SOLUTION MANUAL
608070 _ISM_ThomasCalc_WeirHass_ttl.qxd:harsh_569709_ttl
9/3/09
3:11 PM
Page 1
INSTRUCTOR’S
SOLUTIONS MANUAL
SINGLE VARIABLE
Collin County Community College
WILLIAM ARDIS
THOMAS’ CALCULUS
TWELFTH EDITION
BASED ON THE ORIGINAL WORK BY
George B. Thomas, Jr.
Massachusetts Institute of Technology
AS
REVISED BY
Maurice D. Weir
Naval Postgraduate School
Joel Hass
University of California, Davis
608070 _ISM_ThomasCalc_WeirHass_ttl.qxd:harsh_569709_ttl
9/3/09
3:11 PM
Page 2
This work is protected by United States copyright laws and is provided solely
for the use of instructors in teaching their courses and assessing student
learning. Dissemination or sale of any part of this work (including on the
World Wide Web) will destroy the integrity of the work and is not permitted. The work and materials from it should never be made available to
students except by instructors using the accompanying text in their
classes. All recipients of this work are expected to abide by these
restrictions and to honor the intended pedagogical purposes and the needs of
other instructors who rely on these materials.
The author and publisher of this book have used their best efforts in preparing this book. These efforts
include the development, research, and testing of the theories and programs to determine their
effectiveness. The author and publisher make no warranty of any kind, expressed or implied, with regard
to these programs or the documentation contained in this book. The author and publisher shall not be
liable in any event for incidental or consequential damages in connection with, or arising out of, the
furnishing, performance, or use of these programs.
Reproduced by Pearson Addison-Wesley from electronic files supplied by the author.
Copyright © 2010, 2005, 2001 Pearson Education, Inc.
Publishing as Pearson Addison-Wesley, 75 Arlington Street, Boston, MA 02116.
All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or
transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise,
without the prior written permission of the publisher. Printed in the United States of America.
ISBN-13: 978-0-321-60807-9
ISBN-10: 0-321-60807-0
1 2 3 4 5 6 BB 12 11 10 09
PREFACE TO THE INSTRUCTOR
This Instructor's Solutions Manual contains the solutions to every exercise in the 12th Edition of THOMAS' CALCULUS
by Maurice Weir and Joel Hass, including the Computer Algebra System (CAS) exercises. The corresponding Student's
Solutions Manual omits the solutions to the even-numbered exercises as well as the solutions to the CAS exercises (because
the CAS command templates would give them all away).
In addition to including the solutions to all of the new exercises in this edition of Thomas, we have carefully revised or
rewritten every solution which appeared in previous solutions manuals to ensure that each solution
ì conforms exactly to the methods, procedures and steps presented in the text
ì is mathematically correct
ì includes all of the steps necessary so a typical calculus student can follow the logical argument and algebra
ì includes a graph or figure whenever called for by the exercise, or if needed to help with the explanation
ì is formatted in an appropriate style to aid in its understanding
Every CAS exercise is solved in both the MAPLE and MATHEMATICA computer algebra systems. A template showing
an example of the CAS commands needed to execute the solution is provided for each exercise type. Similar exercises within
the text grouping require a change only in the input function or other numerical input parameters associated with the problem
(such as the interval endpoints or the number of iterations).
For more information about other resources available with Thomas' Calculus, visit http://pearsonhighered.com.
TABLE OF CONTENTS
1 Functions 1
1.1
1.2
1.3
1.4
Functions and Their Graphs 1
Combining Functions; Shifting and Scaling Graphs 8
Trigonometric Functions 19
Graphing with Calculators and Computers 26
Practice Exercises 30
Additional and Advanced Exercises 38
2 Limits and Continuity 43
2.1
2.2
2.3
2.4
2.5
2.6
Rates of Change and Tangents to Curves 43
Limit of a Function and Limit Laws 46
The Precise Definition of a Limit 55
One-Sided Limits 63
Continuity 67
Limits Involving Infinity; Asymptotes of Graphs 73
Practice Exercises 82
Additional and Advanced Exercises 86
3 Differentiation 93
3.1
3.2
3.3
3.4
3.5
3.6
3.7
3.8
3.9
Tangents and the Derivative at a Point 93
The Derivative as a Function 99
Differentiation Rules 109
The Derivative as a Rate of Change 114
Derivatives of Trigonometric Functions 120
The Chain Rule 127
Implicit Differentiation 135
Related Rates 142
Linearizations and Differentials 146
Practice Exercises 151
Additional and Advanced Exercises 162
4 Applications of Derivatives 167
4.1
4.2
4.3
4.4
4.5
4.6
4.7
Extreme Values of Functions 167
The Mean Value Theorem 179
Monotonic Functions and the First Derivative Test 188
Concavity and Curve Sketching 196
Applied Optimization 216
Newton's Method 229
Antiderivatives 233
Practice Exercises 239
Additional and Advanced Exercises 251
5 Integration 257
5.1
5.2
5.3
5.4
5.5
5.6
Area and Estimating with Finite Sums 257
Sigma Notation and Limits of Finite Sums 262
The Definite Integral 268
The Fundamental Theorem of Calculus 282
Indefinite Integrals and the Substitution Rule 290
Substitution and Area Between Curves 296
Practice Exercises 310
Additional and Advanced Exercises 320
6 Applications of Definite Integrals 327
6.1
6.2
6.3
6.4
6.5
6.6
Volumes Using Cross-Sections 327
Volumes Using Cylindrical Shells 337
Arc Lengths 347
Areas of Surfaces of Revolution 353
Work and Fluid Forces 358
Moments and Centers of Mass 365
Practice Exercises 376
Additional and Advanced Exercises 384
7 Transcendental Functions 389
7.1
7.2
7.3
7.4
7.5
7.6
7.7
7.8
Inverse Functions and Their Derivatives 389
Natural Logarithms 396
Exponential Functions 403
Exponential Change and Separable Differential Equations 414
^
Indeterminate Forms and L'Hopital's
Rule 418
Inverse Trigonometric Functions 425
Hyperbolic Functions 436
Relative Rates of Growth 443
Practice Exercises 447
Additional and Advanced Exercises 458
8 Techniques of Integration 461
8.1
8.2
8.3
8.4
8.5
8.6
8.7
Integration by Parts 461
Trigonometric Integrals 471
Trigonometric Substitutions 478
Integration of Rational Functions by Partial Fractions 484
Integral Tables and Computer Algebra Systems 491
Numerical Integration 502
Improper Integrals 510
Practice Exercises 518
Additional and Advanced Exercises 528
9 First-Order Differential Equations 537
9.1
9.2
9.3
9.4
9.5
Solutions, Slope Fields and Euler's Method 537
First-Order Linear Equations 543
Applications 546
Graphical Solutions of Autonomous Equations 549
Systems of Equations and Phase Planes 557
Practice Exercises 562
Additional and Advanced Exercises 567
10 Infinite Sequences and Series 569
10.1 Sequences 569
10.2 Infinite Series 577
10.3 The Integral Test 583
10.4 Comparison Tests 590
10.5 The Ratio and Root Tests 597
10.6 Alternating Series, Absolute and Conditional Convergence 602
10.7 Power Series 608
10.8 Taylor and Maclaurin Series 617
10.9 Convergence of Taylor Series 621
10.10 The Binomial Series and Applications of Taylor Series 627
Practice Exercises 634
Additional and Advanced Exercises 642
TABLE OF CONTENTS
10 Infinite Sequences and Series 569
10.1 Sequences 569
10.2 Infinite Series 577
10.3 The Integral Test 583
10.4 Comparison Tests 590
10.5 The Ratio and Root Tests 597
10.6 Alternating Series, Absolute and Conditional Convergence 602
10.7 Power Series 608
10.8 Taylor and Maclaurin Series 617
10.9 Convergence of Taylor Series 621
10.10 The Binomial Series and Applications of Taylor Series 627
Practice Exercises 634
Additional and Advanced Exercises 642
11 Parametric Equations and Polar Coordinates 647
11.1
11.2
11.3
11.4
11.5
11.6
11.7
Parametrizations of Plane Curves 647
Calculus with Parametric Curves 654
Polar Coordinates 662
Graphing in Polar Coordinates 667
Areas and Lengths in Polar Coordinates 674
Conic Sections 679
Conics in Polar Coordinates 689
Practice Exercises 699
Additional and Advanced Exercises 709
12 Vectors and the Geometry of Space 715
12.1
12.2
12.3
12.4
12.5
12.6
Three-Dimensional Coordinate Systems 715
Vectors 718
The Dot Product 723
The Cross Product 728
Lines and Planes in Space 734
Cylinders and Quadric Surfaces 741
Practice Exercises 746
Additional Exercises 754
13 Vector-Valued Functions and Motion in Space 759
13.1
13.2
13.3
13.4
13.5
13.6
Curves in Space and Their Tangents 759
Integrals of Vector Functions; Projectile Motion 764
Arc Length in Space 770
Curvature and Normal Vectors of a Curve 773
Tangential and Normal Components of Acceleration 778
Velocity and Acceleration in Polar Coordinates 784
Practice Exercises 785
Additional Exercises 791
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
14 Partial Derivatives 795
14.1
14.2
14.3
14.4
14.5
14.6
14.7
14.8
14.9
14.10
Functions of Several Variables 795
Limits and Continuity in Higher Dimensions 804
Partial Derivatives 810
The Chain Rule 816
Directional Derivatives and Gradient Vectors 824
Tangent Planes and Differentials 829
Extreme Values and Saddle Points 836
Lagrange Multipliers 849
Taylor's Formula for Two Variables 857
Partial Derivatives with Constrained Variables 859
Practice Exercises 862
Additional Exercises 876
15 Multiple Integrals 881
15.1
15.2
15.3
15.4
15.5
15.6
15.7
15.8
Double and Iterated Integrals over Rectangles 881
Double Integrals over General Regions 882
Area by Double Integration 896
Double Integrals in Polar Form 900
Triple Integrals in Rectangular Coordinates 904
Moments and Centers of Mass 909
Triple Integrals in Cylindrical and Spherical Coordinates 914
Substitutions in Multiple Integrals 922
Practice Exercises 927
Additional Exercises 933
16 Integration in Vector Fields 939
16.1
16.2
16.3
16.4
16.5
16.6
16.7
16.8
Line Integrals 939
Vector Fields and Line Integrals; Work, Circulation, and Flux 944
Path Independence, Potential Functions, and Conservative Fields 952
Green's Theorem in the Plane 957
Surfaces and Area 963
Surface Integrals 972
Stokes's Theorem 980
The Divergence Theorem and a Unified Theory 984
Practice Exercises 989
Additional Exercises 997
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
CHAPTER 1 FUNCTIONS
1.1 FUNCTIONS AND THEIR GRAPHS
1. domain œ (_ß _); range œ [1ß _)
2. domain œ [0ß _); range œ (_ß 1]
3. domain œ Ò2ß _); y in range and y œ È5x 10
! Ê y can be any positive real number Ê range œ Ò!ß _).
4. domain œ (_ß 0Ó Ò3, _); y in range and y œ Èx2 3x
5. domain œ (_ß 3Ñ Ð3, _); y in range and y œ
Ê3 t!Ê
4
3t
4
3t,
! Ê y can be any positive real number Ê range œ Ò!ß _).
now if t 3 Ê 3 t ! Ê
4
3t
!, or if t 3
! Ê y can be any nonzero real number Ê range œ Ð_ß 0Ñ Ð!ß _).
6. domain œ (_ß %Ñ Ð4, 4Ñ Ð4, _); y in range and y œ
2
% t 4 Ê 16 Ÿ t 16 ! Ê
nonzero real number Ê range œ Ð_ß
#
"'
18 Ó
Ÿ
2
t2 16
2
t2 16 ,
2
t2 16
now if t % Ê t2 16 ! Ê
2
!, or if t % Ê t 16 ! Ê
2
t2 16
!, or if
! Ê y can be any
Ð!ß _).
7. (a) Not the graph of a function of x since it fails the vertical line test.
(b) Is the graph of a function of x since any vertical line intersects the graph at most once.
8. (a) Not the graph of a function of x since it fails the vertical line test.
(b) Not the graph of a function of x since it fails the vertical line test.
#
9. base œ x; (height)# ˆ #x ‰ œ x# Ê height œ
È3
#
x; area is a(x) œ
"
#
(base)(height) œ
"
#
(x) Š
È3
# x‹
œ
È3
4
x# ;
perimeter is p(x) œ x x x œ 3x.
10. s œ side length Ê s# s# œ d# Ê s œ
d
È2
; and area is a œ s# Ê a œ
"
#
d#
11. Let D œ diagonal length of a face of the cube and j œ the length of an edge. Then j# D# œ d# and
D# œ 2j# Ê 3j# œ d# Ê j œ
d
È3
. The surface area is 6j# œ
6d#
3
12. The coordinates of P are ˆxß Èx‰ so the slope of the line joining P to the origin is m œ
ˆx, Èx‰ œ ˆ m"# ,
#
œ 2d# and the volume is j$ œ Š d3 ‹
Èx
x
œ
"
Èx
$Î#
œ
(x 0). Thus,
"‰
m .
13. 2x 4y œ 5 Ê y œ "# x 54 ; L œ ÈÐx 0Ñ2 Ðy 0Ñ2 œ Éx2 Ð "# x 54 Ñ2 œ Éx2 4" x2 54 x œ É 54 x2 54 x 25
16
œ É 20x
2
20x 25
16
œ
È20x2 20x 25
4
14. y œ Èx 3 Ê y2 3 œ x; L œ ÈÐx 4Ñ2 Ðy 0Ñ2 œ ÈÐy2 3 4Ñ2 y2 œ ÈÐy2 1Ñ2 y2
œ Èy4 2y2 1 y2 œ Èy4 y2 1
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
d$
3È 3
25
16
.
2
Chapter 1 Functions
15. The domain is a_ß _b.
16. The domain is a_ß _b.
17. The domain is a_ß _b.
18. The domain is Ð_ß !Ó.
19. The domain is a_ß !b a!ß _b.
20. The domain is a_ß !b a!ß _b.
21. The domain is a_ß 5b Ð5ß 3Ó Ò3, 5Ñ a5, _b 22. The range is Ò2, 3Ñ.
23. Neither graph passes the vertical line test
(a)
(b)
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 1.1 Functions and Their Graphs
24. Neither graph passes the vertical line test
(a)
(b)
Ú xyœ" Þ
Ú yœ1x Þ
or
or
kx yk œ 1 Í Û
Í Û
ß
ß
Ü x y œ " à
Ü y œ " x à
25.
x
y
0
0
1
1
27. Faxb œ œ
2
0
26.
x
y
0
1
1
0
2
0
"
, x0
28. Gaxb œ œ x
x, 0 Ÿ x
4 x2 , x Ÿ 1
x2 2x, x 1
29. (a) Line through a!ß !b and a"ß "b: y œ x; Line through a"ß "b and a#ß !b: y œ x 2
x, 0 Ÿ x Ÿ 1
f(x) œ œ
x 2, 1 x Ÿ 2
Ú
Ý 2, ! Ÿ x "
Ý
!ß " Ÿ x #
(b) f(x) œ Û
Ý
Ý 2ß # Ÿ x $
Ü !ß $ Ÿ x Ÿ %
30. (a) Line through a!ß 2b and a#ß !b: y œ x 2
"
Line through a2ß "b and a&ß !b: m œ !& # œ
x #, 0 x Ÿ #
f(x) œ œ "
$ x &$ , # x Ÿ &
f(x) œ œ
œ "$ , so y œ "$ ax 2b " œ "$ x $ !
! Ð"Ñ œ
" $
%
#! œ #
(b) Line through a"ß !b and a!ß $b: m œ
Line through a!ß $b and a#ß "b: m œ
"
$
&
$
$, so y œ $x $
œ #, so y œ #x $
$x $, " x Ÿ !
#x $, ! x Ÿ #
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
3
4
Chapter 1 Functions
31. (a) Line through a"ß "b and a!ß !b: y œ x
Line through a!ß "b and a"ß "b: y œ "
Line through a"ß "b and a$ß !b: m œ !"
$" œ
Ú
x
" Ÿ x !
"
!xŸ"
f(x) œ Û
Ü "# x $#
"x$
"
#
(b) Line through a2ß 1b and a0ß 0b: y œ 12 x
Line through a0ß 2b and a1ß 0b: y œ 2x 2
Line through a1ß 1b and a3ß 1b: y œ 1
32. (a) Line through ˆ T# ß !‰ and aTß "b: m œ
faxb œ (b)
"!
TaTÎ#b
œ "# , so y œ "# ax "b " œ "# x Ú
1
2x
faxb œ Û 2x 2
Ü 1
$
#
2 Ÿ x Ÿ 0
0xŸ1
1xŸ3
œ T# , so y œ T# ˆx T# ‰ 0 œ T# x "
!, 0 Ÿ x Ÿ T#
#
T
T x ", # x Ÿ T
Ú
A,
Ý
Ý
Ý
Aß
faxb œ Û
Aß
Ý
Ý
Ý
Ü Aß
! Ÿ x T#
T
# Ÿx T
T Ÿ x $#T
$T
# Ÿ x Ÿ #T
33. (a) ÚxÛ œ 0 for x − [0ß 1)
(b) ÜxÝ œ 0 for x − (1ß 0]
34. ÚxÛ œ ÜxÝ only when x is an integer.
35. For any real number x, n Ÿ x Ÿ n ", where n is an integer. Now: n Ÿ x Ÿ n " Ê Ðn "Ñ Ÿ x Ÿ n. By
definition: ÜxÝ œ n and ÚxÛ œ n Ê ÚxÛ œ n. So ÜxÝ œ ÚxÛ for all x − d .
36. To find f(x) you delete the decimal or
fractional portion of x, leaving only
the integer part.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 1.1 Functions and Their Graphs
37. Symmetric about the origin
Dec: _ x _
Inc: nowhere
38. Symmetric about the y-axis
Dec: _ x !
Inc: ! x _
39. Symmetric about the origin
Dec: nowhere
Inc: _ x !
!x_
40. Symmetric about the y-axis
Dec: ! x _
Inc: _ x !
41. Symmetric about the y-axis
Dec: _ x Ÿ !
Inc: ! x _
42. No symmetry
Dec: _ x Ÿ !
Inc: nowhere
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
5
6
Chapter 1 Functions
43. Symmetric about the origin
Dec: nowhere
Inc: _ x _
44. No symmetry
Dec: ! Ÿ x _
Inc: nowhere
45. No symmetry
Dec: ! Ÿ x _
Inc: nowhere
46. Symmetric about the y-axis
Dec: _ x Ÿ !
Inc: ! x _
47. Since a horizontal line not through the origin is symmetric with respect to the y-axis, but not with respect to the origin, the
function is even.
48. faxb œ x& œ
"
x&
and faxb œ axb& œ
"
a x b&
œ ˆ x"& ‰ œ faxb. Thus the function is odd.
49. Since faxb œ x# " œ axb# " œ faxb. The function is even.
50. Since Òfaxb œ x# xÓ Á Òfaxb œ axb# xÓ and Òfaxb œ x# xÓ Á Òfaxb œ axb# xÓ the function is neither even nor
odd.
51. Since gaxb œ x$ x, gaxb œ x$ x œ ax$ xb œ gaxb. So the function is odd.
52. gaxb œ x% $x# " œ axb% $axb# " œ gaxbß thus the function is even.
53. gaxb œ
"
x# "
54. gaxb œ
x
x# " ;
55. hatb œ
"
t ";
œ
"
axb# "
œ gaxb. Thus the function is even.
gaxb œ x#x" œ gaxb. So the function is odd.
hatb œ
"
t " ;
h at b œ
"
" t.
Since hatb Á hatb and hatb Á hatb, the function is neither even nor odd.
56. Since l t$ | œ l atb$ |, hatb œ hatb and the function is even.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 1.1 Functions and Their Graphs
57. hatb œ 2t ", hatb œ 2t ". So hatb Á hatb. hatb œ 2t ", so hatb Á hatb. The function is neither even nor
odd.
58. hatb œ 2l t l " and hatb œ 2l t l " œ 2l t l ". So hatb œ hatb and the function is even.
59. s œ kt Ê 25 œ kÐ75Ñ Ê k œ
"
3
Ê s œ 3" t; 60 œ 3" t Ê t œ 180
60. K œ c v# Ê 12960 œ ca18b2 Ê c œ 40 Ê K œ 40v# ; K œ 40a10b# œ 4000 joules
61. r œ
62. P œ
k
s
Ê6œ
k
v
k
4
Ê k œ 24 Ê r œ
Ê 14.7 œ
k
1000
24
s ;
10 œ
24
s
Ê k œ 14700 Ê P œ
Êsœ
14700
v ;
12
5
23.4 œ
14700
v
Êvœ
24500
39
¸ 628.2 in3
63. v œ f(x) œ xÐ"% 2xÑÐ22 2xÑ œ %x$ 72x# $!)x; ! x 7Þ
64. (a) Let h œ height of the triangle. Since the triangle is isosceles, AB # AB # œ 2# Ê AB œ È2Þ So,
#
h# "# œ ŠÈ2‹ Ê h œ " Ê B is at a!ß "b Ê slope of AB œ " Ê The equation of AB is
y œ f(x) œ x "; x − Ò!ß "Ó.
(b) AÐxÑ œ 2x y œ 2xÐx "Ñ œ 2x# #x; x − Ò!ß "Ó.
65. (a) Graph h because it is an even function and rises less rapidly than does Graph g.
(b) Graph f because it is an odd function.
(c) Graph g because it is an even function and rises more rapidly than does Graph h.
66. (a) Graph f because it is linear.
(b) Graph g because it contains a!ß "b.
(c) Graph h because it is a nonlinear odd function.
x
#
67. (a) From the graph,
(b)
x
#
1
x 0:
x
#
x 0:
x
2
4
x
1
Ê
4
x
x
#
1
4
x
Ê x − (2ß 0) (%ß _)
1 4x 0
#
2x8
0 Ê x 2x
0 Ê
(x4)(x2)
#x
0
(x4)(x2)
#x
0
Ê x 4 since x is positive;
1
4
x
0 Ê
x# 2x8
2x
0 Ê
Ê x 2 since x is negative;
sign of (x 4)(x 2)
ïïïïïðïïïïïðïïïïî
2
%
Solution interval: (#ß 0) (%ß _)
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
7
8
Chapter 1 Functions
3
2
x 1 x 1
3
2
x 1 x 1
68. (a) From the graph,
(b) Case x 1:
Ê x − (_ß 5) (1ß 1)
Ê
3(x1)
x 1
2
Ê 3x 3 2x 2 Ê x 5.
Thus, x − (_ß 5) solves the inequality.
Case 1 x 1:
3
x 1
2
x 1
3(x1)
x 1
Ê
2
Ê 3x 3 2x 2 Ê x 5 which is true
if x 1. Thus, x − (1ß 1) solves the
inequality.
3
Case 1 x: x1 x2 1 Ê 3x 3 2x 2 Ê x 5
which is never true if 1 x, so no solution here.
In conclusion, x − (_ß 5) (1ß 1).
69. A curve symmetric about the x-axis will not pass the vertical line test because the points ax, yb and ax, yb lie on the same
vertical line. The graph of the function y œ faxb œ ! is the x-axis, a horizontal line for which there is a single y-value, !,
for any x.
70. price œ 40 5x, quantity œ 300 25x Ê Raxb œ a40 5xba300 25xb
71. x2 x2 œ h2 Ê x œ
h
È2
œ
È2 h
2 ;
cost œ 5a2xb 10h Ê Cahb œ 10Š
È2 h
2 ‹
10h œ 5hŠÈ2 2‹
72. (a) Note that 2 mi = 10,560 ft, so there are È800# x# feet of river cable at $180 per foot and a10,560 xb feet of land
cable at $100 per foot. The cost is Caxb œ 180È800# x# 100a10,560 xb.
(b) Ca!b œ $"ß #!!ß !!!
Ca&!!b ¸ $"ß "(&ß )"#
Ca"!!!b ¸ $"ß ")'ß &"#
Ca"&!!b ¸ $"ß #"#ß !!!
Ca#!!!b ¸ $"ß #%$ß ($#
Ca#&!!b ¸ $"ß #()ß %(*
Ca$!!!b ¸ $"ß $"%ß )(!
Values beyond this are all larger. It would appear that the least expensive location is less than 2000 feet from the
point P.
1.2 COMBINING FUNCTIONS; SHIFTING AND SCALING GRAPHS
1. Df : _ x _, Dg : x
2. Df : x 1
Rf œ Rg : y
0 Ê x
0, Rf g : y
1 Ê Df
g
1, Dg : x 1
È2, Rfg : y
œ Dfg : x
0 Ê x
1. Rf : _ y _, Rg : y
1. Therefore Df
g
œ Dfg : x
0, Rf g : y
1, Rfg : y
1.
0
3. Df : _ x _, Dg : _ x _, DfÎg : _ x _, DgÎf : _ x _, Rf : y œ 2, Rg : y
RfÎg : 0 y Ÿ 2, RgÎf : "# Ÿ y _
4. Df : _ x _, Dg : x
0 , DfÎg : x
5. (a) 2
(d) (x 5)# 3 œ x# 10x 22
(g) x 10
0
0, DgÎf : x
0; Rf : y œ 1, Rg : y
(b) 22
(e) 5
(h) (x# 3)# 3 œ x% 6x# 6
1,
1, RfÎg : 0 y Ÿ 1, RgÎf : 1 Ÿ y _
(c) x# 2
(f) 2
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 1.2 Combining Functions; Shifting and Scaling Graphs
6. (a) "3
(d)
(b) 2
"
x
(c)
(e) 0
(g) x 2
(h)
(f)
"
"
x 1 1
œ
x
x
"
#
1
x"
x#
œ
"
x 1
3
4
1œ
x
x1
7. af‰g‰hbaxb œ fagahaxbbb œ faga4 xbb œ fa3a4 xbb œ fa12 3xb œ a12 3xb 1 œ 13 3x
8. af‰g‰hbaxb œ fagahaxbbb œ fagax2 bb œ fa2ax2 b 1b œ fa2x2 1b œ 3a2x2 1b 4 œ 6x2 1
9. af‰g‰hbaxb œ fagahaxbbb œ fˆgˆ 1x ‰‰ œ fŠ 1 1 % ‹ œ fˆ 1 x 4x ‰ œ É 1 x 4x " œ É 15x4x"
x
2
10. af‰g‰hbaxb œ fagahaxbbb œ fŠgŠÈ2 x‹‹ œ f
ŠÈ2 x‹
2
ŠÈ2 x‹
œ fˆ $ x ‰ œ
1
2x
2 x
$ x 2
3 $2 xx
8 3x
7 2x
œ
11. (a) af‰gbaxb
(d) a j‰jbaxb
(b) a j‰gbaxb
(e) ag‰h‰f baxb
(c) ag‰gbaxb
(f) ah‰j‰f baxb
12. (a) af‰jbaxb
(d) af‰f baxb
(b) ag‰hbaxb
(e) a j‰g‰f baxb
(c) ah‰hbaxb
(f) ag‰f‰hbaxb
g(x)
f(x)
(f ‰ g)(x)
(a)
x7
Èx
Èx 7
(b)
x2
3x
3(x 2) œ 3x 6
(c)
x#
Èx 5
Èx# 5
(d)
x
x1
x
x1
"
x1
"
x
1
13.
(e)
(f)
"
x
gaxb"
g ax b
œ
x
x (x1)
œx
x
"
x
x
"
lx "l .
14. (a) af‰gbaxb œ lgaxbl œ
(b) af‰gbaxb œ
x
x 1
x
x 1 1
x
x"
œ
Ê"
"
g ax b
œ
x
x"
Ê"
x
x"
œ
"
g ax b
Ê
"
x"
œ
"
gaxb ß so
gaxb œ x ".
(c) Since af‰gbaxb œ Ègaxb œ lxl, gaxb œ x .
(d) Since af‰gbaxb œ fˆÈx‰ œ l x l, faxb œ x# . (Note that the domain of the composite is Ò!ß _Ñ.)
#
The completed table is shown. Note that the absolute value sign in part (d) is optional.
gaxb
faxb
af‰gbaxb
"
"
lxl
x"
lx "l
x"
x#
Èx
x"
x
Èx
#
x
15. (a) faga1bb œ fa1b œ 1
(d) gaga2bb œ ga0b œ 0
x
x"
lxl
lxl
(b) gafa0bb œ ga2b œ 2
(e) gafa2bb œ ga1b œ 1
(c) fafa1bb œ fa0b œ 2
(f) faga1bb œ fa1b œ 0
16. (a) faga0bb œ fa1b œ 2 a1b œ 3, where ga0b œ 0 1 œ 1
(b) gafa3bb œ ga1b œ a1b œ 1, where fa3b œ 2 3 œ 1
(c) gaga1bb œ ga1b œ 1 1 œ 0, where ga1b œ a1b œ 1
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
9
10
Chapter 1 Functions
(d) fafa2bb œ fa0b œ 2 0 œ 2, where fa2b œ 2 2 œ 0
(e) gafa0bb œ ga2b œ 2 1 œ 1, where fa0b œ 2 0 œ 2
(f) fˆgˆ "# ‰‰ œ fˆ #" ‰ œ 2 ˆ #" ‰ œ 5# , where gˆ "# ‰ œ "# 1 œ "#
17. (a) af‰gbaxb œ fagaxbb œ É 1x 1 œ É 1 x x
ag‰f baxb œ gafaxbb œ
1
Èx 1
(b) Domain af‰gb: Ð_, 1Ó Ð0, _Ñ, domain ag‰f b: Ð1, _Ñ
(c) Range af‰gb: Ð1, _Ñ, range ag‰f b: Ð0, _Ñ
18. (a) af‰gbaxb œ fagaxbb œ 1 2Èx x
ag‰f baxb œ gafaxbb œ 1 kxk
(b) Domain af‰gb: Ò0, _Ñ, domain ag‰f b: Ð_, _Ñ
(c) Range af‰gb: Ð0, _Ñ, range ag‰f b: Ð_, 1Ó
19. af‰gbaxb œ x Ê fagaxbb œ x Ê
g ax b
g ax b 2
œ x Ê gaxb œ agaxb 2bx œ x † gaxb 2x
Ê gaxb x † gaxb œ 2x Ê gaxb œ 1 2x
x œ
2x
x1
20. af‰gbaxb œ x 2 Ê fagaxbb œ x 2 Ê 2agaxbb3 4 œ x 2 Ê agaxbb3 œ
21. (a) y œ (x 7)#
(b) y œ (x 4)#
22. (a) y œ x# 3
(b) y œ x# 5
x6
2
3 x6
Ê gaxb œ É
2
23. (a) Position 4
(b) Position 1
(c) Position 2
(d) Position 3
24. (a) y œ (x 1)# 4
(b) y œ (x 2)# 3
(c) y œ (x 4)# 1
(d) y œ (x 2)#
25.
26.
27.
28.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 1.2 Combining Functions; Shifting and Scaling Graphs
29.
30.
31.
32.
33.
34.
35.
36.
37.
38.
39.
40.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
11
12
Chapter 1 Functions
41.
42.
43.
44.
45.
46.
47.
48.
49.
50.
51.
52.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 1.2 Combining Functions; Shifting and Scaling Graphs
53.
54.
55. (a) domain: [0ß 2]; range: [#ß $]
(b) domain: [0ß 2]; range: [1ß 0]
(c) domain: [0ß 2]; range: [0ß 2]
(d) domain: [0ß 2]; range: [1ß 0]
(e) domain: [2ß 0]; range: [!ß 1]
(f) domain: [1ß 3]; range: [!ß "]
(g) domain: [2ß 0]; range: [!ß "]
(h) domain: [1ß 1]; range: [!ß "]
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
13
14
Chapter 1 Functions
56. (a) domain: [0ß 4]; range: [3ß 0]
(b) domain: [4ß 0]; range: [!ß $]
(c) domain: [4ß 0]; range: [!ß $]
(d) domain: [4ß 0]; range: ["ß %]
(e) domain: [#ß 4]; range: [3ß 0]
(f) domain: [2ß 2]; range: [3ß 0]
(g) domain: ["ß 5]; range: [3ß 0]
(h) domain: [0ß 4]; range: [0ß 3]
58. y œ a2xb# 1 œ %x# 1
57. y œ 3x# 3
59. y œ "# ˆ" "‰
x#
œ
"
#
"
#x#
60. y œ 1 "
axÎ$b#
œ1
61. y œ È%x 1
62. y œ 3Èx 1
#
63. y œ É% ˆ x# ‰ œ "# È16 x#
64. y œ "$ È% x#
65. y œ " a3xb$ œ " 27x$
66. y œ " ˆ x# ‰ œ " $
*
x#
x$
)
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 1.2 Combining Functions; Shifting and Scaling Graphs
67. Let y œ È#x " œ faxb and let gaxb œ x"Î# ,
"Î#
"Î#
haxb œ ˆx " ‰ , iaxb œ È#ˆx " ‰ , and
#
#
"Î#
jaxb œ ’È#ˆx "# ‰ “ œ faBb. The graph of
haxb is the graph of gaxb shifted left
"
#
unit; the
graph of iaxb is the graph of haxb stretched
vertically by a factor of È#; and the graph of
jaxb œ faxb is the graph of iaxb reflected across
the x-axis.
68. Let y œ È" x
#
œ faxbÞ Let gaxb œ axb"Î# ,
haxb œ ax #b"Î# , and iaxb œ
œ È" x
#
"
È # a x
#b"Î#
œ faxbÞ The graph of gaxb is the
graph of y œ Èx reflected across the x-axis.
The graph of haxb is the graph of gaxb shifted
right two units. And the graph of iaxb is the
graph of haxb compressed vertically by a factor
of È#.
69. y œ faxb œ x$ . Shift faxb one unit right followed by a
shift two units up to get gaxb œ ax "b3 #.
70. y œ a" Bb$ # œ Òax "b$ a#bÓ œ faxb.
Let gaxb œ x$ , haxb œ ax "b$ , iaxb œ ax "b$ a#b,
and jaxb œ Òax "b$ a#bÓ. The graph of haxb is the
graph of gaxb shifted right one unit; the graph of iaxb is
the graph of haxb shifted down two units; and the graph
of faxb is the graph of iaxb reflected across the x-axis.
71. Compress the graph of faxb œ
of 2 to get gaxb œ
unit to get haxb œ
"
#x . Then
"
#x ".
"
x
horizontally by a factor
shift gaxb vertically down 1
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
15
16
Chapter 1 Functions
72. Let faxb œ
œ
"
#
ŠxÎÈ#‹
"
x#
and gaxb œ
"œ
#
x#
"œ
"
#
’Š"ÎÈ#‹B“
"
#
Š B# ‹
"
"Þ Since
È# ¸ "Þ%, we see that the graph of faxb stretched
horizontally by a factor of 1.4 and shifted up 1 unit
is the graph of gaxb.
$
73. Reflect the graph of y œ faxb œ È
x across the x-axis
$
to get gaxb œ Èx.
74. y œ faxb œ a#xb#Î$ œ Òa"ba#bxÓ#Î$
œ a"b#Î$ a#xb#Î$ œ a#xb#Î$ . So the graph
of faxb is the graph of gaxb œ x#Î$ compressed
horizontally by a factor of 2.
75.
76.
77. *x# #&y# œ ##& Ê
x#
&#
y#
$#
œ"
78. "'x# (y# œ ""# Ê
x#
#
È
Š (‹
y#
%#
œ"
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 1.2 Combining Functions; Shifting and Scaling Graphs
79. $x# ay #b# œ $ Ê
x#
"#
a y #b #
#
ŠÈ$‹
80. ax "b# #y# œ % Ê
œ"
Ê
83.
x#
"'
#
ŠÈ#‹
y#
*
y a#b‘#
#
ŠÈ$‹
#
#
82. 'ˆx $# ‰ *ˆy "# ‰ œ &%
81. $ax "b# #ay #b# œ '
ax " b #
x a"b‘#
##
#
œ"
Ê
’xˆ $# ‰“
$#
ˆy "# ‰#
#
ŠÈ'‹
œ"
œ " has its center at a!ß !b. Shiftinig 4 units
left and 3 units up gives the center at ah, kb œ a%ß $b.
#
x a4b‘#
ay 3#3b œ "
4#
a y $b #
œ ". Center, C, is a%ß
3#
So the equation is
Ê
ax % b #
4#
$b, and
major axis, AB, is the segment from a)ß $b to a!ß $b.
84. The ellipse
x#
%
y#
#&
œ " has center ah, kb œ a!ß !b.
Shifting the ellipse 3 units right and 2 units down
produces an ellipse with center at ah, kb œ a$ß #b
and an equation
ax 3 b#
%
y a#b‘#
#&
œ ". Center,
C, is a3ß #b, and AB, the segment from a$ß $b to
a$ß (b is the major axis.
85. (a) (fg)(x) œ f(x)g(x) œ f(x)(g(x)) œ (fg)(x), odd
(b) Š gf ‹ (x) œ
f(x)
g(x)
œ
f(x)
g(x)
œ Š gf ‹ (x), odd
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
y#
#
È
Š #‹
œ"
17
18
Chapter 1 Functions
(c) ˆ gf ‰ (x) œ
(d)
(e)
(f)
(g)
(h)
(i)
g(x)
f(x)
œ
g(x)
f(x)
œ ˆ gf ‰ (x), odd
f # (x) œ f(x)f(x) œ f(x)f(x) œ f # (x), even
g# (x) œ (g(x))# œ (g(x))# œ g# (x), even
(f ‰ g)(x) œ f(g(x)) œ f(g(x)) œ f(g(x)) œ (f ‰ g)(x), even
(g ‰ f)(x) œ g(f(x)) œ g(f(x)) œ (g ‰ f)(x), even
(f ‰ f)(x) œ f(f(x)) œ f(f(x)) œ (f ‰ f)(x), even
(g ‰ g)(x) œ g(g(x)) œ g(g(x)) œ g(g(x)) œ (g ‰ g)(x), odd
86. Yes, f(x) œ 0 is both even and odd since f(x) œ 0 œ f(x) and f(x) œ 0 œ f(x).
87. (a)
(b)
(c)
(d)
88.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 1.3 Trigonometric Functions
1.3 TRIGONOMETRIC FUNCTIONS
1. (a) s œ r) œ (10) ˆ 451 ‰ œ 81 m
radians and
51
4
1 ‰
3. ) œ 80° Ê ) œ 80° ˆ 180°
œ
41
9
2. ) œ
s
r
œ
101
8
œ
51
4
1 ‰
(b) s œ r) œ (10)(110°) ˆ 180°
œ
1
)
231
0
1
#
s
r
œ
30
50
31
4
"
È2
È"
2
sin )
0
cos )
1
tan )
0
È3
0
und.
"
und.
"
È3
und.
0
1
und.
È 2
1
#
und.
È23
sec )
csc )
0
"
"
0
"
und.
7. cos x œ 45 , tan x œ 34
9. sin x œ È8
3
, tan x œ È8
"
6.
È2
3#1
)
1'
sin )
"
cos )
!
"
#
tan )
und.
È 3
cot )
!
È"3
sec )
und.
#
csc )
"
È23
8. sin x œ
2
È5
10. sin x œ
12
13
13.
14.
period œ 1
13
È
#3
12. cos x œ , cos x œ
"
È2
&1
'
"
#
È
#3
È"3
"
È"3
È 3
"
È 3
2
È3
È2
È23
#
È2
#
"#
È3
#
"
È5
, tan x œ 12
5
È3
#
, tan x œ
"
È3
period œ 41
16.
period œ 2
m
‰ ¸ 34°
œ 0.6 rad or 0.6 ˆ 180°
1
11. sin x œ È"5 , cos x œ È25
15.
551
9
Ê s œ (6) ˆ 491 ‰ œ 8.4 in. (since the diameter œ 12 in. Ê radius œ 6 in.)
È
#3
"#
cot )
œ
ˆ 180°
‰ œ 225°
1
4. d œ 1 meter Ê r œ 50 cm Ê ) œ
5.
1101
18
period œ 4
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
1
%
"
È2
19
20
Chapter 1 Functions
17.
18.
period œ 6
period œ 1
19.
20.
period œ 21
period œ 21
21.
22.
period œ 21
period œ 21
23. period œ 1# , symmetric about the origin
24. period œ 1, symmetric about the origin
25. period œ 4, symmetric about the s-axis
26. period œ 41, symmetric about the origin
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 1.3 Trigonometric Functions
27. (a) Cos x and sec x are positive for x in the interval
ˆ 12 , 12 ‰; and cos x and sec x are negative for x in the
intervals ˆ 321 , 12 ‰ and ˆ 12 , 321 ‰. Sec x is undefined
when cos x is 0. The range of sec x is
(_ß 1] ["ß _); the range of cos x is ["ß 1].
(b) Sin x and csc x are positive for x in the intervals
ˆ 321 , 1‰ and a!, 1b; and sin x and csc x are negative
for x in the intervals a1, !b and ˆ1, 321 ‰. Csc x is
undefined when sin x is 0. The range of csc x is
(_ß 1] [1ß _); the range of sin x is ["ß "].
28. Since cot x œ
"
tan x
, cot x is undefined when tan x œ 0
and is zero when tan x is undefined. As tan x approaches
zero through positive values, cot x approaches infinity.
Also, cot x approaches negative infinity as tan x
approaches zero through negative values.
29. D: _ x _; R: y œ 1, 0, 1
30. D: _ x _; R: y œ 1, 0, 1
31. cos ˆx 1# ‰ œ cos x cos ˆ 1# ‰ sin x sin ˆ 1# ‰ œ (cos x)(0) (sin x)(1) œ sin x
32. cos ˆx 1# ‰ œ cos x cos ˆ 1# ‰ sin x sin ˆ 1# ‰ œ (cos x)(0) (sin x)(1) œ sin x
33. sin ˆx 1# ‰ œ sin x cos ˆ 1# ‰ cos x sin ˆ 1# ‰ œ (sin x)(0) (cos x)(1) œ cos x
34. sin ˆx 1# ‰ œ sin x cos ˆ 1# ‰ cos x sin ˆ 1# ‰ œ (sin x)(0) (cos x)(1) œ cos x
35. cos (A B) œ cos (A (B)) œ cos A cos (B) sin A sin (B) œ cos A cos B sin A (sin B)
œ cos A cos B sin A sin B
36. sin (A B) œ sin (A (B)) œ sin A cos (B) cos A sin (B) œ sin A cos B cos A (sin B)
œ sin A cos B cos A sin B
37. If B œ A, A B œ 0 Ê cos (A B) œ cos 0 œ 1. Also cos (A B) œ cos (A A) œ cos A cos A sin A sin A
œ cos# A sin# A. Therefore, cos# A sin# A œ 1.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
21
22
Chapter 1 Functions
38. If B œ 21, then cos (A 21) œ cos A cos 21 sin A sin 21 œ (cos A)(1) (sin A)(0) œ cos A and
sin (A 21) œ sin A cos 21 cos A sin 21 œ (sin A)(1) (cos A)(0) œ sin A. The result agrees with the
fact that the cosine and sine functions have period 21.
39. cos (1 x) œ cos 1 cos B sin 1 sin x œ (1)(cos x) (0)(sin x) œ cos x
40. sin (21 x) œ sin 21 cos (x) cos (21) sin (x) œ (0)(cos (x)) (1)(sin (x)) œ sin x
41. sin ˆ 3#1 x‰ œ sin ˆ 3#1 ‰ cos (x) cos ˆ 3#1 ‰ sin (x) œ (1)(cos x) (0)(sin (x)) œ cos x
42. cos ˆ 3#1 x‰ œ cos ˆ 3#1 ‰ cos x sin ˆ 3#1 ‰ sin x œ (0)(cos x) (1)(sin x) œ sin x
œ sin ˆ 14 13 ‰ œ sin
44. cos
111
1#
45. cos
1
12
œ cos ˆ 13 14 ‰ œ cos
46. sin
51
1#
œ sin ˆ 231 14 ‰ œ sin ˆ 231 ‰ cos ˆ 14 ‰ cos ˆ 231 ‰ sin ˆ 14 ‰ œ Š
21 ‰
3
È2
1
8
œ
1 cos ˆ 281 ‰
#
œ
1
#
49. sin#
1
1#
œ
1 cos ˆ 211# ‰
#
œ
1
#
3
4
Ê sin ) œ „
52. sin2 ) œ cos2 ) Ê
sin2 )
cos2 )
1
4
œ cos
47. cos#
51. sin2 ) œ
cos
#
È3
#
È3
2
œ
1
3
cos
cos
21
3
1
4
sin
sin
1
4
cos ˆ 14 ‰ sin
1
3
1
3
È2
È3
# ‹Š # ‹
71
1#
œ cos ˆ 14 1
4
È2
ˆ"‰
# ‹ #
43. sin
œŠ
sin
1
3
21
3
œŠ
Š
È2
ˆ "‰
# ‹ #
sin ˆ 14 ‰ œ ˆ "# ‰ Š
Š
È2
# ‹
œ
È2
È3
# ‹Š # ‹
Š
È3
È2
# ‹Š # ‹
2 È2
4
48. cos#
51
1#
œ
1‰
1 cos ˆ 10
1#
#
œ
2 È3
4
50. sin#
31
8
œ
1 cos ˆ 681 ‰
#
Ê tan2 ) œ 1 Ê tan ) œ „ 1 Ê ) œ 14 ,
31 51 71
4 , 4 , 4
cos2 )
cos2 )
œ
È3
È2
# ‹ Š # ‹
œ
Ê ) œ 13 ,
È 6 È 2
4
È 2 È 6
4
1 È 3
2È 2
œ
ˆ "# ‰ Š
œ
œ
1 Š
È3 ‹
#
#
1 Š
#
È2 ‹
#
È2
# ‹
œ
œ
œ
1 È 3
2È 2
2 È3
4
2 È2
4
21 41 51
3 , 3 , 3
53. sin 2) cos ) œ 0 Ê 2sin ) cos ) cos ) œ 0 Ê cos )a2sin ) 1b œ 0 Ê cos ) œ 0 or 2sin ) 1 œ 0 Ê cos ) œ 0 or
sin ) œ "# Ê ) œ 12 , 321 , or ) œ 16 , 561 Ê ) œ 16 , 12 , 561 , 321
54. cos 2) cos ) œ 0 Ê 2cos2 ) 1 cos ) œ 0 Ê 2cos2 ) cos ) 1 œ 0 Ê acos ) 1ba2cos ) 1b œ 0
Ê cos ) 1 œ 0 or 2cos ) 1 œ 0 Ê cos ) œ 1 or cos ) œ "# Ê ) œ 1 or ) œ 13 , 531 Ê ) œ 13 , 1, 531
55. tan (A B) œ
sin (AB)
cos (AB)
œ
sin A cos Bcos A cos B
cos A cos Bsin A sin B
œ
sin A cos B
cos A sin B
cos A cos B cos A cos B
sin A sin B
cos A cos B
cos A cos B cos A cos B
œ
tan Atan B
1tan A tan B
56. tan (A B) œ
sin (AB)
cos (AB)
œ
sin A cos Bcos A cos B
cos A cos Bsin A sin B
œ
sin A cos B
cos A sin B
cos A cos B cos A cos B
sin A sin B
cos A cos B
cos A cos B cos A cos B
œ
tan Atan B
1tan A tan B
57. According to the figure in the text, we have the following: By the law of cosines, c# œ a# b# 2ab cos )
œ 1# 1# 2 cos (A B) œ 2 2 cos (A B). By distance formula, c# œ (cos A cos B)# (sin A sin B)#
œ cos# A 2 cos A cos B cos# B sin# A 2 sin A sin B sin# B œ 2 2(cos A cos B sin A sin B). Thus
c# œ 2 2 cos (A B) œ 2 2(cos A cos B sin A sin B) Ê cos (A B) œ cos A cos B sin A sin B.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 1.3 Trigonometric Functions
58. (a) cosaA Bb œ cos A cos B sin A sin B
sin ) œ cosˆ 1# )‰ and cos ) œ sinˆ 1# )‰
Let ) œ A B
sinaA Bb œ cos’ 1# aA Bb“ œ cos’ˆ 1# A‰ B“ œ cos ˆ 1# A‰ cos B sin ˆ 1# A‰ sin B
œ sin A cos B cos A sin B
(b) cosaA Bb œ cos A cos B sin A sin B
cosaA aBbb œ cos A cos aBb sin A sin aBb
Ê cosaA Bb œ cos A cos aBb sin A sin aBb œ cos A cos B sin A asin Bb
œ cos A cos B sin A sin B
Because the cosine function is even and the sine functions is odd.
59. c# œ a# b# 2ab cos C œ 2# 3# 2(2)(3) cos (60°) œ 4 9 12 cos (60°) œ 13 12 ˆ "# ‰ œ 7.
Thus, c œ È7 ¸ 2.65.
60. c# œ a# b# 2ab cos C œ 2# 3# 2(2)(3) cos (40°) œ 13 12 cos (40°). Thus, c œ È13 12 cos 40° ¸ 1.951.
61. From the figures in the text, we see that sin B œ hc . If C is an acute angle, then sin C œ hb . On the other hand,
if C is obtuse (as in the figure on the right), then sin C œ sin (1 C) œ hb . Thus, in either case,
h œ b sin C œ c sin B Ê ah œ ab sin C œ ac sin B.
a # b # c #
2ab
By the law of cosines, cos C œ
and cos B œ
a # c # b #
.
2ac
Moreover, since the sum of the
interior angles of a triangle is 1, we have sin A œ sin (1 (B C)) œ sin (B C) œ sin B cos C cos B sin C
#
#
#
#
#
#
b c
c b ˆ h ‰
h ‰
œ ˆ hc ‰ ’ a 2ab
a2a# b# c# c# b# b œ
“ ’ a 2ac
“ b œ ˆ 2abc
ah
bc
Ê ah œ bc sin A.
Combining our results we have ah œ ab sin C, ah œ ac sin B, and ah œ bc sin A. Dividing by abc gives
h
sin A
sin C
sin B
bc œ ðóóóóóóóñóóóóóóóò
a œ c œ b .
law of sines
62. By the law of sines,
sin A
#
œ
sin B
3
œ
È3/2
c .
By Exercise 61 we know that c œ È7. Thus sin B œ
3È 3
2È 7
¶ 0.982.
63. From the figure at the right and the law of cosines,
b# œ a# 2# 2(2a) cos B
œ a# 4 4a ˆ "# ‰ œ a# 2a 4.
Applying the law of sines to the figure,
Ê
È2/2
a
œ
È3/2
b
sin A
a
œ
sin B
b
Ê b œ É 3# a. Thus, combining results,
a# 2a 4 œ b# œ
3
#
a# Ê 0 œ
"
#
a# 2a 4
Ê 0 œ a# 4a 8. From the quadratic formula and the fact that a 0, we have
aœ
4È4# 4(1)(8)
#
œ
4 È 3 4
#
¶ 1.464.
64. (a) The graphs of y œ sin x and y œ x nearly coincide when x is near the origin (when the calculator
is in radians mode).
(b) In degree mode, when x is near zero degrees the sine of x is much closer to zero than x itself. The
curves look like intersecting straight lines near the origin when the calculator is in degree mode.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
23
24
Chapter 1 Functions
65. A œ 2, B œ 21, C œ 1, D œ 1
66. A œ "# , B œ 2, C œ 1, D œ
"
#
67. A œ 12 , B œ 4, C œ 0, D œ
68. A œ
L
21 ,
"
1
B œ L, C œ 0, D œ 0
69-72. Example CAS commands:
Maple
f := x -> A*sin((2*Pi/B)*(x-C))+D1;
A:=3; C:=0; D1:=0;
f_list := [seq( f(x), B=[1,3,2*Pi,5*Pi] )];
plot( f_list, x=-4*Pi..4*Pi, scaling=constrained,
color=[red,blue,green,cyan], linestyle=[1,3,4,7],
legend=["B=1","B=3","B=2*Pi","B=3*Pi"],
title="#69 (Section 1.3)" );
Mathematica
Clear[a, b, c, d, f, x]
f[x_]:=a Sin[21/b (x c)] + d
Plot[f[x]/.{a Ä 3, b Ä 1, c Ä 0, d Ä 0}, {x, 41, 41 }]
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 1.3 Trigonometric Functions
69. (a) The graph stretches horizontally.
(b) The period remains the same: period œ l B l. The graph has a horizontal shift of
"
#
period.
70. (a) The graph is shifted right C units.
(b) The graph is shifted left C units.
(c) A shift of „ one period will produce no apparent shift. l C l œ '
71. (a) The graph shifts upwards l D lunits for D !
(b) The graph shifts down l D lunits for D !Þ
72. (a) The graph stretches l A l units.
(b) For A !, the graph is inverted.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
25
26
Chapter 1 Functions
1.4 GRAPHING WITH CALCULATORS AND COMPUTERS
1-4.
The most appropriate viewing window displays the maxima, minima, intercepts, and end behavior of the graphs and
has little unused space.
1. d.
2. c.
3. d.
4. b.
5-30.
For any display there are many appropriate display widows. The graphs given as answers in Exercises 530
are not unique in appearance.
5. Ò2ß 5Ó by Ò15ß 40Ó
6. Ò4ß 4Ó by Ò4ß 4Ó
7. Ò2ß 6Ó by Ò250ß 50Ó
8. Ò1ß 5Ó by Ò5ß 30Ó
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 1.4 Graphing with Calculators and Computers
9. Ò4ß 4Ó by Ò5ß 5Ó
10. Ò2ß 2Ó by Ò2ß 8Ó
11. Ò2ß 6Ó by Ò5ß 4Ó
12. Ò4ß 4Ó by Ò8ß 8Ó
13. Ò1ß 6Ó by Ò1ß 4Ó
14. Ò1ß 6Ó by Ò1ß 5Ó
15. Ò3ß 3Ó by Ò0ß 10Ó
16. Ò1ß 2Ó by Ò0ß 1Ó
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
27
28
Chapter 1 Functions
17. Ò5ß 1Ó by Ò5ß 5Ó
18. Ò5ß 1Ó by Ò2ß 4Ó
19. Ò4ß 4Ó by Ò0ß 3Ó
20. Ò5ß 5Ó by Ò2ß 2Ó
21. Ò"!ß "!Ó by Ò'ß 'Ó
22. Ò&ß &Ó by Ò#ß #Ó
23. Ò'ß "!Ó by Ò'ß 'Ó
24. Ò$ß &Ó by Ò#ß "!Ó
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 1.4 Graphing with Calculators and Computers
25. Ò0Þ03ß 0Þ03Ó by Ò1Þ25ß 1Þ25Ó
26. Ò0Þ1ß 0Þ1Ó by Ò3ß 3Ó
27. Ò300ß 300Ó by Ò1Þ25ß 1Þ25Ó
28. Ò50ß 50Ó by Ò0Þ1ß 0Þ1Ó
29. Ò0Þ25ß 0Þ25Ó by Ò0Þ3ß 0Þ3Ó
30. Ò0Þ15ß 0Þ15Ó by Ò0Þ02ß 0Þ05Ó
31. x# #x œ % %y y# Ê y œ # „ Èx# #x ).
The lower half is produced by graphing
y œ # Èx# #x ).
32. y# "'x# œ " Ê y œ „ È" "'x# . The upper branch
is produced by graphing y œ È" "'x# .
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
29
30
Chapter 1 Functions
33.
34.
35.
36.
37.
38Þ
39.
40.
CHAPTER 1 PRACTICE EXERCISES
1. The area is A œ 1 r# and the circumference is C œ #1 r. Thus, r œ
2. The surface area is S œ %1 r# Ê r œ ˆ %S1 ‰
surface area gives S œ %1 r# œ %1 ˆ $%V1 ‰
"Î#
#Î$
C
#1
#
Ê A œ 1ˆ #C1 ‰ œ
C#
%1 .
$ $V
. The volume is V œ %$ 1 r$ Ê r œ É
%1 . Substitution into the formula for
.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Chapter 1 Practice Exercises
3. The coordinates of a point on the parabola are axß x# b. The angle of inclination ) joining this point to the origin satisfies
the equation tan ) œ
4. tan ) œ
rise
run
œ
h
&!!
x#
x
œ x. Thus the point has coordinates axß x# b œ atan )ß tan# )b.
Ê h œ &!! tan ) ft.
5.
6.
Symmetric about the origin.
Symmetric about the y-axis.
7.
8.
Neither
Symmetric about the y-axis.
9. yaxb œ axb# " œ x# " œ yaxb. Even.
10. yaxb œ axb& axb$ axb œ x& x$ x œ yaxb. Odd.
11. yaxb œ " cosaxb œ " cos x œ yaxb. Even.
12. yaxb œ secaxb tanaxb œ
13. yaxb œ
axb% "
axb$ #axb
œ
x% "
x$ #x
sinaxb
cos# axb
œ
sin x
cos# x
œ sec x tan x œ yaxb. Odd.
%
"
œ xx$ #
x œ yaxb. Odd.
14. yaxb œ axb sinaxb œ axb sin x œ ax sin xb œ yaxb. Odd.
15. yaxb œ x cosaxb œ x cos x. Neither even nor odd.
16. yaxb œ axbcosaxb œ x cos x œ yaxb. Odd.
17. Since f and g are odd Ê faxb œ faxb and gaxb œ gaxb.
(a) af † gbaxb œ faxbgaxb œ ÒfaxbÓÒgaxbÓ œ faxbgaxb œ af † gbaxb Ê f † g is even
(b) f 3 axb œ faxbfaxbfaxb œ ÒfaxbÓÒfaxbÓÒfaxbÓ œ faxb † faxb † faxb œ f 3 axb Ê f 3 is odd.
(c) fasinaxbb œ fasinaxbb œ fasinaxbb Ê fasinaxbb is odd.
(d) gasecaxbb œ gasecaxbb Ê gasecaxbb is even.
(e) lgaxbl œ lgaxbl œ lgaxbl Ê lgl is evenÞ
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
31
32
Chapter 1 Functions
18. Let faa xb œ faa xb and define gaxb œ fax ab. Then gaxb œ faaxb ab œ faa xb œ faa xb œ fax ab œ gaxb
Ê gaxb œ fax ab is even.
19. (a) The function is defined for all values of x, so the domain is a_ß _b.
(b) Since l x l attains all nonnegative values, the range is Ò#ß _Ñ.
20. (a) Since the square root requires " x !, the domain is Ð_ß "Ó.
(b) Since È" x attains all nonnegative values, the range is Ò#ß _Ñ.
21. (a) Since the square root requires "' x#
!, the domain is Ò%ß %Ó.
(b) For values of x in the domain, ! Ÿ "' x# Ÿ "', so ! Ÿ È"' x# Ÿ %. The range is Ò!ß %Ó.
22. (a) The function is defined for all values of x, so the domain is a_ß _b.
(b) Since $#x attains all positive values, the range is a"ß _b.
23. (a) The function is defined for all values of x, so the domain is a_ß _b.
(b) Since #ex attains all positive values, the range is a$ß _b.
24. (a) The function is equivalent to y œ tan #x, so we require #x Á
k1
#
for odd integers k. The domain is given by x Á
k1
%
for
odd integers k.
(b) Since the tangent function attains all values, the range is a_ß _b.
25. (a) The function is defined for all values of x, so the domain is a_ß _b.
(b) The sine function attains values from " to ", so # Ÿ #sina$x 1b Ÿ # and hence $ Ÿ #sina$x 1b " Ÿ ". The
range is Ò3ß 1Ó.
26. (a) The function is defined for all values of x, so the domain is a_ß _b.
&
(b) The function is equivalent to y œ È
x# , which attains all nonnegative values. The range is Ò!ß _Ñ.
27. (a) The logarithm requires x $ !, so the domain is a$ß _b.
(b) The logarithm attains all real values, so the range is a_ß _b.
28. (a) The function is defined for all values of x, so the domain is a_ß _b.
(b) The cube root attains all real values, so the range is a_ß _b.
29. (a)
(b)
(c)
(d)
Increasing because volume increases as radius increases
Neither, since the greatest integer function is composed of horizontal (constant) line segments
Decreasing because as the height increases, the atmospheric pressure decreases.
Increasing because the kinetic (motion) energy increases as the particles velocity increases.
30. (a) Increasing on Ò2, _Ñ
(c) Increasing on a_, _b
(b) Increasing on Ò1, _Ñ
(d) Increasing on Ò "# , _Ñ
31. (a) The function is defined for % Ÿ x Ÿ %, so the domain is Ò%ß %Ó.
(b) The function is equivalent to y œ Èl x l, % Ÿ x Ÿ %, which attains values from ! to # for x in the domain. The
range is Ò!ß #Ó.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Chapter 1 Practice Exercises
33
32. (a) The function is defined for # Ÿ x Ÿ #, so the domain is Ò#ß #Ó.
(b) The range is Ò"ß "Ó.
33. First piece: Line through a!ß "b and a"ß !b. m œ
Second piece: Line through a"ß "b and a#ß !b. m
faxb œ œ
" x, ! Ÿ x "
# x, " Ÿ x Ÿ #
34. First piece: Line through a!ß !b and a2ß 5b. m œ
Second piece: Line through a2ß 5b and a4ß !b. m
faxb œ 10 5
2 x,
5x
2 ,
!"
"
"! œ " œ
"
"
œ !# " œ "
" Ê y œ x " œ " x
œ " Ê y œ ax "b " œ x # œ # x
5!
5
5
2! œ 2 Ê y œ 2x
!5
5
œ 4 2 œ 2 œ 52 Ê
y œ 52 ax 2b 5 œ 52 x 10 œ 10 !Ÿx2
(Note: x œ 2 can be included on either piece.)
2ŸxŸ4
35. (a) af‰gba"b œ faga"bb œ fŠ È"" # ‹ œ fa"b œ
(b) ag‰f ba#b œ gafa#bb œ gˆ "2 ‰ œ
(c) af‰f baxb œ fafaxbb œ fˆ "x ‰ œ
"
É "# #
"
"Îx
œ
"
È#Þ&
"
"
œ"
or É &#
œ x, x Á !
(d) ag‰gbaxb œ gagaxbb œ gŠ Èx" # ‹ œ
"
"
É Èx # #
œ
% x#
È
É " #È x #
$
36. (a) af‰gba"b œ faga"bb œ fˆÈ
" "‰ œ fa!b œ # ! œ #
$
(b) ag‰f ba#b œ faga#bb œ ga# #b œ ga!b œ È
!"œ"
(c) af‰f baxb œ fafaxbb œ fa# xb œ # a# xb œ x
$
$
$
È
(d) ag‰gbaxb œ gagaxbb œ gˆÈ
x "‰ œ É
x""
#
37. (a) af‰gbaxb œ fagaxbb œ fˆÈx #‰ œ # ˆÈx #‰ œ x, x #.
ag‰f baxb œ fagaxbb œ ga# x# b œ Èa# x# b # œ È% x#
(b) Domain of f‰g: Ò#ß _ÑÞ
Domain of g‰f: Ò#ß #ÓÞ
(c) Range of f‰g: Ð_ß #ÓÞ
Range of g‰f: Ò!ß #ÓÞ
%
38. (a) af‰gbaxb œ fagaxbb œ fŠÈ" x‹ œ ÉÈ" x œ È
" x.
ag‰f baxb œ fagaxbb œ gˆÈx‰ œ É" Èx
(b) Domain of f‰g: Ð_ß "ÓÞ
Domain of g‰f: Ò!ß "ÓÞ
39.
y œ faxb
(c) Range of f‰g: Ò!ß _ÑÞ
Range of g‰f: Ò!ß "ÓÞ
y œ af‰f baxb
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
5x
2
34
Chapter 1 Functions
40.
41.
42.
The graph of f# (x) œ f" akxkb is the same as the
graph of f" (x) to the right of the y-axis. The
graph of f# (x) to the left of the y-axis is the
reflection of y œ f" (x), x 0 across the y-axis.
43.
It does not change the graph.
44.
Whenever g" (x) is positive, the graph of y œ g# (x)
œ kg" (x)k is the same as the graph of y œ g" (x).
When g" (x) is negative, the graph of y œ g# (x) is
the reflection of the graph of y œ g" (x) across the
x-axis.
Whenever g" (x) is positive, the graph of y œ g# (x) œ kg" (x)k
is the same as the graph of y œ g" (x). When g" (x) is
negative, the graph of y œ g# (x) is the reflection of the
graph of y œ g" (x) across the x-axis.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Chapter 1 Practice Exercises
45.
46.
The graph of f# (x) œ f" akxkb is the same as the
graph of f" (x) to the right of the y-axis. The
graph of f# (x) to the left of the y-axis is the
reflection of y œ f" (x), x 0 across the y-axis.
Whenever g" (x) is positive, the graph of
y œ g# (x) œ kg" (x)k is the same as the graph of
y œ g" (x). When g" (x) is negative, the graph of
y œ g# (x) is the reflection of the graph of
y œ g" (x) across the x-axis.
47.
48.
The graph of f# (x) œ f" akxkb is the same as the
graph of f" (x) to the right of the y-axis. The
graph of f# (x) to the left of the y-axis is the
reflection of y œ f" (x), x 0 across the y-axis.
49. (a) y œ gax 3b (c) y œ gaxb
(e) y œ 5 † gaxb
"
#
The graph of f# (x) œ f" akxkb is the same as the
graph of f" (x) to the right of the y-axis. The
graph of f# (x) to the left of the y-axis is the
reflection of y œ f" (x), x 0 across the y-axis.
(b) y œ gˆx 3# ‰ 2
(d) y œ gaxb
(f) y œ ga5xb
50. (a) Shift the graph of f right 5 units
(b) Horizontally compress the graph of f by a factor of 4
(c) Horizontally compress the graph of f by a factor of 3 and a then reflect the graph about the y-axis
(d) Horizontally compress the graph of f by a factor of 2 and then shift the graph left "# unit.
(e) Horizontally stretch the graph of f by a factor of 3 and then shift the graph down 4 units.
(f) Vertically stretch the graph of f by a factor of 3, then reflect the graph about the x-axis, and finally shift the
graph up "4 unit.
51. Reflection of the grpah of y œ Èx about the x-axis
followed by a horizontal compression by a factor of
1
2 then a shift left 2 units.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
35
36
Chapter 1 Functions
52. Reflect the graph of y œ x about the x-axis, followed
by a vertical compression of the graph by a factor
of 3, then shift the graph up 1 unit.
53. Vertical compression of the graph of y œ
1
x2
by a
factor of 2, then shift the graph up 1 unit.
54. Reflect the graph of y œ x1Î3 about the y-axis, then
compress the graph horizontally by a factor of 5.
55.
56.
period œ 1
57.
period œ 41
58.
period œ 2
period œ 4
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Chapter 1 Practice Exercises
59.
60.
period œ 21
period œ 21
1
3
61. (a) sin B œ sin
œ
b
c
œ
b
#
Ê b œ 2 sin
1
3
œ 2Š
È3
# ‹
œ È3. By the theorem of Pythagoras,
a# b# œ c# Ê a œ Èc# b# œ È4 3 œ 1.
1
3
(b) sin B œ sin
œ
b
c
œ
2
c
Ê cœ
2
sin 13
œ È23 œ
Š ‹
#
4
È3
#
. Thus, a œ Èc# b# œ ÊŠ È43 ‹ (2)# œ É 34 œ
62. (a) sin A œ
a
c
Ê a œ c sin A
(b) tan A œ
a
b
Ê a œ b tan A
63. (a) tan B œ
b
a
Ê aœ
(b) sin A œ
a
c
Ê cœ
64. (a) sin A œ
a
c
(c) sin A œ
a
c
b
tan B
œ
a
sin A
È c # b #
c
65. Let h œ height of vertical pole, and let b and c denote the
distances of points B and C from the base of the pole,
measured along the flatground, respectively. Then,
tan 50° œ hc , tan 35° œ hb , and b c œ 10.
Thus, h œ c tan 50° and h œ b tan 35° œ (c 10) tan 35°
Ê c tan 50° œ (c 10) tan 35°
Ê c (tan 50° tan 35°) œ 10 tan 35°
tan 35°
Ê c œ tan10
50°tan 35° Ê h œ c tan 50°
œ
10 tan 35° tan 50°
tan 50°tan 35°
¸ 16.98 m.
66. Let h œ height of balloon above ground. From the figure at
the right, tan 40° œ ha , tan 70° œ hb , and a b œ 2. Thus,
h œ b tan 70° Ê h œ (2 a) tan 70° and h œ a tan 40°
Ê (2 a) tan 70° œ a tan 40° Ê a(tan 40° tan 70°)
70°
œ 2 tan 70° Ê a œ tan 240°tantan
70° Ê h œ a tan 40°
œ
2 tan 70° tan 40°
tan 40°tan 70°
¸ 1.3 km.
67. (a)
(b) The period appears to be 41.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
2
È3
.
37
38
Chapter 1 Functions
(c) f(x 41) œ sin (x 41) cos ˆ x#41 ‰ œ sin (x 21) cos ˆ x# 21‰ œ sin x cos
since the period of sine and cosine is 21. Thus, f(x) has period 41.
x
#
68. (a)
(b) D œ (_ß 0) (!ß _); R œ [1ß 1]
(c) f is not periodic. For suppose f has period p. Then f ˆ #"1 kp‰ œ f ˆ #"1 ‰ œ sin 21 œ 0 for all
integers k. Choose k so large that
"
#1
kp "
1
Ê 0
"
(1/21)kp
1. But then
f ˆ #"1 kp‰ œ sin Š (1/#1")kp ‹ 0 which is a contradiction. Thus f has no period, as claimed.
CHAPTER 1 ADDITIONAL AND ADVANCED EXERCISES
1. There are (infinitely) many such function pairs. For example, f(x) œ 3x and g(x) œ 4x satisfy
f(g(x)) œ f(4x) œ 3(4x) œ 12x œ 4(3x) œ g(3x) œ g(f(x)).
2. Yes, there are many such function pairs. For example, if g(x) œ (2x 3)$ and f(x) œ x"Î$ , then
(f ‰ g)(x) œ f(g(x)) œ f a(2x 3)$ b œ a(2x 3)$ b
"Î$
œ 2x 3.
3. If f is odd and defined at x, then f(x) œ f(x). Thus g(x) œ f(x) 2 œ f(x) 2 whereas
g(x) œ (f(x) 2) œ f(x) 2. Then g cannot be odd because g(x) œ g(x) Ê f(x) 2 œ f(x) 2
Ê 4 œ 0, which is a contradiction. Also, g(x) is not even unless f(x) œ 0 for all x. On the other hand, if f is
even, then g(x) œ f(x) 2 is also even: g(x) œ f(x) 2 œ f(x) 2 œ g(x).
4. If g is odd and g(0) is defined, then g(0) œ g(0) œ g(0). Therefore, 2g(0) œ 0 Ê g(0) œ 0.
5. For (xß y) in the 1st quadrant, kxk kyk œ 1 x
Í x y œ 1 x Í y œ 1. For (xß y) in the 2nd
quadrant, kxk kyk œ x 1 Í x y œ x 1
Í y œ 2x 1. In the 3rd quadrant, kxk kyk œ x 1
Í x y œ x 1 Í y œ 2x 1. In the 4th
quadrant, kxk kyk œ x 1 Í x (y) œ x 1
Í y œ 1. The graph is given at the right.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Chapter 1 Additional and Advanced Exercises
6. We use reasoning similar to Exercise 5.
(1) 1st quadrant: y kyk œ x kxk
Í 2y œ 2x Í y œ x.
(2) 2nd quadrant: y kyk œ x kxk
Í 2y œ x (x) œ 0 Í y œ 0.
(3) 3rd quadrant: y kyk œ x kxk
Í y (y) œ x (x) Í 0 œ 0
Ê all points in the 3rd quadrant
satisfy the equation.
(4) 4th quadrant: y kyk œ x kxk
Í y (y) œ 2x Í 0 œ x. Combining
these results we have the graph given at the
right:
7. (a) sin# x cos# x œ 1 Ê sin# x œ 1 cos# x œ (1 cos x)(1 cos x) Ê (1 cos x) œ
Ê
1cos x
sin x
œ
sin# x
1cos x
sin x
1cos x
(b) Using the definition of the tangent function and the double angle formulas, we have
#
tan ˆ x# ‰ œ
sin# ˆ x# ‰
cos# ˆ #x ‰
œ
"
x ‹‹
cos Š2 Š #
#
"cos Š2 Š #x ‹‹
#
œ
1cos x
1cos x
.
8. The angles labeled # in the accompanying figure are
equal since both angles subtend arc CD. Similarly, the
two angles labeled ! are equal since they both subtend
arc AB. Thus, triangles AED and BEC are similar which
) b
implies ab c œ 2a cos
a c
Ê (a c)(a c) œ b(2a cos ) b)
Ê a# c# œ 2ab cos ) b#
Ê c# œ a# b# 2ab cos ).
9. As in the proof of the law of sines of Section 1.3, Exercise 61, ah œ bc sin A œ ab sin C œ ac sin B
Ê the area of ABC œ "# (base)(height) œ "# ah œ "# bc sin A œ "# ab sin C œ "# ac sin B.
10. As in Section 1.3, Exercise 61, (Area of ABC)# œ
œ
"
4
(base)# (height)# œ
"
4
a# h # œ
"
4
a# b# sin# C
a# b# a" cos# Cb . By the law of cosines, c# œ a# b# 2ab cos C Ê cos C œ
Thus, (area of ABC)# œ
œ
"
4
"
16
"
4
a# b# a" cos# Cb œ
#
Š4a# b# aa# b# c# b ‹ œ
"
16
"
4
a# b# Œ" Š a
#
b# c#
‹
#ab
#
œ
a# b#
4
a# b# c#
2ab
Š" .
aa # b # c # b
4a# b#
#
‹
ca2ab aa# b# c# bb a2ab aa# b# c# bbd
"
ca(a b)# c# b ac# (a b)# bd œ 16
c((a b) c)((a b) c)(c (a b))(c (a b))d
a
b
c
a
b
c
a
b
c
a
b
c
œ ˆ # ‰ ˆ # ‰ ˆ # ‰ ˆ # ‰‘ œ s(s a)(s b)(s c), where s œ a#bc .
œ
"
16
Therefore, the area of ABC equals Ès(s a)(s b)(s c) .
11. If f is even and odd, then f(x) œ f(x) and f(x) œ f(x) Ê f(x) œ f(x) for all x in the domain of f.
Thus 2f(x) œ 0 Ê f(x) œ 0.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
39
40
Chapter 1 Functions
f(x) f((x))
œ f(x) #f(x) œ E(x) Ê E
#
even function. Define O(x) œ f(x) E(x) œ f(x) f(x) #f(x) œ f(x) #f(x) . Then
O(x) œ f(x) #f((x)) œ f(x)# f(x) œ Š f(x) #f(x) ‹ œ O(x) Ê O is an odd function
12. (a) As suggested, let E(x) œ
f(x) f(x)
#
Ê E(x) œ
is an
Ê f(x) œ E(x) O(x) is the sum of an even and an odd function.
(b) Part (a) shows that f(x) œ E(x) O(x) is the sum of an even and an odd function. If also
f(x) œ E" (x) O" (x), where E" is even and O" is odd, then f(x) f(x) œ 0 œ aE" (x) O" (x)b
(E(x) O(x)). Thus, E(x) E" (x) œ O" (x) O(x) for all x in the domain of f (which is the same as the
domain of E E" and O O" ). Now (E E" )(x) œ E(x) E" (x) œ E(x) E" (x) (since E and E" are
even) œ (E E" )(x) Ê E E" is even. Likewise, (O" O)(x) œ O" (x) O(x) œ O" (x) (O(x))
(since O and O" are odd) œ (O" (x) O(x)) œ (O" O)(x) Ê O" O is odd. Therefore, E E" and
O" O are both even and odd so they must be zero at each x in the domain of f by Exercise 11. That is,
E" œ E and O" œ O, so the decomposition of f found in part (a) is unique.
13. y œ ax# bx c œ a Šx# ba x b#
4a# ‹
b#
4a
c œ a ˆx b ‰#
2a
b#
4a
c
(a) If a 0 the graph is a parabola that opens upward. Increasing a causes a vertical stretching and a shift
of the vertex toward the y-axis and upward. If a 0 the graph is a parabola that opens downward.
Decreasing a causes a vertical stretching and a shift of the vertex toward the y-axis and downward.
(b) If a 0 the graph is a parabola that opens upward. If also b 0, then increasing b causes a shift of the
graph downward to the left; if b 0, then decreasing b causes a shift of the graph downward and to the
right.
If a 0 the graph is a parabola that opens downward. If b 0, increasing b shifts the graph upward
to the right. If b 0, decreasing b shifts the graph upward to the left.
(c) Changing c (for fixed a and b) by ?c shifts the graph upward ?c units if ?c 0, and downward ?c
units if ?c 0.
14. (a) If a 0, the graph rises to the right of the vertical line x œ b and falls to the left. If a 0, the graph
falls to the right of the line x œ b and rises to the left. If a œ 0, the graph reduces to the horizontal
line y œ c. As kak increases, the slope at any given point x œ x! increases in magnitude and the graph
becomes steeper. As kak decreases, the slope at x! decreases in magnitude and the graph rises or falls
more gradually.
(b) Increasing b shifts the graph to the left; decreasing b shifts it to the right.
(c) Increasing c shifts the graph upward; decreasing c shifts it downward.
15. Each of the triangles pictured has the same base
b œ v?t œ v(1 sec). Moreover, the height of each
triangle is the same value h. Thus "# (base)(height) œ
"
#
bh
œ A" œ A# œ A$ œ á . In conclusion, the object sweeps
out equal areas in each one second interval.
16. (a) Using the midpoint formula, the coordinates of P are ˆ a# 0 ß b# 0 ‰ œ ˆ #a ß b# ‰ . Thus the slope
of OP œ
?y
?x
œ
b/2
a/2
œ
b
a
.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Chapter 1 Additional and Advanced Exercises
(b) The slope of AB œ
b 0
0 a
41
œ ba . The line segments AB and OP are perpendicular when the product
#
of their slopes is " œ ˆ ba ‰ ˆ ba ‰ œ ba# . Thus, b# œ a# Ê a œ b (since both are positive). Therefore, AB
is perpendicular to OP when a œ b.
17. From the figure we see that 0 Ÿ ) Ÿ
cos ) œ
and AB œ AD œ 1. From trigonometry we have the following: sin ) œ
sin )
cos ) .
œ CD, and tan ) œ
œ
We can see that:
w
"
area ˜AEB area sector DB area ˜ADC Ê # aAEbaEBb "# aADb2 ) "# aADbaCDb
AE
AB
œ AE, tan ) œ
1
2
CD
AD
EB
AE
Ê "# sin ) cos ) "# a"b2 ) "# a"batan )b Ê "# sin ) cos ) "# ) " sin )
# cos )
18. af‰gbaxb œ fagaxbb œ aacx db b œ acx ad b and ag‰f baxb œ gafaxbb œ caax bb d œ acx cb d
Thus af‰gbaxb œ ag‰f baxb Ê acx ad b œ acx bc d Ê ad b œ bc d. Note that fadb œ ad b and
gabb œ cb d, thus af‰gbaxb œ ag‰f baxb if fadb œ gabb.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
EB
AB
œ EB,
42
Chapter 1 Functions
NOTES:
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
CHAPTER 2 LIMITS AND CONTINUITY
2.1 RATES OF CHANGE AND TANGENTS TO CURVES
1. (a)
?f
?x
œ
f(3) f(2)
3#
2. (a)
?g
?x
œ
g(1) g(1)
1 (1)
3. (a)
?h
?t
œ
h ˆ 341 ‰ h ˆ 14 ‰
1
31
4 4
œ
?g
?t
œ
g(1) g(0)
10
(2 1) (2 1)
10
4. (a)
5.
?R
?)
œ
R(2) R(0)
20
6.
?P
?)
œ
P(2) P(1)
21
7. (a)
?y
?x
œ
?y
?x
œ
?y
?x
œ
?y
?x
œ
?y
?x
œ
?y
?x
œ
?y
?x
œ
œ
œ
28 9
1
œ
œ
œ
1 1
2
œ 19
(b)
?f
?x
œ
f(1) f(")
1 (1)
œ
20
#
œ1
œ0
(b)
?g
?x
œ
g(0)g(2)
0(2)
œ
04
#
œ 2
(b)
?h
?t
œ
h ˆ 1# ‰ h ˆ 16 ‰
11
#
6
œ
?g
?t
œ
g(1) g(1)
1 (1)
œ
1 1
1
#
È 8 1 È 1
#
œ 14
œ 12
3"
#
œ
(8 16 10)(" % &)
1
ˆa2 h b2 3 ‰ ˆ 2 2 3 ‰
h
œ
(b)
0 È3
1
3
œ
3 È 3
1
(2 1) (2 ")
#1
œ0
œ1
œ22œ0
4 4h h2 3 1
h
œ
4h h2
h
œ 4 h. As h Ä 0, 4 h Ä 4 Ê at Pa2, 1b the slope is 4.
(b) y 1 œ 4ax 2b Ê y 1 œ 4x 8 Ê y œ 4x 7
8. (a)
ˆ 5 a1 h b 2 ‰ ˆ 5 1 2 ‰
h
œ
5 1 2h h2 4
h
œ
2h h2
h
œ 2 h. As h Ä 0, 2 h Ä 2 Ê at Pa1, 4b the
slope is 2.
(b) y 4 œ a2bax 1b Ê y 4 œ 2x 2 Ê y œ 2x 6
9. (a)
ˆa2 h b2 2 a 2 h b 3 ‰ ˆ 2 2 2 a 2 b 3 ‰
h
œ
4 4h h2 4 2h 3 a3b
h
œ
2h h2
h
œ 2 h. As h Ä 0, 2 h Ä 2 Ê at
Pa2, 3b the slope is 2.
(b) y a3b œ 2ax 2b Ê y 3 œ 2x 4 Ê y œ 2x 7.
10. (a)
ˆa1 h b2 4 a 1 h b ‰ ˆ 1 2 4 a 1 b ‰
h
œ
1 2h h2 4 4h a3b
h
œ
h2 2h
h
œ h 2. As h Ä 0, h 2 Ä 2 Ê at
Pa1, 3b the slope is 2.
(b) y a3b œ a2bax 1b Ê y 3 œ 2x 2 Ê y œ 2x 1.
11. (a)
a2 h b 3 2 3
h
œ
8 12h 4h2 h3 8
h
œ
12h 4h2 h3
h
œ 12 4h h2 . As h Ä 0, 12 4h h2 Ä 12, Ê at
Pa2, 8b the slope is 12.
(b) y 8 œ 12ax 2b Ê y 8 œ 12x 24 Ê y œ 12x 16.
12. (a)
2 a1 h b3 ˆ 2 1 3 ‰
h
œ
2 1 3h 3h2 h3 1
h
œ
3h 3h2 h3
h
œ 3 3h h2 . As h Ä 0, 3 3h h2 Ä 3, Ê at
Pa1, 1b the slope is 3.
(b) y 1 œ a3bax 1b Ê y 1 œ 3x 3 Ê y œ 3x 4.
13. (a)
a1 hb3 12a1 hb ˆ13 12a"b‰
h
2
œ
1 3h 3h2 h3 12 12h a11b
h
œ
9h 3h2 h3
h
œ 9 3h h2 . As h Ä 0,
9 3h h Ä 9 Ê at Pa1, 11b the slope is 9.
(b) y a11b œ a9bax 1b Ê y 11 œ 9x 9 Ê y œ 9x 2.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
44
Chapter 2 Limits and Continuity
14. (a)
?y
?x
œ
a2 h b 3 3 a 2 h b 2 4 ˆ 2 3 3 a 2 b 2 4 ‰
h
2
8 12h 6h2 h3 12 12h 3h2 4 0
h
œ
œ
3h2 h3
h
œ 3h h2 . As h Ä 0,
3h h Ä 0 Ê at Pa2, 0b the slope is 0.
(b) y 0 œ 0ax 2b Ê y œ 0.
15. (a)
?p
?t
Slope of PQ œ
Q
650 225
20 10
650 375
20 14
650 475
20 16.5
650 550
20 18
Q" (10ß 225)
Q# (14ß 375)
Q$ (16.5ß 475)
Q% (18ß 550)
œ 42.5 m/sec
œ 45.83 m/sec
œ 50.00 m/sec
œ 50.00 m/sec
(b) At t œ 20, the sportscar was traveling approximately 50 m/sec or 180 km/h.
16. (a)
Slope of PQ œ
Q
Q" (5ß 20)
Q# (7ß 39)
Q$ (8.5ß 58)
Q% (9.5ß 72)
80 20
10 5
80 39
10 7
80 58
10 8.5
80 72
10 9.5
?p
?t
œ 12 m/sec
œ 13.7 m/sec
œ 14.7 m/sec
œ 16 m/sec
(b) Approximately 16 m/sec
17. (a)
(b)
?p
?t
œ
174 62
2004 2002
œ
112
#
œ 56 thousand dollars per year
(c) The average rate of change from 2001 to 2002 is ??pt œ
62 27
20022 2001 œ 35 thousand dollars per year.
111 62
The average rate of change from 2002 to 2003 is ??pt œ 2003
2002 œ 49 thousand dollars per year.
So, the rate at which profits were changing in 2002 is approximatley "# a35 49b œ 42 thousand dollars
18. (a) F(x) œ (x 2)/(x 2)
x
1.2
F(x)
4.0
?F
?x
?F
?x
?F
?x
œ
?g
?x
?g
?x
œ
œ
œ
1.1
3.4
1.01
3.04
1.001
3.004
1.0001
3.0004
1
3
4.0 (3)
œ 5.0;
1.2 1
3.04 (3)
œ 4.04;
1.01 1
3.!!!% (3)
œ 4.!!!%;
1.0001 1
?F
?x
?F
?x
œ
œ
3.4 (3)
œ 4.4;
1.1 1
3.004 (3)
œ 4.!!%;
1.001 1
È
g(2) g(1)
œ #21" ¸ 0.414213
21
È1 h"
g(1 h) g(1)
(1 h) 1 œ
h
?g
?x
œ
g(1.5) g(1)
1.5 1
(b) The rate of change of F(x) at x œ 1 is 4.
19. (a)
œ
œ
È1.5 "
0.5
¸ 0.449489
(b) g(x) œ Èx
1h
È1 h
ŠÈ1 h 1‹ /h
1.1
1.04880
1.01
1.004987
1.001
1.0004998
1.0001
1.0000499
1.00001
1.000005
1.000001
1.0000005
0.4880
0.4987
0.4998
0.499
0.5
0.5
(c) The rate of change of g(x) at x œ 1 is 0.5.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
per year.
Section 2.1 Rates of Change and Tangents to Curves
(d) The calculator gives lim
hÄ!
20. (a) i)
ii)
(b)
f(3) f(2)
32
œ
f(T) f(2)
T#
œ
""
3
#
1
" "
T #
T#
T
f(T)
af(T) f(2)b/aT 2b
"
œ
6
1
È1 h"
h
œ "# .
œ "6
#TT
T#
2
#T
œ
45
œ
2T
#T(T 2)
2.1
0.476190
0.2381
œ
2T
#T(2 T)
2.01
0.497512
0.2488
œ #"T , T Á 2
2.001
0.499750
0.2500
2.0001
0.4999750
0.2500
2.00001
0.499997
0.2500
2.000001
0.499999
0.2500
(c) The table indicates the rate of change is 0.25 at t œ 2.
" ‰
(d) lim ˆ #T
œ 4"
TÄ#
NOTE: Answers will vary in Exercises 21 and 22.
s
15 0
˜s
20 15
10
˜s
30 20
21. (a) Ò0, 1Ó: ˜
˜t œ 1 0 œ 15 mphà Ò1, 2.5Ó: ˜t œ 2.5 1 œ 3 mphà Ò2.5, 3.5Ó: ˜t œ 3.5 2.5 œ 10 mph
(b) At Pˆ "# , 7.5‰: Since the portion of the graph from t œ 0 to t œ 1 is nearly linear, the instantaneous rate of change
will be almost the same as the average rate of change, thus the instantaneous speed at t œ
"
#
is
15 7.5
1 0.5
œ 15 mi/hr.
At Pa2, 20b: Since the portion of the graph from t œ 2 to t œ 2.5 is nearly linear, the instantaneous rate of change will
20
be nearly the same as the average rate of change, thus v œ 20
2.5 2 œ 0 mi/hr. For values of t less than 2, we have
Slope of PQ œ
Q
?s
?t
15 20
1 2 œ 5 mi/hr
19 20
1.5 2 œ 2 mi/hr
19.9 20
1.9 2 œ 1 mi/hr
Q" (1ß 15)
Q# (1.5ß 19)
Q$ (1.9ß 19.9)
Thus, it appears that the instantaneous speed at t œ 2 is 0 mi/hr.
At Pa3, 22b:
s
Q
Slope of PQ œ ?
?t
35 22
43
30 22
3.5 3
23 22
3.1 3
Q" (4ß 35)
Q# (3.5ß 30)
Q$ (3.1ß 23)
Slope of PQ œ
Q
œ 13 mi/hr
Q" (2ß 20)
œ 16 mi/hr
Q# (2.5ß 20)
œ 10 mi/hr
Q$ (2.9ß 21.6)
20 22
2 3 œ 2 mi/hr
20 22
2.5 3 œ 4 mi/hr
21.6 22
2.9 3 œ 4 mi/hr
Thus, it appears that the instantaneous speed at t œ 3 is about 7 mi/hr.
(c) It appears that the curve is increasing the fastest at t œ 3.5. Thus for Pa3.5, 30b
s
Q
Slope of PQ œ ?
Q
Slope of PQ œ
?t
35 30
4 3.5 œ 10 mi/hr
34 30
3.75 3.5 œ 16 mi/hr
32 30
3.6 3.5 œ 20 mi/hr
Q" (4ß 35)
Q# (3.75ß 34)
Q$ (3.6ß 32)
˜A
˜t
œ
10 15
30
(b) At Pa1, 14b:
Q
Q" (2ß 12.2)
Q# (1.5ß 13.2)
Q$ (1.1ß 13.85)
¸ 1.67
gal
day à
Ò0, 5Ó:
Q# (3.25ß 25)
Q$ (3.4ß 28)
Slope of PQ œ
˜A
˜t
?A
?t
12.2 14
2 1 œ 1.8 gal/day
13.2 14
1.5 1 œ 1.6 gal/day
13.85 14
1.1 1 œ 1.5 gal/day
œ
3.9 15
50
¸ 2.2
gal
day à Ò7,
10Ó:
?s
?t
22 30
3 3.5 œ 16 mi/hr
25 30
3.25 3.5 œ 20 mi/hr
28 30
3.4 3.5 œ 20 mi/hr
Q" (3ß 22)
Thus, it appears that the instantaneous speed at t œ 3.5 is about 20 mi/hr.
22. (a) Ò0, 3Ó:
?s
?t
˜A
˜t
Q
Q" (0ß 15)
Q# (0.5ß 14.6)
Q$ (0.9ß 14.86)
œ
0 1.4
10 7
¸ 0.5
Slope of PQ œ
gal
day
?A
?t
15 14
0 1 œ 1 gal/day
14.6 14
0.5 1 œ 1.2 gal/day
14.86 14
0.9 1 œ 1.4 gal/day
Thus, it appears that the instantaneous rate of consumption at t œ 1 is about 1.45 gal/day.
At Pa4, 6b:
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
46
Chapter 2 Limits and Continuity
Q
Q" (5ß 3.9)
Q# (4.5ß 4.8)
Q$ (4.1ß 5.7)
Slope of PQ œ
3.9 6
54
4.8 6
4.5 4
5.7 6
4.1 4
?A
?t
Q
œ 2.1 gal/day
Q" (3ß 10)
œ 2.4 gal/day
Q# (3.5ß 7.8)
œ 3 gal/day
Q$ (3.9ß 6.3)
Slope of PQ œ
10 6
3 4 œ 4 gal/day
7.8 6
3.5 4 œ 3.6 gal/day
6.3 6
3.9 4 œ 3 gal/day
Thus, it appears that the instantaneous rate of consumption at t œ 1 is 3 gal/day.
At Pa8, 1b:
Q
Slope of PQ œ ??At
Q
Slope of PQ œ
Q" (9ß 0.5)
Q# (8.5ß 0.7)
Q$ (8.1ß 0.95)
0.5 1
9 8 œ 0.5 gal/day
0.7 1
8.5 8 œ 0.6 gal/day
0.95 1
8.1 8 œ 0.5 gal/day
Q" (7ß 1.4)
4.8 7.8
4.5 3.5 œ 3 gal/day
6 7.8
4 3.5 œ 3.6 gal/day
7.4 7.8
3.6 3.5 œ 4 gal/day
Q" (2.5ß 11.2)
Q# (7.5ß 1.3)
Q$ (7.9ß 1.04)
?A
?t
?A
?t
1.4 1
7 8 œ 0.6 gal/day
1.3 1
7.5 8 œ 0.6 gal/day
1.04 1
7.9 8 œ 0.6 gal/day
Thus, it appears that the instantaneous rate of consumption at t œ 1 is 0.55 gal/day.
(c) It appears that the curve (the consumption) is decreasing the fastest at t œ 3.5. Thus for Pa3.5, 7.8b
s
Q
Slope of PQ œ ??At
Q
Slope of PQ œ ?
?t
Q" (4.5ß 4.8)
Q# (4ß 6)
Q$ (3.6ß 7.4)
Q# (3ß 10)
Q$ (3.4ß 8.2)
Thus, it appears that the rate of consumption at t œ 3.5 is about 4 gal/day.
11.2 7.8
2.5 3.5 œ 3.4 gal/day
10 7.8
3 3.5 œ 4.4 gal/day
8.2 7.8
3.4 3.5 œ 4 gal/day
2.2 LIMIT OF A FUNCTION AND LIMIT LAWS
1. (a) Does not exist. As x approaches 1 from the right, g(x) approaches 0. As x approaches 1 from the left, g(x)
approaches 1. There is no single number L that all the values g(x) get arbitrarily close to as x Ä 1.
(b) 1
(c) 0
(d) 0.5
2. (a) 0
(b) 1
(c) Does not exist. As t approaches 0 from the left, f(t) approaches 1. As t approaches 0 from the right, f(t)
approaches 1. There is no single number L that f(t) gets arbitrarily close to as t Ä 0.
(d) 1
3. (a) True
(d) False
(g) True
(b) True
(e) False
(c) False
(f) True
4. (a) False
(d) True
(b) False
(e) True
(c) True
5.
x
lim
x Ä 0 kx k
x
kx k
does not exist because
x
kx k
œ
x
x
œ 1 if x 0 and
approaches 1. As x approaches 0 from the right,
x
kx k
x
kxk
œ
x
x
œ 1 if x 0. As x approaches 0 from the left,
approaches 1. There is no single number L that all the
function values get arbitrarily close to as x Ä 0.
6. As x approaches 1 from the left, the values of
"
x 1
become increasingly large and negative. As x approaches 1
from the right, the values become increasingly large and positive. There is no one number L that all the function
values get arbitrarily close to as x Ä 1, so lim x" 1 does not exist.
xÄ1
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 2.2 Limit of a Function and Limit Laws
47
7. Nothing can be said about f(x) because the existence of a limit as x Ä x! does not depend on how the function
is defined at x! . In order for a limit to exist, f(x) must be arbitrarily close to a single real number L when
x is close enough to x! . That is, the existence of a limit depends on the values of f(x) for x near x! , not on the
definition of f(x) at x! itself.
8. Nothing can be said. In order for lim f(x) to exist, f(x) must close to a single value for x near 0 regardless of the
xÄ0
value f(0) itself.
9. No, the definition does not require that f be defined at x œ 1 in order for a limiting value to exist there. If f(1) is
defined, it can be any real number, so we can conclude nothing about f(1) from lim f(x) œ 5.
xÄ1
10. No, because the existence of a limit depends on the values of f(x) when x is near 1, not on f(1) itself. If
lim f(x) exists, its value may be some number other than f(1) œ 5. We can conclude nothing about lim f(x),
xÄ1
xÄ1
whether it exists or what its value is if it does exist, from knowing the value of f(1) alone.
11.
lim (2x 5) œ 2(7) 5 œ 14 5 œ 9
x Ä (
12. lim ax# 5x 2b œ (2)# 5(2) 2 œ 4 10 2 œ 4
xÄ#
13. lim 8(t 5)(t 7) œ 8(6 5)(6 7) œ 8
tÄ'
14.
lim ax$ 2x# 4x 8b œ (2)$ 2(2)# 4(2) 8 œ 8 8 8 8 œ 16
x Ä #
15. lim
x3
œ
x Ä # x6
17.
23
26
16. lim# 3s(2s 1) œ 3 ˆ 23 ‰ 2 ˆ 23 ‰ 1‘ œ 2 ˆ 43 1‰ œ
5
8
sÄ
$
lim 3(2x 1)# œ 3(2(1) 1)# œ 3(3)# œ 27
x Ä "
y2
18. lim
#
y Ä # y 5y 6
19.
œ
œ
22
(2)# 5(#) 6
œ
4
4 10 6
œ
4
#0
œ
"
5
%
lim (5 y)%Î$ œ [5 (3)]%Î$ œ (8)%Î$ œ ˆ(8)"Î$ ‰ œ 2% œ 16
y Ä $
20. lim (2z 8)"Î$ œ (2(0) 8)"Î$ œ (8)"Î$ œ 2
zÄ!
21. lim
3
22. lim
È5h 4 2
h
h Ä ! È3h 1 1
hÄ0
œ
5
È4 2
23. lim
œ
x5
#
x Ä & x 25
24.
œ
3
È3(0) 1 1
œ lim
hÄ0
œ
3
È1 1
È5h 4 2
h
†
œ
3
2
È5h 4 2
È5h 4 2
œ lim
a5h 4b 4
h Ä 0 hŠÈ5h 4 2‹
œ lim
5h
h Ä 0 hŠÈ5h 4 2‹
œ lim
5
4
x5
œ lim
x3
x Ä & (x 5)(x 5)
lim
#
x Ä $ x 4x 3
œ lim
œ lim
x3
1
x Ä & x5
x Ä $ (x 3)(x 1)
œ lim
œ
"
55
œ
"
10
1
œ
"
3 1
x Ä $ x 1
5
h Ä 0 È5h 4 2
œ "2
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
2
3
48
25.
Chapter 2 Limits and Continuity
x# 3x "0
x5
lim
x Ä &
(x 5)(x 2)
x5
œ lim
x Ä &
(x 5)(x 2)
x2
26. lim
x# 7x "0
x#
œ lim
27. lim
t# t 2
t# 1
t Ä " (t 1)(t 1)
xÄ#
tÄ"
t# 3t 2
#
t Ä " t t 2
29.
lim
$
#
x Ä # x 2x
lim
2x 4
5y$ 8y#
1
x 1
x
x Ä 1 1
32. lim
xÄ0
33. lim
1
xc1
u% "
y# (5y 8)
œ lim
uÄ1
œ lim
4x x#
x Ä % 2 Èx
œ lim
36. lim
lim
x Ä "
œ
(v 2) av# 2v 4b
Èx 3
x Ä * ˆÈ x 3 ‰ ˆ È x 3 ‰
x1
x Ä 1 Èx 3 2
37. lim
œ
Èx# 12 4
x2
xÄ2
x Ä "
œ lim
xÄ2
(x 2)(x 2)
x2
x Ä 2 È x # 5 3
lim
x Ä 2
8
16
1 ‰
x1
œ lim
xÄ1
œ lim
uÄ1
au# "b (u 1)
u# u 1
"
œ lim
(x 2)(x 2)
444
(4)(8)
œ
x1
x2
œ 2
4
3
12
32
œ
3
8
xÄ%
œ lim ŠÈx 3 #‹ œ È4 2 œ 4
xÄ1
ax # 8 b *
œ lim
x Ä 1 (x 1) ŠÈx# 8 $‹
2
33
œ
(x 2) ŠÈx# 12 4‹
x Ä 2 Èx# 12 4
œ
2
1
œ lim x ˆ2 Èx‰ œ 4(2 2) œ 16
(x 1) ˆÈx 3 #‰
(x 3) 4
xÄ1
x Ä 1 È x # ) $
œ
œ
"
6
œ lim
(x 1) ŠÈx# 8 $‹
œ "3
ax# 12b 16
œ lim
x Ä 2 (x 2) ŠÈx# 12 4‹
œ
4
È16 4
ax 2b ŠÈx# 5 3‹
È x# 5 3
x2
x Ä 2
œ lim
"
È9 3
œ
x Ä * Èx 3
(1 1)(1 1)
111
œ
#
v Ä # (v 2) av 4b
x ˆ2 È x ‰ ˆ 2 È x ‰
2 Èx
xÄ%
œ lim
œ
v# 2v 4
œ lim
œ lim
œ lim
2
x Ä 1 ax 1bax 1b
x Ä 2 ŠÈx# 5 3‹ ŠÈx# 5 3‹
ax 2b ŠÈx# 5 3‹
œ #"
œ lim x1 œ 1
ŠÈx# 12 4‹ ŠÈx# 12 4‹
x Ä 2 (x 2) ŠÈx# 12 4‹
œ
œ
ŠÈx# 8 $‹ ŠÈx# 8 $‹
lim
(x 1)(x 1)
lim
x Ä 1 (x 1) ŠÈx# ) $‹
lim
5y 8
(x 1) ˆÈx 3 2‰
È
ˆ
x 3 #‰ ˆ È x 3 #‰
xÄ1
39. lim
40.
œ 21
œ lim
È x# 8 3
x1
œ lim
2
4
xÄ1
#
v Ä # (v 2)(v 2) av 4b
x(4 x)
x Ä % 2 Èx
œ
2
œ 13
1
œ lim Š ax 12x
bax 1b † x ‹ œ lim
au# "b (u 1)(u 1)
au# u 1b (u 1)
œ lim
Èx 3
x9
xÄ*
x
xÄ1
œ lim
35. lim
b 1b b ax c 1b
c 1bax b 1b
ax
3
#
1 2
1 2
#
y Ä ! 3y 16
xÄ1
œ
œ
#
x Ä # x
œ lim ˆ 1 x x †
ax
œ lim
t2
t Ä " t 2
œ lim
1cx
x
12
11
œ
œ lim
œ lim
x Ä 1 x1
%
v Ä # v 16
t2
2(x 2)
v$ 8
34. lim
xÄ#
œ lim
œ lim
$
u Ä 1 u 1
38.
t Ä " (t 2)(t 1)
#
#
y Ä ! y a3y 16b
x b1 1
x
œ lim (x 5) œ 2 5 œ 3
t Ä " t1
(t 2)(t 1)
œ lim
x Ä &
œ lim
#
x Ä # x (x 2)
%
#
y Ä 0 3y 16y
31. lim
(t 2)(t 1)
œ lim
28.
30. lim
xÄ#
œ lim (x 2) œ & # œ 7
œ
œ
œ lim
"
2
ax 2b ŠÈx# 5 3‹
x Ä 2
È9 3
4
ax # 5 b 9
œ 23
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 2.2 Limit of a Function and Limit Laws
41.
2 È x# 5
x3
x Ä 3
lim
œ
Š2 Èx# 5‹ Š2 Èx# 5‹
œ lim
(x 3) Š2 Èx# 5‹
x Ä 3
9 x#
lim
œ
x Ä 3 (x 3) Š2 Èx# 5‹
4x
x Ä 4 5 È x# 9
œ lim
x Ä 4 Š5 Èx# 9‹ Š5 Èx# 9‹
a4 xb Š5 Èx# 9‹
xÄ4
x Ä 3 (x 3) Š2 Èx# 5‹
a4 xb Š5 Èx# 9‹
œ lim
42. lim
16 x#
x Ä 3 (x 3) Š2 Èx# 5‹
(3 x)(3 x)
lim
œ lim
(4 x)(4 x)
xÄ4
œ lim
3x
x Ä 3 2 È x # 5
œ
6
2 È4
œ
3
2
a4 xb Š5 Èx# 9‹
25 ax# 9b
xÄ4
a4 xb Š5 Èx# 9‹
œ lim
4 ax # 5 b
œ lim
œ lim
xÄ4
5 È x# 9
4x
œ
5 È25
8
œ
5
4
2
43. lim a2sin x 1b œ 2sin 0 1 œ 0 1 œ 1
44. lim sin2 x œ Š lim sin x‹ œ asin 0b2 œ 02 œ 0
45. lim sec x œ
46. lim tan x œ
xÄ0
xÄ0
47. lim
xÄ0
1 x sin x
3cos x
1
lim
x Ä 0 cos x
œ
œ
1 0 sin 0
3cos 0
1
cos 0
œ
œ
1
1
xÄ0
œ1
100
3
œ
xÄ0
xÄ0
sin x
lim
x Ä 0 cos x
œ
sin 0
cos 0
œ
0
1
œ0
1
3
48. lim ax2 1ba2 cos xb œ a02 1ba2 cos 0b œ a1ba2 1b œ a1ba1b œ 1
xÄ0
49. x Ä
lim1Èx 4 cosax 1b œ x Ä
lim1Èx 4 † x Ä
lim1cosax 1b œ È1 4 † cos 0 œ È4 1 † 1 œ È4 1
50. lim È7 sec2 x œ É lim a7 sec2 xb œ É7 lim sec2 x œ È7 sec2 0 œ É7 a1b2 œ 2È2
xÄ0
xÄ0
xÄ0
51. (a) quotient rule
(c) sum and constant multiple rules
(b) difference and power rules
52. (a) quotient rule
(c) difference and constant multiple rules
(b) power and product rules
53. (a) xlim
f(x) g(x) œ ’xlim
f(x)“ ’ x lim
g(x)“ œ (5)(2) œ 10
Äc
Äc
Äc
(b) xlim
2f(x) g(x) œ 2 ’xlim
f(x)“ ’ xlim
g(x)“ œ 2(5)(2) œ 20
Äc
Äc
Äc
(c) xlim
[f(x) 3g(x)] œ xlim
f(x) 3 xlim
g(x) œ 5 3(2) œ 1
Äc
Äc
Äc
lim
f(x)
f(x)
5
5
xÄc
(d) xlim
œ lim f(x)
lim g(x) œ 5(2) œ 7
Ä c f(x) g(x)
x
54. (a)
(b)
(c)
(d)
55. (a)
(b)
Äc
x
Äc
lim [g(x) 3] œ lim g(x) lim 3 œ $ $ œ !
xÄ%
xÄ%
xÄ%
lim xf(x) œ lim x † lim f(x) œ (4)(0) œ 0
xÄ%
xÄ%
xÄ%
#
lim [g(x)] œ ’ lim g(x)“ œ [3]# œ 9
xÄ%
#
g(x)
x Ä % f(x) 1
lim
xÄ%
œ
Ä%
lim g(x)
x
lim f(x) lim 1
xÄ%
xÄ%
œ
3
01
œ3
lim [f(x) g(x)] œ lim f(x) lim g(x) œ 7 (3) œ 4
xÄb
xÄb
xÄb
lim f(x) † g(x) œ ’ lim f(x)“ ’ lim g(x)“ œ (7)(3) œ 21
xÄb
xÄb
xÄb
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
49
50
Chapter 2 Limits and Continuity
lim 4g(x) œ ’ lim 4“ ’ lim g(x)“ œ (4)(3) œ 12
(c)
xÄb
xÄb
xÄb
xÄb
xÄb
7
3
œ 37
lim [p(x) r(x) s(x)] œ lim p(x) lim r(x) lim s(x) œ 4 0 (3) œ 1
56. (a)
x Ä #
x Ä #
x Ä #
x Ä #
lim p(x) † r(x) † s(x) œ ’ lim p(x)“ ’ lim r(x)“ ’ lim s(x)“ œ (4)(0)(3) œ 0
(b)
x Ä #
x Ä #
x Ä #
x Ä #
lim [4p(x) 5r(x)]/s(x) œ ’4 lim p(x) 5 lim r(x)“ ‚ lim s(x) œ [4(4) 5(0)]/3 œ
(c)
x Ä #
57. lim
hÄ!
x Ä #
(1 h)# 1#
h
œ lim
hÄ!
(2 h)# (2)#
h
hÄ!
58. lim
59. lim
hÄ!
ˆ #" h ‰ ˆ "# ‰
h
hÄ!
È7 h È7
h
hÄ!
61. lim
1 2h h# 1
h
œ lim
hÄ!
44hh# 4
h
hÄ!
œ lim
[3(2 h) 4] [3(2) 4]
h
60. lim
œ
xÄb
lim f(x)/g(x) œ lim f(x)/ lim g(x) œ
(d)
œ lim
2
2 h "
œ lim
h(2 h)
h
œ lim
hÄ!
hÄ!
œ lim (h 4) œ 4
hÄ!
2 (2 h)
2h(# h)
hÄ!
h ŠÈ7 h È7‹
h
œ lim
œ "4
h Ä ! h(4 2h)
œ lim
(7 h) 7
h Ä ! h ŠÈ7 h È7‹
œ lim
h
h Ä ! h ŠÈ7hÈ7‹
œ lim
È3(0 h) 1 È3(0) 1
h
hÄ!
œ lim
3
h Ä ! È3h 1 1
œ
œ lim
ŠÈ3h 1 "‹ ŠÈ3h 1 "‹
h ŠÈ3h 1 "‹
hÄ!
(3h 1) "
œ lim
h Ä ! h ŠÈ3h 1 1 ‹
œ lim
3h
h Ä ! h ŠÈ3h 1 "‹
3
#
63. lim È5 2x# œ È5 2(0)# œ È5 and lim È5 x# œ È5 (0)# œ È5; by the sandwich theorem,
xÄ!
xÄ!
lim f(x) œ È5
xÄ!
64. lim a2 x# b œ 2 0 œ 2 and lim 2 cos x œ 2(1) œ 2; by the sandwich theorem, lim g(x) œ 2
xÄ!
65. (a)
xÄ!
lim Š1 xÄ!
x#
6‹
œ1
0
6
xÄ!
œ 1 and lim 1 œ 1; by the sandwich theorem, lim
(b) For x Á 0, y œ (x sin x)/(2 2 cos x)
lies between the other two graphs in the
figure, and the graphs converge as x Ä 0.
66. (a)
lim Š "# xÄ!
lim
xÄ!
1cos x
x#
x#
24 ‹
œ lim
1
xÄ! #
lim
x#
x Ä ! #4
œ
"
#
x sin x
x Ä ! 22 cos x
xÄ!
0œ
"
#
and lim
"
xÄ! #
œ1
œ "# ; by the sandwich theorem,
œ "# .
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
"
h Ä ! È 7 h È 7
"
#È 7
62. lim
"6
3
œ lim (2 h) œ 2
h(h 4)
h
ŠÈ7 h È7‹ ŠÈ7 h È7‹
hÄ!
x Ä #
œ3
œ lim
2h
hÄ!
œ lim
3h
hÄ! h
x Ä #
Section 2.2 Limit of a Function and Limit Laws
(b) For all x Á 0, the graph of f(x) œ (1 cos x)/x#
lies between the line y œ "# and the parabola
yœ
"
#
x# /24, and the graphs converge as x Ä 0.
67. (a) f(x) œ ax# *b/(x 3)
x
3.1
f(x)
6.1
2.9
5.9
x
f(x)
3.01
6.01
3.001
6.001
3.0001
6.0001
3.00001
6.00001
3.000001
6.000001
2.99
5.99
2.999
5.999
2.9999
5.9999
2.99999
5.99999
2.999999
5.999999
The estimate is lim f(x) œ 6.
x Ä $
(b)
(c) f(x) œ
x# 9
x3
œ
(x 3)(x 3)
x3
œ x 3 if x Á 3, and lim (x 3) œ 3 3 œ 6.
x Ä $
68. (a) g(x) œ ax# #b/ Šx È2‹
x
g(x)
1.4
2.81421
1.41
2.82421
1.414
2.82821
1.4142
2.828413
1.41421
2.828423
1.414213
2.828426
(b)
(c) g(x) œ
x# 2
x È2
œ
Šx È2‹ Šx È2‹
Šx È2‹
œ x È2 if x Á È2, and
69. (a) G(x) œ (x 6)/ ax# 4x 12b
x
5.9
5.99
G(x)
.126582 .1251564
x
G(x)
6.1
.123456
6.01
.124843
5.999
.1250156
6.001
.124984
lim
x Ä È#
5.9999
.1250015
6.0001
.124998
Šx È2‹ œ È2 È2 œ 2È2.
5.99999
.1250001
6.00001
.124999
5.999999
.1250000
6.000001
.124999
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
51
52
Chapter 2 Limits and Continuity
(b)
(c) G(x) œ
x6
ax# 4x 12b
œ
x6
(x 6)(x 2)
œ
"
x#
70. (a) h(x) œ ax# 2x 3b / ax# 4x 3b
x
2.9
2.99
h(x)
2.052631
2.005025
x
h(x)
3.1
1.952380
3.01
1.995024
"
if x Á 6, and lim
x Ä ' x 2
œ
"
' 2
œ "8 œ 0.125.
2.999
2.000500
2.9999
2.000050
2.99999
2.000005
2.999999
2.0000005
3.001
1.999500
3.0001
1.999950
3.00001
1.999995
3.000001
1.999999
(b)
(c) h(x) œ
x# 2x 3
x# 4x 3
œ
(x 3)(x 1)
(x 3)(x 1)
œ
x1
x1
71. (a) f(x) œ ax# 1b / akxk 1b
x
1.1
1.01
f(x)
2.1
2.01
.9
1.9
x
f(x)
.99
1.99
if x Á 3, and lim
x1
x Ä $ x1
œ
31
31
œ
4
#
œ 2.
1.001
2.001
1.0001
2.0001
1.00001
2.00001
1.000001
2.000001
.999
1.999
.9999
1.9999
.99999
1.99999
.999999
1.999999
(b)
(c) f(x) œ
x# "
kx k 1
(x 1)(x 1)
1
œ (x x1)(x
1)
(x 1)
œ x 1, x 0 and x Á 1
, and lim (1 x) œ 1 (1) œ 2.
x Ä 1
œ 1 x, x 0 and x Á 1
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 2.2 Limit of a Function and Limit Laws
72. (a) F(x) œ ax# 3x 2b / a2 kxkb
x
2.1
2.01
F(x)
1.1
1.01
1.9
.9
x
F(x)
1.99
.99
2.001
1.001
2.0001
1.0001
2.00001
1.00001
2.000001
1.000001
1.999
.999
1.9999
.9999
1.99999
.99999
1.999999
.999999
(b)
(c) F(x) œ
x# 3x 2
2 kx k
(x 2)(x 1)
œ (x 2)(x# x")
2x
73. (a) g()) œ (sin ))/)
)
.1
g())
.998334
,
x 0
, and lim (x 1) œ 2 1 œ 1.
x Ä #
œ x 1, x 0 and x Á 2
.01
.999983
.001
.999999
.0001
.999999
.00001
.999999
.000001
.999999
.1
.998334
.01
.999983
.001
.999999
.0001
.999999
.00001
.999999
.000001
.999999
74. (a) G(t) œ (1 cos t)/t#
t
.1
G(t)
.499583
.01
.499995
.001
.499999
.0001
.5
.00001
.5
.000001
.5
.1
.499583
.01
.499995
.001
.499999
.0001
.5
.00001
.5
.000001
.5
)
g())
lim g()) œ 1
)Ä!
(b)
t
G(t)
lim G(t) œ 0.5
tÄ!
(b)
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
53
54
Chapter 2 Limits and Continuity
75. xlim
f(x) exists at those points c where xlim
x% œ xlim
x# . Thus, c% œ c# Ê c# a1 c# b œ 0 Ê c œ 0, 1, or 1.
Äc
Äc
Äc
Moreover, lim f(x) œ lim x# œ 0 and lim f(x) œ lim f(x) œ 1.
xÄ!
xÄ!
x Ä 1
xÄ1
76. Nothing can be concluded about the values of f, g, and h at x œ 2. Yes, f(2) could be 0. Since the conditions of the
sandwich theorem are satisfied, lim f(x) œ 5 Á 0.
xÄ#
lim f(x) lim 5
%
xÄ%
œ xÄlim
x lim 2 œ
f(x)5
x Ä % x 2
77. 1 œ lim
xÄ%
lim f(x) 5
xÄ%
%#
xÄ%
xÄ%
78. (a) 1 œ lim
f(x)
x#
lim f(x)
lim f(x)
œ xÄlim# x# œ xÄ %#
Ê
(b) 1 œ lim
f(x)
x#
œ ’ lim
x Ä #
x Ä #
xÄ
#
x Ä #
79. (a) 0 œ 3 † 0 œ ’ lim
xÄ#
Ê lim f(x) 5 œ 2(1) Ê lim f(x) œ 2 5 œ 7.
xÄ%
lim f(x) œ 4.
x Ä #
f(x)
lim x" “
x “ ’x Ä
#
œ ’ lim
x Ä #
f(x) ˆ " ‰
x “ #
Ê
lim
x Ä #
f(x)
x
œ 2.
f(x) 5
x # “ ’xlim
Ä#
5
(x 2)“ œ lim ’Š f(x)
x # ‹ (x 2)“ œ lim [f(x) 5] œ lim f(x) 5
f(x) 5
x # “ ’xlim
Ä#
(x 2)“ Ê lim f(x) œ 5 as in part (a).
xÄ#
Ê lim f(x) œ 5.
xÄ#
xÄ#
xÄ#
(b) 0 œ 4 † 0 œ ’ lim
xÄ#
80. (a) 0 œ 1 † 0 œ ’ lim
f(x)
# “ ’ lim
xÄ! x
xÄ!
(b) 0 œ 1 † 0 œ
81. (a)
lim x sin
xÄ!
(b) 1 Ÿ sin
82. (a)
"
x
’ lim f(x)
# “ ’ lim
xÄ! x
xÄ!
"
x
xÄ#
#
x“ œ ’ lim
f(x)
#
xÄ! x
x“ œ
lim ’ f(x)
x#
xÄ!
#
“ ’ lim x# “ œ lim ’ f(x)
x# † x “ œ lim f(x). That is, lim f(x) œ 0.
xÄ!
† x“ œ
xÄ!
lim f(x) .
xÄ! x
That is,
xÄ!
lim f(x)
xÄ! x
œ 0.
œ0
Ÿ 1 for x Á 0:
x 0 Ê x Ÿ x sin
"
x
Ÿ x Ê lim x sin
"
x
œ 0 by the sandwich theorem;
x 0 Ê x
"
x
x Ê lim x sin
"
x
œ 0 by the sandwich theorem.
x sin
xÄ!
xÄ!
lim x# cos ˆ x"$ ‰ œ 0
xÄ!
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
xÄ!
Section 2.3 The Precise Definition of a Limit
(b) 1 Ÿ cos ˆ x"$ ‰ Ÿ 1 for x Á 0 Ê x# Ÿ x# cos ˆ x"$ ‰ Ÿ x# Ê lim x# cos ˆ x"$ ‰ œ 0 by the sandwich
xÄ!
theorem since lim x# œ 0.
xÄ!
83-88. Example CAS commands:
Maple:
f := x -> (x^4 16)/(x 2);
x0 := 2;
plot( f(x), x œ x0-1..x0+1, color œ black,
title œ "Section 2.2, #83(a)" );
limit( f(x), x œ x0 );
In Exercise 85, note that the standard cube root, x^(1/3), is not defined for x<0 in many CASs. This can be
overcome in Maple by entering the function as f := x -> (surd(x+1, 3) 1)/x.
Mathematica: (assigned function and values for x0 and h may vary)
Clear[f, x]
f[x_]:=(x3 x2 5x 3)/(x 1)2
x0= 1; h = 0.1;
Plot[f[x],{x, x0 h, x0 h}]
Limit[f[x], x Ä x0]
2.3 THE PRECISE DEFINITION OF A LIMIT
1.
Step 1:
Step 2:
kx 5k $ Ê $ x 5 $ Ê $ 5 x $ 5
$ 5 œ 7 Ê $ œ 2, or $ 5 œ 1 Ê $ œ 4.
The value of $ which assures kx 5k $ Ê 1 x 7 is the smaller value, $ œ 2.
Step 1:
Step 2:
kx 2k $ Ê $ x 2 $ Ê $ # x $ 2
$ 2 œ 1 Ê $ œ 1, or $ 2 œ 7 Ê $ œ 5.
The value of $ which assures kx 2k $ Ê 1 x 7 is the smaller value, $ œ 1.
Step 1:
Step 2:
kx (3)k $ Ê $ x $ $ Ê $ 3 x $ 3
$ 3 œ 7# Ê $ œ "# , or $ $ œ "# Ê $ œ 5# .
2.
3.
The value of $ which assures kx (3)k $ Ê 7# x "# is the smaller value, $ œ "# .
4.
Step 1:
¸x ˆ 3# ‰¸ $ Ê $ x 3
#
$ Ê $ 3
#
x$
3
#
Step 2:
œ 7# Ê $ œ #, or $ 3# œ "# Ê $ œ 1.
The value of $ which assures ¸x ˆ 3# ‰¸ $ Ê 7# x "# is the smaller value, $ œ ".
Step 1:
¸x "# ¸ $ Ê $ x $ 3
#
5.
"
#
$ Ê $ "
#
x$
"
#
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
55
56
Chapter 2 Limits and Continuity
Step 2:
"
or $ #" œ 47 Ê $ œ 14
.
"¸
4
¸
The value of $ which assures x # $ Ê 9 x $ "
#
œ
4
9
Ê $œ
"
18 ,
4
7
is the smaller value, $ œ
"
18 .
6.
Step 1:
Step 2:
kx 3k $ Ê $ x 3 $ Ê $ 3 x $ 3
$ $ œ 2.7591 Ê $ œ 0.2409, or $ $ œ 3.2391 Ê $ œ 0.2391.
The value of $ which assures kx 3k $ Ê 2.7591 x 3.2391 is the smaller value, $ œ 0.2391.
7. Step 1:
Step 2:
kx 5k $ Ê $ x 5 $ Ê $ 5 x $ 5
From the graph, $ 5 œ 4.9 Ê $ œ 0.1, or $ 5 œ 5.1 Ê $ œ 0.1; thus $ œ 0.1 in either case.
8. Step 1:
Step 2:
kx (3)k $ Ê $ x 3 $ Ê $ 3 x $ 3
From the graph, $ 3 œ 3.1 Ê $ œ 0.1, or $ 3 œ 2.9 Ê $ œ 0.1; thus $ œ 0.1.
9. Step 1:
Step 2:
kx 1k $ Ê $ x 1 $ Ê $ 1 x $ 1
9
7
From the graph, $ 1 œ 16
Ê $ œ 16
, or $ 1 œ 25
16 Ê $ œ
10. Step 1:
Step 2:
kx 3k $ Ê $ x 3 $ Ê $ 3 x $ 3
From the graph, $ 3 œ 2.61 Ê $ œ 0.39, or $ 3 œ 3.41 Ê $ œ 0.41; thus $ œ 0.39.
11. Step 1:
kx 2k $ Ê $ x 2 $ Ê $ 2 x $ 2
From the graph, $ 2 œ È3 Ê $ œ 2 È3 ¸ 0.2679, or $ 2 œ È5 Ê $ œ È5 2 ¸ 0.2361;
thus $ œ È5 2.
Step 2:
12. Step 1:
Step 2:
9
16 ;
thus $ œ
kx (1)k $ Ê $ x 1 $ Ê $ 1 x $ 1
From the graph, $ 1 œ thus $ œ
È5 2
# .
È5
#
Ê $œ
È5 2
#
¸ 0.1180, or $ 1 œ 13. Step 1:
Step 2:
kx (1)k $ Ê $ x 1 $ Ê $ 1 x $ 1
7
16
From the graph, $ 1 œ 16
9 Ê $ œ 9 ¸ 0.77, or $ 1 œ 25 Ê
14. Step 1:
¸x "# ¸ $ Ê $ x Step 2:
7
16 .
From the graph, $ thus $ œ 0.00248.
"
#
œ
"
# 1
2.01
$ Ê $ Ê $œ
1
2
"
#
x$
"
#.01
"
#
¸ 0.00248, or $ "
#
œ
È3
#
9
25
Ê $œ
2 È3
#
œ 0.36; thus $ œ
1
1.99
Ê $œ
1
1.99
¸ 0.1340;
9
25
"
#
œ 0.36.
¸ 0.00251;
15. Step 1:
Step 2:
k(x 1) 5k 0.01 Ê kx 4k 0.01 Ê 0.01 x 4 0.01 Ê 3.99 x 4.01
kx 4k $ Ê $ x 4 $ Ê $ 4 x $ 4 Ê $ œ 0.01.
16. Step 1:
k(2x 2) (6)k 0.02 Ê k2x 4k 0.02 Ê 0.02 2x 4 0.02 Ê 4.02 2x 3.98
Ê 2.01 x 1.99
kx (2)k $ Ê $ x 2 $ Ê $ 2 x $ 2 Ê $ œ 0.01.
Step 2:
17. Step 1:
Step 2:
¹Èx 1 "¹ 0.1 Ê 0.1 Èx 1 " 0.1 Ê 0.9 Èx 1 1.1 Ê 0.81 x 1 1.21
Ê 0.19 x 0.21
kx 0k $ Ê $ x $ . Then, $ œ !Þ"* Ê $ œ !Þ"* or $ œ !Þ#"; thus, $ œ 0.19.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 2.3 The Precise Definition of a Limit
18. Step 1:
Step 2:
¸Èx "# ¸ 0.1 Ê 0.1 Èx "# 0.1 Ê 0.4 Èx 0.6 Ê 0.16 x 0.36
¸x "4 ¸ $ Ê $ x 4" $ Ê $ 4" x $ 4" .
Then, $ 19. Step 1:
Step 2:
20. Step 1:
Step 2:
21. Step 1:
Step 2:
22. Step 1:
Step 2:
57
"
4
œ 0.16 Ê $ œ 0.09 or $ "
4
œ 0.36 Ê $ œ 0.11; thus $ œ 0.09.
¹È19 x $¹ " Ê " È19 x $ 1 Ê 2 È19 x % Ê 4 19 x 16
Ê % x 19 16 Ê 15 x 3 or 3 x 15
kx 10k $ Ê $ x 10 $ Ê $ 10 x $ 10.
Then $ 10 œ 3 Ê $ œ 7, or $ 10 œ 15 Ê $ œ 5; thus $ œ 5.
¹Èx 7 4¹ 1 Ê " Èx 7 % 1 Ê 3 Èx 7 5 Ê 9 x 7 25 Ê 16 x 32
kx 23k $ Ê $ x 23 $ Ê $ 23 x $ 23.
Then $ 23 œ 16 Ê $ œ 7, or $ 23 œ 32 Ê $ œ 9; thus $ œ 7.
¸ "x 4" ¸ 0.05 Ê 0.05 "
x
"
4
0.05 Ê 0.2 "
x
0.3 Ê
kx 4k $ Ê $ x 4 $ Ê $ 4 x $ 4.
2
2
Then $ % œ 10
3 or $ œ 3 , or $ 4 œ 5 or $ œ 1; thus $ œ 3 .
10
#
x
10
3
or
10
3
x 5.
kx# 3k !.1 Ê 0.1 x# 3 0.1 Ê 2.9 x# 3.1 Ê È2.9 x È3.1
¹x È3¹ $ Ê $ x È3 $ Ê $ È3 x $ È3.
Then $ È3 œ È2.9 Ê $ œ È3 È2.9 ¸ 0.0291, or $ È3 œ È3.1 Ê $ œ È3.1 È3 ¸ 0.0286;
thus $ œ 0.0286.
23. Step 1:
Step 2:
kx# 4k 0.5 Ê 0.5 x# 4 0.5 Ê 3.5 x# 4.5 Ê È3.5 kxk È4.5 Ê È4.5 x È3.5,
for x near 2.
kx (2)k $ Ê $ x 2 $ Ê $ # x $ 2.
Then $ # œ È4.5 Ê $ œ È4.5 # ¸ 0.1213, or $ # œ È3.5 Ê $ œ # È3.5 ¸ 0.1292;
thus $ œ È4.5 2 ¸ 0.12.
24. Step 1:
Step 2:
25. Step 1:
Step 2:
¸ "x (1)¸ 0.1 Ê 0.1 "
x
1 0.1 Ê 11
10 "
x
9
10
10
10
10
Ê 10
11 x 9 or 9 x 11 .
kx (1)k $ Ê $ x 1 $ Ê $ " x $ ".
"
10
"
Then $ " œ 10
9 Ê $ œ 9 , or $ " œ 11 Ê $ œ 11 ; thus $ œ
"
11 .
kax# 5b 11k " Ê kx# 16k 1 Ê " x# 16 1 Ê 15 x# 17 Ê È15 x È17.
kx 4k $ Ê $ x 4 $ Ê $ % x $ %.
Then $ % œ È15 Ê $ œ % È15 ¸ 0.1270, or $ % œ È17 Ê $ œ È17 % ¸ 0.1231;
thus $ œ È17 4 ¸ 0.12.
26. Step 1:
Step 2:
27. Step 1:
Step 2:
¸ 120
¸
x 5 " Ê " 120
x
&1 Ê 4
120
x
6 Ê
"
4
x
120
"
6
Ê 30 x 20 or 20 x 30.
kx 24k $ Ê $ x 24 $ Ê $ 24 x $ 24.
Then $ 24 œ 20 Ê $ œ 4, or $ 24 œ 30 Ê $ œ 6; thus Ê $ œ 4.
kmx 2mk 0.03 Ê 0.03 mx 2m 0.03 Ê 0.03 2m mx 0.03 2m Ê
0.03
2 0.03
m x2 m .
kx 2k $ Ê $ x 2 $ Ê $ 2 x $ 2.
0.03
0.03
Then $ 2 œ 2 0.03
m Ê $ œ m , or $ 2 œ # m Ê $ œ
0.03
m .
In either case, $ œ
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
0.03
m .
58
Chapter 2 Limits and Continuity
kmx 3mk c Ê c mx 3m c Ê c 3m mx c 3m Ê 3 28. Step 1:
kx 3k $ Ê $ x 3 $ Ê $ $ B $ $.
Then $ $ œ $ mc Ê $ œ mc , or $ $ œ $ mc Ê $ œ
Step 2:
¸(mx b) ˆ m# b‰¸ - Ê c mx m# c Ê c ¸x "# ¸ $ Ê $ x "# $ Ê $ "# x $ "# .
29. Step 1:
Step 2:
Then $ "
#
œ
"
#
c
m
Ê $œ
c
m,
or $ "
#
œ
"
#
c
m
c
m.
m
#
Ê $œ
c
m
x 3
In either case, $ œ
c
m.
c
m.
"
#
In either case, $ œ
c
m.
mx c m
#
Ê
c
m
c
m
x
"
#
c
m.
k(mx b) (m b)k 0.05 Ê 0.05 mx m 0.05 Ê 0.05 m mx 0.05 m
0.05
Ê 1 0.05
m x" m .
30. Step 1:
kx 1k $ Ê $ x 1 $ Ê $ " x $ ".
0.05
0.05
Then $ " œ " 0.05
m Ê $ œ m , or $ " œ " m Ê $ œ
Step 2:
0.05
m .
In either case, $ œ
0.05
m .
31. lim (3 2x) œ 3 2(3) œ 3
xÄ3
ka3 2xb (3)k 0.02 Ê 0.02 6 2x 0.02 Ê 6.02 2x 5.98 Ê 3.01 x 2.99 or
2.99 x 3.01.
0 k x 3k $ Ê $ x 3 $ Ê $ $ x $ $ .
Then $ $ œ 2.99 Ê $ œ 0.01, or $ $ œ 3.01 Ê $ œ 0.01; thus $ œ 0.01.
Step 1:
Step 2:
32.
lim (3x #) œ (3)(1) 2 œ 1
x Ä 1
k(3x 2) 1k 0.03 Ê 0.03 3x 3 0.03 Ê 0.01 x 1 0.01 Ê 1.01 x 0.99.
kx (1)k $ Ê $ x 1 $ Ê $ " x $ 1.
Then $ " œ 1.01 Ê $ œ 0.01, or $ " œ 0.99 Ê $ œ 0.01; thus $ œ 0.01.
Step 1:
Step 2:
33. lim
x# 4
x Ä # x#
34.
35.
œ lim
xÄ#
#
(x 2)(x 2)
(x 2)
œ lim (x 2) œ # # œ 4, x Á 2
xÄ#
(x 2)(x 2)
(x 2)
Step 1:
¹Š xx 24 ‹
Step 2:
Ê 1.95 x 2.05, x Á 2.
kx 2k $ Ê $ x 2 $ Ê $ 2 x $ 2.
Then $ 2 œ 1.95 Ê $ œ 0.05, or $ 2 œ 2.05 Ê $ œ 0.05; thus $ œ 0.05.
lim
x Ä &
x# 6x 5
x5
4¹ 0.05 Ê 0.05 œ lim
x Ä &
(x 5)(x 1)
(x 5)
% 0.05 Ê 3.95 x 2 4.05, x Á 2
œ lim (x 1) œ 4, x Á 5.
x Ä &
(x 5)(x ")
(x 5)
Step 1:
#
¹Š x x 6x5 5 ‹
Step 2:
Ê 5.05 x 4.95, x Á 5.
kx (5)k $ Ê $ x 5 $ Ê $ & x $ &.
Then $ & œ 5.05 Ê $ œ 0.05, or $ & œ 4.95 Ê $ œ 0.05; thus $ œ 0.05.
(4)¹ 0.05 Ê 0.05 4 0.05 Ê 4.05 x 1 3.95, x Á 5
lim È1 5x œ È1 5(3) œ È16 œ 4
x Ä $
Step 1:
¹È1 5x 4¹ 0.5 Ê 0.5 È1 5x 4 0.5 Ê 3.5 È1 5x 4.5 Ê 12.25 1 5x 20.25
Step 2:
Ê 11.25 5x 19.25 Ê 3.85 x 2.25.
kx (3)k $ Ê $ x 3 $ Ê $ $ x $ $.
Then $ $ œ 3.85 Ê $ œ 0.85, or $ $ œ 2.25 Ê 0.75; thus $ œ 0.75.
36. lim
4
xÄ# x
Step 1:
œ
4
#
œ2
¸ 4x 2¸ 0.4 Ê 0.4 4
x
2 0.4 Ê 1.6 4
x
2.4 Ê
10
16
x
4
10
24
Ê
10
4
x
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
10
6
or
5
3
x 25 .
Section 2.3 The Precise Definition of a Limit
Step 2:
59
kx 2k $ Ê $ x 2 $ Ê $ # x $ #.
Then $ # œ 53 Ê $ œ "3 , or $ # œ #5 Ê $ œ "# ; thus $ œ 3" .
37. Step 1:
Step 2:
k(9 x) 5k % Ê % 4 x % Ê % 4 x % 4 Ê % % x 4 % Ê % % x 4 %.
kx 4k $ Ê $ x 4 $ Ê $ % x $ %.
Then $ 4 œ % 4 Ê $ œ %, or $ % œ % % Ê $ œ %. Thus choose $ œ %.
38. Step 1:
k(3x 7) 2k % Ê % 3x 9 % Ê 9 % 3x * % Ê 3 Step 2:
39. Step 1:
Step 2:
40. Step 1:
Step 2:
41. Step 1:
Step 2:
42. Step 1:
Step 2:
43. Step 1:
Step 2:
x 3 3% .
kx 3k $ Ê $ x 3 $ Ê $ 3 x $ 3.
Then $ 3 œ $ 3% Ê $ œ 3% , or $ 3 œ 3 3% Ê $ œ 3% . Thus choose $ œ 3% .
¹Èx 5 2¹ % Ê % Èx 5 2 % Ê 2 % Èx 5 2 % Ê (2 %)# x 5 (2 %)#
Ê (2 %)# & x (2 %)# 5.
kx 9k $ Ê $ x 9 $ Ê $ 9 x $ 9.
Then $ * œ %# %% * Ê $ œ %% %# , or $ * œ %# %% * Ê $ œ %% %# . Thus choose
the smaller distance, $ œ %% %# .
¹È4 x 2¹ % Ê % È4 x 2 % Ê 2 % È4 x 2 % Ê (2 %)# % x (2 %)#
Ê (2 %)# x 4 (2 %)# Ê (2 %)# % x (2 %)# %.
k x 0k $ Ê $ x $ .
Then $ œ (2 %)# 4 œ %# %% Ê $ œ %% %# , or $ œ (2 %)# 4 œ 4% %# . Thus choose
the smaller distance, $ œ 4% %# .
For x Á 1, kx# 1k % Ê % x# " % Ê " % x# " % Ê È1 % kxk È1 %
Ê È" % x È1 % near B œ ".
kx 1k $ Ê $ x 1 $ Ê $ " x $ ".
Then $ " œ È1 % Ê $ œ " È1 %, or $ 1 œ È" % Ê $ œ È" % 1. Choose
$ œ min š" È1 %ß È1 % "›, that is, the smaller of the two distances.
For x Á 2, kx# 4k % Ê % x# 4 % Ê 4 % x# 4 % Ê È4 % kxk È4 %
Ê È4 % x È4 % near B œ 2.
kx (2)k $ Ê $ x 2 $ Ê $ 2 x $ 2.
Then $ 2 œ È% % Ê $ œ È% % #, or $ # œ È% % Ê $ œ # È% %. Choose
$ œ min šÈ% % #ß # È% %› .
¸ "x 1¸ % Ê % "
x
"% Ê "% "
x
"% Ê
"
1%
%
"%,
"
1%.
x
kx 1k $ Ê $ x 1 $ Ê " $ x " $ .
Then " $ œ " " % Ê $ œ " " " % œ " % % , or " $ œ " " % Ê $ œ
Choose $ œ
44. Step 1:
%
3
"
"%
"œ
%
"%.
the smaller of the two distances.
¸ x"# "3 ¸ % Ê % "
x#
"
3
% Ê
"
3
% "
x#
"
3
% Ê
1 3%
3
"
x#
1 $%
3
3
È $.
Ê É 1 3 $% kxk É " 3 $% , or É " 3 $% x É "$
% for x near
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Ê
3
" $%
x# 3
" $%
60
Chapter 2 Limits and Continuity
Step 2:
¹x È3¹ $ Ê $ x È3 $ Ê È3 $ x È3 $ .
Then È3 $ œ É " 3 $% Ê $ œ È3 É " 3 $% , or È3 $ œ É " 3 $% Ê $ œ É " 3 $% È3.
Choose $ œ min šÈ3 É " 3 $% ß É " 3 $% È3›.
45. Step 1:
Step 2:
46. Step 1:
Step 2:
47. Step 1:
#
¹Š xx 3* ‹ (6)¹ % Ê % (x 3) 6 %, x Á 3 Ê % x 3 % Ê % $ x % $.
kx (3)k $ Ê $ x 3 $ Ê $ $ x $ 3.
Then $ $ œ % $ Ê $ œ %, or $ $ œ % $ Ê $ œ %. Choose $ œ %.
#
¹Š xx 11 ‹ 2¹ % Ê % (x 1) 2 %, x Á 1 Ê " % x " %.
kx 1k $ Ê $ x 1 $ Ê " $ x " $ .
Then " $ œ " % Ê $ œ %, or " $ œ " % Ê $ œ %. Choose $ œ %.
x 1: l(4 2x) 2l % Ê ! 2 2x % since x 1Þ Thus, 1 x
Step 2:
48. Step 1:
Step 2:
%
#
x !;
1. Thus, " Ÿ x 1 6% .
1: l(6x 4) 2l % Ê ! Ÿ 6x 6 % since x
kx 1k $ Ê $ x 1 $ Ê " $ x 1 $ .
Then 1 $ œ " #% Ê $ œ #% , or " $ œ 1 6% Ê $ œ 6% . Choose $ œ 6% .
x !: k2x 0k % Ê % 2x ! Ê #% x 0;
x 0: ¸ x# !¸ % Ê ! Ÿ x #%.
k x 0k $ Ê $ x $ .
Then $ œ #% Ê $ œ #% , or $ œ #% Ê $ œ #%. Choose $ œ #% .
49. By the figure, x Ÿ x sin
"
x
Ÿ x for all x 0 and x
x sin
then by the sandwich theorem, in either case, lim x sin
xÄ!
50. By the figure, x# Ÿ x# sin
"
x
"
x
"
x
x for x 0. Since lim (x) œ lim x œ 0,
xÄ!
œ 0.
xÄ!
Ÿ x# for all x except possibly at x œ 0. Since lim ax# b œ lim x# œ 0, then
by the sandwich theorem, lim x# sin
xÄ!
"
x
xÄ!
œ 0.
xÄ!
51. As x approaches the value 0, the values of g(x) approach k. Thus for every number % 0, there exists a $ !
such that ! kx 0k $ Ê kg(x) kk %.
52. Write x œ h c. Then ! lx cl $ Í $ x c $ , x Á c Í $ ah cb c $ , h c Á c
Í $ h $ , h Á ! Í ! lh !l $ .
Thus, limfaxb œ L Í for any % !, there exists $ ! such that lfaxb Ll % whenever ! lx cl $
x Äc
Í lfah cb Ll % whenever ! lh !l $ Í lim fah cb œ L.
hÄ!
53. Let f(x) œ x# . The function values do get closer to 1 as x approaches 0, but lim f(x) œ 0, not 1. The
function f(x) œ x# never gets arbitrarily close to 1 for x near 0.
xÄ!
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 2.3 The Precise Definition of a Limit
61
54. Let f(x) œ sin x, L œ "# , and x! œ 0. There exists a value of x (namely, x œ 16 ) for which ¸sin x "# ¸ % for any given
% 0. However, lim sin x œ 0, not "# . The wrong statement does not require x to be arbitrarily close to x! . As another\
xÄ!
example, let g(x) œ sin "x , L œ #" , and x! œ 0. We can choose infinitely many values of x near 0 such that sin
you can see from the accompanying figure. However, lim sin
xÄ!
"
x
"
x
œ
"
#
as
fails to exist. The wrong statement does not require all
values of x arbitrarily close to x! œ 0 to lie within % 0 of L œ "# . Again you can see from the figure that there are also
infinitely many values of x near 0 such that sin "x œ 0. If we choose % ¸sin "x #" ¸ % for all values of x sufficiently near x! œ 0.
#
55. kA *k Ÿ 0.01 Ê 0.01 Ÿ 1 ˆ x# ‰ 9 Ÿ 0.01 Ê 8.99 Ÿ
Ê
2É 8.99
1
ŸxŸ
2É 9.01
1
1 x#
4
"
4
we cannot satisfy the inequality
Ÿ 9.01 Ê
4
1
(8.99) Ÿ x# Ÿ
4
1
(9.01)
or 3.384 Ÿ x Ÿ 3.387. To be safe, the left endpoint was rounded up and the right
endpoint was rounded down.
56. V œ RI Ê
(120)(10)
51
V
R
ŸRŸ
œ I Ê ¸ VR 5¸ Ÿ 0.1 Ê 0.1 Ÿ
(120)(10)
49
120
R
5 Ÿ 0.1 Ê 4.9 Ÿ
120
R
Ÿ 5.1 Ê
10
49
R
1#0
10
51
Ê
Ê 23.53 Ÿ R Ÿ 24.48.
To be safe, the left endpoint was rounded up and the right endpoint was rounded down.
57. (a) $ x 1 0 Ê " $ x 1 Ê f(x) œ x. Then kf(x) 2k œ kx 2k œ 2 x 2 1 œ 1. That is,
kf(x) 2k 1 "# no matter how small $ is taken when " $ x 1 Ê lim f(x) Á 2.
xÄ1
(b) 0 x 1 $ Ê " x " $ Ê f(x) œ x 1. Then kf(x) 1k œ k(x 1) 1k œ kxk œ x 1. That is,
kf(x) 1k 1 no matter how small $ is taken when " x " $ Ê lim f(x) Á 1.
xÄ1
(c) $ x 1 ! Ê " $ x 1 Ê f(x) œ x. Then kf(x) 1.5k œ kx 1.5k œ 1.5 x 1.5 1 œ 0.5.
Also, ! x 1 $ Ê 1 x " $ Ê f(x) œ x 1. Then kf(x) 1.5k œ k(x 1) 1.5k œ kx 0.5k
œ x 0.5 " 0.5 œ 0.5. Thus, no matter how small $ is taken, there exists a value of x such that
$ x 1 $ but kf(x) 1.5k "# Ê lim f(x) Á 1.5.
xÄ1
58. (a) For 2 x 2 $ Ê h(x) œ 2 Ê kh(x) 4k œ 2. Thus for % 2, kh(x) 4k
matter how small we choose $ 0 Ê lim h(x) Á 4.
% whenever 2 x 2 $ no
(b) For 2 x 2 $ Ê h(x) œ 2 Ê kh(x) 3k œ 1. Thus for % 1, kh(x) 3k
matter how small we choose $ 0 Ê lim h(x) Á 3.
% whenever 2 x 2 $ no
xÄ#
xÄ#
(c) For 2 $ x 2 Ê h(x) œ x# so kh(x) 2k œ kx# 2k . No matter how small $ 0 is chosen, x# is close to 4
when x is near 2 and to the left on the real line Ê kx# 2k will be close to 2. Thus if % 1, kh(x) 2k %
whenever 2 $ x 2 no mater how small we choose $ 0 Ê lim h(x) Á 2.
xÄ#
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
62
Chapter 2 Limits and Continuity
59. (a) For 3 $ x 3 Ê f(x) 4.8 Ê kf(x) 4k 0.8. Thus for % 0.8, kf(x) 4k
3 $ x 3 no matter how small we choose $ 0 Ê lim f(x) Á 4.
xÄ$
% whenever
(b) For 3 x 3 $ Ê f(x) 3 Ê kf(x) 4.8k 1.8. Thus for % 1.8, kf(x) 4.8k
no matter how small we choose $ 0 Ê lim f(x) Á 4.8.
% whenever 3 x 3 $
(c) For 3 $ x 3 Ê f(x) 4.8 Ê kf(x) 3k 1.8. Again, for % 1.8, kf(x) 3k
no matter how small we choose $ 0 Ê lim f(x) Á 3.
% whenever $ $ x 3
xÄ$
xÄ$
60. (a) No matter how small we choose $ 0, for x near 1 satisfying " $ x " $ , the values of g(x) are
near 1 Ê kg(x) 2k is near 1. Then, for % œ "# we have kg(x) 2k "# for some x satisfying
" $ x " $ , or ! kx 1k $ Ê
lim g(x) Á 2.
x Ä 1
(b) Yes, lim g(x) œ 1 because from the graph we can find a $ ! such that kg(x) 1k % if ! kx (1)k $ .
x Ä 1
61-66. Example CAS commands (values of del may vary for a specified eps):
Maple:
f := x -> (x^4-81)/(x-3);x0 := 3;
plot( f(x), x=x0-1..x0+1, color=black,
# (a)
title="Section 2.3, #61(a)" );
L := limit( f(x), x=x0 );
# (b)
epsilon := 0.2;
# (c)
plot( [f(x),L-epsilon,L+epsilon], x=x0-0.01..x0+0.01,
color=black, linestyle=[1,3,3], title="Section 2.3, #61(c)" );
q := fsolve( abs( f(x)-L ) = epsilon, x=x0-1..x0+1 ); # (d)
delta := abs(x0-q);
plot( [f(x),L-epsilon,L+epsilon], x=x0-delta..x0+delta, color=black, title="Section 2.3, #61(d)" );
for eps in [0.1, 0.005, 0.001 ] do
# (e)
q := fsolve( abs( f(x)-L ) = eps, x=x0-1..x0+1 );
delta := abs(x0-q);
head := sprintf("Section 2.3, #61(e)\n epsilon = %5f, delta = %5f\n", eps, delta );
print(plot( [f(x),L-eps,L+eps], x=x0-delta..x0+delta,
color=black, linestyle=[1,3,3], title=head ));
end do:
Mathematica (assigned function and values for x0, eps and del may vary):
Clear[f, x]
y1: œ L eps; y2: œ L eps; x0 œ 1;
f[x_]: œ (3x2 (7x 1)Sqrt[x] 5)/(x 1)
Plot[f[x], {x, x0 0.2, x0 0.2}]
L: œ Limit[f[x], x Ä x0]
eps œ 0.1; del œ 0.2;
Plot[{f[x], y1, y2},{x, x0 del, x0 del}, PlotRange Ä {L 2eps, L 2eps}]
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 2.4 One-Sided Limits
2.4 ONE-SIDED LIMITS
1. (a) True
(e) True
(i) False
(b) True
(f) True
(j) False
(c) False
(g) False
(k) True
(d) True
(h) False
(l) False
2. (a) True
(e) True
(i) True
(b) False
(f) True
(j) False
(c) False
(g) True
(k) True
(d) True
(h) True
3. (a)
lim f(x) œ
x Ä #b
2
#
" œ #, lim c f(x) œ $ # œ "
xÄ#
(b) No, lim f(x) does not exist because lim b f(x) Á lim c f(x)
xÄ#
xÄ#
xÄ#
(c) lim c f(x) œ 4# 1 œ 3, lim b f(x) œ 4# " œ $
xÄ%
xÄ%
(d) Yes, lim f(x) œ 3 because 3 œ lim c f(x) œ lim b f(x)
xÄ%
xÄ%
xÄ%
4. (a)
lim f(x) œ
x Ä #b
2
#
œ 1, lim c f(x) œ $ # œ ", f(2) œ 2
xÄ#
(b) Yes, lim f(x) œ 1 because " œ lim b f(x) œ lim c f(x)
xÄ#
xÄ#
xÄ#
(c)
lim c f(x) œ 3 (1) œ 4, lim b f(x) œ 3 (1) œ 4
x Ä "
x Ä "
(d) Yes, lim f(x) œ 4 because 4 œ
x Ä "
lim
x Ä "c
f(x) œ
lim
x Ä "b
f(x)
5. (a) No, lim b f(x) does not exist since sin ˆ "x ‰ does not approach any single value as x approaches 0
xÄ!
(b) lim c f(x) œ lim c 0 œ 0
xÄ!
(c)
xÄ!
lim f(x) does not exist because lim b f(x) does not exist
xÄ!
xÄ!
6. (a) Yes, lim b g(x) œ 0 by the sandwich theorem since Èx Ÿ g(x) Ÿ Èx when x 0
xÄ!
(b) No, lim c g(x) does not exist since Èx is not defined for x 0
xÄ!
(c) No, lim g(x) does not exist since lim c g(x) does not exist
xÄ!
xÄ!
7. (aÑ
lim f(x) œ " œ lim b f(x)
xÄ1
(c) Yes, lim f(x) œ 1 since the right-hand and left-hand
(b)
x Ä 1c
xÄ1
limits exist and equal 1
8. (a)
(b)
lim f(x) œ 0 œ lim c f(x)
xÄ1
x Ä 1b
(c) Yes, lim f(x) œ 0 since the right-hand and left-hand
xÄ1
limits exist and equal 0
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
63
64
Chapter 2 Limits and Continuity
9. (a) domain: 0 Ÿ x Ÿ 2
range: 0 y Ÿ 1 and y œ 2
(b) xlim
f(x) exists for c belonging to
Äc
(0ß 1) ("ß #)
(c) x œ 2
(d) x œ 0
10. (a) domain: _ x _
range: " Ÿ y Ÿ 1
(b) xlim
f(x) exists for c belonging to
Äc
(_ß 1) ("ß ") ("ß _)
(c) none
(d) none
11.
x Ä !Þ&c
lim
13.
x Ä #b
14.
x Ä 1c
15.
h Ä !b
lim
2
0.5 2
È3
É 3/2
É xx É
1 œ
0.5 1 œ
1/2 œ
lim
x Ä 1b
"
1
È0 œ !
É "1 É xx # œ
# œ
5‰
ˆ x x 1 ‰ ˆ 2x
ˆ 2 ‰ 2(2) 5
ˆ1‰
x# x œ 2 1 Š (2)# (2) ‹ œ (2) 2 œ 1
lim
ˆ x " 1 ‰ ˆ x x 6 ‰ ˆ 3 7 x ‰ œ ˆ 1 " 1 ‰ ˆ 1 1 6 ‰ ˆ 3 7 1 ‰ œ ˆ "# ‰ ˆ 71 ‰ ˆ 27 ‰ œ 1
lim
Èh# 4h 5 È5
h
œ lim b
hÄ!
16.
12.
lim
h Ä !c
(b)
18. (a)
(b)
ah# 4h 5b 5
h ŠÈh# 4h 5 È5‹
È6 È5h# 11h 6
h
œ lim c
hÄ!
17. (a)
œ lim b Š
hÄ!
x Ä #c
lim
x Ä 1b
lim
x Ä 1c
œ lim c Š
hÄ!
6 a5h# 11h 6b
lim
lim
œ lim b
hÄ!
h ŠÈ6 È5h# 11h 6‹
x Ä #b
kx 2 k
x 2
(x 3)
œ
kx 2 k
x2
œ
lim
lim
x Ä #c
È2x (x 1)
kx 1 k
È2x (x 1)
kx 1 k
œ lim b
xÄ1
h(5h 11)
h ŠÈ6 È5h# 11h 6‹
(x 3)
x Ä #b
œ
œ
04
È5 È5
œ
2
È5
È5h# 11h 6
È6 È5h# 11h 6
È
‹ Š È66 ‹
h
È5h# 11h 6
x Ä #b
lim
h(h 4)
h ŠÈh# 4h 5 È5‹
œ lim c
hÄ!
œ
(x 3)
Èh# 4h 5 È5
È # 4h 5 È5
‹ Š Èhh# ‹
h
4h 5 È5
(x2)
(x#)
(0 11)
È6 È6
11
œ 2È
6
akx 2k œ ax 2b for x 2b
(x 3) œ a(2) 3b œ 1
(x 3) ’ (x(x#2)
) “
lim
œ
x Ä #c
akx 2k œ (x 2) for x 2b
(x 3)(1) œ (2 3) œ 1
È2x (x 1)
(x 1)
akx 1k œ x 1 for x 1b
œ lim b È2x œ È2
xÄ1
œ lim c
xÄ1
È2x (x 1)
(x 1)
akx 1k œ (x 1) for x 1b
œ lim c È2x œ È2
xÄ1
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 2.4 One-Sided Limits
19. (a)
) Ä $b
20. (a)
t Ä %b
Ú) Û
)
lim
œ1
3
3
lim at ÚtÛb œ 4 4 œ 0
21. lim
sin È2)
È 2)
22. lim
sin kt
t
23. lim
sin 3y
4y
)Ä!
tÄ!
yÄ!
24.
œ
tan 2x
x
xÄ!
25. lim
2t
27. lim
xÄ!
)Ä!
œ lim c ˆ "3 †
hÄ!
sin 2x ‰
ˆ cos
2x
x
xÄ!
œ lim
œ 2 lim
t
sin t
t Ä ! ˆ cos t ‰
x csc 2x
cos 5x
œ
3h ‰
sin 3h
x x cos x
2
3
lim at ÚtÛb œ 4 3 œ 1
" ‰
cos 5x
xÄ!
œ
"
3
Œ
œ
"
lim
)Ä!c
(where ) œ kt)
(where ) œ 3y)
3
4
"
3
œ
sin )
)
"
‹ Š lim
x Ä ! cos 2x
xÄ!
œ Š #" lim
†1œ
2 sin 2x
#x ‹
t
Ä!
"
3
(where ) œ 3h)
œ1†2œ2
t
"
‹ Š lim cos 5x ‹
x Ä ! sin 2x
xÄ!
6x# cos x
œ lim ˆ sin xxcos x œk†1œk
tÄ!
x Ä ! sin x sin 2x
x Ä ! sin x cos x
t Ä %c
œ
œ 2 Š lim cos t‹ Œ lim" sin t œ 2 † " † " œ 2
t cos t
sin t
tÄ!
xÄ!
(b)
Ú) Û
)
lim
3
sin )
4 )lim
Ä! )
œ
œ Š lim
sin 2x
x Ä ! x cos 2x
œ 2 lim
xÄ!
)Ä!
"
"
sin 3h
3 h lim
Ä !c ˆ 3h ‰
œ
œ lim
œ lim ˆ sinx2x †
sin )
)
œ k lim
sin 3y
3
4 ylim
Ä ! 3y
28. lim 6x# (cot x)(csc 2x) œ lim
29. lim
k sin )
)
œ lim
3 sin 3y
"
4 ylim
3y
Ä!
œ
h
t Ä ! tan t
k sin kt
kt
tÄ!
) Ä $c
(where x œ È2))
œ1
sin x
x
xÄ!
œ lim
lim
h Ä !c sin 3h
26. lim
œ lim
(b)
2x
œ lim ˆ3 cos x †
x
sin x
xÄ!
x cos x ‰
sin x cos x
œ lim ˆ sinx x †
xÄ!
†
œ ˆ #" † 1‰ (1) œ
2x ‰
sin 2x
" ‰
cos x
"
#
œ3†"†1œ3
lim
x
x Ä ! sin x
œ lim Š sin" x ‹ † lim ˆ cos" x ‰ lim Š sin" x ‹ œ (1)(1) 1 œ 2
xÄ!
30. lim
xÄ!
xÄ!
x
x# x sin x
#x
1 cos )
) Ä ! sin 2)
31. lim
œ lim
)Ä!
)Ä!
34. lim
sin (sin h)
sin h
sin )
) Ä ! sin 2)
36. lim
sin 5x
x Ä ! sin 4x
œ
œ lim
)Ä!
œ lim
)Ä!
sin )
)
sin )
)
)Ä!
"# (1) œ 0
1 cos2 )
a2sin ) cos )ba1 cos )b
œ lim
)Ä!
sin2 )
a2sin ) cos )ba1 cos )b
œ lim
xÄ!
xa1 c cos xb
9x2
sin2 3x
9x2
œ lim
1 c cos x
9x
2
x Ä ! ˆ sin3x3x ‰
œ
" lim ˆ 1
9
x
Ä!
cos x ‰
x
2
lim ˆ sin3x3x ‰
xÄ!
œ
"
9 a0 b
12
œ0
œ 1 since ) œ sin h Ä 0 as h Ä 0
2) ‰
#)
5x
œ lim ˆ sin
sin 4x †
4x
5x
)Ä!
œ lim
"
#
œ 1 since ) œ 1 cos t Ä 0 as t Ä 0
sin )
œ lim ˆ sin
2) †
xÄ!
"# ˆ sinx x ‰‰ œ 0 œ0
0
a2ba2b
xa1 cos xb
sin2 3x
xÄ!
sin(1 cos t)
1cos t
x
a1 cos )ba1 cos )b
a2sin ) cos )ba1 cos )b
œ lim
33. lim
35. lim
"
#
xÄ!
œ lim
x x cos x
sin2 3x
xÄ!
hÄ!
œ lim ˆ #x sin )
a2cos )ba1 cos )b
32. lim
tÄ!
xÄ!
œ
"
# )lim
Ä!
† 54 ‰ œ
ˆ sin) ) †
5
4 xlim
Ä!
2) ‰
sin 2)
ˆ sin5x5x †
œ
"
#
4x ‰
sin 4x
†1†1œ
œ
5
4
"
#
†1†1œ
5
4
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
65
66
Chapter 2 Limits and Continuity
37. lim ) cos ) œ 0 † 1 œ 0
)Ä!
38. lim sin ) cot 2) œ lim sin )
)Ä!
)Ä!
tan 3x
39. lim
x Ä ! sin 8x
œ
3
8 xlim
Ä!
40. lim
yÄ!
œ
sin 3x
œ lim ˆ cos
3x †
xÄ!
" ‰
sin 8x
œ lim sin )
)Ä!
œ lim
xÄ!
3
8
†1†1†1œ
sin 3y sin 4y cos 5y
y cos 4y sin 5y
yÄ!
œ lim
) cot 4)
2
2
) Ä ! sin ) cot 2)
42. lim
)Ä!
sin )
cos )
3)
2
) cos
sin 3)
œ lim
)Ä!
2
lim 4) cos2 4) cos )
) Ä ! cos 2) sin 4)
œ lim
cos 4)
sin 4)
2 2)
2
sin ) cos
sin2 2)
"
sin 8x
cos 2)
) Ä ! 2 cos )
†
8x
3x
œ
1
2
† 83 ‰
3
8
yÄ!
sin ) sin 3)
2
) Ä ! ) cos ) cos 3)
)
œ lim
sin 4y
cos 5y
3†4†5y
œ lim Š siny3y ‹ Š cos
4y ‹ Š sin 5y ‹ Š 3†4†5y ‹
cos 5y ˆ 3†4 ‰
lim Š sin3y3y ‹ Š sin4y4y ‹ Š sin5y5y ‹ Š cos
4y ‹ 5
yÄ!
tan )
2
) Ä ! ) cot 3)
cos 2)
2sin ) cos )
sin 3x
œ lim ˆ cos
3x †
ˆ cos"3x ‰ ˆ sin3x3x ‰ ˆ sin8x8x ‰ œ
sin 3y cot 5y
y cot 4y
41. lim
œ
cos 2)
sin 2)
œ1†1†1†1†
œ
12
5
œ lim ˆ sin) ) ‰ˆ sin3)3) ‰ˆ cos ) 3cos 3) ‰ œ a1ba1bˆ 13†1 ‰ œ 3
)Ä!
) cos 4) sin2 2)
2
2
) Ä ! sin ) cos 2) sin 4)
œ lim
12
5
) cos 4) a2sin ) cos )b2
2
2
) Ä ! sin ) cos 2) sin 4)
œ lim
) cos 4) ˆ4sin2 ) cos2 )‰
2
2
) Ä ! sin ) cos 2) sin 4)
œ lim
4) cos )
4) cos )
œ lim ˆ sin4)4) ‰Š coscos
‹ œ lim Š sin14) ‹Š coscos
‹ œ ˆ 11 ‰Š 11†12 ‹ œ 1
2 2)
2 2)
2
)Ä!
2
)Ä!
2
4)
43. Yes. If lim b f(x) œ L œ lim c f(x), then xlim
f(x) œ L. If lim b f(x) Á lim c f(x), then xlim
f(x) does not exist.
Äa
Äa
xÄa
xÄa
xÄa
xÄa
44. Since xlim
f(x) œ L if and only if lim b f(x) œ L and lim c f(x) œ L, then xlim
f(x) can be found by calculating
Äc
Äc
xÄc
xÄc
lim b f(x).
xÄc
45. If f is an odd function of x, then f(x) œ f(x). Given lim b f(x) œ 3, then lim c f(x) œ $.
xÄ!
xÄ!
46. If f is an even function of x, then f(x) œ f(x). Given lim c f(x) œ 7 then
xÄ#
can be said about
lim
x Ä #c
lim
x Ä #b
f(x) œ 7. However, nothing
f(x) because we don't know lim b f(x).
xÄ#
47. I œ (5ß 5 $ ) Ê 5 x & $ . Also, Èx 5 % Ê x 5 %# Ê x & %# . Choose $ œ %#
Ê lim Èx 5 œ 0.
x Ä &b
48. I œ (% $ ß %) Ê % $ x 4. Also, È% x % Ê % x %# Ê x % %# . Choose $ œ %#
Ê lim È% x œ 0.
x Ä %c
49. As x Ä 0 the number x is always negative. Thus, ¹ kxxk (1)¹ % Ê ¸ xx 1¸ % Ê 0 % which is always
true independent of the value of x. Hence we can choose any $ 0 with $ x ! Ê
2
¸ x 2
¸
50. Since x Ä # we have x 2 and kx 2k œ x 2. Then, ¹ kxx
2 k " ¹ œ x 2 " % Ê 0 %
which is always true so long as x #. Hence we can choose any $ !, and thus # x # $
2
Ê ¹ kxx
2k "¹ % . Thus,
x 2
lim
x Ä #b kx2k
x
lim
x Ä ! c kx k
œ 1.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
œ 1.
Section 2.5 Continuity
51. (a)
(b)
lim
x Ä %!!b
67
ÚxÛ œ 400. Just observe that if 400 x 401, then ÚxÛ œ 400. Thus if we choose $ œ ", we have for any
number % ! that 400 x 400 $ Ê lÚxÛ 400l œ l400 400l œ ! %.
lim c ÚxÛ œ 399. Just observe that if 399 x 400 then ÚxÛ œ 399. Thus if we choose $ œ ", we have for any
x Ä %!!
number % ! that 400 $ x 400 Ê lÚxÛ 399l œ l399 399l œ ! %.
(c) Since lim b ÚxÛ Á lim c ÚxÛ we conclude that lim ÚxÛ does not exist.
x Ä %!!
x Ä %!!
52. (a)
x Ä %!!
lim f(x) œ lim b Èx œ È0 œ 0; ¸Èx 0¸ % Ê % Èx % Ê ! x %# for x positive. Choose $ œ %#
xÄ!
Ê lim b f(x) œ 0.
x Ä !b
xÄ!
(b)
lim f(x) œ lim c x# sin ˆ x" ‰ œ 0 by the sandwich theorem since x# Ÿ x# sin ˆ x" ‰ Ÿ x# for all x Á 0.
x Ä !c
xÄ!
Since kx# 0k œ kx# 0k œ x# % whenever kxk È%, we choose $ œ È% and obtain ¸x# sin ˆ "x ‰ 0¸ %
if $ x 0.
(c) The function f has limit 0 at x! œ 0 since both the right-hand and left-hand limits exist and equal 0.
2.5 CONTINUITY
1. No, discontinuous at x œ 2, not defined at x œ 2
2. No, discontinuous at x œ 3, " œ lim c g(x) Á g(3) œ 1.5
xÄ$
3. Continuous on [1ß 3]
4. No, discontinuous at x œ 1, 1.5 œ lim c k(x) Á lim b k(x) œ !
xÄ"
xÄ"
5. (a) Yes
(b) Yes,
(c) Yes
(d) Yes
6. (a) Yes, f(1) œ 1
lim
x Ä "b
f(x) œ 0
(b) Yes, lim f(x) œ 2
xÄ1
(c) No
(d) No
7. (a) No
(b) No
8. ["ß !) (!ß ") ("ß #) (#ß $)
9. f(2) œ 0, since lim c f(x) œ 2(2) 4 œ 0 œ lim b f(x)
xÄ#
xÄ#
10. f(1) should be changed to 2 œ lim f(x)
xÄ1
11. Nonremovable discontinuity at x œ 1 because lim f(x) fails to exist ( limc f(x) œ 1 and lim b f(x) œ 0).
xÄ"
xÄ1
xÄ"
Removable discontinuity at x œ 0 by assigning the number lim f(x) œ 0 to be the value of f(0) rather than f(0) œ 1.
xÄ!
12. Nonremovable discontinuity at x œ 1 because lim f(x) fails to exist ( lim c f(x) œ 2 and lim b f(x) œ 1).
xÄ"
xÄ1
xÄ"
Removable discontinuity at x œ 2 by assigning the number lim f(x) œ 1 to be the value of f(2) rather than f(2) œ 2.
xÄ#
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
68
Chapter 2 Limits and Continuity
13. Discontinuous only when x 2 œ 0 Ê x œ 2
14. Discontinuous only when (x 2)# œ 0 Ê x œ 2
15. Discontinuous only when x# %x $ œ ! Ê (x 3)(x 1) œ 0 Ê x œ 3 or x œ 1
16. Discontinuous only when x# 3x 10 œ 0 Ê (x 5)(x 2) œ 0 Ê x œ 5 or x œ 2
17. Continuous everywhere. ( kx 1k sin x defined for all x; limits exist and are equal to function values.)
18. Continuous everywhere. ( kxk " Á 0 for all x; limits exist and are equal to function values.)
19. Discontinuous only at x œ 0
20. Discontinuous at odd integer multiples of 1# , i.e., x = (2n ") 1# , n an integer, but continuous at all other x.
21. Discontinuous when 2x is an integer multiple of 1, i.e., 2x œ n1, n an integer Ê x œ
n1
# ,
n an integer, but
continuous at all other x.
22. Discontinuous when
1x
#
is an odd integer multiple of 1# , i.e.,
1x
#
œ (2n 1) 1# , n an integer Ê x œ 2n 1, n an
integer (i.e., x is an odd integer). Continuous everywhere else.
23. Discontinuous at odd integer multiples of 1# , i.e., x = (2n 1) 1# , n an integer, but continuous at all other x.
24. Continuous everywhere since x% 1
and are equal to the function values.
1 and " Ÿ sin x Ÿ 1 Ê 0 Ÿ sin# x Ÿ 1 Ê 1 sin# x
1; limits exist
25. Discontinuous when 2x 3 0 or x 3# Ê continuous on the interval 3# ß _‰ .
26. Discontinuous when 3x 1 0 or x "
3
Ê continuous on the interval 3" ß _‰ .
27. Continuous everywhere: (2x 1)"Î$ is defined for all x; limits exist and are equal to function values.
28. Continuous everywhere: (2 x)"Î& is defined for all x; limits exist and are equal to function values.
29. Continuous everywhere since lim
xÄ3
30. Discontinuous at x œ 2 since
x2 x 6
x3
œ lim
xÄ3
ax 3bax 2b
x3
œ lim ax 2b œ 5 œ ga3b
xÄ3
lim faxb does not exist while fa2b œ 4.
x Ä 2
31. xlim
sin (x sin x) œ sin (1 sin 1) œ sin (1 0) œ sin 1 œ 0, and function continuous at x œ 1.
Ä1
32. lim sin ˆ 1# cos (tan t)‰ œ sin ˆ 1# cos (tan (0))‰ œ sin ˆ 1# cos (0)‰ œ sin ˆ 1# ‰ œ 1, and function continuous at t œ !.
tÄ!
33. lim sec ay sec# y tan# y 1b œ lim sec ay sec# y sec# yb œ lim sec a(y 1) sec# yb œ sec a(" ") sec# 1b
yÄ1
yÄ1
yÄ1
œ sec 0 œ 1, and function continuous at y œ ".
34. lim tan 14 cos ˆsin x"Î$ ‰‘ œ tan 14 cos (sin(0))‘ œ tan ˆ 14 cos (0)‰ œ tan ˆ 14 ‰ œ 1, and function continuous at x œ !.
xÄ!
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 2.5 Continuity
35. lim cos ’ È19 13 sec 2t “ œ cos ’ È19 13 sec 0 “ œ cos
tÄ!
1
È16
œ cos
1
4
œ
È2
# ,
and function continuous at t œ !.
36. lim1 Écsc# x 5È3 tan x œ Écsc# ˆ 16 ‰ 5È3 tan ˆ 16 ‰ œ Ê4 5È3 Š È"3 ‹ œ È9 œ 3, and function continuous at
xÄ
'
x œ 1' .
37. g(x) œ
x# 9
x3
(x 3)(x 3)
(x 3)
œ
38. h(t) œ
t# 3t 10
t#
39. f(s) œ
s$ "
s# 1
40. g(x) œ
œ
œ
œ x 3, x Á 3 Ê g(3) œ lim (x 3) œ 6
xÄ$
(t 5)(t 2)
t#
as# s 1b (s 1)
(s 1)(s 1)
x# 16
x# 3x 4
œ
œ t 5, t Á # Ê h(2) œ lim (t 5) œ 7
tÄ#
s# s "
s1 ,
œ
(x 4)(x 4)
(x 4)(x 1)
œ
x4
x1
s Á 1 Ê f(1) œ lim Š s
sÄ1
#
s1
s1 ‹
4‰
, x Á 4 Ê g(4) œ lim ˆ xx 1 œ
xÄ%
œ
3
#
8
5
41. As defined, lim c f(x) œ (3)# 1 œ 8 and lim b (2a)(3) œ 6a. For f(x) to be continuous we must have
xÄ$
xÄ$
6a œ 8 Ê a œ 43 .
42. As defined,
lim
x Ä #c
g(x) œ 2 and
4b œ 2 Ê b œ "# .
lim
x Ä #b
g(x) œ b(2)# œ 4b. For g(x) to be continuous we must have
43. As defined, lim c f(x) œ 12 and lim b f(x) œ a# a2b 2a œ 2a# 2a. For f(x) to be continuous we must have
xÄ#
xÄ#
12 œ 2a# 2a Ê a œ 3 or a œ 2.
44. As defined, lim c g(x) œ
b
b1
xÄ0
0b
b1
œ
b
b1
œ b Ê b œ 0 or b œ 2.
45. As defined,
lim
x Ä 1 c
f(x) œ 2 and
and lim b g(x) œ a0b2 b œ b. For g(x) to be continuous we must have
xÄ0
lim
x Ä 1 b
f(x) œ aa1b b œ a b, and
lim f(x) œ aa1b b œ a b and
x Ä 1c
lim f(x) œ 3. For f(x) to be continuous we must have 2 œ a b and a b œ 3 Ê a œ
x Ä 1b
5
#
and b œ "# .
46. As defined, lim c g(x) œ aa0b 2b œ 2b and lim b g(x) œ a0b2 3a b œ 3a b, and
xÄ0
xÄ0
lim g(x) œ a2b2 3a b œ 4 3a b and lim b g(x) œ 3a2b 5 œ 1. For g(x) to be continuous we must
xÄ0
x Ä 2c
have 2b œ 3a b and 4 3a b œ 1 Ê a œ 3# and b œ 3# .
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
69
70
Chapter 2 Limits and Continuity
47. The function can be extended: f(0) ¸ 2.3.
48. The function cannot be extended to be continuous at
x œ 0. If f(0) ¸ 2.3, it will be continuous from the
right. Or if f(0) ¸ 2.3, it will be continuous from the
left.
49. The function cannot be extended to be continuous
at x œ 0. If f(0) œ 1, it will be continuous from
the right. Or if f(0) œ 1, it will be continuous
from the left.
50. The function can be extended: f(0) ¸ 7.39.
51. f(x) is continuous on [!ß "] and f(0) 0, f(1) 0
Ê by the Intermediate Value Theorem f(x) takes
on every value between f(0) and f(1) Ê the
equation f(x) œ 0 has at least one solution between
x œ 0 and x œ 1.
52. cos x œ x Ê (cos x) x œ 0. If x œ 1# , cos ˆ 1# ‰ ˆ 1# ‰ 0. If x œ 1# , cos ˆ 1# ‰ for some x between 1
#
and
1
#
1
#
0. Thus cos x x œ 0
according to the Intermediate Value Theorem, since the function cos x x is continuous.
53. Let f(x) œ x$ 15x 1, which is continuous on [4ß 4]. Then f(4) œ 3, f(1) œ 15, f(1) œ 13, and f(4) œ 5.
By the Intermediate Value Theorem, f(x) œ 0 for some x in each of the intervals % x 1, " x 1, and
" x 4. That is, x$ 15x 1 œ 0 has three solutions in [%ß 4]. Since a polynomial of degree 3 can have at most 3
solutions, these are the only solutions.
54. Without loss of generality, assume that a b. Then F(x) œ (x a)# (x b)# x is continuous for all values of
x, so it is continuous on the interval [aß b]. Moreover F(a) œ a and F(b) œ b. By the Intermediate Value
Theorem, since a a # b b, there is a number c between a and b such that F(x) œ a # b .
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 2.5 Continuity
55. Answers may vary. Note that f is continuous for every value of x.
(a) f(0) œ 10, f(1) œ 1$ 8(1) 10 œ 3. Since $ 1 10, by the Intermediate Value Theorem, there exists a c
so that ! c 1 and f(c) œ 1.
(b) f(0) œ 10, f(4) œ (4)$ 8(4) 10 œ 22. Since 22 È3 10, by the Intermediate Value
Theorem, there exists a c so that 4 c 0 and f(c) œ È3.
(c) f(0) œ 10, f(1000) œ (1000)$ 8(1000) 10 œ 999,992,010. Since 10 5,000,000 999,992,010, by the
Intermediate Value Theorem, there exists a c so that ! c 1000 and f(c) œ 5,000,000.
56. All five statements ask for the same information because of the intermediate value property of continuous functions.
(a) A root of f(x) œ x$ 3x 1 is a point c where f(c) œ 0.
(b) The points where y œ x$ crosses y œ 3x 1 have the same y-coordinate, or y œ x$ œ 3x 1
Ê f(x) œ x$ 3x 1 œ 0.
(c) x$ 3x œ 1 Ê x$ 3x 1 œ 0. The solutions to the equation are the roots of f(x) œ x$ 3x 1.
(d) The points where y œ x$ 3x crosses y œ 1 have common y-coordinates, or y œ x$ 3x œ 1
Ê f(x) œ x$ 3x 1 œ !.
(e) The solutions of x$ 3x 1 œ 0 are those points where f(x) œ x$ 3x 1 has value 0.
57. Answers may vary. For example, f(x) œ
sin (x 2)
x2
is discontinuous at x œ 2 because it is not defined there.
However, the discontinuity can be removed because f has a limit (namely 1) as x Ä 2.
58. Answers may vary. For example, g(x) œ
"
x1
has a discontinuity at x œ 1 because lim g(x) does not exist.
x Ä "
Š lim c g(x) œ _ and lim b g(x) œ _.‹
x Ä "
x Ä "
59. (a) Suppose x! is rational Ê f(x! ) œ 1. Choose % œ "# . For any $ 0 there is an irrational number x (actually
infinitely many) in the interval (x! $ ß x! $ ) Ê f(x) œ 0. Then 0 kx x! k $ but kf(x) f(x! )k
œ 1 "# œ %, so x lim
f(x) fails to exist Ê f is discontinuous at x! rational.
Äx
!
On the other hand, x! irrational Ê f(x! ) œ 0 and there is a rational number x in (x! $ ß x! $ ) Ê f(x)
œ 1. Again x lim
f(x) fails to exist Ê f is discontinuous at x! irrational. That is, f is discontinuous at
Äx
!
every point.
(b) f is neither right-continuous nor left-continuous at any point x! because in every interval (x! $ ß x! ) or
(x! ß x! $ ) there exist both rational and irrational real numbers. Thus neither limits lim f(x) and
x Ä x!
lim f(x) exist by the same arguments used in part (a).
x Ä x
!
60. Yes. Both f(x) œ x and g(x) œ x g ˆ "# ‰
œ0 Ê
f(x)
g(x)
"
#
are continuous on [!ß "]. However
is discontinuous at x œ
f(x)
g(x)
is undefined at x œ
"
#
since
"
#.
61. No. For instance, if f(x) œ 0, g(x) œ ÜxÝ, then h(x) œ 0 aÜxÝb œ 0 is continuous at x œ 0 and g(x) is not.
62. Let f(x) œ
œ
"
x1
"
(x 1) 1
œ
and g(x) œ x 1. Both functions are continuous at x œ 0. The composition f ‰ g œ f(g(x))
"
x
is discontinuous at x œ 0, since it is not defined there. Theorem 10 requires that f(x) be
continuous at g(0), which is not the case here since g(0) œ 1 and f is undefined at 1.
63. Yes, because of the Intermediate Value Theorem. If f(a) and f(b) did have different signs then f would have to
equal zero at some point between a and b since f is continuous on [aß b].
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
71
72
Chapter 2 Limits and Continuity
64. Let f(x) be the new position of point x and let d(x) œ f(x) x. The displacement function d is negative if x is
the left-hand point of the rubber band and positive if x is the right-hand point of the rubber band. By the
Intermediate Value Theorem, d(x) œ 0 for some point in between. That is, f(x) œ x for some point x, which is
then in its original position.
65. If f(0) œ 0 or f(1) œ 1, we are done (i.e., c œ 0 or c œ 1 in those cases). Then let f(0) œ a 0 and f(1) œ b 1
because 0 Ÿ f(x) Ÿ 1. Define g(x) œ f(x) x Ê g is continuous on [0ß 1]. Moreover, g(0) œ f(0) 0 œ a 0 and
g(1) œ f(1) 1 œ b 1 0 Ê by the Intermediate Value Theorem there is a number c in (!ß ") such that
g(c) œ 0 Ê f(c) c œ 0 or f(c) œ c.
66. Let % œ
kf(c)k
#
0. Since f is continuous at x œ c there is a $ 0 such that kx ck $ Ê kf(x) f(c)k %
Ê f(c) % f(x) f(c) %.
If f(c) 0, then % œ "# f(c) Ê
"
#
"
#
If f(c) 0, then % œ f(c) Ê
f(c) f(x) 3
#
3
#
f(c) f(x) f(c) Ê f(x) 0 on the interval (c $ ß c $ ).
"
#
f(c) Ê f(x) 0 on the interval (c $ ß c $ ).
67. By Exercises 52 in Section 2.3, we have xlim
faxb œ L Í lim fac hb œ L.
Äc
hÄ0
Thus, faxb is continuous at x œ c Í xlim
faxb œ facb Í lim fac hb œ facb.
Äc
hÄ0
68. By Exercise 67, it suffices to show that lim sinac hb œ sin c and lim cosac hb œ cos c.
hÄ0
hÄ0
Now lim sinac hb œ lim asin cbacos hb acos cbasin hb‘ œ asin cbŠ lim cos h‹ acos cbŠ lim sin h‹
hÄ0
hÄ0
hÄ0
hÄ0
By Example 11 Section 2.2, lim cos h œ " and lim sin h œ !. So lim sinac hb œ sin c and thus faxb œ sin x is
hÄ0
continuous at x œ c. Similarly,
hÄ0
hÄ0
lim cosac hb œ lim acos cbacos hb asin cbasin hb‘ œ acos cbŠ lim cos h‹ asin cbŠ lim sin h‹ œ cos c.
hÄ0
hÄ0
Thus, gaxb œ cos x is continuous at x œ c.
hÄ0
69. x ¸ 1.8794, 1.5321, 0.3473
70. x ¸ 1.4516, 0.8547, 0.4030
71. x ¸ 1.7549
72. x ¸ 1.5596
73. x ¸ 3.5156
74. x ¸ 3.9058, 3.8392, 0.0667
75. x ¸ 0.7391
76. x ¸ 1.8955, 0, 1.8955
hÄ0
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 2.6 Limits Involving Infinity; Asymptotes of Graphs
2.6 LIMITS INVOLVING INFINITY; ASMYPTOTES OF GRAPHS
1. (a)
(c)
(e)
(g)
lim f(x) œ 0
(b)
xÄ2
lim
x Ä 3 c
f(x) œ 2
lim
x Ä 3 b
f(x) œ 2
(d) lim f(x) œ does not exist
xÄ3
lim b f(x) œ 1
(f)
lim f(x) œ does not exist
(h) x lim
f(x) œ 1
Ä_
xÄ0
xÄ0
lim f(x) œ _
x Ä 0c
(i) x Ä
lim
f(x) œ 0
_
2. (a)
(c)
lim f(x) œ 2
(b)
lim f(x) œ 1
(d) lim f(x) œ does not exist
xÄ4
x Ä 2c
xÄ2
lim f(x) œ _
x Ä 3 b
(g) lim f(x) œ _
(e)
x Ä 3
(i)
lim f(x) œ 3
x Ä 2b
lim f(x) œ _
(f)
x Ä 3 c
lim
(h)
x Ä 0b
(k) x lim
f(x) œ 0
Ä_
lim f(x) œ _
lim f(x) œ does not exist
(j)
x Ä 0c
xÄ0
(l) x Ä
lim
f(x) œ 1
_
Note: In these exercises we use the result
"
lim
mÎn
xÄ „_ x
Theorem 8 and the power rule in Theorem 1:
lim
xÄ „_
œ 0 whenever
ˆ xm"În ‰ œ
lim
(b) 3
4. (a) 1
(b) 1
5. (a)
"
#
(b)
"
#
6. (a)
"
8
(b)
"
8
7. (a) 53
(b)
10. 3") Ÿ
11.
lim
tÄ_
12. r Ä
lim_
0. This result follows immediately from
ˆ x" ‰mÎn œ Š
"
lim
‹
xÄ „_ x
mÎn
(b) 53
3
4
9. "x Ÿ
m
n
xÄ „_
3. (a) 3
8. (a)
f(x) œ _
sin 2x
x
Ÿ
"
x
cos )
3)
Ÿ
"
3)
2 t sin t
t cos t
Ê x lim
Ä_
Ê
13. (a) x lim
Ä_
2x 3
5x 7
lim
) Ä _
œ lim
2
t
tÄ_
r sin r
2r 7 5 sin r
sin 2x
x
œrÄ
lim_
œ x lim
Ä_
œ 0 by the Sandwich Theorem
cos )
3)
œ 0 by the Sandwich Theorem
1 ˆ sint t ‰
1 ˆ cost t ‰
œ
1 ˆ sinr r ‰
2 7r 5 ˆ sinr r ‰
2 3x
5 7x
3
4
œ
2
5
010
10
œ 1
œrÄ
lim_
10
200
œ
(b)
"
#
2
5
(same process as part (a))
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
œ 0mÎn œ 0.
73
74
Chapter 2 Limits and Continuity
2 Š x7$ ‹
$
2x 7
14. (a) x lim
œ x lim
Ä _ x$ x# x 7
Ä_
(b) 2 (same process as part (a))
"
x
x"#
15. (a) x lim
Ä_
x1
x# 3
œ x lim
Ä_
1 x3#
16. (a) x lim
Ä_
3x 7
x# 2
œ x lim
Ä_
1 x2#
17. (a) x lim
Ä_
7x$
x$ 3x# 6x
18. (a) x lim
Ä_
x$
20. (a) x lim
Ä_
9
#
2x%
x7#
œ x lim
Ä_
œ x lim
Ä_
10x& x% 31
x'
19. (a) x lim
Ä_
(b)
"
4x 1
3
x
œ2
œ0
(b) 0 (same process as part (a))
œ0
(b) 0 (same process as part (a))
œ(
7
1 3x x6#
"
x$
4
x"$
x#
1
œ x lim
Ä_
2
(b) 7 (same process as part (a))
œ!
x"# x31'
1
10
x
œ x lim
Ä_
9x% x
5x# x 6
1 "x x"# x7$
(b) 0 (same process as part (a))
œ0
(b) 0 (same process as part (a))
9 x"$
5
x#
x"$ x6%
œ
9
#
(same process as part (a))
2x$ 2x 3
3x$ 3x# 5x
21. (a) x lim
Ä_
2 x2# x3$
œ x lim
Ä_
œ 23
3 3x x5#
(b) 23 (same process as part (a))
x %
x% 7x$ 7x# 9
22. (a) x lim
Ä_
"
1 7x x7# x9%
œ x lim
Ä_
œ 1
(b) 1 (same process as part (a))
8
8
3
3
3
x2
x2
É 8x
23. x lim
œ Êx lim
œ É 82 00 œ È4 œ 2
2x2 x œ x lim
Ä_
Ä _ Ê 2 1x
Ä _ 2 1x
2
24. x Ä
lim
Šx x1‹
_ 8x2 3
2
1 Î3
œxÄ
lim
_Œ
5
1
" 1x x12
8
x
3
x2
1 Î3
œ Œx Ä
lim
_
5
1
" 1x x12
8
3
x2
5
x
1 Î3
œ ˆ " 8 0 0 0 ‰
1 Î3
œ ˆ "8 ‰
1 Î3
œ
x2
x2
‰ œ_
25. x Ä
lim
lim
œ Œx Ä
lim
œ ˆ 01_
Š 1x ‹ œ x Ä
0
_ x2 7x
_Π1 7x _ 1 7x 3
1
5
1
5
5
x
x
x2
x2
É x 5x œ x lim
26. x lim
œ Êx lim
œ É 1 0 0 0 0 œ È0 œ 0
Ä _ x3 x 2
Ä _Ê 1 x12 x23
Ä _ 1 x12 x23
2
27. x lim
Ä_
2Èx xc"
3x 7
29. x Ä
lim
_
30. x lim
Ä_
œ x lim
Ä_
3 xÈ
5 x
È
3 xÈ
5 x
È
x "x %
x #x $
Š
œxÄ
lim
_
œ x lim
Ä_
2
‹ Š x"# ‹
x"Î#
3 7x
œ0
1 xÐ"Î&Ñ Ð"Î$Ñ
1 xÐ"Î&Ñ Ð"Î$Ñ
x x"#
1 x"
œxÄ
lim
_
28. x lim
Ä_
" ‹
1 Š #Î"&
x
" ‹
1 Š #Î"&
2 Èx
2 Èx
œ x lim
Ä_
2
‹"
x"Î#
2
Š "Î#
‹1
x
Š
œ 1
œ1
x
œ_
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
"
#
Section 2.6 Limits Involving Infinity; Asymptotes of Graphs
2x&Î$ x"Î$ 7
x)Î& 3x Èx
31. x lim
Ä_
3 x 5x 3
È
2x x#Î$ 4
32. x Ä
lim
_
33. x lim
Ä_
È x2 1
x1
34. x Ä
lim
_
œ x lim
Ä_
œxÄ
lim
_
œ x lim
Ä_
È x2 1
x1
1
"
x#Î$
2
È x 2 1 ÎÈ x 2
a x 1 b ÎÈ x 2
œxÄ
lim
_
" 7
x"*Î"&
x)Î&
3
"
x$Î&
x""Î"!
2x"Î"& 5 3x
"
x"Î$
È
3
œ 5# È
x
4x
œ x lim
Ä_
È x 2 1 ÎÈ x 2
a x 1 bÎ È x 2
œ_
È a x 2 1 bÎx 2
ax 1bÎx
œ x lim
Ä_
È 1 1 Îx 2
a1 1 Îx b
È a x 2 1 bÎ x 2
È1 0
a1 0 b
œ
È 1 1 Îx 2
œxÄ
lim
œ x lim
œ
_ ax 1bÎaxb
Ä _ a 1 1 Î x b
œ1
È1 0
a1 0b
a x 3 bÎ x
a x 3 bÎ x
a1 3 Îx b
x3
35. x lim
œ x lim
œ x lim
œ x lim
œ
Ä _ È4x2 25
Ä _ È4x2 25ÎÈx2
Ä _ Èa4x2 25bÎx2
Ä _ È4 25Îx2
ˆ4 3x3 ‰ÎÈx6
4 3x
36. x Ä
lim
œxÄ
lim
œxÄ
lim
_ Èx6 9
_ Èx6 9ÎÈx6
_
3
"
œ_
37.
lim
x Ä !b 3x
39.
lim
x Ä #c x 2
41.
lim
x Ä )b x8
3
2x
4
43. lim
#
x Ä ( (x7)
œ _
œ _
œ_
lim
"Î$
x Ä !b 3x
46. (a)
lim
"Î&
x Ä !b x
4
47. lim
#Î&
xÄ! x
49.
51.
52.
lim
x Ä ˆ 1# ‰
œ lim
4
#
x Ä ! ax"Î& b
œ x lim
Ä_
ˆ 4 Îx 3 3‰
È 1 9 Îx 6
Š positive
positive ‹
40.
lim
x Ä $b x 3
Š negative
positive ‹
42.
lim
x Ä &c 2x10
positive
Š positive
‹
44. lim
œ_
"
3x
"
#
x Ä ! x (x1)
2
(b)
lim
"Î$
x Ä !c 3x
(b)
lim
"Î&
x Ä !c x
48. lim
"
#Î$
xÄ! x
tan x œ _
50.
œ3
œ_
positive
Š negative
‹
2
a0 3 b
È1 0
"
#
positive
Š negative
‹
lim
x Ä !c 2x
5
œ
œ
œ _
38.
œ_
2
ˆ4 3x3 ‰Îˆx3 ‰
Èax6 9bÎx6
positive
Š positive
‹
œ_
2
45. (a)
a1 0 b
È4 0
2
œ 1
œ_
Š negative
negative ‹
œ _
negative
Š positive
†positive ‹
œ _
œ _
œ lim
"
#
x Ä ! ax"Î$ b
œ_
lim sec x œ _
x Ä ˆ #1 ‰
lim (1 csc )) œ _
)Ä!
lim (2 cot )) œ _ and lim c (2 cot )) œ _, so the limit does not exist
)Ä!
) Ä !b
"
œ lim b
xÄ#
"
(x2)(x2)
œ_
Š positive"†positive ‹
"
œ lim c
xÄ#
"
(x2)(x2)
œ _
Š positive†"negative ‹
53. (a)
lim
#
x Ä # b x 4
(b)
lim
#
x Ä # c x 4
(c)
lim
#
x Ä #b x 4
(d)
lim
#
x Ä #c x 4
"
œ
lim
x Ä #b (x2)(x2)
"
œ _
Š positive†"negative ‹
"
œ
lim
x Ä #c (x2)(x2)
"
œ_
Š negative"†negative ‹
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
75
76
Chapter 2 Limits and Continuity
54. (a)
lim
#
x Ä "b x 1
(b)
lim
#
x Ä "c x 1
(c)
lim
#
x Ä "b x 1
(d)
lim
#
x Ä "c x 1
x
œ lim b
xÄ"
x
(x1)(x1)
œ_
positive
Š positive
†positive ‹
x
œ lim c
xÄ"
x
(x1)(x1)
œ _
positive
Š positive
†negative ‹
x
œ
lim
x Ä "b (x1)(x1)
x
œ_
negative
Š positive
†negative ‹
x
œ
lim
x Ä "c (x1)(x1)
x
œ _
negative
Š negative
†negative ‹
55. (a)
lim
x Ä !b #
x#
"
x
œ 0 lim b
xÄ!
"
x
œ _
"
Š negative
‹
(b)
lim
x Ä !c #
x#
"
x
œ 0 lim c
xÄ!
"
x
œ_
"
Š positive
‹
(c)
lim
#
x Ä $È2
(d)
lim
x Ä 1 #
56. (a)
x#
x#
lim
x Ä #b
"
x
"
x
œ
x# 1
2x 4
x# 1
2#Î$
#
œ
"
#
(d)
lim
x Ä !c 2x 4
œ
(b)
(c)
(d)
(e)
58. (a)
x# 3x 2
x$ 2x#
lim b
x# 3x 2
x$ 2x#
lim
xÄ#
#
x 3x 2
x$ 2x#
x 3x 2
x$ 2x#
lim
œ lim c
xÄ#
lim
x Ä #b
(c)
x Ä 0c
(d)
x Ä "b
(e)
lim
x Ä !b x(x #)
x# 1
2x 4
lim
x Ä #c
œ _
positive
Š negative
‹
œ0
(x 2)(x 1)
x# (x 2)
œ _
(x 2)(x 1)
x# (x 2)
œ lim b
xÄ#
(x 2)(x 1)
x# (x 2)
œ lim c
xÄ#
œ lim
œ lim
(x 2)(x 1)
x# (x 2)
œ _
xÄ!
lim
x# 3x 2
x$ 4x
lim
x# 3x 2
x$ 4x
x"
2†0
#4
(x 2)(x 1)
x# (x 2)
xÄ#
œ lim b
xÄ#
x# 3x 2
x$ 4x
(b)
œ
œ lim
x# 3x 2
x$ 4x
lim
x Ä #b
and
œ lim b
xÄ#
x 3x 2
x$ 2x#
#
xÄ!
œ lim b
xÄ!
#
lim
x Ä #c
(x 1)(x 1)
2x 4
(b)
"
4
lim b
xÄ#
3
#
Š positive
positive ‹
œ lim b
xÄ"
xÄ!
œ 2"Î$ 2"Î$ œ 0
œ_
lim
x Ä "b 2x 4
57. (a)
"
#"Î$
ˆ "1 ‰ œ
(c)
x# 1
œ
xÄ#
(x 2)(x ")
x(x #)(x 2)
(x 2)(x ")
œ lim c
xÄ!
œ lim b
xÄ"
(x 2)(x ")
x(x #)(x 2)
(x 2)(x ")
x(x #)(x 2)
œ
x1
x#
œ
"
4
,xÁ2
x1
x#
œ
"
4
,xÁ2
x1
x#
œ
"
4
,xÁ2
†negative
Š negative
positive†negative ‹
œ lim b
xÄ#
lim
x Ä #b x(x #)(x 2)
†negative
Š negative
positive†negative ‹
(x 1)
x(x #)
œ
(x 1)
lim
x Ä #b x(x #)
œ lim c
xÄ!
œ lim b
xÄ"
œ_
(x 1)
x(x #)
œ
negative
Š positive
†positive ‹
x"
negative
Š negative
†positive ‹
œ_
œ
"
8
œ_
(x 1)
x(x #)
œ _
lim
x Ä !c x(x #)
"
#(4)
0
(1)(3)
negative
Š negative
†positive ‹
negative
Š negative
†positive ‹
œ0
so the function has no limit as x Ä 0.
lim 2 59. (a)
t Ä !b
60. (a)
t Ä !b
61. (a)
x Ä !b
(c)
x Ä "b
3 ‘
t"Î$
œ _
"
lim t$Î&
7‘ œ _
lim
2 lim
lim
"
’ x#Î$
2
“
(x 1)#Î$
œ_
lim
"
’ x#Î$
2
“
(x 1)#Î$
œ_
(b)
t Ä !c
(b)
t Ä !c
lim
"
’ x#Î$
2
“
(x 1)#Î$
œ_
(b)
x Ä !c
lim
"
’ x#Î$
2
“
(x 1)#Î$
œ_
(d)
x Ä "c
"
t$Î&
3 ‘
t"Î$
œ_
7‘ œ _
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 2.6 Limits Involving Infinity; Asymptotes of Graphs
lim
"
’ x"Î$
1
“
(x 1)%Î$
œ_
(b)
x Ä !c
lim
"
’ x"Î$
1
“
(x 1)%Î$
œ _
(d)
x Ä "c
62. (a)
x Ä !b
(c)
x Ä "b
lim
"
’ x"Î$
1
“
(x 1)%Î$
œ _
lim
"
’ x"Î$
1
“
(x 1)%Î$
œ _
63. y œ
"
x1
64. y œ
"
x1
65. y œ
"
#x 4
66. y œ
3
x3
67. y œ
x3
x2
68. y œ
2x
x1
œ1
"
x#
69. Here is one possibility.
œ#
2
x1
70. Here is one possibility.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
77
78
Chapter 2 Limits and Continuity
71. Here is one possibility.
72. Here is one possibility.
73. Here is one possibility.
74. Here is one possibility.
75. Here is one possibility.
76. Here is one possibility.
77. Yes. If x lim
Ä_
f(x)
g(x)
œ 2 then the ratio of the polynomials' leading coefficients is 2, so x Ä
lim
_
f(x)
g(x)
œ 2 as well.
78. Yes, it can have a horizontal or oblique asymptote.
79. At most 1 horizontal asymptote: If x lim
Ä_
f(x)
lim
x Ä _ g(x)
f(x)
g(x)
œ L, then the ratio of the polynomials' leading coefficients is L, so
œ L as well.
Èx 9 Èx 4
80. x lim
Š Èx 9 Èx 4‹ œ x lim
’Èx 9 Èx 4“ † ’ Èx 9 Èx 4 “ œ x lim
Ä_
Ä_
Ä_
5
Èx
5
0
œ x lim
œ x lim
œ 11 œ 0
9
4
Ä _ Èx 9 Èx 4
Ä_
ax 9 b a x 4 b
Èx 9 Èx 4
É1 x É1 x
È 2
È 2
81. x lim
Š Èx2 25 Èx2 "‹ œ x lim
’Èx2 25 Èx2 "“ † ’ Èx2 25 Èx2 " “ œ x lim
Ä_
Ä_
Ä_
x 25 x "
œ x lim
Ä_
26
Èx2 25 Èx2 "
œ x lim
Ä_
26
x
É1 x252 É1 x12
œ
0
11
œ0
È 2
ˆx 3 ‰ ˆ x ‰
x
82. x Ä
lim
lim
lim
Š Èx2 3 x‹ œ x Ä
’Èx2 3 x“ † ’ Èxx2 33 “œxÄ
_
_
_ Èx2 3 x
x
3
È x2
3x
3
œxÄ
lim
œ
lim
œ
lim
œ 1 0 1 œ 0
2
È
3
x
_
x Ä _ É1 2 È
x Ä _ É1 32 1
x 3x
x
x
x2
2
2
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
ˆx2 25‰ ˆx2 "‰
Èx2 25 Èx2 "
Section 2.6 Limits Involving Infinity; Asymptotes of Graphs
È
ˆ4x2 ‰ ˆ4x2 3x 2‰
4x 3x 2
83. x Ä
lim
lim
lim
Š 2x È4x2 3x 2‹ œ x Ä
’2x È4x2 3x 2“ † ’ 2x
“œxÄ
_
_
_ 2x È4x2 3x 2
2x È4x2 3x 2
c3x b 2
c3x b 2
3 x2
Èx2
3x 2
cx
œxÄ
lim
œxÄ
lim
œxÄ
lim
œxÄ
lim
_ 2x È4x2 3x 2
_ È2x É4 3 22
_ 2x É4 3 22
_ 2 É4 3 22
x
x
x
cx
x
x
x
x2
œ 3202 œ 43
2
È 2
84. x lim
Š È9x2 x 3x‹ œ œ x lim
’È9x2 x 3x“ † ’ È9x2 x 3x “ œ x lim
Ä_
Ä_
Ä_
9x x 3x
œ x lim
Ä_
x
È9x2 x 3x
œ x lim
Ä_
xx
2
É 9x2
x
xx2 3x
x
1
É9 "x 3
œ x lim
Ä_
œ
1
33
ˆ9x2 x‰ ˆ9x2 ‰
È9x2 x 3x
œ "6
È 2
È 2
85. x lim
Š Èx2 3x Èx2 2x‹ œ x lim
’ Èx2 3x Èx2 2x“ † ’ Èx2 3x Èx2 2x “ œ x lim
Ä_
Ä_
Ä_
x 3x x 2x
5x
5
5
5
œ x lim
œ x lim
œ 11 œ #
È 2
3
2
Ä_ È 2
Ä_
x 3x x 2x
É1 x É1 x
Èx# x Èx# x œ lim ’Èx# x Èx# x“ † ’ Èx# x Èx# x “ œ lim
86. x lim
È x# x È x# x
Ä_
xÄ_
xÄ_
2x
2
2
œ x lim
œ x lim
œ 11 œ 1
È #
"
"
Ä_ È #
Ä_
x x
ˆx2 3x‰ ˆx2 2x‰
Èx2 3x Èx2 2x
x x
ax # x b a x # x b
È x# x È x# x
É1 x É1 x
87. For any % 0, take N œ 1. Then for all x N we have that kf(x) kk œ kk kk œ 0 %.
88. For any % 0, take N œ 1. Then for all y N we have that kf(x) kk œ kk kk œ 0 %.
"
x#
89. For every real number B 0, we must find a $ 0 such that for all x, 0 kx 0k $ Ê
"
x#
Ê
B ! Í
"
x#
"
x#
#
B0 Í x "
B
"
ÈB
Í kxk . Choose $ œ
"
ÈB
, then 0 kxk $ Ê kxk xÄ!
B ! Í lxl B" . Choose $ œ B" . Then ! kx 0k $ Ê lxl "
B
Ê
"
lx l
"
lx l
2
(x 3)#
B ! Í
2
(x 3)#
$ œ É B2 , then 0 kx 3k $ Ê
B0 Í
2
(x 3)#
(x 3)
2
#
"
B
Í (x 3)# B 0 so that lim
2
#
x Ä $ (x 3)
2
B
B. Now,
x Ä ! lx l
2
(x 3)#
Now,
#
B ! Í (x 5) Ê kx 5k "
ÈB
Ê
"
(x 5)#
"
B
Í kx 5k B so that lim
"
"
ÈB
#
x Ä & (x 5)
. Choose $ œ
œ _.
B.
Í ! kB $k É B2 . Choose
œ _.
92. For every real number B 0, we must find a $ 0 such that for all x, 0 kx (5)k $ Ê
1
(x 5)#
"
B so that lim
91. For every real number B 0, we must find a $ 0 such that for all x, 0 kx 3k $ Ê
Now,
"
ÈB
B so that lim x"# œ _.
90. For every real number B 0, we must find a $ 0 such that for all x, ! kx 0k $ Ê
"
lx l
B. Now,
"
ÈB
1
(x 5)#
B.
. Then 0 kx (5)k $
œ _.
93. (a) We say that f(x) approaches infinity as x approaches x! from the left, and write lim c f(x) œ _, if
x Ä x!
for every positive number B, there exists a corresponding number $ 0 such that for all x,
x! $ x x! Ê f(x) B.
(b) We say that f(x) approaches minus infinity as x approaches x! from the right, and write lim b f(x) œ _,
x Ä x!
if for every positive number B (or negative number B) there exists a corresponding number $ 0 such
that for all x, x! x x! $ Ê f(x) B.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
79
80
Chapter 2 Limits and Continuity
(c) We say that f(x) approaches minus infinity as x approaches x! from the left, and write lim f(x) œ _,
x Ä x!
if for every positive number B (or negative number B) there exists a corresponding number $ 0 such
that for all x, x! $ x x! Ê f(x) B.
94. For B 0,
"
x
B 0 Í x B" . Choose $ œ B" . Then ! x $ Ê 0 x 95. For B 0,
"
x
B 0 Í x" B 0 Í x Ê B" x Ê
96. For B !,
"
x#
"
x
B so that lim c
xÄ!
"
x
"
B
Ê
"
x#
"
x#
Ê
99. y œ
101. y œ
"
x#
B ! so that lim b
xÄ#
œx1
x# %
x"
B so that lim b
xÄ!
"
x
œ _.
"
B
Í x 2 B" Í x 2 B" . Choose $ œ B" . Then
"
x#
B 0 so that lim c
xÄ#
"
x#
œ _.
B Í ! x 2 B" . Choose $ œ B" . Then # x # $ Ê ! x # $ Ê ! x 2 "
1 x#
"
x"
œx"
$
x"
œ _.
B Í 1 x# "
#B . Then " $ x " Ê
"
1 x# B for ! x 1 and
x#
x"
"
x
œ _.
B Í x " # B Í (x 2) 98. For B 0 and ! x 1,
$
Ê
Í B" x. Choose $ œ B" . Then $ x !
2 $ x 2 Ê $ x 2 ! Ê B" x 2 0 Ê
97. For B 0,
"
B
"
B
Í (" x)(" x) B" . Now
$ x 1 0 Ê " x $ x near 1 Ê
"
lim
#
x Ä "c " x
"
#B
1x
#
1 since x 1. Choose
Ê (" x)(" x) B" ˆ 1 # x ‰ B"
œ _.
100. y œ
x# "
x1
œx"
102. y œ
x2 "
#x %
œ #" x " #
x1
$
#x %
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
"
B
Section 2.6 Limits Involving Infinity; Asymptotes of Graphs
103. y œ
x# 1
x
105. y œ
x
È 4 x#
œx
107. y œ x#Î$ "
x
104. y œ
x$ 1
x#
106. y œ
"
È 4 x#
œx
"
x#
108. y œ sin ˆ x# 1 1 ‰
"
x"Î$
109. (a) y Ä _ (see accompanying graph)
(b) y Ä _ (see accompanying graph)
(c) cusps at x œ „ 1 (see accompanying graph)
110. (a) y Ä 0 and a cusp at x œ 0 (see the accompanying
graph)
(b) y Ä 32 (see accompanying graph)
(c) a vertical asymptote at x œ 1 and contains the point
Š1,
3
‹
3
2È
4
(see accompanying graph)
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
81
82
Chapter 2 Limits and Continuity
CHAPTER 2 PRACTICE EXERCISES
1. At x œ 1:
Ê
f(x) œ
lim
x Ä "c
lim
x Ä "b
f(x) œ 1
lim f(x) œ 1 œ f(1)
x Ä 1
Ê f is continuous at x œ 1.
At x œ 0: lim c f(x) œ lim b f(x) œ 0 Ê lim f(x) œ 0.
xÄ!
xÄ!
xÄ!
But f(0) œ 1 Á lim f(x)
xÄ!
Ê f is discontinuous at x œ 0.
If we define fa!b œ !, then the discontinuity at x œ ! is
removable.
At x œ 1: lim c f(x) œ 1 and lim f(x) œ 1
xÄ"
Ê lim f(x) does not exist
xÄ"
xÄ1
Ê f is discontinuous at x œ 1.
2. At x œ 1:
Ê
f(x) œ 0 and
lim
x Ä "
lim
x Ä "
f(x) œ 1
lim f(x) does not exist
x Ä "
Ê f is discontinuous at x œ 1.
At x œ 0: lim f(x) œ _ and lim f(x) œ _
xÄ!
Ê lim f(x) does not exist
xÄ!
xÄ!
Ê f is discontinuous at x œ 0.
At x œ 1: lim f(x) œ lim f(x) œ 1 Ê lim f(x) œ 1.
xÄ"
xÄ1
xÄ"
But f(1) œ 0 Á lim f(x)
xÄ1
Ê f is discontinuous at x œ 1.
If we define fa"b œ ", then the discontinuity at x œ " is
removable.
3. (a)
(b)
(c)
(d)
(e)
(f)
lim a3fatbb œ 3 lim fatb œ 3(7) œ 21
t Ä t!
t Ä t!
#
lim afatbb# œ Š lim fatb‹ œ a(b# œ 49
t Ä t!
t Ä t!
lim afatb † gatbb œ lim fatb † lim gatb œ (7)(0) œ 0
t Ä t!
t Ä t!
lim fatb
t Ä t! g(t)7
Ät
t Ä t!
lim fatb
œ
Ät
t
œ
!
lim agatb 7b
t
t
!
Ät
lim fatb
Ät
t
!
Ät
lim gatb lim 7
t
!
!
œ
7
07
œ1
lim cos agatbb œ cos Š lim gatb‹ œ cos ! œ 1
t Ä t!
t Ä t!
lim kfatbk œ ¹ lim fatb¹ œ k7k œ 7
t Ä t!
t Ä t!
(g) lim afatb gatbb œ lim fatb lim gatb œ 7 0 œ 7
t Ä t!
(h)
4. (a)
(b)
(c)
(d)
t Ä t!
lim Š " ‹
t Ä t! fatb
œ
"
lim fatb
t
Ät
t Ä t!
"
7
œ
!
œ 71
lim g(x) œ lim g(x) œ È2
xÄ!
xÄ!
lim ag(x) † f(x)b œ lim g(x) † lim f(x) œ ŠÈ2‹ ˆ "# ‰ œ
xÄ!
xÄ!
xÄ!
lim af(x) g(x)b œ lim f(x) lim g(x) œ
xÄ!
"
lim
x Ä ! f(x)
œ
"
lim f(x)
xÄ!
xÄ!
œ
"
"
#
œ2
xÄ!
"
#
È2
#
È2
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Chapter 2 Practice Exercises
(e)
(f)
"
#
lim ax f(x)b œ lim x lim f(x) œ 0 xÄ!
xÄ!
f(x)†cos x
x 1
xÄ!
lim
xÄ!
lim f(x)† lim cos x
œ
xÄ!
xÄ!
lim x lim 1
xÄ!
xÄ!
œ
ˆ "# ‰ (1)
01
œ
83
"
#
œ #"
5. Since lim x œ 0 we must have that lim (4 g(x)) œ 0. Otherwise, if lim (% g(x)) is a finite positive
xÄ!
xÄ!
xÄ!
’ 4xg(x) “
’ 4xg(x) “
œ _ and lim b
œ _ so the limit could not equal 1 as
xÄ!
x Ä 0. Similar reasoning holds if lim (4 g(x)) is a finite negative number. We conclude that lim g(x) œ 4.
number, we would have lim c
xÄ!
xÄ!
6. 2 œ lim
x Ä %
xÄ!
’x lim g(x)“ œ lim x † lim
xÄ!
x Ä %
’ lim g(x)“ œ 4 lim
xÄ!
x Ä %
(since lim g(x) is a constant) Ê lim g(x) œ
xÄ!
xÄ!
2
%
x Ä %
œ #" .
’ lim g(x)“ œ 4 lim g(x)
xÄ!
xÄ!
7. (a) xlim
faxb œ xlim
x"Î$ œ c"Î$ œ facb for every real number c Ê f is continuous on a_ß _b.
Äc
Äc
(b) xlim
gaxb œ xlim
x$Î% œ c$Î% œ gacb for every nonnegative real number c Ê g is continuous on Ò!ß _Ñ.
Äc
Äc
"
c#Î$
"
c"Î'
(c) xlim
haxb œ xlim
x#Î$ œ
Äc
Äc
(d) xlim
kaxb œ xlim
x"Î' œ
Äc
Äc
œ hacb for every nonzero real number c Ê h is continuous on a_ß !b and a_ß _b.
œ kacb for every positive real number c Ê k is continuous on a!ß _b
8. (a) - ˆˆn "# ‰1ß ˆn "# ‰1‰, where I œ the set of all integers.
n−I
(b) - an1ß an 1b1b, where I œ the set of all integers.
n−I
(c) a_ß 1b a1ß _b
(d) a_ß !b a!ß _b
9.
(a)
x# 4x 4
x2
lim
x Ä !c x(x 7)
(b)
10. (a)
#
x# x
œ lim
1
x# (x 1)
x# x
x # a#
x % a%
œ xlim
Äa
13. lim
(x h)# x#
h
œ lim
(x h)# x#
h
xÄ!
œ lim
xÄ!
x
œ _ and lim b
xÄ!
" Èx
ax # a # b
ax # a # b a x # a # b
hÄ!
œ lim
xÄ!
1
x# (x 1)
2 (2 x)
2x(# x)
œ lim
#
x x
"
"
œ
œ
œ
0
2(9)
œ0
, x Á 0 and x Á 1.
œ _.
, x Á 0 and x Á 1. The limit does not
1
lim
#
x Ä "b x (x 1)
"
#
x Ä 0 x (x 1)
&
%
$
x Ä ! x 2x x
#
x Ä " x (x 1)
"
x # a#
œ xlim
Äa
œ lim
œ _ Ê lim
œ lim
x2
x Ä # x(x 7)
x1
x Ä 1 1 Èx
ax# 2hx h# b x#
h
xÄ!
, x Á 2, and lim
#
x Ä ! x (x 1)(x 1)
œ lim
ax# 2hx h# b x#
h
x2
œ lim
œ _ and
x Ä 1 ˆ1 È x ‰ ˆ 1 È x ‰
, x Á 2; the limit does not exist because
x Ä # x(x 7)
x(x 1)
"
12. xlim
Äa
" "
x
#
x(x 1)
lim
#
x Ä "c x (x 1)
œ lim
#
œ lim
$
#
x Ä " x ax 2x 1b
1 Èx
1x
15. lim
(x 2)(x 2)
œ lim
11. lim
14. lim
œ _
$
#
x Ä ! x ax 2x 1b
lim
&
%
$
x Ä " x 2x x
hÄ!
x2
x(x 7)
x2
x Ä ! x(x 7)
x Ä # x(x 7)(x #)
œ lim
lim &
%
$
x Ä ! x 2x x
xÄ1
œ lim
œ _ and lim b
xÄ!
x 4x 4
exist because
(x 2)(x 2)
x Ä ! x(x 7)(x 2)
lim $
#
x Ä # x 5x 14x
Now lim c
xÄ!
(b)
œ lim
lim $
#
x Ä ! x 5x 14x
œ _.
"
#
"
#a #
œ lim (2x h) œ 2x
hÄ!
œ lim (2x h) œ h
xÄ!
"
x Ä ! 4 #x
œ "4
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
84
Chapter 2 Limits and Continuity
(# x)$ 8
x
16. lim
xÄ!
ax$ 6x# 12x 8b 8
x
œ lim
xÄ!
ˆx1Î3 1‰
x1Î3 1
œ lim ˆÈx 1‰
È
x
1
xÄ1
xÄ1
œ 1 1 1 1 1 œ 23
17. lim
18.
tan 2x
œ lim
sin 2x
x Ä ! cos 2x
lim csc x œ limc
x1
20.
x Ä 1c
21.
xÄ1
22.
xÄ1
ax 1bˆÈx 1‰
œ lim
2Î3
1Î3
x Ä 1 ax 1bax x 1b
Èx 1
œ lim
2Î3
1Î3
x Ä 1 x x 1
1
sin x
†
1x ‰ˆ 1x ‰ˆ 2x ‰
œ lim ˆ sin2x2x ‰ˆ cos
cos 2x
sin 1x
1x œ 1 † 1 † 1 †
cos 1x
sin 1x
xÄ!
2
1
œ
2
1
œ_
lim sin ˆ x2 sin x‰ œ sin ˆ 12 sin 1‰ œ sin ˆ 12 ‰ œ 1
lim cos2 ax tan xb œ cos2 a1 tan 1b œ cos2 a1b œ a1b2 œ 1
23. lim
xÄ0
24. lim
xÄ0
8x
3sin x x
œ lim
xÄ0
cos 2x 1
sin x
2x 1
œ lim ˆ cossin
†
x
xÄ0
8
3 sinx x 1
œ
8
3 a1 b 1
œ4
cos 2x 1 ‰
cos 2x 1
œ lim
xÄ0
"Î$
lim [4 g(x)]"Î$ œ 2 Ê ’ lim b 4 g(x)“
x Ä !b
xÄ!
lim
x Ä È&
27. lim
xÄ1
28.
ˆx2Î3 x1Î3 1‰ˆÈx 1‰
ˆÈx 1‰ax2Î3 x1Î3 1b
ˆx1Î3 4‰ˆx1Î3 4‰
ˆx1Î3 4‰ˆx1Î3 4‰ ˆx2Î3 4x1Î3 16‰ˆÈx )‰
x2Î3 16
œ lim
œ lim
† ˆÈx )‰ax2Î3 4x1Î3 16b
È
È
Èx 8
x
8
x
8
x Ä 64
x Ä 64
x Ä 64
ˆx1Î3 4‰ˆÈx )‰
ax 64bˆx1Î3 4‰ˆÈx )‰
4 4ba8 8b
8
œ lim ax 64bax2Î3 4x1Î3 16b œ lim x2Î3 4x1Î3 16 œ a16
16 16 œ 3
x Ä 64
x Ä 64
x Ä ! tan 1x
26.
xÄ!
lim
19. lim
25.
†
œ lim ax# 6x 12b œ 12
"
x g(x)
3x# 1
g(x)
xÄ1
5 x#
œ
(x g(x)) œ
lim
x Ä È&
œ2 Ê
"
#
œ lim
xÄ0
sin2 2x
sin xacos 2x 1b
œ lim
xÄ0
4sin x cos2 x
cos 2x 1
œ0 Ê
Ê È5 lim
x Ä È5
g(x) œ
"
#
Ê
lim
x Ä È5
g(x) œ
"
#
È5
xÄ1
lim g(x) œ _ since lim a5 x# b œ 1
x Ä #
x Ä #
lim f(x) œ lim c
x Ä "c
x Ä "
lim
x Ä "c
x ax # 1 b
x# 1
œ
lim
x Ä "c
x ax # 1 b
kx # 1 k
x œ 1, and
#
lim f(x) œ lim b xkaxx# 11k b œ lim b
x Ä "b
x Ä "
x Ä "
œ lim (x) œ (1) œ 1. Since
x ax # 1 b
a x # "b
x Ä 1
lim f(x) Á lim b f(x)
x Ä "c
x Ä "
Ê
lim f(x) does not exist, the function f cannot be
x Ä 1
extended to a continuous function at x œ 1.
At x œ 1:
lim f(x) œ lim c
x Ä "c
xÄ"
x ax # 1 b
kx # 1 k
x ax # 1 b
kx # 1 k
œ lim c
xÄ"
x ax # 1 b
x# "
x ax # 1 b
ax # 1 b
œ lim c (x) œ 1, and
xÄ"
lim f(x) œ lim b
œ lim b
œ lim b x œ 1. Again lim f(x) does not exist so f
xÄ1
xÄ"
xÄ"
xÄ1
cannot be extended to a continuous function at x œ 1 either.
x Ä "b
œ
4a0ba1b2
11
lim 4 g(x) œ 8, since 2$ œ 8. Then lim b g(x) œ 2.
xÄ!
x Ä !b
œ _ Ê lim g(x) œ 0 since lim a3x# 1b œ 4
lim
x Ä # Èg(x)
29. At x œ 1:
œ2 Ê
cos2 2x 1
sin xacos 2x 1b
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
œ0
Chapter 2 Practice Exercises
30. The discontinuity at x œ 0 of f(x) œ sin ˆ "x ‰ is nonremovable because lim sin
xÄ!
"
x
does not exist.
31. Yes, f does have a continuous extension to a œ 1:
"
define f(1) œ lim xxÈ
œ 43 .
%
x
xÄ1
32. Yes, g does have a continuous extension to a œ 1# :
)
5
g ˆ 1# ‰ œ lim1 45)cos
#1 œ 4 .
)Ä #
33. From the graph we see that lim h(t) Á lim h(t)
tÄ!
tÄ!
so h cannot be extended to a continuous function
at a œ 0.
34. From the graph we see that lim c k(x) Á lim b k(x)
xÄ!
xÄ!
so k cannot be extended to a continuous function
at a œ 0.
35. (a) f(1) œ 1 and f(2) œ 5 Ê f has a root between 1 and 2 by the Intermediate Value Theorem.
(b), (c) root is 1.32471795724
36. (a) f(2) œ 2 and f(0) œ 2 Ê f has a root between 2 and 0 by the Intermediate Value Theorem.
(b), (c) root is 1.76929235424
# $
#x $
x
37. x lim
œ x lim
œ
Ä _ &x (
Ä _ & (x
x
39. x Ä
lim
_
#
%x )
$x $
#!
&!
ˆ" œxÄ
lim
_ $x
œ
#
#
&
%
$x#
#
#x $
38. x Ä
lim
œxÄ
lim
_ &x# (
_ & ) ‰
$x$
$
x#
(
x#
œ
#!
&!
œ!!!œ!
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
œ
#
&
85
86
Chapter 2 Limits and Continuity
"
"
x#
40. x lim
œ x lim
œ
Ä _ x # (x "
Ä _ " (x x"#
!
"!!
œ!
#
%
x (x
x(
41. x Ä
lim
œxÄ
lim
œ _
_ x 1
_ " "x
$
x x
x"
42. x lim
œxÄ
lim
œ_
Ä _ "#x$ "#)
_ "# "#)
x$
sin x
"
sin x
43. x lim
Ÿ x lim
œ ! since int x Ä _ as x Ä _ Êx lim
œ !.
Ä _ gx h
Ä _ gx h
Ä _ gx h
44.
lim
)Ä_
45. x lim
Ä_
cos ) "
)
Ÿ lim
#
) Ä _)
x sin x #Èx
x sin x
#Î$
œ ! Ê lim
)Ä_
œ x lim
Ä_
cos ) "
)
" sinx x È#x
" sinx x
"
œ
&Î$
x x
" x
46. x lim
œ x lim
#x œ
Ä _ x#Î$ cos# x
Ä _Œ " cos#Î$
œ !.
"!!
"!
"!
"!
œ"
œ"
x
47. (a) y œ
(b) y œ
x2 4
x3
is undefined at x œ 3: lim c xx 34 œ _ and lim b xx 34 œ _, thus x œ 3 is a vertical asymptote.
xÄ3
xÄ3
2
x2 x 2
x2 2x 1
2
is undefined at x œ 1: lim c
xÄ1
x2 x 2
x2 2x 1
œ _ and lim b
xÄ1
x2 x 2
x2 2x 1
œ _, thus x œ 1 is a vertical
asymptote.
(c) y œ
x2 x '
x2 2x 8
is undefined at x œ 2 and 4: lim
x x'
2
œ lim b
x Ä %
lim
2
x Ä %b x 2x 8
48. (a) y œ
1 x2
1 x2
x2 " : x lim
Ä _ x2 "
x3
x4
x2 x '
2
x Ä 2 x 2x 8
x3
œ lim
x Ä 2 x4
œ 56 ; lim c
x Ä %
x2 x '
x2 2x 8
œ lim c
x Ä %
x3
x4
œ_
œ _. Thus x œ 4 is a vertical asymptote.
1
1
x2
œ x lim
Ä _ 1
1
x2
œ
1
1
1
1
1x
œ 1 and x Ä
lim
œ lim x2
œ
_ x2 " x Ä _ 1 x12
œ
10
È1 0
2
1
1
œ 1, thus y œ 1 is a
horizontal asymptote.
(b) y œ
Èx 4
Èx 4
Èx 4 : x lim
Ä _ Èx 4
(c) y œ
È x2 4
È x2 4
: x lim
x
x
Ä_
œx Ä
lim
_
É1 x42
1
œ
œ x lim
Ä_
È1 0
1
1 È4x
É1 B4
œ x lim
Ä_
œ
1
1
É1 x42
1
œ
œ 1 , thus y œ 1 is a horizontal asymptote.
È1 0
1
œ 1 and x lim
Ä _
thus y œ
1
2
"
3
œx Ä
lim_
É1 x42
È
9
"
3
x
xx
2
is a horizontal asymptote.
0.1
0.7943
œx Ä
lim
_
É1 x42
x
cx
0.01
0.9550
0.001
0.9931
0.0001
0.9991
1
9
x 9
0
"
x2
É 9x
and x Ä
lim
lim
œ É 19 21 œ
0 œ 3,
_
x Ä _ Ê 9 x12
CHAPTER 2 ADDITIONAL AND ADVANCED EXERCISES
1. (a)
x
x2
œ 1, thus y œ 1 and y œ 1 are horizontal asymptotes.
x 9
x 9
0
x2
(d) y œ É 9x
lim É 9x
lim
œ É 91 21:
21 œ
0 œ
xÄ_
x Ä _ Ê 9 x12
2
È x2 4
x
0.00001
0.9999
Apparently, lim b xx œ 1
xÄ!
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Chapter 2 Additional and Advanced Exercises 87
(b)
2. (a)
x
ˆ "x ‰"ÎÐln xÑ
Apparently,
10
100
1000
0.3679
0.3679
0.3679
"ÎÐln xÑ
lim ˆ " ‰
xÄ_ x
œ 0.3678 œ
"
e
(b)
3.
lim L œ lim c L! É" v Ä cc
vÄc
v#
c#
lim v
œ L! É1 vÄcc # œ L! É1 #
c#
c#
œ0
The left-hand limit was needed because the function L is undefined if v c (the rocket cannot move faster
than the speed of light).
4. (a) ¹
Èx
#
1¹ 0.2 Ê 0.2 Èx
#
1 0.2 Ê 0.8 Èx
#
1.2 Ê 1.6 Èx 2.4 Ê 2.56 x 5.76.
(b) ¹
Èx
#
1¹ 0.1 Ê 0.1 Èx
#
1 0.1 Ê 0.9 Èx
#
1.1 Ê 1.8 Èx 2.2 Ê 3.24 x 4.84.
5. k10 (t 70) ‚ 10% 10k 0.0005 Ê k(t 70) ‚ 10% k 0.0005 Ê 0.0005 (t 70) ‚ 10% 0.0005
Ê 5 t 70 5 Ê 65° t 75° Ê Within 5° F.
6. We want to know in what interval to hold values of h to make V satisfy the inequality
lV "!!!l œ l$'1h "!!!l Ÿ "!. To find out, we solve the inequality:
**!
l$'1h "!!!l Ÿ "! Ê "! Ÿ $'1h "!!! Ÿ "! Ê **! Ÿ $'1h Ÿ "!"! Ê $'
1 Ÿ hŸ
"!"!
$'1
Ê )Þ) Ÿ h Ÿ )Þ*. where 8.8 was rounded up, to be safe, and 8.9 was rounded down, to be safe.
The interval in which we should hold h is about )Þ* )Þ) œ !Þ" cm wide (1 mm). With stripes 1 mm wide, we can expect
to measure a liter of water with an accuracy of 1%, which is more than enough accuracy for cooking.
7. Show lim f(x) œ lim ax# 7b œ ' œ f(1).
xÄ1
xÄ1
Step 1: kax 7b 6k % Ê % x# 1 % Ê 1 % x# 1 % Ê È1 % x È1 %.
#
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
88
Chapter 2 Limits and Continuity
Step 2: kx 1k $ Ê $ x 1 $ Ê $ " x $ ".
Then $ " œ È1 % or $ " œ È1 %. Choose $ œ min š1 È1 %ß È1 % 1› , then
0 kx 1k $ Ê kax# (b 6k % and lim f(x) œ 6. By the continuity test, f(x) is continuous at x œ 1.
xÄ1
8. Show lim" g(x) œ lim"
xÄ
xÄ
%
"
2x
œ 2 œ g ˆ 4" ‰ .
%
Step 1: ¸ #"x 2¸ % Ê % #"x # % Ê # % #"x # % Ê
Step 2: ¸B "4 ¸ $ Ê $ x 4" $ Ê $ 4" x $ 4" .
Then $ Choose $ œ
"
4
œ
"
4 #%
%
4(#%)
Ê $œ
"
4
"
4 #%
œ
%
4(2 %)
, or $ "
4
œ
, the smaller of the two values. Then 0 ¸x
By the continuity test, g(x) is continuous at x œ
"
4
"
4 #% Ê
4" ¸ $
"
4#%
x
"
4 #%
¸ #"x "
4#%
.
"
4
%
4(2 %)
$œ
œ
Ê
2¸ % and lim"
.
xÄ
%
"
#x
œ 2.
.
9. Show lim h(x) œ lim È2x 3 œ " œ h(2).
xÄ#
xÄ#
Step 1: ¹È2x 3 1¹ % Ê % È2x 3 " % Ê " % È2x 3 " % Ê
(1 %)# $
#
x
(" %)# 3
.
#
Step 2: kx 2k $ Ê $ x 2 $ or $ # x $ #.
(" % )# $
Ê $œ
#
(" % Ñ # $
(" %Ñ# "
#œ
#
#
Then $ # œ
#
Ê $œ
œ%
#
(" %)# $
œ " (1# %)
#
#
%# . Choose $ œ %
œ%
%#
#,
%#
#
, or $ # œ
(" %)# $
#
the smaller of the two values . Then,
! kx 2k $ Ê ¹È2x 3 "¹ %, so lim È2x 3 œ 1. By the continuity test, h(x) is continuous at x œ 2.
xÄ#
10. Show lim F(x) œ lim È9 x œ # œ F(5).
xÄ&
xÄ&
Step 1: ¹È9 x 2¹ % Ê % È9 x # % Ê 9 (2 %)# x * (# %)# .
Step 2: 0 kx 5k $ Ê $ x & $ Ê $ & x $ &.
Then $ & œ * (# %)# Ê $ œ (# %)# % œ %# #%, or $ & œ * (# %)# Ê $ œ % (# %)# œ %# #%.
Choose $ œ %# #%, the smaller of the two values. Then, ! kx 5k $ Ê ¹È9 x #¹ %, so
lim È9 x œ #. By the continuity test, F(x) is continuous at x œ 5.
xÄ&
11. Suppose L" and L# are two different limits. Without loss of generality assume L# L" . Let % œ
"
3
(L# L" ).
Since x lim
f(x) œ L" there is a $" 0 such that 0 kx x! k $" Ê kf(x) L" k % Ê % f(x) L" %
Äx
!
Ê "3 (L# L" ) L" f(x) "
3
(L# L" ) L" Ê 4L" L# 3f(x) 2L" L# . Likewise, x lim
f(x) œ L#
Ä x!
so there is a $# such that 0 kx x! k $# Ê kf(x) L# k % Ê % f(x) L# %
Ê "3 (L# L" ) L# f(x) 3" (L# L" ) L# Ê 2L# L" 3f(x) 4L# L"
Ê L" 4L# 3f(x) 2L# L" . If $ œ min e$" ß $# f both inequalities must hold for 0 kx x! k $ :
4L" L# 3f(x) 2L" L#
Ê 5(L" L# ) 0 L" L# . That is, L" L# 0 and L" L# 0,
L" %L# 3f(x) 2L# L" a contradiction.
12. Suppose xlim
f(x) œ L. If k œ !, then xlim
kf(x) œ xlim
0 œ ! œ ! † xlim
f(x) and we are done.
Äc
Äc
Äc
Äc
%
If k Á 0, then given any % !, there is a $ ! so that ! lx cl $ Ê lfaxb Ll l5l
Ê lkllfaxb Ll %
Ê lkafaxb Lb| % Ê lakfaxbb akLbl %. Thus, xlim
kf(x) œ kL œ kŠxlim
f(x)‹.
Äc
Äc
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Chapter 2 Additional and Advanced Exercises 89
13. (a) Since x Ä 0 , 0 x$ x 1 Ê ax$ xb Ä 0 Ê
lim f ax$ xb œ lim c f(y) œ B where y œ x$ x.
yÄ!
x Ä !b
(b) Since x Ä 0 , 1 x x$ 0 Ê ax$ xb Ä 0 Ê
(c) Since x Ä 0 , 0 x% x# 1 Ê ax# x% b Ä 0 Ê
lim f ax$ xb œ lim b f(y) œ A where y œ x$ x.
yÄ!
x Ä !c
lim f ax# x% b œ lim b f(y) œ A where y œ x# x% .
yÄ!
x Ä !b
(d) Since x Ä 0 , 1 x 0 Ê ! x% x# 1 Ê ax# x% b Ä 0 Ê
lim f ax# x% b œ A as in part (c).
x Ä !b
14. (a) True, because if xlim
(f(x) g(x)) exists then xlim
(f(x) g(x)) xlim
f(x) œ xlim
[(f(x) g(x)) f(x)]
Äa
Äa
Äa
Äa
œ xlim
g(x) exists, contrary to assumption.
Äa
"
x
(b) False; for example take f(x) œ
and g(x) œ x" . Then neither lim f(x) nor lim g(x) exists, but
xÄ!
lim (f(x) g(x)) œ lim ˆ "x x" ‰ œ lim 0 œ 0 exists.
xÄ!
xÄ!
xÄ!
xÄ!
(c) True, because g(x) œ kxk is continuous Ê g(f(x)) œ kf(x)k is continuous (it is the composite of continuous
functions).
1, x Ÿ 0
Ê f(x) is discontinuous at x œ 0. However kf(x)k œ 1 is
(d) False; for example let f(x) œ œ
1, x 0
continuous at x œ 0.
x# "
15. Show lim f(x) œ lim
x Ä 1 x 1
x Ä 1
(x 1)(x ")
(x 1)
œ lim
x Ä 1
Define the continuous extension of f(x) as F(x) œ œ
œ #, x Á 1.
x# 1
x1 ,
2
x Á "
. We now prove the limit of f(x) as x Ä 1
, x œ 1
exists and has the correct value.
#
Step 1: ¹ xx 1" (#)¹ % Ê % (x 1)(x ")
(x 1)
# % Ê % (x 1) # %, x Á " Ê % " x % ".
Step 2: kx (1)k $ Ê $ x 1 $ Ê $ " x $ ".
Then $ " œ % " Ê $ œ %, or $ " œ % " Ê $ œ %. Choose $ œ %. Then ! kx (1)k $
#
Ê ¹ xx 1" a#b¹ % Ê
lim F(x) œ 2. Since the conditions of the continuity test are met by F(x), then f(x) has a
x Ä 1
continuous extension to F(x) at x œ 1.
16. Show lim g(x) œ lim
xÄ$
xÄ$
x# 2x 3
2x 6
œ lim
xÄ$
(x 3)(x ")
2(x 3)
œ #, x Á 3.
#
Define the continuous extension of g(x) as G(x) œ œ
x 2x 3
2x 6 ,
2
xÁ3
. We now prove the limit of g(x) as
, xœ3
x Ä 3 exists and has the correct value.
Step 1: ¹ x
#
2x 3
#x 6
2¹ % Ê % (x 3)(x ")
2(x 3)
# % Ê % x"
#
# % , x Á $ Ê $ #% x $ #% .
Step 2: kx 3k $ Ê $ x 3 $ Ê $ $ x $ $.
Then, $ $ œ $ #% Ê $ œ #%, or $ $ œ $ #% Ê $ œ #%. Choose $ œ #%. Then ! kx 3k $
Ê ¹x
#
2x 3
2x 6
2¹ % Ê lim
xÄ$
(x 3)(x ")
#(x 3)
œ 2. Since the conditions of the continuity test hold for G(x),
g(x) can be continuously extended to G(x) at B œ 3.
17. (a) Let % ! be given. If x is rational, then f(x) œ x Ê kf(x) 0k œ kx 0k % Í kx 0k %; i.e., choose
$ œ %. Then kx 0k $ Ê kf(x) 0k % for x rational. If x is irrational, then f(x) œ 0 Ê kf(x) 0k %
Í ! % which is true no matter how close irrational x is to 0, so again we can choose $ œ %. In either case,
given % ! there is a $ œ % ! such that ! kx 0k $ Ê kf(x) 0k %. Therefore, f is continuous at
x œ 0.
(b) Choose x œ c !. Then within any interval (c $ ß c $ ) there are both rational and irrational numbers.
If c is rational, pick % œ #c . No matter how small we choose $ ! there is an irrational number x in
(c $ ß c $ ) Ê kf(x) f(c)k œ k0 ck œ c c
#
œ %. That is, f is not continuous at any rational c 0. On
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
90
Chapter 2 Limits and Continuity
the other hand, suppose c is irrational Ê f(c) œ 0. Again pick % œ #c . No matter how small we choose $ !
there is a rational number x in (c $ ß c $ ) with kx ck œ kxk c
#
œ% Í
œ % Ê f is not continuous at any irrational c 0.
If x œ c 0, repeat the argument picking % œ
nonzero value x œ c.
18. (a) Let c œ
c
#
kc k
#
œ
c
# .
x
c
#
Then kf(x) f(c)k œ kx 0k
3c
#.
Therefore f fails to be continuous at any
m
n
be a rational number in [0ß 1] reduced to lowest terms Ê f(c) œ "n . Pick % œ
"
#n
œ %. Therefore f is discontinuous at x œ c, a rational number.
"
#n .
No matter how
small $ ! is taken, there is an irrational number x in the interval (c $ ß c $ ) Ê kf(x) f(c)k œ ¸0 "n ¸
œ
"
n
(b) Now suppose c is an irrational number Ê f(c) œ 0. Let % 0 be given. Notice that
number reduced to lowest terms with denominator 2 and belonging to [0ß 1];
denominator 3 belonging to [0ß 1];
"
4
and
[0ß 1]; etc. In general, choose N so that
"
N
3
4
with denominator 4 in [0ß 1];
"
3
and
" 2 3
5, 5, 5
2
3
and
"
#
is the only rational
the only rationals with
4
5
with denominator 5 in
% Ê there exist only finitely many rationals in [!ß "] having
denominator Ÿ N, say r" , r# , á , rp . Let $ œ min ekc ri k : i œ 1ß á ß pf . Then the interval (c $ ß c $ )
contains no rational numbers with denominator Ÿ N. Thus, 0 kx ck $ Ê kf(x) f(c)k œ kf(x) 0k
œ kf(x)k Ÿ N" % Ê f is continuous at x œ c irrational.
(c) The graph looks like the markings on a typical ruler
when the points (xß f(x)) on the graph of f(x) are
connected to the x-axis with vertical lines.
19. Yes. Let R be the radius of the equator (earth) and suppose at a fixed instant of time we label noon as the
zero point, 0, on the equator Ê 0 1R represents the midnight point (at the same exact time). Suppose x"
is a point on the equator “just after" noon Ê x" 1R is simultaneously “just after" midnight. It seems
reasonable that the temperature T at a point just after noon is hotter than it would be at the diametrically
opposite point just after midnight: That is, T(x" ) T(x" 1R) 0. At exactly the same moment in time
pick x# to be a point just before midnight Ê x# 1R is just before noon. Then T(x# ) T(x# 1R) 0.
Assuming the temperature function T is continuous along the equator (which is reasonable), the Intermediate
Value Theorem says there is a point c between 0 (noon) and 1R (simultaneously midnight) such that
T(c) T(c 1R) œ 0; i.e., there is always a pair of antipodal points on the earth's equator where the
temperatures are the same.
#
#
#
#
"
20. xlim
f(x)g(x) œ xlim
af(x) g(x)b‹ Šxlim
af(x) g(x)b‹ “
’af(x) g(x)b af(x) g(x)b “ œ "% ’Šxlim
Äc
Äc %
Äc
Äc
œ "% ˆ$# a"b# ‰ œ #.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Chapter 2 Additional and Advanced Exercises 91
21. (a) At x œ 0: lim r (a) œ lim
aÄ!
œ lim
1 (" a)
aÄ!
a Ä ! a ˆ" È1 a‰
At x œ 1:
(b) At x œ 0:
lim
a Ä "b
œ
r (a) œ
" È1 a
a
1
" È1 0
œ lim c
aÄ!
1 (" a)
a ˆ" È1 a‰
"
#
œ
aÄ!
1 (1 a)
lim
a Ä "b a ˆ1 È1 a‰
lim r (a) œ lim c
aÄ!
a Ä !c
È1 a
œ lim Š " a
" È1 a
a
œ lim c
aÄ!
a
œ lim
a Ä 1 a ˆ" È1 a‰
È1 a
œ lim c Š " a
aÄ!
a
a ˆ 1 È 1 a ‰
" È1 a
‹ Š " È1 a ‹
œ lim c
aÄ!
œ
"
" È0
œ1
" È1 a
‹ Š " È1 a ‹
"
œ _ (because the
" È1 a
"
œ _ (because the
" È1 a
denominator is always negative); lim b r (a) œ lim b
aÄ!
aÄ!
is always positive). Therefore, lim r (a) does not exist.
denominator
aÄ!
At x œ 1:
lim
a Ä "b
r (a) œ
lim
a Ä "b
1 È 1 a
a
œ
lim
"
a Ä 1b " È1 a
œ1
(c)
(d)
22. f(x) œ x 2 cos x Ê f(0) œ 0 2 cos 0 œ 2 0 and f(1) œ 1 2 cos (1) œ 1 # 0. Since f(x) is
continuous on [1ß !], by the Intermediate Value Theorem, f(x) must take on every value between [1 #ß #].
Thus there is some number c in [1ß !] such that f(c) œ 0; i.e., c is a solution to x 2 cos x œ 0.
23. (a) The function f is bounded on D if f(x) M and f(x) Ÿ N for all x in D. This means M Ÿ f(x) Ÿ N for all x
in D. Choose B to be max ekMk ß kNkf . Then kf(x)k Ÿ B. On the other hand, if kf(x)k Ÿ B, then
B Ÿ f(x) Ÿ B Ê f(x) B and f(x) Ÿ B Ê f(x) is bounded on D with N œ B an upper bound and
M œ B a lower bound.
(b) Assume f(x) Ÿ N for all x and that L N. Let % œ L # N . Since x lim
f(x) œ L there is a $ ! such that
Äx
!
0 kx x! k $ Ê kf(x) Lk % Í L % f(x) L % Í L Í
LN
#
f(x) 3L N
# .
But L N Ê
LN
#
LN
#
f(x) L LN
#
N Ê N f(x) contrary to the boundedness assumption
f(x) Ÿ N. This contradiction proves L Ÿ N.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
92
Chapter 2 Limits and Continuity
(c) Assume M Ÿ f(x) for all x and that L M. Let % œ
Ê L
ML
#
f(x) L b, then a b
24. (a) If a
ML
#
3L M
#
Í
f(x)
ML
# . As in part (b), 0 kx L
M
M, a contradiction.
#
0 Ê ka bk œ a b Ê max Öaß b× œ
ab
#
ka b k
#
If a Ÿ b, then a b Ÿ 0 Ê ka bk œ (a b) œ b a Ê max Öaß b× œ
œ
2b
#
xÄ0
ab
#
sina" cos xb
x
œ lim
lim b sinsinÈxx œ
xÄ0
†
sin x
xÄ0
sinasin xb
x
xÄ0
sinax# xb
x
xÄ0
œ lim
29. lim
sinax# %b
x2
œ lim
30. lim
sinˆÈx $‰
x9
28. lim
xÄ2
xÄ9
sin x
" cos x
xÄ0
œ lim
ka b k
#
.
sina" cos xb
" cos x
lim b sinB x †
xÄ0
27. lim
œ lim
xÄ0 x
26.
ab
ab
2a
# # œ # œ a.
ka b k
ab
œ a # b b # a
# #
œ
œ b.
(b) Let min Öaß b× œ
25. lim œ
x! k $
sinasin xb
sin x
†
" cos x
x
Èx
sin Èx
†
†
œ lim
sin x
x
x
Èx
xÄ0
sinasin xb
sin x
sinax# %b
x# %
† ax 2b œ lim
xÄ9
xÄ0
sinˆÈx $‰
Èx $
† lim
" cos# x
x Ä 0 xa" cos xb
"
Èx $
† lim
sin x
xÄ0 x
œ " † lim
sin# x
x Ä 0 xa" cos xb
œ " † " œ ".
sinax# xb
#
x Ä 0 x x
† lim ax "b œ " † " œ "
sinax# %b
x# %
† lim ax 2b œ " † % œ %
xÄ2
†
sina" cos xb
" cos x
œ " † lim b sin"Èx † lim b Èx œ " † ! † ! œ !.
x Ä 0 Š Èx ‹ x Ä 0
† ax "b œ lim
œ lim
œ lim
œ " † ˆ #! ‰ œ !.
sinax# xb
#
x Ä 0 x x
xÄ2
" cos x
" cos x
†
œ lim
xÄ9
xÄ0
xÄ2
sinˆÈx $‰
Èx $
† lim
"
x Ä 9 Èx $
œ"†
"
'
œ
"
'
31. Since the highest power of x in the numerator is 1 more than the highest power of x in the denominator, there is an oblique
asymptote. y œ
32. As x Ä „ _,
2x3Î2 2x 3
Èx 1
1
x
œ 2x 3
Èx 1 ,
thus the oblique asymptote is y œ 2x.
Ä 0 Ê sinˆ 1x ‰ Ä 0 Ê 1 sinˆ 1x ‰ Ä 1, thus as x Ä „ _, y œ x x sinˆ 1x ‰ œ xˆ1 sinˆ 1x ‰‰ Ä x;
thus the oblique asymptote is y œ x.
33. As x Ä „ _, x2 1 Ä x2 Ê Èx2 1 Ä Èx2 ; as x Ä _, Èx2 œ x, and as x Ä _, Èx2 œ x; thus the
oblique asymptotes are y œ x and y œ x.
34. As x Ä „ _, x 2 Ä x Ê Èx2 2x œ Èxax 2b Ä Èx2 ; as x Ä _, Èx2 œ x, and as x Ä _, Èx2 œ x;
asymptotes are y œ x and y œ x.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
CHAPTER 3 DIFFERENTIATION
3.1 TANGENTS AND THE DERIVATIVE AT A POINT
1. P" : m" œ 1, P# : m# œ 5
2. P" : m" œ 2, P# : m# œ 0
3. P" : m" œ 5# , P# : m# œ "#
4. P" : m" œ 3, P# : m# œ 3
5. m œ lim
hÄ!
c4 (" h)# d a4 (1)# b
h
a1 2h h# b1
h
hÄ!
œ lim
œ lim
hÄ!
h(# h)
h
œ 2;
at ("ß $): y œ $ #(x (1)) Ê y œ 2x 5,
tangent line
6. m œ lim
hÄ!
c(1 h 1)# 1d c(" ")# 1d
h
h#
œ lim
hÄ! h
œ lim h œ 0; at ("ß "): y œ 1 0(x 1) Ê y œ 1,
hÄ!
tangent line
È
2È 1 h 2È 1
œ lim 2 1 h h 2
h
hÄ!
hÄ!
4(1 h) 4
œ lim
œ lim È1 2h 1
h Ä ! 2h ŠÈ1 h 1‹
hÄ!
7. m œ lim
†
2È 1 h 2
2È 1 h #
œ 1;
at ("ß #): y œ 2 1(x 1) Ê y œ x 1, tangent line
8. m œ lim
hÄ!
"
( 1 h)#
( "")#
h
a2h h# b
#
h Ä ! h(1 h)
œ lim
1 (1 h)#
#
h Ä ! h(1h)
2h
lim
# œ 2;
h Ä ! (1 h)
œ lim
œ
at ("ß "): y œ 1 2(x (1)) Ê y œ 2x 3,
tangent line
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
94
Chapter 3 Differentiation
(2 h)$ (2)$
h
9. m œ lim
hÄ!
8 12h 6h# h$ 8
h
œ lim
hÄ!
œ lim a12 6h h# b œ 12;
hÄ!
at (2ß 8): y œ 8 12(x (2)) Ê y œ 12x 16,
tangent line
10. m œ lim
(
h
hÄ!
œ
"2
8(8)
hÄ!
hÄ!
at ˆ#ß "8 ‰ : y œ 8" Ê yœ
11. m œ lim
hÄ!
x
"
#,
12 6h h#
8(2 h)$
œ lim
3
œ 16
;
3
16
8 (# h)$
8h(# h)$
œ lim
a12h 6h# h$ b
8h(# h)$
œ lim
hÄ!
"
"
# h)$ ( #)$
3
16 (x
(2))
tangent line
c(2 h)# 1d 5
h
œ lim
hÄ!
a5 4h h# b 5
h
hÄ!
at (2ß 5): y 5 œ 4(x 2), tangent line
12. m œ lim
hÄ!
c(" h) 2(1 h)# d (1)
h
œ lim
hÄ!
h(4 h)
h
œ lim
a1 h 2 4h 2h# b 1
h
3
(3
h
h) 2
3
h
hÄ!
œ lim
hÄ!
(3 h) 3(h 1)
h(h 1)
h Ä ! h(h 1)
at ($ß $): y 3 œ 2(x 3), tangent line
14. m œ lim
hÄ!
8
(2
h)#
2
h
hÄ!
(2 h)$ 8
h
hÄ!
œ lim
a8 12h 6h# h$ b 8
h
hÄ!
œ lim
16. m œ lim
hÄ!
c(1 h)$ 3(1 h)d 4
h
hÄ!
at ("ß %): y 4 œ 6(t 1), tangent line
17. m œ lim
hÄ!
È4 h 2
h
œ lim
hÄ!
œ "4 ; at (%ß #): y 2 œ
18. m œ lim
hÄ!
œ
"
È9 3
È(8 h) 1 3
h
"
4
È4 h 2
h
†
hÄ!
h a12 6h h# b
h
hÄ!
a1 3h 3h# h$ 3 3hb 4
h
œ lim
œ lim
œ lim
at (2ß )): y 8 œ 12(t 2), tangent line
È4 h 2
È4 h 2
œ 3;
œ 2;
8 2 a4 4h h# b
h(2 h)#
hÄ!
8 2(2 h)#
#
h Ä ! h(2 h)
œ lim
at (2ß 2): y 2 œ 2(x 2)
15. m œ lim
2h
œ lim
h(3 2h)
h
œ lim
at ("ß "): y 1 œ 3(x 1), tangent line
13. m œ lim
œ %;
œ lim
2h(4 h)
h(2 h)#
œ
8
4
œ 2;
œ 12;
œ lim
hÄ!
(4 h) 4
h Ä ! h ŠÈ4 h #‹
h a6 3h h# b
h
œ lim
œ 6;
h
h Ä ! h ŠÈ4 h #‹
œ
"
È4 #
(x 4), tangent line
œ lim
hÄ!
È9 h 3
h
œ 6" ; at (8ß 3): y 3 œ
19. At x œ 1, y œ 5 Ê m œ lim
hÄ!
"
6
†
È9 h 3
È9 h 3
œ lim
(9 h) 9
h Ä ! h ŠÈ9 h 3‹
œ lim
h
h Ä ! h ŠÈ9 h 3‹
(x 8), tangent line
5(" h)# 5
h
œ lim
hÄ!
5 a1 2h h# b 5
h
œ lim
hÄ!
5h(2 h)
h
œ 10, slope
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 3.1 Tangents and the Derivative at a Point
c1 (2 h)# d (3)
h
20. At x œ 2, y œ 3 Ê m œ lim
hÄ!
"
#
21. At x œ 3, y œ
Ê m œ lim
"
h) 1
(3
#"
h1
h 1
22. At x œ 0, y œ 1 Ê m œ lim
hÄ!
hÄ!
2 (2 h)
2h(2 h)
œ lim
h
hÄ!
œ lim
hÄ!
(1)
h
a1 4 4h h# b 3
h
hÄ!
hÄ!
h
œ lim
œ lim
h(4 h)
h
œ 4, slope
œ "4 , slope
h Ä ! 2h(2 h)
(h 1) (h ")
h(h 1)
œ lim
œ lim
2h
h Ä ! h(h 1)
œ 2, slope
c(x h)# 4(x h) 1d ax# 4x 1b
h
hÄ!
a2xh h# 4hb
lim
œ lim (2x h 4) œ 2x
h
hÄ!
hÄ!
23. At a horizontal tangent the slope m œ 0 Ê 0 œ m œ lim
ax# 2xh h# 4x 4h 1b ax# 4x 1b
h
hÄ!
œ lim
œ
4;
2x 4 œ 0 Ê x œ 2. Then f(2) œ 4 8 1 œ 5 Ê (2ß 5) is the point on the graph where there is a
horizontal tangent.
c(x h)$ 3(x h)d ax$ 3xb
h
24. 0 œ m œ lim
hÄ!
œ lim
hÄ!
3x# h 3xh# h$ 3h
h
ax$ 3x# h 3xh# h$ 3x 3hb ax$ 3xb
h
œ lim
hÄ!
œ lim a3x# 3xh h# 3b œ 3x# 3; 3x# 3 œ 0 Ê x œ 1 or x œ 1. Then
hÄ!
f(1) œ 2 and f(1) œ 2 Ê ("ß 2) and ("ß 2) are the points on the graph where a horizontal tangent exists.
25. 1 œ m œ lim
"
h) 1
(x
x " 1
h
hÄ!
œ lim
hÄ!
(x 1) (x h 1)
h(x 1)(x h 1)
h
œ lim
h Ä ! h(x 1)(x h 1)
œ (x " 1)#
Ê (x 1)# œ 1 Ê x# 2x œ 0 Ê x(x 2) œ 0 Ê x œ 0 or x œ 2. If x œ 0, then y œ 1 and m œ 1
Ê y œ 1 (x 0) œ (x 1). If x œ 2, then y œ 1 and m œ 1 Ê y œ 1 (x 2) œ (x 3).
26.
"
4
Èx h Èx
œ m œ lim
œ lim
h
y œ 2 "4 (x 4) œ
hÄ!
f(2 h) f(2)
h
x
4
Èx h Èx
h
hÄ!
h Ä ! h ŠÈx h Èx‹
27. lim
œ lim
h
hÄ!
œ
"
#È x
. Thus,
"
4
œ
†
Èx h Èx
Èx h Èx
"
#Èx
(x h) x
œ lim
h Ä ! h ŠÈx h Èx‹
Ê Èx œ 2 Ê x œ 4 Ê y œ 2. The tangent line is
1.
œ lim
hÄ!
a100 4.9(# h)# b a100 4.9(2)# b
h
4.9 a4 4h h# b 4.9(4)
h
œ lim
hÄ!
œ lim (19.6 4.9h) œ 19.6. The minus sign indicates the object is falling downward at a speed of 19.6 m/sec.
hÄ!
28. lim
hÄ!
f(10 h) f(10)
h
hÄ!
1(3 h)# 1(3)#
h
hÄ!
f(3 h) f(3)
h
hÄ!
œ lim
f(2 h) f(2)
h
hÄ!
œ lim
29. lim
30. lim
3(10 h)# 3(10)#
h
œ lim
hÄ!
41
3
3 a20h h# b
h
œ lim
hÄ!
œ 60 ft/sec.
1 c9 6h h# 9d
h
hÄ!
œ lim
(2 h)$ 431 (2)$
h
œ lim
41
3
hÄ!
31. At ax0 , mx0 bb the slope of the tangent line is lim
hÄ!
œ lim 1(6 h) œ 61
hÄ!
c12h 6h# h$ d
h
œ lim
hÄ!
amax0 hb bb am x0 bb
ax 0 h b x 0
41
3
c12 6h h# d œ 161
œ lim
hÄ!
mh
h
The equation of the tangent line is y am x0 bb œ max x0 b Ê y œ mx b.
32. At x œ 4, y œ
1
È4
È
œ
"
#
œ lim – 22hÈ44hh †
hÄ!
and m œ lim
hÄ!
2 È4 h
2 È4 h —
È4
1
h
h
"#
œ lim –
hÄ!
È4
1
h
h
"#
†
2È 4 h
2È 4 h —
œ lim m œ m.
hÄ!
È
œ lim Š 22hÈ44hh ‹
hÄ!
hb
h
œ lim È 4 a4 È
œ lim È
2h 4 hŠ2 4 h‹ 2h 4 hŠ2 È4 h‹ hÄ!
hÄ!
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
95
96
Chapter 3 Differentiation
1
1
œ lim È
œ È 1 È œ 16
2 4 hŠ2 È4 h‹ 2 4Š2 4‹
hÄ!
f(0 h) f(0)
h
hÄ!
33. Slope at origin œ lim
h# sin ˆ "h ‰
h
hÄ!
œ lim
œ lim h sin ˆ "h ‰ œ 0 Ê yes, f(x) does have a tangent at
hÄ!
the origin with slope 0.
g(0 h) g(0)
h
34. lim
hÄ!
œ lim
hÄ!
h sin ˆ "h ‰
h
œ lim sin h" . Since lim sin
hÄ!
hÄ!
"
h
does not exist, f(x) has no tangent at
the origin.
35.
lim
h Ä !c
f(0 h) f(0)
h
lim f(0 h)h f(0)
hÄ!
36.
œ lim c
hÄ!
1 0
h
œ _, and lim b
hÄ!
f(0 h) f(0)
h
10
h
œ lim b
hÄ!
œ _ Ê yes, the graph of f has a vertical tangent at the origin.
œ _, and lim b U(0 h)h U(0) œ lim b
hÄ!
hÄ!
does not have a vertical tangent at (!ß ") because the limit does not exist.
lim
h Ä !c
œ _. Therefore,
U(0 h) U(0)
h
œ lim c
hÄ!
01
h
11
h
œ 0 Ê no, the graph of f
37. (a) The graph appears to have a cusp at x œ 0.
(b)
lim c
hÄ!
f(0 h) f(0)
h
œ lim c
hÄ!
h#Î& 0
h
œ lim c
hÄ!
"
h$Î&
œ _ and lim b
hÄ!
"
h$Î&
œ _ Ê limit does not exist
Ê the graph of y œ x#Î& does not have a vertical tangent at x œ 0.
38. (a) The graph appears to have a cusp at x œ 0.
(b)
lim
h Ä !c
f(0 h) f(0)
h
œ lim c
hÄ!
h%Î& 0
h
œ lim c
hÄ!
"
h"Î&
œ _ and lim b
hÄ!
"
h"Î&
œ _ Ê limit does not exist
Ê y œ x%Î& does not have a vertical tangent at x œ 0.
39. (a) The graph appears to have a vertical tangent at x œ !.
(b)
lim
hÄ!
f(0 h) f(0)
h
œ lim
hÄ!
h"Î& 0
h
œ lim
"
%Î&
hÄ! h
œ _ Ê y œ x"Î& has a vertical tangent at x œ 0.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 3.1 Tangents and the Derivative at a Point
40. (a) The graph appears to have a vertical tangent at x œ 0.
(b)
lim
hÄ!
f(0 h) f(0)
h
œ lim
hÄ!
h$Î& 0
h
"
œ lim
#Î&
hÄ! h
œ _ Ê the graph of y œ x$Î& has a vertical tangent at x œ 0.
41. (a) The graph appears to have a cusp at x œ 0.
(b)
lim c
hÄ!
f(0 h) f(0)
h
œ lim c
hÄ!
4h#Î& 2h
h
œ lim c
hÄ!
4
h$Î&
2 œ _ and lim b
hÄ!
4
h$Î&
#œ_
Ê limit does not exist Ê the graph of y œ 4x#Î& 2x does not have a vertical tangent at x œ 0.
42. (a) The graph appears to have a cusp at x œ 0.
(b)
lim
hÄ!
f(0 h) f(0)
h
œ lim
hÄ!
h&Î$ 5h#Î$
h
œ lim h#Î$ hÄ!
5
h"Î$
œ 0 lim
y œ x&Î$ 5x#Î$ does not have a vertical tangent at x œ !.
5
"Î$
hÄ! h
does not exist Ê the graph of
43. (a) The graph appears to have a vertical tangent at x œ 1
and a cusp at x œ 0.
(b) x œ 1:
lim
hÄ!
(1 h)#Î$ (1 h 1)"Î$ "
h
œ lim
hÄ!
(1 h)#Î$ h"Î$ "
h
œ _
Ê y œ x#Î$ (x 1)"Î$ has a vertical tangent at x œ 1;
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
97
98
Chapter 3 Differentiation
x œ 0:
lim
hÄ!
f(0 h) f(0)
h
œ lim
hÄ!
h#Î$ (h 1)"Î$ (1)"Î$
h
"
œ lim ’ h"Î$
hÄ!
(h ")"Î$
h
h" “
does not exist Ê y œ x#Î$ (x 1)"Î$ does not have a vertical tangent at x œ 0.
44. (a) The graph appears to have vertical tangents at x œ 0 and
x œ 1.
(b) x œ 0:
lim
hÄ!
f(0 h) f(0)
h
œ lim
hÄ!
h"Î$ (h 1)"Î$ (")"Î$
h
œ _ Ê y œ x"Î$ (x 1)"Î$ has a
vertical tangent at x œ 0;
x œ 1:
lim
hÄ!
f(1 h) f(1)
h
œ lim
hÄ!
(1 h)"Î$ (" h 1)"Î$ 1
h
œ _ Ê y œ x"Î$ (x 1)"Î$ has a
vertical tangent at x œ ".
45. (a) The graph appears to have a vertical tangent at x œ 0.
(b)
lim b
hÄ!
f(0 h) f(0)
h
œ lim b
xÄ!
Èh 0
h
œ lim
"
h Ä ! Èh
È kh k 0
f(0 h) f(0)
h
œ lim c
œ lim c
h
hÄ!
hÄ!
Ê y has a vertical tangent at x œ 0.
lim
h Ä !c
œ _;
È kh k
kh k
œ lim c
hÄ!
"
È kh k
œ_
46. (a) The graph appears to have a cusp at x œ 4.
(b)
lim b
f(4 h) f(4)
h
œ lim b
hÄ!
Èk4 (4 h)k 0
h
lim
f(4 h) f(4)
h
œ lim c
Èk4 (4 h)k
h
hÄ!
h Ä !c
hÄ!
œ lim b
hÄ!
œ lim c
hÄ!
È kh k
h
È kh k
lhl
œ lim b
hÄ!
œ lim c
hÄ!
"
Èh
"
È kh k
œ _;
œ _
Ê y œ È% x does not have a vertical tangent at x œ 4.
47-50. Example CAS commands:
Maple:
f := x -> x^3 + 2*x;x0 := 0;
plot( f(x), x=x0-1/2..x0+3, color=black,
title="Section 3.1, #47(a)" );
q := unapply( (f(x0+h)-f(x0))/h, h );
# part (a)
# part (b)
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 3.2 The Derivative as a Function
L := limit( q(h), h=0 );
# part (c)
sec_lines := seq( f(x0)+q(h)*(x-x0), h=1..3 );
# part (d)
tan_line := f(x0) + L*(x-x0);
plot( [f(x),tan_line,sec_lines], x=x0-1/2..x0+3, color=black,
linestyle=[1,2,5,6,7], title="Section 3.1, #47(d)",
legend=["y=f(x)","Tangent line at x=0","Secant line (h=1)",
"Secant line (h=2)","Secant line (h=3)"] );
Mathematica: (function and value for x0 may change)
Clear[f, m, x, h]
x0 œ p;
f[x_]: œ Cos[x] 4Sin[2x]
Plot[f[x], {x, x0 1, x0 3}]
dq[h_]: œ (f[x0+h] f[x0])/h
m œ Limit[dq[h], h Ä 0]
ytan: œ f[x0] m(x x0)
y1: œ f[x0] dq[1](x x0)
y2: œ f[x0] dq[2](x x0)
y3: œ f[x0] dq[3](x x0)
Plot[{f[x], ytan, y1, y2, y3}, {x, x0 1, x0 3}]
3.2 THE DERIVATIVE AS A FUNCTION
1. Step 1: f(x) œ 4 x# and f(x h) œ 4 (x h)#
f(x h) f(x)
h
Step 2:
œ
c4 (x h)# d a4 x# b
h
œ
a4 x# 2xh h# b 4 x#
h
œ
2xh h#
h
œ
h(2x h)
h
œ 2x h
Step 3: f w (x) œ lim (2x h) œ 2x; f w ($) œ 6, f w (0) œ 0, f w (1) œ 2
hÄ!
2. F(x) œ (x 1)# 1 and F(x h) œ (x h 1)# " Ê Fw (x) œ lim
œ lim
hÄ!
hÄ!
ax# 2xh h# 2x 2h 1 1b ax# 2x 1 1b
h
w
w
œ lim
w
œ 2(x 1); F (1) œ 4, F (0) œ 2, F (2) œ 2
3. Step 1: g(t) œ
"
t#
and g(t h) œ
"
Step 2:
"
# #
g(t h) g(t)
œ (t h)h t
h
2t h)
2t h
œ h(
(t h)# t# h œ (t h)# t#
Step 3: gw (t) œ lim
2t h
# #
h Ä ! (t h) t
4. k(z) œ
1 z
#z
and k(z h) œ
œ
œ
"
2z#
œ
Œ
2t
t# †t#
1 (z h)
2(z h)
(" z)(z h)
lim (1 z h)z
#(z h)zh
hÄ!
2xh h# 2h
h
œ lim (2x h 2)
hÄ!
"
(t h)#
œ
œ
hÄ!
c(x h 1)# 1d c(x 1)# 1d
h
t# (t h)#
(t h)# †t# h
œ
2
t$
t# at# 2th h# b
(t h)# †t# †h
œ
œ
2th h#
(t h)# t# h
2
; gw (1) œ 2, gw (2) œ "4 , gw ŠÈ3‹ œ 3È
3
Ê kw (z) œ lim
hÄ!
Š
"
(z h)
" z
#(z h) #z ‹
#
z h z# zh
lim z z zh
2(z h)zh
hÄ!
h
œ lim
h
h Ä ! 2(z h)zh
œ lim
"
h Ä ! #(z h)z
; kw (") œ "# , kw (1) œ "# , kw ŠÈ2‹ œ 4"
5. Step 1: p()) œ È3) and p() h) œ È3() h)
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
99
100
Chapter 3 Differentiation
Step 2:
p() h) p())
h
œ
œ
È3() h) È3)
h
3h
h ŠÈ3) 3h È3)‹
Step 3: pw ()) œ lim
œ
ŠÈ3) 3h È3)‹
h
œ
3
È3) 3h È3)
3
œ
h Ä ! È3) 3h È3)
†
œ
3
È 3) È 3)
3
2È 3 )
ŠÈ3) 3h È3)‹
ŠÈ3) 3h È3)‹
; pw (1) œ
œ
(3) 3h) 3)
h ŠÈ3) 3h È3)‹
, pw (3) œ "# , pw ˆ 32 ‰ œ
3
2È 3
3
#È2
È2s 2h 1 È2s 1
h
hÄ!
6. r(s) œ È2s 1 and r(s h) œ È2(s h) 1 Ê rw (s) œ lim
œ lim
ŠÈ2s h 1 È2s 1‹
h
hÄ!
œ lim
ŠÈ2s 2h 1 È2s 1‹
†
œ lim
2h
œ
"
È2s 1
; rw (0) œ 1, rw (1) œ
2
"
È3
, rw ˆ #" ‰ œ
hÄ!
6x# h 6xh# 2h$
h
hÄ!
9. s œ r(t) œ
œ lim
t
2t1
Š
dr
ds
œ
2
2È2s 1
2 ax$ 3x# h 3xh# h$ b 2x$
h
hÄ!
2(x h)$ 2x$
h
hÄ!
œ lim
œ lim
œ lim a6x# 6xh 2h# b œ 6x#
hÄ!
œ lim
th
2(th)1
and r(t h) œ
(t b h)(2t b 1) c t(2t b 2h b 1)
‹
(2t b 2h b 1)(2t b 1)
h
hÄ!
œ lim
œ
2t# t 2ht h 2t# 2ht t
(2t 2h 1)(2t 1)h
hÄ!
"
"
(2t 1)(2t 1) œ (2t 1)#
dv
dt
œ lim
hÄ!
œ lim
10.
2
È2s 1 È2s 1
3
2
2
3
2
2
3
2
ˆas hb3 2as hb2 3‰ ˆs3 2s2 3‰
œ lim s 3s h 3sh h 2s h 4sh h 3 s 2s 3
h
hÄ!
hÄ!
2
2
3
2
hˆ3s2 3sh h2 4s h‰
lim 3s h 3sh hh 4sh h œ lim
œ lim a3s2 3sh h2 4s hb œ 3s2 2s
h
hÄ!
hÄ!
hÄ!
8. r œ s3 2s2 3 Ê
œ
dy
dx
h a6x# 6xh 2h# b
h
œ lim
œ
"
È2
7. y œ f(x) œ 2x$ and f(x h) œ 2(x h)$ Ê
œ lim
h Ä ! h ŠÈ2s 2h 1 È2s 1‹
h Ä ! È2s 2h 1 È2s 1
h Ä ! h ŠÈ2s 2h 1 È2s 1‹
(2s 2h 1) (2s 1)
œ lim
ŠÈ2s 2h 1 È2s 1‹
’(t h) hÄ!
ht# h# t h
h(t h)t
hÄ!
œ lim
Œ
" “ ˆt " ‰
t
h
h
œ lim
11. p œ f(q) œ
t
"
Èq 1
h
hÄ!
(t h)(2t 1) t(2t 2h 1)
(2t 2h 1)(2t 1)h
h
h Ä ! (2t 2h 1)(2t 1)h
œ lim
hÄ!
t# ht 1
h Ä ! (t h)t
and f(q h) œ
œ lim
œ
h
t
" "
t
h
h
t# 1
t#
"
È(q h) 1
Ê
h Ä ! (2t 2h 1)(2t 1)
Š
œ lim
h(t
h)t t (t
(t h)t
dp
dq
h)
‹
h
hÄ!
œ1
"
œ lim
"
t#
œ lim
Š È(q
hÄ!
"
h)
1
‹ Š Èq"
1
‹
h
Èq 1 Èq h 1
h Ä ! hÈ q h 1 È q 1
h
hÄ!
t ‰
Š 2(t bt bh)hb 1 ‹ ˆ 2t b
1
œ lim
ds
dt
œ lim
œ lim
Èq b 1 c Èq b h b 1
Èq b h b 1 Èq b 1 Ê
œ
ˆÈ q 1 È q h 1 ‰ ˆ È q 1 È q h 1 ‰
1) (q h 1)
† ˆÈq 1 Èq h 1‰ œ lim hÈq h 1(qÈq 1 ˆÈ q 1 È q h 1 ‰
h Ä ! h Èq h 1 Èq 1
hÄ!
h
"
lim
œ lim Èq h 1 Èq 1 ˆÈq 1 Èq h 1‰
h Ä ! h È q h 1 È q 1 ˆÈ q 1 È q h 1 ‰
hÄ!
"
"
œ
È q 1 È q 1 ˆÈ q 1 È q 1 ‰
2(q 1) Èq 1
dz
dw
œ lim
œ lim
œ
12.
Š È3(w " h) 2 h
hÄ!
œ lim
hÄ!
"
È3w 2 ‹
ŠÈ3w 2 È3w 3h 2‹
hÈ3w 3h 2 È3w 2
œ lim
œ lim
È3w 2 È3w 3h 2
h Ä ! hÈ3w 3h 2 È3w 2
†
ŠÈ3w2È3w3h2‹
ŠÈ3w 2 È3w 3h 2‹
3
h Ä ! È3w 3h 2 È3w 2 ŠÈ3w 2 È3w 3h 2‹
œ
œ
œ lim
(3w 2) (3w 3h 2)
h Ä ! hÈ3w 3h 2 È3w 2 ŠÈ3w 2 È3w 3h 2‹
3
È3w 2 È3w 2 ŠÈ3w 2 È3w 2‹
3
2(3w 2) È3w 2
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 3.2 The Derivative as a Function
13. f(x) œ x œ
x(x h)# 9x x# (x h) 9(x h)
x(x h)h
œ
h(x# xh 9)
x(x h)h
"
#x
14. k(x) œ
w
hÄ!
œ
œ
16.
dy
dx
œ
œ
x# 9
x#
œ
9
9
(x b h) “ ’x x “
’(x h) h
x$ 2x# h xh# 9x x$ x# h 9x 9h
x(x h)h
œ
x# xh 9
h Ä ! x(x h)
and k(x h) œ
x# h xh# 9h
x(x h)h
œ
œ1
9
x#
; m œ f w (3) œ 0
Š # "x h k(x h) k(x)
œ
lim
h
h
hÄ!
hÄ!
h
"
"
lim
œ lim (2 x)(# x h) œ (2 x)# ;
h Ä ! h(2 x)(2 x h)
hÄ!
"
2 (x h)
Ê kw (x) œ lim
$
#
#
$
#
#
$
#
c(t h)$ (t h)# d at$ t# b
œ lim at 3t h 3th h b h at 2th h b t t
h
hÄ!
hÄ!
#
#
#
#
$
#
lim 3t h 3th hh 2th h œ lim h a3t 3th h h 2t hb œ lim a3t# 3th hÄ!
hÄ!
hÄ!
¸
3t# 2t; m œ ds
œ
5
dt tœ"
#
" ‹
x
œ lim
ax b hb b 3
1 c ax b hb
œ lim
b3
1x c
x
h
hÄ!
œ lim
17. f(x) œ
4h
b h b 3ba1 c xb c ax b 3ba1 c x c hb
a1 c x c hba1 c xb
h
œ lim
8 ŠÈx 2 Èx h 2‹
hÈ x h 2 È x 2
4
†
8
È(x h) 2
ŠÈx 2 Èx h 2‹
œ
8h
hÈx h 2 Èx 2 ŠÈx 2 Èx h 2‹
œ
8
Èx 2 Èx 2 ŠÈx 2 Èx 2‹
œ
œ
œ
h# 2t hb
x h 3 x2 xh 3x x 3 x2 3x xh 3h
h a1 x h b a 1 x b
4
; dy ¹
a1 xb2 dx xœ2
f(x h) f(x)
h
Ê
ŠÈx 2 Èx h 2‹
œ lim
hÄ!
h Ä ! a1 x hba1 xb
and f(x h) œ
8
Èx 2
ax
œ lim
hÄ!
h Ä ! ha1 x hba1 xb
œ
f(x h) f(x)
h
Ê
9
(x h)
; f w (x) œ lim
"
16
k (2) œ ds
dt
x# xh 9
x(x h)
œ
(# x) (2 x h)
h(2 x)(2 x h)
œ lim
15.
and f(x h) œ (x h) 9
x
œ
œ
4
a3 b 2
4
9
È(x b h) c 2 Èx c 2
8
œ
8
h
8[(x 2) (x h 2)]
hÈx h 2 Èx 2 ŠÈx 2 Èx h 2‹
8
Ê f w (x) œ lim
h Ä ! Èx h 2 Èx 2 ŠÈx 2 Èx h 2‹
4
(x 2)Èx 2
; m œ f w (6) œ
4
4È 4
œ "# Ê the equation of the tangent
line at (6ß 4) is y 4 œ "# (x 6) Ê y œ "# x $ % Ê y œ "# x (.
ˆ1 È4 (z h)‰ Š1 È4 z‹
18. gw (z) œ lim
h
hÄ!
œ
h
hÄ!
(4 z h) (4 z)
lim
h Ä ! h ŠÈ4 z h È4 z‹
"
œ "#
2È 4 3
"# z $# # Ê w
ŠÈ4 z h È4 z‹
œ lim
œ
h
lim
h Ä ! h ŠÈ4 z h È4 z‹
†
ŠÈ4 z h È4 z‹
ŠÈ4 z h È4 z‹
"
œ lim
h Ä ! ŠÈ4 z h È4 z‹
œ
"
2È 4 z
m œ gw (3) œ
Ê the equation of the tangent line at ($ß #) is w 2 œ "# (z 3)
Êwœ
œ "# z (# .
19. s œ f(t) œ 1 3t# and f(t h) œ 1 3(t h)# œ 1 3t# 6th 3h# Ê
a1 3t# 6th 3h# b a1 3t# b
h
hÄ!
œ lim
20. y œ f(x) œ " œ lim
" "
x
21. r œ f()) œ
hÄ!
h
h
hÄ!
œ lim
x
"
x
2
È4 )
œ lim (6t 3h) œ 6t Ê
hÄ!
and f(x h) œ 1 œ lim
h
h Ä ! x(x h)h
œ
œ lim
Ê
dy
dx
"
h Ä ! x(x h)
œ
œ lim
"
x#
"
3
Ê
È œ
dy
dx ¹x= 3
2
È4 () h)
Ê
dr
d)
œ lim
hÄ!
f(t h) f(t)
h
œ6
f(x h) f(x)
h
hÄ!
œ lim
œ lim
Š1 x
" ‹ Š1 " ‹
h
x
h
hÄ!
f() h) f())
œ lim
h
hÄ!
hÄ!
È
È
2È4 ) #È% ) h Š2 % ) 2 4 ) h‹
lim
† È
Š2 4 ) #È4 ) h‹
h Ä ! hÈ 4 ) È 4 ) h
and f() h) œ
2È 4 ) 2È 4 ) h
hÈ 4 ) È 4 ) h
"
xh
ds ¸
dt t=c"
ds
dt
È4 c ) c h È4 c )
2
2
h
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
;
101
102
Chapter 3 Differentiation
4(% )) 4(% ) h)
œ lim
h Ä ! 2hÈ4 ) È4 ) h ŠÈ4 ) È4 ) h‹
œ
2
(4 )) Š2È4 )‹
œ
"
(4 ))È4 )
Ê
dr ¸
d) )œ!
œ lim
2
h Ä ! È4 ) È4 ) h ŠÈ4 ) È% ) h‹
œ
"
8
22. w œ f(z) œ z Èz and f(z h) œ (z h) Èz h Ê
œ lim
Šz h Èz h‹ ˆz Èz‰
hÄ!
h
œ 1 lim
(z h) z
h Ä ! h ŠÈz h Èz‹
h Èz h Èz
h
hÄ!
œ lim
œ 1 lim
"
h Ä ! Èz h Èz
"
dw
dz
œ lim
hÄ!
œ lim –1 hÄ!
œ"
"
2È z
Ê
f(z h) f(z)
h
Èz h Èz
h
dw ¸
dz zœ4
œ
†
ŠÈz h Èz‹
ŠÈz h Èz‹ —
5
4
"
fazb faxb
a x #b a z # b
xz
"
z #x #
23. f w axb œ zlim
œ zlim
œ zlim
œ zlim
œ zlim
œ
Äx zx
Ä x zx
Ä x az xbaz #bax #b
Ä x az xbaz #bax #b
Ä x az #bax #b
ˆz2 3z 4‰ ˆx2 3x 4‰
"
ax #b #
fazb faxb
z 3z x 3x
z x 3z 3x
24. f w axb œ zlim
œ zlim
œ zlim
œ zlim
zx
zx
zx
Ä x zx
Äx
Äx
Äx
az xbaz xb 3‘
az xbaz xb 3az xb
az xb 3‘ œ 2x 3
œ zlim
œ zlim
œ zlim
zx
zx
Äx
Äx
Äx
z
2
2
2
2
x
gazb gaxb
z a x "b x a z " b
z x
"
zc" x "
25. gw axb œ zlim
œ zlim
œ zlim
œ zlim
œ zlim
œ
Äx zx
Äx zx
Ä x az xbaz "bax "b
Ä x az xbaz "bax "b
Ä x az "bax "b
g az b g a x b
26. gw axb œ zlim
œ zlim
Äx zx
Äx
ˆ" Èz‰ˆ" Èx‰
zx
œ zlim
Äx
Èz Èx
zx
†
Èz Èx
Èz Èx
"
a x "b #
zx
"
œ zlim
œ zlim
œ
Ä x az x bˆÈ z È x ‰
Ä x Èz Èx
"
#È x
27. Note that as x increases, the slope of the tangent line to the curve is first negative, then zero (when x œ 0),
then positive Ê the slope is always increasing which matches (b).
28. Note that the slope of the tangent line is never negative. For x negative, f#w (x) is positive but decreasing as x increases.
When x œ 0, the slope of the tangent line to x is 0. For x 0, f#w (x) is positive and increasing. This graph matches (a).
29. f$ (x) is an oscillating function like the cosine. Everywhere that the graph of f$ has a horizontal tangent we expect f$w to be
zero, and (d) matches this condition.
30. The graph matches with (c).
31. (a) f w is not defined at x œ 0, 1, 4. At these points, the left-hand and right-hand derivatives do not agree.
For example, lim c
xÄ!
f(x) f(0)
x0
œ slope of line joining (%ß 0) and (!ß #) œ
"
#
but lim b
xÄ!
line joining (0ß 2) and ("ß 2) œ 4. Since these values are not equal, f w (0) œ
f(x) f(0)
x0
f(0)
lim f(x)x 0
xÄ!
(b)
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
œ slope of
does not exist.
Section 3.2 The Derivative as a Function
32. (a)
103
(b) Shift the graph in (a) down 3 units
33.
(b) The fastest is between the 20th and 30th days;
slowest is between the 40th and 50th days.
34. (a)
35. Answers may vary. In each case, draw a tangent line and estimate its slope.
‰F
(a) i) slope ¸ 1.54 Ê dT
ii) slope ¸ 2.86 Ê
dt ¸ 1.54 hr
iii) slope ¸ 0 Ê
dT
dt
¸ 0‰ hrF
iv) slope ¸ 3.75
dT
‰F
dt ¸ 2.86 hr
‰F
Ê dT
dt ¸ 3.75 hr
(b) The tangent with the steepest positive slope appears to occur at t œ 6 Ê 12 p.m. and slope ¸ 7.27 Ê
The tangent with the steepest negative slope appears to occur at t œ 12 Ê 6 p.m. and
‰F
slope ¸ 8.00 Ê dT
dt ¸ 8.00 hr
(c)
36. Answers may vary. In each case, draw a tangent line and estimate the slope.
lb
(a) i) slope ¸ 20.83 Ê dW
ii) slope ¸ 35.00 Ê
dt ¸ 20.83 month
iii) slope ¸ 6.25 Ê
dW
dt
dW
dt
lb
¸ 35.00 month
lb
¸ 6.25 month
(b) The tangentwith the steepest positive slope appears to occur at t œ 2.7 months. and slope ¸ 7.27
lb
Ê dW
dt ¸ 53.13 month
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
dT
dt
¸ 7.27‰ hrF .
104
Chapter 3 Differentiation
(c)
37. Left-hand derivative: For h 0, f(0 h) œ f(h) œ h# (using y œ x# curve) Ê
œ lim c
hÄ!
h# 0
h
œ lim c h œ 0;
hÄ!
Right-hand derivative: For h 0, f(0 h) œ f(h) œ h (using y œ x curve) Ê
œ lim b
hÄ!
Then lim c
hÄ!
h0
h
lim
h Ä !c
œ lim b 1 œ 1;
hÄ!
f(0 h) f(0)
h
Á lim b
hÄ!
f(0 h) f(0)
h
lim
h Ä !b
œ lim c 0 œ 0;
hÄ!
f(1 h) f(1)
h
lim
h Ä !c
Right-hand derivative: When h !, 1 h 1 Ê f(1 h) œ 2(1 h) œ 2 2h Ê
Then lim c
hÄ!
(2 2h)2
h
œ lim b
hÄ!
f(1 h) f(1)
h
2h
h
hÄ!
È1 h "
h
œ lim c
lim
h Ä !b
ŠÈ1 h "‹
h
hÄ!
†
ŠÈ1 h "‹
ŠÈ1 h 1‹
œ lim c
hÄ!
lim
h Ä !c
Then lim c
hÄ!
(2h 1) "
h
f(1 h) f(1)
h
40. Left-hand derivative:
lim
h Ä !c
Right-hand derivative:
œ lim b
hÄ!
Then lim c
hÄ!
h
h(1 h)
œ lim b 2 œ 2;
hÄ!
f(1 h) f(")
h
lim b
hÄ!
œ lim b
hÄ!
f(1 h) f(1)
h
f(1 h) f(1)
h
Á lim b
hÄ!
(1 h) "
h ŠÈ1 h "‹
œ lim c
hÄ!
Á lim b
hÄ!
"
È1 h 1
lim
h Ä !b
Ê the derivative f w (1) does not exist.
(1 h) "
h
œ lim c
hÄ!
f(1 h) f(")
h
"
1h
f(1 h) f(1)
h
f(" h) f(1)
h
Right-hand derivative: When h 0, 1 h 1 Ê f(1 h) œ 2(1 h) 1 œ 2h 1 Ê
œ lim b
hÄ!
œ lim c 1 œ 1;
hÄ!
Š 1 " h "‹
œ lim b
hÄ!
h
œ lim b
hÄ!
Š
1 (1 h)
1 h ‹
h
œ 1;
f(1 h) f(1)
h
Ê the derivative f w (1) does not exist.
41. f is not continuous at x œ 0 since lim faxb œ does not exist and fa0b œ 1
xÄ!
42. Left-hand derivative:
Right-hand derivative:
Then lim c
hÄ!
g(h) g(0)
h
lim
h Ä !c
g(h) g(0)
h
lim
h Ä !b
œ lim b
hÄ!
22
h
Ê the derivative f w (1) does not exist.
39. Left-hand derivative: When h 0, 1 h 1 Ê f(1 h) œ È1 h Ê
œ lim c
œ lim c
hÄ!
œ lim b 2 œ 2;
hÄ!
f(1 h) f(1)
h
Á lim b
hÄ!
f(0 h) f(0)
h
Ê the derivative f w (0) does not exist.
38. Left-hand derivative: When h !, 1 h 1 Ê f(1 h) œ 2 Ê
œ lim b
hÄ!
f(0 h) f(0)
h
œ lim c
hÄ!
g(h) g(0)
h
œ lim b
hÄ!
g(h) g(0)
h
h1Î3 0
h
h
2Î3
œ lim c
hÄ!
0
h
1
h2Î3
œ lim b
hÄ!
œ +_;
1
h1Î3
œ +_;
œ _ Ê the derivative gw (0) does not exist.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
œ #" ;
f("h)f(1)
h
Section 3.2 The Derivative as a Function
43. (a) The function is differentiable on its domain $ Ÿ x Ÿ 2 (it is smooth)
(b) none
(c) none
44. (a) The function is differentiable on its domain # Ÿ x Ÿ 3 (it is smooth)
(b) none
(c) none
45. (a) The function is differentiable on $ Ÿ x 0 and ! x Ÿ 3
(b) none
(c) The function is neither continuous nor differentiable at x œ 0 since lim c f(x) Á lim b f(x)
xÄ!
xÄ!
46. (a) f is differentiable on # Ÿ x 1, " x 0, 0 x 2, and 2 x Ÿ 3
(b) f is continuous but not differentiable at x œ 1: lim f(x) œ 0 exists but there is a corner at x œ 1 since
x Ä 1
œ 3 and lim b f(" h)h f(1) œ 3 Ê f w (1) does not exist
hÄ!
hÄ!
(c) f is neither continuous nor differentiable at x œ 0 and x œ 2:
at x œ 0, lim c f(x) œ 3 but lim b f(x) œ 0 Ê lim f(x) does not exist;
lim c
f(1 h) f(")
h
xÄ!
xÄ0
xÄ!
at x œ 2, lim f(x) exists but lim f(x) Á f(2)
xÄ#
xÄ#
47. (a) f is differentiable on " Ÿ x 0 and 0 x Ÿ 2
(b) f is continuous but not differentiable at x œ 0: lim f(x) œ 0 exists but there is a cusp at x œ 0, so
f(0 h) f(0)
h
hÄ!
f w (0) œ lim
xÄ!
does not exist
(c) none
48. (a) f is differentiable on $ Ÿ x 2, 2 x 2, and 2 x Ÿ 3
(b) f is continuous but not differentiable at x œ 2 and x œ 2: there are corners at those points
(c) none
49. (a) f w (x) œ lim
hÄ!
f(x h) f(x)
h
œ lim
hÄ!
(x h)# ax# b
h
œ lim
hÄ!
x# 2xh h# x#
h
œ lim (2x h) œ 2x
hÄ!
(b)
(c) yw œ 2x is positive for x 0, yw is zero when x œ 0, yw is negative when x 0
(d) y œ x# is increasing for _ x 0 and decreasing for ! x _; the function is increasing on intervals
where yw 0 and decreasing on intervals where yw 0
f(x h) f(x)
h
hÄ!
50. (a) f w (x) œ lim
œ lim
hÄ!
Š xc"
h h
1
x ‹
œ lim
hÄ!
x (x h)
x(x h)h
œ lim
"
h Ä ! x(x h)
œ
"
x#
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
105
106
Chapter 3 Differentiation
(b)
(c) yw is positive for all x Á 0, yw is never 0, yw is never negative
(d) y œ "x is increasing for _ x 0 and ! x _
51. (a) Using the alternate formula for calculating derivatives: f w (x) œ zlim
Äx
œ
$
$
lim z x
z Ä x 3(z x)
œ
az# zx x# b
lim (z x)3(z
x)
zÄx
œ
#
#
lim z zx3 x
zÄx
f(z) f(x)
zx
#
w
$
Š z3 œ zlim
Äx
œ x Ê f (x) œ x
x$
3 ‹
zx
#
(b)
(c) yw is positive for all x Á 0, and yw œ 0 when x œ 0; yw is never negative
(d) y œ
x$
3
is increasing for all x Á 0 (the graph is horizontal at x œ 0) because y is increasing where yw 0; y is
never decreasing
52. (a) Using the alternate form for calculating derivatives: f w (x) œ zlim
Äx
œ
%
%
lim z x
z Ä x 4(z x)
œ
$
xz# x# z x$ b
lim (z x) az 4(z
x)
zÄx
œ
f(z) f(x)
zx
$
#
#
$
lim z xz 4 x z x
zÄx
œ zlim
Äx
$
Œ
z%
4
x%
4 zx
w
œ x Ê f (x) œ x$
(b)
(c) yw is positive for x 0, yw is zero for x œ 0, yw is negative for x 0
(d) y œ
x%
4
is increasing on 0 x _ and decreasing on _ x 0
#
#
#
a2(x h)# 13(x h) 5b a2x# 13x 5b
œ lim 2x 4xh 2h 13x h13h 5 2x 13x 5
h
hÄ!
hÄ!
#
lim 4xh 2hh 13h œ lim (4x 2h 13) œ 4x 13, slope at x. The slope is 1 when
hÄ!
hÄ!
53. yw œ lim
œ
4x 13 œ "
Ê 4x œ 12 Ê x œ 3 Ê y œ 2 † 3# 13 † 3 5 œ 16. Thus the tangent line is y 16 œ (1)(x 3)
Ê y œ x "$ and the point of tangency is (3ß 16).
54. For the curve y œ Èx, we have yw œ lim
hÄ!
œ lim
"
h Ä ! Èx h Èx
œ
"
#Èx
ŠÈx h Èx‹
h
†
ŠÈx h Èx‹
ŠÈx h Èx‹
œ lim
(x h) x
h Ä ! ŠÈx h Èx‹ h
. Suppose ˆ+ß Èa‰ is the point of tangency of such a line and ("ß !) is the point
on the line where it crosses the x-axis. Then the slope of the line is
Èa 0
a (1)
œ
Èa
a1
which must also equal
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 3.2 The Derivative as a Function
"
;
2È a
using the derivative formula at x œ a Ê
exist: its point of tangency is ("ß "), its slope is
Èa
a1
œ
"
#È a
œ
"
Ê 2a œ a 1 Ê a œ 1.
#Èa
"
# ; and an equation of the line is
Thus such a line does
y1œ
"
#
(x 1)
Ê y œ "# x "# .
55. Yes; the derivative of f is f w so that f w (x! ) exists Ê f w (x! ) exists as well.
56. Yes; the derivative of 3g is 3gw so that gw (7) exists Ê 3gw (7) exists as well.
57. Yes, lim
g(t)
t Ä ! h(t)
can exist but it need not equal zero. For example, let g(t) œ mt and h(t) œ t. Then g(0) œ h(0)
œ 0, but lim
g(t)
t Ä ! h(t)
œ lim
tÄ!
mt
t
œ lim m œ m, which need not be zero.
tÄ!
58. (a) Suppose kf(x)k Ÿ x# for " Ÿ x Ÿ 1. Then kf(0)k Ÿ 0# Ê f(0) œ 0. Then f w (0) œ lim
œ lim
hÄ!
f(h) 0
h
œ lim
hÄ!
f(h)
h .
For khk Ÿ 1, h# Ÿ f(h) Ÿ h# Ê h Ÿ
hÄ!
f(h)
h
f(0 h) f(0)
h
Ÿ h Ê f w (0) œ lim
hÄ!
f(h)
h
œ0
by the Sandwich Theorem for limits.
(b) Note that for x Á 0, kf(x)k œ ¸x# sin "x ¸ œ kx# k ksin xk Ÿ kx# k † 1 œ x# (since " Ÿ sin x Ÿ 1). By part (a),
f is differentiable at x œ 0 and f w (0) œ 0.
59. The graphs are shown below for h œ 1, 0.5, 0.1. The function y œ
y œ Èx so that
"
#È x
œ lim
hÄ!
Èx h Èx
h
"
2È x
. The graphs reveal that y œ
is the derivative of the function
Èx h Èx
h
gets closer to y œ
"
#È x
as h gets smaller and smaller.
60. The graphs are shown below for h œ 2, 1, 0.5. The function y œ 3x# is the derivative of the function y œ x$ so
that 3x# œ lim
hÄ!
(xh)$ x$
h
. The graphs reveal that y œ
(xh)$ x$
h
gets closer to y œ 3x# as h
gets smaller and smaller.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
107
108
Chapter 3 Differentiation
61. The graphs are the same. So we know that
for f(x) œ kxk , we have f w (x) œ
kx k
x
.
62. Weierstrass's nowhere differentiable continuous function.
63-68. Example CAS commands:
Maple:
f := x -> x^3 + x^2 - x;
x0 := 1;
plot( f(x), x=x0-5..x0+2, color=black,
title="Section 3.2, #63(a)" );
q := unapply( (f(x+h)-f(x))/h, (x,h) );
# (b)
L := limit( q(x,h), h=0 );
# (c)
m := eval( L, x=x0 );
tan_line := f(x0) + m*(x-x0);
plot( [f(x),tan_line], x=x0-2..x0+3, color=black,
linestyle=[1,7], title="Section 3.2 #63(d)",
legend=["y=f(x)","Tangent line at x=1"] );
Xvals := sort( [ x0+2^(-k) $ k=0..5, x0-2^(-k) $ k=0..5 ] ):
# (e)
Yvals := map( f, Xvals ):
evalf[4](< convert(Xvals,Matrix) , convert(Yvals,Matrix) >);
plot( L, x=x0-5..x0+3, color=black, title="Section 3.2 #63(f)" );
Mathematica: (functions and x0 may vary) (see section 2.5 re. RealOnly ):
<<Miscellaneous`RealOnly`
Clear[f, m, x, y, h]
x0= 1 /4;
f[x_]:=x2 Cos[x]
Plot[f[x], {x, x0 3, x0 3}]
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 3.3 Differentiation Rules
q[x_, h_]:=(f[x h] f[x])/h
m[x_]:=Limit[q[x, h], h Ä 0]
ytan:=f[x0] m[x0] (x x0)
Plot[{f[x], ytan},{x, x0 3, x0 3}]
m[x0 1]//N
m[x0 1]//N
Plot[{f[x], m[x]},{x, x0 3, x0 3}]
3.3 DIFFERENTIATION RULES
1. y œ x# 3 Ê
œ
dy
dx
2. y œ x# x 8 Ê
dy
dx
3. s œ 5t$ 3t& Ê
œ
ds
dt
ax# b d
dx
x$ x Ê
4
3
d
dt
a5t$ b 6. y œ
x$
3
x#
#
d
dt
d# y
dx#
"
4
œ x# x 7. w œ 3z# z" Ê
dw
dz
œ 6z$ z# œ
x
4
8. s œ 2t" 4t# Ê
ds
dt
10. y œ 4 2x x$ Ê
11. r œ
" #
3 s
5# s" Ê
dy
dx
dr
ds
24
)$
48
)&
d# s
dt#
d# w
dz#
œ
d# y
dx#
a15t# b d
dt
a15t% b œ 30t 60t$
6
z$
2
t#
œ 2x 1 0 œ 2x 1
"
z#
d# w
dz#
Ê
d# s
dt#
Ê
8
t$
œ # 3x% œ # œ 23 s$ 5# s# œ
dr
d)
d
dt
œ 126z& 42z 42
œ 18z% 2z$ œ
œ 4t$ 24t% œ
œ 12x 10 10x$ œ 12x 10 dy
dx
12. r œ 12)" 4)$ )% Ê
œ
Ê
œ 2t# 8t$ œ
9. y œ 6x# 10x 5x# Ê
œ 2
œ 8x
dy
dx
d# y
dx#
œ#
a3t& b œ 15t# 15t% Ê
Ê
d# y
dx#
œ 21z' 21z# 42z Ê
dw
dz
œ 4x# 1 Ê
dy
dx
(3) œ 2x 0 œ #B Ê
œ 2x 1 0 œ 2x 1 Ê
4. w œ 3z( 7z$ 21z# Ê
5. y œ
d
dx
3
x%
2
3s$
Ê
d# y
dx#
5
2s#
Ê
10
x$
Ê
d# y
dx#
œ 0 12x& œ
d# r
ds#
œ 12)# 12)% 4)& œ
18
z%
4
t$
12
)%
2
z$
24
t%
œ 12 0 30x% œ 12 30
x%
12
x&
œ 2s% 5s$ œ
12
)#
4
)&
Ê
2
s%
d# r
d) #
5
s$
œ 24)$ 48)& 20)'
20
)'
13. (a) y œ a3 x# b ax$ x 1b Ê yw œ a3 x# b †
#
#
d
dx
$
ax$ x 1b ax$ x 1b †
%
d
dx
a3 x# b
#
œ a3 x b a3x 1b ax x 1b (2x) œ 5x 12x 2x 3
(b) y œ x& 4x$ x# 3x 3 Ê yw œ 5x% 12x# 2x 3
14. (a) y œ (2x 3) a5x# 4xb Ê yw œ (2x 3)(10x 4) a5x# 4xb (2) œ 30x# 14x 12
(b) y œ (2x 3) a5x# 4xb œ 10x$ 7x2 12x Ê yw œ 30x# 14x 12
15. (a) y œ ax# 1b ˆx 5 "x ‰ Ê yw œ ax# 1b †
d
dx
ˆx 5 "x ‰ ˆx 5 x" ‰ †
d
dx
ax# 1b
œ ax# 1b a1 x# b ax 5 x" b (2x) œ ax# 1 1 x# b a2x# 10x 2b œ 3x# 10x 2 (b) y œ x$ 5x# 2x 5 "
x
Ê yw œ 3x# 10x 2 "
x#
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
"
x#
109
110
Chapter 3 Differentiation
16. y œ a1 x2 b ˆx3Î4 x3 ‰
(a) yw œ a1 x2 b † ˆ 34 x1Î4 3x4 ‰ ˆx3Î4 x3 ‰ a2xb œ
(b) y œ x3Î4 x3 x11Î4 x1 Ê yw œ
3
4x1Î4
2x 5
3x 2 ; use the quotient rule: u œ 2x 5 and
(3x 2)(2) (2x 5)(3)
4 6x 15
œ 6x (3x
œ (3x192)#
(3x 2)#
#)#
17. y œ
œ
4 3x
3x2 x ;use the quotient
ˆ3x2 x‰a3b a4 3xba6x1b
a3x2 xb2
18. y œ
œ
t# "
t# t 2
œ
at "bat "b
at #bat "b
21. v œ (1 t) a1 t# b
22. w œ
x5
2x 7
Ê ww œ
Ès "
Ès 1
23. f(s) œ
NOTE:
d
ds
"
œ
œ
9x2 3x 18x2 21x 4
a3x2 xb2
ˆÈ s ‰ œ
24. u œ
5x "
#È x
25. v œ
1 x 4È x
x
Ê
du
dx
œ
œ
1 t
1t#
t"
t2,
Ê
"
#È s
ˆ È s "‰ Š
"
ax # 1 b a x # x 1 b
#
œ
1
x2
vuw uvw
v#
v œ x 0.5 Ê uw œ 2x and vw œ 1 Ê gw (x) œ
at #ba"b at "ba"b
at 2 b 2
2x 7 2x 10
(2x 7)#
œ
"
"
Ès ‹ ˆÈs 1‰ Š #Ès ‹
#
#
vuw uvw
v#
9x2 24x 4
a3x2 xb2
a1 t# b (") (1 t)(2t)
a1 t# b#
ˆÈ s 1 ‰
œ
œ
œ
t#t"
at 2 b 2
œ
" t# 2t 2t#
a1 t# b#
œ
œ
vuw uvw
v#
"
at 2 b 2
t# 2t "
a1 t# b#
17
(2x 7)#
ˆ È s "‰ ˆ È s 1 ‰
2 È s ˆÈ s 1 ‰
#
œ
"
È s ˆÈ s 1‰#
from Example 2 in Section 3.2
4x
x Š1 œ
5x 1
4x$Î#
È x ‹ ˆ1 x 4 È x ‰
2
x#
26. r œ 2 Š È" È)‹ Ê rw œ 2 )
27. y œ
dv
dt
ˆ2Èx‰ (5) (5x 1) Š È" ‹
x
Ê vw œ
œ
t Á " Ê f w (t) œ
(2x 7)(1) (x 5)(2)
(2x 7)#
Ê f w (s) œ
11 7Î4
4 x
rule: u œ 4 3x and v œ 3x2 x Ê uw œ 3 and vw œ 6x 1 Ê yw œ
x# 4
#
x0.5 ; use the quotient rule: u œ x 4 and
#
#
#
#
(x 0.5)(2x) ax 4b (")
x 4
x4
œ 2x (xx 0.5)
œ x(x #
(x 0.5)#
0.5)#
20. f(t) œ
v œ 3x 2 Ê uw œ 2 and vw œ 3 Ê yw œ
19. g(x) œ
œ
3
x4
3
x34
4x1Î4
11 7Î4
x12
4 x
È)(0) 1 Š
)
"
È) ‹
#
œ
2È x "
x#
"
#È ) "
œ )$Î#
"
)"Î#
; use the quotient rule: u œ 1 and v œ ax# 1b ax# x 1b Ê uw œ 0 and
vw œ ax 1b (2x 1) ax# x 1b (2x) œ 2x$ x# 2x 1 2x$ 2x# 2x œ 4x$ 3x# 1
Ê
dy
dx
œ
vuw uvw
v#
28. y œ
(x 1)(x 2)
(x 1)(x #)
29. y œ
"
#
30. y œ
"
1 #0
x% 3
#
œ
œ
0 1 a4x$ 3x# 1b
ax # 1 b # a x # x 1 b #
x# 3x 2
x# 3x 2
Ê yw œ
œ
4x$ 3x# 1
ax # 1 b# a x # x 1 b #
ax# 3x 2b (2x 3) ax# 3x 2b (2x 3)
(x 1)# (x 2)#
œ
6x# 12
(x 1)# (x 2)#
œ
6 ax # 2b
(x 1)# (x 2)#
x# x Ê yw œ 2x$ 3x 1 Ê yww œ 6x# 3 Ê ywww œ 12x Ê yÐ%Ñ œ 12 Ê yÐnÑ œ 0 for all n
x& Ê yw œ
"
#4
x% Ê yww œ
"
6
x$ Ê ywww œ
"
#
x# Ê yÐ%Ñ œ x Ê yÐ&Ñ œ 1 Ê yÐnÑ œ 0 for all n
5
6
31. y œ ax 1bax2 3x 5b œ x3 2x2 8x 5 Ê yw œ 3x2 4x 8 Ê yww œ 6x 4 Ê ywww œ 6 Ê yÐnÑ œ 0 for all
n 4
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 3.3 Differentiation Rules
111
32. y œ a4x3 3xba2 xb œ 4x4 8x3 3x2 6x Ê yw œ 16x3 24x2 6x 6 Ê yww œ 48x2 48x 6
Ê ywww œ 96x 48 Ê yÐ%Ñ œ 96 Ê yÐnÑ œ 0 for all n 5
33. y œ
x $ 7
x
34. s œ
t# 5t 1
œ 1 5t t"# œ 1 5t"
t#
#
'
d s
$
6t% œ "!
dt# œ 10t
t$ t%
Ê
œ x# 7x" Ê
() " ) a ) # ) 1 b
)$
35. r œ
36. u œ
Ê
œ
dy
dx
)$ "
)$
"
)$
œ"
t# Ê
Ê
œ 1 )$ Ê
dr
d)
d# w
dz#
"Ê
œ 2z$ 0 œ 2z$ œ
"
3
1
#
%
Ê
40. p œ
Ê
d# p
dq#
œ
"
6
#" q% 5q' œ
q# 3
(q 1)$ (q 1)$
dp
dq
q' q# 3q% 3
12q%
œ
"
6
œ
"
#q %
q# "
1#
q# d# p
dq#
&
t#
#
t$
$
)%
Ê
d# r
d) #
z Ê
dw
dz
œ z# 0 1 œ z# 1
œ 12)& œ
"#
)&
œ 1 x$
8
3
"
4
dw
dz
œ 4z$ 0 œ 4z$ Ê
4" q% Ê
dp
dq
œ
"
6
d# w
dz#
œ 12z#
q 6" q$ q& œ
"
6
q
"
6q$
"
q&
&
q6
q# 3
aq$ 3q# 3q 1b aq$ 3q# 3q 1b
œ "# q# œ #"q# Ê
"%
x$
#
z$
"
1#
x
x%
3 z œ z" 38. w œ (z 1)(z 1) az# 1b œ az# 1b az# 1b œ z% 1 Ê
3
39. p œ Š q12q
‹ Š q q$ 1 ‹ œ
œ 2 14x$ œ # œ 0 $)% œ $)% œ
#
$
%
ax # x b a x # x 1 b
œ x(x 1) axx% x "b œ x axx% 1b œ x x% x œ
x%
#
du
%
œ 3x% œ $
Ê ddxu# œ 12x& œ "#
dx œ 0 3x
x&
x%
"
z#
d# y
dx#
œ 0 5t# 2t$ œ 5t# 2t$ œ
ds
dt
37. w œ ˆ 13z3z ‰ (3 z) œ ˆ "3 z" 1‰ (3 z) œ z" œ
(
x#
œ 2x 7x# œ #x œ q$ œ
œ
q# 3
2q$ 6q
œ
q# 3
2q aq# 3b
œ
"
#q
œ
"
#
q"
"
q$
41. u(0) œ 5, uw (0) œ 3, v(0) œ 1, vw (0) œ 2
d
d
(a) dx
(uv) œ uvw vuw Ê dx
(uv)¸ x = 0 œ u(0)vw (0) v(0)uw (0) œ 5 † 2 (1)(3) œ 13
(b)
d
dx
ˆ vu ‰ œ
vuw uvw
v#
Ê
d
dx
(c)
d
dx
d
dx
ˆ uv ‰
œ
uvw vuw
u#
Ê
d
dx
(d)
(7v 2u) œ 7vw ˆ vu ‰¸
x=0
ˆ uv ‰¸
x=0
d
2uw Ê dx
œ
œ
v(0)uw (0) u(0)vw (0)
(v(0))#
u(0)vw (0) v(0)uw (0)
(u(0))#
(")(3) (5)(2)
(1)#
(5)(2) (1)(3)
œ
(5)#
w
w
œ
œ 7
œ
7
25
(7v 2u)¸ x = 0 œ 7v (0) 2u (0) œ 7 † 2 2(3) œ 20
42. u(1) œ 2, uw (1) œ 0, v(1) œ 5, vw (1) œ 1
d
(a) dx
(uv)¸ x = 1 œ u(1)vw (1) v(1)uw (1) œ 2 † (1) 5 † 0 œ 2
(b)
d
dx
ˆ vu ‰¸
x=1
œ
(c)
d
dx
d
dx
ˆ uv ‰¸
x=1
œ
(d)
v(1)uw (")u(1)vw (1)
(v(1))#
u(1)vw (")v(1)uw (1)
(u(1))#
w
œ
œ
5†02†(1)
(5)#
2†(1)5†0
(2)#
œ
2
25
œ 12
(7v 2u)¸ x = 1 œ 7v (1) 2uw (1) œ 7 † (1) 2 † 0 œ 7
43. y œ x$ 4x 1. Note that (#ß ") is on the curve: 1 œ 2$ 4(2) 1
(a) Slope of the tangent at (xß y) is yw œ 3x# 4 Ê slope of the tangent at (#ß ") is yw (2) œ 3(2)# 4 œ 8. Thus the slope
of the line perpendicular to the tangent at (#ß ") is "8 Ê the equation of the line perpendicular to the tangent line at
(#ß ") is y 1 œ "8 (x 2) or y œ 8x 45 .
(b) The slope of the curve at x is m œ 3x# 4 and the smallest value for m is 4 when x œ 0 and y œ 1.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
112
Chapter 3 Differentiation
(c) We want the slope of the curve to be 8 Ê yw œ 8 Ê 3x# 4 œ 8 Ê 3x# œ 12 Ê x# œ 4 Ê x œ „ 2. When x œ 2,
y œ 1 and the tangent line has equation y 1 œ 8(x 2) or y œ 8x 15; when x œ 2, y œ (2)$ 4(2) 1
œ 1, and the tangent line has equation y 1 œ 8(x 2) or y œ 8x 17.
44. (a) y œ x$ 3x 2 Ê yw œ 3x# 3. For the tangent to be horizontal, we need m œ yw œ 0 Ê 0 œ 3x# 3 Ê 3x# œ 3
Ê x œ „ 1. When x œ 1, y œ 0 Ê the tangent line has equation y œ 0. The line perpendicular to this line at
("ß !) is x œ 1. When x œ 1, y œ 4 Ê the tangent line has equation y œ 4. The line perpendicular to this
line at ("ß %) is x œ 1.
(b) The smallest value of yw is 3, and this occurs when x œ 0 and y œ 2. The tangent to the curve at (!ß 2)
has slope 3 Ê the line perpendicular to the tangent at (!ß 2) has slope "3 Ê y 2 œ "3 (x 0) or
yœ
45. y œ
"
3
4x
x# 1
x 2 is an equation of the perpendicular line.
Ê
dy
dx
œ
ax# 1b(4) (4x)(2x)
ax # 1 b #
œ
4x# 4 8x#
ax # 1 b#
œ
4 ax# "b
ax # 1 b#
. When x œ 0, y œ 0 and yw œ
4(0 1)
1
œ %, so the
tangent to the curve at (!ß !) is the line y œ 4x. When x œ 1, y œ 2 Ê yw œ 0, so the tangent to the curve at ("ß 2) is the
line y œ 2.
46. y œ
8
x# 4
Ê yw œ
ax# 4b(0) 8(2x)
ax # 4 b #
œ
16x
ax # 4 b#
. When x œ 2, y œ 1 and yw œ
16(2)
a2 # 4 b#
œ "# , so the tangent
line to the curve at (2ß ") has the equation y 1 œ "# (x 2), or y œ x# 2.
47. y œ ax# bx c passes through (!ß !) Ê 0 œ a(0) b(0) c Ê c œ 0; y œ ax# bx passes through ("ß #)
Ê 2 œ a b; yw œ 2ax b and since the curve is tangent to y œ x at the origin, its slope is 1 at x œ 0
Ê yw œ 1 when x œ 0 Ê 1 œ 2a(0) b Ê b œ 1. Then a b œ 2 Ê a œ 1. In summary a œ b œ 1 and c œ 0 so
the curve is y œ x# x.
48. y œ cx x# passes through ("ß !) Ê 0 œ c(1) 1 Ê c œ 1 Ê the curve is y œ x x# . For this curve,
yw œ 1 2x and x œ 1 Ê yw œ 1. Since y œ x x# and y œ x# ax b have common tangents at x œ 0,
y œ x# ax b must also have slope 1 at x œ 1. Thus yw œ 2x a Ê 1 œ 2 † 1 a Ê a œ 3
Ê y œ x# 3x b. Since this last curve passes through ("ß !), we have 0 œ 1 3 b Ê b œ 2. In summary,
a œ 3, b œ 2 and c œ 1 so the curves are y œ x# 3x 2 and y œ x x# .
49. y œ 8x 5 Ê m œ 8; faxb œ 3x2 4x Ê f w axb œ 6x 4; 6x 4 œ 8 Ê x œ 2 Ê fa2b œ 3a2b2 4a2b œ 4 Ê a2, 4b
50. 8x 2y œ 1 Ê y œ 4x "
#
Ê m œ 4; gaxb œ 13 x3 32 x2 1 Ê g w axb œ x2 3x; x2 3x œ 4 Ê x œ 4 or x œ 1
Ê ga4b œ 13 a4b3 32 a4b2 1 œ 53 , ga1b œ 13 a1b3 32 a1b2 1 œ 56 Ê ˆ4, 53 ‰ or ˆ1, 56 ‰
51. y œ 2x 3 Ê m œ 2 Ê m¼ œ 12 ; y œ
x
x2
Ê „ 2 œ x 2 Ê x œ 4 or x œ 0 Ê if x œ 4, y œ
52. m œ
y8
x3;
ax 2ba1b xa1b
2
œ ax ; 2
ax 2 b 2
2 b2 ax 2 b 2
4
0
4 2 œ 2, and if x œ 0, y œ 0 2
Ê yw œ
faxb œ x2 Ê f w axb œ 2x; m œ f w axb Ê
y8
x3
œ 2x Ê
x2 8
x3
œ 12 Ê 4 œ ax 2b2
œ 0 Ê a4, 2b or a0, 0b.
œ 2x Ê x2 8 œ 2x2 6x Ê x2 6x 8 œ 0
Ê x œ 4 or x œ 2 Ê fa4b œ 4 œ 16, fa2b œ 2 œ 4 Ê a4, 16b or a2, 4b.
2
2
53. (a) y œ x$ x Ê yw œ 3x# 1. When x œ 1, y œ 0 and yw œ 2 Ê the tangent line to the curve at ("ß !) is
y œ 2(x 1) or y œ 2x 2.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 3.3 Differentiation Rules
113
(b)
(c)
y œ x$ x
Ê x$ x œ 2x 2 Ê x$ 3x 2 œ (x 2)(x 1)# œ 0 Ê x œ 2 or x œ 1. Since
y œ 2x 2 y œ 2a2b 2 œ 6; the other intersection point is (2ß 6)
54. (a) y œ x$ 6x# 5x Ê yw œ 3x# 12x 5. When x œ 0, y œ 0 and yw œ 5 Ê the tangent line to the curve at
(0ß 0) is y œ 5x.
(b)
(c)
y œ x$ 6x# 5x
$
#
$
#
#
Ê x 6x 5x œ 5x Ê x 6x œ 0 Ê x (x 6) œ 0 Ê x œ 0 or x œ 6.
y œ 5x
Since y œ 5a6b œ $!, the other intersection point is (6ß 30).
55. lim
xÄ1
56.
x50 1
x1
lim
x Ä 1
œ 50 x49 ¹
x2Î9 1
x1
xœ1
œ 92 x7Î9 ¹
œ 50 a1b49 œ 50
x œ 1
œ
2
9a1b7Î9
œ 92
57. gw axb œ œ
2x 3
a
x0
, since g is differentiable at x œ 0 Ê lim b a2x 3b œ 3 and lim c a œ a Ê a œ 3
x0
xÄ0
xÄ0
58. f w axb œ œ
a
x 1
, since f is differentiable at x œ 1 Ê lim b a œ a and lim c a2bxb œ 2b Ê a œ 2b, and
2bx x 1
x Ä 1
x Ä 1
since f is continuous at x œ 1 Ê lim b aax bb œ a b and lim c abx2 3b œ b 3 Ê a b œ b 3
x Ä 1
x Ä 1
Ê a œ 3 Ê 3 œ 2b Ê b œ 32 .
59. Paxb œ an xn an" xn" â a# x# a" x a! Ê P w axb œ nan xn" an "ban" xn# â #a# x a"
60. R œ M# ˆ C# M‰
3
œ
C
#
61. Let c be a constant Ê
M# "3 M$ , where C is a constant Ê
dc
dx
œ0 Ê
d
dx
(u † c) œ u †
dc
dx
c†
du
dx
dR
dM
œ CM M#
œu†0c
du
dx
œc
du
dx
. Thus when one of the
functions is a constant, the Product Rule is just the Constant Multiple Rule Ê the Constant Multiple Rule is
a special case of the Product Rule.
62. (a) We use the Quotient rule to derive the Reciprocal Rule (with u œ 1):
d
dx
ˆ "v ‰ œ
v†0 1† dv
dx
v#
œ
"† dv
dx
v#
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
œ v"# †
dv
dx
.
114
Chapter 3 Differentiation
(b) Now, using the Reciprocal Rule and the Product Rule, we'll derive the Quotient Rule:
d ˆ"‰
" du
dx v v † dx (Product Rule)
dv
v du
dx u dx
, the Quotient Rule.
v#
œu†
œ
63. (a)
d
dx
(uvw) œ
d
dx
w
dv
œ u † ˆ v#1 ‰ dx
((uv) † w) œ (uv) dw
dx w †
d
dx
" du
v dx
(uv) œ uv
œ uvww uv w uw vw
d
d
%
(b) dx
au" u# u$ u% b œ dx
aau" u# u$ b u% b œ au" u# u$ b du
dx u%
du$
du#
du" ‰
%
ˆ
œ u" u# u$ du
dx u% u" u# dx u$ u" dx u$ u# dx
dw
dx
d
dx
(Reciprocal Rule) Ê
w ˆu
dv
dx
v
a u" u# u$ b Ê
d
dx
du ‰
dx
d
dx
œ uv
d
dx
d ˆ
"‰
dx u † v
du
u dv
dx v dx
v#
ˆ vu ‰ œ
ˆ vu ‰ œ
dw
dx
wu
dv
dx
wv
du
dx
a u" u# u$ u% b
(using (a) above)
du$
du#
du"
%
Ê
au" u# u$ u% b œ u" u# u$ du
dx u" u# u% dx u" u$ u% dx u# u$ u% dx
œ u" u# u$ u%w u" u# u$w u% u" u#w u$ u% u"w u# u$ u%
d
Generalizing (a) and (b) above, dx
au" âun b œ u" u# âun" unw u" u# âun# unw " un
d
dx
(c)
64.
d
m
b
dx ax
65. P œ
Ê
œ
œ
d ˆ 1 ‰
dx xm
xm †01ˆm†xm 1 ‰
ax m b 2
œ
m†xm
x2m
1
œ m † xm12m œ m † xm1
nRT
an#
Vnb V# . We are holding T constant, and a, b, n, R are
#
#
(V nb)†0 (nRT)(1)
dP
2an#
V (0) aVa#anb# b (2V) œ (VnRT
dV œ
(Vnb)#
nb)# V$
66. Aaqb œ
km
q
cm hq
#
á u"w u# âun
œ akmbq" cm ˆ h# ‰q Ê
dA
dq
also constant so their derivatives are zero
œ akmbq# ˆ h# ‰ œ km
q# h
#
Ê
d# A
dt#
œ #akmbq$ œ
#km
q$
3.4 THE DERIVATIVE AS A RATE OF CHANGE
1. s œ t# $t #, 0 Ÿ t Ÿ #
(a) displacement œ ?s œ s(#) s(0) œ !m #m œ # m, vav œ
(b) v œ
aœ
ds
dt œ #t
d# s
dt# œ #
?s
?t
œ
Ê a(0) œ # m/sec# and a(#) œ # m/sec#
changes direction at t œ
$
#.
2. s œ 't t# , ! Ÿ t Ÿ '
(a) displacement œ ?s œ s(') s(0) œ ! m, vav œ
aœ
œ " m/sec
$ Ê kv(0)k œ l$l œ $ m/sec and kv(#)k œ 1 m/sec;
(c) v œ 0 Ê #t $ œ 0 Ê t œ $# . v is negative in the interval ! t (b) v œ
#
#
ds
dt œ ' d# s
dt# œ #
?s
?t
œ
!
'
$
#
and v is positive when
$
#
t # Ê the body
œ ! m/sec
#> Ê kv(0)k œ l 'l œ ' m/sec and kv(')k œ l'l œ ' m/sec;
Ê a(0) œ # m/sec# and a(') œ # m/sec#
(c) v œ 0 Ê ' #t œ 0 Ê t œ $. v is positive in the interval ! t $ and v is negative when $ t ' Ê the body
changes direction at t œ $.
3. s œ t$ 3t# 3t, 0 Ÿ t Ÿ 3
(a) displacement œ ?s œ s(3) s(0) œ 9 m, vav œ
(b) v œ
ds
dt
#
?s
?t
œ
9
3
œ 3 m/sec
œ 3t 6t 3 Ê kv(0)k œ k3k œ 3 m/sec and kv(3)k œ k12k œ 12 m/sec; a œ
#
#
d# s
dt#
œ 6t 6
Ê a(0) œ 6 m/sec and a(3) œ 12 m/sec
(c) v œ 0 Ê 3t# 6t 3 œ 0 Ê t# 2t 1 œ 0 Ê (t 1)# œ 0 Ê t œ 1. For all other values of t in the
interval the velocity v is negative (the graph of v œ 3t# 6t 3 is a parabola with vertex at t œ 1 which
opens downward Ê the body never changes direction).
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 3.4 The Derivative as a Rate of Change
4. s œ
t%
4
115
t$ t# , 0 Ÿ t Ÿ $
(a) ?s œ s($) s(0) œ
$
m, vav œ
?s
?t
œ
*
%
$
œ
$
%
m/sec
(b) v œ t 3t 2t Ê kv(0)k œ 0 m/sec and kv($)k œ ' m/sec; a œ 3t# 6t 2 Ê a(0) œ 2 m/sec# and
a($) œ "" m/sec#
(c) v œ 0 Ê t$ 3t# 2t œ 0 Ê t(t 2)(t 1) œ 0 Ê t œ 0, 1, 2 Ê v œ t(t 2)(t 1) is positive in the interval
for 0 t 1 and v is negative for 1 t 2 and v is positive for # t $ Ê the body changes direction at
t œ 1 and at t œ #.
5. s œ
25
t#
#
*
%
5t , 1 Ÿ t Ÿ 5
(a) ?s œ s(5) s(1) œ 20 m, vav œ
(b) v œ
50
t$
a(5) œ
4
25
(c) v œ 0 Ê
6. s œ
25
t5
5
t#
20
4
œ 5 m/sec
Ê kv(1)k œ 45 m/sec and kv(5)k œ
"
5
m/sec; a œ
150
t%
10
t$
Ê a(1) œ 140 m/sec# and
m/sec#
50 5t
t$
œ 0 Ê 50 5t œ 0 Ê t œ 10 Ê the body does not change direction in the interval
, % Ÿ t Ÿ 0
(a) ?s œ s(0) s(4) œ 20 m, vav œ 20
4 œ 5 m/sec
(b) v œ
a(0)
(c) v œ
25
(t 5)# Ê kv(4)k œ 25
œ 25 m/sec#
0 Ê (t255)# œ 0 Ê v is
m/sec and kv(0)k œ " m/sec; a œ
50
(t5)$
Ê a(4) œ 50 m/sec# and
never 0 Ê the body never changes direction
7. s œ t$ 6t# 9t and let the positive direction be to the right on the s-axis.
(a) v œ 3t# 12t 9 so that v œ 0 Ê t# 4t 3 œ (t 3)(t 1) œ 0 Ê t œ 1 or 3; a œ 6t 12 Ê a(1)
œ 6 m/sec# and a(3) œ 6 m/sec# . Thus the body is motionless but being accelerated left when t œ 1, and
motionless but being accelerated right when t œ 3.
(b) a œ 0 Ê 6t 12 œ 0 Ê t œ 2 with speed kv(2)k œ k12 24 9k œ 3 m/sec
(c) The body moves to the right or forward on 0 Ÿ t 1, and to the left or backward on 1 t 2. The
positions are s(0) œ 0, s(1) œ 4 and s(2) œ 2 Ê total distance œ ks(1) s(0)k ks(2) s(1)k œ k4k k2k œ 6 m.
8. v œ t# 4t 3 Ê a œ 2t 4
(a) v œ 0 Ê t# 4t 3 œ 0 Ê t œ 1 or 3 Ê a(1) œ 2 m/sec# and a(3) œ 2 m/sec#
(b) v 0 Ê (t 3)(t 1) 0 Ê 0 Ÿ t 1 or t 3 and the body is moving forward; v 0 Ê at 3bat 1b 0
Ê " t 3 and the body is moving backward
(c) velocity increasing Ê a 0 Ê 2t 4 0 Ê t 2; velocity decreasing Ê a 0 Ê 2t 4 0 Ê ! Ÿ t 2
9. sm œ 1.86t# Ê vm œ 3.72t and solving 3.72t œ 27.8 Ê t ¸ 7.5 sec on Mars; sj œ 11.44t# Ê vj œ 22.88t and
solving 22.88t œ 27.8 Ê t ¸ 1.2 sec on Jupiter.
10. (a) v(t) œ sw (t) œ 24 1.6t m/sec, and a(t) œ vw (t) œ sw w (t) œ 1.6 m/sec#
(b) Solve v(t) œ 0 Ê 24 1.6t œ 0 Ê t œ 15 sec
(c) s(15) œ 24(15) .8(15)# œ 180 m
(d) Solve s(t) œ 90 Ê 24t .8t# œ 90 Ê t œ
30 „ 15È2
#
¸ 4.39 sec going up and 25.6 sec going down
(e) Twice the time it took to reach its highest point or 30 sec
11. s œ 15t "# gs t# Ê v œ 15 gs t so that v œ 0 Ê 15 gs t œ 0 Ê gs œ
15
t
. Therefore gs œ
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
15
20
œ
3
4
œ 0.75 m/sec#
116
Chapter 3 Differentiation
12. Solving sm œ 832t 2.6t# œ 0 Ê t(832 2.6t) œ 0 Ê t œ 0 or 320 Ê 320 sec on the moon; solving
se œ 832t 16t# œ 0 Ê t(832 16t) œ 0 Ê t œ 0 or 52 Ê 52 sec on the earth. Also, vm œ 832 5.2t œ 0
Ê t œ 160 and sm (160) œ 66,560 ft, the height it reaches above the moon's surface; ve œ 832 32t œ 0
Ê t œ 26 and se (26) œ 10,816 ft, the height it reaches above the earth's surface.
13. (a) s œ 179 16t# Ê v œ 32t Ê speed œ kvk œ 32t ft/sec and a œ 32 ft/sec#
(b) s œ 0 Ê 179 16t# œ 0 Ê t œ É 179
16 ¸ 3.3 sec
È
É 179
(c) When t œ É 179
16 , v œ 32
16 œ 8 179 ¸ 107.0 ft/sec
14. (a)
lim1 v œ lim1 9.8(sin ))t œ 9.8t so we expect v œ 9.8t m/sec in free fall
)Ä
(b) a œ
#
dv
dt
)Ä
#
œ 9.8 m/sec#
(b) between 3 and 6 seconds: $ Ÿ t Ÿ 6
(d)
15. (a) at 2 and 7 seconds
(c)
16. (a) P is moving to the left when 2 t 3 or 5 t 6; P is moving to the right when 0 t 1; P is standing
still when 1 t 2 or 3 t 5
(b)
17. (a)
(c)
(e)
(f)
190 ft/sec
at 8 sec, 0 ft/sec
From t œ 8 until t œ 10.8 sec, a total of 2.8 sec
Greatest acceleration happens 2 sec after launch
(g) From t œ 2 to t œ 10.8 sec; during this period, a œ
(b) 2 sec
(d) 10.8 sec, 90 ft/sec
v(10.8) v(2)
10.8 2
¸ 32 ft/sec#
18. (a) Forward: 0 Ÿ t 1 and 5 t 7; Backward: 1 t 5; Speeds up: 1 t 2 and 5 t 6;
Slows down: 0 Ÿ t 1, 3 t 5, and 6 t 7
(b) Positive: 3 t 6; negative: 0 Ÿ t 2 and 6 t 7; zero: 2 t 3 and 7 t 9
(c) t œ 0 and 2 Ÿ t Ÿ 3
(d) 7 Ÿ t Ÿ 9
19. s œ 490t# Ê v œ 980t Ê a œ 980
(a) Solving 160 œ 490t# Ê t œ
4
7
sec. The average velocity was
s(4/7) s(0)
4/7
œ 280 cm/sec.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 3.4 The Derivative as a Rate of Change
(b) At the 160 cm mark the balls are falling at v(4/7) œ 560 cm/sec. The acceleration at the 160 cm mark
was 980 cm/sec# .
17
(c) The light was flashing at a rate of 4/7
œ 29.75 flashes per second.
20. (a)
(b)
21. C œ position, A œ velocity, and B œ acceleration. Neither A nor C can be the derivative of B because B's derivative
is constant. Graph C cannot be the derivative of A either, because A has some negative slopes while C has only
positive values. So, C (being the derivative of neither A nor B) must be the graph of position. Curve C has both
positive and negative slopes, so its derivative, the velocity, must be A and not B. That leaves B for acceleration.
22. C œ position, B œ velocity, and A œ acceleration. Curve C cannot be the derivative of either A or B because
C has only negative values while both A and B have some positive slopes. So, C represents position. Curve C
has no positive slopes, so its derivative, the velocity, must be B. That leaves A for acceleration. Indeed, A is
negative where B has negative slopes and positive where B has positive slopes.
23. (a) c(100) œ 11,000 Ê cav œ
#
11,000
100
œ $110
(b) c(x) œ 2000 100x .1x Ê cw (x) œ 100 .2x. Marginal cost œ cw (x) Ê the marginal cost of producing 100
machines is cw (100) œ $80
(c) The cost of producing the 101st machine is c(101) c(100) œ 100 201
10 œ $79.90
24. (a) r(x) œ 20000 ˆ1 "x ‰ Ê rw (x) œ
w
(b) r a"!"b œ $"Þ*'Þ
(c) x lim
rw (x) œ x lim
Ä_
Ä_
20000
x#
20000
x#
, which is marginal revenue. rw a"!!b œ
20000
100#
œ $#Þ
œ 0. The increase in revenue as the number of items increases without bound
will approach zero.
25. b(t) œ 10' 10% t 10$ t# Ê bw (t) œ 10% (2) a10$ tb œ 10$ (10 2t)
(a) bw (0) œ 10% bacteria/hr
(b) bw (5) œ 0 bacteria/hr
w
%
(c) b (10) œ 10 bacteria/hr
26. Q(t) œ 200(30 t)# œ 200 a900 60t t# b Ê Qw (t) œ 200(60 2t) Ê Qw (10) œ 8,000 gallons/min is the rate
the water is running at the end of 10 min. Then
Q(10) Q(0)
10
œ 10,000 gallons/min is the average rate the water flows
during the first 10 min. The negative signs indicate water is leaving the tank.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
117
118
Chapter 3 Differentiation
27. (a) y œ 6 ˆ1 t ‰#
1#
œ 6 Š1 (b) The largest value of
value of
dy
dt
dy
dt
t
6
t#
144 ‹
Ê
dy
dt
œ
t
12
1
is 0 m/h when t œ 12 and the fluid level is falling the slowest at that time. The smallest
is 1 m/h, when t œ 0, and the fluid level is falling the fastest at that time.
(c) In this situation,
Ÿ 0 Ê the graph of y is
dy
dt
always decreasing. As
dy
dt
increases in value,
the slope of the graph of y increases from 1
to 0 over the interval 0 Ÿ t Ÿ 12.
28. (a) V œ
4
3
1 r$ Ê
(b) When r œ 2,
dV
dr
dV
dr
œ 41 r # Ê
dV ¸
dr r=2
œ 41(2)# œ 161 ft$ /ft
œ 161 so that when r changes by 1 unit, we expect V to change by approximately 161. Therefore
when r changes by 0.2 units V changes by approximately (161)(0.2) œ 3.21 ¸ 10.05 ft$ . Note that
V(2.2) V(2) ¸ 11.09 ft$ .
29. 200 km/hr œ 55 59 m/sec œ
t œ 25, D œ
10
9
#
(25) œ
500
9 m/sec,
6250
9 m
and D œ
10 #
9 t
30. s œ v! t 16t# Ê v œ v! 32t; v œ 0 Ê t œ
v!
32
Ê Vœ
20
9
t. Thus V œ
500
9
Ê
; 1900 œ v! t 16t# so that t œ
Ê v! œ È(64)(1900) œ 80È19 ft/sec and, finally,
80È19 ft
sec
†
60 sec
1 min
†
60 min
1 hr
†
1 mi
5280 ft
v!
32
20
9
tœ
500
9
Ê t œ 25 sec. When
Ê 1900 œ
v!#
3#
v!#
64
¸ 238 mph.
31.
v œ 0 when t œ 6.25 sec
v 0 when 0 Ÿ t 6.25 Ê body moves right (up); v 0 when 6.25 t Ÿ 12.5 Ê body moves left (down)
body changes direction at t œ 6.25 sec
body speeds up on (6.25ß 12.5] and slows down on [0ß 6.25)
The body is moving fastest at the endpoints t œ 0 and t œ 12.5 when it is traveling 200 ft/sec. It's moving slowest at
t œ 6.25 when the speed is 0.
(f) When t œ 6.25 the body is s œ 625 m from the origin and farthest away.
(a)
(b)
(c)
(d)
(e)
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 3.4 The Derivative as a Rate of Change
32.
(a) v œ 0 when t œ
3
#
sec
(b) v 0 when 0 Ÿ t 1.5 Ê body moves left (down); v 0 when 1.5 t Ÿ 5 Ê body moves right (up)
(c) body changes direction at t œ 3# sec
(d) body speeds up on ˆ 3# ß &‘ and slows down on !ß 3# ‰
(e) body is moving fastest at t œ 5 when the speed œ kv(5)k œ 7 units/sec; it is moving slowest at t œ
3
#
when the
speed is 0
(f) When t œ 5 the body is s œ 12 units from the origin and farthest away.
33.
6 „ È15
3
6 È15
t
3
(a) v œ 0 when t œ
sec
(b) v 0 when
6 È15
3
Ê body moves left (down); v 0 when 0 Ÿ t 6 È15
3
or
6 È15
3
tŸ4
Ê body moves right (up)
6 „ È15
sec
3
6 È15
6 È15
Š 3 ß #‹ Š 3 ß %“
(c) body changes direction at t œ
(d) body speeds up on
È15
and slows down on ’0ß 6 3
È15
‹ Š#ß 6 3
‹.
(e) The body is moving fastest at t œ 0 and t œ 4 when it is moving 7 units/sec and slowest at t œ
(f) When t œ
6È15
3
the body is at position s ¸ 6.303 units and farthest from the origin.
34.
(a) v œ 0 when t œ
6 „ È15
3
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
6„È15
3
sec
119
120
Chapter 3 Differentiation
È
6 È15
or 6 3 15 t Ÿ 4 Ê body is moving left (down); v 0 when
3
È
6 È15
t 6 3 15 Ê body is moving right (up)
3
È
body changes direction at t œ 6 „ 3 15 sec
È
È
È
È
body speeds up on Š 6 3 15 ß #‹ Š 6 3 15 ß %“ and slows down on ’!ß 6 3 15 ‹ Š#ß 6 3 15 ‹
(b) v 0 when 0 Ÿ t (c)
(d)
(e) The body is moving fastest at 7 units/sec when t œ 0 and t œ 4; it is moving slowest and stationary at t œ
(f) When t œ
6 È15
3
6 „ È15
3
the position is s ¸ 10.303 units and the body is farthest from the origin.
3.5 DERIVATIVES OF TRIGONOMETRIC FUNCTIONS
1. y œ 10x 3 cos x Ê
2. y œ
3
x
5 sin x Ê
3. y œ x2 cos x Ê
dy
dx
œ
dy
dx
5
(cos x) œ 10 3 sin x
(sin x) œ
d
dx
œ Èx sec x tan x dy
dx
5. y œ csc x 4Èx 7 Ê
"
x#
3
x#
d
dx
3
x#
5 cos x
œ x2 asin xb 2x cos x œ x2 sin x 2x cos x
4. y œ Èx sec x 3 Ê
6. y œ x# cot x œ 10 3
dy
dx
Ê
dy
dx
dy
dx
œ csc x cot x œ x#
œ x# csc# x 2x cot x sec x
2È x
d
dx
0 œ Èx sec x tan x 4
#È x
(cot x) cot x †
d
dx
sec x
2È x
0 œ csc x cot x ax# b 2
x$
2
Èx
œ x# csc# x (cot x)(2x) 2
x$
2
x$
sin x
2
7. faxb œ sin x tan x Ê f w axb œ sin x sec2 x cos x tan x œ sin x sec2 x cos x cos
x œ sin xasec x 1b
8. gaxb œ csc xcot x Ê gw axb œ csc xacsc2 xb acsc xcot xbcot x œ csc3 x csc x cot2 x œ csc xacsc2 x cot2 xb
9. y œ (sec x tan x)(sec x tan x) Ê
#
dy
dx
œ (sec x tan x)
d
dx
(sec x tan x) (sec x tan x)
#
d
dx
(sec x tan x)
œ (sec x tan x) asec x tan x sec xb (sec x tan x) asec x tan x sec xb
œ asec# x tan x sec x tan# x sec$ x sec# x tan xb asec# x tan x sec x tan# x sec$ x tan x sec# xb œ 0.
ŠNote also that y œ sec# x tan# x œ atan# x 1b tan# x œ 1 Ê
10. y œ (sin x cos x) sec x Ê
œ (sin x cos x)
dy
dx
d
dx
sin# x cos x sin x cos# x cos x sin x
cos# x
œ
"
cos# x
d
dx (sin x cos x)
(sin x cos x) sin x
x sin x
cos cos
cos# x
x
œ sec# x
ŠNote also that y œ sin x sec x cos x sec x œ tan x 1 Ê
11. y œ
œ
Ê
dy
dx
œ
(1 cot x)
csc# x csc# x cot x csc# x cot x
(1 cot x)#
12. y œ
œ
cot x
1 cot x
cos x
1 sin x
Ê
dy
dx
œ
sin x sin# x cos# x
(1 sin x)#
(1 sin x)
œ
d
dx
œ
œ 0.‹
(sec x) sec x
œ (sin x cos x)(sec x tan x) (sec x)(cos x sin x) œ
œ
dy
dx
(cot x) (cot x)
(1 cot x)#
d
dx
(1 cot x)
œ
dy
dx
œ sec# x.‹
(1 cot x) acsc# xb (cot x) acsc# xb
(1 cot x)#
csc# x
(1 cot x)#
d
(cos x) (cos x) dx
(1 sin x)
b (cos x) acos xb
œ (1 sin x) a(1sin xsin
(1 sin x)#
x)#
(1 sin x)
sin x 1
"
(1 sin x)# œ (1 sin x)# œ 1 sin x
d
dx
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 3.5 Derivatives of Trigonometric Functions
13. y œ
4
cos x
"
tan x
14. y œ
cos x
x
x
cos x
œ 4 sec x cot x Ê
Ê
œ
dy
dx
œ 4 sec x tan x csc# x
dy
dx
x(sin x) (cos x)(1)
x#
15. y œ x# sin x 2x cos x 2 sin x Ê
121
(cos x)(1) x(sin x)
cos# x
œ
x sin x cos x
x#
cos x x sin x
cos# x
dy
dx
œ ax# cos x (sin x)(2x)b a(2x)(sin x) (cos x)(2)b 2 cos x
dy
dx
œ ax# (sin x) (cos x)(2x)b a2x cos x (sin x)(2)b 2(sin x)
œ x# cos x 2x sin x 2x sin x 2 cos x 2 cos x œ x# cos x
16. y œ x# cos x 2x sin x 2 cos x Ê
œ x# sin x 2x cos x 2x cos x 2 sin x 2 sin x œ x# sin x
17. faxb œ x3 sin x cos x Ê f w axb œ x3 sin xasin xb x3 cos xacos xb 3x2 sin x cos x œ x3 sin2 x x3 cos2 x 3x2 sin x cos x
18. gaxb œ a2 xbtan2 x Ê gw axb œ a2 xba2 tan x sec2 xb a1btan2 x œ 2a2 xbtan x sec2 x tan2 x
œ 2a2 xbtan xa sec2 x tan xb
19. s œ tan t t Ê
1 csc t
1 csc t
21. s œ
œ
Ê
œ sec# t 1
ds
dt
ds
dt
20. s œ t# sec t 1 Ê
sin t
1 cos t
Ê
ds
dt
œ 2t sec t tan t
(1 csc t)(csc t cot t) (" csc t)(csc t cot t)
(1 csc t)#
œ
csc t cot t csc# t cot t csc t cot t csc# t cot t
(1 csc t)#
22. s œ
ds
dt
(1 cos t)(cos t) (sin t)(sin t)
(1 cos t)#
œ
23. r œ 4 )# sin ) Ê
2 csc t cot t
(1 csc t)#
œ
dr
d)
œ ˆ) #
24. r œ ) sin ) cos ) Ê
dr
d)
d
d)
œ
cos t cos# t sin# t
(1 cos t)#
œ
cos t "
(1 cos t)#
œ 1 "cos t œ
"
cos t 1
(sin )) (sin ))(2))‰ œ a)# cos ) 2) sin )b œ )() cos ) # sin ))
œ () cos ) (sin ))(1)) sin ) œ ) cos )
dr
25. r œ sec ) csc ) Ê d)
œ (sec ))(csc ) cot )) (csc ))(sec ) tan ))
"
"
cos
)
sin ) ‰
"
"
#
#
œ ˆ cos ) ‰ ˆ sin ) ‰ ˆ sin ) ‰ ˆ sin" ) ‰ ˆ cos" ) ‰ ˆ cos
) œ sin# ) cos# ) œ sec ) csc )
26. r œ (1 sec )) sin ) Ê
27. p œ & "
cot q
œ
sin q cos q
cos q
Ê
dp
dq
œ
dp
dq
dp
dq
tan q
1 tan q
31. p œ
q sin q
q2 1
Ê
Ê
dp
dq
œ sec# q
œ (1 csc q)(sin q) (cos q)(csc q cot q) œ (sin q 1) cot# q œ sin q csc# q
(cos q)(cos q sin q) (sin q cos q)(sin q)
cos# q
cos# q cos q sin q sin# q cos q sin q
cos# q
30. p œ
œ
œ (" sec )) cos ) (sin ))(sec ) tan )) œ (cos ) ") tan# ) œ cos ) sec# )
œ 5 tan q Ê
28. p œ (1 csc q) cos q Ê
29. p œ
dr
d)
œ
"
cos# q
œ sec# q
(1 tan q) asec# qb (tan q) asec# qb
(1 tan q)#
dp
dq
œ
œ
ˆq2 1‰aq cos q sin qa1bb aq sin qba2qb
aq 2 1 b 2
œ
sec# q tan q sec# q tan q sec# q
(1 tan q)#
œ
œ
sec# q
(1 tan q)#
q3 cos q q2 sin q q cos q sin q 2q2 sin q
aq 2 1 b 2
q3 cos q q2 sin q q cos q sin q
aq 2 1 b 2
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
122
Chapter 3 Differentiation
aq sec qbˆ3 sec2 q‰ a3q tan qbaq sec q tan q sec qa1bb
aq sec qb2
3
2
ˆ
3q sec q q sec q 3q sec q tan q 3q sec q q sec q tan2 q sec q tan q‰
q sec3 q 3q2 sec q tan q q sec q tan2 q sec q tan q
32. p œ
œ
3q tan q
q sec q
Ê
dp
dq
œ
œ
aq sec qb2
aq sec qb2
33. (a) y œ csc x Ê yw œ csc x cot x Ê yww œ a(csc x) acsc# xb (cot x)(csc x cot x)b œ csc$ x csc x cot# x
œ (csc x) acsc# x cot# xb œ (csc x) acsc# x csc# x 1b œ 2 csc$ x csc x
(b) y œ sec x Ê yw œ sec x tan x Ê yww œ (sec x) asec# xb (tan x)(sec x tan x) œ sec$ x sec x tan# x
œ (sec x) asec# x tan# xb œ (sec x) asec# x sec# x 1b œ 2 sec$ x sec x
34. (a) y œ 2 sin x Ê yw œ 2 cos x Ê yww œ 2(sin x) œ 2 sin x Ê ywww œ 2 cos x Ê yÐ%Ñ œ 2 sin x
(b) y œ 9 cos x Ê yw œ 9 sin x Ê yww œ 9 cos x Ê ywww œ 9(sin x) œ 9 sin x Ê yÐ%Ñ œ 9 cos x
35. y œ sin x Ê yw œ cos x Ê slope of tangent at
x œ 1 is yw (1) œ cos (1) œ "; slope of
tangent at x œ 0 is yw (0) œ cos (0) œ 1; and
slope of tangent at x œ 3#1 is yw ˆ 3#1 ‰ œ cos 3#1
œ 0. The tangent at (1ß !) is y 0 œ 1(x 1),
or y œ x 1; the tangent at (0ß 0) is
y 0 œ 1(x 0), or y œ x; and the tangent at
ˆ 31
‰
# ß 1 is y œ 1.
36. y œ tan x Ê yw œ sec# x Ê slope of tangent at x œ 13
is sec# ˆ 13 ‰ œ 4; slope of tangent at x œ 0 is sec# (0) œ 1;
and slope of tangent at x œ
1
3
is sec# ˆ 13 ‰ œ 4. The tangent
at ˆ 13 ß tanˆ 13 ‰‰ œ Š 13 ß È3‹ is y È3 œ 4ˆx 13 ‰ ;
the tangent at (0ß 0) is y œ x; and the tangent at ˆ 13 ß tan ˆ 13 ‰‰
œ Š 13 ß È3‹ is y È3 œ 4 ˆx 13 ‰ .
37. y œ sec x Ê yw œ sec x tan x Ê slope of tangent at
x œ 13 is sec ˆ 13 ‰ tan ˆ 13 ‰ œ 2È3 ; slope of tangent
is sec ˆ 14 ‰ tan ˆ 14 ‰ œ È2 . The tangent at the point
ˆ 1 ß sec ˆ 1 ‰‰ œ ˆ 1 ß #‰ is y 2 œ #È3 ˆx 1 ‰ ;
at x œ
1
4
3
3
3
the tangent at the point
3
ˆ 14 ß sec ˆ 14 ‰‰
œ
Š 14 ß È2‹
is y È2
œ È2 ˆx 14 ‰ .
38. y œ 1 cos x Ê yw œ sin x Ê slope of tangent at
È
x œ 13 is sin ˆ 13 ‰ œ #3 ; slope of tangent at x œ
‰ œ 1. The tangent at the point
is sin ˆ 31
#
31
#
ˆ 13 ß " cos ˆ 13 ‰‰ œ ˆ 13 ß 3# ‰
È
is y 3# œ #3 ˆx 13 ‰ ; the tangent at the point
ˆ 3#1 ß " cos ˆ 3#1 ‰‰ œ ˆ 3#1 ß 1‰ is y 1 œ x 3#1
39. Yes, y œ x sin x Ê yw œ " cos x; horizontal tangent occurs where 1 cos x œ 0 Ê cos x œ 1 Ê x œ 1
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 3.5 Derivatives of Trigonometric Functions
40. No, y œ 2x sin x Ê yw œ 2 cos x; horizontal tangent occurs where 2 cos x œ 0 Ê cos x œ #. But there
are no x-values for which cos x œ #.
41. No, y œ x cot x Ê yw œ 1 csc# x; horizontal tangent occurs where 1 csc# x œ 0 Ê csc# x œ 1. But there
are no x-values for which csc# x œ 1.
42. Yes, y œ x 2 cos x Ê yw œ 1 2 sin x; horizontal tangent occurs where 1 2 sin x œ 0 Ê 1 œ 2 sin x
Ê "# œ sin x Ê x œ 16 or x œ 561
43. We want all points on the curve where the tangent
line has slope 2. Thus, y œ tan x Ê yw œ sec# x so
that yw œ 2 Ê sec# x œ 2 Ê sec x œ „ È2
Ê x œ „ 14 . Then the tangent line at ˆ 14 ß "‰ has
equation y 1 œ 2 ˆx 14 ‰ ; the tangent line at
ˆ 14 ß "‰ has equation y 1 œ 2 ˆx 14 ‰ .
44. We want all points on the curve y œ cot x where
the tangent line has slope 1. Thus y œ cot x
Ê yw œ csc# x so that yw œ 1 Ê csc# x œ 1
Ê csc# x œ 1 Ê csc x œ „ 1 Ê x œ 1# . The
tangent line at ˆ 1# ß !‰ is y œ x 12 .
2 cos x ‰
45. y œ 4 cot x 2 csc x Ê yw œ csc# x 2 csc x cot x œ ˆ sin" x ‰ ˆ 1 sin
x
(a) When x œ 1# , then yw œ 1; the tangent line is y œ x w
1
#
2.
(b) To find the location of the horizontal tangent set y œ 0 Ê 1 2 cos x œ 0 Ê x œ
then y œ % È3 is the horizontal tangent.
46. y œ 1 È2 csc x cot x Ê yw œ È2 csc x cot x csc# x œ ˆ sin" x ‰ Š
1
3
È2 cos x 1
‹
sin x
(a) If x œ 14 , then yw œ 4; the tangent line is y œ 4x 1 4.
(b) To find the location of the horizontal tangent set yw œ 0 Ê È2 cos x 1 œ 0 Ê x œ
xœ
31
4 ,
radians. When x œ 13 ,
31
4
radians. When
then y œ 2 is the horizontal tangent.
47. lim sin ˆ "x #" ‰ œ sin ˆ #" #" ‰ œ sin 0 œ 0
xÄ2
48.
lim
x Ä 16
49. lim1
)Ä
6
È1 cos (1 csc x) œ É1 cos ˆ1 csc ˆ 1 ‰‰ œ È1 cos a1 † a2bb œ È2
sin ) "#
) 16
6
œ
d
d) asin )b¹)œ 1
6
œ cos )¹
)œ 16
œ cosˆ 16 ‰ œ
È3
2
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
123
124
Chapter 3 Differentiation
tan ) 1
) 14
50. lim1
)Ä
4
œ
d
d) atan )b¹)œ 1
œ sec2 )¹
4
œ sec2 ˆ 14 ‰ œ 2
)œ 14
1 ‰
‘
ˆ 1 ‰
‘
ˆ 1 ‰‘ œ sec 1 œ 1
51. lim sec cos x 1 tan ˆ 4 sec
x 1 œ sec 1 1 tan 4 sec 0 1 œ œ sec 1 tan 4
xÄ!
x ‰
ˆ 1 tan 0 ‰
ˆ 1‰
52. lim sin ˆ tan1xtan
2 sec x œ sin tan 0 2 sec 0 œ sin # œ 1
xÄ!
53. lim tan ˆ1 tÄ!
sin t ‰
t
œ tan Š1 lim
tÄ!
1) ‰
54. lim cos ˆ sin
) œ cos Š1 lim
)
sin t
t ‹
‹
) Ä ! sin )
)Ä!
œ tan (1 1) œ 0
"
œ cos Œ1 †
lim
sin )
)
)Ä!
"
œ cos ˆ1 † 1 ‰ œ 1
dv
da
ˆ1‰
55. s œ 2 2 sin t Ê v œ ds
dt œ 2 cos t Ê a œ dt œ 2 sin t Ê j œ dt œ 2 cos t. Therefore, velocity œ v 4
œ È2 m/sec; speed œ ¸v ˆ 1 ‰¸ œ È2 m/sec; acceleration œ a ˆ 1 ‰ œ È2 m/sec# ; jerk œ j ˆ 1 ‰ œ È2 m/sec$ .
4
56. s œ sin t cos t Ê v œ
velocity œ
v ˆ 14 ‰
58.
œ cos t sin t Ê a œ
ds
dt
œ 0 m/sec; speed œ
sin# 3x
x#
xÄ!
57. lim f(x) œ lim
xÄ!
4
¸v ˆ 14 ‰¸
4
œ sin t cos t Ê j œ
dv
dt
œ 0 m/sec; acceleration œ
œ cos t sin t. Therefore
œ È2 m/sec# ; jerk œ j ˆ 1 ‰ œ 0 m/sec$ .
da
dt
a ˆ 14 ‰
4
œ lim 9 ˆ sin3x3x ‰ ˆ sin3x3x ‰ œ 9 so that f is continuous at x œ 0 Ê lim f(x) œ f(0) Ê 9 œ c.
xÄ!
xÄ!
lim g(x) œ lim c (x b) œ b and lim b g(x) œ lim b cos x œ 1 so that g is continuous at x œ 0 Ê lim c g(x)
xÄ!
xÄ!
xÄ!
xÄ!
œ lim b g(x) Ê b œ 1. Now g is not differentiable at x œ 0: At x œ 0, the left-hand derivative is
x Ä !c
xÄ!
d
dx
(x b)¸ x = 0 œ 1, but the right-hand derivative is
(cos x)¸ x=0 œ sin 0 œ 0. The left- and right-hand
d
dx
derivatives can never agree at x œ 0, so g is not differentiable at x œ 0 for any value of b (including b œ 1).
59.
d***
dx***
d%
dx%
(cos x) œ sin x because
(cos x) œ cos x Ê the derivative of cos x any number of times that is a
multiple of 4 is cos x. Thus, dividing 999 by 4 gives 999 œ 249 † 4 3 Ê
œ
$
d
dx$
#%*†%
d
’ dx
#%* % (cos x)“ œ
†
60. (a) y œ sec x œ
Ê
d
dx
"
cos x
$
d
dx$
Ê
d***
dx***
(cos x)
(cos x) œ sin x.
dy
dx
œ
(cos x)(0) (1)(sin x)
(cos x)#
œ
(sin x)(0) (1)(cos x)
(sin x)#
œ
sin x
cos# x
sin x ‰
œ ˆ cos" x ‰ ˆ cos
x œ sec x tan x
(sec x) œ sec x tan x
(b) y œ csc x œ
Ê
d
dx
d
dx
Ê
dy
dx
cos x
sin# x
œ
" ‰ ˆ cos x ‰
œ ˆ sin
x
sin x œ csc x cot x
(csc x) œ csc x cot x
(c) y œ cot x œ
Ê
"
sin x
cos x
sin x
Ê
dy
dx
#
œ
(sin x)(sin x) (cos x)(cos x)
(sin x)#
œ
sin# xcos# x
sin# x
œ
"
sin# x
œ csc# x
(cot x) œ csc x
61. (a) t œ 0 Ä x œ 10 cosa0b œ 10 cm; t œ
cm
(b) t œ 0 Ä v œ 10 sina0b œ 0 sec
;tœ
1
3
1
3
Ä x œ 10 cosˆ 13 ‰ œ 5 cm; t œ 341 Ä x œ 10 cosˆ 341 ‰ œ 5È2 cm
cm
cm
Ä v œ 10 sinˆ 13 ‰ œ 5È3 sec
; t œ 341 Ä v œ 10 sinˆ 341 ‰ œ 5È2 sec
62. (a) t œ 0 Ä x œ 3 cosa0b 4 sina0b œ 3 ft; t œ
1
2
Ä x œ 3 cosˆ 12 ‰ 4 sinˆ 12 ‰ œ 4 ft;
t œ 1 Ä x œ 3 cosa1b 4 sina1b œ 3 ft
ft
(b) t œ 0 Ä v œ 3 sina0b 4 cosa0b œ 4 sec
;tœ
t œ 1 Ä v œ 3 sina1b 4 cosa1b œ 4
1
2
Ä v œ 3 sinˆ 12 ‰ 4 cosˆ 12 ‰ œ 3
ft
sec ;
ft
sec
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 3.5 Derivatives of Trigonometric Functions
125
63.
As h takes on the values of 1, 0.5, 0.3 and 0.1 the corresponding dashed curves of y œ
closer and closer to the black curve y œ cos x because
d
dx
(sin x) œ
is true as h takes on the values of 1, 0.5, 0.3 and 0.1.
sin x
lim sin (x h)
h
hÄ!
sin (x h) sin x
h
get
œ cos x. The same
64.
cos (x h) cos x
h
cos x
lim cos (x h)
œ
sin x.
h
hÄ!
As h takes on the values of 1, 0.5, 0.3, and 0.1 the corresponding dashed curves of y œ
get
closer and closer to the black curve y œ sin x because
The
d
dx
(cos x) œ
same is true as h takes on the values of 1, 0.5, 0.3, and 0.1.
65. (a)
The dashed curves of y œ
sinax hb sinax hb
#h
are closer to the black curve y œ cos x than the corresponding dashed
curves in Exercise 63 illustrating that the centered difference quotient is a better approximation of the derivative of
this function.
(b)
The dashed curves of y œ
cosax hb cosax hb
#h
are closer to the black curve y œ sin x than the corresponding dashed
curves in Exercise 64 illustrating that the centered difference quotient is a better approximation of the derivative of
this function.
66. lim
hÄ!
k0 h k k 0 h k
2h
œ lim
xÄ!
k h k k hk
2h
œ lim 0 œ 0 Ê the limits of the centered difference quotient exists even
hÄ!
though the derivative of f(x) œ kxk does not exist at x œ 0.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
126
Chapter 3 Differentiation
67. y œ tan x Ê yw œ sec# x, so the smallest value
yw œ sec# x takes on is yw œ 1 when x œ 0;
yw has no maximum value since sec# x has no
largest value on ˆ 1# ß 1# ‰ ; yw is never negative
since sec# x
1.
68. y œ cot x Ê yw œ csc# x so yw has no smallest
value since csc# x has no minimum value on
(!ß 1); the largest value of yw is 1, when x œ 1# ;
the slope is never positive since the largest
value yw œ csc2 x takes on is 1.
69. y œ
sin x
x appears to cross the y-axis at y œ 1, since
lim sin x œ 1; y œ sinx2x appears to cross the y-axis
xÄ! x
at y œ 2, since lim sinx2x œ 2; y œ sinx4x appears to
xÄ!
cross the y-axis at y œ 4, since lim sinx4x œ 4.
xÄ!
However, none of these graphs actually cross the y-axis
since x œ 0 is not in the domain of the functions. Also,
lim
xÄ!
sin 5x
x
sin (3x)
x
œ 5, lim
xÄ!
œ k Ê the graphs of y œ
yœ
sin kx
x
œ 3, and lim
sin kx
x
xÄ!
yœ
sin 5x
x ,
sin (3x)
,
x
and
approach 5, 3, and k, respectively, as
x Ä 0. However, the graphs do not actually cross the
y-axis.
70. (a)
sin h
h
h
1
0.01
0.001
0.0001
lim
hÄ!
sin h°
h
ˆ sinh h ‰ ˆ 180
‰
1
.99994923
1
1
1
.017452406
.017453292
.017453292
.017453292
œ lim
xÄ!
1 ‰
sin ˆh† 180
h
œ lim
1
180
hÄ!
1 ‰
sin ˆh† 180
1
180 †h
œ lim
)Ä!
1 sin )
180
)
œ
1
180
() œ h †
1
180 )
(converting to radians)
(b)
cos h1
h
h
1
0.01
0.001
0.0001
lim
hÄ!
0.0001523
0.0000015
0.0000001
0
cos h1
h
(c) In degrees,
œ 0, whether h is measured in degrees or radians.
d
dx
(sin x) œ lim
hÄ!
œ lim ˆsin x †
hÄ!
cos h 1 ‰
h
sin (x h) sin x
h
lim ˆcos x †
hÄ!
1 ‰
œ (sin x)(0) (cos x) ˆ 180
œ
1
180
œ lim
hÄ!
sin h ‰
h
(sin x cos h cos x sin h) sin x
h
œ (sin x) † lim ˆ cos hh 1 ‰ (cos x) † lim ˆ sinh h ‰
hÄ!
hÄ!
cos x
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 3.6 The Chain Rule
(d) In degrees,
œ lim
hÄ!
d
dx
hÄ!
(cos x)(cos h 1) sin x sin h
h
œ (cos x) lim ˆ
hÄ!
(e)
d#
dx#
d#
dx#
cos (x h) cos x
h
(cos x) œ lim
cos h 1 ‰
h
œ lim
hÄ!
œ lim ˆcos x †
hÄ!
127
(cos x cos h sin x sin h) cos x
h
cos h 1 ‰
h
lim ˆsin x †
hÄ!
sin h ‰
h
1 ‰
1
(sin x) lim ˆ sinh h ‰ œ (cos x)(0) (sin x) ˆ 180
œ 180
sin x
hÄ!
1 ‰#
ˆ 180
d$
dx$
(sin x) œ
d
dx
1
ˆ 180
(cos x) œ
d
dx
1
1 ‰#
ˆ 180
sin x‰ œ ˆ 180
cos x;
cos x‰ œ
sin x;
(sin x) œ
d$
dx$
d
dx
#
$
1 ‰
1 ‰
sin x‹ œ ˆ 180
cos x;
Š ˆ 180
(cos x) œ
d
dx
#
$
1 ‰
1 ‰
cos x‹ œ ˆ 180
sin x
Š ˆ 180
3.6 THE CHAIN RULE
1. f(u) œ 6u 9 Ê f w (u) œ 6 Ê f w (g(x)) œ 6; g(x) œ
"
#
x% Ê gw (x) œ 2x$ ; therefore
dy
dx
œ f w (g(x))gw (x) œ 6 † 2x$ œ 12x$
2. f(u) œ 2u$ Ê f w (u) œ 6u# Ê f w (g(x)) œ 6(8x 1)# ; g(x) œ 8x 1 Ê gw (x) œ 8; therefore
œ 6(8x 1)# † 8 œ 48(8x 1)#
dy
dx
œ f w (g(x))gw (x)
3. f(u) œ sin u Ê f w (u) œ cos u Ê f w (g(x)) œ cos (3x 1); g(x) œ 3x 1 Ê gw (x) œ 3; therefore
œ f w (g(x))gw (x)
dy
dx
œ (cos (3x 1))(3) œ 3 cos (3x 1)
4. f(u) œ cos u Ê f w (u) œ sin u Ê f w (g(x)) œ sin ˆ 3x ‰ ; g(x) œ
‰ œ "3 sin ˆ 3x ‰
œ sin ˆ 3x ‰ † ˆ "
3
x
3
Ê gw (x) œ "3 ; therefore
dy
dx
œ f w (g(x))gw (x)
5. f(u) œ cos u Ê f w (u) œ sin u Ê f w (g(x)) œ sin (sin x); g(x) œ sin x Ê gw (x) œ cos x; therefore
dy
dx
œ f w (g(x))gw (x) œ (sin (sin x)) cos x
6. f(u) œ sin u Ê f w (u) œ cos u Ê f w (g(x)) œ cos (x cos x); g(x) œ x cos x Ê gw (x) œ 1 sin x; therefore
dy
dx
œ f w (g(x))gw (x) œ (cos (x cos x))(1 sin x)
7. f(u) œ tan u Ê f w (u) œ sec# u Ê f w (g(x)) œ sec# (10x 5); g(x) œ 10x 5 Ê gw (x) œ 10; therefore
dy
dx
œ f w (g(x))gw (x) œ asec# (10x 5)b (10) œ 10 sec# (10x 5)
8. f(u) œ sec u Ê f w (u) œ sec u tan u Ê f w (g(x)) œ sec ax# 7xb tan ax# 7xb ; g(x) œ x# 7x
Ê gw (x) œ 2x 7; therefore
dy
dx
œ f w (g(x))gw (x) œ (2x 7) sec ax# 7xb tan ax# 7xb
9. With u œ (2x 1), y œ u& :
dy
dx
œ
dy du
du dx
œ 5u% † 2 œ 10(2x 1)%
10. With u œ (4 3x), y œ u* :
dy
dx
œ
dy du
du dx
œ 9u) † (3) œ 27(4 3x))
11. With u œ ˆ1 x7 ‰ , y œ u( :
12. With u œ ˆ x# 1‰ , y œ u"! :
dy
dx
œ
dy
dx
œ
#
13. With u œ Š x8 x "x ‹ , y œ u% :
14. With u œ 3x2 4x 6, y œ u1Î2 :
)
œ 7u) † ˆ "7 ‰ œ ˆ" x7 ‰
dy du
du dx
dy du
du dx
dy
dx
dy
dx
œ
œ
œ 10u"" † ˆ "# ‰ œ 5 ˆ x# 1‰
dy du
du dx
dy du
du dx
œ 4u$ † ˆ x4 1 "‰
x#
""
#
$
œ 4 Š x8 x "x ‹ ˆ x4 1 œ "# u1Î2 † a6x 4b œ
3x 2
È3x2 4x6
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
"‰
x#
128
Chapter 3 Differentiation
15. With u œ tan x, y œ sec u:
16. With u œ 1 "x , y œ cot u:
17. With u œ sin x, y œ u$ :
œ
dy
dx
dy
dx
œ
dy du
du dx
dy
dx
œ
dy du
du dx
18. With u œ cos x, y œ 5u% :
dy
dx
œ
19. p œ È3 t œ (3 t)"Î# Ê
3
20. q œ È
2r r# œ a2r r# b
21. s œ
œ
4
31
4
1
sin 3t 4
51
dp
dt
"Î3
œ
œ acsc# ub ˆ x"# ‰ œ x"# csc# ˆ1 x" ‰
œ 3u# cos x œ 3 asin# xb (cos x)
dy du
du dx
Ê
cos 5t Ê
œ (sec u tan u) asec# xb œ (sec (tan x) tan (tan x)) sec# x
dy du
du dx
"
#
dq
dr
œ a20u& b (sin x) œ 20 acos& xb (sin x)
(3 t)"Î# †
œ
"
3
d
dt
a2r r# b
(3 t) œ "# (3 t)"Î# œ
2 Î3
†
d
dr
a2r r# b œ
ds
dt
œ
cos 3t †
d
dt
(3t) ds
dt
œ cos ˆ 3#1t ‰ †
d
dt
ˆ 3#1t ‰ sin ˆ 3#1t ‰ †
4
31
4
51
(sin 5t) †
d
dt
"
3
"
2È 3 t
a2r r# b
(5t) œ
4
1
2 Î3
2 2r
3a2rr# b2Î3
(2 2r) œ
cos 3t 4
1
sin 5t
(cos 3t sin 5t)
22. s œ sin ˆ 3#1t ‰ cos ˆ 3#1t ‰ Ê
œ 321 ˆcos 3#1t sin 3#1t ‰
23. r œ (csc ) cot ))" Ê
dr
d)
24. r œ 6(sec ) tan ))3Î2 Ê
œ (csc ) cot ))#
dr
d)
d
d)
d
dt
ˆ 3#1t ‰ œ
(csc ) cot )) œ
œ 6 † 3# asec ) tan )b1Î#
d
d) asec
31
2
cos ˆ 3#1t ‰ csc ) cot ) csc# )
(csc ) cot ))#
œ
31
2
sin ˆ 3#1t ‰
csc ) (cot ) csc ))
(csc ) cot ))#
œ
csc )
csc ) cot )
) tan )b œ 9Èsec ) tan )asec ) tan ) sec2 )b
d
d
# d
%
%
#
#
25. y œ x# sin% x x cos# x Ê dy
xb cos# x †
dx œ x dx asin xb sin x † dx ax b x dx acos
d
d
œ x# ˆ4 sin$ x dx
(sin x)‰ 2x sin% x x ˆ2 cos$ x † dx
(cos x)‰ cos# x
d
dx
(x)
œ x# a4 sin$ x cos xb 2x sin% x xa a2 cos$ xb (sin x)b cos# x
œ 4x# sin$ x cos x 2x sin% x 2x sin x cos$ x cos# x
d ˆ"‰
x d
$
$
asin& xb sin& x † dx
x 3 dx acos xb cos x †
œ "x a5 sin' x cos xb asin& xb ˆ x"# ‰ 3x a a3 cos# xb (sin x)b acos$ xb ˆ 3" ‰
26. y œ
"
x
sin& x x
3
cos$ x Ê yw œ
œ 5x sin' x cos x "
x#
" d
x dx
sin& x x cos# x sin x "
3
ˆ x3 ‰
cos$ x
"
" ‰"
7
d
(
'
ˆ
Ê dy
21 (3x 2) 4 #x#
dx œ 21 (3x 2) † dx (3x 2) 7
" ‰# ˆ " ‰
"
'
'
ˆ
#
21 (3x 2) † 3 (1) 4 #x#
x$ œ (3x 2) $
x Š4 "# ‹
27. y œ
œ
d
dx
(1) ˆ4 " ‰#
#x #
†
d
dx
ˆ4 " ‰
#x#
#x
%
28. y œ (5 2x)$ "8 ˆ x2 1‰ Ê
dy
dx
$
$
œ 3(5 2x)% (2) 84 ˆ x2 1‰ ˆ x2# ‰ œ 6(5 2x)% ˆ x"# ‰ ˆ x2 1‰
$
œ
6
(5 2x)%
Š 2x 1‹
x#
29. y œ (4x 3)% (x 1)$ Ê
%
dy
dx
œ (4x 3)% (3)(x 1)% †
%
$
$
d
dx
(x 1) (x 1)$ (4)(4x 3)$ †
%
œ (4x 3) (3)(x 1) (1) (x 1) (4)(4x 3) (4) œ 3(4x 3) (x 1)
œ
$
(4x 3)
(x 1)%
c3(4x 3) 16(x 1)d œ
%
d
dx
$
(4x 3)
16(4x 3) (x 1)$
$
(4x 3) (4x 7)
(x 1)%
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 3.6 The Chain Rule
'
30. y œ (2x 5)" ax# 5xb Ê
&
2 ax# 5xb
(2x 5)#
œ 6 ax# 5xb dy
dx
&
'
œ (2x 5)" (6) ax# 5xb (2x 5) ax# 5xb (1)(2x 5)# (2)
'
d ˆ
d
31. h(x) œ x tan ˆ2Èx‰ 7 Ê hw (x) œ x dx
tan ˆ2x"Î# ‰‰ tan ˆ2x"Î# ‰ † dx
(x) 0
d ˆ "Î# ‰
"
# ˆ "Î# ‰
"Î# ‰
#ˆ È ‰
ˆ
ˆ
† dx 2x
œ x sec 2x
tan 2x
œ x sec 2 x † È tan 2Èx‰ œ Èx sec# ˆ2Èx‰ tan ˆ2Èx‰
x
d ˆ
d
32. k(x) œ x# sec ˆ "x ‰ Ê kw (x) œ x# dx
sec x" ‰ sec ˆ x" ‰ † dx
ax# b œ x# sec ˆ x" ‰ tan ˆ x" ‰ †
œ x# sec ˆ "x ‰ tan ˆ x" ‰ † ˆ x"# ‰ 2x sec ˆ x" ‰ œ 2x sec ˆ x" ‰ sec ˆ x" ‰ tan ˆ x" ‰
33. faxb œ È7 x sec x Ê f w axb œ "# a7 x sec xb1Î2 ax † asec x tan xb asec xb † "b œ
34. gaxb œ
œ
tan 3x
ax 7 b 4
ax 7b4 ˆsec2 3x†$‰ atan 3xb4ax 7b3 †1
ax 7b4 ‘2
Ê gw axb œ
x sec x tan x sec x
#È7x sec x
ax 7b3 ˆ$ax 7bsec2 3x 4tan 3x‰
ax 7 b 8
ˆ$ax 7bsec2 3x 4tan 3x‰
ax 7 b5
#
35. f()) œ ˆ 1 sincos) ) ‰ Ê f w ()) œ 2 ˆ 1 sincos) ) ‰ †
œ
œ
ˆ x" ‰ 2x sec ˆ x" ‰
d
dx
(2 sin )) acos ) cos# ) sin# )b
(1 cos ))$
sin 3t ‰
36. g(t) œ ˆ 1 3 2t
"
(2 sin )) (cos ) 1)
(1 cos ))$
œ
3 2t
1 sin 3t
œ
37. r œ sin a)# b cos (2)) Ê
d
d)
Ê gw (t) œ
ˆ 1 sincos) ) ‰ œ
œ
†
(1 cos ))(cos )) (sin ))(sin ))
(1 cos ))#
2 sin )
(1 cos ))#
a1 sin 3tba2b a3 2tba3 cos 3tb
a1 sin 3tb2
œ sin a)# b (sin 2))
dr
d)
2 sin )
1 cos )
d
d)
œ
2 2sin 3t 9 cos 3t 6t cos 3t
a1 sin 3tb2
(2)) cos (2)) acos a)# bb †
d
d)
a) # b
œ sin a)# b (sin 2))(2) (cos 2)) acos a)# bb (2)) œ 2 sin a)# b sin (#)) 2) cos (2)) cos a)# b
38. r œ Šsec È)‹ tan ˆ ") ‰ Ê
dr
d)
œ )"# sec È) sec# ˆ ") ‰ 39. q œ sin Š Ètt 1 ‹ Ê
œ cos Š Ètt 1 ‹ †
2
t1
dq
dt
41. y œ sin# (1t 2) Ê
"
#È )
tan ˆ ") ‰ sec È) tan È) œ Šsec È)‹ ”
œ cos Š Ètt 1 ‹ †
dq
dt
Èt 1 40. q œ cot ˆ sint t ‰ Ê
œ Šsec È)‹ ˆ sec# ") ‰ ˆ )"# ‰ tan ˆ ") ‰ Šsec È) tan È)‹ Š
È
t
t
1
d
dt
Èt 1 (1)t †
d
dt
sec# ˆ )" ‰
)# •
ˆÈ t 1 ‰
ˆÈ t 1 ‰
#
1) t
œ cos Š Ètt 1 ‹ Š 2(t
‹ œ Š 2(tt1)2$Î# ‹ cos Š Ètt 1 ‹
2(t 1)$Î#
œ csc# ˆ sint t ‰ †
d
dt
ˆ sint t ‰ œ ˆcsc# ˆ sint t ‰‰ ˆ t cos tt# sin t ‰
œ 2 sin (1t 2) †
dy
dt
Š Ètt 1 ‹ œ cos Š Ètt 1 ‹ †
tan È) tan ˆ ") ‰
#È )
"
‹
#È )
d
dt
sin (1t 2) œ 2 sin (1t 2) † cos (1t 2) †
d
dt
(1t 2)
œ 21 sin (1t 2) cos (1t 2)
42. y œ sec# 1t Ê
dy
dt
œ (2 sec 1t) †
43. y œ (1 cos 2t)% Ê
44. y œ ˆ1 cot ˆ #t ‰‰
œ
csc# ˆ #t ‰
$
ˆ1 cot ˆ t ‰‰
#
#
dy
dt
Ê
d
dt
(sec 1t) œ (2 sec 1t)(sec 1t tan 1t) †
œ 4(1 cos 2t)& †
dy
dt
œ 2 ˆ1 cot ˆ #t ‰‰
d
dt
$
d
dt
(1t) œ 21 sec# 1t tan 1t
(1 cos 2t) œ 4(1 cos 2t)& (sin 2t) †
†
dˆ
dt 1
cot ˆ #t ‰‰ œ 2 ˆ1 cot ˆ #t ‰‰
$
d
dt
(2t) œ
8 sin 2t
(1 cos 2t)&
† ˆcsc# ˆ #t ‰‰ †
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
dˆt‰
dt #
129
130
Chapter 3 Differentiation
45. y œ at tan tb10 Ê
46. y œ ˆt3Î4 sin t‰
œ
4 Î3
œ t1 asin tb4Î3 Ê
dy
dt
œ t1 ˆ 34 ‰asin tb1Î3 cost t2 asin tb4Î3 œ
4asin tb1Î3 cos t
3t
asin tb4Î3
t2
asin tb1Î3 a4t cos t 3cos tb
3t2
3
2
47. y œ Š t3 t 4t ‹ Ê
4‰
48. y œ ˆ 3t
5t 2
œ
œ 10at tan tb9 at † sec2 t 1 † tan tb œ 10 t9 tan9 tat sec2 t tan tb œ 10 t10 tan9 t sec2 t 10 t9 tan10 t
dy
dt
5
2
2
dy
dt
œ 3Š t3 t 4t ‹ †
dy
dt
4‰
œ 5ˆ 3t
5t 2
Ê
ˆt3 4t‰a2tb t2 ˆ3t2 4‰
6
at3 4tb2
†
a5t 2b†3 a3t 4b†5
a5t 2b2
œ
3t4
at3 4tb2
†
2t4 8t2 3t4 4t2
at3 4tb2
6
2‰
œ 5ˆ 5t
†
3t 4
œ
15t 6 15t 20
a5t 2b2
3t4 ˆt4 4t2 ‰
t4 at2 4b4
3t2 ˆt2 4‰
at2 4b4
œ
6
5t 2b
œ 5 aa3t
†
4 b6
26
a5t 2b2
130a5t 2b4
a3t 4b6
49. y œ sin acos (2t 5)b Ê
dy
dt
œ cos (cos (2t 5)) †
d
dt
cos (2t 5) œ cos (cos (2t 5)) † (sin (2t 5)) †
d
dt
(2t 5)
œ 2 cos (cos (2t 5))(sin (2t 5))
ˆ
ˆ t ‰‰ †
50. y œ cos ˆ5 sin ˆ 3t ‰‰ Ê dy
dt œ sin 5 sin 3
œ 53 sin ˆ5 sin ˆ 3t ‰‰ ˆcos ˆ 3t ‰‰
d
dt
ˆ5 sin ˆ 3t ‰‰ œ sin ˆ5 sin ˆ 3t ‰‰ ˆ5 cos ˆ 3t ‰‰ †
$
dy
% ˆ t ‰‘# d † dt 1
dt œ 3 1 tan
1#
" ‘
% ˆ t ‰‘# $ˆ t ‰
#ˆ t ‰
tan 1#
tan 1# sec 1# † 1# œ 1 51. y œ 1 tan% ˆ 1t# ‰‘ Ê
œ 12 1 52. y œ
"
6
$
c1 cos# (7t)d Ê
"Î#
Ê
œ "# a1 cos at# bb
"Î#
53. y œ a1 cos at# bb
dy
dt
2 cos ŒÉ1 Èt
É1 Èt†2Èt
#
tan% ˆ 1t# ‰‘ tan$ ˆ 1t# ‰ sec# ˆ 1t# ‰‘
#
d
dt
tan ˆ 1t# ‰‘
#
c1 cos# (7t)d † 2 cos (7t)(sin (7t))(7) œ 7 c1 cos# (7t)d (cos (7t) sin (7t))
3
6
"
#
#
tan% ˆ 1t# ‰‘ œ 3 1 tan% ˆ 1t# ‰‘ 4 tan$ ˆ 1t# ‰ †
a1 cos at# bb
"Î#
†
d
dt
a1 cos at# bb œ
#
"
#
a1 cos at# bb
"Î#
ˆsin at# b †
d
dt
a t# b ‰
dy
dt
œ 4 cos ŒÉ1 Èt †
d
dt
"
ŒÉ1 Èt œ 4 cos ŒÉ1 Èt †
# É 1 È t
†
d
dt
ˆ1 Èt‰
cos ŒÉ1 Èt
œ
55. y œ tan2 asin3 tb Ê
œ
ˆ 3t ‰
at b
asin at# bb † 2t œ È1t sin
cos at# b
54. y œ 4 sin ŒÉ1 Èt Ê
œ
œ
dy
dt
d
dt
Ét tÈt
dy
dt
56. y œ cos4 asec2 3tb Ê
œ 2 tanasin3 tb † sec2 asin3 tb † a3sin2 t † acos tbb œ 6 tanasin3 tbsec2 asin3 tbsin2 t cos t
dy
dt
œ 4 cos3 asec2 a3tbbasinasec2 a3tbb † 2 aseca3tbbaseca3tb tana3tb † 3bb
œ 24 cos3 asec2 a3tbbsinasec2 a3tbbsec2 a3tb tana3tb
4
57. y œ 3ta2t2 5b Ê
dy
dt
3
58. y œ Ê3t É2 È1 t Ê
œ
"
#Ê3tÉ2È1t
3 4
3
3
œ 3t † 4a2t2 5b a4tb 3 † a2t2 5b œ 3a2t2 5b ’16t2 2t2 5“ œ 3a2t2 5b a18t2 5b
dy
dt
"
#É2È1t
1 Î2
œ "# Œ3t É2 È1 t
†
1
#È 1 t œ
"
#Ê3tÉ2È1t
"
Œ3 # Š2 È1 t‹
12È1tÉ2È1t "
4È1tÉ2È1t
œ
1 Î2
"
# a1
tb1Î2 a1b
12È1tÉ2È1t "
8È1tÉ2È1tÊ3tÉ2È1t
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 3.6 The Chain Rule
$
#
#
59. y œ ˆ1 "x ‰ Ê yw œ 3 ˆ1 x" ‰ ˆ x"# ‰ œ x3# ˆ1 x" ‰ Ê yww œ ˆ x3# ‰ †
#
œ ˆ x3# ‰ ˆ2 ˆ1 x" ‰ ˆ x"# ‰‰ ˆ x6$ ‰ ˆ1 x" ‰ œ
œ x6$ ˆ1 x" ‰ ˆ1 2x ‰
60. y œ ˆ1 Èx‰
Ê yww œ
œ
"
#
œ
"
#x
61. y œ
"
#
"
Ê yw œ ˆ1 Èx‰
œ
ˆ1 x" ‰ ˆ "# x"Î# ‰ œ
"
#
6
x$
ˆ1 x" ‰# œ
ˆ1 x" ‰# ˆ1 x" ‰# †
ˆ1 Èx‰$ Š
ˆ x3# ‰
ˆ1 x" ‰ ˆ x" 1 x" ‰
ˆ1 Èx‰# x"Î#
#
"
#È x
$
x" ˆ1 Èx‰ “ œ
"
#
1‹ œ
"
#x
"
#
x" ˆ1 Èx‰
ˆ1 Èx‰$ Š 3# $
"# x"Î# ˆ1 Èx‰ 1‘
"
‹
#Èx
cot (3x 1) Ê yw œ 9" csc# (3x 1)(3) œ 3" csc# (3x 1) Ê yww œ ˆ 32 ‰ (csc (3x 1) †
2
3
d
dx
#
$
’ˆ1 Èx‰ ˆ "# x$Î# ‰ x"Î# (2) ˆ1 Èx‰ ˆ "# x"Î# ‰“
$Î# ˆ
1 Èx‰
’ "
# x
"
9
#
6
x%
d
dx
6
x$
131
csc (3x 1)(csc (3x 1) cot (3x 1) †
d
dx
csc (3x 1))
d
dx
#
(3x 1)) œ 2 csc (3x 1) cot (3x 1)
62. y œ 9 tan ˆ x3 ‰ Ê yw œ 9 ˆsec# ˆ x3 ‰‰ ˆ 3" ‰ œ 3 sec# ˆ x3 ‰ Ê yww œ 3 † 2 sec ˆ x3 ‰ ˆsec ˆ x3 ‰ tan ˆ x3 ‰‰ ˆ "3 ‰ œ 2 sec# ˆ 3x ‰ tan ˆ 3x ‰
63. y œ xa2x 1b4 Ê yw œ x † 4a2x 1b3 a2b 1 † a2x 1b4 œ a2x 1b3 a8x a2x 1bb œ a2x 1b3 a10x 1b
Ê yww œ a2x 1b3 a10b 3a2x 1b2 a2ba10x 1b œ 2a2x 1b2 a5a2x 1b 3a10x 1bb œ 2a2x 1b2 a40x 8b
œ 16a2x 1b2 a5x 1b
5
4
5
4
4
64. y œ x2 ax3 1b Ê yw œ x2 † 5ax3 1b a3x2 b 2xax3 1b œ xax3 1b ’15x3 2ax3 1b“ œ ax3 1b a17x4 2xb
4
3
3
Ê yww œ ax3 1b a68x3 2b 4ax3 1b a3x2 ba17x4 2xb œ 2ax3 1b ’ax3 1ba34x3 1b 6x2 a17x4 2xb“
3
œ 2ax3 1b a136x6 47x3 1b
65. g(x) œ Èx Ê gw (x) œ
"
#È x
Ê g(1) œ 1 and gw (1) œ
therefore, (f ‰ g)w (1) œ f w (g(1)) † gw (1) œ 5 †
"
#
œ
"
u#
; f(u) œ u& 1 Ê f w (u) œ 5u% Ê f w (g(1)) œ f w (1) œ 5;
5
#
66. g(x) œ (1 x)" Ê gw (x) œ (1 x)# (1) œ
Ê f w (u) œ
"
#
"
(1x)#
Ê g(1) œ
"
#
w
and gw (1) œ
"
4
; f(u) œ 1 Ê f w (g(1)) œ f w ˆ #" ‰ œ 4; therefore, (f ‰ g)w (1) œ f (g(1))gw (1) œ 4 †
67. g(x) œ 5Èx Ê gw (x) œ
Ê f w (g(1)) œ f w (5) œ 5
Ê g(1) œ 5 and gw (1) œ #5
#È x
1
1
# ˆ1‰
10 csc
# œ 10 ; therefore,
"
4
"
u
œ1
1‰
; f(u) œ cot ˆ 110u ‰ Ê f w (u) œ csc# ˆ 110u ‰ ˆ 10
œ
1
10
csc# ˆ 110u ‰
1
(f ‰ g)w (1) œ f w (g(1))gw (1) œ 10
† 5# = 14
68. g(x) œ 1x Ê gw (x) œ 1 Ê g ˆ "4 ‰ œ 14 and gw ˆ 4" ‰ œ 1; f(u) œ u sec# u Ê f w (u) œ 1 2 sec u † sec u tan u
œ 1 2 sec# u tan u Ê f w ˆg ˆ "4 ‰‰ œ f w ˆ 14 ‰ œ 1 2 sec# 14 tan 14 œ 5; therefore, (f ‰ g)w ˆ 4" ‰ œ f w ˆg ˆ 4" ‰‰ gw ˆ 4" ‰ œ 51
69. g(x) œ 10x# x 1 Ê gw (x) œ 20x 1 Ê g(0) œ 1 and gw (0) œ 1; f(u) œ
œ
2u# 2
au # 1 b #
Ê f w (u) œ
au# 1b(2) (2u)(2u)
au # 1 b #
Ê f w (g(0)) œ f w (1) œ 0; therefore, (f ‰ g)w (0) œ f w (g(0))gw (0) œ 0 † 1 œ 0
"
2
w
x# 1 Ê g (x) œ x$ Ê g(1) œ 0 and
4(u 1)
1 ‰ (u 1)(1) (u 1)(1)
2 ˆ uu œ 2(u(u1)(2)
1 †
(u 1)#
1)$ œ (u 1)$
w
w
w
70. g(x) œ
œ
2u
u # 1
#
1‰
1‰
gw (1) œ 2; f(u) œ ˆ uu Ê f w (u) œ 2 ˆ uu 1
1
Ê f w (g(1)) œ f w (0) œ 4; therefore,
(f ‰ g) (1) œ f (g(1))g (1) œ (4)(2) œ 8
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
d
du
1‰
ˆ uu 1
132
Chapter 3 Differentiation
71. y œ fagaxbb, f w a3b œ 1, gw a2b œ 5, ga2b œ 3 Ê y w œ f w agaxbbg w axb Ê y w ¹
x œ2
œ a1b † 5 œ 5
72. r œ sinafatbb, fa0b œ 13 , f w a0b œ 4 Ê
73. (a) y œ 2f(x) Ê
dy
dx
œ 2f w (x) Ê
(b) y œ f(x) g(x) Ê
(c) y œ f(x) † g(x) Ê
dr
dt
œ cosafatbb † f w atb Ê
œ f w (x) gw (x) Ê
dy
dx
dy
dx
œ
g(x)f w (x) f(x)gw (x)
[g(x)]#
(e) y œ f(g(x)) Ê
dy
dx
œ f w (g(x))gw (x) Ê
(d) y œ
f(x)
g(x)
Ê
"
#
Ê
dy
dx ¹ x=2
dy
dx
œ
(g) y œ (g(x))# Ê
dy
dx
œ 2(g(x))$ † gw (x) Ê
Ê
dy
dx ¹ x=2
œ
"
#
(f(x))"Î# † f w (x) œ
"Î#
Ê
"
#
# "Î#
dy
dx
a(f(2))# (g(2)) b
74. (a) y œ 5f(x) g(x) Ê
(b) y œ f(x)(g(x))$ Ê
f (x)
#Èf(x)
œ
dy
dx ¹ x=3
dy
dx ¹ x=2
œ
"
3
f (2)
#Èf(2)
dy
dx ¹ x=1
ˆ "3 ‰
#È 8
œ
œ
37
6
(3) œ 1
œ
"
6È 8
"
1 #È 2
œ
œ 2(g(3))$ gw (3) œ 2(4)$ † 5 œ
"Î#
œ
È2
24
5
3#
a2f(x) † f w (x) 2g(x) † gw (x)b
"
#
a8# 2# b
"Î#
ˆ2 † 8 †
"
3
2 † 2 † (3)‰ œ 3È517
œ 5f w (1) gw (1) œ 5 ˆ 3" ‰ ˆ 38 ‰ œ 1
œ f(x) a3(g(x))# gw (x)b (g(x))$ f w (x) Ê
dy
dx
(2) ˆ "3 ‰ (8)(3)
##
w
a2f(2)f w (2) 2g(2)gw (2)b œ
œ 5f w (x) gw (x) Ê
dy
dx
Ê
a(f(x))# (g(x))# b
œ
œ f(3)gw (3) g(3)f w (3) œ 3 † 5 (4)(21) œ 15 81
œ f w (g(2))gw (2) œ f w (2)(3) œ
w
(f) y œ (f(x))"Î# Ê
(h) y œ a(f(x))# (g(x))# b
dy
dx ¹ x=3
g(2)f w (2) f(2)gw (2)
[g(2)]#
œ
dy
dx ¹ x=2
2
3
œ f w (3) gw (3) œ 21 5
dy
dx ¹ x=3
œ f(x)gw (x) g(x)f w (x) Ê
dy
dx
œ cosafa0bb † f w a0b œ cosˆ 13 ‰ † 4 œ ˆ "# ‰ † 4 œ 2
dr
dt ¹tœ0
œ 2f w (2) œ 2 ˆ "3 ‰ œ
dy
dx ¹ x=2
œ f w aga2bbgw a2b œ f w a3b † 5
dy
dx ¹ x = 0
œ $f(0)(g(0))# gw (0) (g(0))$ f w (0)
œ 3(1)(1)# ˆ 3" ‰ (1)$ (5) œ 6
(c) y œ
œ
f(x)
g(x) 1
Ê
(g(x) 1)f w (x) f(x) gw (x)
(g(x) 1)#
œ
dy
dx
(4") ˆ "3 ‰(3) ˆ 83 ‰
(41)#
Ê
dy
dx ¹ x = 1
(g(1) 1)f w (1) f(1)gw (1)
(g(1) 1)#
œ
œ1
(d) y œ f(g(x)) Ê
dy
dx
œ f w (g(x))gw (x) Ê
dy
dx ¹ x = 0
œ f w (g(0))gw (0) œ f w (1) ˆ "3 ‰ œ ˆ "3 ‰ ˆ 3" ‰ œ 9"
(e) y œ g(f(x)) Ê
dy
dx
œ gw (f(x))f w (x) Ê
dy
dx ¹ x = 0
œ gw (f(0))f w (0) œ gw (1)(5) œ ˆ 83 ‰ (5) œ 40
3
(f) y œ ax"" f(x)b
#
Ê
œ 2 ax"" f(x)b
dy
dx
$
a11x"! f w (x)b Ê
dy
dx ¹ x=1
œ 2(1 f(1))$ a11 f w (1)b
"
‰
œ 2(1 3)$ ˆ11 "3 ‰ œ ˆ 42$ ‰ ˆ 32
3 œ 3
(g) y œ f(x g(x)) Ê
dy
dx
œ f w (x g(x)) a1 gw (x)b Ê
dy
dx ¹ x = 0
œ f w (0 g(0)) a1 gw (0)b œ f w (1) ˆ1 "3 ‰
œ ˆ "3 ‰ ˆ 43 ‰ œ 49
75.
ds
dt
œ
ds
d)
†
d)
dt :
s œ cos ) Ê
76.
dy
dt
œ
dy
dx
†
dx
dt :
y œ x# 7x 5 Ê
77. With y œ x, we should get
(a) y œ
(b)
u
5
7 Ê
y œ 1 "u Ê
"
œ "
u# † (x 1)#
dy
du
dy
du
œ
œ
"
5
œ sin ) Ê
ds
d)
dy
dx
dy
dx
œ 2x 7 Ê
œ
dy
dx ¹ x = 1
œ 9 so that
dy
dt
œ
ds
dt
dy
dx
œ
†
dx
dt
ds
d)
†
d)
dt
œ9†
œ 1†5œ 5
"
3
œ3
œ 1 for both (a) and (b):
; u œ 5x 35 Ê
"
u#
œ sin ˆ 3#1 ‰ œ 1 so that
ds ¸
d) ) = 321
; u œ (x 1)
"
a(x 1) " b#
†
"
(x 1)#
"
du
dx
œ 5; therefore,
Ê
du
dx
œ (x 1)# †
dy
dy
dx œ du
#
†
œ (x 1) (1) œ
"
(x 1)#
du
"
dx œ 5
"
(x 1)# ;
† 5 œ 1, as expected
therefore
dy
dx
œ
œ 1, again as expected
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
dy
du
†
du
dx
Section 3.6 The Chain Rule
78. With y œ x$Î# , we should get
(a) y œ u$ Ê
dy
du
as expected.
(b) y œ Èu Ê
œ
dy
dx
3
#
x"Î# for both (a) and (b):
œ 3u# ; u œ Èx Ê
dy
du
œ
"
#È u
du
dx
; u œ x$ Ê
du
dx
"
#È x
; therefore,
dy
dx
œ
dy
du
†
du
dx
œ 3u# †
œ 3x# ; therefore,
dy
dx
œ
dy
du
†
du
dx
œ
œ
"
#Èu
"
#È x
#
œ 3 ˆÈx‰ †
† 3x# œ
"
#È x $
"
#Èx
† 3x# œ
3
#
œ
3
#
Èx,
x"Î# ,
again as expected.
2
2
1‰
1‰
1‰
79. y œ ˆ xx and x œ 0 Ê y œ ˆ 00 œ a1b2 œ 1. yw œ 2ˆ xx 1
1
1 †
yw ¹
xœ0
œ
4 a0 1 b
a0 1 b 3
œ
4
13
ax 1b†1 ax 1b†1
ax 1 b 2
1b
2
œ 2 aaxx 1 b ax 1 b 2 œ
œ 4 Ê y 1 œ 4ax 0b Ê y œ 4x 1
80. y œ Èx2 x 7 and x œ 2 Ê y œ Éa2b2 a2b 7 œ È9 œ 3. y w œ "# ax2 x 7b
y w¹
xœ2
œ
2 a2 b 1
2 É a2 b 2 a 2 b 7
81. y œ 2 tan ˆ 14x ‰ Ê
(a)
dy
dx ¹ x = 1
œ
1
#
dy
dx
4 ax 1 b
ax 1 b 3
œ
3
6
œ
"
#
œ ˆ2 sec#
1 Î2
a2x 1b œ
2x 1
2 È x 2 x 7
Ê y 3 œ "# ax 2b Ê y œ "# x 2
1x ‰ ˆ 1 ‰
4
4
œ
1
#
sec#
1x
4
sec# ˆ 14 ‰ œ 1 Ê slope of tangent is 2; thus, y(1) œ 2 tan ˆ 14 ‰ œ 2 and yw (1) œ 1 Ê tangent line is
given by y 2 œ 1(x 1) Ê y œ 1x 2 1
(b) yw œ 1# sec# ˆ 14x ‰ and the smallest value the secant function can have in # x 2 is 1 Ê the minimum
value of yw is 1# and that occurs when 1# œ 1# sec# ˆ 14x ‰ Ê 1 œ sec# ˆ 14x ‰ Ê „ 1 œ sec ˆ 14x ‰ Ê x œ 0.
82. (a) y œ sin 2x Ê yw œ 2 cos 2x Ê yw (0) œ 2 cos (0) œ 2 Ê tangent to y œ sin 2x at the origin is y œ 2x;
y œ sin ˆ x# ‰ Ê yw œ "# cos ˆ x# ‰ Ê yw (0) œ "# cos 0 œ "# Ê tangent to y œ sin ˆ x# ‰ at the origin is
y œ "# x. The tangents are perpendicular to each other at the origin since the product of their slopes is 1.
(b) y œ sin (mx) Ê yw œ m cos (mx) Ê yw (0) œ m cos 0 œ m; y œ sin ˆ mx ‰ Ê yw œ m" cos ˆ mx ‰
Ê yw (0) œ m" cos (0) œ m" . Since m † ˆ m" ‰ œ 1, the tangent lines are perpendicular at the origin.
(c) y œ sin (mx) Ê yw œ m cos (mx). The largest value cos (mx) can attain is 1 at x œ 0 Ê the largest value
yw can attain is kmk because kyw k œ km cos (mx)k œ kmk kcos mxk Ÿ kmk † 1 œ kmk . Also, y œ sin ˆ mx ‰
ˆ x ‰¸ Ÿ ¸ m" ¸ ¸cos ˆ mx ‰¸ Ÿ km" k Ê the largest value yw can attain is ¸ m" ¸ .
Ê yw œ m" cos ˆ mx ‰ Ê kyw k œ ¸ "
m cos m
(d) y œ sin (mx) Ê yw œ m cos (mx) Ê yw (0) œ m Ê slope of curve at the origin is m. Also, sin (mx) completes
m periods on [0ß 21]. Therefore the slope of the curve y œ sin (mx) at the origin is the same as the number
of periods it completes on [0ß 21]. In particular, for large m, we can think of “compressing" the graph of
y œ sin x horizontally which gives more periods completed on [0ß 21], but also increases the slope of the
graph at the origin.
83. s œ A cos (21bt) Ê v œ
ds
dt
œ A sin (21bt)(21b) œ 21bA sin (21bt). If we replace b with 2b to double the
frequency, the velocity formula gives v œ 41bA sin (41bt) Ê doubling the frequency causes the velocity to
# #
double. Also v œ #1bA sin (21bt) Ê a œ dv
dt œ 41 b A cos (21bt). If we replace b with 2b in the
acceleration formula, we get a œ 161# b# A cos (41bt) Ê doubling the frequency causes the acceleration to
$ $
quadruple. Finally, a œ 41# b# A cos (21bt) Ê j œ da
dt œ 81 b A sin (21bt). If we replace b with 2b in the jerk
formula, we get j œ 641$ b$ A sin (41bt) Ê doubling the frequency multiplies the jerk by a factor of 8.
21
21
21 ‰
84. (a) y œ 37 sin 365
(x 101)‘ 25 Ê yw œ 37 cos 365
(x 101)‘ ˆ 365
œ
741
365
21
cos 365
(x 101)‘ .
The temperature is increasing the fastest when yw is as large as possible. The largest value of
21
21
cos 365
(x 101)‘ is 1 and occurs when 365
(x 101) œ 0 Ê x œ 101 Ê on day 101 of the year
( µ April 11), the temperature is increasing the fastest.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
133
134
Chapter 3 Differentiation
(b) yw (101) œ
741
365
21
cos 365
(101 101)‘ œ
85. s œ (" 4t)"Î# Ê v œ
v œ 2(" 4t)
"Î#
ds
dt
œ
Ê aœ
dv
dt
86. We need to show a œ
œ
k
2È s
† kÈs œ
87. v proportional to
œ 2sk$Î# †
dx
dt
88. Let
k
Ès
kT
2
k
#
cos (0) œ
741
365
¸ 0.64 °F/day
(1 4t)"Î# (4) œ 2(1 4t)"Î# Ê v(6) œ 2(" % † 6)"Î# œ
"
#
œ † 2(1 4t)
is constant: a œ
dv
dt
œ
$Î#
dv
ds
(4) œ 4(1 4t)
†
ds
dt
and
dv
ds
œ
"
Ès
Ê vœ
k
Ès
for some constant k Ê
dv
ds
m/sec;
Ê a(6) œ 4(1 4 † 6)$Î# œ 14#5 m/sec#
ˆkÈs‰ œ
k
2È s
œ 2sk$Î# . Thus, a œ
#
œ f(x). Then, a œ
dT
dL
d
ds
$Î#
2
5
Ê aœ
dv
ds
†
†
ds
dt
ds
dt
œ
dv
ds
†v
which is a constant.
œ k# ˆ s"# ‰ Ê acceleration is a constant times
89. T œ 21É Lg Ê
œ
#
dv
dt
"
#
741
365
dv
dt
œ 21 †
œ
"
#É Lg
dv
dx
†
dx
dt
œ
†
"
g
œ
1
gÉ Lg
dv
dx
† f(x) œ
œ
1
ÈgL
d
dx
"
s#
dT
du
œ
dv
ds
œ
dv
ds
†v
so a is inversely proportional to s# .
ˆ dx
‰
dt † f(x) œ
. Therefore,
dv
dt
œ
d
dx
(f(x)) † f(x) œ f w (x)f(x), as required.
dT
dL
†
dL
du
œ
1
ÈgL
† kL œ
1 kÈ L
Èg
œ
"
#
† 21kÉ Lg
, as required.
90. No. The chain rule says that when g is differentiable at 0 and f is differentiable at g(0), then f ‰ g is
differentiable at 0. But the chain rule says nothing about what happens when g is not differentiable at 0 so
there is no contradiction.
91. As h Ä 0, the graph of y œ
sin 2(xh)sin 2x
h
approaches the graph of y œ 2 cos 2x because
lim
hÄ!
sin 2(xh)sin 2x
h
œ
d
dx
(sin 2x) œ 2 cos 2x.
92. As h Ä 0, the graph of y œ
cos c(x h)# dcos ax# b
h
#
approaches the graph of y œ 2x sin ax b because
cos c(x h)# dcos ax# b
h
hÄ!
lim
œ
d
dx
ccos ax# bd œ 2x sin ax# b.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 3.7 Implicit Differentiation
93. (a)
(b)
df
dt
œ 1.27324 sin 2t 0.42444 sin 6t 0.2546 sin 10t 0.18186 sin 14t
(c) The curve of y œ
approximates y œ
df
dt
dg
dt
the best when t is not 1, 1# , 0, 1# , nor 1.
94. (a)
(b)
dh
dt
œ 2.5464 cos (2t) 2.5464 cos (6t) 2.5465 cos (10t) 2.54646 cos (14t) 2.54646 cos (18t)
(c)
3.7 IMPLICIT DIFFERENTIATION
1. x# y xy# œ 6:
Step 1:
Šx#
Step 2:
x#
dy
dx
Step 3:
dy
dx
dy
dx
ax# 2xyb œ 2xy y#
Step 4:
dy
dx
œ
y † 2x‹ Šx † 2y
2xy
dy
dx
dy
dx
y# † 1‹ œ 0
œ 2xy y#
2xy y#
x# 2xy
2. x$ y$ œ 18xy Ê 3x# 3y#
dy
dx
œ 18y 18x
dy
dx
Ê a3y# 18xb
dy
dx
œ 18y 3x# Ê
dy
dx
œ
6y x#
y# 6x
3. 2xy y# œ x y:
Step 1:
Š2x
Step 2:
2x
dy
dx
dy
dx
2y‹ 2y
2y
dy
dx
dy
dx
dy
dx
œ1
dy
dx
œ 1 2y
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
135
136
Chapter 3 Differentiation
dy
dx
dy
dx
Step 3:
Step 4:
(2x 2y 1) œ " 2y
œ
1 2y
2x 2y 1
4. x$ xy y$ œ 1 Ê 3x# y x
dy
dx
3y#
œ 0 Ê a3y# xb
dy
dx
dy
dx
œ y 3x# Ê
dy
dx
y 3x#
3y# x
œ
5. x# (x y)# œ x# y# :
Step 1:
x# ’2(x y) Š1 Step 2:
2x# (x y)
Step 3:
dy
dx
dy
dx
Step 4:
dy
dx
dy
dx ‹“
2y
(x y)# (2x) œ 2x 2y
dy
dx
œ 2x 2x# (x y) 2x(x y)#
c2x# (x y) 2yd œ 2x c1 x(x y) (x y)# d
œ
œ
2x c1 x(x y) (x y)# d
2x# (x y) 2y
œ
dy
dx
x c1 x(x y) (x y)# d
y x# (x y)
œ
x a1 x# xy x# 2xy y# b
x# y x$ y
x 2x$ 3x# y xy#
x# y x$ y
6. (3xy 7)# œ 6y Ê 2(3xy 7) † Š3x
Ê
dy
dx
dy
dx
3y‹ œ 6
[6x(3xy 7) 6] œ 6y(3xy 7) Ê
(x 1) (x 1)
(x 1)#
dy
dx
dy
dx
Ê 2(3xy 7)(3x)
y(3xy 7)
œ x(3xy
7) 1 œ
dy
dx
6
dy
dx
œ 6y(3xy 7)
#
3xy 7y
1 3x# y 7x
7. y# œ
x"
x1
Ê 2y
8. x3 œ
2x y
x 3y
Ê x4 3x3 y œ 2x y Ê 4x3 9x2 y 3x3 y w œ 2 y w Ê a3x3 1by w œ 2 4x3 9x2 y
Ê yw œ
œ
dy
dx
œ
Ê
2
(x 1)#
dy
dx
œ
"
y(x 1)#
2 4x3 9x2 y
3x3 1
9. x œ tan y Ê 1 œ asec# yb
dy
dx
Ê
dy
dx
œ
"
sec# y
œ cos# y
dy
dy
dy
#
#
#
10. xy œ cot axyb Ê x dy
dx y œ csc (xy)Šx dx y‹ Ê x dx x csc (xy) dx œ y csc (xy) y
Ê
dy dx x
x csc# (xy)‘ œ y csc# (xy) "‘ Ê
11. x tan (xy) œ ! Ê 1 csec# (xy)d Šy x
œ
1
x sec# (xy)
y
x
œ
cos# (xy)
x
y
x
œ
dy
dx ‹
œ
y csc# (xy) "‘
x" csc# (xy)‘
œ yx
œ 0 Ê x sec# (xy)
dy
dx
œ 1 y sec# (xy) Ê
dy
dx
œ
" y sec# (xy)
x sec# (xy)
cos# (xy) y
x
12. x4 sin y œ x3 y2 Ê 4x3 (cos y)
dy
dx
œ 3x2 y2 x3 † 2y
13. y sin Š "y ‹ œ 1 xy Ê y ’cos Š y" ‹ † (1)
dy
dx
dy
dx
’ "y cos Š "y ‹ sin Š y" ‹ x“ œ y Ê
"
y#
dy
dx
†
dy
dx “
œ
dy
dx
Ê acos y 2x3 yb
sin Š y" ‹ †
dy
dx
œ x
y
"y cos Š "y ‹ sin Š "y ‹ x
œ
dy
dx
dy
dx
œ 3x2 y2 4x3 Ê
dy
dx
œ
3x2 y2 4x3
cos y 2x3 y
y Ê
y #
y sin Š "y ‹ cos Š "y ‹ xy
14. x cosa2x 3yb œ y sin x Ê x sina2x 3yba2 3y w b cosa2x 3yb œ y cos x y w sin x
Ê 2x sina2x 3yb 3x y w sina2x 3yb cosa2x 3yb œ y cos x y w sin x
Ê cosa2x 3yb 2x sina2x 3yb y cos x œ asin x 3x sina2x 3ybby w Ê y w œ
15. )"Î# r"Î# œ 1 Ê
16. r 2È) œ
3
#
"
#
)"Î# "# r"Î# †
)#Î$ 34 )$Î% Ê
dr
d)
dr
d)
œ0 Ê
dr
d)
"
’ #È
“œ
r
)"Î# œ )"Î$ )"Î% Ê
"
#È )
dr
d)
Ê
dr
d)
œ
2È r
2È )
cosa2x 3yb 2x sina2x 3yb y cos x
sin x 3x sina2x 3yb
Èr
œÈ
)
œ )"Î# )"Î$ )"Î%
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 3.7 Implicit Differentiation
17. sin (r )) œ
"
#
Ê [cos (r ))] ˆr )
18. cos r cot ) œ r ) Ê (sin r)
dr
d)
œ0Ê
dr ‰
d)
dr
d)
[) cos (r ))] œ r cos (r )) Ê
csc# ) œ r )
Ê
dr
d)
19. x# y# œ 1 Ê 2x 2yyw œ 0 Ê 2yyw œ 2x Ê
y(1) xyw
y#
Ê yww œ
œ
20. x#Î$ y#Î$ œ 1 Ê
2
3
y x Š xy ‹
y#
since yw œ xy Ê
x"Î$ 23 y"Î$
œ0 Ê
dy
dx
dy
dx
d# y
dx#
œ
"
3
2x 2
2y
21. y# œ x# 2x Ê 2yyw œ 2x 2 Ê yw œ
d# y
dx#
œ yww œ
d# y
dx#
Ê
y# (x 1)#
y$
ww
œy œ
y # x #
y$
d# y
dx#
d
dx
,
y# a" y# b
y$
œ
dr
d)
#
csc )
œ rsin
r)
ayw b œ
œ
dy
dx
œ )r , cos (r )) Á 0
d
dx
Š xy ‹
"
y$
œ yx
"Î$
"Î$
"Î$
œ ˆ yx ‰
y (x 1)yw
y#
; then yww œ
"
y1
œ
y (x 1) Š x y 1 ‹
y#
œ (y 1)" ; then yww œ (y 1)# † yw
"
(y 1)$
œ yww œ
23. 2Èy œ x y Ê y"Î# yw œ 1 yw Ê yw ˆy"Î# 1‰ œ 1 Ê
dy
dx
œ yw œ
"
y "Î# 1
Èy
Èy 1
œ
; we can
"
#
differentiate the equation yw ˆy"Î# 1‰ œ 1 again to find yww : yw ˆ y$Î# yw ‰ ˆy"Î# 1‰ yww œ 0
Ê ˆy"Î# 1‰ yww œ
"
#
w # $Î#
cy d y
Ê
d# y
dx#
"
#
œ yww œ
#
Œy
"
$Î#
"Î# 1 y
ay "Î# 1b
"
$
2y$Î# ay "Î# 1b
œ
œ
"
$
# ˆ1 È y ‰
24. xy y# œ 1 Ê xyw y 2yyw œ 0 Ê xyw 2yyw œ y Ê yw (x 2y) œ y Ê yw œ
œ
(x 2y)yw y(1 2yw )
(x 2y)#
œ
2y(x 2y) 2y#
(x 2y)$
œ
y
œ
y
(x 2y) ’ (x 2y) “ y ’1 2 Š (x 2y) ‹“
(x 2y)#
2y# 2xy
(x 2y)$
œ
œ
"
(x 2y)
y
(x2y)
;
d# y
dx#
œ yww
cy(x 2y) y(x 2y) 2y# d
(x 2y)#
2y(x y)
(x 2y)$
#
25. x$ y$ œ 16 Ê 3x# 3y# yw œ 0 Ê 3y# yw œ 3x# Ê yw œ xy# ; we differentiate y# yw œ x# to find yww :
# ww
w
w
w #
# ww
y y y c2y † y d œ 2x Ê y y œ 2x 2y cy d
œ
2xy$ 2x%
y&
Ê
d# y
dx# ¹ (2ß2)
œ
32 32
32
ww
Ê y œ
2x 2y Š
y#
26. xy y# œ 1 Ê xyw y 2yyw œ 0 Ê yw (x 2y) œ y Ê yw œ
since yw k (0
ß
1)
œ "# we obtain yww k (0ß
1)
œ
27. y# x# œ y% 2x at (#ß ") and (#ß 1) Ê 2y
Ê
dy
dx
a2y 4y$ b œ 2 2x Ê
#
x#
‹
y#
œ
2x y#
2x%
y$
œ 2
(2) ˆ "# ‰ (1)(0)
4
dy
dx
œ
x"
#y $ y
Ê
;
"Î$
x"Î$ †ˆ "3 y #Î$ ‰ Œ y"Î$ y"Î$ ˆ "3 x #Î$ ‰
x
x#Î$
22. y# 2x œ 1 2y Ê 2y † yw 2 œ 2yw Ê yw (2y 2) œ 2 Ê yw œ
œ (y 1)# (y 1)" Ê
d# y
dx#
23 y"Î$ ‘ œ 23 x"Î$ Ê yw œ
x1
y
œ
[sin r )] œ r csc# ) Ê
œ yw œ xy ; now to find
x"Î$ †ˆ 3" y #Î$ ‰ yw y"Î$ ˆ "3 x #Î$ ‰
œ
x#Î$
"Î$
y
" "Î$ %Î$
"
x
œ 3x
%Î$ 3y"Î$ x#Î$
3 y
x#Î$ y"Î$ r cos (r ))
) cos (r ))
œ
dy
dx
Differentiating again, yww œ
Ê
dr
d)
dr
d)
y
(x2y)
Ê yww œ
(x 2y) ayw b (y) a1 2yw b
(x 2y)#
œ 4"
dy
dx 2x
dy
dx ¹ ( 2ß1)
œ 4y$
dy
dx
2 Ê 2y
œ 1 and
dy
dx ¹ ( 2ß 1)
dy
dx
4y$
dy
dx
œ 2 2x
œ1
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
;
137
138
Chapter 3 Differentiation
#
28. ax# y# b œ (x y)# at ("ß !) and ("ß 1) Ê 2 ax# y# b Š2x 2y
Ê
and
dy
dx
c2y ax# y# b (x y)d œ 2x ax# y# b (x y) Ê
dy
dx ¹ (1ßc1)
dy
dx
dy
dx ‹
œ
œ 2(x y) Š1 2x ax# y# b (x y)
2y ax# y# b (x y)
dy
dx ‹
Ê
dy
dx ¹ (1ß0)
œ 1
œ1
29. x# xy y# œ 1 Ê 2x y xyw 2yyw œ 0 Ê (x 2y)yw œ 2x y Ê yw œ
(a) the slope of the tangent line m œ yw k (2 3) œ
ß
Ê the tangent line is y 3 œ
7
4
(b) the normal line is y 3 œ 47 (x 2) Ê y œ 47 x 7
4
2x y
2y x
;
(x 2) Ê y œ
7
4
x
"
#
29
7
30. x# y# œ 25 Ê 2x 2yyw œ 0 Ê yw œ xy ;
(a) the slope of the tangent line m œ yw k (3
4)
ß
œ xy ¹
(3ß 4)
œ
Ê the tangent line is y 4 œ
3
4
3
4
(x 3) Ê y œ
3
4
x
25
4
(b) the normal line is y 4 œ 43 (x 3) Ê y œ 43 x
31. x# y# œ 9 Ê 2xy# 2x# yyw œ 0 Ê x# yyw œ xy# Ê yw œ yx ;
(a) the slope of the tangent line m œ yw k ( 1ß3) œ yx ¸ ( 1ß3) œ 3 Ê the tangent line is y 3 œ 3(x 1) Ê y œ 3x 6
(b) the normal line is y 3 œ "3 (x 1) Ê y œ 3" x 8
3
32. y# 2x 4y " œ ! Ê 2yyw 2 4yw œ 0 Ê 2(y 2)yw œ 2 Ê yw œ
"
y#
;
(a) the slope of the tangent line m œ yw k ( 2ß1) œ 1 Ê the tangent line is y 1 œ 1(x 2) Ê y œ x 1
(b) the normal line is y 1 œ 1(x 2) Ê y œ x 3
33. 6x# 3xy 2y# 17y 6 œ 0 Ê 12x 3y 3xyw 4yyw 17yw œ 0 Ê yw (3x 4y 17) œ 12x 3y
12x 3y
Ê yw œ 3x
4y 17 ;
(a) the slope of the tangent line m œ yw k ( 1ß0) œ
Ê yœ
6
7
x
"2x 3y
3x 4y 17 ¹ ( 1ß0)
œ
6
7
Ê the tangent line is y 0 œ
6
7
(x 1)
6
7
(b) the normal line is y 0 œ 76 (x 1) Ê y œ 76 x 7
6
34. x# È3xy 2y# œ 5 Ê 2x È3xyw È3y 4yyw œ 0 Ê yw Š4y È3x‹ œ È3y 2x Ê yw œ
(a) the slope of the tangent line m œ yw k ŠÈ3 2‹ œ
ß
È3y 2x
¹
4y È3x ŠÈ3ß2‹
œ 0 Ê the tangent line is y œ 2
(b) the normal line is x œ È3
35. 2xy 1 sin y œ 21 Ê 2xyw 2y 1(cos y)yw œ 0 Ê yw (2x 1 cos y) œ 2y Ê yw œ
(a) the slope of the tangent line m œ yw k ˆ1 12 ‰ œ
ß
2y
2x 1 cos y ¹ ˆ1ß 1 ‰
2
y
1
#
œ 1# (x 1) Ê y œ 1# x 1
(b) the normal line is y 1
#
œ
2
1
(x 1) Ê y œ
2
1
x
2
1
2y
2x 1 cos y
œ 1# Ê the tangent line is
1
#
36. x sin 2y œ y cos 2x Ê x(cos 2y)2yw sin 2y œ 2y sin 2x yw cos 2x Ê yw (2x cos 2y cos 2x)
œ sin 2y 2y sin 2x Ê yw œ
sin 2y 2y sin 2x
cos 2x 2x cos 2y
(a) the slope of the tangent line m œ yw k ˆ 14
ß
1‰
2
œ
;
sin 2y 2y sin 2x
cos 2x 2x cos 2y ¹ ˆ 1 ß 1 ‰
4 2
y
1
#
œ 2 ˆx 14 ‰ Ê y œ 2x
(b) the normal line is y 1
#
œ "# ˆx 14 ‰ Ê y œ "# x œ
1
1
#
œ 2 Ê the tangent line is
51
8
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
;
È3y 2x
4y È3x
;
Section 3.7 Implicit Differentiation
37. y œ 2 sin (1x y) Ê yw œ 2 [cos (1x y)] † a1 yw b Ê yw [1 2 cos (1x y)] œ 21 cos (1x y) Ê yw œ
(a) the slope of the tangent line m œ yw k (1 0) œ
ß
21 cos (1x y)
1 2 cos (1x y) ¹(1ß0)
y 0 œ 21(x 1) Ê y œ 21x 21
(b) the normal line is y 0 œ #"1 (x 1) Ê y œ 2x1 139
21 cos (1x y)
1 # cos (1x y)
;
œ 21 Ê the tangent line is
"
#1
38. x# cos# y sin y œ 0 Ê x# (2 cos y)(sin y)yw 2x cos# y yw cos y œ 0 Ê yw c2x# cos y sin y cos yd
2x cos# y
2x# cos y sin y cos y
œ 2x cos# y Ê yw œ
;
(a) the slope of the tangent line m œ yw k (0 1) œ
ß
2x cos# y
2x# cos y sin y cos y ¹ (0ß1)
œ 0 Ê the tangent line is y œ 1
(b) the normal line is x œ 0
39. Solving x# xy y# œ 7 and y œ 0 Ê x# œ 7 Ê x œ „ È7 Ê ŠÈ7ß !‹ and ŠÈ7ß !‹ are the points where the
curve crosses the x-axis. Now x# xy y# œ 7 Ê 2x y xyw 2yyw œ 0 Ê (x 2y)yw œ 2x y
2 È 7
y
2x y
È
È
Ê yw œ 2x
x 2y Ê m œ x 2y Ê the slope at Š 7ß !‹ is m œ È7 œ 2 and the slope at Š 7ß !‹ is
È
m œ 2È77 œ 2. Since the slope is 2 in each case, the corresponding tangents must be parallel.
40. xy 2x y œ 0 Ê x
dy
dx
y2
dy
dx
œ0 Ê
dy
dx
œ
y2
1x
; the slope of the line 2x y œ 0 is 2. In order to be
parallel, the normal lines must also have slope of 2. Since a normal is perpendicular to a tangent, the slope of
2
"
the tangent is "# . Therefore, y1 x œ # Ê 2y 4 œ 1 x Ê x œ 3 2y. Substituting in the original equation,
y(3 2y) 2(3 2y) y œ 0 Ê y# 4y 3 œ 0 Ê y œ 3 or y œ 1. If y œ 3, then x œ 3 and
y 3 œ 2(x 3) Ê y œ 2x 3. If y œ 1, then x œ 1 and y 1 œ 2(x 1) Ê y œ 2x 3.
41. y% œ y# x# Ê 4y$ yw œ 2yyw 2x Ê 2 a2y$ yb yw œ 2x Ê yw œ y x2y$ ; the slope of the tangent line at
È3
"
È È
È
Š 43 ß #3 ‹ is y x2y$ ¹ È3 È3 œ È3 4 6È3 œ " 4 3 œ # " 3 œ 1; the slope of the tangent line at Š 43 ß #" ‹
#
4
Œ
is
x
y2y$ ¹
Œ
È3
4
ß
œ
1
2
È3
"
#
4
28
4
ß
#
2
œ
2È 3
42
8
œ È3
42. y# (2 x) œ x$ Ê 2yyw (2 x) y# (1) œ 3x# Ê yw œ
œ
4
#
y# 3x#
2y(2 x)
; the slope of the tangent line is m œ
œ 2 Ê the tangent line is y 1 œ 2(x 1) Ê y œ 2x 1; the normal line is y 1 œ "# (x 1) Ê y œ "# x 43. y% 4y# œ x% 9x# Ê 4y$ yw 8yyw œ 4x$ 18x Ê yw a4y$ 8yb œ 4x$ 18x Ê yw œ
œ
y# 3x#
2y(2 x) ¹ (1ß1)
x a2x# 9b
y a2y# 4b
œ m; (3ß 2): m œ
(3)(18 9)
2(8 4)
œ 27
8 ; ($ß #): m œ
27
8
; (3ß #): m œ
27
8
ß
5
4
(b) yw œ 0 Ê
and yw k (2 4) œ
ß
#
3y x
y# 3x
4
5
œ
2x$ 9x
2y$ 4y
; (3ß #): m œ 27
8
44. x$ y$ 9xy œ 0 Ê 3x# 3y# yw 9xyw 9y œ 0 Ê yw a3y# 9xb œ 9y 3x# Ê yw œ
(a) yw k (4 2) œ
4x$ 18x
4y$ 8y
3
#
9y 3x#
3y# 9x
œ
3y x#
y# 3x
;
œ 0 Ê 3y x# œ 0 Ê y œ
x#
3
#
$
#
Ê x$ Š x3 ‹ 9x Š x3 ‹ œ 0 Ê x' 54x$ œ 0
Ê x$ ax$ 54b œ 0 Ê x œ 0 or x œ $È54 œ 3 $È2 Ê there is a horizontal tangent at x œ 3 $È2 . To find the
corresponding y-value, we will use part (c).
(c)
dx
dy
œ0 Ê
y# 3x
3y x#
$
œ 0 Ê y# 3x œ 0 Ê y œ „ È3x ; y œ È3x Ê x$ ŠÈ3x‹ 9xÈ3x œ 0
3
3
Ê x$ 6È3 x$Î# œ 0 Ê x$Î# Šx$Î# 6È3‹ œ 0 Ê x$Î# œ 0 or x$Î# œ 6È3 Ê x œ 0 or x œ È
108 œ 3 È
4.
Since the equation x$ y$ 9xy œ 0 is symmetric in x and y, the graph is symmetric about the line y œ x. That is, if
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
140
Chapter 3 Differentiation
(aß b) is a point on the folium, then so is (bß a). Moreover, if yw k (a b) œ m, then yw k (b a) œ
ß
ß
"
m
. Thus, if the folium has a
horizontal tangent at (aß b), it has a vertical tangent at (bß a) so one might expect that with a horizontal tangent at
3
3
3
3
3
xœÈ
54 and a vertical tangent at x œ 3 $È4, the points of tangency are ŠÈ
54ß 3 È
4‹ and Š3 È
4ß È
54‹,
respectively. One can check that these points do satisfy the equation x$ y$ 9xy œ 0.
45. x# 2xy 3y# œ 0 Ê 2x 2xyw 2y 6yyw œ 0 Ê yw (2x 6y) œ 2x 2y Ê yw œ
xy
3y x ¹ (1ß1)
line m œ yw k (1 1) œ
ß
xy
3y x
Ê the slope of the tangent
œ 1 Ê the equation of the normal line at (1ß 1) is y 1 œ 1(x 1) Ê y œ x 2. To find
where the normal line intersects the curve we substitute into its equation: x# 2x(2 x) 3(2 x)# œ 0
Ê x# 4x 2x# 3 a4 4x x# b œ 0 Ê 4x# 16x 12 œ 0 Ê x# 4x 3 œ 0 Ê (x 3)(x 1) œ 0
Ê x œ 3 and y œ x 2 œ 1. Therefore, the normal to the curve at (1ß 1) intersects the curve at the point (3ß 1).
Note that it also intersects the curve at (1ß 1).
46. Let p and q be integers with q 0 and suppose that y œ Èxp œ xpÎq . Then yq œ xp . Since p and q are integers and
q
d
q
dx ay b
xp 1ap pÎqb œ qp
assuming y is a differentiable function of x,
œ
p
q
†
xp
axpÎq b
1
q
47. y# œ x Ê
1
œ
p
q
†
dy
dx
œ
"
#y
xp 1
xp pÎq
œ
p
q
†
œ
p1
Ê qyq 1 dy
Ê
dx œ px
d
p
dx ax b
dy
dx
pxp
qyq
œ
œ
p
q
xp
yq
†
1
1
a p Îq b 1
†x
. If a normal is drawn from (aß 0) to (x" ß y" ) on the curve its slope satisfies
Ê y" œ 2y" (x" a) or a œ x" "# . Since x"
1
1
"
#
0 on the curve, we must have that a
y" 0
x" a
œ 2y"
. By symmetry, the two
Èx
Èx
points on the parabola are ˆx" ß Èx" ‰ and ˆx" ß Èx" ‰ . For the normal to be perpendicular, Š x" "a ‹ Š a x"" ‹ œ 1
Ê
x"
(a x" )#
œ 1 Ê x" œ (a x" )# Ê x" œ ˆx" "
#
#
2x
w
48. 2x# 3y# œ 5 Ê 4x 6yyw œ 0 Ê yw œ 2x
3y Ê y k (1 1) œ 3y ¹
3x#
2y
Ê yw k (1 1) œ
ß
3x#
2y ¹ (1ß1)
œ
and yw k (1
3
#
and y" œ „ #" . Therefore, ˆ 4" ß „ #" ‰ and a œ
œ 23 and yw k (1
(1ß1)
ß
y# œ x$ Ê 2yyw œ 3x# Ê yw œ
"
4
x" ‰ Ê x " œ
ß
1)
œ
3x#
2y ¹ (1ß 1)
ß
1)
œ 2x
3y ¹
(1ß 1)
œ
2
3
; also,
œ #3 . Therefore the
tangents to the curves are perpendicular at (1ß 1) and (1ß 1) (i.e., the curves are orthogonal at these two points of
intersection).
49. (a) x2 y2 œ 4, x2 œ 3y2 Ê a3y2 b y2 œ 4 Ê y2 œ 1 Ê y œ „ 1. If y œ 1 Ê x2 a1b2 œ 4 Ê x2 œ 3
Ê x œ „ È3. If y œ 1 Ê x2 a1b2 œ 4 Ê x2 œ 3 Ê x œ „ È3.
x2 y2 œ 4 Ê 2x 2y dy
dx œ 0 Ê m1 œ
At ŠÈ3ß 1‹: m1 œ
œ
dy
dx
È3
1
dy
dx
œ
At ŠÈ3ß 1‹: m1 œ
dy
dx
œ
At ŠÈ3ß 1‹: m1 œ
dy
dx
1
È3
2
Ê x œ 1 Š
dy
Ê 1 œ 23 y dx
Ê m2 œ
At Š 14 ß
È3
2 ‹:
At Š 14 ß m1 œ
È3
2 ‹:
dy
dx
m1 œ
dy
dx
È3 2
2 ‹
œ
œ
dy
dx
œ
œ
œ È3 and m2 œ
dy
dx
œ
ŠÈ3‹
a 1 b
œ È3 and m2 œ
3
4
È3
3
œ
È 3
3 a1 b
œ
dy
dx
Êyœ „
È
33
œ
Ê m1 † m2 œ
È3
3
ŠÈ3‹
3 a 1 b
È3
2 .
œ
dy
dx
Ê m1 † m2 œ ŠÈ3‹Š
œ
If y œ
œ 14 . x œ 1 y2 Ê 1 œ 2y dy
dx Ê m1 œ
x
3y
È3
3 ‹
œ 1
È
ŠÈ3‹Š 33 ‹
Ê m1 † m2 œ ŠÈ3‹Š
È3
3
È3
2
dy
dx
œ 1
È3
3 ‹
Ê m1 † m2 œ ŠÈ3‹Š
Êxœ1Š
2
È3
2 ‹
œ 1
È3
3 ‹
œ 1
œ 41 . If
1
œ 2y
and x œ 13 y2
3
2y
1
2ŠÈ3Î2‹
œ
È3
3 a1 b
œ
dy
dx
(b) x œ 1 y2 , x œ 13 y2 Ê ˆ 13 y2 ‰ œ 1 y2 Ê y2 œ
yœ
dy
dx
È3
3 a 1 b
œ È3 and m2 œ
ŠÈ3‹
œ
œ xy and x2 œ 3y2 Ê 2x œ 6y dy
dx Ê m2 œ
œ È3 and m2 œ
È
a13b
At ŠÈ3ß 1‹: m1 œ
dy
dx
œ È13 and m2 œ
1
2ŠÈ3Î2‹
œ
1
È3
dy
dx
and m2 œ
œ
dy
dx
3
2ŠÈ3Î2‹
œ
œ
3
2ŠÈ3Î2‹
3
È3
Ê m1 † m2 œ Š È13 ‹Š È33 ‹ œ 1
œ È33 Ê m1 † m2 œ Š È13 ‹Š È33 ‹ œ 1
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
3
4
.
Section 3.7 Implicit Differentiation
50. y œ 13 x b, y2 œ x3 Ê
Ê
x4
4
dy
dx
dy
œ 13 and 2y dx
œ 3x2 Ê
51. xy$ x# y œ 6 Ê x Š3y#
dy
dx ‹
a4 b 2
2
3x2
2y
2
Ê ˆ 13 ‰Š 3x
2y ‹ œ 1 Ê
y$ x#
dy
dx
2xy œ 0 Ê
$
#
#
x
œ 3xy
y$ 2xy ; thus
dx
dy
a0b2
2
x2
2
appears to equal
"
dy
dx
dx
dy
dy
dx
2
œ ! and ˆ 13 ‰Š 3x
2y ‹ œ 1 is
a3xy# x# b œ y$ 2xy Ê
x# y Š2x
dx
dy ‹
œ0 Ê
dx
dy
2
2
œ y Ê Š x2 ‹ œ x3
œ 8. At a4ß 8b, y œ 31 x b Ê 8 œ 31 a4b b Ê b œ
y 2xy
$
#
#
$
œ 3xy
# x# ; also, xy x y œ 6 Ê x a3y b y
dx
dy
œ
œ x3 Ê x4 4x3 œ 0 Ê x3 ax 4b œ 0 Ê x œ 0 or x œ 4. If x œ 0 Ê y œ
indeterminant at a0, 0b. If x œ 4 Ê y œ
Ê
dy
dx
dy
dx
œ
28
3 .
y$ 2xy
3xy# x#
ay$ 2xyb œ 3xy# x#
. The two different treatments view the graphs as functions
symmetric across the line y œ x, so their slopes are reciprocals of one another at the corresponding points
(aß b) and (bß a).
52. x$ y# œ sin# y Ê 3x# 2y
œ
3x#
2 sin y cos y 2y
appears to equal
dy
dx
œ (2 sin y)(cos y)
; also, x$ y# œ sin# y Ê 3x#
"
dy
dx
dx
dy
dy
dx
Ê
dy
dx
(2y 2 sin y cos y) œ 3x# Ê
2y œ 2 sin y cos y Ê
dx
dy
œ
2 sin y cos y 2y
3x#
dy
dx
œ
3x#
2y 2 sin y cos y
; thus
dx
dy
. The two different treatments view the graphs as functions symmetric across the line
y œ x so their slopes are reciprocals of one another at the corresponding points (aß b) and (bß a).
53-60. Example CAS commands:
Maple:
q1 := x^3-x*y+y^3 = 7;
pt := [x=2,y=1];
p1 := implicitplot( q1, x=-3..3, y=-3..3 ):
p1;
eval( q1, pt );
q2 := implicitdiff( q1, y, x );
m := eval( q2, pt );
tan_line := y = 1 + m*(x-2);
p2 := implicitplot( tan_line, x=-5..5, y=-5..5, color=green ):
p3 := pointplot( eval([x,y],pt), color=blue ):
display( [p1,p2,p3], ="Section 3.7 #57(c)" );
Mathematica: (functions and x0 may vary):
Note use of double equal sign (logic statement) in definition of eqn and tanline.
<<Graphics`ImplicitPlot`
Clear[x, y]
{x0, y0}={1, 1/4};
eqn=x + Tan[y/x]==2;
ImplicitPlot[eqn,{ x, x0 3, x0 3},{y, y0 3, y0 3}]
eqn/.{x Ä x0, y Ä y0}
eqn/.{ y Ä y[x]}
D[%, x]
Solve[%, y'[x]]
slope=y'[x]/.First[%]
m=slope/.{x Ä x0, y[x] Ä y0}
tanline=y==y0 m (x x0)
ImplicitPlot[{eqn, tanline}, {x, x0 3, x0 3},{y, y0 3, y0 + 3}]
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
141
142
Chapter 3 Differentiation
3.8 RELATED RATES
1. A œ 1r# Ê
2. S œ 41r# Ê
3. y œ 5x,
dx
dt
dx
dt
dS
dt
dy
dt
7. x2 y2 œ 25,
dy
dt
œ 5 dx
dt Ê
dy
dt
œ 5a2b œ 10
œ 2x dx
dt ; when x œ 1 Ê
dy
dt
œ5Ê
dx
dt
dr
dt
dy
dx
œ 2 Ê 2 dx
dt 3 dt œ 0 Ê 2 dt 3a2b œ 0 Ê
dy
dt
œ3Ê
6. x œ y3 y,
dr
dt
œ 81r
œ2Ê
4. 2x 3y œ 12,
5. y œ x2 ,
œ 21r
dA
dt
œ 3y2 dy
dt dx
dt
dy
dt ;
dL
dt
dx
dt
œ 1,
a5ba1b a12ba3b
œ
dx
dt
œ 3a2b2 a5b a5b œ 55
dy
dy
œ 2 Ê 2x dx
dt 2y dt œ 0; when x œ 3 and y œ 4 Ê 2a3ba2b 2a4b dt œ 0 Ê
4 dy
"
2 2 dy
3 dx
27 , dt œ # Ê 3x y dt 2x y dt œ 0;
2
3
dx
9
3a2b2 ˆ 13 ‰ ˆ "# ‰ 2a2b ˆ 13 ‰ dx
dt œ 0 Ê dt œ 2
Ê
œ3
œ 2a1ba3b œ 6
when y œ 2 Ê
8. x2 y3 œ
9. L œ Èx2 y2 ,
dy
dt
dx
dt
Éa5b2 a12b2
10. r s2 v3 œ 12,
dr
dt
Ê 4 2a1ba3b œ
dy
dt
œ3Ê
dL
dt
œ
when x œ 2 Ê a2b2 y3 œ
1
Š2x dx
dt
2È x2 y2
2y dy
dt ‹ œ
dy
dt
Ê y œ 13 . Thus
4
27
dy
x dx
dt y dt
È x2 y2 ;
when x œ 5 and y œ 12
31
13
œ 4,
ds
dt
3a2b2 dv
dt
œ 3 Ê
œ0Ê
dr
ds
dt 2s dt
dv
1
dt œ 6
2
3
3v2 dv
dt œ 0; when r œ 3 and s œ 1 Ê a3b a1b v œ 12 Ê v œ 2
dx
m
dS
dx
dS
m2
dt œ 5 min Ê dt œ 12x dt ; when x œ 3 Ê dt œ 12a3ba5b œ 180 min
2
m
dV
dV
m3
2 dx
x3 , dx
dt œ 5 min Ê dt œ 3x dt ; when x œ 3 Ê dt œ 3a3b a5b œ 135 min
11. (a) S œ 6x2 ,
(b) V œ
12. S œ 6x2 ,
Ê
dV
dt
dS
dt
2
in
œ 72 sec
Ê
œ 3a3b2 a2b œ 54
13. (a) V œ 1r# h Ê
dV
dt
dV
dt
(c) V œ 1r# h Ê
dS
dt œ
in3
sec
dx
12x dx
dt Ê 72 œ 12a3b dt Ê
œ 1 r#
œ
14. (a) V œ "3 1r# h Ê
dh
dt
1r# dh
dt
dV
dt œ
" # dh
2
3 1r dt 3 1rh
dx
dt
œ2
in
sec ;
V œ x3 Ê
(b) V œ 1r# h Ê
21rh
dV
dt
" # dh
3 1r dt
dr
dt
(b) V œ "3 1r# h Ê
œ
15. (a)
dV
dt
dV
dt
dR
dt
"
œ 1 volt/sec
(b) dI
dt œ 3 amp/sec
" ˆ dV
dI ‰
" ˆ dV
V dI ‰
‰
ˆ dR ‰ Ê dR
œ R ˆ dI
Ê dR
dt I dt
dt œ I
dt R dt
dt œ I
dt I dt
ˆ " ‰‘ œ ˆ #" ‰ (3) œ 3# ohms/sec, R is increasing
œ "# 1 12
# 3
16. (a) P œ RI# Ê
dP
dt
œ I#
(b) P œ RI# Ê 0 œ
dP
dt
dR
dt
2RI
œ I#
dR
dt
œ 3x2 dx
dt ; when x œ 3
œ 21rh
dV
dt
dV
dt
(d)
dV
dt
dr
dt
dr
dt
(c)
(c)
œ 32
œ 23 1rh
dr
dt
dI
dt
2RI
dI
dt
Ê
dR
dt
œ 2RI
I#
dI
dt
œ
2 ˆ PI ‰ dI
I#
dt
œ 2P
I$
dI
dt
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 3.8 Related Rates
17. (a) s œ Èx# y# œ ax# y# b
"Î#
Ê
ds
dt
œ
x
dx
Èx# y# dt
(b) s œ Èx# y# œ ax# y# b
"Î#
Ê
ds
dt
œ
x
dx
Èx# y# dt
(c) s œ Èx# y# Ê s# œ x# y# Ê 2s
ds
dt
œ 2x
ds
dt
œ
x
dx
Èx# y# z# dt
y
dy
Èx# y# z# dt
(b) From part (a) with
dx
dt
œ0 Ê
(c) From part (a) with
ds
dt
œ 0 Ê 0 œ 2x
19. (a) A œ
(c) A œ
"
#
"
#
ab sin ) Ê
ab sin ) Ê
20. Given A œ 1r# ,
(b)
(c)
œ
œ
"
#
"
#
œ
ab cos )
ab cos )
d)
dt
d)
dt
2y
dy
dt
2z
œ
Ê
dy
dt
dx
dt
œ yx
dy
dt
dz
dt
z
dz
Èx# y# z# dt
2z
dy
dt
Ê
dz
dt
(b) A œ
"# b sin )
z dz
x dt
ab sin ) Ê
dA
dt
dx
dt
"
#
"# a sin )
da
dt
œ 0.01 cm/sec, and r œ 50 cm. Since
dr
dt
2y
dx
dt
2y
dx
dt
z
dz
Èx# y# z# dt
y
dy
Èx# y# z# dt
dx
dt
œ 2x
ds
dt
Ê 2s † 0 œ 2x
dy
dt
œ 21r
dA
dt
y dy
x dt
œ0
œ
"
#
ab cos )
d)
dt
"# b sin )
da
dt
db
dt
dr
dt
, then
dA ¸
dt r=50
" ‰
œ 21(50) ˆ 100
œ 1 cm# /min.
dw
dt œ 2 cm/sec, j œ 12 cm and w œ 5 cm.
dA
dj
dA
#
A œ jw Ê dt œ j dw
dt w dt Ê dt œ 12(2) 5(2) œ 14 cm /sec, increasing
dj
dw
P œ 2j 2w Ê dP
dt œ 2 dt 2 dt œ 2(2) 2(2) œ 0 cm/sec, constant
"Î#
"
dj ‰
#
# "Î# ˆ
D œ Èw# j# œ aw# j# b
Ê dD
2w dw
Ê dD
dt œ # aw j b
dt 2j dt
dt
21. Given
(a)
dj
dt
dA
dt
dA
dt
ds
dt
y
dy
Èx# y# dt
2y
dx
dt
18. (a) s œ Èx# y# z# Ê s# œ x# y# z# Ê 2s
Ê
143
œ 2 cm/sec,
(5)(2) (12)(2)
È25 144
22. (a) V œ xyz Ê
œ
dj
w dw
dt j dt
Èw# j#
œ 14
13 cm/sec, decreasing
œ yz
dV
dt
dx
dt
xz
xy
dy
dt
dz
dt
Ê
dV ¸
dt (4ß3ß2)
œ (3)(2)(1) (4)(2)(2) (4)(3)(1) œ 2 m$ /sec
dx
(b) S œ 2xy 2xz 2yz Ê dS
dt œ (2y 2z) dt (2x 2z)
¸
Ê dS
œ (10)(1) (12)(2) (14)(1) œ 0 m# /sec
dt
dy
dt
(2x 2y)
dz
dt
(4ß3ß2)
"Î#
(c) j œ Èx# y# z# œ ax# y# z# b Ê
Ê
23. Given:
dj ¸
dt (4ß3ß2)
dx
dt
dj
dt
œ
x
dx
Èx# y# z# dt
y
dy
Èx# y# z# dt
z
dz
Èx# y# z# dt
œ Š È429 ‹ (1) Š È329 ‹ (2) Š È229 ‹ (1) œ 0 m/sec
œ 5 ft/sec, the ladder is 13 ft long, and x œ 12, y œ 5 at the instant of time
(a) Since x# y# œ 169 Ê
dy
dt
œ xy
dx
dt
‰
œ ˆ 12
5 (5) œ 12 ft/sec, the ladder is sliding down the wall
(b) The area of the triangle formed by the ladder and walls is A œ
is changing at
(c) cos ) œ
x
13
"
#
xy Ê
dA
dt
œ ˆ "# ‰ Šx
dy
dt
y
dx
dt ‹ .
The area
#
[12(12) 5(5)] œ 119
# œ 59.5 ft /sec.
Ê sin )
24. s# œ y# x# Ê 2s
"
#
ds
dt
œ 2x
d)
dt
œ
"
13
dx
dt
2y
†
dx
dt
dy
dt
Ê
Ê
ds
dt
d)
dt
"
œ 13 sin
) †
dx
dt
œ
"
s
dy
dt ‹
Šx
dx
dt
y
œ ˆ 5" ‰ (5) œ 1 rad/sec
Ê
ds
dt
œ
"
È169
[5(442) 12(481)] œ 614 knots
25. Let s represent the distance between the girl and the kite and x represents the horizontal distance between the girl and kite
Ê s# œ (300)# x# Ê
ds
dt
œ
x dx
s dt
œ
400(25)
500
œ 20 ft/sec.
"
#
26. When the diameter is 3.8 in., the radius is 1.9 in. and dr
dt œ 3000 in/min. Also V œ 61r Ê
$
ˆ " ‰
Ê dV
dt œ 121(1.9) 3000 œ 0.00761. The volume is changing at about 0.0239 in /min.
dV
dt
œ 121r
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
dr
dt
144
Chapter 3 Differentiation
27. V œ
"
3
3r
4h
#
3 1r h, h œ 8 (2r) œ 4 Ê r œ 3 Ê
dh ¸
90
ˆ 9 ‰
dt h = 4 œ 1614# (10) œ 2561 ¸ 0.1119
dr
4 dh
4 ˆ 90 ‰
15
r œ 4h
3 Ê dt œ 3 dt œ 3 2561 œ 321
(a)
(b)
28. (a) V œ
"
3
1r# h and r œ
Ê Vœ
15h
#
"
3
"
3
Vœ
#
‰ hœ
1 ˆ 4h
3
1
3
y# (3R y) Ê
y œ 8 we have
"
1441 (6)
œ
dy
dt
1
3
œ
dV
dt
#
751h$
4
œ
161h# dh
9
dt
2251h# dh
4
dt
œ
dV
dt
Ê
dh ¸
dt h = 5
œ
4(50)
2251(5)#
8
2251
œ
¸ 0.0849 m/sec œ 8.49 cm/sec
Ê
dy
dt
dy
dt
" dV
dt
œ 13 a6Ry 3y# b‘
Ê at R œ 13 and
m/min
(b) The hemisphere is on the circle r (13 y)# œ 169 Ê r œ È26y y# m
(c) r œ a26y y# b
5
2881
œ
30. If V œ
4
3
"Î#
Ê
"
#
œ
dr
dt
#
Ê
4
151
c2y(3R y) y# (1)d
œ
dV
dt
¸ 0.1492 m/sec œ 14.92 cm/sec
‰ hœ
1 ˆ 15h
#
"
241
Ê
m/sec œ 11.19 cm/sec
¸ 0.0113 m/min œ 1.13 cm/min
dr
15 dh
dr ¸
8 ‰
ˆ 15 ‰ ˆ 225
(b) r œ 15h
# Ê dt œ # dt Ê dt h = 5 œ #
1 œ
29. (a) V œ
161h$
27
a26y y# b
"Î#
(26 2y)
dy
dt
Ê
dr
dt
œ
13 y
dy
È26y y# dt
Ê
dr ¸
dt y = 8
œ
13 8
È26†8 64
ˆ #"
‰
41
m/min
1r$ , S œ 41r# , and
dV
dt
œ kS œ 4k1r# , then
dV
dt
œ 41r#
Ê 4k1r# œ 41r#
dr
dt
dr
dt
Ê
dr
dt
œ k, a constant.
Therefore, the radius is increasing at a constant rate.
4
dV
dV
dr
$
$
# dr
3 1r , r œ 5, and dt œ 1001 ft /min, then dt œ 41r dt Ê dt
dr
#
dt œ 81(5)(1) œ 401 ft /min, the rate at which the surface area
31. If V œ
œ 81r
œ 1 ft/min. Then S œ 41r# Ê
dS
dt
is increasing.
32. Let s represent the length of the rope and x the horizontal distance of the boat from the dock.
(a) We have s# œ x# 36 Ê
dx ¸
dt s = 10
œ
(b) cos ) œ
d)
dt
Ê
6
r
œ
10
È10# 36
œ
s ds
x dt
œ
s
ds
Ès# 36 dt
. Therefore, the boat is approaching the dock at
(2) œ 2.5 ft/sec.
Ê sin )
6
8 ‰
10# ˆ 10
dx
dt
d)
dt
œ r6#
dr
dt
Ê
d)
dt
œ
6
dr
r# sin ) dt
. Thus, r œ 10, x œ 8, and sin ) œ
8
10
3
† (2) œ 20
rad/sec
33. Let s represent the distance between the bicycle and balloon, h the height of the balloon and x the horizontal
distance between the balloon and the bicycle. The relationship between the variables is s# œ h# x#
" ˆ dh
dx ‰
"
Ê ds
Ê ds
dt œ s h dt x dt
dt œ 85 [68(1) 51(17)] œ 11 ft/sec.
34. (a) Let h be the height of the coffee in the pot. Since the radius of the pot is 3, the volume of the coffee is
dh
dh
" dV
10
V œ 91h Ê dV
dt œ 91 dt Ê the rate the coffee is rising is dt œ 91 dt œ 91 in/min.
(b) Let h be the height of the coffee in the pot. From the figure, the radius of the filter r œ
œ
$
1h
1#
, the volume of the filter. The rate the coffee is falling is
35. y œ QD" Ê
dy
dt
œ D"
dQ
dt
QD#
dD
dt
œ
"
41
(0) 233
(41)#
(2) œ
dh
dt
466
1681
œ
4 dV
1h# dt
œ
4
#5 1
h
#
Ê Vœ
"
3
1r# h
(10) œ 581 in/min.
L/min Ê increasing about 0.2772 L/min
36. Let P(xß y) represent a point on the curve y œ x# and ) the angle of inclination of a line containing P and the
origin. Consequently, tan ) œ
#
and cos )kx=3 œ
#
x
y # x #
œ
#
3
9 # 3 #
y
x
œ
Ê tan ) œ
"
10
, we have
x#
d)
#
x œ x Ê sec ) dt
d) ¸
dt x=3 œ 1 rad/sec.
œ
dx
dt
Ê
d)
dt
œ cos# )
dx
dt
. Since
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
dx
dt
œ 10 m/sec
Section 3.8 Related Rates
37. The distance from the origin is s œ Èx# y# and we wish to find
œ
(5)(1) (12)(5)
È25 144
ds ¸
dt (5ß12)
"
#
œ
ax# y# b
d)
dt
œ
dx
dt
Š2x
2y
dy
dt ‹¹ (5ß12)
œ 5 m/sec
38. Let s œ distance of car from foot of perpendicular in the textbook diagram Ê tan ) œ
Ê
"Î#
145
cos# ) ds
132 dt
;
ds
dt
d)
dt
œ 264 and ) œ 0 Ê
Ê sec# )
s
13#
d)
dt
œ
" ds
13# dt
œ 2 rad/sec. A half second later the car has traveled 132 ft
right of the perpendicular Ê k)k œ 14 , cos# ) œ "# , and
ds
dt
œ 264 (since s increases) Ê
d)
dt
œ
ˆ "# ‰
132
(264) œ 1 rad/sec.
39. Let s œ 16t# represent the distance the ball has fallen, h the
distance between the ball and the ground, and I the distance
between the shadow and the point directly beneath the ball.
Accordingly, s h œ 50 and since the triangle LOQ and
#
triangle PRQ are similar we have I œ 5030h
h Ê h œ 50 16t
and I œ
Ê
30 a50 16t# b
50 a50 16t# b
dI ¸
dt t= 12
œ
30 Ê
1500
16t#
dI
dt
œ 1500
8t$
œ 1500 ft/sec.
40. When x represents the length of the shadow, then tan ) œ
given that
d)
dt
31
#000
œ 0.27° œ
Ê sec# )
80
x
rad/min. At x œ 60, cos ) œ
d)
dt
œ 80
x#
#
x sec
¸
Ê ¸ dx
dt œ ¹
80
3
5
#
dx
dt
Ê
) d)
dt ¹¹ Š d) =
dt
dx
dt
31
2000
x# sec# ) d)
80
dt
œ
and sec ) = 35 ‹
œ
31
16
. We are
ft/min
¸ 0.589 ft/min ¸ 7.1 in./min.
41. The volume of the ice is V œ
4
3
1r$ 43 14$ Ê
thickness of the ice is decreasing at
5
721
œ 41r#
dV
dt
Ê
dr
dt
dr ¸
dt r=6
œ
5
721
in./min when
in/min. The surface area is S œ 41r# Ê
#
œ 10
3 in /min, the outer surface area of the ice is decreasing at
10
3
dS
dt
œ 81r
dr
dt
in# /min.
œ 10 in$ /min, the
5 ‰
¸ œ 481 ˆ 72
Ê dS
dt
1
dV
dt
r=6
42. Let s represent the horizontal distance between the car and plane while r is the line-of-sight distance between the car and
r
dr
ds ¸
5
plane Ê 9 s# œ r# Ê ds
dt œ È #
dt Ê dt r=5 œ È16 (160) œ 200 mph Ê speed of plane speed of car
r 9
œ 200 mph Ê the speed of the car is 80 mph.
43. Let x represent distance of the player from second base and s the distance to third base. Then
#
#
(a) s œ x 8100 Ê 2s œ 2x
Ê
ds
60
È
Ê s œ 30 13 and dt œ 30È13 (16) œ
ds
dt
(b) sin )" œ
Ê
(c)
d)"
dt
†
90
s
Ê cos )"
œ
90
Š30È13‹ (60)
œ xs dx
dt
32
È13 ¸
ds
dt
d)"
dt
90
s†x
d)"
dt
90
œ s# cos
)" †
ds
dt
8.875 ft/sec
œ 90
s# †
ds
dt
Ê
32
† ŠÈ
‹œ
13
8
65
rad/sec; cos )# œ
90
œ s# cos
)" †
xÄ!
œ ˆ x# 908100 ‰
dx
dt
d)#
dt
90
s
œ
ds
dt
œ s90†x †
Ê sin )#
90
Š30È13‹ (60)
s
Ê
lim d)#
x Ä ! dt
"
6
rad/sec;
œ
"6
d)#
dt
œ
90
s# sin )#
†
ds
dt
ds
dt
. Therefore, x œ 60 and s œ 30È13
d)#
dt
œ 90
s# †
ds
dt
Ê
d)#
dt
œ
90
s# sin )#
†
ds
dt
8
32
† ŠÈ
rad/sec.
‹ œ 65
13
90
ˆ xs ‰ † ˆ dx
‰ ˆ 90 ‰ ˆ dx
‰ ˆ
‰
œ ˆs90
#† x ‰ †
dt œ s#
dt œ x# 8100
œ lim ˆ x# 908100 ‰ (15) œ
œ 16 ft/sec
. When the player is 30 ft from first base, x œ 60
. Therefore, x œ 60 and s œ 30È13 Ê
œ
ds
dt
d)"
dt
dx
dt
dx
dt
dx
dt
Ê lim
xÄ!
d)"
dt
ˆ xs ‰ ˆ dx
‰ ˆ 90 ‰ ˆ dx
‰
œ Š s90
#† x ‹
dt œ s#
dt
s
rad/sec
44. Let a represent the distance between point O and ship A, b the distance between point O and ship B, and D the distance
" da
db
db
da ‘
between the ships. By the Law of Cosines, D# œ a# b# 2ab cos 120° Ê dD
dt œ #D 2a dt 2b dt a dt b dt .
When a œ 5,
da
dt
œ 14, b œ 3, and
db
dt
œ 21, then
dD
dt
œ
413
2D
where D œ 7. The ships are moving
dD
dt
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
œ 29.5 knots apart.
146
Chapter 3 Differentiation
3.9 LINEARIZATION AND DIFFERENTIALS
1. f(x) œ x$ 2x 3 Ê f w (x) œ 3x# 2 Ê L(x) œ f w (2)(x 2) f(2) œ 10(x 2) 7 Ê L(x) œ 10x 13 at x œ 2
2. f(x) œ Èx# 9 œ ax# 9b
"Î#
Ê f w (x) œ ˆ "# ‰ ax# 9b
œ 45 (x 4) 5 Ê L(x) œ 45 x 3. f(x) œ x "
x
9
5
"Î#
(2x) œ
x
È x# 9
Ê L(x) œ f w (4)(x 4) f(4)
at x œ 4
Ê f w (x) œ 1 x# Ê L(x) œ f(1) f w (1)(x 1) œ # !(x 1) œ #
4. f(x) œ x"Î$ Ê f w (x) œ
"
$x#Î$
"
1#
Ê L(x) œ f w (8)ax a8bb fa8b œ
(x 8) 2 Ê L(x) œ
"
1#
x
4
3
5. f(x) œ tan x Ê f w axb œ sec2 x Ê Laxb œ fa1b f w a1bax 1b œ 0 1ax 1b œ x 1
6. (a) f(x) œ sin x Ê f w axb œ cos x Ê Laxb œ fa0b f w a0bax 0b œ x Ê Laxb œ x
(b) f(x) œ cos x Ê f w axb œ sin x Ê Laxb œ fa0b f w a0bax 0b œ 1 Ê Laxb œ 1
(c) f(x) œ tan x Ê f w axb œ sec2 x Ê Laxb œ fa0b f w a0bax 0b œ x Ê Laxb œ x
7. f(x) œ x# 2x Ê f w (x) œ 2x 2 Ê L(x) œ f w (0)(x 0) f(0) œ 2(x 0) 0 Ê L(x) œ 2x at x œ 0
8. f(x) œ x" Ê f w (x) œ x# Ê L(x) œ f w (1)(x 1) f(1) œ (1)(x 1) 1 Ê L(x) œ x 2 at x œ 1
9. f(x) œ 2x# 4x 3 Ê f w (x) œ 4x 4 Ê L(x) œ f w (1)(x 1) f(1) œ 0(x 1) (5) Ê L(x) œ 5 at x œ 1
10. f(x) œ 1 x Ê f w (x) œ 1 Ê L(x) œ f w (8)(x 8) f(8) œ 1(x 8) 9 Ê L(x) œ x 1 at x œ 8
3
11. f(x) œ È
x œ x"Î$ Ê f w (x) œ ˆ "3 ‰ x#Î$ Ê L(x) œ f w (8)(x 8) f(8) œ
12. f(x) œ
x
x1
Ê L(x) œ
Ê f w (x) œ
"
4
x
"
4
(1)(x 1) (")(x)
(x 1)#
œ
"
(x 1)#
"
1#
(x 8) 2 Ê L(x) œ
Ê L(x) œ f w (1)(x 1) f(1) œ
"
4
(x 1) "
1#
x
4
3
at x œ 8
"
#
at x œ 1
13. f w axb œ ka" xbk" . We have fa!b œ " and f w a!b œ k. Laxb œ fa!b f w a!bax !b œ " kax !b œ " kx
'
14. (a) faxb œ a" xb' œ " axb‘ ¸ " 'axb œ " 'x
(b) faxb œ
#
" x
"
œ #" axb‘
(c) faxb œ a" xb
"Î#
¸ " ˆ "# ‰x œ " (d) faxb œ È2 x# œ È#Š" x#
#‹
(e) faxb œ a% $xb"Î$ œ %"Î$ ˆ" (f) faxb œ ˆ" ¸ #" a"baxb‘ œ # #x
" ‰2Î$
#x
"Î#
x
#
¸ È#Š" $x ‰"Î$
%
" x#
# #‹
¸ %"Î$ ˆ" 2Î$
œ ’" ˆ # " x ‰“
œ È#Š" " $x ‰
$ %
x#
%‹
œ %"Î$ ˆ" x% ‰
¸ " $# ˆ # " x ‰ œ " #
' $x
15. (a) (1.0002)&! œ (1 0.0002)&! ¸ 1 50(0.0002) œ 1 .01 œ 1.01
3
(b) È
1.009 œ (1 0.009)"Î$ ¸ 1 ˆ " ‰ (0.009) œ 1 0.003 œ 1.003
3
16. f(x) œ Èx 1 sin x œ (x 1)"Î# sin x Ê f w (x) œ ˆ "# ‰ (x 1)"Î# cos x Ê Lf (x) œ f w (0)(x 0) f(0)
œ 3 (x 0) 1 Ê Lf (x) œ 3 x 1, the linearization of f(x); g(x) œ Èx 1 œ (x 1)"Î# Ê gw (x)
#
#
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 3.9 Linearization and Differentials
œ ˆ "# ‰ (x 1)"Î# Ê Lg (x) œ gw (0)(x 0) g(0) œ
w
w
"
#
(x 0) 1 Ê Lg (x) œ
"
#
x 1, the linearization of g(x);
h(x) œ sin x Ê h (x) œ cos x Ê Lh (x) œ h (0)(x 0) h(0) œ (1)(x 0) 0 Ê Lh (x) œ x, the linearization of
h(x). Lf (x) œ Lg (x) Lh (x) implies that the linearization of a sum is equal to the sum of the linearizations.
17. y œ x$ 3Èx œ x$ 3x"Î# Ê dy œ ˆ3x# #3 x"Î# ‰ dx Ê dy œ Š3x# 3
‹
2È x
dx
"Î#
"Î#
"Î#
18. y œ xÈ1 x# œ x a1 x# b
Ê dy œ ’(1) a1 x# b (x) ˆ "# ‰ a1 x# b
(2x)“ dx
œ a1 x# b
"Î#
#
a1 2x# b
È 1 x#
Ê dy œ Š (2) a1 a1xb x# b(2x)(2x)
‹ dx œ
#
19. y œ
2x
1 x #
20. y œ
2È x
3 ˆ1 È x ‰
Ê dy œ
ca1 x# b x# d dx œ
œ
2x"Î#
3 a1 x"Î# b
"
#
3 È x ˆ1 È x ‰
Ê dy œ Š
dx
2 2x#
a1 x # b#
dx
x "Î# ˆ3 ˆ1 x"Î# ‰‰ 2x"Î# ˆ #3 x "Î# ‰
9 a1 x"Î# b
#
‹ dx œ
3x "Î# 3 3
#
9 a1 x"Î# b
dx
dx
21. 2y$Î# xy x œ 0 Ê 3y"Î# dy y dx x dy dx œ 0 Ê ˆ3y"Î# x‰ dy œ (1 y) dx Ê dy œ
1y
3 È y x
22. xy# 4x$Î# y œ 0 Ê y# dx 2xy dy 6x"Î# dx dy œ 0 Ê (2xy 1) dy œ ˆ6x"Î# y# ‰ dx
Ê dy œ
6È x y#
2xy 1
dx
23. y œ sin ˆ5Èx‰ œ sin ˆ5x"Î# ‰ Ê dy œ ˆcos ˆ5x"Î# ‰‰ ˆ 5# x"Î# ‰ dx Ê dy œ
5 cos ˆ5Èx‰
2È x
dx
24. y œ cos ax# b Ê dy œ csin ax# bd (2x) dx œ 2x sin ax# b dx
$
$
$
25. y œ 4 tan Š x3 ‹ Ê dy œ 4 Šsec# Š x3 ‹‹ ax# b dx Ê dy œ 4x# sec# Š x3 ‹ dx
26. y œ sec ax# 1b Ê dy œ csec ax# 1b tan ax# 1bd (2x) dx œ 2x csec ax# 1b tan ax# 1bd dx
27. y œ 3 csc ˆ1 2Èx‰ œ 3 csc ˆ1 2x"Î# ‰ Ê dy œ 3 ˆcsc ˆ1 2x"Î# ‰‰ cot ˆ1 2x"Î# ‰ ˆx"Î# ‰ dx
Ê dy œ È3 csc ˆ1 2Èx‰ cot ˆ1 2Èx‰ dx
x
28. y œ 2 cot Š È"x ‹ œ 2 cot ˆx"Î# ‰ Ê dy œ 2 csc# ˆx"Î# ‰ ˆ #" ‰ ˆx$Î# ‰ dx Ê dy œ
"
È x$
csc# Š È"x ‹ dx
29. f(x) œ x# 2x, x! œ 1, dx œ 0.1 Ê f w (x) œ 2x 2
(a) ?f œ f(x! dx) f(x! ) œ f(1.1) f(1) œ 3.41 3 œ 0.41
(b) df œ f w (x! ) dx œ [2(1) 2](0.1) œ 0.4
(c) k?f df k œ k0.41 0.4k œ 0.01
30. f(x) œ 2x# 4x 3, x! œ 1, dx œ 0.1 Ê f w (x) œ 4x 4
(a) ?f œ f(x! dx) f(x! ) œ f(.9) f(1) œ .02
(b) df œ f w (x! ) dx œ [4(1) 4](.1) œ 0
(c) k?f df k œ k.02 0k œ .02
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
dx
147
148
Chapter 3 Differentiation
31. f(x) œ x$ x, x! œ 1, dx œ 0.1 Ê f w (x) œ 3x# 1
(a) ?f œ f(x! dx) f(x! ) œ f(1.1) f(1) œ .231
(b) df œ f w (x! ) dx œ [3(1)# 1](.1) œ .2
(c) k?f df k œ k.231 .2k œ .031
32. f(x) œ x% , x! œ 1, dx œ 0.1 Ê f w (x) œ 4x$
(a) ?f œ f(x! dx) f(x! ) œ f(1.1) f(1) œ .4641
(b) df œ f w (x! ) dx œ 4(1)$ (.1) œ .4
(c) k?f df k œ k.4641 .4k œ .0641
33. f(x) œ x" , x! œ 0.5, dx œ 0.1 Ê f w (x) œ x#
(a) ?f œ f(x! dx) f(x! ) œ f(.6) f(.5) œ "3
" ‰
(b) df œ f w (x! ) dx œ (4) ˆ 10
œ 52
(c) k?f df k œ ¸ "3 25 ¸ œ
"
15
34. f(x) œ x$ 2x 3, x! œ 2, dx œ 0.1 Ê f w (x) œ 3x# 2
(a) ?f œ f(x! dx) f(x! ) œ f(2.1) f(2) œ 1.061
(b) df œ f w (x! ) dx œ (10)(0.10) œ 1
(c) k?f df k œ k1.061 1k œ .061
35. V œ
4
3
1r$ Ê dV œ 41r!# dr
36. V œ x$ Ê dV œ 3x!# dx
37. S œ 6x# Ê dS œ 12x! dx
38. S œ 1rÈr# h# œ 1r ar# h# b
Ê
dS
dr
œ
1 a r # h # b 1 r#
È r# h #
"Î#
Ê dS œ
, h constant Ê
1 a2r#! h# b
Ér#! h#
dS
dr
œ 1 ar# h# b
"Î#
1r † r ar# h# b
"Î#
dr, h constant
39. V œ 1r# h, height constant Ê dV œ 21r! h dr
40. S œ 21rh Ê dS œ 21r dh
41. Given r œ 2 m, dr œ .02 m
(a) A œ 1r# Ê dA œ 21r dr œ 21(2)(.02) œ .081 m#
1‰
(b) ˆ .08
41 (100%) œ 2%
42. C œ 21r and dC œ 2 in. Ê dC œ 21 dr Ê dr œ
œ 21(5) ˆ 1" ‰ œ 10 in.#
"
1
Ê the diameter grew about
43. The volume of a cylinder is V œ 1r# h. When h is held fixed, we have
dV
dr
2
1
in.; A œ 1r# Ê dA œ 21r dr
œ #1rh, and so dV œ #1rh dr. For h œ $! in.,
r œ ' in., and dr œ !Þ& in., the volume of the material in the shell is approximately dV œ #1rh dr œ #1a'ba$!ba!Þ&b
œ ")!1 ¸ &'&Þ& in$ .
44. Let ) œ angle of elevation and h œ height of building. Then h œ $!tan ), so dh œ $!sec# ) d). We want ldhl !Þ!%h,
&1
&1
sin )
which gives: l$!sec# ) d)l !Þ!%l$!tan )l Ê cos"# ) ld)l !Þ!%
cos ) Ê ld)l !Þ!%sin ) cos ) Ê ld)l !Þ!%sin "# cos "#
œ !Þ!" radian. The angle should be measured with an error of less than !Þ!" radian (or approximatley !Þ&( degrees),
which is a percentage error of approximately !Þ('%.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 3.9 Linearization and Differentials
45. The percentage error in the radius is
(a) Since C œ 21 r Ê
ˆ21 dr
‰
dt
21 r
œ
ˆ21 r dr
‰
dt
1 r2
œ
‚ 100 Ÿ 2%.
dC
dr
dt œ 21 dt . The percentage
ˆ dr
‰
dt
r ‚ 100 Ÿ 2%.
‚ 100 œ
(b) Since A œ 1 r2 Ê
ˆ dr
‰
dt
r
‚ 100 œ 2
ˆ dr
‰
dt
r
(a) Since S œ 6x Ê
dS
dt
œ
ˆ dx ‰
2 dtx
œ
œ 12x
dx
dt .
ˆ dx
‰
dt
x
‚ 100 Ÿ 0.5%.
The percentage error in the cube's surface area is
œ 3x2 dx
dt . The percentage error in the cube's volume is
dV
dt
Ê k31h# dhk Ÿ
of h is
ˆ dS
‰
dt
S
ˆ12x dx
‰
dt
6x2
‚ 100 œ
‚ 100
ˆ dV
‰
dt
V
‚ 100 œ
ˆ3x2 dx
‰
dt
x3
‚ 100
‚ 100 Ÿ 3a0.5%b œ 1.5%
47. V œ 1h$ Ê dV œ 31h# dh; recall that ?V ¸ dV. Then k?Vk Ÿ (1%)(V) œ
"
3
‚ 100
‚ 100 Ÿ 2a0.5%b œ 1%
(b) Since V œ x3 Ê
ˆ dx ‰
3 dtx
ˆ dA
‰
dt
A
‚ 100
‚ 100 Ÿ 2a2%b œ 4%.
46. The percentage error in the edge of the cube is
2
ˆ dC
‰
dt
C
error in calculating the circle's circumference is
œ 21 r dr
dt . The percentage error in calculating the circle's area is given by
dA
dt
149
(1) a1h$ b
100
Ê kdhk Ÿ
"
300
(1) a1h$ b
100
Ê kdVk Ÿ
(1) a1h$ b
100
h œ ˆ 3" %‰ h. Therefore the greatest tolerated error in the measurement
%.
#
48. (a) Let Di represent the interior diameter. Then V œ 1r# h œ 1 ˆ D#i ‰ h œ
1D#i h
4
#
" ‰ 5 1 Di
dV œ 51Di dDi . Recall that ?V ¸ dV. We want k?Vk Ÿ (1%)(V) Ê kdVk Ÿ ˆ 100
Š # ‹
Ê 51Di dDi Ÿ
1D#i
40
Ê
dDi
Di
51D#i
#
1Di#
œ 40
and h œ 10 Ê V œ
Ê
Ÿ 200. The inside diameter must be measured to within 0.5%.
(b) Let De represent the exterior diameter, h the height and S the area of the painted surface. S œ 1De h Ê dS œ 1hdDe
dDe
Ê dS
S œ De . Thus for small changes in exterior diameter, the approximate percentage change in the exterior diameter
is equal to the approximate percentage change in the area painted, and to estimate the amount of paint required to
within 5%, the tanks's exterior diameter must be measured to within 5%.
49. Given D œ 100 cm, dD œ 1 cm, V œ
œ”
50. V œ
œ
10% 1
#
10' 1
6
4
3
1 D$
200
#
• a10 %b œ ”
1 r$ œ
4
3
$
10' 1
#
10' 1
6
1 ˆ D# ‰ œ
Ê kdVk Ÿ
1 D$
200
4
3
$
1 ˆ D# ‰ œ
1 D$
6
Ê dV œ
1
#
D# dD œ
1
#
(100)# (1) œ
10% 1
#
. Then
dV
V
(100%)
• % œ 3%
1 D$
6
Ê dV œ
#
Ê ¹ 1D# dD¹ Ÿ
1 D#
#
3 ‰ 1D
dD; recall that ?V ¸ dV. Then k?Vk Ÿ (3%)V œ ˆ 100
Š 6 ‹
$
1 D$
#00
Ê kdDk Ÿ
D
100
œ (1%) D Ê the allowable percentage error in
measuring the diameter is 1%.
51. W œ a b
g
œ a bg" Ê dW œ bg# dg œ bgdg
Ê
#
dWmoon
dWearth
œ
b dg
‹
(5.2)#
b dg
Š # ‹
(32)
Š
#
32 ‰
œ ˆ 5.2
œ 37.87, so a change of
gravity on the moon has about 38 times the effect that a change of the same magnitude has on Earth.
52. (a) T œ 21 Š Lg ‹
"Î#
Ê dT œ 21ÈL ˆ "# g$Î# ‰ dg œ 1ÈL g$Î# dg
(b) If g increases, then dg 0 Ê dT 0. The period T decreases and the clock ticks more frequently. Both
the pendulum speed and clock speed increase.
(c) 0.001 œ 1È100 ˆ980$Î# ‰ dg Ê dg ¸ 0.977 cm/sec# Ê the new g ¸ 979 cm/sec#
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
150
Chapter 3 Differentiation
53. E(x) œ f(x) g(x) Ê E(x) œ f(x) m(x a) c. Then E(a) œ 0 Ê f(a) m(a a) c œ 0 Ê c œ f(a). Next
f(x) m(x a) c
f(a)
œ 0 Ê xlim
œ 0 Ê xlim
’ f(x)x xa
a m“ œ 0 (since c œ f(a))
Äa
Äa
Ê f w (a) m œ 0 Ê m œ f w (a). Therefore, g(x) œ m(x a) c œ f w (a)(x a) f(a) is the linear approximation,
as claimed.
we calculate m: xlim
Äa
E(x)
xa
54. (a) i. Qaab œ faab implies that b! œ faab.
ii. Since Qw axb œ b" #b# ax ab, Qw aab œ f w aab implies that b" œ f w aab.
iii. Since Qww axb œ #b# , Qww aab œ f ww aab implies that b2 œ
In summary, b! œ faab, b" œ f w aab, and b2 œ
ww
f aa b
# .
ww
f aa b
# .
(b) faxb œ a" xb" ; f w axb œ "a" xb# a"b œ a" xb# ; f ww axb œ #a" xb$ a"b œ #a" xb$
Since fa!b œ ", f w a!b œ ", and f ww a!b œ #, the coefficients are b! œ ", b" œ ", b# œ ## œ ". The quadratic
approximation is Qaxb œ " x x# .
(c)
As one zooms in, the two graphs quickly become
indistinguishable. They appear to be identical.
(d) gaxb œ x" ; gw axb œ "x# ; gww axb œ #x$
Since ga"b œ ", gw a"b œ ", and gww a"b œ # , the coefficients are b! œ ", b" œ ", b# œ
#
#
#
œ ". The quadratic
approximation is Qaxb œ " ax "b ax "b .
As one zooms in, the two graphs quickly become
indistinguishable. They appear to be identical.
(e) haxb œ a" xb"Î# ; hw axb œ "# a" xb"Î# ; hww axb œ "% a" xb$Î#
Since ha!b œ ", hw a!b œ "# , and hww a!b œ "% , the coefficients are b! œ ", b" œ "# , b# œ
approximation is Qaxb œ " x
#
#
x
8
"%
2
œ "8 . The quadratic
.
As one zooms in, the two graphs quickly become
indistinguishable. They appear to be identical.
(f) The linearization of any differentiable function uaxb at x œ a is Laxb œ uaab uw aabax ab œ b! b" ax ab, where
b! and b" are the coefficients of the constant and linear terms of the quadratic approximation. Thus, the linearization
for faxb at x œ ! is " x; the linearization for gaxb at x œ " is " ax "b or # x; and the linearization for haxb at
x œ ! is " x# .
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Chapter 3 Practice Exercises
151
55-58. Example CAS commands:
Maple:
with(plots):
a:= 1: f:=x -> x • 3 x • 2 2*x;
plot(f(x), x=1..2);
diff(f(x),x);
fp := unapply (ww ,x);
L:=x -> f(a) fp(a)*(x a);
plot({f(x), L(x)}, x=1..2);
err:=x -> abs(f(x) L(x));
plot(err(x), x=1..2, title = #absolute error function#);
err(1);
Mathematica: (function, x1, x2, and a may vary):
Clear[f, x]
{x1, x2} = {1, 2}; a = 1;
f[x_]:=x3 x2 2x
Plot[f[x], {x, x1, x2}]
lin[x_]=f[a] f'[a](x a)
Plot[{f[x], lin[x]}, {x, x1, x2}]
err[x_]=Abs[f[x] lin[x]]
Plot[err[x], {x, x1,x 2}]
err//N
After reviewing the error function, plot the error function and epsilon for differing values of epsilon (eps) and delta (del)
eps = 0.5; del = 0.4
Plot[{err[x], eps},{x, a del, a del}]
CHAPTER 3 PRACTICE EXERCISES
1. y œ x& 0.125x# 0.25x Ê
2. y œ 3 0.7x$ 0.3x( Ê
3. y œ x$ 3 ax# 1# b Ê
4. y œ x( È7x "
1 1
Ê
dy
dx
œ 3x# 3(2x 0) œ 3x# 6x œ 3x(x 2)
dy
dx
œ 7x' È7
œ 2(x 1) a2x# 4x 1b
6. y œ (2x 5)(4 x)" Ê
œ 3(4 x)#
$
8. y œ Š1 csc )
#
)#
4‹
#
œ 5x% 0.25x 0.25
œ 2.1x# 2.1x'
dy
dx
5. y œ (x 1)# ax# 2xb Ê
7. y œ a)# sec ) 1b Ê
dy
dx
dy
dx
œ (x 1)# (2x 2) ax# 2xb (2(x 1)) œ 2(x 1) c(x 1)# x(x 2)d
dy
dx
œ (2x 5)(1)(4 x)# (1) (4 x)" (2) œ (4 x)# c(2x 5) 2(4 x)d
dy
d)
Ê
#
œ 3 a)# sec ) 1b (2) sec ) tan ))
dy
d)
œ 2 Š1 csc )
#
)# ˆ csc ) cot )
4‹
#
#) ‰ œ Š1 csc )
#
)#
4 ‹ (csc
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
) cot ) ))
152
Chapter 3 Differentiation
9. s œ
Èt
1 Èt
Ê
ds
dt
œ
10. s œ
"
Èt 1
Ê
ds
dt
œ
ˆ1 Èt‰†
"
sin# x
"
ˆ1 Èt‰
ˆÈ t 1 ‰
#
"
Èt ‹
#
œ
13. s œ cos% (1 2t) Ê
14. s œ cot$ ˆ 2t ‰ Ê
ds
dt
ˆ1 Èt‰ Èt
2Èt ˆ1 Èt‰
#
"
#
#Èt ˆ1 Èt‰
"
#
2 È t ˆÈ t 1 ‰
dy
dx
œ (2 csc x)(csc x cot x) 2( csc x cot x) œ (2 csc x cot x)(1 csc x)
œ 4 cos$ (1 2t)(sin (1 2t))(2) œ 8 cos$ (1 2t) sin (1 2t)
ds
dt
œ 3 cot# ˆ 2t ‰ ˆcsc# ˆ 2t ‰‰ ˆ t#2 ‰ œ
15. s œ (sec t tan t)& Ê
œ
œ (4 tan x) asec# xb (2 sec x)(sec x tan x) œ 2 sec# x tan x
dy
dx
œ csc# x 2 csc x Ê
2
sin x
œ
#
ˆÈt 1‰ (0) 1 Š
11. y œ 2 tan# x sec# x Ê
12. y œ
"
Èt Èt Š #Èt ‹
#
ds
dt
16. s œ csc& a1 t 3t# b Ê
6
t#
cot# ˆ 2t ‰ csc# ˆ 2t ‰
œ 5(sec t tan t)% asec t tan t sec# tb œ 5(sec t)(sec t tan t)&
ds
dt
œ 5 csc% a1 t 3t# b acsc a1 t 3t# b cot a1 t 3t# bb (1 6t)
œ 5(6t 1) csc& a1 t 3t# b cot a1 t 3t# b
"
#
) cos ) sin )
È2) sin )
17. r œ È2) sin ) œ (2) sin ))"Î# Ê
dr
d)
œ
(2) sin ))"Î# (#) cos ) 2 sin )) œ
18. r œ 2)Ècos ) œ 2) (cos ))"Î# Ê
dr
d)
œ 2) ˆ "# ‰ (cos ))"Î# (sin )) 2(cos ))"Î# œ
) sin )
Ècos )
2Ècos )
2 cos ) ) sin )
Ècos )
œ
19. r œ sin È2) œ sin (2))"Î# Ê
20. r œ sin Š) È) 1‹ Ê
œ cos (2))"Î# ˆ "# (2))"Î# (2)‰ œ
œ cos Š) È) 1‹ Š1 cos È2)
È 2)
"
‹
2È ) 1
œ
2È)"1
#È ) "
œ
"
#
22. y œ 2Èx sin Èx Ê
dy
dx
"
2
œ 2Èx ˆcos Èx‰ Š 2È
‹ ˆsin Èx‰ Š 2È
‹ œ cos Èx x
x
x# csc
2
x
Ê
x# ˆcsc
2
x
cot x2 ‰ ˆ x#2 ‰ ˆcsc x2 ‰ ˆ "# † 2x‰ œ csc
cos Š) È) 1‹
dy
dx
21. y œ
"
#
dr
d)
dr
d)
2
x
cot
2
x
x csc
2
x
sin Èx
Èx
dy
"Î#
sec (2x)# tan (2x)# (2(2x) † 2) sec (2x)# ˆ "# x$Î# ‰
dx œ x
8x"Î# sec (2x)# tan (2x)# "# x$Î# sec (2x)# œ "# x"Î# sec (2x)# c16 tan (2x)# x# d or #x"$Î# seca#xb2 16x# tana2xb#
23. y œ x"Î# sec (2x)# Ê
œ
24. y œ Èx csc (x 1)$ œ x"Î# csc (x 1)$
Ê
dy
dx
œ x"Î# acsc (x 1)$ cot (x 1)$ b a3(x 1)# b csc (x 1)$ ˆ "# x"Î# ‰
œ 3Èx (x 1)# csc (x 1)$ cot (x 1)$ or
"
csc(x
#È x
csc (x 1)$
2È x
œ
"
#
Èx csc (x 1)$ x" 6(x 1)# cot (x 1)$ ‘
1)$ c1 6x(x 1)# cot (x 1)$ d
25. y œ 5 cot x# Ê
26. y œ x# cot 5x Ê
dy
dx
œ 5 acsc# x# b (2x) œ 10x csc# ax# b
dy
dx
œ x# acsc# 5xb (5) (cot 5x)(2x) œ 5x# csc# 5x 2x cot 5x
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
"‘
Chapter 3 Practice Exercises
153
27. y œ x# sin# a2x# b Ê
dy
dx
œ x# a2 sin a2x# bb acos a2x# bb (4x) sin# a2x# b (2x) œ 8x$ sin a2x# b cos a2x# b 2x sin# a2x# b
28. y œ x# sin# ax$ b Ê
dy
dx
œ x# a2 sin ax$ bb acos ax$ bb a3x# b sin# ax$ b a2x$ b œ 6 sin ax$ b cos ax$ b 2x$ sin# ax$ b
29. s œ ˆ t 4t 1 ‰
30. s œ
#
Ê
"
15(15t 1)$
œ 2 ˆ t 4t 1 ‰
$
(4t)(1)
Š (t 1)(4)
‹ œ 2 ˆ t 4t 1 ‰
(t 1)#
"
œ 15
(15t 1)$ Ê
#
Èx
ds
dt
31. y œ Š x 1 ‹ Ê
dy
dx
#
2È x
32. y œ Š 2Èx 1 ‹ Ê
dy
dx
(x 1)#
2È x
œ 2 Š 2È x 1 ‹ #
"Î#
33. y œ É x x# x œ ˆ1 "x ‰
Ê
dy
dx
œ
34. y œ 4xÉx Èx œ 4x ˆx x"Î# ‰
œ ˆx Èx‰
"Î#
#
’2x Š1 "
‹
#È x
"Î#
(x 1) 2x
(x 1)$
œ
ˆ2Èx 1‰ Š È" ‹ ˆ2Èx‰ Š È" ‹
x
x
ˆ2 È x 1 ‰
#
ˆ1 "x ‰"Î# ˆ x"# ‰ œ "
#
4
(t 1)#
"
œ 15
(3)(15t 1)% (15) œ
ds
dt
"
(x 1) Š #È
‹ ˆÈx‰ (1)
x
Èx
œ 2 Šx1‹ †
$
Ê
dy
dx
œ
"
"Î#
"Î#
3
(15t 1)%
1x
(x 1)$
#x # É 1 œ 4x ˆ "# ‰ ˆx x"Î# ‰
4 ˆx Èx‰“ œ ˆx Èx‰
œ
œ (t 8t$1)
4Èx Š È"x ‹
ˆ2 È x 1 ‰ $
œ
"
x
ˆ1 "# x"Î# ‰ ˆx x"Î# ‰"Î# (4)
ˆ2x Èx 4x 4Èx‰ œ
35. r œ ˆ cossin) ) 1 ‰ Ê
dr
d)
)) (sin ))(sin ))
œ 2 ˆ cossin) ) 1 ‰ ’ (cos ) 1)(cos
“ œ 2 ˆ cossin) ) " ‰ Š cos
(cos ) 1)#
(2 sin )) (1 cos ))
(cos ) 1)$
œ
2 sin )
(cos ) ")#
dr
d)
1 ‰ (1 cos ))(cos )) (sin ) ")(sin ))
œ 2 ˆ 1sin )cos
“œ
) ’
(1 cos ))#
œ
#
1 ‰
36. r œ ˆ 1sin )cos
Ê
)
œ
4
ˆ2 È x 1‰$
2(sin ) ")
(1 cos ))$
#
6x 5Èx
É x Èx
) cos ) sin# )
‹
(cos ) ")#
acos ) cos# ) sin# ) sin )b
2(sin ) 1)(cos ) sin ) 1)
(1 c os ))$
37. y œ (2x 1) È2x 1 œ (2x 1)$Î# Ê
dy
dx
œ
3
#
(2x 1)"Î# (2) œ 3È2x 1
38. y œ 20(3x 4)"Î% (3x 4)"Î& œ 20(3x 4)"Î#! Ê
39. y œ 3 a5x# sin 2xb
40. y œ a3 cos$ 3xb
$Î#
"Î$
Ê
Ê
dy
dx
dy
dx
dy
dx
" ‰
œ 20 ˆ 20
(3x 4)"*Î#! (3) œ
œ 3 ˆ 3# ‰ a5x# sin 2xb
œ "3 a3 cos$ 3xb
%Î$
&Î#
[10x (cos 2x)(2)] œ
a3 cos# 3xb (sin 3x)(3) œ
3
(3x 4)"*Î#!
9(5x cos 2x)
a5x# sin 2xb&Î#
3 cos# 3x sin 3x
a3 cos$ 3xb%Î$
2
41. xy 2x 3y œ 1 Ê axyw yb 2 3yw œ 0 Ê xyw 3yw œ 2 y Ê yw (x 3) œ 2 y Ê yw œ yx 3
42. x# xy y# 5x œ 2 Ê 2x Šx
Ê
dy
dx
œ
dy
dx
dy
dx
dy
dx
5œ!Êx
dy
dx
2y
dy
dx
œ 5 2x y Ê
dy
dx
(x 2y) œ 5 2x y
5 2x y
x 2y
43. x$ 4xy 3y%Î$ œ 2x Ê 3x# Š4x
Ê
y‹ 2y
dy
dx
ˆ4x 4y"Î$ ‰ œ 2 3x# 4y Ê
4y‹ 4y"Î$
dy
dx
œ
dy
dx
œ 2 Ê 4x
dy
dx
4y"Î$
dy
dx
œ 2 3x# 4y
2 3x# 4y
4x 4y"Î$
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
154
Chapter 3 Differentiation
44. 5x%Î& 10y'Î& œ 15 Ê 4x"Î& 12y"Î&
"
#
45. (xy)"Î# œ 1 Ê
(xy)"Î# Šx
46. x# y# œ 1 Ê x# Š2y
47. y# œ
Ê 2y
x
x 1
x‰
48. y# œ ˆ 11 x
"Î#
dy
dx
dy
dx ‹
y# (2x) œ 0 Ê 2x# y
Ê y% œ
"x
1x
49. p$ 4pq 3q# œ 2 Ê 3p#
Ê
dp
dq
œ
dp
dq
Ê
Ê 4y$
dp
dq
dy
dx
œ x"Î# y"Î# Ê
dy
dx
œ 2xy# Ê
dy
dx
œ yx
Ê
dy
dx
œ
6q œ 0 Ê 3p#
dp
dq
dy
dx
œ
œ
(1 x)(1) (1 x)Ð")
(" x)#
Ê
$Î#
dp
dq ‹
Ê 1 œ 3# a5p# 2pb
dy
dx
œ yx
4q
dp
dq
œ 6q 4p Ê
dp
dq
a3p# 4qb œ 6q 4p
&Î#
Š10p
dp
dq
2
dp
dq ‹
Ê 23 a5p# 2pb
&Î#
œ
dp
dq
(10p 2)
#
œ a5p3(5p 2p1)b
2r sin 2s sin 2s
cos 2s
œ
œ
(2r 1)(sin 2s)
cos 2s
53. (a) x$ y$ œ 1 Ê 3x# 3y#
Ê
d# y
dx#
œ
(b) y# œ 1 Ê
d# y
dx#
2
x
œ
2xy# a2yx# b Š
y%
Ê 2y
dy
dx
2xy x# Š
y# x%
œ
"
‹
yx#
œ
x
y
Ê
d# y
dx#
œ
dy
dx
x#
‹
y#
œ
y(1) x
y#
(cos 2s) œ 2r sin 2s 2 sin s cos s
dr
ds
2xy# y%
dy
dx
1 2s œ 0 Ê
dr
ds
d# y
dx#
œ
#
œ yx# Ê
dy
dx
2x%
y
œ
"
yx#
œ
(2s 1) œ 1 2s 2r Ê
y# (2x) ax# b Š2y
dr
ds
œ
" 2s 2r
2s 1
dy
dx ‹
y%
2xy$ 2x%
y&
Ê
dy
dx
œ ayx# b
"
Ê
dy
dx
œ
d# y
dx#
œ ayx# b
#
’y(2x) x#
dy
dx “
2xy# 1
y$ x%
dy
dx
dy
dx
œ0 Ê
œ
dr
ds
œ (2r 1)(tan 2s)
dr ‰
ds
Ê
2
x#
54. (a) x# y# œ 1 Ê 2x 2y
dy
dx
œ x" y Ê
&Î#
52. 2rs r s s# œ 3 Ê 2 ˆr s
(b)
dy
dx
"
2y$ (1 x)#
dr ‰
51. r cos 2s sin# s œ 1 Ê r(sin 2s)(2) (cos 2s) ˆ ds
2 sin s cos s œ 0 Ê
dr
ds
"
œ "3 x"Î& y"Î& œ 3(xy)
"Î&
dy
dx
"
#y(x 1)#
dy
dx
4 Šp q
œ 4x"Î& Ê
dy
dx
6q 4p
3p# 4q
50. q œ a5p# 2pb
Ê
y‹ œ 0 Ê x"Î# y"Î#
dy
dx
(x 1)(1) (x)(1)
(x 1)#
œ
œ 0 Ê 12y"Î&
dy
dx
œ 0 Ê 2y
œ
y x Š xy ‹
y#
œ
œ 2x Ê
dy
dx
y# x#
y$
œ
"
y$
x
y
(since y# x# œ 1)
55. (a) Let h(x) œ 6f(x) g(x) Ê hw (x) œ 6f w (x) gw (x) Ê hw (1) œ 6f w (1) gw (1) œ 6 ˆ "# ‰ a%b œ (
(b) Let h(x) œ f(x)g# (x) Ê hw (x) œ f(x) a#g(x)b gw (x) g# (x)f w (x) Ê hw (0) œ #f(0)g(0)gw (0) g# (0)f w (0)
œ #(1)(1) ˆ "# ‰ (1)# ($) œ #
(c) Let h(x) œ
f(x)
g(x) 1
Ê hw (x) œ
(g(x) 1)f (x) f(x)g (x)
(g(x) 1)#
w
w
w
w
(& 1) ˆ "# ‰ 3 a%b
(g(1) ")f (1) f(1)g (1)
œ
(g(1) 1)#
(& 1)#
"
"
w
w
w
f (g(0))g (0) œ f (1) ˆ # ‰ œ ˆ # ‰ ˆ "# ‰ œ "%
w
w
w
w
Ê hw (1) œ
(d) Let h(x) œ f(g(x)) Ê hw (x) œ f (g(x))g (x) Ê hw (0) œ
w
w
œ
&
"#
(e) Let h(x) œ g(f(x)) Ê hw (x) œ gw (f(x))f w (x) Ê hw (0) œ g (f(0))f (0) œ g (1)f (0) œ a%b ($) œ "#
(f) Let h(x) œ (x f(x))$Î# Ê hw (x) œ 3# (x f(x))"Î# a1 f w (x)b Ê hw (1) œ 3# (1 f(1))"Î# a1 f w (1)b
œ 3# (1 3)"Î# ˆ1 "# ‰ œ *#
(g) Let h(x) œ f(x g(x)) Ê hw (x) œ f w (x g(x)) a1 gw (x)b Ê hw (0) œ f w (g(0)) a1 gw (0)b
œ f w (1) ˆ1 "# ‰ œ ˆ "# ‰ ˆ $# ‰ œ $%
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Chapter 3 Practice Exercises
"
#È x
56. (a) Let h(x) œ Èx f(x) Ê hw (x) œ Èx f w (x) f(x) †
"
#
(b) Let h(x) œ (f(x))"Î# Ê hw (x) œ
(f(x))"Î# af w (x)b Ê hw (0) œ
(c) Let h(x) œ f ˆÈx‰ Ê hw (x) œ f w ˆÈx‰ †
"
#È x
"
œ 5" (3) ˆ #" ‰
#È 1
"
"Î#
(2) œ 3"
# (9)
Ê hw (1) œ È1 f w (1) f(1) †
"
#
(f(0))"Î# f w (0) œ
"
#È 1
w
Ê hw (1) œ f w ŠÈ1‹ †
"
5
œ
†
"
#
œ
œ 13
10
"
10
(d) Let h(x) œ f(1 5 tan x) Ê hw (x) œ f w (1 5 tan x) a5 sec# xb Ê h (0) œ f w (1 5 tan 0) a5 sec# 0b
œ f w (1)(5) œ "5 (5) œ 1
(2 cos x)f (x) f(x)(sin x)
f(0)(0)
Ê hw (0) œ (2 1)f(2(0)
œ 3(9 2) œ
(2 cos x)#
1)#
h(x) œ 10 sin ˆ 1#x ‰ f # (x) Ê hw (x) œ 10 sin ˆ 1#x ‰ a2f(x)f w (x)b f # (x) ˆ10 cos ˆ 1#x ‰‰ ˆ 1# ‰
hw (1) œ 10 sin ˆ 1# ‰ a2f(1)f w (1)b f # (1) ˆ10 cos ˆ 1# ‰‰ ˆ 1# ‰ œ 20(3) ˆ "5 ‰ ! œ 12
(e) Let h(x) œ
(f) Let
Ê
57. x œ t# 1 Ê
dy
dt
œ
dy
dx
†
"Î$
œ 2 au# 2ub
ds ¸
du u=2
Ê
"Î$
"
3
5; thus
œ ’2 a2# 2(2)b
cos É8 sin ˆs 16 ‰ 2
2É8 sin ˆs Ê
œ
dt
du
59. r œ 8 sin ˆs 16 ‰ Ê
œ
œ
dw ¸
ds s = 0
1‰
and
2 sin (0)
3 1
Ê
œ
"Î$
5“ ˆ 23 ‰ a2# 2(2)b
; thus,
d# y
dx#
œ
œ
"3
8#Î$
œ
"
3
7b
#Î$
2
3
d ) ‰‰
dt
(2)) œ
(1 7)#Î$ œ
d# y
dx#
œ
†
dw
dr
dy
dx
"
6
2
3
d)
dt
"
3
2
#
3 au
"Î$
#Î$
2ub
dr ¸
dt t = 0
d)
dt
#Î$
œ
1b
x#Î$ 3" y#Î$
dy
ˆ #Î$ ‰ ˆ 23
dx ‹ y
#
ax#Î$ b
2ub
#Î$
#
dw
dr
dy
dx
† 8 cos ˆs 16 ‰‘
d)
dt
œ
) #
2)t1
; r œ a)# 7b
œ
ˆ 6" ‰ (1)
œ
a3y# 1b œ 2 sin x Ê
dy
dx
œ
2 sin x
3y# 1
Ê
dy
dx ‹
x "Î$ ‰
Ê
dy
dx
#Î$
œ yx#Î$ Ê
d# y
dx# ¹ (8ß8)
œ
dy
dx ¹ (8ß8)
œ 1;
dy
dx
œ
y#Î$
x#Î$
ˆ8#Î$ ‰ 23 †8 "Î$ †(1)‘ ˆ8#Î$ ‰ ˆ 23 †8 "Î$ ‰
8%Î$
"
6
"
"
f(t h) f(t)
2t 1 (2t 2h 1)
"
"
œ #(t h) 1h #t 1 œ (2t
2t 1 and f(t h) œ #(t h) 1 Ê
h
2h 1)(2t 1)h
f(t h) f(t)
2h
2
w
œ lim (2t 2h 21)(#t 1)
(2t 2h 1)(2t 1)h œ (2t 2h 1)(2t 1) Ê f (t) œ hlim
h
Ä!
hÄ!
#
#
(2t 1)
63. f(t) œ
œ
œ
œ
"
6
œ #"
œ0 Ê
"Î$
d) ¸
dt t= 0, ) = 1
#
dy
dx
9
#
œ È3
d) ¸
dt t = 0
†
œ 2t 5
"
œ cos ˆÈr 2‰ Š #È
‹
r
(2)t 1) œ )# Ê
dr ¸
d) t = 0
ds
dt
(u 1)
; now t œ 0 and )# t ) œ 1 Ê ) œ 1 so that
œ 2 sin x Ê
a3y#
È3 ‹
2È4
) a)# 7b
Ê
dy
dx
œ
(u 1); s œ t# 5t Ê
(2 1) œ 2 ˆ2 † 8"Î$ 5‰ ˆ8#Î$ ‰ œ 2(2 † 2 5) ˆ 4" ‰ œ
# É8 sinˆ s 16 ‰
(cos 0)(8) Š
#Î$
5“ ˆ 32 ‰ au#
cos ŠÉ8 sin ˆs 16 ‰ 2‹
œ0 Ê
(3 1)(2 cos 0) (2 sin 0)(6†0)
(3 1)#
œ
œ
dr
ds
a3y# 1b (2 cos x) (2 sin x) Š6y
ˆx#Î$ ‰ Š 23 y "Î$
4
œ
dw
ds
6
62. x"Î$ y"Î$ œ 4 Ê
Ê
(2u 2) œ
œ ’2 au# 2ub
dt
du
2É8 sin ˆ 16 ‰
œ 0;
d# y
dx# ¹ (0ß1)
#Î$
œ
ds
dt
†
œ 3(cos 2x)(2) œ 6 cos 2x œ 6 cos a2t# 21b œ 6 cos a2t# b ; thus,
œ 8 cos ˆs 16 ‰ ; w œ sin ˆÈr 2‰ Ê
dr
ds
61. y$ y œ 2 cos x Ê 3y#
œ
au# 2ub
32
œ 6 cos (0) † 0 œ 0
dy
dt ¹ t = 0
cos ŠÉ8 sin ˆ 16 ‰ 2‹†8 cos ˆ 16 ‰
dr
"
#
d ) œ 3 a)
dr ¸
2
d) ) = 1 œ 3
dy
dx
w
ds
du
60. )# t ) œ 1 Ê ˆ)# t ˆ2)
Ê
w
œ 2t; y œ 3 sin 2x Ê
dx
dt
œ 6 cos a2t# b † 2t Ê
dx
dt
58. t œ au# 2ub
Ê
Ê hw (x) œ
f(x)
2 cos x
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
dy
dx ¹ (0ß1)
1
1
œ 1
155
156
Chapter 3 Differentiation
64. g(x) œ 2x# 1 and g(x h) œ 2(x h)# 1 œ 2x# 4xh 2h# 1 Ê
œ
4xh 2h#
h
g(x h) g(x)
h
œ 4x 2h Ê gw (x) œ lim
hÄ!
g(x h) g(x)
h
œ
a2x# 4xh 2h# 1b a2x# 1b
h
œ lim (4x 2h) œ 4x
hÄ!
65. (a)
lim f(x) œ lim c x# œ 0 and lim b f(x) œ lim b x# œ 0 Ê lim f(x) œ 0. Since lim f(x) œ 0 œ f(0) it
xÄ!
xÄ!
xÄ!
xÄ!
xÄ!
follows that f is continuous at x œ 0.
(c) lim c f w (x) œ lim c (2x) œ 0 and lim b f w (x) œ lim b (2x) œ 0 Ê lim f w (x) œ 0. Since this limit exists, it
(b)
x Ä !c
xÄ!
xÄ!
xÄ!
follows that f is differentiable at x œ 0.
xÄ!
xÄ!
66. (a)
lim f(x) œ lim c x œ 0 and lim b f(x) œ lim b tan x œ 0 Ê lim f(x) œ 0. Since lim f(x) œ 0 œ f(0), it
xÄ!
xÄ!
xÄ!
xÄ!
xÄ!
follows that f is continuous at x œ 0.
(c) lim c f w (x) œ lim c 1 œ 1 and lim b f w (x) œ lim b sec# x œ 1 Ê lim f w (x) œ 1. Since this limit exists it
(b)
x Ä !c
xÄ!
xÄ!
xÄ!
follows that f is differentiable at x œ 0.
xÄ!
xÄ!
67. (a)
lim f(x) œ lim c x œ 1 and lim b f(x) œ lim b (2 x) œ 1 Ê lim f(x) œ 1. Since lim f(x) œ 1 œ f(1), it
xÄ"
xÄ"
xÄ"
xÄ"
xÄ"
follows that f is continuous at x œ 1.
(c) lim c f w (x) œ lim c 1 œ 1 and lim b f w (x) œ lim b 1 œ 1 Ê lim c f w (x) Á lim b f w (x), so lim f w (x) does
(b)
x Ä "c
xÄ"
xÄ"
xÄ"
not exist Ê f is not differentiable at x œ 1.
68. (a)
xÄ"
xÄ"
xÄ"
lim f(x) œ lim c sin 2x œ 0 and lim b f(x) œ lim b mx œ 0 Ê lim f(x) œ 0, independent of m; since
xÄ!
xÄ!
xÄ!
xÄ!
f(0) œ 0 œ lim f(x) it follows that f is continuous at x œ 0 for all values of m.
x Ä !c
xÄ!
(b)
xÄ1
lim f w (x) œ lim c (sin 2x)w œ lim c 2 cos 2x œ 2 and lim b f w (x) œ lim b (mx)w œ lim b m œ m Ê f is
x Ä !c
xÄ!
xÄ!
xÄ!
xÄ!
xÄ!
differentiable at x œ 0 provided that lim c f w (x) œ lim b f w (x) Ê m œ 2.
xÄ!
xÄ!
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Chapter 3 Practice Exercises
69. y œ
x
#
"
#x 4
"
#
œ
x (2x 4)" Ê
dy
dx
œ
Ê (2x 5)(2x 3) œ 0 Ê x
"
2x
Ê
Ê xœ „
"
#
œ1
dy
dx
2
(2x)#
2(2x 4)# ; the slope of the tangent is 3# Ê 3# œ
"
#
2(2x 4)#
"
#
#
#
(2x 4)# Ê (2x 4) œ 1 Ê 4x 16x 16 œ 1 Ê 4x 16x 15 œ 0
œ 5# or x œ 3# Ê ˆ 5# ß 49 ‰ and ˆ 3# ß 4" ‰ are points on the curve where the slope
Ê 2 œ 2(2x 4)# Ê 1 œ
70. y œ x "
#
"
"
#x# ; the slope of the tangent is 3 Ê 3 œ 1 #x#
" "‰
# ß # are points on the curve where the slope is 3.
œ1
Ê ˆ "# ß "# ‰ and ˆ
71. y œ 2x$ 3x# 12x 20 Ê
dy
dx
Ê 2œ
œ 6x# 6x 12; the tangent is parallel to the x-axis when
dy
dx
"
#x #
157
is 3# .
Ê x# œ
"
4
œ0
Ê 6x# 6x 12 œ 0 Ê x# x 2 œ 0 Ê (x 2)(x 1) œ 0 Ê x œ 2 or x œ 1 Ê (#ß !) and ("ß #7) are
points on the curve where the tangent is parallel to the x-axis.
72. y œ x$ Ê
œ 3x# Ê
dy
dx
œ 12; an equation of the tangent line at (#ß )) is y 8 œ 12(x 2)
dy
dx ¹ ( 2ß 8)
Ê y œ 12x 16; x-intercept: 0 œ 12x 16 Ê x œ 43 Ê ˆ 43 ß !‰ ; y-intercept: y œ 12(0) 16 œ 16 Ê (0ß 16)
73. y œ 2x$ 3x# 12x 20 Ê
dy
dx
œ 6x# 6x 12
(a) The tangent is perpendicular to the line y œ 1 x
24
when
dy
dx
œ Š ˆ" " ‰ ‹ œ 24; 6x# 6x 12 œ 24
#4
Ê x# x 2 œ 4 Ê x# x 6 œ 0 Ê (x 3)(x 2) œ 0 Ê x œ 2 or x œ 3 Ê (#ß 16) and ($ß 11) are
x
points where the tangent is perpendicular to y œ 1 24
.
dy
È
(b) The tangent is parallel to the line y œ 2 12x when dx œ 12 Ê 6x# 6x 12 œ 12 Ê x# x œ 0
Ê x(x 1) œ 0 Ê x œ 0 or x œ 1 Ê (!ß 20) and ("ß () are points where the tangent is parallel to
y œ È2 12x.
74. y œ
1 sin x
x
Ê
dy
dx
œ
x(1 cos x) (1 sin x)(1)
x#
Ê m" œ
dy
dx ¹ x=1
œ
1 #
1#
œ 1 and m# œ
dy
1#
dx ¹ x=c1 1#
œ 1. Since m" œ m"# the
tangents intersect at right angles.
75. y œ tan x, 1# x of y
yœ
Ê
1
#
Ê
dy
dx
œ sec# x; now the slope
œ is "# Ê the normal line is parallel to
"
#
x# when dy
dx œ 2. Thus, sec x œ 2 Ê cos# x œ 2
„"
cos# x œ "# Ê cos x œ È
Ê x œ 14 and x œ 14
2
x
#
for 1# x 1
#
Ê ˆ 14 ß 1‰ and ˆ 14 ß "‰ are points
where the normal is parallel to y œ x# .
76. y œ 1 cos x Ê
dy
dx
œ sin x Ê
dy
dx ¹ ˆ 1 ß1‰
œ 1
2
Ê the tangent at ˆ 1# ß 1‰ is the line y 1 œ ˆx 1# ‰
Ê y œ x 1# 1; the normal at ˆ 1# ß 1‰ is
y 1 œ (1) ˆx 1# ‰ Ê y œ x 77. y œ x# C Ê
"
#
œ
ˆ "# ‰#
dy
dx
1
#
1
œ 2x and y œ x Ê
CÊCœ
dy
dx
œ 1; the parabola is tangent to y œ x when 2x œ 1 Ê x œ
"
4
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
"
#
Êyœ
"
#
; thus,
158
Chapter 3 Differentiation
78. y œ x$ Ê
dy
dx
œ 3x# Ê
dy
dx ¹ x œ a
œ 3a# Ê the tangent line at aaß a$ b is y a$ œ 3a# (x a). The tangent line
intersects y œ x$ when x$ a$ œ 3a# (x a) Ê (x a) ax# xa a# b œ 3a# (x a) Ê (x a) ax# xa 2a# b œ 0
Ê (x a)# (x 2a) œ 0 Ê x œ a or x œ 2a. Now
dy
dx ¹ x œ c2a
œ 3(2a)# œ 12a# œ 4 a3a# b, so the slope at
x œ 2a is 4 times as large as the slope at aaß a$ b where x œ a.
79. The line through a0ß 3b and a5ß 2b has slope m œ
y œ x 3; y œ
c
x1
Ê
dy
dx
c
(x 1)# ,
œ
3 (2)
05
œ 1 Ê the line through a0ß 3b and a5ß 2b is
so the curve is tangent to y œ x 3 Ê
Ê (x 1)# œ c, x Á 1. Moreover, y œ
c
x1
intersects y œ x 3 Ê
#
dy
dx
œ 1 œ
c
(x 1)#
œ x 3, x Á 1
c
x 1
Ê c œ (x 1)(x 3), x Á 1. Thus c œ c Ê (x 1) œ (x 1)(x 3) Ê (x 1)[x 1 (x 3)]
œ !, x Á 1 Ê (x 1)(2x 2) œ 0 Ê x œ 1 (since x Á 1) Ê c œ 4.
80. Let Šbß „ Èa# b# ‹ be a point on the circle x# y# œ a# . Then x# y# œ a# Ê 2x 2y
Ê
dy
dx ¹ x œ b
œ
b
„È a # b #
y Š „ Èa# b# ‹ œ
Ê normal line through Šbß „ Èa# b# ‹ has slope
„È a # b #
b
(x b) Ê y … Èa# b# œ
„È a # b #
b
„È a # b #
b
œ0 Ê
dy
dx
dy
dx
œ xy
Ê normal line is
x … Èa# b# Ê y œ „
È a# b #
b
x
which passes through the origin.
81. x# 2y# œ 9 Ê 2x 4y
œ "4 x 9
4
5
#
œ0 Ê
dy
dx
x
œ 2y
Ê
œ "4 Ê the tangent line is y œ 2 "4 (x 1)
dy
dx ¹ (1ß2)
and the normal line is y œ 2 4(x 1) œ 4x 2.
82. x$ y# œ 2 Ê 3x# 2y
œ 3# x dy
dx
dy
dx
œ0 Ê
œ
dy
dx
3x#
2y
Ê
dy
dx ¹ (1ß1)
and the normal line is y œ 1 32 (x 1) œ
83. xy 2x 5y œ 2 Ê Šx
y‹ 2 5
dy
dx
dy
dx
œ0 Ê
(x 5) œ y 2 Ê
Ê the tangent line is y œ 2 2(x 3) œ 2x 4 and the normal line is y œ 2 84. (y x)# œ 2x 4 Ê 2(y x) Š dy
dx 1‹ œ 2 Ê (y x)
Ê the tangent line is y œ 2 34 (x 6) œ
85. x Èxy œ 6 Ê 1 "
#Èxy
Šx
dy
dx
3
4
x
dy
dx
œ 1 (y x) Ê
y 2
x 5
dy
dx
œ
Ê
1
#
(x 3) œ "# x 7# .
dy
dx
œ
1yx
yx
3
2
x"Î# 3y"Î#
y œ 4 "4 (x 1) œ 4" x 17
4
87. x$ y$ y# œ x y Ê ’x$ Š3y#
Ê
dy
dx
dy
dx
œ0 Ê
dy
dx ¹ (3ß2)
Ê
dy
dx
y œ 2Èxy Ê
dy
dx
œ
2Èxy y
x
Ê
œ2
dy
dx ¹ (6ß2)
dy
dx ¹ (4ß1)
Ê the tangent line is y œ 1 54 (x 4) = 54 x 6 and the normal line is y œ " 45 (x 4) œ
86. x$Î# 2y$Î# œ 17 Ê
(x 1)
œ
and the normal line is y œ 2 43 (x 6) œ 43 x 10.
5
#
y‹ œ 0 Ê x
3
#
x 3" .
2
3
dy
dx
œ #3 Ê the tangent line is y œ 1 dy
dx
œ
x"Î#
2y"Î#
Ê
dy
dx ¹ (1ß4)
4
5
œ
x
5
4
11
5
.
œ "4 Ê the tangent line is
and the normal line is y œ 4 4(x 1) œ 4x.
dy
dx ‹
y$ a3x# b“ 2y
a3x$ y# 2y 1b œ 1 3x# y$ Ê
dy
dx
œ
dy
dx
œ1
1 3x# y$
3x$ y# 2y 1
Ê
dy
dx
Ê 3x$ y#
dy
dx ¹ (1ß1)
dy
dx
2y
œ 24 , but
Therefore, the curve has slope "# at ("ß ") but the slope is undefined at ("ß 1).
dy
dx
dy
dx ¹ (1ßc1) is
dy
dx
œ " 3x# y$
undefined.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
3
4
Chapter 3 Practice Exercises
88. y œ sin (x sin x) Ê
dy
dx
œ [cos (x sin x)](1 cos x); y œ 0 Ê sin (x sin x) œ 0 Ê x sin x œ k1,
k œ 2, 1, 0, 1, 2 (for our interval) Ê cos (x sin x) œ cos (k1) œ „ 1. Therefore,
dy
dx
œ 0 and y œ 0 when
1 cos x œ 0 and x œ k1. For #1 Ÿ x Ÿ 21, these equations hold when k œ 2, 0, and 2 (since
cos (1) œ cos 1 œ 1). Thus the curve has horizontal tangents at the x-axis for the x-values 21, 0, and 21
(which are even integer multiples of 1) Ê the curve has an infinite number of horizontal tangents.
89. B œ graph of f, A œ graph of f w . Curve B cannot be the derivative of A because A has only negative slopes
while some of B's values are positive.
90. A œ graph of f, B œ graph of f w . Curve A cannot be the derivative of B because B has only negative slopes
while A has positive values for x 0.
91.
92.
93. (a) 0, 0
(b) largest 1700, smallest about 1400
94. rabbits/day and foxes/day
sin x
95. lim
#
x Ä ! 2x x
96. lim
3x tan 7x
#x
97. lim
sin r
98. lim
sin (sin ))
)
xÄ!
r Ä ! tan 2r
)Ä!
Ê lim
)Ä!
99.
lim c
) Ä ˆ1‰
xÄ!
"
(#x 1) “
œ lim ˆ 3x
2x sin 7x ‰
2x cos 7x
œ lim ’ˆ sinx x ‰ †
xÄ!
œ lim ˆ sinr r †
rÄ!
101.
lim b
)Ä!
œ lim
xÄ!
lim
œ
1 2 cot# )
5 cot# ) 7 cot ) 8
1cos )
)#
œ lim b
)Ä!
x sin x
x Ä ! 2(1 cos x)
œ lim
)Ä!
2 sin# ˆ #) ‰
)#
sin (sin ))
sin )
"
ˆ 27 ‰
†
‹œ
œ ˆ "# ‰ (1) ˆ 1" ‰ œ
3
#
ˆ1 † 1 † 27 ‰ œ 2
"
#
. Let x œ sin ). Then x Ä 0 as ) Ä 0
œ1
lim c
) Ä ˆ1‰
œ lim
cos 2r
sin 2r
r Ä ! ˆ 2r ‰
sin x
x
sin 7x
7x
xÄ!
† "# ‰ œ ˆ "# ‰ (1) lim
2
x sin x
x Ä ! 2 2 cos x
)Ä!
lim Š cos"7x †
)Ä!
4 tan# ) tan ) 1
tan# ) &
œ (1)(1)(1) œ 1
102.
3
#
)Ä!
sin (sin ))
sin )
lim
œ
(sin )) ˆ sin ) ‰
œ lim Š sinsin
œ lim
) ‹
)
2
100.
2r
tan 2r
œ (1) ˆ "1 ‰ œ 1
Š4 tan" ) tan"# ) ‹
Š" tan5# ) ‹
Š cot"# ) 2‹
Š5 cot7 ) cot8# ) ‹
œ
œ
(4 0 0)
(1 0)
(0 2)
(5 0 0)
œ lim
x sin x
# x
x Ä ! 2 ˆ2 sin ˆ # ‰‰
œ lim ’
)Ä!
sin ˆ #) ‰
ˆ #) ‰
†
sin ˆ #) ‰
ˆ #) ‰
œ4
œ 52
†
x x
œ lim ’ sin## ˆ# x ‰ †
xÄ!
#
sin x
x “
† "# “ œ (1)(1) ˆ "# ‰ œ
ˆx‰
œ lim ’ sin #ˆ x ‰ †
xÄ!
#
"
#
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
ˆ #x ‰
sin ˆ #x ‰
†
sin x
x “
159
160
103.
Chapter 3 Differentiation
tan x
x
lim
xÄ!
œ lim
)Ä!
104.
œ lim ˆ cos" x †
tan )
)
œ 1; let ) œ tan x Ê ) Ä 0 as x Ä 0 Ê lim g(x) œ lim
xÄ!
(tan x)
œ lim ’ tantan
†
x
tan (tan x)
x Ä ! sin (sin x)
105. (a) S œ 21r# 21rh and h constant Ê
(b) S œ 21r# 21rh and r constant Ê
(d) S constant Ê
dh
dt
(b) r constant Ê
dr
dt
108. V œ s$ Ê
109.
dR"
dt
dr
dt
œ 3s# †
ds
dt
œ 1 ohm/sec,
R# œ 50 ohms Ê
" dR
(30)# dt
110.
dR
dt
œ
"
(75)#
"
R
œ
dZ
dt
œ
dr
dt
" dV
3s# dt
Ê (2r ; so s œ 20 and
"
R
œ
"
R"
"
R#
dV
dt
Ê
dA
dt
œ 2x
dx
dt
2y
Ê D
dD
dt
Ê
dr
dt
œ
r dh
2rh dt
1 r#
dr
Èr# h# “ dt
œ
"
È5
œ (21)(10) ˆ 12 ‰ œ 40 m# /sec
œ 1200 cm$ /min Ê
" dR
R# dt
œ
" dR"
R"# dt
ds
dt
" dR#
R## dt
œ
"
3(20)#
(1200) œ 1 cm/min
. Also, R" œ 75 ohms and
Therefore, from the derivative equation,
Ê
dR
dt
5625 ‰
œ (900) ˆ 5000
5625†5000 œ
œ
dy
dt
dh
dt
dh
dt
dh
1rh
Èr# h# dt
(10)(3)(20)(2)
È10# 20#
œ 10 m/sec and
dr
dt 21r
h) dr
dt œ r
œ ’1Èr# h# dr
dt
œ 2 ohms/sec; Z œ ÈR# X# Ê
dx
dt
dr
dt
;
œ 12 m/sec Ê
"
"
75 50 Ê R œ 30 ohms.
"
" ‰
ˆ "
(50)# (0.5) œ 5625 5000
111. Given
dD
dt
ds
dt
(using the result of #105);
œ 1. Therefore, to make f continuous at the origin,
œ (41r 21h)
dr
dt
dh
dt
1Èr# h#
dr
1 r#
Èr# h# “ dt
; so r œ 10 and
Ê
dr
dt
sin x
x Ä ! sin (sin x)
œ (41r 21h)
dr ‰
dt
21r
dr
dt
œ 1 † lim
dX
dt
œ 3 ohms/sec and
X œ 20 ohms Ê
5
tan (tan x)
tan x
1rh
dh
Èr# h# dt
œ
dS
dt
1r# dr
dt
È r# h #
œ
"
cos x “
lim )
) Ä ! sin )
1Èr# h#
œ 0.5 ohm/sec; and
dR#
dt
(1) dy
dt
dS
dt
œ0 Ê
œ 21 r
dA
dt
dV
dt
œ0 Ê
dh ‰
ˆr dr
dt h dt
È r# h #
œ ’1Èr# h# dS
dt
(c) In general,
œ 1r †
dS
dt
(a) h constant Ê
œ 21r dh
dt
#1 ˆr dh
h
dt
dr
dt
†
œ 41r dr
dt 21 h
dS
dt
dS
dt
œ 0 Ê 0 œ (41r 21h)
dS
dt
106. S œ 1rÈr# h# Ê
œ 41r
dS
dt
œ
sin x
lim
x Ä ! sin (sin x)
define f(0) œ 1.
(c) S œ 21r# 21rh Ê
sin x
sin (sin x)
xÄ!
let ) œ sin x Ê ) Ä 0 as x Ä 0 Ê
107. A œ 1r# Ê
xÄ!
œ 1. Therefore, to make g continuous at the origin, define g(0) œ 1.
lim f(x) œ lim
xÄ!
sin x ‰
x
xÄ!
dZ
dt
œ
dX
R dR
dt X dt
È R # X#
9(625)
50(5625)
œ
"
50
œ 0.02 ohm/sec.
so that R œ 10 ohms and
¸ 0.45 ohm/sec.
œ 5 m/sec, let D be the distance from the origin Ê D# œ x# y# Ê 2D
œx
œ (3)(10) (%)(5) Ê
dy
dx
dt y dt
dD
10
dt œ 5 œ
dD
dt
. When (xß y) œ ($ß %), D œ É$# a%b# œ & and
2. Therefore, the particle is moving away from the origin at 2 m/sec
(because the distance D is increasing).
#
œ 11 units/sec. Then D# œ x# y# œ x# ˆx$Î# ‰
È3# 3$ œ 6 and substitution in the
œ x(2 3x) dx
dt ; x œ 3 Ê D œ
112. Let D be the distance from the origin. We are given that
œ x# x$ Ê 2D
dD
dt
œ 2x
dx
dt
3x#
dx
dt
derivative equation gives (2)(6)(11) œ (3)(2 9)
113. (a) From the diagram we have
(b) V œ
"
3
#
1r h œ
"
3
1 ˆ 52
#
10
h
h‰ h œ
œ
4
r
41 h$
75
Ê rœ
Ê
dV
dt
2
5
œ
dx
dt
Ê
dD
dt
dx
dt
œ 4 units/sec.
h.
41h# dh
25 dt
, so
dV
dt
œ 5 and h œ 6 Ê
dh
dt
125
œ 144
1 ft/min.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Chapter 3 Practice Exercises
114. From the sketch in the text, s œ r) Ê
Therefore,
ds
dt
œ 6 ft/sec and r œ 1.2 ft Ê
ds
dt
115. (a) From the sketch in the text,
Ê)œ0Ê
d)
dt
d)
dt
œr
d)
dt
)
dr
dt
. Also r œ 1.2 is constant Ê
dr
dt
œ0 Ê
dx
dt
œ sec# )
œr
d)
dt
œ (1.2)
d)
dt
œ 0.6 rad/sec and x œ tan ). Also x œ tan ) Ê
œ asec 0b (0.6) œ 0.6. Therefore the speed of the light is 0.6 œ
3
5
d)
dt ;
at point A, x œ 0
km/sec when it reaches
point A.
(b)
(3/5) rad
sec
†
1 rev
21 rad
116. From the figure,
a
r
†
60 sec
min
œ
b
BC
œ
18
1
Ê
a
r
revs/min
œ
b
È b # r#
. We are given
that r is constant. Differentiation gives,
"
r
†
da
dt
œ
‰
ŠÈb# r# ‹ ˆ db
dt (b) Š È
b # r#
b œ 2r and
Ê
œ
da
dt
db
dt
b
‰
‹ ˆ db
dt
b# r#
. Then,
œ 0.3r
Ô È(2r)# r# (0.3r) (2r) É2r( #0.3r)# ×
(2r)
r
Ù
œ rÖ
# #
(2r) r
Õ
È3r# (0.3r) 4r# (0.3r)
3r#
È
3r
œ
Ø
a3r# b (0.3r) a4r# b (0.3r)
3 È 3 r#
œ
0.3r
3È 3
œ
r
10È3
m/sec. Since
da
dt
is positive, the distance OA is increasing
when OB œ 2r, and B is moving toward O at the rate of 0.3r m/sec.
117. (a) If f(x) œ tan x and x œ 14 , then f w (x) œ sec# x,
f ˆ 14 ‰ œ 1 and f w ˆ 14 ‰ œ 2. The linearization of
f(x) is L(x) œ 2 ˆx 14 ‰ (1) œ 2x 1 2
#
.
(b) If f(x) œ sec x and x œ 14 , then f w (x) œ sec x tan x,
f ˆ 1 ‰ œ È2 and f w ˆ 1 ‰ œ È2. The
4
4
linearization of f(x) is L(x) œ È2 ˆx 14 ‰ È2
œ È2x 118. f(x) œ
"
1 tan x
È2(% 1)
.
4
Ê f w (x) œ
sec# x
(1 tan x)#
. The linearization at x œ 0 is L(x) œ f w (0)(x 0) f(0) œ 1 x.
119. f(x) œ Èx 1 sin x 0.5 œ (x 1)"Î# sin x 0.5 Ê f w (x) œ ˆ "# ‰ (x 1)"Î# cos x
Ê L(x) œ f w (0)(x 0) f(0) œ 1.5(x 0) 0.5 Ê L(x) œ 1.5x 0.5, the linearization of f(x).
120. f(x) œ
œ
2
1 x
2
(1 x)#
È1 x 3.1 œ 2(1 x)" (1 x)"Î# 3.1 Ê f w (x) œ 2(1 x)# (1) "# (1 x)"Î#
"
2È 1 x
.
œ 5 rad/sec
#
dx
dt
ds
dt
161
Ê L(x) œ f w (0)(x 0) f(0) œ 2.5x 0.1, the linearization of f(x).
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
162
Chapter 3 Differentiation
121. S œ 1 rÈr# h# , r constant Ê dS œ 1 r † "# ar# h# b
Ê dS œ
"Î#
#h dh œ
1rh
Èr# h# dh.
Height changes from h! to h! dh
1 r h! adhb
Ér# h#!
122. (a) S œ 6r# Ê dS œ 12r dr. We want kdSk Ÿ (2%) S Ê k12r drk Ÿ
12r#
100
Ê kdrk Ÿ
r
100
. The measurement of the
edge r must have an error less than 1%.
#
3r dr
‰
(b) When V œ r$ , then dV œ 3r# dr. The accuracy of the volume is ˆ dV
V (100%) œ Š r$ ‹ (100%)
r ‰
œ ˆ 3r ‰ (dr)(100%) œ ˆ 3r ‰ ˆ 100
(100%) œ 3%
123. C œ 21r Ê r œ
dV œ
C#
21 #
C
21
, S œ 41 r # œ
C#
1
, and V œ
4
3
1 r$ œ
C$
61 #
. It also follows that dr œ
"
#1
dC, dS œ
2C
1
dC and
dC. Recall that C œ 10 cm and dC œ 0.4 cm.
0.2
ˆ drr ‰ (100%) œ ˆ 0.2
‰ ˆ 2101 ‰ (100%) œ (.04)(100%) œ 4%
(a) dr œ 0.4
21 œ 1 cm Ê
1
8
1 ‰
ˆ dS
‰
ˆ 8 ‰ ˆ 100
(b) dS œ 20
(100%) œ 8%
1 (0.4) œ 1 cm Ê
S (100%) œ 1
10#
21 #
#
(0.4) œ
20
1#
‰
ˆ 20 ‰ 61
cm Ê ˆ dV
V (100%) œ 1# Š 1000 ‹ (100%) œ 12%
124. Similar triangles yield
35
h
œ
(c) dV œ
Ê dh œ 120a# da œ 15
6
120
a#
Ê h œ 14 ft. The same triangles imply that 20h a œ 6a Ê h œ 120a" 6
" ‰
2
‰ ˆ „ 1"# ‰ œ ˆ "#!
‰ˆ „ "#
da œ ˆ 120
œ „ 45
¸ „ .0444 ft œ „ 0.53 inches.
a#
"&#
CHAPTER 3 ADDITIONAL AND ADVANCED EXERCISES
1. (a) sin 2) œ 2 sin ) cos ) Ê
#
#
d
d)
Ê cos 2) œ cos ) sin )
(b) cos 2) œ cos# ) sin# ) Ê
(sin 2)) œ
d
d)
d
d)
(cos 2)) œ
(2 sin ) cos )) Ê 2 cos 2) œ 2[(sin ))(sin )) (cos ))(cos ))]
d
d)
acos# ) sin# )b Ê 2 sin 2) œ (2 cos ))(sin )) (2 sin ))(cos ))
Ê sin 2) œ cos ) sin ) sin ) cos ) Ê sin 2) œ 2 sin ) cos )
2. The derivative of sin (x a) œ sin x cos a cos x sin a with respect to x is cos (x a) œ cos x cos a sin x sin a, which
is also an identity. This principle does not apply to the equation x# 2x 8 œ 0, since x# 2x 8 œ 0 is not an identity:
it holds for 2 values of x (2 and 4), but not for all x.
3. (a) f(x) œ cos x Ê f w (x) œ sin x Ê f ww (x) œ cos x, and g(x) œ a bx cx# Ê gw (x) œ b 2cx Ê gww (x) œ 2c;
also, f(0) œ g(0) Ê cos (0) œ a Ê a œ 1; f w (0) œ gw (0) Ê sin (0) œ b Ê b œ 0; f ww (0) œ gww (0) Ê cos (0) œ 2c
Ê c œ "# . Therefore, g(x) œ 1 "# x# .
(b) f(x) œ sin (x a) Ê f w (x) œ cos (x a), and g(x) œ b sin x c cos x Ê gw (x) œ b cos x c sin x; also, f(0) œ g(0)
Ê sin (a) œ b sin (0) c cos (0) Ê c œ sin a; f w (0) œ gw (0) Ê cos (a) œ b cos (0) c sin (0) Ê b œ cos a.
Therefore, g(x) œ sin x cos a cos x sin a.
(c) When f(x) œ cos x, f www (x) œ sin x and f Ð%Ñ (x) œ cos x; when g(x) œ 1 "# x# , gwww (x) œ 0 and gÐ%Ñ (x) œ 0. Thus
f www (0) œ 0 œ gwww (0) so the third derivatives agree at x œ 0. However, the fourth derivatives do not agree since
f Ð%Ñ (0) œ 1 but gÐ%Ñ (0) œ 0. In case (b), when f(x) œ sin (x a) and g(x) œ sin x cos a cos x sin a, notice that
f(x) œ g(x) for all x, not just x œ 0. Since this is an identity, we have f ÐnÑ (x) œ gÐnÑ (x) for any x and any positive
integer n.
4. (a) y œ sin x Ê yw œ cos x Ê yww œ sin x Ê yww y œ sin x sin x œ 0; y œ cos x Ê yw œ sin x
Ê yww œ cos x Ê yww y œ cos x cos x œ 0; y œ a cos x b sin x Ê yw œ a sin x b cos x
Ê yww œ a cos x b sin x Ê yww y œ (a cos x b sin x) (a cos x b sin x) œ 0
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Chapter 3 Additional and Advanced Exercises
163
(b) y œ sin (2x) Ê yw œ 2 cos (2x) Ê yww œ 4 sin (2x) Ê yww 4y œ 4 sin (2x) 4 sin (2x) œ 0. Similarly,
y œ cos (2x) and y œ a cos (2x) b sin (2x) satisfy the differential equation yw w 4y œ 0. In general,
y œ cos (mx), y œ sin (mx) and y œ a cos (mx) b sin (mx) satisfy the differential equation yww m# y œ 0.
5. If the circle (x h)# (y k)# œ a# and y œ x# 1 are tangent at ("ß #), then the slope of this tangent is
m œ 2xk (1 2) œ 2 and the tangent line is y œ 2x. The line containing (hß k) and ("ß #) is perpendicular to
ß
y œ 2x Ê
k2
h1
œ "# Ê h œ 5 2k Ê the location of the center is (5 2kß k). Also, (x h)# (y k)# œ a#
Ê x h (y k)yw œ 0 Ê 1 ayw b# (y k)yw w œ 0 Ê yww œ
w
1 ay b#
ky
w
. At the point ("ß #) we know
ww
y œ 2 from the tangent line and that y œ 2 from the parabola. Since the second derivatives are equal at ("ß #)
we obtain 2 œ
1 (2)#
k#
Ê kœ
9
#
#
. Then h œ 5 2k œ 4 Ê the circle is (x 4)# ˆy 9# ‰ œ a# . Since ("ß #)
lies on the circle we have that a œ
5È 5
2
.
6. The total revenue is the number of people times the price of the fare: r(x) œ xp œ x ˆ3 x ‰#
, where
40
x ‰ ˆ
x ‰
40
3 40
" ‰
dr
x ‰#
x ‰ˆ
dr
‘
0 Ÿ x Ÿ 60. The marginal revenue is dx
œ ˆ3 40
2x ˆ3 40
40
Ê dx
œ ˆ3
2x
40
x
x
dr
œ 3 ˆ3 40 ‰ ˆ1 40 ‰ . Then dx œ 0 Ê x œ 40 (since x œ 120 does not belong to the domain). When 40 people
are on the bus the marginal revenue is zero and the fare is p(40) œ ˆ3 7. (a) y œ uv Ê
dy
dt
œ
du
dt
x ‰#
40 ¹ x œ 40
œ $4.00.
v u dv
dt œ (0.04u)v u(0.05v) œ 0.09uv œ 0.09y Ê the rate of growth of the total production is
9% per year.
(b) If
du
dt
œ 0.02u and
dv
dt
œ 0.03v, then
dy
dt
œ (0.02u)v (0.03v)u œ 0.01uv œ 0.01y, increasing at 1% per year.
8. When x# y# œ 225, then yw œ xy . The tangent
line to the balloon at (12ß 9) is y 9 œ
Ê yœ
4
3
4
3
(x 12)
x 25. The top of the gondola is 15 8
œ 23 ft below the center of the balloon. The intersection of y œ 23 and y œ 43 x 25 is at the far
right edge of the gondola Ê 23 œ
Ê xœ
3
#
4
3
x 25
. Thus the gondola is 2x œ 3 ft wide.
9. Answers will vary. Here is one possibility.
10. s(t) œ 10 cos ˆt 14 ‰ Ê v(t) œ
10
(a) s(0) œ 10 cos ˆ 14 ‰ œ È
ds
dt
œ 10 sin ˆt 14 ‰ Ê a(t) œ
dv
dt
œ
d# s
dt#
œ 10 cos ˆt 14 ‰
2
(b) Left: 10, Right: 10
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
164
Chapter 3 Differentiation
(c) Solving 10 cos ˆt 14 ‰ œ 10 Ê cos ˆt 14 ‰ œ 1 Ê t œ 341 when the particle is farthest to the left.
Solving 10 cos ˆt 14 ‰ œ 10 Ê cos ˆt 14 ‰ œ 1 Ê t œ 14 , but t 0 Ê t œ 21 41 œ 741 when the particle
is farthest to the right. Thus, v ˆ 341 ‰ œ 0, v ˆ 741 ‰ œ 0, a ˆ 341 ‰ œ 10, and a ˆ 741 ‰ œ 10.
(d) Solving 10 cos ˆt 14 ‰ œ 0 Ê t œ
11. (a) s(t) œ 64t 16t# Ê v(t) œ
ds
dt
1
4
Ê v ˆ 14 ‰ œ 10, ¸v ˆ 14 ‰¸ œ 10 and a ˆ 14 ‰ œ !.
œ 64 32t œ 32(2 t). The maximum height is reached when v(t) œ 0
Ê t œ 2 sec. The velocity when it leaves the hand is v(0) œ 64 ft/sec.
(b) s(t) œ 64t 2.6t# Ê v(t) œ ds
dt œ 64 5.2t. The maximum height is reached when v(t) œ 0 Ê t ¸ 12.31 sec.
The maximum height is about s(12.31) œ 393.85 ft.
12. s" œ 3t$ 12t# 18t 5 and s# œ t$ 9t# 12t Ê v" œ 9t# 24t 18 and v# œ 3t# 18t 12; v" œ v#
Ê 9t# 24t 18 œ 3t# 18t 12 Ê 2t# 7t 5 œ 0 Ê (t 1)(2t 5) œ 0 Ê t œ 1 sec and t œ 2.5 sec.
13. m av# v#! b œ k ax#! x# b Ê m ˆ2v
substituting
dx
dt
œv Ê m
dv
dt
dv ‰
dt
œ k ˆ2x
dx ‰
dt
Ê m
dv
dt
2x ‰
œ k ˆ 2v
dx
dt
Ê m
dv
dt
œ kx ˆ "v ‰
dx
dt
œ 2At B Ê v ˆ t" # t# ‰ œ 2A ˆ t" # t# ‰ B œ A at" t# b B is the
instantaneous velocity at the midpoint. The average velocity over the time interval is vav œ
œ
Bt# Cb aAt#"
t# t"
. Then
œ kx, as claimed.
14. (a) x œ At# Bt C on ct" ß t# d Ê v œ
aAt##
dx
dt
Bt" Cb
œ
at# t" b cA at# t" b Bd
t# t"
#
?x
?t
œ A at# t" b B.
(b) On the graph of the parabola x œ At Bt C, the slope of the curve at the midpoint of the interval
ct" ß t# d is the same as the average slope of the curve over the interval.
15. (a) To be continuous at x œ 1 requires that lim c sin x œ lim b (mx b) Ê 0 œ m1 b Ê m œ 1b ;
xÄ1
xÄ1
(b) If yw œ œ
cos x, x 1
is differentiable at x œ 1, then lim c cos x œ m Ê m œ 1 and b œ 1.
xÄ1
m, x 1
16. faxb is continuous at ! because lim
xÄ!
œ
x ‰ ˆ 1 cos x ‰
lim ˆ 1 xcos
#
1 cos x
xÄ!
œ
" cos x
x
#
lim ˆ sinx x ‰
xÄ!
œ ! œ fa!b. f w (0) œ lim
f(x) f(0)
x0
ˆ 1 "cos x ‰
w
xÄ!
œ
"
#
œ lim
xÄ!
1 c cos x
0
x
x
. Therefore f (0) exists with value
"
#
.
17. (a) For all a, b and for all x Á 2, f is differentiable at x. Next, f differentiable at x œ 2 Ê f continuous at x œ 2
Ê lim c f(x) œ f(2) Ê 2a œ 4a 2b 3 Ê 2a 2b 3 œ 0. Also, f differentiable at x Á 2
xÄ2
Ê f w (x) œ œ
a, x 2
. In order that f w (2) exist we must have a œ 2a(2) b Ê a œ 4a b Ê 3a œ b.
2ax b, x 2
Then 2a 2b 3 œ 0 and 3a œ b Ê a œ
3
4
and b œ
9
4
.
(b) For x #, the graph of f is a straight line having a slope of
$
%
and passing through the origin; for x
is a parabola. At x œ #, the value of the y-coordinate on the parabola is
$
#
#, the graph of f
which matches the y-coordinate of the point
on the straight line at x œ #. In addition, the slope of the parabola at the match up point is
$
%
which is equal to the
slope of the straight line. Therefore, since the graph is differentiable at the match up point, the graph is smooth there.
18. (a) For any a, b and for any x Á 1, g is differentiable at x. Next, g differentiable at x œ 1 Ê g continuous at
x œ 1 Ê lim b g(x) œ g(1) Ê a 1 2b œ a b Ê b œ 1. Also, g differentiable at x Á 1
x Ä "
Ê gw (x) œ œ
a, x 1
. In order that gw (1) exist we must have a œ 3a(1)# 1 Ê a œ 3a 1
3ax# 1, x 1
Ê a œ "# .
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Chapter 3 Additional and Advanced Exercises
165
(b) For x Ÿ ", the graph of g is a straight line having a slope of "# and a y-intercept of ". For x ", the graph of g is
a cubic. At x œ ", the value of the y-coordinate on the cubic is
$
#
which matches the y-coordinate of the point
on the straight line at x œ ". In addition, the slope of the cubic at the match up point is "# which is equal to the
slope of the straight line. Therefore, since the graph is differentiable at the match up point, the graph is smooth there.
19. f odd Ê f(x) œ f(x) Ê
20. f even Ê f(x) œ f(x) Ê
d
dx
(f(x)) œ
(f(x)) œ
d
dx
d
dx
d
dx
(f(x)) Ê f w (x)(1) œ f w (x) Ê f w (x) œ f w (x) Ê f w is even.
(f(x)) Ê f w (x)(1) œ f w (x) Ê f w (x) œ f w (x) Ê f w is odd.
21. Let h(x) œ (fg)(x) œ f(x) g(x) Ê hw (x) œ x lim
Äx
œ x lim
Äx
œ
!
f(x) g(x) f(x) g(x! ) f(x) g(x! ) f(x! ) g(x! )
x x!
g(x! )
f(x! ) x lim
’ g(x)x x! “
Ä x!
!
h(x) h(x! )
x x!
œ x lim
Äx
!
f(x) g(x) f(x! ) g(x! )
x x!
f(x! )
!)
œ x lim
’f(x) ’ g(x)x xg(x
““ x lim
’g(x! ) ’ f(x)x x! ““
Äx
Äx
!
!
w
g(x! ) f (x! ) œ 0 †
g(x! )
lim ’ g(x)x x! “
x Ä x!
!
w
g(x! ) f (x! ) œ g(x! ) f w (x! ), if g is
continuous at x! . Therefore (fg)(x) is differentiable at x! if f(x! ) œ 0, and (fg)w (x! ) œ g(x! ) f w (x! ).
22. From Exercise 21 we have that fg is differentiable at 0 if f is differentiable at 0, f(0) œ 0 and g is continuous at 0.
(a) If f(x) œ sin x and g(x) œ kxk , then kxk sin x is differentiable because f w (0) œ cos (0) œ 1, f(0) œ sin (0) œ 0
and g(x) œ kxk is continuous at x œ 0.
(b) If f(x) œ sin x and g(x) œ x#Î$ , then x#Î$ sin x is differentiable because f w (0) œ cos (0) œ 1, f(0) œ sin (0) œ 0
and g(x) œ x#Î$ is continuous at x œ 0.
(c) If f(x) œ 1 cos x and g(x) œ $Èx, then $Èx (1 cos x) is differentiable because f w (0) œ sin (0) œ 0,
f(0) œ 1 cos (0) œ 0 and g(x) œ x"Î$ is continuous at x œ 0.
(d) If f(x) œ x and g(x) œ x sin ˆ "x ‰ , then x# sin ˆ x" ‰ is differentiable because f w (0) œ 1, f(0) œ 0 and
sin ˆ "x ‰
lim x sin ˆ "x ‰ œ lim
xÄ!
"
x
xÄ!
œ lim
tÄ_
sin t
t
œ 0 (so g is continuous at x œ 0).
23. If f(x) œ x and g(x) œ x sin ˆ "x ‰ , then x# sin ˆ x" ‰ is differentiable at x œ 0 because f w (0) œ 1, f(0) œ 0 and
lim x sin ˆ "x ‰ œ lim
xÄ!
sin ˆ "x ‰
"
x
xÄ!
œ lim
tÄ_
sin t
t
œ 0 (so g is continuous at x œ 0). In fact, from Exercise 21,
hw (0) œ g(0) f w (0) œ 0. However, for x Á 0, hw (x) œ x# cos ˆ "x ‰‘ ˆ x"# ‰ 2x sin ˆ x" ‰ . But
lim hw (x) œ lim cos ˆ "x ‰ 2x sin ˆ x" ‰‘ does not exist because cos ˆ x" ‰ has no limit as x Ä 0. Therefore,
xÄ!
xÄ!
the derivative is not continuous at x œ 0 because it has no limit there.
24. From the given conditions we have f(x h) œ f(x) f(h), f(h) 1 œ hg(h) and lim g(h) œ 1. Therefore,
hÄ!
f(xh) f(x)
h
hÄ!
w
f w (x) œ lim
f(x) f(h) f(x)
h
hÄ!
œ lim
œ lim f(x) ’ f(h)h 1 “ œ f(x) ’ lim g(h)“ œ f(x) † 1 œ f(x)
hÄ!
Ê f w (x) œ f(x) and f axbexists at every value of x.
hÄ!
25. Step 1: The formula holds for n œ 2 (a single product) since y œ u" u# Ê
dy
dx
œ
du"
dx
u# u"
du#
dx
Step 2: Assume the formula holds for n œ k:
y œ u" u# âuk Ê
du#
duk
dx u$ âuk á u" u# âuk-1 dx
d(u" u# âuk )
If y œ u" u# âuk ukb1 œ au" u# âuk b ukb1 , then dy
ukb1 u" u# âuk dudxkb1
dx œ
dx
dukb1
du#
duk ‰
"
œ ˆ du
dx u# u$ âuk u" dx u$ âuk â u" u# âukc1 dx ukb1 u" u# âuk dx
dukb1
du#
duk
"
œ du
dx u# u$ âukb1 u" dx u$ â ukb1 â u" u# âukc1 dx ukb1 u" u# âuk dx .
dy
dx
œ
du"
dx
u# u$ âuk u"
.
Thus the original formula holds for n œ (k1) whenever it holds for n œ k.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
.
166
Chapter 3 Differentiation
26. Recall ˆ mk ‰ œ
œ
m!
m!
m!
m!
ˆm‰
ˆm‰ ˆ m ‰
k! (m k)! . Then 1 œ 1! (m 1)! œ m and k k 1 œ k! (m k)! (k 1)! (m k 1)!
m! (k 1) m! (m k)
(m 1)!
ˆm1‰
œ (k m!1)!(m(m 1)k)! œ (k 1)! ((m
(k 1)! (m k)!
1) (k 1))! œ k 1 . Now, we prove
Leibniz's rule by mathematical induction.
Step 1: If n œ 1, then
Step 2:
d(uv)
dv
du
dx œ u dx v dx . Assume that the statement is true for n œ k, that is:
#
k #
k "
dk (uv)
dk u
dk " u dv
dk v
ˆk‰ d u d v
ˆ k ‰ du d v
dxk œ dxk v k dxk " dx 2 dxk # dx# á k 1 dv dxk " u dxk .
kb"
k
k "
k
(uv)
d
dk u dv
dk" u d# v
ddxk u" v ddxuk dv
‘
If n œ k 1, then d dx(uv)
œ dx
Š d dx
k "
k ‹ œ
dx ’k dxk dx k dxk" dx# “
’ˆ k2 ‰
dk " u d# v
dxk " dx#
ˆ k2 ‰
dk # u d$ v
dxk # dx$ “
á ’ˆ k k 1 ‰
d# u dk " v
dx# dxk "
ˆ kk 1 ‰
du dk u
dx dxk
v“
k"
#
dk v
dkb" u ‘
dk " u
dk u dv
ˆ k1 ‰ ˆ k2 ‰‘ ddxk"u ddxv# á
dxk u dxk " œ dxk " v (k 1) dxk dx du dk v
dkb" v
dk " u
dk u dv
dk " u d# v
ˆ k k 1 ‰ ˆ kk ‰‘ dx
ˆ k 2 1 ‰ dx
k "
dxk u dxk" œ dxk" v (k 1) dxk dx dx#
du dk v
dkb" v
ˆ k k 1 ‰ dx
dxk u dxk " .
du
dx
á
Therefore the formula (c) holds for n œ (k 1) whenever it holds for n œ k.
27. (a) T# œ
41 # L
g
(b) T# œ
41 # L
g
ÊLœ
T# g
41 #
ÊTœ
#1 È
L;
Èg
ÊLœ
a1 sec# ba32.2 ft/sec# b
41 #
dT œ
#1
Èg
†
"
dL
#È L
Ê L ¸ 0.8156 ft
œ
1
ÈLg dL;
dT œ
1
Èa!Þ)"&' ftba32.2 ft/sec# b a!Þ!"
ftb ¸ 0.00613 sec.
(c) Since there are 86,400 sec in a day, we have a0.00613 secba86,400 sec/dayb ¸ 529.6 sec/day, or 8.83 min/day; the
clock will lose about 8.83 min/day.
28. v œ s$ Ê
dv
dt
#
œ $s# ds
dt œ ka's b Ê
ds
dt
œ #k. If s! œ the initial length of the cube's side, then s" œ s! #k
Ê #k œ s! s" . Let t œ the time it will take the ice cube to melt. Now, t œ
œ
"
"Î$
" ˆ $% ‰
s!
#k
œ
s!
s ! s "
œ
av! b"Î$
"Î$
av! b ˆ $% v! ‰
"Î$
¸ "" hr.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
CHAPTER 4 APPLICATIONS OF DERIVATIVES
4.1 EXTREME VALUES OF FUNCTIONS
1. An absolute minimum at x œ c# , an absolute maximum at x œ b. Theorem 1 guarantees the existence of such
extreme values because h is continuous on [aß b].
2. An absolute minimum at x œ b, an absolute maximum at x œ c. Theorem 1 guarantees the existence of such
extreme values because f is continuous on [aß b].
3. No absolute minimum. An absolute maximum at x œ c. Since the function's domain is an open interval, the
function does not satisfy the hypotheses of Theorem 1 and need not have absolute extreme values.
4. No absolute extrema. The function is neither continuous nor defined on a closed interval, so it need not fulfill
the conclusions of Theorem 1.
5. An absolute minimum at x œ a and an absolute maximum at x œ c. Note that y œ g(x) is not continuous but
still has extrema. When the hypothesis of Theorem 1 is satisfied then extrema are guaranteed, but when the
hypothesis is not satisfied, absolute extrema may or may not occur.
6. Absolute minimum at x œ c and an absolute maximum at x œ a. Note that y œ g(x) is not continuous but still
has absolute extrema. When the hypothesis of Theorem 1 is satisfied then extrema are guaranteed, but when
the hypothesis is not satisfied, absolute extrema may or may not occur.
7. Local minimum at a"ß !b, local maximum at a"ß !b
8. Minima at a#ß !b and a#ß !b, maximum at a!ß #b
9. Maximum at a!ß &b. Note that there is no minimum since the endpoint a#ß !b is excluded from the graph.
10. Local maximum at a$ß !b, local minimum at a#ß !b, maximum at a"ß #b, minimum at a!ß "b
11. Graph (c), since this the only graph that has positive slope at c.
12. Graph (b), since this is the only graph that represents a differentiable function at a and b and has negative slope at c.
13. Graph (d), since this is the only graph representing a funtion that is differentiable at b but not at a.
14. Graph (a), since this is the only graph that represents a function that is not differentiable at a or b.
15. f has an absolute min at x œ 0 but does not have an absolute
max. Since the interval on which f is defined, 1 x 2,
is an open interval, we do not meet the conditions of
Theorem 1.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
168
Chapter 4 Applications of Derivatives
16. f has an absolute max at x œ 0 but does not have an absolute
min. Since the interval on which f is defined, 1 x 1, is
an open interval, we do not meet the conditions of
Theorem 1.
17. f has an absolute max at x œ 2 but does not have an absolute
min. Since the function is not continuous at x œ 1, we do
not meet the conditions of Theorem 1.
18. f has an absolute max at x œ 4 but does not have an absolute
min. Since the function is not continuous at x œ 0, we do
not meet the conditions of Theorem 1.
19. f has an absolute max at x œ
xœ
31
2 .
1
2
and an absolute min at
Since the interval on which f is defined,
0 x 21, is an open interval, we do not meet the
conditions of Theorem 1.
20. f has an absolute max at x œ 0 and an absolute min at
x œ 12 and x œ 1. Since f is continuous on the closed
interval on which it is defined, 1 Ÿ x Ÿ 21, we do meet
the conditions of Theorem 1.
21. f(x) œ
2
3
x 5 Ê f w (x) œ
f(2) œ 19
3 ,
2
3
Ê no critical points;
f(3) œ 3 Ê the absolute maximum
is 3 at x œ 3 and the absolute minimum is 19
3 at
x œ 2
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 4.1 Extreme Values of Functions
22. f(x) œ x 4 Ê f w (x) œ 1 Ê no critical points;
f(4) œ 0, f(1) œ 5 Ê the absolute maximum is 0
at x œ 4 and the absolute minimum is 5 at x œ "
23. f(x) œ x# 1 Ê f w (x) œ 2x Ê a critical point at
x œ 0; f(1) œ 0, f(0) œ 1, f(2) œ 3 Ê the absolute
maximum is 3 at x œ 2 and the absolute minimum is 1
at x œ 0
24. f(x) œ % x# Ê f w (x) œ 2x Ê a critical point at
x œ 0; f(3) œ 5, f(0) œ 4, f(1) œ 3 Ê the absolute
maximum is 4 at x œ 0 and the absolute minimum is 5
at x œ 3
25. F(x) œ x"# œ x# Ê Fw (x) œ 2x$ œ
2
x$
, however
x œ 0 is not a critical point since 0 is not in the domain;
F(0.5) œ 4, F(2) œ 0.25 Ê the absolute maximum is
0.25 at x œ 2 and the absolute minimum is 4 at
x œ 0.5
26. F(x) œ "x œ x" Ê Fw (x) œ x# œ
"
x#
, however
x œ 0 is not a critical point since 0 is not in the domain;
F(2) œ "# , F(1) œ 1 Ê the absolute maximum is 1 at
x œ 1 and the absolute minimum is
27. h(x) œ $Èx œ x"Î$ Ê hw (x) œ
"
3
"
#
at x œ 2
x#Î$ Ê a critical point
at x œ 0; h(1) œ 1, h(0) œ 0, h(8) œ 2 Ê the absolute
maximum is 2 at x œ 8 and the absolute minimum is 1
at x œ 1
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
169
170
Chapter 4 Applications of Derivatives
28. h(x) œ 3x#Î$ Ê hw (x) œ #x"Î$ Ê a critical point at
x œ 0; h(1) œ 3, h(0) œ 0, h(1) œ 3 Ê the absolute
maximum is 0 at x œ 0 and the absolute minimum is 3
at x œ 1 and at x œ 1
29. g(x) œ È4 x# œ a4 x# b
Ê gw (x) œ
"
#
a4 x# b
"Î#
"Î#
(2x) œ
x
È 4 x#
Ê critical points at x œ 2 and x œ 0, but not at x œ 2
because 2 is not in the domain; g(2) œ 0, g(0) œ 2,
g(1) œ È3 Ê the absolute maximum is 2 at x œ 0 and the
absolute minimum is 0 at x œ 2
30. g(x) œ È5 x# œ a& x# b
a5 x# b
(2x)
x
"
w
Ê g (x) œ ˆ # ‰ œ È # Ê critical points at x œ È5
"Î#
"Î#
&x
and x œ 0, but not at x œ È5 because È5 is not in the
domain; f ŠÈ5‹ œ 0, f(0) œ È5
Ê the absolute maximum is 0 at x œ È5 and the absolute
minimum is È5 at x œ 0
31. f()) œ sin ) Ê f w ()) œ cos ) Ê ) œ
1
#
is a critical point,
is not a critical point because #1 is not interior to
the domain; f ˆ #1 ‰ œ 1, f ˆ 1# ‰ œ 1, f ˆ 561 ‰ œ "#
but ) œ
1
#
Ê the absolute maximum is 1 at ) œ
minimum is 1 at ) œ
1
#
1
#
and the absolute
32. f()) œ tan ) Ê f w ()) œ sec# ) Ê f has no critical points in
1‰
ˆ 1
3 ß 4 . The extreme values therefore occur at the
‰ œ È3 and f ˆ 14 ‰ œ 1 Ê the absolute
endpoints: f ˆ 1
3
maximum is 1 at ) œ 14 and the absolute
minimum is È3 at ) œ 1
3
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 4.1 Extreme Values of Functions
171
33. g(x) œ csc x Ê gw (x) œ (csc x)(cot x) Ê a critical point
at x œ 1# ; g ˆ 13 ‰ œ È23 , g ˆ 1# ‰ œ 1, g ˆ 231 ‰ œ È23 Ê the
absolute maximum is
at x œ
2
È3
absolute minimum is 1 at x œ
1
3
and x œ
21
3 ,
and the
1
#
34. g(x) œ sec x Ê gw (x) œ (sec x)(tan x) Ê a critical point at
x œ 0; g ˆ 13 ‰ œ 2, g(0) œ 1, g ˆ 16 ‰ œ È23 Ê the absolute
maximum is 2 at x œ 13 and the absolute minimum is 1
at x œ 0
35. f(t) œ 2 ktk œ # Èt# œ # at# b
Ê f w (t) œ "# at# b
"Î#
"Î#
(2t) œ Èt # œ kttk
t
Ê a critical point at t œ 0; f(1) œ 1,
f(0) œ 2, f(3) œ 1 Ê the absolute maximum is 2 at t œ 0
and the absolute minimum is 1 at t œ 3
36. f(t) œ kt 5k œ È(t 5)# œ a(t 5)# b
œ
"
#
a(t 5)# b
"Î#
(2(t 5)) œ
t5
È(t 5)#
"Î#
œ
Ê f w (t)
t5
kt 5 k
Ê a critical point at t œ 5; f(4) œ 1, f(5) œ 0, f(7) œ 2
Ê the absolute maximum is 2 at t œ 7 and the absolute
minimum is 0 at t œ 5
37. f(x) œ x%Î$ Ê f w (x) œ
4
3
x"Î$ Ê a critical point at x œ 0; f(1) œ 1, f(0) œ 0, f(8) œ 16 Ê the absolute
maximum is 16 at x œ 8 and the absolute minimum is 0 at x œ 0
38. f(x) œ x&Î$ Ê f w (x) œ
5
3
x#Î$ Ê a critical point at x œ 0; f(1) œ 1, f(0) œ 0, f(8) œ 32 Ê the absolute
maximum is 32 at x œ 8 and the absolute minimum is 1 at x œ 1
39. g()) œ )$Î& Ê gw ()) œ
3
5
)#Î& Ê a critical point at ) œ 0; g(32) œ 8, g(0) œ 0, g(1) œ 1 Ê the absolute
maximum is 1 at ) œ 1 and the absolute minimum is 8 at ) œ 32
40. h()) œ 3)#Î$ Ê hw ()) œ 2)"Î$ Ê a critical point at ) œ 0; h(27) œ 27, h(0) œ 0, h(8) œ 12 Ê the absolute
maximum is 27 at ) œ 27 and the absolute minimum is 0 at ) œ 0
41. y œ x2 6x 7 Ê y w œ 2x 6 Ê 2x 6 œ 0 Ê x œ 3. The critical point is x œ 3.
42. faxb œ 6x2 x3 Ê f w axb œ 12x 3x2 Ê 12x 3x2 œ 0 Ê 3xa4 xb œ 0 Ê x œ 0 or x œ 4. The critical pointss are
x œ 0 and x œ 4.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
172
Chapter 4 Applications of Derivatives
43. faxb œ xa4 xb3 Ê f w axb œ x3a4 xb2 a1b‘ a4 xb3 œ a4 xb2 3x a4 xb‘ œ a4 xb2 a4 4xb
œ 4a4 xb2 a1 xb Ê 4a4 xb2 a1 xb œ 0 Ê x œ 1 or x œ 4. The critical points are x œ 1 and x œ 4.
44. gaxb œ ax 1b2 ax 3b2 Ê g w axb œ ax 1b2 † 2ax 3ba1b 2ax 1ba1b † ax 3b2
œ 2ax 3bax 1bax 1b ax 3b‘ œ 4ax 3bax 1bax 2b Ê 4ax 3bax 1bax 2b œ 0 Ê x œ 3 or x œ 1 or
x œ 2. The critical points are x œ 1, x œ 2, and x œ 3.
45. y œ x2 2
x
Ê y w œ 2x 2
x2
œ
2x3 2
x2
2x3 2
x2
Ê
œ ! Ê 2x3 2 œ ! Ê x œ 1;
2x3 2
x2
œ undefined Ê x2 œ 0 Ê x œ 0.
The domain of the function is a_, 0b a0, _b, thus x œ 0 is not in the domain, so the only critical point is x œ 1.
46. faxb œ
Ê f w axb œ
x2
x2
ax 2†b2x x2 a1b
ax 2 b 2
œ
x2 4x
ax 2 b 2
Ê
x2 4x
ax 2 b 2
œ ! Ê x2 4x œ ! Ê x œ 0 or x œ 4;
x2 4x
ax 2 b 2
œ undefined
2
Ê ax 2b œ 0 Ê x œ 2. The domain of the function is a_, 2b a2, _b, thus x œ 2 is not in the domain, so the only
critical points are x œ 0 and x œ 4
47. y œ x2 32Èx Ê y w œ 2x 16
Èx
œ
2x3Î2 16
Èx
Ê
2x3Î2 16
Èx
œ ! Ê 2x3Î2 16 œ 0 Ê x œ 4;
2x3Î2 16
Èx
œ undefined
Ê Èx œ 0 Ê x œ 0. The critical points are x œ 4 and x œ 0.
48. gaxb œ È2x x2 Ê g w axb œ
1x
È2x x2
Ê
1x
È2x x2
œ 0 Ê 1 x œ 0 Ê x œ 1;
1x
È2x x2
œ undefined Ê È2x x2 œ 0
2x x2 œ 0 Ê x œ 0 or x œ 2. The critical points are x œ 0, x œ 1, and x œ 2.
49. Minimum value is 1 at x œ 2.
50. To find the exact values, note that yw œ 3x2 2,
which is zero when x œ „ É 23 . Local maximum at
ŠÉ 23 ß 4 4È 6
9 ‹
¸ a0Þ816ß 5Þ089b; local
minimum at ŠÉ #$ ß % %È '
* ‹
¸ a0.816ß 2.911b
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 4.1 Extreme Values of Functions
51. To find the exact values, note that that yw œ 3x2 2x 8
œ a3x 4bax 2b, which is zero when x œ 2 or x œ %$ .
‰
Local maximum at a2ß 17b; local minimum at ˆ %$ ß %"
#(
52. Note that yw œ 5x# ax 5bax 3b, which is zero at
x œ 0, x œ 3, and x œ 5. Local maximum at a3, 108b;
local minimum at a5, 0b; a0, 0b is neither a maximum nor
a minimum.
53. Minimum value is 0 when x œ " or x œ ".
54. Note that yw œ
Èx 2
Èx ,
which is zero at x œ 4 and is
undefined when x œ 0. Local maximum at a0, 0b;
absolute minimum at a4, 4b
55. The actual graph of the function has asymptotes at x œ „ ",
so there are no extrema near these values. (This is an
example of grapher failure.) There is a local minimum at
a!ß "b.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
173
174
Chapter 4 Applications of Derivatives
56. Maximum value is 2 at x œ ";
minimum value is 0 at x œ " and x œ $.
"
# at x œ "à
"# as x œ ".
57. Maximum value is
minimum value is
"
# at x œ 0à
"# as x œ 2.
58. Maximum value is
minimum value is
59. yw œ x#Î$ a"b #$ x"Î$ ax #b œ
crit. pt.
x œ %&
xœ!
derivative
!
undefined
&x %
$ x
$È
extremum
local max
local min
value
"#
"Î$
œ "Þ!$%
#& "!
0
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 4.1 Extreme Values of Functions
60. yw œ x#Î$ a#xb #$ x"Î$ ax# %b œ
crit. pt.
x œ "
xœ!
xœ"
derivative
!
undefined
!
extremum
minimum
local max
minimum
"
a #xb a"bÈ%
#È % x #
x# a% x# b
% #x #
œÈ
È % x#
% x#
61. yw œ x
œ
crit. pt.
x œ #
x œ È #
x œ È#
xœ#
)x# )
$ x
$È
derivative
undefined
!
!
undefined
value
$
0
$
x#
extremum
local max
minimum
maximum
local min
value
!
#
#
!
62. yw œ x# #È$" x a 1b #xÈ$ x
œ
x# a%xba$ xb
#È $ x
crit. pt.
xœ0
x œ "#
&
xœ$
œ
_5x# "#x
#È $ x
derivative
!
!
undefined
extremum
minimum
local max
minimum
value
!
"%%
"Î#
¸ %Þ%'#
"#& "&
!
extremum
minimum
value
#
#, x "
63. yw œ œ
", x "
crit. pt.
xœ"
64. yw œ œ
crit. pt.
xœ!
xœ"
derivative
undefined
",
# #x,
x!
x!
derivative
undefined
!
extremum
local min
local max
value
$
%
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
175
176
Chapter 4 Applications of Derivatives
2x 2,
65. yw œ œ
2x 6,
crit. pt.
x œ 1
xœ1
xœ3
x1
x1
derivative
!
undefined
!
extremum
maximum
local min
maximum
value
5
1
5
"% x# "# x "&
% , xŸ"
x$ 'x# )x,
x"
w
w
#
if x ", and limc f a" hb œ ". Also, f axb œ $x "#x ) if x ", and
66. We begin by determining whether f w axb is defined at x œ ", where faxb œ œ
Clearly, f w axb œ "# x "
#
hÄ!
limb f w a" hb œ ". Since f is continuous at x œ ", we have that f w a"b œ ". Thus,
hÄ!
f w axb œ œ
"# x "# ,
$x "#x ) ,
#
Note that "# x But # #È $
$
crit. pt.
x œ "
x ¸ $Þ"&&
"
#
xŸ"
x"
œ ! when x œ ", and $x# "#x ) œ ! when x œ
¸ !Þ)%& ", so the critical points occur at x œ " and x œ
derivative
!
!
extremum
local max
local min
È
"# „ È"## %a$ba)b
œ "# „' %)
#a$b
È
# # $ $ ¸ $Þ"&&.
œ#„
#È$
$ .
value
4
¸ $Þ!(*
67. (a) No, since f w axb œ #$ ax #b"Î$ , which is undefined at x œ #.
(b) The derivative is defined and nonzero for all x Á #. Also, fa#b œ ! and faxb ! for all x Á #.
(c) No, faxb need not have a global maximum because its domain is all real numbers. Any restriction of f to a closed
interval of the form Òa, bÓ would have both a maximum value and minimum value on the interval.
(d) The answers are the same as (a) and (b) with 2 replaced by a.
x$ *x, x Ÿ $ or ! Ÿ x $
$x$ *, x $ or ! x $
w
.
Therefore,
f
a
x
b
œ
.
œ
x$ *x, $ x ! or x $
$x$ *, $ x ! or x $
No, since the left- and right-hand derivatives at x œ !, are * and *, respectively.
No, since the left- and right-hand derivatives at x œ $, are ") and "), respectively.
No, since the left- and right-hand derivatives at x œ $, are ") and "), respectively.
The critical points occur when f w axb œ ! (at x œ „ È$) and when f w axb is undefined (at x œ ! and x œ „ $). The
68. Note that faxb œ œ
(a)
(b)
(c)
(d)
minimum value is ! at x œ $, at x œ !, and at x œ $; local maxima occur at ŠÈ$ß 'È$‹ and ŠÈ$ß 'È$‹.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 4.1 Extreme Values of Functions
69. Yes, since f(x) œ kxk œ Èx# œ ax# b
"Î#
Ê f w (x) œ
"
#
ax# b
"Î#
(2x) œ
x
ax# b"Î#
œ
x
kx k
177
is not defined at x œ 0. Thus it is
not required that f w be zero at a local extreme point since f w may be undefined there.
70. If f(c) is a local maximum value of f, then f(x) Ÿ f(c) for all x in some open interval (aß b) containing c. Since f is even,
f(x) œ f(x) Ÿ f(c) œ f(c) for all x in the open interval (bß a) containing c. That is, f assumesa local maximum at
the point c. This is also clear from the graph of f because the graph of an even function is symmetric about the y-axis.
71. If g(c) is a local minimum value of g, then g(x) g(c) for all x in some open interval (aß b) containing c. Since g is odd,
g(x) œ g(x) Ÿ g(c) œ g(c) for all x in the open interval (bß a) containing c. That is, g assumes a local
maximum at the point c. This is also clear from the graph of g because the graph of an odd function is symmetric about
the origin.
72. If there are no boundary points or critical points the function will have no extreme values in its domain. Such functions do
indeed exist, for example f(x) œ x for _ x _. (Any other linear function f(x) œ mx b with m Á 0 will do as
well.)
73. (a) Vaxb œ "'!x &#x# %x$
Vw axb œ "'! "!%x "#x# œ %ax #ba$x #!b
The only critical point in the interval a!ß &b is at x œ #. The maximum value of Vaxb is 144 at x œ #.
(b) The largest possible volume of the box is 144 cubic units, and it occurs when x œ # units.
74. (a) f w axb œ $ax# #bx c is a quadratic, so it can have 0, 1, or 2 zeros, which would be the critical points of f. The
function faxb œ x$ $x has two critical points at x œ " and x œ ". The function faxb œ x$ " has one critical point
at x œ !Þ The function faxb œ x$ x has no critical points.
(b) The function can have either two local extreme values or no extreme values. (If there is only one critical point, the
cubic function has no extreme values.)
75. s œ "# gt# v! t s! Ê
ds
dt
œ gt v! œ ! Ê t œ
2
Thus sŠ vg! ‹ œ "# gŠ vg! ‹ v0 Š vg! ‹ s0 œ
76.
Now satb œ s0 Í tˆ gt2 v0 ‰ œ 0 Í t œ 0 or t œ
s0 s0 is the maximum height over the interval 0 Ÿ t Ÿ
œ ! Ê tan t œ " Ê t œ
never negative) Ê the peak current is #È# amps.
dI
dt
œ #sin t #cos t, solving
v!2
2g
v!
g.
dI
dt
1
%
2v0
g .
2v0
g .
n1 where n is a nonnegative integer (in this exercise t is
77. Maximum value is 11 at x œ 5;
minimum value is 5 on the interval Ò3ß 2Ó;
local maximum at a5ß 9b
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
178
Chapter 4 Applications of Derivatives
78. Maximum value is 4 on the interval Ò5ß 7Ó;
minimum value is 4 on the interval Ò2ß 1Ó.
79. Maximum value is 5 on the interval Ò3ß _Ñ;
minimum value is 5 on the interval Ð_ß 2Ó.
80. Minimum value is 4 on the interval Ò"ß $Ó
81-86. Example CAS commands:
Maple:
with(student):
f := x -> x^4 - 8*x^2 + 4*x + 2;
domain := x=-20/25..64/25;
plot( f(x), domain, color=black, title="Section 4.1 #81(a)" );
Df := D(f);
plot( Df(x), domain, color=black, title="Section 4.1 # 81(b)" )
StatPt := fsolve( Df(x)=0, domain )
SingPt := NULL;
EndPt := op(rhs(domain));
Pts :=evalf([EndPt,StatPt,SingPt]);
Values := [seq( f(x), x=Pts )];
Maximum value is 2.7608 and occurs at x=2.56 (right endpoint).
%
Minimum value $ is -6.2680 and occurs at x=1.86081 (singular point).
Mathematica: (functions may vary) (see section 2.5 re. RealsOnly ):
<<Miscellaneous `RealOnly`
Clear[f,x]
a = 1; b = 10/3;
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 4.2 The Mean Value Theorem
179
f[x_] =2 2x 3 x2/3
f'[x]
Plot[{f[x], f'[x]}, {x, a, b}]
NSolve[f'[x]==0, x]
{f[a], f[0], f[x]/.%, f[b]//N
In more complicated expressions, NSolve may not yield results. In this case, an approximate solution (say 1.1 here)
is observed from the graph and the following command is used:
FindRoot[f'[x]==0,{x, 1.1}]
4.2 THE MEAN VALUE THEOREM
1. When f(x) œ x# 2x 1 for 0 Ÿ x Ÿ 1, then
2. When f(x) œ x#Î$ for 0 Ÿ x Ÿ 1, then
3. When f(x) œ x "
x
for
"
#
fa1b fa0b
10
Ÿ x Ÿ 2, then
fa3b fa1b
31
5. When f(x) œ x3 x2 for 1 Ÿ x Ÿ 2, then
¸ 1.22 and
1 È7
3
œ f w acb Ê 3 œ 2c 2 Ê c œ #" .
œ f w acb Ê 1 œ ˆ 32 ‰ c"Î$ Ê c œ
fa2b fa1Î2b
2 1 Î2
4. When f(x) œ Èx 1 for 1 Ÿ x Ÿ 3, then
1 È7
3
fa1b fa0b
10
œ f w acb Ê 0 œ " œ f w acb Ê
fa2b fa1b
2 a1b
È2
#
œ
"
c#
8
#7 .
Ê c œ 1.
"
#È c 1
Ê c œ #3 .
œ f w acb Ê 2 œ 3c2 2c Ê c œ
1 „ È7
.
3
¸ 0.549 are both in the interval 1 Ÿ x Ÿ 2.
x3 2 Ÿ x Ÿ 0
g a 2 b
w
w
w
2
w
, then ga22ba
2b œ g acb Ê 3 œ g acb. If 2 Ÿ x 0, then g (x) œ 3x Ê 3 œ g acb
x2 ! x Ÿ 2
Ê 3c2 œ 3 Ê c œ „ 1. Only c œ 1 is in the interval. If ! x Ÿ 2, then gw (x) œ 2x Ê 3 œ g w acb Ê 2c œ 3 Ê c œ 32 .
6. When g(x) œ œ
7. Does not; f(x) is not differentiable at x œ 0 in ("ß 8).
8. Does; f(x) is continuous for every point of [0ß 1] and differentiable for every point in (0ß 1).
9. Does; f(x) is continuous for every point of [0ß 1] and differentiable for every point in (0ß 1).
10. Does not; f(x) is not continuous at x œ 0 because lim c f(x) œ 1 Á 0 œ f(0).
xÄ!
11. Does not; f is not differentiable at x œ 1 in (2ß 0).
12. Does; f(x) is continuous for every point of [0ß 3] and differentiable for every point in (0ß 3).
13. Since f(x) is not continuous on 0 Ÿ x Ÿ 1, Rolle's Theorem does not apply:
lim f(x) œ lim c x œ 1 Á 0 œ f(1).
x Ä 1c
xÄ1
14. Since f(x) must be continuous at x œ 0 and x œ 1 we have lim b f(x) œ a œ f(0) Ê a œ 3 and
xÄ!
lim c f(x) œ lim b f(x) Ê 1 3 a œ m b Ê 5 œ m b. Since f(x) must also be differentiable at
xÄ1
xÄ1
x œ 1 we have lim c f w (x) œ lim b f w (x) Ê 2x 3k x=1 œ mk x=1 Ê 1 œ m. Therefore, a œ 3, m œ 1 and b œ 4.
xÄ1
xÄ1
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
180
Chapter 4 Applications of Derivatives
15. (a) i
ii
iii
iv
(b) Let r" and r# be zeros of the polynomial P(x) œ xn an-1 xn-1 á a" x a! , then P(r" ) œ P(r# ) œ 0.
Since polynomials are everywhere continuous and differentiable, by Rolle's Theorem Pw (r) œ 0 for some r
between r" and r# , where Pw (x) œ nxn-1 (n 1) an-1 xn-2 á a" .
16. With f both differentiable and continuous on [aß b] and f(r" ) œ f(r# ) œ f(r$ ) œ 0 where r" , r# and r$ are in [aß b],
then by Rolle's Theorem there exists a c" between r" and r# such that f w (c" ) œ 0 and a c# between r# and r$
such that f w (c# ) œ 0. Since f w is both differentiable and continuous on [aß b], Rolle's Theorem again applies and
we have a c$ between c" and c# such that f ww (c$ ) œ 0. To generalize, if f has n1 zeros in [aß b] and f ÐnÑ is
continuous on [aß b], then f ÐnÑ has at least one zero between a and b.
17. Since f ww exists throughout [aß b] the derivative function f w is continuous there. If f w has more than one zero in [aß b], say
f w (r" ) œ f w (r# ) œ 0 for r" Á r# , then by Rolle's Theorem there is a c between r" and r# such that f ww (c) œ 0, contrary to
f ww 0 throughout [aß b]. Therefore f w has at most one zero in [aß b]. The same argument holds if f ww 0 throughout [aß b].
18. If f(x) is a cubic polynomial with four or more zeros, then by Rolle's Theorem f w (x) has three or more zeros, f ww (x) has 2 or
more zeros and f www (x) has at least one zero. This is a contradiction since f www (x) is a non-zero constant when f(x) is a cubic
polynomial.
19. With f(2) œ 11 0 and f(1) œ 1 0 we conclude from the Intermediate Value Theorem that f(x) œ x% 3x 1
has at least one zero between 2 and 1. Then 2 x 1 Ê ) x$ 1 Ê 32 4x$ 4
Ê 29 4x$ 3 1 Ê f w (x) 0 for 2 x 1 Ê f(x) is decreasing on [#ß 1] Ê f(x) œ 0 has exactly one
solution in the interval (#ß 1).
20. f(x) œ x$ 4
x#
7 Ê f w (x) œ 3x# 8
x$
0 on (_ß 0) Ê f(x) is increasing on (_ß 0). Also, f(x) 0 if x 2 and
f(x) 0 if 2 x 0 Ê f(x) has exactly one zero in (_ß !).
21. g(t) œ Èt Èt 1 4 Ê gw (t) œ
"
#È t
"
2Èt1
0 Ê g(t) is increasing for t in (!ß _); g(3) œ È3 2 0 and
g(15) œ È15 0 Ê g(t) has exactly one zero in (!ß _)Þ
22. g(t) œ
"
"t
È1 t 3.1 Ê gw (t) œ
"
("t)#
"
2 È 1 t
0 Ê g(t) is increasing for t in (1ß 1); g(0.99) œ 2.5 and
g(0.99) œ 98.3 Ê g(t) has exactly one zero in (1ß 1).
23. r()) œ ) sin# ˆ 3) ‰ 8 Ê rw ()) œ 1 32 sin ˆ 3) ‰ cos ˆ 3) ‰ œ 1 "3 sin ˆ 23) ‰ 0 on (_ß _) Ê r()) is increasing on
(_ß _); r(0) œ 8 and r(8) œ sin# ˆ 83 ‰ 0 Ê r()) has exactly one zero in (_ß _).
24. r()) œ 2) cos# ) È2 Ê rw ()) œ 2 2 sin ) cos ) œ 2 sin 2) 0 on (_ß _) Ê r()) is increasing on (_ß _);
r(#1) œ 41 cos (#1) È2 œ 41 1 È2 0 and r(21) œ 41 1 È2 0 Ê r()) has exactly one zero in
(_ß _).
5 Ê rw ()) œ (sec ))(tan )) )3% 0 on ˆ!ß 1# ‰ Ê r()) is increasing on ˆ!ß 1# ‰ ; r(0.1) ¸ 994 and
r(1.57) ¸ 1260.5 Ê r()) has exactly one zero in ˆ!ß 1# ‰ .
25. r()) œ sec ) "
)$
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 4.2 The Mean Value Theorem
181
26. r()) œ tan ) cot ) ) Ê rw ()) œ sec# ) csc# ) 1 œ sec# ) cot# ) 0 on ˆ!ß 1# ‰ Ê r()) is increasing on ˆ0ß 1# ‰ ;
r ˆ 14 ‰ œ 14 0 and r(1.57) ¸ 1254.2 Ê r()) has exactly one zero in ˆ!ß 1# ‰ .
27. By Corollary 1, f w (x) œ 0 for all x Ê f(x) œ C, where C is a constant. Since f(1) œ 3 we have C œ 3 Ê f(x) œ 3 for
all x.
28. g(x) œ 2x 5 Ê gw (x) œ 2 œ f w (x) for all x. By Corollary 2, f(x) œ g(x) C for some constant C. Then
f(0) œ g(0) C Ê 5 œ 5 C Ê C œ 0 Ê f(x) œ g(x) œ 2x 5 for all x.
29. g(x) œ x# Ê gw (x) œ 2x œ f w (x) for all x. By Corollary 2, f(x) œ g(x) C.
(a) f(0) œ 0 Ê 0 œ g(0) C œ 0 C Ê C œ 0 Ê f(x) œ x# Ê f(2) œ 4
(b) f(1) œ 0 Ê 0 œ g(1) C œ 1 C Ê C œ 1 Ê f(x) œ x# 1 Ê f(2) œ 3
(c) f(2) œ 3 Ê 3 œ g(2) C Ê 3 œ 4 C Ê C œ 1 Ê f(x) œ x# 1 Ê f(2) œ 3
30. g(x) œ mx Ê gw (x) œ m, a constant. If f w (x) œ m, then by Corollary 2, f(x) œ g(x) b œ mx b where b is a constant.
Therefore all functions whose derivatives are constant can be graphed as straight lines y œ mx b.
31. (a) y œ
x#
#
C
(b) y œ
32. (a) y œ x# C
x$
3
C
(b) y œ x# x C
33. (a) yw œ x# Ê y œ
"
x
C
(b) y œ x "
x
35. (a) y œ "# cos 2t C
(c) y œ cos 2t 2 sin
(c) y œ 5x (b) y œ 2 sin
t
#
C
"
x
C
(b) y œ 2Èx C
"
#
"
#
x%
4
(c) y œ x$ x# x C
C
x"Î# Ê y œ x"Î# C Ê y œ Èx C
(c) y œ 2x# 2Èx C
34. (a) yw œ
(c) y œ
C
t
#
C
(b) yw œ )"Î# Ê y œ
36. (a) y œ tan ) C
2
3
)$Î# C
(c) y œ
2
3
)$Î# tan ) C
37. f(x) œ x# x C; 0 œ f(0) œ 0# 0 C Ê C œ 0 Ê f(x) œ x# x
38. g(x) œ "x x# C; 1 œ g(1) œ "1 (1)# C Ê C œ 1 Ê g(x) œ x" x# 1
39. r()) œ 8) cot ) C; 0 œ r ˆ 14 ‰ œ 8 ˆ 14 ‰ cot ˆ 14 ‰ C Ê 0 œ 21 1 C Ê C œ 21 1
Ê r()) œ 8) cot ) 21 1
40. r(t) œ sec t t C; 0 œ r(0) œ sec (0) 0 C Ê C œ 1 Ê r(t) œ sec t t 1
41. v œ
ds
dt
œ *Þ)t & Ê s œ %Þ*t# &t C; at s œ "! and t œ ! we have C œ "! Ê s œ %Þ*t# &t "!
42. v œ
ds
dt
œ $#t # Ê s œ "'t# #t C; at s œ % and t œ
43. v œ
ds
dt
œ sina1tb Ê s œ 1" cosa1tb C; at s œ ! and t œ ! we have C œ
44. v œ
ds
dt
œ 12 cosˆ #1t ‰ Ê s œ sinˆ #1t ‰ C; at s œ " and t œ 1# we have C œ " Ê s œ sinˆ #1t ‰ "
"
#
we have C œ " Ê s œ "'t# #t "
"
1
Êsœ
" cosa1tb
1
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
182
Chapter 4 Applications of Derivatives
45. a œ $# Ê v œ $#t C" ; at v œ #! and t œ ! we have C" œ #! Ê v œ $#t #! Ê s œ "'t# #!t C# ; at s œ & and
t œ ! we have C# œ & Ê s œ "'t# #!t &
46. a œ 9.8 Ê v œ 9.8t C" ; at v œ $ and t œ ! we have C" œ $ Ê v œ *Þ)t $ Ê s œ %Þ*t# $t C# ; at s œ ! and
t œ ! we have C# œ ! Ê s œ %Þ*t# $t
47. a œ %sina#tb Ê v œ #cosa#tb C" ; at v œ # and t œ ! we have C" œ ! Ê v œ #cosa#tb Ê s œ sina#tb C# ; at s œ $
and t œ ! we have C# œ $ Ê s œ sina#tb $
Ê v œ 1$ sinˆ $1t ‰ C" ; at v œ ! and t œ ! we have C" œ ! Ê v œ 1$ sinˆ $1t ‰ Ê s œ cosˆ $1t ‰ C# ; at
s œ " and t œ ! we have C# œ ! Ê s œ cosˆ $1t ‰
48. a œ
*
ˆ $t ‰
1# cos 1
49. If T(t) is the temperature of the thermometer at time t, then T(0) œ 19° C and T(14) œ 100° C. From the Mean Value
Theorem there exists a 0 t! 14 such that
T(14) T(0)
14 0
œ 8.5° C/sec œ Tw (t! ), the rate at which the temperature was
changing at t œ t! as measured by the rising mercury on the thermometer.
50. Because the trucker's average speed was 79.5 mph, by the Mean Value Theorem, the trucker must have been going that
speed at least once during the trip.
51. Because its average speed was approximately 7.667 knots, and by the Mean Value Theorem, it must have been going that
speed at least once during the trip.
52. The runner's average speed for the marathon was approximately 11.909 mph. Therefore, by the Mean Value Theorem, the
runner must have been going that speed at least once during the marathon. Since the initial speed and final speed are both 0
mph and the runner's speed is continuous, by the Intermediate Value Theorem, the runner's speed must have been 11 mph
at least twice.
53. Let d(t) represent the distance the automobile traveled in time t. The average speed over 0 Ÿ t Ÿ 2 is
w
Value Theorem says that for some 0 t! 2, d (t! ) œ
d(2) d(0)
#0 .
d(2) d(0)
#0 .
The Mean
w
The value d (t! ) is the speed of the automobile at time t!
(which is read on the speedometer).
54. aatb œ vw atb œ "Þ' Ê vatb œ "Þ't C; at a!ß !b we have C œ ! Ê vatb œ "Þ't. When t œ $!, then va$!b œ %) m/sec.
55. The conclusion of the Mean Value Theorem yields
56. The conclusion of the Mean Value Theorem yields
"
b
"a
ba
b # a#
ba
b‰
œ c"# Ê c# ˆ a ab
œ a b Ê c œ Èab.
œ 2c Ê c œ
a b
# .
57. f w (x) œ [cos x sin (x 2) sin x cos (x 2)] 2 sin (x 1) cos (x 1) œ sin (x x 2) sin 2(x 1)
œ sin (2x 2) sin (2x 2) œ 0. Therefore, the function has the constant value f(0) œ sin# 1 ¸ 0.7081
which explains why the graph is a horizontal line.
58. (a) faxb œ ax #bax "bxax "bax #b œ x& &x$ %x is one possibility.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 4.2 The Mean Value Theorem
183
(b) Graphing faxb œ x& &x$ %x and f w axb œ &x% "&x# % on Ò$ß $Ó by Ò(ß (Ó we see that each x-intercept of
f w axb lies between a pair of x-intercepts of faxb, as expected by Rolle's Theorem.
(c) Yes, since sin is continuous and differentiable on a _ß _b.
59. faxb must be zero at least once between a and b by the Intermediate Value Theorem. Now suppose that faxb is zero twice
between a and b. Then by the Mean Value Theorem, f w axb would have to be zero at least once between the two zeros of
faxb, but this can't be true since we are given that f w axb Á ! on this interval. Therefore, faxb is zero once and only once
between a and b.
60. Consider the function kaxb œ faxb gaxb. kaxb is continuous
and differentiable on Òa, bÓ, and since kaab œ faab gaab and
kabb œ fabb gabb, by the Mean Value Theorem, there must
be a point c in aa, bb where kw acb œ !. But since
kw acb œ f w acb gw acb, this means that f w acb œ gw acb, and c is a
point where the graphs of f and g have tangent lines with the
same slope, so these lines are either parallel or are the same
line.
61. f w axb Ÿ 1 for 1 Ÿ x Ÿ 4 Ê faxb is differentiable on 1 Ÿ x Ÿ 4 Ê f is continuous on 1 Ÿ x Ÿ 4 Ê f satisfies the
conditions of the Mean Value Theorem Ê
Ê fa4b fa1b Ÿ 3
62. 0 f w axb "
#
fa4b fa1b
41
œ f w acb for some c in 1 x 4 Ê f w acb Ÿ 1 Ê
fa4b fa1b
3
Ÿ1
for all x Ê f w axb exists for all x, thus f is differentiable on a1, 1b Ê f is continuous on Ò1, 1Ó
Ê f satisfies the conditions of the Mean Value Theorem Ê
fa1b fa1b
1 a1b
œ f w acb for some c in Ò1, 1Ó Ê 0 fa1b fa1b
2
Ê 0 fa1b fa1b 1. Since fa1b fa1b 1 Ê fa1b 1 fa1b 2 fa1b, and since 0 fa1b fa1b
we have fa1b fa1b. Together we have fa1b fa1b 2 fa1b.
63. Let fatb œ cos t and consider the interval Ò0, xÓ where x is a real number. f is continuous on Ò0, xÓ and f is differentiable on
a0, xb since f w atb œ sin t Ê f satisfies the conditions of the Mean Value Theorem Ê
Ò0, xÓ Ê
cos x 1
x
œ sin c. Since 1 Ÿ sin c Ÿ 1 Ê 1 Ÿ sin c Ÿ 1 Ê 1 Ÿ
Ê x Ÿ cos x 1 Ÿ x Ê lcos x 1l Ÿ x œ l x l. If x 0, 1 Ÿ
cos x 1
x
cos x 1
x
Ÿ 1 Ê x
faxb fa0b
x a0 b
œ f w acb for some c in
Ÿ 1. If x 0, 1 Ÿ
cos x 1
cos x 1
x
x
Ê x Ÿ cos x 1 Ÿ x Ê axb Ÿ cos x 1 Ÿ x Ê lcos x 1l Ÿ x œ l x l. Thus, in both cases, we have
lcos x 1l Ÿ l x l. If x œ 0, then lcos 0 1l œ l1 1l œ l0l Ÿ l0l, thus lcos x 1l Ÿ l x l is true for all x.
64. Let f(x) œ sin x for a Ÿ x Ÿ b. From the Mean Value Theorem there exists a c between a and b such that
sin b sin a
sin a
sin a ¸
œ cos c Ê 1 Ÿ sin bb Ÿ 1 Ê ¸ sin bb Ÿ 1 Ê ksin b sin ak Ÿ kb ak.
ba
a
a
65. Yes. By Corollary 2 we have f(x) œ g(x) c since f w (x) œ gw (x). If the graphs start at the same point x œ a,
then f(a) œ g(a) Ê c œ 0 Ê f(x) œ g(x).
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Ÿ1
"
#
184
Chapter 4 Applications of Derivatives
66. Assume f is differentiable and lfawb faxbl Ÿ lw xl for all values of w and x. Since f is differentiable, f w axb exists and
f w axb œ lim
wÄx
fawb faxb
wx
using the alternative formula for the derivative. Let gaxb œ lxl, which is continuous for all x.
By Theorem 10 from Chapter 2, ¸f w axb¸ œ º lim
wÄx
lfawb faxbl Ÿ lw xl for all w and x Ê
¸f w axb¸ œ lim
wÄx
lfawb faxbl
lw x l
fawb faxb
wx º
lfawbfaxbl
lw x l
faxb
œ lim ¹ fawwb x ¹ œ wlim
wÄx
Äx
lfawb faxbl
lw x l .
Since
Ÿ 1 as long as w Á x. By Theorem 5 from Chapter 2,
Ÿ lim 1 œ 1 Ê ¸f w axb¸ Ÿ 1 Ê 1 Ÿ f w axb Ÿ 1.
wÄx
67. By the Mean Value Theorem we have
we have f(b) f(a) 0 Ê f w (c) 0.
f(b) f(a)
ba
œ f w (c) for some point c between a and b. Since b a 0 and f(b) f(a),
68. The condition is that f w should be continuous over [aß b]. The Mean Value Theorem then guarantees the
existence of a point c in (aß b) such that
w
f(b) f(a)
ba
w
œ f w (c). If f w is continuous, then it has a minimum and
maximum value on [aß b], and min f Ÿ f (c) Ÿ max f w , as required.
69. f w (x) œ a1 x% cos xb
"
Ê f ww (x) œ a1 x% cos xb
#
a4x$ cos x x% sin xb
#
œ x$ a1 x% cos xb (4 cos x x sin x) 0 for 0 Ÿ x Ÿ 0.1 Ê f w (x) is decreasing when 0 Ÿ x Ÿ 0.1
f(0.1) "
0.1
Ê min f w ¸ 0.9999 and max f w œ 1. Now we have 0.9999 Ÿ
Ÿ 1 Ê 0.09999 Ÿ f(0.1) 1 Ÿ 0.1
Ê 1.09999 Ÿ f(0.1) Ÿ 1.1.
70. f w (x) œ a1 x% b
"
Ê f ww (x) œ a1 x% b
#
a4x$ b œ
4x$
$
a1 x % b
0 for 0 x Ÿ 0.1 Ê f w (x) is increasing when
0 Ÿ x Ÿ 0.1 Ê min f w œ 1 and max f w œ 1.0001. Now we have 1 Ÿ
f(0.1) 2
0.1
Ÿ 1.0001
Ê 0.1 Ÿ f(0.1) 2 Ÿ 0.10001 Ê 2.1 Ÿ f(0.1) Ÿ 2.10001.
71. (a) Suppose x 1, then by the Mean Value Theorem
(b)
f(1)
Mean Value Theorem f(x)x 0
1
f(x) f(1)
Yes. From part (a), lim c x 1
xÄ1
f(x) f(1)
x1
0 Ê f(x) f(1). Suppose x 1, then by the
Ê f(x) f(1). Therefore f(x)
1 for all x since f(1) œ 1.
f(x) f(1)
x1
Ÿ 0 and lim b
0. Since f w (1) exists, these two one-sided
xÄ1
limits are equal and have the value f w (1) Ê f w (1) Ÿ 0 and f w (1) 0 Ê f w (1) œ 0.
72. From the Mean Value Theorem we have
f(b) f(a)
ba
œ f w (c) where c is between a and b. But f w (c) œ 2pc q œ 0
has only one solution c œ #qp . (Note: p Á 0 since f is a quadratic function.)
4.3 MONOTONIC FUNCTIONS AND THE FIRST DERIVATIVE TEST
1. (a) f w (x) œ x(x 1) Ê critical points at 0 and 1
(b) f w œ ± ± Ê increasing on (_ß !) and ("ß _), decreasing on (!ß ")
!
"
(c) Local maximum at x œ 0 and a local minimum at x œ 1
2. (a) f w (x) œ (x 1)(x 2) Ê critical points at 2 and 1
(b) f w œ ± ± Ê increasing on (_ß #) and ("ß _), decreasing on (2ß ")
#
"
(c) Local maximum at x œ 2 and a local minimum at x œ 1
3. (a) f w (x) œ (x 1)# (x 2) Ê critical points at 2 and 1
(b) f w œ ± ± Ê increasing on (2ß 1) and ("ß _), decreasing on (_ß 2)
#
"
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 4.3 Monotonic Functions and the First Derivative Test
(c) No local maximum and a local minimum at x œ 2
4. (a) f w (x) œ (x 1)# (x 2)# Ê critical points at 2 and 1
(b) f w œ ± ± Ê increasing on (_ß 2) (#ß ") ("ß _), never decreasing
#
"
(c) No local extrema
5. (a) f w (x) œ (x 1)(x 2)(x 3) Ê critical points at 2, 1 and 3
(b) f w œ ± ± ± Ê increasing on (2ß 1) and ($ß _), decreasing on (_ß 2) and ("ß $)
#
"
$
(c) Local maximum at x œ 1, local minima at x œ 2 and x œ 3
6. (a) f w (x) œ (x 7)(x 1)(x 5) Ê critical points at 5, 1 and 7
(b) f w œ ± ± ± Ê increasing on (5ß 1) and (7ß _), decreasing on (_ß 5) and ("ß 7)
&
"
(
(c) Local maximum at x œ 1, local minima at x œ 5 and x œ 7
7. (a) f w (x) œ
x 2 ax 1 b
ax 2 b
Ê critical points at x œ 0, x œ 1 and x œ 2
w
(b) f œ )( ± ± Ê increasing on a_ß 2b and a1ß _b, decreasing on a2ß 0b and a0ß 1b
2
0
1
(c) Local minimum at x œ 1
8. (a) f w (x) œ
ax 2bax 4b
ax 1bax 3b
Ê critical points at x œ 2, x œ 4, x œ 1, and x œ 3
w
(b) f œ ± )( ± )( Ê increasing on a_ß 4b, a1ß 2b and a3ß _b, decreasing on
1
3
4
2
a4ß 1b and a2ß 3b
(c) Local maximum at x œ 4 and x œ 2
9. (a) f w (x) œ 1 w
4
x2
œ
x2 4
x2
Ê critical points at x œ 2, x œ 2 and x œ 0.
(b) f œ ± )( ± Ê increasing on a_ß 2b and a2ß _b, decreasing on a2ß 0b and a0ß 2b
0
2
2
(c) Local maximum at x œ 2, local minimum at x œ 2
10. (a) f w (x) œ 3 6
Èx
œ
3È x 6
Èx
Ê critical points at x œ 4 and x œ 0
(b) f w œ ( ± Ê increasing on a4ß _b, decreasing on a0ß 4b
0
4
(c) Local minimum at x œ 4
11. (a) f w (x) œ x"Î$ (x 2) Ê critical points at x œ 2 and x œ 0
(b) f w œ ± )( Ê increasing on (_ß 2) and (0ß _), decreasing on (2ß 0)
0
2
(c) Local maximum at x œ 2, local minimum at x œ 0
12. (a) f w (x) œ x"Î# (x 3) Ê critical points at x œ 0 and x œ 3
(b) f w œ ( ± Ê increasing on (3ß _), decreasing on (0ß 3)
0
3
(c) No local maximum and a local minimum at x œ 3
13. (a) f w (x) œ asin x 1ba2cos x 1b, 0 Ÿ x Ÿ 21 Ê critical points at x œ 12 , x œ 231 , and x œ 431
(b) f w œ Ò ± ± ± Ó Ê increasing on ˆ 231 ß 431 ‰, decreasing on ˆ0ß 12 ‰, ˆ 12 ß 231 ‰ and ˆ 431 ß 21‰
1
21
41
0
21
2
3
3
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
185
186
Chapter 4 Applications of Derivatives
(c) Local maximum at x œ
41
3
and x œ 0, local minimum at x œ
21
3
and x œ 21
14. (a) f w (x) œ asin x cos xbasin x cos xb, 0 Ÿ x Ÿ 21 Ê critical points at x œ 14 , x œ 341 , x œ 541 , and x œ 741
(b) f w œ Ò ± ± ± ± Ó Ê increasing on ˆ 14 ß 341 ‰ and ˆ 541 ß 741 ‰, decreasing on ˆ0ß 14 ‰,
1
51
71
31
0
21
4
4
ˆ 341 ß 541 ‰ and ˆ 741 ß 21‰
4
(c) Local maximum at x œ 0, x œ
4
31
4
and x œ
71
4 ,
local minimum at x œ 14 , x œ
51
4
and x œ 21
15. (a) Increasing on a2ß 0b and a2ß 4b, decreasing on a4ß 2b and a0ß 2b
(b) Absolute maximum at a4ß 2b, local maximum at a0ß 1b and a4ß 1b; Absolute minimum at a2ß 3b, local minimum at
a2ß 0b
16. (a) Increasing on a4ß 3.25b, a1.5ß 1b, and a2ß 4b, decreasing on a3.25ß 1.5b and a1ß 2b
(b) Absolute maximum at a4ß 2b, local maximum at a3.25ß 1b and a1ß 1b; Absolute minimum at a1.5ß 1b, local
minimum at a4ß 0b and a2ß 0b
17. (a) Increasing on a4ß 1b, a0.5ß 2b, and a2ß 4b, decreasing on a1ß 0.5b
(b) Absolute maximum at a4ß 3b, local maximum at a1ß 2b and a2ß 1b; No absolute minimum, local minimum at
a4ß 1band a0.5ß 1b
18. (a) Increasing on a4ß 2.5b, a1ß 1b, and a3ß 4b, decreasing on a2.5ß 1b and a1ß 3b
(b) No absolute maximum, local maximum at a2.5ß 1b, a1ß 2b and a4ß 2b; No absolute minimum, local minimum at
a1ß 0b and a3ß 1b
19. (a) g(t) œ t# 3t 3 Ê gw (t) œ 2t 3 Ê a critical point at t œ 3# ; gw œ ± , increasing on
$Î#
ˆ_ß 3# ‰ , decreasing on ˆ 3# ß _‰
3
21
3
(b) local maximum value of g ˆ 3# ‰ œ 21
4 at t œ # , absolute maximum is 4 at t œ #
20. (a) g(t) œ 3t# 9t 5 Ê gw (t) œ 6t 9 Ê a critical point at t œ
3
#
; gw œ ± , increasing on ˆ_ß 32 ‰ ,
$Î#
decreasing on ˆ 3# ß _‰
(b) local maximum value of g ˆ 3# ‰ œ
47
4
at t œ 3# , absolute maximum is
47
4
at t œ
3
#
21. (a) h(x) œ x$ 2x# Ê hw (x) œ 3x# 4x œ x(4 3x) Ê critical points at x œ 0, 43
Ê hw œ ± ± , increasing on ˆ0ß 43 ‰ , decreasing on (_ß !) and ˆ 43 ß _‰
!
%Î$
4
(b) local maximum value of h ˆ 43 ‰ œ 32
27 at x œ 3 ; local minimum value of h(0) œ 0 at x œ 0, no absolute extrema
22. (a) h(x) œ 2x$ 18x Ê hw (x) œ 6x# 18 œ 6 Šx È3‹ Šx È3‹ Ê critical points at x œ „ È3
Ê hw œ | | , increasing on Š_ß È3‹ and ŠÈ$ß _‹ , decreasing on ŠÈ$ß È3‹
È$
È $
(b) a local maximum is h ŠÈ3‹ œ 12È3 at x œ È3; local minimum is h ŠÈ3‹ œ 12È3 at x œ È3, no absolute
extrema
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 4.3 Monotonic Functions and the First Derivative Test
23. (a) f()) œ 3)# 4)$ Ê f w ()) œ 6) 12)# œ 6)(1 2)) Ê critical points at ) œ 0,
"
#
187
Ê f w œ ± ± ,
!
"Î#
increasing on ˆ0ß "# ‰ , decreasing on (_ß !) and ˆ "# ß _‰
(b) a local maximum is f ˆ "# ‰ œ 4" at ) œ #" , a local minimum is f(0) œ 0 at ) œ 0, no absolute extrema
24. (a) f()) œ 6) )$ Ê f w ()) œ 6 3)# œ 3 ŠÈ2 )‹ ŠÈ2 )‹ Ê critical points at ) œ „ È2 Ê
f w œ ± ± , increasing on ŠÈ2ß È2‹, decreasing on Š_ß È2‹ and ŠÈ2ß _‹
È#
È #
(b) a local maximum is f ŠÈ2‹ œ 4È2 at ) œ È2, a local minimum is f ŠÈ2‹ œ %È2 at ) œ È2, no absolute
extrema
25. (a) f(r) œ 3r$ 16r Ê f w (r) œ 9r# 16 Ê no critical points Ê f w œ , increasing on (_ß _), never
decreasing
(b) no local extrema, no absolute extrema
26. (a) h(r) œ (r 7)$ Ê hw (r) œ 3(r 7)# Ê a critical point at r œ 7 Ê hw œ ± , increasing on
(
(_ß 7) ((ß _), never decreasing
(b) no local extrema, no absolute extrema
27. (a) f(x) œ x% 8x# 16 Ê f w (x) œ 4x$ 16x œ 4x(x 2)(x 2) Ê critical points at x œ 0 and x œ „ 2
Ê f w œ ± ± ± , increasing on (#ß !) and (#ß _), decreasing on (_ß 2) and (!ß #)
#
!
#
(b) a local maximum is f(0) œ 16 at x œ 0, local minima are f a „ 2b œ 0 at x œ „ 2, no absolute maximum; absolute
minimum is 0 at x œ „ 2
28. (a) g(x) œ x% 4x$ 4x# Ê gw (x) œ 4x$ 12x# )x œ 4x(x 2)(x 1) Ê critical points at x œ 0, 1, 2
Ê gw œ ± ± ± , increasing on (0ß 1) and (#ß _), decreasing on (_ß 0) and (1ß #)
!
"
#
(b) a local maximum is g(1) œ 1 at x œ 1, local minima are g(0) œ 0 at x œ 0 and g(2) œ 0 at x œ 2, no absolute
maximum; absolute minimum is 0 at x œ 0, 2
29. (a) H(t) œ
3 %
# t
t' Ê Hw (t) œ 6t$ 6t& œ 6t$ (1 t)(" t) Ê critical points at t œ 0, „ 1
Ê Hw œ ± ± ± , increasing on (_ß 1) and (0ß 1), decreasing on ("ß 0) and ("ß _)
"
!
"
(b) the local maxima are H(1) œ "# at t œ 1 and H(1) œ "# at t œ 1, the local minimum is H(0) œ 0 at t œ 0, absolute
maximum is
"
#
at t œ „ 1; no absolute minimum
30. (a) K(t) œ 15t$ t& Ê Kw (t) œ 45t# 5t% œ 5t# (3 t)(3 t) Ê critical points at t œ 0, „ 3
Ê Kw œ ± ± ± , increasing on (3ß 0) (0ß 3), decreasing on (_ß 3) and (3ß _)
$
!
$
(b) a local maximum is K(3) œ 162 at t œ 3, a local minimum is K(3) œ 162 at t œ 3, no absolute extrema
31. (a) f(x) œ x 6Èx 1 Ê f w (x) œ 1 3
Èx 1
œ
Èx 1 3
Èx 1
Ê critical points at x œ 1 and x œ 10
Ê f w œ ( ± , increasing on a10, _b, decreasing on a1, 10b
1
10
(b) a local minimum is fa10b œ 8, a local and absolute maximum is fa1b œ 1, absolute minimum of 8 at x œ 10
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
188
Chapter 4 Applications of Derivatives
32. (a) g(x) œ 4Èx x2 3 Ê gw (x) œ
2
Èx
2 2x3Î2
Èx
2x œ
Ê critical points at x œ 1 and x œ 0
Ê g w œ Ð ± , increasing on a0, 1b, decreasing on a1, _b
0
1
(b) a local minimum is fa0b œ 3, a local maximum is fa1b œ 6, absolute maximum of 6 at x œ 1
33. (a) g(x) œ xÈ8 x# œ x a8 x# b
"Î#
Ê gw (x) œ a8 x# b
"Î#
x ˆ "# ‰ a) x# b
"Î#
(2x) œ
2(2 x)(2 x)
ÊŠ2È2 x‹ Š2È2 x‹
Ê critical points at x œ „ 2, „ 2È2 Ê gw œ ( ± ± Ñ
, increasing on (#ß #), decreasing on
#
#
#È#
#È #
Š#È2ß #‹ and Š#ß #È2‹
(b) local maxima are g(2) œ 4 at x œ 2 and g Š2È2‹ œ 0 at x œ 2È2, local minima are g(2) œ 4 at
x œ 2 and g Š2È2‹ œ 0 at x œ 2È2, absolute maximum is 4 at x œ 2; absolute minimum is 4 at x œ 2
34. (a) g(x) œ x# È5 x œ x# (5 x)"Î# Ê gw (x) œ 2x(5 x)"Î# x# ˆ "# ‰ (5 x)"Î# (1) œ
5x(4x)
2 È 5 x
Ê critical points at
x œ 0, 4 and 5 Ê gw œ ± ± Ñ , increasing on (0ß 4), decreasing on (_ß !) and (%ß &)
&
!
%
(b) a local maximum is g(4) œ 16 at x œ 4, a local minimum is 0 at x œ 0 and x œ 5, no absolute maximum; absolute
minimum is 0 at x œ 0, 5
35. (a) f(x) œ
x# 3
x#
Ê f w (x) œ
2x(x 2) ax# 3b (1)
(x 2)#
œ
(x 3)(x ")
(x #)#
Ê critical points at x œ 1, 3
Ê f w œ ± )( ± , increasing on (_ß 1) and ($ß _), decreasing on ("ß #) and (#ß $),
#
"
$
discontinuous at x œ 2
(b) a local maximum is f(1) œ 2 at x œ 1, a local minimum is f(3) œ 6 at x œ 3, no absolute extrema
36. (a) f(x) œ
x$
3x# 1
Ê f w (x) œ
3x# a3x# 1b x$ (6x)
a3x# 1b#
œ
3x# ax# 1b
a3x# 1b#
Ê a critical point at x œ 0
Ê f w œ ± , increasing on (_ß !) (!ß _), and never decreasing
!
(b) no local extrema, no absolute extrema
37. (a) f(x) œ x"Î$ (x 8) œ x%Î$ 8x"Î$ Ê f w (x) œ
4
3
x"Î$ 83 x#Î$ œ
w
4(x 2)
3x#Î$
Ê critical points at x œ 0, 2
Ê f œ ± )( , increasing on (#ß !) (!ß _), decreasing on (_ß 2)
!
#
(b) no local maximum, a local minimum is f(2) œ 6 $È2 ¸ 7.56 at x œ 2, no absolute maximum; absolute
minimum is 6 $È2 at x œ 2
38. (a) g(x) œ x#Î$ (x 5) œ x&Î$ 5x#Î$ Ê gw (x) œ
5
3
x#Î$ 10
3
x"Î$ œ
5(x 2)
$
3È
x
Ê critical points at x œ 2 and
w
x œ 0 Ê g œ ± )( , increasing on (_ß 2) and (!ß _), decreasing on (2ß !)
!
#
(b) local maximum is g(2) œ 3 $È4 ¸ 4.762 at x œ 2, a local minimum is g(0) œ 0 at x œ 0, no absolute extrema
39. (a) h(x) œ x"Î$ ax# 4b œ x(Î$ 4x"Î$ Ê hw (x) œ
x œ 0,
„2
È7
7
3
x%Î$ 43 x#Î$ œ
ŠÈ7x 2‹ ŠÈ7x #‹
$ #
3È
x
Ê critical points at
2
Ê hw œ ±
)( ± , increasing on Š_ß È
‹ and Š È27 ß _‹ , decreasing on
7
!
#ÎÈ(
#ÎÈ(
2
ß !‹ and Š!ß È27 ‹
ŠÈ
7
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 4.3 Monotonic Functions and the First Derivative Test
2
(b) local maximum is h Š È
‹œ
7
$
24 È
2
7(Î'
¸ 3.12 at x œ
2
È7 ,
the local minimum is h Š È27 ‹ œ $
24 È
2
7(Î'
189
¸ 3.12, no absolute
extrema
40. (a) k(x) œ x#Î$ ax# 4b œ x)Î$ 4x#Î$ Ê kw (x) œ
8
3
x&Î$ 83 x"Î$ œ
8(x 1)(x 1)
$
3È
x
Ê critical points at
w
x œ 0, „ 1 Ê k œ ± )( ± , increasing on ("ß !) and ("ß _), decreasing on (_ß 1)
!
"
"
and (!ß 1)
(b) local maximum is k(0) œ 0 at x œ 0, local minima are k a „ 1b œ 3 at x œ „ 1, no absolute maximum; absolute
minimum is 3 at x œ „ 1
41. (a) f(x) œ 2x x# Ê f w (x) œ 2 2x Ê a critical point at x œ 1 Ê f w œ ± ] and fa1b œ 1 and fa2b œ 0
2
1
a local maximum is 1 at x œ 1, a local minimum is 0 at x œ 2.
(b) There is an absolute maximum of 1 at x œ 1; no absolute minimum.
(c)
42. (a) f(x) œ (x 1)# Ê f w (x) œ 2(x 1) Ê a critical point at x œ 1 Ê f w œ ± ] and f(1) œ 0, f(0) œ 1
!
"
Ê a local maximum is 1 at x œ 0, a local minimum is 0 at x œ 1
(b) no absolute maximum; absolute minimum is 0 at x œ 1
(c)
43. (a) g(x) œ x# 4x 4 Ê gw (x) œ 2x 4 œ 2(x 2) Ê a critical point at x œ 2 Ê gw œ [ ± and
"
#
g(1) œ 1, g(2) œ 0 Ê a local maximum is 1 at x œ 1, a local minimum is g(2) œ 0 at x œ 2
(b) no absolute maximum; absolute minimum is 0 at x œ 2
(c)
44. (a) g(x) œ x# 6x 9 Ê gw (x) œ 2x 6 œ 2(x 3) Ê a critical point at x œ 3 Ê gw œ [ ± and
%
$
g(4) œ 1, g(3) œ 0 Ê a local maximum is 0 at x œ 3, a local minimum is 1 at x œ 4
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
190
Chapter 4 Applications of Derivatives
(b) absolute maximum is 0 at x œ 3; no absolute minimum
(c)
45. (a) f(t) œ 12t t$ Ê f w (t) œ 12 3t# œ 3(2 t)(2 t) Ê critical points at t œ „ 2 Ê f w œ [ ± ± $
#
#
and f(3) œ 9, f(2) œ 16, f(2) œ 16 Ê local maxima are 9 at t œ 3 and 16 at t œ 2, a local minimum is
16 at t œ 2
(b) absolute maximum is 16 at t œ 2; no absolute minimum
(c)
46. (a) f(t) œ t$ 3t# Ê f w (t) œ 3t# 6t œ 3t(t 2) Ê critical points at t œ 0 and t œ 2
Ê f w œ ± ± ] and f(0) œ 0, f(2) œ 4, f(3) œ 0 Ê a local maximum is 0 at t œ 0 and t œ 3, a
$
!
#
local minimum is 4 at t œ 2
(b) absolute maximum is 0 at t œ 0, 3; no absolute minimum
(c)
x$
3
2x# 4x Ê hw (x) œ x# 4x 4 œ (x 2)# Ê a critical point at x œ 2 Ê hw œ [ ± and
!
#
h(0) œ 0 Ê no local maximum, a local minimum is 0 at x œ 0
(b) no absolute maximum; absolute minimum is 0 at x œ 0
(c)
47. (a) h(x) œ
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 4.3 Monotonic Functions and the First Derivative Test
191
48. (a) k(x) œ x$ 3x# 3x 1 Ê kw (x) œ 3x# 6x 3 œ 3(x 1)# Ê a critical point at x œ 1
Ê kw œ ± ] and k(1) œ 0, k(0) œ 1 Ê a local maximum is 1 at x œ 0, no local minimum
!
"
(b) absolute maximum is 1 at x œ 0; no absolute minimum
(c)
49. (a) faxb œ È25 x2 Ê f w axb œ
x
È25 x2
Ê critical points at x œ 0, x œ 5, and x œ 5
w
Ê f œ Ð ± Ñ , fa5b œ 0, fa0b œ 5, fa5b œ 0 Ê local maximum is 5 at x œ 0; local minimum of 0 at
5
5
0
x œ 5 and x œ 5
(b) absolute maximum is 5 at x œ 0; absolute minimum of 0 at x œ 5 and x œ 5
(c)
50. (a) faxb œ Èx2 2x 3, 3 Ÿ x _ Ê f w axb œ
2x 2
Èx2 2x 3
Ê only critical point in 3 Ÿ x _ is at x œ 3
w
Ê f œ [ , fa3b œ 0 Ê local minimum of 0 at x œ 3, no local maximum
3
(b) absolute minimum of 0 at x œ 3, no absolute maximum
(c)
51. (a) gaxb œ
x2
x2 1 ,
0 Ÿ x 1 Ê g w axb œ
x2 4x 1
ax 2 1 b 2
Ê only critical point in 0 Ÿ x 1 is x œ 2 È3 ¸ 0.268
È
Ê g w œ [ l ), gŠ2 È3‹ œ 4È3 3 6 ¸ 1.866 Ê local minimum of
0
1
0.268
maximum at x œ 0.
È
(b) absolute minimum of 4È3 3 6 at x œ 2 È3, no absolute maximum
È3
4È 3 6
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
at x œ 2 È3, local
192
Chapter 4 Applications of Derivatives
(c)
52. (a) gaxb œ
x2
4 x2 ,
2 x Ÿ 1 Ê g w axb œ
Ê only critical point in 2 x Ÿ 1 is x œ 0
8x
a4 x 2 b 2
w
Ê g œ ( l ], ga0b œ 0 Ê local minimum of 0 at x œ 0, local maximum of
1
2
0
(b) absolute minimum of 0 at x œ 0, no absolute maximum
(c)
1
3
at x œ 1.
53. (a) faxb œ sin 2x, 0 Ÿ x Ÿ 1 Ê f w axb œ 2cos 2x, f w (x) œ 0 Ê cos 2x œ 0 Ê critical points are x œ 14 and x œ 341
Ê f w œ [ ± ± ] , fa0b œ 0, fˆ 14 ‰ œ 1, fˆ 341 ‰ œ 1, fa1b œ 0 Ê local maxima are 1 at x œ 14 and 0
1
1
31
0
4
4
at x œ 1, and local minima are 1 at x œ
w
31
4
and 0 at x œ 0.
(b) The graph of f rises when f 0, falls when f w 0,
and has local extreme values where f w œ 0. The function
f has a local minimum value at x œ 0 and x œ 341 , where
the values of f w change from negative to positive. The
function f has a local maximum value at x œ 1 and
x œ 14 , where the values of f w change from positive to
negative.
54. (a) faxb œ sin x cos x, 0 Ÿ x Ÿ 21 Ê f w axb œ cos x sin x, f w (x) œ 0 Ê tan x œ 1 Ê critical points are x œ 341 and
x œ 741 Ê f w œ [ ± ± ] , fa0b œ 1, fˆ 341 ‰ œ È2, fˆ 741 ‰ œ È2, fa21b œ 1 Ê local maxima are
71
31
0
21
È2 at x œ
4
4
31
4
and 1 at x œ 21, and local minima are È2 at x œ
w
w
71
4
and 1 at x œ 0.
(b) The graph of f rises when f 0, falls when f 0,
and has local extreme values where f w œ 0. The function
f has a local minimum value at x œ 0 and x œ 741 , where
the values of f w change from negative to positive. The
function f has a local maximum value at x œ 21 and
x œ 341 , where the values of f w change from positive to
negative.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 4.3 Monotonic Functions and the First Derivative Test
55. (a) faxb œ È3cos x sin x, 0 Ÿ x Ÿ 21 Ê f w axb œ È3sin x cos x, f w (x) œ 0 Ê tan x œ
xœ
1
6
and x œ
71
6
1
È3
193
Ê critical points are
Ê f w œ [ ± ± ] , fa0b œ È3, fˆ 16 ‰ œ 2, fˆ 761 ‰ œ 2, fa21b œ È3 Ê local
1
71
0
21
6
6
maxima are 2 at x œ
1
6
and È3 at x œ 21, and local minima are 2 at x œ
w
w
71
6
and È3 at x œ 0.
(b) The graph of f rises when f 0, falls when f 0,
and has local extreme values where f w œ 0. The function
f has a local minimum value at x œ 0 and x œ 761 , where
the values of f w change from negative to positive. The
function f has a local maximum value at x œ 21 and
x œ 16 , where the values of f w change from positive to
negative.
56. (a) faxb œ 2x tan x, 12 x x œ 14 and x œ
maximum is
1
2
1
4
1
2
Ê f w axb œ 2 sec2 x, f w (x) œ 0 Ê sec2 x œ 2 Ê critical points are
Ê f w œ ( ± ± ) , fˆ 14 ‰ œ 12 1, fˆ 14 ‰ œ 1 12 Ê local
1
1
12
14
2
4
1 at x œ 14 , and local minimum is 1 (b) The graph of f rises when f w 0, falls when f w 0,
and has local extreme values where f w œ 0. The function
f has a local minimum value at x œ 14 , where the values
1
2
at x œ 14 .
of f w change from negative to positive. The function f
has a local maximum value at x œ 14 , where the values
of f w change from positive to negative.
57. (a) f(x) œ
x
#
2 sin ˆ x# ‰ Ê f w (x) œ
Ê f w œ [ ± ]
!
#1
#1Î$
cos ˆ x# ‰ , f w (x) œ 0 Ê cos ˆ x# ‰ œ "# Ê a critical point at x œ 231
and f(0) œ 0, f ˆ 231 ‰ œ 13 È3, f(21) œ 1 Ê local maxima are 0 at x œ 0 and 1
"
#
at x œ 21, a local minimum is 13 È3 at x œ 231
(b) The graph of f rises when f w 0, falls when f w 0,
and has a local minimum value at the point where f w
changes from negative to positive.
58. (a) f(x) œ 2 cos x cos# x Ê f w (x) œ 2 sin x 2 cos x sin x œ 2(sin x)(1 cos x) Ê critical points at x œ 1, 0, 1
Ê f w œ [ ± ] and f(1) œ 1, f(0) œ 3, f(1) œ 1 Ê a local maximum is 1 at x œ „ 1, a local
1
1
!
minimum is 3 at x œ 0
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
194
Chapter 4 Applications of Derivatives
(b) The graph of f rises when f w 0, falls when f w 0,
and has local extreme values where f w œ 0. The
function f has a local minimum value at x œ 0, where
the values of f w change from negative to positive.
59. (a) f(x) œ csc# x 2 cot x Ê f w (x) œ 2(csc x)(csc x)(cot x) 2 acsc# xb œ 2 acsc# xb (cot x 1) Ê a critical
point at x œ 14 Ê f w œ ( ± ) and f ˆ 14 ‰ œ 0 Ê no local maximum, a local minimum is 0 at x œ 14
1
!
1Î%
w
(b) The graph of f rises when f 0, falls when f w 0,
and has a local minimum value at the point where
f w œ 0 and the values of f w change from negative to
positive. The graph of f steepens as f w (x) Ä „ _.
60. (a) f(x) œ sec# x 2 tan x Ê f w (x) œ 2(sec x)(sec x)(tan x) 2 sec# x œ a2 sec# xb (tan x 1) Ê a critical point
at x œ 14 Ê f w œ ( ± ) and f ˆ 14 ‰ œ 0 Ê no local maximum, a local minimum is 0 at x œ 14
1Î#
1Î#
1Î%
w
(b) The graph of f rises when f 0, falls when f w 0,
and has a local minimum value where f w œ 0 and the
values of f w change from negative to positive.
61. h()) œ 3 cos ˆ #) ‰ Ê hw ()) œ 3# sin ˆ #) ‰ Ê hw œ [ ] , (!ß $) and (#1ß 3) Ê a local maximum is 3 at ) œ 0,
!
#1
a local minimum is 3 at ) œ 21
62. h()) œ 5 sin ˆ #) ‰ Ê hw ()) œ
minimum is 0 at ) œ 0
63. (a)
5
#
cos ˆ #) ‰ Ê hw œ [ ] , (!ß 0) and (1ß 5) Ê a local maximum is 5 at ) œ 1, a local
1
!
(b)
(c)
(d)
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 4.3 Monotonic Functions and the First Derivative Test
64. (a)
(b)
(c)
(d)
65. (a)
(b)
66. (a)
(b)
195
67. The function faxb œ x sinˆ 1x ‰ has an infinite number of local maxima and minima. The function sin x has the following
properties: a) it is continuous on a_, _b; b) it is periodic; and c) its range is Ò1, 1Ó. Also, for a 0, the function
a range of Ð_, aÓ Òa, _Ñ on
’ 1a , 1a “.
In particular, if a œ 1, then
1
x
Ÿ 1 or
1
x
1
x
has
1 when x is in Ò1, 1Ó. This means
sinˆ 1x ‰ takes on the values of 1 and 1 infinitely many times in times on the interval Ò1, 1Ó, which occur when
1
1
31
51
2
2
2
ˆ1‰
x œ „ 2 , „ 2 , „ 2 ,. . . . Ê x œ „ 1 , „ 31 , „ 51 , . . . . Thus sin x has infinitely many local maxima and minima
in the interval Ò1, 1Ó. On the interval Ò0, 1Ó, 1 Ÿ sinˆ 1x ‰ Ÿ 1 and since x 0 we have x Ÿ x sinˆ 1x ‰ Ÿ x. On the
interval Ò1, 0Ó, 1 Ÿ sinˆ 1x ‰ Ÿ 1 and since x 0 we have x x sinˆ 1x ‰ x. Thus faxb is bounded by the lines y œ x
and y œ x. Since sinˆ 1x ‰ oscillates between 1 and 1 infinitely many times on Ò1, 1Ó then f will oscillate between y œ x
and y œ x infinitely many times. Thus f has infinitely many local maxima and minima. We can see from the graph (and
verify later in Chapter 7) that lim x sinˆ 1x ‰ œ 1 and lim x sinˆ 1x ‰ œ 1. The graph of f does not have any absolute
xÄ_
xÄ_
maxima., but it does have two absolute minima.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
196
Chapter 4 Applications of Derivatives
68. f(x) œ a x# b x c œ a ˆx# ba x‰ c œ a Šx# ba x b#
4a# ‹
b#
4a
c œ a ˆx b ‰#
2a
b# 4ac
4a
ˆ 2ab ß _‰
, a parabola whose
vertex is at x œ . Thus when a 0, f is increasing on
and decreasing on ˆ_ß
when a 0,
b
b
w
f is increasing on ˆ_ß #a ‰ and decreasing on ˆ #a ß _‰ . Also note that f (x) œ 2ax b œ 2a ˆx #ba ‰ Ê for
b
2a
b ‰
#a ;
a 0, f w œ | ; for a 0, f w œ ± .
bÎ2a
bÎ2a
69. f(x) œ a x# b x Ê f w axb œ 2a x b, fa1b œ 2 Ê a b œ 2, f w a1b œ 0 Ê 2a b œ 0 Ê a œ 2, b œ 4
Ê f(x) œ 2x# 4x
70. f(x) œ a x3 b x# c x d Ê f w axb œ 3a x2 2b x c, fa0b œ 0 Ê d œ 0, fa1b œ 1 Ê a b c d œ 1,
f w a0b œ 0 Ê c œ 0, f w a1b œ 0 Ê 3a 2b c œ 0 Ê a œ 2, b œ 3, c œ 0, d œ 0 Ê f(x) œ 2x3 3x#
4.4 CONCAVITY AND CURVE SKETCHING
1. y œ
x$
3
x#
#
2. y œ
x%
4
2x# 4 Ê yw œ x$ 4x œ x ax# 4b œ x(x 2)(x 2) Ê yww œ 3x# 4 œ ŠÈ3x 2‹ ŠÈ3x 2‹ . The
Ê yw œ x# x 2 œ (x 2)(x 1) Ê yww œ 2x 1 œ 2 ˆx "# ‰ . The graph is rising on
(_ß 1) and (#ß _), falling on ("ß #), concave up on ˆ "# ß _‰ and concave down on ˆ_ß "# ‰ . Consequently,
a local maximum is 3# at x œ 1, a local minimum is 3 at x œ 2, and ˆ "# ß 34 ‰ is a point of inflection.
2x "
3
graph is rising on (2ß 0) and (#ß _), falling on (_ß #) and (!ß #), concave up on Š_ß È23 ‹ and Š È23 ß _‹ and
concave down on Š È23 ß È23 ‹ . Consequently, a local maximum is 4 at x œ 0, local minima are 0 at x œ „ 2, and
2
16
Š È23 ß 16
9 ‹ and Š È3 ß 9 ‹ are points of inflection.
3
4
ax# 1b
#Î$
Ê yw œ ˆ 34 ‰ ˆ 23 ‰ ax# 1b
"Î$
minima are 0 at x œ „ 1; yww œ ax# 1b
"Î$
(2x) œ x ax# 1b
"Î$
, yw œ ) ( ± )( "
"
!
Ê the graph is rising on ("ß !) and ("ß _), falling on (_ß ") and (!ß ") Ê a local maximum is 34 at x œ 0, local
3. y œ
(x) ˆ 3" ‰ ax# 1b
%Î$
(2x) œ
x # 3
3
$
É
ax # 1 b %
,
yww œ ± ) ( )( ± Ê the graph is concave up on Š_ß È3‹ and ŠÈ3ß _‹, concave
"
"
È$
È $
$
È
down on ŠÈ3ß È3‹ Ê points of inflection at Š „ È3ß $ % ‹
%
x#Î$ ax# 1b, yw œ ± )( ± !
"
"
Ê the graph is rising on (_ß 1) and ("ß _), falling on (1ß ") Ê a local maximum is 27
7 at x œ 1, a local
4. y œ
9
14
x"Î$ ax# 7b Ê yw œ
3
14
x#Î$ ax# 7b 9
14
x"Î$ (2x) œ
3
#
ww
&Î$
minimum is 27
ax# 1b 3x"Î$ œ 2x"Î$ x&Î$ œ x&Î$ a2x# 1b ,
7 at x œ 1; y œ x
yww œ )( Ê the graph is concave up on (!ß _), concave down on (_ß !) Ê a point of inflection at (!ß !).
!
5. y œ x sin 2x Ê yw œ 1 2 cos 2x, yw œ [ ± ± ] Ê the graph is rising on ˆ 13 ß 13 ‰ , falling
#1Î$ 1Î$
#1Î$
1Î$
on ˆ #31 ß 13 ‰ and ˆ 13 ß 231 ‰ Ê local maxima are 231 13 È3
#
at x œ 13 and
21
3
È3
#
È3
#
at x œ 231 and
1
3
È3
#
at x œ
1
3
21
3
, local minima are
; yww œ 4 sin 2x, yww œ [ ± ± ± ]
Ê the
!
#1Î$ 1Î#
#1Î$
1Î#
graph is concave up on ˆ 1# ß !‰ and ˆ 1# ß 231 ‰ , concave down on ˆ 231 ß 1# ‰ and ˆ!ß 1# ‰ Ê points of inflection at
at x œ
ˆ 1# ß 1# ‰ , (!ß !), and ˆ 1# ß 1# ‰
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 4.4 Concavity and Curve Sketching
197
6. y œ tan x 4x Ê yw œ sec# x 4, yw œ ( ± ± )
Ê the graph is rising on ˆ 12 ß 13 ‰ and
1
1Î# 1Î$
Î#
1Î$
ˆ 13 ß 1# ‰ , falling on ˆ 13 ß 13 ‰ Ê a local maximum is È3 431 at x œ 13 , a local minimum is È3 431 at x œ 13 ;
yww œ 2(sec x)(sec x)(tan x) œ 2 asec# xb (tan x), yww œ ( ± ) Ê the graph is concave up on ˆ0ß 1# ‰ ,
!
1Î#
1Î#
1
ˆ
‰
concave down on 2 ß 0 Ê a point of inflection at (0ß 0)
7. If x 0, sin kxk œ sin x and if x 0, sin kxk œ sin (x)
œ sin x. From the sketch the graph is rising on
ˆ 3#1 ß 1# ‰ , ˆ!ß 1# ‰ and ˆ 3#1 ß #1‰ , falling on ˆ21ß 3#1 ‰ ,
ˆ 1# ß !‰ and ˆ 1# ß 3#1 ‰ ; local minima are 1 at x œ „ 3#1
and 0 at x œ !; local maxima are 1 at x œ „
1
#
and 0 at
x œ „ #1; concave up on (#1ß 1) and (1ß #1), and
concave down on (1ß 0) and (!ß 1) Ê points of inflection
are (1ß !) and (1ß !)
8. y œ 2 cos x È2 x Ê yw œ 2 sin x È2, yw œ [ ±
± ± ] Ê rising on
1
$1Î#
$1Î%
1Î%
&1Î%
ˆ 341 ß 14 ‰and ˆ 541 ß 3#1 ‰ , falling on ˆ1ß 341 ‰ and ˆ 14 ß 541 ‰ Ê local maxima are 2 1È2 at x œ 1, È2 È2
at x œ 14 and 31#
at x œ
31
# ,
È
at x œ 341 and È2 514 2 at x œ 541 ;
Ê concave up on ˆ1ß 1# ‰ and ˆ 1# ß 3#1 ‰ , concave down on
and local minima are È2 yww œ 2 cos x, yww œ [ ± ± ]
1
$1Î#
1Î#
1Î#
ˆ 1# ß 1# ‰ Ê points of inflection at Š 1# ß
È 21
# ‹
and Š 1# ß 31 È 2
4
1È2
4
È 21
# ‹
9. When y œ x# 4x 3, then yw œ 2x 4 œ 2(x 2) and
yww œ 2. The curve rises on (#ß _) and falls on (_ß #).
At x œ 2 there is a minimum. Since yww 0, the curve is
concave up for all x.
10. When y œ ' 2x x# , then yw œ # 2x œ 2(" x) and
yww œ 2. The curve rises on (_ß 1) and falls on
(1ß _). At x œ 1 there is a maximum. Since yw w 0, the
curve is concave down for all x.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
198
Chapter 4 Applications of Derivatives
11. When y œ x$ 3x 3, then yw œ 3x# 3 œ 3(x 1)(x 1)
and yww œ 6x. The curve rises on (_ß 1) ("ß _) and
falls on (1ß 1). At x œ 1 there is a local maximum and at
x œ 1 a local minimum. The curve is concave down on
(_ß 0) and concave up on (!ß _). There is a point of
inflection at x œ 0.
12. When y œ x(6 2x)# , then yw œ 4x(6 2x) (' 2x)#
œ 12(3 x)(" x) and yww œ 12(3 x) 12(" x)
œ 24(x 2). The curve rises on (_ß ") ($ß _) and
falls on ("ß $). The curve is concave down on (_ß #) and
concave up on (#ß _). At x œ 2 there is a point of
inflection.
13. When y œ 2x$ 6x# 3, then yw œ 6x# 12x
œ 6x(x 2) and yww œ 12x 12 œ 12(x 1). The
curve rises on (!ß #) and falls on (_ß 0) and (#ß _).
At x œ 0 there is a local minimum and at x œ 2 a local
maximum. The curve is concave up on (_ß ") and
concave down on ("ß _). At x œ 1 there is a point of
inflection.
14. When y œ 1 9x 6x# x$ , then yw œ 9 12x 3x#
œ $(x 3)(B 1) and yww œ 12 6x œ 6(x 2).
The curve rises on ($ß ") and falls on (_ß 3) and
("ß _). At x œ 1 there is a local maximum and at
x œ 3 a local minimum. The curve is concave up on
(_ß 2) and concave down on (#ß _). At x œ 2
there is a point of inflection.
15. When y œ (x 2)$ 1, then yw œ 3(x 2)# and
yww œ 6(x 2). The curve never falls and there are no
local extrema. The curve is concave down on (_ß #)
and concave up on (#ß _). At x œ 2 there is a point
of inflection.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 4.4 Concavity and Curve Sketching
16. When y œ 1 (x 1)$ , then yw œ 3(x 1)# and
yww œ 6(x 1). The curve never rises and there are
no local extrema. The curve is concave up on (_ß 1)
and concave down on ("ß _). At x œ 1 there is a
point of inflection.
17. When y œ x% 2x# , then yw œ 4x$ 4x œ 4x(x 1)(x 1)
and yww œ 12x# 4 œ 12 Šx "
È 3 ‹ Šx
"
È3 ‹ .
The curve
rises on ("ß !) and ("ß _) and falls on (_ß 1) and (!ß ").
At x œ „ 1 there are local minima and at x œ 0 a local
maximum. The curve is concave up on Š_ß È"3 ‹ and
Š È"3 ß _‹ and concave down on Š È"3 ß È"3 ‹ . At x œ
„"
È3
there are points of inflection.
18. When y œ x% 6x# 4, then yw œ 4x$ 12x
œ 4x Šx È3‹ Šx È3‹ and yww œ 12x# 12
œ 12(x 1)(x 1). The curve rises on Š_ß È3‹
and Š!ß È3‹ , and falls on ŠÈ3ß !‹ and ŠÈ3ß _‹ . At
x œ „ È3there are local maxima and at x œ 0 a local
minimum. The curve is concave up on ("ß ") and concave
down on (_ß 1) and ("ß _). At x œ „ 1 there are points
of inflection.
19. When y œ 4x$ x% , then yw œ 12x# 4x$ œ 4x# ($ x) and
yww œ 24x 12x# œ 12x(2 x). The curve rises on a_ß $b
and falls on a$ß _b. At x œ 3 there is a local maximum, but
there is no local minimum. The graph is concave up on
a!ß #b and concave down on a_ß !b and a#ß _b. There are
inflection points at x œ 0 and x œ 2.
20. When y œ x% 2x$ , then yw œ 4x$ 6x# œ 2x# (2x 3) and
yww œ 12x# 12x œ 12x(x 1). The curve rises on
ˆ 3# ß _‰ and falls on ˆ_ß 32 ‰ . There is a local
minimum at x œ 3# , but no local maximum. The curve is
concave up on (_ß 1) and (!ß _), and concave down on
(1ß 0). At x œ 1 and x œ 0 there are points of inflection.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
199
200
Chapter 4 Applications of Derivatives
21. When y œ x& 5x% , then yw œ 5x% 20x$ œ 5x$ (x 4) and
yww œ 20x$ 60x# œ 20x# (x 3). The curve rises on
(_ß !) and (%ß _), and falls on (!ß %). There is a local
maximum at x œ 0, and a local minimum at x œ 4. The
curve is concave down on (_ß 3) and concave up on
(3ß _). At x œ 3 there is a point of inflection.
%
%
$
22. When y œ x ˆ x# 5‰ , then yw œ ˆ x# 5‰ x(4)ˆ x# 5‰ ˆ "# ‰
$
ww
‰
ˆx
‰# ˆ "# ‰ ˆ 5x
‰
œ ˆ x# 5‰ ˆ 5x
# 5 , and y œ 3 # 5
# 5
$
#
ˆ x# 5‰ ˆ 5# ‰ œ 5 ˆ x# 5‰ (x 4). The curve is rising
on (_ß #) and (10ß _), and falling on (#ß 10). There is a
local maximum at x œ 2 and a local minimum at x œ 10.
The curve is concave down on (_ß %) and concave up on
(%ß _). At x œ 4 there is a point of inflection.
23. When y œ x sin x, then yw œ " cos x and yww œ sin x.
The curve rises on (!ß 21). At x œ 0 there is a local and
absolute minimum and at x œ 21 there is a local and absolute
maximum. The curve is concave down on (!ß 1) and concave
up on (1ß #1). At x œ 1 there is a point of inflection.
24. When y œ x sin x, then yw œ " cos x and yww œ sin x.
The curve rises on (!ß 21). At x œ 0 there is a local and
absolute minimum and at x œ 21 there is a local and absolute
maximum. The curve is concave up on (!ß 1) and concave
down on (1ß #1). At x œ 1 there is a point of inflection.
25. When y œ È3x 2 cos x, then yw œ È3 2 sin x and
yww œ 2 cos x. The curve is increasing on ˆ!ß 431 ‰ and
ˆ 531 ß 21‰, and decreasing on ˆ 431 ß 531 ‰. At x œ 0 there
is a local and absolute minimum, at x œ
maximum, at x œ
51
3
41
3
there is a local
there is a local minimum, and and at
x œ 21 there is a local and absolute maximum. The curve
is concave up on ˆ!ß 12 ‰ and ˆ 321 ß 21‰, and is concave
down onˆ 12 ß 321 ‰. At x œ 12 and x œ 321 there are points
of inflection.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 4.4 Concavity and Curve Sketching
26. When y œ 43 x tan x, then yw œ
sec2 x and
4
3
yww œ 2 sec2 x tan x. The curve is increasing on ˆ 16 ß 16 ‰,
and decreasing on ˆ 12 ß 16 ‰ and ˆ 16 ß 12 ‰. At x œ 16
there is a local minimum, at x œ
1
6
there is a local
maximum,there are no absolute maxima or absolute minima.
The curve is concave up on ˆ 12 ß 0‰, and is concave
down onˆ0ß 12 ‰. At x œ 0 there is a point of inflection.
27. When y œ sin x cos x, then yw œ sin2 x cos2 x œ cos 2x
and yww œ 2 sin 2x. The curve is increasing on ˆ0ß 14 ‰ and
ˆ 341 ß 1‰, and decreasing on ˆ 14 ß 341 ‰. At x œ 0 there is a
local minimum, at x œ
maximum, at x œ
31
4
1
4
there is a local and absolute
there is a local and absolute minimum,
and at x œ 1 there is a local maximum. The curve is concave
down on ˆ0ß 12 ‰, and is concave up onˆ 12 ß 1‰. At x œ 12
there is a point of inflection.
28. When y œ cos x È3sin x, then yw œ sin x È3cos x
and yww œ cos x È3sin x. The curve is increasing on
ˆ0ß 13 ‰ and ˆ 431 ß 21‰, and decreasing on ˆ 13 ß 431 ‰. At
x œ 0 there is a local minimum, at x œ
and absolute maximum, at x œ
41
3
1
3
there is a local
there is a local and
absolute minimum, and at x œ 21 there is a local maximum.
The curve is concave down on ˆ0ß 561 ‰ and ˆ 11'1 ß 21‰,
and is concave up on ˆ 561 ß 11'1 ‰. At x œ 561 and x œ 1161
there are points of inflection.
29. When y œ x"Î& , then yw œ
"
5
4 *Î&
x%Î& and yww œ 25
x
.
The curve rises on (_ß _) and there are no extrema.
The curve is concave up on (_ß !) and concave down
on (!ß _). At x œ 0 there is a point of inflection.
30. When y œ x#Î& , then yw œ
2
5
6 )Î&
x$Î& and yww œ 25
x
.
The curve is rising on (0ß _) and falling on (_ß !). At
x œ 0 there is a local and absolute minimum. There is
no local or absolute maximum. The curve is concave
down on (_ß !) and (!ß _). There are no points of
inflection, but a cusp exists at x œ 0.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
201
202
Chapter 4 Applications of Derivatives
31. When y œ
yww œ
x
È x2 1 ,
3x
. The
ax2 1b5Î2
then yw œ
1
ax2 1b3Î2
and
curve is increasing on a_ß _b.
There are no local or absolute extrema. The curve is
concave up on a_ß !b and concave down on a!ß _b.
At x œ 0 there is a point of inflection.
32. When y œ
È 1 x2
2x 1 ,
then yw œ
ax 2 b
a2x 1b2 È1 x2
and
4x3 12x2 7
. The curve is decreasing on
a2x 1b3 a1 x2 b3Î2
ˆ1ß "# ‰ and ˆ "# ß 1‰. There are no absolute extrrema,
yww œ
there is a local maximum at x œ 1 and a local minimum
at x œ 1. The curve is concave up on a1ß 0.92b and
ˆ "# ß 0.69‰, and concave down on ˆ0.92ß "# ‰ and
a0.69ß 1b. At x ¸ 0.92 and x ¸ 0.69 there are points of
inflection.
33. When y œ 2x 3x#Î$ , then yw œ 2 2x"Î$ and
yww œ 23 x%Î$ . The curve is rising on (_ß !) and
("ß _), and falling on (!ß "). There is a local maximum
at x œ 0 and a local minimum at x œ 1. The curve is
concave up on (_ß !) and (!ß _). There are no
points of inflection, but a cusp exists at x œ 0.
34. When y œ 5x#Î& 2x, then yw œ 2x$Î& 2 œ 2 ˆx$Î& 1‰
and yww œ 65 x)Î& . The curve is rising on (0ß 1) and
falling on (_ß 0) and ("ß _). There is a local minimum
at x œ 0 and a local maximum at x œ 1. The curve is
concave down on (_ß !) and (!ß _). There are no
points of inflection, but a cusp exists at x œ 0.
35. When y œ x#Î$ ˆ #5 x‰ œ
5 #Î$
x&Î$ , then
# x
yw œ 53 x"Î$ 53 x#Î$ œ 53 x"Î$ (1 x) and
"Î$
yww œ 59 x%Î$ 10
œ 95 x%Î$ (1 2x).
9 x
The curve is rising on (!ß ") and falling on (_ß !) and
("ß _). There is a local minimum at x œ 0 and a local
maximum at x œ 1. The curve is concave up on ˆ_ß "# ‰
and concave down on ˆ "# ß !‰ and (0ß _). There is a point
of inflection at x œ "# and a cusp at x œ 0.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 4.4 Concavity and Curve Sketching
36. When y œ x#Î$ (x 5) œ x&Î$ 5x#Î$ , then
"Î$
yw œ 53 x#Î$ 10
œ 35 x"Î$ (x 2) and
3 x
yww œ
10
9
x"Î$ 10
9
x%Î$ œ
10
9
x%Î$ (x 1). The curve
is rising on (_ß !) and (#ß _), and falling on (!ß #).
There is a local minimum at x œ 2 and a local maximum
at x œ 0. The curve is concave up on ("ß 0) and (!ß _),
and concave down on (_ß 1). There is a point of
inflection at x œ 1 and a cusp at x œ 0.
37. When y œ xÈ8 x# œ x a8 x# b
# "Î#
yw œ a 8 x b
# "Î#
œ a8 x b
œ
2x ax 12b
É a8 x # b $
, then
# "Î#
(x) ˆ "# ‰ a8 x b
a8 2x b œ
$#
(#x)
2(2 x)(2 x)
#
yww œ ˆ "# ‰a8 x# b
#
"Î#
ÊŠ2È2 x‹ Š2È2 x‹
and
(2x)a8 2x# b a8 x# b
#"
(4x)
. The curve is rising on (#ß #), and falling
on Š2È2ß 2‹ and Š#ß 2È2‹ . There are local minima
x œ 2 and x œ 2È2, and local maxima at x œ 2È2 and
x œ 2. The curve is concave up on Š2È2ß !‹ and
concave down on Š!ß #È2‹ . There is a point of inflection
at x œ 0.
38. When y œ a2 x# b
$Î#
, then yw œ ˆ 3# ‰ a2 x# b
"Î#
(2x)
œ 3xÈ2 x# œ 3xÊŠÈ2 x‹ ŠÈ2 x‹ and
yww œ (3) a2 x# b
œ
"Î#
6(" x)(1 x)
ÊŠÈ2 x‹ ŠÈ2 x‹
(3x) ˆ "# ‰ a2 x# b
"Î#
(2x)
. The curve is rising on
ŠÈ2ß !‹ and falling on Š!ß È2‹ . There is a local
maximum at x œ 0, and local minima at x œ „ È2. The
curve is concave down on ("ß ") and concave up on
ŠÈ2ß "‹ and Š"ß È2‹ . There are points of inflection at
x œ „ 1.
39. When y œ È16 x# , then yw œ
yww œ
16
a16 x# b3Î2
x
È16 x#
and
. The curve is rising on a4ß 0b and falling
on a0ß 4b. There is a local and absolute maximum at x œ 0
and local and absolute minima at x œ 4 and x œ 4. The
curve is concave down on a4ß 4b. There are no points
of inflection.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
203
204
Chapter 4 Applications of Derivatives
40. When y œ x# #x , then yw œ 2x ww
y œ2
4
x3
œ
2x3 4
x3 . The
#
x#
œ
2x3 #
x#
and
curve is falling on a_ß 0b
and a0ß 1b, and rising on a1ß _b. There is a local minimum
at x œ 1. There are no absolute maxima or absolute minima.
3
The curve is concave up on Š_ß È
2‹ and a0, _b, and
3
concave down on ŠÈ
2, 0‹ . There is a point of
3
inflection at x œ È
2.
x# 3
x # , then
(x 3)(x 1)
and
(x 2)#
41. When y œ
œ
yww œ
yw œ
2x(x 2) ax# 3b (")
(x2)#
(2x 4)(x 2)# ax# 4x 3b# (x 2)
(x 2)%
œ
2
(x 2)$
.
The curve is rising on (_ß ") and ($ß _), and falling on
("ß #) and (#ß $). There is a local maximum at x œ 1 and a
local minimum at x œ 3. The curve is concave down on
(_ß #) and concave up on (#ß _). There are no points
of inflection because x œ 2 is not in the domain.
3
42. When y œ È
x3 1, then yw œ
yww œ
2x
. The
ax3 1b5Î3
x#
ax3 1b2Î3
and
curve is risng on a_ß 1b,
a1, 0b, and a0ß _b. There is are no local or absolute
extrema. The curve is concave up on a_ß 1b and
a0, _b, and concave down on a1, 0b . There are points of
inflection at x œ 1 and x œ 0.
43. When y œ
yww œ
8x
x2 4 ,
then yw œ
16xˆx 12‰
. The
ax 2 4 b 3
2
8 ˆx 2 4 ‰
ax 2 4 b 2
and
curve is fallng on a_ß 2b
and a2ß _b, and is rising on a2ß 2b. There is a local and
absolute minimum at x œ 2, and a local and absolute
maximum at x œ 2. The curve is concave down on
Š_ß 2È3‹ and Š0, 2È3‹, and concave up on
Š2È3, 0‹and Š2È3, _‹. There are points of inflection at x œ 2È3, x œ 0, and x œ 2È3.
y œ 0 is a horizontal asymptote.
44. When y œ
yww œ
5
x4 5 ,
then yw œ
100x2 ˆx4 3‰
. The
ax 4 5 b 3
20x3
ax 4 5 b 2
and
curve is risng on a_ß 0b,
and is falling on a0ß _b. There is a local and
absolute maximum at x œ 0, and there is no local or
aboslute minimum. The curve is concave up on
4
4
4
4
4
3‹ and ŠÈ
3, _‹, and concave down on ŠÈ
3, 0‹ and Š0, È
3‹. There are points of inflection at x œ È
3
Š_ß È
4
and x œ È
3. There is a horizontal asymptote of y œ 0.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 4.4 Concavity and Curve Sketching
205
x# 1, kxk 1
, then
1 x# , kxk 1
2x, kxk 1
2, kxk "
yw œ œ
and yww œ œ
. The
2x, kxk "
#, kxk "
45. When y œ kx# 1k œ œ
curve rises on ("ß !) and ("ß _) and falls on (_ß 1)
and (0ß 1). There is a local maximum at x œ 0 and local
minima at x œ „ 1. The curve is concave up on (_ß 1)
and ("ß _), and concave down on ("ß "). There are no
points of inflection because y is not differentiable at x œ „ 1 (so there is no tangent line at those points).
Ú x# 2x, x 0
46. When y œ kx 2xk œ Û 2x x# , 0 Ÿ x Ÿ 2 , then
Ü x# 2x, x 2
Ú 2x 2, x 0
Ú 2, x 0
w
ww
2
2x,
0
x
2
y œÛ
, and y œ Û 2, 0 x 2 .
Ü 2x 2, x 2
Ü 2, x 2
#
The curve is rising on (!ß 1) and (#ß _), and falling on
(_ß !) and ("ß #). There is a local maximum at x œ 1 and local minima at x œ 0 and x œ 2. The curve is concave up
on (_ß !) and (#ß _), and concave down on (!ß #). There are no points of inflection because y is not
differentiable at x œ 0 and x œ 2 (so there is no tangent at those points).
47. When y œ Èkxk œ Èx , x 0
, then
È x , x 0
Ú
"
x $Î#
, x0
, x0
#È x
y œ Û "
and yww œ (x)4 $Î#
.
,
x
0
, x0
Ü 2 È x
4
w
Since lim c yw œ _ and lim b yw œ _ there is a
xÄ!
xÄ!
cusp at x œ 0. There is a local minimum at x œ 0, but no local maximum. The curve is concave down on (_ß !)
and (!ß _). There are no points of inflection.
48. When y œ Èkx 4k œ Ú
w
y œÛ
Ü
"
2È x 4
"
#È 4 x
, x4
Èx 4 , x 4
, then
È4 x , x 4
ww
and y œ , x4
(x 4) $Î#
4
(4 x) $Î#
4
, x4
, x4
.
Since lim c yw œ _ and lim b yw œ _ there is a cusp
xÄ4
xÄ4
at x œ 4. There is a local minimum at x œ 4, but no local
maximum. The curve is concave down on (_ß %) and (%ß _). There are no points of inflection.
49. yw œ 2 x x# œ (1 x)(# x), yw œ ± ± "
#
Ê rising on ("ß #), falling on (_ß 1) and (#ß _)
Ê there is a local maximum at x œ 2 and a local minimum
at x œ 1; yww œ 1 2x, yww œ ± "Î#
"‰
ˆ
Ê concave up on _ß # , concave down on ˆ "# ß _‰ Ê a point of inflection at x œ
"
#
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
206
Chapter 4 Applications of Derivatives
50. yw œ x# x 6 œ (x 3)(x 2), yw œ ± ± #
$
Ê rising on (_ß #) and (3ß _), falling on (2ß 3)
Ê there is a local maximum at x œ 2 and a local
minimum at x œ 3; yww œ 2x 1, yww œ ± "Î#
"
ˆ
‰
Ê concave up on # ß _ , concave down on ˆ_ß "# ‰
Ê a point of inflection at x œ
"
#
51. yw œ x(x 3)# , yw œ ± ± Ê rising on
!
$
(!ß _), falling on (_ß !) Ê no local maximum, but there
is a local minimum at x œ 0; yww œ (x 3)# x(2)(x 3)
œ 3(x 3)(x 1), yww œ ± ± Ê concave
"
$
up on (_ß ") and ($ß _), concave down on ("ß $) Ê points of inflection at x œ 1 and x œ 3
52. yw œ x# (2 x), yw œ ± ± Ê rising on
!
#
(_ß #), falling on (2ß _) Ê there is a local maximum at
x œ 2, but no local minimum; yww œ 2x(2 x) x# (1)
œ x(4 3x), yww œ ± ± Ê concave up
!
%Î$
on ˆ!ß 43 ‰ , concave down on (_ß !) and ˆ 43 ß _‰ Ê points of inflection at x œ 0 and x œ
4
3
53. yw œ x ax# 12b œ x Šx 2È3‹ Šx 2È3‹ ,
yw œ ±
± ± Ê rising on
!
#È$
#È $
Š2È3ß !‹ and Š#È3ß _‹ , falling on Š_ß #È3‹
and Š!ß #È3‹ Ê a local maximum at x œ 0, local minima at x œ „ 2È3 ; yww œ 1 ax# 12b xa2xb œ 3ax 2bax 2b,
yww œ ± ± Ê concave up on (_ß #) and (#ß _), concave down on (#ß #) Ê points of inflection
#
#
at x œ „ 2
54. yw œ (x 1)# (2x 3), yw œ ± ± "
$Î#
3
3‰
ˆ
‰
ˆ
Ê rising on # ß _ , falling on _ß # Ê no local
maximum, a local minimum at x œ 3# ;
yww œ 2(x 1)(2x 3) (x 1)# (2) œ 2(x 1)(3x 2),
yww œ ± ± Ê concave up on
"
#Î$
2
ˆ_ß 3 ‰ and ("ß _), concave down on ˆ 23 ß "‰ Ê points of inflection at x œ 23 and x œ 1
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 4.4 Concavity and Curve Sketching
207
55. yw œ a8x 5x# b (4 x)# œ x(8 5x)(% x)# ,
yw œ ± ± ± Ê rising on ˆ!ß 85 ‰ ,
!
%
)Î&
8
ˆ
falling on (_ß !) and 5 ß _‰ Ê a local maximum at
xœ
8
5
, a local minimum at x œ 0;
yww œ (8 10x)(4 x)# a8x 5x# b (2)(% x)(1) œ 4(4 x) a5x# 16x 8b ,
yww œ ±
)#È'
&
8 2È 6 8 2È 6
Š 5 ß 5 ‹
±
)#È'
&
± Ê concave up on Š_ß 8 52
%
and (4ß _) Ê points of inflection at x œ
8 „ 2È 6
5
È6
‹ and Š 8 52
È6
ß %‹ , concave down on
and x œ 4
56. yw œ ax# 2xb (x 5)# œ x(x 2)(x 5)# ,
yw œ ± ± ± Ê rising on (_ß !) and
!
#
&
a2ß _b, falling on a!ß 2b Ê a local maximum at x œ 0, a local
minimum at x œ 2;
yww œ a2x 2bax 5b# 2ax# 2xbax 5b
œ 2ax 5ba2x# 8x 5b ,
yww œ ± ± ± Ê concave up on
&
%È'
%È'
#
4 È 6 4 È 6
Š # ß 2 ‹
#
È
È
and a5ß _b, concave down on Š_ß %2 6 ‹ and Š 4# 6 ß 5‹ Ê points of inflection at x œ
57. yw œ sec# x, yw œ ( ) Ê rising on ˆ 1# ß 1# ‰ ,
1Î#
1Î#
never falling Ê no local extrema; yww œ 2(sec x)(sec x)(tan x)
œ 2 asec# xb (tan x), yww œ ( ± ) Ê concave
!
1Î#
1Î#
1‰
1
ˆ
ˆ
‰
up on !, # , concave down on # ß ! , ! is a point of
inflection.
58. yw œ tan x, yw œ ( ± ) Ê rising on ˆ0ß 1# ‰ ,
!
1Î#
1Î#
1
falling on ˆ # ß !‰ Ê no local maximum, a local minimum
at x œ 0; yww œ sec# x, yww œ ( ) Ê concave up
1Î#
1Î#
1
1
on ˆ # ß # ‰ Ê no points of inflection
59. yw œ cot
)
#
, yw œ ( ± ) Ê rising on (!ß 1),
1
!
#1
falling on (1ß #1) Ê a local maximum at ) œ 1, no local
minimum; yww œ "# csc# #) , yww œ ( ) Ê never
!
#1
concave up, concave down on (!ß #1) Ê no points of
inflection
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
4 „È 6
2
and x œ 5
208
Chapter 4 Applications of Derivatives
60. yw œ csc#
)
#
, yw œ ( ) Ê rising on (!ß 21), never
!
#1
falling Ê no local extrema;
yww œ 2 ˆcsc #) ‰ ˆcsc #) ‰ ˆcot #) ‰ ˆ "# ‰
œ ˆcsc# #) ‰ ˆcot #) ‰, yww œ ( ± )
1
!
#1
Ê concave up on (1ß #1), concave down on (!ß 1)
Ê a point of inflection at ) œ 1
61. yw œ tan# ) 1 œ (tan ) 1)(tan ) 1),
yw œ ( | ± ) Ê rising on
1Î#
1Î%
1Î#
1Î%
1
1
1
1
ˆ # ß 4 ‰ and ˆ 4 ß # ‰ , falling on ˆ 14 ß 14 ‰
Ê a local maximum at ) œ 14 , a local minimum at ) œ 14 ;
yww œ 2 tan ) sec# ), yww œ ( ± )
!
1Î#
1Î#
1‰
ˆ
Ê concave up on !ß # , concave down on ˆ 1# ß !‰
Ê a point of inflection at ) œ 0
62. yw œ 1 cot# ) œ (" cot ))(1 cot )),
yw œ ( | ± ) Ê rising on ˆ 14 ß 341 ‰ ,
1
!
1Î%
$1Î%
1
3
1
falling on ˆ0ß 4 ‰ and ˆ 4 ß 1‰ Ê a local maximum at
)œ
ww
31
4 ,
a local minimum at ) œ
#
ww
1
4
;
y œ 2(cot )) acsc )b, y œ ( ± )
1
!
1Î#
1
1
Ê concave up on ˆ!ß # ‰ , concave down on ˆ # ß 1‰
Ê a point of inflection at ) œ
1
#
63. yw œ cos t, yw œ [ ± ± ] Ê rising on
!
#1
1Î#
$1Î#
ˆ!ß 1# ‰ and ˆ 3#1 ß 21‰ , falling on ˆ 1# ß 3#1 ‰ Ê local maxima at
tœ
1
#
and t œ 21, local minima at t œ 0 and t œ
yww œ sin t, yww œ [ ± ]
1
!
#1
Ê concave up on (1ß #1), concave down
on (!ß 1) Ê a point of inflection at t œ 1
31
#
;
64. yw œ sin t, yw œ [ ± ] Ê rising on (!ß 1),
1
!
#1
falling on (1ß 21) Ê a local maximum at t œ 1, local
minima at t œ 0 and t œ 21; yww œ cos t,
yww œ [ ± ± ] Ê concave up on ˆ!ß 1# ‰
!
#1
1Î#
$1Î#
3
1
and ˆ # ß #1‰ , concave down on ˆ 1# ß 3#1 ‰ Ê points
of inflection at t œ
1
#
and t œ
31
#
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 4.4 Concavity and Curve Sketching
65. yw œ (x 1)#Î$ , yw œ ) ( Ê rising on
"
(_ß _), never falling Ê no local extrema;
yww œ 23 (x 1)&Î$ , yww œ ) ( "
Ê concave up on (_ß 1), concave down on ("ß _)
Ê a point of inflection and vertical tangent at x œ 1
66. yw œ (x 2)"Î$ , yw œ )( Ê rising on (2ß _),
#
falling on (_ß #) Ê no local maximum, but a local
minimum at x œ 2; yww œ 13 (x 2)%Î$ ,
yww œ )( Ê concave down on (_ß 2) and
#
(#ß _) Ê no points of inflection, but there is a cusp at
xœ2
67. yw œ x#Î$ (x 1), yw œ )( ± Ê rising on
!
"
("ß _), falling on (_ß ") Ê no local maximum, but a
local minimum at x œ 1; yww œ "3 x#Î$ 32 x&Î$
"
3
x&Î$ (x 2), yww œ ± )( !
#
Ê concave up on (_ß 2) and (!ß _), concave down on
(#ß !) Ê points of inflection at x œ 2 and x œ 0, and a
vertical tangent at x œ 0
œ
68. yw œ x%Î& (x 1), yw œ ± )( Ê rising on
!
"
("ß 0) and (!ß _), falling on (_ß ") Ê no local
maximum, but a local minimum at x œ 1;
yww œ "5 x%Î& 54 x*Î& œ "5 x*Î& (x 4),
yww œ )( ± Ê concave up on (_ß 0) and
!
%
(4ß _), concave down on (0ß 4) Ê points of inflection at
x œ 0 and x œ 4, and a vertical tangent at x œ 0
69. yw œ œ
#x, x Ÿ 0 w
, y œ ± Ê rising on
2x, x 0
!
(_ß _) Ê no local extrema; yww œ œ
2, x 0
,
2, x 0
yww œ )( Ê concave up on (!ß _), concave
!
down on (_ß !) Ê a point of inflection at x œ 0
70. yw œ œ
x# , x Ÿ 0 w
, y œ ± Ê rising on
x# , x 0
!
(!ß _), falling on (_ß !) Ê no local maximum, but a
2x, x Ÿ 0
local minimum at x œ 0; yww œ œ
,
2x, x 0
yww œ ± Ê concave up on (_ß _)
!
Ê no point of inflection
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
209
210
Chapter 4 Applications of Derivatives
71. The graph of y œ f ww (x) Ê the graph of y œ f(x) is concave
up on (!ß _), concave down on (_ß !) Ê a point of
inflection at x œ 0; the graph of y œ f w (x)
Ê yw œ ± ± Ê the graph y œ f(x) has
both a local maximum and a local minimum
72. The graph of y œ f ww (x) Ê yww œ ± Ê the
graph of y œ f(x) has a point of inflection, the graph of
y œ f w (x) Ê yw œ ± ± Ê the graph of
y œ f(x) has both a local maximum and a local minimum
73. The graph of y œ f ww (x) Ê yww œ ± ± Ê the graph of y œ f(x) has two points of inflection, the
graph of y œ f w (x) Ê yw œ ± Ê the graph of
y œ f(x) has a local minimum
74. The graph of y œ f ww (x) Ê yww œ ± Ê the
graph of y œ f(x) has a point of inflection; the graph of
y œ f w (x) Ê yw œ ± ± Ê the graph of
y œ f(x) has both a local maximum and a local minimum
75. y œ
2x# x 1
x# 1
76. y œ
x# 49
x# 5x 14
œ1
5
x2
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 4.4 Concavity and Curve Sketching
77. y œ
x% "
x#
79. y œ
"
x# 1
œ x# #
2
81. y œ xx# 1 œ " 83. y œ
x#
x1
78. y œ
x# 4
2x
œ
80. y œ
x#
x# 1
œ"
"
x# 1
"
x# 1
82. y œ
x# 4
x# 2
œ"
2
x# 2
"
x1
84. y œ xx 14 œ 1 x "
x#
œx1
#
x
#
2
x
3
x1
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
211
212
Chapter 4 Applications of Derivatives
85. y œ
x# x 1
x1
87. y œ
x$ 3x# 3x 1
x# x 2
89. y œ
91. y œ
93.
œx
"
x1
86. y œ x
#
x1
x1
88. y œ
x$ x 2
x x#
x
x# 1
90. y œ
x1
x # ax 2 b
8
x# 4
92. y œ
4x
x# 4
Point
P
Q
R
S
T
yw
!
œx4
5x 7
x# x #
œ x "
x1
œ x 1 2x 2
x x#
yww
!
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 4.4 Concavity and Curve Sketching
94.
213
95.
96.
97. Graphs printed in color can shift during a press run, so your values may differ somewhat from those given here.
(a) The body is moving away from the origin when kdisplacementk is increasing as t increases, 0 t 2 and
6 t 9.5; the body is moving toward the origin when kdisplacementk is decreasing as t increases, 2 t 6
and 9.5 t 15
(b) The velocity will be zero when the slope of the tangent line for y œ s(t) is horizontal. The velocity is zero
when t is approximately 2, 6, or 9.5 sec.
(c) The acceleration will be zero at those values of t where the curve y œ s(t) has points of inflection. The
acceleration is zero when t is approximately 4, 7.5, or 12.5 sec.
(d) The acceleration is positive when the concavity is up, 4 t 7.5 and 12.5 t 15; the acceleration is
negative when the concavity is down, 0 t 4 and 7.5 t 12.5
98. (a) The body is moving away from the origin when kdisplacementk is increasing as t increases, 1.5 t 4,
10 t 12 and 13.5 t 16; the body is moving toward the origin when kdisplacementk is decreasing as t
increases, 0 t 1.5, 4 t 10 and 12 t 13.5
(b) The velocity will be zero when the slope of the tangent line for y œ s(t) is horizontal. The velocity is zero
when t is approximately 0, 4, 12 or 16 sec.
(c) The acceleration will be zero at those values of t where the curve y œ s(t) has points of inflection. The
acceleration is zero when t is approximately 1.5, 6, 8, 10.5, or 13.5 sec.
(d) The acceleration is positive when the concavity is up, 0 t 1.5, 6 t 8 and 10 t 13.5, the
acceleration is negative when the concavity is down, 1.5 t 6, 8 t 10 and 13.5 t 16.
99. The marginal cost is
dc
dx
which changes from decreasing to increasing when its derivative
d# c
dx#
is zero. This is a
point of inflection of the cost curve and occurs when the production level x is approximately 60 thousand units.
100. The marginal revenue is
dy
dx
d# y
dx# is positive Ê the curve is concave up
#
when ddxy# 0 Ê the curve is concave down
and it is increasing when its derivative
Ê ! t 2 and 5 t 9; marginal revenue is decreasing
Ê 2 t 5 and 9 t 12.
101. When yw œ (x 1)# (x 2), then yww œ 2(x 1)(x 2) (x 1)# . The curve falls on (_ß 2) and rises on
(#ß _). At x œ 2 there is a local minimum. There is no local maximum. The curve is concave upward on (_ß ") and
ˆ 53 ß _‰ , and concave downward on ˆ"ß 53 ‰ . At x œ 1 or x œ 53 there are inflection points.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
214
Chapter 4 Applications of Derivatives
102. When yw œ (x 1)# (x 2)(x 4), then yww œ 2(x 1)(x 2)(x 4) (x 1)# (x 4) (x 1)# (x 2)
œ (x 1) c2 ax# 6x 8b ax# 5x 4b ax# 3x 2bd œ 2(x 1) a2x# 10x 11b. The curve rises on
(_ß 2) and (%ß _) and falls on (#ß %). At x œ 2 there is a local maximum and at x œ 4 a local minimum. The
È3 5 È3
ß # ‹
curve is concave downward on (_ß ") and Š 5 2
È3
Š 5 #
ß _‹ . At x œ 1,
5 È3
#
and
5 È3
#
È3
and concave upward on Š1ß 5 #
‹ and
there are inflection points.
103. The graph must be concave down for x 0 because
f ww (x) œ x"# 0.
104. The second derivative, being continuous and never zero, cannot change sign. Therefore the graph will always
be concave up or concave down so it will have no inflection points and no cusps or corners.
105. The curve will have a point of inflection at x œ 1 if 1 is a solution of yww œ 0; y œ x$ bx# cx d
Ê yw œ 3x# 2bx c Ê yw w œ 6x 2b and 6(1) 2b œ 0 Ê b œ 3.
106. (a) f(x) œ ax# bx c œ a ˆx# ba x‰ c œ a Šx# ba B b#
4a# ‹
b#
4a
c œ a ˆx b
b
whose vertex is at x œ 2a
Ê the coordinates of the vertex are Š 2a
ß b
#
b ‰#
#a
b# 4ac
4a
a parabola
4ac
4a ‹
(b) The second derivative, f ww (x) œ 2a, describes concavity Ê when a 0 the parabola is concave up and
when a 0 the parabola is concave down.
107. A quadratic curve never has an inflection point. If y œ ax# bx c where a Á 0, then yw œ 2ax b and
yww œ 2a. Since 2a is a constant, it is not possible for yww to change signs.
108. A cubic curve always has exactly one inflection point. If y œ ax$ bx# cx d where a Á 0, then
yw œ 3ax# 2bx c and yww œ 6ax 2b. Since 3ab is a solution of yww œ 0, we have that yww changes its sign
b
b
at x œ 3a
and yw exists everywhere (so there is a tangent at x œ 3a
). Thus the curve has an inflection
b
point at x œ 3a
. There are no other inflection points because yww changes sign only at this zero.
109. y ww œ ax 1bax 2b, when y ww œ 0 Ê x œ 1 or x œ 2; y ww œ ± l Ê points of inflection at x œ 1
1
2
and x œ 2
110. y ww œ x2 ax 2b3 ax 3b, when y ww œ 0 Ê x œ 3, x œ 0, or x œ 2; y ww œ ± l l Ê points of
0
2
3
inflection at x œ 3 and x œ 2
111. y œ a x3 b x2 c x Ê y w œ 3a x2 2b x c and y ww œ 6a x 2b; local maximum at x œ 3
Ê 3a a3b2 2ba3b c œ 0 Ê 27a 6b c œ 0; local mimimum at x œ 1 Ê 3a a1b2 2ba1b c œ 0
Ê 3a 2b c œ 0; point of inflection at a1, 11b Ê aa1b3 ba1b2 ca1b œ 11 Ê a b c œ 11 and
6aa1b 2b œ 0 Ê 6a 2b œ 0. Solving 27a 6b c œ 0, 3a 2b c œ 0, a b c œ 11, and 6a 2b œ 0
Ê a œ 1,b œ 3, and c œ 9 Ê y œ x3 3x2 9x
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 4.4 Concavity and Curve Sketching
112. y œ
x2 a
bxc
2
a3 b a b
b x2 2c x a b
à local maximum at x œ 3 Ê ba3baba32c
œ
ab x c b 2
b c b2
ba1b2 2ca1b a b
a1b2 a
œ 0 Ê b 2c a b œ 0 and ba1b c œ 2 Ê
ab a 1 b c b 2
Ê yw œ
a1 ,2b Ê
0 Ê 9b 6c a b œ 0; local minimum at
a 2b 2c œ 1. Solving
9b 6c a b œ 0, b 2c a b œ 0, and a 2b 2c œ 1 Ê a œ 3,b œ 1, and c œ 1 Ê y œ
113. If y œ x& 5x% 240, then yw œ 5x$ (x 4) and
yww œ 20x# (x 3). The zeros of yw are extrema, and
there is a point of inflection at x œ $Þ
114. If y œ x$ 12x# , then yw œ 3x(x 8) and
yww œ 6(x 4). The zeros of yw and yww are
extrema and points of inflection, respectively.
115. If y œ
4
5
x& 16x# 25, then yw œ 4x ax$ 8b and
yww œ 16 ax$ 2b . The zeros of yw and yww are
extrema and points of inflection, respectively.
116. If y œ
x%
4
$
x$
3
#
215
4x# 12x 20, then
yw œ x x )x "# œ (x 3)(x 2)# Þ
So y has a local minimum at x œ $ as its only extreme
value. Also yww œ $x# #x ) œ (3x 4)(x 2) and there
are inflection points at both zeros, %$ and 2, of yww .
117. The graph of f falls where f w 0, rises where f w 0,
and has horizontal tangents where f w œ 0. It has local
minima at points where f w changes from negative to
positive and local maxima where f w changes from
positive to negative. The graph of f is concave down
where f ww 0 and concave up where f ww 0. It has an
inflection point each time f ww changes sign, provided a
tangent line exists there.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
x2 3
x1 Þ
216
Chapter 4 Applications of Derivatives
118. The graph f is concave down where f ww 0, and concave
up where f ww 0. It has an inflection point each time
f ww changes sign, provided a tangent line exists there.
4.5 APPLIED OPTIMIZATION
1. Let j and w represent the length and width of the rectangle, respectively. With an area of 16 in.# , we have
that (j)(w) œ 16 Ê w œ 16j" Ê the perimeter is P œ 2j 2w œ 2j 32j" and Pw (j) œ 2 w
Solving P (j) œ 0 Ê
2(j 4)(j 4)
j#
32
j#
œ
2 aj# 16b
j#
.
œ 0 Ê j œ 4, 4. Since j 0 for the length of a rectangle, j must be 4 and
w œ 4 Ê the perimeter is 16 in., a minimum since Pww (j) œ
16
j$
0.
2. Let x represent the length of the rectangle in meters (! x %) Then the width is % x and the area is
Aaxb œ xa% xb œ %x x# . Since Aw axb œ % #x, the critical point occurs at x œ #. Since, Aw axb ! for ! x # and
Aw axb ! for # x %, this critical point corresponds to the maximum area. The rectangle with the largest area measures
# m by % # œ # m, so it is a square.
Graphical Support:
3. (a) The line containing point P also contains the points (!ß ") and ("ß !) Ê the line containing P is y œ 1 x
Ê a general point on that line is (xß 1 x).
(b) The area A(x) œ 2x(1 x), where 0 Ÿ x Ÿ 1.
(c) When A(x) œ 2x 2x# , then Aw (x) œ 0 Ê 2 4x œ 0 Ê x œ "# . Since A(0) œ 0 and A(1) œ 0, we conclude
that A ˆ "# ‰ œ "# sq units is the largest area. The dimensions are " unit by "# unit.
4. The area of the rectangle is A œ 2xy œ 2x a12 x# b ,
where 0 Ÿ x Ÿ È12 . Solving Aw (x) œ 0 Ê 24 6x# œ 0
Ê x œ 2 or 2. Now 2 is not in the domain, and since
A(0) œ 0 and A ŠÈ12‹ œ 0, we conclude that A(2) œ 32
square units is the maximum area. The dimensions are 4 units
by 8 units.
5. The volume of the box is V(x) œ x(15 2x)(8 2x)
œ 120x 46x# 4x$ , where 0 Ÿ x Ÿ 4. Solving Vw (x) œ 0
Ê 120 92x 12x# œ 4(6 x)(5 3x) œ 0 Ê x œ 53
or 6, but 6 is not in the domain. Since V(0) œ V(4) œ 0,
$
V ˆ 53 ‰ œ #%&!
#( ¸ *" in must be the maximum volume of
the box with dimensions
14
3
‚
35
3
‚
5
3
inches.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 4.5 Applied Optimization
217
6. The area of the triangle is A œ "# ba œ b# È400 b# , where
b#
0 Ÿ b Ÿ 20. Then dA œ " È400 b# db
œ
#
200 b
È400 b#
#
2È400 b#
œ 0 Ê the interior critical point is b œ 10È2.
When b œ 0 or 20, the area is zero Ê A Š10È2‹ is the
maximum area. When a# b# œ 400 and b œ 10È2, the
value of a is also 10È2 Ê the maximum area occurs when
a œ b.
7. The area is A(x) œ x(800 2x), where 0 Ÿ x Ÿ 400.
Solving Aw (x) œ 800 4x œ 0 Ê x œ 200. With
A(0) œ A(400) œ 0, the maximum area is
A(200) œ 80,000 m# . The dimensions are 200 m by 400 m.
8. The area is 2xy œ 216 Ê y œ
108
x
. The amount of fence
needed is P œ 4x 3y œ 4x 324x" , where ! x;
dP
324
#
dx œ 4 x# œ 0 Ê x 81 œ 0 Ê the critical points are
0 and „ 9, but 0 and 9 are not in the domain. Then
Pww (9) 0 Ê at x œ 9 there is a minimum Ê the
dimensions of the outer rectangle are 18 m by 12 m
Ê 72 meters of fence will be needed.
9. (a) We minimize the weight œ tS where S is the surface area, and t is the thickness of the steel walls of the tank. The
surface area is S œ x# %xy where x is the length of a side of the square base of the tank, and y is its depth. The
ˆ # #!!! ‰. Treating the
volume of the tank must be &!!ft$ Ê y œ &!!
x# . Therefore, the weight of the tank is waxb œ t x x
‰ . The critical value is at x œ "!. Since www a"!b œ tˆ# %!!!
‰ !,
thickness as a constant gives ww axb œ tˆ#x #!!!
x#
"!$
there is a minimum at x œ "!. Therefore, the optimum dimensions of the tank are "! ft on the base edges and & ft
deep.
(b) Minimizing the surface area of the tank minimizes its weight for a given wall thickness. The thickness of the steel
walls would likely be determined by other considerations such as structural requirements.
10. (a) The volume of the tank being ""#& ft$ , we have that yx# œ ""#& Ê y œ ""#&
x# . The cost of building the tank is
""#&
$$(&!
#
w
caxb œ &x $!xˆ x# ‰, where ! x. Then c axb œ "!x x# œ ! Ê the critical points are ! and "&, but ! is not
in the domain. Thus, cww a"&b ! Ê at x œ "& we have a minimum. The values of x œ "& ft and y œ & ft will
minimize the cost.
(b) The cost function c œ &ax# %xyb "!xy, can be separated into two items: (1) the cost of the materials and labor to
fabricate the tank, and (2) the cost for the excavation. Since the area of the sides and bottom of the tanks is ax# %xyb,
it can be deduced that the unit cost to fabricate the tanks is $&/ft# . Normally, excavation costs are per unit volume of
‰ #
excavated material. Consequently, the total excavation cost can be taken as "!xy œ ˆ "!
x ax yb. This suggests that the
unit cost of excavation is
$"!Îft#
x
where x is the length of a side of the square base of the tank in feet. For the least
expensive tank, the unit cost for the excavation is
$"!Îft#
"& ft
œ
$!Þ'(
ft$
œ
$")
yd$ .
The total cost of the least expensive tank is
$$$(&, which is the sum of $#'#& for fabrication and $(&! for the excavation.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
218
Chapter 4 Applications of Derivatives
11. The area of the printing is (y 4)(x 8) œ 50.
‰
Consequently, y œ ˆ x 50
8 4. The area of the paper is
50
A(x) œ x ˆ x 8 4‰ , where 8 x. Then
50
‰
Aw (x) œ ˆ x 50
8 4 x Š (x 8)# ‹ œ
4(x 8)# 400
(x 8)#
œ0
Ê the critical points are 2 and 18, but 2 is not in the
domain. Thus Aww (18) 0 Ê at x œ 18 we have a minimum
Therefore the dimensions 18 by 9 inches minimize the
amount minimize the amount of paper.
12. The volume of the cone is V œ
V(y) œ
1
3
1
3
a9 y# b (y 3) œ
"
3
1r# h, where r œ x œ È9 y# and h œ y 3 (from the figure in the text). Thus,
a27 9y 3y# y$ b Ê Vw (y) œ
points are 3 and 1, but 3 is not in the domain. Thus Vww (1) œ
volume of V(1) œ
1
3
321
3
(8)(4) œ
13. The area of the triangle is A()) œ
cubic units.
ab sin )
#
, where 0 ) 1.
Solving A ()) œ 0 Ê
œ 0 Ê ) œ 1# . Since Aww ())
)
œ ab sin
Ê Aww ˆ 1# ‰ 0, there is a maximum at ) œ 1# .
#
w
ab cos )
#
1
#
3 a9 6y 3y b œ 1(1 y)(3 y). The critical
1
3 (' 6(1)) 0 Ê at y œ 1 we have a maximum
14. A volume V œ 1r# h œ 1000 Ê h œ
1000
1 r#
. The amount of
material is the surface area given by the sides and bottom of
#
the can Ê S œ 21rh 1r# œ 2000
r 1r , 0 r. Then
dS
dr
œ 2000
r# 21r œ ! Ê
are 0 and
d# S
dr#
rœ
10
$
È
1
1r$ 1000
r#
œ 0. The critical points
, but 0 is not in the domain. Since
œ
4000
r$ #1 0, we
10
cm and h œ 1000
$
È
1 r#
1
have a minimum surface area when
œ
10
$
È
1
cm. Comparing this result to
the result found in Example 2, if we include both ends of the
can, then we have a minimum surface area when the can is
shorter-specifically, when the height of the can is the same as
its diameter.
15. With a volume of 1000 cm and V œ 1r# h, then h œ
A œ 8r# 21rh œ 8r# 2000
r
. Then Aw (r) œ
but r œ 0 results in no can. Since Aww (r) œ 16
16. (a) The base measures "! #x in. by
"&#x
#
1000
1r# . The amount of aluminum used per can is
8r$ 1000
16r 2000
œ 0 Ê the critical points are 0 and 5,
r# œ 0 Ê
r#
1000
0
we
have
a
minimum
at r œ 5 Ê h œ 40
r$
1 and h:r œ 8:1.
in., so the volume formula is Vaxb œ
xa"!#xba"&#xb
#
œ #x$ #&x# (&x.
(b) We require x !, #x "!, and #x "&. Combining these requirements, the domain is the interval a!ß &b.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 4.5 Applied Optimization
219
(c) The maximum volume is approximately 66.02 in.$ when x ¸ "Þ*' in.
(d) Vw axb œ 'x# &!x (&. The critical point occurs when Vw axb œ !, at x œ
œ
ww
#& „ &È(
, that
'
&! „ Éa&!b# %a'ba(&b
# a 'b
œ
&! „ È(!!
"#
is, x ¸ "Þ*' or x ¸ 'Þ$(. We discard the larger value because it is not in the domain. Since
V axb œ "#x &!, which is negative when x ¸ "Þ*' , the critical point corresponds to the maximum volume. The
maximum volume occurs when x œ
#& &È(
'
¸ "Þ*', which comfimrs the result in (c).
17. (a) The "sides" of the suitcase will measure 24 2x in. by 18 2x in. and will be 2x in. apart, so the volume formula is
Vaxb œ 2xa24 2xba18 2xb œ 8x$ 168x# 862x.
(b) We require x !, 2x 18, and 2x 12. Combining these requirements, the domain is the interval a!ß *b.
(c) The maximum volume is approximately 1309.95 in.$ when x ¸ 3Þ3* in.
(d) Vw axb œ #%x# $$'x )'% œ #%ax# "%x $'b. The critical point is at x œ
"% „ Éa"%b# %a"ba$'b
# a" b
œ
"% „ È&#
#
œ ( „ È"$, that is, x ¸ $Þ$* or x ¸ "!Þ'". We discard the larger value because it is not in the domain. Since
Vww axb œ #%a#x "%b which is negative when x ¸ $Þ$*, the critical point corresponds to the maximum volume. The
maximum value occurs at x œ ( È"$ ¸ $Þ$*, which confirms the results in (c).
(e) )x$ "')x# )'#x œ ""#! Ê 8ax$ #"x# "!)x "%!b œ ! Ê )ax #bax &bax "%b œ !. Since "% is not in
the fomain, the possible values of x are x œ # in. or x œ & in.
(f) The dimensions of the resulting box are #x in., a#% #xb in., and a") #xb. Each of these measurements must be
positive, so that gives the domain of a!ß *b.
18. If the upper right vertex of the rectangle is located at axß % cos !Þ& xb for ! x 1, then the rectangle has width #x and
height % cos !Þ&x, so the area is Aaxb œ )x cos !Þ&x. Solving Aw axb œ ! graphically for ! x 1, we find that
x ¸ #Þ#"%. Evaluating #x and % cos !Þ&x for x ¸ #Þ#"%, the dimensions of the rectangle are approximately %Þ%$ (width) by
"Þ(* (height), and the maximum area is approximately (Þ*#$.
19. Let the radius of the cylinder be r cm, ! r "!. Then the height is #È"!! r# and the volume is
Varb œ #1r# È"!! r# cm$ . Then, Vw arb œ #1r# Š " ‹a#rb Š#1È"!! r# ‹a#rb
È"!! r#
œ
#1r$ %1ra"!! r# b
È"!! r#
œ
#1ra#!! $r# b
È"!! r# .
É $# . Since Vw arb ! for
The critical point for ! r "! occurs at r œ É #!!
$ œ "!
! r "!É #$ and Vw arb ! for "!É #$ r "!, the critical point corresponds to the maximum volume. The
dimensions are r œ "!É #$ ¸ )Þ"' cm and h œ
#!
È$
¸ ""Þ&& cm, and the volume is
%!!!1
$È $
¸ #%")Þ%! cm$ .
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
220
Chapter 4 Applications of Derivatives
20. (a) From the diagram we have 4x j œ 108 and V œ x# j.
The volume of the box is V(x) œ x# (108 4x), where
0 Ÿ x 27. Then
Vw (x) œ 216x 12x# œ 12x(18 x) œ 0
Ê the critical points are 0 and 18, but x œ 0 results in
no box. Since Vww (x) œ 216 24x 0 at x œ 18 we
have a maximum. The dimensions of the box are
18 ‚ 18 ‚ 36 in.
#
(b) In terms of length, V(j) œ x# j œ ˆ 1084 j ‰ j. The graph
indicates that the maximum volume occurs near j œ 36,
which is consistent with the result of part (a).
21. (a) From the diagram we have 3h 2w œ 108 and
V œ h# w Ê V(h) œ h# ˆ54 #3 h‰ œ 54h# 3# h$ .
Then Vw (h) œ 108h 9# h# œ
9
#
h(24 h) œ 0
Ê h œ 0 or h œ 24, but h œ 0 results in no box. Since
Vww (h) œ 108 9h 0 at h œ 24, we have a maximum
volume at h œ 24 and w œ 54 3# h œ 18.
(b)
22. From the diagram the perimeter is P œ 2r 2h 1r,
where r is the radius of the semicircle and h is the
height of the rectangle. The amount of light transmitted
proportional to
A œ 2rh "4 1r# œ r(P 2r 1r) 4" 1r#
œ rP 2r# 34 1r# . Then
dA
dr
œ P 4r 3# 1r œ 0
(4 1)P
2P
4P
21 P
8 31 Ê 2h œ P 8 31 8 31 œ 8 31 .
Therefore, 2rh œ 4 8 1 gives the proportions that admit the
#
most light since ddrA# œ 4 3# 1 0.
Ê rœ
23. The fixed volume is V œ 1r# h 23 1r$ Ê h œ
V
1 r#
2r
3
, where h is the height of the cylinder and r is the radius
of the hemisphere. To minimize the cost we must minimize surface area of the cylinder added to twice the
8
#
surface area of the hemisphere. Thus, we minimize C œ 21rh 41r# œ 21r ˆ 1Vr# 2r3 ‰ 41r# œ 2V
r 3 1r .
Then
œ
dC
dr
œ 2V
r# 4V"Î$
1"Î$ †3#Î$
16
3
2†3"Î$ †V"Î$
3†#†1"Î$
1r œ 0 Ê V œ
œ
8
3
‰
1r$ Ê r œ ˆ 3V
81
3"Î$ †2†4†V"Î$ 2†3"Î$ †V"Î$
3†#†1"Î$
‰
œ ˆ 3V
1
"Î$
"Î$
. From the volume equation, h œ
. Since
d# C
dr#
œ
4V
r$
16
3
V
1 r#
1 0, these
dimensions do minimize the cost.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
2r
3
Section 4.5 Applied Optimization
24. The volume of the trough is maximized when the area of the cross section is maximized. From the diagram
the area of the cross section is A()) œ cos ) sin ) cos ), 0 ) 1# . Then Aw ()) œ sin ) cos# ) sin# )
œ a2 sin# ) sin ) 1b œ (2 sin ) 1)(sin ) 1) so Aw ()) œ 0 Ê sin ) œ
sin ) Á 1 when 0 ) 1
#
w
. Also, A ()) 0 for 0 ) 1
6
w
and A ()) 0 for
1
6
"
#
or sin ) œ 1 Ê ) œ
)
1
#
1
6
because
. Therefore, at ) œ
1
6
there is a maximum.
25. (a) From the diagram we have: AP œ x, RA œ ÈL x# ,
PB œ 8.5 x, CH œ DR œ 11 RA œ 11 ÈL x# ,
QB œ Èx# (8.5 x)# , HQ œ 11 CH QB
œ 11 ’11 ÈL x# Èx# (8.5 x)# “
#
#
#
œ ÈL x# Èx# (8.5 x)# , RQ œ RH HQ
#
œ (8.5)# ŠÈL x# Èx# (8.5 x)# ‹ . It
#
#
#
#
follows that RP œ PQ RQ Ê L# œ x# ŠÈL# x# Èx# (x 8.5)# ‹ (8.5)#
Ê L# œ x# L# x# 2ÈL# x# È17x (8.5)# 17x (8.5)# (8.5)#
Ê 17# x# œ 4 aL# x# b a17x (8.5)# b Ê L# œ x# œ
4x$
4x 17
(b) If f(x) œ
2x$
2x 8.5
œ
$
4x
4x17
51
8 ,
is minimized, then L# is minimized. Now f w (x) œ
51
8
. Thus L# is minimized when x œ
cylinder is formed, x œ 21r Ê r œ
Then Vww (x)
17x$
17x (8.5)#
œ
17x$
‰#
17x ˆ 17
#
4x# (8x 51)
(4x 17)#
Ê f w (x) 0 when x 51
8
51
8 .
then L ¸ 11.0 in.
26. (a) From the figure in the text we have P œ 2x 2y Ê y œ
Ê V(x) œ
œ
.
and f w (x) 0 when x (c) When x œ
17# x#
4 c17x (8.5)# d
x
#1
P
#
x. If P œ 36, then y œ 18 x. When the
and h œ y Ê h œ 18 x. The volume of the cylinder is V œ 1r# h
x)
18x# x$
. Solving Vw (x) œ 3x(12
œ0 Ê
41
41
œ 13 ˆ3 x# ‰ Ê Vww (12) 0 Ê there is a
x œ 0 or 12; but when x œ 0, there is no cylinder.
maximum at x œ 12. The values of x œ 12 cm and
y œ 6 cm give the largest volume.
(b) In this case V(x) œ 1x# (18 x). Solving Vw (x) œ 31x(12 x) œ 0 Ê x œ 0 or 12; but x œ 0 would result in
no cylinder. Then Vww (x) œ 61(6 x) Ê Vww (12) 0 Ê there is a maximum at x œ 12. The values of
x œ 12 cm and y œ 6 cm give the largest volume.
27. Note that h# r# œ $ and so r œ È$ h# . Then the volume is given by V œ 1$ r# h œ 1$ a$ h# bh œ 1h 1$ h$ for
dV
#
#
! h È$, and so dV
dh œ 1 1r œ 1a" r b. The critical point (for h !) occurs at h œ ". Since dh ! for
! h ", and dV ! for " h È$, the critical point corresponds to the maximum volume. The cone of greatest
dh
volume has radius È# m, height "m, and volume
28. Let d œ Éax 0b2 ay 0b2 œ Èx2 y2 and
2
#1
$
x
a
m$ .
y
b
œ 1 Ê y œ ba x b. We can minimize d by minimizing
2
2b2
a2 x
a b2
a2 b2
is the critical point Ê y œ ba Š a2ab b2 ‹ b œ
D œ ˆÈx2 y2 ‰ œ x2 ˆ ba x b‰ Ê D w œ 2x 2ˆ ba x b‰ˆ ba ‰ œ 2x Ê 2Šx b2
a2 x
b2
a ‹
œ0Êxœ
2
2b2
a .
Dw œ 0
a2 b
a2 b2 .
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
221
222
Chapter 4 Applications of Derivatives
D ww œ 2 line
x
a
y
b
2b2
a2
Ê D ww Š a2ab b2 ‹ œ 2 2
2b2
a2
a2 b
a2 b2 ‹
is the point on the
œ 1 that is closest to the origin.
29. Let Saxb œ x 1x , x 0 Ê S w axb œ 1 we only consider x œ 1. S ww axb œ
30. Let Saxb œ
S ww axb œ
2
0 Ê the critical point is local minimum Ê Š a2ab b2 ,
2
x3
2
x3
1
x2
ww
œ
Ê S a1b
x2 1
x2 1
w
x2 . S axb œ 0 Ê x2 œ
œ 123 0 Ê local minimum
0 Ê x2 1 œ 0 Ê x œ „ 1. Since x 0,
when x œ 1
4x2 , x 0 Ê S w axb œ x12 8x œ 8x x2 1 . S w axb œ 0 Ê 8x x2 1 œ 0 Ê 8x3 1 œ 0 Ê x œ "# .
8 Ê S ww ˆ "# ‰ œ a1Î22b3 8 0 Ê local minimum when x œ "# .
3
1
x
3
31. The length of the wire b œ perimeter of the triangle circumference of the circle. Let x œ length of a side of the
equilateral triangle Ê P œ 3x, and let r œ radius of the circle Ê C œ 21r. Thus b œ 3x 21 r Ê r œ b 213x . The area of
the circle is 1 r2 and the area of an equilateral triangle whose sides are x is "# axbŠ
given by A œ
Aw œ 0 Ê
È3 2
4 x
È3
2 x
3b
21
P œ 3Š È33b
‹œ
19
œb
9b
È3 1 9
1 r2 œ
9
21 x
œ
È3 2
4 x .
Thus, the total area is
È3 2
È
È
ab 3xb2
Ê Aw œ 23 x 231 ab 3xb œ 23 x 23b1
4 x 41
È3
3b
9
ww
È3 1 9 . A œ 2 21 0 Ê local minimum at the critical point.
2
1 ˆ b 213x ‰ œ
œ0Êxœ
9
21 x
m is the length of the trianglular segment and C œ 2 1ˆ b 213x ‰ œ b 3x
9b
È3 1 9
È3 1 b
È3 1 9
œ
È3 2
4 x
È3
2 x‹
m is the length of the circular segment.
32. The length of the wire b œ perimeter of the square circunference of the circle. Let x œ length of a side of the square
Ê P œ 4x, and let r œ radius of the circle Ê C œ 21r. Thus b œ 4x 21 r Ê r œ b 214x . The area of the circle is
1 r2 and the area of a square whose sides are x is x2 . Thus, the total area is given by A œ x2 1 r2
2
œ x2 1 ˆ b 214x ‰ œ x2 ab 4xb2
41
Ê Aw œ 2x 8
18 x, Aw œ 0 Ê 2x 2b
1 1x œ 0
0 Ê local minimum at the critical point. P œ 4ˆ 4 b 1 ‰ œ 4 4b
1 m is the length of the square
Aww œ 2 18
segment and C œ 2 1ˆ b 214x ‰ œ b 4x œ b Êxœ
b
41.
4b
41
4
2 1 ab
œ
4xb œ 2x b1
41
2b
1
m is the length of the circular segment.
33. Let ax, yb œ ˆx, 43 x‰ be the coordinates of the corner that intersects the line. Then base œ 3 x and height œ y œ 43 x, thus
the area of therectangle is given by A œ a3 xbˆ 43 x‰ œ 4x 43 x2 , 0 Ÿ x Ÿ 3. Aw œ 4 83 x, Aw œ ! Ê x œ 32 . Aww œ 43
Ê Aww ˆ 32 ‰ 0 Ê local maximum at the critical point. The base œ 3 3
2
œ
3
2
and the height œ 43 ˆ 32 ‰ œ 2.
34. Let ax, yb œ Šx, È9 x2 ‹ be the coordinates of the corner that intersects the semicircle. Then base œ 2x and height œ y
œ È9 x2 , thus the area of the inscribed rectangle is given by A œ a2xbÈ9 x2 , 0 Ÿ x Ÿ 3. Then
2
È
2ˆ9 x2 ‰ 2x2
Aw œ 2È9 x2 a2xb x œ
œ 18 4x , Aw œ 0 Ê 18 4x2 œ 0 Ê x œ „ 3 2 , only x œ
È 9 x2
È 9 x2
È 4 x2
2
3È 2
2
lies in
0 Ÿ x Ÿ 3. A is continuous on the closed interval 0 Ÿ x Ÿ 3 Ê A has an absolute maxima and absolute minima.
È
È
Aa0b œ 0, Aa3b œ 0, and AŠ 3 2 ‹ œ Š3È2‹Š 3 2 ‹ œ 9 Ê absolute maxima. Base of rectangle is 3È2 and height
2
is
2
3È 2
2 .
35. (a) f(x) œ x# (b) f(x) œ x# a
x
a
x
Ê f w (x) œ x# a2x$ ab , so that f w (x) œ 0 when x œ 2 implies a œ 16
Ê f ww (x) œ 2x$ ax$ ab , so that f ww (x) œ 0 when x œ 1 implies a œ 1
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 4.5 Applied Optimization
223
36. If f(x) œ x$ ax# bx, then f w (x) œ 3x# 2ax b and f ww (x) œ 6x 2a.
(a) A local maximum at x œ 1 and local minimum at x œ 3 Ê f w (1) œ 0 and f w (3) œ 0 Ê 3 2a b œ 0 and
27 6a b œ 0 Ê a œ 3 and b œ 9.
(b) A local minimum at x œ 4 and a point of inflection at x œ 1 Ê f w (4) œ 0 and f ww (1) œ 0 Ê 48 8a b œ 0
and 6 2a œ 0 Ê a œ 3 and b œ 24.
37. (a) satb œ "'t# *'t ""# Ê vatb œ sw atb œ $#t *'. At t œ !, the velocity is va!b œ *' ft/sec.
(b) The maximum height ocurs when vatb œ !, when t œ $. The maximum height is sa$b œ #&' ft and it occurs at t œ $
sec.
(c) Note that satb œ "'t# *'t ""# œ "'at "bat (b, so s œ ! at t œ " or t œ (. Choosing the positive value
of t, the velocity when s œ ! is va(b œ "#) ft/sec.
38.
Let x be the distance from the point on the shoreline nearest Jane's boat to the point where she lands her boat. Then she
needs to row È% x# mi at 2 mph and walk ' x mi at 5 mph. The total amount of time to reach the village is
faxb œ
È % x#
#
'x
&
have: #È%x x# œ
"
&
hours (! Ÿ x Ÿ '). Then f w axb œ
"
"
# #È% x# a#xb
"
&
œ
"& . Solving f w axb œ !, we
x
#È% x#
Ê &x œ #È% x# Ê #&x# œ %a% x# b Ê #"x# œ "' Ê x œ „
%
È#" .
We discard the negative
value of x because it is not in the domain. Checking the endpoints and critical point, we have fa!b œ #Þ#,
fŠ È%#" ‹ ¸ #Þ"#, and fa'b ¸ $Þ"'. Jane should land her boat
%
È#"
¸ !Þ)( miles down the shoreline from the point
nearest her boat.
39.
)
x
œ
h
x #(
Êhœ)
œ Ɉ) #"' ‰#
x
#"'
x
and Laxb œ Éh# ax #(b#
ax #(b# when x
minimized when faxb œ ˆ) #"' ‰#
x
!. Note that Laxb is
ax #(b# is
minimized. If f w axb œ !, then
‰ˆ #"'
‰
#ˆ) #"'
x
x# #ax #(b œ !
"(#) ‰
x$
Ê ax #(bˆ" œ ! Ê x œ #( (not acceptable
since distance is never negative or x œ "#. Then La"#b œ È#"*( ¸ %'Þ)( ftÞ
40. (a) s" œ s# Ê sin t œ sin ˆt 13 ‰ Ê sin t œ sin t cos
Êtœ
1
3
1
3
sin
1
3
"
#
cos t Ê sin t œ
41
3
or
sin t (b) The distance between the particles is s(t) œ ks" s# k œ ¸sin t sin ˆt 13 ‰¸ œ
Ê sw (t) œ
then s(0) œ
Šsin t È3 cos t‹ Šcos t È3 sin t‹
2 ¹sin t È3 cos t¹
È3
#
since
d
dx
kxk œ
x
kx k
"
#
È3
#
cos t Ê tan t œ È3
¹sin t È3 cos t¹
Ê critical times and endpoints are 0, 13 ,
, s ˆ 13 ‰ œ 0, s ˆ 561 ‰ œ 1, s ˆ 431 ‰ œ 0, s ˆ 1161 ‰ œ 1, s(21) œ
È3
#
51 41 111
6 , 3 , 6 ,
21;
Ê the greatest distance between the
particles is 1.
(c) Since sw (t) œ
Šsin t È3 cos t‹ Šcos t È3 sin t‹
2 ¹sin t È3 cos t¹
we can conclude that at t œ
1
3
and
41 w
3 , s (t)
has cusps and the distance
between the particles is changing the fastest near these points.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
224
Chapter 4 Applications of Derivatives
41. I œ
k
d2 ,
let x œ distance the point is from the stronger light source Ê 6 x œ distance the point is from the other light
source. The intensity of illumination at the point from the stronger light is I1 œ
point from the weaker light is I2 œ
light Ê k1 œ 8k2 . Ê I1 œ
8k2
x2 .
k2
.
a6 x b 2
The total intensity is given by I œ I1 I2 œ
3
3
16a6 xb3 k2 2x3 k2
and I w œ 0 Ê 16a6x3 ax6bk2xb3 2x k2 œ 0 Ê
x 3 a6 x b 3
6k2
2
Ê I ww a4b œ 48k
44 a6 4b4 0 Ê local minimum. The point
v20
g sin 2!
œ
4v20
g
Ê
dR
d!
œ
2v20
g cos 2!
and
dR
d!
and intensity of illumination at the
Since the intensity of the first light is eight times the intensity of the second
œ
42. R œ
k1
x2 ,
œ0Ê
2v20
g cos 2!
8k2
x2
k2
a6 xb2
2
Ê I w œ 16k
x3 2k2
a6 x b 3
16a6 xb3 k2 2x3 k2 œ 0 Ê x œ 4 m. I ww œ
48k2
x4
6k2
a6 x b 4
should be 4 m from the stronger light source.
œ 0 Ê ! œ 14 .
0 Ê local maximum. Thus, the firing angle of ! œ
1
4
d2 R
d !2
œ
4v20
g sin 2!
Ê
d2 R
d ! 2 ¹! œ 1
œ
4v20
ˆ1‰
g sin 2 4
4
œ 45‰ will maximize the range R.
43. (a) From the diagram we have d# œ 4r# w# . The strength of the beam is S œ kwd# œ kw a4r# w# b . When
r œ 6, then S œ 144kw kw$ . Also, Sw (w) œ 144k 3kw# œ 3k a48 w# b so Sw (w) œ 0 Ê w œ „ 4È3 ;
Sww Š4È3‹ 0 and 4È3 is not acceptable. Therefore S Š4È3‹ is the maximum strength. The dimensions
of the strongest beam are 4È3 by 4È6 inches.
(b)
(c)
Both graphs indicate the same maximum value and are consistent with each other. Changing k does not
change the dimensions that give the strongest beam (i.e., do not change the values of w and d that produce
the strongest beam).
44. (a) From the situation we have w# œ 144 d# . The stiffness of the beam is S œ kwd$ œ kd$ a144 d# b
#
#
where 0 Ÿ d Ÿ 12. Also, Sw (d) œ 4kd a108 d b Ê critical points at 0, 12, and 6È3. Both d œ 0 and
"Î#
,
È144 d#
d œ 12 cause S œ 0. The maximum occurs at d œ 6È3. The dimensions are 6 by 6È3 inches.
(b)
(c)
Both graphs indicate the same maximum value and are consistent with each other. The changing of k has
no effect.
45. (a) s œ 10 cos (1t) Ê v œ 101 sin (1t) Ê speed œ k101 sin (1t)k œ 101 ksin (1t)k Ê the maximum speed is
101 ¸ 31.42 cm/sec since the maximum value of ksin (1t)k is 1; the cart is moving the fastest at t œ 0.5 sec, 1.5 sec,
2.5 sec and 3.5 sec when ksin (1t)k is 1. At these times the distance is s œ 10 cos ˆ 1# ‰ œ 0 cm and
a œ 101# cos (1t) Ê kak œ 101# kcos (1t)k Ê kak œ 0 cm/sec#
(b) kak œ 101# kcos (1t)k is greatest at t œ 0.0 sec, 1.0 sec, 2.0 sec, 3.0 sec and 4.0 sec, and at these times the
magnitude of the cart's position is ksk œ 10 cm from the rest position and the speed is 0 cm/sec.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 4.5 Applied Optimization
225
46. (a) 2 sin t œ sin 2t Ê 2 sin t 2 sin t cos t œ 0 Ê (2 sin t)(1 cos t) œ 0 Ê t œ k1 where k is a positive integer
(b) The vertical distance between the masses is s(t) œ ks" s# k œ ˆas" s# b# ‰
Ê sw (t) œ ˆ "# ‰ a(sin 2t 2 sin t)# b
œ
4(2 cos t 1)(cos t ")(sin t)(cos t 1)
ksin 2t 2 sin tk
"Î#
Ê the greatest distance is
3È 3
#
3È 3
#
at t œ
21
3
ds
dt
œ
ds ¸
dt t=1
"
#
a(12 12t)# 64t# b
"Î#
and
1,
41
3 ,
21; then s(0) œ 0,
3È 3
2 ,
s(21) œ 0
41
3
"Î#
[2(12 12t)(12) 128t] œ
208t 144
È(12 12t)# 64t#
Ê
ds ¸
dt t=0
œ 12 knots and
œ 8 knots
(c) The graph indicates that the ships did not see
each other because s(t) 5 for all values of t.
(e)
21
3 ,
"Î#
2(cos 2t cos t)(sin 2t 2 sin t)
ksin 2t 2 sin tk
, s(1) œ 0, s ˆ 431 ‰ œ ¸sin ˆ 831 ‰ 2 sin ˆ 431 ‰¸ œ
47. (a) s œ È(12 12t)# (8t)# œ a(12 12t)# 64t# b
(b)
œ a(sin 2t 2 sin t)# b
(2)(sin 2t 2 sin t)(2 cos 2t 2 cos t) œ
Ê critical times at 0,
s ˆ 231 ‰ œ ¸sin ˆ 431 ‰ 2 sin ˆ 231 ‰¸ œ
"Î#
lim ds
t Ä _ dt
œ
(208t 144)#
É lim 144(
" t)# 64t#
tÄ_
(d) The graph supports the conclusions in parts (b)
and (c).
Š208 œ Ë lim
#
144
t ‹
#
t Ä _ 144 Š " 1‹ 64
t
#
œ É 144208 64 œ È208 œ 4È13
which equals the square root of the sums of the squares of the individual speeds.
48. The distance OT TB is minimized when OB is
a straight line. Hence n! œ n" Ê )" œ )# .
49. If v œ kax kx# , then vw œ ka 2kx and vww œ 2k, so vw œ 0 Ê x œ
v
ww
ˆ #a ‰
œ 2k 0. The maximum value of v is
ka
4
#
a
#
. At x œ
a
#
there is a maximum since
.
50. (a) According to the graph, yw a!b œ !.
(b) According to the graph, yw aLb œ !.
(c) ya!b œ !, so d œ !. Now yw axb œ $ax# #bx c, so yw a!b œ ! implies that c œ !. There fore, yaxb œ ax$ bx# and
yw axb œ $ax# #bx. Then yaLb œ aL$ bL# œ H and yw aLb œ $aL# #bL œ !, so we have two linear
equations in two unknowns a and b. The second equation gives b œ $aL
# . Substituting into the first equation, we have
aL$ $aL$
#
œ H, or
aL$
#
œ H, so a œ # LH$ . Therefore, b œ $ LH# and the equation for y is
$
#
yaxb œ # LH$ x$ $ LH# x# , or yaxb œ H’#ˆ Lx ‰ $ˆ Lx ‰ “.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
226
Chapter 4 Applications of Derivatives
51. The profit is p œ nx nc œ n(x c) œ ca(x c)" b(100 x)d (x c) œ a b(100 x)(x c)
œ a (bc 100b)x 100bc bx# . Then pw (x) œ bc 100b 2bx and pww (x) œ 2b. Solving pw (x) œ 0 Ê x œ
At x œ
c
#
ww
50 there is a maximum profit since p (x) œ 2b 0 for all x.
c
#
50.
52. Let x represent the number of people over 50. The profit is p(x) œ (50 x)(200 2x) 32(50 x) 6000
œ 2x# 68x 2400. Then pw (x) œ 4x 68 and pww œ 4. Solving pw (x) œ 0 Ê x œ 17. At x œ 17 there is a
maximum since pww (17) 0. It would take 67 people to maximize the profit.
53. (a) A(q) œ kmq" cm h# q, where q 0 Ê Aw (q) œ kmq# h
#
œ
hq# 2km
2q#
and Aww (q) œ 2kmq$ . The
ww É 2km
É 2km
É 2km
critical points are É 2km
h , 0, and
h , but only
h is in the domain. Then A Š
h ‹ 0 Ê at
q œ É 2km
h there is a minimum average weekly cost.
(b) A(q) œ
(kbq)m
q
cm h# q œ kmq" bm cm h# q, where q 0 Ê Aw (q) œ 0 at q œ É 2km
h as in (a).
Also Aww (q) œ 2kmq$ 0 so the most economical quantity to order is still q œ É 2km
h which minimizes the
average weekly cost.
54. We start with caxb œ the cost of producing x items, x !, and
c ax b
x
œ the average cost of producing x items, assumed to
be differentiable. If the average cost can be minimized, it will be at a production level at which
Ê
x c w ax b c a x b
x#
d c ax b
dx Š x ‹
œ ! (by the quotient rule) Ê x cw axb caxb œ ! (multiply both sides by x# ) Ê cw axb œ
œ!
c ax b
x
where cw axb is
the marginal cost. This concludes the proof. (Note: The theorem does not assure a production level that will give a
minimum cost, but rather, it indicates where to look to see if there is one. Find the production levels where the average cost
equals the marginal cost, then check to see if any of them give a mimimum.)
55. The profit p(x) œ r(x) c(x) œ 6x ax$ 6x# 15xb œ x$ 6x# 9x, where x 0. Then pw (x) œ 3x# 12x 9
œ 3(x 3)(x 1) and pw w (x) œ 6x 12. The critical points are 1 and 3. Thus pww (1) œ 6 0 Ê at x œ 1 there is a
local minimum, and pww (3) œ ' 0 Ê at x œ 3 there is a local maximum. But p(3) œ 0 Ê the best you can do is
break even.
56. The average cost of producing x items is caxb œ
c ax b
x
œ x# #!x #!ß !!! Ê c w axb œ #x #! œ ! Ê x œ "!, the
only critical value. The average cost is ca"!b œ $"*ß *!! per item is a minimum cost because c ww a"!b œ # !.
57. Let x œ the length of a side of the square base of the box and h œ the height of the box. V œ x2 h œ 48 Ê h œ
total cost is given by C œ 6 † x 4a4 † x hb œ 6x 2
3
C w œ 0 Ê 12x x2 768
x œ 4 Ê h œ 48
42 œ
2
‰
16xˆ 48
x2
ww
œ 6x œ 0 Ê 12x3 768 œ 0 Ê x œ 4; C œ 12 2
3 and Ca4b œ 6a4b 768
4
w
x 0 Ê C œ 12x 2
768
x ,
1536
x2
Ê C ww a4b œ 12 1536
42
768
x2
œ
48
x2 .
12x3 768
x2
The
0 Ê local minimum.
œ 288 Ê the box is 4 ft ‚ 4 ft ‚ 3 ft, with a minimum cost of $288
58. Let x œ the number of $10 increases in the charge per room, then price per room œ 50 10x, and the number of rooms
filled each night œ 800 40x Ê the total revenue is Raxb œ a50 10xba800 40xb œ 400x2 6000x 40000,
ww
0 Ÿ x Ÿ 20 Ê Rw axb œ 800x 6000; Rw axb œ 0 Ê 800x 6000 œ 0 Ê x œ 15
2 ; R axb œ 800
‰
ˆ 15 ‰
Ê Rww ˆ 15
2 œ 800 0 Ê local maximum. The price per room is 50 10 2 œ $125.
59. We have
dR
dM
œ CM M# . Solving
d# R
dM#
œ C #M œ ! Ê M œ C# . Also,
d$ R
dM$
œ # ! Ê at M œ
maximum.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
C
#
there is a
Section 4.5 Applied Optimization
227
60. (a) If v œ cr! r# cr$ , then vw œ 2cr! r 3cr# œ cr a2r! 3rb and vww œ 2cr! 6cr œ 2c ar! 3rb . The solution of
vw œ 0 is r œ 0 or 2r3! , but 0 is not in the domain. Also, vw 0 for r 2r3! and vw 0 for r 2r3! Ê at r œ 2r3!
there is a maximum.
(b) The graph confirms the findings in (a).
61. If x 0, then (x 1)#
then
0 Ê x# 1
#
#
#
#
Š a a 1 ‹ Š b b 1 ‹ Š c c 1 ‹ Š d d " ‹
62. (a) f(x) œ
x
È a# x#
Ê f w (x) œ
aa # x # b
2x Ê
"Î#
x # aa # x # b
aa # x # b
Ê gw (x) œ
ab# (d x)# b (d x)#
ab# (d x)# b$Î#
ab# (d x)# b
"Î#
œ
a# x# x#
aa# x# b$Î#
œ
a#
aa# x# b$Î#
0
"Î#
"Î#
(d x)# ab# (d x)# b
b# (d x)#
b #
0 Ê g(x) is a decreasing function of x
ab# (d x)# b$Î#
dt
Since c" , c# 0, the derivative dx is an increasing function of x (from part (a)) minus a decreasing
dt
d# t
" w
" w
w
function of x (from part (b)): dx
œ c"" f(x) c"# g(x) Ê dx
# œ c" f (x) c# g (x) 0 since f (x) dt
gw (x) 0 Ê dx
is an increasing function of x.
œ
(c)
dx
Èb# (d x)#
2. In particular if a, b, c and d are positive integers,
16.
Ê f(x) is an increasing function of x
(b) g(x) œ
x# 1
x
œ
0 and
63. At x œ c, the tangents to the curves are parallel. Justification: The vertical distance between the curves is
D(x) œ f(x) g(x), so Dw (x) œ f w (x) gw (x). The maximum value of D will occur at a point c where Dw œ 0. At
such a point, f w (c) gw (c) œ 0, or f w (c) œ gw (c).
64. (a) f(x) œ 3 4 cos x cos 2x is a periodic function with period 21
(b) No, f(x) œ 3 4 cos x cos 2x œ 3 4 cos x a2 cos# x 1b œ 2 a1 2 cos x cos# xb œ 2(1 cos x)#
Ê f(x) is never negative.
65. (a) If y œ cot x È2 csc x where 0 x 1, then yw œ (csc x) ŠÈ2 cot x csc x‹. Solving yw œ 0 Ê cos x œ
Ê x œ 14 . For 0 x 1
4
we have yw 0, and yw 0 when
1
4
x 1. Therefore, at x œ
value of y œ 1.
(b)
The graph confirms the findings in (a).
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
1
4
0
"
È2
there is a maximum
228
Chapter 4 Applications of Derivatives
66. (a) If y œ tan x 3 cot x where 0 x 1
#
, then yw œ sec# x 3 csc# x. Solving yw œ 0 Ê tan x œ „ È3
Ê x œ „ 13 , but 13 is not in the domain. Also, yww œ 2 sec# x tan x 6 csc# x cot x 0 for all 0 x Therefore at x œ 1 there is a minimum value of y œ 2È3.
1
2
.
3
(b)
The graph confirms the findings in (a).
#
#
67. (a) The square of the distance is Daxb œ ˆx $# ‰ ˆÈx !‰ œ x# #x *% , so Dw axb œ #x # and the critical
point occurs at x œ ". Since Dw axb ! for x " and Dw axb ! for x ", the critical point corresponds to the
minimum distance. The minimum distance is ÈDa"b œ
È&
# .
(b)
The minimum distance is from the point ˆ $# ß !‰ to the point a"ß "b on the graph of y œ Èx, and this occurs at the
value x œ " where Daxb, the distance squared, has its minimum value.
68. (a) Calculus Method:
The square of the distance from the point Š"ß È$‹ to Šxß È"' x# ‹ is given by
#
Daxb œ ax "b# ŠÈ"' x# È$‹ œ x# #x " "' x# #È%) $x# $ œ #x #! #È%) $x# .
Then Dw axb œ # "
#
†
#
È%) $x# a'xb
œ #
'x
È%) $x# .
Solving Dw axb œ ! we have: 'x œ #È%) $x#
Ê $'x# œ %a%) $x# b Ê *x# œ %) $x# Ê "#x# œ %) Ê x œ „ #. We discard x œ # as an extraneous solution,
leaving x œ #. Since Dw axb ! for % x # and Dw axb ! for # x %, the critical point corresponds to the
minimum distance. The minimum distance is ÈDa#b œ #.
Geometry Method:
The semicircle is centered at the origin and has radius %. The distance from the origin to Š"ß È$‹ is
#
Ê"# ŠÈ$‹ œ #. The shortest distance from the point to the semicircle is the distance along the radius
containing the point Š"ß È$‹. That distance is % # œ #.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 4.6 Newton's Method
229
(b)
The minimum distance is from the point Š"ß È$‹ to the point Š#ß #È$‹ on the graph of y œ È"' x# , and this
occurs at the value x œ # where Daxb, the distance squared, has its minimum value.
4Þ6 NEWTON'S METHOD
1. y œ x# x 1 Ê yw œ 2x 1 Ê xnb1 œ xn Ê x# œ
2
3
4
9
23 1
4
3 1
Ê x# œ 2 Ê x# œ
42"
4 1
œ
2
3
4 6 9
129
œ
x#n xn 1
# xn 1
2
3
"
#1
#"7 1 1
"
3 3
œ 3" "
90
6
5
1296
6
625 5 3
864
1
125
œ 2 Ê x# œ 2 œ
6
5
16 2 3
32 1
1296 750 1875
4320 625
œ 2 11
31
1 "4 "
œ #"
#1
20 25 4
29
œ #5 1"# œ 12
1 #
Ê x# œ "# œ
¸ 2.41667
"
1#
5
4
113
2000
œ
2500113
2000
œ
2387
#000
6. From Exercise 5, xnb1 œ xn œ 54 625512
2000
œ 54 111
# 1
œ 2
"
3
œ "3
œ
6
5
171
4945
; x! œ 1 Ê x " œ 1 œ
5763
4945
1 1 3
4 1
œ
6
5
¸ 1.16542; x! œ 1 Ê x" œ 1 "13
4 1
œ 51
31 ¸ 1.64516
2xn x#n 1
2 2xn
; x! œ 0 Ê x" œ 0 x%n 2
4xn$ ; x!
œ 1 Ê x" œ 1 "2
4
œ
5
4
00"
#0
œ "#
44"
#4
œ
5
#
Ê x# œ
5
4
Ê x# œ
5
#
625
256 2
125
16
œ
5
4
5 25
4 1
#5
625512
2000
¸ 1.1935
x%n 2
4xn$
113
2000
; x! œ 0 Ê x" œ 0 œ 15# ¸ .41667; x! œ 2 Ê x" œ 2 5. y œ x% 2 Ê yw œ 4x$ Ê xnb1 œ xn œ
2
3
¸ .61905; x! œ 1 Ê x" œ 1 x$n 3xn 1
3xn# 3
x%n xn 3
4xn$ 1
4. y œ 2x x# 1 Ê yw œ 2 2x Ê xnb1 œ xn 5
#
œ
29
œ 90
¸ 0.32222
3. y œ x% x 3 Ê yw œ 4x$ 1 Ê xnb1 œ xn Ê x# œ
13
21
111
#1
¸ 1.66667
5
3
2. y œ x$ 3x 1 Ê yw œ 3x# 3 Ê xnb1 œ xn Ê x# œ "3 œ
; x! œ 1 Ê x " œ 1 ; x! œ 1 Ê x" œ 1 "2
4
œ 1 "
4
œ 54 Ê x# œ 54 625
256 2
125
16
¸ 1.1935
7. f(x! ) œ 0 and f w (x! ) Á ! Ê xnb1 œ xn f axn b
f axn b
w
gives x" œ x! Ê x# œ x! Ê xn œ x! for all n
0. That is, all of
the approximations in Newton's method will be the root of f(x) œ 0.
8. It does matter. If you start too far away from x œ
x! œ 0.5, for instance, leads to x œ 1
#
1
#
, the calculated values may approach some other root. Starting with
as the root, not x œ
1
#
.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
230
Chapter 4 Applications of Derivatives
9. If x! œ h 0 Ê x" œ x! œh
Èh
Š
f(x! )
f w (x! )
Èh
Š È" ‹
2
f(h)
f w (h)
œ h ŠÈh‹ Š2Èh‹ œ h;
"
‹
#Èh
if x! œ h 0 Ê x" œ x! œ h œh
f(x! )
f w (x! )
œ h f(h)
f w (h)
œ h ŠÈh‹ Š2Èh‹ œ h.
h
"Î$
10. f(x) œ x"Î$ Ê f w (x) œ ˆ "3 ‰ x#Î$ Ê xnb1 œ xn " xn #Î$
ˆ ‰ xn
3
œ 2xn ; x! œ 1 Ê x" œ 2, x# œ 4, x$ œ 8, and
x% œ 16 and so forth. Since kxn k œ 2lxnc1 l we may conclude
that n Ä _ Ê kxn k Ä _.
11. i) is equivalent to solving x$ $x " œ !.
ii) is equivalent to solving x$ $x " œ !.
iii) is equivalent to solving x$ $x " œ !.
iv) is equivalent to solving x$ $x " œ !.
All four equations are equivalent.
12. f(x) œ x 1 0.5 sin x Ê f w (x) œ 1 0.5 cos x Ê xnb1 œ xn 13. f(x) œ tan x 2x Ê f w (x) œ sec# x 2 Ê xnb1 œ xn xn 1 0.5 sin xn
1 0.5 cos xn
tan axn b 2xn
sec# axn b
; if x! œ 1.5, then x" œ 1.49870
; x! œ 1 Ê x" œ 1.2920445
Ê x# œ 1.155327774 Ê x16 œ x17 œ 1.165561185
14. f(x) œ x% 2x$ x# 2x 2 Ê f w (x) œ 4x$ 6x# 2x 2 Ê xnb1 œ xn x%n 2xn$ xn# 2xn 2
4xn$ 6xn# 2xn 2
if x! œ 0.5, then x% œ 0.630115396; if x! œ 2.5, then x% œ 2.57327196
15. (a) The graph of f(x) œ sin 3x 0.99 x# in the window
2 Ÿ x Ÿ 2, 2 Ÿ y Ÿ 3 suggests three roots.
However, when you zoom in on the x-axis near x œ 1.2,
you can see that the graph lies above the axis there.
There are only two roots, one near x œ 1, the other
near x œ 0.4.
(b) f(x) œ sin 3x 0.99 x# Ê f w (x) œ 3 cos 3x 2x
Ê xnb1 œ xn sin (3xn ) 0.99xn#
3 cos (3xn ) 2xn
and the solutions
are approximately 0.35003501505249 and
1.0261731615301
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
;
Section 4.6 Newton's Method
16. (a) Yes, three times as indicted by the graphs
(b) f(x) œ cos 3x x Ê f w (x)
œ 3 sin 3x 1 Ê xnb1
œ xn cos a3xn b xn
3 sin a3xn b 1
; at
approximately 0.979367,
0.887726, and 0.39004 we have
cos 3x œ x
17. f(x) œ 2x% 4x# 1 Ê f w (x) œ 8x$ 8x Ê xnb1 œ xn 2x%n 4x#n 1
8xn$ 8xn
; if x! œ 2, then x' œ 1.30656296; if
x! œ 0.5, then x$ œ 0.5411961; the roots are approximately „ 0.5411961 and „ 1.30656296 because f(x) is
an even function.
18. f(x) œ tan x Ê f w (x) œ sec# x Ê xnb1 œ xn tan axn b
sec# axn b
; x! œ 3 Ê x" œ 3.13971 Ê x# œ 3.14159 and we
approximate 1 to be 3.14159.
19. From the graph we let x! œ 0.5 and f(x) œ cos x 2x
Ê xnb1 œ xn cos axn b 2xn
sin axn b 2
Ê x" œ .45063
Ê x# œ .45018 Ê at x ¸ 0.45 we have cos x œ 2x.
20. From the graph we let x! œ 0.7 and f(x) œ cos x x
Ê xnb1 œ xn xn cos axn b
1 sin axn b
Ê x" œ .73944
Ê x# œ .73908 Ê at x ¸ 0.74 we have cos x œ x.
21. The x-coordinate of the point of intersection of y œ x2 ax 1b and y œ
Ê x3 x2 1
x
Ê xnb1 œ xn 1
x
œ 0 Ê The x-coordinate is the root of faxb œ x3 x2 xn xn x1n
3x2n 2xn 12
x
3
2
is the solution of x2 ax 1b œ
1
x
Ê f w axb œ 3x2 2x 1
x2 .
1
x
Let x0 œ 1
Ê x" œ 0.83333 Ê x2 œ 0.81924 Ê x3 œ 0.81917 Ê x7 œ 0.81917 Ê r ¸ 0.8192
n
22. The x-coordinate of the point of intersection of y œ Èx and y œ 3 x2 is the solution of Èx œ 3 x2
1
Ê Èx 3 x2 œ 0 Ê The x-coordinate is the root of faxb œ Èx 3 x2 Ê f w axb œ 2È
2x. Let x0 œ 1
x
Ê xnb1 œ xn Èxn 3 x2n
Èxn 2xn
1
2
Ê x" œ 1.4 Ê x2 œ 1.35556 Ê x3 œ 1.35498 Ê x7 œ 1.35498 Ê r ¸ 1.3550
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
231
232
Chapter 4 Applications of Derivatives
23. If f(x) œ x$ 2x 4, then f(1) œ 1 0 and f(2) œ 8 0 Ê by the Intermediate Value Theorem the equation
x$ 2x 4 œ 0 has a solution between 1 and 2. Consequently, f w (x) œ 3x# 2 and xnb1 œ xn x$n 2xn 4
3x#n 2
.
Then x! œ 1 Ê x" œ 1.2 Ê x# œ 1.17975 Ê x$ œ 1.179509 Ê x% œ 1.1795090 Ê the root is approximately
1.17951.
24. We wish to solve 8x% 14x$ 9x# 11x 1 œ 0. Let f(x) œ 8x% 14x$ 9x# 11x 1, then
f w (x) œ 32x$ 42x# 18x 11 Ê xnb1 œ xn x!
1.0
0.1
0.6
2.0
8x%n 14x$n 9x#n 11xn 1
3#xn$ 42xn# 18xn 11
.
approximation of corresponding root
0.976823589
0.100363332
0.642746671
1.983713587
25. f(x) œ 4x% 4x# Ê f w (x) œ 16x$ 8x Ê xib1 œ xi faxi b
f axi b
w
œ xi xi$ xi
.
%x#i #
Iterations are performed using the
procedure in problem 13 in this section.
(a) For x! œ # or x! œ !Þ), xi Ä " as i gets large.
(b) For x! œ !Þ& or x! œ !Þ#&, xi Ä ! as i gets large.
(c) For x! œ !Þ) or x! œ #, xi Ä " as i gets large.
(d) (If your calculator has a CAS, put it in exact mode, otherwise approximate the radicals with a decimal value.)
For x! œ x! œ
È21
7
È
721
or x! œ or x! œ
È
721
È21
7 ,
Newton's method does not converge. The values of xi alternate between
as i increases.
26. (a) The distance can be represented by
#
D(x) œ É(x 2)# ˆx# "# ‰ , where x
0. The
distance D(x) is minimized when
#
f(x) œ (x 2)# ˆx# "# ‰ is minimized. If
#
f(x) œ (x 2)# ˆx# "# ‰ , then
f w (x) œ 4 ax$ x 1b and f ww (x) œ 4 a3x# 1b 0.
Now f w (x) œ 0 Ê x$ x 1 œ 0 Ê x ax# 1b œ 1
Ê x œ x#"1 .
(b) Let g(x) œ
"
x # 1
Ê xnb1 œ xn x œ ax# 1b
" xn Πx#
n 1
Î
2xn Ñ
#
x# 1 1
ÏŠ
‹
n
"
#
x Ê gw (x) œ ax# 1b (2x) 1 œ
2x
ax # 1 b #
1
; x! œ 1 Ê x% œ 0.68233 to five decimal places.
Ò
27. f(x) œ (x 1)%! Ê f w (x) œ 40(x 1)$* Ê xnb1 œ xn axn 1b%!
40 axn 1b$*
39xn "
40
œ
. With x! œ 2, our computer
gave x)( œ x)) œ x)* œ â œ x#!! œ 1.11051, coming within 0.11051 of the root x œ 1.
28. Since s œ r ) Ê 3 œ r ) Ê ) œ 3r . Bisect the angle ) to obtain a right tringle with hypotenuse r and opposite side
of length 1. Then sin
farb œ sinˆ 2r3 ‰ 1
r
)
2
œ
1
r
Ê sin
ˆ 3r ‰
2
œ
1
r
Ê sinˆ 2r3 ‰ œ
1
r
Ê sin
3
2r
Ê f w arb œ 2r32 cosˆ 2r3 ‰ r12 ; r0 œ 1 Ê rn 1 œ rn Ê r2 œ 1.00282 Ê r3 œ 1.00282 Ê r ¸ 1.0028 Ê ) œ
3
1.00282
1
r
œ 0. Thus the solution r is a root of
sinˆ 2r3n ‰ r1n
32 cosˆ 2r3n ‰ 2rn
1
r2
n
Ê r1 œ 1.00280
¸ 2.9916
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 4.7 Antiderivatives
4.7 ANTIDERIVATIVES
1. (a) x#
(b)
x$
3
(c)
x$
3
x# x
2. (a) 3x#
(b)
x)
8
(c)
x)
8
3x# 8x
3. (a) x$
(b) x3
4. (a) x#
(b) x4 $
$
(c) x3 x# 3x
#
x$
3
(c)
x #
#
x#
2
x
5. (a)
"
x
(b)
5
x
(c) 2x 6. (a)
"
x#
(b)
"
4x#
(c)
x%
4
(c)
2
3
Èx$ 2Èx
(c)
3
4
x%Î$ 3# x#Î$
5
x
"
#x #
7. (a) Èx$
(b) Èx
8. (a) x%Î$
(b)
9. (a) x#Î$
(b) x"Î$
(c) x"Î$
10. (a) x"Î#
(b) x"Î#
(c) x$Î#
11. (a) cos (1x)
(b) 3 cos x
(c)
12. (a) sin (1x)
(b) sin ˆ 1#x ‰
(c) ˆ 12 ‰ sin ˆ 1#x ‰ 1 sin x
13. (a) tan x
(b) 2 tan ˆ x3 ‰
‰
(c) 23 tan ˆ 3x
#
14. (a) cot x
‰
(b) cot ˆ 3x
#
(c) x 4 cot (2x)
15. (a) csc x
(b)
"
5
csc (5x)
(c) 2 csc ˆ 1#x ‰
16. (a) sec x
(b)
4
3
sec (3x)
(c)
17.
' (x 1) dx œ
19.
'
ˆ3t# #t ‰ dt œ t$ 21.
'
a2x$ 5x 7b dx œ
23.
'
ˆ x"# x# 3" ‰ dx œ ' ˆx# x# 3" ‰ dx œ
24.
'
ˆ "5 2
x$
x#
#
"
#
x#Î$
xC
t#
4
C
"
#
x% 5# x# 7x C
x "
1
2x‰ dx œ ' ˆ 5" 2x$ 2x‰ dx œ
"
5
x$
3
cos (1x)
1
2
1
cos (3x)
sec ˆ 1#x ‰
18.
' (5 6x) dx œ 5x 3x# C
20.
'
Š t# 4t$ ‹ dt œ
22.
'
a1 x# 3x& b dx œ x "3 x$ #" x' C
#
3" x C œ x" #
x Š 2x# ‹ 2x#
#
Cœ
x
5
x$
3
t$
6
"
x#
t% C
x
3
C
x# C
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
233
234
Chapter 4 Applications of Derivatives
x#Î$
25.
' x"Î$ dx œ
27.
'
ˆÈx $Èx‰ dx œ ' ˆx"Î# x"Î$ ‰ dx œ
28.
'
Š
29.
'
Š8y 30.
'
Èx
#
Š "7 2
3
2
Èx ‹
2
‹
y"Î%
1
‹
y&Î%
Cœ
3
#
x#Î$ C
' x&Î% dx œ
26.
x$Î#
dx œ ' ˆ "# x"Î# 2x"Î# ‰ dx œ
dy œ ' ˆ8y 2y"Î% ‰ dy œ
dy œ ' ˆ "7 y&Î% ‰ dy œ
"
7
8y#
#
x%Î$
3
#
"
#
Cœ
4
3
2
3
x "Î%
"4
Cœ
4
% x
È
C
x$Î# 43 x%Î$ C
$Î#
"Î#
#
#
Šx3 ‹ 2 Šx" ‹ C œ
"
3
x$Î# 4x"Î# C
$Î%
2 Š y 3 ‹ C œ 4y# 83 y$Î% C
4
"Î%
y Š y 1 ‹ C œ
4
y
7
4
y"Î%
C
31.
'
2x a1 x$ b dx œ ' a2x 2x# b dx œ
32.
'
x$ (x 1) dx œ ' ax# x$ b dx œ
33.
'
tÈtÈt
t#
34.
'
4 È t
t$
35.
' 2 cos t dt œ 2 sin t C
36.
' 5 sin t dt œ 5 cos t C
37.
' 7 sin 3) d) œ 21 cos 3) C
38.
' 3 cos 5) d) œ 35 sin 5) C
39.
' 3 csc# x dx œ 3 cot x C
40.
' sec3# x dx œ tan3 x C
41.
'
42.
'
43.
' a4 sec x tan x 2 sec# xb dx œ 4 sec x 2 tan x C
44.
'
45.
' asin 2x csc# xb dx œ "# cos 2x cot x C
46.
' (2 cos 2x 3 sin 3x) dx œ sin 2x cos 3x C
47.
'
1 cos 4t
#
dt œ ' ˆ "# "
#
cos 4t‰ dt œ
"
#
t "# ˆ sin4 4t ‰ C œ
t
2
sin 4t
8
C
48.
'
1 cos 6t
#
dt œ ' ˆ "# "
#
cos 6t‰ dt œ
"
#
t "# ˆ sin6 6t ‰ C œ
t
2
sin 6t
12
C
49.
' a1 tan# )b d) œ ' sec# ) d) œ tan ) C
50.
' a2 tan# )b d) œ ' a1 1 tan# )b d) œ ' a1 sec# )b d) œ ) tan ) C
51.
' cot# x dx œ ' acsc# x 1b dx œ cot x x C
$Î#
dt œ ' Š t4$ csc ) cot )
#
"
2
dt œ ' Š t t# t"Î#
t# ‹
t"Î#
t$ ‹
2x#
#
x "
1
"
2 Š x1 ‹ C œ x# #
Š x# ‹ C œ "x dt œ ' ˆt"Î# t$Î# ‰ dt œ
t"Î#
"
#
2
x
C
"
#x #
C
"Î#
Š t " ‹ C œ 2Èt #
t
dt œ ' ˆ4t$ t&Î# ‰ dt œ 4 Š #
‹ Š t 3 ‹ C œ t2# #
$Î#
#
d) œ "# csc ) C
acsc# x csc x cot xb dx œ #" cot x "
#
2
5
sec ) tan ) d) œ
2
5
2
Èt
C
2
3t$Î#
C
sec ) C
csc x C
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 4.7 Antiderivatives
52.
' a1 cot# xb dx œ ' a1 acsc# x 1bb dx œ ' a2 csc# xb dx œ 2x cot x C
53.
' cos ) (tan ) sec )) d) œ ' (sin ) 1) d) œ cos ) ) C
54.
'
csc )
csc ) sin )
55.
d
dx
2)
C‹ œ
Š (7x28
56.
d
dx
Š (3 x3 5)
57.
d
dx
ˆ "5 tan (5x 1) C‰ œ
58.
d
dx
ˆ3 cot ˆ x 3 " ‰ C‰ œ 3 ˆcsc# ˆ x 3 " ‰‰ ˆ "3 ‰ œ csc# ˆ x 3 " ‰
59.
d
dx
#
ˆ x"
‰
œ
1 C œ (1)(1)(x 1)
)
)‰
'
‰ ˆ sin
d) œ ' ˆ csc csc
) sin )
sin ) d) œ
%
"
4(7x 2)$ (7)
28
d) œ '
# (3)
‹ œ (3x 5)#
asec# (5x 1)b (5) œ sec# (5x 1)
#
"
(x 1)#
x#
#
d
dx
Š x# sin x C‹ œ
(b) Wrong:
d
dx
(x cos x C) œ cos x x sin x Á x sin x
(c) Right:
d
d)
d
d)
(b) Right:
(c) Right:
2x
#
d
dx
ˆ xx 1 C‰ œ
x#
#
cos x œ x sin x $
Š sec3 ) C‹ œ
3 sec# )
3
ˆ "# tan# ) C‰ œ "# (2 tan )) sec# ) œ tan ) sec# )
ˆ "# sec# ) C‰ œ "# (2 sec )) sec ) tan ) œ tan ) sec# )
$
3(2x 1)# (2)
3
Š (2x 3 1) C‹ œ
(b) Wrong:
d
dx
a(2x 1)$ Cb œ 3(2x 1)# (2) œ 6(2x 1)# Á 3(2x 1)#
d
dx
ax# x Cb
(b) Wrong:
d
dx
Šax# xb
65. Right:
66. Wrong:
d
dx
Š
"Î#
"Î#
œ
"
#
ax# x Cb
C‹ œ
$
"
#
ax# xb
"
Œ 3 ŠÈ2x 1‹ C œ
d ˆ x 3 ‰3
dx Š x 2
d
dx
œ 2(2x 1)# Á (2x 1)#
a(2x 1)$ Cb œ 6(2x 1)#
64. (a) Wrong:
(c) Right:
"
(x 1)#
(sec ) tan )) œ sec$ ) tan ) Á tan ) sec# )
d
dx
d
dx
œ
cos x Á x sin x
63. (a) Wrong:
(c) Right:
(x 1)(") x(1)
(x 1)#
(x cos x sin x C) œ cos x x sin x cos x œ x sin x
d
d)
62. (a) Wrong:
sin x 60.
61. (a) Wrong:
d
dx
d) œ ' sec# ) d) œ tan ) C
"
cos# )
œ (7x 2)$
C‹ œ Š (3x 35)
"
5
"
1sin# )
sinˆx2 ‰
x
3‰
C‹ œ 3ˆ xx 2
C‹ œ
d
dx
"Î#
"Î#
(2x 1) œ
(2x 1) œ
2x 1
2 È x# x C
2x 1
2 È x# x
ˆ 3" (2x 1)$Î# C‰ œ
2 ax 2b†1 ax 3b†1
ax 2 b
2
x†cosˆx2 ‰a2xb sinˆx2 ‰†1
x2
œ
3
6
2
Á
Á È2x 1
(2x 1)"Î# (2) œ È2x 1
3b
5
œ 3 aaxx œ
2 b2 ax 2 b 2
2x2 cosˆx2 ‰ sinˆx2 ‰
x2
Á È2x 1
15ax 3b2
ax 2 b 4
x cosˆx2 ‰ sinˆx2 ‰
x2
67. Graph (b), because
dy
dx
œ 2B Ê y œ x# C. Then y(1) œ 4 Ê C œ 3.
68. Graph (b), because
dy
dx
œ B Ê y œ "# x# C. Then y(1) œ 1 Ê C œ
3
#
.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
235
236
Chapter 4 Applications of Derivatives
69.
dy
dx
œ 2x 7 Ê y œ x# 7x C; at x œ 2 and y œ 0 we have 0 œ 2# 7(2) C Ê C œ 10 Ê y œ x# 7x 10
70.
dy
dx
œ 10 x Ê y œ 10x 71.
dy
dx
œ
"
x#
C; at x œ 0 and y œ 1 we have 1 œ 10(0) x œ x# x Ê y œ x" Ê y œ x" 72.
x#
#
x#
#
"
#
or y œ x" x#
#
0#
#
C Ê C œ 1 Ê y œ 10x x#
#
C; at x œ 2 and y œ 1 we have 1 œ 2" "
#
2#
#
x#
#
1
C Ê C œ "#
dy
dx
œ 9x# 4x 5 Ê y œ 3x$ 2x# 5x C; at x œ 1 and y œ 0 we have 0 œ 3(1)$ 2(1)# 5(1) C
dy
dx
œ $x#Î$ Ê y œ
Ê C œ 10 Ê y œ 3x$ 2x# 5x 10
73.
$x"Î$
"
$
C œ *; at x œ 9x"Î$ C; at x œ " and y œ & we have & œ *(")"Î$ C Ê C œ %
Ê y œ 9x"Î$ %
"
#È x
œ
"
#
x"Î# Ê y œ x"Î# C; at x œ 4 and y œ 0 we have 0 œ 4"Î# C Ê C œ 2 Ê y œ x"Î# 2
74.
dy
dx
œ
75.
ds
dt
œ 1 cos t Ê s œ t sin t C; at t œ 0 and s œ 4 we have 4 œ 0 sin 0 C Ê C œ 4 Ê s œ t sin t 4
76.
ds
dt
œ cos t sin t Ê s œ sin t cos t C; at t œ 1 and s œ 1 we have 1 œ sin 1 cos 1 C Ê C œ 0
Ê s œ sin t cos t
77.
dr
d)
œ 1 sin 1) Ê r œ cos (1)) C; at r œ 0 and ) œ 0 we have 0 œ cos (10) C Ê C œ " Ê r œ cos (1)) 1
78.
dr
d)
œ cos 1) Ê r œ
79.
dv
dt
œ
80.
dv
dt
œ 8t csc# t Ê v œ 4t# cot t C; at v œ 7 and t œ
"
#
"
1
sin(1)) C; at r œ 1 and ) œ 0 we have 1 œ
sec t tan t Ê v œ
"
#
sec t C; at v œ 1 and t œ 0 we have 1 œ
1
#
Ê v œ 4t# cot t 7 1#
81.
d# y
dx#
Ê
œ 2 6x Ê
dy
dx
œ 2x 3x# C" ; at
dy
dx
#
"
1
dy
dx
sin (10) C Ê C œ " Ê r œ
"
#
sec (0) C Ê C œ
"
#
"
1
Ê vœ
sin (1)) 1
"
#
sec t "
#
#
we have 7 œ 4 ˆ 1# ‰ cot ˆ 1# ‰ C Ê C œ 7 1#
œ 4 and x œ 0 we have 4 œ 2(0) 3(0)# C" Ê C" œ 4
œ 2x 3x 4 Ê y œ x# x$ 4x C# ; at y œ 1 and x œ 0 we have 1 œ 0# 0$ 4(0) C# Ê C# œ 1
Ê y œ x# x$ 4x 1
82.
d# y
dx#
œ0 Ê
dy
dx
œ C" ; at
dy
dx
œ 2 and x œ 0 we have C" œ 2 Ê
dy
dx
œ 2 Ê y œ 2x C# ; at y œ 0 and x œ 0 we
have 0 œ 2(0) C# Ê C# œ 0 Ê y œ 2x
83.
d# r
dt#
œ
d# s
dt#
œ
2
t$
œ 2t$ Ê
dr
dt
œ t# C" ; at
dr
dt
œ 1 and t œ 1 we have 1 œ (1)# C" Ê C" œ 2 Ê
dr
dt
œ t# 2
Ê r œ t" 2t C# ; at r œ 1 and t œ 1 we have 1 œ 1" 2(1) C# Ê C# œ 2 Ê r œ t" 2t 2 or
r œ "t 2t 2
84.
3t
8
Ê
ds
dt
œ
3t#
16
C" ; at
s œ 4 and t œ 4 we have 4 œ
ds
dt
$
4
16
œ 3 and t œ 4 we have 3 œ
C# Ê C# œ 0 Ê s œ
3(4)#
16
C" Ê C" œ 0 Ê
ds
dt
œ
3t#
16
$
t
16
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Ê sœ
t$
16
C# ; at
Section 4.7 Antiderivatives
85.
d$ y
dx$
œ6 Ê
Ê
dy
dx
d# y
dx#
#
œ 6x C" ; at
œ 3x 8x C# ; at
d$ )
dt$
œ0 Ê
#
œ 8 and x œ 0 we have 8 œ 6(0) C" Ê C" œ 8 Ê
œ 0 and x œ 0 we have 0 œ 3(0)# 8(0) C# Ê C# œ 0 Ê
d# y
dx# œ 6x 8
dy
#
dx œ 3x 8x
$
#
Ê y œ x 4x C$ ; at y œ 5 and x œ 0 we have 5 œ 0$ 4(0)# C$ Ê C$ œ 5 Ê y œ x 4x 5
86.
$
dy
dx
d# y
dx#
237
d# )
dt#
œ C" ; at
d# )
dt#
d# )
d)
dt# œ 2 Ê dt
"
d)
#
dt œ 2t # Ê ) œ t
) œ t# "# t È2
œ 2 and t œ 0 we have
have "# œ 2(0) C# Ê C# œ "# Ê
È2 œ 0# " (0) C$ Ê C$ œ È2 Ê
#
œ "# and t œ 0 we
"# t C$ ; at ) œ È2 and t œ 0 we have
œ 2t C# ; at
d)
dt
87. yÐ%Ñ œ sin t cos t Ê ywww œ cos t sin t C" ; at ywww œ 7 and t œ 0 we have 7 œ cos (0) sin (0) C" Ê C" œ 6
Ê ywww œ cos t sin t 6 Ê yww œ sin t cos t 6t C# ; at yww œ 1 and t œ 0 we have
1 œ sin (0) cos (0) 6(0) C# Ê C# œ 0 Ê yww œ sin t cos t 6t Ê yw œ cos t sin t 3t# C$ ; at
yw œ 1 and t œ 0 we have 1 œ cos (0) sin (0) 3(0)# C$ Ê C$ œ 0 Ê yw œ cos t sin t 3t#
Ê y œ sin t cos t t$ C% ; at y œ 0 and t œ 0 we have 0 œ sin (0) cos (0) 0$ C% Ê C% œ 1
Ê y œ sin t cos t t$ 1
88. yÐ%Ñ œ cos x 8 sin (2x) Ê ywww œ sin x 4 cos (2x) C" ; at ywww œ 0 and x œ 0 we have
0 œ sin (0) % cos (2(0)) C" Ê C" œ 4 Ê ywww œ sin x 4 cos (2x) 4 Ê yww œ cos x 2 sin (2x) 4x C# ;
at yww œ 1 and x œ 0 we have 1 œ cos (0) 2 sin (2(0)) 4(0) C# Ê C# œ 0 Ê yww œ cos x 2 sin (2x) 4x
Ê yw œ sin x cos (2x) 2x# C$ ; at yw œ 1 and x œ 0 we have 1 œ sin (0) cos (2(0)) 2(0)# C$ Ê C$ œ 0
Ê yw œ sin x cos (2x) 2x# Ê y œ cos x "# sin (2x) 23 x$ C% ; at y œ 3 and x œ 0 we have
3 œ cos (0) "
#
sin (2(0)) 23 (0)$ C% Ê C% œ 4 Ê y œ cos x "
#
sin (2x) 23 x$ 4
89. m œ yw œ 3Èx œ 3x"Î# Ê y œ 2x$Î# C; at (*ß 4) we have 4 œ 2(9)$Î# C Ê C œ 50 Ê y œ 2x$Î# 50
90. Yes. If F(x) and G(x) both solve the initial value problem on an interval I then they both have the same first derivative.
Therefore, by Corollary 2 of the Mean Value Theorem there is a constant C such that F(x) œ G(x) C for all x. In
particular, F(x! ) œ G(x! ) C, so C œ F(x! ) G(x! ) œ 0. Hence F(x) œ G(x) for all x.
91.
dy
dx
œ 1 34 x"Î$ Ê y œ ' ˆ1 34 x"Î$ ‰ dx œ x x%Î$ C; at (1ß 0.5) on the curve we have 0.5 œ 1 1%Î$ C
Ê C œ 0.5 Ê y œ x x%Î$ 92.
dy
dx
œ x 1 Ê y œ ' (x 1) dx œ
Ê yœ
93.
dy
dx
"
#
x#
#
x
x#
#
x C; at (1ß 1) on the curve we have 1 œ
(")#
#
(1) C Ê C œ "#
"
#
œ sin x cos x Ê y œ ' (sin x cos x) dx œ cos x sin x C; at (1ß 1) on the curve we have
" œ cos (1) sin (1) C Ê C œ 2 Ê y œ cos x sin x 2
94.
dy
dx
œ
"
#È x
1 sin 1x œ
"
#
x"Î# 1 sin 1x Ê y œ ' ˆ #" x"Î# sin 1x‰ dx œ x"Î# cos 1x C; at (1ß #) on the
curve we have 2 œ 1"Î# cos 1(1) C Ê C œ 0 Ê y œ Èx cos 1x
95. (a)
ds
dt
œ 9.8t 3 Ê s œ 4.9t# 3t C; (i) at s œ 5 and t œ 0 we have C œ 5 Ê s œ 4.9t# 3t 5;
displacement œ s(3) s(1) œ ((4.9)(9) 9 5) (4.9 3 5) œ 33.2 units; (ii) at s œ 2 and t œ 0 we have
C œ 2 Ê s œ 4.9t# 3t 2; displacement œ s(3) s(1) œ ((4.9)(9) 9 2) (4.9 3 2) œ 33.2 units;
(iii) at s œ s! and t œ 0 we have C œ s! Ê s œ 4.9t# 3t s! ; displacement œ s(3) s(1)
œ ((4.9)(9) 9 s! ) (4.9 3 s! ) œ 33.2 units
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
238
Chapter 4 Applications of Derivatives
(b) True. Given an antiderivative f(t) of the velocity function, we know that the body's position function is
s œ f(t) C for some constant C. Therefore, the displacement from t œ a to t œ b is (f(b) C) (f(a) C)
œ f(b) f(a). Thus we can find the displacement from any antiderivative f as the numerical difference
f(b) f(a) without knowing the exact values of C and s.
96. a(t) œ vw (t) œ 20 Ê v(t) œ 20t C; at (0ß 0) we have C œ 0 Ê v(t) œ 20t. When t œ 60, then v(60) œ 20(60) œ 1200
97. Step 1:
d# s
dt#
œ k Ê
#
k Š t# ‹
sœ
Step 2:
ds
dt
d# s
dt#
œ k Ê
Ê
ds
dt
œ kt C" ; at
ds
dt
‰
k ˆ 88
k
#
#
#
œ kt 44 Ê s œ #
Ê s œ kt# 44t. Then
Ê 968
k 1936
k
œ 45 Ê
#
kt
#
kt#
#
ds
dt
œ kt 88 Ê
88t
88
k
‰ Ê 242 œ (88)
88 ˆ 88
k
2k œ ' k dt œ kt C; at
ds
dt
œ 88 and t œ 0 we have C" œ 88 Ê
88t C# ; at s œ 0 and t œ 0 we have C# œ 0 Ê s œ œ 0 Ê 0 œ kt 88 Ê t œ
Step 3: 242 œ
98.
ds
dt
m
sec .
ds
dt
(88)#
k
Ê 242 œ
(88)#
2k
Ê k œ 16
œ 44 when t œ 0 we have 44 œ k(0) C Ê C œ 44
#
44t C" ; at s œ 0 when t œ 0 we have 0 œ k(0)
# 44(0) C" Ê C" œ 0
ds
44
dt œ 0 Ê kt 44 œ 0 Ê t œ k
968
968
ft
k œ 45 Ê k œ 45 ¸ 21.5 sec2 .
99. (a) v œ ' a dt œ ' ˆ15t"Î# 3t"Î# ‰ dt œ 10t$Î# 6t"Î# C;
‰
and s ˆ 44
k œ
ds
dt
‰
k ˆ 44
k
#
#
‰
44 ˆ 44
k œ 45
(1) œ 4 Ê 4 œ 10(1)$Î# 6(1)"Î# C Ê C œ 0
Ê v œ 10t$Î# 6t"Î#
(b) s œ ' v dt œ ' ˆ10t$Î# 6t"Î# ‰ dt œ 4t&Î# 4t$Î# C; s(1) œ 0 Ê 0 œ 4(1)&Î# 4(1)$Î# C Ê C œ 0
Ê s œ 4t&Î# 4t$Î#
100.
d# s
dt#
œ 5.2 Ê
ds
dt
œ 5.2t C" ; at
ds
dt
œ 0 and t œ 0 we have C" œ 0 Ê
ds
dt
œ 5.2t Ê s œ 2.6t# C# ; at s œ 4
4
and t œ 0 we have C# œ 4 Ê s œ 2.6t# 4. Then s œ 0 Ê 0 œ 2.6t# 4 Ê t œ É 2.6
¸ 1.24 sec, since t 0
101.
d# s
dt#
œa Ê
ds
dt
œ ' a dt œ at C;
when t œ 0 Ê s! œ
a(0)#
#
ds
dt
œ v! when t œ 0 Ê C œ v! Ê
v! (0) C" Ê C" œ s! Ê s œ
at#
#
102. The appropriate initial value problem is: Differential Equation:
s œ s! when t œ 0. Thus,
Ê
ds
dt
ds
dt
œ'
œ gt v! . Thus s œ '
"
#
Thus s œ gt# v! t s!.
ds
dt
œ at v! Ê s œ
at#
#
v! t C" ; s œ s!
v! t s!
d# s
dt#
œ g with Initial Conditions:
g dt œ gt C" ; ds
dt (0) œ v! Ê v! œ (g)(0) agt v! b dt œ "# gt# v! t C# ; s(0) œ s! œ "#
ds
dt
œ v! and
C" Ê C" œ v!
(g)(0)# v! (0) C# Ê C# œ s!
103 106 Example CAS commands:
Maple:
with(student):
f := x -> cos(x)^2 + sin(x);
ic := [x=Pi,y=1];
F := unapply( int( f(x), x ) + C, x );
eq := eval( y=F(x), ic );
solnC := solve( eq, {C} );
Y := unapply( eval( F(x), solnC ), x );
DEplot( diff(y(x),x) = f(x), y(x), x=0..2*Pi, [[y(Pi)=1]],
color=black, linecolor=black, stepsize=0.05, title="Section 4.7 #103" );
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Chapter 4 Practice Exercises
239
Mathematica: (functions and values may vary)
The following commands use the definite integral and the Fundamental Theorem of calculus to construct the solution
of the initial value problems for exercises 103 - 105.
Clear[x, y, yprime]
yprime[x_] = Cos[x]2 Sin[x];
initxvalue = 1; inityvalue = 1;
y[x_] = Integrate[yprime[t], {t, initxvalue, x}] inityvalue
If the solution satisfies the differential equation and initial condition, the following yield True
yprime[x]==D[y[x], x] //Simplify
y[initxvalue]==inityvalue
Since exercise 106 is a second order differential equation, two integrations will be required.
Clear[x, y, yprime]
y2prime[x_] = 3 Exp[x/2] 1;
initxval = 0; inityval = 4; inityprimeval = 1;
yprime[x_] = Integrate[y2prime[t],{t, initxval, x}] inityprimeval
y[x_] = Integrate[yprime[t], {t, initxval, x}] inityval
Verify that y[x] solves the differential equation and initial condition and plot the solution (red) and its derivative (blue).
y2prime[x]==D[y[x], {x, 2}]//Simplify
y[initxval]==inityval
yprime[initxval]==inityprimeval
Plot[{y[x], yprime[x]}, {x, initxval 3, initxval 3}, PlotStyle Ä {RGBColor[1,0,0], RGBColor[0,0,1]}]
CHAPTER 4 PRACTICE EXERCISES
1. No, since f(x) œ x$ 2x tan x Ê f w (x) œ 3x# 2 sec# x 0 Ê f(x) is always increasing on its domain
cos x
2. No, since g(x) œ csc x 2 cot x Ê gw (x) œ csc x cot x 2 csc# x œ sin
#x 2
sin# x
œ sin"# x (cos x 2) 0
Ê g(x) is always decreasing on its domain
3. No absolute minimum because x lim
(7 x)(11 3x)"Î$ œ _. Next f w (x) œ
Ä_
(11 3x)"Î$ (7 x)(11 3x)#Î$ œ
(11 3x) (7 x)
(11 3x)#Î$
œ
4(1 x)
(11 3x)#Î$
Ê x œ 1 and x œ
11
3
are critical points.
Since f w 0 if x 1 and f w 0 if x 1, f(1) œ 16 is the absolute maximum.
4. f(x) œ
ax b
x# 1
Ê f w (x) œ
We require also that f(3)
w
#a$x "bax $b
ax # 1 b #
#
a ax# 1b 2x(ax b)
ab
œ aaxax#2bx
1 b#
ax # 1 b#
œ 1. Thus " œ 3a8b Ê 3a b œ
w
"
; f w (3) œ 0 Ê '%
(*a 'b a) œ ! Ê &a $b œ !.
). Solving both equations yields a œ 6 and b œ 10. Now,
so that f œ ± ± ± ± . Thus f w changes sign at x œ $ from
1
1
3
1/3
positive to negative so there is a local maximum at x œ $ which has a value f(3) œ 1.
f (x) œ
5. Yes, because at each point of [!ß "Ñ except x œ 0, the function's value is a local minimum value as well as a
local maximum value. At x œ 0 the function's value, 0, is not a local minimum value because each open
interval around x œ 0 on the x-axis contains points to the left of 0 where f equals 1.
6. (a) The first derivative of the function f(x) œ x$ is zero at x œ 0 even though f has no local extreme value at x œ 0.
(b) Theorem 2 says only that if f is differentiable and f has a local extreme at x œ c then f w (c) œ 0. It does not
assert the (false) reverse implication f w (c) œ 0 Ê f has a local extreme at x œ c.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
240
Chapter 4 Applications of Derivatives
7. No, because the interval 0 x 1 fails to be closed. The Extreme Value Theorem says that if the function is continuous
throughout a finite closed interval a Ÿ x Ÿ b then the existence of absolute extrema is guaranteed on that interval.
8. The absolute maximum is k1k œ 1 and the absolute minimum is k0k œ 0. This is not inconsistent with the Extreme Value
Theorem for continuous functions, which says a continuous function on a closed interval attains its extreme values on that
interval. The theorem says nothing about the behavior of a continuous function on an interval which is half open and half
closed, such as Ò"ß "Ñ, so there is nothing to contradict.
9. (a) There appear to be local minima at x œ 1.75
and 1.8. Points of inflection are indicated at
approximately x œ 0 and x œ „ 1.
(b) f w (x) œ x( 3x& 5x% 15x# œ x# ax# 3b ax$ 5b. The pattern yw œ ± ± ± ± 3
!
È
È$
È $
5
3
indicates a local maximum at x œ È5 and local minima at x œ „ È3 .
(c)
10. (a) The graph does not indicate any local
extremum. Points of inflection are indicated at
approximately x œ $% and x œ 1.
(b) f w (x) œ x( 2x% 5 10
x$
œ x$ ax$ 2b ax( 5b . The pattern f w œ )( ± ± indicates
7
3
!
È
È
5
2
3
7
a local maximum at x œ È
5 and a local minimum at x œ È
2.
(c)
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Chapter 4 Practice Exercises
241
11. (a) g(t) œ sin# t 3t Ê gw (t) œ 2 sin t cos t 3 œ sin (2t) 3 Ê gw 0 Ê g(t) is always falling and hence must
decrease on every interval in its domain.
(b) One, since sin# t 3t 5 œ 0 and sin# t 3t œ 5 have the same solutions: f(t) œ sin# t 3t 5 has the same
derivative as g(t) in part (a) and is always decreasing with f(3) 0 and f(0) 0. The Intermediate Value Theorem
guarantees the continuous function f has a root in [$ß 0].
12. (a) y œ tan ) Ê
dy
d)
œ sec# ) 0 Ê y œ tan ) is always rising on its domain Ê y œ tan ) increases on every interval
in its domain
(b) The interval 14 ß 1‘ is not in the tangent's domain because tan ) is undefined at ) œ
1
#
. Thus the tangent need not
increase on this interval.
13. (a) f(x) œ x% 2x# 2 Ê f w (x) œ 4x$ 4x. Since f(0) œ 2 0, f(1) œ 1 0 and f w (x) 0 for 0 Ÿ x Ÿ 1, we
may conclude from the Intermediate Value Theorem that f(x) has exactly one solution when 0 Ÿ x Ÿ 1.
È
(b) x# œ 2 „ 4 8 0 Ê x# œ È3 1 and x 0 Ê x ¸ È.7320508076 ¸ .8555996772
#
14. (a) y œ
x
x1
$
Ê yw œ
"
(x 1)# w
#
0, for all x in the domain of
x
x1
Êyœ
x
x1
is increasing in every interval in its domain.
$
(b) y œ x 2x Ê y œ 3x 2 0 for all x Ê the graph of y œ x 2x is always increasing and can never have a
local maximum or minimum
15. Let V(t) represent the volume of the water in the reservoir at time t, in minutes, let V(0) œ a! be the initial amount and
V(1440) œ a! (1400)(43,560)(7.48) gallons be the amount of water contained in the reservoir after the rain, where
24 hr œ 1440 min. Assume that V(t) is continuous on [!ß 1440] and differentiable on (!ß 1440). The Mean Value Theorem
says that for some t! in (!ß 1440) we have Vw (t! ) œ
V(1440) V(0)
1440 0
œ
a! (1400)(43,560)(7.48) a!
1440
œ
456,160,320 gal
1440 min
œ 316,778 gal/min. Therefore at t! the reservoir's volume was increasing at a rate in excess of 225,000 gal/min.
16. Yes, all differentiable functions g(x) having 3 as a derivative differ by only a constant. Consequently, the
d
difference 3x g(x) is a constant K because gw (x) œ 3 œ dx
(3x). Thus g(x) œ 3x K, the same form as F(x).
x
1
x
1
x 1 œ 1 x 1 Ê x 1 differs from x 1
(x 1) x(1)
d ˆ x ‰
d ˆ " ‰
œ (x " 1)# œ dx
dx x 1 œ
(x 1)#
x1 .
17. No,
18. f w (x) œ gw (x) œ
2x
ax # 1 b #
by the constant 1. Both functions have the same derivative
Ê f(x) g(x) œ C for some constant C Ê the graphs differ by a vertical shift.
19. The global minimum value of
"
#
occurs at x œ #.
20. (a) The function is increasing on the intervals Ò$ß #Ó and Ò"ß #Ó.
(b) The function is decreasing on the intervals Ò#ß !Ñ and Ð!ß "Ó.
(c) The local maximum values occur only at x œ #, and at x œ #; local minimum values occur at x œ $ and at x œ "
provided f is continuous at x œ !.
21. (a) t œ 0, 6, 12
(b) t œ 3, 9
(c) 6 t 12
(d) 0 t 6, 12 t 14
22. (a) t œ 4
(b) at no time
(c) 0 t 4
(d) 4 t 8
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
242
Chapter 4 Applications of Derivatives
23.
24.
25.
26.
27.
28.
29.
30.
31.
32.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Chapter 4 Practice Exercises
243
33. (a) yw œ 16 x# Ê yw œ ± ± Ê the curve is rising on (%ß %), falling on (_ß 4) and (%ß _)
%
%
Ê a local maximum at x œ 4 and a local minimum at x œ 4; yww œ 2x Ê yww œ ± Ê the curve
!
is concave up on (_ß !), concave down on (!ß _) Ê a point of inflection at x œ 0
(b)
34. (a) yw œ x# x 6 œ (x $)(x 2) Ê yw œ ± ± Ê the curve is rising on (_ß 2) and ($ß _),
#
$
falling on (#ß $) Ê local maximum at x œ 2 and a local minimum at x œ 3; yww œ 2x 1
Ê yww œ ± Ê concave up on ˆ "# ß _‰ , concave down on ˆ_ß "# ‰ Ê a point of inflection at x œ "#
"Î#
(b)
35. (a) yw œ 6x(x 1)(x 2) œ 6x$ 6x# 12x Ê yw œ ± ± ± Ê the graph is rising on ("ß !)
"
!
#
and (#ß _), falling on (_ß 1) and (!ß #) Ê a local maximum at x œ 0, local minima at x œ 1 and
x œ 2; yww œ 18x# 12x 12 œ 6 a3x# 2x 2b œ 6 Šx yww œ ±
on
±
"È(
$
1 È7 1 È7
Š 3 ß 3 ‹
"È(
$
1 È7
3 ‹ Šx
1 È7
3 ‹
È7
Ê the curve is concave up on Š_ß 1 3
Ê points of inflection at x œ
Ê
È7
‹ and Š 1 3
ß _‹ , concave down
1 „ È7
3
(b)
36. (a) yw œ x# (6 4x) œ 6x# 4x$ Ê yw œ ± ± Ê the curve is rising on ˆ_ß #3 ‰, falling on ˆ #3 ß _‰
!
$Î#
3
ww
#
Ê a local maximum at x œ # ; y œ 12x 12x œ 12x(" x) Ê yww œ ± ± Ê concave up on
!
"
(!ß "), concave down on (_ß !) and ("ß _) Ê points of inflection at x œ 0 and x œ 1
(b)
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
244
Chapter 4 Applications of Derivatives
37. (a) yw œ x% 2x# œ x# ax# 2b Ê yw œ ± ± ± Ê the curve is rising on Š_ß È2‹ and
!
È#
È #
ŠÈ2ß _‹ , falling on ŠÈ2ß È2‹ Ê a local maximum at x œ È2 and a local minimum at x œ È2 ;
yww œ 4x$ 4x œ 4x(x 1)(x 1) Ê yww œ ± ± ± Ê concave up on ("ß 0) and ("ß _),
"
!
"
concave down on (_ß 1) and (0ß 1) Ê points of inflection at x œ 0 and x œ „ 1
(b)
38. (a) yw œ 4x# x% œ x# a4 x# b Ê yw œ ± ± ± Ê the curve is rising on (2ß 0) and (0ß 2),
#
!
#
falling on (_ß 2) and (#ß _) Ê a local maximum at x œ 2, a local minimum at x œ 2; yww œ 8x 4x$
œ 4x a2 x# b Ê yww œ ± ± ± Ê concave up on Š_ß È2‹ and Š0ß È2‹ , concave
!
È#
È #
down on ŠÈ2ß 0‹ and ŠÈ2ß _‹ Ê points of inflection at x œ 0 and x œ „ È2
(b)
39. The values of the first derivative indicate that the curve is rising on (!ß _) and falling on (_ß 0). The slope of the curve
approaches _ as x Ä ! , and approaches _ as x Ä 0 and x Ä 1. The curve should therefore have a cusp and
local minimum at x œ 0, and a vertical tangent at x œ 1.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Chapter 4 Practice Exercises
40. The values of the first derivative indicate that the curve is rising on ˆ!ß "# ‰ and ("ß _), and falling on (_ß !)
and ˆ "# ß "‰ . The derivative changes from positive to negative at x œ "# , indicating a local maximum there. The
slope of the curve approaches _ as x Ä 0 and x Ä 1 , and approaches _ as x Ä 0 and as x Ä 1 ,
indicating cusps and local minima at both x œ 0 and x œ 1.
41. The values of the first derivative indicate that the curve is always rising. The slope of the curve approaches _
as x Ä 0 and as x Ä 1, indicating vertical tangents at both x œ 0 and x œ 1.
È33
42. The graph of the first derivative indicates that the curve is rising on Š!ß 17 16
on (_ß !) and
xœ
17 È33
16
È
È
Š 17 16 33 ß 17 16 33 ‹
Ê a local maximum at x œ
17 È33
16
È33
‹ and Š 17 16
ß _‹ , falling
, a local minimum at
. The derivative approaches _ as x Ä 0 and x Ä 1, and approaches _ as x Ä 0 ,
indicating a cusp and local minimum at x œ 0 and a vertical tangent at x œ 1.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
245
246
Chapter 4 Applications of Derivatives
43. y œ
x1
x3
45. y œ
x# 1
x
œx
47. y œ
x$ 2
#x
œ
49. y œ
x# 4
x# 3
œ1
œ1
x#
#
4
x3
"
x
"
x
"
x# 3
44. y œ
2x
x5
œ2
46. y œ
x# x 1
x
48. y œ
x% 1
x#
œ x# 50. y œ
x#
x# 4
œ1
10
x5
œx1
"
x
"
x#
4
x# 4
51. (a) Maximize f(x) œ Èx È36 x œ x"Î# (36 x)"Î# where 0 Ÿ x Ÿ 36
Ê f w (x) œ
"
#
x"Î# "# (36 x)"Î# (1) œ
È36 x Èx
#Èx È36 x
Ê derivative fails to exist at 0 and 36; f(0) œ 6,
and f(36) œ 6 Ê the numbers are 0 and 36
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Chapter 4 Practice Exercises
(b) Maximize g(x) œ Èx È36 x œ x"Î# (36 x)"Î# where 0 Ÿ x Ÿ 36
Ê gw (x) œ
"
#
x"Î# "# (36 x)"Î# (1) œ
È36 x Èx
#Èx È36 x
Ê critical points at 0, 18 and 36; g(0) œ 6,
g(18) œ 2È18 œ 6È2 and g(36) œ 6 Ê the numbers are 18 and 18
52. (a) Maximize f(x) œ Èx (20 x) œ 20x"Î# x$Î# where 0 Ÿ x Ÿ 20 Ê f w (x) œ 10x"Î# 3# x"Î#
œ
20 3x
#È x
œ 0 Ê x œ 0 and x œ
œ
40È20
3È 3
Ê the numbers are
20
3
20
3
‰ É 20
ˆ
are critical points; f(0) œ f(20) œ 0 and f ˆ 20
3 œ
3 20 and
40
3
.
(b) Maximize g(x) œ x È20 x œ x (20 x)"Î# where 0 Ÿ x Ÿ 20 Ê gw (x) œ
Ê È20 x œ
"
#
Ê xœ
the numbers must be
"
#
53. A(x) œ
79
4
and
79
4 .
"
4 .
The critical points are x œ
79
4
2È20 x 1
#È20 x
‰
and x œ 20. Since g ˆ 79
4 œ
(2x) a27 x# b for 0 Ÿ x Ÿ È27
Ê Aw (x) œ 3(3 x)(3 x) and Aw w (x) œ 6x.
The critical points are 3 and 3, but 3 is not in the
domain. Since Aw w (3) œ 18 0 and A ŠÈ27‹ œ 0,
the maximum occurs at x œ 3 Ê the largest area is
A(3) œ 54 sq units.
54. The volume is V œ x# h œ 32 Ê h œ 32
x# . The
32 ‰
#
ˆ
surface area is S(x) œ x 4x x# œ x# 128
x ,
where x 0 Ê Sw (x) œ
2(x 4) ax# 4x 16b
x#
Ê the critical points are 0 and 4, but 0 is not in the
domain. Now Sw w (4) œ 2 256
4$ 0 Ê at x œ 4 there
is a minimum. The dimensions 4 ft by 4 ft by 2 ft
minimize the surface area.
#
55. From the diagram we have ˆ h# ‰ r# œ ŠÈ3‹
Ê r# œ
12h#
4
#
. The volume of the cylinder is
#
V œ 1r h œ 1 Š 12 4 h ‹ h œ
#
1
4
0 Ÿ h Ÿ 2È3 . Then Vw (h) œ
a12h h$ b , where
31
4
20 ‰
3
(2 h)(2 h)
Ê the critical points are 2 and 2, but 2 is not in
the domain. At h œ 2 there is a maximum since
Vw w (2) œ 31 0. The dimensions of the largest
cylinder are radius œ È2 and height œ 2.
56. From the diagram we have x œ radius and
y œ height œ 12 2x and V(x) œ "3 1x# (12 2x), where
0 Ÿ x Ÿ 6 Ê Vw (x) œ 21x(4 x) and Vw w (4) œ 81. The
critical points are 0 and 4; V(0) œ V(6) œ 0 Ê x œ 4
gives the maximum. Thus the values of r œ 4 and
h œ 4 yield the largest volume for the smaller cone.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
81
4
œ0
and g(20) œ 20,
247
248
Chapter 4 Applications of Derivatives
‰ , where p is the profit on grade B tires and 0 Ÿ x Ÿ 4. Thus
57. The profit P œ 2px py œ 2px p ˆ 40510x
x
Pw (x) œ
2p
(5 x)#
ax# 10x 20b Ê the critical points are Š5 È5‹, 5, and Š5 È5‹ , but only Š5 È5‹ is in
the domain. Now Pw (x) 0 for 0 x Š5 È5‹ and Pw (x) 0 for Š5 È5‹ x 4 Ê at x œ Š5 È5‹ there
is a local maximum. Also P(0) œ 8p, P Š5 È5‹ œ 4p Š5 È5‹ ¸ 11p, and P(4) œ 8p Ê at x œ Š5 È5‹ there
is an absolute maximum. The maximum occurs when x œ Š5 È5‹ and y œ 2 Š5 È5‹ , the units are
hundreds of tires, i.e., x ¸ 276 tires and y ¸ 553 tires.
58. (a) The distance between the particles is lfatbl where fatb œ cos t cosˆt 1% ‰. Then, f w atb œ sin t sinˆt 1% ‰.
Solving f w atb œ ! graphically, we obtain t ¸ "Þ"(), t ¸ %Þ$#!, and so on.
Alternatively, f w atb œ ! may be solved analytically as follows. f w atb œ sin’ˆt 1) ‰ 1) “ sin’ˆt 1) ‰ 1) “
œ ’sinˆt 1) ‰cos 1) cosˆt 1) ‰sin 1) “ ’sinˆt 1) ‰cos 1) cosˆt 1) ‰sin 1) “ œ #sin 1) cosˆt 1) ‰
so the critical points occur when cosˆt 1) ‰ œ !, or t œ
$1
)
k1. At each of these values, fatb œ „ cos $)1
¸ „ !Þ('& units, so the maximum distance between the particles is !Þ('& units.
(b) Solving cos t œ cos ˆt 1% ‰ graphically, we obtain t ¸ #Þ(%*, t ¸ &Þ)*!, and so on.
Alternatively, this problem can be solved analytically as follows.
cos t œ cos ˆt 1% ‰
cos’ˆt 1) ‰ 1) “ œ cos’ˆt 1) ‰ 1) “
cosˆt 1) ‰cos 1) sinˆt 1) ‰sin 1) œ cosˆt 1) ‰cos 1) sinˆt 1) ‰sin 1)
#sin ˆt 1) ‰sin 1) œ !
sin ˆt 1) ‰ œ !
tœ
The particles collide when t œ
(1
)
(1
)
k1
¸ #Þ(%*. (plus multiples of 1 if they keep going.)
59. The dimensions will be x in. by "! #x in. by "' #x in., so Vaxb œ xa"! #xba"' #xb œ %x$ &#x# "'!x for
! x &. Then Vw axb œ "#x# "!%x "'! œ %ax #ba$x #!b , so the critical point in the correct domain is x œ #.
This critical point corresponds to the maximum possible volume because Vw axb ! for ! x # and Vw axb ! for
2 x &. The box of largest volume has a height of 2 in. and a base measuring 6 in. by 12 in., and its volume is 144 in.$
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Chapter 4 Practice Exercises
Graphical support:
60. The length of the ladder is d" d# œ 8 sec ) 6 csc ). We
wish to maximize I()) œ 8 sec ) 6 csc ) Ê Iw ())
œ 8 sec ) tan ) 6 csc ) cot ). Then Iw ()) œ 0
Ê 8 sin$ ) 6 cos$ ) œ 0 Ê tan ) œ
3
È
6
#
Ê
3
3
3
d" œ 4 É 4 È
36 and d# œ È
36 É4 È
36
Ê the length of the ladder is about
3
3
3
36‹ É4 È
36 œ Š4 È
36‹
Š4 È
$Î#
¸ "*Þ( ft.
61. g(x) œ 3x x$ 4 Ê g(2) œ 2 0 and g(3) œ 14 0 Ê g(x) œ 0 in the interval [#ß 3] by the Intermediate
Value Theorem. Then gw (x) œ 3 3x# Ê xnb1 œ xn 3xn x$n 4
33xn#
; x! œ 2 Ê x" œ 2.22 Ê x# œ 2.196215, and
so forth to x& œ 2.195823345.
62. g(x) œ x% x$ 75 Ê g(3) œ 21 0 and g(4) œ 117 0 Ê g(x) œ 0 in the interval [$ß %] by the Intermediate
Value Theorem. Then gw (x) œ 4x$ 3x# Ê xnb1 œ xn x%n x$n 75
4xn$ 3xn#
; x! œ 3 Ê x" œ 3.259259
Ê x# œ 3.229050, and so forth to x& œ 3.22857729.
63.
' ax$ 5x 7b dx œ
64.
' Š8t$ t# t‹ dt œ 8t4% t6$ t## C œ 2t% t6$ t## C
65.
' ˆ3Èt t4# ‰ dt œ ' ˆ3t"Î# 4t# ‰ dt œ 3t$Î# 4t1" C œ 2t$Î# 4t C
66.
' Š #È" t t3% ‹ dt œ ' ˆ #" t"Î# 3t% ‰ dt œ #" Œ t"Î# (3t3)$ C œ Èt t"$ C
x%
4
5x#
#
7x C
#
Š 3# ‹
"
#
67. Let u œ r 5 Ê du œ dr
' ar dr5b
#
œ'
du
u#
œ ' u# du œ
u "
1
C œ u" C œ ar " 5b C
68. Let u œ r È2 Ê du œ dr
'
6 dr
$
Šr È2‹
œ 6'
dr
$
Šr È2‹
œ 6'
du
u$
#
œ 6' u$ du œ 6 Š u# ‹ C œ 3u# C œ 3
#
ŠrÈ2‹
C
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
249
250
Chapter 4 Applications of Derivatives
69. Let u œ )# 1 Ê du œ 2) d) Ê
'
3)È)# 1 d) œ ' Èu ˆ #3 du‰ œ
70. Let u œ 7 )2 Ê du œ 2) d) Ê
'È)
d) œ '
7 ) 2
"
Èu
ˆ #" du‰ œ
"
#
x$ a 1 x % b
"Î%
du œ ) d)
3
#
"
#
$Î#
$Î#
' u"Î# du œ 3# Œ u$Î#
C œ a ) # 1b C
3 C œ u
#
du œ ) d)
"Î#
' u"Î# du œ #" Œ u"Î#
C œ È7 )2 C
" C œ u
#
71. Let u œ 1 x% Ê du œ 4x$ dx Ê
'
"
#
"
4
du œ x$ dx
dx œ ' u"Î% ˆ "4 du‰ œ
"
4
$Î%
" $Î%
' u"Î% du œ 4" Œ u$Î%
C œ 3" a1 x% b C
3 C œ 3 u
4
72. Let u œ 2 x Ê du œ dx Ê du œ dx
' (2 x)$Î& dx œ ' u$Î& ( du) œ ' u$Î& du œ u
)Î&
Š 85 ‹
73. Let u œ
'
sec# 10s
"
10
Ê du œ
s
10
C œ 85 u)Î& C œ 85 (2 x))Î& C
ds Ê 10 du œ ds
ds œ ' asec# ub (10 du) œ 10 ' sec# u du œ 10 tan u C œ 10 tan
74. Let u œ 1s Ê du œ 1 ds Ê
"
1
s
10
C
du œ ds
' csc# 1s ds œ ' acsc# ub ˆ 1" du‰ œ 1" ' csc# u du œ 1" cot u C œ 1" cot 1s C
75. Let u œ È2 ) Ê du œ È2 d) Ê
' csc È2) cot È2) d) œ '
)
3
76. Let u œ
'
sec
)
3
tan
77. Let u œ
'
Ê du œ
x
4
)
3
"
3
"
È2
du œ d)
(csc u cot u) Š È"2 du‹ œ
"
È2
(csc u) C œ È"2 csc È2) C
d) Ê 3 du œ d)
d) œ ' (sec u tan u)(3 du) œ 3 sec u C œ 3 sec
Ê du œ
"
4
)
3
C
dx Ê 4 du œ dx
2u ‰
dx œ ' asin# ub (4 du) œ ' 4 ˆ 1 cos
du œ 2' (1 cos 2u) du œ 2 ˆu #
œ 2u sin 2u C œ 2 ˆ x4 ‰ sin 2 ˆ x4 ‰ C œ #x sin x# C
sin#
x
4
78. Let u œ
'
cos#
79. y œ '
x
#
x
#
Ê du œ
"
#
dx œ ' a1 x# b dx œ x x" C œ x Ê C œ 1 Ê y œ x "
x
81.
dr
dt
œ ' Š15Èt 3
Èt ‹
"
3
"
x
"
#
Cœ
x
#
C; y œ 1 when x œ 1 Ê 1 1
1
C œ 1
sin 2u
#
sin x C
1
#
80. y œ ' ˆx "x ‰ dx œ ' ˆx# 2 y œ 1 when x œ 1 Ê
C
dx Ê 2 du œ dx
2u ‰
dx œ ' acos# ub (2 du) œ ' 2 ˆ 1 cos
du œ ' (1 cos 2u) du œ u #
x# "
x#
sin 2u ‰
#
2
1
1
"‰
x#
dx œ ' ax# 2 x# b dx œ
C œ 1 Ê C œ 3" Ê y œ
$
x
3
x$
3
2x x" C œ
2x dt œ ' ˆ15t"Î# 3t"Î# ‰ dt œ 10t$Î# 6t"Î# C;
Ê 10(1)$Î# 6(1)"Î# C œ 8 Ê C œ 8. Thus
dr
dt
dr
dt
"
x
x$
3
2x "
x
C;
"
3
œ 8 when t œ 1
œ 10t$Î# 6t"Î# 8 Ê r œ ' ˆ10t$Î# 6t"Î# 8‰ dt
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Chapter 4 Additional and Advanced Exercises
œ 4t&Î# 4t$Î# 8t C; r œ 0 when t œ 1 Ê 4(1)&Î# 4(1)$Î# 8(1) C" œ 0 Ê C" œ 0. Therefore,
r œ 4t&Î# 4t$Î# 8t
82.
d# r
dt#
Ê
œ ' cos t dt œ sin t C; rw w œ 0 when t œ 0 Ê sin 0 C œ 0 Ê C œ 0. Thus,
dr
dt
œ ' sin t dt œ cos t C" ; rw œ 0 when t œ 0 Ê 1 C" œ 0 Ê C" œ 1. Then
d# r
dt# œ sin t
dr
dt œ cos t 1
Ê r œ ' (cos t 1) dt œ sin t t C# ; r œ 1 when t œ 0 Ê 0 0 C# œ 1 Ê C# œ 1. Therefore,
r œ sin t t 1
CHAPTER 4 ADDITIONAL AND ADVANCED EXERCISES
1. If M and m are the maximum and minimum values, respectively, then m Ÿ f(x) Ÿ M for all x − I. If m œ M
then f is constant on I.
3x 6, 2 Ÿ x 0
has an absolute minimum value of 0 at x œ 2 and an absolute
9 x# , 0 Ÿ x Ÿ 2
maximum value of 9 at x œ 0, but it is discontinuous at x œ 0.
2. No, the function f(x) œ œ
3. On an open interval the extreme values of a continuous function (if any) must occur at an interior critical
point. On a half-open interval the extreme values of a continuous function may be at a critical point or at the
closed endpoint. Extreme values occur only where f w œ 0, f w does not exist, or at the endpoints of the interval.
Thus the extreme points will not be at the ends of an open interval.
4. The pattern f w œ ± ± ± ± indicates a local maximum at x œ 1 and a local
"
#
$
%
minimum at x œ 3.
5. (a) If yw œ 6(x 1)(x 2)# , then yw 0 for x 1 and yw 0 for x 1. The sign pattern is
f w œ ± ± Ê f has a local minimum at x œ 1. Also yww œ 6(x 2)# 12(x 1)(x 2)
"
#
œ 6(x 2)(3x) Ê yw w 0 for x 0 or x 2, while yww 0 for 0 x 2. Therefore f has points of inflection
at x œ 0 and x œ 2. There is no local maximum.
(b) If yw œ 6x(x 1)(x 2), then yw 0 for x 1 and 0 x 2; yw 0 for " x 0 and x 2. The sign
sign pattern is yw œ ± ± ± . Therefore f has a local maximum at x œ 0 and
"
!
#
È7
local minima at x œ 1 and x œ 2. Also, yww œ ") ’x Š 1 $
1 È 7
$
x
1 È 7
$
È7
‹“ ’x Š 1 $
‹“ , so yww 0 for
and yww 0 for all other x Ê f has points of inflection at x œ
6. The Mean Value Theorem indicates that
f(6) f(0)
60
1 „È 7
$
.
œ f w (c) Ÿ 2 for some c in (0ß 6). Then f(6) f(0) Ÿ 12 indicates the
most that f can increase is 12.
7. If f is continuous on [aß c) and f w (x) Ÿ 0 on [aß c), then by the Mean Value Theorem for all x − [aß c) we have
f(c) f(x)
cx
f(c). Also if f is continuous on (cß b] and f w (x)
Ÿ 0 Ê f(c) f(x) Ÿ 0 Ê f(x)
all x − (cß b] we have
f(x) f(c)
xc
0 Ê f(x) f(c)
0 Ê f(x)
f(c). Therefore f(x)
8. (a) For all x, (x 1)# Ÿ 0 Ÿ (x 1)# Ê a1 x# b Ÿ 2x Ÿ a1 x# b Ê "# Ÿ
(b) There exists c − (aß b) such that
Ê kf(b) f(a)k Ÿ
"
#
c
1 c#
œ
f(b) f(a)
ba
Ê
f(a)
¹ f(b)b a ¹
œ ¸ 1 c c# ¸ Ÿ
"
#
0 on (cß b], then for
f(c) for all x − [aß b].
x
1 x#
Ÿ
"
#
, from part (a)
kb ak .
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
.
251
252
Chapter 4 Applications of Derivatives
9. No. Corollary 1 requires that f w (x) œ 0 for all x in some interval I, not f w (x) œ 0 at a single point in I.
10. (a) h(x) œ f(x)g(x) Ê hw (x) œ f w (x)g(x) f(x)gw (x) which changes signs at x œ a since f w (x), gw (x) 0 when
x a, f w (x), gw (x) 0 when x a and f(x), g(x) 0 for all x. Therefore h(x) does have a local maximum at x œ a.
(b) No, let f(x) œ g(x) œ x$ which have points of inflection at x œ 0, but h(x) œ x' has no point of inflection
(it has a local minimum at x œ 0).
" a
bc#
11. From (ii), f(1) œ
lim
xÄ „_
x"
#
x Ä „ _ bx cx #
f(x) œ
1 "x
x
c 2x
xÄ „_
dy
dx
œ ! and if c œ 0,
œ 3x# 2kx 3 œ 0 Ê x œ
13. The area of the ?ABC is A(x) œ
w
where 0 Ÿ x Ÿ 1. Thus A (x) œ
1 "x
2
x Ä „ _ bx c x
"
1 x
then lim
2
x Ä „ _ bx x
œ
lim
lim
12.
œ 0 Ê a œ 1; from (iii), either 1 œ x lim
f(x) or 1 œ x Ä
lim
f(x). In either case,
Ä_
_
lim
2k „ È4k# 36
6
"
#
œ " Ê b œ 0 and c œ ". For if b œ ", then
œ
2
x
xÄ „_
œ „ _. Thus a œ 1, b œ 0, and c œ 1.
Ê x has only one value when 4k# 36 œ 0 Ê k# œ 9 or k œ „ 3.
(2) È1 x# œ a1 x# b
x
È 1 x#
1 x"
lim
"Î#
,
Ê 0 and „ 1 are
critical points. Also A a „ 1b œ 0 so A(0) œ 1 is the
maximum. When x œ 0 the ?ABC is isosceles since
AC œ BC œ È2 .
f (c h) f (c)
œ f ww (c) Ê for % œ "# kf ww (c)k 0
h
hÄ0
Ê ¹ f (ch)h f (c) f ww (c)¹ "# kf ww (c)k . Then f w (c) œ
w
14. lim
w
w
w
3
#
f ww (c) ww
0 Ê "# kf ww (c)k f (c h)
f ww (c) "# kf ww (c)k . If f ww (c) 0, then
h
f (c h)
"# f ww (c) 0; likewise if f ww (c) 0, then 0 "#
h
w
Ê f ww (c) "# kf ww (c)k Ê
there exists a $ 0 such that 0 khk $
w
w
f (c h)
h
"
#
f ww (c) w
kf ww (c)k
kf ww (c)k œ f ww (c)
f ww (c) w
f (c h)
h
w
3
#
f ww (c).
(a) If f (c) 0, then $ h 0 Ê f (c h) 0 and 0 h $ Ê f (c h) 0. Therefore, f(c) is a local
maximum.
(b) If f ww (c) 0, then $ h 0 Ê f w (c h) 0 and 0 h $ Ê f w (c h) 0. Therefore, f(c) is a local
minimum.
15. The time it would take the water to hit the ground from height y is É 2y
g , where g is the acceleration of gravity. The
product of time and exit velocity (rate) yields the distance the water travels:
È64(h y) œ 8 É 2 ahy y# b
D(y) œ É 2y
g
g
"Î#
, 0 Ÿ y Ÿ h Ê Dw (y) œ 4 É 2g ahy y# b
# "Î#
are critical points. Now D(0) œ 0, D ˆ h# ‰ œ 8 É g2 Šhˆ h# ‰ ˆ h# ‰ ‹
the hole is at y œ
h
#
ba
h
œ
(h 2y) Ê 0,
tan " 1 ha tan "
a
h
œ
h tan " a
h a tan "
ba
h ;
tan (" )) œ
tan " tan )
1 tan " tan )
. Solving for tan " gives tan " œ
; and tan ) œ
bh
h# a(b a)
a
h
. These equations
or
#
ah a(b a)b tan " œ bh. Differentiating both sides with respect to h gives
2h tan " ah# a(b a)b sec# "
d"
dh
œ b. Then
d"
dh
h
#
and h
œ 4hÉ g2 and D(h) œ 0 Ê the best place to drill
.
16. From the figure in the text, tan (" )) œ
give
"Î#
bh
œ 0 Ê 2h tan " œ b Ê 2h Š h# a(b
a) ‹ œ b
Ê 2bh# œ bh# ab(b a) Ê h# œ a(b a) Ê h œ Èa(a b) .
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Chapter 4 Additional and Advanced Exercises
253
17. The surface area of the cylinder is S œ 21r# 21rh. From
the diagram we have Rr œ H H h Ê h œ RH R rH and
S(r) œ 21r(r h) œ 21r ˆr H r HR ‰
œ 21 ˆ1 HR ‰ r# 21Hr, where 0 Ÿ r Ÿ R.
Case 1: H R Ê S(r) is a quadratic equation containing
the origin and concave upward Ê S(r) is maximum at
r œ R.
Case 2: H œ R Ê S(r) is a linear equation containing the
origin with a positive slope Ê S(r) is maximum at
r œ R.
Case 3: H R Ê S(r) is a quadratic equation containing the origin and concave downward. Then
dS
H‰
dS
H‰
RH
ˆ
ˆ
dr œ 41 1 R r 21H and dr œ 0 Ê 41 1 R r 21H œ 0 Ê r œ 2(H R) . For simplification
we let r‡ œ
RH
2(H R)
.
(a) If R H 2R, then 0 H 2R Ê H 2(H R) Ê r*= 2(HRH
R) R. Therefore, the maximum occurs at the
right endpoint R of the interval 0 Ÿ r Ÿ R because S(r) is an increasing function of r.
(b) If H œ 2R, then r‡ œ
2R#
2R
œ R Ê S(r) is maximum at r œ R.
(c) If H 2R, then 2R H 2H Ê H 2(H R) Ê
S(r) is a maximum at r œ r‡ œ
RH
2(H R)
H
2(H R)
1 Ê
RH
2(H R)
R Ê r‡ R. Therefore,
.
Conclusion: If H − (0ß 2R], then the maximum surface area is at r œ R. If H − (2Rß _), then the maximum is at
r œ r‡ œ 2(HRH
R) .
18. f(x) œ mx 1 If f Š È"m ‹
"
x
Ê f w (x) œ m "
x#
and f w w (x) œ
0, then Èm 1 Èm œ 2Èm 1
2
x$
0 when x 0. Then f w (x) œ 0 Ê x œ
0 Ê m
"
4
"
Èm
yields a minimum.
. Thus the smallest acceptable value for m is
"
4
.
19. (a) The profit function is Paxb œ ac exbx aa bxb œ ex# ac bbx a. Pw axb œ #ex c b œ !
Ê x œ c#eb . Pww axb œ #e ! if e ! so that the profit function is maximized at x œ c #e b .
(b) The price therefore that corresponds to a production level yeilding a maximum profit is
p¹
xœ c#eb
œ c eˆ c #e b ‰ œ
c b
#
dollars.
#
(c) The weekly profit at this production level is Paxb œ eˆ c #e b ‰ ac bbˆ c #e b ‰ a œ
ac b b #
%e
#
a.
(d) The tax increases cost to the new profit function is Faxb œ ac exbx aa bx txb œ ex ac b tbx a.
bc
cbt
ww
Now Fw axb œ #ex c b t œ ! when x œ t #
#e . Since F axb œ #e ! if e !, F is maximized
e œ
when x œ c #be t units per week. Thus the price per unit is p œ c eˆ c #be t ‰ œ c #b t dollars. Thus, such a tax
increases the cost per unit by
cbt
#
The x-intercept occurs when
"
x
cb
#
œ
t
#
dollars if units are priced to maximize profit.
20. (a)
$œ!Ê
"
x
œ $ Ê x œ $" .
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
254
Chapter 4 Applications of Derivatives
(b) By Newton's method, xn" œ xn faxn b
f ax n b .
w
Here f w axn b œ x#
n œ
"
x#n .
"
xn $
So xn" œ xn "
x#
n
œ xn Š x"n $‹x#n
œ xn xn $x#n œ #xn $xn# œ xn a# $xn b.
21. x" œ x! and
a
q "
x!
fax! b
f w ax ! b
q
qx!q x!q a
qxq! "
x a
œ x! ! q " œ
qx
!
with weights m! œ
In the case where x! œ
a
xq! "
q"
q
q
x aq "b a
œ ! q "
œ x! Š q q " ‹ a
Š"‹
xq! " q
qx!
and m" œ "q .
we have xq! œ a and x" œ
a
Š q q " ‹ qa " Š q" ‹
xq! "
x!
œ
so that x" is a weighted average of x!
a
Š q q "
xq! "
q" ‹ œ
a
.
xq! "
#
dy
d y
22. We have that ax hb# ay hb# œ r# and so #ax hb #ay hb dy
dx œ ! and # # dx #ay hb dx# œ ! hold.
dy
x y dx
dy .
" dx
dy
Thus #x #y dy
dx œ #h #h dx , by the former. Solving for h, we obtain h œ
#
d y
equation yields # # dy
dx #y dx# #Œ
dy
x y dx
dy " dx
œ !. Dividing by 2 results in " Substituting this into the second
dy
dx
#
y ddxy# Œ
dy
x y dx
dy " dx
œ !.
23. (a) aatb œ sww atb œ k ak !b Ê sw atb œ kt C" , where sw a!b œ )) Ê C" œ )) Ê sw atb œ kt )). So
satb œ
#
kt
#
kt#
#
))t C# where sa!b œ ! Ê C# œ ! so satb œ
))t œ "!!. Solving for t we obtain t œ
kŠ )) È))# #!!k
‹
k
))#
#!!
so that k œ
)) œ ! or kŠ )) )) „ È))# #!!k
.
k
È))# #!!k
‹
k
kt#
#
))t. Now satb œ "!! when
At such t we want sw atb œ !, thus
)) œ !. In either case we obtain ))# #!!k œ !
¸ $)Þ(# ft/sec# .
(b) The initial condition that sw a!b œ %% ft/sec implies that sw atb œ kt %% and satb œ
w
The car is stopped at a time t such that s atb œ kt %% œ ! Ê t œ
‰
sˆ %%
k
œ
k ˆ %% ‰#
#
k
‰
%%ˆ %%
k
œ
%%#
#k
œ
*')
k
œ
‰
*')ˆ #!!
))#
%%
k .
kt#
#
%%t where k is as above.
At this time the car has traveled a distance
œ #& feet. Thus halving the initial velocity quarters
stopping distance.
24. haxb œ f # axb g# axb Ê hw axb œ #faxbf w axb #gaxbgw axb œ #faxbf w axb gaxbgw axb‘ œ #faxbgaxb gaxbafaxbb‘
œ # † ! œ !. Thus haxb œ c, a constant. Since ha!b œ &, haxb œ & for all x in the domain of h. Thus ha"!b œ &.
25. Yes. The curve y œ x satisfies all three conditions since
dy
dx
œ " everywhere, when x œ !, y œ !, and
d# y
dx#
œ ! everywhere.
26. yw œ $x# # for all x Ê y œ x$ #x C where " œ "$ # † " C Ê C œ % Ê y œ x$ #x %.
27. sww atb œ a œ t# Ê v œ sw atb œ
maximum for this t‡ . Since satb
t‡ œ a$Cb"Î$ . So
ÊCœ
a%bb$Î%
$ .
a$Cb"Î$ ‘%
12
t$
w
‡
‡
$ C. We seek v! œ s a!b œ C. We know that sat b œ b for some t and s is at a
%
%
œ 12t Ct k and sa!b œ ! we have that satb œ 12t Ct and also sw at‡ b œ ! so that
Ca$Cb"Î$ œ b Ê a$Cb"Î$ ˆC Thus v! œ sw a!b œ
a%bb$Î%
$
œ
#È# $Î%
.
$ b
$C ‰
"#
œ b Ê a$Cb"Î$ ˆ $%C ‰ œ b Ê $"Î$ C%Î$ œ
28. (a) sww atb œ t"Î# t"Î# Ê vatb œ sw atb œ #$ t$Î# #t"Î# k where va!b œ k œ
(b) satb œ
% &Î#
"& t
%
%$ t$Î# %$ t k# where sa!b œ k# œ "&
. Thus satb œ
%
# $Î#
#t"Î#
$ Ê vatb œ $ t
% &Î#
%
%$ t$Î# %$ t "&
.
"& t
29. The graph of faxb œ ax# bx c with a ! is a parabola opening upwards. Thus faxb
one real value of x. The solutions to faxb œ ! are, by the quadratic equation
#
%b
$
%$ Þ
! for all x if faxb œ ! for at most
#
#b „ Éa#bb %ac
.
#a
Thus we require
#
a#bb %ac Ÿ ! Ê b ac Ÿ !.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Chapter 4 Additional and Advanced Exercises
30. (a) Clearly faxb œ aa" x b" b# Þ Þ Þ aan x bn b# ! for all x. Expanding we see
faxb œ aa#" x# #a" b" x b"# b Þ Þ Þ aan# x# #an bn x bn# b
œ aa#" a## Þ Þ Þ an# bx# #aa" b" a# b# Þ Þ Þ an bn bx ab"# b## Þ Þ Þ bn# b
Thus aa" b" a# b# Þ Þ Þ #
an bn b aa#" a## Þ Þ Þ an# bab"# b## an bn b# Ÿ aa#" a## Þ Þ Þ an# bab"# b## Þ Þ Þ bn# b Ÿ
Þ Þ Þ bn# b.
#
!.
! by Exercise 29.
Thus aa" b" a# b# Þ Þ Þ (b) Referring to Exercise 29: It is clear that faxb œ ! for some real x Í b %ac œ !, by quadratic formula.
Now notice that this implies that
faxb œ aa" x b" b# Þ Þ Þ aan x bn b#
œ aa#" a## Þ Þ Þ an# bx# #aa" b" a# b# Þ Þ Þ an bn bx ab"# b## Þ Þ Þ bn# b œ !
Í aa" b" a# b# Þ Þ Þ an bn b# aa#" a## Þ Þ Þ an# bab"# b## Þ Þ Þ bn# b œ !
Í aa" b" a# b# Þ Þ Þ an bn b# œ aa#" a## Þ Þ Þ an# bab"# b## Þ Þ Þ bn# b
But now faxb œ ! Í ai x bi œ ! for all i œ "ß #ß Þ Þ Þ ß n Í ai x œ bi œ ! for all i œ "ß #ß Þ Þ Þ ß n.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
255
256
Chapter 4 Applications of Derivatives
NOTES
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
CHAPTER 5 INTEGRATION
5.1 AREA AND ESTIMATING WITH FINITE SUMS
1. faxb œ x#
Since f is increasing on Ò!ß "Ó, we use left endpoints to obtain
lower sums and right endpoints to obtain upper sums.
"
#
iœ!
$
"
#
œ "# Š!# ˆ "# ‰ ‹ œ
#
"
4
œ 4" Š!# ˆ 4" ‰ ˆ #" ‰ ˆ 4$ ‰ ‹ œ
(a) ˜x œ
"!
#
œ
"
#
and xi œ i˜x œ
i
#
Ê a lower sum is !ˆ #i ‰ †
(b) ˜x œ
"!
%
œ
"
%
and xi œ i˜x œ
i
%
Ê a lower sum is !ˆ 4i ‰ †
(c) ˜x œ
"!
#
œ
"
#
and xi œ i˜x œ
i
#
Ê an upper sum is !ˆ #i ‰ †
(d) ˜x œ
"!
%
œ
"
%
and xi œ i˜x œ
i
%
Ê an upper sum is !ˆ 4i ‰ †
2. faxb œ x$
iœ!
2
iœ1
%
#
"
)
#
#
#
#
"
#
#
œ "# Šˆ "# ‰ +1# ‹ œ
#
"
4
#
#
#
œ 4" Šˆ 4" ‰ ˆ #" ‰ ˆ 4$ ‰ +1# ‹ œ
iœ"
"
%
†
(
)
"
%
$! ‰
† ˆ "'
œ
œ
(
$#
&
)
"&
$#
Since f is increasing on Ò!ß "Ó, we use left endpoints to obtain
lower sums and right endpoints to obtain upper sums.
"
$
iœ!
$
"
#
œ "# Š!$ ˆ "# ‰ ‹ œ
$
"
4
œ 4" Š!$ ˆ 4" ‰ ˆ #" ‰ ˆ 4$ ‰ ‹ œ
œ
"
#
and xi œ i˜x œ
i
#
Ê a lower sum is !ˆ #i ‰ †
(b) ˜x œ
"!
%
œ
"
%
and xi œ i˜x œ
i
%
Ê a lower sum is !ˆ 4i ‰ †
(c) ˜x œ
"!
#
œ
"
#
and xi œ i˜x œ
i
#
Ê an upper sum is !ˆ #i ‰ †
(d) ˜x œ
"!
%
œ
"
%
and xi œ i˜x œ
i
%
Ê an upper sum is !ˆ 4i ‰ †
(a) ˜x œ
"!
#
iœ!
2
iœ1
%
iœ"
$
"
"'
$
$
$
$'
#&'
$
"
#
$
œ "# Šˆ "# ‰ +1$ ‹ œ
$
"
4
$
$
$
œ 4" Šˆ 4" ‰ ˆ #" ‰ ˆ 4$ ‰ +1$ ‹ œ œ
"
#
†
*
)
œ
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
œ
*
'%
"!!
#&'
œ
*
"'
#&
'%
258
Chapter 5 Integration
3. faxb œ
"
x
Since f is decreasing on Ò1ß 5Ó, we use left endpoints to obtain
upper sums and right endpoints to obtain lower sums.
#
(a) ˜x œ
&"
#
œ # and xi œ " i˜x œ " #i Ê a lower sum is ! x"i † # œ #ˆ "$ "& ‰ œ
(b) ˜x œ
&"
%
œ 1 and xi œ " i˜x œ " i Ê a lower sum is
(c) ˜x œ
&"
#
œ # and xi œ " i˜x œ " #i Ê an upper sum is ! x"i † # œ #ˆ" $" ‰ œ
(d) ˜x œ
&"
%
œ 1 and xi œ " i˜x œ " i Ê an upper sum is
4. faxb œ % x#
iœ"
%
!"
xi
iœ"
"
† " œ "ˆ #" iœ!
$
!"
xi
iœ!
"
$
† " œ "ˆ" "
#
&" ‰ œ
"
%
"'
"&
"
$
((
'!
)
$
"% ‰ œ
#&
"#
Since f is increasing on Ò#ß !Ó and decreasing on Ò!ß #Ó, we use
left endpoints on Ò#ß !Ó and right endpoints on Ò!ß #Ó to obtain
lower sums and use right endpoints on Ò#ß !Ó and left endpoints
on Ò!ß #Ó to obtain upper sums.
(a) ˜x œ
# a#b
#
œ # and xi œ # i˜x œ # #i Ê a lower sum is # † ˆ% a#b# ‰ # † a% ## b œ !
(b) ˜x œ
# a#b
%
œ " and xi œ # i˜x œ # i Ê a lower sum is !ˆ% axi b# ‰ † " !ˆ% axi b# ‰ † "
"
%
iœ!
iœ$
œ "ˆˆ% a#b# ‰ ˆ% a"b# ‰ a% "# b a% ## b‰ œ '
(c) ˜x œ
# a#b
#
œ # and xi œ # i˜x œ # #i Ê a upper sum is # † ˆ% a!b# ‰ # † a% !# b œ "'
(d) ˜x œ
# a#b
%
œ " and xi œ # i˜x œ # i Ê a upper sum is !ˆ% axi b# ‰ † " !ˆ% axi b# ‰ † "
#
$
iœ"
iœ#
œ "ˆˆ% a"b# ‰ a% !# b a% !# b a% "# b‰ œ "%
5. faxb œ x#
œ
"
#
Using 4 rectangles Ê ˜x œ " % ! œ
Ê "% ˆfˆ ") ‰ fˆ $) ‰ fˆ &) ‰ fˆ () ‰‰
"
%
Using 2 rectangles Ê ˜x œ
#
#
œ "# Šˆ "% ‰ ˆ $% ‰ ‹ œ
#
#
"!
$#
#
œ
"!
#
Ê "# ˆfˆ "% ‰ fˆ $% ‰‰
&
"'
#
œ "% Šˆ ") ‰ ˆ $) ‰ ˆ &) ‰ ˆ () ‰ ‹ œ
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
#"
'%
Section 5.1 Area and Estimating with Finite Sums
6. faxb œ x$
œ
"
#
Using 4 rectangles Ê ˜x œ " % ! œ
Ê "% ˆfˆ ") ‰ fˆ $) ‰ fˆ &) ‰ fˆ () ‰‰
"
%
$
$
œ "# Šˆ "% ‰ ˆ $% ‰ ‹ œ
$
$
$
$
(
œ "% Š " $ )&
‹œ
$
7. faxb œ
"
x
"!
#
Using 2 rectangles Ê ˜x œ
#)
# † '%
%*'
% † )$
œ
œ
Using 2 rectangles Ê ˜x œ
œ #ˆ "# "% ‰ œ $#
Ê "# ˆfˆ "% ‰ fˆ $% ‰‰
(
$#
"#%
)$
&"
#
œ
$"
"#)
œ # Ê #afa#b fa%bb
Using 4 rectangles Ê ˜x œ & % " œ "
Ê "ˆfˆ $# ‰ fˆ &# ‰ fˆ (# ‰ fˆ *# ‰‰
œ "ˆ #$ 8. faxb œ % x#
#
&
#
(
#* ‰ œ
"%))
$†&†(†*
Using 2 rectangles Ê ˜x œ
œ #a$ $b œ "#
259
œ
# a#b
#
%*'
&†(†*
œ
%*'
$"&
œ # Ê #afa"b fa"bb
Using 4 rectangles Ê ˜x œ # %a#b œ "
Ê "ˆfˆ $# ‰ fˆ "# ‰ fˆ "# ‰ fˆ $# ‰‰
#
#
#
#
œ "ŠŠ% ˆ $# ‰ ‹ Š% ˆ "# ‰ ‹ Š% ˆ "# ‰ ‹ Š% ˆ $# ‰ ‹‹
œ "' ˆ *% † # "% † #‰ œ "' "!
# œ ""
9. (a) D ¸ (0)(1) (12)(1) (22)(1) (10)(1) (5)(1) (13)(1) (11)(1) (6)(1) (2)(1) (6)(1) œ 87 inches
(b) D ¸ (12)(1) (22)(1) (10)(1) (5)(1) (13)(1) (11)(1) (6)(1) (2)(1) (6)(1) (0)(1) œ 87 inches
10. (a) D ¸ (1)(300) (1.2)(300) (1.7)(300) (2.0)(300) (1.8)(300) (1.6)(300) (1.4)(300) (1.2)(300)
(1.0)(300) (1.8)(300) (1.5)(300) (1.2)(300) œ 5220 meters (NOTE: 5 minutes œ 300 seconds)
(b) D ¸ (1.2)(300) (1.7)(300) (2.0)(300) (1.8)(300) (1.6)(300) (1.4)(300) (1.2)(300) (1.0)(300)
(1.8)(300) (1.5)(300) (1.2)(300) (0)(300) œ 4920 meters (NOTE: 5 minutes œ 300 seconds)
11. (a) D ¸ (0)(10) (44)(10) (15)(10) (35)(10) (30)(10) (44)(10) (35)(10) (15)(10) (22)(10)
(35)(10) (44)(10) (30)(10) œ 3490 feet ¸ 0.66 miles
(b) D ¸ (44)(10) (15)(10) (35)(10) (30)(10) (44)(10) (35)(10) (15)(10) (22)(10) (35)(10)
(44)(10) (30)(10) (35)(10) œ 3840 feet ¸ 0.73 miles
12. (a) The distance traveled will be the area under the curve. We will use the approximate velocities at the
midpoints of each time interval to approximate this area using rectangles. Thus,
D ¸ (20)(0.001) (50)(0.001) (72)(0.001) (90)(0.001) (102)(0.001) (112)(0.001) (120)(0.001)
(128)(0.001) (134)(0.001) (139)(0.001) ¸ 0.967 miles
(b) Roughly, after 0.0063 hours, the car would have gone 0.484 miles, where 0.0060 hours œ 22.7 sec. At 22.7
sec, the velocity was approximately 120 mi/hr.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
260
Chapter 5 Integration
13. (a) Because the acceleration is decreasing, an upper estimate is obtained using left end-points in summing
acceleration † ?t. Thus, ?t œ 1 and speed ¸ [32.00 19.41 11.77 7.14 4.33](1) œ 74.65 ft/sec
(b) Using right end-points we obtain a lower estimate: speed ¸ [19.41 11.77 7.14 4.33 2.63](1)
œ 45.28 ft/sec
(c) Upper estimates for the speed at each second are:
t
0
1
2
3
4
5
v
0
32.00 51.41 63.18 70.32
74.65
Thus, the distance fallen when t œ 3 seconds is s ¸ [32.00 51.41 63.18](1) œ 146.59 ft.
14. (a) The speed is a decreasing function of time Ê right end-points give an lower estimate for the height (distance)
attained. Also
t
0
1
2
3
4
5
v
400
368
336
304
272
240
gives the time-velocity table by subtracting the constant g œ 32 from the speed at each time increment
?t œ 1 sec. Thus, the speed ¸ 240 ft/sec after 5 seconds.
(b) A lower estimate for height attained is h ¸ [368 336 304 272 240](1) œ 1520 ft.
15. Partition [!ß #] into the four subintervals [0ß 0.5], [0.5ß 1], [1ß 1.5], and [1.5ß 2]. The midpoints of these
subintervals are m" œ 0.25, m# œ 0.75, m$ œ 1.25, and m% œ 1.75. The heights of the four approximating
1
125
343
$
$
rectangles are f(m" ) œ (0.25)$ œ 64
, f(m# ) œ (0.75)$ œ 27
64 , f(m$ ) œ (1.25) œ 64 , and f(m% ) œ (1.75) œ 64
Notice that the average value is approximated by
œ
"
length of [!ß#]
†”
"
#
$
$
$
$
’ˆ 4" ‰ ˆ #" ‰ ˆ 43 ‰ ˆ #" ‰ ˆ 45 ‰ ˆ #" ‰ ˆ 47 ‰ ˆ #" ‰“ œ
$"
"'
approximate area under
• . We use this observation in solving the next several exercises.
curve f(x) œ x$
16. Partition [1ß 9] into the four subintervals ["ß $], [3ß &], [&ß (], and [(ß *]. The midpoints of these subintervals are
m" œ 2, m# œ 4, m$ œ 6, and m% œ 8. The heights of the four approximating rectangles are f(m" ) œ "# ,
f(m# ) œ "4 , f(m$ ) œ 6" , and f(m% ) œ 8" . The width of each rectangle is ?x œ 2. Thus,
Area ¸ 2 ˆ "# ‰ 2 ˆ 4" ‰ 2 ˆ 6" ‰ 2 ˆ 8" ‰ œ
Ê average value ¸
25
1#
area
length of ["ß*]
œ
ˆ 25
‰
12
8
œ
25
96 .
17. Partition [0ß 2] into the four subintervals [0ß 0.5], [0.5ß 1], [1ß 1.5], and [1.5ß 2]. The midpoints of the subintervals
are m" œ 0.25, m# œ 0.75, m$ œ 1.25, and m% œ 1.75. The heights of the four approximating rectangles are
"
#
f(m" ) œ
œ
"
#
"
#
sin#
1
4
œ
"
#
"
#
œ 1, and f(m% ) œ
œ 1, f(m# ) œ
"
2
sin#
71
4
œ
sin#
"
#
Š È"2 ‹ œ 1. The width of each rectangle is ?x œ #" . Thus,
31
4
œ
"
#
"
#
œ 1, f(m$ ) œ
"
2
sin#
51
4
œ
"
#
Š È"2 ‹
#
"
2
#
Area ¸ (1 1 1 1) ˆ "# ‰ œ 2 Ê average value ¸
area
length of [0ß2]
œ
2
#
œ 1.
18. Partition [0ß 4] into the four subintervals [0ß 1], [1ß 2ß ], [2ß 3], and [3ß 4]. The midpoints of the subintervals
are m" œ "# , m# œ #3 , m$ œ 5# , and m% œ 7# . The heights of the four approximating rectangles are
f(m" ) œ 1 Šcos Š
%
1 ˆ "# ‰
4 ‹‹
œ 1 ˆcos ˆ 18 ‰‰ œ 0.27145 (to 5 decimal places),
f(m# ) œ 1 Šcos Š
%
1 ˆ 3# ‰
4 ‹‹
œ 1 ˆcos ˆ 381 ‰‰ œ 0.97855, f(m3 ) œ 1 Šcos Š
%
œ 0.97855, and f(m% ) œ 1 Šcos Š
%
%
1 ˆ 7# ‰
4 ‹‹
%
1 ˆ #5 ‰
4 ‹‹
œ 1 ˆcos ˆ 581 ‰‰
%
%
œ 1 ˆcos ˆ 781 ‰‰ œ 0.27145. The width of each rectangle is
?x œ ". Thus, Area ¸ (0.27145)(1) (0.97855)(1) (0.97855)(1) (0.27145)(1) œ 2.5 Ê average
2.5
5
value ¸ lengtharea
of [0ß4] œ 4 œ 8 .
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 5.1 Area and Estimating with Finite Sums
261
19. Since the leakage is increasing, an upper estimate uses right endpoints and a lower estimate uses left
endpoints:
(a) upper estimate œ (70)(1) (97)(1) (136)(1) (190)(1) (265)(1) œ 758 gal,
lower estimate œ (50)(1) (70)(1) (97)(1) (136)(1) (190)(1) œ 543 gal.
(b) upper estimate œ (70 97 136 190 265 369 516 720) œ 2363 gal,
lower estimate œ (50 70 97 136 190 265 369 516) œ 1693 gal.
(c) worst case: 2363 720t œ 25,000 Ê t ¸ 31.4 hrs;
best case: 1693 720t œ 25,000 Ê t ¸ 32.4 hrs
20. Since the pollutant release increases over time, an upper estimate uses right endpoints and a lower estimate
uses left endpoints:
(a) upper estimate œ (0.2)(30) (0.25)(30) (0.27)(30) (0.34)(30) (0.45)(30) (0.52)(30) œ 60.9 tons
lower estimate œ (0.05)(30) (0.2)(30) (0.25)(30) (0.27)(30) (0.34)(30) (0.45)(30) œ 46.8 tons
(b) Using the lower (best case) estimate: 46.8 (0.52)(30) (0.63)(30) (0.70)(30) (0.81)(30) œ 126.6 tons,
so near the end of September 125 tons of pollutants will have been released.
#
21. (a) The diagonal of the square has length 2, so the side length is È#. Area œ ŠÈ#‹ œ #
(b) Think of the octagon as a collection of 16 right triangles with a hypotenuse of length 1 and an acute angle measuring
#1
1
"' œ ) .
Area œ "'ˆ " ‰ˆsin 1 ‰ˆcos 1 ‰ œ % sin 1 œ #È# ¸ #Þ)#)
#
)
)
%
(c) Think of the 16-gon as a collection of 32 right triangles with a hypotenuse of length 1 and an acute angle measuring
#1
1
$# œ "' .
Area œ $#ˆ " ‰ˆsin 1 ‰ˆcos 1 ‰ œ ) sin 1 œ #È# ¸ $Þ!'"
#
"'
"'
)
(d) Each area is less than the area of the circle, 1. As n increases, the area approaches 1.
22. (a) Each of the isosceles triangles is made up of two right triangles having hypotenuse 1 and an acute angle measuring
#1
1
1 ‰ˆ
ˆ " ‰ˆ
cos 1n ‰ œ "# sin #n1 .
#n œ n . The area of each isosceles triangle is AT œ # # sin n
(b) The area of the polygon is AP œ nAT œ
n
#
sin
#1
n ,
n
nÄ_ #
so lim
sin
#1
n
œ lim 1 †
nÄ_
sin #n1
ˆ #n1 ‰
œ1
(c) Multiply each area by r# .
AT œ "# r# sin #n1
AP œ n# r# sin
lim AP œ 1r
#
#1
n
nÄ_
23-26. Example CAS commands:
Maple:
with( Student[Calculus1] );
f := x -> sin(x);
a := 0;
b := Pi;
plot( f(x), x=a..b, title="#23(a) (Section 5.1)" );
N := [ 100, 200, 1000 ];
# (b)
for n in N do
Xlist := [ a+1.*(b-a)/n*i $ i=0..n ];
Ylist := map( f, Xlist );
end do:
for n in N do
# (c)
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
262
Chapter 5 Integration
Avg[n] := evalf(add(y,y=Ylist)/nops(Ylist));
end do;
avg := FunctionAverage( f(x), x=a..b, output=value );
evalf( avg );
FunctionAverage(f(x),x=a..b,output=plot);
# (d)
fsolve( f(x)=avg, x=0.5 );
fsolve( f(x)=avg, x=2.5 );
fsolve( f(x)=Avg[1000], x=0.5 );
fsolve( f(x)=Avg[1000], x=2.5 );
Mathematica: (assigned function and values for a and b may vary):
Symbols for 1, Ä , powers, roots, fractions, etc. are available in Palettes (under File).
Never insert a space between the name of a function and its argument.
Clear[x]
f[x_]:=x Sin[1/x]
{a,b}={1/4, 1}
Plot[f[x],{x, a, b}]
The following code computes the value of the function for each interval midpoint and then finds the average. Each
sequence of commands for a different value of n (number of subdivisions) should be placed in a separate cell.
n =100; dx = (b a) /n;
values = Table[N[f[x]], {x, a dx/2, b, dx}]
average=Sum[values[[i]],{i, 1, Length[values]}] / n
n =200; dx = (b a) /n;
values = Table[N[f[x]],{x, a + dx/2, b, dx}]
average=Sum[values[[i]],{i, 1, Length[values]}] / n
n =1000; dx = (b a) /n;
values = Table[N[f[x]],{x, a dx/2, b, dx}]
average=Sum[values[[i]],{i, 1, Length[values]}] / n
FindRoot[f[x] == average,{x, a}]
5.2 SIGMA NOTATION AND LIMITS OF FINITE SUMS
2
1. !
kœ1
3
2. !
kœ1
6k
k1
œ
6(1)
11
6(2)
21
œ
6
2
k1
k
œ
11
1
21
2
31
3
12
3
œ7
œ0
1
2
2
3
œ
7
6
4
3. ! cos k1 œ cos (11) cos (21) cos (31) cos (41) œ 1 1 1 1 œ 0
kœ1
5
4. ! sin k1 œ sin (11) sin (21) sin (31) sin (41) sin (51) œ 0 0 0 0 0 œ 0
kœ1
3
5. ! (1)kb1 sin
kœ1
1
k
œ (1)"" sin
1
1
(1)#" sin
1
#
(")$" sin
1
3
œ 01
È3
#
œ
È3 2
#
4
6. ! (1)k cos k1 œ (1)" cos (11) (1)# cos (21) (1)$ cos (31) (1)% cos (41)
kœ1
œ (1) 1 (1) 1 œ 4
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 5.2 Sigma Notation and Limits of Finite Sums
6
7. (a) ! 2kc1 œ 2"" 2#" 2$" 2%" 2&" 2'" œ 1 2 4 8 16 32
kœ1
5
(b) ! 2k œ 2! 2" 2# 2$ 2% 2& œ 1 2 4 8 16 32
kœ0
4
(c) ! 2k1 œ 2"" 2!" 2"" 2#" 2$" 2%" œ 1 2 4 8 16 32
kœ
"
All of them represent 1 2 4 8 16 32
6
8. (a) ! (2)k
œ (2)"" (2)#" (2)$" (2)%" (2)&" (2)'" œ 1 2 4 8 16 32
1
kœ1
5
(b) ! (1)k 2k œ (1)! 2! Ð")" 2" (1)# 2# (1)$ 2$ (1)% 2% (1)& 2& œ 1 2 4 8 16 32
kœ0
3
(c) ! (1)k1 2k2 œ Ð")#" 2## (")"" 2"# (")!" 2!# (1)"" 2"# (")#" 2##
kœ 2
(1)$" 2$# œ 1 2 4 8 16 32;
(a) and (b) represent 1 2 4 8 16 32; (c) is not equivalent to the other two
4
(")k "
k1
9. (a) !
kœ2
2
(b) !
kœ0
1
(c) !
kœ
"
(1)# "
21
œ
(")k
k1
œ
(1)!
01
(")k
k2
œ
(1) "
1 2
(")$ "
31
(")"
11
(")!
02
(")#
21
(")% "
41
œ 1 œ1
(")"
12
"
#
œ 1 "
#
"
#
"
3
"
3
"
3
(a) and (c) are equivalent; (b) is not equivalent to the other two.
4
10. (a) ! (k 1)# œ (1 1)# (2 1)# (3 1)# (4 1)# œ 0 1 4 9
kœ1
3
(b) ! (k 1)# œ (1 1)# (0 1)# (1 1)# (2 1)# (3 1)# œ 0 1 4 9 16
kœ 1
"
(c) ! k# œ (3)# (2)# (1)# œ 9 4 1
kœ 3
(a) and (c) are equivalent to each other; (b) is not equivalent to the other two.
6
4
11. ! k
4
12. ! k#
kœ1
13. !
kœ1
5
5
15. ! (1)k1
14. ! 2k
kœ1
kœ1
n
kœ1
"
k
5
16. ! (1)k
kœ1
n
17. (a) ! 3ak œ 3 ! ak œ 3(5) œ 15
kœ1
n
(b) !
kœ1
n
bk
6
œ
"
6
kœ1
n
! bk œ
kœ1
"
6
(6) œ 1
n
n
kœ1
n
kœ1
n
kœ1
n
kœ1
n
kœ1
n
kœ1
(c) ! (ak bk ) œ ! ak ! bk œ 5 6 œ 1
(d) ! (ak bk ) œ ! ak ! bk œ 5 6 œ 11
n
(e) ! (bk 2ak ) œ ! bk 2 ! ak œ 6 2(5) œ 16
kœ1
kœ1
"
#k
kœ1
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
k
5
263
264
Chapter 5 Integration
n
n
kœ1
n
kœ1
n
18. (a) ! 8ak œ 8 ! ak œ 8(0) œ 0
n
n
kœ1
kœ1
(c) ! (ak 1) œ ! ak ! 1 œ 0 n œ n
kœ1
10
19. (a) ! k œ
kœ1
10(10 1)
#
n
(b) ! 250bk œ 250 ! bk œ 250(1) œ 250
kœ1
n
n
kœ1
n
kœ1
kœ1
kœ1
(d) ! (bk 1) œ ! bk ! 1 œ " n
10
(b) ! k# œ
œ 55
kœ1
10(10 1)(2(10) 1)
6
œ 385
13(13 1)(2(13) 1)
6
œ 819
#
10
(c) ! k$ œ ’ 10(10# 1) “ œ 55# œ 3025
kœ1
13
20. (a) ! k œ
kœ1
13(13 1)
#
13
(b) ! k# œ
œ 91
kœ1
#
13
(c) ! k$ œ ’ 13(13# 1) “ œ 91# œ 8281
kœ1
7
7
kœ1
kœ1
6
6
6
kœ1
kœ1
kœ1
6
6
6
kœ1
kœ1
kœ1
5
21. ! 2k œ 2 ! k œ 2 Š 7(7 # ") ‹ œ 56
23. ! a3 k# b œ ! 3 ! k# œ 3(6) 24. ! ak# 5b œ ! k# ! 5 œ
22. !
kœ1
6(6 ")(2(6) 1)
6
6(6 ")(2(6) 1)
6
1k
15
1
15
œ
5
!kœ
1
15
kœ1
Š 5(5 # 1) ‹ œ 1
œ 73
5(6) œ 61
5
5
5
5
kœ1
kœ1
kœ1
kœ1
7
7
7
7
kœ1
kœ1
kœ1
kœ1
1)
25. ! k(3k 5) œ ! a3k# 5kb œ 3 ! k# 5 ! k œ 3 Š 5(5 1)(2(5)
‹ 5 Š 5(5 # 1) ‹ œ 240
6
1)
26. ! k(2k 1) œ ! a2k# kb œ 2 ! k# ! k œ 2 Š 7(7 1)(2(7)
‹
6
5
27. !
k$
225
kœ1
kœ1
#
7
$
5
Œ! k œ
7
28. Œ! k !
kœ1
kœ1
k$
4
"
2 #5
7
5
5
kœ1
kœ1
$
! k $ Œ! k œ
#
œ Œ! k kœ1
"
4
"
#25
#
7(7 1)
#
œ 308
$
Š 5(5 # 1) ‹ Š 5(5 # 1) ‹ œ 3376
#
7
! k$ œ Š 7(7 1) ‹ #
kœ1
7
"
4
#
Š 7(7 # 1) ‹ œ 588
500
29. (a) ! 3 œ 3a7b œ 21
(b) ! 7 œ 7a500b œ 3500
kœ1
kœ1
264
262
kœ3
jœ1
(c) Let j œ k 2 Ê k œ j 2; if k œ 3 Ê j œ 1 and if k œ 264 Ê j œ 262 Ê ! 10 œ ! 10 œ 10a262b œ 2620
36
28
28
28
kœ9
jœ1
jœ1
jœ1
30. (a) Let j œ k 8 Ê k œ j 8; if k œ 9 Ê j œ 1 and if k œ 36 Ê j œ 28 Ê ! k œ ! a j 8b œ ! j !8
œ
28a28 1b
2
)a28b œ 630
17
15
kœ3
jœ1
(b) Let j œ k 2 Ê k œ j 2; if k œ 3 Ê j œ 1 and if k œ 17 Ê j œ 15 Ê ! k2 œ ! a j 2b2
15
15
15
15
jœ1
jœ1
jœ1
jœ1
œ ! a j2 4j 4b œ ! j2 ! 4j ! 4 œ
15a15 1ba2a15b 1b
6
4†
15a15 1b
2
4a15b
œ 1240 480 60 œ 1780
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 5.2 Sigma Notation and Limits of Finite Sums
71
(c) Let j œ k 17 Ê k œ j 17; if k œ 18 Ê j œ 1 and if k œ 71 Ê j œ 54 Ê !kak 1b
kœ3
54
54
54
54
54
jœ1
jœ1
jœ1
jœ1
œ ! a j 17baa j 17b 1b œ ! a j2 33j 272b œ ! j2 ! 33j ! 272
jœ1
œ
54a54 1ba2a54b 1b
6
33 †
54a54 1b
2
272a54b œ 53955 49005 14688 œ 117648
n
n
31. (a) ! 4 œ 4n
(b) ! c œ cn
kœ1
n
n
n
kœ1
kœ1
kœ1
kœ1
(c) ! ak 1b œ ! k ! 1 œ
n an 1 b
2
nœ
n
n
32. (a) ! ˆ 1n 2n‰ œ ˆ 1n 2n‰n œ 1 2n2
kœ1
n
(c) !
kœ1
k
n2
œ
1 n an 1 b
n2
2
œ
n2 n
2
(b) !
kœ1
c
n
œ
c
n
†nœc
n1
2n
33. (a)
(b)
(c)
34. (a)
(b)
(c)
35. (a)
(b)
(c)
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
265
266
Chapter 5 Integration
36. (a)
(b)
(c)
37. kx" x! k œ k1.2 0k œ 1.2, kx# x" k œ k1.5 1.2k œ 0.3, kx$ x# k œ k2.3 1.5k œ 0.8, kx% x$ k œ k2.6 2.3k œ 0.3,
and kx& x% k œ k3 2.6k œ 0.4; the largest is lPl œ 1.2.
38. kx" x! k œ k1.6 (2)k œ 0.4, kx# x" k œ k0.5 (1.6)k œ 1.1, kx$ x# k œ k0 (0.5)k œ 0.5,
kx% x$ k œ k0.8 0k œ 0.8, and kx& x% k œ k1 0.8k œ 0.2; the largest is lPl œ 1.1.
39. faxb œ " x#
Let ˜x œ
"!
n
œ
n
!a" c#i b " œ
n
$
œ
n
n$
# $n n"#
'
. Thus,
nÄ_
Let ˜x œ
# '
$!
n
œ
"
n#
$
n
n
!#ci ˆ $ ‰ œ ! 'i †
n
n
iœ"
iœ"
n
Thus,
41. faxb œ x# "
lim ! 'i
nÄ_ iœ" n
Let ˜x œ
$!
n
n
œ"
œ
n
#( ! #
i
n
iœ"
nÄ_ iœ1
œ"
"
$
œ
#
$
and ci œ i˜x œ
$
n
œ
")
n#
n
!i œ
iœ"
†
#
œ lim *n n# *n
nÄ_
œ
$
n
n
and ci œ i˜x œ
†nœ
$i
n.
")
n#
$
n
iœ"
$
n
#n$ $n# n
'n$
lim !a" ci# b n"
The right-hand sum is
n an " b
#
†
œ
*n# *n
n#
œ lim ˆ* n* ‰ œ *.
!ac#i "b $ œ !Šˆ $i ‰# "‹ $ œ
n
n
n
iœ"
iœ1
n
$
n
n
nan "ba#n "b
'n $
n
!an# i# b
"
n$
n
iœ1
œ"
œ lim Œ" 40. faxb œ #x
!Š" ˆ i ‰# ‹ œ
"! #
i
n$
iœ1
œ"
and ci œ i˜x œ ni . The right-hand sum is
n
"
n
iœ1
n
"
n
nÄ_
$i
n.
$
n
#( nan "ba#n "b
‹
n$ Š
'
The right-hand sum is
n
!Š *i## "‹
n
iœ"
$
*
") #(
*a#n$ $n# nb
n n#
$œ
$Þ Thus,
#n $
#
n
#(
") n n*#
lim !ac#i "b $n œ lim Œ
$ œ
#
nÄ_ iœ"
nÄ_
œ
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
* $ œ "#.
Section 5.2 Sigma Notation and Limits of Finite Sums
42. faxb œ $x#
"!
n
Let ˜x œ
n
œ
"
n
and ci œ i˜x œ ni . The right-hand sum is
n
!$c#i ˆ " ‰ œ !$ˆ i ‰# ˆ " ‰ œ
n
iœ"
#n$ $n# n
#n$
œ
œ lim Œ
nÄ_
43. faxb œ x x# œ xa" xb
n
iœ"
n
# $n n"#
#
œ
# $n n"#
#
œ
"!
n
"
n
Let ˜x œ
œ
n
#
#
$
n$
n
iœ"
n
. Thus, lim !$c#i ˆ "n ‰
nÄ_ iœ"
œ ".
and ci œ i˜x œ ni . The right-hand sum is
n
n
"!
i
n#
iœ"
44. faxb œ $x #x#
iœ"
œ
" n a n "b
‹
n# Š
#
œ
" "n
#
$ nan "ba#n "b
‹
n$ Š
'
! i# œ
!aci ci# b " œ !Š i ˆ i ‰# ‹ " œ
n
n
n
n
iœ"
œ lim ”Š
nÄ_
" "n
# ‹
Œ
Let ˜x œ
"!
n
"
n
œ
# $n n"#
'
• œ
"
#
#
'
n
$!
i
#
n
iœ"
n
#! #
i
$
n
iœ"
$n# $n
#n#
#n# $n "
$n#
!a$ci #c#i b " œ !Š $i #ˆ i ‰# ‹ " œ
n
n
n
n
iœ"
iœ"
œ
$ n a n "b
‹
n# Š
#
œ
$ $n
#
n#$ Š nan "ba' #n "b ‹
n
# $ "
$
nÄ_ iœ"
œ lim ”Š
$ $n
# ‹
Œ
Let ˜x œ
"!
n
n
"
n
œ
# $n n"#
$
iœ"
iœ"
œ
#n# an# 2n "b
4n4
n
œ
• œ
$
#
#
$
œ
"$
' .
and ci œ i˜x œ ni . The right-hand sum is
!a2c3i b " œ !Š2ˆ i ‰3 ‹ " œ
n
n
n
Thus,
œ
. Thus, lim !a$ci #c#i b "n
n#
n
œ &' .
and ci œ i˜x œ ni . The right-hand sum is
n
n
n
"! #
i
n$
iœ"
n
45. faxb œ 2x3
#
$
#
" nan "ba#n "b
Š
‹ œ n #n# n #n '$nn$ n
'
n$
n
# $n n"#
. Thus, lim !aci c#i b "n
'
nÄ_ iœ"
nÄ_
267
n# 2n "
#n #
n
#! 3
i
n4
iœ"
œ
œ
" #n n"#
#
" # "
lim !a2c3i b "n œ lim ” n# n#
nÄ_ iœ"
nÄ_
# n a n "b
Š # ‹
n4
.
"
• œ #.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
2
268
Chapter 5 Integration
46. faxb œ x2 x3
Let ˜x œ
! a"b
n
n
œ
"
n
and ci œ " i˜x œ " ni . The rightn
2
$
hand sum is !ac#i ci3 b "n œ !Šˆ" ni ‰ ˆ" ni ‰ ‹ n"
iœ1
n
œ ! Š2 œ
iœ1
n
!2
n
iœ"
œ 2n anb n# 2n "
4n#
4 6n n#2
1 4
5n 5
#n
4n2 6n#
3n2
œ lim ”2 5
#
5
n
œ2
nÄ_
3
2
n
œ2
"
n#
5 5n
#
•œ2
5
2
4 6n n#2
3
1 2n 4
œ
7
12 .
4
3
1
4
5i
n
n
5!
i
n2
iœ"
%i
n2
2
n
%! 2
i
n3
iœ"
5 n an "b
‹
n2 Š
#
"
n#
i
"
n3 ‹ n
3
iœ1
n
œ !Š n2 iœ1
n
1! 3
i
n4
iœ"
5i
n2
% nan "ba#n "b
‹
n3 Š
'
%i2
n3
i3
n4 ‹
1 nan "b
‹
n4 Š
#
2
n
. Thus, lim !ac#i ci3 b n"
nÄ_ iœ"
5.3 THE DEFINITE INTEGRAL
1.
'02 x# dx
2.
'"! 2x$ dx
3.
'(& ax# 3xb dx
4.
'"% "x dx
5.
'#$ 1 " x dx
6.
'0" È4 x# dx
7.
' ! Î% (sec x) dx
8.
'0 Î% (tan x) dx
1
1
9. (a)
(c)
(e)
(f)
10. (a)
(b)
(c)
(d)
(e)
(f)
11. (a)
(c)
"
&
'#2 g(x) dx œ 0
(b) ' g(x) dx œ ' g(x) dx œ 8
&
"
&
&
2
'"2 3f(x) dx œ 3'"2 f(x) dx œ 3(4) œ 12
(d) ' f(x) dx œ ' f(x) dx ' f(x) dx œ 6 (4) œ 10
#
"
"
&
&
&
'" [f(x) g(x)] dx œ '" f(x) dx '" g(x) dx œ 6 8 œ 2
'"& [4f(x) g(x)] dx œ 4 '"& f(x) dx '"& g(x) dx œ 4(6) 8 œ 16
'"* 2f(x) dx œ 2 '"* f(x) dx œ 2(1) œ 2
'(* [f(x) h(x)] dx œ '(*f(x) dx '(* h(x) dx œ 5 4 œ 9
'(* [2f(x) 3h(x)] dx œ 2 '(* f(x) dx 3 '(* h(x) dx œ 2(5) 3(4) œ 2
'*"f(x) dx œ '"* f(x) dx œ (1) œ 1
'"( f(x) dx œ '"* f(x) dx '(* f(x) dx œ 1 5 œ 6
'*( [h(x) f(x)] dx œ '(* [f(x) h(x)] dx œ '(* f(x) dx '(* h(x) dx œ 5 4 œ 1
'"2 f(u) du œ '"2 f(x) dx œ 5
'#" f(t) dt œ '"2 f(t) dt œ 5
(b)
(d)
'"2 È3 f(z) dz œ È3 '"2 f(z) dz œ 5È3
'"2 [f(x)] dx œ '"2 f(x) dx œ 5
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 5.3 The Definite Integral
'!$ g(t) dt œ '$! g(t) dt œ È2
'$! [g(x)] dx œ '$! g(x) dx œ È2
12. (a)
(c)
(b)
'"$ h(r) dr œ '"$ h(r) dr '"" h(r) dr œ 6 0 œ 6
"
$
$
' h(u) du œ Œ ' h(u) du œ ' h(u) du œ 6
$
"
"
14. (a)
(b)
15. The area of the trapezoid is A œ
"
#
(5 2)(6) œ 21 Ê
œ 21 square units
"
#
(3 1)(1) œ 2 Ê
"
#
(B b)h
"
#
(B b)h
'# ˆ #x 3‰ dx
%
16. The area of the trapezoid is A œ
œ
(d)
'$! g(u) du œ '$! g(t) dt œ È2
'$! Èg(r)2 dr œ È"2 '$! g(t) dt œ Š È"2 ‹ ŠÈ2‹ œ 1
'$% f(z) dz œ '!% f(z) dz '!$ f(z) dz œ 7 3 œ 4
'%$ f(t) dt œ '$% f(t) dt œ 4
13. (a)
œ
(b)
'"Î#
$Î#
(2x 4) dx
œ 2 square units
17. The area of the semicircle is A œ
œ
9
#
1 Ê
"
#
1r# œ
"
#
1(3)#
'$$ È9 x# dx œ 9# 1 square units
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
269
270
Chapter 5 Integration
18. The graph of the quarter circle is A œ
œ 41 Ê
"
4
1 r# œ
"
4
1(4)#
'%! È16 x# dx œ 41 square units
19. The area of the triangle on the left is A œ
"
#
bh œ
œ 2. The area of the triangle on the right is A œ
œ
"
#
Ê
(1)(1) œ
"
#.
(2)(2)
bh
Then, the total area is 2.5
'# kxk dx œ 2.5 square units
"
20. The area of the triangle is A œ
Ê
"
#
"
#
"
#
bh œ
'" a1 kxkb dx œ 1 square unit
"
21. The area of the triangular peak is A œ
"
#
(2)(1) œ 1
"
#
bh œ
"
#
(2)(1) œ 1.
The area of the rectangular base is S œ jw œ (2)(1) œ 2.
Then the total area is 3 Ê
'"" a2 kxkb dx œ 3 square units
22. y œ 1 È1 x# Ê y 1 œ È1 x#
Ê (y 1)# œ 1 x# Ê x# (y 1)# œ 1, a circle with
center (!ß ") and radius of 1 Ê y œ 1 È1 x# is the
upper semicircle. The area of this semicircle is
A œ "# 1r# œ "# 1(1)# œ 1# . The area of the rectangular base
is A œ jw œ (2)(1) œ 2. Then the total area is 2 Ê
'"" Š1 È1 x# ‹ dx œ 2 1# square units
1
#
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 5.3 The Definite Integral
23.
'!b x2 dx œ "# (b)( b2 ) œ b4
24.
'!b 4x dx œ "# b(4b) œ 2b#
25.
'ab 2s ds œ "# b(2b) "# a(2a) œ b# a#
26.
'ab 3t dt œ "# b(3b) "# a(3a) œ 3# ab# a# b
(b)
'02 È4 x2 dx œ "4 1a2b2 ‘ œ 1
#
27. (a)
'22 È4 x2 dx œ "# 1a2b2 ‘ œ 21
28. (a)
'01 Š3x È1 x2 ‹ dx œ '01 3x dx '01 È1 x2 dx œ "# a1ba3b‘ 4" 1a1b2 ‘ œ 14 3#
(b)
271
'01 Š3x È1 x2 ‹ dx œ '01 3x dx '01 3x dx '11 È1 x2 dx œ "# a1ba3b‘ "# a1ba3b‘ 2" 1a1b2 ‘ œ 12
È#
#
29.
'"
31.
'1#1 ) d) œ (2#1)
33.
'0
35.
'!"Î# t# dt œ ˆ 3‰
œ
37.
'a#a x dx œ (2a)#
39.
'!
3
È
7
x dx œ
ŠÈ2‹
#
#
(1)#
#
œ
1#
#
œ
31 #
#
"
#
30.
'!Þ&#Þ& x dx œ (2.5)#
32.
'È& # # r dr œ Š5È#2‹
x dx œ
3
7‹
ŠÈ
œ
3
#
7
3
34.
'!!Þ$ s# ds œ (0.3)3
"
24
36.
'!1Î# )# d) œ ˆ 3‰
a#
#
œ
3a#
#
38.
'a
$
b‹
ŠÈ
3
œ
b
3
40.
'!$b x# dx œ (3b)3
(0.5)#
#
#
$
1 $
#
È$a
$
x# dx œ
#
È
$
#
" $
#
$
È
b
œ3
#
ŠÈ2‹
#
œ 24
œ 0.009
œ
1$
#4
#
x dx œ
ŠÈ3a‹
#
$
a#
#
œ a#
œ 9b$
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
272
Chapter 5 Integration
41.
'$" 7 dx œ 7(1 3) œ 14
43.
'!2 (2t 3) dt œ 2 '"" t dt '!2 3 dt œ 2 ’ 2#
44.
'!
45.
'#" ˆ1 #z ‰ dz œ '#" 1 dz '#" #z dz œ '#" 1 dz "# '"# z dz œ 1[1 2] "# ’ 2# 1# “ œ " "# ˆ 3# ‰ œ 74
46.
'$! (2z 3) dz œ '$! 2z dz '$! 3 dz œ 2 '!$ z dz '$! 3 dz œ 2 ’ 3#
47.
'"# 3u# du œ 3 '"# u# du œ 3 ”'!# u# du '!" u# du• œ 3 Š’ 23
48.
'"Î#" 24u# du œ 24 '"Î#"
49.
'!# a3x# x 5b dx œ 3 '!# x# dx '!# x dx '!# 5 dx œ 3 ’ 23
50.
'"! a3x# x 5b dx œ '!" a3x# x 5b dx œ ”3 '!" x# dx '!" x dx '!" 5 dx•
#
È2
Št È2‹ dt œ
È2
'!
'!2 5x dx œ 5 '!2 x dx œ 5 ’ 2#
#
42.
t dt È2
'!
È2 dt œ
0#
#“
ŠÈ2‹
#
0#
#—
#
$
'!"
u# du '!"Î#
0$
3“
$
0$
3‹
51. Let ?x œ
b0
n
œ
b
n
#
Š 1# 0#
#‹
$
’ "3 $
u# du— œ 24 ” 13 $
œ ’3 Š 13 5(1 0)“ œ ˆ 3# 5‰ œ
0$
3“
0#
#“
#
3[0 3] œ 9 9 œ 0
0$
3 “‹
ˆ "# ‰$
3 •
#
’ 2# $
œ 3 ’ 23 œ 24 ’
0#
#“
ˆ 78 ‰
3
1$
3“
5[2 0] œ (8 2) 10 œ 0
7
#
x# œ 2?xß á ß xn " œ (n 1)?x, xn œ n?x œ b.
Let the ck 's be the right end-points of the subintervals
Ê c" œ x" , c# œ x# , and so on. The rectangles
defined have areas:
f(c" ) ?x œ f(?x) ?x œ 3(?x)# ?x œ 3(?x)$
f(c# ) ?x œ f(2?x) ?x œ 3(2?x)# ?x œ 3(2)# (?x)$
f(c$ ) ?x œ f(3?x) ?x œ 3(3?x)# ?x œ 3(3)# (?x)$
ã
f(cn ) ?x œ f(n?x) ?x œ 3(n?x)# ?x œ 3(n)# (?x)$
n
Then Sn œ ! f(ck ) ?x œ ! 3k# (?x)$
kœ1
n
kœ1
1)
œ 3(?x)$ ! k# œ 3 Š bn$ ‹ Š n(n 1)(2n
‹
6
$
kœ1
œ
$
b
#
ˆ2 3
n
"‰
n#
Ê
'!b 3x# dx œ n lim
Ä_
b$
#
ˆ2 3
n
"‰
n#
œ 3 ˆ 37 ‰ œ 7
“œ7
and let x! œ 0, x" œ ?x,
n
œ 10
È2 ’È2 0“ œ 1 2 œ 1
#
u# du œ 24 –
0#
#“
3(2 0) œ 4 6 œ 2
#
–
œ b$ .
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 5.3 The Definite Integral
52. Let ?x œ
b0
n
œ
and let x! œ 0, x" œ ?x,
b
n
x# œ 2?xß á ß xn " œ (n 1)?x, xn œ n?x œ b.
Let the ck 's be the right end-points of the subintervals
Ê c" œ x" , c# œ x# , and so on. The rectangles
defined have areas:
f(c" ) ?x œ f(?x) ?x œ 1(?x)# ?x œ 1(?x)$
f(c# ) ?x œ f(2?x) ?x œ 1(2?x)# ?x œ 1(2)# (?x)$
f(c$ ) ?x œ f(3?x) ?x œ 1(3?x)# ?x œ 1(3)# (?x)$
ã
f(cn ) ?x œ f(n?x) ?x œ 1(n?x)# ?x œ 1(n)# (?x)$
n
n
Then Sn œ ! f(ck ) ?x œ ! 1k# (?x)$
kœ1
n
kœ1
1)
œ 1(?x)$ ! k# œ 1 Š bn$ ‹ Š n(n 1)(2n
‹
6
$
kœ1
œ
1b
6
$
ˆ2 53. Let ?x œ
3
n
b0
n
"‰
n#
œ
b
n
Ê
'!b 1x# dx œ n lim
Ä_
1 b$
6
ˆ2 3
n
"‰
n#
œ
1 b$
3 .
and let x! œ 0, x" œ ?x,
x# œ 2?xß á ß xn " œ (n 1)?x, xn œ n?x œ b.
Let the ck 's be the right end-points of the subintervals
Ê c" œ x" , c# œ x# , and so on. The rectangles
defined have areas:
f(c" ) ?x œ f(?x) ?x œ 2(?x)(?x) œ 2(?x)#
f(c# ) ?x œ f(2?x) ?x œ 2(2?x)(?x) œ 2(2)(?x)#
f(c$ ) ?x œ f(3?x) ?x œ 2(3?x)(?x) œ 2(3)(?x)#
ã
f(cn ) ?x œ f(n?x) ?x œ 2(n?x)(?x) œ 2(n)(?x)#
n
n
Then Sn œ ! f(ck ) ?x œ ! 2k(?x)#
kœ1
n
kœ1
œ 2(?x)# ! k œ
kœ1
œ b# ˆ1 "n ‰ Ê
54. Let ?x œ
b0
n
œ
#
2 Š bn# ‹ Š n(n 2 1) ‹
'!b 2x dx œ n lim
Ä_
b
n
b# ˆ1 n" ‰ œ b# .
and let x! œ 0, x" œ ?x,
x# œ 2?xß á ß xn " œ (n 1)?x, xn œ n?x œ b.
Let the ck 's be the right end-points of the subintervals
Ê c" œ x" , c# œ x# , and so on. The rectangles
defined have areas:
"
#
‰
f(c" ) ?x œ f(?x) ?x œ ˆ ?x
# 1 (?x) œ # (?x) ?x
f(c# ) ?x œ f(2?x) ?x œ ˆ 2?# x 1‰ (?x) œ "# (2)(?x)# ?x
f(c$ ) ?x œ f(3?x) ?x œ ˆ 3?# x 1‰ (?x) œ
"
#
(3)(?x)# ?x
f(cn ) ?x œ f(n?x) ?x œ ˆ n?# x 1‰ (?x) œ
"
#
(n)(?x)# ?x
ã
n
n
kœ1
kœ1
Then Sn œ ! f(ck ) ?x œ ! ˆ "# k(?x)# ?x‰ œ
œ
"
4
b# ˆ1 1n ‰ b Ê
'! ˆ x# 1‰ dx œ n lim
Ä_
b
"
#
n
n
kœ1
kœ1
(?x)# ! k ?x ! 1 œ
ˆ 4" b# ˆ1 n" ‰ b‰ œ
"
4
"
#
Š bn# ‹ Š n(n 2 1) ‹ ˆ bn ‰ (n)
#
b# b.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
273
274
Chapter 5 Integration
55. av(f) œ Š È3" 0 ‹
œ
È$
'!
"
È3
È$
'!
x# dx "
È3
ŠÈ3‹
3
56. av(f) œ ˆ 3 " 0 ‰
È$
'!
"
È3
$
œ
ax# 1b dx
"
È3
1 dx
ŠÈ3 0‹ œ 1 1 œ 0.
'!$ Š x# ‹ dx œ 3" ˆ #" ‰ '!$ x# dx
#
$
#
œ "6 Š 33 ‹ œ 3# ; x# œ 3# .
'!" a3x# 1b dx œ
"
"
œ 3 ' x# dx ' 1 dx œ 3 Š 13 ‹ (1 0)
!
!
57. av(f) œ ˆ 1 " 0 ‰
$
œ #.
'!" a3x# 3b dx œ
"
"
œ 3 ' x# dx ' 3 dx œ 3 Š 13 ‹ 3(1 0)
!
!
58. av(f) œ ˆ 1 " 0 ‰
$
œ #.
'!$ (t 1)# dt
$
$
$
œ 3" ' t# dt 32 ' t dt 3" ' 1 dt
!
!
!
59. av(f) œ ˆ 3 " 0 ‰
œ
"
3
$
#
Š 33 ‹ 32 Š 3# 0#
#‹
3" (3 0) œ 1.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 5.3 The Definite Integral
60. av(f) œ Š 1 1(2) ‹
'#" at# tb dt
'#" t# dt 3" '#" t dt
"
#
œ "3 ' t# dt 3" '
t# dt 3" Š 1#
!
!
œ
"
3
#
œ
"
3
$
Š 13 ‹ 3" Š (32) ‹ $
61. (a) av(g) œ Š 1 "(1) ‹
"
#
œ
3
#
(2)#
# ‹
.
'"" akxk 1b dx
'"! (x 1) dx "# '!" (x 1) dx
!
!
"
"
œ "# ' x dx "# ' 1 dx "# ' x dx "# ' 1 dx
"
"
!
!
œ
"
#
#
œ "# Š 0# (1)#
# ‹
#
"# (0 (1)) "# Š 1# 0#
#‹
"# (1 0)
œ "# .
'"$ akxk 1b dx œ #" '"$ (x 1) dx
$
$
œ "# ' x dx "# ' 1 dx œ "# Š 3# 1# ‹ "# (3 1)
"
"
(b) av(g) œ ˆ 3 " 1 ‰
#
#
œ 1.
(c) av(g) œ Š 3 "(1) ‹
œ
"
4
"
4
'"$ akxk 1b dx
'"" akxk 1b dx 4" '"$ akxk 1b dx
"
4
(see parts (a) and (b) above).
62. (a) av(h) œ Š 0 "(1) ‹
'"0 kxk dx œ '"0 (x) dx
œ
œ
(1 2) œ
'"0 x dx œ 0#
#
(1)#
#
œ "# .
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
275
276
Chapter 5 Integration
(b) av(h) œ ˆ 1 " 0 ‰
#
œ Š "# '0" kxk dx œ '0" x dx
0#
#‹
œ "# .
(c) av(h) œ Š 1 "(1) ‹
'"" kxk dx
'"0 kxk dx '0" kxk dx
œ
"
#
Œ
œ
"
#
ˆ "# ˆ "# ‰‰ œ "# (see parts (a) and (b)
above).
ba
n and let ck be the right
n ab a b
× and ck œ a kabn ab .
n
63. Consider the partition P that subdivides the interval Òa, bÓ into n subintervals of width ˜x œ
endpoint of each subinterval. So the partition is P œ Öa, a n
n
kœ"
kœ"
b
We get the Riemann sum ! fack b˜x œ ! c †
this expression remains cab ab. Thus,
ba
n
ba
n ,
n
c ab a b !
"
n
kœ"
œ
#a b a b
,
n
a
œ
c ab a b
n
...,a
† n œ cab ab. As n Ä _ and mPm Ä !
'a c dx œ cab ab.
64. Consider the partition P that subdivides the interval Ò0, 2Ó into n subintervals of width ˜x œ
right endpoint of each subinterval. So the partition is P œ Ö0,
n
n
kœ"
kœ"
‰
Riemann sum ! fack b˜x œ ! ’2ˆ 2k
n 1“ †
As n Ä _ and mPm Ä ! the expression
œ
2
n
4 an "b
n
2
n
2
n,
2†
2
n,
n
! ˆ 4k 1‰ œ
n
kœ"
...,n†
8
n2
n
!k kœ"
2
n
n
!1 œ
kœ"
8
n2
†
n
n
œ
n
ba ! #
a
n Œ
kœ"
kœ"
n
#a a b a b !
k
n
kœ"
œ ab aba# aab ab# †
n"
n
kœ"
n
ab a b ! #
k n#
kœ"
#
ab a b $
'
†
œ
b a
n
n an " b
#
2
n
and let ck be the
2
n
œ
2
n
2k
n .
We get the
†nœ
4 an " b
n
2.
'02 a2x 1b dx œ 6.
65. Consider the partition P that subdivides the interval Òa, bÓ into n subintervals of width ˜x œ
We get the Riemann sum ! fack b˜x œ ! c#k ˆ b n a ‰
œ
œ 2× and ck œ k †
2
n
2 has the value 4 2 œ 6. Thus,
endpoint of each subinterval. So the partition is P œ
20
n
ba
n
and let ck be the right
Öa, a b n a , a #abn ab , . . ., a nabn ab × and ck œ a kabn ab .
n
n
#
#
#
œ b n a ! Ša kabn ab ‹ œ bn a ! Ša# #akabn ab k abn# ab ‹
kœ"
kœ"
† na# an "ba#n "b
n#
#a a b a b #
n#
†
n a n "b
#
ab a b $
n$
†
nan "ba#n "b
'
$
"
" n"
ab ab$ # n n#
†
" '
"
ab a b $
†#
'
b
$
$
x# dx œ b$ a$ .
a
œ ab aba# aab ab# †
As n Ä _ and mPm Ä ! this expression has value ab aba# aab ab# † " œ ba# a$ ab# #a# b a$ "$ ab$ $b# a $ba# a$ b œ
b$
$
a$
$.
Thus,
'
66. Consider the partition P that subdivides the interval Ò1, 0Ó into n subintervals of width ˜x œ
the right endpoint of each subinterval. So the partition is P œ Ö1, 1 1
n,
1 2 †
1
n,
0 a 1 b
n
. . ., 1 n †
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
œ
1
n
1
n
and let ck be
œ 0× and
Section 5.3 The Definite Integral
ck œ 1 k †
œ
1
n
n
kœ"
n
kœ"
2
œ 1 kn . We get the Riemann sum ! fack b˜x œ ! Šˆ1 kn ‰ ˆ1 kn ‰ ‹ †
k
n
1
n
! Š1 kœ"
3 an " b
2n
œ 2 n
1
n
n
2
ˆ kn ‰ ‹ œ 2n ! 1 2k
n
kœ"
an "ba#n "b
.
'n 2
3
n2
!k 1
n3
kœ"
n
! k2 œ 2 † n n
kœ"
3
n2
n an " b
#
†
As n Ä _ and mPm Ä ! this expression has value 2 '01 ax x# bdx œ 56 .
3
2
1
n3
the right endpoint of each subinterval. So the partition is P œ Ö1, 1 ck œ 1 k †
œ
3
n
n
! Š3 kœ"
œ 18 3
n
18k
n
36an "b
n
œ 1 27k2
n2
3k
n .
n
n
kœ"
n
kœ"
3
n,
1 2 †
We get the Riemann sum ! fack b˜x œ ! Š3ˆ1 2
6k
n
27an "ba#n "b
.
2n2
1‹ œ
18
n
n
!1 kœ"
72
n2
!k kœ"
81
n3
n
! k2 œ
kœ"
18
n
†n
3
n,
3k ‰2
n
72
n2
†
1
n
nan "ba#n "b
'
œ 56 . Thus,
1
3
67. Consider the partition P that subdivides the interval Ò1, 2Ó into n subintervals of width ˜x œ
†
2 a 1 b
n
. . ., 1 n †
2ˆ1 n an " b
#
277
œ
3
n
3k ‰
n
81
n3
†
3
n
and let ck be
œ 2× and
1‹ †
3
n
nan "ba#n "b
'
As n Ä _ and mPm Ä ! this expression has value 18 36 27 œ 9. Thus,
'21 a3x# 2x 1bdx œ 9.
68. Consider the partition P that subdivides the interval Ò1, 1Ó into n subintervals of width ˜x œ
the right endpoint of each subinterval. So the partition is P œ Ö1, 1 ck œ 1 k †
œ
2
n
œ 1 2
n
n
! Š1 kœ"
6k
n
œ 2n † n 12
n#
†
œ 2 6 †
" "n
"
12k2
n2
n an " b
#
4†
'1 x3 dx œ 0.
2k
n .
24
n$
n
2
n,
1 2 †
n
We get the Riemann sum ! fack b˜x œ ! c3k ˆ 2n ‰ œ
kœ"
8k3
n3 ‹
†
# $n n"#
"
n
n
kœ"
kœ"
œ n2 Œ! 1 n6 ! k nan "ba#n "b
'
4†
" 2n n12
1
16
n4
†
12
n2
nn
Š a # "b ‹
n
! k2 kœ"
2
kœ"
n
8 ! 3
k 3
n
kœ"
œ 2 6 †
n"
n
2
n
4†
#
n,
1 a 1 b
n
. . ., 1 n †
n
! ˆ1 kœ"
œ
2
n
2
n
and let ck be
œ 1× and
2k ‰3
n
an "ba#n "b
n#
4†
an "b2
n2
. As n Ä _ and mPm Ä ! this expression has value 2 6 8 4 œ 0.
1
Thus,
ba
n and let ck be the
#
a
b
a
b
n
a
b
a
b
endpoint of each subinterval. So the partition is P œ Öa, a b n a , a n , . . . , a n œ b× and
n
n
n
3
ck œ a kabn ab . We get the Riemann sum ! fack b˜x œ ! c3k ˆ b n a ‰ œ b n a ! Ša kabn ab ‹
kœ"
kœ"
kœ"
n
n
n
2
3
2 n
3 n
2
2
3
2
œ b n a ! Ša3 3a kanb ab 3ak anb2 ab k abn3 ab ‹ œ b n a Œ ! a3 3a abn ab ! k 3aabn2 ab ! k2 ab n3ab ! k3 kœ"
kœ"
kœ"
kœ"
kœ"
2
#
$
4
2
3a
b
a
n
n
"
3a
b
a
n
n
"
#
n
"
b
a
n
n
"
a
b
a
b
a
b
a
ba
b
a
b
a
b
œ b n a † na3 † # n$
†
n4 † Š # ‹
n#
'
69. Consider the partition P that subdivides the interval Òa, bÓ into n subintervals of width ˜x œ
3a2 ab ab#
2
a ab a b $
2
4
2
an "ba#n "b
ab 4 ab † an n2"b
n#
2
1
#
$
2
# $n n"#
" "
ab ab4 " n n2
œ ab aba3 3a ab2 ab † " n aab 2 ab †
†
. As n
"
4
1
b
#
4
2
4
4
4
ab aba3 3a ab2 ab aab ab$ ab 4 ab œ b4 a4 . Thus,
x3 dx œ b4
a
œ ab aba3 †
n"
n
right
†
'
Ä _ and mPm Ä ! this expression has value
a4
4.
10
n
70. Consider the partition P that subdivides the interval Ò0, 1Ó into n subintervals of width ˜x œ
right endpoint of each subinterval. So the partition is P œ Ö0, 0 n
n
kœ"
kœ"
We get the Riemann sum ! fack b˜x œ ! a3ck c3k bˆ 1n ‰ œ
1
n
n
1
n,
02†
! Š3 †
kœ"
k
n
1
n,
. . ., 0 n †
1
n
n
œ
1
n
and let ck be the
œ 1× and ck œ 0 k †
3
ˆ kn ‰ ‹ œ 1n Œ 3n ! k kœ"
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
1
n3
n
! k3 kœ"
1
n
œ kn .
278
Chapter 5 Integration
œ
3
n2
†
n an " b
#
3
2
has value
1
4
1
n4
2
† Š nan # "b ‹ œ
œ 54 . Thus,
3
2
†
n"
n
1
4
†
a n "b 2
n2
œ
3
2
" "n
"
†
" 2n n12
1
†
1
4
. As n Ä _ and mPm Ä ! this expression
'01 a3x x3 bdx œ 54 .
71. To find where x x# 0, let x x# œ 0 Ê x(1 x) œ 0 Ê x œ 0 or x œ 1. If 0 x 1, then 0 x x# Ê a œ 0
and b œ 1 maximize the integral.
72. To find where x% 2x# Ÿ 0, let x% 2x# œ 0 Ê x# ax# 2b œ 0 Ê x œ 0 or x œ „ È2. By the sign graph,
0 0 0 , we can see that x% 2x# Ÿ 0 on ’È2ß È2“ Ê a œ È2 and b œ È2
!
È#
È #
minimize the integral.
73. f(x) œ
"
1 x#
is decreasing on [0ß 1] Ê maximum value of f occurs at 0 Ê max f œ f(0) œ 1; minimum value of f occurs
at 1 Ê min f œ f(1) œ
"
1 1#
œ
"
#
. Therefore, (1 0) min f Ÿ
"
#
That is, an upper bound œ 1 and a lower bound œ
74. See Exercise 73 above. On [0ß 0.5], max f œ
(0.5 0) min f Ÿ
min f œ
Then
"
4
"
1 1#
2
5
'0
0.5
"
1 0#
Ÿ
'00.5 1 " x
#
dx '0.5" 1 " x
#
dx Ÿ
75. 1 Ÿ sin ax# b Ÿ 1 for all x Ê (1 0)(1) Ÿ
"
#
œ 1, min f œ
Ÿ
2
5
'0
'0.5" 1 " x
#
dx Ÿ (1 0) max f Ê
"
#
'0" 1 " x
Ÿ
#
dx Ÿ 1.
.
f(x) dx Ÿ (0.5 0) max f Ê
œ 0.5. Therefore (1 0.5) min f Ÿ
'0" 1 " x
2
5
Ê
0.5
#
"
1 (0.5)#
"
1 x#
œ 0.8. Therefore
dx Ÿ
"
#
. On [0.5ß 1], max f œ
dx Ÿ (1 0.5) max f Ê
13
20
Ÿ
'0" 1 " x
#
dx Ÿ
9
10
"
4
Ÿ
"
1 (0.5)#
'0.5" 1 1 x
#
dx Ÿ
œ 0.8 and
2
5
.
.
'0" sin ax# b dx Ÿ (1 0)(1) or '0"sin x# dx Ÿ 1
Ê
'0"sin x# dx cannot
equal 2.
76. f(x) œ Èx 8 is increasing on [!ß "] Ê max f œ f(1) œ È1 8 œ 3 and min f œ f(0) œ È0 8 œ 2È2 .
Therefore, (1 0) min f Ÿ
77. If f(x)
Then b
'0" Èx 8 dx Ÿ (1 0) max f
0 on [aß b], then min f
a Ê ba
0 and max f
0 Ê (b a) min f
Ê 2È 2 Ÿ
'0" Èx 8 dx Ÿ 3.
0 on [aß b]. Now, (b a) min f Ÿ
0 Ê
'ab f(x) dx
0.
78. If f(x) Ÿ 0 on [aß b], then min f Ÿ 0 and max f Ÿ 0. Now, (b a) min f Ÿ
b
a Ê ba
79. sin x Ÿ x for x
Ê
0 Ê (b a) max f Ÿ 0 Ê
Ê
'ab f(x) dx Ÿ (b a) max f.
Then
b
'0" (sin x x) dx Ÿ 0 (see Exercise 78) Ê '0" sin x dx '0" x dx Ÿ 0
'0" sin x dx Ÿ Š 1# 0# ‹ Ê '0" sin x dx Ÿ "# . Thus an upper bound is "# .
0 Ê sin x x Ÿ 0 for x
'0" sin x dx Ÿ '0" x dx
'a f(x) dx Ÿ 0.
'ab f(x) dx Ÿ (b a) max f.
0Ê
#
#
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 5.3 The Definite Integral
80. sec x
1
x#
#
on ˆ 1# ß 1# ‰ Ê sec x Š1 since [0ß 1] is contained in ˆ 1# ß 1# ‰ Ê
Ê
is
x#
#‹
0 on ˆ 1# ß 1# ‰ Ê
'0" ’sec x Š1 x# ‹“ dx
'0" sec x dx '0" Š1 x# ‹ dx
#
'0" sec x dx '0" 1 dx "# '0" x# dx Ê '0" sec x dx
#
0Ê
$
0 (see Exercise 77)
'0" sec x dx '0" Š1 x# ‹
(1 0) "# Š 13 ‹ Ê
#
'0" sec x dx
7
6.
Thus a lower bound
7
6.
'ab f(x) dx is a constant K.
b
b
b
Ê ' av(f) dx œ (b a)K œ (b a) † b " a ' f(x) dx œ ' f(x) dx.
a
a
a
81. Yes, for the following reasons: av(f) œ
"
ba
'ab av(f) dx œ 'ab K dx
Thus
œ K(b a)
82. All three rules hold. The reasons: On any interval [aß b] on which f and g are integrable, we have:
(a) av(f g) œ
"
ba
'ab [f(x) g(x)] dx œ b " a ”'ab f(x) dx 'ab g(x) dx• œ b " a 'ab f(x) dx b " a 'ab g(x) dx
œ av(f) av(g)
(b) av(kf) œ
(c) av(f) œ
"
ba
"
ba
'ab kf(x) dx œ b " a ”k 'ab f(x) dx• œ k ” b " a 'ab f(x) dx• œ k av(f)
'ab f(x) dx Ÿ b " a 'ab g(x) dx since f(x) Ÿ g(x) on [aß b], and b " a 'ab g(x) dx œ av(g).
Therefore, av(f) Ÿ av(g).
83. (a) U œ max" ?x max# ?x á maxn ?x where max" œ f(x" ), max# œ f(x# ), á , maxn œ f(xn ) since f is
increasing on [aß b]; L œ min" ?x min# ?x á minn ?x where min" œ f(x! ), min# œ f(x" ), á ,
minn œ f(xnc1 ) since f is increasing on [aß b]. Therefore
U L œ (max" min" ) ?x (max# min# ) ?x á (maxn minn ) ?x
œ (f(x" ) f(x! )) ?x (f(x# ) f(x" ))?x á (f(xn ) f(xnc1 )) ?x œ (f(xn ) f(x! )) ?x œ (f(b) f(a)) ?x.
(b) U œ max" ?x" max# ?x# á maxn ?xn where max" œ f(x" ), max# œ f(x# ), á , maxn œ f(xn ) since f
is increasing on[aß b]; L œ min" ?x" min# ?x# á minn ?xn where
min" œ f(x! ), min# œ f(x" ), á , minn œ f(xnc1 ) since f is increasing on [aß b]. Therefore
U L œ (max" min" ) ?x" (max# min# ) ?x# á (maxn minn ) ?xn
œ (f(x" ) f(x! )) ?x" (f(x# ) f(x" ))?x# á (f(xn ) f(xnc1 )) ?xn
Ÿ (f(x" ) f(x! )) ?xmax (f(x# ) f(x" )) ?xmax á (f(xn ) f(xnc1 )) ?xmax . Then
U L Ÿ (f(xn ) f(x! )) ?xmax œ (f(b) f(a)) ?xmax œ kf(b) f(a)k ?xmax since f(b) f(a). Thus
lim (U L) œ lim (f(b) f(a)) ?xmax œ 0, since ?xmax œ lPl .
lPl Ä 0
279
lPl Ä 0
84. (a) U œ max" ?x max# ?x á maxn ?x where
max" œ f(x! ), max# œ f(x" ), á , maxn œ f(xnc" )
since f is decreasing on [aß b];
L œ min" ?x min# ?x á minn ?x where
min" œ f(x" ), min# œ f(x# )ß á , minn œ f(xn )
since f is decreasing on [aß b]. Therefore
U L œ (max" min" ) ?x (max# min# ) ?x
á (maxn minn ) ?x
œ (f(x! ) f(x" )) ?x (f(x" ) f(x# ))?x
á (f(xn " ) f(xn )) ?x œ (f(x! ) f(xn )) ?x
œ (f(a) f(b)) ?x.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
280
Chapter 5 Integration
(b) U œ max" ?x" max# ?x# á maxn ?xn where max" œ f(x! ), max# œ f(x" ), á , maxn œ f(xn " ) since
f is decreasing on[aß b]; L œ min" ?x" min# ?x# á minn ?xn where
min" œ f(x" ), min# œ f(x# ), á , minn œ f(xn ) since f is decreasing on [aß b]. Therefore
U L œ (max" min" ) ?x" (max# min# ) ?x# á (maxn minn ) ?xn
œ (f(x! ) f(x" )) ?x" (f(x" ) f(x# ))?x# á (f(xn " ) f(xn )) ?xn
Ÿ (f(x! ) f(xn )) ?xmax œ (f(a) f(b) ?xmax œ kf(b) f(a)k ?xmax since f(b) Ÿ f(a). Thus
lim (U L) œ lim kf(b) f(a)k ?xmax œ 0, since ?xmax œ lPl .
lPl Ä 0
lPl Ä 0
85. (a) Partition 0ß 1# ‘ into n subintervals, each of length ?x œ
x# œ 2?x, á , xn œ n?x œ
1
#n
with points x! œ 0, x" œ ?x,
Since sin x is increasing on 0ß 1# ‘ , the upper sum U is the sum of the areas
1
#.
of the circumscribed rectangles of areas f(x" ) ?x œ (sin ?x)?x, f(x# ) ?x œ (sin 2?x) ?x, á , f(xn ) ?x
œ (sin n?x) ?x. Then U œ (sin ?x sin 2?x á sin n?x) ?x œ ”
œ”
1 cos ˆˆn " ‰ 1 ‰
cos 4n
1
# 2n
1
• ˆ #n ‰
# sin 4n
1 cos ˆ 1 1 ‰‰
1 ˆcos 4n
#
4n
1
4n sin 4n
œ
1 cos ˆ 1 1 ‰
cos 4n
#
4n
sin 1
Š 14n ‹
4n
'!
1Î#
(b) The area is
œ
cos ?#x cosˆ ˆn #" ‰ ?x‰
• ?x
# sin ?#x
sin x dx œ n lim
Ä_
1 cos ˆ 1 1 ‰
cos 4n
#
4n
sin 1
Š 14n ‹
œ
1 cos 1#
1
œ 1.
4n
n
86. (a) The area of the shaded region is !˜xi † mi which is equal to L.
iœ"
n
(b) The area of the shaded region is !˜xi † Mi which is equal to U.
iœ"
(c) The area of the shaded region is the difference in the areas of the shaded regions shown in the second part of the figure
and the first part of the figure. Thus this area is U L.
n
n
iœ"
iœ"
87. By Exercise 86, U L œ !˜xi † Mi !˜xi † mi where Mi œ maxÖfaxb on the ith subinterval× and
n
n
iœ"
iœ"
mi œ minÖfaxb on the ith subinterval×. Thus U L œ !aMi mi b˜xi !% † ˜xi provided ˜xi $ for each
n
n
iœ"
iœ"
i œ "ß Þ Þ Þ , n. Since !% † ˜xi œ % !˜xi œ %ab ab the result, U L %ab ab follows.
88. The car drove the first 150 miles in 5 hours and the
second 150 miles in 3 hours, which means it drove 300
miles in 8 hours, for an average of 300
8 mi/hr
œ 37.5 mi/hr. In terms of average values of functions,
the function whose average value we seek is
30, 0 Ÿ t Ÿ 5
v(t) œ œ
, and the average value is
50, 5 1 Ÿ 8
(30)(5) (50)(3)
8
œ 37.5.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 5.3 The Definite Integral
89-94. Example CAS commands:
Maple:
with( plots );
with( Student[Calculus1] );
f := x -> 1-x;
a := 0;
b := 1;
N :=[ 4, 10, 20, 50 ];
P := [seq( RiemannSum( f(x), x=a..b, partition=n, method=random, output=plot ), n=N )]:
display( P, insequence=true );
95-98. Example CAS commands:
Maple:
with( Student[Calculus1] );
f := x -> sin(x);
a := 0;
b := Pi;
plot( f(x), x=a..b, title="#95(a) (Section 5.3)" );
N := [ 100, 200, 1000 ];
# (b)
for n in N do
Xlist := [ a+1.*(b-a)/n*i $ i=0..n ];
Ylist := map( f, Xlist );
end do:
for n in N do
# (c)
Avg[n] := evalf(add(y,y=Ylist)/nops(Ylist));
end do;
avg := FunctionAverage( f(x), x=a..b, output=value );
evalf( avg );
FunctionAverage(f(x),x=a..b,output=plot);
# (d)
fsolve( f(x)=avg, x=0.5 );
fsolve( f(x)=avg, x=2.5 );
fsolve( f(x)=Avg[1000], x=0.5 );
fsolve( f(x)=Avg[1000], x=2.5 );
89-98. Example CAS commands:
Mathematica: (assigned function and values for a, b, and n may vary)
Sums of rectangles evaluated at left-hand endpoints can be represented and evaluated by this set of commands
Clear[x, f, a, b, n]
{a, b}={0, 1}; n =10; dx = (b a)/n;
f = Sin[x]2 ;
xvals =Table[N[x], {x, a, b dx, dx}];
yvals = f /.x Ä xvals;
boxes = MapThread[Line[{{#1,0},{#1, #3},{#2, #3},{#2, 0}]&,{xvals, xvals dx, yvals}];
Plot[f, {x, a, b}, Epilog Ä boxes];
Sum[yvals[[i]] dx, {i, 1, Length[yvals]}]//N
Sums of rectangles evaluated at right-hand endpoints can be represented and evaluated by this set of commands.
Clear[x, f, a, b, n]
{a, b}={0, 1}; n =10; dx = (b a)/n;
f = Sin[x]2 ;
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
281
282
Chapter 5 Integration
xvals =Table[N[x], {x, a dx, b, dx}];
yvals = f /.x Ä xvals;
boxes = MapThread[Line[{{#1,0},{#1, #3},{#2, #3},{#2, 0}]&,{xvals dx,xvals, yvals}];
Plot[f, {x, a, b}, Epilog Ä boxes];
Sum[yvals[[i]] dx, {i, 1,Length[yvals]}]//N
Sums of rectangles evaluated at midpoints can be represented and evaluated by this set of commands.
Clear[x, f, a, b, n]
{a, b}={0, 1}; n =10; dx = (b a)/n;
f = Sin[x]2 ;
xvals =Table[N[x], {x, a dx/2, b dx/2, dx}];
yvals = f /.x Ä xvals;
boxes = MapThread[Line[{{#1,0},{#1, #3},{#2, #3},{#2, 0}]&,{xvals dx/2, xvals dx/2, yvals}];
Plot[f, {x, a, b},Epilog Ä boxes];
Sum[yvals[[i]] dx, {i, 1, Length[yvals]}]//N
5.4 THE FUNDAMENTAL THEOREM OF CALCULUS
1.
'c (2x 5) dx œ cx# 5xd#! œ a0# 5(0)b a(2)# 5(2)b œ 6
2.
'c ˆ5 x# ‰ dx œ ’5x x4 “ %
0
2
4
3.
#
$
3
'
2
0
xax 3b dx œ
'
2
0
5.
3
'c ax2 2x 3b dx œ ’ x3
3
1
'
4
0
Š3x x$
4‹
#
dx œ ’ 3x# 'c ax$ 2x 3b dx œ ’ x4
2
6.
%
2
7.
'
8.
'
'
1Î3
9.
'
1
10.
'
31Î4
11.
'
1Î3
12.
1
%
x%
16 “ !
1
3
œ Š a23b œ
3 a2 b 2
2 ‹
133
4
3
Š a03b 3
1
#
#
#
3 a0 b 2
2 ‹
œ 10
3
œ Š a13b a1b# 3a1b‹ Š (31) (1)# 3(1)‹ œ
œ Š 3(4)
# x# 3x“
2
3x2
2 “!
(3)#
4 ‹
4%
16 ‹
3
#
Š 3(0)
# (0)%
16 ‹
œ8
%
œ Š 24 2# 3(2)‹ Š (42) (2)# 3(2)‹ œ 12
%
"
$
ˆx# Èx‰ dx œ ’ x3 32 x$Î# “ œ ˆ "3 32 ‰ 0 œ 1
32
1
$#
x'Î& dx œ 5x"Î& ‘ " œ ˆ #5 ‰ (5) œ
0
1Î$
2 sec# x dx œ [2 tan x]!
5
#
œ ˆ2 tan ˆ 13 ‰‰ (2 tan 0) œ 2È3 0 œ 2È3
a1 cos xb dx œ [x sin x]1! œ a1 sin 1b a0 sin 0b œ 1
1Î4
0
x# 3x“
Š5(3) !
0
0
4#
4‹
ax2 3xb dx œ ’ x3 1
4.
œ Š5(4) $1Î%
csc ) cot ) d) œ [csc )]1Î% œ ˆcsc ˆ 341 ‰‰ ˆcsc ˆ 14 ‰‰ œ È2 ŠÈ2‹ œ 0
1Î$
4 sec u tan u du œ [4 sec u]!
œ 4 sec ˆ 13 ‰ 4 sec 0 œ 4(2) 4(1) œ 4
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
20
3
Section 5.4 The Fundamental Theorem of Calculus
13.
14.
'
0
" cos 2t
#
1Î2
'
dt œ
0
ˆ" 1Î2 #
2t
'c ÎÎ " cos
dt œ '
#
1 3
1Î3
1 3
1Î3 #
œ ˆ "# ˆ 13 ‰ '
1Î4
15.
'
1Î6
16.
0
0
"
4
"
#
cos 2t‰ dt œ "# t ˆ" "
#
'
1Î4
0
sin 2t‘ 1Î# œ ˆ "# (0) cos 2t‰ dt œ "# t sin 2 ˆ 13 ‰‰ ˆ #" ˆ 13 ‰ tan# x dx œ
!
"
4
"
4
1Î6
0
sin 2 ˆ 13 ‰‰ œ
1
6
"
4
sin 231 1
6
"
4
sin ˆ 321 ‰ œ
1
3
œ ˆtan ˆ 14 ‰ 14 ‰ atan a0b 0b œ 1 asec2 x 2sec x tan x tan2 xbdx œ
'
1Î6
0
'
1Î8
1 Î8
sin 2x dx œ ’ "# cos 2x“
!
0
œ ˆ "# cos 2ˆ 18 ‰‰ ˆ "# cos 2a0b‰ œ
a2sec2 x 2sec x tan x 1bdx
1
6
2
2 È 2
4
1
#
1 3
1 3
œ Š4 tan ˆ 14 ‰ 1
ˆ 14 ‰ ‹
Š4 tan ˆ 13 ‰ 1
ˆ 13 ‰ ‹
œ (4(1) 4) Š4 ŠÈ3‹ 3‹ œ 4È3 3
19.
'"" (r 1)# dr œ '"" ar# 2r 1b dr œ ’ r3 r# r“ " œ Š (31)
20.
'È (t 1) at# 4b dt œ 'È at$ t# 4t 4b dt œ ’ t4 t3 2t# 4t“ÈÈ$
$
"
È3
È3
3
œ
%
ŠÈ3‹
4
$
ŠÈ3‹
2 ŠÈ3‹ 4È3 3
'È" Š u#
22.
'cc y y 2y dy œ 'cc ay2 2y2 b dy œ ’ y3
(
2
3
'
È2
1
$
(1)# (1)‹ Š 13 1# 1‹ œ 38
$
%
#
21.
1
$
3
%
23.
È3
4
'cÎ Î% ˆ4 sec# t t1 ‰ dt œ ' Î Î% a4 sec# t 1t# b dt œ 4 tan t 1t ‘ 11Î%Î$
1
18.
sin 2 ˆ 1# ‰‰ œ 14
1
4
œ [2 tan x 2sec x x]1!Î6 œ ˆ2 tanˆ 16 ‰ 2secˆ 16 ‰ ˆ 16 ‰‰ a2 tan 0 2sec 0 0b œ 2È3 17.
"
4
sin 2t‘ 1Î$
1 Î4
'
sin 2(0)‰ ˆ "# ˆ 1# ‰ 1Î$
"
4
asec# x 1bdx œ [tan x x]!
asec x tan xb2 dx œ
"
4
"
u& ‹
du œ
'È" Š u#
(
2
)
u
u& ‹ du œ ’ 16
1
5
3
3
3
s# È s
s#
ds œ
'
1
È2
ŠÈ3‹
4
$
ŠÈ3‹
3
$
#
2 ŠÈ3‹ 4 ŠÈ3‹ œ 10È3
)
"
"
4u% “È#
)
1
œ Š 16
c1
"
4(1)% ‹
ŠÈ2‹
16
"
œ Š ac31b ac21b ‹ Š ac33b ac23b ‹ œ
c3
3
2y1 “
ˆ1 s$Î# ‰ ds œ ’s È#
2
“
Ès
"
œ È 2 3
2
É È2 Š1 2
È1 ‹
%
4 ŠÈ2‹
œ4
3
22
3
œ È2 2$Î% 1
4
œ È2 È
81
24.
'
8
ˆx1Î3 1‰ˆ2 x2Î3 ‰
x1Î3
1
dx œ
'
8
2x1Î3 x 2 x2Î3
x1Î3
1
dx œ
'
1
8
ˆ2 x2Î3 2x1Î3 x1Î3 ‰ dx œ
2x 35 x5Î3 3x2Î3 34 x4Î3 ‘3 œ Š2a8b 35 a8b5Î3 3a8b2Î3 34 a8b4Î3 ‹ Š2a1b 35 a1b5Î3 3a1b2Î3 34 a1b4Î3 ‹
1
œ 137
20
25.
'
1
sin 2x
1Î2 2 sin x
dx œ
'
1
2 sin x cos x
1Î2 2 sin x
dx œ
'
1
1Î2
1
cos x dx œ ’sin x“
1Î2
œ asin a1bb ˆsin ˆ 12 ‰‰ œ 1
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
283
284
26.
Chapter 5 Integration
'
1Î3
0
'
œ
'
acos x sec xb2 dx œ
1Î3
0
ˆ 12 cos 2x 5
2
1Î3
0
'
acos2 x 2 sec2 xbdx œ
1Î3
0
ˆ cos 2x2 1 2 sec2 x‰dx
1Î3
sec2 x‰dx œ ’ 14 sin 2x 52 x tan x“
!
œ ˆ 14 sin 2ˆ 13 ‰ 52 ˆ 13 ‰ tanˆ 13 ‰‰ ˆ 14 sin 2a0b 52 a0b tana0b‰ œ
'c% kxk dx œ ' !% kxk dx '!
4
27.
28.
'
1
"
! #
acos x kcos xk b dx œ
1
#
œ sin
29. (a)
!
d
dx
30. (a)
'
(b)
d
dx
31. (a)
'
(b)
d
dt
32. (a)
'
'
1Î#
!
"
# (cos
' !% x dx '!
x cos x) dx 4
#
x dx œ ’ x# “
'
1
"
1Î# #
!
#
%
%
#
’ x# “ œ Š 0# !
(cos x cos x) dx œ
'
1Î#
!
(4)#
# ‹
Èx
cos t dt œ [sin t]! œ sin Èx sin 0 œ sin Èx Ê
Èx
'
Œ
!
sin x
d
dx
Œ
'
Èx
!
cos t dt œ
d ˆÈ ‰‰
cos t dt œ ˆcos Èx‰ ˆ dx
x œ ˆcos Èx ‰ ˆ "# x"Î# ‰ œ
3t# dt œ ct$ d "
sin x
1
'
Œ
t%
sin x
1
Œ'
t%
!
tan )
!
d
d)
œ sin$ x 1 Ê
d
dx
Œ
'
sin x
1
3t# dt œ
d
dx
d
dx
ˆsin Èx‰ œ cos Èx ˆ "# x"Î# ‰
cos Èx
2È x
asin$ x 1b œ 3 sin# x cos x
'
t%
!
t%
u"Î# du œ 23 u$Î# ‘ ! œ
2
3
at% b
$Î#
0œ
2 '
3 t
Ê
d
dt
Œ'
Œ
'
t%
!
Èu du œ
ˆ 23 t' ‰ œ 4t&
d
dt
Èu du œ Èt% ˆ dtd at% b‰ œ t# a4t$ b œ 4t&
)
sec# y dy œ [tan y]tan
œ tan (tan )) 0 œ tan (tan )) Ê
!
'
Œ
!
tan )
d
d)
tan )
!
sec# y dy œ
d
d)
(tan (tan )))
sec# y dy œ asec# (tan ))b ˆ dd) (tan ))‰ œ asec# (tan ))b sec# )
33. y œ
'
35. y œ
'È sin t# dt œ '
x
!
È1 t# dt Ê
'
x2
2
sin t3 dt Ê
'
2
x
34. y œ
sin t# dt Ê
œx†
dy
dx
'
x2
dy
dx
œ
d
dx Œ
2
'
1
x
"
t
dt Ê
dy
dx
œ
"
x
,x0
#
d ˆÈ ‰‰
œ Šsin ˆÈx‰ ‹ ˆ dx
x œ (sin x) ˆ "# x"Î# ‰ œ 2sinÈxx
sin t3 dt 1 †
'
x2
2
sin t3 dt œ x † sin ax# b
3 d
#
dx ax b
'
2
x2
sin t3 dt
sin t3 dt
x
#
#
dy
dx
œ È1 x#
x2
'c t t 4 dt '
1
Èx
!
x
œ 2x# sin x6 37. y œ
dy
dx
!
36. y œ x
œ 16
1Î#
œ asec# (tan ))b sec# )
(b)
0#
#‹
cos x dx œ [sin x]!
d
3t# dt œ a3 sin# xb ˆ dx
(sin x)‰ œ 3 sin# x cos x
Èu du œ
!
#
Š 4# cos Èx
2È x
œ
(b)
kxk dx œ 9È 3
8
sin 0 œ 1
Èx
'
4
51
6
3
t#
t# 4
dt Ê
x#
x# 4
x#
x# 4
œ0
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 5.4 The Fundamental Theorem of Calculus
'
38. y œ Œ
x
0
39. y œ
'
40. y œ
'
sin x
dt
È1 t#
!
tan x
dt
1 t#
!
, kxk Ê
'
3
10
at3 "b dt Ê
dy
dx
1
#
dy
dx
Ê
œ 3Œ
dy
dx
0
x
"
È1 sin# x
œ
'
10
d
at3 "b dt dx
Œ
d
ˆ dx
(sin x)‰ œ
0
x
10
10
'
at3 "b dt œ 3ax3 "b Œ
"
Ècos# x
(cos x) œ
cos x
kcos xk
œ
cos x
cos x
"
d
‰ ˆ dx
œ ˆ 1 tan
(tan x)‰ œ ˆ sec"# x ‰ asec# xb œ 1
#x
'$# ax# 2xbdx '#! ax# 2xbdx '!# ax# 2xbdx
$
œ ’ x3 x# “
#
$
$
’ x3 x# “
!
$
#
$
’ x3 x# “
#
!
$
œ ŠŠ (32) (2)# ‹ Š (33) (3)# ‹‹
$
ŠŠ 03 0# ‹ Š (32) (2)# ‹‹
$
$
$
ŠŠ 23 2# ‹ Š 03 0# ‹‹ œ
28
3
42. 3x# 3 œ 0 Ê x# œ 1 Ê x œ „ 1; because of symmetry about
the y-axis, Area œ 2 Œ
'!" a3x# 3bdx '"# a3x# 3bdx
"
#
2 Š cx$ 3xd ! cx$ 3xd " ‹ œ 2 c aa1$ 3(1)b a0$ 3(0)bb
aa2$ 3(2)b a1$ 3(1)bd œ 2(6) œ 12
43. x$ 3x# 2x œ 0 Ê x ax# 3x 2b œ 0
Ê x(x 2)(x 1) œ 0 Ê x œ 0, 1, or 2;
Area œ
'!" ax$ 3x# 2xbdx '"# ax$ 3x# 2xbdx
"
%
%
œ ’ x4 x$ x# “ ’ x4 x$ x# “
!
œ
%
Š 14
$
#
1 1 ‹
%
%
Š 04
$
#
"
#
0 0 ‹
%
’Š 24 2$ 2# ‹ Š 14 1$ 1# ‹“ œ
10
at3 "b dt
œ 1 since kxk 41. x# 2x œ 0 Ê x(x 2) œ 0 Ê x œ 0 or x œ 2; Area
œ
0
x
"
#
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
1
#
285
286
Chapter 5 Integration
44. x"Î$ x œ 0 Ê x"Î$ ˆ1 x#Î$ ‰ œ 0 Ê x"Î$ œ 0 or
1 x#Î$ œ 0 Ê x œ 0 or 1 œ x#Î$ Ê x œ 0 or
1 œ x# Ê x œ 0 or „ 1;
Area œ œ
'c! ˆx"Î$ x‰dx '!" ˆx"Î$ x‰dx '") ˆx"Î$ x‰dx
"
%Î$
’ 34
x
!
x#
# “ "
œ ’Š 34 (0)%Î$ ’ 34 x%Î$ 0#
#‹
"
x#
# “!
’ 43 x%Î$ (1)#
# ‹“
Š 34 (1)%Î$ ’Š 34 (1)%Î$ 1#
#‹
Š 34 (0)%Î$ 0#
# ‹“
’Š 34 (8)%Î$ 8#
#‹
Š 34 (1)%Î$ 1#
# ‹“
œ
"
4
"
4
ˆ2! $
4
#" ‰ œ
)
x#
# “"
83
4
45. The area of the rectangle bounded by the lines y œ 2, y œ 0, x œ 1, and x œ 0 is 21. The area under the curve
y œ 1 cos x on [0ß 1] is
'!
1
(1 cos x) dx œ [x sin x]!1 œ (1 sin 1) (0 sin 0) œ 1. Therefore the area of
the shaded region is 21 1 œ 1.
46. The area of the rectangle bounded by the lines x œ 16 , x œ
"
#
51
6 ,
y œ sin
ˆ 561 16 ‰ œ 13 . The area under the curve y œ sin x on 16 ß 561 ‘ is
œ ˆcos
51 ‰
6
È3
# ‹
ˆcos 16 ‰ œ Š
È3
#
'
1
6
œ
51Î6
1Î6
"
#
œ sin
51
6
, and y œ 0 is
&1Î'
sin x dx œ [cos x]1Î'
œ È3. Therefore the area of the shaded region is È3 13 .
47. On 14 ß 0‘ : The area of the rectangle bounded by the lines y œ È2, y œ 0, ) œ 0, and ) œ 14 is È2 ˆ 14 ‰
œ
1È2
4
. The area between the curve y œ sec ) tan ) and y œ 0 is 'c
!
1Î4
sec ) tan ) d) œ [sec )]!1Î%
œ (sec 0) ˆsec ˆ 14 ‰‰ œ È2 1. Therefore the area of the shaded region on 14 ß !‘ is
1È2
4
On 0ß 14 ‘ : The area of the rectangle bounded by ) œ 14 , ) œ 0, y œ È2, and y œ 0 is È2 ˆ 14 ‰ œ
under the curve y œ sec ) tan ) is
of the shaded region on !ß 14 ‘ is
È
'
1Î4
!
1È2
4
1Î%
sec ) tan ) d) œ [sec )]!
œ sec
1
4
Š È 2 1‹ .
1È2
4
. The area
sec 0 œ È2 1. Therefore the area
ŠÈ2 1‹ . Thus, the area of the total shaded region is
È
1È2
#
Š 1 4 2 È2 1‹ Š 1 4 2 È2 1‹ œ
.
48. The area of the rectangle bounded by the lines y œ 2, y œ 0, t œ 14 , and t œ 1 is 2 ˆ1 ˆ 14 ‰‰ œ 2 area under the curve y œ sec# t on 14 ß !‘ is
under the curve y œ 1 t# on [!ß "] is
'c
!
1Î4
"
$
œ
dt 3 œ 0 3 œ 3 Ê (d) is a solution to this problem.
dy
dx
œ
50. y œ
'c sec t dt 4
Ê
dy
dx
œ sec x and y(1) œ
51. y œ
'
sec t dt 4 Ê
dy
dx
œ sec x and y(0) œ
x
1
!
x
and y(1) œ
'
"
t
dt 3 Ê
"
x
1
1
Thus, the total
. Therefore the area of the shaded region is ˆ2 1# ‰ '
"
1 t
$
5
3
49. y œ
x
$
!
2
3
'
!
. The
!
ˆ 1‰
sec# t dt œ [tan t]
1Î% œ tan 0 tan 4 œ 1. The area
'! a1 t# b dt œ ’t t3 “ " œ Š1 13 ‹ Š0 03 ‹ œ 32 .
area under the curves on 14 ß "‘ is 1 1
#
'cc sec t dt 4 œ 0 4 œ 4
1
1
!
5
3
œ
"
3
1
#
.
Ê (c) is a solution to this problem.
sec t dt 4 œ 0 4 œ 4 Ê (b) is a solution to this problem.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 5.4 The Fundamental Theorem of Calculus
52. y œ
'
53. y œ
'
x
"
" t
x
#
55. Area œ
dt 3 Ê
dy
dx
œ
"
x
and y(1) œ
'
"
"
t
"
dt 3 œ 0 3 œ 3 Ê (a) is a solution to this problem.
sec t dt 3
'c ÎÎ
b 2
b 2
54. y œ
$
4h ˆ b# ‰
3b#
œ ˆ bh
# ˆ bh
# bh ‰
6
Œh ˆ # ‰ b
bh ‰
6
œ bh x
"
È1 t# dt 2
bÎ2
4h ˆ #b ‰
3b#
œ
bh
3
'
4hx$
3b# “ bÎ2
ˆh ˆ 4h
‰ # ‰ dx œ ’hx b# x
œ Œhˆ #b ‰ $
2
3
bh
56. k 0 Ê one arch of y œ sin kx will occur over the interval 0ß 1k ‘ Ê the area œ
œ "k cos ˆk ˆ 1k ‰‰ ˆ k" cos (0)‰ œ
57.
dc
dx
œ
58. r œ
"
#È x
'
$
!
œ
Š2 "
#
x"Î# Ê c œ
'
dx œ 2
'
2
(x 1)# ‹
x
!
$
!
'
1Îk
!
sin kx dx œ "
k
œ t"Î# ‘ 0 œ Èx; c(100) c(1) œ È100 È1 œ $9.00
x
"
(x 1)# ‹
$
dx œ 2 x ˆ x11 ‰‘ ! œ 2 ’Š3 "
(3 1) ‹
Š0 "
(0 1) ‹“
œ 2 3 "4 1‘ œ 2 ˆ2 4" ‰ œ 4.5 or $4500
59. (a) t œ 0 Ê T œ 85 3È25 0 œ 70‰ F; t œ 16 Ê T œ 85 3È25 16 œ 76‰ F;
t œ 25 Ê T œ 85 3È25 25 œ 85‰ F
(b) average temperatuve œ
œ
1
25 Š85a25b
1
25 0
2a25 25b
' 25 Š85 3È25 t‹ dt œ 251 ’85t 2a25 tb3Î2 “ 25
!
0
3 Î2
‹
1
25 Š85a0b
2a25 0b
3Î2
‰
‹ œ 75 F
3
60. (a) t œ 0 Ê H œ È0 1 5a0b1Î3 œ 1 ft; t œ 4 Ê H œ È4 1 5a4b1Î3 œ È5 5È
4 ¸ 10.17 ft;
1
Î
3
t œ 8 Ê H œ È8 1 5a8b œ 13 ft
(b) average height œ
œ
61.
'
62.
'
x
1
!
x
1 2
8 Š 3 a8
1b
3 Î2
1
80
' 8 ŠÈt 1 5 t1Î3 ‹ dt œ 18 ’ 23 at 1b3Î2 154 t4Î3 “ 8
15
4
!
0
a8b
4Î3
‹
f(t) dt œ x# 2x 1 Ê f(x) œ
f(t) dt œ x cos 1x Ê f(x) œ
63. f(x) œ 2 '#
x
"
9
1t
d
dx
1 2
8 Š 3 a0
'
d
dx
'
!
1
x
x
1b
3Î2
f(t) dt œ
d
dx
4Î3
15
‹
4 a0b
œ
29
3
¸ 9.67 ft
ax# 2x 1b œ 2x 2
f(t) dt œ cos 1x 1x sin 1x Ê f(4) œ cos 1(4) 1(4) sin 1(4) œ 1
dt Ê f w (x) œ 1 (x9 1) œ
9
x 2
L(x) œ 3(x 1) f(1) œ 3(x 1) 2 œ 3x 5
Ê f w (1) œ 3; f(1) œ 2 '#" " 1 9 t dt œ 2 0 œ 2;
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
1 Îk
cos kx‘ !
2
k
" "Î#
dt
# t
Š1 287
288
Chapter 5 Integration
64. g(x) œ 3 '
1
x#
sec (t 1) dt Ê gw (x) œ asec ax# 1bb (2x) œ 2x sec ax# 1b Ê gw (1) œ 2(1) sec a(1)# 1b
#
a"b
"
œ 2; g(1) œ 3 '
sec (t 1) dt œ 3 ' sec (t 1) dt œ 3 0 œ 3; L(x) œ 2(x (1)) g(1)
1
1
œ 2(x 1) 3 œ 2x 1
65. (a)
(b)
(c)
(d)
(e)
(f)
(g)
True: since f is continuous, g is differentiable by Part 1 of the Fundamental Theorem of Calculus.
True: g is continuous because it is differentiable.
True, since gw (1) œ f(1) œ 0.
False, since gww (1) œ f w (1) 0.
True, since gw (1) œ 0 and gww (1) œ f w (1) 0.
False: gww (x) œ f w (x) 0, so gww never changes sign.
True, since gw (1) œ f(1) œ 0 and gw (x) œ f(x) is an increasing function of x (because f w (x) 0).
66. Let a œ x0 x1 x2 â xn œ b be any partition of Òa, bÓ and let F be any antiderivative of f.
n
(a) !Faxi b Faxi1 b‘
iœ1
œ Fax1 b Fax0 b‘ Fax2 b Fax1 b‘ Fax3 b Fax2 b‘ â Faxn1 b Faxn2 b‘ Faxn b Faxn1 b‘
œ Fax0 b Fax1 b Fax1 b Fax2 b Fax2 b â Faxn1 b Faxn1 b Faxn b œ Faxn b Fax0 b œ Fabb Faab
(b) Since F is any antiderivative of f on Òa, bÓ Ê F is differentiable on Òa, bÓ Ê F is continuous on Òa, bÓ. Consider any
subinterval Òxi1 , xi Ó in Òa, bÓ, then by the Mean Value Theorem there is at least one number ci in Ðxi1 , xi Ñ such that
n
Faxi b Faxi1 b‘ œ Fw aci baxi xi1 b œ faci baxi xi1 b œ faci b?xi . Thus Fabb Faab œ !Faxi b Faxi1 b‘
iœ1
n
œ !faci b?xi .
iœ1
n
(c) Taking the limit of Fabb Faab œ !faci b?xi we obtain
Ê Fabb Faab œ 'a faxb dx
iœ1
lim aFabb Faabb œ
mPmÄ0
n
lim Œ!faci b?xi mPmÄ0
iœ1
b
67-70. Example CAS commands:
Maple:
with( plots );
f := x -> x^3-4*x^2+3*x;
a := 0;
b := 4;
F := unapply( int(f(t),t=a..x), x );
# (a)
p1 := plot( [f(x),F(x)], x=a..b, legend=["y = f(x)","y = F(x)"], title="#67(a) (Section 5.4)" ):
p1;
dF := D(F);
# (b)
q1 := solve( dF(x)=0, x );
pts1 := [ seq( [x,f(x)], x=remove(has,evalf([q1]),I) ) ];
p2 := plot( pts1, style=point, color=blue, symbolsize=18, symbol=diamond, legend="(x,f(x)) where F '(x)=0" ):
display( [p1,p2], title="81(b) (Section 5.4)" );
incr := solve( dF(x)>0, x );
# (c)
decr := solve( dF(x)<0, x );
df := D(f);
# (d)
p3 := plot( [df(x),F(x)], x=a..b, legend=["y = f '(x)","y = F(x)"], title="#67(d) (Section 5.4)" ):
p3;
q2 := solve( df(x)=0, x );
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 5.4 The Fundamental Theorem of Calculus
pts2 := [ seq( [x,F(x)], x=remove(has,evalf([q2]),I) ) ];
p4 := plot( pts2, style=point, color=blue, symbolsize=18, symbol=diamond, legend="(x,f(x)) where f '(x)=0" ):
display( [p3,p4], title="81(d) (Section 5.4)" );
71-74. Example CAS commands:
Maple:
a := 1;
u := x -> x^2;
f := x -> sqrt(1-x^2);
F := unapply( int( f(t), t=a..u(x) ), x );
dF := D(F);
# (b)
cp := solve( dF(x)=0, x );
solve( dF(x)>0, x );
solve( dF(x)<0, x );
d2F := D(dF);
# (c)
solve( d2F(x)=0, x );
plot( F(x), x=-1..1, title="#71(d) (Section 5.4)" );
75.
Example CAS commands:
Maple:
f := `f`;
q1 := Diff( Int( f(t), t=a..u(x) ), x );
d1 := value( q1 );
76.
Example CAS commands:
Maple:
f := `f`;
q2 := Diff( Int( f(t), t=a..u(x) ), x,x );
value( q2 );
67-76. Example CAS commands:
Mathematica: (assigned function and values for a, and b may vary)
For transcendental functions the FindRoot is needed instead of the Solve command.
The Map command executes FindRoot over a set of initial guesses
Initial guesses will vary as the functions vary.
Clear[x, f, F]
{a, b}= {0, 21}; f[x_] = Sin[2x] Cos[x/3]
F[x_] = Integrate[f[t], {t, a, x}]
Plot[{f[x], F[x]},{x, a, b}]
x/.Map[FindRoot[F'[x]==0, {x, #}] &,{2, 3, 5, 6}]
x/.Map[FindRoot[f'[x]==0, {x, #}] &,{1, 2, 4, 5, 6}]
Slightly alter above commands for 75 - 80.
Clear[x, f, F, u]
a=0; f[x_] = x2 2x 3
u[x_] = 1 x2
F[x_] = Integrate[f[t], {t, a, u(x)}]
x/.Map[FindRoot[F'[x]==0,{x, #}] &,{1, 2, 3, 4}]
x/.Map[FindRoot[F''[x]==0,{x,#}] &,{1, 2, 3, 4}]
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
289
290
Chapter 5 Integration
After determining an appropriate value for b, the following can be entered
b = 4;
Plot[{F[x], {x, a, b}]
5.5 INDEFINTE INTEGRALS AND THE SUBSTITUTION RULE
"
2
1. Let u œ 2x 4 Ê du œ 2 dx Ê
' 2a 2x 4b dx œ '
5
2u5
"
2
du œ ' u5 du œ
"
7
2. Let u œ 7x 1 Ê du œ 7 dx Ê
' 7È7x 1 dx œ ' 7a7x 1b
' 2xa x2 5b
4x3
dx
ax 4 1 b 2
u6 C œ
1
2È x
dx œ ' ˆ1 Èx‰
"
4
"
3
sin 3x dx œ '
"
3
x sin a2x b dx œ '
sec 2t tan 2t dt œ '
"
4
C
"
2
'
"
10
u4 du œ
1
Èx
"
2
1
x4 1
C
du œ a3x 2b dx
u5 C œ
"
10
a 3x2 4xb
5
C
dx
dx œ ' u1Î3 2 du œ 2' u1Î3 du œ 2 †
3
4
u4Î3 C œ
3
2
ˆ1 Èx‰4Î3 C
du œ dx
"
4
"
#
du œ x dx
sin u du œ 4" cos u C œ 4" cos 2x# C
"
#
du œ dt
sec u tan u du œ
10. Let u œ 1 cos 2t Ê du œ
ˆ1 cos
a7x 1b3Î2 C
sin u du œ 3" cos u C œ 3" cos 3x C
#
t ‰#
#
2
3
du œ x3 dx
dx Ê 2 du œ
1 Î3 1
Èx
9. Let u œ 2t Ê du œ 2 dt Ê
'
u3Î2 C œ
3
u4 "2 du œ
8. Let u œ 2x# Ê du œ 4x dx Ê
'
2
3
2
7. Let u œ 3x Ê du œ 3 dx Ê
'
C
œ ' 4x3 ax4 1b dx œ ' 4 u2 4" du œ ' u2 du œ u1 C œ
6. Let u œ 1 Èx Ê du œ
'
6
dx œ ' 2 u4 2" du œ ' u4 du œ 3" u3 C œ 3" a x2 5b
' a3x 2ba3 x2 4xb4 dx œ '
' ˆ1ÈÈxx‰
a 2x 4b
du œ x dx
5. Let u œ 3x2 4x Ê du œ a6x 4bdx œ 2a3x 2b dx Ê
1Î3
"
6
du œ dx
"
2
4. Let u œ x4 1 Ê du œ 4x3 dx Ê
'
"
6
dx œ ' 7u1Î2 "7 du œ ' u1Î2 du œ
1 Î2
3. Let u œ x2 5 Ê du œ 2x dx Ê
4
du œ dx
"
#
sin
t
#
"
#
sec u C œ
dt Ê 2 du œ sin
ˆsin #t ‰ dt œ ' 2u# du œ
2
3
u$ C œ
t
2
2
3
"
#
sec 2t C
dt
ˆ1 cos #t ‰$ C
11. Let u œ 1 r$ Ê du œ 3r# dr Ê 3 du œ 9r# dr
' È9r
#
dr
1 r$
œ ' 3u"Î# du œ 3(2)u"Î# C œ 6 a1 r$ b
"Î#
C
12. Let u œ y% 4y# 1 Ê du œ a4y$ 8yb dy Ê 3 du œ 12 ay$ 2yb dy
'
12 ay% 4y# 1b ay$ 2yb dy œ ' 3u# du œ u$ C œ ay% 4y# 1b C
#
$
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 5.5 Indefinite Integrals and the Substitution Rule
13. Let u œ x$Î# 1 Ê du œ
'
x"Î# dx Ê
Èx sin# ˆx$Î# 1‰ dx œ '
14. Let u œ "x Ê du œ
'
3
#
"
x#
2
3
2
3
du œ Èx dx
sin# u du œ
2
3
ˆ #u "
4
sin 2u‰ C œ
"
3
ˆx$Î# 1‰ "
6
sin ˆ2x$Î# 2‰ C
dx
cos# ˆ x" ‰ dx œ ' cos# aub du œ
"
œ 2x
4" sin ˆ x2 ‰ C
"
x#
'
cos# aub du œ ˆ u# "
4
"
sin 2u‰ C œ 2x
"
4
sin ˆ x2 ‰ C
15. (a) Let u œ cot 2) Ê du œ 2 csc# 2) d) Ê "# du œ csc# 2) d)
'
csc# 2) cot 2) d) œ '
"
#
#
#
u du œ "# Š u# ‹ C œ u4 C œ 4" cot# 2) C
(b) Let u œ csc 2) Ê du œ 2 csc 2) cot 2) d) Ê "# du œ csc 2) cot 2) d)
'
csc# 2) cot 2) d) œ ' "# u du œ "# Š u# ‹ C œ u4 C œ 4" csc# 2) C
#
16. (a) Let u œ 5x 8 Ê du œ 5 dx Ê
'
dx
È5x 8
œ'
"
5
Š È"u ‹ du œ
"
#
(b) Let u œ È5x 8 Ê du œ
'
dx
È5x 8
œ'
2
5
du œ
2
5
"
5
'
"
5
#
du œ dx
u"Î# du œ
"
5
ˆ2u"Î# ‰ C œ
(5x 8)"Î# (5) dx Ê
uCœ
2
5
2
5
du œ
2
5
u"Î# C œ
2
5
È5x 8 C
dx
È5x8
È5x 8 C
17. Let u œ 3 2s Ê du œ 2 ds Ê "# du œ ds
'
È3 2s ds œ ' Èu ˆ " du‰ œ " ' u"Î# du œ ˆ " ‰ ˆ 2 u$Î# ‰ C œ " (3 2s)$Î# C
#
#
#
3
3
18. Let u œ 5s 4 Ê du œ 5 ds Ê
'
"
È5s 4
ds œ '
"
Èu
ˆ 5" du‰ œ
"
5
"
5
'
du œ ds
u"Î# du œ ˆ 5" ‰ ˆ2u"Î# ‰ C œ
2
5
È5s 4 C
19. Let u œ 1 )# Ê du œ 2) d) Ê "# du œ ) d)
'
&Î%
4
4
)È
1 ) # d) œ ' È
u ˆ "# du‰ œ "# ' u"Î% du œ ˆ "# ‰ ˆ 45 u&Î% ‰ C œ 25 a1 )# b C
20. Let u œ 7 3y# Ê du œ 6y dy Ê "# du œ 3y dy
'
$Î#
3yÈ7 3y# dy œ ' Èu ˆ "# du‰ œ "# ' u"Î# du œ ˆ "# ‰ ˆ 23 u$Î# ‰ C œ 3" a7 3y# b C
21. Let u œ 1 Èx Ê du œ
'
"
È x ˆ" È x ‰ #
dx œ '
2 du
u#
"
2È x
dx Ê 2 du œ
œ 2u C œ
22. Let u œ 3z 4 Ê du œ 3 dz Ê
'
"
3
2
1 È x
dx
C
du œ dz
cos (3z 4) dz œ ' (cos u) ˆ "3 du‰ œ
23. Let u œ 3x 2 Ê du œ 3 dx Ê
'
"
3
"
Èx
"
3
' cos u du œ 3" sin u C œ 3" sin (3z 4) C
du œ dx
sec# (3x 2) dx œ ' asec# ub ˆ "3 du‰ œ
"
3
'
sec# u du œ
"
3
tan u C œ
"
3
tan (3x 2) C
24. Let u œ tan x Ê du œ sec# x dx
'
tan# x sec# x dx œ ' u# du œ
"
3
u$ C œ
"
3
tan$ x C
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
291
292
Chapter 5 Integration
25. Let u œ sin ˆ x3 ‰ Ê du œ
'
r$
18
1 Ê du œ
r#
6
$
sec# ˆ x# ‰ dx Ê 2 du œ sec# ˆ x# ‰ dx
"
4
tan) ˆ x# ‰ C
dr Ê 6 du œ r# dr
r % Š7 r&
10
$
r&
10 ‹
'
'
$
Ê du œ "# r% dr Ê 2 du œ r% dr
dr œ ' u$ (2 du) œ 2 ' u$ du œ 2 Š u4 ‹ C œ "# Š7 %
29. Let u œ x$Î# 1 Ê du œ
'
sin' ˆ x3 ‰ C
r
r
r# Š 18
1‹ dr œ ' u& (6 du) œ 6 ' u& du œ 6 Š u6 ‹ C œ Š 18
1‹ C
&
28. Let u œ 7 '
"
#
"
#
tan( ˆ x# ‰ sec# ˆ x# ‰ dx œ ' u( (2 du) œ 2 ˆ 8" u) ‰ C œ
27. Let u œ
'
cos ˆ x3 ‰ dx Ê 3 du œ cos ˆ x3 ‰ dx
sin& ˆ x3 ‰ cos ˆ x3 ‰ dx œ ' u& (3 du) œ 3 ˆ 6" u' ‰ C œ
26. Let u œ tan ˆ x# ‰ Ê du œ
'
"
3
3
#
x"Î# dx Ê
2
3
r&
10 ‹
%
C
du œ x"Î# dx
x"Î# sin ˆx$Î# 1‰ dx œ ' (sin u) ˆ 23 du‰ œ
2
3
'
sin u du œ
2
3
(cos u) C œ 23 cos ˆx$Î# 1‰ C
30. Let u œ csc ˆ v # 1 ‰ Ê du œ "# csc ˆ v # 1 ‰ cot ˆ v # 1 ‰ dv Ê 2 du œ csc ˆ v # 1 ‰ cot ˆ v # 1 ‰ dv
'
csc ˆ v # 1 ‰ cot ˆ v # 1 ‰ dv œ ' 2 du œ 2u C œ 2 csc ˆ v # 1 ‰ C
31. Let u œ cos (2t 1) Ê du œ 2 sin (2t 1) dt Ê "# du œ sin (2t 1) dt
'
sin (2t 1)
cos# (2t 1)
dt œ ' #"
du
u#
œ
"
#u
Cœ
"
# cos (2t 1)
C
32. Let u œ sec z Ê du œ sec z tan z dz
'
sec z tan z
Èsec z
33. Let u œ
'
"
t#
"
t
dz œ '
"
Èu
du œ ' u"Î# du œ 2u"Î# C œ 2Èsec z C
1 œ t" 1 Ê du œ t# dt Ê du œ
"
Èt
"
)#
" "Î#
# t
dt Ê 2 du œ
sin
"
)
"
)
cos
Ê du œ ˆcos ") ‰ ˆ )"# ‰ d) Ê du œ
"
)
cos È)
È) sin# È)
d) œ '
"
È)
dt
t$ a1 t% b dt œ ' u$ ˆ "4 du‰ œ
$
cos
"
‹
#È )
"
)
"
)
d)
C
d) Ê 2 du œ
"
È)
cot È) csc È) d)
cot È) csc È) d) œ ' 2 du œ 2u C œ 2 csc È) C œ 37. Let u œ 1 t% Ê du œ 4t$ dt Ê
'
"
)#
d) œ ' u du œ #" u# C œ #" sin#
36. Let u œ csc È) Ê du œ Šcsc È) cot È)‹ Š
'
"
Èt
cos ˆÈt 3‰ dt œ ' (cos u)(2 du) œ 2 ' cos u du œ 2 sin u C œ 2 sin ˆÈt 3‰ C
35. Let u œ sin
'
dt
cos ˆ "t 1‰ dt œ ' (cos u)(du) œ ' cos u du œ sin u C œ sin ˆ "t 1‰ C
34. Let u œ Èt 3 œ t"Î# 3 Ê du œ
'
"
t#
"
4
"
4
2
sin È)
du œ t$ dt
ˆ 4" u% ‰ C œ
"
16
%
a 1 t% b C
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
C
Section 5.5 Indefinite Integrals and the Substitution Rule
38. Let u œ 1 '
"
É2
x#
"
x
"
x
40. Let u œ 1 "
x#
Ê du œ
É x x& 1 dx œ '
39. Let u œ 2 '
"
x
"
x#
dx
É x x 1 dx œ '
"
x#
Ê du œ
Ê du œ
2
x3
#
'É
3
x11
x3
#
3
3
x3
Ê du œ
dx œ '
1É
x4
9
x4
3
x3
x3
"
#
dx Ê
#
41. Let u œ 1 É1 4
3
x2
È x3 1
du œ
"
9
dx Ê
du œ
dx œ '
dx œ '
1
É1
x4
"
3
u$Î# C œ
2
3
1
x3
2
3
u$Î# C œ
2
3
ˆ1 "x ‰$Î# C
2
3
ˆ2 "x ‰$Î# C
dx
dx œ ' Èu
42. Let u œ x3 1 Ê du œ 3x2 dx Ê
' É x x 1 dx œ '
dx œ ' Èu du œ ' u"Î# du œ
"
x
dx
' x1 É x x 1 dx œ ' x1 É1 x"
3
"
x#
dx œ ' Èu du œ ' u"Î# du œ
"
x#
293
"
#
1
x4
du œ
"
#
' u"Î# du œ 13 u$Î# C œ 13 ˆ1 x" ‰$Î# C
#
dx
3
x3
dx œ ' Èu
"
9
du œ
"
9
' u"Î# du œ 272 u$Î# C œ 272 ˆ1 x3 ‰$Î# C
3
du œ x2 dx
1 "
Èu 3
du œ
"
3
' u"Î# du œ 23 u1Î# C œ 23 ax3 1b$Î# C
43. Let u œ x ". Then du œ dx and x œ u ". Thus ' xax "b10 dx œ ' au "bu10 du œ ' au11 u10 b du
œ
1 12
12 u
1 11
11 u
Cœ
1
12 ax
"b12 1
11 ax
"b11 C
44. Let u œ 4 x. Then du œ 1 dx and a1b du œ dx and x œ 4 u. Thus ' x È4 xdx œ ' a4 ubÈu a1bdu
œ ' a4 ubˆu1Î2 ‰du œ ' au3Î2 4u1Î2 b du œ 25 u5Î2 83 u3Î2 C œ 25 a4 xb5Î2 83 a4 xb3Î2 C
45. Let u œ " x. Then du œ 1 dx and a1b du œ dx and x œ 1 u. Thus ' ax 1b2 a" xb5 dx
œ ' a2 ub2 u5 a1b du œ ' au7 4u6 4u5 b du œ 18 u8 47 u7 23 u6 C
œ 18 a" xb8 47 a" xb7 23 a" xb6 C
46. Let u œ x 5. Then du œ dx and x œ u 5. Thus ' ax 5bax 5b1Î3 dx œ ' au 10bu1Î3 du œ ' ˆu4Î3 10u1Î3 ‰ du
œ 37 u7Î3 15 4Î3
2 u
C œ 37 ax 5b7Î3 15
2 ax
5b4Î3 C
47. Let u œ x# ". Then du œ #x dx and "# du œ x dx and x# œ u ". Thus ' x$ Èx# " dx œ ' au "b "# Èu du
œ
"
#
' au$Î# u"Î# bdu œ "# ’ #& u&Î# #$ u$Î# “ C œ "& u&Î# "$ u$Î# C œ "& ax# "b&Î# "$ ax# "b$Î# C
48. Let u œ x3 " Ê du œ 3x# dx and x3 œ u ". So ' 3B& Èx3 " dx œ ' au "bÈu du œ ' au$Î# u"Î# bdu
œ #& u&Î# #$ u$Î# C œ #& ax$ "b
49. Let u œ x2 4 Ê du œ 2x dx and
œ "4 u2 C œ 4" ax2 4b
2
&Î#
"
#
#$ ax$ "b
$Î#
C
du œ x dx . Thus '
x
ax 2 4 b 3
dx œ ' ax2 4b x dx œ ' u3 "# du œ
3
"
#
' u3 du
C
50. Let u œ x 4 Ê du œ dx and x œ u 4 . Thus '
x
ax 4 b 3
dx œ ' ax 4b3 x dx œ ' u3 au 4bdu œ ' au2 4u3 b du
œ u1 2u2 C œ ax 4b1 2ax 4b2 C
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
294
Chapter 5 Integration
51. (a) Let u œ tan x Ê du œ sec# x dx; v œ u$ Ê dv œ 3u# du Ê 6 dv œ 18u# du; w œ 2 v Ê dw œ dv
'
18 tan# x sec# x
dx œ
a2 tan$ xb#
œ 2 6 u$ C
$
'
'
18u#
du œ
a2 u $ b#
6
2 tan
$x C
#
#
œ
6 dv
(2 v)#
œ'
6 dw
w#
œ 6 ' w# dw œ 6w" C œ # 6 v C
(b) Let u œ tan x Ê du œ 3 tan x sec x dx Ê 6 du œ 18 tan# x sec# x dx; v œ 2 u Ê dv œ du
'
dx œ '
18 tan# x sec# x
a2 tan$ xb#
œ'
6 du
(2 u)#
6 dv
v#
6
œ v6 C œ 2 6 u C œ # tan
$x C
(c) Let u œ 2 tan$ x Ê du œ 3 tan# x sec# x dx Ê 6 du œ 18 tan# x sec# x dx
'
dx œ '
18 tan# x sec# x
a2 tan$ xb#
6 du
u#
6
œ u6 C œ 2 tan
$x C
52. (a) Let u œ x 1 Ê du œ dx; v œ sin u Ê dv œ cos u du; w œ 1 v# Ê dw œ 2v dv Ê
'
"
#
Èw dw œ
"
3
w$Î# C œ
"
3
a 1 v# b
$Î#
Cœ
"
3
a1 sin# ub
$Î#
Cœ
"
#
#
(b) Let u œ sin (x 1) Ê du œ cos (x 1) dx; v œ 1 u Ê dv œ 2u du Ê
'
È1 sin# (x 1) sin (x 1) cos (x 1) dx œ ' u È1 u# du œ '
œ ˆ "# ˆ 23 ‰ v$Î# ‰ C œ
"
3
v$Î# C œ
"
3
a1 u # b
$Î#
Cœ
"
3
(c) Let u œ 1 sin (x 1) Ê du œ 2 sin (x 1) cos (x 1) dx Ê
'
È1 sin# (x 1) sin (x 1) cos (x 1) dx œ '
œ
"
3
a1 sin# (x 1)b
$Î#
œ
"
6
dr œ ' Š
cos Èu
ˆ"
È u ‹ 1#
sin È)
É) cos$ È)
d) œ
4
Écos È)
'
sin È)
È) Écos$ È)
Èv dv œ '
$Î#
"
#
"
1#
du œ sin (x 1) cos (x 1) dx
"
#
u"Î# du œ
du‰ œ ' (cos v) ˆ 6" dv‰ œ
"
‹
#È )
d) œ '
d) Ê 2 du œ
2 du
u$Î#
$
"
#
ˆ 23 u$Î# ‰ C
"
6
sin v C œ
"
#È u
"
6
du Ê
sin Èu C
sin È)
È)
d)
œ 2 ' u$Î# du œ 2 ˆ2u"Î# ‰ C œ
4
Èu
C
"
#
u% C œ
"
#
%
a3t# 1b C;
(3 1)% C Ê 3 œ 8 C Ê C œ 5 Ê s œ
"
#
%
a3t# 1b 5
56. Let u œ x# 8 Ê du œ 2x dx Ê 2 du œ 4x dx
"Î$
dx œ ' u"Î$ (2 du) œ 2 ˆ 3# u#Î$ ‰ C œ 3u#Î$ C œ 3 ax# 8b
y œ 0 when x œ 0 Ê 0 œ 3(8)
57. Let u œ t 1
1#
#Î$
#
C Ê C œ 12 Ê y œ 3 ax 8b
#Î$
#Î$
C;
12
Ê du œ dt
s œ ' 8 sin# ˆt "
1 #È u
C
"
#
du œ (2r 1) dr; v œ Èu Ê dv œ
s œ ' 12t a3t# 1b dt œ ' u$ (2 du) œ 2 ˆ "4 u% ‰ C œ
y œ ' 4x ax# 8b
dv œ
v"Î# dv
55. Let u œ 3t# 1 Ê du œ 6t dt Ê 2 du œ 12t dt
"
#
C
dv œ u du
C
s œ 3 when t œ 1 Ê 3 œ
$Î#
a1 sin# (x 1)b
sin È3(2r 1)# 6 C
54. Let u œ cos È) Ê du œ Šsin È)‹ Š
œ
Èu du œ '
"
3
C
53. Let u œ 3(2r 1)# 6 Ê du œ 6(2r 1)(2) dr Ê
(2r 1) cos È3(2r 1)# 6
È3(2r 1)# 6
"
#
"
#
a1 sin# (x 1)b
#
'
dw œ v dv
È1 sin# (x 1) sin (x 1) cos (x 1) dx œ ' È1 sin# u sin u cos u du œ ' vÈ1 v# dv
œ'
'
"
#
dt œ ' 8 sin# u du œ 8 ˆ u# "4 sin 2u‰ C œ 4 ˆt 11# ‰ 2 sin ˆ2t 16 ‰ C;
s œ 8 when t œ 0 Ê 8 œ 4 ˆ 11# ‰ 2 sin ˆ 16 ‰ C Ê C œ 8 13 1 œ 9 13
Ê s œ 4ˆt 11# ‰ 2 sin ˆ2t 16 ‰ 9 13 œ 4t 2 sin ˆ2t 16 ‰ 9
1‰
1#
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
"
6
du
Section 5.5 Indefinite Integrals and the Substitution Rule
58. Let u œ
1
4
) Ê du œ d)
r œ ' 3 cos# ˆ 14 )‰ d) œ ' 3 cos# u du œ 3 ˆ u# sin 2u‰ C œ 3# ˆ 14 )‰ 34 sin ˆ 1# 2)‰ C;
C Ê C œ 1# 43 Ê r œ 3# ˆ 14 )‰ 43 sin ˆ 1# 2)‰ 1# when ) œ 0 Ê 18 œ 381 43 sin 1#
Ê r œ 3# ) 34 sin ˆ 1# 2)‰ 18 43 Ê r œ
rœ
1
8
1
#
59. Let u œ 2t ds
dt
295
3
2
)
3
4
"
4
cos 2) 1
8
3
4
3
4
Ê du œ 2 dt Ê 2 du œ 4 dt
œ ' 4 sin ˆ2t 1# ‰ dt œ ' (sin u)(2 du) œ 2 cos u C" œ 2 cos ˆ2t 1# ‰ C" ;
at t œ 0 and
œ 100 we have 100 œ 2 cos ˆ 1# ‰ C" Ê C" œ 100 Ê
ds
dt
œ 2 cos ˆ2t 1# ‰ 100
ds
dt
Ê s œ ' ˆ2 cos ˆ2t 1# ‰ 100‰ dt œ ' (cos u 50) du œ sin u 50u C# œ sin ˆ2t 1# ‰ 50 ˆ2t 1# ‰ C# ;
at t œ 0 and s œ 0 we have 0 œ sin ˆ 1# ‰ 50 ˆ 1# ‰ C# Ê C# œ 1 251
Ê s œ sin ˆ2t 1# ‰ 100t 251 (1 251) Ê s œ sin ˆ2t 1# ‰ 100t 1
60. Let u œ tan 2x Ê du œ 2 sec# 2x dx Ê 2 du œ 4 sec# 2x dx; v œ 2x Ê dv œ 2 dx Ê
dy
dx
œ ' 4 sec# 2x tan 2x dx œ ' u(2 du) œ u# C" œ tan# 2x C" ;
at x œ 0 and
dy
dx
œ 4 we have 4 œ 0 C" Ê C" œ 4 Ê
Ê y œ ' asec# 2x 3b dx œ ' asec# v 3b ˆ "# dv‰ œ
at x œ 0 and y œ 1 we have 1 œ
"
#
dy
dx
"
#
"
#
dv œ dx
œ tan# 2x 4 œ asec# 2x 1b 4 œ sec# 2x 3
tan v 3# v C# œ
"
#
(0) 0 C# Ê C# œ 1 Ê y œ
"
#
tan 2x 3x C# ;
tan 2x 3x 1
61. Let u œ 2t Ê du œ 2 dt Ê 3 du œ 6 dt
s œ ' 6 sin 2t dt œ ' (sin u)(3 du) œ 3 cos u C œ 3 cos 2t C;
at t œ 0 and s œ 0 we have 0 œ 3 cos 0 C Ê C œ 3 Ê s œ 3 3 cos 2t Ê s ˆ 1# ‰ œ 3 3 cos (1) œ 6 m
62. Let u œ 1t Ê du œ 1 dt Ê 1 du œ 1# dt
v œ ' 1# cos 1t dt œ ' (cos u)(1 du) œ 1 sin u C" œ 1 sin (1t) C" ;
at t œ 0 and v œ 8 we have 8 œ 1(0) C" Ê C" œ 8 Ê v œ
ds
dt
œ 1 sin (1t) 8 Ê s œ ' (1 sin (1t) 8) dt
œ ' sin u du 8t C# œ cos (1t) 8t C# ; at t œ 0 and s œ 0 we have 0 œ 1 C# Ê C# œ 1
Ê s œ 8t cos (1t) 1 Ê s(1) œ 8 cos 1 1 œ 10 m
63. All three integrations are correct. In each case, the derivative of the function on the right is the integrand on
the left, and each formula has an arbitrary constant for generating the remaining antiderivatives. Moreover,
sin# x C" œ 1 cos# x C" Ê C# œ 1 C" ; also cos# x C# œ cos#2x "# C# Ê C$ œ C# "# œ C" "# .
64. (a) Š
"
60
"
‹
0
'
0
1Î60
"Î'!
Vmax sin 1201t dt œ 60 Vmax ˆ 120" 1 ‰ cos (1201t)‘ !
œ V#1max [1 1] œ 0
(b) Vmax œ È2 Vrms œ È2 (240) ¸ 339 volts
(c)
'
0
œ
1Î60
aVmax b# sin# 1201t dt œ aVmax b#
aVmax b#
#
t ˆ 240" 1 ‰ sin
"Î'!
2401t‘ !
'
œ
1Î60
0
ˆ 1 cos# 2401t ‰ dt œ
aVmax b#
#
aVmax b#
#
œ V#max
1 [cos 21 cos 0]
'
0
1Î60
(1 cos 2401t) dt
"
ˆ 60
ˆ 240" 1 ‰ sin (41)‰ ˆ0 ˆ #40" 1 ‰ sin (0)‰‘ œ
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
aVmax b#
1#0
296
Chapter 5 Integration
5.6 SUBSTITUTION AND AREA BETWEEN CURVES
1. (a) Let u œ y 1 Ê du œ dy; y œ 0 Ê u œ 1, y œ 3 Ê u œ 4
'
3
0
Èy 1 dy œ
'
4
1
%
u"Î# du œ 23 u$Î# ‘ " œ ˆ 23 ‰ (4)$Î# ˆ 23 ‰ (1)$Î# œ ˆ 23 ‰ (8) ˆ 23 ‰ (1) œ
14
3
(b) Use the same substitution for u as in part (a); y œ 1 Ê u œ 0, y œ 0 Ê u œ 1
'c Èy 1 dy œ '
0
1
1
0
"
u"Î# du œ 23 u$Î# ‘ ! œ ˆ 23 ‰ (1)$Î# 0 œ
2
3
2. (a) Let u œ 1 r# Ê du œ 2r dr Ê "# du œ r dr; r œ 0 Ê u œ 1, r œ 1 Ê u œ 0
'
1
0
r È1 r# dr œ
'
0
!
"# Èu du œ 3" u$Î# ‘ " œ 0 ˆ 3" ‰ (1)$Î# œ
1
"
3
(b) Use the same substitution for u as in part (a); r œ 1 Ê u œ 0, r œ 1 Ê u œ 0
'c r È1 r# dr œ '
1
1
0
0
"# Èu du œ 0
3. (a) Let u œ tan x Ê du œ sec# x dx; x œ 0 Ê u œ 0, x œ
'
1Î4
0
'
tan x sec# x dx œ
1
"
#
1#
#
u du œ ’ u# “ œ
!
0
0œ
1
4
Ê uœ1
"
#
(b) Use the same substitution as in part (a); x œ 14 Ê u œ 1, x œ 0 Ê u œ 0
'c Î
0
1 4
'
tan x sec# x dx œ
0
#
1
u du œ ’ u# “
!
œ0
"
"
#
œ "#
4. (a) Let u œ cos x Ê du œ sin x dx Ê du œ sin x dx; x œ 0 Ê u œ 1, x œ 1 Ê u œ 1
'
1
0
$
$
' 3u# du œ cu$ d "
" œ (1) a(1) b œ 2
1
3 cos# x sin x dx œ
1
(b) Use the same substitution as in part (a); x œ 21 Ê u œ 1, x œ 31 Ê u œ 1
'
31
3 cos# x sin x dx œ
21
'
1
1
3u# du œ 2
5. (a) u œ 1 t% Ê du œ 4t$ dt Ê
'
1
0
'
$
2
"
1 4
t$ a1 t% b dt œ
"
4
du œ t$ dt; t œ 0 Ê u œ 1, t œ 1 Ê u œ 2
%
#
u
u$ du œ ’ 16
“ œ
"
2%
16
1%
16
œ
15
16
(b) Use the same substitution as in part (a); t œ 1 Ê u œ 2, t œ 1 Ê u œ 2
'c
1
'
$
t$ a1 t% b dt œ
1
2
"
4
2
u$ du œ 0
6. (a) Let u œ t# 1 Ê du œ 2t dt Ê
'
È7
0
t at# 1b
"Î$
dt œ
'
8
1
"
#
"
#
du œ t dt; t œ 0 Ê u œ 1, t œ È7 Ê u œ 8
)
u"Î$ du œ ˆ "# ‰ ˆ 34 ‰ u%Î$ ‘ " œ ˆ 38 ‰ (8)%Î$ ˆ 38 ‰ (1)%Î$ œ
45
8
(b) Use the same substitution as in part (a); t œ È7 Ê u œ 8, t œ 0 Ê u œ 1
'cÈ
0
7
t at# 1b
"Î$
dt œ
'
1
"
8 #
u"Î$ du œ 7. (a) Let u œ 4 r# Ê du œ 2r dr Ê
' a4 5rr b
1
1
# #
dr œ 5
'
5
5
"
#
"
#
'
1
8
"
#
u"Î$ du œ 45
8
du œ r dr; r œ 1 Ê u œ 5, r œ 1 Ê u œ 5
u# du œ 0
(b) Use the same substitution as in part (a); r œ 0 Ê u œ 4, r œ 1 Ê u œ 5
'
0
1
5r
a4 r# b #
dr œ 5
'
4
5
"
#
&
u# du œ 5 "# u" ‘ % œ 5 ˆ "# (5)" ‰ 5 ˆ "# (4)" ‰ œ
"
8
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 5.6 Substitution and Area Between Curves
8. (a) Let u œ 1 v$Î# Ê du œ
'
1
10Èv
a1 v$Î# b
0
#
dv œ
'
2
"
u#
1
3
#
v"Î# dv Ê
ˆ 20
‰
3 du œ
'
20
3
du œ 10Èv dv; v œ 0 Ê u œ 1, v œ 1 Ê u œ 2
20
3
2
20 "
1‘
"‘#
u# du œ 20
3 u " œ 3 # 1 œ
1
297
10
3
(b) Use the same substitution as in part (a); v œ 1 Ê u œ 2, v œ 4 Ê u œ 1 4$Î# œ 9
'
4
10Èv
#
1 a1 v$Î# b
dv œ
'
9
20 " ‘ *
20 ˆ "
1‰
20 ˆ
7 ‰
ˆ 20
‰
3 du œ 3 u # œ 3 9 2 œ 3 18 œ
"
u#
2
70
#7
9. (a) Let u œ x# 1 Ê du œ 2x dx Ê 2 du œ 4x dx; x œ 0 Ê u œ 1, x œ È3 Ê u œ 4
'
È3
4x
È x# 1
0
dx œ
'
4
'
du œ
2
1 Èu
4
1
%
2u"Î# du œ 4u"Î# ‘ " œ 4(4)"Î# 4(1)"Î# œ 4
(b) Use the same substitution as in part (a); x œ È3 Ê u œ 4, x œ È3 Ê u œ 4
È3
'cÈ
4x
3 È x# 1
'
dx œ
4
du œ 0
2
Èu
4
10. (a) Let u œ x% 9 Ê du œ 4x$ dx Ê
'
1
x$
0 È x% 9
dx œ
'
10
9
"
4
du œ x$ dx; x œ 0 Ê u œ 9, x œ 1 Ê u œ 10
"!
"
4
u"Î# du œ 4" (2)u"Î# ‘ * œ
"
#
(10)"Î# #" (9)"Î# œ
È10 3
#
(b) Use the same substitution as in part (a); x œ 1 Ê u œ 10, x œ 0 Ê u œ 9
'c
0
x$
1 È x% 9
dx œ
'
9
"
10 4
u"Î# du œ '
10
9
"
4
"
3
11. (a) Let u œ 1 cos 3t Ê du œ 3 sin 3t dt Ê
'
1Î6
0
(1 cos 3t) sin 3t dt œ
'
1
"
3
0
'
1Î6
(1 cos 3t) sin 3t dt œ
12. (a) Let u œ 2 tan
'c
0
1Î2
t
#
t
#
dt œ
"
3
1
"
#
Ê du œ
ˆ2 tan #t ‰ sec#
'
2
2
1
'c
1Î2
ˆ2 tan #t ‰ sec#
t
#
'
dt œ 2
1
'
0
cos z
È4 3 sin z
dz œ
'
4
4
"
Èu
"
6
#
#
"
t
#
(1)# 6" (0)# œ
Ê u œ 1, t œ
u du œ ’ 3" Š u# ‹ “ œ
dt Ê 2 du œ sec#
1
3
1
6
Ê u œ 1 cos
1
#
œ1
"
6
Ê u œ 1 cos 1 œ 2
"
6
(2)# 6" (1)# œ
t
#
dt; t œ
1
#
"
2
Ê u œ 2 tan ˆ 41 ‰ œ 1, t œ 0 Ê u œ 2
#
u (2 du) œ cu# d " œ 2# 1# œ 3
1
#
3
Ê u œ 1, t œ
1
#
Ê uœ3
$
u du œ cu# d " œ 3# 1# œ 8
"
3
13. (a) Let u œ 4 3 sin z Ê du œ 3 cos z dz Ê
21
!
1
6
(b) Use the same substitution as in part (a); t œ
1Î2
"
#
sec#
'
du œ sin 3t dt; t œ 0 Ê u œ 0, t œ
u du œ ’ 3" Š u# ‹ “ œ
(b) Use the same substitution as in part (a); t œ
1Î3
3 È10
#
u"Î# du œ
du œ cos z dz; z œ 0 Ê u œ 4, z œ 21 Ê u œ 4
ˆ 3" du‰ œ 0
(b) Use the same substitution as in part (a); z œ 1 Ê u œ 4 3 sin (1) œ 4, z œ 1 Ê u œ 4
'c
1
cos z
1 È4 3 sin z
dz œ
'
4
4
"
Èu
ˆ 3" du‰ œ 0
14. (a) Let u œ 3 2 cos w Ê du œ 2 sin w dw Ê "# du œ sin w dw; w œ 1# Ê u œ 3, w œ 0 Ê u œ 5
'c
0
sin w
#
1Î2 (3 2 cos w)
dw œ
'
5
3
u# ˆ #" du‰ œ
"
#
&
cu" d $ œ
"
#
"
ˆ "5 "3 ‰ œ 15
(b) Use the same substitution as in part (a); w œ 0 Ê u œ 5, w œ
'
!
1Î2
sin w
(3 2 cos w)#
dw œ
'
5
3
u# ˆ #" du‰ œ
"
#
5
' u# du œ
1
#
Ê uœ3
"
15
3
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
298
Chapter 5 Integration
15. Let u œ t& 2t Ê du œ a5t% 2b dt; t œ 0 Ê u œ 0, t œ 1 Ê u œ 3
'
1
0
'
Èt& 2t a5t% 2b dt œ
0
16. Let u œ 1 Èy Ê du œ
'
4
dy
#
1 2 È y ˆ1 È y ‰
œ
'
3
"
#
2 u
3
$
u"Î# du œ 23 u$Î# ‘ ! œ
; y œ 1 Ê u œ 2, y œ 4 Ê u œ 3
dy
2È y
'
du œ
(3)$Î# 23 (0)$Î# œ 2È3
2
3
3
2
$
u# du œ cu" d # œ ˆ 13 ‰ ˆ 12 ‰ œ
"
6
17. Let u œ cos 2) Ê du œ 2 sin 2) d) Ê "# du œ sin 2) d); ) œ 0 Ê u œ 1, ) œ
'
1Î6
!
cos$ 2) sin 2) d) œ
18. Let u œ tan ˆ 6) ‰ Ê du œ
'
31Î2
1
'
1Î2
1
"
6
'
'
cot& ˆ 6) ‰ sec# ˆ 6) ‰ d) œ
1
#
u$ du œ ’ 2" Š u# ‹“
"Î#
"
œ
1
"
%
"
#
4 ˆ 1# ‰
'
!
"
È3
"
4(1)#
,)œ
31
#
9
1
"
4
Ê u œ tan
"
'
1
1
4
œ1
œ 12
%
du œ sin t dt; t œ 0 Ê u œ 5 4 cos 0 œ 1, t œ 1 Ê u œ 5 4 cos 1 œ 9
5u"Î% ˆ "4 du‰ œ
0
"
#
3
4
È3 ‹
'
(1 sin 2t)$Î# cos 2t dt œ
œ
u
3
3
3
&
È3 u (6 du) œ ’6 Š 4 ‹“ "ÎÈ$ œ 2u% ‘ "ÎÈ$ œ 2(1)% # Š "
'
5
4
9
1
*
u"Î% du œ 54 ˆ 45 u&Î% ‰‘ " œ 9&Î% 1 œ $&Î# "
20. Let u œ 1 sin 2t Ê du œ 2 cos 2t dt Ê "# du œ cos 2t dt; t œ 0 Ê u œ 1, t œ
1Î4
Ê u œ cos 2 ˆ 16 ‰ œ
1Î
5 (5 4 cos t)"Î% sin t dt œ
!
1Î2
sec# ˆ 6) ‰ d) Ê 6 du œ sec# ˆ 6) ‰ d); ) œ 1 Ê u œ tan ˆ 16 ‰ œ
19. Let u œ 5 4 cos t Ê du œ 4 sin t dt Ê
1
'
u$ ˆ "# du‰ œ "#
1
6
" ˆ 2 &Î# ‰‘ !
2 5 u
"
"# u$Î# du œ 1
4
Ê uœ0
œ ˆ 15 (0)&Î# ‰ ˆ 15 (1)&Î# ‰ œ
"
5
21. Let u œ 4y y# 4y$ 1 Ê du œ a4 2y 12y# b dy; y œ 0 Ê u œ 1, y œ 1 Ê u œ 4(1) (1)# 4(1)$ 1 œ 8
'
1
!
a4y y# 4y$ 1b
#Î$
a12y# 2y 4b dy œ
'
8
1
)
u#Î$ du œ 3u"Î$ ‘ " œ 3(8)"Î$ 3(1)"Î$ œ 3
22. Let u œ y$ 6y# 12y 9 Ê du œ a3y# 12y 12b dy Ê
'
1
!
ay$ 6y# 12y 9b
23. Let u œ )$Î# Ê du œ
'
È1
3
!
1Î2
1
œ
"
#
'
È) cos# ˆ)$Î# ‰ d) œ
"
t
"
4
1
!
2
3
'
9
du œ ay# 4y 4b dy; y œ 0 Ê u œ 9, y œ 1 Ê u œ 4
%
"
3
u"Î# du œ 3" ˆ2u"Î# ‰‘ * œ
2
3
(4)"Î# 32 (9)"Î# œ
2
3
(2 3) œ 32
3
du œ È) d); ) œ 0 Ê u œ 0, ) œ È
1# Ê u œ 1
cos# u ˆ 23 du‰ œ 23 ˆ #u "
4
1
sin 2u‰‘ ! œ
2
3
ˆ 1# "
4
sin 21‰ 32 (0) œ
1
3
Ê du œ t# dt; t œ 1 Ê u œ 0, t œ #" Ê u œ 1
t# sin# ˆ1 "t ‰ dt œ
ay# 4y 4b dy œ
)"Î# d) Ê
#
24. Let u œ 1 'c
3
#
"Î#
4
"
3
'!
1
sin# u du œ ˆ u2 "
4
"
sin 2u‰‘ ! œ ’ˆ #" "
4
sin (2)‰ ˆ #0 "
4
sin 0‰“
sin 2
25. Let u œ 4 x# Ê du œ 2x dx Ê "# du œ x dx; x œ 2 Ê u œ 0, x œ 0 Ê u œ 4, x œ 2 Ê u œ 0
Aœ
'c
0
2
xÈ4 x# dx %
œ 23 u$Î# ‘ ! œ
2
3
'
2
!
xÈ4 x# dx œ (4)$Î# 23 (0)$Î# œ
'
4
!
"# u"Î# du '
4
0
"# u"Î# du œ 2
'
4
"
! #
u"Î# du œ
16
3
26. Let u œ 1 cos x Ê du œ sin x dx; x œ 0 Ê u œ 0, x œ 1 Ê u œ 2
'
!
1
(1 cos x) sin x dx œ
'
!
2
#
#
u du œ ’ u2 “ œ
!
2#
#
0#
#
œ2
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
'
!
4
u"Î# du
Section 5.6 Substitution and Area Between Curves
299
27. Let u œ 1 cos x Ê du œ sin x dx Ê du œ sin x dx; x œ 1 Ê u œ 1 cos (1) œ 0, x œ 0 Ê u œ 1 cos 0 œ 2
Aœ
'c
0
1
3 (sin x) È1 cos x dx œ '
2
!
28. Let u œ 1 1 sin x Ê du œ 1 cos x dx Ê
Because of symmetry about x œ 1# , A œ 2
œ
'
!
1
3u"Î# (du) œ 3
"
1
'c
'
2
!
#
u"Î# du œ 2u$Î# ‘ ! œ 2(2)$Î# 2(0)$Î# œ 2&Î#
du œ cos x dx; x œ 1# Ê u œ 1 1 sin ˆ 1# ‰ œ 0, x œ 0 Ê u œ 1
0
1
1Î2 #
(cos x) (sin (1 1 sin x)) dx œ 2
'
1
(" cos 2x)
#
!
dx œ
'
"
#
1
!
(1 cos 2x) dx œ
"
#
x 30. For the sketch given, a œ 13 , b œ 13 ; f(t) g(t) œ
Aœ
œ
"
#
1
!
1
#
(sin u) ˆ 1" du‰
sin u du œ [cos u]1! œ (cos 1) (cos 0) œ 2
1 cos 2x
;
#
29. For the sketch given, a œ 0, b œ 1; f(x) g(x) œ 1 cos# x œ sin# x œ
Aœ
'
'c ÎÎ ˆ "# sec# t 4 sin# t‰ dt œ "# '
1 3
1Î3
1 3
1Î3
1Î3
1Î3
1 3
1Î3
'c Î sec# t dt 2'
"
#
"
#
"
#
œ
'
1Î3
1Î3
sin# t dt œ
"
#
"
#
'
1Î3
1Î3
1Î$
sin 2t
# ]1Î$
1Î$
[tan t]1Î$ 2[t 1
#
[(1 0) (0 0)] œ
sec# t a4 sin# tb œ
sec# t dt 4
(1 cos 2t) dt œ
sin 2x ‘ 1
#
!
sec# t 4 sin# t;
'
sec# t dt 4
œ È3 4 †
1
3
1Î3
(" cos 2t)
#
1Î3
È3 œ
dt
41
3
31. For the sketch given, a œ 2, b œ 2; f(x) g(x) œ 2x# ax% 2x# b œ 4x# x% ;
Aœ
'c
2
$
2
a4x# x% b dx œ ’ 4x3 #
x&
5 “ #
œ ˆ 32
3 32 ‰
5
ˆ 32 ‰‘ œ
32
3 5
64
5
œ
(" 0)
4
œ
"
3
64
3
320192
15
œ
128
15
32. For the sketch given, c œ 0, d œ 1; f(y) g(y) œ y# y$ ;
Aœ
'
1
!
ay# y$ b dy œ
'
1
!
y# dy '
!
1
$
"
%
"
(" 0)
3
y$ dy œ ’ y3 “ ’ y4 “ œ
!
!
"
4
œ
"
1#
33. For the sketch given, c œ 0, d œ 1; f(y) g(y) œ a12y# 12y$ b a2y# 2yb œ 10y# 12y$ 2y;
Aœ
'
1
!
a10y# 12y$ 2yb dy œ
'
‰
œ ˆ 10
3 0 (3 0) (1 0) œ
!
1
10y# dy '
!
1
12y$ dy '
1
!
"
"
"
$‘
12 % ‘
2 #‘
2y dy œ 10
3 y ! 4 y ! # y !
4
3
34. For the sketch given, a œ 1, b œ 1; f(x) g(x) œ x# a2x% b œ x# 2x% ;
Aœ
'c ax# 2x% b dx œ ’ x3
1
$
1
"
2x&
5 “ "
œ ˆ 3" 52 ‰ 3" ˆ 52 ‰‘ œ
35. We want the area between the line y œ 1, 0 Ÿ x Ÿ 2, and the curve y œ
(formed by y œ x and y œ 1) with base 1 and height 1. Thus, A œ
œ ˆ2 8 ‰
1#
"
#
œ2
2
3
"
#
œ
'
!
2
2
3
x#
4,
4
5
œ
10 12
15
œ
22
15
738?= the area of a triangle
Š1 x#
4‹
dx "# (1)(1) œ ’x #
x$
1# “ !
"
#
5
6
36. We want the area between the x-axis and the curve y œ x# , 0 Ÿ x Ÿ 1 :6?= the area of a triangle (formed by x œ 1,
x y œ 2, and the x-axis) with base 1 and height 1. Thus, A œ
'
!
1
$
"
x# dx "# (1)(1) œ ’ x3 “ !
"
#
œ
"
3
"
#
œ
5
6
37. AREA œ A1 A2
A1: For the sketch given, a œ 3 and we find b by solving the equations y œ x# 4 and y œ x# 2x
simultaneously for x: x# 4 œ x# 2x Ê 2x# 2x 4 œ 0 Ê 2(x 2)(x 1) Ê x œ 2 or x œ 1 so
b œ 2: f(x) g(x) œ ax# 4b ax# 2xb œ 2x# 2x 4 Ê A1 œ
'cc a2x# 2x 4b dx
2
3
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
300
Chapter 5 Integration
$
œ ’ 2x3 2x#
#
4x“
#
$
‰
œ ˆ 16
3 4 8 (18 9 12) œ 9 œ
16
3
11
3 ;
A2: For the sketch given, a œ 2 and b œ 1: f(x) g(x) œ ax# 2xb ax# 4b œ 2x# 2x 4
'c a2x# 2x 4b dx œ ’ 2x3
1
Ê A2 œ $
2
œ 23 1 4 16
3
x# 4x“
"
‰
œ ˆ 23 1 4‰ ˆ 16
3 48
#
4 8 œ 9;
Therefore, AREA œ A1 A2 œ
11
3
9œ
38
3
38. AREA œ A1 A2
A1: For the sketch given, a œ 2 and b œ 0: f(x) g(x) œ a2x$ x# 5xb ax# 3xb œ 2x$ 8x
Ê A1 œ
'c a2x$ 8xb dx œ ’ 2x4
0
%
2
!
8x#
# “ #
œ 0 (8 16) œ 8;
A2: For the sketch given, a œ 0 and b œ 2: f(x) g(x) œ ax# 3xb a2x$ x# 5xb œ 8x 2x$
Ê A2 œ
'
2
0
#
a8x 2x$ b dx œ ’ 8x2 Therefore, AREA œ A1 A2 œ 16
#
2x%
“
4 !
œ (16 8) œ 8;
39. AREA œ A1 A2 A3
A1: For the sketch given, a œ 2 and b œ 1: f(x) g(x) œ (x 2) a4 x# b œ x# x 2
Ê A1 œ
'cc ax# x 2b dx œ ’ x3
1
$
2
x#
#
2x“
"
#
œ ˆ "3 "
#
2‰ ˆ 83 4‰ œ
4
2
7
3
"
#
œ
143
6
œ
1"
6 ;
"
#
œ 9# ;
A2: For the sketch given, a œ 1 and b œ 2: f(x) g(x) œ a4 x# b (x 2) œ ax# x 2b
Ê A2 œ 'c
2
$
1
ax# x 2b dx œ ’ x3 x#
#
2x“
#
"
œ ˆ 83 4
#
4‰ ˆ 13 1
2
2‰ œ 3 8 A3: For the sketch given, a œ 2 and b œ 3: f(x) g(x) œ (x 2) a4 x# b œ x# x 2
Ê A3 œ
'
3
$
x#
#
ax# x 2b dx œ ’ x3 2
Therefore, AREA œ A1 A2 A3 œ
11
6
9
#
$
2x“ œ ˆ 27
3 #
9
#
ˆ9 9
#
6‰ ˆ 38 83 ‰ œ 9 5
6
œ
4
2
4‰ œ 9 9
#
38 ;
49
6
40. AREA œ A1 A2 A3
$
A1: For the sketch given, a œ 2 and b œ 0: f(x) g(x) œ Š x3 x‹ Ê A1 œ
"
3
'c
0
2
ax$ 4xb dx œ
"
3
%
’ x4 2x# “
!
x
3
for x:
xœ
f(x) g(x) œ
œ
"
3
x
3
x
3
Ê
x$
3
xœ0 Ê
4
3
x
3
43 x œ
"
3
ax$ 4xb
œ 0 3" (4 8) œ 34 ;
#
A2: For the sketch given, a œ 0 and we find b by solving the equations y œ
x$
3
x$
3
œ
x$
3
x and y œ
x
3
simultaneously
(x 2)(x 2) œ 0 Ê x œ 2, x œ 0, or x œ 2 so b œ 2:
$
Š x3 x‹ œ 3" ax$ 4xb Ê A2 œ 3"
'
0
2
ax$ 4xb dx œ
"
3
'
0
2
a4x x$ b œ
(8 4) œ 43 ;
$
A3: For the sketch given, a œ 2 and b œ 3: f(x) g(x) œ Š x3 x‹ Ê A3 œ
"
3
'
3
2
ax$ 4xb dx œ
"
3
Therefore, AREA œ A1 A2 A3 œ
$
%
’ x4 2x# “ œ
#
4
3
4
3
25
12
œ
"
3
x
3
œ
"
3
ax$ 4xb
ˆ 81
‰ ˆ 16
‰‘ œ
4 2†9 4 8
3225
1#
œ
"
3
ˆ 81
‰
4 14 œ
19
4
41. a œ 2, b œ 2;
f(x) g(x) œ 2 ax# 2b œ 4 x#
Ê Aœ
"
3
'c a4 x# bdx œ ’4x x3 “ #
2
$
#
2
8‰
œ 2 † ˆ 24
3 3 œ
œ ˆ8 83 ‰ ˆ8 83 ‰
32
3
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
25
12 ;
’2x# #
x%
4 “!
Section 5.6 Substitution and Area Between Curves
42. a œ 1, b œ 3;
f(x) g(x) œ a2x x# b (3) œ 2x x# 3
'c a2x x# 3b dx œ ’x# x3
3
Ê Aœ
œ ˆ9 $
3x“
1
9‰ ˆ1 27
3
1
3
43. a œ 0, b œ 2;
f(x) g(x) œ 8x x% Ê A œ
#
œ ’ 8x2 #
x&
5 “!
œ 16 32
5
"
3
3‰ œ 11 œ
'
2
0
œ
$
"
32
3
a8x x% b dx
80 32
5
œ
48
5
44. Limits of integration: x# 2x œ x Ê x# œ 3x
Ê x(x 3) œ 0 Ê a œ 0 and b œ 3;
f(x) g(x) œ x ax# 2xb œ 3x x#
Ê Aœ
œ
27
#
'
3
0
#
a3x x# b dx œ ’ 3x2 27 18
#
9œ
œ
$
x$
3 “!
9
#
45. Limits of integration: x# œ x# 4x Ê 2x# 4x œ 0
Ê 2x(x 2) œ 0 Ê a œ 0 and b œ 2;
f(x) g(x) œ ax# 4xb x# œ 2x# 4x
Ê Aœ
'
2
0
œ 16
3 $
a2x# 4xb dx œ ’ 2x
3 œ
16
#
32 48
6
œ
#
4x#
2 “!
8
3
46. Limits of integration: 7 2x# œ x# 4 Ê 3x# 3 œ 0
Ê 3(x 1)(x 1) œ 0 Ê a œ 1 and b œ 1;
f(x) g(x) œ a7 2x# b ax# 4b œ 3 3x#
Ê Aœ
'c a3 3x# b dx œ 3 ’x x3 “ "
1
$
"
1
œ 3 ˆ1 "3 ‰ ˆ1 3" ‰‘ œ 6 ˆ 32 ‰ œ 4
47. Limits of integration: x% 4x# 4 œ x#
Ê x% 5x# 4 œ 0 Ê ax# 4b ax# 1b œ 0
Ê (x 2)(x 2)(x 1)(x 1) œ ! Ê x œ 2, 1, 1, 2;
f(x) g(x) œ ax% 4x# 4b x# œ x% 5x# 4 and
g(x) f(x) œ x# ax% 4x# 4b œ x% 5x# 4
Ê Aœ
'
1
2
'cc ax% 5x# 4bdx 'c ax% 5x# 4bdx
1
1
ax% 5x# 4bdx
&
œ ’ x5 œ ˆ "5 œ
1
2
60
5
5
3
5x$
3
4x“
"
&
’ x5 #
4‰ ˆ 32
5 60
3
œ
300180
15
40
3
5x$
3
4x“
8‰ ˆ 5" 5
3
"
"
&
’ 5x 5x$
3
4‰ ˆ 5" 4x“
5
3
#
"
4‰ ˆ 32
5 40
3
8‰ ˆ 5" œ8
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
5
3
4‰
301
302
Chapter 5 Integration
48. Limits of integration: xÈa# x# œ 0 Ê x œ 0 or
Èa# x# œ 0 Ê x œ 0 or a# x# œ 0 Ê x œ a, 0, a;
Aœ
œ
"
#
œ
"
3
'c xÈa# x# dx '
0
a
a
0
’ 23 aa# x# b
# $Î#
aa b
$Î# !
“
"
3
a
xÈa# x# dx
"# ’ 23 aa# x# b
# $Î#
’ aa b
$Î# a
“
!
2a$
3
“œ
49. Limits of integration: y œ Èkxk œ Èx, x Ÿ 0
and
Èx, x 0
5y œ x 6 or y œ x5 65 ; for x Ÿ 0: Èx œ x5 65
Ê 5Èx œ x 6 Ê 25(x) œ x# 12x 36
Ê x# 37x 36 œ 0 Ê (x 1)(x 36) œ 0
Ê x œ 1, 36 (but x œ 36 is not a solution);
for x 0: 5Èx œ x 6 Ê 25x œ x# 12x 36
Ê x# 13x 36 œ 0 Ê (x 4)(x 9) œ 0
Ê x œ 4, 9; there are three intersection points and
Aœ
'c ˆ x 5 6 Èx‰dx '
0
1
4
0
#
œ ’ (x 106) 23 (x)$Î# “
œ ˆ 36
10 25
10
!
ˆ x 5 6 Èx‰dx ˆÈ x ’ (x 106) 23 x$Î# “ ’ 23 x$Î# 23 ‰ ˆ 100
10 2
3
† 4$Î# !
36
10
0‰ ˆ 32 † 9$Î# 50. Limits of integration:
x# 4, x Ÿ 2 or x
y œ kx# 4k œ œ
4 x# , 2 Ÿ x Ÿ 2
for x Ÿ 2 and x
4
9
%
#
"
'
2: x# 4 œ
x#
2
x6‰
5
dx
*
(x 6)#
10 “ %
225
10
2
3
† 4$Î# 100 ‰
10
œ 50
10 20
3
2
4
Ê 2x# 8 œ x# 8 Ê x# œ 16 Ê x œ „ 4;
for 2 Ÿ x Ÿ 2: 4 x# œ
x#
#
4 Ê 8 2x# œ x# 8
Ê x# œ 0 Ê x œ 0; by symmetry of the graph,
Aœ2
'
0
2
#
’Š x2 4‹ a4 x# b“dx 2
œ 2 ˆ 8# 0‰ 2 ˆ32 64
6
'
4
2
#
$
#
’Š x2 4‹ ax# 4b“dx œ 2 ’ x2 “ 2 ’8x 16 68 ‰ œ 40 !
56
3
œ
%
x$
6 “#
64
3
51. Limits of integration: c œ 0 and d œ 3;
f(y) g(y) œ 2y# 0 œ 2y#
Ê Aœ
'
0
3
$
$
2y# dy œ ’ 2y3 “ œ 2 † 9 œ 18
!
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
œ
5
3
Section 5.6 Substitution and Area Between Curves
52. Limits of integration: y# œ y 2 Ê (y 1)(y 2) œ 0
Ê c œ 1 and d œ 2; f(y) g(y) œ (y 2) y#
'c ay 2 y# b dy œ ’ y#
2
Ê Aœ
#
1
2y œ ˆ 4# 4 83 ‰ ˆ "# 2 3" ‰ œ 6 8
3
"
#
#
y$
3 “ "
2
"
3
œ
9
#
53. Limits of integration: 4x œ y# 4 and 4x œ 16 y
Ê y# 4 œ 16 y Ê y# y 20 œ 0 Ê
(y 5)(y 4) œ 0 Ê c œ 4 and d œ 5;
#
f(y) g(y) œ ˆ 164y ‰ Š y 44 ‹ œ
"
4
Ê Aœ
y$
3
'c ay# y 20b dy
5
4
y#
#
œ
"
4
’
œ
"
4
"
4
ˆ 125
3 189
ˆ 3 œ
y# y20
4
20y“
&
%
25
‰
100
4" ˆ 64
2
3
9
243
‰
180
œ
2
8
80‰
16
#
54. Limits of integration: x œ y# and x œ 3 2y#
Ê y# œ 3 2y# Ê 3y# œ 3 Ê 3(y 1)(y 1) œ 0
Ê c œ 1 and d œ 1; f(y) g(y) œ a3 2y# b y#
œ 3 3y# œ 3 a1 y# b Ê A œ 3
"
y$
3 “ "
œ 3 ’y œ 3 ˆ1 "‰
3
'c a1 y# b dy
1
1
3 ˆ1 3" ‰
œ 3 † 2 ˆ1 "3 ‰ œ 4
55. Limits of integration: x œ y# y and x œ 2y# 2y 6
Ê y# y œ 2y# 2y 6 Ê y# y 6 œ 0
Ê ay 3bay 2b œ 0 Ê c œ 2 and d œ 3;
f(y) g(y) œ ay# yb a2y# 2y 6b œ y# y 6
Ê Aœ
'c ay# y 6b dy œ ’ y3
3
$
2
œ ˆ9 18‰ ˆ 83 2 12‰ œ
9
2
"# y2 6y“
3
2
125
6
56. Limits of integration: x œ y#Î$ and x œ 2 y%
Ê y#Î$ œ 2 y% Ê c œ 1 and d œ 1;
f(y) g(y) œ a2 y% b y#Î$
Ê Aœ
œ ’2y 'c ˆ2 y% y#Î$ ‰ dy
1
1
y&
5
53 y&Î$ “
"
"
œ ˆ2 "5 35 ‰ ˆ2 œ 2 ˆ2 "5 35 ‰ œ 12
5
"
5
35 ‰
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
303
304
Chapter 5 Integration
57. Limits of integration: x œ y# 1 and x œ kyk È1 y#
Ê y# 1 œ kyk È1 y# Ê y% 2y# 1 œ y# a1 y# b
Ê y% 2y# 1 œ y# y% Ê 2y% 3y# 1 œ 0
Ê a2y# 1b ay# 1b œ 0 Ê 2y# 1 œ 0 or y# 1 œ 0
"
#
Ê y# œ
or y# œ 1 Ê y œ „
„È 2
#
È2
#
or y œ „ 1.
are not solutions Ê y œ „ 1;
for 1 Ÿ y Ÿ 0, f(x) g(x) œ yÈ1 y# ay# 1b
Substitution shows that
œ 1 y# y a1 y# b
Aœ2
"Î#
, and by symmetry of the graph,
'c ’1 y# y a1 y# b"Î# “ dy
0
1
'c a1 y# b dy 2 'c y a1 y# b"Î# dy œ 2 ’y y3 “ !
œ2
0
0
1
$
"
1
œ 2 (! 0) ˆ1 " ‰‘
3
ˆ 23
# $Î#
2 ˆ "# ‰ ” 2 a1 3y b •
!
"
0‰ œ 2
58. AREA œ A1 A2
Limits of integration: x œ 2y and x œ y$ y# Ê
y$ y# œ 2y Ê y ay# y 2b œ y(y 1)(y 2) œ 0
Ê y œ 1, 0, 2:
for 1 Ÿ y Ÿ 0, f(y) g(y) œ y$ y# 2y
Ê A1 œ
'c
0
1
œ 0 ˆ "4 "
3
%
y$
3
ay$ y# 2yb dy œ ’ y4 1‰ œ
y# “
!
"
5
12 ;
for 0 Ÿ y Ÿ 2, f(y) g(y) œ 2y y$ y#
'! a2y y$ y# b dy œ ’y# y4
2
Ê A2 œ
Ê ˆ4 16
4
%
#
y$
3 “!
38 ‰ 0 œ 38 ;
Therefore, A1 A2 œ
5
12
8
3
œ
37
12
59. Limits of integration: y œ 4x# 4 and y œ x% 1
Ê x% 1 œ 4x# 4 Ê x% 4x# 5 œ 0
Ê ax# 5b (x 1)(x 1) œ 0 Ê a œ 1 and b œ 1;
f(x) g(x) œ 4x# 4 x% 1 œ 4x# x% 5
Ê Aœ
'c a4x# x% 5b dx œ ’ 4x3
1
1
œ ˆ 43 "
5
5‰ ˆ 43 "
5
$
5‰ œ 2 ˆ 43 "
x&
5
5x“
"
5
5‰ œ
"
104
15
60. Limits of integration: y œ x$ and y œ 3x# 4
Ê x$ 3x# 4 œ 0 Ê ax# x 2b (x 2) œ 0
Ê (x 1)(x 2)# œ 0 Ê a œ 1 and b œ 2;
f(x) g(x) œ x$ a3x# 4b œ x$ 3x# 4
Ê Aœ
'c ax$ 3x# 4b dx œ ’ x4
œ ˆ 16
4 24
3
2
%
1
8‰ ˆ 41 " 4‰ œ
3x$
3
4x“
#
"
27
4
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 5.6 Substitution and Area Between Curves
61. Limits of integration: x œ 4 4y# and x œ 1 y%
Ê 4 4y# œ 1 y% Ê y% 4y# 3 œ 0
Ê Šy È3‹ Šy È3‹ (y 1)(y 1) œ 0 Ê c œ 1
0; f(y) g(y) œ a4 4y# b a1 y% b
and d œ 1 since x
'c a3 4y# y% b dy
1
œ 3 4y# y% Ê A œ
4y$
3
œ ’3y "
y&
5 “ "
1
œ 2ˆ3 4
3
5" ‰ œ
56
15
#
62. Limits of integration: x œ 3 y# and x œ y4
#
Ê 3 y# œ y4 Ê
3y#
4
3œ0 Ê
3
4
(y 2)(y 2) œ 0
#
Ê c œ 2 and d œ 2; f(y) g(y) œ a3 y# b Š y4 ‹
y#
4‹
œ 3 Š1 œ 3 ˆ2 8 ‰
12
'c Š1 y4 ‹ dy œ 3 ’y 1y# “ #
2
Ê Aœ3
ˆ 2 #
$
#
2
8 ‰‘
12
œ 3 ˆ4 16 ‰
12
œ 12 4 œ 8
63. a œ 0, b œ 1; f(x) g(x) œ 2 sin x sin 2x
Ê Aœ
'
1
0
(2 sin x sin 2x) dx œ 2 cos x cos 2x ‘ 1
2
!
œ 2(1) "# ‘ ˆ2 † 1 "# ‰ œ 4
64. a œ 13 , b œ 13 ; f(x) g(x) œ 8 cos x sec# x
Ê Aœ
œ Š8 †
'c
1Î3
1Î3
È3
#
1Î$
a8 cos x sec# xb dx œ [8 sin x tan x] 1Î$
È3
#
È3‹ Š8 †
È3‹ œ 6È3
‰
65. a œ 1, b œ 1; f(x) g(x) œ a1 x# b cos ˆ 1x
#
Ê Aœ
œ ˆ1 "
3
'c 1 x# cos ˆ 1#x ‰‘ dx œ ’x x3
1
$
1
12 ‰ ˆ1 "
3
12 ‰ œ 2 ˆ 23 12 ‰ œ
2
1
4
3
sin ˆ 1#x ‰“
"
"
4
1
66. A œ A1 A2
a" œ 1, b" œ 0 and a# œ 0, b# œ 1;
f" (x) g" (x) œ x sin ˆ 1#x ‰ and f# (x) g# (x) œ sin ˆ 1#x ‰ x
Ê by symmetry about the origin,
A" A# œ 2A" Ê A œ 2
œ 2 ’ 12 cos ˆ 1#x ‰ "
x#
# “!
œ 2 ˆ 12 "# ‰ œ 2 ˆ 4211 ‰ œ
'
0
1
sin ˆ 1x
‰
‘
# x dx
œ 2 ˆ 12 † 0 "# ‰ ˆ 12 † 1 0‰‘
4 1
1
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
305
306
Chapter 5 Integration
67. a œ 14 , b œ 14 ; f(x) g(x) œ sec# x tan# x
Ê Aœ
œ
œ
'c
1Î4
1Î4
1Î4
'c
1Î4
'c
1Î4
1Î4
asec# x tan# xb dx
csec# x asec# x 1bd dx
1Î%
1 † dx œ [x]1Î% œ
1
4
ˆ 14 ‰ œ
1
#
68. c œ 14 , d œ 14 ; f(y) g(y) œ tan# y a tan# yb œ 2 tan# y
œ 2 asec# y 1b Ê A œ
'c
1Î4
1Î4
2 asec# y 1b dy
1Î%
œ 2[tan y y]1Î% œ 2 ˆ1 14 ‰ ˆ1 14 ‰‘
œ 4 ˆ1 14 ‰ œ 4 1
69. c œ 0, d œ 1# ; f(y) g(y) œ 3 sin yÈcos y 0 œ 3 sin yÈcos y
Ê Aœ3
'
1Î2
0
1Î#
sin yÈcos y dy œ 3 23 (cos y)$Î# ‘ !
œ 2(0 1) œ 2
"Î$
‰
70. a œ 1, b œ 1; f(x) g(x) œ sec# ˆ 1x
3 x
Ê Aœ
'c sec# ˆ 13x ‰ x"Î$ ‘ dx œ 13 tan ˆ 13x ‰ 43 x%Î$ ‘ ""
1
1
œ Š 13 È3 34 ‹ ’ 13 ŠÈ3‹ 34 “ œ
6È 3
1
71. A œ A" A#
Limits of integration: x œ y$ and x œ y Ê y œ y$
Ê y$ y œ 0 Ê y(y 1)(y 1) œ 0 Ê c" œ 1, d" œ 0
and c# œ 0, d# œ 1; f" (y) g" (y) œ y$ y and
f# (y) g# (y) œ y y$ Ê by symmetry about the origin,
A" A# œ 2A# Ê A œ 2
œ 2 ˆ "# 4" ‰ œ
'
1
0
#
ay y$ b dy œ 2 ’ y# "
#
"
y%
4 “!
72. A œ A" A#
Limits of integration: y œ x$ and y œ x& Ê x$ œ x&
Ê x& x$ œ 0 Ê x$ (x 1)(x 1) œ 0 Ê a" œ 1, b" œ 0
and a# œ 0, b# œ 1; f" (x) g" (x) œ x$ x& and
f# (x) g# (x) œ x& x$ Ê by symmetry about the origin,
A" A# œ 2A# Ê A œ 2
œ 2 ˆ "4 6" ‰ œ
"
6
'
0
1
%
ax$ x& b dx œ 2 ’ x4 "
x'
6 “!
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 5.6 Substitution and Area Between Curves
73. A œ A" A#
Limits of integration: y œ x and y œ
"
x#
$
Ê xœ
"
x# ,
307
xÁ0
Ê x œ 1 Ê x œ 1 , f" (x) g" (x) œ x 0 œ x
Ê A" œ
'
1
0
#
"
x dx œ ’ x2 “ œ "# ; f# (x) g# (x) œ
'
œ x# Ê A# œ
A œ A" A# œ
!
2
1
"
#
and b œ
Ê Aœ
œŠ
È2
#
#
"
#
œ1
1
4
Ê aœ0
f(x) g(x) œ cos x sin x
'
1Î4
0
0
"
"
‘
x# dx œ "
x " œ # 1 œ #;
74. Limits of integration: sin x œ cos x Ê x œ
1
4;
"
x#
1Î%
(cos x sin x) dx œ [sin x cos x]!
È2
# ‹
(0 1) œ È2 1
75. (a) The coordinates of the points of intersection of the
line and parabola are c œ x# Ê x œ „ Èc and y œ c
(b) f(y) g(y) œ Èy ˆÈy‰ œ 2Èy Ê the area of the
lower section is, AL œ
œ2
'
0
c
'
0
c
[f(y) g(y)] dy
Èy dy œ 2 23 y$Î# ‘ ! œ
c
4
3
c$Î# . The area of the
entire shaded region can be found by setting c œ 4: A œ ˆ 43 ‰ 4$Î# œ 43†8 œ 32
3 . Since we want c to divide the region
32
4
$Î#
into subsections of equal area we have A œ 2AL Ê 3 œ 2 ˆ 3 c ‰ Ê c œ 4#Î$
(c) f(x) g(x) œ c x# Ê AL œ
œ
4
3
Èc
Èc
c
c
'cÈ [f(x) g(x)] dx œ 'cÈ ac x# b dx œ ’cx x3 “ È
$
c
cÈc
œ 2 ’c$Î# c$Î# . Again, the area of the whole shaded region can be found by setting c œ 4 Ê A œ
condition A œ 2AL , we get
4
3
$Î#
c
œ
#Î$
Ê cœ4
32
3
32
3 .
c$Î#
3 “
From the
as in part (b).
76. (a) Limits of integration: y œ 3 x# and y œ 1
Ê 3 x# œ 1 Ê x# œ 4 Ê a œ 2 and b œ 2;
f(x) g(x) œ a3 x# b (1) œ 4 x#
Ê Aœ
'c a4 x# b dx œ ’4x x3 “ #
2
$
#
32
3
2
œ ˆ8 83 ‰ ˆ8 83 ‰ œ 16 16
3
œ
(b) Limits of integration: let x œ 0 in y œ 3 x#
Ê y œ 3; f(y) g(y) œ È3 y ˆÈ3 y‰
œ 2(3 y)"Î#
Ê Aœ2
'c (3 y)"Î# dy œ 2 'c (3 y)"Î# (1) dy œ (2) ’ 2(3 3y)
œ ˆ 43 ‰ (8) œ
3
1
3
1
$Î#
“
$
"
œ ˆ 43 ‰ 0 (3 1)$Î# ‘
32
3
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
308
Chapter 5 Integration
77. Limits of integration: y œ 1 Èx and y œ
Ê 1 Èx œ
2
Èx
2
Èx
, x Á 0 Ê Èx x œ 2 Ê x œ (2 x)#
Ê x œ 4 4x x# Ê x# 5x 4 œ 0
Ê (x 4)(x 1) œ 0 Ê x œ 1, 4 (but x œ 4 does not
satisfy the equation); y œ È2x and y œ x4 Ê È2x œ x4
Ê 8 œ xÈx Ê 64 œ x$ Ê x œ 4.
Therefore, AREA œ A" A# : f" (x) g" (x) œ ˆ1 x"Î# ‰ Ê A" œ
'
œ ˆ1 "8 ‰ 0 œ
2
3
1
0
œ ˆ4 † 2 "
x#
8 “!
ˆ1 x"Î# x4 ‰ dx œ ’x 23 x$Î# 16 ‰
8
37
24 ; f# (x)
g# (x) œ 2x"Î# ˆ4 "8 ‰ œ 4 œ
15
8
17
8 ;
x
4
x
4
Ê A# œ
'
1
4
ˆ2x"Î# 4x ‰ dx œ ’4x"Î# Therefore, AREA œ A" A# œ
37
24
17
8
œ
3751
24
œ
88
24
%
x#
8 “"
œ
11
3
78. Limits of integration: (y 1)# œ 3 y Ê y# 2y 1
œ 3 y Ê y# y 2 œ 0 Ê (y 2)(y 1) œ 0
Ê y œ 2 since y 0; also, 2Èy œ 3 y
Ê 4y œ 9 6y y# Ê y# 10y 9 œ 0
Ê (y 9)(y 1) œ 0 Ê y œ 1 since y œ 9 does not
satisfy the equation;
AREA œ A" A#
f" (y) g" (y) œ 2Èy 0 œ 2y"Î#
Ê A" œ 2
Ê A# œ
'
1
0
'
1
2
"
$Î#
y"Î# dy œ 2 ’ 2y3 “ œ 34 ; f# (y) g# (y) œ (3 y) (y 1)#
!
#
c3 y (y 1)# d dy œ 3y "# y# "3 (y 1)$ ‘ " œ ˆ6 2 3" ‰ ˆ3 Therefore, A" A# œ
4
3
7
6
œ
œ
15
6
"
#
80. A œ
'
a
b
2f(x) dx '
a
b
'
0
a
aa# x# b dx œ 2 a# x "3 x$ ‘ ! œ 2 Ša$ a
a$
$
Š 4a3 ‹
(2a) aa# b œ a$ ; limit of ratio œ lim b
aÄ!
'
f(x) dx œ 2
a
b
0‰ œ 1 "
3
"
#
œ 67 ;
5
2
79. Area between parabola and y œ a# : A œ 2
Area of triangle AOC:
"
#
f(x) dx '
a
b
f(x) dx œ
'
a
b
œ
3
4
a$
3‹
0œ
4a$
3 ;
which is independent of a.
f(x) dx œ 4
81. The lower boundary of the region is the line through the points azß 1 z2 b and Šz 1ß 1 az 1b2 ‹. The equation of this
line is y a1 z2 b œ
ˆ 1 az 1 b 2 ‰ ˆ 1 z 2 ‰
ax
z1z
z1
The area of theregion is given by
œ
'
z
1b œ a2z 1bax 1b Ê y œ a2z 1bx az2 z 1b.
aa1 x2 b aa2z 1bx az2 z 1bbbdy
' z ax2 a2z 1bx z2 zbdy œ 13 x3 "# a2z 1bx2 az2 zbx‘ zz 1
1
z
œ Š 13 az 1b3 "# a2z 1baz 1b2 az2 zbaz 1b‹ ˆ 13 z3 "# a2z 1bz2 az2 zbz‰ œ 16 . No matter where we
choose z, the area of the region bounded by y œ 1 x2 and the line through the points azß 1 z2 b and
Šz 1ß 1 az 1b2 ‹ is always 16 .
82. It is sometimes true. It is true if f(x) g(x) for all x between a and b. Otherwise it is false. If the graph of f
lies below the graph of g for a portion of the interval of integration, the integral over that portion will be
negative and the integral over [aß b] will be less than the area between the curves (see Exercise 71).
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 5.6 Substitution and Area Between Curves
"
#
83. Let u œ 2x Ê du œ 2 dx Ê
'
3
sin 2x
x
1
'
dx œ
6
ˆ "# du‰ œ
sin u
ˆ u# ‰
2
du œ dx; x œ 1 Ê u œ 2, x œ 3 Ê u œ 6
'
6
du œ cF(u)d '# œ F(6) F(2)
sin u
u
2
84. Let u œ 1 x Ê du œ dx Ê du œ dx; x œ 0 Ê u œ 1, x œ 1 Ê u œ 0
'
1
0
'
f(1 x) dx œ
0
1
f(u) ( du) œ '
0
1
f(u) du œ
'
1
f(u) du œ
0
'
1
0
f(x) dx
85. (a) Let u œ x Ê du œ dx; x œ 1 Ê u œ 1, x œ 0 Ê u œ 0
'c f(x) dx œ '
0
f odd Ê f(x) œ f(x). Then
1
0
1
f(u) ( du) œ
'
1
œ 3
(b) Let u œ x Ê du œ dx; x œ 1 Ê u œ 1, x œ 0 Ê u œ 0
f even Ê f(x) œ f(x). Then
'c f(x) dx œ '
0
1
1
0
0
f(u) ( du) œ '
f(u) ( du) œ
1
0
f(u) du œ
'
0
1
'
1
0
f(u) du œ '
0
1
f(u) du
f(u) du œ 3
'c f(x) dx when f is odd. Let u œ x Ê du œ dx Ê du œ dx and x œ a Ê u œ a and x œ !
Ê u œ !. Thus ' f(x) dx œ ' f(u) du œ ' f(u) du œ ' f(u) du œ ' f(x) dx.
c
Thus ' f(x) dx œ ' f(x) dx ' f(x) dx œ ' f(x) dx ' f(x) dx œ !.
c
c
'c sin x dx œ [cos x]1Î#1Î# œ cos ˆ 1# ‰ cos ˆ 1# ‰ œ ! ! œ !.
0
86. (a) Consider
a
0
0
a
0
a
a
0
a
a
a
a
a
a
0
a
0
0
a
0
0
1/2
(b)
1/2
87. Let u œ a x Ê du œ dx; x œ 0 Ê u œ a, x œ a Ê u œ 0
Iœ
'
' f(u)f(af(au) duu) œ ' f(x)f(af(ax) dxx)
dx
f(x)f(ax)
' f(x)f(af(ax) dxx) œ ' f(x)
' dx œ [x]! œ a 0 œ a.
I I œ ' f(x)f(x)
f(ax) f(ax) dx œ
a
0
f(x) dx
f(x)f(ax)
œ
'
0
a
f(au)
f(au)f(u)
a
Ê
a
0
0
a
a
a
a
0
0
Therefore, 2I œ a Ê I œ
88. Let u œ
'
a
( du) œ
xy
x
"
t
xy
t
dt œ
a
#
0
.
t
Ê du œ xy
t# dt Ê xy du œ
'
y
1
u" du œ 0
'
y
1
"
u
du œ
'
y
1
"
t
"
u
dt Ê u" du œ
du œ
'
1
y
"
t
"
t
dt; t œ x Ê u œ y, t œ xy Ê u œ 1. Therefore,
dt
89. Let u œ x c Ê du œ dx; x œ a c Ê u œ a, x œ b c Ê u œ b
' cc
b c
a c
90. (a)
f(x c) dx œ
'
a
b
f(u) du œ
'
a
b
f(x) dx
(b)
(c)
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
309
310
Chapter 5 Integration
91-94. Example CAS commands:
Maple:
f := x -> x^3/3-x^2/2-2*x+1/3;
g := x -> x-1;
plot( [f(x),g(x)], x=-5..5, legend=["y = f(x)","y = g(x)"], title="#91(a) (Section 5.6)" );
q1 := [ -5, -2, 1, 4 ];
# (b)
q2 := [seq( fsolve( f(x)=g(x), x=q1[i]..q1[i+1] ), i=1..nops(q1)-1 )];
for i from 1 to nops(q2)-1 do
# (c)
area[i] := int( abs(f(x)-g(x)),x=q2[i]..q2[i+1] );
end do;
add( area[i], i=1..nops(q2)-1 );
# (d)
Mathematica: (assigned functions may vary)
Clear[x, f, g]
f[x_] = x2 Cos[x]
g[x_] = x3 x
Plot[{f[x], g[x]}, {x, 2, 2}]
After examining the plots, the initial guesses for FindRoot can be determined.
pts = x/.Map[FindRoot[f[x]==g[x],{x, #}]&, {1, 0, 1}]
i1=NIntegrate[f[x] g[x], {x, pts[[1]], pts[[2]]}]
i2=NIntegrate[f[x] g[x], {x, pts[[2]], pts[[3]]}]
i1 i2
CHAPTER 5 PRACTICE EXERCISES
1. (a) Each time subinterval is of length ?t œ 0.4 sec. The distance traveled over each subinterval, using the
midpoint rule, is ?h œ "# avi vib1 b ?t, where vi is the velocity at the left endpoint and vib1 the velocity at
the right endpoint of the subinterval. We then add ?h to the height attained so far at the left endpoint vi to
arrive at the height associated with velocity vib1 at the right endpoint. Using this methodology we build
the following table based on the figure in the text:
t (sec) 0 0.4 0.8 1.2 1.6 2.0 2.4 2.8 3.2 3.6 4.0 4.4 4.8 5.2 5.6
6.0
v (fps) 0 10 25 55 100 190 180 165 150 140 130 115 105 90
76
65
h (ft)
0 2
9
25 56 114 188 257 320 378 432 481 525 564 592 620.2
t (sec)
v (fps)
h (ft)
6.4
50
643.2
6.8
37
660.6
7.2
25
672
7.6
12
679.4
8.0
0
681.8
NOTE: Your table values may vary slightly from ours depending on the v-values you read from the graph.
Remember that some shifting of the graph occurs in the printing process.
The total height attained is about 680 ft.
(b) The graph is based on the table in part (a).
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Chapter 5 Practice Exercises
311
2. (a) Each time subinterval is of length ?t œ 1 sec. The distance traveled over each subinterval, using the
midpoint rule, is ?s œ "# avi vib1 b ?t, where vi is the velocity at the left, and vib1 the velocity at the
right, endpoint of the subinterval. We then add ?s to the distance attained so far at the left endpoint vi
to arrive at the distance associated with velocity vib1 at the right endpoint. Using this methodology we
build the table given below based on the figure in the text, obtaining approximately 26 m for the total
distance traveled:
t (sec)
0
1
2
3
4
5
6
7
8
9
10
v (m/sec)
0
0.5
1.2
2
3.4
4.5
4.8
4.5
3.5
2
0
s (m)
0
0.25
1.1
2.7
5.4
9.35
14
18.65 22.65 25.4 26.4
(b) The graph shows the distance traveled by the
moving body as a function of time for
0 Ÿ t Ÿ 10.
10
!
3. (a)
kœ1
10
ak
4
œ
"
4
10
"
4
! ak œ
kœ1
(2) œ #"
10
10
kœ1
10
!
(d)
kœ1
kœ1
ˆ 5#
10
bk ‰ œ !
5
#
kœ1
20
20
kœ1
20
kœ1
! ˆ" #
(c)
kœ1
20
kœ1
10
! bk œ
kœ1
2bk ‰
7
20
œ !
kœ1
20
"
#
2
7
kœ1
kœ1
kœ1
kœ1
5
#
(10) 25 œ 0
(b)
20
! bk œ
kœ1
20
kœ1
kœ1
"
#
20
20
20
kœ1
kœ1
kœ1
! (ak bk ) œ ! ak ! bk œ 0 7 œ 7
(20) 27 (7) œ 8
kœ1
5. Let u œ 2x 1 Ê du œ 2 dx Ê
5
1
(2x 1)"Î# dx œ
'
9
1
3
1
x ax# 1b
7. Let u œ
x
2
"Î$
dx œ
'
8
0
"
#
du œ dx; x œ 1 Ê u œ 1, x œ 5 Ê u œ 9
*
u"Î# ˆ "# du‰ œ u"Î# ‘ " œ 3 1 œ 2
6. Let u œ x# 1 Ê du œ 2x dx Ê
"
#
du œ x dx; x œ 1 Ê u œ 0, x œ 3 Ê u œ 8
)
u"Î$ ˆ "# du‰ œ 38 u%Î$ ‘ ! œ
3
8
(16 0) œ 6
Ê 2 du œ dx; x œ 1 Ê u œ 1# , x œ 0 Ê u œ 0
'c cos ˆ x# ‰ dx œ 'c
0
0
1
1Î2
(cos u)(2 du) œ [2 sin u]!1Î# œ 2 sin 0 2 sin ˆ 1# ‰ œ 2(0 (1)) œ 2
8. Let u œ sin x Ê du œ cos x dx; x œ 0 Ê u œ 0, x œ
0
10
! aak 2b œ ! ak ! 2 œ 0 2(20) œ 40
(d)
'
10
10
! 3ak œ 3 ! ak œ 3(0) œ 0
4. (a)
'
10
! (bk 3ak ) œ ! bk 3 ! ak œ 25 3(2) œ 31
! (ak bk 1) œ ! ak ! bk ! " œ 2 25 (1)(10) œ 13
(c)
'
(b)
1Î2
(sin x)(cos x) dx œ
'
0
1
#
"
u du œ ’ u2 “ œ
!
1
#
Ê uœ1
"
#
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
312
Chapter 5 Integration
'c f(x) dx œ "3 'c 3 f(x) dx œ 3" (12) œ 4
(b) ' f(x) dx œ ' f(x) dx ' f(x) dx œ 6 4 œ 2
c
c
c
' g(x) dx œ 'c g(x) dx œ 2
'
'
(d)
(1 g(x)) dx œ 1
g(x) dx œ 1(2) œ 21
c
c
'c Š f(x) 5 g(x) ‹ dx œ 5" 'c f(x) dx 5" 'c g(x) dx œ 5" (6) 5" (2) œ 85
2
9. (a)
2
2
5
2
2
2
(c)
5
5
(e)
'
'
'
10. (a)
2
5
5
5
2
2
2
2
0
(c)
0
2
(e)
' 7 g(x) dx œ "7 (7) œ 1
f(x) dx œ ' f(x) dx œ 1
[g(x) 3 f(x)] dx œ ' g(x) dx 3'
g(x) dx œ
2
"
7
(b)
0
2
(d)
0
2
0
5
2
0
2
0
'
'
2
2
2
5
5
2
2
2
1
2
0
g(x) dx œ
'
0
2
g(x) dx È2 f(x) dx œ È2
'
0
2
'
0
1
g(x) dx œ 1 2 œ 1
f(x) dx œ È2 (1) œ 1È2
f(x) dx œ 1 31
11. x# 4x 3 œ 0 Ê (x 3)(x 1) œ 0 Ê x œ 3 or x œ 1;
Area œ
'
1
0
ax# 4x 3b dx "
$
'
1
3
ax# 4x 3b dx
$
œ ’ x3 2x# 3x“ ’ x3 2x# 3x“
!
œ
$
’Š "3
$
"
#
2(1) 3(1)‹ 0“
$
$
’Š 33 2(3)# 3(3)‹ Š 13 2(1)# 3(1)‹“
œ ˆ "3 1‰ 0 ˆ 3" 1‰‘ œ
12. 1 x#
4
8
3
œ 0 Ê 4 x# 0 Ê x œ „ 2;
Area œ
'c Š1 x4 ‹ dx '
2
œ ’x 2
#
x$
12 “ #
2$
12 ‹
œ ’Š2 3
#
2
’x Š1 x#
4‹
dx
$
x$
12 “ #
Š2 (2)$
12 ‹“
œ 43 ˆ 43 ‰‘ ˆ 34 43 ‰ œ
’Š3 3$
12 ‹
Š2 2$
12 ‹“
13
4
13. 5 5x#Î$ œ 0 Ê 1 x#Î$ œ 0 Ê x œ „ 1;
Area œ
'c ˆ5 5x#Î$ ‰ dx '
1
1
1
8
ˆ5 5x#Î$ ‰ dx
"
)
œ 5x 3x&Î$ ‘ " 5x 3x&Î$ ‘ "
œ ˆ5(1) 3(1)&Î$ ‰ ˆ5(1) 3(1)&Î$ ‰‘
ˆ5(8) 3(8)&Î$ ‰ ˆ5(1) 3(1)&Î$ ‰‘
œ [2 (2)] [(40 96) 2] œ 62
14. 1 Èx œ 0 Ê x œ 1;
Area œ
'
0
1
ˆ1 Èx‰ dx '
1
4
ˆ1 Èx‰ dx
"
%
œ x 23 x$Î# ‘ ! x 23 x$Î# ‘ "
œ ˆ1 23 (1)$Î# ‰ 0‘ ˆ4 23 (4)$Î# ‰ ˆ1 23 (1)$Î# ‰‘
‰ "‘
œ "3 ˆ4 16
3 3 œ 2
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Chapter 5 Practice Exercises
"
x# ,
15. f(x) œ x, g(x) œ
œ
'
2
1
"‰
x#
ˆx '
œ
Š 42
2
1
Šx '
b
[f(x) g(x)] dx
a
#
#
dx œ ’ x# "x “ œ ˆ #4 #" ‰ ˆ #" 1‰ œ 1
"
"
Èx
16. f(x) œ x, g(x) œ
œ
a œ 1, b œ 2 Ê A œ
"
Èx ‹dx
, a œ 1, b œ 2 Ê A œ
#
œ ’ x# 2Èx“
'
b
a
[f(x) g(x)] dx
#
"
7 4 È 2
#
2È2‹ ˆ "# 2‰ œ
'
#
17. f(x) œ ˆ1 Èx‰ , g(x) œ 0, a œ 0, b œ 1 Ê A œ
œ
'
1
ˆ1 2x"Î# x‰ dx œ ’x 43 x$Î# 0
"
x#
# “!
#
x%
#
"
x(
7 “!
œ1
"
#
"
7
œ
b
a
œ1
18. f(x) œ a1 x$ b , g(x) œ 0, a œ 0, b œ 1 Ê A œ
œ ’x '
a
b
'
[f(x) g(x)] dx œ
4
3
"
#
œ
"
6
1
0
ˆ1 Èx‰# dx œ
(6 8 3) œ
[f(x) g(x)] dx œ
'
0
1
œ2
'
0
3
'
d
c
[f(y) g(y)] dy œ
y# dy œ
2
3
'
#
a1 x$ b dx œ
'
1
0
9
14
3
0
a2y# 0b dy
$
cy$ d ! œ 18
20. f(y) œ 4 y# , g(y) œ 0, c œ 2, d œ 2
Ê Aœ
œ ’4y '
c
d
[f(y) g(y)] dy œ
#
y$
3 “ #
œ 2 ˆ8 8‰
3
œ
'
0
1
ˆ1 2Èx x‰ dx
"
6
19. f(y) œ 2y# , g(y) œ 0, c œ 0, d œ 3
Ê Aœ
313
'c a4 y# b dy
2
2
32
3
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
a1 2x$ x' b dx
314
Chapter 5 Integration
y#
4
21. Let us find the intersection points:
y2
4
œ
Ê y# y 2 œ 0 Ê (y 2)(y 1) œ 0 Ê y œ 1
or y œ 2 Ê c œ 1, d œ 2; f(y) œ
Ê Aœ
'
d
c
y2
4
2
#
1
œ
"
4
'c ay 2 y# b dy œ 4" ’ y#
œ
"
4
ˆ 4# 4 38 ‰ ˆ "# 2 3" ‰‘ œ
#
1
y#
4
'c Š y 4 2 y4 ‹ dy
[f(y) g(y)] dy œ
2
, g(y) œ
2y 9
8
y# 4
4
22. Let us find the intersection points:
#
y$
3 “ "
œ
y 16
4
Ê y# y 20 œ 0 Ê (y 5)(y 4) œ 0 Ê y œ 4
or y œ 5 Ê c œ 4, d œ 5; f(y) œ
Ê Aœ
'
d
c
[f(y) g(y)] dy œ
y 16
4
, g(y) œ
y# 4
4
'c Š y 416 y 4 4 ‹ dy
5
#
4
œ
"
4
'c ay 20 y# b dy œ "4 ’ y#
œ
"
4
"
4
125 ‰
ˆ 25
‰‘
ˆ "#6 80 64
# 100 3
3
ˆ 9# 180 63‰ œ 4" ˆ 9# 117‰ œ 8" (9 234) œ
œ
5
#
20y 4
23. f(x) œ x, g(x) œ sin x, a œ 0, b œ
Ê Aœ
'
b
a
#
[f(x) g(x)] dx œ
œ ’ x# cos x“
1Î%
!
#
œ Š 31# '
1Î4
(x sin x) dx
1
24. f(x) œ 1, g(x) œ ksin xk , a œ 1# , b œ
Ê Aœ
œ
'c
œ2
0
'
b
a
[f(x) g(x)] dx œ
(1 sin x) dx 1Î2
1Î2
'
0
'
0
1Î2
243
8
1
4
0
È2
# ‹
&
y$
3 “ %
'c
1Î2
1Î2
1
2
a1 ksin xkb dx
(1 sin x) dx
1Î#
(1 sin x) dx œ 2[x cos x]!
œ 2 ˆ 1# 1‰ œ 1 2
25. a œ 0, b œ 1, f(x) g(x) œ 2 sin x sin 2x
Ê Aœ
'
0
1
(2 sin x sin 2x) dx œ 2 cos x cos 2x ‘ 1
#
!
œ 2 † (1) "# ‘ ˆ2 † 1 "# ‰ œ 4
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Chapter 5 Practice Exercises
315
26. a œ 13 , b œ 13 , f(x) g(x) œ 8 cos x sec# x
'c
1Î3
Ê Aœ
œ Š8 †
1Î3
È3
#
1Î$
a8 cos x sec# xb dx œ [8 sin x tan x]1Î$
È3‹ Š8 †
È3
#
È3‹ œ 6È3
27. f(y) œ Èy, g(y) œ 2 y, c œ 1, d œ 2
'
Ê Aœ
œ
'
1
2
d
c
[f(y) g(y)] dy œ
'
2
1
Èy (2 y)‘ dy
ˆÈy 2 y‰ dy œ ’ 23 y$Î# 2y œ Š 43 È2 4 2‹ ˆ 23 2 "# ‰ œ
4
3
#
y#
# “"
È2 7
6
œ
8 È 2 7
6
28. f(y) œ 6 y, g(y) œ y# , c œ 1, d œ 2
Ê Aœ
'
œ ’6y y#
#
œ4
c
7
3
d
[f(y) g(y)] dy œ
"
#
#
y$
3 “"
œ
'
2
1
a6 y y# b dy
œ ˆ12 2 38 ‰ ˆ6 24 14 3
6
œ
"
#
3" ‰
13
6
29. f(x) œ x$ 3x# œ x# (x 3) Ê f w (x) œ 3x# 6x œ 3x(x 2) Ê f w œ ± ± !
#
Ê f(0) œ 0 is a maximum and f(2) œ 4 is a minimum. A œ 30. A œ
'
a
0
ˆa"Î# x"Î# ‰# dx œ
œ a# ˆ1 4
3
"# ‰ œ
a#
6
'
a
0
&Î$
A# œ
"
y#
“
# !
œ
"
10
(6 8 3) œ
'
0
1
%
&Î$
1
Ê the total area is A" A# œ
!
a
x#
# “0
œ a# 34 Èa † aÈa a#
6
ˆy#Î$ y‰ dy
!
y#
# “ "
œ
$
‰
ax$ 3x# b dx œ ’ x4 x$ “ œ ˆ 81
4 27 œ
; the area below the x-axis is
'c ˆy#Î$ y‰ dy œ ’ 3y5
0
0
3
ˆa 2Èa x"Î# x‰ dx œ ’ax 43 Èa x$Î# 31. The area above the x-axis is A" œ
œ ’ 3y5 '
11
10
6
5
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
a#
#
27
4
316
Chapter 5 Integration
32. A œ
'
1Î4
0
'
31Î2
51Î4
(cos x sin x) dx '
51Î4
1Î4
(sin x cos x) dx
1Î%
(cos x sin x) dx œ [sin x cos x]!
&1Î%
$1Î#
[ cos x sin x]1Î% [sin x cos x]&1Î%
œ ’Š
È2
#
È2
# ‹
(0 1)“ ’Š
’(1 0) Š
33. y œ x# 34. y œ
'
x
0
'
x
1
"
t
È2
#
dt Ê
dy
dx
È2
# ‹“
xœ0 Ê yœ
'
0
0
È2
# ‹
È2
#
Š
œ
8È 2
#
2 œ 4È2 2
"
x
Ê
d# y
dx#
œ 2x ˆ1 2Èsec t‰ dt Ê
È2
#
œ2
"
x#
; y(1) œ 1 œ 1 2Èsec x Ê
dy
dx
d# y
dx#
ˆ1 2Èsec t‰ dt œ 0 and x œ 0 Ê
dy
dx
36. y œ
'c È2 sin# t dt 2 so that dydx œ È2 sin# x; x œ 1
dt 3 Ê
dy
dx
œ
;xœ5 Ê yœ
sin x
x
5
5
sin t
t
1
1
"
t
dt œ 1 and yw (1) œ 2 1 œ 3
œ 1 2Èsec 0 œ 3
'
sin t
t
5
'
œ 2 ˆ "# ‰ (sec x)"Î# (sec x tan x) œ Èsec x (tan x);
35. y œ
x
'
È2
# ‹“
dt 3 œ 3
x
Ê yœ
1
'cc È2 sin# t dt 2 œ 2
1
1
37. Let u œ cos x Ê du œ sin x dx Ê du œ sin x dx
' 2(cos x)"Î# sin x dx œ ' 2u"Î# ( du) œ 2 ' u"Î# du œ 2 Š u"Î#" ‹ C œ 4u"Î# C œ 4(cos x)"Î# C
#
38. Let u œ tan x Ê du œ sec# x dx
' (tan x)$Î# sec# x dx œ ' u$Î# du œ ˆu"Î#"‰ C œ 2u"Î# C œ (tanx)2 "Î# C
#
39. Let u œ 2) 1 Ê du œ 2 d) Ê
"
#
du œ d)
' [2) 1 2 cos (2) 1)] d) œ ' (u 2 cos u) ˆ "# du‰ œ u4
#
#
œ ) ) sin (2) 1) C, where C œ C" 40. Let u œ 2) 1 Ê du œ 2 d) Ê
'Š
œ
41.
42.
"
#
"
È 2 ) 1
"
#
2 sec# (#) 1)‹ d) œ
"
4
sin u C" œ
(2)1)#
4
sin (2) 1) C"
is still an arbitrary constant
du œ d)
' Š È"u 2 sec# u‹ ˆ #" du‰ œ #" ' ˆu"Î# 2 sec# u‰ du
"Î#
Š u " ‹ "# (2 tan u) C œ u"Î# tan u C œ (2) 1)"Î# tan (2) 1) C
#
' ˆt 2t ‰ ˆt 2t ‰ dt œ ' ˆt# t4 ‰ dt œ ' at# 4t# b dt œ t3$ 4 Š t 1" ‹ C œ t3$ 4t C
#
t #
' (t1)t%#1 dt œ ' t#t%2t dt œ ' ˆ t"# t2$ ‰ dt œ ' at# 2t$ b dt œ (t 1)" 2 Š #
‹ C œ "t t"# C
43. Let u œ #t$Î# Ê du œ $Èt dt Ê "$ du œ Èt dt
' Èt sin ˆ#t$Î# ‰dt œ "$ ' sin u du œ "$ cos u C œ "$ cosˆ#t$Î# ‰ C
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Chapter 5 Practice Exercises
44. Let u œ " sec ) Ê du œ sec ) tan) d) Ê
317
' sec ) tan) È" sec ) d) œ ' u"Î# du œ #$ u$Î# C œ #$ a" sec )b$Î# C
'c a3x# 4x 7b dx œ cx$ 2x# 7xd "" œ c1$ 2(1)# 7(1)d c(1)$ 2(1)# 7(1)d œ 6 (10) œ 16
1
45.
1
46.
'
47.
'
48.
'
49.
'
1
0
"
a8s$ 12s# 5b ds œ c2s% 4s$ 5sd ! œ c2(1)% 4(1)$ 5(1)d 0 œ 3
2
4
#
1 v
27
1
'
dv œ
2
#
4v# dv œ c4v" d " œ ˆ #4 ‰ ˆ 14 ‰ œ 2
1
#(
x%Î$ dx œ 3x"Î$ ‘ " œ 3(27)"Î$ ˆ3(1)"Î$ ‰ œ 3 ˆ "3 ‰ 3(1) œ 2
4
dt
1 tÈt
œ
'
4
œ
dt
t$Î#
1
'
4
1
%
50. Let x œ 1 Èu Ê dx œ
'
4
ˆ1 Èu‰"Î#
Èu
1
'
du œ
3
2
2
È4
t$Î# dt œ 2t"Î# ‘ " œ
"
#
u"Î# du Ê 2 dx œ
du
Èu
(2)
È1
œ1
; u œ 1 Ê x œ 2, u œ 4 Ê x œ 3
$
x"Î# (2 dx) œ 2 ˆ 23 ‰ x$Î# ‘ # œ
4
3
ˆ3$Î# ‰ 43 ˆ2$Î# ‰ œ 4È3 83 È2 œ
4
3
Š3È3 2È2‹
51. Let u œ 2x 1 Ê du œ 2 dx Ê 18 du œ 36 dx; x œ 0 Ê u œ 1, x œ 1 Ê u œ 3
'
'
1
36 dx
$
0 (2x1)
œ
3
# $
$
9 ‘
ˆ 9 ‰ ˆ 9 ‰
18u$ du œ ’ "8u
2 “ œ u # " œ 3 # 1 # œ 8
"
1
52. Let u œ 7 5r Ê du œ 5 dr Ê "5 du œ dr; r œ 0 Ê u œ 7, r œ 1 Ê u œ 2
'
1
dr
3
#
0 È(7 5r)
œ
'
1
0
(7 5r)#Î$ dr œ
'
2
#
u#Î$ ˆ 5" du‰ œ 5" 3u"Î$ ‘ ( œ
7
&Î#
"
8
!
#
$Î%
53. Let u œ 1 x#Î$ Ê du œ 23 x"Î$ dx Ê #3 du œ x"Î$ dx; x œ
'
1
1Î8
œ
x"Î$ ˆ1 x#Î$ ‰
$Î#
'
dx œ
0
3Î4
u$Î# ˆ #3 du‰ œ ’ˆ 23 ‰ Š u 5 ‹“
1Î2
0
x$ a1 9x% b
" ˆ 25 ‰
œ 18
16
"Î#
$Î#
dx œ
'
0
1
sin# 5r dr œ
56. Let u œ 4t '
1Î%
0
œ
1
8
1
4
"
36
#Î$
œ
3
4
, x œ 1 Ê u œ 1 1#Î$ œ 0
!
&Î#
œ 53 u&Î# ‘ $Î% œ 53 (0)&Î# ˆ 53 ‰ ˆ 43 ‰
'
51
0
"
16
"
16
"
5
"Î#
#
œ
1
8
1Î4
"
%
Ê u œ 1 9 ˆ #" ‰ œ
25
16
#&Î"'
" "Î# ‘
œ 18
u
"
du œ dr; r œ 0 Ê u œ 0, r œ 1 Ê u œ 51
Ê du œ 4 dt Ê
'
#&Î"'
"
#
"
90
asin# ub ˆ "5 du‰ œ
31Î4
du œ x$ dx; x œ 0 Ê u œ 1, x œ
"
"
u$Î# ˆ 36
du‰ œ ’ 36
Š u " ‹“
"
ˆ 18
(1)"Î# ‰ œ
cos# ˆ4t 14 ‰ dt œ
25Î16
1
55. Let u œ 5r Ê du œ 5 dr Ê
'
Ê u œ 1 ˆ 8" ‰
27È3
160
54. Let u œ 1 9x% Ê du œ 36x$ dx Ê
'
3
3
7È
2‹
ŠÈ
3
5
"
4
"
5
2u sin 2u ‘ &1
4
!
œ ˆ 1# sin 101 ‰
#0
du œ dt; t œ 0 Ê u œ 14 , t œ
acos# ub ˆ "4 du‰ œ
"
4
u2 sin 2u ‘ $1Î%
4
1Î%
œ
ˆ0 1
4
"
4
sin 0 ‰
20
Ê uœ
Š 381 œ
1
#
31
4
sin ˆ 3#1 ‰
‹
4
4" Š 18 Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
sin ˆ 1# ‰
‹
4
318
Chapter 5 Integration
'
1Î$
57.
'
31Î4
58.
0
1Î$
sec# ) d) œ [tan )]!
1Î4
31
1
1
3
tan 0 œ È3
$1Î%
csc# x dx œ [cot x]1Î% œ ˆ cot
59. Let u œ
'
œ tan
cot#
x
6
x
6
Ê du œ
dx œ
'
"
6
ˆ cot 14 ‰ œ 2
dx Ê 6 du œ dx; x œ 1 Ê u œ 16 , x œ 31 Ê u œ
1Î2
1Î6
31 ‰
4
'
6 cot# u du œ 6
1Î2
1
#
1Î#
acsc# u 1b du œ [6(cot u u)]1Î' œ 6 ˆ cot
1Î6
1
#
1# ‰ 6 ˆcot
1
6
16 ‰
œ 6È3 21
60. Let u œ
'
1
0
tan#
)
3
)
3
Ê du œ
d) œ
'
1
0
"
3
d) Ê 3 du œ d); ) œ 0 Ê u œ 0, ) œ 1 Ê u œ
ˆsec#
)
3
1‰ d) œ
'
1Î3
0
1
3
1Î$
3 asec# u 1b du œ [3 tan u 3u]!
œ 3 tan
1
3
3 ˆ 13 ‰‘ (3 tan 0 0)
œ 3È3 1
'c
sec x tan x dx œ [sec x]!1Î$ œ sec 0 sec ˆ 13 ‰ œ 1 2 œ 1
'
csc z cot z dz œ [csc z]1Î% œ ˆ csc
0
61.
62.
1Î3
31Î4
1Î4
$1Î%
31 ‰
4
ˆ csc 14 ‰ œ È2 È2 œ 0
63. Let u œ sin x Ê du œ cos x dx; x œ 0 Ê u œ 0, x œ
'
1Î2
0
'
5(sin x)$Î# cos x dx œ
1
0
1
#
Ê uœ1
"
"
5u$Î# du œ 5 ˆ 25 ‰ u&Î# ‘ ! œ 2u&Î# ‘ ! œ 2(1)&Î# 2(0)&Î# œ 2
64. Let u œ 1 x# Ê du œ 2x dx Ê du œ 2x dx; x œ 1 Ê u œ 0, x œ 1 Ê u œ 0
'c
1
1
'
2x sin a1 x# b dx œ
0
sin u du œ 0
0
65. Let u œ sin 3x Ê du œ 3 cos 3x dx Ê
'c ÎÎ
1 2
15 sin% 3x cos 3x dx œ
1 2
'
1
1
"
3
du œ cos 3x dx; x œ 1# Ê u œ sin ˆ 3#1 ‰ œ 1, x œ
15u% ˆ "3 du‰ œ
'
1
1
1
#
"
5u% du œ cu& d " œ (1)& (1)& œ 2
66. Let u œ cos ˆ x# ‰ Ê du œ "# sin ˆ x# ‰ dx Ê 2 du œ sin ˆ x# ‰ dx; x œ 0 Ê u œ cos ˆ 0# ‰ œ 1, x œ
'
21Î3
0
cos% ˆ x# ‰ sin ˆ x# ‰ dx œ
'
1
1Î2
$
u% (2 du) œ ’2 Š u3 ‹“
67. Let u œ 1 3 sin# x Ê du œ 6 sin x cos x dx Ê
'
1Î2
0
3 sin x cos x
È1 3 sin# x
dx œ
'
4
1
"
Èu
ˆ #" du‰ œ
'
4
1
68. Let u œ 1 7 tan x Ê du œ 7 sec# x dx Ê
Ê u œ 1 7 tan
'
0
1Î4
#
sec x
(1 7 tan x)#Î$
1
4
"
#
Ê u œ sin ˆ 3#1 ‰ œ 1
"Î#
"
œ
2
3
ˆ "# ‰$ 32 (1)$ œ
2
3
"Î#
#
"
1
#
21
Ê u œ cos Š #3 ‹ œ
(8 1) œ
du œ 3 sin x cos x dx; x œ 0 Ê u œ 1, x œ
%
21
3
14
3
Ê u œ 1 3 sin#
%
"
#
u"Î# du œ ’ 2" Š u " ‹“ œ u"Î# ‘ " œ 4"Î# 1"Î# œ 1
"
7
du œ sec# x dx; x œ 0 Ê u œ 1 7 tan 0 œ 1, x œ
1
4
œ8
dx œ
'
1
8
"
u#Î$
ˆ 7" du‰ œ
'
1
8
"
7
"Î$
)
3
"
)
u#Î$ du œ ’ 7" Š u " ‹“ œ 37 u"Î$ ‘ " œ
3
7
(8)"Î$ 37 (1)"Î$ œ
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
"
#
3
7
1
#
œ4
Chapter 5 Practice Exercises
69. Let u œ sec ) Ê du œ sec ) tan ) d); ) œ 0 Ê u œ sec 0 œ 1, ) œ
'
1Î3
0
tan )
È2 sec )
"
È2
œ
'
d) œ
"Î#
#
#
"
'
1Î3
0
’ ˆu " ‰ “ œ ’
1Î3
sec ) tan )
sec ) tan )
d) œ
È2 (sec ))$Î#
sec ) È2 sec )
0
#
2
2
2
È2u “ œ È2(2) Š È2(1) ‹ œ
"
cos Èt
2È t
70. Let u œ sin Èt Ê du œ ˆcos Èt‰ ˆ "# t"Î# ‰ dt œ
#
1
4
tœ
'
1 Î4
#
Ê u œ sin
71. (a) av(f) œ
'
"
1 (1)
'c
"
k (k)
(b) av(f) œ
1
"
1Î2 Èu
1
1
'c
k
k
(2 du) œ 2
"
30
73. favw œ
"
#k
0
3
0
0
"
ba
'
a
b
1
"
È2 u$Î#
du œ
"
È2
'
1
1
3
2
œ2
u$Î# du
È2 1
dt Ê 2 du œ
cos Èt
Èt
dt; t œ
1#
36
Ê u œ sin
1
6
œ
"
#
,
a
"
b a
"
b a
"
#
’ mx2 bx“
Èa x"Î# dx œ
[f(x)]ab œ
"
’ mx2 bx“
È3 x"Î# dx œ
0
f w (x) dx œ
"
u"Î# du œ 4Èu‘ "Î# œ 4È1 4É #" œ 2 Š2 È2‹
#
(mx b) dx œ
3
1
1Î2
"
#
a
(b) yav
'
(mx b) dx œ
' È3x dx œ "3 '
œ a " 0 ' Èax dx œ "a '
72. (a) yav œ
'
Ê u œ sec
œ1
dt œ
cos Èt
Ét sin Èt
1# Î36
1
#
d) œ
2
1
3
319
k
ck
#
#
m(1)
b(1)‹“ œ
’Š m(1)
2 b(1)‹ Š #
œ
"
#
œ
"
#k
#
"
#
#
m(k)
b(k)‹“ œ
’Š m(k)
2 b(k)‹ Š #
(2b) œ b
"
#k
(2bk) œ b
È3
3
23 x$Î# ‘ $ œ
!
È3
3
23 (3)$Î# 23 (0)$Î# ‘ œ
È3
3
Š2È3‹ œ 2
Èa
a
32 x$Î# ‘ a œ
!
Èa
a
ˆ 32 (a)$Î# 32 (0)$Î# ‰ œ
Èa
a
ˆ 32 aÈa‰ œ
[f(b) f(a)] œ
f(b) f(a)
ba
2
3
a
so the average value of f w over [aß b] is the
slope of the secant line joining the points (aß f(a)) and (bß f(b)), which is the average rate of change of f over [aß b].
74. Yes, because the average value of f on [aß b] is
and the average value of the function is
"
#
'
a
"
ba
'
a
b
f(x) dx. If the length of the interval is 2, then b a œ 2
b
f(x) dx.
75. We want to evaluate
"
$'& !
'
$'&
!
f(x) dx œ
"
$'&
'
$'&
!
#1
Œ$(sin” $'& ax "!"b• #&dx œ
#1
Notice that the period of y œ sin” $'&
ax "!"b• is
length 365. Thus the value of
76.
"
'(&#!
œ
'#
'(&
!
$(
$'&
'
$'&
!
"
'&& Œ”)Þ#(a'(&b
#'a'(&b
#†"!&
#1
$'&
"Þ)(a'(&b
$†"!&
$
'
!
$'&
#1
sin” $'&
ax "!"b•dx #&
$'&
'
$'&
!
dx
œ $'& and that we are integrating this function over an iterval of
#1
ax "!"b•dx sin” $'&
a)Þ#( "!& a#'T "Þ)(T# bbdT œ
#
#1
$(
$'&
"
'&& ”)Þ#(T
• ”)Þ#(a#!b #&
$'&
#'T#
#†"!&
#'a#!b
#†"!&
#
'
!
$'&
dx is
"Þ)(T$
$†"!& •
"Þ)(a#!b
$†"!&
$
$(
$'&
†!
#&
$'&
† $'& œ #&.
'(&
#!
• ¸
"
'&& a$(#%Þ%%
"'&Þ%!b
œ &Þ%$ œ the average value of Cv on [20, 675]. To find the temperature T at which Cv œ &Þ%$, solve
&Þ%$ œ )Þ#( "!& a#'T "Þ)(T# b for T. We obtain "Þ)(T# #'T #)%!!! œ !
ÊTœ
#' „ Éa#'b# %a"Þ)(ba#)%!!!b
È#"#%**'
œ #' „ $Þ(%
.
#a"Þ)(b
‰
So T œ $)#Þ)# or T œ $*'Þ(#. Only T œ $*'Þ(# lies in the
interval [20, 675], so T œ $*'Þ(# C.
77.
dy
dx
œ È2 cos$ x
78.
dy
dx
œ È2 cos3 a7x2 b †
d
2
dx a7x b
œ 14xÈ2 cos3 a7x2 b
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
320
Chapter 5 Integration
79.
dy
dx
œ
d
dx Œ
80.
dy
dx
œ
d
dx Œ
' x $ ' t dt œ $'x
%
1
%
'sec# x t " " dt œ dxd Œ'#sec x t " " dt œ sec "x " dxd asec xb œ sec" xsectan xx
#
#
#
#
81. Yes. The function f, being differentiable on [aß b], is then continuous on [aß b]. The Fundamental Theorem of
Calculus says that every continuous function on [aß b] is the derivative of a function on [aß b].
82. The second part of the Fundamental Theorem of Calculus states that if F(x) is an antiderivative of f(x) on [aß b], then
'
a
b
f(x) dx œ F(b) F(a). In particular, if F(x) is an antiderivaitve of È1 x% on [0ß 1], then
'
0
1
È1 x% dx
œ F(1) F(0).
83. y œ
'x1 È1 t# dt œ '1x È1 t# dt
84. y œ
'
0
"
#
cos x 1 t
dt œ '
cos x
0
"
1 t#
dt Ê
Ê
dy
dx
œ
d
dx
”
dy
dx
œ
d
dx
”
'
'1x È1 t# dt• œ dxd ”'1x È1 t# dt• œ È1 x#
cos x
0
"
d
‰ ˆ dx
œ ˆ 1 cos
(cos x)‰ œ ˆ sin"# x ‰ ( sin x) œ
#x
"
1 t#
"
sin x
d
dt• œ dx
”
'
cos x
0
"
1 t#
dt•
œ csc x
85. We estimate the area A using midpoints of the vertical intervals, and we will estimate the width of the parking lot on each
interval by averaging the widths at top and bottom. This gives the estimate
A ¸ "& † ˆ ! # $' $' # &% &% # &" &" #%*Þ& %*Þ&# &% &% #'%Þ% '%Þ% # '(Þ& '(Þ&# %# ‰
A ¸ &*'" ft# . The cost is Area † ($2.10/ft# ) ¸ a5961 ft# b a$2.10/ft# b œ $12,518.10 Ê the job cannot be done for $11,000.
86. (a) Before the chute opens for A, a œ 32 ft/sec# . Since the helicopter is hovering, v! œ 0 ft/sec
Ê v œ ' 32 dt œ 32t v! œ 32t. Then s! œ 6400 ft Ê s œ ' 32t dt œ 16t# s! œ 16t# 6400.
At t œ 4 sec, s œ 16(4)# 6400 œ 6144 ft when A's chute opens;
(b) For B, s! œ 7000 ft, v! œ 0, a œ 32 ft/sec# Ê v œ ' 32 dt œ 32t v! œ 32t Ê s œ ' 32t dt
œ 16t# s! œ 16t# 7000. At t œ 13 sec, s œ 16(13)# 7000 œ 4296 ft when B's chute opens;
(c) After the chutes open, v œ 16 ft/sec Ê s œ ' 16 dt œ 16t s! . For A, s! œ 6144 ft and for B,
s! œ 4296 ft. Therefore, for A, s œ 16t 6144 and for B, s œ 16t 4296. When they hit the ground,
4296
s œ 0 Ê for A, 0 œ 16t 6144 Ê t œ 6144
16 œ 384 seconds, and for B, 0 œ 16t 4296 Ê t œ 16
œ 268.5 seconds to hit the ground after the chutes open. Since B's chute opens 58 seconds after A's opens
Ê B hits the ground first.
CHAPTER 5 ADDITIONAL AND ADVANCED EXERCISES
1. (a) Yes, because
'
0
1
f(x) dx œ
(b) No. For example,
'
0
1
(b) True:
'
0
1
7f(x) dx œ
"
7
(7) œ 1
"
8x dx œ c4x# d ! œ 4, but
'
0
1
"
È8x dx œ ’2È2 Š x$Î#
œ
3 ‹“
!
#
' f(x) dx œ ' f(x) dx œ 3
'c [f(x) g(x)] dx œ 'c f(x) dx 'c g(x) dx œ 'c f(x) dx '
2
2. (a) True:
"
7
5
4È 2
3
ˆ1$Î# 0$Î# ‰ œ
4È 2
3
Á È4
5
2
5
5
5
2
2
2
2
2
2
5
f(x) dx 'c g(x) dx œ 4 3 2 œ 9
5
2
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Chapter 5 Additional and Advanced Exercises
'c f(x) dx œ 4 3 œ 7 2 œ 'c g(x) dx
(c) False:
5
5
2
2
œ
"
a
'
0
sin ax
a
x
sin ax
a
'
0
'
d
Πdx
œ cos ax
0
'
x
2
2
5
0
2
x
x
0
0
x
0
0
'
f(t) cos at dt sin ax
f(t) cos at dt 0
5
x
f(t) cos at dt x
5
' f(t) sin ax cos at dt "a ' f(t) cos ax sin at dt
cos ax '
' f(t) cos at dt
f(t) sin at dt Ê dy
a
dx œ cos ax Œ
"
a
f(t) sin a(x t) dt œ
x
'c [f(x) g(x)] dx 0 Ê 'c [g(x) f(x)] dx 0.
Ê ' [g(x) f(x)] dx 0 which is a contradiction.
c
Ê
On the other hand, f(x) Ÿ g(x) Ê [g(x) f(x)]
3. y œ
x
0
f(t) sin at dt '
(f(x) cos ax) sin ax
sin ax
a
cos ax
a
x
'
d
Πdx
x
f(t) sin at dt
0
f(t) sin at dt 0
cos ax
a
(f(x) sin ax)
' f(t) cos at dt sin ax ' f(t) sin at dt. Next,
d y
' f(t) cos at dt (cos ax) Πdxd ' f(t) cos at dt a cos ax '
dx œ a sin ax
Ê
x
œ cos ax
dy
dx
x
0
x
#
#
0
x
0
'
d
(sin ax) Πdx
a cos ax
'
0
x
0
x
0
4. x œ
'
"
#
œ
'
x
0
"
0 È1 4t#
'
d
dx
(x) œ
Š dy
dx ‹ Ê
"Î#
x
0
f(t) cos at dt dt Ê
"
È14y#
a1 4y# b
x
0
'
'
d
dx
x
0
y
0
'
"
È1 4t#
dy
4y Š dx
‹
(8y) Š dy
dx ‹ œ
È1 4y#
0
'
x
0
x
f(t) cos at dt a cos ax
'
0
x
f(t) sin at dt f(x).
f(t) cos at dt f(x)
f(t) sin at dt œ f(x). Note also that yw (0) œ y(0) œ 0.
dt œ
œ È1 4y# . Then
dy
dx
f(t) sin at dt
f(t) cos at dt (cos ax)f(x) cos ax
f(t) sin at dt a sin ax
cos ax
a
x
0
f(t) sin at dt (sin ax)f(x) sin ax œ a sin ax
y
Ê 1œ
'
f(t) sin at dt œ a sin ax
Therefore, yww a# y œ a cos ax
a# Πsinaax
321
œ
'
d
dy
”
d# y
dx#
œ
4y ˆÈ1 4y# ‰
È1 4y#
y
0
"
È1 4t#
d
dx
ˆÈ1 4y# ‰ œ
dt• Š dy
dx ‹ from the chain rule
œ 4y. Thus
d# y
dx#
ˆÈ1 4y# ‰ Š dy
dx ‹
d
dy
œ 4y, and the constant of
proportionality is 4.
5. (a)
'
0
x#
f(t) dt œ x cos 1x Ê
Ê f ax# b œ
(b)
'
0
f(x)
d
dx
cos 1x 1x sin 1x
.
2x
$
t# dt œ ’ t3 “
f(x)
!
œ
"
3
'
x#
0
f(t) dt œ cos 1x 1x sin 1x Ê f ax# b (2x) œ cos 1x 1x sin 1x
cos 21 21 sin 21
4
Thus, x œ 2 Ê f(4) œ
(f(x))$ Ê
"
3
"
4
œ
3
(f(x))$ œ x cos 1x Ê (f(x))$ œ 3x cos 1x Ê f(x) œ È
3x cos 1 x
3
3
Ê f(4) œ È
3(4) cos 41 œ È
12
6.
'
a
0
f(x) dx œ
a#
#
a
#
sin a Ê f(a) œ Fw (a) œ a 7.
'
1
b
"
#
1
#
cos a. Let F(a) œ
sin a a
#
cos a f(x) dx œ Èb# 1 È2 Ê f(b) œ
d
db
1
#
'
a
0
f(t) dt Ê f(a) œ Fw (a). Now F(a) œ
sin a Ê f ˆ 1# ‰ œ
'
b
1
f(x) dx œ
"
#
side of the equation is:
d
dx
”
'
0
x
f(u)(x u) du• œ
d
dx
”
' ”'
d
dx
0
'
0
x
ab# 1b
x
8. The derivative of the left side of the equation is:
1
#
0
u
"
#
1
#
sin
"Î#
d
dx
#
#
(2b) œ
f(t) dt• du• œ
f(u) x du ˆ1‰
'
0
'
0
cos
b
È b# 1
1
#
a#
#
1
#
a
#
sin a sin
1
#
Ê f(x) œ
œ
1
#
1
#
cos a
"
#
1
#
x
È x# 1
x
f(t) dt; the derivative of the right
x
u f(u) du
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
œ
"
#
322
Chapter 5 Integration
œ
d
dx
œ
'
'
”x
0
x
f(u) du• '
d
dx
x
0
u f(u) du œ
'
x
0
d
f(u) du x ” dx
dy
dx
0
x
f(u) du• xf(x) œ
'
0
x
f(u) du xf(x) xf(x)
x
0
f(u) du. Since each side has the same derivative, they differ by a constant, and since both sides equal 0
when x œ 0, the constant must be 0. Therefore,
9.
'
' ”'
x
0
0
u
f(t) dt• du œ
'
0
x
f(u)(x u) du.
œ 3x# 2 Ê y œ ' a3x# 2b dx œ x$ 2x C. Then (1ß 1) on the curve Ê 1$ 2(1) C œ 1 Ê C œ 4
Ê y œ x$ 2x 4
10. The acceleration due to gravity downward is 32 ft/sec# Ê v œ ' 32 dt œ 32t v! , where v! is the initial
velocity Ê v œ 32t 32 Ê s œ ' (32t 32) dt œ 16t# 32t C. If the release point, at t œ !, is s œ 0, then
C œ 0 Ê s œ 16t# 32t. Then s œ 17 Ê 17 œ 16t# 32t Ê 16t# 32t 17 œ 0. The discriminant of this
quadratic equation is 64 which says there is no real time when s œ 17 ft. You had better duck.
11.
'c f(x) dx œ 'c x#Î$ dx '
œ
œ
œ
12.
3
0
8
8
3
4 dx
0
35 x&Î$ ‘ ! [4x]!$
)
ˆ0 35 (8)&Î$ ‰ (4(3)
36
5
0) œ
'c f(x) dx œ 'c Èx dx '
3
0
4
4
3
$
$
œ 23 (x)$Î# ‘ % ’ x3 4x“
œ 0 ˆ (4)
2
3
œ
13.
'
3œ
7
3
g(t) dt œ
'
16
3
2
0
#
1
0
$Î# ‰‘
t dt "
$
’ Š 33
'
1
2
12
ax# 4b dx
0
!
96
5
!
4(3)‹ 0 “
sin 1t dt
#
œ ’ t2 “ 1" cos 1t‘ "
!
œ ˆ "# 0‰ 1" cos 21 ˆ 1" cos 1‰‘
"
#
œ
14.
'
0
2
2
1
h(z) dz œ
'
0
1
È1 z dz '
1
2
(7z 6)"Î$ dz
"
#
3
œ 23 (1 z)$Î# ‘ ! 14
(7z 6)#Î$ ‘ "
œ 23 (1 1)$Î# ˆ 23 (1 0)$Î# ‰‘
3
14
(7(2) 6)#Î$ 3 ‰
55
œ 23 ˆ 67 14
œ 42
3
14
(7(1) 6)#Î$ ‘
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Chapter 5 Additional and Advanced Exercises
'c f(x) dx œ 'cc dx 'c a1 x# b dx '
2
15.
1
2
1
2
1
"
x$
3 “ "
œ [x]"
# ’x 1$
3‹
16.
ˆ 23 ‰ 4 2 œ
2
3
'c h(r) dr œ 'c r dr '
2
0
1
1
#
œ ’ r2 “
!
"
"
#
’2(2) 2(1)“
13
3
a1 r# b dr '
2
dr
1
"
!
2
3
Š Š1 1œ
17. Ave. value œ
œ
(1)$
3 ‹•
Š1 ’r r3 “ [r]#"
(1)#
# ‹
œ "# 1
0
$
œ Š0 2 dx
1
[2x]#"
œ a1 (2)b ”Š1 œ1
2
"
ba
1$
3‹
0‹ a2 1b
7
6
'
b
a
#
f(x) dx œ
"
#0
#
'
2
0
#
’Š 1# 0‹ Š 2# 2‹ Š 1# 1‹“ œ
18. Ave. value œ
"
ba
'
b
a
f(x) dx œ
"
30
'
3
0
"
#
f(x) dx œ
”
'
1
0
x dx '
2
1
(x 1) dx• œ
"
#
"
#
#
’ x2 “ #" ’ x2 x“
!
#
"
"
#
f(x) dx œ
"
3
”
'
0
1
dx '
1
2
'
0 dx 19. Let f(x) œ x& on [0ß 1]. Partition [0ß 1] into n subintervals with ?x œ
2
10
n
3
dx• œ
"
3
[1 0 0 3 2] œ
œ "n . Then "n , n2 , á ,
_
&
n
n
2
3
are the
right-hand endpoints of the subintervals. Since f is increasing on [0ß 1], U œ ! Š nj ‹ ˆ n" ‰ is the upper sum for
œ
j 1
_
&
! Š j ‹ ˆ " ‰ œ lim
f(x) œ x& on [0ß 1] Ê n lim
n
n
Ä_
nÄ_
jœ1
œ
'
1
'
"
x& dx œ ’ x6 “ œ
!
0
"
n
&
&
&
’ˆ n" ‰ ˆ n2 ‰ á ˆ nn ‰ “ œ n lim
’1
Ä_
&
2& á n&
“
n'
"
6
20. Let f(x) œ x$ on [0ß 1]. Partition [0ß 1] into n subintervals with ?x œ
10
n
œ "n . Then "n , n2 , á ,
_
$
n
n
are the
right-hand endpoints of the subintervals. Since f is increasing on [0ß 1], U œ ! Š nj ‹ ˆ n" ‰ is the upper sum for
_
$
!
f(x) œ x on [0ß 1] Ê n lim
Ä_
jœ1
œ
'
0
1
%
"
x$ dx œ ’ x4 “ œ
!
œ
j 1
Š nj ‹
$
ˆ "n ‰
œ
lim "
nÄ_ n
$
’ˆ n" ‰
ˆ n2 ‰$
$
á ˆ nn ‰ “ œ n lim
’1
Ä_
$
2$ á n$
“
n%
"
4
21. Let y œ f(x) on [0ß 1]. Partition [0ß 1] into n subintervals with ?x œ
10
n
œ "n . Then "n , n2 , á ,
_
right-hand endpoints of the subintervals. Since f is continuous on [!ß 1], !
œ
j 1
_
! f Š j ‹ ˆ " ‰ œ lim
y œ f(x) on [0ß 1] Ê n lim
n
n
Ä_
nÄ_
jœ1
"
22. (a) n lim
[2 4 6 á 2n] œ n lim
Ä _ n#
Ä_
on [0ß 1] (see Exercise 21)
"
n
"
n
f Š nj ‹ ˆ n" ‰
4
n
6
n
á 2n ‘
n
œ
'
0
1
are the
is a Riemann sum of
f ˆ n" ‰ f ˆ 2n ‰ á f ˆ nn ‰‘ œ
2n n
n
'
0
1
f(x) dx
"
2x dx œ cx# d! œ 1, where f(x) œ 2x
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
323
324
Chapter 5 Integration
"
(b) n lim
c1"& 2"& á n"& d œ n lim
Ä _ n"'
Ä_
"&
f(x) œ x on [0ß 1] (see Exercise 21)
"&
"&
"&
’ˆ n1 ‰ ˆ n2 ‰ á ˆ nn ‰ “ œ
"
n
'
'
1
0
"'
"
"
16 ,
x"& dx œ ’ x16 “ œ
!
where
1
"
" (c) n lim
sin 1n sin 2n1 á sin nn1 ‘ œ
sin n1 dx œ 1" cos 1x‘ ! œ 1" cos 1 ˆ 1" cos 0‰
Ä_ n
0
œ 12 , where f(x) œ sin 1x on [0ß 1] (see Exercise 21)
"
(d) n lim
c1"& 2"& á n"& d œ Šn lim
Ä _ n"(
Ä_
" ‰
œ 0 ˆ 16
œ 0 (see part (b) above)
"
n"&
(e) n lim
Ä_
c1"& 2"& á n"& d œ n lim
Ä_
œ Šn lim
n‹ Šn lim
Ä_
Ä_
"
n"'
"
"
n ‹ Šn lim
Ä _ n"'
c1"& 2"& á n"& d‹ œ Šn lim
Ä_
"
n‹
'
1
0
x"& dx
c1"& 2"& á n"& d
n
n"'
c1"& 2"& á n"& d‹ œ Šn lim
n‹
Ä_
'
1
0
x"& dx œ _ (see part (b) above)
23. (a) Let the polygon be inscribed in a circle of radius r. If we draw a radius from the center of the circle (and
the polygon) to each vertex of the polygon, we have n isosceles triangles formed (the equal sides are equal
to r, the radius of the circle) and a vertex angle of )n where )n œ 2n1 . The area of each triangle is
An œ
" #
# r
sin )n Ê the area of the polygon is A œ nAn œ
nr#
#
(b) n lim
A œ n lim
Ä_
Ä_
sin
21
n
œ n lim
Ä_
n 1 r#
21
sin
21
n
nr#
#
nr#
21
# sin n .
2
1
sin ˆ n ‰
#
ˆ 2n1 ‰ œ a1r b
sin )n œ
œ n lim
a1 r # b
Ä_
lim
2 1 În Ä 0
sin ˆ 2n1 ‰
ˆ 2n1 ‰
œ 1 r#
24. Partition [!ß 1] into n subintervals, each of length ?x œ 1n with the points x! œ 0, x" œ n" , x# œ n2 , á , xn œ nn œ 1.
The inscribed rectangles so determined have areas
#
#
#
f(x! ) ?x œ (0)# ?x, f(x" ) ?x œ ˆ "n ‰ ?x, f(x# ) ?x œ ˆ n2 ‰ ?x, á , f(xnc1 ) œ ˆ n n 1 ‰ ?x. The sum of these areas
#
#
#
#
is Sn œ Š0# ˆ "n ‰ ˆ n2 ‰ á ˆ n n 1 ‰ ‹ ?x œ Š "n# lim
nÄ_
'
ga$b œ '
25. (a) ga"b œ
(b)
#
Sn œ n lim
Š "n$ Ä_
1
1
$
1
(c) ga"b œ
2#
n$
á (n ")#
n$ ‹
œ
'
1
0
2#
n#
x# dx œ
á 1$
3
(n 1)# "
n# ‹ n
œ
1#
n$
2#
n$
á (n ")#
n$ .
Then
œ 3" .
fatb dt œ !
fatb dt œ "# a#ba"b œ "
' fatb dt œ ' fatb dt œ "% a1 ## b œ 1
1
1
1
1
(d) gw axb œ faxb œ ! Ê x œ $, ", $ and the sign chart for gw axb œ faxb is
relative maximum at x œ ".
(e) gw a"b œ fa"b œ # is the slope and ga"b œ
± ± ± . So g has a
3
1
3
' fatb dt œ 1, by (c). Thus the equation is y 1 œ #ax "b
1
1
y œ #x # 1 .
(f) gww axb œ f w axb œ ! at x œ " and gww axb œ f w axb is negative on a$ß "b and positive on a"ß "b so there is an
inflection point for g at x œ ". We notice that gww axb œ f w axb ! for x on a"ß #b and gww axb œ f w axb ! for x on
a#ß %b, even though gww a#b does not exist, g has a tangent line at x œ #, so there is an inflection point at x œ #.
(g) g is continuous on Ò$ß %Ó and so it attains its absolute maximum and minimum values on this interval. We saw in (d)
that gw axb œ ! Ê x œ $, ", $. We have that
ga$b œ
' $ fatb dt œ '$" fatb dt œ 1##
1
#
œ #1
' fatb dt œ !
$
ga$b œ ' fatb dt œ "
%
ga%b œ ' fatb dt œ " "# † " † " œ "#
ga"b œ
1
1
1
1
Thus, the absolute minimum is #1 and the absolute maximum is !. Thus, the range is Ò#1ß !Ó.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Chapter 5 Additional and Advanced Exercises
325
'x cos 2t dt " œ sin x ' x cos 2t dt " Ê yw œ cos x cosa2xb; when x œ 1 we have
1
yw œ cos 1 cosa21b œ " " œ #. And yww œ sin x 2sina2xb; when x œ 1, y œ sin 1 ' cos 2t dt "
x
1
26. y œ sin x 1
œ ! ! " œ ".
27. f(x) œ
'
28. f(x) œ
'
œ
x
"
t
1/x
dt Ê f w (x) œ
sin x
"
1 t#
"
sin x
cos x
"
cos x
'ÈÈ sin t# dt
30. f(x) œ
'
y
x
xb3
y
dx ‰
d ˆ " ‰‰
ˆ dx
Š "" ‹ ˆ dx
œ
x
x
"
x
x ˆ x"# ‰ œ
"
x
"
x
œ
2
x
"
d
"
d
‰ ˆ dx
‰ ˆ dx
dt Ê f w (x) œ ˆ 1 sin
(sin x)‰ ˆ 1 cos
(cos x)‰ œ
#x
#x
29. g(y) œ
2
"
x
cos x
cos# x
sin x
sin# x
#
#
d ˆ
d ˆ
Èy‰‹ œ
Ê gw (y) œ Šsin ˆ2Èy‰ ‹ Š dy
2Èy‰‹ Šsin ˆÈy‰ ‹ Š dy
sin 4y
Èy
sin y
2È y
d
‰
t(5 t) dt Ê f w (x) œ (x 3)(& (x 3)) ˆ dx
(x 3)‰ x(5 x) ˆ dx
dx œ (x 3)(2 x) x(5 x)
œ 6 x x# 5x x# œ 6 6x. Thus f w (x) œ 0 Ê 6 6x œ 0 Ê x œ 1. Also, f ww (x) œ 6 ! Ê x œ 1 gives a
maximum.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
326
Chapter 5 Integration
NOTES
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
CHAPTER 6 APPLICATIONS OF DEFINITE INTEGRALS
6.1 VOLUMES USING CROSS-SECTIONS
(diagonal)#
#
1. A(x) œ
œ
ˆ È x ˆ È x ‰ ‰ #
#
œ 2x; a œ 0, b œ 4;
V œ 'a A(x) dx œ '0 2x dx œ cx# d ! œ 16
b
4
1(diameter)#
4
2. A(x) œ
œ
%
1 c a2 x # b x # d
4
#
œ
1c2 a1 x# bd
4
#
œ 1 a1 2x# x% b ; a œ 1, b œ 1;
V œ 'a A(x) dx œ 'c1 1 a1 2x# x% b dx œ 1 ’x 23 x$ b
1
"
x&
5 “ "
#
œ 21 ˆ1 2
3
"5 ‰ œ
161
15
#
3. A(x) œ (edge)# œ ’È1 x# ŠÈ1 x# ‹“ œ Š2È1 x# ‹ œ 4 a1 x# b ; a œ 1, b œ 1;
V œ 'a A(x) dx œ 'c1 4a1 x# b dx œ 4 ’x b
1
#
(diagonal)#
#
4. A(x) œ
œ
œ
#
Š2È1 x# ‹
V œ 'a A(x) dx œ 2'c1 a1 x# b dx œ 2 ’x 1
"
#
5. (a) STEP 1) A(x) œ
#
"
x$
3 “ "
(side) † (side) † ˆsin 13 ‰ œ
STEP 2) a œ 0, b œ 1
œ 8 ˆ1 "3 ‰ œ
16
3
#
’È1 x# ŠÈ1 x# ‹“
b
"
x$
3 “ "
"
#
œ 2 a1 x# b; a œ 1, b œ 1;
œ 4 ˆ1 "3 ‰ œ
8
3
† Š2Èsin x‹ † Š2Èsin x‹ ˆsin 13 ‰ œ È3 sin x
STEP 3) V œ 'a A(x) dx œ È3 '0 sin x dx œ ’È3 cos x“ œ È3(1 1) œ 2È3
1
1
b
!
(b) STEP 1) A(x) œ (side)# œ Š2Èsin x‹ Š2Èsin x‹ œ 4 sin x
STEP 2) a œ 0, b œ 1
STEP 3) V œ 'a A(x) dx œ '0 4 sin x dx œ c4 cos xd 1! œ 8
1
b
#
6. (a) STEP 1) A(x) œ 1(diameter)
œ 14 (sec x tan x)# œ
4
1 sin x ‘
#
œ 4 sec x asec# x 1b 2 cos
#x
STEP 2) a œ 13 , b œ
1Î3
b
1
4
’2È3 asec# x tan# x 2 sec x tan xb
1
3
STEP 3) V œ 'a A(x) dx œ 'c1Î3
œ
1
4
1
3
1
4
ˆ2 sec# x 1 2 sin x ‰
cos# x
2 Š ˆ "" ‰ ‹ Š2È3 #
1
3
1
3
STEP 3) V œ 'a A(x) dx œ 'c1Î3 ˆ2 sec# x 1 b
1Î3
2 tan x x 2 ˆ cos" x ‰‘1Î$
1Î$
2 Š ˆ "" ‰ ‹‹“ œ
#
(b) STEP 1) A(x) œ (edge)# œ (sec x tan x)# œ ˆ2 sec# x 1 2
STEP 2) a œ 13 , b œ
1
4
dx œ
2 sin x ‰
cos# x
1
4
Š4È3 21
3 ‹
sin x ‰
cos# x
dx œ 2 Š2È3 13 ‹ œ 4È3 21
3
7. (a) STEP 1) A(x) œ alengthb † aheightb œ a6 3xb † a10b œ 60 30x
STEP 2) a œ 0, b œ 2
STEP 3) V œ 'a A(x) dx œ
b
'02 a60 30xb dx œ c60x 15x2 d 2! œ a120 60b 0 œ 60
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
328
Chapter 6 Applications of Definite Integrals
(b) STEP 1) A(x) œ alengthb † aheightb œ a6 3xb † Š 20 2a26 3xb ‹ œ a6 3xba4 3xb œ 24 6x 9x2
STEP 2) a œ 0, b œ 2
STEP 3) V œ 'a A(x) dx œ '0 a24 6x 9x2 bdx œ c24x 3x# 3x3 d ! œ a48 12 24b 0 œ 36
b
2
2
8. (a) STEP 1) A(x) œ "# abaseb † aheightb œ ˆÈx x2 ‰ † a6b œ 6Èx 3x
STEP 2) a œ 0, b œ 4
STEP 3) V œ 'a A(x) dx œ
b
"
#
(b) STEP 1) A(x) œ
'04 ˆ6x1Î2 3x‰ dx œ 4x3Î2 32 x2 ‘ 4! œ a32 24b 0 œ 8
2
‰ œ
† 1ˆ diameter
2
STEP 2) a œ 0, b œ 4
STEP 3) V œ 'a A(x) dx œ
b
1
4
9. A(y) œ
(diameter)# œ
1
4
#
ŠÈ5y# 0‹ œ
d
&
#
!
"
#
10. A(y) œ
1
4
(leg)(leg) œ
† 1Š
2
51
4
51
4
œ
1
2
†
x x3Î2 14 x2
4
œ 18 ˆx x3Î2 41 x2 ‰
41 x2 ‰ dx œ "# x2 52 x5Î2 1 3‘ 4
12 x !
œ 18 ˆ8 64
5
16 ‰
3
18 a0b œ
1
"&
y% ;
y% dy
a2& 0b œ 81
"
#
#
È1 y# ˆÈ1 y# ‰‘ œ
V œ 'c A(y) dy œ 'c1 2a1 y# b dy œ 2 ’y d
Èx x2 2
‹
2
'04 ˆx x3Î2
1
8
c œ 0, d œ 2; V œ 'c A(y) dy œ '0
œ ’ˆ 541 ‰ Š y5 ‹“ œ
"
#
1
"
y$
3 “ "
"
#
#
ˆ2È1 y# ‰ œ 2 a1 y# b ; c œ 1, d œ 1;
œ 4 ˆ1 "3 ‰ œ
8
3
11. The slices perpendicular to the edge labeled 5 are triangles, and by similar triangles we have
b
h
œ
4
3
Ê h œ 34 b. The
equation of the line through a5, 0b and a0, 4b is y œ 45 x 4, thus the length of the base œ 45 x 4 and the
6 2
height œ 34 ˆ 45 x 4‰ œ 35 x 3.Thus Aaxb œ "# abaseb † aheightb œ "# ˆ 45 x 4‰ † ˆ 35 x 3‰ œ 25
x 12
5 x6
6 2
and V œ 'a Aaxb dx œ '0 ˆ 25
x b
5
12
5 x
5
2 3
6‰ dx œ 25
x 65 x2 6x‘ 0 œ a10 30 30b 0 œ 10
12. The slices parallel to the base are squares. The cross section of the pyramid is a triangle, and by similar triangles we have
b
h
œ
3
5
2
Ê b œ 35 h. Thus Aayb œ abaseb2 œ ˆ 35 y‰ œ
9 2
25 y
Ê V œ 'c Aayb dy œ '0
d
5
9 2
25 y
3 3‘ 5
dy œ 25
y 0 œ 15 0 œ 15
13. (a) It follows from Cavalieri's Principle that the volume of a column is the same as the volume of a right
prism with a square base of side length s and altitude h. Thus, STEP 1) A(x) œ (side length)# œ s# ;
STEP 2) a œ 0, b œ h; STEP 3) V œ 'a A(x) dx œ '0 s# dx œ s# h
b
h
(b) From Cavalieri's Principle we conclude that the volume of the column is the same as the volume of the prism
described above, regardless of the number of turns Ê V œ s# h
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 6.1 Volumes Using Cross-Sections
14. 1) The solid and the cone have the same altitude of 12.
2) The cross sections of the solid are disks of diameter
x ˆ x# ‰ œ x# . If we place the vertex of the cone at the
origin of the coordinate system and make its axis of
symmetry coincide with the x-axis then the cone's cross
sections will be circular disks of diameter
x
ˆ x‰ x
4 4 œ # (see accompanying figure).
3) The solid and the cone have equal altitudes and identical
parallel cross sections. From Cavalieri's Principle we
conclude that the solid and the cone have the same
volume.
15. R(x) œ y œ 1 œ 1 ˆ2 4
2
16. R(y) œ x œ
3y
#
x
#
8 ‰
12
#
Ê V œ '0 1[R(x)]# dx œ 1'0 ˆ1 x# ‰ dx œ 1'0 Š1 x 2
2
2
x#
4‹
dx œ 1 ’x x#
#
21
3
œ
‰ dy œ 1'
Ê V œ '0 1[R(y)]# dy œ 1'0 ˆ 3y
#
0
2
17. R(y) œ tan ˆ 14 y‰ ; u œ
1
4
#
2
y Ê du œ
1
4
2
9
4
#
y# dy œ 1 34 y$ ‘ ! œ 1 †
3
4
dy Ê 4 du œ 1 dy; y œ 0 Ê u œ 0, y œ 1 Ê u œ
#
x$
12 “ !
† 8 œ 61
1
4
;
#
1Î%
V œ '0 1[R(y)]# dy œ 1'0 tan ˆ 14 y‰‘ dy œ 4 '0 tan# u du œ 4 '0 a1 sec# ub du œ 4[u tan u]!
1
1Î4
1
1Î4
œ 4 ˆ 14 1 0‰ œ 4 1
1
#
18. R(x) œ sin x cos x; R(x) œ 0 Ê a œ 0 and b œ
œ 1'0
1Î2
xœ
1
#
(sin x cos x)# dx œ 1 '0
1Î2
Ê u œ 1‘ Ä V œ 1'0
1
"
8
(sin 2x)#
4
are the limits of integration; V œ '0
1Î2
dx; u œ 2x Ê du œ 2 dx Ê
sin# u du œ
1
8
#u "
4
sin
1
2u‘ !
œ
1
8
du
8
œ
dx
4
1[R(x)]# dx
; x œ 0 Ê u œ 0,
ˆ 1# 0‰ 0‘ œ
1#
16
19. R(x) œ x# Ê V œ '0 1[R(x)]# dx œ 1 '0 ax# b dx
2
2
œ 1 '0 x% dx œ 1 ’ x5 “ œ
2
#
&
!
#
321
5
20. R(x) œ x$ Ê V œ '0 1[R(x)]# dx œ 1'0 ax$ b dx
2
2
œ 1 '0 x' dx œ 1 ’ x7 “ œ
2
(
#
!
#
1281
7
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
329
330
Chapter 6 Applications of Definite Integrals
21. R(x) œ È9 x# Ê V œ 'c3 1[R(x)]# dx œ 1 'c3 a9 x# b dx
3
$
x$
3 “ $
œ 1 ’9x 3
œ 21 9(3) 27 ‘
3
œ 2 † 1 † 18 œ 361
22. R(x) œ x x# Ê V œ '0 1[R(x)]# dx œ 1'0 ax x# b dx
1
1
œ 1'0 ax# 2x$ x% b dx œ 1 ’ x3 1
$
œ 1 ˆ 13 "
#
5" ‰ œ
1
30
(10 15 6) œ
23. R(x) œ Ècos x Ê V œ '0
1Î2
1Î#
œ 1 csin xd !
2x%
4
1
30
#
"
x&
5 “!
1[R(x)]# dx œ 1'0 cos x dx
1Î2
œ 1(1 0) œ 1
1Î4
1Î4
24. R(x) œ sec x Ê V œ 'c1Î4 1[R(x)]# dx œ 1 ' 1Î4 sec# x dx
1Î%
œ 1 ctan xd 1Î% œ 1[1 (1)] œ 21
25. R(x) œ È2 sec x tan x Ê V œ
œ1
'01Î4 1[R(x)]# dx
'01Î4 ŠÈ2 sec x tan x‹# dx
œ 1 '0 Š2 2È2 sec x tan x sec# x tan# x‹ dx
1Î4
œ 1 Œ'0 2 dx 2È2 '0 sec x tan x dx 1Î4
1Î%
œ 1 Œ[2x]!
1Î4
1Î%
2È2 [sec x]!
$
’ tan3 x “
'01Î4 (tan x)# sec# x dx
1Î%
!
œ 1 ’ˆ 1# 0‰ 2È2 ŠÈ2 1‹ "3 a1$ 0b“ œ 1 Š 1# 2È2 11
3 ‹
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 6.1 Volumes Using Cross-Sections
26. R(x) œ 2 2 sin x œ 2(1 sin x) Ê V œ '0 1[R(x)]# dx
1Î2
œ 1 '0 4(1 sin x)# dx œ 41 '0 a1 sin# x 2 sin xb dx
1Î2
1Î2
œ 41'0 1 "# (1 cos 2x) 2 sin x‘ dx
1Î2
œ 41'0 ˆ 3# 1Î2
cos 2x
2
2 sin x‰
1Î#
œ 41 3# x sin42x 2 cos x‘ !
œ 41 ˆ 341 0 0‰ (0 0 2)‘ œ 1(31 8)
27. R(y) œ È5 y# Ê V œ 'c1 1[R(y)]# dy œ 1 'c1 5y% dy
1
1
"
œ 1 cy& d " œ 1[1 (1)] œ 21
28. R(y) œ y$Î# Ê V œ '0 1[R(y)]# dy œ 1'0 y$ dy
2
2
#
%
œ 1 ’ y4 “ œ 41
!
29. R(y) œ È2 sin 2y Ê V œ '0 1[R(y)]# dy
1Î2
œ 1'0 2 sin 2y dy œ 1 c cos 2yd !
1Î2
1Î#
œ 1[1 (1)] œ 21
30. R(y) œ Écos
1y
4
Ê V œ 'c2 1[R(y)]# dy
0
œ 1 'c2 cos ˆ 14y ‰ dy œ 4 sin
0
31. R(y) œ
2
y1
1y ‘ !
4 #
œ 4[0 (1)] œ 4
Ê V œ '0 1[R(y)]# dy œ 41 '0
3
3
"
ay 1 b 2
dy
$
œ 41 ’ y 1 1 “ œ 41 4" a1b‘ œ 31
!
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
331
332
Chapter 6 Applications of Definite Integrals
È2y
y # 1
32. R(y) œ
Ê V œ '0 1[R(y)]# dy œ 1'0 2y ay# 1b
1
1
#
dy;
#
cu œ y 1 Ê du œ 2y dy; y œ 0 Ê u œ 1, y œ 1 Ê u œ 2d
Ä V œ 1'1 u# du œ 1 "u ‘ " œ 1 #" (1)‘ œ
2
#
1
#
33. For the sketch given, a œ 1# , b œ 1# ; R(x) œ 1, r(x) œ Ècos x; V œ 'a 1 a[R(x)]# [r(x)]# b dx
b
œ 'c1Î2 1(1 cos x) dx œ 21'0 (1 cos x) dx œ 21[x sin x]!
1Î2
1Î2
1Î#
œ 21 ˆ 1# 1‰ œ 1# 21
34. For the sketch given, c œ 0, d œ 14 ; R(y) œ 1, r(y) œ tan y; V œ 'c 1 a[R(y)]# [r(y)]# b dy
d
œ 1'0 a1 tan# yb dy œ 1 '0 a2 sec# yb dy œ 1[2y tan y]!
1Î4
1Î4
35. r(x) œ x and R(x) œ 1 Ê V œ
œ '0 1 a1 x# b dx œ 1 ’x 1
1Î%
œ 1 ˆ 1# 1‰ œ
1#
#
1
'01 1 a[R(x)]# [r(x)]# b dx
"
x$
3 “!
œ 1 ˆ1 "3 ‰ 0‘ œ
21
3
36. r(x) œ 2Èx and R(x) œ 2 Ê V œ '0 1 a[R(x)]# [r(x)]# b dx
1
œ 1'0 (4 4x) dx œ 41’x 1
"
x#
# “!
œ 41 ˆ1 "# ‰ œ 21
37. r(x) œ x# 1 and R(x) œ x 3
Ê V œ 'c1 1 a[R(x)]# [r(x)]# b dx
2
œ 1'c1 ’(x 3)# ax# 1b “ dx
2
#
œ 1 'c1 cax# 6x 9b ax% 2x# 1bd dx
2
œ 1 'c1 ax% x# 6x 8b dx
2
&
œ 1 ’ x5 x$
3
œ 1 ˆ 32
5 8
3
#
6x#
#
8x“
24
#
16‰ ˆ 5" "
"
3
6
#
‰
ˆ 5†30533 ‰ œ
8‰‘ œ 1 ˆ 33
5 3 28 3 8 œ 1
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
1171
5
Section 6.1 Volumes Using Cross-Sections
38. r(x) œ 2 x and R(x) œ 4 x#
Ê V œ 'c1 1 a[R(x)]# [r(x)]# b dx
2
œ 1'c1 ’a4 x# b (2 x)# “ dx
2
#
œ 1 'c1 ca16 8x# x% b a4 4x x# bd dx
2
œ 1'c1 a12 4x 9x# x% b dx
2
œ 1 ’12x 2x# 3x$ œ 1 ˆ24 8 24 #
x&
5 “ "
32 ‰
5
ˆ12 2 3 5" ‰‘ œ 1 ˆ15 33 ‰
5
œ
1081
5
39. r(x) œ sec x and R(x) œ È2
Ê V œ 'c1Î4 1 a[R(x)]# [r(x)]# b dx
1Î4
œ 1 'c1Î4 a2 sec# xb dx œ 1[2x tan x]1Î%
1Î4
1Î%
œ 1 ˆ 1# 1‰ ˆ 1# 1‰‘ œ 1(1 2)
40. R(x) œ sec x and r(x) œ tan x
Ê V œ '0 1 a[R(x)]# [r(x)]# b dx
1
œ 1 '0 asec# x tan# xb dx œ 1 '0 1 dx œ 1[x]!" œ 1
1
1
41. r(y) œ 1 and R(y) œ 1 y
Ê V œ '0 1 a[R(y)]# [r(y)]# b dy
1
œ 1'0 c(1 y)# 1d dy œ 1 '0 a1 2y y# 1b dy
1
1
œ 1 '0 a2y y# b dy œ 1 ’y# "
y$
3 “!
1
œ 1 ˆ1 "3 ‰ œ
41
3
42. R(y) œ 1 and r(y) œ 1 y Ê V œ '0 1 a[R(y)]# [r(y)]# b dy
1
œ 1'0 c1 (1 y)# d dy œ 1'0 c1 a1 2y y# bd dy
1
1
œ 1'0 a2y y# b dy œ 1 ’y# 1
"
y$
3 “!
œ 1 ˆ1 "3 ‰ œ
21
3
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
333
334
Chapter 6 Applications of Definite Integrals
43. R(y) œ 2 and r(y) œ Èy
Ê V œ '0 1 a[R(y)]# [r(y)]# b dy
4
œ 1'0 (4 y) dy œ 1 ’4y 4
%
y#
2 “!
œ 1(16 8) œ 81
44. R(y) œ È3 and r(y) œ È3 y#
È3
Ê V œ '0
È3
œ 1 '0
$
œ 1 ’ y3 “
1 a[R(y)]# [r(y)]# b dy
È3
c3 a3 y# bd dy œ 1'0
È$
!
y# dy
œ 1È3
45. R(y) œ 2 and r(y) œ 1 Èy
Ê V œ '0 1 a[R(y)]# [r(y)]# b dy
1
œ 1'0 ’4 ˆ1 Èy‰ “ dy
1
#
œ 1 '0 ˆ4 1 2Èy y‰ dy
1
œ 1 '0 ˆ3 2Èy y‰ dy
1
œ 1 ’3y 43 y$Î# œ 1 ˆ3 "
y#
# “!
"# ‰ œ 1 ˆ 18683 ‰ œ
4
3
71
6
46. R(y) œ 2 y"Î$ and r(y) œ 1
Ê V œ '0 1 a[R(y)]# [r(y)]# b dy
1
#
œ 1'0 ’ˆ2 y"Î$ ‰ 1“ dy
1
œ 1'0 ˆ4 4y"Î$ y#Î$ 1‰ dy
1
œ 1 '0 ˆ3 4y"Î$ y#Î$ ‰ dy
1
œ 1 ’3y 3y%Î$ "
3y&Î$
5 “!
œ 1 ˆ3 3 53 ‰ œ
31
5
47. (a) r(x) œ Èx and R(x) œ 2
Ê V œ '0 1 a[R(x)]# [r(x)]# b dx
4
œ 1'0 (4 x) dx œ 1 ’4x 4
(b) r(y) œ 0 and R(y) œ y#
%
x#
# “!
œ 1(16 8) œ 81
Ê V œ '0 1 a[R(y)]# [r(y)]# b dy
2
œ 1'0 y% dy œ 1 ’ y5 “ œ
2
&
#
!
321
5
#
(c) r(x) œ 0 and R(x) œ 2 Èx Ê V œ '0 1 a[R(x)]# [r(x)]# b dx œ 1'0 ˆ2 Èx‰ dx
4
œ 1'0 ˆ4 4Èx x‰ dx œ 1 ’4x 4
8x$Î#
3
4
%
x#
“
# !
œ 1 ˆ16 64
3
16 ‰
#
œ
81
3
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 6.1 Volumes Using Cross-Sections
(d) r(y) œ 4 y# and R(y) œ 4 Ê V œ '0 1 a[R(y)]# [r(y)]# b dy œ 1 '0 ’16 a4 y# b “ dy
2
2
œ 1 '0 a16 16 8y# y% b dy œ 1 '0 a8y# y% b dy œ 1 ’ 83 y$ 2
2
48. (a) r(y) œ 0 and R(y) œ 1 #
y&
“
5 !
#
œ 1 ˆ 64
3 32 ‰
5
œ
2241
15
y
#
Ê V œ '0 1 a[R(y)]# [r(y)]# b dy
2
#
œ 1'0 ˆ1 y# ‰ dy œ 1'0 Š1 y 2
2
y#
#
œ 1 ’y #
y$
12 “ !
œ 1 ˆ# (b) r(y) œ 1 and R(y) œ 2 4
2
8 ‰
12
y#
4‹
œ
dy
21
3
y
#
#
Ê V œ '0 1 a[R(y)]# [r(y)]# b dy œ 1 '0 ’ˆ2 y# ‰ 1“ dy œ 1 '0 Š4 2y 2
2
œ 1'0 Š3 2y 2
y#
4‹
dy œ 1 ’3y y# 2
#
y$
12 “ !
œ 1 ˆ6 4 8 ‰
12
œ 1 ˆ2 23 ‰ œ
y#
4
1‹ dy
81
3
49. (a) r(x) œ 0 and R(x) œ 1 x#
Ê V œ 'c1 1 a[R(x)]# [r(x)]# b dx
1
œ 1 'c1 a1 x# b dx œ 1 'c1 a1 2x# x% b dx
1
1
#
2x$
3
œ 1 ’x "
x&
5 “ "
103 ‰
œ 21 ˆ 1515
œ
œ 21 ˆ1 2
3
15 ‰
161
15
(b) r(x) œ 1 and R(x) œ 2 x# Ê V œ 'c1 1 a[R(x)]# [r(x)]# b dx œ 1 'c1 ’a2 x# b 1“ dx
1
1
#
œ 1 'c1 a4 4x# x% 1b dx œ 1'c1 a3 4x# x% b dx œ 1 ’3x 43 x$ 1
œ
21
15
1
(45 20 3) œ
561
15
"
x&
5 “ "
œ 21 ˆ3 4
3
15 ‰
2
3
15 ‰
(c) r(x) œ 1 x# and R(x) œ 2 Ê V œ 'c1 1 a[R(x)]# [r(x)]# b dx œ 1 'c1 ’4 a1 x# b “ dx
1
1
#
œ 1 'c1 a4 1 2x# x% b dx œ 1'c1 a3 2x# x% b dx œ 1 ’3x 23 x$ 1
œ
21
15
1
(45 10 3) œ
641
15
"
x&
5 “ "
œ 21 ˆ3 50. (a) r(x) œ 0 and R(x) œ hb x h
Ê V œ '0 1 a[R(x)]# [r(x)]# b dx
b
#
œ 1 '0 ˆ hb x h‰ dx
b
œ 1'0 Š hb# x# b
#
$
x
œ 1h# ’ 3b
# x#
b
2h#
b
x h# ‹ dx
b
x“ œ 1h# ˆ b3 b b‰ œ
!
1 h# b
3
#
(b) r(y) œ 0 and R(y) œ b ˆ1 yh ‰ Ê V œ '0 1 a[R(y)]# [r(y)]# b dy œ 1b# '0 ˆ1 yh ‰ dy
h
œ 1b# '0 Š1 h
2y
h
y#
h# ‹
dy œ 1b# ’y y#
h
h
h
y$
3h# “ !
œ 1b# ˆh h 3h ‰ œ
1 b# h
3
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
335
336
Chapter 6 Applications of Definite Integrals
51. R(y) œ b Èa# y# and r(y) œ b Èa# y#
Ê V œ 'ca 1 a[R(y)]# [r(y)]# b dy
a
œ 1 'ca ’ˆb Èa# y# ‰ ˆb Èa# y# ‰ “ dy
#
a
#
œ 1 'ca 4bÈa# y# dy œ 4b1'ca Èa# y# dy
a
a
1a#
#
œ 4b1 † area of semicircle of radius a œ 4b1 †
œ 2a# b1#
52. (a) A cross section has radius r œ È#y and area 1r# œ #1y. The volume is '0 #1ydy œ 1 cy# d ! œ #&1.
&
(b) Vahb œ ' Aahbdh, so
dV
dh
œ Aahb. Therefore
For h œ %, the area is #1a%b œ )1, so
dh
dt
œ
dV
dt
"
)1
œ
dV
dh
†
œ Aahb †
dh
dt
$
$
)1
† $ units
sec œ
hca
53. (a) R(y) œ Èa# y# Ê V œ 1'ca aa# y# b dy œ 1 ’a# y a$
3“
œ 1 ’a# h "3 ah$ 3h# a 3ha# a$ b dV
$
dt œ 0.2 m /sec
dV
#
dh œ 101h 1h
(b) Given
Ê
and a œ 5 m, find
Ê
dV
dt
œ
dV
dh
†
dh
dt
œ 1 Ša# h †
so
dh
dt
œ
"
A ah b
†
dV
dt .
units$
sec .
hca
y$
3 “ ca
h$
3
dh
dt ,
&
œ 1 ’a# h a$ h# a ha# ‹ œ
(h a)$
3
Ša$ a$
3 ‹“
1h# (3a h)
3
#
From part (a), V(h) œ 1h (153 h) œ 51h# 13h
dh ¸
0.2
"
"
œ 1h(10 h) dh
dt Ê dt hœ4 œ 41(10 4) œ (201)(6) œ 1#01 m/sec.
dh ¸
dt hœ4 .
$
54. Suppose the solid is produced by revolving y œ 2 x about
the y-axis. Cast a shadow of the solid on a plane parallel to
the xy-plane.
Use an approximation such as the Trapezoid Rule, to
#
estimate 'a 1cRaybd# dy ¸ ! 1Œ #k ˜y.
n
b
d^
kœ"
55. The cross section of a solid right circular cylinder with a cone removed is a disk with radius R from which a disk of radius
h has been removed. Thus its area is A" œ 1R# 1h# œ 1 aR# h# b . The cross section of the hemisphere is a disk of
#
radius ÈR# h# . Therefore its area is A# œ 1 ŠÈR# h# ‹ œ 1 aR# h# b . We can see that A" œ A# . The altitudes of
both solids are R. Applying Cavalieri's Principle we find
Volume of Hemisphere œ (Volume of Cylinder) (Volume of Cone) œ a1R# b R "3 1 aR# b R œ
56. R(x) œ
œ
1
144
x
1#
È36 x# Ê V œ ' 1[R(x)]# dx œ 1'
0
0
’12x$ 6
'
x&
5 “!
œ
1
144
Š12 † 6$ 6
6&
5‹
œ
1 †6 $
144
x#
144
ˆ12 a36 x# b dx œ
36 ‰
5
1
144
2
3
1 R$ .
'06 a36x# x% b dx
1 ‰ ˆ 6036 ‰
œ ˆ 196
œ
144
5
361
5
cm$ . The plumb bob will
weigh about W œ (8.5) ˆ 3651 ‰ ¸ 192 gm, to the nearest gram.
c7
c7
57. R(y) œ È256 y# Ê V œ 'c16 1[R(y)]# dy œ 1'c16 a256 y# b dy œ 1 ’256y œ 1 ’(256)(7) 7$
3
Š(256)(16) 16$
3 ‹“
$
œ 1 Š 73 256(16 7) 16$
3 ‹
(
y$
3 “ "'
œ 10531 cm$ ¸ 3308 cm$
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 6.2 Volume Using Cylindrical Shells
337
58. (a) R(x) œ kc sin xk , so V œ 1'0 [R(x)]# dx œ 1'0 (c sin x)# dx œ 1'0 ac# 2c sin x sin# xb dx
1
1
œ 1'0 ˆc# 2c sin x '1
1
(b)
1
1cos 2x ‰
dx œ 1 0 ˆc# "# 2c sin x cos#2x ‰ dx
#
1
œ 1 ˆc# "# ‰ x 2c cos x sin42x ‘ ! œ 1 ˆc# 1 1# 2c 0‰ (0 2c 0)‘ œ 1 ˆc# 1 1# 4c‰ . Let
2
V(c) œ 1 ˆc# 1 1# 4c‰ . We find the extreme values of V(c): dV
dc œ 1(2c1 4) œ 0 Ê c œ 1 is a critical
#
#
point, and V ˆ 12 ‰ œ 1 ˆ 14 1# 18 ‰ œ 1 ˆ 1# 14 ‰ œ 1# 4; Evaluate V at the endpoints: V(0) œ 1# and
#
#
V(1) œ 1 ˆ 3# 1 4‰ œ 1# (4 1)1. Now we see that the function's absolute minimum value is 1# 4,
taken on at the critical point c œ 12 . (See also the accompanying graph.)
#
From the discussion in part (a) we conclude that the function's absolute maximum value is 1# , taken on at
the endpoint c œ 0.
(c) The graph of the solid's volume as a function of c for
0 Ÿ c Ÿ 1 is given at the right. As c moves away from
[!ß "] the volume of the solid increases without bound.
If we approximate the solid as a set of solid disks, we
can see that the radius of a typical disk increases without
bounds as c moves away from [0ß 1].
59. Volume of the solid generated by rotating the region bounded by the x-axis and y œ faxb from x œ a to x œ b about the
x-axis is V œ 'a 1[f(x)]# dx œ 41, and the volume of the solid generated by rotating the same region about the line
b
'ab 1[f(x) 1]# dx œ 81. Thus 'ab 1[f(x) 1]# dx 'ab 1[f(x)]# dx œ 81 41
b
b
b
b
Ê 1'a a[f(x)]# 2f(x) " [f(x)]# b dx œ 41 Ê 'a a2f(x) "b dx œ 4 Ê 2'a f(x) dx 'a dx œ 4
b
b
Ê 'a f(x) dx "# ab ab œ 2 Ê 'a f(x) dx œ 4 #b a
y œ 1 is V œ
60. Volume of the solid generated by rotating the region bounded by the x-axis and y œ faxb from x œ a to x œ b about the
x-axis is V œ 'a 1[f(x)]# dx œ 61, and the volume of the solid generated by rotating the same region about the line
b
'ab 1[f(x) 2]# dx œ 101. Thus 'ab 1[f(x) 2]# dx 'ab 1[f(x)]# dx œ 101 61
b
b
b
b
Ê 1'a a[f(x)]# 4f(x) 4 [f(x)]# b dx œ 41 Ê 'a a4f(x) 4b dx œ 4 Ê 4'a f(x) dx 4'a dx œ 4
b
b
Ê 'a f(x) dx ab ab œ 1 Ê 'a f(x) dx œ 1 b a
y œ 2 is V œ
6.2 VOLUME USING CYLINDRICAL SHELLS
1. For the sketch given, a œ 0, b œ 2;
shell ‰
shell
V œ 'a 21 ˆ radius
Š height
‹ dx œ '0 21x Š1 b
2
x#
4‹
dx œ 21'0 Šx x#
4‹
dx œ 21'0 Š2x 2
x$
4‹
#
dx œ 21 ’ x# œ 21 † 3 œ 61
2. For the sketch given, a œ 0, b œ 2;
shell ‰
shell
V œ 'a 21 ˆ radius
Š height
‹ dx œ '0 21x Š2 b
2
3. For the sketch given, c œ 0, d œ È2;
È2
shell ‰
shell
V œ 'c 21 ˆ radius
Š height
‹ dy œ '0
d
2
È2
21y † ay# b dy œ 21'0
x$
4‹
#
x%
16 “ !
dx œ 21 ’x# %
y$ dy œ 21 ’ y4 “
È#
!
œ 21 ˆ 4# #
x%
16 “ !
16 ‰
16
œ 21(4 1) œ 61
œ 21
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
338
Chapter 6 Applications of Definite Integrals
4. For the sketch given, c œ 0, d œ È3;
È3
È3
shell ‰
shell
V œ 'c 21 ˆ radius
Š height
‹ dy œ '0 21y † c3 a3 y# bd dy œ 21 '0
d
%
y$ dy œ 21 ’ y4 “
È3
!
œ
91
#
5. For the sketch given, a œ 0, b œ È3;
È3
shell ‰
shell
V œ 'a 21 ˆ radius
Š height
‹ dx œ '0 21x † ŠÈx# 1‹ dx;
b
’u œ x# 1 Ê du œ 2x dx; x œ 0 Ê u œ 1, x œ È3 Ê u œ 4“
Ä V œ 1'1 u"Î# du œ 1 23 u$Î# ‘ " œ
%
4
21
3
ˆ4$Î# 1‰ œ ˆ 231 ‰ (8 1) œ
141
3
6. For the sketch given, a œ 0, b œ 3;
shell ‰
shell
V œ 'a 21 ˆ radius
Š height
‹ dx œ '0 21x Š Èx9x
‹ dx;
$9
b
3
cu œ x$ 9 Ê du œ 3x# dx Ê 3 du œ 9x# dx; x œ 0 Ê u œ 9, x œ 3 Ê u œ 36d
Ä V œ 21 '9 3u"Î# du œ 61 2u"Î# ‘ * œ 121 ŠÈ36 È9‹ œ 361
$'
36
7. a œ 0, b œ 2;
shell ‰
shell
V œ 'a 21 ˆ radius
Š height
‹ dx œ '0 21x x ˆ x2 ‰‘ dx
b
2
œ '0 21x# †
2
3
#
dx œ 1 '0 3x# dx œ 1 cx$ d ! œ 81
2
#
8. a œ 0, b œ 1;
shell ‰
shell
V œ 'a 21 ˆ radius
Š height
‹ dx œ '0 21x ˆ2x x2 ‰ dx
b
1
œ 1 '0 2 Š 3x# ‹ dx œ 1 ' 3x# dx œ 1 cx$ d ! œ 1
1
1
#
"
0
9. a œ 0, b œ 1;
shell ‰
shell
V œ 'a 21 ˆ radius
Š height
‹ dx œ '0 21x c(2 x) x# d dx
b
1
œ 21'0 a2x x# x$ b dx œ 21 ’x# 1
œ 21 ˆ1 "
3
4" ‰ œ 21 ˆ 12 124 3 ‰ œ
x$
3
101
12
œ
"
x%
4 “!
51
6
10. a œ 0, b œ 1;
shell ‰
shell
V œ 'a 21 ˆ radius
Š height
‹ dx œ '0 21x ca2 x# b x# d dx
b
1
œ 21'0 x a2 2x# b dx œ 41'0 ax x$ b dx
1
#
œ 41 ’ x# 1
"
x%
4 “!
œ 41 ˆ "2 4" ‰ œ 1
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 6.2 Volume Using Cylindrical Shells
11. a œ 0, b œ 1;
shell ‰
shell
V œ 'a 21 ˆ radius
Š height
‹ dx œ '0 21x Èx (2x 1)‘ dx
b
1
"
œ 21'0 ˆx$Î# 2x# x‰ dx œ 21 25 x&Î# 23 x$ "# x# ‘ !
1
œ 21 ˆ 25 2
3
15 ‰
"# ‰ œ 21 ˆ 12 20
œ
30
71
15
12. a œ ", b œ 4;
shell ‰
shell
V œ 'a 21 ˆ radius
Š height
‹ dx œ '1 21x ˆ 32 x"Î# ‰ dx
b
4
œ 31'1 x"Î# dx œ 31 23 x$Î# ‘ " œ 21 ˆ4$Î# "‰
%
4
œ 21(8 1) œ 141
sin x, 0 x Ÿ 1
0xŸ1
Ê xf(x) œ œ
; since sin 0 œ 0 we have
0, x œ 0
x, x œ 0
sin x, 0 x Ÿ 1
Ê xf(x) œ sin x, 0 Ÿ x Ÿ 1
xf(x) œ œ
sin x, x œ 0
13. (a) xf(x) œ œ
x†
sin x
x ,
shell ‰
shell
(b) V œ 'a 21 ˆ radius
Š height
‹ dx œ '0 21x † f(x) dx and x † f(x) œ sin x, 0 Ÿ x Ÿ 1 by part (a)
1
b
Ê V œ 21'0 sin x dx œ 21[ cos x]1! œ 21( cos 1 cos 0) œ 41
1
tan# x
x ,
tan# x, 0 x Ÿ 1/4
0 x Ÿ 14
Ê xg(x) œ œ
; since tan 0 œ 0 we have
0, x œ 0
x † 0, x œ 0
tan# x, 0 x Ÿ 1/4
Ê xg(x) œ tan# x, 0 Ÿ x Ÿ 1/4
xg(x) œ œ
tan# x, x œ 0
14. (a) xg(x) œ œ
x†
shell ‰
shell
(b) V œ 'a 21 ˆ radius
Š height
‹ dx œ '0 21x † g(x) dx and x † g(x) œ tan# x, 0 Ÿ x Ÿ 1/4 by part (a)
1Î4
b
Ê V œ 21'0 tan# x dx œ 21'0 asec# x 1b dx œ 21[tan x x]!
1Î4
1Î4
1Î%
œ 21 ˆ1 14 ‰ œ
41 1 #
#
15. c œ 0, d œ 2;
shell ‰
shell
V œ 'c 21 ˆ radius
Š height
‹ dy œ '0 21y Èy (y)‘ dy
d
2
œ 21'0 ˆy$Î# y# ‰ dy œ 21 ’ 2y5
2
&Î#
&
œ 21 ” 25 ŠÈ2‹ œ
161
15
2$
3•
È
#
y$
3 “!
œ 21 Š 8 5 2 83 ‹ œ 161 Š
È2
5
3" ‹
Š3È2 5‹
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
339
340
Chapter 6 Applications of Definite Integrals
16. c œ 0, d œ 2;
shell ‰
shell
V œ 'c 21 ˆ radius
Š height
‹ dy œ '0 21y cy# (y)ddy
d
2
œ 21'0 ay$ y# b dy œ 21 ’ y4 2
%
œ 161 ˆ 56 ‰ œ
401
3
#
y$
3 “!
œ 161 ˆ 42 3" ‰
17. c œ 0, d œ 2;
shell ‰
shell
V œ 'c 21 ˆ radius
Š height
‹ dy œ '0 21y a2y y# bdy
d
2
œ 21'0 a2y# y$ b dy œ 21 ’ 2y3 2
$
œ 321 ˆ "3 4" ‰ œ
321
12
œ
81
3
#
y%
4 “!
œ 21 ˆ 16
3 "6 ‰
4
18. c œ 0, d œ 1;
shell ‰
shell
V œ 'c 21 ˆ radius
Š height
‹ dy œ '0 21y a2y y# ybdy
d
1
œ 21'0 y ay y# b dy œ 21'0 ay# y$ b dy
1
1
$
œ 21 ’ y3 "
y%
4 “!
œ 21 ˆ 13 "4 ‰ œ
1
6
19. c œ 0, d œ 1;
shell ‰
shell
V œ 'c 21 ˆ radius
Š height
‹ dy œ 21'0 y[y (y)]dy
d
1
œ 21'0 2y# dy œ
1
41
3
"
cy$ d ! œ
41
3
20. c œ 0, d œ 2;
shell ‰
shell
V œ 'c 21 ˆ radius
Š height
‹ dy œ '0 21 yˆy y2 ‰dy
d
2
œ 21 '0
2
y2
2 dy
2
œ 13 c y3 d ! œ
81
3
21. c œ 0, d œ 2;
shell ‰
shell
V œ 'c 21 ˆ radius
Š height
‹ dy œ '0 21y c(2 y) y# d dy
d
2
œ 21 '0 a2y y# y$ b dy œ 21 ’y# 2
œ 21 ˆ4 8
3
16 ‰
4
œ
1
6
y$
3
(48 32 48) œ
#
y%
4 “!
161
3
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 6.2 Volume Using Cylindrical Shells
341
22. c œ 0, d œ 1;
shell ‰
shell
V œ 'c 21 ˆ radius
Š height
‹ dy œ '0 21y c(2 y) y# d dy
d
1
œ 21'0 a2y y# y$ b dy œ 21 ’y# 1
œ 21 ˆ1 1
3
14 ‰ œ
1
6
(12 4 3) œ
y$
3
51
6
"
y%
4 “!
shell ‰ shell
23. (a) V œ 'a 21 ˆ radius
Š height ‹dx œ '0 21 x a3xbdx œ 61'0 x2 dx œ 21 cx3 d ! œ 161
b
2
2
2
2
shell ‰ shell
(b) V œ 'a 21 ˆ radius
Š height ‹dx œ '0 21 a4 xb a3xbdx œ 61'0 a4x x2 bdx œ 61 2x2 13 x3 ‘ ! œ 61ˆ8 83 ‰ œ 321
b
2
2
2
shell ‰ shell
(c) V œ 'a 21 ˆ radius
Š height ‹dx œ '0 21 ax 1b a3xbdx œ 61'0 ax2 xbdx œ 61 13 x3 12 x2 ‘ ! œ 61ˆ 83 2‰ œ 281
b
2
2
6
shell ‰ shell
(d) V œ 'c 21 ˆ radius
Š height ‹dy œ '0 21 y ˆ2 13 y‰dy œ 21'0 ˆ2y 13 y2 ‰dy œ 21 y2 19 y3 ‘ ! œ 21a36 24b œ 241
d
6
6
shell ‰ shell
(e) V œ 'c 21 ˆ radius
Š height ‹dy œ '0 21 a7 yb ˆ2 13 y‰dy œ 21'0 ˆ14 d
6
6
13
3 y
13 y2 ‰dy œ 21 14y 13 2
6 y
6
19 y3 ‘ !
œ 21a84 78 24b œ 601
6
shell ‰ shell
(f) V œ 'c 21 ˆ radius
Š height ‹dy œ '0 21 ay 2b ˆ2 13 y‰dy œ 21'0 ˆ4 43 y 13 y2 ‰dy œ 21 4y 23 y2 19 y3 ‘ !
d
6
6
œ 21a24 24 24b œ 481
shell ‰ shell
24. (a) V œ 'a 21 ˆ radius
Š height ‹dx œ '0 21 x a8 x3 bdx œ 21'0 a8x x4 bdx œ 21 4x2 15 x5 ‘ ! œ 21ˆ16 b
2
2
2
32 ‰
5
œ
961
5
shell ‰ shell
(b) V œ 'a 21 ˆ radius
Š height ‹dx œ '0 21 a3 xb a8 x3 bdx œ 21'0 a24 8x 3x3 x4 bdx
b
2
2
2
œ 21 24x 4x2 34 x4 15 x5 ‘ ! œ 21ˆ48 16 12 32 ‰
5
œ
2641
5
shell ‰ shell
(c) V œ 'a 21 ˆ radius
Š height ‹dx œ '0 21 ax 2b a8 x3 bdx œ 21'0 a16 8x 2x3 x4 bdx
b
2
2
2
œ 21 16x 4x2 12 x4 15 x5 ‘ ! œ 21ˆ32 16 8 32 ‰
5
œ
3361
5
shell ‰ shell
(d) V œ 'c 21 ˆ radius
Š height ‹dy œ '0 21 y † y1Î3 dy œ 21'0 y4Î3 dy œ
d
8
8
61
7
y7Î3 ‘ 8 œ
!
61
7 a128b
œ
7681
7
8
shell ‰ shell
(e) V œ 'c 21 ˆ radius
Š height ‹dy œ '0 21 a8 yb y1Î3 dy œ 21'0 ˆ8y1Î3 y4Î3 ‰dy œ 21 6y4Î3 37 y7Î3 ‘ !
d
8
œ 21ˆ96 384 ‰
7
œ
8
5761
7
8
shell ‰ shell
(f) V œ 'c 21 ˆ radius
Š height ‹dy œ '0 21 ay 1b y1Î3 dx œ 21'0 ˆy4Î3 y1Î3 ‰dy œ 21 37 y7Î3 34 y4Î3 ‘ !
d
8
‰
œ 21ˆ 384
7 12 œ
8
9361
7
shell ‰ shell
25. (a) V œ 'a 21 ˆ radius
Š height ‹dx œ 'c1 21 a2 xb ax 2 x2 bdx œ 21'c1 a4 3x2 x3 bdx œ 21 4x x3 14 x4 ‘ 1
b
2
œ 21a8 8 4b 21ˆ4 1 14 ‰ œ
2
271
2
shell ‰ shell
(b) V œ 'a 21 ˆ radius
Š height ‹dx œ 'c1 21 ax 1b ax 2 x2 bdx œ 21'c1 a2 3x x3 bdx œ 21 2x 32 x2 14 x4 ‘ 1
b
.
2
2
œ 21a4 6 4b 21ˆ2 3
2
14 ‰ œ
2
271
2
shell ‰ shell
(c) V œ 'c 21 ˆ radius
Š height ‹dy œ '0 21 yˆÈy ˆÈy‰‰dy '1 21 yˆÈy ay 2b‰dy
d
4
1
œ 41'0 y3Î2 dy 21'1 ˆy3Î2 y2 2y‰dy œ
4
1
œ
81
5 a1b
21ˆ 64
5 64
3
16‰ 21ˆ 52 1
3
81
5
1‰ œ
y5Î2 ‘ 1 21 52 y5Î2 31 y3 y2 ‘ 4
!
1
721
5
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
2
342
Chapter 6 Applications of Definite Integrals
shell ‰ shell
(d) V œ 'c 21 ˆ radius
Š height ‹dy œ '0 21 a4 ybˆÈy ˆÈy‰‰dy '1 21 a4 ybˆÈy ay 2b‰dy
d
4
1
œ 41'0 ˆ4Èy y3Î2 ‰dy 21'1 ˆy2 y3Î2 6y 4Èy 8‰dy
4
1
1
4
œ 41 83 y3Î2 25 y5Î2 ‘ ! 21 13 y3 25 y5Î2 3y2 83 y3Î2 8y‘ 1
64
64
‰
ˆ1 2
œ 41ˆ 83 25 ‰ 21ˆ 64
3 5 48 3 32 21 3 5 3 8‰ œ
8
3
1081
5
shell ‰ shell
26. (a) V œ 'a 21 ˆ radius
Š height ‹dx œ 'c1 21 a1 xb a4 3x2 x4 bdx œ 21'c1 ax5 x4 3x3 3x2 4x 4bdx
b
1
1
1
œ 21 16 x6 15 x5 34 x4 x3 2x2 4x‘ 1 œ 21ˆ 16 1
5
3
4
1 2 4‰ 21ˆ 16 1
5
1 2 4‰ œ
3
4
shell ‰ shell
4 y ˆÈ
4 y‰‰dy ' 21 y É 4 y ŠÉ 4 y ‹ dy
(b) V œ 'c 21 ˆ radius
Š height ‹dy œ '0 21 yˆÈ
”
•
3
3
1
d
œ 41'0 y5Î4 dy 1
œ
161
9
œ
161
9
4
1
y9Î4 ‘ 1 !
41
È3
41
È3
41
È
È 3 Š8 3
'14 yÈ4 ydy cu œ 4 y Ê y œ 4 u Ê du œ dy; y œ 1 Ê u œ 3, y œ 4 Ê u œ 0d
'30 a4 ubÈu du œ 1691 a1b È41 '03 ˆ4Èu u3Î2 ‰du œ 1691 È41 38 u3Î2 52 u5Î2 ‘ 30
3
18 È
3‹
5
161
9
œ
881
5
œ
3
8721
45
shell ‰
shell
27. (a) V œ 'c 21 ˆ radius
Š height
‹ dy œ '0 21y † 12 ay# y$ b dy œ 241 '0 ay$ y% b dy œ 241 ’ y4 d
1
œ 241 ˆ 14 15 ‰ œ
241
20
œ
1
"
y&
5 “!
%
61
5
shell ‰
shell
(b) V œ 'c 21 ˆ radius
Š height
‹ dy œ '0 21(1 y) c12 ay# y$ bd dy œ 241'0 (1 y) ay# y$ b dy
d
1
1
œ 241'0 ay# 2y$ y% b dy œ 241 ’ y3 1
$
y%
2
"
y&
5 “!
œ 241 ˆ "3 1
2
" ‰
15 ‰ œ 241 ˆ 30
œ
41
5
shell ‰
shell
(c) V œ 'c 21 ˆ radius
Š height
‹ dy œ '0 21 ˆ 85 y‰ c12 ay# y$ bd dy œ 241 '0 ˆ 85 y‰ ay# y$ b dy
d
1
œ 241'0 ˆ 85 y# 1
œ
241
12
13
5
1
8 $
y$ y% ‰ dy œ 241 ’ 15
y 13
20
y% "
y&
5 “!
8
œ 241 ˆ 15
13
20
241
60
15 ‰ œ
(32 39 12)
œ 21
shell ‰
shell
(d) V œ 'c 21 ˆ radius
Š height
‹ dy œ '0 21 ˆy 25 ‰ c12 ay# y$ bd dy œ 241'0 ˆy 25 ‰ ay# y$ b dy
d
1
1
2 $
œ 241'0 ˆy$ y% 25 y# 25 y$ ‰ dy œ 241'0 ˆ 25 y# 35 y$ y% ‰ dy œ 241 ’ 15
y 1
1
2
œ 241 ˆ 15
3
20
15 ‰ œ
241
60
(8 9 12) œ
241
12
2
%
œ 21 ’ y4 #
y'
24 “ !
%
2'
24 ‹
œ 21 Š 24 #
%
œ 321 ˆ 4" 4 ‰
24
dy œ '0 21y Šy# 2
y#
# ‹“
2
œ 21 '0 Š2y# 2
y%
2
y$ y&
4‹
#
$
dy œ 21 ’ 2y3 %
y&
10
y%
4
#
y'
#4 “ !
2
œ 21'0 Š5y# 54 y% y$ 2
y&
4‹
#
$
dy œ 21 ’ 5y3 %
5y&
20
y%
4
shell ‰
shell
(d) V œ 'c 21 ˆ radius
Š height
‹ dy œ '0 21 ˆy 58 ‰ ’ y# Š y4 d
2
œ 21'0 Šy$ 2
y&
4
85 y# 5
32
#
%
y% ‹ dy œ 21 ’ y4 %
y'
#4
5y$
#4
81
3
y%
4‹
œ 21 ˆ 16
3 œ
32
10
16
4
64 ‰
24
dy œ '0 21(5 y) Šy# #
y'
#4 “ !
y#
# ‹“
2
2
y#
# ‹“
dy œ 21'0 Šy$ dy œ '0 21(2 y) Šy# y#
# ‹“
shell ‰
shell
(c) V œ 'c 21 ˆ radius
Š height
‹ dy œ '0 21(5 y) ’ y# Š y4 d
y%
4‹
2 ‰
œ 321 ˆ 4" 6" ‰ œ 321 ˆ 24
œ
shell ‰
shell
(b) V œ 'c 21 ˆ radius
Š height
‹ dy œ '0 21(2 y) ’ y# Š y4 d
"
y&
5 “!
y% œ 21
shell ‰
shell
28. (a) V œ 'c 21 ˆ radius
Š height
‹ dy œ '0 21y ’ y# Š y4 d
3
20
2
œ 21 ˆ 40
3 160
20
16
4
dy œ '0 21 ˆy 58 ‰ Šy# 2
#
5y&
160 “ !
œ 21 ˆ 16
4 64
24
40
24
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
81
5
y%
4‹
64 ‰
24
dy
dy
œ 81
y%
4‹
160 ‰
160
dy
œ 41
y&
4‹
dy
561
5
Section 6.2 Volume Using Cylindrical Shells
343
shell ‰
shell
29. (a) About x-axis: V œ 'c 21 ˆ radius
Š height
‹dy
d
œ '0 21yˆÈy y‰dy œ 21'0 ˆy$Î# y# ‰dy
1
1
"
œ 21 #& y&Î# "$ y$ ‘ ! œ 21ˆ #& "$ ‰ œ
#1
"&
shell ‰
shell
About y-axis: V œ 'a 21 ˆ radius
Š height
‹dx
b
œ '0 21xax x# bdx œ 21'0 ax2 x3 bdx
1
1
$
œ 21’ x$ "
x%
% “!
œ 21ˆ "$ "% ‰ œ
1
'
(b) About x-axis: Raxb œ x and raxb œ x# Ê V œ 'a 1Raxb# raxb# ‘dx œ '0 1cx# x% ddx
b
$
œ 1’ x$ "
x&
& “!
œ 1ˆ "$ "& ‰ œ
1
#1
"&
About y-axis: Rayb œ Èy and rayb œ y Ê V œ 'c 1Rayb# rayb# ‘dy œ '0 1cy y2 ddy
d
#
œ 1’ y# "
y$
$ “!
œ 1ˆ "# "$ ‰ œ
1
1
'
#
30. (a) V œ 'a 1Raxb# raxb# ‘dx œ 1'0 ’ˆ #x #‰ x# “dx
%
b
œ 1'0 ˆ $% x# #x %‰dx œ 1’ x% x# %x“
%
$
œ 1a"' "' "'b œ "'1
%
!
shell ‰
shell
(b) V œ 'a 21 ˆ radius
Š height
‹dx œ '0 #1xˆ x# # x‰dx
%
b
œ '0 #1xˆ# x# ‰dx œ #1'0 Š#x %
%
œ #1’x# %
x$
' “!
œ #1ˆ"' '% ‰
'
x#
# ‹dx
$#1
$
œ
shell ‰
shell
(c) V œ 'a 21 ˆ radius
Š height
‹dx œ '0 #1a% xbˆ x# # x‰dx œ '0 #1a% xbˆ# x# ‰dx œ #1'0 Š) %x %
b
œ #1’)x #x# %
x$
“
' !
%
œ #1ˆ$# $# '% ‰
'
%
x#
# ‹dx
'%1
$
œ
#
(d) V œ 'a 1Raxb# raxb# ‘dx œ 1'0 ’a) xb# ˆ' #x ‰ “dx œ 1'0 ’a'% "'x x# b Š$' 'x x% ‹“dx
%
b
%
#
1'0 ˆ $% x# "!x #)‰dx œ 1’ x% &x# #)x“ œ 1"' a&ba"'b a(ba"'b‘ œ 1a$ba"'b œ %)1
%
%
$
!
shell ‰
shell
31. (a) V œ 'c 21 ˆ radius
Š height
‹ dy œ '1 21y(y 1) dy
d
2
œ 21'1 ay# yb dy œ 21 ’ y3 2
$
#
y#
# “"
œ 21 ˆ 83 42 ‰ ˆ "3 #" ‰‘
œ 21 ˆ 73 2 "# ‰ œ 13 (14 12 3) œ
51
3
shell ‰
shell
(b) V œ 'a 21 ˆ radius
Š height
‹ dx œ '1 21x(2 x) dx œ 21'1 a2x x# b dx œ 21 ’x# b
2
2
œ 21 ˆ4 83 ‰ ˆ1 3" ‰‘ œ 21 ˆ 12 3 8 ‰ ˆ 3 3 " ‰‘ œ 21 ˆ 34 32 ‰ œ
41
3
shell ‰
shell
' ˆ 203 ‰
(c) V œ 'a 21 ˆ radius
Š height
‹ dx œ '1 21 ˆ 10
3 x (2 x) dx œ 21 1
b
2
#
" $‘
8 #
ˆ 40
œ 21 20
3 x 3 x 3 x " œ 21
3 2
32
3
38 ‰ ˆ 20
3 8
3
16
3
#
x$
3 “"
x x# ‰ dx
3" ‰‘ œ 21 ˆ 33 ‰ œ 21
shell ‰
shell
(d) V œ 'c 21 ˆ radius
Š height
‹ dy œ '1 21(y 1)(y 1) dy œ 21'1 (y 1)# œ 21 ’ (y31) “ œ
d
2
2
$
#
"
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
21
3
344
Chapter 6 Applications of Definite Integrals
shell ‰
shell
32. (a) V œ 'c 21 ˆ radius
Š height
‹ dy œ '0 21yay# 0b dy
d
2
œ 21'0 y$ dy œ 21 ’ y4 “ œ 21 Š 24 ‹ œ 81
2
%
#
%
!
shell ‰
shell
(b) V œ 'a 21 ˆ radius
Š height
‹ dx
b
œ '0 21x ˆ2 Èx‰ dx œ 21'0 ˆ2x x$Î# ‰ dx
4
4
%
2 †2 &
5 ‹
œ 21 x# 25 x&Î# ‘ ! œ 21 Š16 œ 21 ˆ16 64 ‰
5
21
5
œ
321
5
(80 64) œ
shell ‰
shell
(c) V œ 'a 21 ˆ radius
Š height
‹ dx œ '0 21(4 x) ˆ2 Èx‰ dx œ 21'0 ˆ8 4x"Î# 2x x$Î# ‰ dx
b
4
4
%
œ 21 8x 83 x$Î# x# 25 x&Î# ‘ ! œ 21 ˆ32 64
3
16 64 ‰
5
œ
21
15
(240 320 192) œ
21
15
shell ‰
shell
(d) V œ 'c 21 ˆ radius
Š height
‹ dy œ '0 21(2 y) ay# b dy œ 21 '0 a2y# y$ b dy œ 21 ’ 23 y$ d
2
œ 21 ˆ 16
3 16 ‰
4
œ
321
12
2
81
3
(4 3) œ
(112) œ
2241
15
#
y%
4 “!
shell ‰
shell
33. (a) V œ 'c 21 ˆ radius
Š height
‹ dy œ '0 21yay y$ b dy
d
1
œ '0 21 ay# y% b dy œ 21 ’ y3 1
œ
$
41
15
"
y&
“
5 !
œ 21 ˆ "3 5" ‰
shell ‰
shell
(b) V œ 'c 21 ˆ radius
Š height
‹ dy
d
œ '0 21(1 y) ay y$ b dy
1
œ 21 '0 ay y# y$ y% b dy œ 21 ’ y# 1
#
y$
3
y%
4
"
y&
5 “!
œ 21 ˆ "# "
3
"
4
5" ‰ œ
21
60
(30 20 15 12) œ
71
30
shell ‰
shell
34. (a) V œ 'c 21 ˆ radius
Š height
‹dy
d
œ '0 21y c1 ay y$ bddy
1
œ 21 '0 ay y# y% b dy œ 21 ’ y# 1
#
œ 21 ˆ "# œ
"
3
5" ‰ œ
21
30
y$
3
"
y&
5 “!
(15 10 6)
111
15
(b) Use the washer method:
V œ 'c 1 cR# (y) r# (y)d dy œ '0 1 ’1# ay y$ b “ dy œ 1 '0 a1 y# y' 2y% b dy œ 1 ’y d
1
œ 1 ˆ1 "
3
"
7
52 ‰ œ
1
105
1
#
(105 35 15 42) œ
y$
3
y(
7
971
105
"
2y&
5 “!
(c) Use the washer method:
V œ 'c 1 cR# (y) r# (y)d dy œ '0 1 ’c1 ay y$ bd 0“ dy œ 1'0 ’1 2 ay y$ b ay y$ b “ dy
d
1
œ 1'0 a1 y# y' 2y 2y$ 2y% b dy œ 1 ’y 1
œ
1
210
(70 30 105 2 † 42) œ
1
#
y$
3
y(
7
y# #
y%
#
1211
210
"
2y&
5 “!
œ 1 ˆ1 "
3
"
7
1
"
#
25 ‰
shell ‰
shell
(d) V œ 'c 21 ˆ radius
Š height
‹ dy œ '0 21(1 y) c1 ay y$ bd dy œ 21 '0 (1 y) a1 y y$ b dy
d
1
1
œ 21'0 a1 y y$ y y# y% b dy œ 21'0 a1 2y y# y$ y% b dy œ 21 ’y y# 1
œ 21 ˆ 1 1 1
"
3
"
4
5" ‰ œ
21
60
(20 15 12) œ
231
30
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
y$
3
y%
4
"
y&
5 “!
Section 6.2 Volume Using Cylindrical Shells
shell ‰
shell
35. (a) V œ 'c 21 ˆ radius
Š height
‹dy œ '0 21y ˆÈ8y y# ‰ dy
d
2
œ 21'0 Š2È2 y$Î# y$ ‹ dy œ 21 ’ 4 5 2 y&Î# 2
È
#
y%
4 “!
&
œ 21 4È2†ŠÈ2‹
2%
4
5
81
5
œ 21 † 4 ˆ 85 1‰ œ
$
œ 21 Š 4†52 (8 5) œ
4 †4
4 ‹
241
5
shell ‰
shell
(b) V œ 'a 21 ˆ radius
Š height
‹ dx œ '0 21x ŠÈx b
4
&
œ 21 Š 2†52 4%
3# ‹
'
œ 21 Š 25 2)
32 ‹
œ
1†2(
160
x#
8‹
dx œ 21'0 Šx$Î# (32 20) œ
4
1†2* †3
160
œ
1†2% †3
5
œ
x$
8‹
dx œ 21 ’ 25 x&Î# 481
5
shell ‰
shell
36. (a) V œ 'a 21 ˆ radius
Š height
‹ dx
b
œ '0 21x ca2x x# b xd dx
1
œ 21 '0 x ax x# b dx œ 21'0 ax# x$ b dx
1
$
1
œ 21 ’ x3 "
x%
4 “!
œ 21 ˆ "3 4" ‰ œ
1
6
shell ‰
shell
(b) V œ 'a 21 ˆ radius
Š height
‹ dx œ '0 21(1 x) ca2x x# b xd dx œ 21'0 (1 x) ax x# b dx
b
1
1
œ 21 '0 ax 2x# x$ b dx œ 21 ’ x2 32 x$ 1
"
x%
“
4 !
#
œ 21 ˆ 12 2
3
"4 ‰ œ
21
1#
(6 8 3) œ
37. (a) V œ 'a 1 cR# (x) r# (x)d dx œ 1 '1Î16 ˆx"Î# 1‰ dx
b
1
"
œ 1 2x"Î# x‘"Î"' œ 1 (2 1) ˆ2 †
œ 1 ˆ1 7 ‰
16
œ
"
4
" ‰‘
16
91
16
shell ‰
shell
(b) V œ 'a 21 ˆ radius
Š height
‹ dy œ '1 21y Š y"% b
2
œ 21'1 ˆy$ 2
y ‰
16
dy œ 21 ’ 12 y# œ 21 ˆ "8 8" ‰ ˆ #" œ
21
32
91
16
(8 1) œ
" ‰‘
3#
d
2
œ 1 "3 y$ œ
1
48
y ‘#
16 "
" ‰
32
"
16 ‹
dy
"
œ 1 ˆ 24
8" ‰ ˆ 3" (2 6 16 3) œ
dy
#
y#
32 “ "
œ 21 ˆ 4" 38. (a) V œ 'c 1 cR# (y) r# (y)d dy œ '1 1 Š y"% "
16 ‹
" ‰‘
16
111
48
shell ‰
shell
(b) V œ 'a 21 ˆ radius
Š height
‹ dx œ '1Î4 21x Š È"x "‹ dx
b
1
œ 21 '1Î4 ˆx"Î# x‰ dx œ 21 ’ 23 x$Î# 1
œ 21 ˆ 23 "# ‰ ˆ 23 †
"
8
" ‰‘
3#
"
x#
“
2 "Î%
œ 1 ˆ 43 1 "
6
" ‰
16
œ
1
48
(4 † 16 48 8 3) œ
111
48
39. (a) H3=5 : V œ V" V#
V" œ 'a 1[R" (x)]# dx and V# œ 'a 1[R# (x)]# with R" (x) œ É x 3 2 and R# (x) œ Èx,
b"
b#
"
#
a" œ 2, b" œ 1; a# œ 0, b# œ 1 Ê two integrals are required
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
1
6
%
x%
“
32 !
345
346
Chapter 6 Applications of Definite Integrals
(b) [ +=2/<: V œ V" V#
V" œ 'a 1 a[R" (x)]# [r" (x)]# b dx with R" (x) œ É x 3 2 and r" (x) œ 0; a" œ 2 and b" œ 0;
b"
"
b
V# œ 'a
#
1 a[R# (x)]# [r# (x)]# b dx with R# (x) œ É x 3 2 and r# (x) œ Èx; a# œ 0 and b# œ 1
#
Ê two integrals are required
shell ‰
shell
shell
(c) W2/66: V œ 'c 21 ˆ radius
Š height
‹ dy œ 'c 21y Š height
‹ dy where shell height œ y# a3y# 2b œ 2 2y# ;
d
d
c œ 0 and d œ 1. Only 98/ integral is required. It is, therefore preferable to use the =2/66 method.
However, whichever method you use, you will get V œ 1.
40. (a) H3=k: V œ V" V# V$
Vi œ 'c 1[Ri (y)]# dy, i œ 1, 2, 3 with R" (y) œ 1 and c" œ 1, d" œ 1; R# (y) œ Èy and c# œ 0 and d# œ 1;
di
i
R$ (y) œ (y)"Î% and c$ œ 1, d$ œ 0 Ê three integrals are required
(b) [ +=2/<: V œ V" V#
Vi œ 'c 1a[Ri (y)]# [ri (y)]# b dy, i œ 1, 2 with R" (y) œ 1, r" (y) œ Èy, c" œ 0 and d" œ 1;
di
i
R# (y) œ 1, r# (y) œ (y)"Î% , c# œ 1 and d# œ 0 Ê two integrals are required
shell ‰
shell
shell
(c) W2/66: V œ 'a 21 ˆ radius
Š height
‹dx œ 'a 21xŠ height
‹dx, where shell height œ x# ax% b œ x# x% ,
b
b
a œ 0 and b œ 1 Ê only one integral is required. It is, therefore preferable to use the =2/66 method.
However, whichever method you use, you will get V œ 561 .
41. (a) V œ 'a 1 cR# axb r# axbd dx œ 'c4 1 ”ŠÈ25 x2 ‹ a3b# •dx œ 1'c4 c25 x2 9d dx œ 1'c4 a16 x2 bdx
b
4
%
4
64 ‰
64 ‰
2561
ˆ
3 1 64 3 œ 3
5001
3 Ê Volume of portion removed
œ 116x 13 x3 ‘ 4 œ 1ˆ64 (b) Volume of sphere œ 43 1a5b3 œ
È" 1
shell ‰
shell
42. V œ 'a 21 ˆ radius
Š height
‹ dx œ '1
b
#
œ
4
5001
3
2561
3
œ
2441
3
21 x sinax2 1b dx; Òu œ x2 1 Ê du œ 2x dx; x œ 1 Ê u œ 0,
x œ È" 1 Ê u œ 1Ó Ä 1'0 sin u du œ 1 ccos ud 1! œ 1a1 1b œ 21
1
r
shell ‰
shell
43. V œ 'a 21 ˆ radius
Š height
‹ dx œ '0 21 x ˆ hr x h‰dx œ 21'0 ˆ hr x2 h x‰dx œ 21 3rh x3 h2 x2 ‘ !
b
r
2
œ 21Š r3h r2 h
# ‹
r
œ 13 1 r2 h
shell ‰
shell
44. V œ 'c 21 ˆ radius
Š height
‹dy œ '0 21 y”Èr2 y2 ˆÈr2 y2 ‰•dy œ 41'0 yÈr2 y2 dy
r
d
r
Òu œ r2 y2 Ê du œ 2y dy; y œ 0 Ê u œ r2 , y œ r Ê u œ 0] Ä 21'r2 Èu du œ 21'0 u1Î2 du
0
œ
2
4 1 3 Î2 ‘ r
3 u
!
œ
r2
41 3
3 r
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 6.3 Arc Lengths
6.3 ARC LENGTHS
1.
dy
dx
œ
"
3
† 3# ax# 2b
"Î#
† 2x œ Èax# 2b † x
Ê L œ '0 È1 ax# 2b x# dx œ '0 È1 2x# x% dx
$
3
œ '0 Éa1 x# b# dx œ '0 a1 x# b dx œ ’x $
œ3
2.
dy
dx
œ
3
#
œ 12
27
3
Èx Ê L œ ' É1 49 x dx; u œ 1 49 x
0
4
Ê du œ
dx Ê
9
4
du œ dx; x œ 0 Ê u œ 1; x œ 4
4
9
Ê u œ 10d Ä L œ '1 u"Î# ˆ 49 du‰ œ
10
œ
3.
dx
dy
8
27
œ y# #
"
4y#
%
Ê Š dx
dy ‹ œ y 3
œ '1 Éy% 3
"
#
œ '1 ÊŠy# 3
$
œ ’ y3 dx
dy
œ
4
9
23 u$Î# ‘ "!
"
Š10È10 1‹
Ê L œ '1 É1 y% 4.
$
x$
“
3 !
$
"
#
"
16y%
"
4y# ‹
$
y "
“
4 "
#
"
#
"
16y%
"
#
"
16y%
dy
dy
dy œ '1 Šy# 3
" ‰
1#
œ ˆ 27
3 "
4y# ‹
dy
ˆ 3" 4" ‰ œ 9 #
y"Î# "# y"Î# Ê Š dx
dy ‹ œ
"
4
"
1#
"
3
"
4
œ9
(1 4 3)
1#
œ9
(2)
1#
œ
53
6
Šy 2 y" ‹
Ê L œ '1 Ê1 "4 Šy 2 y" ‹ dy
9
œ '1 Ê "4 Šy 2 y" ‹ dy œ '1
9
œ
"
#
9
"
#
"
Èy ‹
Ê ŠÈ y *
$Î#
$
"
dx
dy
dy
'19 ˆy"Î# y"Î# ‰ dy œ "# 23 y$Î# 2y"Î# ‘ *"
œ ’ y 3 y"Î# “ œ Š 33 3‹ ˆ "3 1‰ œ 11 5.
#
œ y$ "
4y$
#
'
Ê Š dx
dy ‹ œ y Ê L œ '1 É1 y' 2
œ '1 Éy' 2
œ '1 Šy$ 2
œ Š 16
4 "
2
y $
4 ‹
"
(16)(2) ‹
"
16y'
"
2
"
16y'
"
#
32
3
dy
2
%
œ
"
16y'
dy œ '1 ÊŠy$ dy œ ’ y4 "
3
y $
4 ‹
#
dy
#
y #
“
8 "
ˆ "4 "8 ‰ œ 4 "
32
"
4
"
8
œ
128184
32
œ
123
32
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
347
348
6.
Chapter 6 Applications of Definite Integrals
dx
dy
œ
y#
#
"
#y #
#
Ê Š dx
dy ‹ œ
"
4
ay% 2 y% b
Ê L œ '2 É1 "4 ay% 2 y% b dy
3
œ '2 É "4 ay% 2 y% b dy
3
7.
œ
"
#
'23 Éay# y# b# dy œ "# '23 ay# y# b dy
œ
"
#
’ y3 y" “ œ
dy
dx
$
$
"
#
#
"‰
ˆ 27
ˆ 8 " ‰‘ œ
3 3 3 #
#
#Î$
œ x"Î$ "4 x"Î$ Ê Š dy
dx ‹ œ x
Ê L œ '1 É1 x#Î$ 8
œ '1 Éx#Î$ 8
"
#
x #Î$
16
"
#
x #Î$
16
"
#
"
#
ˆ 26
3 8
3
#" ‰ œ
"
#
ˆ6 #" ‰ œ
13
4
x #Î$
16
dx
dx
#
œ '1 Ɉx"Î$ "4 x"Î$ ‰ dx œ '1 ˆx"Î$ "4 x"Î$ ‰ dx
8
8
)
œ 34 x%Î$ 38 x#Î$ ‘ " œ
œ
8.
dy
dx
3
8
2x%Î$ x#Î$ ‘ )
"
3
8
ca2 † 2% 2# b (2 1)d œ
œ x# 2x 1 œ (1 x)# #
2
œ '0 É(1 x)% "
#
œ '0 Ê’(1 x)# 2
œ '0 ’(1 x)# 2
"
#
(1x) %
16
#
(1x) #
“
4
(1x) #
“
4
(1x) %
16
99
8
"
"
4 (1x)#
%
Ê Š dy
dx ‹ œ (1 x) Ê L œ '0 É1 (1 x)% 2
(32 4 3) œ
œ x# 2x 1 4
(4x4)#
"
"
4 (1x)#
3
8
"
#
"
16(1x)%
dx
dx
dx
dx; cu œ 1 x Ê du œ dx; x œ 0 Ê u œ 1, x œ 2 Ê u œ 3d
Ä L œ '1 ˆu# "4 u# ‰ du œ ’ u3 "4 u" “ œ ˆ9 3
$
$
"
9.
dx
dy
" ‰
1#
ˆ 3" 4" ‰ œ
108143
12
œ
106
12
œ
53
6
#
%
œ Èsec% y 1 Ê Š dx
dy ‹ œ sec y 1
1Î4
1Î4
Ê L œ 'c1Î4 È1 asec% y 1b dy œ ' 1Î4 sec# y dy
1Î%
œ ctan yd 1Î% œ 1 (1) œ 2
10.
dy
dx
#
%
œ È3x% 1 Ê Š dy
dx ‹ œ 3x 1
c1
c1
Ê L œ 'c2 È1 a3x% 1b dx œ 'c2 È3 x# dx
$
œ È3 ’ x3 “
"
#
œ
È3
3
c1 (2)$ d œ
È3
3
(" 8) œ
7È 3
3
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 6.3 Arc Lengths
11. (a)
dy
dx
#
(b)
#
œ 2x Ê Š dy
dx ‹ œ 4x
Ê L œ 'c1 Ê1 Š dy
dx ‹ dx
#
2
œ 'c1 È1 4x# dx
2
(c) L ¸ 6.13
12. (a)
dy
dx
#
(b)
%
œ sec# x Ê Š dy
dx ‹ œ sec x
Ê L œ 'c1Î3 È1 sec% x dx
0
(c) L ¸ 2.06
13. (a)
dx
dy
#
(b)
#
œ cos y Ê Š dx
dy ‹ œ cos y
Ê L œ '0 È1 cos# y dy
1
(c) L ¸ 3.82
14. (a)
dx
dy
#
œ È1y y# Ê Š dx
dy ‹ œ
1Î2
Ê L œ 'c1Î2 É1 œ 'c1Î2 a1 y# b
1Î2
"Î#
y#
a1 y # b
y#
1 y#
(b)
1Î2
dy œ ' 1Î2 É 1 " y# dy
dy
(c) L ¸ 1.05
15. (a) 2y 2 œ 2
dx
dy
#
#
Ê Š dx
dy ‹ œ (y 1)
(b)
Ê L œ 'c1 È1 (y 1)# dy
3
(c) L ¸ 9.29
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
349
350
Chapter 6 Applications of Definite Integrals
16. (a)
dy
dx
#
(b)
#
#
œ cos x - cos x + x sin x Ê Š dy
dx ‹ œ x sin x
Ê L œ '0 È1 x# sin# x dx
1
(c) L ¸ 4.70
17. (a)
dy
dx
#
(b)
#
œ tan x Ê Š dy
dx ‹ œ tan x
# x cos# x
Ê L œ '0 È1 tan# x dx œ '0 É sin cos
dx
#x
1Î6
œ '0
1Î6
1Î6
œ '0 sec x dx
1Î6
dx
cos x
(c) L ¸ 0.55
18. (a)
dx
dy
#
(b)
#
œ Èsec# y 1 Ê Š dx
dy ‹ œ sec y 1
Ê L œ 'c1Î3 È1 asec# y 1b dy
1Î4
1Î4
1Î4
œ 'c1Î3 ksec yk dy œ ' 1Î3 sec y dy
(c) L ¸ 2.20
#
"
19. (a) Š dy
dx ‹ corresponds to 4x here, so take
So y œ Èx from ("ß ") to (4ß 2).
dy
dx
"
.
#È x
as
Then y œ Èx C and since ("ß ") lies on the curve, C œ 0.
(b) Only one. We know the derivative of the function and the value of the function at one value of x.
#
20. (a) Š dx
dy ‹ corresponds to
So y œ
"
y%
here, so take
dy
dx
"
y# .
as
Then x œ y" C and, since (!ß ") lies on the curve, C œ 1
"
"x.
(b) Only one. We know the derivative of the function and the value of the function at one value of x.
21. y œ '0 Ècos2t dt Ê
x
œ '0
1Î4
œ Ècos2x Ê L œ '0
1Î4
Ê1 ’Ècos2x“ dx œ '0
#
1Î4
È1 cos2x dx œ '
1Î4
0
È2cos2 x dx
È2cos x dx œ È2csin xd 1Î4 œ È2sinˆ 1 ‰ È2sina!b œ 1
0
4
22. y œ ˆ1 x2Î3 ‰
3 Î2 È 2
, 4
œ 'È2Î4 É1 1
dy
dx
ŸxŸ1Ê
1 x2Î3
dx
x2Î3
œ 32 a1b2Î3 32 Š
È2 2Î3
4 ‹
dy
dx
1Î2
œ 32 ˆ1 x2Î3 ‰ ˆ 23 x1Î3 ‰ œ œ 'È2Î4 É1 1
x2Î3
œ
3
4
1
3
2
32 ˆ 12 ‰ œ
ˆ1 x2Î3 ‰1Î2
x1Î3
1
1 dx œ 'È2Î4 É x12Î3 dx œ 'È2Î4 x11Î3 dx œ 'È2Î4 x1Î3 dx œ
1
1
#
Ê L œ 'È2Î4 Ë1 ” a1 xx1Î3 b • dx
1
Ê total length œ 8ˆ 34 ‰ œ 6
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
2Î3 1Î2
3
2
"
x2Î3 ‘ È
2Î4
Section 6.3 Arc Lengths
23. y œ 3 2x, 0 Ÿ x Ÿ 2 Ê
dy
dx
351
œ 2 Ê L œ '0 É1 a2b# dx œ '0 È5 dx œ ’È5 x“ œ 2È5.
2
2
2
0
d œ Éa2 0b a3 a1bb œ 2È5
2
2
24. Consider the circle x2 y2 œ r2 , we will find the length of the portion in the first quadrant, and multiply our result by 4.
y œ Èr2 x2 , 0 Ÿ x Ÿ r Ê
œ 4'0
r
r
È r2 x 2
dx œ 4r'0
r
25. 9x2 œ yay 3b2 Ê
Ê dx œ
2
œ ’ ay 4y1b 1“dy2 œ
26. 4x2 y2 œ 64 Ê
#
Ê L œ 4'0 Ë1 ” È 2 x 2 • dx œ 4'0 É1 r x
r
œ
d
dy ’yay
2
y2 2y 1 4y
dy2
4y
2
r
x2
r2 x 2
dx œ 4'0 É r2 r x2 dx
r
dx
3b2 “ Ê 18x dy
œ 2yay 3b ay 3b2 œ 3ay 3bay 1b Ê
bay 1b
ds2 œ dx2 dy2 œ ’ ay 36x
dy“ dy2 œ
d
2
dx ’4x
y2 “ œ
2
œ dx2 ’ 4x
y dx“ œ dx 16x2
2
y2 dx
27. È2 x œ '0 Ê1 Š dy
dt ‹ dt, x
#
x
x
È r2 x 2
2
dx
È r2 x 2
d
2
dy ’9x “
ay 3bay 1b
dy;
6x
œ
dy
dx
œ
ay 3 b2 ay 1 b 2
dy2
36x2
dy2 œ
dx
dy
œ
ay 3bay 1b
6x
ay 3 b2 ay 1b 2
dy2
4yay3b2
dy2
ay 1 b 2
2
4y dy
d
dx ’64“
Ê 8x 2y dy
dx œ 0 Ê
œ Š1 16x2
2
y2 ‹dx
œ
dy
dx
œ
4x
y
y2 16x2
dx2
y2
œ
4x2 64 16x2
dx2
y2
#
dy
0 Ê È2 œ Ê1 Š dx
‹ Ê
dy
dx
Ê dy œ
4x
y dx;
ds2 œ dx2 dy2
œ
20x2 64
dx2
y2
œ
4
2
y2 a5x
16bdx2
œ „ 1 Ê y œ f(x) œ „ x C where C is any real
number.
28. (a) From the accompanying figure and definition of the
differential (change along the tangent line) we see that
dy œ f w (xkc1 ) ˜ xk Ê length of kth tangent fin is
È( ˜ xk )# (dy)# œ È( ˜ xk )# [f w (xkc1 ) ˜ xk ]# .
n
n
! (length of kth tangent fin) œ lim ! È( ˜ xk )# [f w (xk 1 ) ˜ xk ]#
(b) Length of curve œ n lim
Ä_
nÄ_
kœ1
! È1 [f w (xk 1 )]# ˜ xk œ ' È1 [f w (x)]# dx
œ n lim
Ä_
a
n
kœ1
b
kœ1
4
2
29. x2 y2 œ 1 Ê y œ È1 x2 ; P œ Ö0, 14 , 12 , 34 , 1× Ê L ¸ ! Éaxi xi1 b2 ayi yi1 b2 œ ʈ 14 0‰ Š
kœ1
ʈ 1# 1 ‰2
4
È
Š 23
È15 2
4 ‹
ʈ 43 1 ‰2
2
È
Š 47
È3 2
2 ‹
2
ʈ1 34 ‰ Š0 30. Let ax1 , y1 b and ax2 , y2 b, with x2 x1 , lie on y œ m x b, where m œ
#
y2 y1
x2 x1 ,
then
dy
dx
È7 2
4 ‹
œ m Ê L œ 'x È1 m# dx
x2
1
#
#
2
ax 2 x 1 b
1
ax2 x1 b œ Éax2 x1 b# ay2 y1 b# .
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
2
1‹
¸ 1.55225
ax 2 x 1 b a y 2 y 1 b
y1
œ È1 m# c xd xx21 œ È1 m# ax2 x1 b œ Ê1 Š yx22 ax2 x1 b
x1 ‹ ax2 x1 b œ Ê
ax x b #
É ax 2 x 1 b# a y 2 y 1 b #
È15
4
352
Chapter 6 Applications of Definite Integrals
31. y œ 2x3Î2 Ê
#
œ 3x1Î2 ; Laxb œ '0 É1 3t1Î2 ‘ dt œ '0 È1 9t dt; Òu œ 1 9t Ê du œ 9dt, t œ 0 Ê u œ 1,
x
dy
dx
t œ x Ê u œ 1 9x] Ä
32. y œ
x3
3
x2 x 1
4x 4
1
9
È
'119x Èu du œ 272 u3Î2 ‘ 119x œ 272 a1 9xb3Î2 272 ; La1b œ 272 a10b3Î2 272 œ 2Š10 2710 1‹
Ê
dy
dx
Laxb œ '0 Ë1 ”at 1b2 x
œ '0 Ê 16at 1b
x
œ '0
x
4at1b4 1
4 at 1 b 2
Ä '1
x 1
La1b œ
8
3
4
œ x2 2x 1 #
1
4 at 1 b 2 •
16at 1b8 8at 1b4 "
16at 1b4
dt œ '0 ”at 1b2 x
x
x
1
12
œ
1
;
4 ax 1 b 2
#
4
b 8 at 1b
dt œ '0 Ê 16at 116
at 1 b 4
x
1
4 at 1 b 2
8
4
"
x
dt œ '0 Ê
x
4at1b4 1‘#
16at 1b4
4at1b4 1‘#
16at 1b4
dt
dt
•dt; Òu œ t 1 Ê du œ dt, t œ 0 Ê u œ 1, t œ x Ê u œ x 1Ó
x 1
1
8
œ ax 1b2 1
dt œ '0 Ë1 ” 4a4tat 1b 1
dt œ '0 Ê1 b2 •
1 2
1 3
1 1
2
”u 4 u •du œ 3 u 4 u ‘ 1
1
4 ax 1 b 2
œ Š 31 ax 1b3 1
4 ax 1 b ‹
ˆ 13 14 ‰ œ 13 ax 1b3 59
24
33-38. Example CAS commands:
Maple:
with( plots );
with( Student[Calculus1] );
with( student );
f := x -> sqrt(1-x^2);a := -1;
b := 1;
N := [2, 4, 8 ];
for n in N do
xx := [seq( a+i*(b-a)/n, i=0..n )];
pts := [seq([x,f(x)],x=xx)];
L := simplify(add( distance(pts[i+1],pts[i]), i=1..n ));
T := sprintf("#33(a) (Section 6.3)\nn=%3d L=%8.5f\n", n, L );
P[n] := plot( [f(x),pts], x=a..b, title=T ):
end do:
display( [seq(P[n],n=N)], insequence=true, scaling=constrained );
L := ArcLength( f(x), x=a..b, output=integral ):
L = evalf( L );
# (b)
# (a)
# (c)
33-38. Example CAS commands:
Mathematica: (assigned function and values for a, b, and n may vary)
Clear[x, f]
{a, b} = {1, 1}; f[x_] = Sqrt[1 x2 ]
p1 = Plot[f[x], {x, a, b}]
n = 8;
pts = Table[{xn, f[xn]}, {xn, a, b, (b a)/n}]/ / N
Show[{p1,Graphics[{Line[pts]}]}]
Sum[ Sqrt[ (pts[[i 1, 1]] pts[[i, 1]])2 (pts[[i 1, 2]] pts[[i, 2]])2 ], {i, 1, n}]
NIntegrate[ Sqrt[ 1 f'[x]2 ],{x, a, b}]
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
1
4 ax 1 b
1
12 ;
Section 6.4 Areas of Surfaces of Revolution
6.4 AREAS OF SURFACES OF REVOLUTION
1.
(a)
dy
dx
#
(b)
%
œ sec# x Ê Š dy
dx ‹ œ sec x
Ê S œ 21'0
1Î4
(tan x) È1 sec% x dx
(c) S ¸ 3.84
2. (a)
dy
dx
#
(b)
2
œ 2x Ê Š dy
dx ‹ œ 4x
Ê S œ 21'0 x# È1 4x# dx
2
(c) S ¸ 53.23
3. (a) xy œ 1 Ê x œ
Ê S œ 21'1
2
"
y
"
y
Ê
dx
dy
#
dx
œ y"# Ê Š dy
‹ œ
"
y%
(b)
È1 y% dy
(c) S ¸ 5.02
4. (a)
dx
dy
#
#
œ cos y Ê Š dx
dy ‹ œ cos y
(b)
Ê S œ 21'0 (sin y) È1 cos# y dy
1
(c) S ¸ 14.42
#
5. (a) x"Î# y"Î# œ 3 Ê y œ ˆ3 x"Î# ‰
"Î# ‰ ˆ
ˆ
Ê dy
"# x"Î# ‰
dx œ 2 3 x
#
"Î# ‰
ˆ
Ê Š dy
dx ‹ œ 1 3x
(b)
#
#
#
Ê S œ 21'1 ˆ3 x"Î# ‰ É1 a1 3x"Î# b dx
4
(c) S ¸ 63.37
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
353
354
Chapter 6 Applications of Definite Integrals
#
"Î# ‰
ˆ
œ 1 y"Î# Ê Š dx
dy ‹ œ 1 y
dx
dy
6. (a)
#
(b)
#
Ê S œ 21 '1 ˆy 2Èy‰ É1 a1 y"Î# b dx
2
(c) S ¸ 51.33
#
(b)
#
œ tan y Ê Š dx
dy ‹ œ tan y
dx
dy
7. (a)
Ê S œ 21'0 Š'0 tan t dt‹ È1 tan# y dy
1Î3
y
œ 21'0 Š'0 tan t dt‹ sec y dy
1Î3
y
(c) S ¸ 2.08
#
(b)
#
œ Èx# 1 Ê Š dy
dx ‹ œ x 1
dy
dx
8. (a)
È5
Ê S œ 21'1 Š'1 Èt# 1 dt‹ È1 ax# 1b dx
È5
x
œ 21'1 Š'1 Èt# 1 dt‹ x dx
x
(c) S ¸ 8.55
9. y œ
œ
Ê
x
#
1È5
#
dy
dx
' ˆ x ‰ É1 œ "# ; S œ 'a 21y Ê1 Š dy
dx ‹ dx Ê S œ 0 21 #
"
4
dx œ
1È5
#
'04 x dx
%
!
Ê x œ 2y Ê
x
#
4
#
’ x# “ œ 41È5; Geometry formula: base circumference œ 21(2), slant height œ È4# 2# œ 2È5
Ê Lateral surface area œ
10. y œ
#
b
dx
dy
"
#
(41) Š2È5‹ œ 41È5 in agreement with the integral value
#
È
È '
È #
'
œ 2; S œ 'c 21x Ê1 Š dx
dy ‹ dy œ 0 21 † 2y 1 2 dy œ 41 5 0 y dy œ 21 5 cy d !
#
d
2
2
#
œ 21È5 † 4 œ 81È5; Geometry formula: base circumference œ 21(4), slant height œ È4# 2# œ 2È5
Ê Lateral surface area œ " (81) Š2È5‹ œ 81È5 in agreement with the integral value
#
11.
dy
dx
'
œ "# ; S œ 'a 21yÊ1 Š dy
dx ‹ dx œ 1 21
#
b
3
(x 1)
#
É1 ˆ "# ‰# dx œ
1È5
#
'13 (x 1) dx
œ
1È5
#
#
’ x# x“
$
"
1È5
#
È
ˆ 9# 3‰ ˆ "# 1‰‘ œ 1 # 5 (4 2) œ 31È5; Geometry formula: r" œ "# "# œ 1, r# œ 3# "# œ 2,
œ
slant height œ È(2 1)# (3 1)# œ È5 Ê Frustum surface area œ 1(r" r# ) ‚ slant height œ 1(1 2)È5
œ 31È5 in agreement with the integral value
12. y œ
x
#
"
#
Ê x œ 2y 1 Ê
dx
dy
dx
œ 2; S œ 'c 21x Ê1 Š dy
‹ dy œ '1 21(2y 1)È1 4 dy œ 21È5 '1 (2y 1) dy
d
#
2
#
œ 21È5 cy# yd " œ 21È5 [(4 2) (1 1)] œ 41È5; Geometry formula: r" œ 1, r# œ 3,
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
2
Section 6.4 Areas of Surfaces of Revolution
slant height œ È(2 1)# (3 1)# œ È5 Ê Frustum surface area œ 1(1 3)È5 œ 41È5 in agreement with
the integral value
13.
dy
dx
#
x#
3
œ
’u œ 1 x%
9
Ê S œ '0
2
x%
9
Ê Š dy
dx ‹ œ
Ê du œ
4
9
x œ 0 Ê u œ 1, x œ 2 Ê u œ
Ä S œ 21 '1
25Î9
14.
œ
1
3
dy
dx
œ
1 2
"
4 du œ # 3
1 ˆ 12527 ‰
œ 98811
3
27
#
Ê S œ '3Î4 21Èx É1 15Î4
15Î4
15.
œ
’ˆ 15
4
œ
41
3
(8 1) œ
dy
dx
" (2 2x)
# È2x x#
œ
ˆ 34
dx;
dx;
#&Î*
u$Î# ‘ "
"
4x
dx
$Î#
dx œ 21 ’ 23 ˆx 4" ‰ “
"
4
" ‰$Î#
4
41
3
x$
9
du œ
x%
9
"
4x
x"Î# Ê Š dy
dx ‹ œ
œ 21'3Î4 Éx É1 25 ‘
9
u"Î# †
ˆ 125
‰
27 1 œ
"
#
"
4
x$ dx Ê
21 x$
9
" ‰$Î#
“
4
"&Î%
$Î%
41
3
$
’ˆ 42 ‰
Ê Š dy
dx ‹ œ
(1 x)#
2x x#
œ
1“
281
3
œ
#
1x
È2x x#
Ê S œ '0 5 21È2x x# É1 1Þ5
Þ
œ 21'0 5 È2x x#
È2x x#
1Þ5
(1 x)#
2x x#
1 2x x#
È2x x#
Þ
dx
dx
œ 21'0 5 dx œ 21[x]"Þ&
!Þ& œ 21
1Þ5
Þ
16.
dy
dx
"
2È x 1
œ
#
dy
Ê Š dx
‹ œ
"
4(x 1)
Ê S œ '1 21Èx 1 É1 5
œ 21'1 É(x 1) 5
œ 21 ’ 23 ˆx 17.
œ
41
3
œ
1
6
dx
dy
"
4
&
5 ‰$Î#
“
4
"
981
6
dx
dx œ 21'1 Éx 5
œ
41
3
‰$Î# ˆ 94 ‰$Î# “ œ
’ˆ 25
4
(125 27) œ
"
4(x 1)
œ
5
4
dx
$Î#
$Î#
’ˆ5 45 ‰ ˆ1 45 ‰ “
41
3
$
Š 52$ 3$
2$ ‹
491
3
%
'
œ y# Ê Š dx
dy ‹ œ y Ê S œ 0
#
1
u œ 1 y% Ê du œ 4y$ dy Ê
"
4
21 y$
3
È1 y% dy;
du œ y$ dy; y œ 0
Ê u œ 1, y œ 1 Ê u œ 2d Ä S œ '1 21 ˆ "3 ‰ u"Î# ˆ 4" du‰
2
œ
1
6
'12 u"Î# du œ 16 32 u$Î# ‘ #" œ 19 ŠÈ8 1‹
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
355
356
Chapter 6 Applications of Definite Integrals
18. x œ ˆ "3 y$Î# y"Î# ‰ Ÿ 0, when 1 Ÿ y Ÿ 3. To get positive
area, we take x œ ˆ "3 y$Î# y"Î# ‰
Ê
dx
dy
#
œ "# ˆy"Î# y"Î# ‰ Ê Š dx
dy ‹ œ
"
4
ay 2 y" b
Ê S œ '1 21 ˆ "3 y$Î# y"Î# ‰ É1 4" ay 2 y" b dy
3
œ 21'1 ˆ 3" y$Î# y"Î# ‰ É 4" ay 2 y" b dy
3
Éay"Î# y "Î# b#
œ 21'1 ˆ "3 y$Î# y"Î# ‰
3
œ 1'1 ˆ "3 y# 3
2
3
dx
dy
œ
"
È 4 y
œ 41 '0
15Î4
œ
20.
dx
dy
81
3
œ
3
$
y#
3
$
y“ œ 1 ˆ 27
9 "
161
9
#
Ê S œ '0
15Î4
"
4 y
Ê Š dx
dy ‹ œ
21 † 2È4 y É1 È5 y dy œ 41 23 (5 y)$Î# ‘ "&Î% œ 831 ’ˆ5 !
Š 5È 5 "
È2y 1
dy œ 1'1 y"Î# ˆ 3" y 1‰ Šy"Î# y 1‰ dy œ 1 ’ y9 œ 19 (18 1 3) œ
19.
#
5È 5
8 ‹
œ
81
3
#
Š 40
"
2y1
Ê Š dx
dy ‹ œ
41 È 2
3
"
œ 21È2 23 y$Î# ‘ &Î) œ
È 5 5 È 5
‹
8
œ
1
$Î#
3
3‰ ˆ "9 9
3
"
4y
15 ‰$Î#
4
"
3
dy œ 41'0
15Î4
1‰‘ œ 1 ˆ3 "
9
"
3
1‰
È(4 y) 1 dy
$Î#
5$Î# “ œ 831 ’ˆ 45 ‰ 5$Î# “
“œ
41 È 2
3
Š1 #
dy œ 21'5Î8 È(2y 1) 1 dy œ 21'5Î8 È2 y"Î# dy
1
"
2y 1
5È 5
‹
8È 8
41 È 2
3
œ
21. S œ 21'1Î2 È2y 1 Ê1 Š È2y" 1 ‹ dy œ 21 '1Î2 È2y 1 É1 1
dy œ 1 '1 ˆ 3" y 1‰ (y 1) dy
351È5
3
Ê S œ '5Î8 21È2y 1 É1 ’1$Î# ˆ 85 ‰
"
‹
y"Î#
1
"
2y1
1
È
5
Š 8†2 8†22È
2
È5
‹œ
1
12
Š16È2 5È5‹
dy œ 21'1Î2 È2y 1 É 2y2y1 dy
1
1
3
œ 21 '1Î2 È2y dy œ 2È2 1 '1Î2 Èy dy œ 2È2 1 23 y3Î2 ‘1Î2 œ 2È2 1 ”Š 23 È13 ‹ Œ 23 Ɉ "# ‰ • œ 2È2 1 Š 23 1
1
È
œ 2È2 1 Š 2 È21 ‹ œ
3 2
22. y œ
"
3
ax# 2b
21
3
Š2È2 1‹
È2
Ê dy œ xÈx# 2 dx Ê ds œ È1 a2x# x% b dx Ê S œ 21'0 x È1 2x# x% dx
$Î#
È2
È2
œ 21'0 xÉax# 1b# dx œ 21'0
23. ds œ Èdx# dy# œ ÊŠy$ œ ÊŠy$ "
4y$ ‹
#
dy œ Šy$ "
4y$ ‹
#
"
4y$ ‹
È2
x ax# 1b dx œ 21'0 ax$ xb dx œ 21 ’ x4 1 dy œ ÊŠy' "
#
%
"
16y' ‹
dy; S œ '1 21y ds œ 21'1 y Šy$ 2
2
#
&
1 dy œ ÊŠy' "‰
"‰
ˆ " " ‰‘ œ 21 ˆ 31
œ 21 ’ y5 4" y" “ œ 21 ˆ 32
5 8 5 4
5 8 œ
"
24. y œ cos x Ê
dy
dx
"
#
œ 21 ˆ 44 22 ‰ œ 41
"
16y' ‹
dy
dy œ 21'1 ˆy% "4 y# ‰ dy
2
(8 † 31 5) œ
2531
20
#
È1 sin# x dx
'
œ sin x Ê Š dy
dx ‹ œ sin x Ê S œ 21 c1Î2 (cos x)
25. y œ Èa# x# Ê
#
dy
dx
œ
"
#
aa# x# b
Ê S œ 21'ca Èa# x# É1 a
21
40
"
4y$ ‹
È#
x#
# “!
"Î#
x#
aa # x # b
1Î2
(2x) œ
x
È a# x#
#
Ê Š dy
dx ‹ œ
x#
aa # x # b
dx œ 21'ca Èaa# x# b x# dx œ 21'ca a dx œ 21a[x]ca a
a
a
œ 21a[a (a)] œ (21a)(2a) œ 41a#
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
"
‹
3È 2
Section 6.4 Areas of Surfaces of Revolution
26. y œ
œ
r
h
21 r
h
x Ê
dy
dx
#
É h h# r
#
œ
r
h
#
Ê Š dy
dx ‹ œ
Ê S œ 21 '0
h
r#
h#
r
h
x É1 r#
h#
dx œ 21'0
h
r
h
#
#
x É h h# r dx
'0h x dx œ 2h1r Èh# r# ’ x# “ h œ 2h1r Èh# r# Š h# ‹ œ 1rÈh# r#
#
#
#
#
0
#
#
#
È16# y#
27. The area of the surface of one wok is S œ 'c 21x Ê1 Š dx
dy ‹ dy. Now, x y œ 16 Ê x œ
#
d
Ê
dx
dy
œ
c7
y
È16# y#
#
Ê Š dx
dy ‹ œ
c7
; S œ 'c16 21È16# y# É1 y#
16# y#
y#
16# y#
c7
dy œ 21'c16 Èa16# y# b y# dy
œ 21'c16 16 dy œ 321 † 9 œ 2881 ¸ 904.78 cm# . The enamel needed to cover one surface of one wok is
V œ S † 0.5 mm œ S † 0.05 cm œ (904.78)(0.05) cm$ œ 45.24 cm$ . For 5000 woks, we need
5000 † V œ 5000 † 45.24 cm$ œ (5)(45.24)L œ 226.2L Ê 226.2 liters of each color are needed.
28. y œ Èr# x# Ê
œ 21'a
abh
œ 21'a
œ "#
2x
È r# x #
œ
Èar# x# b x# dx œ 21r'
a
29. y œ ÈR# x# Ê
abh
dy
dx
dy
dx
œ "#
2x
È R # x#
abh
œ
x#
r# x # ;
S œ 21 'a
abh
Èr# x# É1 x#
r# x #
dx
dx œ 21rh, which is independent of a.
abh
30. (a) x# y# œ 45# Ê x œ È45# y# Ê
45
Ê Š dx
dy ‹ œ
#
x
È R # x#
ÈaR# x# b x# dx œ 21R '
a
S œ 'c22Þ5 21 È45# y# É1 #
x
È r# x #
y#
45# y#
dx
Ê Š dy
‹ œ
x#
R # x# ;
S œ 21'a
abh
ÈR# x# É1 x#
R # x#
dx œ 21Rh
dx
dy
œ
y
È45# y#
dy œ 21 '
45
22Þ5
#
Ê Š dx
dy ‹ œ
y#
45# y#
;
Èa45# y# b y# dy œ 21 † 45'
45
22Þ5
dy
œ (21)(45)(67.5) œ 60751 square feet
(b) 19,085 square feet
31. (a) An equation of the tangent line segment is
(see figure) y œ f(mk ) f w (mk )(x mk ).
When x œ xkc1 we have
r" œ f(mk ) f w (mk )(x5 1 mk )
œ f(mk ) f w (mk ) ˆ ?#xk ‰ œ f(mk ) f w (mk )
when x œ xk we have
r# œ f(mk ) f w (mk )(x5 mk )
k
œ f(mk ) f w (mk ) ?x
# ;
(b) L#k œ (?xk )# (r# r" )#
?x k
#
;
#
ˆf w (mk ) ?#xk ‰‘
œ (?xk )# [f w (mk )?xk ]# Ê Lk œ È(?xk )# [f w (mk )?xk ]# , as claimed
(c) From geometry it is a fact that the lateral surface area of the frustum obtained by revolving the tangent
œ (?xk )# f w (mk )
?x k
#
line segment about the x-axis is given by ?Sk œ 1(r" r# )Lk œ 1[2f(mk )] Éa?xk b# [f w (mk )?xk ]#
using parts (a) and (b) above. Thus, ?Sk œ 21f(mk ) È1 [f w (mk )]# ?xk .
! ?Sk œ lim ! 21f(mk ) È1 [f w (mk )]# ?xk œ ' 21f(x) È1 [f w (x)]# dx
(d) S œ n lim
Ä_
nÄ_
a
kœ1
kœ1
n
32. y œ ˆ1 x#Î$ ‰
$Î#
n
Ê
dy
dx
œ
Ê S œ 2'0 21 ˆ1 x#Î$ ‰
1
3
#
b
ˆ1 x#Î$ ‰"Î# ˆ 32 x"Î$ ‰ œ $Î#
ˆ1x#Î$ ‰"Î#
x"Î$
#
Ê Š dy
dx ‹ œ
1x#Î$
x#Î$
œ
"
x#Î$
$Î#
"
É1 ˆ x#Î$
1‰ dx œ 41'0 ˆ1 x#Î$ ‰ Èx#Î$ dx
1
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
1
dx
357
358
Chapter 6 Applications of Definite Integrals
$Î#
œ 41'0 ˆ1 x#Î$ ‰ x"Î$ dx; u œ 1 x#Î$ Ê du œ 23 x"Î$ dx Ê 32 du œ x"Î$ dx;
1
!
x œ 0 Ê u œ 1, x œ 1 Ê u œ 0d Ä S œ 41'1 u$Î# ˆ 3# du‰ œ 61 52 u&Î# ‘ " œ 61 ˆ0 52 ‰ œ
0
121
5
6.5 WORK AND FLUID FORCES
1. The force required to stretch the spring from its natural length of 2 m to a length of 5 m is F(x) œ kx. The work done
by F is W œ '0 F(x) dx œ k '0 x dx œ
3
3
k
#
$
cx# d ! œ
9k
# .
This work is equal to 1800 J Ê
2. (a) We find the force constant from Hooke's Law: F œ kx Ê k œ
Ê kœ
F
x
800
4
9
#
k œ 1800 Ê k œ 400 N/m
œ 200 lb/in.
(b) The work done to stretch the spring 2 inches beyond its natural length is W œ '0 kx dx œ 200 '0 x dx œ 200 ’ x# “
2
2
#
œ 200(2 0) œ 400 in † lb œ 33.3 ft † lb
(c) We substitute F œ 1600 into the equation F œ 200x to find 1600 œ 200x Ê x œ 8 in.
#
!
3. We find the force constant from Hooke's law: F œ kx. A force of 2 N stretches the spring to 0.02 m Ê 2 œ k † (0.02)
N
4N
Ê k œ 100 m
. The force of 4 N will stretch the rubber band y m, where F œ ky Ê y œ Fk Ê y œ 100
N Ê y œ 0.04 m
m
œ 4 cm. The work done to stretch the rubber band 0.04 m is W œ '0
0Þ04
œ
(100)(0.04)#
#
kx dx œ 100 '0
0Þ04
#
x dx œ 100 ’ x# “
!Þ!%
!
œ 0.08 J
4. We find the force constant from Hooke's law: F œ kx Ê k œ
F
x
Ê kœ
90
1
Ê k œ 90
N
m.
The work done to stretch the
‰
spring 5 m beyond its natural length is W œ '0 kx dx œ 90 '0 x dx œ 90 ’ x# “ œ (90) ˆ 25
# œ 1125 J
5
5
#
&
!
5. (a) We find the spring's constant from Hooke's law: F œ kx Ê k œ
œ
F
x
21,714
8 5
œ
21,714
3
Ê k œ 7238
lb
in
(b) The work done to compress the assembly the first half inch is W œ '0 kx dx œ 7238 '0 x dx œ 7238 ’ x# “
0Þ5
#
œ (7238) (0.5)
# œ
(7238)(0.25)
#
1Þ0
1Þ0
Þ
Þ
#
"Þ!
!Þ&
œ
7238
#
c1 (0.5)# d œ
6. First, we find the force constant from Hooke's law: F œ kx Ê k œ
compresses the scale x œ
this far is W œ '0
1Î8
#
!Þ&
!
¸ 905 in † lb. The work done to compress the assembly the second half inch is:
W œ '0 5 kx dx œ 7238 '0 5 x dx œ 7238 ’ x# “
"
8
0Þ5
in, he/she must weigh F œ kx œ
#
kx dx œ 2400 ’ x# “
"Î)
!
œ
2400
2†64
œ
F
x
2,400 ˆ 8" ‰
œ 18.75 lb † in. œ
(7238)(0.75)
#
150
" ‰
ˆ 16
¸ 2714 in † lb
œ 16 † 150 œ 2,400
lb
in .
If someone
œ 300 lb. The work done to compress the scale
25
16
ft † lb
7. The force required to haul up the rope is equal to the rope's weight, which varies steadily and is proportional to x, the
length of the rope still hanging: F(x) œ 0.624x. The work done is: W œ '0 F(x) dx œ '0 0.624x dx œ 0.624 ’ x# “
50
50
#
œ 780 J
&!
!
8. The weight of sand decreases steadily by 72 lb over the 18 ft, at 4 lb/ft. So the weight of sand when the bag is x ft off the
ground is Faxb œ "%% %x. The work done is: W œ 'a F(x) dx œ '0 a"%% %xbdx œ c144x 2x# d ! œ 1944 ft † lb
b
18
")
9. The force required to lift the cable is equal to the weight of the cable paid out: F(x) œ (4.5)(180 x) where x
is the position of the car off the first floor. The work done is: W œ '0
180
F(x) dx œ 4.5'0
180
(180 x) dx
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 6.5 Work and Fluid Forces
œ 4.5 ’180x ")!
x#
# “!
180#
# ‹
œ 4.5 Š180# œ
4.5†180#
#
359
œ 72,900 ft † lb
10. Since the force is acting toward the origin, it acts opposite to the positive x-direction. Thus F(x) œ xk# . The work done
b
is W œ 'a xk# dx œ k 'a x"# dx œ k x" ‘ a œ k ˆ b" "a ‰ œ
b
b
k(a b)
ab
11. Let r œ the constant rate of leakage. Since the bucket is leaking at a constant rate and the bucket is rising at a constant rate,
the amount of water in the bucket is proportional to a#! xb, the distance the bucket is being raised. The leakage rate of
the water is 0.8 lb/ft raised and the weight of the water in the bucket is F œ 0.8a#! xb. So:
W œ '0 0.8a#! xb dx œ 0.8 ’20x 20
#!
x#
# “!
œ 160 ft † lb.
12. Let r œ the constant rate of leakage. Since the bucket is leaking at a constant rate and the bucket is rising at a constant rate,
the amount of water in the bucket is proportional to a#! xb, the distance the bucket is being raised. The leakage rate of
the water is 2 lb/ft raised and the weight of the water in the bucket is F œ 2a#! xb. So:
W œ '0 2a#! xb dx œ 2 ’20x 20
#!
x#
# “!
œ 400 ft † lb.
Note that since the force in Exercise 12 is 2.5 times the force in Exercise 11 at each elevation, the total work is also 2.5
times as great.
13. We will use the coordinate system given.
(a) The typical slab between the planes at y and y ?y has
a volume of ?V œ (10)(12) ?y œ 120 ?y ft$ . The force
F required to lift the slab is equal to its weight:
F œ 62.4 ?V œ 62.4 † 120 ?y lb. The distance through
which F must act is about y ft, so the work done lifting
the slab is about ?W œ force ‚ distance
œ 62.4 † 120 † y † ?y ft † lb. The work it takes to lift all
20
the water is approximately W ¸ ! ?W
0
20
œ ! 62.4 † 120y † ?y ft † lb. This is a Riemann sum for
0
the function 62.4 † 120y over the interval 0 Ÿ y Ÿ 20. The work of pumping the tank empty is the limit of these sums:
W œ '0 62.4 † 120y dy œ (62.4)(120) ’ y# “
20
#
#!
!
‰ œ (62.4)(120)(200) œ 1,497,600 ft † lb
œ (62.4)(120) ˆ 400
#
5 ‰
(b) The time t it takes to empty the full tank with ˆ 11
–hp motor is t œ
W
†lb
250 ftsec
œ
1,497,600 ft†lb
†lb
250 ftsec
œ 5990.4 sec œ 1.664 hr
Ê t ¸ 1 hr and 40 min
(c) Following all the steps of part (a), we find that the work it takes to lower the water level 10 ft is
W œ '0 62.4 † 120y dy œ (62.4)(120) ’ y# “
10
#
œ 1497.6 sec œ 0.416 hr ¸ 25 min
(d) In a location where water weighs 62.26
"!
!
‰ œ 374,400 ft † lb and the time is t œ
œ (62.4)(120) ˆ 100
#
lb
ft$ :
a) W œ (62.26)(24,000) œ 1,494,240 ft † lb.
b) t œ 1,494,240
œ 5976.96 sec ¸ 1.660 hr Ê t ¸ 1 hr and 40 min
250
In a location where water weighs 62.59
lb
ft$
a) W œ (62.59)(24,000) œ 1,502,160 ft † lb
b) t œ 1,502,160
œ 6008.64 sec ¸ 1.669 hr Ê t ¸ 1 hr and 40.1 min
250
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
W
†lb
250 ftsec
360
Chapter 6 Applications of Definite Integrals
14. We will use the coordinate system given.
(a) The typical slab between the planes at y and y ?y has
a volume of ?V œ (20)(12) ?y œ 240 ?y ft$ . The force
F required to lift the slab is equal to its weight:
F œ 62.4 ?V œ 62.4 † 240 ?y lb. The distance through
which F must act is about y ft, so the work done lifting
the slab is about ?W œ force ‚ distance
20
œ 62.4 † 240 † y † ?y ft † lb. The work it takes to lift all the water is approximately W ¸ ! ?W
10
20
œ ! 62.4 † 240y † ?y ft † lb. This is a Riemann sum for the function 62.4 † 240y over the interval
10
10 Ÿ y Ÿ 20. The work it takes to empty the cistern is the limit of these sums: W œ '10 62.4 † 240y dy
20
#
œ (62.4)(240) ’ y# “
(b) t œ
W
†lb
275 ftsec
œ
#!
œ (62.4)(240)(200 50) œ (62.4)(240)(150) œ 2,246,400 ft † lb
"!
2,246,400 ft†lb
275
¸ 8168.73 sec ¸ 2.27 hours ¸ 2 hr and 16.1 min
(c) Following all the steps of part (a), we find that the work it takes to empty the tank halfway is
W œ '10 62.4 † 240y dy œ (62.4)(240) ’ y# “
15
#
Then the time is t œ
W
†lb
275 ftsec
œ
936,000
#75
"&
"!
œ (62.4)(240) ˆ 225
# 100 ‰
#
‰ œ 936,000 ft.
œ (62.4)(240) ˆ 125
#
¸ 3403.64 sec ¸ 56.7 min
lb
ft$ :
(d) In a location where water weighs 62.26
a) W œ (62.26)(240)(150) œ 2,241,360 ft † lb.
b) t œ 2,241,360
œ 8150.40 sec œ 2.264 hours ¸ 2 hr and 15.8 min
275
‰ œ 933,900 ft † lb; t œ 933,900
c) W œ (62.26)(240) ˆ 125
#
#75 œ 3396 sec ¸ 0.94 hours ¸ 56.6 min
lb
ft$
In a location where water weighs 62.59
a) W œ (62.59)(240)(150) œ 2,253,240 ft † lb.
b) t œ 2,253,240
œ 8193.60 sec œ 2.276 hours ¸ 2 hr and 16.56 min
275
‰ œ 938,850 ft † lb; t œ 938,850
c) W œ (62.59)(240) ˆ 125
#
275 ¸ 3414 sec ¸ 0.95 hours ¸ 56.9 min
#
15. The slab is a disk of area 1x# œ 1ˆ y# ‰ , thickness ˜y, and height below the top of the tank a"! yb. So the work to pump
#
the oil in this slab, ˜W, is 57a"! yb1ˆ y# ‰ . The work to pump all the oil to the top of the tank is
W œ '0
10
571
#
4 a"!y
y$ bdy œ
571
4
$
’ "!$y "!
y%
% “!
œ 11,8751 ft † lb ¸ 37,306 ft † lb.
#
16. Each slab of oil is to be pumped to a height of 14 ft. So the work to pump a slab is a"% yba1bˆ y# ‰ and since the tank is
half full and the volume of the original cone is V œ "$ 1r# h œ "$ 1a&# ba"!b œ
with half the volume the cone is filled to a height y,
œ
571 "%y$
4 ’ $
$
È
&!!
y%
“
% !
#&!1
'
#&!1
$
ft3 , half the volume œ
$
È
$
œ $" 1 y% y Ê y œ È
&!! ft. So W œ '0
#
&!!
#&!1
'
571
#
4 a"%y
ft3 , and
y$ b dy
¸ 60,042 ft † lb.
#
‰ ?y
17. The typical slab between the planes at y and and y ?y has a volume of ?V œ 1(radius)# (thickness) œ 1 ˆ 20
#
œ 1 † 100 ?y ft$ . The force F required to lift the slab is equal to its weight: F œ 51.2 ?V œ 51.2 † 1001 ?y lb
Ê F œ 51201 ?y lb. The distance through which F must act is about (30 y) ft. The work it takes to lift all the
30
30
kerosene is approximately W ¸ ! ?W œ ! 51201(30 y) ?y ft † lb which is a Riemann sum. The work to pump the
0
0
tank dry is the limit of these sums: W œ '0 51201(30 y) dy œ 51201 ’30y 30
¸ 7,238,229.48 ft † lb
$!
y#
# “!
‰ œ (5120)(4501)
œ 51201 ˆ 900
#
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 6.5 Work and Fluid Forces
18. (a) Follow all the steps of Example 5 but make the substitution of 64.5
W œ '0
8
œ
64.51
4
64.51†8$
3
(10 y)y# dy œ
64.51
4
$
’ 10y
3 %
y
4
)
“ œ
!
64.51
4
$
Š 103†8 lb
ft$
%
8
4
lb
ft$ .
for 57
361
Then,
1‰
‰
a8$ b ˆ 10
‹ œ ˆ 64.5
4
3 2
œ 21.51 † 8$ ¸ 34,582.65 ft † lb
(b) Exactly as done in Example 5 but change the distance through which F acts to distance ¸ (13 y) ft. Then
W œ '0
8
571
4
(13 y)y# dy œ
571
4
$
’ 13y
3 œ (191) a8# b (7)(2) ¸ 53,482.5 ft † lb
)
y%
4 “!
œ
571
4
$
Š 133†8 8%
4‹
‰
œ ˆ 5741 ‰ a8$ b ˆ 13
3 2 œ
571†8$ †7
3 †4
#
19. The typical slab between the planes at y and y?y has a volume of about ?V œ 1(radius)# (thickness) œ 1 ˆÈy‰ ?y ft$ .
#
The force F(y) required to lift this slab is equal to its weight: F(y) œ 73 † ?V œ 731 ˆÈy‰ ?y œ 731 y ?y lb. The
distance through which F(y) must act to lift the slab to the top of the reservoir is about a4 yb ft, so the work done is
approximately ?W ¸ 731 y a4 yb?y ft † lb. The work done lifting all the slabs from y œ 0 ft to y œ 4 ft is
n
approximately W ¸ ! 731 yk a4 yk b?y ft † lb. Taking the limit of these Riemann sums as n Ä _, we get
kœ0
4
4
4
1
‰ 2336
W œ '0 731 y a4 ybdy œ 731'0 a4y y2 bdy œ 731 2y# 13 y$ ‘ 0 œ 731ˆ32 64
ft † lb.
3 œ
3
20. The typical slab between the planes at y and y?y has a volume of about ?V œ (length)(width)(thickness)
œ ˆ2È25 y2 ‰a10b ?y ft$ . The force F(y) required to lift this slab is equal to its weight: F(y) œ 53 † ?V
œ 53ˆ2È25 y2 ‰a10b ?y œ 1060È25 y2 ?y lb. The distance through which F(y) must act to lift the slab to the
level of 15 m above the top of the reservoir is about a20 yb ft, so the work done is approximately
?W ¸ 1060È25 y2 a20 yb?y ft † lb. The work done lifting all the slabs from y œ 5 ft to y œ 5 ft is
n
approximately W ¸ ! 1060É25 y2k a20 yk b?y ft † lb. Taking the limit of these Riemann sums as n Ä _, we get
kœ0
W œ 'c5 1060È25 y2 a20 ybdy œ 1060'c5 a20 ybÈ25 y2 dy œ 1060”'c5 20 È25 y2 dy 'c5 y È25 y2 dy•
5
5
5
5
To evaluate the first integral, we use we can interpret 'c5 È25 y2 dy as the area of the semicircle whose radius is 5, thus
5
'c55 20È25 y2 dy œ 20'c55 È25 y2 dy œ 20 "# 1a5b2 ‘ œ 2501. To evaluate the second integral let u œ 25 y2
5
0
Ê du œ 2y dy; y œ 5 Ê u œ 0, y œ 5 Ê u œ 0, thus 'c5 y È25 y2 dy œ "# '0 Èu du œ 0. Thus,
1060”'c5 20 È25 y2 dy 'c5 y È25 y2 dy• œ 1060a2501 0b œ 2650001 ¸ 832522 ft † lb.
5
5
21. The typical slab between the planes at y and y?y has a volume of about ?V œ 1(radius)# (thickness)
#
œ 1 ˆÈ25 y# ‰ ?y m$ . The force F(y) required to lift this slab is equal to its weight: F(y) œ 9800 † ?V
#
œ 98001 ˆÈ25 y# ‰ ?y œ 98001 a25 y# b ?y N. The distance through which F(y) must act to lift the
slab to the level of 4 m above the top of the reservoir is about (4 y) m, so the work done is approximately
?W ¸ 98001 a25 y# b (4 y) ?y N † m. The work done lifting all the slabs from y œ 5 m to y œ 0 m is
0
approximately W ¸ ! 98001 a25 y# b (4 y) ?y N † m. Taking the limit of these Riemann sums, we get
c5
W œ 'c5 98001 a25 y# b (4 y) dy œ 98001 'c5 a100 25y 4y# y$ b dy œ 98001 ’100y 0
0
œ 98001 ˆ500 25†25
#
4
3
† 125 625 ‰
4
25
#
y# 34 y$ ¸ 15,073,099.75 J
22. The typical slab between the planes at y and y?y has a volume of about ?V œ 1(radius)# (thickness)
#
œ 1 ˆÈ100 y# ‰ ?y œ 1 a100 y# b ?y ft$ . The force is F(y) œ 56ft$lb † ?V œ 561 a100 y# b ?y lb. The
distance through which F(y) must act to lift the slab to the level of 2 ft above the top of the tank is about
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
!
y%
4 “ &
362
Chapter 6 Applications of Definite Integrals
(12 y) ft, so the work done is ?W ¸ 561 a100 y# b (12 y) ?y lb † ft. The work done lifting all the slabs
10
from y œ 0 ft to y œ 10 ft is approximately W ¸ ! 561 a100 y# b (12 y) ?y lb † ft. Taking the limit of these
0
Riemann sums, we get W œ '0 561 a100 y b (12 y) dy œ 561'0 a100 y# b (12 y) dy
10
10
#
œ 561'0 a1200 100y 12y# y$ b dy œ 561 ’1200C 10
œ 561 ˆ12,000 10,000
#
4 † 1000 10,000 ‰
4
100y#
#
12y$
3
"!
y%
4 “!
œ (561) ˆ12 5 4 5# ‰ (1000) ¸ 967,611 ft † lb.
It would cost (0.5)(967,611) œ 483,805¢ œ $4838.05. Yes, you can afford to hire the firm.
23. F œ m
œ
"
#
dv
dt
œ mv
dv
dx
#
by the chain rule Ê W œ 'x mv
x#
m cv# (x# ) v (x" )d œ
24. weight œ 2 oz œ
dv
dx
"
"
#
weight
32
œ
"
8
œ
3#
"
#56
26. weight œ 1.6 oz œ 0.1 lb Ê m œ
0.1 lb
32 ft/sec#
œ
ft
27. v1 œ 0 mph œ 0 sec
, v2 œ 153 mph œ 224.4
W œ 'x Faxb dx œ
"
28. weight œ 6.5 oz œ
"
#
6.5
16
mv##
"
dv ‰
dx
dx œ m "# v# (x)‘ x"
x#
"
slugs; W œ ˆ #" ‰ ˆ #56
slugs‰ (160 ft/sec)# ¸ 50 ft † lb
hr
1 min
5280 ft
25. 90 mph œ 901 hrmi † 601 min
† 60
sec † 1 mi œ 132 ft/sec; m œ
0.3125 lb ‰
#
W œ ˆ "# ‰ ˆ 32
ft/sec# (132 ft/sec) ¸ 85.1 ft † lb
x#
x#
mv## "# mv"# , as claimed.
lb; mass œ
2
16
dx œ m'x ˆv
"
#
mv1#
lb Ê m œ
œ
"
3 #0
ft
sec ;
œ
0.3125
32
slugs;
slugs; W œ ˆ "# ‰ ˆ 3"#0 slugs‰ (280 ft/sec)# œ 122.5 ft † lb
2 oz œ 0.125 lb Ê m œ
2
"ˆ " ‰
# 256 a224.4b
6.5
(16)(32)
0.3125 lb
32 ft/sec#
2
"ˆ " ‰
# 256 a!b
0.125 lb
32 ft/sec#
œ
"
256
slugs;
œ 98.35 ft-lb.
6.5
slugs; W œ ˆ "# ‰ Š (16)(32)
slugs‹ (132 ft/sec)# ¸ 110.6 ft † lb
29. We imagine the milkshake divided into thin slabs by planes perpendicular to the y-axis at the points of a partition of the
interval [!ß (]. The typical slab between the planes at y and y ?y has a volume of about?V œ 1(radius)# (thickness)
#
œ 1 ˆ y 1417.5 ‰ ?y in$ . The force F(y) required to lift this slab is equal to its weight: F(y) œ
4
9
?V œ
41
9
#
ˆ y 1417.5 ‰ ?y oz.
The distance through which F(y) must act to lift this slab to the level of 1 inch above the top is about (8 y) in. The work
#
b
done lifting the slab is about ?W œ ˆ 491 ‰ ay 1417.5
a8 yb?y in † oz. The work done lifting all the slabs from y œ 0 to
#
7
y œ 7 is approximately W œ !
0
41
9†14# (y
17.5)# (8 y) ?y in † oz which is a Riemann sum. The work is the limit of these
sums as the norm of the partition goes to zero: W œ '0
7
œ
41
9†14#
œ
41
9†14#
'07 a2450 26.25y 27y# y$ bdy œ 9†4141
7%
4
’
$
9†7 30. Work œ '6 370 000
35ß780ß000
ß
ß
1000 MG
r#
26.25
#
41
9†14# ay
%
#
17.5b# a8 ybdy
’ y4 9y$ 26.25
#
y# 2450y“
(
!
#
† 7 2450 † 7“ ¸ 91.32 in † oz
dr œ 1000 MG '6 370 000
35ß780ß000
ß
ß
"
œ (1000) a5.975 † 10#% b a6.672 † 10"" b Š 6,370,000
dr
r#
$&ß()!ß!!!
œ 1000 MG "r ‘ 'ß$(!ß!!!
"
35,780,000 ‹
¸ 5.144 ‚ 10"! J
31. To find the width of the plate at a typical depth y, we first find an equation for the line of the plate's
right-hand edge: y œ x 5. If we let x denote the width of the right-hand half of the triangle at depth y, then
x œ 5 y and the total width is L(y) œ 2x œ 2(5 y). The depth of the strip is (y). The force exerted by the
c2
c2
water against one side of the plate is therefore F œ 'c5 w(y) † L(y) dy œ 'c5 62.4 † (y) † 2(5 y) dy
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 6.5 Work and Fluid Forces
c2
œ 124.8 'c5 a5y y# b dy œ 124.8 5# y# "3 y$ ‘ & œ 124.8 ˆ 5# † 4 œ
(124.8) ˆ 105
#
#
117 ‰
3
œ
(124.8) ˆ 315 6 234 ‰
"
3
† 8‰ ˆ 5# † 25 "
3
363
† 125‰‘
œ 1684.8 lb
32. An equation for the line of the plate's right-hand edge is y œ x 3 Ê x œ y 3. Thus the total width is
L(y) œ 2x œ 2(y 3). The depth of the strip is (2 y). The force exerted by the water is
F œ 'c3 w(2 y)L(y) dy œ 'c3 62.4 † (2 y) † 2(3 y) dy œ 124.8'c3 a6 y y# b dy œ 124.8 ’6y 0
0
œ (124.8) ˆ18 9
#
0
y#
#
!
y$
3 “ $
‰
9‰ œ (124.8) ˆ 27
# œ 1684.8 lb
strip
33. (a) The width of the strip is Layb œ 4, the depth of the strip is a10 yb Ê F œ 'a w † Š depth
‹Faybdy
b
œ '0 62.4a10 yba4bdy œ 249.6'0 a10 ybdy œ 249.6’10y 3
3
3
y#
“
# 0
œ 249.6ˆ30 92 ‰ œ 6364.8 lb
strip
(b) The width of the strip is Layb œ 3, the depth of the strip is a10 yb Ê F œ 'a w † Š depth
‹Faybdy
b
œ '0 62.4a10 yba3bdy œ 187.2'0 a10 ybdy œ 187.2’10y 4
4
4
y#
# “0
œ 187.2a40 8b œ 5990.4 lb
strip
34. The width of the strip is Layb œ 2È25 y2 , the depth of the strip is a6 yb Ê F œ 'a w † Š depth
‹Faybdy
b
œ '0 62.4a6 ybˆ2È25 y2 ‰dy œ 124.8'0 a6 ybÈ25 y2 dy œ 124.8”'0 6 È25 y2 dy '0 y È25 y2 dy•
5
5
5
5
To evaluate the first integral, we use we can interpret '0 È25 y2 dy as the area of a quarter circle whose radius is 5, thus
5
'05 6È25 y2 dy œ 6'05 È25 y2 dy œ 6 4" 1a5b2 ‘ œ 7521 . To evaluate the second integral let u œ 25 y2
Ê du œ 2y dy; y œ 0 Ê u œ 25, y œ 5 Ê u œ 0, thus '0 y È25 y2 dy œ "# '25 Èu du œ
5
25
œ 13 u3Î2 ‘ 0 œ
125
3 .
0
Thus, 124.8”'0 6 È25 y2 dy '0 y È25 y2 dy• œ 124.8ˆ 7521 5
5
125 ‰
3
"
#
'025 u1Î2 du
¸ 9502.7 lb.
35. Using the coordinate system of Exercise 32, we find the equation for the line of the plate's right-hand edge to be
y œ 2x 4 Ê x œ y # 4 and L(y) œ 2x œ y 4. The depth of the strip is (1 y).
(a) F œ 'c4 w(1 y)L(y) dy œ 'c4 62.4 † (1 y)(y 4) dy œ 62.4 'c4 a4 3y y# b dy œ 62.4 ’4y 0
0
œ (62.4) ’(4)(4) (3)(16)
#
(b) F œ (64.0) ’(4)(4) (3)(16)
#
0
64
3 “
œ (62.4) ˆ16 24 64
3 “
œ
(64.0)(120 64)
3
64 ‰
3
œ
(62.4)(120 64)
3
œ 1164.8 lb
¸ 1194.7 lb
36. Using the coordinate system given, we find an equation for
the line of the plate's right-hand edge to be y œ 2x 4
Ê x œ 4#y and L(y) œ 2x œ 4 y. The depth of the
strip is (1 y) Ê F œ '0 w(1 y)(4 y) dy
1
œ 62.4'0 ay# 5y 4b dy œ 62.4 ’ y3 1
œ (62.4) ˆ "3 $
5
#
5y#
#
4‰ œ (62.4) ˆ 2 156 24 ‰ œ
4y“
"
!
(62.4)(11)
6
œ 114.4 lb
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
3y#
#
!
y$
3 “ %
364
Chapter 6 Applications of Definite Integrals
37. Using the coordinate system given in the accompanying
figure, we see that the total width is L(y) œ 63 and the depth
of the strip is (33.5 y) Ê F œ '0 w(33.5 y)L(y) dy
33
œ '0
33
64 ‰
† (33.5 y) † 63 dy œ ˆ 12
(63)'0 (33.5 y) dy
$
33
64
1 #$
64 ‰
œ ˆ 12
(63) ’33.5y $
œ
(64)(63)(33)(67 33)
(#) a12$ b
$$
y#
# “!
‰ ’(33.5)(33) œ ˆ 641#†63
$
33#
# “
œ 1309 lb
38. Using the coordinate system given in the accompanying
figure, we see that the right-hand edge is x œ È1 y#
so the total width is L(y) œ 2x œ 2È1 y# and the depth
of the strip is (y). The force exerted by the water is
therefore F œ 'c1 w † (y) † 2È1 y# dy
0
œ 62.4'c1 È1 y# d a1 y# b œ 62.4 ’ 23 a1 y# b
0
39. (a) F œ ˆ62.4
lb ‰
ft3 a8
$Î# !
“
"
œ (62.4) ˆ 23 ‰ (1 0) œ 41.6 lb
ftba25 ft2 b œ 12480 lb
strip
(b) The width of the strip is Layb œ 5, the depth of the strip is a8 yb Ê F œ 'a w † Š depth
‹Faybdy
b
œ '0 62.4a8 yba5bdy œ 312'0 a8 ybdy œ 312’8y 5
5
5
y#
“
# 0
œ 312ˆ40 25 ‰
2
œ 8580 lb
(c) The width of the strip is Layb œ 5, the depth of the strip is a8 yb, the height of the strip is È2 dy
5ÎÈ2
strip
Ê F œ 'a w † Š depth
‹Faybdy œ '0
b
40
œ 312È2 Š È
2
25
4 ‹
5ÎÈ2
62.4a8 yba5bÈ2 dy œ 312È2 '0
a8 ybdy œ 312È2 ’8y œ 9722.3
40. The width of the strip is Layb œ 34 Š2È3 y‹, the depth of the strip is a6 yb, the height of the strip is
2È 3
strip
Ê F œ 'a w † Š depth
‹Faybdy œ '0
b
œ
93.6
È
È3 ’12y 3
5 ÎÈ 2
y#
“
# 0
3y2 y2 È3 62.4a6 yb † 34 Š2È3 y‹ È23 dy œ
2È 3
y3
“
3 0
œ
93.6
È3 Š72
93.6
È3
'02
È3
2
È3
dy
Š12È3 6y 2yÈ3 y2 ‹dy
36 12È3 8È3 ‹ ¸ 1571.04 lb
41. The coordinate system is given in the text. The right-hand edge is x œ Èy and the total width is L(y) œ 2x œ 2Èy.
(a) The depth of the strip is (2 y) so the force exerted by the liquid on the gate is F œ '0 w(2 y)L(y) dy
1
"
œ '0 50(2 y) † 2Èy dy œ 100 '0 (2 y)Èy dy œ 100'0 ˆ2y"Î# y$Î# ‰ dy œ 100 43 y$Î# 25 y&Î# ‘ !
1
1
1
‰
œ 100 ˆ 43 25 ‰ œ ˆ 100
15 (20 6) œ 93.33 lb
2‰
(b) We need to solve 160 œ '0 w(H y) † 2Èy dy for h. 160 œ 100 ˆ 2H
3 5 Ê H œ 3 ftÞ
1
42. Suppose that h is the maximum height. Using the coordinate system given in the text, we find an equation for
the line of the end plate's right-hand edge is y œ 5# x Ê x œ 52 y. The total width is L(y) œ 2x œ 54 y and the
depth of the typical horizontal strip at level y is (h y). Then the force is F œ '0 w(h y)L(y) dy œ Fmax ,
h
where Fmax œ 6667 lb. Hence, Fmax œ w'0 (h y) † 45 y dy œ (62.4) ˆ 45 ‰'0 ahy y# b dy
h
#
œ (62.4) ˆ 45 ‰ ’ hy# h
y$
3 “0
$
œ (62.4) ˆ 45 ‰ Š h# h
h$
3‹
3
max ‰
ˆ 54 ‰ ˆ F10.4
œ (62.4) ˆ 45 ‰ ˆ "6 ‰ h$ œ (10.4) ˆ 45 ‰ h$ Ê h œ É
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 6.6 Moments and Centers of Mass
3
ˆ 54 ‰ ˆ 6667
‰
σ
10.4 ¸ 9.288 ft. The volume of water which the tank can hold is V œ
Height œ h and
"
#
(Base) œ
2
5
h Ê Vœ
ˆ 25
#‰
h
#
"
#
365
(Base)(Height) † 30, where
#
(30) œ 12h ¸ 12(9.288) ¸ 1035 ft$ .
43. The pressure at level y is p(y) œ w † y Ê the average
pressure is p œ
#
œ ˆ wb ‰ Š b# ‹ œ
'0b p(y) dy œ b" '0b w † y dy œ b" w ’ y# “ b
#
"
b
0
wb
#
. This is the pressure at level
b
#
, which
is the pressure at the middle of the plate.
44. The force exerted by the fluid is F œ '0 w(depth)(length) dy œ '0 w † y † a dy œ (w † a)'0 y dy œ (w † a) ’ y# “
b
œ
#
w Š ab# ‹
b
b
#
b
0
‰
œ ˆ wb
# (ab) œ p † Area, where p is the average value of the pressure.
45. When the water reaches the top of the tank the force on the movable side is 'c2 (62.4) ˆ2È4 y# ‰ (y) dy
0
œ (62.4)'c2 a4 y# b
0
"Î#
(2y) dy œ (62.4) ’ 23 a4 y# b
$Î# !
“
#
œ (62.4) ˆ 23 ‰ ˆ4$Î# ‰ œ 332.8 ft † lb. The force
compressing the spring is F œ 100x, so when the tank is full we have 332.8 œ 100x Ê x ¸ 3.33 ft. Therefore the
movable end does not reach the required 5 ft to allow drainage Ê the tank will overflow.
46. (a) Using the given coordinate system we see that the total
width is L(y) œ 3 and the depth of the strip is (3 y).
Thus, F œ '0 w(3 y)L(y) dy œ '0 (62.4)(3 y) † 3 dy
3
3
œ (62.4)(3)'0 (3 y) dy œ (62.4)(3) ’3y 3
$
y#
# “!
œ (62.4)(3) ˆ9 9# ‰ œ (62.4)(3) ˆ 9# ‰ œ 842.4 lb
(b) Find a new water level Y such that FY œ (0.75)(842.4 lb) œ 631.8 lb. The new depth of the strip is (Y y) and Y is
the new upper limit of integration. Thus, FY œ '0 w(Y y)L(y) dy œ 62.4'0 (Y y) † 3 dy
Y
œ (62.4)(3)'0 (Y y) dy œ (62.4)(3) ’Yy Y
Y
y#
# “0
Y
œ (62.4)(3) ŠY# Y#
# ‹
#
œ (62.4)(3) Š Y# ‹ . Therefore,
2FY
È6.75 ¸ 2.598 ft. So, ?Y œ 3 Y ¸ 3 2.598 ¸ 0.402 ft ¸ 4.8 in
Y œ É (62.4)(3)
œ É 1263.6
187.2 œ
6.6 MOMENTS AND CENTERS OF MASS
1. Since the plate is symmetric about the y-axis and its density is
constant, the distribution of mass is symmetric about the y-axis
and the center of mass lies on the y-axis. This means that
x œ 0. It remains to find y œ MMx . We model the distribution of
mass with @/<>3-+6 strips. The typical strip has center of mass:
#
(µ
x ßµ
y ) œ Šxß x 4 ‹ , length: 4 x# , width: dx, area:
#
dA œ a4 x# b dx, mass: dm œ $ dA œ $ a4 x# b dx. The moment of the strip about the x-axis is
#
µ
C dm œ Š x #4 ‹ $ a4 x# b dx œ #$ a16 x% b dx. The moment of the plate about the x-axis is Mx œ ' µ
C dm
œ 'c2 #$ a16 x% b dx œ
2
$
#
’16x #
x&
“
5 #
œ
$
#
’Š16 † 2 2&
5‹
Š16 † 2 2&
5 ‹“
œ
$ †2
#
ˆ32 32 ‰
5
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
œ
128$
5 .
The mass of the
366
Chapter 6 Applications of Definite Integrals
plate is M œ ' $ a4 x# b dx œ $ ’4x #
x$
“
3 #
32$
3 .
œ 2$ ˆ8 83 ‰ œ
Therefore y œ
Mx
M
œ
$
Š 128
5 ‹
Š 323$ ‹
œ
12
5 .
The plate's center of
‰
mass is the point (xß y) œ ˆ!ß 12
5 .
2. Applying the symmetry argument analogous to the one in
Exercise 1, we find x œ 0. To find y œ MMx , we use the
@/<>3-+6 strips technique. The typical strip has center of
#
mass: (µ
x ßµ
y ) œ Šxß 25 x ‹ , length: 25 x# , width: dx,
#
#
area: dA œ a25 x bdx, mass: dm œ $ dA œ $ a25 x# b dx.
The moment of the strip about the x-axis is
#
µ
y dm œ Š 25 # x ‹ $ a25 x# b dx œ
œ 'c5 #$ a25 x# b dx œ
5
#
œ $ † 625 ˆ5 œ 2$ Š5$ 10
3
5$
3‹
$
#
'c55
$
#
a25 x# b dx. The moment of the plate about the x-axis is Mx œ ' µ
y dm
#
$
#
a625 50x# x% b dx œ
’625x 50
3
x$ &
x&
5 “ &
œ 2 † #$ Š625 † 5 50
3
† 5$ 1‰ œ $ † 625 † ˆ 38 ‰ . The mass of the plate is M œ ' dm œ 'c5 $ a25 x# b dx œ $ ’25x œ
5
4
3
$ † 5$ . Therefore y œ
Mx
M
œ
$ †5% † ˆ 83 ‰
$ †5$ †ˆ 43 ‰
5&
5‹
&
x$
“
3 &
œ 10. The plate's center of mass is the point (xß y) œ (!ß 10).
3. Intersection points: x x# œ x Ê 2x x# œ 0
Ê x(2 x) œ 0 Ê x œ 0 or x œ 2. The typical @/<>3-+6
#
strip has center of mass: (µ
x ßµ
y ) œ Šxß ax x b (x) ‹
#
#
œ Šxß x# ‹ , length: ax x# b (x) œ 2x x# , width: dx,
area: dA œ a2x x# b dx, mass: dm œ $ dA œ $ a2x x# b dx.
The moment of the strip about the x-axis is
#
µ
y dm œ Š x# ‹ $ a2x x# b dx; about the y-axis it is µ
x dm œ x † $ a2x x# b dx. Thus, Mx œ ' µ
y dm
œ '0 ˆ #$ x# ‰ a2x x# b dx œ #$ '0 a2x$ x% b dx œ #$ ’ x# 2
2
%
#
x&
5 “!
œ #$ Š2$ œ 45$ ; My œ ' µ
x dm œ '0 x † $ a2x x# b dx œ $ '0 a2x# x$ b œ $ ’ 23 x$ 2
2
M œ ' dm œ '0 $ a2x x# b dx œ $ '0 a2x x# b dx œ $ ’x# 2
2
œ ˆ 43$ ‰ ˆ 43$ ‰ œ 1 and y œ
Mx
M
#
x$
3 “!
#
x%
4 “!
2&
5‹
œ #$ † 2$ ˆ1 45 ‰
œ $ Š2 †
œ $ ˆ4 38 ‰ œ
4$
3
2$
3
2%
4‹
œ
. Therefore, x œ
œ ˆ 45$ ‰ ˆ 43$ ‰ œ 53 Ê (xß y) œ ˆ1ß 53 ‰ is the center of mass.
4. Intersection points: x# 3 œ 2x# Ê 3x# 3 œ 0
Ê 3(x 1)(x 1) œ 0 Ê x œ 1 or x œ 1. Applying the
symmetry argument analogous to the one in Exercise 1, we
find x œ 0. The typical @/<>3-+6 strip has center of mass:
#
#
#
(µ
x ßµ
y ) œ Šxß 2x ax 3b ‹ œ Šxß x 3 ‹ ,
#
#
length: 2x# ax# 3b œ 3 a1 x# b, width: dx,
area: dA œ 3 a1 x# b dx, mass: dm œ $ dA œ 3$ a1 x# b dx.
The moment of the strip about the x-axis is
µ
y dm œ 3 $ ax# 3b a1 x# b dx œ 3 $ ax% 3x# x# 3b dx œ
#
œ
3
#
$ 'c1 ax% 2x# 3b dx œ
1
#
3
#
&
$ ’ x5 2x$
3
3x“
"
"
œ
3
#
3
#
$ ax% 2x# 3b dx; Mx œ ' µ
y dm
† $ † 2 ˆ 5" 2
3
45 ‰
3‰ œ 3$ ˆ 310
œ 325$ ;
15
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
$ †2%
1#
My
M
œ
4$
3
;
Section 6.6 Moments and Centers of Mass
M œ ' dm œ 3$ 'c1 a1 x# b dx œ 3$ ’x "
x$
3 “ "
1
œ 3$ † 2 ˆ1 3" ‰ œ 4$ . Therefore, y œ
Mx
M
367
œ 5$††$32†4 œ 58
Ê (xß y) œ ˆ0ß 85 ‰ is the center of mass.
5. The typical 29<3D98>+6 strip has center of mass:
$
(µ
x ßµ
y ) œ Š y y ß y‹ , length: y y$ , width: dy,
#
area: dA œ ay y$ b dy, mass: dm œ $ dA œ $ ay y$ b dy.
The moment of the strip about the y-axis is
$
#
µ
x dm œ $ Š y y ‹ ay y$ b dy œ $ ay y$ b dy
#
œ
$
#
#
ay# 2y% y' b dy; the moment about the x-axis is
1
$
µ
y dm œ $ y ay y$ b dy œ $ ay# y% b dy. Thus, Mx œ ' µ
y dm œ $ '0 ay# y% b dy œ $ ’ y3 My œ ' µ
x dm œ
$
#
'01 ay# 2y% y' b dy œ #$ ’ y3
$
œ $ '0 ay y$ b dy œ $ ’ y# 1
œ
#
"
y%
4 “!
2y&
5
œ $ ˆ "# 4" ‰ œ
$
4
"
y(
7 “!
œ
$
#
ˆ "3 . Therefore, x œ
2
5
7" ‰ œ
$
#
œ $ ˆ "3 "5 ‰ œ
15 ‰
ˆ 35 3†42
œ
5†7
4$ ‰ ˆ 4 ‰
œ ˆ 105
$ œ
My
M
"
y&
5 “!
16
105
2$
15
;
4$
105
; M œ ' dm
Mx
M
2$ ‰ ˆ 4 ‰
œ ˆ 15
$
and y œ
16 8 ‰
Ê (xß y) œ ˆ 105
ß 15 is the center of mass.
8
15
6. Intersection points: y œ y# y Ê y# 2y œ 0
Ê y(y 2) œ 0 Ê y œ 0 or y œ 2. The typical
29<3D98>+6 strip has center of mass:
#
#
(µ
x ßµ
y ) œ Š ay yb y ß y‹ œ Š y ß y‹ ,
#
2
#
#
length: y ay yb œ 2y y , width: dy,
area: dA œ a2y y# b dy, mass: dm œ $ dA œ $ a2y y# b dy.
The moment about the y-axis is µ
x dm œ #$ † y# a2y y# b dy
œ #$ a2y$ y% b dy; the moment about the x-axis is µ
y dm œ $ y a2y y# b dy œ $ a2y# y$ b dy. Thus,
Mx œ ' µ
y dm œ '0 $ a2y# y$ b dy œ $ ’ 2y3 2
œ '0
2
$
#
$
a2y$ y% b dy œ
œ $ ’y# #
y$
3 “!
$
#
%
’ y2 œ $ ˆ4 83 ‰ œ
#
y&
“
5 !
4$
3
œ
$
#
ˆ8 #
y%
4 “!
16$
1#
ˆ 405 32 ‰ œ
4$
5
; M œ ' dm œ '0 $ a2y y# b dy
œ
$
#
My
M
œ ˆ 45$ ‰ ˆ 43$ ‰ œ
32 ‰
5
. Therefore, x œ
œ
(4 3) œ
4$
3
; My œ ' µ
x dm
16 ‰
4
œ $ ˆ 16
3 2
3
5
and y œ
Mx
M
œ ˆ 43$ ‰ ˆ 43$ ‰ œ 1
Ê (xß y) œ ˆ 35 ß "‰ is the center of mass.
7. Applying the symmetry argument analogous to the one used
in Exercise 1, we find x œ 0. The typical @/<>3-+6 strip has
center of mass: (µ
x ßµ
y ) œ ˆxß cos# x ‰ , length: cos x, width: dx,
area: dA œ cos x dx, mass: dm œ $ dA œ $ cos x dx. The
moment of the strip about the x-axis is µ
y dm œ $ † cos# x † cos x dx
2x ‰
œ #$ cos# x dx œ #$ ˆ 1 cos
dx œ 4$ (1 cos 2x) dx; thus,
#
1Î2
Mx œ ' µ
y dm œ 'c1Î2 4$ (1 cos 2x) dx œ
1Î#
œ $ [sin x]1Î# œ 2$ . Therefore, y œ
Mx
M
œ
$
4
x $1
4 †# $
œ
sin 2x ‘ 1Î#
#
1Î#
1
8
œ
$
4
ˆ 1# 0‰ ˆ 1# ‰‘ œ
$1
4
1Î2
; M œ ' dm œ $ ' 1Î2 cos x dx
Ê (xß y) œ ˆ!ß 18 ‰ is the center of mass.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
368
Chapter 6 Applications of Definite Integrals
8. Applying the symmetry argument analogous to the one used
in Exercise 1, we find x œ 0. The typical vertical strip has
#
center of mass: (µ
x ßµ
y ) œ Šxß sec# x ‹ , length: sec# x, width: dx,
area: dA œ sec# x dx, mass: dm œ $ dA œ $ sec# x dx. The
#
moment about the x-axis is µ
y dm œ Š sec x ‹ a$ sec# xb dx
#
1Î4
sec% x dx. Mx œ 'c1Î4 µ
y dm œ
$
#
' 11ÎÎ44 sec% x dx
œ
$
#
œ
$
#
'c11ÎÎ44 atan# x 1b asec# xb dx œ #$ ' 11ÎÎ44 (tan x)# asec# xb dx #$ ' 11ÎÎ44 sec# x dx œ 2$ ’ (tan3x) “ 1Î4
œ
$
2
3" ˆ 3" ‰‘ #$ [1 (1)] œ
$
1 Î4
Therefore, y œ
Mx
M
œ ˆ 43$ ‰ ˆ 2"$ ‰ œ
2
3
$
3
$ œ
4$
3
1Î%
#$ [tan x]1Î%
; M œ ' dm œ $ 'c1Î4 sec# x dx œ $ [tan x]1Î4 œ $ [1 (1)] œ 2$ .
1Î4
1Î4
Ê (xß y) œ ˆ!ß 32 ‰ is the center of mass.
9. Since the plate is symmetric about the line x œ 1 and its
density is constant, the distribution of mass is symmetric
about this line and the center of mass lies on it. This means
that x œ 1. The typical @/<>3-+6 strip has center of mass:
#
#
#
(µ
x ßµ
y ) œ Šxß a2x x ba2x 4xb ‹ œ Šxß x 2x ‹ ,
#
#
#
#
#
length: a2x x b a2x 4xb œ 3x 6x œ 3 a2x x# b ,
width: dx, area: dA œ 3 a2x x# b dx, mass: dm œ $ dA
œ 3$ a2x x# b dx. The moment about the x-axis is
#
µ
y dm œ 3 $ ax# 2xb a2x x# b dx œ 3 $ ax# 2xb dx
#
#
œ $ ax 4x 4x b dx. Thus, Mx œ ' µ
y dm œ '0
%
3
#
œ $
3
2
&
Š 25
$
%
2 2
#
4
3
&
$ ax% 4x$ 4x# b dx œ 23 $ ’ x5 x% 34 x$ “
10 ‰
† 2 ‹ œ $ † 2 ˆ 25 1 23 ‰ œ 3# $ † 2% ˆ 6 15
œ 85$ ; M œ ' dm
15
$
œ '0 3$ a2x x# b dx œ 3$ ’x# 2
3
2
3
#
#
x$
3 “!
#
!
%
œ 3$ ˆ4 83 ‰ œ 4$ . Therefore, y œ
Mx
M
œ ˆ 85$ ‰ ˆ 4"$ ‰ œ 25
Ê (xß y) œ ˆ1ß 25 ‰ is the center of mass.
10. (a) Since the plate is symmetric about the line x œ y and
its density is constant, the distribution of mass is
symmetric about this line. This means that x œ y. The
typical @/<>3-+6 strip has center of mass:
È
#
(µ
x ßµ
y ) œ Šxß 9 x ‹ , length: È9 x# , width: dx,
#
area: dA œ È9 x# dx,
mass: dm œ $ dA œ $ È9 x# dx.
The moment about the x-axis is
È
#
µ
y dm œ $ Š 9# x ‹ È9 x# dx œ
$
#
a9 x# b dx. Thus, Mx œ ' µ
y dm œ '0
3
$
#
a9 x# b dx œ
$
#
’9x $
x$
“
3 !
(27 9) œ 9$ ; M œ ' dm œ ' $ dA œ $ ' dA œ $ (Area of a quarter of a circle of radius 3) œ $ ˆ 941 ‰ œ
4 ‰
Therefore, y œ MMx œ (9$ ) ˆ 91$
œ 14 Ê (xß y) œ ˆ 14 ß 14 ‰ is the center of mass.
œ
$
#
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
91$
4
.
Section 6.6 Moments and Centers of Mass
369
(b) Applying the symmetry argument analogous to the one
used in Exercise 1, we find that x œ 0. The typical
vertical strip has the same parameters as in part (a).
3
Thus, M œ ' µ
y dm œ ' $ a9 x# b dx
c3 #
x
œ #'
3
$
0 #
a9 x# b dx œ 2(9$ ) œ 18$ ;
M œ ' dm œ ' $ dA œ $ ' dA
œ $ (Area of a semi-circle of radius 3) œ $ ˆ 921 ‰ œ 91$
2 . Therefore, y œ
4‰
ˆ
as in part (a) Ê (xß y) œ 0ß 1 is the center of mass.
Mx
M
2 ‰
œ (18$ ) ˆ 91$
œ
4
1
, the same y
11. Since the plate is symmetric about the line x œ y and its
density is constant, the distribution of mass is symmetric
about this line. This means that x œ y. The typical @/<>3-+6
strip has
È
#
center of mass: (µ
x ßµ
y ) œ Šxß 3 9 x ‹ ,
#
length: 3 È9 x# , width: dx,
area: dA œ Š3 È9 x# ‹ dx,
mass: dm œ $ dA œ $ Š3 È9 x# ‹ dx.
The moment about the x-axis is
µ
y dm œ $
Š3 È9 x# ‹ Š3 È9 x# ‹
#
dx œ
$
#
c9 a9 x# bd dx œ
$ x#
#
dx. Thus, Mx œ '0
3
$ x#
#
dx œ
$
6
$
cx$ d ! œ
#
9$
#
equals the area of a square with side length 3 minus one quarter the area of a disk with radius 3 Ê A œ 3 œ
9
4
9$
4
(4 1) Ê M œ $ A œ
(4 1). Therefore, y œ
œ ˆ 9#$ ‰ ’ 9$(44 1) “ œ
Mx
M
2
41
. The area
19
4
Ê (xß y) œ ˆ 4 2 1 ß 4 2 1 ‰ is the
center of mass.
12. Applying the symmetry argument analogous to the one used
in Exercise 1, we find that y œ 0. The typical @/<>3-+6 strip
has center of mass: (µ
x ßµ
y ) œ Œxß
length:
"
x$
ˆ x"$ ‰ œ
2
x$
"
x$
x"$
#
œ (xß 0),
, width: dx, area: dA œ
2
x$
dx,
2$
x$
mass: dm œ $ dA œ dx. The moment about the y-axis is
a
µ
x dm œ x † 2x$$ dx œ 2x$# dx. Thus, My œ ' µ
x dm œ '1 2x$# dx
œ 2$ x" ‘ " œ 2$ ˆ "a 1‰ œ
a
xœ
My
M
2$ (a1)
a
œ ’ 2$(aa 1) “ ’ $ aa#a 1b “ œ
#
13. Mx œ ' µ
y dm œ '1
2
Š x2# ‹
#
2a
a1
; M œ ' dm œ '1
a
2$
x$
dx œ $ x"# ‘ " œ $ ˆ a"# 1‰ œ
a
$ aa# 1b
a#
. Therefore,
‰
Ê (xß y) œ ˆ a 2a
x œ 2.
1 ß 0 . Also, a lim
Ä_
† $ † ˆ x2# ‰ dx
œ '1 ˆ x"# ‰ ax# b ˆ x2# ‰ dx œ '1
2
2
2
x#
dx œ 2'1 x# dx
2
#
œ 2 cx" d " œ 2 ˆ "# ‰ (1)‘ œ 2 ˆ "# ‰ œ 1;
My œ ' µ
x dm œ '1 x † $ † ˆ x2# ‰ dx
2
œ '1 x ax# b ˆ x2# ‰ dx œ 2'1 x dx œ 2 ’ x# “
2
2
#
#
"
œ 2 ˆ2 "# ‰ œ 4 1 œ 3; M œ ' dm œ '1 $ ˆ x2# ‰ dx œ '1 x# ˆ x2# ‰ dx œ 2'1 dx œ 2[x]"# œ 2(2 1) œ 2. So
xœ
My
M
œ
3
#
and y œ
Mx
M
œ
"
#
2
2
2
Ê (xß y) œ ˆ 3# ß "# ‰ is the center of mass.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
370
Chapter 6 Applications of Definite Integrals
14. We use the @/<>3-+6 strip approach:
1
#
M œ'µ
y dm œ ' ax x b ax x# b † $ dx
x
#
0
'0 ax# x% b † 12x dx
1
"
#
œ
œ 6'0 ax$ x& b dx œ 6 ’ x4 1
%
œ 6 ˆ "4 6" ‰ œ
6
4
1œ
"
#
"
x'
6 “!
;
My œ ' µ
x dm œ '0 x ax x# b † $ dx œ '0 ax# x$ b † 12x dx œ 12'0 ax$ x% b dx œ 12 ’ x4 1
œ
12
#0
xœ
œ
My
M
3
5
œ
1
1
%
; M œ ' dm œ '0 ax x# b † $ dx œ 12'0 ax# x$ b dx œ 12 ’ x3 1
3
5
and y œ
Mx
M
œ
"
#
1
$
"
x%
4 “!
"
x&
5 “!
œ 12 ˆ "4 5" ‰
œ 12 ˆ 3" 4" ‰ œ
12
12
œ 1. So
Ê ˆ 53 ß "# ‰ is the center of mass.
shell ‰
shell
15. (a) We use the shell method: V œ 'a 21 ˆ radius
Š height
‹ dx œ '1 21x ’ È4x Š È4x ‹“ dx œ 161'1
b
4
%
œ 161'1 x"Î# dx œ 161 23 x$Î# ‘ " œ 161 ˆ 23 † 8 32 ‰ œ
4
321
3
4
(8 1) œ
(b) Since the plate is symmetric about the x-axis and its density $ (x) œ
"
x
x
Èx
dx
2241
3
is a function of x alone, the distribution of its
mass is symmetric about the x-axis. This means that y œ 0. We use the vertical strip approach to find x:
4
4
4
%
My œ ' µ
x dm œ '1 x † ’ È4x Š È4x ‹“ † $ dx œ '1 x † È8x † x" dx œ 8'1 x"Î# dx œ 8 2x"Î# ‘ " œ 8(2 † 2 2) œ 16;
4
M œ ' dm œ '1 ’ È4x Š È
‹“ † $ dx œ 8'1 Š È"x ‹ ˆ "x ‰ dx œ 8'1 x$Î# dx œ 8 2x"Î# ‘ " œ 8[1 (2)] œ 8.
x
4
So x œ
My
M
œ
16
8
4
%
4
œ 2 Ê (xß y) œ (2ß 0) is the center of mass.
(c)
‘
16. (a) We use the disk method: V œ 'a 1R# (x) dx œ '1 1 ˆ x4# ‰ dx œ 41'1 x# dx œ 41 x" ‘ " œ 41 "
4 (1)
b
4
4
%
œ 1[1 4] œ 31
(b) We model the distribution of mass with vertical strips: Mx œ ' µ
y dm œ '1
4
2
œ 2'1 x$Î# dx œ 2 ’ È
x dm œ '1 x †
“ œ 2[1 (2)] œ 2; My œ ' µ
x
%
4
4
"
2 16
3
2‘
3
xœ
My
M
œ
2
x
ˆ 2x ‰
2
† ˆ x2 ‰ † $ dx œ '1
4
2
x#
† Èx dx
† $ dx œ 2'1 x"Î# dx œ 2 ’ 2x3 “ œ
4
4
4
4
' dm œ ' x2 † $ dx œ 2' Èxx dx œ 2' x"Î# dx œ 2 2x"Î# ‘ %" œ 2(4 2) œ 4.
1
1
1
œ 28
3 ;Mœ
ˆ 28
‰
7
3
4 œ 3 and
yœ
Mx
M
œ
2
4
œ
"
#
Ê (xß y) œ ˆ 73 ß "# ‰ is the center of mass.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
$Î#
%
"
So
Section 6.6 Moments and Centers of Mass
(c)
17. The mass of a horizontal strip is dm œ $ dA œ $ L dy, where L is the width of the triangle at a distance of y above
its base on the x-axis as shown in the figure in the text. Also, by similar triangles we have
Ê Lœ
b
h
(h y). Thus, Mx œ ' µ
y dm œ '0 $ y ˆ bh ‰ (h y) dy œ
h
œ
$b
h
Š h# $
h$
3‹
œ
$b
h
#
h#
2‹
Šh œ $ bh# ˆ "# 3" ‰ œ
œ
$ bh
2
. So y œ
Mx
M
$ bh#
6
œ
$b
h
#
ˆ 2 ‰
Š $bh
6 ‹ $ bh
œ
h
3
œ
hy
h
'0h ahy y# b dy œ $hb ’ hy#
; M œ ' dm œ '0 $ ˆ hb ‰ (h y) dy œ
h
L
b
$b
h
#
h
y$
3 “!
'0h ah yb dy œ $hb ’hy y2 “ h
Ê the center of mass lies above the base of the
triangle one-third of the way toward the opposite vertex. Similarly the other two sides of the triangle can be
placed on the x-axis and the same results will occur. Therefore the centroid does lie at the intersection of the
medians, as claimed.
18. From the symmetry about the y-axis it follows that x œ 0.
It also follows that the line through the points (!ß !) and
(!ß $) is a median Ê y œ "3 (3 0) œ 1 Ê (xß y) œ (!ß ").
19. From the symmetry about the line x œ y it follows that
x œ y. It also follows that the line through the points (!ß !)
and ˆ "# ß "# ‰ is a median Ê y œ x œ 23 † ˆ "# 0‰ œ 3"
Ê (xß y) œ ˆ "3 ß 3" ‰ .
20. From the symmetry about the line x œ y it follows that
x œ y. It also follows that the line through the point (!ß !)
and ˆ #a ß #a ‰ is a median Ê y œ x œ 32 ˆ #a 0‰ œ "3 a
Ê (xß y) œ ˆ 3a ß 3a ‰ .
21. The point of intersection of the median from the vertex (0ß b)
to the opposite side has coordinates ˆ!ß #a ‰
Ê y œ (b 0) † "3 œ 3b and x œ ˆ #a !‰ † 32 œ 3a
Ê (xß y) œ ˆ 3a ß 3b ‰ .
22. From the symmetry about the line x œ
a
#
it follows that
xœ
It also follows that the line through the points
ˆ #a ß !‰ and ˆ #a ß b‰ is a median Ê y œ "3 (b 0) œ 3b
a
#.
Ê (xß y) œ ˆ #a ß b3 ‰ .
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
#
!
371
372
Chapter 6 Applications of Definite Integrals
"
#
23. y œ x"Î# Ê dy œ
x"Î# dx
"
4x
Ê ds œ È(dx)# (dy)# œ É1 Mx œ $ '0 Èx É1 2
œ $ '0 Éx 2
"
4
dx œ
"
4x
dx
2$
3
$Î#
’ˆx 4" ‰ “
dx ;
#
!
$Î#
$Î#
’ˆ2 "4 ‰ ˆ 4" ‰ “
œ
2$
3
œ
2$ ˆ 9 ‰$Î#
3 ’ 4
ˆ 4" ‰
$Î#
2$
3
“œ
"‰
ˆ 27
8 8 œ
13$
6
24. y œ x$ Ê dy œ 3x# dx
Ê dx œ É(dx)# a3x# dxb# œ È1 9x% dx;
Mx œ $ '0 x$ È1 9x% dx;
1
"
36
[u œ 1 9x% Ê du œ 36x$ dx Ê
du œ x$ dx;
x œ 0 Ê u œ 1, x œ 1 Ê u œ 10]
Ä Mx œ $ '1
10
"
36
u"Î# du œ
$
36
32 u$Î# ‘ "! œ
"
$
54
ˆ10$Î# 1‰
25. From Example 4 we have Mx œ '0 a(a sin ))(k sin )) d) œ a# k'0 sin# ) d) œ
'01 (1 cos 2)) d) œ a#k ) sin#2) ‘ !1
1
1
1
1
œ a #k1 ; My œ '0 a(a cos ))(k sin )) d) œ a# k '0 sin ) cos ) d) œ a#k csin# )d ! œ 0; M œ '0 ak sin ) d) œ ak[ cos )]1!
1
1
#
a# k
#
#
#
œ 2ak. Therefore, x œ
My
M
œ 0 and y œ
Mx
M
#
" ‰
œ Š a 2k1 ‹ ˆ 2ak
œ
a1
4
Ê ˆ!ß a41 ‰ is the center of mass.
26. Mx œ ' µ
y dm œ '0 (a sin )) † $ † a d)
1
œ '0 aa# sin )b a1 k kcos )kb d)
1
œ a# '0 (sin ))(1 k cos )) d)
1Î2
a# '1Î2 (sin ))(1 k cos )) d)
1
œ a# '0 sin ) d) a# k'0 sin ) cos ) d) a# '1Î2 sin ) d) a# k '1Î2 sin ) cos ) d)
1Î2
1Î2
1Î#
#
1
a# k ’ sin# ) “
œ a# [ cos )]!
#
#
k ˆ "#
1Î#
1
#
a# [ cos )]11Î# a# k ’ sin# ) “
!
1Î#
#
"‰
#
œ a [0 (1)] a
0‰ a [(1) 0] a k ˆ0 œ a# a#k a# 1
1
M œ'µ
x dm œ ' (a cos )) † $ † a d) œ ' aa# cos )b a1 k kcos )kb d)
y
#
1
0
#
a# k
#
œ 2a# a# k œ a# (2 k);
0
1Î2
1
œ a '0 (cos ))(1 k cos )) d) a# '1Î2 (cos ))(1 k cos )) d)
#
œ a#
1Î2
'01Î2 cos ) d) a# k '
0
#
œ a [sin
1Î#
) ]!
œ a# (1 0) a# k
#
a# k
#
) 2) ‰
2) ‰
ˆ 1 cos
d) a# '1Î2 cos ) d) a# k'1Î2 ˆ 1 cos
d)
#
#
sin 2) ‘ 1Î#
#
!
1
a# [sin )]11Î# 1
a# k
#
ˆ 1# 0‰ (! 0)‘ a# (0 1) ) a# k
#
sin 2) ‘ 1
#
1Î#
(1 0) ˆ 1# 0‰‘ œ a# a# k1
4
a# M œ '0 $ † a d) œ a'0 (1 k kcos )k) d) œ a '0 (1 k cos )) d) a'1Î2 (1 k cos )) d)
1
1Î#
œ a[) k sin )]!
œ
a1
#
1
1Î2
1
a[) k sin )]11Î# œ a ˆ 1# k‰ 0‘ a (1 0) ˆ 1# k‰‘
ak a ˆ 1# k‰ œ a1 2ak œ a(1 2k). So x œ
My
M
œ 0 and y œ
Mx
M
œ
a# (2 k)
a(1 #k)
œ
a(2 k)
1 #k
ka ‰
Ê ˆ0ß 2a
1 #k is the center of mass.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
a# k1
4
œ 0;
Section 6.6 Moments and Centers of Mass
373
27. faxb œ x 6, gaxb œ x2 , faxb œ gaxb Ê x 6 œ x2
Ê x2 x 6 œ 0 Ê x œ 3, x œ 2; $ œ 1
M œ 'c2 cax 6b x2 ddx œ "# x2 6x 13 x3 ‘2
3
3
œ ˆ 92 18 9‰ ˆ2 12 83 ‰ œ
xœ
1
125Î6
125
6
3
6 '
'c32 xcax 6b x2 ddx œ 125
cx2 6x x3 ddx
c2
œ
6 1 3
125 3 x
œ
6 ˆ
125 9
3
3x2 14 x4 ‘2
27 81 ‰
4
6 ˆ 8
125 3
12 4‰ œ 12 ; y œ
3 1 3
1 5 ‘3
2
125 3 x 6x 36x 5 x 2
Ê ˆ 12 , 4‰ is the center of mass.
œ
œ
3 ˆ
125 9
1
125Î6
54 108 3
3 '
'c32 12 ’ax 6b2 ax2 b2 “dx œ 125
cx2 12x 36 x4 ddx
c2
243 ‰
5
3 ˆ 8
125 3
24 72 32 ‰
5
œ4
28. faxb œ 2, gaxb œ x2 ax 1b, faxb œ gaxb Ê 2 œ x2 ax 1b
Ê x3 x2 2 œ 0 Ê x œ 1; $ œ 1
M œ '0 c2 x2 ax 1bd dx œ '0 c2 x3 x2 d dx
1
1
1
œ 2x "4 x4 13 x3 ‘0 œ ˆ2 xœ
1
17Î12
œ
12 2
17 x
œ
12 ˆ
17 1
œ
6 17 4x
"
4
13 ‰ 0 œ
17
12
'01 xc2 x2 ax 1bddx œ 1217 '01 c2x x4 x3 ddx
1
15 x5 14 x4 ‘0
1
5
14 ‰ 0 œ
33
85 ;
yœ
1
17Î12
1
17 x7 13 x6 15 x5 ‘0 œ
'01 12 ’22 ax2 ax 1bb2 “dx œ 176 '01 c4 x6 2x5 x4 ddx
6 ˆ
17 4
1
7
1
3
15 ‰ 0 œ
698
595
Ê ˆ 33
85 ,
698 ‰
595
is the center of mass.
29. faxb œ x2 , gaxb œ x2 ax 1b, faxb œ gaxb Ê x2 œ x2 ax 1b
Ê x3 2x2 œ 0 Ê x œ 0, x œ 2; $ œ 1
M œ '0 cx2 x2 ax 1bddx œ '0 c2x2 x3 ddx
2
2
2
‰
œ 23 x3 "4 x4 ‘0 œ ˆ 16
3 4 0œ
xœ
1
4 Î3
'02 xcx2 x2 ax 1bddx œ 43 '02 c2x3 x4 ddx
2
œ 34 12 x4 5" x5 ‘0 œ 34 ˆ8 yœ
1
4 Î3
4
3
32 ‰
5
0 œ 56 ;
'02 12 ’ax2 b2 ax2 ax 1bb2 “dx œ 38 '02 c2x5 x6 ddx œ 38 13 x6 7" x7 ‘20 œ 38 ˆ 643 "728 ‰ 0 œ 78
Ê ˆ 65 , 87 ‰ is the center of mass.
30. faxb œ 2 sin x, gaxb œ 0, x œ 0, x œ 21; $ œ 1;
M œ '0 c2 sin xddx œ c2x cos xd201
21
œ a41 1b a0 1b œ 41
xœ
1
41
'021 xc2 sin x 0ddx œ 411 '021 c2x x sin xddx
'021 2x dx 411 '021 x sin xdx
œ
1
41
œ
1
2 21
4 1 cx d 0
œ
1
2
4 1 a 41 b
0
œ
1
81
œ
1
81 c4x
1
41 csin x
x cos xd201
1
4 1 a0
21b 0 œ
21 1
2 ;
yœ
1
41
'021 21 ’a2 sin xb2 a0b2 “dx œ 811 '021 c4 4 sin x sin2 xddx
2x ‘
'021 c4 4 sin xddx 811 '021 csin2 xddx œ 811 '021 c4 4 sin xddx 811 '021 1 cos
dx
2
4cos x d201 1
161
'021 dx 1611 '021 cos 2x dx [u œ 2x Ê du œ 2dx, x œ 0 Ê u œ 0, x œ 21 Ê u œ 41]
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
374
Chapter 6 Applications of Definite Integrals
Ä
œ
21
21
1
1
1
81 c4x 4cos xd0 161 cxd0 321
1
1
1
81 a81 4b 81 a0 4b 161 a21b
'041 cos u du œ 811 c4x 4cos xd201 1611 cxd201 3211 csin ud401
00œ
9
8
Ê ˆ 212 1 , 89 ‰ is the center of mass.
31. Consider the curve as an infinite number of line segments joined together. From the derivation of arc length we have that
the length of a particular segment is ds œ Éadxb# adyb# . This implies that Mx œ ' $ y ds, My œ ' $ x ds and
M œ ' $ ds. If $ is constant, then x œ
My
M
' x ds
œ ' ds œ
' x ds
length
and y œ
Mx
M
' y ds
œ ' ds œ
' y ds
length
.
32. Applying the symmetry argument analogous to the one used in Exercise 1, we find that x œ 0. The typical vertical strip
x#
a
has center of mass: (µ
x ßµ
y ) œ Œxß 2 4p , length: a œ $ Ša œ
$
#
x#
4p ‹
’a# x 2
#Èpa
x&
80p# “ c2 pa
È
‰
œ 2a# $ Èpa ˆ 64
80 œ
œ 2$ Š2aÈpa œ
3
5
Èpa
dx. Thus, Mx œ ' µ
y dm œ 'c2Èpa "# Ša œ 2 † #$ ’a# x 8a# $Èpa
5
2$ paÈpa
12p ‹
2
x&
80p# “ 0
Èpa
x#
4p ,
x#
4p ‹ Ša
œ $ Š2a# Èpa Èpa
; M œ ' dm œ $ 'c2Èpa Ša 2
œ 4a$ Èpa ˆ1 width: dx, area: dA œ Ša 4 ‰
12
x#
4p ‹
x#
4p ‹ $
dx œ
2& p# a# Èpa
‹
80p#
$
#
dx, mass: dm œ $ dA
x
'c22ÈÈpapa Ša# 16p
‹ dx
%
#
œ 2a# $ Èpa ˆ1 2
16 ‰
80
"6 ‰
œ 2a# $ Èpa ˆ 8080
Èpa
œ 2 † $ ’ax È
dx œ $ ’ax x$
12p “ c2 pa
8a$ Èpa
3
. So y œ
œ 4a$ Èpa ˆ 121#4 ‰ œ
x#
4p ‹
Mx
M
œŠ
2
Èpa
x$
12p “ !
8a# $ Èpa
3
‹ Š 8a$È
5
pa ‹
a, as claimed.
33. The centroid of the square is located at (#ß #). The volume is V œ (21) ayb (A) œ (21)(2)(8) œ 321 and the surface area is
S œ (21) ayb (L) œ (21)(2) Š4È8‹ œ 32È21 (where È8 is the length of a side).
34. The midpoint of the hypotenuse of the triangle is ˆ 3# ß 3‰
Ê y œ 2x is an equation of the median Ê the line
y œ 2x contains the centroid. The point ˆ 3# ß $‰ is
3È 5
#
units from the origin Ê the x-coordinate of the
#
centroid solves the equation Ɉx 3# ‰ (2x 3)#
œ
È5
#
Ê ˆx# 3x 94 ‰ a4x# 12x 9b œ
5
4
Ê 5x# 15x 9 œ 1
Ê x# 3x 2 œ (x 2)(x 1) œ 0 Ê x œ 1 since the centroid must lie inside the triangle Ê y œ 2. By the
Theorem of Pappus, the volume is V œ (distance traveled by the centroid)(area of the region) œ 21 a5 xb "# (3)(6)‘
œ (21)(4)(9) œ 721
35. The centroid is located at (#ß !) Ê V œ (21) axb (A) œ (21)(2)(1) œ 41#
36. We create the cone by revolving the triangle with vertices
(0ß 0), (hß r) and (hß 0) about the x-axis (see the accompanying
figure). Thus, the cone has height h and base radius r. By
Theorem of Pappus, the lateral surface area swept out by the
hypotenuse L is given by S œ 21yL œ 21 ˆ #r ‰ Èh# r#
œ 1rÈr# h# . To calculate the volume we need the position
of the centroid of the triangle. From the diagram we see that
the centroid lies on the line y œ
r
2h
#
x. The x-coordinate of the centroid solves the equation É(x h)# ˆ 2hr x #r ‰
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 6.6 Moments and Centers of Mass
œ
"
3
Éh# r#
4
#
#
#
#
Ê Š 4h4h# r ‹ x# Š 4h 2h r ‹ x inside the triangle Ê y œ
r
2h
xœ
r
3.
39. V œ 21 yA Ê
4
3
2a ‰‘
(1a)
1
2a
1,
2 ar# 4h# b
9
œ0 Ê xœ
2h
3
or
21 ˆ 3r ‰‘ ˆ "#
4h
3
Ê xœ
hr‰ œ
"
3
2h
3 ,
since the centroid must lie
1r# h.
and by symmetry x œ 0
œ 21a# (1 2)
1ab# œ a21yb ˆ 1#ab ‰ Ê y œ
40. V œ 213A Ê V œ 21 ˆa By the Theorem of Pappus, V œ
37. S œ 21 y L Ê 41a# œ a21yb (1a) Ê y œ
38. S œ 213 L Ê 21 ˆa r#
4
375
4a ‰‘ 1a#
Š # ‹
31
œ
4b
31
and by symmetry x œ 0
1a$ (31 4)
3
41. V œ 213 A œ (21)(area of the region) † (distance from the centroid to the line y œ x a). We must find the distance from
4a ‰
ˆ0ß 31
to y œ x a. The line containing the centroid and perpendicular to y œ x a has slope 1 and contains the point
4a
ˆ!ß 31 ‰ . This line is y œ x 34a1 . The intersection of y œ x a and y œ x 34a1 is the point ˆ 4a 613a1 ß 4a 613a1 ‰ . Thus,
#
the distance from the centroid to the line y œ x a is Ɉ 4a 613a1 ‰ ˆ 34a1 Ê V œ (21) Š
È2 (4a 3a1)
#
‹ Š 1#a ‹
61
œ
4a
61
3a1 ‰#
61
œ
È2 (4a 3a1)
61
È2 1a$ (4 31)
6
‰
42. The line perpendicular to y œ x a and passing through the centroid ˆ!ß 2a
1 has equation y œ x of the two perpendicular lines occurs when x a œ x #
2a
1
centroid to the line y œ x a is Ɉ 2a 2 1a 0‰ ˆ 2a 2 1a Ê xœ
2a ‰#
#
œ
2a a1
21
a(21)
È 21
Ê yœ
2a a1
21 .
2a
1.
The intersection
Thus the distance from the
. Therefore, by the Theorem of Pappus the
1 )
surface area is S œ 21 ’ a(2
“ (1a) œ È21a# (2 1).
È
21
43. If we revolve the region about the y-axis: r œ a, h œ b Ê A œ 12 ab, V œ 13 1 a2 bß and 3 œ x. By the Theorem of Pappus:
1
a
1
1
2
2
ˆ1 ‰
3 1 a b œ 21 x 2 ab Ê x œ 3 ; If we revolve the region about the x-axis: r œ b, h œ a Ê A œ 2 ab, V œ 3 1 b aß and
3 œ y. By the Theorem of Pappus: 13 1 b2 a œ 21 y ˆ 12 ab‰ Ê y œ
b
3
Ê ˆ 3a , 3b ‰ is the center of mass.
44. Let Oa0, 0b, Paa, cb, and Qaa, bb be the vertices of the given triangle. If we revolve the region about the x-axis: Let R be
the point Raa, 0b. The volume is given by the volume of the outer cone, radius œ RP œ c, minus the volume of the inner
cone, radius œ RQ œ b, thus V œ 13 1 c2 a 13 1 b2 a œ 13 1 aac2 b2 b, the area is given by the area of triangle OPR minus
area of triangle OQR, A œ "# ac "# ab œ "# aac bb, and 3 œ y. By the Theorem of Pappus: 13 1 aac2 b2 b
œ 21 y ’ "# aac bb“ Ê y œ
cb
3 ;
If we revolve the region about the y-axis: Let S and T be the points Sa0, cb and Ta0, bb,
respectively. Then the volume is the volume of the cylinder with radius OR œ a and height RP œ c, minus the sum of the
volumes of the cone with radius œ SP œ a and height œ OS œ c and the portion of the cylinder with height œ OT œ b and
radius œ TQ œ a with a cone of height œ OT œ b and radius œ TQ œ a removed. Thus
V œ 1 a2 c ’ 13 1 a2 c ˆ1 a2 , 13 1 a2 b‰“ œ 23 1 a2 c 23 1 a2 b œ 23 1 a2 aa bb. The area of the triangle is the same as
before, A œ "# ac "# ab œ "# aac bb, and 3 œ x. By the Theorem of Pappus: 23 1 a2 aa bb œ 21 x ’ "# aac bb“
Êxœ
2aaa bb
3 ac b b
aa b b
Ê Š 2a
3 ac b b ,
cb
2 ‹
is the center of mass.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
376
Chapter 6 Applications of Definite Integrals
CHAPTER 6 PRACTICE EXERCISES
#
1. A(x) œ 14 (diameter)# œ 14 ˆÈx x# ‰
œ 14 ˆx 2Èx † x# x% ‰ ; a œ 0, b œ 1
Ê V œ 'a A(x) dx œ
b
#
œ
1
4
œ
1
4†70
x&
5 “!
’ x# 74 x(Î# È3
4
"
#
ˆ "# 4
7
5" ‰
91
280
È3
4
ˆ2Èx x‰#
ˆ4x 4xÈx x# ‰ ; a œ 0, b œ 4
È3
4
b
œ
È3
4
œ
32È3
4
ˆ1 1
4
8
5
'04 ˆ4x 4x$Î# x# ‰ dx
%
x$
“
3 !
’2x# 58 x&Î# 3. A(x) œ
1
4
1
4
œ
(side)# ˆsin 13 ‰ œ
Ê V œ 'a A(x) dx œ
œ
'01 ˆx 2x&Î# x% ‰ dx
(35 40 14) œ
2. A(x) œ
œ
1
4
"
œ
8È 3
15
32 ‰ œ
(diameter)# œ
1
4
È3
4
8†32
5
ˆ32 (15 24 10) œ
64 ‰
3
8È 3
15
(2 sin x 2 cos x)#
† 4 asin# x 2 sin x cos x cos# xb
œ 1(1 sin 2x); a œ
1
4
,bœ
51
4
Ê V œ 'a A(x) dx œ 1 '1Î4 (1 sin 2x) dx
51Î4
b
œ 1 x cos 2x ‘ &1Î%
#
1Î%
œ 1 ’Š 541 cos 5#1
# ‹
Š 14 cos 1#
# ‹“
œ 1#
#
#
%
4. A(x) œ (edge)# œ ŒŠÈ6 Èx‹ 0 œ ŠÈ6 Èx‹ œ 36 24È6 Èx 36x 4È6 x$Î# x# ;
a œ 0, b œ 6 Ê V œ 'a A(x) dx œ '0 Š36 24È6 Èx 36x 4È6 x$Î# x# ‹ dx
b
6
œ ’36x 24È6 † 23 x$Î# 18x# 4È6 † 25 x&Î# œ 216 576 648 5. A(x) œ
(diameter)# œ
1
4
72 œ 360 Š2Èx x#
4‹
#
1728
5
œ
1
4
œ
œ 216 16 † È6 È6 † 6 18 † 6# 58 È6 È6 † 6# 1800 1728
5
Š4x x&Î# œ
72
5
x%
16 ‹ ;
a œ 0, b œ 4 Ê V œ 'a A(x) dx
b
'04 Š4x x&Î# 16x ‹ dx œ 14 ’2x# 27 x(Î# 5x†16 “ % œ 14 ˆ32 32 † 87 25 † 32‰
œ
1
4
œ
321
4
%
ˆ1 6. A(x) œ
œ
1
4
1728
5
'
x$
3 “!
È3
4
"
#
8
7
52 ‰ œ
81
35
&
(35 40 14) œ
(edge)# sin ˆ 13 ‰ œ
È3
4
!
721
35
2Èx ˆ2Èx‰‘#
ˆ4Èx‰# œ 4È3 x; a œ 0, b œ 1
Ê V œ 'a A(x) dx œ '0 4È3 x dx œ ’2È3 x# “
b
1
"
!
œ 2È3
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
6$
3
Chapter 6 Practice Exercises
377
7. (a) .3=5 7/>29. :
V œ 'a 1R# (x) dx œ 'c1 1 a3x% b dx œ 1 'c1 9x) dx
b
1
1
#
"
œ 1 cx* d " œ 21
(b) =2/66 7/>29. :
shell ‰
shell
V œ 'a 21 ˆ radius
Š height
‹ dx œ '0 21x a3x% b dx œ 21 † 3'0 x& dx œ 21 † 3 ’ x6 “ œ 1
b
1
1
'
!
Note: The lower limit of integration is 0 rather than 1.
(c) =2/66 7/>29. :
shell ‰
shell
V œ 'a 21 ˆ radius
Š height
‹ dx œ 21'c1 (1 x) a3x% b dx œ 21 ’ 3x5 b
"
1
&
(d) A+=2/< 7/>29. :
"
x'
2 “ "
œ 21 ˆ 35 "# ‰ ˆ 35 "# ‰‘ œ
121
5
R(x) œ 3, r(x) œ 3 3x% œ 3 a1 x% b Ê V œ 'a 1 cR# (x) r# (x)d dx œ 'c1 1 ’9 9 a1 x% b “ dx
b
1
œ 91 'c1 c1 a1 2x% x) bd dx œ 91 'c1 a2x% x) b dx œ 91 ’ 2x5 1
1
8. (a) A+=2/< 7/>29. :
R(x) œ
, r(x) œ
4
x$
"
#
&
"
x*
9 “ "
b
21†13
5
œ
261
5
2
(b) =2/66 7/>29. :
V œ 21'1 x ˆ x4$ "# ‰ dx œ 21 ’4x" 2
(c) =2/66 7/>29. :
"
#
#
x#
4 “"
16
5
4" ‰ œ
1
20
(2 10 64 5) œ
b
2
x
(d) A+=2/< 7/>29. :
#
x#
4 “"
571
#0
œ 21 ˆ 4# 1‰ ˆ4 4" ‰‘ œ 21 ˆ 54 ‰ œ
shell ‰
shell
V œ 21'a ˆ radius
Š height
‹ dx œ 21'1 (2 x) ˆ x4$ "# ‰ dx œ 21'1 ˆ x8$ 4
x
œ 181 25 "9 ‘ œ
#
#
#
&
Ê V œ 'a 1cR# (x) r# (x)d dx œ '1 1 ’ˆ x4$ ‰ ˆ "# ‰ “ dx œ 1 16
x4 ‘ "
5 x
"‰
"
ˆ 16 " ‰‘ œ 1 ˆ 10
œ 1 ˆ 5†16
32 # 5 4
œ 21 ’ x4# #
2
4
x#
51
#
1 x# ‰ dx
œ 21 (1 2 2 1) ˆ4 4 1 4" ‰‘ œ
31
#
V œ 'a 1cR# (x) r# (x)d dx
b
#
œ 1 '1 ’ˆ 7# ‰ ˆ4 2
dx
œ
491
4
161'1 a1 2x$ x' b dx
œ
491
4
161 ’x x# œ
491
4
491
4
491
4
161 ˆ2 4" 5†"3# ‰ ˆ1 1 5" ‰‘
"
161 ˆ 4" 160
5" ‰
œ
œ
9.
4 ‰#
x$ “
2
161
160
#
x &
5 “"
(40 1 32) œ
491
4
711
10
œ
1031
20
(a) .3=5 7/>29. :
#
V œ 1 '1 ŠÈx 1‹ dx œ 1'1 (x 1) dx œ 1 ’ x# x“
#
5
5
‰ ˆ"
‰‘ œ 1 ˆ 24
‰
œ 1 ˆ 25
# 5 # 1
# 4 œ 81
&
"
(b) A+=2/< 7/>29. :
R(y) œ 5, r(y) œ y# 1 Ê V œ 'c 1 cR# (y) r# (y)d dy œ 1 'c2 ’25 ay# 1b “ dy
d
2
œ 1'c2 a25 y% 2y# 1b dy œ 1 'c2 a24 y% 2y# b dy œ 1 ’24y 2
2
#
y&
5
32 y$ “
#
#
œ 21 ˆ24 † 2 Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
32
5
2
3
† 8‰
378
Chapter 6 Applications of Definite Integrals
œ 321 ˆ3 2
5
"3 ‰ œ
321
15
(45 6 5) œ
10881
15
(c) .3=5 7/>29. :
R(y) œ 5 ay# 1b œ 4 y#
Ê V œ 'c 1R# (y) dy œ 'c2 1 a4 y# b dy
d
2
#
œ 1 'c2 a16 8y# y% b dy
2
œ 1 ’16y 8y$
3
œ 641 ˆ1 2
3
#
y&
5 “ #
"5 ‰ œ
œ 21 ˆ32 641
15
64
3
(15 10 3) œ
32 ‰
5
5121
15
10. (a) =2/66 7/>29. :
shell ‰
shell
V œ 'c 21 ˆ radius
Š height
‹ dy œ '0 21y Šy d
4
œ 21'0 Šy# 4
œ
21
1#
† 64 œ
y$
4‹
$
dy œ 21 ’ y3 321
3
%
y%
16 “ !
y#
4‹
dy
œ 21 ˆ 64
3 64 ‰
4
(b) =2/66 7/>29. :
shell ‰
shell
V œ 'a 21 ˆ radius
Š height
‹ dx œ '0 21x ˆ2Èx x‰ dx œ 21'0 ˆ2x$Î# x# ‰ dx œ 21 ’ 45 x&Î# b
4
œ 21 ˆ 45 † 32 64 ‰
3
œ
4
1281
15
%
x$
3 “!
(c) =2/66 7/>29. :
shell ‰
shell
V œ 'a 21 ˆ radius
Š height
‹ dx œ '0 21(4 x) ˆ2Èx x‰ dx œ 21'0 ˆ8x"Î# 4x 2x$Î# x# ‰ dx
b
4
$Î#
œ 21 ’ 16
2x# 54 x&Î# 3 x
œ 641 ˆ1 45 ‰ œ
641
5
4
%
x$
3 “!
œ 21 ˆ 16
3 † 8 32 (d) =2/66 7/>29. :
shell ‰
shell
V œ 'c 21 ˆ radius
Š height
‹ dy œ '0 21(4 y) Šy d
4
œ 21'0 Š4y 2y# 4
y$
4‹
dy œ 21 ’2y# 32 y$ y#
4‹
%
y%
16 “ !
4
5
† 32 64 ‰
3
œ 641 ˆ 34 1 dy œ 21'0 Š4y y# y# 4
œ 21 ˆ32 2
3
4
5
y$
4‹
32 ‰
dy
† 64 16‰ œ 321 ˆ2 8
3
1‰ œ
321
3
11. .3=5 7/>29. :
R(x) œ tan x, a œ 0, b œ
1
3
Ê V œ 1 '0 tan# x dx œ 1'0 asec# x 1b dx œ 1[tan x x]!
1Î3
12. .3=5 7/>29. :
1Î3
1Î$
V œ 1'0 (2 sin x)# dx œ 1 '0 a4 4 sin x sin# xb dx œ 1'0 ˆ4 4 sin x 1
1
œ 1 4x 4 cos x x
#
sin 2x ‘ 1
4
!
1
œ 1 ˆ41 4 1
#
0‰ (0 4 0 0)‘ œ
œ
1cos 2x ‰
dx
#
9
1
1 ˆ # 8‰ œ 1#
1 Š3È31‹
3
(91 16)
13. (a) .3=5 7/>29. :
V œ 1'0 ax# 2xb dx œ 1'0 ax% 4x$ 4x# b dx œ 1 ’ x5 x% 34 x$ “ œ 1 ˆ 32
5 16 2
œ
161
15
2
#
(6 15 10) œ
#
&
!
161
15
32 ‰
3
(b) A+=2/< 7/>29. :
V œ '0 1’1# ax# 2x "b “ dx œ '0 1 dx '0 1 ax "b% dx œ #1 ’1
2
2
#
2
(c) =2/66 7/>29. :
#
ax"b&
& “!
œ #1 1 †
shell ‰
shell
V œ 'a 21 ˆ radius
Š height
‹ dx œ 21'0 (2 x) c ax# 2xbd dx œ 21'0 (2 x) a2x x# b dx
b
2
2
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
#
&
œ
)1
&
Chapter 6 Practice Exercises
œ 21'0 a4x 2x# 2x# x$ b dx œ 21'0 ax$ 4x# 4xb dx œ 21 ’ x4 34 x$ 2x# “ œ 21 ˆ4 2
œ
21
3
2
#
%
81
3
(36 32) œ
32
3
!
8‰
(d) A+=2/< 7/>29. :
V œ 1 '0 c2 ax# 2xbd dx 1'0 2# dx œ 1'0 ’4 4 ax# 2xb ax# 2xb “ dx 81
2
2
#
2
#
œ 1'0 a4 4x# 8x x% 4x$ 4x# b dx 81 œ 1'0 ax% 4x$ 8x 4b dx 81
2
2
#
&
‰
œ 1 ’ x5 x% 4x# 4x“ 81 œ 1 ˆ 32
5 16 16 8 81 œ
!
1
5
(32 40) 81 œ
721
5
401
5
œ
321
5
14. .3=5 7/>29. :
V œ 21'0 4 tan# x dx œ 81'0 asec# x 1b dx œ 81[tan x x]!
1Î4
1Î4
1Î%
œ 21(4 1)
15. The material removed from the sphere consists of a cylinder
and two "caps." From the diagram, the height of the cylinder
#
is 2h, where h# ŠÈ$‹ œ ## , i.e. h œ ". Thus
#
Vcyl œ a#hb1ŠÈ$‹ œ '1 ft$ . To get the volume of a cap,
use the disk method and x# y# œ ## : Vcap œ '" 1x# dy
2
œ '" 1a% y# bdy œ 1’%y 2
œ 1ˆ8 83 ‰ ˆ% 3" ‰‘ œ
#
y3
3 “"
&1
3
Vremoved œ Vcyl #Vcap œ '1 ft$ . Therefore,
"!1
3
œ
#)1
3
ft$ .
16. We rotate the region enclosed by the curve y œ É12 ˆ1 4x# ‰
121
and the x-axis around the x-axis. To find the
11Î2
volume we use the .3=5 method: V œ 'a 1R# (x) dx œ 'c11Î2 1 ŠÉ12 ˆ1 b
œ 121'c11Î2 Š1 11Î2
4x#
121 ‹
œ 1321 ˆ1 "3 ‰ œ
17. y œ x"Î# x$Î#
3
Ê
dx œ 121 ’x 2641
3
dy
dx
œ
""Î#
4x$
363 “ ""Î#
ˆ4 2
3
18. x œ y#Î$ Ê
"
#
#
x"Î# "# x"Î# Ê Š dy
dx ‹ œ
œ '1
8
† 8‰ ˆ2 23 ‰‘ œ
È9y#Î$ 4
3y"Î$
œ
"
4
"
3
5
12
12 Š1 4x#
121 ‹
dx
#
4
ˆ2 14 ‰
3
œ
4y #Î$
9
"
#
ˆx"Î# x"Î# ‰ dx œ
"
#
2x"Î# 23 x$Î# ‘ %
"
10
3
dx
Ê L œ '1 Ê1 Š dy
‹ dy œ '1 É1 #
8
8
4
9y#Î$
dy
'18 È9y#Î$ 4 ˆy"Î$ ‰ dy; u œ 9y#Î$ 4 Ê du œ 6y"Î$ dy; y œ 1 Ê u œ 13,
y œ 8 Ê u œ 40d Ä L œ
19. y œ
"
#
#
dy œ
11Î2
ˆ x" 2 x‰ Ê L œ ' É1 4" ˆ x" 2 x‰ dx
1
4
y"Î$ Ê Š dx
dy ‹ œ
2
3
11Î2
$
4
dx
dy
dx œ 1 '
œ 881 ¸ 276 in$
4
"
#
#
4 ‰ 11
ˆ 4 ‰ ˆ 11
‰ “ œ 1321 ’1 ˆ 363
œ 241 ’ 11
Š 4 ‹“
2 363
#
#
Ê L œ '1 É 4" ˆ x" 2 x‰ dx œ '1 É 4" ax"Î# x"Î# b dx œ '1
œ
4x# ‰
121 ‹
x'Î& 58 x%Î& Ê
dy
dx
"
18
œ
'1340 u"Î# du œ 18" 32 u$Î# ‘ %!
œ #"7 40$Î# 13$Î# ‘ ¸ 7.634
"$
"
#
#
dy
x"Î& "# x"Î& Ê Š dx
‹ œ
"
4
ˆx#Î& 2 x#Î& ‰
#
Ê L œ '1 É1 "4 ax#Î& 2 x#Î& b dx Ê L œ '1 É 4" ax#Î& 2 x#Î& b dx œ ' É 4" ax"Î& x"Î& b dx
32
32
32
1
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
379
380
Chapter 6 Applications of Definite Integrals
œ '1
32
œ
"
48
20. x œ
"
#
ˆx"Î& x"Î& ‰ dx œ
(1260 450) œ
"
1#
y$ "
y
Ê
" %
œ '1 É 16
y 2
"
#
œ
1710
48
" 5
# 6
285
8
$#
x'Î& 45 x%Î& ‘ " œ
"
y#
#
dx
dy
œ
"
4
"
y%
dy œ '1 ÊŠ 4" y# y# 2
8
œ ˆ 12
"# ‰ ˆ 1"# 1‰ œ
7
1#
"
#
21. S œ 'a 21y Ê1 Š dy
dx ‹ dx;
dy
dx
œ
#
b
"
16
Ê Š dx
dy ‹ œ
œ
"
y# ‹
#
"
#
ˆ 65 † 2' y% "
#
"
y%
5
4
† 2% ‰ ˆ 56 54 ‰‘ œ
"
#
ˆ 315
6 " %
Ê L œ '1 Ê1 Š 16
y 2
dy œ '1 Š 4" y# 2
"
y# ‹
dy œ ’ 1"# y$ y" “
"
#
75 ‰
4
"
y% ‹
dy
#
"
13
12
#
"
È2x 1
Ê Š dy
dx ‹ œ
"
#x 1
Ê S œ '0 21È2x 1 É1 3
"
#x 1
dx
2
È ' Èx 1 dx œ 2È21 2 (x 1)$Î# ‘ $ œ 2È21 † 2 (8 1) œ
œ 21'0 È2x 1 É 2x
2x1 dx œ 2 21 0
3
3
!
3
3
22. S œ 'a 21y Ê1 Š dy
dx ‹ dx;
#
b
œ
1
6
dy
dx
%
'
œ x# Ê Š dy
dx ‹ œ x Ê S œ 0 21 †
#
1
x$
3
È1 x% dx œ
1
6
281È2
3
'01 È1 x% a4x$ b dx
'01 È1 x% d a1 x% b œ 16 ’ 32 a1 x% b$Î# “ " œ 19 ’2È2 1“
!
23. S œ 'c 21x Ê1 Š dx
dy ‹ dy;
#
d
dx
dy
ˆ "# ‰ (4 2y)
È4y y#
œ
œ
2y
È4y y#
#
Ê 1 Š dx
dy ‹ œ
4y y# 4 4y y#
4y y#
œ
4
4y y#
Ê S œ '1 21 È4y y# É 4y 4 y# dy œ 41'1 dx œ 41
2
2
24. S œ 'c 21x Ê1 Š dx
dy ‹ dy;
#
d
œ 1'2 È4y 1 dy œ
6
1
4
dx
dy
œ
1
2È y
#
Ê 1 Š dx
dy ‹ œ 1 32 (4y 1)$Î# ‘ ' œ
#
1
6
(125 27) œ
1
6
"
4y
œ
4y 1
4y
(98) œ
Ê S œ '2 21Èy †
6
È4y 1
È4y
dy
491
3
25. The equipment alone: the force required to lift the equipment is equal to its weight Ê F" (x) œ 100 N.
The work done is W" œ 'a F" (x) dx œ '0 100 dx œ [100x]%!
! œ 4000 J; the rope alone: the force required
b
40
to lift the rope is equal to the weight of the rope paid out at elevation x Ê F# (x) œ 0.8(40 x). The work
done is W# œ 'a F# (x) dx œ '0 0.8(40 x) dx œ 0.8 ’40x b
40
the total work is W œ W" W# œ 4000 640 œ 4640 J
%!
x#
# “!
œ 0.8 Š40# 40#
# ‹
œ
(0.8)(1600)
#
œ 640 J;
26. The force required to lift the water is equal to the water's weight, which varies steadily from 8 † 800 lb to
8 † 400 lb over the 4750 ft elevation. When the truck is x ft off the base of Mt. Washington, the water weight is
x‰
F(x) œ 8 † 800 † ˆ 2†24750
œ (6400) ˆ1 †4750
œ '0
4750
6400 ˆ1 x ‰
9500
dx œ 6400 ’x œ 22,800,000 ft † lb
x ‰
9500
lb. The work done is W œ 'a F(x) dx
%(&!
x#
2†9500 “ !
b
œ 6400 Š4750 4750#
4†4750 ‹
œ ˆ 34 ‰ (6400)(4750)
27. Force constant: F œ kx Ê 20 œ k † 1 Ê k œ 20 lb/ft; the work to stretch the spring 1 ft is
W œ '0 kx dx œ k'0 x dx œ ’20 x# “ œ 10 ft † lb; the work to stretch the spring an additional foot is
1
1
#
"
!
#
W œ '1 kx dx œ k '1 x dx œ 20 ’ x# “ œ 20 ˆ 4# "# ‰ œ 20 ˆ 3# ‰ œ 30 ft † lb
2
2
#
"
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Chapter 6 Practice Exercises
28. Force constant: F œ kx Ê 200 œ k(0.8) Ê k œ 250 N/m; the 300 N force stretches the spring x œ
œ
300
250
381
F
k
œ 1.2 m; the work required to stretch the spring that far is then W œ '0 F(x) dx œ '0 250x dx
1Þ2
1Þ2
œ [125x# ]!"Þ# œ 125(1.2)# œ 180 J
29. We imagine the water divided into thin slabs by planes
perpendicular to the y-axis at the points of a partition of the
interval [0ß 8]. The typical slab between the planes at y and
y ?y has a volume of about ?V œ 1(radius)# (thickness)
#
œ 1 ˆ 54 y‰ ?y œ 25161 y# ?y ft$ . The force F(y) required to
lift this slab is equal to its weight: F(y) œ 62.4 ?V
œ
(62.4)(25)
16
1y# ?y lb. The distance through which F(y)
must act to lift this slab to the level 6 ft above the top is
about (6 8 y) ft, so the work done lifting the slab is about ?W œ
(62.4)(25)
16
1y# (14 y) ?y ft † lb. The work done
lifting all the slabs from y œ 0 to y œ 8 to the level 6 ft above the top is approximately
8
W¸!
!
(62.4)(25)
16
1y# (14 y) ?y ft † lb so the work to pump the water is the limit of these Riemann sums as the norm of
the partition goes to zero: W œ '0
8
$
œ (62.4) ˆ 25161 ‰ Š 14
3 †8 8%
4‹
(62.4)(25)
(16)
(62.4)(25)1
16
1y# (14 y) dy œ
'08 a14y# y$ b dy œ (62.4) ˆ 25161 ‰ ’ 143 y$ y4 “ )
%
!
¸ 418,208.81 ft † lb
30. The same as in Exercise 29, but change the distance through which F(y) must act to (8 y) rather than (6 8 y). Also
change the upper limit of integration from 8 to 5. The integral is:W œ '0
5
œ (62.4) ˆ 25161 ‰'0 a8y# y$ b dy œ (62.4) ˆ 25161 ‰ ’ 83 y$ 5
&
y%
4 “!
(62.4)(25)1
16
y# (8 y) dy
œ (62.4) ˆ 25161 ‰ Š 83 † 5$ 5%
4‹
31. The tank's cross section looks like the figure in Exercise 29 with right edge given by x œ
#
slab has volume ?V œ 1(radius)# (thickness) œ 1 ˆ #y ‰ ?y œ
F(y) œ 60 †
1
4
y œ y# . A typical horizontal
y# ?y. The force required to lift thisslab is its weight:
y# ?y. The distance through which F(y) must act is (2 10 y) ft, so the work to pump the liquid is
W œ 60'0 1a12 ybŠ y4 ‹dy œ 151 ’ 12y
3 10
22,5001 ft†lb
275 ft†lb/sec
1
4
5
10
¸ 54,241.56 ft † lb
#
$
"!
y%
4 “!
œ 22,5001 ft † lb; the time needed to empty the tank is
¸ 257 sec
32. A typical horizontal slab has volume about ?V œ (20)(2x)?y œ (20) ˆ2È16 y# ‰ ?y and the force required to lift this
slab is its weight F(y) œ (57)(20) ˆ2È16 y# ‰ ?y. The distance through which F(y) must act is (6 4 y) ft, so the
work to pump the olive oil from the half-full tank is W œ 57'c4 (10 y)(20) ˆ2È16 y# ‰ dy
0
"Î#
œ 2880 'c4 10È16 y# dy 1140'c4 a16 y# b (2y) dy
0
0
œ 22,800 † (area of a quarter circle having radius 4) 23 (1140) ’a16 y# b
œ 335,153.25 ft † lb
$Î# !
“
%
œ (22,800)(41) 48,640
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
382
Chapter 6 Applications of Definite Integrals
33. Intersection points: 3 x# œ 2x# Ê 3x# 3 œ 0
Ê 3(x 1)(x 1) œ 0 Ê x œ 1 or x œ 1. Symmetry
suggests that x œ 0. The typical @/<>3-+6 strip has
#
#
#
center of mass: (µ
x ßµ
y ) œ Šxß 2x a3 x b ‹ œ Šxß x 3 ‹ ,
#
#
#
#
#
length: a3 x b 2x œ 3 a1 x b, width: dx,
area: dA œ 3 a1 x# b dx, and mass: dm œ $ † dA
œ 3$ a1 x# b dx Ê the moment about the x-axis is
µ
y dm œ
œ
3
#
$ ax# 3b a1 x# b dx œ
3
#
&
2x$
3
$ ’ x5 "
x$
3 “ "
œ 3$ ’x 3x“
"
"
3
#
$ ax% 2x# 3b dx Ê Mx œ ' µ
y dm œ
œ 3$ ˆ 5" 2
3
3$
15
3‰ œ
œ 6$ ˆ1 "3 ‰ œ 4$ Ê y œ
Mx
M
œ
(3 10 45) œ
32$
5 †4 $
œ
8
5
32$
5
3
#
$ 'c1 ax% 2x# 3b dx
"
; M œ ' dm œ 3$ 'c1 a1 x# b dx
"
. Therefore, the centroid is (xß y) œ ˆ!ß 85 ‰ .
34. Symmetry suggests that x œ 0. The typical @/<>3-+6
#
strip has center of mass: (µ
x ßµ
y ) œ Šxß x# ‹ , length: x# ,
width: dx, area: dA œ x# dx, mass: dm œ $ † dA œ $ x# dx
Ê the moment about the x-axis is µ
y dm œ #$ x# † x# dx
œ
$
#
x% dx Ê Mx œ ' µ
y dm œ
$
#
'c22 x% dx œ 10$ cx& d ##
35. The typical @/<>3-+6 strip has: center of mass: (µ
x ßµ
y )
œ Œxß
#
4 x4
#
, length: 4 area: dA œ Š4 œ $ Š4 x#
4‹
x#
4 ‹dx,
width: dx,
mass: dm œ $ † dA
dx Ê the moment about the x-axis is
#
Š4 x4 ‹
µ
y dm œ $ †
x#
4,
#
Š4 x#
4‹
dx œ
$
#
Š16 moment about the y-axis is µ
x dm œ $ Š4 œ
$
2
’16x %
x&
5†16 “ !
$
#
œ
64 64 ‘
5
œ
x%
16 ‹
x#
4‹
œ
16†$ †3
32†$
œ
3
2
and y œ
Mx
M
œ
dx. Thus, Mx œ ' µ
y dm œ
4
4
My
M
x$
4‹
† x dx œ $ Š4x ; My œ ' µ
x dm œ $ '0 Š4x 128$
5
œ $ (32 16) œ 16$ ; M œ ' dm œ $ '0 Š4 Ê xœ
dx; the
128†$ †3
5†32†$
x#
4‹
œ
dx œ $ ’4x 12
5
%
x$
12 “ !
x$
4‹
dx œ $ ’2x# œ $ ˆ16 64 ‰
1#
œ
$
#
'04 Š16 16x ‹ dx
%
%
x%
16 “ !
32$
3
‰
. Therefore, the centroid is (xß y) œ ˆ 3# ß 12
5 .
36. A typical 29<3D98>+6 strip has:
#
center of mass: (µ
x ßµ
y ) œ Š y # 2y ß y‹ , length: 2y y# ,
width: dy, area: dA œ a2y y# b dy, mass: dm œ $ † dA
œ $ a2y y# b dy; the moment about the x-axis is
µ
y dm œ $ † y † a2y y# b dy œ $ a2y# y$ b ; the moment
#
about the y-axis is µ
x dm œ $ † ay 2yb † a2y y# b dy
#
œ a4y y b dy Ê Mx œ ' µ
y dm œ $ '0 a2y# y$ b dy
$
#
#
œ $ ’ 23 y$ œ
$
#
yœ
ˆ 43†8
Mx
M
2
%
œ
#
y%
4 “!
32 ‰
5
4†$ †3
3 † 4 †$
œ
œ $ ˆ 23 † 8 32$
15
16 ‰
4
œ $ ˆ 16
3 16 ‰
4
œ
$ †16
12
œ
4$
3
; My œ ' µ
x dm œ
$
#
'02 a4y# y% b dy œ #$ ’ 34 y$ y5 “ #
$
; M œ ' dm œ $ '0 a2y y# b dy œ $ ’y# y3 “ œ $ ˆ4 83 ‰ œ
2
#
!
&
!
4$
3
Ê xœ
œ 1. Therefore, the centroid is (xß y) œ ˆ 58 ß 1‰ .
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
My
M
œ
$ †32†3
15†$ †4
œ
8
5
and
Chapter 6 Practice Exercises
383
37. A typical horizontal strip has: center of mass: (µ
x ßµ
y )
#
œ Š y #2y ß y‹ , length: 2y y# , width: dy,
area: dA œ a2y y# b dy, mass: dm œ $ † dA
œ (1 y) a2y y# b dy Ê the moment about the
x-axis is µ
y dm œ y(1 y) a2y y# b dy
#
œ a2y 2y$ y$ y% b dy
œ a2y# y$ y% b dy; the moment about the y-axis is
#
µ
x dm œ Š y 2y ‹ (1 y) a2y y# b dy œ " a4y# y% b (1 y) dy œ
#
Ê Mx œ ' µ
y dm œ '0 a2y# y$ y% b dy œ ’ 23 y$ 2
œ
16
60
œ
"
#
"
#
#
(20 15 24) œ
$
Š 4†32 2% 2&
5
(11) œ
4
15
2'
6‹
2
44
40
œ
11
10
y$
3
; My œ ' µ
x dm œ '0
2
œ 4 ˆ 43 2 œ '0 a2y y# y$ b dy œ ’y# ‰ ˆ 38 ‰ œ
œ ˆ 44
15
44
15
y%
4
4
5
œ ˆ4 8
3
16 ‰
4
œ ˆ 16
3 16
4
32 ‰
5
œ 16 ˆ "3 a4y# 4y$ y% y& b dy œ
86 ‰ œ 4 ˆ2 45 ‰ œ
#
y%
4 “!
"
#
#
y&
5 “!
a4y# 4y$ y% y& b dy
œ
8
3
24
5
"
#
"
4
25 ‰
’ 43 y$ y% y&
5
; M œ ' dm œ '0 (1 y) a2y y# b dy
#
y'
6 “!
2
Ê xœ
My
M
‰ ˆ 83 ‰ œ
œ ˆ 24
5
9
5
and y œ
Mx
M
‰
. Therefore, the center of mass is (xß y) œ ˆ 95 ß 11
10 .
3
3
38. A typical vertical strip has: center of mass: (µ
x ßµ
y ) œ ˆxß 2x3$Î# ‰ , length: x$Î#
, width: dx, area: dA œ x$Î#
dx,
µ
3
3
3
9$
mass: dm œ $ † dA œ $ †
dx Ê the moment about the x-axis is y dm œ
†$
dx œ
dx; the moment about
x$Î#
3
the y-axis is µ
x dm œ x † $ x$Î#
dx œ
(a) Mx œ $ '1
9
M œ $ '1
9
(b) Mx œ '1
9
"
#
3
x$Î#
x
#
9$
#
ˆ x9$ ‰ dx œ
3$
x"Î#
#x$Î#
# *
20$
9
’ x# “ œ
"
3 ‰
; My œ $ '1 x ˆ x$Î#
dx œ 3$ 2x"Î# ‘ " œ 12$ ;
*
*
9
#
*
9
dx œ 6$ x"Î# ‘ " œ 4$ Ê x œ
ˆ x9$ ‰ dx œ
2x$
x$Î#
dx.
My
M
œ
12$
4$
œ 3 and y œ
Mx
M
œ
ˆ 209$ ‰
4$
œ
5
9
*
3 ‰
3 ‰
x" ‘ * œ 4; My œ ' x# ˆ $Î#
dx œ 2x$Î# ‘ " œ 52; M œ '1 x ˆ x$Î#
dx
"
x
1
œ 6 x"Î# ‘ " œ 12 Ê x œ
9
My
M
œ
13
3
and y œ
Mx
M
9
œ
"
3
strip
39. F œ 'a W † Š depth
‹ † L(y) dy Ê F œ 2 '0 (62.4)(2 y)(2y) dy œ 249.6'0 a2y y# b dy œ 249.6 ’y# b
2
2
#
y$
3 “!
œ (249.6) ˆ4 83 ‰ œ (249.6) ˆ 43 ‰ œ 332.8 lb
strip
40. F œ 'a W † Š depth
‹ † L(y) dy Ê F œ '0 75 ˆ 56 y‰ (2y 4) dy œ 75'0 ˆ 53 y 5Î6
b
5Î6
10
3
2y# 4y‰ dy
7
7 #
2 $ ‘ &Î'
50 ‰
25 ‰
125 ‰‘
#‰
ˆ 18
œ 75 '0 ˆ 10
dy œ 75 10
ˆ 67 ‰ ˆ 36
ˆ 32 ‰ ˆ 216
3 3 y 2y
3 y 6 y 3 y ! œ (75)
5Î6
œ (75) ˆ 25
9 175
216
250 ‰
3†#16
‰
œ ˆ 9†75
#16 (25 † 216 175 † 9 250 † 3) œ
strip
41. F œ 'a W † Š depth
‹ † L(y) dy Ê F œ 62.4'0 (9 y) Š2 †
b
4
%
œ 62.4 6y$Î# 25 y&Î# ‘ ! œ (62.4) ˆ6 † 8 2
5
Èy
2 ‹
(75)(3075)
9†#16
¸ 118.63 lb.
dy œ 62.4'0 ˆ9y"Î# 3y$Î# ‰ dy
4
‰ (48 † 5 64) œ
† 32‰ œ ˆ 62.4
5
(62.4)(176)
5
œ 2196.48 lb
strip
42. Place the origin at the bottom of the tank. Then F œ '0 W † Š depth
‹ † L(y) dy, h œ the height of the mercury column,
h
strip depth œ h y, L(y) œ 1 Ê F œ '0 849(h y) " dy œ (849)'0 (h y) dy œ 849’hy h
œ
849 #
2 h .
Now solve
849 #
2 h
h
h
y#
# “!
œ 849 Šh# h#
#‹
œ 40000 to get h ¸ 9.707 ft. The volume of the mercury is s2 h œ 12 † 9.707 œ 9.707 ft$ Þ
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
384
Chapter 6 Applications of Definite Integrals
CHAPTER 6 ADDITIONAL AND ADVANCED EXERCISES
1. V œ 1 'a cf(x)d# dx œ b# ab Ê 1'a cf(t)d# dt œ x# ax for all x a Ê 1 [f(x)]# œ 2x a Ê f(x) œ É 2x1 a
b
x
2. V œ 1 '0 [f(x)]# dx œ a# a Ê 1 '0 [f(t)]# dt œ x# x for all x a Ê 1[f(x)]# œ 2x 1 Ê f(x) œ É 2x1 1
a
x
3. s(x) œ Cx Ê '0 È1 [f w (t)]# dt œ Cx Ê È1 [f w (x)]# œ C Ê f w (x) œ ÈC# 1 for C
x
1
Ê f(x) œ '0 ÈC# 1 dt k. Then f(0) œ a Ê a œ 0 k Ê f(x) œ '0 ÈC# 1 dt a Ê f(x) œ xÈC# 1 a,
x
where C
x
1.
4. (a) The graph of f(x) œ sin x traces out a path from (!ß !) to (!ß sin !) whose length is L œ '0 È1 cos# ) d).
!
The line segment from (0ß 0) to (!ß sin !) has length È(! 0)# (sin ! 0)# œ È!# sin# !. Since the
shortest distance between two points is the length of the straight line segment joining them, we have
!
immediately that ' È1 cos# ) d) È!# sin# ! if 0 ! Ÿ 1 .
#
0
(b) In general, if y œ f(x) is continuously differentiable and f(0) œ 0, then '0 È1 [f w (t)]# dt È!# f # (!)
!
for ! 0.
5. We can find the centroid and then use Pappus' Theorem to calculate the volume. faxb œ x, gaxb œ x2 , faxb œ gaxb
1
Ê x œ x2 Ê x2 x œ 0 Ê x œ 0, x œ 1; $ œ 1; M œ '0 cx x2 ddx œ "# x2 13 x3 ‘0 œ ˆ "# 13 ‰ 0 œ
1
1
6
xœ
1
1 Î6
'01 xcx x2 ddx œ 6'01 cx2 x3 ddx œ 6 31 x3 41 x4 ‘10 œ 6ˆ 31 41 ‰ 0 œ 21
yœ
1
1 Î6
'01 12 ’x2 ax2 b2 “dx œ 3'01 cx2 x4 ddx œ 3 13 x3 15 x5 ‘10 œ 3ˆ 13 15 ‰ 0 œ 25 Ê The centroid is ˆ 12 , 25 ‰.
3 is the distance from ˆ 12 , 25 ‰ to the axis of rotation, y œ x. To calculate this distance we must find the point on y œ x that
also lies on the line perpendicular to y œ x that passes through ˆ 12 , 25 ‰. The equation of this line is y 25 œ 1ˆx 12 ‰
Êxyœ
9
10 .
9
3 œ Ɉ 20
The point of intersection of the lines x y œ
1 ‰2
2
9
ˆ 20
2 ‰2
5
œ
1
.
10È2
9
and y œ x is ˆ 20
,
9
10
Thus V œ 21Š 101È2 ‹ˆ 16 ‰ œ
9 ‰
20 .
Thus,
1
.
30È2
6. Since the slice is made at an angle of 45‰ , the volume of the wedge is half the volume of the cylinder of radius
height 1. Thus, V œ
"
ˆ " ‰2
# ’1 # a1b
“œ
"
#
and
1
8.
7. y œ 2Èx Ê ds œ É "x 1 dx Ê A œ '0 2Èx É "x 1 dx œ
3
4
3
(1 x)$Î# ‘ $ œ
!
28
3
8. This surface is a triangle having a base of 21a and a height of 21ak. Therefore the surface area is
"
# #
# (21a)(21ak) œ 21 a k.
d# x
dt#
9. F œ ma œ t# Ê
œaœ
t#
m
Ê vœ
x œ 0 when t œ 0 Ê C" œ 0 Ê x œ
W œ ' F dx œ '0
Ð12mhÑ"Î%
œ
(12mh)$Î#
18m
œ
F(t) †
12mh†È12mh
18m
œ
2h
3
dx
dt
dx
t$
dt œ 3m C; v œ 0 when t œ 0 Ê
t%
"Î%
.
12m . Then x œ h Ê t œ (12mh)
dt œ '0
Ð12mhÑ"Î%
† 2È3mh œ
4h
3
t# †
$
t
3m
dt œ
"
3m
'
’ t6 “
Ð12mh)"Î%
0
Cœ0 Ê
dx
dt
œ
t$
3m
The work done is
" ‰
œ ˆ 18m
(12mh)'Î%
È3mh
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Ê xœ
t%
12m
C" ;
Chapter 6 Additional and Advanced Exercises
10. Converting to pounds and feet, 2 lb/in œ
†
2 lb
1 in
œ 24 lb/ft. Thus, F œ 24x Ê W œ '0
1Î2
12 in
1 ft
24x dx
"Î#
"
"
‰
œ c12x# d ! œ 3 ft † lb. Since W œ "# mv!# "# mv"# , where W œ 3 ft † lb, m œ ˆ 10
lb‰ ˆ 3# ft/sec
#
"
œ 320
slugs, and v" œ 0 ft/sec, we have 3 œ ˆ #" ‰ ˆ 3#"0 v#! ‰ Ê v!# œ 3 † 640. For the projectile height,
s œ 16t# v! t (since s œ 0 at t œ 0) Ê ds
dt œ v œ 32t v! . At the top of the ball's path, v œ 0 Ê
#
and the height is s œ 16 ˆ 3v#! ‰ v! ˆ 3v#! ‰ œ
v!#
64
3†640
64
œ
385
tœ
v!
3#
œ 30 ft.
11. From the symmetry of y œ 1 xn , n even, about the y-axis for 1 Ÿ x Ÿ 1, we have x œ 0. To find y œ MMx , we
n
use the vertical strips technique. The typical strip has center of mass: (µ
x ßµ
y ) œ ˆxß 1 2 x ‰ , length: 1 xn ,
width: dx, area: dA œ a1 xn b dx, mass: dm œ 1 † dA œ a1 xn b dx. The moment of the strip about the
1
1
n #
n #
"
nb1
2n b 1
x-axis is µ
y dm œ a1 x b dx Ê M œ ' a1 x b dx œ 2' " a1 2xn x2n b dx œ x 2x x ‘
#
œ1
2
n1
"
#n 1
œ
x
c1 #
(n 1)(2n 1) 2(2n ") (n 1)
(n 1)(#n 1)
œ
0 #
2n# 3n 1 4n 2 n 1
(n 1)(#n 1)
Also, M œ 'c1 dA œ 'c1 a1 xn b dx œ 2 '0 a1 xn b dx œ 2 x 1
yœ
Mx
M
œ
1
#
†
2n
(n 1)(2n 1)
1
(n 1)
2n
œ
xn b 1 ‘ "
n1 !
n1
œ
2n#
(n 1)(#n 1)
œ 2 ˆ1 " ‰
n1
#n 1 !
.
œ
2n
n1.
Therefore,
Ê ˆ!ß #n n 1 ‰ is the location of the centroid. As n Ä _, y Ä
n
2n 1
the limiting position of the centroid is ˆ!ß
"
#
so
"‰
# .
12. Align the telephone pole along the x-axis as shown in the
accompanying figure. The slope of the top length of pole is
9 ‰
ˆ 14.5
"
81 81
œ 8"1 † 40
† (14.5 9) œ 815.5
†40
40
11 ‰
y œ 891 8111†80 x œ 8"1 ˆ9 80
x is an
œ
11
81†80 .
Thus,
equation of the
line representing the top of the pole. Then,
My œ 'a x † 1y# dx œ 1 '0 x 8"1 ˆ9 b
40
11
80
#
x‰‘ dx
b
11 ‰#
'040 x ˆ9 80
x dx; M œ 'a 1y# dx
40
40
11 ‰‘#
‰# dx.
œ 1 '0 8"1 ˆ9 80
x
dx œ 64"1 '0 ˆ9 11
80 x
œ
"
641
Thus, x œ
My
M
¸
129,700
5623.3
¸ 23.06 (using a calculator to compute
the integrals). By symmetry about the x-axis, y œ 0 so the center of mass is about 23 ft from the top of the pole.
13. (a) Consider a single vertical strip with center of mass (µ
x ßµ
y ). If the plate lies to the right of the line, then
µ
µ b) $ dA Ê the plate's first moment
the moment of this strip about the line x œ b is (x b) dm œ (x
about x œ b is the integral ' (x b)$ dA œ ' $ x dA ' $ b dA œ My b$ A.
(b) If the plate lies to the left of the line, the moment of a vertical strip about the line x œ b is
ab µ
x b dm œ ab µ
x b $ dA Ê the plate's first moment about x œ b is ' (b x)$ dA œ ' b$ dA ' $ x dA
œ b$ A My .
14. (a) By symmetry of the plate about the x-axis, y œ 0. A typical vertical strip has center of mass:
(µ
x ßµ
y ) œ (xß 0), length: 4Èax, width: dx, area: 4Èax dx, mass: dm œ $ dA œ kx † 4Èax dx, for some
a
proportionality constant k. The moment of the strip about the y-axis is M œ ' µ
x dm œ ' 4kx# Èax dx
y
œ 4kÈa'0 x
a
&Î#
dx œ
4kÈa 27
x
(Î# ‘ a
0
œ 4ka
"Î#
† a
2
7
(Î#
œ
œ 4kÈa'0 x$Î# dx œ 4kÈa 25 x&Î# ‘ 0 œ 4ka"Î# † 25 a&Î# œ
a
Ê (xß y) œ
a
ˆ 5a
‰
7 ß0
0
8ka%
7
. Also, M œ ' dm œ '0 4kxÈax dx
8ka$
5
. Thus, x œ
a
My
M
œ
8ka%
7
†
5
8ka$
œ
5
7
is the center of mass.
y#
#
#
a
(b) A typical horizontal strip has center of mass: (µ
x ßµ
y ) œ Œ 4a # ß y œ Š y 8a4a ß y‹ , length: a width: dy, area: Ša œ 'c2a y kyk Ša 2a
a
y#
4a ‹
y#
4a ‹
dy, mass: dm œ $ dA œ kyk Ša dy œ 'c2a y# Ša 0
y#
4a ‹
y#
4a ‹
dy '0 y# Ša 2a
dy. Thus, Mx œ ' µ
y dm
y#
4a ‹
dy
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
y#
4a
,
386
Chapter 6 Applications of Definite Integrals
œ 'c2a Šay# 0
%
32a&
20a
œ 8a3 dy '0 Šay# 2a
y%
4a ‹
8a%
3
y%
4a ‹
œ 0; My œ ' µ
x dm œ 'c2a Š y
2a
32a&
#0a
'c2a2a kyk ay# 4a# b Š 4a 4a y ‹ dy œ 32a"
#
!
y&
#0a “ #a
dy œ ’ 3a y$ #
4a#
8a ‹
#
"
8a
œ
"
3 #a #
'c02a a16a% y y& b dy 32a" '02a a16a% y y& b dy œ 3#"a
œ
"
32a#
’8a% † 4a# "
32a#
2a
œ
#
#
64a'
6 “
œ
"
16a#
#
’8a% y# 32a'
3 ‹
Š32a' œ
#a
œ2†
#
16a%
4 ‹
#
Š2a † 4a "
#a
œ
!
y'
6 “ #a
1
3#a#
’8a% y# † 32 a32a' b œ
4
3
#a
y'
6 “!
a% ;
'c2a2a kyk a4a# y# b dy
"
4a
%
"
4a
dy
"
16a#
'c02a a4a# y y$ b dy 4a" '02a a4a# y y$ b dy œ 4a" ’2a# y# y4 “ !
"
4a
yœ
c2a
’8a% † 4a# M œ ' dm œ 'c2a kyk Š 4a 4ay ‹ dy œ
y#
4a ‹
kyk Ša ' kyk a16a% y% b dy
#
#
#a
y&
#0a “ !
2a
œ
64a'
6 “
’ 3a y$ %
%
$
a8a 4a b œ 2a . Therefore, x œ
"
4a
’2a# y# œ ˆ 34 a% ‰ ˆ 2a"$ ‰ œ
My
M
#a
y%
4 “!
2a
3
and
œ 0 is the center of mass.
Mx
M
15. (a) On [0ß a] a typical @/<>3-+6 strip has center of mass: (µ
x ßµ
y ) œ Šx,
È b# x # È a# x#
‹,
#
length: Èb# x# Èa# x# , width: dx, area: dA œ ŠÈb# x# Èa# x# ‹ dx, mass: dm œ $ dA
œ $ ŠÈb# x# Èa# x# ‹ dx. On [aß b] a typical @/<>3-+6 strip has center of mass:
È #
#
(µ
x ßµ
y ) œ Šxß b # x ‹ , length: Èb# x# , width: dx, area: dA œ Èb# x# dx,
mass: dm œ $ dA œ $ Èb# x# dx. Thus, Mx œ ' µ
y dm
œ '0
a
"
#
ŠÈb# x# Èa# x# ‹ $ ŠÈb# x# Èa# x# ‹ dx 'a
b
"
#
Èb# x# $ Èb# x# dx
œ
$
#
'0a cab# x# b aa# x# bd dx #$ 'ab ab# x# b dx œ #$ '0a ab# a# b dx #$ 'ab ab# x# b dx
œ
$
#
cab# a# b xd ! #$ ’b# x œ
$
#
aab# a$ b #$ Š 23 b$ ab# a
b
x$
3 “a
œ
a$
3‹
$
#
b$
3‹
cab# a# b ad #$ ’Šb$ œ
$ b$
3
$ a$
3
œ $ Šb
$
a$
3 ‹;
a$
3 ‹“
Š b# a My œ ' µ
x dm
œ '0 x$ ŠÈb# x# Èa# x# ‹ dx 'a x$ Èb# x# dx
a
b
œ $ '0 x ab# x# b
a
œ
$
#
”
2 ab # x # b
3
$Î#
"Î#
dx $ '0 x aa# x# b
a
a
$ 2 aa
• #”
#
x# b
3
$Î#
# $Î#
#
œ ’ab a b
# $Î#
ab b
a
dx $ 'a x ab# x# b
$ 2 ab
• #”
b
#
!
0
$
3
"Î#
# $Î#
$
3
“ ’0 aa b
x# b
3
"Î#
$Î#
•
a
$
3
“ ’0 ab# a# b
#
#
$Î#
We calculate the mass geometrically: M œ $ A œ $ Š 14b ‹ $ Š 14a ‹ œ
œ
$ ab $ a $ b
3
yœ
(b) lim
œ
Mx
M
4
b Ä a 31
†
4
$1 ab# a# b
#
#
4 aa abb b
31(ab)
Ša
#
ab b
ab
#
œ
4
31
$
$
a
Š bb# a# ‹ œ
dx
b
4 (b a) aa# ab b# b
31
(b a)(b a)
“œ
$1
4
$ b$
3
$ a$
3
œ
$ ab $ a $ b
3
ab# a# b . Thus, x œ
œ Mx ;
My
M
œ
4 aa# ab b# b
31(a b)
2a
1
2a ‰
Ê (xß y) œ ˆ 2a
1 ß 1 is the limiting
; likewise
.
‹ œ ˆ 341 ‰ Š a
#
a# a#
‹
aa
#
œ ˆ 341 ‰ Š 3a
2a ‹ œ
position of the centroid as b Ä a. This is the centroid of a circle of radius a (and we note the two circles
coincide when b œ a).
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Chapter 6 Additional and Advanced Exercises
16. Since the area of the traingle is 36, the diagram may be
labeled as shown at the right. The centroid of the triangle is
ˆ 3a , 24
‰
a . The shaded portion is 144 36 œ 108. Write
ax, yb for the centroid of the remaining region. The centroid
of the whole square is obviously a6, 6b. Think of the square
as a sheet of uniform density, so that the centroid of the
square is the average of the centroids of the two regions,
weighted by area:
'œ
$'ˆ 3a ‰ "!)axb
"%%
and ' œ
‰
$'ˆ 24
a "!)ayb
"%%
which we solve to get x œ ) a
*
and y œ
)a a " b
.
a
Set
x œ 7 in. (Given). It follows that a œ *, whence y œ
œ
7 "*
'%
*
in. The distances of the centroid ax, yb from the other sides are easily computed. (Note that if we set y œ 7 in.
above, we will find x œ 7 "* .)
17. The submerged triangular plate is depicted in the figure
at the right. The hypotenuse of the triangle has slope 1
Ê y (2) œ (x 0) Ê x œ (y 2) is an equation
of the hypotenuse. Using a typical horizontal strip, the fluid
strip
strip
pressure is F œ ' (62.4) † Š depth
‹ † Š length
‹ dy
c2
c2
œ 'c6 (62.4)(y)[(y 2)] dy œ 62.4 'c6 ay# 2yb dy
$
œ 62.4 ’ y3 y# “
#
‰‘
œ (62.4) ˆ 83 4‰ ˆ 216
3 36
'
‰
œ (62.4) ˆ 208
3 32 œ
(62.4)(112)
3
¸ 2329.6 lb
18. Consider a rectangular plate of length j and width w.
The length is parallel with the surface of the fluid of
weight density =. The force on one side of the plate is
F œ ='cw (y)(j) dy œ =j ’ y# “
0
#
!
w
œ
=jw#
#
. The
average force on one side of the plate is Fav œ
œ
=
w
#
’ y# “
!
w
œ
=w
#
. Therefore the force
=
w
'c0w (y)dy
=jw#
#
‰
œ ˆ =w
# (jw) œ (the average pressure up and down) † (the area of the plate).
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
387
388
Chapter 6 Applications of Definite Integrals
NOTES:
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
CHAPTER 7 TRANSCENDENTAL FUNCTIONS
7.1 INVERSE FUNCTIONS AND THEIR DERIVATIVES
1. Yes one-to-one, the graph passes the horizontal line test.
2. Not one-to-one, the graph fails the horizontal line test.
3. Not one-to-one since (for example) the horizontal line y œ 2 intersects the graph twice.
4. Not one-to-one, the graph fails the horizontal line test.
5. Yes one-to-one, the graph passes the horizontal line test
6. Yes one-to-one, the graph passes the horizontal line test
7. Not one-to-one since the horizontal line y œ 3 intersects the graph an infinite number of times.
8. Yes one-to-one, the graph passes the horizontal line test
9. Yes one-to-one, the graph passes the horizontal line test
10. Not one-to-one since (for example) the horizontal line y œ 1 intersects the graph twice.
11. Domain: 0 x Ÿ 1, Range: 0 Ÿ y
13. Domain: 1 Ÿ x Ÿ 1, Range: 1# Ÿ y Ÿ
12. Domain: x 1, Range: y 0
1
#
14. Domain: _ x _, Range: 1# y Ÿ
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
1
#
390
Chapter 7 Transcendental Functions
15. Domain: 0 Ÿ x Ÿ 6, Range: 0 Ÿ y Ÿ 3
16. Domain: 2 Ÿ x Ÿ 1, Range: 1 Ÿ y 3
17. The graph is symmetric about y œ x.
(b) y œ È1 x# Ê y# œ 1 x# Ê x# œ 1 y# Ê x œ È1 y# Ê y œ È1 x# œ f " (x)
18. The graph is symmetric about y œ x.
yœ
"
x
Ê xœ
"
y
Ê yœ
"
x
œ f " (x)
19. Step 1: y œ x# 1 Ê x# œ y 1 Ê x œ Èy 1
Step 2: y œ Èx 1 œ f " (x)
20. Step 1: y œ x# Ê x œ Èy, since x Ÿ !.
Step 2: y œ Èx œ f " (x)
21. Step 1: y œ x$ 1 Ê x$ œ y 1 Ê x œ (y 1)"Î$
Step 2: y œ $Èx 1 œ f " (x)
22. Step 1: y œ x# 2x 1 Ê y œ (x 1)# Ê Èy œ x 1, since x
Step 2: y œ 1 Èx œ f
"
1 Ê x œ 1 Èy
(x)
23. Step 1: y œ (x 1)# Ê Èy œ x 1, since x
1 Ê x œ È y 1
Step 2: y œ Èx 1 œ f " (x)
24. Step 1: y œ x#Î$ Ê x œ y$Î#
Step 2: y œ x$Î# œ f " (x)
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 7.1 Inverse Functions and Their Derivatives
25. Step 1: y œ x& Ê x œ y"Î&
5
Step 2: y œ È
x œ f " (x);
Domain and Range of f " : all reals;
&
f af " (x)b œ ˆx"Î& ‰ œ x and f " (f(x)) œ ax& b
"Î&
œx
"Î%
œx
26. Step 1: y œ x% Ê x œ y"Î%
Step 2: y œ %Èx œ f " (x);
Domain of f " : x
f af
"
(x)b œ ˆx
0, Range of f " : y
"Î% ‰%
œ x and f
"
0;
(f(x)) œ ax% b
27. Step 1: y œ x$ 1 Ê x$ œ y 1 Ê x œ (y 1)"Î$
Step 2: y œ $Èx 1 œ f " (x);
Domain and Range of f " : all reals;
$
f af " (x)b œ ˆ(x 1)"Î$ ‰ 1 œ (x 1) 1 œ x and f " (f(x)) œ aax$ 1b 1b
28. Step 1: y œ
"
#
x
7
#
Ê
"
"
#
xœy
7
#
Step 2: y œ
"
x#
Ê x# œ
"
Èx
œ f " (x)
"
y
Ê xœ
7
#
30. Step 1: y œ
"
x$
"
x"Î$
"
Step 2: y œ
Domain of f
f af " axbb œ
"
y
Ê xœ
Step 2: y
32. Step 1: y œ
œ
"
Š "x ‹
œ x since x 0
"
y"Î$
: x Á 0, Range of f " : y Á 0;
"
$
ax "Î$ b
œ
"
x "
œ x and f " afaxbb œ ˆ x"$ ‰
x3
x 2 Ê yax 2b
3
1
œ 2x
axb;
x1 œ f
"
f af " axbb œ
"
É x"#
3 "
"
σ
(x);
x œ f
31. Step 1: y œ
Domain of f
œx
"
Èy
Šx‹
Ê x$ œ
"Î$
œ x and f " (f(x)) œ 2 ˆ "# x 7# ‰ 7 œ (x 7) 7 œ x
Domain of f " : x 0, Range of f " : y 0;
f af " (x)b œ "" # œ "" œ x and f " (f(x)) œ
Š Èx ‹
œ ax$ b
Ê x œ 2y 7
Step 2: y œ 2x 7 œ f (x);
Domain and Range of f " : all reals;
f af " (x)b œ "# (2x 7) 7# œ ˆx 7# ‰ 29. Step 1: y œ
"Î$
"Î$
œ ˆ x" ‰
"
œx
œ x 3 Ê x y 2y œ x 3 Ê x y x œ 2y 3 Ê x œ
: x Á 1, Range of f " : y Á 2;
b3‰
ˆ 2x
xc1 3
b3‰
ˆ 2x
xc1 2
Èx
Èx 3
œ
a2x 3b 3ax 1b
a2x 3b 2ax 1b
œ
5x
5
œ x and f " afaxbb œ
3
2ˆ xx b
c2‰ 3
3
ˆ xx b
c2‰ 1
œ
2y 3
y1
2 ax 3 b 3 a x 2 b
ax 3 b ax 2 b
œ
5x
5
œx
Ê yˆÈx 3‰ œ Èx Ê yÈx 3y œ Èx Ê yÈx Èx œ 3y Ê x œ Š y 3y
1‹
2
2
‰ œ f 1 a xb ;
Step 2: y œ ˆ x 3x
1
Domain of f " : Ð_, 0Ó a1, _b, Range of f " : Ò0, 9Ñ a9, _b;
f af " axbb œ
f
"
2
Ɉ x 3x
c1‰
2
Ɉ x 3x
c1‰ 3
afaxbb œ ; If x 1 or x Ÿ 0 Ê
Èx
3Š È x c 3 ‹
Èx
Š Èx c 3 ‹ 1 3x
x1
0Ê
2
Ɉ x 3x
c1‰
2
Ɉ x 3x
c1‰ 3
œ
3x
xc1
3x
xc1 3
œ
3x
3x 3ax 1b
œ
2
œ
9x
ˆÈ x ˆÈ x 3 ‰‰ 2
œ
9x
9
œx
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
3x
3
œ x and
391
392
Chapter 7 Transcendental Functions
33. Step 1: y œ x2 2x, x Ÿ 1 Ê y 1 œ ax 1b2 , x Ÿ 1 Ê Èy 1 œ x 1, x Ÿ 1 Ê x œ 1 Èy 1
Step 2: y œ 1 Èx 1 œ f 1 axb;
Domain of f " : Ò1, _Ñ, Range of f " : Ð_, 1Ó;
2
f af " axbb œ Š1 Èx 1‹ 2Š1 Èx 1‹ œ 1 2Èx 1 x 1 2 2Èx 1 œ x and
f " afaxbb œ 1 Èax2 2xb 1, x Ÿ 1 œ 1 Éax 1b2 , x Ÿ 1 œ 1 lx 1l œ 1 a1 xb œ x
34. Step 1: y œ a2x3 1b
3 x
Step 2: y œ É
5
1
2
1 Î5
Ê y5 œ 2x3 1 Ê y5 1 œ 2x3 Ê
y5 1
2
3 y
œ x3 Ê x œ É
5
1
2
œ f 1 ax b ;
Domain of f " : a_, _b, Range of f " : a_, _b;
3 x
f af " axbb œ Œ2ŠÉ
5
1
2 ‹
3
1 Î5
1
œ Š 2Š x
5
1
2 ‹
1‹
1 Î5
œ aax5 1b 1b
1 Î5
œ ax5 b
1 Î5
œ x and
5
1Î5
’a2x3 1b “ 1
3
f " afaxbb œ Ê
2
3 a2x
σ
35. (a) y œ 2x 3 Ê 2x œ y 3
Ê x œ y# 3# Ê f " (x) œ
(c)
df ¸
dx xœ 1
36. (a) y œ
"
5
œ 2,
df c"
dx ¹ xœ1
x7 Ê
"
5
œ
(c)
œ
" df c"
5 , dx ¹ xœ$%Î&
(c)
œ 4,
df c"
dx ¹ xœ3
38. (a) y œ 2x# Ê x# œ
Ê xœ
(c)
df ¸
dx xœ&
"
È2
"
#
3
3
#
"
#
"
(b)
(x) œ 5x 35
œ5
œ
(b)
5
4
x
4
"
4
(b)
y
Èy Ê f
3 2x
σ
2 œ x
(b)
x
#
37. (a) y œ 5 4x Ê 4x œ 5 y
Ê x œ 54 y4 Ê f " (x) œ
df ¸
dx xœ1Î#
1 b 1
2
xœy7
Ê x œ 5y 35 Ê f
df ¸
dx xœ 1
3
"
(x) œ
È x#
œ 4xk xœ5 œ 20,
df c"
dx ¹ xœ&0
œ
"
#È 2
x"Î# ¹
xœ50
œ
"
#0
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 7.1 Inverse Functions and Their Derivatives
$
3
3
39. (a) f(g(x)) œ ˆÈ
x‰ œ x, g(f(x)) œ È
x3 œ x
w
#
w
(b)
w
(c) f (x) œ 3x Ê f (1) œ 3, f (1) œ 3;
gw (x) œ 3" x#Î$ Ê gw (1) œ 3" , gw (1) œ
"
3
(d) The line y œ 0 is tangent to f(x) œ x$ at (!ß !);
the line x œ 0 is tangent to g(x) œ $Èx at (0ß 0)
40. (a) h(k(x)) œ
"
4
ˆ(4x)"Î$ ‰$ œ x,
k(h(x)) œ Š4 †
(c) hw (x) œ
w
k (x) œ
x$
4‹
"Î$
(b)
œx
3x#
w
w
4 Ê h (2) œ 3, h (2)
4
#Î$
Ê kw (2) œ "3 ,
3 (4x)
œ 3;
kw (2) œ
(d) The line y œ 0 is tangent to h(x) œ
x$
4
"
3
at (!ß !);
the line x œ 0 is tangent to k(x) œ (4x)"Î$ at
(!ß !)
œ 3x# 6x Ê
41.
df
dx
43.
df "
dx ¹ x œ 4
df c"
dx ¹ x œ f(3)
df "
dx ¹ x œ f(2)
œ
45. (a) y œ mx Ê x œ
"
m
œ
(b) The graph of y œ f
46. y œ mx b Ê x œ
y
m
"
df
dx
º
œ
xœ2
"
df
dx
œ
º
xœ3
"
ˆ 3" ‰
œ3
y Ê f " (x) œ
"
"
9
œ
"
m
œ 2x 4 Ê
42.
df
dx
44.
dg "
dx ¹x œ 0
b
m
dg "
dx ¹ x œ f(0)
œ
"
dg
dx
º
œ
xœ0
"
df
dx
º
œ
xœ5
œ
"
6
"
2
x
(x) is a line through the origin with slope
œ
df "
dx ¹ x œ f(5)
Ê f " (x) œ
"
m
x
b
m;
"
m.
the graph of f " (x) is a line with slope
"
m
and y-intercept mb .
47. (a) y œ x 1 Ê x œ y 1 Ê f " (x) œ x 1
(b) y œ x b Ê x œ y b Ê f " (x) œ x b
(c) Their graphs will be parallel to one another and lie on
opposite sides of the line y œ x equidistant from that
line.
48. (a) y œ x 1 Ê x œ y 1 Ê f " (x) œ 1 x;
the lines intersect at a right angle
(b) y œ x b Ê x œ y b Ê f " (x) œ b x;
the lines intersect at a right angle
(c) Such a function is its own inverse.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
393
394
Chapter 7 Transcendental Functions
49. Let x" Á x# be two numbers in the domain of an increasing function f. Then, either x" x# or
x" x# which implies f(x" ) f(x# ) or f(x" ) f(x# ), since f(x) is increasing. In either case,
f(x" ) Á f(x# ) and f is one-to-one. Similar arguments hold if f is decreasing.
50. f(x) is increasing since x# x" Ê
"
3
x# 5
6
"
3
x" 56 ;
df
dx
œ
"
3
51. f(x) is increasing since x# x" Ê 27x$# 27x"$ ; y œ 27x$ Ê x œ
df
dx
œ 81x# Ê
df "
dx
œ
" ¸
81x# 13 x"Î$
œ
"
9x#Î$
œ
"
9
df c"
dx
Ê
"
3
œ
df
dx
œ 24x# Ê
dx
œ
" ¸
24x# 12 Ð1 xÑ"Î$
œ
œ3
y"Î$ Ê f " (x) œ
"
3
x"Î$ ;
x#Î$
52. f(x) is decreasing since x# x" Ê 1 8x$# 1 8x"$ ; y œ 1 8x$ Ê x œ
df c"
"
ˆ "3 ‰
"
6(" x)#Î$
"
#
(1 y)"Î$ Ê f " (x) œ
"
#
(1 x)"Î$ ;
œ "6 (1 x)#Î$
53. f(x) is decreasing since x# x" Ê (1 x# )$ (1 x" )$ ; y œ (1 x)$ Ê x œ 1 y"Î$ Ê f " (x) œ 1 x"Î$ ;
df
dx
œ 3(1 x)# Ê
df c"
dx
œ
"
3(1 x)# ¹ 1cx"Î$
&Î$
54. f(x) is increasing since x# x" Ê x#
df
dx
œ
5
3
x#Î$ Ê
df c"
dx
œ
5
3
"
¹
x#Î$ x$Î&
œ
"
3x#Î$
œ "3 x#Î$
&Î$
x" ; y œ x&Î$ Ê x œ y$Î& Ê f " (x) œ x$Î& ;
œ
3
5x#Î&
œ
3
5
x#Î&
55. The function g(x) is also one-to-one. The reasoning: f(x) is one-to-one means that if x" Á x# then f(x" ) Á f(x# ), so
f(x" ) Á f(x# ) and therefore g(x" ) Á g(x# ). Therefore g(x) is one-to-one as well.
56. The function h(x) is also one-to-one. The reasoning: f(x) is one-to-one means that if x" Á x# then f(x" ) Á f(x# ), so
"
"
f(x" ) Á f(x# ) , and therefore h(x" ) Á h(x# ).
57. The composite is one-to-one also. The reasoning: If x" Á x# then g(x" ) Á g(x# ) because g is one-to-one. Since
g(x" ) Á g(x# ), we also have f(g(x" )) Á f(g(x# )) because f is one-to-one. Thus, f ‰ g is one-to-one because
x" Á x# Ê f(g(x" )) Á f(g(x# )).
58. Yes, g must be one-to-one. If g were not one-to-one, there would exist numbers x" Á x# in the domain of g with
g(x" ) œ g(x# ). For these numbers we would also have f(g(x" )) œ f(g(x# )), contradicting the assumption that
f ‰ g is one-to-one.
59. (g ‰ f)(x) œ x Ê g(f(x)) œ x Ê gw (f(x))f w (x) œ 1
60. W(a) œ 'f(a) 1 ’af " (y)b a# “ dy œ 0 œ 'a 21x[f(a) f(x)] dx œ S(a); Ww (t) œ 1’af " (f(t))b a# “ f w (t)
f(a)
a
#
#
œ 1 at# a# b f w (t); also S(t) œ 21f(t)'a x dx 21'a xf(x) dx œ c1f(t)t# 1f(t)a# d 21'a xf(x) dx Ê Sw (t)
t
t
t
œ 1t# f w (t) 21tf(t) 1a# f w (t) 21tf(t) œ 1 at# a# b f w (t) Ê Ww (t) œ Sw (t). Therefore, W(t) œ S(t) for all t − [aß b].
61-68. Example CAS commands:
Maple:
with( plots );#63
f := x -> sqrt(3*x-2);
domain := 2/3 .. 4;
x0 := 3;
Df := D(f);
# (a)
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 7.1 Inverse Functions and Their Derivatives
395
plot( [f(x),Df(x)], x=domain, color=[red,blue], linestyle=[1,3], legend=["y=f(x)","y=f '(x)"],
title="#61(a) (Section 7.1)" );
q1 := solve( y=f(x), x );
# (b)
g := unapply( q1, y );
m1 := Df(x0);
# (c)
t1 := f(x0)+m1*(x-x0);
y=t1;
m2 := 1/Df(x0);
# (d)
t2 := g(f(x0)) + m2*(x-f(x0));
y=t2;
domaing := map(f,domain);
# (e)
p1 := plot( [f(x),x], x=domain, color=[pink,green], linestyle=[1,9], thickness=[3,0] ):
p2 := plot( g(x), x=domaing, color=cyan, linestyle=3, thickness=4 ):
p3 := plot( t1, x=x0-1..x0+1, color=red, linestyle=4, thickness=0 ):
p4 := plot( t2, x=f(x0)-1..f(x0)+1, color=blue, linestyle=7, thickness=1 ):
p5 := plot( [ [x0,f(x0)], [f(x0),x0] ], color=green ):
display( [p1,p2,p3,p4,p5], scaling=constrained, title="#63(e) (Section 7.1)" );
Mathematica: (assigned function and values for a, b, and x0 may vary)
If a function requires the odd root of a negative number, begin by loading the RealOnly package that allows Mathematica
to do this. See section 2.5 for details.
<<Miscellaneous `RealOnly`
Clear[x, y]
{a,b} = {2, 1}; x0 = 1/2 ;
f[x_] = (3x 2) / (2x 11)
Plot[{f[x], f'[x]}, {x, a, b}]
solx = Solve[y == f[x], x]
g[y_] = x /. solx[[1]]
y0 = f[x0]
ftan[x_] = y0 f'[x0] (x-x0)
gtan[y_] = x0 1/ f'[x0] (y y0)
Plot[{f[x], ftan[x], g[x], gtan[x], Identity[x]},{x, a, b},
Epilog Ä Line[{{x0, y0},{y0, x0}}], PlotRange Ä {{a,b},{a,b}}, AspectRatio Ä Automatic]
69-70. Example CAS commands:
Maple:
with( plots );
eq := cos(y) = x^(1/5);
domain := 0 .. 1;
x0 := 1/2;
f := unapply( solve( eq, y ), x ); # (a)
Df := D(f);
plot( [f(x),Df(x)], x=domain, color=[red,blue], linestyle=[1,3], legend=["y=f(x)","y=f '(x)"],
title="#70(a) (Section 7.1)" );
q1 := solve( eq, x );
# (b)
g := unapply( q1, y );
m1 := Df(x0);
# (c)
t1 := f(x0)+m1*(x-x0);
y=t1;
m2 := 1/Df(x0);
# (d)
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
396
Chapter 7 Transcendental Functions
t2 := g(f(x0)) + m2*(x-f(x0));
y=t2;
domaing := map(f,domain);
# (e)
p1 := plot( [f(x),x], x=domain, color=[pink,green], linestyle=[1,9], thickness=[3,0] ):
p2 := plot( g(x), x=domaing, color=cyan, linestyle=3, thickness=4 ):
p3 := plot( t1, x=x0-1..x0+1, color=red, linestyle=4, thickness=0 ):
p4 := plot( t2, x=f(x0)-1..f(x0)+1, color=blue, linestyle=7, thickness=1 ):
p5 := plot( [ [x0,f(x0)], [f(x0),x0] ], color=green ):
display( [p1,p2,p3,p4,p5], scaling=constrained, title="#70(e) (Section 7.1)" );
Mathematica: (assigned function and values for a, b, and x0 may vary)
For problems 69 and 70, the code is just slightly altered. At times, different "parts" of solutions need to be used, as in the
definitions of f[x] and g[y]
Clear[x, y]
{a,b} = {0, 1}; x0 = 1/2 ;
eqn = Cos[y] == x1/5
soly = Solve[eqn, y]
f[x_] = y /. soly[[2]]
Plot[{f[x], f'[x]}, {x, a, b}]
solx = Solve[eqn, x]
g[y_] = x /. solx[[1]]
y0 = f[x0]
ftan[x_] = y0 f'[x0] (x x0)
gtan[y_] = x0 1/ f'[x0] (y y0)
Plot[{f[x], ftan[x], g[x], gtan[x], Identity[x]},{x, a, b},
Epilog Ä Line[{{x0, y0},{y0, x0}}], PlotRange Ä {{a, b}, {a, b}}, AspectRatio Ä Automatic]
7.2 NATURAL LOGARITHMS
1. (a) ln 0.75 œ ln
(b) ln
4
9
"
#
3
4
œ ln 3 ln 4 œ ln 3 ln 2# œ ln 3 2 ln 2
œ ln 4 ln 9 œ ln 2# ln 3# œ 2 ln 2 2 ln 3
3
œ ln 1 ln 2 œ ln 2
(d) ln È
9 œ "3 ln 9 œ
(e) ln 3È2 œ ln 3 ln 2"Î# œ ln 3 "# ln 2
(f) ln È13.5 œ " ln 13.5 œ " ln 27 œ " aln 3$ ln 2b œ " (3 ln 3 ln 2)
(c) ln
#
2. (a) ln
"
125
#
(e) ln 0.056 œ ln
7
125
3
#
ln 3# œ
2
3
ln 3
#
(b) ln 9.8 œ ln
49
5
œ ln 7# ln 5 œ 2 ln 7 ln 5
(d) ln 1225 œ ln 35# œ 2 ln 35 œ 2 ln 5 2 ln 7
ln 7
œ ln 7 ln 5$ œ ln 7 3 ln 5
3. (a) ln sin ) ln ˆ sin5 ) ‰ œ ln "
#
#
œ ln 1 3 ln 5 œ 3 ln 5
(c) ln 7È7 œ ln 7$Î# œ
(c)
#
"
3
sin )
Š sin5 ) ‹ œ ln 5
(f)
ln 35 ln
ln 25
"
7
œ
ln 5 ln 7 ln 7
# ln 5
œ
"
#
#
" ‰
(b) ln a3x# 9xb ln ˆ 3x
œ ln Š 3x 3x 9x ‹ œ ln (x 3)
#
ln a4t% b ln 2 œ ln È4t% ln 2 œ ln 2t# ln 2 œ ln Š 2t# ‹ œ ln at# b
4. (a) ln sec ) ln cos ) œ ln [(sec ))(cos ))] œ ln 1 œ 0
(b) ln (8x 4) ln 2# œ ln (8x 4) ln 4 œ ln ˆ 8x 4 4 ‰ œ ln (2x 1)
$
"Î$
")
(c) 3 ln Èt# 1 ln (t 1) œ 3 ln at# 1b ln (t 1) œ 3 ˆ "3 ‰ ln at# 1b ln (t 1) œ ln Š (t (t1)(t
‹
1)
œ ln (t 1)
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 7.2 Natural Logarithms
1 ‰
5. y œ ln 3x Ê yw œ ˆ 3x
(3) œ
7. y œ ln at# b Ê
œ ˆ t"# ‰ (2t) œ
dy
dt
œ ln 3x" Ê
9. y œ ln
3
x
10. y œ ln
10
x
dy
dx
œ ln 10x" Ê
11. y œ ln () 1) Ê
13. y œ ln x$ Ê
dy
dt
17. y œ
x%
4
ln x dy
dx
œ ˆ ) " 1 ‰ (1) œ
dy
d)
4
18. y œ ax2 ln xb Ê
Ê
"
)1
dy
dt
14. y œ (ln x)$ Ê
d
dt
(ln t) œ (ln t)# œ (ln t)"Î# "# t(ln t)"Î# †
x%
4
4x$
16
œ
t ˆ "t ‰ (ln t)(1)
t#
dy
dt
œ
ln x
1 ln x
Ê yw œ
(1 ln x) ˆ "x ‰ (ln x) ˆ "x ‰
(1 ln x)#
22. y œ
x ln x
1 ln x
Ê yw œ
(1ln x) ˆln x x† "x ‰ (x ln x) ˆ x" ‰
(1ln x)#
œ
t ˆ "t ‰ (" ln t)(1)
t#
23. y œ ln (ln x) Ê yw œ ˆ ln"x ‰ ˆ "x ‰ œ
24. y œ ln (ln (ln x)) Ê yw œ
"
ln (ln x)
25. y œ )[sin (ln )) cos (ln ))] Ê
œ x$ ln x
†
œ
œ
"
x
œ lnt# t
lnxx lnxx
(1 ln x)#
œ
œ
"
x(1 ln x)#
(" ln x)# ln x
(1 ln x)#
œ1
ln x
(1 ln x)#
"
x ln x
d
dx
dy
d)
" 1 ln t
t#
(ln (ln x)) œ
"
ln (ln x)
†
"
ln x
†
d
dx
(ln x) œ
"
x (ln x) ln (ln x)
œ [sin (ln )) cos (ln ))] ) cos (ln )) †
œ sin (ln )) cos (ln )) cos (ln )) sin (ln )) œ 2 cos (ln ))
26. y œ ln (sec ) tan )) Ê
dy
d)
"
xÈ x 1
"
#
œ ln x œ
"
#
3(ln x)#
x
t(ln t)c"Î#
#t
(ln t) œ (ln t)"Î# 1 ln t
t#
21. y œ
1x
1x
(ln x) œ
2x ln x‰ œ 4x6 aln xb3 ax 2x ln xb œ 4x7 aln xb3 8x7 aln xb4
Ê
ln
d
dx
1
x
" ln t
t
"
#
œ 3(ln x)# †
"
)1
3
œ 4ax2 ln xb ˆx2 †
20. y œ
28. y œ
œ ˆ #) " 2 ‰ (2) œ
œ (ln t)# 2 ln t
2t ln t
t
d
dt
dy
dx
dy
d)
ln t
t
27. y œ ln
3
2t
"
x
œ x$ ln x 19. y œ
dy
dt
" ‰ ˆ 3 "Î# ‰
œ ˆ t$Î#
œ
# t
†
dy
dx
dy
dx
12. y œ ln (2) 2) Ê
3
x
œ (ln t)# 2t(ln t) †
Ê
dy
dt
œ ˆ 10x" " ‰ a10x# b œ x"
"
#(ln t)"Î#
x%
16
8. y œ ln ˆt$Î# ‰ Ê
2
t
"
x
œ ˆ 3x" " ‰ a3x# b œ x"
16. y œ tÈln t œ t(ln t)"Î# Ê
œ (ln t)"Î# " ‰
6. y œ ln kx Ê yw œ ˆ kx
(k) œ
œ ˆ x"$ ‰ a3x# b œ
dy
dx
15. y œ t(ln t)# Ê
"
x
œ
sec ) tan ) sec# )
sec ) tan )
œ
sec )(tan ) sec ))
tan ) sec )
"
)
sin (ln )) † ") ‘
œ sec )
1) x
3x 2
ln (x 1) Ê yw œ x" #" ˆ x " 1 ‰ œ 2(x
2x(x 1) œ 2x(x 1)
cln (1 x) ln (1 x)d Ê yw œ
"
#
1 " x ˆ 1 " x ‰ (1)‘ œ
"
#
1x1x
’ (1
x)(" x) “ œ
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
"
1 x#
397
398
Chapter 7 Transcendental Functions
29. y œ
1 ln t
1 ln t
Ê
dy
dt
(1ln t) ˆ "t ‰ (1 ln t) ˆ t" ‰
œ
(1 ln t)#
30. y œ Éln Èt œ ˆln t"Î# ‰
œ
"
#
ˆln t"Î# ‰"Î# †
"
t"Î#
Èsin ) cos )
1 2 ln )
œ
"
#
dy
d)
"
#
œ
œ
œ
dy
dt
"
sec (ln ))
"
#
t
t
ˆln t"Î# ‰"Î# †
œ
t
(1 ln t)#
2
t(1 ln t)#
ˆln t"Î# ‰ œ
"
#
ˆln t"Î# ‰"Î# †
sec (ln )) tan (ln ))
sec (ln ))
†
d
d)
d
dt
"
t"Î#
†
d
dt
ˆt"Î# ‰
†
d
d)
(sec (ln ))) œ
œ
dy
d)
"
#
(ln )) œ
)
ˆ cos
sin ) tan (ln ))
)
sin ) ‰
cos )
2
)
1 # ln )
4
)(1 2 ln )) “
&
&
t
"
4tÉln Èt
x 1b
33. y œ ln Š aÈ
‹ œ 5 ln ax# 1b 1x
34. y œ ln É (x(x2)1)#! œ
"
#
" ln t " ln t
(ln sin ) ln cos )) ln (1 2 ln )) Ê
’cot ) tan ) #
Ê
† #" t"Î# œ
31. y œ ln (sec (ln ))) Ê
32. y œ ln
"Î#
œ
"
#
ln (1 x) Ê yw œ
5†2x
x# 1
[5 ln (x 1) 20 ln (x 2)] Ê yw œ
"
#
#" ˆ 1 " x ‰ (1) œ
ˆ x 5 1 20 ‰
x#
œ
5
#
10x
x# 1
"
#(1 x)
4(x 1)
’ (x(x2)1)(x
2) “
2
œ 5# ’ (x 3x1)(x
#) “
35. y œ 'x#Î2 ln Èt dt Ê
x#
36. y œ 'Èx ln t dt Ê
3 x
È
œ
3 x
ln È
3
3È x2
dy
dx
dy
dx
œ Šln Èx# ‹ †
3
œ ˆln È
x‰ †
d
dx
d
dx
#
ax# b Šln É x# ‹ †
3
ˆÈ
x‰ ˆln Èx‰ †
d
dx
d
dx
#
Š x# ‹ œ 2x ln kxk x ln
kx k
È2
3
ˆÈx‰ œ ˆln È
x‰ ˆ 3" x#Î$ ‰ ˆln Èx‰ ˆ #" x"Î# ‰
ln Èx
2È x
37.
2
'cc32 x" dx œ cln kxkd #
$ œ ln 2 ln 3 œ ln 3
38.
'c01 3x3# dx œ cln k3x 2kd !" œ ln 2 ln 5 œ ln 52
39.
' y 2y25 dy œ ln ky# 25k C
40.
' 4r8r5 dr œ ln k4r# 5k C
41.
42.
#
#
'01 2sincost t dt œ cln k2 cos tkd !1 œ ln 3 ln 1 œ ln 3; or let u œ 2 cos t Ê du œ sin t dt with t œ 0
1
3
Ê u œ 1 and t œ 1 Ê u œ 3 Ê '0 2sincost t dt œ '1 u" du œ cln kukd $" œ ln 3 ln 1 œ ln 3
'01Î3 14 4sincos) ) d) œ cln k1 4 cos )kd !1Î$ œ ln k1 2k œ ln 3 œ ln "3 ; or let u œ 1 4 cos ) Ê du œ 4 sin ) d)
1Î3
c1
"
with ) œ 0 Ê u œ 3 and ) œ 13 Ê u œ 1 Ê '0 14 4sincos) ) d) œ 'c3 u" du œ cln kukd "
$ œ ln 3 œ ln 3
43. Let u œ ln x Ê du œ
'1
2
2 ln x
x
dx œ '0
ln 2
'2
dx
x ln x
œ 'ln 2
ln 4
"
u
dx; x œ 1 Ê u œ 0 and x œ 2 Ê u œ ln 2;
2u du œ cu# d 0 œ (ln 2)#
ln 2
44. Let u œ ln x Ê du œ
4
"
x
"
x
dx; x œ 2 Ê u œ ln 2 and x œ 4 Ê u œ ln 4;
#
4‰
ln 2
ˆ 2 ln 2 ‰ œ ln 2
du œ cln ud lnln 42 œ ln (ln 4) ln (ln 2) œ ln ˆ ln
ln 2 œ ln Š ln 2 ‹ œ ln ln 2
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 7.2 Natural Logarithms
45. Let u œ ln x Ê du œ
'2
4
dx
x(ln x)#
'2
dx; x œ 2 Ê u œ ln 2 and x œ 4 Ê u œ ln 4;
œ 'ln 2 u# du œ "u ‘ ln 2 œ ln"4 ln 4
ln 4
46. Let u œ ln x Ê du œ
16
"
x
dx
2xÈln x
œ
'ln 2
ln 16
"
#
"
x
"
ln #
œ ln"## "
ln 2
œ 2 ln" # "
ln #
"
# ln 2
œ
œ
"
ln 4
dx; x œ 2 Ê u œ ln 2 and x œ 16 Ê u œ ln 16;
u"Î# du œ u"Î# ‘ ln 2 œ Èln 16 Èln 2 œ È4 ln 2 Èln 2 œ 2Èln 2 Èln 2 œ Èln 2
ln 16
47. Let u œ 6 3 tan t Ê du œ 3 sec# t dt;
t
' 6 3sec
' duu œ ln kuk C œ ln k6 3 tan tk C
3 tan t dt œ
#
48. Let u œ 2 sec y Ê du œ sec y tan y dy;
' sec# ysectanyy dy œ ' duu œ ln kuk C œ ln k2 sec yk C
49. Let u œ cos
x
#
Ê du œ "# sin
sin
'01Î2 tan x# dx œ '01Î2 cos
x
#
dx Ê 2 du œ sin
1Î
50. Let u œ sin t Ê du œ cos t dt; t œ
'11ÎÎ42 cot t dt œ '11ÎÎ42
51. Let u œ sin
)
3
cos t
sin t
Ê du œ
'1Î2 2 cot 3) d) œ '1Î2
1
1
4
du
u
dx; x œ 0 Ê u œ 1 and x œ
È2
œ c2 ln kukd 11Î
Ê uœ
"
È2
and t œ
1
#
"
dt œ '1ÎÈ2 du
u œ cln kukd "ÎÈ# œ ln
1
"
3
1 2 cos
sin
È2
dx œ 2 '1
x
#
x
#
x
#
)
3
cos
)
3
)
3
d) Ê 6 du œ 2 cos
È3Î2
d) œ 6 '1Î2
du
u
)
3
"
È2
œ 2 ln
1
#
Ê uœ
œ 2 ln È2 œ ln 2
"
È2
d) ; ) œ
È3Î2
œ ln È2
1
#
"
#
Ê uœ
È3
#
and ) œ 1 Ê u œ
53.
'
œ'
dx
2Èx 2x
dx
;
2 È x ˆ1 È x ‰
du
u
È2
œ 2 cln kukd 11Î
let u œ 1 Èx Ê du œ
"
#È x
dx;
È3
#
;
ln "# ‹ œ 6 ln È3 œ ln 27
52. Let u œ cos 3x Ê du œ 3 sin 3x dx Ê 2 du œ 6 sin 3x dx; x œ 0 Ê u œ 1 and x œ
È2
;
Ê u œ 1;
œ 6 cln kukd 1Î2 œ 6 Šln
'01Î12 6 tan 3x dx œ '01Î12 6cossin3x3x dx œ 2 '11Î
"
È2
œ 2 ln
"
È2
1
1#
Ê uœ
"
È2
;
ln 1 œ 2 ln È2 œ ln 2
' 2Èx ˆdx1 Èx‰ œ '
du
u
œ ln kuk C
œ ln ¸1 Èx¸ C œ ln ˆ1 Èx‰ C
54. Let u œ sec x tan x Ê du œ asec x tan x sec# xb dx œ (sec x)(tan x sec x) dx Ê sec x dx œ
'
sec x dx
Èln (sec x tan x)
œ'
du
uÈln u
œ ' (ln u)"Î# †
55. y œ Èx(x 1) œ (x(x 1))"Î# Ê ln y œ
Ê yw œ ˆ "# ‰ Èx(x 1) ˆ x" " ‰
x 1
œ
"
#
56. y œ Èax# 1b (x 1)# Ê ln y œ
Ê yw œ Èax# +1b (x 1)# ˆ x# x 1 57. y œ É t t 1 œ ˆ t t 1 ‰
Ê
dy
dt
œ
"
#
"Î#
É t t 1 ˆ "t Ê ln y œ
" ‰
t1
œ
"
#
"
#
"
#
"
u
du œ 2(ln u)"Î# C œ 2Èln (sec x tan x) C
ln (x(x 1)) Ê 2 ln y œ ln (x) ln (x 1) Ê
Èx(x 1) (2x 1)
2x(x 1)
œ
w
y
y
"
#
œ
#
ˆ x#2x 1 #
œ Èax# 1b (x 1)# ’ axx#x1b (xx 1)1 “ œ
[ln t ln (t 1)] Ê
É t t 1 ’ t(t " 1) “ œ
2y
y
w
2x "
2Èx(x 1)
cln ax# 1b 2 ln (x 1)d Ê
" ‰
x1
du
u ;
" dy
y dt
œ
"
#
ˆ "t 2 ‰
x1
a2x# x 1b kx 1k
Èx# 1 (x 1)
" ‰
t1
"
2Èt (t 1)$Î#
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
œ
"
x
"
x 1
399
400
Chapter 7 Transcendental Functions
"
#
58. y œ É t(t 1 1) œ [t(t 1)]"Î# Ê ln y œ
Ê
dy
dt
"
œ "# É t(t 1 1) ’ t(t2t 1)
“œ
dy
d)
"
#
ln () 3) ln (sin )) Ê
dy
d)
#
)
œ (tan )) È2) 1 Š sec
tan ) "
#) 1 ‹
"
#
œ asec# )b È2) 1 62. y œ
Ê
dy
dt
œ t(t 1)(t+2) ˆ "t "
t(t 1)(t 2)
dy
dt
" dy
y d)
œ
"
#() 3)
cos )
sin )
œ
"
t1
" ‰
t#
"t "
t1
œ
sec# )
tan )
ˆ #" ‰ ˆ #) 2 1 ‰
œ
"
t
"
t1
"
t#
t(t 2) t(t 1)
œ t(t 1)(t 2) ’ (t 1)(t t(t2)1)(t
“ œ 3t# 6t 2
2)
" dy
y dt
Ê ln y œ ln 1 ln t ln (t 1) ln (t 2) Ê
"
t(t 1)(t 2)
" dy
y d)
ln (2) 1) Ê
tan )
È 2) 1
" dy
y dt
61. y œ t(t 1)(t 2) Ê ln y œ ln t ln (t 1) ln (t 2) Ê
Ê
" ‰
t1
œ È) 3 (sin )) ’ 2() " 3) cot )“
60. y œ (tan )) È2) 1 œ (tan ))(2) 1)"Î# Ê ln y œ ln (tan )) Ê
œ #" ˆ "t 2t 1
2 at# tb$Î#
59. y œ È) 3 (sin )) œ () 3)"Î# sin ) Ê ln y œ
Ê
" dy
y dt
[ln t ln (t 1)] Ê
" ‘
t#
œ
"
t(t 1)(t #)
œ "t "
t1
"
t#
t(t 2) t(t 1)
’ (t 1)(t t(t2)1)(t
“
2)
#
œ at$3t3t#6t2t2b#
63. y œ
)5
) cos )
64. y œ
) sin )
Èsec ) Ê ln y
dy
) sin ) ˆ "
d) œ Èsec ) )
Ê
65. y œ
Ê
Ê ln y œ ln () 5) ln ) ln (cos )) Ê
œ ln ) ln (sin )) cot ) "
#
"!
"
#
"!
3 x(x 2) "
ˆ
Ê yw œ 3" É
x# 1 x "
3
"
)5
" dy
y d)
"
)
œ ’ ") 1b 2
3
ln (x 1) Ê
[10 ln (x 1) 5 ln (2x 1)] Ê
(x 1) ˆ 5
Ê yw œ É (2x
x1 1)&
3 x(x 2)
67. y œ É
x# 1 Ê ln y œ
ln (sec )) Ê
œ
sin )
cos )
cos )
sin )
Ê
dy
d)
5 ‰ ˆ "
œ ˆ ) )cos
)
)5 "
)
tan )‰
(sec ))(tan ))
2 sec ) “
tan )‰
xÈ x# 1
Ê ln y œ ln x "# ln ax# (x 1)#Î$
È #
yw œ x(x x1)#Î$1 ’ "x x# x 1 3(x 2 1) “
(x 1)
66. y œ É (2x
1)& Ê ln y œ
"
#
" dy
y d)
yw
y
yw
y
œ
œ
"
x
5
x1
x
x# 1
2
3(x 1)
5
2x 1
5 ‰
2x 1
cln x ln (x 2) ln ax# 1bd Ê
"
x#
x(x 1)(x 2)
3
68. y œ É
ax# 1b (2x 3) Ê ln y œ
"
3
x(x 1)(x 2) ˆ "
3
Ê yw œ "3 É
ax# 1b (2x 3) x yw
y
œ
"
3
ˆ x" "
x#
2x ‰
x# 1
2x ‰
x# 1
cln x ln (x 1) ln (x 2) ln ax# 1b ln (2x 3)d
"
x1
"
x#
2x
x# 1
2 ‰
2x 3
sin x
1
w
w
69. (a) f(x) œ ln (cos x) Ê f w (x) œ cos
x œ tan x œ 0 Ê x œ 0; f (x) 0 for 4 Ÿ x 0 and f (x) 0 for
0 x Ÿ 13 Ê there is a relative maximum at x œ 0 with f(0) œ ln (cos 0) œ ln 1 œ 0; f ˆ 14 ‰ œ ln ˆcos ˆ 14 ‰‰
œ ln Š È"2 ‹ œ #" ln 2 and f ˆ 13 ‰ œ ln ˆcos ˆ 13 ‰‰ œ ln
xœ
1
3
with
f ˆ 13 ‰
"
#
œ ln 2. Therefore, the absolute minimum occurs at
œ ln 2 and the absolute maximum occurs at x œ 0 with f(0) œ 0.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 7.2 Natural Logarithms
sin (ln x)
x
(b) f(x) œ cos (ln x) Ê f w (x) œ
401
"
#
œ 0 Ê x œ 1; f w (x) 0 for
Ÿ x 1 and f w (x) 0 for 1 x Ÿ 2
Ê there is a relative maximum at x œ 1 with f(1) œ cos (ln 1) œ cos 0 œ 1; f ˆ "# ‰ œ cos ˆln ˆ "# ‰‰
œ cos ( ln 2) œ cos (ln 2) and f(2) œ cos (ln 2). Therefore, the absolute minimum occurs at x œ
x œ 2 with f ˆ "# ‰ œ f(2) œ cos (ln 2), and the absolute maximum occurs at x œ 1 with f(1) œ 1.
"
#
and
70. (a) f(x) œ x ln x Ê f w (x) œ 1 "x ; if x 1, then f w (x) 0 which means that f(x) is increasing
(b) f(1) œ 1 ln 1 œ 1 Ê f(x) œ x ln x 0, if x 1 by part (a) Ê x ln x if x 1
71.
'15 (ln 2x ln x) dx œ '15 ( ln x ln 2 ln x) dx œ (ln 2)'15 dx œ (ln 2)(5 1) œ ln 2% œ ln 16
72. A œ
'c01Î4 tan x dx '01Î3 tan x dx œ ' 01Î4 cossinxx dx '01Î3 cossinxx dx œ cln kcos xkd !1Î% cln kcos xkd !1Î$
"
È2 ‹
œ Šln 1 ln
ˆln
"
#
ln 1‰ œ ln È2 ln 2 œ
73. V œ 1'0 Š Èy2 1 ‹ dy œ 41 '0
#
3
3
74. V œ 1 '1Î6 cot x dx œ 1 '1Î6
1Î2
1Î2
"
y 1
cos x
sin x
3
#
ln 2
dy œ 41 cln ky 1kd $! œ 41(ln 4 ln 1) œ 41 ln 4
1Î#
dx œ 1 cln (sin x)d 1Î' œ 1 ˆln 1 ln "# ‰ œ 1 ln 2
75. V œ 21'1Î2 x ˆ x"# ‰ dx œ 21 '1Î2 x" dx œ 21 cln kxkd #"Î# œ 21 ˆln 2 ln #" ‰ œ 21(2 ln 2) œ 1 ln 2% œ 1 ln 16
2
2
76. V œ 1 '0 Š Èx9x
‹ dx œ 271'0 dx œ 271 cln ax$ 9bd ! œ 271(ln 36 ln 9) œ 271(ln 4 ln 9 ln 9)
$9
#
3
3
$
œ 271 ln 4 œ 541 ln 2
77. (a) y œ
x#
8
œ '4
#
ln x Ê 1 ayw b# œ 1 ˆ 4x x" ‰ œ 1 Š x 4x 4 ‹ œ Š x 4x 4 ‹ Ê L œ '4 É1 ayw b# dx
#
8 #
x 4
4x
#
#
#
8
dx œ '4 ˆ x4 "x ‰ dx œ ’ x8 ln kxk“ œ (8 ln 8) (2 ln 4) œ 6 ln 2
8
)
#
%
#
(b) x œ ˆ y4 ‰ 2 ln ˆ y4 ‰ Ê
dx
dy
œ
y
8
2
y
'
Ê L œ '4 Ê1 Š dx
dy ‹ dy œ 4
#
12
Ê 1
12 #
y 16
8y
Š dx
dy ‹
#
#
œ 1 Š y8 2y ‹ œ 1 Š y
#
y
dy œ '4 Š y8 2y ‹ dy œ ’ 16
2 ln y“
12
#
16
8y ‹
"#
%
#
œ Šy
#
16
8y ‹
#
œ (9 2 ln 12) (1 2 ln 4)
œ 8 2 ln 3 œ 8 ln 9
78. L œ '1 É1 2
"
x#
dx Ê
dy
dx
œ
"
x
Ê y œ ln kxk C œ ln x C since x 0 Ê 0 œ ln 1 C Ê C œ 0 Ê y œ ln x
" ‰ ˆ"‰
79. (a) My œ '1 x ˆ "x ‰ dx œ 1, Mx œ '1 ˆ 2x
x dx œ
2
Ê xœ
2
My
M
œ
"
ln 2
¸ 1.44 and y œ
Mx
M
œ
ˆ "4 ‰
ln 2
"
#
'12 x"
#
" ‘
dx œ 2x
œ 4" , M œ '1
"
#
2
"
x
dx œ cln kxkd #" œ ln 2
¸ 0.36
(b)
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
402
Chapter 7 Transcendental Functions
80. (a) My œ '1 x Š È"x ‹ dx œ '1 x"Î# dx œ
16
œ
16
'
cln kxkd "'
" œ ln 4, M œ 1
16
"
#
"
Èx
2
3
"
x$Î# ‘ "' œ 42; Mx œ ' Š È
‹ Š È"x ‹ dx œ
"
2 x
1
16
"'
dx œ 2x"Î# ‘ " œ 6 Ê x œ
My
M
œ 7 and y œ
Mx
M
"
#
œ
'116 x" dx
ln 4
6
"
(b) My œ '1 x Š È"x ‹ Š È4x ‹ dx œ 4'1 dx œ 60, Mx œ '1 Š 2È
‹ Š È"x ‹ Š È4x ‹ dx œ #'1 x$Î# dx
x
16
16
16
œ 4 x"Î# ‘" œ 3, M œ '1 Š È"x ‹ Š È4x ‹ dx œ 4'1
"'
yœ
Mx
M
œ
16
16
"
x
16
dx œ c4 ln kxkd "'
" œ 4 ln 16 Ê x œ
My
M
œ
15
ln 16
and
3
4 ln 16
81. faxb œ lnax3 1b, domain of f: a1, _b Ê f w axb œ
3x2
x3 1 ;
f w axb œ 0 Ê 3x2 œ 0 Ê x œ 0, not in the domain;
f w axb œ undefined Ê x3 1 œ 0 Ê x œ 1, not in domain. On a1, _b, f w axb 0 Ê f is increasing on a1, _b
Ê f is one-to-one
82. gaxb œ Èx2 ln x, domain of g: x 0.652919 Ê g w axb œ
2x 1x
2Èx2 ln x
œ
2x2 1
;
2xÈx2 ln x
g w axb œ 0 Ê 2x2 1 œ 0 Ê no real
solutions; g w axb œ undefined Ê 2xÈx2 ln x œ 0 Ê x œ 0 or x ¸ 0.652919, neither in domain. On x 0.652919,
g w axb 0 Ê g is increasing for x 0.652919 Ê g is one-to-one
83.
dy
dx
84.
d# y
dx#
œ1
"
x
at ("ß 3) Ê y œ x ln kxk C; y œ 3 at x œ 1 Ê C œ 2 Ê y œ x ln kxk 2
œ sec# x Ê
dy
dx
œ tan x C and 1 œ tan 0 C Ê
dy
dx
œ tan x 1 Ê y œ ' (tan x 1) dx
œ ln ksec xk x C" and 0 œ ln ksec 0k 0 C" Ê C" œ 0 Ê y œ ln ksec xk x
85. (a) L(x) œ f(0) f w (0) † x, and f(x) œ ln (1 x) Ê f w (x)k xœ0 œ
ww
(b) Let faxb œ lnax "b. Since f axb œ
"
ax"
b#
" ¸
1x xœ0
œ 1 Ê L(x) œ ln 1 1 † x Ê L(x) œ x
! on Ò!ß !Þ"Ó, the graph of f is concave down on this interval and the
largest error in the linear approximation will occur when x œ !Þ". This error is !Þ" lna"Þ"b ¸ !Þ!!%'* to five
decimal places.
(c) The approximation y œ x for ln (1 x) is best for smaller
positive values of x; in particular for 0 Ÿ x Ÿ 0.1 in the
graph. As x increases, so does the error x ln (1 x).
From the graph an upper bound for the error is
0.5 ln (1 0.5) ¸ 0.095; i.e., kE(x)k Ÿ 0.095 for
0 Ÿ x Ÿ 0.5. Note from the graph that 0.1 ln (1 0.1)
¸ 0.00469 estimates the error in replacing ln (1 x) by
x over 0 Ÿ x Ÿ 0.1. This is consistent with the estimate
given in part (b) above.
86. For all positive values of x,
d ln a
x
dx c
dœ
1
a
x
† xa2 œ 1x and
d
ln a
dx c
ln x d œ 0 1
x
œ 1x . Since ln xa and ln a ln x have
the same derivative, then ln xa œ ln a ln x C for some constant C. Since this equation holds for all positve values of x,
it must be true for x œ 1 Ê ln 1x œ ln 1 ln x C œ 0 ln x C Ê ln 1x œ ln x C. By part 3 we know that
ln 1x œ ln x Ê C œ 0 Ê ln xa œ ln a ln x.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 7.3 Exponential Functions 403
(b) yw œ
87. (a)
cos x
asin x .
Since lsin xl and lcos xl are less than
or equal to 1, we have for a "
"
"
w
a" Ÿ y Ÿ a" for all x.
Thus, lim yw œ ! for all x Ê the graph of y looks
aÄ_
more and more horizontal as a Ä _.
88. (a) The graph of y œ Èx ln x appears to be concave
upward for all x 0.
(b) y œ Èx ln x Ê yw œ
"
#È x
"
x
Ê yww œ 4x"$Î# "
x#
œ
"
x#
Š
Èx
4
1‹ œ 0 Ê Èx œ 4 Ê x œ 16.
Thus, yww 0 if 0 x 16 and yww 0 if x 16 so a point of inflection exists at x œ 16. The graph of
y œ Èx ln x closely resembles a straight line for x 10 and it is impossible to discuss the point of
inflection visually from the graph.
7.3 EXPONENTIAL FUNCTIONS
1. (a) e0Þ3t œ 27 Ê ln e0Þ3t œ ln 3$ Ê (0.3t) ln e œ 3 ln 3 Ê 0.3t œ 3 ln 3 Ê t œ 10 ln 3
(b) ekt œ "# Ê ln ekt œ ln 2" œ kt ln e œ ln 2 Ê t œ lnk2
t
(c) eÐln 0Þ2Ñt œ 0.4 Ê ˆeln 0Þ2 ‰ œ 0.4 Ê 0.2t œ 0.4 Ê ln 0.2t œ ln 0.4 Ê t ln 0.2 œ ln 0.4 Ê t œ
ln 0.4
ln 0.2
2. (a) e0Þ01t œ 1000 Ê ln e0Þ01t œ ln 1000 Ê (0.01t) ln e œ ln 1000 Ê 0.01t œ ln 1000 Ê t œ 100 ln 1000
"
(b) ekt œ 10
Ê ln ekt œ ln 10" œ kt ln e œ ln 10 Ê kt œ ln 10 Ê t œ lnk10
(c) eÐln 2Ñt œ
Èt
3. e
"
#
t
Ê ˆeln 2 ‰ œ 2" Ê 2t œ 2" Ê t œ 1
Èt
œ x# Ê ln e
#
œ ln x# Ê Èt œ 2 ln x Ê t œ 4(ln x)#
#
#
4. ex e2x1 œ et Ê ex 2x1 œ et Ê ln ex 2x1 œ ln et Ê t œ x# 2x 1
5. y œ e5x Ê yw œ e5x
d
dx
6. y œ e2xÎ3 Ê yw œ e2xÎ3
7. y œ e57x Ê yw œ e57x
#
(5x) Ê yw œ 5e5x
d
dx
d
dx
ˆ 2x
‰ Ê yw œ
3
2
3
e2xÎ3
(5 7x) Ê yw œ 7e57x
#
8. y œ eˆ4Èxx ‰ Ê yw œ eˆ4Èxx ‰
d
dx
ˆ4Èx x# ‰ Ê yw œ Š È2 2x‹ eˆ4Èxx# ‰
x
9. y œ xex ex Ê yw œ aex xex b ex œ xex
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
404
Chapter 7 Transcendental Functions
10. y œ (1 2x) e2x Ê yw œ 2e2x (1 2x)e2x
d
dx
(2x) Ê yw œ 2e2x 2(1 2x) e2x œ 4xe2x
11. y œ ax# 2x 2b ex Ê yw œ (2x 2)ex ax# 2x 2b ex œ x# ex
12. y œ a9x# 6x 2b e3x Ê yw œ (18x 6)e3x a9x# 6x 2b e3x
(3x) Ê yw œ (18x 6)e3x 3 a9x# 6x 2b e3x
d
dx
œ 27x# e3x
13. y œ e) (sin ) cos )) Ê yw œ e) (sin ) cos )) e) (cos ) sin )) œ 2e) cos )
14. y œ ln ˆ3)e) ‰ œ ln 3 ln ) ln e) œ ln 3 ln ) ) Ê
#
15. y œ cos Še) ‹ Ê
dy
d)
16. y œ )$ e#) cos 5) Ê
#
œ sin Še) ‹
dy
d)
d
d)
dy
d)
#
"
)
œ
#
#
Še) ‹ œ Š sin Še) ‹‹ Še) ‹
œ a3)# b ˆe#) cos 5)‰ a)$ cos 5)b e#)
œ )# e#) (3 cos 5) 2) cos 5) 5) sin 5))
17. y œ ln a3tet b œ ln 3 ln t ln et œ ln 3 ln t t Ê
dy
dt
œ
"
t
d
d)
#
#
a)# b œ 2)e) sin Še) ‹
1t
t
dy
dt
œ 1 ˆ sin" t ‰
d
dt
(sin t) œ 1 cos t
sin t
cos t sin t
sin t
19. y œ ln
e)
1 e)
20. y œ ln
È)
1 È)
œ ln e) ln ˆ1 e) ‰ œ ) ln ˆ1+e) ‰ Ê
œ ln È) ln Š1 È)‹ Ê
"
"
œ Š È" ‹ Š È
‹ Š 1 "È) ‹ Š #È
‹œ
)
# )
)
21. y œ eÐcos tln tÑ œ ecos t eln t œ tecos t Ê
22. y œ esin t aln t# 1b Ê
23.
d
d)
(2)) 5(sin 5)) ˆ)$ e#) ‰
1œ
18. y œ ln a2et sin tb œ ln 2 ln et ln sin t œ ln 2 t ln sin t Ê
œ
1
'0ln x sin et dt
dy
dt
œ Š È" ‹
)
Š1 È)‹ È)
dy
dt
2) Š1 È)‹
œ
œ 1 ˆ 1 " e) ‰
d
d)
d
dt
ˆ 1 e) ‰ œ 1 ŠÈ)‹ Š 1 "È) ‹
"
#) Š1 È)‹
œ ecos t tecos t
d
d)
œ
d
d)
e)
1 e)
œ
d
dx
24. y œ 'e4Èx ln t dt Ê yw œ aln e2x b †
Š1 È ) ‹
"
#) a1)"Î# b
(cos t) œ (1 t sin t) ecos t
(ln x) œ
d
dx
sin x
x
ae2x b Šln e4Èx ‹ †
d
dx
Še4Èx ‹ œ (2x) a2e2x b ˆ4Èx‰ Še4Èx ‹ †
œ 4xe2x 4Èx e4Èx Š È2x ‹ œ 4xe2x 8e4Èx
25. ln y œ ey sin x Ê Š y" ‹ yw œ ayw ey b (sin x) ey cos x Ê yw Š y" ey sin x‹ œ ey cos x
Ê yw Š 1 yey sin x ‹ œ ey cos x Ê yw œ
y
26. ln xy œ exy Ê ln x ln y œ exy Ê
Ê yw Š 1 ye
y
xby
‹œ
xex b y "
x
"
1 e)
œ esin t (cos t) aln t# 1b 2t esin t œ esin t aln t# 1b (cos t) 2t ‘
Ê yw œ ˆsin eln x ‰ †
e2x
dy
d)
dy
d)
Ê yw œ
yey cos x
1 yey sin x
"
x
Š y" ‹ yw œ a1 yw b exy Ê yw Š y" exy ‹ œ exy y axex b y "b
x a1 yex b y b
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
"
x
d
dx
ˆ4Èx‰
Section 7.3 Exponential Functions 405
27. e2x œ sin (x 3y) Ê 2e2x œ a1 3yw b cos (x 3y) Ê 1 3yw œ
28. tan y œ ex ln x Ê asec# yb yw œ ex 29.
' ae3x 5ex b dx œ e3
31.
"
x
Ê yw œ
2e2x
cos (x 3y)
Ê 3yw œ
2e2x
cos (x 3y)
1 Ê yw œ
2e2x cos (x 3y)
3 cos (x 3y)
axex "b cos# y
x
30.
' a2ex 3e2x b dx œ 2ex #3 e2x C
'lnln23 ex dx œ cex d lnln 32 œ eln 3 eln 2 œ 3 2 œ 1
32.
'0ln 2 ex dx œ cex d 0 ln 2 œ e! eln 2 œ 1 2 œ 1
33.
' 8eÐx1Ñ dx œ 8eÐx1Ñ C
34.
' 2eÐ2x1Ñ dx œ eÐ2x1Ñ C
35.
'lnln49 exÎ2 dx œ 2exÎ2 ‘ lnln 94 œ 2 eÐln 9ÑÎ2 eÐln 4)Î2 ‘ œ 2 ˆeln 3 eln 2 ‰ œ 2(3 2) œ 2
36.
'0ln 16 exÎ4 dx œ 4exÎ4 ‘ ln0 16 œ 4 ˆeÐln 16ÑÎ4 e0 ‰ œ 4 ˆeln 2 1‰ œ 4(2 1) œ 4
3x
5ex C
37. Let u œ r"Î# Ê du œ "# r"Î# dr Ê 2 du œ r"Î# dr;
' eÈÈrr dr œ ' er"Î# † r"Î# dr œ 2 ' eu du œ 2eu C œ 2er"Î# C œ 2eÈr C
38. Let u œ r"Î# Ê du œ "# r"Î# dr Ê 2 du œ r"Î# dr;
' eÈcÈrr dr œ ' er"Î# † r"Î# dr œ 2 ' eu du œ 2er"Î# C œ 2eÈr C
39. Let u œ t# Ê du œ 2t dt Ê du œ 2t dt;
' 2tet
#
dt œ ' eu du œ eu C œ et C
#
"
4
40. Let u œ t% Ê du œ 4t$ dt Ê
'
%
t$ et dt œ
41. Let u œ
' ex#
1Îx
"
x
"
4
du œ t$ dt;
' eu du œ 4" et% C
Ê du œ x"# dx Ê du œ
dx;
dx œ ' eu du œ eu C œ e1Îx C
42. Let u œ x# Ê du œ 2x$ dx Ê
' e x$Î #
1 x
"
x#
dx œ ' ex † x$ dx œ
#
"
#
"
#
du œ x$ dx;
' eu du œ "# eu C œ "# ex # C œ "# e1Îx# C
43. Let u œ tan ) Ê du œ sec# ) d); ) œ 0 Ê u œ 0, ) œ
'0
1 Î4
ˆ1 etan ) ‰ sec# ) d) œ '
1Î4
0
1
4
Ê u œ 1;
sec# ) d) '0 eu du œ ctan )d 0
1
1 Î4
ceu d "! œ tan ˆ 14 ‰ tan (0)‘ ae" e! b
œ (1 0) (e 1) œ e
44. Let u œ cot ) Ê du œ csc# ) d); ) œ
'1Î4 ˆ1 ecot ) ‰ csc# ) d) œ '1Î4
1 Î2
1Î2
1
4
Ê u œ 1, ) œ
1
2
Ê u œ 0;
csc# ) d) '1 eu du œ c cot )d 1Î4 ceu d !" œ cot ˆ 12 ‰ cot ˆ 14 ‰‘ ae! e" b
0
1 Î2
œ (0 1) (1 e) œ e
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
406
Chapter 7 Transcendental Functions
45. Let u œ sec 1t Ê du œ 1 sec 1t tan 1t dt Ê
' esec Ð1tÑ sec (1t) tan (1t) dt œ 1" ' eu du œ e1
u
œ sec 1t tan 1t dt;
du
1
Cœ
esec a1tb
1
C
46. Let u œ csc (1 t) Ê du œ csc (1 t) cot (1 t) dt;
' ecsc Ð1tÑ csc (1 t) cot (1 t) dt œ ' eu du œ eu C œ ecsc Ð1tÑ C
47. Let u œ ev Ê du œ ev dv Ê 2 du œ 2ev dv; v œ ln
1
6
Ê u œ 16 , v œ ln
1
#
Ê u œ 1# ;
'lnlnÐÐ11ÎÎ62ÑÑ 2ev cos ev dv œ 2 '11ÎÎ62 cos u du œ c2 sin ud 11ÎÎ26 œ 2 sin ˆ 1# ‰ sin ˆ 16 ‰‘ œ 2 ˆ1 "# ‰ œ 1
#
#
48. Let u œ ex Ê du œ 2xex dx; x œ 0 Ê u œ 1, x œ Èln 1 Ê u œ eln 1 œ 1;
Èln 1
'0
2xex cos Šex ‹ dx œ '1 cos u du œ csin ud 1" œ sin (1) sin (1) œ sin (1) ¸ 0.84147
#
1
#
49. Let u œ 1 er Ê du œ er dr;
' 1 e e
r
dr œ '
x
dx œ '
r
50.
' 1 " e
x
let u œ e
"
u
du œ ln kuk C œ ln a1 er b C
ecx
ecx 1
dx;
1 Ê du œ ex dx Ê du œ ex dx;
' ecec 1 dx œ ' "u du œ ln kuk C œ ln aex 1b C
x
x
51.
dy
dt
œ et sin aet 2b Ê y œ ' et sin aet 2b dt;
dy
dt
œ et sec# a1et b Ê y œ ' et sec# a1et b dt;
let u œ et 2 Ê du œ et dt Ê y œ ' sin u du œ cos u C œ cos aet 2b C; y(ln 2) œ 0
Ê cos ˆeln 2 2‰ C œ 0 Ê cos (2 2) C œ 0 Ê C œ cos 0 œ 1; thus, y œ 1 cos aet 2b
52.
let u œ 1et Ê du œ 1et dt Ê 1" du œ et dt Ê y œ 1" ' sec# u du œ 1" tan u C
œ 1" tan a1et b C; y(ln 4) œ 12 Ê 1" tan ˆ1eln 4 ‰ C œ 12 Ê 1" tan ˆ1 † 4" ‰ C œ 12
Ê 1" (1) C œ
53.
d# y
dx#
œ 2ex Ê
dy
dx
2
1
Ê C œ 13 ; thus, y œ
œ 2ex C; x œ 0 and
3
1
dy
dx
"
1
tan a1et b
œ 0 Ê 0 œ 2e! C Ê C œ 2; thus
dy
dx
x
Ê y œ 2ex 2x C" ; x œ 0 and y œ 1 Ê 1 œ 2e! C" Ê C" œ 1 Ê y œ 2e
54.
d# y
dy
" 2t
" #
2t
Ê dy
dt# œ 1 e
dt œ t # e C; t œ 1 and dt œ 0 Ê 0 œ 1 # e C Ê
dy
" 2t
" #
" #
" 2t
ˆ" #
‰
dt œ t # e # e 1 Ê y œ # t 4 e # e 1 t C" ; t œ 1 and y œ
"
" #
" #
" 2t
"
#
Ê C" œ # 4 e Ê y œ # t 4 e ˆ # e 1‰ t ˆ #" 4" e# ‰
#
58. y œ 2s Ê
dy
ds
dy
ds
2x 1 œ 2 aex xb 1
"
#
e# 1; thus
1 Ê " œ
œ 5Ès (ln 5) ˆ "# s"Î# ‰ œ Š 2lnÈ5s ‹ 5Ès
#
"
#
4" e# #" e# 1 C"
56. y œ 3cx Ê y w œ 3cx (ln 3)(1) œ 3cx ln 3
55. y œ 2x Ê y w œ 2x ln 2
57. y œ 5Ès Ê
Cœ
œ 2ex 2
#
œ 2s (ln 2)2s œ aln 2# b Šs2s ‹ œ (ln 4)s2s
#
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 7.3 Exponential Functions 407
59. y œ x1 Ê y w œ 1xÐ1 1Ñ
61. y œ (cos ))
È2 Ê
62. y œ (ln ))1 Ê
60. y œ t1
œ (1 e) t
dy
dt
œ 1(ln ))Ð1 1Ñ ˆ ") ‰ œ
1(ln ))Ð1 1Ñ
)
63. y œ 7sec ) ln 7 Ê
dy
d)
œ a7sec ) ln 7b(ln 7)(sec ) tan )) œ 7sec ) (ln 7)# (sec ) tan ))
64. y œ 3tan ) ln 3 Ê
dy
d)
œ a3tan ) ln 3b(ln 3) sec# ) œ 3tan ) (ln 3)# sec# )
65. y œ 2sin 3t Ê
dy
dt
œ a2sin 3t ln 2b(cos 3t)(3) œ (3 cos 3t) a2sin 3t b (ln 2)
66. y œ 5c cos 2t Ê
dy
dt
67. y œ log2 5) œ
ln 5)
ln #
œ a5c cos 2t ln 5b(sin 2t)(2) œ (2 sin 2t) a5c cos 2t b (ln 5)
Ê
ln x
ln 4
70. y œ
x ln e
ln #5
ln x#
ln 4
œ
ln x
2 ln 5
œ
ln (1 ) ln 3)
ln 3
2
ln x
ln 4
œ ˆ ln"# ‰ ˆ 5") ‰ (5) œ
dy
d)
68. y œ log3 (1 ) ln 3) œ
69. y œ
x
# ln 5
ln x
ln 4
Ê
œ3
ln x
2 ln 5
x ‰
71. y œ x3 log10 x œ x3 ˆ lnln10
œ
1‰
73. y œ log3 Šˆ xx ‹œ
1
ln 3
dy
dx
œ
"
x1
"
x1
74. y œ log5 Ɉ 3x7x 2 ‰
œ
"
#
ln 7x "
#
ln 5
œ
ln 3
ln 3
cos )
(sin ))(ln 7)
77. y œ log10 ex œ
78. y œ
) †5 )
2 log5 )
œ
ln ex
ln 10
ln# r
(ln 3)(ln 9)
dy
dr
Ê
1
(ln 3) ln Š xx b
c1‹
ln 3
†
1
x
3x2 ln x‰ œ
1
2
ln 10 x
"
ˆ"‰
œ ’ (ln 3)(ln
9) “ (2 ln r) r œ
Ðln 5ÑÎ2
œ log5 ˆ 3x7x 2 ‰
œ
dy
dx
œ
7
2†7x
dy
d)
ln ˆ 3x7x 2 ‰
Ðln 5ÑÎ2
ln 5
3
2†(3x 2)
œ
œ ˆ ln#5 ‰ ”
(3x 2) 3x
2x(3x 2)
œ
ln ˆ 3x7x 2 ‰
ln 5
1
2
ln 10 x
3x2 log10 x
2 ln r
r(ln 3)(ln 9)
x
ln 10
Ê yw œ
"
#
ln ˆ 3x7x # ‰
ln ) ‰
ˆ ln ) ‰‘ ˆ ) ln" 7 ‰ œ sin (log7 )) œ sin ˆ ln
7 ) cos ln 7
œ
Ê yw œ
•œ
"
x(3x 2)
) †5 )
)
2 ln
ln 5
x
3x2 lnln10
œ
1‰
œ ln ˆ xx 1 œ ln (x 1) ln (x 1)
ln (sin )) ln (cos )) ln e) ln 2)
)) ) ) ln 2
œ ln (sin )) ln (cos
ln 7
ln 7
sin )
"
ln 2
ˆ " ‰
(cos ))(ln 7) ln 7 ln 7 œ ln 7 (cot ) tan ) 1 ln
)
œ
1 ˆ 3
ln 10 x
x1
2x ln 5
2
(x 1)(x 1)
ln (3x 2) Ê
)‰
76. y œ log7 ˆ sin e) #cos
œ
)
dy
d)
"
1 ) ln 3
3
x ln 4
ln x Ê y w œ
œ
ln ) ‰
75. y œ ) sin (log7 )) œ ) sin ˆ ln
Ê
7
Ê
œ ˆ ln"3 ‰ ˆ 1 )" ln 3 ‰ (ln 3) œ
œ ˆ # ln" 5 ‰ (x ln x) Ê y w œ ˆ # ln" 5 ‰ ˆ1 "x ‰ œ
1
3
ln 10 x
1
ln ˆ xx b
c1‰
dy
d)
"
) ln #
Ê yw œ
ln x
ln 4
72. y œ log3 r † log9 r œ ˆ lnln 3r ‰ ˆ lnln 9r ‰ œ
Ê
e
È
œ È2 (cos ))Š 2c1‹ (sin ))
dy
d)
dy
d)
Ê
e
"
ln 7
cos (log7 ))
2)
"
ln 10
) ‰ˆ
)
)
ˆ2 ln
‰ ˆ ) ‰ˆ ) ln1 5 ‰
ln 5 )†5 ln 5 5 a1b )†5
) ‰2
ˆ2 ln
ln 5
œ
5) ln 5a2 log5 )ba) ln 5 1b 5)
ln 5a2 log5 )b2
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
408
Chapter 7 Transcendental Functions
79. y œ 3log2 t œ 3Ðln tÑÎÐln 2Ñ Ê
80. y œ 3 log8 (log2 t) œ
83.
t ln Šˆeln 3 ‰
sin t
‹
ln 3
3 ln (log2 t)
ln 8
ln 8 ln ˆtln 2 ‰
ln #
81. y œ log2 a8tln 2 b œ
82. y œ
œ c3Ðln tÑÎÐln 2Ñ (ln 3)d ˆ t ln" 2 ‰ œ
dy
dt
t ln ˆ3sin t ‰
ln 3
œ
Ê
dy
dt
3 ln 2 (ln 2)(ln t)
ln #
œ
œ
3 ln ˆ lnln 2t ‰
ln 8
œ
t(sin t)(ln 3)
ln 3
"
t
alog2 3b 3log2 t
"
ˆ " ‰
œ ˆ ln38 ‰ ’ (ln t)/(ln
2) “ t ln # œ
œ 3 ln t Ê
œ t sin t Ê
dy
dt
œ
dy
dt
3
t(ln t)(ln 8)
œ
"
t(ln t)(ln #)
"
t
œ sin t t cos t
' 5x dx œ ln5 5 C
x
84. Let u œ 3 3x Ê du œ 3x ln 3 dx Ê ln13 du œ 3x dx;
' 3 3 3 dx œ ln13 ' u1 du œ ln13 lnlul C œ lnl3ln33 l C
x
x
x
85.
'01 2c
)
'c2
0
86.
5c) d) œ
'
0
ˆ " ‰)
2 5
'1
d) œ –
88. Let u œ x"Î# Ê du œ
"Î#
"
#
"
#
ln Š "# ‹
"
ln Š "# ‹
!
ln Š 5" ‹ —
#
x2ax b dx œ '1 ˆ "# ‰ 2u du œ
x
!
)
2
'14 È2Èx dx œ '14 2x
— œ
Š "5 ‹
87. Let u œ x# Ê du œ 2x dx Ê
È2
"
)
"
#‹
ln Š "# ‹
1
Š
)
d) œ '0 ˆ "# ‰ d) œ –
œ
"
#
œ
ln Š "# ‹
"
2(ln 1 ln 2)
c#
"
ln Š 5" ‹
œ
#
Š 5" ‹
ln Š 5" ‹
œ
"
ln Š "5 ‹
(1 25) œ
'0
24
ln 1 ln 5
"
#
du œ x dx; x œ 1 Ê u œ 1, x œ È2 Ê u œ 2;
"
#
ln2 # ‘ # œ ˆ 2 ln" 2 ‰ a2# 2" b œ
"
x"Î# dx Ê 2 du œ
dx
Èx
Ðu 1Ñ #
"
7cos t sin t dt œ '1 7u du œ 0
7 ‘!
ln 7 "
u
œ ˆ ln"7 ‰ a7! 7b œ
90. Let u œ tan t Ê du œ sec# t dt; t œ 0 Ê u œ 0, t œ
'01Î4 ˆ 3" ‰tan t sec# t dt œ '01 ˆ 3" ‰u du œ –
91. Let u œ x2x Ê ln u œ 2x ln x Ê
" du
u dx
"
u
Š "3 ‹
ln Š 3" ‹
24
ln 5
; x œ 1 Ê u œ 1, x œ 4 Ê u œ 2;
† x"Î# dx œ 2'1 2u du œ ’ 2ln # “ œ ˆ ln"# ‰ a2$ 2# b œ
2
œ
"
ln #
u
89. Let u œ cos t Ê du œ sin t dt Ê du œ sin t dt; t œ 0 Ê u œ 1, t œ
1Î2
"
# ln 2
œ
1
4
1
#
4
ln #
Ê u œ 0;
6
ln 7
Ê u œ 1;
!
"
"
"
"
— œ ˆ ln 3 ‰ ’ˆ 3 ‰ ˆ 3 ‰ “ œ
2
3 ln 3
!
œ 2 ln x (2x) ˆ x" ‰ Ê
du
dx
œ 2u(ln x 1) Ê
x œ 2 Ê u œ 2% œ 16, x œ 4 Ê u œ 4) œ 65,536;
"
#
du œ x2x (1 ln x) dx;
'24 x2x (1 ln x) dx œ "# '1665 536 du œ "# cud 6516 536 œ "# (65,536 16) œ 65,520
œ 32,760
#
ß
2
2
92. Let u œ 1 2x Ê du œ 2x a2xbln 2 dx Ê
'
93.
'
2
x 2x
dx
1 2x2
3x
œ
È3 dx œ
ß
1
2 ln 2 du
2
œ 2x x dx
' 1u du œ 2 ln1 2 lnlul C œ lnŠ12 ln 22
x2
1
2 ln 2
3xŠ 3b1‹
È 3 1
È
C
‹
C
94.
'
È
xŠ 2c1‹ dx œ
È
x 2
È2
C
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 7.3 Exponential Functions 409
95.
'03 ŠÈ2 1‹ xÈ2 dx œ ’xŠÈ2
97.
'
98.
$
"‹
È2
“ œ 3Š
96.
!
x ‰ ˆ"‰
dx œ ' ˆ lnln10
x dx; u œ ln x Ê du œ
log10 x
x
Ä
"‹
"
x
'1e xÐln 2Ñ1 dx œ lnx # ‘ e1 œ e
ln 2
1ln 2
ln 2
ln 2
œ
"
ln #
' ˆ lnln10x ‰ ˆ x" ‰ dx œ ln"10 ' u du œ ˆ ln"10 ‰ ˆ #" u# ‰ C œ 2(lnlnx)10 C
#
'14 logx x dx œ '14 ˆ lnln #x ‰ ˆ x" ‰ dx; u œ ln x Ê du œ x" dx; x œ 1 Ê u œ 0, x œ 4 Ê u œ ln 4‘
4
x‰ ˆ"‰
' ln 4 ˆ ln"# ‰ u du œ ˆ ln"# ‰ #" u# ‘ ln0 4 œ ˆ ln"# ‰ #" (ln 4)# ‘ œ (ln2 ln4)# œ (lnln4)4 œ ln 4
Ä '1 ˆ ln
ln #
x dx œ 0
2
#
4
x
"
"
"
ln x ‰
#‘ %
#
#
#
#
#
'14 ln 2 log
' 4 ln x
"
dx œ '1 ˆ lnx2 ‰ ˆ ln
# dx œ 1 x dx œ # (ln x) " œ # c(ln 4) (ln 1) d œ # (ln 4) œ # (2 ln 2) œ 2(ln 2)
x
2
100.
'1e 2 ln 10 x(log
101.
'02 logx (x # 2) dx œ ln"# '02 cln (x 2)d ˆ x " # ‰ dx œ ˆ ln"# ‰ ’ (ln (x # 2)) “ # œ ˆ ln"# ‰ ’ (ln#4)
10
x)
dx œ '1
e
(ln 10)(2 ln x)
(ln 10)
ˆ x" ‰ dx œ c(ln x)# d e1 œ (ln e)# (ln 1)# œ 1
#
2
#
!
#
œ ˆ ln"# ‰ ’ 4(ln# 2) 102.
2 1
ln 2
dx‘
#
99.
œ
'110Î10 log
10
(10x)
x
dx œ
(ln 2)#
# “
"0
ln 10
œ
3
#
(ln 2)#
# “
ln 2
"!
'110Î10 cln (10x)d ˆ 10x" ‰ dx œ ˆ ln"010 ‰ ’ (ln (10x))
“
#0
#
"Î"!
#
œ ˆ ln"010 ‰ ’ (ln #100)
0
(ln 1)#
# “
#
œ ˆ ln"010 ‰ ’ 4(ln#010) “ œ # ln 10
103.
'09 2 logx (x1 1) dx œ ln210 '09 ln (x 1) ˆ x " 1 ‰ dx œ ˆ ln210 ‰ ’ (ln (x# 1)) “ * œ ˆ ln210 ‰ ’ (ln 210)
104.
'23 2 logx (x1 1) dx œ ln22 '23 ln (x 1) ˆ x" 1 ‰ dx œ ˆ ln22 ‰ ’ (ln (x#1)) “ $ œ ˆ ln22 ‰ ’ (ln22)
105.
'
#
10
!
#
2
#
#
dx
x log10 x
‰ ˆ x" ‰ dx œ (ln 10) ' ˆ ln"x ‰ ˆ x" ‰ dx; u œ ln x Ê du œ
œ ' ˆ lnln10
x
Ä (ln 10) ' ˆ ln"x ‰ ˆ "x ‰ dx œ (ln 10) '
'
107.
'1ln x "t dt œ cln ktkd ln1 x œ ln kln xk ln 1 œ ln (ln x), x 1
108.
'1e "t dt œ cln ktkd e1
109.
'11/x "t dt œ cln ktkd 1"Îx œ ln ¸ x" ¸ ln 1 œ aln 1 ln kxkb ln 1 œ ln x, x 0
110.
"
ln a
dx
x ‰#
x ˆ ln
ln 8
œ (ln 8)#
'
(ln x) #
x
(ln ")#
# “
(ln ")#
# “
œ ln 10
œ ln 2
dx‘
du œ (ln 10) ln kuk C œ (ln 10) ln kln xk C
106.
dx
x (log8 x)#
œ'
"
u
"
x
#
(ln x) "
1
dx œ (ln 8)#
#
C œ (lnln 8)x C
x
x
œ ln ex ln 1 œ x ln e œ x
'1x "t dt œ ln"a ln ktk‘ x1 œ lnln xa lnln 1a œ loga x, x 0
111. y œ (x 1)x Ê ln y œ ln (x 1)x œ x ln (x 1) Ê
w
y
y
œ ln (x 1) x †
"
(x 1)
Ê yw œ (x 1)x x x 1 ln (x 1)‘
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
410
Chapter 7 Transcendental Functions
112. y œ x2 x2x Ê y x2 œ x2x Ê lnay x2 b œ ln x2x œ 2x ln x Ê
w
w
1
w
y x 2 ay
2xb œ 2x †
1
x 2x
2 † ln x œ 2 2ln x
Ê y 2x œ ay x ba2 2ln xb Ê y œ aax x b x ba2 2ln xb 2x œ 2ax x x2x ln xb
2
t
2
2x
2
113. y œ ˆÈt‰ œ ˆt"Î# ‰ œ ttÎ# Ê ln y œ ln ttÎ# œ ˆ #t ‰ ln t Ê
Ê
dy
dt
t
" dy
y dt
œ ˆ #" ‰ (ln t) ˆ #t ‰ ˆ "t ‰ œ
"
#
ln t
#
t
œ ˆÈt‰ ˆ ln# t "# ‰
114. y œ tÈt œ tˆt
"Î# ‰
Ê ln y œ ln tˆt
œ ˆt"Î# ‰ (ln t) Ê
"Î# ‰
" dy
y dt
115. y œ (sin x)x Ê ln y œ ln (sin x)x œ x ln (sin x) Ê
116. y œ xsin x Ê ln y œ ln xsin x œ (sin x)(ln x) Ê
w
y
y
w
y
y
œ ˆ #" t"Î# ‰ (ln t) t"Î# ˆ "t ‰ œ
ln t2
2È t
Ê
t2
œ Š ln2È
‹ tÈ t
t
dy
dt
x‰
œ ln (sin x) x ˆ cos
Ê yw œ (sin x)x cln (sin x) x cot xd
sin x
œ (cos x)(ln x) (sin x) ˆ x" ‰ œ
sin x x (ln x)(cos x)
x
Ê yw œ xsin x ’ sin x x(lnx x)(cos x) “
117. y œ sin xx Ê y w œ cos xx
w
d
x
dx ax b;
w
if u œ xx Ê ln u œ ln xx œ x ln x Ê
w
u
u
œ x†
1
x
1 † ln x œ 1 ln x
Ê u œ x a1 ln xb Ê y œ cos x † x a1 ln xb œ x cos x a1 ln xb
x
x
x
118. y œ (ln x)ln x Ê ln y œ (ln x) ln (ln x) Ê
x
w
y
y
x
œ ˆ "x ‰ ln (ln x) (ln x) ˆ ln"x ‰
d
dx
(ln x) œ
ln (ln x)
x
"
x
Ê yw œ Š ln (ln xx) " ‹ (ln x)ln x
119. f(x) œ ex 2x Ê f w (x) œ ex 2; f w (x) œ 0 Ê ex œ 2 Ê x œ ln 2; f(0) œ 1, the absolute maximum; f(ln 2) œ 2 2 ln 2
¸ 0.613706, the absolute minimum; f(1) œ e 2 ¸ 0.71828, a relative or local maximum since f ww (x) œ ex is always
positive.
120. The function f(x) œ 2esin ÐxÎ2Ñ has a maximum whenever sin
x
#
œ 1 and a minimum whenever sin
œ 1. Therefore the
x
#
maximums occur at x œ 1 2k(21) and the minimums occur at x œ 31 2k(21), where k is any integer. The maximum
is 2e ¸ 5.43656 and the minimum is 2e ¸ 0.73576.
121. faxb œ x ex Ê f w axb œ x ex a1b ex œ ex x ex Ê f ww axb œ ex ax ex a1b ex b œ x ex 2ex
(a) f w axb œ 0 Ê ex x ex œ ex a1 xb œ 0 Ê ex œ 0 or 1 x œ 0 Ê x œ 1, fa1b œ a1be1 œ 1e ; using second
derivative test, f ww a1b œ a1be1 2e1 œ 1e 0 Ê absolute maximum at ˆ1, 1e ‰
(b) f ww axb œ 0 Ê x ex 2ex œ ex ax 2b œ 0 Ê ex œ 0 or x 2 œ 0 Ê x œ 2, fa2b œ a2be2 œ
f ww a1b 0 and f ww a3b œ e3 a3 2b œ e13 0 Ê point of inflection at ˆ2, e22 ‰
122. faxb œ
œ
ex
1 e2x
Ê f w axb œ
ˆ1 e2x ‰ex ex ˆ2e2x ‰
a1 e2x b2
œ
ex e3x
a1 e2x b2
Ê f ww axb œ
2
e2 ;
since
ˆ1 e2x ‰2 ˆex 3e3x ‰ ˆex e3x ‰2ˆ1 e2x ‰ˆ2e2x ‰
2
2
’a1 e2x b “
ex ˆ1 6e2x e4x ‰
a1 e2x b3
(a) f w axb œ 0 Ê ex e3x œ 0 Ê ex a1 e2x b œ 0 Ê e2x œ 1 Ê x œ 0; fa0b œ
e0
1 e2a0b
œ 12 ;
2
f w axb œ undefined Ê a1 e2x b œ 0 Ê e2x œ 1 Ê no real solutions. Using the second derivative test,
f ww a0b œ
e0 ˆ1 6e2a0b e4a0b ‰
a1 e2a0b b
3
œ
4
8
0 Ê absolute maximum at ˆ0, 12 ‰
(b) f ww axb œ 0 Ê ex a1 6e2x e4x b Ê ex œ 0 or 1 6e2x e4x œ 0 Ê e2x œ
Êxœ
lnŠ32È2‹
2
or x œ
lnŠ32È2‹
2
. f
lnŠ32È2‹
2
œ
É 3 2È 2
4 2È 2
and f
a6b „ È36 4
2
lnŠ32È2‹
2
œ
œ 3 „ 2È2,
É 3 2È 2
4 2È 2
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
;
Section 7.3 Exponential Functions 411
since f ww a1b 0, f ww a0b 0, and f ww a1b 0 Ê points of inflection at lnŠ32È2‹
2
123. f(x) œ x# ln
"
x
É 3 2È 2
,
4 2È 2
"
x
x# Š "" ‹ ax# b œ 2x ln
x
Since x œ 0 is not in the domain of f, x œ e"Î# œ
"
e
2
É 3 2È 2
,
4 2È 2
and
.
Ê f w (x) œ 2x ln
Therefore, f Š È"e ‹ œ
lnŠ32È2‹
ln Èe œ
"
e
"
#e
ln e"Î# œ
"
Èe
"
x
x œ x(2 ln x 1); f w (x) œ 0 Ê x œ 0 or ln x œ "# .
. Also, f w (x) 0 for 0 x ln e œ
"
#e
"
Èe
"
Èe
and f w (x) 0 for x is the absolute maximum value of f assumed at x œ
"
Èe
.
.
124. f(x) œ (x 3)# ex Ê f w (x) œ 2(x 3) ex (x 3)# ex
œ (x 3) ex (2 x 3) œ (x 1)(x 3) ex ; thus
f w (x) 0 for x 1 or x 3, and f w (x) 0 for
1 x 3 Ê f(1) œ 4e ¸ 10.87 is a local maximum and
f(3) œ 0 is a local minimum. Since f(x) 0 for all x,
f(3) œ 0 is also an absolute minimum.
125.
'0ln 3 ae2x ex b dx œ ’ e#
126.
'02 ln 2 ˆexÎ2 exÎ2 ‰ dx œ 2exÎ2 2exÎ2 ‘ 20 ln 2 œ ˆ2eln 2 2e ln 2 ‰ a2e! 2e! b œ (4 1) (2 2) œ 5 4 œ 1
2x
127. L œ '0 É1 ex
4
128. S œ 21'0 ˆ e
ecy ‰
#
1
ln 2
œ 21'0
ln 2
œ
œ
1
#
1
#
ˆe
y
y
dx Ê
ecy ‰
#
ex “
œ
dy
dx
ln 3
0
exÎ2
#
!
œ Š e # eln 3 ‹ Š e# e! ‹ œ ˆ 9# 3‰ ˆ "# 1‰ œ
2 ln 3
8
#
2œ2
Ê y œ exÎ2 C; y(0) œ 0 Ê 0 œ e0 C Ê C œ 1 Ê y œ exÎ2 1
É1 ˆ ey #ecy ‰# dy œ 21 '
0
ln 2
Ɉ ey #ecy ‰# dy œ 21 '0ln 2 ˆ e
y
ˆe
y
ecy ‰
#
ecy ‰#
#
É1 "4 ae2y 2 e2y b dy
dy œ
1
#
'0ln 2 ae2y 2 e2y bdy
"# e2y 2y "# e2y ‘ ln 2 œ 1# ˆ "# e2 ln 2 2 ln 2 "# e2 ln 2 ‰ ˆ "# 0 "# ‰‘
0
ˆ "# † 4 2 ln 2 "# † 4" ‰ œ 1# ˆ2 8" 2 ln 2‰ œ 1 ˆ 15
‰
16 ln 2
129. y œ "# aex ex b Ê
œ '0 É e4 1
2x
"
#
dy
dx
ec2x
4
2
œ "# aex ex b; L œ '0 É1 ˆ "# aex ex b‰ dx œ '0 É1 1
ln 3
œ 'ln 2
4x
2e2x 1 4e2x
ae2x 1b2
ln 3 x
e ecx
ex ecx
"
#
ec2x
4
dy
dx
œ
ex
ex 1
ex
ex 1
œ
2ex
e2x 1 ;
dy
dx
œ
dx
x
2
L œ 'ln 2 É1 ˆ e2x2e 1 ‰ dx œ 'ln 2 É1 ln 3
ln 3
ln 3
4x
2x
ln 3
2x
ln 3 2x
e 1
2
e2x 1
ln 3 e2x b 1
dx œ 'ln 2
dx; ’let u œ ex ex Ê du œ aex ex bdx, x œ ln 2 Ê u œ eln 2 eln 2 œ 2 131. y œ ln cos x Ê
1Î4
1
1b
dx œ 'ln 2 É e ae2x2e 1b2 1 dx œ 'ln 2 Ê aaee2x dx œ 'ln 2
1 b2
Ê u œ eln 3 eln 3 œ 3 œ '0
e2x
4
2
dx œ '0 Ɉ "# aex ex b‰ dx œ '0 "# aex ex b dx œ "# cex ex d 10 œ "# ˆe 1e ‰ 0 œ
1
130. y œ lnaex 1b lnaex 1b Ê
œ 'ln 2 É e
1
sin x
cos x
1
3
1
2
ex
e2x c 1
ex
4e2x
ae2x 1b2
œ 32 , x œ ln 3
8 Î3
œ tan x; L œ '0
1Î4
1Î4
sec x dx œ cln lsec x tan xld 0
É1 atan xb2 dx œ '
0
1Î4
È1 tan2 x dx œ '
œ ˆln lsecˆ 14 ‰ tan ˆ 14 ‰l‰ a0b œ lnŠÈ2 1‹
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
1Î4
0
dx
dx
‰
œ 83 “ Ä '3Î2 1u du œ cln luld 3Î2 œ lnˆ 83 ‰ lnˆ 32 ‰ œ lnˆ 16
9
8Î3
e2 1
2e
Èsec2 x dx
412
Chapter 7 Transcendental Functions
132. y œ ln csc x Ê
dy
dx
csc x cot x
csc x
œ
œ cot x; L œ '1Î6 É1 acot xb2 dx œ '1Î6 È1 cot2 x dx œ '1Î6 Ècsc2 x dx
1Î4
1Î4
1Î4
œ '1Î6 csc x dx œ cln lcsc x cot xld 1Î6 œ ˆln lcscˆ 14 ‰ cot ˆ 14 ‰l‰ ˆln lcscˆ 16 ‰ cot ˆ 16 ‰l‰
1 Î4
1 Î4
È
œ lnŠÈ2 1‹ lnŠ2 È3‹ œ lnŠ 2È2 13 ‹
133. (a)
d
dx
(x ln x x C) œ x †
(b) average value œ
134. average value œ
"
2 1
'1
e
"
e1
"
x
ln x 1 0 œ ln x
ln x dx œ
"
e1
cx ln x xd e1 œ
"
e1
[(e ln e e) (1 ln 1 1)] œ
"
e 1
(e e 1) œ
'12 "x dx œ cln kxkd #" œ ln 2 ln 1 œ ln 2
135. (a) f(x) œ ex Ê f w (x) œ ex ; L(x) œ f(0) f w (0)(x 0) Ê L(x) œ 1 x
(b) f(0) œ 1 and L(0) œ 1 Ê error œ 0; f(0.2) œ e0 2 ¸ 1.22140 and L(0.2) œ 1.2 Ê error ¸ 0.02140
(c) Since yww œ ex 0, the tangent line
approximation always lies below the curve y œ ex .
Thus L(x) œ x 1 never overestimates ex .
Þ
136. (a) y œ ex Ê yww œ ex 0 for all x Ê the graph of y œ ex is always concave upward
(b) area of the trapezoid ABCD 'ln a ex dx area of the trapezoid AEFD Ê
ln b
'ln a ex dx Š e
ln b
M œ eÐln a
ln bÑÎ2
ln a
eln b
‹ (ln
#
b ln a). Now
"
#
'ln a ex dx Š e
ln a
e
#
ln b
(AB CD)(ln b ln a)
(AB CD) is the height of the midpoint
since the curve containing the points B and C is linear Ê eÐln a
ln b
"
#
ln bÑÎ2
(ln b ln a)
‹ (ln b ln a)
'ln a ex dx œ cex d lnln ba œ eln b eln a œ b a, so part (b) implies that
ln b
(c)
eÐln aln bÑÎ2 (ln b ln a) b a Š e
Ê eln aÎ2 † eln bÎ2 137. A œ 'c2 1 2xx# dx œ 2'0
2
2
Ä A œ 2'1
5
138.
1
A œ '1 2Ð1
xÑ
"
u
ba
ln b ln a
2x
1 x#
ab
#
ln a
eln b
‹ (ln
#
b ln a) Ê eÐln aln bÑÎ2 Ê Èeln a Èeln b ba
ln b ln a
ab
#
ba
ln b ln a
Ê Èab ab
#
ba
ln b ln a
dx; cu œ 1 x# Ê du œ 2x dx; x œ 0 Ê u œ 1, x œ 2 Ê u œ 5d
du œ 2 cln kukd &" œ 2(ln 5 ln 1) œ 2 ln 5
"
x
"
#‹
ln Š "# ‹
1
Š
x
dx œ 2 '1 ˆ "# ‰ dx œ 2 –
—
œ ln2# ˆ #" 2‰ œ ˆ ln2# ‰ ˆ 3# ‰ œ
3
ln #
"
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
ab
#
"
e1
Section 7.3 Exponential Functions 413
139. From zooming in on the graph at the right, we estimate
the third root to be x ¸ 0.76666
140. The functions f(x) œ xln 2 and g(x) œ 2ln x appear to
have identical graphs for x 0. This is no accident,
because xln 2 œ eln 2 ln x œ aeln 2 bln x œ 2ln x .
†
141. (a) f(x) œ 2x Ê f w (x) œ 2x ln 2; L(x) œ a2! ln 2b x 2! œ x ln 2 1 ¸ 0.69x 1
(b)
142. (a) f(x) œ log3 x Ê f w (x) œ
"
x ln 3
, and f(3) œ
ln 3
ln 3
Ê L(x) œ
"
3 ln 3
(x 3) ln 3
ln 3
œ
x
3 ln 3
"
ln 3
1 ¸ 0.30x 0.09
(b)
143. (a) The point of tangency is apß ln pb and mtangent œ
"
p
since
dy
dx
œ x" . The tangent line passes through a!ß !b Ê the
equation of the tangent line is y œ "p x. The tangent line also passes throughapß ln pb Ê ln p œ "p p œ " Ê p œ e, and
the tangent line equation is y œ "e x.
(b)
d# y
dx#
œ x"# for x Á ! Ê y œ ln x is concave downward over its domain. Therefore, y œ ln x lies below the graph of
y œ "e x for all x !, x Á e, and ln x x
e
for x !, x Á e.
(c) Multiplying by e, e ln x x or ln x x.
e
(d) Exponentiating both sides of ln xe x, we have eln x ex , or xe ex for all positive x Á e.
(e) Let x œ 1 to see that 1e e1 . Therefore, e1 is bigger.
e
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
414
Chapter 7 Transcendental Functions
144. Using Newton's Method: faxb œ lnaxb " Ê f w axb œ
"
x
Ê xn" œ xn lnaxn b"
"
x8
Ê xn" œ xn ’# lnaxn b“.
Then, x1 œ 2, x2 œ 2.61370564, x3 œ 2.71624393, and x& œ 2.71828183. Many other methods may be used. For
example, graph y œ ln x " and determine the zero of y.
7.4 EXPONENTIAL CHANGE AND SEPARABLE DIFFERENTIAL EQUATIONS
1. (a) y œ ecx Ê y w œ ecx Ê 2y w 3y œ 2 aecx b 3ecx œ ecx
(b) y œ ecx ec3xÎ2 Ê y w œ e x 3# e 3xÎ2 Ê 2y w 3y œ 2 ˆe x 3# e 3xÎ2 ‰ 3 ae x e 3xÎ2 b œ e x
(c) y œ e x Ce 3xÎ2 Ê y w œ e x 3# Ce 3xÎ2 Ê 2y w 3y œ 2 ˆe x 3# Ce 3xÎ2 ‰ 3 ae x Ce 3xÎ2 b œ e
2. (a) y œ "x Ê y w œ
"
x#
(b) y œ x " 3 Ê y w œ
(c) y œ
3. y œ
"
xC
'1x
"
x
et
t
Ê yw œ
x
#
œ ˆ x" ‰ œ y#
"
(x 3)#
#
œ ’ (x " 3) “ œ y#
#
"
(x C)#
œ x " C ‘ œ y#
dt Ê yw œ x"# '1
x t
e
t
dt ˆ x" ‰ ˆ ex ‰ Ê x# y w œ '1
x t
e
x
dt ex œ x Š x"
t
'1x et
t
dt‹ ex œ xy ex
Ê x# y w xy œ ex
4. y œ
"
È 1 x%
'1x È1 t% dt
$
Ê y w œ Š 12xx% ‹ Š È
"
1 x%
Ê y w œ #" –
4x$
$
ŠÈ1 x% ‹
'1x È1 t% dt‹ 1
—
'1x È1 t% dt È "
1 x%
$
ŠÈ 1 x% ‹
Ê y w œ Š 12xx% ‹ y 1 Ê y w 2x$
1 x%
†yœ1
5. y œ ecx tan" a2ex b Ê y w œ ecx tan" a2ex b ecx ’ 1 a"2ex b# “ a2ex b œ ecx tan" a2ex b Ê y w œ y 2
1 4e2x
Ê yw y œ
2
1 4e2x
; y( ln 2) œ ecÐ
ln 2Ñ
tan" a2e
ln 2
2
1 4e2x
b œ 2 tan" 1 œ 2 ˆ 14 ‰ œ
1
#
#
#
#
#
#
6. y œ (x 2) ecx Ê y w œ ecx ˆ2xecx ‰ (x 2) Ê y w œ ecx 2xy; y(2) œ (2 2) ec2 œ 0
7. y œ cosx x Ê y w œ x sin xx# cos x Ê y w œ sinx x "x ˆ cosx x ‰ Ê y w œ sinx x 1/2)
y ˆ 1# ‰ œ cos(1(/2)
œ0
8. y œ
x
ln x
9. 2Èxy
Ê
2
3
Ê yw œ
dy
dx œ
$Î#
y
ln x x Š "x ‹
(ln x)#
Ê yw œ
"
ln x
"
(ln x)#
Ê x# y w œ
x#
ln x
x#
(ln x)#
y
x
Ê xy w œ sin x y Ê xy w y œ sin x;
Ê x# y w œ xy y# ; y(e) œ
e
ln e
œ e.
1 Ê 2x"Î# y"Î# dy œ dx Ê 2y"Î# dy œ x"Î# dx Ê ' 2y"Î# dy œ ' x"Î# dx Ê 2 ˆ 23 y$Î# ‰ œ 2x"Î# C"
x"Î# œ C, where C œ
"
#
C"
10.
dy
dx
œ x# Èy Ê dy œ x# y"Î# dx Ê y"Î# dy œ x# dx Ê ' y"Î# dy œ ' x# dx Ê 2y"Î# œ
11.
dy
dx
œ excy Ê dy œ ex ecy dx Ê ey dy œ ex dx Ê
12.
dy
dx
œ 3x# ey Ê dy œ 3x# ey dx Ê ey dy œ 3x# dx Ê ' ey dy œ ' 3x# dx Ê ey œ x3 C Ê ey x3 œ C
' ey dy œ ' ex dx
x$
3
C Ê 2y"Î# "3 x$ œ C
Ê ey œ ex C Ê ey ex œ C
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 7.4 Exponential Change and Separable Differential Equations
13.
dy
dx
œ Èy cos# Èy Ê dy œ ˆÈy cos# Èy‰ dx Ê
side, substitute u œ Èy Ê du œ
1
2Èy dy
sec# Èy
Èy dy
Ê 2 du œ
1
Èy dy,
œ dx Ê '
415
œ ' dx. In the integral on the left-hand
sec# Èy
Èy dy
and we have ' sec# u du œ ' dx Ê 2 tan u œ x C
Ê x 2 tan Èy œ C
14. È2xy dy
dx œ 1 Ê dy œ
Ê È2
15. Èx
y3/2
3
2
dy œ
"
#
dy
dx
œ
ey eÈ x
Èx
Ê dy œ
hand side, substitute u œ Èx Ê du œ
ey eÈ x
Èx dx
"
#È x
Ê ecy œ 2eÈx C, where C œ C1
16. asec xb
dy
dx
1
Èx
dx Ê È2 y1/2 dy œ x1/2 dx Ê È2 ' y1/2 dy œ ' x1/2 dx
3
C1 Ê È2 y3/2 œ 3Èx 32 C1 Ê È2 ˆÈy‰ 3Èx œ C, where C œ 32 C1
x1/2
œ eybÈx Ê
dy
dx
dx Ê È2Èydy œ
1
È2xy
œ eybsin x Ê
dy
dx
eÈ x
Èx
Ê ecy dy œ
dx Ê 2 du œ
"
Èx
dx Ê ' ecy dy œ '
dx, and we have
eÈ x
Èx
dxÞ In the integral on the right-
' ecy dy œ 2 ' eu du Ê ecy œ 2eu C1
œ eybsin x cos x Ê dy œ aey esin x cos xbdx Ê ecy dy œ esin x cos x dx
Ê ' ecy dy œ ' esin x cos x dx Ê ecy œ esin x C1 Ê ecy esin x œ C, where C œ C1
17.
dy
dx
œ 2xÈ1 y2 Ê dy œ 2xÈ1 y2 dx Ê
dy
È 1 y2
œ 2x dx Ê '
dy
È 1 y2
œ ' 2x dx Ê sin" y œ x# C since kyk "
Ê y œ sinax2 Cb
18.
dy
dx
œ
e2x c y
ex b y
2y
Ê dy œ
e2x c y
ex b y dx
Ê dy œ
e2x ecy
ex ey dx
œ
Ê e2y dy œ ex dx Ê ' e2y dy œ ' ex dx Ê
ex
e2y dx
e2y
#
œ ex C1
Ê e 2ex œ C where C œ 2C1
2 3
2
2
2 3
19. y2 dy
dx œ 3x y 6x Ê y dy œ 3x ay 2bdx Ê
20.
dy
dx
œ x y 3x 2y 6 œ ay 3bax 2b Ê
y2
y3 2
1
y 3 dy
dy œ 3x2 dx Ê '
œ ax 2bdx Ê '
y2
y3 2
dy œ ' 3x2 dx Ê 31 lnly3 2l œ x3 C
1
y 3 dy
œ ' ax 2bdx
Ê lnly 3l œ "# x2 2x C
21.
1 dy
x dx
2
2
2
œ yex 2Èy ex œ ex ˆy 2Èy‰ Ê
Ê'
22.
dy
dx
1
dy
È y ˆÈ y 2 ‰
1
y 2Èy dy
ey
1 ey dy
2
2
2
1
ecy 1 dy
ln 0.99
1000
2
1
ecy 1 dy
œ ' aex 1bdx
¸ 0.00001
ln (0.9)
0.00001
6Þ05
¸ 2.389 millibars
900 ‰
(c) 900 œ 1013eÐ 0Þ121Ñh Ê 0.121h œ ln ˆ 1013
Ê hœ
dy
dt
œ aex 1bdx Ê '
¸ 10,536 years
œ kp Ê p œ p! ekh where p! œ 1013; 90 œ 1013e20k Ê k œ
(b) p œ 1013e
25.
2
2
(b) 0.9 œ eÐ 0Þ00001)t Ê (0.00001)t œ ln (0.9) Ê t œ
(c) y œ y! eÐ20ß000Ñk ¸ y! e 0Þ2 œ y! (0.82) Ê 82%
dp
dh
œ ' x ex dx
œ ' aex 1bdx Ê lnl1 ey l œ ex x C Ê lna1 ey b œ ex x C
23. (a) y œ y! ekt Ê 0.99y! œ y! e1000k Ê k œ
24. (a)
1
y 2Èy dy
œ ' x ex dx Ê 2 lnlÈy 2l œ "# ex C Ê 4 lnlÈy 2l œ ex C Ê 4 lnˆÈy 2‰ œ ex C
œ ex y ex ey 1 œ aey 1baex 1b Ê
Ê'
œ x ex dx Ê '
ln (90) ln (1013)
20
ln (1013) ln (900)
0.121
¸ 0.121
¸ 0.977 km
œ 0.6y Ê y œ y! e 0Þ6t ; y! œ 100 Ê y œ 100e 0Þ6t Ê y œ 100e 0Þ6 ¸ 54.88 grams when t œ 1 hr
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
416
Chapter 7 Transcendental Functions
26. A œ A! ekt Ê 800 œ 1000e10k Ê k œ
ln (0.8)
10
Ê A œ 1000eÐln (0Þ8ÑÎ10Ñt , where A represents the amount of sugar that
remains after time t. Thus after another 14 hrs, A œ 1000eÐln Ð0Þ8ÑÎ10Ñ24 ¸ 585.35 kg
27. L(x) œ L! e
kx
Ê
L!
#
œ L! e
Ê ln
18k
one-tenth of the surface value,
L!
10
"
#
œ 18k Ê k œ
ln 2
18
¸ 0.0385 Ê L(x) œ L! e 0Þ0385x ; when the intensity is
œ L! ec0Þ0385x Ê ln 10 œ 0.0385x Ê x ¸ 59.8 ft
28. V(t) œ V! e tÎ40 Ê 0.1V! œ V! e tÎ40 when the voltage is 10% of its original value Ê t œ 40 ln (0.1) ¸ 92.1 sec
29. y œ y! ekt and y! œ 1 Ê y œ ekt Ê at y œ 2 and t œ 0.5 we have 2 œ e0Þ5k Ê ln 2 œ 0.5k Ê k œ
Therefore, y œ eÐln 4Ñt Ê y œ e24 ln 4 œ 424 œ 2.81474978 ‚ 1014 at the end of 24 hrs
ln 2
0.5
œ ln 4.
30. y œ y! ekt and y(3) œ 10,000 Ê 10,000 œ y! e3k ; also y(5) œ 40,000 œ y! e5k . Therefore y! e5k œ 4y! e3k
Ê e5k œ 4e3k Ê e2k œ 4 Ê k œ ln 2. Thus, y œ y! eÐln 2Ñt Ê 10,000 œ y! e3 ln 2 œ y! eln 8 Ê 10,000 œ 8y!
Ê y! œ 10,000
œ 1250
8
31. (a) 10,000ekÐ1Ñ œ 7500 Ê ek œ 0.75 Ê k œ ln 0.75 and y œ 10,000eÐln 0Þ75Ñt . Now 1000 œ 10,000eÐln 0Þ75Ñt
0.1
Ê ln 0.1 œ (ln 0.75)t Ê t œ lnln0.75
¸ 8.00 years (to the nearest hundredth of a year)
(b) 1 œ 10,000eÐln 0Þ75Ñt Ê ln 0.0001 œ (ln 0.75)t Ê t œ
ln 0.0001
ln 0.75
¸ 32.02 years (to the nearest hundredth of a year)
32. (a) There are (60)(60)(24)(365) œ 31,536,000 seconds in a year. Thus, assuming exponential growth,
P œ 257,313,431ekt and 257,313,432 œ 257,313,431eÐ14kÎ31ß536ß000Ñ Ê ln Š 257,313,432
257,313,431 ‹ œ
14k
31,536,000
Ê k ¸ 0.0087542
(b) P œ 257,313,431eÐ0.0087542Ña"&b ¸ 293,420,847 (to the nearest integer). Answers will vary considerably with the
number of decimal places retained.
33. 0.9P! œ P! ek Ê k œ ln 0.9; when the well's output falls to one-fifth of its present value P œ 0.2P!
0.2
Ê 0.2P! œ P! eÐln 0Þ9Ñt Ê 0.2 œ eÐln 0Þ9Ñt Ê ln (0.2) œ (ln 0.9)t Ê t œ ln
ln 0.9 ¸ 15.28 yr
34. (a)
dp
dx
"
œ 100
p Ê
dp
p
"
"
œ 100
dx Ê ln p œ 100
x C Ê p œ eÐ 0Þ01xCÑ œ eC e 0Þ01x œ C" e 0Þ01x ;
p(100) œ 20.09 Ê 20.09 œ C" eÐ 0Þ01ÑÐ100Ñ Ê C" œ 20.09e ¸ 54.61 Ê p(x) œ 54.61e 0Þ01x (in dollars)
(b) p(10) œ 54.61eÐ 0Þ01ÑÐ10Ñ œ $49.41, and p(90) œ 54.61eÐ 0Þ01ÑÐ90Ñ œ $22.20
(c) r(x) œ xp(x) Ê rw (x) œ p(x) xpw (x);
pw (x) œ .5461e 0Þ01x Ê rw (x)
œ (54.61 .5461x)e 0Þ01x . Thus, rw (x) œ 0
Ê 54.61 œ .5461x Ê x œ 100. Since rw 0
for any x 100 and rw 0 for x 100, then
r(x) must be a maximum at x œ 100.
35. A œ A! ekt and A! œ 10 Ê A œ 10 ekt , 5 œ 10 ekÐ24360Ñ Ê k œ
then 0.2Ð10Ñ œ 10 e0.000028254t Ê t œ
36. A œ A! ekt and
Êtœ
37. y œ y! e
ln 0.05
0.00499
kt
"
#
A! œ A! e139k Ê
"
#
ln 0.2
0.000028454
œ e139k Ê k œ
ln (0.5)
24360
¸ 0.000028454 Ê A œ 10 e0.000028454t ,
¸ 56563 years
ln (0.5)
139
¸ 0.00499; then 0.05A! œ A! ec0Þ00499t
¸ 600 days
œ y! e ÐkÑÐ3ÎkÑ œ y! e
3
œ
y!
e$
y!
20
œ (0.05)(y! ) Ê after three mean lifetimes less than 5% remains
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 7.4 Exponential Change and Separable Differential Equations
38. (a) A œ A! eckt Ê
(b)
"
k
œ ec2Þ645k Ê k œ
"
#
417
¸ 0.262
ln 2
#.645
¸ 3.816 years
ln 2 ‰
ln 2 ‰
(c) (0.05)A œ A exp ˆ 2.645
t Ê ln 20 œ ˆ 2.645
t Ê tœ
2.645 ln 20
ln #
¸ 11.431 years
39. T Ts œ (T! Ts ) e kt , T! œ 90°C, Ts œ 20°C, T œ 60°C Ê 60 20 œ 70e
10k
Ê
4
7
œe
10k
Êkœ
ln ˆ 74 ‰
10
¸ 0.05596
(a) 35 20 œ 70ec0Þ05596t Ê t ¸ 27.5 min is the total time Ê it will take 27.5 10 œ 17.5 minutes longer to reach 35°C
(b) T T œ (T T ) e kt , T œ 90°C, T œ 15°C Ê 35 15 œ 105e 0Þ05596t Ê t ¸ 13.26 min
s
!
s
!
s
40. T 65° œ (T! 65°) e kt Ê 35° 65° œ (T! 65°) e 10k and 50° 65° œ (T! 65°) e 20k . Solving
30° œ (T! 65°) e 10k and 15° œ (T! 65°) e 20k simultaneously Ê (T! 65°) e 10k œ 2(T! 65°) e 20k
ln 2
Ê e10k œ 2 Ê k œ ln102 and 30° œ T!e10k65° Ê 30° e10 ˆ 10 ‰ ‘ œ T! 65° Ê T! œ 65° 30° ˆeln 2 ‰ œ 65° 60° œ 5°
41. T Ts œ (T! Ts ) eckt Ê 39 Ts œ (46 Ts ) ec10k and 33 Ts œ (46 Ts ) ec20k Ê
33Ts
46Ts
œ ec20k œ aec10k b# Ê
33Ts
46Ts
39Ts
46Ts
œ ec10k and
#
Ts
#
#
œ Š 39
46Ts ‹ Ê (33 Ts )(46 Ts ) œ (39 Ts ) Ê 1518 79Ts Ts
œ 1521 78Ts T#s Ê Ts œ 3 Ê Ts œ 3°C
42. Let x represent how far above room temperature the silver will be 15 min from now, y how far above room temperature the
silver will be 120 min from now, and t! the time the silver will be 10°C above room temperature. We then have the
following time-temperature table:
time in min.
0
20 (Now) 35
140
t!
temperature
Ts 70° Ts 60° Ts x Ts y Ts 10°
" ‰
T Ts œ (T! Ts ) eckt Ê (60 Ts ) Ts œ c(70 Ts ) Ts d ec20k Ê 60 œ 70ec20k Ê k œ ˆ 20
ln ˆ 67 ‰ ¸ 0.00771
(a) T Ts œ (T! Ts ) ec0Þ00771t Ê (Ts x) Ts œ c(70 Ts ) Ts d e Ð0Þ00771ÑÐ35Ñ Ê x œ 70e 0Þ26985 ¸ 53.44°C
(b) T Ts œ (T! Ts ) e 0Þ00771t Ê (Ts y) Ts œ c(70 Ts ) Ts d e Ð0Þ00771ÑÐ140Ñ Ê y œ 70e 1Þ0794 ¸ 23.79°C
(c) T Ts œ (T! Ts ) e 0Þ00771t Ê (Ts 10) Ts œ c(70 Ts ) Ts d e Ð0Þ00771Ñ t! Ê 10 œ 70e 0Þ00771t!
"
‰ ln ˆ "7 ‰ œ 252.39 Ê 252.39 20 ¸ 232 minutes from now the
Ê ln ˆ "7 ‰ œ 0.00771t! Ê t! œ ˆ 0.00771
silver will be 10°C above room temperature
43. From Example 4, the half-life of carbon-14 is 5700 yr Ê
Ê (0.445)c! œ c! e 0Þ0001216t Ê t œ
ln (0.445)
0.0001216
"
#
c! œ c! eckÐ5700Ñ Ê k œ
ln 2
5700
¸ 0.0001216 Ê c œ c! e 0Þ0001216t
¸ 6659 years
44. From Exercise 43, k ¸ 0.0001216 for carbon-14.
(a) c œ c! e 0Þ0001216t Ê (0.17)c! œ c! e 0Þ0001216t Ê t ¸ 14,571.44 years Ê 12,571 BC
(b) (0.18)c! œ c! e 0Þ0001216t Ê t ¸ 14,101.41 years Ê 12,101 BC
(c) (0.16)c! œ c! e 0Þ0001216t Ê t ¸ 15,069.98 years Ê 13,070 BC
45. From Exercise 43, k ¸ 0.0001216 for carbon-14 Ê y œ y0 e0.0001216t . When t œ 5000
Ê y œ y0 e0.0001216a5000b ¸ 0.5444y0 Ê yy0 ¸ 0.5444 Ê approximately 54.44% remains
46. From Exercise 43, k ¸ 0.0001216 for carbon-14. Thus, c œ c! e 0Þ0001216t Ê (0.995)c! œ c! e 0Þ0001216t
Êtœ
ln (0.995)
0.0001216
¸ 41 years old
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
418
Chapter 7 Transcendental Functions
^
7.5 INDETERMINATE FORMS AND L'HOPITAL'S
RULE
^
1. l'Hopital:
lim
x2
œ
"
#x ¹xœ#
^
2. l'Hopital:
lim
sin 5x
x
œ
5 cos 5x
¹
1
xœ!
#
x Ä 2 x 4
xÄ0
^
3. l'Hopital:
lim
xÄ_
œ lim
3x#
œ
x2
lim #
x Ä 2 x 4
9.
t$ 4t 15
#
t Ä 3 t t 12
œ lim
xÄ!
#x # $ x
x$ x 1
"
œ lim
7.
x Ä 2 #x
sin x
2x
3
11
œ lim
œ
3t# 4
t Ä 3 2t 1
œ
3
11
3(3)# 4
2(3) 1
5x3 2x
7x3 3
œ x lim
Ä_
"5x2 2
21x2
œ x lim
Ä_
30x
42x
12. x lim
Ä_
x 8x#
12x# 5x
1 "6x
24x 5
œ x lim
Ä_
16
24
15. lim
sin t#
t
xÄ0
œ x lim
Ä_
œ lim
16x
sin x x
x$
œ lim
cos x "
3x#
2) 1
lim
) Ä 1Î2 cos (21 ))
18.
1
) Ä 1Î3 sin ˆ) 3 ‰
19.
lim
) Ä 1Î2 1 cos 2)
œ
3) 1
1 sin )
œ
xÄ!
œ5†1œ5
5x# 3x
7x# 1
1cos x
x#
œ x lim
Ä_
5
7
3
x
"
œ
x#
5
7
ax "bax# x "b
16
x Ä 0 cos x
œ lim
xÄ0
2
œ x lim
Ä_
3
lim
1
) Ä 1Î3 cos ˆ) 3 ‰
cos )
lim
) Ä 1Î2 2 sin 2)
œ
#x# 3x
x$ x 1
œ ! or x lim
Ä_
lim
x Ä 5
x# 25
x5
œ x lim
Ä_
œ lim
2x
x Ä 5 1
#
x
"
"
x#
3
x#
30
42
œ
5
7
sin 5t
2t
16
1
tÄ0
5 cos 5t
2
œ
5
2
œ 16
œ lim
2
sin ˆ 3#1 ‰
œ lim
xÄ0
cos x
6
œ "6
œ 2
œ3
sin )
lim
) Ä 1Î2 4 cos 2)
œ
"
(4)(1)
œ
"
x$
œ 10
œ 23
œ
sin x
6x
œ
xb ˆ " cos x ‰
œ lim ” a" xcos
2
" cos x •
xÄ!
"
#
tÄ0
lim
) Ä 1Î2 sin (21 ))
œ
"
4
2
x Ä 1 ax "ba4x + 4x + 3b
14. lim
œ lim
x Ä 0 sin x
xÄ0
%
'x
œ0
8x#
17.
lim
acos t# b (2t)
1
tÄ0
x Ä 0 cos x 1
16. lim
œ
#
t Ä 1 12t 1
œ lim
œ
œ 23
7
11. x lim
Ä_
tÄ0
"
x Ä 2 x#
œ lim
or lim
8.
3t#
13. lim
x $ 1
"
#
œ
"
4
œ lim
$
t Ä 1 4t t 3
sin 5x
5x
or x lim
Ä_
$
x Ä 1 4x x3
œ x lim
Ä_
t$ 1
10. lim
5
7
or lim
cos x
2
xÄ!
%x 3
$x # "
œ x lim
Ä_
œ lim
lim
œ
œ
10
14
œ lim ”ˆ sinx x ‰ˆ sinx x ‰ˆ " "cos x ‰• œ
xÄ!
2
x Ä ! x a" cos xb
^
6. l'Hopital:
lim
xÄ_
œ x lim
Ä_
#
x Ä 1 12x 1
œ 5 lim
5x Ä 0
œ lim
3
11
1 cos x
x#
sin# x
œ lim
x Ä 2 ax #bax #b
sin 5x
x
xÄ0
x$ 1
ax # x " b
xÄ!
œ 5 or lim
x2
œ lim
#
x Ä 2 x 4
"0x 3
14x
2
x Ä 1 a4x + 4x + 3b
x2
or lim
œ x lim
Ä_
$
x Ä 1 4x x 3
^
5. l'Hopital:
lim
"
4
5x# 3x
7x# 1
^
4. l'Hopital:
lim
œ lim
œ
"
4
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
œ
!
"
œ!
^
Section 7.5 Indeterminate Forms and L'Hopital's
Rule
x"
20. lim
x Ä 1 ln x sin (1x)
x#
21. lim
ln (csc x)
1 #
x Ä 1 Î 2 ˆx ˆ # ‰‰
tÄ0
t(1 cos t)
t sin t
24. lim
t sin t
t Ä 0 1 cos t
"
x
"
1 cos (1x)
œ
œ
lim
(1 cos t) t(sin t)
1 cos t
œ lim
tÄ0
sin t t cos t
sin t
œ lim
tÄ0
œ lim
2x
lim
tÄ0
tÄ0
œ
26.
ˆ 1 x‰ tan x œ
lim
lim
x Ä Ð1 Î 2 Ñ c #
x Ä Ð1 Î 2 Ñ c
ˆ 1# x‰
cot x
œ
œ lim
ˆ "# ‰) 1
)
)Ä0
œ lim
28. lim
29. lim
x 2x
x
x Ä 0 2 1
30. lim
3x "
x
x Ä 0 2 1
)Ä0
ˆln ˆ "# ‰‰ ˆ "# ‰)
1
)Ä0
œ lim
3x ln 3
x
x Ä 0 2 ln 2
32. x lim
Ä_
log2 x
log3 (x 3)
œ x lim
Ä_
34.
ln (x1)
x‰
ˆ ln
ln #
œ x lim
Ä_
"
1
3‰
œ ˆ ln
ln # x lim
Ä_
œ
ln ax# 2xb
ln x
œ lim b
xÄ!
lim b
ln aex "b
ln x
œ lim b
xÄ!
œ lim
Èay a# a
y
yÄ0
œ lim
yÄ0
36. lim
œ lim
tÄ0
1 (1 0)
1
œ
" ‰
ˆ sin
x œ
lim
" ‰
ˆ csc
œ
#x
x Ä Ð1 Î 2 Ñ c
œ
1#
#
œ
"
#
cos t cos t cos t t sin t
cos t
œ
11"0
1
œ2
lim
x Ä Ð1 Î 2 Ñ c
csc# x
#
"
1
œ 1
lim
x Ä Ð1 Î 2Ñ c
sin# x œ 1
œ ln 3
œ
1 †2 ! 0
(ln 2)†2!
œ
ˆx"1‰
ˆ "x ‰
3‰
œ ˆ ln
ln # x lim
Ä_
Š ln (xln 3 3) ‹
"
ln #
ln 3
ln #
œ (ln 2) x lim
Ä_
x‰
ˆ ln
ln #
œ
œ (ln 2) x lim
Ä_
ln x
ln (x 3)
x
x 1
œ (ln 2) x lim
Ä_
ln 3 ‰
œ ˆ ln
# x lim
Ä_
ˆ "x ‰
ˆx " 3‰
1
1
œ ln 2
œ ˆ llnn 3# ‰ x lim
Ä_
Š x2x
#
2
‹
2x
ˆ x" ‰
œ lim b
xÄ!
2x# 2x
x# 2x
4x 2
2x 2
œ lim b
xÄ!
œ lim b
xÄ!
2
#
œ1
x
È5y 25 5
y
35. lim
lim
x Ä 1 Î2
ln 3
ln #
lim b
xÄ!
3! †ln 3
2! †ln 2
œ
a3! b (ln 3)(1)
1
œ2
œ ln ˆ "# ‰ œ ln 1 ln 2 œ ln 2
(1) a2x b (x)(ln 2) a2x b
(ln 2) a2x b
xÄ0
ln (x1)
log2 x
xÄ!
œ
œ lim
31. x lim
Ä_
33.
3sin ) (ln 3)(cos ))
1
œ
cos t (cos t t sin t)
cos t
œ lim
ˆx 1# ‰
cos x
3sin ) "
)
cot x
2
1#
sin t (sin t t cos t)
sin t
œ lim
ˆx 1# ‰ sec x œ
lim
lim
x Ä Ð1 Î 2 Ñ c
x Ä Ð1 Î 2 Ñ c
)Ä0
œ
1
x Ä 1 Î 2 2 ˆ x ˆ # ‰‰
25.
27. lim
2
#
x Ä 0 sec x
x Ä 0 tan x
x cot x ‰
ˆ csccsc
x
ˆx ˆ 1# ‰‰
2
x Ä 1 Î2
œ
"
11
œ lim
2x
sec x tan x
x Ä 0 ˆ sec x ‰
lim
23. lim
xÄ1
œ lim
x Ä 0 ln (sec x)
22.
œ lim
yÄ0
Š exec 1 ‹
ˆ "x ‰
œ lim b
xÄ!
(5y 25)"Î# 5
y
aay a# b
y
yÄ0
"Î#
a
xex
ex 1
œ lim
yÄ0
œ lim b
xÄ!
ex xex
ex
ˆ "# ‰ (5y 25) "Î# (5)
1
ˆ "# ‰ aay a# b "Î# (a)
1
yÄ0
œ lim
‰
37. x lim
[ln 2x ln (x 1)] œ x lim
ln ˆ x 2x
1 œ ln Šx lim
Ä_
Ä_
Ä_
œ
10
1
œ1
œ lim
5
y Ä 0 2È5y 25
œ lim
a
y Ä 0 2Èay a#
2x
x1‹
œ ln Šx lim
Ä_
œ
"
#
œ "# , a 0
2
1‹
œ ln 2
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
x3
x
œ3
419
420
38.
Chapter 7 Transcendental Functions
lim
x Ä !b
(ln x ln sin x) œ lim b ln ˆ sinx x ‰ œ ln Š lim b
xÄ!
xÄ!
2
xb
39. lim b lnaln
asin xb œ lim b
x Ä0
40.
2aln xbˆ 1x ‰
cos x
sin x
x Ä0
lim
x Ä !b
ˆ 3x x " " ‰
sin x
basin xb
aln xb
œ lim b 2alnxxcos
œ lim b ’ 2cos
x
x †
x Ä0
x Ä0
x) x
œ lim b Š (3x x1)(sin
‹ œ lim b
sin x
xÄ!
xÄ!
3 cos x (3x 1)( sin x)
œ lim b Š 3 cos xcos
‹œ
x cos x x sin x
xÄ!
41.
lim b ˆ x " 1 xÄ1
" ‰
ln x
lim
x Ä !b
"
(0 1) 1
#
cos ) "
#
eh (" h)
h#
hÄ0
45.
et t#
lim t
t Ä _ e 1
sin )
œ lim
44. lim
eh "
h Ä 0 #h
et 2t
et
tÄ_
x sin x
xÄ0 x tan x
œ lim
ae x 1 b 2
xÄ0 x sin x
œ lim
48. lim
) sin ) cos )
)Ä0 tan ) )
49. lim
50. lim
x Ä0
eh
hÄ0 #
x#
ex
œ lim
"
tÄ_
œ x lim
Ä_
cos x
sin x
"
#
œ lim
2 ae x 1 be x
x
xÄ0 cos x sin x
œ lim
et
œ lim
t
tÄ_ e
œ x lim
Ä_
2x
ex
(sin x)(cos x)
cos x‰ œ lim b Š (1 cos x) sin
‹
x
xÄ!
œ 1
œ
et 2
et
œ lim
1 cos x
2
xÄ0 x sec x tan x
2
ex
œ1
œ0
sin x
2
2
xÄ0 2x sec x tan x 2sec x
2e2x 2ex
xÄ0 x cos x sin x
1 sin2 ) cos2 )
sec2 ) "
)Ä0
œ lim
sin 3x 3x x2
sin x sin 2x
œ3
œ1
cos )
e)
)Ä0
œ lim
œ lim
010
1
œ lim
)
) Ä 0 e 1
46. x lim
x# ex œ x lim
Ä_
Ä_
47. lim
6
#
œ _ † 1 œ _
œ #"
(csc x cot x cos x) œ lim b ˆ sin" x xÄ!
)
) Ä 0 e )1
œ
œ ln 1 œ 0
3 sin x (3x 1)(cos x) 1
sin x x cos x
"
x sin x
œ lim b Š sin x cos
‹œ
cos x
xÄ!
43. lim
3 3 (1)(0)
110
sin x
x “
"
cos x ‹
(x 1)
1x
x
œ lim b Š ln(xx1)(ln
x) ‹ œ lim b Š (ln x) (x 1) ˆ "x ‰ ‹ œ lim b Š (x ln x) x 1 ‹
xÄ1
xÄ1
xÄ1
"
œ lim b Š (ln x 1) 1 ‹ œ
xÄ1
42.
œ ln Š lim b
xÄ!
x
sin x ‹
0
2
œ0
4e2x 2ex
xÄ0 x sin x 2cos x
œ lim
2sin2 )
2
)Ä0 tan )
œ lim
3cos 3x 3 2x
xÄ0 2sin x cos 2x cos x sin 2x
œ
œ lim
2
2
œ1
œ lim 2 cos2 ) œ 2
)Ä0
3cos 3x 3 2x
xÄ0 sin x cos 2x sin 3x
œ lim
œ
9sin 3x 2
xÄ0 2sin x sin 2x cos x cos 2x 3cos 3x
œ lim
51. The limit leads to the indeterminate form 1_ . Let faxb œ x1ÎÐ1 xÑ Ê ln faxb œ ln ax1ÎÐ1 xÑ b œ
lim
x Ä 1b
ln faxb œ lim b
xÄ1
ln x
1x
œ lim b
xÄ1
ˆ "x ‰
1
lim
ln faxb œ lim b
xÄ1
ln x
x1
œ lim b
xÄ1
ˆ "x ‰
lim ln faxb œ
xÄ_
œ x lim
Ä_
œ
1
2
. Now
1
ln x
x1.
"
e
Now
œ 1. Therefore lim b x1ÎÐx 1Ñ œ lim b faxb œ lim b eln faxb œ e" œ e
xÄ1
xÄ1
xÄ1
53. The limit leads to the indeterminate form _! . Let faxb œ (ln x)1Îx Ê ln faxb œ ln (ln x)1Îx œ
x)
lim ln (ln
x
xÄ_
2
4
œ 1. Therefore lim b x1ÎÐ1 xÑ œ lim b faxb œ lim b eln faxb œ e" œ
xÄ1
xÄ1
xÄ1
52. The limit leads to the indeterminate form 1_ . Let faxb œ x1ÎÐx 1Ñ Ê ln faxb œ ln ax1ÎÐx 1Ñ b œ
x Ä 1b
ln x
1x
œ
ˆ x ln" x ‰
1
ln (ln x)
x .
Now
œ 0. Therefore x lim
(ln x)1Îx œ x lim
faxb œ x lim
eln faxb œ e! œ 1
Ä_
Ä_
Ä_
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
^
Section 7.5 Indeterminate Forms and L'Hopital's
Rule
54. The limit leads to the indeterminate form 1_ . Let faxb œ (ln x)1ÎÐx eÑ Ê ln faxb œ
œ lim b
xÄe
ˆ x ln" x ‰
1
œ lim b
xÄe
ln (ln x)
x e
ln (ln x)
x e
421
œ lim ln faxb
xÄe
œ "e . Therefore (ln x)1ÎÐx eÑ œ lim b faxb œ lim b eln faxb œ e"Îe
xÄe
xÄe
x
55. The limit leads to the indeterminate form 0! . Let faxb œ xc1Îln x Ê ln faxb œ ln
ln x œ 1. Therefore
lim
xÄ!
x1Îln x œ lim
eln faxb œ e" œ
faxb œ lim
xÄ!
xÄ!
"
e
56. The limit leads to the indeterminate form _! . Let faxb œ x1Îln x Ê ln faxb œ
ln x
ln x
œ x lim
faxb œ x lim
e1n faxb œ e" œ e
Ä_
Ä_
œ 1. Therefore x lim
x1Îln x
Ä_
ln (1 2x)
2 ln x
57. The limit leads to the indeterminate form _! . Let faxb œ (1 2x)1ÎÐ2 ln xÑ Ê ln faxb œ
ln (1 2x)
Ê x lim
ln faxb œ x lim
œ x lim
2 ln x
Ä_
Ä_
Ä_
ln faxb
"Î#
œ x lim
faxb œ x lim
e
œe
Ä_
Ä_
x
1 2x
œ x lim
Ä_
"
#
œ
"
#
. Therefore x lim
(1 2x)1ÎÐ2 ln xÑ
Ä_
58. The limit leads to the indeterminate form 1_ . Let faxb œ aex xb1Îx Ê ln faxb œ
ln aex xb
x
Ê lim ln faxb œ lim
xÄ0
xÄ0
ex 1
œ lim
x
x Ä 0 e x
ln aex xb
x
œ 2. Therefore lim aex xb1Îx œ lim faxb œ lim eln faxb œ e#
xÄ0
xÄ0
xÄ0
59. The limit leads to the indeterminate form 0! . Let faxb œ xx Ê ln faxb œ x ln x Ê ln faxb œ
œ lim b ln faxb œ lim b
xÄ!
xÄ!
ln x
ˆ "x ‰
ˆ "x ‰
œ lim b
xÄ!
Š x"# ‹
ln x
ˆ "x ‰
œ lim b (x) œ 0. Therefore lim b xx œ lim b faxb
xÄ!
xÄ!
xÄ!
œ lim b eln faxb œ e! œ 1
xÄ!
x
60. The limit leads to the indeterminate form _! . Let faxb œ ˆ1 "x ‰ Ê ln faxb œ
œ
c#
Š cx " ‹
lim b 1xx#
xÄ!
"
1 x"
œ lim b
xÄ!
œ lim b eln faxb œ e! œ 1
xÄ!
œ lim b
xÄ!
x
x1
ln a1xc" b
x "
Ê
lim
x Ä !b
ln faxb
x
œ 0. Therefore lim b ˆ1 x" ‰ œ lim b faxb
xÄ!
xÄ!
x
x
2‰
2‰
2‰
61. The limit leads to the indeterminate form 1_ . Let faxb œ ˆ xx Ê ln faxb œ ln ˆ xx œ x ln ˆ xx 1
1
1 Ê lim ln faxb
2‰
œ lim x ln ˆ xx 1 œ lim Š
xÄ_
xÄ_
2
ln ˆ xx b
c1‰
1
x
‹ œ lim Š
ln ax 2b ln ax 1b
1
x
xÄ_
‹œ
2
œ lim Š ax 23xbax 1b ‹ œ lim ˆ 2x6x 1 ‰ œ lim ˆ 62 ‰ œ 3. Therefore,
xÄ_
xÄ_
xÄ_
62. The limit leads to the indeterminate form _! . Let faxb œ Š xx 21 ‹
2
ln Š xx bb21 ‹
2
Ê lim ln faxb œ lim 1x ln Š xx 21 ‹ œ lim
2
xÄ_
xÄ_
x2 4x 1
3
2
xÄ_ x 2x x 2
œ lim
xÄ_
2x 4
2
xÄ_ 3x 4 x 1
œ lim
x
œ lim
2
xÄ_ 6 x 4
1
63.
x
lim ˆ x 2 ‰
xÄ_ x 1
lim faxb œ
xÄ_
2
ln ˆx2 1‰ ln ax 2b
x
xÄ_
3
œ
Ê ln faxb œ ln Š xx 21 ‹
œ lim
œ 0. Therefore,
œ xlim
Œ
Ä_
œ lim
2x
x2 b 1
xÄ_
1 Îx
lim Š xx 21 ‹
2
xÄ_
1Îx
x b1 2
1
ax
xÄ_
c3
b 2bax c 1b
x12
lim eln faxb œ e3
xÄ_
œ x1 ln Š xx 21 ‹
2
x2 4x 1
2
xÄ_ ax 1bax 2b
œ lim
œ lim faxb œ lim eln faxb œ e0 œ 1
xÄ_
xÄ_
2
x
lim x2 ln x œ limb Œ ln1 x œ limb Œ x2 œ limb Š 2x
‹ œ limb Š 3x2 ‹ œ 0
2
3
x Ä0
x Ä0
x Ä0
x Ä0
x Ä0 b
x
x
2
64.
1 Îx
1
lim Πx b 2 1x c 1
xÄ_
x2
1
lim x aln xb2 œ limb Š aln1xb ‹ œ limb Œ
x Ä0 b
x Ä0
x
x Ä0
2aln xb 1x
x12 2
2
œ limb Š 2ln1x ‹ œ limb Œ x1 œ limb Š 2xx ‹ œ limb a2xb œ 0
2
x Ä0
x Ä0
x Ä0
x Ä0
x
x
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
422
65.
Chapter 7 Transcendental Functions
lim x tanˆ 12 x‰ œ limb Š cotˆ 1x x‰ ‹ œ limb Š csc2 ˆ11 x‰ ‹ œ
x Ä0 b
x Ä0
x Ä0
2
2
1
1
œ1
1
66.
ln x ‰
x
ˆ sin xxtan x ‰ œ lim Š sin x sec
lim sin x † ln x œ limb ˆ csc
x œ limb Š csc x cot x ‹ œ limb b
x Ä0 b
x Ä0
67. x lim
Ä_
È9x 1
Èx 1
68.
Èx
Èsin x
lim b
xÄ!
70.
lim
x Ä !b csc x
cot x
71. x lim
Ä_
lim b
xÄ!
ex
x ex
lim
x
ec1Îx
œ lim b
xÄ!
e1Îx
1
x
9
1
x Ä0
2
x cos x tan x
‹
1
œ
0
1
œ0
œ È9 œ 3
"
œ1
lim
x Ä 1Î2c sin x
œ lim b cos x œ 1
xÄ!
ˆ 23 ‰x 1
x
1 ˆ 43 ‰
2
ex c x
x
œ x lim
Ä_
œ Éx lim
Ä_
x‰
ˆ cos" x ‰ ˆ cos
sin x œ
x‰
ˆ cos
sin x
ˆ sin" x ‰
œxÄ
lim
_
x Ä0
œ É 1" œ 1
sin x
x
x Ä 1 Î2 c
œ x lim
Ä_
2x 4x
5x 2x
2
73. x lim
Ä_
"
lim
xÄ!b
œ lim b
xÄ!
2x 3x
3x 4x
72. x Ä
lim
_
74.
œ
sec x
lim
x Ä 1Î2c tan x
9x 1
x1
œ Éx lim
Ä_
œÊ
69.
x Ä0
œ0
x
1 ˆ 42 ‰
ˆ 52 ‰x 1
1 2x
ˆ 52 ‰x 1
œxÄ
lim
_
exax c 1b
x
œ x lim
Ä_
x12
2x 2
x Ä 0 2x cos x
œ 1
œ_
œ lim b e1Îx œ _
xÄ!
75. Part (b) is correct because part (a) is neither in the
76. Part (b) is correct; the step lim
10
01
exax c 1b a2x 1b
1
œ x lim
Ä_
e1Îx Š x12 ‹
œ lim b
xÄ!
œ
0
0
nor
œ lim
2
_
_
x Ä 0 # sin x
^
form and so l'Hopital's
rule may not be used.
in part (a) is false because lim
2x 2
x Ä 0 2x cos x
is not an
indeterminate quotient form.
77. Part (d) is correct, the other parts are indeterminate forms and cannot be calculated by the incorrect arithmetic
78. (a) We seek c in a#ß !b so that
f ac b
g ac b
w
fa!b fa#b
ga!b ga#b
œ
w
œ
!#
!%
œ #" . Since f w acb œ " and gw acb œ #c we have that
"
#c
œ #"
Ê c œ ".
(b) We seek c in aaß bb so that
Êcœ
œ
fabb faab
g ab b g a a b
f ac b
g ac b
œ
fa$b fa!b
$ !
ga$b ga!b œ * !
È
œ " $ $( .
w
w
ba
# .
(c) We seek c in a!ß $b so that
c# %
#c
f ac b
g ac b
œ "$ Ê c œ
w
w
" „ È$(
$
Êc
œ
ba
b # a#
œ
"
ba.
Since f w acb œ " and gw acb œ #c we have that
œ lim
xÄ0
27 sin 3x
30x
œ lim
xÄ0
81 cos 3x
30
œ
27
10
œ
œ "$ . Since f w acb œ c# % and gw acb œ #c we have that
79. If f(x) is to be continuous at x œ 0, then lim f(x) œ f(0) Ê c œ f(0) œ lim
xÄ0
"
#c
xÄ0
9x 3 sin 3x
5x$
œ lim
xÄ0
.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
9 9 cos 3x
15x#
"
ba
^
Section 7.5 Indeterminate Forms and L'Hopital's
Rule
80.
limˆ tanx32x a
x2
x Ä0
sin bx ‰
x
œ limŠ tan 2x axx3 x
2
sin bx
x Ä0
‹ œ limŠ 2sec
2
x Ä0
2x a bx2 cos bx 2x sin bx
‹will
3x2
lima2sec2 2x a bx2 cos bx 2x sin bxb œ a 2 œ 0 Ê a œ 2; limŠ 2sec
x Ä0
x Ä0
œ limŠ 8sec
œ
x Ä0
16 6b
6
2
2x tan 2x b2 x2 sin bx 4bx cos bx 2sin bx
‹
6x
œ limŠ 32sec
x Ä0
2
2
be in
0
0
form if
2x 2 bx2 cos bx 2x sin bx
‹
3x2
2x tan2 2x 16sec4 2x b3 x2 cos bx 6b2 x sin bx 6b cos bx
6
‹
œ 0 Ê 16 6b œ 0 Ê b œ 38
81. (a)
(b) The limit leads to the indeterminate form _ _:
È #
lim Šx Èx# x‹ œ x lim
Šx Èx# x‹Š x Èx# x ‹ œ x lim
Š
xÄ_
Ä_
Ä_
x
œ x lim
Ä_
82.
"
" É" "
x
œ
"
" È" !
lim ŠÈx2 1 Èx‹ œ lim xŠ
_
_
xÄ
xÄ
x x
x # ax # x b
‹
x È x# x
œ x lim
Ä_
x
x È x# x
œ "#
È x2 1
x
Èx
x ‹
œ lim xŠÉ x x2 1 È xx2 ‹ œ lim xŠÉ1 2
_
_
xÄ
xÄ
1
x2
É 1x ‹ œ _
83. The graph indicates a limit near 1. The limit leads to the
2x# (3x 1) Èx 2
x 1
xÄ1
#
$Î#
"Î#
4x 9# x"Î# lim 2x 3xx 1 x 2 œ lim
1
xÄ1
xÄ1
4 9# #"
45
œ 1 œ 1
1
indeterminate form 00 : lim
œ
œ
"
#
x "Î#
x
84. (a) The limit leads to the indeterminate form 1_ . Let f(x) œ ˆ1 "x ‰ Ê ln f(x) œ x ln ˆ1 x" ‰ Ê x lim
ln f(x)
Ä_
œ x lim
Ä_
ln ˆ1 "x ‰
ˆ "x ‰
œ x lim
Ä_
ln a1 x " b
x "
#
œ x lim
Ä_
Š 1 xx " ‹
x #
œ x lim
Ä_
"
1 ˆ "x ‰
œ
"
10
œ1
ˆ1 "x ‰x œ lim f(x) œ lim eln fÐxÑ œ e" œ e
Ê x lim
Ä_
xÄ_
xÄ_
ˆ1 "x ‰x
(b)
x
10
100
1000
10,000
100,000
2.5937424601
2.70481382942
2.71692393224
2.71814592683
2.71826823717
Both functions have limits as x approaches
infinity. The function f has a maximum but
no minimum while g has no extrema. The limit
of f(x) leads to the indeterminate form 1_ .
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
423
424
Chapter 7 Transcendental Functions
" ‰x
x#
(c) Let f(x) œ ˆ1 Ê ln f(x) œ x ln a1 x# b
c#
Š
2x
$
#‹
#
ln a1 x b
1x
2x
Ê x lim
ln f(x) œ x lim
œ x lim
œ x lim
œ x lim
x "
Ä_
Ä_
Ä _ x #
Ä _ ax $ x b
Ä_
x
"
ln
f
Ð
x
Ñ
!
ˆ1 x# ‰ œ lim f(x) œ lim e
Therefore x lim
e
1
œ
œ
Ä_
xÄ_
xÄ_
ln a1 rkc" b
k "
k
85. Let f(k) œ ˆ1 kr ‰ Ê ln f(k) œ
œ lim
œ lim
rk
k Ä _ kr
r
kÄ_ 1
86. (a) y œ x1Îx Ê ln y œ
œ r. Therefore
Ê
ln x
x
w
y
y
œ x lim
Ä_
4
6x
œ 0.
#
Š 1 rkrk " ‹
ln a1 rk " b
œ lim
œ lim 1 rrk "
"
k
k #
kÄ_
kÄ_
kÄ_
r ‰k
ln fÐkÑ
ˆ
lim 1 k œ lim f(k) œ lim e
œ er .
kÄ_
kÄ_
kÄ_
Ê
lim
ˆ x" ‰ (x) ln x
x#
œ
4x
a3x# 1b
Ê yw œ ˆ 1 x#ln x ‰ ax1Îx b . The sign pattern is
yw œ ± ± which indicates a maximum value of y œ e1Îe when x œ e
e
!
w
ˆ " ‰ ax# b 2x ln x
#
#
(b) y œ x1Îx Ê ln y œ lnx#x Ê yy œ x
Ê yw œ ˆ 1 x2$ln x ‰ ax1Îx b . The sign pattern is
x%
yw œ ± ± which indicates a maximum of y œ e1Î2e when x œ Èe
!
Èe
(c) y œ x1Îx Ê ln y œ
ˆ "x ‰ axn b (ln x) ˆnxnc1 ‰
x2n
ln x
xn
œ
Ê yw œ
xnc1 (1 n ln x)
x2n
† x1Îx . The sign pattern is
n
yw œ ± ± which indicates a maximum of y œ e1Îne when x œ È
e
n
!
È
e
n
n
ˆeln x ‰1Îx œ lim eÐln xÑÎxn œ exp Š lim
(d) x lim
x1Îx œ x lim
Ä_
Ä_
xÄ_
xÄ_
n
87. (a) y œ x tanˆ 1x ‰, lim ˆx tanˆ 1x ‰‰ œ lim Š
xÄ
œ lim Š
xÄ
_
tanˆ 1x ‰
1
x
_
_
xÄ
‹ œ lim xÄ_
sec2 ˆ 1x ‰Š x12 ‹
Š x12 ‹
tanˆ 1x ‰
1
x
‹ œ lim xÄ_
n
ln x
xn ‹
sec2 ˆ 1x ‰Š x12 ‹
Š x12 ‹
ˆ " ‰‹ œ e! œ 1
œ exp Šx lim
Ä _ nxn
sec2 ˆ 1x ‰ œ 1; lim ˆx tanˆ 1x ‰‰
œ xlim
Ä_
xÄ_
lim sec2 ˆ 1x ‰ œ 1 Ê the horizontal asymptote is y œ 1 as x Ä _ and as
œ xÄ
_
x Ä _.
(b) y œ
3x e2x
2x e3x ,
e
lim Š 3x
2x e3x ‹ œ
2x
_
xÄ
2e2x
lim Š 32 3e3x ‹
xÄ
œ
_
œ
3
2
h Ä0
2x
_
xÄ
2x
lim Š 4e
9e3x ‹ œ
_
xÄ
3x e
lim ˆ 9e4x ‰ œ 0; lim Š 2x
e3x ‹
2x
_
_
xÄ
xÄ
Ê the horizontal asymptotes are y œ 0 as x Ä _ and y œ
lim fa0 hhb fa0b œ lim e
88. f w a0b œ
2e
lim Š 32 3e3x ‹ œ
h Ä0
c1Îh2 0
h
œ lim e
h Ä0
1Îh2
h
1
3
2
as x Ä _.
1
œ limŠ e1hÎh2 ‹ œ lim 1Îh2 h2 2 œ limŠ 2e1hÎh2 ‹
Š 3 ‹
h Ä0
h Ä0 e
h Ä0
h
œ limŠ h2 e1Îh ‹ œ 0
2
h Ä0
89. (a) We should assign the value 1 to f(x) œ (sin x)x to
make it continuous at x œ 0.
(b) ln f(x) œ x ln (sin x) œ
œ lim
x
#
x Ä 0 tan x
œ lim
ln (sin x)
ˆ "x ‰
2x
#
x Ä 0 sec x
Ê
lim ln f(x) œ lim b
x Ä !b
xÄ!
ln (sin x)
ˆ "x ‰
œ lim b
xÄ!
ˆ sin" x ‰ (cos x)
Š x"# ‹
œ 0 Ê lim f(x) œ e! œ 1
xÄ0
(c) The maximum value of f(x) is close to 1 near the point x ¸ 1.55 (see the graph in part (a)).
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 7.6 Inverse Trigonometric Functions
(d) The root in question is near 1.57.
90. (a) When sin x 0 there are gaps in the sketch. The width
of each gap is 1.
(b) Let f(x) œ (sin x)tan x Ê ln f(x) œ (tan x) ln (sin x)
Ê
œ
ˆ sin" x ‰ (cos x)
csc# x
lim
x Ä 1 Î2 c
Ê
ln f(x) œ
lim
x Ä 1 Î2 c
lim
œ
ln (sin x)
cot x
cos x
lim
x Ä 1Î2c ( csc x)
œ0
f(x) œ e! œ 1. Similarly,
lim
x Ä 1 Î2 c
x Ä 1 Î2 b
lim
x Ä 1 Î2 c
f(x) œ e! œ 1. Therefore,
lim
x Ä 1 Î2
f(x) œ 1.
(c) From the graph in part (b) we have a minimum of about 0.665 at x ¸ 0.47 and the maximum is about 1.491 at
x ¸ 2.66.
7.6 INVERSE TRIGONOMETRIC FUNCTIONS
1. (a)
1
4
(b) 13
3. (a) 16
(c)
1
6
(b)
1
4
(c) 13
2. (a) 14
(b)
1
3
(c) 16
4. (a)
1
6
(b) 14
(c)
1
3
(c)
1
6
(c)
21
3
5. (a)
1
3
(b)
31
4
(c)
1
6
6. (a)
1
4
(b) 13
7. (a)
31
4
(b)
1
6
(c)
21
3
8. (a)
31
4
(b)
9. sin Šcos"
È2
# ‹
œ sin ˆ 14 ‰ œ
"
È2
11. tan ˆsin" ˆ "# ‰‰ œ tan ˆ 16 ‰ œ È"3
1
6
10. sec ˆcos" #" ‰ œ sec ˆ 13 ‰ œ 2
12. cot Šsin" Š
È3
# ‹‹
œ cot ˆ 13 ‰ œ È"3
sin" x œ
1
#
14.
15. x lim
tan" x œ
Ä_
1
#
16. x Ä
lim
tan" x œ 1#
_
17. x lim
sec" x œ
Ä_
1
#
18. x Ä
lim
sec" x œ x Ä
lim
cos" ˆ "x ‰ œ
_
_
13.
lim
x Ä 1c
19. x lim
csc" x œ x lim
sin" ˆ "x ‰ œ 0
Ä_
Ä_
lim
x Ä 1 b
cos" x œ 1
1
#
20. x Ä
lim
csc" x œ x Ä
lim
sin" ˆ "x ‰ œ 0
_
_
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
425
426
Chapter 7 Transcendental Functions
21. y œ cos" ax# b Ê
dy
dx
œ
23. y œ sin" È2t Ê
dy
dt
œ
È2
#
Ê1 ŠÈ2t‹
25. y œ sec" (2s 1) Ê
dy
ds
26. y œ sec" 5s Ê
5
k5sk È(5s)# 1
dy
ds
œ
27. y œ csc" ax# 1b Ê
28. y œ csc" ˆ x# ‰ Ê
dy
dx
œ
œ
2x
È 1 x%
22. y œ cos" ˆ "x ‰ œ sec" x Ê
œ
È2
È1 2t#
24. y œ sin" (1 t) Ê
œ
dy
dt
#
30. y œ sin" ˆ t3# ‰ œ csc" Š t3 ‹ Ê
œ
"
kx k É x
#
œ
34. y œ tan" (ln x) Ê
dy
dx
œ
35. y œ csc" aet b Ê
36. y œ cos" aet b Ê
dy
dt
œ
dy
dt
Š
"
1
x#
‹
tanc" x
4
ˆ 2t ‰
œ
œ
ˆ "x ‰
1 (ln x)#
3
Š "# ‹ tc"Î#
1 at"Î# b
dy
dt
#
œ
s#
È 1 s#
œ
et
e t
É 1 ae t b #
"
È 1 s#
œ
œ
"
k2s 1k Ès# s
2t
%
t# É t 9
9
œ
6
t Èt% 9
Š "# ‹ (t 1)c"Î#
1 c(t 1)"Î# d
#
œ
"
2Èt 1 (1 t 1)
œ
"
2tÈt 1
"
x c1 (ln x)# d
œ
"
Èe2t 1
œ
e t
È1 e
œ È1 s# 38. y œ Ès# 1 sec" s œ as# 1b
"
È2t t#
"
#Èt(1 t)
2t
"Î#
37. y œ sÈ1 s# cos" s œ s a1 s# b cos" s Ê
œ È1 s# œ
"
atan " xb a1x# b
ke t k É a e t b # 1
œ
"
È1 (1 t)#
œ
2
kx k È x # 4
3
œ #
œ
# #
¹ t ¹ ÊŠ t ‹ 1
32. y œ cot" Èt 1 œ cot" (t 1)"Î# Ê
dy
dx
"
kx k È x # 1
"
È1 t#
dy
dt
dy
dt
œ
œ
2x
ax# 1b Èx% 2x#
œ
4
3
33. y œ ln atan" xb Ê
œ
2x
kx # 1 k É a x # 1 b # 1
¸ x# ¸ Ɉ x# ‰# 1
31. y œ cot" Èt œ cot" t"Î# Ê
2
k2s 1k È4s# 4s
dy
dt
dy
dx
"
ksk È25s# 1
œ
Š "# ‹
29. y œ sec" ˆ "t ‰ œ cos" t Ê
œ
2
k2s 1k È(2s 1)# 1
œ
dy
dx
œ
2x
É 1 ax # b#
"Î#
s# 1
È 1 s#
sec" s Ê
dy
ds
œ
dy
dx
œ a1 s# b
1 s# s# 1
È 1 s#
"Î#
œ
œ ˆ "# ‰ as# 1b
s ˆ "# ‰ a1 s# b
"Î#
(2s) "
È 1 s#
2s#
È 1 s#
"Î#
(2s) "
ks k È s # 1
œ
s
È s# 1
s ks k 1
ks k È s # 1
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
"
k sk È s # 1
Section 7.6 Inverse Trigonometric Functions
"Î#
39. y œ tan" Èx# 1 csc" x œ tan" ax# 1b csc" x Ê
œ
"
x È x# 1
"
kx k È x # 1
1
#
tan" ax" b tan" x Ê
41. y œ x sin" x È1 x# œ x sin" x a1 x# b
x
È 1 x#
x
È 1 x#
'È"
44.
'È
"
1 4x#
46.
' 9 "3x
47.
'
48.
'
dx œ '
#
dx œ
"
È2
"
#
œ
49.
'01 È4 ds
50.
'03
4 s#
È2Î4
"
#
ŠÈ17‹ x#
'
"
3
œ'
dx
xÈ25x# 2
dx
xÈ5x# 4
du
uÈ u# 2
œ'
du
uÈ u# 4
dx œ
"
kx k È x # 1
"Î#
Ê
dy
dx
œ0
xc#
1 ax " b #
œ sin" x x Š È
dy
dx
"
1 x#
"
‹
1 x#
œ
"
x# 1
"
1 x#
ˆ #" ‰ a1 x# b
œ0
"Î#
(2x)
2x
4
x#
tan" ˆ #x ‰ x –
Š "# ‹
1 ˆ #x ‰
#
—œ
2x
x# 4
tan" ˆ #x ‰ 2x
4 x#
œ tan" ˆ #x ‰
"
#
'03
'02
œ ’ È"2 †
"
È8
œ
È2
1 u #
dx œ
"
È17
tan"
"
3È 3
"
È2
, where u œ 2x and du œ 2 dx
x
È17
C
tan" Š Èx3 ‹ C œ
È3
9
tan" Š Èx3 ‹ C
5x
sec" ¹ È
¹C
2
, where u œ È5x and du œ È5 dx
"
#
sec" ¹
"
œ 4 sin" #s ‘ ! œ 4 ˆsin"
œ
' È du
"
#
, where u œ 5x and du œ 5 dx
sec" ¸ u# ¸ C œ
"
È2
#
œ
dy
dx
dx œ
#
sec" ¹ Èu2 ¹ C œ
ds
È9 4s#
'02 8 dt2t
(2x)
#
1 ’ax# 1b"Î# “
sin" (2x) C
"
#
ŠÈ3‹ x#
3
œ "# sin" u3 ‘ 0
51.
"
#
sin" u C œ
#
œ
' È1 2(2x)
"
#
dx œ
"
#
œ
' 17 " x
"Î#
dx œ sin" ˆ x3 ‰ C
9 x#
45.
Š "# ‹ ax# 1b
œ sin" x
42. y œ ln ax# 4b x tan" ˆ x# ‰ Ê
43.
œ
œ 0, for x 1
40. y œ cot" ˆ x" ‰ tan" x œ
œ sin" x dy
dx
È2Î4
È2Î2
du
8 u#
tan"
du
È 9 u#
œ
"
#
È5x
# ¹
"
#
C
sin" 0‰ œ 4 ˆ 16 0‰ œ
21
3
, where u œ 2s and du œ 2 ds; s œ 0 Ê u œ 0, s œ
Šsin"
È2
#
sin" 0‹ œ
"
#
ˆ 14 0‰ œ
3È 2
4
Ê uœ
3È 2
#
1
8
, where u œ È2t and du œ È2 dt; t œ 0 Ê u œ 0, t œ 2 Ê u œ 2È2
u
È8 “
#È #
!
œ
"
4
Štan"
2È 2
È8
tan" 0‹ œ
"
4
atan" 1 tan" 0b œ
"
4
ˆ 14 0‰ œ
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
1
16
427
428
52.
Chapter 7 Transcendental Functions
'c22
œ ’ È"3 †
53.
È2Î2
'cc1
È
'c22È33
"
È3
œ
dt
4 3t#
"
#
tan" u# “
È2
œ 'c2
dy
yÈ4y# 1
È #
œ csec" kukd #
54.
È
#
œ
56.
œ
#
' È4 6 (rdr 1)
' 2 (xdx 1)
œ
58.
œ
59.
œ'
"
È2
"
3
"
#
"
#
†
u
È2
'c11ÎÎ22 12cos(sin) d)))
#
'11ÎÎ64
sec" ¸ u# ¸
'0ln
È3
¸ 5u ¸
œ2'
1
"
3
œ 'È3
1
È3
œ '1
œ 11#
1
3
È2
3
Ê u œ È 2
œ 11#
sin" 2(r 1) C
, where u œ r 1 and du œ dr
"
È2
tan" Š xÈ1 ‹ C
2
tan" (3x 1) C
du
, where u œ x
uÈu# 25
"
C œ 5 sec" ¸ x5 3 ¸ C
1 and du œ 2 dx
3 and du œ dx
, where u œ sin ) and du œ cos ) d); ) œ 1# Ê u œ ", ) œ
du
1 u# ,
where u œ cot x and du œ csc# x dx; x œ
œ tan" 1 tan" È3 œ 14 du
1 u#
È$
œ ctan" ud "
1
4
Ê u œ È 2
1
#
Êuœ"
œ 2 atan" 1 tan" (1)b œ 2 14 ˆ 14 ‰‘ œ 1
"
c tan" ud È$
ex dx
1 e2x
1
3
, where u œ 3y and du œ 3 dy; y œ 32 Ê u œ 2, y œ du
, where u œ 2x uÈ u# 4
"
C œ 4 sec" ¸ 2x # 1 ¸ C
du
1 u#
1
"
tan" ud "
csc# x dx
1 (cot x)#
È2
#
, where u œ 3x 1 and du œ 3 dx
#
uCœ
sec
1
4
1
3È 3
, where u œ 2(r 1) and du œ 2 dr
Cœ
' 1 duu
"
13 ˆ 13 ‰‘ œ
, where u œ x 1 and du œ dx
#
"
5
"
#È 3
, where u œ 2y and du œ 2 dy; y œ 1 Ê u œ 2, y œ ' (x 3)È(xdx 3) 25 œ '
œ
63.
"
3
"
tan
œ c2
62.
tan"
’tan" È3 tan" ŠÈ3‹“ œ
C œ 6 sin" ˆ r# 1 ‰ C
du
2 u#
œ
#
du
È 4 u#
#
œ
61.
u
#
3
#
' (2x 1)Èdx(2x 1) 4 œ "# '
œ
60.
du
È 1 u#
œ6'
#
#
' 1 (3xdx 1)
'
"
#È 3
œ sec" ¹È2¹ sec" k2k œ
sin" u C œ
3
#
œ 6 sin"
57.
3
#
œ
du
uÈ u# 1
2
È #
' È1 34(rdr 1)
#È$
œ sec" ¹È2¹ sec" k2k œ
œ csec" kukd #
55.
#È $
du
uÈ u# 1
È2
'c2Î32Î3 yÈ9ydy 1 œ '
, where u œ È3t and du œ È3 dt; t œ 2 Ê u œ 2È3, t œ 2 Ê u œ 2È3
du
4 u#
1
3
œ
1
6
Ê u œ È3 , x œ
1
4
Ê uœ1
1
1#
, where u œ ex and du œ ex dx; x œ 0 Ê u œ 1, x œ ln È3 Ê u œ È3
œ tan" È3 tan" 1 œ
1
3
1
4
œ
1
1#
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 7.6 Inverse Trigonometric Functions
64.
'1e
1Î%
œ c4
65.
' Èy1 dy y
œ
%
"
#
œ
66.
œ 4'0
1Î4
du
1 u# , where u œ ln t and
1Î%
tan" ud ! œ 4 ˆtan" 14 tan" 0‰
4 dt
t a1 ln# tb
"
#
' È du
1 u#
"
t
dt; t œ 1 Ê u œ 0, t œ e1Î4 Ê u œ
œ 4 tan"
1
4
1
4
, where u œ y# and du œ 2y dy
sin" u C œ
' Èsec1 ytandy y œ ' È du
#
#
du œ
429
"
#
sin" y# C
, where u œ tan y and du œ sec# y dy
1 u#
œ sin" u C œ sin" (tan y) C
67.
'È
68.
' È dx
69.
'01
œ'
dx
x# 4x 3
2x x#
œ'
dx
È1 ax# 4x 4b
œ'
œ'
dx
È1 ax# 2x 1b
dx
È1 (x 2)#
dx
È1 (x 1)#
œ sin" (x 2) C
œ sin" (x 1) C
œ 6 '1 È4 at#dt 2t 1b œ 6 '1 È2# dt(t 1)# œ 6 sin" ˆ t # 1 ‰‘ "
0
6 dt
È3 2t t#
0
!
œ 6 sin" ˆ "# ‰ sin" 0‘ œ 6 ˆ 16 0‰ œ 1
70.
'11Î2
6 dt
È3 4t 4t#
dt
1 ‰‘
œ 3'1Î2 È4 a4t2#dt 4t 1b œ 3 '1Î2 È2# 2 (2t
œ 3 sin" ˆ 2t #
"Î#
1)#
1
1
œ 3 sin" ˆ "# ‰ sin" 0‘ œ 3 ˆ 16 0‰ œ
"
1
#
71.
' y dy2y 5 œ ' 4 y dy 2y 1 œ ' # (ydy 1)
72.
' y 6ydy 10 œ ' 1 ay dy 6y 9b œ ' 1 (ydy 3)
73.
'12 x 8 2xdx 2 œ 8'12 1 ax dx 2x 1b œ 8'12 1 (xdx 1)
74.
'24
75.
#
#
#
#
#
#
œ
#
#
œ tan" (y 3) C
#
2 dx
x# 6x 10
œ 2'2
4
dx
1 ax# 6x 9b
œ 2'2
4
tan" ˆ y # 1 ‰ C
"
#
#
œ 8 ctan" (x 1)d " œ 8 atan" 1 tan" 0b œ 8 ˆ 14 0‰ œ 21
#
dx
1 (x 3)#
%
œ 2 ctan" (x 3)d # œ 2 ctan" 1 tan" (1)d œ 2 14 ˆ 14 ‰‘ œ 1
' xx 44 dx œ ' x x 4 dx ' x 4 4 dx; ' x x 4 dx œ #" ' u1 du where u œ x# 4 Ê du œ 2x dx Ê #" du œ x dx
Ê ' xx 44 dx œ 12 lnax# 4b 2 tan1 ˆ x2 ‰ C
#
#
#
#
#
76.
' t t 6t # 10 dt œ ' at t3b # 1 dt ’Let w œ t 3 Ê w 3 œ t Ê dw œ dt“ Ä ' ww 1" dw œ ' w w " dw ' w 1 " dw;
#
#
#
#
#
' w w " dw œ #" ' 1u du where u œ w# 1 Ê du œ 2w dw Ê #" du œ w dw Ê ' w w " dw ' w 1 " dw
#
œ
77.
#
1
#
2 lnaw
1
1b tan awb C œ
1 ˆ
2 ln at
1
3b 1‰ tan at 3b C œ
#
1
#
2 lnat
6t 10b tan1 at 3b C
' x x 2x 9 1 dx œ ' a1 2xx 109 bdx œ ' dx ' x 2x 9 dx 10' x 1 9 dx; ' x 2x 9 dx œ ' u1 du where u œ x# 9
1 ˆ x ‰
Ê du œ 2x dx Ê ' dx ' x 2x 9 dx 10' x 1 9 dx œ x lnax# 9b 10
3 tan
3 C
#
#
#
#
#
78.
#
#
#
#
' t 2tt 13t 4 dt œ ' at 2 2tt 12 bdt œ ' at 2bdt ' t 2t 1 dt 2' t 1 1 dt; ' t 2t 1 dt œ ' u1 du where u œ t# 1
1
1 2
#
1
Ê du œ 2t dt Ê ' at 2bdt ' t 2t 1 dt 2' t 1 dt œ 2 t 2t lnat 1b 2 tan atb C
3
2
#
#
#
#
#
#
#
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
430
79.
Chapter 7 Transcendental Functions
'
œ sec
80.
'
sinc" x
' Èe
1 x#
cosc" x
' Èe
1 x#
"x
1x
84.
du
,
uÈ u# 1
where u œ x 1 and du œ dx
œ'
"
uÈ u# 1
du, where u œ x 2 and du œ dx
kx 2k C
dx
È 1 x#
C
c" x
dx
È 1 x#
C
u# du, where u œ sin" x and du œ
$
asin " xb
3
Cœ
œ'
dx
(x 2)È(x 2)# 1
dx
È 1 x#
C
' È1tan cx x dx œ ' u"Î# du, where u œ tan" x and du œ 1 dxx
"
#
œ
85.
u$
3
œ'
dx œ ' eu du, where u œ cos" x and du œ
' aÈsin " xb## dx œ '
œ
dx
(x 2)Èx# 4x 4 1
"
dx œ ' eu du, where u œ sin" x and du œ
œ eu C œ ecos
83.
dx
(x 1)È(x 1)# 1
kx 1k C
kuk C œ sec
œ eu C œ esin
82.
œ'
dx
(x 2)Èx# 4x 3
"
œ'
dx
(x 1)Èx# 2x 1 1
"
kuk C œ sec
œ sec
81.
œ'
dx
(x 1)Èx# 2x
"
2
3
u$Î# C œ
2
3
atan" xb
' atanc" yb"a1 y#b dy œ '
$Î#
Cœ
"
‹
y#
tan " y
Š
1
2
3
dy œ '
#
$
Éatan" xb C
"
u
du, where u œ tan" y and du œ
dy
1 y#
œ ln kuk C œ ln ktan" yk C
86.
'
"
"
asinc" yb È1 y#
dy œ
' Ésin " y# dy œ ' u" du, where u œ sin" y and du œ È1dy y#
1
y
œ ln kuk C œ ln ksin" yk C
87.
'È22
sec# asec " xb
xÈ x# 1
dx œ '1Î4 sec# u du, where u œ sec" x and du œ
1Î3
1Î$
œ ctan ud 1Î4 œ tan
88.
'22ÎÈ3
cos asecc" xb
xÈ x# 1
"
' eÈsinc e
x
" x
1 e#x
tanc1 Èx
‹
3
1
6
sin
Èxax 1b ”ˆtanc1 Èx‰2 9•
œ 23 tan1 Š
90.
1
3
; x œ È2 Ê u œ
1
4
,xœ2 Ê uœ
1
3
dx
xÈ x# 1
;xœ
1
6
,xœ2 Ê uœ
1
3
œ È3 1
1Î3
1Î$
'
1
4
tan
dx œ '1Î6 cos u du, where u œ sec" x and du œ
œ csin ud 1Î' œ sin
89.
1
3
dx
xÈ x# 1
œ
2
È3
Ê uœ
È3 "
#
dx œ 2'
"
u2 9 du
where u œ tan1 Èx Ê du œ
1
1
dx
2
1 + ˆÈ x ‰ 2 È x
Ê 2du œ
C
dx œ ' u du where u œ sin" ex Ê du œ
1
x
È1 e#x e dx
œ "# asin" ex b C
2
sinc" 5x
x
xÄ0
91. lim
œ lim
ŠÈ
xÄ0
5
1
25x#
1
‹
œ5
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
1
dx
a 1 x bÈ x
Section 7.6 Inverse Trigonometric Functions
92.
È x# 1
sec " x
lim b
xÄ1
"Î#
ax # 1 b
sec " x
œ lim b
xÄ1
tanc" a2x " b
x "
93. x lim
x tan" ˆ 2x ‰ œ x lim
Ä_
Ä_
2 tanc" 3x#
7x#
xÄ0
94. lim
95.
96.
lim
x Ä0
lim
tanc1 x2
x sinc1 x
e tanc1 ex
e2x x
98.
œ
x
’tanc1 ˆÈx‰“
ˆe2x 3‰
4ae2x 1b2 —
x Ä0
xÈ x 1
sinc1 ˆx2 ‰
asinc1 xb2
œ limb Š
x Ä0
x Ä0
x
1 x#
œ lim
Èx 1
‹œ
2x
1 c x2
"
#
ln a1 x# b "
x a1 x # b
tanc" x
x
tan " x
x# ‹
œ x lim
Ä_
x #
œ
œ lim b x kxk œ 1
xÄ1
6
7
2Š3x4 1‹
Š1
Š1
Ñ
2 0 1
12
Ó
œ
c0 2 œ
2
3Î2
1 0 3Î2
Ò
a
2
x4 ‹
x2
x2‹
a
b
b
œ limb
x Ä0
È
œ1
ex tanc1 ex 2
œ lim
e2x Še2x b 3‹
2
Še2x b 1‹
4e2x
xÄ_
ˆ1 3ec2x ‰
4aex ecx b2 —
tanc1 ˆÈx‰
xa1 b xb
3x b 2
2Èx b 1
2
2
2e2x
Še2x b 1‹
4e2x
x
œ2
2
14x #
œ00œ0
È bx
2tanc1 ˆÈx‰
œ limb Š a3x 2b ÈxÈx 1 ‹ œ limb 12x2 xb113x
b2 x Ä0
x Ä0
1
a
b
2ÈxÈx
b1
œ1
È
œ limb Œ 2asinc11xcbx4 1
È
x Ä0
œ
#
#‹
2x
1 4x
xÄ_
c1
2
2
1
2x
xÄ_
x
2Èx b 1
x#
(2x)
ex tanc1 ex e2xe b 1 œ lim – tan4ex e È 1 x2 È 1 x2
‹
1 x2 xÈ1 x2 sinc1 x
99. If y œ ln x œ Š "x 6
È
Š
%
x Ä 0 7 a1 9x b
tanc1 ˆÈx‰ Èxa11 b xb
œ limb
2
lim Š 2
xÄ0b a12x 13x 2bÈx 1
limb
œ lim
ex tanc1 ex e2xe b 1
lim
2e2x 1
xÄ_
2
œ
x
k k
œ x lim
Ä_
2x
c1
x Ä0
Œ
"Î#
"
Î
œ lim Œ x 1 sin1 x œ lim Ð
x Ä0
x Ä0
È1 x2
Ï
xÄ_
limb
12x
‹
9x%
14x
xÄ0
œ lim – tan4ex e 97.
1
Š "# ‹ ax# 1b
2x
1 x4
x
xÄ_
Š
œ lim
œ lim b
xÄ1
431
1
1
Š c1 x
‹ œ limb Œ sinc1 x†
œ xlim
2
Ä0b sin x È1 x
x Ä0
1
È1 b x2 È1 c x2 È1 x2
x
1
œ
œ1
C, then dy œ – x" dx œ
x
1 x#
Š
x
1
x a1 x# b x$ x atan " xba1 x# b
x # a1 x # b
x#
‹ tan " x
x#
dx œ
— dx
tan " x
x#
dx,
which verifies the formula
100. If y œ
x%
4
cos" 5x 5
4
'È
x%
1 25x#
%
dx, then dy œ ’x$ cos" 5x Š x4 ‹ Š È
5
‹
1 25x#
45 Š È
x%
‹“
1 25x#
dx
œ ax$ cos" 5xb dx, which verifies the formula
#
101. If y œ x asin" xb 2x 2È1 x# sin" x C, then
c"
#
dy œ ’asin" xb 2x asin xb 2 2x sin" x 2È1 x# Š
È 1 x#
È 1 x#
"
È 1 x # ‹“
#
dx œ asin" xb dx, which verifies
the formula
102. If y œ x ln aa# x# b 2x 2a tan" ˆ xa ‰ C, then dy œ –ln aa# x# b #
2x#
a# x#
2
2
#
1 Š x# ‹ —
dx
a
#
x
#
#
œ ’ln aa# x# b 2 Š aa# x# ‹ 2“ dx œ ln aa x b dx, which verifies the formula
103.
dy
dx
œ
"
È 1 x#
Ê dy œ
dx
È 1 x#
Ê y œ sin" x C; x œ 0 and y œ 0 Ê 0 œ sin" 0 C Ê C œ 0 Ê y œ sin" x
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
432
104.
Chapter 7 Transcendental Functions
dy
dx
œ
"
x# 1
1 Ê dy œ ˆ 1 " x# 1‰ dx Ê y œ tan" (x) x C; x œ 0 and y œ 1 Ê 1 œ tan" 0 0 C
Ê C œ 1 Ê y œ tan" (x) x 1
105.
dy
dx
œ
œ1
106.
dy
dx
œ
"
Ê dy œ Èdx#
Ê y œ sec"
xÈ x# 1
x x 1
13 œ 231 Ê y œ sec" (x) 231 , x 1
"
1 x#
2
È 1 x#
Ê dy œ Š 1 " x# kxk C; x œ 2 and y œ 1 Ê 1 œ sec" 2 C Ê C œ 1 sec" 2
2
È 1 x# ‹
dx Ê y œ tan" x 2 sin" x C; x œ 0 and y œ 2
Ê 2 œ tan" 0 2 sin" 0 C Ê C œ 2 Ê y œ tan" x 2 sin" x 2
107. (a) The angle ! is the large angle between the wall and the right end of the blackboard minus the small angle between
x ‰
the left end of the blackboard and the wall Ê ! œ cot" ˆ 15
cot" ˆ 3x ‰ .
(b)
d!
dt
œ
1
15
x ‰
1 ˆ 15
2
1
3
1 ˆ 3x ‰
2
œ 22515 x2 œ
3
9 x2
540 12x2
d!
a225 x2 ba9 x2 b ; dt
œ 0 Ê 540 12x2 œ 0 Ê x œ „ 3È5
È
È
Since x 0, consider only x œ 3È5 Ê !Š3È5‹ œ cot" Š 3155 ‹ cot" Š 3 3 5 ‹ ¸ 0.729728 ¸ 41.8103‰ . Using
the first derivative test,
d!
dt ¹xœ1
œ
132
565
0 and
d!
dt ¹xœ10
132
‰
œ (!)
5 0 Ê local maximum of 41.8103 when
x œ 3È5 ¸ 6.7082 ft.
108. V œ 1'0 c2# (sec y)# d dy œ 1 c4y tan yd !
1Î3
1Î$
œ 1 Š 431 È3‹
109. V œ ˆ "3 ‰ 1r# h œ ˆ 3" ‰ 1(3 sin ))# (3 cos )) œ 91 acos ) cos$ )b, where 0 Ÿ ) Ÿ
Ê
dV
d)
œ 91(sin )) a1 3 cos# )b œ ! Ê sin ) œ 0 or cos ) œ „
"
È3
1
#
Ê the critical points are: 0, cos" Š È"3 ‹ , and
cos" Š È"3 ‹ ; but cos" Š È"3 ‹ is not in the domain. When ) œ 0, we have a minimum and when ) œ cos" Š È"3 ‹
¸ 54.7°, we have a maximum volume.
‰
110. 65° (90° " ) (90° !) œ 180° Ê ! œ 65° " œ 65° tan" ˆ 21
50 ¸ 65° 22.78° ¸ 42.22°
111. Take each square as a unit square. From the diagram we have the following: the smallest angle ! has a
tangent of 1 Ê ! œ tan" 1; the middle angle " has a tangent of 2 Ê " œ tan" 2; and the largest angle #
has a tangent of 3 Ê # œ tan" 3. The sum of these three angles is 1 Ê ! " # œ 1
Ê tan" 1 tan" 2 tan" 3 œ 1.
112. (a) From the symmetry of the diagram, we see that 1 sec" x is the vertical distance from the graph of y œ sec" x to
the line y œ 1 and this distance is the same as the height of y œ sec" x above the x-axis at x;
i.e., 1 sec" x œ sec" (x).
(b) cos" (x) œ 1 cos" x, where 1 Ÿ x Ÿ 1 Ê cos" ˆ "x ‰ œ 1 cos" ˆ "x ‰, where x 1 or x Ÿ 1
Ê sec" (x) œ 1 sec" x
113. sin" (1) cos" (1) œ
1
#
0œ
1
#
; sin" (0) cos" (0) œ 0 1
#
œ
1
#
; and sin" (1) cos" (1) œ 1# 1 œ 1# .
If x − ("ß 0) and x œ a, then sin" (x) cos" (x) œ sin" (a) cos" (a) œ sin" a a1 cos" ab
œ 1 asin" a cos" ab œ 1 1# œ 1# from Equations (3) and (4) in the text.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 7.6 Inverse Trigonometric Functions
Ê tan ! œ x and tan " œ
114.
1
#
115. csc" u œ
sec" u Ê
acsc" ub œ
d
dx
d
dx
"
x
Ê
1
#
"
x
œ ! " œ tan" x tan"
ˆ 1# sec" u‰ œ 0 du
dx
ku k È u # 1
œ
433
.
du
dx
ku k È u # 1
, ku k 1
116. y œ tan" x Ê tan y œ x Ê
Ê asec# yb
œ
"
1 x#
dy
dx
œ1 Ê
d
d
dx (tan y) œ dx (x)
"
"
#
sec# y œ È
Š 1 x# ‹
œ
dy
dx
, as indicated by the triangle
117. f(x) œ sec x Ê f w (x) œ sec x tan x Ê
df "
dx ¹xœb
œ
"
df
dx ¹x œ f
"abb
œ
"
secasec " bbtanasec " bb
Since the slope of sec" x is always positive, we the right sign by writing
1
#
118. cot" u œ
tan" u Ê
acot" ub œ
d
dx
d
dx
du
dx
ˆ 1# tan" u‰ œ 0 119. The functions f and g have the same derivative (for x
xœ
"
.
lx l È x # "
du
1 u#
0), namely
d
"
dx sec
"
.
b Š„ È b # " ‹
œ
œ 1 dxu#
"
Èx (x 1)
. The functions therefore differ
by a constant. To identify the constant we can set x equal to 0 in the equation f(x) œ g(x) C, obtaining
1‰
" È
sin" (1) œ 2 tan" (0) C Ê 1# œ 0 C Ê C œ 1# . For x 0, we have sin" ˆ xx x
1 œ 2 tan
120. The functions f and g have the same derivative for x 0, namely
"
1 x#
1
#
.
. The functions therefore differ by a
constant for x 0. To identify the constant we can set x equal to 1 in the equation f(x) œ g(x) C, obtaining
sin" Š È" ‹ œ tan" 1 C Ê
2
È3
1
4
œ
1
4
C Ê C œ 0. For x 0, we have sin"
È3
121. V œ 1 'cÈ3Î3 Š È1" x# ‹ dx œ 1 ' È3Î3
#
œ 1 13 ˆ 16 ‰‘ œ
r
È#
r ÎÈ 2
œ '0
œ
dy
dx
x
È r2 x 2 ;
(in other words, the length of
É r2 r x2 dx œ '
2
rÎÈ2
0
r
Èr2 x2 dx
œ r sin1 Š È12 ‹ 0 œ rˆ 14 ‰ œ
123. (a) A(x) œ
1
4
œ tan"
È3
dx œ 1 ctan" xd È3Î3 œ 1 ’tan" È3 tan" Š
"
x
.
È3
3 ‹“
1#
#
122. Consider y œ Èr2 x2 Ê
to x œ
"
1 x#
"
È x# 1
(diameter)# œ
1
4
"
1r
4 .
"
1 x#
dy
dx
is undefined at x œ r and x œ r, we will find the length from x œ 0
rÎÈ2
of a circle) Ê L œ '0
r ÎÈ 2
œ r sin1 ˆ xr ‰‘ 0
rÎÈ2
Ê1 Š È 2 x 2 ‹ dx œ '0
2
r x
É1 È
œ r sin1 Š rÎ r 2 ‹ r sin1 a0b
The total circumference of the circle is C œ 8L œ 8ˆ 14r ‰ œ 21 r.
#
’ È1" x# Š È1" x# ‹“ œ
œ 1 ctan" xd " œ (1)(2) ˆ 14 ‰ œ
(b) A(x) œ (edge)# œ ’ È
1
8
Since
1#
#
Š È
#
"
‹
“
#
1x
œ
4
1 x#
1
1 x#
Ê V œ 'a A(x) dx œ 'c1
b
1
1 dx
1 x#
Ê V œ 'a A(x) dx œ 'c1 14dxx#
b
1
"
œ 4 ctan" xd " œ 4 ctan" (1) tan" (1)d œ 4 14 ˆ 14 ‰‘ œ 21
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
x2
r2 x2 dx
434
Chapter 7 Transcendental Functions
124. (a) A(x) œ
1
4
È2Î2
œ 'cÈ2Î2
(b) A(x) œ
(diameter)# œ
1
È 1 x#
(diagonal)#
2
1
4
Š
2
%
È
1 x#
#
0‹ œ
1
4
ŠÈ
È2Î2
4
‹
1 x#
dx œ 1 csin" xd È2Î2 œ 1 ’sin" Š
œ
È
1
2
Š
2
%
È
1 x#
#
0‹ œ
2
È 1 x#
È2
# ‹
œ
1
È 1 x#
Ê V œ 'a A(x) dx
sin" Š
b
È2
# ‹“
œ 1 14 ˆ 14 ‰‘ œ
È2Î2
Ê V œ 'a A(x) dx œ 'cÈ2Î2
b
2
È 1 x#
1#
#
dx
2Î2
œ 2 csin" xd È2Î2 œ 2 ˆ 14 † 2‰ œ 1
125. (a) sec" 1.5 œ cos"
(c) cot" 2 œ
1
#
tan
"
1.5
"
" ‰
(b) csc" (1.5) œ sin" ˆ 1.5
¸ 0.72973
¸ 0.84107
2 ¸ 0.46365
126. (a) sec" (3) œ cos" ˆ "3 ‰ ¸ 1.91063
(c) cot" (2) œ
1
#
" ‰
(b) csc" 1.7 œ sin" ˆ 1.7
¸ 0.62887
tan" (2) ¸ 2.67795
127. (a) Domain: all real numbers except those having
the form 1# k1 where k is an integer.
Range: 1# y 1
#
(b) Domain: _ x _; Range: _ y _
The graph of y œ tan" (tan x) is periodic, the
graph of y œ tan atan" xb œ x for _ Ÿ x _.
128. (a) Domain: _ x _; Range: 1# Ÿ y Ÿ
1
#
(b) Domain: " Ÿ x Ÿ 1; Range: " Ÿ y Ÿ 1
The graph of y œ sin" (sin x) is periodic; the
graph of y œ sin asin" xb œ x for " Ÿ x Ÿ 1.
129. (a) Domain: _ x _; Range: 0 Ÿ y Ÿ 1
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 7.6 Inverse Trigonometric Functions
(b) Domain: 1 Ÿ x Ÿ 1; Range: " Ÿ y Ÿ 1
The graph of y œ cos" (cos x) is periodic; the
graph of y œ cos acos" xb œ x for " Ÿ x Ÿ 1.
130. Since the domain of sec" x is (_ß 1] ["ß _), we
have sec asec" xb œ x for kxk 1. The graph of
y œ sec asec" xb is the line y œ x with the open
line segment from ("ß ") to ("ß ") removed.
131. The graphs are identical for y œ 2 sin a2 tan" xb
œ 4 csin atan" xbd ccos atan" xbd œ 4 Š È
œ
4x
x# 1
x
‹ Š Èx#" 1 ‹
x# 1
from the triangle
132. The graphs are identical for y œ cos a2 sec" xb
œ cos# asec" xb sin# asec" xb œ
œ
2 x #
x#
"
x#
x# 1
x#
from the triangle
133. The values of f increase over the interval ["ß 1] because
f w 0, and the graph of f steepens as the values of f w
increase towards the ends of the interval. The graph of f
is concave down to the left of the origin where f ww 0,
and concave up to the right of the origin where f ww 0.
There is an inflection point at x œ 0 where f ww œ 0 and
f w has a local minimum value.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
435
436
Chapter 7 Transcendental Functions
134. The values of f increase throughout the interval (_ß _)
because f w 0, and they increase most rapidly near the
origin where the values of f w are relatively large. The
graph of f is concave up to the left of the origin where
f ww 0, and concave down to the right of the origin
where f ww 0. There is an inflection point at x œ 0
where f ww œ 0 and f w has a local maximum value.
7.7 HYPERBOLIC FUNCTIONS
#
1. sinh x œ 34 Ê cosh x œ È1 sinh# x œ É1 ˆ 34 ‰ œ É1 coth x œ
2. sinh x œ
sech x œ
3. cosh x œ
œ
8
17
"
tanh x
coth x œ
œ
, and csch x œ
4
5
Ê cosh x œ È1 sinh# x œ É1 4
3
"
cosh x
17
15
œ
13
5
3
5
, and csch x œ
"
sinh x
œ
16
9
"
sin x
œ É 25
9 œ
25
œ É 16
œ
5
4
œ
17
8
5
3
, tanh x œ
œ
15
17
, and csch x œ
"
sinh x
œ
É 144
, x 0 Ê sinh x œ Ècosh# x 1 œ É 169
25 1 œ
25 œ
"
tanh x
œ
13
12
5. 2 cosh (ln x) œ 2 Š e
6. sinh (2 ln x) œ
"
cosh x
, sech x œ
ln x
ecln x
‹
#
e2 ln x ec2 ln x
#
7. cosh 5x sinh 5x œ
œ
e5x e
#
9. (sinh x cosh x)% œ ˆ e
x
5x
ecx
#
œ
5
13
œ eln x "
eln x
#
eln x eln x
#
#
, and csch x œ
œ
e5x e
#
ex ecx ‰%
#
5x
œ 35 ,
sinh x
cosh x
œ
ˆ 43 ‰
ˆ 53 ‰
4
5
, coth x œ
"
tanh x
œ
5
4
8
15
, tanh x œ
sinh x
cosh x
œ
8 ‰
ˆ 15
ˆ 17
‰
15
œ
,
3
4
"
cosh x
, sech x œ
ˆ 34 ‰
ˆ 54 ‰
sinh x
cosh x
œ 34
#
"
tanh x
œ
, tanh x œ
17 ‰
289
64
, x 0 Ê sinh x œ Ècosh# x 1 œ Ɉ 15
1 œ É 225
1 œ É 225
œ
, coth x œ
4. cosh x œ
"
cosh x
œ 35 , sech x œ
9
16
œx
Šx# x"# ‹
#
"
sinh x
œ
15
8
12
5
, tanh x œ
sinh x
cosh x
œ
ˆ 12
‰
5
ˆ 13
‰
5
œ
12
13
,
5
12
"
x
œ
x% "
#x#
œ e5x
8. cosh 3x sinh 3x œ
e3x e
#
3x
e3x e
#
3x
œ e3x
œ aex b% œ e4x
10. ln (cosh x sinh x) ln (cosh x sinh x) œ ln acosh# x sinh# xb œ ln 1 œ 0
11. (a) sinh 2x œ sinh (x x) œ sinh x cosh x cosh x sinh x œ 2 sinh x cosh x
(b) cosh 2x œ cosh (x x) œ cosh x cosh x sinh x sin x œ cosh# x sinh# x
12. cosh# x sinh# x œ ˆ e
œ
"
4
!
a4e b œ
13. y œ 6 sinh
14. y œ
"
#
x
3
"
4
x
ecx ‰#
#
ˆe
x
ecx ‰#
#
œ
"
4
caex ex b aex ex bd caex ex b aex ex bd œ
"
4
a2ex b a2ex b
(4) œ 1
Ê
dy
dx
œ 6 ˆcosh x3 ‰ ˆ "3 ‰ œ 2 cosh
sinh (2x 1) Ê
dy
dx
œ
"
#
x
3
[cosh (2x 1)](2) œ cosh (2x 1)
15. y œ 2Èt tanh Èt œ 2t"Î# tanh t"Î# Ê
dy
dt
œ sech# ˆt"Î# ‰‘ ˆ "# t"Î# ‰ ˆ2t"Î# ‰ ˆtanh t"Î# ‰ ˆt"Î# ‰ œ sech# Èt Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
tanh Èt
Èt
Section 7.7 Hyperbolic Functions 437
16. y œ t# tanh
"
t
œ t# tanh t" Ê
17. y œ ln (sinh z) Ê
œ
dy
dz
œ csech# at" bd at# b at# b (2t) atanh t" b œ sech#
dy
dt
œ coth z
cosh z
sinh z
19. y œ (sech ))(1 ln sech )) Ê
18. y œ ln (cosh z) Ê
dy
dz
œ
"
t
2t tanh
sinh z
cosh z
"
t
œ tanh z
dy
d)
) tanh ) ‰
œ ˆ sech
(sech )) ( sech ) tanh ))(1 ln sech ))
sech )
dy
d)
) coth ) ‰
œ (csch )) ˆ csch
(1 ln csch ))( csch ) coth ))
csch )
œ sech ) tanh ) (sech ) tanh ))(1 ln sech )) œ (sech ) tanh ))[1 (1 ln sech ))] œ (sech ) tanh ))(ln sech ))
20. y œ (csch ))(1 ln csch )) Ê
œ csch ) coth ) (1 ln csch ))(csch ) coth )) œ (csch ) coth ))(1 1 ln csch )) œ (csch ) coth ))(ln csch ))
21. y œ ln cosh v "
#
tanh# v Ê
"
#
coth# v Ê
dy
dv
œ
dy
dv
œ
sinh v
cosh v
ˆ "# ‰ (2 tanh v) asech# vb œ tanh v (tanh v) asech# vb
cosh v
sinh v
ˆ "# ‰ (2 coth v) a csch# vb œ coth v (coth v) acsch# vb
œ (tanh v) a1 sech# vb œ (tanh v) atanh# vb œ tanh$ v
22. y œ ln sinh v œ (coth v) a1 csch# vb œ (coth v) acoth# vb œ coth$ v
23. y œ ax# 1b sech (ln x) œ ax# 1b ˆ eln x 2ec ln x ‰ œ ax# 1b ˆ x 2xc" ‰ œ ax# 1b ˆ x#2x1 ‰ œ 2x Ê
24. y œ a4x# 1b csch (ln 2x) œ a4x# 1b ˆ eln 2x 2e
25. y œ sinh" Èx œ sinh" ˆx"Î# ‰ Ê
dy
dx
É1 ax"Î# b#
26. y œ cosh" 2Èx 1 œ cosh" ˆ2(x 1)"Î# ‰ Ê
27. y œ (1 )) tanh" ) Ê
dy
d)
dy
dx
œ
œ
"
#È x È 1 x
dy
d)
œ
dy
dx
œ4
"
#Èx(1 x)
(2) Š "# ‹ (x 1)c"Î#
œ
Éc2(x 1)"Î# d# 1
œ (1 )) ˆ 1 " )# ‰ (1) tanh" ) œ
28. y œ a)# 2)b tanh" () 1) Ê
œ2
2
4x
#
‰ œ a4x# 1b Š 2x (2x)
" ‹ œ a4x 1b ˆ 4x# 1 ‰ œ 4x Ê
ln 2x
Š "# ‹ xc"Î#
œ
dy
dx
"
1)
"
Èx 1 È4x 3
œ
"
È4x# 7x 3
tanh" )
œ a)# 2)b ’ 1 ()"1)# “ (2) 2) tanh" () 1) œ
) # 2)
) # 2 )
(2) 2) tanh" () 1)
œ (2) 2) tanh" () 1) 1
29. y œ (1 t) coth" Èt œ (1 t) coth" ˆt"Î# ‰ Ê
30. y œ a1 t# b coth" t Ê
dy
dt
œ
dy
dx
"
È 1 x#
’x Š
32. y œ ln x È1 x# sech" x œ ln x a1 x# b
dy
dx
"
x
œ
"
33. y œ csch
a1 x# b
ˆ "# ‰)
"Î#
Š
œ (1 t) –
Š "# ‹ tc"Î#
1 at"Î# b
#
"
ˆt"Î# ‰ œ
"
È 1 x#
— (1) coth
"
‹
xÈ 1 x#
"
‹
xÈ 1 x#
"Î#
(1) sech" x“ œ
"
È 1 x#
Ê
coth" Èt
sech" x œ sech" x
sech" x
ˆ "# ‰ a1 x# b
"Î#
(2x) sech" x œ
"
x
"
x
x
È 1 x#
sech" x œ
)
dy
d)
"
#È t
œ a1 t# b ˆ 1 " t# ‰ (2t) coth" t œ 1 2t coth" t
31. y œ cos" x x sech" x Ê
Ê
dy
dt
œ
’ln Š "# ‹“ Š "# ‹
)
Š "# ‹
) #
"
Ë 1 ”Š # ‹
•
œ ln (1) ln (2)
œ
#)
Ê1 Š "# ‹
ln 2
#)
Ê1 Š "# ‹
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
x
È 1 x#
sech" x
438
Chapter 7 Transcendental Functions
34. y œ csch" 2) Ê
dy
d)
œ
35. y œ sinh" (tan x) Ê
œ
dy
dx
36. y œ cosh" (sec x) Ê
(ln 2) 2)
2 ) É 1 a2 ) b
œ
dy
dx
ln 2
È1 2#)
œ
sec# x
Èsec# x
(sec x)(tan x)
Èsec# x 1
œ
(sec x)(tan x)
Ètan# x
(b) If y œ sin" (tanh x) C, then
x#
#
œ
sec# x
È1 (tan x)#
37. (a) If y œ tan" (sinh x) C, then
38. If y œ
#
dy
dx
dy
dx
œ
œ
œ
sec# x
ksec xk
œ
œ
ksec xk ksec xk
ksec xk
(sec x)(tan x)
ktan xk
œ ksec xk
œ sec x, 0 x 1
#
cosh x
cosh x
1 sinh# x œ cosh# x œ sech x, which verifies the formula
#
sech x
sech# x
È1 tanh# x œ sech x œ sech x, which verifies the formula
sech" x "# È1 x# C, then
œ x sech" x dy
dx
x#
#
Š
"
‹
xÈ 1 x#
2x
4È 1 x#
œ x sech" x, which verifies the
formula
x# "
#
39. If y œ
coth" x 40. If y œ x tanh" x 41.
'
sinh 2x dx œ
œ
42.
'
sinh
x
5
cosh u
#
"
#
'
"
#
x
#
C, then
'
44.
'
'
dy
dx
Cœ
cosh 2x
#
4 cosh (3x ln 2) dx œ
tanh
x
7
sinh u C œ
dx œ 7 '
" ˆ " ‰
# ‹ 1 x#
"
#
œ x coth" x, which verifies the formula
sinh u du, where u œ 2x and du œ 2 dx
C
dx œ 5 ' sinh u du, where u œ
4
3
#
œ tanh" x x ˆ 1 " x# ‰ #" ˆ 12xx# ‰ œ tanh" x, which verifies the formula
x
5
x
5
"
5
and du œ
sinh u
cosh u
4
3
4
3
'
dx
C
6 cosh ˆ x# ln 3‰ dx œ 12 ' cosh u du, where u œ
œ 12 sinh u C œ 12 sinh ˆ x# ln 3‰ C
œ
45.
œ x coth" x Š x
ln a1 x# b C, then
œ 5 cosh u C œ 5 cosh
43.
dy
dx
x
#
ln 3 and du œ
"
#
dx
cosh u du, where u œ 3x ln 2 and du œ 3 dx
sinh (3x ln 2) C
du, where u œ
x
7
and du œ
"
7
dx
œ 7 ln kcosh uk C" œ 7 ln ¸cosh x7 ¸ C" œ 7 ln ¹ e
xÎ7
e xÎ7
¹
#
C" œ 7 ln ¸exÎ7 exÎ7 ¸ 7 ln 2 C"
œ 7 ln kexÎ7 e xÎ7 k C
46.
'
coth
)
È3
d) œ È 3 '
cosh u
sinh u
du, where u œ
œ È3 ln ksinh uk C" œ È3 ln ¹sinh
œ È3 ln ¹e)Î
47.
'
È$
e)Î
È$
)
È3
and du œ
)
È3 ¹
d)
È3
C" œ È3 ln ¹ e
¹ È3 ln 2 C" œ È3 ln ¹e)Î
È3
È$ e )ÎÈ$
)Î
#
e)Î
È3
¹ C"
¹C
sech# ˆx "# ‰ dx œ ' sech# u du, where u œ ˆx "# ‰ and du œ dx
œ tanh u C œ tanh ˆx "# ‰ C
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 7.7 Hyperbolic Functions 439
48.
'
csch# (5 x) dx œ ' csch# u du, where u œ (5 x) and du œ dx
œ ( coth u) C œ coth u C œ coth (5 x) C
49.
'
dt œ 2 ' sech u tanh u du, where u œ Èt œ t"Î# and du œ
sech Èt tanh Èt
Èt
dt
2È t
œ 2( sech u) C œ 2 sech Èt C
50.
'
csch (ln t) coth (ln t)
t
dt œ ' csch u coth u du, where u œ ln t and du œ
dt
t
œ csch u C œ csch (ln t) C
51.
x
'lnln24 coth x dx œ 'lnln24 cosh
' 15Î8
sinh x dx œ 3Î4
"
u
"&Î)
¸
¸3¸
¸ 15 4 ¸
du œ cln kukd $Î% œ ln ¸ 15
8 ln 4 œ ln 8 † 3 œ ln
where u œ sinh x, du œ cosh x dx, the lower limit is sinh (ln 2) œ
limit is sinh (ln 4) œ
52.
eln 4 ec ln 4
#
œ
4 Š "4 ‹
#
œ
17Î8
sinh 2x
" '
'0ln 2 tanh 2x dx œ '0ln 2 cosh
2x dx œ # 1
eln 2 ec ln 2
#
œ
2 Š "# ‹
5
#
,
œ
3
4
and the upper
ln ˆ 17
‰
‘
8 ln 1 œ
"
#
ln
#
15
8
"
u
du œ
"
#
"(Î)
cln kukd "
"
#
œ
17
8
, where
u œ cosh 2x, du œ 2 sinh (2x) dx, the lower limit is cosh 0 œ 1 and the upper limit is cosh (2 ln 2) œ cosh (ln 4)
œ
53.
eln 4 ec ln 4
#
c2 ln 2
#
'0ln 2 4ec
)
ln 2‹ Š e
#
e
2 ln 2
#
ec)
‹
#
c ln 2
d) œ 'c ln 4 ae2) 1b d) œ e# )‘ c ln 4
"
ln 4‹ œ ˆ 8" ln 2‰ ˆ 32
ln 4‰ œ
‹ Š0 )
e0
# ‹“
ec)
‹
#
ln 2
2)
3
32
ln 2 2 ln 2 œ
d) œ 2 '0 a1 ec2) b d) œ 2 ’) ln 2
œ 2 ˆln 2 "
8
"# ‰ œ 2 ln 2 "
4
3
3#
ln 2
ec#)
# “0
ln 2
1 œ ln 4 3
4
'c11ÎÎ44 cosh (tan )) sec# ) d) œ ' 11 cosh u du œ csinh ud "" œ sinh (1) sinh (1) œ Š e #e " ‹ Š e " # e" ‹
"
e" , where u œ tan ), du œ sec# ) d), the lower limit is tan ˆ 14 ‰ œ 1 and the upper
'01Î2 2 sinh (sin )) cos ) d) œ 2'01 sinh u du
"
œee
œ 2 ccosh ud "! œ 2(cosh 1 cosh 0) œ 2 Š e #e
c"
1‹
2, where u œ sin ), du œ cos ) d), the lower limit is sin 0 œ 0 and the upper limit is sin ˆ 1# ‰ œ 1
'12 cosht(ln t) dt œ '0ln 2 cosh u du œ csinh ud ln0 2 œ sinh (ln 2) sinh (0) œ e
u œ ln t, du œ
58.
c2 ln 4
ln 2
ee "e "e
œe
#
1
limit is tan ˆ 4 ‰ œ 1
57.
17
8
sinh ) d) œ '0 4ec) Š e
œ
56.
œ
)
œ 2 ’Šln 2 55.
#
'cclnln42 2e) cosh ) d) œ 'cclnln42 2e) Š e
œ Še
54.
4 Š "4 ‹
œ
"
t
ln 2
ec ln 2
#
0œ
2
#
"
#
œ
3
4
, where
dt, the lower limit is ln 1 œ 0 and the upper limit is ln 2
2
Èx
#
"
'14 8 cosh
dx œ 16'1 cosh u du œ 16 csinh ud #" œ 16(sinh 2 sinh 1) œ 16 ’Š e #e ‹ Š e #e ‹“
Èx
#
œ 8 ae# e# e e" b , where u œ Èx œ x"Î# , du œ
"
#
x"Î# dx œ
dx
2È x
, the lower limit is È1 œ 1 and the upper
limit is È4 œ 2
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
440
59.
60.
Chapter 7 Transcendental Functions
'c0ln 2 cosh# ˆ x# ‰ dx œ 'c0ln 2 cosh#x " dx œ "# 'c0ln 2 (cosh x 1) dx œ "# csinh x xd c0 ln 2
œ
"
#
[(sinh 0 0) (sinh ( ln 2) ln 2)] œ
œ
"
#
ˆ1 "
4
ln 2‰ œ
3
8
"
#
ln 2 œ
"
#
’(0 0) Š e
c ln 2 eln 2
#
ln 2‹“ œ
"
#
–
Š "# ‹ 2
#
ln 2—
ln È2
3
8
'0ln 10 4 sinh# ˆ x# ‰ dx œ '0ln 10 4 ˆ cosh#x 1 ‰ dx œ 2'0ln 10 (cosh x 1) dx œ 2 csinh x xd ln0 10
œ 2[(sinh (ln 10) ln 10) (sinh 0 0)] œ eln 10 ec ln 10 2 ln 10 œ 10 5
25
61. sinh" ˆ 1#5 ‰ œ ln Š 12
É 144
1‹ œ ln ˆ 32 ‰
63. tanh" ˆ "# ‰ œ
"
#
65. sech" ˆ 35 ‰ œ ln Š
67. (a)
'02
È3
dx
È 4 x#
"
10
2 ln 10 œ 9.9 2 ln 10
62. cosh" ˆ 35 ‰ œ ln Š 35 É 25
9 1‹ œ ln 3
(1/2)
ln 3
ln Š 11 (1/2) ‹ œ #
64. coth" ˆ 45 ‰ œ
1È1 (9/25)
‹
(3/5)
66. csch" Š È"3 ‹ œ ln È3 2
œ sinh" x# ‘ 0
œ ln 3
È3
"
#
ln Š (9/4)
(1/4) ‹ œ
"
#
ln 9 œ ln 3
È4/3
Š1/È3‹ œ ln ŠÈ3 2‹
œ sinh" È3 sinh 0 œ sinh" È3
(b) sinh" È3 œ ln ŠÈ3 È3 1‹ œ ln ŠÈ3 2‹
68. (a)
'01Î3 È 6 dx
œ 2'0
1
1 9x#
dx
È a# u# ,
where u œ 3x, du œ 3 dx, a œ 1
"
œ c2 sinh" ud ! œ 2 asinh" 1 sinh" 0b œ 2 sinh" 1
(b) 2 sinh" 1 œ 2 ln Š1 È1# 1‹ œ 2 ln Š1 È2‹
69. (a)
'52Î4
#
"
1 x#
dx œ ccoth" xd &Î% œ coth" 2 coth"
(b) coth" 2 coth"
70. (a)
'01Î2
71. (a)
'13ÎÎ513
"
#
"
#
œ
ln 3 ln ˆ 9/4
‰‘ œ
1/4
"Î#
"
1 x #
(b) tanh"
5
4
dx œ ctanh" xd !
"
#
œ
(1/2)
ln Š 11 (1/2) ‹ œ
dx
xÈ1 16x#
œ '4Î5
12Î13
œ tanh"
"
#
œ c sech" ud 4Î5 œ sech"
(b) sech"
12
13
sech"
œ ln Š 13 72. (a)
(b)
73. (a)
'12
"
#
dx
xÈ 4 x#
ˆcsch"
'01
È169 144
‹
1#
"
3
tanh" 0 œ tanh"
"
#
where u œ 4x, du œ 4 dx, a œ 1
12
13
œ ln Š
4
5
ln
ln 3
du
,
u È a# u#
12Î13
"
#
"
#
5
4
sech"
4
5
1È1 (12/13)#
‹
(12/13)
ln Š 5 È25 16
‹
4
ln Š
1È1 (4/5)#
‹
(4/5)
œ ln ˆ 5 4 3 ‰ ln ˆ 1312 5 ‰ œ ln 2 ln
#
œ "# csch" ¸ x# ¸‘ " œ "# ˆcsch" 1 csch" "# ‰ œ
"
#
csch" 1‰ œ
cos x
È1 sin# x
dx œ '0
0
"
#
’ln Š2 "
È 1 u#
È5/4
(1/2) ‹
ln Š1 È2‹“ œ
"
#
ˆcsch"
"
#
"
#
5
ln Š 21 ‹
È2
3
#
œ ln ˆ2 † 23 ‰ œ ln
4
3
csch" 1‰
È
!
du œ csinh" ud ! œ sinh" 0 sinh" 0 œ 0, where u œ sin x, du œ cos x dx
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 7.7 Hyperbolic Functions 441
(b) sinh" 0 sinh" 0 œ ln Š0 È0 1‹ ln Š0 È0 1‹ œ 0
74. (a)
'1e xÈ1 dx(ln x)
œ
œ '0
1
#
"
csinh" ud !
du
È a# u#
, where u œ ln x, du œ
"
x
dx, a œ 1
œ sinh" 1 sinh" 0 œ sinh" 1
(b) sinh" 1 sinh" 0 œ ln Š1 È1# 1‹ ln Š0 È0# 1‹ œ ln Š1 È2‹
f(x) f(x)
and O(x) œ f(x) #f(x) .
#
f(x) f((x))
œ f(x) #f(x) œ E(x)
#
75. Let E(x) œ
E(x) œ
Then E(x) O(x) œ
f(x) f(x)
#
Ê E(x) is even, and O(x) œ
f(x) f(x)
œ 2f(x)
#
# œ f(x).
f(x) f((x))
œ f(x) #f(x)
#
Also,
œ O(x)
Ê O(x) is odd. Consequently, f(x) can be written as a sum of an even and an odd function.
f(x) œ
f(x) f(x)
#
because
Thus, if f is even f(x) œ
f(x) f(x)
œ
#
2f(x)
# 0 and
if f is odd, f(x) œ 0 76. y œ sinh" x Ê x œ sinh y Ê x œ
Ê ey œ
2x „ È4x# 4
#
f(x) f(x)
#
2f(x)
#
0 if f is even and f(x) œ
ey ecy
#
Ê 2x œ ey "
ey
because
f(x) f(x)
#
œ 0 if f is odd.
Ê 2xey œ e2y 1 Ê e2y 2xey 1 œ 0
Ê ey œ x Èx# 1 Ê sinh" x œ y œ ln Šx Èx# 1‹ . Since ey 0, we cannot
choose ey œ x Èx# 1 because x Èx# 1 0.
É gk
77. (a) v œ É mg
k tanhŒ
m t Ê
dv
dt
# É gk
# É gk
É gk
œ É mg
k ”sech Œ
m t•Œ
m œ g sech Œ
m t.
# É gk
# É gk
#
Thus m dv
dt œ mg sech Œ
m t œ mgŒ" tanh Œ
m t œ mg kv . Also, since tanh x œ ! when x œ !, v œ !
when t œ !.
(b)
mg
mg
mg
É kg
lim v œ lim É mg
lim tanh ΃ kg
k tanh Œ
m t œ É k
m t œ É k (1) œ É k
tÄ_
tÄ_
tÄ_
160
(c) É 0.005
œ É 160,000
œ
5
400
È5
œ 80È5 ¸ 178.89 ft/sec
78. (a) s(t) œ a cos kt b sin kt Ê
#
ds
dt
œ ak sin kt bk cos kt Ê
œ ak# cos kt bk# sin kt
œ k (a cos kt b sin kt) œ k s(t) Ê acceleration is proportional to s. The negative constant k#
implies that the acceleration is directed toward the origin.
(b) s(t) œ a cosh kt b sinh kt Ê
#
d# s
dt#
ds
dt
œ ak sinh kt bk cosh kt Ê
d# s
dt#
œ ak# cosh kt bk# sinh kt
œ k# (a cosh kt b sinh kt) œ k# s(t) Ê acceleration is proportional to s. The positive constant k# implies
that the acceleration is directed away from the origin.
79. V œ 1'0 acosh# x sinh# xb dx œ 1'0 1 dx œ 21
2
2
80. V œ 21 '0
ln
81. y œ
"
#
È3
sech# x dx œ 21 ctanh xd ln0
È3
È3 Š1/È3‹
œ 21 – È
cosh 2x Ê yw œ sinh 2x Ê L œ '0
œ ’ "# Š e
ln
2x
ec2x
‹“
#
0
ln
È5
œ
"
4
ˆ5 "5 ‰ œ
È5
3 Š1/È3‹ —
œ1
È1 (sinh 2x)# dx œ '
0
ln
È5
ln
cosh 2x dx œ "# sinh 2x‘ 0
6
5
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
È5
442
Chapter 7 Transcendental Functions
82. (a) x lim
tanh x œ x lim
Ä_
Ä_
ex ecx
ex ecx
(b) x Ä
lim
tanh x œ x Ä
lim
_
_
œ x lim
Ä_
ex ecx
ex ecx
ex e1x
ex e1x
ex e1x
ex e1x
œxÄ
lim
_
e ecx
ˆex e1x ‰
ˆex e1x ‰
œ x lim
Ä_
œxÄ
lim
_
†
1
ex
1
ex
œ x lim
Ä_
ˆex e1x ‰
ˆex e1x ‰
†
ex
ex
1
1 e2x
1
1 e2x
œ
10
10
e2x 1
e2x 1
œxÄ
lim
_
œ1
01
01
œ
œ 1
e ˆ e 2e1x ‰ œ _ 0 œ _
(c) x lim
sinh x œ x lim
œ x lim
œ x lim
2
2
Ä_
Ä_
Ä_
Ä_ 2
x
cx
ex ecx
ˆ e e2 ‰ œ 0 _ œ _
(d) x Ä
lim
sinh x œ x Ä
lim
œxÄ
lim
2
_
_
_ 2
1
ex
x
x
x
1
ex
1
ex
(e) x lim
sech x œ x lim
Ä_
Ä_
2
ex ecx
œ x lim
Ä_
2
ex e1x
†
(f) x lim
coth x œ x lim
Ä_
Ä_
ex ecx
ex ecx
œ x lim
Ä_
ex e1x
ex e1x
œ x lim
Ä_
lim coth x œ lim b
xÄ0
(g)
x Ä 0b
lim coth x œ lim c
xÄ0
(h)
x Ä 0c
ex ecx
ex ecx
ex ecx
ex ecx
(i) x Ä
lim
csch x œ x Ä
lim
_
_
83. (a) y œ
H
w
œ x lim
Ä_
2
ex
œ
1
1 e2x
ˆex e1x ‰
ˆex e1x ‰
†
1
ex
1
ex
œ!
0
10
œ x lim
Ä_
œ lim b
xÄ0
ex e1x
ex e1x
†
ex
ex
œ lim b
xÄ0
e2x 1
e2x 1
œ _
œ lim c
xÄ0
e ex †
ex
ex
œ lim c
xÄ0
e2x 1
e2x 1
œ _
2
ex ecx
‰
cosh ˆ w
H x Ê tan 9 œ
x
1
ex
1
ex
œxÄ
lim
_
dy
dx
2
ex e1x
†
x
e
ex
œxÄ
lim
_
2ex
e2x 1
œ
1
1 e2x
1
1 e2x
0
01
œ
10
10
œ1
œ!
‰ w
ˆ w ‰‘ œ sinh ˆ w
‰
œ ˆH
w
H sinh H x
H x
(b) The tension at P is given by T cos 9 œ H Ê T œ H sec 9 œ HÈ1 tan# 9 œ HÉ1 ˆsinh
w
H
#
x‰
‰
ˆH‰
ˆw ‰
œ H cosh ˆ w
H x œ w w cosh H x œ wy
"
a
84. s œ
œ
"
a
sinh ax Ê sinh ax œ as Ê ax œ sinh" as Ê x œ
Èsinh# ax 1 œ
"
a
Èa# s# 1 œ És# 85. To find the length of the curve: y œ
"
a
œ a"# sinh ax‘ 0 œ
"
a#
"
a
cosh ax œ
Ècosh# ax
"
a#
b
b
b
"
a
sinh" as; y œ
cosh ax Ê yw œ sinh ax Ê L œ '0 È1 (sinh ax)# dx
Ê L œ '0 cosh ax dx œ "a sinh ax‘ 0 œ
b
"
a
"
a
sinh ab. The area under the curve is A œ '0
b
"
a
sinh ab œ ˆ "a ‰ ˆ "a sinh ab‰ which is the area of the rectangle of height
cosh ax dx
"
a
and length L
as claimed, and which is illustrated below.
86. (a) Let the point located at (cosh uß 0) be called T. Then A(u) œ area of the triangle ?OTP minus the area
under the curve y œ Èx# 1 from A to T Ê A(u) œ
(b) A(u) œ
œ
"
#
"
#
cosh u sinh u '1
cosh u
cosh# u (c) Aw (u) œ
"
#
"
#
Èx# 1 dx Ê Aw (u) œ
sinh# u sinh# u œ
Ê A(u) œ
u
#
"
#
"
#
cosh u sinh u "
#
'1cosh u Èx# 1 dx.
acosh# u sinh# ub ŠÈcosh# u 1‹ (sinh u)
acosh# u sinh# ub œ ˆ "# ‰ (1) œ
"
#
C, and from part (a) we have A(0) œ 0 Ê C œ 0 Ê A(u) œ
u
#
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Ê u œ 2A
Section 7.8 Relative Rates of Growth
443
7.8 RELATIVE RATES OF GROWTH
lim x 3 œ lim e"x œ 0
x Ä _ ex
xÄ_
#
x$ sin# x
x cos x
2x
(b) slower, lim
œ lim 3x 2 sin
œ lim 6x 2excos 2x œ lim 6 4esin
œ 0 by the
x
ex
ex
xÄ_
xÄ_
xÄ_
xÄ_
2
6 4 sin 2x
10
2
"0
Sandwich Theorem because ex Ÿ
Ÿ ex for all reals and lim ex œ 0 œ lim
ex
xÄ_
x Ä _ ex
1. (a) slower,
Š "# ‹ x "Î#
"Î#
Èx
lim
œ lim xex œ lim
x Ä _ ex
xÄ_
xÄ_
4x
4 ‰x
ˆ
(d) faster, lim ex œ lim
œ _ since
xÄ_
xÄ_ e
(c) slower,
ex
œ
lim
xÄ_
"
#Èx ex
œ0
1
4
e
x
(e) slower,
lim
xÄ_
(f) slower, lim
xÄ_
(g) same,
lim
xÄ_
2. (a) slower,
lim
xÄ_
(b) slower,
lim
xÄ_
Š "x ‹
ex
(c) slower,
xÎ2
e
ex
x
lim
xÄ_
œ È0 œ 0
x
lim ˆ 3 ‰ œ 0 since
x Ä _ 2e
œ lim ex"Î2 œ 0
xÄ_
œ
œ
ex
lim
xÄ_
lim
xÄ_
ex
Š e# ‹
(h) slower,
œ
Š 3# ‹
lim
xÄ_
œ
log10 x
ex
lim
xÄ_
10x% 30x 1
ex
x ln x x
ex
œ
œ
lim
xÄ_
È 1 x%
ex
"
#
œ
œ
"
xex
"
#
ln x
(ln 10) ex
lim
xÄ_
lim
xÄ_
1
3
2e
œ
40x$ 30
ex
x (ln x 1)
ex
"
x
lim
xÄ_
œ
œ
œ
(ln 10) ex
lim
xÄ_
lim
xÄ_
lim
xÄ_
"20x#
ex
œ
240x
ex
lim
xÄ_
ln x 1 x Š "x ‹
ex
"
(ln 10)xex
œ
lim
xÄ_
œ0
œ
240
ex
lim
xÄ_
ln x 1 1
ex
œ
œ0
lim
xÄ_
œ0
œ É lim
xÄ_
1 x%
e2x
œ É lim
xÄ_
4x$
2e2x
œ É lim
xÄ_
12x#
4e2x
œ É lim
xÄ_
24x
8e2x
œ É lim
xÄ_
x
Š5‹
x
#
5
lim
œ lim ˆ #5e ‰ œ 0 since 2e
1
x Ä _ ex
xÄ_
ecx
"
(e) slower, lim
œ lim e2x œ 0
x Ä _ ex
xÄ_
x
(f) faster, lim xeex œ lim x œ _
xÄ_
xÄ_
(g) slower, since for all reals we have 1 Ÿ cos x Ÿ 1 Ê e" Ÿ ecos x Ÿ e" Ê
(d) slower,
e "
e"
ec"
ex
Ÿ
ecos x
ex
lim
œ 0 œ lim ex , so by the Sandwich Theorem we conclude that lim
x Ä _ ex
xÄ_
xÄ_
x 1
(h) same, lim e ex œ lim eÐx "x 1Ñ œ lim "e œ "e
xÄ_
xÄ_
xÄ_
3. (a) same,
lim
xÄ_
(b) faster, lim
xÄ_
(c) same,
lim
xÄ_
(d) same,
lim
xÄ_
x# 4x
x#
x& x#
x#
lim 2x 4 œ lim
x Ä _ 2x
xÄ_
œ lim ax$ 1b œ _
xÄ_
œ
È x% x$
x#
(x 3)#
x#
œ É lim
xÄ_
œ
lim
xÄ_
x% x$
x%
2(x 3)
#x
œ
Ÿ
e
cos x
ex
œ1
2
#
œ É lim ˆ1 "x ‰ œ È1 œ 1
xÄ_
lim
xÄ_
2
#
œ1
Š"‹
x
lim x ln x œ lim lnxx œ lim
œ0
x Ä _ x#
xÄ_
xÄ_ 1
x
x
#
x
2) 2
(f) faster, lim 2x# œ lim (ln2x
œ lim (ln 2)# 2 œ _
xÄ_
xÄ_
xÄ_
$ x
(g) slower, lim x xe# œ lim exx œ lim e"x œ 0
xÄ_
xÄ_
xÄ_
#
(h) same, lim 8x
œ
lim
8
œ
8
x Ä _ x#
xÄ_
(e) slower,
# È
x x
" ‰
lim
œ lim ˆ1 x$Î#
œ1
x Ä _ x#
xÄ_
#
(b) same, lim "0x
œ lim 10 œ 10
x Ä _ x#
xÄ_
x# e x
"
(c) slower, lim
œ
lim
œ0
x Ä _ x#
x Ä _ ex
4. (a) same,
ln x
ex
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
e"
ex
and also
œ0
24
16e2x
444
Chapter 7 Transcendental Functions
#
Š ln x ‹
#
ln 10
lim log10 x œ lim
œ ln"10 lim
x Ä _ x#
x Ä _ x#
xÄ_
$
#
(e) faster, lim x x#x œ lim (x 1) œ _
xÄ_
xÄ_
(d) slower,
œ
2 ln x
x#
2
ln 10
lim
xÄ_
Š x" ‹
2x
"
ln 10
œ
lim
xÄ_
"
x#
œ0
x
"
Š 10
‹
lim
œ lim 10"x x# œ 0
x Ä _ x#
xÄ_
x
x
(g) faster, lim (1.1)
œ
lim (ln 1.1)(1.1)
œ lim
#x
x Ä _ x#
xÄ_
xÄ_
#
‰
(h) same, lim x x100x
œ lim ˆ1 100
œ
1
#
x
xÄ_
xÄ_
(f) slower,
5. (a) same,
lim
xÄ_
log3 x
ln x
(b) same,
lim
xÄ_
ln 2x
ln x
(c) same,
lim
xÄ_
(d) faster,
lim
xÄ_
(e) faster,
(f) same,
œ
œ
ln Èx
ln x
x
Š ln
ln 3 ‹
lim
xÄ_
ln x
ˆ #2x ‰
ˆ x" ‰
lim
xÄ_
œ
œ
lim
xÄ_
lim
xÄ_
x
ln x
œ
lim
xÄ_
lim
xÄ_
5 ln x
ln x
œ
Š"‹
lim
xÄ_
"
ln 3
œ
"
ln 3
"
#
œ
"
#
œ_
œ1
Š "# ‹ ln x
lim
xÄ_
Èx
ln x
œ
(ln 1.1)# (1.1)x
#
œ
ln x
lim
xÄ_
Š "# ‹ x "Î#
x"Î#
ln x
œ
lim
xÄ_
"
Š "x ‹
œ
lim x œ _
xÄ_
Š x" ‹
œ
lim
xÄ_
x
#Èx
œ
Èx
#
lim
xÄ_
œ_
lim 5 œ 5
xÄ_
x
lim
œ lim x ln" x œ 0
x Ä _ ln x
xÄ_
x
x
(h) faster, lim lne x œ lim ˆe" ‰ œ lim xex œ _
xÄ_
xÄ_ x
xÄ_
(g) slower,
6. (a) same,
lim
xÄ_
(b) same,
lim
xÄ_
lim
xÄ_
(d) slower,
lim
xÄ_
lim
xÄ_
(g) slower,
lim
xÄ_
lim
xÄ_
œ
lim
xÄ_
œ
lim
xÄ_
ln x
"
‹
x#
ln x
ecx
ln x
œ
œ
"
ln #
œ
ln x
lim
xÄ_
Š È"x ‹
Š
Š lnlnx2 ‹
lim
xÄ_
œ
x2 ln x
ln x
lim
xÄ_
(f) slower,
(h) same,
œ
log10 10x
ln x
(c) slower,
(e) faster,
#
log2 x#
ln x
Š lnln10x
10 ‹
ln x
œ
"
ˆÈx‰ (ln x)
"
x# ln x
lim
xÄ_
"
ln 10
ln x#
ln x
lim
xÄ_
"
ex ln x
ln (ln x)
ln x
œ
lim
xÄ_
ln (2x5)
ln x
œ
lim
xÄ_
"
ln #
ln 10x
ln x
œ
lim
xÄ_
"
ln 10
2 ln x
ln x
lim
xÄ_
œ
"
ln #
"0
Š 10x
‹
Š x" ‹
lim 2 œ
xÄ_
œ
"
ln 10
2
ln #
lim 1 œ
xÄ_
"
ln 10
œ0
œ0
lim ˆ x 2‰ œ Š lim
x Ä _ ln x
xÄ_
lim
xÄ_
œ
x
ln x ‹
2 œ lim
xÄ_
"
Š "x ‹ 2 œ Š lim x‹ 2 œ _
xÄ_
œ0
"/x
Š ln
x‹
Š x" ‹
Š 2x2 5 ‹
Š x" ‹
œ
lim
xÄ_
œ
"
ln x
œ0
2x
2x5
lim
xÄ_
œ
lim
xÄ_
2
#
œ
lim 1 œ 1
xÄ_
x
7.
e
lim
œ lim exÎ2 œ _ Ê ex grows faster than exÎ2 ; since for x ee we have ln x e and lim (lnexx)
x Ä _ exÎ2 x Ä _
xÄ_
x
x
x
œ lim ˆ lne x ‰ œ _ Ê (ln x)x grows faster than ex ; since x ln x for all x 0 and lim (lnxx)x œ lim ˆ lnxx ‰
xÄ_
xÄ_
xÄ_
œ _ Ê xx grows faster than (ln x)x . Therefore, slowest to fastest are: exÎ2 , ex , (ln x)x , xx so the order is d, a, c, b
8.
2)
lim (ln 2) œ lim (ln (ln 2))(ln
œ lim (ln (ln 2))# (ln 2) œ (ln (ln# 2))
lim (ln 2)x œ 0
#x
x Ä _ x#
xÄ_
xÄ_
xÄ_
#
Ê (ln 2)x grows slower than x# ; lim x2x œ lim (ln2x#)2x œ lim (ln 2)2 # #x œ 0 Ê x# grows slower than 2x ;
xÄ_
xÄ_
xÄ_
x
x
lim 2ex œ lim ˆ 2e ‰ œ 0 Ê 2x grows slower than ex . Therefore, the slowest to the fastest is: (ln 2)x , x# , 2x
xÄ_
xÄ_
and ex so the order is c, b, a, d
x
x
x
#
x
#
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 7.8 Relative Rates of Growth
lim x œ 1
xÄ_ x
(b) false; lim x x 5 œ 1" œ 1
xÄ_
(c) true; x x 5 Ê xx 5 1 if x 1 (or sufficiently large)
9. (a) false;
(d) true; x 2x Ê
(e) true;
lim
xÄ_
(f) true;
x ln x
x
(g) false;
È x# 5
x
(h) true;
Šx " 3‹
10. (a) true;
Š x" ‹
Š "x (b) true;
œ lim
xÄ0
œ1
œ
"‹
x#
œ
"
x
(e) true;
Š "x "
‹
x#
"
Šx‹
œ1
(f) true;
lim
xÄ_
ln (ln x)
(g) true; ln x (h) false;
lim
xÄ_
x
ex
Š "x ‹
œ
Š #2x ‹
œ1
2 if x 1 (or sufficiently large)
lim 1 œ 1
xÄ_
6 if x 1 (or sufficiently large)
5
x
lim ˆ1 "x ‰ œ 1
xÄ_
œ
and
2 cos x
#
x
x
e Ä
Ÿ
3
#
if x is sufficiently large
0 as x Ä _ Ê 1 lim ln x œ lim
xÄ_ x
xÄ_
œ 1 if x is sufficiently large
x ln x
x#
ln x
ln x
x5
x
"
Èx
œ1
2 if x 1 (or sufficiently large)
(d) true; 2 cos x Ÿ 3 Ê
ex x
ex
Èx
x
1 if x 1 (or sufficiently large)
œ1
lim
xÄ_
œ0
lim
xÄ_
È(x 5)#
x
x
x3
"
ex
1
ln x
x
ln x
ln 2x
Š "x ‹
(c) false;
ex
e2x
lim
xÄ_
1 if x 1 (or sufficiently large)
x
2x
œ
ln x
ln ax# 1b
œ
lim
xÄ_
Š "x ‹
Š
2x
x#
1
‹
œ
Š "x ‹
1
x
ex
2 if x is sufficiently large
œ0
lim
xÄ_
x# "
#x #
œ
lim ˆ " xÄ_ #
lim
xÄ_
f(x)
g(x)
œLÁ0 Ê
lim
xÄ_
f(x)
L¹ 1 if x is sufficiently large Ê L 1 ¹ g(x)
f(x)
g(x)
L1 Ê
f(x)
g(x)
11. If f(x) and g(x) grow at the same rate, then
Ê f œ O(g). Similarly,
g(x)
f(x)
" ‰
#x#
g(x)
f(x)
œ
œ
"
L
"
#
Á 0. Then
Ÿ kLk 1 if x is sufficiently large
Ÿ ¸ L" ¸ 1 Ê g œ O(f).
12. When the degree of f is less than the degree of g since in that case
lim
xÄ_
f(x)
g(x)
œ 0.
f(x)
lim
œ 0 when the degree of f is smaller
x Ä _ g(x)
(the ratio of the leading coefficients) when the degrees are the same.
13. When the degree of f is less than or equal to the degree of g since
than the degree of g, and
lim
xÄ_
f(x)
g(x)
œ
a
b
14. Polynomials of a greater degree grow at a greater rate than polynomials of a lesser degree. Polynomials of the
same degree grow at the same rate.
15.
lim
xÄ_
œ
16.
ln (x ")
ln x
lim
xÄ_
lim
xÄ_
œ
x
x 999
ln (x a)
ln x
œ
lim
xÄ_
Šx " 1‹
Š x "999 ‹
œ lim xx 1 œ lim "1 œ 1 and lim ln (xlnx999) œ lim
"
xÄ_
xÄ_
xÄ_
x Ä _ Š x" ‹
Šx‹
œ1
lim
xÄ_
Šx " a‹
Š x" ‹
œ
lim
xÄ_
x
x a
œ
lim
xÄ_
"
1
œ 1. Therefore, the relative rates are the same.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
445
446
17.
18.
19.
Chapter 7 Transcendental Functions
È10x 1
È x 1
lim
œ É lim "0xx 1 œ È10 and lim
œ É lim x x 1 œ È1 œ 1. Since the growth rate
Èx
xÄ_
xÄ_
x Ä _ Èx
xÄ_
È
È
is transitive, we conclude that 10x 1 and x 1 have the same growth rate ˆthat of Èx‰ .
È x% x
x#
È
%
$
x x
œ É lim x x% x œ 1 and lim
œ É lim x x% x œ 1. Since the growth rate is
x#
xÄ_
xÄ_
xÄ_
È
%
%
$
È
transitive, we conclude that x x and x x have the same growth rate athat of x# b .
lim
xÄ_
lim
xÄ_
xn
ex
œ
%
nxnc1
ex
lim
xÄ_
œá œ
n!
ex
lim
xÄ_
%
$
œ 0 Ê xn œ o aex b for any non-negative integer n
20. If p(x) œ an xn an1 xn1 á a" x a! , then
nc1
n
lim p(x) œ an lim xex an1 lim xex á
x Ä _ ex
xÄ_
xÄ_
x
"
a" lim ex a! lim ex where each limit is zero (from Exercise 19). Therefore, lim p(x)
œ0
xÄ_
xÄ_
x Ä _ ex
x
Ê e grows faster than any polynomial.
21. (a)
lim
xÄ_
x1În
ln x
œ
lim
xÄ_
xÐ1 nÑÎn
n ˆ "x ‰
œ ˆ "n ‰
lim x1În œ _ Ê ln x œ o ˆx1În ‰ for any positive integer n
xÄ_
'
(b) ln ae17ß000ß000 b œ 17,000,000 Še"(‚"! ‹
1Î10'
œ e"( ¸ 24,154,952.75
(c) x ¸ 3.430631121 ‚ 10"&
(d) In the interval c3.41 ‚ 10"& ß 3.45 ‚ 10"& d we have
ln x œ 10 ln (ln x). The graphs cross at about
3.4306311 ‚ 10"& .
22.
lim
xÄ_
œ
ln x
an xn anc1 xn 1 á a" x a!
lim
xÄ_
lim Š ln x ‹
xÄ_ xn
anc1
Ša n x á a"
xn 1
a!
xn
Ê ln x grows slower than any non-constant polynomial (n
23. (a)
lim
nÄ_
n log2 n
n alog2 nb#
œ
slower than n (log2 n)# ;
Š "n ‹
1“
œ
lim
xÄ_
"
aa n b a n x n b
œ0
1)
(b)
œ 0 Ê n log2 n grows
n log2 n
n$Î#
lim
nÄ_
1Îx
lim ’
xÄ_ nxn
an
œ
lim
nÄ_
n
Š ln
ln 2 ‹
n"Î#
"
lim
œ ln2# lim n"Î#
œ0
n Ä _ ˆ "# ‰ n "Î#
nÄ_
$Î#
Ê n log2 n grows slower than n . Therefore, n log2 n
grows at the slowest rate Ê the algorithm that takes
O(n log2 n) steps is the most efficient in the long run.
œ
"
ln #
"
log2 n
lim
nÄ_
‹
œ
#
24. (a)
lim
nÄ_
(log2 n)#
n
œ
lim
nÄ_
œ
2
(ln 2)#
œ
2(ln n) Š "n ‹
(ln 2)#
lim
nÄ_
than n;
lim
nÄ_
œ
lim
nÄ_
n
Š ln
ln 2 ‹
n"Î#
œ
Š "n ‹
1
œ
lim
nÄ_
lim
nÄ_
(ln n)#
n(ln 2)#
(b)
ln n
n
œ 0 Ê (log2 n)# grows slower
"
ln #
Š"‹
n
2
(ln 2)#
(log2 n)#
Èn log2 n
œ
n
Š ln
ln 2 ‹
œ
lim
nÄ_
lim
nÄ_
log2 n
Èn
ln n
n"Î#
n
"
lim
œ ln2# lim n"Î#
œ 0 Ê (log2 n)# grows slower than Èn log2 n. Therefore (log2 n)# grows
x Ä _ ˆ "# ‰ n "Î#
nÄ_
at the slowest rate Ê the algorithm that takes O a(log2 n)# b steps is the most efficient in the long run.
œ
"
ln #
lim
nÄ_
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Chapter 7 Practice Exercises
25. It could take one million steps for a sequential search, but at most 20 steps for a binary search because
2"* œ 524,288 1,000,000 1,048,576 œ 2#! .
26. It could take 450,000 steps for a sequential search, but at most 19 steps for a binary search because
2") œ 262,144 450,000 524,288 œ 2"* .
CHAPTER 7 PRACTICE EXERCISES
1. y œ 10ecxÎ5 Ê
3. y œ
"
4
"
16
xe4x dy
dx
œ (10) ˆ 5" ‰ e xÎ5 œ 2e xÎ5
dy
dx
œ
"
4
Ê
dy
dx
œ x# ca2x# b e
e4x Ê
cx a4e4x b e4x (1)d 4. y œ x# ec2Îx œ x# e
2x "
5. y œ ln asin# )b Ê
dy
d)
œ
2(sin ))(cos ))
sin# )
6. y œ ln asec# )b Ê
dy
d)
œ
2(sec ))(sec ) tan ))
sec# )
#
#
7. y œ log2 Š x# ‹ œ
ln Š x# ‹
ln #
8. y œ log5 (3x 7) œ
9. y œ 8ct Ê
dy
dt
11. y œ 5x$Þ' Ê
12. y œ È2 x
Ê
dy
dx
ln (3x7)
ln 5
2. y œ È2 e
œ
Ê
2 cos )
sin )
œ
"
ln #
2x "
"
16
de
2x "
(2x) œ (2 2x)e
dy
dx
Š x# ‹ œ
x
#
œ 8ct (ln 8)(1) œ 8ct (ln 8)
œ 2e 2Îx (1 x)
3
(ln 5)(3x7)
10. y œ 92t Ê
dy
dt
œ 92t (ln 9)(2) œ 92t (2 ln 9)
œ 5(3.6)x#Þ' œ 18x#Þ'
dy
dx
È2 Ê
dy
dx
œ ŠÈ2‹ ŠÈ2‹ xŠ
È 2 1‹
œ 2xŠ
È 2 1‹
Ê y w œ # ln" x ˆ "# ‰ ln (ln x)‘ 2 (ln x)xÎ2 œ (ln x)xÎ2 ln (ln x) 15. y œ sin" È1 u# œ sin" a1 u# b
œ
2x "
2
(ln 2)x
w
y
y
œ (x 2)xb2 cln (x 2) 1d
u
uÈ 1 u#
È2x œ 2eÈ2x
œ 2 tan )
"
È 1 u#
"Î#
Ê
dy
du
œ
"
#
a1 u # b
1 Î2
" ‰
œ (x 2) ˆ x#
(1) ln (x 2)
w
14. y œ 2(ln x)xÎ2 Ê ln y œ ln c2(ln x)xÎ2 d œ ln (2) ˆ x# ‰ ln (ln x) Ê
œ
œ ŠÈ 2 ‹ Š È 2 ‹ e
œ 2 cot )
13. y œ (x 2)xb2 Ê ln y œ ln (x 2)xb2 œ (x 2) ln (x 2) Ê
Ê
dy
dx
a4e4x b œ xe4x 4" e4x 4" e4x œ xe4x
œ ˆ ln"5 ‰ ˆ 3x37 ‰ œ
dy
dx
È2x Ê
y
y
ˆ"‰
œ 0 ˆ x# ‰ ’ lnx x “ (ln (ln x)) ˆ "# ‰
" ‘
ln x
(2u)
#
"Î#
Ê1 ’a1 u# b “
œ
u
È 1 u # È 1 a1 u # b
œ
u
ku k È 1 u #
,0u1
16. y œ sin" Š È"v ‹ œ sin" v"Î# Ê
17. y œ ln acos" xb Ê y w œ
Š È c"
1
x#
cosc" x
‹
dy
dv
œ
œ
"
#v
c$Î#
É1 av "Î# b#
œ
"
2v$Î# È1 v "
œ
"
2v$Î# É v
1
v
œ
È v
2v$Î# Èv 1
œ
"
È1 x# cos " x
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
"
2vÈv 1
447
448
Chapter 7 Transcendental Functions
18. y œ z cos" z È1 z# œ z cos" z a1 z# b
œ cos" z z
È 1 z#
z
È 1 z#
"Î#
Ê
dy
dz
œ cos" z dy
dt
20. y œ a1 t# b cot" 2t Ê
œ 2t cot" 2t a1 t# b ˆ 1 24t# ‰
œ tan" t t ˆ 1 " t# ‰ ˆ #" ‰ ˆ "t ‰ œ tan" t 21. y œ z sec" z Èz# 1 œ z sec" z az# 1b
œ
z
kz k È z # 1
z
È z# 1
ˆ "# ‰ a1 z# b
"Î#
(2z)
œ cos" z
19. y œ t tan" t ˆ "# ‰ ln t Ê
dy
dt
z
È 1 z#
sec" z œ
1z
È z# 1
"Î#
Ê
dy
dz
œ zŠ
"
‹
kz k È z # 1
t
1 t#
"
2t
asec" zb (1) #" az# 1b
"Î#
(2z)
sec" z, z 1
22. y œ 2Èx 1 sec" Èx œ 2(x 1)"Î# sec" ˆx"Î# ‰
Ê
Š " ‹ xc"Î#
23. y œ csc" (sec )) Ê
"x
24. y œ a1 x# b etan
25. y œ
sec " Èx
œ 2 –ˆ "# ‰ (x 1)"Î# sec" ˆx"Î# ‰ (x 1)"Î# È#xÈx 1 — œ 2 Š 2Èx 1 dy
dx
2 ax# 1b
Ècos 2x
dy
d)
œ
sec ) tan )
ksec )k Èsec# ) 1
Ê y w œ 2xetan
"x
tan
#
2 ax # 1 b
Ècos 2x
"! 3x 4
"! 3x 4
É
26. y œ É
2x 4 Ê ln y œ ln
2x 4 œ
"
10
"x
a1 x# b Š e1 x# ‹ œ 2xetan
Ê ln y œ ln Š 2Èaxcos2x1b ‹ œ ln (2) ln ax# 1b Ê y w œ ˆ x#2x 1 tan 2x‰ y œ
Ê yw œ
)
œ ktan
tan )k œ 1, 0 ) ˆ 3x 3 4 " ‰
x2 y
"
#
"
#x ‹
œ
sec " Èx
Èx 1
"x
etan
"x
ln (cos 2x) Ê
cln (3x 4) ln (2x 4)d Ê
"! 3x 4
ˆ " ‰ ˆ 3x 3 4 œÉ
2x 4 10
yw
y
œ0
2x
x # 1
2 sin 2x)
ˆ #" ‰ (cos
2x
yw
y
œ
"
10
ˆ 3x 3 4 2 ‰
2x 4
" ‰
x2
&
1)(t 1)
dy
"
27. y œ ’ (t(t 2)(t 3) “ Ê ln y œ 5 cln (t 1) ln (t 1) ln (t 2) ln (t 3)d Ê Š y ‹ Š dt ‹
œ 5 ˆ t " 1 28. y œ
Ê
"
t1
2u2u
Èu# 1 Ê ln y
dy
2u2u ˆ "
du œ Èu# 1 u
È)
29. y œ (sin ))
Ê
"
t#
dy
d)
" ‰
t3
Ê
dy
dt
œ ln 2 ln u u ln 2 ln 2 &
1)(t 1)
ˆ t " 1 œ 5 ’ (t(t 2)(t 3) “
"
#
"
t1
"
t#
dy
ln au# 1b Ê Š "y ‹ Š du
‹œ
"
u
" ‰
t3
ln 2 #" ˆ u#2u 1 ‰
u ‰
u# 1
È) ˆ cos ) ‰ " )"Î# ln (sin ))
Ê ln y œ È) ln (sin )) Ê Š "y ‹ Š dy
d) ‹ œ
sin )
#
œ (sin ))
È)
ŠÈ) cot ) ln (sin ))
‹
2È )
30. y œ (ln x)1Îln x Ê ln y œ ˆ ln"x ‰ ln (ln x) Ê
w
y
y
œ ˆ ln"x ‰ ˆ ln"x ‰ ˆ x" ‰ ln (ln x) ’ (ln"x)# “ ˆ x" ‰
ln (ln x)
Ê y w œ (ln x)1Îln x ’ 1 x(ln
x)# “
31.
"
x
1
#
ˆ x#2x 1 tan 2x‰
"
10
' ex sin aex b dx œ ' sin u du, where u œ ex and du œ ex dx
œ cos u C œ cos aex b C
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Chapter 7 Practice Exercises
32.
' et cos a3et 2b dt œ 3" ' cos u du, where u œ 3et 2 and du œ 3et dt
"
3
œ
33.
449
"
3
sin u C œ
sin a3et 2b C
' ex sec# aex 7b dx œ ' sec# u du, where u œ ex 7 and du œ ex dx
œ tan u C œ tan aex 7b C
34.
' ey csc aey 1b cot aey 1b dy œ ' csc u cot u du, where u œ ey 1 and du œ ey dy
œ csc u C œ csc aey 1b C
35.
' asec# xb etan x dx œ ' eu du, where u œ tan x and du œ sec# x dx
œ eu C œ etan x C
36.
' acsc# xb ecot x dx œ ' eu du, where u œ cot x and du œ csc# x dx
œ eu C œ ecot x C
' c1
37.
'c11
38.
'1e Èlnx x dx œ '01 u"Î# du, where u œ ln x, du œ "x dx; x œ 1
"
"
"
3x 4 dx œ 3 c7 u du, where u œ 3x 4, du œ 3 dx; x œ 1 Ê
"
"
ln 7
œ "3 cln kukd "
( œ 3 cln k1k ln k7kd œ 3 [0 ln 7] œ 3
œ
39.
23 u$Î# ‘ "
!
œ 23 1$Î# 23 0$Î# ‘ œ
Ê u œ 0, x œ e Ê u œ 1
2
3
sin ˆ ‰
"
'01 tan ˆ x3 ‰ dx œ '01 cos
' 1Î2 "
ˆx‰
ˆx‰
ˆ ‰ dx œ 3 1 u du, where u œ cos 3 , du œ 3 sin 3 dx; x œ 0
x
3
x
3
Ê uœ
"Î#
œ 3 cln kukd "
40.
'11ÎÎ64
œ 3 ln ¸ "# ¸ ln k1k‘ œ 3 ln
2 cot 1x dx œ 2'1Î6
1Î4
cos 1x
sin 1x
dx œ
2
1
È2
'11ÎÎ2
"
u
"
#
œ
'04
2t
t# 25
2
1
È
cln kukd 11ÎÎ2 2 œ
c9
dt œ 'c25
"
u
2
1
œ ln 2 œ ln 8
du, where u œ sin 1x, du œ 1 cos 1x dx; x œ
’ln ¹ È"2 ¹ ln ¸ #" ¸“ œ
2
1
ln 1 "
#
"Î#
43.
'
tan (ln v)
v
2
1
#" ln 2‘ œ
ln 2
1
"
u
9
25
du, where u œ 1 sin t, du œ cos t dt; t œ 1# Ê u œ 2, t œ
1
6
Ê uœ
œ ln ¸ "# ¸ ln k2k‘ œ ln 1 ln 2 ln 2 œ 2 ln 2 œ ln 4
dv œ ' tan u du œ '
sin u
cos u
du, where u œ ln v and du œ
"
v
dv
œ ln kcos uk C œ ln kcos (ln v)k C
44.
'
"
v ln v
dv œ '
"
u
Ê uœ
du, where u œ t# 25, du œ 2t dt; t œ 0 Ê u œ 25, t œ 4 Ê u œ 9
'c11ÎÎ62 1 cossint t dt œ '21Î2
œ cln kukd #
"
6
"
È2
ln 2 ln 1 ln 2‘ œ
œ cln kukd *
#& œ ln k9k ln k25k œ ln 9 ln 25 œ ln
42.
Ê u œ 1, x œ 1
"
#
$
Ê uœ
41.
u œ 7, x œ 1 Ê u œ 1
du, where u œ ln v and du œ
"
v
dv
œ ln kuk C œ ln kln vk C
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
"
#
"
#
,xœ
"
4
450
45.
Chapter 7 Transcendental Functions
'
(ln x)c$
x
œ
46.
'
u #
#
ln (x 5)
x5
œ
47.
dx œ ' u$ du, where u œ ln x and du œ
'
"
r
"
#
C œ (ln x)
#
"
x
dx
C
dx œ ' u du, where u œ ln (x 5) and du œ
#
Cœ
u
#
cln (x5)d
2
#
"
x 5
dx
C
csc# (1 ln r) dr œ ' csc# u du, where u œ 1 ln r and du œ
"
r
dr
œ cot u C œ cot (1 ln r) C
48.
'
cos (1 ln v)
v
dv œ ' cos u du, where u œ 1 ln v and du œ "v dv
œ sin u C œ sin (1 ln v) C
49.
'
#
œ
50.
'
"
#
x3x dx œ
"
# ln 3
'
3u du, where u œ x# and du œ 2x dx
a3u b C œ
"
# ln 3
#
Š3x ‹ C
2tan x sec# x dx œ ' 2u du, where u œ tan x and du œ sec# x dx
œ
"
ln #
2tan x
ln #
a2u b C œ
51.
'17
52.
"
"
È
'132 5x" dx œ 5" '132 x" dx œ 5" cln kxkd $#
" œ 5 aln 32 ln 1b œ 5 ln 32 œ ln Š 32‹ œ ln 2
53.
15
'14 ˆ x8 #"x ‰ dx œ #" '14 ˆ 4" x x" ‰ dx œ #" 8" x# ln kxk‘ %" œ #" ˆ 168 ln 4‰ ˆ 8" ln 1‰‘ œ 16
#" ln 4
ln È4 œ
15
16
ˆln 8 3
#
15
16
ln 2
12‰ œ
2
3
ˆln 8 21 ‰
#
œ
2
3
(ln 8) 7 œ ln ˆ8#Î$ ‰ 7 œ ln 4 7
'cc21 eÐx1Ñ dx œ '10 eu du, where u œ (x 1), du œ dx; x œ 2
œ ce
'c0ln 2
d !"
e2w dw œ
œ
57.
dx œ 3 cln kxkd (" œ 3 aln 7 ln 1b œ 3 ln 7
#
2
3
u
56.
"
x
'18 ˆ 3x2 x8 ‰ dx œ 23 '18 ˆ "x 12x# ‰ dx œ 23 cln kxk 12x" d )" œ 23 ˆln 8 128 ‰ (ln 1 12)‘
œ
55.
7
3
x
&
œ
54.
dx œ 3'1
C
"
#
!
Ê u œ 1, x œ 1 Ê u œ 0
"
œ ae e b œ e 1
"
#
'ln0Ð1Î4Ñ eu du, where u œ 2w, du œ 2 dw; w œ ln 2
ceu d 0ln Ð1Î4Ñ œ
"
#
ce! eln Ð1Î4Ñ d œ
"
#
ˆ1 4" ‰ œ
Ê u œ ln "4 , w œ 0 Ê u œ 0
3
8
'1ln 5 er a3er 1b$Î# dr œ "3 '416 u$Î# du, where u œ 3er 1, du œ 3er dr; r œ 0
Ê u œ 4, r œ ln 5 Ê u œ 16
"'
œ 23 u"Î# ‘ % œ 23 ˆ16"Î# 4"Î# ‰ œ ˆ 23 ‰ ˆ 4" #" ‰ œ ˆ 23 ‰ ˆ 4" ‰ œ
58.
'0ln 9 e
)
"
6
ae) 1b"Î# d) œ '0 u"Î# du, where u œ e) 1, du œ e) d); ) œ 0 Ê u œ 0, ) œ ln 9 Ê u œ 8
8
œ
2
3
u$Î# ‘ ) œ
!
2
3
ˆ8$Î# 0$Î# ‰ œ
2
3
ˆ2*Î# 0‰ œ
2""Î#
3
œ
32È2
3
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Chapter 7 Practice Exercises
59.
'1e "x (1 7 ln x)"Î$ dx œ 7" '18 u"Î$ du, where u œ 1 7 ln x, du œ x7 dx, x œ 1
œ
60.
'ee
#
"
xÈln x
3 #Î$ ‘ )
14 u
"
œ
3
14
3 ‰
ˆ8#Î$ 1#Î$ ‰ œ ˆ 14
(4 1) œ
dx œ 'e (ln x)"Î#
e#
Ê u œ 1, x œ e Ê u œ 8
9
14
dx œ '1 u"Î# du, where u œ ln x, du œ
2
"
x
451
"
x
dx; x œ e Ê u œ 1, x œ e# Ê u œ 2
#
œ 2 u"Î# ‘ " œ 2 ŠÈ2 1‹ œ 2È2 2
61.
'13 [ln (vv 11)]
dv œ '1 [ln (v 1)]#
3
#
"
v1
dv œ 'ln 2 u# du, where u œ ln (v 1), du œ
ln 4
"
v 1
dv;
v œ 1 Ê u œ ln 2, v œ 3 Ê u œ ln 4;
œ
62.
"
3
"
3
$ ln 4
cu d ln 2 œ
$
$
c(ln 4) (ln 2) d œ
"
3
$
$
c(2 ln 2) (ln 2) d œ
(ln 2)$
3
(8 1) œ
7
3
(ln 2)$
'24 (1 ln t)(t ln t) dt œ '24 (t ln t)(1 ln t) dt œ '24lnln24 u du, where u œ t ln t, du œ ˆ(t) ˆ "t ‰ (ln t)(1)‰ dt
œ (1 ln t) dt; t œ 2 Ê u œ 2 ln 2, t œ 4
Ê u œ 4 ln 4
œ
63.
"
#
cu# d 2 ln 2 œ
4 ln 4
"
#
c(4 ln 4)# (2 ln 2)# d œ
c(8 ln 2)# (2 ln 2)# d œ
(2 ln 2)#
#
(16 1) œ 30 (ln 2)#
'18 log) ) d) œ ln"4 '18 (ln )) ˆ ") ‰ d) œ ln"4 '0ln 8 u du, where u œ ln ), du œ ") d), ) œ 1
4
œ
64.
"
#
'1e
"
# ln 4
cu # d
ln 8
!
"
ln 16
œ
d) œ '1
8(ln 3)(log3 ))
)
c(ln 8)# 0# d œ
e
8(ln 3)(ln ))
)(ln 3)
#
#
(3 ln 2)#
4 ln 2
œ
Ê u œ 0, ) œ 8 Ê u œ ln 8
9 ln 2
4
d) œ 8 '1 (ln )) ˆ ") ‰ d) œ 8'0 u du, where u œ ln ), du œ
e
1
"
)
d) ;
) œ 1 Ê u œ 0, ) œ e Ê u œ 1
œ
65.
'c33ÎÎ44 È
"
4 cu # d !
6
9 4x#
œ 4 a1 0 b œ 4
3Î4
3Î2
dx œ 3 ' 3Î4 È3# 2 (2x)# dx œ 3' 3Î2 È #" # du, where u œ 2x, du œ 2 dx;
3 u
x œ 34 Ê u œ 3# , x œ
Ê uœ
3
4
$Î#
3
#
œ 3 sin" ˆ u3 ‰‘ $Î# œ 3 sin" ˆ "# ‰ sin" ˆ "# ‰‘ œ 3 16 ˆ 16 ‰‘ œ 3 ˆ 13 ‰ œ 1
66.
'c11ÎÎ55
6
È4 25x#
œ
67.
'c22
3
4 3t#
6
5
dx œ
"
sin
6
5
' 11ÎÎ55
ˆ u2 ‰‘ "
"
dt œ È3 'c2
5
È2# (5x)#
œ
6
5
"
sin
È3
È3
œ È3 "# tan" ˆ u2 ‰‘ c2È3 œ
'È33
69.
'
"
3 t#
dt œ 'È3
"
yÈ4y# 1
3
"
#
ŠÈ3‹ t#
dy œ '
' 11 È
ˆ "# ‰
"
sin
"
2# u#
È3
ˆ
dt œ È3 'c2È3
2
#
ŠÈ3t‹
2
68.
6
5
du, where u œ 5x, du œ 5 dx;
x œ 15 Ê u œ 1, x œ
2
2#
dx œ
"
2# u#
œ sec" kuk C œ sec
œ
6
5
16
ˆ
1 ‰‘
6
œ
6
5
ˆ 13 ‰
œ
Ê uœ1
21
5
du, where u œ È3t, du œ È3 dt;
t œ 2 Ê u œ 2È3, t œ 2 Ê u œ 2È3
È3
#
’tan" ŠÈ3‹ tan" ŠÈ3‹“ œ
dt œ ’ È"3 tan" Š Èt 3 ‹“
2
(2y)È(2y)# 1
"
" ‰‘
#
1
5
dy œ '
"
uÈ u# 1
3
È3 œ
"
È3
È3
#
13 ˆ 13 ‰‘ œ
Štan" È3 tan" 1‹ œ
"
È3
1
È3
ˆ 13 14 ‰ œ
du, where u œ 2y and du œ 2 dy
k2yk C
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
È 31
36
452
Chapter 7 Transcendental Functions
70.
'
71.
'È2Î23Î3
dy œ 24 '
24
yÈy# 16
"
kyk È9y# 1
"
yÈ y# 4#
dy œ 24 ˆ 4" sec" ¸ 4y ¸‰ C œ 6 sec" ¸ 4y ¸ C
3
'È2
dy œ 'È2Î3 k3yk È(3y)
# 1 dy œ
2Î3
2
"
ku k È u # 1
du, where u œ 3y, du œ 3 dy;
yœ
œ csec" ud È2 œ ’sec" 2 sec" È2“ œ
2
72.
È È5
'cc2/È65Î
"
kyk È5y# 3
È6ÎÈ5
dy œ 'c2/È5
1
3
1
4
È6
#
dy œ 'c2
#
È5y ÊŠÈ5y‹ ŠÈ3‹
Ê u œ È2, y œ
2
3
Ê uœ2
1
12
œ
È5
È2
3
"
u Ê u #
#
du,
ŠÈ3‹
È
œ ’ È"3 sec" ¹ Èu3 ¹“
73.
'
dx œ '
"
È2x x#
cÈ6
c2
where u œ È5y, du œ È5 dy; y œ È25 Ê u œ 2, y œ È56 Ê u œ È6
œ
"
È1 ax# 2x 1b
1
È3
’sec" È2 sec"
dx œ '
"
È1 (x 1)#
2
È3 “
dx œ '
œ
"
È3
"
È 1 u#
ˆ 14 16 ‰ œ
"
È3
3121 21 ‘
1#
œ
1
12È3
È 3 1
36
œ
du, where u œ x 1 and
du œ dx
"
"
œ sin
74.
'È
u C œ sin
dx œ '
"
x# 4x 1
(x 1) C
"
È3 ax# 4x 4b
dx œ '
"
#
ÊŠÈ3‹ (x 2)#
dx œ '
"
du
#
ÊŠÈ3‹ u#
where u œ x 2 and du œ dx
œ sin" Š Èu3 ‹ C œ sin" Š xÈ32 ‹ C
75.
'cc21 v 4v2 5 dv œ 2 'cc21 1 av " 4v 4b dv œ 2 'cc21 1 (v" 2)
#
#
"
œ 2 ctan
76.
77.
'c11
3
4v# 4v 4
dv œ
"
œ 2 atan
3
4
'c11
2u
’ È23 tan" Š È
‹“
3
œ
3
4
œ
È 31
4
'
"
ud !
"
(t 1)Èt# 2t 8
3
4
$Î#
"Î#
dt œ '
Šv#
dv œ 2'0
1
#
"
1 u#
du,
where u œ v 2, du œ dv; v œ 2 Ê u œ 0, v œ 1 Ê u œ 1
1
ˆ
0b œ 2 4 0‰ œ 1#
1 tan
"
"
v 4" ‹
dv œ
3
4
'c11
Œ
È3
#
#
"
Šv #
"
#‹
dv œ
3
4
'c31ÎÎ22
Œ
È3
#
"
du
#
u#
where u œ v "# , du œ dv; v œ 1 Ê u œ "# , v œ 1 Ê u œ
œ
È3
#
’tan" È3 tan" Š È"3 ‹“ œ
"
(t 1)Èat# 2t 1b 9
dt œ '
È3
#
"
(t 1)È(t 1)# 3#
13 ˆ 16 ‰‘ œ
dt œ '
"
uÈ u# 3#
È3
#
ˆ 261 16 ‰ œ
È3
#
†
1
#
du
where u œ t 1 and du œ dt
œ
78.
'
"
3
"
sec
"
(3t 1)È9t# 6t
¸ 3u ¸ C œ
dt œ '
"
3
"
sec
¸ t31 ¸ C
"
(3t 1)Èa9t# 6t 1b 1
dt œ '
"
(3t 1)È(3t 1)# 1#
dt œ
"
3
'
"
uÈ u# 1
du
where u œ 3t 1 and du œ 3 dt
œ
"
3
"
sec
ku k C œ
"
3
"
sec
k3t 1k C
79. 3y œ 2y1 Ê ln 3y œ ln 2y1 Ê y(ln 3) œ (y 1) ln 2 Ê (ln 3 ln 2)y œ ln 2 Ê ˆln 3# ‰ y œ ln 2 Ê y œ
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
ln 2
ln Š 3# ‹
3
#
Chapter 7 Practice Exercises
453
80. 4y œ 3y2 Ê ln 4y œ ln 3y2 Ê y ln 4 œ ay 2b ln 3 Ê 2 ln 3 œ aln 3 ln 4by Ê aln 12by œ 2 ln 3
9
Ê y œ lnln12
x#
9
81. 9e2y œ x# Ê e2y œ
#
#
82. 3y œ 3 ln x Ê ln 3y œ ln (3 ln x) Ê y ln 3 œ ln (3 ln x) Ê y œ
#
ln Š x9 ‹ œ ln É x9 œ ln ¸ x3 ¸ œ ln kxk ln 3
ln 3 ln (ln x)
ln 3
œ
ln (3 ln x)
ln 3
#
"
#
Ê ln e2y œ ln Š x9 ‹ Ê 2y(ln e) œ ln Š x9 ‹ Ê y œ
83. ln (y 1) œ x ln y Ê eln Ðy1Ñ œ eÐxln yÑ œ ex eln y Ê y 1 œ yex Ê y yex œ 1 Ê y a1 ex b œ 1 Ê y œ
84. ln (10 ln y) œ ln 5x Ê eln Ð10 ln yÑ œ eln 5x Ê 10 ln y œ 5x Ê ln y œ
85. lim
x # $x %
x"
87. xlim
Ä1
tan x
x
xÄ"
89. lim
œ
sin# x
#
x Ä ! tanax b
90. lim
sinamxb
x Ä ! sinanxb
91. lim
x Ä 1Î#c
92.
lim
x Ä !b
œ lim
xÄ"
tan 1
1
#x $
"
œ5
b
x Ä " x "
œ!
x Ä ! x sin x
# #
x Ä ! #x sec ax b
œ lim
xÄ!
m cosamxb
n cosanxb
sina#xb
œ lim
# #
x Ä ! #x sec ax b
œ
cosa$xb
x Ä 1Î#c cosa(xb
Èx sec x œ lim
x Ä !b
Èx
cos x
œ
!
"
" cos x
sin x
94. lim ˆ x"% #
xÄ!
"‰
x#
# #
#
# #
x Ä ! #x a#sec ax btanax b†#xb #sec ax b
$sina$xb
œ lim
x Ä 1Î#c (sina(xb
œ lim
sin x
x Ä ! cos x
œ
œ lim Š " x%x ‹ œ lim a" x# b †
xÄ!
œ
"
""
œ
#
! #†"
œ"
xÄ!
!
"
"
x%
œ!
"
œ lim a" x# b œ lim
%
xÄ! x
xÄ!
œ"†_œ_
Èx# x "Èx# x
È x# x " È x# x
x x
Notice that x œ Èx# for x ! so this is equivalent to
œ x lim
Ä_
#x
#
Éx
$
x
"
" É x # x
x#
x#
x
$
œ x lim
Ä_
96. x lim
Š x#x " x#x " ‹ œ lim_
Ä_
xÄ
"#
"
œ x lim
œ x lim
œ!
Ä _ #%x
Ä _ #x
œ
$
(
95. x lim
ŠÈx# x " Èx# x‹ œ x lim
Š È x # x " È x # x‹ †
Ä_
Ä_
#x "
œ x lim
È #
Ä_ È #
x x"
œ
x Ä ! " cos x
œ!
93. lim acsc x cot xb œ lim
xÄ!
a
b
sec# x
œ lim
#cosa#xb
œ lim
œ
m
n
seca(xbcosa$xb œ lim
xÄ!
axa "
b "
x Ä " bx
œ lim
tan x
88. lim
#sin x†cos x
œ lim
xa "
86. lim
Ê eln y œ exÎ2 Ê y œ exÎ2
x
#
# x"
œ È"# È" œ "
"
É" x x"# É" x"
x $ ax # " b x $ a x # " b
ax# "bax# "b
œ x lim
Ä_
97. The limit leads to the indeterminate form 00 : lim
10x 1
x
98. The limit leads to the indeterminate form 00 : lim
3) 1
)
xÄ0
)Ä0
œ lim
)Ä0
œ x lim
Ä_
(ln 10)10x
1
œ lim
xÄ0
2sin x 1
x
x Ä 0 e 1
99. The limit leads to the indeterminate form 00 : lim
#x $
x% "
(ln 3)3)
1
œ lim
xÄ0
'x #
%x $
œ x lim
Ä_
"#x
"#x#
œ ln 10
œ ln 3
2sin x (ln 2)(cos x)
ex
œ ln 2
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
"
#
"
1 ex
454
Chapter 7 Transcendental Functions
2c sin x "
ex 1
100. The limit leads to the indeterminate form 00 : lim
xÄ0
œ lim
xÄ0
5 5 cos x
101. The limit leads to the indeterminate form 00 : lim
œ lim
x
x Ä 0 e x1
x sin x2
tan3 x
102. The limit leads to the indeterminate form 00 : lim
xÄ0
œ lim
6x cos x2 4x3 sin x2
3
2
2
x Ä 0 12tan x sec x 6tan x sec x
6x cos x2 4x3 sin x2
œ lim
103. The limit leads to the indeterminate form 00 :
tÄ!
œ lim
xÄ4
1 sin (21x)
ex 4 1
21# cos (21x)
ex 4
œ lim
xÄ4
106. The limit leads to the indeterminate form
lim
y Ä !
œ5
5 cos x
ex
2x2 cos x2 sin x2
3tan4 x 3tan2 x
œ lim
xÄ0
4
2
2
2
2
x Ä 0 60tan x sec x 54tan x sec x 6sec x
œ
lim
t Ä !
œ lim
xÄ4
Š1 1 b2 2t ‹
2t
œ
6
6
œ1
œ _
21(sin 1x)(cos 1x)
ex 4 1
œ 21 #
105. The limit leads to the indeterminate form 00 :
œ
œ ln 2
ˆ6 8x4 ‰cos x2 24x2 sin x2
œ lim
sin# (1x)
x 43x
e
xÄ4
104. The limit leads to the indeterminate form 00 : lim
2x2 cos x2 sin x2
3tan2 x sec2 x
xÄ0
t ln (1 2t)
t#
lim œ lim
xÄ0
5 sin x
ex 1
xÄ0
œ lim
5
3
x Ä 0 12tan x 18tan x 6tan x
2c sin x (ln 2)( cos x)
ex
Š et "t ‹ œ lim Š e "
t ‹ œ lim t
lim
t Ä !
_
_:
t
tÄ!
lim e1Îy ln y œ
yÄ!
tÄ!
ln y
eyc"
lim yÄ!
œ
lim yÄ!
et
1
œ1
y "
e y
" ˆy #‰
Š eyyc" ‹ œ 0
1‰
107. Let f(x) œ ˆ eex 1
ln x
1‰
1‰
Ê ln f(x) œ ln x ln ˆ eex ln f(x) œ x lim
ln x ln ˆ eex 1 Ê x lim
1 ; this is limit is currently of
Ä_
Ä_
x
1
exÎ2 e xÎ2
ˆx‰
the form 0 † _. Before we put in one of the indeterminate forms, we rewrite eex 1 œ exÎ2 e xÎ2 œ coth 2 ; the limit is
x
x
ln cothˆ x2 ‰
lim ln x ln cothˆ x2 ‰ œ x lim
xÄ_
Ä_
Î
Ð
œ x lim
Ä_
Ï
csch2 Š x2 ‹
cothŠ x2 ‹
1
ln x
x
; the limit leads to the indeterminate form 00 : x lim
Ä_
ˆ "# ‰ Ñ
1 2 ˆ x1 ‰
aln xb
ln cothˆ x2 ‰
1
ln x
2
2xaln xbˆ x ‰ aln xb
a b
aln xb
Ó
œ x lim
Š 2 sinhxˆ xln‰ xcoshˆ x ‰ ‹ œ x lim
Š xsinh
‹ œ x lim
Š
‹
x
cosh x
Ä_
Ä
_
Ä
_
2
2
Ò
2
2
aln xb
œ x lim
Š 2ln xcosh
‹ œ x lim
Š
x
Ä_
Ä_
1
2
2ˆ 1x ‰ 2aln xbˆ 1x ‰
‹
sinh x
2
2ln x ‰
ˆ 2xsinh
œ x lim
Š x cosh xx sinh x ‹
x œ x lim
Ä_
Ä_
2
1 ‰ln x
ˆ
‰ œ 0 Ê lim ˆ eex œ x lim
œ x lim
eln fÐxÑ œ e0 œ 1
1
Ä _ x2 cosh x x sinh x
xÄ_
Ä_
x
x
108. Let f(x) œ ˆ1 3x ‰ Ê ln f(x) œ x ln ˆ1 3x ‰ Ê
indeterminate form
109. (a) x lim
Ä_
log2 x
log3 x
_
_:
lim
x Ä !b
œ x lim
Ä_
c#
Š c3x " ‹
1 3x
x
Š ln
ln 2 ‹
x
Š ln
ln 3 ‹
œ x lim
Ä_
xc#
œ lim b
xÄ!
œ x lim
Ä_
x#
x # 1
ln 3
ln 2
œ x lim
Ä_
(b) x lim
Ä_
x
x Š "x ‹
(c) x lim
Ä_
(d) x lim
Ä_
ˆ x ‰
xex
ex
100
œ x lim
xe x œ x lim
Ä _ 100x
Ä _ 100
x
tan " x œ _ Ê faster
2x
#x
œ
lim
xÄ!
3x
x3
csc " x
Š "x ‹
œ x lim
Ä_
sin " ax " b
x "
xÄ!
œ0 Ê
lim
x Ä !b
the limit leads to the
ˆ1 3x ‰x œ lim eln fÐxÑ œ e! œ 1
x Ä !b
Ê same rate
ln 3
ln #
œ x lim
" œ 1 Ê same rate
Ä_
œ _ Ê faster
Š x
(e) x lim
Ä_
ln a1 3xc" b
;
x "
ln f(x) œ lim
œ x lim
Ä_
Ê1 Šx
#‹
x #
" ‹#
œ x lim
Ä_
"
"
Ê 1 Š x# ‹
œ 1 Ê same rate
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Chapter 7 Practice Exercises
$
10x 2x
ex
(c) x lim
Ä_
Š "x ‹
(b)
"
‹
x%
"
Š #‹
x
Š
"
x#
œ x lim
Ä_
tan " ax " b
x "
"
"
lim sin x a#x b
xÄ_
"
x#
œ1
"
‹
x%
"
Š %‹
x
"ec2x
#
Š ex b2ecx ‹
ecx
œ x lim
Ä_
œ x lim
Ä_
œ x lim
Ä_
œ x lim
Ä_
œ
"
#
Ê same rate
"# ‰ œ
60x4
ex
Š x
1x
Ê same rate
œ x lim
Ä_
#
#‹
x #
x
É1
"
#
ˆx
#
œ x lim
Ä_
"‰# 2x $
2
ecx aex ecx b
60
ex
œ 0 Ê slower
"
1 "#
œ 1 Ê same rate
x
œ x lim
Ä_
x
2É1 x"#
œ _ Ê faster
ˆ 2 ‰ œ 2 Ê same rate
œ x lim
Ä _ 1 ec2x
Ÿ 2 for x sufficiently large Ê true
œ x# 1 M for any positive integer M whenever x ÈM Ê false
œ x lim
Ä_
x
x ln x
(c) x lim
Ä_
30x 4x
ex
œ x lim
Ä_
sech x
ecx
#
œ x lim
Ä_
œ
Š "# ‹
x
(f) x lim
Ä_
x
sinc" Š "x ‹
(e) x lim
Ä_
"
x#
#
tanc" Š " ‹
(d) x lim
Ä_
Š
œxÄ
lim_
3cx
ˆ 23 ‰x œ 0 Ê slower
2cx œ x lim
Ä_
ln 2x
ln 2 ln x
ˆ ln 2
œ x lim
ln x# œ x lim
Ä _ 2 (ln x)
Ä _ # ln x
110. (a) x lim
Ä_
(b) x lim
Ä_
111. (a)
aex ecx b
#e x
œxÄ
lim_
sinh x
ex
(f) x lim
Ä_
"
1
"
x
œ 1 Ê the same growth rate Ê false
Š x" ‹
(e)
(f)
112. (a)
tan " x
1
cosh x
ex
œ
"
‹
x%
Š "# "% ‹
x
x
Š
Š
x
(c) x lim
Ä_
(d) lnln2x
x œ
(f)
113.
df
dx
"
x # 1
œ
"
%‹
Ÿ 1 if x 0 Ê true
ˆ " ‰ œ 0 Ê true
œ x lim
Ä _ x# 1
"
‹
x%
cos " Š "x ‹
1
Ÿ
ˆ1‰
#
114. y œ f(x) Ê y œ 1 f w (x) œ x"# Ê
"
#
"
Šx
"
x œ fÐln 2Ñ
"
x
1‹
Ê
true
2 Ê true
if x 1 Ê true
if x 0 Ê true
c"
œ ex 1 Ê Š dfdx ‹
1
#
œ
1
a1 ec2x b Ÿ
f af " (x)b œ 1 œ 0 Ê grows slower Ê true
1) œ 1 if x 0 Ê true
Š "x ‹
ln x
œ
lim
œ0 Ê
x1
xÄ_ 1
ln 2
ln x 1 Ÿ 1 1 œ 2 if x
secc" x
œ
1
sinh x
"
œ
ex
#
"
ln x
œ x lim
Ä_
ˆ "x ‰
1
# for all x Ê true
"
c2x b Ÿ " (1 # a1 e
#
Ÿ
x
(b) x lim
Ä _ Š "# (e)
œ x lim
Ä_
ln (ln x)
ln x
(d) x lim
Ä_
– ln x —
"
x
œ
"
df
Š dx
‹
x œ ln 2
œy1 Ê xœ
œ 1 (x 1) œ x;
df "
dx ¹ fÐxÑ
œ
"
Ê Š dfdx ‹
"
y 1
df c"
dx ¹ fÐxÑ
œ
x œ fÐln 2Ñ
œ
"
aex 1bx œ ln 2
Ê f " (x) œ
"
(x 1)# ¹ fÐxÑ
œ
"
x 1
œ
"
#1
œ
; f " (f(x)) œ
"
’Š1 x" ‹1“
#
"
3
"
Š1 "x ‹1
œ x# ;
"
f w (x)
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
œ
"
Š x" ‹
œ x and
455
456
Chapter 7 Transcendental Functions
2 ‰
115. y œ x ln 2x x Ê y w œ x ˆ 2x
ln (2x) 1 œ ln 2x;
"
#
"
w
# and y 0 for
x "# Ê relative minimum of "# at x œ "# ; f ˆ #"e ‰ œ "e
and f ˆ #e ‰ œ 0 Ê absolute minimum is "# at x œ "# and
the absolute maximum is 0 at x œ #e
solving y w œ 0 Ê x œ
; y w 0 for x 116. y œ 10x(2 ln x) Ê y w œ 10(2 ln x) 10x ˆ "x ‰
œ 20 10 ln x 10 œ 10(1 ln x); solving y w œ 0
Ê x œ e; y w 0 for x e and y w 0 for x e
Ê relative maximum at x œ e of 10e; y ! on Ð!ß e# Ó and
y ae# b œ 10e# (2 2 ln e) œ 0 Ê absolute minimum is 0
at x œ e# and the absolute maximum is 10e at x œ e
117. A œ '1
e
dx œ '0 2u du œ cu# d ! œ 1, where u œ ln x and du œ
1
2 ln x
x
118. (a) A" œ '10
20
(b) A" œ 'ka
kb
119. y œ ln x Ê
dy
dx
120. y œ 9ecxÎ3 Ê
Ê
dx ¸
dt xœ9
dx œ cln kxkd #!
"! œ ln 20 ln 10 œ ln
"
x
"
x
kb
dx œ cln kxkd ka
œ ln kb ln ka œ ln
"
x
œ
dy
dx
Š
xœ
;
dA
dx
dA
dx
œ
dx
dt
"
4
(dy/dt)
(dy/dx)
œ ˆ "x ‰ Èx œ
Ê
dx
dt
œ
œ ln 2, and A# œ '1
2
œ ln
kb
ka
dx; x œ 1 Ê u œ 0, x œ e Ê u œ 1
"
Èx
b
a
Ê
Š "4 ‹ È9 y
3e xÎ3
#
Ê
dy
dt ¹ e#
dx œ cln kxkd #" œ ln 2 ln 1 œ ln 2
œ
"
e
b
"
x
dx œ cln kxkd ab œ ln b ln a
m/sec
; x œ 9 Ê y œ 9e$
#
"
È2
dA
dx
and
"
Èe
œ
œ
"
x#
dA
dx
#
0 for 0 x "
È2
ln x
2x$Î#
œ
dA
dx
œ 0 Ê 1 2x# œ 0
Ê absolute maximum of
"
È2
e"Î# œ
ln x
x#
œ
1ln x
x#
. Solving
dA
dx
at
œ 0 Ê 1 ln x œ 0 Ê x œ e;
ln e
e
œ
"
e
at x œ e units long and y œ
2 ln x
2xÈx
Ê y ww œ 34 x&Î# (2 ln x) "# x&Î# œ x&Î# ˆ 34 ln x 2‰ ;
solving y w œ 0 Ê ln x œ 2 Ê x œ e# ; y w 0 for x e# and
and y w 0 for x e# Ê a maximum of 2e ; y ww œ 0
Ê ln x œ 83 Ê x œ e)Î$ ; the curve is concave down on
ˆ0ß e)Î$ ‰ and concave up on ˆe)Î$ ß _‰; so there is an
inflection point at ˆe)Î$ ß
"
È2e
units high.
0 for x e Ê absolute maximum of
"
xÈ x
"
x
œ ln b ln a, and A# œ 'a
œ ecx (x)(2x) ecx œ ecx a1 2x# b . Solving
ln x
x
and dA
dx
Ê yw œ
œ
dy
dt
20
10
"
x
Èe$ Èe$ 1 ¸ 5 ft/sec
units long by y œ e
ln x
Èx
Ê
dy dx
dx dt
"Î#
0 for x e
123. (a) y œ
œ
0 for x 122. A œ xy œ x ˆ lnx#x ‰ œ
dA
dx
9
e$
3
‹
e$
#
"
È2
dy
dt
Š "4 ‹ É9 œ
"
È2
;
œ 3e xÎ3 ;
121. A œ xy œ xecx Ê
Ê xœ
"
) ‰
.
$e%Î$
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
"
e#
units high.
Chapter 7 Practice Exercises
#
#
#
(b) y œ ex Ê y w œ 2xex Ê y ww œ 2ex 4x# ex
457
#
#
œ a4x# 2bex ; solving y w œ 0 Ê x œ 0; y w 0 for
x 0 and y w 0 for x 0 Ê a maximum at x œ 0 of
e! œ 1; there are points of inflection at x œ „ È" ; the
2
curve is concave down for "
È2
x
"
È2
and concave
up otherwise.
(c) y œ (1 x) ecx Ê y w œ ecx (1 x) ecx œ xecx
Ê y ww œ ecx xecx œ (x 1) ecx ; solving y w œ 0
Ê xecx œ 0 Ê x œ 0; y w 0 for x 0 and y w 0
for x 0 Ê a maximum at x œ 0 of (1 0) e! œ 1;
there is a point of inflection at x œ 1 and the curve is
concave up for x 1 and concave down for x 1.
124. y œ x ln x Ê y w œ ln x x ˆ "x ‰ œ ln x 1; solving y w œ 0
Ê ln x 1 œ 0 Ê ln x œ 1 Ê x œ e" ; y w 0 for
x e" and y w 0 for x e" Ê a minimum of e" ln e"
œ "e at x œ e" . This minimum is an absolute minimum
since yww œ
125.
dy
dx
"
x
is positive for all x !.
œ Èy cos2 Èy Ê
126. y w œ
3yax1b2
y 1
Ê
dy
Èy cos2 Èy
ay 1 b
y dy
127. yy w œ secay2 bsec2 x Ê
œ dx Ê 2tanÈy œ x C Ê y œ ˆtan1 ˆ x 2 C ‰‰
2
œ 3ax 1b2 dx Ê y ln y œ ax 1b3 C
œ sec2 x dx Ê
y dy
secay2 b
sinˆy2 ‰
2
œ tan x C Ê sinay2 b œ 2tan x C1
sin x
128. y cos2 axb dy sin x dx œ 0 Ê y dy œ cos
2 axb dx Ê
y2
2
œ cos1axb C Ê y œ „ É cosa2xb C1
129.
œ exy2 Ê ey dy œ eax2b dx Ê ey œ eax2b C. We have ya0b œ 2, so e2 œ e2 C Ê C œ 2e2 and
e œ eax2b 2e2 Ê y œ lnˆeax2b 2e2 ‰
130.
dy
dx
dy
dx
y
œ
y ln y
1 x2
Ê etan
Ê
c1 a0bC
dy
y ln y
œ
Ê lnaln yb œ tan1 axb C Ê y œ ee
dx
1 x2
tanc1 axbbC
. We have ya0b œ e2 Ê e2 œ ee
œ 2 Ê tan1 a0b C œ ln 2 Ê 0 C œ ln 2 Ê C œ ln 2 Ê y œ ee
131. x dy ˆy Èy‰dx œ 0 Ê
dy
ˆy È y ‰
œ
dx
x
tanc1 a0bbC
tanc1 axbbln 2
Ê 2lnˆÈy 1‰ œ ln x C. We have ya1b œ 1 Ê 2 lnŠÈ1 1‹ œ ln 1 C
Ê 2 ln 2 œ C œ ln 22 œ ln 4. So 2 lnˆÈy 1‰ œ ln x ln 4 œ lna4xb Ê lnˆÈy 1‰ œ "# lna4xb œ lna4xb1/2
Ê eln
ˆ È y 1 ‰
132. y2 dx
dy œ
So
y3
3
1/2
œ elna4xb Ê Èy 1 œ 2Èx Ê y œ ˆ2Èx 1‰
ex
e2x 1
Ê
e2x 1
ex dx
œ ex ex 1
3
œ
dy
yc2
Ê
y3
3
2
œ ex ex C. We have ya0b œ 1 Ê
a1 b 3
3
œ e0 e0 C Ê C œ 13 .
Ê y3 œ 3aex ex b 1 Ê y œ c3aex ex b 1 d1/3
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
458
Chapter 7 Transcendental Functions
133. Since the half life is 5700 years and A(t) œ A! ekt we have
Ê kœ
ln (0.5)
5700
Ê ln (0.1) œ
A!
#
"
#
œ A! e5700k Ê
œ e5700k Ê ln (0.5) œ 5700k
ln Ð0Þ5Ñ
. With 10% of the original carbon-14 remaining we have 0.1A! œ A! e 5700
ln (0.5)
5700
t Ê tœ
(5700) ln (0.1)
ln (0.5)
t
ln Ð0Þ5Ñ
Ê 0.1 œ e 5700
t
¸ 18,935 years (rounded to the nearest year).
134. T Ts œ (To Ts ) eckt Ê 180 40 œ a220 40b eckÎ4 , time in hours, Ê k œ 4 ln ˆ 79 ‰ œ 4 ln ˆ 97 ‰ Ê 70 40
œ a220 40b ec4 ln Ð9Î7Ñ t Ê t œ
¸ 1.78 hr ¸ 107 min, the total time Ê the time it took to cool from 180° F to
ln 6
4 ln ˆ 97 ‰
70° F was 107 15 œ 92 min
x ‰
135. ) œ 1 cot" ˆ 60
cot" ˆ 35 œ 30 ’ 60# 2 x# "
30# (50 x)# “ ;
x ‰
30 ,
solving
d)
dx
100 20È17 is not in the domain;
0 x 50 Ê
d)
dx
d)
dx
œ
1
Š 60
‹
1ˆ
x ‰#
60
Š
"
30 ‹
cx‹
1 Š 5030
#
œ 0 Ê x# 200x 3200 œ 0 Ê x œ 100 „ 20È17, but
d)
dx
0 for x 20 Š5 È17‹ and
0 for 20 Š5 È17‹ x 50
Ê x œ 20 Š5 È17‹ ¸ 17.54 m maximizes )
136. v œ x# ln ˆ "x ‰ œ x# (ln 1 ln x) œ x# ln x Ê
Ê 2 ln x 1 œ 0 Ê ln x œ maximum at x œ e
1Î2
;
r
h
Ê x œ ec1Î2 ;
"
#
œ 2x ln x x# ˆ "x ‰ œ x(2 ln x 1); solving
dv
dx
dv
dx
1Î2
œ x and r œ 1 Ê h œ e
1Î2
0 for x e
and
È
œ e ¸ 1.65 cm
dv
dx
0 for x e
1Î2
dv
dx
œ0
Ê a relative
CHAPTER 7 ADDITIONAL AND ADVANCED EXERCISES
lim c '0
b
1.
2.
bÄ1
lim "
xÄ_ x
"
È 1 x#
dx œ lim c csin" xd 0 œ lim c asin" b sin" 0b œ lim c asin" b 0b œ lim c sin" b œ
bÄ1
bÄ1
bÄ1
bÄ1
b
'0x tan" t dt œ x lim
Ä_
tanc" x
œ x lim
Ä_
3. y œ ˆcos Èx‰
1
1Îx
œ "# lim b
xÄ!
"
#x
œ
"Î# sec# Èx
" x "Î#
#
4. y œ ax ex b2Îx Ê ln y œ
Ê x lim
ax e b
Ä_
ˆ " 5. x lim
Ä _ n1
"
n#
x
_
ˆ_
form‰
1
#
Ê ln y œ
x 2Îx
'0x tanc" t dt
"
x
ln ˆcos Èx‰
x
ln ˆcos Èx‰ and lim b
xÄ!
1Îx
lim b ˆcos Èx‰ œ e 1Î2 œ
œ "# Ê
xÄ!
2 ln ax ex b
x
Ê x lim
ln y œ x lim
Ä_
Ä_
y
#
œ x lim
e
œ
e
Ä_
á " ‰
#n
œ lim b
xÄ!
2 a1 e x b
x ex
1
6. x lim
Ä_
"
1x
"
n
œ
"
#
lim
x Ä !b
tan Èx
Èx
"
Èe
œ x lim
Ä_
2ex
1 ex
œ x lim
Ä_
2ex
ex
œ2
1
ˆ"‰
œ x lim
ˆ n" ‰ – 1 " — á ˆ n" ‰ – 1 " —
Ä _ n – 1 Š"‹ —
12Š ‹
1nŠ ‹
n
which can be interpreted as a Riemann sum with partitioning ?x œ
œ '0
sin Èx
2Èx cos Èx
n
"
n
n
ˆ " Ê x lim
Ä _ n1
"
n#
á " ‰
#n
dx œ cln (1 x)d "! œ ln 2
ˆ n" ‰ eÐ1ÎnÑ ˆ n" ‰ e2Ð1ÎnÑ á ˆ n" ‰ enÐ1ÎnÑ ‘ which can be interpreted as a
ce1În e2În á ed œ x lim
Ä_
Riemann sum with partitioning ?x œ
"
n
Ê x lim
Ä_
"
n
ce1În e2În á ed œ '0 ex dx œ cex d "! œ e 1
1
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
1
#
Chapter 7 Additional and Advanced Exercises
t
7. A(t) œ '0 ecx dx œ cecx d t0 œ 1 ect , V(t) œ 1'0 ec2x dx œ 1# ec2x ‘ 0 œ
t
(a)
(b)
(c)
8. (a)
t
lim A(t) œ lim a1 ect b œ 1
tÄ_
tÄ_
lim V(t)
t Ä _ A(t)
limb V(t)
A(t)
œ lim
tÄ_
œ limb
tÄ!
tÄ!
1
#
ˆ1 ec2t ‰
1 ect
1
#
ˆ1 ec2t ‰
1 ect
œ 0;
lim loga 2 œ lim c
aÄ1
ln 2
ln a
œ _;
lim loga 2 œ lim b
aÄ1
ln 2
ln 1
œ _;
lim loga 2 œ a lim
Ä_
ln 2
ln a
œ0
aÄ_
9. A" œ '1
e
dx œ
2 log2 x
x
e
#
x)
œ ’ (ln
2 ln # “ œ
1
"
# ln 2
2
ln 2
"
1 x#
#
lim
x Ä !c
a1 ec2t b
a1 ect b œ 1
(b)
4
x
1
dx œ
2
ln 4
'1e lnxx dx
Ê A" : A# œ 2 : 1
"
1 x#
Š
"
‹
x#
Š1 x"# ‹
œ 0 Ê tan" x tan" ˆ x" ‰ is a
1
#
for x 0; it is 1# for
0 since tan" x tan" ˆ "x ‰ is odd. Next
lim b tan" x tan" ˆ "x ‰‘ œ ! 1# œ 1#
xÄ!
1
#
œ limb
tÄ!
#
constant and the constant is
and
a1 ect b a1 ect b
a1 ect b
'1e lnxx dx œ ’ (lnlnx)2 “ e œ ln"# ; A# œ '1e 2 log4
10. y œ tan" x tan" ˆ "x ‰ Ê yw œ
"
1 x#
1
œ limb
tÄ!
ln 2
ln a
a Ä 1b
x
1
#
lim loga 2 œ lim b
aÄ!
a Ä !b
a Ä 1c
œ
œ
1
#
459
the
ˆtan" x tan" ˆ "x ‰‰ œ 0 ˆ 1# ‰ œ 1#
11. ln xax b œ xx ln x and ln axx bx œ x ln xx œ x# ln x; then, xx ln x œ x# ln x Ê axx x# bln x œ ! Ê xx œ x# or ln x œ !Þ
x
ln x œ ! Ê x œ "; xx œ x# Ê x ln x œ 2 ln x Ê x œ 2. Therefore, xax b œ axx bx when x œ 2 or x œ ".
x
12. In the interval 1 x 21 the function sin x 0
Ê (sin x)sin x is not defined for all values in that
interval or its translation by 21.
13. f(x) œ egÐxÑ Ê f w (x) œ egÐxÑ gw (x), where gw (x) œ
14. (a)
(c)
df
dx
df
dx
œ
2 ln ex
ex
x
1 x%
Ê f w (2) œ e! ˆ 1 2 16 ‰ œ
(b) f(0) œ '1
1
† ex œ 2x
#
2 ln t
t
2
17
dt œ 0
#
œ 2x Ê f(x) œ x C; f(0) œ 0 Ê C œ 0 Ê f(x) œ x Ê the graph of f(x) is a parabola
15. (a) g(x) h(x) œ 0 Ê g(x) œ h(x); also g(x) h(x) œ 0 Ê g(x) h(x) œ 0 Ê g(x) h(x) œ 0
Ê g(x) œ h(x); therefore h(x) œ h(x) Ê h(x) œ 0 Ê g(x) œ 0
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
460
Chapter 7 Transcendental Functions
f(x) f(x)
#
f(x) f(x)
#
(b)
cfE (x) fO (x)d b fE (x) fO (x)‘
œ fE (x) fO (x) # fE (x) fO (x) œ fE (x);
#
cfE (x) fO (x)d cfE (x) fO (x)d
œ fE (x) fO (x) # fE (x) fO (x) œ fO (x)
#
œ
œ
(c) Part b Ê such a decomposition is unique.
16. (a) g(0 0) œ
g(0) g(0)
1 g(0) g(0)
#
Ê c1 g# (0)d g(0) œ 2g(0) Ê g(0) g$ (0) œ 2g(0) Ê g$ (0) g(0) œ 0
Ê g(0) cg (0) 1d œ 0 Ê g(0) œ 0
g(x) b g(h)
’
“ g(x)
g(x h) g(x)
g(x) g# (x) g(h)
œ lim 1 c g(x) g(h)
œ lim g(x) g(h)
h
h
h
c
1
g(x) g(h)d
hÄ0
hÄ0
hÄ0
g(h)
1 g# (x)
lim ’ h “ ’ 1 g(x) g(h) “ œ 1 † c1 g# (x)d œ 1 g# (x) œ 1 [g(x)]#
hÄ0
(b) gw (x) œ lim
œ
dy
dx
(c)
œ 1 y# Ê
dy
1 y#
"
Ê C œ 0 Ê tan
œ dx Ê tan" y œ x C Ê tan" (g(x)) œ x C; g(0) œ 0 Ê tan" 0 œ 0 C
(g(x)) œ x Ê g(x) œ tan x
17. M œ '0
1
"
2
"
xd ! œ 1#
1 x# dx œ 2 ctan
ln 2
ln 4
ˆ 1 ‰ œ 1 ; y œ 0 by symmetry
#
œ
and My œ '0
1
2x
1 x#
"
dx œ cln a1 x# bd ! œ ln 2 Ê x œ
My
M
"
18. (a) V œ 1 '1Î4 Š #È
‹ dx œ
x
1
4
'14Î4 x" dx œ 14 cln kxkd %"Î% œ 14 ˆln 4 ln 4" ‰ œ 14 ln 16 œ 14 ln a2% b œ 1 ln 2
"
(b) My œ '1Î4 x Š #È
‹ dx œ
x
1
2
63
'14Î4 x"Î# dx œ 3" x$Î# ‘ %"Î% œ ˆ 83 24" ‰ œ 64#4 1 œ 24
;
#
4
4
"
"
Mx œ '1Î4 "# Š #È
‹ Š 2È
‹ dx œ
x
x
4
M œ '1Î4
4
yœ
Mx
M
œ
"
#È x
ˆ "#
'14Î4
"
x
%
dx œ 8" ln kxk‘ "Î% œ
dx œ '1Î4 "2 x"Î# dx œ x"Î# ‘ "Î% œ 2 %
4
ln
œ
ln 2
3
b csc ) ‰
r%
Ê
2‰ ˆ 32 ‰
19. (a) L œ k ˆ a bR%cot ) %
1
8
#
dL
d)
#
œ k Š b csc
R%
)
"
#
œ
3
#
b csc ) cot )
‹;
r%
%
"
8
ln 16 œ
"
#
ln 2;
; therefore, x œ
solving
%
dL
d)
My
M
‰ ˆ 23 ‰ œ
œ ˆ 63
24
21
1#
œ
7
4
and
œ0
%
Ê r b csc ) bR csc ) cot ) œ 0 Ê (b csc )) ar csc ) R cot )b œ 0; but b csc ) Á 0 since
)Á
1
#
Ê r% csc ) R% cot ) œ 0 Ê cos ) œ
r%
R%
%
Ê ) œ cos" Š Rr % ‹ , the critical value of )
%
(b) ) œ cos" ˆ 56 ‰ ¸ cos" (0.48225) ¸ 61°
20. In order to maximize the amount of sunlight, we need to maximize the angle ) formed by extending the two red line
segments to their vertex. The angle between the two lines is given by ) œ 1 a)1 a1 )# bb. From trig we have
1 ˆ 200 ‰
tan )1 œ 450350 x Ê )1 œ tan1 ˆ 450350 x ‰ and tan a1 )# b œ 200
x Ê a1 )# b œ tan
x
‰
Ê ) œ 1 a)1 a1 )# bb œ 1 tan1 ˆ 450350 x ‰ tan1 ˆ 200
x
Ê
d)
dx
d)
dx
œ
œ0Ê
1
2
1 ˆ 450350c x ‰
†
350
a450 xb2
350
a450 xb2 122500
1
‰2
1 ˆ 200
x
200
x2 40000
‰œ
† ˆ 200
x2
350
a450 xb2 122500
200
x2 40000
œ 0 Ê 200Ša450 xb2 122500‹ œ 350ax2 40000b
Ê 3x2 3600x 1020000 œ 0 Ê x œ 600 „ 100È70. Since x 0, consider only x œ 600 100È70.
Using the first derivative test,
d)
dx ¹x œ 100
œ
9
3500
0 and
d)
dx ¹x œ 400
œ
9
5000
0 Ê local max when
x œ 600 100È70 ¸ 236.67 ft.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
CHAPTER 8 TECHNIQUES OF INTEGRATION
8.1 INTEGRATION BY PARTS
1. u œ x, du œ dx; dv œ sin
'
x sin
x
#
dx œ 2x cos
x
#
x
#
dx, v œ 2 cos
' ˆ2 cos
"
1
2. u œ ), du œ d); dv œ cos 1) d), v œ
'
) cos 1) d) œ
3.
sin 1) '
)
1
"
1
x‰
#
x
#
;
dx œ 2x cos ˆ x# ‰ 4 sin ˆ x# ‰ C
sin 1);
sin 1) d) œ
)
1
sin 1) "
1#
cos 1) C
cos t
ÐÑ
t# ïïïïî
ÐÑ
2t ïïïïî
ÐÑ
2 ïïïïî
sin t
cos t
sin t
0
4.
'
t# cos t dt œ t# sin t 2t cos t 2 sin t C
'
x# sin x dx œ x# cos x 2x sin x 2 cos x C
sin x
ÐÑ
x# ïïïïî
ÐÑ
2x ïïïïî
ÐÑ
2 ïïïïî
cos x
sin x
cos x
0
5. u œ ln x, du œ
dx
x ;
dv œ x dx, v œ
'1 x ln x dx œ ’ x#
2
6. u œ ln x, du œ
'1 x
e
ln x“ '1
#
#
$
ln x dx œ
dx
x ;
2 #
x
"
#
x#
#
dx
x
dv œ x$ dx, v œ
%
’ x4
ln x“ '
e
1
e %
x
;
x%
4
dx
1 4 x
#
#
œ 2 ln 2 ’ x4 “ œ 2 ln 2 "
3
4
œ ln 4 3
4
;
œ
e%
4
%
e
x
’ 16
“ œ
1
3e% 1
16
7. u œ x, du œ dx ; dv œ ex dx, v œ ex ;
' x ex dx œ x ex ' ex dx œ x ex ex C
8. u œ x, du œ dx ; dv œ e3x dx, v œ 13 e3x ;
' x e3x dx œ x3 e3x 13 ' e3x dx œ x3 e3x 19 e3x C
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
462
Chapter 8 Techniques of Integration
ex
9.
ÐÑ
x# ïïïïî
ÐÑ
2x ïïïïî
ÐÑ
2 ïïïïî
ex
ex
ex
'
0
x# ex dx œ x# ex 2x ex 2 ex C
e2x
10.
ÐÑ
x# 2x 1 ïïïïî 12 e2x
ÐÑ
2x 2
ïïïïî 14 e2x
ÐÑ
2
ïïïïî 18 e2x
' ax# 2x 1be2x dx œ 12 ax# 2x 1be2x 14 a2x 2be2x 14 e2x C
0
œ ˆ 12 x# 32 x 54 ‰e2x C
11. u œ tan" y, du œ
dy
1 y #
; dv œ dy, v œ y;
' tan" y dy œ y tan" y ' a1ydyy b œ y tan" y #" ln a1 y# b C œ y tan" y ln È1 y# C
#
12. u œ sin" y, du œ
dy
È 1 y#
; dv œ dy, v œ y;
' sin" y dy œ y sin" y ' Èy1 dy y
#
œ y sin" y È1 y# C
13. u œ x, du œ dx; dv œ sec# x dx, v œ tan x;
' x sec# x dx œ x tan x ' tan x dx œ x tan x ln kcos xk C
14.
' 4x sec# 2x dx; [y œ 2x]
' y sec# y dy œ y tan y ' tan y dy œ y tan y ln ksec yk C
Ä
œ 2x tan 2x ln ksec 2xk C
ex
15.
ÐÑ
x$ ïïïïî
ÐÑ
3x# ïïïïî
ÐÑ
6x ïïïïî
ÐÑ
6 ïïïïî
0
ex
ex
ex
ex
'
x$ ex dx œ x$ ex 3x# ex 6xex 6ex C œ ax$ 3x# 6x 6b ex C
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 8.1 Integration by Parts
ecp
16.
ÐÑ
ïïïïî
ÐÑ
4p$ ïïïïî
ÐÑ
12p# ïïïïî
ÐÑ
24p ïïïïî
ÐÑ
24 ïïïïî
p%
ecp
ecp
ecp
ecp
ecp
'
0
p% ecp dp œ p% ecp 4p$ ecp 12p# ecp 24pecp 24ecp C
œ ap% 4p$ 12p# 24p 24b ecp C
ex
17.
ÐÑ
x# 5x ïïïïî ex
ÐÑ
2x 5 ïïïïî ex
ÐÑ
2
ïïïïî ex
' ax# 5xb ex dx œ ax# 5xb ex (2x 5)ex 2ex C œ x# ex 7xex 7ex C
0
œ ax# 7x 7b ex C
er
18.
ÐÑ
r# r 1 ïïïïî er
ÐÑ
2r 1
ïïïïî er
ÐÑ
2
ïïïïî er
0
' ar# r 1b er dr œ ar# r 1b er (2r 1) er 2er C
œ car# r 1b (2r 1) 2d er C œ ar# r 2b er C
ex
19.
x&
5x%
20x$
60x#
120x
120
0
ÐÑ
ïïïïî ex
ÐÑ
ïïïïî ex
ÐÑ
ïïïïî ex
ÐÑ
ïïïïî ex
ÐÑ
ïïïïî ex
ÐÑ
ïïïïî ex
' x& ex dx œ x& ex 5x% ex 20x$ ex 60x# ex 120xex 120ex C
œ ax& 5x% 20x$ 60x# 120x 120b ex C
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
463
464
Chapter 8 Techniques of Integration
e4t
20.
ÐÑ
t# ïïïïî
ÐÑ
2t ïïïïî
ÐÑ
2 ïïïïî
"
4
e4t
"
16
e4t
"
64
e4t
' t# e4t dt œ t4 e4t 162t e4t 642 e4t C œ t4 e4t 8t e4t 3"# e4t C
#
0
#
œ Š t4 t
8
#
"
4t
3# ‹ e
C
21. I œ ' e) sin ) d); cu œ sin ), du œ cos ) d); dv œ e) d), v œ e) d Ê I œ e) sin ) ' e) cos ) d);
cu œ cos ), du œ sin ) d); dv œ e) d), v œ e) d Ê I œ e) sin ) Še) cos ) ' e) sin ) d)‹
œ e) sin ) e) cos ) I Cw Ê 2I œ ae) sin ) e) cos )b Cw Ê I œ
"
#
ae) sin ) e) cos )b C, where C œ
w
C
#
is
Ê 2I œ "# ecy (sin y cos y) Cw Ê I œ "4 ecy (sin y cos y) C œ e 4 (sin 2x cos 2x) C, where C œ
C
#
another arbitrary constant
22. I œ ' ecy cos y dy; cu œ cos y, du œ sin y dy; dv œ ecy dy, v œ ecy d
Ê I œ ecy cos y ' aecy b (sin y) dy œ ecy cos y ' ecy sin y dy; cu œ sin y, du œ cos y dy;
dv œ ecy dy, v œ ecy d Ê I œ ecy cos y Šecy sin y ' aey b cos y dy‹ œ ecy cos y ecy sin y I Cw
Ê 2I œ ecy (sin y cos y) Cw Ê I œ
"
#
aecy sin y ecy cos yb C, where C œ
23. I œ ' e2x cos 3x dx; u œ cos 3x; du œ 3 sin 3x dx, dv œ e2x dx; v œ
is another arbitrary constant
e2x ‘
Ê
' e2x sin 3x dx; u œ sin 3x, du œ 3 cos 3x, dv œ e2x dx; v œ "# e2x ‘
I œ "# e2x cos 3x 3# Š "# e2x sin 3x 3# ' e2x cos 3x dx‹ œ "# e2x cos 3x 34 e2x sin 3x 94 I Cw
Ê
13
4
Ê Iœ
24.
"
#
w
C
#
"
#
e2x cos 3x Iœ
"
#
3
#
e2x cos 3x 34 e2x sin 3x Cw Ê
e2x
13
(3 sin 3x 2 cos 3x) C, where C œ
4
13
Cw
' ec2x sin 2x dx; [y œ 2x] Ä "# ' ecy sin y dy œ I; cu œ sin y, du œ cos y dy; dv œ ecy dy, v œ ecyd
Ê I œ "# Šecy sin y ' ecy cos y dy‹ cu œ cos y, du œ sin y; dv œ ecy dy, v œ ecy d
Ê I œ "# ecy sin y "# Šecy cos y ' aecy b ( sin y) dy‹ œ "# ecy (sin y cos y) I Cw
c2x
#
25.
' eÈ3sb9 ds; ” 3s 92 œ x
ds œ
2
3
' xex dx œ
2
3
3
x dx •
Ä
' ex † 23 x dx œ 23 ' xex dx; cu œ x, du œ dx; dv œ ex dx, v œ ex d ;
Šxex ' ex dx‹ œ
2
3
axex ex b C œ
2
3
ŠÈ3s 9 eÈ3sb9 eÈ3sb9 ‹ C
26. u œ x, du œ dx; dv œ È1 x dx, v œ 23 È(1 x)$ ;
'01 xÈ1 x dx œ 23 È(1 x)$ x‘ "! 23 '01 È(1 x)$ dx œ 23 25 (1 x)&Î# ‘ "! œ 154
27. u œ x, du œ dx; dv œ tan# x dx, v œ ' tan# x dx œ '
œ tan x x;'0
1Î3
œ
1
3
1Î$
x tan# x dx œ cx(tan x x)d !
ŠÈ3 13 ‹ ln
"
#
1#
18
œ
1È3
3
ln 2 sin# x
cos# x
dx œ '
" cos# x
cos# x
'0 (tan x x) dx œ
1Î3
1
3
dx œ '
dx
cos# x
' dx
ŠÈ3 13 ‹ ’ln kcos xk 1#
18
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
1Î$
x#
# “!
w
Section 8.1 Integration by Parts
28. u œ ln ax x# b, du œ
œ x ln ax x# b '
; dv œ dx, v œ x; ' ln ax x# b dx œ x ln ax x# b '
(2x 1) dx
x x #
œ x ln ax x# b '
(2x 1) dx
x 1
u œ ln x
29.
' sin (ln x) dx; Ô du œ "x dx ×
œ
"
#
Ä
Õ dx œ eu du Ø
cx cos (ln x) x sin (ln x)d C
u œ ln z
30.
' z(ln z)# dz; Ô du œ "z dz ×
"
2
e2u
"
4
e2u
"
8
e2u
'
0
u# e2u du œ
œ
† x dx
dx œ x ln ax x# b 2x ln kx 1k C
From Exercise 21, ' (sin u) eu du œ eu ˆ sin u # cos u ‰ C
' eu † u# † eu du œ ' e2u † u# du;
Ä
Õ dz œ eu du Ø
e2u
ÐÑ
u# ïïïïî
ÐÑ
2u ïïïïî
ÐÑ
2 ïïïïî
31.
' (sin u) eu du.
2(x 1) "
x 1
2x "
x(x 1)
465
#
z
4
u#
#
#
e2u #u e2u "4 e2u C œ
e2u
4
c2u# 2u 1d C
c2(ln z) 2 ln z 1d C
' x sec x# dx ’Let u œ x# , du œ 2x dx Ê 12 du œ x dx“ Ä ' x sec x# dx
œ
1
2
' sec u du œ 12 lnlsec u tan ul C
œ 12 lnlsec x# tan x# l C
32.
' cosÈÈx x dx ’Let u œ Èx, du œ 2È1 x dx Ê 2du œ È1x dx“ Ä ' cosÈÈx x dx
33.
' xaln xb# dx; Ô du œ x" dx ×
u œ ln x
Õ dx œ eu du Ø
e2u
ÐÑ
u# ïïïïî
ÐÑ
2u ïïïïî
ÐÑ
2 ïïïïî
"
2
e2u
"
4
e2u
"
8
e2u
'
0
Ä
34.
' xaln" xb
#
' lnx x dx
2
36.
' alnxxb
3
x#
4
dx ’Let u œ ln x, du œ
35. u œ ln x, du œ
1
x
dx; dv œ
œ lnx x '
"
x#
1
x2
' eu † u# † eu du œ ' e2u † u# du;
u# e2u du œ
œ
1
x
œ 2' cos u du œ 2 sin u C œ 2 sinÈx C
u#
#
#
e2u #u e2u "4 e2u C œ
c2(ln x) 2 ln x 1d C œ
dx“ Ä '
"
xaln xb#
dx œ '
1
u2
x#
2 aln
e2u
4
c2u# 2u 1d C
xb # x#
2 ln
x
x#
4
C
du œ 1u C œ ln1x C
dx, v œ x1 ;
dx œ lnx x dx ’Let u œ ln x, du œ
1
x
1
x
C
dx“ Ä '
aln xb3
x
dx œ ' u3 du œ 41 u4 C œ 41 aln xb4 C
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
466
37.
Chapter 8 Techniques of Integration
' x3 ex
4
dx ’Let u œ x4 , du œ 4x3 dx Ê 14 du œ x3 dx“ Ä ' x3 ex dx œ
4
3
1
4
' eu du
4
œ 14 eu C œ 14 ex C
3
38. u œ x3 , du œ 3x2 dx; dv œ x2 ex dx, v œ 13 ex ;
' x5 ex
3
dx œ ' x3 ex x2 dx œ 13 x3 ex 3
3
1
3
' ex
3
3
39. u œ x2 , du œ 2x dx; dv œ Èx2 1 x dx, v œ 13 ax2 1b
' x3 Èx2 1 dx
40.
œ
1 2 2
3 x ax
1b
3Î2
1
3
3
3x2 dx œ 13 x3 ex 13 ex C
' ax2 1b
3Î2
3Î2
;
3Î2
œ
1
3
2x dx œ 13 x2 ax2 1b
' x2 sin x3 dx ’Let u œ x3 , du œ 3x2 dx Ê 13 du œ x2 dx“ Ä ' x2 sin x3 dx
2
2
15 ax
1b
' sin u du
5Î2
C
œ 13 cos u C
œ 13 cos x3 C
41. u œ sin 3x, du œ 3cos 3x dx; dv œ cos 2x dx, v œ 12 sin 2x ;
' sin 3x cos 2x dx œ 12 sin 3x sin 2x 32 ' cos 3x sin 2x dx
u œ cos 3x, du œ 3sin 3x dx; dv œ sin 2x dx, v œ 12 cos 2x ;
' sin3x cos 2x dx œ 12 sin 3x sin 2x 32 ” 12 cos 3x cos 2x 32 ' sin3x cos 2x dx•
œ 12 sin 3x sin 2x 34 cos 3x cos 2x 9
4
' sin 3x cos 2x dx Ê 54 ' sin 3x cos 2x dx œ 12 sin 3x sin 2x 34 cos 3x cos 2x
Ê ' sin 3x cos 2x dx œ 25 sin 3x sin 2x 35 cos 3x cos 2x C
42. u œ sin 2x, du œ 2cos 2x dx; dv œ cos 4x dx, v œ 14 sin 4x ;
' sin 2x cos 4x dx œ 14 sin 2x sin 4x 12 ' cos 2x sin 4x dx
u œ cos 2x, du œ 2sin 2x dx; dv œ sin 4x dx, v œ 14 cos 4x ;
' sin 2x cos 4x dx œ 14 sin 2x sin 4x 12 ” 14 cos 2x cos 4x 12 ' sin 2x cos 4x dx•
œ 14 sin 2x sin 4x 18 cos 2x cos 4x 1
4
' sin 2x cos 4x dx Ê 34 ' sin 2x cos 4x dx œ 14 sin 2x sin 4x 18 cos 2x cos 4x
Ê ' sin 2x cos 4x dx œ 13 sin 2x sin 4x 16 cos 2x cos 4x C
43.
' ex sin ex dx ’Let u œ ex , du œ ex dx“ Ä ' ex sin ex dx
44.
'
45.
' cosÈx dx; Ö dy œ 2È" x dx Ù
eÈ x
Èx
dx ’Let u œ Èx, du œ
Ô
y œ Èx
1
2È x
×
dx Ê 2du œ
Ä
1
Èx
œ ' sin u du œ cos u C œ cos ex C
dx“ Ä '
eÈ x
Èx
dx œ 2' eu du œ 2eu C œ 2eÈx C
' cos y 2y dy œ ' 2y cos y dy;
Õ dx œ 2y dy Ø
u œ 2y, du œ 2 dy; dv œ cos y dy, v œ sin y ;
' 2y cos y dy œ 2y sin y ' 2 sin y dy œ 2y sin y 2 cos y C œ 2Èx sin Èx 2 cos Èx C
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 8.1 Integration by Parts
Ô
46.
y œ Èx
×
' Èx eÈx dx; Ö dy œ 2È" x dx Ù
'
Ä
467
y ey 2y dy œ ' 2y2 ey dy;
Õ dx œ 2y dy Ø
ey
ÐÑ
2y# ïïïïî ey
ÐÑ
4y ïïïïî ey
ÐÑ
4
ïïïïî ey
' 2y2 ey dy œ 2y# ey 4y ey 4 ey C œ 2x eÈx 4Èx eÈx 4eÈx C
0
47.
sin 2)
ÐÑ
)# ïïïïî "2 cos 2)
ÐÑ
2) ïïïïî "4 sin 2)
ÐÑ
2 ïïïïî "8 cos 2)
'01Î2 )# sin 2) d) œ ’ )#
#
0
œ ’
48.
1#
8
† (1) 1
4
cos 2) †0
"
4
)
#
sin 2) "
4
cos 2)“
† (1)“ 0 0 "
4
1Î#
!
† 1‘ œ
1#
8
"
#
1# 4
8
œ
cos 2x
ÐÑ
x$ ïïïïî "2 sin 2x
ÐÑ
3x# ïïïïî "4 cos 2x
ÐÑ
6x ïïïïî "8 sin 2x
ÐÑ
"
6 ïïïïî 16
cos 2x
'01Î2 x$ cos 2x dx œ ’ x#
$
0
œ
49. u œ sec" t, du œ
dt
tÈt# 1
'22ÎÈ3 t sec" t dt œ ’ t#
#
œ
51
9
’ "# Èt# 1“
50. u œ sin" ax# b , du œ
È2
'01Î
1$
’ 16
†0
; dv œ t dt, v œ
sec" t“
#
#ÎÈ$
œ
2x dx
È 1 x %
#
#
’È1 x% “
!
œ
†0
1
1#
3
8
œ ˆ2 †
"# ŠÈ3 É 43 1‹ œ
"ÎÈ#
1
1#
dt
tÈt# 1
; dv œ 2x dx, v œ x# ;
"ÎÈ#
31
8
3x#
4
cos 2x 3x
4
sin 2x 3
8
cos 2x“
† (1)“ 0 0 0 3
8
1Î#
!
#
1
† 1‘ œ 316
;
'2ÎÈ3 Š t# ‹
2x sin" ax# b dx œ cx# sin" ax# bd !
œ
t#
#
† (1) 2
#ÎÈ$
51
9
31 #
16
sin 2x '0
É 34 1 œ
È2
1Î
x# †
51
9
1
3
2
3
† 16 ‰ '2ÎÈ3
"# ŠÈ3 2x dx
È 1 x%
2
È3
3 ‹
œ
œ ˆ "# ‰ ˆ 16 ‰ '0
t dt
2Èt# 1
51
9
È3
3
œ
51 3È 3
9
È 2 d ˆ1 x % ‰
1Î
2È 1 x%
16È312
1#
51. (a) u œ x, du œ dx; dv œ sin x dx, v œ cos x;
S" œ '0 x sin x dx œ [x cos x]!1 '0 cos x dx œ 1 [sin x]1! œ 1
1
1
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
3
4
œ
3 a4 1 # b
16
468
Chapter 8 Techniques of Integration
(b) S# œ '1 x sin x dx œ ”[x cos x]#11 '1 cos x dx• œ c31 [sin x]1#1 d œ 31
21
21
(c) S$ œ '21 x sin x dx œ [x cos x]$#11 '21 cos x dx œ 51 [sin x]$#11 œ 51
31
31
Ðn 1Ñ1
(d) S8" œ (1)nb1 'n1
x sin x dx œ (1)nb1 c[x cos x]Ðnn1 1Ñ1 [sin x]Ðnn1 1Ñ1 d
œ (1)nb1 c(n 1)1(1)n n1(1)nb1 d 0 œ (2n 1)1
52. (a) u œ x, du œ dx; dv œ cos x dx, v œ sin x;
31Î2
1‰
S" œ '1Î2 x cos x dx œ ”[x sin x]311Î2Î2 '1Î2 sin x dx• œ ˆ 31
# # [cos x]1Î2 œ 21
31Î2
31Î2
(b) S# œ '31Î2 x cos x dx œ [x sin x]&$11ÎÎ22 '31Î2 sin x dx œ 5#1 ˆ 3#1 ‰‘ [cos x]&$11ÎÎ22 œ 41
51Î2
51Î2
(c) S$ œ '51Î2 x cos x dx œ ”[x sin x](&11ÎÎ22 '51Î2 sin x dx• œ ˆ 7#1 71Î2
71Î2
Ð2n1Ñ1Î2
51 ‰
#
[cos x](&11ÎÎ22 œ 61
Ð2n1Ñ1Î2
n1Ñ1Î2
n'
(d) Sn œ (1)n 'Ð2n 1Ñ1Î2 x cos x dx œ (1)n ”[x sin x]Ð#
sin x dx•
Ð2n 1Ñ1Î2 Ð2n 1Ñ1Î2
œ (1)n ’ (2n# 1)1 (1)n 53. V œ '0
ln 2
(2n1)1
#
n 1Ñ1Î2
(1)nc1 “ [cos x]Ð#
Ð2n1Ñ1Î2 œ
"
#
(2n1 1 2n1 1) œ 2n1
21(ln 2 x) ex dx œ 21 ln 2 '0 ex dx 21'0 xex dx
ln 2
ln 2
œ (21 ln 2) cex d ln0 2 21 Œcxex d ln0 2 '0 ex dx
ln 2
œ 21 ln 2 21 ˆ2 ln 2 cex d ln0 2 ‰ œ 21 ln 2 21 œ 21(1 ln 2)
54. (a) V œ '0 21xecx dx œ 21 Œcxecx d "! '0 ecx dx
1
1
œ 21 Š "e cecx d "! ‹ œ 21 ˆ "e œ 21 "
e
1‰
41
e
(b) V œ '0 21(1 x)ecx dx; u œ 1 x, du œ dx; dv œ ecx dx,
1
v œ ecx ; V œ 21 ”c(1 x) aecx bd "! '0 ecx dx•
1
œ 21 ’[0 1(1)] cecx d "! “ œ 21 ˆ1 55. (a) V œ '0 21x cos x dx œ 21 Œ[x sin x] !
1Î2
"
e
1Î#
1‰ œ
21
e
'0 sin x dx
1Î2
1Î#
œ 21 Š 1# [cos x] ! ‹ œ 21 ˆ 1# 0 1‰ œ 1(1 2)
(b) V œ '0 21 ˆ 1# x‰ cos x dx; u œ
1Î2
1Î#
V œ 21 ˆ 1# x‰ sin x‘ !
1
#
x, du œ dx; dv œ cos x dx, v œ sin x;
21'0 sin x dx œ 0 21[ cos x] !
1Î2
1Î#
œ 21(0 1) œ 21
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 8.1 Integration by Parts
469
56. (a) V œ '0 21x(x sin x) dx;
1
sin x
ÐÑ
x# ïïïïî cos x
ÐÑ
2x ïïïïî sin x
ÐÑ
2 ïïïïî cos x
Ê V œ 21'0 x# sin x dx œ 21 cx# cos x 2x sin x 2 cos xd ! œ 21 a1# 4b
1
0
1
(b) V œ '0 21(1 x)x sin x dx œ 21# '0 x sin x dx 21 '0 x# sin x dx œ 21# [x cos x sin x]1! a21$ 81b
1
1
1
œ 81
57. (a) A œ '1 ln x dx œ ’x ln x“ '1 dx
e
e
e
1
e
œ ae ln e 1 ln 1b ’x“ œ e ae 1b œ 1
1
(b) V œ '1 1aln xb dx œ 1Š’x aln xb2 “ '1 2 ln x dx‹
e
e
2
e
1
œ 1”Šealn eb 1aln 1b ‹ Š’2x ln x“ '1 2 dx‹•
2
e
2
e
1
e
œ 1”e Ša2e ln e 2a1b ln 1b ’2x“ ‹•
1
œ 1”e a2e a2e 2bb• œ 1ae 2b
e
(c) V œ '1 21ax 2b ln x dx œ 21'1 ax 2b ln x dx œ 21Œ”ˆ 12 x2 2x‰ln x• '1 ˆ 12 x 2‰ dx
e
e
e
1
e
œ 21Œˆ 12 e2 2e‰ln e ˆ 12 2‰ln 1 ”ˆ 14 x2 2x‰• œ 21ˆˆ 12 e2 2e‰ ˆˆ 14 e2 2e‰ 94 ‰‰ œ 12 ae2 9b
1
(d) M œ '1 ln x dx œ 1 (from part (a)); x œ
e
1
1
'1 x ln x dx œ ” 12 x2 ln x•
œ 12 e2 Š 14 e2 14 a1b2 ‹ œ 14 ae2 1b; y œ
e
1
1
œ
œ
1
2 ae
e
1
2e 2e 2b œ
2b Ê axß yb œ
1
58. (a) A œ '0 tan1 x dx œ ”x tan1 x• '0
1
1
0
e
e
1 † aln 1b ‹ Œ”2x ln x• '1 2 dx œ
2
1
2 ae
1
'1 12 x dx œ Š 12 e2 ln e 12 a1b2 ln 1‹ ” 14 x2 •
'1 12 aln xb2 dx œ 12 Œ”x aln xb2 •
e
2
1
2 ŒŠe aln eb
e
2
Š e 4 1 ß e 2 2 ‹
1
2 Še
e
1
'1 2 ln x dx
e
e
Ša2e ln e 2a1b ln "b ’2x“ ‹‹
1
is the centroid.
x
1 x2 dx
1
œ atan1 1 0b 12 ”lna1 x2 b•
œ
1
4
21 aln 2 ln 1b œ
(b) V œ '0 21 x tan1 x dx
1
4
0
21 ln 2
1
1
œ 21” x2 tan1 x• "
#
'01 1 x x dx
œ 21Œ 12 tan1 1 0 "
#
'01 ˆ1 1 1 x ‰dx œ 21 18
2
0
œ
21ˆ 18
"ˆ
# 1
1 ‰‰
4
2
2
œ
2
1
#" ”x tan1 x• œ 21ˆ 18 #" a1 tan1 1 a0 0bb‰
0
1a1 2b
2
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
e
1
470
Chapter 8 Techniques of Integration
59. av(y) œ
œ
"
1
"
#1
'021 2ect cos t dt
ect ˆ sin t # cos t ‰‘ #1
!
(see Exercise 22) Ê av(y) œ
"
#1
a1 ec21 b
'021 4ect (sin t cos t) dt
21
21
œ 12 '0 ect sin t dt 12 '0 ect cos t dt
60. av(y) œ
œ
œ
2
1
2
1
"
#1
ect ˆ sin t# cos t ‰ ect ˆ sin t # cos t ‰‘ #1
!
cect sin td #!1 œ 0
61. I œ ' xn cos x dx; cu œ xn , du œ nxn" dx; dv œ cos x dx, v œ sin xd
Ê I œ xn sin x ' nxn" sin x dx
62. I œ ' xn sin x dx; cu œ xn , du œ nxn" dx; dv œ sin x dx, v œ cos xd
Ê I œ xn cos x ' nxn" cos x dx
63. I œ ' xn eax dx; u œ xn , du œ nxn" dx; dv œ eax dx, v œ "a eax ‘
ÊIœ
xn eax ax
a e
n
a
' xn" eax dx, a Á !
64. I œ ' aln xbn dx; ’u œ aln xbn , du œ
naln xbn "
x
dx; dv œ " dx, v œ x“
Ê I œ xaln xbn ' naln xbn" dx
65.
'ab ax ab faxb dx; ’u œ x a, du œ dx; dv œ faxb dx, v œ 'bx fatb dt œ 'xb fatb dt“
b
œ ”ax ab'b fatb dt• 'a Š'b fatb dt‹ dx œ Œab ab'b fatb dt aa ab'b fatb dt 'a Œ'x fatb dt dx
x
b
x
b
a
b
b
a
œ 0 'a Œ'x fatb dt dx œ 'a Œ'x fatb dt dx
b
66.
b
b
b
' È1 x2 dx; ’u œ È1 x2 , du œ È x
œ x È1 x2 '
1 x2
x 2
È 1 x2
dx œ x È1 x2 '
œ x È1 x2 ' È1 x2 dx '
Ê ' È1 x2 dx œ x È1 x2 '
Ê ' È1 x2 dx œ
x
2
dx; dv œ dx, v œ x“
È1 x2 1
2
"
È 1 x2
"
È 1 x2
'È"
" x2 "
È 1 x2
dx œ x È1 x2 Š'
" x2
È 1 x2
dx '
"
È 1 x2
dx‹
dx
dx ' È1 x2 dx Ê 2' È1 x2 dx œ x È1 x2 '
1 x2
dx C
67.
' sin" x dx œ x sin" x ' sin y dy œ x sin" x cos y C œ x sin" x cos asin" xb C
68.
' tan" x dx œ x tan" x ' tan y dy œ x tan" x ln kcos yk C œ x tan" x ln kcos atan" xbk C
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
"
È 1 x2
dx
Section 8.2 Trigonometric Integrals
69.
471
' sec" x dx œ x sec" x ' sec y dy œ x sec" x ln ksec y tan yk C
œ x sec" x ln ksec asec" xb tan asec" xbk C œ x sec" x ln ¹x Èx# 1¹ C
70.
' log2 x dx œ x log2 x ' 2y dy œ x log2 x ln2 # C œ x log2 x lnx# C
y
71. Yes, cos" x is the angle whose cosine is x which implies sin acos" xb œ È1 x# .
72. Yes, tan" x is the angle whose tangent is x which implies sec atan" xb œ È1 x# .
73. (a)
' sinh" x dx œ x sinh" x ' sinh y dy œ x sinh" x cosh y C œ x sinh" x cosh asinh" xb C;
check: d cx sinh" x cosh asinh" xb Cd œ ’sinh" x x
È 1 x#
œ sinh" x dx
(b)
' sinh" x dx œ x sinh" x '
œ x sinh" x a1 x# b
"Î#
"
‹
1 x#
dx œ x sinh" x "Î#
C“ œ ’sinh" x x
È 1 x#
dx
' a1 x# b"Î# 2x dx
x
È 1 x# “
dx œ sinh" x dx
' tanh" x dx œ x tanh" x ' tanh y dy œ x tanh" x ln kcosh yk C œ x tanh" x ln kcosh atanh" xbk C;
check: d cx tanh" x ln kcosh atanh" xbk Cd œ ’tanh" x œ tanh" x (b)
"
#
"
È 1 x# “
C
check: d ’x sinh" x a1 x# b
74. (a)
x ŠÈ
sinh asinh" xb
x
1 x#
x ‘
1 x#
"
#
sinh atanh " xb
"
cosh atanh " xb 1 x# “
dx
dx œ tanh" x dx
' tanh" x dx œ x tanh" x ' 1 x x
check: d x tanh" x x
1 x#
dx œ x tanh" x #" ' 12xx# dx œ x tanh" x #" ln k1 x# k C
ln k1 x# k C‘ œ tanh" x 1 x x# 1 x x# ‘ dx œ tanh" x dx
#
8.2 TRIGONOMETRIC INTEGRALS
1.
' cos 2x dx œ "# ' cos 2x † 2dx œ "# sin 2x C
2.
'01 3 sin x3 dx œ 9'01 sin x3 †
3.
' cos3 x sin x dx œ ' cos3 x asin xbdx œ 14 cos4 x C
4.
' sin4 2x cos 2x dx œ "# ' sin4 2x cos 2x † 2dx œ 101 sin5 2x C
5.
' sin3 x dx œ ' sin2 x sin x dx œ ' a1 cos2 xb sin x dx œ ' sin x dx ' cos2 x sin x dx œ cos x 13 cos3 x C
6.
' cos3 4x dx œ ' cos2 4x cos 4x dx œ 14 ' a1 sin2 4xb cos 4x † 4dx œ 14 ' cos 4x † 4dx 14 ' sin2 4x cos 4x † 4dx
œ 14 sin 4x 7.
1
3
12 sin 4x
1
3 dx
1
œ 9’cos x3 “ œ 9ˆcos 13 cos 0‰ œ 9ˆ "# 1‰ œ
0
9
#
C
' sin5 x dx œ ' asin# xb# sin x dx œ ' a" cos# xb# sin x dx œ ' a" 2cos# x cos4 xbsin x dx
œ ' sin x dx ' 2cos# x sin x dx ' cos4 x sin x dx œ cos x #3 cos3 x 51 cos5 x C
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
472
8.
Chapter 8 Techniques of Integration
'01 sin5 ˆ x2 ‰dx (using Exercise 7) œ '01 sinˆ x2 ‰dx '01 2cos# ˆ x2 ‰sinˆ x2 ‰dx '01 cos4 ˆ x2 ‰sinˆ x2 ‰dx
1
œ #cos ˆ x2 ‰ 3% cos3 ˆ 2x ‰ 5# cos5 ˆ 2x ‰‘ 0 œ a!b ˆ# %
$
&# ‰ œ
"'
"&
9.
' cos3 x dx œ ' acos# xbcos x dx œ ' a" sin# xbcos x dx œ ' cos x dx ' sin# x cos x dx œ sin x 13 sin3 x C
10.
'01/6 3cos5 3x dx œ '01/6 acos# 3xb# cos 3x † 3dx œ '01/6 a" sin# 3xb# cos 3x † 3dx œ '01/6 a" #sin# 3x sin% 3xbcos 3x † 3dx
œ '0 cos 3x † 3dx #'0 sin# 3x cos 3x † 3dx '0 sin% 3x cos 3x † 3dx œ ’sin 3x # sin33x 1/6
œ ˆ" 11.
1/6
2
$
"& ‰ a!b œ
1/6
3
1Î6
sin5 3x
5 “0
)
"&
' sin3 x cos3 x dx œ ' sin3 x cos2 x cos xdx œ ' sin3 x a1 sin2 xbcos x dx œ ' sin3 x cos x dx ' sin5 x cos x dx
œ 14 sin4 x 16 sin6 x C
12.
13.
' cos3 2x sin5 2x dx œ "# ' cos3 2x sin5 2x † 2dx œ "# ' cos 2x cos2 2x sin5 2x † 2dx œ "# ' a1 sin2 2xb sin5 2x cos 2x † 2dx
1
1
œ "# ' sin5 2x cos 2x † 2dx "# ' sin7 2x cos 2x † 2dx œ 12
sin6 2x 16
sin8 2x C
2x
' cos2 x dx œ ' 1 cos
dx œ "# ' a1 cos 2xbdx œ "# ' dx "# ' cos 2x dx œ "# ' dx 4" ' cos 2x † 2dx
2
œ "# x 4" sin 2x C
14.
1/2
1/2
1/2
1/2
1/2
2x
'01/2 sin2 x dx œ '01/2 1 cos
dx œ "# '0 a1 cos 2xbdx œ "# '0 dx "# '0 cos 2x dx œ "# '0 dx 4" '0 cos 2x † 2dx
2
1 Î2
œ "# x "4 sin 2x‘ 0
15.
œ ˆ #" ˆ 12 ‰ 4" sin 2ˆ 12 ‰‰ ˆ #" a0b 4" sin 2a0b‰ œ ˆ 14 0‰ a0 0b œ
'01/2 sin7 y dy œ '01/2 sin6 y sin y dy œ '01/2 a" cos2 yb$ sin y dy œ '01/2 sin y dy $'01/2 cos2 y sin y dy
$'0 cos4 y sin y dy '0 cos6 y sin y dy œ ’cos y $ cos3 y $ cos5 y 1/2
16.
1/2
3
18.
19.
3
&
1Î#
cos( y
( “0
œ a!b ˆ" " $
&
"( ‰ œ
' 7cos7 t dt (using Exercise 15) œ 7’' cos t dt $' sin2 t cos t dt $' sin4 t cos t dt ' sin6 t cos t dt“
&
œ 7Šsin t $ sin3 t $ sin5 t 17.
1
4
sin( t
( ‹
C œ 7sin t 7sin3 t 21
&
5 sin t
sin( t C
1
1
1
1
4x
#x ‰ #
'01 )sin4 x dx œ )'01 ˆ " cos
dx œ #'0 a" #cos #x cos# #xbdx œ #'0 dx #'0 cos #x † #dx #'0 " cos
dx
#
#
1
1
1
1
"
œ c#x #sin #xd 0 '0 dx '0 cos 4x dx œ #1 x # sin 4x‘ 0 œ #1 1 œ $1
' )cos4 21x dx œ )' ˆ " cos# 41x ‰# dx œ #' a" #cos 41x cos# 41xbdx œ #' dx %' cos 41x dx #' " cos# )1x dx
œ 3' dx %' cos 41x dx ' cos )1x dx œ 3x 1" sin 41 x )"1 sin )1 x C
#x ‰ˆ " cos #x ‰
4x ‰
' 16 sin# x cos# x dx œ 16' ˆ " cos
dx œ %' a" cos# 2xbdx œ %' dx %' ˆ " cos
dx
#
#
2
œ 4x 2' dx 2' cos 4x dx œ 4x 2x "2 sin 4x C œ 2x 2" sin 4x C œ 2x sin 2x cos 2x C
œ 2x 2sin x cos x a2cos2 x 1b C œ 2x 4sin x cos3 x 2sin x cos x C
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
"'
$&
Section 8.2 Trigonometric Integrals
20.
473
1
1
1
1
#y ‰# ˆ " cos #y ‰
'01 8 sin4 y cos# y dy œ 8'01 ˆ " cos
dy œ '0 dy '0 cos #y dy '0 cos# #y dy '0 cos$ #y dy
#
#
1
1
1
1
1
1
4y ‰
"'
"'
#
'
'
œ y 2" sin 2y‘ 0 '0 ˆ " cos
dy
a
"
sin
#
y
b
cos
#
y
dy
œ
1
dy
cos
4y
dy
cos #y dy
#
# 0
# 0
0
0
1
1
sin
2y
'0 sin# #y cos #y dy œ 1 ’ "# y ") sin 4y "# sin 2y "# † 3 “ œ 1 1# œ 1#
3
0
21.
'
22.
'01/2 sin2 2) cos3 2) d) œ '01/2 sin2 2)a" sin2 2)bcos 2) d) œ '01/2 sin2 2) cos 2) d) '01/2 sin4 2) cos 2) d)
8cos3 2) sin 2) d) œ 8ˆ "# ‰ cos4 2) C œ cos4 2) C
4
œ ’ "# †
sin3 2)
3
"
#
†
1/2
sin5 2)
5 “0
œ!
23.
'021 É " #cos x dx œ '021 ¹ sin x# ¹dx œ '021 sin x# dx œ #cos x# ‘ 20 1 œ # # œ %
24.
'01 È" cos 2x dx œ '01 È# lsin 2x ldx œ '01 È# sin 2x dx œ ’È#cos 2x“ 1 œ È# È# œ #È#
0
25.
'01 È" sin# t dt œ '01 l cos t ldt œ '01/2 cos t dt '11/2 cos t dt œ csin td 10 /2 csin td 11/2 œ " ! ! " œ #
26.
'01 È" cos# ) d) œ '01 l sin ) ld) œ '01 sin ) d) œ ccos )d 10 œ " " œ #
27.
'11ÎÎ32 È sin x
2
" cos x
1 Î2
dx œ '1Î3
1Î2
È" cos x
sin2 x
È" cos x È" cos x
dx œ '1Î3
1Î2
œ '1Î3 sin x È" cos x dx œ ’ 23 a" cos xb3Î2 “
1 Î2
1 Î3
œ
28.
É 32
1Î2
sin2 x È" cos x
Èsin2 x
2
" sin x
'511Î6 Ècos x
" sin x
dx œ '0
1Î6
cos x
È" sin x
dx
dx œ '51Î6
'
" sin x
œ'
1
'
1
1
4 È
cos4 x È" sin x
x " sin x
cos4 x È" sin x
dx œ 51Î6 cos È
dx
È" sin x È" sin x dx œ 51Î6
È" sin2 x
cos2 x
1
1
1
cos4 x È" sin x
dx œ 51Î6 cos3 x È" sin x dx œ 51Î6 cos xa" sin2 xb È" sin x dx
cos x
5 1 Î6
1
1
51Î6 cos x È" sin x dx 51Î6 cos x sin2 x È" sin x dx; u2 Èu du
4
'
'
'
'
’Let u œ 1 sin x Ê u 1 œ sin x Ê du œ cos x dx, x œ
1
œ ’ 23 a" sin xb3Î2 “
Š 23 a"
sin 1b
3Î2
51
6
Ê u œ 1 sinˆ 561 ‰ œ 32 , x œ 1 Ê u œ 1 sin 1 œ 1 “
'3Î2 au 1b2 Èu du œ ’ 23 a" sin xb3Î2 “
1
1
5 1 Î6
œ
2
" sin x
œ 2É" sin ˆ 16 ‰ 2È" sin 0 œ 2É 1# 2È1 œ 2 È2
0
œ
dx
3 Î2
3 Î2
3Î2
œ 23 ˆ" cosˆ 12 ‰‰ 23 ˆ" cos ˆ 13 ‰‰ œ 32 32 ˆ 23 ‰
'01Î6 È" sin x dx œ '01Î6 È" 1 sin x ÈÈ" sin x dx œ '01Î6 ÈÈ" sin x dx œ '01Î6 ÈÈcos x
1Î6
2ˆ
3 "
3 Î2
œ Š 23 23 ˆ 32 ‰ ‹ ˆ 27 30.
dx œ '1Î3
2
3
œ ’2a" sin xb1Î2 “
29.
sin2 x È" cos x
È" cos2 x
5 1 Î6
'3Î2 ˆu5Î2 2u3Î2 Èu‰ du
1
1
3Î2
sin ˆ 561 ‰‰ ‹ 72 u7Î2 54 u5Î2 32 u3Î2 ‘3Î2
4
5
7Î2
5Î2
3Î2
5Î2
7Î2
23 ‰ Š 27 ˆ 32 ‰ 45 ˆ 32 ‰ 23 ˆ 32 ‰ ‹ œ 45 ˆ 32 ‰ 27 ˆ 32 ‰ '17Î12Î12 È1 sin 2x dx œ '17Î12Î12 È1 1sin 2x ÈÈ1 sin 2x dx œ '17Î12Î12 ÈÈ1 sin 2x dx œ '17Î12Î12 ÈÈcos 2x
2
œ '1Î2
71Î12
cos 2x
È1 sin 2x
1 sin 2x
71Î12
dx œ ’È1 sin 2x“
1 Î2
1 sin 2x
2
1 sin 2x
18
35
dx
œ É1 sin 2ˆ 7121 ‰ É1 sin 2ˆ 12 ‰ œ É "# 1 œ 1 Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
1
È2
œ
È2 1
È2
474
Chapter 8 Techniques of Integration
31.
'01/2 )È" cos 2) d) œ '01/2 )È# l sin ) l d) œ È#'01/2 ) sin ) d) œ È# c)cos ) sin )d 10 /2 œ È#a"b œ È#
32.
'11 a" cos# tb$Î# dt œ '11 asin# tb$Î# dt œ '11 ¸ sin$ t¸ dt œ '!1 sin$ t dt '!1 sin$ t dt œ '!1 a" cos# tbsin t dt
'! a" cos# tbsin t dt œ '1 sin t dt '1 cos# t sin t dt '! sin t dt '! cos# t sin t dt œ ’cos t 1
’cos t 33.
34.
!
1
cos3 t
3 “!
œ ˆ" "
$
!
" "$ ‰ ˆ" 1
"
$
" "$ ‰ œ
1
!
cos3 t
3 “ 1
)
$
' sec2 x tan x dx œ ' tan x sec2 x dx œ 12 tan 2 x C
' sec x tan2 x dx œ ' sec x tan x tan xdx; u œ tan x, du œ sec2 x dx, dv œ sec x tan x dx, v œ sec x;
œ sec x tan x ' sec3 x dx œ sec x tan x ' sec2 x sec xdx œ sec x tan x ' atan2 x 1bsec xdx
œ sec x tan x Š' tan2 x sec xdx ' sec xdx‹ œ sec x tan x lnlsec x tan xl ' tan2 x sec xdx
Ê ' sec x tan2 x dx œ sec x tan x lnlsec x tan xl ' tan2 x sec xdx
Ê 2' tan2 x sec xdx œ sec x tan x lnlsec x tan xl Ê ' tan2 x sec xdx œ "# sec x tan x "# lnlsec x tan xl C
35.
36.
' sec3 x tan x dx œ ' sec2 x sec x tan x dx œ 13 sec 3 x C
' sec3 x tan3 x dx œ ' sec2 x tan2 x sec x tan x dx œ ' sec2 xasec2 x 1bsec x tan x dx
œ ' sec4 x sec x tan x dx ' sec2 x sec x tan x dx œ 15 sec 5 x 13 sec 3 x C
37.
' sec2 x tan2 x dx œ ' tan2 x sec2 x dx œ 13 tan 3 x C
38.
' sec4 x tan2 x dx œ ' sec2 x tan2 x sec2 x dx œ ' atan2 x 1btan2 x sec2 x dx œ ' tan4 x sec2 x dx ' tan2 x sec2 x dx
œ 15 tan 5 x 13 tan 3 x C
39.
'!1/3 2 sec$ x dx; u œ sec x, du œ sec x tan x dx, dv œ sec# x dx, v œ tan x;
'!1/3 2 sec$ x dx œ c2 sec x tan xd !1Î$ #'!1/3 sec x tan2 x dx œ # † " † ! # † # † È$ #'!1/3 sec x asec2 x "bdx
œ %È$ #'1/3 sec$ x dx #'1/3 sec x dx; 2'1/3 2 sec$ x dx œ %È$ c#ln l sec x + tan xld !1Î$
!
!
!
2'1/3 2 sec$ x dx œ %È$ #ln l " + !l #ln l # È$ l œ %È$ # ln Š# È$‹
!
'!1/3 2 sec$ x dx œ #È$ ln Š#
40.
È $‹
' ex sec$ aex bdx; u œ secaex b, du œ secaex btanaex bex dx, dv œ sec# aex bex dx, v œ tanaex b.
' ex sec$ aex b dx œ secaex btanaex b ' secaex btan# aex bex dx
œ secaex btanaex b ' secaex basec# aex b "bex dx
œ secaex btanaex b ' sec$ aex bex dx ' secaex bex dx
2' ex sec$ aex b dx œ secaex btanaex b ln¸secaex b tanaex b¸ C
' ex sec$ aex b dx œ "# ˆsecaex btanaex b ln¸secaex b tanaex b¸‰ C
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 8.2 Trigonometric Integrals
41.
475
' sec4 ) d) œ ' a" tan2 )bsec2 ) d) œ ' sec2 ) d) ' tan2 ) sec2 ) d) œ tan ) 13 tan3 ) C
œ tan ) 13 tan )asec2 ) 1b C œ 13 tan ) sec2 ) 23 tan ) C
42.
'
3sec4 a3xb dx œ ' a" tan2 a3xbbsec2 a3xb3dx œ ' sec2 a3xb3dx ' tan2 a3xb sec2 a3xb3dx œ tan a3xb 13 tan3 a3xb C
43.
'11/4/2 csc4 ) d) œ '11/4/2 a" cot# )bcsc# ) d) œ '11/4/2 csc# ) d) '11/4/2 cot# ) csc# ) d) œ ’cot ) cot3 ) “ 1Î2
$
œ a!b ˆ" "$ ‰ œ
44.
45.
1/4
%
$
' sec6 x dx œ ' sec4 x sec2 x dx œ ' atan2 x 1b2 sec2 x dx œ ' atan4 x 2tan2 x 1bsec2 x dx
œ ' tan4 x sec2 x dx 2' tan2 x sec2 x dx ' sec2 x dx œ 15 tan 5 x 23 tan 3 x tan x C
' 4 tan3 x dx œ 4'
asec# x "btan x dx œ 4' sec# x tan x dx 4' tan x dx œ % tan# x % ln lsec xl C
#
œ 2 tan# x 4 ln lsec xl C œ 2 tan# x 2 ln lsec2 xl C œ 2 tan# x 2 ln a1 tan2 xb C
46.
'11/4/4 6 tan4 x dx œ 6'11/4/4
asec# x "btan2 x dx œ 6'1/4 sec# x tan2 x dx 6'1/4 tan2 x dx
1/4
1/4
œ 6'1/4 sec2 x tan2 x dx 6'1/4 asec2 x 1bdx œ ’' tan$ x “
1/4
œ #a" a"bb c'tan
47.
1 Î4
xd 1Î4
1/4
$
1Î4
c'xd 1Î4
49.
œ % 'a" a"bb 1/4
$1
#
$1
#
1/4
œ $1 )
1
4
atan2 x 1b
2
atan2 x 1b lnlsec xl C œ 14 tan4 x "# tan2 x lnlsec xl C
' cot6 2x dx œ ' cot4 2x cot2 2x dx œ ' cot4 2x acsc2 2x 1b dx œ ' cot4 2x csc2 2x dx ' cot4 2x dx
œ ' cot4 2x csc2 2x dx ' cot2 2x cot2 2x dx œ ' cot4 2x csc2 2x dx ' cot2 2xacsc2 2x 1bdx
œ ' cot4 2x csc2 2x dx ' cot2 2x csc2 2x dx ' cot2 2x dx
œ ' cot4 2x csc2 2x dx ' cot2 2x csc2 2xdx ' acsc2 2x 1bdx
1
œ ' cot4 2x csc2 2x dx ' cot2 2x csc2 2xdx ' csc2 2x dx ' dx œ 10
cot5 2x 16 cot3 2x "# cot 2x x C
'11/6/3 cot3 x dx œ '11/6/3 acsc2 x " bcot x dx œ '11/6/3 csc2 x cot x dx '11/6/3 cot x dx œ ’ cot# x ln l csc xl“ 1Î3
#
1Î6
œ "# ˆ "$ $‰ Šln È#$ ln #‹ œ
50.
1 Î4
6'1/4 sec# x dx 6'1/4 dx
' tan5 x dx œ ' tan4 x tan x dx œ ' asec2 x 1b2 tan x dx œ ' asec4 x 2sec2 x 1btan x dx
œ ' sec4 x tan x dx 2' sec2 x tan x dx ' tan x dx œ ' sec3 x sec x tan x dx 2' sec x sec x tan x dx ' tan x dx
œ 14 sec4 x sec2 x lnlsec xl C œ
48.
1Î4
%
$
lnÈ$
' 8 cot4 t dt œ 8' acsc2 t " bcot2 t dt œ 8' csc2 t cot2 t dt 8' cot2 t dt œ 83 cot3 t 8' acsc2 t " bdt
œ 83 cot3 t 8 cot t 8t C
51.
' sin 3x cos 2x dx œ "# ' asin x sin 5xb dx œ "# cos x 10" cos 5x C
52.
' sin 2x cos 3x dx œ "# ' asinaxb sin 5xb dx œ "# ' asin x sin 5xb dx œ "# cos x 10" cos 5x C
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
476
Chapter 8 Techniques of Integration
53.
'11 sin 3x sin 3x dx œ "# '11 acos ! cos 6xb dx œ "# '11 dx "# '11 cos 6x dx œ "# x "#" sin 6x‘ 11 œ 1# 1# ! œ 1
54.
'!1Î2 sin x cos x dx œ "# '!1Î2 asin ! sin 2xb dx œ "# '!1Î2 sin 2x dx œ "% ccos 2xd 1! Î# œ "% a" "b œ "#
55.
' cos 3x cos 4x dx œ "# ' acosaxb cos 7xb dx œ "# ' acos x cos 7xb dx œ 2" sin x 14" sin 7x C
56.
'11ÎÎ22 cos 7x cos x dx œ "# '11ÎÎ22 acos 6x cos 8xb dx œ 2" 6" sin 6x 8" sin 8x‘ 1Î12Î2 œ 0
57.
58.
59.
60.
2)
' sin2 ) cos 3) d) œ ' 1 cos
cos 3) d) œ "# ' cos 3) d) "# ' cos 2) cos 3) d)
2
œ "# ' cos 3) d) "# ' "# acosa 2 3b) cosa 2 3b)b d) œ "# ' cos 3) d) 4" ' acosa)b
1
œ "# ' cos 3) d) 4" ' cos ) d) 4" ' cos 5) d) œ 16 sin 3) 4" sin ) 20
sin 5) C
cos 5)b d)
' cos2 2) sin ) d) œ ' a2cos2 ) 1b2 sin ) d) œ ' a4cos4 ) 4cos2 ) 1b sin ) d)
œ ' 4cos4 ) sin ) d) ' 4cos2 ) sin ) d) ' sin ) d) œ 45 cos5 ) 43 cos3 ) cos ) C
' cos3 ) sin 2) d) œ ' cos3 ) a2sin ) cos )b d) œ 2' cos4 ) sin ) d) œ 25 cos5 ) C
' sin3 ) cos 2) d) œ ' sin2 ) cos 2) sin ) d) œ ' a1 cos2 )ba2cos2 ) 1bsin ) d)
œ ' a2cos4 ) 3cos2 ) 1bsin ) d) œ 2' cos4 ) sin ) d) 3' cos2 ) sin ) d) ' sin ) d)
œ 25 cos5 ) cos3 ) cos ) C
61.
62.
' sin ) cos ) cos 3) d) œ "# ' 2sin ) cos ) cos 3) d) œ "# ' sin 2) cos 3) d) œ "# ' "# asina2 3b) sina2 3b)b d)
1
œ "4 ' asina)b sin 5)b d) œ 4" ' asin ) sin 5)b d) œ 4" cos ) 20
cos 5) C
' sin ) sin 2) sin 3) d) œ ' "# acosa1 2b) cosa1 2b)b sin 3) d) œ "# ' acosa)b cos 3)b sin 3) d)
œ "# ' sin 3) cos ) d) "# ' sin 3) cos 3) d) œ "# ' "# asina3 1b) sina3 1b)bd) 4" ' 2sin 3) cos 3) d)
œ "4 ' asin 2) sin 4)bd) 4" ' sin 6) d) œ 4" ' asin 2) sin 4)bd) 4" ' sin 6) d)
œ 18 cos 2) 63.
'
sec3 x
tan x
dx œ '
1
16 cos 4)
sec2 x sec x
tan x
dx œ '
6) C
1
24 cos
atan2 x 1bsec x
tan x
dx œ '
tan2 x sec x
tan x
dx '
sec x
tan x
dx œ ' tan x sec x dx ' csc x dx
œ sec x lnlcsc x cot xl C
64.
'
sin3 x
cos4 x
dx œ '
tan2 x
csc x
dx œ '
sin# x sin x
cos4 x
dx œ '
a1cos# xb sin x
cos4
x
dx œ '
sin x
cos4 x
dx '
œ ' sec2 x sec x tan x dx ' sec x tan x dx œ 13 sec3 x sec x C
65.
'
sin2 x
sin x dx
cos2 x
œ'
a1cos# xb
cos2 x
sin x dx œ '
cos# x sin x
cos4 x
1
sin x dx
cos2 x
'
dx œ ' sec3 x tan x dx ' sec x tan x dx
cos# x
sin x dx
cos2 x
œ ' sec x tan x dx ' sin x dx
œ sec x cos x C
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 8.2 Trigonometric Integrals
66.
67.
'
dx œ '
cot x
cos2 x
cos x
sin x
dx œ '
1
cos2 x
†
2
2sin x cos x
dx œ '
2
sin 2x
477
dx œ ' csc 2x 2dx œ lnl csc 2x cot 2x l C
2x
' x sin2 x dx œ ' x 1 cos
dx œ "# ' x dx "# ' x cos 2x dx ’u œ x, du œ dx, dv œ cos 2x dx, v œ "# sin 2x“
2
œ "4 x2 #" ’ #" x sin 2x ' #" sin 2x dx“ œ 4" x2 4" x sin 2x 18 cos 2x C
68.
' x cos3 x dx œ ' x cos2 x cos x dx œ ' xa1 sin2 xb cos x dx œ ' x cos x dx ' x sin2 x cos x dx;
' x cos x dx œ x sin x ' sin x dx œ x sin x cos x;
’u œ x, du œ dx, dv œ cos x dx, v œ sin x“
' x sin2 x cos x dx
œ 13 x sin3 x ' 13 sin3 x dx;
’u œ x, du œ dx, dv œ sin2 x cos x dx, v œ 13 sin3 x“
' a1 cos2 xb sin x dx œ 13 x sin3 x 13 ' sin x dx 13 ' cos2 x sin x dx œ 13 x sin3 x 13 cos x 19 cos3 x;
Ê ' x cos x dx ' x sin2 x cos x dx œ ax sin x cos xb ˆ 13 x sin3 x 13 cos x 19 cos3 x‰ C
œ 13 x sin3 x 1
3
œ x sin x 13 x sin3 x 23 cos x 19 cos3 x C
69. y œ lnasec xb; y w œ
sec x tan x
sec x
œ tan x;ay w b# œ tan# x; '!
1Î4
È" tan# x dx œ '
1Î4
!
lsec xl dx œ clnlsec x tan xld !1/4
œ lnŠÈ# "‹ lna! "b œ lnŠÈ# "‹
"
70. M œ '1Î4 sec x dx œ clnlsec x tan xld 1/41Î4 œ lnŠÈ# "‹ ln lÈ# "l œ ln È## "
1Î4
yœ
È
È '
ln È# " 1Î4
#"
1Î4
"
sec# x
#
Ê ax, yb œ Œ!ß Šln
dx œ
1Î4
"
ctan xd 1Î4 œ
È
#
"
#ln È
#"
È#" "
È#" ‹
71. V œ 1'! sin# x dx œ 1'!
1
1
"
È a" a"bb
#ln È# "
#"
1
#
dx œ
È#
#
#
1
1Î%
csin 2xd !
È#
#
1Î%
$1Î%
csin 2xd 1Î% È#
#
73. M œ '0 ax cos xbdx œ ’ "# x2 sin x“
0
1
21 2
csin 2xd 1$1Î% œ
21
21
xœ
#"
'!1 dx 1# '!1 cos 2x dx œ 1# cxd 1! 14 csin 2xd 1! œ 1# a1 !b 14 a! !b œ 1#
72. A œ '! È" cos 4x dx œ '! È# lcos 2xldx œ È# '!
œ
"
È
ln È# "
" cos 2x
2
1
œ
cos 2x dx È# '1Î% cos 2x dx È# '$1Î% cos 2x dx
È#
#
$1Î%
a" !b È#
#
1
a" "b È#
#
a ! "b œ È # È # œ # È #
œ Š "# a21b2 sin a21b‹ Š "# a0b2 sin a0b‹ œ 212 ;
'021 xax cos xbdx œ 211 '021 ax2 x cos xbdx œ 211 '021 x2 dx 211 '021 x cos xdx
2
2
2
’u œ x, du œ dx, dv œ cos x dx, v œ sin x“
0 '0 sin xdx
21
1
3
612 ’x “0
œ
41
3
œ
1
41 2
'021 ax2 2x cos x cos2 xbdx œ 411 '021 x2 dx 211 '021 x cos xdx 411 '021 cos2 x dx
21
1
212 Œ’x sin x“0
'0 sin xdx œ
œ
21
1
cos
x
’
“
2
21
0
œ
41
3
21
1
212 acos 21
1
3
612 a81
cos 0b œ
2
41
3
0b 0œ
2
41
3 ;
1
212 Œ21 sin 21
yœ
1
21 2
21
'021 "# ax cos xb2 dx
2
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
478
Chapter 8 Techniques of Integration
œ
21
1
3
1212 ’x “0
œ
21
3
1Î3
œ 1'!
1 Î3
œ
'!
1Î3
1
2
dx 1 Î3
œ
21
3
1Î3
0
12
6
0
21
1
812 ’x“0
sin# x dx 1'!
1Î3
21
cos x“
1asin x sec xb2 dx œ 1'!
œ 12 ’ x“
œ
1
212 ’x sin x
21
1
1612 ’sin 2x“0
74. V œ '!
1
2
'!
1Î3
'021 cos 2x2 1 dx œ 231 0 811 '021 cos 2x dx 811 '021 dx
2
0
1Î3
cos 2x dx 21ˆlnlsec
1Î3
0
1
41
œ
81 2 3
121
2
Ê The centroid is Š 431 ,
81 2 3
121 ‹.
asin# x 2sin x sec x sec2 xb dx
2tan x dx 1'!
14 ’ sin 2x“
1È3
8
1
41 2
sec2 x dx œ 1'!
1Î3
1
3l
1 cos 2x
2
lnlsec 0l‰ 1ˆtan
1
3
1Î3
dx 21’ lnlsec xl“
0
1Î3
1’ tan x“
0
tan 0‰
21 ln 2 1È3 œ 12 ˆ 13 0‰ 14 ˆsin 2ˆ 13 ‰ sin 2a0b‰ 21 ln 2 1È3
1Š41 21È3 48 ln 2‹
21 ln 2 1È3 œ
24
8.3 TRIGONOMETRIC SUBSTITUTIONS
1. x œ 3 tan ), 1# ) 1# , dx œ
3 d)
cos# )
, 9 x# œ 9 a1 tan# )b œ 9 sec# ) Ê
ˆbecause cos ) 0 when 1# ) 1# ‰ ;
' È dx
œ 3'
' È 3 dx
; c3x œ ud Ä
' È du
œ'
3.
'22
#
œ #" tan" #x ‘ # œ
4.
'02 8 dx2x
5.
'03Î2 È dx
6.
'01Î2
9 x#
2.
1 9x#
1 u#
dx
4 x#
#
œ
9 x#
È2
"
#
cos ) d)
3 cos# )
dt
cos# t asec tb
'02 4dxx
#
œ'
'
d)
cos )
œ ln ksec ) tan )k Cw œ ln ¹
du
È 1 u#
; u œ tan t, 1# t , du œ
œ
"
#
"
#
tan" 1 #" tan" #x ‘ # œ
!
; ct œ 2xd Ä
'01Î2
È2
"
#
"
#
25
#
8. t œ
"
3
1
3ksec )k
œ
kcos )k
3
œ
cos )
3 ;
x3 ¹ Cw œ ln ¹È9 x# x¹ C
, È1 u# œ lsec tl œ sec t ;
"
#
tan" (1) œ ˆ #" ‰ ˆ 14 ‰ ˆ #" ‰ ˆ 14 ‰ œ
ˆ #" tan" 1 sin" 0 œ
dt
È1 t#
1
6
"
#
0œ
È2
1Î
œ csin" td 0
a) sin ) cos )b C œ
25
#
sin ), 1# ) 1# , dt œ
"
3
’sin" ˆ 5t ‰ 1
4
tan" 0‰ œ ˆ #" ‰ ˆ #" ‰ ˆ 14 ‰ 0 œ
1
16
1
6
œ sin"
7. t œ 5 sin ), 1# ) 1# , dt œ 5 cos ) d), È25 t# œ 5 cos );
' È25 t# dt œ ' (5 cos ))(5 cos )) d) œ 25 ' cos# ) d) œ 25 '
œ
dt
cos# t
œ
œ ' sec t dt œ ln ksec t tan tk C œ ln ¹Èu# 1 u¹ C œ ln ¹È1 9x# 3x¹ C
$Î#
œ sin" 3x ‘ ! œ sin"
2 dx
È1 4x#
1
#
È 9 x#
3
"
È 9 x#
È
ˆ 5t ‰ Š 255 t# ‹“
"
È2
sin" 0 œ
1 cos 2)
#
Cœ
25
#
1
4
0œ
d) œ 25 ˆ #) sin" ˆ 5t ‰ 1
4
sin 2) ‰
4
tÈ25 t#
#
C
C
cos ) d), È1 9t# œ cos );
' È1 9t# dt œ "3 ' (cos ))(cos )) d) œ 3" ' cos# ) d) œ 6" a) sin ) cos )b C œ 6" ’sin" (3t) 3tÈ1 9t# “ C
9. x œ
'È
7
#
sec ), 0 ) 1# , dx œ
dx
4x# 49
œ'
ˆ 7#
sec ) tan )‰ d)
7 tan )
7
#
œ
sec ) tan ) d), È4x# 49 œ È49 sec# ) 49 œ 7 tan );
"
#
' sec ) d) œ "# ln ksec ) tan )k C œ "# ln ¹ 2x7 È4x7# 49 ¹ C
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 8.3 Trigonometric Substitutions
10. x œ
sec ), 0 ) 1# , dx œ
3
5
' È 5 dx
25x# 9
œ'
sec ) tan ) d), È25x# 9 œ È9 sec# ) 9 œ 3 tan );
3
5
œ ' sec ) d) œ ln ksec ) tan )k C œ ln ¹ 5x
3 5 ˆ 35 sec ) tan )‰ d)
3 tan )
È25x# 9
¹
3
C
11. y œ 7 sec ), 0 ) 1# , dy œ 7 sec ) tan ) d), Èy# 49 œ 7 tan );
) tan )) d)
' Èy y 49 dy œ ' (7 tan ))(77 sec
œ 7 ' tan# ) d) œ 7 ' asec# ) 1b d) œ 7(tan ) )) C
sec )
#
œ 7’
Èy# 49
7
sec" ˆ y7 ‰“ C
12. y œ 5 sec ), 0 ) 1# , dy œ 5 sec ) tan ) d), Èy# 25 œ 5 tan );
sec ) tan )) d)
' Èyy 25 dy œ ' (5 tan ))(5
œ "5 '
125 sec )
#
$
œ
"
10
$
a) sin ) cos )b C œ
"
10
tan# ) cos# ) d) œ
’sec" ˆ y5 ‰ Š
"
5
Èy# 25
‹ Š 5y ‹“
y
'
sin# ) d) œ
Cœ’
sec "
10
"
10
ˆ 5y ‰
' (1 cos 2)) d)
Èy# 25
“
#y #
C
13. x œ sec ), 0 ) 1# , dx œ sec ) tan ) d), Èx# 1 œ tan );
'
dx
x# È x# 1
œ'
sec ) tan ) d)
sec# ) tan )
œ'
d)
sec )
œ sin ) C œ
È x# 1
x
C
14. x œ sec ), 0 ) 1# , dx œ sec ) tan ) d), Èx# 1 œ tan );
'
2 dx
x$ È x# 1
œ'
2 tan ) sec ) d)
sec$ ) tan )
2) ‰
œ 2 ' cos# ) d) œ 2 ' ˆ 1 cos
d) œ ) sin ) cos ) C
#
#
œ ) tan ) cos# ) C œ sec" x Èx# 1 ˆ "x ‰ C œ sec" x È x# 1
x#
C
15. u œ 9 x# Ê du œ 2x dx Ê "# du œ x dx;
' Èx dx
9 x#
œ "# '
1
Èu du
œ Èu C œ È9 x# C
16. x œ 2 tan ), 1# ) 1# , dx œ 2sec# ) d) , 4 x# œ 4sec# )
' 4xdxx
2
#
œx
œ'
ˆ4tan2 )‰a2sec# )bd)
4sec# )
1 ˆ x ‰
2 tan
2 C
œ ' 2 tan2 ) d) œ 2' asec# ) 1bd) œ 2' sec# ) d) 2' d) œ 2 tan ) 2) C
17. x œ 2 tan ), 1# ) 1# , dx œ
' Èx
$
dx
x# 4
œ'
œ 8 Œ
œ 8'
a8 tan$ )b (cos )) d)
cos# )
ct œ cos )d Ä 8 '
È x# 4
#
t# 1
t%
2 d)
cos# )
, Èx# 4 œ
sin$ ) d)
cos% )
œ 8'
2
cos ) ;
acos# ) 1b ( sin )) d)
cos% )
dt œ 8' ˆ t"# t"% ‰ dt œ 8 ˆ "t ax # 4 b
8 †3
$Î#
Cœ
"
3
ax# 4b
$Î#
" ‰
3t$
;
C œ 8 Š sec ) sec$ )
3 ‹
C
4Èx# 4 C œ 13 ax# 8bÈx# 4 C
18. x œ tan ), 1# ) 1# , dx œ sec# ) d), Èx# 1 œ sec );
'
dx
x# È x# 1
œ'
sec# ) d)
tan# ) sec )
œ'
cos ) d)
sin# )
œ sin" ) C œ
È x # 1
x
C
19. w œ 2 sin ), 1# ) 1# , dw œ 2 cos ) d), È4 w# œ 2 cos );
'
8 dw
w# È 4 w#
œ'
8†2 cos ) d)
4 sin# )†2 cos )
œ 2'
d)
sin# )
œ 2 cot ) C œ
2 È 4 w #
w
C
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
479
480
Chapter 8 Techniques of Integration
20. w œ 3 sin ), 1# ) 1# , dw œ 3 cos ) d), È9 w# œ 3 cos );
' È9w w
#
#
dw œ '
)
' acsc# ) 1b d)
œ ' cot# ) d) œ ' Š 1 sinsin
# ) ‹ d) œ
#
3 cos )†3 cos ) d)
9 sin# )
œ cot ) ) C œ È 9 w#
w
sin" ˆ w3 ‰ C
21. u œ 5x Ê du œ 5dx, a œ 6
1
10
" ˆ u ‰
" ˆ 5x ‰
' 36 100
' 1
' 1
25x dx œ 20 a6b a5xb 5dx œ 20 a u du œ 20 † 6 tan
6 C œ 3 tan
6 C
#
#
2
#
2
22. u œ x2 4 Ê du œ 2x dx Ê "# du œ x dx
' x Èx# 4 dx œ "# ' Èu du œ 13 u3Î2 C œ 13 ax# 4b3Î2 C
23. x œ sin ), 0 Ÿ ) Ÿ 13 , dx œ cos ) d), a1 x# b
È3Î2
'0
4x# dx
a1 x# b$Î#
œ '0
1Î3
1Î$
œ cos$ );
)
'0 asec# ) 1b d)
œ 4 '0 Š 1 coscos
# ) ‹ d) œ 4
1Î3
4 sin# ) cos ) d)
cos$ )
24. x œ 2 sin ), 0 Ÿ ) Ÿ 16 , dx œ 2 cos ) d), a4 x# b
'01
dx
a4 x# b$Î#
œ '0
1Î6
2 cos ) d)
8 cos$ )
œ
$Î#
dx
ax# 1b$Î#
œ'
sec ) tan ) d)
tan$ )
#
œ'
cos ) d)
sin# )
x# dx
ax# 1b&Î#
œ'
sec# )†sec ) tan ) d)
tan& )
27. x œ sin ), 1# ) ' a1 xxb
# $Î#
'
dx
œ'
' a1 xxb
%
29. x œ
'
"
3
œ'
œ'
8 ˆ "# sec# )‰ d)
sec% )
6 dt
a9t# 1b#
œ'
sec )‰ d)
sec% )
#
x
x# 1
$Î#
C
œ tan& );
x$
3 ax# 1b$Î#
"Î#
È 1 x# &
‹
x
C
œ cos );
œ ' cot# ) csc# ) d) œ cot3 ) C œ "3 Š
$
"
#
C
œ cos$ );
&
È 1 x# $
‹
x
C
#
sec# ) d), a4x# 1b œ sec% );
œ 4 ' cos# ) d) œ 2() sin ) cos )) C œ 2 tan" 2x tan ), 1# ) 1# , dt œ
6 ˆ "3
œ tan$ );
d) œ 3 sin" $ ) C œ , dx œ cos ) d), a1 x# b
cos )†cos ) d)
sin% )
&Î#
"
4È 3
œ ' cot% ) csc# ) d) œ cot5 ) C œ "5 Š
tan ), 1# ) 1# , dx œ
8 dx
a4x# 1b#
30. t œ
'
"
#
dx
1
#
cos )
sin% )
, dx œ cos ) d), a1 x# b
cos$ )†cos ) d)
sin' )
28. x œ sin ), 1# ) # "Î#
1
#
œ'
$Î#
œ sin" ) C œ È
26. x œ sec ), 0 ) 12 , dx œ sec ) tan ) d), ax# 1b
'
œ 8 cos$ );
'01Î6 cosd) ) œ "4 ctan )d 1! Î' œ È123 œ
"
4
25. x œ sec ), 0 ) 12 , dx œ sec ) tan ) d), ax# 1b
'
1Î3
#
41
3
œ 4È3 œ 4 ctan ) )d !
$Î#
"
3
4x
a4x# 1b
C
sec# ) d), 9t# 1 œ sec# );
œ 2 ' cos# ) d) œ ) sin ) cos ) C œ tan" 3t 3t
a9t# 1b
C
31. u œ x2 1 Ê du œ 2x dx Ê "# du œ x dx
' x x 1 dx œ ' ax x x 1 bdx œ ' x dx ' x x 1 dx œ #" x2 #" ' u1 du œ #" x2 #" ln lul C œ #" x2 #" ln lx2 1l C
$
2
2
2
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 8.3 Trigonometric Substitutions
32. u œ 25 %x2 Ê du œ 8x dx Ê "8 du œ x dx
' 25 x %x dx œ 8" ' u1 du œ 8" ln lul C œ 8" ln a25 %x2 b C
2
33. v œ sin ), 1# ) 1# , dv œ cos ) d), a1 v# b
'
œ'
v# dv
a1 v# b&Î#
sin# ) cos ) d)
cos& )
œ ' tan# ) sec# ) d) œ
34. r œ sin ), 1# ) 1# ;
' a1 r rb
# &Î#
)
œ'
dr
&Î#
œ cos& );
tan$ )
3
Cœ
"
3
'0
et dt
Èe2t 9
tan " Ð4Î3Ñ
œ 'tanc" Ð1Î3Ñ
œ ln ˆ 53 43 ‰ ln Š
(
tan " Ð4Î3Ñ
œ 'tan
3 tan )†sec# ) d)
tan )†3 sec )
È10
3
" Ð1Î3Ñ
37.
'11ÎÎ124
2 dt
Èt 4tÈt
'11ÎÈ3
2 du
1 u#
tan " Ð4Î3Ñ
#
)
(tan )) Š sec
tan ) ‹ d)
; ’u œ 2Èt, du œ
œ '1Î6
1Î4
C
È 1 r# (
“
r
C
d), Èe2t 9 œ È9 tan# ) 9 œ 3 sec );
"
3" ‹ œ ln 9 ln Š1 È10‹
œ 'tanc" Ð3Î4Ñ
et dt
a1 e2t b$Î#
sec# )
tan )
$
Ð4Î3Ñ
sec ) d) œ cln ksec ) tan )kd tan
tanc" Ð1Î3Ñ
36. Let et œ tan ), t œ ln (tan )), tan" ˆ 34 ‰ Ÿ ) Ÿ tan" ˆ 43 ‰ , dt œ
'lnlnÐ3Ð4ÎÎ43Ñ Ñ
v
‹
1 v#
œ ' cot' ) csc# ) d) œ cot7 ) C œ "7 ’
cos& )†cos ) d)
sin) )
35. Let et œ 3 tan ), t œ ln (3 tan )), tan" ˆ "$ ‰ Ÿ ) Ÿ tan" ˆ %$ ‰, dt œ
ln 4
ŠÈ
2 sec# ) d)
sec# )
tan " Ð4Î3Ñ
sec$
)
œ 'tan
" Ð3Î4Ñ
"
Èt
dt“ Ä '1ÎÈ3
2 du
1 u#
1
sec# )
tan )
d), 1 e2t œ 1 tan# ) œ sec# );
"
Ð4Î3Ñ
cos ) d) œ csin )d tan
œ
tanc" Ð$Î%Ñ
; u œ tan ),
1Î%
œ c2)d 1Î' œ 2 ˆ 14 16 ‰ œ
1
6
4
5
3
5
œ
"
5
Ÿ ) Ÿ 14 , du œ sec# ) d), 1 u# œ sec# );
1
6
38. y œ etan ) , 0 Ÿ ) Ÿ 14 , dy œ etan ) sec# ) d), È1 (ln y)# œ È1 tan# ) œ sec );
'1e yÈ1 dy(ln y)
œ '0
1Î4
#
etan ) sec# )
etan ) sec )
d) œ '0 sec ) d) œ cln ksec ) tan )kd !
1Î4
1Î%
œ ln Š1 È2‹
39. x œ sec ), 0 ) 1# , dx œ sec ) tan ) d), Èx# 1 œ Èsec# ) 1 œ tan );
'
dx
xÈ x# 1
œ'
sec ) tan ) d)
sec ) tan )
œ ) C œ sec" x C
40. x œ tan ), dx œ sec# ) d), 1 x# œ sec# );
' x dx1 œ ' secsec) )d) œ ) C œ tan" x C
#
#
#
41. x œ sec ), dx œ sec ) tan ) d), Èx# 1 œ Èsec# ) 1 œ tan );
' x dx œ ' sec )†sec ) tan ) d) œ ' sec# ) d) œ tan ) C œ Èx# 1 C
È x# 1
tan )
42. x œ sin ), dx œ cos ) d), 1# ) ' È dx
1 x#
œ'
cos ) d)
cos )
1
#
;
œ ) C œ sin" x C
43. Let x2 œ tan ), 0 Ÿ ) 1# , 2x dx œ sec2 ) d) Ê x dx œ "# sec2 ) d); È1 x4 œ È1 tan2 ) œ sec )
'Èx
1 x4
dx œ
"
#
)
"'
"
"
2
4
È
' sec
sec ) d) œ # sec )d) œ # lnlsec ) tan )l C œ # lnl 1 x x l C
2
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
481
482
Chapter 8 Techniques of Integration
44. Let ln x œ sin ), 1# Ÿ ) 0 or 0 ) Ÿ 1# , 1x dx œ cos ) d), É1 aln xb2 œ cos )
)
' É1xlnalnx xb dx œ ' cos
' 1 sinsin) ) d) œ ' csc ) d) ' sin ) d) œ
sin ) d) œ
2
2
œ ln» ln1x É1aln xb2
ln x
2
2
» É1 aln xb C œ ln»
1 É1aln xb2
»
ln x
lnlcsc ) cot )l cos ) C
É1 aln xb2 C
45. Let u œ Èx Ê x œ u2 Ê dx œ 2u du Ê ' É 4 x x dx œ ' É 4 u2u 2u du œ 2' È4 u2 du;
2
1
#
u œ 2 sin ), du œ 2 cos ) d), 0 ) Ÿ
, È4 u2 œ 2 cos )
2' È4 u2 du œ 2' a2 cos )b a2 cos )b d) œ 8' cos2 ) d) œ 8'
'
œ 4) 2 sin 2) C œ 4) 4 sin ) cos ) C œ 4 sin1 ˆ u2 ‰ œ 4 sin1 Š
œ 4 sin1 Š
Èx
2 ‹
1 cos 2)
d) œ 4
2
È
2
4ˆ u2 ‰Š 42u ‹ C
d) 4' cos 2) d)
Èx
2 ‹
ÈxÈ4 x C
È4x x2 C
46. Let u œ x3Î2 Ê x œ u2Î3 Ê dx œ 23 u1Î3 du
' È 1 x x dx œ ' É
3
u2Î3
ˆ 2 u1Î3 ‰du
3
1 au2Î3 b 3
œ'
u1Î3 ˆ 2 ‰
È1 u# 3u1Î3 du
œ
2
3
'È1
1 u#
du œ
2
3
sin1 u C œ
2
3
sin1 ˆx3Î2 ‰ C
47. Let u œ Èx Ê x œ u2 Ê dx œ 2u du Ê ' ÈxÈ1 x dx œ ' u È1 u2 2u du œ 2' u2 È1 u2 du;
u œ sin ), du œ cos ) d), 1 ) Ÿ 1 , È1 u2 œ cos )
#
#
4)
2' u2 È1 u2 du œ 2' sin2 ) cos ) cos ) d) œ 2' sin2 ) cos2 ) d) œ 12 ' sin2 2) d) œ 12 ' 1 cos
d)
2
1'
1'
1
1
1
1
1
1
œ 4 d) 4 cos 4) d) œ 4 ) 16 sin 4) C œ 4 ) 8 sin 2) cos 2) C œ 4 ) 4 sin ) cos ) a2cos2 ) 1b C
œ 14 ) 12 sin ) cos3 ) 14 sin ) cos ) C œ 14 sin1 u 12 u a1 u2 b
œ 1 sin1 Èx 1 Èx a1 xb3Î2 1 Èx È1 x C
4
2
3Î2
14 u È1 u2 C
4
48. Let w œ Èx 1 Ê w2 œ x 1 Ê 2w dw œ dx Ê '
w œ sec ), dx œ sec ) tan ) d), 0 ) 1
#
Èx 2
Èx 1 dx
œ'
È w2 1
2w dw
w
œ 2' Èw2 1 dw
, Èw2 1 œ tan )
2' Èw2 1 dw œ 2' tan ) sec ) tan ) d); u œ tan ), du œ sec2 ) d), dv œ sec ) tan ) d), v œ sec )
2' tan ) sec ) tan ) d) œ 2 sec ) tan ) 2' sec3 ) d) œ 2 sec ) tan ) 2' sec2 ) sec )d)
œ 2 sec ) tan ) 2' atan2 ) 1bsec ) d) œ 2 sec ) tan ) 2Š' tan2 ) sec ) d) ' sec ) d)‹
œ 2sec ) tan ) 2lnlsec ) tan )l 2' tan2 ) sec ) d) Ê 2' tan2 ) sec ) d) œ sec ) tan ) lnlsec ) tan )l C
œ w Èw2 1 lnlw Èw2 1l C œ Èx 1 Èx 2 lnlÈx 1 Èx 2l C
49. x
dy
dx
œ Èx# 4; dy œ Èx# 4
Ä yœ'
œ 2’
(2 tan ))(2 sec ) tan )) d)
2 sec )
È x# 4
#
dx
x
;yœ'
1
È x# 4
x
Ô x œ 2 sec ), 0 ) # ×
dx; Ö dx œ 2 sec ) tan ) d) Ù
Õ Èx# 4 œ 2 tan ) Ø
œ 2' tan# ) d) œ 2 ' asec# ) 1b d) œ 2(tan) )) C
sec" ˆ x# ‰“ C; x œ 2 and y œ 0 Ê 0 œ 0 C Ê C œ 0 Ê y œ 2 ’
È x# 4
#
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
sec" x# “
Section 8.3 Trigonometric Substitutions
50. Èx# 9
œ 1, dy œ
dy
dx
;yœ'
dx
È x# 9
1
Ô x œ 3 sec ), 0 ) # ×
dx
Ö dx œ 3 sec ) tan ) d) Ù Ä y œ '
È x# 9 ;
Õ Èx# 9 œ 3 tan ) Ø
œ ' sec ) d) œ ln ksec ) tan )k C œ ln ¹ x3 Ê y œ ln ¹ x3 51. ax# 4b
yœ'
' x dx 4 œ #3 tan" #x C; x œ 2 and y œ 0
3 dx
x # 4 ; y œ 3
œ 3# tan" ˆ x# ‰
Ê C œ 381 Ê y
52. ax# 1b
# dy
dx
œ Èx# 1, dy œ
sec# ) d)
sec$ )
3
È 9 x#
3
A œ '0
1Î2
54.
x2
a2
y2
b2
#
dx
ax# 1b$Î#
; x œ tan ), dx œ sec# ) d), ax# 1b
È x# 1
œ 3'0
1Î2
œ 1 Ê y œ „ bÉ1 1
#
a
œ 2ab'0
x2
a2
dx œ 4b'0
1Î2
d) 2ab'0
1 Î2
1Î2
55. (a) A œ '0
1 Î2
cos# ) d) œ
x2
a2 ;
1Î#
c) sin ) cos )d !
3
#
A œ 4'0 bÉ1 a
œ ’x sin1 x“
0
(b) M œ '0
1 Î2
1 Î2
0
'0
sin1 x dx œ
œ sec$ );
x
È x# 1
x2
a2
x
È 1 x2
œ
a
x2
a2
dx
œ cos ), x œ 0 œ a sin ) Ê ) œ 0, x œ a œ a sin ) Ê ) œ 12 “
cos2 ) d) œ 4ab'0
1Î2
1Î2
ab’sin 2)“
0
1
È1 x2 dx,
dv œ dx, v œ x“
1 Î2
dx œ œ ˆ "# sin1 "# !‰ ’È1 x2 “
0
1 6È3 12
;
1#
xœ
1
È
1 6 3 12
1#
12
’ " x2
1 6È3 12 Œ #
1 Î2
sin1 x“
0
"
#
'01Î2
x2
È 1 x2
1 cos2)
d)
#
œ 2abˆ 12 0‰ abasin 1 sin 0b œ 1ab
'01Î2 x sin1 x dx œ
œ
1 6È3 12
1#
12
1 6È3 12
’u œ sin1 x, du œ
œ
C; x œ 0 and y œ 1
31
4
dx œ 4b'0 É1 1Î2
cos2) d) œ 2ab’ )“
1 Î2
x2
a2
cos ) aa cos )b d) œ 4ab'0
sin1 x dx ’u œ sin1 x, du œ
1 Î2
Cœ
tan" 1 C
, dx œ 3 cos ) d), È9 x# œ È9 9 sin# ) œ 3 cos );
’x œ a sin ), 1# Ÿ ) Ÿ 1# , dx œ a cos ) d), É1 4b'0 É1 tan )
sec )
$Î#
3
#
1
x
dx; x œ 3 sin ), 0 Ÿ ) Ÿ
3 cos )†3 cos ) d)
3
Ê 0œ
31
8
œ ' cos ) d) œ sin ) C œ tan ) cos ) C œ
Ê 1œ0C Ê yœ
53. A œ '0
C; x œ 5 and y œ ln 3 Ê ln 3 œ ln 3 C Ê C œ 0
È x# 9
¹
3
œ 3, dy œ
dy
dx
È x# 9
¹
3
3 sec ) tan ) d)
3 tan )
'01Î2 x sin1 x dx
1
È1 x2 dx,
dv œ x dx, v œ "# x2 “
dx
’x œ sin ), 1# ) 1# , dx œ cos ) d), È1 x2 œ cos ), x œ 0 œ sin ) Ê ) œ 0, x œ
œ
2
12
Š"ˆ"‰
1 6È3 12 Œ # #
œ
12
1
1 6È3 12 Œ 48
œ
1
12
1 6È3 12 Œ 48
’ 4) "8 sin 2)“
"
#
sin1 ˆ "# ‰ 0‹ '01Î6
1 cos 2)
2
"
#
'01Î6
d) œ
1 Î6
0
œ
sin2 )
cos )
cos ) d) œ
12
1
1 6È3 12 Œ 48
3È 3 1
;
4Š1 6È3 12‹
"
4
12
1
1 6È3 12 Œ 48
"
#
"
#
œ sin ) Ê ) œ 16 “
'01Î6 sin2 ) d)
'01Î6 d) "4 '01Î6 cos 2) d)
yœ
1
È
1 6 3 12
1#
1
2
'01Î2
’u œ asin xb , du œ
"
1 2
# asin xb dx
2 sinc1 x
È1 x2 dx,
dv œ dx, v œ x“
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
483
484
Chapter 8 Techniques of Integration
œ
2
xasin1 x dxb •
6
1 6È3 12 ”
1 Î2
'0
1 Î2
2x sinc1 x
È1 x2 dx
0
’u œ sin1 x, du œ
œ
œ
2
6
Š " ˆsin1 ˆ "# ‰‰
1 6È3 12 #
0‹ ”2È1 x2 sin1 x•
Œ2É1 ˆ "# ‰ sin1 ˆ "# ‰ 0 ’2x“
0
56. V œ '0 1ŠÈx tan1 x‹ dx œ 1'0 x tan1 x dx
2
1
1
œ 1Œ’ "# x2 tan1 x“ 0
œ 1Œ 18 "
#
"
#
1
'01
x2
1 x2
'0
1 Î2
0
1 Î2
2
6
12
1 6È3 12 Œ 72
1 Î2
1
È1 x2 dx,
œ
"
#
2È 1 x2
È1 x2 dx
1
1 x2 dx,
1‹ œ
12 121È3 72
12Š1 6È3 12‹
'01 ˆ 1 1 1 x ‰dx œ 1Œ 18 #" '01 ˆ 1 1 1 x ‰dx
2
2
2
0
x3 È1 x2 dx œ 13 x2 a1 x2 b
3Î2
1
3
'
a1 x2 b
3Î2
2
x3 È1 x2 dx œ ' x2 È1 x2 x dx œ
œ 13 a1 x2 b
3 Î2
5" a1 x2 b
(c) Trig substitution: x œ sin ),
1
#
5Î2
"#
3Î2
2x dx œ 13 x2 a1 x2 b
(b) Substitution: u œ 1 x Ê x œ 1 u Ê du œ 2x dx Ê
'
1È3
6
'01 dx "# '01 1 1 x dx œ 1Œ 18 ’ #" x #" tan1 x“1 œ 1ˆ 18 ˆ #" #" tan1 1 0 0‰‰ œ 1a14 2b
2
'
dv œ x dx, v œ #" x2 “
57. (a) Integration by parts: u œ x2 , du œ 2x dx, dv œ x È1 x2 dx, v œ 13 a1 x2 b
'
v œ 2È1 x2 “
2x
È1 x2 dx