The-calculus-with-analytic-geometry-Louis-Leithold

Lo**ryotl '20. "' f du f ---::r J u\/a+n f t l - {Ll *, - ir a > o [-1r" "' 16lT \/i + lal +Tn l\/a | - lZr^n_r l!-a L" i.J '-' b(2n - 3) f \/;T6 itu 2a(n - l) n(n - 7\yn-r J ,nytfifi q6 ifa <o r| @-_a - du J un-r la * bu iflul<a * I#F:+'"\#l ,4.Ih:]tan-'1*, [ 1 t " r , r , -a, L + c ,u.Ih:*'nl=;l*c= la if lul> a iflul<a l l c o t r , -a, L + c i f l u l > a la Tql"c {&) = )v.cerrt.Dn $,non,ol 7"C "*[n-k In formulas27 through38,we may replace ln(z * t/FRl by sinh-lf, lnlu * \trt41 by cosh-'f, .t"l l a + t f i4l , bY s i n h -' f, I +{F+l+c ,, I#:rnlu ,8. t/itTV du:l 1ffif I + $U" + vFEA + c f_ 2e. uz\/uz+ 02d" -- t (2u' + a2) r/uz J :; to. I ry -" : \fr'* nz . tffil "lLPl*, \FtA I/FTF , ,-r-. tAA z 'lf- f f : - T * au hlu+vTfr\+c zz.1 u'z ilu J ffi:i u 1n2 t/F tF -;hlu " I#:sin-l *. nt I f- J uz1ffi- uzo":t tr. f J (uz-+ szytrz o":t (2u,+'Saz)lirE7 +{r'r1"+ tFTTl+c d" qR I :-LLc *' J (uz-rszlstz -raz1/ffi - * tffiv1 + c '42. - ' J1 i*' 1/o:utur:l tF7+f ,uI+:_YT*, +c tfi, - a2-4 sec-, .. 1'l ,t. =-: I *. ' 3s. lgl " " ' JtL:1r".-, uy@-- oz a"-lal "in-,I*, (2u,- sz1 \/F= uz+ sin-' I*, f; t/F=F au: {a,77 - a tnl'* Yal u u 1 : {a'7 u' - a cosh-l *, X L? 1tlF=Tau:_l;'-u' -". u u2 J _ s---in-lL+c a I *, 44. t % : - t t J t/az-uz 2 45. l L : - 1 a J u!a2-u2 i \m+o: 2 n u t "' - sq'){az - uz + aAa ? . _, Ir , r sin-l ;* 4sI#:#+c *. ,nlo*ffi! I du: -[ 47. oz - uz)stz I sin-rL+c I : -+cosh-l 1* ' 46. f \E-u' du - J uztfaz - u2 , -, azu Fsftrl*,:Cont4futing,Lwu- n3 49. 50. .- f a u- \ t r a u - F d, u : T I a' (r-+)*, t / 2 a u - u z + 7 c^ ô-s- ,- r 55. 166 \ffire ["rmdu:2u2-T-to' vsr '4 .t + J lffiu-uz uz du trtr [ t't- r- a u a (" |l')- *acos-'(r-g) +c \ \ER a/ +3o-' cos-l *tcos-l(r-)*, sl I ry * acos-' (r-:)+ c : tffi -:)*, -cos-,(1 sz. Iry:-rY Forms Containing Trigonometric f. f 59.f sinudu cosu+C f 60. f cosudu:sinu*C J : -u - a du f T,-/-' * J (2au - u2)st2 , -. a2!2au - uz s 8 .1 (Zau v'Lr' , J u2)3t2 atffi=E Functions r c o t 2u d u : - c o t u - u + C I -rilcosu+t J t t : - l , i n n - r u c o s t l+ n n" 1 l . ,ui r ^r 'r- ' u d u 73. n J Jsinnut 72. f f 6 1 . I t a t u d u : l n l s e c u l+ C J 62. | .o. u du- lnlsin ul + c J f 63. I t". u du - lnl secu * tan ul + C r : lnltan(*n + iu)l+ C u du- lnlcsct/ - cot ul + C 64. [ J "r. - lnltan iul + C 65.Iru.'udu:tanu*C J f 56.f csc2udu J r J cotu+C secil tanudu: secil+C 58. f.r. u cotu du J f sin2 u du : Lrtt- *sin 2u * C 69. I f 70. I cos' u du: *u * *sin 2u * C J 71. @- ''' J 57. du udu. -++c -T)*, s 3I # : c o s - r ( 1 rF,o . f Jffi:- f J t a n zu d u : t a n t t - u + C u du u d u : + c o s ' - ra s i n * + J . o ' " - 2 " Jcosn u ilu u du:#tan'-r 7s. "- Itann-2 J.ur," u ilu 76. .ot"u du: -i+cotn-' " - I cotu-z / 74. 77.Jr".' u du:#secn-2 u du u tan"+#/'"t'-2 78. csc'u du #csc'-z I 7s. ri^ mu sinnudu-- -W / t""-' u du u cot"+ #J + c W+ --n)usin (m * n)u , sr\(m tt Lc - f d, u : - ; f f i + @ cosnu 80. lcosmu 81. f J f 82. I a s i n m uc o s n ud u - cos(rn * n)u - @c o s ( m u sin u du: sin u - tt cos ll + C f il cos u du: cosu * u sin z + C I 84. I u'sin z du - 2u srnu * (2 u2) cos u * C J 83. - n)u r- L .L t WITH ANALYTICGEOMETRY edition g third ( ( I { { \ ! \ i t LoulsLelthold UNIVERSITY OF SOUTHERN CALIFORNIA HARPER & ROW, PUBLISHERS New York, Hagerstoutn,SanFrancisco,London Sponsoring Editor: GeorgeJ. Telecki Project Editor: Karen A. fudd Designer: Rita Naughton Production Supervisor:FrancisX. Giordano Compositor: ProgressiveTypographers Printer and Binder: Kingsport Press Art Studio: J & R TechnicalServicesInc. Chapter opening art: "Study Light,, by patrick Caulfield THE CALCULUS WITH ANALYTIC GEOMETRY, Copyright @ 1.968,1,972,t976 by Louis Leithold Edition All rights reserved. Printed in the United States of America. No part of this book may be used or reproduced in any manner whatsoever without written permission except in the case of brief quotations embodied in critical articles and reviews. For information address Harper & Row, Publishers, Inc., 10 East 53rd street, New york, N.y. 10022. Library of Congress Cataloging in Publication Data Leithold, Louis. The calculus, with analytic geometry. Includes index. L. Calculus. 2. QA303.L428 7975b ISBN 0-06-043951-3 Geometry, Analytic, L Title. 515'.1s 75-26639 To Gordon Marc Gontents Preface Chapter 1 REAL NUMBERS, INTRODUCTION TO ANALYTIC GEOMETRY, AND FUNCTIONS page 1 Chapter 2 LIMITS AND CONTINUIT} paSe oc Chapter 3 THE DERIVATIVE page110 )cu, 1.1 1.2 1.3 1.4 1.5 1.6 L.7 1.8 2 Sets,Real Numbers, and Inequalities 14 Absolute Value 2L The Number Plane and Graphs of Equations 28 Formula Distance Formula and Midpoint 33 Equations of a Line 43 The Circle 48 Functions and Their Graphs Function Notation, Operations on Functions, and Types of Func56 tions 2.1, 2.2 2.3 2.4 2.5 2.6 66 The Limit of a Function 74 Theorems on Limits of Functions 85 One-Sided Limits 88 Infinite Limits Continuity of a Function at a Number 1-0L Theorems on Continuity 97 1-1-i3.1 The Tangent Line 11,5 3.2 InstantaneousVelocity in Rectilinear Motion L213.3 The Derivative of a Function L26 3.4 Differentiability and Continuity 3.5 Some Theorems on Differentiation of Algebraic Functions L38 3.6 The Derivative of a Composite Function 3.7 The Derivative of the Power Function for Rational Exponents 1-45 3.8 Implicit Differentiation 1.50 3.9 The Derivative as a Rate of Change 154 3.10 RelatedRates 157 3.1L Derivatives of Higher Order 1.30 L42 Viii CONTENTS -T t I \ Chapter 4 TOPICS ON LIMITS, CONTINUITY, AND THE DERIVATIVE page164 Chapter 5 ADDITIONAT APPLICATIONS OF THE DERIVATIVE page204 Chapter 5 THE DIFFERENTIAL AND ANTIDIFFERENTIATION page243 Chapter 7 THE DEFINITE INTEGRAL page275 Chapter 8 APPLICATIONS OF THE DEFINITE INTEGRAL page323 4.L 4.2 4.3 4.4 4.5 4.6 Limits at Infinity 765 Horizontal and Vertical Asymptotes L71 Additional Theorems on Limits of Functions 1.74 Continuity on an Interval 177 Maximum and Minimum Values of a Function 19L Applications Involving an Absolute Extremum on a Closed Interval L89 4.7 Rolle's Theorem and the Mean-Value Theorem lgs 5.1 Increasing and Decreasing Functions and the First-Derivative Test 205 5.2 The second-Derivative Test for Relative Extrema 2i.1 5.3 Additional Problems Involving Absolute Extrema 213 5.4 Concavity and Points of Inflection 220 5.5 Applications to Drawing a Sketchof the Graph of a Function 227 5.6 An Application of the Derivative in Economics 280 5.1 The Differential 244 6.2 Differential Formulas 249 5.3 The Inverse of Differentiation 253 6.4 Differential Equations with Variables Separable 261 6.5 Antidifferentiation and Rectilinear Motion 265 6.6 Applications of Antidifferentiation in Economics 269 7.L 7.2 7.3 7.4 7.5 7.6 The Sigma Notation 276 Area 28L The Definite Integral 288 Properties of the Definite Integral 2gG The Mean-Value Theorem for Integrals 306 The Fundamental Theorem of the Calculus SLi. 8.1 Area of a Region in a Plane 324 8.2 Volume of a Solid of Revolution: Circular-Disk and Circular-Ring Methods 330 8.3 Volume of a Solid of Revolution: Cylindrical-Shell Method 3J6 8.4 Volume of a Solid Having Known Parallel Plane Sections 341 8.5 Work 344 8.6 Liquid Pressure 348 8.7 Center of Mass of a Rod 35L 8.8 Center of Mass of a Plane Region 356 8.9 Center of Mass of a Solid of Revolution 366 8.10 Length of Arc of a Plane Curve 372 CONTENTS Chapter 9 LOGARITHMIC AND EXPONENTIAL FUNCTIONS page38L ChaPterL0 TRIGONOMETRIC FUNCTIONS page430 382 9.1 The Natural Logarithmic Function 9.2 The Graph of the Natural Logarithmic Function 395 9.3 The Inverse of a Function 405 9.4 The Exponential Function 9.5 Other Exponential and Logarithmic Functions 420 9 . 6 Laws of Growth and DecaY 391' 4L4 431 10.1 The Sine and Cosine Functions 438 t0.2 Derivatives of the Sine and Cosine Functions 447 10.3 Integrals Involving Powers of Sine and Cosine 452 Functions and Cosecant Secant, Cotangent, Tangent, 10.4 The of a to the Slope Function Tangent the of 10.s An Application 10.5 L0.7 10.8 10.9 461, Line Integrals Involving the Tangent, Cotangent, Secant, and Co466 secant 47L InverseTrigonometricFunctions 477 Functions Trigonometric Derivatives of the Inverse 484 Functions Integrals Yielding Inverse Trigonometric 492 Introduction 493 Integration by Parts 498 Integration by Trigonometric Substitution Integration of Rational Functions by Partial Fractions. CasesL and 504 2: The Denominator Has Otly Linear Factors 11.5 Integration of Rational Functions by Partial Fractions. Cases3 and 51.2 4: The Denominator Contains Quadratic Factors 516 LL.6' Integration ofRational Functions of Sine and Cosine 51-9 Ll..7 MiscellaneousSubstitutions 521, 1.L.8 The Trapezoidal Rule 'n.9 526 Simpson'sRule Chapter LL 1 1 . 1 TECHNIQUES OF LL.2 INTEGRATION 1 1 . 3 page 497 Lt.4 536 Chapter L2 Lz.L The Hyperbolic Functions 543 Functions Hyperbolic Inverse The I2,2 HYPERBOTIC FUNCTIONS page535 12.3 Integrals Yielding Inverse Hyperbolic Functions 555 Chapter13 13.1 The Polar Coordinate System 560 Polar in Coordinates Equations Graphs of T3.2 POLAR COORDINATES page 554 13.3 Intersection of Graphs in Polar Coordinates 567 57113.4 Tangent Lines of Polar Curves 573 13.5 Area of a Region in Polar Coordinates 579 Chapter 1.4 LA.l The Parabola 583 Axes of L4.2 Translation THE CONIC SECTIONS page 578 588 14.3 Some Properties of Conics 592 L4.4 Polar Equations of the Conics 548 X CONTENTS I {I L4.5 14.5 L4.7 14.8 :L I N i 1 chapter L5 INDETERMINATE FORMS, IMPROPER INTEGRALS, AND TAYLOR'S FORMULA 628 Page Cartesian Equations of the Conics The Ellipse 606 The Hyperbola 613 Rotation of Axes 620 Sgg 15.1 The IndeterminateForm 0/0 629 1,5.2 Other IndeterminateForms 636 15.3 Improper Integralswith Infinite Limits of Integration lS.4 Other Improper Integrals 647 15.5 Taylor's Formula 6s7 Chapter 15 1,5.1 Sequences 5G0 INFINITE SERIES 1'6.2 Monotonic and Bounded Sequence s 667 page65e 16.g Infinite Series of Constant Terms 673 1,6.4 Infinite Seriesof Positive Terms 6g4 1,6.5 The Integral Test 694 15.6 Infinite series of Positive and Negative Terms L6 7 Power Series 707 L6.8 Differentiation of Power Series 713 16.9 Integration of Power Series 722 15.10Taylor Series 729 L6.ll The Binomial Series 738 Chapter 17 VECTORS IN THE PLANE AND PARAMETRIC EQUATIONS page745 Chapter 18 VECTORS IN THREEDIMENSIONAL SPACE AND SOLID ANALYTIC GEOMETRY page 810 77.1, 77.2 L7.3 17.4 77.5 77.6 t7.7 I7.8 641 697 Vectors in the Plane 746 Properties of Vector Addition and Scalar Multiplication 751 Dot Product 756 vector-Valued Functions and Parametric Equations 763 Calculus of Vector-Valued Function s 772 Length of Arc 779 Plane Motion 785 The Unit Tangent and Unit Normal Vectors and Arc Length as a Parameter 792 L7.9 Curvature 796 17.10 Tangential and Normal Components of Acceleration g04 18.1 R3,The Three-Dimensional Number Space 8L1 L8.2 Vectors in Three-Dimensional Space 818 18.3 The Dot Product in V, 825 18.4 Planes 829 18.5 Lines in R3 836 L8.6 Cross Product 842 1,8.7 Cylinders and Surfaces of Revolution BS2 18.8 Quadric Surfaces 858 18.9 Curves in R3 864 18.10 Cylindrical and Spherical Coordinates 872 CONTENTS Chapter 19 CALCULUS DIFFERENTIAL OF FUNCTIONS OF SEVERAL VARIABLES page 880 Chapter 20 DIRECTIONAL DERIVATIVES, GRADIENTS, APPTICATIONS OF PARTIAL DERIVATIVES, AND LINE INTEGRALS page 944 ChaPter2| MULTIPTE INTEGRATION page 1001 APPENDIX pageA-1 88L L9.']-. Functions of More Than One Variable 889 lg.2 Limits of Functions of More Than one Variable 900 tg.3 Continuity of Functions of More Than One Variable 905 19.4 Partial Derivatives 91'3 19.5 Differentiability and the Total Differential 926 19.6 The Chain Rule 934 19.7 Higher-Order Partial Derivatives 20.L 20.2 20.3 20.4 20.5 20.6 20.7 945 Directional Derivatives and Gradients 953 Surfaces to Normals Tangent Planes and 956 Variables Two Extrema of Functions of Some Applications of Partial Derivatives to Economics 975 Obtaining a Function from Its Gradient 981 Line Integrals 989 Line Integrals Independent of the Path 21.1, 2'1..2 21.3 21.4 21.5 21.6 21.7 L002 The Double Integral 1008 Evaluation of Double Integrals and Iterated Integrals 1'0L6 Center of Mass and Moments of Inertia 1'022 The Double Integral in Polar Coordinates 1028 Area of a Surface 1034 The Triple Integral L039 The Triple Integral in Cylindrical and Spherical Coordinates A-2 Table 1. Powers and Roots A-3 Table 2 Natural Logarithms A-5 Table 3 Exponential Functions A-L2 Table 4 Hyperbolic Functions A-13 Table 5 Trigonometric Functions A-1-4 Table 6 Common Logarithms A-15 Table 7 The Greek Alphabet ANSWERS TO ODD-NUMBERED EXERCISES INDEX 4.45 A.1.7 967 ACKNOWLEDGMENTS Reviewers of rheCatcutu Professor William D. Professor Archie D. l Professor Phillip Cla Professor Reuben W ProfessorJacob Golil ProfessorRobert K. Goodrich, University of Colorado Professor Albert Herr, Drexel University ProfessorJamesF. Hurley, University of Connecticut Professor Gordon L. Miller, Wisconsin State Universitv Professor William W. Mitchell, Jr., phoenix College Professor Roger B. Nelsen, Lewis and Clark Colle-ge Professor Robert A. Nowlan, southern connectictit state college Sister Madeleine Rose, Holy Names College Professor George W. Schultz, St. petersbuig Junior College Professor Donald R. Sherbert, Universitv of Illinois Professor]ohn V_adney,Fulton-Montgomery community college Professor David Whitman, San Diego State College Production Staff at Harper & Row GeorgeTelecki, MathematicsEditor Karen fudd, Project Editor Rita Naughton, Designer Assistantsfor Answers to Exercises JacquelineDewar, Loyola Marymount University Ken Kast, Logicon, Inc. |ean Kilmer, West Covina Unified SchoolDistrict Cover and ChapterOpenins Artist Patrick Cadlfield,Londin, England To these _peopleand to all the users of the first and second editions who have suggested changes,I expressmy deep appreciation. L. L. rl] FTETAGE This third edition of THE CALCULUS WITH ANALYTIC GEOMETRY, like the other two, is designed for prospective mathematics majors as well as for students whose primary interest is in engineering, the physical sciences, or nontechnical fields. A knowledge of high-school algebra and geometry is assumed. The text is available either in one volume or in two parts: Part I consists of the first sixteen chapters, and Part II comprises Chapters 16 through 21 (Chapter L6 on Infinite Series is included in both parts to make the use of the two-volume set more flexible). The material in Part I consists of the differential and integral calculus of functions of a single variable and plane analytic geometrlz, and it may be covered in a one-year course of nine or ten semester hours or twelve quarter hours. The second part is suitable for a course consisting of five or six semester hours or eight quarter hours. It includes the calculus of several variables and a treatment of vectors in the plane, as well as in three dimensions, with a vector approach to solid analytic geometry. The objectives of the previous editions have been maintained. I have endeavored to achieve a healthy balance between the presentation of elementary calculus from a rigorous approach and that from the older, intuitive, and computational point of view. Bearing in mind that a textbook should be written for the student, I have attempted to keep the presentation geared to a beginner's experience and maturity and to leave no step unexplained or omitted. I desire that the reader be aware that proofs of theorems are necessary and that these proofs be well motivated and carefully explained so that they are understandable to the student who has achieved an average mastery of the preceding sections of the book. If a theorem is stated without proof ,I have generally augmented the discussion by both figures and examples, and in such cases I have always stressed that what is presented is an illustration of the content of the theorem and is not a proof. Changes in the third edition occur in the first five chapters. The first xlv PREFACE section of Chapter I has been rewritten to give a more detailed exposition of the real-number system. The introduction to analytic geometry in this chapter includes the traditional material on straight lines as well as that of the circle, but a discussion of the parabola is postponed to Chapter 14, The Conic Sections. Functions are now introduced in Chapter 1. I have defined a function as a set of ordered pairs and have used this idea to point up the concept of a function as a correspondence between sets of real numbers. The treatment of limits and continuity which formerly consisted of ten sections in Chapter 2 is now in fwo chapters (2 and 4), with the chapter on the derivative placed between them. The concepts of limit and continuity are at the heart of any first course in the calculus. The notion of a limit of a function is first given a step-by-step motivation, which brings the discussion from computing the value of a function near a number, through an intuitive treatment of the limiting process, up to a rigorous epsilon-delta definition. A sequence of examples progressively graded in difficulty is included. All the limit theorems are stated, and some proofs are presented in the text, while other proofs have been outlined in the exercises. In the discussion of continuity, I have used as examples and counterexamples "common, everyday" functions and have avoided those that would have little intuitive meaning. In Chapter 3, before giving the formal definition of a derivative, I have defined the tangent line to a curve and instantaneous velocity in rectilinear motion in order to demonstrate in advance that the concept of a derivative is of wide application, both geometrical and physical. Theorems on differentiation are proved and illustrated by examples. Application of the derivative to related rates is included. Additional topics on limits and continuity are given in Chapter 4. Continuity on a closed interwal is defined and discussed, followed by the introduction of the Extreme-Value Theorem, which involves such functions. Then the Extreme-Value Theorem is used to find the absolute extrema of functions continuous on a closed interval. Chapter 4 concludes with Rolle's Theorem and the Mean-Value Theorem. Chapter 5 gives additional applications of the derivative, including problems on curve sketching as well as some related to business and economics. The antiderivative is treated in Chapter 6. I use the term "antidifferentiation" instead of indefinite integration, but the standard notation ! f (x) dx is retained so that you are not given a bizarre new notation that would make the reading of standard references difficult. This notation will suggest that some relation must exist between definite integrals, introduced in Chapter 7, and antiderivatives, but I see no harm in this as long as the presentation gives the theoretically proper view of the definite integral as the limit of sums. Exercises involving the evaluation of definite integrals by finding limits of sums are given in Chapter 7 to stress that this is how they are calculated. The introduction of the definite inte- PREFACE gral follows the definition of the measure of the area under a curye as a limit of sums. Elementary properties of the definite integral are derived and the fundamental theorem of the calculus is proved. It is emphasized that this is a theorem, and an important one, because it provides us with an alternative to computing limits of sums. It is also emphasized that the definite integral is in no sense some special type of antiderivative. In Chapter 8 I have given numerous applications of definite integrals. The presentation highlights not only the manipulative techniques but also the fundamental principles involved. In each application, the definitions of the new terms are intuitively motivated and explained. The treatment of logarithmic and exponential functions in Chapter 9 is the modern approach. The natural logarithm is defined as an integral, and after the discussion of the inverse of a function, the exponential function is defined as the inverse of the natural logarithmic function. An irrational power of a real number is then defined. The trigonometric functions are defined in Chapter 10 as functions assigning numbers to numbers. The important trigonometric identities are derived and used to obtain the formulas for the derivatives and integrals of these functions. Following are sections on the differentiation and integration of the trigonometric functions as well as of the inverse trigonometric functions. Chapter LL, on techniques of integration, involves one of the most important computational aspects of the calculus. I have explained the theoretical backgrounds of each different method after an introductory motivation. The mastery of integration techniques depends upon the examples, and I have used as illustrations problems that the student will certainly meet in practice, those which require patience and persistence to solve. The material on the approximation of definite integrals includes the statement of theorems for computing the bounds of the error involved in these approximations. The theorems and the problems that go with them, being self-contained, can be omitted from a course if the instructor so wishes. A self-contained treatment of hyperbolic functions is in Chapter 12. This chapter may be studied immediately following the discussion of the circular trigonometric functions in Chapter L0, if so desired. The geometric interpretation of the hyperbolic functions is postponed until Chapter L7 because it involves the use of parametric equations. Polar coordinates and some of their applications are given in Chapter 13. In Chapter 1.4,conics are treated as a unified subject to stress their natural and close relationship to each other. The parabola is discussed in the first two sections. Then equations of the conics in polar coordinates are treated, and the cartesian equations of the ellipse and the hyperbola are derived from the polar equations. The topics of indeterminate forms, improper integrals, and Taylor's formula, and the computational techniques involved, are presented in Chapter L5. I have attempted in Chapter 16 to give as complete a treatment of xvr PREFACE infinite series as is feasible in an elementary calculus text. In addition to the customary computational material, I have included the proof of the equivalence of convergence and boundedness of monotonic sequences based on the completeness property of the real numbers and the proofs of the computational processes involving differentiation and integration of power series. The first five sections of Chapter L7 on vectors in the plane can be taken up after Chapter 5 if it is desired to introduce vectors earlier in the course. The approach to vectors is modern, and it serves both as an introduction to the viewpoint of linear algebra and to that of classical vector analysis. The applications are to physics and geometry. Chapter 18 treats vectors in three-dimensional space, and, if desired, the topics in the first three sections of this chapter may be studied concurrently with the corresponding topics in Chapter 17. Limits, continuity, and differentiation of functions of several variables are considered in Chapter L9. The discussion and examples are applied mainly to functions of two and three variables; however, statements of most of the definitions and theorems are extended to functions of n variables. In Chaptet 20, a section on directional derivatives and gradients is followed by a section that shows the application of the gradient to finding an equation of the tangent plane to a surface. Applications of partial derivatives to the solution of extrema problems and an introduction to Lagrange multipliers are presented, as well as a section on applications of partial derivatives in economics. Three sections, new in the third edition, are devoted to line integrals and related topics. The double integral of a function of two variables and the triple integral of a function of three variables, along with some applications to physics, engineering, and geometry/ are given in Chapter 2/... New to this edition is a short table of integrals appearing on the front and back endpapers. However, as stated in Chapter L1, you are advised to use a table of integrals only after you have mastered integration. Louis Leithold Realnumberq,introduction to analytic geometry and functions REAL NUMBERS,INTRODUCTION TO ANALYTICGEOMETRY. AND FUNCTIONS 1.1 SETS, REAL NUMBERS, AND INEQUALITIES The idea of "set" is used extensively in mathematics and is such a basic concept that it is not given a formal definition. We can say that a set is a collection of objects, and the objects in a set are called the elements of a set. We may speak of the set of books in the New York Public Library, the set of citizens of the United States, the set of trees in Golden Gate Park, and so on. In calculus, we are concerned with the set of real numbers. Before discussing this set, we introduce some notation and definitions. We want every set to be weII defined; that is, there should be some rule or property that enables one to decide whether a given object is or is not an element of a specific set. A pair of braces { } used with words or symbols can describe a set. If S is the set of natural numbers less than 6,we can write the set S as {L,2,3, 4,5} We can also write the set S as {r, such that r is a natural number less than 6} where the symbol. "x" is called a"variable." A aariable is a symbol used to represent any element of a given set. Another way of writing the above set S is to use what is called set-builder notation, where a vertical bar is used in place of the words "such that." Using set-builder notation to describe the set S, we have {rlr is a natural number less than 6} which is read "the set of all r such that r is a natural number less than G." The set of natural numbers will be denoted by N. Therefore, we may write the set N as {7,2,3, I .l where the three dots are used to indicate that the list goeson and on with no last number. With set-builder notation the set N may be written as {rlr is a natural number}. The symbol " e " is used to indicate that a specificelementbelongs to a set. Hence, we may write 8 € N, which is read "8 is an elementof N.,, The notation a,b e S indicates that both a andb are elementsof S. The symbol fi is read "is not an elementof." Thus, we read * € w as "+ is not an elementof N." We denote the set of all integers by /. Becauseevery element of N is also an element of / (that is, every natural number is an integer),we say that N is a "subset" of /, written N E /. :',.t Definition -J The set S is a subsetof the set T, written S e T,if and only if every element of S is also an element of T.If , in addition, there is at least one element of T 1.1 SETS,REAL NUMBERS,AND INEQUALITIES which is not an elementof S, then S is a propersubsetof T, and it is writtenS C T. Observe from the definition that every set is a subsetof itself, but a set is not a proper subset of itself. In Definition 1.1.1,the "if and only if" qualification is used to combine two statements:(i) "the set S is a subset of the set T if every element of S is also an element of T"; and (ii) "the set S is a subset of set T only if every element of S is also an element of 7," which is logically equivalent to the statement"if. S is a subset of.T, then every elementof S is also an element of.T." o rLLUsrRArroN1: Let N be the set of natural numbers and let M be the set denoted by {rlr is a natural number less than 1,0}.Becauseevery element of M is also an element of N, M is a subset of N and we write M e N. Also, there is at least one element of N which is not an element of M, and so M is a proper subsetof N and we may write M C N. Furthermore, because{5} is the set consisting of the number 5, {6} C M, which statesthat the set consisting of the single element 5 is a Proper subset of the set M. We may also write 5 € M, which statesthat the number 6 is o an element of the set M. Consider the set {xlzx * L:0, and x e I}. This set containsno elements because there is no integer solution of the equation 2x * 1.: 0. Such a set is called the "empty set" or the "null set." 1.t.2 Definition The empty set (or null sef) is the set that contains no elements.The emPty set is denoted by the symbol A. The concept of "subset" may be used to define what is meant by two sets being "equal." 1.1.3 Definition Two setsA and B are said to be equal,written A: andB e A. B, if and only if A e B Essentially, this definition states that the two sets A and B are equal if and only if every element of A is an element of B and every element of B is an element of A, that is, if the sets A and B have identical elements. There are two operations on sets that we shall find useful as we proceed. These operations are given in Definitions 1.1.4 and 1.1.5. 1.1.4 Definition Let A and B be two sets. The union of A and B, denoted by A U B and read "A union 8," is the set of all elements that are in A or in B or in both A and B. REAL NUMBERS,INTRODUCTION TO ANALYTICGEOMETRY,AND FUNCTIONS E x A M P LE1: Le t A : { 2 , 4 , 6 ,8 , 10,LzI, B : {1, 4, 9, '1.5} , and C - { 2 , 1 0 } .F i n d (a)AuB (c)BUC (b)Auc (d)AuA 1..1..5 Definition nxelvrpr.E 2: If A, B, and C are the sets defined in Example1, find @)AnB (c)BnC SOLUTION: (a)Au (b)Au (c)BU (d)Au BC: C: A: 1 ,, 21, , 6 } {1,2,4,6,9,9, "1.0 { 2 , 4 , 6 , 8 , 1 , 0 1, 2 } { 1 , 2 ,4 , 9 , 1 , 0 , 1 , 6 } {2, 4, 6, 8, I0,12} : 4 Let A and B be two sets.The intersectionof A and B, denoted by A n B and tead " A intersection8," is the set of all elementsthat are in both A and B. SOLUTION: ( a )A n g : { 4 } ( c )B f i C : A (b)An C: {2, 1.0} ( d ) A n A : { 2 , 4 , 6 , 8 , t 0 , 7 2 1: 4 (b)Anc @)AnA The real number system consists of a set of elements called real numbers and two operations called addition and multiplication The set of real numbers is denoted by Rt. The operation of addition is denoted by the symbol "*", and the operation of multiplication is denoted by the symbol',.',. rf a, b c Rt, a * b denotes the sum of a and b, and a - b (or ab) denotes their product. We now present seven axioms that give laws governing the operations addition of and multiplication on the set Rl. The word axiom is used to indicate a formal statement that is assumed to be true without proof. 1.1.6Axiom (Closure and UniquenessLaws) 1.1.7 Axiom rf a ,b e R t, then si b number. i s a uni que real number, andab i s a unique r eal If a, b e Rr, then (CommutatiaeLaws) a*b:b*a L.L.8 Axiom (AssociatiaeLaws) 1."1,.9 Axiom and ab:ba If. a, b, c € Rl, then a + (b* c) : (a + b) + c and a(bc): (ab)c l f a , b , c C R 1 ,t h e n (DistributiaeLaw) a(b+c):ab*ac 1..L.10 Axiom There exist two distinct real numbers 0 and 1 such that for anv real num- (Existence of IdentityElements) ber A, ail:a J and A.t:a AND INEOUALTTIES 1.1 SETS,REAL NI.JMBERS, L.1.11 Axiom For every real number a, there exists a real number called the negatiae lnaerse) of a (or additiae inaerse of a), denoted by -a (read "the negative of a"), or Additiae (Existence of Negatiae such that a*(-s):Q '1,.1.12 Axiom For every real number a, except 0, there exists a real number called the (Existenceof Reciprocalor reciprocat of a (or multiplicatiae inaerse of a), denoted by a-t, such that Multiplicatiae lnuerse) a ' a-r,: t Axioms 1,.1,.6through 1,.I.L2 are called field axioms because if these axioms are satisfied by a set of elements, then the set is called afield under the two operations involved. Hence, the set R1 is a field under addition and multiplication. For the set / of integers, each of the axioms t.1.6 is not satisfied (for instance, through 1.1.11 is satisfied, but Axiom L.1,.1,2 it /). Therefore, the set of inverse no multiplicative has the integer 2 multiplication. and field under addition a integers is not Definition 1..1..13 lf a, b € Rl, the operation of subtraction assigns to a andb areal number, denoted by a - b (read "a minus b"), called the differenceof a and b, where (1) a-b:A*(-b) Equality (1) is read "A minus b equals a plus the negatle 'i,.1.14 Definition of-b." I f . a , b G R l , a n d b * 0 , t h e o p e r a t i o no f d i a i s i o na s s i g n s t o a a n d b a r c a l number, denoted by o + b (read "a divided by b"), called the quotient of a and b, where a:b:A'b-r Other notations for the quotient of a and b ate ! b and alb By using the field axioms and Definitions 1.1.13 and 1.1.14, we can derive properties of the real numbers from which follow the familiar algebraic operations as well as the techniques of solving equations, factoring, and so forth. In this book we are not concerned with showing how such properties are derived from the axioms. Properties that can be shown to be logical consequences of axioms are theorems. In the statement of most theorems there are two parts: the "if" part, called the hypothesis, and the "then" part, called the conclusion. The argument verifying a theorem is a proof. A proof consists of showing that the conclusion follows from the assumed truth of the hypothesis. REAL NUMBERS,INTRODUCTION TO ANALYTICGEOMETRY, AND FUNCTIONS The concept of a real number being "positive" lowing axiom. is given in the fol- L.1..L5Axiom In the set of real numbers there exists a subset called the positiae numbers (OrderAxiom) such that (i) if a € Rr, exactly one of the following three statements holds: -a i s posi ti ve. a: 0 a i s posi ti ve (ii) the sum of two positive numbers is positive. (iii) the product of two positive numbers is positive. Axiom 1.1.15 is called the order axiom because it enables us to order the elements of the set Rr. In Definitions 'J,.7.77 and 1.1.18 we use this axiom to define the relations of "greater than" and "less than" on Rr. The negatives of the elements of the set of positive numbers form the set of "negative" numbers, as given in the following definition. 1*1.16Definition The real number a is negatiae iI and only if -a is positive. From Axiom 1.1.15and Definition 7.1..76it follows that a real number is either a positive number, a negative number, or zero. Ary real number can be classified as a rqtional number or an irrational number. A rational number is any number that can be expressed as the ratio of two integers. That is, a rational number is a number of the form plq, where p a:nd,q are integers and q + 0.The rational numbers consist of the following: The integers (positive, negative, and zero) . , -5, -4, -3, -2, -L, 0, 1, 2, 3, 4, 5, . The positive and negative fractions such as +-#+ The positive and negative terminating decimar.ssuch as -0.003251 : 3,251 L,000,000 2s6:ffi The positive and negative nonterminatingrepeatingdecimalssuch as 0.333. :+ -0549s49s49...:-ft The real numbers which are not rational numbers are called irrational numbers. These are positive and negative nonterminating, nonrepeating decimals, for example, rt: t.732. n : 3.L4'1,59. tan 140o: -0.839L. The field axioms do not imply any ordering of the real numbers. That is, by means of the field axioms alone we cannot state that 1 is less than 1.1 SETS,REAL NUMBERS,AND INEQUALITIES 2,2isless than 3, and so on. However, we have introduced the order axiom (Axiom 1.1.15),and because the set Rr of real numbers satisfies the order axiom and the field axioms, we say that Rl is an ordered field. We use the concept of a positive number given in the order axiom to define what we mean by one real number being "less than" another. 11*\ 7 D e fi n i ti o n If a ,b € R t, th e n a i s l essthan b (w ri tten a< b) positive. i f and onl y i f b-ars and 5 is positive. (b) o rt,r,usrRArIbrs 2: (a) 3 b) than a; with symbols we write if andonlyif b isless albifandonlyifb<a l.l.l9 Definition The symbols = (" is less than or equal to") and = (" is greater than or equal to") are defined as follows: (1 ) a - b i f and onl y i f ei ther a b if. and only if either a > b ot a: bT h e s ta te m entsa 1b, a) b, a < b, anda = b are cal l ed i nequal i ti es. -4, -3> -6, Some examplesare 2 <7, -5 < 6, -S 1-4,"].'4> 8,2> -10 -7.In < > b are called strict inequalparticular, a 1b and a 5 = 3, -b anda= b are called nonstrict inequalities.. ities, whereas a The following theorems can be proved by using L.L.6 through 1.1'.19. "1..1.20Theorem 1 . 7 . 2 1T h e o r e m (i) (ii) (1ii) (iv) a a a a > < > < 0 0 0 0 if if if if If.a<bandb1c, and and and and only only only only if if if if a is positivq. a is negative. -a < 0. -a > 0. thena1c. o rLLUsrRArroN 3: 3 < 5 and 6 <'1.4; so 3 < 14. 1 . 1 . 2 2T h e o r e m lf a< b,then a* c<'b+ o rLLUsrRATroN4: 41 1.1.23 Theorem c,and a-c<b- cif c isanyrealnumber. s o4 * 5 < 7 * 5 ; a n d 4 - 5 < 7 - 5 . If. a < b and c I d, then a * c < b + d. 5: 2 <6 and -3 < 1; so 2+ (-3) < 6 + L. o rLLUsrRArIoN REALNUMBERS,INTRODUCTION TO ANALYTICGEOMETRY, AND FUNCTIONS 1.1.24 Theorem lf a < b, and c is any positive number, then ac I o TLLUSTRATToN6: 2 I L.l.25 Theorem bc. 5; so 2 . 4 < s rf a < b, and c is any negative number, then ac ) bc. o rLLUSrRArroN7: 2 < 5; so 2(-4) > 5(-4). 1 , . 1 . 2 T6 h e o r e m I f 0 < a < b and0(c( d,thenac<bd. o r L L U S r R A r r o N 80: < 4 < 7 and0 < 8 < 9;so4(8) <7(9). Theorem 1'.1.24states that if both members of an inequality are multiplied by a positive number, the direction of the inequality remains unchanged, whereas Theorem 1.1.25 states that if both members of an inequality are multiplied by a negative number, the direction of the inequality is reversed. Theorems 7.7.24 and 1.1..25also hold for division, because dividing both members of an inequality by a number d is equivalent to multiplying both members by lld. To illustrate the type of proof that is usually given, we present a proof of Theorem 1.L.23: By hypothesis a I b. Then b - a is positive (by Definition 1..1,.17). By hypothesis c I d. Then d - c is positive (by Definition 1.1.17). H e n ce, (b - a) + (d- c) i s posi ti ve (by A xi om 1.1.15(i i )). T h erefore, (b t d) - (a * c) i s posi ti ve (by A xi oms 1.1.gand l. l. T and Definition 1.1.13). Therefore, A * c b and b ) c, then a ) c. . r L L U s r R A r r o N 9 : 8> 4 a n d 4 > - 2 ; ljl..28 Theorem so8 )-2. rf a > b , then a* c > b + c, and a- c > b- . c i f c i s any real num ber . o rLLUSrRArroN 10: 3 = -5; so 3 - 4 > -5 - 4. 1.1.29 Theorem If a > b and c ) d, then a I c > b + d. o r L L U S r R A r r otN ' 1 , : 7> 2 a n d 3 L.1.30Theorem >-5; so7*3>2+(-5). If a > b and if c is any positive number, then ac ) bc. o rLLUsrRArroN 12: -3 > -7; so (-3) > eDA. o 1.1 SETS,REAL NUMBERS,AND INEQUALITIES 1.1.31Theorem lf. a > b and if c is any negative number, then ac I bc. o rLLUsrRArroN 13: -3 > -7; so (-3)(-4) < (-7)(-4). o 1.1.32Theorem If a > b > }and c > d > 0, then ac > bd. o r L L U S r R A r r o1N4 : 4 > 3 > 0 a n d 7 n 79 tz24 lllttrttllrll -4t -2 0 | zt 4 F i g u r e1 . 1 . 1 >5 ) 0; so 4(7)>3(6). o So far we have required the set Rl of real numbers to satisfy the field axioms and the order axiom, and we have stated that because of this requirement Rr is an ordered field. There is one more condition that is imposed upon the set R1. This condition is called the axiom of completeness (Axiom 16.2.5). We defer the statement of this axiom until Section 16.2 because it requires some terminology that is best introduced and discussed later. However, we now give a geometric interpretation to the set of real numbers by associating them with the points on a horizontal line, called an axis. The axiom of completeness guarantees that there is a oneto-one correspondence between the set Rl and the set of points on an axis. Refer to Figure 1.1.1. A point on the axis is chosen to represent the number 0. This point is called the origin A unit of distance is selected. Then each positive number x is represented by the point at a distance of r units to the right of the origin, and each negative number x is represented by the point at a distance of -x units to the left of the origin (it should be noted that if r is negative, then -r is positive). To each real number there corresponds a unique point on the axis, and with each point on the axis there is associated only one real number; hence, we have a one-to-one correspondence between Rl and the points on the axis. So the points on the axis are identified with the numbers they represent, and we shall use the same symbol for both the number and the point representing that number on the axis. We identify Rl with the axis, and we call Rl the real number line. We see that a b if and only if the point representing a is to the right of the point representing b. For instance, the number 2 is less than the number 5 and the point 2 is to the left of the point 5. We could also write 5 ) 2 and say that the point 5 is to the right of the point 2. A numberx is between a andbif a ( rand x <b. Wecanwritethis as a continued inequality as follows: alxlb (2) The set of all numbers x satisfying the continued inequality (2) is called an "open interyal." 1.L.33 Definition The open interaal from a to b, denoted by (o, b), is defined by (a,b):{rlalx<b} 10 REALNUMBERS, INTRODUCTION TO ANALYTICGEOMETRY, AND FUNCTIONS The "closed interval" from 4 to b is the open interval (a, ,) together with the two endpoints c and b. 1.1.34Definition - The closedinteraalfrom a to b, denoted by lo, bl, is defined by b,bl:{xla-x-b} a F i g u r e1 . 1 . 2 b a Figure 1,.1,.2 illustrates the open interval (a,b) and Fig. 1.1.3illustrates the closed interval la, bl. The "interval half-open on the lef.t" is the open interval (a, b) together with the right endpoint b. F i g u r e1 . 1 . 3 " 1.1.35 Definition The interaal half-open on the left, denoted by (o, bl, is defined by (a,bl:{xla<x-b} we define an "interval half-open on the right" in a similar way. '1,.1.36 Definition The interaal half-open on the right, denoted by lo, b), is defined by la,b):{rl ab Figure 1'.1.4illustrates the interval (a,bl and Fig. 1.1.5 illustrates the interval la, b). we shall use the symbol +@ ("positive infinity") and the symbol -m ("negative infinity"); however, care must be taken not to confuse these symbols with real numbers, for they do not obey the properties of the real numbers F i g u r e1 . 1 . 4 f--L ab F i g u r e1 . 1 . 5 7.1.37 Definition (i) (a, +-) : (ii) (-.o, b): (iii) la, +*7 : (iv) (-m, bl: (v) (-*, a **) {xlx > {xlx < {xlx > {xlx : a} b} a} b} Rl Figure 1,.1,.6illustrates the interval (a, *a), and Fig. r.7.T illustrates the interval (-*, b). Note that (-m,1oo) denotes the set of all real numbers. For each of the intervals (a,b), [a, b], lq, b), and (a, b] the numbers a F i g u r e1 . 1 . 6 b F i g u r e1 . 1 . 7 a=x<b} 1.1 SETS,REAL NUMBERS,AND INEQUALITIES 11 of its endpoints, and a closed interval can be regarded as one which conis considered tains all of its endpoints. Consequently, the intewalla,ao). to be a closed interval because it contains its only endpoint a. Similarly, (--, b] is a closed interval, whereas (a, **) and (-oo, b) are oPen. The intervals la, b) and (a, bl are neither open nor closed. The interval (--, *-) has no endpoints, and it is considered both oPen and closed. Intervals are used to represent "solution sets" of inequalities in one variable. The solution set of such an inequality is the set of all numbers that satisfy the inequality. EXAMPLE 3: Find the solution set of the inequality 2*3x<5r*8 Illustrate it on the real number line. solurroN: lf x is a number such that 2*3x<5r*8 then 2 t 3x - 2 < 5x * 8 - 2 (by Theorem 1'.1'.22) or, equivalently, 3x<5xI6 Then, adding -5r to both membersof this inequality, we have -2x < 6 (by Theorem7.7.22) Dividing on both sides of this inequality by -2 and reversing the direction of the inequality, we obtain r ) -3 (by Theorem1.1.25) What we have proved is that if 2*3x<5r*8 then x>-3 Each of the steps is reversible; that is, if we start with x>-3 we multiply on each side by -2, reversethe direction of the inequality, and obtain -2x<6 Then we add 5x and 2 to both membersof the inequality, in which case we get 2i3x<5x*8 12 REAL NUMBERS;INTRODUCTION TO ANALYTICGEOMETRY,AND FUNCTIONS -30 F i g u r e1 . 1 . 8 Therefore, we can conclude that 2*3x<5r*8 i f a n d o n l y i fx > - 3 So the solutionsetof the given inequalityis the interval(-3, +-;, which is illustrated in Fig. 1.1.8. ExAMPLE4: Find the solution set of the inequality 4<3x-2<'1,0 Illustrate it on the real number line. 0 F i g u r e1 . 1 . 9 rxauprn 5: Find the solution set of the inequality 722 x Illustrate it on the real number line. solurroN: Adding 2 to each member of the inequalit|, we obtain 6<3x<12 Dividing each member by 3, we get 21x<4 Each step is reversible; so the solution set is the interval (2,41, as is illusstrated in Fig. 7.1,.9. solurroN: we wish to multiply both members of the inequality by x. However, the direction of the inequality that results will depend upon whether x is positive or negative.So we must consider two cases. Case7: r is positive;that is, x ) 0. Multiplying on both sides by *, we obtain 7>2x Dividing on both sidesby 2,we get E > x or, equivalently, x < t Therefore,since the abovestepsare reversible,the solution set of Case1 is {xlx > 0} n {xlx < E} ot, equivalently, {rl0 < 7s<E}, which is the interval (0, g). Cqse2: x is negative;that is, x ( 0. Multiplying on both sides by * and reversing the direction of the inequality, we find 7 <2x Dividing on both sides by 2,we have E < x or, equivalently, x > E Again, becausethe above stepsare reversible,the solution set of Case2 is {rl*< 0} n {x>g}:A. From Cases1 and 2 we concludethat the solution set of the given inequality is the open interval (0,2), which is illustratedin Fig. 1.1.10. 1.1 SETS,REAL NUMBERS,AND INEQUALITIES 13 EXAMPLE6: Find the solution set of the inequality x=14 x-5 Illustrate it on the real number line. 0 F i g u r e1 , 1 .11 solurroN: To multiply both members of the inequality by x-3, must consider two cases,as in Example5. Case'l: x- 3 > 0; that is, x ) 3. Multiplying on both sides of the inequality by x- 3, we get we x14x-12 Adding -4x to both members,we obtain -3x < -12 Dividing on both sidesby-3 and reversingthe direction of the inequality, we have x) 4 Thus the solution set of Case1 is {rlx > 3} n {xlr > 4} or, equivalently, {xlx > 4}, which is the interval (4, +oo;. Case2: x-3 < 0; thatis,x <-3. Multiplying on both sides by * - 3 and reversing the direction of the inequ ality,we have x) 4x-L2 or, equivalently, -3x > -12 or, equivalently, x14 Therefore,.x must be less than 4 and also less than 3. Thus, the solution set of Case2 is the interval (-*, 3). If the solution sets for Cases 1 and 2 are combined, we obtain (--, 3) U (4, **), which is illustrated in Fig. 1.1.11. ExavrplE 7: Find the solution set of the inequality (r+3)(x+4) > 0 solurroN: The inequality will be satisfied when both factors have the s a m e s i g n , t h a t i isf ,x t 3 > 0 a n d x + 4 > 0 , o r i f x t 3 < 0 a n d r + 4 < 0 . Let us consider the two cases. C a s e l - : x t 3 > 0 a n dx t 4 x>-3 ) 0.Thatis, and x)-4 Thus, both inequalities hold if x > -3, which is the interval (-3, +*;. C a s e 2 : x - 1 3 < 0 a n dx * 4 < 0 . T h a t i s , and x 1-4 x 1-3 Both inequalities hold if x < -4, which is the interval (-*,-4). If we combine the solution setsfor Cases1 and 2,we have (-*,-4) (-3, +-;. U 14 REAL NUMBERS,INTRODUCTION TO ANALYTICGEOMETRY,AND FUNCTIONS Exercises7.L In Exercises1 through 10,list the elementsof the given set if A: { 0 , 3 ,6 , 9 } . {0,2,4,6,8}, B : { 1 , z , 4 , B } , C : { 1 , i , s , 7 , 9 } , a n d D : I,.AUB 2.CUD 4.CnD 5.BUD 6. AUC 8.AnC g. 7.BnD 3. ANB @n D)u (Bnc) 10.(AuB)n(cuD) -ilP';3,t"i l1"t*T"-:11 ,T:tyo.n 1.'1..5x+2>x-6 ? 1 4 . 3 x - 5- 2x-3>5 1,6.2 < 5 - 3 x i14 ,f8.I .9 -7 { r : i1 - 3 > Z x "x r) 20r i-L - x = 1. \-r 2 1 ,*. > 4 22. f < g 23.(x-3)(r+s)>0 2 4 . x 2- 3 r * 26.**3r*1>0 27.4fI9x<9 28. 2 f - 6 x * 3 < 0 29. x*'j,x ? " -n' 2 - r - 3 * x 31,. 32. 33. Prove Theorem 1,.1.27. 94. Prove Theorem 1,.1,.22. 17.2>-3-3x>-7 35. Prove 36. lf 2> 0 <ll 25. l - x - 2 x 2 > 0 74 3x-7 3-2x Theorems 1.1,.24and 1.1.25. n > b = O,prove that a2 > W. 37. lf. a and b are nonnegative numbers, and az ) br, prove that a > b. 38. Prove that if a > 0 and b > 0, then A2: b2 if and only if a: 39.Pncve that if b > a > 0 and c ) 0, then a* c b + t' b. 40. Prove that if x 1 !, then r < t(x + y) < y. a b 1.2 ABSOLUTE VALUE 1.2|t Definition The conceptof the "absolute value" ofa number is used in some important definitions in the study of calculus. Furtherrnore, you will need.tb work with inequalities involving "absolute value." The absoluteaalue of x, denoted by lrl, is defined by ifx>0 l*l-** ltl :-t l 0 l: 0 ifx<o 1.2ABSOLUTE VALUE 1 5 l-u - a: la br| Thus, the absolute value of a positive number or zero is equal to the number itself. The absolute value of a nggative numloer is the corresponding positive number because the negative of a negative number is positive. . I L L U s T R A T I1o:Nl 3 l : 3 ; l - 5 1 : - ( - 5 ) : 5 ; l,-b=la-bl1 lb F igure1.2.1 -L4l: l-61:-(-6):6. 18 We see from the definition that the absolute value of a number is either a positive number or zero;that is, it is nonnegative. In terms of geometry, the absolute value of a number x is its distance from 0, without regard to direction. In general, lo - bl is the distance between a and b without regard to direction, that is, without regard to which is the larger number. Refer to Fig. 1..2.1'. We have the following properties of absolute values. 1.2.2 Theorem l"l < a lf. and only if -a 1x I a,where a > "1..2.3 Corollary Irl = a if. and only if -a < )c < a, where a > \.2.4 Theorem ltl > a if andonly if x > a or x I -a,where 7.2.5 Corollary ltl = a if andonly if x > a or x s -a,where 0. 0. a > 0. a ) 0. The proof of a theorem that has ant"if and only if " qualification requires two parts, as illustrated in the following proof of Theorem 1.2.2. plnr l.: Prove that lrl < a if.-a < x I a,where a ) O.Here,we haveto consider two cases:.x > 0 and r < 0. Casel: x=0. Then ltl : x. Becauser < a,we concludethat lrl < a. C a s e 2 :x < 0 . and obtain Then lrl : -r. Because-a < x, we aPPlyTheorem L.1,.25 a)-x or, equivalently,-x < a.But because-x: lrl, we have l*l < o. In both cases,then, wherea>0 <n if -a<x(a, -a 1x I A, where a > 0. Here we panr 2: Prove that lrl < a only if must show that whenever the inequality ltl < a holds, the inequality -a < x 1a also holds. Assume that lrl < a and consider the two cases x>0andr(0. CaseT: x>0. Then ltl : r. Because lxl < a,we conclude that r < a. Also, because -a 10 a ) 0, it follows from Theorem 1.1.31 that-a < 0. Thus/ we have <x<a,ot-a<x1a. C a s e 2 :x < 0 . Then ltl : Because ltl < a, we conclude that -x ( a. Also, be- 16 REAL NUMBERS,INTRODUCTION TO ANALYTICGEOMETRY, AND FUNCTIONS cause x I 0, it follows from Theorem l.r.zs that 0 < -x. Therefore,we have -a < 0 ( -r < a, or -a <-x < a, which by applying Theorem 1..7.25 yields-a < x 1a. In both cases, onlyif -a<x1n, l*l<o wherea>0 I The proof of Theorcm7.2.4is left as an exercise(seeExercise27). The following examples illustrate the solution of equations and inequalitiesinvolving absolutevalues. EXAMPLE l-: Solve for x: lar+ 2l:5. solurroN: This equation will be satisfiedif either 3x*2:5 or -(3r*2):5 Considering each equation separatellr w€ have x: L and x: -& which are the two solutions to the given equation. uxluplr 2: Solve for x: lz*- rl: l4x+ 31. solurroN: This equation will be satisfiedif either 2x-L-4x*3 or 2x-1--Gx+j\ Solving the first equation, w€ have x: -2; solving the second,we get x: -*, thus giving us two solutions to the original equation. Exarnrpt,n3: Solve f.or x: l 5 r + 4 1: - 3 . rxr.vrpu 4: Find the solution set of the inequality l x - 5 1< + Illustrate it on the real number line. soLUTroN: Because the absolute value of a number may never be negative, this equation has no solution. solurroN: If r is a number such that lr-51 <a then -4 < x-S < 4 (by Theorem' 1,.2.2) Adding 5 to each member of the preceding inequality, we obtain 7<x<9 Because each step is reversible, we can conclude that lx-51 < ifandonlyif L<x<9 So, the solution set of the given inequality is (1, 9), which is illustrated in Fig. 1,.2.2. 1.2ABSOLUTE VALUE 17 EXAMPLE 5: Find the solution set of the inequatity =n l3=-,,*l l2+xl-' By Corollary solurroN '1..2.3, the given inequality is equivalent to - ! ,- -< . 3 =. 2 ' - 4 2* x If we multiply by 2 * x, we must consider two cases,depending uPon whether 2 t x is positive or negative. C a s e ' l - :2 * r > 0 o r x > . - 2 . Then we have -4(2*r) =3-2x=4(2+x) or, equivalently, -8- 4x <3-2x <8*4x So, if x ) -2, then also-8 - 4x < 3- 2x andS-2x < 8 + 4x.We solve these two inequalities. The first inequality is -8- 4x <3-2x Adding 2x * 8 to both members gives -2x < LL Dividing both members by .2 and reversing the inequality sign, we obtain x>-# The second inequality is 3-2x<8t4x Adding -4x - 3 to both members gives -6x <5 Dividing both members by -5 and reversing the inequality sign, we obtain x>-E Therefore, if x ) -2, x>-#andr>-8. then the inequality holds if and only if Becauseall three inequalities r ) -2, )c > -+, and r > -B must be satisfied by the same value of x, we have x > -8, or the interval [-8, +-). Case2: 2*x(0orx<-2. Thus, we have -4(2*r) =3-2x>4(2*x) or, equivalently, -8-4x>3-2x>814x 18 REALNUMBERS,INTROEUCTION TO ANALYTICGEOMETRY, AND FUNCTIONS Considering the left inequality, we have -8-4x>3-2x or, equivalently, -2x >'l-.'1, or, equivalently, rs-+ From the right inequality we have 3-2x>8*4x or, equivalently, -5x>5 or, equivalently, r s-8 Therefore,if x 1-2, and r = -8. the original inequality holds if and only if x = -t Because all three inequalities must be satisfied by the same value of x, we have x =-+, or the interval (-*, -u, l. Combining the solution sets of Cases 1 and 2, we have as the solution set(-o, -+l u [-8, **). rxaprprr 6: Find the solution set of the inequality l 3 r + 2 1> 5 soLUrIoN' By Theorem L.2.4,the given inequality is equivarent to 3x*2>5 or 3x*2<-s (1) That is, the givgn inequality will be satisfied if either of the inequalities in (1) is satisfied. Considering the first inequality, we have 3x*2>5 or, €guivalently, x)l Therefore, every number in the interval (1, +oo; is a solution. From the second inequality, we have 3xt2<-5 or, equivalently, x<-rs Hence, every number in the interval (-*, -+) is a solution. 1.2 ABSOLUTEVALUE The solution set of the given inequality is therefore (1,+*;. 19 _+)u You may recall from algebra that the symbol {a, where A > 0, is defined as the unique nonnegfltloe number x such that x2: A. We read tfi as "the principal square root of fl." For example, fr:2 \6:0 G:E NorE: tfr + -2; -2 is a square root of.4, bfi {4 denotesonly t]nepositiae square root of 4. Becausewe are concerned only with real numbers in this book, \/i is not defined lf. a < 0. From the definition of fr, it follows that G:lxl -{5zl':3. 5 and .... For example, {*: The following theorems about absolute value will be useful later. 1.2.6 Theorem lf. a and b are any numbers, then l a b l :l o l' l u l Expressedin words, this equation states that the absolute value of the product of two numbers is the product of the absolute values of the two numbers. PROOF: labl:'@ :{ffi :G'\E : l o l' l b l 7.2.7 Theorem lf a is any number and b is any number except 0, That is, the absolute value of the quotient of two numbers is the quotient of the absolute values of the two numbers. The proof of Theorem 1.2.7 is left as an exercise (see Exercise 28). 1..2.8Theorem The Triangle lnequality If. a and b are any numbers, then la+bl=lal+ lbl REALNUMBERS,]NTRODUCTION TO ANALYTICGEOMETRY, AND FUNCTIONS pRooF: By Definition '1..2.'1., either a : or a: _lol; thus -lol=a-lal Furthermore, -lbl =b-lbl (3) From inequalities (2) and (3) and Theorem 1.1.23, we have - ( l a l + l b l )= a + b < + lal lul Hence, from Corollary '/...2.3, it follows that l a + b l= l a l + l b l I Theorem r.2.8 has two important and prove. 1.2.9 Corollary corollaries which we now If a and b are any numbers, then lo-bl =lal+ lbl pRooF:lo- bl: la * (-b)l = lol+ l(-b) l: lal+ 1.2.10 Corollary If a and b arc any numbers, then l o l l b ls l a - b l pRooF: lol: l(r- b) + bl = lo - bl + lbl; thus, subtractinglbl both membersof the inequality,we have l o l - l b l= l o - b l '.Exercises1,.2** st , T -tarL, In Exercises1 through L0, solve fo, J l. l4x* 3l:7 4 . 1 4 +3 r l: 1 7.lTxl:4-x ,olffil:n Q ) l s t - 3 1: l 3 r + 5 1 g.2x*3:lar+51 3. 15- 2xl:11 6 . l x - 2 l: 1 3- 2 x l g.l'+31 :s zl lx- In ExercisesL1.through 14, find all the values of r for which the number is real. At.i s+ ,:: ,j state .t v ' Uttt.-l . 1.3 THE NUMBERPLANEAND GRAPHSOF EQUATIONS 21 >),23 fi-')fitu5 i tn In Exercises15 through 26, find the solution set of the given inequality, and illustrate the solution on the real number line. 1 5 .l r + 4 1 < 7 1 8 . 1 6- 2 x l > 7 !. lr+41= lzx-61 -) -a' l.1 3o + x- s l -r2=l 1 27. Prcve Theorem 1.2.4. 16. l2x-sl < 3 1 9 .l 2 x - s l > 3 22. l3rl > 16- }xl v*?l .n 2s.llffil 1 7 .l 3 x - 4 1- z 2 0 .1 3 + 2 x l < l a - r l 23.le- zxl =- l4xl rc I l1_l =1,-rr 26. !ffiI 28. Prove Theorem 1'.2.7. In Exercises29 through 32, solve for r and use absolute value bars to write the answer. 29. 31.. 12'8') thenlc-bl = l4l +lbl.(HINI:Writea-basa'+(-b) anduseTheorem 33. prcvethatif , andb areany nurnberg, 34. prcve that if a and b are any numben, then lal lbl < la bl. (Hrr.rr:Let lal: l(a b) + bl, and useTheorem1.2.8.) 35. What singleinequalityis equivatentto the followingtwo inequalities:s>b+ ca d a> b- c? 1.3 THE NUMBER PLANE AND GRAPHS OF EQUATIONS writing them in parentheses with a comma separating them as (I/ y). Note that the ordered pair (3, 7) is different from the ordered pair (7,3). Definition 1..3.1. The set of all ordered pairs of real numbers is called the number Plane' and each ordered Pafi (x, D is called a point in the number plane. The number plane is denoted by R2. fust as we can identify Rl with points on an axis (a one-dimensional space), we can identify R2 with points in a geometric plane (a two-dimendional space). The method we use with R2 is the one attributed to the French mathematician Rene Descartes (1595-1650), who is credited with the invention of analytic geometry in L637. A horizontal line is chosen in the geometric plane and is called the r axis. A vertical line is chosen and is called lhe y axis. The point of intersection of the r axis and tlne y axis is called the origin and is denoted by the letter O. A unit of length is chosen (usually the unit length on each axis is the same). We establish the positive direction on the r axis to the right of the origin, and the positive direction on the y axis above the origin. REAL NUMBERS,INTRODUCTION TO ANALYTICGEOMETRY, AND FUNCTIONS F i g u r e1 . 3 . 1 We now associate an ordered pair of real numbers (x, y) with a point P in the geometric plane. The distance of P from the y axis (considered as positive if P is to the right of the y axis and negative if P is to the left of the y axis) is called the abscissa(or x coordinate) of P and is denoted by *.The distance of P from the r axis (considered as positive if P is above r the x axis and negative if P is below the x axis) is called the ordinate (or y coordinate)of P and is denoted by y. The abscissa and the ordinate of a point are called the rectqngular cartesian coordinatesof the point. There is a one-to-one correspondence between the points in a geometric plane 7) and R2; that is, with each point there corresponds a unique ordered pair (x,y), and with each ordered pafu (x, y) there is associated only one point. This one-to-one correspondence is called a rectangular cartesian coordinate system, Figure 1.3.1 illustrates a rectangular cartesian coordinate system with some points plotted. The x and y axes are called the coordinateaxes,They divide the plane into four parts, called quadrants, The first quadrant is the one in which the abscissa and the ordinate are both positive, that is, the upper right quadrant. The other quadrants are numbered in the counterclockwise direction, with the fourth, for example, being the lower right quadrant. Because of the one-to-one correspondence, we identify R2 with the geometric plane. For this reason we call an ordered pair (x, y) a point. Similarly, we refer to a "line" in R2 as the set of all points corresponding to a line in the geometric plane, and we use other geometric terms for sets of points in R2. Consider the equation A:x2-2 (1) where (x,y) is a point inR2. we call this an equation inR2. By u solution of this equation, we mean an ordered pair of numbers, one for r and one for y, which satisfies the equation. Foi example, if r is re p l a c e d by 3 i n E q. (1), w e see that A :7; thus, x:3 and,y:7 const itutes a solution of this equation. If any number is substituted for x in the right side of Eq. (1), we obtain a corresponding value for y.It is seen, then, that Eq. (1) has an unlimited number of solutions. Table 1.3.1 gives a few such solutions. T a b l e1 .3 .7 x l0 a:x2-zl-z F i g u r e1 . 3 . 2 L2 -1 3 -3-4 4-1.-Z 2 7 74 -1 2 7 14 If we plot the points having as coordinates the number pairs (x,y) satisfying Eq. (1), we have a sketch of the graph of the equation. In Fig. 1'.3.2we have plotted points whose coordinates are the number pairs ob- 1.3 THE NUMBERPLANEAND GRAPHSOF EQUATIONS tained frorn Table 1.3.1. These points are connected by u smooth curve. Ary point (x, y) on this curye has coordinates satisfying Eq. (L). Also, the coordinates of any point not on this curve do not satisfy the equation. We have the following general definition. 1.g.2 Definition The graph of an equation in I€ is the set of all points (x, y) in R2 whose coordinates are numbers satisfying the equation. We sometimes call the graph of an equation the locus of the equation. The graph of an equation in R2 is also called a curae. Unless otherrryise stated, an equation with two unknowns, x and y, is considered an equation in R2. sxelrpr-n L: Draw a sketch of the graph of the equation' Y'-x-2:0 Q) soLUrIoN: Solving Eq. (2) for !, we have (3) y:!\tX+2 Equations (3) are equivalent to the two equations (4 ) Y:lx+2 (s) Y-_\'x-2 The coordinates of all points that satisfy Eq. (3) will satisfy either Eq. (4) or (5), and the coordinates of any point that satisfieseither Eq. (4) or (5) will satisfy Eq. (3). Table 1.3.2 gives some of these values of x and y. Table1-.3.2 x 00 v \/2 -\/2 1 \/i "1. 2 -\/5 2 2 -2 3 3-L-1-2 \/5 -\tr 1, -L 0 Note that for any value of r <-2 there is no real value for y. Also, for each value of x > -2 there are two values for y. A sketch of the graph of Eq. (2) is shown in Fig. 1.3.3.The graph is a parabola. REAL NUMBERS,INTRODUCTION TO ANALYTICGEOMETRY, AND FUNCTIONS 2: Draw sketchesof ExAMPLE the graphs of the equations y:\/x+2 and soLUrIoN: Equation (5) is the same as Eq. (a). The value of y is nonnegative; hence, the graph of Eq. (6) is the upper half of the graph of Eq. (3). A sketch of this graph is shown in Fig. 1..3.4. (5) Similarly, the graph of the equation Y:-\E+2 Y:-1/Yt'2 a sketch of which is shown in Fig. 1.3.5, is the lower half of the parabola o f F i g . 1 .3.3. F i g u r e1 . 3 . 4 F i g u r e1 . 3 . 5 rxltrpr.n 3: Draw a sketch of the graph of the equation soLUTroN: From the definition have y:lx*31 0) y:x*3 if of the absolute value of a number, we r*3>0 and y--(x+3) if r*3<0 or, equivalently, y:x*3 if x>-3 and y--(x+3) if x1-3 Table 1.3.3 gives some values of x and y satisfying Eq. (Z). T a b l e1 .3.3 Figure1.3.6 x 0 L 2 3 v 3 4 5 6 -L 2 -2 1 -3 0 -4 1, -5 2 -6 -7 3 -8 4 5 A sketch of the graph of Eq. (7) is shown in Fig. 1,.9.6. rxetvrplr 4: Draw a sketch of the graph of the equation (x-2y+3)(V-x'):0 (8) soLUTroN: By the property of real numbers that ab:O a : 0 o r b :0, w e have from E q. (8) x-2y*3:0 if and -9 6 1.3 THE NUMBERPLANE AND GRAPHSOF EQUATIONS y-xz:o (10) The coordinates of all points that satisfy Eq. (8) will satisfy either Eq. (9) or Eq. (L0), and the coordinates of any point that satisfies either Eq. (9) or (10) will satisfy Eq. (8). Therefore, the graph of Eq. (8) will consist of the graphs of Eqs. (9) and (10).Table 1..3.4gives some values of r and y satisfying Eq. (9), and Table 1.3.5 gives some values of x and y satisfying Eq. (10).A sketchof the graph of Eq. (8) is shown in Fig. 7.3.7. L.3.4 Table F i g u r e1 . 3 . 7 x 0L23-1-2-3-4-5 v 821r31,+o-+-1 T a b l e1 .3 .5 1.3.3 Definition x 01,23-7-2-3 v 0L49149 An equation of a graph is an equation which is satisfied by the coordinates of those, and only those, points on the graph. is an equation whose graph consists of o ILLUSTRATIoN1: In R2, y:8 those points having an ordinate of 8. This is a line which is parallel to the o x axis, and 8 units above the r axis. In drawing a sketch of the graph of an equation, it is often helpful to consider properties of symmetry of a graph. 1.3.4 Definition (3'2\o F i g u r e1 . 3 . 8 Two points P and Q are said to be symmetric with respect to a line if and only if the line is the perpendicular bisector of the line segment PQ. Two points P and Q are said to be symmetric with respectto a thirdpoint if and only if the third point is the midpoint of the line segment PQ. o rLLUsrRArroN 2: The points (3,2) and (3,-2) are symmetric with respect to the x axis, the points (3, 2) and (-3, 2) are symmetric with respect to the y axis, and the points (3,2) and (-3, -2) are symmetricwith o respect to the origin (see Fig. 1.3.8). In general, the points (x, y) and (x, -y) are symmetric with respect to the x axis, the points (r, y) and (-x, y) are symmetric with respect to the y axLS,and the points (x,y) and (-r, -y) arc symmetric with respect to the origin. REAL NUMBERS,INTRODUCTION TO ANALYTICGEOMETRY, AND FUNCTIONS 1.3.5 Definition are symmetric with respect to R. From Definition 1.3.5 it follows that if a point (x, y) is on a graph which is symmetric with respect to the x axis, then the point (x,-y) also must be on the graph. And, if both the points (x,y) and (x,-y) are on the graph, then the graph is symmetric with respect to the x axis. Therefore, the coordinates of the point (x,-y) as well as (x, y) must satisfy an equation of the graph. Hence, we may conclude that the graph of an equation in r and y is symmetric with respect to the r axis if and only if an equivalent equation is obtained when y is replaced by -y in the equation. We have thus proved part (i) in the following theorem. The proofs of parts (ii) and (iii) are similar. L.3.6 Theorem (Testst'or Symmetry) The graph of an equation in r and y is (i) symmetric with respect to the x axis if and only if an equivalent equation is obtained when y is replaced by -y in the equation; (ii) symmetric with respect to the y axis if and only if an equivalent equation is obtained when x is replaced by -x in the equation; (iii) symmetric with respect to the origin if and only if an equivalent equation is obtained when r is replaced by -x and y is replaced by -y in the equation. The graph in Fig. 1.3.2 is symmetric with respect to the y axis, and for Eq. (1) an equivalent equation is obtained when x is replaced by -r. In Example 1 we have Eq. (2) for which an equivalent equation is obtained when y is replaced by-y, and its graph sketched inFig. L.3.3 is symmetric with respect to the r axis. The following example gives a graph which is symmetric with respect to the origin. rxlrvrpr.r 5: Draw a sketch of the graph of the equation xy:1 solurroN: We see that if in Eq. (1,1)x is replaced by -x and y is replaced by -y,an equivalent equation is obtained; hence,by Theorem 1.3.6(iii) the graph is symmetric with respect to the origin. Table L.3.6 gives some (11) values of x and y satisfying Eq. (11). T a b l eL .3.6 1.3 THE NUMBERPLANEAND GRAPHSOF EQUATIONS 'l.lx. We seethat as r increasesthrough From Eq. (11) we obtain y : positive values and gets closer and positive values, y decreasesthrough positive values,y increasesthrough closerto zero.As x decreasesthrough positive values and gets larger and larger. As I increasesthrough negative values (i.e.,r takeson the values-4, -3, -2, -t, -+, etc.),y takes on negative values having larger and larger absolute values. A sketch of the graph is shown in Fig. 1..3.9. F i g u r e1 . 3 . 9 1.3 Exercises In Exercises 1 through 6, plot the dven Point P and such of the following PoinB as may aPPly: (a) The point O suitr ttrit ttre tnJ through Q and P is perpendicular to the r axis and is bisected by it. Give the coordinates of Q. (b) The poiniR such that the line through P and R is perpmdicular to and i6 bisected by the y axis. Give the coordinates of R. (c) The point s such that the line through P and 5 iE bisected by the origrn. Give the_coordin"F-"-9f s.by the als' line though the origin 1ai fn" poi"t T such that the line *Eough P and 7 is perpendicular to and is bisected of T. coordinates bisecting the first and thtd quadrants. Give the 1 . P ( 1, - 2 ) 4. P(-2, -2) 2. P(-2,2) 5 . P ( - 1, - 3 ) 3. P(2,2) 5 . P ( 0 ,- 3 ) In Exercises7 through2S, draw a sketch of the graph of the equation. 7'Y:2x*5 l0.y:-nffi 1 3 .x : - J 15.y:-lx+21 1 9 .4 f * 9 Y 2 : 3 6 22. Y' : 41cs 25. xz*y',:O 2 8 . ( y 2- x * z ) ( V + t E - 8'Y:4x-3 11.y':x-3 'I.. y2 * 1 7 .y : l r l - 5 2 0 . 4 x 2- 9 Y ' : 3 6 L4. x: 23. 4x2- Az:0 2 6 .( 2 x + V - l ) ( 4 V + r 2 ) : 0 Y: \81 y:5 y:lx-51 y--lxl+z A:4x3 3x2- L3xy- L0y2: g xa- 5x2yI 4y2: g +):O 29. Draw a sketch of the graph of each of the following equations: (a) V: lzx (b\ v : -{2x (c) y' :2x 30. Draw a sketch of the graph of each of the following equations: (a) Y: !-2x (b\ y : -!-2x (c\ y' : -2N REAL NUMBERS,INTRODUCTION TO ANALYTICGEOMETRY, AND FUNCTIONS 3L. Draw a sketch of the graph of each of the following equations: ( a )r * 3 y : 0 (c) r' - 9y' :0 (b)r-3y:0 32. (a) Write an equation whose gtaPh is the r axis. (b) Write an equation whose graph is the y axis. (c) Write an equafion whose graph is the set of all points on either the .x axis or the y axis. 33. (a) Write a,n__equ{iol whose graph consists of all points having an abscissa of 4. O) Write an equation whose graph consistE of all points having an ordinate of -3. 34. Prove that a graPh that i5 symmetric with respect to both coordinate axes is also symmetric with rcspect to the origin. 35' Pro-eethat a graPh that is symmeFic with rcspect to any two perpendicular lines is also sym.rretric with respect to their point of intersection- 1.4 DISTANCE FORMULA AND MIDPOINT FORMULA If A is the point (x,.,y) and B is the point (xr,y) (i.e., A and B havethe same ordinate but different abscissas), then the directed distance from A to B, denoted by AB, is defined ds x2 - x1. o ILLUSTRATToN 1: Refer to Fig. 1.4.1(a),(b), and (c). u B(r,z) A(4,2) AB:'/.,4 (b) Figur1 e.4.1 If .4 is the point (3, 4) and B is the point (9, 4), then AB :9 - 3: 6. rf A is the point (-8, 0) and B is the point (6,0), then B :6 - (-s) : 14. If A is the point (4, 2) and B is the point (L,2), then AB : 1.- 4: -3. we see that AB is positive if B is to the right of A, and AB is negative if B is to the left of A. o v D(-2,4) C(-2, -g) CD: -6 (a) Figure 1.4.2 CD:7 (b) If C is the point (x,yr) and D is the point (x, yr), then the directed distance from C to D, denoted by CD, is defined as!z!r. o rLLUsrRArroN 2: Refer to Fig. La.2@) and (b). If C is the point (L, -2) and D is the point (1, -8), then CD : -g -J-2)--6. If C is the point (-2, -3) and D is the point (-2,4), then (-3) :7. Tl r.enumber C p i s posi ti vei f D i s above C , and CD C D :4 is negative if D is below C. o We consider a directed distance AB as the signed distance traveled by u particle that starts at A(rr, y) and travels to B(x2, !). ln such a case, the abscissa of the particle changes from 11 to x2, dfld we use the notation Ar ("delta x") to denote this change; that is, A,x: xz- xt Therefore, AB: Lx. 1.4 DISTANCEFORMULAAND MIDPOINTFORMULA Pr(x", vz) It is important to note that the symbol Ar denotes the difference between the abscissa of B and the abscissa of A, and it doesnot mean "delta multiplied by x." Similarly, if we consider a particle moving along a line parallel to the y axis from a point C(x, yr) to a point D(x, yr), then the ordinate of the particle changes from Ar to Az.We denote this change by Ly ot Ly: az- ar Thus, CD: Ly. Now let Pr(r1, yr) and Pr(xr, a) be any two points in the plane. we wish to obtain a formula for finding the nonrtegative distance between these two points. We shatl denote this distanceby lPtPrl. W" use absolute-value bars becausewe are concernedonly with the length, which is a nonnegative number, of the line segmentbetween the two points P1and Pr. To derive the formula, we note that lffil is the length of the hypotThis is illustrated in Fig. 1.4.3for P1and of a right triangle PLMP2. "rrrrr" quadrant. first are in the Pr, both of which we have theorem, Using the Pythagorean F i g u r e1 . 4 . 3 +llvl' EF"f: lArl2 So lF-nl:\MV+WP That is, (1) lF:P,l: Formula (1) holds for all possible positions of Pt and P, in all four qqadrants. The length of the hypotenuse will always be lFfrl, and the lengths of the two legs will always be [Ar[ and lAyl (seeExercisesL and 2). We state this result as a theorem. '1,.4.1 Theorem The undirected distance between the two points Pr(xr, Ar) and Pr(xr, yr) is given by EXAMPLEl.: If a point P(x, U) is such that its distance from A(3,2) is always twice its distance from B(-4, L), find an equation which the coordinates of P must satisfy. solurroN: From the statement of the problem lPTl:2lPBl Using formula (1), we have @:z@ Squaring on both sides, we have x 2- 6 x + 9 + y ' - 4 y I 4 : 4(xz*8r or, equivalently, 3 x 2* 3 y ' * 3 8 r - 4 y + 5 5 : 0 * t6 + y2- 2y * 1) REAL NUMBERS,INTRODUCTION TO ANALYTICGEOMETRY, AND FUNCTIONS ExAMPLE2: Show that the triangle with verticesat A(-2, 4), B(-5, L), and C(-6,5) is isosceles. SOLUTION: The triangle is shown in Fig. L.4.4. lBel: @ : \ 4 + 1 6 : f r : @:\tr6+1:{lT la-e; IEAI:W : \ / N : j f i Therefore, A(-2,4) Hence, triangle ABC is isosceles. B(-5, i.) Figure1.4.4 nxervrrrn3: Prove analytically that the lengths of the diagonals of a rectangleare equal. C(0, b) (0,0) F i g u r e1 . 4 . 5 B(a,b) SOLUTION:Draw a generalrectangle.Becausewe can choosethe coordinate axes anywhere in the plane, and becausethe choice of the position of the axes does not affect the truth of the theorem, we take the origin at one vertex, the x axis along one side, and the y axis along another side. This procedure simplifies the coordinates of the vertices on the two axes. Referto Fig. 1..4.5. Now the hypothesis and the conclusion of the theorem can be stated. Hypothesis:OABC is a rectanglewith diagonalsOB and AC. Conclusion: IO-B| : lrel . l o B:lf f i : r t + 6 2 l A C: lW : \ f r ] A t Therefore, lml : ld-cl Let P, (xr, yr) and P, (xr, yr) be the endpoints of a line segment. We shall denote this line segment by PtPr. This is not to be confused with the notation PrPr, which denotes the directed distance from P, to Pr. That is, PrP, denotes a number, whereas PrP2is a line segment. Let P(x, y) be the midpoint of the line segment P,Pr. Refer to Fig. L.4.6. In Fig. I.4.6 we see that triangles P'RP andPTPrare congruent. Therefore, l-PtRl: lPTl, and so r - xr: xz- x, giving us 1.4 DISTANCEFORMULAAND MIDPOINTFORMULA Similarly, lRPl : lTPzl.Then A Ar: Az- A, andtherefore (3) Hence, the coordinates of the midpoint of a line segment are, respectively, the average of the abscissasand the average of the ordinates of the endpoints of the line segment. P z( x z ,v z ) P(x,y) u__ R(t, yt) T(xr, Y) __JS(rr,yr) F i g u re1 .4 .6 In the derivation of formulas (2) and (3) it was assumed that x, ) xt and y, ) yr.The same formulas are obtained by using any orderings of these numbers (see Exercises 3 and 4). nxaurr,E 4: Prove analytically that the line segments ioining the midpoints of the opposite sides of any quadrilateral bisect each other. B(b,c) F i g u r e1 . 4 . 7 Draw a general quadrilateral. Take the origin at one vertex solurroN: and the x axis along one side. This method simplifies the coordinates of the two vertices on the r axis. See Fig. L.4.7. Hypothesis: OABC is a quadrilateral. M is the midpoint of OA, N is the midpoint of CB, R is the midpoint of OC, and S is the midpoint of AB. Conclusion:MN and RS bisect each other. pRooF: To prove that two line segments bisect each other, we show that they have the same midpoint. Using formulas (2) and (3), we obtain the coordinates of M, N, R, and S. M is the point Ga, 0), N is the point ( + ( b + d ) , t r ( c * e ) ) ,R i s t h e p o i n t 1 ! d , t e ) , a n d S i s t h e p o i n t G @ + b ) , i c ) . + d). T h e a b s c i s s ao f t h e m i d p o i n t o f M N i s i l L a + + ( b + d ) l : * ( a + b : e). The ordinate of the midpoint of MN is +[0 + Lk + e)f ik + Therefore, the midpoint of MN is the point (ifu + b + d), ik + e)). + d). T h e a b s c i s s ao f t h e m i d p o i n t o f R S i s i l i a + * @ + b ) l : * ( a + b The ordinate of the midpoint of RS is tlie + *cf : Ik + e). Therefore, the midpoint of RS is the point (ifu + b + d), i(c + e)). Thus, the midpoint of MN is the same point as the midpoint of RS. I Therefore, MN and RS bisect each other. 92 REAL NUMBERS,INTRODUCTION TO ANALYTICGEOMETRY, AND FUNCTIONS Exercises 7.4 1. Derive distance formula (1) if Pr is in the third quadrant and P, is in the second quadrant. Draw a figure. 2. Derive distance formula (1) if Pr is in the second quadrant and P, is in the fourth quadrant. Draw a figure. 3. Derive midpoint formulas (2) and (3) if Pr is in the first quadrant and p, is in the third quadrant. 4. Derive midpoint formulas (2) and (3) iI4 (rr, yr) and &(rr, yr) are both in the second quadrant and x, > xrand,y, > yr. 5. Find the length of the medians of the rriangle having vertices A(2,3),8(3,-g), and C(-1,-1). 5. Find the midpoints of the diagonals of the quadrilateral whose vertices are (0,0), (0,4), (3,5), and (3, 1). (.\Prove that the trianSle with vertices A(3,-6), B(8,-2), and C(-1, -1) is a right triangle. Find the areaof the triangle. \'-l(HrNr: Use the converseof the Pythagoreantheorem.) 8. Prove that the Points A(6, -lS) , B(-2, 2), C(13, 10), and D(21, -5) are the vertices of a square. Find the tength of a diagonal. 9. By using distance formula (1), prove that the points (-3, 2) , (1, -2), and (9, -10) lie on a line. 10. If one end of a line segment is the point (-4, 2) and the midpoint is (3, -L), find the coordinates of the other end of the line segment. 11. The abscissa of a Point iB -6, and its distance from the point (1, 3) is V-74. Find the ordinate of the point. 12. Determine whether or not the poifis (14,7) , (2, 2) , and,(-4, -1) lie on a line by using distance formula (1). /;:\ '\3.llf two vertices of an equilateral hiangle are (-4,3) and (0,0), find the third vertex. d Findan equation that must be satisfied by the coordinatesof any point that is equidistant from the two points (-3, 2) and (4,6). 15. Find an equation that must be satisfied by the coordinates of any point whose distance from the point (5, 3) is atways two units Feater than it8 distance frcm the point (-4, -2). 16. Giver the two Points d(-3,4) andB(2,5), find the coordinates of a point P on the line through A and B such thatp is (a) twice as far from 4 as frcm B, and (b) twice as far frcm B as from _A. 17. Find the coordinates of the three points that divide the tine segment ftom l(-5, 3) to 8(6, g) into four equal parts. 18. If rr. and r, are positiv€_intgg:ers,-provethat the coordinates of the point p(r, y), which divides the line segment plp, : r,/rr-are given by in the ratio rr/r2-that i3, IP-,,PUlPrEI ,: (t" - rt)x, + rrxu fz"t2 ur4 , : (r, - ,r)y, + ,rv" In Exercises 19 through 23, use the formulas of Exercise 18 to find the coordinates oI point p. 19' The Point P is on the line segment between points Pr (1 ,3) ?flrd,P26,2) and is three times as far from P, as it b ftom pr. 20. The Point P is on the line segmentbetween points Pl(1 ,3) and Pr(6,2) and is three times as far from Pr as it is frcmpr. 21. The Point P i8 on the line through P' and P, and is three times as far ftom Pr(6, 2) as it is from P, (1,3) but is not between Pr and Pr. 22. The Point P is on the Une though P, and P, and is three times as far from P1(1, 3) as it is ftom Pr(6,2) but is not between P1and P2. 23. The point P is on the line through Pr(-3,5) and Pr(-L,2) so that ppr : + . p.,pr. 24. Find an equation whose graph is the circle that is the set of all points that are at a distance of 4 units from the point (1, 3). 1.5 EQUATIONSOF A LINE 25. (a) Find an equation whose graph consist8 of all points equidistant frcm the Points (-1,2) and (3,4). (b) Draw a skekh of the gnph of the equation found in (a)' 26. prove analytically that the sum of the squares of the distanc$ of any Point from two opposite vertic€s of any rcctangle is equal to the sum of the squares of its distances from the other two verticeE' 22. prcve analytically that the line segment ioining the midpoints of two opposite sides of any quadrilateral and the line segment joining the midpoints of the diagonala of the quadrilateral bieect each other' 28. prove analytically that the midpoint of the hypotenuse of any right triangle is equidistant frcm each of the three vettices. 29. Prove analytically that if the lengths of two of the medians of a triangle arc equal, the triangle is isoeceles. 1..sEQUATIONS OF A LINE Letl be a nonverticalline and Pr(x' U) andP"(xr, A)be any two distinct points on l. Figure L.5.1shows such a line. In the figure, R is the point Qc,Ur),and the points Pr, Pu and R are vertices of a right triangle; furthermore, P,.R: xz-xr and R-Pr: Az-Ar The number Az-Ar gives the measure of the change in the ordinate from Pr to Pr, and it may be positive, negative, or zero.The number x2- xl gives the measureof the change in the abscissafrom P, to Pr, and it may be positive or negative.The number x, - rr may not be zero becaus?x2 * r, since the line I is not vertical. For all choices of the points P, and P, on l, the quotient R(rz,yJ Az- Ur Xz- xr , is constanq this quotient is called the "slope" of the line. Following is the formal definition. F i g u r e1 . 5 . 1 1.S.1 Definition If Pr(xr, Ar) and Pr(xr, yr) arc any two distinct points on line l, which is not parallel to the y axis, then the slopeof l, denoted by m, is given by ; : y EXz * -at P, (xr, y, R(xr,Yr) Pr(7r,V,) (1) ?Cr In Eq. (L), x, * x1 since / is not parallel to the y axis. The value of m computed from Eq.(t) is independent of the choice of the two pointr_P, and P2 on L To show this, suppose we choose two different points, tr(n, yr) and Pr(ir,yr), and compute a number nr from Ee. (1). ar *m- :_- A - zxz- xt 7z,Ai x We shall show tlnilt n - m. Refer to Fig. 1.5.2. Triangles PrRPz and P1RP2 are similar, so the lengths of corresponding sides are proportional. Therefore, F i g u r e1 . 5 . 2 V:- Y_t- Az Xz- Xr or m:m - Xz- Ar Xt 34 REAL NUMBERS,INTRODUCTION T O ANALYTICGEOMETRY, AND FUNCTIONS Hence, we conclude that the value of m computedfrom Eq. (1) is the same number no matter what two points on I are selected. In Fig. 1.5.2, x2 ) x1, Uz) Ar, x, > n, and y2 > fr. The discussion above is valid for any ordering of these pairs of numbers since Definition 1.5.1holds for any ordering. In sec. 1.4 we defined Ly:uz-ur and Ar:xz rr. substituting thesevalues into Eq. (L), we have F i g u r e1 . 5 . 3 Multiplying on both sidesof this equationby Lx,we obtain Ly: mLx e) It is seen from Eq. (2) that if we consider a particle moving along line l, the change in the ordinate of the particle is proportional to the change in the abscissa, and the constant of proportionality is the slope of the line. If the slope of a line is positive, then as the abscissa of a point on the line increases, the ordinate increases. Such a line is shown in Fig. 1.5.3. In Fig. "1..5.4, we have a line whose slope is negative. For this line, as the abscissa of a point on the line increases, the ordinate decreases. Note that if the line is parallel to the x axis, then Uz: Ur and so m:0. If the line is parallel to the y axis, xz: xr, thus, Eq. (1) is meaningless because we cannot divide by zerc. This is the reason that lines paiallel to the y axis, or vertical lines, are excluded in Definition 1.5.L. We Jay that a vertical line does not have a slope. F i g u r e1 . 5 . 4 . rLLUSrRArroNl.: Let I be the line through the points P,,(2,3) and p2(4,7). The slope of I, by Definition "!..5.'!., is given by *:7n!:z L Pr $ , 9 ) Referto Fig. 1.5.5.rf P(x,y) and Q@+ L*,y + Lil are any two points on I, then L A_ . , Ax-' Ay :2 P:-(2,3) P+(-1 F i g u r e1 . 5 . 5 A,x Thus, if a particle is moving along the line l, the change in the ordinate is twice the change in the abscissa. That is, if the particle is at pr(4,7) and the abscissa is increased by one unit, then the ordinate is increased by two units, and the particle is at the point Pr(5,9). Similarly, if the particle is at Pt(2,3) and the abscissa is decreased by three units, then the ordinate is decreased by six units, and the particle is at Pr(-L, -3). o Since two points Pr(xr, yr) and Pr(xr, y) determine a unique line, we 1.5 EQUATIONS OF A LINE should be able to obtain an equation of the line through these two points. Consider P(x,y) any point on the line. We want an equation that is satisfied by r and y if and only rf P(x, D is on the line through P1(rr, yr) and Pr(xr, Ar).We distinguish two cases. C a s e7 : x z : x r . In this case the line through P, and P, is parallel to the y axis, and all points on this line have the same abscissa. So P(x,,U) ts any point on the line if and only if (3) X: Xr Equation (3) is an equation of a line parallel to the y axis. Note that this equation is independent of y; that is, the ordinate may have any value whatsoever, and the point P(x,y) is on the line whenever the abscissais rr. Case2: x2 * x1. The slope of the line through Pt and P, is given by Ar ry:Uzxz- xr lf P(x, g) is any point on the line except (\, given by (4) !r), the slope is also At 1--A x- xr (s) The point P will be on the line through P, and Pz if and only if the value of m fuomEq. (4) is the same as the value of m ftom Eq. (5), that is, if and only if U A t- A z - A r X- Xr Xz- Xr Multiplying on both sides of this equation by (x - xr),we obtain (6) Equation (5) is satisfied by the coordinates of P, as well as by the coordinates of any other point on the line through P1 and Pr. Equation (6) is called tkte two-point fonn of an equation of the line. It gives an equation of the line if two points on the line are known. -3) o rLLUsrRArroN 2: An equation of the line through the two points (5, and (-2,3) is y-(-3):*(r-5) y+3--2@-6) 3x*4Y:6 REALNUMBERS,INTRODUCTION TO ANALYTICGEOMETRY, AND FUNCTIONS If in Eq. (6) we replace(y, - !r)l(x, - x) by *, we get I i.:r )i' I.' + l- o : t I '; ... } i Yfl:gqt{,&Es'*"atu e) Equation (7) is called the point-slope form of an equation of the line. It gives an equation of the line if a point Pr(xr, a) on the line and the slope m of the line are known. . rLLUSrRArroN 3: An equation of the line through the point (-4,-5) and having a slope of 2 is y - (-s) :Zlx - (-4)l 2x-y*3:0 o If we choosethe particular point (0, b) (i.e., the point where the line intersectsthe y axis) for the point (h, A) in Eq. (7), we have y-b:m(x-0) or, equivalently, +b (8) The number b, which is the ordinate of the point where the line intersects the y axis, is called the y intercept of the line. Consequently, Eq. (8) is called the slope-intercept form of an equation of the line. This is especially important because it enables us to find the slope of a _form line from its equation. It is also important becauseit expressesttre y coordinate explicitly in terms of the r coordinate. rxevrpln L: Given the line having the equation 3r I 4y :7, find the slope of the line. sol.urroN: Solving the equation f.or !, we have ,:-fix*I Comparing this equation with Therefore, the slope is -*. Eq. (8), we see that and b:*. Another form of an equation of a line is the one involving the intercepts of a line. We define the r intercept of a line as the abscissa of the point at which the line intersects the r axis. The r intercept is denoted by o. If the r intercept a and the y intercept b arc given, we have two points (a, 0) and (0, b) on the line. Applying Eq. (6), the two-point form, we have h- 0 . Y-o:6ik- -aY:bx-ab b x* a A : a b a) OF A LINE 1.5 EOUATIONS by ab, if. a * 0 and b + 0, we obtain Dividing (e) Equation (9) is called the intercept fonn of an equation of the line. Obviously it does not apply to a line through the origin, becausefor such a line both a and b arc zero. nxlupln 2: The point (2,3) bisects that portion of a line which is cut off by the coordinate axes.Find an equation of the line. Refer to Fig. 1.5.6. lf. a is the r intercept of the line and b is the y intercept of the line, then the point (2, 3) is the midPoint of the line segment joining (a, 0) and (0, b). By the midpoint formulas/ we have soLUrIoN: A andt:ry atl A:4 and b:6 The intercept form, Eq. (9), gives us x +!-t 4'6 or, equivalently, F i g u r e1 . 5 . 6 3x*2y-12 t.5.2 Theorem The graph of the equation (10) Ax*By*C:0 where A, B, and C are constants and where not both A and B are zero, is a straight line. PRooF: Consider the two cases B + 0 and B : 0. CaseL: B+0. Because B + 0, we divide on both sides of Eq. (10) by B and obtain AC (11) y--i*-, Equation (11) is an equation of a straight line becauseit is in the slope-- -CIB. intercept form, where m: -AlB and b C a s e2 : B : 0 . BecauseB:0, Ax*C:0 we may concludethat A * 0 and thus have REAL NUMBERS,INTRODUCTION TO ANALYTICGEOMETRY, AND FUNCTIONS or/ equivalently, *:- aC (12) Equation (12) is in the form of Eq. (3), and so the graph is a straight line parallel to the y axis. This completes the proof. r Because the graph of Eq. (10) is a straight line, it is called a linear equation.Equation (10) is the general equation of the first degree in x andy. Because two points determine a line, to draw a sketch of the graph of a straight line from its equation we need only determine the cooidit utet of two points on the line, plot the two points, and then draw the line. Aty two points will suffice, but it is usually convenient to plot the two points where the line intersects the two axes (which are given by the intercepts). rxeupm 3: Given line lr, having the equation 2x - 3y : 12, and line Ir, having the equation 4x * 3y:6, draw a sketch of each of the lines. Then find the coordinates of the point of intersection of l, and Ir. solurroN: To draw a sketch of the graph of lr, we find the intercepts a and b. In the equation of Ir, we substitute 0 for r and get b - -4.In the equation of Ir, we substitute 0 for y and get a: 6. Similarly, we obtain the intercepts a and b for lr, and for l, we have a:8 and,b: 2. The two lines are plotted in Fig. 1..5.7. To find the coordinates of the point of intersection of /, and Ir, we solve the two equations simultaneously. Because the point must lie on both lines, its coordinates must satisfy both equations. If both equations are put in the slope-intercept form, we have a:?x-4 and -tx*2 Eliminating y gives 3x-4:-gx*2 2x-L2:-4x#6 x:3 So y:BG)_4 F i g u r e1 . 5 . 7 Therefore, the point of intersection is (3, -2). 1'5'3 Theorem If /r and l, are two distinct nonvertical lines having slopes m, and mr, respectively, then /, and l, are parallel if and only if mr: //t2. PRooF: Let an equation of /, be : tftrx * br, and let an equation of l, A be y : rnzx * br. Because there is an "if and only if " qualiiication, the proof consists of two parts. panr 1: Prove that l, and I, are parallel if ftir: tnz. 1.5E QU A TIONOF S A LIN E Assume that I, and I, are not parallel. Let us show that this assumPtion leads to a contradiction. If I, and I, are not parallel, then they intersect. Catl this point of intersection P(*o, /o).The coordinates of P must satisfy the equations of 11and Ir, and so we have ls: But mr: ls: !o: fti2xo* b2 !o: nttfio* b2 rnlxs* b1 and rrt2rwhich gives mfis * b1 and b2,both from which it follows thatbl: bz.Thus, becauserrtr: mrandbl: lines 11and /, have the same equation , A : mrx * br, and so the lines are the same. But this contradicts the hypothesis that l, and I, are distinct lines. Therefore, our assumption is false. So we conclude that /r and 12 are parallel. pARr 2: Prove that l, and I, are parallel only if mt: lftz. Here we must show that if Ir and l, are parallel, then rflt: rtrz. Assume that m, * mr. Solving the equations for \ and I, simultaneously, we get, upon eliminating V, m t x * b r : m 2 x* b 2 from which it follows that ( m , - m r ) x : b z- b t Because we have assumed that m, * *r, this gives - - _ _b r - b , L _ t7\- ffiz Hence, 11and l, have a point of intersection, which contradicts the hypothesis that l, and 12are parallel. So our assumption is false, and thereI fore mr: nflz. In Fig. 1.5.8, the two lines Ir and l, are perpendicular. We state and prove the following theorem on the slopes of two perpendicular lines. 1.5.4 Theorem If neither line /, nor line l, is vertical, then Ir and 12are perpendicular if and only if the product of their slopes is -1.. That is, if m, is the slope of 11and m, is the slope of. Ir, then Ir and I, are perpendicular if and only if n tfl tz - -1 . PROOF: - -1' Panr 1: Prove that l, and 12are perPendicular only if mtm2 Let L, be the line through the origin parallel to l, and let L, be the line through the origin parallel to Ir. See Fig. 1.5.8. Therefore,by Theorem 1..5.3,the slope of line L, ts m, and the slope of line L, is m2.Because neither REAL NUMBERS,INTRODUCTION TO ANALYTICGEOMETRY,AND FUNCTIONS L, nor L, rs vertical, these two lines intersect the line x: ! at points p, and Pr, respectively.The abscissaof both P, and P, is L. Let be the ordi/ nate of Pt. Since Lr contains the points (0,0) and (1.,y) and its slopeis mr, we have from Definition 1.5.1 ttL: a-0 i= and so y = mr.Similarly, the ordinate of P, is shown to be mr. Applying the Pythagorean theorem to right triangle prop2, we get l o t r ,+; zl @ f : l P , k l , F i g u r e1 . 5 . 8 (1 3 ) By applyingthe distanceformula,we obtain l o - 4 l r : ( 1- 0 ) ,f - ( m r - 0 ) ,: 7 * m ] l O t r l r : ( 1- o ) , * ( m r - o ) , : r * m z 2 l F J r P : ( 1 - 1 ) , * ( m r - m r ) ,- n t r z- 2 m r m r t m r , Substituting into Eq. (13)gives (l + mrz)+ (1 + m22)- ftizz- 2mrm,I.mrt 2: -2mrffiz tltinz - -1 penr 2: Prove that I, and l, are perpendicular if.mrm" - -1. Starting with lt|fitz - -'1, we can reverse the steps of the proof of Part 1 in order to obtain lO-&l'+ l1trl': lPEl'z from which it follows, from the converse of the Pythagorean theorem, that Lt andL, are pe{pendicular; hence, /, and l, are perpendicular. I Theorem I.5.4 states that if two lines are perpendicular and neither one is vertical, the slope of one of the lines is the negative reciprocal of the slope of the other line. o rLLUsrRArroN 4: If line /t is perpendicular to line I, and the slope of /, is 3, then the slope of /, must be -9. o sxarvrpr,n 4: Prove by meahs of slopes that the four points A(5,2), B(9,6), C(4,g), and DQ,4) are the verticesof a rectangle. solurroN: See Fig. 1.5.9. Let 6-2 ntr: the slope of AB - 8 - 6 8-6 m, J the slope of BC - 4- I 1.5 EQUATIONSOF A LINE /fi.:the sloPe of DC:ft: C(4,8) B(8,6) 41 '' rn.:the slopeof.AD:=:-+. Because D Tnr: mr, AB ll DC. /n2: ma,BC ll AD. ntrttrz--'1, AB L BC. F i g u r e1 . 5 . 9 Therefore, quadrilateral ABCD has opposite sides parallel and two adjacent sides perpendicular, and we conclude that ABCD is a rectangle. EXAMPLE5: Given the line / having the equation Because the required line is perpendicular to line l, its slope solurroN: must be the negative reciprocal of the slope of l. We find the slope of lby putting its equation into the slope-intercept form. Solving the given' equation for y, we obtain 2x*3Y-5:0 find an equation of the line perpendicular to line I and passing through the point Aet,3). y--tx** Therefore, the slope of I is -3, and the slope of the required line is 9. gecausewe also know that the required line contains the point (-L, 3), we use the point-slope form, which gives y-3:$(x*t) 2y-6:3x*3 3x-2Y+9:0 1-.5 Exercises In Exercises L through 4, find the slope of the line through the given points. 1 . ( 2 ,- 3 ) , ( - 4 , 3 ) 2 . ( 5 , 2 ), ( - 2 , - 3 ) 3.(+,+),eE,&) 4 . ( - 2 . T , 0 . 3 ), ( 2 . 3 , 1 . 4 ) In Exercises5 through14, find an equation of the line satisfying the given conditions. 5. The slope is 4 and through the point (2,-3). 6. Through the two points (3, L) and (-5,4). 7. Through the point (-3, -4) and parallel to the y axis. 8. Through the point (I,-7) and parallelto the x axis. 9. The x intercept is -3, and the y intercept is 4. TO ANALYTICGEOMETRY. REAL NUMBERS,INTRODUCTION AND FUNCTIONS 10. Through (1, 4) and parallel to the line whose equation is 2r - 5y + 7:0. 11. Through (-2, -5) and having a slope of l/5. 12. Through the origin and bisecting the angle between the axes in the first and third quadranb, 13. Through the origin and bisecting the angle between the axes in the second and fou h quadrants. 14. The slope is -2, and the r intercept is 4. the points (1, 3) and (2, -2), and put the equation in the intercept form. equation of the line through the points (3, -5) and (1, -2), and put the equation in the slope-interceptform. dg. Fitrd "tr 17. Show by means of slopes that the points (-4, -1) , (3, +) , (8, -4) , and (2, -9) are the vertices of a trapezoid. 15. Find an equation of the line thouth 18. Thrce consecutivevertices of a parallelogramare (-4, 1), (2,3),and (8,9). Find the coordinatesof the fou h vertex. 19. For each of the following sets of three point6, detemine by means of slopes if the points are on a line: (a) (2, 3), (-4,-7), (s,8);(b) (-3,6), (3,2), (e,-2); (cl (2,-7), (7,7), (3,t); and (d) (1,6), (1,2), (-s,-4). 20. Prove by means of slopes that the three points A(3, 1), B(6,0), and C(4, ) are the vertices of a right triangle, and find the area of the triangle. /\2T. Given the line I having the equation 2y - 3r : 4 and the point P(1, -3), find (a) an equation of the line through P " and perpendicul to l/(b) the shortest distance from P to line l. \ 22. lf A,B, C, and D arc cdhstarts, show that (a) the lines Ax + By+ C:0 a dAx+By+D:0 are parallel and (b) the ljmesAr+ By + C:0 and Bx- Ay * D:0 areperpendicular. 23. Given the line l, having the equation Ar + By + C: O, B + 0, find (a) the slope, (b) the y intercept, (c) the x intercept, (d) an equation of the line through the origin perpendicular to L 24. Find an equation of the line which has equal intercepts and which passesthrouth the point (8, -6). 25. Find equations o{ the three medians of the triangle having vertices A(3, -2) , BG, a), and C(-l,1), they m€et in a point. and prove that 26. Find equations of the perpendicular bisectors of the sides of the triande having vertices A(-l , -3) , B (5, -3), and C(5,5), and prove that they meet in a point. 27. Find an equation of each of the lines through the point (3,2), which forms with the coordinateaxesa triangle of area12. 28. Let il be the line having the equation.r4'r* Bly+ C1:0,andlet I'be the line having the equationArx+ 82! + Ca:0. If ll is not parallel to L and iJ & is any constant,the equation A'x 'l Bry I C' * k(A"x * Bry * C") :0 represents an unlimited number of lines. Prove that eadr of these lines contains the point of intersection df lr and t. 29. Given an equation ot It is 2x 'f 3y - 5: 0 and an equation of l, is 3x + 5y - 8 : 0, by using Exercise 28 and without finding the coordinates of the point of intersection of ,r and lr, find an equation of the line through this point of intersection and (a) passing through the point (1, 3); (b) parallel to the r axis; (c) parallel to the y axis; (d) having slope-2; (e) perpendicular to the line having the eqtatio\ 2x + y :7; (f) forming an isoscelestriangle with the coordinate axes. 30. Find an equation of each straight line that is perpendicular to the line having the equation 5r - y: I and that forms with the coordinate axes a triantle having an area of measure 5. 31. Prove analytically that the diatonals of a rhombus are perpendicular. 32. Prove analytically that the line segments joining conEecutivemidpoints of the 6ide8 of any quadrilateral form a parallelogram. 1.6THECIBCLE 43 33. Prove analytically that the diagonals of a parallelogram bisect each other. 34. Prove analytically that if the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram. 1.5 THE CIRCLE The simplest curve that is the graph of a quadratic equation in two variables is the "citcle," which we now define. '1,.6.1 Definition 1.6.2 Theorem A circle is the set of all points in a plane equidistant from a fixed point. The fixed point is called the center of the circle, and the measure of the constant equal distance is called the radius of the circle. The circle with center at the point C(h, k) and radius r has as an equation '; l -,: i : ;,(x *'h}P, * (g1= k)2,* 7zi'1 (1) pRooF: The point P(x, y) lies on the circle if and only if l n c l :, that is, if and only if @:r This is true if and only if (x-h)'+ (y-k)z--tz which is Eq. (1). Equation (1) is satisfied by the coordinates of those and only those points which lie on the given circle. Hence, (1) is an equation I of the circle. From Definition L.3.2, it follows that the graph of Eq. (1) is the circle with center at (h, k) and radius r. If the center of the circle is at the origin, then h: k: 0; therefore, its equation is x' + y' : 12. ' :-LJ-' C(2, - 3) o rLLUsrRArroN 1: Figure 1.6.1 shows the circle with center at (2, -3) and radius equal to 4. For this circle h:2,k:-3, and r:4. We obtain an equation of the circle by substituting these values into Eq. (1) and we obtain (x-2)'+ly - (-3)12:42 (x-2)'+ (y+3)':t6 Squaring and then combining terms, we have x2-4x+4+y'*5y*9:t6 F i g u r e1 . 6 . 1 )c'+y'-4x+6y-3:0 REALNUMBERS,INTRODUCTION TO ANALYTICGEOMETRY, AND FUNCTIONS EXAMPLE1: Given the equation x'+ y'* 6x-2y - 15: o prove that the graph of this equation is a circle, and find its center and radius. solurroN: The given equation may be written as (xr+6x)t(yr-2y):ts Completing the squares of the terms in parentheses by adding 9 and 1 on both sides of the equation, we have (x' * 6x*9) + (y' - 2y I1) : 15+ 9 + 1, (r*3)2+(V-1)':25 Comparing this equation with Eq. (1), we see that this is an equation of a circle with its center at (-3, I) and a radius of 5. In Eq. (L), removing parenthesesand combining terms gives x ' + y ' - Z h x- Z k y+ ( h , + k 2- r r ) : 0 Q) Equation (2) is of the form )e+y'*Dx+Ey*F:0 (3) -2h, -2k, : : : where D E and F h2* k2 f. Equation (3) is called the general form of an equation of a circle, whereas Eq. (1) is called the center-rsdiusform. Becauseevery circle has a center and a radius, its equation can be put in the center-radius form, and hence into the general form. We now consider the question of whether or not the graph of every equation of the form x'+y,*Dx*Ey*F:0 is a circle. To determine this, we shall attempt to write this equation in the center-iadius form. We rewrite the equation as x'+y,*Dx*Ey--F and completethe squaresof the terms in parenthesesby adding]nDzand iE' on both sides, thus giving us --F + +Dz* iE, (x' * Dxt +D')+ (y' * Ey++E'z) or, €guivalently, (x + LD)2+ (y + +E)2: i(D' + Ez- 4F) ( 4) Equation (a) is in the form of Eq. (1) if and only if +(D,+E2-4F):rz We now consider three cases,namely, (D'+ E2- 4F) as positive, zero, and negative. Case7: (D, + E2- 4F) > 0. Then r': *(D' + E2- 4F), and so Eq. (a) is an equation of a circle having a radius equal to it/FE= 4F and its center at (-*D,-+E). 1.6 THE CIRCLE C a s e2 : D z + E z- 4 F : 0 . Equation ( ) is then of the form (x + tD)z + (Y * iE)z :0 (5) - -L2D and Becausethe only real values of r and y satisfying Eq. (5) are v -+E). Comparing Eq. (5) with y:-+E, the graph is the point (-+D, and r - 0. Thus, this point can be Eq. (1),we see that h:-+D, k:-iE, called a point-circle. Case3: (Dt + E2- 4F) < 0. Then Eq. (4) has a negative number on the right side and the sum of the squaresof two real numbers on the left side. There are no real values of r and y that satisfy such an equation; consequently,we say the graph is the empty set. Before stating the results of these three casesas a theorem, we observe that an equation of the form Axz*Ay'+Dx*Ey* F:0 w h e r eA + 0 (6) can be written in the form of Eq. (3) by dividing by A, thereby obtaining F r c ' + y ' +Dt r *, *E= A,U +i:0 Equation (6) is a special case of the general equation of the second degree: Axz * Bxy * Cy' t Dx * Ey * F : 0 in which the coefficients of x2 and yz are equal and which has no xy term. We have, then, the following theorem. 1.6.3 Theorem The graph of any second-degree equation in R2 in r and U, in which the coefficients of x2 and y2 are equal and in which there is no xy term, is either a circle, a point-circle, or the empty set. o rLLUSrRArroN 2: The equation 2x2*2y'*LZx-By+31:0 is of the form (6), and therefore its graph is either a circle, a point-circle, or the empty set. If the equation is put in the form of Eq. (1), we have x'+y't6x-4y-F#:o ( x ' + 6 x )+ ( y ' - 4 y ) : - T ( x ' * 6 x* 9 ) + ( y ' - 4 y + 4 ) : - # + (r*3)2 + (Y-2)':-E Therefore,the graphis the empty set. 9+ 4 REAL NUMBERS,INTRODUCTION TO ANALYTICGEOMETRY, AND FUNCTIONS ExAMPLE2: Find an equation of the circle through the three points A(4,5), B(3, -2), and c(1,,-4). soLUrIoN: The general form of an equation of the circle is x'+y,tDx-fEy*F:0 Because the threepoints A, B,and C must lie on the circle,the coordinates of these points must satisfy the equation. So we have L6+25+4D+5E+F:0 9+ 4+3D-2E*F:0 1+16+ D-4E*F:0 or, €guivalently, 4D+ 5E+F:-47 3D-2E+F:-L3 D - 4 E* F : - 1 7 Solving these three equations simultaneously, we get D:7 E:-5 F:-44 Thus, an equation of the circle is x'+y'*7x-5y-44:0 In the following example we have a line which is tangent to a circle. The definition of the tangent line to a general curve at a specific point is given in Sec. 3.L. However, for a circle we use the definition from plane geometry which states that a tangent line at a point P on the circle is the line intersecting the circle at only the point P. nxeuprr 3: Find an equation of the circle with its center at the point C(1,6) and tangent to the line / having the equation x-v-1:0. soLUrIoN: See Fig. L.6.2. Given that h: L and k: 6, if we find r, we can obtain an equation of the circle by using the center-radius form. Let I, be the line through C and the point P, which is the point of tangency of line I with the circle. r: lPCl Hence, we must find the coordinates of P. We do this by finding an equation of /, and then finding the point of intersection of /1 with /. Since 11 is along a diameter of the circle, and / is tangent to the circle , I, is perpendicular to l. Because the slope of / is L, the slope of 11is -L. Therefore, using the point-slope form of an equation of a line, we obtain as an equation of I, y-5--1(r-1) 1,6THE CIRCLE or, equivalently, x+Y-7:0 Solving this equation simultaneouslywith the given equation of l, namely, X- A- L:0 we get x: 4 and y: 3. Thus, P is the point (4, 3). Therefore, ,:l-PCl:ffi or, equivalently, r: !18 So, an equation of the circle is (x-l)'+ (Y-6)': (\ft)' or, equivalently, F i g u r e1 . 6 . 2 x ' + y ' - 2 x - 1 2 y* L 9 : 0 1.6 /- )Exercises In Exercises 1 through 4, find an equation of the circle with center at C and radius r. Write the equation in both the centerradius forrr and the gmeral form. ' , . c ( Q ,0 ) , r : 8 r . c 1 4 , - 3 ) ,r : S 3. C(-5, -12), f :3 4. C(-l,l),r:2 In Exercises5 through 10, find an equation of the circle satisfying the given condition$. 5. Center is at (1, 2\ and through the point (3, -1). 5. Center is at (-2,5) and tangent to the line x:7. is at (-3, -5) and tangent to the line l}x t 5y - 4:0. C"n /Z) w "r 8. Through the three points (2, 8) , (7, 3), and (-2,0). 9. Tangent to the line 3x * y * 2:0 at (-L, 1) and through the point (3, 5). 10. Tangent to the line 3x * 4y - 16:0 at (4,1.) and with a radius of 5. (Two possiblecircles.) In ExercisesL1 through 14, find the center and radius of each circle, and draw a sketch of the graph. ll. f*y'-6x-8y*9:0 13. 3f * 3y' * 4y - 7 :0 12.2f*2y'-2x*2y*7:0 1 4 .* * y ' - 10r-lOy+25:0 In ExercisesL5 through 20, determine whether the graph is a circle, a point-circle, or the empty set. 16. 4* l- 4y' * 24x - 4y t L : 0 1 5. 12* y'- 2x * LO y* L9:0 1 7 .f * y ' - 10r* 6y*36:0 1 8 .x 2 * y ' * 2 x - 4 y *5:0 48 REALNUMBERS, INTRODUCTION TO ANALYTICGEOMETRY, AND FUNCTIONS 19. *+ f -{f.x+36v-719:0 20. W 1 9 f I 6 x - 6 y + 5 - 0 21. Find an equation of the comrnon chord_of the two cirdes f* - 6y - 12:0 and f* 2yrg:o. 3F+rlt l+gr(rrrrvr: If the coordinates of a Point gatisfy two different equations, then thjcoordinates also satisfy the difierence of the two equatione.) 22. Find the points of intersection of the two cirdes in Exercise 21. i\ Find an equation of the line which is tangent to the circle .d + y, - 4t + 6y - 12: O at the point (5, 1). 24. Findanequationofeachofthetwolineshavingslope-jgwhicharetangenttothecirclef*l*2x-W-g:0. I 25. From the origin, dbrda of the cirde t' * I * 4r: 0 are drawn. Pr€ve that the set of midpoinb of these chords is a circle. 26. Prove analytically that a line from th€ center of any circle biEecting any drord is perpendicular to the chord. 27. Prcve analytically that an angle inscribed in a semicircle is a right angle. 28. Given the line y: flr + b tangent to the circle rP * y, = r2, find an equation involving m, b, and t 1.7 FUNCTIONS AND We intuitively consider y to be a function of r if there is some rule by THEIR GRAPHS which a unique value is assignedto y by a correspondingvalue of x. Familiar examples of such relationships are given by equations such as y:2f+5 (1) Y:tP-g (2) and It is not necessary that r and y be related by an equation in order for a functional relationship to exist between them, as shown in the following illustration. o ILLUsTRATIoN1,: lf y is the number of cents in the postage of a domestic first class letter, and if r is the number of ounces in the weight of the letter, then y is a function of x. For this functional relationship, there is no equation involving r and /; however, the relationship between r and y may be given by means of a table, such as Table 1.7.L. . T a b l e1 . 7 . 1 r: number of ounces in the weight of the letter x <3 y: number of cents in first classpostage The formal definition 1.7.1 Definition makes the concept of a function precise. A function is a set of ordered pairs of numbers (r, fi in which no two A N D T H E I RG R A P H S 1.7 FUNCTIONS distinct ordered pairs have the same first number. The set of all possible values of r is called the domain of the function, and the set of all possible values of y is called the range of the function. In Definition 1.7.1, the restriction that no two distinct ordered pairs can have the same first number assures us that y is unique for a specific value of r. :2x2+ 5} f : { ( x ,y ) l y The numbers x and,y are called aariables.Because for the function / values are assigned to x and because the value of y is dependent upon the choice of x, we call x the independent aariable andy the dependentaariable' The domain of the function is the set of all possible values of the independent variable, and the range of the function is the set of all possible values of the dependent variable. For the function / under consideration, the domain ii tne set of all real numbers which can be denoted with interval notation as (-m, +€). The smallest value that y can assume is 5 (when The range of f is then the set of all positive numbers greater than r:0). or equal to 5, which is [5, +m). o rLLUsrRArroN 2: Let g be the function which is the set of all ordered pairs (x,y) defined by Eq. (2); that is, g : { @ ,Y ) l y : t / T vY Because the numbers are confined to real numbers, ! is a function of r only for x > 3 or x <-3 (orsimply lrl = 3) because forany rsatisfying either of these inequalities a unique value of y ts determined. However, if x is in the intenial (-9, g), a square root of a negative number is obtained, and hence no real number y exists. Therefore, we must restrict r, and so g : { G , y ) l y : t / 7 - g a n dl r l > 3 } The domain of g is (-o, -31 U [3, *-), and the range of g is [0, +-;' o It should be stressed that in order to have a function there must be exactly one aalue of the dependent variable for a value of the independent variable in the domain of the function. 1.7.2 Definition If f is a function, then the graph of f is the set of all points (x,y) in R2 for which (x,y) is an ordered Pafu in f . REALNUMBERS,INTRODUCTION TO ANALYTIC GEOMETRY, AND FUNCTIONS Hence, the graph of a function is a curve which is the set of all points in R2 whose cartesian coordinates are given by the ordered pairs of numbers (x, y). Because for each value of r in the domain of the function there corresPonds a unique value of y, no vertical line can intersect the graph of the function at more than one point. o rLLUSrRArroN 3: Let f : {(x, y)ly : \6=}. A sketch of the graph of f is shown in Fig. 1.7.1,.The domain of is the set of all real numbers less / than or equal to 5, which is (-*,5], and the range of is the set of all non_ / negative real numbers, which is [0, +oo). o ' rLLUSrRArroN 4: Let I be the function which is the set of all ordered pairs (x,y) such that F i g u r e1 . 7 . 1 -3 y:[ 1 4 if x <-1. if -1 < x <2 if2<x The domain of g is (-m, *m), while the range of g consists of the three numbers -3, 1, and 4. A sketch of the graph is shown in Fig. 1,.7.2. o observe in Fig. r.z.z that there is a break at x:-L and another at x: 2. we say that g is discontinuous at -7 and 2. Continuous and discontinuous functions are considered in Sec. 2.5. o TLLUSTRATToN 5: Consider the set { ( x ,y ) l * ,+ y , : 2 5 } This set of ordered pairs is not a function because for any x in the interval (-5, 5) there are two ordered pairs having that number as a first element. EXAMPLE Let h : { ( x ,y ) l y: l * l } solurroN: The domain of h is (--, +*), and the range of h is [0, +*;. A sketch of the graph of h is shown in Fig. 1,.7.9. v Find the domain and range of h, and draw a sketch of the graph of h. F i g u r e1 . 7 . 3 1.7 FUNCTIONSAND THEIRGRAPHS 51 EXAMPLE2: Let F be the function which is the set of all ordered pairs (x, Y) such that (3x-2 U:l* A sketchof the graph of F is shown in Fig. l.7.A.The domain of F is (-m , **), and the range of F is (--, +m). SOLUTION: if.x<"1' ifL<x Find the domain and range of-F, and draw a sketch of the graPh of F. F i g u r e1 . 7 . 4 nxnvrrr,n 3: Let G be the function which is the set of all ordered pairs (x, Y) such that u :" x2-9 x- soLUrIoN: A sketch of the graph is shown in Fig. 1'.7.5-Because a value for y is determined for each valrue of r except x: 3, the domain of G conboth the numerator and sists of all real numbers except 3. When i:3, undefined. denominator are zero, and 0/0 is - 3) (r + 3), we obtain Factoring the numerator into (r J Find the domain and range of' G, and draw a sketch of the graPh of.G. Y- (x-3)(r+3) e1l or y: x* 3, provided that r # 3.In other words, the function G consists of all ordered pairs (x,y) such that y-x+3 and x*3 The range of G consists of all real numbers excePt6. The graph consists of all points on the line y : x * 3 except the point (3, 6). F i g u r e1 . 7 . 5 REAL NUMBERS,INTRODUCTION TO ANALYTICGEOMETBY, AND FUNCTIONS 4: Let H be the funcEXAMPLE tion which is the set of all ordered pairs (x, y) such that lx*3 A:lz ifx*3 ifx_3 solurroN: A sketch of the graph of this function is shown in Fig. 1.2.6. The graph consists of the point (9, 2) and all points on the hne y - x * 3 exceptthe point (3,5). Function H is defined for all valuesof x, and.therefore the domain of H is (-co,+*). The range of H consistsof all real numbers except6. u Find the domain and range of H, and draw a sketch of the graph of H. F i g u r e1 . 7 . 6 nxenapr,n5: Let d be the function which is the set of all ordered pairs (x,y) such that ,r , _ ( x 2 * 3 x - 4 ) ( f - 9 ) (x'+x-12)(r+3) Find the domain and range of O, and draw a sketch of the graph of 0. v soLUrIoN: A sketch of the graph of this function is shown in Fig. 7.7.7. Factoring the numerator and denominator, we obtain ,' , _ (x * 4) (r- 1)(x - 3)(x + 3) (x*4)(x-3)(r+3) We see that the denominator is zero f.or x: -4,-3, and 3; therefore, j, @ is undefined for these three values of x. For values of.x * -4,-g,or we may divide numerator and denominator by the common factors and obtain !:x-I if x -3, ot 3 Therefore, the domain of f is the set of all real numbers except, -3, and 3, and the range of Q is the set of all real numbers except those values of (x - 1,) obtained by replacing r by -4,-3, or 3, that is, all real numbers except -5,-4, and 2. The graph of this function is the straight line y - x - 1, with the points (-4, -5), (-3 -4), and (3, 2) deleted. F i g u r e1 . 7 . 7 1.7 FUNCTIONSAND THEIRGRAPHS EXAMPLE6: Let f be the function which is the set of all ordered pairs (x,y) such that (x' A:lz tfx*2 soLUrIoN: A sketch of the graph of / is shown in Fig. 1'7'8' The graph : consistsof the point (2, 7) and all points on the parabola A x2, except the point (2,A)."Function/ is defined for all values of.x, and so the domain of / is (-*, +*). The range of f consistsof all nonnegirtivereal numbers. ifx-2 Find the domain and runge of t', and draw a sketch of the graPh of f. F i g u r e1 . 7 . 8 7' Let h be the funcnx.s.lrpr-n tion which is the set of all ordered pairs (x,y) such that I x-L a:lzx+t If x<3 if3<x The domain solurroN: A sketchof the graph of.h ts shown in Fig. 1,.7.9. of h is (-*, +oo).The values of y are either less than 2 ot greaterthan or equal to 7. So the range of h is (-*,2) U 17,+*7 or, equivalently,all real numbers not in 12,7). u Find the domain and range of h, and draw a sketch of the graPh of.h. F i g u r e1 . 7 . 9 REAL NUMBERS,INTRODUCTION TO ANALYTICGEOMETRY, AND FUNCTIONS rxaupr.n 8: Let 8: { @,y)ly: t/i64} Find the domain and range of g, and draw a sketch of the graph of 8. solurroN: Because\RFT is not a rcal number when x(x - 2) < 0, the d-omainof g consistsof the values of x for which x(x - 2) > 0. This in_ equality will be satisfied when one of the following . x> 0 a n d r - 2 > 0 ; o r x < } a n d x - 2 < 0 . two cases holds: CaseT: r>0andr-Z>0. That is, x>0 and x>2 Both inequalities hold if x > 2, which is the interval [2, +*;. Case2: x<0andx-2<0. That is, x<0 and x<2 F i g u r e1 . 7 . 1 0 In the next illustration, the symbol [r]] is used. [r]l is defined as the greatest integer which is less than or equal to r; that is, [*n-n ff n<tc<n*l where n is an integer This function is called thegreatestinteger function It follows that [1]l : 1, [1.3n:1, [+]l:0, [-4.2n:-5, [-8]l :-8, [9.8n:9, and.so on. o rLLUSrRArroN 6: Ler F be the greatest integer function; that is, F: { (x , y ) l y : fi rn]. A sketch of the graph of F i s show n i n Fi g. 1.7. 11.The following values are used to draw the sketch. 5 4 2 -4<x<-3 -5-4 -3-Z-7 -1 -2 -3 -4 -J F i g u r e1 . 7 . 1 lrn flxn: firn : [x]l - -5<x<-4 3 -D 2345 -3<x1-2 -2<x<-'j, -L<x( -s -4 -3 -z 0 [x]J : -t 0<x< 1. fxn: o 7<x< 2 l[r]l : 1 2<x1 3 [t]l : 2 3<x< 4 ["n: 3 4<x1 5 [r]l - 4 A N D T H E I RG R A P H S 1.7 FUNCTIONS The domain of F is the set of all real numbers. The range of F consists of all the integers. SOLUTION: EXAMPLE9: Let G:{(x,y)ly:firn-x] Draw a sketch of the graPh of G. State the domain and range of G. v 0; ["]l : : L fitn : 2 ; [tJl fixn:-1; If 0 < x 1!, If L < x <-2, If 2 < x 13, lf-l<x--o' so so so so G(r) :-x' G(x):"1- x' G(r) :2- x' G(x):-l-x' and so on. A sketchof the graph is shown in Fig. L.7.L2.The domain of G is the set of all real numbers,and the rangeof G is (-1,0] F i g u r e1 . 7 . 1 2 1 s. 7 t . ' l Exercise the gfaPh of the function' In Exercises 1 through 10, find. the domain and range of the given function, and draw a sketch of 3. F: {(x,y)ly:3x' - 6} 2.g:{(x,y)ly:x'+2} y ) l y: 3 x - 1 } 1. : f {(x, 4. G: {(x, y)ly : t/iT17 7 . g : { @ ,y ) l y : t / P - + l ( S . h : { ( x ,y ) l y: f t x - + } 8. H: {(x,y\ly:l*-31} 6 .f : { ( x ,y ) l V : { a ; F } 9 . 6 : { G ,y ) 1:, l 3 x+ 2 l } 4t'-11 r o .F : I e , y ) t v : 2 * r J Find the domain In Exercises11 through 34, the function is the set of all ordered pairs (r, y) satisfying the given equation the functlon' and range of the function, ard draw a sketch of the graPh of It G:Y:l2 (-2 t> ifr<3 if 3<x [x*5 1 , 4g. : V : \ \ t r 5 = 7 Lr-5 1 . -:.@ - ( "r ;+ t ) ( f I /n. t :, a ifr<-5 if -s <x<5 if5<t_, \ +3r-10) i, 2 0 .h : u : { F 6 4 23.h: y: ' f*5x2-6x-30 x*5 26g . :y:lxl 'lr-11 l-4 ifx<-2 1f-2<x<2 12.h:y:l-l t 3 if.Z<x (x2-4 15.H:v:lir_i if3<x ( v 2 r -e r - 4 ) ( f zL. 8iu: ifx*2 if x :2 fzx-t lo [6x*7 ro'8:Y:I4-x ifr<3 18.G:V:ffi . 1 5 .I i ' : -5x*6) xr - 2x2 *_2 if'x=-2 if-2<x 1 9 .f : y : 22.f: y: x*3 24.F:y:T 25.f: y: lxl+ lr - 1l 2 7 .G : y : x - f l t n 28F . :!:[x+2n xa*x3-9x2-3rf18 REAL NUMBERS,INTRODUCTION TO ANALYTICGEOMETRY, AND FUNCTIONS 29.h:y: [x'n 30.H: y: lxl + flrn 32.f: U: (x- [rn)' 3 3 .f : U : 2 + ( - 1 ) " , w h e r en : [,xn 1.8 FUNCTIONNOTATION, If f is the function having as its domain values of x and as its range values OPERATIONS ON FUNCTIONS, of y ,the symb ol t'@) (read "f oI r") denotes the particular value of y which AND TYPESOF FUNCTIONS corresponds to the value of x. o ILLUSTRATToN 1: In Illustration 3 of Sec. "1..7, f : { ( x ,y ) l y: l i - , } Thus, we may write f (x) - \6=. Becausewhen )c: l, ti=:2, we write f(1):2. Similarly,f(-6): \/TT, t'e): t6, andso on. . When defining a function, the domain of the independent variable must be given either explicitty or implicitly. For instance,if we are given that f is defined by f(x):3x2-5x*2 it is implied that r can be any real number. However, if we are given that / is defined by f (x):3x2-5x*Z I < x < 10 then the domain of / consists of all real numbers between and including 1 and 10. Similarly, if g is defined by the equation / \ 5x-2 8(x/: x+4 it is implied that x # -4, becausethe quotient is undefined for x: hence, the domain of g is the set of all real numbers except--4. If we are given -4; h(x) - \/e -E it is implied that r is in the closed interval -3 < x = J, because t is undefined (i.e., not a real number) for x ) 3 or tr < -3. So the domain of h is [-3, 3], and the range of h is [0, 3]. Exevrpr.n1: Given that / is the function defined by f (x) - tsz I3x - 4, find: (a) f (0); (b) f (2); (c) f (h); (d) f Qh); (e) f (2x); (t) f(x + h);(d f @)+ f(h). SOLUTION: (a) /(0) :02 * 3 . 0 - 4 - - 4 (b) f(2):221-3' 2-4:6 (c)f(h):h2+3h -4 AND TYPESOF FUNCTIONS ON FUNCTIONS, 1.8 FUNCTIONNOTATION,OPERATIONS ( d ) f ( z h ) : ( 2 h ) 2+ 3 ( 2 h ) - t - 4 h 2+ 6 h - 4 (e) f(zx) : (zx)z * 3(2x) - 4 - 4x2t 6x - 4 ( f ) f ( x + h ) : ( r + h ) ' * 3 ( . r+ h ) - 4 :x2*2hx*h2*3x+3h-4 : ) c zI ( 2 h + 3 ) x + ( h ' + 3 h - 4 ) 4) (g)/(r)+f(h)::9'lrtJ* ,nll:I':3r- SOLUTION: s G + h ) - g , k ) \MT6=1-\ffi h h \tri1)6ffi=+ 6ffi58=uNWT.+tF) h+0 (3x+3h-L)- (3x-L) Vgr-1) h(\M.+ : t6_1) 3h h(\triT5T=+ ..ffi) 3 16;agy-r+ \E- In the second step of this solution, the numerator and denominator were multiplied by the conjugate of the numerator in order to rationalize the numerator, and this gave a common factor of h tn the numerator and the denominator. We now consider operations (i.e., addition, subtraction, multiplication, division) on functions. The functions obtained from these operathe sum, the difference, the product, and the quotient of tions-called defined as follows. the original functions-are L.8.1 Definition Given the two functions f and g: (i) their sltm, denoted by f + g, is the function defined bV U +SXr) :f(x)+g(x); (ii) their difference,denoted by the function defined bv (f-g(x):f(x)-B@); (iii) their product,denotedby f '9, is the function definedbV (/' gXr) :f(x)'B@); (iv) their quotient,denoted by flg, is the function defined by (flil@) : f (x)lg(x). TO ANALYTICGEOMETRY, REAL NUMBERS,INTRODUCTION AND FUNCTIONS In each case the domain of the resulting function consists of those values of x common to the domains of.f and g, with the exception that in case(iv) the values of x for which g(r) - 0 are excluded. rxeupr.n 3: Given that / is the function defined by f(x):\/x+Landgistne function defined by 4, find: (a)(f + gXr); B@): {r (b)(/ - s)(r);(c)(/ ' s)(x); G) (flS)(x).In each case/determine the domain of the resulting function. SOLUTION: ( a )( f + i l ( r ) : \ E + 1 + \ t x - 4 ( b )( f - i l k ) : \ / x + I - . / x - 4 ( c )( f . i l ( r ) : I x + 1 . { x - 4 \/# r (d) (fls)(r)- \tr-4 The domain of f is 11,, **), and the domain of g is [4, +*;. So in parts (a), (b), and (c) the domain of the resulting function is [4, *m). In part (d) the denominator is zero when x: 4; thus, 4 is excluded from the domain, and the domain is thereforc (4, +oo). To indicate the product of a function / multiplied by itself, or f . f , we write f 2. For example,if f is defined by f (x) : 3x, then f it the function defined by f '(x): (3x)(3r):9x2. In addition to combining two functions by the operations given in Definition L.8.L,we shall consider the "composite function" of two given functions. 1.8.2 Definition Given the two functions f and g, the composite function, denoted by f " g, is defined by (/"s)(r): fk?)) and the domain of f . g is the set of all numbers r in the domain of g such that g(x) is in the domain of f . rxervrrr.n 4: Given that / is defined by f (x) : \G and g is defined by g(x) :2x - 3, find F(x) if F : f o g, and find the domain of F. rxarvrprE5: Given that f is defined by f (r): lx and g is defined by S@)- f -'/.., find: (a)f"f;(b)g"g; (c)/.9;(d) g"f. Also find the domain of the composite function in each part. soLUrIoN:F(r): ff " S)(r): f@(x)) :f(2x_3) : f2x-3 The domain of g is (-m, *oo), and the domain of / is [0, +m). So the domain of F is the set of real numbers for which 2x - 3 > 0 or, equivalently, [9, **). solurroN: The domain of / is [0, *-), and the domain of g is (-*, ( a )( f " f ) ( x ) : f ( f ( x ) ) : f ( { - * ) : { T x : 9 i The domain of f " f i" [0, +*;. (b) (g" g)(r):Sk(r)): g(*' - 1): (x'-l)'-"1,:xa-2x2 The domain of I ' I is (-co, +*). +oo). AND TYPESOF FUNCTIONS ON FUNCTIONS, 1.8 FUNCTIONNOTATION,OPERATIONS ( c )( f " i l ( r ) : f k G D : f e - 1 ) : { P - r The domain of f " git (--,-1] not in (-1, 1). ( d ) ( s " i l ( x ) : S U ( x ) ): 8 ( { i ) : U [1, +*; or, equivalently,all r ( f r ) ' - 1- x - L 'l' is deThe domain of I " f is [0, +*). Note that even though x '1..8.2, fined for all values of x, the domain of I " f , by Definition is the set of all numbers I in the domain of / such that f (x) is in the domain of g. 1.8.3 Definition (i) A function / is said to be an eaen function if for every r in the domain of f , f(-x): /(r). (ii) A function / is said to be an odd function if for every r in the -f(x). domain of f , f(-x): In both parts (i) and (ii) it is understood that -r is in the domatn of f whenever x is. . ILLUSTnATTON 2: 3(-x)n - 2(-x)' * 7 :3xa (a) If f(x): 3xa- 2x2+ 7, then f(-x): 2x2* Z : f(x).Therefore, f is an even function. (b) If g(r) : 2x5I Sxs- 8x, then S(-x) : 2(-x)5 * 5(-r)B - 8(-x) : -2x5 - 5x3* 8r : -(2xs * 5x3- 8x): -g(r). Thus, g is an odd function. *5(-r)3 - (-x)' * 8: ( c ) I f h ( x ) : 2 x a * 5 r 3 - x 2* 8 , h ( - x ) : 2 ( - x ) n 2xa 5x3 x2 * 8. We see that the function h is neither even o nor odd. From the definition of an even function and Theorem 1.3.6(ii) it follows that the graph of an even function is symmetric with respect to tlne y axis. From Theorem 1.3.5(iii) and the definition of an odd function, we see that the graph of an odd function is symmetric with respect to the origin. If the range of a function / consists of only one number, then / is called a constant function So if f(x): c, and if c is any real number, then / is a constant function and its graph is a straight line parallel to the r axis at a directed distance of c units from the x axis. If a function f is defined by f ( x ) : a s x n* a 1 x 6 nI - ra 2 x n -*z' + a n - 1 xI a n where n is a nonnegative integer, and ao, ar, . . . , an ate real numbers (ao * 0), then / is called a polynomial function of degree n. Thus, the function / defined by f(x):3x5-x2+7x-r is a polynomial function of degree 5. REALNUMBERS, INTRODUCTION TO ANALYTIC GEOMETRY. AND FUNCTIONS If the degree of a polynomial function is 7, then the function is called a linear function; if the degree is 2, the function is called a quad- 'r!!,"':::_: if the degreeis 3' thefunctionis careda cubicfunc- The function / defined by f (x) : 3x * 4 is a linear function. The function g defined by S@) :5x2 - 8x * 1 is a quadratic function. The function h defined by h(x):8x3 - x * 4 is a cubic function. o If the degree of a polynomial function is zero, the function is a constant function. The general linear function is defined by f(x):mx*b where m and b are constants and m # 0. The graph of this function is a straight line having m as its slope and b aSits y intercept. The particular linear function defined by f(x):x 1S called the identity function, The general qLadratic function is defined by f(x):af*bxrc where a, b, and c are constantsand a * 0. If a function can be expressedas the quotient of two polynomial functions, the function is called a rational function For example, the function / defined by -s-x'za5 : \--/ Jf(r) _9 f is a rational function, for which the domain is the set of all real numbers except 3 and -3. An algebraicfunction ts a function formed by a finite number of algebraic operations on the identity function and the constant function. These algebraic operations include addition, subtraction, multiplication, division, raising to powers, and extracting roots. An example of an algebraic function is the function f defined by f(x) : In addition to algebraicfunctions, transcendental functionsare considered in elementary calculus.Examplesof transcendentalfunctions are trigonometric functions, logarithmic functions, and exponential functions, which are discussed in later chapters. ANDTYPESOF FUNCTIONS ON FUNCTIONS, OPERATIONS NOTATION, 1.8 FUNCTION Exercises7.8 / 1.)Given f (x) :Zxz + 5x - 3, fnd: (a) f(-z) (c) /(o) (d) /(3) (e)/(h + 1) G)f @'- s) (h\ f(x+h) (i) /(r) + f(h) 2. Givengk) :3x2 - 4, find: (a) g(-4) (b) (e)g(x - h) (f) (c) 8(+) s@)- s(h) (d) g(3xz- a) 8e) (s)s ( x + hh) - s ( x ), h + 0 F(x) : \/zxT 3, i3-..,iGiven (a) F(-1 ) (b) F(4) (d) F(30) (e) F(2r + 3) 4. Given G(x) : lET7, frnd: (a) G(-2) (b ) G(0 ) (c) c(+) (d) G(+) (e \ G(Z x ' z -l ) (f) G(r+h)-G(x) h+0 h 5. Given ilrl o"*o t\x):1x [1 if t:o find:(a)/(1); (b)/(-t); (4 f G); @) f|a); @)f (-r), (t) f(x + 7); (g)l(f); (n)/(-f). l a r r l - i r l- a 6. Given l(r) expressf(t) without absolute-value bars if (a) f >0;(b)-3 =t <0; (c) f <-3. g arc defined. In each problem define the following functions and determine In Exercises 7 through fz, tn" for,"tlon, 1 "nd ( a ) f f u n c t i o n : + g ; (b)f -g;G)f .S;@)flg;Grglh(Df "g;Grg"f. o f t h e r e s u l t i n g thedomain 7.f(x)-x-5;S@):x2-7 8. f(x) : tfi; g(r) : xz+ 1 s.f(*):fit sQ):* 10./(x): \/i -'2, SG):1 rr. f (x) : \/iz - r; 8@) : {i] 12.f (x) : lxl;g(r) : lr - 3l L3. For each of the following functions, determine whether f is even, odd, or neither. (a) /(x) :2xa - 3xz I 1 (d) /(r) : .16- L tt3 G)fQ):ffi (b) /(x) :Sxs - 7x ( c )/ ( s ) : s 2 * 2 s t 2 (e ) f (t) :5 t7 * ' 1 ., (f) f(x): lxl v-1 (h)/(r):ffi 1.4.Prove that if f and g arc both odd functions, then (/ * g) and (f - il are also odd functions. d2 REALNUMBERS, INTBODUCTION TO ANALYTICGEOMETRY, AND FUNCTIONS 15. Prove that if I and g are both odd functions, then / . g and /g are both even functions. 16. Prove that any function can be expressed as the sum of an even function and an odd function by writing f (r) :trlf (x)+ l(-x)l + *If k) - l(-x)l and showint that the function having Iunction valuestr[l(r) + l(-r)] is an even function and the function having function values{ff(r) - f(-r)l is an odd function. 17. Usethe result of Exercise16to exprcssthe functions defined by the followint equationsasthe sumof an evenfunction a n da no d df u n c t i o n( :a ) l ( x ) : * + 2 ; b ) f k ) : t C - L ( c r f ( x ) : f + x a- x + 3 ;( d )l ( r ) : 1 l x ,( e )f ( r ) : ( x - r \ l ( x + 7 ) ; (f)l(r) : lrl + lr - 11. 18. There is one function that is both even and odd. What is it? 19. Determine whether the comPositefunction I " g is odd or even in each of the following cases:(a)land g are both even; (b) I and I are both odd; (c) / is even ard g i8 odd; (d) I is odd and g is even. 20. The function I is defned by g(r) =f. Define a function f such that (/"g)(r):r if (a) r > 0; (b) r < 0. In Exercises 21 through 34, draw a sketch of the graph of the given function. In Exercises 23 through 34, the function U is the unit step function defined in Exertise 21 and the function sgn is the signum function defined in Exercise 22. 21.tI isthetunction defined by,(t : {l if l: 3 This function U is called the u t step f nctiofl. 22. The signum functiotr (w sign functioz), denoted by sgn, is defined by sgnr:l l-1 0 [ 1 ifx<0 itx:0 ifr>0 sgn r is read "signum of x." 2 3 .f ( x ) : U ( x - 1 ) 2a. gQ): U(r) - U(x - 1) 2 5 . g ( x ) : s g n ( x+ 1 ) - s g n ( x- 1 ) 26. f(x): 2 7 .h ( x ) : x ' U ( x ) 2 9 . G ( x ) : ( r f 1 , ). U ( x + 1 ) - x . l l ( x ) 2 8 .F ( x ) : ( r * 1 ) . U ( x + 1 ) 30.F(r) -x-2sgnx 31,.h(x):sgnr-U(x) 32.f(x):sgnx'U(x+1,) 3 3 .g ( r ) : s g n x * x . U ( x ) 3 4 .G ( r ) : s g n ( U ( r ) ) bi. fi"a formulas tor (f " g)(r) if ifr<0 [0 if0<x<1 f(x):]2x 1.0 itx>7 lr and g(x):]Lx tt ifr<0 if0<x<1 ifx>1. 36. Find formulas for (g " fXr) for the functions of Exercise 35. 37. Find formulas for (f . g)(r) if [0 f(*):lf [0 ifr<O if0<x<'l.. ifx>1 [1, and g(x):lZx Ll ifr<0 if0<x<1, ifx]l sgn x2- sgnx REVIEWEXERCISES 63 v2,find two functionsg for which (/ " S)(r) : 4x2-12x*9. f i n d t w o f u n c t i o n sg f o r w h i c h ( f " i l ( r ) : x 2 - 4 x 1 5 . 39. I t f ( x ) : x 2 * 2 x 1 2 , 38. It f(x): (Chapter1,) ReaieutExercises In Exercises 1 thrcugh 6, find the solution set of the given inequality, and illustrate the solution on the real number line. 1.8<5r*4<10 ?7 4 .x++<4 + r-5 ,#.,L 3. Zxz* r < 3 s. 13+Sxl-9 6- ' lz-gxl 1 l3+rl=a 7. Define the foUowing sets of points by either ar equation or an inequality: (a) the point cide (3, -5); (b) the set of all points whose distancefrom the point (3, -5) is lessthan 4; (c) the set of all points whose distancefrom the point (3, -5) is at least 5. 8. Ptove that the points (1, -l) , (3, 2), and (7, 8) ar€ collinear in two ways: (a) by usint the distance formula; (b) by ueing slopes. 9. Find equations of the lines passing through the origin that are tangent to the circle having its center at (2, 1) and a radius of 2. L0. Prov€ that the quaddlateral having vertices at (1, 2); (5, -1\ , (71,7) , and (7, 10) is a rectangle. and 5t - y + ,l:0 are equations of the same line. 12. Shovvthat the tdangle with vertices at (-8, 1), (-f, -6) and (2, 4) is isoscelesand find its area. 1 3. Two vertices of a parallelogram are at (-3, 4) and (2, 3) and its center is at (0, -1). Find the other two vertices. 14. Two opposite verticrs of a square are at (3, -4) and (9, -4). Find the other two vertices. 1 1 . Determine the values of & and h ifSr + ky + 2:0 15. DetemrineallvaluesoftforwNchthegraphsofthetwoequations**f:kandx+y:fintersect. t6. Prcve that iI r i8 any real number, lrl < f + 1. 1 7 . Find an equation of the circle circumscribedabout the triangle having sideson the lines r - 3y * 2 :0,3x and2r+y-3:0. - 2y * 6:0, 18. Find an equation of the circle having as its diameter the common chord of the two circles rl + y'z+ 2x - 2y - 74: O andt'+!P-4r+4y-2:O. 19. Find an equation of the line through the point of intersection of the lines 5r + 6y - 4: Oa d'x- 3y*2:0andperpendicular to the line x - 4y - 20:0 without finding the point of intersection of the two lines. (nrtr: See Exercise 28 in Exercisee1.5.) +20:0, x+2y-15=0, and 3r- y * 10: 0. 20. The sides of a parallelogramare on the lines r+ 2y- 10:0,3r-y Find the equations of the diagonals urithout finding the vertic$ of the parallelogram, (rrrm: See Exercise 28 in Exerciees1.5.) In Exercises 21 through 24, the function is the set of all ordered pairs (r, y) satisfying the given equation. Find the domain and range o{ the function, and draw a sketch of the graph of the function. 2 1 .f .: y : \ E x + 5 2 2g. : y : # 64 REALNUMBERS,INTRODUCTION TO ANALYTICGEOMETRY, AND FUNCTIONS | *+3 24. F: "al : 1\/9= 7 I x-3 ifx(-3 i f-3 < x < 3 if3<r In Exetcises 25 through 28, for the given functions f and g, define the following functions and determine the domain of the resultingtunction: (a\ f+ g; (b)f - gt 1c)f'g; (d) flg; (e)Slf; (D f. S; G) S" f. E. f ( x ) : x z- 4a n d g (r) : 4 x - 3 . 26. f(x) : \/x+ 2and g(r) : f * 4. 27. f ( x ) : t l( x - 3 ) a n d t@ ) : x l k + 1 ,). 28. f (x) : t/i and S@) : Uxr. 29. Prove that r sgn r: l.xl. 30. Provethat the two lines.Afi+Bg +C1:O andAzx + B2V+C,:0 are parallelif and only if ArBr-Are:0. 31. Prove analytically that the three medians of any t angle meet in a point. 32. Plt)ve analytically that the line segment joining the midpoints of any two sides of a triangle is panllel to the third side and that its length is one-half the length of the third side. 33. Prove analytically that if the diagonals of a rectangle 6re perpendicular, then the rectangle is a square. 34. Prove analytically that the set of points equidistant from two given points is the perpendicular bisector of the line segment joining the two points. 35. In a triangle the point of intersection of the medians, the point of intersection of the altitudes, and the center of the circumscribed cirde are collinear. Find these three points and prove they are collinear for the triangle having vertices at (2,8), (5,-L), and (6,5\. lcontinuity LIMITSAND CONTINUITY 2.1 THE LIMIT OF A FUNCTION Consider the function / defined by the equation (2x+3)(x-1) lt \l -x. )\ _: n -1;- (1) / is defined for all values of x except x: l. Furthermore, if x * r, the numerator and denominator can be divided by (r - 1) to obtain f(x):2x*3 x*t e) T a b l e2 . 1 . 1 x 0 0.25 0.5 0.75 0.9 0.99 0.999 0.gggg o.ggggg f(x) -2x*3 (x + 1,) 3 3.5 4 4.5 4.8 4.98 4.998 4.9998 4.gggg8 .r approach L, through values that are greater than 1; that is, let r take on the values2, 1.75,'1,.5, 1.25,7.L,7.01, 1.001,,1.0001, L.0000L,and so on. Referto Table 2.1.2. Table 2.7.2 x 2 L.75 1.5 1,.25 1.1 1.01 1.001 1.0001 1.00001 f(r):2xi3 (x+I) 7 6.5 6.0 5.5 5.2 5.02 5.002 5.0002 5.00002 We seefrom both tables that as r gets closerand closerto l,/(r) gets closerand closerto 5; and the closerr is to 1, the closer (x) is to 5. For inf stance,from Table2.1.7,when x:0.9, f (x): 4.8; that is, when r is 0.j. less than 1, f (x) is 0.2lessthan 5. When x : 0.999,f(x) : 4.998;that is, when r is 0.001less than 1,f(r) is 0.002less than 5. Furthermore,when x:0.9999, f(x) : 0.49998;that is, when x is 0.0001lessthan r, f(x) is 0.0002lessthan 5. Table2.1..2shows that when x: 1,.r,f(x):5.2, that is, when r is 0.1 greaterthan 7, f(*) is 0.2 greaterthan 5. when r: r.001, f(x): 5.002;that is, when r is 0.001greater than L, f (x) is 0.002greater than 5. when r: L.0001, f(*),:5.0002; that is, when r is 0.0001greaterthan l,f (x) is 0.0002 greater than 5. Therefore,from the two tables, we see that when r differs from l by -f 0.001(i.e., r :0.999 or x: 1.001),/(r) differs from 5 by -10.002[i.e., f !!).:4.998 or f (x):5.0021. And when r differs from l by -+0.0001, /(r) differs from 5 bv +0.0002. 2.1THELIMITOFA FUNCTION67 Now, looking at the situation another wdlt we consider the values of f (x\ first. We see that we can make the value of f (x) as close to 5 as we please by taking x close enough to 1. Another way of saying this is that *" .ut make the absolute value of the difference between f (x\ and 5 as small as we please by making the absolute value of the difference between r and L small enough. That is, lf (x) 5l can be made as small as we pleaseby making l, - Ll small enough. A more precise way of noting this is by using two symbols for these small differences.The symbols usually used are € (epsilon) and 6 (delta)So we state that If(x) - 5 | will be less than any given positive number e whenever lr - Ll is less than some appropriatelychosenpositive number that lf(*)-51 <e w h e n e v e r0 < l r - 1 1 (3) <6 We seefrom the two tablesthat lf (x)- 5l :0.2 when lx So,given e:0.2, we take 6:0.L and statethat 1l:0.1. - 5l < 0.2 whenever 0 < lt - 1l < 0.1 This is statement(3),with e: 0.2 and 6 : 0.1. Also, lf (*) - 5l : 0.002when lx - Ll : 0.00t.Hence,if e : 0.002,we take 6: 0.001,and then lf (x) l f ( x ) - 5 1 < 0 . 0 0 2 w h e n e v e r0 < l r - l l <0.001 This is statement(3),with e: 0.002and 6: 0.001. Similarly, if e: 0.0002,we take D : 0.0001and statethat lf (x) - 5l < 0.0002 whenever 0 < l*-11 < 0.0001 This is statement(3),with e: 0.0002and 6: 0.000L. We could go on and give € any small positive value, and find a suitable value for 6 such thai lf(x) - 5l will be less than e whenever lx- 1,1 i s l e s st h a n 6 a n d x # t ( o r l r - 1 1 > 0 ) .N o w , b e c a u s e f o r a n y e) 0 w e can tind a E > 0 such that If(x) -51 < e whenever 0 < lx- 1l < 6, we state that the limit of f (x) as r approaches1 is equal to 5 or, exPressedin symbols, (4) trm/(r):s 1-6 F i g u r e2 . 1 . 1 1+E You will note that we state 0 < lr - 11.This condition is imposed becausewe are concernedonly with values of f (x) for x closeto L, but not f.orx: L. As a matter of fact,this function is not defined for x:L. Let us see what this means geometrically for the particular function defined by Eq. (1). Figure 2.1..1illustrates the geometric significance of LIMITSAND CONTINUIW € and 6. We see that /(x) on the vertical axis will lie between 5 - e and 5 * e whenever x on the horizontal axis lies between 1 - 6 and 1 * 6; or lf(x)-51 <. whenever 0(l*-11 <a Another way of stating this is that f (x) on the vertical axis can be restricted to lie between 5 - e and 5 -l eby restricting r on the horizontal axis to lie between L - 6 and 1 + 6. Note that the values of e are chosen arbitrarily and can be as small as desired, and that the value of a 6 is dependent on the e chosen. We should also point out that the smaller the value of e, the smaller will be the corresponding value of 6. Summing up for this example, we state that lim f (*):5 because for any e ) 0, however small, there exists a 6 ) 0 such that lf(*)-51 <e w h e n e v e r0 < l r - 1 1 < D We are now in a position to define the limit of a function in general. 2.1.1 Definition Let f be a function which is defined at every number in some open interval 1 containing a, except possibly at the number a itself. The timit of f (x) ns x approachesa is L, written as ti2f(c):L (s) if for any € ) 0, however small, there exists a E ) 0 such that lf!)-Ll <e whenever 0<lr-al <6 (6) In words, Definition 2.1.L states that the function values f (x) approach a limit L as x approaches a number a if the absolute value of the difference between /(x) and L can be made as small as we please by taking x sufficiently near a,but not equal to a. It is important to realize that in the above definition nothing is mentioned about the value of the function when x: a. That is, it is not necessary that the function be defined for x: A in order for the lim /(x) to exist. In particular, we saw in our example that ,. (2x*3)(r-1) TTGI)__c but that (2x+ 3) (x - 1,) (r- 1) is not defined f.or x: L. However, the first sentence in Definition 2.1.1 requires that the function of our example be defined at all numbers except f. in some open interval containing 1. LIMIT OF A FUNCTION ExAMPLEl-: Let the function / be defined by the equation f(x):4x-1' Given lim f(r) --ll, find a D for e: 0.0Lsuchthat lf(*) - 111< 0.01 SOLUTION: ' l f ( * )- 1 1 1 : l @ x -1 ) 1 1 1 - l4x- 121 -Alx-31 Therefore/ we want 4lx- 3l < 0.01 whenever 0 ( lr - 3l < 6 or, equivalently, whenever 0 < lr-31 < 6 lr - 3l < 0.0025 whenever 0 < <D If we take 6: 0.0025,we have l(ax- 1) - 111< 0.01 whenever0 ( < 0.0025 (7) It is important to realize that in this example any positive number less than 0.0025 can be used in place of 0.0025 as the required 6. That is, if 0 < y < 0.0025 and statement (7) holds, then - 1) - 111< 0.01 whenever 0 ( l*- 3l < y (8) because every number r satisfying the inequality 0 < satisfiesthe inequality 0 < lx - 3l < 0.0025. < 7 also l(+* The solution of Example L consisted of finding a 6 for a specific e. If for any € we can find a 6 that will work, we shall have established that the value of the limit is LL. We do this in the next example. nxervrrr.n2: Using Definition z.L.L,prove that L'T tnt - 1): 1.1 solurroN: We must showthat for any€ ) 0 thereexistsa E > 0 suchthat r <lt-31 <D l(+x-1)-111 <e wheneve0 From Example L, we note that l@x-1)-11l:lax-Dl Therefore, we want alx-31<e w h e n e v e r0 < <6 w h e n e v e r0 < l x - 3 1 <6 or, equivalently, l"-31 <|e So if we choose6: *e, we have alx- 3l < 46 whenever 0 < lx- 3l < 6 LIMITSAND CONTINUIry or/ equivalently, al*- 3l< 4 *e whenever 0 ( <6 or, equivalently, 4lx- 3l < . giving us l ( a r- 1 ) whenever <€ 0 ( whenever 0( <6 <6 if 6: *e. This provesthat lim (4x-- 1) : 11. In particular, if e : 0.0L, then we take E : 0.01/4, or 0.0025,which correspondsto our result in Example 1. A.y positive number 6' ( *e can be used.in place of ie as the required 6 in this example. rxauprr 3: Using Definition 2.1.7, prove that limx2:4 solurroN: wemustshowthatforany€) 0thereexistsa6 ) 0suchthat l * ' - 4 1 < , w h e n e v e r0 < l x - Z l < a Factoring, we get lx,-41:lx-2l .lx+zl we want to show that ll - 4l is small when r is closeto 2. To do this, we first find an upper bound for the factor + zl.lf x is closeto 2, we lx know that the factor I* - 2l is small, and that the factor * 2l is closeto 4. Belx causewe are considering values of r close to 2, we can concernourselves with only those values of x for which l* - zl < 1; that is, we are requiring the 6, for which we are looking, to be lessthan or equal to 1. The inequalit! l x - 2 1< r is equivalent to -L < x-2<'!, which is equivalent to 1,<x<3 or, equivalently, 3<x+2<s This means that if lx - 2l < 1, then 3 < l" + 2l < 5; therefore,we have .lx+Zl < l*'-41:lx-Zl fr- 2l .S wheneverlx_21 <1, Now we want l* 2l . 5 ( e or, equivalently, l* - 2l < *e Thus, if we chooseD to be the smallerof 1 and *e, then whenever lx- zl 2.1THELIMITOF A FUNCTION < 6, it follows that lr - 2l < *e and lx + 2l < 5 (becausethis is true when ' that lx 2l < 1) and so lr2 4l < (*e1 (5). Therefore, we conclude l r ' - 4 1 < e w h e n e v e r0 < l t - 2 1 < 6 if E is the smaller of the two numbers L and *e, which we write as 6 - min(1, *e). 4: Using Definition ExAMPLE prove that 2.1.'/.,, 8 lim*:2 t-7 t- solurroN: We must show that for any€ > 0 there existsaD ) 0 suchthat ls'=-21 I I -l <e v-3 c Now w h e n e v e r0 ( l t - z l <a :l#l t*-21 :# . ).m =lt-71 We wish to show that l8/(f - 3) 2l is small when f is close to 7. We - 31.By requiring proceedto find some upper bound for the fraction zlv ln" A for which we are looking to be less than or equal to 1",we can say that whenever lt - Zl ( 6, then certainly lt 7l < L. The inequality It-zl <t is equivalent to -L < t-7 <L which is equivalent to 3< t- 3< 5 Thus, if lf - 7l < 1.,then 3 < lt-31< s we haveshown wheneverlt - 7l < l,lt - 3l > 3, and because Therefore, that 18/(t 3) - 2l: lt - 7l ' zllt - 3',,we have l*-21 : V-71 h< lr-71<L lr- rl .? whenever We want then V-71 'A < € or, equivalently,V-71 <*e. Consequently, we take 6 as the smaller of L and Be, which assuresus that whenever LIMITSAND CONTINUITY I t - z l < 6 , t h e nl t - z l 3 (becausethis is true when whenever 0 ( lt_ zl < D, and where 6: min(!, te). we have therbfore proved that lim l8l(t - 3) f : 2. The following theorem states that a function cannot approach two different limits at the same time. It is called, a uniquenesstheirem because it guarantees that if the limit of a function exists, it is unique. 2.'1,.2Theorem U fT fk): L, and li^ f(r): Lr, then Lr.: Lz. pRooF: we shall assume that L, * L, and show that this assumption leads to a contradiction. Because lt : Lr, it follows from Definition f@ 2.1'.1that for any € > 0 there exists a 6r ) 0 such that r (lr-al lf(x)-Ltl <e wheneve0 <6, e) Also, because f <*l : Lr, we know that thereexistsa 6z) 0 suchthat l\ l f ( * ) - L r l < . w h e n e v e 0r ( l * - o l < 0 , (10) Now, writing Lr- L, as L, - f(x) + f (x) - L, and applying the triangle inequality,we have lL,- L,l : l[t, - f(x)l + [f(x) - L"]l < l L ,- f ( x ) l+ V @ )- L , l (11) so from (9), (10), and (11) we may conclude that for any € ) 0 there exists a 6r ) 0 and a 6z ) 0 such that lLr-Lrl< e*e w h e n e v e r0 < l * - - o l < 8 , a n d 0 < l x - a l < 6 , ( t z ) If 6 is the smaller of E, and 6r, then E = Er and I S 6r, and (12)statesthat for any e ) 0 there exists a D > 0 such that w h e n e v e r0 ( (13) l*-ol <6 : However, if we take e tlL, hl, then (13) states that there exists a 6>0suchthat - Lrl < lt, lLr- Lrl whenever 0 ( lr- al < D (14) lLr-Lrl<2e 2.1 THE LIMIT OF A FUNCTION 73 Because(1a) is a contradiction, our assumption is false. So Lt : Lr, and I the theorem is proved. 2.1 Exercises In Exercises1 though 8, we are given l(r)' a' and L' as well as lim l(x) : L' Deternine a number 6 for the given e such that ll(r) - Ll ( e whenever0 ( h- al < 6. ' 3' lin (3-ttr):7,e :o'o2 2. lim (4r-5):3;€:0.m1 t : ' t t ' ^ ( Z r + l ) : t 0 ;e : 0 . 0 1 n . o i ' , r * u t , : - 8 ; e : 0 . 0 @ ' t ' ; ; : * -t-8s ; , : o . o o s e'-t-'"r6:z;':o'oos t--2 q*-1 ar-A. -4; € = o,o1 8. lim ff=t' : 2; 6: 9.91 '7. tm I-:: nB5x- L ;:-, x+2 In Exercises9 through 29, establishthe lirrit by using Definition 2.1.1;that i8, for any € > 0, find a 6 > 0, such that l f ( r \ - L l ( e w h e n e v e 0r < l r - c l < 6 ' ri.#:6 J0. lim (7-2r):11 9 , { m ( 5 x - 3 ) :' Z 'r r . ,-s x-3 .-1 ,7;i' p. \^{:=i2:+ ,-t I-1 + . J9"$,*:e !?,urn*-r , / r ' . -------= t 2 - - ; -lim ;r-r 15. -76. - ; -nm=:2 ;x-3 1 18. lim ----:-==-l 19. lim (f -3r):10 21. lim (5-r-f):-1 22. 6- 11-4:9 t r--1x + 3 17.lim-!--;:2 ".rr-4 20' Itm (f +2x-l):7 5+E zg' llm r/r+5-:g '-1 t-t]25-I r--a 11 24. lim =#:) r _ 1v 5 - r z 25. Prove that lim f: at i4 cis any Positive number' 26. Prcve that lim f :a2if 27. Prove that litrr t/i: ais any negative number. la if a is a y Positive number. 28. Prove that u'm Vi :'Va. 29. Prove that iI lim l(r) (mxr: qs- bs: (a - b\ (d + ab + 8).) exists and is L, e* liT ll(r) | exists and is lll. 30. Prove that, if ^t) : 8(t) for all values of r except r : a, then lim ir) : t(d if the limits exist. ItT 31. Prcve that, if l(x) : g(r) for all values of x except r: n, then if lim 8(x) does not exist, then lim /(r) does not exist. (Hrlfn Show that the assumPtion that lim l(x) does exist leads to a contadiction.) 74 LIMITS ANDCONTINUITY 2.2 THEOREMS ON LIMITS In order to find limits of functions in a straightforward manner, we shall OF FUNCTIONS need some theorems. The proofs of the theorems are based on Definition 2.1'.1.These theorems, as well as other theorems on limits of functions appearing in later sections of this chapter, will be labeled "limit theorems" and will be so designated as they are presented. 2.2.1.Limit theorem 1 If m and b are any constants, tu(*r+b)-ma*b PRooF: To prove this theorem, we use Definition2.1,.1. For any e ) 0, we must prove that there exists a D ) 0 such that w h e n e v e r0 ( l * - o l < 6 (1) l(**+b)-(ma*b)l<e C a s e ' l -m : * 0. B e c a u sl e ( m x + b )- ( m a * a ) l : l * * - m a l - l * l . l * - a l , w e w a n t to find a D ) 0 for any € > 0 suchthat w h e n e v e r0 < l r - a l < 6 l*l .l*-al<e or,becausem*0, l*- ol.* lml w h e n e v e rO ( l r - a l <6 e) Statement (2) will hold if we take E : ellml, and thereforewe conclude that l(** + b) - (ma* b)l < e whenever 0 < <6 if 6:+l m l This proves the theorem for Case L. C a s e2 : m : 0 . I f m : 0 , t h e n l ( * * + b ) - ( m a* b ) l : O f o r a l l values of x. So we take 6 to be any positive number, and statement (L) holds. This proves the theorem for Case2. I . rLLUsrRArroN1: From Limit theorem 1, it follows that lim(3r+5)-3.2+5 t-2 2.2.2 Limit theorem 2 If c is a constant, then for any number a, lim c: c t-A pRooF: This follows immediately from Limit theorem 1 by taking m : andb-c. 0 I E;,. 2,2 THEOREMSON LIMITSOF FUNCTIONS 2.2.3 Limit theorem 3 lim x: 75 a I-A PRooF: This also follows immediately from Limit theorem 1 by taking I tn:Landb:0. O ILLUSTRATION 2: From Limit theorem 2, limT -7 t-5 and from Limit theorem 3, ]1 2.2.4 Limit theorem 4 '--6 If lim f (x) : L and lim g(r) : M, then t-A E-A }l tf@) I g(r)l : L -+M pRooF: We shall Prove this theorem using the plus sign. Given tuf('):r (3) *l 8G):M (4 ) we wish to prove that (5) l i m [ / ( x )+ S ( r ) l : L + M To prove Eq. (5), we must use Definition 2.1.1;.thatis, for any e ) 0 we must prove that there exists a 6 ) 0 such that llf(x)+g(r)l-(L+M)l <e w h e n e v e r0 ( l * - o l <6 (6 ) Becausewe are given Eq. (3), we know from the definition of a limit that for ie > 0 there exists a 6r ) 0 such that lf(*)-Ll<*, w h e n e v e r0 ( l r - a l <6' Similarly, from Eq. (4), for ie ) 0 there exists a Dz) 0 such that -Ml<Le w h e n e v e r0 < l * - o l ( 6 , lS(") Now, let 6 be the smaller of the two numbers S1and 6r. Therefore, E = Dt and E < 62.So we can say lf (x) Ll < *e whenever 0 ( lr al < 6 lg(r)-Ml<te w h e n e v e r0 ( l r - o l <6 LIMITSAND CONTINUITY Hence, we have l t / ( r )+ s ( r ) l - ( L+ M ) l: l ( / ( r )- L )+ ( s ( r )- M ) l : j:'?;:'..53;I'o.,x-a, <6 ln this warlr we have obtained statement (6), thereby proving that lr lf(x)+S(r)l:L+M The proof of Limit theorem4 using the minus sign is left as an exercise (see Exercise 24). I Limit theorem 4 can be extended to any finite number of functions. 2.2.5Limit theorem5 If lim fr(x) : Lr, lim fr(x) : Lr, ., "-irf and fu f r ( r ) : L , , ,t h e n -,f,(x)J : L,* L,* t iia t" lf,(x) . -{-Ln This theorem may be prove dby applying Limit theorem 4 and mathematical induction (see Exercise 25). 2.2.5 Limit theorem 6 If lim f(*): L and lim g(x) : M, then limf(x)'Sk):L.M The proof of this theorem is more sophisticated than those of the preceding theorems, and it is often omitted from a beginning calculus text. We have outlined the proof in Exercises 27 and 28. o rLLUSrRArroN3: From Limit theorem r, tiT x:3, and from Limit theorem L, lim (2x + 1) - 7. Thus, from Limit theorem 6, we have lim x(2r + 1) : lim r ' lim (2x + I) t-3 t-3 J'-3 Limit theorem 6 also can be extended to any finite number of functions by applying mathematical induction. 2.2.7 Limit theorem 7 If lim fr(x):Lr,limfr(x):Lr, lim [/r G)fr(x) . . , andlimfn(x): . f"(x)f : L,L, Ln The proof is left as an exercise (see Exercise 29). L,,,then 2.2 THEOREMSON LIMITSOF FUNCTIONS 2.2.8 Limit theorem 8 U Q) : L and n is any positive integer, then lT f tiq t/(r) l" : L The proof follows immediately from Limit theorem 7 by taking ' ' _ L, f"(x) and L:Lr-Lr: f(x):f'(x) -f2(x):' -3. Therefore, 4: From Limit theorem L, lim (5r * 7): . TLLUSTRATIoN from Limit theorem 8, it follows that lim (5r + 7)n: I lim (5x + 7)la &--2 : (-3)n :81. Z.Z.gLimit theorem 9 If lim f(x): L and tr1 S(x) : M, and M # 0, then n#:h is also frequently omitted from The proof, based on Definition 2.L."1., a beginning calculus text. However, a proof is given in Sec.2.6. Although we are postponing the proof, we apply this theorem when necessary. o rLLUsrRArroN5: From Limit theoremr, tj1 x:4, and from Limit theorem 1.,hm (-7x + 1) - -27. Therefore,from Limit theorem 9, lim r s-4 lim (-7x + t) t-4 _4 -27 __ 4 27 2.2.70Limit theorem L0 tt tji f @) : L, then n{f@:{L if L > 0 and rais any positive integer, or if L < 0 and n ts a positive odd integer. A proof of this theorem is given in Sec.2.6, and as is the casewith the preceding theorem, we apply it when necessary. F LIMITSAND CONTINUIW o ILLUSTRATToN 6: From the result of Illustration 5 and Limit theorem 1.0, it follows that lim e-4 J?# L- Following are some examplesillustrating the application of the above _ theorems. To indicate the limit theorem being ,s"d, we use the abbreviation "L.T." followed by the theorem number; for example, ,'L.T. 2,, refers to Limit theorem 2. EXAMPLE L: lT Find (* +7x-5) SOLUTION: lim (f *7x- 5) : lim f * lim 7x-lim 5 .1'-3 and, when applicable, indicate the limit theorems which are being used. (1.r.s) I-B : lim x . Iim r* t-3 ."-3 lim 7 . limr- lim 5 :r-S r_g -3'3+7.3-5 t-B (L.T.6) G.T.3andL.T.2) :9*2L-5 It is important, at this point, to realize that the limit in Example 1 was evaluated by direct application of the theorems on limits. For the f u n c t i o n/ d e f i n e db y f ( x ) : x 2 i r x - S , w e s e et h a t f ( g ) : g z + 7 . g - 5 - 25, which is the same as lim (x' + 7x - 5). It is not always true that we havet_i1/(r) : f (a) (seeExample4). In ExampleL, rim (x) :/(3) f cause the function / is continuous at x:3. continuous functions in Sec. 2.5. be- we discuss the meaning of rxatvrpr,n 2: Find tFzxc+3 VT+s and when applicable indicate the limit theorems being used. (1.T.10) lim (r3 * 2x t 3) (L.r.e) lim 12 * lim 5 t-2 u-2 (L.r.s) 2.2 THEOREMSON LIMITSOF FUNCTIONS imx)3+lim2'limr*lim3 (L.T.7 and L.T. 8) (L.T. 3 and L.T. 2) 8+4+3 EXAMPLE 3: Find -27 r. x3 llm-^ x-B x- J and, when applicable, indicate the limit theoremsbeing used' solurroN: Here we have a more difficult problem since Limit theorem 9 - 3) because lim (r - 3) - g. cannot be applied to the quotient (rt - 27)l(x However, factoring the numerator, we obtain -xs : : 27 - (x - 3) (x'z* 3x -f 9) x-3 x-3 This quotient is (r2*3x+9) if x*3 (sinceif x* 3 we can divide the numerator and denominator by (r - 3)). When evaluatingItT [(xt - 27)l(x - 3)], we are consideringvaluesof r close to 3, but not equal to 3. Therefore, it is possible to divide the numerator and denominator by (x- 3). The solution to this problem takes the following form: xs_27 .. ttfr-3:itTT : lT (r-3)(f+3x*9) (x' i 3x + 9) : lim : (lim r)2 + 1.8 dividing numerator and denominator by (, - 3) since x * 3 (1.r.4) (L.T. 8 and L.T. 1) I-3 :32 * L8 (L.r.3) :27 Note that in Example 3 (rt - 27)l(r - 3) is not defined when x:3, but lim [(rt - 27)l(x - 3)l exists and is equal to 27. NY 80 LIMITSAND CONTINUITY ExAMPLE4: Given that / is the function defined by 'f[(cx ): (* -3 li ifx*4 ifx:4 soLUrIoN: when evaluating lim f (*), we are considering values o f x closeto 4 but not equalto 4. Tf;il, we have lim /(r) : lim (x - 3) find lim f (x). :t (1.r.1) I n E x a m p l e4 l ; r y f @ ) : t b u t f ( 4 ) : 5 ' t h e r e f o r e ,t T f @ ) + f ( ) . This is an example of a function which is discontinuousat x:4. In terms of geometry, this means that there is a break in the graph of the function at the point where x: 4 (seeFig. 2.2.1).The graprr or ttre function consistsof the isolatedpoint (4,5) and the straight line whose equa_ tion is y : x - 3, with the point (4, L) deleted. There are two limits that we need in order to prove Limit theorems 9 and 10 in Sec.2.6. They are given in the following two theorems. Figure2.2.1 2.2.11Theorem If. a is any number except zero, ,. 1 llm-:t-aX 1 a pRooF: We need to consider two cases: a > 0 and a < 0. We prove the theorem if a > 0 and leave the proof for a 1 O as an exercise (see Exercise 30). so, if a is any positive number, from Definition 2.L.j, we must show that for any € ) 0 there exists a 6 > 0 such that Ir 1l li-il <e lx al Now w h e n e v e r0 < I r - o l < a 1 1 t l ___ l ta_ - x l lx- al li-d- I * I:loll|* and because a ) 0, we have : tx- ot-L 11-ll al alxl lx (6) We proceed to find some upper bound for the fraction Llalxl. If we require the D for which we are looking to be less than or equal to La, then certainly whenever lx- al < 6, we know l*- ol < ia. The inequality l*-ol<*a 2.2 THEOREMSON LIMITSOF FUNCTIONS 81 is equivalent to ba<x-a<ia which is equivalent to ia < x <Ea or,becausea)0, t a < l x l< 8 a Hence,whenever lx - al < ta,l"l > ta. So from Eq' (6) we have -v-'t'h,<lx-ath-tx-'t # l1-;l whenever l*- rl <ta We want then l, - ol - (2ta') < e or, equivalentlY,l* ol < La'e.Hence, we take D as the smaller of ta andiaze, which assuresus that whenever >ta (becausethis is true when l*- ol < 6, then lr- ol <*a'e and Jrf - ol < ia).This gives us l, :tx-'t'h<ia'e h, li-*l or Il -t- - l 1 l < lx al € whenever 0 < lx - ,l ( 6, and where 6 - min(# a, iaze). Hence, we have I proved that lim (U x) : Ll a if a is any positive number' The proof of the next theorem rnakes use of the following formula, where n is any positive integer. A n - b n : ( a - b ) ( s " - r a o n - 2 f ua+n - 3 b 2 + . . + a 2 b n - 3a* b " - z + b " - r ) Q ) The proof of Eq. (7) follows. a ( a n - rl a " - z b + ' ' ' * a b " - z+ b " - ' ) : a nl a " - r b* " ' b ( s n - r* a n - z b + ' ' * a b " - z+ b " - r ) - a ' - r b + ' ' + 0 ; 2 b n*- 2a b " - r ( 8 ) + f l i z b n*- za b - r * b subtracting terms of Eq. (9) from terms of Eq. (8) gives us Eq. (7). 2.2.12 Theorem lf n is a positive integer, then lim ,trx: \fr (9) ilv LIMITSANDCONTINUITY if either (i) a is any positive number or (ii) a is a negative number or zero and n is odd. pRooF: We prove part (i) and leave the proof of paft (ii) as an exercise (seeExercise31). Letting a be any positive number and using Definition 2.1.1, we must show that for any € ) 0 there exists a 6 ) 0 such that w h e n e v e r0 < l r - a l liT-{il<, <6 To expressl{G {al in terms of lx- al, we use Eq. (z). we have l{i- q' t." ).1(x-' t" )"(7rtn)n-zsrtn .-' * + ' {i l :l (xtt" Wl l"-rl Applying l{i- Eq,. (7) to the numerator, we have ffil: lx- rl l'{"-rlt" + /(n-z)tnartn + a artng@-2)tn + A@-r)tnl (10) We wish to find an upper bound for the fraction on the right side of Eq. (10).If we require the 6 for which we are looking to be less than ot "q.rilis to a, then whenever lx- al < E, we know that lr- rl < a, which equivalent to -a < x- a < a or 0<r<2a So, whenever lx - al ( a, then x ) 0. Therefore, if in the denominator of the fraction on the right side of Eq. (10)the x is replacedby 0, we obtain'/..I a@-t)in,which is greater than the fraction appearing in Eq. (10). Flence, whenever lx - al 1 a, we have l\Ti- til .l*- ol # - ol < a@--r)tne. we want, then, l* - ol ' 1,fa@-r1rnI e or, equivalently, l, Thus, we take 6 as the smaller of a and 6@-r)tnr,which assures us that whenever lx- al < 6, then l*- al q s(n-r)rne and r ) 0. Therefore, l{i - {Gl -lx-al ly(r-r)tn | 1(n-z)lnortnJ a yrtng(n-z)tn+ A@-L)tnl "l[t_ 2,2 THEOREMSON LIMITSOF FUNCTIONS 1 < a@ -r)tne. + A\n-r)ln :e whenever 0 < l* - ol < 6, where 6 : min( a, 6(n-r)tne).We have therefore I proved that lim ffi - \fr, if a is any positive number. 7: From part (i) of Theorem2.2.'l'2 it follows thatlim li: o TLLUSTRATIoN 8-5 rG and \im '{i: 2; frompart (ii) it follows that lim {i: -3. We do not n'-27 tr-16 consider a limit such as lim Vi because fi is not defined on any oPen interval containin g - 4. 2.2 Exercises In Exercises 1 through 17, find the value of the limit. and when applicable indicate rl l used. I 1 . l i m (x't 2x - 1. \ 4. - "t. lrfl ;; , 2xtL t-3x*4 d' Irm tt' 10. r-o ,1f,--r c - tE (y' - 2y' * 3y - 4) 3. u3+8 6. 5. trm-u--zVtz ---------;- ,. x2+5x+6 rlm--------.;-; f -x-12 13. lim :t:, 3x2-l7x*20 4x'z- zsx+'J6 r--E ]'-T,!9. 1 " t'i ig. L2. r, . s (nrwr: Rationalize the numerator.) X r s _ x z- r -* 1 0 L 6 .l i m - - - : ^ x--z x'+ cx I 2 2 x 3 - S x 2- 2 x - 3 i'"7 ' i i m # 13x2I 4x - 3 4x3 ;:; i:r s h o w t h a t l i mf ( x ) : f ( 2 ) . 1 8 .I f f ( x ) : x 2 + 5 x - 3 , 1 9. If F(r) : 2x 3*7x - 1, s h o w th a t l i m F (r) : F (-1 ). {,\t "! 20. If g@) : (x' - 0lQ - 2) , show that lim 8(r) : 4 but that g(2) is not defined. 21,.lf h(x) : 6trT9 - 3)lx, show that lim h(x) : * but that h(0) is not defined. the limit theorems being ..uF 84 LIMITS AND CONTINUIry 22. Given that f is the function defined by f ( x ) : {[?r ' (a) Find lim/(r), . -t^:\ t-2 1 i rx r 2 i f .x : 2 and show that lim f(x) + f(Z). x-2 (b)- praw a sketch of the graph of f . 23. Given that / is the function defined by (-z-9 ifx*-j f ( x ) : lLi 4 ifr:-3 (a) Find,li1 /(r), and show that,lim f (x) + /(-gl. (b) Draw a sketch of the graph of /. 24. Using Definition 2.'1,.1, prove that if lim /(r) : 1 and lim g(x) : 1'4 then tq t/(r)- s?)l: L- M 25- Prove Limit theorem 5 by applying 26. Prove that if lim /(r) Limit theorem 4 and mathematical induction. : L, then lim (/(x) - L) : 0. 27. Using Definition 2.'/-..L, prove that if lim /(x) : 1 and lim g(r) : g then tim /(r) ' g(r) : 0 (rflNT:Inordertoprovethatliml(.r)'g(r):0,wemustshowthatforanye>0thereexistsa6>0suchthat < 6 . F i " " t s h o w t h a t t h e r e i s a d 1> 0 s u c h t h a t l / ( x ) l < 1 + ll(r)'sk)l<ewhenever0<lr-al lll whenever 0 < lr- al < 61,by applying Definition 2.1.1to lim l(x) : L,withe : t and 6 : 6,, and then use the triangle inequality. Then show that ther€ is a& > o-suchthat kr"lil.ltr * lrl) whenever 0 < lx- al < 8", by applying Definition 2.1.1 to lim g(r) : 0. By taking 6 as the smalhr of the two nunbers A and 6r, the theorem i. p-""ay' 28. Prove Limit theot€m 6: If lim /(r) = L and lim g(r) = M, then rimU(x)'s(xll:L.M ' (ym, _writg-{1) 8e)=[l(x) cises26 and,27.) -t]s(r)+tk(r) - Ml+L.M. Apply Limit theorem5 and the resultsof Exer- 29. ProveLinit theorcm 7 by applying Limit theor€m 6 and mathematicalinduction. 3). ProveTheorem2.2.1.1if a < 0. 31. ProveTheorem2.2.12(ii). "*ry,". 2.3 ONE-SIDEDLIMITS 2.3 ONE-SIDED LIMITS Definition 2.9."1. 85 f @) we are concernedwith values of r in an oPen lrT interval containing abut not at a itself, that is, at values of r closeto a and either greater than a or less than a. However,_supPose,for example, Because/(r) doesnot that we have the function / for which f (x) : {R' containing 4. Hence, interval on any open < not defined 4, is x exist if f is restricted to values x if. 4. However, lim {x consider cannot we greater than 4, the"utlrJ of fr-x - +can be made as closeto 0 as we please by taking r sufficiently closeto 4 but greater than 4.ln such a caseas this, we let r approach 4 from the right and consider the one-sidedlimit from the right or the right-handlimit, which we now define. When considering Let f be a function which is defined at every number in some open ina from the right, is L, terval (a, c).Then the limit of f (x), as x npproaches written : L |'T,/t') if for any € ) 0, however small, there exists a 6 > 0 such that lf(x) x- -Ll <. w h e n e v e r0 < x - a 1 6 (1) Note that in statement (1) there are no absolute-valuebars around a s i n c e x - a ) 0, becausex ) a. It follows from Definition 2.3.1 that hm Vr - 4:0 I-4+ If when considering the limit of a function the independent variable r is restricted to values less than a number a,we say that x a|Ptoachesa from the left; the limit is called the one-sidedlimit from the left ot the lefthandlimit. 2.9.2 Definition Let f be a function which is defined at every number in some open ina from the left, is L, written terval (d, a).Then thelimit of f (x), asx approaches '"^tk):L if for any € > 0, however small, there exists a D ) 0 such that -Ll <. w h e n e v e r- D < x - a 1 0 lf (*) limit, or the undirectedlimit, We can refer ,o I,,T f @) as the tuso-sided to distinguish it from the one-sided limits. Limit theorems t-10 given in Sec. 2.2 remain unchanged when 4 a" is replacedby "x + a*" ot "x + a-." "x LIMITSAND CONTINUITY EXAMPLEL: Let f be defined by f-t ifr<o r : s g n f(x): J 0 if r: 0 [ 1 if0<r solurroN: A sketch of the graph is shown in Fig. 2 . 3 . 7 l.i m - -1 /(r) since, if r is any number less than 0, f(x) has the value -1. Similarly, tt1/(r) : *1. v (a) Draw a sketch of the graph of f . (b) Determine Iim f (x) it it exists. (c) Determine lim f (x) if it exists. F i g u r e2 . 3 . 1 In the aboveexamplelim /(r) * Lim (x). Becausethe left-hand limit f the right-hand limit J# r",o, say that the rwo-sided limit, 1nd "qdi,';e ttl, does not exist.The concept of the two-sided limit failing to exist f ftq becausethe two one-sided limits are unequal is stated in the following theorem. 2'3'3 Theorem @) exists and is equal to L if and only if nf lin_r/(r) and lim /(r) both exist and both are equal to L. The proof of this theorem is left as an exercise(seeExercise16). Let g be defined by (t-t g ( r ): { I ' l1 ifx#0 solurroN: lim g(r) : lim (-r) ifx:0 (a) Draw a sketch of the graph of g. (b) Find lim g(r) if it exists. A sketch of the graph is shown in Fig. Z.g.Z. : 0 and lim g(r) lim g(r) : lim lim r: Jr-o+ 0 J-O+ Therefore,by Theorem 2.9.9lim g(r) exists and is equal to 0. Note that 8(0): 2, which has no effectJ"ii- S@). Figure2.3.2 LIMITS 2.3 ONE-SIDED Let h be defined by EXAMPLE3: (4- xz . h ( x ): 1 ' i f .x = t , 1zr xz if.L < x sol,urroN: A sketch of the graph is shown in Fig. 2-3.3. lim h(r) - lim (4 - f) :3 t-l- (a) Draw a sketch of the graph of h. (b) Find each of the following limits if they exist: lim h (r), r-l- tim h(r) : lim (Z + f7 :3 t-l+ Therefore, by Theorem 2.3.3 lim h(r) exists and is equal to 3. Note that lim h(r), lim h(r). c-l Figure2.3.3 2.3 Exercises In Exercises 1 through 10, draw a skekh oI the graph and find the indicated limit if it exists; if the limit does not odst, give the reason. [ 2 if x<1'l r. f (x): 1-1 if x :1[; (a)tim /(r); (b) lim f (x); (c)lim /(r) r-rr'r '-1+ L-3 1f.| < xJ 2.g(s): {;l: (c) st'l (u) ii'-;;1}; (a),riq.s(s); "!1_s(s);111 s.h(x): {?t!i lf ; : l}, ,n I*nr,t' or }:Th(x\;(c)rimft(x) a. Fk): tt'1,tu) F(x);(c)rimF(r) l{ - r, il;:2,}; @llT }11'_ (2r+3 s. f (r) : li 17 = 2r ifr<1l if r: r [; (a) trmf (r); (b) lim f (r); (c)lim /(r) if 7 < r) r''- r-t- r'r y-1 11 \1't5 ,1 LIMITSAND CONTINUITY [3 +t2 if r<-2] 6 . 8(f):19. - f., :f"r=-21, it-Z < t) Ul (a).lims(r);(b)lim g(f);(c)lims(f) t--2+ t--z-" t--2 F(x): lx- sl;f") F(x);trl F(x);(c)limF(r) l'-T l'_? 8. f (x):3 + l2x- ah(a)tryf @; (b) (x);(c)rr\ f (x) l'* f 7. 9. f(x) :f; l-l (a)rim f (x); (b) lim /(r); (c)lim f(x) I r-O+ ft-O- t-O 10. The absolute value of the signum function (see Exercise 22 in Exercises 1.g). (") rl; to) lsgnrl; rc) lsgnrl. 1'_T"lsgn llT. I'i 1 1 . F is the function of Exercise30 in Exercises1.g. Find, if they exist: (") F(x);(b) F(x);G) F(r). 11[ ]1T1T 12. lt is the functionof Exercise31 in Exer€ises 1.8.Find, if they exis* (a) lim h(r), (b) lim_ft(r); (c)lim l(r). 13. Find, il they exisr (a) liq [r]l; (b) tim [r]l; (c) tim l[rj. ,-2a_2 14' Let G(t) : ["n + [a - x]1.Draw a sketchof the graphof G.Find,if theyexish(a)rim G(r);(b) lim G(r);(c)lim G(r). 15. Given if x=l xl:[f +g if rsl . . , a. tr**\r_- 1l f, [x*1 if r>1-'if r>1 (a) show that lim /(r) and lim /(r) both exist but ar€ not equal, and hencelim /(r) doesnot exist. r_l- - I-l+ t_l (b) Show that lim g(x) and lim g(r) both exist but are not equal, and hence u lim g(r) mg I- 1t-l+ .ifil. -E\rl. (g) Find formulastor ; } k J _ r . r r r q a v l r t . u r q r l ( r l I \ L t f(x) --7(G)"Frove ( \--.- -.1'r"r .X 7'{ * r l "I * f - j . - '1: , 'fr, ),:. '. i",,,t , _ . * , ' , . ' , Ii .l "i 'f. i .: : that lim f (x) .g(r) existsby' sfiowing that tim /(x) . S(r) : Iim (x) . S(r). fr-l ,_r_ does not exist. t-l ' l Ir- -l +1 + L5. Prove Theorem2.3.9. 17. Let rf \( 'x ) : l ( - l' i f x < 0 L 1 ifx>0 show that lim /(x) does not exist but that lim l/(r)l does exist. .z'--0 r-O 2.4 INFINITE LIMITS Let f be the tunction defined by i@'):& A sketch of the graph of this function is We investigate the function values of / when .r is close to 2. Letting 2.4 INFINITELIMITS x approach 2 from the right, w€ have the values of f (x) given in Table 2.4.L. From this table we intuitively see that as r gets closer and closer T a b l e2 .4 .7 x 3 t T 3 t2 I 5 4 27 48 2l TO 300 20r T0o- 30,000 2001 T0T-6 3,000,000 to 2 through values greater than 2, f (*) increaseswithout bound; in other words, we can make f (x) as large as we please by taking r close enough to 2. To indicate that /(x) increases without bound as x approaches 2 through values greater than 2, we write Figure2.4.1 ? l i m f t : -r - * .r-z* (I Z) If we let r approach 2 from the left, we have the values of f (x) given in Table 2.4.2.'We intuitively see from this table that as r gets closer and T a b l e2 .4 .2 x 1 3 .) B 4 3 t2 27 48 D l9 10 300 199 T00. 30,000 1999 TIITT' 3,000,000 closer to 2, through values less than 2, f (x) increaseswithout bound; so we write f i m $ : *-c c ,-z- \X z)- Therefore/as )c approaches2 fuomeither the right or the left, f (x) increases without bound, and we write h m j -T: *-z \x *oo We have the following definition. 2.4.1 Definition Let f be a function which is defined at every number in some open interval I containing a, exceptpossibly at the number a itself. As x approaches a, f (x) increaseswithout bound,which is written lim /(r) I-A : *co (1) LIMITSAND CONTINUITY if foranynumberN > 0 thereexistsa 6 > 0 suchthat/(r) > Nwhenever 0 < l r - a l< D . In words, Definition 2.4.1states that we can make f (x) as large as we please (i.e., greater than any positive number N) by taking r sufficiently close to a. NorE: It should be stressed again (as in Sec. 1.1) that *m is not a real number; hence, when we write it does not have the same |2fO:*a, meaning ur l,ti f@) : L, where L is a real number. Equation (1) can be read as "the limit of f(x) as r approaches a is positive infinity." In such a case the limit does not exist, but the symbolism r'+@" indicates the behavior of the function values f(x) as x gets closer and closer to a. In an analogous manner, we can indicate the behavior of a function whose function values decrease without bound. To lead up to this, we consider the function g defined by the equation -3 /\ g\x):lJ--2]l A sketchof the graph of this function is in Fig.2.a.2. The function valuesgiven by g(r):-31(x-Z)'are the negativesof the function valuesgiven by f (x)-31(x-2)'. So for the functiong as r approaches 2, either from the right or the left, g(r) decreaseswithout bound, and we write Figure 2.4.2 -3 #W:-q) In general,we have the following definition. 2.4.2 Definition Let / be a function which is defined at every number in some open interval / containing a, exceptpossibly at the number a itself.As x approaches a, f(x) decreases without bound,which is written rim f(x) ( 2) t-A if for any number N < 0 t h e r e e x i s t s a 6 ) 0 s u c h t h a t / ( x ) < N w h e n e v e r <6. 0<lr-al NorE: Equation (2) can be read as "the limit of f (x) as r approaches a is negative infinity," noting again that the limit does not exist and the symbolisrrr tt-@tr indicates only the behavior of the function values as r approaches a. We can consider one-sided limits which are "infinite." In particular, lim /(x) : *m if f is defined at every number in some open interval (a, c) and if for any numberN > 0 thereexistsa D ) 0 suchthat/(x) > N when- 2.4 INFINITELIMITS 91 a < 6. Similar definitions can be given if lim f(x):an lim /(x) _ -@, and lim f (x) : -a. "-o* *o* suppose ,t"i n is the function defined by the equation ever 0 < x- ,2x h(x):- (3) A sketch of the graph of this function is in Fig.2.4.3. By referring to Figs. 2.4.I, 2.4.2, and 2.4.3, note the difference in the behavior of the function whose graph is sketched in Fig. 2.4.3 from the functions of the other two figures. We see that 2x lrfll r-t X- Figure2.4.3 : --r, L and 2x lim -;: L 2-vX- aco That is, for the function defined by Eq. (3), as r approaches1 through values less than L, the function values decreasewithout bound, and as x approachesL through values greater than L, the function values increase without bound. Before giving some examples,we need two limit theorems involving "infinite" limits. 2.4.3 Limit theorem 11 If r is any positive integer, then (i) lim *: ,t-O+ ** 4 ...\ ,. L [-o if r is odd (tt, lt_TF: l+.o if r ts even pRooF: We prove part (i). The proof of part (ii) is analogous and is left as an exercise (see Exercise 19). We must show that for any N > 0 there existsa6)0suchthat 1 ;tN whenever 0(r<6 or, equivalently, because r > 0 and N ) 0, 1 t"aft w h e n e v e r0 ( r < 6 or, equivalently, since r ) 0, x< whenever 0(x(6 92 LIMITSAND CONTINUITY The above statement holds if E: 1 ;t N w henever (l/Nr;tt'. We conclude that 0 ( x ( 6,i f 6: /1\r/r (N / I o ILLusrRArroN L: From Limit theorem 11(i), it follows that h m 4: t-O* )P * oo and 1i m 1: ,.o* f * cc From Limit theorem 11(ii), we have rim 4: r-o-f, -@ and hm 1: t-o- X* *oo 2.4.4 Limit theorem 1.2 lf. a is any real number, and if lim /(x) : 0 and lim g(r) : c, where c is a t-A constant not equal to 0, then (0 if c ) 0 andif f(x)- 0throughpositivevaluesof f(x), 'jT#:*m (ii) if c ) 0 andif f (x)- 0 throughnegativevaluesof f (x), Ir#:-co (iii) if c < 0 andif f(x) - 0throughpositivevalues of.f(x), n#:-m (iv) if c < 0 and if f(x) - 0 through negative values of f(x), n8:*oo The theorem is also valid if "x + a" is replaced by "x + a*" ot "x + a-." pRooF: We prove part (i) and leave the proofs of the other parts as exercises (see Exercises 20-22). To prove that ,. , q(r) lim i*: *co " -" l \x) we must show that for anv N ) 0 there exists a D ) 0 such that #>N w h e n e v eor < l r - o l < o (4) Sincelim g(r) - c > 0,by taking e: tc in Definition 2.1.1,it follows that t-A 2.4 INFINITELIMITS 93 there exists a 0r ) 0 such that lg(r) - cl < trc whenever 0 <lr-al <0, By applying Theorcm 1.2.2 to the above inequality, it follows that there existsaDr)0suchthat -tc<Sk) <6t c < * c w h e n e v e r0 < l r - r l or, equivalently, *c<g(x)<tc w h e n e v e r0 ( l * - o l <Et So there exists a Dr ) 0 such that (5) whenever 0 < lx - al < 6' Now lt^ f (*): 0. Thus, for any € ) 0, there exists a 6z ) 0 such that g(x) > ic l/(r)l<e w h e n e v e r0 < l r - a l <6' Since f (x) is approaching zero through positive values of f (x), we can remove the absolute-value bars around /(x); hence, for any € ) 0 there existsaEz)0suchthat f(x)<, w h e n e v e r0 < l * - o l <0, (5) From statements(5) and (6) we can conclude that for any € > 0 there exist a Dr ) 0 and a 6z ) 0 such that g_2rit f(x) ' Hence, if e : € w h e n e v e r0 < l r - a l cl2N and E: &) t--r p *t N:N M <6, and 0<lr-al<6, min(Dt, Er), then w h e n e v e r0 < l r - - o l < S which is statement (4). Hence, part (i) is proved. t SOLUTION: x2+x+2 ,. (u)llTm:l,TIm x2*x*2 The limit of the numerator rs L4, which can be easily verified. lim (r-3)(r+ 1): lim (r-3)' lim (r+ 1) The limit of the denominator is 0, and the denominator is approaching 0 through positive values. Then applying Limit theorem Lz(i), we obtain 94 LIMITSAND CONTINUITY )c2+x+2 (r-3)(r+t) As in part (a), the limit of the numerator is L4. li? ('- 3)(r + 1): llT ('- 3) ' lim (r + 1) =0'4:0 In this case, the limit of the denominator is zero, but the denominator is approaching zero through negative values. Applying Limit theorem lz(ii), we have s o r , u r r o N : ( a )B e c a u sxe+ 2 + ,x - 2 ,rlilh. ^ \--------E - 4--- , :: , -rlli:i. - @x-2* x-2 *-* : lim I-2+ ) 0,andso.r- Z: {@-2Y. Thus, !(x-2\2 \/x-2\/x+2 !x-2 lx-2 Yx*2 ,r ::'-Tffi The limit of the numerator is 2. The limit of the denominator is 0, and the denominator is approaching 0 through positive values. Therefore, by Limit theorem 12(i) it follows that ,. {7-4 IIIIIt-2+ X- ^ Z --T€ (b) Because x+2-, Therefore, \/4=7 : r l l f.. ,. r: rill:r------t-z- x-'2 x-2<0, and so x-2--(2-x):-\/Q=T \/r- r-2--V2Y2* ,r :j'_T:ffi \/r+ x x Y2- x x The limit of the numerator is2. The limit of the denominator is 0, and the denominator is approaching 0 through negative values. Hence, by Limit theorem lz(ii), we have \/4=T l r m . 7 - : -- € Z .x-2- X 2.4 INFINITELIMITS ExAMPLE 3: sor,urroN: lim [r]l :3., Therefore, lim ([rl - 4): -t. Furthermore, (t - 4) : 0, and x - 4 is approaching 0 through negative values. l:? Hence, from Limit theorem lZ(iv), we have Find: u* l[4];4 x-4 r-4- u * f lXr- n+ l 4 : * o r-4- Remember that because *co and -oo are not numbers, the Limit theorems 1-10 of Sec.2.2 do not hold for "infinite" limits. However/ we do have the following properties regarding such limits. The proofs are left as exercises(seeExercises23-25). (i) If lim /(r) : *@, and lim g(r) : c, where c is any constant, then 2.4.5 Theorem t^lf(x) +g(r)I:+* (ii) If lim /(r) : -Q, and lim 8@) : c, where c is any constant, then t-A r^ lf (x)+ g(r)l: -a" is replaced by "x+ The theorem is valid if. "x+ fl*" ot "x+ A-." :1, it follows o rLLUSrRArroN 2: Because lim +:- ''t' *m and lim + 2-2+x * 2 + s-2+x from Theorem 2.4.5(i) that ,' 1 I r r m tI _ _ _ _ _ + _ + l : * m i-i;Lx-2' x*21 2.4.6Theorem If l;af{x):*.o and . fiB@): c, where c is any constantexcept0, then (i) if c ) 0,I\f O>. s@): *oo. (ii) if c I 0,l\f(x). g(r) : -o. The theorem is valid if. "x -> a" is replaced by . rr,r,usrRArrorq 3: q l i m . . * f -u : * o 5) r_s tx r r. x*4 and lim *-s x- . --7 4 Therefore, from Theorem 2.4.6(ii), we have S i. I ' x*Af 'iT LG-3F x-41:-€' 96 LIMITSAND CONTINUITY 2.4.7 Theorem If t;yf@)_-oo and limg(x):c, where c is any constant except 0, then (i ) i f c ) O,Li ^ f (x)' S (r) : -oo. (i i ) i f c I 0,l t^ f (x)' g(r): T h e th e orem i s val i d i f " x+ * o. a" i s repl aced by " )c+ a* " ot" x+ a- . " o ILLUsrRArroN 4: In Example 2(b) we showed {44 ,llm. -----: -oo ;:;- x-2 Furthermore, ,. x-3 llm -- _ ;:;--x + 2 t 4 Thus, from Theorem 2.4.7(ii), it follows that .lfm . l$=7 l--:;:;-L x-2 Exercises 2.4 ,,Q In Exercises1 through"l.4, evaluatethe limit. 1. lim -x2-a+X ' 4 t+) 4.rimffi ,. t/s+* /. rlm .r*o- X' '\R 10. lim r-4- X'- 4 Dm(i-+) 16. Prove that lim 17. Prove that lim ? ffi: -) ffi: *o by using Definition 2.4.1. -@ by using Definition 2.4.2. 18. Use Definition 2.4.1,to prove that jlils+'l :*,. 15-rl x - 3 1| : x*21 r€ o AT A NUMBER OF A FUNCTION 2.5 CONTINUIry 19. Prove Theorem 2.4.3(ii). 20. Prove Theorem 2.4.4(1i). 21. Prove Theorem 2.4.4(iii). 22. Prove Theorem Z.a.a(iv\. 23. Prove Theorem 2.4.5. 24. Prove Theorem 2.4.6. 25. Prove Theorem 2.4.7. 2.5 CONTINUITY OF A FUNCTION AT A NUMBER v In Sec.2.1 we considered the function / defined by the equation ,.,-.\ (2x * 3) (x - 1) I\x): x_.1 We noted that / is defined for all values of r except 1. A sketch of the graph consisting of all points on the line y :2x * 3 except (1, 5) is shown in Fig. 2.5.1.There is a break in the graph at the point (1,5), and we state at the number 1.. that the function f is discontinuous for instance, the function is defined for all If we define f(l):2, values of x, but there is still a break in the graph, and the function is still however, there is no break in discontinuous at L. If we define f(l):5, the graph, and the function / is said to be continuousat all values of x. We have the following definition. 2.5.1 Definition The function / is said to be continuousat the number a if and only if the following three conditions are satisfied: (i) f (a) exists. (ii) /(x) exists. L'i (iii) lim f (x): f (o). If one or more of these three conditions fails to hold at a, the function f at a. is said to be discontinuous We now consider some illustrations of discontinuous functions. In each illustration we draw a sketch of the graph, determine the points where there is a break in the graph, and show which of the three conditions in Definition 2.5.1fails to hold at each discontinuity. L: Let / be defined as follows: . rLLUSrRArroN ?r. [(2x+3)(l-1) ,-1 f(x):j lz ifx*t if x:'1. A sketch of the graph of this function is given in Fig. 2.5.2.We see that there is a break in the graph at the point where x : 1, and so we investi- LIMITSAND CONTINU]TY gate there the conditions of Definition 2.5.1.. f (1) :2; thercfore,condition (i) is satisfied. LtT fttl Itm f (*): :5; therefore,condition (ii) is satisfied. 5, but f (L) :2; therefore,condition (iii) is not satisfied. We conclude that f is discontinuous at L. o Note that if in Illustration I f (l) is defined to be 5, then lim /(r) : f (1) and / would be continuous at 1.. o rLLUsrRArroN2: Let f be defined bv 1 ,f,\( xt \ : +x _ 2 Figure2.5.3 A sketch of the graph of f is given in Fig. 2.s.2. There is a break in the graph at the point where x : 2, and so we investigate there the conditions of Definition 2.5.1.. f (2) is not defined; therefore, condition (i) is not satisfied. We conclude that f is discontinuous at 2. o o rLLUsrRArroN 3: Let g be defined by rJ 8(r): l[3 Figure2.5.4 if x # 2 ifx-2 A sketch of the graph of g is shown in Fig. 2.5.4.Investigatingthe three conditions of Definition 2.5.1,at 2 we have the following. S(2):3; therefore,condition (i) is satisfied. : *mi therefore, condition (ii) is not liT g(x) : -m, and ]i1 S(r) satisfied. Thus, g is discontinuousat 2. . . rLLUsrRArroN4: Let h be defined by 7., [3*x n\x):tg_, if.x="]. ifj,<x A sketchof the graph of h is shown in Fig. 2.5.5.Becausethere is a break in the graph at the point where x:1, we investigate the conditions of Definition 2.5.1at L. We have the following. h(t1 :4; therefore,condition (i) is satisfied. lim h(r) : lim (3 + r1 :4 t-7lim h(r) - lim (3- x):2 a-1" Figure2.5.5 ,t-l+ AT A NUMBER 99 OF A FUNCTION 2.5 CONTINUIW Because limh(x) +limh(r), we conclude that lim h(x) does not exis| lr-l+ t-l- t'l therefore, condition (ii) fails to hold at 1. Hence, h is discontinuous at 1. 5: LCt F bC dCfiNCdbY o ILLUSTRATION F(r) Figure2.5.6 flr-sl L2 ifx*3 i f .x : 3 A sketch of the graph of F is shown in Fig. 2.5.5. We investigate the three conditions of Definition 2.5.1at the point where x: 3. We have F(3) :2; therefore, condition (i) is satisfied. nttl : 0. So lim F(r) exists and is 0; therefore, tttl : 0 and l,T_ l.T condition (ii) is satisfied. tim F(r) :0 but F(3) :2; therefore,condition (iii) is not satisfied. c'3 Therefore, F is discontinuous at 3. o It should be apparent that the geometric notion of a break in the graph at a certain point is synonymous with the analytic concept of a function being discontinuous at a certain value of the independent variable. After Illustration L we mentioned that it f(1) had been defined to be 5, then / would be continuous at 1. This illustrates the concept of.aremovable discontinuity. In general, suppose that f is a function which is discontinuous at the number a,but for which f(tl exists. Then either f(a) + Tq fOl or else /(a) does not exist. Such a discontinuity is called a refi movable discontinuity because if f wete redefined at a so that f(a) l;a f@), / becomes continuous at a.If. the discontinuity is not removable it is called an essentialdiscontinuita. nxervrpr.El-: In each of the Illustrations 1 through 5, determine if the discontinuity is removable or essential. solurroN: In Illustration 1 the function is discontinuous at 1 but lilf (x): 5. By redefining f0 :5, we have lim /(x) : f (t), and so the discontir-r nuity is removable. In Illustration 2, the function / is discontinuous at 2. lim /(r) does not r'2 exist; hence, the discontinuity is essential. In Illustration 3, lim g(r) does not exist, and so the discontinuity is I-2 essential. In Illustration 4, the function h is discontinuousbecauselim h(r) does t-r not exist, and so again the discontinuity is essential. F(x):0, but F(3): 2, and so F is discontinuousat In IllustrationU, TT 3. However, if F(3) is redefined to be 0, then the function is continuous at 3, and so the discontinuity is removable. lOO LIMITSAND CONTINUITY Exercises 2.5 y ,1 In Exercises 1 through 22, draw a sketch of the graph of the function; then by observing wher€ there are breals in the graph, detemrine the values of the independent variable at which the fun*ion is discontin;ous and show why Definition 2.5.1 is not satisfied at each discontinuitv. 4 l . f .( ,x )\ : t , fx+ -4 Lt irx#4 i f .x : 4 ra_16 a.h@):fr . r+3 7.h(x):Vfrj_6 I t+x 1 0 .H ( r ) : l 2 - x l2x-1 x2+x-6 3. F(r) : x*3 6. f(x): f-2x2-llx*12 x2-5x*4 e.f(x):l t1, if x < 0 0 if x :0 I r if 0 < x v2-4 8.g(r):;d if x<-2 it-Z1x<2 if21x 13. f(x\:ffi 't [t-+*-t !6.f(x\:l ,c+2 irx#-Z 2 . g ( x.) : llx' + * 2 if x: -2 1.0 (x2+x-6 ifx*-3 5.s(r):{ r+3 ifx:-3 Lt + sl i f x + - t 1 , 1f ,( .x ) : l Z x + 5 1 1 2 .G k \ : f_r-? r a .F ( r ) : l l ' - 3 T i r x # 3 ifx:3 L0 1 s s. ( r ) : . 18 - 3 r {l2r LJ f2x*j I x*3 if.x:-* iIx<l ifl<x<-2 if.Z < x 1f'x*-2 ifx:-2 l-3 17. The greatest inteter function. (See lllustration 5 in Sec. 1.7.) 18. The unit step function, (SeeExercise21 in Exercises1.8.) 19. The signum function. (See Exercise 22 in Exercises 1.8.) 20. The function of Exercise 24 in Exercises 1.8. 21.. The function of Exercise 33 in Exercises 1.8. 22. The function of Exercise 32 in Exercises 1.8. In Exercises 23 through 30, prove that the function is discontinuous at the number c. Then detemrine if the discontinuity is rcrrovable or eaeential. If the discontinuity is removable, define l(a) so that the discontinuity is removed. 23. f(x):'ft, o:l 2a. f (s): {# [0 - '' 2s. f (x): {f ifx#3) ifx:3J;a:c ir s *-u}, ,: -g ifs:-5) r,\ [*-ax-+3 26.f(x):l x-3 [5 ze.f0):#,a:o i f x * g )- l ; a : 3 if x :3) ON CONTINUITY101 2,6 THEOREMS ( t z - L- i f t < 2 I . " : , a: zs.f(t): ll {rt riJ, z 3 0 .f ( x ) a:0 31. Let / be the function defined bY rt'r- [rXl 1 'f ( x ): lt1 tr-[x+lnl if fixn is even if flrnisodd Draw a sketch of the graph of l. At what numbers is f discontinuous? 2.6 THEOREMS ON CONTINUITY 2.6.t Theorem By applying Definition 2.5.1. and limit theorems, we have the following theorem about functions which are continuous at a number. lf f and,g are two functions which are continuous at the number a, then (i) (ii) (iiD (iv) / + g is continuous at a. / 8 is continuous at a. /' I is continuous at a. 0' flg is continuous at a, provided that g(a) * To illustrate the kind of proof required for each part of this theorem, we prove part (i): Because f and,g are continuous at a, from Definition 2.5.1 we have (1) lim/(r):f(a) 8-& and ljT st'l : s(a) Q) Therefore, from Eqs. (1) and (2) and Limit theoretrr 4, we have (3) l i m [ / ( r ) + s ( r ) l : f f u )+ g ( a ) fi'A Equation (3) is the condition that / *g is continuous at a, which furnishes the proof of (i). The proofs of parts (ii), (iii), and (iv) are left as exercises (see Exercises 1-3). 2.6.2 Theorem A polynomial function is continuous at every number. To prove this theorem, consider the polynomial function / defined by f ( x ) : b s x "* b 1 Y n -*r b 2 x " - *2' ' + b n - r x* b n b 0* 0 where n is a nonnegative integer, and bo, br, . , bn are real numbers' By successive applications of limit theorems, w€ can show that if a is 102 LIMITSAND CONTINUITY any number II tt*l : boan* brqn-r t bran-z+ . . + bn_ra* b, from which it follows that Iim f(x) : f(a) thus establishing the theorem. The details of the proof are left as an exercise (see Exercise 4). 2'6'3 Theorem A rational function is continuous at every number in its domain. PRooF: If is a rational function, it can be expressed as the quotient of / two polynomial functions. So f can be defined by o( v\ /f (\ x \: 9h ()x*)a where g and h are two polynomial functions, and the domain of consists / of all numbers except those for which h(x): g. If a is any number in the domain of I then h(a) + 0; and so by Limit theorem 9, l i m g(r) limf(x):-im h(x) .r-o ( 4) I-A Becauseg and h are polynomial functions, by Theorem 2.6.2 they are continuous at A, and so lim g(r) : g(a) and lim h(x): h(a). Consequently, t-A from Eq. (4) we have o! l i m f ('"" r ') : { \h(a) i-i;' Thus, we can conclude that / is continuous domain. at every number in its I Definition 2.5.'l'statesthat the function / is continuous at the number a iI f(a) exists,if l:\f(x) exists,and if ri^ f(r): (s) f(q) Definition2.7.1 statesthat lim f (x):Lif x'a 6 > 0 such that lf(x)-Ll <e w h e n e v e r0 < 1 " - a l for any € ) 0 there existsa <6 Applying this definition to Eq. (5), we have lim /(r) : f (a) if for any 2.6THEOREMS ONCONTINUITY 103 € > 0 there exists a 6 > 0 such that lf(x)-f(a)l <e w h e n e v e r0 ( l x - o l <D (5) If f is continuous at at we know thatf (a) exists;thus, in statement(6) it is not necessarythat lr - al > 0 becausewhen x: a, statement(6)obviously holds. We have, then, the following definition of continuity of a function by using e and 6 notation. 2.6.4Dehnition The function f is said to be continuous at the number a if / is defined on some open interval containing a and if for any € > 0 there exists a 6 > 0 such that lf(x)-f(a)l <e wheneverlx-41 <6 This alternate definition is used in proving the followirtg important theorem regarding the limit of a composite function. 2.6.5 Theorem If lim g(r) - b and if the function / is continuous at b, (/'s)@):f(b) or, equivalently, rim f (g(x) ) : /(lim S(r) ) PRooF: Because / is continuous at b, we have the following statement from Definition2.6.4. For any e1 ) 0 there exists a 61 ) 0 such that (7) <6' whenever lf@-f(Ull<., lY-bl Because lim g(x) : b, for any 6, ) 0 there exists a 62 > 0 such that lg(r) - bl < 6, whenever 0 < lx- al < 6, (8) Whenever 0 < l* - ol < Dr,we replacey in statement(7) by g(x) and obtain the following: For any €r > 0 there exists a 6r ) 0 such that - (b) < ., whenever lS(t) - bl < D' (e) lf QQ)) | f From statements (9) and (8), we conclude that for any er ) 0 there exists a 6z ) 0 such that ( l* - ol < D' IfGGD f(b) | < ., whenever 0 from which it follows that lim/(s(r)):/(b) 104 LIMITSAND CONTINUITY or, equivalently, ) : /(tims(r)) fif G<*) T An immediate application of Theorem 2.6.5 is in proving Limit theorems 9 and 10, the proofs of which were deferred until now. We restate these theorems and prove them, but before the statement of each theorem a special case is given as an example. EXAMpLE1: By using Theorem 2.6.5 and not Limit theorem 9 (limit of a quotient), find solurroN: Let the functions / and g be defined by a n d g ( r ) : x 2- t 2 x t S f(x) :4x-3 We consider /(r) lSk) as the product of /(r) and U S@) and use Limit theorem 6 (limit of a product). First of all, though, we must find lim Llg(x) 4x-3 trTfr]x+s r. and this we do by considering l,lg(x) as a composite function value. If h is the function defined by h(x) - 1lx, then the composite function h " B is the function defined by h(g(x)11: llg(x). Now : t"tg(x'+ 2x-r 5) : 13 }T sftl To use Theorem 2.6.5,h must be continuous at L3, which follows from Theorem 2.2.L7.We have, then, r.:- 1' ,. |\Nzx+s:f'3m 1, : t"T h(sk)) - n\\S(r)) (by Theorem2.6.5) : h(13) :+ Therefore, by using Limit theorem G,we have - 3)' t"'3\4x Ig v+fi 4x-Z ,. r;+ affi: -5.* - 1 75 Limit theorem 9 nf@: 7.fk\L lfm ---: L and tf ]1g-g(;r): M and,M + 0, then _ M "-i 8@) PRooF: Let h be the function defined by h(x):.l.lx. Then the composite 2.6 THEOREMSON CONTINUITY 105 function h " I is defined by h(S@D: Ug@).The function h is continuous everywhere except at 0, which follows from Theorem 2.2.7t. Hence, 1 r. lim ,,:limh(S@)) t-a *-" 8\x) - h(lim S(r) ) (by Theorem2.6.5) ,,;; -h Applying Limit theorem 6 and the above result, we have r. L r. t / \ 'lim r. lim f(x) ^ ' t-e' ,-" 8\x) fk) lim *: *-" 8\x) _L M rxavrrr.n 2: BY using Theorem 2.6.5and not Limit theorem L0, find 11^flffi solurroN; Let the functions / and h be defined by and h(fi:ffi f(x):3x-5 5 The compositefunction h. f is defined by h(f(x)) :9gxf (x) *g '1.6; lrtg tgr - 5): so, to use Theorem 2.5.5 for the composite function h " f ,h must be continuousat 76, which follows from Theorem2.2.12(1). The solution, then, is as follows. fim,vex - 5: lim h(f (x)) - nQ)) (x)) f (by rheorem2.6.5) : h(L6) :% -/ Limit theoremLo U Q) : L, then fT f lim1@:\/L if. L > 0 and n is any positive integer, or if L < 0 and n is a positive odd integer. 106 LIMITSAND CONTINUITY pRooF: Let h be the function defined by h(t) : Vr. Then the composite function h . f ts defined by h(f (x)) : \yf6. From Theorem 2.2.t2 it follows that h is continuous at L If L > 0 and n is any positive integer or if L < 0 and n is a positive odd integer. Therefore, we have lim tf@: lim h(f (x)) : ,-ri;i f @)) (byrheorem2.6.5) : h(L) :"W Another application of Theorem 2.6.5is in proving thal a continuous function of a continuousfunction is continuozs.This is now stated as a theorem. 2.6.(, Theorem. If the function g is continuous at a and the function is continuous at / g@),then the compositefunction / . g is continuous at a. pROoF: Becauseg is continuous at a we have gf'l : sk) LT (10) Now / is continuous at g(a); thus, we can apply Theorem 2.G.5 to the composite function f o g, thereby giving us lim (/. g) (r) : lim f (g(x)) :",r;*s(r)) : f (s(a)) (by Eq. (10)) : (f '8) (a) which proves that f . g is continuous at a. Theorem 2.6.6 enables us to determine the numbers for which a particular function is continuous. The following example illustrates this. EXAMPLE 3: Given h(x): \FT' determinethe valuesof x tor which h is continuous. s o r T - r r r o NI:f . g ( x ) : 4 xz andf (x): \6, then h(x): (f " il(r). Because g is a polynomial function, g is continuous everywhere. Furthermore/ / is continuous at every positive number. Therefore, by Theorem 2.6.6,ft is continuous at every number x for which g(x) > 0, that is, when 4 - xz > 0. Hence, h is continuous at every number in the open interval (-2,2). 2.6 THEOREMSON CONTINUITY 1O7 In Sec.4.4 (Continuity on an interval) we extend the concept of continuity to include continuity at an endpoint of a closed interva.l. We define "right-hand continuity" and "left-hand continuity" and with these definitions we show that the function h of Example3 is continuous on the ciosed interval [-2, 2]. Exercises 2.6 1. Prove Theorem 2.6.1(ii). 2. Prove Theorem 2.6.1(iii). 3. Prove Theorem 2.6.1(iv). 4. Prove Theorem 2.6.2, showing step by steP which limit theorems are used. In Exercises 5 through 10, show the application of Theorem 2.6.5 to find the limit. 7. lim l{/3y + + U'-4 10. rim €=F J-3 Y t 4 In Exercises11.through 26, determine all values of.x f.or which the given function is continuous. Indicate which theorems you apply. w3-1 1 1 ' f. ( x ) : 1 , 2f.( x ) : ( x - 5 ) ' ( r ' + 4 ) ' x 2 ( x+ 3 ) ' -J-2 x 2+ 2 x - 8 x3* 6* t 5x -'1,2 1 3 .g ( x ) : ; d 16. SQ\ : 7/P a 4 14.h(x): +x3-7x-6 1 5 . F (x ) : 1 7 .f ( x ) : f f - ls. F(x): {L6 -E t9.h(x\: lfi . tT4 20.f(x): {ffi 2 1 ,f.( x ) : l r - 5 l 22.g(x\ : l9 - xzl + x-'\t2)s zs.f (x)- yz(:s-z 2a.G(x): ffi,- 15 It +4 1)"' 2 5 .g ( x ) : [ \ R \ 26.F(x)- 1 -r + [r]l - [1 - r] Prove that if the function / is continuous at t, then lim f (t h) : f (t). prove that if f is continuous at a and g is discontinuous at a, then f + g is discontinuous at a. If( fr'r:E'lf;:3 . and g o (\ -r -) l: { ^ tt ifr<o tr iff>0 prove that f and I ar€ both discontinuous at 0 but that the Product f ' I is continuous at 0' 30. Give an example of two functionE that are both discontinuous at a number a' but the sum of the two functions fu continuous at 4. LIMITSANO CONTINUITY 108 31. Give an example to show that the product of tr,yo functions f and g may be continuous at a number c where / is continuous at 4 but I is discontinuous at 4. 32. If the function g is continuous at a number c and the function f is discontinuous at 4, is it possible for the quotient of the two functions, /9, to be continuous at 4? Prove )'our answer. (Chapter2) ReaiewExercises In Exercises 1 through 4, evaluate the limit, and when applicable indicate the limit theoremsbeing used. \/9-t-3 1. lim (3* - 4r * 5) J. t-2 rrlrr - ,-o t- n ---14 Jlltih In Exercises5 through 8, draw a sketch of the graph and discuss the continuity of the function. x-3 5. 6' 8G): 7. 8.F(r):# 4-x ln Exercises9 through 14, establish the limit by using Definition 2.1.1; that is, for any € > 0 find a E ) 0 such that l f ( x ) - L l < . w h e n e v e r 0< l x - a l < 6 . 9. limr(8 - 3x) :14 10. lim (2x, - x - 6) :9 11.lim l./T-3:7 t-4 3 12. , : _ _ n 2 v- t : - 1 15. Prove that lim {Fl:0 .^ ,. t'' 5*3r I T 2 ** , : 2 1,4. lim 13: -$ I--2 by showing that for any € } 0 there exists a E > 0 such that \/F4 ( e whenever 0 < t - 2 < 6. 1,6.If f (x) : ( lxl - x) lx, evaluate: (a) lim f (x); (b) lim f(x); (c) lim f (x). .T-0+ "u-0 In Exercises 17 through 24, evaluate the limit if it exists. ,+7t+ t j- , (nrm: Write --t 24. ri*[',4_-$4E .r-l+ X' L 25. D r a w a s k e t c h o f t h e g r a p ho f f i t f ( x ) : [ 7 - x " n a n d - 2 < r c < 2 . ( a ) D o e sl i m / ( x ) e x i s t ? ( b ) I s / c o n t i n u o u s a t 0 ? REVIEWEXERCISES 109 26. Dlaw a sketchof the graph of g if g@) : (x- 1)[r]l,and 0 < x < 2. (a) Does ItT Sttl exist? (b) Is g continuous at L? 27. Give an example of a function for which lim l/(r) | exists but lim /(r) does not exist. r-0 J-0 28. Give an example of a function f that is discontinuous at I for whidr (a) lim l(r) exists but /(1) does not exiEq (b) /(1) exists but tim l(r) does not exist; (c) lin /(r) and l(1) both exist but are not equal. r-l 29. Letlbe the tunctiondefinedby f1 if r is an inteser [0 if r is not an integer (a) Draw a sketchof the graph of f. (b) For what valuesof c doestim f(r) extut?(c)At what numbersis f cutinuous? rNhy? 30. If the function g is continuous at a and f is continuous at g(a), is the composite function f 'g continuous at a? ( b ) S h o w t h a t t h e c o n v e r s e ot h f etheoremin thenlti/(r+h):|tT/(r-h). 31. ( a )P r o v e t h a t i f ItT/(r+h):f(r) (a)isnottruebygivinganexampleofafunctionforwhichtjryf(:+h):lim/(r-h)but{iq/(x+h)+f(x). ' : 32. If the domain of f is the set of all real nu.urbers,and I is continuousat 0, prove that if l(c * b) f(a) f(b\ for all a and b, then f is continuousat every numb€r. 33. If the domain of I is the set of all real numbers and / is continuous at 0, Prove that tl f(a + b) = l(a) + f(b) tor dl a and b, thin / is continuous at every number. The derivative 3.1THETANGENT LINE 111 3.1 THE TANGENT LINE Many of the important problems in calculus depend on the problem of finding the tangent line to a given curye at a specific point on the curue. If the curve is a circle, we know from plane geometry that the tangent line at a point P on the circle is defined as the line intersecting the circle at only one point P. This definition does not suffice for a curve in general. For example, in Fig. 3.1.1 the line which we wish to be the tangent line to the curve at point P intersectsthe curye at another point Q. In this section we arrive at a suitable definition of the tangent line to the graph of a function at a point on the graph. We proceed by considering how we should define the slope of the tangent line at a point, becauseif we know the slope of a line and a point on the line, the line is determined. Q(xr, f(xr)) 1) left side of P. Let us denote the difference of the abscissasof Q and P by Ar so that At: xz- xr P ( x 1 ,f @ 1 ) ) Ar may be either positive or negative. The slope of the secantline PQ then is given by f(xr) f(x') Ittpe: LX provided that line PQ is not vertical. Becaus€ 12 : Xt * At, we can write the above equation as T|flpe: f(x'+Ar)-f(x') LX Now, think of point P as being tixed, and move point Q along the curve toward P; that is, Q approaches P. This is equivalent to stating Ar approaches zero the line PQ approaches the line through P which is p*uU"t to the y axis.In this case we would want the tangent line to the graph at P to be the line r : xr.The preceding discussion leads us to the following definition. 3.1.1 Definition If the function f is continuous at )cu then the tangent line to the graph of 112 THE DERIVATIVE / at the point P(*r, /(r,)) is (i) the line through P having slope m(xr), given by -"_f(xr*Ar) -f(*r) rrc*ixrl*fiat% ar*o (1) AX if this limit exists; (ii) the line x : x, if ,t^f(xt+ L{) - f(xr):*oo or-oo If neither (i) nor (ii) olDefinr,r"^"r:.1 holds, then there is no tangent line to the graph of f at the point p(xr, (xr)). f EXAMPLE1: Find the slope of the tangent line to the curve a: x2 4x * 3 at the point (xr, yt). s o l -u rro N , f(x) - x2- 4x* 3; therefore, f(xr) - xr2- 4xr* J, : (r, Ax) * Lx), 4(xr+ A r) * 3. From E q. (l ).w e have f(x r+ lim f(xr+Ax)-f(xr) m(x): Ar-O : A,x -f Lx)z- 4(x, -t A,x)* 3f lxrz - 4x, * jl lim l(x, Ar ar-0 : lim Ax At:-O : Iim Ar-0 2xrA,x*(A,x)z_-4Ax A^x BecauseAx * 0, we can divide the numerator and the denominator bv Ar and obtain m(xr) : lim ( 2 x 1 - tA x - 4 ) m(x):2h-4 (2) 3.1 THE TANGENTLINE 113 T a b l e3 . 7 . 1 2 1 0 -1 3 4 5 -1 0-2 3-4 8-5 02 34 85 0 F i g u r e3 . 1 . 3 2: Find an equation of EXAMPLE the tangent line to the curve of Example L at the Point (4, 3). solurroN: Because the slope of the tangent line at any point (xt, Ut) is given by m(xr):2h- 4 - 4:4' the slope of the tangent l i n e a t t h e p o i n t ( 4 , 3 ) L s m ( 4 ) : 2 ( 4 ) Therefore, an equation of the desired line, if we use the point-slope form, is y-3:4(x-4) 4x-y-L3-0 g.l.z Definition nxevrpr-n 3: Find the normal line to 3 which y : {F theline 5x*3Y- an equation of the curye is Parallel to 4:0. The normal line to a curve at a given point is the line perpendicular to the tangent line at that Point. write its soLUrIoN: Let I be the given line. To find the slope of l, we is which form, equation in the slope-intercept y--Zx++ -2, and the slope of the desired normal line Therefore, the slope of I is -2 becausethe two lines are parallel' is also To find the slope of the tangent line to the€iven curve at any point -3, and we have (xr, Ar),we apply plfinition 3.l.iwith f (x) \F m(x): lrP. To evaluate this limit, we rationaltze the numerator. m(xr): lr"T. 114 THE DERIVATIVE Dividing numerator and denominator by Ax (since A,x * 0), we obtain m(x,): [1 1_ 3 :--L 2Y x, - 3 Because the normal line at a point is perpendicular to the tangent line at that point, the product of their slopes is -1.. Hence, the slope of the normal line at (x1, U) is given by -z\/xr - 3 As shown above, the slope of the desired line is -2. so we solve the equation -z\/xt-3--2 giving us xt: 4 Therefore, the desired line is the line through point (4, l) on the curve and has a slope of -2. using the point-slope form oi ur,'equation of a line, we obtain y-L--2@-a) F i g u r e3 . 1 , 4 2x*y-9:0 to Fig. 3.L.4, which shows a sketch of the curve together with - _Refer the line /, the normal line pN at (4,1), and the tangent line pi at (4, r). Exercises 3.1 In Exercises 1 through 8, find the slope of ,the-tangent line to the $aph at-the point (xr, y1). Make a table of values of r, y, and m at vadous Points on the graphl and includeln the taule all ioiits wheteih" grifl, ha, h-irorriJ turrg"r,t. o."* skekh of the graph. " " l'Y:9-* 2.y:x2-6x*9 3.y:7-5x-x2 4. y: lx2 5.y:x3-3x 6.y:x3-xcz-r*L0 7.y:4x3-13x2 I4x-3 8'Y:\/x+t In Exercises 9 through 18, find an equation of the tangent line and an equation of the normal line to the given curve at the indicated point Draw a skekh of the curve together with the resulting tangent line and normal line. " 9.y:x.'-1x-5;(-2,7) 12.y :2x - x3;(-2, 4) 15. y : 6 |; 13.Y: \/q4i; R (3, 2) 18.y:Vfi; 1 0 . y : x z * 2x * 1,;(1, 4) e4,5) 15.y:-+; (4,-4) e3,2\ 1 1 .y : t x 3 ;( 4 , 8 ) 14. y: \/+i=; (3, 3) 17.y : {i; @,2) MOTION .II5 VELOCIryIN BECTILINEAR 3.2 INSTANTANEOUS + 3 that is Parallelto the line 8r - y * 3: 0' that is parallel to the linezx+78y-9=o' 20. Find an equation of the nomral line to the curve y:t3-3r is PerPendicularto the line x+2y-77:0' equation of the tangent line to the curve y: \/b=-1that f|' Fird "r, -2) that is tantent to the curve y: r'- 7' 22. Find an equation of each line through the point (3, -6) that is tangent to the curve y: 3xu- 8' 23. Find an equation of each lin€ through th€ Point (2, 24. Prcve analytically that therc is no line through the Point (1, 2) that is talSent to the curve y: 4 x,. 19. Find an equation of the tantent line to the curve y:21 3.2 INSTANTANEOUS VELOCITY IN RECTILINEAR MOTION Consider a particle moving along a straight line. Such a motion is called rectilinear motion. One diiection is chosen arbitrarily as positive, and the opposite direction is negative. For simplicity, w€ assll'me that the motion of the particle is along a horizontal line, with distance to the right as positirru u.d distance to the left as negative. Select some point on the line and denote it by the letter O. Let / be the function determining the directed distance of the particle from O at any particular time. To be more specific, let s feet be the directed distance of the particle from O at t r".o1d, of time. Then / is the function defined by the equation (1) s: f (t) which gives the directed distance from the point O to the particle at a particular instant of time. Equation (1) is called an equationof motion of' the Particle. O ILLUSTRATION 1: LCt s--P+2t-3 Then, when t:0, s: -3; therefore,the particle is 3 ft to the left of point O when f : 0. When t: I, s : 0; so the particle is at point O at l- sec. When t: 2, s : 5; so the particle is 5 ft to the right of point O at 2 sec. When t:3, s: !2; so the particleis 12 ft to the right of point O at 3 sec. Figure g.z.L illustrates the various positions of the particle for specific values of f. the particle moves from the Between the time fot t: L and t-3, point where s : 0 to the point where s:12; thus, in the 2-sec interval ihe change in the directed distance from O is L2 ft. The average velocity of the pirti.t" is the ratio of the change in the directed distance from a 116 THEDERIVATIVE fixed point to the change in the time. So the number of feet per second in the average velocity of the particle from t: ! to f : 3 is la: 6. From f : 0 to f : 2, the change in the directed distance from O of the particle is 8 ft, and so the number of feet per second in the average velocity of the particle ' in this 2-sec interval is $: 4. . In Illustration 1, the average velocity of the particle is obviously not constanU and the average velocity supplies no specific information about the motion of the particle at any particular instant. For example, if a person is driving a car a distance of 70 miles in the same direction, and it takes him 2 hr, we say his average velocity in traveling that distance is 35 mi/hr. However, from this information we cannot determine the speedometer reading of the car at any particular time in the 2-hr penod. The speedometer reading at a specific time is referred to as the iistantaneous aelocity, The following discussion enables us to arrive at a definition of what is meant by "instantaneous velocity.,, I9t Eq. (1) define s (the number of feet in the directed distance of the particle from point O) as a function of f (the number of seconds in the time)' When t: tr, s: sr. The change in tire directed distance from O is (s - st) ft over the interv al of time (t - tr) sec, and the number of feet per second in the average velocity of the particle over this interval of time is given by s-sr t-t, or, because s: f(t) - f(t,) t- f (t) and st: f (t), the average verocity is found from tt Now the shorter the interval is from t, to t, the closer the average velocity will be to what we would intuitively think of as the instantaneous veloclty at tr. For example, if the speedometer reading of a car as it passes a point Pt is 40 mi/hr and if a point p is, for instance, 100 ft from pr, then the average velocity of the car as it travels this 100 ft will very likely be close to 40 mi/hr because the variation of the velocity of the car along this short stretch is probably slight. Now, if the distance irorn p, to p were shortened to 50 ft, the average velocity of the car in this interyal would be even closer to the speedometer reading of the car as it passes pr. We can continue this process, and the speedometer reading ui ê, ."n b" represented as the limit of the average velocity between P, ind pas p approaches pr. We have, then, the following definition. 3.2.7 Definition rf f is a function given by the equation s: f(t) (1) and a particle is moving along a straight line so that s is the number of MOTION 117 VELOCIWIN RECTILINEAR 3.2 INSTANTANEOUS units in the directed distance of the particle from a fixed point on the line at f units of time, then the instantaneousuelocity of the particle at f1 units of time is u(f1) units of velocity where if this limit exists. Because t + tr, we can write (3) t:h+At and conclude that "t -+ tl' is equivalent to "Lt + 0" (4) From Eqs. (2) and (3) and statement (4), we obtain the following expression for a(tt): + Ll +af)- (f') (5) if this limit exists. Formula (5) can be substituted for formula (2) in the definition of instantaneous velocity. The instantaneous velocity may be either positive or negative, depending on whether the particle is moving along the line in the positive -or the negative direction. When the instantaneous velocity is zero, the particle is at rest. The speed of a particle at any time is defined as the absolute value of the instantaneous velocity. Hence, the speed is a nonnegative number. The terms "speed" and "instantaneous velOcity" are often cOnfused. It should be noted that the speed indicates only how fast the particle is moving, whereas the instantaneous velocity also tells us the direction of motion. rxavrrr.r L: A Particle is moving along a straight line according to the equation of motion s:2t3-4t2+2t-L Determine the intervals of time when the particle is moving to the right and when it is moving to the left. Also determine the instant when the particle reverses its direction. s o l u r r o N f U ) : Z t B - 4 t 2+ 2 t - t ' Applying formula (2) in the definition of the instantaneousvelocity of the particle al tr, we have a(t): lim f t-tt @ -- f - ( t ) L Ll - (zttg- Attz+ 2h - 1) : Iim (2t3 4t2J 2t 1) t-tt 2(t3 - tf) - 4(t2 - ttz) + 2(t - t) : lim t-tr 2 ( t- t ) l ( t z + t h + t f ) - 2 ( t+ t ) + L f 118 THEDERIVATIVE Dividing the numerator and the denominator by t - t, (becauset + t), we obtain a(tr) : lim z(tz + ttt + tr, - 2t - 2tr + 't) I _tt : 2 ( t r 2 * t r ,I t r ,- 2 t ,- 2 t 1 + l ) :2(3tt2-4h+1,) : 2 ( 3 t , - 1 )( f r - 1 ) I T a b l e3 . 2 . 1 tr- 1 f'<* L -l Ll-g zr(fr) is negative, and the particle is moving to the left "l' 11t, Table3.2.2 -1 0 + 1 2 -g -L2 1,6 -#o -1 310 0 -* a(tr) is positive, and the particle is moving to the right u(tr) is zero, and the particle is changing direction from right to left *<f, tr: Conclusion -10-5o510 Figure3.2.2 o(fr) is zero, and the particle is changing direction from left to right zr(fr) is positive, and the particle is moving to the right IN RECTILINEAR MOTION 119 VELOCITY 3.2 INSTANTANEOUS In the above example, u(ft) can also be found by applying formula (5). (See Exercise 1.) EXAMPLE2: A ball is thrown vertically upward from the ground with an initial velocitY of 64 ftlsec.If the positive direction of the distance from the starting point is up, the equation of motion is solurroN: a(t) is the number of feet per second in the instantaneous velocity of the ball at f, sec. where f (t) : -16t2 + 64t s: f (t) Applying formula (2), we find that a(tr\:tr-Tfry s: -L 6t z + 64t If f is the number of seconds in the time which has elaPsed since the ball was thrown, and s is the number of feet in the distance of the ball from the starting Point at f sec, find: (a) the instantaneous velocity of the ball at the end of L sec; (b) the instantaneous velocitY of the ball at the end of 3 sec; (c) how many seconds it takes the ball to reach its highest point; (d) how high the ball will go; (e) the sPeed of the ball at the end of 1 sec and at the end of 3 sec; (f ) how many seconds it takes the ball to reach the ground; (g) the instantaneous velocity of the ball when it reaches the ground. At the end of L sec is the ball rising or falling? At the end of 3 sec is the ball rising or falling? : lim -16t2 + 54t- (-L6tf + 64t) t - t-tt : lim tt -1'6(t2- tf) + 64(t - t) t - t-tt tt Dividing the numerator and the denominator by t h (becauset + t), we obtain o(t'): tt-T [-16(r + 11)+ 64]: -32h + 54 ( a ) o ( 1 ) : - 3 2 ( 1 ) + 5 4 : 3 2 ; s o a t t h e e n d o f L s e c t h e b a l li s r i s i n g with an instantaneousvelocity of 32 ft/sec. (b) u(3) : -g2(3) + 64: -32; so at the end of 3 secthe baII is falling with an instantaneousvelocity of -32 ftlsec. (c) The ball reaches its highest point when the direction of motion changes,that is, when o(tt): 0. Settinga(tt) - 0, we obtain -32' * 54:0 Thus, h:2 (d) when t:2, s: 64; therefore,the ball reachesa highest point of 64 ft above the starting Point. (e) lr(tr)lis the number of feet per secondin the speedof theball at f1 sec. l a ( 1 )l : 3 2 a n d l u ( 3 )l : 3 2 (f) The ball will reach the ground when s: 0. Setting 5: 0, we have -15t2 + 64t: 0, from which we obtain f : 0 and t: 4. Therefore,the ball will reach the ground in 4 sec. -64; when the ball reachesthe ground, its instantaneous G) u(4) : velocitv is -64 ftlsec. 1 20 THE DERIVATIVE Table 3.2.3gives values of s and a for some specificvalues of f. The motion of the ball is indicated in Fig. 3.2.3.The motion is assumedto be in a straight vertical line, and the behavior of the motion is indicated to the left of the line. (r:S) T a b l e3 . 2 . 3 0 ,I 1 2 3 , 4 G:zt 0 28 48 64 48 28 0 64 48 32 0 -32 -48 -64 (t:4) Figure3.2.3 Exercises3.2 t 1. Apply formula (5) to find 2,(t1) for the rectilinear motion in Example 1. 2. Apply formula (5) to find r(tr) {or the rectilinear motion in Example 2. In Exercises 3-through 8, a particle is rnoving along a horizontal lirle according to the given equation of motion, rgher€ s ft b the directed distance of the pa*ide fron a pointb at I sec. the instantineo"" .iG,iiis-eclt find r, sec; and ttren find z(f,) for t}le pa icular value of t, given. "lfoatf 3 . s : 3 P+ l ; t r : 3 .11, o' s: 4 t ;h : , 4. s:8-t2; fr:5 5. s: GT; ,52 ,J,':m)h:2 tr: j 8. s: Vl-+Z\:6 In Exerogel 9 thmugh 12, the motion of a particle is along a horizontal line according !o the given equation of motion, where s ft is the directed distance of.the particle from a point o at t sec. The positive direction is to the right. Determine the intervals of time when the Particle is moving to the ight and when it is moving to the left. Also deterErine when the Partide rcveres its direction. Show the behavior of the iotion by a figure similar to Fig. g,2.2, choosing values of, at random but induding the values of f when the particle reverses it; direction. . . 9 .s : t 3 + 3 t 2 - g t + 4 "r I..' L 0 .s : 2 t 3 - g t z - l z t + g l+t s: 1-,+t r 1) c: t- L+t2 $. rf an obiect fa s ftom rest, its equation of motion i6 s : -161e, where f is the number of seconds in the tirne that has ' d"Pl"l since the.obiect left the starting point, s is the number of feet in tre digtance oith" oi;o rro gr" Ii I Point at , 8ec, and the Positive direction is upward. If a stone b dropped from a bui in! zleiirigh, ri"rd: (a)"r"*ir,g the instantaneou6 velocity of the stone I sec after it is dropped; (b) the initantaneous velociti of the stine 2 sec after it is 3.3 THE DERIVATIVE OF A FUNCTION 121 dropped; (c) how long it takes the stone to reach the ground; (d) the in8tantaneous velocity of the stone when it reaches the ground. is thrown vedically upward from the ground with an initial velocity of 32 ft/sec, the equation of motion i8 hi. It u ',., "t*" -l32t, where t sec is the time that has elapsed since the Etonewas thrown, s ft is the distance of the stone s: -l6P from the starting point at t sec, and the positive direction is upward. Find: (a) the average velocity of the 6tone during thetimeinterval:*-t-*(b)theinstantareousvelocityofthestoneat*secandat*sec;(c)thesPeedofthestone at I sec and at I sec; (d) the average velocity of the stone during the time interval: t = t = *, (e) how many seconds it rvill take the stone to reach the highest point; (f ) how high the stone will go; (g) how many seconds it will take the stone to reach the gFound; (h) the instantaneous velocity of the stone when it readles the ground. Show the behavior of the motion by a figur€ similar to Fig. 3.2.3. 15. If a ball is given a push so that it has an initial velocity ot 24 ftlsec down a certain inclined plane, then s : 241* 1012, where s ft is the distance of the ball from the starting point at t Bec and the Positive direction i3 down the inclined plane. (a) What is the instantaneous velocity of the batl at I, sec? (b) How long does it take for the velocity to increase to tl8 ft/sec? : 16, A rocket i6 fired vertically upward, and it is s ft above the ground t sec after being fired, where s 560t 16tsand the long it takes for positive dircction is upward. Find (a) the velocity of the rocket 2 sec after being fired, and (b) how the rocket to reach its maximum height. 3.3 THE DERIVATIVE OF A FUNCTION In Sec. 3.1 the slope of the tangent line to the graph of y: point (xr, f (x')) is defined by 1 \ ftt\xrl : ,. lrm ar-o fkt+Ar)-f(xt) f (x) at the (1) Lx if this limit exists. In Sec. 3.2 we learned that if a particle is moving along a straight line so that its directed distance s units from a fixed Point at f units of time is given by r : /(f), then if r:(tr) units of velocity is the instantaneous velocity of the particle at f1 units of time, a(t):lli'ry (2) The limits in formulas (1) and (2) are of the same form. This type of limit occurs in other problems too, and it is given a sPecific name. 3.3.1 Definition ',such that The deriuatiae of the function / is that function, denoted by f its value at any number x in the domain of / is given by yi+5xYtgry, (3) '(x) is read "f prime of x.") if this limit exists. (f is read "f prime," and f '(x) is D"f(x), which is read Another symbol that is used instead of f "the derivative of f of x with resPectto x." If y - f (x), then f '(x) is the derivative of y with resPectto x, and we use the notation D*A. The notaliony'is also used for the derivative of y 122 THE DERIVATIVE with respect to an independent variable, if the independent variable is understood. If r, is a particular number in the domain of f, then ( 4) if this limit exists. Comparing formulas (1) and (4), note that the slope of the tangent line to the graph of y: f (x) at the point (xr, (xr)) is precisely f the derivative of / evaluated at xr. If a particle is moving along a straight line according to the equation of motion s_ /(/), then upon comparing formulas (2) utra 1+;, it is seen that a(tt) in the definition of instantaneous velocity of the particle at f, is the derivative of f evaluated at t, or, equivalently, the deiivative of s with respect to f evaluated at fr. EXAMPLE1: Given f (x): hcz+ L2, find the derivative of. f. SOLUTION: If x is any number in the domain of f , ftom Eq. (3) we obtain ,,* f(x+Ar)-f(x1 f ' ( x ) - i}l [3(r + Lx)' + 121- Gr, + t2) ,:_ IIITI AX : lim 3x2l6x A,x* 3(Ar)' + 12- Jxz L2 A,x Ar_o : lim 6x A'x* 3(A'x)2 : Lx Ar-o 111 (6xi 3 Ar) :5X '(x): Therefore, the derivative of f is the function 6r. The /' defined by f domain of f is the set of all real numbers, which is the same as the domain of f. In Sec. 3.2 we showed that the limit t:,nfW is equivalent to ,, f(t) - f(t') rrr-n--t-tr t - tt Therefore, an alternative formula to (4) for f,(rr) is given by (r) - f (x,) 3.3 THE DERIVATIVE OF A FUNCTION EXAMPLE 2: For the function / of Example 1, find the derivative of f at 2 in three ways: (a) apply formula ( ); (b) apply formula (5); (c) substitute 2 for r in the expressionfor f '(x) in Example1. (a) Applying SOLUTION: f ' ( 2 ): 129 formula (4), we have lim Ar-0 lim A.z-0 lim Ar-O 1 !.__ :um - - 12 A ,x+ 3(A x)' z ^fi : (n+'^', :LZ (b) From formula (5) we obtain f,(2):LTry : (3x' + tZ) - 24 ,,,llr-I1 ,'...' :llmx-2 ,*:,;, x- /. (x-2)(x+2) -Jlllrl- :r;; (.+2; :12 ' (c) Because,from ExampleL, f ' (x) : 6x, we obtain f (2) : 12. If the function / is given by the equation y: f(x), we can let Ly:f(*+Lx)-f(x) '(x), so that from formula (3) we have and writs DrA in place of f (6) A derivative is sometimes indicated by the notation dyldx. But we avoid this symbolism until we have defined what is meant by dY and dx. 124 THE DERIVATIVE EXAMPLE 3: Given 2* x Y: g= find D;y. SOLUTION: DrA: li- & or-u Ar : lim An-0 : lim A.r-0 Ar-0 5 (3-x-Ar)(3-r) lim Ar-0 5 : rxavrpln 4: find /'(r). Given /(x) ( 2 + x + A x ) l G - x - A x )- ( 2 + x ) l G - x ) A,x 5Ax Ar(3-x-Ar)(3-r) : lim : f(x+Lx)-f(x) Ax - r;' 13, SOLUTION: f@ + a'x) f(x), f' (x): lim . ( x * A r ) z t -s x 2 t 3 o"-oT:it"Tof We rationalize the numerator in order to obtain a common factor of Ar in the numerator and the denominator; this yields f'(x): ltl : lim Ar-0 (x*Lx)z-f Ar[ (r + Ax) nt7+ (x + Ar) zts*zre -; *er1 _ r:* :ili : lim As-o : lim Ar-0 x2 * Zx(A,x) * (Lx)z - x2 2x(A,x)* (Ar), Axl(x + Ar) nt7+ (x * Ar) zt%czts + x4t3l 2x* Ax (x * Ar) 4t3* (x + Ar) 2tlx2t3+ x4t3 2x X4l3+)C2l3X2l3+X413 3.3 THE DERIVATIVEOF A FUNCTION 125 _2x w _2 Sxus -B -6 - 4-zO Note that /'(0) does not exist even though f is continuous at 0. A sketch of the graph of / is shown in Fig. 3.3.1. F i g u r e3 , 3 . 1 If for the function of Example 4 we evaluate ,l i. m ^ -ft(' x r t L x ) f ( x r ) Ia t r l : O Lx Ar-o+ we have .. ufii ;;:i f(0 + Ar) - f(0) ,. (Ar;"' - O A'x ;;:;. Ax -:lilll If we evaluate ,,_ -f(x, + Ax) f(r') lilfl Aa-o- Ax at .r1:0 we obtain + Ar) - f(0) _ _@ ,._ r l m - -1(0 *" .#lude that the tangent line to the graph of f attheorigin ,n"*ir", is the y axrs. Example4 shows thatf ' (x) can exist for somevalues of r in the domain of f but fail to exist for other values of r in the domain of f .W" have the following definition. 3.3.2 Definition The function / is said to be differentiable at xt if.f '(r1) exists. From Definition 3.3.2it follows that the function of Example4 is differentiable at every number except 0. 3.3.3 Definition A function is said to be differentiableif it is differentiable at every number in its domain. Exercises 3.3 In ExercisesL through L0, tind f '(r) for the given function by applying formula (3) of this section. l.f@):4**5r*3 2.f(x):vz 3.f(x):{i 126 THEDERIVATIVE a.f@): v3;E 1 7.f(x\:ax' 5 .f ( , ) : ; - , 8./(r):# x r0.f(x):ffii In Exercises L1through16,find f '(a) for the given valueoI a by applyingformula(a) of this section. i!. ltrl -1- x2;a:t 1 , 2f (. x \ : f i , a : 2 u.f(x):+-L;a:4 15.f (x) : 1/fi; ) r J .f ( x \ : f i ; a : 6 n: s 1 . 6f.( x ) : ] * x*x2;a:-j In Exercises 17 through 22, find f '(a) for the given value of a by applying formula (5) of this section. 1 7 .f ( x ) : 3 r * 2 ; a : - 3 L 8 .f ( x ) - x 2 - x * 4 ; a : 4 1 9 .f ( x ) - 2 - - x 3 ; a : - 2 20. f (x) : \/TT ei; a:7 2r.f (x): !+TE, a: l 2 2 .'fY ( xx) : + - x ; a - - 8 'l zs. Givenl(r):gi=T, findl'(r). Is / di{ferentiableat 1?Draw a sketchof the graphof I : +/(E=T),, fil:.dl'(r). Is I differenriableat f ? Dnw a sketchot the gaph of u. ctven l(x) l. 25. If g is continuousat a andl(r) : (r - a)g(r), find l,(c). (nrr.rr:UseEq. (5).) . Let f be a function whose domain is the set of all real numbers andf(c * b) : f(s) . f(b) for all a and D.Furthermore, supposethat/(0) : 1 andl'(0) odsts.Provethat/, (r) existsfor all r and that l,(r) : l'(O) . /(r). 27. If I is differentiable at a, prove that ,. f(a + Ax)- f(a - Ax) / {4) : rlm --2l;(rruvr:l(c * Ax) - f (a - Lr) : f (a + Ax)- f (a) + f (a) - f (d - Ax).) 3.4 DIFFERENTIABITITY The function of Example 4 of Sec. 3.3 is continuous at the number zero AND CONTINUITY but is not differentiable there. Hence, it may be concluded that continuity of a function at a number does not imply differentiability of the function at that number. However, differentiability does imply continuiry, which is given by the next theorem. 3.4.'1,Theorem If a function f is differentiable dt xy then / is continuous at xr PRooF: To prove that f is continuous aI xr, we must show that the three conditions of Definition 2.5.1, hold there. That is, w€ must show that (i) /(r') exists; (ii) lim /(r) exists; and (iii) lim f (x): f (xr'1. By hypothesis,/ is differentiableat )+ Therefore,f ' (xr)exists.Because by formula (5) of Sec. 3.3 f '(*r): lim f @;::f;! (1) ANDCONTINUITY 127 3.4 DIFFERENTIABILIW we conclude fhat f(xr) must exist; otherwise, the above limit has no meaning. Therefore, condition (i) holds at xt. Now, let us consider t.ty,lf(x) - f(x')l We can write r f'r)-f(r')l ''fr] t/(r)-f(x,)l:IT, L('-x') ]1m (2) Because (r- rr): o and,,+,fff: l1T, f'(xr) we apply the theorem on the limit of a product (Theorem 2.2.5\ to the right side of Eq. (2), and we have r(r) - f (xr) - n - f(x,)l: lim (x- x,) ]lt'H !\,lf(*) 0' f' (x') so that l.r\,Ufr)-f(x')l:o Then we have t"\'f(*) : ;*' r',' lli'r,,,, o',,'r'-i r* r*o'u',' :0*f(xr) which givesus lim/(r):f(x') (3) &-Ir From Eq. (3) we may conclude that conditions (ii) and (iii) for continuity t of f at 11 hold. Therefore, the theorem is proved. As previously stated, Example 4 of Sec. 3.3 shows that the converse of the above theorem is not true. Before giving an additional example of a function which is continuous at a number but which is not differentiable there, the concept of a one-sidedderiaatiae is introduced. g.4.2 Defirrition If the function / is defined at rr, then the deriaatiae from the right of f at 11, denoted by f '*(xr), is defined by (4) 128 THE DERIVATIVE or, equivalently, ., fk) -f(xr) ' - , ' ' , . : tim fitrr} ,', rt=.rr+ X-Xt (5) if the limit exists. 3.4.3 Definition If the function / is defined at rr, then the deriaatiuefrom the left of f at xr, denoted by f -(xr), is defined by lim f k' f -(x,): ai,-o_ + L9-- f (x) t)r (6) or, equivalently, f '-(x,): lim f k), f=!x) -, X 7r-nr- Xt (7) if the limit exists. EXAMTLE 1: Let f be the function defined by (,t*-1 ifx<3 f(x): t-i _, if3< r (a) Draw a sketch of the graph of f . (b) Prove that / is continuous at 3. (c) Find f '-(3) and f 'ap). (d) Is / differentiable at 3? v solurroN: (a) A sketchof the graph is shown in Fig. 9.4.1,. (b) To Prove that f is continuous at 3, we verify the three conditions for continuity at a number: (i) /(3) :5 (ii) Iim f(x) : lim ( 2 x - 1 ) : 5 &-3- !E-3- lim /(r) : lim ( 8 - r ) : 5 t-3+ t-3+ Therefore, lim /(x) : 5 fi-3 (iii) lim f(x) : f(3) E-3 Becauseconditions (i), (ii), and (iii) all hold at 3, f is continuous at 3. (c)/t(3):^"1 # ,. urrr- lzQ+Ar)-1.l -5 AX A..r-o- : lim Ax-o- Figure3.4.1 6t2-A'x-6 LX 2 A,x -;llm d,r-o- AX 1. :lim2 Ae:-O- -2 AND CONTINUITY 129 3.4 DIFFERENTIABILIW f f(3+Lx)-f(3) '+(3)= lim L,x Ar-0+ : [8 - (3 + Ax)] - s Ar lim Ar-0+ : lim 8 - 3 - A r - 5 Ar-.0+ :lim+ -Av Ar-0,* L)C : lim (-1) AJr ' 0+ --l (d) Because ,. utl- Af -0- f ( 3 +a r ) - / ( 3 ) Lx we conclude that -/(3) ,tlm . -f G + A r ) A?..n LX does not exist. Hence, / is not differentiable at 3. However, the ilerivative from the left and the derivative from the right both exist at 3. The function in Example 1 giveb us another illustration of a function which is continuous at a number but not differentiable there. 3.4 Exercises conIn Exercises1 through 14, do each of the following: (a) Draw a sketchof the graph of the function; (b) determine if f is tinuous at :r; (c) find l1(r,) and l1(rr) if they exi6u (d) determine if I is differcntiable at Ir. ( 3-2x if x<2 ( n r *' 1 if.x<-4 7 L . tf\k*\' : 4 l_x_6 xt: -4 : ifx>_4 2.f(x):ls*-i' irx>2 xt:2 3 . f ( x ) : l r - 3 l ;x , : 3 a.fk):L*lx+2lr;lh:-2 ?t s.f(x):trl1 6.f(x):lbll; == 3 ifr<o if r>o rr:0 xr:0 (r2 i.f < 7 . f ( x ) : F r , i rxr r o0 rr:0 ^ ., \, tf-+ 6. l\x): ltffi xt:2 Lf.x<2 ifx>2 130 THE DERIVATIVE e. f (x): {tr-a . ifx<t 1,1'. f (x) : Y-x+'/-.;xt: -7 ( ^ - 6 x' l i l-4-x, xt:3 ifx<-l ifx>-1, xl ,: -l xt: I 1 3 .rf \( .x ) : l l'2 1 0 ./f \( x ): { ^ , (-7-2x .t {1i'_ ,1, ir x > 12.f(x) : (x - 2)-2; xr:2 ifr<3 if r>3 M . f (/ x\ ) : { (-*2t3 ifx<0 -: ifr>0 lxr," Ir:0 15. Given l(r) : lrl, draw a sketch of the graph of I Prove that is continuous at 0. Prove that is not differentiabte / at O, f but that f'(r) : lrllx lor all x + 0. (HrNr: Let lrl : r/r1) t0 Given l(r) : \g' Prove that neither the derivative from the right at -3 nor the derivative from the left at 3 exist. Draw a sketch of the graph. 17. Given l(r) : d/r, ptove that fi(o) exist8 and find its value. Draw a sketch of the graph. (l- i')3t2, Prove that /'(r) exists for all values of r in the open interval (-1, 18 Givm/(r): 1), and that both /-i(-1) and fl(1) exist. Draw a sketdr of the graph ofl on [-1, l]. ^..a f 19-rFind the values of a and b so thatl'(1) exists if ' itx<r ,r,.,: I.f l a x +b i t x > 7 z0' Given l(r) : [tn, find /'(xr) if xr is not an integer. Prove by applying Theorem 3.4.1 that l,(r,) does not exist if ,1 is an inteter. lf x, is arl integer, what can you say about f1(x,) li(r,)Z 21. G i v e nf ( x ) : ( r - 1 ) [ x ] . D r a w a s k e t c h o ft h e g r a p ho f f o r x i""a n [ 0 , 2 ] . F i n d :( a ) f_ ( r ) ; ( b )f l ( l ) ; ( c ) f , ( r ) . f 22. Given f@):{0., lf;;3 where n ]s a Positive integer. (a) For what values of zr is differentiable for all values of x? (b) For what values of n is / /' continuous for all values of x? 2 3. G iv enf ( x ) : s g n r. (a ) P ro v eth a t/i (0 ):+ o o and,f' -(0):+ * . (b) provethat l i m /' (r) : 0 and l i m f ' (r) : 0. ' 'O+ "r 24. Let the function / be defined by ( g ( x )- g ( a ) . . tr x f a f (x) : l=--T:;if x: a Lg'(a) J.-O- Prove that if g'@) exists,/ is continuous at a. 3'5 SOME THEOREMS ON DIFFERENTIATION oF ALGEBRAIC FUNCTIONS The operation of finding the derivative of a function is called differentiation, which can be peiformed by applying Definition 3.3.1. However, because this proc"r, i, usually rather lengthy, we state and prove some theorems which enable us to find the derivative of certain functions more easily. These theorems are proved by applying Definition 3.3.1. Following the proof of each theorem, we state the corresponding formula of differentiation. OF ALGEBRAICFUNCTIONS 131 3.5 SOME THEOREMSON DIFFERENTIATION 3.5.1 Theorem If c is a constant and if f (x): c for all x, then f'(r) - o PROOF: f'(x):ltg'ry :lt'T,T limO , I :Q ffiffiffi (1) The deriaatiaeof a constantis zero. . rLLUsrRArIoN1: If f(x):5, then f(x):o g.5.2 Theorem lt n is a positive integer and if f (x): ro, then f'(x) : nYn-r PROOF: (tv+Lx)-f(x) f'(x):llq'tr :lt"T,ry Applying the binomial theorem to (r +Ar)n, wQ have xn + nxn-tO, *W ,n-z([y)2 * * nx(Lx)"-l * (Ar)n lim f'(r): ' Ar-0 : lim nxn-r Lx*n(n;l) "4 2l ,o-z1yx\z+ . . . *nx(Ax)"-t+ (Ar)' Aa-0 Dividing the numerator and the denominator by Ax, we have f,(x):l,*|n**_,*w*n-z1Lx)+...*nx(Lx)n+(Ar),-'] 132 THE DERIVATIVE Every term, except the first, has a factor of A,x;therefore, every term, except the first, approacheszero as Ar approacheszero. So we obtain I f '(x) nx6n-r (2) o ILLUSTRATToN2: (a) lf f (x) : x8, then f'(x) : 8x7 (b) If f(x): x, then ' f'(x): L ro :L'1. -1 3'5'3 Theorem If f is a function, c is a constant, and g is the function defined by 8(r)-c.f(x) then if f '(x) exists, g'(r)-c'f'(*) PROOF: 8'k):lt*ry :l:Try : lin lf@+ Ar) - f(x)1 o"lorL:j :c.t.-/(r+AI)-/(r) Arr-.o A)C - cf'(x) i ffi##iF;rn**l (3) The deriaatiaeof a constanttimes a function is the constanttimesthe deriaatiaeof the function if this deriaatiae exists. By combining Theorems 3.5.2 and 3.5.3 we obtain the following result: If f (x) : ocn, where n is a positive integer and c is a constant, ' f (x): sn26n-r (4) FUNCTIONS 133 OF ALGEBRAIC ON DIFFERENTIATION 3.5 SOMETHEOREMS o rLLUSrRArroN3: If f(x) :5x7, then :5 ' 7xG f'(x) : 3516 3.5.4 Theorem If / and g are functions and if h is the function defined by h(x):f(x)+s@) theniI f ' (x) andg'(r) exist, h'(x):1'(x)*g'(r) PROOF: h'(x): : : h ( x* L x )- h ( x ) ll* t/(r + ar) + s(r +Ar)l [/(r) + s(r)] lr,T. Ax)- s(r)l l'"T. :1'"T'ry*hlry - 7 '@ )* 8 ' ( r ) I (s) The deriuative of the sum of two functions is the sum of their deriratiaes if these deriaatiaesexist. The result of the preceding theorem can be extended to any finite number of functions, by mathematical induction, and this is stated as another theorem. 3.5.5 Theorem The derivative of the sum of a finite number of functions is equal to the sum of their derivatives if these derivatives exist. From the preceding theorems the derivative of any polynomial func' tion can be found easilY. Exevrpr,r1: Given f(x):7xa-2x3*8x*5 hnd f ' (x). SOLUTION: f' (x) : D*(7xa 2x3* 8x + 5) : D*(7xa)+ D*(-Zx") * D*(8r) + D'(5) :28xs-6x2+8 DERIVATIVE 3.5.5 Theorem If f andg are functions and if ft is the function defined by h ( x ): f ( x ) s ! ) then if.f '(x) and g'(x) exist, h ' ( x ) : f ( x ) g '( r ) + g ( x ) f ' ( x ) PROOF: h'(x):l'nW . _- lim fG + A'x)' S@+ A,x) f(x) g(x) A,x We subtract and add f (x * Ar) . S(r) in the numerator, therebv grvlng us h'(x\ : Ln . g(r+Ar)-g(x)+g(r).f(x+ t4- r<xr11 Ax) H kt'+ ' 8(r+ ar)- 8(r)] m kr'* Ar) : lim f (x+ Ar). lim A.r- 0 Ar- 0 Ax .trrn /(x + AI)-/(x) * lim s@) " A,r .0 Ar. .0 ax Because / is differentiable at x, by Theorem 3.4.1, i" continuous at r; f therefore, lim f(x+ Ar) : f(*). Also, + AI).- g(r) _ 6R,(x) ,o, 8(r \ Ar Ar-o HW:f'(x) and : llq st'l 8(r) thus giving us h ' ( x ) : f ( x ) g '( x ) + g ( x ) f , ( x ) r OF ALGEBRAICFUNCTIONS 135 3.5 SOME THEOREMSON DIFFERENTIATION the deriaatiaeof the secondfunctionplus the secondfunction timesthe deriaatiae of tlte first functionif thesederiaatiaesexist. EXAMPLE2: h(x) : Given SOLUTION: (2*t - 4x2)(3x5 * x') find h' (r). h' (x) : (2x3- 4x',)('l.Sxa+ 2x) * (3r5 t x',)(6x2 8x) - (30r?- 50rG* 4xa- 8rt) * (18r?-24xG * 5xa- 8rt) :48x7-84x6*1014-L6xB In Example 2, note that if we multiply first and then perform the differentiation, the same result is obtained. Doing this, we have h (x ) :6 x B - L2x7* Zxs- 4xa "Thus, h' (x) : g.5.7 Theorem 'l'5x3 48x7- 84x6* L0ra - If f andg are functions and if h is the function defined by f( v\ h(x\ : wherexU) * o 8\x) thenrf f ' (x) andg'(r) exist, o ( * ) f , ( * )- f @ ) SG ) h,(x) : { [s (r) ], PROOF: h'(x):lt1W f (x + A,x)_ I@- :l'"T'ry :l'n Subtracting and adding f (x) 'g(r) h,(x):Inf in the numerator, we obtain 136 THEDERIVATIVE e+4i3. : l:l'r'r llt lli rr'r llt e.r*Fs!{ ]t"qst'l H s(r * Ar) _g(x)' f'(x)- f(x)- g'@) s ( r )' s ( r ) _s@)f'(x)-f(x)s'(x) [s(x)l (7) The deriaatiaeof the quotientof two functionsis the fraction haaingas its denominatorthe squareof the original denominator,and as its numerator the denominatortimes the deriuatiaeof the numeratorminus the numerator timesthe deriaatiaeof the denominatorif thesederiaatiaesexist. nxaupln 3: Given SOLUTION: h,(x) - k2 ax+ r)-(6!') (2r1! D Qx a) 2x3+4 h(x) : (x'-4x*1,)z xz- 4x *'1, find h'(x). 3.5.8 Theorem rf f (x) - x-n, where -n is a negative integer and.x * 0, then ' (x) - -nx-n-r f PRooF: If -n is a negative integer, then n is apositive integer. We write f(x) -1 1 Applying Theorem 3.5.7,we have -n.0-1.nxn-r _nxn-r ' f (x) : nxauprn 4: Given f(x\ :4 hnd f ' (x). ffi - -nx'|i-r-2it : -T solurroN, f (x) :3x-5. Hence, ? f '(x): 3(-5r-u): -15r-6: - g )c6 OF ALGEBRAICFUNCTIONS 137 3.5 SOME THEOREMSON DIFFERENTIATION If r is any positive or negative integer it follows from Theorems 3.5.2 and 3.5.8 that D*(xn) : rxr-r and from Theorems 3.5.2,3.5.3,and 3.5.8 we obtain D*(cxr') + s77;r-r Exercises3.5 In ExercisesL throu gh 26, differentiate the given function by applying the theorems of this section. 3. f(x) - $74- xa 2. f(x) :3xa - 5x2* 1 l.f(x):x3-3x2,I5x-2 a. g(x) : )c7- 2x5+Sxs - 7x 5. F(f) : t.to- it' 6. H(x):*r3-x*2 7. a(r) : t.nr' 8.C(y):y'ot7y"-y"+l 9. F(x):x2*Zr+ 10. f(x) : - 5 * x-2 I 4x-a .x4 13.f(s) : t6(s'- s') L6.g(x) : (4xz+ 3)' r s f. ( x ) : h 2 2 .g ( x ) : xa*2x2*5r*L zs.f (x):'# (3r- 1) 1 1 .g ( r ) : 35 12.H(x) - ,r+ j ) * 1,a.S@) : (2x2+ 5) (ar - 1) #+).x+l 15.f(x) : (2xa- 1) (5x3* 6x) 17.H\x):'ffi 18.F(y)-'# 20.f (x) : (x' - 3x * 2) (2x3+ 1') 2r.h(x):{* r*, 2a-fG):m -3-R. 23.f(x):;"fr v3+1 * r) 26.g(x): ?fr (x'- 2x-1 \,22. V f, g, andh are functionsand if O(x):f(x.)'8e) 'ft(r), prove that it f'(x)' 8'(r)' and l'(r) exist,d'(r): l(i):S@).h'(x)+f(x).g'(x).h(r)+f'(x).gG).h(r).(rru.rr:ApplyTheor€m3.5.6twice.) Use the result of Exercide27 to differentiate the functions in Exercises28 through 31' 2 8 .f ( x ) : ( * + s ) ( 2 x - s ) ( s x+ 2 ) 3 0 .g ( r ) : ( 3 r s+ r - 3 )( r + 3 ) ( f - s ) 2 9 .h ( x ) : ( 3 x + 2 ) ' ( f - 1 ' ) 3 1 .d ( x ) : ( 2 f + x + r ) z 32. Find arl equation of the tangentline to the curve y : 8/ (f + 4) at the Point (2, 1)' 33. Find an equationof eachof the lines throughthe point (-1,2) which is a tangentline to the curvey: (r-1)/(r*3). +6r+4, which is parallelto the line 34. Find an equation of each of the tangmt lines to the curve 3y:x"-3f 2x-y+3:0. - : 35. Find an equation of eachof the normal lines to the cur-sey: it 4x which is Parallelto the line x + 8y 8 0. 36. An obiectis moving alonga straightline accordingto the equationof motion s:3r/(f + 9), with t > 0'where6 ft i8 the directed distan& of the obje* from the starting point atJ 8€c.(a) What is the instantaneousvelocity of the object at ,1 sec?(b) What is the instantaneousvelocity at 1 iec? (c) At what time i5 the instantaneousveloclty zeto? 138 THEDERIVATIVE 3.6 THE DERMTIVE OF A COMPOSITE FUNCTION Suppose that y is a function of u and u, it'r.turn, is a function of t. For example,let Y : f ( u ): u 5 (1) u:8@):2x3-5x2+4 e) and Equations (1) and (2) together define y as a function of x because if we replace u in (1) by the right side of (2) we have A : h ( x ) : f \ e ) 1 : ( 2 x B _ 5 x+24 ) ' where h ts a composite function which was defined previously (refer to Definition 1.8.2). We now state and prove a theorem for finding the derivative of a composite function. This theorem is known as the chain rule. 3.6.1 Theorem If y is a function of u, defined by y: f (u), and Duy exists, and if u is a ChainRute function of x, defined by r: g@i, ind'Dru exists, tiren y is afunction of r and Dry exists and is given by D rA: D uy ' D ru pRooF: From Eq. (6) of Sec. 3.3, we have Duu:Iin#, Hence, when l\"l difference is small (close to zero but not equal to zero) the Au fi'- o"v is numerically small. Denoting the difference by n, we have A,u n:ft,-DuA ifLu#0 The above equation defines T as a function of Lu, provided that Lu * 0. Letting 11: F(Lu), we obtain Att F(Az):ffi- Dua if Lu * o (3) Equation (3) defines F(Aa) provided that Az * 0. In a later part of this proof , we want the function F to be continuous at 0. Thus, *" d"fi.re F(0) to be lim F(Ar.r).From Eq. (3) we have : H-'î\n,, Hllnro') llT OF A COMPOSITEFUNCTION 3.6 THE DERIVATIVE 139 Duu- D"a : Hence, we define F(0) : 0, and we have , lN-D",u F ( A u )- ' l A u " u r L0 irLu*o if\u-0 Solving Eq. (3) for LY, we obtain Ly: DuU' Lu * Lu ' F(A'u) (4 ) 1f Lu * 0 Note that Eq. (4) also holds rf Lu:0 becausewe have Ly:0 (remember t h a tL y : f ( u + A u ) - f ( u ) ) . Dividing on both sides of Eq. (4) by Ar, where Ax * 0, we obtain La = *:Dua' Att htt 6+Ô'r(Au) Taking the limit on both sides of the above equation as Ar approaches zero and applying limit theorems, we have F(Au) Hence, DrA: DuA ' Dru * Dru ' Lim F(Lu) (5) Ar-0 We now show that lim F(Az):0 by making use of Theorem 2.6-5.We Atr-O first expressAtt as a function of Ax. Because rim * ar-o AI : D,u it follows that when lAxl is small and Ax * 0, Lul Lr differs from D "u by a small number which dePends on Ar, which we call d(Ar). Hence, we write Lu ffi-- D,u* 0(Ar) if a'x* o and multiplying by Lx we obtain Lu:D*il' Lx* d(Ar) 'Ar if Lx # 0 The above equation expressesAz as a function of Ar. Calling this function G, we have G(Ar) : D,il' Ar * d(Ar) ' Ar ifA,x*0 THEDERIVATIVE 140 H e n c e , F(A z) : F(G(A 1) ), and lim F(Az) : l''1 F(G(Ax)) :0 and F is continuous at 0 (rememberthat we made Because ln "(Ax) it so), we can apply Theorem2.6.5,to the right side of the aboveequation and we have lim F(Au): F(l',Tc(Ax)) : F(0) -0 So in Eq. (5) if we replace F(Az) by 0, we obtain olim D,A:DuA.Dru*Dru.0 Thus, D*A: Duy . Dru I In the following example, we apply the chain rule to the function given at the beginning of this section. EXAMPLEL: ,: Given ( 2x s- Sx z+ 4 )t soLUTroN: Considering y as a function of z, where u is a function of x, we have U : tts find D,y. w here u:2xr - S xzI 4 Therefore, from the chain rule, DrA : DuA . Dru: 5ua(6x2- 10x) : 5(2x3 - 5x2* 4)a(6x2- 10r) rxavrprn 2: f(x) : Given 4x3*5x2-7xt8 solurroN: to obtain write f (x) : (4x3 + 5x2- zx * 8)-t, and apply the chain - -1,(4xB * 5x2- Zx * 8)-2(1,2x2* llx - T\ f' (x) -1.2x2-1,0x*7 (4x"*5x2-7x+g)2 nx,q.upr.n 3: Given f(x):e#)^ find f '(x). solurroN: Applying the chain rule, we have (3r - 1)(2J- (2{ + 1)(3) ,(x\ /f \ / : 4- PJI l)3 (3r-1), \3r-1l OF A COMPOSITEFUNCTION 3.6 THE DERIVATIVE ExAMPLE 4: Given : (3x2t 2)'(x' - 5r)' f(x) find f '(x). 141 Consider f as the product of the two functionsg and h, where : (x2- Sx)B 8(r) : (3x2* 2)z and h(x) Using Theorem 3.5.5 for the derivative of the product of two functions, we have ' ( x ) : g @ ) h ' ( x )+ h ( x ) g ' ( x ) f We find h' (x) and g'(r) by the chain rule, thus giving us i.J 'i i i! t ' ( x ) : ( 3 x z* Z ) ' I g k ' 5x)'(2x- 5)l + (x'- 5r)3[2(3r'+ 2)(6r)] f : 3 ( 3 f * 2 ) ( x z- 5 r ) ' [ ( 3 x 2* 2 ) ( 2 x - 5 ) + 4 x ( f - 5 r ) ] : 3(3x2* 2) (x2- 5r)'16x" - L1xzt 4x- 10 * 4xs- 20x21 solurroN: : 3(3f * 2) (x2- 5x)'(10r3- 35xz* 4x- 10) 3.6 Exercises In Exercises1 through20, find the derivative of the given function. 2. f(x): (10- Sx)n 1. F(r) : (x'* 4r - 5)3 a. S(r) : (zta* 8rz + 1)' 7. h(u) : ( 3 u 2* 5 ) 3 ( 3 u _1 ) ' 5 . f ( x ) : ( x* 4 ) - ' 8. f(x): (4xzr7)2(2x3+ 1)' /)t2 + l\2 ' 1_2. g(t): (fr,,117 1;-r 10.f (x): (f - n*-z1z1xz* i'-r-l. f(il: 13.f(x):*fu M h(x):Gj!#*6)' (F)'' - 4) : t6.f(il : (y*s)'(sy+ L)2(3y2 -r7.f(z) ffi, Tg.G(x):ffi g. f(t) : (2t4- 7tB+ 2t - l)2 5. H(z) : (2"- 322* t1-e 9. S (r) : (2x - 5)-' (4x * 3\-2 (\t + s)' 15.f(r) : (r?+ 7)3(1r 18. g(r) : (2x - 9)t(x" t 4x - 5)3 - 20.F(x):ffi- g)-2 21. A particle i5 moving along a straight line according to the equation of motion s: [(t2 l)/(rc + 1)]" with t = 0, velocity of the instantaneous What i3 the (a) t sec. the origin at particle frorn oif the where s ft is the dire-cted distance particle at t1 sec? (b) Whai is the instantaneous velocity of the partide at 1 sec? (c) What is the instantaneous velocity at t sec? at each of the followint Points: (0'l.)' (1'&)' (2'+)' 22. Find an equation of the tangent line to the curve y:2/(4-r)r (g,2), (5,2), (6, *). Draw J sketch of the graph and setments of the tangent lines at the given points. - 2x - 4)2 at the point (3.:2). 23. Find an equation of the normal line to the curve y : ?l(x' Find an equation of the tangent line to the curve y: (x'- 4)'l(3x 5)2 at the point (1, *). ' (x ' ); (b ) g ' (r). 25. Gi ve nf ( x ) : 13and S @ ) : f (f).F i n d : (a )f Fi nd the deri vati veof f " g i n tw o w ays: (a) by fi rst fi ndi n g a n d g (r): (x + 7 )l (r-1 ). 26. Gi ven f( u) : u2*S u*5 U. il(r) and then finding U " 8)'@); (b) by using the chain rule. a i ? 142 THE DEFIIVATIVE 27. Suppose that I and g are two functio,nssuch that (i) g'(rl) and l'(g(r,)) terval containing xr, g(r) - g(rr) + 0. Then exist and (ii) for all x * 11in some open rn- ( / " 9 ) ( r )- ( / . 9 ) ( r , )_ ( / . s ) ( r ) - ( / . g ) ( r , ) g ( r )- s ( x , ) . x- xr x- xr s(r) - g(x') (6) (a) Provethat as , - *r, gk) - gk) and hencethat (f . s)'(rr): f' (g(xr))g'(xr) thus simplifying the proof of the chain rule under the additional hypothesis (ii); (b) Show that the proof of the chain rule given in part (a) applies it f(u): LP arrdg(x) : xs,but that it ioes not apply'ii (u) : u2 and,gtr) : sgn r. f 28. Use the chain rule to Prove that (a) the derivative of an even function is an odd function, and (b) the derivative of an odd function is an even function, provided that these derivatives exist. 29.use the result of Exercise28(a) to Prove that if g is an even function and g'G) exists, then if h(x) : (f " s) @) and,f is differentiableeverywhere,h'(0) : 0. 30. Supposethat/ and g are functionssuch thatf' (x): llx and,f(S@)):r. prove that if g,@) existstheng,(r):S(r). 31.luppose that y is a function of a, ts is a function oI u, and u is a function of.x, and,thatthe derivative sDrA,Duv,and, Dru all exist. Prove the chain rule for three functions: D,A : (D,y) (D"a) (D"u) 3.7 THE DERIVATIVE oF THE The tunction defined by / POWER FUNCTION FOR x' I\x) RATTONAL EXpONENTS (1) is called the poaw function In Sec. 3.5 we obtained the following formula for the derivative of this function when r is a positive or negatirre irrt"gen f ' (x ) : p6r-r ( 2) We now show that this formula holds when r is a rational number, with certain stipulations when x: 0. We first consider x * 0, and r : Llq, where q is a positive integer. Equation (1) then can be written f (x) : 26tro From Definition 3.3.1, we have f,(x):111 w (3) To evaluate the limit in Eq. (3) we must rationalize the numerator. To do this, we use formula (7) inSec.2.2, which is an-bn: (a-b)(a"-t so we rationalize * a " - z b+ a n - s b + z . the numerator + a z b n - i* a b " - z+ b " - r ) of the fraction (4) in Eq. (3) by applying '.qw; FORRATIONAL EXPONENTS1 4 3 FUNCTION OF THEPOWER 3.7 THEDERIVATIVE Eq. (4), where a: (r + Ar)tto, b: x7rt8,and n: q. So we multiPly the numerator and denominator by + ' + [ (r + Ar) rtqf(q-2)xrtq [ (r + Lx)rta1<o-1) + (fto)(a-rt + (r * Ar) rro(vrta)(s-z) We have, then, from Eq. (3), f lim '(x) equals [- '( r + A x ) 'u c - x v q ] [ ( x + A r ) ( c - r l r a *( r * L,xl(x*Ax)(a-trto+ (xl 1; L4v"'orrtq+ '' Lx){a-zttqxrtq+' '+ '+ x@-r\tql xG-\tql (s) and now applying Eq. (4) to the numerator we get (r * Ar)nts )csts, which is Ar. So from (5) we obtain t, f ' ( x ) : Al ri-m 0 Ar[ (r * Ar)tc-t)ta a ,* * Ar)ta-z)ta2srto + xg-r)tqf 1 : lim (r * Ar)(q-l)/s + (r * Ar)ta-z)tqxrts+ Ar-0 f a(a-r)/o 1 - y(a-t)tu j xg{s-tyu t, + x9-r)tq Because there are exactlY q terms in the denominator of the above fraction, we have 1 ' f'\x):fuo qx' or, equivalently, ,(x) 'fL l - 1 ! 26ua-t (5) which is formula (2) with r - Llq. Now, in Eq. (1) with x # o,let r: plq, whete p is any nonzero integer, and.q is any positive integer; that is, r is any rational number except zero. Then Eq. (1) is written as f (x): frq or, equivalently, f(x) : (xrraln Because p is either a positive or negative integer, we have, from the chain rule and Theorems 3.5.2 and 3.5.8, ' (x ) : ' P (x tto)e-r D r(xrtq) f Applying formula (5) for D,(x''o), we get ' '+ xuq-l 'q f (x) : P(x'to)'-' "'sjr' 144 THEDEBIVATIVE f f ,(x\ -P q ,nh-uq+uq-l '(x) -L vnrc-t 'qs (7) 1* Formula (7) is the same as forrnula (2) with r: plq. l f r:0, and x * 0, E q. (1) becomesf (x): r0: 1..In thi s case / , ( x) : 0, which can be written as f ' (r): 0 . r0-1. Therefore, formula (Z) holds if r: 0 with x * 0. We have therefore shown that formula (2) holds when r is any rational number and x * 0. Now 0 is in the domain of the power functionf if and only if r is a positive number because when r = 0, /(0) is not defined. Hence, we wish to determine for what positive values of r, f ' (0) will be given by formula (2). We must exclude the numbers in the interval (0, 1.1because for those values of r, x'-r is not a real number when r: 0. Suppose, then, that r ) 1. By the definition of a derivative f '(o):tjTH: When r ) l,lim :f -0 ljT''-' r'-r exists and equals 0 provided that r is a number such that x"-1 is defined on some open interval containing 0. For example, if r: i, x''-1: x:ttz-r:. vtt2,which is not defined on any open interval containing 0 (since r1l2 does not exist when r < 0). However, if. r x,-l - x,ts-l : x2t3, which "a, is defined on every open interval containing 0. Hence, we conclude that formula (2) gives the derivative of the power function when x: 0 provided that r is a number for which r'-r is dlfined on some oPen interval containing 0. Thus, we have proved the following theorem. 3.7.1,Theorem rf f is the power function, where r is any rational number (i.e., (x) : x,), f then f '(x) : 716r-l For this formula to give f '(0), r must be a number such that x'-1 is defined on some open interval containing 0. An immediate consequence of Theorem 3.7.1 and the chain rule is the following theorem. 3.7.2 Theorem rf f and g are functions such that /(x) : number, and if g'@) exists, then [g(x)]', where r is any rational ' f (x): rlB@)l'-'g'G) For this formula to give f '(0), r must be a number such that [g(r)]"-t defined on some open interval containing 0. is -. OF THE POWERFUNCTIONFOR RATIONALEXPONENTS 3.7 THE DERIVATIVE EXAMPLE 1: Given f (x) :4VP findf '(x). solurroN, f (x):4xzt3. Applying Theorem 3.7.1,we find f '(x) - l,' :9 I ilxzt}-rl t-r,, 38 3*u3 8 :5ffi Exauprn 2: Given h(x)-\ffiB find h'(r). solurroN h(x): (ZxB- 4x * 5;trz. Applying Theorem 3.7.2,we have - 4) h' (x) - $(Zxs- 4x* 5)-t'215x2 3x2-2 EXAMPLE Given soLUrIoN: Writing the given fraction as a product we have g@):xl(3xz_L)-us Using Theorems 3.5.6 and 3.7.2, we have g ' G) :3 x 2 ( 3f - L)-tts - i (3,x2 1' )-e' a16x)(xB ) : x2(3xz- L)-ais;3(3xz- t) - 2x2f - 3)-- *(7xz(3 f L)ri s 3.7 Exercises In Exercises1 through 18, find the derivative of the given function. 2'f(s):ffi 3. s(r) : 4.h(t):{F1 5.f(x)-{16112+5x-rt2 6. sfu) : (y, * 3)tralrs- L)ttz 7. F(x):9Fffi s.g(x):{@ffi 1. f(x): ( 3 x* 5 ; z B 10. f(x) - Ja2t3- 6xtt3 a xs-rtz \F 1 1F . (r):? '1,9. 12. ' .'-f;F- 146 THE DERIVATIVE M. f (s): ( s nl S s z* l ) - z r s 17.f(x): \/s+m 19. Find an equation of the tangent line to the cur.ve y : {f + 9 at the point (4, 5). 20. Find an equation of the tansent rine to the {urve y: (6- zx)Ltxat each of the folrowing points: (-1, 2), (L,.{y4), (3,01, 15,-94r. (7, -2). Driw a skekh of the gt"irr se-jirents of the tangent rines at the given ponts. ""a 21. Find an equation of the normal line to the cuwe y : a1/l$ ap at the origin. 2, Find, aa equation of the tantent line to the curve y:1/$?r=Twhich is perpendicular to th e line 12x_7y I2_O. 23' Alr obiect i8 moYing along a straitht line according_to the equation of motion s : l4FT3, with t > 0. Find the values of I for which the measure of rhe instantaneous vilocity is 0; (b) U (c) Z. ?a) 24'Givenf(u):llu2andg(r):{xl{2filal,findthederivativeof/"gintwoways:(a)byfirstfinding(l.g)(r) and then finding U " d,e); (b) by using the chain rule. In Exercises 25 through 28, find the derivative of the given function. (nrwr: jal : fii.y 2 5 .f ( x ) : l x z- 4 l 2 7 .g ( x ) : l r l ' 2 6 .g ( x ) : r l r l 28. h(x): VltlE 29. Supposeg(x) : lf (x) l. prove that if ,(x) and g,(x) exist,then ,(x)1. f lg,(r) | : lf : 30' Supposethat 8(r) \/FE and ft(x) : f kQ)) where / is differentiableat 3. prove that h,(0; : g. 31. If g and h are functions and if / is the function defined by f(x): [s(x)]'[h(x)]' where r and s are rationar numbers, prove that if g,@) and,h, (r) exist '(x): [ g ( r ) ] , - , l h ( x ) 1 s - r [ r. h ( x ) g ,( r ) + s . g ( r ) h , ( x ) ] f In Exercises32 through 35, use the result of Exercise3L to find the derivative 32. g(x) : (4x + 3)rt2(4 - x2)rt3 u. f(r):(#)'{r*n;,,, 3.8 IMPLICIT DIFFERENTIATION of the given function. 33. f (x): ( 3 x - t 2 ) a ( x 2- l ) z r s 35. F(f) : ( t ' - 2 t + 1 ) z t z ( 1 2 +f + 5 ) 1 / 3 If f - {(x, y)ly : g*" * 5x + 1}, then the equation a:3x2*5x*L ( 1) defines the function / explicitly. However, not all functions are defined explicitly. For example, if we have the equation f-2x-3yu*y'-y' e) we cannot solve for y in terms of r; however, there may exist one or more functions / such that if y: f(x), Eq. (2) is satisfied, that is, such that the equation x B- 2 x : i l f ( x ) l u+ [ / ( x ) ] ' - l f ( x ) 1 , .aie9s . 147 DIFFERENTIATION 3.8 IMPLICIT is true for all values of r in the domain of f . In this case we state that the function / is defined implicitly by the given equation. Using the assumption that Eq. (2) defines y as one or more differentiable functions of x, we can find the derivative of y with respect to xby the process called implicit differentiation, which we now do. The left side of Eq. (2) is a function of x, and the right side is a function of y. Let F be the function defined by the left side of (2), and let G be the function defined by the right side of (2). Thus, F(r) - x6 - 2x (3) G(y)-3yu*y"-Y' (4) and where y is a function of.x, saY, Y: f(x) So we write Eq. (2) as (s) F(r) - G(f(x)) Equation (5) is satisfied by all values of x in the domain of / for which exists. Glf(x)l ' Then for all values of x fot which / is differentiable, we have D,lxu - 2xl - D,l3yu* y' - y'l (6) The derivative on the left side,of Eq. (5) is easily found, and we have V) D*lxu-2xl:6x5-2 we find the derivative on the right side of Eq. (5) by the chain rule, giving us D'lSyu l y'- a'l: t}y' ' D*a * Syn' D,V - 2a ' D,a (8) substituting the values from (7) and (8) into (6), we obtain 6x5- 2: (LB!' t Syn- 2y)D"y Solving for D rY, we get D,U 6x5-2 L8y5*Syn-2y Equation (2) is a special type of equation involvrng x and y because it can be written so that all the terms involving x are on one side of the equation and all the terms involving y are on the other side. In the following illustration we use the method of implicit differentiation to findD*y from a more general tyPe of equation. 148 THE DERIVATIVE o rLLUSrRArroN 1: Consider the equation 3*nY'-7xy":4-8U (9) and assumethat thereexist one or more differentiablefunctions such that / if y: f (x), Eq. (9) is satisfied. Differentiating on both sides of Eq. (9) (bearing in mind that y is one or more differentiable functions of r), and applying the theorems for the derivative of a product, the derivative of a power, and the chain rule, we obtain 72x3y2 * 3xa(2yD*a)- 7yt - 7x(3y2D,y) - 0 - gD,a Solving for Dry, we have D,y(6xay - 27xy2+ 8) : 7yt - l2xsy2 DrA:ffi o Remember that we assumed that both Eqs. (2) and (9) define y as one or more differentiable functions of x. It may be that an equation in x and y does not imply the existence of any real-valued function, as is the case for the equation x'+y'*4:0 which is not satisfied by any real values of r and y. Furtherrnore,it is possible that an equation in I and V may be satisfied by many different functions, some of which are differentiable and some of which are not. A general discussion is beyond the scope of this book, but can be found in an advancedcalculustext. In subsequentdiscussions,when we statethat an equation in x and y definesy implicitly as a function of x, we assume that one or more of these functions is differentiable. Example 3, which follows, illustrates the fact that implicit differentiation gives the derivative of every differentiable function defined by the given equation. EXAMPLE (x * Y)'- Given ( x - y ) 2 : x a* A a find D,y. solurroN: Differentiating implicitly with respect to x, we have 2 ( x * V ) 0 * D , y ) - 2 ( +- V ) ( t - D * y ) : A x s - r A y B D , y from whicir we obtain 2x * 2y + (2x + 2y)D"y - 2x + 2y * (2x - 2y)D,y : 4xB* yBD,y D"y(Ax-4y'):4x3-4y D rxevrrrE 2: Find an equation of the tangent line to the curye f t y":9 at the point (1,,2). solurroN: xg-tl ,!:;=T Differentiating implicitly with respect to x, we obtain 3x2*3y2D,y:g I 3 . 8 I M P L I C I TD I F F E R E N T I A T I O N1 4 9 Hence, v2 D,u:-+ @r y" Therefore,at the point (L, 2), is then An equation of the tangent line y-2:-I(x-1) (a) Differentiating implicitly, we find ExAMPLE3: Given the equation solurroN: x' * y' :9, find: (a) D"Y bY -X 2x t 2YD*Y: 0 and so D-Y: implicit differentiation; @) rwo functions defined bY the equa(b) Solving the given equation for y, we obtain tion; (c) the derivative of each of and Y : -!9 - xz the functions obtained in Part (b) y -- ttr=p by explicit differentiation. Let fl be the function for which (d) Verify that the result obtained in part (a) agreeswith the results fr(x):6zT obtained in Part (c). and f, be the function for which frv ' 'i,'.,4 f,(x): -\E=i' ;+Y i * ,J,- r' 'r/ '' *J "'l (c) Because/, (r) : (9 - xr)tt', bY using the chain rule we obtain - x')-trz1-2x) : f ,.'(x) i(9 I C,,t ii.1f f t/9 - x' r0' Similarly, we get -x f''\x):ffi; fr(x), where f ,(x) (d) For a: -- \tr=T, we have from part (c) --x f " ( \"/ -t) .,I y {r4 which agrees with the answer in part (a). Fot y: -fr=E, we have from Part (c) fr'(x) : -r678 v/ & __L v which also agrees with the answer in part (a)' : fr(x), where fr(x) 150 THEDERIVATIVE Exercises 3.8 In Exercises L through L6, find D,y by implicit 1 , .f + A 2 : 1 6 , i T , ' * : '- + ? Y \.J XC Zy 7. \/i * t/y:4 differentiation. (z., zr'y * 3xys: 5 3 . x s+ y t : 8 x y s.l+1:t xy 6.x-4u:x 8.y+{xy:g*" 9. ,'y': v x2i A2 t O .y t f f i t x f f y : y 11..(x * y)'- (x - A)": x" + y" 1 2 . ( 2 xt 3 ) n : 3 y n 13. Y : 2 1 x 2 x-y 1.4.t/y + VV + {y: 15.l*V + 2x: fr 1 6. fyt: * xn - yn In Exercises 77 through 20, consider y as the independent variable and, find, Dox. 1 7. xn * yn: L2x2y 18.y:2x3 - 5x t9. x " y * 2 y n - x a : 0 2 0 .y f r - x t / y : 9 2L. Find an equation of the tangent line to the curve "L6xaI at Aa:32 the point (1,2). 22. There are two lines through the point (-r, g) which are tangent to the curve x2+4y2-4x-8y*3:0 Find an equation of each of these lines. Prove that the sum of the r and y intercepts of any tangent line to the curve {tz a yuz: If. x"y*: (r * y)n+*, prove that r . DrA: A. 2 5 -y 2 : X t - g ; l r : g 2 7 .x 2 - y ' : 9 ; x r : - 5 29.f * y' - 2x - 4y - 4- 0; xr:'!. 3.9 THE DERIVATIVE AS A RATE OF CHANGE ku2 isconstant and equal to k. 2 6 .f + \ e : Z l ; x t : A 28. y' - x2:' ).6; x1- -3 30. x2* 4y, * 6x - 40y* 9Z : 0; xr: -2 The concept of velocity in rectilinear motion coffesponds to the more general concept of instantaneous rate of change. For eiample, ifa particle is moving along a straight line according to the-equation of motion s: f(t), we have seen that the velocity of the particle at f units of time is determined by the derivative of s .with respect to f. Because velocity can be interpreted as a rate of change of distance per unit change in time, we see that the derivative of s with respect to f is the rate of c"hange of s per unit change in f. In a simirar way, if a quantity y is a function of a quantity x, wemay exPress the rate of change of y per unit change in r. The discussion is analogous to the discussions of the slope of a tangent line to a graph and AS A RATEOF CHANGE 151 3.9 THE DERIVATIVE the instantaneous velocity of a particle moving along a straight line. If the functional relationship between y and r is given by Y: f(x) f (x,+ Lx) f (x) :Ly Lx A,x (1) If the limit of this quotient exists as Ax e 0, this limit is what we intuitively think of as the instantaneous rate of change of.y per unit change in r at lr. Accordingly, we have the following definition. 3.9|t Definition rate of changeof y per unit changein x at xt lf y - f (x), tirreinstantaneous derivative of y with respect to r at xr, if it the it7'drj or, equivalently, exists there. The instantaneousrate of changeof y per unit change in r may be interpreted as the change in y causedby a change of one unit in x if the rate of changeremains constant.To illustrate this geometrically,letl'(rr) be the instantaneousrate of change of.y per unit changein x at xr' Then if we multi ply f' (rr) by Ar (the change in x), we have the change that would occur t" y it'the point (x,y) were to move along the tangent line at (x1,y) of the graph of y:/(r). 3ee Fig. 3.9.1,.The averagerate of change o{ y per unil.h"r,g" i" r is given by ihe fraction in Eq. (L), and if this is multiplied by Ax, we have :Lu n. ^*- Ly Figure3.9.1 which is the actual change in y caused by a change of Ar in r when the point (x,y) moves along the graPh. sxevlprE L: LetV cubic inches be the volume of a cube having an edge of length e inches. Find the average rate of change of the volume per inch change in the length of the edge as e changes from (a) 3.00 to 3.20; (b) 3.00 to 3.10; (c) 3.00 to 3.01. (d) What is the instantaneous rate of change of the volume Per inch change in the length of the edge when e:3? Because the formula for finding the volume of a cube is solurroN: : e3. Then the average rate v: e3, let f be the function defined by f (e) from e, to et * Ae is changes e e as in change unit of chang e of V per f (e'+ A'e) f (e') A,e - 0.2,and'=s@: 9Jf i, A,e (a)or-fllfi/Q) (b)e,: 3,A,e:0.1; ".a : g}}9 :# : 28.8 :'# : 27'e 152 THE DERIVATIVE ( 3 ' 0 1 ) 3- 3 3 - 0 ' 2 7 1 ' : 27 (c) €r:3, A,e:0.01, undl(3'01.)- /(3) 't 0.0L 0.01 0J01 In part (a) we see that as the length of the edge of the cube changes from 3.00 inches to 3.20 inches, the change in the volume is S.7T cubic inches and the average rate of change of the volume is 28.8 cubic inches per inch change in the length of the edge. There are similar interpretations of parts (b) and (c). (d) The instantaneous rate of change of V per unit change in e at 3 is /'(3). '(e) :3e2 f Hence, f ' ( 3 ): 2 7 Therefore, when the length of the edge of the cube is 3 inches, the instantaneous rate of change of the volume is 27 cubic inches per inch change in the length of the edge. ExAMPLE 2: The annual gross earnings of a particular corporation f years after January I,'1974, is p millions of dollars and P:€F+2t+10 Find: (a) the rate at which the gross earnings were growing January 7, 1976; (b) the rate at which the gross earnings should be growing January 1,1980. solurroN: (a) On January '1.,'j.976, DtP:tt+z hence, we find Dp when ,,r],:, -3+ 2:3.G s9.9" I^!!?rv 1,1976,the gross earnings were growing at the rate of 3.6 million dollars per year. (b) On January 1,,lg10, t: 6 and. I Dtpl :+*Z:6.8 Jt:e Therefore, on January l, 7980,the gross earnings should be growing at the rate of 6.8 million dollars per year. The results of Example 2 are meaningful only if they are compared to the actual earnings of the corporation. For eximple, ir o. fanuary 1, 1975, it was found that the earnings of the coqporation for the year lg14 had been 3 million dollars, then the rate of growth on fanuary 7, !9T6, of 3.4 million dollars annually would have b-een excellent. Howev er, if the earnings in 7974 had been 300 million dollars, then the growth rate on fanuary 1,"1,976,would have been very poor. The measire used to comPare the rate of change with the amount of the quantity which is being changed is called the relatiue rate. 3.9.2 Definition y -- f (r), the relatiae rate of changeof y per unit changein x at xris given lf by f'(xr)lf(xr) or, equivalently, O.VIV' evaluated at xJ xr. 3.9 THE DERIVATIVE AS A RATE OF CHANGE 153 If the relative rate is multiplied by 100, we have the percent rate of change. nxaupln 3: Find the relative rate of growth of the gross earnings on Januaryt,'1.976,and January l, L980,for the corPoration of Example2. solurroN: (a) When t:2, P : +(4) + 2(2) + L0 : L5.6.Hence,on Januaty 1, 1976,the relative rate of growth of the corporation's annual gross earningswas : : 23.'m ry1,:,:ffi 0.231 (b) When t:6, p: ?(36)+ 2(6) + 10 :36.4. Therefore,on January 1980, the relative rate of growth of the corporation's annual gross earnings should be '!,, %P.l : 6'R P lt:e 967: 0'1'87:'l'8'7% Note that the growth rate of 6.8 million dollars for ]anuary 1, 1980, is greater than the 3.6 million dollars for January 1,1976; however, the rela'J.8.7Vo for January 1, 1980, is less than the relative tive growth rate of growth rate of 23.LVofor January L, L976. 3.9 Exercises l. If A in., is the area of a square and s in, i5 the length of a side of the square, find the average rate of change of A with respectto s as s changesfrom (a) 4.00to 4.50;(b) 4.00to 4.30;(c) 4.00to 4.10.(d) What is the instantaneousrate of change of d vrith respect to s when s is 4.00? 2. Supposea right-circular cylinder has a constant height of 10.00in. If V in.3 is the volume of the right-circular cylinder, ani r in. is the radius of its base, find the average rate of change of V with rcsPect to / as t changes from (a) 5.00 to 5.40; (b) 5.00 to 5.10;(c) 5.00 to 5.01.(d) Find the instantaneousrate of change of y with resPectto r when r is 5.00. 3. Let r be the reciprocal of a number r. Find the instantaneous rate of drante of / with resPect to t and the relative rate of change of r per unit change in n when z is (a) 4 and (b) 10' 4. Let s be the principal square root of a number l. Find the instantaneous rate of change of s with resPect to r and the relative rate of drange of s Per unit change in x when r is (a) 9 and (b) a. 5. If water is being drained from a swimming pool and y gal is the volume of water in the Pool I min after the draining starts, where y-= 250(4O- l)1, find (a) the averagerate at which the water leaves the pool during the first 5 min, and (b) how faet the water ie flowing out of the pool 5 min after the draining starts. 6, The supply equation for a certain kind of pencil is r: 3f I 2p whete p cents fu the price P,erPencil when 10mt PenAt are suppfea. (a) Find the average rate of drange of the supply per I cent drange in the Pdce when the Price is increased &om 10 cents to 11 cents. (b) Find the instantareous (or marginal) rate of drange of the suPPly Per 1 cent change in the price whm the price is 10 cents. - O.Zf. 7. The profit of a rctail store is 100y dollars when r dolLars are spent daily on advertiBing a\d V : 25,0O+ Xx if the currmt increased Use the derivative to detemrine ii it would be profitable for the daily advertising budget to be (b) $100 daily advertising budget is (a) $60 ard 154 THE DERIVATIVE 8' A balloon maintains the shape of a spherc as it is being inflated. Find the rate of change of the surface area with respect to the radius at the instant when the radius is 2 in. 9. In an electdc circuit, if E volt8 is the electromotive force, R ohms is the resistance, and I amperes is the current, Ohm,8 l,aw states that IR: E. Assuming that E is constant, show that R decreasesat a rate that is-proportional to the inverse square of L 10. Boyle's lgw fo1 tle expansion of a gas is PV: C, where P is the number of pounds per square unit of pressure, y is -change the numbet of cubic units in the_volume of the gas, and C is a constant. Fina the insiantaneous rate of of the volune per change of one pound per square unit in the pressure when p : 4 and V: g. 11' A bomber is flying parallel to the ground at an altitude of 2 mi and at a speed of 4l mi/rrin. If the bomber flies directly over a target, at what rate is the line-of-Eight distance between the bomber and the target changing 20 sec later? 12. At 8 a.M. a ship sailing due north at 24 knots (nauticat miles per hour) is at a point P. At 10 A.M. a second ship sailing due east at 32 knots i8 at P. At what rate is the distance between the two ships changing at (a) 9 e.r.,r.and @!1 e.r,a.? 3.10 RELATED RATES ' ExAMrLE1: A ladder ZSft long is leaning against a veftical wall. If the bottom of the ladder is pulled horizont ally away from the wall at 3 ftlsec, how fast is the top of the ladder sliding down the wall, when the bottom is 15 ft from the wall? There are many problems concerned with the rate of change of two or more related variables with respect to time, in which it is not necessary to express each of these variables directly as functions of time. For examPle, supPose that we are given an equation involving the variables r and y, and that both x and y arc functions of a third variable f, where t sec denotes time. Then, because the rate of change of r with respect to f and the rate of change of y with respect to f are given by D,x aid. D,y, resPectively, we differentiate on both sides of the given equation with respect to t by applying the chain rule and proceed as below. SoLUTIoN: Let f : the number of seconds in the time that has elapsed since the ladder started to slide down the wall; a: the number of feet in the distance from the ground to the top of the ladder at f sec; r: the number of feet in the distance from the bottom of the ladder to the wall at f sec. See Fig. 3.10.1.Because the bottom of the ladder is pulled horizontally away from the wall at 3 ftlsec,D.x:3. We wish to find Dry whenx:.1_5. From the Pythagorean theorem, we have U2:625 - xz (1) Because r and y are functions of t, we differentiate on both sides of Eq. (t) with respect to f and obtain 2y Dty : -2x Dtx giving us Dta: -1o,* u F i g u r e3 . 1 0 . 1 When x: lS,it follows from Eq. (1) thaty : (2) 2f.BecauseDg:3, we 3.10 RELATEDRATES 155 get from (2) o,rf":,o Therefore, the top of the ladder is sliding down the wall at the rate of 2* ftlsec when the bottom is 15 ft from the wall. (The significance of the minus sign is that y is decreasing as f is increasing.) EXAMPLE2: A tank is in the form of an inverted cone, having an altitude of L6 ft and a base radius of 4 ft. Water is flowing into the.tank at the rate of 2 ftglri;rir. How fast is the water level rising when the water is 5 ft deep? Let f : the number of minutes in the time that has elapsed since water started to flow into the tank; the number of feet in the height of the water level at h f min; r: the number of feet in the radius of the surface of the water at f min; V: the number of cubic feet in the volume of water in the tank at f min. At anv time, the volume of water in the tank may be expressed in terms of the volume of a cone (see Fig. 3.1'0.2). solurroN: y: (3) lnrzh V, r, and h are all functions of f. Becausewater is flowing into the tank at the rate of 2 ft}lmtn, D1V: 2.We wish to find Dfu when h: 5. To express r in terms of.h, we have from similar triangles r:ih ;:+ Substituting this value of r into formula (3), we obtain V : in(*h)'(h) 3.10. 2 F ig ure or Y : |gnhg ( 4) Differentiating on both sides of Eq. (a) with respect to f, we get D1V: L*rhz D1h Substituting 2 for D1V and solving for D1h, we obtain D -, h : 1 ?) Tn- Therefore, *o,']o:, :# We conclude that the water level is rising at the rate of 32125znftlmin when the water is 5 ft deeP. 156 THE DERIVATIVE EXAMPLE3: Two cars, one going due east at the rate of 37.5 mi/hr and the other going due south at the rate of 30.0 mi/hr, are traveling toward an intersection of the two roads. At what rate are the two cars approaching each other at the instant when the first car is 400 ft and the second car is 300 ft from the intersection? solurroN: Refer to Fig. 3.10.3, where the point P is the intersection of the two roads. Let x: the number of feet in the distance of the first car from P at f sec; y: the number of feet in the distance of the second car from P at f sec; z: the number of feet in the distance between the two cars at f sec. Because the first car is approaching P at the rate of jT.S mi/nr : 37.5' rcuftlsec:55 ft/sec and because r is decreasing as f is increasing, .# f t lsec: i t fo l l o w s that D fi :-55. S i mi l arl y, because 30 mi /hr:30 4 4 ttl s e c, D tA :-44. W e w i sh to fi nd D rz w hen tr:400 and U: 200. From the Pythagorean theorem we have z 2 : x 2* y 2 (5) Differentiating on both sides of Eq. (5) with respect to f, we obtain 2z DP:2x Dg I 2U D,Y and so (5) F i g u r e3 . 1 0 . 3 when x: 400 and y: 300,it follows from Eq. (5) that z: 500.In Eq. (6), substitutingDrx: -55, DtA: -44, x: 400,y : 300, and,z: 500,we get I : (4oo)(-5s) + (3oo)(-44) :_- t 0n, .4 D,zl 500 Jz:soo Therefore,at the instant in question the cars are approachingeachother at the rate of 70.4 ftlsec. Exercises 3.10 1. A kite is llying at a height of 40 ft. A boy flying it so that it is moving hodzontally at a rate of 3 ft/sec. If the string is taut, at what rate is the stdnt being paid! out when the length of the i-t ing releasei is 50 ft? 2 A spherical balloon is being inflated so that its volume is increasing at the rate of 5 ftslmin. At what rate is the diameter incteasing when the diameter is 12 ft? 3' A spherical snowball i5 being made so that its volume is incr€asing at the rate of 8 ft3lmin. Find the rate at which the radius is increasing when the snowball is 4 ft in diameter. 4' suPPose that when the diameter is 6 ft the snowball in Exercise 3 stopped growing and started to melt at the rate of I ft3lmin. Find the rate at which the radius is changing when the radiu; is tft. 5' Sand is being dropped at the rate of 10 ft3lmin onto a conical pile. ft the height of the pile is always twice the base radius, at what rate is the height increasing when the pile is a fi nign? 6. l_llght is hunt 15 ft above a 6tr-aight horizontal path. ff a man 6 ft tall is walking away ftom the litht at therateof 5 ft/eec, how {ast is his shadow lengthening? 3.11 DERIVATIVES OF HIGHERORDER 157 7. .In Exercise 6 at what rate is the tip of the man's shadow moving? / b. A man 6 ft tall is walking toward a building at the rate of 5 ft/sec. If there is a light on the ground 50 ft from the buitding, how fast is the man's shadow on the building growing shotter when he is 30 ft from the building? 9. A water tank in the form of an inverled cone ie being emptied at the rate of 6 ft8/min, The dtitude of the cone i6 24 ft, and the base radius is 12 ft. Find how fa6t the water level is lowerint when the water is 10 ft deep. 10. A trough is 12 ft long and its ends are in the forrr of inverted isoscelea triangles having an altitude of 3 ft and a bas€ of 3 ft. Water is flowing into the trough at the rate of 2 ft3/min. How fa6t is the water level rising when the water is 1,ft deep? 11, Boyle's law for the expansion of gas ig PV: C, where P is the number of pounds per square unit of pressure, y is the number of cubic units of volume of the gas, and C is a constant. At a certain instant the pr€sEure is 3000 lb/fP, the vol: ume is 5 fC, and the volume is increasing at the rate of 3 fP/min. Find the rate of change of the pressure at this instant. 12. The adiabatic law (no gain or loss of heat) for the expansion of air is PIl ' : C, where P is the number of pounds per square unit of pressure, y is the number of cubic unit8 of volume, and C is a constant. At a specific instant, the pressure is 40 lb/in.2 ard is increasing at the rate of 8 lb/in.'2 each second. What i6 the rate of change of volume at this instant? ' 13, An autourobile traveung at a rate of 30 ft/sec is approaching an intersection. When the automobile is 120 ft from the i' intersection, a truck traveling at'the rate of atoft/sec crosseqthe inter8ection, The automobile and the truck are on roads that are at right angles to each other. How fast are the automobile and the truck separating 2 sec after the truck leaveg the intersection? 14. A man on a dock is pnlling in a boat at the rate of 50 ft/min by means of a rope attached to the boat at water level. If the man's hands are 16 ft above the water level, how fast is the boat approadring the dock when the amount of rope out is 20 ft? 15. A ladder 20 ft long is leaning against an embankment inclined 50' to the horizontat. If the bottom of the ladder i8 being uroved horizontally toward the embar*ment at L fvsec, how fast is the top of the ladder moving when the bottom is 4 ft from the embankment? 16. A horizontal trough is 16 ft lon& and its ends are isosceles trapezoids with an altitude of 4 ft, a lower base of 4 ft, and an upper base oI 5 ft. Water is being poured into the trough at the rate of l0 felmin. How fast is the water level rising when the water i6 2 ft deep? 17. In Exercise 16 if the water level is decreasing at the rate of I ft/min when the water i8 3 ft deeP, at what rate is water being drawn from the trough? 18. Water is being pour€d at the rate of 8 ft3/min into a tank in the forrr of a cone. The cone is 20 ft deep and 10 ft in diameter at the top. If there iB a leak in the bottom, and the water level is rising at the rate of 1 in./rnin, when the water is 16 ft deep, how fast is the watet leaking? OF 3.11 DERMTIVES HIGHER ORDER lf f ' rs the derivative of the function I then /' is also a function, and it is the first deriaatiae of.f . It is sometimes referred to as the first deriaed function. If the derivative of f ' exists, it is called the secondderiaatiae of f , or the second derived function, and can be denotedby f" (read as "f double prime"). Similarly, we define the third deriaatiae of f , or the third derived function, as the first derivative of f " if it exists. We denote the third derivative of f by f "' (read as "f triple prime"). The nth deriaatiae of the function /, where n is a positive integer greater than L, is the first derivative of the (n - l)st derivative of /. We denote the nth derivative of f by f{"t. Thus, if. f{"t denotes the nth derived THEDERIVATIVE 158 function, we can denote the function / itself by fo.Another symbol for the nth derivative of f is D *"f . If the function / is defined by the equation y: f(x), we can denote the nth derivative of f by D,"y. EXAMPLE 1: Find all the derivatives of the function / defined by f(x):8ra*5r3- SOLUTION: f'(x) :32x8 * LSxz 2x f" (x) :96x2 * 3ox 2 x2+7 f "'(x) :192x + 30 f+t(x) : tg2 / ( 5 ) ( r ): 0 n>S f{n)(y)-0 Becausef '(x) gives the rate of change of f (x) per unit change in x, ' '(r) per f " (x), being the derivative of f (x), gives the rate of change of f unit change in r. The second derivative f " (x) is expressed in units of f'(x) per unit of x, which is units of f(x) per unit of x, per unit of r. For example, in straight-line motion, it f (t) feet is the distance of a particle from the origin at f seconds, then f '(t) feet per second is the velocity of the particle at f seconds, and f " (t) feet per second per second is the instantaneous rate of change of the velocity at f seconds. In physics, the instantaneous rate of change of the velocity is called the instantaneousacceleration Therefore, if a particle is moving along a straight line according to the equation of motion s : f (t), where the instantaneous velocity at f sec is given by a ftisec and the instantaneous acceleration is given by a ftlsecz, then a is the first derivative of z with respect to t or, equivalently, the second derivative of s with respect to f; that is, z t:.DtS rxavrprn 2: A particle is moving along a straight line according to the equation of motion '1 4t ,, :; -5 A : D rts: D tzs SOLUTION: r,t:D6:f i-'T7-- 2-'t+t where s ft is the directed distance of the particle from the origin at f sec. If u ftlsec is the instantaneous velocity at f sec and a ft/secz is the instantaneous acceleration at f sec, find f, s, and a when a: 0. and A: Dp: Se tti n g a :0, *' , t ! = , . (t + 1.)2 r -.,- 1 D' -t 2 w e have ( r + 1 ) 3 _8 _ (f + 1)3 o or (f + 1)t: 8 8 (t+1)t 3.11DERIVATIVES OF HIGHER ORDER 159 from which the only real value of t is obtained from the principal cube root of 8, so that t*l:2 When t: or I ,:|(1)'+ #:2+ ll: 7 lf (x, y) is any point on the graph of y : /(x), then D"y gives the slope of the tangent line to the graph at the point (x, y).Thus, D"'A is the rate of changeof the slopeof the tangentline with respectto r at the point (x,y). rxavrprr 3: Let m(x) be the slope of the tangent line to the c u r v eA : x 3 - 2 x 2 * r a t t h e point (x,y).Find the instantaneous rate of change of.m per unit change in r at the point Q,2). '1,. soLUTroN m: Dr! : 3xz - 4x * The instantaneous rate of change of. m per unit change in x is given by Drm or, equivalently, Dr'A. D rffi - D r' A :6x - 4 At the point (2,2), D,'A:8. is 8 times the change in r. Hence, at the point (2,2), the change in m Further applications of the second derivative are its uses in the second-derivative test for relative extrema (Sec. 5.2) and the sketching of the graph of a function (Secs. 5.4 and 5.5). An important application of other higher-ordered derivatives is to determine infinite series as shown in Chapter 76. The following example illustrates how the second derivative is found for functions defined implicitly. rxlvrprs 4: Given 4xz * 9Y2:36 find D,2y by implicit differentiation. sot.urroN: Differentiating implicitly with respect to x, we have 8x *'l,8yD"y - 0 so that Dr!:-fr (1) To find Dr'A, we find the derivative of a quotient and keep in mind that y Lsa function of r. So we'have (2) THE DERIVATIVE Substituting the value of D,y from Eq. (1) into Eq. (2), we get -36y'- l6x2 8Ly" Thus, -L(gyz * 4x2) Dr'y:ft (3) Becauseany values of r and y satisfying Eq. (3) must also satisfy the original equation, we can replace (9y'* 4xz) by 36 and obtain Drra:-#:-# Exercises 3.71 In Exercises1 through 10, find the first and second derivative of the function defined by the given equation. t.f(x):x5-2x3*x 2. F(x) :7xs - 8x2 3. S(s) :2sa - 4s3* 7s- 7 4. G(t):f3- P+t 5. f(x): \Eq 6.h(V):gffi 7. F(x): xztG-5x 8.g(t):lr+W 1 1 9 .G \ x ) : f f i r z 2- \n ru./(I) : rT yE 1 1 .F i n d D , 3 y i fA : f -2f*r-5. 12. Find Dfs if s: @T.. {3. Find D,"f (x)' it f(x) : i: (1-x)" "' ' ..1 ,, o-/ffIn. Find /ra)(r) if /(r) : i=T. 15. Find DraY if. A : x7t2- 2x5t2+ xrt2 15. Find D,su if u: aG2.. -') 17. Given x" + y": 1, show that Drzy:;118. Given xtg * yLt2:2, show that D"'y : v $. 19. Given y' + l:4' (a is a constant),find D,? in simplest form. 20. Given F* a2f = FV (a allrdb are constants), find D"!r in simplest fomr. 2t. Find the slope of the tangent line at each point of the graph of y : t' + f - 3f wnet€ the rate of change of the slope iB zerc, In B<ercises 22 and 23, a particle ie moving along a straight line according to the given equation of motion, where s ft i8 BEVIEWEXERCISES 161 the direct€d distance of the particle from the origin at t 6ec. If ftlsec is the velocity and c ftlseC is the acceleration of the " partide at t sec, find ? and 4 in terms of l. Also find when the accelemtion is zero and the intervals of tine when the particle is moving toward the origin and $'hen it is moving away ftom the origin. 23. s: *tn - ztg+ 4t2 22. s: t3- 9t2+ l5t In Exercises 24 through 27, a patnde is moving along a Btraight Iine according to the Biven equation of motion, wherc s ft is the directed distance of the particle from the origin at t sec. Find the tiure when the instantan€ous acceleration is zeto, and then find the directed distance of the particle from the origin and the instantaneous velocity at this instant. >o 2 ss. : f f i - z u f , f 2 4 .s = 2 t 3 - 6 t 2 + 3 t - 4 , f > 0 - {2?. s: $Ssrz + 2trt2,f > 0 2 6 .s : g t t + 2 \ / 2 t + l , t > 0 I n E xe rci ses 28t hr ough31 .,fi n d fo rm u l al os rf' (r) a n d f" (x ) andstatethedomai nsof f' and.f" (x' t.rx# o zB. f(x):IFTi f x : 0 L0 ze. f(x): ff 3 0 .f ( x ) : l r l ' 31. lf; : 3 32. For the function of Exercise30 find f "' (x) when it exists. 33. For the function of Exercise31 find f "'(x) when it exists. 34. Show that if xA:1, then D,zy ' Dozx: 4. expressh"(r) in terms of the derivatives offand g. 35. fi f', g', f", and8" exist and if tt:/"9, 36. If / and g are two functions such that their first and secondderivatives exist and if h is the function definedby h(x): ' f(x) sG) ' Prove that n" G\ : f(x) . g" (x) + 2f'(x) ' g'(x) + f" (x) ' g(x) 37. lI V = t, 38. If whete z is any positive integer, Prove by mathematical induction that D'?: n! 1, - L- 2X prcve by mathematical induction that D nt : t,rl -------:-i::- (l - 2x't"*l Reaiew Exercises(Chapter3) In Exercises1.through 14, find D"y. +2 l.y:fra 2. 3 . 4 x z* 4 y ' - U 3 : 0 4.y:\GV+\Fx 5. 6. v: x2 (x*2)2(4r-5) 162 THE DERIVATIVE 7.y:{tr t4 8.*y'*2y": x-2U 10. XztB* yzts: g2t3 lL. y: xffi t4-y: -7*., lrt* (xn*x)'ft 13.Y:f* l ) s t z ( x z- 4 ) r r z (x'- 15. A parttcle is movint in a straight line according to the equation of motion s: r3- llP + 24t + 100 where s ft i8 the direct€d distance of the particle fiom the starting point at , sec. (a) The partide is at the starting poht when t: 0. For what other values of t is the particle at the starting point? (b) Determine the velocity of the partide at each instant that it is at the startint point, and interpret the sign of the velocity in each case. + 4.C - r that have the slope l. 17. Find equations of the tangent line and normal line to the curve 2/ * 2y3 - 9xy : g at the point (2, 1). 16. Find equations of the tangent lines to the curve y:2f 18. An obiect is sliding down an hclined plane according to the equation of motion s : 121,* 5t wherc s ft is the dirccted distance of the obiect ftom the top I sec after stafting. (a) Find the velocity 3 sec after the start. (b) Find the initial velocity. 19. Using only the definition of a derivative find l'(r) it f (x') : \/-4x - 3. 20. Using only the definition of a derivative find /'(5) it f(x) : 19iiT1. 2 7 - F i n d f, ' ( x ) i f l ( r ) : ( l r + 1 1 - l r l ) , . 2 2 . F i n df , ( x \ i I f ( x ) : x - [ x \ . : 23. Given /(r) lrl3. 1a)Draw a sketch of the graph of f (b) Find lirn l(r) if ir existe. (c) Find l,(0) if it exists. 24. Given l(r) : t' sgn x. (a) Discuss the differentiability of. f. (b) Is f' continuous on its domain? 25.Findl'(-3) if /(r): [r + ]lV9r. 26.Findt,(-3) iff(r):(lrl 27- Ptove tha! the line tantent to the curve y:-y' point, and find this point - x)fl-sx. + Zf + r at the point (1, 2) is also tangent to the curve at another 28. Find an equation of the normal line to the eurve r - y: at the point 13, l). fETl 29. A ball is thrown vertically uPward from the top of a houee 112 ft high. Its equation of motion is s : -15f + l)5t wherc s ft is the dirccted distance of the ball from the starting point at t sec. Find: (a) the in8tantaneous velocity of the batl at 2_sec; O) how high the ball will go; (c) how long it takes for the ball to reach the ground; (d) the instantaneous velocity of the ball when it reaches the ground. 30. Prove that the tangent lines to the curves 4Vs- t1y - x + sy:g and,/ - 4y3+'Sxt y:0 at the origin are perpmdicular. 31. Giv€n (qf+b ((t\ :1 1 Ei if r =l ifx>1 Find the values of c and, 8o that/'(1) exists. 32. Suppose IVJ: (f lot"+bt +, if x<1 if , =t Find the values of a, ,, and c so that f"(1) exi8ts. 33. A sltip leavea a Port at 12 noon and havels due west at 20 knots. At 12 noon the next day, a second ship leaves the same REVIEWEXERCISES 163 port and travels northwest at 15 knots. How fast are the two ships separating when the second ship has traveled 90 nautical miles? 34. A particle is moving in a straight line according to the equation of motion s: liT bF, whete a and D are positive constants. Prove that the measure of the acceleration of the particle is inveBely proportional to f for any ,. 35. A funnel in the form of a cone is 10 in. across the top and 8 in. deep. Water is flowing into the funnel at the rate of 12 in.3/sec,and out at the rate of4 in.8/sec.How fast is the surfaceof the waterrising when it i8 5 in. deep? 36. As the last car of a train passes under a bridge, an automobile cro8sesthe bridge on a rcadway pe4rendicular to the track and 30 ft above it. The train is traveling at the rate of 80 ft/sec and the automobile is traveling at the rate of 40 ftl6ec. How fast are the train and the automobile separating after 2 sec? 37. A partide is moving along the cuwe rl - / : 9, and the ordinate of the point of its position is inqeasing at the rate of 3 units per second.How fast is the abscissachanging at the point (5, -4)? 38. Supposey is the number of workers in the labor force needed to produce r units of a certain commodity, and x: 4yz. If the production of the commodity this year is 250,0Q0units and the prcduction is increa6ing at the rate of 18,000units per year, what is the current rate at which the labor force should be increased? 39. Using Boyle's law for the expansion of gas (Exercise 10 in Exercises 3.9), find the rate of change of the pressur€ of a certain gas at the instant when the pressure is 8 lb/in.? and the volume in 7(X)ff if the volume of the gas is increasint at the rate of 3 ff/min. 40. If the two functionE f and 8 arc differ€ntiable at the number .rr, is the composite function f. g necessarily differentiabl€ at 11?If your answer G yes, prove it. If your answer is no, tive a connters(ample. + lrl and g(r) : tr - *lrl. Prove that neith€r f'(0) norg'(0) exists but that (l " g)'(0) does exist. 41. Supposel(r):3r 42. Give an example of two functions f and g for which I is differentiable at g(0), 8 is not differentiable at 0, and l . g is diff€r€ntiable at 0. 43. Give an example of two functions f and g for whidr I is not diffet€ntiable at g(0), I is differentiable at 0, and I . g is differentiable at 0. ,14. In Exercise 27 of Exercises 3.3, you are to prcve that if / is differentiable at 4, then + Lt) - f(a - Ax) .. f(s ----------i;t (4, : lrm Show by using the absolute value function that it is possible for the limit in the above equation to exist even though f'(c) does not exist. 45. ll f' (x) exists, prove that ,^4k,): tl?) : fe) _ xl.,(x) 46. Let I and g be two functions whose domains are the set of all real numbers. Furthermore, suppose that (i) 8(r) : x/(r) + 1; (ii) g(a + D) : gb) ' 8@) lor all a and.b; (iii) lim /(r) : 1. Prove that 8'(r) : 8(x). 47. The remainder theorem of elenientary algebra states that iI P(r) is a polynomial in .r and r is any real number, then thert is a polynomial Q(r) such that P(r) = Q(i) (r - r) + P(r). What is lim O(r)? tl8. Suppose g(r) : ll(r) l. If lt't(a) exists and /(r) {^t(*):ffif@G) + 0, prove that Topicsonlimits, continuity and the derivative 4.1 LIMITSAT INFINITY 1 6 5 4.1 LIMITS AT INFINITY Consider the function / defined by the equation v )'v2 rf \(-x- /\ : +X z+ . 1 , A sketch of the graph of this function is shown in Fig. 4.L.7.Let x take on the values 0, L, 2, 3, 4, 5, 10, L00, 1000, and so on, allowing x to increase without bound. The corresponding functionvalues are given inTable 4.L.1.. F i g u r e4 . 1 . 1 T a b l e4 . 1 . 1 10 100 1000 x 0 1 2 3 f(x) û .t g 5 18 32 50 200 2o,ooo 2,000,000 10 rr % 101 1o"oo1 lpoo"ool 4 5 We see from Table 4.1..1thatas r increasesthrough positive values, the function values f (x) get closer and closer to 2. In particular, when x:4, A2x2n322 z- r: x2+l:z- 17:O Therefore, the difference between 2 and /(r) is # when x: 100, 2f 4. When 20,000 z - x z + r - zA- L o , o o L10,00L : Hence, the difference between 2 and /(r) is 21t0,001when r: L00. Continuing on, we intuitively see that we can make the value of f (x'1 as close to 2 as we please by taking r large enough. In other words, we can make the difference between 2 andf (x) as small as we please by taking r large enough; or going a step further, for any e ) 0, however small, we can find a number N ) 0 such that If(x) - 2l < e whenever r ) N. When an independent variable r is increasing without bound through positive values, we write "x + *q." From the illustrative example above, then, we can say that hm4:2 r-*oftL In general, we have the following definition. 4.'1,.1,Definition Let f be a function which is defined at every number in some intenral (a, +c,c'1. The limit of f (x), as x increaseswithout bound, is L, written (1) Jltf?):t if for any € > 0, however small, there exists a number N ) 0 such that lf(*)-Ll <e w h e n e v e rr ) N 166 TOPICSON LIMITS,CONTINU|ry,AND THE DERIVATIVE NorE: When we write x + *m, it does not have the same meaning as, for instance, x --+ 1000. The symbolism "x -->*a" indicates the behavior of the variable x. However, w€ can read Eq. (1) as "the limit of f(x) as x approaches positive infinity is L," bearing in mind this note. Now let us consider the same function and let r take on the values 0, -L, -2, -3, -4, -5, -10, -L00, -1000, and so on, allowing x to decrease through negative values without bound. Table 4.7.2 gives the corresponding function values of f (x). Table 4.7.2 0 -1 ') v2 f(*):ft+I -2 -3 -4 -5 8183250 5 10 tr% -10 -100 -1000 200 20,000- 2,000,000 101 L0,001. 1,000,001 Observe that the function values are the same for the negative numbers as for the corresponding positive numbers. So we intuitively see that as x decreases without bound, f (x) approaches 2, and formally we say that for any e ) 0, however small, we can find a number N < 0 such that lf (x) 2l < , whenever x < N. Using the symbolism "x -->-@" to denote that the variable r is decreasing without bound, we write 2x2 1. jl*i+-2 In general, we have the following definition. 4.1.2 Definition Let f be a function which is defined at every number in some interval (-*, a). The limit of f(x), as x decreasesraithoutbound,is L, written f?):t ,t1T- Q) if for any € ) 0, however small, there exists a number N < 0 such that -Ll <. whenever r(N lf(*) NorE: As in the note following Definition4.r.L, the symbolism "x + only indicates the behavior of the variable x, but we can read Eq. "the limit of f (x) as r approaches negative infinity is L." Limit theorems 2, 4, 5, 6,7, 8, 9, and 10 given in Sec. 2.2 and theorems 11 and 12 given in Sec. 2.4 remain unchanged when "x + replaced by r-++co or x--->-oo. We have the following additional theorem. 4.1.3 Limit theorem 1,3 If r is any positive integer, then (t) j: o "t1T_ (ii),11a4:o -6" (2) as Limit a" is limit 4.1 LIMITSAT INFINITY pRooF oE (i): To prove part (i), we must show that Definition 4.1,.1, holds tor f (x): tlx' and L: 0; that is, we must show that for any e ) 0 there exists a number N ) 0 such that lr I l+-01 <e I lx' w h e n e v e rr ) N or, equivalently, 1 lrl"=Z wheneverrlN or, equivalently, since r ) 0, /1\rb whenever r)N ltl t(;/ In order for the above to hold, take N : (lle)rtr. Thus, we can concludethat rr | /l\ur l+ - 0l < e whenever x ) N, if N: l: I ter I lx' r This proves part (i). The proof of part (ii) is analogous and is left as an exercise (see Exercise 22). rxlvrplr r. rm 'j-+@ L: Find 4x-3 ;J= L* | J and when applicable indicate the limit theorems being used. sol,urroN: To use Limit theorem 1.3,we divide the numerator and the denominator by r, thus giving us 4x-3 "11 ,. 4-3lx zrc+s:"t1t7-+i|jlim (4 - 3lx) - (L.r. e) t-lo lim (2 + slx) t-l@ lim 4_ t-+6 Iim (3/r) t-*o j1t2+"lit $tx') lim 4- lim 3' lim (Ux\ lim 2 * lim 5 ' lim (Ux) J| -*@ _4-3.0 2+5.0 -2 (L.r.4) Jf-j-o (L.r. 6) t-*@ (L.T.2 and L.T. 13) 168 TOPICS ON LIMITS, CONTINUITY, ANDTHEDERIVATIVE EXAMPLE2: Find r. 2x2-r*5 ,'14 4,,*l and when applicable indicate the limit theorems being used. soLUrIoN: To use Limit theorem 13, we divide the numerator and the denominator by the highest power of x occurring in either the numerator or denominator, which in this case is 13. So we have r:* 2x2-x*5_ rj_ 2lx-'l,lxz*Slx} "'14 Axs-L:"'1a@ lim (2lx - tlx2 + Slx") - t'-@ (L.r.e) (4- ttxs) Ji* lim (zlx) - lim (Ux') * lim (5/r') 04-@ I+-@ lim 4 - (1.r.5) t+-@ lim (Ux') Q--6 2 ' l i m (ux) - lim (Llf) * lim 5 . lim (Llx") I--@ I--@ 4 - lim (Ux') !t--6 (1.r.6) fi--@ 2. 0- 0 + 5 . 0 (L.T.2 and L. T. 13) 4-0 EXAMPLE ,. riii:t Find -- 3x*4 r-l.a V2* - 5 and when applicable indicate the limit theorems being used. soLUrIoN: We divide the numerator and the denominator of the fraction by r. In the denominator we let r: t/7 since we are considering only positive values of r. r:_ 3x*4 3+4lx ,:- "'11ffi,:,t1tffilw lim G + alx) lim 1D=W (L.r.e) I-lq I--lq lim G + alx) t-*a lim ,(L.r. 10) (2 - 5lx2) f-*m ]g3*_li1 @tx) .T-*€ (1.r.4) t-*a lim 3* lim 4. lim (Tlx) t-*a Vlim 2- I-l@ ,f,-*co lim 5. lim (tlfl (L.r. 6) t-'ta 3+4.0 (L.T.2 andL.T.13) 4.1 LIMITSAT INFINITY 1 6 9 The function is the same as the one in Example 3; however, solurroN. -x or, since we are considering negative values of x, in this case #: equivalen tly, -t/P : r. Hence, in the first step when we divide numerin the denominator, and we ator and denominator bv x, we let x :-!* have ,. 3xt4 ,:-- Ja6-s:*'11 3*4lx ffi lim (3 + alx) t--@ lim 1-t/ffi) lim 3* lim (alx) Jf--@ c--@ -@ The remaining steps are similar to those in the solirtion of Example 3, and in the denominator we obtain -tfZ instead of t/-2. Hence, the limit is 4trt. "Infinite" limits at infinity can be considbred. There are formal definitions for each of the following. l i m /(r) : * m l i m /(r) : *m t--@ t-l@ l i m /(r) : - m lim /(x) : -@ iX--@ fi-l@ For exampte, f@): *m if the function/is defined on some interval "lit (a, +*1 and if for any number N > 0 there exists an M > 0 such that f (x) > N whenever r > M. The other definitions are left as an exercise (seeExercise17). Find EXAMpLE 5: ,2 lim-* rr-*o X t L Dividing solurroN; obtain the numerator Y2 = *rlim t.1 t w x ' lim _*: r - * c ox f L and the denominatot by x2, we 1 Evaluating the limit of the denominator, we have l i m f t * +f ) : H m1X * r i m4 :o*o:o T r-+- \I / r-+* r-+* Therefore, the limit of the denominator is 0, and the denominator is approaching 0 through positive values. The limit of the numerator is 1, and so by Limit theorem lz(i) (2.4.4) it follows that ryz lim _*: r-*a X+ L *cc 170 TOPICSON LIMITS,CONTINUIW,AND THE DERIVATIVE EXAMPLE 6: soLU'oN, ,rit#-*litf,ffi Find ,l r.m . -2 x - * We consider the limits of the numerator and the denominator separately. x-*@ 5X + 5 /)\t lim I'r-+o 1l: \I / lim /q* 4): f/ r-+* \I lim:,-+* lim 1:0-1:-1 X r-+cc tim !+ ti* i": o * o: o f . . r - + c oI *-+* Therefore, we have the limit of a quotient in which the limit of the numerator is - L and the limit of the denominator,is 0, where the denominator is approaching 0 through positive values. By Limit theorem 1.2(iii) it follows that r. llm r-+€ 2x- xz --:-fr, JX + 5 Exercises 4.1 In ExercisesL through'l'4, find the limits, and when applicable indicate the limit theorems being used. 4 a. r. u mr-+a 2x*l JX - ,^ . _r .l l f l4's32J* 3- I L {FT4 *-Zx*5 .,' X,i.T x s + x + 1 5. trm u**o y+4 ,. 4f+2f-5 '',11 w to. ,{1 fffi x- x) ^ ,. 3xa-7f +2 o' 2x4+ l "1T. 11. lim (\7F; - fF + 1) 9. lim ,3-*@ 12. lim [- \FT *o r s .n ^ ?5 "y -* 34 y-a- For the functions defined in Ex€rrises 15 and 16, prove that lim l(x) :1 by applying Definition 4.1.1; that is, for any e ) 0, show that there exists a number N > 0 such that lf (*l - 1l < . whenever x ) N. r s .f ( x ) : * 16.f(x):H 17. Give a definition for each of the following: (a) lim /(x) - -@; (b) lim f (x): *o; ,f--6 18. Prove tlut lim (f -4):+€ (c) lim /(r) : -oo f--@ by showing that for anyN > 0 there exists anM > 0 such t}at (f -4) > N whenever x> M. 19. Prove that lim (6 - I - t') : -6 by applying the definition in Exercise 17(a). l 4.2 HORIZONTALAND VERTICALASYMPTOTES 171 20. Prove part (i) of Limit theorem 12 (2.4.4) if "x + a" is replaced by "x + *a." 21. Prove that ,._ 8r*3 _n j1i 2'- 1 = by showing that for any e ) 0 there exists a number N < 0 such that - l <. lry_+l lzx-1 whenever r ( N. 22. Prove part (ii) of Limit theorem 13 (4.1.3). fr, c fne function / is defined bY f(x):]n#* Draw a sketch of the graph of f . At what values of r is / discontinuous? 4.2 HORIZONTAL AND An aid in drawing the sketch of the graph of a function is to find, if there VERTICAL ASYMPTOTES are any, the "horizontal and vertical asymptotes" of the graph. Consider the function / defined by v (1) r asymptote. Figure4.2.1 y A sketch of the graph of.f is in Fig. 4.2.1,.Any line parallel to and above the x axis will intersect this graph in two points: one point to the left of the line x: a and one point to the right of this line. Thus, for any k > 0, no matter how large, the line y: k will intersect the graph of f in two points; the distance of these two points from the line x: a gets smaller and smaller as k gets larger and larger. The line r : A is called a "vertical asymptote" of the graph of /. Following is the definition of a vertical 4.2.1 Definition The line x: A is said to be a aerticalasymptoteof the graph of the function f if at least one of the following statementsis true: (i) lim f ( x ) : 4 a t-d+ (ii) lim 8'd+ (iii) lim tr-A- f(x):-* f@)-a*^. (iv) lim f ( x ) : - a . fi'A- For the function defined by Eq. (1), both parts (i) and (iii) of the above definition are true. If g is the function defined by Figure 4.2.2 . q\ (. ,x/ ) : o ---1 (X _ A), 172 TOPICSON LIMITS,CONTINUITY,AND THE DERIVATIVE then both parts (ii) and (iv) are true, and the line r : A isa vertical asymptote of the graph of g. This is shown in Fig. 4.2.2. A "hotizontal asymptote" of a graph is a line parallel to the x axis. 4.2.2 Definition The line A: b is said to be a horizontalasymptoteof the graph of the function f tf at least one of the following statementsis true: (i) ,lit f @):u. (ii) lim f (x):to. ExAMPLE L: Find the horizontal asymptotes of the graph of the function / defined by r/ r I\x):ffi I and draw a sketch of the graph. solurroN: First we consider,lim f (x), and we have lirfe)-J1*# To evaluate this limit we write x: t/x' (x > 0, because r -> +co) and then divide the numerator and the denominator, under the radical sign, by xr. x t:--llm ....--=: r-*co Vxz * 1 1. --- \/P ltm .-J; Vf + | : lim ,T-l@ (Ltf) -1 l Therefore,by Definition a.2.2(1), the line y: L is a horizontal asymptote. Now we consider (x); in this case we write x- - \/P because ,lim f if x --+-@, x < 0. So we have -l./& lim/(r):,tiaffi :limt--a t I * Llxz (uf) --1 Figure4.2.3 Accordingly,by Definition 4.2.2(ii), the line y - -1 is a horizontal asymptote. A sketch of the graph is in Fig.4.2.2. Exavprn 2: Find the vertical and horizontal asymptotes of the graph of the equation solurroN: y:t2 Solving the given equation for y, we obtain 4.2 HORIZONTALAND VERTICALASYMPTOTES xy' - 2y' - 4x -- 0, and draw a Equation (2) defines two functions: sketch of the graph. U v : f ,(x) where fi is defined by f ,(x) : *2 {- and - -ztlh A: fr(x) wheref2 is definedby fr(x) The graph of the given equation is composed of the graphs of the two functions f, and fr. The domains of the two functions consist of those > 0. By using the result of Example 8, values of x f.or which xl(x-z) we see that the domain of /r and fz is )c:2, excluding 1.7, and Sec. (-*,0] u (2,+m). Figure 4.2.4 Now, consider /r necause Ir2f,(x): ri^.21#: *oo by Definition 4.2.1,(i)the line x - 2 is a vertical asymptote of the graph of ft' 2rl#:,t11 zyftO:z ,tit f,@):*t1t Figure4.2.5 so by Definition 4.2.2(1)the line y : 2 is a horizontal asymptoteof the graph of ft. similarly,li1; fr@):2. A sketch of the graph of f is shown in Fig. 4.2.4. Hence, by Definition 4.2.1,(ii)the line x: graph of fz. 2 is a vertical asymptote of the so by Definition a.2.2(1)the line y : -2 is a horizontal asymptoteof the graph of fr. -2. A sketchof the graph of fris shown in Fig.4.2.5. Also, 11* fr(x): The graph of the given equation is the union of the graphs of.ft and fz, and a sketchis shown in Fig. 4.2.6. 4.2 Exercises In Exercises I through 14, find the horizontal and vertical asymptotes of the graph of the function defined by the givm equation, and draw a sketch of the graPh. l.f(x):# 2. f (x):;, _a 3. g(x): _? G;zy 174 TOPICS ON LIMITS, CONTINU|ry, ANDTHEDERIVATIVE 4. F(x): x2*8x*-16 4x2 s .f ( x ) : # x _ 6 8./(r):;a / . h \ x ): 7 ; J 10.s(r):ffi6 /Ly2 6. G(r): 6x2*11.r-10 y2 11.F(x):A;+ A.f(x):w#fr 13.tf(x)-+r\, \/x"_2 In ExercisesL5 through 21, find the horizontal and vertical asymptotes of the graph of the given equation, and draw a sketch of the graph. 1 5 . 3 x y- 2 x - 4 y - 3 : 0 '/..8. Zxyz* 4y, - Jx:0 21. x2y * 6xy - f + 2x * 9y * 3 : 0 4'3 ADDITIONAL THEOREMS ON LIMITS OF FUNCTIONS 4'3'1'Theorem 1 6 .Z x y* 4 x - 3 y * 6 : 0 1 9 .( y ' - 1 ) ( x - 3 ) : 6 17. x2y2- x2+ 4yr:0 20. xyzI 3y, - 9r: 0 We now discuss five theorems which are needed to prove some rmportant theorems in later sections.After the statementof each theorem, a graphicalillustration is given. If liry /(x) existsand is positive, then there is an open interval containing a such that f (x) > 0 for every x # a in the interval. . rLLUsrRArroN1: Consider the function defined bv f f(x) : 2 x - l A sketchof the graph of / is in Fig. 4.9.r. Becausetiq /(4 : L, and 1 > 0, according to Theorem 4.3.1 there is an open interval containing 3 such that /(r) ) 0 for every x * 3 in the interval. such an interval is (2, 4). Actually, any open interval (a,b)for which i = a< 3 and b > gwill do. o pRooFoF rHEoRa''4.3.7: Let L: tjT f@).By hypothesis,L > 0. Apply_ ing Definition z.r.r and taking e:!L, we know thereis a 6 > 0 such that lf (*) - Ll < +f whenever 0 < l* - ol < a (1) Also, lf (x)- LI < lL is equivalentto -*L < f (x)- L < tf (refer to Theorcm'/-..2.2),which in turn is equivalent to Figure4,3.1 tt<f(x) <*t (2) Also, 0 < l" - al < 6 is equivalentto -D < x- a < 6 but x # a, which 4.3 ADDITIONALTHEOREMSON LIMITSOF FUNCTIONS 175 in turn is equivalent to but a-61x1a*D x*a which is equivalent to stating that r is in the open interval (a - 6, a * 6) but x * a (3) From statements (2) and (3), we can replace (1) by the statement +L < f(x) <Et when r is in the open intenral (a- 6, a * E) butx * a Since L > 0,we have the conclusion that/(r) the open interval (a - 6,4 * 6). 4.3.2 Theorem > 0 for every x * a in tt tti /(r) exists and is negative, there is an open interval containing a such that /(x) < 0 for every x * a in the interval. The proof of this theorem is similar to the proof of Theorem4.3.Land is left as an exercise(SeeExerciseL). . rLLUsrRATroN 2: Let 6-x , \ g e c ): g _ , h - -4 ( 0; hence, Figure 4.3.2 shows a sketch of the graph of 8. ljT Sttl by Theorem 4.3.2there is an oPen interval containing 2 such that g(r) < 0 for every x # 2 in the interval. Such an interval is (8,3). Any oPen interval o (a,b) for which t = a <2 and} < b = 6 will suffice. theotem. The following theoremis sometimesreferredto as the squeeze Figure4.3,2 4.3.3 Theorem Suppose that the functions f , g, and h are defined on some oPen interval / containinga exceptpossiblyat a itself, and that /(x) < g(r) 'h(x) for all x in I for which x # a. Also suPPosethat lim f (x) andr11\r(r) both exist and are equal to L. Then li* S(t) also exists and is equal to L. The proof is left as an exercise(seeExercise2). o rLLUsrRArIoN3: Consider the functions f , 8, and h defined by -4(x - 2)''+ 3 f (x) : ^/-\_(x-2)(x2-4x1-7) 8\X):x_2 h(x):4(x-2)2+3 The graphs of the functions / and h are parabolas having their vertex at 176 TOPICSON LIMITS,CONTINUIW,AND THE DERIVATIVE fore satisfied, and it follows that lim g(r) - 3. c-2 o Figure 4.3.3 4.3.4 Theorem Supposethat the function / is defined on some open interval / containing a, except possibly at a. Also suppose that there is some number M fir which thereisaE ) 0suchthat/(x) <. llrlwhenever0( lr- ol < 6.Then, tt f Q) existsand is equal to L, L < M. IrT o TLLUSTRATIoN 4: Figure 4.3.4shows a sketch of the graph of a function / satisfying the hypothesis of Theorem 4.9.4. From the figure we see that /(1) is not defined, but / is defined on the open interval G, il exceptat 1.. Furthermorc,f(x) s I whenever 0 < lr - 1l < *. Thus, we can conclude from Theorem 4.3.4that if lim /(r) exists and is L, then L = *. From the figure we observe that th"rJii an L and it is 2. . PRooFoF rHEoRnrv4.3.4: We assumethat M < L and show that this as- ON AN INTERVAL 177 4.4 CONTINUITY sumption leads to a contradiction.If M < L, there is some e ) 0 such that M + e: L. Because fif lf(x)-Ll <e <O: L, thereexistsa 6r ) 0 suchthat w h e n e v e r0 < l r - a l <Dt which is equivalent to L-e<f(x)<L*e w h e n e v e r0 < l x - a l <Dt ReplaceL by M + e; it follows that there exists a 6r ) 0 such that (M *e)-e<f(x) w h e n e v e r0 ( l * - o l <6' or, equivalently, M<f(x) w h e n e v e r0 ( l * - o l <Et (4) But, by hypothesis, there is a 6 such that (s) f (x) = tvt whenever 0 < l* ol < 6 Statements(4) and (5) contradict each other. Hence, our assumption that I . herefore,L'M. M < L is falseT Figure4.3.4 4.3.5 Theorem Supposethat the function / is defined on some open interval I containing a, exceptpossibly at a. Also suppose that there is some number M fot which there is a a ) 0 such that /(x) > M whenever 0 < l*- ol < E. > Then tt /(r) exists and is equal to L, L M' lT The proof is left as an exercise(SeeExercise3)' 5: Figure 4.3.4 also illustrates Theorem 4.3.5.From the o rLLUSrRArroN - 1l < i; and, becauseas figure we see that /(r) = I whenever 0 < lt (*, B) exceptat L, Theointerval pieviously stated, f is dehnedon the open o L > +' L, then rem 4.3.5statesthat if lim /(r) exists and is 4.3 Exercises 1. Prove Theorem 4.3.2. - cl ts sufficiently small, and € > 0, the following inequalities 2. prove Theorem 4.3.3. (rrrNr: Ftust show that when lr m u s t h o l d : L - e < f ( x ) < I * e , a n d L - e < h ( x )< L * ' e ' \ 3. Prove Theorem 4.3.5. and that l'(0) : 0. (IIrNr: UEeTheo4. Ler f be a function such that ll(r) | = I for all r. Prove that f is differentiable at 0 rem 4.3.3.) ON 4.4 CONTINUITy INTERVAL AN In Example 3 of Sec. 2.6 we showed that the function h, for which is continuous at every number in the oPen interval h(x): {4-P, 178 TOPICSON LIMITS, CONTINUIry, AND THEDERIVATIVE ' 4'4'l Definition ExAMPLE1: If f (x) : U (x - g) , on what open intervals is f continuous? (-2, 2). Because of this fact, we say that h is continuous on the open interval (-2, 2).Following is the general definition of continuity on an open interval. A function is said to be continuoLtson an open interaalif and only if it is continuous at every number in the open interval. solurroN: The function / is continuous at every number except3. Hence, by Definition 4.4.1,/ is continuous on every open interval which does not contain the number 3. We refer again to the function hfor which h(x) - {E4.We know that h is continuous on the open interval (-2,2). However, becauseh is not defined on any oPen interval containing either -2 or 2, we cannot consider lim h(r) or lim h(x).Hence, to discussthe question of the r--2 r-2 continuity of h on the closedinterval [-2,21, we must extend.the concept continuity to include continuity at an endpoint of a closed interval. _of We do this by first defining right-handcontinuity and,left-handcontinuity. 4'4'2 Definition The function / is said to be continuous from the right at the numbera if. and only if the following three conditions are satisfied: (t) f (a) exists. (ii) exists. liT /(r) (iii) lim f(x):f(o). 4.4.3 Definition Th9 fungtion / is said to be continuous from the teft at the numbera if and. only if the following three conditions are satisfied: (i) f(a) exists. Gt)I'T /(x) exists. (iii) lim f(x):f(a). 4.4.4 Definition rxaupr.n 2: Prove that the function h, f.or which h(x): l/4- xr, is continuouson the closedinterval l-2,21. A function whose domain includes the closed interval la, bl is said to be continuous on fa,b] if and only if it is continuous on the open interval (a, b), as well as continuous from the right at a and,continuous from the left at b. soLUTroN: The function h is continuouson the open interval (-2,2) and \/4-7:0: "ti1. h(-2) 4.4 CONTINUIW ON AN INTERVAL tim GE 179 - o: h(2) I-2- Thus, by Definition 4.4.4,h is continuous on the closedinterval l-2,2f . 4.4.5 Definition nxavrprr 3: Given / is the function defined bY f(x) : determine whether f is continuous or discontinuous on each of the following intervals: (-3,2),?3,21,[-3,2),(-3,21. (i) A function whose domain includes the interval half-open on the right la, b) is said to be continuous on la, b) if and only if it is continuous on the open interval (a, b) and continuous from the rig}:rt at a. (ii) A function whose domain includes the interval half-open on the left (a, bl is said to be continuouson (a,bl if and only if it is continuous on the oPen interval (a,b) and continuous from the left at b. solurroN: We first determine the domain of /. Th" domain of / is the is nonnegative. Thus, any set of all numbers for which (2-x)l(3*x) of this fraction the denominator and numerator the which r for values of changes numerator The domain. the from excluded are signs have opposite -3. We : I : when sign changes denominator artd the 2, I sign when *ik" use of Table 4.4.L to determine when the fraction is positive, negative, zero,or undefined, from which we are able to determine the values of.x for which /(x) exists. The domain of / is then the interual half-open on the left (-3, 21. The function / is continuous on the oPen interval (-3,z).It is continuous from the left at 2 because t2-' l'_t{ffi:s:f(2) -3 because However, f i" not continuous from the right at b-- ,ri1.rl'#:*rc We concludethat f is continuous on (-3, 2] and discontinuouson [-3,2] and [-3, 2). Table4.4.7 2- x 3*x 2- x 3+x f(x) x<-3 x: -3 + 5 0 undefined does not exist -3 1x + + + + x:2 0 5 0 0 2<x does not exist + does not exist 180 TOPICS ON LIMITS, CONTINUITY, ANDTHEDERIVATIVE ExAMrLE4: Given g(r) : firn and L < x < 3. Discussthe continuity of g. solurroN: Figure 4.4.1,shows a sketch of the graph of g. Supposethat a is any number in (1, 2). Then f f u n : 1 , a n d l i m [t]l : L. Hence, if t 1 u . 2, fig(r) limg(r):lim : g(u), and so g is continuouson [r]l :f (r,2). :g(1) Therefore,g is continuous from the right at 1..But lim g(r) : lim firn : t I-2- r-2' and g(2) :2. So 3 is not continuous from the left at 2. The function g is thereforecontinuous on the interval half-open on the right [1,2). similarly, g is continuous at every number u for *ni.r, 2 < u < j. At 2, g is continuous from the right since lim g(r) : g(z), but g is not Figure4.4.1 continuous from the left at 3 becauselim Siij.:2 g is also continuous on the interv ul tZ., ii. and g(3) : 3. Hence, Exercises4.4 In Exercises 1 through 16, determine whether the function is continuous or discontinuous on each of the indicated intervals. , t. f (i : ;i-s; $, 7)' l-6, 4J.(-@,o), (-s, +-), I-s, +-), t-10,-s). r)-1 z. f(r):Flt (0' 41,(-2,2), (--,-zl, (2,+@),l-4,4f, (-2,21. 3. s(r) : t@-- s;1--,-s1, (--, -31, (3,+oo),[s,+6), (_3,3). a . f @ ) : [ x n ; e r , b , G ,D , 0 , , 2 ) , 1 1 , 2 )( .7 , 2 ] . tr-ll s. f(t)-fr (-@,r),(--,1), t-r,tl, (-1,+e),(1,+@). (2x-3 rrx<-2 l 6. h(x): ] | - 5 lr,-2- x - 1|l, (--, r), (-2, +*), (-2.'t), t-2, 1), l-2. Ll l3-x ifl<r I 7.f(r): \/4- '*. (-2,2),t-2,21, t-2,2),(-2,21,(-@,-21,(2,+-). 6=- e. t('l : ,,l'r:i e2,2), l-2.21,l-2,2), (-2,21,(--,-2), 12,+-). e'r{il:6fu;?1,s), 10./(r) : ITIE=E; to_J n. c(x):li_|; (-r,31. t-1,31, t-1,3), (-1,3), t-1,31,t-1,3), (-1,31. (--.-3). (-3,3),t_3.31, (3,41, l_s,s),ls.41, t4,+*),(4,+_). 4 . 5 M A X I M U MA N D M I N I M U MV A L U E SO F A F U N C T I O N 1 8 1 t z + T ( - * , - 5 ) , (, - * , - 5 1, 1 - 5 , - 2 ) , 1 - 2 , 5 1 , 1 - 2 ,(5-)2, , 5 1(,- 2 , 5 ) ,[ s , + @ )( ,5 , + - ; ' t z .S @ ) :l # + ; In Exercises 13 through 20, determine the intervals on which the given function is continuous. 1 6 .F ( x ) : 1 3f .( x ) : h tFE Vr+5 17. The function of Exercise25 in Exercises1.8. 1.8. 18. The functionof Exercise26 in Exercises 1.7. 31 in Exercises 19. The functionof Exerciee 20. The function of Exercise34 in Exercises1.7. In Exercises21 through 28, functions f and g are defined. In eachexetcisedefine I " g, and determine all valuesof r for which f . I is continuous. 21.f (x) : x3;g(x) : li 22. f(x) : x2;8(x) : x2- 3 2s.f(x): t7;s(x): # 2 a .f @ ) : : . f x ; 8 @ ) : x + " l ' 25. f (x): fi, s@): {i 26.f (x) : fl-xi8(x) : \tr + 1 v*1 sG): t/i 2 7f.( x ) : f i , 2s.f (x\ : \ETI; g@): {x (-oo, +6) and In Exercise8 29 through 32, find the values of the constants c and lr that male the function continuous on function' resulting the graph of draw a sketch of the lfx<2 if.2 < x zsf(x):{*:T il;2i fx gr.f(x):lcx*k l-2x ifr<L ifl<x14 if4<x 3o/(r): 32.f(x): ifx<-2 if-z<x<L if1.<r 33. Given that f is defined by f(x):{f,[;]il|-=::i If g is continuous on [c, b), and ft is continuous on [b, c], can we conclude thatf is continuous on [4, c] ? If your answ_er is-yes, prove it. If your answer is no, what additionat condition or conditions would assute continuity of f on la' c)? U. 1I f is the function defined by /(x) : 9!;, terval [0, +@). where n and n are positive integers, prove that f is continuous on the in- 4.5 MAXIMUM AND We have seen that the geometrical interpretation of the derivative of a MINIMUM VALUES function is the slope of the tangent line to the graph of a function at a OF A FUNCTION point. This fact enables us to apply derivatives as an aid in sketching graphs. For example, the derivative may be used to determine at what points the tangent line is horizontal; these are the points where the derivative is zero.Also, the derivative may be used to find the intervals for TOPICSON LIMITS,CONTINUIW,AND THE DERIVATIVE which the graph of a function lies above the tangent line and the intervals for which the graph lies below the tangent line. Beforeapplying the derivative to draw sketchesof graphs, we need some definitionr u.,d theorems. 4.5.1 Definition The function f is said to have a relatiaemaximumaalue at c if there exists an open interval containing c, on which / is defined, such that /(c) > f (x) for all I in this interval. Figures 4.5.1,and 4.s.2 each show a sketch of a portion of the graph of a function having a relative maximum value at c. x 4.5.2 Definition Figure4.5.2 The function / is said to have a relatioe minimum aalue at c if there exists an open interval containing c, on which is defined, such that (c) = (x) / f f for all r in this interval. Figures 4.5.3 and 4.5.4 each show a sketch of a portion of the graph of a function having a relative minimum value at c. ac Figure4.5.3 Figure 4.5.4 If the function / has either a relative maximum or a relative minimum value at c, then / is said to have a relatiue extremum at c. (The plurals of maximum and minimum are maxima and minima; the plural of extremum is extrema.) The following theorem enables us to locate the possible values of c for which there is a relative extremum. 4.5 MAXIMUMAND MINIMUMVALUESOF A FUNCTION 183 4.5.3 Theorem lf f(x) exists for all values of x in the open interval (a,b) and if / has a relative extremum at c, where a < c 1 b, then It f' (c) exists,f' (c) : 0. pRooF: The proof will be given for the casewhen / has a relative minimum value at c. tf f '(c) exists,from formula (5) of Sec.3.3 we have (1) f,(,):tjTfff Because / has a relative minimum existsaD)0suchthat value at c, by Definition 4.5.2, there w h e n e v e r(o l x - c l < 6 f(x)-f(c) =o If r is approachingc from the right, x - c ) 0, and therefore f(c) = o f(x) x-c whenevero ( r - c ( D By Theorem 4.3.5,if the limit exists, ,. fk)-f(c) lrrll->U n-c+ )C - - n (2) C Similarly, if x is approaching c from the left, x - c ( 0, and therefore f(x) f(c) -o x-c whenever-6 < x- c < o so that by Theorem 4.3.4,if the limit exists, Ilm- ;-c- f(x) - f(c) x- c Sv (3) Becausef '(c) exists, the limits in inequalities (2) and (3) must be '(c). So from (2) we have equal, and both must be equal to f f'(c) =o (4) and from (3), f'(c)'o (5) Becauseboth (4) and (5) are taken to be true, we conclude that f ' ( c ): o which was to be proved. The proof for the case when / has a relative maximum value at c is I similar and is left as an exercise (see Exercise 37). The geometrical interpretation of Theorem 4.5.3 is that if / has a rela- 184 TOPICS ON LIMITS, CONTINUITY, ANDTHEDERIVATIVE tive extremum at c and if f ' (c) exists,then the graph of y : /(x) must have a horizontal tangent line at the point where x: c. If / is a differentiable function, then the only possible values of x for which f canhave a relative extremum are thosefor which f ' (x): 0. However, f ' (x) can be equal to zero for a specific value of x, and yet f may not have a relative extremum there, as shown in the following illustration. 1: Consider the function f defined bv r rLLUSrRArroN f(x):(r-1;' A sketch of the graph of this function is shown in Fig. 4.5.5. ,(x): f 3 ( x - L ) 2 ,a n d s o f ' ( 1 ) : 0 . H o w e v e r f, ( x ) < 0 , i f x . - - ' 1 ,a, n d / ( r ) j 0 , if. x > 1,.So / does not have a relative extremum at L. o A function / may have a relative extremum at a number and f, may fail to exist there. This situation is shown in Illustration 2. Figure4.5.5 o rLLUSrRArroN2: Let the function f be defined as follows: f(x): {';_: lf; .=i x A sketch of the graph of this function is shown in Fig. 4.5.6.The function / has a relative maximum value at 3. The derivativ" irorr, the left at 3 is given by f -(3) : 2, and the derivative from the right at 3 is given by f *(3) - -1. Therefore,we concludethat /'(3) does notlxist o Illustration 2 demonstrateswhy the condition ,'f ,(c) exists,,must be included in the hypothesis of Theorem 4.5.3. In summary, then, if. a function / is defined at a number c, anecessary condition fot f to have a relative extremum there is that either ,(c) :'O f or f '(c) does not exist. But we have noted that this condition is not sufficient. Figure4.5.6 4.5.4 Definition If c is a number in the domain of the function and if either , (c): f f f'(c) does not exist, then c is called a critical number of f. 0 or Because of this definition and the previous discussion, we can conclude that a necessary condition for a function to have a relative extremum at a number c is for c to be a critical number. ExAMPLE1: Find the critical numbers of the function / defined by f (x) - x4ts+ 4xrt3. SoLUTIoN:f,(x)-!*,,,+t*_,,":t*_,,'1x*|):w '(x): 0 when r_ -L, and f '(x) doesnot existwhen t: 0. Both-1 f and 0 are in the domain of f therefore, the critical numbers of f are -L and 0. AND MINIMUM VALUES 4.5 MAXIMUM OF A FUNCTION 185 We are frequently concerned with a function which is defined on a given interval, and we wish to find the largest or smallest function value on the interual. These intervals can be either closed, open, or closed at one end and open at the other. The greatest function value on an interval is called the "absolute maximum vallJe," and the smallest function value on an interval is called the "absolute minimum value." Following are the precise definitions. 4.5.5 Definition The function / is said to have an absolute maximum aalue on an interaal if there is some number c in the interval such that /( c) > f (x) tor all r in the interval. In such a case, f(c) is the absolute maximum value of / on the interval. 4.5.6 Definition The function / is said to have an absoluteminimum aalue on an interaal if there is some number c in the interval such that /(c) = f (x) for all r in the interual. In such a case, f(c) is the absolute minimum value of / on the interval. An absolute extremum of. a function on an interval is either an absolute maximum value or an absolute minimum value of the function on the interval. A function may or may not have an absolute extremum on a given interval. In each of the following illustrations, a function and an interval are given, and we determine the absolute extrema of the function on the interval if there are any. o rLLUsrRArroN 3: Suppose / is the function defined by Figure4.5.7 f(x) -2x A sketchof the graphof / on ll, 4) is shownin Fig.4.5.7.The function/ has an absoluteminimum valueof 2 on 11,4).Thereis no absolutemaxilim /(r) : 8, but /(r) is alwayslessthan mum valueof / on lL, 4),because 8 on the given interval. o rLLUsrRArIoN 4: Consider the function / defined by f(x):-* A sketch of the graph of f on (-3,2f is shown in Fig. 4.5.8.The function / has an absolute maximum value of 0 on (-3' 21. There is no absolute minimum value of / on (-3,21, because lim /(r) -9,but, f(x) is always t'-Bo greater than -9 on the given interval. o TLLUSTRATIoN 5: The function / defined by Figure4.5.8 f(x) : L- x2 186 TOPICSON LIMITS,CONTINUITY,AND THE DERIVATIVE has neither an absolute maximum value nor an absolute minimum value on (-1, 1). A sketch of the graph of / on (-1, 1) is shown in Fig. 4.s.9. Observe that ,t1T. f @) : -a and l_tT_fOl: *oo o o rLLUSrRArroN 6: Let f be the function defined by f(x): {X,*_rr* *, ifx<1 'l-, if <x A sketchof the graph of f on [-5, 4] is shown in Fig. 4.5.10.The absolute maximum value of / on [-5, 4] occurs at ']-.,and /(1) :2; the absolute minimum value of / on l-5, 4l occursat -5, and /(-5) : -4. Note that / has a relative maximum value at 1 and a relative minimum value at 3. Also, note that 1 is a critical number of f because/' doesnot exist at 1, and 3 is a critical number of f because/'(3) : 0. o Figure4.5.9 F i g u re 4 .5.10 . rLLUSrRArroN 7: The function f defined bv 1 tf(x) \,/ : x_3 has neither an absolute maximum value nor an absolute minimum value o n [ L , 5 ] . s e e F i g . 4 . 5 . 1 1f o r a s k e t c h o f t h e g r a p h o f /.,11p f@):-a; so /(x) can be made less than any negative number by taking (3 - r) > 0 -- *a) ro and less than a suitable positive 6. Also, 1ry can be Ir1/tr) made greater than any positive number by taking (x - 3) > 0 and less than a suitable positive D. We may speak of an absolute extremum of a function when no interval is specified. In such a case we are referring to an absolute extremum of the function on the entire domain of the function. F i g u r e4 . 5 . 1 1 4.5.7 Definition /(c) is said to be the absolute maximum aalue of the function f if c is in the domain of f and it f (c) > f (*) for all varues of r in the domain of f. 4.5 MAXIMUMAND MINIMUMVALUESOF A FUNCTION 187 4.5.8 Definition /(c) is said to be the absoluteminimumaalueof the function f if c is in the domain of / and if f (c) = f (x) for all values of r in the domain of /. . rLLUsrRArIoN 8: The graph of the function / defined by f(x):x2-4x*8 is a parabola, and a sketch is shown in Fig. 4.5.12. The lowest point of the parabola is at (2, 4), and the parabola oPens upward. The function has an absolute minimum value of 4 at 2. There is no absolute maxio mum value of /. Referring back to Illustrations 3-8, we see that the only case in which there is both an absolute maximum function value and an absolute minimum function value is in Illustration 5, where the function is continuous on the closed interval l-5,4]. In the other illustrations, either we do not have a closed interval or we do not have a continuous function. If a function is continuous on a closed interval, there is a theorem, called the extreme-oalue theorem, whrch assures us that the function has both an absolute maximum value and an absolute minimum value on the interval. The proof of this theorem is beyond the scope of this book, but we can state it without proof. You are referred to an advanced calculus text for the F i g u r e4 . 5 . 1 2 proof. 4.5.9 Theorem Extreme-Vglue Theorem If the function / is continuous on the closed interval la, bf , then / has an absolute maximum value and an absolute minimum value on la, blAn absolute extremum of a function continuous on a closed interval must be either a relative extremum or a function value at an endpoint of the interval. Because a necessary condition for a function to have a relative extremum at a number c is for c to be a critical number, we can determine the absolute maximum value and the absolute minimum value of a continuous function f on a closed interval la,bl by the following procedure: (1) (2) (3) . rxeuplr 2: Given f(x):x3+x2-x*l find the absolute extrema of f on [_2,+1. find the function values at the critical numbers of.f on la, bl; find the values of f(a) and f(b); the largest of the values from steps (1) and (2) is the absolute maximum value, and the smallest of the values from steps (L) and (2) is the absolute minimum value. sor,urroN: Because/ is continuous on l-2, +1,the extreme-valuetheorem applies. To find the critical numbers of f *e first find /': '(x) :3x2 * 2x -'l' f ' will f (x) exists for all real numbers, and so the only critical numbers of / 188 TOPICSON LIMITS,CONTINUIW,AND THE DERIVATIVE be the valuesof x for which f '(x):0. ( 3 x - 1 ) ( x* 1 ) : s Settingf ' (x) - 0, we have from which we obtain and The critical numbers of f are-1 and *, and each of these numbers is in the given closed interval l-2, +l.We find the function values at the critical numbers and at the endpoints of the interval, which are given in Table 4.5.7. The absolute maximum value of f on l-2, +l is therefore 2, which occurs at-L, and the absoluteminimum value of /on l-2,+l is-1, which occurs at the left endpoint -2. A sketch of the graph of this function on [-2, +l is shown in Fig. 4.5.73. I Table4.5.1 F i g u r e4 . 5 . 1 3 EXAMPLE 3: Given f (x) : (x - 21rr" find the absolute extrema of f on [1,5]. solurroN: applies. Because / is continuous on Il, sl, the extreme-value theorem rf ' (\ "x/ ) : *3J( x - 2-1-u a There is no value of x for which f'(x):0. However,becausef,(x) does not exist at 2, we conclude that 2 is a critical number of , so that the f absolute extrema occur either at 2 ot at one of the endpoints of the interval. The function values at these numbers are given in tabte 4.s.2. From the table we conclude that the absolute minimum value of / ot [L5] is 0, occurring at2, and the absolutemaximum value of /on [1,5J is179, occurringat 5-A sketchof the graph of this function on [1,5] is shown in Fig. 4.5.14. Table4.5.2 Exercises4.5 In ExercisesI through 10, find the critical numbers of the given function. 1. f (x) : x3I7x2 - 5x 2. f (x) :2x3 - 2x2- 1.6x-f 1. 3. f(x) : xa * 4x3- 2x2- 12x INVOLVINGAN ABSOLUTEEXTREMUM ON A CLOSEDINTERVAL 189 4.6 APPLICATIONS a. f@) : v7t3+ x4tl- 3xrt3 7. f (x): 1 0 .f t x ) : (* - A)ztz 6. f(x) : xa* 1L13* 34* + 75x- 2 x 5.f(x)-x6t5-12rrts 8. f (x\: (r' - 3f * 4;tte e.flx):74 r*1 f _Sx+4 In Exercises 11 through 24, find the absolute extrema of the given function on the given interval' if there are any' and find the values of r at which the absolute extrema occur, Draw a eketch of the graph of the function on the interval. ,l 1,2.f (x) : x2- 2x -l 4; (-*, **) 73,f (x):|; ?L3J 1'a. f k) :;; [2,3) 15.f (x) : \/t4; [-3, +-1 16. f(x):fi, 4 17. f (x) : ---^\,,; 12,51 J)' 18./(r) : \84'; e2,2) 1 9 f. ( x ) : l x - 4 l + r ; ( 0 , 6 ) r]1,'f (x) - 4 - 3x;(-1, 2f 1 \x- [2 ] irx* s[; t3,s] ifr:5J 20. f (x) : 14- 12l;(-co, 1m; z r .f ( x ) : l r - 5 L2 23.f (x) : )c- [rn; (1,3) 2 a .f @ ) : U ( x )- U ( x - 1 ) ; ( - 1 , 1 ) ?3,2) 22f(x):{r *'' ;i:=_l};e, t In Exercises25 through 36, find the absolute maximum value and the absolute minimum value of the given function on the indicated intervul by the method used in Examples2 and 3 of this section. Draw a sketch of the graph of the function on the interval. 26. f (x): rs * 3x2- 9x; l-4, 4l 2 5.f (x): 13* 5x - 4; [ - 3, -1 ] 'l'6; 28. f (x) : x4- 8x2+ 16; l-1, 4l 27. f (x) : x4- 8x2+ l-4, 0l 30. f (x) : 74- 8xzI 16; l-3 ' 2l 2 9 .f ( x ) : x 4- 8 x 2* 1 6 ; 1 0 , 3 f 31.f (x): f* ,, l-1, 2) t-l-6 32.f (x): fi; l-s,2l 3 3 .f ( x ) : ( r * 1 . ) 2 t 3 ; l - 2 , " 1 . 1 3 a .f @ ) - L - 35f (x): {'J-t lf;'=; :: t},r-3,3r (x-31't'' l-5,41 36f (x): {rl- [; I ?]; ;l-:-=:=oa];r-0,o1 37. Prove Theorem 4.5.3 for the case when f has a relative maximum value at c. 4.S APPLICATIONS INVOLVING AN ABSOLUTE EXTREMUM ON A CLOSED INTERVAL We consider some problerns in whidr the solution is air absolute extremum oI a function on a dosed interval. Use is made of the extremevalue theorem, which assures us that both an absolute maximum value and an absolute minimum value of a function exist on a closed interval if the function is continuous on that closed interval. The procedure is illustrated by some examples. TOPICSON LIMITS, CONTINUITY, AND THEDERIVATIVE L: A cardboard box EXAMPLE manufacturer wishes to make open boxes from pieces of cardboard 12 in. square by cutting equal squares from the four corners and turning up the sides. Find the length of the side of the square to be cut out in order to obtain a box of the largest possible volume. i ft N I N ri rn. r in. (L2 -zr) in. solurroN: Let x: the number of inches in the length of the side of the square to be cut out; V: the number of cubic inches in the volume of the box. The number of inches in the dimensions of the box are then r, (12 - 2x), and (12- 2r). Figurc 4.6.t representsa given piece of cardboard, and Fig. 4.6.2representsthe box. The volume of the box is the product of the three dimensions, and so V is a function of x, and we write V(x) : xc(Lz- 2x)(12 - 2x) (1) lf. x : 0, l/ :0, and if.x: 6, V :0. The value of r that we wish to find is in the closedinterval l0,6l.BecauseV is continuous on the closedinterval 10, 61,it follows from the extreme-valuetheorem that V has an absolute maximum value on this interval. We also know that this absolute maximum value of V must occur either at a critical number or at an endpoint of the interval. To find the critical numbers of V, we findV, (r), and then find the values of x for which either V' (x) : 0 or V, (x) does not exist. From Eq. (1), we obtain V (x) : LMx - 48x2* 4xg Thus, V' (x) : lM - 96x * l,xz Figure4.6.1 V' (x) exists for all values of r. Setting V, (r) - 0, we have l2(*-8r*12):g from which we obtain x: 6 and x:2 k-trz -2r) in. Figure 4.6.2 The critical numbers of V are2 and 5, both of which are in the closed interval [0, 6]. The absolute maximum value of v on [0, 6] must occur at either a critical number or at an endpoint of the interval. BecauseV(0) - g and v(6) : 0, while v(2) : 128,we concludethat the absolutemaximum value of V on [0, 6] is L28,occurring at 2. Therefore, the largest possible volume is l2g in.B, and this is obtained when the length of the side of the square cut out is 2 in. We should emphasize that in Example 1 the existence of an absolute maximum value of I/ is guaranteed by the extreme-value theorem. In the following example, the existence of an absolute minimum value is guaranteed by the same theorem. nxaprprr 2: An island is at point 4,6 miles offshore from S O L U T I O N : Refer to Fig. 4.6.3.Let P be the point on the beach where the man lands. Therefore, the man rows from A to P and walks from p to c. 4.6 APPLICATIONSINVOLVINGAN ABSOLUTEEXTREMUMON A CLOSED INTERVAL the nearest point B on a straight beach. A store is at point C,7 miles down the beach from B. If a man can row at the rate of 4 mi/hr and walk at the rate of 5 mi/hr, where should he land in order to go from the island to the store in the least possible time? 191 the number of miles in the distance from B to P. the number of hours in the time it takes the man to make the trip from A to C. Then T: the number of hours in the time to go from A to P plus the number of hours in the time to go from P to C. Because time is obtained by dividing distance by rate, we have Let r: T: : _5 lAPl , -, 4 lPCl (2) From Fig. 4.6.3,we seethat triangle ABP. Therefore, is the length of the hypotenuse of right lAPl: \/F+ 35 We also see from the figure that lPe | : 7 - x. So from Eq. (2) T can be expressedas a function of x, and we have 6 miles \456,7-x T(x):--+? Becausethe distance from B to C is 7 miles and becauseP can be any point on the line segment BC, we know that r is in the closed interual Figure4.6.3 n. 10, We wish to find the value of x for which T has an absolute minimum value on [0, Zl. Because T is a continuous function of x on [0, 71,we know that such a value exists. The critical numbers of T are found by first computing T (x): ?- "\*'' (r):-L-L 4\/i, +-56 5 T' (x) existsfor all valuesof x. Setting T'(r) - 0 and solving for x, we have (3) 'zi::2':,:*' 5x:4\m 36 x2:64 r: +8 The number -8 is an extraneous root of Eq. (3), and 8 is not in the interval l0,7l.Therefore, there are no critical numbers of T in [0, 7]. The absolute minimum value of T on 10,7) must therefore occur at an endpoint of the interval. Computing T(0) and T(7), we get T(0): i3 and T(7): +\65 192 TOPICSON LIMITS, AND THEDERIVATIVE CONTINUIW, Since +\65 < H, the absolute minimum value of T on [0, 7] is +\/85, occurring when x:7. Therefore, in order for the man to go from the island to the store in the least possible time, he should row directly there and do no walking. EXAMPLE3: A rectangular field is to be fenced off along the bank of a river and no fence is required along the river. If the material for the fence costs $4 per running foot for the two ends and $6 per running foot for the side parallel to the river, find the dimensions of the field of largest possible area that can be enclosed with $1800 worth of fence. soLUrIoN: Let r: the number of feet in the length of an end of the field; a: the number of feet in the length of the side parallel to the river; A: the number of square feet in the area of the field. Hence, A: )cA (4) Since the cost of the material for each end is g4 per running foot and the length of an end is x ft, the total cost for the fence for each end is 4r dollars. Similarly, the total cost of the fence for the third side is 5y dollars. We have, then, 4x * 4x * 6y :1800 (s) To expressA in terms of a single variable, we solve Eq. (5) for y in terms of x and substitute this value into Eq. (4), yielding A as a function of x, and A ( x ) : r ( 3 0 0- t x ) (6) If y - 0, x: 225,and if x: 0, A :300. Becauseboth r and y must be nonnegative, the value of x that will make A an absolute maximum is in the closed interval 10, 2251.BecauseA is continuous on the closed interval 10,2251,we conclude from the extreme-valuetheorem that A has an absolute maximum value on this interval. From Eq,.(6),we havei A ( x ) : 3 0 0 r- t x ' Hence, A ' ( x ) : 3 0 0- t x BecauseA'(x) exists for all x, the critical numbers of A will be found bv settingA'(x):0, which gives x:'l,t2t The only critical number of A is ILz+, which is in the closed interval 10,2251.Thus, the absolute maximum value of A must occur at either 0, tl2+, or 225.Because,4(0) :0 and A(225):0, while A(1,1,2t):16,875, we conclude that the absolute maximum value of A on [0, 22Slis L6,BT| occurring when x: LL2i and y: L50 (obtained from Eq. (5) by substi- INVOLVINGAN ABSOLUTEEXTREMUMON A CLOSEDINTERVAL 193 4.6 APPLICATIONS ExAMPLE4: In the planning of a coffee shop it is estimated that if there are places for 40 to 80 people, the weekly profit will be $8 per place. However, if the seating capacity is above 80 places, the weekly profit on each place will be decreasedbY 4 cents times the number of Placesabove 80. What should be the seating capacity in order to Yield the greatestweekly profit? SoLUrroN: Let r: the number of placesin the seating capacity; P: the number of dollars in the total weekly profit. The value of P depends uPon r. and it is obtained by multiplying r by the number of dollars in the profit per place. When 40 < r < 80, $8 is the profit per place, and so P :8x. However, when r ) 80, the numthus giving ber of dollars in the profit per place is [8-0.04(r-80)], P - r[8 - 0.04(r - 80)] : 11.20x- 0-04x2.so we have if40<r<80 ^,.. : [8r Y\x) ttt.zox- o.oLxzifSo< x < z8o x - 0.04x2:0 The upper bound of 280 for x is obtainedby noting that 1.L.20 when x:280i and when t > 280,1L.20r- 0.04x2is negative. Even though x, by definition, is an integer, to have a continuous function we let x take on all real values in the intewall40,280l. Note that there is continuity at 80 because llt P(r): l11_t': 640 : jt1. P(x): l1T. Gt.20x 0.04x2\ 640 :540: from which it follows that the two-sided limit lim P(r) P(80). So P is continuous on the closed interval 140,2801andthe extreme-value theorem guaranteesan absolute'maximum value of P on this interval. :lL'20 When 40 < r ( 80,P'(r):8, and when 80 < x 1280, P'(x) - 0.08r.P'(80) doesnot exist sincePl(80) - 8 and Pi(80) : 4.80.Setting P'(x) : 0, we have 11.20-0.08r-0 x:'1.40 The critical numbers of P are then 80 and 1.40.Evaluating P(x) at the endpoints of the intenral [40,280]and at the critical numbers, we have P(40) : 920,P(ao;: 540,P(140):784, and P(280):0. The absolutemaximum value of.P, then, is 784 occurring when x:140The seating capacity should be L40places,which gives a total weekly profit of$784. rxavpr,n 5: Find the dimensions of the right-circular cYlinder of greatest volume which can be inscribed in a right-circular cone with a radius of 5 in. and a height of L2 in. solurroN: Let r: h: v: the number of inches in the radius of the cylinder; the number of inches in the height of the cylinder; the number of cubic inches in the volume of the cylinder. the rylinder inscribed in the cone, and illustrates 4.6.4 Figure through the axis of the cone. plane a section 4.6.5illustrates 194 TOPICSON LIMITS,CONTINUITY, AND THE DERIVATIVE lf r:0 and h: t2, we have a degeneratecylinder, which is the axis of the cone. rf r:5 and h: O,we alsohave a degeneratecylinder, which is a diameter of the base of the cone. We conclude that r is in the closed interval [0, 5] and h is in the closed interval 10,tzl. The following formula expressesv in terms of r and h: V: nrzh (7) To expressV in terms of a single variable we need another equation involving r and h. From Fig. 1.6.s, and using similar triangles,w€ have 12-h:p r5 Figure4.6,4 Thus, , 50-L2r ":--- Substituting from Eg. (8) into formula (7), we obtain V as a function of r and write I V(r): #n(Sf - d) N r N F{ rln 5 in. Figure4.6.5 ( 8) with r in [0, 5] (e) Because V is continuous on the closed interval [0, 5], it follows from the extreme-value theorem that V has an absolute maximum value on this interval. The values of r and h that give this absolute maximum value for V are the numbers we wish to find. V'(r):En(l}r-3.rz) To find the critical numbers of V,we set V,(r):0 and solve for r: r(1,0-3r):g from which we obtain and BecauseV'(r) exists for all values of r,rhe only critical numbers of Vare 0 and *, both of which are in the closedintervil [0,5J.The absolutemaximum value of v on [0, 5] must occur at either 0, +, or 5. From Eq. (9) we obtain v(0) -0,v(+):#r, and y(5):0. we thereforeconcludethat the absolute maximum value of.v is a8szr.,and this occurs when r: *. When r: #, we find from Eq. (8) that h: 4. . ^Thtt, lh" greatestvolume of an inscribed cylinder in the given cone is "8n"t in.3, which occurs when the radius is # in. and the heiiht is 4 in. Exercises 4.6 1. Find the area of the largest rectangle having a perimeter of 200 ft. THEOREM 1 9 5 4.7 ROLLE'STHEOREMAND THE MEAN-VALUE 2. Find the area of the largest isosceles triangle having a perimeter of 18 in. 3 . A manufacturer of tin boxe8 wishes to make use of pieces of tin rvith dimeNions 8 in. by 15 in. by cutting equal Equar€8 from the four comers and tuming up the sides. Find the length of the side of the squarc to be cut out if an oPen box having the largest possible volume is to be obtained from each Piece of tin. 4 . A r€ctangular plot of ground is to be endosed by a fmce and then divided down the middle by another fence' If the fmce down thi middle costs $1.pet running foot and the other fence costs $2.50 Per running foot, find the dimen8ions of the plot of latgest possible arca that can be mcloEed with $al8oworth of fence. 5 . Points /{ and B are opposite each other on shoresof a Etraightriver that is 3 mi wide. Point C i9 on the sameshoreas B but 6 mi down the river from 8. A telephone courpany wishes to lay a cable from A to C. If the cost per nile of the cable is 25% nore under the water than it is on land, what line of eable would be least exp€nsive for the comPany? 6. golve Exercisa 5 if point C i8 only 3 mi down the river from B. 7. Solve Example 2 of this section if the store i8 9 mi down the beach from Point B. : u1/o7*_7 8 . Example 2 and Exercises 5, 6, and.7 are special cases of the following more general problem. Let f(r) milimum value of absolute for the > > that in order r, 0. Show f to occur at a + o(i- t), where r is in [0, b] and tl < na b\/F=e' must be satisfied: (0, inequality following b) the nuurber in the open interval 9 . Find the dimensions of the right-circular cylinder of greatest lateral sur{ace area that can be inscribed in a sPherc with a radius of 6 in. 10. Find the dimeneions of the right-circular cylinder of greatest volume that can be inscribed in a sphere with a radius of 6 in. point on the 1 1 .Given the cirrJe haviirg the equation f + 4 : 9, 6tt4 ,", the shortest distance from the Point (4' 5) to a the circle' on (4, 5) to a from the longist distance Point circle, and (b) the Point more than 8OOitems are Produced each week. The Profit item if not each profit of on a $20 12. A manufacturer can make to have declreases2 C€nts per item over 800. How many items shpuld the manufacturer produce eadl week in order the grcatest Profit? 13. A school-sponsored trip will cost each student $15 if not morc than 150 student8 make the lriP; however, the cost Per student wiil be reduced Qcents for each student in excessof 150. How many students should make the trip in order fur the school to receive the Large8tSross income? L4. Solve Exerciee 13 iI the reduction Per student in excess of 150 i5 7 cent8. 15. A pdvate club charges annual memhrship dues of $lfi).per qrember less 50 cents for each member over 600 and plus S0'cents for eactr mJmber less than 600. How many members would give the club the most revenue from annual dues? 15. Suppoae a weight is to be held 10 ft below a horizontal line AB by a wire in the shape of a Y If the Points A and B art 8 ft lpa*, wttai is the shortest total length of wire that can be used? 12. -' A piece of wire 10 ft long ie cut into two pieces. One piece i8 bent into the sh4pe of a circle and th€ other into the shaPe i; shouli the wire be cui eo that (a)-the combined area of the two fiSures i8 as small as possible and ;;;;;;;(b) the combined area of the two figues is as large as Possible? lg, Solve Exercise 17 if one piece of wire is bent into the shape of an equilateral tdangle and the other Piece is bent into the ehape of a square. 4.7 ROLLE,S THEOREM AND TIIE MEAN-VALUE THEOREM . Letl be a function which is continuous on the closed interval [4, b], differ€ntiable on the open interval (a, b), and such thatf(a):0 andl(b):0. The French mathematician Michel Rolle (1652-1779) proved that if a 196 TOPICSON LIMITS,CONTINUIW,AND THE DERIVATIVE x function / satisfies these conditions, there is at least one number c between a and b for which f '(c) : 0. Let us see what this means geometrically. Figure 4.7.1,shows a sketch of the graph of a function / that satisfies the conditions in the preceding ParagtaPh. We intuitively see that there is at least one point on the curve between the points (a,0) and (b,0) where the tangent line is parallelto the r axis; that is, the slope of the tangent line is zero. This situation is illustrated in Fig. 4.7.7 at the point P. So the abscissa of P would be the c such thaLf ' (c): 0. The function, whose graph is sketched in Fig. 4.7,1, not only is differentiable on the open interval (a, b) but also is differentiable at the end- It is necessary, however, that the function be continuous at the endpoints of the interval to guarantee a horizontal tangent line at an interior point' Figure 4.7.3 shows a sketch of the graph of a function that is continuous on the interval [a, b) but discontinuous at b; the function is differentiable on the oPen interval (a, b), and the function values are zero at both a and b. However, there is no point at which the graph has a horizontal tangent line. We now state and prove Rolle,s theorem. Figure 4.7.3 4,7.'1,Theorem Rolle's Theorem Let f be a function such that (i) it is continuous on the closed interval la, bl; (ii) it is differentiable on the open interval (a, b); (iii) f ( a ): f ( b ): 0 . Then there is a number c in the open interval (a, b) such that f ' ( c ): o PRooF: We consider two cases. Case1-: f (x) :0 for all r in la, bl. Then f '(x): 0 for all r in (a, b); therefore, any number between a and b can be taken f.or c. Case2: /(r) is not zero for some value of r in the open interval (a,b). Because / is continuous on the closed interval fa, bl, we know by _ Theorem 4.5.9 that f has an absolute maximum value on la, bl and an a b s o l u te mi ni mum val ue on l a, bl . B y hypothesi s, (r):-f(b):0. Fur _ f thermore , f (x) is not zero f.or some x in (a, b). Hence, we can conclude that / will have either a positive absolute maximum value at some c, in (a, b) or a negative absolute minimum value at some c, in (a, b), or both. Thus, THEOREM 197 ANDTHEMEAN-VALUE THEOREM 4.7 ROLLE'S tot c: Cr, or c: cz as the case may be, we have an absolute extremum at an interior point of the interval la, bl.Therefore, the absolute extrernum ' f (c) is also a relative extremum , and because f (c) exists by hypothesis, it follows from Theorem 4.5.3 that f' (c) :0. This proves the theorem. I It should be noted that there may be more than one number in the open interval (a, b) for which the derivative of / is zero. This is illustrated geometricalty in Fig. 4.7.4, where there is a horizontal tangent line at the ' point where X: Crand also at the point where *: ,r, so that both f (ct) :0 '(tr) :0. a n df The converse of Rolle's theorem is not true. That is, we cannot conc l u d e th a t i f a functi on/i s such thatf' (c) -0, w i th a 1c ( b, then th e conditions (i), (ii), and (iii) must hold. (See Exercise 40.) Figure 4.7.4 EXAMPLE Given f(x):4x3-9x verify that conditions (i), (ii), and (iii) of the hyoothesis of Rolle's theorem are satisfied for each of the following intervals: [-8, 01, [0,3r],and [-8, B].Then find a suitable value for c in each of these intervals for which f ' ( c ): o ' sor,urroN: f'(x):12*-9; f'(x) existsfor all valuesof x, and so / is differentiable on (-oo, 1m) and thereforecontinuous on (-*, +*). Conditions (i) and (ii) of Rolle's theorem thus hold on any interval. To determine on which intervals condition (iii) holds, we find the values of x fot which f (x) - 0. Settingf (x) :0, we have Ax(f - *) :0 which gives us - _ rL- _3 2 -_ 'r- 3 2 and b:0, we see that Rolle's theorem holds on [-8, 0]. Taking a:-E Rolle's theorem holds on [0, srland I-9, gl. Similarly, To find the suitable values for c, we set /'(r) - 0 and get - 9:0 12x2 which gives us x - -+\E and *: *15 -tfi. In the Therefore, in the interv al l-ar,0l a suitable choice for c is gl, possiare two there the interval interval lO, we take c: +\/3.In l-9, *i bilities for c: either -+\/g or t{5. We apply Rolle's theorem to prove one of the most important theorems in calculus-that known as the mean-aalue theorem (or law of the mean). The mean-value theorem is used to prove many theorems of both differential and integral calculus. You should become thoroughly familiar with the content of this theorem. 4.7.2 Theorem Let f be a function such that Mean-Value Theorem (i) it is continuous on the closed interval la, bl; (ii) it is differentiable on the oPen interval (a,b). 198 TOPICS ON LIMITS, CONTINUIry, ANDTHEDERIVATIVE Then there is a number c in the open interval (a,b) such that -f(u) - f(') ' f'(c) 0-a r r n l_ f @ ) f'(c):tff Refer to Fig. 4.7.5. By taking the r axis along the line segment AB,we observethat the mean-value theorem is a generalization of Rolle's theorem, which is used in its proof. p,(b, f(b)) pRooF: An equation of the line through A and B in Fig. 4.7.s is y-f(a):fffk_a) or, equivalently, f ( b )- f ( a ) ( x a )+ f ( a ) Y:'-ff Figure4.7.5 Now if F(r) measuresthe verticaldistancebetweena point (x,/(r)) on lhe graph of the function / and the corresponding point on the secant line through A and B, then F ( r ) :f ( x ) - f % P k - a ) - f ( o ) (1) We show that this function F satisfies the three conditions of the hypothesis of Rolle's theorem. But (u|--lfu) F'(x): f '(x)-f THEOREM199 ANDTHEMEAN-VALUE THEOREM 4.7ROLLE'S Thus F'(c)- 1'G)- f(u)r=fJ'l Therefore, there is a number c rn (a, b) such that o:f,(c)_fW or, equivalently, r,,.\ f(b)-f(a) !f l\ c- ,t : - - - - b_a which was to be proved. 2: Given nxarurPr,n f(x):x3-5x2-3x verify that the hlryothesis of the mean-value theorem is satisfied for a: 1 and b : 3. Then find all numbers c in the oPen interval (1,3) such that f'(c\:fW solurroN: Because/ is a polynomial function, / is continuous and differentiable for all values of r. Therefore,the hypothesis of the mean-value theorem is satisfied for any a and b. - Lor - 3 f' (x) :3xz -z and (3) : -22 f f (t) : Hence, _-27 _(-7):_l.o 2 3-1 ' Settingf (c): -1.0, we obtain 3c2-10c-3:-L0 3c2-10c*7:0 (3c-7)(c-1):0 which gives us c:& and c:t Because1.is not in the open interval (1,3), the only possible value for c is 6. nxavrpr.n3: Given solurroN: A sketch of the graph of f is shown in Fig. 4.7.6. f (x) : xzrs draw a sketch of the graPh of f . Show that there is no number c in the open interval (-2,2) such that Figure4.7.6 TOPICS ON LIMITS, CONTINUlry, ANDTHEDERIVATIVE Ic ,\/c^)\:--f (22=) -l f ( - 2 ) Which condition of the hypothesis of the mean-value theorem fails to hold for f when a: -2 a n db : 2 ? f (2)- f(-2) 4u3_ 4u3 2- (-2) 4 -0 There is no number c for which 2l3crts: e. The function / is continuous on the closedinterval l-2,21; however, is not differentiable on the open interval (-2,2) because / f,bl does not exist. Therefore, condition (ii) of the hypothesis of the mean-value theorem fails to hold for f when a: -2 and,b : 2. Exercises 4.7 In Exercises1 through 4, verify that conditions (i), (ii), and (iii) of the hypothesis of Rolle's theorem are satisfied by the given function on the indicated interval' Then find a suitable value for i that satisfies the conclusion of Rolle,s theorem. r.f(x)-*-4x*3; : [1,3] 2. f(x) x3 2xz x * 2; lt, Zl 3. f(x) : x3- 2xz- x * 2; l-1, 2f a. f Q): .r3- 1.6x;l-4,01 5' For the function f defined bv f(r) : Ltt l 12* - x- 3, determine three sets of values for 4 and D so that conditions (i), (ii)' and (iii) of the hypotheiis of Rolle's *t r*L rina a suttabt".,,ar." to" ir, of tlr" th*" op.r. intervale (a, b) for which /, (c) : 0. "",itii"llir-",' " """t In 6 thtough 13, verify that the hypothesis of the mean-value theorem ,Exercises is satisfied for the given function on the indicated interval. Then find a ;uitable valu-e for c that sati"ri"" ti" *".ro.ior, of tle mean-value theorcm. 6.f(x):xs**-x;l-2,tl 7 . f (x ) : xz l 2x - l ; [0, 1] 9. f (x) = x2t3; [0, 1] 1,0.f (x) : ffifi' -]; 12. f(x):ff, 1,J. f(x):ffi,?r,4f 12,61 l-6, 8l 8.f(x)-tc-r+!_y[*,s] 1,r.f (x): \ETz; 14,61 In Ex€rcises 14 through 23, (a) draw a sketch olthe_graph of the given function on the indicated interval; (b) test the three 6:T {itt) .oj 1ty hypothesis of _nol-e,itheorem ind determine which conditions are satisfied and rdhich, fi:l'l: !i)1 iia, if any, are not satisfied; and (c) if the three conditions in part (b) are satisfied, detenni i ne a point at which there is a horizontal tangent line. M. f (x) : y3t4- 2xtt4. 10, 4l 15. f(x) : y4t! - 3xrt3. [0, 3] 16. f(x):'#;F+,sl if x=21. 1 7 .f ( x ) : [ x + l-3'71 -3 1,8. r(x): {t::X il,;:l.}; ?2, +l lz * irz<;J; THEOREM 4.7 ROLLE'STHEOREMAND THE MEAN.VALUE fx ' - 5x * 4 2 0f.( x ) : l ,-L L0 i rx * , ] , ifx:U t r ,n l 2 2 .f ( x ) : | - l r l ; [ - 1 , 1 ] 201 21,', f (x): {:.-t ;i::'*}, ?r, +) 23.f (x) : 19- 4x'l; l-*, *l 2a. lt f(x): (2r - 7)[Qx- 4) and if c:1 and b:2, show that there is no number c in the oPen interval (c, b) that satis- fiis the conclusion of the mean-value theorem. Which part of the hypothesis of the mean-value theorerr fails to hold? Draw a sketch of the graph of I on [1, 2]. The geometric interprctation of the mean-value theorem is that for a suitable c in the oPen interval- (4, b), the tangent (a,l(a)), and (b' f(b)).ln tine io the curve y =l(r) -30, at the point (c,l(c)) is parallel to the secantline through the points find a vllue of c satisfying the conclusion of the mean-value theorem, draw a sketch of the graph Exercises 25 through on the dosed interv al la , bl , and show the tangent line and secant line. 2 : 2 7 .f ( x \ : =;a:3.'!.,b:3.2 2 6 .f ( x ) : x 2 ) a : 2 , b : 4 79''f(x) x2;a: 3, b 5 x- 3 28.f(x) : x - t; a : "l'0,b : 26 Z !. f(x ) : x s- 9 x * l ; a: -3, b : 4 30. f (x) : ) fi; a: 3.01,b : 3.02 For each of the functions in Exercises 31 through 34, there is no number c in the open interval (a, ,) that satisfies the conclusion of the mean-value theorem. In each exercise, det€mrine which pafi of the hypothesis of the mean-value theorem fails to hold. Draw a skekh of the 6taph ot y : f(x) and the line through the points (a, f(t)) and (b, f(b)). ffJ f ( x ): a: \, b:6 {, - sy,, n i f @ ) : 3 ( r - 4 ) u e 'a : - 4 , b : 5 3 2f.( x ) : ' # a:1,b:2 3afQ): {?;:tr.lf; : '.},,: -!, b:5 -6*+4t-7. Prove by Rolle's theorcm that the equation 4ti-6f : t - 2xr * ?l - x, then l'(t):4f 35, '' If f(r) (0, 1). open interval + +t* t:O has at least one r€al root in the 36, prove by Rolle,g theorem that the equation f + 2r + c = O,where c is any constant, cannot have more than one rcal root 37. Use Rolle,s theorem to prove that the equation4rl + 3d * 3r 2 : 0 has exactly one root that lies in the interval (0, 1). (0, 1) that is a root of the equation. Tlten assumethat there is number in least one is at that ihere First show (HrNr: more than one root of the equation in (0, 1) and show that this leads to a contradiction ) 38. Suppose s : /(t) is an equation of motion of a particle moving in a sttaight line where f_satisfies the-hypothesis of the meiir-value iheorem. Show that the condusion of the mean-value theorcm assures us that there will be some instant durint any time intel\'al when the instantan€ou8 velocity will equal the average velocity during that time interval. 39. Supposethat the tunction f is continuous on [a, D] and /'(r) : 1 for all r in (4, ]). Prove thatf(x): x- a + f(a) lot all x in [a, bl. ,l(). The converse of Rolle's theorem is not true, Make up an example of a function for which the condusion of Rolle's theorem is true and for which (a) condition (i) i8 not satisfied but conditions (ii) and (iii) are Eatisfied; (b) condition (D is not satisfied but conditions (i) and (iii) are satisfied; (c) condition (iii) is not satisfied but conditions (i) and (ii) are satisfied. Draw a skekh of the graph showing the horizontal tangent Une for each case. 41., Use Rolle,s theorem to prove that if every polynomial of the fourth degree has at most Iour real_roots, then every Poly(HINr; Assume a polynomial of the fifth degree has six real roots nomial of the fifth degree has at most tiiJ""it "oote. contradiction.) leads to a that this and show 42. Use the method of Exercise 41 and mathematical induction to prove that a polynomial of the nth degree has at most z real rcots. 202 TOPICSON LIMITS,CONTINUIW,AND THE DERIVATIVE ReaiewExercises (Chapter4) In Exercises1 through4, evaluate the limit. r' ,. 8r3-Sf+3 :1 2r+?x-4 3. tr- ffi .r--€ 2x - 4 'A' , tr1. 1 4. 3x2-4x-l 6f +x'z+4 :im O/F+t-\/F+4) In Exercises 5 and 6, find the horizontal and vertical asymptotes o{ the graph of the function defined by the grven equatron, and draw a sketch of the graph. s.f(x):# In Exercises 7 thmugh 10, functions I and g are defined. In each exercise, define . g, and determine all values of x for which I is continuous. f"g 7. f(x) : t/-x - 3; g(r) : x * 2 9. f(x): sgn x; gQ) - x2- | : li 8.f (x):;ft, s@) 10. f (x): sgn x; gQ) : xz- ,c In Exetcises 11 thrcugh 16, find the absolute extrema of the given function on the given interval, if there are any, and find the values of t at which the absolute extt€ma ocour. Draw a sketch of the graph oi the function on the interval. rr. f (x): \6 + r; [-5,+*1 1 3f.( x ) : l e- * l ; [ - 2 , 3 ] 1 5. f ( x ) : x 4- 12;z+ 3 5 ; [0 ,5 ] 12.f(x): t$- rz; (-3, 3) 1,a.f Q) : x4 -'l.2xz * 36; l-2,61 1 6rf\.( x' ) : f : : 1 lxz1_4 =':1I;?2,21 :i f' .L :< 'x < 2 ) 17. Find two nonnegative numbers whose sum is 12 such that the sum of their squares is an absolute minimum. 18. A pieceof wire 80 in. long is bent to forrr a rectangle. Find the dimensions of the rectangle so that its area is as large as possible. 19. In order for a packageto be mailed by parcel post, the sum of the length and girth (the perimeter of a cro6ssection)must not be greater than 100 in. If a package is to be in the shape of a rectangulir box with a square cross section, find the dimensions of the package having the greatest possible volume that can be sent by parcel post. 20. If lis_ the function defined byJ(r) : l2x - 4l - 6, ttenl(-l) Rolle's theorem does not hold, : l(5) : 0. However, /, never has the value 0. Show why 21,.(a) lf. f is a polynomial tunction and f(a): f(b) : f'(a):/,(D) : 0, prove that there are at least two numbers in the open interval (s, b) that are loot8 of the equation f,'(x) : O- (b) Show that the function defined by the equation - a)' satisfiespart (a,. f(t) : $ 22. If I is a polynomial function, ehow that between any two consecutive roots of the equation l,(r) one root of the equation l(r) : 0. : 0 there is at most 23. SuPPose that / and I ar€ two func'tions that satisfy the hypothe8is of the mean-value theorem on [a, D]. Furthermore, - g<a) tot all r in the suppose that /'(.r)_lg'(r) for all r in the open interval (s, b). prcve that t'e) - g(r.):1k) dosed interval [c, D]. 24. Let I and 8 be two functions that ate differentiable at every number in the closed interval [a, b]. Suppose further that REVIEWEXERCISES 203 : gla) and f(b) : g,(b). Prove that there exists a number c in the open interval (4, b) such that l' (c) : 8' (c) ' f(a) (HrNr: Consider the function f - g') zs. lI f(x) : zttlP= 6i4 , find the horizontal and vertical asymptotes of the graph of l, and draw a sketch of the graph. 26. Let [3x + 6a l f x- 3< - <3 x < 3 f(x):l3ax-7b if lx - 12b i f 3 < x a sketch of the Find the values of the constants a and D that make the function f continuous on (-€, +'o), and draw gnph of l. of a straight 27. Two towns A and B are to tet their water suPPly from the EamePumPing station to be located on_the bank mi aPartand A B are 20 to and river that is 15 mi from town.A and 10 mi fromiown B. If the points on the river nearest of least amount that the located so be A and B are on the same side of the river, where should the pumping station piping is required? table on the entire 28. A manufacturer offers to deliver to a dealer 300 tables at $90 Per table and to reduce the Price Per possible transaction largest in the order by Z5 cents for each additional table over 3fl). Find the dolar total involved these circumstances' between the manufactuter and the dealer under at every number 29. Give an example of a function whose domain is the set of all real numbers and that (a) i6 continuous from the discontinuous left at I but the from (c) continuous 0 and 1; number except at every except 0; (b) continuous right at 1. Additional applications of the derivative TEST AND THE FIRST-DERIVATIVE FUNCTIONS AND DECREASING 5.1 INCREASING 205 function f 5.1 INCREASING AND Suppose that Fig. 5.1.1.representsa sketch of the graph of a we have sketch this drawing In )crl. interval lxr, DECREASING FUNCTIONS foi itt r in the closed rr]. on [rr, AND THE FIRST. assumed that f is continuous DERIVATIVE TEST : f(x) G(xr, yr) D ( x + , Yn ) F 1 x 6y, 6 ) B ( x z, y z ) A(xt, y) E(ru,Yt) C(xt, Yt) 5 .' 1 .1 F i g u re Figure 5.1.1 shows that as a point moves along the curve from A to B, the function values increase as the abscissa increases, and that as a point moves along the curve from B to C, the function values decrease as the abscissa increases. we say, then, that / is "increasing" on the closed interval lxr, xr] and that f is"d,ecreasing" on the closed interval lxr, x"l' Following are the precise definitions of a function increasing or decreasing on an interval. 5.1.L Definition A function / defined on an irrterval is said to be increasing on that interval if and only if f (*r) < f (xr) whenevet x, I )c, where r, and )cz ate any numbers in the interval' The function of Fig. 5.1.1 is increasing on the following closed intervals: lxr, xrf; lxr, xnl; lxr, xuf; lxu, xrl; lxr, xrls.1,2 Definition A function / defined on an interval is said to be decressingon that interval if and only if f (*r) > f (rr) whenevet xt I x, where rr and x2 are any numbers in the interval. The function of Fig. 5.1.1 is decreasing on the following closed intervals: lxr, xt); fxn, x"f. ADDITIONALAPPLICATIONS OF THE DERIVATIVE If a function / is either increasing on an interval or decreasing on an interval, then I is said to be monotonicon the interval. Before stating a theorem that gives a test for determining if a given function is monotonic on an interval, let us see what is happining geometrically. Referring to Fig. 5.1.1,we observe that when tfre itope oittre tangent line is positive the function is increasing, and when the slope of the tangent-line is negative the function is decreasing. Because /,i:) is the slope of the tangent line to the cuwe y: f(x), iJ increasing when / f'(x) - 0, and f is decreasing when f'(x) < 0. Also, becausel'(rj is the rate of changeof the function valuesl(r) with respectto r, whenl,(x) > 0, the function values are increasing as r increases; and when /,(r) < 0, the function values are decreasing as .r increases. We have the iollowing theorem. . 5.1.3 Theorem Let the function f be continuous on the closed interval la, bl and differentiable on the open interval (a, b): (i) { f'@) > 0 for all x n (a, b), then f is increasing on [a, b]; (ii) if l'(r) < 0 for all x in (a, b), then / is decreasingon [a, b]. pRooF oF (i): Let r, and r, be any two numbers in [a, b] such that r, < :r. Then / is continuous on [rr, rr] and differentiable on (xr, rr). From the mean-value theorem it follows that there is some number c in (xr, x2) such that 1' 1r'':f(x,) - f(x') Because:, a rr, tn"^ rz- rr )0. Atso, f,(c) > 0 by hypothesis. Theretore f(x) - f(rr) > 0, and so f(xr) > f(xr). We have shown, then, that f(xr) < f(h) whenever x, < xr, where x., and r, are any numbers in the interval- [4, b]. Therefore, by Definition 5.1.1, it follows ti.rat is increasing / on fa, bl. The proof of part (ii) is similar and is left as an exercise(see Exer_ cise 34). I An immediate application of Theorem 5.1.1,is in the Droof of what is known as the first-deriaatioetestfor relafiaeextremaof a funcuon. 5.1'4 Theorem Let the function f be continuous at all points of the opm interval (a, b) Fitst-Derioatioe Test Extrena containing the number c, and suppose that lor Relatioe l' exists at'aii points of (a, b) except possibly at c: (1) if f' (x) > 0 for all values of r in some open intewal having c as its right endpoint, and if f'(x) < O for all values of r in iome open interval having c as its left endpoint, then f has a relative maximum value at c; TEST ANDTHEFIRST-DERIVATIVE FUNCTIONS AND DECREASING 5 . 1 INCREASING 207 (ii) if f '(x) < 0 for all values of r in some oPen interval having c as its right endpoint, and if f ' (x) ) 0 for all values of x in some open interval having c as its left endpoint, then /"has a relative minimum value at c. F i g u r e5 . 1 . 2 F i g u r e5 . 1 . 3 pRooF or (i): Let (d, c) be the interval having c as its right endpoint for which /'(r) > 0 for all r in the interval. It follows from Theorem 5.1.3(i) that / is increasing on ld, c). Let (c, e) be the interval having c as '(x) < 0 for all x in the interval. By Theorem its left endpoint for which f 5.1.3(ii) / is decreasing on lc, ef. Because/is increasing onld, c], it follows from DefinitionS.l..L that if x, is tnld, c] and xr* c, then f(xt) < f (c) .Also, because / is decreasing on fc, ef , it follows from Definition1.'1..2 that if x, is in [c, e] and xr* c, then f(c) > f(rr). Therefore, from Definition 4.5.L, / has a relative maximum value at c. The proof of part (ii) is similar to the proof of part (i), and it is left I as an exercise (see Exercise 35). The first-derivative test for relative extrema essentially states that if '(x) changes algebraic sign from positive to is continuous at c and f / negative as r increases through the number c, then / has a relative maxi'(x) changes algebraic sign from negative to posimum value at c; and tf f tive as r increases through c, then / has a relative minimum value at c. Figures 5.1.2 and 5.1.3 illustrate parts (i) and (ii), respectively, of Theorem 5.1,.4when /'(c) exists. Figure 5.1,.4shows a sketch of the graph '(c) of a function / that has a relative maximum value at a numbet c,but f d o e s n o te x i s t; how ever,f' (x) ) 0w hen x l c,andf' (r) < 0w hen x ) c . In Fig. 5.1.5, we show a sketch of the graph of a function / for which c is (r) <0whenxlc; acriticalnumber,andf'(x) < 0when xlc,andf' c. at a relative extremum not have does / F i g u r e5 . 1 . 4 F i g u r e5 . 1 . 5 Further illustrations of Theorem 5.1,.4occur in Fig. 5.1.1. At x, and xo the function has a relative maximum value, and at 13 and rr the function ADDITIONALAPPLICATIONS OF THEDERIVATIVE has a relative minimum value; even though x6 is a critical number for the function, there is no relative extremum at 16. In summary, to determine the relative extrema of a function f: (1) Find f '(x). (2) Find the critical numbers of f, that is, the values of x for which ' (x): 0 or for w hi ch f ,(x) does not exi st. f (3) Apply the first-derivative test (Theorem 5.1.4). The following examples illustrate this procedure. EXAMPLE Given f(x):x3-5x2*9x*L find the relative extrema of f by applying the first-derivative test. Determine the values of x at which the relative extrema occur, as well as the intervals on which / is increasing and the intervals on which / is decreasing.Draw a sketch of the graph. solurroN, f' (r) - Bx2-I2x*9. '(x) exists for all values of r. Settingf , (r) : 0, we have f 3x2-'1,2x*9:0 3(r-3)(r-1):0 which gives us and Thus, the critical numbers of f are 1 and 3. To determine whether has a / relative extremum at either of these numbers, we apply the first-derivative test. The results are summarized in Table 5.1.1. We see from the table that 5 is a relative maximum value of occur/ ring at L, and 1 is a relative minimum value of occurring at 3. A sketch f of the graph is shown in Fig. 5.1.6. T a b l e5 . 1 . 1 f'(x) Exeupln 2: Given (-2-4 ifx<3 ,f(x) t \ - - /: l*^ if3<x Lg_r find the relative extrema of f by applying the first-derivative test. Conclusion x <'/.. x: "L / has a relative maximum value t<x<3 / is decreasing x--3 / has a relative minimum 3<x / is increasing / is increasing value s o r u r r o x : I f x 1 3 , f ' ( x ) : 2 x . I f x ) 3 , , ( x )- - 1 . B e c a u s e f f_(j):6 and f *(3):-L, we conclude that f' (3) does not exist. Therefore,3 is a critical number of /. Becausef'(x):0 when x:0, it follows that 0 is a critical number of f . Applying the first-derivative test, we summ arize the results in Table 5.L.2. A sketch of the graph is shown in Fig. S.I.Z. AND DECREASING FUNCTIONSAND THE FIRST-DERIVATIVE TEST 5 . 1INCREASING Determine the values of x at which the relative extrema occur, as well as the intervals on which / is increasing and the intervals on which / is decreasing. Draw a sketch of the graph. T a b l e5 . 1 . 2 (r) I x:0 I -+ 0<r<31 : x4ts+ 4'cu3 f (x) find the relative extrema of /, determine the values of x at which the relative extrema occur/ and determine the intervals on which / is increasing and the intervals on which / is decreasing. Draw a sketch of the graph. I 3<x | '(x) - r(0 x:3 f 5 Conclusion /isdecreasing 0 f has a relative minimum value + /isincreasing does not exist - / has a relative maximum value /isdecreasing + 1). SOLUTIoN f '(x)- Lxrts* $v-us: $v'2t3(x '(r) - 0 when x: -t, Becausef ' (x) doesnot exist when x: 0, andf -1 and 0. We apply the first-derivative test the critical numbers of f are and summarizethe results in Table 5.1.3.A sketchof the graph is shown in Fig. 5.1.8. T a b l e5 . L . 3 (r) x1-l x: -"L '(x) Conclusion / is decreasing -3 0 / has a relative minimum value -1<r<0 + / is increasing r:0 does not exist 0<x + / does not have a relative extremum at x: 0 / is increasing 210 ADDITIONALAPPLICATIONS OF THE DERIVATIVE Exercises 5.'L In Fxetcises 1 through 30, do each oI the following: (a) find the relative extrema of by applying the first-derivauve test / -on O) deteflnine the values of t at whidt the relative extrema occur; (c) determine tit" i"te*ar which I is increasing; (d) determine the intervals on which / is decreasing; (e) draw a sketch of the graph. r.f(x)-f -4x-r 2 . f ( x ) : 1 3- 9 x zl 1 5 r - 5 3. f (x) :2x3 - x2+ 3x -"1. a.fQ):x4+4x 5.f(x):x5- 6.f(x):2x+fi 1 5r3-20x-2 7.f(x): \/x-ffi1 y - ' ) 8 . 'f ( x ) : " x-r z 9 . f ( x ) : 2x t / 3 - x 1 1 .f ( x ) : ( 1 , - x ) z ( l * r ) ' 10.f(x): x\/fr L 2 .f ( x ) : ( r * 2 ) ' ( x- 1 ) ' 13. f ( x ) : 2 - 3 ( x - A ) z t s 1 , af .@ ) : l - 15. f(x):{T:: itr;=: if x<-2 1 7 ./f \(.x ) : [ 2 1 + s Lx"+l if_2<x ((*-)-)r-g ifr < 5 te. (x): 1 , 6 .tf \( x /) : (r- 1;'r' ( q-2x .l^" [3x-10 if x<3 if 3<x (r * 5)'z if x < -4 1 , 8 .f ( x ) : { , : - (x -f 1)2 if.-4 < x rLz ir5 < r --. 20../ f(x\ \--': [!tr=T*W U2_ xz if r < -3 if _3 < r [3x+5 ifx<-t 2 1 ,f .( x ) : ] x 2 * ' t i f - 1 < x < 2 (r-6 2 2 .f ( x ) : J - \ [ 4 = l t = ) r 120-2x ifx<6 if 6<r<10 if10<r 112-(x*S)2 2a.fQ):J5-r ifx<-3 if-3(x<-L f tiL *rr | 7- x if 2 < x ((r+9)r-s 23.f(x): t-t/tr= 1@ t(x-2)r-r i f x < -7 if -z < . x = 0 if0< x tvmo--T=ry if -1 < r 2 5 .f ( x ) : y s t 4 + 1 0 x t t 4 26. f(x):y;tl-l1x2ts 2 7 .f ( x ) : ( r * l ) 2 t 3 ( x - 2 ) r r c ' 29. f(x) - rrrs(x* 41-ue' 28. f(x): (4x- 6)trt(2x- a)ztt 30.f (x): ( r - a ) z r s* , 31. Find 4 and b so that the function defined byl(r):13+ af* b will have a relative extrernum at (2,3). 32. Find 4, ,, and c so that the funcdon defined by l(r) : af + bx+ c will have a relative maximum value of Zatlandthe graph oI y : f(x) will go through the point (2, -2). 33. Find c, b, c, and d so that th€ function defined byf(t) : af ibf f u I dwill have relative exhema at (1,2) and (2,3). 34. Prove Theorem 5.1.3(ii). prcve 35. Theorem 5.1.4(ii). 36. Given that the function f is continu""s 1o. values o{r,/(3) :2,f'(x) < 0 ifx < 3, and,f'(x) >0 if r > 3, draw a sketch of a Possible graph of I in each of_the. "l following casis, where the additional condition is satisfied: (a) is con/, t i n u o u sa t 3 ; ( b ) / ' ( x ) : - 1 . i l t < 3 , a n d f , ( x ) : 1 i f r > 3 ; ( c ) h 1 f , e ) : - 1 , I i - m /,(r) :1,andf,(a) + f''6j;lo * U. 32 /(1.- r)1 whe.re P gd 4 are positive integers greater than 1, prove each of the folowing: (a) if p is even, 9I* {1,1 : / has a relative minimum value at 0; (b) if 4 is evm, / has a relative rrinimum value at 1; (c) I has a Ghtive rraxirnum value at pl(p + q) whether p and 4 are odd or even. TESTFORRELATIVE 5,2THESECOND-DERIVATIVE EXTREMA 211 38. Prcvethat if I is incrcasingon [a, b] andif 8 is increasingon lf(a'1,l(b\1, thenrt g. f existson [a, b], g . I is increasing on fa, bl. 39. The functionI is inqeasingon the intervalI. Prcve:(a) if g(x): -f(x\, then 3 is decreasingon 4 (b) il h(x):1'lf(t) andf(x) > 0 on I, then ft is decreasingon I. 40. The functionf i8 differcntiableat eachnumberin the closedinterval [a, b]. Provethat if f'(a) ' f'(b) < 0, therei8 a number c in the opn interval (c, b) such that l'(c) : 0. 5.2 THE SECOND- In Sec. 5.1. we learned how to determine whether a function f has a relaDERIVATIVE TEST FOR tive maximum value or a relative minimum value at a critical number c ' RELATIVE EXTREMA by checking the algebraic sign of f ut numbers in intervals to the left and right of c. Another test for relative extrema is one that involves only the critical numb er c, and often it is the easier test to apply. It is called the second-deriuatiae test for relatiae extrema, and it is stated in the following theorem. and let f ' exist for all values of r in some open interval containing c. Then if f " (c) exists and '(c):0, 5.2.1Theorem Let c be a critical number of a function / at which f Second-DeriuatiueTest for Relatiae Extremct (i) if f "(c) ( 0, then / has a relative maximum value at c; (ii) if f "(c) ) 0, then / has a relative minimum value at c. pRooF or (i): By hypothesis, f " (c) exists and is negative; so we have f,,(c):}Tfff.o Therefore, by Theorem 4.3.2, there is an oPen interval I containing c such that . f'(x)-f'(c) <o x- c (1) for every x * c in the interval. Let I' be the open interval containing all values of r in I for which X 1 c; therefore, c is the right endpoint of the open interval /'. Let l" be the open interval containing all values of x in I for which x ) c; so c is the left endpoint of the open interval 1". Then if r is in I' , (x - c) ( 0, and it follows from inequality (1) that lf'(x) ',( x ) > f ' ( c ) . I f - x i s i n 1 " , ( x c ) > f ' ( c ) l > 0 o r , e q u i v a l e n t l yf '(x) '(c)l < 0 or, equivalently, 0, and it follows from (1) that lf f < f'(x) f'(r). '(r) > 0, and if But becausef '(c) : 0, we concludethat if x is in l' , f '(r) '(r) < 0. Therefore,f changesalgebraicsign from positive r is in 1" , f to negativeas r increasesthrough c, and so, by Theorem5.L.4,/hasa relative maximum value at c. The proof of part (ii) is similar and is left as an exercise (see ExerI cise L8). 212 ADDITIONAL APPLICATIONS OF THEDERIVATIVE EXAMPLE Given SOLUTION: f(x):xa*$x3-4x2 find the relative maxima and the relative minima of f by applying the second-derivativetest. f'(x) :4xs * 4xz 8x f"(x):12x2*8r-8 Settingf '(x):0, we have 4 x ( x+ 2 )( x - 1 ) which gives x:L ff:0 Thus, the critical numbers of f are-2,0, and L. We determine whether or not there is a relative extremum at any of these critical numbers by finding the sign of the second derivative there. The results are summ arized. in Table 5.2.1. T a b l e5 .2 .1 f'(x) -+ 0 0 0 / has a relative maximum value 0 / has a relative minimum value _5 x f" (*) f(x) 3 T Conclusion t' has a relative minimum value rf f " (c) : 0, as well as f '(r) :0, nothing can be concluded regarding a relative extremum of f at c. The following three illustrations lustify thii statement. .IL L U ST R A TToN7: If f(x)- x4, then f ' (x):4x3 and ,,(x):' !,2x2. Thus, f '(0), and ''(0) all have the value zero. By applying the first-derivative f (0), f f test we see that / has a relative minimum value at 0. A sketch of the graph of / is shown in Fig. 5.2.1. o Figure5.2.1 Figure5.2.2 o IL L U S TR A TION 2: If 8(x):-xn, then g' (r) :-4x3 and g,,(r): - 12x2. Hence, g(0) : g'(0) : g"(0) : 0. In this case,g has a rerative maximum value at 0, as can be seen by applying the first-derivative test. A sketch of the graph of g is shown in Fig. 5.2.2. o . rLLUsrRArroN 3: If h(x) : r3, then h'(*) : 3x2 and,h" (*) : 6x; so h(0) : h'(0) : h" (0): 0. The function h does not have a relative extremum at 0 because if r < 0, h(x) < h(0); and if x ) o, h(x) > h(0). A sketch of the graph of h rs shown in Fig. 5.2.9. o Figure5.2,3 In Illustrations 7,2, and 3 we have examples of three functions, each of which has zero for its second derivative at a number for which its first derivative is zero; yet one function has a relative minimum value at that 5.3 ADDITIONALPROBLEMSINVOLVINGABSOLUTEEXTREMA 213 number, another function has a relative maximum value at that number, and the third function has neither a relative maximum value nor a relative minimum value at that number. 5.2 Exercises In Exercises 1 through 14, find the relative o(trema of the given function by using the second-derivative test, if it can be applied. If the second-derivative test cannot be applied, us€ the first-derivative test. 2.s@):13-sx*6 5. f(x) : (x - 4)2 8 . f(* ) : x (x - 1)' 1.f(x):3x2-2x*l a. h@) :2xs - %c2+ 27 7. G(x) : ( r - 3) a 1 0.f(x): x\/R 13.F ( r ) : 6 x r t 3 - x z t l 11'.f(x) - lx6rtz+ x-rtz 3.f(x):-4x3+3f *'8r 6. G(x): (x * 2)3 9. h(x): xfr$ Ov2 1 2 .S U ) : ; * ; - 4)' M. G(x) - xszts(a 15.G i v e n f ( x ) : x 3 + s r x * 5 , p r o v e t h a t ( a ) i f r ) 0 , / h a s n o r e l a t i v e e x t r e m a ; ( b ) i f r / - 0 , / h a s b o t h a r e l a t i v e m a x i m u m value and a relative minimum value. 16. Givenf(r):x'-1.r+ k, where r > 0 and r + 1, prove that (a) if 0 < r< 1,/hasa relativemaximumvalue at 1; (b) if r > 1.,f ha8 a relative minimwn valu€ at l. 77. Given f(x) : f * rr-1, prove that regardless of the value of r, f has a relative qrinimum value and no rclative maximurn value. 18. Prove Theorem 5.2.1(ii). 5.3 ADDITIONAL The extreme-value thmrem (Iheorem 4.5.9) guarantees an absolute maximum value and an absolute minimum value for a function which is continuous on a dosed interval. We now consider some functions defned on intervals for which the extreme-value theorem does not apply and which may or may not have absolute extrema. Given solurroN: f is continuous on the interval [0, 6) becausethe only discontinuity of f is at 6, which is not in the interval. t, t ^"\_zx(ffi _ xz L2x* 27 _ (r 3) (r 9) PROBLEMS INVOLVING ABSOLUTE EXTREMA EXAMpLE l,: f(x\: t\re, yz_27 x_5 find the absolute extrema of f on the interval [0, 5) if there are any. I\x) @: 1r_5y::-ffiy ' '(r) : 0 when x: 3 or 9; so f (*) exists for all values of r in 10,6), and f the only critical number of / in the interval [0, 5) is 3. The first-derivative test is applied to determine if f has a relative extremum at 3. The results are summarizedin Table 5.3.1. Because/ has a relative maximum value at 3, and f is increasing on the interval [0, 3) and decreasingon the interval (3,6), we conclude that 214 ADDITIONALAPPLICATIONS OF THE DERIVATIVE on [0, 6) f has an absolute maximum value at 3, and it is f(3), which is 6. Noting that lim f (x) : -m, w€ conclude that there is no absolute minimum value of / on [0, 6). T a b l e5 .3 . 1 Conclusion EXAMPLE Given f ( x \ : r r(-X., * t vrz * 6)2 find the absolute extrema of f on (0, +m; if there are any. 0<r(3 / is increasing x:3 / has a relative maximum value 3<r<6 / is decreasing solurroN: f'(x) l l f is continuous for all values of x. -"*'* -r Q)? 4r'Q'* A --k2 e2*6)a * 6) * 4x2- _3x'- 6 1xriffi:ft7y f'g) existsfor all valuesof x. Settingf'(x):0, we obtain x:-+{2; so V2 is the only critical number of f in (0, +*;. The first-derivative test is applied at f2, and the results are summarizedin Table s.g.2. Because/ has a relative minimum value at f2 and because is de/ creasingon (0, {21 and increasinAon ({2,**), we conclude that f has an absolute minimum value at f1 on (0, +oo).The absolute minimum value is-#{2. There is no absolutemaximum value of f on (0, +*;. T a b l e5 . 3 . 2 f(x) f'(r) 0<r<\/2 x: \/2 \/2 < r ( *rc nxl.trapr-n3: Find the shortest distance from the point A(2, l) to a point on the parabola A : x', and find the point on the parabola that is closest to A. Conclusion / is decreasing _#\/i o + / has a relative minimum value / is increasing soLUTroN: Let z: the number of units in the distance from the point AQ, +) to a point P(x, U) on the parabola. Refer to Fig. 5.3.L. (1) Because P(x, y) is on the parabola, its coordinates satisfy the equation A: x2. Substituting this value of y into Eq. (1), we obtain z as a function of x and write z(x):@ 5.3 ADDITIONALPROBLEMSINVOLVINGABSOLUTEEXTREMA z(x)- \E=ET+ 215 (2) BecauseP can be any point on the parabola, x is any real number. We wish to find the absoluteminimum value of z on (-*, +oo). -tt2:Zf' z' (x) : t(4x3- 4) (xn- 4x+i +) 4t t, y) A(r'+) x Figure5.3.1 - U t'" + - + tl \F@ z' (x) exists for all values of r because the denominator is z(x), which is never zero. Consequently, the critical numbers are obtained by setting z'(x): 0. The only real solution is 1; thus, L is the only critical number of z. We apply the first-derivative test and summarize the results in Table 5.3.3. T a b l e5 .3 .3 z(x) z'(x) -@<x1l x:L Ilxafo Conclusion z is decreasing +f5 0 z has a relative minimum value + z is increasing Because z has a relative minimum value at l, and becausez is decreasing on (-m, L) and increasing_on (L, +*;, w€ concludethat z has an absolute minimum value of tt/S at L. So the point on the parabola closest to A is (1, t). We can show that the point AQ, +) lies on the normal line at (L, t) of the graph of the parabola. Becausethe slope of the tangent line at any point (x, y) of the parabola is 2x, the slope of the tangent line at (1, 1) is 2. Therefore, the slope of the normal line at (1, 1) is -*, which is the same as the slope of the line through A(2, *) and (1, 1). In the next two examples, where the second-derivative test is used to determine the relative extrema, we use the following theorem to determine the absolute extrema. S.3.1 Theorem Let the function f b" continuous on the interval I containing the number c. It f(c) is a relative extremum of f on / and c is the only number in I for which / has a relative extremum, then /(c) is an absolute extremurnof / on I. Furthermore, (i) if /(c) is a relative maximum value of / on I, then /(c) is an absolute maximum value of f on I; 216 ADDITIONALAPPLICATIONS OF THE DERIVATIVE (ii) if /(c) is a relative minimum value of / on I, then /(c) is an absolute minimum value of f on I. PRooF: We prove part (i). The proof of part (ii) is similar. Because/(c) is a relative maximum value of on I, then by Definition / 4.5.1. there is an open interval I where I c I, and where / contains c, such that f(c) = f(x) for all x e I Because c is the only number in I for which f has a relative maximum value, it follows that f(c)>f(k) ifkelandk*c (3) In order to show that /(c) is an absolute maximum value of on l, f we show that if d is any number other than c in l, then (c) > f f(d). we assumethat f(c) -f(d) e) and show that this assumption leads to a contradiction. Because d + c, then either c 1 d or d < c. we consider the case that c < d (the proof is similarifd<c). Because/ is continuous on 1, then / is continuous on the closed interval lc, d]. Therefore, by Theorem 4.5.9, / has an absolute minimum value on fc, dl. Assume this absolute minimum value occurs at e where c < e < d. From inequality (3) it follows that e # c, and,from inequalities (3) and (4) it follows that e * d. Consequently, c 1 e 1 d, and hence / has a relative minimum value at e. But this statement contradicts the hypothesis that c is the only number in I for which has a relative extre/ mum. Thus, our assumption that /(c) = f (d) is false. Therefore,f (c) > (d) f if d e I and d * c, and consequently /(c) is an absolute maximum value offonl. I nxavpr.n 4: A closed box with a square base is to have a volume of 2000 cubic inches. The material for the top and bottom of the box is to cost $3 per square inch, and the material for the sides is to cost $L.50 per square inch. If the cost of the material is to be the least, find the dimensions of the box. solurroN: Let r: the number of inches in the length of a side of the square base; A : the number of inches in the depth of the box; C: the number of dollars in the cost of the material. The total number of square inches in the combined area of the top and bottom is 2x2, and for the sides it is 4xy; so we have C: 3 (2xr) + E(4xV) (5) Because the volume of the box is the product of the area of the base and the depth, we have x'Y:2ooo 5.3ADDITIONAL PROBLEMS INVOLVING ABSOLUTE EXTREMA 217 solving Eq. (6) for y in terms of r and substituting into Eq. (5) we get c: 6xz+ L2'ooo xQ) x is in the interval (0, **), and C is a function of r, which is continuous on (0, +oo).From Eq. (Z) we obtain DrC: l2x - 12,000 (8) xc2 DrC does not exist when x:0, but 0 is not in (0, +*). Hence, the only critical numbers will be those obtained by setting D,C: 0. The only real solution is L0; thus, 10 is the only critical number. To determine if x: 10 makes C a relative minimum we apply the second-derivative test. From Eq. (8) it follows that Dr'C: !2 +24'q00 x3 The results of the second-derivative test are summarized in Table 5.9.4. Table5.3.4 r: L0 D rC D *,C Conclusion 0 + C has a relative minimum value From Eq. (7) we see that C is a continuous function of x on (0, *m). Because the one and only relative extremum of C on (0, +m; is at x: 10, it follows from Theorem 5.3.1(ii) that this relative minimum value of C is the absolute minimum value of C. Hence, we conclude that the total cost of the material will be the least when the side of the square base is 10 in. and the depth is 20 in. In the preceding examples and in the exercisesof Sec.4.6, the variable for which we wished to find an absolute extremum was expressed as a function of only one variable. Sometimes this procedure is either too difficult or too laborious, or occasionally even impossible. Often the given information enables us to obtain two equations involving three variables. Instead of eliminating one of the variables, it may be more advantageous to differentiate implicitly. The following example illustrates this method. nxarvrlr 5: If a closed tin can of specific volume is to be in the form of a right-circular cylinder, find the ratio of the height to the base radius if the least amount of material is to be used in its manufacture. 8:-, ' . We wish to find a relationship between the height and the solurroN: base radius of the right-circular cylinder in order for the total surface area to be an absolute minimum for a fixed volume. Therefore, we consider the volume of the cylinder a constant. Let v: the number of cubic units in the volume of a cylinder (a constant). We now define the variables: 218 OF THE DERIVATIVE ADDITIONALAPPLICATIONS r - the number of units in the base radius of the cylinder; 0 ( r < +e; h : the number of units in the height of the cylinder; 0 < h < +oc; S - the number of square units in the total surface area of the cylinder. We have the following equations: S:2nr2 (e) * 2nrh V : rrzh (10) BecauseV is a constant,we could solveEq. (10)for either r or h in terms of the other and substitute into Eq. (9), which will give us S as a function of one variable. The alternative method is to consider S as a function of two variables r and h; however, r and h arenot independent of eachother. That is, if we choose r as the independent variable, then S depends on r; also,h depends on r. Differentiating S and V with respect to r and bearing in mind that h is a function of. r, we have D,S: (1 1 ) 4nr * 2nh * 2nr D,h and DrV :2nrh (12) * nrz Drh Because V is a constant, DrV :0; therefore, from Eq. (12) we have Z n r h* n r 2 D r h : O with r # 0, and we can divide by r and solve f.or Drh, thus obtaining D,h: -; )h (13) Substituting from Eq. (13)into Eq. (11),we obtain tD,s:zolzr+h+r(+)) DrS: 2n(2r - h) To find when S has a relative minimum obtain 2r * h: 0, which gives us (14) value, we set D r S : 0 and r: th To determine if this relationship between r and h makes S a relative minimum, w€ apply the second-derivative test. Then from Eq. (14) we find that D,'S:2n(2-D,h) (1s) 5.3 ADDITIONALPROBLEMSINVOLVINGABSOLUTEEXTREMA 2 1 9 Substituting from Eq. (13)into (15),we get D,2 5 : 2 r olrL e)l:2n(r.+) The results ofrf the second-derivative test are summarized in Table 5.3.5. Table5.3.5 r:*h I I DrS Dr'S 0 + Conclusion Shasarelativeminimumvalue From Eqs. (9) and (10) we see that S is a continuous function of r on (0, +"o;. Because the one and only relative extremum of s on (0, *co) is at r: +h, we conclude from Theorem 5.3.1(ii) that S has an absolute minimum value at r: ih. Therefore, the total surface area of the tin can will be least for a specific volume when the ratio of the height to the base radius is 2. Exercises5.3 In Exercises 1 through8, find the absolute extrema of the given function on the given interval if there are any. L. f (x) : x2)(-3,2l 2 . g (x ): 1 3* 2 x2- 4x * L; ( - 3 , 2 ) 3.F(r):#, a.f@):fi3,?4,-rl 5. S(r) : 4x2- 2x * 1; (-* , + @ ) 6. G(x) : (x - S )zrs'(--, * * ) 7. f (x) : B.f(x):ffi,F*,4) k"; +-; 4yrr,[0, V4,41 9. A rectangular field, having an area of 27C0ydz, is to be enclosed by a fence, and an additional fence is to be used to divide the field down the middle. If the cost of the fence down the middle is 92 per rururing yard, and the fence alont the sides costs $3 per running yard, find the dimensions of the field so that the cost of the fencing will be the least. 10. A rectangular open tank i8 to have a squarc base, and its volume is to be 125 yd8. The cost per square yard for the bottom is $8 and for the sides is $4. Find the dimensione of the tank in order for the cost oflhe material to be the leaet. 1 1 . A box manufacturer is to produce a closed box of specific volume whose base is a rectangle having a length that is three times its width. Find the most economical dimensions. t2. Solve Exercise 11 if the box is to have an open top. 13.A funnel of specific volume is to be in the shape of a right-circular cone. Find fhe ratio of the height to the base radius if the least amount of material is to be used in its manufacture. 't4. A Norman window consists of a rectangle surmounted by a semicircle. Find the shape of such a window admit the most light for a given perimeter. that will 22O ADDITIONAL APPLICATIONS OF THE DERIVATIVE 15, Solve Exercise 14 if the window is such that the Bemicircle transmits only half as much light per square unit of area as the rectantle. 16, The strength of a rectantulat beam is proportional to the breadth and the square of its depth. Find the dimensions of the strong*t beam that can be cut from a log in the shape of a right-circular cylinder of radius c in. 17. The stifbnees of a Ectangular beam fu proportional to the breadth and the cube of its depth. I4rhat is the shape ofthe stiffest beam that can be cut from a log in the shape of a dght-circular cylinder? L8. A page of print is to contain 24 in.z of printed area, a margin of 1+in. at the top and bottom, and a rrargin of 1 in. at the eides. What are the dimeruions oJ the snallest pate that wou.ld fi.ll tlese requirements? 19. A one-story building having a rectangular floor space of 13,200 fe is to be constructed where a 22-ft easerrent is re. quircd in the front and back and a 15-ft easement iE required on eadr side, Find the dimensions of the lot havint the least area on which this building can be located. 20. Find the point on the curve yr - t':1 that is do8est to the point (2,0). 21. A direct current generator has an electromotive force of E volts and an intemal rcsistance of / ohms, E and r are mnstants. If R ohms is the external resistance, the total resistance is (r * R) ohms and iI P wafts is the power, then -^ ER (r+x1r What odemal resi8tance will consume the most powe!? 22. A ritht-cfuculal cone is to be inscribed in a sphere of givm radius. Find the ratio of the altitude to the bage radius of the cone of largest possible volume. 23. A right-circular cone is to be circumacribed about a sphere of given radius. Find the ratio of the altitude to the base radius of the cone of least possible volume. 24. Prove by the method of this section that the shortest distance from the point P1(r1, yr) to the line l, having the equation Ar+_By+ C: O,is lA4*Byl*' CV\/TTF. (nrnr: If sis the number oi uniis trom p. to a point p[.r, y) on t, then s will be an absolute ininimum when d is an absolute minimum.) 5'4 CONCAVITY AND POINTS OF INFLECTION Figure 5.4.1,shows a sketch of the graph of a function / whose first and second derivatives exist on the closJd interval lxr, xrl. Becauseboth and f ' ' f are differentiable there, / and f are continuous on lxr, xzl. If we consider a point P moving along the graph of Fig. s.4.l from A to G, then the position of P varies as we increasex from r, to xr. As P B(x Uz) G(xy vr) C(x", y t) A ( xr , y r ) F ( x u ,V 6 ) E(xr,y t) Figure5.4.1 D(xn,vn) 5.4 CONCAVITY AND POINTSOF INFLECTION 22'I moves along the graph from A to B, the slope of the tangent line to the graph is positive and is decreasing; that is, the tangent line is turning clockwise, and the graph lies below the tangent line. When the point F is at B, the slope of the tangent line is zero and is still decreasing. As P moves along the graph from B to C, the slope of the tangent line is negative and is still decreasing; the tangent line is still tuming clockwise, and the graph is below its tangent line. We say that the graph is "concave downward" from A to c. As P moves along the graph from c to D, the slope of the tangent line is negative and is increasing; thatis, the tangent line is turning counterclockwise, and the graph is above its tangent line. At D, the slope of the tangent line is zero and is still increasing. From D to E, the slope of the tangent line is positive and increasing; the tangent line is still turning counterclockwise, and the graph is above its tangent line. we say that the graph is "concave upward" from c to E. At the point c, the graph changes from concave downward to concave upward. Point C is called a "point of inflection." we have the following definitions. 5.4J1'Definition The graph of a function / is said to be concaveupward at the point (c,f (c)) if f '(c) exists and if there is an open interval 1 containing c such that for all values of x * c in I the point (x,f (x)) on the graph is above the tangent line to the graph at (c, f (c)). Definition The graph of a function / is said to be concaae downward at the point (c,f (c)) if f'(c) exists and if there is an open interval lcontaining c such that for all values of x * c in l the point (x,f (x)) on the graph is below the tangent line to the graph at (c, f (t)). l) o ILLUSTRATToN 1: Figure 5.4.2 shows a sketch of a portion of the graph of a function / that is concave upward at the point (c, f(c)), and Fig. s.4.9 shows a sketch of a portion of the graph of a function / that is concave downward at the point (c,f(c)). o Figure5.4.2 The graph of the function f of Fig. 5.4.1,is concave downward at all points (x, f(x)) for which x is in either of the following open intervals: (xt, x") or (xr, x6). Similarly, the graph of the function / in Fig. 5.4.i. is concave upward at all points (x, f(x)) for which r is in either (xs, xr) or (xu, xr). The following theorem gives a test for concavity. Figure5.4.3 5.4.3 Theorem Let f be a function which is differentiable taining c. Then on some open interval con- (i ) i f f" (c ) ) 0, the graph of /i s concaveupw ard at (c,f(t)); (ii) if f " (c) ( 0, the graph of / is concave downward at (c, f (c)). OF THE DERIVATIVE ADDITIONALAPPLICATIONS PRooF or (i): f,,(c):*Tfry Becausef"(c)>0, ,. f'(x)- f'(cl rmJ)0 tr-c x- c Then, by Theorem 4.3.1, there is an open interval I containing c such that f'(*)*f'(c) >o (r) x- c foreveryx*cinl. Now consider the tangent line to the graph of f at the point (c, f (c)). An equation of this tangent line is y : f ( c ) + f ' ( c )( x- c ) (c' f(c)) iQ(',f(')) Figure5.4.4 (2) Let x be a number in the interval I such that r * c, and let Q be the point on the graph of / whose abscissa is r. Through Q draw a line parallel to the y axis, and let T be the point of intersection of this line with the tangent line (see Fig. 5.4.4). To prove that the graph of / is concave upward at (c, f (c)), we must show that the point Q is above the point T or, equivalently, that the directed distance TQ > 0 for all values of x * c in 1. TQ equals the ordinate of Q minus the ordinate of T. The ordinate of Q is f (x), and the ordinate of T is obtained from Eq. (2); so we have r Q : f k ) - l f ( ' ) + f ' ( r ) ( x* c ) l r Q : l f ( x )- f ( c ) J- f ' ( c )( r - c ) (3) From the mean-value theorem there exists some number d between x and c such that f , ( d )- f ( x ) f ( c ) ' x- c That is, f(x)-f(c):f'(d)(x-c) for some d between x and c Substituting from Eq. (4) into (3), we have TQ- f'(d)(x- c)- f'(c)(x- c) TQ: k-c)lf'(d)-f'(c)l Because d is between x and c, d is in the interval /, and so by taking x: (s) d 5.4 CONCAVIryAND POINTSOF INFLECTION in inequality (1), we obtain f ' ( d ) _ -f ' ( r ) > d- c o (6) To prove that TQ ) 0, we show that both of the factors on the right side of Eq. (5) have the same sign. rf (x - c) > 0, then x ) c. And because d is between r and c, then d > c; therefore, from inequality (6), lf'(d)-f_'(c)l >0. If (r-c) <0, then x1c, andso d<ritherefore, (d) - f' (r)l < 0. We concludethat(r- c) and lf, (d) {rom (5),lf' f, (c)l have the same sign; therefore, re, is a positive number. Thus, the gruph of f is concave upward at (c, f(c)), which is what we wished to Figure5.4.5 Prove. The proof of part (ii) is similar and is left as an exercise (see Exerctse 27). t { T -t{ i Figure5.4.6 5.4.4 Definition The converse of Theorem 5.4.i is not true. For example, if / is the function defined by f(x) :1s4, the graph of / is concave upward at the point (0, 0) but f" (0) :0 (see Fig. s.2.1). Accordingly, a sufficient condition for the graph of a function / to be concave upward at the point (c, f (c)) is that f " (c) ) 0, but this is not a necessary condition. similarly, a sufficient-but not a necessary-condition that the graph of a function ,,(c) < 0. / be concave downward at the point (c, f(c)) is that f If there is a point on the graph of a function at which the sense of concavity changes, then the graph crosses its tangent line at this point, as shown in Figs. 5.4.5, 5.4.6, and 5.4.7. Such a point is called a "point of inflection." The point (c, f (c)) is apoint of inflection of the graph of the function/if the graph has a tangent line there, and if there exists an open interval / containing c such that if r is in /, then either ( i ) f " ( r ) < 0 i f x ( c a n d .f " ( x ) > 0 i f x ) c ; o r (ii) f"(r) > 0 if x ( c and f" (x) < 0 if x ) c o rLLUsrRArroN 2: Figure 5.4.5 illustrates a point of inflection where condition (i) of Definition 5.4.4 holds; in this case,the graph is concave downward at points immediately to the left of the point of inflection, and the graph is concave upward at points immediately to the right of the point of inflection. Condition (ii) is illustrated in Fig. 5.4.6, where the sense of concavity changes from upward to downward at the point of inflection. Figure 5.4.7 is another illustration of condition (i), where the sense of concavity changes from downward to upward at the point of inflection. Note that in Fig. 5.4.7 there is a horizontal tangent line at the point of inflection. . OF THEDERIVATIVE APPLICATIONS ADDITIONAL For the graph in Fig. 5.4.1, there are points of inflection at C, E, and F. Definition 5.4.4 indicates nothing about the value of the second derivative of f at a point of inflection. The following theorem states that if the second derivative exists al a point of inflection, it must be zero there. 5.4.5 Theorem If the function / is differentiable on some open interval containing c, and fi (c, f(c)) is a point of inflection of the graph of f, then if f" (c) exists, /" (c)- o' pRooF: Let g be the function such that g(r) - 1' k); then g'(x) : f " (x). Because(c, f (c)) is a point of inflection of the graph of /, then f " (x) changessign at c and so 8'(r) changessign at c. Therefore,by the firstderivative,test(Theorem5.1.4),g has a relative extremum at c, and c is a critical number of g. Becauseg' G) : f " (c) and sinceby hypothesisf " (c) exists, it follows that g'k) exists.Therefore,by Theorem4.5.3 g'G) :0 and f" (c) : 0, which is what we wanted to prove. I The converse of Theorem 5.4.5 is not true. That is, if the second derivative of a function is zero at a number c, it is not necessarily true that the graph of the function has a point of inflection at the point where x: c. This fact is shown in the following illustration. o rLLUsrRArroN 3: Consider the function / defined by f (x) : x4.f '(x) : 4x3 a n d f" (x):L2x2. Further, f " (0):0; but becausef" (x) > 0 i f x ( 0 and f"(x) > 0 if x) 0, the graph is concave upward at points on the graph immediately to the left of (0, 0) and at points immediately to the right of (0, 0). Consequently, (0, 0) is not a point of inflection. In Illustration 1 of Section 5.2 we showed that this function / has a relative minimum value at zero. Furthermore, the graph is concave upward at the point (0, 0) (see F i g . 5 .2 .1). o The graph of a function may have a point of inflection at a point, and the second derivative may fail to exist there, as shown in the next illustration. If / is the function defined by f (x) : tr1l3,then * $y-l tl f' (x ) : !x6-zt? and f" (x) : . rLLUSrRerrox 4: r ,rr6,r* rc /f\ -/l Figure5.4.8 I 4-- (r) < 0. f " ( 0 ) d o e s n o t e x i s t ;b u t i f x < 0 , f " ( x ) > 0 , a n d t f x ) 0 , f " Hence, / has a point of inflection at (0, 0). A sketch of the graph of this function is shown in Fig. 5.4.8. Note that for this function /'(0) also fails to exist. The tangentline to the graph at(0,0) is the y axis. . In drawing a sketch of a graph having points of inflection, it is helpful to draw a segment of the tangent line at a point of inflection. Such a tangent line is called an inflectional tangent. 5.4 CONCAVITYAND POINTSOF INFLECTION EXAMPLE 1: For the function in Example1 of Sec.5.1, find the points of inflection of the graph of the function, and determine where the graph is concaveupward and where it is concave downward. SOLUTION: f(x):x3-6x2*9x*1' f ' ( x ) : 3 x 2 r 2 x* 9 f " ( x ): 5 x - L 2 f " (x) existsfor all valuesof x; so the only possiblepoint of inflection is where f " (x): 0, which occurs at x:2. To determine whether there is a point of inflection at x:2, we must check to see if f " (x) changes sign; at the same time, we determine the concavity of the graph for the respective intervals. The results are summanzed in Table 5.4.1. Table5.4,1 f (x) -oo< f '(x) Conclusion graph is concave downward x12 x:2 -3 3 2{x(foo Figure5.4.9 f " (x) 0 graph has a point of inflection + graph is concave upward In Example L of Sec.5.1 we showed that / has a relative maximum value at L and a relative minimum value at 3. A sketch of the graph showing a segmentof the inflectional tangent is shown in Fig. 5.4.9. Exruprn 2: If f (x) : (1 - 2x)t, find the points of inflection of the graph of f and determine where the graph is concave upward and where it is concave downward. Draw a sketch of the graph of f. v SOLUTION: f '(x) - -6(1. * 2x)2 - 2x) f " (x) :24(l Becausef " (x) exists for all values of x, the only possible point of inflection is where f " (x) : 0, that is, at x: *.By using the results summarized in Table 5.4.2,we see that /"(x) changessign from rt+" to "-" at x: L, and so the graph has a point of inflection there. Note also that becausef '(+): 0, the graph has a horizontal tangent line at the point of inflection. A sketchof the graph is shown in Fig. 5.4.10. Table 5.4.2 f (x) '(x) f " (x) Conclusion -m< x<i T graph is concave upward x:* 0 graph has a point of inflection i<x(foo F i g u r e5 . 4 . 1 0 f graph is concave downward ADDITIONALAPPLICATIONS OF THE DERIVATIVE 5.4 Exercises ln Exercises I through 10, determine whele the Faph of the given function is concave upward, where it is concave downward, and find the point8 of inflection if there are any. L.f(x\:x3+9x 3. F(r) : x4- 8xB+ 24xz 2.sG):.r3+3x2-3x-3 c.g(rr:7q 6 .G k ) : (x' , ^+? *41tn a. f Q) :"l.6xa * 32xs* 24x2- 5x - 2o F/rI 8. F(r) : (2x - 6)srz* t (*z if.x<l lo' f(x): U , _ 4 x 2* T x _ 3 i f r > l 7.f(x):(x-))rrs (*2 :ff<0 9 . f/ (\ x ) : 1 ^ :' l-* " ifx>0 17. Il f(x) : sp a 65, detennine 4 and D so that the graph o{ / will have a point of inflection at (1, 2). 12. Itf(x):sf abf * cr, determine a, b, and c ao that the gaph of I will have a point of inflection at (1,2) and so that the slope of the inflectional tangent there w.ill be -2. 13. Ilf(r): af tb*l cr* d, detemdne a, b, c, a\d. d 80 that I will have a rel,ativeextremum at (0, g) and so that the graph ofl will have a point of inflection at (1, -1). 7a. fif(x): al +bf +cf + dr+ e, deternine the values of a, b, c, d, and.e so the graph of will have a point of inflec/ tion at (1, -1), have the origin on it, and be symmetric with respect to the y axis. In Exercises 15 through 24, draw a skekh of a portion of the graph ot a function through the point where r: c if the I given conditions are satisfied. If the conditions are incomplete or inconsistent, explain. It is assumed that is continuous f on some open interval containing c. 1 5 .f ' ( x ) > 0 r t x < c ; f ' ( x ) < 0 ifr > c;f"(x) < 0itx < c;f"(r)< 0if x> c. 1 6 .f ' ( x ) > 0 i t x < c ; f ' ( x ) > 0 i f r > c ; f " ( x ) > 0 i t x < c ; f , , ( t ) < 0 i f r > c . 7 7 .f " ( c ) : o rf t ( c ) : o ; 1 " ( x ) > 0 i f r < c ; 1 " ( x ) > o i I x > c . 1 8 .f ' ( c ) : 0 ; f ' ( x ) > 0 i f r < c;f"(x) > 0ifr>c. 79. f' (c) 0; f' (x) < 0 if r < c; f" (x) > 0 iI x > c. m . f " ( c ): 0 ;f ' ( c ): * ; 1 " ( r ) > 0 i f r < c ; f , , ( x ) < o i t r > c . 21.l'(c) doesnot exist;f" (r\ >0itr < c;f,(x\ > 0 ifr > c. 22. l'(c) doesnot exisf /" (c) doesnot eist f" (x) < 0 ifr < c; f,'(r) > 0 ifr > c. 23. lim f'(x) - +co,lim l'(t) : o; f" (x\ > 0 ifr < c;f"(x) < 0 ifr > c. 24. lim f'(x):*@j limf' (r): -*;f" (x) > 0 if r < c;f" (x) > 0 if x > c. E. Draw a Bketchof the graph of a functionf for which l(x), f'(r), andf,,(r) exist and are positive for all r. 26. II f(t\ :3f + xlxl, prove that l" (0) does not exist but the graph of f is concaveupward everyrvhere. 27. Prove Theorem 5.4.3(ii). 28. SuPPos€ that I is a function for which l" (x) exists for all values of r in some open interval I and that at a number c in I, f" (r) : O and'l"' (c) €xists and is not zero. Prove that the point (c, l(c) ) is a point of inflection oI the graph of /. (HrNr; The proof is similar to the prcof of the second-derivative test (Theorem 5.2.1.).) TO DRAWING A SKETCHOF THE GRAPHOF A FUNCTION 5.5 APPLICATIONS 5.5 APPLICATIONS TO DRAWING A SKETCH OF THE GRAPH OF A FUNCTION nxanaptnL: Given f(x):x3-3**3 find: the relative extrema of f the points of inflection of the graph of f the intervals on which f is increasing;the intervals on which / is decreasing;where the graph is concaveupward; where the graph is concavedownward; and the slope of any inflectional tangent. Draw a sketch of the graph. We now apply the discussionsin Secs.5.'1.,5.2,and 5.4 to drawing a sketch of the graph of a function. If we are given f (x) and wish to draw a sketchof the graph of.f ,we proceedas follows. First, findf' (r) and f"(x). Then the critical numbers of f arc the values of r in the domain of f for which either f '(x) does not exist or f ' (r):0. Next, apply the first-derivative test (Theorem 5.1.4)or the second-derivative test (Theorem 5.2.1)to determine whether we have at a critical number a-relativemaximum value, a relative minimum v3lue, or neither. To determine the intervals on which '(x) is positive; to de/ is increasing, we find the values.of.x for which f termine the intervals on which / is decreasing,we find the values of r for which f '(x) is negative. In determining the intervals on which / is monotonic, we also check the critical numbers at which / does not have a relative extremum. The values of r forwhich f"(*):0 or f"(x) does not exist give us the possible points of inflection, and we check to seeif f " (x) changessign at eachof these values of r to determine whether we actually have a point of inflectioh. The values of r for which f " (x) is positive and those for which f" (r) is negative will give us points at which the graph is concaveupward and points at which the graph is concavedownward. It is also helpful to find the slope of each inflectional tangent. It is suggestedthat all the information so obtained be incorporated into a table, as illustrated in the following examples. S e t t i n gf ' ( x ) : 0 , w e o b sol,urroN f' (r)-3x2-5x; f" (r):5x-5. tain r : 0 and x: 2. Setting f " (x) : 0, we obtain x: t. In making the table,we considerthe points at which x:0, x:L, and r:2, and the intervals excluding these values of.x: -oo<r<0 0<r<1. t<x<2 2<xqfm Using the information in Table 5.5.1 and plotting a few obtain the sketch of the graph shown in Fig. 5.5.1. Figure5.5.1 ADDITIONALAPPLICATIONS OF THE DERIVATIVE T a b l e5 . 5 . 1 f (x) f '(x) -@<x<0 x:0 / has a relative maximum value; graph is concave downward 0<r<1 / is decreasing; graph is concave downward -3 1<x<2 x-2 21x(*m f (x) : Sxzt7 )c5t3 find: the relative extrem a of f; the points of inflection of the graph of f, the intervals on which / is increasing; the intervals on which / is decreasing; where the graph is concave upward; where the graph is concave downward; and the slope of any inflectional tangent. Draw a sketch of the graph. Conclusion / is increasing; graph is concave downward x:1 nxarvrplr 2: Given f " (x) / is decreasing; graph has a point of inflection / is decreasing; graph is concave upward -1 / has a relative minimum value; graph is concave upward + / is increasing; graph is concave upward SOLUTION: f'(x)' - +x._rts and f" (x) : _-ff1s=ats ' f (x) doesnot exist when x : 0. Settingf ' (x): 0, we obtain x: 2. Therefore, the critical numbers of f are0 and z. f " (r) doesnot existwhen r : 0. F i g u r e5 . 5 . 2 5.5 APPLICATIONS TO DRAWINGA SKETCHOF THE GRAPHOF A FUNCTION setting f " (x): 0, we obtain x: -'1.. In making the table, we consider the points at which r: -L, x:0, and x:2, and the following intervals: -m< x<-1 -1 <.t<0 0<x<2 21x({m A sketch of the graph, drawn from the information in Table 5.5.2 and by plotting a few points, is shown in Fig. 5.5.2. Table5.5.2 f(x) f'(x) f" (x) -oo<r(-1 / is decreasing; graph is concave upward x: -l -) / is decreasing; graph has a point of inflection is decreasing; graph is / concave downward -1 <r<0 0 l:0 does not exist + 0<x<2 gV-+: 4.8 x:2 Conclusion 21x({m does not exist / has a relative minimum value / is increasing; graph is concave downward / has a relative maximum value; graph is concave downward / is decreasing; graph is concave downward 5.5 Exercises For each of the following functions find: the relative extrema of t the point8 of inflection of the graph ol f; the intervals on which f is inseasin& the intervals on which f is decreasing; where the graph is concave upward; where the traph is concave downward; the slope of any hflectional tangent. Draw a sketch of the graph. 2.f(x):x3 lx2-5x 1. f (x) :2)c3 - 6x -f L 3.f(x):)ca-2x3 -: a. f k) 1 3+ 5 x 2+ 3x- 4 3xa* 2x3 5 . /(r) 6. f(x) : 2x3- tx' - l2x * 1 7.f(x):2s4- 3x3*3*+l 8. /(r) : v4- 4x3+ L6x 9. f (x) : ixn - *r$ - x2+'l.. (v2 ifr<0 (-# ifx<0 x): {1 1 __ . 1.tf. (.\--, 1 2 .!f \( x') : l 10. f (x) :3xa * 4x3* 6x2- 4 : ,n l /x ," i f x > 0 Lxr if x>0 (-va 1v3. .r f\k- - \/ : l L .* [rn (tt*-11t i.fx<0 : if x > 0 ra.f Q): t?)^_,l- ifx<l if x > I ls. /(r) : (r * 1)*(r -2)' ADDITIONAL APPLICATIONS OF THE DERIVATIVE 1 6 .f ( x ) : * ( x + 4 ) ' 1 9 ,f ( x ) - 3 x 2 t 3 - 2 x 1 7 .f ( x ) : 3 x E * S x a 20- f(x) : t7rt3- , 2 3 .f ( x ) : 2 * ( x - 3 ) t r a 18. f(x):3xs * 5r3 X. f (x):2 * (r - 31'r' 28.f(x):3 * (x + r7'r' 26. f (x) :2 * (x - 31z' 2 9 .f ( x ) : 7 2 \ P y 2 1 .f ( x ) : 7 r t 3 a 2 r t t s 2a. f @) :2 t (x - 31nrc (x+1;uru 2 7 .f ( x ) : 3 * 30. f(x) v11{- rz 3t.f(x):H 32. f(x):h 33.f(x): 22.f(x):Jall-4* (r * 2) \R U.f(x):ffi 5'6 AN APPLICATION OF THE DERIVATIVE IN ECoNoMIcS In economicsthe variation of one quantity with respectto another can be described by either an aaerageconceptor a margi;a/ concept. The average concePtexPressesthe variation of one quantity over a specified range of values of a second quantity, whereas the marginal concept is the instantaneous change in the first quantity that results from a very small unit change in the second quantity. For example, in our discussion of rectilinear motion (Sec.3.2) we considered both an averageconcept and a marginal concept. The averagevelocity in traveling a distanceof J miles in f hours is s/f mi/hr, whereas the instantaneous velocity is a marginal concePt as it gives the rate of change of s with respect to a small unit change in f at a particular instant of time. We begin our examplesin economics with the definitions of averagecost and marginal cosi. It should be clear that to define a marginal concept precisely we must use the notion of a limit, and this will lead to the derivative. Supposethat C(r) dollars is the total cost of producing r units of a commodity. The function C is called a total cost function. In normal circumstances,r and C(r) are nonnegative. If zero is in the domain of C, C(0) is called the oaerheadcost of production. Note that since r representsthe number of units of a commodity, r must be a nonnegative inieger. However, t9 apply the calculuswe assumethat r is a nonnegative reil number (and then round off any noninteger values of x to th" ,rearestinteger), thus giving us the continuity requirements for the function C. The aaeragecost of producing each unit of a commodity is obtained by dividing the total cost by the number of units produced. Letting e(r) be the number of dollars in the averagecost, *e hane C(x) Q (r), xand Q is called an aaeragecostfunction. Nory,,let us suPPosethat the number of units in a particular output is rt and this is changed by A,x.Then the changein the total cost is given by C(xt + Ar) - C(xt), and the averagechangl in the total cost wiih re- 5.6 AN APPLICATIONOF THE DERIVATIVEIN ECONOMICS 231 spect to the change in the number of units produced is given by C(xr*Ax)-C(r,) Ax (1) Economists use the term marginal cost for the limit of the quotient in (1) as At approacheszero, provided the limit exists. Being the derivative of C at xr, this limit gives us the following definition. 5.5.1 Definition If C(r) is the number of dollars in the total cost of producing r units of a commodity, then the marginalcost,when *:rr., is given by C'(rt) if it exists. The function C' is called the marginqlcostfunction. In the above definition , C'(xr) can be interpreted as the rate of change of the total cost per unit change in the quantity produced, when 11 units are produced. o rLLUsrRArroNL: Supposethat C(r) is the number of dollars in the total cost of producing x picture frames (x > 10) and C ( x ) : 1 5 * 8 r +x.r y (a) The marginal cost function is C' and :8 -# C'(x) (b) The marginal cost when I : 20 is C'(20) and C ' ( 2 0-) 8 - ' # : 8 - 0 . s 0: 7 . 5 0 (c) The number of dollars in the cost of producing the twenty-first frame is C(21) - C(20) and C(21) - C(20) :192-52 - 1'85:7.52 Observe discrepancy of change of approximate frame. that the answers in parts (b) and (c) differ by 0.02. This occurs because the marginal cost is the instantaneous rate C(r) with respect to a unit change in r. Hence, C'(20) is the number of dollars in the cost of producing the twenty-first . Reasoning similar to that preceding Definition following definition. 5.5.2 Definition 5.5.1 leads to the lf Q@) is the number of dollars in the average cost of producing one unit of r units of a commodity, then the marginal average cost, when x: x1,is given by Q'(r1) if it exists, and Q' is called the marginal aaerage cost function. ADDITIONALAPPLICATIONS OFTHEDERIVATIVE The graphs of the total cost function, the marginal cost function, and the averagecost function are called the total costcurae (labeled TC), the marginalcost curae (labeled MC), and the nrreragecost curae (labeled AC), respectively. Q'(r1) gives the slope of the tangent line to the averagecost curve at the point where x: xr. Figure5.6,1 o rlr,usrRArroN 2: Consider a linear total cost function. C(x):mx*b Figure5.6.2 Note that b rePresentsthe overheadcost (the total cost when.r: 0). The marginal cost is given by C'(r) : m. If Q is the averagecost function, Qk) : m * blx, and the marginal averagecostis given by e,k) : -blxr. Refer to Fig. 5.6.1'for sketchesof the total cost curve and the average cost curve. The total cost curve is a segment of a straight line in the first quadrant having slope m and y-intercept b.The averagecost curve is a branch of an equilateral hyperbola in the first quadrant having as horizontal asymptote the line y : m. BecauseQ'(r) is always negatirre, the averagecost function is always decreasinB,and as r increases,the value of Q@) gets closer and closer to m. . o rtlustRATIoN suppose that we have a quadratic total cost function C(r) -- ax2* bx * c where a and c are positive. Here c is the number of dollars in the overhead cost. The total cost curve is a parabola opening upward. Because c'(x):2Ax*b, a critical number of c is-blza. We distinguish two cases:b>0andb<0. Figure5.6.3 rxeivrr.r 1: Supposethat C(x) dollars is the total cost of producing 100r units of a commodity and C(r) : Lxz- 2x * 5. Find the function giving (a) the average cost; (b) the marginal cost; (c) the marginal averagecost. (d) Find the absolute minimum average unit cost. (e) Draw sketchesof the total cost curve, the average Case7: b > 0. -bl2a is then either negative or zero, and the vertex of the parabola is either to the left of the y axis or on the y axis. Hence, the domain of C is the set of all nonnegative numbers. A sketch of TC for which b > 0 is shown in Fig. 5.6.2. Case2: b < O.-bl2a is positive; so the vertexof the parabolais to the right of the y axis, and the domain of C is restricted to numbers in the interval l-bl2a, +oo).A sketchof TC for which b < 0 is shown in Fig. s.G.g. . SOLUTION: (a) If Q is the average cost function, (_(*) 1 ^ , 5 QQ):Y:i.r-2*; (b) The marginal cost function is C,, and c'(x) 5.6 AN APPLICATIONOF THE DERIVATIVEIN ECONOMICS cost curve, and the marginal cost curve on the same set of axes. (c) Q' is the marginal average cost function, and Q',k):+-* (d) Setting Q' @): 0, we obtain \[0 as a critical number of e, and Q(V10) : VLO- / - L.L6.BecauseQ" k) :101x3, Q" (\m) ) 0, and so Q has a relative minimum value of approximately '1,.L6at x: {fr. From the equation definirg Q(r) we seethat Q is continuous on (0, +oo). Becausethe only relative extremum of Q on (0, aoo) is at x: \m, it follows from Theorem5.3.1(ii)that Q has an absoluteminimumvalue there. when x: t/fr : 3.L6,100r : 3L6,and so we concludethat the absolute minimum averageunit cost is $1.15when 3L6 units are produced. The sketchesof the curves TC, AC, and MC are shown in Fig. s.6.4. Figure5.6.4 In Fig. 5.6.4, note that the lowest point on curve AC occurs at the point of intersection of the curves AC and MC, and this is the point where the marginal average cost is zero. This is true in general and it follows from the fact that the value of r which causes the marginal average cost to be zero is a critical number of the function Q. That is, since Q(r) : C(x)lx, Q',k) - *r_,(x)_ 4v and so Q'k): C(x) )c2 0 when xC'(x)- C(x) : 0 or, equivalently, when "r r) c'(x):aYou should note the economic significance that when the marginal cost and the average cost are equal, the commodity is being produced at the very lowest average unit cost. In normal circumstances, when the number of units of the commodity produced is large, the marginal cost will be eventually increasing or zero; hence, C" (x) > 0 for r greater than some positive number N. So unless C" (x) : 0, the graph of the total cost function is concave upward tor x > N. However, the marginal cost may decrease for some values of x; hence, for these values of x, C" (x) < 0, and therefore the graph of the total cost function will be concave downward for these values of x. The following example involving a cubic cost function illustrates the case in which the concavity of the graph of the total cost function changes. 234 ADDITIONALAPPLICATIONSOF THE DERIVATIVE 2: Draw a sketch of ExAMPLE the graph of the total cost function C for which C(r):x3-6x2* 13r*1' Determine where the graph is concaveupward and where it is concavedownward. Find any points of inflection and an equation of any inflectional tangent. Draw a segment of the inflectional tangent. SOLUTION: C'(x) :3xz - 12x+ 13 C"(x):6x-t2 C'(x) can be written as 3(r - 2)2 + 1. Hence, C'(x) is never zero. C" (x): 0 when r: 2. To determine the concavity of the graph for the intervals (0, 2) and (2, am) and if the graph has a point of inflection at x:2, we use the resultssummarizedin Table 5.6.1. T a b l e5 . 6 . 1 C(x) C'(r) C" (x) Conclusion 0<x<2 x:2 graph has a point of inflection 2<x(fm graph is concaveupward An equation of the inflectional tangent is x - y + 9: 0. A sketch of the graph of the total cost function together with a segment of the inflectional tangent is shown in Fig. 5.6.5. Figure5.6.5 Consider now an economic sifuation in which the variables are the price and quantity of the commodity demanded. Let p dollars be the price of one unit of the commodity and x be the number of units of the commodity. Upon reflection, it should seem reasonablethat the amount of the commodity demanded in the marketplaceby consumersdepends on the price of the commodity. As the price falls, consumers generally demand more of the commodity. Should the price rise, the opposite occurs. Consumers demand less. Thus, if p, and p2 are the number of dollars in the prices of x, and r, units, respectively, of a commodity, then pt ) pz if and only if x, 1 x, (2) An equation giving the relationship between the amount, given by x, of a commodity demanded and the price, given by p, is called a demand equation.If the equation is solved for p, we have P: P(x) The function P is called the pricefunction and P(r) is the number of dollars in the unit price for which r units of the commodity are demanded at that price. Becauseof statement (2),P is a decreasingfunction. In a normal eco- 5.6 AN APPLICATIONOF THE DERIVATIVEIN ECONOMICS nomic situation, r and P(r) are nonnegative numbers. Even though r should be an integer (becauser is the number of units of a commodity) we assumeonly that r is a real number in order that P may be a continuous function so that we can apply the calculus. Another function important in economicsis the total reaenuefunction, which we denote by R, and R(x) - xP(x) where P(x) dollars is the price of each unit and r is the number of units sold. When x * 0, from the above equation we obtain R(r) _ p(r) x which shows that the revenue per unit (the average revenue) and the price per unit are equal. 5.6.3 Definition If R(r) is the number of dollars in the total revenue obtained when x units of a commodity are demanded, then the marginal reaenue,when *: *r, is given by R'(rt) , if.it exists.The function R' is called the marginalreaenue function. R'(rr) can be positive, negative, or zero, and it can be interpreted as the rate of changeof the total revenueper unit changein the demand when rt units are demanded. The graphs of the functions R and R' are called the total rer)enuecurae(IabeledTR) and the marginalreaenuecurve(labeled MR), respectively.The graph of the demand equation is called the demand curTJe. ExAMPLE3: The demand equation for a particular commodity is p' * x - 12: 0. Find the price function, the total revenue function, and the marginal revenue function. Draw sketches of the demand curve, the total revenue curve, and the marginal revenue curye on the same set of axes. solurroN: If the demand equation is solved for p,we find p : !\EV. Since P(r) > 0, we have P(r):Ex R(x) : x{FV R'(r) : A-3x 2\/12 - x Setting R'(r) :0, we obtain r: 8. Using the information in Table 5.6.2, we seethat the required sketchesare drawn as shown in Fig. 5.6.6. T a b l e5 . 6 . 2 x P(x) R(r) R'(r) 0\n0\m. 33e8 82t60 11. 1 12 0 L1 0 -g does not exist ADDITIONALAPPLICATIONS OF THE DERIVATIVE demand curve Figure5.6.6 we have seen that for a given demand equation the amount demanded by the consumer depends only on the price of the commodity. Under a mlnopoly (which means that there is only one producer of a certain commodity), price and, hence, demand can be controlled by regulating the quantity of the commodity produced. The producer undei a monopoly is called a monopolist,and he wishes to control the quantity produced and, hence,the price per unit so that the profit will be as largl as possible. The profil earned by u business is the difference between the total revenue and the total cost. That is, if 5(x) dollars is the profit obtained by producing r units of a commodity, then 5(r) - R(x) - C(r) (3) where R(x) dollars is the total revenue and C(x) dollars is the total cost. S is calledthe profit function.From Eq. (3),if we differentiatewith respect to x, we obtain S ' ( r ) : R ' ( r ) - C '( x ) (4) S" (x) : R" (r) - C" (x) (5) and It follows from Eq. (4) that s'(r) > 0 if and only if R'(r) > c, (x); therefore, the profit is increasing if and only if the marginal revenue is greater than the marginal cost. Let us determine what level of production is necessary to obtain the greatestprofit. S has a relative maximum function value at a number x for which S'(r) : 0 and S" (x) ( 0. From Eqs. ( ) and (5) 5.6 AN APPLICATION OF THE DERIVATIVE IN ECONOMICS (r, R(r)) Figure5.6.7 4: Supposethat the ExAMPLE demand equation for a certain commodityis P:4 - 0.0002r, where x is the number of units produced each week and p dollars is the price of each unit. The number of dollars in the total cost of producing x units is 600 + 3x. If the weekly profit is to be as large as possible, find: (a) the number of units that should be produced each week; (b) the price of each unit; (c) the weekly profit. we observe that this will occur at a value of x for which the marginal revenue equals the marginal cost and R" (r) < C" (x). The values of r will be restricted to a closed interval, in which the left endpoint is 0 (since r = 0) and the right endpoint (the largest permissible value of x) is determined from the demand equation. At each of the endpoints there will be no profit, and so the absolute maximum value of 5 occurs at a value of r where S has a relative maximum value. To illustrate this geometrically, refer to Fig. 5.6.7, where both the x total revenue curve and the total cost curye are drawn on the same set of axes. The vertical distance between the two curves for a particular value of r is 5(r), which gives the profit corresponding to that value of x. When this vertical distance is largest, S(r) is an absolute maximum. This is the distance an in the figure where A and B are the points on the two curves where the tangent lines are parallel, and hence C'(x) - R'(r). sor,urroN: The price function is given by P (r) - 4 - 0.0002x,and r is in the closed interval [0, 20,000]becauser and P(r) must be nonnegative. R(r) : xP(x), and so R(x) :4x - 0.000212 r in [0, 20,000] (6) The cost function is given by C ( x ) : 6 0 0* 3 r (7) If S(r) dollarsis the profit, S(r): S(r):r-0.0002f -600 R(r) - C(x), and so r in [0,20,000] From Eqs. (6) and (7) we obtain R'(r) - 4- 0.0004x c'(r) : 3 Also, R"(r) : -0.0004 and C"(x):0. 4 - 0.0004x: 3 EquatingR'(r) and C'(r) we get x:2500 BecauseR" (r) is always less than C" (x) , S" (x): R" (x) - C" (x) < 0, and so we conclude that 5(2500) is an absolute maximum value. (Note that S(x) is negative for r:0 and r:20,000.) P(2500):3.50 and s(2500):650. Therefore, to have the greatest weekly profit, 2500 units should be produced each week to be sold at $3.50each for a total profit of $550. 238 ADDITIONALAPPLICATIONS OF THE DERIVATIVE ExAMPLE 5: Solve Example 4 if a tax of 20 cents is levied by the govemment on the monopolist for each unit produced. soLUrIoN: With the added tax, the total cost function is now given by C(x):(500*3r) *0.20x and so the marginal cost function is given by C'(x):3.20 EquatingR'(r) and C'(r), we get 4 - 0.0004x:3.20 As in Example4 it follows that when x: 2000,S has an absolute maximum value on [0, 20,000]. From P(x) - 4- 0.0002r and S(x) : O.gx - 0.0002x2 - 600,we have P(2000): 3.60and S(2000): 200. Therefore, if the tax of 20 cents per unit is levied, only 2000units should be produced each week to be sold at $3.60 in order to attain a maximum total weekly profit of $200. It is interesting to note that in comparing the results of Examples 4 and 5, the entire 20 cents rise should not be passedon to the consumer to achieve the greatestweekly profit. That is, it is most profitable to raise the unit price by only L0 cents. The economic significance of this result is that consumers are sensitive to price changes, which prohibits the monopolist from passing on the tax completely to the consumer. A further note of interest in the two examplesis how the fixed costs of a company do not affect the determination of the number of units to be produced or the unit price such that maximum profit is obtained. The fixed costof a company is the cost that does not change as the company's output changes.Regardlessof whether anything is produced, the fixed cost must be met. In Examples 4 and 5, becauseC(r) : d00+ 3r and 600 + 3.2x, respectively, the fixed cost is $600. If the 600 in the expressions for C(r) is replacedby any constantk, C'(x) is not affected;hence, the value of x for which the marginal cost equals the marginal revenue is not affected by any such change. Of course, a change in fixed cost affects the unit cost and hence the actual profit; however, if a company is to have the greatestprofit possible, a change in its fixed cost will not affect the number of units to be produced nor the price per unit. Exercises5.6 1. The number of dollars in the total cost of manufacturing .x watches in a certain plant is given by C(r) :1500 + 30r + 20/r' Find (a) the matginal cost function, (b) the marginal cost when x: 40, and (c) the cost of manufacturing the forty-first watch. 2. lf C(r) dollars is the total cost of manufacturing r toys and C(r) : 110 + 4r * 0.02f, find (a) the marginal cost function, (b) the marginal cost when r: 10, and (c) the cost of manufaduring the eleventh toy. 5.6 AN APPLICATION OF THE DERIVATIVE IN ECONOMICS 239 3. Suppose a liquid is produced by a certain dremical process and the total cost function C is given by C(x) :514rrr, where C(r) dollars is the total cost of producing t gallons of the liquid. Find (a) the marginal cost when 16 gal are prcduced and (b) the number of gallons produced when the marginal cost is 40 cents per gal. 4. The number of dollaB in the total cost of prcducing x units of a certain commodity is C(r) : & + 3x + 9{2t. Fi^d, (a) the marginal cost when 50 units are produced and (b) the nurnber of units produced when the marginal cost is $4.50. 5. T'he number of dollars in the total cost of prcducing r units of a commodity is C(r) : ae14r * * Find the function grving (a) the average cost, (b) the marginal cost, and (c) the marginal average cost (d) Find the absolute ninimum average unit cost. (e) Draw sketches of the total cost, average cost, and marginal cost curves on the same Bet of axes. Verify that the average cost and marginal cost are equal when the average cost has it8 least value. - 6t + 4, find (a) the average cost 6. II C(x) dollars is the total cost of producing t units of a commodity and C(x):3f function, (b) the marginal cost function, arld (c) the marginal average cost function. (d) What is the range of C? (e) Find the absolute minimum average unit cost. (f) Draw sketches of the total cost, average cost, and marginal cost curves on the same set of axes. Verify that the average cost and marginal cost are equal when the average cost has its least value. 7. The total cost function C is given by C(x) = lat - 2a2+ 5x + 2. (a) Determine the range of C. (b) Find the marginal co9t function, (c) Find the interval on whidr the martinal cost function is decreaaing and the interval on which it is incr€asing. (d) Draw a sketch of the graph of the total cost function; detemdne where the graph is concave upward and where it is concave downward, and find the points of inflection and an equation of any inflectional tangent. 8. If C(r) dollars b the total cost of producing t units of a commodity and C(r) = 2f - 8r + 18, find (a) the domain and range of C, (b) the average cost function, (c) the absolute minimum average unit cost, and (d) the marginal cost tunction. (e) Draw sketches of the total cost, average cost, and marginal cost curves on the same set of axe8. 9. The fixed overhead expense of a manufactuer of childrcnls toys is $4fi) per week, and other costs amount to $3 fot each toy ploduced. Find (a) the total cost function, (b) the average cost function, and (c) the marginal cost function. (d) Show thaf there is no absolute minimum average unit cost. (e) What is the smallest number of toys that mu8t be Produced so that the average cost per toy is le8s than $3.42?(f) Draw sketches of the graphs of the functions in (a), (b), and (c) on the 6ame Bet of axes. 10. The number of hundreds of dollars in the total cost of producing 100xradios per day in a certain factory is C (x) : 4x + 5' Find (a) the average cost function, (b) the marginal cost function, and (c) the martinal average cost function. (d) Show that there i8 no absolute minimum average urlit cost. (e) What is the smallest number of radios that the factory must produce in a day so that the averagecost per radio is lessthan $7?(f) Draw sketchesof the total cost, averageco6t, and martinal cost curves on the same set of axes. 11. If the demand equation for a particular commodlty is 3r I 4p : 12, find (a) the price function, O) the total r€venue function, and (c) the marginal revenue function, Draw sketches of the demand, total revenue, and marginal revenue curves on the same set of axes. Verify that the marginal revenue curve intersects the t axis at the Point whose abscissa is the value of r for whidr the total levenue iB greate8t and that the demand curve interEects the t axis at the Point whose absci66a is twice that. 12. The derrand equation for a particular commodity is pf + ? - 18:0 wherc p dollars is the price per unit when 100r units are demanded. Find (a) the price function, (b) the total revenue function, and (c) the marginal revenue function. (d) Find the absolute maximum total revenue. 13. Follow the instructiona of Exercise 12 if the demand equation is t' + Pt - 36: 0' 14. Let R(x) dollars be the total revenue obtained when r rrnits of a comrnodity are demanded and R(r):2+31i1, where I is in the closed interval [2, 17]. Find (a) the demand equation, O) the Price function, and (c) the marginal revenue function. (d) Find the absolute maxtnum total levenue. (e) Draw skekhes of the demand, total revenue, and marginal revenue curves on the Sameset of axe8. 1.5.The demand equation for a certain commodity is r + p: 14, where r is the numbet of units Produced daily andp is the number of hundreds of dollars in the price of each unii. The number of hundre& of dollars in the total cost of Pro- 240 ADDITIONALAPPLICATIONS OF THE DERIVATIVE ducing x units is given by C(x) : * - 2x I 2, and r is in the closed interval [1, 1a]. (a) Find the profit function and drav, a sketch of its graph. (b) On a set of axes different from that in (a), draw sketches of the total revenue and total cost curves and show the Seometrical interpretation of the profit function. (c) Find the maximum daily profit. (d) Find the marginal tevenue and rrarginal cost functions. (e) Draw sketches of the graphs of the marginal revinue and margi nal cost functions on the same set of axes and show that they intersect at the point for which the value of r makes the prcfit a maximum, 16. Follow the instructions of Exercise15 if the demand equation is f + p :32 and C(x) :5r. 17' The demand equation for a ce*ain commodity is p : (x - 8),, and the total cost function is given by C(r) : l8t - f , where C(r) dollars is the total cost when x units are purchased. (a) Detemine the permissible vaiues of r. (b) Find margin-al revenue and marginal co6t functions. (c) Find the value oI r which yields the maximum piofit. (d) Draw $e sketches of the marginal revenue and marginal cost functions on the same set of axes. 18. A monoPotst d€termines that if C(r) cents is the total cost of producing x units of a certain commodity, then C(r) : 25r + 20/000. The demand equation is x + sOp: 5000, where x units are demanded each week when the unit Drice tj.P If the weekly profit is to be maximized, find (a) the numbe! of units that should be produced each week, -"".b: O) the price of each unit, and (c) the weekly profit. 19. solve Exercise 18 if the govemment levies a tax on the monopolist of 10 cents per unit produced. 20, Solve Exercise18 if the govemment imposes a 10% tax based upon the consume/s price, 21. For the monopoligt of Exerciee 18, determine the amount of tax that shoutd be levied by the government on each unit prcduced in order fo. the tax revenue received by the government to be maxirnized. 22' Find the maximum tax revenue that can be received by the government if an additive tax for each unit produced is leyied on a monopolist for which th_edeaTd equation is xllp:75, where.' units are demanded when p dollars is the price of one unit, and c(r) :3x r 1e0,where c(r) dollars is the total cost of producint r units. 23. The de-mand equation for a certain commodity pmduced by a monopolist is p : a - 6r, ,ne total cost, C(r) dollars, of Prcducing :r units is detennined by C1r; : r a dx, where a, b, c, aid, d arc positive constants. "rrd If the government levies a tax on_the monopolist of f dollars per unit produced, show that in order ior the monopolist to nJximize his profits he should pass on to the consumer only one-half of the tax; that is, he should increase his unit price bv it dollars. ReaiewExercises (Chnpter5) 1. Find the shortest distance from the point (*, 0) to the curve y: t[. 2. Prove that among all the rectangleshaving a given perimeter, the square has the greatestarea. 3. Prove that among all the rectangles of a given area, the square has the least perimeter. 4. The demand in a cerlain market for a particular kind of breakfastcereal is given by the demand equation p, + 25p 2000: 0, where P cents is the price of one box and x thousandso{ boxes is the quantity demandedper week. If the current Price of the cerealis 40 cents per box and the pdce per box is increasing at th€ rate of 0.2 cent each week, find the rate of change in the demand. 5. Two Particles start theit motion at the same time. one particle is moving along a horizontal line and rts equation of motion i6 t = t: - 2t, where r ft is the directed distance of the particle from the o;igin at t sec.The otherparticle is moving along a vedical line that intersectsthe horizontal line at the odgin, and its equation of motion is y : p - 2, where y ft is the directed distance of the partide from the odgin at t sec. Find when the directed distance Letween the two particles is least, and their velocities at that time. REVIEWEXERCISES 241 the intervals on which f In Exercises 5 through 9, find: the relative extremi of f the points of inflection of the traph of t glaph i6 concave downwhete the upward; graph is concave the tf," i.tlrvals on which I is decreasing witere i'i""r""ri"g graph' of the Draw a sketch tangmt. ward; the sl--opeof any inflectional 6.f(x):(x*2)at3 8. /(r) : x\/FT 7.f(x): (r- 4)'(x+2)' 9 . f ( x ) : ( r - 3 ) s r*r t 10. If I\x): x*L xr+1 gxaPh. prove that the graph of I ha8 three points of inflection that are collinear. DrarY a Eketch of the 11. If l(r) : 1111,show that the graph of I has a Point oI inflection at the origin' and only xo where r is a positive integer. Prove that the gnph of I has a point oJ inflection at th€ origin if at 0. value relative minimum has a is wen, , > 1. Furthirmore show that if I / Jd int"g". "nd (f * ar)!, where p is a rational numb er and'p * 0, prove that if P < ]the graph ofl has two Point8 of inflectg. fif(x\: ti6n and if p > * the graph of f has no Pointg of inflection' what can you condude, if an''ll1ing, about (a) the L4. Suppose the glaph of a function has a point of inflection at l: c. of continuity (c) c; the of at f" at c? continuity of / ai c; (b) the continuity /' P is the numbef of millions 15. Suppose c is the number of millions of dollars in the caPitalization of a certain colPoration, - 0.004c'. r its capitalization is incleasing at the rate of : P 0-.05c profit, and annual of dollars in the corpofation's profit if its current caPitalization is (a) $3 million $400,000per year, find the rate of chanie of the corporation's annual and (b) $6 million. when r aPartmentsare rented and 16. A property development company rents each apartment at p dollars per month is zero? revenue marginal How many apartments must be rented before the p :2(;\/ffi=E -!2. Let fe\: ;; f; about a right-circular cylinder L7 . Find the dimensions of the right-circular cone of leastvolume that can be circumscribed of radius r in. and altitude h in. radius to the measure of the altitude for a 1g. A tent is to be in the shape of a cone. Find the mtio of the measure of the tent of given volume to require the least material' :l18' whete P dollars is the Price per unit when 19. The demand equation for a particular commodity is (P + 4)(r + 3) the marginal revenue function' lg0r units are demanded. Find (a) the price functi;n, 16) the total rwenue function, (c) and margiral revenue revenue, total the demand, of (d) Find the absolute maxrsrum iotal revenue. (e) Draw sketches cuw$ on the same set of axes. $ and the total cost function is given by c(r) : 2r 1' 20. The demand equation for a certain commodity i8 p + 2V?J: -(c) Find the functions. cost marginal permissible values of r. iul ritta the marginal revenue and t"l-oJ""rr""'ae value of r which yields the maximum profit' 21. The demand equation for a certain cpmmodity is l16px:70s - 2. 1.03r* 18 ' 10'3f 6f unit, and I > 100. The number of where r is the number oI units produced weekly and p dollars is the price of each ' : 10k-1' Find the number of units * 11 24 dolLars in the avenge cost of producing each unit is given by QQ) "\7for the weekly unit in oldet price of each Profit to be marirnized. that should be produced each week and the 242 ADDITIONALAPPLICATIONS OF THE DEBIVATIVE 22. When 1.000rboxes of a c€rtain*ind of material are produced, the-nurnber of dollars in the total cost of production is givenbv c(r) - 135r1i3 + ft50.Find (a)the marginicost when8000br""r ifi'r,*uu. ot uor"" produced whm the marginal cost "i"-p-au""a?ffi Grer thousand boxes) is g20. 23' A ladder is to rcach over a fence ' ft high to a watl ur ft behind the fence. Find the length of the shortest ladder that may be used. 24. Dtaw a sketch of the graph of.a function on the interval I in eachcase: (a) I is the open interval (0, 2) ard/is f con_ tinuous on r' At 1, I has a relative maximum value but /'(1) does not exirt. 1iy i i, th" to, zl. rrru ror,""io"eJiiiu*"i -i',iio- ua,,"Jt7;;f 0.6ilil; Hll,Ti,""tiffi?ffifrTff""j,1'.but theabsorute olenintervar (0.z), 25. (a) It t'@):3lrl + alx - tl, prove that/has an absolute minimum value of 3. (b) If g(r) =alll + 3lr _ 11,prove that 8 ha8 an absolute minimum value o_f3. (c) rl h(tc): alxl + tp - il, where d > 0 and > b o, plove that , has an abso_ lute minimum value that is the smalter of the t*,o ,rr.-t"""'o f. "r,i 26' rl f(x): lxl" ' lr - tlb, where c and b are positive rational numbers, prove that has a relative maximum / value of L'bbl@.+ bl+b. The differential and antidifferentiation THE DIFFERENTIAL AND ANTIDIFFERENTIATION 6.1 THE DIFFERENTIAL Supposethat the function / is defined by Y:f(x) Then, when f ,(x) exists, A/ '(x) lim /f \ - ar;-o A'r ( 1) where Ly:f(x+ Lx) - f(x). From (1),itfollows thatfor any€ ) 0there existsa6>0suchthat l + - f ' '( x ) 1 I .. lAr w h e n e v e( r ol A r l< 6 which is equivalent to l\l,-f' (x) Arl lA;:<. whenever0( lArl <D This means that lAy -f'(*) Arl is small comparedto lAxl. That is, for a sufficiently small lArl ,f '(x) Ax is a good approximation to the value of Ay. using the symbol :, meaning "is approximatelyequal to,,, we say then that Ly:f,(x)Ax if lArl is sufficiently small. The right side of the above expression is defined to be the ,,differential,, of y. 6'L'1 Definition If the function / is defined by y : (x), then the differentialof y, d,enotedby f dy, is given by (1) where r is in the domain of ' and,Ar is an arbitrary increment f of x. This concept of the differential involves a special type of function of two variables (a detailed study of such functions will be girr"t in Chapter 19)' The notation df may be used to represent this funltion, where the symbol df is tegallud as a single entity. The variable r can be any number in the domain of f ' , andLx can be any number whatsoever.To state that df is a function of the two independent variablesr and Ar means that to each ordered paft (x, Ar) in the domain of df therc correspondsone and only one number in the range of df, and this number .u., b" represented by df (*, Ax), so that df (r, Ar) : f '(x) Ax (2) ComparingEqs.(1) and (2) we seethat when y : (x), dy and df (x, A,x) f 6.1 THE DIFFERENTIAL 245 are two different notations f.orf '(x) Lx. We shall use the "dy" symbolism in subsequentdiscussions. o T L L U S T R A Tt I: oINf y : 4 x 2 - r , t h e nf ( * ) : A x z - x , a n d s o / ' ( r ) : 8 r - 1 " . Therefore,from Definition 6.1.7, dy: (8r-1) Ax In particular, tf.x:2, then dy : 1'5Ax. o When y : f (r), Definition 6.1,.Lindicates what is meant by dy, the differential of the dependent variable. We also wish to define the differential of the independent variable, or dx. To arrive at a suitable definition for dx that is consistent with the definition of dy, we consider the identity '(x): L and function, which is the function / defined by f (x) - r. Then f ' : A : x; so dy L Ar : Ar. Becausey x, we want dx to be equal to dy for this function we want dx: Ax. It is this is, for that this particular function; definition. following to the leads us that reasoning 6.1.2 Definition If the function /is defined by y : f (x), then the differentialof x, denotedby dx, ts given by (3) where Ar is an arbitrary increment domain of f'. of. x, and r is any number in the From Eqs. (1) and (3) we obtain (4) $ Dividing on both sides of the above by dx, we have #:f'k) tfdx*0 (5) Equation (5) expresses the derivative as the quotient of two differentials. The notation dyldr often denotes the derivative of y wlth respect to x-a symbolism we have avoided until now. nxlvpln L: Given y--4x'-3x*I find Ly, dy, and Ly - dy for (a) any )cand Ax; (b) r :2, Lx: 0.1; ( c )x : 2 , A x : 0 . 0 1 ;( d ) x : 2 , Ar: 0.001. solurroN: (a) Becausey :4x2 - 3x * L, we have y + Ly :4(x * Ar)2 - 3(x * Ar) * 1' (4x' - 3x * 1) + Ly : 4x2* 8x A,x* 4(A'x)2- 3x- 3 Ar * 1 Ly : (8r - 3) Ar * A(A,x)z Also, dy: f'(x) dx THE DIFFERENTIAL AND ANTIDTFFERENTIATION Thus, (8r - 3) dx :- (8r - 3) Ar Ay-dy-A(A,x)z The resultsfor parts (b), (c),yd (d) aregiven in Table5.1.-l,,where : Ly (8r - 3) Ar + 4(A,x)zand,dy: (8r - 3) Ar. T a b l e6 . 1 . 1 1..34 0.1304 0.013004 0.04 0.0004 0.000004 Note from Table 5-1.1,that the closer Ax is to zero, the smaller is the difference between Ly and dy. Furthermore, observe that for each value of Ax, the corresponding value of A,y - dy is smaller than the value of Ar. More generally, dy is an approximation of ay when Ar is small, and the approximation is of better accuracy than the size of Ar. For a fixed value of. x, say xo, dy - f '(xs) dx (6) That is, dy is a linear function of dx; consequently dy is usually easier to compute than Ay (this was seen in Example i;. Because (xo + Ar) - /(ro) : f Ly, we have f(xo+Ar): f(x) + Ly Thus, (ro * Ar, f(xo+ ar))Q R(re * Ax, flr) + dy) AT, M (xo+ Ax,f (xJ) f(xo+ Ax) - f(x) + dy (7) We illustrate our results in Fig. 5.1.1.The equation of the curve in the figure is y : /(r). The line pT is tangent to the curye at p(xs, (xo));Ax and f dx ate equal and are represented by the directed distarr..-pfr, where M is the point (ro + a'x,f (xs))._w" ret Q be the point (ro + ax, (xs* Ar)), and the directed distance MA is Ly or, equivalently, (x, f f f @o) L!4_ The slop=f PT is f '(x): dyldx. Aiso, the slope of'pi'i; ffiJm, becausePM: dx, we have dy: MR and Re: Ay - dy. Note that ""a the smaller the value of dx (i.e., the closer the point e i to tire point p), then the smaller will be the value of Ay - dy (i.e., the smaller wilibe the length of the line segmentRQ). An equation of the tangent line pT is y-f(ro)+f'(x)(r-ro) 6.1 THE DIFFERENTIAL 247 Thus, if y is the ordinate of R, then (8) y: f(xi + dy comparing Eqs. (7) and (8), we see that when using /(xo) + dy to ap' proximate the ,ruir" of f:1xo* Ar) we are approximating the ordinate ff tn" point Q(ro * A,x,i@o + Ar)) on the curve by the ordinate of the point filxo + Lx, f (xs)+ dV'l on the line that is tangent to the curve at P(ro,/(ro)). Observethat in Fig. 6.1.1,the graph is concaveupward' In Exercise30 you arc asked to draw a similar figure if the graph is concave downward' (*): 2: Find an aPProximate soLUrIoN: Consider the function / defined by f Hence, value for \l/28 without using 3r \/x tables. EXAMPLE {x and let y : f (x)' and dy: f'(x) dx :i*ut7 , ax The nearest perfect cube to 28 is 27. Thus, we comPute dY with and dx: Ax : L. dy:#wrrt By applyingformula(7)with xs: 27, f(27+ 1) : f(27) +* 1.,and dy : #, we have iln+7: W+# i/2s: g+ ;, Therefor",{28:3.037. : From Table 1 in the Appendix, we get t/28 3.037.Thus, the apdecimal three Places. proximation is accurateto rxarrpr.n 3: Find the approximate volume of a sPherical shell whose inner radius is 4 in. and whose thickness is t in. We consider the volume of the spherical shell as an increment solurroN: of the volume of a sPhere. Let r: the number of inches in the radius of a sphere; V: the number of cubic inches in the volume of a sphere; LV: the number of cubic inches in the volume of a spherical shell. Y: $nrs Substitutingr so that = 4 and dr: dV - 4rr(4)z # :4n dV : 4nr2 dr # into the above, we obtain 248 THE DIFFERENTIAL AND ANTIDIFFERENTIATION Therefore, LV : 4rt, and we conclude that the volume of the spherical shell is approximately 4n in.3. Exercises 5.1 In ExercisesL through6, find (a) Ly; (b) dy; (c) Ay - dV. l' Y: x" 2. y:4x2 - 3x* 1 4.y:h 5. Y:2f + 3* In Exercises7 through 12, find for the given values: qa) ay; (b) dv; (c) Ay - dy. 7.y:*-3x;x:2; Ax:0.03 8. y : x2- 3x; x: -l; Ar : 0.02 10.y : xs* 7; x: -7; Ax: 0.1 1 1 .y : i , 1 ,:2; 3. v : \E 6. v: 9.y: 72.y : Ax: 0.01 I Yx 1 3 * L ;x : 1 ; A r : - 0 . 5 1 ir, x: -3; Ar: -0.1 In Exercises 13 through 20, use differentials to find an approximate value for the given quantity. Express each answer to three significant digits. 13.\/m 14.v.s $. vn 12.fd00b% 18. \/M. 19. 1 YL2O 16. \/{2 20. 1 ffiF 21. n:asureTent of an edge of a cube is found to be 15 in. with a possible error of 0.01 in. Using differ€ntials find lle aPproxrmate tne er()r in comPutins ftom this measuremmt (a) the volume; (b) the area of one of the faces. 22' The altitude of a right-circular con€ is tvdce the radius of the base. The altitude is measured as 12 in., with a possible error of 0.005 in, Find the approdmate error in the calcul,atedvolume of the cone. 23' An-open rylindrical tank is to have an out8ide coating oI thickness I in. If the inner radius is 6 ft and the altih.rde is 10 ft, find by differentials the apprcximate amount of coating material to be used. 24 A metal box in the form of a cube is to have an interior volune of 64 in.s. The six sides are to be madeof metaltin. thick. If the cost of the metal io be used js I cents per cubic inch, use differentiats to find the app;imate cost of the metal to be u6ed in the manufacturc of the box. 25 If the possible error in the measurement of the volurne of a gas is 0.1 fte and the allowable error in the pressure is 0.001Clb/ff' find the size of the smallest container for which !o/e's law (Exerase rO in p*erc-ises i.ry no6". ' A conh'actor agrees to paint on both sides-of 1000_ circular signs each of radius 3 ft. Upon receiving the signs, it is discover€d that the radius i8 I in' too large. use differentials to- find the approximat" p'.i*"i it of pilni ttrat wltt be needed. "*"'"" 27' The measure of the electrical reefoblT o{." wire is proportional to the measure of its length and inversely proportional to the square of the measure of its diameter. Suppose the resistance of a wite of givei length is computed from a measurement of the diaureter l4iith a possible 2% i:rror. Find the possible percent Jror in th-e computed value of the resistance. * the for.?ne c-oTplete swing of a simple pendulum of length I ft, then 4?l t: gp, whereg: 32.2.A clock y ls]e ij: "a Pendulum naung of lmgth 1 fi gains 5 min eadr day, Find the approximate amount by whictr the pendulum should b€ Lengthen€d in order to correct the inaccuracv. 6.2 DIFFERENTIAL FORMULAS 29. For the adiabatic Law for the expansion of air of Exercise12 in Sec.3.10,prove that dPJP:-1.4 dVlV. 30. Draw a figure sirnilar to Fig. 6.1.1, if the graph is concave downward. Lrdicate the line setments whose lengths represmt the following quantities: A:, Ay, dt, and dy. 6.2 DIFFERENTIAL FORMULAS Supposethat y is a function of r and that r, in turn, is a function of a third variable f; that is, y:f(x) and x-gG) (1 ) The two equations in (1) together define y as a function of f. For example, suPPosethat y : r! and x:2tz - 1..Combining these two equations,we get y - (2t2- 1)t. In general,if the two equationsin (1) are combined,we obtain v: f k(t)) (2) The derivative of y with respect to f can be found by the chain rule, which yields DtA : D;y Dp (3) Equation (3) expressesDg as a function of r and f becauseD"y is a function of r and Dg is a function of f. o rLLUsrRATroN 1,:If.y: rl and x:2t2 - 1.,then DtA : Dry DP :3x2(4t) -- LLxzt o BecauseEq. (2) defines y as a function of the independent variable t, the differential of y is obtained from Definition 6.1.1: dy - D,y dt (4) Equation (4) expressesdy as a function of t and dt. Substituting from (3) into (4), we get dy: D,y D6 dt (s) Now because r is a function of the independent variable t, Definition 5.1,.Lcan be applied to obtain the differential of.x, and we have dx - Dtx dt (6 ) Equation (5) expressesdr as a function of t and df. From (5) and (5), we get dY: DrY dx (7 ) You should bear in mind that in Eq. (4 dy is a function of f and dt, and that '(r), we have dx is a function of t and dt. Tf.D;y in Eq. (7) is replacedby f dy: f'(x) dx (8) THE DIFFERENTIAL AND ANTIDIFFERENTIATION Equation (8) resembles Eq. (a) of Sec. 6.1. However, in that equation r is the independent variable, and dy is expressed in terms of r and dx, whereas in Eq. (8) f is the independent variable, and both dy and dx are expressed in terms of f and dt. Thus, we have the following theorem. 6.2.L Theorem If y : /(r), then when f '(x) exists, dy: f'(x) dx whether or not r is an independent variable. If on both sides of Eq. (8) we divide by dx (provided dx + 0), w€ obtain f ' ( x ): -dY dx dx*O (e) Equation (9) statesthat if y: f (r), then f '(x) is the quotient of the two differentials dy and dx, even though r may not be an independent variable. We now proceed to write the chain rule for differentiation by expressingthe derivativesas quotients of differentials.If y: f (u) andDuy exists,and if u : g(x) and D, rzexists,then the chain rule gives D r A : D u yD * u (10) But D*y:dyldx rf dx*0; Duy:dyldu rf du*0; dx * 0. From (10),therefore,we have andD*u:duldx if x-{#){#) if du#0and dx*0 Observe the convenient form the chain rule takes with this notation. The German mathematician Gottfried Wilhelm Leibniz (7646-1716) was the first to use the notation dyldx for the derivative of y with respect to r. The concept of a derivative was introduced, in the seventeenth century, almost simultaneously by Leibniz and Sir Isaac Newton (1,642-7727), who u/ere working independently. Leibniz probably thought of dx and dy as small changes in the variables r and y and of the derivative of y with respect to I as the ratio of dy to dx as dy and dx become small. The concept of a limit as we know it today was not known to Leibniz. Corresponding to the Leibniz notation dyldx for the first derivative of y with respect to x, we have the symbol dzyldxz for the second derivative of y withrespect to x. However, d2yldxz must not be thought of as a quotient because in this text the differential of a differential is not considered. Similarly, d"yldx" is a notation for the nth derivative of. y with respect to r. The symbolism f ' , f ", f " ', and so on, for the successive 6.2 DIFFERENTIAL FORMULAS 251 derivatives of a function / was introduced by the French mathematician Joseph Louis Lagrange (1736-1,813) in the eighteenth century. Earlier in Chapter 3, we derived formulas for finding derivatives. These formulas are now stated using the Leibniz notation. Along with the formula for the derivative, we shall give a corresponding formula for the differential. In these formulas, u and T are functions of. x, and it is understood that the formulas hold if Dru and Dra exist. When c aPPears, it is a constant. r dax! r ) : o ,, !#_ I ' d ( c ): g Il' d(x") - nrn-r dx nlcn-1 ( " u :) , 4 u l d dx ax Ill'd(cu):cdu ( u: *Ea-) E :d, ,, rU , -dT *d, IV'd(u*v):du*da (Ya):u4+o4 v d dx flx ilc __ Yr . /u\ dt-t \al _ dx du an-u a u2 run-,d!- -, !#- da d, V'd(ua):uda*adu vr o(+):a d u - u d a VII' d(u") - nun-r du The operation of differentiation is extended to include the process of : finding ttre differential as well as finding the derivative.If y f (x) , dy can ' be found either by applying formulas I'-VII' or by finding f (x) and multiplying itbY dx. rxluplr Y: find dy. 1: Given Applying formula VI', we obtain solurroN: \m a,., y -: z;+l (zx+ D a(F+ t) - {V + t a2x+ t) n From formula VII' d6/V+ L ) : t ( x ' * 1 ) - r r z2 x d x : x ( x 2* L ) - r r z6 * (L2) and d(2x*L)-24, Substituting values from (12) and (13) into (L1'),we get (1 3) AND ANTIDIFFERENTIATION THE DIFFERENTIAL x ( Z x* t ) ( r ' z* t ) - t r z d x - 2 ( x z* t ) r r zO , a) -y. _: _ (2x2+ x) dx 2oc2+ 1,) dx ( 2 x* L ) ' ( x ' * 1 ; t r z - EXAMPLE 2: Given Zx'y'- 3x3* 5y" I 6xy2:5 where r and y arefunctions of a third variable t, find dyldx by finding the differential term by term. x-2 ,(2x]-7)z{x;4 1*^ solurroN: This is a problem in implict differentiation.Taking the differential term by term, we get 4xy2 dx * 4xzy dy - 9x2dx * 15y2dy * 6y2 dx * 1,2xydy : O Dividing by dx, if dx * 0, we have (4x'y t 15yz+ 12xy) #: -4*y' * 9x2- 6y' dy_: 9xz- 6y2- 4xy2 dx 4x2y11,5y2* I2xy Exercises 6.2 In Exercises1 through 8, find dy. l.y: +, 3. yx : (3x,-2x*l)3 2'Y:\/4-F 3.y:*g2x+3 xr+2 5- . y : 6,y: /.y:ffi 2x lx_l Vr+1 (x+2)ttlbc-2)2t3 8. y: $x+ 4f/71 In Exercises9 through 76, x and y are functions of a third variable f. Find dyldx by finding the differential term by term (seeExample2). 9. 3xs* 4yz-- 48 L 0 . 8 x 2- ! 2 : 3 2 11.t/i+ {y:+ 1 2 . L x z y- 3 x y " * 6 y 2 : 1 1 3 .x a - 3 x " y * 4 x y 3 * y n : 2 .1,4. L5. 3r3 - x'y * 2xy2- y3 - 3x2* A2:'!. In Exercises17 through 24, find dyldt. '1.6. xzI y': +,C+ V 17. y : 3r 3- 5x 2i l ; x : P - 1 . 1 8 .y : 19. y : x2- 3x t- 1; 2s: \/F=j + 4 x2t3 I yzt:t _ fl2t3 4v-) x : ( 2 t- 1 ) ' ff; 20.y: S6;- 1;x: t/2tTJ 21. y - x2- 5x * t;r : s3- 2s * 1; s : t/T+-t 22.y:ft,r:ffi;s:tz-4t+5 23. x3- 3x'y * y" : 5; xs: 4tz + 1. 24. 3xzy- 4xy' l7y": 0; 2x3- 3xtzI t3:'J. 6.3 THEINVERSE OF DIFFERENTIATION 253 5.3 THE INVERSE OF You are already familiar with inuerse operations. Addition and subDIFFERENTIATION traction are inverse operations; multiplication and division are also inverse operations, as well as raising to powers and extracting roots. The inverse operation of differentiation is called antidifferentiation. 5.3.L Definition A function F is called an antideriaatiae of a function f on an interval I if F' (x) : f (x) for every value of r in /. o rL L U s rR A rIo N 1: If F i s defi ned by F(x):4xs]-x2+ 5, then F' (x)l2x2 * 2x. Thus,if f is the function defined by f (x) : L2f * 2x, we state that f is the derivative of F and that F is an antiderivative of f .lf G is the function defined by G(r) - 4xB* x2 - t7, then G is also an antiderivative of f because G'(r) : l2x2 * 2x. Actually any function whose function value is given by 4x3* x2 * C, where C is any constant, is an antiderivao tive of /. In general, if a function F is an antiderivative of a function / on an interval I and if G is defined by G(r):F(r)+C where C is an arbitrary constant, then G' (x) : F' (x) : f (x) and G is also an antiderivative of / on the interval I. We now proceed to prove that if F is any particular antiderivative of / on an interval /, then all possible antiderivatives of / on I are defined by F(x) + C, where C is an arbitrary constant. First, two preliminary theorems are needed. 5.3.2 Theorem If f is a function such that f ,(x) : 0 for all values of r in the interval I, then / is constant on I. pRooF: Let us assume that is not constant on the interval I. Then there / exist two distinct numbers rr and x, in I whetl x1( r, such that / (xt) * : 0 for all r in I, then f' (x) - 0 for all r f (xr). Because,by hypothesis, f' (x) in the closed interval lxr, xr]. Hence, / is differentiable at all x in lxr, x2l and/ is continuous on [xr, rz]. Therefore, the hypothesis of the meanvalue theorem is satisfied, and we conclude that there is a number c, with x, 1c 1*r, such that (1) '(c): 0, and But because/'(r) : 0 for all x in the interval lxr, rr], then f : from Eq. (1) it follows that f (rt) f (xr). Yet our assumption was that (*t) (xr). and so / is constanton f, a contradiction, Hence, there is + f f which is what we wished to prove. AND ANTIDIFFERENTIATION THE DIFFERENTIAL 6.3.3 Theorem If f andg are two functions such that f , (x) : g' (x) for all valuesof r in the interval I, then there is a constant K such that f(x) :g(r) + K for all r in I pRooF: Let h be the function defined on I b v h(x): f (x)- SG) so that for all values of r in l we have h'(x):f' (r) -g'(x) But, by hypothesis,f '(r) : g'k) for all values of r in /. Therefore, h'(x) : g for all valuesof r in I Thus, Theorem 6.3.2 applies to the function h, and there is a constant K such that h(x) : Y for all values of x in I Replacingh(x) by f (x) - S Q), w e have for all values of r in / f (x) : g(x) + K and the theorem is proved. The next theorem follows immediately from Theorem 6.9.2. 6'3'4 Theorem If F is any particular antiderivative of / on an interval l, then the most general antiderivative of / on I is given by F ( r )+ C e) where C is an arbitrary constant,and all antiderivatives of on/can be obf tained from (2) by assigning particular values to C. PRooF: Let G be any antiderivative of on f G'(x) : f (x) on / then Because F is a particular antiderivative of / on .1,we have F'(x):f(x) onl @ From Eqs. (3) and (4), it follows that G'(x) -- F'(x) on I Therefore, from Theorem 6.9.9, there is a constant K such that G(x) : F(r) + K for all x in I Because G is any antiderivative of / on 1, it follows that all antiderivatives of f can be obtained from F(x) + c, where c is an arbitrary constant. Hence, the theorem is proved. I 6.3 THEINVERSE OF DIFFERENTIATION 255 If F is an antiderivative of f, then F'(x) : f (x), and so d(F(x)):f (x) dx Antidifferentiation is the process of finding the most general antiderivative of a given function. The symbol t J denotes the operation of antidifferentiation, and we write f (s) I f (r) dx: F(r) + C J where F'(r) : f (x) or, equivalently, d(F(x)) : f (x) dx From Eqs. (5) and (6) / we can write f (6) v) ld(F(x)):F(x)+c J Equation (7) statesthat when we antidifferentiate the differential of a function we obtain that function plus an arbitrary constant. So we can think of the f symbol for antidifferentiation as meaning that operation which is the inverse of the operation denoted by d for finding the differential. Becauseantidifferentiation is the inverse operation of differentiation, we can obtain antidifferentiation formulas from differentiation formulas. We use the following formulas that can be proved from the corresponding formulas for differentiation. 6 . 3 . 5F o r m u l a l I d * - x f C 6.9.6Formula 2 I af(x) dx: alf@) dx,whete a is a constant Formula 2 states that to find an antiderivative of a constant times a function, first find an antiderivative of the function, and then multiply it by the constant. 6.s.7Formula3 I lf'(x) + f"(x)f dx: If'k) dx* [fr(x) dx Formula 3 states that to find an antiderivative of the sum of two functions, find an antiderivative of each of the functions separatelY, and then add the results. It is understood that both functions must be defined on the same interval. Formula 3 can be extended to any finite number of functions. Combining Formula 3 with Formula 2, we have Formula 4. AND ANTIDIFFERENTIATION THE DIFFERENTIAL 5 . 3 . 8F o r m u l a4 6.3.9 Formula 5 + c " f " ( x ) fd x I [ c t f t ( x )+ c r f r ( x )* ' ' ' : h[ f t(x) dx * cr[fr(x) dx + " :; ' + c"[ f"(x) dx 'tr'*? d**dfo'+ c irn* *r As stated above, these formulas follow from the corresponding formulas for finding the differential. Following is the proof of Formula 5: / ryn+1 "!.)xn (n I _\ 7 d l - : - + C l : \ '!)#: d x n*1-a \/rJ-, :xndx Applications examples. EXAMPLE 1: Evaluate f | (st +5) dx J of the above formulas are illustrated in the following SOLUTION: rfr | (3r + 5) dx: 3 | x dxt 5 | dx (by Formula4) JJJ /vz :3 (i \z \ + C , I + 5 ( r + C r ) ( b y F o r m u l a s5 a n d 1 ) / : Ex'* 5r + 3Cr+ 5c2 Because3Cl + 5C2 is an arbitrary constant, it may be denoted by C, and so our answer is *x2+5xtC This answer may be checked by finding the derivative. Doing this, we have D * ( z r x*25 r + C ) : 3 r * 5 rxauplr 2: Evaluate solurroNt I vP dx: | ,',, 4* JJ Igpa* : J )c213+r 3.. , + C (from Formula 5) : P,xst3 + C nxawrrr,n 3: Evaluate I(++S\a. solurroN ' I $. +) dx-- a-^ * *-rr+)dx [ y-{-r I y-l r I t a 14+ | -4+1.'-*+r ' 6.3 THE INVERSEOF DIFFERENTIATION r-3,x314,^ - J 4^ l 3 l \ . :- 1 =q ! * t , n a g 3x3' 3^ Many antiderivatives cannot be found directly by applying formulas. However, sometimes it is possible to find an antiderivative by the formulas after changing the variable. 2: Suppose that we wish to find . TLLUSTRATTON f_ J If we make the substitution u: f J (8) zxt/t + x'zdx "L* x2, then du:2x dx, and (8) becomes uttzdu which by Formula 5 gives tustz* C Then, replacingu by (1 + r2), we have as our result o 3 ( 1+ r ? ) B I Z + C fustification of the procedure used in Illustration 2 is provided by the following theorem, which is analogousto the chain rule for differentiation, and hence may be called the chain rule for antidifferentiation. 6.g.'l,0Theorem Letg be a differentiable function of r, and let the range of g be an interval I. ChninRule Supposethat / is a function defined on / and that F is an antiderivativeof / forAntidifterentiation on /. Then if u: gG) , ff I f G k D y . l - )d x : JI' f f u ) d u : F ( u )+ c - F ( s ( r ) )+ c J' u: g(r), then z is in PRooF: Because tive of f on l, it follows that and because F is an antideriva- d F ( u \- . / 'f'(\ u ) du Irut F ( u )+ C Also, dF(sk)) _ dF(u) dx dx (10) ( 1 1) THE DIFFERENTIAL AND ANTIDIFFERENTIATION Applying the chain rule for differentiation to the right side of Eq. (11), obtain d F ( g ( x ) )_ d F ( u ) . d u dx du dx Substituting from Eq. (9) into (12) gives dF(sk))_ .. ' Edu tlu) dx Because u: gQ), from Eq. (13) we have d F ( s . @:) ) f e(r)) . 8,(r) dx from which we conclude that r ' 8 ' G )d x : F ( 8 ( r )+) c J f k@) Becauseu: g(x), we have F(g(x))+C:F(u)+C Therefore,from (10), (I4), and (15), we have fr dx: J f G @) g ' ( r ) J f O t d u : F ( u )+ C : F ( g ( x ) )+ C which is what we wished to prove. I As a particular case of Theorem 6.3.10, from Formula 5 we have the generalized power formula for antiderivatives, which we state as Formula 5. 6.3.1,1Formula 6 If g is a differentiable function, then if u: ffi;ffi gG) , [*{*1"*'+c Tff' where n # --L. Examples4, 5, and 5 illustrate the application of Formula 6. rxavrprr r_ 4: Evaluate I t/3x+4 dx J soLUrIoN: To apply Formula6,we makethe substitution u:Bx * 4; then du : 3 dx, or tdu: dx. Hence, Ia,+t dx-[u,,,+ : + / u' Itzdr' t 6.3 THE INVERSEOF DIFFERENTIATION 259 1 u 3- +1a 2 , ^ -a3l\/ J2 : $ystz* C : g(Jx * 41trz* , The details of the solution of Example 4 can be shortened by not explicitly stating the substitution of z. The solution then takes the following form. JJ dx - + [ fg* -r 41r,g o* (3x*4)3t2 *" _ 1 . ---=- - - fi(3x * 41zrz * , EXAMPLE 5: Evaluate f I t(s + 3tz)8dt J sor-urroN: Since d(5 + 3t2)- 6t dt, we write rf I r ( s + g t 2 ) 8 d t : i | ( 5 + 3 t ' z ) 8 5dtt JJ ( 5 + 3 P ) ' g+ c _ - 6! . 9 : #(5 + 3t2)e+ c rxavrpr.r 6: Evaluate f - l l\Y7-4x3dx J r-f xrr/7=B J dx: -# : -*. nxenaprn 7: f_ Evaluate I x2t/'j.,+ x dx J J (7 - 4x3)u5(-\hc2)dx e - xs)o/s-r-g then t:2:'L * x. Hence, r : sor,urroN: Let a : GT; dx:2u fut. Making these substitutions,we have .a . (2a da) | *r ,6,, * dx: | @r- 1), JJ :2 _ J[ a6da - 4 | t:adu + 2 | a2da J J :?a7-fu'+fut+C : ?(t * x)ztz- +(1 * x)srz+ ?(1 * x)zrza , and AND ANTIDIFFERENTIATION THE DIFFERENTIAL 260 The answers to each of the above examples can be checked by finding the derivative (or the differential) of the answer. o TLLUSTRATToN 3: In Example 5, we have r I t(S + 3tz)Bdt : # (5 + 3tz)e+ C J Checking by differentiation gives D r [ # ( 5 + 3 t 2 ) e ]: ; b ' 9 ( 5 + 3 t 2 ) 8' 6 t :f(5+3t2)t . o TLLUSTRATToN 4: In Example 7, we have f_ I x 2 ! 1 , * x d x : + ( 1 + x ) z n - * ( 1 * x ) s r z + 3 ( 1* x ) s r z * , J Checking by differentiation gives D"l+(L* x)zrz-e(1* x)srz+3(1* x)erz1 : ( 1 * x ) s r -z2 ( l * x ) s r z+ ( 1 * x ) t r z : ( 1 * x ) r r z[ ( 1 + x ) r - 2 ( t + r ) + 1 ] : ( 1+ _ A : . l L + 2 x * x z - 2 - 2 x + t l -*zfay Exercises6.3 In Exercises L through26, of your answer. find the most general antiderivative. In Exercises 1 through 10, check by finding the derivative f 1,. 3xadx J 4. f J@*+bx*c) z. to. f J f J (x'- 4x+ 41ut4, F- t.JV3-xxzdx f_ 8I%+d, dx 3. 22. V3 + s(s* 1,)2ds J "Iwh ,n. t. xa!3x3-5 dx ,,I# ,, t . f_ J I (r, * 3)rn*so* ,'\ffi,, r dt JG-zt+t2) 6 J[t:#1 vy ^-_ xV(4- x2)2dx B[J;4# ,u. ' IG***u) * dx ,l(*-#\* IUr"-3xz*6x-r) o, r_ ,'\/x3 -'l".dx J f- J !5r*r dr f 15. x'zG- xz)rdx J " Itlr.,1 | Q't 2x) dx -'' J !x3 1- 3x2+ 1 24. f J {zt't llrrzssO, ,u[ff* 2T.Evaluate!(2xtl)sdxbytwomethods:(a)Expand(2r*1)gbythebinomialtheorem,andapplyForrrulasl,4,and in (a) and (b)' 5; (b) maie the substituiion u = 2x + L ixplain the difference in appearanceof the answersobtained o = V7- 1. 2g. EvaluateJ \/r - | * d.rby two methods:(a) Make the substitution u: x 7; (b) rrake the substitution 29. Letf(x):1 for all x in (-r, 1) andlet Fl. il-1 < r s0 8(rr:11 if0<r<1 f or all r i n (-7 ,1 ) a n d 8 ' (r) :0 Th en f'( x ) : 0 (-1, 1). Why doesn't Theorem6.3.3hold? 3 0. L et f(x) : s gn r and F (x ): l rl .Sh o w th a t F ' (r): w h enever3' existsin (-1, L). However,f(x) + g(r) + K forr in of /on (-m , +-)? Explain. f(x ) i tx * O.Is F an anti deri vati ve 31. Let f (x): lrl and let F be definedby (-lf ifr<o F('):l+o ifr>o Show that F is an antiderivative of / on (-m, +oo)' 1.8) does not have an antiderivative on (-6' +@)' (HIvr: 32. show that the unit step function u (Exercise 21 in Exercises is obtained by showing that it fouows from Assume that U has an antiderivative F on (--, +-), and a contradiction thegrean-valuetheoregrthatthereexistsan.umber'&suchthatF('):'+kifr>0,andF(r):kifr<0.) 5.4 DIFFERENTIAL If F is a function defined by the equation EQUATIONS WITH a:F(x) vARIABLES sEPARABLE and f is the derivative of F, then fu--r(x\ ''\ (1) (2) dx and F is an antiderivative of /' Writing Eq' (2) using differentials, we have (3) dy: f(x) dx Equations (2) and (3) are very simple types of differential,equations' They aie differential equations of the firstorder (becauseonly derivatives To of the first order are involved) for which the variables are separable' : the that so G(r) which for G functions A solve Eq. (3) we must find all G are equatioi is satisfied. So if F is an antiderivative of /, all functions is' if That constant' arbitrary : an C is where C, F(x) + aefined by G(r) solution complete the called is what (x) then dx, d(G(x)) d(F(x) + C)'- f of Eq. (3) is given bY y-F(x)+C (4) THE DIFFERENTIAL AND ANTIDIFFERENTIATION Equation (4) represents a family of functions depending on an arbitrary constant C. This is called a one-parameter amily. The graphs of these f functions form a one-parameter famity of curves in the plane, ind through any particular point (xr,yr) there passes just one curye of the famiiy. o ILLUSTRATTON 1: Suppose we wish to find the complete solution of the differential equation dy -2x dx (5) The most general antiderivative for the left side of Eq. (5) is (y * C,) and the most general antiderivative of 2x is (r2 * Cr).Thus, we have A * Ct: x 2+ C , Because(C" - Cr) is an arbitrary constant if C, and C1 are arbitrary, we can replace (Cz - Cr) by C, thereby obtaining 1 Q: CC- F i g u r e6 . 4 . 1 o -1 -4 Y:x2+C (6) which is the complete solution of the given differential equation. Equation (6) represents a one-parameter family of functions. Figure 6.4.1' shows sketches of the graphs of the functions corresponding to C:-4, C:-L,C:0, C:1, and C:Z. Often in problems involving differential equations it is desired -. to find particular solutions which titirfy certain .onditior.,s called boundary conditions ot initiql conditians. For example, if differential equation (3) is given as well as the boundary condition thaty: y, whe rr x: xllthen after the complete solution (4) is found, if r and y in $) are replaced by r, and Au a particular value of C is determined. When this value of C is substituted back into Eq. (4), a particular solution is obtained. ' rLLusrRArIoN 2,: To find the particular solution of differential equation (5) satisfying the boundary condition that y: 6 when : x Z,we substitute th e s e v a l u es i n E q. (6) and sol ve for c, gi vi ng 6:4+ c, or c:2. substituting this value of C in (6), we obtain U:x2+2 which is the particular solution desired. Another type of differential equation is dra ffi:ftx) (7) Equation (71isa differential equation of the secondorder,Two succes. sive antidifferentiations are necessaryto solve Eq. (T), and two arbitrary constantsoccur in the complete solution. The complete solution of Eq. 1i1 thereforerePresentsa two-parameter family of functions,and the graphsof these functions form a two-parameter family of curves in the plane. EQUATIONSWITH VARIABLESSEPARABLE 6.4 DIFFERENTIAL L: Find the complete EXAMPLE solution of the differential equation d2u *:4x*3 ax' solurroN: Becausedzyldx'* dy'ldr, the given equation can be written AS du' ax Writing this in differential form, we have dy': (4x+3) dx Antidifferentiatin 8, w e have rf ldy':l(4x+3)dx J Jfrom which we get i , , A' :2x2 * 3x* Ct Becausey' -- dyldx, we make this substitution in the above equation and get +ax: 2 x 2 + 3 x * c , Using differentials, we have dU: (2x2+ 3x * Cr) dx Antidifferentiatin g, w e obtain ff I dv- l (zx'* 3r * c,) dx J- J from which we get A:3x3+hcz*Crxi_C, which is the comPlete solution. BxauprE 2: Find the Particular solution of the differential equation in ExamPle L for which A :2 and y' - -3 when tc: L. soLUrIoN: Because y' : 2x2+ 3x * C1,we substitute-3 for y' andl for x' - -8. Substituting this value of Ct into the giving -3 - 2 + g * Cr, or Cr complete solution gives U:&x3+tx'z-8x*C, Becausey :2 when x: L, we substitute these values in the above equag- 8 + C2,fromwhich we obtaif|Cz: atz.The partiction and get2:3+ ular solution desired, then, is A:3x3+*x2-8r++ THE DIFFERENTIAL AND ANTIDIFFERENTIATION form A more general type of differential equation of the first order is of the -dv : dx f(x) sQ) (8) For such equations,the variablescan be separatedby multiplyirrg on both sidesof the equationby g(y) dx; thus, Eq.(g) can be writienias g(y) dy : f (x) dx The following example gives us an equation of this type. ExAMPLE3: The slope of the tangent line to a curve at any point (x, y) on the curye is equal to 3*'y'. Find an equation of the curve if it contains the point (2, l) . G";i.h" ,"" soLUTroN: Because the (x, y) is the value of the derivative at that point, we have l du r' : 3x'u' a)c This is a first-order differential equation, in which the varia plescan be separated,and we have du u' errtiAifferentiating, we have rr lY-'dy-l3x'dx JJ from which we obtain -;:1 x3+ c T-his rrt'r'DrD is 4rr an equarlon equation or of a one-parameter one-parameter tamily family of curyes. curves. To To fin{ the the particpartic- ular curve of this family which containsthe point (2,1),*; ;;br;itute 2 for -LI _: Og T* Lc,, Ifrom r-and 1 for Jy,' which r v L . L - L . gives us LrD rOm 6rvsD W which nl we- obthin - - "f' C: -9. Therefore, Therefore, the the required required curve curve has has the the equation equation -+:r-e Exercises 6,4 ln Exercises 1 through 8, find the complete solution of the given differential equation. -dv du - L. d x : 3 x z* 2 x 7 Z. #: ( 3 x+ 1 ) t 3. du E:3xy2 AND RECTILINEAB MOTION 6.5 ANTIDIFFERENTIATION ds dt:5Vs , - dy d2a 7.-?+:5x'l7 8. #: 4. J . d2u dr du 3xt/TT! t/i+ 6' dx:fr- 1 axv x v YZx-3 In Exercises 9 through 14, for each of the differential equations find the particular solution determined by the given boundarv conditions. f. n du i:x2-2x-4;y:-6 dx 4da ;:t;y:-z tO #: w h e nx : 3 ( x* 1 )( r + 2 ) ;y : - + w h e nx : - 3 t.H:ffi;Y:owhen w h e nx : 4 ,!2,, y : -1andy': -2 whenx: -1 s. ffi: ut )- 3x)2; dzU u |rt:Azx-t;v:2ar.dv':5 x:L whenx:3 2r - 3. Find an 15. The point (3, 2) is on a cuwe, and at any point (n y) on the curve the tangent line has a slope equal to equation of the curve. ) an equation 16! The slope of the tangent line at any point (r, y) on a curve is 3Vi. If the Point (9, 4) is on the curve, find -.cf the curve. :2 - 4x Find an equation of 17j'fn" poir,t" (-1, 3) and (0, 2) are on a curve, and at ary Point (r, y) on the curve D ,'y -' the curve. -1: y) on the curve, D"'zy : 5a, 18. An equation of the tangent line to a curve at the point (1, 3) is y a 2. If at any Point (t, find an equation of the curve. and an equation of the tangent line to the curve at the Point (1, 1) is 19. At any point (r, y) on a (j(rwe,Dx2y:1-i, ! : 2 t- Find an equation of the curve' the slope of the inflectional tangent 20. At any point (r, y) on a cuwe, D"? : 2, and (1, 3) is a Point of inflection at which is -2. Find an equation of the curve. of another one-parameter 21. The equation f :4oy represents a one-parametef family of parabolas. Find an equation and the tangent lines to the two it through family cuwe oi eadr y) there is a point (rl at any family of curves such-thai any Point (r, y), not on the line at of the tangent the sloPe that (nrr.rr: First show p"rp".rdlcrrlu". u"e point at this curves y axis, of the paraboLa of the given family through that point is 2ylx ) y": a"' 22. Solve Exercise 21 if the given one-Parameter family of cuwes has the equation '" + 6.5 ANTIDIFFERENTIATION AND RECTILINEAR MOTION We learned in Secs.3.2 and 3.11 that in considering the motion of a particle along a straight line, when an equation of motion, s: f (t), is given, then the instantaneous velocity and the instantaneous acceleration can be determined from the equations u:Dls-f'(t) and a:Dtzs:Dtu-f"(t) Hence, if we are given o or fl aS a function of f , as well aS Some boundary conditions, it is possible to determine the equation of motion by antidifferentiation. This is illustrated in the following example. THE DIFFERENTIAL AND ANTIDIFFERENTIATION ExAMPLE 1.: A particle is moving on a straight line; s is the number of feet in the directed distance of the particle from the origin at f sec of time, u is the number of ftlsec in the velocity of the particle at f sec, and a is the number of ft/sec2 in the acceleration of the particle . I andv :3, a t f s e c .l f n : 2 t and s : 4 when t: t, express u, and s as functions of f. soLUTroN: Becausea: D{s, we have d'-Jt-1 dtr which, expressed in differential form, gives dzs: (2t - t) dt Antidifferentiatin g, we have rr IJ J d u : I e t - r ) d t from which we get a:P-t+Cl Substitutinga: 3 and t: (1) L, we have 3:1,-L+Cl :il*rT*€ cr :3. Therefore, substitutingthis valueof c, intf Eq. (1),we u:t2-f+3 (2) which expresses? as a function of f. Now, letting u : D 7sin Eq. e), *e have +d t :e LP - t + Z and writing this in differential form gives ds:(tr_f+3)dt Antidifferentiating, we have rt dt: | (tr-t+s) dt I JJ s:*ft-ttr+gt+Cz (3) Letting s:4 and f : f. in Eq. (3) and solving for Cr, we o$tain Cz: t. Therefore, by substituting 6 fot C, in Eq. (3f we have s exfressed as a function of t, which is s:*fa-+tr+gt++ If an object is moving fueelyin a vertical line and is leing pulled toward the earth by a force of gravity, the accelerationof grorlty, denoted by g ftfsecz,varies with the distanceof the object from ti" dur,t"r of the earth. However, for small changesof distancesthe acceleratiqnof gravity AND RECTILINEAR MOTION 6.5 ANTIDIFFERENTIATION is almost constant, and an approximate value of g, if the object is near sea level, is 32. EXAMrLE 2: A stone is thrown vertically upward from the ground with an initial velocity of L28 ftlsec. If the only force considered is that attributed to the acceleration of gravity, find how high the stone will rise and the sPeed with which it will strike the ground. Also, find how long it will take for the stone to strike the ground. so1,urroN: The motion of the stone is illustrated in Fig. 6.5.1.The positive direction is taken as upward. Let t: the number of seconds in the time that has elapsed since the stone was thrown; s: the number of feet in the distance of the stone from the ground at / sec of time; u : the number of feet per second in the velocity of the stone at f sec of time; lrl: the number of feet per second in the speed of the stone at f sec of time. The stone will be at its highest point when the velocity is zero. Let s be the particular value of s when o : 0. When the stone strikes the ground, s : 0 . L e t f a n d a b e t h e p a r t i c u l a r v a l u e s o ff a n d p w h e n 5 : 0 , a n d t + 0 . Table 6.5.1 is a table of boundary conditions for this problem. The positive direction of the stone from the starting point is taken as upward. Because the only acceleration is due to gravity, which is in the -32 ft/sec2. downward direction, the acceleration has a constant value of Because the acceleration is given by the derivative of o with respect to f, r'T{:=: we have da dt -32 Writing this in differential form, we have da : -32 dt f,:7 1r:o lo: a F i g u r e6 . 5 . 1 Antidifferentiatingr w€ have I a,: [ -2,a' from which we obtain a:-32t*Ct Becauseu : L28 when t : 0, we substitute these values in the above and get C1:'l'28. Therefore, we have a: -32t + L28 (4) Because u is the derivative of s with respect to f, then a : dsldt, and so T a b l e6 . 5 . 1 t s 7) OT 0s0 I280a dc -32t + 128 :dr: which in differential form is ds: (-32t + L28) dt THE DIFFERENTIAL AND ANTIDIFFERENTIATION Antidifferentiating gives I*:l (-gzt + I28) dt from which we obtain s: -L6t2 + 128t* C, B e c a u sse: 0 w h e n t : A , we get Cz:0, and substituting 0 for C2in the above equation gives s: -L6t2 + L28t Substituting,tfor t and 0 for s, we get 0: -L6l(i- 8) from which we obtain l: 0 and f : g. However, the value 0 occurswhen the stone is thrown; so we concludethat it takes8 secfor the stone to strike the ground. To obtain a, substitute 8 for f and a for a in Eq. (4), which gives 6: (-32X8) + 128,from which we obtain a: -L29. So l7l : L2g.There_ fore, the stone strikes the ground with a speed of 1.2gftlsec. To find i we first find the value of t for which a is 0. From Eq. (4), we obtain t:4 when u:0.In Eq. (5) we substitute4 for f and s fbr s and obtain s:256. Thus, the stone will rise 256 ft. Exercises6,5 In Exercises 1 through 4, a particle is moving on a slraight line, s ft is the_directed distance of the paticle from the origin at I 8ec of time, a ft/eec is rhe velocity of ttre palticle ati sec, and4 ft/s€c ;;;;;";1;;;;ir,1*p'"1"t "*,.". 1. a:5-Zt; a:2 and s:0 when t:0. Express? and s in terms of ,. 2. a-3tf,; p:f 4rd s:1when t:1. Expressz and s in terns of t. 3. c=8@; rr:20whens:1. Find an equationinvolving - o "rrd r. /"rr.rr, \ dt dsdt "=+:4*:"@\ - ds) Find an equation involving u and s. (nrr.rr:SeeHint for Exercise 3.) In Exercises 5 throush 10' the orilv force considered is that due to the acceleration of gravity, which we tak e as 32 ftlsec in the downward dir=ectiorr4. a:2s + L u:2 when s:1. 5' A stone i: vertically upward from the gound with an initial velocity of 20 ft/eec. lf*Y" How long will it take the t9--etrike the ground, and with what speid witt it strike? How tong r,"ilt th. :tot-rc g.lrrj and how high wil it go? "t"""i" "p-"ra, 6' A ball.is dropped from the top of the was_hington monument, which is 555 ft high. How long will it take the ball to readt the groutld, and with what speed will it strike the glound? 7' A man in a balloon droDs his binoculars when it is l5o ft above the gound and rising at the rate of l0 ttlsec. How long will it take the bin&dare to strike the gouna, -fr"t i, if,Ar Jpeea on i-paci ".a IN ECONOMICS 269 OF ANTIDIFFERENTIATION 6.6 APPLICATIONS 8 . A stone is thrown vertically upward from the top of a house 60 ft above the ground with an initial velocity of 40 ftlsec. At what time will the stone reach its greatest height, and what is its greatest height? How long will it take the stone to pass the top of the house on its way down , and what is its velocity at that instant? How long will it take the stone to strike the ground, and with what velocity does it strike the ground? 9. A ball is thrown vertically upward with an initial velocity of 40 ftlsec from a point 20 ft above the ground. lf a ftlsec is the velocity of the ball when it is s ft from the starting point, express o in terms of s. What is the velocity of the ball when it is 36 ft from the ground and rising? (Hrnr: See Hint for Exeicise 3.) 1 0 . If a ball is rolled across level ground with an initial velocity of 20 ftlsec and if the speed of the ball is decreasing at the rate of 6 ftlsecz due to friction, how far will the ball roll? 1 1 . If the driver of an automobile wishes to increase his speed from 20 milhr to 50 mi/hr while traveling a distance of 528 ft, what constant acceleration should he maintain? L2, What constant acceleration (negative) will enable a driver to decrease his speed from 60 miihr to 20 mi/hr while traveling a distance of 300 ft? 13. If the brakes are applied on a car traveling 50 mi/hr and the brakes can give the car a constant negative acceleration of 20 ftlsecz,how long^will it take the car to come to a stop? How far will the car travel before stopping? a downward L4. A ball started upward from the bottom of an inclined plane with an initial velocity of 5 ftlsec. If there is acceleration of 4 ftlsecz, how far up the plane will the ball go before rolling down? greatest speed it may 15. If the brakes on a car can give the car a constant negative acceleration of 20 ft|secz, what is the be going if it is necessary [o be able to stop the car within 80 ft after the brake is applied? and the marginal 6.6 APPLICATIONS OF In Sec. 5.6 we learned that the marginal cost function function and the ANTIDIFFERENTIATION revenue function are the first derivatives of the total cost function IN ECONOMICS total revenue function, respectively. Hence, if the marginal cost and the marginal revenue function are known, the total cost function and the total ,errerrnu function are obtained from them by antidifferentiation. In finding the total cost function from the marginal cost function, the arbitrary constant can be evaluated if we know either the fixed cost (i.e., the cost when no units are produced) or the cost of production of a specific number of units of the commodity. Because it is generally true that the total revenue is zero when the number of units produced is zero, this fact may be used to evaluate the arbitrary constant when finding the total revenue function from the marginal revenue function. nxarvrprE 1: If the marginal revenue is given by 27 - 12x * x2, hnd the total revenue function and the demand equation. Also determine the permissible values of x. Draw sketches of the demand curve, the total revenue curve, and the marginal revenue curve on the same set of axes. solurroN: If R is the total revenue function, the marginal revenue func' tion is R', and R'(r) :27 - 12xI x2 Thus, ntt) : / (27 - LZx* x2) dx :27x-6x2**r3*C THE DIFFERENTIAL AND ANTIDIFFERENTIATION BecauseR(0) R(r) :27x - 6x2+ 1#t J5 If P is the price function, R(x) : xP(r), and so P(r) :27 - 6x + *x2 30 25 20 we get C : 0. Hence, demand curve 15 10 5 o -5 -10 F i g u r e6 . 6 . 1 rxavrprE 2: The marginal cost function C' is given by C'(r) : 4x - 8. If the costof producing 5 units is $20,find the total cost function. Determine the permissible values of r and draw sketchesof the total cost curve and the marginal cost curve on the same set of axes. If we let p dollars be the price of one unit of the commodity when r units are demanded,and becausep : P(x), the demand equation is 3p :81 - 1'8x* f To determine the permissible values of x, we use the facts that r > 0, p > 0, and P is a decreasingfunction. Because P'(x) 3x P is_decreasing when x <9 (i.e., when p'(x) < 0). Also, when x:9, : 0, and so the permissible values of r are the numbers in the closedinP terval [0,9]. The required sketchesare shown in Fig. 6.6.t. solurroN: The marginal cost must be nonnegative. Hence, 4x- 8 = 0, and therefore the permissible v a l u e so f . x a r e x > 2 . C'(x):4x-8 Therefore, r C(x):1 (4x-8) dx J :2tc2-8x*k BecauseC(5) : 20, we get k: I0. Hence, C(r) :2x2 - 8x+ 10 (x = 2) The required sketchesare shown in Fig. b.6.2. v Figure6.6.2 IN ECONOMICS OF ANTIDIFFERENTATION 6.6 APPLICATIONS 3: After experiEXAMPLE mentation, a certain manufacturer determined that if r units of a Particular commodity are produced per week, the marginal cost is given by 0.3r - 17,where the Production cost is in dollars.If the selling price of the commodity is fixed at $19per unit, and the fixed costis $100per week, find the maximum weekly profit that can be obtained. Let C(r) : the number of dollars in the total production cost of r units; : R(r) the number of dollars in the total revenue ob- solurroN: tained by selling r units; the number of dollars in the profit obtained by selling x units. Becauser units are sold at $19 per unit, we have s(x)- R(r) : l9x The marginal revenue then is given by R'(r) : t9 C'. is the marginal cost function, and we are given that C'(r):0.3r-11 The profit will be a maximum when marginal revenue equals marginal costlsee Sec.5.5).Equating R'(r) and C'(x),ws get x:100. So 1'00units should be produced each week for maximum profit. (0.3r - 11,)dx C(r) - LLx * k : 0.1.512 Becausethe fixed cost is $100, C(O;: 100, and hence fore, C(r) :0.15yz- \Lx + L00 L00. There- SettingS(r) - R(r) - C(x), we have - ILx + 100) S(r) : 19x- (0.15x2 - -0.1512+ 30x- 100 Hence, 5(100) :1400. Therefore, the maximum weekly profit is $1400, which is obtained if L00 units are produced weekly' nxlrvrpln 4: A business firm has made an analysis of its Production facilities and its Personnel. With the present equiPment and number of workers, the firm can produce 3000 units Per daY. It was estimated that without anY change in investment, the rate of change of the number of units produced per day with resPect to a Let U: solurroN: +: dx the number of units produced per day' Then Bo- 6xttz from which we get dy: (80 - 5*rrz) dx Antidifferentiating U :8 0 x gives us - 4 xgt2+ C 272 THE DIFFERENTIAL AND ANTIDIFFERENTIATION change in the number of additional workers is 80 - 6x1t2,where r is the number of additional workers. Find the daily production if 25 workers are added to the labor force. BecauseU :300A when x: A, we get C : 3000.Hence, y -80x- 4x3t2 + 3000 Letting Azsbe the value of y when r : 25, we have Azs:2000 - 500+ 3000 Therefore,4500units are produced per day if the labor force is increased by 25 workers. Exercises 6.6 1' Find an equation of the total revenue cule of company iI its slope at any point is given by 12 - 3r and p(4) = 6, I where P is the price function. Draw sketches of tlte toLl revenue curve and the derrand curve on the game set of axes. 2' The rate oJ change of the slope the totakost curve of a particular company is the constant 2, and the total cost curve _of contains the points (2,72) and (3, 18). Find the total cost function. 3' Find the demand equation for a commodity for which the marginal revenue function is given by 10/(r + 5), - 4. 4. The rrarginal cost function i8 given by 3llET-q,.If the fixed cost is zero, find the total cost function. 5. The urarginal revenue function is given by 16 - 3rp. Find the permissible values of t, the total revenue function, and the demand equation. Draw sketches of the demand curve, tire total revenue curye,-u"a ,f," _"[in"f ."venue curye on the Bameset of axes. 6. Find the total revenue function, the demard equation, and the permissible values of x if the marginal revenuefunction is given by frf - 10r + 12. Also draw sketches of the dem*a trt"1t-,a-""ri"r" revenue curve on the same set of axes. "ir*", "ir*i,"""i"ri.ri--ginal 7. The marginal cost function iB given by 3rl + 8r + 4, and the fixed cost is g6. If C(r) dollars is the total cost of , units, find the total cost function, and draw iketches of the total cost curve and the marginal cost curve on the same set of axes. 8' The marginal cost function c' is defined by C'(r) :6r, where c(r) is the number of hundreds of do ars in the total cost of r hundred units of a certain-comm'odity. u ttt"'*"i J irb units is g2000, find the total cost function and the fixed cost. Draw sketches of the totar cost curve ana trr" -*6"J cost culve on the same set of axes. 9' The cost of a certain piece of machinery is $700 and its value depreciates with time according to the formula D,y: -500(' + l)-'z where v dollars is its value t years after its purchase.what is its value three years afte! its purchase? 10' suPpose that a particular company estimates it' growth in income ftom sales by the formula D,s : 2(, - 1)r/3, where s millions of dollars i8 the incone from saLs r y""* i"ii". r trr" g-"" ii..o^" iro,,'-it ye-,, s"t"" is "'ossexpected gross irr.o*Jlrorn 98 million, what should be ihe "'lrri"r,t ,"i". t*o years from now? ReoiewExercises (Chapter6) In Exercises 1 through 5, find the most general antiderivative. ,L. f Vxt , J iF-ax +. ,f$-g)a, f_ Jk3*x)!x2*3dx 5. f_ J \/4x* 3(* + 1,)dx 3. 6. REVIEWEXERCISES 273 In Exercises 7 through 10, find the complete solution of the given differential equation. 7. x'tt ' 9: tlx fu' - 1)' 9. y' dx* Y zdY : dY du "' dx R -:- 10. d2u #: xIL \/x lZxz- 30x 11.If.y:80r-'J.6x2,findthedifferenceAy-dVif(a)r:2,Lx:0.1;(b)r:4,Lx:-0.2. 12. Use differentials to find an apProximatevalue of fl126. -1) is on the curve. Find an 1 3. The slope of the tangent line at any point (x, y) on a curve is L0 4x and the point (1, equation of the curve. f and an equation of the tangent line to the curve at the Point (1, -1) is L4. At any point (r, y) on a curve D,"y:4: 2r 3y 3. Find an equation of the curve. 1 5. ll xs * yB- 3xy?-f l: 0, find dy at the point (1, 1) in fu: 16. Find d.yld.tiI y: rs - 2x + 1 and xl + ts - 3l: 0. 17. Fnd dyl dt it 8f + 27f - 4xy : S and F * fx - f : 4. O.1,- 18. Evaluate J (.ra)04d dr as Juo du and as l4x2r dx and compale the results. 19. Evaluate J(ri * l)2* dx by two methods: (a) Make the substitution u: f'l l; (b) first expand (f * 1)'. ComPare the results and exptain the difference in the appearance of the answers obtained in (a) and (b). 20. Find the particular solution of the differential equation f dy : y3 dx fot whiclj. y: 1 when r = 4. 21,. FindtheparticularsolutionofthedifferentialequationD'2y:\rQltrforwhichy:3andy':2whent:4' 22. A ball is threwn vertically upward from the top of a house 64 ft above the ground, and the initial velocity is r18ft/sec. How long will it take the ball to reach its greatest height, and what iB it6 greatest height? How long will it take the ball to strike the ground and with what velocity does it strike the ground? 23. Suppose the ball in Exercise 22 is thrown downward with an initial velocity of 48 ftlsec. How long will it take the ball to strike the ground and with what velocity does it strike the 8lound? 24. Suppose the ball in Exercise 22 is dropped from the top of the house. Determine how long it takes for the ball to stiike the ground and the velocity when it strikes the ground. 25. Neglecthg air resistance, if an obiect is dropped from an airplane at a height of 30,000 ft above the ocean, how long would it take the obiect to strike the water? 26. Suppose a bullet is lired direcfly downward from the airplane in Exetcise 25 with a muzzle velocity of 2500 ft/sec. If air resistance is neglected, how long would it take the bullet to reach the ocean? 27. A container in the form of a cube having a volume of 1000in.3 is to be made by using 6 equal squares of material costing 2 cents per square inch. How accurately must the side of each square be nade so that the total cost of the material shau be corect to within 50 cents? 28. The measure of the radius of a right-circular cone is * times the measure of the altitude. How accurately must the altitude be measured if the error in the computed volume is not to exceed 3%? 's 29. If t sec is the time for one complet€ swing of a simple pendulurn of length I 11,gt"t1 4721: gP, where 8: 32.2. What the effect upon the tim€ if an eror of 0.01 ft is made in measuring the lentth of the pendulun? 30. An automobile traveung at a constant speed of 60 mi/hr along a straight'highway fails to stoP at a stoP sign. If 3 sec l,ater a highway patrol car starts from rest from the stop sign and maintains a constant acceleration of 8 ft/geC, how long will it take the patrol car to overtake the automobile, and how far from the stoP sign will this occur? Also deteF mine th€ speed of the Pabol car when it overtakes the automobile. 271 THE DIFFERENTIAL AND ANTIDIFFERENTIATION 37. If l(x) = r * lr - tl, and Ftr\:[x lt'-x+t if r<1 if r>1 show that F i8 an antiderivative of I on (-c., +ce). 32. Let / and I be two tunctions sucl_rth31_f91-alt il and g,(r) : *l(r). Furthermore, suppose I !_-, +-), f,(!:g(r) thatl(0) : 0 andS(0) : 1. Provethar_[(r)]'+.[g(r)]': t. lnrwr: Consi'derth" r.i iur". r'"ii'C wiere r(r) : t/(r)1, artd c(:) : -[g(r)]2 and show rhat,F,(x) : c,G) for all x.i 33' The marginal revenue function is given by abl(x * b)2- c. Find the total revenue function and the demand equation. 34' A company has determined that the marginal cost function for the production of a particular commodity is given by C'(r) : 125'tr-1gt - " where C(x) dotlars i, *"'"or, ing 15 units? ot p-ao"ing r units of the commodity. If the fixed cost is $250,what is the cost of produc- The definite integral THE DEFINITE INTEGRAL 7'1 THE SIGMA NOTATION In this chapter we are concemed with the sums of many terms, and so a notation called the sigma notation is introduced to faciliiate writing these sums. This notation involves the use of the symbol ), the capital silma of the Greek alphabet, which correspondsto our letter S. Some examplesof the sigma notation are given in the following illustration. o rLLUsrRATroN 1: 5 \F:72+22f32+42+52 (3t+2): t3(-2)+21+ t3(-1)+ 2l + 13'0 + zl 1>= - ' + [ 3 . 1 + 2 J+ 1 3 . 2 + 2 ] : ( - 4 )+ ( - 1 )+ 2 + s + 8 n *ns )it:Ls+23+33+ j:1 };:f++.+*t*+.t We have, in general, n F ( i ): F ( m )+ F ( m+ 1 ) + F ( m* 2 ) + + F(n) i:m (1) where m and n are integers, and m < n. The right side of formula (1) consists of the sum of (n - m * 1) terms, the first of which is obtained by replacing i by m in F(i), the second by rePlacing iby (m * 1) in F(i), and so on, until the last term is obtained bv replacing i by n in F(i). The number m is called the lower timit of the sum, and.n is called the upper limit. The symbol i is called the index of summation. rt is a ,,dummy,, symbol because any other letter can be used. For example, 5 k2:32+42+52 k:3 is equivalent to 5 i2:32+42+52 i:3 rLLUSrRArroN 2: By using the formula of Eq. (1), we have $ t r _ 3 r _ 4 r _ 5 r , 6'' ei+ 1 3+1 4+']..'5+1 ' 6+']-. Sometimes the terms of a sum involve subscripts, as shown in the next illustration. 7.1 THE SIGMA NOTATION 3: O ILLUSTRATION n '*A' )a':A'+A2+' j:' gbs pûon: 4b++ 5b5+ 6b6+ 7b7+ 8bB+ a 2fQ) A x : f ( x , ) A r * f ( x r )A x * f ( x r )A r * f ( * ) A r * f ( x ' ) L x . ,n" n"tt"wing propertiesof the sigmanotation are useful,and they are easily proved. cn where c is any constant ,: 7.'t.1,Property 1, f The proof is left as an exercise(seeExercise9). property2 7..t..2 ii : l r. F(i): r i ftO wherec is any constant PROOF: n F(1)+c'F(2)*c'F(3)+ T c'F(l) LI ic'F(n) i:l : c [ F ( l ) + F ( 2 )+ F ( 3 )+ ' ' ' +F(n)] =r ; r 1 ; 1 i:r 7.1..3Property 3 i trrr)+ c(t)l:2 rtil +; cr;l .L/ - i:1 L/ i:l .? i:7 The proof is left as an exercise (see Exercise 10). Property 3 can be extended to the sum of any number of functions. 7.1.4 Property 4 n t F ( t )- F ( ,- 1 ) l - - F ( n )- F ( 0 ) > i:1 PROOF: n - F ( 1 ) ] + [ r ( 3 )- F ( 2 ) ] + ' tZ-/ t- r t i l - F ( i - 1 ) l : [ F ( 1 )- F ( 0 ) ] + [ F ( 2 ) i:1 + l F ( n - 1 ) - F ( n - 2 ) l + l F ( n ) - F ( n - 1 )l THE DEFINITEINTEGRAL : - F ( 0 ) + [ F ( 1 )_ r ( r ) ] + [ F ( 2 )_ r 1 z ; 1 + . . . + l F ( n - 1 ) - F ( n- 1 ) l + F ( n ) :-F(0) +0+0+ . . . + 0+ F(n) : F(n) - F(0) I The following formulas, which are numbered for future reference, are also useful. n 7.'l.,.5Formula L . n(n*71 ,:T i:r n 7.1.6 Formula 2 . r-_b n ( n * 1 ) ( 2 n + 1 ) i:r n 7.'1,.7Formula 3 ,, r - -f l- '-(4n-+ 7 ) ' i:r 'n 7.'1,.8Formula 4 i:1 n(n r t) (6nBt 9n2+ n - 1) ,+_ 30 Formulas 1 through 4 can be proved with or without using mathematical induction. The next illustration shows how Formula L can be proved without using mathematical induction. The proof of Form ula 1,by mathematical induction is left as an exercise (see Exercise 11). . ILLUSTRATToN 4: We prove Formula 1. n ) l:1,+2+3+ . . . -t (n-t) *n and n )i:n*(n-1)+ (n-2)+.. +2+t ,it*" add thesetwo equationsterm by term, the left side is n 2>i ""0 .;;he right side are n terms, each having the value (n + 1). Hence, n 2 > i : ( n + 1 ) + ( n * L ) + ( n+ 1 ) + ' i:r :n(n*t) Therefore, S\ : , _ n ( n,)+ t ) z.,l i=l ' '+(n+I) nterms 7.1 THE SIGMA NOTATION EXAMrLE1: Prove Formula 2bY mathematical induction. solurroN: We wish to Prove that *,, _n(n+L)(Zn+I) >r: 6 'L.The left sideis then2 ,': l'. When ,rrr,ln" formulais verifiedfor n: '2'3-li6- l. Theren : L , t h er i g h ts i d ei s [ 1 ( 1+ 1 ) ( 2+ D ) t e - ( 1 fore, the formula is true when n: 1. Now we assume that the formula is true for n: k, where k is any positive integer; and with this assumption we wish to prove that the formula is also true for n:k + 1. If the formula is true for tt: k, we have k S ./-/ i , : i:1 Wh e n n : k+l f k(k + 1,)(2k+ 1,) 6 (2) k * L , w e have i r : 1 2 * 2 2 + 9 2 + . . . + k 2+ ( k + 1 ) ' i:l k :)i2+(k+1)' + L) (k 1), (by applyingEq. (2)) _ k(k + L)_(zk + + 6 _ k ( k+ L )( 2 k+ L ) + 6 ( k+ L ) ' z 6 -_- ( k + 1 )[ k ( 2 k+ 1 ) + 6 ( k+ 1 ) ] _ (k+L)(zk'z+7k+6) 5 _ (k+1)(k+2)(2k+3) 6 - ( / c +1 ) [ ( k + 1 ) + 1 ] [ 2 ( k +1 ) + 1 ] 6 Therefore, the formula is true f.ot n: k * L. We have proved that the formula holds f.or n: 1, and we have also proved that when the formula holds f.or n: k, the formula also holds for n: k * 1. Therefore, it follows that the formula holds when n is any positive integer. A proof of Formula 2, without using mathematical induction, is left as an exercise. The proofs of Formulas 3 and 4 are also left as exercises (see Exercises L2 to 76). THE DEFINITE INTEGRAL ExAMpLE 2: n Evaluate soLUTroN: From Property 4, where F(i) : 4i,1t follows that n (4'- 4i-')- 4 - 4 0 > i:r (4' - 4i-') t .{-J i:r :4n-I nxaivrPrn3: Evaluate SOLUTION: n i(3i- 2) > i:r n r(3t- 2) > i:r (3i2 - 2i) by using Properties 1-4 and Formulas 1-4. (3i') :3 +> (-2i) i'- Z2 i > i:7 (by Property 3) (by Property2) n(n+ 7) 2 (by Formulas2 and 1 ) 2nBl3nzIn-2n2-2n 2 2n3*n2-n Exercises7.1 In Exercises 1 through 8, find the given sum. 6" 1 .> ( s i - 2 ) i=l t $ 2.>(i+1.)' i:r 2 KFz -i . + , -( - t ; u * r ) , k F 3 U ,D-,'' 3. it ? i-1, s1 6. z ' - l + i 2 3 b 8 " '. y t I ,ro-rk+3 9. Prove Property 1 (7.1,.1,). 10. Prove Property 3 (7.1.3). 11. Prove Formula 7 (7.1.5)by mathematicalinduction. L2. Prove Formula 2 (7.1.6\without using mathematicalinduction. (HrNr: iB- (i - 7)":3i ' - 3i + L, so that nn ) t ; '- ( i - 1 ) ' l : ) tgt,- 3t+ 1) On the left side of the above equation, use Property 4; on the right side, use Properties '1,,Z, and 3 and Formula 1.) 7.2 AREA 281 13. Prove Formula 3 (7.1.7)without using mathematicalinduction. (HrNr: ia- (i - qa:4p - rO * 4i- 1, and use a method similar to the one for Exercise12.) 14. Prove Formula 4 (7.1.8)without using mathematical induction (seehints for Exercises12 and 13 above). 15. Prove Formula 3 (7,1.4 by mathematical induction. 16, Prove Formula 4 (7.1.8)by mathematicalinduction. In Exercises17 through 25, evaluatethe indicated sum by using Properties 1 thlough 4 and Formulas 1 through 4. t 20 1 8 .> 3 i ( i 2 + 2 ) 2zi(i-1,) i:r n n 20. > (2k-t-zk) 19.>(10i*'-10,) k:r 40 22.>3/T+r -\Dl-l --'f:,Lk k+L) ,r.''flff-=+-l ,t 24iz(i i:r n - z\ 24.>\i(L+i2) i--l lr n 25. > [(3-o - 3o)'- (3t'-t+ 3-k-r)2] f /i\21r12 2 6 . P r o v e :) l t - { - } | \nl J ,?nL :2} n ? = ,[ ' - (*)']''' *, Prope rty4 toshowthat ,.i;Use ) F ( n )- F ( 1 )- r ( 0 ) > ttt, + 1)- F(i- 1)l : F(nr-1'+ Lli:1 (b) Prove that n t ,1J t(t+1)'-(i-' 1 ) ' l: ( n * 1 , ) 3 * n 3 - L i:r (c) Prove that nnn ) t t ; + 1 ) ' - ( i - 1 ) ' l : 2 ( 0 i '+ 2 ) : 2 n* 6 ) i2 i:r (d) Using the results of parts (b) and (c), prove Formula 2. 28. Use the method of Exercise 27 to prove Formula 3. n )t' n 29. If X:El - , pr ov e n t ha t ) (x , - X)' :2 i:r 7.2 AREA n i=l r. r,' - X 2 * , i--L We use the word mensure extensively throughout the book. A measure refers to a number (no units are included). For example, if the area of a triangle is L0 irt.z, we say that the measure of the area of the triangle is 10. You probably have an intuitive idea of what is meant by the measure of the area of certain geometrical figures; it is a number that in some way THE DEFINITEINTEGRAL Figure7.2.1 Figure7.2.2 gives the size of the region enclosed by the figure. The area of a rectangle is the product of its length and width, and the area of a triangle is half the product of the lengths of the base and the altitude. The area of a polygon can be defined as the sum of the areas of triangles into which it is decomposed, and it can be proved that the area thus obtained is independent of how the polygon is decomposed into triangles (see Fig. 7.2.1). However, how do we define the measure of the area of a region in a plane if the region is bounded by a curve? Are we even certain that such a region has an area? Let us consider a region R in the plane as shown in Fig. 7.2.2. The region R is bounded by the r axis, the lines x: a and x: b, and the curve having the equation y -- f(x), where f is a function continuous on the closed interval [a,b]. For simplicity, we take f (x) > 0 for all x in [a,bl.We wish to assign a number A to be the measure of the area of R. We use a limiting process similar to the one used in defining the area of a circle: The area of a circle is defined as the limit of the areas of inscribed regular polygons as the number of sides increases without bound. We realize intuitively that, whatever number is chosen to represent A, that number must be at least as great as the measure of the area of any polygonal region contained in R, and must be no greater than the measure of the area of any polygonal region containing R. We first define a polygonal region contained in R. Divide the closed interval [a, b] into resubintervals. For simplicity, we shall now take each of these subintervals as being of equal length, for instance, Ar. Therefore, Ar: (b - a)ln. Denote the endpoints of these subintervals by xo, xt, x 2 , . . , x n - r ,r r , w h e r € f f o : f l , x t : a i L x , . . . , x i : A * i L , x , . . , xn-r: a -l (n - t) A,x,xn: b. Let the ith subinterval be denoted by lxr-r, xrf. Because / is.continuous on the closed interval la, bf, it is continuous on each closed subinterval. By the extreme-value theorem (4.5.9), there is a number in each subinterval for which / has an absolute minimum value. In the lth subinterval, let this number be ci, so that /(c) is the absolute minimum value of f on the subinterval fx,-r, r1]. Consider n rectangles, each having a width Ax units and an altitude f (r,) units (see Fig. 7.2.3). Let the sum of the areas of these n rcctangles be given by S, square units; then S":f(cr) Ax*f(rr) Ax*' '+f(c,) Ax.r- * f(c") Lx or, with the sigma notation, n sn:\ f (c) Ax (1) The summation on the right side of Eq. (L) gives the sum of the measuresof the areasof n inscribed rectangles.Thus, however we define A, rt must be such that A= S, 7.2 AREA Figure7.2.3 Figure7.2.4 In Fig. 7.2.3 the shaded region has an area of 5n square units. Now, let n increase. Specifically, multiply nby 2; then the number of rectangles is doubled, and the width of each rectangle is halved. This is illustrated in Fir.7.2.4, showing twice as many rectangles as Fig. 7.2.9. By comparing the two figures, notice that the shaded region in Fig. 7.2.4 appears to THE DEFINITEINTEGRAL approximate the region R more nearly than that of Fig.7.2.3. So the sum of the measures of the areas of the rectangles in Fig. 7.2.4 is closer to the number we wish to represent the measure of the area of R. As n increases, the values of s, found from Eq. (1) increase, and successive values of 5, differ from each other by amounts that become arbitrarily small. This is proved in advanced calculus by a theorem which states that if / is continuous on fa, b], then as n increases without bound, the value of S, given by (1) approaches a limit. It is this limit that we take as the definition of the measure of the area of region R. 7.2.'l' Definition Suppose that the function/ is continuous on the closed interval la,bl,with f (x) > 0 for all x in fa, b), and that R is the region bounded by the curve y - f(x), the x axis, and the lines x: a and r: b. Divide the interval la,bI into n subintervals, each of length Ax: (b - a) ln, and denote the ith subinterval by lxt-t, xil. Then if f(c) is the absolute minimum function value on the ith subinterval, the measure of the area of region R is given by rim ) /(r,) ar A*+€b {*1 (2) . Equation (2) means that, for any e ) 0, there is a number N ) 0 such that -ol l>f,',,a,x . . whenever rz ) N and n is a positive integer. We could take circumscribed rectangles instead of inscribed rectangles. In this case, we take as the measures of the altitudes of the rectangles the absolute maximum value of / on each subinterval. The existence of an absolute maximum value of / on each subinterval is guaranteed by the extreme-value theorem (4.5.9). The corresponding sums of the measures of the areas of the circumscribed rectangles are at least as great as the measure of the area of the region R, and it can be shown that the limit of these sums as n increases without bound is exactly the same as the limit of the sum of the measures of the areas of the inscribed rectangles. This is also proved in advanced calculus. Thus, we could define the measure of the area of the region R by A: n lim 2 f @) Lx (3) wheref (";.;"'.]1" urrotutemaximumvatueof / on lx,-,, x,l. The measure of the altitude of the rectangle in the lth subinterval actually can be taken as the function value of any number in that subinterval, and the limit of the sum of the measures of the areas of the rectangles is the same no matter what numbers are selected. This is also proved in advanced calculus, and later in this chapter we extend the definition of the measure of the area of a region to be the limit of such a sum. 7.2 AREA EXAMPLEL: Find the area of the region bounded by the curve A : x', the r axis, and the line x: 3 by taking inscribed rectangles. solurroN: Figure 7.2.5 shows the region and the ith inscribed rectangle. We apply Definition 7.2.L. Divide the closed interval [0, 3] into n subintervals, each of length A,x: xs: 0, xr: Lx, xz: 2 Ax, . . ., xi: i Lx, . . r x n -r: (n - t) A,x,xn: 3. ^ - rJ,,t, 3-0 nn 3 The function value /(x) -- xz and because / is increasing on [0, 3], the absolute minimum value of f on the ith subinterval lxr-r, r;] is f (xr-r). Therefore, from Eq. (2) n A: lim D fk,-r) Lx Because ::-,-::;- ( 4) 1) Ax and /(x) : ,sz,we have f (x,-,)- [(t 1) Ar]'z Therefore, nn ) f (x,-') Ar: > (t 1)'z(Ar)g 8". ;; : ztn,ro tnut: n n s ( t - L ) ' -Ln:,2 7 n A r : / ( r ' ' ) ) > sL i:r i:r :4[$;,-2$ no t+$ r), '] ,?:, ,?:, f4 and using Formulas 2 and L and Property L from Sec.7.'1., we get ._27 ln(n+t)(2n+L) ,: 2fQ'-')Ax:lirt 27 _2.ry!*"1 2n3*3n2*n-6n2-6n*6n n3 9 2 2n2-3n*L ___v- Then, from Eq. (4), we have (,-1.#) :*(2_0+0) -9 Therefore, the area of the region is 9 square units. THE DEFINITEINTEGRAL 2: Find the areaof the ExAMPLE region in ExampleL by taking circumscribed rectangles. solurroN: we take as the measureof the altitude of the lth rectangle the absolute maximum value of / on the lth subintervaI lxi4, 11] which is f (x,). From Eq. (3), we have n A: lim 2 f k,) tx Because [! r'ir,thenf (x,): (i a,x)z,and so n n Ar: 2f(',) i:r ] L/ i:1 )7n i2(A,x)S: +n"s?- i , l = l 27 l n ( n+ 7 )( 2 n+ 1 ) ' l n3 L6l 2n2*3n*L _____7Therefore, from Eq. (5), we obtain q A: Itm n-+6 , .(r*q+4) \ /r n"/ ( a s in Example1) uxaupru 3: Find the area of the trapezoid which is the region bounded by the line 2x * y : 8, the x axis, and the lines x : 1 and x: 3. Take inscribed rectangles. soLUTroN: The region and the ith inscribed rectangle are shown in Fig. 7.2.6.The closed interval [1, 3] is divided into n subintervals, each of length A,x; xs: L , xr: "1.* Lx, xz: I * 2 A,x, xi:t*i Lx, 1.-t (n - L) L,x,xn 3-1 Solving the equation of the line for y, we obtainy:-2x * B. Therefore, f (x): -2x* 8, and because / is decreasing on [1,3], the absolute minimum value of f on the ith subinterval lxi_r, ri] is f (rr). Because ri: 1 + i Ax a nd f(* ) : -2x* 8, then f(x,) : -2(L+ i A x) * 8 : 6 - zi Ax. Fr om Eq,. (2), we have f(x) A,x lim fl-*@ : lim i tu 2i a,x)ax n-+@ i:l F i g u r e7 . 2 , 6 : n lim ) 2-+@ i:l n i:r I0 Ax - Zi(A,x)'z1 ,,G)',1 I'G)- 7.2 AREA : lim ll-lo 287 l+y-i-E'l we have Using Property 1,(7.1.1)and Formula l, (7.1,.5), : lim ft-A) nl - --\ h-*@ -8 Therefore,the area is 8 square units. Using the formula from plane geometry for the area of a trapezoid,A:ih(b1*b2), where h,br, andb, are, respectively,the number of units in the lengths of the altitude and the two bases,we get A: +(2)(6+ 2): 8, which agreeswith our result. 7,2 Exercises In Exercises 1 through 14, use the method oI this section to find the area of the given region; use inscribed or circumscribed rectantles as indicated. For each exercise, draw a figure showing the redon and the ith rectangle. The region bounded by y: r', the x axis, and the line r = 2; inscribed rectangles, J. 2. The region of Exercise 1; cirormscribed rectangles. ..\ The re6on boun dedby y :2x, 1 and r : 4; circumscribed rectangles. tte r axis, and the lines r: 4. The region of Exercise 3; insclibed rectangles. 5l The region above the r axis and to the right of the line r:1 ''' y=4-f, inscribed rectangles. bounded by the r axis, the line r= 1, and the curve 6, The region of Exercise 5; circumsoibed rectangles. 7i The region lying to the left of the line : = 1 bounded by the curve and lines of Exercise 5; circumscribed rectangles. 8. The region of Exercise 7; inscribed rectangles. 9. The region bounded by y : 3/, dre r axis, and the line r: 1; inscribed rectangles. 10. The region of Exercise 9i circumscribed rectantles. 11. The region bounded by y: ri, the r axis, and the lines r: -L and r: 2; inscribed rectangles. 12. The region of Exercise lli circumscribed rectantles. li.lnre region bou nded by y: ' rectangles. mx, with tt > 0, the r axis, and the lines l: a and r: b, with b ) a > 0; circumscribed 14. The region of Exercbe 13; inscribed rectangles. 15. Use the rrethod of this s€ction to find the ar€a of an isosceles trapezoid whoee bases have measr:res b. and D"and whose altitude has measure l. 284 INTEGRAL THE DEFINITE -4 tox:4lorm 76. The Snph of y:4l.zland the r axis hom x: area of this triangle. a triarrgle. Use the method of this section to lind the In Exercises 17 throtgb 22, tind the area of the region by taking as the measure of the altitude of the ith rectangle l(nr), where z1 is the midpoint of the ith subinterval. (rrrwr: n1: !(4 a 4 1).) 17. The region of Example 1.. 18. The region of Exercise L. 19. The region of Exercise 3. 20. The region of Exercise 5. 21. The region of Exercise 7. 22. The region of Exercise 9. 7.3 THE DEFINITE INTEGRAL In the preceding section, the measure of the area of a region was defined as the following limit: n lim ) f(c1) A,x n++6 (1) i:l To lead up to this definition, we divided the closed interval fa, bl into subintervals of equal length and then took ci as the point in the ith subinterval for which / has an absolute minimum value. We also restricted the function values f (x) to be nonnegative on fa,bl and further required f tobe continuous on la, bl. To define the definite integral, we need to consider a new kind of limiting process/ of which the limit given in (1) is a special case. Let / be a function defined on the closed interval [a, b].Divide this interval into n subintervals by choosing any (n - 1.)intermediate points between a andb. Let t xn-r be the intermediate points so that ro(xr1xr1 1 xn_r I x, The points x6,xr, xz, . r xn-rrx?rarenot necessarilyequidistant.Let Arr be the length of the first subinterval so that Arr : xr - xil let Arr be the length of the secondsubinterval so that Azr - x2- xr; and so forth, so that the length of the lth subinterval is A1r, and a.g: xi A set of all such subintervals of the interval [a, b] is called apartition of the interval [a, b].Let A be such a partition. Figure 7.3.1 illustrates one such partition A, of la, bl. I a:ro I | | I x7 )t2 13 x4 I xn_L l.>r In:o F i g u re7 .3 .1 The partition A contains n subintervals. One of these subintervals is longest; however, there may be more than one such subinterval. The length of the longest subinterval of the partition A, called the norm of the partition, is denoted by llAll. Choose a point in each subinterval of the partition A: Let fr be the point chosen in [xo, xr] so that ro < €, < xr. Let f, be the point chosen in 7.3 THE DEFINITEINTEGRAL v lxr, xrf so that xr s tz = x2, afld so forth, so that fi is the point chosenin lxu-r,x-], and xr-r 3 tt - x.. Form the sum 0 ' 'tf({") A'"x f ( ( , )A , r * f ( ( , )L , x t ' ' ' + f ( ( , ) L p t ' \ or n > f (€,) Lix \ Su.nl sum is called a Riemannsum,named for the mathematicianGeorg Friedrich Bernhard Riemann (1826-1.866). o ILLUSTRATIoN 1: Suppose/(x) :1.0 - x2, with i = x < 3. We will find the Riemann sum for the function f on [i, e1 for the partition A: xo:i, x.t:1, xz:li, xs:Lt, xq:2*, xs:3, andf1: t, tz:Lt, ts:11, €q:2, t": 21. Figure 7.3.2 shows a sketch of the graph of / on l*, Z1 and the five rectangles,the measuresof whose areasare the terms of the Riemannsum. 5 5 2f tt,l L&: f(€,)A,r * f(il A,rx*f(€') A'r * f(t^) Anr*/((') A'r - 1)+ f@(L*- r+) : fE)(l- +)+ fG)0+ + f (2)(2+-1*)+ f (+)(3- 2+) : (e*)(+)+ (8+)(+)+ (6€)(+)+ (5)(+)+ (2+)(+) : 18# . Thenorm of A is the lengthof the longestsubinterval.Hence,llAll:*. o xot, Figure7.3.2 xrtrl xr\xn l, t, !. €rx Because the function values f (x) are not restricted to nonnegative values, some of the /(fi) could be negative. In such a case/ the geometric interpretation of the Riemann sum would be the sum of the measures of the areas of the rectangles lying above the r axis plus the negatives of the measures of the areas of the rectangles lying below the r axis. This situation is illustrated in Fig.7.3.3. Here v a\ y : f(x) J li IC Arl |.i, la' dt, \ tn t, *rl i4 trv xott \ €z O x2 ,{N A^ 'l : i:f: :tt As i/ \ i {o xo .xs f,l f' I ,/,, t1 b .{rc ;-\ Aa t F i g u r e7 . 3 . 3 fto r,s A,I T I t 1 rto THE DEFINITEINTEGRAL j ftg,l Lix: Ar* A2_ As_A4- Au* Au* A7- As- An- Aro because f(€"), fG), fG), f$r), f$n), and /(f,o) are negative numbers. We are now in a position to definewhat is meantby a function / being "integrable" on the closedinterval la, bl. 7.3.1 Definition Let f be a function whose domain includesthe closedinterval [a, b]. Then / is said to be integrableon fa, bl if there is a number L satisfyingthe condition that, for every e ) 0, there exists a 6 ) 0 such that I Lix- tl ' ' fte,l l) ln for every partition A for which llAll < 6, and for any fi in the closed interval f x r _ r x, 1 l i, : 1 , 2 , . . . , n . In words, Definition7.3.L states that, for a given function / defined on the closed interval la, bl, we can make the values of the Riemann sums as close to L as we please by taking the norms llAllof all partitions A of [a, b] sufficiently small for all possible choices of the numbers f1 for which xi-r s €i = xi.If Definition 7.3.1 holds, we write n (2) lim ) /(f,) A,ix: L llAll-0 t:r The above limiting process is different from that discussed in Chapter the number L in (2) exists if for every e ) 0 there 2. From Definition 7.3.'I.., existsaD)0suchthat lf rtr,l Lix- rlI . . lr-:r' for every partition A for which llAll< 6, and for any f, in the closedinterval , n. f x r - r ,x i l ,i : L , 2 , In Definition 2.1.1we had the following: (3) lim/(r):1 if for every e > 0 there exists a 6 > 0 such that lf(x)- Ll < e w h e n e v e r 0< l r - a l < 6 In limiting process(2), for a particular 6 > 0 there are infinitely many partitions A having norm l[ll < 6. This is analogousto the fact that in limiting process (3), for a given 6 > 0 there are infinitely many values of x for which 0 < lr - al < 6. However, in limiting process(2) , for eachpartition 7.3 THE DEFINITE INTEGRAL 291 A there are infinitely many choices of ti.It is in this respect that the two limiting processes differ. In Chapter 2 (Theorem 2.1,.2) we showed that if the number L in limiting process (3) exists, it is unique. In a similar manner we can show that if there is a number L satisfying Definition 7.3.1, then it is unique. Now we can define the "definite integral,." 7.9.2 Definition If /is a function defined on the closed interval [a,bf, then the definite integral of / from a to b, denoted bV Il f @) dx, is given by u,' fl'f;w'n*t' ,tiil,:/tut (4) if the limit exists. Note that the statement"the function / is integrableon the closedinterval la, bf" is synonymouswith the statement"the definite integral of / from atob exists." In the notation for the definite integral I! f @) dx, f (x) is called the integrand,a is called the lower limit, and b is called the upper limit. T}l.e symbol I J is called an integral sign. The integral sign resembles a capital 5, which is appropriate because the definite integral is the limit of a sum. It is the same symbol we used in Chapter 6 to indicate the operation of antidifferentiation. The reason for the common symbol is that a theorem (7.6.2), called the fundamental theorem of the calculus, enables us to evaluate a definite integral by finding an antiderivative (also called an indefinite integral). A question that now arises is as follows: Under what conditions does a number L satisfying Definition 7.3.Lexis| that is, under what conditions is a function f integrable? An answer to this question is given by the following theorem. 7.3.3 Theorem If a function / is continuous on the closed interval lq, bf, then/ is integ ra b l e o n l a , b l . The proof of this theorem is beyond the scope of this book and is given in advanced calculus texts. The condition that / is continuous on la,bl,while being sufficient to guarantee that/is integrable on la,bl,is not a necessary condition for the existence of [l f @) dx. That is, if / is continuous on la,bT,then Theorem 7.3.3 assures us that Il f G) dx exists; however, THE DEFINITEINTEGRAL it is possible for the integral to exist even if the function is discontinuous at somenumbers in la, b]. The following examplegives a function which is discontinuous and yet integrable on a closed interval. EXAMPLE1: defined by Let f be the function ,,, [o ifx*o l \ x ) : I "Ltr r i f r _ o Letfa, bl be any interval such that a 1 0 < b. Show that/is discontinuous on la, bl and yet integrable on [a, bl. solurroN: Becauselim f (x) :0 + f (0),/is discontinuous at 0 and hence discontinuous on [a, b]. To prove that / is integrable on la, bl we show that Definition 7.3.7 is satisfied. Consider the Riemann sum n ) /(f,) l,' If none of the numbers (r, €r, zero.Supposethat $: 0. Then is zero, then the Riemann sum is n . > /(f,) Aix: I L,x ,r, "u:,iur.ur" {rl s llall lf, rr*,, Hence, If, f rc,l Lic_ol . . whenever llAll< e Comparingthe abovewith DefinitionT.3.'1, where D: e and L:0, that f is integrable on la, bl. we see In Definition7.3.2, the closed interval [a, b] is given, and so we assume Ihat a < b. To consider the definite integral of a function from a ta b / when o s b, or when fl: b, we have the following definitions. 7.3.4 Definition I f a > . , b, then rb J"f(x) dx:it tf f( x) 7.3.5 Definition dx [: f(x) dx exists. It f Q) dx 0 if f (a) exists. At the beginning of this section, we stated that the limit used in Definition 7.2.1to define the measure of the area of a region is a special caseof 7.3 THE DEFINITEINTEGRAL the limit used in Definition7.3.2 to define the definite integral. In the discussion of area, the intervalfa, b] was divided into n subintervals of equal length. Such a partition of the interval [a, bj is called a regular partition.If Ar is the length of each subinterval in a regular partition, then each Lix: Ar, and the norm of the partition is Ar. Making these substitutions in Eq. (4) , we have fbn I f- tr>dx: Alim ) f ((,) Lx Ja r ,o i:l Furthermore, b-a n and Lx So from Eq. (6) lim lL- (8) -t@ and from (7),because b > a and Ar approaches zero through positive valu e s (b e c a u s eA x > 0), lim n: *oo (e) Ar-0 From limits (8) and (9), we concludethat Ar --+0 is equivalent to n -->+@ (10) Thus, we have from Eq. (5) and statement(10), fbn )i : l f (€,)Lx I f t*l dx: nlim Ja _*o (11) It should be remembered that fi can be any point in the ith subinterval lxi-t, xrl. In applications of the definite integral, regular partitions are often used; therefore, formulas (5) and (L1) are especially important. Comparing the limit used in Definition 7.2.L, which gives the measure of the area of a region, with the limit on the right side of Eq. (11), we have in the first case, lim ) f(c) Lx (t2) n-+@ i:l where/(ci) is the absolute minimum function value on lxi-t, x1).In the sec- THE DEFINITEINTEGRAL ond case, we have n rim ) f(€,) Lx n-*q i--t (13) where (1 is any number in [r1-r,r,]. Becausethe function / is continuous on fa,bl, by Theorem 7.9.9, Il f @ dx exists;therefore,this definite integral is the limit of all Riemann sums of / on la, bl including those in (12) and (13).Becauseof this, we redefine the area of a region in a more general way. 7.3.6 Definition f(x) Letthe function/becontinuousonfa, bl and f(x) > 0forall rin la,bl.Let R be the region bounded by the curve y - f (x), the r axis, and the lines x: fr and r : b. Then the measure of the area of region R is given by o'* P,rtr'lo":I)f(x) dx ,,1il, F i g u r e7 . 3 . 4 ExAMpLE2: Find the exact value of the definite integral f3 I x2dx Jt Interpret the result geometrically. The above definition states that if f (x) > 0 for all r in Ia, bl, the definite integral [! f (x) dx can be interpreted geometrically as the measure of the area of the region R shown in Fig. 7.i.4. Equation (11) can be used to find the exact value of a definite integral as illustrated in the following example. soLUrIoN: Consider a regular partition of the closedinterval [1, 3] into n subintervals.Then A^x: 2ln. If we choose ti as the right endpoint of each subinterval, we have € r : L + 1 , € z : 1 * r G ) , t s : L+3G), €i:1*'(1), tn:1 *"(t Becausef (x): x', f c,): (r* +)' : (! j-4)' Therefore,by using Eq. (11) and applying properties and formulas from Sec.7.'1,, we get I' *' dx:lri) : (+)'t" )n lim h-*a +s n"? 1:l (n' t 4ni -l 4i2) 7.3 THE DEFINITEINTEGRAL : lim ll- *6 #l*z 'l., t 4n 295 n n i+4 i'] i:r i:r :Ii hl"'" -,r,+ n ' n ( n * L ) ,- T4l n ( n+ t ) ( 2 n + 1 ) l Z :,li #l**2nst2n2.Wl :.llils*f*8n2 -;+ L 2n + I Jn' :,Illo*f* 8g -, 4i -, 4f g"') :6*O+8+0+0 _83 (1, L) Figure7.3.5 We interpret the result geometrically. Because xz = 0 for all x in the x axis, and the lines [1,3], the region bounded by the curve A:x', : 3 has an area of 83 square units. The region is shown x: 1 and x in Fig. 7.3.5. Exercises7.3 In Exercises1 through 6, iind the Riemannsum for the function on the interval, using the given partition A and the given valuesof $. Draw a sketchof the graph of the function on the given interval, and show the rectanglesthe measureof whose areasare the terms of lhe Riemannsum. (SeeIllustration 1 and Fig. 7,3,2.) 1. f (x) : f , O s r = 3; for A: xo: 0, xr: tr,xz: 7*,rs : 2i, xe: 3; : L ta : 1' &: lt, €n= 2i 2.f(x):f,g <r=3;forA: to:0,h:1,x":ll,xs:2,ra:2*,x5:3;t,:i, 1 2 : 1 , h = 7 1 ,E 4 : 2 1 ,€ r= 2 1 3 .f ( x ) : 1 1 r , 1 < r s 3 i f o r A : x o : 7 , x ' : 7 & , r z : A , x " : 2 4 , r n : t t * : 7 1 , 4 2 : 2 ' t r : 2 t , € ' : 2 1 L f(x):f ,-1 =x=2;fotA to:-'1,xr:-1,xn:l,7r:l,xn:lLxo:2;h:-t,t":0,€e:&':7'€s:li 5 . f ( x ) : f - v a 1 , o = x = 7 ; f o r A : r o : 0 , 1 , : 6 . 2 ,a : 0 . 5 , 4 : 1 . 7 , x n = 7 t 1 ' : o . r , t " : 0 . 4 ' & : 0 . 6 ,e 4 : o - 9 6 . l ( x ) : 1 1 1 r a 2 1 ,- 1 < r s 3 ; f o r A : r o : - 1 , h : - + , x 2 : O ,x " : 1 , a n : l l , x s = 2 , r a : 2 L h : 2 1 , x s : 3 ; h : - L ( " : O ,( " : l , ( , : 1 , & : r i , € 6 : 2 , e 1 =2 t r i, s = 3 In Exercises7 tfuough 14,find l:re exactvalue of the definite integral. Use the method of Example2 of this section. f2 J;J,oo* 8 I:xzdx s' L' x3dx THE DEFINITE INTEGRAL ,o ,r. dx I:,xa I" e,*4x*s) dx t dx t e,*x-6) tn. (4x'- Bxz) dx I:, ".1s: ,[_,(r,* 1,)dx In Exercises 15 through 20, find the exact area of the region in the following way: (a) Express the measure of the area as the limit oI a Riemann sum with regular Partitions; (b) express this limit as a definite integral; (c) evaluate the definite integal by the method of this section and a suitable choice of fr. Draw a figure showing the region. 15. Boundedby the line y:2a- 1,,he r axis,and the lines r:1 and r:5. - 5, *e r axis, and the lines x:4 arrdr:2. 17. Bounded by the culve y:4 the r axis, and the lines r:1 and r:2. "e, 18. Bounded by the curve y : (r + 3)r, the r axis, and the lines r : -3 and .r: 0. 19. Bounded by the c:.rrvey :12 - , - f, the r axis, and the lines .r: -3 and r:2. 20. Bounded by the c'lrrrcy:6r+ f - d, the r axis, and the lines r: -1 and r:3. 16. Bounde{ by the line y:2r 21. ExpresEaEa definite integral: iim j 22. Expressa8 a definite inteFal (8ir/n3). (HrNr: Consider the function I for which l(r) : d.) j Um f71n*i). -1'," (HrNr: Consider the tunction f for which f(x):1lx on [1, 2].) 23. Expressas a definitet.:"O", (n/i:). (HrNr:Considerthe tuncrionI for which l(r) : Ul.) ,g : 24,Letla,blbeanyintervalsuchthat4<0<b.Provethateventhoughtheunitstepfunction(Exerci6e21 in Exercises 1.8) is dtucontinuous on [q, b], it is i^tegrableon [a. b] and t! U(r) dt: b. 25. Prove that the signum function is discontinuous !r sgnx dx:0. on [-1, 1] and yet integrable on 1]. Furthermore show that 26. Prcve that the greatest integer function is discontinuous on [0, *] and yet integrable on [0, sl. Furthermore, show that ftp [xn dx:i. 7.4 PROPERTIES OF THE DEFINITE INTEGRAL 7.4j1, Theorem Evaluating a definite integral from the definition, by actuatly finding the limit of a sum as we did in S,ec.7.3,is usually quite tedious and frequently almost impossible. To establish a much simpler method we first need to develop some properties of the definite integral. First the following two theorems about Riemann sums are needed. If A is any partition of the closed interval [a, bl, then lim llall-o b-a PROOF: n L&-(b-a):(b-a)-(b-a) i:r OF THE DEFINITEINTEGRAL 7.4 PROPERTIES 297 Hence, for any e ) 0 any choice of E > 0 guaranteesthe inequality ln lU I or - (b- o,l . . <s whenever lfAll and so by Definition 7.3.1, n I l i m -) A r r : b - a llAll-o i:1 7.4.2 Theorem If f is defined on the closed interval la, bl and if li* f, f G,)o,, ilA|-0 r:1 exists, where A is any partition of [a, b], then if k is any constant, nll lim ) k/(f,) L#: k lim 2 fG) L,x l l a l l - ot : r l l a l l - o3 r The proof of this theorem is left as an exercise (see Exercise 1). We now state and prove some theorems that give important properties of the definite integral. 7.4.g Theorem If the function / is integrable on the closed interval la, bl and if k is any constant, then pRooF: Because / is integrableon la,r], if ,,t^1il. ClAix exists,and so by Theorem 7.4.2 nn lim >,kf(€,) Lix: k lim 2 fG) t,, Ft -J - l l a l l - oF r -' - llall*o Therefore, fb rb Ja Ja' I kf(x)dx:kl 7.4.4 Theorem f@)dx I If the functions f and 8 are integrable on La,bf, then f + S is integrable on 298 THE DEFINITEINTEGRAL PRooF: The functions f and g are integrable on la, bl; thus, let fb rb J" f @)dx: M and dx: N J" SQ) To prove that / * g is integrable on la, bl and that IP, lf (x) + g(r) f dx : M+ N, we must show that for any € > 0 there exists a 6 > 0 such that ln r + s(f,)la#- (M+ N)l. . t/(f,) l> li--r I t- forall partitions A for which llAll< 6 and for any 4i in lxr_r,xil. Because M:,Jim t fG,) o,* and IY: fim XS(f,) a,t l l a l l- o t : t llall-oi:r it follows that for any e ) 0 there exist a 6, ) 0 and a 6z ) 0 such that .*z and r(ri)Aix-rl.f - Air- utl r,*,, l> l7:r' | lp-, for all partitions A for which llAll< 61 and llAll< 6r, and for any fi in fxr-r, ri]. Therefore,if 5 : min(6r, 6r), then for any e ) 0 Air-,l*l> r(ri)Air-Nl.f ' 2+i.:, l>rte,l le' I l?r"" | (1) for all partitions A for which llAll< 6 and for any fi in lxi_r, xif. By the triangle inequalit!, we have \ /n \l Aix- *) * (=>,t(fi) Air- t)l l(p fre,l l/n = fG,)Lic-rl * lI ,^, s(fi) Air- Nl From inequalities (L) and (2), we have lln m \ I lx* ) s(f,)t,r) - (M+N)l . . l() ftr'l (3) From Property 3 (7.1,.3) of the sigma notation, we have nnn > /(4') A,r* ) s((,) Lix: > t/(f,) + s(f,)l Lix ," ;; ,"bstitutin ,: n"^(4) e)0 ln (4) into'O, *" are able to concludethat for any I +s(f,)lL&- (M+N)l.. l}_,ffrc; for all partitions A for which llAll< 6, where 6: min(6r, 5r) and for any€i OF THE DEFINITEINTEGRAL 7.4 PROPERTIES r2l. This proves in lxia, that f * g is integrable 299 on [a, bl and that dx:| ,o, dx* [' su)a* f rru>+ s(r)I Theorem 7.4.4 can be extended to any number of functions. That is, if the functions fr, fr, . . , fn are all integrable on la, b), then (ft+ fr* ' ' ' + f,) ,s integrable on [a,bf and rb IJ a lf,(x) + f"(x) + . . . + f"(x)l dx : lo f,(x)dx* fuf,(*) dx* ' ' ' + [' f,(x) dx Jct Ja Jo'" The plus sign in the statementof Theorem7.4.4canbe replacedby a -'l'. minus sign as a result of applying Theorem 7.4.3,where k: 7.4.5 Theorem If the function / is integrableon the closedintervals la, bf,la, c], andlc, bf, f rr.tdx:["r,,,tdx* [' fr*>o* wherea<c<b. pRooF: Let A be a partition of la, bl. Form the partition A' of la, bl in the following way. If c is one of the partitioning points of A (i.e., c: xi for some i), then A' is exactly the sameas A. If c is not one of the partitioning points of A but is contained in the subinterval lxr-r, ril, then the partition A' hur as its partitioning points all the partitioning points of A and, in addition, the point c. Therefore, the subintervals of the partition A' are the same as the subintervals of A, with the exception that the subinterval lxi-r, xi] of A is divided into the two subintervals [4-r, c] and lc, xofIf llA'llis the norm of A' and if llAllis the norm of A, then = llall llA'll If in the partition A' the intervalla, cl is divided into r subintervals and the interv il lc, bl is divided into (n - r) subintervals, then the part of the partition A' from a to c gives a Riemann sum of the form r 2fG) ^,x other part of the partition A', from c to b,gives a Riemann sum of ""0;" the form n THE DEFINITEINTEGRAL Using the definition of the definite integral and properties of the sigma notation, we have fbn I .f0) dx: lim ). 'f(f,) A'r Ja l l a l l - . oG i : [> rtc,lA;x* i, /tr,to,'] ,,t^1T^ llall-o ui:r : i:r*l f, rrclAir* ,*il. i,f {t) o'' ,,Til, Since0 < llA'lls llAll,we canreplacelAll- 0 by llA'll- 0, giving us fbrn 'f @ dx: JI a lim > /(6,) - A;x* lim ) l l a , l-lo i ? * r l l a , l l -A o' f (€,)Lt Applying the definition of the definite integral to the right side of the above equation, we have y: f(x) fb fc fb I f@ dx: JIa ' fG) dx* J|e t'Q)dx Ja' o rLLUSrRArroN 1: We interpret Theorem 7.4.5 geometrically. If (x) > 0 f for all x rn fa, b], then Theorem 7.4.5 states that the measure of the area of the region bounded by the curve y : f (x) and the x axis from a to b is equal to the sum of the measures of the areas of the regions from a to c and from c to b. See Fig. 7.4.1. o F i g u r e7 . 4 . 1 The result of Theorem7.4.5 is true for any ordering of the numbers a, b, and c. This is stated as another theorem. 7.4.6 Theorem lf f is integrable on a closed interval containing the three numbers a,b, and c, then Irutdx:f" rul dx* f' f ul n, (5/ regardless of the order of.a, b, and c. PRooF: If a,b, and c are distinct, there are six possible orderings of these t h r e e n u m b e r sa: < b I c , a 1 c < b , b 1 a 1 c , b 1 c 1 a , c 1 a 1 b , a n d c < b < a. The second orderi ng, a I c 1 b,i sTheorem7.4.5.W e makeuse of Theorem 7.4.5 in proving that Eq. (5) holds for the other orderings. Suppose that a < b < c; then from Theorem 7.4.5 we have fb fe fe Ju- Jtr' Ja' I fQ) dx* | f(x) dx: I fG) dx (6) INTEGRAL OF THEDEFINITE 7.4 PROPERTIES From Definition 7.3.4 lc fb dx | f(x) ' I f(x) dx:* Jc Ja' Substituting from (7) into (5), we obtain fb fb fc fb fc fb Ja' Ja' I f(*) dx- JIc ' fk) dx: JIa f(x) dx Ja' Thus, I fU) dx: I f(r) dx* J|e f(x) dx which is the desired result. The proofs for the other four orderings are similar and left as exercises(seeExercises2 to 5). There is also the possibility that two of the three numbers are equal; for example,a: c 1 b. Then rc fa I' f trl dx: J|a /trl dx:0 (by Definition7.3.5) Ja' Al s o , b e c a u s ea : rb c, fb I f @ ) d x : Jl a f ( x ) d x Je' Therefore, fc fb fb I f U ) d x * lJ " ' f @ l d x : o + fJ a 'f ( x ) d x Jo' I which is the desired result. 7.4.T Theorem If k is any constant and tf.f isa function such that /(r) : k for all r in la, bf, then fb rb I f t l . )d x : | r ca x : k ( b a ) Ja' Ja PRooF: From Definition 7.3.2, fbn I f (*) dx: lim -)i : r/({,) A,r Ja' llall-o : n l i m Ll! kAox l l a l l - oi : r : k ,.lim )i : r O,t (by Theorcm7.4.2) llall-0 : k(b - a) (by Theorem7.4.1) THE DEFINITEINTEGRAL 2 : A geometric interpretation of Theorem 7.4.7 when 0 is given en by Fig. 7.4.2. The definite integral fil t dx gives the measure of the area of theel sha ded region, which is a rectangle whose dimensions are k units an ar md (lb7 - Q ) units. . o r ILLUSTRATIO IION 7-: n/ > 7.4.8 Theorem IIff tthe functioons / and 8 are integrable on the closed interval [a, bl and if fk' ) > g(r) foor all xr i n [a, b7,then f( fb f 8 (x) dx I rutdx >- l" pRooF: Because and g are integrableonfa,bl, f [! f (x) dx and ff g(x) dx both exist. Therefore(by Theorem 7.4.3), fb,fbrbfb I f f r l d x - JIa -s ( " )d x : lJ o ' f t * l d x * JIn t - s e ) l d x Jo' Figure7.4.2 rb : J" [f {*) - S G)] dx (by Theorem 7-4.4) LeI h be the function defined by h(x):f(x)-sG) Then h(x) - 0 for all r in la,bl becausef (x) > g(x) for all x in la, bl. We wish to prove that Il lf G) - s?)l dx = 0. fb I lf (x)- s(r)l d x : Ja fbn I h(x) dx: Ja lim ). ft(f,) Air l l a l l' o F r (8) Assume that (e) l i m L,l) h ( { , ) A i x : L ( 0 llAll-0 i=1 Then by Definition 7.3.1, with e:-L, ,(fi) Air- .l . -t l> li:l I t ? ' l there exists a 6 > 0 such that whenever llAll< 6 But because n ln h(t) A,*- L = l t h(t) Aix- Ll > i:r l-t li:l from inequality (10)we ha VC n tt(il A,ix- L < - L 2 i:r wheneverllAll< D (10; OF THE DEFINITEINTEGRAL 7.4 PROPERTIES n h(il A;r ( 0 wheneverllAll< S (1 1 ) i:r But statement(11) is impossible becauseevery h((1) is nonnegative and every L,g 7 0; thus, we have a contradiction to our assumption (9). Therefore, (9) is false, and lim ). ft(f,) A1x > 0 (J2) LJ llall*o t:t From (8) and (L2),we have fb IJ a t/tt) - s(r)I dx> 0 Hence, ["r<o dx- dx>o [' sal and so f(x) g(r) x dx> f sroa' f tal . rLLUSrRArroN 3: Figure 7.4.3 gives a geometric interpretation of Theorem 7.4.8 when f(x) > g(x) > 0 for all r in la,bl. Il f @ dx gives the measure of the area of the region bounded by the curye y: f (x)'the x axis, and the lines x: A and r : b. Ig g@) dxgives the measure of the area of the region bounded by the curve y : 8(x), the r axis, and the lines x: a and r : b.In the figure we see that the first area is greater than the second area. Figure7.4.3 7.4.9 Theorem Supposethat the function / is continuous on the closedinterval la, bl.If m and M are, respectively, the absolute minimum and absolute maximum function values of f on la, bl so that m-f(x)=M m ( b- a ) fora-x=b =r f(x) dx < M(b a) pRooF: Because / is continuous on la, bl, the extreme-value theorem (4.5.9) guarantees the existence of m and M. THE DEFINITE INTEGRAL By Theorem 7.4.7 fb J, * d x : m ( b- a ) (13) , dx: M(b- s) (14) and fb J" Because/ is continuous on fa, bl, itfollows from Theorem 7.3.j that/ is integrable on fa, bl. Then because f (*) > m for all r in [a, b], we have from Theorem 7.4.8 fb fb I f ( * ) d x = lJ n m d x Ja' which from (13) gives fb I f' @ d x > m ( b - a ) Ja (15) Similarly, becauseM = f (x) for all .r in la, bf, it follows from Theorem7.4.8 that f,,,dx>[)roo, which from (14) gives M(b_-a)=-f) fato- (16) Combining inequalities (15) and (16) we have v: f@) Figure7.4.4 nxevrprn1: Apply Theorem7.4.9 to find a smallestand a largest possiblevalue of r4 |r l 2 ( r ' - 6 x 2 * 9 x + 1 . ) d x J m(b-a)=['fildx= M(b - a) o rLLUSrRArroN4: A geometric interpretation of Theorem 7.4.9 is given in Fig.7.4.4, where f (x) > 0 for all x in la, b). The integral t!, (x) dx gives the f measure of the area of the region bounded by the curve y : (x),the r axis, f and the lines x: A and x : b. This area is greater than that of the rectangle whose dimensions are m and (b - a) and less than that of the rectangle whose dimensions are M and (b - a). o solurroN: Referring to Example 1, Sec.5.1, we see that has a relative / minimum value of 1 at x:3 and a relative maximum value of 5 at x:"1.. and f $):5. Hence, the absolute minimum value of / on 1i, 41isI, f e):# and the absolute maximum value is 5. Taking m: '!. and M : 5 in Theorem 7.4,.9,we have L(4-+) = f- (rr- 6xz*9x*t) dx=5(a-i) I rtz OF THE DEFINITEINTEGRAL 7.4 PROPERTIES Use the results of ExamPleL, Sec.5.1. and so t=l r4 (x'-6x2* 9x*t) dx <3rE J rtz In Examplel , 9ec.7.6,the exact value of the definite integral is shown to be 10*?. From Example 3, sec. 5.1, we see that/has a relative minimum solurroN: -3 Furthermore, f (l ):5. H ence, on [-L, 1] the abso a t x :-L. value of lute minimum value of / is -3 and the absolute maximum value is 5. So, taking m: -3 and M : 5 in Theorem 7.4.9, we have Exevrpr-E2: Apply Theorem 7.4.9 to find a smallest and a largest possible value of rr | (*n,t * nrtrs) dx (-3)[1 - (-11] = f' (r4l3+ Ax't")dx <sU - (-1)l J-t J-t Use the results of ExamPle3, Sec.5.L. Therefore, _6 < |rr (*n," + Axttt) dx < t0 Jt g The exact value of the definite integral is as shown in Example 2,Sec'7.6' Exercises7.4 L. Prove Theorem7.4'2result of Theorem7.4.5. In Exercises2 througlnT, Prove that Theotem7.4.6is valid; in each caseuse the 4. c<a<b 3.b<c1a 2.b<a1c 7. a:b < c 6. a < c:b 5. c<b<a largest possible value of the given integral. In Exercises8 thro ugh 17, apPly Theorem 7.4.9 to find a smallest and a , tt. t kxdx r6 ,r. t/3 + x dx J_, f,(xn - vxz* 16)dx 'ur_,fia, 'nr,ffio. 17 . tx2 itx tO. lr J_, (r + 1)ztsdx r0 tt. J_r(xn - 8xz i L6) dx f2 dx tU. J_rt/xz+5 I w-zldx 1 8. Show that if / is continuous on [-L, 2], then dx+ rel ax+ [-' f (x)dx:, dx+ I' f ,r<.1 [' f k) 19. show that I x dx > I' *, dx but f' * a* = Ii o dx. Do not evaluatethe definiteintegrals. 306 THE DEFINITEINTEGRAL 20. If / is continuous on la, b], prove that l['-lfro*l=I:lf(x)ldx (Hrur: (x)l = f (x) s l/(x)l.) 7.5 THE MEAN-VALUE Before stating and proving the mean-value theorem for integrals, w€ THEOREM FOR INTEGRALS discuss an important theorem about a function that is continuous on a closedinterval. It is called the intermediate-aalue theorem,and we need to use it to prove the mean-value theorem for integrals. 7.5.1 Theorem I nt erm edint e-V ahte Tlrcor ent v f(b) k f(a) F i g u r e7 . 5 . 1 y : f(x) If the function / is continuous on the closedinterval [a, bl andif f (a) + f (b), for any number k between (a) and (b) there exists a number c f fhen f between a and b such that f (c) : 16. The proof of this theorem is beyond the scope of this book; it can be found in an advancedcalculustext. However, *" dir..rss the geometricinterpretation of the theorem. In Fig.7.s.1, (0,k) is any point-on the y axis between the points (0,f (a)) and (0,f (b)).TheoremT.s.j.statesthat the line y : k must intersect the curve whose equation is y : (x) at the point f (c'k), where c lies between a and b. Figure 7.5.1.shows ihir it,t"rsection. Note that for some values of k there may be more than one possible value f.ot c. The theorem statesthat there is always at least or" lrulue of.c, but that it is not necessarilyunique. Figure 7.5.2 showsthree possible values of c (cr, cr, and c.r)for a particular k. Theorem 7.s.j,statesthat if the function / is continuous on a closed interval [a, bf, then / assumesevery value between f (a) andf (b) as.r assumesall values between a and.b.The importanceof the continuity of /on la,bl is demonstratedin the following illustration. . ILLUSTRATTON 1: Consider the function I defined by tf \('x- \/ : [ x , - 1 lx" ifo<x<2 if2<x<3 A sketch of the graph of this function is shown in Fig. 7.s.g. The function / is discontinuous at 2, which is in the closed interval I0' 3l;/(0) : -L and f (3):9. rf k is any number between 1 and 4, there is no value of c such that f(c)-k because there are no function values between L and 4. . Figure7.5.2 Another function for which Theorem 7.5.7 does not hold is given in the next illustration. . rLLUSrRArroN 2: Let the function g be defined by 8(r): x- 4 FORINTEGRALS 307 THEOREM 7.5 THEMEAN-VALUE A sketch of the graph of this function is shown in Fig. 7.5.4. The function g is discontinuous at 4, which is in the closed interval -1 and 2, there is -1 and 5l; 8(5) : z.If k is any number between 12, g(2) : : no value of c between 2 and 5 such that g(c) k. In particulat,lf k: 1, then ' 8(6) : L, but 6 is not in the interval 12,51. Figure7.5.4 Figure7,5.3 nxevmn 1: Given the function / defined by f(x):4*3x-x2 2=x<5 verify the intermediate-value theorem if k: L. - -6' s o L U rIo N : f (2 ) : 6; f (5) T o fi n d c , s et f (c): k:1; that i s, 4*3c-c2:t c2-3c-3:0 This gives v- g! \E 2 because this number is outside the interval We reject + (3 - \E) and 12,51.The number* G + {n ) is in the interval [ 2 , 5 ] , rF+!n\:t /\ z / We are now ready to state and prove the mean-value theorem for integrals. 7.5.2 Theorem MeturVslttc Theoretn for lntegrnls If the function / is continuous on the closed interval la, bl, then there exists a number X such that a = X < b, and fb IJ a ' f (*) dx: /(x) (b a) 308 THE DEFINITEINTEGRAL pRooF: Because is continuous on / fa, bl, fuom the extreme-value theorem (4.5.9)/ has an absolute maximum value and an absolute minimum value on [a, b]. Let m be the absolute minimum value occurring at x: x^. Thus, A=x*<b f(x*):m (1) Let M be the absolute maximum value, occurring at x: a<xp1 <b f(x")-M rr. Thus, (2) We have, then, m =f(x) -M forall xtnla,bl From Theorem 7.4.9it follows that m ( b- a ) = [ ' f f * l d x - M ( b - a ) Ja(b - a) andnoting that (b - a) is positive because b ) a,we Dividingby get lb m I fQ) a' <*_ 0-a <M But from Eqs. (1) and (2) , /n: f (x^) and M : (x*), and so we have f fb I f(x) dx =f(x*) f(r^)=eVo (3) From inequalities (3) and the intermediate-value theorem (7.5.7),there is some number X in a closed interval containin g x* and ron such that /(x): or t@) fb | fk)dx:f(x)(b-a) Jo" , Figure7.5.5 a<x=b I . rLLUSrRArroN3: To interpret Theorem 7.5.2 geometrically, we consider f(x) > 0 for all values of x in la, bl. Then tl fG) dx is the measure of the area of the region bounded by the curve whose equation is y : (x), the f x axis, and the lines x: a and x: b (see Fig.7.s.il. Theorem7.s.2 states that there is a number X in la, b] such that the area of the recta ngle AEFB of height /(X) units and width (b - a) units is equal to the area of the r€Bion ADCB. . The value of X is not necessarily unique. The theorem does not provide a method for finding X, but it states that a value of X exists, and 7.5 THE MEAN-VALUETHEOREMFOR INTEGRALS 309 this is used to prove other theorems. In some particular cases we can find the value of X, as is illustrated in the following example. 2: Find the value of X EXAMPLE such that f3 I 'f ( * ) d x - l ( x ) ( 3 - 1 ) Jr if f (x) : )c2.Use the result of Example 2 in Sec.7.3. solurroN: In Example2 of Sec.7.3, we obtained r3 I x 2d x : 8 3 JL Therefore,we wish to find X such that ' (2) :zfL /(x) that is, X2: # Therefore, X: _+_ +\/99 We reject -+\E since it is not in the interval [1, 31,and we have f3 dx- fG\E)(3 - 1) I f<rl ' Jr The value /(X) given by Theorem 7.5.2is called the meanaalue (ot aaerageaalue)of / on the interval la, bl. It is a generalization of the arithmetic mean of a iinite set of numbers. That is, if {f (*r),f (xr), ' ' ' 'f (x")} is a set of n numbers, then the arithmetic mean is given by n 2fk,) i:l n To generalizethis definition, consider a regular partition of the closedinterval [a, bl,which is divided into n subintervals of equal length Ar: (b - a) ln Let f, be any point in the ith subinterval. Form the sum: n /(f,) > i:r ( 4) n The quotient (4) corresPonds to the arithmetic Because Lx: (b - a) ln, we have n: b-a Ax Substituting from (5) into (4), we obtain mean of. n numbers. (s) 310 THE DEFINITEINTEGRAL or, equivalently, tt ) /(f,) a' t:1 b-a Taking the limit as n -> *m (or Ax -- 0), we have, if the limit exists, nfb I /(4')a' l!r_ e!_ b-a ,_;* : | to 1tg)ax = b-a This leads to the following definition. 7 '5'3 Definition If the function / is integrable on the closed interval la, bf, the aaerageaalue of / on [a, b] is rb f{r) a* -J, -F, EXAMPLE3: Find the average value of the function / defined by f (*) : x2 otr the interval [1,,3]. SoLUrroN: In Example 2, Sec. we obtained r3 I x2dx: +q Jr So if A.V. is the average value of f on 11,gl, we have _ -236 - 3-1 4^ _rJ 3 An important application of the averagevalue of a function occursin physics and engineering in connection with the conceptof center of mass. This is discussedin the next chaDter. In economics,Definition 2.5.1can be used to find an averagetotal cost or an averagetotal revenue. o rLLusrRArroN4: If the total cost function C is given by C(r): #, where C(r) thousands of dollars is the total cost of producing 100r units of a certain commodity, then the number of thousandsof dollars in the average total cost, when the number of units produced takes on all values from 1O0to 300, is given by the averagevalue of C on [1, 3]. From Example 3, this is rra:4.33. Hence, the averagetotal cost is $4333. o Exercises7.5 ln Exercises 1 through 8, a function / and a closed interval [a, D] are given. Determine if the intermediate_value theorem holds lor the given value of ft. If the theorem holds, find a number c sucihthat c) : lc. If the theorem does not hold, give the f( reason. Draw a sketch of the curve and the line v : k. THEOREMOF THE CALCULUS 7.6 THE FUNDAMENTAL 311 1. f(x):2 + x - *t la,bl: [0,3]; k: t 2. f({: f a5r-6;la,bf : l-1,2l; k: a 3. f (x) : tfi=P, lq, b1: l-a.5,37;k: 3 a. fk):-vffi=F; fa,bf : f0,81;k: -B 4 s- f {x\ : ;}1; (1*r 6 .f @ : 1 ; - ; (*-+ ?.f@ : lf _; . ,la, bl : [-3, 1]; k: t it-4 =x =-?1 i-i., - 1 l; Io'bl:[-4'r],k=L it-Z-t<t) ri r =l = i '], [s,bl: l-2' 3];k: -1' e. t k) : r;-_ y ts,b): [0,7h k : 2 In Exercises 9 through 13, find the value of X satisfying the mean-value theorem for integrals. For the value of the delinite integral, use the resilts of the conesponding Exercisd 9 though 13 in Exercises 7.3 Draw a figure illustrating the application of the theorem. t. ,, f rdx f Q'*x-6)dx t to. dx I_,x4 tt [:,(r'* f (x,* 4x*s)dx t) dx 14. Find the averagevalue of the function / defined byl(r) : f on the interval [2, 4], and Iind the value of r at which it occurs, Make isketch. Use the rcsult of Exercise 8 of Exercises 7.3 for the value of the definite integral. 15. Suppose a ball is dropped from rest and after t sec its velocity is ? ft/sec. Neglecting air r€sistance, exPressr,'in terms of t as v: f(t) and find the averagevalue ot f on 10,21. (r) : 13.Find the 16. Suppose C(r) hundreds of dollars is the total cost of producing 10r units of a ce*ain commodity, and C 7 in ExerExercise result of from 0 to 20. Use the *hen the number of units produced takes on all values .rr6"ig" tot"l "ort integral. cises7.3 for the value of the definite : V16-7 on the interval [-4, 4]. Diaw a figure. (HrNr: Find t Z. Find the average value of the tunction I defined by /(r) the area of a region endosed by a semicircle.) measure of the value of the definite integral by interpr€ting it as the - f on the interval [0, 7]. Draw a figure. (I{INr: Find the 18. Find the averagevalue of the functionl defined byl(r) = @ it as the of the area of a region enclosed by a quarter-circle.) measure interpreting integral by difinite value of the 19. Show that the intermediate-value theoreqr guarantees that the equation r3 4l * r + 3 : 0 has a rcot between I and 2. 20. Suppose f is a function for whidr 0 = f(x) =l if 0 = r = 1. Pncve that ifl is contrnuous on [0, 1] there is at least one (urNr: If neither 0 nor l qualifies as c, then l(0) > 0 and l(1) < 1- consider the in [0, 1] sudr thatl(c):c. trr.-b"" " : x ar\d apply the intermediate-value theotem to 8 on [0, 1] .) function g for which g@) l(x\ 21..lf f is continuous on [a, b] and Jl lG) dx:o, 7.6 Tp.E FUNDAMENTAL THEOREM OF THE CALCULUS prove that there is at least one number c in [c, b] such thatl(c):0. Historically the basic conceptsof the definite integral were used by the ancient Greeks,principally Archimedes (287-212B.c.),more than 2000years ago,which was many yearsbefore the differential calculuswas discovered. 312 INTEGRAL THE DEFINITE In the seventeenth century, almost simultaneously but working independently, Newton and Leibniz showed how the calculus could be used to find the area of a region bounded by u curve or a set of curves, by evaluating a definite integral by antidifferentiation. The procedure involves what is known as the t'undamentaltheorem of the calculus.Before we state and prove this important theorem, w€ discuss definite integrals having a variable upper limit, and a preliminary theorem. Let f be a function continuous on the closed interval la, bl. Then the value of the definite integral ft f Q) dx depends only on the function f and the numbers a and b, and not on the symbol r, which is used here as the independent variable. In Example 2,Sec.7.3,we found the value of .frtx2dx to be 83. Any other symbol instead of x could have been used; for example, l' ,' or: u2du: r2dr- 8a 1," l,' If / is continuous on the closed interval fn, bl, then by Theorem 7.3.3 and the definition of the definite integrcI, [! f (t) dt exists. We previously stated that if the definite integral exists, it is a unique number. If r is a number rnfa, b], then/is continuous onfa, x] because it is continuous on [a, bl.Consequently, If f ft) df exists and is a unique number whose value depends on x. Therefore, I{ fU) dt defines a function F having as its domain all numbers in the closed interval [a, b] and whose function value at any number x in la, bl is given by _r"f(t) dt F(x) _ f(t) F i g u r e7 . 6 . 1 7.6.1 Theorem J, (1) As a notational observation, if the limits of the definite integral are variables, different symbols are uged for these limits and for the independent variable in the integrand. H8nce, in Eq. (1), because x is the upper limit, we use the letter f as the independent variable in the integ.ur,d. If, in Eq. (1),f (t) > 0 for all values of f in la, bl, then the function value F (x) can be interpreted geometrically as the measure of the area of the region bounded by the curve whose equation i" y: f (t), the f axis, and t h e l i n e s t : a a n d f : x . ( S e e F i g . 7 . 6 . 1 ,N. )o t e t h a t F ( a ) : I t fU) dt,which by Definition 7.3.5 equals 0. We now state and prove an important theorem giving the derivative of a function F defined as a definite integral having a variable upper limit. Let the function / be continuous on the closed interval la, bl and let x be any number in la, bl.If F is the function defined by F(x):f ru>0, then F'(x): f(x) THEOREM OF THECALCULUS 313 7.6THEFUNDAMENTAL (If x: x:b, a, the derivative in (2) may be a derivative from the right, and if the derivative in (2) may be a derivative from the left.) pRooF: Consider two numbers 11 and (r1 + Ar) in [a, b]. Then F(r,): dt f' fG) and F(x,* Ax)- ft) at f'*^* f so that F(x,* Ar)- F(r,): (t) dtdt f' f U) f'*'" f (3) By Theorem7.4.6, f e:r+tt f Jrr+Lx f rt dt: or* J" fG\ J., f(t) J, fQ)dt or, equivalently, rrr+ir f n+ Lr lrt fo dt- J, fG)dt: J*, f(t)dt J, (4) Substituting.from Eq. (4) into (3), we get F(xr-f,ax)- F(x,) : f,'*o" f(t) dt (s) By the mean-value theorem for integrals (7.5.2),there is some number the closed interval bounded by x, and (r, + Ax) such that f xr+L.r J., f (t) dt: 1(X)Ar From Eqs. (5) and (6), we obtain F ( x 1* A x ) - F ( r ' ) : / ( X ) A r or, if we divide by Lx, F(x'*Ax)-F(x'):f(X) r \--/ Ar Taking the limit as Ax approaches zerol we have ,.- F(rr * A{) - F(xr) : ,n";;'ide lim f (X) (7) or,* ,r, i, r' (',i"il'u"r"r-ine lim f (x), recallthatX is in 314 THE DEFINITEINTEGRAL the closedinterval bounded by .rr and x, * Lx, and because : ll.T r,: r, and llq (r, * Ar) r, it follows from the squeeze theorem (4.3.3) that continuous at .x1,w€ have lim /(X) : l:l*:xt.Because/is lim f (X) : f (xt); thus, from Eq. (7) we get F'(xr):f(xr) (8) If the function/ is not defined for values of x less than abut is continuous from the right at a, then in the above argumenl, if. xr: a rn Eq. (T), A,x must approach 0 from the right. Hence, the left side of Eq. (g) will be Fi(rr). Similarly, if f is not defined for values of x greater than b but is continuous from the left at b, then if xr: b inEq. (7), A,x must approach 0 from the left. Hence, we have F'_(xr) on the left side of Eq. (8). Because r, is any number in la, bl, Eq.(8) states what we wished to Prove. I Theorem 7.6.L states that the definite integral t{ f(t) dt,withvariable upper limit r, is an antiderivative of f. 7.6.2 Theorem F t r n d n m e n t aT l heorem of the Cnlutlus Let the function / be continuous on the closed interval la, bl and let g be a function such that g'Q):f(x) o) for all x in fa, bl. Then fb I f U )d t : s f t ) - s ( a ) Ja' (If x: x:b, a, the derivative in (9) may be a derivative from the right , and if the derivative in (9) may be a derivative from the left.) pRooF: If is continuous at all numbers in / [a, bf, we know from Theorem 7.6.t that the definite integral Ii f (t) dt, with variable upper limit r, defines a function F whose derivative on la, bl is Becauseby hypothesis /. 8'(x) : f (x), it follows from Theorem 6.3.3 that fr 8 ( r ) - l fJ < tldt+k a (10) where k is some constant. Letting x: b and r : a, successively, in Eq. (10), we get :T: 8@) f(t) dt + k (11) THEOREMOF THE CALCULUS 315 7.6 THE FUNDAMENTAL and (12) s(a)-f,urdt+k From Eqs. (11) and (L2), we obtain 8(b)-s@\:["fUl ot- dt f fu) But, by DefinitionT.3.5, f Al dt:0, and so we have [" fb s(b)- s@): J, f U)at which is what we wished we Prove. If f is not defined for values of x greater than b but is continuous from the left at b, the derivative in (9) is a derivative from the left, and we have g-@) : F'-(b),from which (11) fotlows. Similarly, if f is not defined for values of x less than a but is continuous from the right at a, then the derivative in (9) is a derivative from the right, and we have 7n(a):F'*(a), from I which (12) follows. We are now in a position to find the exact value of a definite integral by applying Theorcm 7.6.2.In applying the theorem, we denote lD ts(b)- stu)l by s(r)l, o rLLUSrRArroN 1: We apply the fundamental theorem of the calculus to evaluate r3 I x2dx Jr Here, f(x): x2. An antiderivative of.x2 is ixt. From this we choose .Y3 g(x): ? Therefore,from Theorem 7.6.2,we get ["*'dr:+-l':t-* 3l' Jr* 3 :83 Compare this result with that of Example 2, Sec. 7.3. Because of the connection between definite integrals and antiderivatives, we used the integral sign J for the notation J/(r) dx for an an- 316 THE DEFINITEINTEGRAL tiderivative. We now dispense with the terminology of antiderivatives and antidifferentiation and begin to call I f Q) dx the indefiniteintegralof "t' of x, dx." The processof evaluating an indefinite integral or a definite integral is called integration. The difference between an indefinite integral and a definite integral should be emphasized. The indefinite integral Ifk) dx is defined as a function g such that its derivative D*[g(x)]: f (x). However, the definite integral Il f Q) dx is a number whose value dependson the function f and, the numbers a andb, and it is defined as the limit of a Riemann sum. The definition of the definite integral makes no reference to differentiation. The general indefinite integral involves an arbitrary constant; for instance, f*3 I x2dx:++C JJ This arbitrary constantC is calleda constantof integration.lnapplying the fundamental theorem to evaluate a definite integral, we do not need.to include the arbitrary constantC in the expressionforg(r) becausethe fundamental theorem permits us to select any antiderivative, including the one for which C:0. Evaluate SOLUTION: - 6xz* ex+ 1,)itx Ii, o' (xt - 6x2* 9x * 1,) dx: [:,, xsdx-6[^rxzdx *t lnn, :t-6.{*, +*,1:, : ( 6 4 - 1 , 2 8 + 7 2 + 4 -) ( # - * + g + * ) Evaluate SOLUTION: (r4l3 + Axrts) dx: *xzrl, + 4 . t*n,tlt_, :++3- (-++3) (x6ars * n*rrs) dx -_ 76 sor,urroN t :)\ f2 J" Zx't/xt + '1,dx : 3 rz J lFn (ixz) dx 7.6 THE FUNDAMENTAL THEOREMOF THE CALCULUS 317 : 6 ( 8 * L 1 t r z- # ( 0 * L } s r z : + ( 2 7- L ) _ 104 I f_ ExavrprE 4: r3 Evaluate I xt/t+x dx Jo solurroN: u: To evaluate the indefinite integral l.f,+x u2:'!,*x J xVt I x dx we let dx-2udu x:tt2-"!. Substituting, we have r_l I xt/t + x dx: | (u' t)u(2udu) JJ -2 r | (un uz) du J : E u ' - f i u Bt C : ? ( 1 * x ) s t z- ? ( 1 * x ) s t za . , Therefore, the definite integral f3 13 I x \ / 1 ,* x d x : 9 ( 1 * x ) s r z 3 ( 1 * x ) s r zI Jo lo : ?@);tz- z(4Y, - !"(1)5t2 + ?(t1r,, :+!-+-E+3 -_ 116 -T5- Another method for evaluating the definite integral in Example 4 involves changing the limits of the definite integral to values of a. The procedure is shown in the following illustration. Often this second method is shorter and its justification follows immediately from Theorems 6.3.10 and 7.6.2. o TLLUSTRATToN 2: Because u: l.'ffi-x, : and when .r 3, u :2. Thus, we have f3 I xt/t + x dx:2 Jo 12 | (un u2)du Jr 12 :?u5 - &ut I It :91-+q-?+e _ 116 15 we see that when ff: 0, u: L; 318 THE DEFINITEINTEGRAL EXAMPLE 5' Evaluate l x + 2 1d x soLUrIoN: If we let f (x): lr * 21,insteadof finding an antiderivativeof / directly, we write f(x) as ifx>-2 ifx<-2 f(x):l::; Applying Theorem7.4.6,we have r_, 2 lx+ 2lO*: f,-3 (.(- ) c - 2 ) a,+ ( x + 2 ) d x [^, T - r,]-_" : [ - _ x2 '*]^-, *lt* :++ 1.8 :31ExavrprE6: Find the area of the region in the first quadrant bounded by the curve whose equation is y : xfp + 5, the x axis,and the line r:2. Make a sketch. solurroN: see Fig. 7.6.2:The region is shown together with one of the rectangular elements of area. We take a regular partition of the interval [0,2l.The width of each rectangle is Ar units, and the altitude of the ith rectangle is &\mT units, where fi is any number in the ith subinterval. Therefore, the measureof the area of the rectangular element is (1@ + s Ar. The sum of the measuresof the areasof n such rectanglesis __ia 2ttY1iz+5 Ar *ht; y: xy/izT s is a Riemannsum.The limit of this sum asAx approaches zero(or n --> *@) gives the measure of the desired area. The limit of the Riemann sum is a definite integral which we evaluate by the fundamental theorem of the calculus. Let A: the number of square units in the area of the region. Then, n A: lim ) f,{F+s Ar-0 i:t - ilx r2 | x\/x21-S dx JO :+l f2 t/x2*S (2xdx) :+. & (xx2 *_' _ - 1r 3'L ,]: s;srz 3t2l (e 312.- (s)' : i ( 27 " 9 )-s5t t./5) Figure7.6.2 Hence, the area is *(27 - 5 \6) squareunits. 7.6 THEFUNDAMENTAL THEOREM OF THECALCULUS 319 ExRtvtptE 7: solurroN: Let A.V. : the averagevalue of / on [5, 8]. We have,from Definition 7.5.3, Given f(x): xtE- E find the average value of /on the interval [5, 81. 118 A.V.:+ | xlx-A d-5 Letu: lxw h e n x :8 , Jt dx 4 ; t h e n u 2 - *x - 4 ; x : t t z * 4 ; d x : Z u d u . W h e n x : 5 , u : tt:2. Therefore, A.V.:*l r2 -: r f6" J, :- " f6r (urt4)u(2udu) Jr (un* 4u2)du 12 + 6u" lius ), :?l+++-+-*l _ - 4 5466 Exercises7.6 ln Exercises L through 18, evaluate the definite integral by using the fundamental theorem of the calculus. ,': ,. (x'- [ f4 + 1'\dY J, @'- Y2 .ro x dx .r5 ''J,WWA f. ,r. rr tn.J,ffa* f, x"*2x2*x*2 @+1y dx r3 - 3l dx J_,lx Y'J, dx *.J_,(r*1)t/x*3 V3+ lxl dx J_" f4 l,5-r ,t.J_,t/lxl-xdx ,n- J^x2!x-4 ,fr o" l t i w d w Jo (1+ 61s t o q15 n ' 1s du J -@ , ;rY .-'' f3 o' J, 1er'- 1y' M - w2 dut 5./ J_rzw f' (y'* 2V)dy ,{i r3 (3x2I 5x -'t) dx J_, ,. 2x) dx a, x 'o' f'3x"-24x2*48x*5 *z - g* a 16 J_, (nrxr: Divide the numerator by the denominator.) rl l,*('n-+.w) at tt. fr J. {i lT +i{x ax 7 ax L; THEDEFINITE INTEGRAL 320 In Exercises1.9through 22, use Theorem 7.6.1 to find the indicated derivative. fr L9. Dr I t/++ t6 dt r5 20.D*l vt+t4 dt ,; i Joi Jr zr. ur r'x g dt J_. + t, frz 22.D*l Jr V1,+t4dt In Exercises 23 throuth 28, find the area of the region bounded by the given curve and lines. Draw a figure showing the retion and a rectangular element of area. Express the measure of the area as the limit of a Riemann sum and tlen as a definite integral. Evaluate the definite integral by the fundanental theorem of the calculus. "N. y : 4tc- x2; x axis; tc:']",, x: 3 : F. v \/i + l; x axis,y axis;r: 8 x a x i sx; : - 1 , x : 4 2 7 .y : x f f i ; r-!'; 24.y:x2-2x*3;x axis;r--2,x:l 26' Y : x\/xz * 9 ; x axis' Y axis;x: 4 1 2 8 .-Vx: ' + - x ; x a x i s x; : 2 , x : 3 In Exercises 29 through 32, find the average value of the flnction f on the tiven interval [c, b]. In Exercises 29 and 30, find the value of r at lvhich the average value of f occurs and make a sketch, 2 9 .f ( x ) : 8 r - x 2 ;l a , b f : 1 0 , 4 1 30. /(r) - Q- x2; l a, bl : [0,3] 31.f(x) : az1/a g; la, bl: 17,l2l 32.f(x) :3xt/F - 16; l a, bf: 14,5f 33. A ball is dropped from rest and after t secits velocity is z ft/sec. Neglecting air resistance,show that the averaBevelocity during the first +T sec is one-third of the avenge velocity during the next +T sec. 34. A stone is thrown downward with an initial vqlggqgll"o ftlsec. Neglect air rcsbtance. (a) Show that if ftlsec is the velocity of the stone after falling s ft, then o : lpoz I 2gs . (b) Find the average velocity during the first 100"ft of fall if the initial velocity is 60 ftlsec. Clake g: 32 and downward as the positive direction.) 35. The demand equation for a certain commodity is a2+ pt : 100, where p dollars is the price of one unit when 100r units of the commodity are demanded. Find the average total revenue when the number of units demanded takes on all values ftom 600 to 800. 36. Let/be a function whose derivative f is continuous on [a, b]. Find the averagevalue of the slope of the tangent line of the graph of / on [c, b] and give a geometric interpretation of the result ReviewExercises (Chapter7) In Exercises 1 and 2, find the exact value of the definite integral by using the definition of the definite integraL do not use the fundamental theorem of the calculus. fr 1,. | J-z (r' * 2x) dx , [, (x, - 1.)zdx In Exercises 3 and 4, find the sum. r00 41 4. > (\?trI_l- V3l+z) 3. > 2i(i3-1.) i:1 i:r t 5. Provethat) /^ O: (I \2 , and verify the formulalorn:1,2, and,3. 6. Express_asa definite integal and evaluate th€ definite integral:"lim f which f(t) : Vr.) @^[1n"',y (nwr: Considet the function f for REVIEWEXERCISES In Exercises 7 throu gh l6,evaluate the definite integrat by using the fundamental theorem of the calculus. ,- . f3 r. . dt J_"U6-3t) s.f -L Jt \/3x- f7 1 0 .f ' r * f l F + z a x 1 ^-' 11.l'z)cygx+ 4 Jo J-s 12.f =+ Jt \/5 dx û2 tlw^ fs f2 dx ts. I lx - 213 J _s x fr ' - ' J| - , \ / i + T t n ' J r ( x ' * + / ] ) 1 \ / F + o 7 + t A x 1a la4dx L# 16. | J-z *l* - 3l dx L7. Show that each of the following inequalities holds: r-'dx=l-'!;@1,'+=l: h;t')f h=f+ (u) x_g J_, 18. rf [0 f (x): lk [1 ifa<x1c if x: c ifc<x<b prove that f is integrable on la, bf urra fu f k) dx b - c regardlessof the value of k. Ja In Exercises L9 and.2A, find a smallest and a largest possible value of the given integralfl f4 19. I V5 *4x-x2 dx Jr 20.I J-z Y2x3-3x2+l dx In Exercises 21 through 24, find the area of the region bounded by the given curve and lines. DIaw a fiturc showing the region and a rectangularelement of area.Expressthe measureof the area as the limit of a Riemann sum and then as a definite integraL Evaluate the definite integral by the fundamenta.l theorem of the calculus. 22. y : 16- x2;x axis;x: -4, x: 4 x: 3 2L. y 9 x2; x axis; y axis; 23. y : 21, - 1 ; x axis;x : 5, x : 17 24.y: 4 rr-x;xaxis;x: 25. It f (x) : 2gz\[x=T, find the average value of / on 17,1,2] 26. Interpret the mean-value theorem for integrats (7.5.2) in terms of an average function value. 27. A body falls from rest and travels a distance of s ft before striking the ground. If the only force acting is that of gravity, which gives the body an acceleration of g ft/sec2 toward the ground, show that the average value of the velocity, expressed as a function of distance, while traveling this distance is ?\/lgs ft/sec, and that this average velocity is twothirds of the final velocity. 28. Suppose a ball is dropped from rest and after f sec its directed distance from the starting point is s ft and its velocity is tr and find the (a) Express ? as a function of. t as a:f(t) ft/sec. Neglect air resistance. When f : tr, s: s1, ?fld o:uy average value of f on [0, t,]. (b) Express z, as a function of s as 7r: 8(s) and find the average velocity of g on [0, st]. (c) Write the results of parts (a) and (b) in terms of. ,, and determine which average velocity is larger. r - 1,0: 0, where p dollars is the price of one unit when 100x units of the commodity are demanded. Find the average total revenue when the number of units demanded takes on all values from 600 to 800. 29. The demand equation of a certain commodity isp'* 322 THE DEFINITE INTEGRAL 30. (a) Find the average value of the function / defined byl(r) found in part (a), find lirn A. 31. Given F(x): tn1 | Jt : t/ri on the interval tt, rl. (b) If.r{ is the average value ;t dt and r > 0. Prove that F is a constant function by showint that F,(.r):0. (nrnr: Use Theorem 7.6.1afterwriting the given integlal as the difference of two integrals.) 32' Make up an examPleof a discontinuous function fo{ which the mean-valuetheorem for integrals (a) does not hold, and (b) does hold. 33. Let/be continuous on [a,D] ^na f{t) di * 0. Show that for any numbert in (0,1) there is a nurnber cin (a, b) such Jfu o' that f. tb I f(t) dt:k I t'() dt 1"r*i",cor,"ia".*" ,""or." F for whichF(x): [" f(t) d, / [' ruldr andapplythe intermediate-varue theorem.) J"' / J"' Applications of the definite integral APPLICATIONS OF THE DEFINITEINTEGRAL 8.1 AREA OF A REGION IN A PLANE From Definition 7.3.6, if f is a function continuous on the closed interval lo, bl and if f (x) =- 0 for all r in la, bl, then the number of square units in the area of the region bounded by the curve y : f (x), the r axis, and the l i n e s x : a and x: b i s n lim -) - f(f,) Ax -0 t:l llAll which is equal to the definite integral ft f Q) dx. suppose that f (x) < 0 for all x in la, bl. Then eachf (il is a negative number, and so we define the number of square units in the area of the regionboundedby y : f(x), the r axis,and the lines)c: aand x : b to be n lim ->. t-f(f') I Ax llAll'-0 i=l which equals - |fb f(x) dx Ja' nxavrprE 1: Find the areaof the region bounded by the cuwe y : f - 4x, the r axis, and the lines x:Landr:3. soLUrIoN: The region, together with a rectangular element of area, is s h o w n i n Fi g. 8.1.1. If one takes a regular partition of the interval [1, 3], the width of each rectangle is Ar. Because x2- 4x { 0 on [1,3], the altitude of the lth rectangle is - (€r' - a€r) : 4€i - trt.Hence, the sum of the measures of the areas of n rectangles is given by 7Z i=1 The measure of the desired area is given by the limit of this sum as Al approaches 0; so if A square units is the area of the region, we have n hm ) Lx:-0 (afi - t,') Lx i:l (4x - xz) dx :2)c2 (€i, €i'- atr) *rt - T -,.) F i g u r e8 . 1 . 1 Thus, the area of the region is ]2 square units. rxavrprn 2: Find the area of the region bounded by the curve y : xs -2x2 -5x * 5, the r axis, and the lines x: -L andx:2. solurroN: The region is shown in Fig. g.l.z.Let (x) : )cB- 2x2 -sx * 6. f Because f (x) > 0 when x is in the closed interval [-1, 1] and (r) = o f when x is in the closed interval [1, 21, we separate the region into two parts. Let At be the number of square units in the area of the region when 8 . 1 A R E A O F A R E G I O NI N A P L A N E t- of square units in the area of the t ,,'l,f risin J t, dtlnd let A, be the number region w he e n l x: i si i n lr, zl. Then n ( f Ar l'"1)i =l1 f ( 'f,) A1 litl : ( € t, f ( t i ) ) rII'r If_ l' -J' l rr 1( 't -' l (r)t ddx a x3_ 2x2-5x*6) dx F(r, f(x) - 2x2 -5x * 6) dx If A square units is the area of the entire region, then Ar* A, A: : F i g u r e8 . 1 . 2 12 (x'- 2x2-5x * 5) dx G' - 2xz- 5x -f 6) dxJ, J_, , lr l-_ - 3x'- Ex'* U*1 2 - 3x"- Ex' * U* ), L*"' liro l_r: [ ( *- z ' - E + o -) ( ++ ? - B - 5 ) ] - l ( 4 - a f - 1 0+ 1 2 -) ( +- ? - B + a ) l : Gi, fGt)) rt t- (-?9) The area of the region is thereforc lft square units. Now consider two functions/ and g continuous on the closed interval la, bf and such that f (x) > 8(x) for all x in la, bl. We wish to find the area of the region bounded by the two curvesy : f (x) and! :8Q) and the two lines x: a and x : b. Such a situation is shown in Fig. 8.1'3. Take a regular partition of the interval la, bl, with each subinterval having a length of Ar. In each subinterval choose a point f1. Consider the rectangle having altitude lf (€,) - S(6;) ] units and width Ar units. Such a rectangle is shown in Fig. 8.1.3. There are n such rectangles, one associated with each subinterval. The sum of the measures of the areas of these n rectangles is given by the following Riemann sum: (f i, g(ti)) F i g u r e8 . 1 . 3 n -s(fi)l Ax > t/(f') fni":tiiu*ann sum is an approximation to what we intuitively think of as A P P L I C A T I O NOSF T H E D E F I N I T E INTEGRAL the number representing the "measure of the area" of the region. The larger the value of n-or, equivalently, the smaller the value of Ax-the better is this approximation. If A square units is the area of the region, we define n, A : l i m > t / ( f , )* s ( f , ) l A x (1) Because f andg are continuouson la, bl, so alsois (f - g); therefore, the limit in Eq. (1) exists and is equal to the definite integral rb J, tf G) nxavrru 3: Find the area of the region bounded by the curyes A:x2 andy:-x2*4x. y: 8@) (2,4) G' , f ( ti ) - g(x)ldx soLUrIoN: To find the points of intersection of the two curves, we solve the equations simultaneously and obtain the points (0, 0) and (2,4). The region is shown in Fig. 8.I.4. Let /(r) : -)c2 * 4x, and g (x) : 12. Therefore, in the interval l0, Zl the curve y : f (x) is above the curve y : g(x). We draw a vertical rectangular element of area,having altitude lf (€r) - s((o)] units and width Ar units. The measure of the area of this rectangle then is given by Ax. The sum of the measures of the areas of n such lf((r) -S({i)] rectangles is given by the Riemann sum n lf (€,)- s(f,)I Ax y : f(x) i:1 If A square units is the area of the region, then n lim ./-/ \ (f;,9(fr)) Ar-0 f:1 lf G,)- s(fi)I Ar and the limit of the Riemann sum is a definite integral. Hence, lf(x) - s(x)l dx F i g u r e8 . 1 . 4 *t' -Jo l(-x'+ 4x)- xzldx :I: (-2x'* : 4x) dx [- l,x'.-u'f', :_++g_0 -_ 58 The area of the region is $ square units. 8 . 1 A R E A O F A R E G I O NI N A P L A N E EXAMPLE4r Find the area of the region bounded bY the Parabola : x - 5. : Az 2x 2 andthe line y -Z) and (9, 4) . The The two curves intersect at the points (3, solurroN: region is shown in Fig. 8.1.5. Theequationy2:2x-2isequivalenttothetwoequations and y--\Ei=T y:\E-z : , : (9,41 , : g(xl ltkl fr\x) F i g u r e8 . 1 . 5 u : g(r) v :frkl \e, 4, Gi, f Jtil) Gt, f r(Eil), with the first equation giving the upper half of the parabola ald thr sectLx- 2 and ond equation ginittg the bottom half. If we let /r(x): : - \Ex - 2, the equation of the top half of the parabola is.A : f t(x) , f ,(x) and the equation of the bottom half of the parabola is U: fr(x)' If we let gG) : x - 5, the equation of the line is y : S (r) . In Fig. 8.1.6 we see two vertical rectangular elements of area. Each rectangleLas the upper base on the curve A : f r(r). Becausethe base of the frG)] first rectangle is on the curve y: fr(x), the altitude is [/t({,) : y its curve is on the rectangle second the base of 8(r), the units. Because using by problem this solve to wish If we units. gG)l altitude is l/,(f,) vertical rectangular elements of area, we must divide the region into two separate regions, for instance R1 and R2,where R, is the region bounded by and u:fr(x) and the l i ne x:3, and w here R 2 i s th e th e c u rv e s y :fi (r) region bounded by the curyes A : f r(x) and y : 8@) and the line r: 3 (see Fig. 8.1'.7). If A, it the number of square units in the area of region R1, w€ have Ar: rim ) lf'Go) - frG,)l Ar i:l (€i, g({i )) Ae-O _t' lf ,(x) _ J, -_ f ' l \ E - z J' (€', /r(€i )) F i g u r e8 . 1 . 6 fz@)l dx :, It t E x - Z +t/zx-zlax ax l3 : I (Zx- Z)srzI lr :+L y : g(x) y : f ,(x') It Az is the number of square units in the area of region R2, w€ have n Az: lim ) :J; :T: l f ' ( € , )- s ( f , ) l A x [/'(r) - g(x)l dx Y: f 2Gl t\E-2 F i g u r e8 . 1 . 7 - (x -s)) dx A P P L I C A T I O NOSF T H E D E F I N I T E INTEGRAL : - z)'z- +x'+ sxf'" [+Qx :[g-++451-t*-8+ts1 Hence At * Az: # + # : 18. Therefore, the area of the entire region is 1g square units. nxevrprE 5: Find the area of the region in Example 4by taking horizontal rectangular elements of area. (3, - 2l soLUrIoN: Figure 8.1.8 illustrates the region with a horizontal rectangular element of areaIf in the equations of the parabola and the line we solve for x,we have x:iQ'+2) * 2) and r(y) : y + s, the equation of the parabola r : tr(y) Letting QQ) :t(y' : QQ) and the equation of the line as tr: L(y). maybe written as x r*e \9, 4l consider the closed interval l-2,4] on the y axis and take a regullr partition of this interval, each subinterval will have a length of Ly-.In the ith t r ( { i ) ,{ i ) subinterval lyi-r, ai], choose a point f;. Then the lenglh of the ith rectangular element is [r(1,) -d(fr)] units and the width is ay units. The measure of the area of the region can be approximated by ihe Riemann sum: n F i g u r e8 . 1 . 8 and x:y*S i:r [r(f,) - 6G)l Lv If A square units is the area of the region, then n A: lim > tl(f,) - d(f,)l Ly A.tl*0 i:1 Because). and Q arc continuouson l-2,4], so arsois (r-d), limit of the Riemann sum is a definite integral: o:I^,[r(y)- 6U)] dv : I:,l ( y+ 5 )- i ( y ' + 2 ) l d v -t -, :+ l n (-y' + 2y + 8) dy J_, [- *y"* v'+ 8yf_, : + [ - g s + 1 6 + 3 2 )- ( 8+ 4 - 1 , 6 ) l :L8 and the 8.1A R E AOF A R E GION IN A P LA N E 32 9 Comparing the solutions in Examples 4 and 5, we see that in the first casewe have two definite integrals to evaluate, whereas in the second case we have only one. In general, if possible, the rectangular elements of area should be constructed so that a single definite integral is obtained. The following example illustrates a situation where two definite integrals are necessary. EXAMPLE6: Find the area of the region bounded by the two curves A: xB- 6X2* 8r and !: x2 4x. €i , f(ti)) solurroN: The points of intersection of the two curves are (0, 0), (3, and (4,0). The region is shown in Fig. 8.1.9. Let f (r) : xB- 6x2* 8x and g(x) : )c2- 4x. In the interval [0, 3] the curvey: f (r) is abovethe curve y: g(r), and in the interval [3, 4] the curvey: g(x) is abovethe curvey: f (r). So the region must be divided into two separateregionsRt and R2,where Rt is the region bounded by the two curves in the interval [0, 3] and R2is the region bounded by the two curvesin the interval 13, 41. Letting A, squareunits be the areaof R1and A, squareunits be the area of Rr, we have n y:8@) y : f(x) Ar: lim n Az:lim Ar-0 R2 ( f ; , g ( f; ) ) €i,f(ti)) 3, -3) (i,s((;)) F i g u r e8 . 1 . 9 lf (€,)- s(fr)I Ax AJr- 0 i : 1 ) [ S ( f , )- f ( € , ) ] A x i:l so that A, * Or: 6xzI 8x) - (x'- ax)l dx It [(x'+ [^ l@' - 4x) - (r' - 6x2+ 8x)l dx :f Js (x, - Txz* I2x) dx + f^ (-x, t Txz- I2x) dx J3 - u*]^, :lr*^ - &x,+ 6r]',+ * Tx" f+r-- -4T5--rs7-27 -_ 6 7L Therefore, the required area is ff square units. 8.1 Exercises h Exercfues 1 though 20, find the area of the region bounded by the given curves. In each Problem do the following: (a) Draw a figure showing the region and a rectangular element of area; (b) expressthe area of the region as the limit of a Rigmann sum; (c) find the limit in part (b) by evaluating a definite ihtegral by the fundamental theorem of the calculus. t J .x. 2 : - y ; y - - 4 4. yt - -x; x: -2; x: -4 330 APPLICATIONS OF THE DEFINITE INTEGRAL 3. tt + y I 4: 0; / : -8. Take the elemmts of area perpendicular to the y axis. 4, The same region as in Exercise 3. Take the elements of area parallel to the y axis. 'l 5. ir:2y2.x=O,y:-2 d.y":4t;r:O;y:-2 ,).y:2-i".y:-7 8.y:f;y:r' 9. y2:y-1;a:g 1 0 -y : y z ; f : 1 6 - , 1 1 .V : { r ; y : * t 2 .x : 4 - y 2 ; x : 4 - 4 y l\. ys: f;|-gy 4 4:g 1 4 .r y z : y z 7 ; x : 7 t y = l ; y : 4 1 5 .x : y 2 - ) ; y : g - y z 1 6 .t 6 : y z - y ; x : y - y z : : lV. y 2rt 3g 9x; y S 23 tr tB. 3y : as- 2rz - t1x; y : I - ng _ ,r, * * b. y:lxl,y:S-r,x=-r,x:t 2 0y. : l x + 1 1 * l x l , y : 0 , x : - 2 , x - _ 3 Find by integration the area of the triangle having vertices at (S, l), (1, 3), and (-1,_2). T. 22. Find by integation the area of the triangle having vertice8 at (3,4), (2,0), and (0, 1). 23. Findthe areaof the region bounded by the curvc r1- f +2ry -y2:Oandthelirrer:4. (HrNr: Solvethe cubic equation for y in terms of r, and express y as two functions of r,) 24. Find the area of the region bounded by the three curves y: f , x: yB, and x + y:2. 2 6 . F i n d t h e a r e ao f t h e r e g i o n b o u n d e d b y t h e t h r e e c u r v e s y : * , y : g - y r , a n d . 4 r - y + 1 2 : 0 . 26. Find the_areaof the region above the panbola * : 4py and inside the triangle formed by the r axis and the lines y : x * 8P and Y : -a 1gp. 27. (a) Find the area of the region bounded by the curves y2 : Apx and f: of the area in part (a) with respect to p when p :3. 4py. (b) Find the rate of change of the measure 28' Find the rate of drange of the measurc of the area of Exercise 25 with respect to p when p: *. 29. If d square unit6 i8 the area of the region bounded by the parabola y, : 4r and the line : y flr (z > 0), find the rate of change of A with respect to t||, 30. Detersdne n so that the region above the line y : 1nx and below the parabola y = 2x _ .f has an area of 36 square units. 8.2 VOLUME OF A SOLID OF REVOLUTION: CIRCULAR-DISK AND CIRCUTAR.RING METHODS F i g u r e8 . 2 . 1 We must first define what is meant by the "volume" of asolid of revolution; that is, we wish to assign a number,for example V,to what we intuitively think of as the measure of the volume of such a solid. We define the measure of the volume of a right-circular cylinder by mzh, where r and h are, respectively, the number of units in the base radius and altitude. Consider first the case where the axis of revolution is a boundary of C:I R C U L A R - D I SAKN D C I R C U L A R - R I NM GE T H O D S 3 3 1 8 . 2 V O L U M EO F A S O L I DO F R E V O L U T I O N the region that is revolved. Let the function / be continuous on the closed i n te rv a l l a ,b l a nd assume that f (x) > 0 foral l x i n l a,bl . Let R be the region bounded bythe curve y: f (r), the x axis, and the lines x: A and x: b. Let S be the solid of revolution obtained by revolving the region R about the r axis. We proceed to find a suitable definition for the number V which gives the measure of the volume of S. Let A be a partition of the closed interval la, bl given by a:xo1xtlxr( 1 xn-t I xr: b . T h e n w e h a v e n subi nterval sof the form l * r-r,r;], w herei :7,2, ,rt , any Lrx: xi-xi-y Choose being length of the ith subinterval with the point fi, with xi-r s ti s xi, in each subinterval and draw the rectangles having widths A1r units and altitudes f (€r) units. Figure 8.2.3 shows the region R together with the ith rectangle. When the ith rectangle is revolved about the x axis, we obtain a circular disk in the form of a right-circular cylinder whose base radius is given by f ((r)units and whose altitude is given by A'ixunits, as shown in Fig.8.2.4. The measure of the volurne of this circular disk, which we denote by L,1V,is given by (t; 'f(tt)) L,V-- nlf (il)' L,x F i g u r e8 . 2 . 3 (1) Because there are n rectangles, n curclrlardisks are obtained in this way, and the sum of the measures of the volumes of these n circular disks is given by fi:r o,r: fi:r rrlf(€,)l'L,x (2) which is a Riemann sum. The Riemann sum given in Eq. (2) is an approximation to what we intuitively think of as the number of cubic units in the volume of the solid of revolution. The smaller we take the norm llAllof the partition, the larger will be n, and the closer this approximation will be to the number V we wish to assign to the measure of the volume. We therefore define V to be the limit of the Riemann sum in Eq. (2) as llAllapproaches zero. This limit exists because f i" continuous on la, bf, which is true because / is continuous there. We have, then, the following definition. F i g u r e8 . 2 . 4 8.2.1,Definition Let the function / be continuous on the closed interval lo, bl, and assume that f (r) > 0 for all r in la, bl. If S is the solid of revolution obtained by revolving about the r axis the region bounded by the curve y : f (x), the x axis, and the lines x : a and x : b, and if V is the number of cubic units in the volume of S, then v- : ' u|)f' dx f. ,tfG,)l'A,x f ,,t;il, (3) APPLICATIONS OF THE DEFINITEINTEGRAL v A similar definition applies when both the axis of revolution and a boundary of the revolved region is the y axis or any line parallel to either the r axis or the y axis. (tr,€i' o TLLUSTRATIoN 1: We find the volume of the solid of revolution generated when the region bounded by the curve U : xr, the x axis, and the lines x:"1. and x:2 is revolved about the r axis. Refer to Fig. B.z.s, showing the region and a rectangular element of area. Figure 8.2.6 shows an element of volume and the solid of revolution. The measure of the volume of the circular disk is given by L1V: n(€r')' Lix: rr(f A,ix Then Figure8.2.5 v: .Jimf ntf t* llAll-oi:t :, r2 I x4dx Jr : rr(*rt)l' lr -_ 37_ 5ta Therefore, the volume of the solid of revolution is #a cubic units. o Now suPPose that the axis of revolution is not a boundary of the region being revolved. Let f and 8 be two continuous functions on the we wish to find a value for V. Let A be a partition of the interval la, bl, given by A:XolXrlXrl Figure8.2.6 1 xn_, I xr: fu and the lth subinterval fr,-r, ri] has length Lix: xr- xr_.tIn the lth subinterval, choose any €i, with xi-r s €i = xi.Consider the n rectangular elements of area for the region R. See Fig.8.2.7 illustrating the region and the ith rectangle, and Fig. 8.2.8 showing the solid of revolution. When the ith rectangle is revolved about the x axis, a circular ring (or "washe{'), as shown in Fig. 8.2.9, is obtained. The number giving the difference of the measures of the areas of the two circular regions is (rlf ((,)l'rlg(f,)1,) and the thickness is A;x units. Therefore, the measure of the volume of the circular ring is given by L v : r ( l f ( f , ) l ' - [ s ( { , ) ] 'A ) ir The sum of the measures of the volumes of the n circular rings formed by METHODS AND CIRCULAR-RING CIRCULAR-DISK 8.2 VOLUMEOF A SOLID OF REVOLUTION: f(€t) ti, f(t;)) f(x) s(x) F i g u r e8 , 2 . 9 Figure8.2.8 Figure8.2.7 revolving the n rectangular elements of area about the r axis is ) i=L (4) l,Y i:l The number of cubic units in the volume, then, is defined to be the limit of the Riemann sum in Eq. (4) as llAllapproacheszero.The limit exists since , f and g2 are continuous on la, b) because f and I are continuous there. 8.2.2 Definition Let the functions f and I be continuous on the closedinterval la, bl, and assumethatf (x) > g(x) > 0 forall x in la,bf. Then if Vcubic units isthe volume of the solid of revolution generatedby revolving about the r axis and A: gk) and the lines the region bounded by the curves y:/(r) x : n a n dx : b , As before, a similar definition applies when the axis of revolution the y axis or any line parallel to either the r axis or the y axis. Exavrpr.r l.: Find the volume of the solid generated by revolving about the r axis the region bounded by the parabola : x * 3. A : xcz* L and the line y solurroN: The points of intersection are (-1., 2) and (2,5). Figure 8.2.10 shows the region and a rectangular element of area. An element of volume and the solid of revolution are shown in Fig. 8.2.1'L. andg(r) T a k i n gf ( x ) : x * 3 the volume of the circular ring is n(lf G)l'- [ s ( 1 ' ) ] ' )A i x * L, we find that the measure of 334 APPLICATIONS OF THE DEFINITEINTEGRAL ti , fGi) Gi , s ft)) Fi g ur e 8. 2. 10 F i g u r e8 . 2 . 1 1 If V cubic units is the volume of the solid, then n v : lim ) '( lf G)l' - [s(f )],) Aix iIr llall-o :7r If2 (lf(x)l'J-t f2 : T I J-r ls@)fz)dx [ ( r ' + 3 ) ' - ( x ' + t ) ' z 1d x f2 :rT I l-xn-f J-r +6x*sldx : nl-Ex'- *x3* gr'+ grlt L l-r : rl(-+- S + L z+ 1 6 )- ( t + + + 3 - 8 ) l _rL7_ 5tl Therefore, the volume of the solid of revolution is LFr cubic units. AND CIRCULAR-RING METHODS CIRCULAR-DISK 8.2 VOLUMEOF A SOLID OF REVOLUTION: ExAMPLE 2r Find the volume of the solid generatedby revolving about the line x: -4 the region bounded by the two parabolas )c:y-yz andx:A'-3. solurroN: The curyes intersectat the points (-2, -1.) and (-+,il. The region and a rectangular element of area are shown in Fig. P.2.12.Figure 8.2.13shows the solid of revolution as well as an elementof volume, which is a circular ring. F i g u r e8 . 2 . 1 3 Let F(y) - y - y2 and G(y) = y' - 3. The number of cubic units in the volume of the circular ring is L1V: n(14 + F(f')1, - 14+ G(f')l') Lua (G(6r),fi) Thus, v- F i g u r e8 . 2 . 1 2 n + F(€,)f'- 14+ G(€)]') L,y lim -). l l A l l - 0 i : 1 "-([4 -, f3l2 -y')' - (4* y'-g)'l dY |J - r [ ( 4 + y -n I ?2y'-9y'*8y+L5) dy J-r l3l2 13tz - 3Yt* 4Y'* tu,l-, : n ll+rn _ - 875* 32,. The volume of the solid of revolution is then Y*n cubic units. 8.2 Exercises In Exercises 1. through 8, find the volume of the solid of revolution when the given region of Fig. 8.2.1,4is revolved about the indicated line. An equation of the curve in the figure is y2 : *2. 2. OAC about the line AC l. OAC about the r axis 3. OAC about the line BC ' 4. OAC about the y axis 5. OBC about the y axis 6. OBC about the line BC 7. OBC about the line AC 8. OBC about the x axis F i g u r e8 . 2 . 1 4 336 APPLICATIONS OF THE DEFINITEINTEGRAL 9 . Find the volume of the sphere generated by revolving the circle whose equation is 12 * A2: 12 about a diameter. 10.Find by integration the volurne of a right-circular cone of altitude /z units and base radius a units. 1 1 .Find the volume of the solid generated by revolving about the x axis the region bounded by the curve : x3 and the A linesy:0andx:2. 12. Find the volume of the solid generated by revolving the region in Exercise L1,about the y axis. 13. Find the volume of the solid generated by revolving about the line x: -4 the region bounded by that line and the p a r a b o l ax : 4 i 6 y - 2 y 2 . 14. Find the volume of the solid generated by revolving the region bounded by the curves U2: 4x andy: r about the x axis. 1 5. Find the volume of the solid generated by revolving the region of Exercise 14 about the line x: 4. 16. An oil tank in the shape of a sphere has a diameter of 50 ft. How much oil does the tank contain if the depth of the oil is 25 ft? 1 7. A paraboloid of revolution is obtained by revolving the parab ola y2 : 4px about the r axis. Find the volume bounded by a paraboloid of revolution and a plane perpendicular to its axis if the plane is 10 in. from the vertex, and if the plane section of intersection is a circle having a radius of 5 in. 18. Find the volume of the solid generated by revolving about the x axis the region bounded by the loop of the curve whose equation is 2y2 : x(* - 4). 19. Find the volume of the solid generated when the region bounded by one loop of the curve whose equation is xzyz: (x'- 9) (1 - x2) is revolved about the x axis. 20. The region bounded by a pentagon having vertices at (-4,4), (-2,0), (0, 8), (2,0), and (4,4) is revolved about the x axis. Find the volume of the solid generated. 8.3 VOLUME OF A SOLID OF REVOLUTION: CYLINDRICAL-SHELL METHOD Figure8.3.'l In the preceding section we found the volume of a solid of revolution by taking the rectangular elements of area perpendicular to the axis of revolution, and the element of volume was either a circular disk or a circular ring. If a rectangular element of area is parallel to the axis of revolution, then when this element of area is revolved about the axis of revolution, a cylindrical shell is obtained. A cylindrical shell is a solid contained between two rylinders having the same center and axis. Such a cylindrical shell is shown in Fig. 8.3.1. If the cylindrical shell has an inner radius r, units, outer radius r, units, and altitude h units, then its volume V cubic units is given by V: nrzzh- Trrzh (l) Let R be the region bounded by the curye y : f (x), the r axis, and the lines x: a and r : b, where / is continuous on the closed interval la, bl and f(x) > 0 for all x in la, b]; furtherrnore, assume that a > 0. such a region is shown in Fig. 8.3.2. If R is revolved about the y axis, a solid of revolution S is generated. Such a solid is shown in Fig. 8.3.3. To find the volume of S when the rectangular elements of area are taken parallel to the y axis, we proceed in the following manner. Let A be a partition of the closed interval [a, b] given by A:Xo<Il<Xz1 1 xn-t 1 xn: b METHOD 337 CYLTNDRICAL-SHELL OF A SOLIDOF REVOLUTION: 8.3 VOLUME tt^.\ trx) Let mi be the midpoint of the lth subinterval lxr-r, ri]. Then ttti: Lr(xo-,* xi). Consider the rectangle having altitude f (mr) units and width tnlr rectangle is revolved about the y axis, a cylindrical shell is A,* ""itt."if Figure 8.3.3 shows the cylindrical shell generated by the recobtained. tangular element of area. If LiV gives the measure of the volume of this cylindrical shell, we : have, from lormula ( 1.) , whe tE T1: xi-1, 12: xi, and h f (mr) ' o'':i.lr'!',,-;'iir-)',!r'*' F i g u r e8 . 3 . 2 L rV: rr(x i - )ci -)(xr* xr-t: Becausexi- L1V:2nmif Lrx and becaus€ Ii I (m) (2 ) xr-r)f(m) Xi-r:2*r, We have from Eq. (2) (3) Lix lf n rectangular elements of area are revolved about the y axLS,n cylindrical shells are obtained. The sum of the measures of their volumes is given by nn ii : r o,u: ti : l 2nmif(m,) Lix (4) which is a Riemann sum. Then we define the measure of the volume of the solid of revolution to be the limit of the Riemann sum in Eq. (4) as llAll approaches zero' The limit exists because it f is continuous on la, bl, so is the function having function values Znxf (x). Figure8.3.3 8. 3. 1 D e fi n i ti o n the volume of S, then V- n l i m )Ll , Z r r m j ( m ) llall-o i:1 Arr:,, I: xf(x) dx (5) Definition 8.3.1 is consistent with Definitions 8.2.1 and 8.2.2 for the measure of the volume of a solid of revolution for which the definitions apPly. The formula for the measure of the volume of the shell is easily remembered by noticing that 2nmr, f (m), and Lp ate, respectively, the numbers giving the circumference of the circle having as radius the mean of the inner and outer radii of the shell, the altitude of the shell, and the thickness. APPLICATIONS OF THE DEFINITEINTEGRAL ExAMPLEL: The region bounded by the curye A : x2, the r axis, and the line x:2 is revolved about they axis. Find the volume of the solid generated. Take the elements of area parallel to the axis of revolution. solurroN: Figure 8.3.4shows the region and a rectangular element of area. Figure 8.3.5 shows the solid of revolution and the cylindrical shell obtained by revolving the rectangular element of area about the y axis. a: xz (m r, m i') 'i-, L i* Figure8.3.4 Figure8.3.5 The element of volume is a cylindrical shell the measure of whose volume is given by LiV : 2nmi(mr') Aix: 2rmi3 A,ix Thus, v- li* > 2nm13A,6 l l A l l - oi : t 2rl f2 x 3dx JO : 2'(if) f' Jo :8rt Therefore, the volume of the solid of revolution is gzr cubic units. rxavrpr,E 2: The region bounded by the curye A : xz and the lines y : l and r :2is revolved about the line A : -3. Find the volume of the solid generated by taking the rectangular elements of area parallel to the axis of revolution. soLUTroN: The region and a rectangular element of area are shown in Fig. 8 .3 .6 . The equation of the curve is y : 12. Solving for x, we obtain x : !fr. Because x ) 0 for the given region, we have ,: fr. The solid of revolution as well as a cylindrical shell element of volume is shown in Fig. 8.3.7. The outer radius of the cylindrical shell is (y1+ 3) units and the inner radius is (yr-, * 3) units. Hence, the mean of the inner METHOD CYLINDRICAL.SHELL 8.3 VOLUMEOF A SOLID OF REVOLUTION: y:7 (tffi, m; __\_ !g-) Figure8'3'7 Figure8.3.6 and outer radii is (mi* 3) units. Because the altitude and thickness of - {nx) units and Liy units, the cylindrical shell are, resPectively, Q L,V:2n(mi+ 3) (2 - l*) Lry Hence, if V cubic units is the volume of the solid of revolution, n_ Y: |i : l 2 n ( m 1 +3 )( 2 ! m , ) L i l J i m ilail-0 : l -r4 zrr(!+3)(2-frlaV Jr r4 : 2rr I ?y"'' + 2y - Sytrz+ 6) dY Jr f14 : 2n | 4y,,' + 6y I * y' - 2yzrz lr L _ ss_* 5t, Therefore, the volume is Tzi cubic units. 340 APPLICATIONS OF THE DEFINITE INTEGRAL 8.3 Exercises 1-8. Solve Exercises1 through 8 in Sec.8.2 by taking the rectangular elementsparallel to the axis of revolution. In Fig. 8.3.8the region bounded by the r axis, the line r : 1, and the cuwe y : I is denoted by R1;the region bounded bythetwocurvesy:landy,:risdenotedbyRttheregionboundedbytheyaxis,theliney:l,andthecuweyr:ris denoted by Rs.In Exercises9 through 15,find the volume of the solid generatedwhen the indicated region is revolved abodt the given line. Figure8.3.8 9. R1 is revolved about the y axisi the rectangular elementsare parallel to the axis of revolution. 10. Sameas Exercise9; but the rectangularelementsare perpendicular to the axis of revolution, 11. R, is revolved about the r axi6; the rectangular elements are parallel to the axis of revolution. 12. Same as Exercise 11; but the rectantular elements are perpendicular to the axis of revolution. 13. & i8 revolved about the line y:2; the rectanguLarelerrents are paiallel to the axis of revolution. 14. same as Exercise 13; but the rectantular elements are perpendicular to the axis of revolution. 15. R, is revolved about the line r : -2; the rectangular elements are parallel to the axis of revolution. 16. Sameas Exercise15; but the rectangularelements are perpendicular to the axis of revolution. 17. Find the volume of the solid generated if the rcgion bounded by the parabola A, : 4ax (a > 0) and the line r : a is revolved about r: a, 18. Find the volume of the solid genented by revolving the region bounded by the curve li3 * yrl": a2i' about the y axis. 19. Find the volume of the solid generatedby revolving about the y axis the region outside the cuwe y : f and between the Lines y : 2a - 1 ?n\d.y : x + 2. 20. Through a spherical shaped solid of radius 6 in., a hole of radius 2 in. is bored, and the axis of the hole is a diameter of the sphere. Find the volume of the part of the solid that remains. 21. A hole of radius 2\6 in. is bored through the center of a spherical shaped solid of radius 4 in. Find the volume of the portion of the solid cut out. 22. Find the volume of the solid generatedif the region bounded by the parabola y, :4pr about the y axis. and the line r: p is revolved PLANESECTIONS 341 8.4 VOLUME OF A SOLIDHAVINGKNOWNPARALLEL 8.4 VOLUME OF A SOLID Let S be a solid. By a plane sectionof S is meant a plane region formed by HAVING KNOWN PARALLEL the intersection of a plane with S. In Sec.8.2 we learned how to find the PLANE SECTIONS volume of a solid of revolution for which all plane sections perpendicular to the axis of revolution are circular. We now generalize this method to find the volume of a solid for which it is possible to express the area of any plane section perpendicular to a fixed line in terms of the perpendicular distance of the plane section from a fixed point. We first define what we mean when we say that a solid is a "cylinder." 8.4.1. Definition A solid is a right cylinder if it is bounded by two congruent plane regions R1 and R2 lying in parallel planes and by a lateral surface generated by u line segment, having its endpoints on the boundaries of R1 and Rz, which moves so that it is always PerPendicular to the planes of Rt and R2. F i g u r e8 . 4 . 1 Figure8.4.2 Figure8.4.3 Figure 8.4.1.illustrates a right cylinder. The heighf of the cylinder is the perpendicular distance between the planes of Rr and R2, and the base is either R1 or R2. If the base of the right cylinder is a region enclosed by a circle, we have a right-circular cylinder (see Fig. 8.a.4; if the base is a region enclosed by u rectangle, we have a rectangular parallelepiped (see F i g . 8 .a .3 ). If the area of the base of a right cylinder is A square units and the height is h units, then by definition we let the volume of the right cylinder be measured by the product of A and h. As stated above, we are considering solids for which the area of any plane section that is perpendicular to a fixed line is a function of the perpendicular distance of the plane section from a fixed point. The solid 5 of Fig. 8.4.4 is such a solid, and it lies between the planes perpendicular to the r axis at a and b. We represent the number of square units in the area of the plane section of S in the plane perpendicular to the r axis at x by A(x), where A is continuous on la, bl. Let A be a partition of the closed interval la, bl given by a:xo/-xr1xzl A(€ t) Figure 8.4.4 '1xn:6 . , Wehave, then, n subintervals of the form lxr-r, ri], wherei:1,2, any Choose x^. xiLrx: being n,wittr the length of the ith subinterval APPLICATIONS OF THE DEFINITEINTEGRAL number f1, with xi-t s €i a xi, in each subinterval and construct the right cylin-dersof heights A;x units and plane section areasA(tr) square units. The volume of the lth cylinder is A(f,) A;x cubic units. We obtain n right cylinders,and the sum of the measuresof the volumes of thesen cvlinders is given by the Riemann sum n A(() ap (1) i:r This Riemann sum is an approximation of what we intuitively think of as the measure of the volume of S, and the smaller we take the norm llAllof the partition A, the larger will be n, and the closer this approximation will be to the number we wish to assign to the measure of the volume. We have, then, the following definition. 8.4.2 Definition Let S be a solid such that S lies between planes drawn perpendicular to the x axis at a and b. If the measure of the area of the plane section of S drawn perpendicular to the x axis at r is given by A(x) , wherc A is continuous on la, bl, then the measure of the volume of S is given by lim 2 a(f,) Aix: [' A ( x ) d x .0 i=l J(l (2) lllll Note that Definitions 8.2.7 and 8.2.2 are special cases of Definition 8.4.2. If in Eq,. (2) we take A (x): nl f (x)12,w e have E q. (3) of S e c.g. Z. lf w e t a k e A ( x ) : n ( l f ( x ) l ' - [S(x)]'), *" have Eq. (S) of Sec. 8.2. nxervrpr.r 1: If the base of a solid is a circle with a radius of r units and if all plane sections perpendicular to a fixed diameter of the base are squares, find the volume of the solid. solurroN: Take the circle in the xy plane, the center at the origin, and the fixed diameter along the r axis. Therefore, an equation of the circle is 12 I y' : r2. Figure 8.4.5 shows the solid and an element of volume, which is a right cylinder of altitude A1r units and with an area of the base given by l2f (il]2 square units, where /(r) is obtained by solving the equation of thpirc,fe j?, v and setting y : f (x). This computation give f (x) : Yr" - xz. Therefore, if V cubic units is the volume of the solid, "we have lim -tt llall-0 n lzf((,)l' Lix i:r (r'- : +l'"* x 2 )d x 8.4 VOLUMEOF A SOLID HAVINGKNOWN PARALLELPLANESECTIONS F i g u r e8 . 4 . 5 rxeuPrn 2: A wedge is cut from a right-circular cylinder with a radius of r in. bY two Planes, one perpendicular to the axis of the cylinder and the other intersecting the first at an angle of measurement 60oalong a diameter of the circular plane section. Find the volume of the wedge. solurroN: The wedge is shown in Fig. 8.4.6. The xy plane is taken as the plane perpendicular to the axis of the cylinder, and the origin is at the An equation of the circular plane section is then point of plrp"ndicularity. : 12.Every plane section of the wedge perpendicular to the r axis is * * y, a right triangle. An element of volume is,a right cylinder having altitude (€r)12 in.2, where /(r) is obA,r-in., and irea of the base given bV i$lf for y and setting y : f (x), circle tained by solving the equation of the in.3 is the volume of the if V Therefore, \fr'=T. thereby giving f (x) wedge, n_ V: U, - €rr)L,x lim )iVe .0,:r llall _ :+\/g fr | J-r (r, - x2)dx I - i*"1'_, : +\/5lr'* : fi\/3r3 Hence, the volume of the wedge is 3 tEr3 in'3 Figure8.4.6 U4 APPLICATIONS OF THEDEFINITE INTEGBAL Exercises 8.4 l The baseof a solid is a circlehawing a radius of / units. Find the volume of the solid if all plane sectionsperpendicular to a fixed diameter of the base are equilateral triangles, The base of a solid is a circle with a radius of r units, and all pLanesections perpendicular to a fixed diameter of the base are isoscelesright triangles haYing the hypotenuse in the plane of the base. iind the volume of the solid. 3 . Solve Exercise2 if the isoscelesdght t angles have one leg in the plane of the base. 4. Find the volume of a right pyramid having a height of t units and a square base of side n units. 5 . Find the volume of the tebahedmn havint 3 mutually perpendicular faces and three mutually perpendicular edges whose lengths have measuresa, b, and c. 6 . The base of a solid is a cirde with a radius of 4 in. , and each plane section perpendicular to a fixed diameter of the base is an isosceles triangle having an altitude of 10 in. and a chord of the cirde as a base. Find the volume of the solid. 7. The base of a solid is a circle with a radius of 9 in., and each plane section perpendicular to a fixed diameter of the base is a square having a chord of the circle as a diagonal. Find the volume of the solid. 8. Two right-circular cylinde{s, each having a radius of / units, have axes that intersect at right angles. Find the volume of the solid common to the two cylinders. 9 . A wedge is cut ftom a solid in the shapeof a right-circular cylinder with a radius of r in. by a -the plane through a diameter of the base and inclined to the Plane of the base at an angle of measurement 45". Fi;d volume" of the wedge. 10. A wedge is cut from a solid in the shaPeof a right-circular cone having a base radius of 5 ft and an altitude of 20 ft by two half planes through the axis of the cone, The argle between the two pl-aneshas a measurement of 30.. Find the volume of the wedge cut out. 8'5 WORK The "work" done by u force acting on an object is defined in physics as "force times displacement." For example, suppose that an otiect is moving to the right along the r axis from a point a to apoint b, and.a constant force of F lb is acting on the object in the direction of motion. Then if the displacement is measured in feet, (b - a) is the number of feet in the displacement. And if W is the number of foot-pounds of work done by the force, W is defiped by (1) o tttustRArroN '1-.: lI W is the number of foot-pounds in the work necessary to lift a 70-lb weight to a height of 3 ft, then W:70 ' 3:270 . In this section we consider the work done by a variable force, which is a function of the position of the object on which the force is acting. We wish to define what is meant by the term "work" in such a case. suppose that /(x) , where /is continuous on la, bl, is the number of units in the force acting in the direction of motion on an object as it moves to the right along the x axis from point a to pointb. Let A be a partition of 8.5 WORK the closedinterval la, bf : 1 xn-r I xr: a:xolxrlxzl b The ith subinterval is [x1-1, x]; and if xi-, is close to xi, the force is almost constant in this subinterval. If we assume that the force is constant in the ith subinterval and if fi is any point such that ri-, s fi s ri, then if AiW is the number of units of work done on the object as it moves from the point xi-1 Io the point xi, from formula (1) we have A ,W: f (€ ,) (x,- xi -r) Replacin E )ci- xi-r by A;x, w€ have A1W:f (t) L'x and n AiW:2fG,) L,x i:1 (2) i:r The smaller we take the norrn of the partition A, the larger n will be and the closer the Riemann sum in Eq. (2) will be to what we intuitively think of as the measure of the total work done. We therefore define the measure of the total work as the limit of the Riemann sum in Eq. (2). 8.5.1 Definition Let the function / be continuous on the closed interval la, bl and /(r) be the number of units in the force acting on an object at the point / on the r axis. Then if W units is the work done by the force as the obfect moves from a to b, W is given by (3) In the following examplewe use Hooke'slaw, which statesthat if a spring is stretchedx in. beyond its natural length, it is pulled back with a force equal to kx lb, where k is a constant depending on the wire used. nxalvrpn L: A sPring has a natural length of 14 in. If a force of 5 lb is required to keep the sPring stretched 2rrt., how much work is done in stretching the sPring from its natural length to a length of 18 in.? solurroN: Place the spring along the r axis with the origin at the point where the stretching starts (see Fig. 8.5.1). 14in.# T 8 .5 .1 F i g u re Let r: the number of inches the spring is stretched; : the number of pounds in the force acting on the spring r in. f (x) beyond its natural length. APPLICATIONS OF THE DEFINITEINTEGRAL Then, by Hooke's law, f (x) : kr. Because/(2) : 5, we have . 5:k'2 k:E Therefore, f(x):Ex If W: the number of inch-pounds of work done in stretching the spring from its natural length of 'J,4in. to a length of 1g in., we have w: li3 f. f G,)o,* llall-0 t:I f(x) dx Exdx : T x z 1I4 JO :20 Therefore, the work done in stretching the spring is 20 in.-lb. EXAMPLE2: A water tank in the form of an inverted right-circular cone is 20 ft across the top and 15 ft deep. If the surface of the water is 5 ft below the top of the tank, find the work done in pumping the water to the top of the tank. solurroN: Referto Fig. 8.5.2.The positive r axis is chosenin the downward direction becausethe motion is vertical. Take the origin at the top of the tank. We considera partition of the closedinterval [5, 15] on the r ixis and let fi be any point in the ith subinterval l*r-r, r;]. An elementof volume is a circular disk having thickness A,ixft and radiu s (il ft, where the f function / is determined by an equation of the line through the points (0, 10) and (L5, 0) in the form y - f (r). The number of cubic feet in the volume of this elementis given by AîV: nlf (&)1, Lrx.If w is the number of pounds in the weight of L fts of water, then the number of pounds in the weight of this element is wnlf(t)1, A;r, which is the force required to pump the element to the top of the tank. If xi-, is closeto xi, then the distance through which this element moves is approximately& ft.Thus, if A'rWft-lb is the work done in pumping the element to the iop of the tank, AiW is approximately(wrrlf G)l' Lrx) . fii so if w is the number of footpounds of work done, w: . Ji3 f .df G,)1, €i Lix llall-0 f:t : *n Figure8.5.2 fr5 (x)lzxdx J, lf To determine f (x), we find an equation of the line through the points 8.5 WORK (15, 0) and (0, 10) by using the interceptform: *15 + + : 10 1 or y---Bx*Lo T h e r e f o r e , f ( x ) : - & x * L 0 ,a n d w - *n f" ?3x * \o)zxdx Js rr5 : rpir | (6rt - #x't Js toox) dx |:'u)n lt*n - Xqr3* 5012 1ts rJt : + (10,000nar) Therefore,the work doneis 1,0,000rrw19 ft-lb. 8.5 Exercises 1. A spring has a natural length of 8 in. If a force of 20lb stretchesthe spring i in., find the work done in stretching the spring from 8 in. to 11 in. 2. A spring has a natural length of 10 in.. and a 30-lb force stretches it to 11iin. Find the work done in stretching the sprint from 10 in. to 12 in. Then find the work done in stretchint the sPdng from 12 in. to 14 in. 3. A spring has a natural length of 6 in. A 12,000-lb force compressesthe spring to 5* in Find the work done in compressing it from 6 in. to 5 in. Hooke's law holds for compression as well as for extension. 4. A spring has a natural length o{ 6 in. A 12001b force compresses it to 5} in. Find the wotk done in compressing it from 6 in. to 4* in. 5. A swimming pool full of water is in the form of a rectangularparallelepiped5 ft deep, 15 ft wide, and 25 ft long. Find the work required to pump the water in the pool up to a level I ft above the surface of the pool. 6. A trough full of water is 10 ft long, and its cross section is in the shape of an isosceles triangle 2 ft wide acrossthe top and 2 ft high. How much work is done in pumping all the water out of the trcugh over the top? 7. A hemispherical tank with a radius of 5 ft is filed with water to a depth of 4 ft. Find the work done in pumping the water to the top of the tank. 8. A right-circular cylindrical tank with a depth of 12 ft and a radius of 4 ft is half tull of oit weighing 60 lb/ft3. Find the wotk done in pumping the oil to a height 6 ft above the tank. 9. A cable 200 ft lont and weighing 4 lb/ft is hanging vertically down a well. If a weight of 100Ib is susp€nded from the lower end of the cable, find the work done in pulling the cable and weight to the top of the well, 10. A bucket weighing 20lb containing 60 lb of sand is attached to the lower end of a 100 ft long chain that weighs 10lb and is hanging in a deep well. Find the work done in raising the bucket to the top of the well, 11. SolveExercise10 if the sand is leaking out of the bucket at a constantrate and has all leakedout iust as soon as the bucket i8 at the toD of the well. 348 APPLICATIONS OF THE DEFINITEINTEGRAL 12. As a water tank is being raised, water spills out at a constant rate of 2 fC per foot of rise, If the tank originally contained 1000 fe of water, find the work done in raising the tank 20 ft. 13. A tank in the form of a rectangular parallelepiped 6 ft deep,4 ft wide, and 12 ft long is fuJl of oil weighing 50lb/fi3. When one-third of the work necessary to pump the oil to the top of the tank has been done, find by how much the sudace of the oil is lowered. 14. A cylindrical tank 10 ft high and 5 ft in radius is standing on a platform 50 ft high. Find the depth of the water whm onehalf of the work required to fill the tank faom the Found level through a pipe in the bottom has been done. 15. A one horsepower motor can do 550 ft-lb of work per second. If a 0.1.hp motor is used to pump water from a full tank in theshapeofarectangularparallelepiped2ftdeep,2ftwide,and6ftlonttoapoint5ftabovethetopofthetank,how long will it take? 16. A meteorite is a miles from the center of the earth and falls to the surface of the earth. The folce of sraviw is inverselv proportional to the square of the distance of a body from the center of the earth. Find the work doie by gravity if the weight of the meteorite is r, lb at the sudace of the earth. Let R rriles be the Edius of the earth, 8'6 LIQUID PRESSURE Another application of the definite integral in physics is to find the force causedby liquid pressureon a plate submergedin the liquid or on a side of a containerholding the liquid. First of all, supposethat a flat plate is insefted horizontally into a liquid in a container. The weight of the liquid exertsa force on the plate. The force per square unit of area exerted by the liquid on the plate is called the pressureof the liquid. Let w be the number of pounds in the weight of one cubic foot of the liquid and h be the number of feet in the depth of a point below the surfaceof the liqui d. If p is the number of pounds per square foot of pressure exertedby the liquid at the point, then 'u:oi"'rne (1) number of square feet in the area of a flat plate that is submerged horizontally in the liquid, and F is the number of pounds in the force causedby liquid pressure acting on the upper face of the plate, then F:PA ( 2) Substituting from formula (1) into (2) gives us F : whA (3) Note that formula (1) states that the size of the container is immaterial so far as liquid pressure is concerned. For example, at a depth of 5 ft in a swimming pool filled with salt water the pressure is the same as at a depth of 5 ft in the Pacific Ocean, assuming the density of the water is the same. Now suPPose that the plate is submerged vertically in the liquid. Then at points on the plate at different depths the pressure, computed from formula (1), will be different and will be greater at the bottom of the plate than at the top. We now proceed to define the force caused by liquid Pressure when the plate is submerged vertically in the liquid. We use 8.6 LIQUIDPRESSUBE 349 Pascal's principle: At any point in a liquid, the pressure is the same in all directions. In Fig. 8.6.1 let ABCD be the region bounded by the x axis, the lines x: a and r : b, andthe curve y: f (x), where the function /is continuous and/(r) > 0 on the closed interval fa, bl. Choose the coordinate axes so they axis lies along the line of the surface of the liquid. Take the r axis vertical with the positive direction downward. The length of the plate at a depth r ft is given by f (x) ft. Let A be a partition of the closed interval la, bl which divides the interval into n subintervals. Choose a point f; in the ith subinterval, with xrr 3 €i = xr. Draw n hofizontal rectangles. The ith rectangle has a length of f (€r) ft and a width of L,ix ft (see Fig. 8.6.1). If we rotate each rectangular element through an angle of 90o,each element becomes a plate submerged in the liquid at a depth of f6 ft below the surface of the liquid and perpendicular to the region ABCD. Then the force on the rth rectangular element is given by wtrf (t) L# lb. An approximation to F, the number of pounds in the total force on the vertical plate, is given by F i g u r e8 . 6 . 1 n wt,fGi) Air 2 i:r (4) which is a Riemann sum. The smaller we take llAll,the larger n willbe and the closer the approximation given by (4) will be to what we wish to be the measure of the total force. We have, then, the following definition. 8'6'1 Definition Suppose that a flat plate is submerged vertically in a liquid of weight ut pounds per cubic unit. The length of the plate at a depth of r units below the surface of the liquid is /(x) units, where / is continuous on the closed interval la, bl and f(x) > 0 on la, bl. Then F, the number of pounds of force causedby liquid pressureon the plate, is given by t: Air: wxf(x)dx [: 2*rOrri) ,,Til, (s) Exevpr.E 1: A trough having a soLUrIoN: Figure 8.5.2 illustrates one end of the trough together with a trapezoidal cross section is full of rectangularelement of area. An equation of line AB it y :g - *r. Let water. If the trapezoid is 3 ft wide f (x) : B *x.If we rotate the rectangularelement through 90', th" force on at the top,2 ft wide at the bottom, the element is given by 2w(1fG) Lfi lb. If F is the number of pounds in and 2 ft deep, find the total force the total force on the side of the trough, owing to liquid pressure on one n end of the trough. F - lim zwt,f (t) L# 2 H llAll-0 f:t APPLICATIONS OF THE DEFINITEINTEGRAL :2us :Ztll [, oo) d x I,.* - ix) dx lrr'-# ' ] , 2 , O )A ( 2 , l ) Figure8.6.2 Taking w : 62.5, we find that the total force is 291,.2lb. Exavprr 2: The ends of a trough are semicircular regions, each with a radius of 2 ft. Find the force caused by liquid pressure on one end if the trough is full of water. soLUrIoN: Figure 8.5.3 shows one end of the trough together with a rectangular element of area. An equation of the semicircle is xz * yr:4. Solving for y gives y : t[=7, and so let f (x) : {E -7. The force on the rectangular element is given by 2w(j(fr) Arr. so if F pounds is the total force on the side of the trough, F: ,,limf ,*g,f(f,) Air llall.0 t:l :2ut Ir2 xf(x) dx Jo r2 :2'ut I xt/E=P dx JO : -&w(4- xr)r,rl' JO Figure8.6.3 Therefore, the total force is ffut lb. 8.6 Exercises 1. A platein the shaPeof a rectangle is submerged vertically in a tank of water, with the upper edge lying in the surface. If -"ia" the width of the plate is 10 ft and the depth is 8 ft, find the {orce due to liquid presiure ott otr" of the plate. 2. A square plate of side 4 ft is submerged ve ically in a tank of water and its center is 2 ft below the surface.Find the force due to liquid ptessure on one side of the plate. 3. Solve Exercise 2 if the center of the plate is 4 ft below the surface. 4. A Platein the shapeof an isoscelesright triantle is submergedvertically in a tank of water, with one let lying in the surface.The legs are each 6 ft lont, Find the force due to liquid pressureon one side of the plate. 5. A rectantular tank fuIl of water is 2 ft wide and 18 in. deep, Find the force due to liquid pressure on one end of the tank. 6. The,endsof a trough are equilateraltriangles having sides with lengtls of 2 ft. If the water in the trough is 1 ft deep, find the force due to liquid prrssure on one end. 8.7 CENTEROF MASSOF A ROD 351 7. The face of a dam adiacent to the water is vertical, and its shape is in the form of an isosceles triangle 250 ft wide across the top and 100 ft high in the center. If the water is 10 ft deep in the center, find the total force on the dam due to liquid Pressure. 8. An oil tank is in the shape of a right-circular cylinder 4 ft in diameter, and its axis is horizontal. If the tank is half full of oil weighing 50 lb/fe, find the total force on one end due to liquid Pressure. 9. The faceof the gate of a dam is in the shape of an isoscelestriangle 4 ft wide at the top and 3 ft high. If the upper edge of the face of the gate is 15 ft below the surface of the water, find the total force due to liquid pressure on the gate. 10. The face of a gate of a dam is vertical and in the shape of an isoscelestrapezoid 3 ft wide at the top, 4 ft wide at the bottom, and 3 ft high. If the upper base is 20 ft below the surface of the water, find the total force due to liquid pressure on the gate. 11. The face of a dam adjacentto the water is inclined at an angle of 30ofrom the vertical. Tlie shape of the face is a rectangle of width 50 ft and slant height 30 ft. If the dam is full of water, find the total force due to liquid pressure on the face. 12. Solve Exercise11 if the face of the dam is an isoscelestrapezoid 120ft wide at the top, 80 ft wide at the bottom, and with a slant height of.40 ft13. The bottom of a swimming pool is an inclined plane. The pool is 2 ft deep at one end and 8 ft deep at the other. If the width of the pool is 25 ft and the length is 40 ft, find the total force due to liquid pressure on the bottom. 14. If the end of a water tank is in the shapeof a rectangle and the tank is full, show that the measure of the force due to liq' uid pressureon the end is the product of the measureof the areaof the end and the measureof the force at the geometrical center. 8.7 CENTER OF MASS OF A ROD In Sec. 7.5 we leamed that if the function / is continuous on the closed interval l-a,bf', the averagevalue of f on fa, b] is given by I rat dk b:- a An important application of the average value of a function occurs in physics in connection with the concept of centerof mass. To arrive at a definition of "mass," consider a particle that is set into motion along an axis by a force of F lb exerted on the particle. So long as the force is acting on the particle, the velocity of the particle is increasing; that is, the particle has an acceleration.The ratio of the.force to the acceleration is.constant regardless of the nagnitude of the force, and this constant ratio is called the massof the particle. . rLLUsrRArror L: If the acceleration of a certain particle is 10 ft/sec2 when the force is 30Ib, the mass of the particle is 30 lb : - 3lb 10 ftlsecz 1 ftlsec2 Thus, for every 1 ft/secz of acceleration,a force of 3 lb must be exerted on . the particle. When the unit of force is 1 lb and the unit of acceleration is7 ftlsecz, ,.-w 352 APPLICATIONS OF THE DEFINITEINTEGRAL the unit of mass is called one slug. That is, L slug is the mass of a particle whose acceleration is 1. ftlsec2 when the magnitude of the force on the particle is 1 lb. Hence, the particle of Illustration 1 that has an acceleration of 10 ftlsec2 when the force is 30 lb has a mass of 3 slugs. From physics, if W lb is the weight of an object having a mass of m slugs, and g ftlsec2 is the constant of acceleration due to gravity, then W: mB Consider now a horizontal rod, of negligible weight and thickness, placed on the r axis. On the rod is a system of n particles located at points x 1 , x 2 , .. . , l n . T h e i t h p a r t i d e ( i : 1 , 2 , . ,n) is atadirecteddistance 4 ft from the origin and its mass is mi slugs. See Fig.8.7.1. The n number of slugs in the total mass of the system ir ) /ni. We define the i :l F m2 lTl s xs O x4 x,r { t- x,2 x F i g ur e 8 . 7. 1 momentof massof the ith particle with respect to the origin as mixi slug-ft. The moment of mass for the system is defined as the sum of the moments .all of mass of the particles. Hence, If M, slug-ft is the moment of mass of the system with respect to the origin, then Now we wish to find a point f such that if the total mass of the system were concentrated there, its moment of mass with respect to the origin would be equal to the moment of mass of the system with respect to the origin. Then f must satisfy the equation nn t> mi:2 i:l flltXt i :l and so (1) The point r is called the centerof massof the system, and it is the point where the system will balance. The position of the center of mass is indepehd'entof the position of the origin; that is, the location of ttre center of mass relative to the positions of the particles does not change when the origin is changed. 8,7 CENTEROF MASS OF A ROD ExAMPLE L: Given four particles of masses 2, 3,1,, and 5 slugs located on the r axis at the Points having coordinates 5, 2, -3, and -4, respectively, where distance measurement is in feet, find the center of mass of this system. solurroN: lf 7 is the coordinate of the center of mass, we have from formula (1) 7 fi, Thus, the center of mass is # tt to the left of the origin. The preceding discussion is now extended to a rigid horizontal rod having a continuously distributed mass.The rod is said tobehomogeneous if its mass is directly proportional to its length. In other words, if the segment of the rod whose length is Arr ft has a mass of L1mslugs, and'A,1m: k Air, then the rod is homogeneous. The number k is a constant and k slugs/ft is called the.linear dcnsity of the rod. Lim = p((i)Lir il*], , ,,_r€i -;. O tfi ,, L--- Figwe 8.7.2 Suppose that we have a nonhomogeneous rod, in which case the linear density varies along the rod. Let L ft be the length of the rod, and place the rod on the r axis so the left endpoint of the rod is at the origin and the right endpoint is at L. See Fi9.8.7.2. The linear density at any point r on the rod is p(r) slugs/ft, where p is continuous on [0, L]. To find the total mass of the rod we consider a partition A of the closed interval [0, L] into z subintervals. The fth subinterval is [ra-t, ra], and its length is A4x ft. If f6 is any point in lxp1, xrf, ffir approximation to the mass of the part of the rod contained in the ith subinterval is Lam slags, where Arnt: p(t) L* The number of slugs in the total mass of the rod is approximated by nn s A ; t n - t pG) Lfc LJ"Ll i:l (2) i:l The smaller we take the norm of the partition A, the doser the Riemann sum in Eq. (2) will be to what we intuitively think of as the measure of the mass of the rod, and so we define the measure of the mass as the limit of the Riemann sum in Eq. (2). 8.7.1 Definition A rod of length L ft has its left endpoint at the origin. If the number of slugs per foot in the linear density at a point x ft from the origin is p(x), where p is continuous on 10,L), then the total massof the rod is M slugs, APPLICATIONS OF THE DEFINITEINTEGRAL where p(x) dx ExAMPrn 2: The density at any point of a rod 4 ftlong varies directly as the distance from the point to an external point in the line of the rod and 2 ft from an end, where the density is 5 slugs/ft. Find the total mass of the rod. solurroN: Figure 8.7.3shows the rod placed on the r axis. If p(r) is the number of slugs per foot in the density of the rod at the point r ft from the end having the greater density, then P(x):c(6-x) where c is the constant of proportionality. Becausep(4):5, we have 5:2c or c:*. Hence,p(x):E(6x). Therefore,if M slugs is the total mass of the rod, we have from Definition 8.7.1. M : , l i p f Z < r -f , )A , x llAll-0 i=l E(6- x) dx -40 The total mass of the rod is therefore 40 slugs. 6-'* ti x ox.,(E>,46 F i g u r e8 . 7 . 3 we now proceed to define the center of mass of the rod of Definition 8.7.L. However, first we must define the moment of mass of the rod with respect to the origin. ti L rm : p(t;)Aix t ,, ,r_, \; Figure8.7.4 As before, place the rod on the r axis with the left endpoint at the origin and the right endpoint at L. see Fig. 8.7.4.Let A be a partition of [0, L] into n subintervals, with the lth subinterval lxo_r, xolhaving length Aix ft. If f' is any point in lx;r, ri], an approximation to the moment of mass I 8.7 CENTEROF MASS OF A ROD 355 with respect to the origin of the part of the rod contained in the ith subinterval is & Lim slug-ft, where Liftr: pG) L*. The number of slug-feet in the moment of mass of the entire rod is approximated by n n i:l i:l (4) PGt) Lix The smaller we take the norm of the partition A, the closer the Riemann sum in Eq. (4) will be to what we intuitively think of as the measure of the moment of mass of the rod with respect to the origin. We have, then, the followi.g definition. 8.7.2 Definition A rod of length L ft has its left endpoint at the origin and the number of slugs per foot in the linear density at a point x ft from the origin is p (x) , where p is continuous on [0, L]. The moment of mass of the rod with respect to the origin rs Mo slug-ft, where (s) The center of massof the rod is at the point r such that if M slugs is the total mass of the rod, iM : Mo. Thus, from Eqs. (3) and (5) we get (6) EXAMPLE 3: Find the center of mass for the rod in Example 2. soLUrIoN: In Example 2, we found M:4A. E(5 - x), we have r Using Eq. (6) with p(r) : Ex(6- x) dx 40 Theref.ore, the center of mass is at I ft from the end having the greater density. 2: If a rod is of uniform density k slugs/ft, where k is a constant, then from formula (6) we have O ILLUSTRATION kLz 2L kx2 2 L kx A T kL2 0 Thus, the center of mass is at the center of the rod, as is to be expected.o 356 APPLICATIONS OF THE DEFINITEINTEGRAL 8.7 Exercises In Exercises1 through 4, a system of particles is located on the r axis. The number of slugs in the mass of each particle and the coordinate of its position are given. Distance is measured in feet. Find the center of mass of each svstem. l. mr:5 at 2; m2: 6 at 3; ms: 4 at 5; mq: 3 at 8. 2. mr:2 a t - 4 ; m z : 8 a l - 7 ; m s : 4 a t 2 ;m q , : 2 a t 3 . 3. mt: 2 at -3; frrz: 4 at -2; ms: 20 at 4; mn: 70 at 6; ms: 30 at 9. 4. m, : 5 at -7; ttrz: 3 al -2; m": 5 at 0i ffiE: ! at 2; m5: 8 at 10. In Exercises5 through 9, find the total mass of the given rod and the center of mass. 5 . The length of a rod is 9 in. and the linear density of the rod at a point r in. from one end is (4r * 1) slugs/in. 6 . The length of a rod is 3 ft, and the linear density of the rod at a point r ft from one end is (5 * 2x) slugslft. 7. The length of a rod is 10 ft and the measure of the linear density at a point is a linear function of the measureof the distance of the point from the left end of the rod. The linear density at the left end is 2 slugs/ft and at the right end is 3 slugs/ft. 8 . A rod is 10 ft long, and the measure of the linear density at a point is a linear function of the measureof the distance from the center of the rod. The linear density at each end of the rod is 5 slugs/ft and at the center the linear density is 3i slugs/ft. 9 . The measureof the linear density at a point of a rod varies directly as the third power of the measureof the distance of the point from one end. The length of the rod is 4 ft and the linear density is 2 slugs/ft at the center. 10.A rod is 6 ft long and its mass is 24 slugs. If the measure of the linear density at any point of the rod varies directly as the square of the distance of the point from one end, find the largest value of the linear density. L L . The length of a rod is L ft and the center of mass of the rod is at the point *L ft from the left end. If the measureof the linear density at a point is proportional to a power of the measure of the distance of the point from the left end and the linear density at the right end is 20 slugs/ft, find the linear density at a point x ft from the left end. Assume the massis measured in slugs. t2. The total mass of a rod of length L ft is M slugs and the measure of the linear density at a point r ft from the left end is proportional to the measureof the distance of the point from the right end. Show that the linear density at a point on the rod r ft from the left end is 2M(L - x) lLz slugs/ft. 8.8 CENTER OF MASS OF A PLANE REGION Let the massesof z particleslocated at the points (xr,yr), (x",yr) , . . . , (xn, An)in the ry pline be measuredin slugs by *r,;r, . . . , mn, and consider the problem of finding the center of mass of this system. we may imagine the partides being supported by a sheet of negligible weight and negligible thickness and may assumethat each particle has its position at exactly one point. The center of mass is the point where the sheet will balance. To determine the center of mass, we must find two averages:f, which is the average value for the abscissasof the n points, and y', the averagevalue for the ordinates of the z points. we first define the moment of mass of a system of particles with respect to an axis. ' If a particle at a distance d ft from an axis has a mass of.m slugs,then if 8.8 CENTEROF MASS OF A PLANEREGION M, slug-ft is the moment of mass of the particle with respect to the axis, Mt: (1) md If the ith particle, having trraissftti slugs, is located at the point (xi, y) ,Its distance from the y axis is xrft; thus, from formula (1), the moment of massof this particle with respectto the y axis ismixl slug-ft. Similarly, the moment of mass of the particle with respectto the r axis is miylslug-ft-The moment of the system of n particles with respectto the y axis is Mu slug-ft, where (2) and the moment of the system with respect to the x axis Ls Mr slug-ft, where (3) The total mass of the system Ls M slugs, where (4) The center of mass of the system is at the point (f , Y), where i,:i'.u.llilliillffilr::,i :'....it.*.. ...':,.. (s) (6) The point (i, !) canbe interpreted as the point such that, if the total mass M slugs of the system were concentrated there, its moment of mass with respectto the y axis,Mn slug-ft, would be determin edby Mu : Mi , and its moment of mass with respect to the I axis, M" slug-ft, would be determined by Mr: My. SOLUTION: ExAMPLEL: Find the centerof massof the four particleshavi^g Mo: masses 2, 6, 4, and 1.slugslocated at the points (5, -2) , (-2, L) , (0,3), and (4,-L), respectively. M, '. - M 4 t f t i x r - 2 ( 5 ) + 6 ( - 2 ) + 4 ( 0 ) + L( 4 ) : 2 i:7 4 \l .at i:t - s4 Z,r i:t rniAt-2(-2) + 6(1)+ 4(3)+ L(-1) :13 lfii:2 + 6 + 4 + L : 13 358 A P P L I C A T I O NO SF T H E D E F I N I T EI N T E G R A L Therefore, W:a M13JM13 andy-W- 1 3- 1 The center of mass is at (#, L ). 0i, f0)) xi-rTi \/ L;x F i g u r e8 . 8 . 1 f(r) consider now a thin sheet of continuously distributed mass, for example, a piece of paper or a flat strip of tin. we regard such sheetsas being two dimensional and call such a plane region a lamina.In this sectionw-e confine our discussion to homogeneous laminae, that is, laminae having constant area density. Laminae of variable area density are considered in Chapter 2L in connection with applications of multiple integrals. Let a homogeneous lamina of areaA ff have a mass of M slugs. Then if the constant area density is k slugs/ff , M: kA. rf the homogeneous lamina is a rectangle,its center of mass is defined to be at the center of the rectangle. we use this definition to define the center of mass of a more general homogeneous lamina. Let L be the homogeneous lamina whose constant area density is k slugs/ff, and which is bounded by the curve y : (x), the r axis, and the f lines x : a and x : b. The function / is continuous on the closed interval l!' bl, andf (x) > 0 for all r in f-a,bf . see Fig. 8.8.1.Let A be a partition of the interval la, bl inton subintervals.The ith subinterval is [4_r, 4] and A6x: xi - xi-r The midpoint of [r1-1, xi] is yi.Associatedwith eachsubinterval is a rectangular lamina whose width, altitude, and area densitv are given by Luxft, f (yt) tt, and k slugs/ft2, respectively, and whose centlr of mass is at the point (7,, if 0)). The area of the rectangularlamina is f (yr) L,,xft"; hence, kf (y) Ag slugs is its mass. Consequently,if L,iMo slug-ft is the moment of mass of this rectangularelemeniwith respeciti the y axis, LtMo: yftf(y) A1x The sum of the measures of the moments of mass of n such rectangular laminae with respect to the y axis is given by the Riemann sum n kv,f(y,) aic i:l If M, slug-ft is the moment of mass of the lamina L with respectto the y axis, we define Similarly, if AiM, slug-ft is the moment of mass of the ith rectangular lamina with respect to the x axrs, LoM":if(y)kf(y) Aux andthe sumof the measuresof the momentsof massof tt suchrectangular laminaewith respectto the r axis is given by the Riemannsum > +klf(v,)f'L,x (8) Thus, if M* slug-ft is the moment of massof the laminaL with resPect to the r axis, w€ define ' ... . ., ;,. +*ur+*+il*#*t* i+ffi r,, (e) The mass of the lth rectangular lamina is kf (y) Alx slugs, and so the sum of the measuresof the massesof n rectangular laminae is given by n ) kf (y) L1x (10) the point (x, y) , we detine which by using formulas (7) , (9) , and (10) gives b b k furr*l dx Ja and y- fb k I f(x) dx Ja Dividing both the nu merator and denominator by k, we get *- rb | *f(*) dx ttl rb IJ a f@) dx (11) and lf (x)l' L f(x) dx (12) 360 A P P L I C A T I O NO SF T H E D E F I N I T EI N T E G R A L In formulas (11) and (12) the denominator is the number of square units in the area of the region, and so we have expressed a physical problem in terms of a geometric one. That is, r and ! can be considered as the averageabscissaand the averageordinate, respectively, of a geometric region. In such a case,x and y depend only on the region, not on the mass of the lamina. so we refer to the center of mass of a plane region instead of to the center of mass of a homogeneous lamina. In such a case,we call the center of mass the centroidof the region. Instead of moments of mass, we consider moments of the region. we define the moments of the plane region in the above discussion with respect to the r axis and the y axis by the following: If. (I, y) is the centroid of the plane region and tf.M, andMu are defined as above, ExAMPLE2: Find the centroid of the first quadrant region bounded by the curye y' : 4x, the x axis, and the lines N - L and x: 4. soLUrIoN: Let f (x) : 2x1t2. The equation of the curve is then y : /(x). In Fig. 8.8.2, the region is shown together with the lth rectangular element. The centroid of the rectangle is at (7;, *f @).). The areaA square units of the region is given by A- n lim 2 f0) l l a l l o- i : r Lic l,^ f(x) dx r 0i, f 0)) 2xrt2dx Ti Lix F i g u r e8 . 8 . 2 We now computeMo and Mr. Mo: lim f,r,f' (yr) Aix l l a l l -o , ? : r xf(x) dx re;: 8.8 CENTEROF MASS OF A PLANEREGION 361 x(2xrrz1dx t24 5 - M* - li* Zif1,) f(v,) Ln llall-oi:1 : * :t f4 J, lf(x)f' f4 Jr dx 4xdx - xrln lr :L5 Hence, +_Ma rw A _+ Zsg _93 35 and M- r, rA: r 2 84 15 :E:- 45 Therefore,the centroid is at the point (89,#). ExAMPr,n3: Find the centroid of the region bounded bY the curyes y:x2 andy--2x*3. solurroN: The points of intersection of the two curves are (-1,1) and (3, 9). The region is shown in Fig. 8.8.3, together with the ith rectangular element. Let f (x) : f and g(r) : 2x * 3. The centroid of the ith rectangular elementis at the point (71,ilf 0) + S(yr)l ) where 7; is the midpoint of the ith subinterval l-xr-r, xr]. The measure of the area of the region is given by : tim [s(y,)-f(y,)] Lrx i llall-oi:l :f f3 J-t ts(r)-f(x)ldx f3 - | J-r -u 3 lZx+3-xzldx APPLICATIONS OF THE DEFINITEINTEGRAL v : f@lt y : s(*) (3,9) (v t, g(y ,) Q,,tlfrl + s(y,)l) 0i, f0)) (-1,1) Figure 8.8.3 We now compute Mu and M*. n lim ) y,lSQ) - f(y,)l Lix Mo: l l a l l -s , 7 : 1 : f xkk) - f(x)l dx J-r : f x l 2 x + 3- x r l d x :t;'3 M,: lim >+ls4) + f (y)I tsQo)- f (y)l 4x s ,-:1 llallfg :u + (x)]ts(r)-, f (x)l dx J_,[s(r) f -1 -i I lZ, + 3 * x'flzx + 3 - xrf dx J-r -l -r ,:E l t/^-\-tr^/ fB fg J-t :# t l+xr*I2x+9-xnf dx 8 , 8 C E N T E RO F M A S S O F A P L A N ER E G I O N 363 Therefore, +:#-1, -3 .n , : 5i # andy : TM: E _L7 Hence,the centroidis at the point (1, +). If a plane region has an axis of symmetry, the centroid of the region lies on the axis of symmetry. This is now stated and proved as a theorem. 8.8.1 Theorem f(-v,)) f0t)) Lix Aix If the plane region R has the line L as an axis of symmetry, the centroid of R lies on L. pRooF: Choose the coordinate axes so that L is on the y axis and the origin is in the region R. Figure 8.8.4 illustrates an example of this situation. In the figure, R is the region CDE, C is the point (-a,0), E is the point (a,0) , and an equationof the curve CDE is y : f (x). Consider a partition of the interval 10, af .Let y5be the midpoint of the ith subinterval lxo-r, xrl. The moment with resPectto the y axis of the rectangular element having an altitude /(7,) and a width Air is yo[f1) Alr]. Becauseof symmetry, for a similar partition of the interval l-a, 0f there is a corresPondingelement having as its moment with respect to the y axis -ytf 0) A1r.The sum of thesetwo moments is 0; theref.ore,Mo: 0. Because i : Mul A, we concludethat f : 0. Thus, the centroid I of the region R lies on the y axis, which is what was to be proved. F i g u r e8 . 8 . 4 By applying the preceding theorem, we can simplify the problem of finding the centroid of a plane region that can be divided into regions having axes of symmetry. ExAMPLE4: Find the centroid of the region bounded bY the semiand the r axis. circle y - \tr4 soLUTIoN: The region is shown in Fig. 8.8.5. Because the y axLSis an axis of symmetry, w€ conclude that the centroid lies on the y axLS;so .r : 0. The moment of the region with respect to the x axis is given by M*: lim >+l\fy|f'L,x i:1 ;;all-s - 2 . + f G - x ' )d x JO F i g u r e8 . 8 . 5 The area of the region .ts 2n square units; so v- +8 2rr 3rr 364 A P P L I C A T I O NO SF T H E D E F I N I T EINTEGRAL There is a useful relation between the force causedby liquid pressure on a plane region and the location of the centroid of the region. As in sec. 8.6, let ABCD be the region bounded by the r axis, the lines x: a artd x:b, and the cur''ey:f(x), where/is continuousand f(x) > 0 on the closed interval la,bl.The region ABCD can be consideredas a vertical plate immersed in a liquid having weight ar pounds per cubic unit of the liquid (see Fig. 8.8.6).If F lb is the force owing to liquid pressureon the vertical plate, r:,li3 f w{f((1)a1x llAll-0 i:t or, equivalently, Figure 8.8.6 xf(x) dx F_ (13) If i is the abscissaof the centroid of the region ABCD, then i : Mo lA. BecauseMn: Il xf(x) dx, we have I: xf(x) dx A and so I: xf (x) dx - iA (14) Substituting from Eq. (14) into (13), *e obtain F- wIA (ls) Formula (15) states that the total force owing to liquid pressure against a vertical plane region is the sameas it would be if the region were horizontal at a depth f units below the surface of a liquid. . ILLUsrRArroN1: Consider a trough full of water having asends semicircular regions each with a radius of.2 ft. Using the result of Example4, we find that the centroid of the region is at a depth of Sljn ft. Therefore,using formula (15), we see that if F lb is the force on one end of the trough, _8t6 F:tu.*.2r:;. This agreeswith the result found in Example2 of Sec.8.6. . For various simple plane regions, the centroid may be found in a table. when both the areaof the region and the centroid of the region may be obtained directly, formqla (15) is easy to apply and is used in such casesby engineers to find the force caused by liquid pressure. K 8.8 CENTEROF MASS OF A PLANEREGION Exercises 8.8 l,2,andSslugsandlocatedatthepoints (-1,3),(2'1), 1. Findthecenterof massof thethreeparticleshavingmassesof -1), (3, respectivelY' and -2) ' 2. Find the center of mass of the four particles having masses of.2, 3,3, and 4 slugs and located at the points (-1, (1, 3), (0, 5), and (2,l) , respectivelY. prove that the centroid of three particles, having equal masses,in a plane lies at the point of intersection of the medians 3. of the triangle having as vertices the points at which the partides are located. In Exercises4 through 11, find the centroid of the region with the indicated boundaries the Y axis. 4. The parabolax:2y - y' aurrd rP and the r axis. 5. The parabolay:4- 6. The parabola A2: 4x, the y axis, and the line y: 4. and the line Y:4. 8. The lines y : 2x * l, x * Y : 7, and x: 8. 9. The curyes A : f and A : 4x in the first quadrant. 7. The parabola!: f 10. The curves A: * and'A: x3. -4artd.y:2x-f. 1 1 . T h e c u r v e sy : f 12. prove that the distance from the centroid of a triangle to any side of the triangle is equal to one-third the length of the altitude to that side. 13. If the centroid of the region bounded by the parabola!':4px value of a. and the line r: a is to be at the point (p, 0), find the 14. Solve Exercise4 of Sec.8.6 by using formula (15) of this section. 15. Solve Exercise5 of Sec.8.5 by using formula (15) of this section. 16. The face of a dam adjacent to the water is vertical and is in the shapeof an isoscelestrapezoid 90 ft wide at the top, 50 ft wide at the bottom, and 20 ft high. Use forrrula (15) of this section to find the total force due to liquid Pressureon the face of the dam. 17. Find the moment about the lower base of the trapezoid of the force in Exercise L5. L8. Solve Exercise 6 of Sec. 8.5 by using formula (15) of this section. L9. Find the center of mass of the lamina bounded by the parabola 2y' - 18 - 3r and the y axis if the area density at any point (x,y) is \ffi slugs/ftz. fr(x) 20. Solve ExerciseL9 if the area density at any point (x, y) is r slugs/fP. 2 L . Let R be the regionboundedby the curvesy: fr(r) and y: fr(x) (seeFig. 8.8.7).lf A is the measureof the areaof R and if y is the ordinate of the centroid of R, prove that the measureof the volume,V, of the solid of revolution obtained by revolving R about the r axis is given by V - 2nyA fr(x) F i g u r e8 . 8 . 7 Stating this formula in words we have: If a plane regionis revoloeil abouta line in its plane that doesnot cut the region,then the measureof the aolumeof the solid 366 OF THE DEFINITE INTEGRAL APPLICATIONS of rettolutiongenetateilis equalto the productof the measureof the areaof the regionand the measureof the distancetra,eled by the centroidof the region. The above statement is known as the theoremot' Pappusfor volumes of solids of revolution. 22. Us_ethe theorem of Pappus to find the volume of the torus (doughnut-shaped) generatedby revolving a circle with a radius of r units about a line in its plane at a distance of b units from its center, whereb ) t. 23. Use the theorem of Pappus to find the centroid of the region bounded by a semicircle and its diameter. 24. Use the theorem of Pappus to find the volume of a sphere with a radius of r units. 25. LetR be the region bounded by the semicircle V: moment of R with respect to the line y : -4. t/rt= land the x axis. Use the theorem of pappus to find the 26. If R is the region of Exercise25, use the theorem of Pappus to find the volume of the solid of revolution generated by revolving R about the line x - y : r. (nrrrn: use the result of Exercise24 in sec. 5.3.) l' 8.9 CENTER OF MASS OF A SOLID OF REVOLUTION v:f@) 0i,f0)) xi-tTi)cib Lix Figure8.9.1 To tind the center of mass of a solid, in general we must make use of multiple integration. This procedure is taken up in chapter 21 as an application of multiple integrals. However, if the shape of the solid is thai of a solid of revolution, and its volume density is constant, we find the center of mass by a method similar to the one used to obtain the center of massof a homogeneous lamina. Following is the procedure for finding the center of mass of a homogeneous solid of revolution, with the assumption that the center of mass is on the axis of revolution. We first set up a three-dimensional coordinate system. The r and.y axesare taken as in two dimensions, and the third axis, the z axis, is taken perpendicular to them at the origin. A point in three dimensions is then given by (x, y , z) . The plane containing the r and y axesis called the ry plane, and the xz plane and the yz plane are defined similarly. suppose that the r axis is the axis of revolution. Then under the assumption that the center of mass lies on the axis of revolution, the y andz coordinates of the center of mass are each zero, and so it is only necessary to find the r coordinate, which we call i. To find r we make use of thb moment of the solid of revolution with respect to the yz plane. Let / be a function that is continuous on the closed interval la , bl , and assumethat/(r) > 0 for all r in la, bf. R is the region bounded by the cufite y : f(x), the r axis, and the lines r : a and x : b;S is the homogeneous solid of revolution whose volume density is k slugs/ft', where k is a constant, and which is generated by revolving the region R about the r axis. Take a partition A of the closed interval la, bf , and denote the ith subintervalby lxn-r,4] (with i: 7,2, . . . ,n ). Let 7i be the midpoint of Lxrr, xtf.Form n rectangleshaving altitudes of f (y) ft and baseswhose width is A;r ft. Refer to Fig. 8.9.1, showing the region R and the ith rectangle.If each of the n rectanglesis revolved about the r axis, n circular disks are generated. The ith rectangle generates a circular disk having a radius of f(y) ft and a thicknessof Lp ft; its volume is rrff(y)fz Aaxff , OF MASSOF A SOLIDOF REVOLUTION 8.9 CENTER and its mass is knlf (y)1, Ap slugs.Figure8.9.2shows the solid of revolution S and the ith circular disk' The center of mass of the circular disk lies on the axis of revolution at the center of the disk (yi, 0, 0). The moment of mass of the disk with respect to the yz plarte is then LMo" slug-ft where LrMo,: Tt(knlf (yr)l' Lrx) The sum of the measuresof the moments of mass of the n circular disks with resPectto the yz plane is given by the Riemann sum Zy,t nlf(y)fz L$ (1) The number of slugs-feet in the moment of massof s with respectto theyz plane, denoted by Mu", is then defined to be the limit of the Riemann sum in (1.)as llAllapproacheszero; so we have Figure8.9.2 The volume , V ft3, of S was, defined in Sec.8.2 b v fb v - lim t nlf (yi)12A i r - r T I t f @ ) f ' d x Ja Fr llall-o The mass,M slugs, of solid S is defined by We define the center of mass of S as the point (r, 0, 0) such that +ffi +i=_qi#+++$ffffi Substituting from Eqs. (2) and (4) into (5), we get xlf(x)l' dx lf(x)l' dx 368 APPLICATIONS OF THE DEFINITEINTEGRAL From Eq. (6), we see that the center of mass of a homogeneous solid of revolution depends only on the shape of the solid, not on-its substance. Therefore, as for a homogeneouslamina, we refer to the center of mass as the centroid of the solid of revolution. when we have a homogeneous solid of revolution, instead of the moment of mass we consider the moment of the solid. The moment, Maz, of the solid S with respectto theyz plane is given by (7) Thus, if (I, 0, 0) is the centroid of S, from Eqs. (3), (d) , and,(7) we have ffi ExAMPLE1: Find the centroid of the solid of revolution generated by revolving about the r axis the region bounded by the curve y : x2, the x axis,and the line x=3. SOLUTION: The region and a rectangular element are shown in Fig. 8.9.3. The solid of revolution and an element of volume are shown in Fig . 8.g.4. f(x) : x2 Mor: lim 2 y,nlf (y,)f' L,x n Ilall-o f-:r xlf(x)f' dx v=f@) x5 dx _ryn f(v)) v -,,Til, rilf(vt)f'Nx } f3 I tf?)f'dx Jo -_1r[t*nA* Jo Figure8.9.3 :#n Therefore, , =T n - 5 x - TM : 4, ,q_ 2 Therefore, the centroid is at the point (8, O, 0 ) . F i g u r e8 . 9 . 4 8.9 CENTEROF MASS OF A SOLID OF REVOLUTION 0i, f0)) Figure8.9.6 Figure8.9.5 The cenkoid of a solid of revolution also can be found by the cylindrical-shell method. Let R be the region_bou_nded by the curve > 0 on la, bl, the r axis, and the i : f @),where / is continuous and /(r) of revolution generated by solid the be Let s and, x:b. r:a iirr", of S is then at the point centroid y axis. The the about R revolving (0, , , O).lf the rectangular elements are taken parallel to the y axis, then the Llement of volume is a cylindrical shell. Let the ith rectangle have_a width of L1x- xi - xi_1,u.'.dlit 7abe the midpoint of the interval lxi-r, xtf. The centroid of the rylindrical shell obtained by revolving this rectangle about the y axis is at the center of the cylindrical shell, which is the point (0,tf Q), 0). Figure 8.9.5shows the region R and a rectangular element of area,and Fig.8.9.5 shows the cylindrical shell. The moment, Mrz, of s with respect to the xz Plarreis given bY lf V cubic units is the volume of S, v- lim f ,nr,f(y,) \x Ilall- o t--r xf(x) dx i::; APPLICATIONS OF THE DEFINITEINTEGRAL I EXAMPLE 2: Use the cylindricalshell method to find the centroid of the solid of revolution generated by revolving the region in Example L about the y axrs. solurroN: Figure 8.9.7 shows th e region and a rectangular element of area. The solid of revolution and a cylindrical-shell element of volume are shown in Fig. 8.9.8. M,,: ,lip 2 +ffvt)2nyi(yt)A$ llAll-0i-t f3 v:f@) (3,9) =rr I xlf (x)f' dx JO -7Tft"d* JO =Tn V - lim f, rorrf (y,) Lix llall-oFi r3 ,2n I xf(x) dx 0 i, f0 i)) JO :2tr ft *, d* JO :#n Q,,tftv,l) Therefore, 2!3n M y_f:#_3 aix Figure8.9.7 Hence, the centroid is at the point ( 0 ,3 , 0 ) . Figure8.9.8 8.9 CENTEROF MASS OF A SOLID OF REVOLUTION 371 ExAMPLE3: Solve Example L by the cylindrical-shell method. Figure 8.9.9 shows the region and a rectangular element of solurroN: area, and Fig. 8.9.10 shows the cylindrical-shell element of volume obtained by reiolving the rectangle about the r axis. The centroid of the ,7.'n),0,0)' The cylindricalshellis it its center,i"hi.h is the point (+(3 + (f, 0,0)' is at centroidof the solid of revolution v (3,9) M,, - 6l -- lim i +(3+ 6,1 2nyo(3 Lil l l a l l *o 7 : t fs : n l" y(3+ frl(3 - {vl dy JO fs :n | (gv-yz) dy JO (tE , v,) (*tt + tfr) ,y) Ai v :#n AiA (3, Yi ) Finding V by the cylindrical-shell method, we have v- A i-t lim $ 2nyo(3- ,6) Lra l l a l l -o ? t fs 2n I V(3 - y't') dY JO #n F i g u r e8 . 9 . 9 _ M.," *v2 t:-:- x 5 and the centroid is at (8,0,0). The result agreeswith that of Examplet. t 8.9 Exercises revolving the given region about the inIn Exercises1 through 16, findthe centroid of the solid of revolution generatedby dicated line. elements perpendicular to the 1. The region bounded by y : 4x f and the x axis, about the y axis. Take the rectangular axis of revolution. 2. Same as Exercise 1, but take the rectangular elements parallel to the axis of revolution. 972 APPLICATIONS OF THE DEFINITEINTEGRAL 3. The region bounded by x * 2y :2, the x axis, and the y axis, about the r axis. Take the rectangular elementsperpendicular to the axis of revolution. Same as Exercise3, but take the rectangular elements parallel to the axis of revolution. The region bounded by y' : r3 and x - 4, about the ff axis. Take the rectangular elements perpendicular to the axis of revolution. Same as Exercise5, but take the rectangularelementsparallel to the axis of revolution. The region bounded by y : x3,x - z, and the r axis, about the lin e x: 2. Take the rectangular elements perpendicular to the axis of revolution. 8. Same as Exercise7, but take the rectangular elements parallel to the axis of revolution. 9. The region bounded by xny: l, y :1, and y :4, about the y axis. 10. The region in Exercise9, about the r axis. 11. The region bounded by the lines y : x, ! : 2x, and.x * y : 6, aboutthe y axis. 12' The region bounded by the portion of the circle r3 -t y' :4 in the first quadrant, the portion of the line 2r - y : 4 in the fourth quadrant, and the y axis, about the y axis. L3. The region bounded by y: rP and !: x + 2, aboutthe line y:4. 1.4.The region bounded by y": 4x and :16 - 4x, about the r axis. !2 L5. The region bounded by y : \/W , the x axis, and the line r: p, aboul the line r: p. 15. The region of Exercise 15, about the line y:2p. L7. Find the centroid of the right-circular cone of altitude ft units and radius r units. L8. Find the center of mass of the solid of revolution of Exercise3 if the measure of the volume density of the solid at any point is equal to a constant k times the measure of the distance of the point from tn" y, pi*". 19. Find the center of mass of the solid of revolution of Exercise5 if the measure of the volume density of the solid at any point is equal to a constant k times the measure of the distance of the point from the yz p?ane. 20. that a cylindrical hole with a radius of r units is bored through a solid wooden hemisphere luppose of radius 2r units, so that the axis of the cylinder is the same as the axis of ttre hemifrhere. Find the centroii of the solid remaining. 8.L0 LENGTH oF ARc OF A PLANE CURVE Let the function f be continuous on the closed interval la, bl. Consider the graph of the equati on y : f (x) of which a sketch is shown in Fig. 9 . 1 0L. . b : f(x) F i g u r e8 . 1 0 . 1 8 . 1 0 L E N G T HO F A R C O F A P L A N EC U R V E 373 The portion of the curve from the point A(a, f (a)) to the point B(b, f(b)) is called an arc. We wish to assigna number to what we intuitively think of as the length of such an arc. If the arc is a line segmentfrom the point (xr, yr) to the point (rr, Ur), we know from the formula for the distance between two points that its length is given We use this formula for defining the length of av ffi. an arc in general. Recallfrom geometry that the circumference of a circle is defined as the limit of the perimeters of regular polygons inscribed in the circle. For other curves we proceed in a similar way. Let A be a partition of the closed interval la, b) fotmed by dividing the interval into r subintervals by choosing any (n - 1) intermediate numbers between a and b. Let X0: a, and xn:b, and let xt , Xz, Xs, 1 xn-r I xr. Xn-r be the intermediate numbers so that Xo I Xr I xz I Then the fth subinterval is [ri-1, x1]; and its length, denoted by Air, is . . . ,2. Thenif llAllisthenormofthepartition x1- xi-lrwherei:!,2,3, = A, each Ap llAll. u 8.10.2 Figure Associatedwith eachpoint (4, 0) on the r axis is a point Prktu f k)) on the curve. Draw a line segmentfrom eachpoint Pi-l to the next point P, as shown in Fig.8.10.2 The length of the line segmentfrom Pi-l to Pi is denoted UV lFr-Fr I and is given by the distance formula lPt-t.tr,l: xr-r)' + (1) A*)' The sum of the lengths of the line segments is l p a l + l p - l p+r l p r 4 l+ . . . + l n , - p , l + " ' + lPr-rPnl which can be written in sigma notation as lP,-rPil (2) i:l It seems plausible that if n is sufficiently large, the sum in (2) will be "close to" what we would intuitively think of as the length of the arc AB. So we define the length of the arc as the limit of the sum in (2) as the norrn 374 APPLICATIONS OF THEDEFINITE INTEGRAL of A approacheszero, in which casen increaseswithout bound. we have, then, the following definition. 8.L0.L Definition If the function / is continuous on the closed interval la, bl and,if there exists a number L having the following property: for any e ) 0 there is a 6>0suchthat l-r. _ | l ) l l P , _ , P -, | L l < e l - " ' ' I li:l I for every partition A of the interual la, bf forwhich llAll< 6, then we write n t: ,lip > lF;El llAll-0 i:r (3) and L is called the length of the arc of fhe curve y : (x) from the point f A(a, f(a)) to the point B(b, f(U11. If the limit in Eq. (3) exists,the arc is said tobe rectifiable. we derive a formula for finding the length L of an arc that is rectifiable. The derivation requires that the derivative of be continuous on f la,bf; such a function is said to be "smooth,, on lo,bl. 8.1.0.2Definition A function / is said tobe smootfton an interval I if f , is continuous on I. (xi,yi) Ui-t : LiU (xi-r,Ai-t) Figure8.10.3 Refer now to Figure 8.10.3. If. pi_r has coordinates (4_1, yi_1) and.p, has coordinates (xv y), then the length of the chord n*rr, is given by formula (1). Lettin g h - xrr : A# and Ui - Ai_r - lPr-rPrl: AtA, we have (4) or/ equivalently,becauseAg # 0, lPr-tP,l: (Air) (s) CURVE 375 OFARCOFA PLANE 8.10LENGTH Becausewe required that f ' be continuous on [4, b], the hypothesis of the mean-value theorem (4.7.2) is satisfied by f , and so there is a number z1in the open intervd (xr-r, r1) such that ' (6) f (xt) f (xt-') : f (z) (x,- xr-r) BecauseL$: f(x) * f (xr-t) and A6r- x1 ri-1, from Eq. (6) we have +r!_:f, (zr) Lrx (7) Substituting from Eq. (7) into (5), we get (8) lml--ffiLix where x*r 1 zi 1 xi. For each i from L to n there is an equation of the form (8). nn (e) i :l i:l Taking the limit on both sides of Eq. (9) as llAllapproadres zeto,we obtain n-n- lim ). lP;F;l : lim ) Vt * lf'(z)12 LP llAll-0 i:l (10) llAll-0 t=l if this limit exists. To show that the limit on the right side of Eq. (10) exists, let g be the function defined by s@): \ET(@F Because/' is continuous on [4, bl, g is continuous on la, bf. Therefore, . ,n, b e c a u sxe* r 1 ? , 1 l x y f o t i : L , 2 , n-n \ s@)t* li- ) \trTTfGJP Ap: ,lim llAll-0 l=1 llAll-0 i=l or, equivalently, (11) ll B e c a u sge( x ) : f f i , f r o mE q . ( 1 1 )w e h a v e rimtffiLrx:f'rydx \ o'J 116-11*of:r (12) Jo Substituting from Eq. (L2) into (10),we get rim>Finl : ['ry Jo '=r llall-o dx (13) 376 A P P L I C A T I O NO SF T H E D E F I N I T EI N T E G R A L Then from Eqs. (3) and (13) we obtain rb L-l ffidx Ja In this way, we have proved the followirg 8.L0.3 Theorem (14) theorem. If the functi on f and its derivativ e ' are continuous on the closed interval f fa, bf , then the length of arc of the curve y: f (x) from the point (a, (a)) f to the point (b, f (b)) is given by we also have the following theorem, which gives the length of the arc of a curve when r is expressedas a function of.y. g.10.4 Theorem If the function F and its derivative F' are continuous on the closed interval lr'lf ,thenthelglgat of arcof thecurvex:F(y) fromthepoint (F(c),c) to the point (F(d) , d) is given by The proof of rheorem 8.10.4is identical with that of rheorem g.10.3; here we interchange x and y as well as the functions f and F, The definite integral obtained when applying Theorem g.10.3 or Theorem 8.10.4is often difficult to evaluate.Becauseour techniques of integration have so far been limited to integration of powers, we are further restricted in finding equations of curves for which we can evaluate the resulting definite integrals to find the length of an arc. EXAMPLE1.:Find the length of the arc of the curye y : x2t3from the p o i n t ( 1 , 1 ) t o (8, 4) by using Theorem 8.10.3. '(x): solurroN: See Fig.8.10.4. Because f ( x ) : ) c 2 t 3f, Theorem 8.L0.3we have &X-rtt. From v v: To evalu ate this definite integral, let u - gxztT* 4; then du:6x-1t| W h e n x : l , u - L 3 ; when x - 8, Lt- 40. Therefore, F i g u r e8 , 1 0 . 4 uLtz du dx. 8 . 1 0 L E N G T HO F A R C O F A P L A N EC U R V E ExAMPrn 2: Find the length of the arc in Example L by using Theorem 8.L0.4. soLUTroN: Because y - x2t3and r Lettin g F (y) - At'', we have 377 0, we solve for x and obtain x - y3t2 F ' ( Y )- E Y u z Then, from Theorem 8.'1.0.4,we have fq L- | Vr+*y dy JT -rftMdv -, v J, -- l f 14 rg | 3 (a + 9y)3/2| lr L - ;7 (40tr' - 133t2) : 7.6 is continuous on fa,bl, then the definite integral If f dt is a function of r; and it gives the length of the arc tf \trTV@P of the cuwey:f(x) fromthe point(a,f(a)) to thepoint (x,f(x)),where r is any number in the closed interval la, bf .Let s denote the length of this arc; thus, s is a function of x, and we have ar s(x):f Vl+tf'G)12dt Ja we have From Theorem7.6.'1., s'(r): ttt+TfTdP or, becauses'1r;: dsldxandf'(x): dyldx, Multiplying by dx, we obtain dsSimilarly, if we are talking about the length of arc of the curve r:80) (15) 378 APPLICATIONS OF THE DEFINITEINTEGRAL f r o m ( S ( c ), c ) t o ( S Q ) , y ) , w e h a v e ds- dy (16) Squaring on both sides of Eq. (15) gives ffiffi ffilF $#rl (r7) From Eq. (r7) we get the geometric interpretation of ds, which is shown in Fig. 8.10.5. Q@ * Ar, V * tV) Figure8.10.5 In Fig. 8.10.5, line T is tangent to the cur.rtey: f (x) at the pointP. lFMl= Ax: dx; lMSl: Ay;lMRl: dV;lpRl:}r; ih" l"ngthof a r cP Q : 4 r . Exercises 8.10 1. Find the length of the arc of the curve 9yz : 4rs from the origin to the point (g, 2\/t) . 2. Find the length of the arc of the curve *: (2V + 3)s from (1, -t) ro (Z\/V, D. 3. Find the length of the arc of the curve 8y : * * 2x-2 fuom the point where r: 1 to the point where r: 2. 4. Use Theorem 8.10.3to find the length of the arc of the curve ys:gf from the point (1,2) to the point (27,7g). 5. Solve Exercise4 by using Theorem 8.10.4. $ Find the entire length of the arc of the curve x2ts+ y2t3: 1 from the point where x : * to the point where r: 1. E Find the length of the arc of the curve y: *(* *21srz from the point where r:0 to the pointwhere r:3. 8. Find the length of the curve 6ry: yo * 3 from the point where y: I to the point where y: !. BEVIEWEXERCISES 379 1 in the first quadrant from r : [a to x: a. 10. Find the length of the curve 9yz : 4(t * rP)a in the first quadrant from x : 0 to r :2\/2. 11. Find the length of the curve 9y' : x(x - 3)' in the first quadrant from x: L to x:3. 12. Find the length of the curve 9y': x'(2x * 3) in the secondquadrant from r: -1 to r: 0. 9. Find the length of the cuwe (xla)2t} t (ylb)'t": (Chapter8) ReoiewExercises 1. Find the area of the region bounded by the loop of the curve y2: *(4- x). 2. The region in the first quadrant bounded by the curyes r : y2 and,x : ya is revolved about'the y axis Find the volume of the solid generated. 3. The base of a solid is the region bounded by the parabola !2:8x and the line r: every plane section perpendicular to the axis of the base is a square. 8. Find the volume of the solid if 4. A container has the sameshape and dimensions as a solid of revolution formed by revolving about the y axis the region in the first quadrant bounded by the parabola *: 4py, the y axis, and the line y : p. If the container is full of water, find the work done in pumping all the water up to a point 3p ft above the top of the container. 5. The sur{aceof a tank is the sameas that of a paraboloid of revolution which is obtained by revolving the parabola y : f about the r axis. The vertex of the parabola is at the bottom of the tank and the tank is 35 ft high. If the tank is filled with water to a depth of.20 ft, find the work done in pumping all of the water out over the top. 6. A plate in the shape of a region bounded by the parabola * : 6y and the line 2y: 3 is Placed in a water tank with its vertex downward and the line in the surfaceof the water. Find the force due to liquid Pressureon one side of the plate' 7. Find the area of the region bounded by the curves y: lx- 11,V: f - 2x, the y axis, and the line r: 2. 8. Find the area of the region bounded by the curves y: lxl + lr 1l and y: x + t, : lx - 21, the r axis, and the 9. Find the volume of the solid generated by revolving the region bounded by the curve y lines l: I and r: 4 about the r axis. 10. Find the length of arc of the curve ayz: f from the origin to (4a,8a). 11. Find the length of the curve 6y': x(x - 2)2 fuom (2, 0) to (8' A\tr) ' (2,2),(2'4)'and(-1,2). 12. Findthecenterofmassofthefourparticleshavingequalmasseslocatedatthepoints(3,0), -l) ' and (5, 2). Find 13. Tfuee particles having masses5, 2, andS slugs are located, respectively, at the points (-1, 3), (2, mass. the center of 14. The length of a rod is 8 in. and the linear density of the rod at a point r in. from the left end is 216 + 1 slugs/in. Find the total mass of the rod and the center of mass. 15. Find the centroid of the region bounded by the parabola !2: x and the line y : x - 2. 16. Find the centroid of the region bounded by the loop of the curve !2: * - x3. 12. The length of a rod is 4 ft and the linear density of the rod at a point r ft from the left end is (3r * 1) slugs/ft. Find the total mass of the rod and the center of mass. 18. Give an example to show that the centroid of a plane region is not necessarily a point within the region. 19. Find the volume of the solid generatedby revolving about the lin e y : -'I..the region bounded by the line 2y : r * 3 and outside the curves!2 I x:0 and.y2- 4x:0. 20. Use integration to find the volume of a segment of a sphere if the sphere has a radius of r units and the altitude of the segment is h units. 380 APPLICATIONS OF THE DEFINITE INTEGRAL 2 L . A church steepleis 30 ft high and every horizontal plane section is a square having sides of length one-tenth of the distance of the plane section from the top of the steeple. Find the volume of the steeple. 22. A trough full of water is 5 ft long, and its cross section is in the shape of a semicircle with a diameter oI 2It at the top. How much work is required to pump the water out over the top? 23. A water tank is in the shapeof a hemisphere surmounted by a right-circular cylinder. The radius of both the hemisphere and the cylinder is 4 ft and the altitude of the cylinder is 8 ft. If the tank is full of water, find the work necessaryto empty the tank by pumping it through an outlet at the top of the tank. 24. A force of 500lb is required to compress a spring whose natural length is 10 in. to a length of 9 in. Find the work done to compress the spring to a length of 8 in. 25. A cable is 20 ft long and weighs 2lblft, and is hanging vertically from the top of a pole. Find the work done in raising the entire cable to the top of the pole. 26. The work necessaryto stretch a spring from 9 in. to L0 in. is I times the work necessaryto stretch it from 8 in. to 9 in. What is the natural length of the spring? 27. Find the length of the curve 9x2tt+ Ayzrz- 36 in the second quadrant from r: -1 to r: -*. 28. A semicircular plate with a radius of 3 ft is submerged vertically in a tank of water, with its diameter lying in the surface. Use formula (15) of Sec.8.8 to find the force due to liquid pressure on one side of the plate. 29. A cylindrical tank is half full of gasoline weighing 42lblfts.If the axis is horizontal and the diameter is 6 ft, find the force on an end due to liquid pressure. 30. Find the centroid of the region bounded above by the parabola4*:36-9y andbelow by the r axis. 3 1 . Use the theorem of Pappus to find the volume of a right-circular cone with a base radius of r units and an altitude of h units. 32. Theregionboundedbythe parabolal: of the solid of revolution formed. 4y,thex axis,andthe liner:4 is revolvedaboutthey axis.Findthe centroid 33. Find the center of mass of the region of Exercise30 if the measure of the area density at any point (x, y) is l[ - V 34. Find the center of mass of the solid of revolution of Exercise32 if the measureof the volume density of the solid at any point is equal to the measure of the distance of the point from the rz plane. Logarithmlc and rl functions FUNCTIONS AND EXPONENTIAL LOGARITHMIC 9.I THE NATURAL LOGARITHMIC FUNCTION The definition of the logarithmic function that you encountered in algebra is based on exponents. The laws of logarithms are then proved from corresponding laws of exponents. One such law of exponents is Ac . Aa: gc*u (1) If the exponents, r and ! t altl positive integers and a is any real number, (1) follows from the definition of a positive integer exponent and mathematical induction. If the exponents are allowed to be any integers, either positive, negative, ot zeto, and,a # 0, then (1) will hold if a zero exponent and a negative integer exponent are defined by ao: I and n>0 If the exponents are rational numbers and a > 0, then Eq. (1) holds when a^tn is defined bv antn: (%)* It is not quite so simple to define a'when r is an ilrational number. For example, what is meant by 4fi? The definition of the logarithmic function, as given in elementary algebra, is based on the assumption that aa exists if a is any positive number and x is any real number. This definition states that the equation 4t: N where a is any positive number except 1 and N is any positive number, can be solved for r, and r is uniquely determined by x: loio N In elementary algebra,logarithms are used mainly as an aid to computation, and for such purposes the number a (called the base)is taken as 10. The following laws of logarithms are proved from the laws of exponents: Law 1 logoMN:logo M * logoN .M : loga M- log" N Law 2 tog, f Law 3 logo1:0 Law 4 logoM:nlogoM Law 5 logoa:l In this chapter we define the logarithmic function by using calculus and prove the laws of logarithms by means of this definition. Then the exponential function is defined in terms of the logarithmic function. This definition enables us to define a' when r is any real number and a * 0. LOGARITHMIC 9.1THENATURAL FUNCTION383 The laws of exponents will then be proved if the exponent is any real number. Let us recall the formula | +n+l *C I t"dt:*= n+ L n#-t J Consider the function This formula does not hold when n:-1. defined by the equation A: t-r, where f is positive. A sketch of the graph of this equation is shown in Fig. 9.1.1. LetR be the region bounded above by the curve y : Uf, below by the f axis, on the left by the line t : l, and on the right by the line f : r, where r is greater than 1. This region R is shown in Fig. 9.1.1.The measureof the areaof R is a function of r; call it A(x) and define it as a definite integral by F i g u r e9 . 1 . 1 (2) A(x) Now consider the integral in (2) if 0 < r ( 1. From Definition 7.3.4, we have f +dt:-t:|0, Then the integral I: Glt) df represents the measure of the area of the region bounded above by the curve y: Ut, below by the t axis, on the left by the line t:x, and on the right by the line f :1. So the integral li $lt) df is then the negative of the measure of the area of the region shown in Fig.9.1.2. rf.x:L, the integral li Glt) df becomesIl Olt) dt, which equals 0 by Definition7.3.5.In this casethe left and right boundariesof the region are the same and the measure of the area is 0. Thus, we see that the integral t{ Glt) dt for x } 0 can be interpreted s in terms of the measure of the area of a region. Its value depends on r and is therefore a function of r. This integral is used to define the "nafural logarithmic function." Figure 9.1.2 9.1.1 Definition The natural logarithmicfunction is the function defined by r>0 The domain of the natural logarithmic function is the set of all tive numbers. We read ln x as "the natural logarithm of x." The natural logarithmic function is differentiable becauseby applying Theorem 7.5.1we have D"(ln r) 384 FUNCTIONS ANDEXPONENTIAL LOGARITHMIC Therefore, D"(ln i:1 (3) From formula (3) and the chain rule,7f.u is a differentiable function of x, and u(x) > 0, then *#*g##;r{f*'-, ., (4) ffi'D.ua EXAMPLE 1: Given y-ln(3r2-5x+8) fu__ - r dx tind dyldx. ExAMPw 2: solurroN: Applying formula (4), we get Given y : l n [ ( 4 x '+ 3 ) ( 2 x- 1 ) ] SOLUTION: D"U find D *U. 3x2 6x + 8 '(6x-6)- 6x-5 3x2- 6x + 8 Applying formula (4) , we get 7 (4x' + 3) (2x- 1) [8r (2x - 1) + 2(4xz+ 3) ] 24x2-8x+6 (4x'+ 3) (2x- 1) SOLUTION: dy From formula (4) , we have 1 (x+1) -)c It should be emphasizedthat when using formula (4), u(x) must be positive; that is, a number in the domain of the derivative must be in the domain of the given natural logarithmic function. o rLLUsrRArroxL: In ExampleL, the domain of the given natural logarithmic function is the set of all real numbers because3x2- 6x * 8 > 0 for all r. This can be seen from the fact that the parabola having equation - 6x * 8 has its vertex at (1,5) and opensupward. Hence,(6x - 6)l A :3f (3x' 6r * 8) is the derivative for all values of x. In Example 2, because (4x2t 3)(2x- 1) > 0 only when x > i, the domain of the given natural logarithmic function is the interval (*, +-). FUNCTION 9.1 THE NATURALLOGARITHMIC Therefore, it is understood that fraction (5) x>i. Becausexl(x * 1) > 0 when either x 1-1 the natural logarithmic function in Example and so 1.lx(x-t 1) is the derivative if either x < is the derivative only if or r > 0, the domain of 3 is (-oo, -1) U (0, **), -1 or r ) 0. o We show that the natural logarithmic function obeys the laws of logarithms as stated earlier. However, first we need a preliminary theorem, which we state and prove. positive numbers, then lf a 9.1.2 Theorem 1 idt PRooF: In the integral ["u (Llt) dt, make the substitution t: aur then u : 1, and when t - ab, u - b. Therefore, we dt-adu. When t:a, have f'lo':I:# @ du) : ['Lau u Jr -1,+" , I F i g u r e9 . 1. 3 o TLLUsTRATIoN2: In terms of geometrY, Theorem 9."1..2states that the measure of the area of the region shown in Fig. 9.1.3 is the same as the measure of the area of the region shown in Fig.9."1..4. o If we take x -'/-. in Definition 9.1.1.,we have l n 1 : [ ' I ta , Jr The right side of the above is zero by Definition t 7.3.5. So we have (6) lnL-0 Equation (5) corresponds to Law 3 of logarithms, given earlier. The following three theorems give some properties of the natural logarithmic function. F i g u r e9 . 1. 4 9.L.3 Theorem If a and b arc any Positive numbers, then ln (ab):ln a+ lnb 386 FUNCTIONS LOGARITHMICANDEXPONENTIAL PRooF:From Definition 9.1.1, rn(ab):f"oIL at Jr which, from Theorem7.4.6, gives tn(ab): I:+ at+l"' I at Applying Theorem 9.1.2 to the second integral on the right side of the above, we obtain tn(ab):f+at+lolat t t Jr Jt and so from Definition 9.1.1 we have ln(ab\ :ltt 9.1.4 Theorem n * ln b I lL a and b are any positive numbers, then lnl:ln a -lnb pRooF: Becauseo: (alb) . b, we have ln4:hf+.b) \al Applying Theorem 9.1.3 to the right side of the above equation, we get l na : t n l + n a ubtractingln b on both sidesof the aboveequation,w e obtain ln4:ha-lnb b 9.1.5 Theorem rf. n is any positive number and r is any rational number, then lna':rlna pRooF: From formula (4), if. r is any rational number and r ) 0, we have 1 D"(lnr):*.rrr-r-!* and D,(rlnr):!* 9.1 THE NATURALLOGARITHMIC FUNCTION Therefore, D'(ln x') : D,(r ln x) From the above equation the derivatives of ln x'and rln x are equal, and so it follows from Theorem 5.3.3 that there is a constant K such that lnx":rlnx*K forallr>O (7\ To determine K, substitute 1 for r in Eq. (7) and we have ln 1": r ln 1* K But, by Eq. (5), ln 1 : 0; hence,K : 0. ReplacingK by0 in Eq. (7),we get lrtx':rlnx forallr>0 from which it follows that if x: fl, where a is any positive number, then lna':rlna I Note that Theorems 9.1.3,9.1.4,and 9.1.5are the same properties as the laws of logarithms 1.,2, and 4, respectively, given earlier. . rLLUsrRArroN3: In Example 2, if Theorem 9.1.3is applied before finding the derivative, we have y : ln(Axz + 3) * ln (2x - 1) (8) Now the domain of the function defined by the above equation is the interval (4, aoo), which is the same as the domain of the given function. From Eq. (8) we get L'tU : 8x2 Ef+ gT E=T and combining the fractions gives 8x(2x- 1,)* 2(4x2* 3) un t A_.:_@ which is the same as the first line of the solution of Example 2. . o rLLUSrnerroN 4: If we apply Theorem 9.1,.4before finding the derivative in Example 3, we have lnY:ln x- ln(r+1) (e) ln r is defined only when x ) 0, and ln(x * 1) is defined only when x > -1. Therefore, the domain of the function defined by Eq. (9) is the interval (0, 1o). But the domain of the function given in Example 3 consists of the two intervals (--, -1) and (0, +-;. Finding D*y from Eq. (9), we have x+"1, 388 FUNCTIONS LOGARITHMICAND EXPONENTIAL but remember here that r must be greater than 0, whereas in the solution of Example 3 values of r less than -1 are also included. o Illustration 4 shows the care that must be taken when applying Theorems 9.1..3,9.L.4,and 9.1.5to natural logarithmic functions. ExAMPLE4: y:ln(2x find D *A. Given - l)t solurroN: From Theorem 9.1.5we have y :ln(2x - l)t: 3 ln(2x - 1) Note that ln(2x - 1)3 and 3ln(2x - 1) both have the samedomain, x ) E. Applying formula (4) gives DrA:3 . 1 ,rr1 . '_ r- 6 2x_ | In the di scutssron ror ) n tth ] Iflt tr follo,1 find this by.Utsin S rg fo: rla ((4), )ffn nu fon (l l' D,(ln lrll))- ' D)*"'(ln 1 ve need to make use of D,(ln lrl). $o is substitutedfor l*l,and so we ha{re tfr,2) w* r ' D " (({x' , ) .f 1 {x' x x2 Hence, D"(ln lxl):, 1 (10) From formula (10) and the chain rule, if u is a differentiable function of x, we get (11) The followirg example logarithmic function, given plify the work involved in volving products, quotients, soLUrIoN: illustrates how the properties oT the natural in Theorems 9.'1,.3,9."1,.4, and 9.'1..5,can simdifferentiating complicated expressions inand powers. From the given equation ' f?E' = l:, lf'FTl lvl:lr(r*D\mllr+|[ml FUNCTION 9.1 THE NATURALLOGARITHMIC Taking the natural logarithm and applying the properties of logarithms, we obtain l r ,l y l: * l n l r + 1 l- l n l r * 2 l - + l n l r + 3 1 Differentiating on both sides implicitly with respect to r and applying formula (11), we get 1, ^ I-'rY : I 11 1 2 / -1 rr + 1 \- T Ix 5 - +2-m Multiplying on both sides by Ur we obtain DrA: y 2 ( x+ 2 ) ( r + 3 ) - 6 ( r * 1 ) ( r + 3 ) - 3 ( r * 1 ) ( x+ 2 ) 5(x+1)(r+2)(r+3) Replacing y by its given value, we obtain D rA equal to Z x z* 1 0 r + 1 2 - 6 x z - 2 4 x - 1 8 -3x2-9x-6 6(x+1)(r+2)(r+3) -7x2*23x-12 Dra: 6 ( x + 1 ) 2 t 3 ( *x 2 ) ' ( r * 3 ) s r z (r2) we have, for n any real number, (13) Evaluate SOLUTION: f xzdx I f 3x2dx Jm:3Jffi:3'^ EXAMprn 7: Evaluate I:#'. 1r +C lr3+11 sor,urroN: Because(x' * 2) l(x + 1) is an improper fraction, we divide the numerator by the denominator and obtain x2*2 4, '+1I-x-I+I+1 3 390 LOGARITHMICAND EXPONENTIAL FUNCTIONS Therefore, Q-1.#) dx I#dx:l: l: :txr-x*3lnlr+11 -2-2+3ln3-3ln -3In3-3 L 0 -3In3 The answer in the above example also can be written asln27 because, by Theorem 9.'1,.5,3In 3 : ln 33. ExAMPr,r 8: Evaluate SOLUTION: f hr , | -ax Jx LCt LI : then du : dxlx; so we have ln I+dx-1"d t-t I " r + C: + ( h r )z+ C Exercises 9.1 In Exercises1 through 10, differentiate the given function and simplify the result. L f (x): ln(l + 4x') 2 .f ( x ) : l n \ M \r '\$; \E3. f(x): ln 4. S(r) : ln(ln r) 5. h(x) : ln (x2ln x) 7.f(x):ln 6.f(x):ln|/# 8. f (x): f4nT --xln(r* \ffi)U.G(r) lr3+11 O F ( r )- \ E T -h(1 + lm) In Exercises11 through 16, findD"y by logarithmic differentiation. xlffi 4A LZ.y:ffi ffi 14.1t: \' (x * l)zrs L6. y - (5x - 4) (x, + 3) (3rt - 5) In Exercises L7 and L8, fin d D ry by implicit differentiation. (b. m!*xy-l W, lnxy*x*y-z In Exercises 19 through x 26, evaluatethe indefinite integral. \R 9.2 THEGRAPHOF THENATURALLOGARITHMICFUNCTION 21Jt+x L n x fdx J s-zx \q' ntlffi 5 - nt z,\A + ' f g + 2 * a* J In Exercises27 through 30, evaluatethe definite integral. 2l dx 27. [5 Jt i2-5 29. 31.P r o v e t h a t l n r n : n ' l n x b y f i r s t s h o w i n g t h a t l n t n a n d n ' l n x d i f f e r b y a c o n s t a n t a n d t h e n f i n d t h e c o n s t a n t b y taking r: 1. ln r ) 0 and 1- ln r- 32. Prove that x-7- Ux < 0 for all r > 0 and x * l,thus establishingthe inequality r-1.hr(x-l x 1. (nrnr: Letf(x):x-1-lnrandg(x):t-lnx-llxanddeterminethesigns forallr>0andr# g'(r) on the intervals (0, 1) and (1, +o;.; off'(x) and 33. Use the result of Exercise32 to prove lt* t-o h(l* r) : t 1C which 34. Establish the limit of Exercise33 by using the definition of the derivative to find F'(0) for the function F for F(r) : ln(l + r). rt rc (Jlt) dt by using Theorem 7.4.8.) (ln r)/r : 0. (mNr: First prove ttrat ol \/i) o, = 35. prove J. J. "lif_ J 1 J l g.Z TlflIE GRAPH OF THE NATURAL LOGARITHMIC FUNCTION To draw a sketch of the graph of the natural logarithmic function, we must first consider some Properties of this function' Let / be the function defined by f(r):lnt: J[ r' It o , r>o The domain of / is the set of all positive numbers. The function / is differentiable for all r in its domain, and f'(x):+ (1) Because/ is differentiable for all x > 0, / is continuous for all r > 0. From (1) weconclude thatf'(r) > 0 forall x) 0, andtherefore/is anincreasing function. 392 FUNCTIONS AND EXPONENTIAL LOGARITHMIC 1 f"(*):-* Q) From (2) it follows that f" (r) < 0 when x > 0. Therefore, the graph of y : f (x) is concavedownward at every point. Because/(1) : ln L : 0, the r intercept of the graph is 1. f(2):ln L t To determine an approximate numerical value for ln 2, the definite integral |l[lt) df is interpreted as the number of square units in the area of the region shown in Fig.9.2.1. From Fig. 9.2.1 we see that ln 2 is between the measuresof the areas of the rectangles, each having a base of length 1 unit and altitudes of lengths * unit and 1 unit; that is, 0.5< ln2 < 1 F i g ur e 9 . 2 . 1 (3) An inequality can be obtained analytically, using Theorem 7.4.g,by proceedingas follows.Letf (t) :Llt andg(t) : *. Then/(t) > g(t) for all f in [1,2]. Becausef and g are continuous on 11.,21,they are integrable on f7,2f, and we have from Theorem 7.4.8 f+ dt> or, equivalently, l n2 = + (4) Similarly, it f (t) : l,lt and h(t'S: t, then h(t) = f (t) for all t in [1, 2]. Becauseh and / are continuous on l'1,,21,they are integrable there; and again using Theorem 7.4.8, we obtain or/ equivalently, 1 > ln? (s) Combining (4) and (5), we get 0.5<ln2 < 1 (5) The number 0.5 is a lower bound of.ln 2 and 1 is an upper bound of ln 2. In a similar manner we obtain a lower and upper bound for the natural logarithm of any positive real number. Later we learn, by applying infinite series, how to compute the natural logarithm of any positive real number to any desired number of decimal places. The value of ln 2 to five decimal places is given by ln 2 - 0.69315 l 9.2 THE GRAPH OF THE NATURALLOGARITHMICFUNCTION 393 Using this value of.ln 2 and applying Theorem 9.1.5,we can find an approximate value for the natural logarithm of any power of 2. In particular, we have : ln4:tn22 2ln2- L.38630 : 2.07945 3ln2= ln f : ln 2-r : -'1. ln 2 - -0.69315 ln | : ln 2-z : -2 ln 2 - -1.38630 ln8:1n23 In the appendix there is a table of natural logarithms, giving the values to four decimal places. We now determine the behavior of the natural logarithmic function tn x. for large values of r by considering lig Becausethe natural logarithmic function is an increasing function, if we take p as any positive number, ln x ) ln 2P x ) 2P whenever V\ From Theorem9.1.5 we have rn2p:pk (8) 2 Substituting from (8) into (7) , we get lnr) pln2 whenever x>2e Because from (6) ,ln 2 > +, we have from the above ln x > +p whenever x Letting p :ZTt, where n ) 0, we have whenever lnr)n x (e) Itfol1owsfrom(9),.bytakingN-2,n,thatforanyn> lnr)n whenever r>N So we may conclude that (10) l" r: *oo "li11 To determine the behavior of the natural logarithmic function for postn r. itive values of r near zero, we investigate ]1p Becauseln r: ln(r-t;-t, we have ln r: -trr 1 x (11) The expression "x + 0+" is equivalent to "Ux -->+@," and so from LOGARITHMIC AND EXPONENTIAL FUNCTIONS (11) we write lim ln r-- lim I-o+ llT-*a tr, 1)c (r2) From (10) we have lim tr, 1: X ll,f-*a *o (13) Therefore,from (L2) and (13) *e get limlnr--@ (ta1 Jl- 0+ From (14) and (10) and the intermediate-value theorem (2.5.1) we conclude that the range of the natural logarithmic function is the set of all real numbers. From (14) we conclude that the graph of the natural logarithmic function is asymptotic to the negative side of the y axis through the fourth quadrant. The properties of the natural logarithmic function are summarized as follows: (i) ( ii) (iii) (iv) (v) (vi) The domain is the set of all positive numbers. The range is the set of all real numbers. The function is increasing on its entire domain. The function is continuous at all numbers in its domain. The graph of the function is concave downward at all points. The graph of the function is asymptotic to the negative side of the y ax;isthrough the fourth quadrant. By using these properties and plotting a few points with a piece of the tangent line at the points, we can draw a sketch of the graph of ttre natural logarithmic function. In Fig.9.2.2 we have plotted the points having abscissasof i, t,1,2,4, and 8. The slope of the tangent line is foundlrom D,(ln x) : Ux. slope: I slope: + slope - 'l,, fI rL2 slope - 2 slope - 4 Figure 9.2.2 1, 2 9 . 3 T H E I N V E R S EO F A F U N C T I O N 9.2 Exercises 1. Draw a sketch of the graph of y : 1tr x by plotting the points having the abscissas*, *, 7,3, and 9,.and use tr 3 1.09861.At each of the five points, find the slope of the tangent line and draw a piece of the tangent line. In Exercises2 through 9, draw a sketch of the graph of the curve having the given equation. S.y:ln(r+l) !. y:rn(-x) 6.Y:ln(x-l) 8' Y:xlnx 4. vux -tr, ! d.x:lny t y:rnlx 7.Y:x-lnx 10. Show the geometric interpretation of the inequality in Exercise32 of Exercises9.1 by drawing on the same pair of axes - Ux, g(r) : In x, and'h(x) : x - 7. sketchesof the graphs of the functions f , g, and lr defined by the equations /(r) : t f { firra an equation of the tangent line to the curve A : ln x at the point whose abscissais 2. f.d pina an equation of the normal line to the curve A:lnx that is parallel to the line x*2y - 1:0. In Exercises13 through 20, find the exactvalue of the number to be found and then give an approximation of this number to three decimal places by using the table of natural logarithms in,the appendix. ff. fina the area of the region bounded by the curve y : 2l(x- 3), the r axis, and the lines r : 4 and la. ft f(x) : Ux, find the averagevalue of f on the interval [1, 5]. 15. Using Boyle's law for the expansion of a gas(seeExercise10 in Exercises3.9); find the averagePressurewith respectto the volume as the volume increasesfrom 4 ff to 8 fts and the pressure is 2000 lb/ft2 when the volume is 4 ff. 'l' 3lx, the r axis, and fe{ fina the volume of the solid of revolution generated when the region bounded by the curve y: the line r: L is revolved about the r axis' tl. lna telegraph cable, the measure of the speed of the signal is proportionalto f ln(Ux), where r is the ratio of the measu.e of the radius of the core of the cable to the measure of the thickness of the cable's winding. Find the value of r for which the speed of the signal is greatest. 18./A particle is moving on a straight line according to the equation of motion s : (t + 1)'zln(f t 1) where s ft is the directed distance of the partide from the starting point at f sec. Find the velocity and accelerationwhen f : 3. 19. The linear density of a rod at a point r ft from one end is 2/(1 * x) slugs/ft. If the rod is 3 ft long, find the mass and center of mass of the rod. 20. A manufacturer of electric generatorsbegan operations on lan. l, 1966.During the first year there were no salesbecause the company concentrated on product development and research.After the first year, the saleshave increased steadily according to the equation y : r ln r, where x is the number of years during which the company has been operating and y is the number of millions of dollars in the salesvolume. (a) Draw a sketch of the graph of the equation. Determine the rate at which the saleswere increasing (b) on Jan. l, L970,and (c) on Jan. 7, 1976. 9.3 THE INVERSE Let us consider the equation OF A FUNCTION x2_u2:4 If this equation is solved for y in Jerms of x, we obtain Y:+\84 (1) LOGARITHMICAND EXPONENTIAL F UNCTIONS Therefore, Eg. (1) does not define y as a function of r becausefor each value of r greater than 2 or less than -2 there are two values of y, and to have a functional relationship, there must be a unique value of the dependent variable for a value of the independent variable. If Eq. (1) is solved for r in terms of y, we obtain x:+{N and so x is not a function of y. Therefore, Eq. (1) neither defines y as a function of r nor x as a function of y. The equation u:x2-t F i g u r e9 . 3 . 1 expressesy as afunction of r becausefor each value of x a unique value of y is determined. As stated in Sec.1.7, no vertical line can intersect the graph of a function in more than one point. In Fig. 9.3.1 we have a sketch of the graph of Eq. (1). We see.thatif a > 2 or a 1-2, the straight line ;r: a intersects the graph in two points. In Fig. 9.8.2 we have a sketch of the graph of Eq. (2). Here we see that any vertical line intersects the graph in only one point. Equation (2) does not define x as a function of y because for each value of y gteatet than -r., two values of r are determined. In some cases,an equation involving x and y will define both y as a function of r and x as a function oly, as shown in the following illustration. o TLLUSTRATToN1.: Consider Figure9.3.2 Q) the equation Y : )c3+'l' (3) If we let f be the function defined by f(x):f+r (4) then Eq. (3) can be written as y : f(x),and y is a function of.x. lf.we solve Eq. (3) for r, we obtain x-w (5) and if we let g be the function defined by sQ)-w then Eq. (5) may be written as x: (6) g Q) , and r is a function of y. o The functions / and g defined by Eqs. (4) and (6) are called inuerse functions.-wealso say that the function g is the inoerseof the function/and that/ is the inaerseof g. The notation f -1 denotes the inverse of function /. Note that in using -1. to denote the inverse of a function, it should not be confused with the exponent -1. we have the following formal definition of the inverse of a function. 9.3 THE INVERSEOF A FUNCTION 9.3.1 Definition If the function / is the set of ordered pairs (x, y), and if there is a function f-r such that : ::.::...::j::l::::.:::i::i.r':....r......:..:,.:,,::::r,:::i::ii:..:r': ,: ': . . . . : : .r.i: .r i::::..::..:1..: .1 . :....!..r. ,:. , . :. .,:.::: i: ::. .:i ir i:::::::!:l:l:ii::::::i::i::i::rli::iil:li;:l;::tji:::::irj::!:::iil]j':::::ii::i::::S:::ii:::: .:.... ::::..:.: i......:::....... :. i :::::i:::;:::i:::i (7) then/-1, which is the set of ordered pairs (y, r), is called the inaerseof the function f , and /and f-r arecalled inaersefunctions. The domain of /-1 is the range of f , and the range of /-1 is the domain of /. Eliminating y between the equations in (7) , we obtain x -- f -'(f (x)), (8) and eliminating r between the same pair of equations, w€ get v:f(f-'(v)) (e) So from Eqs. (8) and (9) we seethat if the inverse of the function/ is the function /-t, then the inverse of the function /-1 is the function /. Becausethe functions / and g defined by Eqs. (4) and (5) are inverse functions, f-r can be written in place of g. We have from (a) and (6) (10) and f-'(x):9i-1 f(x):f+l Sketches of the graphs of functions / and f-| defined in (10) are shown in Figs. 9.3.3and 9.3.4,respectively. We observe that in Fig. 9.3.3the function / is continuous and increasing on its entire domain. The definition of an increasing function (Definition 5.1.1)is satisfiedby f for all valuesof r in its domain. We alsoobserve in Fig. 9.3.3 that a horizontal line intersects the graph of / in only one point. We intuitively suspect that if a function is continuous and increasing, then a horizontal line will intersect the graph of the function in only one point, and so the function will have an inverse. The following theorem verifies this fact. Figure9.3.3 Figure9.3.4 9.3.2 Theorem Supposethat the function / is continuous and increasing on the closedintewalla, bl. Then (i) / has an inverse /-1, which is defined on If @), f (b)l; (ii) /-' is increasineon If @), f (b)l; (iii) /-1 is continuoason lf(a), f(b)|. PRooFor (i): Since/ is continuous on Ia, bl, then if k is any number such that f(a) < k < f(b), by the intermediate-value theorem (7.5.1), there exists a number c in (a, b) such that /(c) : k. So if y is any number in the closed interval lf (a) , f (b)1, there is at least one number x in la, bl such that y : f (x). We wish to show that for each number y there corresponds only one number x. Supposeto a number y, in lf (a),l(b)l there correspond 398 LOGARITHMICAND EXPONENTIAL FUNCTIONS two numbers x, and x, (x, # 12) in [a, b] such that !r: /(rr). Then, we must have f(x,): f @r) and y, : (11) f(xr) Becausewe have assumedx1 * x2, either xr I xz ot x2 < xr. lt q 1 xr, because/ is increasing on la, bl, it follows that /(r1) < f(xr); this contradicts (11). lf. xr1r1, then f(x") < f(x), and,this also contradicts (11). ThelefoJe,our assumption that x, # xris false, and so to eachvalue of y in Lf(a), f (b)l there corresponds exactly one number x in [a, b] such ihat y : f (x). Therefore,/ has an inverse /-r, which is defined for all numbers in ff(a), f(b)|. pRooFor (ii): To prove that /-1 is increasing on lf(a), /(b)], we must show that y, and Az are two numbers in [/(a), /(b)] such that y1 I y2, if then f-l(yr) < f-,(yr). Because/-1 is defined on lf(a), f(b)|, theie exist numbers r, and x2 in fa, bl such that y, : f (xr) ^dyr: f (xr). Therefore, f -tQ) f-'(f (x')) : xr (12) f -t(yr) - (13) and f -'(f (xr)) : x, If x2 < 11, then because / is increasing on fa,b), f (x) < (x1) or, f 'th"n equivalently, azl yt. But ur<uzi therefore, rz cannot be less rr. lf. xr: rr, then because/ is a function, f (x) : f(xr) or, equivalently, Ar: Az,but this also contradicts the fact that Ur 1Az. Therefore, x, # xr. So if r, is not less than r, and x2 * x1,it follows that 11 ( rr; hence, from (12) and (13), f-t.(yr) < f-'(yr). Thus, we have proved that /-l is increasingon [f (a), f (b)1. PRooFor (iii): To prove that ;-t is continuous on the closed interval tfQ), f(b)J, we must show that if r is any number in the open interval (f(a)'f(b))' then/-r is continuousat r, andf-r is continuoirsfrom the right at f (a), and /-1 is continuous from the left at (b). f we prove that 1-t is continuous at any r in the open interval (f(a), f(b)) by showing that Definition 2.6.4 holds at r. We wish to show that, f_orany e > 0 small enough so that f-r(r) - e and f-r(r) * e are both in la, bf, there exists a 6 > 0 such that whenever lV ,l -r(r) : s. Then /(s) : r. Becausefrom (ii), f-t is increasingon ,, .Let f [f(a), f(b)], we concludethat a < s < D. Therefore, a<s-e(s(s*e<b Because/ is increasing on fa, bf, (14) 9.3 THE INVERSEOF A FUNCTION Let E be the smaller of the two numberc r - f(s- e) and f(s * e) - r; so 6 < t - f(s- e) and 6 = /(s * e) - r or, equivalently, /(s-.) <r-D (15) and r+D =f(s*e) WheneverlV- rl (16) <r +6 (17) r-6<y From (14), (15), (15), and (I7), w€ have, whenever ly - ,l f ( a )= f ( s - € ) < y < f ( s * e ) = f ( b ) Because/-t is increasingon If @), f (b)l,it follows from the above that whenever ly ,l f'(f(s - € )) < f' Q) < f' ( fG + e)) or, equivalently, s- € < f'(Y) < s * e whenever lV-rl or, equivalently, -€ < f -'(Y) s or, equivalently, whenever lY-rl <E lf'(y)-f'(r)l<, So1-t is continuouson the openintenral(f (a), f (b)). -1 Theproofsthatf is continuousfrom the right at /(a ) and continuous from the left at f(b) are left as an exercise (see Exercise 24). ',l$ffi'ill ilTfff:;,:-#".?H:rilT#::,"1?i1"f r.,l$if,fi tewal la, bl. 9.3.3 Theorem Supposethat the function / is continuous and decreasingon the closedintewalla, bl. Then -',which is defined o n L f( b ) ,f ( a ) l ; (i) / has an inverse f -r (ii) f is decreasingon lf (b), f (a)l; (iii) f-r is continuouson l,f(b), f (a)1. The proof is similar to the Proof of Theorem 9.3.2 and is left as an exercise (see Exercises 25 to 27). o ILLUsrnATroN 2: We reconsider Eq. (2), y : x2 - "1..As stated earlier in this section, this equation doeS not define .r as a function of y because if we solve for x we obtain x:\F and x:-11y11 (18) t I FUNCTIONS AND EXPONENTIAL LOGARITHMIC So actually Eq. (2) defines r as two distinct functions of y. If in Eq. (2) we restrict r to the closedinterval [0, c] and lety : fr(x) if r is in [0, c], we have f'(x): x2- t and r in [0, c] (19) If in Eq. (2) we restrict r to the closed interval [-c, 0] and let y : fr(x) if.x is in [-c,0], we have fr(x) : x2 I' and r in [-c, 0] (20) The functions f, andf, defined by (19) and (20) are distinct functions becausetheir domains .ue different. If we find the derivatives of f and f2, we obtain 'r(x) : zx and r in [0, c] f and f Lk) : Zx and r in[-c, 0] Because/,is continuous on [0, c] and f'r(x) > 0 for all r in (0, c),by Theorem5.l.3 (i), f is increasingon [0, c]; and so by Theorep !.1.!, /, has an inverse on [-L, c2-ll. The inverse function is given by (2r) Similarly, because/2 is continuous on [-c, 0] and becausef L@) < 0 for all x in (-c,0), by Theorem5.1.3(ii), /, is decreasihgon [-c, 0]. Hence, by Theorem 9.3.3,fzhas an inverse on [-1, ,, - tl, and the inverse function is given by fr-'(y):-\m Figure 9.3.5 (22) Becausethe letter used to represent the independent variable is irrelevant as far as the function is concerned, r could be used instead of u in (21) and (22), and we write fr-'(r) - \m and r in t-l , c2- 1] (23) and Qfu,a) -\ffi fr-t(r) - o Rb, u) and r in t-l , cz- 1] (24) o In Fig. 9.3.5 there are sketchesof the graphs of /, and its inverse fi-l, as defined in (19) and.(23), plotted on the same set of axes. In Fig. 9.3.6 there are sketchesof the graphs of f2 andits inverse/2-1, as defined in (20) and (24), plotted on the same set of axes. It appears intuitively true from Fig.9.3.5 that if the point Q@, o) is on the graph of fi, then the point R(o, u) is on the graph of fr-'. From Fig. 9.3.6 it appears to be true that if the point Q@, a) is on the graph of fz, then the point R(o, u) is on the graph of fr-'. In general, if Q is the point (u, a) and R is the point (2, z), the line 9.3 THE INVERSEOF A FUNCTION segment QR is perpendicular to the line y: t and is bisected by it. We state that the point Q is a reflectionof the point R with respect to the line U : x, and the point R is a reflectionof the point Q with respect to the line : A : x.If r and y are interchanged in the equation y f(x), we obtain the x: (y) equation graph of the the and x: equation , f (y) is said to be a f refleciionof thegraph of the equati ony : f (x) with respectto the line y : 7 Becausethe equation x: f (y) is equivalent to the equation y : fr (x) , we conclude that the graph of the equation y: f-t(x) is a reflection of the graph of the equationy : f (x) with respectto the line y : r. Therefore,if a function / haq an inverse /-1, their graphs are reflections of eachother with respectto the line y: v. EXAMPLEL: Given the function / is defined by f(x):# -t find the inverse f if it exists. Draw a sketchof the graPh of f , and if 1-r exists,draw a sketchof the graph of 1-t on the sameset of axes as the graPh of f . v y - f-'(x) : /'U v:fk) so1urroN: f '(x) : -Sl@- L)'. Because/is continuous for allvalues of r except7, andf' (r) < 0 if x *'l', itfollows fromTheorem9.3.3that/has-an inveise if r is any real number excePt1' To find/-r,lety : f (x), and solve for x, thus giving x: f-t(y).So we have 2x*3 u : -----T " x- L rY-Y:2x*3 x(y-2):y*3 v*3 i: ----=' a-2 Thus, f-'(v): y+3 y-2 or, equivalently, v - f-'(xl ) f-'(x) The domain of f-r is the set of all real numbers excePt2. Sketchesof the graphs of f and f-L are shown in Fig. 9.3.7. Figure 9.3.7 The following theorem bxpressesa relationship between the derivative of a function and the derivative of the inverse of the function if the function has an inverse. 9.3.4 Theorem closedinSupposethat the function / is continuous and monotonic on the ' (r) * Ofor :f bltr'df on!a, (x).If differentiable tew-al1a,bT,andlety /is any r in la, bl, then the derivative of the inverse function /-1, defined by *:1-r(A), is given by 402 LOGARITHMIC AND EXPONENTIAL FUNCTIONS pRooF: Because is continuous and monotonic on / la, b], then by Theorems 9.3.2 and 9.3.3,/ has an inverse which is continuous and monot o n i co n [ f ( a ) , f ( b ) ] ( o r l f ( b ) , f ( a ) l i f f ( b ) < f ( a ) ) . If r is a number in la, bf, LetAr be an increment of.x, Ax # 0, such that x * Lx is also in [a, b]. Then the corresponding increment of.y is given by Ly:f(x*Ar) -f(x) (25) Ly * 0 sinceAr * 0 and/is monotonicon la,bf; that is, either onla,bl f(x+ Lx) < f(x) or f(x+ Lx) > f(x) If r is in [a,b] andy_: f (x),theny is in [/(a) , f (b)] (or lf (b),f (a)l). Also,if x t Ax is in [a, b], theny + Ly is in [/(a) ,f (b)] (or lf (D; @))) f because y + Ly: f(x* Ax)by (25).So x: f-r(y) e6\ )c+ Ar: f_r(y + Ly) It followsfrom (26) and (27) that Ar: f-t(y + Ly)- f-10) (27) and (28) From the definition of a derivative, + Lv) - f -'(y) Dox: lim f-'Q substit"tiJ;"* es) ^::(2s) into the aboveequatior,we set A,x D o x : lim ;;-; f(x*Ar) -f(x) and becauseAr + 0, Dox: lim Ag- 0 f(x*Ar) -f(x) (2e) Lx Beforewe find the limit in (29) we show that under the hypothesis of this theorem "Lx---> 0" is equivalent to ,'Ly -+ 0.,, First we-show that lim Ar: 0. From (28) we have As-0 hm lf-'!+ In at: As-0 Ar-O ty) -f-,(y)l Because/-r is continuous o n l f ( a ) , f ( b ) l ( o r l f ( b ) , f ( a ) l ) , t h el i m i to n the right side of the above equation is zero. So ln Ar-0 (30) 9.3 THE INVERSEOF A FUNCTION that li$ AV:0. From (25) we have Now we demonstrate lim AY: lim lf (x + Lx) - f(x)l Ar-0 At-o Because/ is continuous on Ia, bl, the limit on the right side of the above equation is zeto, and therefore we have -o Jn LY (31) From (30) and (31) it follows that Ar+0 ifandonlyif (32) Ly-0 Thus, applying the limit theorem (regarding the limit of a quotient) to (29) and using (32), we have 1, Dox: Because/ is differentiable on la, bl, the,limit in the denominator of the above is-/'(r) or, equivalently,DrA, and so we have EXAMPLEZ: Show that Theorem g.3.Aholds for the function of Example L. solurroN: lf y: f (x) : (2x + g)/(r - 1), then D,A:f'(*):-C+ at all Because/,(r) exists if.x * 1,/is differentiable and hence continuous # l. x forall is_decreasing L,andsof * < ir, b x # l.F.irth"r*o"",f,(x) b] interval any for holds [a, 9.3.4 Theorem of esis Therefore, tt t ypott " 1. number the which does not contain In the solution of Example 1 we showed that x : f - r ( v\ )r :' Y + 3 y-2 Computing Dux ftom this equation, we get Dox:(f-\'Q):-G+ If in the above equation we let ,: 5 (2x + 3)/(r - 1), we obtain 404 AND EXPONENTIAL FUNCTIONS LOGARITHMIC EXAMPLE3: Given f is the function defined by f (x) : xs * x, determine if / has an inverse. If it does, find the derivative of the inverse function. '(x) :3xz + 1. Therefore, ' (x) ) soLUrIoN: Because 0 for f (x) : x3* x, f f all real numbers, and so f is increasing on its entire domain. Because/ is also continuous on its entire domain, it follows from Theorem 9.3.2 that f has an inverse f-r L e t y - f ( x ) a n d t h e nx - f - ' U ) . S o by Theorem9.3.4, r\17 uax:W- g x z+ l 9.3 Exercises In ExercisesL through 8, find the inverse of the given function, if it exists, and determine its domain. Draw a sketch of the graph of the given function, and if the given function does not have an inverse, show that a horizontal line intersects the graph in more than one point. If the given function has an inverse, draw a sketch of the graph of the inverse function on the same set of axes as the graph of the given function. 1. f(x) : x3 2. f (x) : x2+ 5 3. f (x) 4 . f ( x )- ( x + 2 ) ' 5. f (x) 6. f(x) 7. f (x) - 2lxl+ x 8. f(x) In Exercises9 through 14, perform each of the following steps: (a) Solve the equation for y interms of r and expressy as one or more functions of r; (b) for each of the functions obtained in (a) determine if the function has an inverse, and if it does, determine the domain of the inverse function; (c) use implicit differentiation to find D,y and.Duxand determine the values of r and y for which D,y and.Dux arereciprocals. 9. x2+A,:9 12.9y'- 813-0 10.x 2 - 4 Y ' - t 6 11. 13.2x2-3xy+L-0 1 , 4 . 2 x 2 * 2 y *L : 0 In Exercises 15 through 20, determine if the given function has an inverse, and if it does, determine the domain and range of the inverse function. L 6 .f ( x ) - ( x + 3 ) ' 1 7f.( x )- x 2- + , r ) 0 x 19. f(x) : x5 * xB 20.f(x) *x FUNCTION 9.4 THE EXPONENTIAL 21. Let the function / be defined by f (x): I ttr=V Jo df. Determine if / has an inverse, and if it does, find the derivative of the inverse function. : (r * 5)/(r + k) will be its own inverse' 22. Determine the value of the constant k so that the function d.efined by /(r) rfx < L fx it L < x < 9 23.Givenf (x) -- l*' ifx>9 lzrtfr -t Provethat / has an inversefunctionand find f (x) . 24. Given that the function / is continuous and increasing on the closed intervalla,bl.AssumingTheorem9'3'1(i) prove/-1 is continuour i.o* the right atf(a) and,continuous from the left atl(b). 25. Prove Theorem 9.3.3 (ii). Prove Theorem 9.3.3(i). and (ii), 27. Prove Theorem 9.3.3 (iii). Show that the formula of Theorem 9.3.4 can be written as ( f - ' ) 'v(' r ) :' (#f - t ( r ) ) f 29.use the formula of Exercise 28 to show that (f-')"(r):-ffi 9.4 THE EXPONENTIAL FUNCTION 9.A.1Definition Becausethe natural logarithmic function is increasing on its entire domain, then by Theorem 9.3.2 it has an inverse which is also an increasing function. The inverse of the natural logarithmic function is called the "exponential function," which we now define' The exponentialfunction is the inverse of the natural logarithmic function' and it is defined by (1) The exponential function "exp(r)" is read as "the value of the exponential function at r." 'Becausethe range of the natural logarithmic function is the set of all real numbers, the domain of the exponential function is the set of all real numbers. The range of the exponential function is the set of positive numbers because this is the domain of the natural logarithmic function' Becausethe natural logarithmic function and the exponential function are inverses of each other, it follows from Eqs. (8) and (9) of sec.9.3 that ln(exP(x)) : x (2) -- x exP (ln r) (3) and Because 0 : ln L, we have exp 0 - exP (ln 1) 406 AND EXPONENTIAL FUNCTIONS LOGARITHMIC which from (3) gives us exP0:L We now theorems. 9.4.2 Theorem (4) state some properties of the exponential function AS lf. a and b are any real numbers, then exP(a* b) : exp(a)' exp(b) PRooF: Let A: exp(a), and so from Definition 9.4.1 we have a:ln A Let B: exp(b), and from Definition 9.4.1.it follows that b:lnB From Theorem9.1.3 we have lnA+lnB:lnAB (Z\ Substituting from (5) and (5) into (Z), we obtain a * b:ln AB So exp(a * b) : exp(ln AB) (s) From Eq. (3) it follows that the right side of Eq. (8) is AB, and so we have exp(a*b):An Replacing A and B by their values, we get exp(a* b): exp(a) . exp(b) 9.4.3 Theorem r lI a and b arc any real numbers, exP(a- b) : exp(a) -:- exp(b) The proof is analogousto the proof of Theorem 9.4.2,where Theorem 9.1.3is replacedby Theorem 9.1,.4.rt is left as an exercise(seeExercise1). 9.4.4 Theorem If a is any real number and r is any rational number, then exp(ra): [exp(a)]" pRooF: If in Eq. (g) x : lexp(a)1,,we have lexp (a)] " : exp{lnfexp(a)1,] Applying Theorem 9.1.5to the right side of the above equation, we obtain FUNCTION 9.4 THE EXPONENTIAL [exp@)]r : exP{rln[exP(a)] ] But from Eq. (2),ln[exp @)] - n, and therefore [exp(a)]': T exp(ra) We now wish to define what is meant by a', where a is a positive number and r is an irrational number. To arrive at a reasonabledefinition, consider the casea,, where a ) 0 and r is a rational number. we have from Eq. (3) (e) sr : exp[ln (a')] But by Theorem9."!,.5,lna'- rln a; so from Eq. (9) 61r: exp(r ln a) (10) Becausethe right side of Eq. (10) has a meaning if r is any real number, we use it for our definition. 9.4.5 Definition g.4.6 Theorem lf a is any positive number and r is any real number, we define lf. a is any positive number and r is any real number, lna':xlr':.a PRooF: From Definition 9.4.5, a' : exp(x ln a) Hence, from Definition 9.4.1, we have ln n": xln a I Following is the definition of one of the most important numbers in mathematics. 9.4.7 Definition The number e is defined by the forimula e: exp L The letter "e,, was chosen because of the swiss mathematician and physicist Leonhard Euler (7707-1783)' number; that is, it cannot be exThe number e is a transcend.ental pressed as the root of any polynomial with integer coefficients. The number zr is another example of a transcendentalnumber. The proof that e is transcendentalwas first given inl873,by Charles Hermite, and its value AND EXPONENTIAL FUNCTIONS LOGARITHMIC can be expressed to any required degree of accuracy.In Chapter 16 we leam a method for doing this. The value of e to seven decimal places is 2.71,8281,8. 9.4.8 Theorem ln e: L. PRooF' By Definition 9.4.7, e-expL Hence, ', ln e - ln(exp 1) (11) Becausethe natural logarithmic function and the exponential function are inverse functions, it follows that ln(exP 1) : L (r2) Substituting from (L2) into (11), w€ have lne-L 9.4.9 Theorem exp (x) : st , for all values of x. PRooF' By Definition 9.4.5, st : exp (r ln e) (13) But by Theorem 9.4.8, ln e - 1. Substituting this in (13), we obtain sr : exp (r) From now on, we write e'in place of exp(r), and so from Definition 9.4.1we have (r4) We now derive the formula for the derivative of the exponential function. Let a:e' Then from (14) we have x-lny (ls) Differentiatirg get 1:1 v D"U on both sides of (15) implicitly with respect to x, we FUNCTION 9,4 THE EXPONENTIAL 409 So, D"a- y Replacing y by e*, we obtain ( 16,) D r(e") : sr If.u is adifferentiable function of.x,7t follows from (L6) and the chain rule that (17) It follows that the derivative of the function defined by f(x):ke', where k is a constant, is itself. The only other function we have previously encountered that has this property is the constant function zero; actaally, this is the specialcaseof f(r) : ke' when k: 0. It can be proved that the : ke' most general function which is its own derivative is given by f(x) (seeExercise35). EXAMPLE Given Y : srtrz find +- dyI dx. solurroN; From (17) : H:"ttz (-i) /grlrz x3 From (I7) we obtain the following indefinite integration formula: (18) EXAMptn 2: l4a* J\E Find Let u - t E ; then du r 2Jeudu 2e'+C 2en+C : Becausefrom (14), e' : A if and only if x : ln y, the graph of y t' i" : identical with the graph of tt: ln y. So we can obtain the graph of y et by interchanging the r and y axesin Fig' 9.2-2(seeFig' 9'a'1)' The gruph ol y : et carrbe obtained without referring to the graph of iogarithmic function. Because the range of the exponential natuial the set of all positive numbers, it follows that e' > 0 for all valis the function graph lies entirely above the r axis' D,A : e" ) 0 for all r; the r. So ues of so the function is increasing for all x. D r'y : e' ) 0 for all r; so the graph is concave upward at all Points. 410 FUNCTIONS AND EXPONENTIAL LOGARITHMIC We have the followi^g two limits: (le) lim st--fm X- *@ and e*: o "tlq (20) The proofs of (19) and (20) are left as exercises(see Exercises44 and 45). To plot somespecificpoints use the table in the Appendix giving powers of e. F i g u r e9 . 4 . 1 ExAMPr,u3: Find the area of the region bounded by the curye y : e', thecoordinate axes,and the line x: 2. Y: The region is shown in Fig. 9.4.2. solurroN: tip l lAal l -o f , 1 T er dx e, (2, ezl et e2 e0 : e2- 1 From a table of powers of e, we see that the value of e2to two decimal places is 7 .39. Therefore, the area of the region is approximately G.gg square units. Ei Air Figure 9.4.2 The conclusionsof Theorems9.4.2,9.4.3,and9.4.4are now restatedby using e" in place of exp(r). If. a and b are any real numbers, then ea+b: eo , eb eo-b: €ra: ea + eb (sa)r where r is any rational number. 9,4 THE EXPONENTIAL FUNCTION Writing Eqs. (2) , (3), and (4) by substituting have 411 er in place of exp (x) , we ln(e") : )c - ,lnx X and eo:t ExAMPrn 4: y : t2r*ln find D *A. Given n t2r*ln r : soLUTroN ' t2r trnr : e'" (x). So y - xez' Therefore, D rU : s2r* 2xe2* In Definition9.A.7, the number e was defined as the value of the exponential function at L; that is, e: exp L. To arrive at another way of defining e, we consider the natural logarithmic function f(x) :ln x '(x) :1/r; hence, We know that the derivative of / is given by f L. However, let us apply the definition of the derivative to find f,(t): /'(1). We have ' ): !f ' (\ L : r(1 l i m E !_L*)._lG)_ Lx Ar-o lim Ar-0 _ lim A.r- 0 Therefore, 1 lim ;* ln(1+ Ar) : 1 ar-o AI Replacing Ar by h, we have from the above equation and Theorem9'4.6 (27\ limln(l *h)un-, ft-0 Now, because the exponential function and the natural logarithmic function are inverse functions, we have lim (L * hlun: lim exp[ln(l * h)trn1 h-0 h-O Q2) AND EXPONENTIAL FUNCTIONS LOGARITHMIC Becausethe exponential function is continuous and lim ln(l + h)unexists and equals L as shown in Eq. (21))rwe can apply H"orem right side of Eq. (22) and get lg 2.5.5to the (1+ h)trn-exp h(l + h)',of: exp1 fm Hence, (23) Equation (23) is sometimes used to define e; however, to use this as a definition it is necessaryto prove that the limit exists. Let us consider the function F defined by F(h)-(1 +h)un (24) and determine the function values for some values of h closeto zero. These values are given in Table 9.4.1. Table9.4.1 h 1 0.5 0.05 0.0L 0.001 0.001-0.0L -0.05 -0.5 F(h):(1 +hyrn 2 2.25 2.65 2.702.71,692.7196 2.73 2.79 4 Table9.4.1leadsus to suspectthat lim (1 + lr; ru ir probably a number that lies between 2.7169 and2.T191.As previously mentioned, in Chapter 16 we leam a method for finding the value of e to any desired number of decimal places. Exercises 9,4 1. Prove Theorem9.4.g. A 1. 1 2. Draw a sketch of the graph of y : 3. Draw a sketch of the graph of A - el"l e-s". In Exercises4 throu gh 14, find dyl dx. 4. a:e5' 5. y : g-9tz 8' a:e"* \.trt. y: 1'4. y : *lr-lrna' gnlt/A+rz I 9.4 THE EXPONENTIAL FUNCTION 413 In Exercises1.5through 1.8,find D*y by implicit differentiation. r:. : [F. e' * eu et+u 16. ye'" * xezu- 1 tt[. y'e'* * xyt: I \/ 1.9through26, In Exercises 18. €u: ln(r3 + gy) r evaluate the indefinite rl4't 19. 1 ez-sr dx 20. l tzr+r dx 22. lr"r" 4* 23' J ,-ks"Ydx J J f ,]-t"' integral. J f plr ?26. I#. oZc -EY.J, Jl # d x ex+J In Exercises27 through 30, evaluate the definite integral. 27. 29. fr J. f2 J, -?8. e2dx xea-" dx 30. In Exercises 31 and 32, find. the relative extrema of f , the points of inllection of the graph of.f , the intervals on which / is increasing, the intervals on which / is decreasing, where the gfaph is concave upward, where the graph is concave downward, and the slope of any inflectional tangent. Draw a sketch of the graph of /. 3 1 .f ( x ) : x e - c &}f(*)-e-*' d. fitta an equation of the tangent line to the curve !: e-" that is perpendicular to the line 2r !:5. (0, 1) and (1, e). Cl fitra the area of the region bounded by the curve y: e' and the line through the points : 1. If every plane section per35. A solid has as its basethe region bounded by the curves ! : e' and A : e-" and the line r pendicular to the r axis is a square find the volume of the solid. : 35. Prove that the most generalfunction that is equal to its derivative is given by /(r) : ke". (rrrNr: Let y /(x) and solve the differential equation dVldx: y.) id|* lf p lblftz is the atmospheric pressure at a height of h ft above sea level, then p :2ll6e-o'owo31t". Find the time rate of change of the atmospheric press.rr" outside of an airplane that is 5000 ft high and rising at the rate of 160 ftlsec. 3g. At a certain height the gaugeon an airplane indicates that the atmospheric pressrre is 1500lb/ff. Applying the formula of Exercise37, ipproximat- by differentials how much higher the airplane must rise so that the pressure will be 1480 lblft2. Use differentials to find the aPProxg9. lllft is the length of an iron rod when f degreesis its temperature, then I : 60e0'00001'. l'0. imate increase in I when f increasesfrom 0 to (F. A simple electric circuit containing no condensers,a resistanceof R ohms, and an inductance of L henrys has the electromotive force cut off when the current is 16amperes. The current dies down so that at f secthe current is i amperes and i: loe-<RtLY Show that the rate of dtange of the curent is proportional to the current. 41. A body is moving along a straight line and at f secthe velocity is u ftlsecwhere p: the particle while zr ) 0 after f : 0. es- e2t.Find the distance traveled by 414 LOGARITHMIC AND EXPONENTIAL FUNCTIONS 42. An advertising agency determined statistically that if a breakfast food manufacturer increasesits budget for television commercialsby r thousand dollars there will be an increasein the total profit of 25i?e-0.2s hundred dollirs. What should be the advertising budget increase in order for the manufacturer to hive the greatest profit? What will be the coresponding increase in the company's profit? 43. A tank in the shape of the solid of revolution formed by rotating about the r axis the region bounded by the curve yzx: e and the lines r : 1 and x : 4. If the tank is full of water, find the work done in pumplng all the water to a point 1 ft above the top of the tank. Distance is measured in feet. 44. Prove: lim e":*o,by X-*o showing that for any N>0 there exists an M>0 such that e")N whenever x>M. 45. P r o v e : l i T ; ' e c : O , b Y s h o w i n g t h a t f o r a n y € > 0 t h e r e e x i s t s a n N < 0 s u c h t h a t e s < e w h e n e v e r r ( N . 46. Draw a sketch of the graph of F if F(h) : (7 t h)tth. 9 q NTHFR FXPONFNTIAL AND I OGARITHMIC FTINCTIONS From Definition 9.4.5, we have qx : ,"rln u (1) The expression on the left side of Eq. (1) is called the " exponential function to the base a." Definition If a is any positive number and r is any real number; then the function / defined by f(x) : 6t is called the exponentialfunction to the basea. The exponential function to the base a satisfies the same Properties as the exponential function to tlE-base e. . TLLUSTRATTON 1.: If x and y are any real numbers and n is positive, then from Eq. (1) we have frr7a - gtlnarulna _ grlna*ulna : g(t*u)ln _ a Or*U From Illustration 7 we have the property ar aa - ,r*a We also have the followirg At+fr.u:ar-a (2) properties: (3) 9.5 OTHEREXPONENTIAL AND LOGARITHMIC FUNCTIONS 415 (a")u - ata (4) (ab)' : arbt (s) (6) ao: I The proofs of (3) through (5) are left as exercises (see Exercises 1 to 4). To find the derivative of the exponential function to the base a, we set ac - ecrnaand apply the chain rule. We have At: exlna s c r n aD r ( x l n a ) Dr(a'): : errno(lna) :4'ln a Hence, itu is a differentiable function of r, (7) o TLLUSTRATToN2: If y : D*A:3" Jr2, then from formula ln 3(2x)- z(ln 3)x3"' (7) we have o From (7) we obtain the followi.g indefinite integration formula: (8) Find dx SOLUTION: I \4OT dX: I 103't2dx. Let Lt: Ex; then du :8 dx; thus JJ ? du: dx. We have then ff dx: I to & du | 1or"rz JJ :5 2 1' 0 " r - C lt:r L0 : ExAMPrn 2: Draw sketches of the graPhs of y -- 2" andY :2-* on the same set of axes. Find the area of the region bounded bY these two graphs and the line x: 2. 2 . t03Jft2 31,,L0 -, C The region is soLUrIoN: The required sketches are shown in Fig. 9.5.'1.. shaded in the figure. If A square units is the desired atea, we have A- lim i:r llall-o 416 FUNCTIONS AND EXPONENTIAL LOGARITHMIC (2 - 2-") dx (2, 4) (t i , zE') 2t ln 2 2-r 12 ln? Jo \ ((i,z-ti'1 F i g u r e9 . 5 . 1 We can now define the "logarithmic function to the base a', if a is any positive number other than 1. 9.5.2 Definition '1., If a is any positive number except the logarithmicfunction to the basea is the inverse of the exponential function to the base a; and we write (e) The above is the definition of the logarithmic function to the base a usually given in elementary algebra; however, (9) has meaning for y any real number becauseaa hasbeen precisely defined. It should be noted,that 7fa: e, we have the logarithmic function to the base e, which is the natural logarithmic function. log" x is read as "the logarithm of.x to the base a.', The logarithmic function to the base a obeys the samelaws as the natural logarithmic function. We list them. logo Qy) - logo x + Iogo y logoQ:U):logox-logoy logoL :0 logo xa : y logo x (10) (11) (12) (13) The proofs of (10) through (13) are left as exercises (see Exercises 5 to 8). A relationship between logarithms to the base a and natural logarithms follows easily. Let Y :logo x 9.5 OTHEREXPONENTIAL AND LOGARITHMIC FUNCTIONS 417 Then aa:)c ln aa: ln r ylna--lnr Y- lnr l"a Replacing y by logo x, we obtain ll::r,,., ::rili.iiuiiiniii:ii.:i:i...iiirii:iitixi:liiiii,iiirli.,..ilillir::iiiffiiri ,,*fiffitff't : r ' : : 1 r : : : : : r : : : r : : : t : : : : : . . : : r' ,tj : : : : . : : : : : : : : . : : :i :. :i : .: j: : : : . : : : :. . . : (14) . Equation (14) sometimes is used as the definition of the logarithmic function to the base a. Becausethe natural logarithmic function is continuous at all x > 0, it follows from (14) that the logarithmic function to the base a is continuous at all r ) 0. If in (14) we take x: e,wehave ot, (ls) We now find the derivative of the logarithmic function to the base a. We differentiate on both sides of Eq. (14) with respect to r, and we obtain D , ( l o g o x ) : + D , ( l nw lna r) D*(logox):#'+ (16) Substituting from (15) into (16), we get e D r(logox) :lo8o x (rz7 If u is a differentiable function of x, we have (18) Note that if in (18) we take n - e, we get D*(rog,u) : lo? t D*u u FUNCTIONS LOGARITHMICAND EXPONENTIAL 418 or, equivalently, 1 D"(lnu):!D,u which is the formula we had previously for the derivative of the natural logarithmic function. EXAMPTT 3: Y Given soLUrIoN: Using (11), w€ write : logto x * L y: xz +'/', l o g t o ( r + 1 ) - l o g r o ( x '+ 1 ) From (18) we have find dy ldx. dy e _Iogro e . )y _logn x+1,dx )c2+L'zL Becauser" has been defined for any real number n, we can now prove the formula for finding the derivative of the power function if the exponent is any real number. 9.5.3 Theorem rf n is any real number and the function / is defined by wherer)0 f(x):x" then f'(x) : ny"-r PRooF: Lety: Y rn. Then from (1) : ,nrnr Thus, DrU - ,nrnt ,nln ln r) t -vn'L x - nxn-r I Theorem 9.5.3 enablesus to find the derivative of a variable to a constant power. Previously in this section we leamed how to differentiate a 9.5 OTHER EXPONENTIALAND LOGARITHMICFUNCTIONS 4I9 constant to a variable power. We now consider the derivative of a function whose function value is a variable to a variable power. The method of Iogarithmic differentiationis used and is illustrated in the following example. EXAMPTr 4: Given y - x', find dyldx. solurroN: BecauseA: f , then lyl : lr'1. We take the natural logarithm on both sides of the equation, and we have ln lYl : ltt lr"l or, equivalent$, ltt lYl: r ln lrl Differentiating on both sides of the above equation with respect to r, + * Lx H:a}nl'l + 1) - r'(ln lrl + 1) 9,5 Exercises In Exercises1 through 4, prove the given ProPerty if a is any positive. number and x and y are any real numbers. L. 'A* + Aa- 3. (ab)t : 2. (a")u - ata As-a 4 . a o: ' L arbt 8, prove 5 through In Exercises the given ProPerty if a \s any positive number and x and y arc any positive numbers. 5. logok : !) : logo x - logo y 5. logo(rV) - logo )c+ log" y 7. Iogo xu -- y Log" x 8 . l o g oL : 0 '(x). In Exercises9 through24, find f , ' 9 rf.k \ - 3 5 ' i 1 0 .f ( x ) : $ - B t logrox 1 ' 3 f. ( x ) : ' x ;r' I?. f(x) - (x3+3)2-'* l s .f ( x ) : @ 18. f (x) - vtnx 21.f(x\:1ge' 24. f (x) - (ln x)tn r i i t"t. f(x)-25t34cz 14. f ( x ) : l o g r o # r) ] 16. f (x) - logo[1o8o(log, 17. f (x) : logro[logto(r+ 1) ] 1 9 .f ( x ) : x 6 20. f ( x ) : 22.f(x):(x)"" 23. f (x) : (4e"'1r" xstz FUNCTIONS AND EXPONENTIAL LOGARITHMIC In Exercises 25 through 32, evaluate the indefinite integral. 2s. *" a* [ 27. 28. Iu**,(2xr+L)dx '30. I a"e"dx Ir"rnr(lnr+1)dx 31. *r,'3," dx I 33. Find an equation of the tangent line to the curve y24v- /2x at (4,2). 34. A partide is moving along a straight line according to the equation of motion s - A2r' + B2-ht,where s ft is the directed distance of the particle from the starting point at f sec. Prove that if a ft/sec2is the accelerationat f sec, then a is proportional to s. A company has leamed that when it initiates a new salescampaign the number of salesper day increases.However, the number of extra daily salesper day decreasesas the impact of the campaign wears off. For a specific campaign the company has determined that if S is the number of extra daily salesas a result of the campaign and r is the number of days that have elapsed since the campaign ended, then s: 1000. 3-'/2 Find the rate at which the extradaily salesis decreasingwhen (a) x: 4 and,(b) r: 10. In Exercises 36 and 37,usedifferentialsto find an approximatevalueof the givenlogarithmandexpressthe answerto three decimal places. 36. logro 1..0L5 37. log,o 997 38. Draw sketchesof the graphs of.y : lo916r and y :ln r on the same set of axes. 39. Given:f(x) :t@" * a-"). Prove that f(b * c) * f(b - c) :zf(b)f(c). tl(). A particle moves along a straight line according to the equation of motion s : fll, where s ft is the directed distance of the partide from the starting point at f sec. Find the velocity and accelerationat 2 sec. 9.5 LAWS OF GROWTH AND DECAY The laws of growth and decay provide applications of the exponential function in chemistry, physics, biology, and business. Such a situation would arise when the rate of change of the amount of a substancewith respect to time is proportional to the amount of the substancepresent at a given instant. This would occur in biology, where under certain circumstances the rate of grovWh of a culture of bacteria is proportional to the amount of bacteria present at any given instant. In a chemical reaction, it is often the casethat the rate of the reaction is proportional to the quantity of the substancepresen! for instance, it is known from experiments that the rate of decay of radium is proportional to the amount of radium present at a given instant. An application in business occurswhen interest is compounded continuously. In such cases,if the time is represented by f units, and A units represents the amount of the substancepresent at any time, then dA 7l: Ktt 9.6 LAWS OF GROWTHAND DECAY where k is a constant. If A increasesasf increases,then k > O,and we have lhe law of naturalgrowth.If A decreasesas f increases,then k ( 0, and we have the law of natural decny.In problems involving the law of natural decay, thehalf life of a substanceis the time required for half of it to decay. L: The rate of decay of. EXAMPLE is radium proportional to the amount present at any time. If 50 mg of radium are Presentnow, and its half life is 1.690years,how much radium will be Present100 years from now? Let f : the number of years in the time from now; A : the number of milligrams of radium Present at f years. we have the initial conditions given in Table 9.6.1.The differential equation is solurroN: Separatingthe variables, we obtain 9.6.1 Table t 0 1590 100 A 60 30 Aro, #-kdt Integrating, we have lnlAl:H+e lAl:ekt+e-F'ekt Letting eE: C, we have lAl : Cro', and becauseA is nonnegative we can omit the absolute-value bars around A, thereby giving us A: Cekt BecauseA: 50 when f : 0, we obtain 50 : C. So (1) A - 50ekt - 'J..690, or we get 30 : 6l0e16e0k Because A - 30 when t 0.5 : gr6e0k So ln 0.5 : 1,590k and llr9:5 : -0.0004L0 k - 1690 :-o='929r 1690 Substituting this value of k into Eq. (1), we obtain A - 50e-o.o00410t When t - L00, A Aroo,and we have Aroo:50e-o'0410:57.5 Therefore, there will be 57.6mg of radium present 100 years from now. 422 FUNCTIONS AND EXPONENTIAL LOGARITHMIC ExAMPtn 2: In a certain culture, the rate of growth of bacteria is proportional to the amount present. If there are 1000 bacteria present initially, and the amount doubles in L2 min, how long will it take before there will be 1.,000,000bacteria present? The differential equation is the same as we had in Example above, the general solution is Table9.6.2 t A 0 soLUrIoN: Let f : the number of minutes in the time from now; A: the number of bacteria present at f min. Even though by definition A is a nonnegative integer, we assumethat A can be any nonnegativenumber for A to be a continuous function of f. Table9.6.2 gives the initial conditions. The differential equation is L2 T L,000 2 ,000 L,000,000 hence, ds A - Cekt When t - 0, A: L000;hence, 1,000,which gives A:1000ekt From the condition that A: erzk- 2 2000when f :12, we obtain and so k: #ln2:0.05776 Hence, we have A- 1000eo.o5776t Replacing t by T and A by 1,000,000,we get 1,000,000- 1000e0'05776r e0.05776?:1000 0.057767: ln 1000 ln 1( ,: O ffi:179.6 Therefore, there will be 1,000,000bacteria present in t hr, 59 min, 36 sec. ExAMPrn 3: Newton,s law of cooling states that the rate at which a body changes temperature is proportional to the difference between its temperature and that of the surrounding medium. If a body is in air of temperature 35oand the body cools solurroN: Let f : the number of minutes in the time that has elapsed since the body started to cool; r: the number of degreesin the temperature of the body at f minutes. Table 9.6.3 gives the initial conditions. From Newton's law of cooling, we have dx 7t- k(x - 35) DECAY from 120"to 60oin 40 min, find the temperatureof the body after 100 min. Separating the variables, we obtain dx ffi-kdt T n b l e9 . 6 . 3 t 0 40 x I20 60 kl dt 100 ' rroo kt+ c x: cekt+ 35 Therefore, When x-85ekt+35 When t - 40, x - 60; and we obtain + 35 85e4ok 60: 40k- ln # n:l[il;'::'Lz) x -_ g5e-0.0306t+ 35 :r f f r o o- 8 5 e - 3 ' o 6 1 3 5 : 3 9 Therefore, ExAMPrn4: There are L00 gal of brine in a tank and the brine contains 70lb of dissolved salt. Fresh water runs into the tank at the rate of 3 gal/min, and the mixture, kePt uniform by stirring, runs out at the same rate. How many Pounds of salt are there in the tank at the end of 1.hr? the temperature of the body is 39o after 100 min. Let f : the number of minutes that have elapsed since the water started flowing into the tank; t: the number of pounds of salt in the tank at t min' Because1,00gal of brine are in the tank at all times, at f minutes the number of pounds of salt per gallon is r/100. Three gallons of the mixture run out of the tank each minute, and so the tank loses3(r/100) pounds of salt per minute. BecauseDp is the rate of change of x with respect to f, and r is decreasing as f increases,we have the differential equation solurroN: dt 100 we also have the initial conditions given in Table9.6.4. Separating the variables and integratingr w€ have LOGARITHMIC AND EXPONENTIAL FUNCTIONS T a b l e9 . 6 . 4 t 0 60 x 70 reo I+--o.otIat ln lrl - -0.03f + e x: Ce When x:70, and so C:70. Letting 60 and x - xeo,we have x6o: 70e-1'8 :70(0.L553) - 1'/-,.57 So there are 11.57lb of salt in the tank after L hr. The calculus is often very useful to the economist for evaluating certain business decisions.However, to use the calcuruswe must beconcemed with continuous functions. consider, for example, the following formula which givesA, the number of dollars in the amount after f years, ii P dollars is invested at a rate of 100i percent, compound.edz times per year: A : p (- t\ +- 'Jm- \l ^ ' (2) Let us conceive of a situation in which the interest is continuously compounding; that is, consider formula (2), where we let the number of interest qerlo$s per year increasewithout bound. Then going to the limit in formula (2), we have A: p mtim /r + a)'' _*@ m/ \ which can be written as 1)mtil* A : p t nl it -m *@ l(t+ L\ m/ J (3) Now consider tim /r + -!-\''n z-+o \ ml Letting h : il m, we have mli : 1lh; and because',m -->*o,, is equivalent tO ,rh + 0+r, we have == tim /r + l-\'" (1 * hlttn- ' m/ nli$ n-+o \ 9.6 LAWS OF GROWTHAND DECAY 425 Hence, using Theorem 2.6.5, we have .ri* [(t +h)*'')": : git and so Eq. (3) becomes A - Peit (4) By letting t vary through the set of nonnegative real numbers, we seethat Eq. (4) expressesA as a continuous function of f. Anotherway of looking at the samesituation is to consider an investment of P dollars, which increasesat a rate proportional to its size. This is the law of natural growth. Then if A dollars is the amount at f years, we have lnlAl:H+e A- Cekt When t: 0, A: P, and so C : P. Therefore,we have A: pekt (5) Comparing Eq. (5) withEq. (4), we seethat they are the same if.k: i. So if an investment increasesat a rate proportional to its size, we say that the interest is compoundedcontinuously, and the interest rate is the constant of proportionality. . rLLUSrRArroN1: If P dollars is invested at a rate of 8vo Per year compounded continuously, and A dollars is the amount of the investment at f years, dA ff: o'oee and A- Peo.ost o If in Eq. (4) we take P : L,i:1, and t: l,we getA: e, which gives a justification for the economist's interpretation of the number "e" as the yield on an investment of one dollar for a year at an interest rate of 700Vo compounded continuously. ii't .'ii FUNCTIONS AND EXPONENTIAL LOGARITHMIC EXAMPIE5: If $5000 is borrowed at an interest rate of L2Voper year, compounded continuously, and the loan is to be repaid in one payment at the end of a year, how much must the borrower repay? Also, find the effective rate of interest which is the rate that gives the same amount of interest compounded once a yeat. solurroN: Letting A dollars be the amount to be repaid, and becauseP : 5000,i :0.12, and f : 1, we have from Eq. (4) d: 5gggro'tz : s000(1.1275) :5637.50 Hence, the borrower must rcpay $5637.50.Lettingl be the effective rate of interest, we have 5000(1*i) :5000e0'12 1*i:to'rz i:1.1275 - L : 0.1275 - 12,7570 9;6 Exercises 1. Bacteriagrown in a certain culture increaseat a rate proportional to the amount present. If there are 1000bacteria present initially and the amount doubles in L hr, how many bacteria will there be in 3* hr? 2. In a certain culture where_the rate of growth of bacteria is proportional to the amount present, the number triples in 3 hr, and at the end of 12 hr there were 10 million bacteria. How many bacteria were present initially? 3' In a certain chemical reaction the rate of conversion of a substanceis proportional to the amount of the substancestill untransformed at that time. After 10 min one-third of the original amount of the substancehas been converted, and 20 g has been converted after 15 min. What was the original amount of the substance? 4. Sugar decomPosesin water at a rate proportional to the amount still unchanged. If there were 50 lb of sugar present initiallyandattheendof 5hrthisisreduced ro20 lb,howlongwillittaie u n t i l 9 0 V oo f t h e s u g a r i J d e l o m p o s e d l 5. The rate of natural increase of the population of a certain city is proportional to the population. If the population increasesfrom 40,000to 50,000in 40 years, when will the population be 80,000? 6. Using Newton's law of cooling (see Example 3), if a body in air at a temperature of 0ocools from 200oto 100' in 40 min; how many more minutes will it take for the body to cool to 50.? 7. Under the conditions of Exarnple3, alter how many minutes will the temperature of the body be 45'? 8. When a simple electric circuit, containing no condensersbut having inductance and resistance,has the electromotive force removed, the rate of decreaseof the current is proportional to the current. The current is i amperesf sec after the. cutoff, and i: 4Owhen t: 0. If the current dies down to L5 amperes in 0.01 sec, find i in terms of f. 9. If a thermometer is taken from a room in which the temperature is 75ointo the open, where the temperature is 35", and the reading of the thermometer is 55oafter 30 sec, (a) how long after the removal will the reading be Sb.f 1U)What is the thermometer reading 3 min after the removal? Use Newton's law of cooling (see Example 3). 10. Thirty Percent of a radioactive substancedisappears in 15 years. Find the half life of the substance. REVIEWEXERCISES 427 11. If the half life of radium is 1690years, what percent of the amount present now will be remaining after (a) 100years and (b) 1000years? 12. A tank contains 200 gal of brine in which there are 3 lb of salt per gallon. It is desired to dilute this solution by adding brine containing t lb of salt per gallon, which flows into the tank at the rate of 4 gaUmin and runs out at the same rate. When will the tank contain 1+ lb of salt per gallon? 13. A tank contains 100 gal of fresh water and brine containing2lb of salt per gallon flows into the tank at the rate of 3 gal/min. If the mixture, kept uniform by stirring, flows out at the same rate, how many pounds of salt are there in the tank at the end of 30 min? 14. A loan of $100is repaid in one payment at the end of a year. If the interest rate is 8% compounded continuously, determine (a) the total amount repaid and (b) the effective rate of interest. s 15. If an amount of money invested doubles itself in L0 years at interest compounded continuously, how long will it take for the original amount to triple itself? 15. If the purchasing power of a dollar is decreasingat the rate of 8% annually, cornpounded continuously, how long will it take for the purchasing power to be 50 cents? Professor Willard Libby of University of Califomia at Los Angeles was awarded the Nobel prize in chemistry for discovering a method of determining the date of death of a once-living object. Professor Libby made use of the fact that the tissue of a living organism is composed of two kinds of carbons, a radioactive carbon A and a stable carbon B, in which the ratio of the amount of A to the amount of B is approximately constant. When the organism dies, the law of natural decay applies to A If it is determined that the amount of A in a piece of charcoal is only 15Voof its original amount and the half life of A is 5500 years, when did the tree from which the charcoal came die? (Chapter9) ReaiewExercises In Exercises1 through 8, differentiate the given function. 1 .f ( x ) : ( l nf ) 2 2 .f ( x ) : @ h 3 .f ( x ) : a. f @) - 10-5" 5. /(r) - ifttrr 6. f(x)-ttt( 8. f(x):3"'" (7)t"' integral. the indefinite 14, evaluate 9 through Exercises In r aozc dx lo. I eu'-*1x-l) logro L*x t-x +xz) z. f(x): s. lfi*ax r nlffa, BI# (e" * at*) dx In Exercises15 through L8, evaluate the definite integral. ts. fzfr- ln 17. J,fua* 18 16. x2e8dx "l l 3 , 19. Find D*A if ye' * xea* x * y - 0. '(x)2 0 . l f f ( x ) : l o g ( " ' y (+x 1 ) , f i n d f Jo kb *7)z dx 428 FUNCTIONS AND EXPONENTIAL LOGARITHMIC 2 1 . T h e l i n e a r d e n s i t y o fa r o d a t a p o i n t x f t f r o m o n e e n d i s U ( x * 1 ) center of mass of the rod. 22. Find an equation of the tangent line to the curve y : xs-r af (2,2). s l u g s / f tI.f t h e r o d i s 4 f t l o n g , f i n d t h e m a s s a n d 23. A particle is moving on a straight line wheres ft is the directed distance of the particle from the origin, o ftlsec is the velocityof the particle, and aftlse& is theaccelerationof the particle at f sec.If. a: et * e-tand o:1and s:2 when t: 0, find o and s in terms of l. 24.T'heareaoftheregign,boundedbythecurvey:e-s,thecoordinateaxes,andtheliner:b(b>0),isafunctionofb.lf / is this tunction, find /(b). Also find 1im /(b) . 25. The volume of the solid of revolution obtained by revolving the region in Exercise24 about the r axis is a function of b. If g is this function, find g(b). Also find la S(A). 25. The rate of natural increaseof the population of a certain city is proportional to the population. If the population doubles in 60 years and if the population in 1950was 60,000,estimate the population in the year 2000. 27. The rate of decay of a radioactive substanceis proportional to the amount present. If half of a given deposit of the substance disappears in 1900years, how long will it take for 95Voof the deposit to disappear? 28. Prove that if a rectangle is to have its base on the r axis and two of its vertices on the curve y : s-cz, then the rectangle will have the largest possible area if the two vertices are at the points of inflection of the graph. 29. Givenf (x) : ln lrl and r < 0. Show that / has an inverse function. If g is the inverse functioru find g(r) and the domain of g. In Exercises30 and 31, find the inverse of the given function if there is one, and determine its domain. Draw a sketch of the given function, and if the given function has an inverse, draw a sketch of the graph of the inverse function on the same set of axes. 3 0 .Jf '(\ x ) - ! + 4- 3 2x 3 2 . P r o v e t h a t i fr < 1 , L r r ( r . (nrur: Letf(x):r-lnrandshowthat/isdecreasingon(0,1)andfind/(l).) 33. When a gas undergoesan adiabatic (no gain or loss of heat) expansionor compression,then the rate of changeof the Pressurewith respect to the volume varies directly as the pressure and inversely as the volume. If the pressure is p lb/in.'zwhen the volume is u in.3, and the initial pressure and volume are polblin.z and a6 in.s, show thai pztk: p6ooi. U. ft W in.lb is the work done by a gas expanding against a piston in a cylinder and P lb/in., is the pressure of the gas when the volume of the gas is V in.3, show that if V, in.3 and.V"in.s are the initial and final volumes, respectively, then W:I fvz PN Jvr 35. Supposea piston comPressesa gas in a cylinder from an initial volume of 50 in.s to a volume of 40 in.3. If Boyle's law (Exercise L0 in Exercises3.9) holds, and the initial pressure is 50 lblin.z, find the work done by the piston. (Use the result of Exercise34.) 36. The charge of electricity on a spherical surfaceleaks off at a rate proportional to the charge. Initially, the charge of electricity was 8 coulombs and one-fourth leaks off in 15 min. When will there be only 2 coulombs remaining? 37. How long will it take for an investment to double itself if interest is paid at the rate of 8% compounded continuously? 38. A tank contains 50 gal of salt water with 120lb of dissolved salt. Salt water with 3 lb of salt per gallon flows into the tank at the rate of 2 gallmin and the mixture, kept uniform by stirring, flows out at the same rate. How long will it be before there are 200 lb of salt in the tank? REVIEWEXERCISES rt 39.Find I e-t"t dx if f is any real number. Jo 40. Provethat rf.x> 0, and f" ,o-t dt:L, (1 + h)rm. then ti* x:lim Jt h_O h_0 4 L . Prove that lim 1og,(l * r) : Logoe X t-O (Norr: Comparewith Exercise33 in Exercises9.1.) 42. Prove that li*a'-1:loo c-O X (nrnr: Lety:a'-Tandexpress lim s(y).) (a"-l)lxas a f u n c t i o no f y , s a y g ( y ) ' T h e n s h o w t h a t ! + 0 a s x + 0 ' andfind 43. Use the results of Exercises41 and 42 to Prove that r i mU _ _Lb x'l (HrNr: X,- Write xb-L -_to'n'-L blnx )c-L .bLnx x-l Then let s b ln r and t: x - L.) 44. Prove that \ ean- "l' : fl lim X t-0 eo'and find /'(0) by two methods.) cf(x) for allr where c is a constant, Prove that there is a con45. If the domain of /is the set of allreal numbers and f'(x): function 8 for which 8(x) : f(x)e-co and find 8'(x)') stant k for which f(x) - ke"* for all r. (HrNr: Consider the (nrNr: Let f (x) : 46. Prove that Drn(lnx)-(-l)n-r+ (nrxr: Use mathematical induction' ) each integral' 47. DoExercise 17 in the Review Exercisesof chapter 7 by evaluating Iiigonometric functions 1 0 . 1T H E S I N E A N D C O S I N EF U N C T I O N S 431 10.1.THE SINE AND COSINE FUNCTIONS F i g u r e1 0 . 1 . 1 L0.L.LDefinition In geometry an angleis defined as the union of two rays called the sides, having a cornmon endpoint called the vertex. Any angle is congruent to some angle having its vertex at the origin and one side, called the initial side,lyingon the positive side of the r axis. Such an angle is said to be in standardposition.Figure 10.1.1shows an angle AOB in standard position with AO as the initial side. The other side, OB, is called the terminal side. The angle AAB canbe formed by rotating the side OA to the side OB, arrd under such a rotation the point A moves along the circumference of a circle, having its center at O and radius IOA I to the point B. In the study of trigonometry dealing with problems involving angles of triangles, the measurement of an angle is usually given in degtees. However, in the calculuswe are concernedwith trigonometric functions of real numbers, and we use radian tneasureto define these functions. To define the radian measure of an angle we use the length of an arc of a circle. If such an arc is smaller than a semicircle, it can be considered the by Theorem $aph of a function having a continuous derivative; and so g.r0.z it has length. If the arc is a semicircle or larger, it has a length that is the sum of the lengths of arcs which are smaller than semicircles. Let AOB be an angle in standard position and 16Z'l :1. If s units is the length of the arc oi the circle traveled by point A as the initial side OA is rotated to the terminal side OB, the radianmeasure,f, of angle AOB is given by t: s if the rotation is counterclockwise and t: -s if the rotation is clockwise o rr.r,usrnerrow 1: By using the fact that the measure of the length of the unit circle's circumference is 2tr, we determine the radian measuresof the -+T , ,\r , - n%r , and angles in Fig. 10.1.2a,b, c, d, e, and f. They aretzt , *n , tn, respectively. Ln 2 (d) 10.1.2 Figure In Definition 10.1.1 it is possible that there may be more than one complete revolution in the rotation of. OA' TRIGONOMETRIC FUNCTIONS o ILLUSTRATIoN 2: Figure 10.1.3a shows such an angle whose radian measure is +7r,and Fig. 10.1.3bshows one whose radian measure is -*r. An angle formed by one complete revolution so that oA is coincident with oB has degree measure of 360 and radian measure of 2rr. Hence, there is the following correspondencebetween degreemeasureand radian measure (where the symbol - indicates that the given measurementsare for the same or congruent angles): 350" - 2zr rad or, equivalently, 180'- zr rad From this it follows that 1' - r#ozr rad and 180' 1 rad - 57"78' From this correspondencethe measurement of an angre can be converted from one system of units to the other. F i g u r e1 0 . 1 . 3 o ILLUSTRATToN 3: L62" - 162 . tl1-n rad: r%.zrrad - T n5' - 1m" _ 75o n Table 10.1.1gives the corresponding degreeand radian measuresof certain angles. Table10.1.1 ' t we now define ,n" ,r"" 10'1'2 Definition ; *n ;; tr ""0 "r;"" ', &n "-; tr ;r;r'.; Ezr ",jr """ *o ,5; Eo 2n ,"" ""-0"".Suppose that f is a real number. Placean angle, having radian measure f, in standard position and let point P be at the interseciion of the terminal side of the.angle with the unit circle having its center at the origin. If p is the point (*, y),then the cosinefunctionlJaefinea Uy c o sf : r 10.1 THE SINE AND COSINEFUNCTIONS and the sinefunction is defined by 0, 1) ir/z) F i g u r e1 0 . 1 . 4 sinf:/From the above definition it is seen that sin f and cos f are defined for any value of f. Hence, the domain of the sine and cosine functions is the set of all real numbers. The largest value either function may have is 1, and the smallest value is -1. It will be shown later that the sine and cosine functions are continuous everywhere, and from this it follows that the range of the two functions is [-L, 1]' For certain values of f, the cosine and sine are easily obtained from a figure. From Fig. 10.1.4we see that cos 0: 1 and sin 0: 0, cos*zr: L\/, 0 and sin *z : L, cos rr:-7 and sin 7r:0, and sin 1n: l!2, cosir: Table 10.1'.2gives these values and some cos *zr:0 and sin *rr:-1. others that are frequently used. +\n +\E 1 +\E +\n -+\n -+\E -1 1 +\E t{z z u - 2 , & 1 -1 n r;- An equation of the unit circle having its center at the origin is x2 - cos f and y : sin t, it follows that * y' - L. Because x (1) F i g u r e1 0 . 1 . 5 Note that coszf and sin2 f stand for (cos t)2 and (sin t)2. Equation (1) is called anidentity becauseit is valid for any real number f. Figures L0.1.5and 10.l..6show angleshaving a negative radian measure of lf and corresponding angles having a positive radian measure of t' From these figures we see that . '.i.anii,..., ..,'ffib{.Itli.''l#..'cod.,.*. , sin{*f } +-',,usin:,f (2) These equations hold for any real number f becausethe points where the -t) intersect terminal sides of the angles (having radian measures f and the unit circle have equal abscissasand ordinates that differ only in sign. Hence, Eqs. (2) are identities. From these equations it follows that the cosine function is even and the sine function is odd. From Definition 1.0.1.2the following identities are obtained. t F i g u r e1 0 . 1 . 6 and Bin:(f,*,2rrr) '= ein f (3) The property of cosineand sine statedby Eqs. (3) is called periodicity, which is now defined. TRIGONOMETRIC FUNCTIONS L0.L.3Definition A function f ,t said to be periodicwith periodp * 0 if whenever x is in the domain of /, then x * p is also in the domain of f and From the above definition and Eqs. (3) it is seen that the sine and cosine functions are periodic with period 2n; that is, whenever the value of the independent variable f is increasedby 2n, the value of each of the functions is repeated.It is becauseof the periodicity of the sine and cosine that these functions have important applications in physics and engineering in connection with periodically repetitive phenomena such as wave motion and vibrations. , we now proceed to derive a useful formula becauseother important identities can be obtained from it. The formula we derive is as fblows: (4) P1 (cos a, sin a) x where a and b are any real numbers. Refer to Fig. 10.1.7 showing a unit circle with the points Q ( 1 , o ) , P r ( c o sa, sin a), Pr(cosb, srn b), Pr(cos(a * b), sin (a * b)), and Po(cosa, -sin a) - Using the notatiot 8 to denote the measure of the length of arc from R to S, we have @, O t r r :n , @ r : b , f jP rPr: a, and ffi: l-al: o.Because Gr:@+ffir, w€have C4-b*a F i g u r e1 0 . 1 . 7 and because PnP,- a frr:&+OFr, +b it fotlowsthat (5) From Eqs. (5) and (5) we seethat @, :'pnFr;therefore, the length of the joining the p_oints-Qand Ps is the same as the length of the chord :hg1d joining the points Pn and Pr. squaring the measuresof tfiese lengths, we have loP;|,:l4J'l (7) Using the distance formula, we get IOF;!,: [cos(a + b) - 1], + [sin(a * b) - 0], : cosz(a f b) - 2 cos(a+ b) + L * sin2(a+ b) and because cosz(a * b) + sinz (a * b) : 1,, the above may be written as l84l':2-2cos(a+b) Again applying the distance formula, we have l@rp: (cosb-cos a)2+ (sin b+sin a), (8) I O . 1T H E S I N EA N D C O S I N EF U N C T I O N S 435 : cosz b - 2 cos a cos b + cosz a * sin2 b * 2 sin a sinb* sin2 a and because coszb * sin2 b : 1.and cos2a * sinz a: l, the above becomes (9) cosa cosb * 2 sir.a sinb lF/,l':l-2 Substituting from Eqs. (8) and (9) into (7),we obtain 2 - 2 cos(a+ b) : 2 - 2 cosn cos b * 2 sin a sin b from which follows cos(a * b) : cos a cosb sin a sin b which is formula (4). (a) by substiA formula for cos(a - b) can be obtained from formula tuting -b for b. Doing this, we have cos4 cos(-b) - sin a sin(-b) c o s( a * ( - b ) ) : - -sin b, we get Because cos(-b) - cos b and sin (-b) cssta *',b}"* s'.n Gosl'li'* sin at'lbin h' (10) for all real numberc a NtdbLetting a:Lrr in formula (1'0),we have the equation,cos(Ln-b): : sin *zr : L' it .or J"'-"ol a + ,it Lrr sin b; and'becausecos *rz 0 and - b: c' b".orr,"s cos(*zr- b) : sin b. Now if in this equation we-let $n c and we obtain cos c: sin(*n- c)' We have therefore then b :irrproved the following two identities. -*$r (11) Formulassimilarto(4)and(10)forthesineofthesumanddifference from those we have. A formula for the sine of of two numbers follow ""rity from (11) the sum of two numbers follo*, from (10) and (11)' We have -b) s i n ( a* b ) : c o s ( i z - ( a * b ) ) : c o s ( ( * z t - a ) With (10) it follows that sin(a * b) : cos(tz a) cos b * sin(tzr a) sin b and from (L1) we get $D' for all real numbers a and b. we To obtain a formula for the sine of the difference of two numbers have we write a -b as a+ (-b) and apply formula (12)' Doing this' sin(a * (-b)) : sin a cos(-b) * cos a sin(-b) :-sin b' we obtain and becausecos(-b) : cos b and sin(-b) 436 TRIGONOMETRIC FUNCTIONS (13) :h is valid for all real numb ers a and b. Bylettinga-b- f informulas (L2) and (4) , respectively,we get the (14) (ls) Using the identity sin2 t + cos2 t - I we may rewrite formula (15) as cos2t:2coszt-1, (16) or as c o s2 t : 1 , - 2 s i n 2f (r7) Replacingt by tt in formulas (15) and (lr) and performingsome algebraicmanipulations, we get, respectively, (18) and (1e) (19) are identities because they hold for all real numbers t. By subhacting the terms of formula (4) from the corresponding terms -of formula (10) the following is obtained: sin a sin g:{[-cos(a * b) * cos(a- b)] e0) By adding corresponding terms of formulas (4) and (10) we get cosd cos 6:glcos(a + b) + cos(a- b)l er) and by adding corresponding terms of formulas (12) and (13) we have sin a cos6:f fsin(a + b) + sin(a _ b)] ez) Bylettingc:a * b and d:ab informulas (22), (21),and (20),we obtain, respectively, sinc+ sind-2 sin rycos + (zt1 c o sc + c o s d : 2 cos c + d cos c - d 2 T (24) and c o s c - c o s d : - n -z sln c*4^r^c-d 2 Sln 2 (2s) 10.1 THE SINE AND COSINEFUNCTIONS If the terms of formula (13) are subtracted from the corresponding terms of formula (L2) and a * b and a - b arereplaced by c and il, tespectively, we have P(x,v) Pt (rt , at) s i n c - s i n d - 2 c o s+ sz i2n F i g u r e1 0 . 1. 8 v 7 P( o i 1 / \ _L 4 (26) You are asked to perform the indicated operations in the derivation of some of the above formulas in Exercises7 to 6. When referring to the trigonometric functions with a domain of angle measurements we use the notation f, to denote the measurement of an angle if its degree measure is 0. For example, 45ois the measurementof an angle whose degree measure is 45 or, equivalently, radian measure is *zr' Consider an angle of d|oin standard position on a rectangular cartesian-coordinate system. Chooseany point P, excluding the vertex, on the terminal side of the angle and let its abscissabe r, its ordinate be y, and lOf l : r. Referto Fig. 10.1.8.The ratios xly andy/r ateindependentof the choice of P becauseif the point Pr is chosen instead of P, we see by Fig. 10.1.8that xlr: xlrl and ylr: alrr Becausethe position of the terminal side depends on the angle, these two ratios are functions of the measurement of the angle, and we define cos 0o:x rr and sin 0o:Y (27) Becauseany point P (other than the origin) may be chosen on the terminal side, we could choose the point for which / : 1, and this is the point where the terminal side intersects the unit circle * * y': 1 (see Fig. 10.1.9).Then cos gois the abscissaof the point and sin 0ois the ordinate of the point. This gives the analogy between the sine and cosine of real numbers and those of angle measurements.We have the next definition. .9 Figure10.1 10.1.4 Definition If a de'greesand r radians are measurements for the same angle, then cos co : cos r and sin ao: sin t 1-0.1. Exercises (10). 1. Derive formula (20) by subtracting the terms of formula (4) from the corresponding terms of formula 2. Derive formula (21) by a method similar to that suggestedin Exercise 1. 3. Derive formula (22) by a method similar to that suggestedin Exercise 1. 4. Derive forrrula (23) by using formula (22). 5. Derive formula QD by using formula (21). 6. Derive formula (25) by using formula (20). I. Derive a formula for sin 3t in terms of sin f by using formulas (72), (74), (15), and (1). 438 TRIGONOMETRIC FUNCTIONS 8 . Derive a formula for cos 3f in terms of cos f. Use a method similar to that suggested in Exercise7. 9 . Without using tables, find the value of (a) sin #r and.(b) cos gz. 10. Without using tables, find the value of (a) sin *z and (b) cos *rr. 1 1 .Without using tables, find the value of (a) sin ttn and.(b) cos tzr. T2, Express each of the following in terms of sin f or sin(iz.-t): (a) sin(Ezr-t); (d) cos(Bzrf f). (b) cos(gzr- (c) sin(Bzr+ t); 13. Express the function values of Exercise12 in terms of cos f or cos($z - f). t4. Expresseachof the following in terms of sin f or sin (tn - t): (a) sin(z - t); (b) cos(z - f); (c) sin(z + l); (d) cos(z * f). 15. Express the function values of Exercise 14 in terms of cos f or cos(*a. - f). rc.Find all values of.t for which (a) sin f :0, and (b) cos f :0. 17. Find all values of f for which (a) sin f : 1, and (b) cos f : 1. 18. Find all values of f for which (a) sin f : -1, and (b) cost: -1. 19. Find all values of t for which (a) sin f : *, and (b) cos t: i. 20. Find all values of t for which (a) sin t: -+\/r, and (b) cos f : -Lt/-2. 21. Suppose/ is a function which is periodic with period 2r , and,whose domain is the set of all real numbers prove that / is also periodic with period -22. 22. Prove that the function of Exercise 21 is periodic with period 2nn f.orevery integer n. (Hrr.rr: Use mathematical induction. ) 23. Prove that if f is defined by f (x) - )c- [r]] , then ,t periodic. What is the smallest positive period of f? f 1.O.2DERIVATIVES OF THE SINE AND COSINE FUNCTIONS Before the formula for the derivative of the sine function can be derived we need to know the value of ,. sin f 1iT t Letting f(t): (sin t)lt, we see that /(0) is not defined. However, we prove that lim /(f) exists and is equal to L. 10.2.LTheorem lim t-O PRooF: We first assume that 0 < t < tn. Refel to Fig. 10.2.7,which shows the unit circle 12*yr:1 and the shaded sector BOP, where B is the point (1, 0) and p is the point (cos f, sin f). The area of a circular sector of radius r and central angle of radian measure f is determined by trzt; and so if s square units is the area of sector BOP, s -tt OF THE SINE AND COSINE FUNCTIONS 10.2 DERIVATIVES 439 consider now the triangle BoP, and let K1 square units be the area of this triangle. Hence, K r:t2l F l ' IOB- I:j( sint) ' ( 1) :f sint The line through the points o(0,0) sin f/cos f; therefore, its equation is ( 2) and P(cos f, sin f) has slope sin f x u:----cost " B (L, o) F i g u r e1 0 . 2 . 1 This line intersectsthe line x:1 at the point (1, sin tlcos t), which is the point T in the figure. Letting K, square units be the area of right triangle BOT, we have K,:+lBr |. 16|:+'#l' ! : +.#+ (3) From Fig. 1'0.2.1'we see that K1 (4) , we get Substituting from Eqs. (1), (2), and (3) in inequality L L T s i nt < ; t . ; ' f f i sinf pos: Multiplying eachmember of the above inequality by Zlsin f , which is itive because0 < f < *n, we obtain r1 1' "inT';o;7 Taking the reciProcal of each member of the above inequality and reversing the direction of the inequality signs, we get (s) si+f .1 c o st <\ t \ From the right-hand inequality in the above we have (5) sint<t and from formula (19) in Sec. 10.1 we have I-cosf 2 1', - s ol n e ,' Replacing t by *t in inequality (7) (5) and squarifr1, we obtain sinztt < It' Thus, from (7) and (8) it follows that 1-cosf 2 ,.-Sz -z (8) 440 TRIGONOMETRIC FUNCTIONS which is equivalent to 1 , - t t ' < c o sf From (5) and (9) and because0 1- !*.#.1 iro (10) < t < 0, then 0 If 4n 1 - + e 1 r . s ifn<(1r-) But sin(-f) i f - i n1. t < 0 sin f; thus, the above can be written as 1-|,'.ff (11) From (10) and (11) we conclude that 1-I*.#. 1 ir-+, t 1 - t. Er r.\, 0 (r2) 2 Becauselim (1 - +t') - 1 and lim 1 - 1,, it follows from (I2) and the t-O t-O squeeze theorem r. llmL4 t-O (4.3.5) that sin f _1 L We wish to write the quotient sin 3xlstn 5x in such a way solurroN: '/-.0.2.1. that we can apply Theorem We have, rf x # 0, As r approaches zeto, so do 3x and Sx. Hence, w€ have 1. sin 3r ltm#:lim Jx i-o sin 3r - 1 ,. sr-o 3x and r. 1. sin 5r hm#:lim--1 r-0 5X sin 5r b.llr o 5X Therefore, singx:'l$ ff) rim ;-:o sin 51 5r) 5 hm fg" r-o 5X \ / :1.1 . 1 :g 5 5 FUNCTIONS441 OFTHESINEANDCOSINE 10,2DERIVATIVES From Theorem 10.2.1 we can prove that the sine function and cosine function are continuous at 0. 10.2.2 Theorem the The sine function is continuous at 0. pRooF: We show that the three conditions necessaryfor continuity at a number are satisfied. (i) sin0:0. ( i i ) l i m s i n f : l i m 4 l t' f :lim+ ,-0 r-o t-o t l i m f: 1 ' 0 : 0 . t-o (iii) lim sin.f : sin 0. Therefore, the sine function is continuous at 0' 10.2.3 Therirem I The cosine function is continuous at 0. PROOF: ( i ) c o s0 : 1 . (ii) lim cosf : lim VT- sinu t-o t-o : Vii* (r -i.t) : vIlI-:1. sin4 since cos f ) 0 when Nore: We can replace cos f by \G <0. < t <ir when-tn < and t 0 (iii) lim cos f- cos 0. Therefore, the cosine function is continuous at 0. I The limit in the following theorem is also important. It follows from previous three theorems and limit theorems. the 10.2.4Theorem ltS T{: o PROOF: (1 - cosf)(1 * cost) ,,_ 1 cosf _,,_ :nm@ 111. , -: ,'- k-gos2l)- lS 16+ cosrt :U*''.sin,f_ r--o-f(l * cos f) TRIGONOMETRIC FUNCTIONS :+$,t* l'-t@ By Theorem 10.2.'l.,lLTtri" tlt):1, and becausethe sine and cosine functions are continuous at 0, it follows that Tgor[sin tl(L+ cos f)]: o/(1 + 1) :0. Therefore, L-cosf l'-TT:o ^ I / We now show that the sine function has a derivative. Let f be the function defined by /(r) : sin r From the definition of a derivative, we have f ' '(x) - lim Lt-O - f(x*Ar) -f(x) A,x r:_ sin (x * Ax) - sin r Ilm Atr-o Af - sin r - lim sin r cos(Ar) * cos r sin(Ar) Lx A.r- 0 ,:_ :ttm A.r-o sin rfcos(Ax) - 1] , r:- cosx sin(Ar) rufrrAJr*o LX LX (riq sin')* (*+ cos')g* # 1, (by Theorems10.2.4and 1,0.2.1) Therefore, we have the formula x) : cos r (13) "(sin If u is a differentiable function of x, we have from (13) and the chain D rule (14) The derivative of the cosine function is obtained by making use of (13) and the following identities: cos x - sin (in - x) D "(cos and sin x: r) : D *[sin (in - r) ] cos(tn - x) 10.2 DERIVATIVES OF THE SINE AND COSINEFUNCTIONS 443 : cos(+zr- x) ' D,(*r - x) : sin r(-1.) Therefore, D'(cos r) : -sin r (15) So if. u is a differentiable function of x, it follows from (L5) and the chain rule that (16) ExAMPLE 2: Given SOLUTION: ( 1 - 2 c o sr ) D " ( s i n r ) - s i n r ' D ' ( 1 - 2 c o sr ) -2cosx)' f'(x) (1 (L - 2 cosr) (cosx) - sin r(2 sin r) ( 1 - 2 c o sx ) ' cosr-2(cos2r*sin2r) ( 1 - 2 c o sx ) ' c o sx - 2 ( 1- 2 c o s x Y ExAMPrn3: Given y : (1 + cos3x,)n find dyldx. EXAMPLE du -dx- 4 ( 1+ cos 3x2)t (- sin 3x2)(6r) - -24x sin 3x2(1 + cos 3x')t Given r cos f y c o s x : t find D,A. SOLUTION: solurloN: Differentiating implicitly with respect to )c,we get L ' c o sy * x ( - s i n y ) D * A * D r y ( c o s x ) * y ( - s i n r ) : 0 D * A ( c o s) c - r s i n y ) - y sinx- cosy srnx-cosY Dw * t"r - Y c o sx - x s l n y Geometric problems involving absolute extrema occasionally are more easily solved by using trigonometric functions. The following example illustrates this fact. TRIGONOMETRIC FUNCTIONS EXAMPLE5: A right-circular cylinder is to be inscribed in a sphere of given radius. Find the ratio of the altitude to the base radius of the cylinder having the largest lateral surface area. soLUrIoN: Refer to Fig. 10.2.2.The measure of the constant radius of the sphere is taken as a. Let 0: the number of radians in the angle at the center of the sphere subtended by the radius of the cylinder; r: the number of inches in the radils of the cylinder; h: the number of inches in the altitude of the cylinder; S : the number of square inches in the lateral surface area of the cylinder. '1,0.2.2, From Fig. we seethat r:Asin0 and h-2acos0 Because S : 2rrrh, we have S - 2n(a sin e)Qa cos 0) :2rA2(2 sin g cos 0) :Znaz sin 20 DsS:4na2 cos20 and DszS:-8na2 sin20 Setting Dfi:0, we get c o s2 d : 0 Figure10.2.2 Because 0 < 01trr, 0: i,n We apply the second derivative test for relative extrema and summarize the results in Table 10.2.1,. Table10.2.1 DeS 0:In 0 Dr'S Conclusion S has a relative maximum value Becausethe domain of 0 is the open interval (0,tn),the relative maximum value of S is the absolute maximum value of S. When 0 : *r, r: t\Da, and h : l/-2a.So for the cylinder having the largestlateral surface area,hlr:2. BecauseDr(sin r) : cos r and cos x exists for all values of r, the sine function is differentiable everywhere and therefore continuous everywhere. Similarly, the cosine function is differentiable and continuous everywhere. We now discuss the graph of the sine function. Let f(x): sinr OF THE SINE AND COSINEFUNCTIONS 445 10.2 DERIVATIVES Thenf'(r) : cos xandf" (r) :-sin r. To determinethe relativeextrema, -r2, . . . . If. n is we set f' (x) :0 and get x:$rr * ntr, where rr:0, *7, and [tr:-L; an even integer, then /" (I2n* nn) =-sin(iz * nn\:-sin *nr):-sin *n-l. if n is an odd integer, f "(*o*nn):-sin(*zr Therefore,/ has relative extrema when x : lr * nrr; and if n is an even integer,f has a relative maximum value; and if n is an odd integer, / has a relative minimum value. To determine the points of inflection of the graPh, we set f " (x) :0 . Becausef" (x) changes and obtain x: nnr, where n:0, !L, -r2, . sign at each of these values of x, the graph has a point of inflection at each '(x) : point having these abscissas. At each point of inflection f co's nn:-f].. Therefore, the slopes of the inflectional tangents are either*1 or-1. Furthermore,becausesin(r * 2rr): sin x, the sine function is periodic and has the period 2rr. The absolute maximum value of the sine function is 1, and the absolute minimum value is -1. The graph intersects the r axis at those points where sin r: 0, that is, at the points where x: nzr(n is any integer). From the information obtained above, we draw a sketch of the graph of the sine function; it is shown in Fig. 1.0.2.3' y:sinr Figure 10.2.3 To obtain the graph of the cosine function, wp use the identity cosr: sin(*z * r) Hence,the graph of the cosine which follows from formula (12)of Sec.1.0.1.. function is obtained from the graph of the sine function by translating the y axis*z units to the right (seeFig. 10.2.a). q -lf 2 -- 7r y: Figurc 10.2.4 ,;-$bi' - cos.r 446 TRIGONOMETRIC FUNCTIONS Exercises 10.2 In Exercises 1.through 8, evaluate the limit, if it exists. sin 4r 4 ,. I. lrm 2. lim X 3.-O iin - x !__ 4. lim 1-cosr F ,. J. rlfrr r-o x2 COS I s-nl 2 T.lim = r-o t*'== 1-cos2r ., r. O. uFr 2x2 r-o 1-cos2*r f-A x ,r-0 COS.f In Exercises9 through 24, find the derivative of the given function. 9. f (x) - 3 sin 2r 10.f (x) : cos(3x'+ 1) 1 2 .f ( t ) : c o sP 13.h(t) : 2 cos\tr 15. f (x) : ln sin 5r 16.F(x): sin(lnr) 1 1 .S @ ) : s i n3 x c o s3 r f..,, l + . 8 ( x ): \ f f i r 1 7 .h ( x ) : - ? t nt r 2 + c o st r 18.g(t) : sin(cosf) 20. h(y) : y'22. S(x) : L9. S@) - 2 cos r sin 2r - sin x cos2x yz cosy * 2y sin y + 2 cos y tl 23. h(x) - (cosr)sinr asinc 21,.f ( x ) : (sinxz)a* 24. f ( x ) : (ln sin r)"' In Exercises 25 and 26, use logarithmic differentiation to find dyldx. .rF zc. y: sinrffi 2 a"' - : g V1 -3cosx In Exercises27 artd 28, find D,y by implicit differentiation. *4. V: cos(r- y) 28. cos(r + il : y sin x In Exercises29 through 34, draw a sketch of the graph of the function defined by the given equations. 2 9 .f ( x ) : s i n 2 r 30. 3L. f (x) : lcos3rl rsinr 3 3 .f ( x ) : l lsin(r-tn\ f(x)-zsin3r 32.f (x) - +(1- cosr) if0 <x<tn ifin<x<n . 3 4 .f ( x' ) : l lcosr if-ir<r=0 Lcos(zr-r), if 0<x<rT 35. If a ladder of length 30 ft which is leaning against a wall has its upper end sliding down the wall at the rate of tftlsec, what is the rate of change of the measure of the acute angle maae by ttre ladder wiih the ground when the uppei end is 18 ft above theground? 36. The cross section of a trough has the shape of an inverted isoscelestriangle If the lengths of the equal sides are 15 in., find the size of the vertex angle that will give maximum capacity for the trough. ;p. tf a }oay of weight W lb is dragged along a horizontal floor by means of a force of magnitude F lb and directed at ai angle of 0 radians with the plane of the floor, then F is given by the equation -.: o* ksin0*cos0 where k is a constant and is called the coefficient of friction. Find cos 0 when F is least. 10.3 INTEGRALSINVOLVINGPOWERSOF SINE AND COSINE 447 38. Find the altitude of the right-circular cone of largest possible volume that can be inscribed in a sphere of radius a units. Let 20 be the radian measure of the vertical angle of the cone. A partide moving in a straight line is said to have simple harmonicmotion if the measure of its accelerationis always proportional to the measure of its displacement from a fixed point on the line and its accelerationand displacement are oppositelydirected.Showthatthestraight-linemotionof aparticledescribedbys: Asin2r.ktlBcos2rkt,wheresft is the directed distance of the particle from the origin at f sec, and A, B, and k are constants, is a simple harmonic motion. Show that if a partide is moving along a straight line according to the equation of motion s : a cos(kf * d) (whqre a, k, and 0 are constants) and s ft is the directed distance of the particle from the origin at f sec,then the motion is simple harmonic. (See Exercise39.) 4 L . Given /(r) : x sin(nlx), prove that I has an infinite number of relative extrema. 42. If the domain of the function of Exercise41 is the interval (0, 1], how should/(0) be defined so that / is continuous on the closedinterval [0,1]? 1.0.3INTEGRALS INVOLVING POWERS OF SINE AND COSINE The formulas for the indefinite integral of the sine and cosine functions follow immediately from the corresponding formulas for differentiation. (1) PRooF: Du(-cos u) : sin z. (2) PRooF: EXAMPLE Find Du(sin u) : solurroN: cos y. Let u: ln r. Then du: dxlx; so we have f f cos(lnr) , l--ffrdx:fcosudu JXJ :sinu+C : sin(ln r) + C EXAMPLE 2: Find solurroN: Let u- f sinxdx L - cos r. Then du: f du J1-cosr: J u l " l+ c :ln11 -cosrl +C sin x dx; so 448 TRIGONOMETRIC FUNCTIONS EXAMPIn 3: Find fn | ( 1+ sin r) dx * sin x) dx - x - cos JO :Tr-cos (0 - cos0) :rr+L- 1) :Tr+2 We now consider four casesof integrals involving powers of sine and cosine.The method used in each caseis shown by an illustration. Case1: J sinn u du or J cos, u du, whercn is an odd integer. . rLLUsrRATrow L: ff I cos3r dx: JJ I cos2r (cos x dx) I = | (f - sin2x)(cos x dx) J So .or' r ar: I cos x dx- f ,rr,,x cosx dx Jf J J (3) To evaluatethe second integral on the right side of (3), we let a: sin r; then du: cos r dr. This gives I ff I sin2r (cos r d.x): I u2 ilu JJ =tu3*C1 =$sin3x*C1 Becausethe first integral on the right side of (3) is sin r * Cr, we have f J EXAMPTN 4: / cos3xdx:sinx-*sin3r*C Find sin' x d x (sin2 x)' sin x dx (1 - cos2x)' sin x dx (1-2coszx+ cos4r) sin xdx . 10.3 INTEGRALSINVOLVINGPOWERSOF SINE AND COSINE Therefore, ffrf I sinsr dx: I sinxdx-2 JJJJ 449 | cos2xsinxdx+ | cosar(sinrdr) (4) To evaluate the secondand third integrals in (4), we let a : cos r and du: -sin x dx; and we obtain Isintr dx:-cosx+Z Iu2du- [unau JJJ :-cosx*&u3-!u5IC r * fi cossr - $ cossr * C :-cos Case2: J sinn u du and J cosnu du,wheren is an even integer. The method used in CaseL does not work in this case.We use the following formulas which are obtained from formulas (19) and (18) of Sec. L0.1by substituting 2x for t. . o L-coszx sln- x: 2 s, L + cosz)c c o s -x : 2 O ILLUSTNATION 2: f f r sin'xdx- l1-cos2x rf Jfdx tx - * sin 2x * C EXAMPTN 5: Find SOLUTION: f / f q, \o , : f /1+cos2x\2 , xdx: (cos2 dx x)2dx lcos4 J J (T/ .oro x d x -1 t 4J dx*; 1 r n c ?2vx; vd+xl * i lcos : i x1+ ] s t n1 2 x f 2xdx Jcos2 +1f_t-cos4.x, 4J fax - t x + + s i n2 x * t r + # s i n 4 x * C - 8x + * sin 2x * # sin 4x * C Case3: I sinn r cosmx dx, where at least one of the exponents is odd. The solution of this case is similar to the method used for Case '/-.. . ILLUSTNAUON I 3: r sin3 x cos4 sin2 r cos4r(sin x dx) T R I G O N O M E T R IFCU N C T I O N S : | (1 cos2r) cosax(sin x dx) J ff I cosar sin xdxJJ | cos6r sin xdx :-*cos5x**cos7x*C . Case4: t sinn x cos' x dx, where both m and n are even. The solution of this caseis similar to the method used in Case 2. O ILLUSTRATI f I sin2r J oN 4: cosaxdx:f(ry) 1f (=*)'o* 1f aO J I d* +* O J | cosZxdx - i JL r c o s22x d x -1i Jr c o s2sx d x :* x*+sinzx-+/#0. - 1f (1 - sin22x) cos2x dx i J sin4r : xg - ,r sin2x - x 1,G- 64 L6 - L r cos2x dx+ l f sin22x cos2x dx ; J i J _ x * sin 2x G+T x , sin32x 164g ExAMPtn6: Find I sinn t cosax dx J sin 4r sin 2x__ sin3 2r +6+c sin4r . A Tt64 o soLUTroN: If we make use of the formula sin r cos r : { sin 2x, we have, 1 f I sinnr cosaxdx : i J lsinnzxax : t 6LJ \f /1,- cos 4r\2 , z 1o' 1f :A 7 r dr-i r f cos4xdx+fr cos2 4xdx J J J _tc _sin4r * 1 [ 1*cosStr_ *n 64 7zg 64J 2 10.3INTEGRALS POWERS INVOLVING OF SINEAND COSINE :- x 64 sin4r L28 , sin 4r r28 : - 3x L28 x , sinSr , A '. ' v L28 1024 sin 8r . r. I \/ 1024 The following example illustrates another type of integral involving a product of a sine and a cosine. EXAMPLE7: solurroN: We use formula (22)of Sec.10.1,which is Find r I sin 3x cos 2x dx J s i na c o s b - * s i n ( a - b ) + + s i n ( a + b ) So ff I sin 3r cos2x dx - I t+ sin x +t sin 5rl dx J 10.3 Exercises ir. ,,r. r. ':* ::t',)))!l:J,:: f_ I J In Exercises L throu gh 22, evaluate the indefinite integral. 1. 5. I(3sin / g. / x*2cos x)dx 2. f I (sin 3x * cos2x) dx 4 sinxdx 10. /.oru 7. 1,4. ,ir,' 2t cosa2t dt / L5. ,r Ii##d. 18. L9. 2!. I (sin 3x - sin 2r)2dx / 11. x dx 13. f sint xcoszxdx J / ,i" 4. x eeos'dx / f cos366* *Jrc ,i" r . sin(cos x) dx 3. ,i" 2x cos4x dx / / / 8. ,ir,' x dx .or' tx dx sin' 3f cosz3t dt .o, 4r cos 3x dx 22. [ ,ir, r sin 3r sin 5x dx J In Exercises 23 through 28, evaluate the definite integral. 24. la;it,, _ lr'^,e' sine' dx cossx dx f cosr l2ffidx r cossx dx I L2. I sint x cos' x dx J f 16. {cos z sin3 z dz J T R I G O N O M E T R IFCU N C T I O N S 26. fr | ,ir,- inx dx 27. I sin2 zrx coszrrx dx Jo If z is any positive integer, prove that f" "rn, Jo If n is a positive odd integer, prove that l 28. f rl6 J, sin 2x cos4x dx nx dx : Ln. "o"^* dx:0. In Exercises31 through 33, m and rr are arly integers except zero; show that the given formula is true. ft , [ o i f m #' n 3 1 . I cos nTTxcos mTTxdx - { : :-, "' J_r Lr fl f0 ifm*n 33. I Srnnix srn mTx.dx: 1J-t LI fr 32. I cos nTTx srn mTTxdx - 0 J-r fim-n lIm: n 34. If q coulombs is the chargeof electricity received by a condenserfrom an electric circuit of i amperesat f sec,then 1: Dt1. Supposei:5 sin 50t and 4:0 when t:tr, find the greatestpositive chargeon the condenser. 35. In an electric circuit supPosethat E volts is the electromotive force at f sec and E:2 from f: 0 to f: *zr. sin 3f. Find the averagevalue of E For the electric circuit of Exercise35, find the square root of the averagevalue of E2from f: 0 to t: *r. Find the area of the region bounded by one arch of the sine curve. Find the volume of the solid of revolution generated if the region of Exercise 37 is revolved about the r axis. Find the volume of the solid generatedif the region bounded by the curve y : sin2 r and the r axis from x: 0 to x: r is revolved about the line y: 1. 40. Find the area of the region bounded by the two curves y : sin x andy : cos x between two consecutivepoints of intersection. 10.4 THE TANGENT, COTANGENT, SECANT, AND COSECANT FUNCTIONS 10.4.1 Definition The other trigonometric functions, tangent, cotangent, secant, and cosecant,are defined in terms of the sine ind cosine. The tangenf and secantfunctions are defined by : : : : :i :, , ' ' . . i .f ' +, , S' G :: C,, :#: CSS:ff, for all real numbers r for which cos r # 0. The cotangentartd cosecantfunctions are defined by *"*'**;ffi for all real numbers x fot which sin x * 0. The tangent and secant functions are not defined when cos x:0, which occurls when r isir,tr, orin * nrr,wheren is any positive or nega- 10.4 THE TANGENT,COTANGENT,SECANT,AND COSECANTFUNCTIONS tive integer, or zeto.Therefore, the domain of the tangent and secantfunctions is the set of all real numbers except numbers of the form!-r * nrr, where z is any integer. Similarly, becausecot x and csc r are not defined when sin x: 0, the domain of the cotangent and cosecantfunctions is the set of all real numbers except numbers of the form ntr, where n is any integer. By Theorem 2.6.1.(iv),if the functions / and g are continuous at the number a, then f lg is continuous at a, provided that g(a) + 0. Becausethe sine and cosine functions are continuous at all real numbers, it follows that the tangent, cotangent,secant,and cosecantfunctions are continuous at all numbers in their domain. By using the identity (1) cos2r*sinzr:l and Definition 10.4.1,we obtain two other important identities. If on both sides of (1) we divide by cos2r, when cos x + 0, we get cos2r, sinzt 1 ;oF;"os';:;;t; and becausesin x/cos r: tan r and 1/cos r: secr, we have the identiW (2) By dividing on both sides of (1) by sinz x when sin x + 0, we obtain in a similar way the identity (3) Three other important formulas follow immediately from Definition 10.4.1..They are (4) (5) and (6) Formulas (4), (5), and (6) are valid for all values of x for which the functions are defined, and therefore they are identities. Another formula that we need is one that expressesthe tangent of the difference of two numbers in terms of the tangents of the two numbers. By using formulas (13) and (10)from Sec.1.0.L,we have tan (a - b) : Dividing sin(a - Ut) cos(a - b) sinacosb-cosAsinb c o sa c o s b + s i n a s i n b both the numerator and denominator by cos a cos b, we get T R I G O N O M E T R IFCU N C T I O N S tan(a - b) : sin a cos b cos a cos ? cos a cos b cos a cosb cos a sin b cgs a cgs -b smj sin b cos a cos b sin a sin b cos b cos a ,, , sin a sin b rT 'cos4cosb Therefore, we have the identity tan a-tanb (7) L + tan a tanb Taking a - 0 in formula (7) gives tan(0- b) - tan0-tanb 1+tan0tanb Because tan 0 - 0, we have from the above equation . , , : . : ; ; . . . . , . '.....,:- : : , , . ,,,..,. .,1 ...,, lfffi{*;,b),,*:,'- tan'b (8) Therefore, the tangent is an odd function. Takingb_-b tn formula (7) gives tan a- tan (a - (-b) ) : tan(-b) L+tan atan(-b) and replacing tan(-b) by -tan b, we get the identity (e) The derivatives of the tangent, cotangent, secant, and cosecantfunctions are obtained from those of the sine and cosine functions and differentiation formulas. D "(tan cos r ' D, (sin r) - sin r . Dr(cos r) r) C OS2 T C OS2 T c o s zx + s i n 2 x coszr - L cos,t : sec2r So we have Dr(tan r) : sec2r (10) It u is a differentiable function of r, then from (10) and the chain rule 10,4 THE TANGENT,COTANGENT,SECANT,AND COSECANTFUNCTIONS 455 we obtain (11) ExAMPr.rL: Find f '(x) if f(x)-2tantx-x solurroN: f'(x) -2 secztx(t) - 1 : s e c 2 t x - 1 . If we use identity (2) , this simplifies to ' f (x) - tanztx The formula for the derivative of the cotangent function is obtained in a manner analogous to that for the tangent function. The'result is D"(cot x) : -csc2 x (r2) The derivation of (1,2)is left as an exercise (see Exercise L). From (12) and the chain rule, 7f u is a differentiable function of x, we have (13) D x) : D "(sec -t] r) "l(cos -'(sin r) L ( cosx) - - - - ' SI r n f cos- r _ 1 . sin.r cos I cos I :secxtanx We have, then, the following formula: (14) x) : secx tan x "(sec lf u is a differentiable function of x, then from (14) and the chain rule D (ls) EXAMPLE 2: f(x): Find f s e c a3 r '(x) if SOLUTION: ' ' f (x) 4 sec33x D"(sec 3r) - 4 sec83r(sec 3r tan 3r) (3) : 12 seca3x tan 3x 456 T R I G O N O M E T R IFCU N C T I O N S In a manner similar to that for the secant,the formula for the derivative of the cosecantfunction may be derived, and we obtain (15) D"(csc r) : -csc x cot x So by applying (16) and the chain rule, if u is a differentiable function of r, we have D"(csc a) + *-csc u cot u D*u (r7) The derivation of (16)is left as an exercise(seeExercise2). EXAMPLE Find f '(x) if f (x) cot r csc r SOLUTION: f'(x) .. : :": r,! ;:: ; ;:; ",: =-csc x cotz x ' D"(cot r) x(-csc2 x) cscsr In Definition 10.1.4 we gave the analogy between the sine and cosine of real numbers and those of angle measurements. The following definition gives a similar analogy for the other four trigonometric functions. 10.4.2 Definition If a degrees and r radians are measurements for the same angle, then tan a": tan r EXAMpLE4: An airplane is flying west at 500 ftlsec at an altitude of 4000 ft. The airplane is in a vertical plane with a searchlight on the ground. If the light is to be kept on the plane, how fast is the searchlight revolving when the airplane is due east of the searchlight at an airline distance of 2000 ft? cot a": cot r secao: sec.x csc ao: cscr sor,urroN: Refer to Fig. 10.4.1.The searchlightis at point L, and at a particular instant of time the airplane is at point P. Let f : the number of secondsin the time; r: the number of feet due east in the airline distance of the airplane from the searchlight at time f sec; 0 : the number of radians in the angle of elevation of the airplane at the searchlight at time f sec. given We are Dg: -500. We wish to find D10when x: 2000. tan o - 4ooo x Differentiating on both sides of Eq. (18) with respect to t, we obtain s e c02D , e - - r y D t x Substituting D'rx: -500 in the above and dividing by secz0 D10: (18) 2,000,000 x2 secz 0 10.4 THE TANGENT,COTANGENT, SECANT,AND COSECANTFUNCTIONS Whenx:2000, tan 0:2. Therefore, secz0: L * tan20:5. Substituting thesevaluesinto Eq. (19),we have,when x:2000, Irta: 2,000,000 : 7 4roopoo(s) 10 We concludethat at the given instantthe measurement of the angleis increasing at the rate of rb rad/sec, and this is how fast the searchlight is revolving. L xft Figure 10.4.1 We now consider the graph of the tangent function. Becausefrom Eq. (8) tan(-r) : -tan r, the graph is symmetric with respect to the origin. Furthermore, L__r-. , _.,_ sin(r* t€tnlxfTr):- z) cos(r * zr) sin r cos rr + cos x sin n cos x cos zr - sin r sin zr _ sin r(-1) * cos r(0) cos x(-1) - sin r(0) -sin r _ -cos .r :tanr and so by Definition 10.1.3,the tangent function is periodic with period zr. The tangent function is continuous at all numbers in its domain which is the set of all real numbers except those of the form tn * nzr, where n is any integer. Ffowever, tan r: -roo,and so the lines ,_EEr,, having equations x:Lr-lnn are vertical asymptotes of the graph. If n is any integer, sin nn: 0 and cosnT is either *1 or -L, and so tan nn: 0. Therefore, the graph intersects the r axis at the points (nr, 0). To find the relative extrema of the tangent function and the points of inflection of its graph, we consider the first and second derivatives. So if f(x) : tan x, then f'(x) : sec2r and f" (x) :2 seczr tan x. gives seczr:0. Becauseseczx- 1 for all x, we Setting f'(x):0 concludethat there are no relative extrema.Setting f" (x):0, we obtain 4s8 T R I G O N O M E T R IFCU N C T I O N S 2seczxtan r:0, from which we get tanr:0 becauseseczx # 0. Therefore, f" (r):0 if. x:nn, wheren is any integer. At these values of x, (x) changessign, and so the points of inflection are the points (nn,0), f" which are the points where the graph intersects the r axis. Because '(no) : sec2nnr:'J., the slopesof the inflectional tangentsare L. f Consider now the open interval (-tn, tn) on which the tangent function is defined everywhere. Because/' (r) : sec, r ) 0 for all values of.x, it follows that the tangent is an increasing function on this interval. When -ir < x( 0, x tanr ( 0; hence,the graph is concave f"(*):2secz downward on the open interval (-tn,O). When 0 < x < +7, f" (x) > 0, from which it follows that the graph is concave upward on the open interval (0, *z). T a b l e1 0 . 4 . 1 x tan r 0 *\R - 0.58 1, tB - l.Tg In Table 10.4.1there are some correspondingvalues of r and y satisfying the equation y : tan r. By plotting the points having as coordinates the number pairs (r, y) and using the above information, a sketch of the graph of the tangent function may be drawn, and it is shown in Fig. 1,0.4.2. '1, 2 y: tanr Figure 10.4.2 The graph of the cotangent function can be obtained from the graph of the tangent function by using an identity which will now be proved. sin(tz + r) tan(Ln* r) : cos(*z * r) 10.4 THE TANGENT,COTANGENT,SECANT,AND COSECANTFUNCTIONS sin *zr cos x * cos *n stn x cos*z cosr - sin tzr sin r -_ (1) cosr * (0) sin r (0) cosx- (1) sin r -# : -cot r Therefore, cot x: -tan(*zr * r) From the above identity it follows that the graph of the cotangent function is obtained from the graph of the tangent function by translating the y axis *z units to the right and then taking a reflection of the graph with respect to the r axis. A sketch of the graph of the cotangent function is in Fig. 1,0.4.3. y: cotr Figure 10.4.3 The secant function is periodic with period 2zr because sec(r*2zrl: coslx.+ z7r) L :secr cos .r The domain of the secant function is the set of all real numbers except those of the form*rr * nn, and the function is continuous on its domain. Because lim secr:1o, the graph of the secantfunction has the lines I -fr12+nfr x: tn * nn as vertical asymptotes. There is no intersection of the graph with the r axis because sec r is never zero. The secant function is even because 11 : -----:-: secI sec(-x) cos(-r, cos r 460 TRIGONOMETRIC FUNCTIONS and so the graph of the secantfunction is symmetric with respect to the y axis. lf f(x): sec r/ it follows that f' (r): sec x tan r and f,,(x): sec r(2 tan2 x * 1.).Setting f '(x):0 gives sec x tan x:0. Because secx * 0,f '(t) : 0 when tan r: 0, which is when x: nzr,wheren is any integer. Also, f"(ntr1: secnrr[2tanf nzr+1]. If n is an even integer, a n d i f . n i s a n o d d i n t e g e r ,f " ( n t ) : f"(nr):1'(0*1):1' (-1)(0+ 1):-1. Therefore,when x:nn and n is an even integer,/ has a relative minimum value; and when x: nnTand z is an odd integer, / has a relative maximum value. There are no points of inflection because for all x, f " (x) + 0. Using the above information and plotting a few points, we get a sketch of the graph of the secantfunction shown in Fig. 10.4.4. Because sec(r * tn) :-:-:-uDL,r 1, cos(r * in) 1, -sin r vv then cscx: -sec(r * $r) and so the graph of the cosecant function is obtained from that of the secantfunction by translating the y axis$r units to the right and taking a reflection of the graph with respect to the r axis. A sketch of the graph of the cosecantfunction is in Fig. 10.4.5. y: secr Figure 10.4.4 v F i g u r e1 0 . 4 . 5 Exercises 1-0.4 1. Derivei Dr(cot x) : -csC r. 2. Derive: Dr(csc x) : -csc r cot x. TO THESLOPEOF A LINE OF THE TANGENTFUNCTION 10.5 AN APPLICATION 461 In Exercises 3 through 20, find the derivative of the given function. 3. f(x): 4,a s(r) : ln .rd?* 7. G(r): \ffi sec12 6 . h ( x ) : l n l c o tt r l 9. sQ)- 5. F(r) : ln lsec2rl 8. h(t): sec22t - tanz2t 10. f (x) : sin x tan x secx tan x 13. H(t): 1 2 . 8 ( f ) : 2 s e c\ F ln lsec5r * tan 5rl 18. G(r) logrslcscx - cot rl 15. f(x): cotat - cscat L5. f (x) : + seca2x - sec2x 'Ig. F(r) - (sin r)tanc 1 1 .f ( t ) : c s c ( f + a 1) 2: 14.F(r) : ":': l*xz L 7 .S ( x ) : 3 ' s e cx 20. G(r) - (tan r)" In Exercises2l and 22, use logarithmic differentiation to find D ,y . r sec rVl=6f zly::ft- 22.'y:ffitarf x In Exercises23 through 26, hnd D"y by implicit differentiation. : S, , tan(x-rA) '@ t".' x* cs9y:4 24.cotxy t xy:O 26-csc(x-0 + sec(r* !) : a In Exercises27 through 30, draw a sketch of the graph of the given function. 2 8 .f ( x ) - + s e c 2 x 30.f (x) - Jcsc $rl 27. f (x) : tan 2x 2e.f (x) -- + lcotrJ fi"a an equation of the tangent line to the curve y: sec r at the point (In, t/-Z). Q 32. A radar antenna is located on a ship that is 4 miles from a straight shore and it is rotating at 32 rpm. How fast is the radar beam moving along the shoreline when the beam makes an angle of 45" with the shore? (g3l, A steel girder 27 frlongis moved horizontally along a passageway8 ft wide and into a corridor at right anglesto the pas-' sageway.How wide must the corridor be in order for the girder to go around the comer? Neglect the horizontal width of the girder. 34. If two corridors at right angles to each other are 10 ft and 15 ft wide, respectively, what is the length of the longest steel girder that can be moved horizontally around the comer? Neglect the horizontal width of the girder. 10.5 AN APPLICATION OF THE TANGENT FUNCTION TO THE SLOPE OF A LINE 10.5.1Definition The tangent function can be used in connection with the slope of a straight line. We first define the "angle of inclination" of a line. Tthe angle of inclination oI a line not parallel to the r axis is the smallest angle measured counterdockwise from the positive direction of the r axis to the line. The inclination of a line parallel to the r axis is defined to have measure zero. If a is the degree measure of the angle of inclination of a line, a may be any number in the interval 0 = c ( 180. Figure 10.5.1shows a line L for which 0 ( a ( 90, and Fig. 10.5.2shows one forwhich90 < a < 180. 462 TRIGONOMETRIC FUNCTIONS F i g u r e1 0 . 5 . 1 10.5.2 Theorem F i g u r e1 0 . 5 . 2 If a is the degree measure of the angle of inclination of line L, not parallel to the y axis, then the slope m of L is given by m: tatt q.o PRooF: Refer to Figs. 10.5.1 and'1.0.5.2,which show the given line L whose angle of inclination has degree measure a and whose slope is rn. The line L' that passesthrough the origin and is parallel to L also has slope m and an angle of inclination whose degree measure is c. The point P(cos ao, sin ao), at the intersection of L and the unit circle I * yr:1, lies on L'. And becausethe point (0, 0) also lies on L', it follows from Definition 1.5.1that sin co - 0 *:6;A14:;0;;. sin co : tan 6r- I If the line L is parallel to the y axis, the degree measure of the angle of inclination of L is 90 and tan 90odoes not exist. This is consistent with the fact that a vertical line has no slope. Theorem 10.5.2is used to obtain a formula for finding the angle between two nonvertical intersecting lines. If two lines intersect, two supplementary angles are formed at their point of intersection. To distinguish these two angles, let L2 be the line with the greater angle of inclination of degree measure a, and let L1 be the other line for which the degree measure of its angle of inclination is or. If 0 is the degree measure of the angle between the two lines, then we define 0: e.2- a1 (1) If L, and L, are parallel, then c1 : c, and the angle between the two lines 10.5 AN APPLICATIONOF THE TANGENTFUNCTIONTO THE SLOPE OF A LINE 463 has degreemeasure0. Thus, if Lt and L, are two distinct lines,0 < 0 < 180. Refer to Figs. 10.5.3and 10.5.4. v F i g u r e1 0 . 5 . 3 10.5.4 Figure ./ The following theorem enables us to find 0 when the slopes of L1and Lo ate known. 10.5.3 Theorem Let L, and L, be two nonvertical lines, which intersect and are not perpendicular, and let L, be the line having the greater angle of inclination. Then if. m1is the slope of Lr, m, is the slope of.Lz, and 0 is the degree measure of the angle between Lt and Lr, (2) pRooF: If a1 and d.2are degree measuresof the angles of inclination of Lt and L", respectively, from Eq. (1) we have 0: a2- e.1 and so tan 0o: tan(azo- oro) Applying formula (7) of Sec. 10.4 to the right side of the above, we get - L-- Ao_ tan a2o tan d-:1H"fr;5;7tan c1o I T R I G O N O M E T R IFCU N C T I O N S EXAMPLEL: Find to the nearest degree the measurements of the interior angles of the triangle having vertices B(-2, l) , C(2, 2) , D (-3, 4). solurroN: Let B, y, and 6 be the degreemeasuresof the interior anglesof the triangle at the vertic es B, c, and D , respectively. Let u denote ttre line through B and C,let a denote the line through C and D,let w denote the line through B and D, and let m, m, and mrbe their respectiveslopes. Refer to Fig. 10.5.5. Using the formula for the slope of a line through fwo given points, we get ttlu: i ffio:-? ffi.:-3 To determine B, we observe that line ar has a greater angle of inclination than line u; so in formula (2), tnz: fitw:-3 and trh- ntu: {. The degree measure of the angle between lines u and. a is B. Thus, from formula (2) we have -e-_L tanBo:1ffi:-13 F i g u r e1 0 . 5 . 5 Therefore, F:94 To determine 7, becauseline a has a greater angle of inclination than line u, ffiz: ttr: { and rnr: nru: }. The degree measure of the angle between lines u and a is, by definition, 180- 7. Applying formula (2), then, we get - yo) :#+I tan(180" : -+g Hence, 180- Y:144 so that y:35 To determine 6, becauseline o has a greater angle of inclination than line w, friz: frtr: { and rltr: tnw: -3. The degreemeasureof the angle between lines o and w is then 6. Applying formula (2), we obtain 2 - (-a\ tan 6o: =-'-! ,\ ,el:: 1 + (-f )(-3) 4{ rr Therefore, 6: 50 The sum of the degree measuresof the angles of a triangle must be r.80.we use this to check the results, and we have F+y+6:94+35+50:180 OF THE TANGENTFUNCTIONTO THE SLOPEOF A LINE 10.5 AN APPLICATION 10.5.4 Definition ExAMPLEZ: Find to the nearest degree the measurement of the angle between the cun/es y : f - 2 and y : -2x2 + 10 at their points of intersection. v The anglebetweentwo curoesat a point of intersection is the angle between the tangent lines to the curves at the point. so urroN: Solving the two equations simultaneously,we obtain (2,2) and (-2,2) as the points of intersection.In Fig. 10.5.5sketchesof the two curves are shown together with the tangent lines at one point of intersection. Becauseof the symmetry of the two curves with respect to the y axis, the angle between the curvet at (2,2) has the szunemeasure as the angle between them at (-2 , 2) . Let P be the point (2, 2) , Tt be the tangent line --2x2 * L0 atP. Theline toy: -hasx2 2atP, andTzbe the tangent line to a a greater angle of inclination than Tt. so if 0 is the degreemeasureof z; the angle between T1 arrd T2, - lllz . ^6 tan fl-:TTffi tflr : where mristhe slope of T, and rz, is the slope of Tr. From the equation y :-2x2 L0' + x'- 2,wl get D,A-:2x, and so rttl:4. From the equationy we get DrY:-4x; tlrrtts,ffiz:-8. Hence, P(2,2) tan 0o: -R -j-=:*:0.387 (4X-6, Jr l-----:-+ Therefore,to the nearestdegree,0":21". F i g u r e1 0 . 5 . 6 1-0.5 Exercises (a) I and *; (b) 5 and -*; (c) - 3 and 1. Find tan 0. if 0" isthe measurementof the angle between the lines whose slopesare -rb; (e) -S and -8' *; (d) - * and (a) 5 and -6i (b) -3 and 2. Find, to the nearest degree, the measurementof the angle between the lines whose slopes are -*. -* and 2; (c) t and *; (d) lines that have equa3. Find, to the nearest degree, the measurementsof the interior angles of the triangle formed by the : 8 0. 4y * tions 2r * y 6: 0, 3x y 4: 0, and 3x * (1, 0) , (-3, 2) ' and 4. Find, to the nearestdegree, the measurementsof the interior anglesof the triangle having vertices at (2,3). line having equa5. Find an equation of a line through the point (-1,4) making an angle of radian measure*z with the tion 2r * y 5: 0 (two solutions). -2) making an angle of radian measure tzr with the line having equa6. Find an equation of a line through the poin t (-3 , tion3r-2Y-7:0. T R I G O N O M E T R IFCU N C T I O N S 7' Using Theorem 10.5.3,prove that the hiangle having vertices at (2, 3), (6, 2), and (3, -1) is isosceles. 8' Using Theorem 10.5.3,prove that the triangle having verticesat (-3, 0), (-1, 0), and (-2, \/g)is equilateral. 9' Find the slope of the bisector of the angle at A in the triangle having the verticesA(4, 1); 8(6, S);and C(-1, g). 10' Find the slope of the bisector of the angle at A in the triangle having the vertices A(-3,5); B(2, -4); and C(-1,7). 17' Let A' B' and c be the.vertices of an oblique triangle (no right angle), and let a, B, and.7be the radian measures of the interior angles at vertices A, B, and c, respectiveiy. prove-that ti'n a * tan B * tan 7 : tan a tan B tan y. (Hrwr: First show that tan(a * F) : -tan y.) 12' Find the tangents of the measurementsof the interior angles of the triangle having vertices at (-3, -2), Fe ,3), and (5,1) and checkby applyrng the result of Exercise11. 13' Find the points of intersection of the graphs of the sine and cosine functions and find the measurement of the angle between their tangent lines at their poiints of intersection. 14' Find, to the nearest10 minutes, the measurementof the angle between the curvesy:2f11 and !: * -2 at their points of intersecfion. 10.6 INTEGRALS INVOLVING A formula for the indefinite integral of the tangent function is derived as THE TANGENT, follows. Because COTANGENT, SECANT, t'n u AND COSECANT du f ,ur,u du: I J we let a: J cosu cos u, da: -sin u du, and obtain [ ,ur,u du: - JtU+: -ln lrl+ c J Hence, f u d u - - l n l c o sz l + C Jtan B e c a u s-el n l c o su l : l n l ( c o su ) - t l : l n l s e cu l , w e can also write , '' p : " " "' ' ' ' t,,. , ,"j",' ,x4s, 14,t,il''trn |$ecal + c ';j' ' : ,t,1... ri ,': (1) ..r' :; =-- (2) O ILLUSTRATION "1.: f I tan 3x dx: "tf tan3x (s dx) i J :glnlsecSrl+C o The following formula is derived in a way similar to the derivation of (2). (See Exercise 29.) / .ot u du- tn lsinzl + C (3) ANDCOSECANT 467 SECANT, THETANGENT, COTANGENT, INVOLVING 10.6INTEGRALS To integrate ! sec u ilu, we multiply the numerator and denominator of the integrand by (sec u * tan u), and we have I r"" u au:= It"t l,(i!f !l,.tT') J J@tanu Letu: itu: [ (secz * secrl tanil) ' secuTGT- au J secu * tant. Then 471: (secutarru *seCu) du; andsowehave ["""uAu:[@:;o-lol +c J JO Therefore, (4) The formula for J csc u ilu is derived by multiplying the numerator and denominator of the integran{ by (csc u - cot z). and proceeding as above. This derivation is left as an exercise(seeExercise30).The formula is (s) r t u _ I f ^csc ^ 2x (zdx'l I d * == csc2x dx:; J sin2x J J - f l n l c s c2 x - c o t 2 x l + C The following two indefinite integral formulas follow immediately from the corresponding differentiation formulas. (5) (7) Many integrals involving powers of tangent and secant can be evaluated by applying these two forrnulas and the identity given by formula (2) in Sec.10.4,which is (8) 1 * tanz u: se* u There are similar formulas involving the cotangent and cosecant. (e) TRIGONOMETRIC FUNCTIONS 468 PRooF: Du(-cot u) : - (- csczu): csc2u. (10) pRooF: Du(-csc n) : - (-csc u cot u) : csc u cot u. . rLLUsrRArrou3: ' . + d u * 3f + [ 2 t 9Sms ous u d u : 2 1 Sln'4 sm J J :2 J tl . " ? " ud , sln rl ff | c s c 2u d u * 3 | c s cu c o t u d u JJ :-2cotz-3cscu*C . From formula (3) in Sec. 10.4we have the identity 1 * cof z: (l1.\ csC u Formulas (6) , (7) , and (8) are used to evaluate integrals of the form f I tan^ u secnu ilu J and formulas (9), (10), and (11) are used to evaluateintegrals of the form I I cot' u cscnu ilu J where m andn are positive integers. We distinguish various cases. Case1: t tan" u du or ! cot" u du, wheren is a positive integer. We write tattn u: tanz-2 u tanz u : fala-z z(seC u- 1) and cot' ll : ExAMPrn 1: Find cotn-2 u cotz u cotn-z u(cscz u - l) SOLUTION: f I tant x dx J f f I tant r d x J | t a n r ( s e c 2x J -+tanz r*ln lcos tan x sec2 tan x dx 10.6 INTEGRALSINVOLVINGTHE TANGENT,COTANGENT,SECANT,AND COSECANT ExAMPrn 2: Find SOLUTION: r f f I cota3x dx - I cof 3r(csc23x l) dx I cota3x dx J J J I cot' 3r csC 3x dx | co1 lx dx JJ + (-cot3 3x)- (csc23x - 1) dx I -+cot33x*.*cot3x*x*C Case2: J sec" u du or J csc" u du, where n is a positive even integer. We write secnU: gggn-2u seczy: cscnIt: sgsn-2u csc2u : (tanz u * l)@-z)tz seczu and EXAMPT,N 3: f I Find """2 u SOLUTION: ff csc6x d x (cof u * l)@-z)tz I cscux dx: J -f | (cotz x * L)2 csc2x dx t, r c o t a xc s C x d x + 2 | cof x csc2 x JJ lcot5 x- S c o t 3x - csC x dx c o t ) c +c To integrate J secnu du or ! csc"u dz when n is a positive odd integer, we must use integration by parts, which is discussedin Sec.11.2. Case3: t tar.^ u secnu du or J cot' u cscnu du, wheren is a positive even integer. This caseis illustrated by the following example. EXAMPI.,N 4: / Find tur,' r sec4x dx SOLUTION: / tur,t.r sec4x dx: I t a n s x( t a n z x + 1 ) s e c 2x d x :T t a n T r s e c x2 d x * f t u r , t r s e c 2x d x -_ B1 J t a n sx + * t a n 6 x + C Case4: t tan* u secnu du or cotmu cscnu du, where m Ls a positive odd "[ integer. The following example illustrates this case. 470 T R I G O N O M E T R IFCU N C T I O N S EXAMprr 5: f I Find SOLUTION: tan5r secTx dx I tun'ffseczx dx: t a n ax s e c 6 r s e c x t a n x d x I :I ( s e c 2x - L)2sec6r (secxtanxdx) - - [ seclo.r ( secr tan x dx) fr 2 | s e c s r ( s e cx t a n x d x ) + | s e c 6 r ( s e cx t a n x d x ) JJ _ T r1 s e c l l x - 3 s e c e x * * s e c z x * C Integration by parts (sec. 11.2)is used to integrate J tan- usecou ilu and I cot^ u cscnu du whenz is a positive even integer and n is a positive odd integer. Exercises10.6 In ExercisesL throu gh 22, evaluate r 1. I tan 2x dx {*, J the indefinite integral. f 2. I cot(3r * t) dx J 3. I - cscSxzdx 4. f\ 5, I cotsxdx J ft^,[tan6rseca xdx 13. ry")/ f I ,/ 10. / t"n x dx ,ur,t r sec' x dx (tan 3x * cot 3r) z dx cot'2t dt 7. / .r.' 6xdx 11. cot'3r csca3x dx 15. ,un^tx dx I [ f secr tan r tan(secx) dx I r 8. 12sec2xBdx I r 12. (sec5r * csc5r) 2dx J L6. I t"r,t 3xdx J fdu 20.[ ""i, ,* J cotz x J@ In Exercises 23 through 28, ev,aluatethe definite integral. f rl4 25. I J -nt+ frls 28. I Jo sec6xdx -lun} dx * , SeC .f 30. Derive the formula J csc u du - ln lcscu,Z?.Derive the formula J cot u du: ln lsin ul + C. Find the length of the arc of the curve y : ln c,scr from x - *n' to x: 4, in. 32. Find the area of the region bounded by the curye y : tan2x, the r axis, and the lin e x : ir. cotul + c. T R I G O N O M E T R IFCU N C T I O N S 471 1 0 . 7I N V E R S E 33. Find the volume of the solid of revolution generated if the rigion bounded by the ctrrvey: seC r, the axes,and the line x: Ir is revolved about the x axis. ' if n*o' 34. Prove: cot r cscn, d x : - c s c " x * r , n I'r - " 6t-tann-r-x - [ ann-'xdx - - J I tannr 35. Prove: n-7 J if.nisapositiveintegergreaterthanl. 36. Derive a formula similar to that in Exercise34 fot t tan I secnx dx iI n * 0. 32. Derive a formula similar to that in Exercise35 for J col" xdx,if nis a positive integer Sreaterthan 1. and #tW"canintegrate I seCxtanxilx intwoways asfollowi: ! se9 xtanxilx: J tanr(seCx dx):ttan2x*C ll7\ .u:Iwvers. the two of :tseC appearance the in the difference Explain r * C. sec2x tani dx: ! sec r(sec 2ctan x dx) -ln 39. Prove that J csc u du: lcsc z * cot zl * C' IO.7 INVERSE Consider the equation r: sin y. A sketchof the graph of this equation is the figure, the vertical line r : * is shown; this line interTRIGONOMETRIC in Fig. 1.0.7.1,.In and so on' FUNCTIONS sectsthe graph at the points for which y : +7,8r,,-&zr,-#n, of y as a function not define y does we see,then, that the equation x: sin at in function a graph of the r because every vertical line must intersect have not does function the sine most one point. Hence, we conclude that an inverse function. Refer now to the graph of the sine function (Fig. 10.2.3).We seefrom the figure that the sine is an increasing function on the interval l-Lo,En). This follows from Theorem 5'1.3 because if /(r) : sin r, then /'(r) : cos r > 0 for atl x in (-in, *zr). Thus, even though the sine function does not have an inverse function, it follows from Theorem 9.3'2that the function F for which F(r) : sin t and -tn = x -Lrr (1) does have an inverse function. The domain of F is l-in,Enl and its range The inverseof is [-1, I ]. A sketchof the graph of F is shown in Fig. 1'0;7.2. and is function" sine the "inverse is called (1) by the function defined definition. formal is the Following sin-l. the symbol denoted by -z") - TI\ 6r) Figure F i g u r e1 0 . 7 . 2 472 TRIGONOMETRIC FUNCTIONS 10.7.1 Definition The denoted by sin-l, is defined as follows: if and only if The domain of sin-l is the closedinterval [- "!.,'].], and the range is the closedinterval l-+n, +rrl. A sketch of the graph is shown in Fig. 1,0.7.3. It follows from Detinition I0.7.Lthat sin(sin-t x) - x for r in [-1 , Lf sin-r(sin y) : y for y rn l-in, and (2) I tnf (3) In (1) the domain of F was restrictedto the closedinterval l-h,!v1 so the function would be monotonic on its domain and therefore have an inverse function. However, the sine function has period 2n andis increasing on other intervals, for instance, l-Err, -Bz'] and liln, Enl. Also, the function is decreasingon certain closedintervals, in particular the intervals l-Er, -$z] and ltn, Errf. Any one of these intervals could just as well have been chosen for the domain of the inverse sine function. The choice of the interval l-l:r, ]zr], however, is customary becauseit is the largest interval containing the number 0 on which the function is monotonic. The cosine function does not have an inverse function for the same reason that the sine function doesn't. To define the inverse cosine function, we restrict the cosine to an interval on which the function is monotonic. we choosethe interval 10, rrl on which the cosine is decreasing.so consider the function G defined bv F i g u r e1 0 . 7 . 3 G(r)-cos.r and 0 <x< jr (4) The range of this function G is the closed interval l-1,11, and its domain is the closed interval [0, ,f . A sketch of the graph of G is shown in Fig. 1'0.7.4. Because this function is continuous and decreasing on its domain, it has an inverse function which is called the ,,inverse cosine function" and is denoted by cos-l. 1.0.7.2Definition The The domain of cos-l is the closed interval l-1,1], and the range is the closed interval 10, lrl. A sketch of its graph is shown in Fig. 7O.Z.S. From Definition 1.0.7.2it follows that cos(cos-l x) : x for r in [-1 , lf (s) cos-l(cos y) : y for y rn 10,nl (6) and There is an interesting equation relating sin-r and cos-l that is giv in the following example. 10.7 FUNCTIONS TRIGONOMETRIC Y Figure10.7.4 : 479 c o s- 1 r Figure 10.7.5 Let x be in t-l , Il and Tt): tn - sin-l ExAMpr,nL:' Prove cos-1 x: irr _ sin-l r sorurroN: for lrl < 1,. and so . sin(sin-1x) : sin (En - w) Then s i n _ 1x _ L n _ w and using (2) , we get x-sin(*n-w) Applying (11)in Sec.10.1to the right side of the above equatio(I, we get (7) x-cosa) Because-En = sin-1 r = in, by adding -tn to eachmember we have -rT =-in * sin-l r < 0 Multiplying each mernber of the above inequality by-l direction of the inequality signs, we have 0 =tn and reversing the - sin-l r < Tr Replacing tn - sin-l x by rD, we obtain 0=w<7r From (7), (8), and Definition 10.7.2 it follows that u) - cos-l r for lrl < 1. and replacing w by *n - sin-l x,, we get tn-sin-1 x- c o s - 1r forltl <L which is what we wished to prove. (e) 474 T R I G O N O M E T R IFCU N C T I O N S Note in the above solution that the relation given by (9) depends on choosing the range of the inverse cosine function to be [0, zr]. To obtain the inverse tangent function we first restrict the tangent function to the open interval (-izr, tn) on which the function is continuous and increasing. We let H be the function defined by H(x)-tanff and -in<x<*n (10) The domain of His the open interval (-Lzr,ir), and the range is the set of all real numbers. A sketch of its graph is shown in Fig. 1.0.7.6.B,ecause the function H is continuous and increasing on its domain, it has an inverse function, which is called the "inverse tangent function" and is denoted bv tan-r. F i g u r e1 0 . 7 . 6 10.7.3 Definition Figure 10.7 .7 The inuersetangentfunction, denoted by tan-l, is defined as follows: Y : tan-1 x i f and only if )c- tan y and -tn < y < *n The domain of tan-l is the set of all real numbers, and the rangeis the open interval (-hr, ir). A sketch of its graph is shown in Fig. 10.7.7. Before defining the inverse cotangent function, we refer back to Example 1, in which we proved the relationship between cos-1 and sin-l given by Eq. (9). This equation can be used to define the inverse cosine function, and then it can be proved that the range of cos-l is [0, zr].We use this kind of procedure in discussing the inverse cotangent function. 10.7.4 Definition The inaersecotangentfunction, denoted by cot-l, is defined by cot-r x: tn - tan-l r where x is any real number (11) It follows from the definition that the domain of cot-l is the set of all FUNCTIONS 10.7 INVERSETRIGONOMETRIC 475 real numbers. To obtain the range we write Eq. (11) as tan-l x-tn- cot-l x Because -tn<tan-rx< in by substituting from Eq. (12) into this inequality, we get -in<tzr-cot-l xlin Subtracting tr from each member, we get -7r<-cot-l r(0 and now multiplying each member by -1 and reversing the direction of the inequality signs, we obtain Y : cot 0<cot-rxlr -1r Therefore, the range of the inverse cotangent function is the open interval (0, zr). A sketch of its graph is in Fig. 1'0.7.8' The inverse secant and the inverse cosecantfunctions are defined in terms of cos-l and sin-r, respectively. Figure 10.7.8 denoted by sec-l, is defined by 10.7.5Definition The L0.7.6Definition The inaersecosecantfunction, denoted by csc-1,is defined by it follows that the domain of sec-l is (- n,-!f U From Definition "!.0.7,5 Lzn) U (tn, nf . Note that the numb er +rr is [1, +*), and the range is 10, not in the range because sec trr is not defined. A sketch of the graph of sec-r is shown in Fig. 10.7.9. Y : sec-l r F i g u r e1 0 . 7 . 9 476 FUNCTIONS TRIGONOMETRIC It follows from Defini tlon 1.0.7.6 that the domain of csc-l is (-o, - 11 g 11,+*), and the range is L-*r,0) U (0, *z]. The number 0 is not in the range becausecsc 0 is not defined. A sketch of the graph of csc-r is shown in Fig. 1,0.7.10. Y : ExAMPtn 2: Find the exact value of sec[sin-l (-t) ]. csc-l r F i g u r e1 0 . 7 . 1 0 solurroN: Let t : sin 1(-*). Then sin t: -1. Becausethe range of sin-l is l-in, !nl, and becausesin f is negative, it follows that-tzr < f < 0. Figure 10.7.1,1, shows an angle having radian measuref that satisfiesthese requirements. From the figure we see that sec t:#17, and so we conclude that seclsin-l(-il1: +\/7. Fig ure EXAMPT,T 3: Find the exact value of sin[2 cos- 1 ( - g ) l . solurroN: Let f : cos-l(-$). so we wish to find the exact value of sin 2f. From formula (14) of Sec. 10.1, we have sin 2t:2 sin f cos t Becausef _ cos-l(-g), it follows that cos t - -B andtn < f < identity, sin2 t+ cos2t:1., and because sin t > 0, since f is we obtain sint-1 m-ffi:+ Therefore, from formula (13) we get sin 2t : z(il (- g) from which we conclude that sin[2 cos-l(-g) ] (13) From the (*r, n) , 10.8DERIVATIVES OF THE INVERSE TRIGONOMETRIC FUNCTIONS 477 ExC T C I S C S1 0 . 7 /l).Find the value of each of the following: (p) V sin-r(-t); (h) cos-r(-l); (f) csc-1(-1) e) tan-'(-l); (d) cot-1(-1); (e) sec-1(-1); (t (S);cos-l(!); (.d') cos-l(-*); (; Find the value of each of the following: (a) sin-'(i); (b) sin-l(-*); (e) sec-1(2); (f) csc-l(-2) 3. Given: 3t: sin-t *. Find the exact value of each of the following: (a) cos y; (b) tan y; (c) cot y; (d) sec y; (e\ cscy. 4. Given: y : cos-t(-I). Find the exact value of each of the following: (a) sin y; (b) tan y; (c) cot y; (d) secy; (e) csc y. 5. Given: y : tan-r(-2). Find the exact value of each of the following: (a) sin y; (b) cos y; (c) cot y; (d) sec y; (e) cscy. 6. Given: y : cot-'(-t). Find the exactvalue of each of the following: (a) sin y; (b) cos y; (c) tan y; (d) secy; (e) csc y. 7. Find the exactvalue of each of the following: (a) tan[sin-'(*16;1, (b) sin[tan-l(] \6)1. 8. Find the exactvalue of each of the following: (a) cos[tan-t(-3)]; (b) tan[sec-1(-3)]. 9. Find the exact value of each of the following: (a) sin-lltan 1z]; (b) tan-1[sin(-fu)l; (d) cot-l[csc(-*n)1. (b) sin-r[cos(-*zr)]; 10. Find the exact value of each of the following: (a) cos-1[sin(-fz)]; (d) sin-1[cos3zr]. (c) sec-r[tan(-*n)]; (c) cos-l[sin 3zr]; In Exercises11 through 18, find the exact value of the given quantity' 12' tartl2 sec-l(-t)l 11. cos[2 sin-'(-t)] 13. sinfsin-l 3* cos-l i] '' L5. tan(tan-l *- sin-' 14' cos[sin-1(-+)+ sin-lrl 17. cos(sin-l *- tan-'*) 18. sin[cos-1(-3) + 2 sin-l(-])l 20. Prove:2 tan-1 f - tan-r(-f):in. 15. tanfsec-l** csc-1(-i8)] 19. Prove:cos-1(3/V10)* cos-1(2/r6):*r. In Exercises2L through 28, draw a sketch of the graph of the given equation. 2l.y:isin-l 22'Y:sin-rix 7 A. Y :2 23. y : Ian-r 2x 25.y: 26.y:tsec-t2x c o s - l3 1 28.Y:*csc-tlr 2 7 .y : 2 c o t - r i x 1.0.8DERIVATIVES OF THE INVERSE TRIGONOMETRIC FUNCTIONS tan-t x Becausethe inverse trigonometric functions are continuous and monotonic on their domains, it follows from Theorem 9.3.4 that they have derivatives. To derive the formula for the derivative of the inverse sine function, let Y : sin-l r which is equivalent to x- sin y and y in l-to, Lnl (1) 478 FUNCTIONS TRIGONOMETRIC Differentiating on both sides of (1) with respect to y, we obtain Dux:cosy (2) and y inl-tn,irf From the identity sinz y * coszy: 1, and replacing sin y by & we obtain coszy:l-x2 (3) If. y is in [-*zr, *z'], cos y is nonnegative, and so from (3) it follows that cosy: \R rf y is in l-tn,inl (4) Substitutingfrom (4) into (2), we get Dor:\R FromTheorem9.3.4, DrA: llDox; hence, D ' ( s i n - r{ : ! (5) vI - xz The domain of the derivative of the inverse sine function is the open interval (-1,1). rf.u is a differentiable function of x,we obtain from (5) and the chain rule D * ( s i n - ul ) : # D , u o ILLUSTRATTOTV 1: If y: (6) sin-l dvl)r dx Vt-(f)' 12, then .._:_\..1:_ vr-r a To derive the formula for the derivative of the inverse cosine function, we use Eq. (9) of Sec.10.2,which is cos-r t:tr - sin-l r Differentiating with respect to r, we have D"(cos-l x):Dr(Ln - sin-l r) and so D'(cos-lx):- 7 ffi (7) where r is in (-1, 1). If u is a differentiable function of x,we obtain from (7) and the chain rule 1, (8) 479 FUNCTIONS TRIGONOMETRIC 10.8DERIVATIVES OFTHEINVERSE we now derive the formula for the derivative of the inverse tangent function. If U: tan-l x then x: tan y and y is in (-in, irr) Differentiating on both sides of the above equation with respect to y, we obtain D o x : s e c z A a n d y L Sin (-in, tr) (e) From the identity sec' y - | * tanzy, and replacing tan y by x, we have (10) sec2Y:t*x2 Substituting from (10) into (9) , we get Dox-1i-x2 So from Theorem 9.3.4 we obtain (11) x):# D r(tan -' The domain of the derivative of the inverse tangent function is the set of all real numbers. lf uis a differentiable function of x,we obtain from (11) and the chain rule (12) o rt,r,usrnerroN2: lf f (x): tan-t #, 1 then f'(x) From Detinition ']..0.7.4 we have cot_l )c: in _ tan_l r Differentiating with resPect to x, we obtain the formula D *(cot-l x) : - L L*xz TRIGONOMETRIC FUNCTIONS From this formula and the chain rule, it follows that if u is a differentiable (13) EXAMPLE1: Given SOLUTION: du ! - y : rs cot-t *x ffic 3x2 cot-l find dy ldx. : EXAMPIr 2: Given ln(r*y)-tan-r 3x2 cot-l solurroN: Differentiating implicitly on both sides of the given equation with respect to r, we get (;) find D *A. L+Dr!/ w9, : -y-7sD*U _ x+y y'* x' y'+ )c2 + (y',* x',)D*a: xy * y',- (x',* xy) Dra D*A:ffi Definition 1,0.7.5and formula (8) of this section are used to derive the formula for the derivative of the inverse secantfunction. From Definition "1.0.7.5 we have sec-lx - cos-l (+) for lrl ! vc/ So D 1 "(sec-r I r) 1, (+(-+) )' 1 *21ffi \E l*l *21ffi Therefore, we have D * ( s e cx- r) : + pltF (r4) FUNCTIONS 481 10.8 DERIVATIVES OF THE INVERSETRIGONOMETRIC wherelxl > 1. lf.u is a differentiablefunction of.x, it follows from (14) and the chain rule that u):ffiD*u D*(sec-l (ls) o rLLUSrRArroN3: If f (x) - sec-' 3x, then f ' ( x )_ 1 l3rlffi From (9) of Sec.10.7we have for lxl = 1 cos-lr sin-lr:tn- Replacing xby llx in the above equation, we obtain 1 s'm ' /1\ \-/:Zn - (1) "or-' \ r / ror lrl > 1 From the above equation and Definitions 10.7.5 and L0.7.6 it follows that csc-l x:in - sec-1r for lrl > 1 Equations (16) and (L4) are used to find the derivative of the inverse cosecantfunction. Differentiating on both sides of (15) with respectto x, we get D ' ( c s c - rx ) : - t Wffi G7) where lrl > 1. From (17) and the chain rule, Lf.u is a differentiablefunction of r, we have u) ffiD*u ExAMPLE3: A picture 7 ft high is placed on a wall with its bas e 9 ft above the level of the eye of an observer. How far fuom the wall should the obsenrer stand in order for the angle subtended at his eye by the picture to be the greatest? ( 1 8) sol,urroN: Let x ft be the distance of the observer from the wall,0 be the radian measure of the angle subtended at the observer's eye by the picture, o be the radian measureof the angle subtended at the observer's eye by the portion of the wall above his eye level and below the picture, and F: o + 0. Referto Fig. 10.8.1. We wish to find the value of r that will make d an absolute maximum. Becauser is in the interval (0, *-), the absolute maximum value of 0 will --] FUNCTIONS TRIGONOMETRIC be a relative maximum value. We see from the figure that cotB-+ and c o.t)oc: o B e c a u s e<0B < t n a n d 0 f' "1,6 B-cot-r a - cot-t { |and 9 Substituting these values of p and u in the equation 0 - B - q . , w e g e t x )c e - cot-' : - G - c o t "-r g Differentiating with respect to r gives 1L D*o tf 9r* ,rr* t * (G/ 1+ (;/ +rft L5Lg '1,62*x2' F i g u r e1 0 . 8 . 1 92*x2 Setting D r0 0, we obtain 9(L62* x') - l5(9' + ,c') :0 -7f+9'16(15-9):0 x2:9.75 x: 12 -12 The is rejectedbecauseit is not in the interval (0, +o;. The resultsof the first-derivative test are shown in Table 10.9.1.Becausethe relative maximum value of 0 is an absolute maximum value, we conclude that the observer should stand 12 ft from the wall Table10.8.1 Dr| 0< x<12 + X:'/',2 0 Conclusion d has a relative maximum value 12<x(fm Exercises10.8 L. Derive the formula D"(cot-l -1 x) 2. Derive the formula D*(csc-rx):4 lxl"ffi' TRIGONOMETRIC FUNCTIONS OF THE INVERSE 10.8DERIVATIVES In Exercises3 through 22, find' the derivative of the given function. (3,, (x) : sin-t tx ,4,-8(r) - tan-r 2x f 5. 8(t) : sec-r5t 8. f (x) - csc-r2x ii.:,',1, \{ ' : y sin-t 2y \*---' (y) 7. F ( r ) : c o t - t ? + t " r , - t I x2 f0i f (t) : t2 cos-' f 1 1 .S ( r ) : x . r . - ' I 1 2 ;h ( x ) : t a n - ' , 2 ' , ' , .' L-X- 1 3 .f ( x ) - 4 s i n - 1 i x * x t R l 4 . S ( s ) c: o s' r-: i + ' 1',1",' 16'. f (x) : }n(tan-l 3x) 17. S(r) : cos-l(sin r) 19. h(x) - sec-l r * csc-l r 20. F(r)- 6. F(r) - cos-t{x i I t 15.f(t) - a sin-r ;+ 18.f(x):sin-l \m x*cos-l t sec-r1ffi 22.f (t) - ,',,-' '* I4 Exercises23 through 26, find D,y by implicit differentiation. 21.G(x): I cot-' r + hv{T7- (%'e"*Y:cos-tr 24. ln(sin2 3x) : t" * cot-r y :ta -'y &:rsiny* f 26.sin-r(ry): cos-t(r* Y) iei. l1ight is 3 mi from a straight beach. If the light revolves and makes 2 rpm, find the speed of the spot of light along the beach when the spot is 2 mi from the point on the beach nearest the light. 2g. A ladder 25 ft long is leaning against a vertical wall. If the bottom of the ladder is pulled horizontally away fromthe wall so that the top is s-liding aownit g ftlsec, how fast is the measure of the angle between the ladder and the ground changing when the bottom of the ladder is 15 ft from the wall? 29. A picture 4 ft high is placed on a wall with its base 3 ft above the level of the eye of an observer. If the observer is approactring ttre watt it the rate of.4 fllsec, how fast is the measure of the angle subtended at his eye by the picture changing when the observer is 10 ft from the wall? 30. A man on a dock is pulling in at the rate of 2ftlsec a rowboat by means of a rope. The man's handq are 20 ft above the level of the point where thi rope is attached to the boat. How fast is the measure of the angle of depression of the rope changing when there are 52 ft of rope out? A man is walking at the rate of 5 ftlsec along the diameter of a circular courtyard. A light at one end of a diameter perpendicular to his path castsa shadow on the circular wall. How fast is the shadow moving along the wall when the disiance from the min to the center of the courty atd,is tr , where r ft is the radius of the courtyard? 32, In Exercise31, how far is the man from the center of the courtyard when the speed of his shadow along the wall is 9 ftlsec? 33. A rope is attachedto a weight and it passesover a hook that is 8 ft above the ground. The rope is pulled over the hook at the rate of * ftlsec and drags the weight along level ground. If the length of the rope between the weight and the hook is r ft when the radian measure of the angle between the rope and the floor is 0, find the time rate of change of d in terms ol x. 34. Given: l(r) : tan-'(l/r) - cot-r x. (a) Show that /'(r) : 0 for all r in the domain of /. (b) Prove that there is no constant C for which f (x) : C for all r in its domain. (c) Why doesn't the statement in part (b) contradict Theorem 5.3.2? (Z -"t " 484 TRIGONOMETRIC FUNCTIONS 10.9 INTEGRALS YIELDING INVERSE TRIGONOMETRIC FUNCTIONS From the formulas for the derivatives of the inverse trigonometric functions, we obtain the following indefinite integral formulas: (1) (2) (3) Formulas(1) and (2) follow directlyfrom the formulasfor the derivatives of sin-r z and tan-l u. Formula(3) needsan explanation. If u > 0, then z : lul, andwe have fdufdu Jnffi: J 1"6: s e c - ' uc* : s e c -l zr l* c II u < 0, then u: -lul, and we have f d- u - fdu J u\/u2-7 J -l"lttd= _ f : -ilu J lul\/u2 _ 1, f dFu\ J |ufftrdt= : sec-1(-z) * C : sec-l lul + c Therefore, if u > 0 or u < 0, we obtain (3). we also have the following formulas. (4) (s) (5) These formulas can be proved by finding the derivative of the right side and obtaining the integrand. This is illustrated by proving (Z). D,(sin-'t):#""e) v,- \;/ \+. 10.9INTEGRALY S I E L D I N GI N V E R S E T R I G O N O M E T R IFCU N C T I O N S \tr tfa>0 tfa>0 The proofs of (5) and (6) are left as exercises (see Exercises L and 2). ExAMPLE l,: Evaluate fdx J t/l - 9xz :+ sin-r ++ c EXAMPug2: fdx Jffi Evaluate SOLUTION: df d, I :[ J 3 x ' - 2 x * 5 J 3 ( x -' f x ) + 5 To complete the square of.xz - &x we add *, and because* is multiplied by 3, we actually add * to the denominator, and so we also subtract * from the denominator. Therefore, we have d* d* _f I J ixz-2x+5- J 3(x,-?x++)+5-+ dx dx 1f - r 3 ( x + ) ' + # s J @- +)r+# J :l- solurroN: Because d(xz* 2x * 5) : (2x I 2) dx,we write the numerator as (2x + 2) dx * 5 dx, and expressthe original integral asthe sum of two integrals. d.x I Qx*7\ dx I Qx+2\ dx , - | 'J x'*2x*5 J x'*2x*5 J x'*2x*5 .d 486 TRIGONOMETRIC FUNCTIONS dx : l n l x ' l . 2 x + 5 1*u/ (x*I)'+4 +]t an-l tl1+ :ln(x'*2x*5) 2 NorE: Becausexz * 2x + 5 > 0, for all x, Evaluate :I I 3dx 3dx (x*z)ffi (r*Dffi C l x ' 1 - 2 x + 5: 1x2* 2x + 5. 3dx ( -3 sec-1lx + 2l + C ExAMPrr 5: Find the area of the region in the first quadrant bounded by the curue sot,urroN: Figure 1.0.9. 1, shows the region and a rectangular element of area. If A square units i s the area of the region, w€ have Athe x axis, the y axis, and the line x: l. : lim $= '1 ttiii-o?:'r*€" Aix fdx Jo L*x2 : tan-t r It Jo :tan-l 1-tan-1 :in-0 :in The result of Example 5 states that the number z is four times t6'e number of square units in the area of the region shown in Fig. 10.9.1. Exercises 10.9 In Exercises1 through 5, prove the given formula by showing that the derivative of the right side is equal to the integrand - f --;--:----;: du -7 t"rr-'. u1+ C l. | J A'tU' A A 10.9INTEGRALS YIELDING INVERSE TRIGONOMETRIC FUNCTIONS '-'' du 1 ^ f z. t--------::-sec-rlal +C a J u!ut-a' lal if a>0 Take two cases:a > 0 and z < 0. du ^f 3. | -:-cos-'z*C J \/l- u, Is this formula equivalent to formula (1)? Why? 4' fdu Jr;7: -cot-ru * C Is this formula equivalent to formula (2)? Why? du -f 5. l;ffi:-csc-l lul+c Take two cases:z > 0 and z < 0. Is this formula equivalent to formula (3)? Why? In Exercises5 through 25, evaluate the indefinite integral. 7I# i ,101 .L 13I# 14t4 J Iz,''.t f L 6 .l L J\ffi dx ^fdx ' t ,89.Jffi a -,--- 22. [ = !d* J x2+x*L f c o s zx L7:l " J x2-x*2 i&lJ,'ffi, 6 xdx 1 1Jt LV r 6 - e t ' J xtF \''--" i S I x4+r6 . ,- . , .f - Q + x ) d x J V4-2x-x2 3dx z) a+l ' J G + D f f i - In Exercis es 26 through 33, evaluate the definite integral. . . : 'f r 1 * x 27.' J| o #d* , LtX' 30. f d x J, x'z-4x+13 4". Q, r[L + (ln r)2] \ 34.Find the area of the region bounded by the curve y : 8l (x2 * 4) , the x axis, the y axis, and the line x: 2. :i5. Find the abscissaof the centroid of the region of Exercise34. 36. Find the area of the region bounded by the curves x2: 4ay and y :8a31 (x2 * 4a2), 37. Find the circumference of the circle rP * y' : I by integration. */ t'i FUNCTIONS TRIGONOMETRIC 488 38. A particle moving in a straight line is said to have simple harmonicmotion if the measure of its accelerationis always proportional to the measure of its displacement from a fixed point on the line and its accelerationand displacement are oppositely directed. So if at f sec,s ft is the directed distance of the particle from the origin and o ftlsec is the velocity of the particle, then a differential equation for simple harmonic motion is da -- -kzs dt (7) where k2 is the constant of proportionality and the minus sign indicates that the acceleration is opposite in direction from the displacement.Becausedoldt : (dulds)(dsldt) : o(dolds), Eq. (7) may be written as I'j da E: -kzs (8) (a) Solve Eq. (8) for a to get Tr- *k\m. Note: Take a2k2as the arbitrary constant of integration and justify this choice. (b) Letting a - dsldt in the solution of part (u), we obtain the differential equation i< fi:tktto'- t" Taking f : 0 at the instant when o : 0 (and hence s : a), solve Eq. (9) to obtain s: a coskt (10) (c) Show that the largest value for lsl is c. The number a is called the amplitudeof the motion. (d) The particle will oscillatebetweenthepointswheres:aands:-a.If Tsecisthetimefortheparticletogofrom ato-aandreturn,show that 7: Znlk. The number T is called Ihe periodof the motion. 39. A particle is moving in a straight line according to the equation of motion s:5 - L0 sin2 2t, wherc s ft is the directed distance of the particle from the origin at f sec.Use the result of part (b) of Exercise38 to show that the motion is simple harmonic. Find the amplitude and period of this motion. 40. Show that the motion of Exercise 39 is simple harmonic by showing that differential equation (7) is satisfied. Reaiew Exercises(Chapter 10) In Exercises L throu gh 4, evaluate the given limit, if it exists. I l. .r r' r. llm r-o 2Ljsry sin2 r f. L2 CJL cos 5r 4. lim - tan a 2 x 4*^ lt$ 5. show tlnat r-7r12 COS lX d(Lnsin r) x cot x J(ln rf 6. Prove that lim sin x2 0. r-o X '1.4, In Exercises7 through find the derivative of the function defined by the given equation. 7.f(x)--ffi g. S@) - xcos! 8. f(x) - \63r . lnf 1 0 ..f ,( \t ') - : secf 1 1 . F ( r ) : t a n -' ; ' 12. s(r) : cot SxtTsin2x 13. f(x) - (tanx)rt*' 1 4 . h ( x ) - ( c o sx ) e " REVIEWEXERCISES 489 In Exercises L5 and L6, find D *A by implicit differentiation. y2 t6. sin(r + V) * sin(x - A) - 1 L 5 . c o t - l| + x A z : 0 5y In ExercisesL7 through 24, find the indefinite integral. r 17. I sinaix cosztx dx f 18. J 20. 23. I J IUl-cosxdx (csc 2x - cot 2x) dx 2t. I# sin 3r cos5x dx 24. f sint 2x coss2x dx J r J dx ffi In Exercises25 through 28, evaluate the definite integral. lils 25. I Jo lal6 secs x tans x ilx (x -t 2) dx 20. I Jo tana 2x dx fntz 2 7 . f2 Jrffi ,t.J,,n(sina2x-2sin4x)dx 29. Find the length of the arc of the curve y : ln cos r from the origin to the point (in, -ln Z). 30. Find the averagevalue of the cosine function on the closed interval fa, a * 2nl. 31. Find to the nearest degree the measurementsof the four interior angles of the quadrilateral having vertices at (5, 6) , (-2, 4) , (-2,7) and (3, 1) and verify that the sum is 360'. 32. Findthevolumeof the solidgeneratedifthe regionboundedbythecurve!: revolved about the x axis. sinzrandther axisfromr:0to x: zris 33. Two points A and B are diametrically opposite each other on the shoresof a circularlake 1 mi in diameter. A man desires to go from point A to point B. He can row at the rate of 1* mi/hr and walk at the rate of 5 mi/hr. Find the least amount of time it can take for him to get from point A to point B. 34. Solve Exercise33 if the rates of rowing and walking are, respectively,2 milhr and 4 mi/hr. 35. A partide is moving along a straight line and s: sin(4f * tn') + sin(4f * *r) where s ft is the directed distance of the particle from the origin at t sec. Show that the motion is simple harmonic and find the amplitude of the motion. (SeeExercise 38 in Exercises10.9.) 35. If an equation of motion is s: cos 2f * cos f , prove that the motion is not simple harmonic. fn 37. Evaluate: I lcosx+t\tu. Jo fnl2 38. Evaluate:I Jo lcosx - sin xl dx. 39. Find the area of the region in the first quadrant boundedby the y axis and the curves A: seczx and y :2 tanzx. 40. Find an equation of the normal line to the curve /: cos r at the point (32, -*). 41,.A p a r t i c l e i s m o v i n g a l o n g a s t r a i g h t l i n e a c c o r d i n g t o t h e e q u a t i o n o f m o t i o n s : 5 - 2 c o s z t w h e r e s f t i s t h e d i r e c t e d 490 TRIGONOMETRICFUNCTIONS distance of the particle from the origin at f sec. lf. a fitlsec anda fitlseczare, respectively, the velocity and accelerationof the particle at f sec, find o and a in terms of s. A. ff y ft is the range of a projectile, theS "c onzsin 20 where o6ftlsec is the initial velocity, g ftlse3 is the accelerationdue to gravity, and 0 is the radian measureof the angle that the gun makes with the horizontal. Find the value of 0 that makes the range a maximum. 43. A searchlight is * mi from a straight road and it keeps a light trained on an automobile that is traveling at the constant speedof 50 mi/hr. Find the rate at which the light beam is changing direction (a) when the car is at the point on the road nearest the searchlight and (b) when the car is * mi down the road from this point. 44. A helicopter leaves the ground at a point 800 ft from an observer and rises vertically at 25 fllsec. Find the time rate of change of the measure of the observer's angle of elevation of the helicopter when the helicopter is 500 ft above the ground. 4s. In an electriccircuit let E : f (f) and i - g(f) where E volts and i amperes are, respectively, the electromotive force and current at t sec.If P watts is the averagepower in the interval [0 , T), where T sec is the common period of andg, then f P-+ f fG)su) dt If E - L00sin f and i: A(sin f - *zr), first determine T and then find P. 46. Find the area of the region bounded by the curve y:glt6=7, x - 2\n. 47. Find.the absolute maximum value attained by the function / if (i / constants. the two coordinate axes, and the line : a sin kr * B coskx, where A, B, and k are positive 48. Suppose the function / is defined on the open interval (0, 1) and f (x): x^:?:'l (x - 1) Define f at 0 and L so that / is continuous on the closed interval 49. Prove: Dru (sin x) : sin(x + tnn). (urNr: Use mathematical cos(x + tn) sin r after each differentiation.) L O1, J . induction and the formulas sin(x + trr)_ cos x or l Techniquesof integration 492 TECHNIQUEO S F INTEGRATION 11.1 INTRODUCTION In Chapter 7 the definite integral of a function / from a to b was defined as the limit of a Riemann sum as follows: fbn dx:,,tiil,> f (€,)Lix J"f k) (1) if the limit exists. We stated a theorem, proved in advanced calculus, that the limit on the right side of (1) exists if / is continuous on the dosed intewal fa, bl. The exact value of a definite integral may be calculated by the fundamental theorem of the calculus,provided that we can find an antiderivative of the integrand. The process of finding the most general antiderivative of a given integrand is called indefiniteintegration.We use the term indefinite integral to mean the most general antiderivative of a given integrand. In practice, it is not always possible to find the indefinite integral. That is, we may have a definite integral that exists, but the integrand has no antiderivative that can be expressedin terms of elementary functions. An example of this is [;tz t-t' df (a method for computing the value of this definite integral to any required degreeof accuracyby using infinite series is given in Sec. 1.6.9).However, many of the definite integrals that occur in practice can be evaluated by finding an antiderivative of the integrand. Some methods for doing this were given previously, and additional ones are presented in this chapter. Often you may find it desirable to resort to a tableof integralsinstead of performing a complicatedintegration. (A short table of integrals can be found on the front and back endpapers.)However, it may be necessaryto employ some of the techniquesof integration in order to express the integrand in a form that is found in a table. Therefore,you should acquire proficiency in recognizing which technique to apply to a given integral. Furthermore, development of computational skills is important in all branches of mathematics, and the exercisesin this chapter provide a good training ground. For thesereasonsyou are advised to use a table of integrals only after you have mastered integration. The standard indefinite integration formulas, which we leamed in previous chapters and which are used frequently, are listed below. 1,"du-#+c u+c : ALI + C I l f ( u )+ s @ ) l I un du f du lT: lr n l " l+ c where a is any constant :gu+c f(u)du* udu--cosu+C n+-L u du: sin u + C u du: tan u + C I BY PARTS 11.2 INTEGRATION [ ","' /,". udu:-cotu*C /,". utanudu:secu*C /.,. l s e cu * t a n u l + C udu:lnlcscu-cotul+C d' I cot u du udu:ln - -:--tu-+c Jffi:sln'; wherea>o L+c | =dY ^ -'l' tan-1 l'rr u du: ln lseczl + C J a2+uz t4 u d u : l n l s i nu l + C a fl J uluz - n2 1 sec-r v Y - lYl+ c a lal wherea > o parts. It II.2 INTEGRATION A method of integration that is quite useful is integrationby a product: of BY PARTS depends on the formula for the differential d(ua)-uda*adu or, equivalently, udo-d(ua)-adu \ (1) Integrating on both sides of (1), we have (2) Formula (2) is called the formula for integrntionby pntts. This formula expressesthe integral [u do in terms of another integral, lo du. By a suitable choice of u and do, it may be easier to integrate the second integral than the first. When choosing the substitutions for u and do,we usually want da to be the most complicated factor of the integrand that can be integrated directly and u to be a function whose derivative is a simpler funciion. The method is shown by the following illustrations and examples. o ILLUSTRATTON 1.: We wish to evaluate r I tan-r x dx J Let u: t a n - l r and da - dx. Then TECHNIQUESOF INTEGRATION 494 So from (2) we have f J tan-l x dx : ( t a n x- ,) ( x *C , )- I(r* -xtan-l x*crtan-rx-[+d* C , )# n rR-''JR r dx : x t a n - l x * C , t a n - r x - * l n 1 1+ x r l - C , t a n - r x * C , :xtan-rr-*ln (1,+x2)*C, I In Illustration L observe that the first constant of integration c1 does not appear in the final result. This is true in general, and we prove it as follows: Bv writing u * C, in place of a in formula 2, we have ff J u du: u(a* C,)- + Cr)du J kt :uo*cru-[adu-crIau JJ :uu*cfl-[oa"-cfl :nn- [ o au J Therefore,it is not necessaryto write C1when finding a from do. EXAMPLE 1: Find f I rln xdx J SOLUTION:Let Lt-- ln r and da : x dx. Then ,dxxz au: x2 ,v2 . h r d x - i t n xt I .vz It+ _ 1;I^ '_. d( ,x 2 |tnx txz ln lx EXAMPLE 2: f Find f rsinxdx J xdx tx' '+ C SOLUTTON: Let Lt- r andl d v : s i n r : d x Then du-dx a n d 7 ) - - cos S )c Therefore, we have f I r sin x dx:-x J f c o sx )+ f c o sx;ddx : --ficos. J x + sinx+. C BY PARTS 495 11.2INTEGRATION 2: IN o ILLUSTRATION Example 2, if instead of our choice s of u and du as above, we let and Lt: sin r da:xdx we get and o:tf du:cosxdx So we have f J f r sinx dx:i sin--; 1f J f cosx dx The integral on the right is more complicated than the integral with which we started, thereby indicating that these are not desirable choices for z o and ilo. It may happen that a particular integral may require repeated applications of integration by parts. This is illustrated in the following example. EXAMPTn 3: Find r I xze*dx J - x2 and da - et dx. Then soLUTIoN: Let u and 7):er du-Zxdx We have, then, ff I xre" dx: JJ xzet - 2 | xe' dx We now apply integration r,t-x by Parts to the integral on the right. Let and d0:etdx Then and u:e* dil-dx So we obtain ff | *t" dx-xex- JJ | e'dx :xet-er+C Therefore, f I x"" dx: )cze" zlxe" er + C] J : Nzen- Zxe' * 2e'+ C Another situation that sometimes occurs when using integration by parts is shown in ExamPle4. I TECHNIQUES OF INTEGRATION Find soLUTroN: Let u - e, and du - sin x dx. Then du: sin x dx e* dx and T)-- -cos x Therefore, r I e" sin xdx cos x dx JJ The integral on the right is similar to the first integral, exceptit has cosr in place of sin r. We apply integration by purts again by letting fr:et and d0-cosxdx So dil-e'dx and o-sinx Thus, we have ff t * s i nx d x JI J e'cosx* [e'sin x- | e* sinxdx] Now we have on the right the same integral that we have on the left. so we add I e" sin x dx to both sides of the equation, thus giving us r 2 l e ' s i nx d x J Dividirg e ' c o sx * e " s i n x * e on both sides by 2, we obtain r I t " s i n x d x - + e " ( s i n x - c o sr ) + C J In applying integration by parts to a specific integral one pair of / chrrices fot u and du may work while another pair may nofl We saw this in Illustration 2, and another case occurs in Illustration 3. OI LLUSrRarroN 3: In Example 4, rn the step where we have fr l t " s i nx d x : - e * c o s x * JJ l t * c o sx d x i f v ve evaluate the integral on the right by letting u-cosff di: we get and d0:e*dx -sin x dx and u : er fa-r t" sin x dx Jl J ff I t" sin x dx: JJ e* cosx * (e" cosr * l t" sin x dx) I e* sin x dx BYPARTS 11.2INTEGRATION ,497"" In Sec. 10.8 we noted that in order to integrate odd powers of the secantand cosecantwe use integration by parts. This processis illustrated in Example 5. EXAMPLE 5: soLUTroN: Let Lt- secr and da : sec2x dx. Then du : sec r tan x dx and a - tan x Find f I sect x dx J Therefore, fr xdx=sec xtanxI sec3 JJ rf I secsx dx: secr tan xJJ ffr I s e c tx d x : s e c JJJ xtanx- f s e cx t a n z x d x | secr(seczx- I) dx | s e c tx d x * | s e cx d x Adding I secsx dx to both sidet, yu get r 2 | t".t xdx:secrtanr*ln J l s e cx * t a n x l+ 2 C f x d x - + s e c r t a n) c + * l n l s e cx + t a n r l + C lsecs Integration by parts often is used when the integrand involves logarithms, inverse trigonometric functions, and products. 11..2 Exercises In Exercises L through t6, evaluate the indefinite 1. f Jrnxdx u..! x3'dx I t a n - lx d x lx r L0. sin(ln r) dx J 7. integral. , UJ ,6I#" tA ? I-cos2xdx g,/ 'i"-'\x dx 3. 8. *' stn3x dx I ,. ''ir. I#. 14. /.r.' 6. xdx I-secz I.'tnxdx I e* cosx dx L2. x'e" dx I x dx 1s. /r".'xdx TECHNIQUES OF INTEGRATION In Exercises17 through 24, evaluatethe definite integral. t . [:'" sin 3r cosx dr Jo F\. Jo ,-j L'^ e*asin4xdx cos2x itx {9 '-' f _x2 J-n '-') f ,r,, a, $ Jo ,r. 1""'o-x surr cotxcsc cscx r 4tr itx Jtu zz. fn '"' ,., sec-,t/i dx ,0. [""'' cos\/2x dx Jo e. ,4' r Jo x sin-r x ilx /,$. fi"a the area of the region bounded by the cule y : In x, the r axis, and the line x: ez. ?$. fi"a the volume of the solid generated by revolving the region in Exercise25 about the r axis. {}. ni"a the volume of the solid generated by revolving the region in Exercise25 about the y axis. 28. Find the centroid of the region bounded by the cule y : er, the coordinate axes, and the line : r 3. 29' The linear density of a rod at a point r ft from one end is 2e-t slugs/ft. If the rod is 6 ft long, find the mass and center of mass of the rod. 30' Find the centroid of the solid of revolution obtained by revolving about the x axis the region bounded by the curve y : sin x, the x axis, and the line x: tr. 31' A board is in the shape of a region bounded by a straight line and one arch of the sine curve. If the board is submerged vertically in water so that the straight line is the lowei bound ary 2 ft below the surfaceof the water, find the force on the board due to liquid pressure. 32' A particle is rnoving along a straight line and s ft is the directed distance of the particle from the origin at t sec.rf o f t / s e ci s t h e v e l o c i t ya t f s e c , s : 0 w h e n f : 0 , a n d o . s : t s i n t , f i n d s i n t e r m so f f a n d a l s os w h e n t : t n . 33' A water tank full of water is inthe shape of the solid of revolution formed by rotating about the r axis the region bounded by the cuwe y: e-", the coordinate axes, and the line r: 4. Find the work d6ne in pumping all the water to the top of the tank. Distance is measured in feet. Take the positive r axis vertically downward. 34. The face of a dam is in the shape of one arch of the curve y :100 cos z&ozrrand the surface of the water is at the top of the dam' Find the force due to liquid pressure on the face of the dam. Distance is measured in feet. 35' A manufacturer has discovered that if x hundreds_of units of a particular commodity are produced per week, the marginal iost is determined by x2'r' and the marginal revenue is d'etermined by s . z-ttz,wliere both the production cost thousands of dollars' If the weekly fixed costs amount to $2000,find the *a"i-.r* *"etaf frorit ili:::T:ftTarein 11.3 INTEGRATION By TRIGONOMETRIC SUBSTITUTION If the integrand contains an expression of the form \/ar=, \/FTE, or {F=V, where a } O,iiis often possible to perform the integration by making a trigonometric substitution which reiults in an inte_ gral involving trigonometric functions. we consider each form as a separate case. Case 1: The integrand contains an expression of the form \/F7A, a > 0. we introduce a new variable 0 by letting u: a sir, g, where 0 < 0 =+rif u > 0 and -*rr < g < 0 iIu < 0. Ttrendu: a cos0 d0,and, \/F=7 : \/|r=}Gfr 0:{F6=R6 : a@0 Beggge'-*n= 0 -in,cos 0 - 0; hence,\re 0:cos 0, andwe have \/F=7: a cos0. Because sin 0: ula and.-trr - 0 = t,,it followsthat 0 : sin-r(ula). SUBSTITUTION 499 BY TRIGONOMETRIC 11.3 INTEGRATION soLUrIoN:'Letr:3 sin d, where0 < 0 < tnif. x > 0 and-*rr <0 <OIf' r ( 0. Then dr:3 cos 0 d0 and .\5=7 : t/iTg sin4 : S:'/ coPE : 3 cos 0 Therefore, Iryo,:l+#'? (scosodo) :Icop0d0 \lfr F i g u r e11 . 3 . 1 : [ 1"r., 0-1) d0 r>0 + \/ffi :1.o, o-o+C Becausesin g: *r and -Lr = 0 =8n,0: sin-l(*r)' To find cot 0 refer to Figs.11.3.1 (for r > 0) and 11'3.2 (for x < 0). We see that cot9: V9 - *lr. Therefore, t/g =7 f ..F-- I , --t dx \5=7 _,,__,x c - sin-ri* F i g u r e1 1 . 3 . 2 Case 2: The integrand contains an expression of the form \817, a > 0. we introduce a new varidble 0 by letting u: a tan 0, where 0 < 0 < tn lf.u > 0 and -Lo < e < O if u < }'Then du: a secz0 il9'and \/F 1fr : GTV@o : a{I+ffin: a{@T :sec0' andwehave B e c a u s-ei n < 0 < t n , s e c 0> 1 ; t h u s\ / s @ e -in g: < izr, it followsthat < 0 ula and tan \/7 + fr: sec0. Because 0: tan-r(ula). EXAMPLE 2: r | !x'+ JI EVAIUATC 5 dx solurroN: Letr: tEtan0,where0 < 0 < irif.x = 0and-izr < 0 < 0if r < 0. Then dr: \6 se& 0 d0 and \ffi _m-\6\@_\E sec0 Therefore, sec'ade 0 d0) dx: | \re secO(fi secz | \ffi / JJ Using the resultof Example5 of Sec.1L.2,we have | \m J dx:f r " . 0 t a n0 + * t " P e ce + t a n g l + C : TECHNIQUES OF INTEGRATION r>0 F i g u r e11 . 3 . 3 F i g u r e11 . 3 . 4 we find sec0 from Figs.11.3.3(for x - 0) and 1,1.3.4 (for r < 0), where tarr 0 : xlt6. We see that sec e: \/FTEIVE. Hence, ,5 ax: 2 -tx \ffi -tx \ffi *Bh l\ffi *Bh (lffi *xl -Brr,\6 +c *r) +cl Because\ffi*x)0,wedroptheabsolute-valuebars. case 3:_ The integrand contains an expression of the form t/F1v, a > 0. we introduce a new variable d by letting u: a sec0, where < 0 0 <+r7f.u > a andr < 0 <8rrif u = -r. Th"rrTu: osec0ian 0 it| and \E=7 : \Gr;;A4- dt:1/p@[=j : a{R o Becauseeither 0 < 0 < Er or r, < 0 < tr, tan 0 - Q;thus, VTan??: tan 0, and we have l/FjE : a tan 0. Note in Fig. ft.g.S ifrut > u a, sec d: ula >'1., and 0 < 0 < ia. Hence, 6: ssg-"r(ula) if. u > a. ln Fig.11.3.5,u --flt sec d: ula =-1, and z < 0 <gzr.'Because when u = -atiz < sec-l(ula) = zr, it follows that 0 : 2r _ sec-r(ula). \/u'z-A u\a F i g u r e 11 . 3 . s u1-a F i g u r e11 . 3 . 0 SUBSTITUTION BY TRIGONOMETRIC 11.3 INTEGRATION EXAMPLE 3: sec0,where 0 < d < tn 1f x ) 3 and r < 0 <En if. solurroN: Letr:3 x 1-3. Then dr: 3 sec 0 tan 0 d0 and Evaluate fdx 0 \/F4:ffi64:3ffi0:3tan J *\ffi Hence, dx f 3secgtan0d0 f J /? r""'o'g t"r'ro Jffi: : 1f * I (1 * cos 20) d0:*(0 ** sin 2e) + C :#(e*sin0cos0)*C When x) 3,0 < d < Ln, and we obtain sin 0 and cos 0from Fig.1'1.3.7. When x 1 -3, r < 0 < 8zr,andwe obtain sin 0 and cos 0 from Fig' 11.3.8. In either case,sin 0: {FaT lxand cos0: 3lx.To exPress0 in terms of r -3. When we must distinguish between the two cases: , > 3 and x < - sec-r *x (referto x ) 3, d: sec-l *r. However, when x 1-3, 0:2zr Fig.11.3.8).We have, then, F i g ur e 1 1. 3 . 7 \ffi.3\+c x/ x x 3 {ffi + tfx>3 \ffi ,x rfx<-3 3)c and letting C Cr * *n, x( we get _tx*F 3 -3 Lf+c ifx>3 F i g u r e1 1 . 3 . 8 _tx*F rf+c 3 EXAMPLE 4: Evaluate fdx J t/x2 - 25 tf x<-3 n <0 <*n if solurroN: Letr:5 sec0, where 0 < 0 < Ln lf x > 5 and -5. and Then dx: 5 sec0 tan 0 d0 x < {F =Z r:{25@ 0- 25 :5@6:5tan 0 Therefore, dx -l 5secltanoitl:I sec'de I Stano J ItlP=6-t : l n l s e c d * t a n 0 l+ C To find sec 0 and tan 0 refer to Fig. 11.3.9(for x > 5) and Fig. tt'g.to (for TECHNIQUEO S F INTEGRATION < -5). In either case,sec0 - *r and tan tL*\ffi| l4 Jffi--rnl;--ln lr+ \ffi-l -ln lr+ \ml I tlffi. Wehave,then, *. -ln5+C +C, t r)s F i g u r e11 . 3 . 9 ExAMPr.n 5: Evaluate fz J,@ dx F i g u r e1 1 . 3 . 1 0 solurroN: To evaluatethe indefinite integral I dxl(6- *)ttz,we make the substitution r: \6- sin d. In this casewe can restrict 0 to the interval 0 < 0 < *z becausewe are evaluating a definite integrar for which > r 0 berauser-is in [1,2]. So we have i: {d sin 0, 0 2 e <fur, and,dx: V6 cos 0 d0. Furthennore/ (6 - P1"r'\ (6 - 6sin25t;arz - 6\6 (1 sinz o)ltz - 6\6(cos2 0)3t2 - 6\6 coss 0 Hence, f dx l{6cos0do JG-fft'z:J sG"""'e 7l d0 6J cos20 :i'1 ' f J""coae :$tan0*C Wefind tan d from Fig.L1..3Jt,in which sin 0: xltG and0 < 0 < tr.We NY T R I G O N O M E T R ISCU B S T I T U T I O N 1 1 . 3 I N T E G R A T I OB see that tan e- and so we have xl\ffi, fdx)c, l - - : t f J G - x2)3t2 6t75 - x2 I Therefore, f2dxx12 | - - - l J, rc- x2)3i2-6\ffi F i g u r e1 1 . 3 . 1 1 ), L 11.3 Exercises In Exercises L through L8, evaluate the indefinite integral. 3I 7I 11I lsI 14. (5 - 4x- x2)3t2 h In Exercises L9 through 26, evaluate the definite integral. 22. /, I re. : \/P -9lx', 27. Find the area of the region bounded by the cufve y 28. Find the centroid of the region of Exercise27. 29. Find the length of the arc of the curve A : ln x from r: 1 to x: the x axis, and the line r:5. 3. 30. Find the volume of the solid generated by revolving the region of Exercise27 about the y axis. 31. Find the centroid of the solid of revolution of Exercise 30. 32. The linear density of a rod at a point r ft from one end is li17 ter of mass of the rod. slugs/ft. If the rod is 3 ft long, find the mass and cen- TECHNIQUEO S F INTEGRATION 33. If an gbject is moving vertically due to the forces produced by a spring and gravity, then under certain conditions v2: & - k2s2,where at f sec,s ft is the directed distance of the o-bjectirorith" stlartinjpoint and o ftlsec is the velocity of the object. Find s in terms of f where o is positive. 34. A gate in an irrigation ditch is in the shape of a segment of a circle of radius 4 ft. The top of the gate is horizontal and 3 ft above the lowest point on the gate. If the water level is 2 ft above the top of the gate, find the force on the gate due to the water pressure. 35. A cylindrical pipe is 4 ft in diameter and is.closed at one end by a circular gate which just fits over the pipe. If the pipe contains water to a depth of.3 ft, find the force on the gate due to the water pressua 35. An automobile's gasoline tank is in the shape of a right-circular cylinder of radius 8 in. with a horizontal axis. Find the total force on one end when the gasoline is 12 in. deep and w oz. is the weight of 1 in.B of gasoline. 11.4 INTEGRATION oF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS. CASES 1 AND 2: THE DENOMINATOR HAS ONIY LINEAR FACTORS In Sec.1.8 a rational function was defined as one which can be expressed as the quotient of two polynomial functions. That is, the function H is a rational function if.H(x): P(x)le(x), where p(x) and e(r) are polyno_ mials. We saw previously that if the degree of the numerator is not less than the degree of the denominator, *" liurr" an improper fraction, and in that casewe divide the numerator by the denominator until we obtain a ProPer fraction, one in which the degree of the numerator is less than the degree of the denominator. For example, x4- 10x2* 3x + 1 -)c2-6*3x,-?3 x2-4 x2-4 So if we wish to integrate the problem is reduced to integrating I e'-6) dx* In general, then, we are concerned with the integration of expressions , of the form where the degreeof p(r) is less than the degreeof e(r). To do this it is often necessaryto write p (r) | e @) as the sum of p artial fractions' The denominators of the partial fractions are obtained by factoring Q(r) into a product of linear and quadratic factors. Sometimes it may be difficult to find these factors of Q (x); however, a theorem from advanced algebra states that theoretically this can always be d.one.We state this theorem without proof. 11'.4.1Theorem Any polynomial with real coefficients can be expressed as a product of FRACTIONS. CASES1 AND 2 BY PARTIAL FUNCTIONS 11.4INTEGRATION OF RATIONAL linear and quadratic factors in such a way that each of the factors has real coefficients. After Q(r) has been factored into products of linear and quadratic factors, the method of determining the Partial fractions depends on the nature of these factors. We consider various casesseParately.The results of advanced algebra, which we do not prove here, provide us with the form of the partial fractions in each case. We may assume,without loss of generality, that if Q(r) is a polynomial of the nth degree, then the coefficient of x" is 1 becauseif * Crxn-r+''' I Cn-6* Cn Q(r) : Csxn '!., we divide the numerator and the denominator of the fracthen if Co + tion P(r)/Q(r) by Co. Case1: The factors of Q(r) are all linear and none is repeated. That is, ' ' ' (x- a,) Q(r) : (x- ar)(x- ar) where no two of the a1are identical. In this casewe write lql x-ar' Q(r) ' + . . . + .'h = + x -A x-fln a2 (1) where Ar, Ar, . , An are constants to be determined' Note that we used "=" (tead as "identically equal") instead of.":" ln (1). This is because(1) is an identity for the value of each4,. The following illustration shows how the values of A1 ate found' o TLLUSTRATTON1.: We wish f (r-1) Jm to evaluate dx We factor the denominator, and we have x-L x3-x2-2x x-l So we write x-1 x ( x - 2 )( r + 1 ) Equation (2) is an identity for all x (except x:0,2, (2) -1). From (2) we get x-7: A(x-2)(x+1) tBx(x + 1)+ Cx(x-2) (3) Equation(3) is an identity which is true for all valuesof r including0,2, TECHNIQUES INTEGRATION and -1. We wish to find the constants A, B, and C. Substituting 0 for r in (3), we obtain -7: -2A or A: t Substituting 2 for x in (3), we get 1,:68 or B: t Substituting -'J. for r in (3), we obtain -2:3C or C:-E There is another method for finding the values of A, B,and C. If on the right side of (3) we combine terms, we have x-] : (A+B+C)* + (-A+B- 2C)x-2A (4\ For (4) to be an identity, the coefficientson the left must equal the correspondingcoefficientson the right. Hence, A+B* C:0 -A+B-2C:1 -24--1 Solving these equations simultaneously, we get A:i, -3. Substituting these values in (2), we get x-t x(x-2)(x*1)- _+ , x' + x-2' B: t and C: -3 x*1. So our given integral can be expressedas follows: [ ^' .t -ar:![@ *71 2 J xt-x2-2x-*- ax -2t 6 J x-2J i:f ln lrl+ *ln lr- 2l: * ( 3 l n l r l + l n l r - 2 l- :f,mlwl dx 3J,t+1 Bl n l r * 1 l + g t n C A l n l r * 1 l+ l n c ) . Case2: The factors of Q(x) are all linear, and some are repeated. Suppose that (r - a) is a p-fold factor. Then, corresponding to this factor, there will be the sum of p paftial fractions A, Ar, .+. , --+, (x - a,)o ' (x - a,)o-r ' *-T A o - , G-F -r - Ao i- where Ar, A", . , Ao are constants to be determined. Example 1 following illustrates this caseand the method of determining each 46. 11.4 INTEGRATIONOF RATIONALFUNCTIONSBY PARTIALFRACTIONS.CASES 1 AND 2 507 solurroN: The fraction in the integrand is written as a sum of partial fractions as follows: x}-L _ A Wry:V- , B , C J- D =L rTGTF-1;-2)'' E x-2 (5) Multiplying on both The above is an identity for all r (exceptx:0,2). sides of (5) by the lowest common denominator, we get x3-l = A(x-2)t +Bx(x-2)s * Cxz+D*(x-2) + E*(x-2)z xs- L = A(xs - 5f + l2x - 8) * Bx(xs- 5* + l2x - 8) + C* + Dxs- 2Df + Ef(* - 4x * 4) xs-L = (B* E)xa* (A-58*D-AE)xz + (-6A + r2B+ C - 2D+ 4E)* + (L2A- 8B)r- 8,4 Equatingthe coefficientsof like powersof x, we obtain Bi E:0 A-68+D-4E:L -6A+728+C-zD *4E:0 t2A-88 -0 -84--1 Solvingr we get A:i B:* C:I Therefore, from (5) we have D:1 t r -- _ 3 lJ 16 TECHNIQUES OF INTEGRATION SOLUTION: 1 A _ LB u2-a2 u-a' u*a Multiplying by (u - a) (u * a), *e get 1 - A ( u * a ) + B ( u- a ) L: (A+B)u*Aa-Ba Equating coefficients, we have A+B_O Aa-Ba-l Solving simultaneously, we get ^1rh1 A-+ za and B- 2a ' 2Ln :*rnlu- ol I 2a lnlu+al+C (6) -*k 2a ^ ^ 1l u-+ a l '* c lu ol This is also listed as a formula. (7) FRACTIONS. CASES1 AND 2 FUNCTIONS BY PARTIAL 11.4INTEGRATION OF RATIONAL In chemistry, the law of massaction alfords us an application of integration that leads to the use of partial fractions. Under certain conditions it is found that a substance A reacts with a substance B to form a third substance G in such a way that the rate of change of the amount of C is proportional to the product of the amounts of A and B remaining at any given time. Supposethat initially there are a grams of A and B grams of B and that r grams of A combine with s grams of B to form (r * s) grams of C. If we let r be the number of grams of substanceC present at f units of time, then C contains rxl(r * s) grams of A and sxl(r * s) grams of B' The number of grams of substanceA remaining is then q - rxl(r * s), and the number of grams of substAnce B remaining is F- sxl(t * s). Therefore, the law of mass action gives us @ - nu ( ^. - -tt_\/" - --q{-\ A: \d- /+s/\p--l+s/ where K is the constant of proportionality. This equation can be written as dx ,4rt ,^(+ r A:@f (8) a - . * \ ( ' tst - B' - r ) / /\ Letting k__Krs trff a- r*s ' ot b -: * p a s r Eq. (8) becomes dx ii-k(a-r)(b-x) (e) We can seParate the variables in Eq. (9) and get dx (a-r)(b-x) :kdt lf a : b, then the left side of the above equation can be integrated by the power formula. If a + b, partial fractions can be used for the integration. ExAMPr.n3: A chemical reaction causes a substance Ato combine with a substance B to form a substance C so that the law of mass action is obeyed. If in Eq. (9) a- 8 and b - 6, and 2 g of substance C are formed in L0 min, how many grams of C are formed in 15 min? soLUrIoN: Lettin g x grams be the amount of substance C present at t Equation minutes, we have the initial conditions shown in Table 11,.4.'1,. (9) becomes dx fr- k(8 x)(6 x) Separating the variables we have dx (8- x)(6-x) (10) 510 TECHNIQUES OF INTEGRATION Writing the integrand as the sum of partial fractions gives Table11.4.1 t 0 10 15 x 0 2 Xs 1LB @=3_r+ 6_x from which we obtain 1,: A(6-x) +B(8-r) Substituting 6 for x gives B : t, and substituting 8 for x givesA - -+. Hence,Eg. (10) is written as --rJ8--lld* f +lldr_ zJG:k Jdt Integrating, we have *ln 18-xl-*ln 16-xl+*ln lCl:tt , | 5 - n l ::-_ 2 k t rnl@.-)-l 5-x ffi:Qs-21;t substituting x:0, 6-x 8-r t - 0 in Eq. (11) we get c - *. Hence, _ 3 o_zkt 4v Substitutingx : 2, t: (r2) L0 in Eq. (12), we have 6 : te-zort e- 20k: 9€. Substituting )c- xrs,t - L5 into Eq. (lZ), we get 6-fu- 3 n-sok 3{:4, 4(6- x.'")- (11) 3(e-2ok)stz(g- xrr) 24- 4xrs- 3(8)t/,(8- xG) 24- Axrr:ry (8- )crs) 54 - 32\D xrs:W - 2.6 Therefote, there will be 2.6 g of substance C formed in L5 min. -INTEGRATIONRATIONAL CASES 1 AND FUNCTIONSBY PARTIALFRACTIONS. 11.4 OF 511 L1,,4 Exercises In ExercisesL through "l'6,evaluatethe indefinite integral. l"v ,Î' d x J xz_4 (!x-? ax 44 ' [ x 3 - x 2 - 2 x J .t ,' Jr f f i *dzx 10'r d t J * z . [ =x 2**ix d -6 4' J 5I 8I LLI(2x* 3) (x + 1)' dx 1 4 |. " ' =2 x=5 -' x*nI t a * J .,F | 44x3*30x2*52x+17 rD.J 9xa-5x3-LLx2*4x+4 , dx In Exercises17 through 24, evaluatethe definite integral. (x' - 3) dx /Fr.. 5 ' Jf' r f f i 24. t x2 dx o 2x3*9x2+t2x+4 & . F i n d t h e a r e a o ft h e r e g i o n b o u n d e d b y t h e c u r v e y : ( x - l ) l e - S x * 5 ) , t h e r a x i s , a n d t h e l i n e s x : 4 a r t d ' x : 6 ' 25. Find the centroid of the region in Exercise25. y axis. 27. Find the volume of the solid of revolution generated by revolving the region in Exercise25 about the : 4 x' 28. Find the area of the regfrn in the first quadrant bounded by the curve (r + z)'y 29. Find the centroid of the region of Exercise28. 30. Find the volume of the solid of revolution generated if the region in Exercise28 is revolved about the r axis. 31. Suppose in Example 3 that a : 5 and b : 4 and I g of substanceC is formed in 5 min. How many grams of C are formed in 10 min? substanceCisformedin4min'Howlongwillittake2gof subSflSoppor"inExample3 thata:5and,b:3and1gof stance C to be formed? of the amount of the substance 33. At any instant the rate at whictr a substancedissolves is proportional to the product in solution at that instant and substance pr"r"rrt at that instant and the difference between the concentration of the is mixed with 10 lb of salt material insoluble quantity of A the concentration of the substance in a saturated solution. in 10 rnin and the condissolves salt 5 lb of If gal water. bf 20 containing in a tank is dissolving initially, and the salt min? in 20 centration of salt in a saturated rolotio.. is 3 lb/gal, how much salt will dissolve then 34. A particle is moving along a straight line so that if u ftlsec is the velocity of the particle at f sec, t+3 a:F+3,tTr. Find the distance traveled by the particle from.the time when t: 0 to the time when f : 2' OFINTEGRATION 512 TECHNIQUES INTEGRATION OF Case3: The factors of Q (x) are linear and quadratic, and none of the qua1.1..5 RATIONAL FUNCTIONS dratic factors is repeated. BY PARTIAL FRACTIONS. Corresponding to the quadratic f.actor* * px * 4 in the denominator CASES 3 AND 4z is the partial fraction of the form THE DENOMINATOR Ax-t B CONTAINS **px*q QUADRATIC FACTORS soLUrIoN; The fraction in the integrand is written as a sum of partial fractions as follows: x2-2x-3 : ( x - 1 ) ( x z+ 2 x * 2 ) Ax*B , C x+2x+2T x-"1 (1) Multiplying on both sides of (1) by the lowest common denominator, we have * - 2x- 3 : *-2x-3: (Ax * B) (x - 1,)+ C(* * 2x * 2) (A+C)xz+ (B- A-r2C)x+ (zC-B) Equating coefficients of like powers of r gives Al C:7 B-Al2C:-2 2C-B:-3 Solving for A, B, and C, we obtain A:E B:! C:_t Substituting thesevalues into (1.),we get : x2-2x-3 (x-l)(x'*2x+2) I x2-2x-3 (x-l)(x'*2x+2) tx+E * -+ x2+2x*2', Jc-1 dx xdx 9f dx 4f dx ,7f '5 (2) 5 J x2+2x*2 5J x-t J )c2+2x*2 To integratu I@ dx)l(* -l2x*2), we see that the differential of the denominator is 2(x t 7) dx; so we add and subtract 1 in the numerator, thereby giving us 9f xdx 9f (r+1)dx 5Jm:EJW-TJm 9f dx Substituting from (3) into (2) and combining terms, we get (3) FUNCTIONS FRACTIONS, CASES3 AND 4 11.5INTEGRATION OF RATIONAL BY PARTIAL I 513 x2-2x-3 (x - l ) ( x ' * 2 x + 2 9. L f 2@+1) dx 5 dx x-"1, 9 ln l*'+2x+21 10 lr- 1l zJm l*' + 2x* 2l -?tan-1(r* 1,)- # ln lt 1l + + ln C l c ( x z + z x + z ) n l_ ) (x-1)' I f,tan-l(r*L) | a ILLUSTRATION L: In Example 1, we would have saved some steps if instead of (1) we had expressed the original fraction as x2-2x-3 D ( 2 x + 2 )+ E , ' \-rw (x-I)(x'*2x+2) t . -/ F - - x2*2x*2 + x-1 NorE: We write D(2x * 2) + E instead of Ax * B because 2x*2:D,(**2x*2) Then, solving for D, E, and F, we obtain D: * E=-3 F:-t giving us (4) directly. . Case4: The factors of Q(r) are linear and quadratic, and some of the quadratic factors are repeated. If. x2* px * q is an n-fold quadratic factor of QG) ' then, corresponding to this factor (f + px * q)", we have the sum of the following n partial fractions: Arxi-B, Azx*Bz - + @TV|TT'(x'+px*q)"-' . . - Anx*Bn f*px*q -5x*2)3, o rLLUSrRArroN2: If the denominator contains the factor (f we have, corresponding to this factor, Ax* B , Cx*D @r@ or, more conveniently, A(2x-5) +B -C(Zx-5) *D * E(2r-5) *F ( i , - 5 r * 2 ) t ' ( x ,- 5 x + 2 ) ' ' x 2- 5 x + 2 SOLUTION: x-2 x(x'-4x+5)' A , B(2x-4)+C =;+ffi+ D(Zx-4)+E x:2-4x*5 TECHNIQUES OF INTEGRATION Multiplying on both sidesof (5) by the lowestcommondenominator, we have x - 2 : A(f - 4x * S)2* x(2Bx- 48 + C) + x(xz- 4x * 5) (2Dx- 4D + E) x - 2 = Af + 16A* + 25A- 9Af + tlA* - 40Ax* 28* - 4Bx* Cx * 2Dxa- l2Dxs* Ers* 26D* - 4Ef - 20Dx* SEx x - 2 = ( A + 2 D ) f + ( - 8 A - 7 2 D * E ) x a+ ( 2 6 A + z B+ z 6 D - 4 E ) * + (-40A- 48 + C - 20D* SE)r + 25A Equatingcoefficientsand solving simultaneously,we obtain A:-& B:* C:* D:* E:-* Therefore, (x-2\ dx | I ,@=a*g _ ![ , ( ? x - + )a x = + L [ _ _ _ _ _ d r _ _ - 1 f L 2 x - 4 )d x 2 sJ[ @x*' 5 2 sJ F - + x + s J {P=E+W-E J At=7;*5y' dx - E4 f :-z JE=E+s :-E ) r nlr l- 1 ,lr +i 1 flr Jl1g= .g;iqTp +litn lx2- 4x +51 _+l we evaluate separatelythe integrals in the third and right side of (5). tdxfdx lwffir:Jffi - 2 - tan 0, where 0 dx: s e c z0 d 0 a n d ( x dx [(r- 2)'+ L]' 2 ) ' + r _ t a n 2 e + 1 , : s e c 20 . H e n c e , 515 FRACTIONS. CASES3 AND 4 FUNCTIONS BY PARTIAL OF RATIONAL 11.5INTEGRATION A1 :;+|sinecos0*C1 - 2)' We find sin 0 Becausetan 0 : x - 2 and -tr < 0 < tr, 0 : tan-l(r and cos g from Figs.11.5.1(if. x > 2) and 11.5.2(if x < 2).In either case x)2 v -+ s i n0 vr-r F i g u r e11 . 5 . 1 l/xz - 4x + 5 a n d c o s0 - ffi So we have 1 1 d, tan-'tr- 2)'r'ffi rlz:i + 2)z J I(r I x-2 ' L 7Et=fi77 + cr Thus, ' dx I lT6zr+W:|tan-'(x-2) x-2 *'" ', +Z@-4r*5) V) Now, consideringthe other integralon the right side of.(6),we have fdxldx Jffi:tan-l(r-2) I @ffiT|: (8) +cz Substitutingfrom (7) and (8) into (5), we get f x<z (x-2) dx JWTW F i g ur e 1 1. 5 . 2 5(f r - ll *t'l- 4 x * 5 1 : ; i r1 n Exercises71.5 In Exercises L through L6, evaluate the ',[dx r' J 2x3*x 2. c.Jffi 6. 1 - x - '22 2)+ tan-'(x fu ffi*rtat + f r" l*'- 4x*sl - +tan-l(x- z)+ c - 5 36 taaan- - -r((lxy --r2\ +) + f f i ; ax s- 4f .-+r?\ - TECHNIQUEO S F INTEGRATION 516 9. 13. [ (2*-x+2) dx J x'+zx'+x IJ (4ftsdx + e), (2t!_9*) ,r 0u,' J[ (,l +. 3 x l - 2 x + d3 *) ,trL. ' J[ " Iffi## " /ee;f$f;:e In Exercises17 through 24, evaluate the definite integral. 17. fg+#4 20. ft 9 dx (xr+zx-t) dx zTxs-t f dx (* + 3x + 3) dx 23. ft J" r"+*+x+l " I#h n.l"fffla- "' rol',,*rIo1,*, (x -t l) dx ,n ['t, "' Jo r - | "' J-,(2r -r2x* t)z J, 8'+l d* 1 L 2z ' [l 1 t z * r 1 t fnt2 cosx dx za'J,,usinr+sinsr 25. Find the areaof the region bounded by the curve y:41(xs * 8), the r axis, the y axis, and the line x:2. 26. Find the abscissaof the centroid of the region in Exercise 25. 27. Find the mass of a lamina in the shape of the region of Exercise25 if the area density at any point (x,y) iskf slugslfP. 28. Find the volume of the solid of revolution generated by revolving the region of Exercise25 about the y axis. 29. A particle is moving along a straight line so that if v ftlsec is the velocity of the particle at I sec, then a- t2-t+l (f+2)r(t2+L) Find a formula for the distance traveled by the particle from the time when f : 0 to the time when f : fr. 11..6INTEGRATION OF If an integrand is a rational function of sin x and cos x, Ttcan be reduced to RATIONAL FUNCTIONS a rational function of z by the substitution oF SINE AND COSINE z_ hntx (1) c o s zV - 1 and letting y - +x, we have Using the identity cos 2y : 2 c o sx - 2 c o s 2 L r x -1 - 2.24 -1 sec2+r 1+tanzLx r I*zz I So . (2) In a similar manner, from the identity sin 2y: 2 sin y cosy, we have s i n ) c - 2 s i n t r c o sl x - 2 -2 sin *r cos'tx cos+x , r- tan - --- tt z- 1 sec, tx --- - Z t a n t x 1 + tanz tx \ OF RATIONALFUNCTIONSOF SINE AND COSINE 11.6 INTEGRATION 517 Thus, (3) Becausez-tani,x, dz- * sec2tx dx - +(1 + tan2tx) dx Hence, (4) Find soLurroN: Letting dx l-sin x*cos )c I dx 1-sinxlcosr tan ix and using (2) , (3), and (4) , we obtain :T 1-RT -n I I 2dz t*22 2z ,1-z' L+22 dz r *+-z2 '2 ) - 2 2 * ( I - z ' ) / (1 - 2 l d zdz 2 I 22 - t22z --l lz dz L 1 --zz 1n l n l L1-- r lz l + C tan Lxl+ C ln ln llr1-; te ExAMPtn 2: Find r secxdx I J by using the substitution of this section. s o , , u r l o Nl,s e cx d x : l + c o s x J J using (2) and (4), we get and tan z: Letting tx I A*:l+.!+,i_2t 1'-z'-'J J c o si - J L + 7 + t-22 "!'"1'.4, we have Using formula (7) from Sec. f dz ,11*zl *' 2Jft-rnlffil Therefore, r r d x - r n1l + t a n ? r*l. Jsec' tt-tan#l Equation (5) can be written in another form by noting that 1' tan *n 7 TECHNIQUEO S F INTEGRATION andusing the identity tan(a * B) : (tan a * tan illG we have -tan d tan F). So ta4 *"t,* tat *r,-l +' c f ,"" r dx: ln | = - tan r tl. *n . tan|xl or, equivalently, r IJ q*qr dr - ln lt*(*o + *r)l + C (6) Formula (U) ,, un alternative form of the formula I I secr dx:ln lsecr * tan rl * C *nr"n is obtained by the trick of multiplying the numerator and denominator of the integrand by sec r * tan r. It is worth noting that still another form for J sec x dx canbe obtained as follows: f r l secrdx: l J dx J "o--:J Substituting u : sin x, du: f c-costt:J osxdx f cosxdx l=;fit; cosx dx, we have ' lt+ul *a:hrll+ti"tl"'*c A r : [ -,d- uu r : t t t1l T -l JI r " . r J tr_sm.rl Because-l. < sin r < l for allx,1.*sin rand 1- sin rarenonnegative; hence, the absolute-value bars can be removed and we have | /1*sinr\t/r, secx dx: tn (i:-ffi/ J /i+sinr * c- _:1 _ln 11=16; + c^ (7) Exercises11.6 In Exercises 1 through'1,2, evaluate the indefinite integral. ,r ' [--4Js+4costr J. dx ,..' r 2 l3-fu dx costr-sinx*L o. 8dx 3 c o s2 x * L fdxldx l- J sinr*tanx s.[-Jro* J 6+4secx J tanx-l 1 0 .I d* dx J sinr-tanr 2 s i n x * 2 c o sx * 3 In Exercises L3 through L8, evaluate the definite integral. frtz 1 3 .1 Jo fntz 1 , 6 .| Jo dx 5. *s i n x * 3 dx 3 " _+ c o s 2 x fnta - 8 dx 1 4 .| ro tanx+j. frts 3 dx IT. I Jnt62sin2x*"1, 4I 8I 12I dx cot 2x(1 - cos 2r) 1 1 . 7 M I S C E L L A N E O USSU B S T I T U T I O N S 5 1 9 19. Show that formula (7) of this section is equivalent to the formula J sec r dx: lnlsec r f tan rl * C. (xrNt Multiply the numerator and denominator under the radical sign by (1 + sin r).) 20. By using the substitution z : tam*r, prove that f / 1- c o s r J c s cr d x : l n V 1 a . * ; +c lnlcsc r - cot xl + C. (nrr.rr:Use a method 21. Show that the result in Exercise20 is equivalent to the formula J cscx dr: similar to that suggestedin the hint for Exercise 19.) IT.7 MISCELLANEOUS If an integrand involves fractional powers of a variable r, the integrand SUBSTITUTIONS can be simplified by the substitution x: zn where n is the lowest common denominator of the denominators of the exponents. This is illustrated by the following example. z 5 tdz. S soLUTroN: we let r : z6; then dxt c : 625 f )crtzdx J].*,.:J Dividing - =oL$, l t zt(6zt dz) - t 28 3-, ;dz +1 J z 2+ inator, IO mrin the numerator by th ' d €en( z 2-_1 1 + iz'.3 - - 1 z2+L nave la, z * tan- ,) +C 6 ry7t6 - grst6 + 2XU2- 5Xrt6+ 5r tar fr 7* 5 xrt6+ C No general rule can be given to determine a substitution that will result in a simpler integrand. Sometimes a substitution which does not rationalize the given integrand may still result in a simpler integrand. EXAMPLI. 2: EVAIUATE f I xit/xz+4 dx J SOL Let z- \m. Then x2+ 4, and 2z dz - 2x dx. So we hav rrtffidx fr | (*1r\Eq JJ x dx- I e, - 4)zz(zdz) f -!.622) _ _ dz izz - Ez, + #zt + C I ku gza* J + 5601+ C #ztll1za -'1,6822 * 4) + 5601+ C + 4)2 - 1,68(x2 T+5(x, + 4)srz715(x2 - 48x2+ L28l + C r+E(x' * 4)srzSLlxa 520 TECHNIQUESOF INTEGRATION ExAMPLE3: Evaluate ld* Jxzm sot,urloN: If x - tlz, then dx: -dzlzz. We have, then, -dz d* I -[ J*2ffi by using the reciprocalsubstitution x - Uz. ,' J/_1 \ tre; r \;r) \,17*Z- ifz>0 if z < 0 f l: 1 f f : + . {?l - 2 2 + 6 ) d z , ^ f zdz 2JV27*62-22 J\/27*62-22 dz J\/27*62-z2 :-z-J 1ffifdz f -dz ffi+" "J ffi + 3 s i n - , +b + C 1-g l__ 5 1, \ 1 2+7| - t + 3 ffiLF) - s i n -+r + TJ ffi..r c SIN +JSln Substituting from (2) into (1) we get Id* Jyzffi --r' ^" 1 , - 3 x . 6i*+c if x <o 11.8 THE TRAPEZOIDALRULE ffi 521 lfx>0 rfx<0 11,7 Exercises In Exercises L through t6, evaluate the indefinite integral. r4 .f J f f i ; d x ,l'Jffi o dx ,- f. J f f i dx fdx 72-Jffi Use the reciprocal substitution x: Ilz. t3.DoExercise1,2byusingthesubstitutionffi:z-x. r o .f Jd xf f i In Exercises 17 through 22, evaluate the definite integral. d* 1 Lnl . Jl no f f i f18dx 20' J,uffi 1.1..8THE TRAPEZOTDAL RULE 1 9 .f2J , , , @dx f2 dx /L'L' J, xlx2 * 4x 4 ,.tt | - We have seen previously how many problems can be solved by evaluating definite integrals. In evaluating a definite integral by the fundamental theorem of the calculus,it is necessaryto find an indefinite integral (or antiderivative). There are many functions for which there is no known method for finding an indefinite integral. However, if a function f is.continuous on a closed interval fa,bl, the definite integral I'"f(x) dr exists and is a unique number. In this section and the next we leam two methods value of a definite integral. These methods for computit g "tt "pptoximate can often give us fairly good accuracyand can be used for evaluating a definite integral by electronic computers. The first method is known as the trapezoidalrule. TECHNIQUEO S F INTEGRATION We know that if f is continuous on the closed interval [a, b], the definite integral Io"f(x) dr is the limit of a Riemann sum fbn I f (x) dx: lim ). fG) Lix Ja l l a l l *o 7 : r The geometric interpretation of the Riemann sum n fG) Lix i:l is that it is equal to the sum of the measuresof the areasof the rectangles lying above the r axis plus the negative of the measuresof the areasofthe rectangleslying below the.r axis (seeFig.7.3.3). To approximate the measure of the area of a region let us use trapezoids instead of rectangles. Let us also use regular partitions and function values at equally spaced points. Thus, if we are considering the definite integral Io,f (x) ilx,we divide the interval la, bl into n subintervals, each of length Ax: (b - a) ln. This gives the following (n + 1.) points: xo: a, xr: a * Lx, xr: a * 2 Ax, . , xt: a* i Lx, . . r xn-r: a* (n-l) Ax,xn:b. Then the definite integral I2 f @) dx may be expressedas the sum of z definite integrals as follows: I: f(x) dx f(x) f(x) dx + . . . + [:' f@) dx*. . . + f f@ ) dx ( 1) Jrr_r Jrn_, To interpret (1) geometrically, refer to Fig. 11..8.i.,in which we have takenf(x) > 0forall r in la, bh however, (1) holds for anyfunction which is continuous on la, bf. f(x) Figur1 e1.8.1 Then the integral I t' f @) dx is the measure of the area of the region bounded by the x axis, the lines x : a a n d x : x 1 r a n d the portion of the 11 . 8 T H E T R A P E Z O I D ARL U L E curve from P0to P1.This integral may be approximated by the measure of the area of the trapezoid formed by the lines r : a, x: x1, P6P1,and the x axis. By a formula from geometry, the measureof the areaof 'this trapezoid is iLf@,)+ f(x)l Lx Similarly, the other integrals on the right side of (1) may be approximated by the measure of the area of a trapezoid. Using the symbol "-" fot "is approximately equal to," we have then for the ith integral ftt (*) ilx - tlf (x;,) + f (x,)l Lx J",_,f (2) So, using (2) for each of the integrals on the right side of (1) , we have fb J,f{*) Thus, d x = E l f ( x )+ / ( r ' ) l L x + i [ f @ ' ) + / ( r , )A] r * ' ' + tlf(x"-,) + f (x"-')fLx+ tlf Q"-r)* /(r")l Ar ,r, Formula (3) is known as the trapezoidalrule. ExAMPLE 1: Compute Pdx Joffi by using the trapezoidal rule with n: 6. Expressthe result to three decimal places.Check by finding the exactvalue of the definite integral. Because la, bl - [0, 3] and solurroN: 6, we have ,o:1:0.5 n6 Lx:b Therefore, f' d* Jo LG*x2 - + + 2f(xr)+ f (x)l [/(ro)+zf(x,)* 2f(xr)+ Zf(xr)+ 2f(xn) where f (x) - Il(16 + x').The computationof the sum in brackets in the aboveis shown in Table11.8.1.So, [' j' Jo 6fu: 0'25(0'5427) I'J* Q'l.607 Jorcfu: Rounding the result off to three decimal places, we get I'J' Jorcfu: Q"l'6! We check by finding the exactvalue. We have It d* 1 . _^-1rlt J of f i : a t a n ' Z l o TECHNIQUEO S F INTEGRATION : + tan-l *- * tan-l 0 - +(0) : +(0.6435) - Q.1,6A9 to four decimal places Table11.8.1 f (x,) X1 0 1 2 3 4 5 6 0 0.5 L L.5 2 2.5 3 0.0625 0.0615 0.0588 0.0548 0.0500 0.0450 0.0400 1, 2 2 2 2 2 1, I o.o62si 0.L230 I 0.LL76 I 0."1.096 0.1000 0.0900 0.0400 I - 0.6427 io'r@') i :0 I To consider the accuracyof the approximation of a definite integral by the trapezoidal rule, we prove first that as Ar approaches zero and.n increases without bound, the limit of the approximation by the trapezoidal rule is the exact value of the definite integral. Let T: i Lx[f(x) + zf(x) + . . . -r 2f(x,_,)+ f(x")] Then T :l f( xr ) + f( x) + . . . + f( x") l A^x+ilf( x)- f( x,) l Lx or, equivalently, n r: ) f(x) ax + ilf@) - f(b)l Ax i=1 Therefore, if. n -->*o and At -+ 0, we have lim ttf@) - f(b)) Lx ["] r: aa-o !* i 'f@1)ax* a*_o ?r Aa-o fb : I f @ )d x + 0 Ja' Thus, we can make the difference between Tand the value of the definite integral as small as we please by taking n sufficiently large (and consequently Ar sufficiently small). The following theorem, which is proved in advanced calculus, gives us a method for estimating the error obtained when using the trapez6idal rule. The error is denoted by e7. 11.8 THE TRAPEZOIDAL RULE 1,L.8.LTheorem Let thre funct: Iu rcti10n f 1 be continuous on the closedinterval fa, bl, and f ' andf " )n both exisl ist or 1 [ a , bb l . If on fbb tl f(' ( x ) €yT - d: dl xx'- T ,J/a where3 TI'i is tl he aP rro ximate value of [f f (x) dx found by the trapezoidal th PPI rule, th en threre is3 Ssor then ( me number 4 in la, bl such that €y EXAMPug 2: Find the bounds for the error in the result of Example L. __ 1 t2 ,(b at ) f '" (q) (Ar)' (4) soLUrIoN: We first find the absolute minimum and absolute maximum values of f" (x) on [0, 3]. f ( x ) : ( 1 0+ l ) - t '(x) : -2x(16 * f1-z f - 2(LG+ f):2: 10* - g2)(L6-t *1-t f,,(x):B*(1.6 * f;-s -6x(6xz - 32)(1.6* *1-+ * L2x(1'6* l; -r f "' (x) : :24x(L5- xz)(L6* *1-+ Becausef"'(x) > 0 for all x in the open interval (0, 3), then /"'is increasing on the open interval (0,3). Therefore, the absolute minimum value of f " on [0, 3] is f " (0), and the absolute maximum value of f " on [0, 3] is f " (3). f"(il --h andf"\3)-& T a k i n g n - 0 o n t h e right side of (4), w€ get 3 ( r\L _1 -:r,.\-ru)z 2048 Taking n 3 on the right side of (4), w€ have \L - 3122 L2 \E2s )Z 11 45,000 Therefore, if e1 is the error in the result of Example L, we conclude 1L.8 Exercises In Exercises 1 through 14, compute the approximate value of the given definite integral by the trapezoidal rule for the indicated value of z. Express the result to three decimal places.In ExercisesI through 8, find the exact value of the definite integral and compare the result with the approximation. TECHNIQUESOF INTEGRATION 526 1. ['4,n:S Jr )c' 4. 7. r2 Jrx!4-f fr J, z . f " Jt *lx ' , n : 8 3. ?rrdx 5 . J,ffi;n:5 6. J, dx;n-8 sinr dx;n-5 8. fn J, ln(l * x') d x ;n - 4 \R dx;n-6 tr cos x2 dx; n: LL. [' ," dx; nJo 14. fr J. 4 5 g. r2 Jo*"dx;n-4 f3 J, V l + x 2d x ; n - 6 sin r dx; n [37rtz x Jnn 6 1 . 2f." + i " d x ; n - 6 L*x Jo VI + xsdx;n: 4 In Exercises15 through 20, find. the bounds for the error in the approximation of the indicated exercises. 15. Exercise1 15. Exercise2 12. Exercise3 18. Exercise6 19. Exercise 1.1 20. Exercise L0 21. The integral Io' e-" dx is very important in mathematical statistics. It is called a "probability integral" and it cannot be evaluated exactly in terms of elementary functions. Use the trapezoidal rule with n :5 to find an approximate value and express the result to three decimal places. 22. The region bounded by the curve whose equation is y : g-ttz, the r axis, the y axis, and the line r: 2 is revolved about. the r axis. Find the volume of the solid of revolution generated. Approdmaie the definite integral by the trapezoidal rule to three decimal places, with z :5. 23' Show that the exact value of l& t/+ - f dx ig z. Approximate the definite integral by the trapezoidal rule to three decimal places,with z:8, and compare the value so obtained with the exact value. 24. Show that the exactvalue of.t losdxl (x * 1) is ln 2. Approximate the definite integral by the trapezoidal rule with n : 6 to three decimal places, and compare the value so obtained with the exact value of ln2 as given in a table. 11'9 SIMPSoN'S RULE 1L.9.LTheorem Another method for approximating the value of a definite integral is provided by Simpson'srule (sometimesreferred to as the parabotiirute). For a given partition of the dosed interval [a, bl, simpson,s rule rr,r"liy gives a better approximation than the trapezoidal rule. Howevei, simpson's rule requires more effort to apply. tn the trapezoidal rule, successive points on the graph of y: /(r) are connected by segments of straight lines, whereas in simpson's rule the points are conhectJaby ,"gments of parabolas. Before simpson's rule is developed, we statl an--d pnove a theorem which will be needed. If.Po(xs,Ao),Pr(xr, U), and pr(xr, Ar\ are three noncollinearpoints on the parabola having the equation y: A* + Bx * C, where Ao2 0, Ar > 0, Az > 0, xr : xo * h, and xz : xo * 2h, then the measure of the area of the region bounded by the parabola, the r axis, and the lines l: xo arrdx: x, is given by *h(yo*4yr*yr) (1) 11 . 9 S I M P S O N 'R SU L E ( x o ,y o ) 527 pRooF: The parabola whose equation is y : Ax2 * Bx * C has a vertical 'axis. Refer to Fig. 1'1..9.'1., which shows the region bounded by the parabola, the x axis,and the lines r : ro afld x: xz. BecausePo, Pr, and P, are points on the parabola, their coordinates satisfy the equation of the parabola. So when we replace xlby xs * h, and xrby xs*2h,wehave yr: Axoz* Bxs-l C lr : A(xo + h)2 + B (16* h) + C : A(xs2I 2hxs* h2) + B (xo+ h) + C U z : A ( x o + 2 h ) z* B ( x s + 2 h ) + C : A ( x n z* 4 h x s +4 h 2 )* B ( x e+ 2 h ) + C Therefore, Ao* 4yr* yr- F i g u r e11. 9 . 1 A ( 6 x o ' * 1 2 h x 0+ t h ' ) + B ( 6 x o + 6 h ) + 6 C (2) Now if K square units is the area of the region, then K can be computed by the limit of a Riemann sum, and we have K- lim Atr-0 n sl Ll i:l 2h (At,'* Bt,+ c) L,x (Ax' * Bx + C) dx - $Ax3+ LBx' + - +A(ro* 2h)' 2h)2+ C(ro + 2h) - (*Axr' + +Bxo'* Cxs) - +hlA(6xo'* L2hx0+ th2) + B(6xo+ 6h) + 6Cl (3) Substituting from (2) in (3), we get K- *hlyot 4y, * yrl I Let the function f be continuous on the closed interval [4, b]. Consider a regular partition of the interval la, bl of 2n subintervals (2n is used instead of z becausewe want an even number of subintervals). The length of each subinterval is given by Ar : (b - a) l2n. Let the points on the curve y: f(x) having these partitioning points as abscissasbe denoted by Po(xo,yr), Pt(xt, yr), , Pro(xzn,!zn)) see Fig. 77.9.2,where /(r) = O for all x in fa, bl. We approximate the segmentof the ctfivey : f (x) from Psto P2by the segment of the parabola with its vertical axis through Pr, Pr, and Pr. Then by Theorem 11.9.1the measureof the area of the region bounded by this parabola,the r axis, and the lines r:ro and x: xz,withh: Lx, is given by $ Ar(yo * 4y, * yr) or * Ar[/(ro) + af k) + f (xr)] In a similar manner, we approximate the segment of the curve 528 T E C H N I Q U EO S F INTEGRATION N I N n- A:Xo.Xr X2 i Xs X+ X5 X6 N F 1 I \ N HH F i g ur e 1 1. 9 . 2 , + f1*2n L N y : f (x) from P" to Paby the segment of the parabola with its vertical axis through P2,Ps,and Pn.The measure of the area of the region bounded by this parabola, the x axis, and the lines r: xs and x: xt is given by i Lx(yr* 4ys* y,) or t Axff(xr)+ 4f(xr)+ f(x)l This processis continued until we have n such regions, and the measureof the area of the last region is given by * Axlyro-r*4ar,-'* ar) or t Lx lf(xr^) + +1@r*_r) + f(xr*)l The sum of the measuresof the areas of these regions approximates the measure of the area of the region bounded by thelurv" *io"u uq,rution is y : f(x), the r axis, and the lines r : a andx: b. The measureof the area of this region is given by the definite integral ft fG) dx. so we have as an approximation to the definite integral t Ax[f(x)+af@)+f(x,)l+ t Axlf(x,) +4f(x")+f(x)]+ . . . + * AxLf@zn-) t 4f(x,,-r)+ f(xr^-r)l+g axlf(xr^_r) + 4f(x2,) + f(xr^)l dx -: * Arlf (xo) + a f Q , ) + z f ( x r ) + a f f t s ) + Z f (+ x .) + 2f(xzn_z) + 4f(xrn_J+ f Gill where Lx - (b - a)l\n. Formula (4) is known as Simpson,srule. ExAMPLEL: Use Simpson,s rule to approximate the value of soLUrIoN' *, and Applying Simpson's rule with 2n: 4, we have Ar : +(1 - 1, 12 lf (xi + af@r)+ 2f(xr)+ af@s)+ f (x)l . . G) RULE 1 1 . 9S I M P S O N ' S with 2n: 4. Give the result to four decimal places. The computation of the expression in brackets on the right side of (5) where f(x): U(x + l)' is shown in Table 1'!'.9.'l', Substituting the sum from Table 11'9'1in (5), we get : + : Q.69325+ (8.31905) T a b l e 1 7 . 9. 1 xi 0 t 2 3 41 kt f@) 0 0.25 0.5 0.75 L.00000 0.80000 0.66567 0.57143 0.50000 4 kr'f@i) | L.00000 3.20000 4 L-33334 2 2.28572 4 10.50000 k,f(x) - 8.31e06 i :0 Rounding off the result to four decimal places gives us F dy: J, x+t Q v '.vG' vg g g The exactvalue of y; dxl(x + 1) is found as follows: I:h:'n l r + t l ] , : r n 2 - l n 1: r n 2 From a table of natural logarithms, the value of ln 2 to four decimal places is 0.6931,which agreeswith our aPProximllion in the first three -0.0002' plu."t. And the elror in our aPProximationis In applying simpson,s rule, the larger we take the value of.Zn, the smaller will be the vaiue of. Lx,and so geometrically it seemsevident that the greater will be the accuracyof the approximation, becausea parabola p"rJttg through three points of a curve that are close to each other will be close to the curve throughout the subinterval of width 2 Ar' The following theorem, which is proved in advanced calculus, gives a method for determining the error in applying Simpson's rule' The error is denoted by es. 11.9.2Theorem ' Let the function f be continuous on the closed interval la, bl, and f , f " , ftt' and fcv)all exist on la, bl- lf It fb €s: I f@)dx-S Ja 530 TECHNIQUEO S F INTEGRATION lvhere s is the approximatevalue of [! f (x) dr found by simpson'srule, then there is somenumber 11in fa, bl such that €,s: -rto (b - a)f Gv\ (il (Ax)n EXAMPLN2: Find the bounds for (6) SOLUTION: the effor in Example1. f(x): (x* 1)-' f'(x) f " ( x ) 2 ( x* 1 ) - ' f "' (r) : -5(x * t;-+ f ( i v )( r ) : 2 4 ( x * t ; - s (x) : - 120(x * 1) -u 1<vt _ BecausefF)(x) < 0 for all r in [0,11, fiut is decreasingon [0, 1]. Thus, the absolute minimum value of /(tu) is at the right endpoini 1, and the absolute maximum value of .f(tu)on [0, 1] is at the left endpoint 0. f(iv)(o):24 and /(iv)(l) : * Substituting 0 for q in the right side of.(6), we get -rto (b - a)/rivl(O)(Ar)a :-t+u Q4) (+)^: -r.#o : -0.00052 Substituting 1. for rl in the right side of (6), we have - 1 ( . - -- - \ . r ( i v ) l r \ / ^ - - \ 4 180@ a ) f t v ) ( 1 )( A r ) 4 - - So we concludethat -0.00052=€s <-0.00002 1' m _11_l\ a\z/ (7) The inequ ality (7) agrees with the discussion in Example L regarding the error in the approximation of [or dxl (r + 1) by Simpson's rule because -0.00052 If f (x) is a polynomial of degree three or less, then ltiv)(r) = 0 and therefore€s: 0. In other words, simpson's rule gives an exactresult for a polynomial of the third degree or lower. This statement is geometrically obvious if f(x) is of the second or first degree becausein the first casethe graph of. y : f(x) is a parabola, and in the second case the graph is a straight line (a degenerateparabola). _ We apply Simpson's rule to the definite integral Il fe) dx, where f(-x) is a third-degree polynomial, and take 2n:2. Here'simpson,s rule gives an exactvalue for the definite integral. Because2n:2,-xs: a, x1: SULE 11. 9 S I M P S O N ' R 531 i @ + b), and x2: b; and A,x- + (b - a) . So we have I:f(x)dx-+ lro +4f(+) * f@f Formula (8) is known as the prismoidalformula. In Sec.8.4, we leamed that Il f Q) dx yields the measure of the volume of a solid for which /(r) representsthe measure of the area of a plane section formed by a plane perpendicular to the r axis at a distance r units from the origin. Therefore, lf f (x) is a polynomial of the third degree or less,the prismoidal formula gives the exactmeasureof the volume of such a solid. EXAMPLE3: Find the volunie of a right-circular cone of height h and base radius r by the prismoidal formula. soLUrIoN: Figure 11.9.3illustrates a right-circular cone having its vertex at the origin and its axis along the positive side of the r axis. The area of the plane section fi units from the origin is zr[g(fr)]'zsquareunits, where 8(x) : (rlh)x, which is obtained from an equation of line OA in the xy plane. So if. V cubic units is the volume of the right-circular cone, then lim llallo - 7:r (fi,8(f i )) nlg(r) I2 dx \_/ L iX \_---_. ,, \r - \i-l ,, B nf h2 F i g u r e1 1 . 9 . 3 dx Evaluating t! x2 dx by the prismoidal formula (8) with f(x): a:0, andi@ + b) : th, we have f , b:h, :t,r"h v: $ .I o * n'T * h'): # (2h') 71.9 Exercises In Exercises 1 through 6, approximate the definite integral by Simpson's rule, using the indicated value of 22. Express the answers to three decimal places. Also, find the exact value of the definite integral and compare the results. 2 2n t*.t2n-8 2n-4 6. | ,ir, ' 2n: 6 TECHNIQUES OF INTEGRATION In Exercises 7 throu gh 9, find bounds for the error in the indicated exercise. 7. ExerciseL 8. Exercise2 9. Exercise 5 Each of the definite integrals in Exercises 10 through 15 cannot be evaluated exactly in terms of elementary functions. use Simpson's rule, with the indicated value of 2n, to find an aPProximate value of the given definite integral. Express results to three decimal places. f nl2 Vsin r dx; 2n - G 10. I Jo f2 13. I e-" dx; 2n - 4 Jo In Exercises15 through 19, use the prismoidal formula to find the exact volume of the given solid. 16. A sphere of radius r. 17. A right-circular cylinder of height h and base radius r. 18. A right pyramid of height ft and base a square having a side of length s. 19. A frustrum of a right-circular cone of height ft, and base radii r, and.rr. 20' Show that the exact value of 4 lor \/T=7 dx is n. Then use Simpson's rule with 2n:6to get an approximate value of n tott/l=E dr to three decimii places. Compare the results. 21. Find an approximate value to four decimal places of the definite integral ,[f,/r logro cosx dx, (a) by the prismoidal for_ mula; (b) by simpson's rule, taking ax: #n; (c) by the trapezoidal ,-"r", iin"fi, : io. 22- Find' the area of the region enclosed by the loop of the curve whose equation is y2 : 8x2- x5.Evaluate the definite integral by Simpson,s rule, with 2n:9. ExpresJthe result to three decimal places. ReaiewExercises (Chapter11) In Exercises I through 52, evaluate the indefinite integral. L. I tun'4x cosa 4x dx 4t4 J x2\/a2* x2 T. [ "or' 10t! J ly tx dx +1 13I# 16.I 2. fsL.- 3 a* xs_x 5. r I 8. [ffi J Yx*L 11. r I +t -L ,sI#* a- sin r sin 3x dx f d* \/4- tan-l tfr Ax 14. tVzt - szdt J J \/e, _ 1 314 J J tr[ffib, 2 0J X[ 'Y * e" 6Ih sI#0. 12. .o, g cos20d0 / 15. (sec3x * csc3r)2dx I 18. x"er,dx I 2t. / sir,t 3r cosz3x dx R E V I E WE X E R C I S E S 23I# x3 cos x2 dx 26Iffi^ 28tM du 31I sinb nx dx 34I 5 + 4dxs e c x sin x dx 37I T:r-ffi 40I ffi dx 43I 46I 4x2*x-2 f sin2 2t dt e* dx I+ - 9e2' cot x dx 3 + 2 sin'x 4sIs i n x -dx 2 c s cx dx s5t s8I 2 * 2 s dx * c o s x 61,. t x n l n x dx 1nI T 27. 2s|ffi* 32I# 35I 30. 33. * 38I# 4rIh 44I# 47. l* dx x3-5x2*8x-4 s i n - 1x d x r ln(sinx) dx 50. /.o, s3IY* 36. I I I I I I I I I ett2cos2t dt \6-a , 7t) cscs x dx x5 dx (x' - a.')t 39. tffiat 42. -dt 45. cotz 3x csca3x dx 48. 51. \tr-1 a. yt +L dx xffi cos3t dt sin3fffi 54. dx s6. /r"Q2tL) 57. se. ItffiAx 60. xdx,n>0 52. Irur,"rseca In Exercises 53 through 94, evaluate the definite integral. rr tT- 64'Jr',rl#o* GT f+ Jo \/4+t2 70Iffi 73' 7nl8 2YdY J,,rrcots 5sr@2 xa2x* x * 4 58r 71,. r 74r , l, sins t cos3tdt l4 seca x dx (2" * x2) dx TECHNIQUES OF INTEGRATION 75.["''wry {ax Jo xB dx \ffi 7- 76. 79. t; (ln r) 2 dx 77. logro tE 80. Ax f!n12 xe" cos x2 dx 83. fuz 2x dx S5.J,ffi gg. 85. dx [7'tr2 cosa 3r Jo g1,.l"'' I; x3lx B l,* 82. I Jo (2x' - 2x + 1) dx 89. d, 92. J o 1 2 + 1 3 c o sf fn r Isin x- cosrl dx T lcossrl dx I,', l,'@ay *s1fg 4* I (rrD\m dr ,. ""i n * f 94. | /'=' at Jrn,, L + cos Lzt In Exercises95 and 95, find an approximate value-for the given integral by using (a) the trapezoidal rule, and (b) simpson,s rule. Take four subintervals and express the results to thiee decimir places. fsts 96. I Vl*x7dx JT 97. Find the length of the arc of the parabola yz :6r from x: 6 Io x: 12. 98. Find the area of the region bounded by the curve y: sin-l 2x, theline x: *lVI,, andthe r axis. 99. Find the centroid of the solid of revolution obtained by revolving about the r axis the region bounded by the curve y : e-t, the coordinate axes, and the line r : 1. 100. Find the centroid of the solid obtained by revolving the region of Exercise99 about the y axis. 1 0 1 .The vertical end of a water trough is 3 ft wide_atthe top, 2 ft deep,and has the form of the region bounded by the one arch of the curve-y: 2 sin tm. rfthe trough is fulli water, find rhe force due to liquid pressure on ffi:il:t 1,02.The linear density of a rod 3 ft long at a point x ft fuomone end is ke-s, slugs/ft. Find the mass and center of mass of the rod. 103. Find the centroid of the region bounded by the curve y : x ln x, the x axis, and the line x: e. 104. Find the area of the region enclosedby one loop of the curve * : yo(l _ yr). 105' Two chemicals A and B react to form a chemical C and the rate of change of the amount of c is proportional to the product of the amounts of A and B remaining at any given time. Initialiy there are 50 lb of chemical A and 50 lb of dremical B, and to form 5lb of c,3lb of A and zb oin are required. Aiter t hr, 1.5lbof careformed. (a)If xlbof C are formed at f hr, find an expression for r in terms of t. (b) Fini the amount of C after 3 hr. 106' A tank is in the shape of the solid of revolution f:*d by rotating about the x axis the region bounded by the curve ! : h x, the r axis, and the lines r : e and x: e2.rf.the tank is ruu or water, find the work done in pumping all the water to the top of the tank. Distance is measured in feet. Take the positive x axis vertically downward. Hyperbolicfunctions 536 HYPERBOLIC FUNCTIONS I2.I THE HYPERBOLIC FUNCTIONS l2.l.l Definition certain combinations of e" and e-'appear so frequently in applications of mathematics that they are given special names. Two of these functions are the "hyperbolic sine function" and the "hyperbolic cosine function.,, The function values are related to the coordinates of the points of an equilateral hyperbola in a manner similar to that in which the values of the corresponding trigonometric functions are related to the coordinates of points of a circle (see sec. 1.7.4)Following are the definitions of the hyperbolic sine function and the hyperbolic cosine function. The hyperbolicsinefunction is defined by The domain and range are the set of all real numbers. 12.1.2 Definition The hyperboliccosinefunction LSdefined by The domain is the set of all real numbers and the range is the set of all numbers in the interval [L, + *;. y : sinhr Figure 12.1.1 12.1.3 Definition Sketches of the graphs of the hyperbolic sine function and the hyperbolic cosine function are shown in Figs. 12.1.1and 12.'1,.2, r"rp".tively. The graphs are easily sketched by using values in the table of hyperbolic functions in the Appendix. Note that these functions are not periodic. The remaining four hyperbolic functions may be defined in terms of the hyperbolic sine and hyperbolic cosine functions. you should note that eachsatisfies a relation analogous to one satisfied by corresponding trigonometric functions. The hyperbolictangentfunction, hyperboliccotangent function, hyperbolic secantfunction, and hyperboliccosecant function are defined, resplctively, as follows: (1) (2) (3) y:coshr Figure 12.1.2 (4) 12.1THE HYPERBOLIC FUNCTIONS y:tanhr F i g u r e1 2 . 1 . 3 Lr)].f urt "L"u' A sketch of the graph of the hyperbolic tangent function is shown in Fi9.12.1.3, There are identities that are satisfied by the hyperbolic functions which are similar to those satisfied by the trigonometric functions. Four of the fundamental identities are given in Definition 12.1,.3.The other four fundamental identities are as follows: tanh x -- (5) coth r cosh2 x - sinh2 )c: t (6) 1 - tanhz x: sech2x (7) | - coth2 x: -csch2 r (8) Equation (5) follows immediately from (1) and (2).Following is the proof of (6). cosh2 r- sinh2-: (*)'- (-ti.:l : t(er, * 2eo* e-2, - ezr+ zeo- e-rr) :l Equation (7) can be proved by using the definitions of tanh r and sech r in terms of e' and e-' as in the proof above, or an alternateproof is obtained by using other identities as follows: '/., sinh2r cosh2r-sinh2x I - tann. N: | secn-r c6sh7;-: 6;hr;: Ail2f: The proof of (8) is left as an exercise(seeExercise1). From the eight fundamental identities, it is possible to prove others. Some of these are given in the exercisesthat follow. Other identities, such as the hyperbolic functions of the sum or difference of two numbers and the hyperbolic functions of twice a number or one-half a number, are similar to the coresponding trigonometric identities. Sometimes it is helpful to make use of the following two relations, which follow from Definitions 12.1.1 and 12."1,.2. cosh x + sinh x: er cosh x- e-ff sinh x- We use them in proving the following (e) (10) identity: y sinh(r * A) : sinh r cosh y + cosh x sLnln. From Definition 12.1,.1we have sinh(r*a):ry:y ( 1 1) 538 FUNCTIONS HYPERBOLIC Applying (9) and (10) to the right side of the above equation, we obtain sinh(r * y) : *[(cosh r * sinh x) (coshy * sinh y) - (cosh r - sinh r) (cosh y - sinh y)l Expanding the right side of the above equation and combining terms, formula (11) is obtained. In a similar manner we can prove cosh(x + V) : cosh x cosh y + sinh x sinh y (r2) If in (11) and (LZ) y i" replacedbyx, the followirg two formulas are obtained: sinh 2x - 2 sinh r cosh x (13) cosh 2x - cosh2 x + sinh2 r (14) and Formula (14) combined with the identity (5) gives us two alternate formulas for cosh 2x, which are cosh 2x :2 sinhz x * '/., (15) and cosh2r:2coslfr-1 (16) Solving (15) and (15) for sinh x and cosh r, respectively, and replacing xby Lx, we get .lx ffi Srnn =V 2: (17) 2 and cosh;:tr (18) We do not have a -+ sign on the right side of (18) since the range of the hyperbolic cosinefunction is [ 1,+o;. The detailsof the proofs of Eqs. (12) through (18) are left as exercises(seeExercises2,4, and 5). The formulas for the derivatives of the hyperbolic sine and hyperbolic cosine functions may be obtained by applying Definitions 12.1.1and '1,2.'1,.2 and differentiating the resulting expressionsinvolving exponential functions. For example, D"(cosh x): D, (ry) : =t --sinhr In a similar manner, we find D."(sinh r) : cosh r (see Exercise15). 12.1THE HYPERBOLIC FUNCTIONS To find the derivative of the hyperbolic tangent function, we use some of the identities we have previously proved. D,(tanh x): - cosh2r--si-nh2 x : D, /g+a) -+: cosh'l cosh'r \cosh r/ sech2x The formulas for the derivatives of the remaining hyperbolic functions are as follows: Dr(coth r) : -csch2 r; D"(sech r) : -sech r tanh r; D"(csch r) : -csch x coth r. The proofs of these formulas are left as exercises(seeExercises15 and 16). From the above formulas and the chain rule, if z is a differentiable function of.x, we have the following more general formulas: D D D "(sinh "(cosh "(tanh u) : cosh u D ru (1e) u) : sinh u D *u (20) u) : sech2u D ru (2r) D"(coth u) : -cschz u D*u (22) D"(sech u) : -sech a tanh u D*u (23) D"(csch u) : -csch u coth u D*u (24') You should notice that the formulas for the derivatives of the hyperbolic sine, cosine, and tangent all have a plus sign, whereas the formulas for the derivatives of the hyperbolic cotangent, secant, and cosecantall have a minus sign. Otherwise, the formulas are similar to the corresponding formulas for the derivatives of the trigonometric functions. EXAMPLE 1: Find D ry tf SOLUTION' D rA : -2x SOLUTION: +:_ sech2 (L y - tanh (1 - x') EXAMPLE 2: Find dyldx if du 1, ax slnn x cosh x , cosh x - .-: slnn ff coth .r. y - ln sinh r Formulas (19) through (24) give rise to the following indefinite integration formulas: | ,inh u du- coshu + C J (2s) FUNCTIONS HYPERBOLIC (28) (2e) udu EXAMPTB 3: Evaluate f I sinh3 x coshz x dx J (30) SOLUTION: r I sinh3r coshz x dx J I I I 1 b EXAMPLE 4: Evaluate r I sechax. dx J sinh2 r coshz r (sinh x dx) (cosh2 x - 1) cosh2 r (sinh x dx) cosh4x(sinh xdx) f I cosh2 r (sinh x dx) J c o s h sx - *coshsx* C SOLUTION: r I sechax dx J I I I sech2r (sech2x dx) (1 - tanh2 x ) (sechz x dx) sechz x dx - tan h r - * t a n f I tanh2x (sech2x dx) h3x+C A catenary is the curve formed by a homogeneous flexible cable hanging from two points under its own weight. If the lowest point of the catenary is the point (0, a), it can be shown that an equation of it is y -a cosh(+) \a/ Figure 12.1.4 a> o A sketch of the graph of Eq. (31) is shown in Fig. 12.1.4. (31) 12.1THE HYPERBOLIC FUNCTIONS 541 ExAMPLE5: Find the length of arc of the catenary having Eq. (31)from the point (0, a) to the point (xr, Ar) wh€r€ ff1 > 0. soLUTroN: If. f(x):, *rn (#) then f '(x):a sinh(tJ (tJ: 'i"r,(f) and using Theorem 8.10.3, if L units is the length of the given arc, we have fcr L- | Vl +lf'(x)l'dx JO (32) From formula (6) , L + sinh2 ,c- coshzr, and because cosh(xla)> L, rt follows that @ : cosh and substituting fromn t h to (32) L- I:.o,n(f) (dx - asinh (il - asinh(+) l; tr 0 A I sinh 0 - asinh(+) Exercises12.L In Exercises L through 10/ prove the identities. L. 1- cothzx: -csch2 r 2. cosh(r * U) : cosh r cosh y * sinh x sinh y tanhr*tanhy 3 . tannrr -r v t : -----:--------:1*tanhxtanhy 4. (a) sinh 2r : 2 sinh r cosh r; (b) cosh 2r : cosh2r * sinh2 x : 2 sinhz x * 7 : 2 cosrf x - | 5. (a)cosh#r: 1$; 6 . csch2r:$ sechr cschr (b)sinhlr:=,F 542 FUNCTIONS HYPERBOLIC 7. (a) tanh(-r) : -tanh x; (b) sech(-x) : sechr 8. sinh 3x -3 sinh x * 4 sinhsr 9. cosh3x :4 coshsx - 3 coshx 10. sinhz x - srnhzy: sinh (x + y) sinh (x - y) 11. Prove:(sinh x * cosh x)n: cosh nx * sinh nx, tf n is any positive integer. (ntNr: Use formula (9).) 12. Prove that the hyperbolic sine function is an odd function and the hyperbolic cosine function is an even function. tanF x : s,x L4.prove: 1* I-tannx *'--- L 13. Prove:tanh(ln r) : x2 * "1. 15. Prove: (a) D"(sinh r) : cosh r; (b) D"(coth x) :-csctri r. 15. Prove: (a) D,(sedr x) : -sedr r tanh r; (b) D"(csch r) : -csch r coth r. In ExercisesL7 through 25, find the derivative of the given function. 4v*1 tZ. f (x): tanh aj 1 19. f (x) - sr cosh x 18.g(r) : ln(sinh13) 20. h(x): coth I 21..h(t): 23. G(x): sin-r(tanh ri) 2a. t'Q): ln(coth 3r - csch3r) 2 6 .g ( x ) : ( 6 o s hx ) ' 22. f (r) : tan-l(sinh 12) ln(tanh f) 25. f(x) -tsinhr 2 7 . P r o v e :( a ) J t a n h u d u : l n l c o s hu l + C ; ( b ) J c o t h u d u : l n l s i n h u l + C . 2 8 . P r o v e :J s e c hu d u : 2 t a n - i e u l C . 2 9 . P r o v e :J c s c ha d u : l n ltanh+ul+ C. In Exercises30 through37, evaluatethe indefinite integral. SO.f sintrnx coshs xdx JJJJ Sf. I t".rn r In(coshx) dr tz. I tanhzZxdx Zl. I coshnzxdx u. 3s. I sechnSxdx J so. f s".f, xtanhsxdx J ,r. - " J ["t"h! t/x | *sinh2 f dx J a, In Exercises38 and 39, evaluate the definite integral. 38. [' cosh'rdr J-r 39. [' sinhtxdx Jo 40. Draw a sketch of the graph of the hyperbolic cotangent function. Find equations of the asymptotes. 41.. Draw a sketch of the graph of the hyperbolic secant function. Find the relative extrema of the function, the points of inflection of the graph, the intervals on which the graph is concaveupward, and the intervals on which the graph is concave downward. 42. Prove that a catenary is concave upward at each point. tl3. Find the area of the region bounded by the catenary of Example 5, the y axis, the r axis, and the line r : 11,where 11)0. tl4. Find the volume of the solid of revolution if the region of Exercise43 is revolved about the r axis. 12,2THE INVERSEHYPERBOLIC FUNCTIONS 45. A particle is moving along a straight line according to the equation of rnotion s: e-ctt2(Asinh f * B cosh f) where s ft is the directed distance of the particle from the origin at f sec.If o ftlsec and a ftlsec2are the velocity and acceleration, respectivelp of the particle at f sec, find o and a. Also show that a is the sum of two numbers, one of which is proportional to s and the other is proportional to p. 45. Prove that the hyperbolic sine function is continuous and increasing on its entire domain. 47. Prove that the hyperbolic tangent function is continuous and increasing on its entire domain. 48. Prove that the hyperbolic cosine function is continuous on its entire domain but is not monotonic on its entire domain. Find the intewals on which the function is increasing and the intervals on which the function is decreasing. 12.2 THE INVERSE HYPERBOLIC FUNCTIONS ' 12.2.1 Definition r we see that a From the graph of the hyperbolic sine function (Fig. 12.'1..1), point; line the graph in one and only one therefore, to horizontal intersects number in the range of the function there corresponds one and only ""ch one number in the domain. The hyperbolic sine function is continuous and increasingon its domain (seeExercise46 in Exercises12.1),and so by Theorem 9.3.2 the function has an inverse function. The inaerse hyperbolic sine function, denoted by sinh-l, is defined as follows: A sketch of the graph of the inverse hyperbolic sine function is shown in Fig. 12.2.1'.Itsdomain is the set of all real numbers, and its range is the set of all real numbers. that It follows from Definition 12.2.'1, sinh(sinh-t r) : x (1) sinh-r(sinhy\:y (2) and sinh-l r Figure 12.2.1 we notice that a horizontal line y: k, where k > 1, , From Fig. 1.2.1..2, intersects the graph of the hlryerbolic cosine function in two points. Therefore, for each number greater than L in the range of this function, there correspond two numbers in the domain. So the hyperbolic cosine function does not have an inverse function. However, we define a function F as follows: F(r) : cosh r Figure12.2.2 t ' forx>0 (3) The domain of F is the interval [0, +-;, and the range is the interval A sketch of the graph of F is shown in Fig. 12.2.2.The function F +-;. [1, is continuous and increasing on its entire domain (seeExerciseL), and so by Theorem 9.9.2, F has an inverse function, which we call the "inverse hyperbolic cosine function." FUNCTIONS HYPERBOLIC 12.2.2Definition The inaersehyperboliccosinefunction, denoted by cosh-l, is defined as follows: l: cosh-t r if and only if x: cosh y and y z 0 A sketch of the graph of the inverse hyperbolic cosine function is shown in Fig. 12.2.3.The domain of this function is the interval [1, +oo), and the range is the interval [0, +*;. From Definition12.2.2 we conclude that cosh(cosh-t x) - x tf x (4) cosh-l(coshy):y ify>0 (5) and As with the hyperbolic sine function, a horizontal line intersects each of the graphs of the hyperbolic tangent function, the hyperbolic cotangent function, and the hyperbolic cosecantfunction at one and only one point. For the hyperbolic tangent function, this may be seen in Fig. 12.1.3.Each of the above three functions is continuous and monotonic on its domain; hence, each has an inverse function. Figure12.2,3 12.2.3 Definition The inaersehyperbolictangentfunction, the inverse hyperboliccotangent function,and the inaersehyperboliccosecant function,denoted respectively by tanh-r, coth-l, and csch-r, are defined as follows: A sketch of the graph of the inverse hyperbolic tangent function is in Fi9.12.2.4. The hyperbolic secant function does not have an inverse function. However, we proceed as we did with the hyperbolic cosine function and define a new function which has an inverse function. Let the function G be defined by G (r) : sechr Y : tanh-l x Figure12.2.4 12.2.4 Definition (5) forx>0 It can be shown that G is continuous and monotonic on its entire domain (see Exercise 2); therefore, G has an inverse function, which we call the "inverse hyperbolic secantfunction." The inaersehyperbolicsecantfunction, denoted by sech-l, is defined as follows: y : sech-l r if and only if x - sechy / and y >0 (7) 12.2THEINVERSE HYPERBOLIC FUNCTIONS545 The inverse hyperbolic functions can be expressedin terms of natural logarithms. This should not be surprising because the hyperbolic functions were defined in terms of the exponential function, and the natural logarithmic function is the inverse of the exponential function. Following are these expressions for sinh-l, cosh-l, tanh-r, and coth-l. sinh-l x:ln(x + \F +l) cosh-1r:ln(r+\/P-) 1 vI1 (s) x>I (9) lrl < 1 (10) lrl > 1 (11) tanh-r *:tnffi coth-lx:|lnffi r any real number We prove (9) and leave the proofs of the other three formulas as exercises(see Exercises3 through 5). To prove (9), let V: cosh-r x, x > 1.. Then, by Definition 12.2.2, r : cosh U, A > 0. Applying Definition 12.7.2to cosh y, we get ea + e-u y>0 from which we obtain ,y >0 ezo-2xea+1-0 Solving this equation for e' by the quadratic formula, we obtain € u : 2 * r W2: x t \ F y>o (r2) We know that y >O and r > 1. Therefore,eu> 7. W.hen x:1, : 1. Furthermore,when x )']-., 0 < r- 1 < x+ lF - t : x- \/71 x * l, and so \/i=T < t/x + t . Therefore, \81 \81 < \81 \tr+1 giving us x-'!.1\/r'z--. Hence, whdn x)1,, y-1/p=1 ay Consequently, we can safely reject the minus sign in (12). And since y: cosh-l .r, we have cosh-lr: ln(x + t/P - t) x > ! which is (9). ExAMPLE L: Express each of the following in terms of a natural logarithm: (a) tanh-l(- *); (b) sinh-l 2. SOLUTION: tanh-l (a) From (10), we obtain 3 (-f):+t"i: +'"i:rn+:-h (b) From (8), we get sinh-l 2- ln(2 * \6) HYPERBOLIC FUNCTIONS To obtain a formula for the derivative of the inverse hyperbolic sine functiofl, let y : sinh-l r, and then x - sinh y. Thus, becauseDox: cosh y and DrU : LlDdc, we have Dra: (13) cosh y From the identity cosh2y-sinh2 U:1, it follows that cosh y: VA;nt!=. (Norr: When taking the iquare root, the minus sign is rejectedbecausecosh y > 1.) Hence, becauser: sinh /, cosh y: lF +1. Substituting this in (13), we obtain D *(sinh-x) L :#TT (14) If. u is a differentiable function of r, from (14) and the chain rule it follows that (15) Formula (14) can also be derived by using (8), as follows: -r D x) : D r l n ( r * \m) "(sinh 1 r 'r : x 1ffi x,+\m \m ==: \ffi( r *tffil -:- - *x 1 Yxz + 1 The followitg differentiation formulas can be obtained in analogous ways. Their proofs are left as exercises(see Exercises6 throu each formula u is a differentiable function of x. FUNCTIONS 547 12.2THE INVERSEHYPERBOLIC (le) (20) Find f '(x) if EXAMPLE EXAMPTE 3: "(r Verify Applying formula (L7), w€ get r, / r : -2 I (x, 1,- tanh-l(cos 2x) f(x) D solurroN: stn 2x 2*: "orz -2 stn 2x : - ., ., z csc zx ,inz ,, SOLUTION: cosh-L x - Dr(r cosh-Lx - tffil - cosh-t x * x - cosh -r x The main use of the inverse hyperbolic functions is in connection with integration. This topic is discussedin the next section. Exercises L2.2 1. If F(r) : cosh x and.x = 0, prove that F is continuous and increasing on its entire domain. 2. lt G(x): sechx and x = 0, prove that G is continuous and monotonic on its entire domain. In Exercises3 through 10, prove the indicated formula of this section. 3. Formula (8) 4. Formula (10) 5. Formula (11) 6 . Formula (15) 7. Formula (L7) 8. Formula (18) 9. Formula (19) 1 0 . Formula (20) ln Exercises 11.through L4, express the given quantity in terms of a natural logarithm. n. sinh-r * 1,4. coth-l (-2) 13. tanh-l * 12. cosh-r 3 In Exercises L5 through 28, find the derivative of the given function. 15. f (x) : sinh-r x2 18. h(w) - tanh-t Tt)B 1'5.G(x) : cosh-t tx 21. f (x) : x2 cosh-l x2 24. S(r) : tanh-l(sin 3r) 27. G(x) - x sinh-r x- \ffi 22. h(x) - (sech-lr)2 19. S(x) - coth-t(3r + 1) 25. h(x) - cosh-l(cscx) - x t a n h - rx 2 8 .H ( x ) : l n \ F 1 7 .F ( r ) - tanh-r 4x - csch-r +r2 20.f(r) 23.f(x) : sinh-l(tan x) 26.F(x) - c o t h - l ( c o s h x ) In Exercises29 through 32, draw a sketch of the graph of the indicated function. 29. The inverse hyperbolic tangent function. 30. The inverse hyperbolic cotangent function. 31. The inverse hyperbolic secant function. 32. The inverse hyperbolic cosecant function. HYPERBOLIC FUNCTIONS 1'2.3INTEGRATS YIELDING INVERSE HYPERBOLIC FUNCTIONS The inverse hyperbolic functions can be applied in integration, and sometimes their use shortens the computation considerably. Il should be noted, however, that no new types of integrals are evaluated by their use. Only new forms of the results are obtained. From formula (15) in 9ec.12.2we have . D,(sinh-l u): U;ft7o,u from which we obtain the integration formula Idu I ffi: sinh-ra * C Expressing sinh-r u as a natural logarithm by using formula (g) of sec. '1,2.2, we get (1) Formula (16) in Sec.12.2 is D * ( c o s h -zr) - L n where u ffiu,u from which it follows that [ -!" Jffi:cosh-tu +C u and combining this with formula (9) in Sec.12.2,we obtain (2) Formulas (17) and (18) of Sec.12.2 are, respectively, D*(tanh-lu):#D,u wherelul D,(coth-lD: firD,u wherelul >t and From the above two formulas we get I d, _ftanh-l z*C i f l r l l< 1 J T=7: l"orh-'u * C if lui > t and combining this with formulas (10) and (11) of Sec.12.2 we have (3) I2,g INTEGRALSYIELDINGINVERSEHYPERBOLICFUNCTIONS 84g; We also have the following three formulas. (4) (s) (5) These formulas can be proved by finding the derivative of the right side and obtaining the integrand, or else more directly by using a h1ryerbolic function substitution. For example, the proof of (a) by differentiating the right side is as follows: / r \ 1 1 \ G=o): l D,(sinh-: ffi VF/*' ' -'': Uffi7' i and becausea ) 0, {-a': a, and,we get / .,\ 1 =o): Du(sinh-' ,6ryt To obtain the natural logarithm representation we use formula (8) in Sec.12.2and we have , lu , \/FTF\ :tl-tt--T--l \4a/ :ln(n+tfft)_lna Therefore, ,rl sinh-l i + C: ln(z + t/NV) a - h a* C : ln(z+ I/FTE) + c, whereCl=C-lna. Ttre proof of (5) by a hyperbolic function substitution makes use of the identity cosh2r - sinh2r: L. Lettingu: a cosh r, where r ) 0, then FUNCTIONS HYPERBOLIC du: a sinh r dx, and r: get f du J '\fr'=A- cosh-l(ula).Substitutingin the integrand,we asinhrdr f J \//frcoffi=@ asinhrdr f I \/F Vcoiltt= -J f asinhxdx \/F \ffiE; a. Furtherrnore, Because a ) 0, lF: becausex ) 0, sinh r ) 0, and so \6in6 t: sinh r. We have,then, du I J !u2-a2 f a sinhx dx J asinhx :Ia, J :r*C : cosh_r I*, The natural logarithm representation may be obtained in a way similar to the way we obtained the one for formula (a). The proof of formula (5) is left as an exercise(seeExercise17). Note that formulas (1) through (5) give altemate representations of the integral in question. In evaluating a definite integral in which one of these forms occurs and a table of hyperbolic functions is available, the inverse hyperbolic function representation may be easier to use. Also the inverse hy?erbolic function representation is a less cumbersome one to write for the indefinite integral. Also note that the nafural logarithm form of formula (5) was obtained in Sec. 1,1,.4 by using partial fractions. EXAMPLE ld* Jffi Evaluate sol,urroN: We apply formula (4) after rewriting the denominator. dx 12.3 INTEGRALS YIELDINGINVERSEHYPERBOLIC FUNCTIONS 551 EXAMPTg2: Evaluate dx fto J, !x2 - 25 solurroN: Using formula (5), we get fro dx I -Ja vx4 :'.lro cosh-r * | - cosh-l 2c Jo cosh-l L.2 Using Table 4 in the Appendix, we find cosh-l 2- c o s h - l( L . 2 ) - ' 1 , . 3 2 - 0 . 6 2 : 0 . 7 0 Instead of applying the formulas, integrals of the forms of those in formulas (1) through (6) can be obtained by using a hyperbolic function substitution and proceeding in a manner similar to that usin g a trigonometric substitution. EXAMPTT 3: Evaluatethe integral of ExampleL without using a formula but by using a hlperbolic function substitution. solurroN: as In the solution of Example L, the given integral was rewritten ldx J tk=ffiT If we let x-3:2 sinh u,then d.x:2 cosha du and z - sinh-ltk-g). We have, therefore, dx which agreeswith the result of Example L. 12.3 Exercises In Exercises 1 through 10, express the indefinite integral in terms of an inverse hyperbolic function and as a natural logarithm. FUNCTIONS HYPERBOLIC 1 r ' J| @ d * = ?f J F, f f ?xdx r ^ f o .Jffi ,-J@ ".Jffi = 5 . lf ,dgxx 2 - ! 5 . t =2=5 d -x2 cosxdx d! rffi dx In Exercises LL through 16, evaluate the definite integral and express the answer in terms of a natural logarithm. 13I::'& dx fz l.o ' J - , f f i 17. Prove formula (6) by using a hlperbolic function substitution. L8. Acurvegoesthroughthepoint (0,a),a > 0,andtheslopeatanypointist/f,ia2-l.Provethatthecurveisacatenary. 19. A man wearing a parachute falls out of an airplane and when the parachute opens his velocity is 200 ftlsec. If t, ftlsec is his velocity f sec after the parachute opens, 32ad!.:32484t az Solve this differential equation to obtain t:E (.otn-, #- coth-'ry) ReaiewExercises (Chapter12) In Exercises1.through 3,prove the identities. 1,.(a) coth(-x)- -coth x; (b) csch(-x) : -csch r 2. sinh x * sinh y :2 sinh +(x+ y) cosh t(x - V) ? cosh 2r * cosh 4y _ coth(x + 2y) r. 4. Prove,:f"l j111 coth x - 1; (b) csch x : 0 "t1i In Exercises5 throu gh 9, find the derivative of the given function. 5. f (x) : coshl 2x 6. 8U) : ln sech f2 8. h(x): 9. f (w) : wz sin1.'1-r 2w e"(cosh r * sinh r) T . rf (' \i l' - cosh tz l*sgcha 10. The gudermannian is a function that occurs frequently in mathematics, and.it is defined by gd x: tan-r(sinh r). Show that D'(gd r) : sech r. E.i'" REVIEWEXERCISES 553 In ExercisesL1.through 14, evaluatethe indefinite integral. 11. .tanh tf itx I 13. /sinnxsin xdx In Exercises15 and 15, evaluatethe definite integral. fr 1 . 5 .I s e c h z t d t Jo In Exercisest7 and 18,obtdn the given result by a hlryerbolic function substitution. 1. dx coth /rir,r,-'f) t c. a > o rt . f al \ ljm q i:- A t "/r*rr-'r)+c.a>o dx I 18. al \ l@:=srnn 19. Find the length of the catenary y: cosh x from (ln 2, *) to (In 3, t). 20. The graph of the equation t7-r:cginh-l 'l\-l-'VAry' is called a tractrix. Provethat the slope of the curve at any point (x, y) is -Vl{FT. Polarcoordinates 13.1 THE POLARCOORDINATE SYSTEM 13.1.THE POLAR COORDINATE SYSTEM P(r, 0) Figure F i g u r e1 3 . 1 . 2 - 7 61r F i g u r e1 3 . 1 . 3 F i g u r e1 3 . 1 . 4 F i g u r e1 3 . 1 . 5 P(-n,t,) a.-A-. 1.1 70 F i g u r e1 3 . 1 . 6 Until now we have located a.point in a plane by its rectangular cartesian coordinates. There are other coordinate systems that can be used. Probably the next in importance to the cartesian coordinate system is the polar coordinatesystem.In the cartesian coordinate system, the coordinates are numbers called the abscissa and the ordinate, and these numbers are directed distances from two fixed lines. In the polar coordinate system, the coordinates consist of a distance and the measure of an angle relative to a fixed point and a fixed ray (or half line). The fixed point is called the pole (or origin), and it is designated by the letter "O." The fixed ray is called the polar axis (or polar line), which we label OA. The ny OA is usually drawn horizontally and to the right, and it extendsindefinitely (seeFig. 13.1.1). Let P be any point in the plane distinct from O. Let 0 be the radian measure of the directed angle AOP, positive when measured counters its initial clockwise and negative when measured clockwise, having aas side the ny OA and as its terminal side the ny OP. Then ift r is the undirected distance from O to P (i.e., ,: lOPl\, one set of polar coordinates of P is given by r and 0, and we write these coordinates as (r' 0)' o rLLUsrRArroN1: The point P(4, ta) is determined by first drawing the angle having radian measure $2, having its vertex at the pole and its initial side along the polar axis. Then the point on the terminal side, which is four uriits from the pole, is the point P (see Fig. 13.1.2).Another set of polar coordinatesfor this samepoint is (4,-&rr); see Fig. 13.1.3.Furthermore, the polar coordinates (4, #rr) would also yield the same point, as ' shown in Fig. 13.L.4. Actually the coordinates (4, $n *2nn), where n is any integer, give the same point as (4,trr). So a given point has an unlimited number of sets of polar coordinates. This is unlike the rectangular cartesian coordinate system becausethere is a one-to-one conespondence between the rectangular cartesiancoordinates and the position of points in the plane, whereas there is no such one-to-one correspondence between the polar coordinates and the position of points in the plane. A further example is obtained by considering sets of polar coordinates for the pole. If r: 0 and 0 is any real number, we have the pole, which is designated by (0, 0). We consider polar coordinates for which r is negative. In this case, instead of the point being on the terminal side of the angle, it is on the extension of the terminal side, which is the ray from the pole extending in the direction opposite to the terminal side. So if P is on the extension of the terminal side of the angle of radian measure 0, a set of polar coordinates of P is (r, 0), where r: -lOPl. is the same . rLLUSrRArroN2: The point (-4, -ttr) shown in Fig. 1"3.1'.5 -&r), another set 1. Still and (4, fzr) in Illustration point as (4, *n), (4, of polar coordinatesfor this samepoint is (-4, l*zr); seeFig. 13.1'6. . The angle is usually meastlred in radians; thus, a set of polar coordi- 556 POLARCOORDINATES nates of a point is an ordered pair of real numbers. For eachordered pair of real numbers there is a unique point having this set of polar coordinates. However, we have seen that a particular point can be given by an unlimited number of ordered pairs of real numbers. If the point p is not the pole, and r and0 arerestricted so thatr > 0 and0 < 0 < 2rr,then there is a unique set of polar coordinates for p. EXAMPLE1: (a) plot the point having polar coordinates (3, -3r). Find another set of polar coordinates of this point for which (b) r is negative and 0 (c) r rs positive and 0 (d) r is negative and -2n < 0 < 0. solurroN: (a) The point is plotted by drawing the angle of radian measure -3a in a clockwisedirection from the polar axis. Becauser)0,p is on the terminal side of the angle, three units from the pole; see Fig. 1,3.1,.7a. The answersto (b), (c), and (d) are, respectively,(-3,$r), (3,trl), and (-3, -En).They are illustrated in Fig. 1,3.1.2b, c, and d. 4_ 5" o/ / x -+" PQ,_3I 7r * { p(- 3, \ +c 4 o(_r, (b) (a) -+c (d) F i g u r e1 3 . 1. 7 (,, (*, often we wish to refer to both the rectangular cartesian coordinates and the polar coordinates of a point. To do this, we take the origin of the first system and the pole of the second system coincident, the polar axis as the positive side of the r axis, and the ray for which 0 : in as the positive side of the y axis. suppose that P is a point whose representation in the rectangular cartesian coordinate system is (x, y) and (r, g) is a polar-coordinate representation of P. we distinguish two cases:r > 0 and r < 0.In the first case, if r ) 0, then the point P is on the terminal side of the angle of 0 radians,and r: loPl. Such a caseis shown in Fig. 13.1.g.Then cos d: xl lOPl : xlr and sin 0 : Vllo-Pl : Alr; and so x-rcose F i g u r e1 3 . 1 . 8 and y-rsrn0 (1) In the second case, tf r I 0, then the point P is on the extension of the terminal side and r--lOTl (see Fig. 13.1.9).Then if a is the point (- x , -y), we have -x -X -X x cosg= =l@T So x- rcos0 ;Fl :-,: r (2) 13.1 THE POLARCOORDINATE SYSTEM 557 Also, -y) Q(- x, -a-a-va sino:16T:l6FT:=:; Hence, P F i g u r e1 3 . 1. 9 y:rsin9 (3) Formulas (2) and (3) are the same as the formulas in (L); thus, the formulas (1) hold in all cases. From formulas (1) we can obtain the rectangular cartesiancoordinates of a point when its polar coordinates are known. Also, from the formulas we can obtain a polar equation of a curve if a rectangular cartesian equation is known. To obtain formulas which give a set of polar coordinates of a point when its rectangular cartesian coordinates are known, we square on both sides of each equation in (1) and obtain 'f: rz cosz0 and A2: 12 sinz0 Equating the sum of the left members of the above to the sum of the right members, we have **y':12 cos2 o * r z s i n zo or, equivalently, * * Y': rz(sin2o * coszo) which gives us *+Yz:rz and so (4) Fromthe equationsin (1) and dividinS,we have r s i ne - _ y ,'cosg i or, equivalently, :':: if lt*i+fi i$il1ii.::':::i:i'iiiiiiirii:i;i::i:li':'ii::i (5) ri;1i1r-1i fffitl++i+++ . rr,r.usrRArrorv3: The point whose polar coordinates are (-6, Znt) is plotted in Fig. 13.1.10.We find its rectangular cartesian coordinates. From (L) we have .10 F i g u r e1 3 . 1 x - r c o s0 - -5 cosZn y-rsin0 - -6 stn tn POLAR COORDINATES --3\n Sothe point is (-3 \n,3\n). The graph of an equation in polar coordinates r and 0 consists of all those points and only those points P having at least one pair of coordinates which satisfy the equation. If an equation of a graph is given in polar coordinates, it is called a polar equation to distinguish if from a cartesianequation,which is the term used when an equation is given in rectangular cartesian coordinates. EXAMPLr'2: Given a polar equation of a graph is 12:4stn20 find a cartesian equation. SoLUTIoN: Because y2 - x, + y, and sin 20:2 sin 0 cose - zfulr)(xlr), from (1) we have, upon substituting in the given polar equation, x 2+ Y ' : 4 ( D + ' ; x2* y',:w r2 xz*Y': !!V = x'+y, (x'*y')'-8xy EXAMPTn3: Find (r, 0) tf r > 0 and0 <0<-2n forthepoint whose rectangular cartesian coordinate representation is soLUrIoN: The point e{3, cause r ) 0, we have t\8,-1). From (5), tan 0:-U(-fi), -1) is plotted in Fig. 1g.1.1,1. From (4), be_ r: {lT3:2 o:trr So the point is Q, &r). F i g u r e1 3 . 1 .11 and since r I 0 < $r, wehave SYSTEM 13.1 THE POLARCOORDINATE ExAMPtn 4: Given a cartesian equation of a graPh is x2+y2-4x-0 find a polar equation. soLUrIoN: Substituting x- /cos g and y: rsin 0 in x2+y2-4x-o we have rz cosz 0 + 12 sinz0 - 4r cos0:0 12-4rcos0:0 r ( r - 4 c o sg ) : 0 Therefore, o r r - 4 c o s0 : 0 The graph of r - 0 is the pole. However, the pole is a Point on the g r a p h o f r - 4 cos 0: O becauser:0 when e Lr. Therefore,a polar equation of the graPh is r-0 r-4cos0 The graph of.* * y'- 4r: 0 is a circle.The equationmay be written in the form (x-2)'*y':4 which is an equationof the circlewith centerat (2, 0) and radius 2. L3.1Exercises then find another set of polar coordiIn Exercises1 through 6, plot the point having the given-set of polar coordinatesi natesforthesamepointforwhich(a)r<oan?0<"0<Ztr;b\i>oand-2n<0'0;(c)r(0and-2zr<0<0' 3. (2, Ln) 2. (3, 8zr) 1. (4, in) 6. (2, tn) 5. ?n,fu) 4. (3, szr) give two other sets of polar coordiIn Exercises 7 through 12, plot the point having the given set of polar coordinates; then sign' opposite having r nates of the same point, one with the same Value of r and one with an ,. @,-&n) 10. (-2, tn) 9. (-4, &n) 12. (-3 , -n) 8. ({2, -*n) 1.L.(-2, -*rr) given: (a) (3' r); 13. Find the rectangular cartesiancoordinates of each of the following points whose polar coordinates are -tn); (e)1-z,Irr);(f) (-1, -Sa'). b) ({r,-*"');-(") ?a,Ir); (d)(-2, qartgsil coordinates are given' Take 14. Find a set of polar coordinates for each of the following points whose rectangular -2); r > 0 and0 = 0 < ir. til 0,-r); (u) (-rfs, r); (c)d',-z);(d) (-s,0); (e)(b, (r) (-z'-zt/i)' In Exercises15 through 22, find. a polar equation of the graph having the given cartesian equation. L5. x2+A2:az L6. xB : 4y2 18.x2-A2:15 L9. (x' * y')'- 4(x' - y') 17.Y' : 4(x + 1) 20. 2xy-- a2 I I POLAR COORDINATES 2I. rf+y3-3axy-0 2x 22.Y : x'z+r In Exercises23 through 30, find a cartesian equation of the graph having the given polar equation. 23. 12: 2 sin 20 24. r2 cos 20: 2 6 .r 2 : 0 27. 12 : 4 cos 20 13.2 GRAPHS OF EQUATIONS IN POLAR COORDINATES Let r: f (0) be a polar equation of a curve. We first derive a formula for finding the slope of a tangent line to a polar curve at a point (r, g) on the curve. Consider a rectangular cartesian coordinate system and a polar coordinate system in the same plane and having the positive side of the x axis coincident with the polar axis. In Sec.13.1 we saw that the two sets of coordinates are related by the equations r: 25. 12: cos 0 t0 28.r:2sin30 / cos 0 and tt: r sin 0 r and y can be considered as functions of 0 becauser:f(0).If we differentiate with respect to 0 on both sides of these equations we get, by applying the chain rule, dx Ag: Adr cos 0 d0- r sin 0 and du ffi:sin0 dr cos0 de*r (2) If a is the radian measure of the inclination of the tangent line, then tan o: du f* and if dxlde # 0, we have -dv :dx dy de dx de substituting from (1) and (2) into (g),we obtain s i no # * r c o s g c o s0 # - r s i n o If cos 0 * 0, we divide the numerator and the denominator of the 13,2 GRAPHSOF EQUATIONSIN POLARCOORDINATES right side of ( ) by cos 0 and replace dyI dx by tan a, thereby giving us (5) ExAMPLEL: Find the sloPeof the tangent line to the curue whose equation is r 1'- cos A at the point (1 +\n, Ln). Pt(rt,01) F i g u r e1 3 . 2 . 1 sor,urrow: Sincer: 1 - cos0,drld0: sin 0.Soat thepoint (1 - i!2, in), d.rlil|: it/2 ana tan0:1; so from (5)we obtain (1,)+\/, (1,- i\/r\: _-!_ r-tanc^.:_ i!?g2_+ 11_+\/2)(L) vz- t: t/2 + L If we are to draw a sketch of the graph of a polar equation, it will be helpful to consider properties of symmetry of the graph. In Sec.1..3(Definition L.3.4) we learned that two points P and Q are said to be symmetric with respect to a line if and only if the line is the perpendicular bisector of the line segment PQ, and that two points P and Q are said to be symmetric with respect to a third point if and only if the third point is the midpoint of the line segment PQ. Therefore, the points 1Z'tr) and (2, $z) are symmetric with respect to the Lzr axis and the points 12,tzr) and (2, -lrr) are symmetric with respect to the pole. We also leamed (Definition 1.3.5)that the graph of an equation is symmetric with respect to a line I if and only' if foi every point F on the graph there is a point Q, also on the graph, such that P and Q are symmetric with resPect to l. Similarly, the graph of an equation is symmetric with respect to a point R if and only if for every poit t f on the graph there is a point S, also on the graph, such that P and s aie symmetric with respect to R. We have the following rules of symmetry for graphs of polar equations. RuIe1. If for an equation in polar coordinatesan equivalent equation is obtained when (r, 0) is replacedby either (r, - 0 * 2nn) or (- r, n 0 * Znt)' where n is any integer, the graph of the equation is symmetric with lespect to the polar axis. RuIe2. If for an equation in polar coordinatesan equivalent equation is obtained wh en (r, 0) is replacedby either (r, o 0 4- 2nn) or (- r, 0 * 2nn), integer, the graph of the equation is symmetric with respect where nis aurty axis. the to tqr RuIe3. If for an equation in polar coordinatesan equivalent equation is obtained when (r, 0) is replacedby either (-t, 0 I 2nzr)ot (r, t * 0 * 2nzr)' where n is any integer, the graph of the equation is symmetric with respect to the pole. The proof of Rule 1 is as follows: If the point P(r, d) is a point on the graph of an equation, then the graph is symmetric with respect to the polar axis if there is a point Pr(rr, 01)on the graph so that the polar axis is the perpendicular bisector of the line segment PtP (see Fig. 13.2.1).So 562 POLARCOORDINATES if r, : r, then d, must equal - d * 2nn,where n is an integer. And if.r, : - 7, then d1must be n - 0 * 2nrr. The proofs of Rules 2 and 3 are similar and are omitted. . rr,r,usrRArror,l1: We test for symmetry with respect to the polar axis, the in axis, and the pole, the graph of the equation r:4 cos 20. Using Rule 1 to test for symmetry with respect to the polar axis, we replace (r, 0) by (r, -0) and obtain r : 4 cos(-20), which is equivalent to r: 4 cos20. So the graph is symmetric with respect to the polar axis. Using Rule 2 to test for symmetry with respect to the ir axis,we replace (r, 0) by (r, T-0) and get r:4 cos(2(?r-l)) or, equivalently, r: 4 cos(2n- 2e), which is equivalentto the equation r - 4cos20. Therefore, the graph is symmetric with respect to the lr axis. To test for symmetry with respect to the pole, we replace (r, 0) by (-r, 0) and obtain the equation -/: 4 cos 20, which is not equivalent to the equation. But we must also determine if the other set of coordinates works. We replace(r, 0) by (r, n * 0) and obtain r:4 cos2(tr+ 0)or, equivalently, r : 4 cos(2rr+ 20),which is equivalent to the equation r: 4 cos20. Therefore, the graph is symmetric with respect to the pole. . when drawing a sketch of a graph, it is desirable to determine if is on the graph. This is done by substituting 0 for r and solving lhe nofg for'0. Also, it is advantageousto plot the points for which r has a relative maximum or relative minimum value. As a further aid in plotting, if a curve contains the pole, it is sometimes helpful to find equations of the tangent lines to the graph at the pole. To find the slope of the tangent line at the pole we use formula (5) with r: 0. If drldO * 0, we have tan a: tan 0 (6) From (6) we conclude that the values of 0, where 0 < 0 < zr, which satisfy a polar equation of the curve when /: 0 are the inclinations of the tangent lines to the curve at the pole. So if.er,02, . , 0* are these values of 0, then equations of the tangent lines to the curve at the pole are 0 : 0 r .0, : e r , EXAMpLE2: Draw a sketch of the graph of r:1,-2cos0 , 0: 0* soLUrIoN: Replacing (r , 0) by (r , -0), we obtain an equivalent equation. Thus, the graph is symmetric with respect to the polai axis. - Table 13.2.1gives the coordinates of some points on the graph. From these points we draw half of the graph and the remainder is-drawn from its symmetry with respect to the polar axis. lf r:0, we obtaincos 0:*, and if 0 < 01r, 0:*r.Therefore, the point (0' *n) is on the graph,and an equation of the tangent line there is 0 : *zr. A sketch of the graph is shown in Fig. 13.2.2.rt is called a limagon. 13,2 GRAPHSOF EQUATIONSIN POLARCOORDINATES 563 Tsble 0 0 tn tn tn 3n r -1 L-\tr 0 t 2 Err 1+\E 7r 3 Fig ure The graph of an equation of the form r-s+bcos0 or r-s+bsin0 is a limaqon. r f b If fl:b, the limaqonisacardioid,whichisheart-shaped.Ifa> limaEon has a shape similar to the one in Example 3. EXAMPIn 3: Draw the graph of 3+2stn0 1_ 7T a sketch of solurroN: The graph is symmetric with respect to the in axis because if (r,0) is replacedby (r, r - 0), an equivalent equation is obtained. Table ti.2.2 gives the coordinates of some of the points on the graph. is drawn by plotting the A sketch of the graph is shown in Fig. 13.2.3.It '1,3.2.2 and using the symin Table points whose coordinates are given metry property. T a b l eL 3. 2 . 2 F i g u r e1 3 . 2 . 3 0 0tntnLniT r 3 4 3+\E 5 3 &n *n En 2 3-\tr L 564 POLARCOORDINATES The graph of an equation of the form 'r-acosn0 or r-astnn? is a rose,having n leavestf n is odd and 2nLeavesrf n is even. EXAMPur4: Draw a sketch of the four-leafed rose r-4cos20 1 7r solurroN: In Illustration L we proved that the graph is symmetric with respect to the polar axis, the trr axis, and the pole. Substituting 0 for r in the given equation, we get c o s2 0 : 0 from which we obtain, for 0 < 0 12rr,0:*rr,Zrr,Zr, and #2. Table 13.2.3gives values of r for some values of d from 0 to jz. From these values and the symmetry properties, we draw a sketch of the graph as shown in Fig. 13.2.4. TableL3.2.3 0 0 #n r 4 2\tr tn L+n tn #n tn 2 0 -2 -z\tr -4 Figure 13.2.4 The graph of the equation 0:C where c is any constant, is a straight line through the pole and makes an angle of C radians with the polar axis. The same line is given by the equation 0:C-rnn where n is any integer. general, the polar form of an equation of a line is not so simple -In as the cartesian form. However, if. the line is parallel to either the poiar axis or the *zr axis, the equation is fairly simple. If a line is parallel to the polar axis and contains the point B whose cartesian coordinates are (0, b) and polar coordinates ^t" 1b,fz'), then a cartesian equation is y : b. If we replace y by r sin d, we have rsin0:b which is a polar equation of any line paraller to the polar axis. If b is posi- 13.2 GRAPHSOF EQUATIONSIN POLARCOORDINATES 565 tive, the line is above the polar It b is negative, it is below the polar we have a sketchof the graph of the equa. rLLusrRArroN2: In Fig. 1,3.2.5 ']..3.2.6 g: we have a sketch of the graph of the 3, and in Fig. tion r sin -3. . equation r sin 0: If a line is parallel to the *zr axis or, equivalently, Perpendicular to the polar axis, and goes through the point A whose cartesian coordinates are (a, 0) and polar coordinates are (a, 0), a cartesian equation is r : a. Replacing xby r cos 0, we obtain F i g u r e1 3 . 2 . 5 rcos0:a rsin0: Figure cO tl <b (n o U It which is an equation of any line perpendicular to the polar axis. If. a is positive, the line is to the right of the *zr axis. If a is negative, the line is to the left of the *zr axis. 3: Figure 13.2.7shows a sketch of the graPh of the equao rLLusrRATroN and Fig. \3.2.8shows a sketch of the graPh of the equa0:3, r cos tion 3 . t i o n r c o s0 The graPh of the equation r:c Figure 13.2.7 where C is any constant, is a circle whose center is at the pole and radius is lcl. The samecircle is given by the equation r--c F i g u r e1 3 . 2 . 8 As was the case with the straight line, the general polar equation of a circle is not so simple as the cartesian form. However, there are special casesof an equation of a circle which are worth considering in polar form' If a circle has its center on the polar axis at the point having.polar coord.inates(a, 0) and is tangent to the tr axis, then its radius is lal. The points of iniersection of s.tlh u circle with the polar axis are the pole if P(r, 0) is any point on iO, O; ana the point (2a,0) (seeFig. 1'3.2.9)'Then a right angle. Therefore, is circle the in P inscribed at angle the ihe circle, : equivalentlY, rl2a ot, cos 0 r:2a c o s0 which is a polar equation of the circle having its center on the polar axis or its extension, and tangent to the Ezt axis. lf. a is a positive number, the circle is to the right of the pole. lf. a is a negative number, the circle is to the left of the pole. In a similar manner, it may be shown that r:2b Fig ure sin0 is a polar equation of the circle having its center on the in axis or its ex- 566 POLAR COORDINATES tension, and tangent to the polar axis. If b is positive, the circle is above the pole. If b is negative, the circle is below the pole. ExAMPrn 5: Draw a sketch of the graph of soLUrIoN: When 0: nn, where n is any integer, the graph intersects the polar axis or its extension, and when 0:inrr, where n is any odd integer, the graph intersects the izr axis or its extension. When r: 0, 0: 0; thus, the tangent line to the curve at the pole is the polar axis. A sketch of the graph is shown in Fig. 13.2.1,0. The curve is calleda spiral of Archimedes. 11 ='lf o F i g u r e1 3 . 2 . 1 0 ExercisesL3,2 In ExercisesL through 40, draw a sketch of the graph of the given equation. 1 , .r c o s0 : 4 2. r sin 3. r-4 c o s0 4. 2sin0 -4 5. 0:5 6. 7. r:5 8. r : - 4 9. r sin 0: -4 1 1 .r : - 4 13. r: sin0 eo(ogarithmic spiral) 10. r c o s 0 : - 5 L2. r : - 5 c o sA 1,4.r - eotg(logarithmic spiral) 15. r - ll0 (reciprocalspiral) L6. r: L7. r : 2 sin 30 (three-leafedrose) 1 8 . r : 3 cos20 (four-leafed rose) 19. r : 2 cos 40 (eight-leafedrose) 20. 21,.r : 4 sin 20 (four-leafed rose) 22. r : 3 cos 30 (three-leafed rose) 23. r : 2 + 2 sin 0 (cardioid) 25. r : 4 - 4 cos 0 (cardioid) 24. 20 (spiral of Archimedes) 4 sin 50 (five-leafed rose) 3+3cos 0(cardioid) 26. r : 3 - 3 s i n 0 ( c a r d i o i d ) 13.3 INTERSECTION OF GRAPHSIN POLARCOORDINATES 2 7 .r : 4 - 3 c o s A ( h m a g o n ) 29.r:4+2 28. r:3-4 sin 0 (limagon) 30. r:2 31. rz :9 32. 12 : -4 sin 20 (lemniscate) sin 20 (lemniscate) -25 : 33. 12 cos2e (lemniscate) -35. r 2 sin 0 tan 0 (cissoid) 37. r : 2 sec 0 - 1 (conchoid of Nicomedes) 99. r: cos 0 (limagon) - 3 sin 0 (limagon) 34. 12 : L6 cos 20 (lemniscate) 35. (r - 2)' :80 38. r: 2 csc 0 + 3 (conchoid of Nicomedes) 40.r:2lcos lsin 201 (parabolic spiral) 0l 41 through 44,find an equationof eachof the tangentlines to the given curveat the pole. In Exercises 4 2 .r : 2 s i n 3 0 4 7 .r : 4 c o s 0 l 2 44.rz :9 sin20 43.rz :4 cos20 r15.Find the slopeof the tangentline to the curver: 4 at (4, *n). rt5. Find a polar equationof the tangentline to the curver = -6 sin g at the point (6, *r) . 13.3 INTERSECTION OF GRAPHS IN POLAR COORDINATES EXAMPLE "1.: DrAW sketchesof 20 and r: 2 sin graphs of the r : L, and find the points of intersection. To find the points of intersection of two curves whose equations are in caftesian coordinates, we solve the two equations simultaneously. The common solutions give all the points of intersection. However, because a point has an unlimited number of sets of polar coordinates, it is possible to have as the intersection of two curves a point for which no single pair of polar coordinates satisfies both equations. This is illustrated in the following example. solurroN: The graph of r:2 sin 20 is a four-leafedrose. The graph of r: 1 is the circle with its center at the pole and radius 1..Sketchesof the graphs are shown in Fig.13.3.1.Solving the two equations simultaneously, we have 1 : 2 sin 20 or sin 20 : i. Therefore, we get 20 : tn, *n , En , and *rr 0:#r,&n,tlr, andlln Hence, we obtain the points of intersection (1, $n), 11.,fzr), (1, l$rr) , and that eight points of intersectionare shown. (1,,I12r).Wenotice in Fig. 1.3.3.1 The other four points are obtained if we take another form of the equa-L, which is the tion of the circle r: 1; that is, consider the equation r: same circle. Solving this equation simultaneously with the equation of the four-leafedrose,we have sin 20 : -+.Then we get 20:&n,#n, #n, andTn 0 - #n, #n,1&n, and #n F i g u r e1 3 . 3 . 1 Thus, we have the four points (-L,#n),1-L,#r), (-7,1&n), and (-t,1?r). 568 POLAR COORDINATES Incidentally, (-1 , #n) alsocan be written as (1,,#n), (-1, #n) can be written as (L, # r ) , ( - 1 , # n ) can be written as (1,#n), and (-t ,Hzr) can be written as (1,#n). Because(0, 0) represents the pole for any 0, we determine if the pole is a point of intersection by setting r: 0 in each equation and solving for 0. Often the coordinatesof the points of intersection of two curuescan be found directly from their graphs. However, the following is a general method. If an equation of a curye in polar coordinatesis given by (1) r - f(0) then the same curye is given by (-l)nr:f@*nn) where n is any integer. o rLLUSrRArroN 1: Consider the curyes of Example 1. The graph of r - 2 sin 20 also has the equation (by taking n:'1. in Eq. (2)) (-t)r- 2 sin 2(0irr) or, equivalently, -r:2 sin 20 If we take n : 2 in Eq. Q), the graph of r : 2 sin 20 also has the equation (-7),r: 2 sin 2(0 + 2n) or, equivalently, r:2 s i n2 0 which is the same as the original equation. Taking n aurry other integer, we get either r:2 sin20 or r: -2 sin 20.The graph of the equation r: 1 also has the equation (by taking n: I in Eq. (Z)) (-1;r: 1 or, equivalently, r: -7 Other integer values of n in (z) applied to the equation 1 give either r: I or r - -1. o If we are given the two equations r - f(0) and r - g(0), we obtain all the points of intersection of the graphs of the equations by doing the following. (a) Use (2) to determine all the distinct equations of the two cur:ves: r - f r ( O ,) r : f r ( 0 ), r : f r ( O ,) (3) 13.3 INTERSECTION OF GRAPHSIN POLARCOORDINATES r:8t(0) , r: S r @ ), r : k @ ) , 'r;;ttaneously (4) (b) Solve each equation in (3) with each equation in (a). (c) Check to see if the pole is a point of intersection by setting r: 0 in each equation, thereby giving and s@)-0 f@l-0 (5) If Eqs. (5) each have a solution for 0, not necessarily the same, then the pole lies on both cunres. EXAMPLE2: Find the points of intersection of the two curyes r -- 2 - 2 cos 0 and r - 2 cos 0. Draw sketches of their graphs. sot-urroN: To find other equations of the cur:ve represented by r 2 2 cos 0, we have (-1)r:2- 2 cos(0+n) or, equivalently, -r:2+2cos0 and (-1)'r:2- 2 cos(p+2n) or, equivalently, r-2-2cos0 which is the same as the original equation. In a similar manner, we find other equations of the curve given by r : 2 c o s0 : (-L)r:2 -r: cos(O* r) -2 cos0 r:2 c o s0 which is the same as the original equation. So there are two possible equations for the first curve, r: 2 - 2 cos 0 -r :2 * 2 cos0, and one equation for the secondcurve, r:2 cos0. and Solving simultaneously r: 2 - 2 cos 0 and r: 2 cos 0 gives 2cos0-1-2cos0 4 c o s0 : 2 c o s0 : * Thus, e - En and En, giving the points (1,+zr)and (I,8n). Solving simult a n e o u s l y - r - 2 + 2 cos g and r :2 cos 0 yields 2 + 2 c o s0 - - 2 cos 0 570 POLARCOORDINATES 4 cos0: -2 c o sd : - * Hence, 0 : *rr and 0 : *r, giving the points (-1,, ?ir) and (-1, fz). However, (-1, 3z) is the samepoint as (1, *rr), and f 1, *z) is the samepoint as (1, $r). Checking to see if the pole is on the first cuwe by substituting r: 0 in the equationr:2 - 2 cos d, we have 0 : 2 - 2 c o s0 cos 0 : 1 0:0 Figure 13.3.2 Therefore, the pole lies on the first currre. In a similar fashion, by substifuting r: 0 in r:2 cos0, we get 0:2 c o s0 c o s0 : 0 0:ir or tn So the pole lies on the second curve. Therefore, the points of intersection of the two curves ate (7, |sr), (1, *n\, and the pole. Sketchesof the two curves are shown in Fig. 13.3.2. L3.3 Exercises In Exercises1 through L4, find the points of intersection of the graphs of the given pair of equations. Draw a sketch of each pair of graphs with the same pole and polar axis. .' ' [ 2 r : 3 l t:3sin0 ^t ' l r(:21r : 3 4- '. [ r : 2 c o s 2 0 [r: 2 sin0 u - '. { r : 4 0 Lr: ir 6 :!n " ' . [[ e r: 2 ," [ r : c o s 0 - l lr: cos20 R Jz:l-sind "' [r: cos20 o [r:sin2l " lr: cos20 *cos 0 n.[r':2cos| I r:1 t- .- [' r : t a n ? Lr:4sind ,-^- ' [ r : 4 ( 1 * s i n 0 ) [r(f - sin 0) :3 ,ô^ ' [ r ' s i n 2 0 : 8 I r c o s0 : 2 ^ f t : 2 c o s0 c'lr:2sinO p - -.-{ r : 4 t a n 0 s i n 0 [r:4cosO In Exercises15 and 15, the graph of the given equation intersects itself. Find the points at which this occurs. 'l..6. 15. r: sin *0 r: ! * 2 cos20 13.4TANGENTLINESOF POLARCURVES 13.4 TANGENT LINES In Sec. 13.2. we derived a formula for finding the slope of a tangent line OF POLAR CURVES to a polar curve whose equation is r: f (0).If a radians is the measureof the inclination of the tangent line to the curve at (r,0)' we have (1) Formula (1) is generally complicated to aPply. A simpler formula is obtained by considering the angle behmeenthe line OP and the tangent line. This angle will have radian measure 1 and is measured from the tine OP counterclockwise to the tangent line, 0 < X < 7t. There are two possible cases:o > I and ot I 0. These two casesare and 13.4.2.InFig. 13.4.L,a) 0 and 1:a-0' illustratedin Figs. 1.3.4.1 In Fig. 13.4.2,a 1 0 and 1: T - (0 - a). In each case, tan 1: tan(a - 0) o F i g u r e1 3 . 4 . 1 or, equivalently, tanX: tan a- tan 0 (2) L+tan o-tan0 Substituting the value of tan a from (1,)in (2), we get (?i' # t,G\;'""') : =^"u : tan x t.K" :') tant #* r -tan e #* r tanz0 o#*r-no Hence, Figurc 13.4.2 iiit+il++$fi+f+1iffi (3) more Comparing formula (3) with formula (1), you can see why it is coordinates. desirable to consider 1 instead of a when working with Polar ExAMPLEL: Find 1 for the cardiotdr- a* a sin 0 at the point (Ea, En). soLUTIoN: See Fig. acose '1.3.4.3. Because a * a sin 0, we have 572 POLARCOORDINATES Applying formula (3), we get tan a*asin0 acos0 1+sin0 cos 0 Therefore, at the point (ra, tn), tt/s ft-- v3 t a nX - # : = 3 So X: *n F i g u r e1 3 . 4 , 3 EXAMPLE2: Find to the nearest L0 min the measurement of the smaller angle between the tangent lines to the curyes r - Z cos 20 and r - 3 sin 20 at the point Pefr,t.o). If B is the radian measureof the anglebetweenthe tangentlines at P?\/2,grr),then F: xt- x, So tan B: tan(11- 12) or, equivalently, tan '6 - =tT xt tan x' 1 + tan Xr tany" (4) and tan X2 arcfound from formula (3). For tan yr, r: _ and, Tul,I, drld0: -6 sin 20. So Figure 13.4.4 1 f-A - -n-Awr_ : _3 6c o s 2 0 sin n:-)cot20 When 0 : tzr, tan 1, : -* cot izr : -*. For tan Xz,r:3 sin 20 and,drlilT: 6cos 20. So t a n ; i:,m : | t a n z o When e - *n, tan Xz: t tan LEzr : t. Substituting tan Xr: -* and tan Xz:* in (4), we obtain : 1l + = t a n B--@ --+ -1 4 j cos20 13.5 AREA OF A REGIONIN POLARCOORDINATES The angle of radian measure B has a measurementof 125'50'to the nearest L0 min. So the measurement of the smaller angle between the two tangent lines is 180o- 125'50': 53o10'. 1-3.4 Exercises In Exercises1 through8, find 1 at the point indicated. 2. r : a0; (tra, Ezr) l. r0 : a; (a, l) 4. r:asin*g; (tra,*r) 7. t2 : a2 cos20; la 1 3' r : a sec20;(t/-za,-+r) 6. r:acoslli(iuIa,Jzn) 5. r:02;(*n',tr) \ \U,i") 8' t: a(l- sin 0); (a' r) at the In Exercises9 through 12, hnd a measurement of the angle between the tangent lines of the given pair of curves indicated point of intersection. t {i=i:trt,(*r2a,rn) 'n {;=Lo"u.e;@,*d y: 4 cosô. --2, - ' o&n) "/ n. " ' . t r : 4 c o s 2 0 - 3 -: '\ p. {:=;'"::\:t the pole U:l pair of curvesat all In Exercises13 through 15,find a measurementof the anglebetweenthe tangentlines of the given points of intersection. : 2(1- cos 0) ' 17. Prove that tan 1 : tan *0 at all points of the cardioidt 18. Prove that at each point of the logarithmic spiral r: beoe,1is the same' L9. prove that at the points of intersection of the cardioids r: pe{pendicular for all values of.a arrd b. a(l * sin 0) and r: 20. Prove that at the points of intersection of the fwo curyes r: Pe{Pendicular. 1.3.5AREA OF A REGION IN POLAR COORDINATES a sec2te and r: b(l - sin 0) their tangent lines are b csc2te their tangent lines are In this section a method is developed for finding the area of a region bounded by a curve whose equation is given in polar coordinates and by two lines through the pole. Let the function / be continuous and nonnegative on the closed interval [a, B]. Let R be the region bounded by the curve whose equation is /: /(0) and by the lines 0 : a and 0: B. Then the region R is the region AOB shown in Fig. 13.5.1' ConsiderapartitionAof[a,B]definedbyo:0o<0,<02<..< 574 POLAR COORDINATES . 10n_, 19n:B. Therefore,we have z subintervalsof 0rr1&1. the form f0i-1,01f, where i:'1 ,2, . . . ,n.Let $ be a value of g in the ith subinterval f0i-1,061. SeeFig.13.5.2,wherethe ith subinterval is shown together with 0: f;. The radian measure of the angle between the lines 0: 0*r and 0: 0r is denoted by Lg. The number of square units in the area of the circular sector of radius /(f1) units and central angle of radian measure AaOis given by ElfG)],L,0 There is such a circular sector for each of the n subintervals. The sum of the measuresof the areas of these n circular sectors is F i g u r e1 3 . 5 . 1 +lf(f,)l' a,o+ +lf(t,)12Lzo + + +lf(f,)I2 Lio+ + +lf(€")1,Ln? which can be written, using sigma notation, as n > +tfc)f, L,0 Let llAllbe the norm of the partition A; that is, IlAllis the measureof the largest Lr?.Then if we let A square units be the area of the region R, we define #*ffi .......#=+U a,o I+r, ft$, Figure13.5.2 The limit in (t) is a definite inte gral, and we have #$ EXAMPLE1: Find the area of the region bounded by the graph of r:2+2cos0. (2) The region together with an element of. areais shown in Fig. ::l]t"_"' 13.5.3.Becausethe curve is symmetric with respectto the polar axis, we take the 0 limits from 0 to z which determint the area or tn" region bounded by the curve above the polar axis. Then the area of the entire region is determined by multiplying that area by 2. Thus, if A square units is the measure of the required area, A :2 ,!ip i+ fr *2 l l a l l - o- : -n 1 -4 costi) , Aio r +(2 + 2 cos 0)2 d0 I F i g u r e1 3 . 5 . 3 (1) ( 1 + 2 c o s 0 + c o s z 0 )d 0 sin 0 + te + * sin *]: 13.5 AREA OF A REGIONIN POLARCOORDINATES - 4(n* 0 + in * 0 - 0) :6n Therefore, the area ts 6n square units. ExAMPLE2: Find the area of the region inside the circle r:3 sin0 3 sin0-2- sin0 sin0:Lz and outside the limagon r:2- To find the points of intersection,we set solurroN: So 0:tr sin0 r :, f (0) Li1 ,0:Ei ' t,) $(Eil (g(€;),tr) and 0:8n The curves are sketched and the region is shown together with an element of area in Fig. 13.5.4. - - sin 0, then the equation of the If we let f @):3 sin 0 and g(0) I circle is r: f (0), and the equation of the limagon is r: 8(0). The measure of the area of the element of area is the difference of the measuresof the areas of two circular sectors. tlf G)l' toe- ils?)l' 7 :, g(0) : +([/(fn)]' - [g(f') ]') aoo Ln? The sum of the measuresof the areas of r such elements is given by n A'o > +(t/(f')l'-[g(f,)]') C=1 Hence, if A square units is the area of the region desired, we have F i g u r e1 3 . 5 . 4 A: tisr i +ttrtE lt'- [g(f,)]')Aoo llall-oEi This limit is a definite integral. Instead of taking the limits tr to *rr, we or" thu property of symmetry with respect to the ttr axis and take the limits from Szrto ir and multiply by 2.We have, then, fn12 A:2. 1| ( t / ( 0 ) l ' - l s @ ) l ' zd)e J t16 : f::re sin 0)'l d0 -8 e d0+4 s i n0 d e - 4 f: f:sinz -8 - c o s 2 0 )d o + f:(1 :40-2sin20-4 cos o - Ae'n'' lnrc 576 POLAR COORDINATES sin 20- 4 cos _ (-2 sin 7T- 4 cos (-2 sin tn - 4 cos En) -2.+\E+4.+\n - 3{g Therefore,the area is 3 \n square 13.5 Exercises In Exercises L throu gh 6,find the area of the region enclo sed by the graph of the given equation. 1. r:3cos0 2.r-Z-sin0 3. r-4 c o s3 0 4.r:4sin2t0 5. r2:4sin20 6. r:4 s i n 20 c o s 0 In Exercises 7 through 10, find the area of the region encl sed by one loop of the graph of the given equation. 7. r: 3 cos20 9. r-"1,+3sin0 8. r:a 10. r: sin30 a ( l - 2 c o s0 ) In Exercises LL through "1,4,find the area of the intersection of the regionsenclosedby the graphsof the two given equations. 1 1 [r-2 ^^' lr - 3 - 2 cos 0 sin20 4 rr'^ l r --3 3 cos 2o lr l r' : 4 : s i n 0 12..t lr-4cos0 (-2 - 2 cos20 1,4.{ ' I r- 1 In ExercisesL5 through 18,find the areaof the region which is inside the graph of the first equation and outside the graph of the secondequation. 15. [ r --o l r a ( l - c o s0 ) - 1 7 . ft 2 sin o Lr-sin0*cosg - 2a srn o L6. {r lr-a - 4 stn 20 j.g. r \ " r [r' L t- \n 19. The face of a bow tie is the region enclosed by the graph of the equation 12: 4 cos 20. How much material is necessary to cover the f.aceof the tie? 20. Find the area of the region swept out by the radius vector of the spiral r: not swept out during its first revolution. a0 duing its second revolution which was 21- Find the area of the region swept out by the radius vector of the curve of Exercise20 during its third revolution which was not swept out during its second revolution. ReaiewExercises (Chapter13) In Exercises7 and'2, find a polar equation of the graph having the given cartesian equation. l. x2*y'-9x+8y:0 Z.yn:*(ar-yr) REVIEWEXERCISES In Exercises3 and4, find a cartesian equation of the graph having the given polar equation. 4. r: 3. r-9sinzi0 a tanz 0 3/d (reciprocal spiral) and (b) r: 013(spiral of Archimedes). 5. Show that the equations r: L I sin 0 and /: sin 0 - t have the same graph. y'lios 0f. 7. Dtaw a sketch of the gaph o1 7 : 5. Draw a sketch of the graph of (a) r: 8. Draw a sketctr of the graph of r: {@{E| 9. Find an equation of each of the tangent lines at the pole to the cardioid' r: 3 3 sin 0' 10. Find an equation of each of the tangent lines at the pole to the four-leafed rose r:4 cos20. 11. Find the area of the region inside the graph of r:2a sin 0 and outside the graph of r: a' a cos 0 and 12. Find the area of the intersection of the regions enclosed by the graphs of the two equations /: > 0. a 0), cos r: a(l r: stco(k > 0) as 0 varies from 13. Find the area of the region swept out by the radius vector of the logarithmic spiral 0 to 2tr. :2 sin 20 and outside the graph of the circle r: 1' 14. Find the area of the region inside the graph of the lemniscate 12 : (o , n) . 15. Find a polar equation of the tangent line to the curve r 0 at the point the two given equations' In Exercises16 and 17, find all the points of intersection of the graphs of L6{;:l i;lo,e aF, | ,-2(l+ Lt' lrr-2cos0 cos 0) of points of intersection. Also prove that the 1g. prove that the graphs of r: a0 and.r0: a have an unlimited number and find these points' intersection, tangent lines are perpendicular at only two of these points of each point of intersection of the graphs of the equa19. Find the radian measure of the angle between the tangent lines at -6 cos 0)' cos 0 and,t:2(1 tions r: : point (1, *n)' 20. Find the slope of the tangent line to the curve r 6 cos0 2 atthe * 2 cos 0) and also the area of the region en21. Find the area of the region enclosed by the loop of the limagon r:4(7 closed by the outer part of the limagon' n is a positive integer' 22. Find.the area enclosed by one loop of the curve r: a sin n0, wherc is 23. Prove that the distance between the two points PJh, 0) and'PzQz' 0) @ 24. Find the points of intersection of the graphs of the equations /: tan 0 and r: col 0; (2, tr,) and the line 0 : 0 25. (a) Use the forrrula of Exercise23 to find a polar equation of the parabola having its focus at of the paraboh hJving its focus at (0,2) and the r axis as its directrix' as its directrix. (b) Write a cartesian "qtr"tiotr Compare the results of parts (a) and (b). law of cosinesto 25. Find a polar equation of the circle having its center at .(rs,!6) and a radius of a units. (rur'rr: Apply the (r, (ro, and 0)') 0), pole, at the the triangle hiving vertices 14.1 THE PARABOLA 14.I THE PARABOLA A conic section is a curve of intersection of a plane with a right cirorlar cone of two nappes. There are three types of curr/esthat occur in this way: the parabola, the ellipse (including the circle as a special case), and the h;rperbola. The resulting cuwe depends on the inclination of the axis of the cone to the cutting plane. In this section we study the parabola. Following is the analytic definition of a parabola. 1,4.1.1Definition A parabolais the set of all points in a plane equidistant from a fixed point and a fixed line. The fixed point is called the focus, and the fixed line is called the directrix. We now derive an equation of a parabola from the definition. In order for this equation to be as simple as possible, we choose the r axis as perpendicular to the directrix and containing the focus. The origin is taken as the point on the r axis midway between the focus and the directrix. It should be stressedthat we are choosing the axes (nof the parabola) in a specialway. SeeFig.1'4.1"1" x: -l a(- p,v) .1 14.1 Figure then P is on the parabola if and only if lFTl:lOPl Because and lilPl: t/(.T\YTTW P is on the parabola if and onlY if \/G=pYTT: {GTff 580 THE CONIC SECTIONS By squaring on both sides of the equation, we obtain xz - 2px * p' * y' : )c2* 2px * p' Y' : 4Px This result is stated as a theorem. 14.1.2Theorem An equation of the panbola having its focus at (p,0) and as its directrix the line y:-p is (1) F(p,o) F i gur e 14 . 1. 2 14.1,.3Theorem '!.4.'1..'!,, In Fig. p is positive; p may be negative,however, becauseit is the directed distance Of. Figure 14.1.2shows a parabola for p < 0. From Figs. 14.1.1and 1,4.'1,.2, we seethat for the equation yz :4px the parabola opens to the right if p > 0 and to the leftif p < 0. The point midway between the focus and the directrix on the parabola is called the ztertex.The vertex of the parabolasin Figs. 14.1.1and 14.1.2is the origin. The line through the vertex and the focus is called the axisof the parabola. The axis of the parabolasin Figs. 14.1.1and 1.4.1.2isthe r axis. In the above derivation, if the r axis and the y axis are interchanged, then the focus is at the point F(0, p), and the directrix is the line having the equation y:-p. An equation of this parabola is *:4py, and,thi result is stated as a theorem. An equation of the parabola having its focus at (0, p) and as its directrix the line y : -p is (2) If p ; 0, the parabola opens upward as shown in Fig. 74.1.2; and ifp as shown in Fig. 14."1,.4. In each case the vertex is at the origin, and the y axis is the axis of the parabola. F(0,p) F i g u r e1 4 . 1. 3 Figure 14.1.4 14.1THE PARABOLA 581 When we draw a sketch of the graph of a parabola, it is helpful to draw the chord through the focus, perpendicular to the axis of the parabola. This chord is called the latus rectum of the parabola. The length of the latus rectum is lapl. (SeeExercise1.7.) EXAMPLE1: Find an equation of the parabola having its focus at (0, -3) and as its directrix the line y - 3. Draw a sketch of the graPh. sor,urroN: Since the focus is on the y axis and is also below the directrix, the parabola opens downward, and p: -3. Hence, an equation of the parabola is *:-12y The length of the latus rectum is l4(-3) | : 12 Q, Qt Qt u -c o A sketch of the graph is shown in Fig. 1'4.1.5. Ary point on the parabola is equidistant from the focus and the directrix. In Fig. 74.L5, three such points (Pr, Pr, an'd Pg) are shown, and we have - lre-,| I : lPre-,| tFP,l-lk-a,l lFtr,l IFP-, F i g u r e1 4 . 1 . 5 EXAMPLE 2. Given the Parabola having the equation solurroN: The given equation is of the form of Eq. (1); so 4p-7 u ': 7 x find the coordinates of the focus, an equation of the directrix, and the length of the latus rectum. Draw a sketch of the graPh. opens to the right. The focus is at the point Becausep - -2. The length of the latus F(+, 0). An equation of the directrix is x rectum rs 7. A sketch of the graph is shown in Fig. 1'4.1.6. Directrix F i g u r e1 4 . 1 , 6 THE CONIC SECTIONS L4.1 Exercises For each of the parabolas in ExercisesI thnrugh 8, find the coordinates of the focus, an equation of the directrix, and the length of the latus rectum. Draw a sketch of the curve. 1. x2:4A 4. xz- -I5y 2' a2:6x 3. y' : -8x 5.f +A:0 6.y2- 5r:0 7.2y'-9x:0 8.3x2*4y-0 In Exercises 9 through 16, find an equation of the parabola having the given properties. 9. Focus, (5, 0); directrix, y: -5. 10. Focus, (0, 4); directrix, A: -4. 11. Focus, (0, -2); directrix, y - 2:0. 12. Focus, (-8, O); directrix, 5 - 3r: 0. 13. Focus,(*, 0); directrrx,2xl l:0. 14. Focus, (0, i); directrix,Sy t2:0. 15. Vertex, (0, 0); opens to the leff length of latus rectum: 6. 15. Vertex, (0, 0); opens upward; length of latus rectum: 3. 17. Prove that the length of the latus rectum of a parabola is l4pl. 18' Find an equation of the parabola having its vertex at the origin, the r axis as its axis, and passing through the point (2, -4\. 19' Find an equation of the parabola having its vertex at the origin , the y axis as its axis, and passing through the point (-2, -4). 20' A parabolic arch has a height of 20 ft and a width of 36 ft at the base. If the vertex of the parabola is at the top of the arch, at what height above the base is it 1g ft wide? 21' The cable of a suspension bridge hangs in th9 form of a parabola when the load is uniformly distributed horizontally. The distance between two towers is rsoo ft, the points ;i;;;6; of the cable on the to*"r, arc 220fr above the road- il1til*:T#ffiit:HfiJ?ff31""T"70rt aloveth",oa'd*"v. Findthevertical distance tothecabre rromapoint 22' Assume that water issuing from the end of a horizontal pipe,25 t lb9",u the ground, describesa parabolic curve, the vertex of the parabola being at the end of the pip-e.r, ui u'poit t 8 ft below trr"etine of the pipe, the flow of water has curved outward 10 ft beyond a vertical line throuih th; ;"e iiir" pip", how far beyond this vertical line will the water strike the ground? 23' A reflecting telescopehas a parabolic mirror for which the distance from the vertex to the focus is 30 ft. If the distance acrossthe top of the mirror is il in , how deep is the mirroiai ttre centerz 24' Using Definition 14'1.1,find an equation of the parabola having as its directrix the line a :4 and.asits focus the point (-3, 8). 25' using Definition t4'l'1, hnd' an equation of the parabola having as its directrix the line r:-3 point (2,5). and as its focus the 26' Find all points on the parabola y2: 8r such that the foot of the perpendicular drawn from the point to the directrix, the focus, and the point itself are verrices of an equirateiJil;gil,' 27' Find' an equation of the circle passing through the vertex and the endpoints of the latus rectum of the parabola ,, - -gy. 14,2 TRANSLATIONOF AXES 2g. prove analytically that the circle having as its diameter the latus rectum of a parabola is tangent to the directrix of the parabola. 29. A focal chord of a parabola is a line segment through the focus and with its endpoints on the parabola. If A and B are the endpoints of ifocal chord of a parabola, and ii C is the point of intersection of the directrix with a line thrcugh the vertex and point A, prove that the line through C and B ii parallel to the axis of the parabola. is half the fl). prove that the distance from the midpoint of a focal chord (see Exercise 29\ of. a parabola to the directrix length of the focal chord. 14.2 TRANSLATION OF AXES The shape of a curve is not affected by the position of the coordinate axes; however, an equation of the curve is affected' . rllusrRArrorv 1: If a circle with a radius of 3 has its center at the point (4, -1), then an equation of this circle is (x-4)'+@+1)':e **Y'-8r*2Y*8:0 Howevef, if the origin is at the center, the same circle has a simpler equation, namelY, ' f+Yz:g Ifwemaytakethecoordinateaxesasweplease,theyaregenerally chosen in such a $/ay that the equations will be as simple as possible. Iftheaxesaregiven,however,weoftenwishtofindasimplerequation of a given curve referred to another set of axes' Ingeneral,ifintheplanewithgivenrandyaxes,Itewcoordinate a chosen parallel to the given ones, we say that there has been "re" """s ttanslation of axesin the Plane' y' In particular' let the gi*tett x and'y axes be translated to the x' and that assume Also, axes. axes,hiving origin (h, k)-withrespect to the givgn x' and y' the positive no*b"rs are on the iame side of the origin on the axes as they are on the x and'y axes (see Fig' 1'a'2'1)' (h,k) A, l N, I I I I Figure14.2.1 A point P in the plane, having coordinates (x, y) wilh resPectto the the given coordinate axes; wiil have coordinates (x' , y') with resPectto 584 THE CONIC SECTIONS new axes.To obtain relationships between these two sets of coordinates, we draw a line through P parallel to the y axis and the y' axis, and also a line through P parallel to the r axis and the I' axis. Let the first line intersect the r axis at the point A and the r' axis at the point A', and the second line intersect the y axis at the point B and the y' axisat the point B'. With respect to the r and y axes,the coordinates of P ate (x, y)' the coordinates of A are (r, 0), and the coordinates of A' are (r, k). Because A,p:n-M,wehave y':Y-k or, equivalently, y:y'*k With respect to the x and y axes, the coordinates of B are (0, y), and the coordinates of B' are (h, y). BecauseB'P : BP - BB' , we have X,:X-h or, equivalently, x:x'*h These results are stated as a theorem. 14.2.1Theorem lf (x, y) representsa point P with respectto a given set of axes,and (x' ,y') is a representation of P after the axesare translated to a new origin having coordinates (ft, k) with respect to the given axes, then y-v' +k (1) y'-y -k (2) ot, Equations (1) or (2) are called the equationsof translating the axes. If an equation of a curve is given in x andy, then an equation in r' and y' is obtained by replacing x by (x' -t h) and y by (y' -f k).The graph of the equation in r and y, with respect to the x and y axes,is exactly the same set of points as the graph of the corresponding equation in r' and y' with respect to the x' and y' axes. ExAMPLE1: Given the equation x2 + Llx * 6y + 19: 0, find an equation of the graph with respect to the x' and y' axes after a translation of axes to the new origin (-5, 1). soLUrIoN: A point P, represented by (x, y) with respect to the old axes, has the representation(r', y') with respectto the new axes.Then by Eqs. (1), with h: -5 and k : 1, we have x'-5 and y-y' +1 Substituting these values of r and y into the given equation, we obtain ( x ' - 5 ) , + 1 0 ( r ' , -5 ) + 6 ( y ' , +1 ) + 1 g : 0 14.2 TRANSLATIONOF AXES x'2 - "1.0x' + 25 * l}x'- v--+ x' Figure 14.2.2 50 * 6y' + 6 + 19 - 0 )c'2: -6Y' The graph of this equation with respect to the r' and y' axesis a parabola with its vertex at the origin, opening downward, andwith 4p :-6. The graph with respect to the x and y axes is, then, a parabola having its vertex at (-5, 1), its focus at (-5, -*), and as its directrix the line y : $ (seeFig. 1,4.2.2). The above example illustrates how an equation can be reduced to a simpler form by a suitable translation of axes. In general, equations of the second degree which contain no term involving xy can be simplified by a translation of axes. This is illustrated in the following examPle. EXAMPLE2: Given the equation * 32y + 37: 0, 9xz * 4y' - "1,8x translate the axes so the equation of the graph with respect to the )c' and y' axes contains no first-degree terms. vy' solurroN: We rewrite the given equation: 9(* - 2x) + 4(yz * 8y) :-37 ' Completing the squares of the terms in parenthesesby adding 9 I' and 4 ' 16 on both sides of the equation, we have 9(* - 2x* 1) + +(y' * 8y-r 1'6):-37 + 9 + 64 9 ( x - L ) ' + 4 ( y + 4 ) 2: 3 6 Then, if we let r' 1 and y' : y + 4, we obtain 9x'2* 4y'' :35 : yI 4 From Eq. (2), wesee that the substitutions of r' : x - L andy''l'4.2.3' -4). we In Fig. result in a translation of axes to a new origin of.(1, have a sketch of the graph of the equation in r' and y' with resPect to the r' and v' axes. Figure14.2.3 :# Figure 14.2.4 We shall now apply the translation of axes to finding the general equation of a parabola having its directrix parallel to a coordinate axis and its vertex at the point (h, k). In particular, let the directrix be parallel to the y axis. If the vertex is at point V(h, k), then the directrix and the focus is at the point F(h *p, k). x, has the equation x:h-p, Let the .r' and y' axes be such that the origin O' is at V(h, k) (see Fig.1,4.2.4). An equation of the parabolain Fig. 1'4.2.4with respectto the r' and *u' axesis y'': 4px' THE CONIC SECTIONS To obtain an equation of this parabola with respect to the x and y axes,we replacex' by (x - h) andy' by (y - k) from Eq,.(2),whichgives us (Y-k)':4P(x-h) The axis of this parabola is parallel to the I axis. Similarly, if the directrix of a parabola is parallel to the r axis and the vertex is at V(h, k), then its focus is at F(h, k + p) and the directrix has the equation A : k - p, and an equation of the parabola with respect to the x and y axes is ( x - 7 1 2 : 4 P ( Y- k ) The axis of this parabola is parallel to the y axis. We have proved, then, the following theorem. 1.4.2.2Theorem If p is the directed distance from the vertex to the focus, an equation of the parabola with its vertex at (h, k) and with its axis parallel to the x axis is 'i{.p'.H. ; Aik..ru}*. fr) (3) A parabola with the same vertex and with its axis parallel to the y axis has for an equation (r - h)' : 4P{V- k) EXAMPLE 3: Find an equation of the parcbola having as its directrix the line U : 1,and as its focus the point F(- 3, 7), F(- 3,7) v(- t, +)l (4) soLUrIoN: Since the directrix is parallel to the x axis, the axis will be parallel to the y axis, and the equation will have the form (4). since the vertex v is halfway between the directrix and the focus, v has coordinates (-3, 4).The directed distance from the vertex to the focus is p, and so P:7-4:3 Therefore, an equation is (r * 3)'z:12(y - 4) Squaring and simplifying, we have * * 6 x - ' l - 2 y* 5 7 : 0 Figure14.2.5 ExAMPrn 4: Given the parabola having the equation y'+6x*8y*1:0 A sketch of the graph of this parabolais shown in Fig. 1.4.2.5. sol.urroN: Rewrite the given equation as y ' + 8 y- - 6 x - 1 Completingthe squareof the terms involving y on the left side of this OF AXES 14.2TRANSLATION find the vertex, the focus, an equation of the directrix, an equation of the axis, and the length of the latus rectum, and draw a sketch of the graph. v equation by adding L5 on both sides/ we obtain y'+8y*'1.6:-6x+L5 (y + 4)' 6(x- B) Comparing this equation with (3), we let Directrix h:E k--4 and 4P--6 or P:-g Therefore,the vertex is at ($, -4); an equation of the axis is y:-4;the focus is at (7,-4); an equation of the directrix is x:4; and the length of the latus rectum is 6. A sketch of the graph is shown in Fig. 14'2.6. Figure 14.2.6 1-4.2 Exercises In Exercises1 through 8, find a new equation of the graph of the given equation after a translation of axesto the new origin as indicated. Draw the original and the new axes and a sketch of the graph. 2. x2+ y2- L}x * 4y + 13- 0; (5,-2) L . x 2+ y ' + 6 x * 4 Y : 0 ; ( - 3 , - 2 ) 3.y2-6x*9:0; (8,01 5. xz+ 4y' * 4x * 8y + 4 : 0; (-2, -l) 7 . y - 4 - 2 ( x- 1 ) t ;( 1 ,4 ) 4. y' + 3x - 2y * 7 - 0; (-2, t) 6 . 2 5 x z* y ' - 5 0 r * 2 0 y - 5 0 0 : 0 ; ( 1 , - 1 0 ) 8. (Y+ l)' : 4(x - 2)'; (2' -I) contain no In Exercises9 through 12, translate the axes so that an equation of the graph with respect to the new axeswill graph. the of and a sketch axes new the first-degree terms. Draw the original and 10. l5x2 * 25y2- 32x - l00y - 284 0 9. x2 + 4y' - l5x * 24y + 84- 0 L2. x:2- y' + L4x- 8y- 35 - 0 L1..3x2- 2y' * 6x- 8Y- LL : 0 In Exercises13 through 18, find the vertex, the focus, an equation of the axis, and an equation of the directrix of the given parabola. Draw a sketch of the graph. 1 5 .y ' + 6 x * 1 0 y + 1 9 - 0 1 , 4 .A x z- 8 r * 3 Y- 2 : 0 L 3 .x 2 * 5 x * 4 y * 8 : 0 1 8 .Y : 3 x 2 - 3 x * 3 17. 2y' 4y 3x 16.3y'-8x-l2Y-4-0 In Exercises19 through 28, find,an equation of the parabola having the given properties. Draw a sketch of the graph. 19. Vertexat (2,4); focusat (-3,4). 20. Vertex at (1, -3); directrix, Y: 1. 21. Focusat (-7,7); directtu, Y:3. 22. Focusat (-*,4); directrix, x:-*. 588 THE CONIC SECTIONS 23. Vertex at (3 , -2); axis, r : 3; length of the latus rectum is 5. 24. Axis parallel to the r axis; through the points (I,Z), (S,Z), and (11,4). 25. Vertex at (-4,2); acis,y :2; through the point (0, 6). 25. Directrix, x: -2; a<is, y: 4; length of the latus rectum is 8. : 27. Directrix , x : 4) aocis, .y 4; through the point (9, Z) . 28. Endpoints of the latus rectum are (1, 3) and (7,3). 29. Given the parabola having the equation V : af I bx -f c, with a * 0, frnd.the coordinates of the vertex. 30. Find the coordinates of the focus of the parabola in Exercise29. 31. Find an equation of every parabola containing the points A(-9, -4) and B (5, -4), such that points A and,B are each 5 units from the focus. 32. Given the equation 4xs- l2xz * l2x - 3y - l0: 0, translate the axes so that the equation of the graph with respect to the new axeswill confail no second-degreeterm and no constant term. Draw a sketch of ttre grap-hand the two sets of axes.(rrrNr:Let 2e:x'Ih andy:y,+ k in the given equation.) 33' Given the equation. xs + 3* - y'-+ 3x * 4y - 3 : 0, translate the axes so that the equation of the graph with respect to the new axes will contain_no first-degree term and no constant term. Draw a sketch of the graph irrd tfru two sets of axes (seehint for Exercise32). 34' lf a parabola has its focus at the origin and the x axis is its axis, prove that it must have an equation of the form !2:4kx+4P,k+0. 1.4.3SOME PROPERTIES In Sec' 14'1we stated that a conicsection(or conic)is a curve of intersection OF CONICS of a plane with a right-circular cone of two nappes, and three types of curves of intersection that occur are the paraboli, the ellipse, and,Tie hyperbola. The Greek mathematician Apollonius studied conic sectiohs, in terms of geometry, by using this concept. we studied the parabola in sec. 14.1,where an analytic definition (ri.l.r) of a parabolawas given. In this section, an analytic definition of a conic r""iior, is given and the three types of curves are obtained as special casesof this d-efinition. generator In a consideration of the geometry of conic sections,a cone is regarded as having two nappes, extending indefinitery far in both directions. A portion of a right-circular cone of two nappes is shown in Fig. r4.g.r. upPer naPPe A generator (or element) of the cone is a line lying in the cone, and all -the the generators of a cone contain the point v, calied aertexof the cone. In Fig. 1'4.3.2we have a cone and a cutting plane which is parallel to one-and only one generator of the cone. Thii conic is a parabola.rf the cutting plane is parallel to two generators, it intersects both nappes of the cone and we have a hyperbola(Fig. 1a.3.3).An ellipseis obtained if lower nappe the cutting plane is parallel to no generator, in which case the cutting plane intersects each generator, as in Fig. 1,4.g.4. generator , A special caseof the ellipse is a cirire, which is formed if the cutting plane, which intersects each generator, is also perpendicular to the axis F i g ur e 14 . 3 . 1 of the cone. Degenerate cases of the conic seitions include a point, a SF C O N I C S 1 4 , 3S O M E P R O P E R T I EO F i g ur e 1 4 . 3 . 2 F i g ur e 14 . 3 . 3 F i g u r e1 4 . 3 . 4 straight line, and two intersecting straight lines. A point is obtained if Ure cutting plane contains the vertex of the cone but does not contain a generator. This is a degenerate ellipse. If the cutting plane contains the vertex of the cone and ot ly orr" generator, then a straight line is obtained, and this is a degenerateparabola. A degeneratehlryerbola is formed when the cutting plane contains the vertex of the cone and two generators, thereby giving two intersecting straight lines' There are many applications of conic sections to both pure and applied mathematics. We shall mention a few of them. The orbits of planets and satellites are ellipses. Ellipses are used in making machine gears' Arches of bridges areiometimes ellipticat or parabolic in shape' The path of a projectite Is a parabola if motion is considered to be in a plane and air resisiance is ne-glected.Parabolas are used in the design of parabolic mirrors, searchlighis, and automobile headlights. Hyperbolas are used in combat in "sound tanging" to locate the position of enemy guns by the sound of the firing of ihoi" g,trrr. If a quantity varies inversely as anpt"ssote and volume in Boyle's law for a perfect other quantity, such "s the graph is a hyperbola' temperature, gas at a constant To discuss conics analytically as plane curves, we first state a defini' tion which gives a property common to all conics' L4.3.1 Definition A conic is the set of all points P in a plane such that the undirected distance of P from a fixed point is in a corrstantratio to the undirected distance of P from a fixed line which does not contain the fixed point. The constant ratio in the above definition is called t}re eccentricityof the conic and is denoted by e. The eccentricity e is a nonnegative number becauseit is the ratio of two undirected distances.Actually for nondegenIf erate conics, e ) 0. (Later we see that when e:0, we have a point') is the conic 14.1.1 that and Definitions'1,4.3.1' e:L,we Seeby comparing THE CONIC SECTIONS a parabola. If e < 1, the conic is an ellipse, and if e t L, the conic is a hyperbola. Tlre fixed point, referred to in Definition 14.3.1,is called a focus of the conic, and the fixed line is called the corresponding directrix. We leamed in sec. 14.1that a parabolahas one focus and one directrix. Later we see that an ellipse and a hyperbola each have two foci and two direcfrices, with each focus corresponding to a particular directrix. The line through a focus of a conic perpendicular to its directrix is called the principal axis of. the conic. The points of intersection of the conic and its principal axis are called the oerticesof the conic. From our study of the parabola, we know that a parabola has one vertex. However, both the ellipse and the hyperbola have two vertices. This is proved in the following theorem. 14.3.2 Theorem If e is the eccentricity of a nondegenerate conic, then if e # 'J.,the conic has two vertices; 7f e: 1, the conic has only one vertex. PRooF: Let F denote the focus of the conic and D denote the point of intersection of the directrix and the principal axis. Let d denote the undirected distance between the focus and iti directrix. In Fig. 14.3.5,F is to the right of the directrix, and in Fig.1.4.3.6,F is to the left of-the directrix. Let v denote a vertex of the conic. we wish to show that if e # L, there are two possible vertices V, and if e: L, there is only one vertex V. From Definition 14.3.1,we have F i g u r e1 4 . 3 . 5 l-wl:elfil Removing the absolute-value bars, we obtain Fn:*e(N) (1) Because D, F, and v are all on the principal axis, If F is to the right of D, DF : d. Thus, we have from the above Tv- d+N (2) Substituting from (1) into (2) gives Figure14.3.6 frl- d+ s(fr/) from which we get ,l Dv:fta (3) rf e:1, the minus sign in the above equations must be rejectedbecause we would be dividing by zero. so rf e * !, friro points v are obtained; if e : L, we obtain only one point V. 14.3SOMEPROPERTIES OFCONICS 591 If F is to the left of D, DF: -d. Instead of (3), we get _A ov:1fi from which the same conclusion follows. So the theorem is proved. I The point on the principal axis of an ellipse or a hyperbola that lies halfway between the two vertices is called the centerof the conic. Hence, the ellipse and the hyperbola are called central conicsin contrast to the parabola that has no center becauseit has only one vertex. 14.3.3 Theorem A conic is symmehic with respect to its principal axis. The proof of this theorem is left as an exercise (see Exercise 7). EXAMPLE1: Use Definition "1.4.3.1 to find an equation of the conic whose eccentricity is # and havin g a focus at the point F(1 , -2) with the line 4y * 17 :0 as the corresponding directrix. soLUTroN: Figure 14.3.7shows the focus, the directrix, and the point P (x, y) representing any point on the conic. Letting the point D be the foot of the pe{pendicular from P to the directrix, we have from Defini'1,4.3.'1, tion lgt:4 lPDl 5 and so :+l Tol lPTl Using the distance formula to find the above equatiolt, we get and lfrl and substituting into _ +yl + + l P(*,y) Squaring on both sides of this equation and simplifying, we get x 2 - 2 x * 1 . +y ' + 4 y * 4 - i # 0 ' + + y +i#) 25x2- 50x * 25yz* I00y + L25- L6yza B6y + 289 25x2t 9y'- 50r - 35y:164 Figure14.3.7 Becausee - + < L, the equation is that of an ellipse. 1,4.3 Exercises In Exercises1 through 5, use Definition 14.3.1and the formula of Exercise 24 of Exercises5.3 to find an equation of the conic having the given properties. I . F o c u sa t ( 2 , 0 ) ; 2. Focusat (0, e: i. e:t. 592 THECONICSECTIONS 3. Focusat (-3, 2); directrix:x: li e:3. 4. Focusat (1, 3); directrix:!:8i e: *. 5. FocusaI (4, -3); directrix: 2x - y - 2:0; 6. Focusat (-1, 4); directrix:2x - y i 3:0; e : i. e:2. 7. Prove Theorem 14.3.3. 8. Find an equation whose graph consists of all points P in a plane such that the undirected distance of P from the point (ae' 0) is in a constant ratio e to the undirected distance of P from the line x : al e. Let a2(l - er) :.r b2 and .orr-ridu. the three cases:e : l, e < 7, and,e > 7. 9. Solve Exercise8 if the point is (-ae , 0\ and the line is r : -ale. 14.4 POLAR EQUATIONS OF THE CONICS D x lTPl: 'lRPl F{ u o I-r 'o F i g u r e1 4 . 4 . 1 lOFl Because P is to the right of the directrix, - r because > r 0. So from (1) we have r - e(RP) (1) RP > 0; thus, lRPl: RF. (2) However, RP - m: DO + OQ d * r cos 0. Substituting this expression for RT in e), we get r-e(d*rcosg) (3) In a similar manner, we can derive an equation of a conic if the directrix coffesponding to the focus at the pole is to the right of the focus, and (4) The derivation of ( ) is left as an exercise (see Exercise L). If a focus of a conic is at the pole and the polar axis is parallel to the 14.4 POLAR EQUATIONSOF THE CONICS corresponding directrix, then the tn axis and its extension are along the principal axis. We obtain the following equation: (5) where e and d are, respectively, the eccentricity and the undirected distance between the focus and the coresponding directrix. The plus sign is taken when the directrix corresponding to the focus at the pole is above the focus, and the minus sign is taken when it is below the focus. The derivations of Eqs. (5) are left as exercises(see Exercises2 and 3). We now discuss the graph of Eq. (3) for each of the three cases:e: 1, ( e 1, and e > 1. Similar discussions apply to the graphs of Eqs. (a) and (5)' Case1: e: 1. The conic is a parabola. Equation (3) becomes to axis (6) 1 , - c o s0 P(r,0) Figure 14.4.2 EXAMPLEL: r is not defined when 0: 0; however, r is defined for all other values of 0 for which 0 < 0 < 2zr. Differentiating in (6), we obtain -dsin0 . . - __ "e, (l _ cos 0)2 polaraxis Setting Der:0, for values of 0 in the interval (0,2n) we obtain 0-n' BecauJewhen 0 < 0 < r,Det < 0, and when zr < e <2n'Dsr ) 0' r has an absolute minimum value at 0: n. when 0: n, r: Ld, and the point distance from the focus 1td, rr) is the vertex. Therefore, the undirected point on the parabola is the when is shortest parabola io a point on the the vertex. By Theorem 14.3.9, the parabola is symmetric with respect to its principal axis; by Theorem 14.3.2there is one vertex. Note that the curve does not contain the pole because there is no value of d which will give a value of 0 for r. A sketch of the parabola is shown in Fig. 1.4.4.2. A parabola has its focus at the pole and its vertex at (4,n). Find an equation of the parabola and an equation of the directrix. Draw a sketch of the parabola and the directrix. soLUrIoN: Becausethe focus is at the pole and the vertex is at (4, t), the polar axis and its extension are along the principal axis of the parabola. Furfhermore, the vertex is to the left of the focus, and so the directrix is also to the left of the focus. Hence, an equation of the parabola is of the form of Eq. (3). Becausethe vertex is at (4, n), Ld:4; thus, d:8' The eccentricity e:1, anrdtherefore we obtain the equation r-1-*,0 - -d, and becaused: 8, An equation of the directrix is given by r cos 0 594 THE CONIC SECTIONS we have r cos 0- -8. Figure 14.4.3 shows a sketch of the parabola and the directrix. |n axis a,tr) polar axis fl_ rocus F i g u r e1 4 . 4 . 3 EXAMPLE2: Follow the instructions of Example 1 if the parabola has its focus at the pole and its vertex at (3, *o). polar axis solurroN: The vertex of the parabola is below the focus, and the lrr axis and its extension are along the principal axis. Therefore, the directrix is also below the focus, and an equation of the parabola is of the form of Eq. (5) with the minus sign. The undirected distance from the focus to the vertex is id:3; thus d:6. Therefore,from Eqs. (5) an equation of the parabola is r' - 6 L-sind An equation of the directrix is given by r sin 0:-d. Becaused:6,we have r sin d: -6. sketches of the parabola and the directrix are shown in Fig. 1,4.4.4. Figure 14.4.4 Case2: e When e < L, the denominator of the fractiom in Eq. (3) is never zero, and so r exists for all values of 0. To find the absolute extrema of r on the interval [0,2n), w€ find Der and obtain D6r: -ezd sin 0 (1 - ecos0)2 (7) For 0 in the interval [0, 2n), D er - 0 when 0 - 0 and zr. When -+ 7r < 0<0,Dsr) 0; and when 0 an absolute maximum value of edl (1,- q. When trr < 0 < n, D6r <--0; and when n 14,4 POLAREQUATIONSOF THE CONICS L zT ax's P(r,o 7\ v1 \ of.edl(l* e) when 0: r. The points (edl(l- e), 0) and (edl(1'+ e), n) ate the vertices of the ellipse. We denote them by % and V2, respectively. It follows, then, that the undirected distance from the focus (0,0) to a point on the ellipse is greatest when the point on the ellipse is V1 and smallest when it is Vz. By Theorem L4.3.3,the ellipse is symmetric with respect to its principal axis. There is no value of 0 that will give a value of 0 for r, and so the curve does not contain the pole. A sketch of the ellipse is shown in Fi9.1,4.4.5. Figure 14.4.5 3: Find an equation of EXAMPLE the ellipse having a focus at the pole and vertices at (5, 0) and (2, n). Write an equation of the directrix corresPondingto the focus at the pole. Draw a sketch of the ellipse. solurroN: Becausethe vertices are at (5, 0) and Q,tr), the polar axis and its extension are along the principal axis of the ellipse. The directrix corresponding to the focus at the pole is to the left of the focus becausethe ,r"ti"* closest to this focus is at (2, rr). The required equation is therefore of the form of Eq. (3).The vertex vr is at (5,0) and v2is at Q, n).Tt follows from the discussion of Case 2 that and Solving these two equations simultaneously, we obtain s: $ and d: #' Hence, from Eq. (3) an equation of the ellipse is Xf l.{ U o \ tzo 1 \7, ; n.29 -' --1 ---J-----9.*cosO I2 zraxis l'{ t v1 tVz 20 7, (5, 0) Ir F i g u r e1 4 . 4 . 6 EXAMPLE An equation of a conic is r-# Find the eccentricity, identify the conic, write an equation of the directrix corresPonding to the focus at the pole, find the vertices/ and draw a sketch of the culve. or, equivalently, .20 ' 7-3cos0 f:- Becaused: T, an equation of the directrix corresponding to the focus - -20. A sketch of at the pole is r cos 0 -+ or, equivalently, 3r cos 0 the ellipse is shown in Fig. L4.4-6. sor,urroN: Dividing the numeratorand denominatorof the fraction in the given equationby 3, we obtain '-L+?sin0 THE CONIC SECTIONS are therefore at (1, in) and (5, *o). A sketch of the ellipse is shown in Fig. 14.4.7. I o axis directrix polar axis t (r, 3i ")l v2 F i g ur e 1 4 . 4 . 7 '1,. The conic is a hyperbola. Case3' e > In the derivation of Eq. (3) we assumed that the point p on the conic is to the right of the given directrix and that p is on the terminal side of the angle of 0 radians, thus making r positive. In the previous two cases, when the conic is an ellipse or a parabora,we obtain tlie entire curve with these assumptions. However, whene ) 1, we shall seethat we obtain only one of two branches of the hyperbola when r ) 0, and this branch is to the right of the directrix. There is also a branch to the left of the given directrix; this is obtained by letting / assume negative as well as positive values as d takes on values in the interval l0,2rrl, lf e> 1 in Eq. (3), r is undefined when cos 0:1.1e. There are two values of 0 in [0, 2r) that satisfy this equation. These varues are 0r: cos-l -1 and p 0z: 2n - cos-l I e we investigatethe values of r as d increasesfrom 0 to 91.when 0:0, edl(7 - e), and becausee ) l, r < 0. So the point (edl(1.- e), 0) is the vp{ex the hyperbola. The cosine function is decreasing in the infeft 9f terval (0,0r); therefore,when 0 < 0 < 0r, cos d ) cos 0r:Uior, equiv_ alently, 1 - e cos 0 < 0. Hence, for vJues ot 0 in the intervat (0', 0r), r: edl(7 - e cos 0) < 0. When 0 : er, r is undefined; however, r: P(r,0) polar axis \ Figure 14.4.8 lim r: 0_0r 1i^ --3! 0_0rl-eCOS0 ^:-* we conclude, then, that as 0 increasesfrom 0 to 01,r < 0 and lrl increases .I4.4POLAR EQUATIONS OFTHECONICS 597 without bound. From these values of 0 we obtain the lower half of the left branch of the curye. See Fig. 1,4.4.8.We show later that a hyperbola has two asymptotes. For this hyperbola one of the asymptotes is parallel to the line 0: 01. Now we consider the points on the hyperbola as g increasesthrough valuesbetween 01and 02.When 0, I 0 ( dr)cosg ( cosfi: lleor, equivalently, L - e cos 0 > 0. Therefore, for values of g in the intenral (0r,0r), r)0. lim r: o-of Hm3 o-oiL-ecosa ^:-l* When 0: r, r: edl(L* e); so the point (edl(L+ e), r) is the right vertex of the hyperbola. Finding Dsr from Eq. (3), we get Der: - ezd srn 0 (1 - e cos 0)2 (8) When 0, I 0 ( z, sin 0 > 0; hence, Der 10, from which we conclude that r is decreasing when 0 is in the interval (0v r). Therefore, for values of 0 in the interval (0u rr) we obtain the top half of the right branch of the hyperbola shown in Fig.1.4.4.8. When n < 0 ( d2,sin 0 < 0; so from Eq. (8), D6r ) 0. Hence, r is increasing when O'is in the interval (n, 0r).Furthermore, limr: 0-02- lim= 0-02- L- ed ;:** e COS A Consequently, for values of 0 in the interval (rr, 0r) we have the lower half of the right branch of the hyperbola. The other asymptote of the hyryerbola is parallel to the line 0: 02. The cosine function is increasing in the interval (02, 2n\. So when 0, 1 0 I 2n, cos 0 ) cos 02: 7le or, equivalently, 1 - e cos 0 < 0. Thus, we see from Eq. (3) that for values of 0 in the interval (0r, 2rr), r < 0. Also, €d lim r: lim = =:-* e cos0 ;:;;L ;:;;' It follows from Eq. (8) that Dsr ) 0; hence, r is increasing when 0, < 0 < 2zr. When r < 0 and r is increasing, lrl is decreasing.Thus, for values of 0 in the interval (02,2r) we obtain the upper half of the left branch of the hyperbola, and so the curve of Fig.14.4.8is complete. EXAMPLE5: The polar axis and its extension are along the principal axis of a hyperbola havin I a focus at the pole. The corresponding directrix is to the left of the focus. If the hyperbola contains the point 1L, ?n) and solurroN: An equation of the hyperbolais of the form of Eq. (3) with e:2.We have,then, ' t:- 2d L - 2 c o s0 Because the point (1, &rr) lies on the hyperbola, its coordinates satisfy the equation. Therefore, THE CONIC SECTIONS e : 2, find (a) an equation of the hlperbola; (b) the vertices; (c) the center; and (d) an equation of the directrix corresponding to the focus at the pole. Draw a sketch of the hyperbola. 2d 1- 2(-+) from which we obtain d:1.. r- Hence, arrrequation of the hyperbola is 2 1-2cos0 (e) The vertices are the points on the hyperbola for which 0 :0 and Q'- z'. From Eq. (9) we see that when e - 0, r - -Z;and when 0 - 7r,r : 3. Consequently, the left vertex Vr is at the point (-2, 0), and the right vertex V2 is at the point (?, n). *ou X l.{ I o F. 'o (- 2,o) 1- polar axis Q,*4 Figure 14.4.9 ExAMPLE 6: Find a polar equation of the hyperboli having tn" line r cos e - 4 as the directrix corresponding to a focus at the pole, and e: 8. The center of the hyperbola is the point on the principal axis halfway between the two vertices. This is thi point (t, zr). An equation of the directrix corresponding to the focus at the pole . is given by r cos 0: -d. Becaused: l,ihis equation is r cos 0: _1. As an aid in drawing a sketch of the hyperbola, we first draw the two asymptotes. In our discussion of case 3 we stated that these are lines through the center of the hyperbola that are parallel to the lines g: 0r and d : 02,where g, and 02 arethe values of C in the interval [0, 2n) foi which r is not defined. From Eq. (9), r is not defined when 1 - 2 cos 0: 0. Therefore, 0r:$r and 02:$zr. Figure 14.4.9shows a sketch of the hyperbola as well th_" two asymptotes, and the directrix corresponding "r pole. to the focus at the solurroN. The given directrix is pe{pendicular to the polar axis and is four units to the right of the focus at the pole. Therefore, an equation of the hyperbola is of the form of Eq. (A),which is r' : ed 1 , * e c o s0 Because d - 4 and e - g, tne equation becomes y' : E'4 1.+9cos0 or/ equivalently, t' : 12 2 + 3 cos 0 Exercises 14.4 1' show that an equation oJ a conic having its principal axis along the polar axis and its extension, a focus at the pole, and the corresponding directrix to the right o? *r" fo*, is r: e1l o i e cos e). 2' show that an equation of.a conic having its principal.axis along the lr axis and its extension, a focus at the pole, and the coresponding directrix above the fJcus ii ,:iAl(t+ e sin?). 14,5 CARTESIANEQUATIONSOF THE CONICS 3. Show that an equation of a conic having its principal axis along the +zr axis and its extension, a focus at the pole, and the corresponding directrix below the focus is r : edl(l - e sin d) . 4. Show that the equation r: k cs€ *0, where k is a constant, is a polar equation of'a parabola. In Exercises5 through 14, the equation is that of a conic having a focus at the pole. In each Exercise, (a) find the eccentricity; (b) identify the conic; (c) write an equation of the directrix whidr corresponds to the focus at the pole; (d) draw a sketch of the curve. v" -:- a -v" 11.,: L4. f' : - 2 t-cos0 4 "1,- 3 cos 0 L 5-5sin0 7 3+4cos0 In Exercises15 through 18, find an equation of the conic having a focus at the pole and satisfying the given conditions. L5. Parabola;vertex at (4,*zr). 15. Ellipse; e: * a vertex at (4' t). 1 7 . Hyryerbola;vertices at (1, tn) and (3 ' trn) . is the directrix corresponding to the focus at the pole. g) and above the parabola r : 3l (7 * sin 0) ' 19. Find the area of the region inside the ellipse r : 6l (2- sin the parabola' 20. For the ellipse and parabola of Exercise 19, find the area of the region inside the ellipse and below 18. Hyperbola; e:*r cos 0:9 is 80,000,000miles 21. A comet is moving in a parabolic orbit around the sun at the focus of the parabola.Wh-enthe comet of the orbit. (a) Find axis with the from the sun the line segrnent from the sun to the comet makes an angle oi *z radians sun? to the come an equation of the comit's orbit. (b) How close does the comet the pole is at the sun' 22. T\e orbit of a planet is in the form of an ellipse having the equation r: pl(l t e cos-0) where to 0' respect with from the sun Find the a.rerage measune of the distance of the planet pole, the principal axis 23. Using Definition 14.3.1 find a polar equation of a central conic for which the center is at the a/e. is a directrix pole to the from distance and the its extension, is aloing the polar axis and - cos 0) are 24. Show that the tangent lines at the points of intersection of the parabolas r: al(l * cos 0) and r: b/(1 perpendicular. 14.5 CARTESIAN EQUATIONS In Sec.14.4we learned that a polar equation of a conic is OF THE CONICS ed r:7-ffi7 or, equivalently, r: e(r cos0 * d) (1) A cartesian representation of Eq. (1) can be obtained by replacing r *{7TT and r cos 0 by x. Making these substitutions' we get by *t/NT: e(x* d) 600 THECONICSECTIONS Squaring on both sides of the above equation gives x2 + y' : e2x2* 2e2dx* e2d2 or, equivalently, y'+ x'(L- e 2 )- Z e z d x * e z d z (2) If e - L, Eg. (2) becomes y' :2d(x + +d) (3) If we translate the origin to the point Fta,O) by replacirg xby I - +d and y by y,Eq,. (3) becomes Y' :2dr Equation(4)resembles Eq. (1)of Sec.1,4.1 Y' : 4Px (4) 1,4.2. '1., If. e * we divide on both sides of Eq. (2) by 1- ez and,obtain z ztexd+17" ]eyd.z: 7 _ r^-. R r, Compleling the square for the terms involving r by adding ead2l(l _ e), on both sides of the above equation, we get ( , _- \r e d \ r - -r 1 " . ":_ e z i l z t - e) 1--@ Y- O=W (s) If the origin is translatedto the point (ezdlG - er),0), Eq. (5)becomes P 1]-, . t _ e z r i':=4(l_4yz or/ equivalently, x2-+-L-l wTafit T1?F:-- Now let ezd'lG - r o-1 lft ed tFI (6) e2)': a2, whete a > 0. Then iro<e ?r tre (7\ 14.5 CARTESIANEQUATIONSOF THE CONICS Then Eq. (5) can be written as * lt2 _--:l oz, sz(l_d) Replacing x arrd y by r and y, we get f . --=--'----v' - 1 - -r, oz sz(l_ g) (8) Equation (8) is a standard form of a cartesian equation of a central conic having its principal axis on the x axis. Because(5) is an equation of a conic having a focus at the origin (pole), and (8) is obtained from (5) it follows that the by translating the origin to the point(*dl(l-ez),O), point at the central conic having Eq. (8) has a focus eezdl(l-e2),0). However, from (7) it follows that -ezd [-ae ifO<e<1 o, if e>L l-rz:l Therefore, we conclude that if (8) is an equation of an ellipse (0 < e < 1), there is a focus at the point (-ae,O). If (8) is an equation of a hyperbola (e > 7), there is a focus at (ae,O). In each case the coresponding directrix is d units to the left of the focus. So if the graPh of (8) is an ellipse, the directrix corresponding to the focus at (-ae,0) has as an equation x:-ae- d d - a.(I - Because when 0 x:-ae-a(L- ez)le, this equation becomes e2) e a e Q r - Similarly, if the graph of (8) is a hyperbola, the directrix corresponding to the focus at (ae,0) has as an equation x: ae- d When e ) l, d: a(E - L) le; tttus, ttte above equation of the directrix can be written as a e Hence, we have shown that if (8) is an equation of an ellipse, a focus and and if (8) is an its corresponding directrix are (-ae,0) and x:-ale; equation of a hyperbola, a focus and its coresponding directrix ate (ae,O) and x: a/e. BecauseEq. (8) contains only even Powers of r and y, its graph is symmetric with respect to both the x and y axes. Therefore, if there is syma focus at (-ae,0) having a conesponding directrixof x:-ale,by THE CONIC SECTIONS metry there is also a focus at (ae,0) having a conesponding directrix of x: ale. Similarly, for a focus at (ae,0) and a coresponding directrix of x : al e, thete is also a focus at (- ae, 0) and a coffesponding directrix of x:-a/e.These results Ere summarized in the following theorem. 14.5.1 Theorem The central conic having as an equation x2 1f a2' (e) -r J - I - 4 a2(I-er) where a ) 0, has a focus at (-ae,0), whose coresponding directrix is x:-a/e, and a focus at (ae,0), whose correspondingdirectrix is x: afe. Figures 14.5.1and 14.5.2show sketchesof the graph of Eq. (9) together with the foci and directrices in the respective casesof an ellipse and a hyperbola. v *v e: - a a *:,Q rre Bb A ( ) A ae, a Xr: L{ii .|.4i: (Jii q.ti t+'li / 'ui B, - b F i g ur e 1 4 . 5 . 1 x E (J (U,r tsr -(' Figure14.5.2 Suppose that in Eq. (9) 0 < e causein this caseL - e2) 0, we conclude that \R Therefore,we let b-a\R is a real number. ( 10) Because a ; 0, it follows that b > 0. Substituting from (10) into (9), we obtain (11) Equation (11) is an equation of an ellipse having its principal axis on the r axis. Becausethe vertices are the points of interiection of the ellipse with its principal axis, they are at (-a,0) and (a, 0). The centerof the ellipse is at the origin because it is the point midway between the vertices. Figure 14.5.1shows a sketch of this ellipse. Refer to this figure as you read the next two paragraphs. If we denote the vertic es (- a, 0) and (a, 0) by A' and A, respectively, 14.5 CARTESIANEQUATIONSOF THE CONICS the segmentA' A of the principal axis is called the maior axis of the ellipse, and its length is 2a units. Then we state that a is the number of units in the length of the semimajor axis of the ellipse. The ellipse having Eq. (11) intersectsthe y axis at the points (-b,0) and (b,0), which we denote by B' and B, respectively. The segment B'B is called the minor axis of. the ellipse. Its length is 2b units. Hence, b is the number of units in the length of the semiminor axis of the ellipse. Becausea and b are positive numbers and 0 < e 1L, it follows from Eq. (10)thatb<a. EXAMPLEI: Given the ellipse having the equation -x:2 - 25 1t2 r' L : 1 "1,6 * find the vertices, foci, directrices, eccentricity, and extremities of the minor axis. Draw a sketch of the ellipse and show the foci and directrices. SoLUTIoN: From the equation of the ellipse,we have az:25 andf :16i thus, a : 5 and b : 4. Therefore, the vertices are at the points A'(-5, 0) and A(5, 0), and the extremities of the minor axis are at the points B'(0,-4) from which it follows that and B(0, 4). From (L0) we have 4 :S{T=V, r: $. The foci and the directrices are obtained by applying Theorem 1.4.5.1.Becauseae:3 and ale:ff, we conclude that one focus F is at (3, 0) and its corresponding directrix has the equation x: T; the other focus F' is at (-3, 0) and its corresponding directrix has the equation x: -4. A sketch of the ellipse, the foci F and F' , and the directrices are in Fig. 14.5.3.From the definition of a conic (1'4.3.1)it follows that if P is any point on this ellipse, the ratio of the undirected distance of P from a focus to the undirected distance of P from the corresponding directrix is equal to the eccentricity 9. F i g u r e1 4 . 5 . 3 ExAMPLE2: An arch is in the form of a semiellipse. It is 48 ft wide at the base and has a height of 20 ft. How wide is the arch at a height of L0 ft above the base? soLUrIoN: Figure 14.5.4 shows a sketch of the arch and the coordinate axes which are chosen so that the r axis is along the base and the origin is at the midpoint of the base. Then the ellipse has its principal axis on the r axis, its center at the origin, a: 24, and b :20. An equation of the ellipse is of the form of Eq. (11): * * a' -1 s76 ' 4 0 0 v (t, L0)i Let 27 be the number of feet in the width of the arch at a height of 10 ft above the base. Therefore, the point (t, 10) lies on the ellipse. Thus, *,100 576' 400 -+ -24|+-Zxft+24 *asft# Figure14.5.4 604 THE CONIC SECTIONS and so #:432 i:7216 Hence, at a height of L0 ft above the base the width of the arch is 24\/j tt. Consider now the central conic of Theorem 14.s.1.when e _e2 is, when the conic is a hyperbola. Because e e - L > 0. Therefore, for a hyperbola we let fo-a\F (r2) is a positive number. Also, from (I2), if b: a; and rf e ( 13) _Equation (13) is an equation of a hyperbola having its principal axis on the x axis,its verticesat the points (-a,0) and(a,0;, ind iti centerat the origin. A sketch of this hyperbola is in Fig. r4.s.2.If we denote the vertices (-4, 0) and (a, 0) by A' and A, the segment A'A is called the transaerseaxis of the hyperbola. The length of the transverse axis is 2a units, and so a is the number of units in the length of the semitransverseaxis. Substituting 0 for r in Eq. (13),we obtain y, : -b", which has no real roots. consequenfly, the hyperbola does not intersect the y axis. However, the line segment having extremities at the points (0,-b) and (0, b) is called the conjugnteaxis of the hyperbola, and its length is 2b units. Hence, b is the number of units in the length of the semiconjugate axis. Solving Eq. (13) fot y in terms of r, we obtain v:4\m (r4) we conclude from (14) that if lxl < a, there is no real value of y. Therefore, there are no points (x, y) on the hyperbola for which -a < x < a. We-also see from (14)that if lxl = a, then y has two real values. As previously learned, the hyperbola has two branches. one branch contains the vertex A(a, 0) and extends indefinitely far to the right of A. The other branch contains the vertex A'(-o,0) and extends indefinitelv far to the left of A'. EXAMPLE Given the hyperbola x 2_ y ' 1 g 1,6 SOLUTION: The given equation is of the form of Eq. (I3); thus, a - 3 and b : 4 . T h e vertices are therefore at the points A' ?9, 0) and A(3, 0). The number of units in the length of the transverse axis is 2a : 6, and the number of units in the length of the conjugate axis is 2b : 8. Because 14.5 CARTESIANEQUATIONSOF THE CONICS find the vertices,foci, directrices, eccentricity, and lengths of the transverseand conjugate axes. Draw a sketch of the hyperbola and show the foci and the directrices. from (12), b: a171, we have !': l{/i, and so e: t. Therefore, Ae:S and ale:$. Hence, the foci are at F'(-5,0) and F(5,0). The corresponding directrices have, respectively, the equations r: -9 and 1: g. A sketch of the hyperbola and the foci and directrices are in Fig. 14.5.5. From Definition 14.3.1it follows that if P is any point on this hyperbola, the ratio of the undirected distance of P from a focus to the undirected distance of P from the corresponding directrix is equal to the eccentricity*. F i g u r e1 4 . 5 . 5 EXAMPLE4: Find an equation of the hyperbola having the ends of its conjugate axis at (0,-2) and (0, 2) and an eccentricity equal to *\6. solurroN: Becausethe ends of the conjugate axis are at (0, -2) and (0,2), it follows that b : 2, the transverse axis is along the r axis, and the center is at the origin. Hence, an equation is of the form x2 v' - 1 a2 b2 To find a we use the equation b - a\FT. and letb-2 and e-g\8, thereby yielding 2 - a\F. Therefore, a - \8, and an equation of the hyperbola is x 2_ a ' - , , 5 4-r L4.5 Exercises In Exercises1 through 4, hnd the vertices, foci, directrices, eccentricif, and.ends of the minor axis of the given ellipse. Draw a sketch of the curve and show the foci and the directnces. 2.4x2*9y'-4 4. 3x2* 4y' :9 L. 4xz* 9y' - 36 3. 2x2* 3y' - Lg In Eiercises 5 through 9, find the vertices, foci, directrices, eccentricity, and lengths of the transverseand conjugateaxes of the given hlryerbola. Draw a sketch of the curve and show the foci and the directrices. 5. 9x2- 4y' - 36 6. x2- 9y' :9 7. 9x2- L6y'- 1 8. 25x2- 25yz- 1 THE CONIC SECTIONS In Exercises9 through 14,find an equation of the given conic satisfying the given conditions and draw a sketch of the graph. 9. Hyperbola having verticesat (-2,0) and (2,0) and a conjugateaxis of length 5. 10. Ellipse having foci at (-5,0) and (5,0) and one directrix the line x:-20. 11. Ellipse having verticesat (-9,0) and (*,0) and one focus (8,0). 12. Hyperbola having the ends of its conjulate axis at (0, -3) and (0, 3) and e:2. 13. Hyperbola having its center at the origin, its foci on the x axis, and passing through the points (4, -2) and (7, -6). 14. Ellipse having its center at the origin, its foci on the r axis, the length of the major axis equal to three times the length of the minor axis, and passing through the point (3,3). 15. Find an equation of the tangent line to the ellipse 4f -f 9y2:72 at the point (3, 2). 16. Find an equation of the normal line to the hyperbol a 4x2- 3y2: 24 at the point (3, 2). 17. Find an equation of the hyperbolawhose foci are thevertices of the ellipse 7f *17y2:77 the foci of this ellipse. and whose vertices are 18. Find an equation of the ellipse whose foci are the vertices of the hyperbola l'l,f - 7y, : 77 and whose vertices are. the foci of this hyperbola. 19. The ceiling in a hallway 20 ft wide is in the shape of a semiellipse and is 18 ft high in the center and 12 fthigh at the side walls. Find the height of the ceiling 4 ft from either wall. 20. A football is 12 in. long, and a plane section containing a seam is an ellipse, of which the length of the minor axis is 7 in. Find the volume of the football if the leather is so stiff that every cross section is a square. 21. Solve Exercise20 if every cross section is a circle. 22. The orbit of the earth around the sun is elliptical in shape with the sun at one focus, a semimajor axis of length 92.9 million miles, and an eccentricity of 0.017.Find (a) how close the earth gets to the sun and @f the greatestp'ossible distance between the earth and the sun. 23. The cost of production.of a commodity is $12 less per unit at a point A than it is at a point B and the distance between A and B is 100 miles. Assuming that the route of delivery of the commodity is alonf a straight line, and that the deIivery cost is 20 cents per unit per mile, find the curve at any point of which the commodity can be supplied from either A or B at the same total cost. (nrur: Take points A and B at (-50, 0) and (50, 0), respeciively.) 24. Prove that there is no tangent line to the hyperbol a * - yr: 1 that passesthrough the origin. 25. Find the volume of the-solidof revolution generatedby revolving the region bounded by the hyperbola xzlaz- yzlbz: 1.and the line r: 2a abott the r axis. 25. Find the centroid of the solid of revolution of Exercise25. 27' A rar& has a horizontal axis of length 20 ft and its ends are semiellipses. The width acrossthe top of the tank is 10 ft and the depth is 5 ft. If the tank is full of water, how much work is necessaryto pump all the water to th" top of the tank? 14.6 THE ELLIPSE In Sec. 14.3 we considered ellipses for which the center is at the origin and the principal axis is on the x axis. Now more general equationJ of ellipses will be discussed. If an ellipse has its center at the origin and its principal axis on the y ucis, then an equation of the ellipse is of the form 14.6THE ELLIPSE (1) which is obtained from Eq. (11) of Sec. 14.5 by interchanging x and y. . rLLUsrRArroNL: Becausefor an ellipse a ) b, it follows that the ellipse having the equation * *t-:t 1,6 25 has its foci on the y axis. This ellipse has the same shape as the ellipse of Example f. in Sec.14.5.The vertices are at (0, -5) and (0, 5), and the foci are at (0, -3) and (0, 3), and their corresponding directrices have the . equations y : -+ and y :4f, respectively. If the center of an ellipse is at the point (h, k) rather than at the origin, and if the principal axis is parallel to one of the coordinate axes, then by a translation of axes so that the point (h, k) is the new origin, an equation -*la2 * y'lb': 1 if the principal axis is horizontal, and of the ellipse is and y'la"+-*1bz:1 if the principal axis is vertical. Becausei:x-h y: y - k, these equations becomethe following in x and y: (2) if the (3) if the principal axis is vertical. In Sec. 1.6 we discussed the general equation of the second degree in two variables: Axz * Bxy * Cy' * Dx * Ey + F- 0 (4) where B:0 and A: C. In such a case, the graph of (4) is either a circle a degenerate case of a circle, which is either a point-circle or no real or locus. If we eliminate the fractions and combine terms in Eqs. (2) and (3), we obtain an equation of the form Axz*Cy'*Dx*Ey+F-0 (5) where A+ C 1f.a*b and AC>0.It can be shown bycompleting the squares in r and y that an equation of the form (5) can be put in the form ( x- h ) ' 11 (v- k)' A C (6) CONIC SECTIONS If. AC > 0, then A and C have the same sign. If G has the same sign as A and C, then Eq. (5) can be written in the form of (2) or (3). Thus, the graph of (5) is an ellipse. 2: Suppose we have the equation o ILLUSTRATToT.T 6**9y2-24x-54y*51:0 which can be written as 6(f-4x)+9(y2-6y):-5L Completing the squares in r and y, we get 6(f -4x* 4) +9(y'-6y t 9):-51 +24+81 or, equivalently, 6(x-2)2+9(V-3)2:54 which can be put in the form (x-2)' --_-r (y -3\' , --]-: J? This is lr, un,r",roi of the form of Eq. (6).By dividing on both sidesby 54 we have ( x - 2 ) ' _ ( y---z-: -3)'_., 9 - ' which has the form of Eq. (2). o If in Eq. (5) G has a sign opposite to that of A and C, then (5) is not satisfied by any real values of r and y. Hence, the graph of (5) is the empty set. . rLLUsrRArroN3: Supposethat Eq. (5) is 6**9y2-24x-54y*1L5:0 Then, upon completing the squares in r and y, we get 6(x-2)z a9(! -3)':-115 + 24+8L which can be written as (x-2\' ------i- (y -3)' T, -=___i_: _10 (7) T h i s i s l r , n " f o r - t E q , . ( 6 ) , w h e rGe : - 1 0 , A : 6 , a n d C : 9 . F o r a l l values of r and y the left side of Eq. (7) is nonnegative; hence, the graph of (7) is the empty set. . If G:0 in (6), then the equation is satisfiedby only the point (h, k). 14.6 THE ELLIPSE 609 Therefore, the graph of (5) is a single point, which we call a point-ellipse. o rLLUsrRATroN4: Becausethe equation 6f*9y2-24x-54Y*105:0 can be written as (* ,2)'*(y *+ _ . 9 )_' o ' its graph is the point (2, 3). If the graph of Eq. (5) is a point-ellipse or the emPty set, the graph is said to be degenerate. lf. A: C in (5), we have either a circle or a degeneratecircle, as mentioned above. A circle is a limiting form of an ellipse. This can be shown by considering the formula relating a, b, and.e fot an ellipse: bz: az('[,_d) From this formula, it is seent$at as e approacheszero, b2approaches 42. lf b2: az,Eqs. (2) and (3) become (x- h)'+ (y - k)': az which is an equation of a circle having its center at (h, k) and radius a. We see that the results of Sec.1.6 for a circle are the sameas those obtained for Eq. (5) applied to an elliPse. The results of the above discussion are summarized in the following theorem. 14.5.1Theorem If in the generalsecond-degreeEq. (4) n:0 and AC > 0, then the graph is either an ellipse, a point-ellipse, or the empty set. In addition, if A: C, then the graph is either a circle, a point-circle, or the empty set' Determine the graph of the equati on 25x2* I6Y2 * 150r - L28y - LIL9: Q. soLUrIoN: From Theorem'1,4.5.7, becauseB:0 and ng:(25)(t0): 4O0> 0, the graph is either an ellipse or is degenerate. Completing the squares in r and y, we have 25(x2* 6x + 9) + l6(y', - 8y * 16) - 1119+ 225+ 256 EXAMPLE1: 2 5 ( x+ 3 ) ' + l 6 ( y - 4 ) ' : (x*3)2 -(y-+)':1 100 54 L500 * (8) Equation (8) is of the form of Eq. (3), and so the graPh is an ellipse having its principal axis parallel to the y axis and its center at (-3' 4). 610 T H E C O N I CS E C T I O N S ExAMPLE2: For the ellipse of Example 1.,find the vertices, foci, directrices, eccentricity, and extremities of the minor axis. Draw a sketch of the ellipse and show the foci and the directrices. solurroN: From Eq. (8) it follows that a:'1.0 and b : 8. Becausethe center of the ellipse is at (-3, 4) and the principal axis is vertical, the vertices are at the points V'(-3, -5) and V(-3,l4). The extremitiesof the minor axis are at the points B'(_1t'1.,4)and B(5,4). Becauseb: a{1V, we have 8:101T=7 and, solving for e,weget e: $. Consequently,ae:6 and ale:ry. Therefore, the foci are at the points F'(-3, -2) and F(-3, L0). The corresponding directrices have, respectively, the equations y : -+ and y: 92. A sketch of the ellipse, the foci, and the directricesare in Fig. 14.6.1. v V: directrix 3 (- z,t+) v \ (-3, L0) F ( - 1 1 , 4) B', ?3,4) \t | | r I I r B (s,4) tl r, r,J F, -/ .J (-3, -6) V', y: - T38 directrix F i g u r e1 4 . 6 . 1 EXAMPLE3: Find an equation of the ellipse for which the foci are at (- 8, 2) and (4,2) and the eccentricity is 3. Draw a sketch of the ellipse. soLUrIoN: The center of the ellipse which is halfway between the foci is the point (-2, 2). The distance between the foci of any ellipse is 2ae, and the distance between (-8,2) and (4, 2) is L2. Therefore, we have the equation Zae: 12. Replacing e by *, we get 2a(|) : 12, and so a:9. Because b : a{177, we have b:9lT=@:9lT=4:3rE The principal axis is parallel to the x axis; hence, an equation of the ellipse is of the form of Eq. (2).Because(h, k): (-2,2), a:9, and b : 3V5, the - 14,6THE ELLIPSE 611 required equation is ( x! 2 ) ' 8L * (y _z)'_ 45 1 A sketchof this ellipse is shown in Fig. "1.4.6.2. (- 2,2+ 3\/s) \ ,2) ( - 1 1 , 2) lrrrrltl \ - ___L!.?)____ 7, 2) lttlll/ o ./ (-2,2 - 3\/-s) F i g ur e 1 4 We conclude this section with a theorem that gives an altemative definition of an ellipse. The theorem is based on a characteristic property of the ellipse. 14.6.2 Theorem d' /(-+,vp r -/h o .,,/ ,p, An ellipse can be defined as the set of points such that the sum of the distances from any point of the set to two given points (the foci) is a constant. pRooF: The proof consists of two parts. In the first part we show that the set of points defined in the theorem is an ellipse. In the second part we prove that any ellipse is such a set of points. Refer to Fig 14.6.3 in both parts of the proof. Let the point Pt(rt, Ar) be any point in the given set. Let the foci be the points F'(-ae,0) and F(ae,0), and let the constant sum of the distances be 24. Then lrEl + 1fi1:2o Using the distance formula, we get Y - - A ' L e F i g u r e1 4 . 6 . 3 {G=7*Tt7 * {Q,*i{tT y}:26 \/@=1d"TT : 2a- \/Cn ad"W Squaring on both sides of the above equation, we obtain x( - 2Ae4 * azE * yl : 4a2- Aa\/(it + aAY + y7 * xr' * 2aex1+ a2e2* yr' 7 612 THE CONIC SECTIONS or, equivalently, \/GfidrTt!: a+ exl Squaring on both sides again, we get xr2+ 2aeh I azez* yr' : a2+ 2nex,* dxr2 ('l'- E)x12* Yt': a2(1'- e2) x,2 -: 71,2 -J- ---------!-:-:1 a2(l - 8) Becauseb2: az(l - e), and replacing 11and yrby x and y, respectively, we obtain a2 f 1t2 7+ fr:1 which is the required form of an equation of an ellipse. Now consider the ellipse in Fig. 1.4.6.3;Pr(xr,yr) is any point on the ellipse. Through P, draw a line that is parallel to the r axis and that intersects the directrices d and d' at the points Q and R, respectively. From Definition 14.3.1it follows that lF-ql: elR4l and lFEl: 'ltr'al Therefore, lrEl + lFEl:'(lR4l+ lFrel) (e) RPt and PrQ are both positive because an ellipse lies between its directrices. Hence, :2- (-3):'+ Ro (10) Substituting from (10)into (9),we get l F E l+ l E l : 2 a which proves that an ellipse is a set of points as described in the theorem. I o rLLUsrRArrow 5: The ellipse of Example L has foci at (-3, -2) and (-3, 10), and 2a: 20. Therefore, by Theorem '1.4.6.2 this ellipse can be defined as the set of points such that the sum of the distancesfrom any point of the set to the points (-3, -2) and (-3, 10) is equal to 20. Similarly, the ellipse of Example3 can be defined as the set of points such that the sum of the distancesfrom any point of the set to the points (-8,2) and (4,2) is equal to 18. 14.7THE HYPERBOLA 013 Exercises L4.6 In Exercises1 through 5, find the eccentricity, center, foci, and directrices of each of the given ellipses and draw a sketch

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