INSTRUCTOR’S
SOLUTIONS MANUAL
DUANE KOUBA
University of California, Davis
T HOMAS ’ C ALCULUS
FOURTEENTH EDITION
Based on the original work by
George B. Thomas, Jr
Massachusetts Institute of Technology
as revised by
Joel Hass
University of California, Davis
Christopher Heil
Georgia Institute of Technology
Maurice D. Weir
Naval Postgraduate School
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 from electronic files supplied by the author.
Copyright © 2018, 2014, 2010 Pearson Education, Inc.
Publishing as Pearson, 330 Hudson Street, NY NY 10013
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: 978-0-13-443918-1
ISBN-10: 0-13-443918-X
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 9
Trigonometric Functions 19
Graphing with Software 27
Practice Exercises 32
Additional and Advanced Exercises 40
2 Limits and Continuity 45
2.1
2.2
2.3
2.4
2.5
2.6
Rates of Change and Tangents to Curves 45
Limit of a Function and Limit Laws 49
The Precise Definition of a Limit 59
One-Sided Limits 66
Continuity 72
Limits Involving Infinity; Asymptotes of Graphs 77
Practice Exercises 87
Additional and Advanced Exercises 93
3 Derivatives 101
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 101
The Derivative as a Function 107
Differentiation Rules 118
The Derivative as a Rate of Change 123
Derivatives of Trigonometric Functions 129
The Chain Rule 138
Implicit Differentiation 148
Related Rates 156
Linearization and Differentials 161
Practice Exercises 167
Additional and Advanced Exercises 179
Copyright 2018 Pearson Education, Inc.
iii
4 Applications of Derivatives 185
4.1
4.2
4.3
4.4
4.5
4.6
4.7
Extreme Values of Functions 185
The Mean Value Theorem 195
Monotonic Functions and the First Derivative Test 201
Concavity and Curve Sketching 212
Applied Optimization 238
Newton's Method 253
Antiderivatives 257
Practice Exercises 266
Additional and Advanced Exercises 280
5 Integrals 287
5.1
5.2
5.3
5.4
5.5
5.6
Area and Estimating with Finite Sums 287
Sigma Notation and Limits of Finite Sums 292
The Definite Integral 298
The Fundamental Theorem of Calculus 313
Indefinite Integrals and the Substitution Method 323
Definite Integral Substitutions and the Area Between Curves 329
Practice Exercises 346
Additional and Advanced Exercises 357
6 Applications of Definite Integrals 363
6.1
6.2
6.3
6.4
6.5
6.6
Volumes Using Cross-Sections 363
Volumes Using Cylindrical Shells 375
Arc Length 386
Areas of Surfaces of Revolution 394
Work and Fluid Forces 400
Moments and Centers of Mass 410
Practice Exercises 425
Additional and Advanced Exercises 436
7 Transcendental Functions 441
7.1
7.2
7.3
7.4
7.5
7.6
7.7
7.8
Inverse Functions and Their Derivatives 441
Natural Logarithms 450
Exponential Functions 459
Exponential Change and Separable Differential Equations 473
Indeterminate Forms and L’Hôpital’s Rule 478
Inverse Trigonometric Functions 488
Hyperbolic Functions 501
Relative Rates of Growth 510
Practice Exercises 515
Additional and Advanced Exercises 529
Copyright 2018 Pearson Education, Inc.
iv
8 Techniques of Integration 533
8.1
8.2
8.3
8.4
8.5
8.6
8.7
8.8
8.9
Using Basic Integration Formulas 533
Integration by Parts 546
Trigonometric Integrals 560
Trigonometric Substitutions 569
Integration of Rational Functions by Partial Fractions 578
Integral Tables and Computer Algebra Systems 589
Numerical Integration 600
Improper Integrals 611
Probability 623
Practice Exercises 632
Additional and Advanced Exercises 646
9 First-Order Differential Equations 655
9.1
9.2
9.3
9.4
9.5
Solutions, Slope Fields, and Euler's Method 655
First-Order Linear Equations 664
Applications 668
Graphical Solutions of Autonomous Equations 673
Systems of Equations and Phase Planes 680
Practice Exercises 686
Additional and Advanced Exercises 694
10 Infinite Sequences and Series 697
10.1 Sequences 697
10.2 Infinite Series 709
10.3 The Integral Test 717
10.4 Comparison Tests 726
10.5 Absolute Convergence; The Ratio and Root Tests 736
10.6 Alternating Series and Conditional Convergence 742
10.7 Power Series 752
10.8 Taylor and Maclaurin Series 765
10.9 Convergence of Taylor Series 771
10.10 The Binomial Series and Applications of Taylor Series 779
Practice Exercises 788
Additional and Advanced Exercises 799
Copyright 2018 Pearson Education, Inc.
v
11 Parametric Equations and Polar Coordinates 805
11.1
11.2
11.3
11.4
11.5
11.6
11.7
Parametrizations of Plane Curves 805
Calculus with Parametric Curves 814
Polar Coordinates 824
Graphing Polar Coordinate Equations 829
Areas and Lengths in Polar Coordinates 837
Conic Sections 843
Conics in Polar Coordinates 854
Practice Exercises 864
Additional and Advanced Exercises 875
12 Vectors and the Geometry of Space 881
12.1
12.2
12.3
12.4
12.5
12.6
Three-Dimensional Coordinate Systems 881
Vectors 886
The Dot Product 892
The Cross Product 897
Lines and Planes in Space 904
Cylinders and Quadric Surfaces 913
Practice Exercises 918
Additional and Advanced Exercises 926
13 Vector-Valued Functions and Motion in Space 933
13.1
13.2
13.3
13.4
13.5
13.6
Curves in Space and Their Tangents 933
Integrals of Vector Functions; Projectile Motion 940
Arc Length in Space 949
Curvature and Normal Vectors of a Curve 953
Tangential and Normal Components of Acceleration 961
Velocity and Acceleration in Polar Coordinates 967
Practice Exercises 970
Additional and Advanced Exercises 977
Copyright 2018 Pearson Education, Inc.
vi
14 Partial Derivatives 981
14.1 Functions of Several Variables 981
14.2 Limits and Continuity in Higher Dimensions 991
14.3 Partial Derivatives 999
14.4 The Chain Rule 1008
14.5 Directional Derivatives and Gradient Vectors 1018
14.6 Tangent Planes and Differentials 1024
14.7 Extreme Values and Saddle Points 1033
14.8 Lagrange Multipliers 1049
14.9 Taylor's Formula for Two Variables 1061
14.10 Partial Derivatives with Constrained Variables 1064
Practice Exercises 1067
Additional and Advanced Exercises 1085
15 Multiple Integrals 1091
15.1
15.2
15.3
15.4
15.5
15.6
15.7
15.8
Double and Iterated Integrals over Rectangles 1091
Double Integrals over General Regions 1094
Area by Double Integration 1108
Double Integrals in Polar Form 1113
Triple Integrals in Rectangular Coordinates 1119
Moments and Centers of Mass 1125
Triple Integrals in Cylindrical and Spherical Coordinates 1132
Substitutions in Multiple Integrals 1146
Practice Exercises 1153
Additional and Advanced Exercises 1160
16 Integrals and Vector Fields 1167
16.1
16.2
16.3
16.4
16.5
16.6
16.7
16.8
Line Integrals 1167
Vector Fields and Line Integrals: Work, Circulation, and Flux 1173
Path Independence, Conservative Fields, and Potential Functions 1185
Green's Theorem in the Plane 1191
Surfaces and Area 1199
Surface Integrals 1209
Stokes' Theorem 1220
The Divergence Theorem and a Unified Theory 1227
Practice Exercises 1234
Additional and Advanced Exercises 1244
Copyright 2018 Pearson Education, Inc.
vii
CHAPTER 1 FUNCTIONS
1.1
FUNCTIONS AND THEIR GRAPHS
1. domain (, ); range [1, )
2. domain [0, ); range (, 1]
3. domain [2, ); y in range and y 5 x 10 0 y can be any nonnegative real number range [0, ).
4. domain (, 0] [3, ); y in range and y x 2 3 x 0 y can be any nonnegative real number
range [0, ).
5. domain (, 3) (3, ); y in range and y
4 ,
3t
now if t 3 3 t 0
4
3t
0, or if t 3
3 t 0 3 4 t 0 y can be any nonzero real number range (, 0) (0, ).
6. domain (, 4) ( 4, 4) (4, ); y in range and y
2
4 t 4 16 t 2 16 0 16
2
t 2 16
2 ,
t 2 16
2
now if t 4 t 2 16 0
, or if t 4 t 16 0
2
t 2 16
2
t 2 16
0, or if
0 y can be any nonzero
real number range (, 18 ] (0, ).
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 x; (height)2
2x
2
x 2 height
3
2
x; area is a ( x)
1
2
(base)(height) 12 ( x)
perimeter is p ( x) x x x 3 x.
x
3 2
x ;
4
3
2
10. s side length s 2 s 2 d 2 s d ; and area is a s 2 a 12 d 2
2
11. Let D diagonal length of a face of the cube and the length of an edge. Then 2 D 2 d 2 and
D 2 2 2 3 2 d 2 d . The surface area is 6 2
3
6d 2
3
2
2d 2 and the volume is 3 d3
3/2
3
d .
12. The coordinates of P are x, x so the slope of the line joining P to the origin is m xx 1 ( x 0).
Thus, x, x
1
m2
,
1
m
.
x
25
13. 2 x 4 y 5 y 12 x 54 ; L ( x 0)2 ( y 0)2 x 2 ( 12 x 54 )2 x 2 14 x 2 54 x 16
5 x2
4
5
4
25
x 16
20 x 2 20 x 25
16
20 x 2 20 x 25
4
14. y x 3 y 2 3 x; L ( x 4) 2 ( y 0) 2 ( y 2 3 4)2 y 2 ( y 2 1)2 y 2
y4 2 y2 1 y2
y4 y2 1
Copyright 2018 Pearson Education, Inc.
1
3 3
2
Chapter 1 Functions
15. The domain is (, ).
16. The domain is (, ).
17. The domain is (, ).
18. The domain is (, 0].
19. The domain is (, 0) (0, ).
20. The domain is (, 0) (0, ).
21. The domain is (, 5) (5, 3] [3, 5) (5, ) 22. The range is [2, 3).
23. Neither graph passes the vertical line test
(a)
(b)
Copyright 2018 Pearson Education, Inc.
Section 1.1 Functions and Their Graphs
24. Neither graph passes the vertical line test
(a)
(b)
x y 1
y 1 x
x y 1
or
or
x y 1
y 1 x
25.
x 0 1 2
y 0 1 0
26.
4 x 2 , x 1
27. F ( x)
2
x 2 x, x 1
x 0 1 2
y 1 0 0
1 , x 0
28. G ( x) x
x, 0 x
29. (a) Line through (0, 0) and (1, 1): y x; Line through (1, 1) and (2, 0): y x 2
x, 0 x 1
f ( x)
x 2, 1 x 2
2,
0,
(b) f ( x)
2,
0,
0 x 1
1 x 2
2 x3
3 x 4
30. (a) Line through (0, 2) and (2, 0): y x 2
0 1
Line through (2, 1) and (5, 0): m 5 2
x 2, 0 x 2
f ( x) 1
5
3 x 3 , 2 x 5
1
3
13 , so y 13 ( x 2) 1 13 x 53
Copyright 2018 Pearson Education, Inc.
3
4
Chapter 1 Functions
(b) Line through (1, 0) and (0, 3): m
Line through (0, 3) and (2, 1) : m
3x 3, 1 x 0
f ( x)
2 x 3, 0 x 2
3 0
3, so y 3x 3
0 ( 1)
1 3 4
2 2, so y 2 x 3
20
31. (a) Line through (1, 1) and (0, 0): y x
Line through (0, 1) and (1, 1): y 1
0 1
Line through (1, 1) and (3, 0): m 3 1 21 12 , so y 12 ( x 1) 1 12 x 32
x
1 x 0
f ( x) 1
0 x 1
1
3
1 x 3
2 x 2
(b) Line through (2, 1) and (0, 0): y 12 x
Line through (0, 2) and (1, 0): y 2 x 2
Line through (1, 1) and (3, 1): y 1
1x
2 x 0
2
f ( x) 2 x 2 0 x 1
1
1 x 3
10
32. (a) Line through T2 , 0 and (T, 1): m T (T /2) T2 , so y T2 x T2 0 T2 x 1
0, 0 x T2
f ( x)
T
2
T x 1, 2 x T
A, 0 x T
2
A, T x T
2
(b) f ( x)
3T
A, T x 2
A, 32T x 2T
33. (a) x 0 for x [0, 1)
(b) x 0 for x (1, 0]
34. x x only when x is an integer.
35. For any real number x, n x n 1, where n is an integer. Now: n x n 1 (n 1) x n.
By definition: x n and x n x n. So x x for all real x.
36. To find f(x) you delete the decimal or
fractional portion of x, leaving only
the integer part.
Copyright 2018 Pearson Education, Inc.
Section 1.1 Functions and Their Graphs
37. Symmetric about the origin
Dec: x
Inc: nowhere
38. Symmetric about the y-axis
Dec: x 0
Inc: 0 x
39. Symmetric about the origin
Dec: nowhere
Inc: x 0
0 x
40. Symmetric about the y-axis
Dec: 0 x
Inc: x 0
41. Symmetric about the y-axis
Dec: x 0
Inc: 0 x
42. No symmetry
Dec: x 0
Inc: nowhere
Copyright 2018 Pearson Education, Inc.
5
6
Chapter 1 Functions
43. Symmetric about the origin
Dec: nowhere
Inc: x
44. No symmetry
Dec: 0 x
Inc: nowhere
45. No symmetry
Dec: 0 x
Inc: nowhere
46. Symmetric about the y-axis
Dec: x 0
Inc: 0 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. f ( x) x 5
1
x5
and f ( x) ( x) 5
1
( x )5
15 f ( x). Thus the function is odd.
x
49. Since f ( x) x 2 1 ( x) 2 1 f ( x). The function is even.
50. Since [ f ( x) x 2 x] [ f ( x) ( x) 2 x] and [ f ( x) x 2 x ] [ f ( x) ( x) 2 x] the function is neither
even nor odd.
51. Since g ( x) x3 x, g ( x) x3 x ( x3 x) g ( x). So the function is odd.
52. g ( x) x 4 3 x 2 1 ( x) 4 3( x) 2 1 g ( x), thus the function is even.
53. g ( x)
1
x2 1
54. g ( x)
x ;
x2 1
1
( x )2 1
g ( x). Thus the function is even.
g ( x) 2x
x 1
g ( x). So the function is odd.
55. h(t ) t 1 1 ; h(t ) t 1 1 ; h(t ) 1 1 t . Since h(t ) h(t ) and h(t ) h(t ), the function is neither even nor odd.
Copyright 2018 Pearson Education, Inc.
Section 1.1 Functions and Their Graphs
56. Since |t 3 | |(t )3 |, h(t ) h( t ) and the function is even.
57. h(t ) 2t 1, h(t ) 2t 1. So h(t ) h(t ). h(t ) 2t 1, so h(t ) h(t ). The function is neither even
nor odd.
58. h(t ) 2| t | 1 and h(t ) 2| t | 1 2| t | 1. So h(t ) h(t ) and the function is even.
59. g ( x) sin 2 x; g ( x) sin 2 x g ( x). So the function is odd.
60. g ( x) sin x 2 ; g ( x) sin x 2 g ( x ). So the function is even.
61. g ( x) cos3 x; g ( x) cos3x g ( x). So the function is even.
62. g ( x) 1 cos x; g ( x) 1 cos x g ( x). So the function is even.
63. s kt 25 k (75) k 13 s 13 t ; 60 13 t t 180
64. K c v 2 12960 c(18)2 c 40 K 40v 2 ; K 40(10) 2 4000 joules
65. r ks 6 k4 k 24 r 24
; 10 24
s 12
5
s
s
k k 14700 P 14700 ; 23.4 14700 V
66. P Vk 14.7 1000
V
V
24500
39
628.2 in 3
67. V f ( x ) x (14 2 x )(22 2 x ) 4 x 3 72 x 2 308 x; 0 x 7.
AB
68. (a) Let h height of the triangle. Since the triangle is isosceles, AB
2
2
22 AB 2. So,
2
h 2 12 2 h 1 B is at (0, 1) slope of AB 1 The equation of AB is
y f ( x) x 1; x [0, 1].
(b) A( x) 2 xy 2 x( x 1) 2 x 2 2 x; x [0, 1].
69. (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.
70. (a) Graph f because it is linear.
(b) Graph g because it contains (0, 1).
(c) Graph h because it is a nonlinear odd function.
Copyright 2018 Pearson Education, Inc.
7
8
Chapter 1 Functions
71. (a) From the graph, 2x 1 4x x (2, 0) (4, )
(b) 2x 1 4x 2x 1 4x 0
x2 2 x 8
( x 4)( x 2)
x 0: 2x 1 4x 0
0
0
2x
2x
x 4 since x is positive;
x2 2 x 8
( x 4)( x 2)
x 0: 2x 1 4x 0
0
0
2x
2x
x 2 since x is negative;
sign of ( x 4)( x 2)
Solution interval: (2, 0) (4, )
72. (a) From the graph,
3
x 1
x 2 1 x (, 5) (1, 1)
3( x 1)
(b) Case x 1: x 3 1 x 2 1 x 1 2
3x 3 2 x 2 x 5.
Thus, x (, 5) solves the inequality.
3( x 1)
Case 1 x 1: x 3 1 x 2 1 x 1 2
3 x 3 2 x 2 x 5 which
is true if x 1. Thus, x (1, 1)
solves the inequality.
Case 1 x : x 3 1 x 2 1 3x 3 2 x 2 x 5
which is never true if 1 x,
so no solution here.
In conclusion, x (, 5) (1, 1).
73. A curve symmetric about the x-axis will not pass the vertical line test because the points (x, y) and ( x, y ) lie
on the same vertical line. The graph of the function y f ( x) 0 is the x-axis, a horizontal line for which there
is a single y-value, 0, for any x.
74. price 40 5 x, quantity 300 25x R( x) (40 5 x)(300 25 x)
75. x 2 x 2 h 2 x h
2
2h
;
2
cost 5(2 x) 10h C (h) 10
10h 5h
2h
2
22
76. (a) Note that 2 mi 10, 560 ft, so there are 8002 x 2 feet of river cable at $180 per foot and (10,560 x)
feet of land cable at $100 per foot. The cost is C ( x) 180 8002 x 2 100(10,560 - x).
(b) C (0) $1, 200, 000
C (500) $1,175,812
C (1000) $1,186,512
C (1500) $1, 212, 000
C (2000) $1, 243, 732
C (2500) $1, 278, 479
C (3000) $1,314,870
Values beyond this are all larger. It would appear that the least expensive location is less than 2000 feet
from the point P.
Copyright 2018 Pearson Education, Inc.
Section 1.2 Combining Functions; Shifting and Scaling Graphs
1.2
9
COMBINING FUNCTIONS; SHIFTING AND SCALING GRAPHS
1. D f : x , Dg : x 1 D f g D fg : x 1. R f : y , Rg : y 0, R f g : y 1, R fg : y 0
2. D f : x 1 0 x 1, Dg : x 1 0 x 1. Therefore D f g D fg : x 1.
R f Rg : y 0, R f g : y 2, R fg : y 0
3. D f : x , Dg : x , D f /g : x , Dg /f : x , R f : y 2, Rg : y 1, R f /g : 0 y 2,
Rg /f : 12 y
4. D f : x , Dg : x 0, D f /g : x 0, Dg /f : x 0; R f : y 1, Rg : y 1, R f /g : 0 y 1, Rg /f : 1 y
5. (a) 2
(d) ( x 5)2 3 x 2 10 x 22
(g) x 10
(b) 22
(e) 5
(h) ( x 2 3)2 3 x 4 6 x 2 6
(c) x 2 2
(f ) 2
6. (a) 13
(b) 2
(c)
(d)
(e) 0
1
x
(g) x 2
(h)
(f )
1
1
x 1
1
x 1 2
x 1
x 1
x2
1 1 x
x 1
x 1
3
4
7. ( f g h)( x) f ( g (h( x))) f ( g (4 x)) f (3(4 x)) f (12 3 x) (12 3 x) 1 13 3x
8. ( f g h)( x) f ( g (h( x))) f ( g ( x 2 )) f (2( x 2 ) 1) f (2 x 2 1) 3(2 x 2 1) 4 6 x 2 1
1x f
9. ( f g h)( x) f ( g (h( x))) f g
1
x
1
4
5x 1
f 1 x4 x 1 x4 x 1 1 4 x
2 x 2
f
f
2 x 2 1
2x
3 x
2x
2
3 x
2x
3 3x
8 3x
7 2x
10. ( f g h)( x) f ( g (h( x))) f g
2 x
11. (a) ( f g )( x)
(d) ( j j )( x)
(b) ( j g )( x)
(e) ( g h f )( x)
(c) ( g g )( x)
(f ) (h j f )( x)
12. (a) ( f j )( x)
(d) ( f f )( x)
(b) ( g h)( x)
(e) ( j g f )( x)
(c) (hh)( x)
(f ) ( g f h)( x)
f (x)
( f g )( x )
(a) x 7
x
x7
(b) x 2
3x
13.
g(x)
(c) x 2
3( x 2) 3 x 6
x5
(d)
x
x 1
x
x 1
(e)
1
x 1
1 1x
x2 5
x
x 1
x
1
x 1
x
x ( x 1)
x
x
Copyright 2018 Pearson Education, Inc.
10
Chapter 1 Functions
(f )
1
x
1
x
x
1 .
x 1
g ( x) 1
x x 1
g ( x)
14. (a) ( f g )( x) |g ( x)|
(b) ( f g )( x)
1 g (1x )
x 1 x
x 1
x 1
2
g (1x ) x 1 1 g (1x ) , so g ( x ) x 1.
(c) Since ( f g )( x) g ( x) | x |, g ( x) x .
(d) Since ( f g )( x) f x | x |, f ( x) x 2 . (Note that the domain of the composite is [0, ).)
The completed table is shown. Note that the absolute value sign in part (d) is optional.
g (x)
f (x)
( f g )(x)
1
x 1
| x|
1
x 1
x 1
x 1
x
x
x 1
x2
x
| x|
x
2
| x|
x
15. (a) f ( g (1)) f (1) 1
(d) g ( g (2)) g (0) 0
16. (a)
(b)
(c)
(d)
(e)
(f )
(b) g ( f (0)) g (2) 2
(e) g ( f (2)) g (1) 1
(c) f ( f (1)) f (0) 2
(f) f ( g (1)) f (1) 0
f ( g (0)) f (1) 2 (1) 3, where g (0) 0 1 1
g ( f (3)) g (1) (1) 1, where f (3) 2 3 1
g ( g (1)) g (1) 1 1 0, where g (1) (1) 1
f ( f (2)) f (0) 2 0 2, where f (2) 2 2 0
g ( f (0)) g (2) 2 1 1, where f (0) 2 0 2
12 f 12 2 12 52 , where g 12 12 1 12
f g
17. (a) ( f g )( x) f ( g ( x))
( g f )( x) g ( f ( x))
1 1
x
1
x 1
1 x
x
(b) Domain ( f g ): (, 1] (0, ), domain ( g f ): (1, )
(c) Range ( f g ): (1, ), range ( g f ): (0, )
18. (a) ( f g )( x) f ( g ( x)) 1 2 x x
( g f )( x) g ( f ( x)) 1 | x |
(b) Domain ( f g ): [0, ), domain ( g f ): (, )
(c) Range ( f g ): (0, ), range ( g f ): (, 1]
19. ( f g )( x) x f ( g ( x)) x
g ( x) x g ( x) 2 x g ( x)
g ( x)
x g ( x)
g ( x) 2
1 2xx x2x 1
( g ( x) 2) x x g ( x) 2 x
20. ( f g )( x ) x 2 f ( g ( x)) x 2 2( g ( x))3 4 x 2 ( g ( x ))3
x6
2
21. V V ( s ) V ( s (t )) V (2t 3)
(2t 3) 2 2(2t 3) 3
4t 2 8t 6
Copyright 2018 Pearson Education, Inc.
g ( x) 3
x6
2
Section 1.2 Combining Functions; Shifting and Scaling Graphs
11
22. (a)
x
4
3
2
1
0
1
2
3
4
g ( x)
2
1
0.5
0.2
0
0.2
0.5
1
2
f ( g ( x))
1
1.3
1.6
1.8
2
1.8
1.5
1
0
x
4
3
2
1
0
1
2
3
4
g ( x)
1.5
0.3
0.7
1.5
2.4
2.8
3
2.7
2
f ( g ( x))
0.8
1.9
1.7
1.5
0.7
0.3
0.2
0.5
0.9
(b)
23. (a) y ( x 7) 2
(b) y ( x 4)2
24. (a) y x 2 3
(b) y x 2 5
25. (a) Position 4
(b) Position 1
(c) Position 2
(d) Position 3
26. (a) y ( x 1)2 4
(b) y ( x 2) 2 3
(c) y ( x 4) 2 1
(d) y ( x 2)2
27.
28.
Copyright 2018 Pearson Education, Inc.
12
Chapter 1 Functions
29.
30.
31.
32.
33.
34.
35.
36.
37.
38.
Copyright 2018 Pearson Education, Inc.
Section 1.2 Combining Functions; Shifting and Scaling Graphs
39.
40.
41.
42.
43.
44.
45.
46.
47.
48.
49.
50.
Copyright 2018 Pearson Education, Inc.
13
14
Chapter 1 Functions
51.
52.
53.
54.
55.
56.
57. (a) domain: [0, 2]; range: [2, 3]
(b) domain: [0, 2]; range: [–1, 0]
(c) domain: [0, 2]; range: [0, 2]
(d) domain: [0, 2]; range: [–1, 0]
Copyright 2018 Pearson Education, Inc.
Section 1.2 Combining Functions; Shifting and Scaling Graphs
(e) domain: [–2, 0]; range: [0, 1]
(f ) domain: [1, 3]; range: [0,1]
(g) domain: [–2, 0]; range: [0, 1]
(h) domain: [–1, 1]; range: [0, 1]
58. (a) domain: [0, 4]; range: [–3, 0]
(b) domain: [–4, 0]; range: [0, 3]
(c) domain: [–4, 0]; range: [0, 3]
(d) domain: [–4, 0]; range: [1, 4]
(e) domain: [2, 4]; range: [–3, 0]
(f ) domain: [–2, 2]; range: [–3, 0]
Copyright 2018 Pearson Education, Inc.
15
16
Chapter 1 Functions
(h) domain: [0, 4]; range: [0, 3]
(g) domain: [1, 5]; range: [–3, 0]
59. y 3x 2 3
61. y
1
2
60. y (2 x) 2 1 4 x 2 1
1
1
x2
1
2
62. y 1
1
2 x2
63. y 4 x 1
65. y 4
1
( x /3) 2
1
64. y 3 x 1
2x
2
1
2
66. y 13 4 x 2
16 x 2
67. y 1 (3x )3 1 27 x3
68. y 1
2x
3
, i( x) 2 x 12 , and
1/2
2 x 12 f ( x). The graph of h( x)
h( x) x 12
1/2
1/2
j ( x)
is the graph of g ( x) shifted left 12 unit; the graph
of i ( x) is the graph of h( x) stretched vertically by
a factor of 2; and the graph of j ( x) f ( x) is the
graph of i ( x) reflected across the x-axis.
70. Let y 1 2x f ( x). Let g ( x) ( x)1/2 ,
h( x) ( x 2)1/2 , and i ( x ) 1 ( x 2)1/2
2
1
x
2
3
1 x8
69. Let y 2 x 1 f ( x) and let g ( x) x1/2 ,
9
x2
f ( x ). The graph of g ( x) is the graph
of y x reflected across the x-axis. The graph
of h( x) is the graph of g ( x) shifted right two units.
And the graph of i ( x) is the graph of h( x)
compressed vertically by a factor of 2 .
Copyright 2018 Pearson Education, Inc.
Section 1.2 Combining Functions; Shifting and Scaling Graphs
71. y f ( x) x3 . Shift f ( x) one unit right followed by
a shift two units up to get g ( x) ( x 1)3 2 .
72.
y (1 x)3 2 [( x 1)3 (2)] f ( x).
Let g ( x) x3 , h( x) ( x 1)3 ,
i( x) ( x 1)3 (2),
and j ( x) [( x 1)3 (2)]. The graph of h( x) is the
graph of g ( x) shifted right one unit; the graph of i ( x)
is the graph of h( x) shifted down two units; and the
graph of f ( x) is the graph of i ( x) reflected across
the x-axis.
73. Compress the graph of f ( x) 1x horizontally by a
factor of 2 to get g ( x) 21x . Then shift g ( x)
vertically down 1 unit to get h( x) 21x 1.
74. Let f ( x)
1
x/ 2
2
1
x2
and g ( x)
1
1
2
1/ 2 x
2
x2
1 12 1
x
2
1. Since 2 1.4, we see
that the graph of f ( x ) stretched horizontally by
a factor of 1.4 and shifted up 1 unit is the graph
of g ( x).
75. Reflect the graph of y f ( x) 3 x across the x-axis
to get g ( x ) 3 x .
Copyright 2018 Pearson Education, Inc.
17
18
76.
Chapter 1 Functions
y f ( x) (2 x) 2/3 [(1)(2) x]2/3 (1) 2/3 (2 x)2/3
(2 x)2/3 . So the graph of f ( x) is the graph of
g ( x) x 2/3 compressed horizontally by a factor of 2.
77.
78.
79. (a) ( fg )( x) f ( x) g ( x) f ( x)( g ( x)) ( fg )( x), odd
(b)
(c)
( x)
( x)
( x), odd
f
g
f ( x)
g ( x)
g ( x ) g ( x), odd
g
f
g ( x)
f ( x)
f ( x)
f ( x)
g ( x)
f
g
f
(d) f 2 ( x) f ( x) f ( x) f ( x) f ( x) f 2 ( x), even
(e) g 2 ( x) ( g ( x)) 2 ( g ( x))2 g 2 ( x), even
(f ) ( f g )( x) f ( g ( x)) f ( g ( x)) f ( g ( x)) ( f g )( x), even
(g) ( g f )( x) g ( f ( x)) g ( f ( x)) ( g f )( x), even
(h) ( f f )( x) f ( f ( x)) f ( f ( x)) ( f f )( x), even
(i) ( g g )( x) g ( g ( x )) g ( g ( x)) g ( g ( x )) ( g g )( x ), odd
80. Yes, f ( x) 0 is both even and odd since f ( x) 0 f ( x) and f ( x) 0 f ( x).
81. (a)
(b)
Copyright 2018 Pearson Education, Inc.
Section 1.3 Trigonometric Functions
(c)
19
(d)
82.
1.3
TRIGONOMETRIC FUNCTIONS
s r (10) 45 8 m
1. (a)
2.
s
r
(b) s r (10)(110)
180 11018 559 m
225
108 54 radians and 54 180
49 s (6) 49 8.4 in. (since the diameter 12 in. radius 6 in.)
3. 80 80 180
4. d 1 meter r 50 cm
s
r
23
0
sin
0
23
0
1
cos
1
0
12
1
0
1
3
0
und.
1
5.
tan
cot
und.
sec
1
csc
und.
1
3
2
2
3
34
30
0.6 rad or 0.6 180
50
2
3
4
1
2
2
6.
32
3
6
4
5
6
sin
1
23
12
1
2
1
2
cos
0
1
2
3
2
1
2
23
tan
und.
3
und.
0
1
1
und.
2
cot
0
und.
1
2
sec
und.
2
csc
1
2
Copyright 2018 Pearson Education, Inc.
1
3
3
1
3
1
1
3
1
3
2
3
2
2
2
3
2
3
2
20
Chapter 1 Functions
7. cos x 54 , tan x 34
9. sin x
11. sin x
8
,
3
tan x 8
1 , cos x
5
13.
8. sin x
2
5
2 ,
5
cos x 1
5
, tan x 12
10. sin x 12
13
5
12. cos x
3
,
2
tan x 1
14.
period 4
period
15.
16.
period 4
period 2
18.
17.
period 1
period 6
19.
20.
period 2
period 2
Copyright 2018 Pearson Education, Inc.
3
Section 1.3 Trigonometric Functions
21.
22.
period 2
period 2
23. period 2 , symmetric about the origin
24. period 1, symmetric about the origin
s
3
2
s = tan t
1
2
1
1
0
2
t
1
2
3
26. period 4 , symmetric about the origin
25. period 4, symmetric about the s-axis
27. (a) Cos x and sec x are positive for x in the interval
2 , 2 ; and cos x and sec x are negative for x
in the intervals 32 , 2 and 2 , 32 . Sec x is
undefined when cos x is 0. The range of sec x is
(, 1] [1,); the range of cos x is [1, 1].
Copyright 2018 Pearson Education, Inc.
21
22
Chapter 1 Functions
(b) Sin x and csc x are positive for x in the intervals
32 , and (0, ); and sin x and csc x are
negative for x in the intervals ( , 0) and
, 32 . Csc x is undefined when sin x is 0. The
range of csc x is (, 1] [1, ); the range of
sin x is [1, 1].
28. Since cot x tan1 x , cot x is undefined when tan x 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 1, 0, 1
30. D : x ; R : y 1, 0, 1
31. cos x 2 cos x cos 2 sin x sin 2 (cos x)(0) (sin x)(1) sin x
32. cos x 2 cos x cos 2 sin x sin 2 (cos x)(0) (sin x)(1) sin x
33. sin x 2 sin x cos 2 cos x sin 2 (sin x)(0) (cos x)(1) cos x
34. sin x 2 sin x cos 2 cos x sin 2 (sin x)(0) (cos x)( 1) cos x
35. cos( A B ) cos( A ( B )) cos A cos( B ) sin A sin( B ) cos A cos B sin A( sin B )
cos A cos B sin A sin B
36. sin( A B) sin( A ( B )) sin A cos( B ) cos A sin( B ) sin A cos B cos A( sin B )
sin A cos B cos A sin B
37. If B A, A B 0 cos( A B) cos 0 1. Also cos( A B ) cos( A A) cos A cos A sin A sin A
cos 2 A sin 2 A. Therefore, cos 2 A sin 2 A 1.
38. If B 2 , then cos( A 2 ) cos A cos 2 sin A sin 2 (cos A)(1) (sin A)(0) cos A and
sin( A 2 ) sin A cos 2 cos A sin 2 (sin A)(1) (cos A)(0) sin A . The result agrees with the fact that the
cosine and sine functions have period 2 .
39. cos( x ) cos cos x sin sin x ( 1)(cos x ) (0)(sin x ) cos x
Copyright 2018 Pearson Education, Inc.
Section 1.3 Trigonometric Functions
40. sin(2 x) sin 2 cos( x) cos(2 ) sin( x) (0)(cos( x)) (1)(sin( x)) sin x
41. sin 32 x sin 32 cos( x) cos 32 sin( x) ( 1)(cos x ) (0)(sin( x)) cos x
42. cos 32 x cos 32 cos x sin 32 sin x (0)(cos x ) (1)(sin x) sin x
43. sin 712 sin 4 3 sin 4 cos 3 cos 4 sin 3
2
2
cos 2 cos cos 2 sin sin 2
44. cos 11
12
4
3
4
3
4
3
6
4
3
2
2
2
1
2
2
2
2
3
2
2
2
1
2
2
4
6
12 22 23 22 12 23
cos cos cos sin sin
45. cos 12
3 4
3
4
3
4
46. sin 512 sin
47.
cos 2 8
23 4 sin 23 cos 4 cos 23 sin 4 23 22 12 22 12 23
1 cos
2
28 1
2
2
212 1
3
2
49. sin 2 12
1 cos
51. sin 2
3
4
2
2
2
1 cos
1012 1 23 2
2 2
4
48.
2 3
4
1 cos
50. sin 2 38
2
cos 2 512
2
2
6
8
1 22 2
2
3
4
2
4
sin 23 3 , 23 , 43 , 53
52. sin 2 cos 2
sin 2
cos 2
2
cos 2 tan 2 1 tan 1 4 , 34 , 54 , 74
cos
53. sin 2 cos 0 2sin cos cos 0 cos (2sin 1) 0 cos 0 or 2sin 1 0
cos 0 or sin 12 2 , 32 , or 6 , 56 6 , 2 , 56 , 32
54. cos 2 cos 0 2 cos 2 1 cos 0 2 cos 2 cos 1 0 (cos 1)(2 cos 1) 0
cos 1 0 or 2 cos 1 0 cos 1 or cos 12 or 3 , 53 3 , , 53
55. tan( A B )
sin( A B )
cos( A B )
cos A cos B sin A sin B
sin A cos B cos A cos B
56. tan( A B )
sin( A B )
cos( A B )
cos A cos B sin A sin B
sin A cos B cos A cos B
sin A cos B
cos A cos B
cos A cos B
cos A cos B
sin A cos B
cos A cos B
cos A cos B
cos A cos B
cos A sin B
cos A cos B
sin A sin B
cos A cos B
1 tan A tan B
cos A sin B
cos A cos B
sin A sin B
cos A cos B
1 tan A tan B
tan A tan B
tan A tan B
57. According to the figure in the text, we have the following: By the law of cosines, c 2 a 2 b 2 2ab cos
12 12 2 cos( A B ) 2 2cos( A B) . By distance formula, c 2 (cos A cos B )2 (sin A sin B )2
cos 2 A 2 cos A cos B cos 2 B sin 2 A 2sin A sin B sin 2 B 2 2(cos A cos B sin A sin B ) . Thus
c 2 2 2 cos( A B ) 2 2(cos A cos B sin A sin B ) cos( A B ) cos A cos B sin A sin B .
Copyright 2018 Pearson Education, Inc.
23
24
Chapter 1 Functions
58. (a) cos( A B ) cos A cos B sin A sin B
sin cos 2 and cos sin 2
Let A B
sin( A B ) cos 2 ( A B) cos 2 A B cos 2 A cos B sin 2 A sin B
sin A cos B cos A sin B
(b) cos( A B) cos A cos B sin A sin B
cos( A ( B )) cos A cos( B ) sin A sin( B )
cos( A B) cos A cos( B ) sin A sin( B) cos A cos B sin A( sin B) cos A cos B sin A sin B
Because the cosine function is even and the sine functions is odd.
59. c 2 a 2 b 2 2ab cos C 22 32 2(2)(3) cos(60) 4 9 12 cos(60) 13 12
Thus, c 7 2.65.
12 7.
60. c 2 a 2 b 2 2ab cos C 22 32 2(2)(3) cos(40) 13 12 cos(40). Thus, c 13 12 cos 40° 1.951.
61. From the figures in the text, we see that sin B hc . If C is an acute angle, then sin C bh . On the other hand,
if C is obtuse (as in the figure on the right in the text), then sin C sin( C ) bh . Thus, in either case,
h b sin C c sin B ah ab sin C ac sin B.
a 2 b2 c2
a 2 c 2 b2
and cos B
. Moreover, since the sum of the interior
By the law of cosines, cos C
2 ab
2 ac
angles of triangle is , we have sin A sin( ( B C )) sin( B C ) sin B cos C cos B sin C
hc
h (2a 2 b 2 c 2 c 2 b 2 ) ah ah bc sin A.
2abc
bc
a 2 b2 c 2 a 2 c 2 b2 h
b
2 ab
2 ac
Combining our results we have ah ab sin C, ah ac sin B, and ah bc sin A. Dividing by abc gives
h sin A sin C sin B .
bc
a
c
b
law of sines
62. By the law of sines, sin2 A
sin B
3
3/2
. By
c
Exercise 59 we know that c 7. Thus sin B
3 3
2 7
0.982.
63. From the figure at the right and the law of cosines,
b 2 a 2 22 2(2a) cos B
a 2 4 4a
12 a2 2a 4.
2/2
a
3/2
b
b
sin A
a
sin B
b
3 a. Thus, combining results,
2
Applying the law of sines to the figure,
a 2 2a 4 b 2 32 a 2 0 12 a 2 2a 4 0 a 2 4a 8 . From the quadratic formula and the fact that
a 0, we have a
64. tan hc c
4
42 4(1)( 8)
2
4 34
2
1.464.
h
tan
h tan
tan b h c h h b tan h
b tan
b tan tan h tan h tan
b tan tan h tan h tan
Copyright 2018 Pearson Education, Inc.
Section 1.3 Trigonometric Functions
b tan tan h(tan tan )
h
b tan tan
tan tan
65. sin r r
sin r sin r
sin r r sin r (1 sin )
r 1sin
sin
66. (a) The graphs of y sin x and y 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.
67. A 2, B 2 , C , D 1
68. A 12 , B 2, C 1, D
1
2
69. A 2 , B 4, C 0, D 1
70.
A 2L , B L, C 0, D 0
Copyright 2018 Pearson Education, Inc.
25
26
Chapter 1 Functions
71–74.
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 ["B1", "B3","B2*Pi","B3*Pi"],
title "#71 (Section 1.3)");
Mathematica:
Clear[a, b, c, d, f, x]
f[x_]: a Sin[2/b (x c)] d
Plot[f[x]/.{a 3, b 1, c 0, d 0}, {x, 4, 4 }]
71. (a) The graph stretches horizontally.
(b) The period remains the same: period | B |. The graph has a horizontal shift of 12 period.
72. (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. | C | 6
73. (a) The graph shifts upwards | D | units for D 0
(b) The graph shifts down | D | units for D 0.
Copyright 2018 Pearson Education, Inc.
Section 1.4 Graphing with Software
74. (a) The graph stretches | A| units.
1.4
27
(b) For A 0, the graph is inverted.
GRAPHING WITH SOFTWARE
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 5–30
are not unique in appearance.
5. [ 2, 5] by [ 15, 40]
6. [ 4, 4] by [ 4, 4]
Copyright 2018 Pearson Education, Inc.
28
Chapter 1 Functions
7. [ 2, 6] by [ 250, 50]
8. [ 1, 5] by [ 5, 30]
9. [ 4, 4] by [ 5, 5]
10. [ 2, 2] by [ 2, 8]
11. [ 2, 6] by [ 5, 4]
12. [ 4, 4] by [ 8, 8]
13. [ 1, 6] by [ 1, 4]
14. [ 1, 6] by [ 1, 5]
Copyright 2018 Pearson Education, Inc.
Section 1.4 Graphing with Software
15. [ 3, 3] by [0, 10]
16. [ 1, 2] by [0, 1]
17. [ 5, 1] by [ 5, 5]
18. [ 5, 1] by [ 2, 4]
19. [ 4, 4] by [0, 3]
20. [ 5, 5] by [ 2, 2]
21. [ 10, 10] by [ 6, 6]
22. [ 5, 5] by [ 2, 2]
Copyright 2018 Pearson Education, Inc.
29
30
Chapter 1 Functions
23. [ 6, 10] by [ 6, 6]
24. [ 3, 5] by [ 2, 10]
25. [0.03, 0.03] by [1.25, 1.25]
26. [0.1, 0.1] by [3, 3]
27. [300, 300] by [ 1.25, 1.25]
28. [50, 50] by [0.1, 0.1]
29. [0.25, 0.25] by[0.3, 0.3]
30. [0.15, 0.15] by [ 0.02, 0.05]
31.
x 2 2 x 4 4 y y 2 y 2 x 2 2 x 8.
The lower half is produced by graphing
y 2 x 2 2 x 8.
Copyright 2018 Pearson Education, Inc.
Section 1.4 Graphing with Software
32. y 2 16 x 2 1 y 1 16 x 2 . The upper branch
is produced by graphing y 1 16 x 2 .
33.
34.
35.
36.
37.
38.
200
8
150
6
100
4
50
60
2
64
68
72
76
80
0
1970 1980 1990 2000 2010 2020
Copyright 2018 Pearson Education, Inc.
31
32
Chapter 1 Functions
40.
39.
(in thousands)
300
26
225
22
R
18
T
150
14
75
10
6
1972 1980 1988 1996 2004 2012
41.
2000 2002 2004 2006 2008
42.
1
600
450
0.5
300
1955
1935
1975
1995
150
2015
0
4
6
1. The area is A r 2 and the circumference is C 2 r. Thus, r 2C A
2C
0.5
0
2
8
10
CHAPTER 1 PRACTICE EXERCISES
2
2
C4 .
4S . The volume is V 43 r 3 r 3 34V . Substitution into the formula
2/3
for surface area gives S 4 r 2 4 34V .
2. The surface area is S 4 r 2 r
1/2
3. The coordinates of a point on the parabola are (x, x2). The angle of inclination joining this point to the origin
2
satisfies the equation tan xx x. Thus the point has coordinates ( x, x 2 ) (tan , tan 2 ).
4. tan
rise
run
h h 500 tan ft .
500
6.
5.
Symmetric about the origin.
Symmetric about the y-axis.
Copyright 2018 Pearson Education, Inc.
Chapter 1 Practice Exercises
8.
7.
Neither
Symmetric about the y-axis.
9. y ( x) ( x )2 1 x 2 1 y ( x). Even.
10. y ( x) ( x)5 ( x)3 ( x) x5 x3 x y ( x). Odd.
11. y ( x) 1 cos( x) 1 cos x y ( x). Even.
12. y ( x) sec( x) tan( x)
13. y ( x)
x 4 1
x 3 2 x
sin x
cos 2 x
x4 1
3
x 2x
sin2 x sec x tan x y ( x). Odd.
cos x
x4 1
x3 2 x
y ( x). Odd.
14. y ( x) ( x) sin( x) ( x) sin x ( x sin x) y ( x). Odd.
15. y ( x) x cos( x) x cos x. Neither even nor odd.
16. y ( x) ( x) cos( x) x cos x y ( x). Odd.
17. Since f and g are odd f ( x) f ( x) and g ( x ) g ( x).
(a) ( f g )( x) f ( x) g ( x) [ f ( x)] [ g ( x)] f ( x) g ( x) ( f g )( x) f g is even.
(b) f 3 ( x ) f ( x) f ( x) f ( x) [ f ( x)] [ f x ] [ f ( x)] f ( x) f ( x) f ( x) f 3 ( x) f 3 is odd.
(c) f (sin( x)) f (sin( x)) f (sin( x )) f (sin( x)) is odd.
(d) g (sec( x)) g (sec( x)) g (sec( x)) is even.
(e) | g ( x)| | g ( x)| | g ( x ) | | g | is even.
18. Let f (a x) f (a x) and define g ( x) f ( x a). Then g ( x) f (( x) a) f (a x) f (a x)
f ( x a ) g ( x) g ( x) f ( x a) is even.
19. (a) The function is defined for all values of x, so the domain is (, ).
(b) Since | x | attains all nonnegative values, the range is [2, ).
20. (a) Since the square root requires 1 x 0, the domain is (,1].
(b) Since 1 x attains all nonnegative values, the range is [2, ).
21. (a) Since the square root requires 16 x 2 0, the domain is [4, 4].
(b) For values of x in the domain, 0 16 x 2 16, so 0 16 x 2 4. The range is [0, 4].
Copyright 2018 Pearson Education, Inc.
33
34
Chapter 1 Functions
22. (a) The function is defined for all values of x, so the domain is (, ).
(b) Since 32 x attains all positive values, the range is (1, ) .
23. (a) The function is defined for all values of x, so the domain is (, ).
(b) Since 2e x attains all positive values, the range is (3, ) .
24. (a) The function is equivalent to y tan 2 x, so we require 2 x k2 for odd integers k. The domain is given by
x k4 for odd integers k.
(b) Since the tangent function attains all values, the range is (, ).
25. (a) The function is defined for all values of x, so the domain is (, ).
(b) The sine function attains values from –1 to 1, so 2 2sin (3 x ) 2 and hence 3 2 sin (3x ) 1 1.
The range is [3, 1].
26. (a) The function is defined for all values of x, so the domain is (, ).
5
(b) The function is equivalent to y x 2 , which attains all nonnegative values. The range is [0, ) .
27. (a) The logarithm requires x 3 0, so the domain is (3, ).
(b) The logarithm attains all real values, so the range is (, ).
28. (a) The function is defined for all values of x, so the domain is (, ).
(b) The cube root attains all real values, so the range is (, ).
29.
y 5 ( x 3)( x 1) so the domain (, 1] [3, );
( x 3)( x 1) 0 and can be any positive number,
so the range (, 5].
3x2
x2 4
y 2
31.
y 4sin
32.
y 3cos x 4sin x so the domain (, );
and
so the domain (, ); 0
3x2
x2 4
30.
3 so the range [2, 5).
1x so the domain (, 0) (0, ); if 32 x 2 , then 1 sin 1x 1, so the range [4, 4].
32 42 5 so 3cos x 4sin x 5
53 cos x 54 sin x
5(cos cos x sin sin x) 5cos( x), and
1 cos( x) 1 so the range [ 5, 5].
33. (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.
34. (a) Increasing on [2, )
(c) Increasing on (, )
(b) Increasing on [1, )
(d) Increasing on 12 ,
35. (a) The function is defined for 4 x 4, so the domain is [4, 4].
(b) The function is equivalent to y | x |, 4 x 4, which attains values from 0 to 2 for x in the domain.
The range is [0, 2].
Copyright 2018 Pearson Education, Inc.
Chapter 1 Practice Exercises
36. (a) The function is defined for 2 x 2, so the domain is [2, 2].
(b) The range is [1, 1].
0 1
37. First piece: Line through (0, 1) and (1, 0). m 1 0 11 1 y x 1 1 x
Second piece: Line through (1, 1) and (2, 0). m
1 x, 0 x 1
f ( x)
2 x, 1 x 2
0 1
2 1
11 1 y ( x 1) 1 x 2 2 x
50
52 y 52 x
20
05
0). m 4 2 25 52
38. First piece: Line through (0, 0) and (2, 5). m
Second piece: Line through (2, 5) and (4,
y 52 ( x 2) 5 52 x 10 10 52x
5 x, 0 x 2
2
f ( x)
(Note: x 2 can be included on either piece.)
5x
10 2 , 2 x 4
39. (a) ( f g )(1) f ( g (1)) f 1 f (1) 11 1
1
2
(b) ( g f )(2) g ( f (2)) g 12 11 1 or 52
2 2.5
( f f )( x) f ( f ( x)) f 1x 1/1x x, x 0
2
(c)
(d) ( g g )( x) g ( g ( x)) g
40. (a) ( f g )(1) f ( g (1)) f
1
x2
1
1 2
x2
4
x2
1 2 x 2
3 1 1 f (0) 2 0 2
(b) ( g f )(2) f ( g (2)) g (2 2) g (0) 3 0 1 1
(c) ( f f )( x) f ( f ( x)) f (2 x) 2 (2 x) x
(d) ( g g )( x) g ( g ( x)) g
41. (a) ( f g )( x) f ( g ( x)) f
3 x 1 3 3 x 1 1
x2 2
( g f )( x) g ( f ( x)) g (2 x 2 )
(b) Domain of f g : [2, ).
Domain of g f : [2, 2].
1 x
( g f )( x) g ( f ( x)) g x 1
42. (a) ( f g )( x) f ( g ( x)) f
(b) Domain of f g : (, 1].
Domain of g f : [0, 1].
x2
2
x, x 2 .
2 x2 2
4 x2
(c) Range of f g : (, 2].
Range of g f : [0, 2].
1 x 4 1 x.
x
(c) Range of f g : [0, ).
Range of g f : [0, 1].
Copyright 2018 Pearson Education, Inc.
35
36
43.
Chapter 1 Functions
y ( f f )( x)
y f ( x)
44.
45.
The graph of f 2 ( x) f1 (| x |) is the same as the
graph of f1 ( x ) to the right of the y-axis. The graph
of f 2 ( x) to the left of the y-axis is the reflection of
y f1 ( x), x 0 across the y-axis.
46.
47.
It does not change the graph.
Whenever g1 ( x) is positive, the graph of y
g 2 ( x) g1 ( x) is the same as the graph of y g1 ( x).
When g1 ( x) is negative, the graph of y g 2 ( x) is
the reflection of the graph of y g1 ( x) across the xaxis.
Copyright 2018 Pearson Education, Inc.
Chapter 1 Practice Exercises
48.
37
49.
Whenever g1 ( x) is positive, the graph of y
g 2 ( x) | g1 ( x)| is the same as the graph of y
g1 ( x). When g1 ( x) is negative, the graph of y
g 2 ( x) is the reflection of the graph of y g1 ( x)
across the
x-axis.
Whenever g1 ( x) is positive, the graph of
y g 2 ( x) | g1 ( x)| is the same as graph of
y g1 ( x ). When g1 ( x) is negative, the graph of
y g 2 ( x) is the reflection of the graph of
y g1 ( x) across the x-axis.
51.
50.
The graph of f 2 ( x) f1 (| x |) is the same as the
graph of f1 ( x ) to the right of the y-axis. The graph
of f 2 ( x) to the left of the y-axis is the reflection of
y f1 ( x ), x 0 across the y-axis.
The graph of f 2 ( x) f1 (| x |) is the same as the
graph of f1 ( x) to the right of the y-axis. The graph
of f 2 ( x) to the left of the y-axis is the reflection of
y f1 ( x), x 0 across the y-axis.
52.
The graph of f 2 ( x) f1 (| x |) is the same as the
graph of f1 ( x) to the right of the y-axis. The graph
of f 2 ( x) to the left of the y-axis is the reflection of
y f1 ( x), x 0 across the y-axis.
53. (a) y g ( x 3) 12
(c) y g ( x)
(e) y 5 g ( x)
(b) y g x 23 2
(d) y g ( x)
(f ) y g (5 x)
54. (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 then reflect the graph about the y-axis
Copyright 2018 Pearson Education, Inc.
38
Chapter 1 Functions
(d) Horizontally compress the graph of f by a factor of 2 and then shift the graph left 12 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 14 unit.
55. Reflection of the graph of y x about the x-axis
followed by a horizontal compression by a factor of
1 then a shift left 2 units.
2
56. Reflect the graph of y 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.
57. Vertical compression of the graph of y
factor of 2, then shift the graph up 1 unit.
1
x2
by a
58. Reflect the graph of y x1/3about the y-axis, then
compress the graph horizontally by a factor of 5.
60.
59.
period
period 4
Copyright 2018 Pearson Education, Inc.
Chapter 1 Practice Exercises
61.
62.
period 4
period 2
64.
63.
period 2
period 2
65. (a) sin B sin 3
2
2
b
c
b2 b 2sin 3 2 23 3. By the theorem of Pythagoras,
2
2
2
a b c a c b 4 3 1.
(b) sin B sin 3
b
c
2c c 2 2 4 . Thus, a c 2 b 2
sin 3
3
2
3
(2)
4
3
2
66. (a) sin A
a
c
a c sin A
(b) tan A
a
b
a b tan A
67. (a) tan B
b
a
a tanb B
(b) sin A
a
c
c sina A
68. (a) sin A
a
c
(b) sin A
a
c
c 2 b2
c
69. Let h 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 flat ground, respectively.
Then, tan 50 hc , tan 35 bh , and b c 10.
Thus, h c tan 50and h b tan 35 (c 10) tan 35
c tan 50 (c 10) tan 35
c(tan 50 tan 35) 10 tan 35
c 10 tan 35 h c tan 50
tan 50 tan 35
10 tan 35 tan 50 16.98
tan 50 tan 35
m.
Copyright 2018 Pearson Education, Inc.
2
4
3
2 .
3
39
40
Chapter 1 Functions
70. Let h height of balloon above ground. From the
figure at the right, tan 40 ah , tan 70 bh , and
a b 2. Thus, h b tan 70 h (2 a ) tan 70
and h a tan 40 (2 a ) tan 70 a tan 40
a (tan 40 tan 70) 2 tan 70
2 tan 70
a tan 40
h a tan 40
tan 70
2 tan 70 tan 40 1.3 km.
tan
40 tan 70
71. (a)
(b) The period appears to be 4 .
(c) f ( x 4 ) sin( x 4 ) cos
x 4
2
sin( x 2 ) cos
x
2
2 sin x cos 2x
since the period of sine and cosine is 2 . Thus, f (x) has period 4 .
72. (a)
(b) D ( ,0) (0, ); R [ 1, 1]
21 kp f 21 sin 2 0 for all integers k.
1
. But then f 21 kp sin (1/(21)) kp 0
Choose k so large that 21 kp 1 0
1/(2 ) kp
(c) f is not periodic. For suppose f has period p. Then f
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) 3 x and g ( x) 4 x satisfy
f ( g ( x)) f (4 x) 3(4 x) 12 x 4(3 x) g (3 x) g ( f ( x)).
2. Yes, there are many such function pairs. For example, if g ( x) (2 x 3)3 and f ( x) x1/3, then
( f g )( x) f ( g ( x)) f ((2 x 3)3 ) ((2 x 3)3 )1/3 2 x 3.
3. If f is odd and defined at x, then f ( x) f ( x). Thus g ( x) f ( x) 2 f ( x) 2 whereas
g ( x) ( f ( x) 2) f ( x) 2. Then g cannot be odd because g ( x) g ( x) f ( x) 2 f ( x) 2
4 0, which is a contradiction. Also, g ( x) is not even unless f ( x ) 0 for all x. On the other hand, if f is
even, then g ( x) f ( x) 2 is also even: g ( x) f ( x) 2 f ( x ) 2 g ( x).
4. If g is odd and g(0) is defined, then g (0) g ( 0) g (0). Therefore, 2 g (0) 0 g (0) 0.
Copyright 2018 Pearson Education, Inc.
Chapter 1 Additional and Advanced Exercises
41
5. For (x, y) in the 1st quadrant, | x | | y | 1 x
x y 1 x y 1. For (x, y) in the 2nd
quadrant, | x | | y | x 1 x y x 1
y 2 x 1. In the 3rd quadrant, | x | | y | x 1
x y x 1 y 2 x 1. In the 4th
quadrant, | x | | y | x 1 x ( y ) x 1
y 1. The graph is given at the right.
6. We use reasoning similar to Exercise 5.
(1) 1st quadrant: y | y | x | x |
2 y 2 x y x.
(2) 2nd quadrant: y | y | x | x |
2 y x ( x) 0 y 0.
(3) 3rd quadrant: y | y | x | x |
y ( y ) x ( x) 0 0
all points in the 3rd quadrant
satisfy the equation.
(4) 4th quadrant: y | y | x | x |
y ( y ) 2 x 0 x. Combining
these results we have the graph given at the right:
sin 2 x
7. (a) sin 2 x cos 2 x 1 sin 2 x 1 cos 2 x (1 cos x) (1 cos x) 1 cos x 1 cos x
1 cos x
sin x
(b) Using the definition of the tangent function and the double angle formulas, we have
1 cos 2 x
x
2
1 cos x
2
tan 2 2x 2 2x
1 cos x .
1 cos 2 x
cos 2
2
sin 2
2
8. The angles labeled in the accompanying figure are
equal since both angles subtend arc CD. Similarly, the
two angles labeled α are equal since they both subtend
arc AB. Thus, triangles AED and BEC are similar which
ac
2 a cos b
implies b a c
(a c)(a c) b(2a cos b)
a 2 c 2 2ab cos b 2
c 2 a 2 b 2 2ab cos .
9. As in the proof of the law of sines of Section 1.3, Exercise 61, ah bc sin A ab sin C ac sin B
the area of ABC 12 (base)(height) 12 ah 12 bc sin A 12 ab sin C 12 ac sin B .
10. As in Section 1.3, Exercise 61, (Area of ABC ) 2 14 (base) 2 (height)2
1 a 2 h2
4
14 a 2b 2 sin 2 C
a 2 b2 c 2
14 a 2 b 2 (1 cos 2 C ) . By the law of cosines, c 2 a 2 b 2 2ab cos C cos C
. Thus,
2 ab
a 2 b2 c 2 2
2
2
2 2
a 2b 2 1 ( a b c )
(area of ABC )2 14 a 2b 2 (1 cos2 C ) 14 a 2b 2 1
2 ab
4
4 a 2b 2
1 4a 2 b 2 ( a 2 b 2 c 2 ) 2 1 [(2ab ( a 2 b 2 c 2 )) (2ab ( a 2 b 2 c 2 ))]
16
16
1
16
2
1 [(( a b) c )(( a b) c )(c ( a b))(c (a b))]
[((a b) c )(c (a b) 2 )] 16
2
2
a b c a b c a b c a b c
abc
s ( s a)( s b)( s c), where s
.
2
2
2
2
2
Therefore, the area of ABC equals s ( s a )( s b)( s c) .
Copyright 2018 Pearson Education, Inc.
x
1 sincos
x
42
Chapter 1 Functions
11. If f is even and odd, then f ( x) f ( x) and f ( x) f ( x) f ( x) f ( x) for all x in the domain of f.
Thus 2 f ( x ) 0 f ( x) 0.
12. (a) As suggested, let E ( x)
f ( x) f ( x)
2
E ( x)
function. Define O ( x) f ( x) E ( x ) f ( x)
f ( x) f ( x)
2
f ( x) f ( x)
2
f ( x ) f ( ( x ))
2
f ( x) f ( x)
2
O( x) O
f ( x) f ( x)
2
f ( x) f ( x)
. Then
2
E ( x) E is an even
O( x)
f ( x) f ( x)
2
is an odd function f ( x) E ( x) O( x) is the sum of an even
and an odd function.
(b) Part (a) shows that f ( x) E ( x) O( x) is the sum of an even and an odd function. If also
f ( x) E1 ( x ) O1 ( x), where E1 is even and O1 is odd, then f ( x) f ( x) 0
( E1 ( x ) O1 ( x )) ( E ( x) O( x)) . Thus, E ( x) E1 ( x) O1 ( x) O ( x) for all x in the domain of f (which is
the same as the domain of E E1 and O O1). Now ( E E1 )( x) E ( x) E1 ( x) E ( x) E1 ( x) (since E
and E1 are even) ( E E1 )( x) E E1 is even. Likewise, (O1 O)( x) O1 ( x ) O( x )
O1 ( x ) (O( x)) (since O and O1 are odd) (O1 ( x) O ( x)) (O1 O ) ( x) O1 O is odd.
Therefore, E E1 and O1 O are both even and odd so they must be zero at each x in the domain of f by
Exercise 11. That is, E1 E and O1 O, so the decomposition of f found in part (a) is unique.
2
2
2
2
x b 2 4ba c a x 2ba 4ba c
4a
(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.
13. y ax 2 bx c a x 2
b
a
14. (a) If a 0, the graph rises to the right of the vertical line x b and falls to the left. If a < 0, the graph falls
to the right of the line x b and rises to the left. If a 0, the graph reduces to the horizontal line y c.
As | a | increases, the slope at any given point x x0 increases in magnitude and the graph becomes steeper. As
| a | decreases, the slope at x0 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 vt v (1 sec) . Moreover, the height of each
triangle is the same value h. Thus 12 (base)(height)
12 bh A1 A2 A3 … . In conclusion,
the object sweeps out equal areas in each one
second interval.
Copyright 2018 Pearson Education, Inc.
Chapter 1 Additional and Advanced Exercises
16. (a) Using the midpoint formula, the coordinates of P are
y
x
a0 b0
, 2
2
43
, . Thus the slope
a b
2 2
ba /2
ba .
/2
b0
(b) The slope of AB 0 a ba . The line segments AB and OP are perpendicular when the product of their
of OP
slopes is 1
ba ba ba
2
2
. Thus, b 2 a 2 a b (since both are positive). Therefore, AB is
perpendicular to OP when a b.
17. From the figure we see that 0 2 and AB AD 1. From trigonometry we have the following:
sin . We can see that:
sin EB
EB, cos AE
AE , tan CD
CD, and tan EB
cos
AB
AB
AD
AE
area ADC 1 ( AE )( EB) 1 ( AD)2 1 ( AD) (CD )
area AEB area sector DB
2
2
sin
12 sin cos 12 (1) 2 12 (1)(tan ) 12 sin cos 12 12 cos
2
18. ( f g )( x) f ( g ( x)) a (cx d ) b acx ad b and ( g f )( x) g ( f ( x)) c (ax b) d acx cb d
Thus ( f g )( x ) ( g f )( x) acx ad b acx bc d ad b bc d . Note that f (d ) ad b and
g (b) cb d , thus ( f g )( x) ( g f )( x ) if f (d ) g (b).
Copyright 2018 Pearson Education, Inc.
CHAPTER 2
2.1
LIMITS AND CONTINUITY
RATES OF CHANGE AND TANGENTS TO CURVES
1. (a)
f
x
f (3) f (2)
3 2
2. (a)
g
x
g (3) g (1)
3 1
3. (a)
h
t
g
t
4. (a)
h
2819 19
(b)
f
x
f (1) f ( 1)
1( 1)
3 ( 1)
2
2
(b)
g
x
g (4) g ( 2)
4 ( 2)
34 h 4 11 4
(b)
h
t
g
t
3
4
4
2
g ( ) g (0)
0
(2 1) (2 1)
0
2
5.
R
R (2) R (0)
20
812 1 321 1
6.
P
P (2) P (1)
21
7. (a)
(b)
8. (a)
(b)
(816 10) (1 4 5)
1
y
x
((2 h )2 5) (22 5)
h
y
x
(7(2 h )2 ) (7 22 )
h
(b)
h
22 0 1
8 6 8 0
2 h 6 0
2
6
g ( ) g ( )
( )
3
3
3 3
(2 1) (2 1)
2
0
22 0
2
2
4 4h hh 51 4h h h 4 h. As h 0, 4 h 4 at P (2, 1) the slope is 4.
y ( 1) 4( x 2) y 1 4 x 8 y 4 x 9
2
2
744hh h 3 4 hhh 4 h. As h 0, 4 h 4 at P (2, 3) the slope
is 4.
y 3 ( 4)( x 2) y 3 4 x 8 y 4 x 11
y
((2 h )2 2(2 h ) 3) (22 2(2) 3)
y
((1 h ) 2 4(1 h )) (12 4(1))
y
(2 h )3 23
y
2 (1 h )3 (2 13 )
4 4 h h 2 4 2 h 3( 3)
9. (a) x
h
h
P(2, 3) the slope is 2.
(b) y (3) 2( x 2) y 3 2 x 4 y 2 x 7.
1 2 h h 2 4 4 h ( 3)
2
2h h h 2 h. As h 0, 2 h 2 at
2
10. (a) x
h h 2h h 2. As h 0, h 2 2 at P (1, 3) the
h
h
slope is 2.
(b) y ( 3) ( 2)( x 1) y 3 2 x 2 y 2 x 1.
2
3
2
3
11. (a) x
812h 4hh h 8 12 h 4hh h 12 4h h 2 . As h 0, 12 4h h 2 12, at P (2, 8)
h
the slope is 12.
(b) y 8 12( x 2) y 8 12 x 24 y 12 x 16.
2
3
2
3
12. (a) x
213h h3h h 1 3h 3hh h 3 3h h 2 . As h 0, 3 3h h 2 3, at
h
P(1, 1) the slope is 3.
(b) y 1 ( 3)( x 1) y 1 3 x 3 y 3x 4.
Copyright 2018 Pearson Education, Inc.
45
46
Chapter 2 Limits and Continuity
3
3
2
3
2
3
y
(1 h ) 12(1 h ) (1 12(1)) 13h 3h h 1212 h ( 11)
13. (a) x
9h 3hh h 9 3h h 2 .
h
h
As h 0, 9 3h h 2 9 at P(1, 11) the slope is 9.
(b) y (11) (9)( x 1) y 11 9 x 9 y 9 x 2.
(2 h )3 3(2 h ) 2 4 (23 3(2)2 4)
y
2
3
12 h 3h
14. (a) x
812h 6h h 12
h
h
As h 0, 3h h 2 0 at P (2, 0) the slope is 0.
(b) y 0 0( x 2) y 0.
15. (a)
y
x
1 1
2 h 2
h
As h 0,
(b)
16. (a)
y
y
x
17. (a)
18. (a)
(4 h )
4
2 (4 h ) 2 4
h
1
2 h
41 , at P 2, 21 the slope is 41 .
4 h 2( 2 h ) 1
42hh 12 1h
h 21h 21 h .
2 h
1,
2
at P (4, 2) the slope is
1.
2
y (2) 12 ( x 4) y 2 12 x 2 y 12 x 4
y
x
(4 h ) 4
1
4 hh 4 4hh 2 4 h 2
.
4 h 2 h ( 4 h 2)
4 h 2
1
4 h 2
1
4 2
14 , at P (4, 2) the slope is
1.
4
y 2 14 ( x 4) y 2 14 x 1 y 14 x 1
y
x
7 ( 2 h ) 7 ( 2)
h
As h 0,
(b)
2
3
3h h h 3h h 2 .
2 ( 2 h )
2( 2 h ) 1h 2( 21 h) .
1
2( 2 h )
As h 0,
(b)
40
21 41 ( x (2)) y 12 41 x 12 y 41 x 1
As h 0,
(b)
2
y 3
19. (a)
1
9 h 3
1 ( x (2))
6
Q
Q1 (10, 225)
Q2 (14,375)
Q3 (16.5, 475)
Q4 (18,550)
9 h 3
h
1
9 3
y 3
1 ,
6
9 h 3 9 h 3
h
9 h 3
(9 h ) 9
h ( 9 h 3)
at P (2, 3) the slope is
1 x 1
6
3
Slope of PQ
650 225
2010
650 375
20 14
650 475
20 16.5
650 550
2018
y
1 .
9 h 3
1 .
6
1 x 8
6
3
p
t
42.5 m/sec
45.83 m/sec
50.00 m/sec
50.00 m/sec
(b) At t 20, the sportscar was traveling approximately 50 m/sec or 180 km/h.
20. (a)
Q
Q1 (5, 20)
Q2 (7,39)
Q3 (8.5,58)
Q4 (9.5, 72)
Slope of PQ
p
t
80 20 12 m/sec
10 5
80 39 13.7 m/sec
10 7
80 58 14.7 m/sec
10 8.5
8072 16 m/sec
10 9.5
(b) Approximately 16 m/sec
Copyright 2018 Pearson Education, Inc.
Section 2.1 Rates of Change and Tangents to Curves
21. (a)
47
p
Profit (1000s)
200
160
120
80
40
0
(b)
p
t
2010 2011 2012 2013 2014
Ye ar
174 62
2014 2012
t
112
56 thousand dollars per year
2
(c) The average rate of change from 2011 to 2012 is
The average rate of change from 2012 to 2013 is
p
t
p
t
62 27
20122011
35 thousand dollars per year.
11162
20132012
49 thousand dollars per year.
So, the rate at which profits were changing in 2012 is approximately 12 (35 49) 42 thousand dollars
per year.
22. (a) F ( x) ( x 2)/( x 2)
x
1.2
1.1
F ( x) 4.0
3.4
F
x
F
x
F
x
1.01
3.04
1.001
3.004
4.0 ( 3)
5.0;
1.2 1
3.04 ( 3)
4.04;
1.011
3.0004 ( 3)
4.0004;
1.00011
F
x
F
x
g
x
1.0001
3.0004
1
3
3.4 ( 3)
4.4;
1.11
3.004 ( 3)
4.004;
1.0011
(b) The rate of change of F ( x) at x 1 is 4.
23. (a)
g
x
g
x
g (2) g (1)
2211 0.414213
2 1
g (1 h ) g (1)
1hh 1
(1 h ) 1
g (1.5) g (1)
1.51
1.5 1
0.5
0.449489
(b) g ( x) x
1 h
1 h
1 h 1 /h
1.1
1.01
1.001
1.0001
1.00001
1.000001
1.04880
1.004987
1.0004998
1.0000499
1.000005
1.0000005
0.4880
0.4987
0.4998
0.499
0.5
0.5
(c) The rate of change of g ( x) at x 1 is 0.5.
1 h 1
h
h 0
(d) The calculator gives lim
24. (a) i)
ii)
12 .
11
1
f (3) f (2)
3 2
6
16
3 2
1
1
1 1
2 T
f (T ) f (2)
TT 22 2TT 22T 2T2(TT 2)
T 2
T
2T2(2
21T , T 2
T )
(b) T
f (T )
( f (T ) f (2))/(T 2)
2.1
2.01
2.001
0.476190
0.497512
0.499750
0.2381
0.2488
0.2500
(c) The table indicates the rate of change is 0.25 at t 2.
(d) lim 21T 14
T 2
2.0001
0.4999750
0.2500
2.00001
0.499997
0.2500
NOTE: Answers will vary in Exercises 25 and 26.
0 15 mph; [1, 2.5]: s
25. (a) [0, 1]: st 15
t
10
20 15
2.51
10
mph; [2.5, 3.5]: st
3
Copyright 2018 Pearson Education, Inc.
30 20
3.5 2.5
10 mph
2.000001
0.499999
0.2500
48
Chapter 2 Limits and Continuity
(b) At P
12 , 7.5 : Since the portion of the graph from t 0 to t 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 12 is
157.5 15 mi/hr. At P (2, 20): Since the portion of the graph from t 2 to t 2.5 is nearly linear, the
10.5
20 0 mi/hr.
instantaneous rate of change will be nearly the same as the average rate of change, thus v 20
2.5 2
For values of t less than 2, we have
Q
Q1 (1, 15)
Q2 (1.5, 19)
Q3 (1.9, 19.9)
Slope of PQ
s
t
15 20 5 mi/hr
1 2
19 20 2 mi/hr
1.5 2
19.9 20 1 mi/hr
1.9 2
Thus, it appears that the instantaneous speed at t 2 is 0 mi/hr.
At P(3, 22):
Q
Q1 (4, 35)
Q2 (3.5, 30)
Q3 (3.1, 23)
Slope of PQ
35 22
4 3
30 22
3.53
23 22
3.13
s
t
Q
13 mi/hr
Q1 (2, 20)
16 mi/hr
Q2 (2.5, 20)
10 mi/hr
Q3 (2.9, 21.6)
Slope of PQ
s
t
20 22 2 mi/hr
2 3
20 22 4 mi/hr
2.53
21.6 22 4 mi/hr
2.9 3
Thus, it appears that the instantaneous speed at t 3 is about 7 mi/hr.
(c) It appears that the curve is increasing the fastest at t 3.5. Thus for P(3.5, 30)
Slope of PQ st
Slope of PQ
Q
Q
Q1 (4, 35)
Q2 (3.75, 34)
Q3 (3.6, 32)
3530 10 mi/hr
43.5
34 30 16 mi/hr
3.753.5
32 30 20 mi/hr
3.6 3.5
Q1 (3, 22)
Q2 (3.25, 25)
Q3 (3.4, 28)
s
t
2230 16 mi/hr
33.5
2530 20 mi/hr
3.253.5
2830 20 mi/hr
3.4 3.5
Thus, it appears that the instantaneous speed at t 3.5 is about 20 mi/hr.
gal
26. (a) [0, 3]: At 10315
1.67 day ; [0, 5]: At
0
(b) At P(1, 14):
Q
Q1 (2, 12.2)
Q2 (1.5, 13.2)
Q3 (1.1, 13.85)
Slope of PQ
A
t
12.2 14 1.8 gal/day
21
13.2 14 1.6 gal/day
1.51
13.8514 1.5 gal/day
1.11
3.915
5 0
gal
gal
2.2 day ; [7, 10]: At 0101.4
0.5 day
7
Q
Q1 (0, 15)
Q2 (0.5, 14.6)
Q3 (0.9, 14.86)
Slope of PQ
A
t
1514 1 gal/day
0 1
14.6 14 1.2 gal/day
0.51
14.86 14 1.4 gal/day
0.9 1
Thus, it appears that the instantaneous rate of consumption at t 1 is about 1.45 gal/day.
At P(4, 6):
Slope of PQ At
Q
Slope of PQ A
Q
t
Q1 (5, 3.9)
Q2 (4.5, 4.8)
Q3 (4.1, 5.7)
3.9 6
5 4
4.86
4.5 4
5.7 6
4.1 4
2.1 gal/day
Q1 (3, 10)
2.4 gal/day
Q2 (3.5, 7.8)
3 gal/day
Q3 (3.9, 6.3)
10 6 4 gal/day
3 4
7.86 3.6 gal/day
3.5 4
6.36 3 gal/day
3.9 4
Thus, it appears that the instantaneous rate of consumption at t 1 is 3 gal/day.
Copyright 2018 Pearson Education, Inc.
Section 2.2 Limit of a Function and Limit Laws
At P(8, 1):
Q
Q1 (9, 0.5)
Q2 (8.5, 0.7)
Q3 (8.1, 0.95)
Slope of PQ
Q
A
t
Q1 (7, 1.4)
0.51 0.5 gal/day
9 8
0.7 1 0.6 gal/day
8.58
0.951 0.5 gal/day
8.18
Q2 (7.5, 1.3)
Q3 (7.9, 1.04)
Slope of PQ
49
A
t
1.4 1 0.6 gal/day
7 8
1.31 0.6 gal/day
7.58
1.04 1 0.6 gal/day
7.98
Thus, it appears that the instantaneous rate of consumption at t 1 is 0.55 gal/day.
(c) It appears that the curve (the consumption) is decreasing the fastest at t 3.5. Thus for P(3.5, 7.8)
Slope of PQ st
Q
Slope of PQ At
Q
11.2 7.8 3.4 gal/day
Q1 (2.5, 11.2)
4.8 7.8 3 gal/day
Q1 (4.5, 4.8)
2.53.5
Q2 (4, 6)
Q3 (3.6, 7.4)
4.53.5
6 7.8 3.6 gal/day
4 3.5
7.4 7.8 4 gal/day
3.6 3.5
Q2 (3, 10)
Q3 (3.4, 8.2)
10 7.8 4.4 gal/day
33.5
8.2 7.8 4 gal/day
3.4 3.5
Thus, it appears that the rate of consumption at t 3.5 is about 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 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 0.
(d) 1
3. (a)
(d)
(g)
(j)
True
False
True
True
(b)
(e)
(h)
(k)
4. (a) False
(d) True
(g) False
x
x0 | x |
(c) False
(f) True
(i) True
(b) False
(e) True
(h) True
(c) True
(f) True
(i) False
1 if x 0 and | xx | xx 1 if x 0. As x approaches 0 from the left, | xx |
approaches 1. As x approaches 0 from the right, | xx | approaches 1. There is no single number L that all the
5. lim
does not exist because | xx |
True
False
False
False
x
x
function values get arbitrarily close to as x 0.
1 become increasingly large and negative. As x approaches 1
6. As x approaches 1 from the left, the values of x
1
from the right, the values become increasingly large and positive. There is no number L that all the function
values get arbitrarily close to as x 1, so lim x11 does not exist.
x 1
7. Nothing can be said about f ( x) because the existence of a limit as x x0 does not depend on how the function
is defined at x0 . 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 x0 . That is, the existence of a limit depends on the values of f ( x) for x near x0 , not on the
definition of f ( x) at x0 itself.
Copyright 2018 Pearson Education, Inc.
50
Chapter 2 Limits and Continuity
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
x 0
the value f (0) itself.
9. No, the definition does not require that f be defined at x 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) 5.
x 1
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) 5. We can conclude nothing about lim f ( x),
x 1
x 1
whether it exists or what its value is if it does exist, from knowing the value of f (1) alone.
11.
lim ( x 2 13) ( 3)2 13 9 13 4
x3
12. lim ( x 2 5 x 2) (2)2 5(2) 2 4 10 2 4
x 2
13. lim 8(t 5)(t 7) 8(6 5)(6 7) 8
t 6
14.
15.
16.
17.
lim ( x3 2 x 2 4 x 8) (2)3 2(2)2 4( 2) 8 8 8 8 8 16
x 2
lim 2 x 53
x2 11 x
2(2) 5
11(2)3
23 2 23 1 (8 2) 43 1 (6) 13 2
lim 4 x (3x 4)2 4 12 3 12 4
x 1/2
y2
2 2
(2) 2 5(2) 6
2
( 2) 23 4
y 3
lim
z 4
22. lim
5h 4 2
h
h 0
x 5
2
x 5 x 25
23. lim
2
2
( 2) 25 25
2
4
24 16
z 2 10 42 10 16 10 6
3
3h 1 1
h 0
4
4 1
410
20
6
5
lim (5 y ) 4/3 [5 (3)]4/3 (8)4/3 (8)1/3
21. lim
24.
3
2
y 2 y 5 y 6
20.
9
3
lim (8 3s )(2 s 1) 8 5
t 2/3
18. lim
19.
3
3(0) 1 1
lim
h 0
3 32
1 1
5 h 4 2 5h 4 2
h
5h 4 2
lim
h 0 h
(5h 4) 4
5h 4 2
lim
h 0 h
5h
5h 4 2
x 5
lim ( x 5)(
lim 1 1 1
x 5) x 5 x 5 55 10
x 5
lim 2 x 3
x 3 x 4 x 3
x 3
lim ( x 3)(
lim 1 1 12
x 1) x 3 x 1 31
x 3
Copyright 2018 Pearson Education, Inc.
lim
h0
5
5h 4 2
5
4 2
54
Section 2.2 Limit of a Function and Limit Laws
25.
2
( x 5)( x 2)
lim x x3x510 lim
lim ( x 2) 5 2 7
x 5
x 5
x 5
x 5
x 2 7 x 10
x2
x 2
26. lim
t 2 t 2
2
t 1 t 1
27. lim
( x 5)( x 2)
x2
lim
x2
lim ( x 5) 2 5 3
x 2
(t 2)(t 1)
lim (t 1)(t 1) lim tt12 1112 32
t 1
t 1
28.
2
(t 2)(t 1)
lim t 23t 2 lim (t 2)(t 1) lim tt 22 11 22 13
t
t
2
t 1
t 1
t 1
29.
lim 32 x 42
x 2 x 2 x
30. lim
2( x 2)
lim
lim 22 42 12
x 2 x
2
x 2 x ( x 2)
5 y 3 8 y 2
y 2 (5 y 8)
lim
4
2
y 0 3 y 16 y
816 12
2
y 0 3 y 16
1 x
5 y 8
lim
2
2
y 0 y (3 y 16)
31.
1
lim xx 11 lim x x 1 lim 1xx x11 lim 1x 1
x1
x 1
x1
x1
32.
lim
x 0
1 1
x 1 x 1
x
u 4 1
3
u 1 u 1
33. lim
lim
lim
4 x x2
x 4 2 x
lim
x 1
lim
x 1
x 1
x 3 2
lim
x 2 8 3
x 1
x 1
( x 1)
x 3 2
lim
x 1
lim
39. lim
x 2
x 2 12 4
x 2
( x 1)
x 1
x 2 12 4
lim
x2
x2
x 2 12 4
x 1
x 2 8 3
lim
lim
x 8 3
x 2 8 3
( x 2)
2
x 1
x 2
x 3 2
x 2 8 3
1
9 3
x (2 x )(2 x )
2 x
x 3 2
1
x 3
x 9
x 4
(11)(11)
111
v 2 2v 4
2
v 2 (v 2)( v 4)
lim
lim
2
x 0 ( x 1)( x 1)
lim
v 2 (v 2)(v 2)( v 4)
x 3
x 9 ( x 3)( x 3)
lim
u 2 u 1
u 1
2
x (4 x )
x 4 2 x
(u 2 1)(u 1)
lim
(v 2)( v 2 2v 4)
lim x 3
x 9 x 9
37. lim
38.
2
u 1 (u u 1)(u 1)
lim
2x
1
x 0 ( x 1)( x 1) x
lim
(u 2 1)(u 1)(u 1)
v3 8
4
v 2 v 16
36. lim
x
x 0
34. lim
35.
( x 1) ( x 1)
( x 1)( x 1)
lim
2
2
3 3
4 4 4
(4)(8)
1
6
2
1
2
4
3
12
3
32 8
x 3 2
lim
lim x 2 x 4(2 2) 16
x 4
( x 1)
( x 3) 4
lim
x 1
( x 2 8) 9
x 1 ( x 1)
x 2 8 3
x32 4 24
lim
( x 1)( x 1)
x 1 ( x 1)
x 2 8 3
13
x 2 12 4
x 12 4
4
16 4
lim
( x 2 12) 16
x 2 ( x 2)
1
2
2
x 12 4
lim
( x 2)( x 2)
x 2 ( x 2)
Copyright 2018 Pearson Education, Inc.
x 2 12 4
51
52
40.
Chapter 2 Limits and Continuity
x2
lim
x 2
2
x 5 3
( x 2)
lim
x 2
x 2
2
lim 2 x x3 5
x 3
lim
x 2 5 3
x 5 3
9 3
4
x 5 3
2
x 3
2
2
(3 x )(3 x )
x 3 ( x 3) 2 x 2 5
4 x
42. lim
2
x 4 5 x 9
lim
(4 x ) 5 x 2 9
x 4
(4 x ) 5 x 2 9
(4 x )(4 x )
( x 2)
lim
x 2 5 3
( x 2 5) 9
x 2
lim
( x 2)
x 2 5 3
( x 2)( x 2)
x 2
32
lim
4 ( x 2 5)
3 x
x 3 2 x 2 5
x 3 ( x 3) 2 x 5
lim
x 4 5 x 2 9 5 x 2 9
lim
x 2 5 2 x 2 5
( x 3) 2 x 5
lim
2
x 2 5 3
x 2
lim
41.
lim
x 4
6
2 4
32
(4 x ) 5 x 2 9
x4
lim 5
2
2
25( x 9)
x2 9
4 x
5 25
8
9 x 2
lim
x 3 ( x 3) 2 x 2 5
lim (4 x)5
x4
16 x
x 2 9
2
5
4
2
43. lim (2sin x 1) 2sin 0 1 0 1 1
44.
45. lim sec x lim cos1 x cos1 0 11 1
x 0
x 0
46.
x 0
lim sin 2 x lim sin x (sin 0)2 02 0
x /4
x /4
sin x sin 0 0 0
lim tan x lim cos
x
cos0
1
x /3
x /3
x sin x 1 0 sin 0 1 0 0 1
47. lim 13cos
x
3cos 0
3
3
x 0
48. lim ( x 2 1)(2 cos x) (02 1)(2 cos 0) ( 1)(2 1) ( 1)(1) 1
x 0
49.
lim
x
x 4 cos( x ) lim
50. lim 7 sec2 x
x 0
x 4 lim cos( x ) 4 cos 0 4 1 4
x
x
lim (7 sec2 x) 7 lim sec 2 x 7 sec2 0 7 (1) 2 2 2
x 0
x 0
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) lim f ( x) g ( x) lim f ( x) lim g ( x) (5)( 2) 10
x c
x c
x c
(b) lim 2 f ( x) g ( x) 2 lim f ( x ) lim g ( x) 2(5)(2) 20
x c
x c
x c
(c) lim [ f ( x ) 3 g ( x)] lim f ( x) 3 lim g ( x) 5 3(2) 1
x c
(d)
f ( x)
lim
x c f ( x ) g ( x )
x c
lim f ( x )
x c
lim f ( x ) lim g ( x ) 5(52) 75
xc
x c
x c
Copyright 2018 Pearson Education, Inc.
Section 2.2 Limit of a Function and Limit Laws
53
54. (a) lim [ g ( x) 3] lim g ( x) lim 3 3 3 0
x 4
x4
x 4
(b) lim xf ( x ) lim x lim f ( x) (4)(0) 0
x 4
x4
x4
2
(c) lim [ g ( x)]2 lim g ( x) [3]2 9
x 4
x 4
lim g ( x )
g ( x)
(d) lim f ( x ) 1 limx f (4x ) lim 1 031 3
x 4
x4
x 4
55. (a) lim [ f ( x) g ( x )] lim f ( x) lim g ( x ) 7 (3) 4
x b
x b
x b
(b) lim f ( x) g ( x) lim f ( x ) lim g ( x) (7)(3) 21
x b
x b
xb
(c) lim 4 g ( x) lim 4 lim g ( x) (4)(3) 12
x b
x b x b
(d) lim f ( x)/g ( x) lim f ( x)/ lim g ( x) 73 73
x b
56. (a)
x b
x b
lim [ p ( x) r ( x) s ( x)] lim p ( x) lim r ( x) lim s ( x ) 4 0 ( 3) 1
x 2
x 2
x 2
x 2
(b) lim p ( x) r ( x) s ( x ) lim p ( x ) lim r ( x) lim s ( x) (4)(0)(3) 0
x 2
x 2
x2
x 2
(c) lim [4 p ( x) 5r ( x)]/s ( x) 4 lim p ( x) 5 lim r ( x ) lim s ( x) [4(4) 5(0)]/ 3 16
3
x 2
x 2
x2
x2
(1 h ) 2 12
h
h 0
57. lim
2
h (2 h )
lim 1 2 h h h 1 lim h lim (2 h) 2
h 0
h 0
h 0
( 2 h ) 2 ( 2)2
h
h 0
58. lim
2
h ( h 4)
lim 4 4h h h 4 lim h lim (h 4) 4
h 0
h 0
h 0
[3(2 h ) 4][3(2) 4]
h
h 0
59. lim
60. lim
h 0
61. lim
h 0
21 h 12 lim
2 1
2 h
h 0 2 h
h
7h 7
h
lim 3hh 3
h 0
lim
h 0
lim
7h 7
h
63. lim
x 0
7 h 7
7 h 7
3(0 h ) 1 3(0) 1
lim
h
h 0
h 0
62. lim
2 ( 2 h )
h0 2 h ( 2 h )
lim
h 0 h
3h 1 1
h
lim h(4h2 h) 14
h 0
(7 h ) 7
7h 7
lim
3h 1 1
3h 1 1
5 2 x 2 5 2(0)2 5 and lim
x0
h 0 h
lim
h 0 h
(3h 1) 1
3h 1 1
h
7h 7
lim
h 0 h
lim
h0
3h
3h 1 1
1
7h 7
lim
h0
x 0
2
3
3h11
2 7
32
5 x 2 5 (0) 2 5; by the sandwich theorem, lim f ( x) 5
x 0
64. lim (2 x 2 ) 2 0 2 and lim 2 cos x 2(1) 2; by the sandwich theorem, lim g ( x) 2
x 0
1
x 0
x 1
65. (a) lim 1 x6 1 06 1 and lim 1 1; by the sandwich theorem, lim 2x2sin
cos x
x0
x0
x 0
Copyright 2018 Pearson Education, Inc.
54
Chapter 2 Limits and Continuity
(b) For x 0, y ( x sin x)/(2 2 cos x) lies
between the other two graphs in the figure,
and the graphs converge as x 0.
1
x 0 2
66. (a) lim
x 2 lim
24
1
x 0 2
x2
lim 24
x 0
1
2
0
1
2
1
x0 2
and lim
(b) For all x 0, the graph of f ( x) (1 cos x)/x 2
lies between the line y 12 and the parabola
x 1.
12 ; by the sandwich theorem, lim 1cos
2
2
x 0
x
y 12 x 2 /24, and the graphs converge as
x 0.
67. (a)
f ( x) ( x 2 9)/( x 3)
x
3.1
3.01
3.001
3.0001
3.00001
3.000001
f ( x)
6.1
6.01
6.001
6.0001
6.00001
6.000001
x
2.9
2.99
2.999
2.9999
2.99999
2.999999
5.999
5.9999
5.99999
5.999999
f ( x)
5.9
5.99
The estimate is lim f ( x) 6.
x 3
(b)
2
( x 3)( x 3)
(c) f ( x) xx 39
x 3 if x 3, and lim ( x 3) 3 3 6.
x 3
x 3
68. (a) g ( x) ( x 2 2)/ x 2
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
Copyright 2018 Pearson Education, Inc.
Section 2.2 Limit of a Function and Limit Laws
(b)
2
x
(c) g ( x) x 2
x 2
x
x 2
2 x 2
69. (a) G ( x) ( x 6)/( x 2 4 x 12)
x
5.9
5.99
G ( x ) .126582 .1251564
x
G ( x)
6.1
.123456
6.01
.124843
2 if x 2, and lim
x 2
x 2
2 2 2 2.
5.999
.1250156
5.9999
.1250015
5.99999 5.999999
.1250001 .1250000
6.001
.124984
6.0001
.124998
6.00001
.124999
6.000001
.124999
(b)
(c) G ( x)
x 6
( x 2 4 x 12)
x6
( x 6)(
x 1 2 if x 6, and lim x 1 2 61 2 18 0.125.
x 2)
x 6
70. (a) h( x) ( x 2 2 x 3)/( x 2 4 x 3)
x
h( x )
2.9
2.052631
2.99
2.005025
2.999
2.000500
2.9999
2.000050
2.99999
2.000005
2.999999
2.0000005
x
3.1
h( x) 1.952380
3.01
1.995024
3.001
1.999500
3.0001
1.999950
3.00001
1.999995
3.000001
1.999999
(b)
2
( x 3)( x 1)
(c) h( x) x2 2 x 3 ( x 3)( x 1) xx 11 if x 3, and lim xx 11 3311
x 4 x 3
x 3
4
2
2.
Copyright 2018 Pearson Education, Inc.
55
56
Chapter 2 Limits and Continuity
71. (a) f ( x ) ( x 2 1)/(| x | 1)
x
f ( x)
1.1
2.1
1.01
2.01
1.001
2.001
1.0001
2.0001
1.00001
2.00001
1.000001
2.000001
x
f ( x)
.9
1.9
.99
1.99
.999
1.999
.9999
1.9999
.99999
1.99999
.999999
1.999999
(b)
(c) f ( x)
x 2 1
x 1
( x 1)( x 1) x 1, x 0 and x 1
x 1
, and lim (1 x ) 1 (1) 2.
( x 1)( x 1)
x 1
( x 1) 1 x, x 0 and x 1
72. (a) F ( x) ( x 2 3 x 2)/(2 | x |)
x
F ( x)
2.1
1.1
2.01
1.01
2.001
1.001
2.0001
1.0001
2.00001
1.00001
2.000001
1.000001
x
F ( x)
1.9
.9
1.99
.99
1.999
.999
1.9999
.9999
1.99999
.99999
1.999999
.999999
(b)
(c) F ( x )
x2 3 x 2
2 x
( x 2)( x 1) ,
x0
2 x
, and lim ( x 1) 2 1 1.
( x 2)( x 1)
x 2
x
1,
x
0
and
x
2
2 x
73. (a) g ( ) (sin )/
g ( )
.1
.998334
.1
g ( )
.998334
lim g( ) 1
.01
.999983
.001
.999999
.01
.001
.999983 .999999
.0001
.999999
.00001
.999999
.000001
.999999
.0001
.999999
.00001
.999999
.000001
.999999
0
Copyright 2018 Pearson Education, Inc.
Section 2.2 Limit of a Function and Limit Laws
57
(b)
74. (a) G (t ) (1 cos t )/t 2
t
G (t )
.1
.499583
.01
.499995
.001
.499999
.0001
.5
.00001
.5
.000001
.5
t
.1
G (t ) .499583
lim G (t ) 0.5
.01
.499995
.001
.499999
.0001 .00001 .000001
.5
.5
.5
t 0
(b)
75. lim f ( x) exists at those points c where lim x 4 lim x 2 . Thus, c 4 c 2 c 2 (1 c 2 ) 0 c 0, 1, or 1.
x c
x c
x c
Moreover, lim f ( x) lim x 2 0 and lim f ( x ) lim f ( x) 1.
x 0
x 0
x 1
x 1
76. Nothing can be concluded about the values of f , g , and h at x 2. Yes, f (2) could be 0. Since the conditions
of the sandwich theorem are satisfied, lim f ( x ) 5 0.
x 2
f ( x ) 5
x 4 x 2
77. 1 lim
78. (a) 1 lim
lim f ( x ) lim 5
4
x4
x lim
x lim 2
x 4
f ( x)
2
x 2 x
(b) 1 lim
f ( x)
2
x 2 x
x4
lim f ( x )
x 2
lim x 2
x 2
4 2
lim f ( x )
x 2
4
lim f ( x) 5 2(1) lim f ( x) 2 5 7.
x 4
x 4
lim f ( x) 4.
x 2
f ( x)
f ( x)
lim x lim 1x lim x
x 2
x2 x 2
f ( x ) 5
79. (a) 0 3 0 lim x 2
x 2
lim f ( x) 5 lim
x2
lim f ( x ) 5
x 4
x 2
f ( x)
2.
12 xlim
2 x
lim ( x 2) lim f ( x ) 5 ( x 2) lim [ f ( x ) 5]
x 2
x 2 x 2
x2
f ( x) 5.
f ( x ) 5
(b) 0 4 0 lim x 2 lim ( x 2) lim f ( x) 5 as in part (a).
x 2
x 2
x 2
Copyright 2018 Pearson Education, Inc.
58
Chapter 2 Limits and Continuity
2
f ( x)
80. (a) 0 1 0 lim 2 lim x lim
x0
x 0 x x 0
That is, lim f ( x) 0.
x 0
(b) 0 1 0 lim
x 0
f ( x)
f ( x)
x 2 lim 2 x 2 lim f ( x).
xlim
x 0
0
x0 x
x2
f ( x)
x2
lim x lim f ( x ) x lim f ( x ) . That is, lim f ( x ) 0.
x0 x
x0 x
x 0 x0 x 2
81. (a) lim x sin 1x 0
x 0
(b) 1 sin 1x 1 for x 0:
x 0 x x sin 1x x lim x sin 1x 0 by the sandwich theorem;
x 0
x 0 x x sin 1x x lim x sin 1x 0 by the sandwich theorem.
x 0
82. (a) lim x 2 cos
x 0
(b) 1 cos
2
0
1
x3
1 for x 0 x
1
x3
2
x 2 cos 13 x 2 lim x 2 cos 13 0 by the sandwich theorem since
lim x 0.
x
x 0
x
x 0
83–88.
Example CAS commands:
Maple:
f : x - (x^4 16)/(x 2);
x0 : 2;
plot( f (x), x x0-1..x0 1, color black,
title "Section 2.2, #83(a)" );
limit( f (x), x x 0 );
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 _]: (x 3 x 2 5x 3)/(x 1) 2
x0 1; h 0.1;
Plot[f[x],{x, x0 h, x0 h}]
Limit[f[x], x x0]
Copyright 2018 Pearson Education, Inc.
Section 2.3 The Precise Definition of a Limit
2.3
THE PRECISE DEFINITION OF A LIMIT
1.
Step 1: x 5 x 5 5 x 5
Step 2: 5 7 2,or 5 1 4.
The value of δ which assures x 5 1 x 7 is the smaller value, 2.
2.
Step 1:
Step 2:
x 2 x 2 2 x 2
2 1 1, or 2 7 5.
The value of which assures x 2 1 x 7 is the smaller value, 1.
Step 1:
Step 2:
x (3) x 3 3 x 3
3 72 12 , or 3 12 52 .
3.
The value of which assures x (3) 72 x 12 is the smaller value, 12 .
4.
Step 1:
x 32 x 32 32 x 32
Step 2:
32 72 2, or 32 12 1.
The value of which assures x 32 72 x 12 is the smaller value, 1.
Step 1:
x 12 x 12 12 x 12
1 , or 1 4 1 .
12 94 18
2 7
14
5.
Step 2:
The value of which assures x 12 94 x
4
7
1.
is the smaller value, 18
6.
Step 1:
Step 2:
7. Step 1:
Step 2:
x 3 x 3 3 x 3
3 2.7591 0.2409, or 3 3.2391 0.2391.
The value of which assures x 3 2.7591 x 3.2391 is the smaller value, 0.2391.
x 5 x 5 5 x 5
From the graph, 5 4.9 0.1, or 5 5.1 0.1; thus 0.1 in either case.
Copyright 2018 Pearson Education, Inc.
59
60
Chapter 2 Limits and Continuity
8. Step 1:
Step 2:
x (3) x 3 3 x 3
From the graph, 3 3.1 0.1, or 3 2.9 0.1; thus 0.1.
9. Step 1:
Step 2:
x 1 x 1 1 x 1
9 7 , or 1 25 9 ; thus 7 .
From the graph, 1 16
16
16
16
16
10. Step 1:
Step 2:
x 3 x 3 3 x 3
From the graph, 3 2.61 0.39, or 3 3.41 0.41; thus 0.39.
11. Step 1:
Step 2:
x 2 x 2 2 x 2
From the graph, 2 3 2 3 0.2679, or 2 5 5 2 0.2361;
thus 5 2.
12. Step 1:
x (1) x 1 1 x 1
Step 2:
From the graph, 1
thus
5 2
.
2
5
2
52 2 0.118 or 1 23 22 3 0.1340;
13. Step 1:
Step 2:
x (1) x 1 1 x 1
From the graph, 1 16
97 0.77, or 1 16
9
25
14. Step 1:
x 12 x 12 12 x 12
1 1 1 0.00248, or 1 1 1 1 0.00251;
From the graph, 12 2.01
2 2.01
2 1.99
1.99 2
thus 0.00248.
Step 2:
9
25
9 0.36.
0.36; thus 25
15. Step 1:
Step 2:
( x 1) 5 0.01 x 4 0.01 0.01 x 4 0.01 3.99 x 4.01
x 4 x 4 4 x 4 0.01.
16. Step 1:
(2 x 2) (6) 0.02 2 x 4 0.02 0.02 2 x 4 0.02
4.02 2 x 3.98 2.01 x 1.99
x (2) x 2 2 x 2 0.01.
Step 2:
17. Step 1:
x 1 1 0.1 0.1 x 1 1 0.1 0.9 x 1 1.1 0.81 x 1 1.21
Step 2:
0.19 x 0.21
x 0 x . Then, 0.19 0.19 or 0.21; thus, 0.19.
18. Step 1:
Step 2:
19. Step 1:
Step 2:
x 12 0.1 0.1 x 12 0.1 0.4 x 0.6 0.16 x 0.36
x 14 x 14 14 x 14 .
Then 14 0.16 0.09 or 14 0.36 0.11; thus 0.09.
19 x 3 1 1 19 x 3 1 2 19 x 4 4 19 x 16
4 x 19 16 15 x 3 or 3 x 15
x 10 x 10 10 x 10.
Then 10 3 7, or 10 15 5; thus 5.
Copyright 2018 Pearson Education, Inc.
Section 2.3 The Precise Definition of a Limit
x 7 4 1 1 x 7 4 1 3 x 7 5 9 x 7 25 16 x 32
20. Step 1:
Step 2:
21. Step 1:
Step 2:
22. Step 1:
Step 2:
61
x 23 x 23 23 x 23.
Then 23 16 7, or 23 32 9; thus 7.
14 0.05 0.05 1x 14 0.05 0.2 1x 0.3 10
x 10
or 10
x 5.
2
3
3
x 4 x 4 4 x 4.
Then 4 10
or 23 , or 4 5 or 1; thus 23 .
3
1
x
x 2 3 0.1 0.1 x 2 3 0.1 2.9 x 2 3.1 2.9 x 3.1
x 3 x 3 3 x 3.
Then 3 2.9 3 2.9 0.0291, or 3 3.1 3.1 3 0.0286;
thus 0.0286
23. Step 1:
Step 2:
24. Step 1:
Step 2:
25. Step 1:
Step 2:
26. Step 1:
Step 2:
x 2 4 0.5 0.5 x 2 4 0.5 3.5 x 2 4.5 3.5 x 4.5 4.5 x 3.5,
for x near 2.
x (2) x 2 2 x 2.
Then 2 4.5 4.5 2 0.1213, or 2 3.5 2 3.5 0.1292;
thus 4.5 2 0.12.
11 1 9 10 x 10 or 10 x 10 .
( 1) 0.1 0.1 1x 1 0.1 10
10
11
9
9
11
x
x (1) x 1 1 x 1.
1 ; thus 1 .
Then 1 10
19 , or 1 10
11
9
11
11
1
x
( x 2 5) 11 1 x 2 16 1 1 x 2 16 1 15 x 2 17 15 x 17.
x 4 x 4 4 x 4.
Then 4 15 4 15 0.1270, or 4 17 17 4 0.1231; thus
17 4 0.12.
x 1 30 x 20 or 20 x 30.
5 1 1 120
5 1 4 120
6 14 120
x
x
6
x 24 x 24 24 x 24.
Then 24 20 4, or 24 30 6; thus 4.
120
x
27. Step 1:
Step 2:
x 2
mx 2m 0.03 0.03 mx 2m 0.03 0.03 2m mx 0.03 2m 2 0.03
m
x 2 x 2 2 x 2.
Then 2 2 0.03
0.03
, or 2 2 0.03
0.03
. In either case, 0.03
.
m
m
m
m
m
28. Step 1:
Step 2:
mx 3m c c mx 3m c c 3m mx c 3m 3 mc x 3 mc
x 3 x 3 3 x 3.
Then 3 3 mc mc , or 3 3 mc mc . In either case, mc .
29. Step 1:
(mx b) m2 b c c mx m2 c c m2 mx c m2 12 mc x 12 mc .
x 12 x 12 12 x 12 .
Step 2:
Then 12 12 mc
c,
m
or 12
1
2
mc mc . In either case, mc .
Copyright 2018 Pearson Education, Inc.
0.03 .
m
62
Chapter 2 Limits and Continuity
(mx b) (m b) 0.05 0.05 mx m 0.05 0.05 m mx 0.05 m
30. Step 1:
.
1 0.05
x 1 0.05
m
m
x 1 x 1 1 x 1.
Then 1 1 0.05
0.05
, or 1 1 0.05
m
m
m
Step 2:
0.05 . In
m
either case,
0.05 .
m
31. lim (3 2 x) 3 2(3) 3
x 3
(3 2 x) (3) 0.02 0.02 6 2 x 0.02 6.02 2 x 5.98 3.01 x 2.99 or
2.99 x 3.01.
0 x 3 x 3 3 x 3.
Then 3 2.99 0.01, or 3 3.01 0.01; thus 0.01.
Step 1:
Step 2:
32.
lim (3 x 2) (3)(1) 2 1
x 1
Step 1:
Step 2:
(3 x 2) 1 0.03 0.03 3 x 3 0.03 0.01 x 1 0.01 1.01 x 0.99.
x (1) x 1 1 x 1.
Then 1 1.01 0.01, or 1 0.99 0.01; thus 0.01.
x2 4
x 2 x 2
lim
33. lim
Step 1:
Step 2:
34.
x 2 4.05, x 2
1.95 x 2.05, x 2.
x 2 x 2 2 x 2.
Then 2 1.95 0.05, or 2 2.05 0.05; thus 0.05.
2
( x 5)( x 1)
lim x x 65x 5 lim
lim ( x 1) 4, x 5.
( x 5)
x5
x5
x 5
Step 1:
Step 2:
35.
( x 2)( x 2)
lim ( x 2) 2 2 4, x 2
( x 2)
x 2
x 2
x 2 4 4 0.05 0.05 ( x 2)( x 2) 4 0.05 3.95
x 2
( x 2)
( x 5)( x 1)
( x 5)
4 0.05 4.05 x 1 3.95, x 5
5.05 x 4.95, x 5.
x (5) x 5 5 x 5.
Then 5 5.05 0.05, or 5 4.95 0.05; thus 0.05.
1 5 x 4 0.5 0.5 1 5 x 4 0.5 3.5 1 5 x 4.5 12.25 1 5 x 20.25
Step 1:
Step 2:
4
x2 x
(4) 0.05 0.05
1 5 x 1 5(3) 16 4
lim
x 3
36. lim
x 2 6 x 5
x 5
11.25 5 x 19.25 3.85 x 2.25.
x (3) x 3 3 x 3.
Then 3 3.85 0.85, or 3 2.25 0.75; thus 0.75.
24 2
Step 1:
Step 2:
37. Step 1:
Step 2:
2 0.4 0.4 4x 2 0.4 1.6 4x 2.4 10
4x 10
10
x 10
or 53 x 52 .
16
24
4
6
x 2 x 2 2 x 2.
Then 2 53 13 , or 2 52 12 ; thus 13 .
4
x
(9 x) 5 4 x 4 x 4 4 x 4 4 x 4 .
x 4 x 4 4 x 4.
Then 4 4 , or 4 4 . Thus choose .
Copyright 2018 Pearson Education, Inc.
Section 2.3 The Precise Definition of a Limit
38. Step 1:
Step 2:
(3x 7) 2 3x 9 9 3 x 9 3 3 x 3 3 .
x 3 x 3 3 x 3.
Then 3 3 3 3 , or 3 3 3 3 . Thus choose 3 .
x 5 2 x 5 2 2 x 5 2 (2 ) 2 x 5 (2 )2
39. Step 1:
Step 2:
63
(2 ) 2 5 x (2 ) 2 5.
x 9 x 9 9 x 9.
Then 9 2 4 9 4 2 , or 9 2 4 9 4 2 . Thus choose the smaller
distance, 4 2 .
4 x 2 4 x 2 2 4 x 2 (2 ) 2 4 x (2 ) 2
40. Step 1:
Step 2:
41. Step 1:
Step 2:
(2 ) 2 x 4 (2 )2 (2 )2 4 x (2 )2 4.
x 0 x .
Then (2 ) 2 4 2 4 4 2 , or (2 ) 2 4 4 2 . Thus choose the
smaller distance, 4 2 .
For x 1, x 2 1 x 2 1 1 x 2 1 1 x 1
1 x 1 near x 1.
x 1 x 1 1 x 1.
Then 1 1 1 1 , or 1 1 1 1. Choose
min 1 1 , 1 1 , that is, the smaller of the two distances.
42. Step 1:
Step 2:
For x 2, x 2 4 x 2 4 4 x 2 4 4 x 4 4 x
4 near x 2.
x (2) x 2 2 x 2.
Then 2 4 4 2, or 2 4 2 4 . Choose
min 4 2, 2 4 .
43. Step 1:
Step 2:
44. Step 1:
Step 2:
1
x
1 1x 1 1 1x 1 11 x 11 .
x 1 x 1 1 x 1 .
Then 1 11 1 11 1 , or 1 11 11 1 1 .
Choose 1 , the smaller of the two distances.
1
x2
13 12 13 13
x
Choose min
Step 2:
1
x2
13 133 12 133
x
133 x 2 133 133 x 133 , or 133 x 133 for x near 3.
x 3 x 3 3 x 3 .
Then 3
45. Step 1:
3
13
3 133 , or 3 133 133 3.
3 133 , 133 3 .
(6) ( x 3) 6 , x 3 x 3 3 x 3.
x 2 9
x 3
x (3) x 3 3 x 3.
Then 3 3 , or 3 3 . Choose .
Copyright 2018 Pearson Education, Inc.
64
Chapter 2 Limits and Continuity
2 ( x 1) 2 , x 1 1 x 1 .
x 2 1
x 1
46. Step 1:
Step 2:
47. Step 1:
Step 2:
48. Step 1:
Step 2:
x 1 x 1 1 x 1 .
Then 1 1 , or 1 1 . Choose .
x 1: (4 2 x) 2 0 2 2 x since x 1. Thus, 1 2 x 0;
x 1: (6 x 4) 2 0 6 x 6 since x 1. Thus, 1 x 1 6 .
x 1 x 1 1 x 1 .
Then 1 1 2 2 , or 1 1 6 6 . Choose 6 .
x 0: 2 x 0 2 x 0 2 x 0;
x 0: 2x 0 0 x 2.
x 0 x .
Then 2 2 , or 2 2. Choose 2 .
49. By the figure, x x sin 1x x for all x 0 and x x sin 1x x for x 0. Since lim ( x) lim x 0, then by
x 0
the sandwich theorem, in either case, lim x sin 1x 0.
x 0
x 0
50. By the figure, x 2 x 2 sin 1x x 2 for all x except possibly at x 0. Since lim ( x 2 ) lim x 2 0, then by the
x 0
sandwich theorem, lim x 2 sin 1x 0.
x 0
x 0
51. As x approaches the value 0, the values of g ( x) approach k. Thus for every number 0, there exists a 0
such that 0 x 0 g ( x) k .
52. Write x h c. Then 0 x c x c , x c (h c) c , h c c h ,
h 0 0 h 0 .
Thus, lim f ( x) L for any 0, there exists 0 such that f ( x) L whenever 0 x c
x c
f (h c) L whenever 0 h 0 lim f (h c) L.
h 0
53. Let f ( x) x 2 . The function values do get closer to 1 as x approaches 0, but lim f ( x) 0, not 1. The
x 0
function f ( x) x 2 never gets arbitrarily close to 1 for x near 0.
54. Let f ( x) sin x, L 12 , and x0 0. There exists a value of x (namely x 6 ) for which sin x 12 for any
given 0. However, lim sin x 0, not 12 . The wrong statement does not require x to be arbitrarily close to x0 .
x 0
As another example, let g ( x) sin 1x , L 12 , and x0 0. We can choose infinitely many values of x near 0 such
that sin 1x 12 as you can see from the accompanying figure. However, lim sin 1x fails to exist. The wrong
x0
statement does not require all values of x arbitrarily close to x0 0 to lie within 0 of L 12 . Again you can
see from the figure that there are also infinitely many values of x near 0 such that sin 1x 0. If we choose 14
we cannot satisfy the inequality sin 1x 12 for all values of x sufficiently near x0 0.
Copyright 2018 Pearson Education, Inc.
Section 2.3 The Precise Definition of a Limit
55.
A 9 0.01 0.01
2x
2
65
2
πx
9 0.01 8.99 4 9.01 π4 (8.99) x 2 π4 (9.01)
2 8.99
x 2 9.01
or 3.384 x 3.387. To be safe, the left endpoint was rounded up and
the right endpoint was rounded down.
56. V RI VR I
(120)(10)
51
R
V 5
R
(120)(10)
49
5 0.1 4.9 120
5.1 10
R 10
0.1 0.1 120
49 120 51
R
R
23.53 R 24.48.
To be safe, the left endpoint was rounded up and the right endpoint was rounded down.
57. (a) x 1 0 1 x 1 f ( x) x. Then f ( x) 2 x 2 2 x 2 1 1. That is,
f ( x) 2 1 12 no matter how small is taken when 1 x 1 lim f ( x) 2.
x 1
(b) 0 x 1 1 x 1 f ( x) x 1. Then f ( x) 1 ( x 1) 1 x x 1. That is, f ( x) 1 1
no matter how small is taken when 1 x 1 lim f ( x) 1.
x 1
(c) x 1 0 1 x 1 f ( x) x. Then f ( x) 1.5 x 1.5 1.5 x 1.5 1 0.5.
Also, 0 x 1 1 x 1 f ( x) x 1. Then f ( x) 1.5 ( x 1) 1.5 x 0.5
x 0.5 1 0.5 0.5. Thus, no matter how small is taken, there exists a value of x such that
x 1 but f ( x ) 1.5 12 lim f ( x) 1.5.
x 1
58. (a) For 2 x 2 h( x) 2 h( x) 4 2. Thus for 2, h( x) 4 whenever 2 x 2 no matter
how small we choose 0 lim h( x) 4.
x2
(b) For 2 x 2 h( x) 2 h( x) 3 1. Thus for 1, h( x) 3 whenever 2 x 2 no matter
how small we choose 0 lim h( x) 3.
x2
(c) For 2 x 2 h( x) x 2 so h( x ) 2 x 2 2 . No matter how small 0 is chosen, x 2 is close to 4
when x is near 2 and to the left on the real line x 2 2 will be close to 2. Thus if 1, h( x) 2
whenever 2 x 2 no matter how small we choose 0 lim h( x) 2.
x2
59. (a) For 3 x 3 f ( x) 4.8 f ( x) 4 0.8. Thus for 0.8, f ( x) 4 whenever 3 x 3 no
matter how small we choose 0 lim f ( x) 4.
x 3
(b) For 3 x 3 f ( x) 3 f ( x) 4.8 1.8. Thus for 1.8, f ( x) 4.8 whenever 3 x 3
no matter how small we choose 0 lim f ( x) 4.8.
x 3
(c) For 3 x 3 f ( x) 4.8 f ( x) 3 1.8. Again, for 1.8, f ( x) 3 whenever 3 x 3 no
matter how small we choose 0 lim f ( x) 3.
x 3
Copyright 2018 Pearson Education, Inc.
66
Chapter 2 Limits and Continuity
60. (a) No matter how small we choose 0, for x near 1 satisfying 1 x 1 , the values of g ( x) are
near 1 g ( x) 2 is near 1. Then, for 12 we have g ( x) 2 12 for some x satisfying 1 x 1 ,
or 0 x 1 lim g ( x) 2.
x 1
(b) Yes, lim g ( x ) 1 because from the graph we can find a 0 such that g ( x) 1 if 0 x (1) .
x 1
Example CAS commands (values of del may vary for a specified eps):
61–66.
Maple:
f : x - (x^4-81)/(x-3); x0 : 3;
plot( f (x), x x0-1..x0 1, color black,
title "Section 2.3, #61(a)" );
.
L : limit( f (x), x x0 );
# (a)
# (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: L eps; y2: L eps; x0 1;
f[x _]: (3x 2 (7x 1)Sqrt[x] 5)/(x 1)
Plot[f [x], {x, x0 0.2, x0 0.2}]
L: Limit[f [x], x x0]
eps 0.1; del 0.2;
Plot[{f [x], y1, y2}, {x, x0 del, x0 del}, PlotRange {L 2eps, L 2eps}]
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
Copyright 2018 Pearson Education, Inc.
Section 2.4 One-Sided Limits
3. (a)
67
lim f ( x) 22 1 2, lim f ( x) 3 2 1
x 2
x 2
(b) No, lim f ( x) does not exist because lim f ( x) lim f ( x)
x 2
(c)
4
2
4
2
x 2
x 2
lim f ( x ) 1 3, lim f ( x) 1 3
x 4
x 4
(d) Yes, lim f ( x) 3 because 3 lim f ( x) lim f ( x)
x 4
x 4
4. (a)
lim f ( x)
x 2
2
2
x 4
1, lim f ( x) 3 2 1, f (2) 2
x 2
(b) Yes, lim f ( x) 1 because 1 lim f ( x) lim f ( x )
x 2
x 2
(c)
lim
x 1
f ( x) 3 (1) 4, lim
x 1
x 2
f ( x) 3 (1) 4
(d) Yes, lim f ( x ) 4 because 4 lim
5. (a) No, lim f ( x) does not exist since sin
(b)
f ( x) lim
x 1
x 1
x 0
x 1
f ( x)
1x does not approach any single value as x approaches 0
lim f ( x) lim 0 0
x 0
x 0
(c) lim f ( x) does not exist because lim f ( x) does not exist
x 0
x 0
6. (a) Yes, lim g ( x) 0 by the sandwich theorem since x g ( x) x when x 0
x 0
(b) No, lim g ( x) does not exist since x is not defined for x 0
x 0
(c) No, lim g ( x) does not exist since lim g ( x ) does not exist
x 0
7. (a)
x 0
(b)
lim f ( x) 1 lim f ( x )
x 1
x 1
(c) Yes, lim f ( x) 1 since the right-hand and left-hand
x 1
limits exist and equal 1
8. (a)
(b)
lim f ( x) 0 lim f ( x )
x 1
x 1
(c) Yes, lim f ( x) 0 since the right-hand and left-hand
x 1
limits exist and equal 0
Copyright 2018 Pearson Education, Inc.
68
Chapter 2 Limits and Continuity
9. (a) domain: 0 x 2
range: 0 y 1 and y 2
(b) lim f ( x) exists for c belonging to (0, 1) (1, 2)
x c
(c) x 2
(d) x 0
10. (a) domain: x
range: 1 y 1
(b) lim f ( x) exists for c belonging to
x c
(, 1) (1, 1) (1, )
(c) none
(d) none
11.
13.
14.
15.
lim
x0.5
lim
x 2
lim
x 1
lim
h 0
x 2
x 1
0.5 2
0.51
3/2
1/2
2
h 0
h2 4h 5 5
h
lim
h 0
6 5h 2 11h 6
h
h 2 4 h 5 5
h
h 0 h
lim ( x 3)
x 2
lim ( x 3)
x 2
h ( h 4)
h2 4h 5 5
lim
h 0
(b)
11
1 2
0 0
x11 xx 6 37 x 111 116 371 12 71 72 1
lim
17. (a)
x 1
x2
lim
x 1
2
h 0 h
lim
12.
xx1 2x x5x 221 (2(2)2)(52) (2) 12 1
lim
16.
3
x2
x2
6 (5h 2 11h 6)
2
6 5h 11h 6
h 4h 5
0 4
5 5
h
( h 2 4 h 5) 5
h2 4h 5 5
lim
h 0 h
( x 2)
h (5h 11)
2
6 5h 11h 6
(0 11)
6 6
(|x 2| ( x 2) for x 2)
lim ( x 3) ((2) 3) 1
x2
x2
2
5
6 5h2 11h 6
6 5h2 11h 6
lim ( x 3) ( x 2)
x 2
lim
5 h 0
2
6 5h 2 11h 6
h
h 2 4h 5 5
x 2
( x 2)
(|x 2| ( x 2) for x 2)
lim ( x 3) ( x 2)
x 2
lim ( x 3)(1) (2 3) 1
x 2
Copyright 2018 Pearson Education, Inc.
11
2 6
Section 2.4 One-Sided Limits
18. (a)
2 x ( x 1)
x 1
lim
x 1
(b)
lim
2 x ( x 1)
( x 1)
lim
2x 2
x 1
x 1
2 x ( x 1)
x 1
lim
x 1
lim
x 1
(|x 1| x 1 for x 1)
2 x ( x 1)
( x 1)
(|x 1| ( x 1) for x 1)
lim 2 x 2
x 1
19. (a) If 0 x 2 , then sin x 0, so that lim
x 0
(b) If 2 x 0, then sin x 0, so that lim
x 0
sin x
sin x
sin x
sin x
x lim 1 1
lim sin
sin x
x 0
x 0
lim
x 0
sin x
sin x
lim 1 1
x 0
20. (a) If 0 x 2 , then cos x 1, so that lim
(b) If 2 x 0, then cos x 1,
lim 33 1
3
21. (a)
lim sin 2 lim sinx x 1
2
sin kt
t
25. lim
sin 3 y
4y
y 0
26.
lim
h 0
tan 2 x
x 0 x
2t
t 0 tan t
28. lim
1 lim 3sin 3 y
4 y 0 3 y
lim
h 0
lim
x 0
2 lim
x csc 2 x
x 0 cos 5 x
29. lim
(where x 2 )
x 0
h
sin 3h
27. lim
t 0
3 lim sin 3 y
4 y 0 3 y
13 sin3h3h 13 hlim
0
sin 2 x
cos
2x
lim
x
t
sin t
cos
t
sin 2 x
x 0 x cos 2 x
x 1
x 0 sin 2 x cos 5 x
lim
3 lim sin
4 0
1
sin 3h
3h
(where kt )
3
4
1
13
lim sin
0
(where 3y )
1 1 1
3
3
(where 3h)
2 x 1 2 2
lim cos12 x lim 2sin
x 0
x 0 2 x
cos t 2 lim cos t
2 lim tsin
t
t 0
t 0
1sin t
lim t
t 0
2 1 1 2
2 x lim
1 1 1 (1) 1
12 xlim
2
2
0 sin 2 x x 0 cos 5 x
6 x 2 cos x
x sin 2 x
sin
x 0
30. lim 6 x 2 (cot x)(csc 2 x) lim
x 0
t 4
kt lim k sin k lim sin k 1 k
lim k sin
0
0
t 0 kt
24. lim
t 0
lim (t t ) 4 3 1
(b)
t 4
0
lim 23
3
(b)
lim (t t ) 4 4 0
22. (a)
23.
1cos x lim 1cos x lim 1cos x lim 1 1
(cos x 1)
1 cos x
x 0 cos x 1 x 0
x 0
x 0
cos x 1 lim 1 1
so that lim cos x 1 lim (cos
x 1)
x 0 cos x 1 x 0
x 0
lim 3cos x sinx x sin2 x2 x 3 1 1 3
x 0
Copyright 2018 Pearson Education, Inc.
69
70
Chapter 2 Limits and Continuity
x x cos x
x 0 sin x cos x
31. lim
x
x 0 sin x cos x
lim
x
sinx cos
lim
x cos x
lim sin1 x lim
x 0 x x 0
x 2 x sin x
2x
x 0
1cos
0 sin 2
33. lim
x
x 0 2
lim
32. lim
x 1
x 0 sin x cos x
cos1 x xlim
0
1
sin x
x
x
xlim
0 sin x
(1)(1) 1 2
sinx x 0 12 12 (1) 0
12 12
(1cos )(1 cos )
0 (2sin cos )(1 cos )
lim
1cos 2
(2sin
cos )(1 cos )
0
lim
sin 2
)(1 cos )
(2sin
cos
0
lim
sin
0 0
0 (2 cos )(1 cos ) (2)(2)
lim
34.
lim x x 2cos x
x 0 sin 3 x
lim
sin(1cos t )
1cos t
36. lim
sin(sin h )
sin h
h 0
sin 2 3 x
x 0
34. lim
t 0
x (1cos x )
lim
x (1 cos x )
9 x2
lim
sin 2 3 x
9 x2
x 0
x 0
1 cos x
9x
sin3 x3 x
2
1 lim 1 cos x
9 x0
x
2
sin 3 x
lim 3 x
x 0
1 (0)
9
2
1
0
lim sin 1 since 1 cos t 0 as t 0
0
lim sin 1 since sin h 0 as h 0
0
sin lim sin 2 1 lim sin 2
12 1 1 12
2 0
sin 2
0 sin 2 0 sin 2 2
37. lim
sin 5 x
x 0 sin 4 x
38. lim
sin 5 x 4 x 5
x 0 sin 4 x 5 x 4
lim
54 xlim
sin5 x5x sin4 x4 x 54 11 54
0
39. lim cos 0 1 0
0
cos 2
sin 2
40. lim sin cot 2 lim sin
0
0
tan 3 x
x 0 sin 8 x
41. lim
42. lim
y 0
cos 2
2sin cos
cos 2 1
2
0 2 cos
lim
xlim
sin 3x 1 8 x 3
0 cos 3 x sin 8 x 3 x 8
83 lim cos13 x sin3 x3 x sin8 x8 x 83 1 1 1 83
x0
sin 3 y cot 5 y
y cot 4 y
lim
y 0
tan
0 cot 3
lim
0
1 1 1 1
sin 3 y sin 4 y cos 5 y
y cos 4 y sin 5 y
sin
cos
2 cos 3
sin 3
lim
sin 3 y
3y
y 0
2
0
sin 3 x 1
x 0 cos 3 x sin 8 x
lim
lim
43. lim
lim sin
sin 4 y
4y
lim
y 0
5y
sin 5 y
sin 3 y
y
sin 4 y
cos 4 y
cos 5 y
cos 4 y
34
5
sin sin 3 lim sin
2
0 cos cos 3 0
cos 5 y
sin 5 y
345 y
345 y
12
5
12
5
sin33 cos 3cos 3 (1)(1) 113 3
4
2
cos
cos 4 (4sin 2 cos 2 )
cos 4 (2sin cos )2
cot 4 lim
sin 4
lim 2cos 42sin 2 lim
lim
2
2
2
2
2
2
2
0 sin cos 2 sin 4
0 sin cot 2 0 sin 2 cos 2 0 sin cos 2 sin 4 0 sin cos 2 sin 4
2
44. lim
sin 2
1
cos 4 cos2
lim sin44
lim sin 4
2
0 cos 2 sin 4
0
cos 2 0 4
lim
4 cos 4 cos 2
2
Copyright 2018 Pearson Education, Inc.
111 1
cos 4 cos2 1
cos2 2 1
2
2
Section 2.4 One-Sided Limits
45.
71
x lim 1cos3 x 1 cos3 x lim 1cos 2 3 x lim
sin 2 3 x
lim 1cos3
lim 3 sin 3 x sin 3 x
2
2
1
cos3
2
(1
cos3
)
2
(1
cos3 x ) x 0 2 3 x 1 cos3 x
x
x
x
x
x
x
x 0
x 0
x 0
x 0
3 (1) 0 0 (where 3 x )
lim 32 sin 1sin
cos
2
11
0
46.
2
cos x (cos x 1)
cos x (cos x 1) cos x 1
cos x (cos 2 x 1)
cos x( sin 2 x )
lim cos x 2 cos x lim
lim
cos x 1 lim 2
lim 2
2
2
x
x 0
x 0
x
x 0
cos x
lim sinx x sinx x cos
x 1
x 0
x
x (cos x 1)
x 0
(1)(1) 111
x 0
x (cos x 1)
1
2
47. Yes. If lim f ( x) L lim f ( x), then lim f ( x) L. If lim f ( x ) lim f ( x), then lim f ( x) does not
exist.
x a
xa
x a
x a
x a
x a
48. Since lim f ( x) L if and only if lim f ( x) L and lim f ( x) L, then lim f ( x) can be found by
x c
x c
calculating lim f ( x).
x c
x c
x c
49. If f is an odd function of x, then f ( x) f ( x). Given lim f ( x) 3, then lim f ( x) 3.
x 0
x 0
50. If f is an even function of x, then f ( x) f ( x). Given lim f ( x) 7 then lim
can be said about lim
x 2
x 2
x 2
f ( x ) because we don’t know lim f ( x).
f ( x) 7. However, nothing
x 2
51. I (5, 5 ) 5 x 5 . Also, x 5 x 5 2 x 5 2 . Choose 2 lim
x 5 0.
52. I (4 , 4) 4 x 4. Also, 4 x 4 x 2 x 4 2 . Choose 2 lim
4 x 0.
x 5
x 4
53. As x 0 the number x is always negative. Thus,
x
x
(1)
x
x
1 0 which is always true
x
x 0 x
independent of the value of x. Hence we can choose any 0 with x 0 lim
54. Since x 2 we have x 2 and x 2 x 2. Then,
x2
x2
1
x2
x2
1 0 which is always true so
long as x 2. Hence we can choose any 0, and thus 2 x 2
55. (a)
(b)
lim
x 400
1.
x2
x2
1 . Thus, lim
x 2
x2
x2
1.
x 400. Just observe that if 400 x 401, then x 400. Thus if we choose 1, we have for
any number 0 that 400 x 400 x 400 400 400 0 .
lim x 399. Just observe that if 399 x 400 then x 399. Thus if we choose 1, we have for
x 400
any number 0 that 400 x 400 x 399 399 399 0 .
(c) Since lim x lim x we conclude that lim x does not exist.
x 400
56. (a)
x 400
lim f ( x) lim
x 0
x 0
x 0 0;
x 400
x 0 x 0 x 2 for x positive. Choose 2
lim f ( x) 0.
x 0
Copyright 2018 Pearson Education, Inc.
72
Chapter 2 Limits and Continuity
(b)
lim f ( x) lim x 2 sin 1x 0 by the sandwich theorem since x 2 x 2 sin 1x x 2 for all x 0.
x 0
x 0
Since x 2 0 x 2 0 x 2 whenever x , we choose and obtain x 2 sin 1x 0
if x 0.
(c) The function f has limit 0 at x0 0 since both the right-hand and left-hand limits exist and equal 0.
2.5
CONTINUITY
1. No, discontinuous at x 2, not defined at x 2
2. No, discontinuous at x 3, 1 lim g ( x) g (3) 1.5
x 3
3. Continuous on [1, 3]
4. No, discontinuous at x 1, 1.5 lim k ( x) lim k ( x) 0
x 1
x 1
5. (a) Yes
(b) Yes, lim
(c) Yes
(d) Yes
6. (a) Yes, f (1) 1
x 1
f ( x) 0
(b) Yes, lim f ( x) 2
(c) No
(d) No
7. (a) No
(b) No
x 1
8. [1, 0) (0, 1) (1, 2) (2, 3)
9. f (2) 0, since lim f ( x) 2(2) 4 0 lim f ( x)
x 2
x 2
10. f (1) should be changed to 2 lim f ( x )
x 1
11. Nonremovable discontinuity at x 1 because lim f ( x) fails to exist ( lim f ( x) 1 and lim f ( x) 0).
x 1
x 1
x 1
Removable discontinuity at x 0 by assigning the number lim f ( x) 0 to be the value of f (0) rather
x 0
than f (0) 1.
12. Nonremovable discontinuity at x 1 because lim f ( x) fails to exist ( lim f ( x) 2 and lim f ( x ) 1).
x 1
x 1
x 1
Removable discontinuity at x 2 by assigning the number lim f ( x) 1 to be the value of f (2) rather than
x 2
f (2) 2.
13. Discontinuous only when x 2 0 x 2
14. Discontinuous only when ( x 2)2 0 x 2
15. Discontinuous only when x 2 4 x 3 0 ( x 3)( x 1) 0 x 3 or x 1
16. Discontinuous only when x 2 3 x 10 0 ( x 5)( x 2) 0 x 5 or x 2
17. Continuous everywhere. (|x 1| sin x defined for all x; limits exist and are equal to function values.)
Copyright 2018 Pearson Education, Inc.
Section 2.5 Continuity
73
18. Continuous everywhere. (|x| 1 0 for all x; limits exist and are equal to function values.)
19. Discontinuous only at x 0
20. Discontinuous at odd integer multiples of 2 , i.e., x (2n 1) 2 , n an integer, but continuous at all other x.
21. Discontinuous when 2x is an integer multiple of , i.e., 2 x n , n an integer x n2 , n an integer, but
continuous at all other x.
22. Discontinuous when 2x is an odd integer multiple of 2 , i.e., 2x (2n 1) 2 , n an integer x 2n 1, n an
integer (i.e., x is an odd integer). Continuous everywhere else.
23. Discontinuous at odd integer multiples of 2 , i.e., x (2n 1) 2 , n an integer, but continuous at all other x.
24. Continuous everywhere since x 4 1 1 and 1 sin x 1 0 sin 2 x 1 1 sin 2 x 1; limits exist and are
equal to the function values.
25. Discontinuous when 2 x 3 0 or x 32 continuous on the interval 32 , .
26. Discontinuous when 3 x 1 0 or x 13 continuous on the interval 13 , .
27. Continuous everywhere: (2 x 1)1/3 is defined for all x; limits exist and are equal to function values.
28. Continuous everywhere: (2 x)1/5 is defined for all x; limits exist and are equal to function values.
2
( x 3)( x 2)
29. Continuous everywhere since lim x xx36 lim
lim ( x 2) 5 g (3)
x 3
x 3
x 3
x 3
30. Discontinuous at x 2 since lim f ( x) does not exist while f (2) 4.
x 2
31. lim sin( x sin x) sin( sin ) sin( 0) sin 0, and function continuous at x .
x
32. lim sin( 2 cos(tan t )) sin( 2 cos(tan(0))) sin 2 cos(0) sin 2 1, and function continuous at t 0.
t 0
33. lim sec ( y sec2 y tan 2 y 1) lim sec ( y sec2 y sec2 y ) lim sec (( y 1) sec2 y ) sec ((1 1) sec 2 1)
y 1
y 1
y 1
sec 0 1, and function continuous at y 1.
34. lim tan 4 cos(sin x1/3 ) tan 4 cos(sin(0)) tan 4 cos(0) tan 4 1, and function continuous at x 0.
x 0
35. lim cos
t 0
36. lim
x 6
19 3 sec 2t
cos
cos cos 4 22 , and function continuous at t 0.
16
193 sec 0
csc2 x 5 3 tan x csc2 6 5 3 tan 6 4 5 3 1
at x 6 .
3
9 3, and function continuous
Copyright 2018 Pearson Education, Inc.
74
Chapter 2 Limits and Continuity
2
( x 3)( x 3)
37. g ( x) xx 39 ( x 3) x 3, x 3 g (3) lim ( x 3) 6
x 3
38. h(t )
39.
t 2 3t 10
t 2
(t 5)(t 2)
t 2
t 5, t 2 h(2) lim (t 5) 7
t 2
2
2
3
2
( s s 1)( s 1)
f ( s ) s2 1 ( s 1)( s 1) s ss11 , s 1 f (1) lim s ss11 32
s 1
s 1
2
( x 4)( x 4)
40. g ( x) 2x 16 ( x 4)( x 1) xx14 , x 4 g (4) lim
x 3 x 4
x 4
xx14 85
41. As defined, lim f ( x) (3)2 1 8 and lim (2a )(3) 6a. For f ( x) to be continuous we must have
6a 8 a
x 3
43 .
x 3
42. As defined, lim g ( x) 2 and lim g ( x ) b(2)2 4b. For g ( x) to be continuous we must have
x 2
x 2
4b 2 b 12 .
43. As defined, lim f ( x) 12 and lim f ( x ) a 2 (2) 2a 2a 2 2a. For f ( x) to be continuous we must have
2
x 2
x 2
12 2a 2a a 3 or a 2.
44. As defined, lim g ( x) 0bb1 bb1 and lim g ( x) (0) 2 b b. For g ( x) to be continuous we must have
x 0
x 0
b
b 1
b b 0 or b 2.
45. As defined, lim
x 1
f ( x) 2 and lim
x 1
f ( x) a (1) b a b, and lim f ( x) a (1) b a b and
x 1
lim f ( x) 3. For f ( x) to be continuous we must have 2 a b and a b 3 a 52 and b 12 .
x 1
46. As defined, lim g ( x) a (0) 2b 2b and lim g ( x) (0)2 3a b 3a b, and lim g ( x) (2)2 3a b
x 0
x 0
x 2
4 3a b and lim g ( x) 3(2) 5 1. For g ( x) to be continuous we must have 2b 3a b and 4 3a b 1
x 0
a 32 and b 32 .
47. f ( x) is continuous on [0, 1] and f (0) 0, f (1) 0
by the Intermediate Value Theorem f ( x) takes on
every value between f (0) and f (1) the equation
f ( x) 0 has at least one solution between x 0
and x 1.
48. cos x x (cos x) x 0. If x 2 , cos 2 2 0. If x 2 , cos 2 2 0. Thus cos x x 0 for
some x between 2 and 2 according to the Intermediate Value Theorem, since the function cos x x is
continuous.
49. Let f ( x) x3 15 x 1, which is continuous on [4, 4]. Then f (4) 3, f (1) 15, f (1) 13, and f (4) 5.
By the Intermediate Value Theorem, f ( x) 0 for some x in each of the intervals 4 x 1, 1 x 1, and
Copyright 2018 Pearson Education, Inc.
Section 2.5 Continuity
75
1 x 4. That is, x3 15 x 1 0 has three solutions in [4, 4]. Since a polynomial of degree 3 can have at
most 3 solutions, these are the only solutions.
50. Without loss of generality, assume that a b. Then F ( x) ( x a )2 ( x b)2 x is continuous for all values of x,
so it is continuous on the interval [a, b]. Moreover F (a ) a and F (b) b. By the Intermediate Value Theorem,
since a a 2 b b, there is a number c between a and b such that F ( x) a 2 b .
51. Answers may vary. Note that f is continuous for every value of x.
(a) f (0) 10, f (1) 13 8(1) 10 3. Since 3 10, by the Intermediate Value Theorem, there exists a c so
that 0 c 1 and f (c) .
(b) f (0) 10, f (4) (4)3 8(4) 10 22. Since 22 3 10, by the Intermediate Value Theorem,
there exists a c so that 4 c 0 and f (c) 3.
(c) f (0) 10, f (1000) (1000)3 8(1000) 10 999,992, 010. Since 10 5, 000, 000 999,992, 010, by the
Intermediate Value Theorem, there exists a c so that 0 c 1000 and f (c) 5, 000, 000.
52. All five statements ask for the same information because of the intermediate value property of continuous
functions.
(a) A root of f ( x) x3 3 x 1 is a point c where f (c) 0.
(b) The point where y x3 crosses y 3x 1 have the same y-coordinate, or y x3 3 x 1 f ( x)
x3 3 x 1 0.
(c) x3 3x 1 x3 3 x 1 0. The solutions to the equation are the roots of f ( x ) x3 3 x 1.
(d) The points where y x3 3 x crosses y 1 have common y-coordinates, or y x3 3 x 1 f ( x)
x3 3 x 1 0.
(e) The solutions of x3 3x 1 0 are those points where f ( x ) x3 3 x 1 has value 0.
sin( x 2)
53. Answers may vary. For example, f ( x) x 2 is discontinuous at x 2 because it is not defined there.
However, the discontinuity can be removed because f has a limit (namely 1) as x 2.
54. Answers may vary. For example, g ( x) x11 has a discontinuity at x 1 because lim g ( x) does not exist.
lim g ( x) and lim g ( x) .
x 1
x 1
x 1
55. (a) Suppose x0 is rational f ( x0 ) 1. Choose 12 . For any 0 there is an irrational number x (actually
infinitely many) in the interval ( x0 , x0 ) f ( x) 0. Then 0 |x x0 | but | f ( x) f ( x0 )|
1 12 , so lim f ( x) fails to exist f is discontinuous at x0 rational.
x x0
On the other hand, x0 irrational f ( x0 ) 0 and there is a rational number x in ( x0 , x0 ) f ( x) 1.
Again lim f ( x) fails to exist f is discontinuous at x0 irrational. That is, f is discontinuous at every point.
x x0
(b) f is neither right-continuous nor left-continuous at any point x0 because in every interval ( x0 , x0 ) or
( x0 , x0 ) there exist both rational and irrational real numbers. Thus neither limits lim f ( x) and
x x0
lim f ( x) exist by the same arguments used in part (a).
x x0
f ( x)
56. Yes. Both f ( x) x and g ( x) x 12 are continuous on [0, 1]. However g ( x ) is undefined at x 12 since
f ( x)
g 12 0 g ( x ) is discontinuous at x 12 .
57. No. For instance, if f ( x) 0, g ( x ) x , then h( x) 0 x 0 is continuous at x 0 and g ( x) is not.
Copyright 2018 Pearson Education, Inc.
76
Chapter 2 Limits and Continuity
58. Let f ( x ) x11 and g ( x) x 1. Both functions are continuous at x 0. The composition f g f ( g ( x))
1
1 is discontinuous at x 0, since it is not defined there. Theorem 10 requires that f ( x ) be continuous
( x 1) 1 x
at g (0), which is not the case here since g (0) 1 and f is undefined at 1.
59. 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 [a, b].
60. Let f ( x ) be the new position of point x and let d ( x) 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) 0 for some point in between. That is, f ( x) x for some point x,
which is then in its original position.
61. If f (0) 0 or f (1) 1, we are done (i.e., c 0 or c 1 in those cases). Then let f (0) a 0 and f (1) b 1
because 0 f ( x) 1. Define g ( x) f ( x ) x g is continuous on [0, 1]. Moreover, g (0) f (0) 0 a 0 and
g (1) f (1) 1 b 1 0 by the Intermediate Value Theorem there is a number c in (0, 1) such that
g (c) 0 f (c) c 0 or f (c) c.
f (c )
62. Let 2 0. Since f is continuous at x c there is a 0 such that x c f ( x) f (c)
f (c) f ( x) f (c) .
If f (c) 0, then 12 f (c ) 12 f (c) f ( x) 32 f (c) f ( x) 0 on the interval (c , c ).
If f (c) 0, then 12 f (c) 32 f (c) f ( x) 12 f (c) f ( x) 0 on the interval (c , c ).
63.
By Exercise 52 in Section 2.3, we have lim f ( x ) L lim f (c h ) L.
x c
h0
Thus, f ( x ) is continuous at x c lim f ( x ) f ( c ) lim f ( c h ) f ( c ).
x c
h 0
64. By Exercise 63, it suffices to show that lim sin(c h) sin c and lim cos(c h) cos c.
h 0
h 0
Now lim sin(c h) lim (sin c)(cos h) (cos c)(sin h) (sin c) lim cos h (cos c ) lim sin h .
h 0
h 0
h 0
h 0
By Example 11 Section 2.2, lim cos h 1 and lim sin h 0. So lim sin(c h) sin c and thus f ( x) sin x is
h 0
continuous at x c. Similarly,
h 0
h 0
lim cos(c h) lim (cos c)(cos h) (sin c)(sin h) (cos c) lim cos h (sin c) lim sin h cos c. Thus,
h 0
h 0
h 0
g ( x) cos x is continuous at x c.
h 0
65. x 1.8794, 1.5321, 0.3473
66. x 1.4516, 0.8547, 0.4030
Copyright 2018 Pearson Education, Inc.
Section 2.6 Limits Involving Infinity; Asymptotes of Graphs
67. x 1.7549
68. x 3.5156
69. x 0.7391
70. x 1.8955, 0, 1.8955
2.6
LIMITS INVOLVING INFINITY; ASYMPTOTES OF GRAPHS
1. (a) lim f ( x) 0
(c)
(e)
x 2
(b)
f ( x) 2
(d)
lim f ( x) 1
(f)
lim
x 3
x 0
(g) lim f ( x) does not exist
(i)
(h)
x 0
lim f ( x) 0
(e)
(g)
(i)
f ( x) 2
lim f ( x ) does not exist
x3
lim f ( x)
x 0
lim f ( x) 1
x
x
2. (a) lim f ( x) 2
(c)
lim
x 3
(b)
x 4
lim f ( x) 1
x 2
lim
x 3
lim f ( x ) 3
x 2
(d) lim f ( x) does not exist
f ( x)
(f)
lim f ( x)
(h)
lim f ( x)
(j)
x 3
x 0
(k) lim f ( x) 0
(l)
x
1
m/ n
x x
Note: In these exercises we use the result lim
Theorem 8 and the power rule in Theorem 1: lim
x 2
lim
x 3
f ( x)
lim f ( x)
x 0
lim f ( x) does not exist
x 0
lim f ( x) 1
x
0 whenever mn 0. This result follows immediately from
1
m/n
x x
1
x x
lim
3. (a) 3
(b) 3
4. (a)
(b)
5. (a) 12
(b) 12
6. (a) 18
(b) 18
7. (a) 53
(b) 53
8. (a) 34
(b) 34
m/ n
lim 1x
x
9. 1x sinx2 x 1x lim sinx2 x 0 by the Sandwich Theorem
x
1 lim cos 0 by the Sandwich Theorem
10. 31 cos
3
3
3
11.
2 t sin t
t t cos t
lim
lim
t
010 1
10
2
1 sint t
t
1 cost t
Copyright 2018 Pearson Education, Inc.
m/ n
0m / n 0.
77
78
Chapter 2 Limits and Continuity
r sin r
r 2 r 7 5sin r
12. lim
sinr r lim 10 1
7
sin r
2
r 2 r 5 r r 2 0 0
13. (a)
2 x 3
x 5 x 7
14. (a)
2 x3 7
3
2
x x x x 7
2 3x
lim
lim
1
lim
7
x 5 x
2
5
2 73
x
7
1 1
x 1 x 2 3
lim
lim
(b) 2 (same process as part (a))
15. (a)
16. (a)
17. (a)
lim x21
x x 3
lim
3x 7
2
x x 2
lim
1 1
x x2
x 1
lim
9 x4 x
4
2
x 2 x 5 x x 6
19. (a)
10 x5 x 4 31
x6
x
20. (a)
lim
lim
x
x 3 7 x 2 2
x 2 x 1
22. (a)
(b)
23.
24.
7
2
lim 3 x 35 x 1
x 6 x 7 x 3
8
3
lim 5 x 2 x 5 9
x 3 x 4 x
x
lim
2
lim x 2x 1
8
x 3
x
x
1/3
(b) 7 (same process as part (a))
1
x3
2 52 13 64
x
x
x
10 1 31
x x 2 x6
1
3 x 4 5 x 1 x 3
2
3
x 67 x 3 x
5 x 3 2 x 2 9 x 5
3 x 5 x 4 4
8
lim
x
1 1x 12
x
lim
3
x 8 2
x
, since x n 0, 5x 3 , and the denominator 4.
3
x2
2 1x
1/3
(b)
, since x n 0 and 3x 4 .
5 x 3 2 x 2 9 x 5
5
4
x 3 x x 4
3
(same process as part (a))
(b) 0 (same process as part (a))
lim
x2
2 1x
9
2
, since x n 0 and 3x 4 .
lim
8
(b)
, since x n 0 and x 7 .
3 x 4 5 x 1 x 3
2
3
x 67 x 3 x
x
9
2
, since x n 0 and x 7 .
lim
lim
0
x 7 2 x 1
1
2
1
x x x
5 x8 2 x 3 9
5
x 3 x 4 x
8 x 2 3
2 x2 x
x
9
lim
3 x 7 5 x 2 1
3
x 6 x 7 x 3
lim
lim
7
9
x2
x 7 2 x 1
1
2
x 1 x x
lim
lim
(b) 0 (same process as part (a))
lim
x 3 7 x 2 2
2
x x x 1
(b)
0
x
lim
21. (a)
2
x2
lim
lim
x
(b) 0 (same process as part (a))
7
3
x 1 x
18. (a)
2
0
lim
lim
x
(same process as part (a))
x2
3 7
x x2
x 1
7 x3
3
x
3
x2 6 x
x
3
2
5
(b)
, since x n 0, 5x 3 , and the denominator 4.
8 0
2 0
42
1 1x 12
x
lim
x 8 32
x
1/3
18000
1/3
18
Copyright 2018 Pearson Education, Inc.
1/3
1
2
Section 2.6 Limits Involving Infinity; Asymptotes of Graphs
25.
x 2 5 x
x3 x 2
x
27.
29.
26. lim
1
lim 2 x x
x 3 x 7
3
lim
x 5 x
x5 x
3
x
x 1 x 4
2
3
x x x
30. lim
5
1 5
x x2
lim
1
x
(1/5) (1/3)
lim 1 x (1/5) (1/3)
1
x
x
x2
1 1x
x
lim
32.
3
x 5 x 3
2/3
x 2 x x 4
lim
33.
lim
34.
35.
36.
37.
39.
41.
x
x
x 2 1
x 1
lim
x 3
2
x
4 3 x 3
6
x
x 9
lim
1
3x
lim
3
x 2
x 0
x 2
lim
x 8
lim
x 2 1/ x 2
x
1 21
x /15
lim
x 1 1
x 2 /15
lim 2 x
x 2 x
2 1
1/ 2
lim x
x 2 1
x1/ 2
1
1
2
52
lim
x
lim
2
( x 3)/ x 2
lim
4 x 2 25 / x 2
(4 3 x3 )/ x 6
x 9 / x
( x 2 1)/ x 2
( x 1)/ x
11/ x 2
(1
x 1/ x )
6
x
lim
1 0
(1 0)
11/ x 2
(
x 11/ x )
lim
( x 3)/ x
(4 x 2 25)/ x 2
(43 x3 )/( x3 )
6
lim
( x 2 1)/ x 2
(
x
1)/( x )
x
x 2 1/ x 2
6
6
lim
lim
1
1 0
( 1 0)
(13/ x )
x
4 25/ x 2
( 4/ x3 3)
1 9/ x6
38.
lim
5
2x
lim
1
x 3
( x 9)/ x
positive
positive
positive
negative
40.
negative
positive
42.
positive
positive
44. lim
x 0
x 3
lim
x 5
1
(1 0)
4 0
x
x
4
2
x 7 ( x 7)
43. lim
1 4
x1/3 x
x
2x
x 8
x 2
x 2/3
x ( x 1)/ x
4 x 25
lim
5 3x
lim
10000 0 0
28.
1 7
x19/15 x8/5
3
1
1 3/5
11/10
x
x
1
lim
1 2
x 2 x3
0
2 x1/15
x ( x 1)/ x
lim
x 1
x
lim
1 5
x x2
lim
1
x
lim
31.
x 2 1
x 1
1 2
x 2 x3
2 1
1/ 2 2
lim x 7 x
3 x
x
5/3
1/3
lim 2 x8/5 x 7
x
3
x
x
x
lim
5
1 x
12 x
5
2
lim x 7 lim x 7 01
0
x 1 x
x 1 x
5
3
lim 12 x
x x 7 x
79
(0 3)
1 0
1
2
3
positive
negative
positive
positive
negative
negative
negative
positivepositive
3x
2 x 10
1
2
x 0 x ( x1)
Copyright 2018 Pearson Education, Inc.
80
Chapter 2 Limits and Continuity
45. (a)
lim 21/3
x 0 3 x
46. (a)
2
lim 1/5
x 0 x
4
2/5
x 0 x
4
1/5 2
x 0 ( x )
2
52.
(b)
lim 21/3
x 0 3 x
(b)
2
lim 1/5
x 0 x
1
2/3
x 0 x
48. lim
lim tan x
x
51.
lim
47. lim
49.
50.
lim
2
x
1
1/3 2
x 0 ( x )
lim
sec x
lim (1 csc )
0
lim (2 cot ) and lim (2 cot ) , so the limit does not exist
0
0
53. (a)
lim 21
x 2 x 4
lim
1
( x 2)( x 2)
(b)
lim 21
x 2 x 4
lim
1
( x 2)( x 2)
(c)
lim 21
x 2 x 4
lim
1
( x 2)( x 2)
(d)
lim 21
x 2 x 4
lim
1
( x 2)( x 2)
54. (a)
(b)
(c)
(d)
55. (a)
(b)
(c)
(d)
56. (a)
(c)
(d)
x
2
x 1 x 1
lim 2x
x 1 x 1
lim
lim
lim
lim
x 0
x 0
3
x 2
x2
2
x2
2
x2
2
x2
x 1 2
lim
x 2
x 2
x 2
lim
x
( x 1)( x 1)
lim
x
( x 1)( x 1)
x 1
x 1
lim 2x
x 1 x 1
lim 2x
x 1 x 1
lim
x 2
lim
x 1
lim
x 1
x 2 1
2 x4
1
x
positive
positivenegative
x
( x 1)( x 1)
negative
negativenegative
x 0
22/3
2
1
2
1
negativenegative
positive
positivepositive
0 lim
2
2
1
x
1
positivenegative
negative
positivenegative
x 0
1
x
1
negative
1
x
1
positive
1
21/3
1
positivenegative
x
( x 1)( x 1)
1x 0 lim
1
x
1
positivepositive
1/3
1/3
0
3
2
1
1
x 2
2
( x 1)( x 1)
lim 2xx 14 lim
2 x4
x 1
x 1
2
x
1
1
lim 2 x 4 4
x 0
positive
positive
20
2 4
(b)
lim
x 2
x 2 1
2 x4
0
Copyright 2018 Pearson Education, Inc.
positive
negative
Section 2.6 Limits Involving Infinity; Asymptotes of Graphs
57.
x 2 3 x 2
x3 2 x 2
lim
( x 2)( x1)
lim
x 2 3 x 2
x3 2 x 2
lim
( x 2)( x 1)
lim
x 2 3 x 2
x3 2 x 2
lim
( x 2)( x 1)
(a) lim
x 0
(b)
(c)
x 2
x 2
lim
x 2 3 x 2
3
2
x 0 x 2 x
lim
(c)
(d)
(e)
lim
x 2
lim
x 2
x 2 ( x 2)
x 2
(e) lim
(b)
x 2 ( x 2)
x2
( x 2)( x 1)
( x 2)( x 1)
x 2 ( x 2)
x 0
x 2 3 x 2
x3 4 x
x2
x 2 3 x 2
x3 4 x
lim
x 2
x 2 3 x 2
x3 4 x
lim
x 2 3 x 2
3
x 1 x 4 x
lim
lim
x 0
lim
lim
x 0
x 1
x ( x 2)
and lim
x 0
x 0
x 1
x 1
2
x 2 x
14 , x 2
( x 2)( x 1)
x ( x 2)( x 2)
( x 2)( x 1)
x ( x 2)( x 2)
14 , x 2
x2
lim
x 2
x 0
lim
x 1
( x 1)
x ( x 2)
negative
positivepositive
negative
negativepositive
1
2(4)
( x 1)
x ( x 2)
( x 1)
x ( x 2)
lim
negativenegative
positivenegative
( x 1)
x ( x 2)
lim
( x 2)( x 1)
x ( x 2)( x 2)
x 1
x ( x 2)
x 1
2
x 2 x
lim
( x 2)( x 1)
x ( x 2)( x 2)
lim
14 , x 2
lim
negativenegative
positivenegative
x 1
2
x 2 x
lim
x 2 ( x 2)
x 2
x 2 3 x 2
3
2
x 2 x 2 x
(d) lim
58. (a)
x 0
x 2 ( x 2)
81
0
(1)(3)
negative
negativepositive
negative
negativepositive
0
so the function has no limit as x 0.
59. (a)
60. (a)
3
lim 2 1/3
t
t 0
1 7
lim 3/5
t
t 0
1
lim 2/3
2 2/3
( x 1)
x
1
(c) lim 2/3
2 2/3
x
x
(
1)
x 1
61. (a)
x 0
1
lim 1/3
1 4/3
( x 1)
x
1
(c) lim 1/3
1 4/3
x
x
(
1)
x 1
62. (a)
63. y
x 0
1
x 1
(b)
(b)
3
lim 2 1/3
t
t 0
1 7
lim 3/5
t
t 0
1
lim 2/3
2 2/3
( x 1)
x
1
(d) lim 2/3
2 2/3
x
x
(
1)
x 1
(b)
x 0
1
lim 1/3
1 4/3
( x 1)
x
1
(d) lim 1/3
1 4/3
x
x
(
1)
x 1
(b)
x 0
64. y
1
x 1
Copyright 2018 Pearson Education, Inc.
81
82
Chapter 2 Limits and Continuity
65. y
1
2 x4
67. y
x 3
x2
1
1
x2
69. domain (, ); y in range and y 4
3 x2 ,
x 2 1
0
66. y
3
x 3
68. y
2x
x 1
3x2 3
x 2 1
3x2
2
x x 1
and lim
asymptote is y 7 .
70. domain (, 1) (1, 1) (1, ); y in range and y
lim 22 x
x 1 x 1
2x
2
x 1 x 1
, and lim
;
lim 22 x
x 1 x 1
2x ;
x 2 1
2x
2
x 1 x 1
asymptote is y 0; vertical asymptotes are x 1, x 1
x2 4
;
x
lim
x0
x2 4
x
, lim
x
x2 4
x
1, and lim
x
x2 4
x
3 range [4, 7) ; horizontal
if x 0, then y 0,
and lim
71. domain (, 0) (0, ); y in the range and y
2 x21
y
0,
range (, ); horizontal
4
x x2 4
2
lim 22 x
x x 1
0;
lim
x0
x2 4
x
,
1 range (, 1) (1, ); horizontal
asymptotes are y 1, y 1; vertical asymptote is x 0
72. domain (, 2) (2, ); y in the range and y
x3
x x 8
and lim
3
x3 ;
x3 8
y
24 x 2
x 8
3
2
x3
3
8
x
x 2
0; lim
,
lim
x 2
x3
x3 8
,
1 range (, 1) (1, ); horizontal asymptote is y 1; vertical asymptote is x 2
Copyright 2018 Pearson Education, Inc.
Section 2.6 Limits Involving Infinity; Asymptotes of Graphs
73. Here is one possibility.
74. Here is one possibility.
75. Here is one possibility.
76. Here is one possibility.
77. Here is one possibility.
78. Here is one possibility.
79. Here is one possibility.
80. Here is one possibility.
f ( x)
x g ( x )
81. Yes. If lim
f ( x)
x g ( x )
2 then the ratio the polynomials’ leading coefficients is 2, so lim
2 as well.
82. Yes, it can have a horizontal or oblique asymptote.
f ( x)
x g ( x )
83. At most 1 horizontal asymptote: If lim
f ( x)
g
x ( x )
so lim
L, then the ratio of the polynomials’ leading coefficients is L,
L as well.
Copyright 2018 Pearson Education, Inc.
83
84
84.
Chapter 2 Limits and Continuity
x
x 9 x 4 lim x 9 x 4
x
lim
5
x 9 x 4
lim
x
lim
x 9 x 4
x 9 x 4
5
x
85. lim x 2 25 x 2 1 lim x 2 25 x 2 1
x
x
26
lim
86.
lim
x 2 25 x 2 1
x
lim x 2 3 x lim x 2 3 x
x
x
x
x2
x 1
87.
3
x2
x
x2
2
x 25 x
x
2
1 252 1
x
3x
x 1
3
x2
1
3 x 2
x 2 x 4 x 2 3 x 2
lim
x 2
3 2x
4 3x 22
x
lim 9 x 2 x 3 x lim 9 x 2 x 3 x
x
x
lim
x
xx
9 x2
x
x2 x2
3xx
lim
x
x
30
2 2
9x
2x
x2
90. lim
x
x 3 x x 2 x
1
3 3
x
x 2 x x 2 x lim x 2 x x 2 x
x
2
2 1
lim
11
1
1
x
3
lim
x 2 3 x
x
4 x 2 3 x 2
4x
2
3 x 2
x2
lim
3 x 2 x
4 3x
lim
2
x
x2
(4 x 2 ) (4 x 2 3 x 2)
2 x 4 x2 3 x 2
3 x 2
x
2x
4
3x 22
x
x
43
89. lim x 2 3 x x 2 2 x lim x 2 3x x 2 2 x
x
x
5x
lim
lim
2
2
x
x 2 3 x
lim
x 3 x x
1
9 1x 3
101 0
1
x2
9 x 2 x 3 x
2
x 2 25 x 2 1
101 0
lim
( x 2 25) ( x 2 1)
lim
1 x
( x 2 3) ( x 2 )
lim
3 x x
lim
2
26
x
lim 2 x 4 x 2 3 x 2 lim 2 x 4 x 2 3x 2 2 x
2 x
x
x
lim
88.
x 2 25 x 2 1
x 2 3 x
3
lim
101 0
1 9x 1 4x
x
( x 9) ( x 4)
x x 9 x 4
lim
(9 x 2 x ) (9 x 2 )
9 x 2 x 3 x
x
x
9 x 2 x 3 x
16
x2 3 x x2 2 x
2
x 3 x
5
1 3x 1 2x
lim
x 2 x x
151 52
x x x
2
( x 2 3 x ) ( x 2 2 x )
2
x2 x x2 x
2
lim
lim
x x
x 2 3 x x 2 2 x
( x 2 x ) ( x 2 x )
x2 x x2 x
lim
x
2x
x2 x x2 x
1 x 1 x
91. For any 0, take N 1. Then for all x N we have that f ( x) k k k 0 .
92. For any 0, take N 1. Then for all y N we have that f ( x) k k k 0 .
93. For every real number B 0, we must find a 0 such that for all x, 0 x 0
Now,
1
x2
B 0
1
x2
2
B0 x
21 B so that lim 12 .
x
x0
1
B
x
1 .
B
Choose
1
B
1
x2
B.
, then 0 x x
x
Copyright 2018 Pearson Education, Inc.
1
B
Section 2.6 Limits Involving Infinity; Asymptotes of Graphs
94. For every real number B 0, we must find a 0 such that for all x, 0 x 0
1
x
1 . Choose
B
B0 x
1 . Then 0
B
1
B
x0 x
1
x
1
x
2
( x 3) 2
( x 3)2
2
B 0 2 B1 ( x 3)2
( x 3) 2
2 B 0 so that lim 2 .
2
( x 3) 2
x 3 ( x 3)
B 0
0 x 3
2
B
0
B. Now,
1
x0 x
B so that lim
95. For every real number B 0, we must find a 0 such that for all x, 0 x 3
X
2
( x 3) 2
96. For every real number B 0, we must find a 0 such that for all x, 0 x (5)
Now,
1
( x 5)2
x5
B 0 ( x 5)2
1
B
1
( x 5)2
1
B
x5
1
B
1
2
x 5 ( x 5)
.
B so that lim
. Choose
.
2 . Choose
B
3
1 . Then 0
B
85
B. Now,
2,
B
1
( x 5) 2
B.
then
x (5)
97. (a) We say that f ( x) approaches infinity as x approaches c from the left, and write lim f ( x ) ,
xc
if for every positive number B, there exists a corresponding number 0 such that for all x,
c x c f ( x ) B.
(b) We say that f ( x) approaches minus infinity as x approaches c from the right, and write lim f ( x ) , if
xc
for every positive number B (or negative number B) there exists a corresponding number 0 such that
for all x, c x c f ( x ) B.
(c) We say that f ( x) approaches minus infinity as x approaches c from the left, and write lim f ( x ) , if
x c
for every positive number B (or negative number B) there exists a corresponding number 0 such that
for all x, c x c f ( x ) B.
98. For B 0,
1
x
B0 x
1 . Choose
B
1 . Then 0
B
99. For B 0, 1x B 0 1x B 0 x
1x B so that lim 1x .
1
B
x 0 x
B1 x. Choose
1
B
1
x
B so that lim
1 . Then
B
x 0
1
x
.
x 0 B1 x
x 0
100. For B 0, x 1 2 B x 1 2 B ( x 2) B1 x 2 B1 x 2 B1 . Choose B1 .
Then 2 x 2 x 2 0 B1 x 2 0 x 1 2 B 0 so that lim x 1 2 .
x 2
101. For B 0,
1
x2
1
x 2
B 0 x2
1 . Choose
B
1 .
x 2
x 2
1, 1 2 B 1 x 2
1 x
1 . Then
B
2 x 2 0 x 2 0 x 2
1
B
B 0 so that lim
102. For B 0 and 0 x
1
B
(1 x)(1 x)
Then 1 x 1 x 1 0 1 x
x near 1
lim 1 2
x 1 1 x
.
1
2B
1.
B
Now 12x 1 since x 1. Choose
(1 x)(1 x)
B1 11x
1 1 x
B
2
Copyright 2018 Pearson Education, Inc.
2
1 .
2B
B for 0 x 1 and
86
Chapter 2 Limits and Continuity
103. y
x2
x 1
105. y
x2 4
x 1
107. y
x 2 1
x
109. y
104. y
x 2 1
x 1
x 1 x21
x 1 x31
106. y
x 2 1
2 x4
12 x 1 2 x3 4
x 1x
108. y
x3 1
x2
x
x 1 x11
x
4 x 2
110. y
1
x2
1
4 x 2
Copyright 2018 Pearson Education, Inc.
Chapter 2 Practice Exercises
111. y x 2/3
1
x1/3
112. y sin
x 2 1
113. (a) y (see accompanying graph)
(b) y (see accompanying graph)
(c) cusps at x 1 (see accompanying graph)
114. (a) y 0 and a cusp at x 0 (see the
accompanying graph)
(b) y 32 (see accompanying graph)
(c) a vertical asymptote at x 1 and contains the
point 1,
CHAPTER 2
3
23 4
(see accompanying graph)
PRACTICE EXERCISES
1. At x 1:
lim f ( x) lim f ( x) 1
x 1
x 1
lim f ( x) 1 f (1)
x 1
f is continuous at x 1.
At x 0 :
lim f ( x) lim f ( x) 0
x 0
lim f ( x) 0.
x 0
x 0
But f (0) 1 lim f ( x)
x 0
f is discontinuous at x 0.
If we define f (0) 0, then the discontinuity at
x 0 is removable.
At x 1:
lim f ( x) 1 and lim f ( x) 1
x 1
x 1
lim f ( x ) does not exist
x 1
f is discontinuous at x 1.
Copyright 2018 Pearson Education, Inc.
87
88
Chapter 2 Limits and Continuity
2. At x 1:
lim f ( x) 0 and lim f ( x) 1
x 1
lim f ( x) does not exist
x 1
x 1
f is discontinuous at x 1.
At x 0 :
lim f ( x ) and lim f ( x)
x 0
x 0
lim f ( x) does not exist
x 0
f is discontinuous at x 0.
At x 1:
lim f ( x) lim f ( x) 1 lim f ( x) 1.
x 1
x 1
x 1
But f (1) 0 lim f ( x)
x 1
f is discontinuous at x 1.
If we define f (1) 1, then the discontinuity at
x 1 is removable.
3. (a) lim (3 f (t )) 3 lim f (t ) 3( 7) 21
t t0
t t0
2
(b) lim ( f (t )) 2 lim f (t ) (7) 2 49
t t0
t t0
(c) lim ( f (t ) g (t )) lim f (t ) lim g (t ) ( 7)(0) 0
t t0
f (t )
g
(
t ) 7
t t0
(d) lim
t t0
lim f (t )
t t0
lim ( g (t ) 7)
t t0
t t0
lim f (t )
t t0
lim g (t ) lim 7
t t0
t t0
7
07
1
(e) lim cos ( g (t )) cos lim g (t ) cos 0 1
t t0
t t0
(f)
lim | f (t )| lim f (t ) | 7| 7
t t0
t t0
(g) lim ( f (t ) g (t )) lim f (t ) lim g (t ) 7 0 7
t t0
(h) lim
t t0
1
f (t )
t t0
1
lim f (t )
t t0
t t0
1
7
17
4. (a) lim g ( x) lim g ( x) 2
x 0
x 0
(b) lim ( g ( x) f ( x)) lim g ( x) lim f ( x)
x 0
x 0
x 0
2 12
2
2
(c) lim ( f ( x) g ( x)) lim f ( x) lim g ( x) 12 2
x 0
x 0
(d) lim f 1( x ) lim 1f ( x ) 11 2
x 0
2
x 0
x 0
(e) lim ( x f ( x)) lim x lim f ( x) 0 12 12
x 0
x 0
x 0
(f)
lim
x 0
f ( x )cos x
x 1
lim f ( x ) lim cos x
x 0
x 0
lim x lim 1
x0
x 0
12 (1) 1
0 1
2
5. Since lim x 0 we must have that lim (4 g ( x)) 0. Otherwise, if lim (4 g ( x)) is a finite positive
x 0
number, we would have lim
x 0
x0
4
g
(
x
)
x
x 0
and lim
x 0
4 g ( x )
x
so the limit could not equal 1 as x 0.
Similar reasoning holds if lim (4 g ( x)) is a finite negative number. We conclude that lim g ( x) 4.
x 0
x 0
Copyright 2018 Pearson Education, Inc.
Chapter 2 Practice Exercises
89
6. 2 lim x lim g ( x) lim x lim lim g ( x ) 4 lim lim g ( x) 4 lim g ( x) (since lim g ( x) is a
x0
x 4 x 0
x 4 x 0
x 0
x 4 x 4 x 0
constant) lim g ( x ) 24 12 .
x 0
7. (a) lim f ( x) lim x1/3 c1/3 f (c) for every real number c f is continuous on ( , ).
x c
x c
x c
x c
x c
x c
1
c 2/3
x c
x c
1
c1/ 6
(b) lim g ( x) lim x3/4 c3/4 g (c) for every nonnegative real number c g is continuous on [0, ).
(c) lim h( x) lim x 2/3
(, ).
(d) lim k ( x) lim x 1/6
8. (a)
h(c) for every nonzero real number c h is continuous on (, 0) and
k (c ) for every positive real number c k is continuous on (0, )
n 12 , n 12 , where I the set of all integers.
n I
(b)
(n , (n 1) ), where I the set of all integers.
n I
(c) (, ) ( , )
(d) (, 0) (0, )
x2 4 x 4
( x 2)( x 2)
lim
lim x 2 , x 2; the limit does not exist because
2
x 0 x 5 x 14 x x 0 x ( x 7)( x 2) x 0 x ( x 7)
lim x (xx27) and lim x (xx27)
x 0
x 0
2
x 4 x4
( x 2)( x 2)
0 0
lim 3 2
lim
lim x 2 , x 2, and lim x (xx27) 2(9)
x 2 x 5 x 14 x x 2 x ( x 7)( x 2) x 2 x ( x 7)
x 2
9. (a) lim
(b)
3
x ( x 1)
x2 x
lim 3 2
lim 2 x 1
lim 2 1 , x 0 and x 1.
4
3
x 0 x 2 x x
x 0 x ( x 2 x 1) x 0 x ( x 1)( x 1) x 0 x ( x 1)
2
Now lim 2 1 and lim 2 1 lim 5 x 4x 3 .
x ( x 1)
x ( x 1)
x 0 x 2 x x
x 0
x 0
2
x ( x 1)
x
x
1
lim 5 4 3 lim 3 2
lim 2
, x 0 and x 1. The limit does not
x 1 x 2 x x
x 1 x ( x 2 x 1) x 1 x ( x 1)
1
1
lim 2
and lim 2
.
x 1 x ( x 1)
x 1 x ( x 1)
10. (a) lim
(b)
1 x
x 1 1 x
11. lim
5
1 x
x 1 (1 x )(1 x )
x2 a2
4
4
x a x a
2
2
2
x a ( x a )( x a )
2
( x h)2 x 2
h
h 0
lim
( x h)2 x 2
h
x 0
lim
13. lim
14. lim
lim
x 0
1 1
2 x 2
x
x1/3 1
x 1 x 1
17. lim
1
2
2
xa x a
lim
1
2a 2
lim (2 x h) 2 x
( x 2 2 hx h 2 ) x 2
h
x 0
lim (2 x h) h
2 (2 x )
x 0 2 x (2 x )
h 0
x 0
1
x 0 4 2 x
lim
( x3 6 x 2 12 x 8) 8
x
x 0
lim
14
lim ( x 2 6 x 12) 12
x 0
( x1/3 1) ( x 2/3 x1/3 1)( x 1)
2/3
1/3
x 1 ( x 1) ( x 1)( x x 1)
lim
1
2
( x 2 2 hx h 2 ) x 2
h
h 0
lim
(2 x )3 8
x
x 0
16. lim
lim
( x2 a2 )
lim
12. lim
15.
1
x 1 1 x
lim
exist because
lim
( x 1)( x 1)
2/3
1/3
x 1 ( x 1)( x x 1)
x 1
2/3
1/3
x 1 x x 1
lim
Copyright 2018 Pearson Education, Inc.
11111
2
3
90
18.
Chapter 2 Limits and Continuity
x 2/3 16
x 8
x 64
sin 2 x cos x
cos
2 x sin x
x 0
tan 2 x
tan
x
x 0
lim
19. lim
20.
( x1/3 4)( x1/3 4)
( x1/3 4)( x1/3 4) ( x 2/3 4 x1/3 16)( x 8)
lim
x 8
x 8
( x 8)( x 2/3 4 x1/3 16)
x 64
x 64
1/3
1/3
( x 64) ( x 4) ( x 8)
( x 4) ( x 8)
(4 4) (88)
lim
lim 2/3 1/3
161616 83
2/3
1/3
(
x
64)
(
x
4
x
16)
x
4
x
16
x 64
x 64
lim
lim
1
sin x
lim csc x lim
x
x
21. lim sin
x
lim
x 0
cos x
x
2 x 1 1 1 2 2
sin2 x2 x cos
2 x sin x x
2x sin x sin 2 sin sin 2 1
22. lim cos 2 ( x tan x) cos 2 ( tan ) cos 2 ( ) (1)2 1
x
8x
x 0 3sin x x
8
sin x
x 0 3 x 1
lim
23. lim
24. lim
x 0
cos 2 x 1
sin x
8
3(1) 1
4
cos 2 x 1 cos 2 x 1
sin x
cos 2 x 1
x 0
4sin x cos 2 x
4(0)(1) 2
lim
11 0
x 0 cos 2 x 1
lim
1/3
25.
lim [4 g ( x)]1/3 2 lim 4 g ( x)
x 0
x 0
26.
1
lim
x 5 x g ( x)
3 x 2 1
x 1 g ( x )
27. lim
28.
sin 2 x
cos 2 x 1 lim
xlim
0 sin x (cos 2 x 1) x 0 sin x (cos 2 x 1)
2
2
2 lim 4 g ( x) 8, since 23 8. Then lim g ( x ) 2.
x 0
2 lim ( x g ( x)) 12 5 lim g ( x)
x 5
x 5
x 0
1
2
lim g ( x) 12 5
x 5
lim g ( x) 0 since lim (3 x 2 1) 4
x 1
5 x 2
x 2 g ( x )
x 1
0 lim g ( x) since lim (5 x 2 ) 1
lim
x 2
x 2
29.(a) f (1) 1 and f (2) 5 f has a root between 1 and 2 by the Intermediate Value Theorem.
(b), (c) root is 1.32471795724
30. (a) f (2) 2 and f (0) 2 f has a root between 2 and 0 by the Intermediate Value Theorem.
(b), (c) root is -1.76929235424
31. At x 1:
lim
lim
x 1
f ( x) lim
x 1
x ( x 2 1)
x 1
lim
x 2 1
x ( x 2 1)
x 1
| x 2 1|
lim x 1, and
x 1
f ( x) lim
x ( x 2 1)
x 1
2
| x 1|
lim
2
x 1 ( x 1)
lim ( x) (1) 1. Since lim
x 1
lim
x 1
x ( x 2 1)
x 1
f ( x)
f ( x) lim f ( x) does not exist, the
x1
function f cannot be extended to a continuous
function at x 1.
Copyright 2018 Pearson Education, Inc.
Chapter 2 Practice Exercises
At x 1:
lim f ( x) lim
x 1
x 1
lim
x 1
x ( x 2 1)
2
| x 1|
x ( x 2 1)
x 2 1
lim
x ( x 2 1)
2
x 1 ( x 1)
lim ( x) 1, and lim f ( x) lim
x 1
x 1
x 1
x ( x 2 1)
| x 2 1|
lim x 1.
x 1
Again lim f ( x) does not exist so f cannot be extended to a continuous function at x 1 either.
x 1
32. The discontinuity at x 0 of f ( x) sin
lim sin 1x does not exist.
1x is nonremovable because x
0
33. Yes, f does have a continuous extension at a 1:
define f (1) lim x 41 43 .
x 1 x x
34. Yes, g does have a continuous extension at a 2 :
5 cos
g 2 lim 4 2 54 .
2
35. From the graph we see that lim h(t ) lim h(t )
t 0
t 0
so h cannot be extended to a continuous function
at a 0.
36. From the graph we see that lim k ( x) lim k ( x)
x 0
x 0
so k cannot be extended to a continuous function at
a 0.
Copyright 2018 Pearson Education, Inc.
91
92
Chapter 2 Limits and Continuity
2 x 3
x 5 x 7
37. lim
2 3x
lim
7
x 5 x
39.
x 2 4 x 8
3
x 3 x
40.
lim 2 1
x x 7 x 1
2 0
5 0
1
x 3 x
lim
lim
2
5
2 x 2 3
2
x 5 x 7
38.
4
3x2
8
3 x3
lim
3
2
20
50
x 1
128
x 12 3
lim
x
x2
5 72
x
2
5
000 0
1
41.
43.
44.
x2
7 1
x 1 x 2
x
100 0 0
lim
2
lim x x71 x lim x 17
x
x 1 x
lim sin x
x x
x 4 x3
3
x 12 x 128
42.
lim
lim
x
lim 1 0 since x as x lim sin x 0.
x
x
x
x
lim cos 1 lim 2 0 lim cos 1 0.
sin x
2
x
45.
1 x
x2 x
lim x sin
lim
sin x
x sin x
x
x 1 x
46.
x 2/3 x 1 lim 1 x 5/3 1 0 1
lim 2/3
2
2
x x cos x x 1 cos x 1 0
x2/3
47. (a) y
(b) y
x2 4
x 3
1100 0 1
x2 4
x 3
is undefined at x 3 : lim
x 3
2
and lim
x 3
x 1 x 2 x 1
x 1 x 2 x 1
asymptote.
2
(c) y x2 x 6 is undefined at x 2 and 4: lim
x 2 x 8
2
lim x2 x 6
x 2 x 8
x 4
48. (a)
, thus x 3 is a vertical asymptote.
2
undefined at x 1: lim x2 x 2 and lim x2 x 2 , thus x 1 is a vertical
x x 2 is
x 2 2 x 1
2
x2 4
x 3
x 2 x 6
2
x
x 2 2 x 8
x 3
x2 x 4
lim
x 2 x 6
2
x
2 x 8
x 4
56 ; lim
lim xx34 . Thus x 4 is a vertical asymptote.
x 4
2
1 x 2
x 1 x x 2 1
y 12 x : lim
1
x2
lim
x 1
1
1
x2
1 x 2
2
x x 1
1
1
1 and lim
1
lim
x2
x 1
1
1
x2
1
1
lim
x 4
x 3
x4
1, thus y 1 is a
horizontal asymptote.
(b)
y x 4 :
lim
x4
(c) y
x2 4
x
x
:
lim
x
x 4
x4
lim
x2 4
x
1
4
x
1 0 1, thus y 1 is a horizontal asymptote.
1 0
x 1 4x
lim
x
lim
x
4
1
11 0 1 and lim
x2
1
x
1
4
x2
x
x
lim
x
1
4
x2
1
x2 4
x
lim
1
x
4
x2
x
x2
110 11 1,
thus y 1 and y 1 are horizontal asymptotes.
(d) y
x 2 9 :
9 x 2 1
lim
x
x 2 9
9 x 2 1
lim
x
1
9
thus y 13 is a horizontal asymptote.
9
x2
1
x2
1 0
9 0
13 and lim
x
x 2 9
9 x 2 1
Copyright 2018 Pearson Education, Inc.
lim
x
1
9
x2
9 12
x
1 0
9 0
13 ,
Chapter 2 Additional and Advanced Exercises
49. domain [4, 2) (2, 4]; y in range and y
lim
x 2
16 x 2
x2
x b
, if x 4, then y 0,
lim
x2
16 x 2
x 2
, and
, range (, )
ax 2 4
x b
50. Since lim
16 x 2
x 2
93
2
x
4
vertical asymptote is x b; lim ax
lim x b a 42
x
b
x
x
x
2
4
lim x x b a 42 a horizontal asymptote is y a , lim ax
x
b
x
x
x
x
lim x b a 42 lim xxb a 42 a horizontal asymptote is y a
x
x
x
x
CHAPTER 2
1.
ADDITIONAL AND ADVANCED EXERCISES
lim L lim L0 1
v c
v c
v2
c2
L0 1
lim v 2
v c
2
c
2
L0 1 c 2 0
c
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).
2. (a)
x
2
1 0.2 0.2 2x 1 0.2 0.8 2x 1.2 1.6 x 2.4 2.56 x 5.76.
(b)
x
2
1 0.1 0.1 2x 1 0.1 0.9 2x 1.1 1.8 x 2.2 3.24 x 4.84.
3. |10 (t 70) 104 10| 0.0005 |(t 70) 104 | 0.0005 0.0005 (t 70) 104 0.0005
5 t 70 5 65 t 75 Within 5 F.
4. We want to know in what interval to hold values of h to make V satisfy the inequality
|V 1000| |36 h 1000| 10. To find out, we solve the inequality:
990 h 1010 8.8 h 8.9
|36 h 1000| 10 10 36 h 1000 10 990 36 h 1010 36
36
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 8.9 8.8 0.1 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.
5. Show lim f ( x) lim ( x 2 7) 6 f (1).
x 1
x 1
Step 1: |( x 2 7) 6| x 2 1 1 x 2 1 1 x 1 .
Step 2: | x 1| x 1 1 x 1.
Then 1 1 or 1 1 . Choose min 1 1 , 1 1 , then 0 | x 1|
2
|( x 7) 6| and lim f ( x) 6. By the continuity text, f ( x) is continuous at x 1.
x 1
1
x 14 2 x
6. Show lim g ( x) lim
x 14
1: 21x
2 g 14 .
2 21x 2 2 21x 2 412 x 412 .
Step 2: X 14 x 14 14 x 14 .
Step
Then 14
Choose
1 1 1
, or
4 2
4 4 2
4(2 )
, the smaller of the two values.
4(2 )
14 412 412 14 4(2 ) .
Then 0 x 14 21x 2 and lim 21x 2.
x 1
By the continuity test, g ( x) is continuous at x 14 .
Copyright 2018 Pearson Education, Inc.
4
94
Chapter 2 Limits and Continuity
7. Show lim h( x) lim 2 x 3 1 h(2).
x 2
Step 1:
x2
2 x 3 1 2 x 3 1 1 2 x 3 1
Step 2: | x 2| x 2 or 2 x 2.
(1 )2 3
2
x
(1 )2 3
.
2
2
(1 ) 2 3
(1 ) 2 3 1(1 )2
(1 )2 3
(1 ) 2 3
(1 )2 1
2
2 , or 2
2
2
2
2
2
2
2
2
2
2 . Choose 2 , the smaller of the two values. Then, 0 | x 2| 2 x 3 1 ,
Then 2
so lim 2 x 3 1. By the continuity test, h( x) is continuous at x 2.
x2
8. Show lim F ( x) lim 9 x 2 F (5).
x 5
Step 1:
x 5
9 x 2 9 x 2 9 (2 )2 x 9 (2 ) 2 .
Step 2: 0 | x 5| x 5 5 x 5.
Then 5 9 (2 )2 (2 )2 4 2 2, or 5 9 (2 ) 2 4 (2 ) 2 2 2.
Choose 2 2, the smaller of the two values. Then, 0 | x 5| 9 x 2 , so lim 9 x 2.
x 5
By the continuity test, F ( x) is continuous at x 5.
9. Suppose L1 and L2 are two different limits. Without loss of generality assume L2 L1. Let 13 ( L2 L1 ). Since
lim f ( x) L1 there is a 1 0 such that 0 | x x0 | 1 | f ( x) L1 | f ( x) L1
x x0
13 ( L2 L1 ) L1 f ( x) 13 ( L2 L1 ) L1 4 L1 L2 3 f ( x) 2 L1 L2 . Likewise, lim f ( x) L2 so
x x0
there is a 2 such that 0 | x x0 | 2 | f ( x) L2 | f ( x ) L2
13 ( L2 L1 ) L2 f ( x) 13 ( L2 L1 ) L2 2 L2 L1 3 f ( x) 4 L2 L1 L1 4 L2 3 f ( x) 2 L2 L1.
If min{1 , 2 } both inequalities must hold for 0 | x x0 | :
4 L1 L2 3 f ( x) 2 L1 L2
5( L1 L2 ) 0 L1 L2 . That is, L1 L2 0 and L1 L2 0, a
L1 4 L2 3 f ( x) 2 L2 L1
contradiction.
10. Suppose lim f ( x) L. If k 0, then lim k f ( x) lim 0 0 0 lim f ( x) and we are done. If k 0, then given
x c
x c
x c
x c
any 0, there is a 0 so that 0 | x c | | f ( x) L | | k | | k || f ( x ) L | | k ( f ( x) L)|
|(kf ( x)) (kL)| . Thus lim k f ( x) kL k lim f ( x) .
x c
x c
11. (a) Since x 0 , 0 x3 x 1 ( x3 x) 0 lim f ( x3 x) lim f ( y ) B where y x3 x.
x 0
y 0
(b) Since x 0 , 1 x x 0 ( x x) 0 lim f ( x x) lim f ( y ) A where y x3 x.
3
3
3
x 0
y 0
(c) Since x 0 , 0 x 4 x 2 1 ( x 2 x 4 ) 0 lim f ( x 2 x 4 ) lim f ( y ) A where y x 2 x 4 .
x 0
4
y 0
2
(d) Since x 0 , 1 x 0 0 x x 1 ( x x ) 0 lim f ( x x 4 ) A as in part (c).
4
2
2
x 0
12. (a) True, because if lim ( f ( x) g ( x)) exists then lim ( f ( x) g ( x)) lim f ( x) lim [( f ( x) g ( x)) f ( x)]
x a
lim g ( x) exists, contrary to assumption.
x a
(b) False; for example take f ( x)
x a
x a
and g ( x) 1x . Then neither lim f ( x) nor lim g ( x) exists, but
0 0 exists.
1 1 xlim
x 0 x x
0
lim ( f ( x) g ( x)) lim
x 0
1
x
x a
x 0
Copyright 2018 Pearson Education, Inc.
x 0
Chapter 2 Additional and Advanced Exercises
95
(c) True, because g ( x) | x | is continuous g ( f ( x)) | f ( x)| is continuous (it is the composite of
continuous functions).
1, x 0
(d) False; for example let f ( x)
f ( x) is discontinuous at x 0. However | f ( x)| 1 is
1, x 0
continuous at x 0.
x 2 1
x 1 x 1
13. Show lim f ( x ) lim
x 1
( x 1)( x 1)
x 1 ( x 1)
lim
2, x 1.
x 2 1 , x 1
. We now prove the limit of f ( x) as x 1
Define the continuous extension of f ( x) as F ( x) x 1
2 , x 1
exists and has the correct value.
Step 1:
x 2 1 ( 2)
x 1
( x 1)( x 1)
( x 1)
2 ( x 1) 2 , x 1 1 x 1.
Step 2: | x (1)| x 1 1 x 1.
Then 1 1 , or 1 1 . Choose . Then 0 | x (1)|
2
xx 11 (2) lim F ( x) 2. Since the conditions of the continuity test are met by F ( x), then f ( x) has
x 1
a continuous extension to F ( x) at x 1.
x 2 2 x 3
x 3 2 x 6
14. Show lim g ( x) lim
x 3
lim
x 3
( x 3)( x 1)
2( x 3)
2, x 3.
x 2 2 x 3 , x 3
Define the continuous extension of g ( x) as G ( x) 2 x 6
. We now prove the limit of g ( x) as x 3
,
x3
2
exists and has the correct value.
Step 1:
x 2 2 x 3 2
2 x 6
( x 3)( x 1)
2( x 3)
2
x 1 2 ,
2
x 3 3 2 x 3 2.
Step 2: | x 3| x 3 3 x 3.
Then, 3 3 2 2, or 3 3 2 2. Choose 2. Then 0 | x 3|
x 2 2 x 3 2
2 x 6
( x 3)( x 1)
x 3 2( x 3)
lim
2. Since the conditions of the continuity test hold for G ( x), g ( x) can be
continuously extended to G ( x ) at x 3.
15. (a) Let 0 be given. If x is rational, then f ( x) x | f ( x) 0| | x 0| | x 0| ; i.e., choose .
Then | x 0| | f ( x) 0| for x rational. If x is irrational, then f ( x) 0 | f ( x) 0| 0
which is true no matter how close irrational x is to 0, so again we can choose . In either case, given
0 there is a 0 such that 0 | x 0| | f ( x) 0| . Therefore, f is continuous at x 0.
(b) Choose x c 0. Then within any interval (c , c ) there are both rational and irrational numbers. If c
is rational, pick 2c . No matter how small we choose 0 there is an irrational number x in
(c , c ) | f ( x) f (c)| |0 c | c 2c . That is, f is not continuous at any rational c 0. On the
other hand, suppose c is irrational f (c) 0. Again pick 2c . No matter how small we choose 0
there is a rational number x in (c , c ) with | x c | 2c 2c x 32c . Then | f ( x) f (c)| | x 0|
| x|
c
2
f is not continuous at any irrational c 0.
If x c 0, repeat the argument picking
value x c.
|c|
2
c . Therefore
2
f fails to be continuous at any nonzero
16. (a) Let c mn be a rational number in [0, 1] reduced to lowest terms f (c) 1n . Pick 21n . No matter
how small 0 is taken, there is an irrational number x in the interval (c , c ) | f ( x) f (c)|
0 1n
1
n
1
2n
. Therefore f is discontinuous at x c, a rational number.
Copyright 2018 Pearson Education, Inc.
96
Chapter 2 Limits and Continuity
(b) Now suppose c is an irrational number f (c) 0. Let 0 be given. Notice that 12 is the only rational
number reduced to lowest terms with denominator 2 and belonging to [0, 1]; 13 and 23 the only rationals with
denominator 3 belonging to [0, 1]; 14 and 34 with denominator 4 in [0, 1]; 15 , 52 , 53 and 54 with denominator 5
in [0, 1]; etc. In general, choose N so that
1
N
there exist only finitely many rationals in [0, 1] having
denominator N , say r1, r2 , , rp . Let min {| c ri |: i 1, , p}. Then the interval (c , c )
contains no rational numbers with denominator N . Thus, 0 | x c | | f ( x) f (c )| | f ( x ) 0|
| f ( x)| N1 f is continuous at x c irrational.
(c) The graph looks like the markings on a typical
ruler when the points ( x, f ( x)) on the graph of
f ( x ) are connected to the x-axis with vertical
lines.
17. 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 0 R represents the midnight point (at the same exact time). Suppose x1 is a point
on the equator “just after” noon x1 R 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 ( x1 ) T ( x1 R) 0. At exactly the same moment in time pick x2 to be a point just before
midnight x2 R is just before noon. Then T ( x2 ) T ( x2 R ) 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 R (simultaneously midnight) such that T (c) T (c R ) 0; i.e., there is always a pair
of antipodal points on the earth’s equator where the temperatures are the same.
1 ( f ( x ) g ( x )) 2
x c 4
1 (32 ( 1) 2 ) 2.
4
18. lim f ( x ) g ( x) lim
x c
2
2
( f ( x) g ( x))2 14 lim ( f ( x ) g ( x)) lim ( f ( x) g ( x))
x c
xc
a 0
At x
(b) At x
1(1 a )
1 1 a
lim 1 a1 a 1 1 a lim
1 12
a
1 1 0
1 1 a
a 0 a ( 1 1 a )
a 0
1(1 a )
a
1
1: lim r (a ) lim
lim
1
1 0
a 1
a 1 a ( 1 1 a ) a 1 a ( 1 1 a )
1 (1 a )
a
0: lim r (a ) lim 1 a1 a lim 1 a1 a 1 1 a lim
lim
a ( 1 1 a )
a ( 1 1 a )
1
1
a
a 0
a 0
a 0
a 0
a 0
1
lim
(because the denominator is always negative); lim r (a)
a 0
a 0 1 1 a
1
lim
(because the denominator is always positive).
a 0 1 1 a
19. (a) At x 0: lim r (a ) lim
a 0
Therefore, lim r (a ) does not exist.
a 0
At x 1: lim r (a ) lim
a 1
a 1
1 1 a
a
1
1
1 a
a 1
lim
1
Copyright 2018 Pearson Education, Inc.
Chapter 2 Additional and Advanced Exercises
97
(c)
(d)
20. f ( x) x 2 cos x f (0) 0 2 cos 0 2 0 and f ( ) 2 cos( ) 2 0. Since f ( x ) is
continuous on [ , 0], by the Intermediate Value Theorem, f ( x ) must take on every value between [ 2, 2].
Thus there is some number c in [ , 0] such that f (c) 0; i.e., c is a solution to x 2 cos x 0.
21. (a) The function f is bounded on D if f ( x) M and f ( x) N for all x in D. This means M f ( x) N for
all x in D. Choose B to be max {| M |, | N |}. Then | f ( x)| B. On the other hand, if | f ( x)| B, then
B f ( x) B f ( x ) B and f ( x) B f ( x) is bounded on D with N B an upper bound and
M B a lower bound.
(b) Assume f ( x) N for all x and that L N . Let L 2N . Since lim f ( x) L there is a 0 such that
x x0
0 | x x0 | | f ( x) L | L f ( x ) L L L 2N f ( x) L L 2N L 2N f ( x)
3 L2 N . But L N L 2N N N f ( x) contrary to the boundedness assumption f ( x) N . This
contradiction proves L N .
(c) Assume M f ( x) for all x and that L M . Let M2 L . As in part (b), 0 | x x0 | L M2 L
f ( x) L M2 L
3L M
2
f ( x)
M L
2
M , a contradiction.
| a b | a b a b
2 2 22a a.
2
|a b|
max {a, b} a 2 b 2 a 2 b b 2 a
22. (a) If a b, then a b 0 | a b | a b max {a, b}
If a b, then a b 0 | a b | (a b) b a
(b) Let min {a, b}
23.
sin(1 cos x )
x
lim
sin 2 x
x
cos x )
(1
x 0
lim
lim
x 0
a b | a b | .
2
2
1 lim
sin(1 cos x ) 1cos x 1 cos x
x 1 cos x
x 0 1cos x
x 0
sin x sin x
x
1 cos x
1
02 0.
a b
2
2
sin(1cos x )
x
lim x1(1cos
x
x)
1
cos
cos
x 0
x 0
lim
Copyright 2018 Pearson Education, Inc.
22b b.
98
24.
Chapter 2 Limits and Continuity
lim sin x
x 0 sin x
lim
x 0
sin(sin x )
x
lim
sin( x 2 x )
x
x 0
lim
sin( x 2 4)
x2
x 2
lim
sin( x 3)
x 9
x 9
lim
25. lim
x 0
x
sin x
x
sin x
x 0
26. lim
27. lim
x
x
sin(sin x ) sin x
x
sin x
sin( x 2 x )
2
x x
x 0
28. lim
sin( x 2 4)
2
x 4
x 2
1 lim
x 0
lim
x 0
lim
x 1 1 0 0.
x 0
sin(sin x )
lim sinx x
sin x
x 0
( x 1) lim
sin( x 2 x )
x2 x
x 0
sin( x 2 4)
( x 2) lim
x2 4
x2
sin( x 3)
1
x 3
x 3
x 9
1
sin x
x
lim ( x 1) 1 1 1.
x 0
lim ( x 2) 1 4 4.
x2
sin( x 3)
lim
x 3
x 9
x 9
lim
1 1 1.
1
x 3
1 16 16 .
29. Since the highest power of x in the numerator is 1 more than the highest power of x in the denominator, there is
3/ 2
an oblique asymptote. y 2 x 2 x 3 2 x 3 , thus the oblique asymptote is y 2 x.
x 1
30. As x ,
1
x
x 1
x; thus
0 sin 1x 0 1 sin 1x 1, thus as x , y x x sin 1x x 1 sin 1x
the oblique asymptote is y x.
31. As x , x 2 1 x 2 x 2 1 x 2 ; as x , x 2 x, and as x , x 2 x; thus the oblique
asymptotes are y x and y x.
32. As x , x 2 x x 2 2 x x( x 2) x 2 ; as x , x 2 x, and as x , x 2 x;
asymptotes are y x and y x.
33. Assume 1 a b and
a
x
x
1
x b
a ( x b) x 2 ( x b) x f ( x) a( x b) x 2 ( x b) x 0; f is
continuous for all x-values and f (0) ab 0, f (a b) a 2 (a b)2 a (a b)
a2
a (a b) 2 a b 0.
()
()
Thus, by the Intermediate Value Theorem there is at least one number c, 0 c a b, so that f (c) 0
a(c b) c 2 (c b) c 0 ac c c 1b .
34. (a)
lim
x 0
a bx 1
x
" a01" , so
a 1 0 a 1, then lim
x 0
b
b 2b4
x 0 1 bx 1 2
tan( ax a ) b 2 "tan 0 b 2" "b 2"
lim
0 ,
x 1
0
x 1
1bx 1 1bx 1
x
1bx 1
(1bx ) 1
x 0 x ( 1bx 1)
lim
lim
(b)
so b 2 0 b 2, then lim
x 1
tan( ax a )
x 1
sin a ( x 1)
lim cos aa( x 1) a ( x 1) cosa 0 1 a 3
x 1
35.
1/6 4 4
1/6
1/6
1/6 2
1/6
1/3
2/3
1 lim ( x ) 1 lim ( x 1)( x 1)(( x ) 1) lim ( x 1)( x 1) (2)(2) 4
lim x 1/2
3
1/6 3
1/6
1/6
1/6 2
1/6
1/3
3
3
x 1 1 x
x 1 1 ( x )
x 1 (1 x )(1 ( x ) ( x ) )
x 1 1 x x
Copyright 2018 Pearson Education, Inc.
tan a ( x 1)
lim a a ( x 1)
x 1
Chapter 2 Additional and Advanced Exercises
36.
lim
3x 4 x 4
x
lim
4x
x
x 0
x 0
lim
x 0
(3 x 4) x 4
x
3x 4 x 4
x
x 0
4 lim
37. (a) Domain 0, 1,
3x 4 x 4
(3 x 4) ( x ) 4
lim 22x 2; assume x 43 lim
lim
x
x
x 0
x 0
x 0
does not exist.
1 , 1 , 1 ,
2 3 4
(b) Consider any open interval (a, b) containing c 0. Choose a positive integer N so that
x
(c)
1
N
1
N
b. Then
1
N
b. Then
is in the domain and in the interval (a, b).
lim f ( x) 0
x0
38. (a) Domain 0, 1,
1 , 1 , 1 ,
2 3 4
(b) Consider any open interval (a, b) containing c 0. Choose a positive integer N so that
x
(c)
99
1
N
is in the domain and in the interval (a, b).
lim f ( x) 0
x0
39. (a) Domain
1 , 31 , 21 51 , 41 71 , 61 , 1 21 , 31 41 , 51 61 , 71
1
N
b. Then
2
x 2
2 .
(b) Consider any open interval (a, b) containing c 0. Choose a positive integer N so that
(c)
x N1 is in the domain and in the interval (a, b).
lim f ( x) 0
x0
40. (a)
(b) Let 0 be given. Find 0 so that if 0 x and x is in the domain of g, then
Choose min
(c)
1 , 1
2
2
so that
1
2
x 1 ,
2
1
2
x 2 3 , and
2
2
x 2
lim g ( x) 2 g (0) so g is continuous at x 0.
x 0
(d) Function g is continuous at every point of its domain.
Copyright 2018 Pearson Education, Inc.
2
2x
x 2
2 12
1
2
.
CHAPTER 3
3.1
DERIVATIVES
TANGENTS AND THE DERIVATIVE AT A POINT
1. P1: m1 1, P2 : m2 5
2. P1: m1 2, P2 : m2 0
3. P1: m1 52 , P2 : m2 12
4. P1: m1 3, P2 : m2 3
[4 ( 1 h ) 2 ](4 ( 1)2 )
h
h 0
h (2 h )
lim
2; at (1, 3): y
h 0 h
5. m lim
(1 2 h h 2 ) 1
h
h 0
lim
3 2( x ( 1))
y 2 x 5, tangent line
[(1 h 1) 2 1] [(1 1) 2 1]
h
h 0
6. m lim
h2
h 0 h
lim
lim h 0;
at (1,1) : y 1 0( x 1) y 1, tangent line
2 1 h 2 1
h
h 0
4(1 h ) 4
7. m lim
lim
h0 2 h 1 h 1
h 0
2 1 h 2 2 1 h 2
h
2 1 h 2
2
lim
1;
h 0 1 h 1
lim
h 0
at (1, 2): y 2 1( x 1) y x 1, tangent line
1
8. m lim
h 0
lim
( 1 h )2
h
2
( 2 h h )
2
h0 h ( 1 h )
1
( 1)2
lim
1( 1 h )2
2
h0 h ( 1 h )
2 h
2
h0 ( 1 h )
lim
2; at (1,1):
y 1 2( x (1)) y 2 x 3, tangent line
Copyright 2018 Pearson Education, Inc.
101
102
Chapter 3 Derivatives
( 2 h )3 ( 2)3
h
h 0
2
812 h 6 h 2 h3 8
h
h 0
9. m lim
lim
lim (12 6h h ) 12;
h 0
at (2, 8): y 8 12( x (2)) y 12 x 16,
tangent line
1
10. m lim
( 2 h )3
1
( 2)3
h
h 0
(12 h 6 h 2 h3 )
lim
lim
8 ( 2 h )3
h 0 8 h ( 2 h )
2
lim 126h h3
h 0 8( 2 h )
3
3
h 0 8h ( 2 h )
3
12
8( 8) 16 ;
3 ( x ( 2))
at 2, 18 : y 18 16
3 x 1 , tangent line
y 16
2
[(2 h ) 2 1]5
h
h 0
11. m lim
(5 4 h h 2 ) 5
h (4 h )
lim
h
h 0
h 0 h
lim
[(1 h ) 2(1 h )2 ]( 1)
h
h 0
h ( 3 2 h )
(1 h 2 4 h 2 h 2 ) 1
lim
h
h
h 0
h 0
12. m lim
13. m lim
3 h
3
(3 h ) 2
h 0
h
14. m lim
(2 h )2
8
h 0
h 0
2
h
(2 h )3 8
h
h 0
15. m lim
8 2(2 h )2
lim
h 0 h (2 h )
2
3; at (1, 1) : y 1 3( x 1), tangent line
lim h(h2h1) 2; at (3,3): y 3 2( x 3), tangent line
h 0
lim
8 2(4 4 h h 2 )
h (2 h )2
h 0
(812 h 6 h 2 h3 ) 8
h
h 0
lim
[(1 h )3 3(1 h )] 4
h
h 0
16. m lim
lim
(3 h ) 3( h 1)
h ( h 1)
lim
4; at (2,5): y 5 4( x 2), tangent line
lim
2 h (4 h )
2
h 0 h (2 h )
h (12 6 h h 2 )
h
h 0
lim
(1 3h 3h 2 h3 33h ) 4
h
h 0
lim
48 2; at (2, 2): y 2 2( x 2)
12; at (2,8): y 8 12(t 2), tangent line
h (6 3h h 2 )
h
h 0
lim
6; at (1, 4): y 4 6(t 1), tangent line
(4 h ) 4
h
lim 4hh 2 4 h 2 lim
lim
1 14 ;
4 2
4 h 2 h 0 h 4 h 2 h 0 h 4 h 2
h 0
1
at (4, 2): y 2 4 ( x 4), tangent line
17. m lim
h 0
4 h 2
h
(8 h ) 1 3
lim 9hh 3 9 h 3
h
9 h 3
h 0
h 0
(8,3): y 3 16 ( x 8), tangent line
18. m lim
at
lim
5(1 h ) 3(1 h )2 2
h
h 0
19. At x 1, y 2 m lim
h 0 h
(9 h ) 9
9 h 3
h 3h 2
h
h0
lim
lim
h 0 h
lim
h0
h
9 h 3
h ( 13h )
h
Copyright 2018 Pearson Education, Inc.
1
9 3
1
16 ;
Section 3.1 Tangents and the Derivative at a Point
[( 2 h )3 2( 2 h ) 7]3
h
h 0
h ( h 2 6h 10)
h
h0
20. At x 2, y 3 m lim
21. At x 3, y
1
2
m lim
1
1
(3 h ) 1 2
h 0
22. At x 0, y 1 m lim
2 (2 h )
h 0 2 h (2 h )
lim
h
h 1 ( 1)
h 1
h
h 0
lim
h 0
lim
103
10
h
14 , slope
lim 2h(2
h)
h 0
( h 1) ( h 1)
h ( h 1)
lim h (2hh1) 2
h 0
23. (a) It is the rate of change of the number of cells when t 5. The units are the number of cells per hour.
(b) P (3) because the slope of the curve is greater there.
(c)
6.10(5 h ) 2 9.28(5 h ) 16.43 [6.10(5) 2 9.28(5) 16.43]
61.0h 6.10h 2 9.28h
lim
h 0
h 0
h
h
lim 51.72 6.10h 51.72 52 cells/hr.
P (5) lim
h 0
24. (a) From t 0 to t 3, the derivative is positive.
(b) At t 3, the derivative appears to be 0. From t 2 to t 3, the derivative is positive but decreasing.
[( x h )2 4( x h) 1]( x 2 4 x 1)
h
h 0
25. At a horizontal tangent the slope m 0 0 m lim
( x 2 2 xh h2 4 x 4 h 1) ( x 2 4 x 1)
h
h 0
lim
(2 xh h 2 4 h)
h
h 0
lim
lim (2 x h 4) 2 x 4; 2 x 4 0
h 0
x 2. Then f (2) 4 8 1 5 (2, 5) is the point on the graph where there is
a horizontal tangent.
[( x h )3 3( x h )]( x3 3 x )
h
h 0
26. 0 m lim
2
2
( x3 3 x 2 h 3 xh 2 h3 3 x 3h ) ( x3 3 x )
h
h 0
lim
2
2
2
3
lim 3 x h 3 xhh h 3h
h 0
2
lim (3x 3xh h 3) 3 x 3; 3 x 3 0 x 1 or x 1. Then f (1) 2 and f (1) 2 (1, 2)
h0
and (1, 2) are the points on the graph where a horizontal tangent exists.
1
1
( x h ) 1 x 1
( x 1) ( x h 1)
lim h ( x 1)( x h 1) lim h( x 1)( hx h 1) 1 2 ( x 1)2 1 x 2 2 x 0
( x 1)
h 0
h 0
x ( x 2) 0 x 0 or x 2. If x 0, then y 1 and m 1 y 1 ( x 0) ( x 1). If x 2,
then y 1 and m 1 y 1 ( x 2) ( x 3).
27. 1 m lim
h
h 0
28.
( x h) x
h
m lim x hh x lim x hh x x h x lim
lim
1 .
2 x
x h x h 0 h x h x h 0 h x h x
h 0
h 0
x
1
1
1
x 2 x 4 y 2. The tangent line is y 2 4 ( x 4) 4 1.
Thus, 4
1
4
2 x
f (2 h ) f (2)
h
h 0
29. lim
(100 4.9(2 h ) 2 ) (100 4.9(2) 2 )
h
h 0
lim
4.9(4 4 h h 2 ) 4.9(4)
h
h 0
lim
lim (19.6 4.9h) 19.6.
h 0
The minus sign indicates the object is falling downward at a speed of 19.6 m/sec.
f (10 h ) f (10)
h
h 0
30. lim
f (3 h ) f (3)
h
h 0
31. lim
3(10 h )2 3(10) 2
h
h 0
lim
lim
h 0
(3 h )2 (3)2 )
h
3(20 h h 2 )
h
h 0
lim
lim
h 0
60 ft/sec.
[9 6 h h 2 9]
h
lim (6 h) 6
h 0
Copyright 2018 Pearson Education, Inc.
104
Chapter 3 Derivatives
f (2 h ) f (2)
h
h 0
32. lim
lim
4
3
h 0
(2 h )3 43 (2)3
h
lim
4
[12 h 6 h 2 h3 ]
3
lim 43 [12 6h h 2 ] 16
h
h 0
33. At ( x0 , mx0 b) the slope of the tangent line is lim
h 0
( m ( x0 h ) b ) ( mx0 b )
( x0 h ) x0
mh
lim h lim m m.
h 0
h 0
The equation of the tangent line is y (mx0 b) m( x x0 ) y mx b.
h 0
34. At x 4, y
1
4
12 and m lim
lim
h 0
1 1
4 h 2
h
h 0
1 1
4 h 2
h
4 h
2 4 h
2
lim 2 4 h lim 2 4 h 2 4 h
h 0 2 h 4 h
h 0 2 h 4 h 2 4 h
4 (4 h )
h
1
1
1
16
lim
lim
lim
2 4 2 4
h 0 2 h 4 h 2 4 h h 0 2 h 4 h 2 4 h h 0 2 4 h 2 4 h
f (0 h ) f (0)
h
h 0
35. Slope at origin lim
lim
origin with slope 0.
g (0 h ) g (0)
h
h 0
36. lim
37.
lim
h 0
lim
f (0 h ) f (0)
h
h sin
h
h 0
1h lim h sin
h0
h
lim
1 0
h
h 0
h
h0
, and lim
h 0
lim
h 0
U (0 h ) U (0)
h
lim
h 0
0 1
h
, and lim
h 0
h
f (0 h) f (0)
h
yes, the graph of f has a vertical tangent at the origin.
38.
1h 0 yes, f ( x) does have a tangent at the
1h lim sin 1 . Since lim sin 1 does not exist, f ( x) has no tangent at the origin.
h 0
h 0
h 2 sin
U (0 h ) U (0)
h
vertical tangent at (0, 1) because the limit does not exist.
lim
h 0
lim
h 0
1 0
h
11
h
. Therefore, lim
h 0
f (0 h ) f (0)
h
0 no, the graph of f does not have a
39. (a) The graph appears to have a cusp at x 0.
(b)
lim
h 0
f (0 h ) f (0)
h
2/5
of y x
lim
h 0
h 2/5 0
h
1
3/5
h
h 0
lim
1
3/5
h 0 h
limit does not exist the graph
1
1/5
h 0 h
limit does not exist y x 4/5
and lim
does not have a vertical tangent at x 0.
40. (a) The graph appears to have a cusp at x 0.
(b)
lim
h 0
f (0 h ) f (0)
h
lim
h 0
h 4/5 0
h
1
1/5
h 0 h
lim
does not have a vertical tangent at x 0.
and lim
Copyright 2018 Pearson Education, Inc.
Section 3.1 Tangents and the Derivative at a Point
105
41. (a) The graph appears to have a vertical tangent
at x 0.
f (0 h ) f (0)
h
(b) lim
h 0
lim
h 0
h1/5 0
h
1
4/5
h 0 h
lim
y x1/5 has a vertical tangent at x 0.
42. (a) The graph appears to have a vertical tangent
at x 0.
h3/5 0
h
h 0
f (0 h ) f (0)
h
h 0
lim
(b) lim
1
25
h 0 h
lim
the graph of y x3/5 has a vertical tangent at x 0.
43. (a) The graph appears to have a cusp at x 0.
(b)
lim
h 0
f (0 h ) f (0)
h
lim
h 0
2/5
the graph of y 4 x
4 h 2/5 2 h
h
lim
h 0
4
h3/5
2 and lim
h 0
4
h3/5
2 limit does not exist
2 x does not have a vertical tangent at x 0.
44. (a) The graph appears to have a cusp at x 0.
(b)
f (0 h ) f (0)
h
h 0
5/3
2/3
lim
yx
5x
h5/3 5h 2/3
h
h 0
lim
lim h 2/3
h 0
5
h1/3
0 lim
5
1/3
h 0 h
does not exist the graph of
does not have a vertical tangent at x 0.
Copyright 2018 Pearson Education, Inc.
106
Chapter 3 Derivatives
45. (a) The graph appears to have a vertical tangent
at x 1 and a cusp at x 0.
(b) x 1:
(1 h )2/3 (1 h 1)1/3 1
h
h 0
lim
at x 1;
x 0:
f (0 h ) f (0)
h
h 0
2/3
lim
yx
(1 h ) 2/3 h1/3 1
h
h 0
lim
y x 2/3 ( x 1)1/3 has a vertical tangent
1 ( h 1)1/3 1
lim 1/3
h h does not exist
h 0 h
does not have a vertical tangent at x 0.
h 2/3 ( h 1)1/3 ( 1)1/3
h
h 0
1/3
lim
( x 1)
46. (a) The graph appears to have vertical tangents
at x 0 and x 1.
(b) x 0:
x 1:
f (0 h ) f (0)
h1/3 ( h 1)1/3 ( 1)1/3
lim
y x1/3 ( x 1)1/3 has a vertical tangent at x 0;
h
h
h 0
h 0
f (1 h ) f (1)
(1 h )1/3 (1 h 1)1/3 1
lim
lim
y x1/3 ( x 1)1/3 has a vertical tangent at x 1.
h
h
h 0
h 0
lim
47. (a) The graph appears to have a vertical tangent
at x 0.
(b)
f (0 h ) f (0)
f (0 h ) f (0)
lim hh0 lim 1 ; lim
h
h
h
h 0
h 0
h 0
x 0
1
lim
y has a vertical tangent at x 0.
h 0 |h|
lim
lim
h 0
Copyright 2018 Pearson Education, Inc.
|h| 0
h
lim
h 0
|h|
|h|
Section 3.2 The Derivative as a Function
48. (a) The graph appears to have a cusp at x 4.
(b)
f (4 h ) f (4)
h
lim
h 0
lim
h 0
49–52.
|h|
| h |
lim
h 0
lim 1
h 0 | h |
|4 (4 h )| 0
h
lim
h 0
|h|
h
lim
h 0
1
h
; lim
h 0
f (4 h ) f (4)
h
lim
h 0
y 4 x does not have a vertical tangent at x 4.
Example CAS commands:
Maple:
f : x - x^3 2*x;x0 : 0;
plot( f (x), x x0-1/2..x0 3, color black,
# part (a)
title "Section 3.1, #49(a)" );
q : unapply( (f (x0 h)-f (x0))/h, h );
# part (b)
L : limit( q(h), h 0 );
sec_lines : seq( f(x0) q(h)*(x-x0), h 1..3 );
tan_ line : f(x0) L*(x-x0);
plot( [f(x),tan_line,sec_lines], x x0-1/2..x0 3, color black,
# part (c)
# part (d)
linestyle [1,2,5,6,7], title "Section 3.1, #49(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 p;
f[x_ ]: Cos[x] 4Sin[2x]
Plot[f [x],{x, x0 1, x0 3}]
dq[h_ ]: (f [x0 h] f [x0])/h
m Limit[dq[h], h 0]
ytan: f [x0] m(x x0)
y1: f [x0] dq[1](x x0)
y2: f [x0] dq[2](x x0)
y3: 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:
Step 2:
Step 3:
f ( x) 4 x 2 and f ( x h) 4 ( x h) 2
f ( x h) f ( x)
h
[4 ( x h )2 ](4 x 2 )
(4 x 2 2 xh h 2 ) 4 x 2
2
2 xhh h
h
h
f ( x) lim (2 x h) 2 x; f (3) 6, f (0) 0, f (1) 2
h 0
Copyright 2018 Pearson Education, Inc.
h ( 2 x h )
h
2 x h
|4 (4 h )|
h
107
108
Chapter 3 Derivatives
[( x h 1)2 1][( x 1)2 1]
h
h 0
2. F ( x) ( x 1)2 1 and F ( x h) ( x h 1)2 1 F ( x ) lim
( x 2 2 xh h 2 2 x 2 h 11) ( x 2 2 x 11)
h
h0
lim
2 xh h 2 2 h
h
h 0
lim
F (1) 4, F (0) 2, F (2) 2
g (t )
3. Step 1:
4.
1
t2
lim (2 x h 2) 2( x 1);
h 0
1
(t h ) 2
t 2 ( t h )2
( t h )2 t 2
2
t 2 (t 2 2th h 2 )
h ( 2t h )
h
2th 2 h2
2 2
(t h ) t h
(t h ) 2 t 2 h
(t h ) t h
2t 2 ; g (1) 2, g (2) 1 , g 3 2
4
3 3
t 2 t 2
t3
and g (t h)
Step 2:
g (t h ) g ( t )
h
Step 3:
g (t )
1
( t h )2
12
t
h
2
t
h
lim
2 2
h 0 ( t h ) t
2t h
(t h ) 2 t 2
1 ( z h )
12zz
2( z h )
1( z h )
(1 z h ) z (1 z )( z h )
k ( z ) 12zz and k ( z h) 2( z h) k ( z ) lim
lim
h
2( z h ) zh
h 0
h 0
h
1
1
1
1
lim 2( z h) zh lim 2( z h ) z 2 ; k (1) 2 , k (1) 2 , k
2 14
2z
h 0
h 0
p ( ) 3 and p ( h) 3( h)
5. Step 1:
p ( h ) p ( )
h
Step 2:
h
3( h ) 3
h
3h
3 3h 3
3 3h 3
h
3 3h
3
h0 3 3h 3
3
3 3
3
2 3
; p (1) 3 , p (3) 12 , p 32 3
2 3
2 2
6. r ( s ) 2 s 1 and r ( s h) 2( s h) 1 r ( s ) lim
lim
2 s h 1 2 s 1
h
h 0
lim
h 0 h
2h
2 s 2 h 1 2 s 1
r (0) 1, r (1)
1 ,
3
r
2 s 2 h 1 2 s 1
2 s 2 h 1 2 s 1
6 x 2 h 6 xh 2 2 h3
h
h 0
8. r s 3 2 s 2 3
dr
ds
2
h 0 2 s 2 h 1 2 s 1
2
2 s 1 2 s 1
2
2 2 s 1
1 ;
2 s 1
h (6 x 2 6 xh 2 h 2 )
h
h 0
dy
dx
2( x h )3 2 x3
h
h 0
lim
2( x3 3 x 2 h 3 xh 2 h3 ) 2 x3
h
h 0
lim
lim (6 x 2 6 xh 2h 2 ) 6 x 2
h 0
(( s h )3 2( s h )2 3) ( s 3 2 s 2 3)
h
h 0
lim
h (3s 2 3sh h 2 4 s 2 h )
h
h 0
lim
lim
h 0
s 3 3s 2 h 3sh 2 h3 2 s 2 4 sh 2 h 2 3 s 3 2 s 2 3
h
lim (3s 2 3sh h 2 4 s 2h) 3s 2 2s
h 0
( t h )(2 t 1) t (2t 2 h 1)
t h
t
(2t 2 h 1)(2 t 1)
t and r (t h) t h ds lim 2( t h ) 1 2t 1 lim
2t 1
h
dt
h
2(t h ) 1
h 0
h 0
2
2
(t h )(2t 1) t (2t 2 h 1)
h
2
t
t
2
ht
h
2
t
2
ht
t
lim (2t 2h 1)(2t 1) h lim
lim (2t 2h 1)(2t 1) h lim (2t 2h 11)(2t 1)
(2t 2 h 1)(2t 1) h
h 0
h 0
h 0
h 0
1
1
(2t 1)(2t 1)
(2t1) 2
9. s r (t )
1
2
lim
3s 2 h 3sh 2 h3 4 sh 2 h 2
h
h 0
lim
2 s 2 h 1 2 s 1
h
lim (2 s 2h1)(2s 1)
h0 h 2s 2h1 2s 1
7. y f ( x) 2 x3 and f ( x h) 2( x h)3
lim
h 0
lim
12
(3 3h)3
3
h 3 3h 3
3 3h 3
3
3 3h 3
p ( ) lim
Step 3:
z z 2 zh z h z 2 zh
2( z h ) zh
h 0
lim
Copyright 2018 Pearson Education, Inc.
Section 3.2 The Derivative as a Function
10.
dv
dt
(t h ) (t )
t h
t
lim
lim
h
11.
dp
dq
lim
1
1
h 0
( q h )3/2 q 3/2
h
h 0
( w h )2 1
lim
( q h )1/2 lim
1
w2 1
h 0
h 0
w2 1 ( w h )2 1
2
q [( q h )1/2 q1/2 ]
h
2
lim
h 0
1
2 x
2
h
h 0 h (2 x )(2 x h )
15.
x 2 h xh 2 9 h
x ( x h) h
h ( x 2 xh 9)
x( x h)h
9
h (3t 2 3th h 2 2t h )
h
h 0
lim
16.
dy
dx
lim
( x h) 3 x 3
1 ( x h ) 1 x
h
h 0
8
x2
h 0
;
dy
dx x2
and f ( x h)
8
( x h) 2
4
h 0 (1 x h )(1 x )
lim
17. f ( x)
h 0
h
4
(1 x )2
4
(3) 2
18. g ( z ) lim
8
8
x 2
( x h)2
h
h
lim
lim
h 0
1
9
x2
ds
dt t1
3t 2 h 3th 2 h3 2th h 2
h
5
4h
h 0 h (1 x h )(1 x )
lim
4 z h 4 z
h
8
x 2
x2 x h2
h x h2
f ( x) lim
h 0 x h 2 x 2
x2 x h2
x 2 x h2
8
x2 x h2
the equation of the tangent line at (6, 4) is
4 z h 4 z
lim (4 z h)(4 z )
4 z h 4 z
h 0
h 0 h 4 z h 4 z
h
1
1
1
lim
lim
; m g (3)
12 the equation
2 4 3
2 4 z
h 0 4 z h 4 z
h 0 h 4 z h 4 z
of the tangent line at (3, 2) is w 2 12 ( z 3) w 12 z 32 2 w 12 z 72 .
h0
x 2 9
x2
4
9
(1 4 ( z h ) ) 1 4 z
h 0 h (2 x )(2 x h )
h
x h 3 x 2 xh 3 x x 3 x 2 3 x xh 3h
h (1 x h )(1 x )
h 0
f ( x h) f ( x )
h
x ( x h )2 9 x x 2 ( x h ) 9( x h )
x( x h)h
lim
8[( x 2) ( x h 2)]
8h
h x h2 x 2 x 2 x h2
h x h2 x2 x 2 x h2
8
4
; m f (6) 4 12
( x 2) x 2
4 4
x2 x 2 x 2 x 2
1
1
1
y 4 2 ( x 6) y 2 x 3 4 y 2 x 7.
w
( w2 1)3/2
x 2 xh 9
h 0 x ( x h )
lim
( x h 3)(1 x ) ( x 3)(1 x h )
(1 x h )(1 x )
9
2 1x h 21 x lim (2 x)(2 x h)
lim (3t 2 3th h 2 2t h) 3t 2 2t ; m
lim
q1/2 23 q1/2
2
f ( x) lim
(t 3 3t 2 h 3th 2 h3 ) (t 2 2th h 2 ) t 3 t 2
h
h 0
lim
q1/2
2
2
2
x 2 xh 9 ;
x ( x h)
k ( x h) k ( x )
1
k ( x) lim
lim
h
2 ( x h)
h 0
h 0
1
1 ; k (2) 1
lim (2 x )(2
x h)
16
(2 x ) 2
h 0
[(t h )3 (t h )2 ](t 3 t 2 )
h
h 0
ds
dt
2
( x h) ( x h ) x x
h
and k ( x h)
lim
t2
w 1 ( wh) 1
w 1 w 1 ( w h ) 1
w2 1 ( w h )2 1
h 0 h ( w h )2 1
f ( x h) f ( x)
h
t2
( q h )1/2
q
( q h )1/2 q1/2
m f (3) 0
14. k ( x)
h 0 (t h )t
2 w h
h0 h ( w h )2 1 w2 1 w2 1 ( w h )2 1
2
2
1/2
h ( qhh )
( q h )1/2 lim
lim
13. f ( x) x 9x and f ( x h) ( x h) ( x 9 h)
x3 2 x 2 h xh 2 9 x x3 x 2 h 9 x 9 h
x ( x h) h
q [( q h ) q ]
h [( q h )1/2 q1/2 ]
h0 h ( w h )2 1 w2 1
h0 h ( w h )2 1 w2 1
lim
lim ht h t h lim t ht 1 t 1 1 1
h 0 h (t h )t
h
w2 1 ( w h )2 1
lim
w2 1 ( w2 2 wh h 2 1)
lim
h (t h )t t (t h )
(t h )t
h 0
( q h )( q h )1/2 qq1/2
h
h 0
h
h 0
lim
lim
q [( q h )1/2 q1/2 ][( q h )1/2 q1/2 ]
h [( q h )1/2 q1/2 ]
1
dz
dw
12.
h
h 0
h0
lim
h t 1h 1t
109
Copyright 2018 Pearson Education, Inc.
;
110
Chapter 3 Derivatives
19. s f (t ) 1 3t 2 and f (t h) 1 3(t h)2 1 3t 2 6th 3h 2
(13t 2 6th 3h 2 ) (13t 2 )
h
h 0
lim (6t 3h) 6t ds
dt
lim
h 0
dy
dx
20. y f ( x) 1 1x and f ( x h) 1 x 1 h
h
h 0 x ( x h ) h
lim
21. r f ( )
1
h 0 x ( x h )
lim
2
4
1
x2
f (t h ) f (t )
h
lim
h 0
6
f ( x h) f ( x)
h
h 0
1
3
lim
lim
h 0
f ( h ) f ( )
h
ddr lim
h 0
2
4 ( h )
2 4 2 4 h
and f ( h)
lim 2 4 2 4 h
1 x 1 h 1 1x lim
h
lim
h
h 0
1 1
x xh
h 0
2
2
4 h
4
4(4 ) 4(4 h )
lim
2 4 2 4 h
h 0 2 h 4 4 h 4 4 h
h0 h 4 4 h
2
2
1
lim
ddr
1
(4 ) 4
0 8
(4 ) 2 4
h 0 4 4 h 4 4 h
dy
dx x 3
t 1
ds
dt
h
lim 2 4 2 4 h
h 0 h 4 4 h
22. w f ( z ) z z and f ( z h) ( z h)
zh
dw
dz
f ( z h ) f ( z )
h
h0
lim
lim
z h
z h ( z z )
h
h0
z h z 1 lim ( z h) z 1 lim 1 1 1
lim h z hh z lim 1 z hh z
2 z
h0
h0
h0 z h z
h0 h z h z
z h z
dw
dz
z4
54
f ( z ) f ( x)
zx
zx
23. f ( x) lim
lim
zx
1 1
z2 x2
zx
( x 2) ( z 2)
z
1
lim ( z x )( z 2)( x 2) lim ( z x )( zx2)(
lim
1 2
x 2) z x ( z 2)( x 2)
( x 2)
zx
zx
f ( z ) f ( x)
( z 2 3 z 4) ( x 2 3 x 4)
lim
z
x
zx
zx
zx
( z x ) ( z x ) 3
lim
lim ( z x) 3 2 x 3
zx
zx
zx
24. f ( x) lim
g ( z ) g ( x)
zx
zx
lim
g ( z ) g ( x)
zx
lim
25. g ( x) lim
26. g ( x) lim
zx
zx
zx
z x
z 1 x 1
zx
2
2
2
2
( z x )( z x ) 3( z x )
lim z 3 zz xx 3 x lim z x z 3x z 3 x lim
zx
zx
zx
zx
z ( x 1) x ( z 1)
x
1
lim ( z x )( z 1)( x 1) lim ( z x )(zz1)(
lim
1 2
x 1) z x ( z 1)( x 1)
( x 1)
zx
zx
(1 z ) (1 x )
zx
lim
zx
z x
zx
z x
z x
zx
z x ( z x )( z x )
lim
lim
zx
1
z x
1
2 x
27. Note that as x increases, the slope of the tangent line to the curve is first negative, then zero (when x 0), then
positive the slope is always increasing which matches (b).
28. Note that the slope of the tangent line is never negative. For x negative, f 2 ( x) is positive but decreasing as
x increases. When x 0, the slope of the tangent line to x is 0. For x 0, f 2 ( x) is positive and increasing. This
graph matches (a).
29. f3 ( x) is an oscillating function like the cosine. Everywhere that the graph of f3 has a horizontal tangent we
expect f3 to be zero, and (d) matches this condition.
30. The graph matches with (c).
Copyright 2018 Pearson Education, Inc.
Section 3.2 The Derivative as a Function
111
31. (a) f is not defined at x 0, 1, 4. At these points, the left-hand and right-hand derivatives do not agree. For
f ( x ) f (0)
f ( x ) f (0)
example, lim
slope of line joining (4, 0) and (0, 2) 12 but lim
slope of line
x 0
x 0
x 0
joining (0, 2) and (1, 2) 4. Since these values are not equal, f (0)
(b)
does not exist.
(b) Shift the graph in (a) down 3 units
32. (a)
33.
x 0
f ( x ) f (0)
lim
x 0
x 0
y’
2
1
6
7
8
9
10
11
x
1
2
3
4
5
(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.
(a) i) slope 1.54 dT
1.54 °F
ii) slope 2.86
dt
hr
iii) slope 0
dT
dt
0 °F
hr
iv)
dT 2.86 °F
dt
hr
dT
slope 3.75 dt 3.75 °F
hr
(b) The tangent with the steepest positive slope appears to occur at t 6 12 p.m. and slope 7.27
dT
7.27 hrF . The tangent with the steepest negative slope appears to occur at t 12 6 p.m. and
dt
slope 8.00 dT
8.00 hrF
dt
Copyright 2018 Pearson Education, Inc.
112
Chapter 3 Derivatives
(c)
36. (a) decrease: 2006–2012, increase: 2012–2015
(b) i) $300,000
ii) $190,000
iii) $280,000
(c) i) $35,000/yr
ii) $0/yr
iii) $8,000/yr
(d) during 2008 at $90,000/yr
(e) during 2013 at $68,500/yr
(f)
37. Left-hand derivative: For h 0, f (0 h) f (h) h 2 (using y x 2 curve) lim
h 0
f (0 h ) f (0)
h
2
lim h h0 lim h 0;
h 0
h 0
f (0 h) f (0)
Right-hand derivative: For h 0, f (0 h) f (h) h (using y x curve) lim
h
lim
h 0
h0
h
lim 1 1; Then lim
h 0
h 0
f (0 h ) f (0)
h
lim
h 0
f (0 h ) f (0)
h
38. Left-hand derivative: When h 0, 1 h 1 f (1 h) 2 lim
h 0
h 0
the derivative f (0) does not exist.
f (1 h ) f (1)
h
Right-hand derivative: When h 0, 1 h 1 f (1 h) 2(1 h) 2 2h
2 2 lim 0 0;
h
h 0
h 0
f (1 h ) f (1)
(2 2 h ) 2
lim
lim
h
h
h 0
h 0
lim
lim 2hh lim 2 2;
h 0
h 0
f (1 h ) f (1)
f (1 h ) f (1)
lim
the derivative f (1) does not exist.
Then lim
h
h
h 0
h 0
39. Left-hand derivative: When h 0,1 h 1 f (1 h) 1 h lim
lim
h 0
1 h 1
h
lim (1 h)1 lim
1 h 1 h 0 h 1 h 1 h 0
1 h 1
h 0
1
1 h 1
f (1 h ) f (1)
h
1;
2
Copyright 2018 Pearson Education, Inc.
lim
h 0
1 h 1
h
Section 3.2 The Derivative as a Function
Right-hand derivative: When h 0,1 h 1 f (1 h) 2(1 h) 1 2h 1 lim
(2 h 1) 1
lim 2 2;
h
h 0
h 0
f (1 h ) f (1)
f (1 h ) f (1)
lim
Then lim
h
h
h 0
h 0
h 0
f (1 h ) f (1)
h
lim
f (1 h ) f (1)
h
40. Left-hand derivative: lim
h 0
Then lim
h 0
f (1 h ) f (1)
h
h 0
lim
h 0
(1 h ) 1
h
lim
h 0
f (1 h ) f (1)
h
Right-hand derivative: lim
the derivative f (1) does not exist.
lim
f (1 h ) f (1)
h
h 0
h 0
1 1
1 h
h
lim 1 1;
lim
h 0
1 (1 h )
1 h
h
lim h(1hh ) lim 11h 1;
h 0
h 0
the derivative f (1) does not exist.
41. f is not continuous at x 0 since lim f ( x) does not exist and f (0) 1
x 0
1/3
g ( h ) g (0)
1 ;
lim h h0 lim 2/3
h
h
h 0
h 0
h 0
g ( h ) g (0)
h 2/3 0 lim 1 ;
lim
Right-hand derivative: lim
1/3
h
h
h 0 h
h 0
h 0
g ( h ) g (0)
g ( h ) g (0)
Then lim
lim
the derivative g (0) does
h
h
h 0
h 0
42. Left-hand derivative: lim
f ( h ) f (0)
h
43. Left-hand derivative: lim
h 0
Right-hand derivative: lim
h 0
f ( h ) f (0)
h
not exist.
2
lim hh lim h 0;
h 0
h 0
lim 2 h htan h lim 2 sinh h cos1 h
h 0
h 0
2 (1) (1) 3
the derivative f (0) does not exist
44. Left-hand derivative: lim
h 0
lim
h 0
1 h ( h 2)
h
h 1
g ( h ) g (0)
h
lim
h 0
h h11 ( 1)
h
lim 1h
h 0
h ( h 1) 1 ( h 1)
h 1
2
lim 1h h h 21h
h 0
2
Right-hand derivative: lim
h 0
g ( h ) g (0)
h0
lim
h 0
2 h h3 1 ( 1)
h
lim
h 0
h (2 h 2 )
h
2 and g is continuous at
x 0 since lim g ( x) g (0) 1 the derivative g (0) 2.
h 0
45. (a) The function is differentiable on its domain 3 x 2 (it is smooth)
(b) none
(c) none
46. (a) The function is differentiable on its domain 2 x 3 (it is smooth)
(b) none
(c) none
47. (a) The function is differentiable on 3 x 0 and 0 x 3
(b) none
(c) The function is neither continuous nor differentiable at x 0 since lim f ( x) lim f ( x)
h 0
Copyright 2018 Pearson Education, Inc.
h 0
113
114
Chapter 3 Derivatives
48. (a) f is differentiable on 2 x 1, 1 x 0, 0 x 2, and 2 x 3
(b) f is continuous but not differentiable at x 1: lim f ( x) 0 exists but there is a corner at x 1 since
lim
h 0
f ( 1 h ) f ( 1)
h
3 and lim
h 0
x 1
f ( 1 h ) f ( 1)
3
h
f (1) does not exist
(c) f is neither continuous nor differentiable at x 0 and x 2:
at x 0, lim f ( x) 3 but lim f ( x) 0 lim f ( x) does not exist;
x 0
x 0
x 0
at x 2, lim f ( x ) exists but lim f ( x ) f (2)
x 2
x 2
49. (a) f is differentiable on 1 x 0 and 0 x 2
(b) f is continuous but not differentiable at x 0: lim f ( x) 0 exists but there is a cusp at x 0,
f (0 h ) f (0)
h
h 0
so f (0) lim
(c) none
x 0
does not exist
50. (a) f is differentiable on 3 x 2, 2 x 2, and 2 x 3
(b) f is continuous but not differentiable at x 2 and x 2: there are corners at those points
(c) none
f ( x h) f ( x)
h
h 0
51. (a) f ( x) lim
(b)
( x h)2 ( x 2 )
h
h 0
lim
2
2
2
lim x 2 xhh h x lim (2 x h) 2 x
h 0
h 0
(c) y 2 x is positive for x 0, y is zero when x 0, y is negative when x 0
(d) y x 2 is increasing for x 0 and decreasing for 0 x ; the function is increasing on intervals
where y 0 and decreasing on intervals where y 0
f ( x h) f ( x )
h
h 0
52. (a) f ( x) lim
(b)
lim
h 0
x1h x1 lim x( x h) lim
h
h 0 x ( x h ) h
1
h 0 x ( x h )
1
x2
(c) y is positive for all x 0, y is never 0, y is never negative
(d) y 1x is increasing for x 0 and 0 x
53. (a) Using the alternate formula for calculating derivatives: f ( x)
( z x )( z 2 zx x 2 )
3( z x )
zx
lim
2
x 2 x 2 f ( x ) x 2
lim z zx
3
f ( z ) f ( x)
lim
zx
zx
zx
Copyright 2018 Pearson Education, Inc.
z 3 x3
3 3
lim z x
zx
3
3
lim 3(z zxx )
zx
Section 3.2 The Derivative as a Function
115
(b)
(c) y is positive for all x 0, and y 0 when x 0; y is never negative
3
(d) y x3 is increasing for all x 0 (the graph is horizontal at x 0 ) because y is increasing where y 0; y is
never decreasing
54. (a)
(b)
z 4 x4
4 4
f ( z ) f ( x)
Using the alternate form for calculating derivatives: f ( x) lim
lim
zx
zx zx
zx
3
2
2
3
z 4 x 4 lim ( z x )( z xz x z x ) lim z 3 xz 2 x 2 z x3 x3 f ( x) x3
lim 4(
4(
)
4
)
z
x
z
x
zx
zx
zx
(c) y is positive for x 0, y is zero for x 0, y is negative for x 0
(d) y
x4
4
is increasing on 0 x and decreasing on x 0
(2( x h )2 13( x h ) 5) (2 x 2 13 x 5)
h
h 0
55. y lim
2
2
2
2
lim 2 x 4 xh 2h 13 xh13h 5 2 x 13 x 5 lim 4 xh 2hh 13h
h 0
h 0
lim (4 x 2h 13) 4 x 13, slope at x. The slope is 1 when 4 x 13 1 4 x 12 x 3
h 0
y 2 32 13 3 5 16. Thus the tangent line is y 16 (1)( x 3) y x 13 and the point of
tangency is (3, 16).
56. For the curve y x , we have y lim
xh x
h
h 0
lim
x h 0
xh x
( x h) x
xh
xh x h
lim
h0
1
xh x
1 .
2 x
Suppose a, a is the point of tangency of such a line and (1, 0) is the point on the line where it crosses the
x-axis. Then the slope of the line is
xa
1
2 a
1;
2
a
a 1
1
2 a
a 0
a ( 1)
a
a 1
which must also equal 1 ; using the derivative formula at
2 a
2a a 1 a 1. Thus such a line does exist: its point of tangency is (1, 1), its slope is
and an equation of the line is y 1 12 ( x 1) y 12 x 12 .
57. Yes; the derivative of f is f so that f ( x0 ) exists f ( x0 ) exists as well.
58. Yes; the derivative of 3g is 3 g so that g (7) exists 3 g (7) exists as well.
g (t )
59. Yes, lim h(t ) can exist but it need not equal zero. For example, let g (t ) mt and h(t ) t. Then g (0) h(0) 0,
t0
g (t )
but lim h (t ) lim mt
lim m m, which need not be zero.
t0
t 0
t0 t
Copyright 2018 Pearson Education, Inc.
116
Chapter 3 Derivatives
60. (a) Suppose | f ( x)| x 2 for 1 x 1. Then | f (0)| 02 f (0) 0. Then f (0) lim
h 0
lim
h 0
f (h)0
h
f ( h)
lim h . For | h | 1, h 2 f (h) h 2 h
h 0
f (h)
h
f (0 h ) f (0)
h
f ( h)
h f (0) lim h 0 by the
h0
Sandwich Theorem for limits.
(b) Note that for x 0, | f ( x)| | x 2 sin 1x | | x 2 ||sin 1x | | x 2 | 1 x 2 (since 1 sin x 1). By part (a), f is
differentiable at x 0 and f (0) 0.
61. The graphs are shown below for h 1, 0.5, 0.1 The function y 1 is the derivative of the function y x so
2 x
that 1 lim x hh x . The graphs reveal that y x hh x gets closer to y 1 as h gets smaller and
2 x
2 x
h 0
smaller.
62. The graphs are shown below for h 2,1, 0.5. The function y 3 x 2 is the derivative of the function y x3 so that
( x h )3 x 3
. The
h
h 0
3 x 2 lim
graphs reveal that y
( x h )3 x 3
h
gets closer to y 3 x 2 as h gets smaller and smaller.
63. The graphs are the same. So we know that for
| x|
f ( x) | x |, we have f ( x) x .
Copyright 2018 Pearson Education, Inc.
Section 3.2 The Derivative as a Function
64. Weierstrass’s nowhere differentiable continuous function.
65–70.
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, #65(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 #65(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 #65(f )" );
Mathematica: (functions and x0 may vary) (see section 2.5 re. RealOnly ):
Miscellaneous`RealOnly`
Clear[f, m, x, y, h]
x0 π/4;
f[x_ ]: x 2 Cos[x]
Plot[f[x], {x, x0 3, x0 3}]
q[x_,h_ ]: (f[x h] f[x])/h
m[x_ ]: Limit[q[x, h], h 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}]
Copyright 2018 Pearson Education, Inc.
117
118
Chapter 3 Derivatives
3.3
DIFFERENTIATION RULES
1. y x 2 3
2
d ( x 2 ) d (3) 2 x 0 2 x d y 2
dx
2
dx
dy
dx
dx
d2y
dy
2. y x 2 x 8 dx 2 x 1 0 2 x 1 2 2
dx
d (5t 3 ) d (3t 5 ) 15t 2 15t 4 d 2 s d (15t 2 ) d (15t 4 ) 30t 60t 3
dt
3. s 5t 3 3t 5 ds
dt
dt
dt
dt
dt 2
4. w 3z 7 7 z 3 21z 2
5. y 43 x3 x
6.
dy
dx
dw
dz
d 2w
dz 2
21z 6 21z 2 42 z
4 x2 1
d2y
dx 2
126 z 5 42 z 42
8x
2
3
2
dy
d y
y x3 x2 4x dx x 2 x 14 2 2 x 1
dx
7. w 3 z 2 z 1
dw
dz
8. s 2t 1 4t 2
6 z 3 z 2
ds
dt
2t 2 8t 3
9. y 6 x 2 10 x 5 x 2
dy
dx
6
z3
2
t2
1
z2
8
t3
d 2w
dz 2
d 2s
dt 2
18 z 4 2 z 3 184
z
4t 3 24t 4
12 x 10 10 x 3 12 x 10 103
x
10. y 4 2 x x 3
dy
dx
2 3x 4 2
3
x4
11. r 13 s 2 52 s 1
dr
ds
23 s 3 52 s 2
2
3s 3
d2y
dx 2
5
2s2
d2y
dx 2
0 12 x 5
d 2r
ds 2
4
t3
2
z3
244
t
12 0 30x 4 12 304
x
12
x5
2s 4 5s 3
2
s4
5
s3
2
124 45 d r2 24 3 48 5 20 6
12. r 12 1 4 3 4 ddr 12 2 12 4 4 5 12
2
d
243 485 206
d ( x3 x 1) ( x3 x 1) d (3 x 2 )
13. (a) y (3 x 2 ) ( x3 x 1) y (3 x 2 ) dx
dx
(3 x 2 ) (3 x 2 1) ( x3 x 1) (2 x) 5 x 4 12 x 2 2 x 3
(b) y x5 4 x3 x 2 3x 3 y 5 x 4 12 x 2 2 x 3
14. (a)
(b)
y (2 x 3)(5 x 2 4 x) y (2 x 3)(10 x 4) (5 x 2 4 x)(2) 30 x 2 14 x 12
y (2 x 3)(5 x 2 4 x) 10 x3 7 x 2 12 x y 30 x 2 14 x 12
d ( x 5 1 ) ( x 5 1 ) d ( x 2 1)
15. (a) y ( x 2 1) ( x 5 1x ) y ( x 2 1) dx
x
x dx
( x 2 1) (1 x 2 ) ( x 5 x 1 ) (2 x) ( x 2 1 1 x 2 ) (2 x 2 10 x 2) 3 x 2 10 x 2
(b) y x3 5 x 2 2 x 5 1x y 3x 2 10 x 2
1
x2
Copyright 2018 Pearson Education, Inc.
1
x2
Section 3.3 Differentiation Rules
16. y (1 x 2 )( x3/4 x 3 )
(a) y (1 x 2 ) 34 x 1/4 3 x 4 ( x3/4 x 3 )(2 x)
(b) y x3/4 x 3 x11/4 x 1 y
2 x 5 ; use the quotient rule:
3x2
6 x 4 6 x 15 19
(3 x 2)2
(3 x 2) 2
17. y
3
4x1/ 4
3 3
4 x1/ 4 x 4
3 11 x 7/4 1
4
x4
x2
11
x7/4
4
43 x ; use the quotient rule: u 4 3 x and v 3 x 2 x u
3 x2 x
2
2
2
(3 x 2 x )( 3) (4 3 x )(6 x 1)
9 x 3 x 218 x 2 21x 4 9 x 224 x 2 4
2
2
(3 x x )
(3 x x )
(3 x x )
x 2 4 ; use the quotient rule:
x 0.5
( x 0.5)(2 x ) ( x 2 4)(1)
2 x2 x x2 4
( x 0.5)
20. f (t )
t 2 1
t t 2
2
2
(t 1)(t 1)
(t 2)(t 1)
21. v (1 t ) (1 t 2 ) 1
22. w
x 5
2 x 7
24. u
5 x 1
2 x
26. r 2
27. y
x
t 1 , t
t 2
dv
dt
( s 1)
1
2 s
(1t )
1
2 s
3 and v 6 x 1 y
(3 x 2)(2) (2 x 5)(3)
(3 x 2)2
vu uv
v2
2 x 7 2 x 10
(2 x 7)2
s 1)
(
1
2 s
t 2 t 1
(t 2) 2
1t 2 2t 2t 2
(1t 2 ) 2
vu uv
v2
1
(t 2) 2
t 2 2t 1
(1t 2 )2
17
(2 x 7)2
s 1) ( s 1)
2 s ( s 1) 2
1
s ( s 1) 2
from Example 2 in Section 3.2
5 x1
1
x
4 x3/ 2
(1 x4
2
x
x
x)
2
2 x 1
x2
(0) 1
r 2
1
( s 1)2
4x
x 1
(t 2) 2
2 2
(
(2 x )(5) (5 x 1)
v
(t 2)(1) (t 1)(1)
(1t 2 )( 1) (1t )(2t )
x2 x 4
( x 0.5)2
1 f (t )
(2 x 7) 2
s)
du
dx
25. v 1 x x4
1t
1t 2
f ( s )
d (
ds
NOTE:
(2 x 7)(1) ( x 5)(2)
w
s 1
s 1
23. f ( s )
( x 0.5)
2
vu uv
v2
u x 2 4 and v x 0.5 u 2 x and v 1 g ( x)
19. g ( x)
1
x2
u 2 x 5 and v 3 x 2 u 2 and v 3 y
18. y
1
; use the
( x 2 1)( x 2 x 1)
2
2
1
2
1 1
3/ 2 1/ 2
1
2
quotient rule: u 1 and v ( x 2 1) ( x 2 x 1) u 0 and
dy
v ( x 1)(2 x 1) ( x x 1)(2 x) 2 x3 x 2 2 x 1 2 x3 2 x 2 2 x 4 x3 3 x 2 1 dx vu2uv
v
0 1(4 x3 3 x 2 1)
( x 2 1) 2 ( x 2 x 1)2
28. y
( x 1)( x 2)
( x 1)( x 2)
4 x3 3 x 2 1
( x 2 1)2 ( x 2 x 1)2
x 2 3 x 2
x 2 3 x 2
y
( x 2 3 x 2)(2 x 3) ( x 2 3 x 2)(2 x 3)
2
( x 1) ( x 2)
2
6 x 2 12
( x 1) 2 ( x 2) 2
Copyright 2018 Pearson Education, Inc.
119
6( x 2 2)
( x 1) 2 ( x 2) 2
120
Chapter 3 Derivatives
29.
y 12 x 4 32 x 2 x y 2 x3 3x 1 y 6 x 2 3 y 12 x y (4) 12 y ( n) 0 for all n 5
30.
1 x5 y 1 x 4 y 1 x3 y 1 x 2 y (4) x y (5) 1 y ( n ) 0 for all n 6
y 120
24
6
2
31.
y ( x 1)( x 2)( x 3) y ( x 2)( x 3) ( x 1)( x 3) ( x 1)( x 2) x 2 5 x 6 x 2 2 x 3
x 2 x 2 3x 2 8 x 1 y 6 x 8 y 6 y ( n ) 0 for n 4.
32.
y (4 x 2 3)(2 x ) x 4 x 3 8 x 2 3x 6 x 4 x 4 8 x 3 3x 2 6 x y 16 x 3 24 x 2 6 x 6
y 48 x 2 48 x 6 y 96 x 48 y (4) 96 y ( n) 0 for n 5
x3 7
x
33. y
34. s
dy
t 2 5t 1 1 5 1 1 5t 1 t 2
t t2
t2
d 2 s 10t 3 6t 4 10 6
dt 2
t3 t 4
( 1)( 2 1)
35. r
3
du
dx
37. w
3
( x 2 x )( x 2 x 1)
36. u
x4
0 3x 4
5 2
t2
t3
1 3 ddr 0 3 4 3 4 34 d r2 12 5 12
5
d
2
3
x ( x 1)( x 2 x 1)
x4
4
3 x 43
x
x ( x3 1)
d 2u
dx 2
x4
12 x
4
x 4 x 1 x4 1 x 3
5
x
x
12
x5
133z z (3 z) 13 z 1 1 (3 z) z 1 13 3 z z 1 83 z dwdz z 2 0 1 z 2 1 z 1 1
2
dz
38. p
ds
0 5t 2 2t 3 5t 2 2t 3
dt
1 1 1
3
2
d w
2 z 3 0 2 z 3
2
d2y
x 2 7 x 1 dx 2 x 7 x 2 2 x 72 2 2 14 x 3 2 143
x
dx
x
q 2 3
3
3
2
z3
q 2 3
3
2
3
2
( q 1) ( q 1)
( q 3q 3q 1) ( q 3q 3q 1)
d2p
3
1
q 3
dq 2
q
q 2 3
3
2q 6q
q 2 3
2 q ( q 2 3)
21q 12 q 1 dq 12 q 2 1 2
2q
39. u (0) 5, u (0) 3, v(0) 1, v (0) 2
d (uv ) uv vu d (uv )
(a) dx
u (0)v (0) v(0)u (0) 5 2 (1)(3) 13
dx
x 0
v (0)u (0) u (0) v(0)
( 1)( 3) (5)(2)
d u
vu2uv dx
7
v x0
(v (0))2
( 1)2
v
u (0) v(0) v (0)u (0)
(5)(2) ( 1)( 3)
d v uvvu d v
7
25
(c) dx
2
2
u
dx
u
(u (0))
u
(5)2
x0
d (7v 2u ) 7v 2u d (7v 2u ) |
(d) dx
x 0 7v (0) 2u (0) 7 2 2( 3) 20
dx
(b)
d u
dx v
40. u (1) 2, u (1) 0, v(1) 5, v (1) 1
d (uv ) |
(a) dx
x 1 u (1)v (1) v (1)u (1) 2 ( 1) 5 0 2
(b)
(c)
(d)
u (1)v(1)
50 2( 1)
2
25
v(1)u((1)v(1))
(5)
u (1)v(1) v (1)u(1)
2( 1) 50
d v
12
dx u x 1
(u (1))
(2)
d u
dx v x 1
2
2
2
2
d (7v 2u ) |
x 1 7v (1) 2u (1)
dx
7 (1) 2 0 7
Copyright 2018 Pearson Education, Inc.
dp
Section 3.3 Differentiation Rules
121
41. y x3 4 x 1. Note that (2, 1) is on the curve: 1 23 4(2) 1
(a) Slope of the tangent at ( x, y ) is y 3x 2 4 slope of the tangent at (2, 1) is y (2) 3(2)2 4 8. Thus
the slope of the line perpendicular to the tangent at (2, 1) is 18 the equation of the line perpendicular to
the tangent line at (2, 1) is y 1 18 ( x 2) or y 8x 54 .
(b) The slope of the curve at x is m 3x 2 4 and the smallest value for m is 4 when x 0 and y 1.
(c) We want the slope of the curve to be 8 y 8 3 x 2 4 8 3x 2 12 x 2 4 x 2. When
x 2, y 1 and the tangent line has equation y 1 8( x 2) or y 8 x 15; When x 2,
y (2)3 4(2) 1 1, and the tangent line has equation y 1 8( x 2) or y 8 x 17.
42. (a) y x3 3x 2 y 3 x 2 3. For the tangent to be horizontal, we need m y 0 0 3x 2 3
3x 2 3 x 1. When x 1, y 0 the tangent line has equation y 0. The line perpendicular to
this line at (1, 0) is x 1. When x 1, y 4 the tangent line has equation y 4. The line
perpendicular to this line at (1, 4) is x 1.
(b) The smallest value of y is 3, and this occurs when x 0 and y 2. The tangent to the curve at (0, 2)
has slope 3 the line perpendicular to the tangent at (0, 2) has slope 13 y 2 13 ( x 0) or y 13 x 2
is an equation of the perpendicular line.
43. y
4 x dy
dx
x 2 1
( x 2 1)(4) (4 x )(2 x )
2
( x 1)
2
4 8 x 2
( x 2 1) 2
4x
2
4( x 2 1)
( x 2 1) 2
. When x 0, y 0 and y
4(0 1)
1
4, so the tangent
to the curve at (0, 0) is the line y 4 x. When x 1, y 2 y 0, so the tangent to the curve at (1, 2) is the
line y 2.
44. y
8
x 4
2
y
( x 2 4)(0) 8(2 x )
( x 2 4) 2
curve at (2, 1) has the equation
16 x . When x
( x 2 4)2
y 1 12 ( x 2), or
2, y 1 and y
16(2)
(22 4) 2
12 , so the tangent line to the
y 2x 2.
45. y ax 2 bx c passes through (0, 0) 0 a (0) b(0) c c 0; y ax 2 bx passes through (1, 2)
2 a b; y 2ax b and since the curve is tangent to y x at the origin, its slope is 1 at x 0 y 1
when x 0 1 2a(0) b b 1. Then a b 2 a 1. In summary a b 1 and c 0 so the curve is
y x 2 x.
46. y cx x 2 passes through (1, 0) 0 c(1) 1 c 1 the curve is y x x 2 . For this curve, y 1 2 x
and x 1 y 1. Since y x x 2 and y x 2 ax b have common tangents at x 1, y x 2 ax b must
also have slope 1 at x 1. Thus y 2 x a 1 2 1 a a 3 y x 2 3 x b. Since this last curve
passes through (1, 0), we have 0 1 3 b b 2. In summary, a 3, b 2 and c 1 so the curves are
y x 2 3x 2 and y x x 2 .
47. y 8 x 5 m 8; f ( x) 3 x 2 4 x f ( x) 6 x 4;6 x 4 8 x 2 f (2) 3(2) 2 4(2) 4 (2, 4)
48. 8 x 2 y 1 y 4 x 12 m 4; g ( x) 13 x3 32 x 2 1 g ( x) x 2 3 x; x 2 3 x 4 x 4 or x 1
g (4) 13 (4)3 32 (4)2 1 53 , g (1) 13 (1)3 32 (1)2 1 56 4, 53 or 1, 56
49. y 2 x 3 m 2 m 12 ; y
x
x 2
y
( x 2)(1) x (1)
( x 2) 2
2 x 2 x 4 or x 0 if x 4, y 44 2 2, and if
2 ; 2
( x 2)2 ( x 2)2
x 0, y 00 2
Copyright 2018 Pearson Education, Inc.
12 4 ( x 2)2
0 (4, 2) or (0, 0).
122
Chapter 3 Derivatives
2
y 8
f ( x ) x 2 f ( x) 2 x; m f ( x) x 3 2 x xx 38 2 x x 2 8 2 x 2 6 x x 2 6 x 8 0
x 4 or x 2 f (4) 42 16, f (2) 22 4 (4, 16) or (2, 4).
50. m
y 8
;
x 3
51. F ( x ) f ( x) g ( x), F (1) f (1) g (1) (2)(4) 8, and, F ( x) f ( x) g ( x) f ( x) g ( x)
F (1) f (1) g (1) f (1) g (1) (2)(2) (3)(4) 16 tangent line is y 8 16( x 1) y 16 x 24
52.
F ( x)
f ( x ) 3
,
x g ( x)
F (2)
(2 g (2)) f (2) ( f (2) 3)(1 g (2))
(2 g (2)) 2
F (2)
f (2) 3
2 g (2)
3 3
2 ( 4)
1 and F ( x)
( x g ( x )) f ( x ) ( f ( x ) 3)(1 g ( x ))
( x g ( x )) 2
(2 ( 4))( 1) (33)(11)
(2 ( 4)) 2
6
36
1
6
normal line is y 1 6( x 2)
y 6 x 11
53. (a) y x3 x y 3 x 2 1. When x 1, y 0 and y 2 the tangent line to the curve at (1, 0) is
y 2( x 1) or y 2 x 2.
(b)
(c)
y x3 x
y 2 x 2
x x 2x 2 x 3x 2 ( x 2)(x 1)
3
3
2
0 x 2 or x 1. Since y 2(2) 2 6; the
other intersection point is (2, 6)
54. (a) y x3 6 x 2 5 x y 3x 2 12 x 5. When x 0, y 0 and y 5 the tangent line to the curve at
(0, 0) is y 5 x.
(b)
(c)
y x3 6 x 2 5 x
y 5 x
x 6 x 5x 5x x 6 x
3
2
3
2
0 x 2 ( x 6) 0 x 0 or x 6. Since y 5(6) 30,
the other intersection point is (6, 30).
x50 1
x 1 x 1
55. lim
56.
50 x 49
x 2/9 1
x 1 x 1
lim
57. g ( x)
x 1
92 x 7/9
50(1) 49 50
x 1
2 x 3 x 0
a
x 0 , since
2
9( 1)7/9
92
g is differentiable at x 0 lim (2 x 3) 3 and lim a a a 3
x 0
Copyright 2018 Pearson Education, Inc.
x 0
Section 3.4 The Derivative as a Rate of Change
58. f ( x)
a
x 1
2bx x 1 ,
123
since f is differentiable at x 1 lim a a and lim (2bx) 2b a 2b, and
x 1
2
x 1
since f is continuous at x 1 lim (ax b) a b and lim (bx 3) b 3 a b b 3
a 3 3 2b b
x 1
32 .
x 1
59. P( x) an x n an 1 x n 1 a2 x 2 a1 x a0 P ( x) nan x n 1 (n 1)an 1 x n 2 2a2 x a1
60. R M 2
dR CM M 2
C2 M3 C2 M 2 13 M 3 , where C is a constant dM
d (u c ) u dc c du u 0 c du c du . Thus when one of the functions is a
0 dx
61. Let c be a constant dc
dx
dx
dx
dx
dx
constant, the Product Rule is just the Constant Multiple Rule 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 1):
d 1
dx v
v0 1 dv
dx
v2
(b) Now, using the Reciprocal Rule and the Product Rule, we’ll derive the Quotient Rule:
1 dv
dx
v2
1 dv .
v 2 dx
dxd u 1v u dxd 1v 1v dudx (Product Rule) u v1 dvdx 1v dudx (Reciprocal Rule)
d u u v v u , the Quotient Rule.
dx
v v
v
d u
dx v
2
dv
dx
du
dx
du
dx
2
63. (a)
(b)
dv
dx
2
d (uvw)
dx
d ((uv ) w) (uv ) dw w d (uv ) uv dw w u dv v du uv dw wu dv wv du
dx
dx
dx
dx
dx
dx
dx
dx
dx
uvw uv w u vw
d (u u u u )
dx 1 2 3 4
d u u u u u u u du4 u d u u u
dx
1 2 3 4 1 2 3 dx 4 dx 1 2 3
d u u u u u u u du4 u u u du3 u u du2 u u du1
dx
1 2 3 4 1 2 3 dx 4 1 2 dx 3 1 dx 3 2 dx
(using (a) above)
d u u u u u u u du4 u u u du3 u u u du2 u u u du1
dx
1 2 3 4 1 2 3 dx 1 2 4 dx 1 3 4 dx 2 3 4 dx
u1u2u3u4 u1u2u3 u4 u1u2 u3u4 u1u2 u3u4
(c) Generalizing (a) and (b) above,
64.
d ( x m )
dx
d
dx
1
xm
x m 0 1( m x m 1 )
m 2
(x )
d (u u )
n
dx 1
u1u2 un 1un u1u2 un 2un 1un u1u2 un
m 1
m2xm m x m 1 2 m m x m 1
x
2
an2 . We are holding T constant, and a, b, n, R are also constant so their derivatives are zero
65. P VnRT
nb
dP
dV
66.
3.4
A( q)
V
(V nb )0 ( nRT )(1)
(V nb )
km
q
cm
2
hq
2
V 2 (0) ( an 2 )(2V )
(V 2 )2
nRT
(V nb )2
2an3
2
V
( km ) q 1 cm h2 q dA
( km ) q 2 h2
dq
2
km2 h2 d 2A 2( km) q 3 2km
3
q
dt
THE DERIVATIVE AS A RATE OF CHANGE
1. s t 2 3t 2, 0 t 2
(a) displacement s s (2) s (0) 0 m 2 m 2 m, vav
(b) v
ds
dt
s
t
2t 3 | v(0)| | 3| 3 m/sec and | v(2)| 1 m/sec; a
a (2) 2 m/sec2
2 1 m/sec
2
2
d 2s 2 a (0) 2
dt
Copyright 2018 Pearson Education, Inc.
m/sec 2 and
q
124
Chapter 3 Derivatives
(c) v 0 2t 3 0 t 32 . v is negative in the interval 0 t
body changes direction at t 32 .
2. s 6t t 2 , 0 t 6
(a) displacement s s (6) s (0) 0 m, vav
(b) v
ds
dt
s
t
0
6
3
2
and v is positive when
3
2
t 2 the
0 m/ sec
2
6 2t | v(0)| |6| 6 m/ sec and | v(6)| | 6| 6 m/ sec; a d 2s 2 a(0) 2 m/ sec2 and
dt
a (6) 2 m/ sec2
(c) v 0 6 2t 0 t 3. v is positive in the interval 0 t 3 and v is negative when 3 t 6 the
body changes direction at t 3.
3. s t 3 3t 2 3t , 0 t 3
(a) displacement s s (3) s (0) 9 m, vav
(b) v
ds
dt
s
t
39 3 m/ sec
2
3t 2 6t 3 | v(0)| | 3| 3 m/ sec and | v(3)| | 12| 12 m/ sec; a d 2s 6t 6
dt
a(0) 6 m/ sec2 and a (3) 12 m/ sec 2
(c) v 0 3t 2 6t 3 0 t 2 2t 1 0 (t 1) 2 0 t 1. For all other values of t in the interval
the velocity v is negative (the graph of v 3t 2 6t 3 is a parabola with vertex at t 1 which opens
downward the body never changes direction).
4. s
t4
4
t 3 t 2 , 0 t 3
(a) s s (3) s (0) 94 m, vav
s
t
9
34 34 m/ sec
(b) v t 3 3t 2 2t | v(0)| 0 m/ sec and | v(3)| 6 m/sec; a 3t 2 6t 2 a (0) 2 m/ sec2 and
a (3) 11 m/ sec 2
(c) v 0 t 3 3t 2 2t 0 t (t 2)(t 1) 0 t 0, 1, 2 v 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 2 t 3 the body changes direction at t 1
and at t 2.
5. s
25
t2
5t , 1 t 5
(a) s s (5) s (1) 20 m, vav
(b) v
(c) v
50
t3
20
4
5 m/ sec
52 | v(1)| 45 m/sec and | v(5)| 15 m/ sec; a 150
103
t
t4
t
0 5035t 0 50 5t 0 t 10 the body does not
t
, 4t 0
6. s t25
5
(a) s s (0) s (4) 20 m, vav 20
5 m/sec
4
(b) v
25
(t 5)2
(c) v 0
| v(4)| 25 m/ sec and | v(0)| 1 m/ sec; a
25
(t 5) 2
50
(t 5)3
4 m/sec 2
a (1) 140 m/ sec2 and a (5) 25
change direction in the interval
a (4) 50 m/ sec2 and a (0) 52 m/ sec2
0 v is never 0 the body never changes direction
7. s t 3 6t 2 9t and let the positive direction be to the right on the s -axis.
(a) v 3t 2 12t 9 so that v 0 t 2 4t 3 (t 3)(t 1) 0 t 1 or 3; a 6t 12 a (1) 6 m/ sec2
and a (3) 6 m/ sec2 . Thus the body is motionless but being accelerated left when t 1, and motionless
but being accelerated right when t 3.
(b) a 0 6t 12 0 t 2 with speed | v(2)| |12 24 9| 3 m/sec
(c) The body moves to the right or forward on 0 t 1, and to the left or backward on 1 t 2. The positions
are s (0) 0, s (1) 4 and s (2) 2 total distance | s (1) s (0)| | s (2) s (1)| |4| | 2| 6 m.
Copyright 2018 Pearson Education, Inc.
Section 3.4 The Derivative as a Rate of Change
125
8. v t 2 4t 3 a 2t 4
(a) v 0 t 2 4t 3 0 t 1 or 3 a (1) 2 m/sec 2 and a (3) 2 m/sec 2
(b) v 0 (t 3) (t 1) 0 0 t 1 or t 3 and the body is moving forward; v 0 (t 3)(t 1) 0
1 t 3 and the body is moving backward
(c) velocity increasing a 0 2t 4 0 t 2; velocity decreasing a 0 2t 4 0 0 t 2
9. sm 1.86t 2 vm 3.72t and solving 3.72t 27.8 t 7.5 sec on Mars; s j 11.44t 2 v j 22.88t and
solving 22.88t 27.8 t 1.2 sec on Jupiter.
10 . (a) v(t ) s (t ) 24 1.6t m/sec, and a(t ) v (t ) s (t ) 1.6 m/sec 2
(b) Solve v(t ) 0 24 1.6t 0 t 15sec
(c) s (15) 24(15) .8(15)2 180 m
2
4.39 sec going up and 25.6 sec going down
(d) Solve s (t ) 90 24t .8t 2 90 t 3015
2
(e) Twice the time it took to reach its highest point or 30 sec
. Therefore g s 15
11. s 15t 12 g s t 2 v 15 g s t so that v 0 15 g s t 0 g s 15
t
20
3
4
0.75 m/sec2
12. Solving sm 832t 2.6t 2 0 t (832 2.6t ) 0 t 0 or 320 320 sec on the moon;
solving se 832t 16t 2 0 t (832 16t ) 0 t 0 or 52 52 sec on the earth. Also, vm 832 5.2t 0
t 160 and sm (160) 66,560 ft, the height it reaches above the moon’s surface; ve 832 32t 0
t 26 and se (26) 10,816 ft, the height it reaches above the earth’s surface.
13. (a) s 179 16t 2 v 32t speed | v | 32t ft/sec and a 32 ft/sec2
(b) s 0 179 16t 2 0 t
(c) When t
14. (a)
179 , v
16
179
16
3.3 sec
32 179
8 179 107.0 ft/sec
16
lim v lim 9.8(sin )t 9.8t so we expect v 9.8t m/sec in free fall
2
(b) a
dv
dt
2
9.8 m/sec 2
15. (a) at 2 and 7 seconds
(c)
(b) between 3 and 6 seconds: 3 t 6
(d)
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
Copyright 2018 Pearson Education, Inc.
126
Chapter 3 Derivatives
(b)
17. (a)
(c)
(e)
(f)
(g)
190 ft/sec
(b) 2 sec
at 8 sec, 0 ft/sec
(d) 10.8 sec, 90 ft/sec
From t 8 until t 10.8 sec, a total of 2.8 sec
Greatest acceleration happens 2 sec after launch
v (10.8) v (2)
From t 2 to t 10.8 sec; during this period, a 10.8 2 32 ft/sec 2
18. (a) Forward: 0 t 1 and 5 t 7; Backward: 1 t 5; Speeds up: 1 t 2 and 5 t 6;
Slows down: 0 t 1, 3 t 5, and 6 t 7
(b) Positive: 3 t 6; negative: 0 t 2 and 6 t 7; zero: 2 t 3 and 7 t 9
(c) t 0 and 2 t 3
(d) 7 t 9
19.
s 490t 2 v 980t a 980
(a) Solving 160 490t 2 t 74 sec. The average velocity was
s (4/7) s (0)
4/7
280 cm/sec.
(b) At the 160 cm mark the balls are falling at v(4/7) 560 cm/sec. The acceleration at the 160 cm mark
was 980 cm/sec2.
17 29.75 flashes per second.
(c) The light was flashing at a rate of 4/7
20. (a)
(b)
21. C position, A velocity, and B 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.
Copyright 2018 Pearson Education, Inc.
Section 3.4 The Derivative as a Rate of Change
127
22. C position, B velocity, and A 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.
$110
23. (a) c(100) 11, 000 cav 11,000
100
(b) c( x) 2000 100 x .1x 2 c ( x) 100 .2 x. Marginal cost c ( x) the marginal cost of producing
100 machines is c(100) $80
(c) The cost of producing the 101st machine is c(101) c(100) 100 201
$79.90
10
24. (a) r ( x) 20000 1 1x r ( x)
(b) r (101) $1.96.
(c) lim r ( x) lim
x
20000
x2
x
20000 ,
x2
which is marginal revenue. r (100)
20000
1002
$2.
0. The increase in revenue as the number of items increases without bound will
approach zero.
25. b(t ) 106 104 t 103 t 2 b(t ) 104 (2)(103 t ) 103 (10 2t )
(b) b(5) 0 bacteria/hr
(a) b(0) 104 bacteria/hr
(c) b(10) 104 bacteria/hr
1
26. S ( w) 120
180
w
27. (a) y 6 1 12t
1
80 w
t
6 1 6t 144
2
; S increases more rapidly at lower weights where the derivative is greater.
2
dy
dt
t 1
12
dy
is 0 m/h when t 12 and the fluid level is falling the slowest at that time.
dt
dy
of dt is 1 m/h, when t 0, and the fluid level is falling the fastest at that time.
(b) The largest value of
The smallest value
(c) In this situation,
dy
dt
0 the graph of y is
dy
always decreasing. As dt increases in value,
the slope of the graph of y increases from 1
to 0 over the interval 0 t 12.
28. Q(t ) 200(30 t )2 200(900 60t t 2 ) Q (t ) 200(60 2t ) Q (10) 8, 000 gallons/min is the rate
Q (10) Q (0)
the water is running at the end of 10 min. Then
10, 000 gallons/min is the average rate the water
10
flows during the first 10 min. The negative signs indicate water is leaving the tank.
29. s ( v ) 1.1 0.108v; s (35) 4.88, s (70) 8.66. The units of ds / dv are ft/mph; ds / dv gives, roughly, the
number of additional feet required to stop the car if its speed increases by 1 mph.
4 r 2 dV
4 (2) 2 16 ft 3 /ft
30. (a) V 43 r 3 dV
dr r 2
dr
(b) When r 2, dV
16 so that when r changes by 1 unit, we expect V to change by approximately 16 .
dr
Therefore when r changes by 0.2 units V changes by approximately (16 )(0.2) 3.2 10.05 ft 3 .
Note that V (2.2) V (2) 11.09 ft 3 .
31. 200 km/hr 55 95 m/sec
t 25, D 10
(25)2
9
500
9
6250 m
9
m/sec, and D 10
t 2 V 20
t. Thus V 500
20
t 500
t 25sec. When
9
9
9
9
9
Copyright 2018 Pearson Education, Inc.
128
Chapter 3 Derivatives
v
v2 v2
v0
; 1900 v0 t 16t 2 so that t 320 1900 320 640
32
80 19 ft 60 sec 60 min 1 mi
finally, sec 1 min 1 hr 5280 ft 238 mph.
32. s v0 t 16t 2 v v0 32t ; v 0 t
v0 (64)(1900) 80 19 ft/sec and,
33.
v 0 when t 6.25sec
v 0 when 0 t 6.25 body moves right (up); v 0 when 6.25 t 12.5 body moves left (down)
body changes direction at t 6.25 sec
body speeds up on (6.25, 12.5] and slows down on [0, 6.25)
The body is moving fastest at the endpoints t 0 and t 12.5 when it is traveling 200 ft/sec. It’s moving
slowest at t 6.25 when the speed is 0.
(f ) When t 6.25 the body is s 625 m from the origin and farthest away.
(a)
(b)
(c)
(d)
(e)
34.
(a) v 0 when t 32 sec
(b) v 0 when 0 t 1.5 body moves left (down); v 0 when 1.5 t 5 body moves right (up)
(c) body changes direction at t 32 sec
(d) body speeds up on 32 , 5 and slows down on 0, 32
(e) body is moving fastest at t 5 when the speed | v(5)| 7 units/sec; it is moving slowest at t 32 when
the speed is 0
(f ) When t 5 the body is s 12 units from the origin and farthest away.
35.
(a) v 0 when t
6 15
sec
3
Copyright 2018 Pearson Education, Inc.
Section 3.5 Derivatives of Trigonometric Functions
(b) v 0 when 6 3 15 t 6 3 15 body moves left (down); v 0 when 0 t
body moves right (up)
(c) body changes direction at t
6 15
sec
3
6 3 15 , 4
6 15
3
or 6 3 15 t 4
and slows down on 0, 6 3 15 2, 6 3 15 .
(e) The body is moving fastest at t 0 and t 4 when it is moving 7 units/sec and slowest at t
(d) body speeds up on
6 15
3
(f ) When t
6 15
,2
3
129
6 15
sec
3
the body is at position s 6.303 units and farthest from the origin.
36.
6 15
3
0 when 0 t 6 3 15
(a) v 0 when t
(b) v
or 6 3 15 t 4 body is moving left (down); v 0 when 6 3 15 t
body is moving right (up)
(c) body changes direction at t
6 15
sec
3
6 3 15 , 4
6 15
3
and slows down on 0, 6 3 15 2, 6 3 15
(e) The body is moving fastest at 7 units/sec when t 0 and t 4; it is moving slowest and stationary
(d) body speeds up on
6 15
3
When t 6 3 15
6 15
,2
3
at t
(f)
3.5
DERIVATIVES OF TRIGONOMETRIC FUNCTIONS
1. y 10 x 3cos x
2. y 3x 5sin x
3. y x 2 cos x
dy
dx
dy
dx
4. y x sec x 3
5.
the position is s 10.303 units and the body is farthest from the origin.
dy
dx
d (cos x ) 10 3sin x
10 3 dx
3 5 d
dx
x2
(sin x) 23 5cos x
x
x 2 ( sin x) 2 x cos x x 2 sin x 2 x cos x
dy
dx
x sec x tan x sec x 0
2 x
x sec x tan x sec x
2 x
dy
y csc x 4 x 7 dx csc x cot x 4
2 x
6. y x 2 cot x
1
x2
dy
dx
x2
x 2 csc2 x 2 x cot x 23
d (cot
dx
d ( x2 )
x) cot x dx
2
x3
x 2 csc2 x (cot x)(2 x) 23
x
Copyright 2018 Pearson Education, Inc.
x
130
Chapter 3 Derivatives
sin x sin x(sec 2 x 1)
7. f ( x) sin x tan x f ( x) sin x sec2 x cos x tan x sin x sec 2 x cos x cos
x
8.
x csc x cot x g ( x ) csc x ( csc2 x ) ( csc x cot x ) cot x csc3 x csc x cot 2 x
g ( x ) cos2 x sin1 x cos
sin x
sin x
csc x(csc2 x cot 2 x)
9.
10.
y x sec x
1
dy d
d
1
1
( x ) sec x x (sec x ) 2 sec x x sec x tan x 2
x
dx dx
dx
x
x
dy
dx
d (sec x) sec x d (sin x cos x )
dx
dx
(sin x cos x )sin x cos x sin x
(sin x cos x)(sec x tan x) (sec x)(cos x sin x)
cos x
cos 2 x
2
2
sin x cos x sin x 2cos x cos x sin x 12 sec2 x
cos x
cos x
y (sin x cos x) sec x
(sin x cos x)
Note also that y sin x sec x cos x sec x tan x 1
11.
x
y 1cot
cot x
12.
d (cot x ) (cot x ) d (1 cot x )
(1cot x ) dx
dx
(1cot x ) 2
csc2 x csc2 x cot x csc2 x cot x
(1cot x ) 2
x
y 1cos
sin x
dy
dx
sin x 1
(1sin x )2
dy
dx
(1sin x )
1cos x csc2 x cot x csc2 x
1cos x 2
2
(1sin x )( sin x )(cos x )(cos x )
(1sin x )2
2
2
sin x sin x2cos x
(1sin x )
1
1sin
x
dy
dx
13.
y cos4 x tan1 x 4sec x cot x
14.
y cosx x cosx x
15.
y (sec x tan x) (sec x tan x)
dy
dx
sec2 x.
csc2 x
(1cot x ) 2
d (cos x ) (cos x ) d (1sin x )
(1sin x ) dx
dx
(1sin x )
(1sin x ) 2
dy
dx
4sec x tan x csc2 x
x ( sin x )(cos x )(1)
x2
dy
dx
(cos x )(1) x ( sin x )
cos 2 x
x sin x cos x
x2
cos x x2 sin x
cos x
d (sec x tan x) (sec x tan x) d (sec x tan x)
(sec x tan x) dx
dx
(sec x tan x)(sec x tan x sec2 x) (sec x tan x) (sec x tan x sec2 x)
(sec2 x tan x sec x tan 2 x sec3 x sec2 x tan x) (sec2 x tan x sec x tan 2 x sec3 x tan x sec2 x ) 0.
Note also that y sec
16.
2
x tan 2 x (tan 2 x 1) tan 2 x 1
y x 2 cos x 2 x sin x 2 cos x
dy
dx
dy
dx
0.
( x 2 ( sin x) (cos x)(2 x)) (2 x cos x (sin x)(2)) 2( sin x)
x 2 sin x 2 x cos x 2 x cos x 2sin x 2sin x x 2 sin x
17.
f ( x) x3 sin x cos x f ( x) x3 sin x( sin x ) x3 cos x(cos x) 3x 2 sin x cos x
x3 sin 2 x x3 cos 2 x 3 x 2 sin x cos x
Copyright 2018 Pearson Education, Inc.
Section 3.5 Derivatives of Trigonometric Functions
18. g ( x ) (2 x) tan 2 x g ( x) (2 x) (2 tan x sec2 x) (1) tan 2 x 2(2 x) tan x sec2 x tan 2 x
2(2 x) tan x (sec 2 x tan x)
csc t
21. s 11csc
t
ds
dt
t
22. s 1sin
cos t
sec2 t 1
ds
dt
19. s tan t t
(1csc t )( csc t cot t ) (1 csc t )(csc t cot t )
ds
dt
20. s t 2 sec t 1
(1csc t )2
(1cos t )
2t sec t tan t
2
2
csc t cot t csc t cot t csc2 t cot t csc t cot t 2 csc t cot2 t
(1csc t )
(1 csc t )
(1cos t )(cos t ) (sin t )(sin t )
ds
dt
2
2
2
1
cos t cos t 2sin t cos t 1 2 1cos
cos1t 1
t
(1cos t )
(1cos t )
23. r 4 2 sin ddr 2 dd (sin ) (sin )(2 ) ( 2 cos 2 sin ) ( cos 2sin )
24. r sin cos ddr ( cos (sin )(1)) sin cos
25. r sec csc ddr (sec )( csc cot ) (csc )(sec tan )
21 12 sec2 csc2
sin
1
sin
1
cos1 sin1 cos
sin sin cos cos
cos
26. r (1 sec ) sin ddr (1 sec ) cos (sin ) (sec tan ) (cos 1) tan 2 cos sec2
27. p 5 cot1 q 5 tan q
28. p (1 csc q) cos q
29. p
sin q cos q
cos q
tan q
30. p 1 tan q
31. p
dp
dq
dp
dq
dp
dq
sec2 q
(1 csc q )( sin q ) (cos q )( csc q cot q ) ( sin q 1) cot 2 q sin q csc2 q
(cos q )(cos q sin q ) (sin q cos q )( sin q )
2
cos q
(1 tan q )(sec2 q ) (tan q )(sec 2 q )
(1 tan q )
2
cos 2 q cos q sin q sin 2 q cos q sin q
cos 2 q
sec 2 q tan q sec2 q tan q sec2 q
( q 2 1)( q cos q sin q (1)) ( q sin q )(2 q )
q 2 1
q 3 cos q q 2 sin q q cos q sin q
( q 2 1)2
(1 tan q )
2
1
cos 2 q
sec2 q
sec2 q
(1 tan q )2
q3 cos q q 2 sin q q cos q sin q 2 q 2 sin q
( q 2 1) 2
( q 2 1)2
32. p
q sin q
dp
dq
dp
dq
3q tan q
q sec q
3
dp
dq
( q sec q )(3 sec2 q ) (3q tan q )( q sec q tan q sec q (1))
( q sec q ) 2
2
3q sec q q sec q (3q sec q tan q 3q sec q q sec q tan 2 q sec q tan q )
( q sec q )2
q sec3 q 3q 2 sec q tan q q sec q tan 2 q sec q tan q
33. (a)
(b)
( q sec q )2
y csc x y csc x cot x y ((csc x)( csc2 x) (cot x)( csc x cot x)) csc3 x csc x cot 2 x
(csc x)(csc2 x cot 2 x) (csc x)(csc2 x csc2 x 1) 2 csc3 x csc x
y sec x y sec x tan x y (sec x)(sec2 x) (tan x)(sec x tan x) sec3 x sec x tan 2 x
(sec x)(sec2 x tan 2 x) (sec x)(sec2 x sec2 x 1) 2sec3 x sec x
Copyright 2018 Pearson Education, Inc.
131
132
Chapter 3 Derivatives
34. (a)
y 2 sin x y 2 cos x y 2( sin x) 2sin x y 2 cos x y (4) 2 sin x
(b)
y 9 cos x y 9sin x y 9 cos x y 9( sin x ) 9sin x y (4) 9 cos x
35. y sin x y cos x slope of tangent at x is
y ( ) cos ( ) 1; slope of tangent at x 0 is
y (0) cos (0) 1; and slope of tangent at x 32 is
y ( 32 ) cos 32 0. The tangent at ( , 0) is
y 0 1( x ), or y x ; the tangent at (0, 0) is
y 0 1 ( x 0), or y x; and the tangent at
3 , 1 is y 1.
2
36. y tan x y sec2 x slope of tangent at x 3 is
sec 2 3 4; slope of tangent at x 0 is sec 2 (0) 1; and
slope of tangent at x 3 is sec2 3 4. The tangent
3
at , tan
3
, 3 is y 3 4 x ; the
3
tangent at (0, 0) is y x; and the tangent at
3 , tan 3 3 , 3 is y
3
3 4 x 3 .
37. y sec x y sec x tan x slope of tangent at
x 3 is sec 3 tan 3 2 3; slope of tangent
4 4 2. The tangent at the point
4
3 , sec 3 3 , 2 is y 2 2 3 x 3 ; the
tangent at the point 4 , sec 4 4 , 2 is
y 2 2 x 4 .
at x is sec tan
38. y 1 cos x y sin x slope of tangent at x 3 is
3π2 1.
The tangent at the point 3 , 1 cos 3 3 , 32
is y 32 23 x 3 ; the tangent at the point
32 ,1 cos 32 32 , 1 is y 1 x 32
sin 3
3
; slope
2
of tangent at x 32 is sin
39. Yes, y x sin x y 1 cos x; horizontal tangent occurs where 1 cos x 0 cos x 1 x
40. No, y 2 x sin x y 2 cos x; horizontal tangent occurs where 2 cos x 0 cos x 2. But there are
no x-values for which cos x 2.
41. No, y x cot x y 1 csc2 x; horizontal tangent occurs where 1 csc2 x 0 csc2 x 1. But there are no
x-values for which csc2 x 1.
42. Yes, y x 2 cos x y 1 2 sin x; horizontal tangent occurs where 1 2 sin x 0 1 2sin x
12 sin x x 6 or x 56
Copyright 2018 Pearson Education, Inc.
Section 3.5 Derivatives of Trigonometric Functions
x y
43. Yes, y 3sec
sec x
(3sec x )sec x tan x sec xsec x tan x
(3sec x )2
133
3sec x tan 2x ; horizontal tangent occurs when
(3 sec x )
sec x tan x 0 tan x 0 x 0, x , or x 2 .
44.
x y
No, y 3cos
4sin x
(3 4sin x )( sin x ) cos x( 4cos x )
(3 4sin x )2
2
2
4sin x 4cos x 2 3sin x
(3 4sin x )
4 3sin x ;
(3 4sin x )2
horizontal tangent
occurs when 4 3 sin x 0 sin x 43 . But there are no x-values for which sin x 43 .
45. We want all points on the curve where the tangent
line has slope 2. Thus, y tan x y sec2 x so that
y 2 sec2 x 2 sec x 2 x 4 . Then the
tangent line at 4 , 1 has equation y 1 2 x 4 .
tangent line at 4 , 1 has equation y 1 2 x 4 ; the
46. We want all points on the curve y cot x where the tangent
line has slope 1. Thus y cot x y csc2 x so that
y 1 csc2 x 1 csc2 x 1 csc x 1 x 2 .
The tangent line at 2 , 0 is y x 2 .
47.
1sin2cosx x
y 4 cot x 2csc x y csc2 x 2csc x cot x sin1 x
(a) When x 2 , then y 1; the tangent line is y x 2 2.
(b) To find the location of the horizontal tangent set y 0 1 2 cos x 0 x 3 radians. When x 3 ,
then y 4 3 is the horizontal tangent.
48.
y 1 2 csc x cot x y 2 csc x cot x csc2 x sin1 x
2 cos x 1
sin x
(a) If x 4 , then y 4; the tangent line is y 4 x 4
(b) To find the location of the horizontal tangent set y 0
2 cos x 1 0 x 34 radians.
When x 34 , then y 2 is the horizontal tangent.
49.
50.
lim sin 1x 12 sin 12 12 sin 0 0
x 2
lim
x 6
1 cos( csc x) 1 cos( csc( 6 )) 1 cos( ( 2)) 2
Copyright 2018 Pearson Education, Inc.
134
51.
52.
53.
54.
55.
56.
Chapter 3 Derivatives
lim
sin 12
6 6
dd (sin ) cos cos 6 23
6
6
lim tan 1 dd (tan ) sec2 sec2 4 2
4
4
4 4
1 sec 1 tan 4 sec0
1 sec tan 4 sec ( ) 1
lim sec cos x tan 4 sec
x
x0
0
lim sin tan x 2tansecx x sin tan0tan
sin 2 1
2sec0
x 0
lim tan 1
t 0
sin t
t
tan 1 lim
sin t
t 0 t
tan (1 1) 0
1
cos lim cos
lim cos sin
0
0 sin
lim sin
0
57. s 2 2 sin t v
ds
dt
2 cos t a
dv
dt
cos 1 1
1
2 sin t j
da
dt
2cos t. Therefore, velocity v 4
2 m/sec; speed | v 4 | 2 m/sec; acceleration a 4 2 m/sec2 ; jerk j 4 2 m/sec3 .
58. s sin t cos t v
ds
dt
cos t sin t a
dv
dt
sin t cos t j
da
dt
cos t sin t. Therefore velocity
v 4 0 m/sec; speed v 4 0 m/sec; acceleration a 4 2 m/sec2 ; jerk j 4 0 m/sec3 .
59.
60.
2
lim f ( x) lim sin 23 x lim 9 sin3 x3 x
x 0
x
x 0
x 0
sin3x3x 9 so that f is continuous at
x 0 lim f ( x ) f (0) 9 c.
x 0
lim g ( x) lim ( x b) b and lim g ( x) lim cos x 1 so that g is continuous at x 0 lim g ( x)
x 0
x 0
x 0
x 0
x 0
lim g ( x) b 1. Now g is not differentiable at x 0: At x 0, the left-hand derivative is
x 0
d ( x b)|
x 0 1,
dx
but the right-hand derivative is
d
dx
(cos x)|x 0 sin 0 0. The left- and right-hand
derivatives can never agree at x 0, so g is not differentiable at x 0 for any value of b (including b 1 )
61. (a)
d 999
dx999
(cos x) sin x because
d4
dx 4
(cos x) cos x the derivative of cos x any number of times that is a
multiple of 4 is cos x. Thus, dividing 999 by 4 gives 999 249 4 3
(b)
d 3 d 249 4
dx3 dx 249 4
d 110
dx110
(cos x)
d3
dx3
d2
dx 2
(cos x) cos x because
(cos x) cos x;
d 110
dx110
(cos x)
(cos x) sin x.
d4
dx 4
(cos x) cos x the derivative of cos x any number of times that is a
multiple of 4 is cos x. Thus, dividing 110 by 4 gives 110 27 4 2
d 999
dx999
(sin x) sin x because
d4
dx 4
d 110
dx110
(cos x)
d2
dx 2
d 27 4 (cos x)
dx 27 4
(sin x) sin x the derivative of sin x any
number of times that is a multiple of 4 is sin x. Thus, dividing110 by 4 gives 110 27 4 2
Copyright 2018 Pearson Education, Inc.
Section 3.5 Derivatives of Trigonometric Functions
d 110
dx110
(c)
(sin x)
d 110
dx110
d2
dx 2
d 27 4 (sin x)
dx 27 4
d2
dx 2
(sin x) sin x. Now,
d 110
dx110
135
(sin x 3cos x)
110
(sin x) 3 d 110 (cos x) sin x 3( cos x) 3cos x sin x
d ( x sin x)
dx
dx
x cos x sin x
d2
dx 2
d3
dx3
( x sin x) x sin x 2cos x
d4
dx 4
( x sin x) x sin x 4cos x
d7
dx 7
( x sin x) x cos x 7sin x; let n 2k 1 for k 0, 1, 2, 3,
(1)k x cos x (1)k n sin x
d5
dx5
( x sin x) x cos x 5sin x
d 73
dx 73
( x sin x)
d 2 361 ( x sin x)
d 2 361
( x sin x) x cos x 3sin x
d6
dx 6
( x sin x) x sin x 6cos x
d n ( x sin x ) d 2 k 1 ( x sin x)
dx n
dx 2 k 1
36
36
(1) x cos x (1)
73sin x
x cos x 73sin x
62. (a)
y sec x cos1 x
dy
dx
(cos x )(0) (1)( sin x )
(b)
y csc x sin1 x
dy
dx
(sin x )(0) (1)(cos x )
(c)
sin 2 x cos 2 x
sin 2 x
1
sin 2 x
(cos x )2
(sin x ) 2
csc2 x
sin x sec x tan x
cos
x
cos x 1
x csc x cot x
sin x cos
sin x
sin x
sin2x cos1 x
cos x
2
d (cot x ) csc 2
dx
d (sec x ) sec x tan x
dx
d (csc x ) csc x cot x
dx
x
cm ; t v 10sin
0 sec
3
3
63. (a) t 0 x 10cos(0) 10 cm; t 3 x 10cos 3 5 cm; t 34 x 10cos 34 5 2 cm
(b) t 0 v 10sin(0)
cm ; t 3 v 10sin 3 5 2 cm
5 3 sec
4
4
sec
64. (a) t 0 x 3cos(0) 4sin(0) 3 ft; t 2 x 3cos 2 4sin 2 4 ft;
t x 3cos( ) 4 sin( ) 3 ft
ft ; t v 3sin 4cos 3 ft ;
(b) t 0 v 3sin(0) 4cos(0) 4 sec
2
2
2
sec
ft
t v 3 sin( ) 4 cos( ) 4 sec
65.
As h takes on the values of 1, 0.5, 0.3 and 0.1 the corresponding dashed curves of y
and closer to the black curve y cos x because
takes on the values of 1, 0.5, 0.3 and 0.1.
d (sin x ) lim sin( x h ) sin x
dx
h
h 0
66.
Copyright 2018 Pearson Education, Inc.
sin( x h ) sin x
h
get closer
cos x. The same is true as h
136
Chapter 3 Derivatives
As h takes on the values of 1, 0.5, 0.3, and 0.1 the corresponding dashed curves of y
and closer to the black curve y sin x because
takes on the values of 1, 0.5, 0.3, and 0.1.
d (cos x ) lim cos( x h ) cos x
dx
h
h 0
cos( x h ) cos x
h
get closer
sin x. The same is true as h
67. (a)
The dashed curves of y
sin( x h ) sin( x h )
2h
are closer to the black curve y cos x than the corresponding
dashed curves in Exercise 65 illustrating that the centered difference quotient is a better approximation of
the derivative of this function.
(b)
The dashed curves of y
cos( x h ) cos( x h )
2h
are closer to the black curve y sin x than the corresponding
dashed curves in Exercise 66 illustrating that the centered difference quotient is a better approximation of the
derivative of this function.
68.
lim
h 0
|0 h||0 h|
2h
|h||h|
lim 2 h lim 0 0 the limits of the centered difference quotient exists even though the
x 0
h 0
derivative of f ( x) | x | does not exist at x 0.
69.
y tan x y sec2 x, so the smallest value y sec2 x
takes on is y 1 when x 0; y has no maximum value since
sec2 x has no largest value on 2 , 2 ; y is never negative
since sec2 x 1.
70.
y cot x y csc2 x so y has no smallest value since
csc2 x has no minimum value on (0, ); the largest value
of y is 1, when x 2 ;
the slope is never positive since the largest value y csc2 x
takes on is 1.
Copyright 2018 Pearson Education, Inc.
Section 3.5 Derivatives of Trigonometric Functions
71.
y sinx x appears to cross the y -axis at y 1, since
lim sin x
x 0 x
1; y sinx2 x appears to cross the y -axis at y 2,
since lim
x 0
sin 2 x
x
2; y sinx4 x appears to cross the y -axis at
sin 4 x
x
y 4, since lim
x0
4. However, none of these graphs
actually cross the y -axis since x 0 is not in the domain of
sin 5 x
x 0 x
the functions. Also, lim
sin( 3 x )
x
x 0
5, lim
3, and
sin( 3 x )
lim sinxkx k the graphs of y sinx5 x , y
, and
x
x 0
y
sin kx
x
approach 5, 3, and k, respectively, as x 0.
However, the graphs do not actually cross the y -axis.
sin h
h
sin h
h
72. (a) h
.017452406
.017453292
.017453292
.017453292
1
0.01
0.001
0.0001
sin h
h 0 h
lim
lim
sin h . 180
h
x 0
180
.99994923
1
1
1
lim 180 sin h 180 lim 180 sin
h 0
180
.h
0
h 180
180
(converting to radians)
cos h 1
(b) h
h
0.0001523
1
0.01
0.001
0.0001
cos h 1
h 0 h
lim
0.0000015
0.0000001
0
0, whether h is measured in degrees or radians.
(c) In degrees,
d (sin x ) lim sin( x h ) sin x
dx
h
h 0
lim sin x
h 0
cos h 1
h
lim
cos h 1
h
h 0
sin h
h
dx
(cos x) lim
h 0
cos h 1
(cos x) lim
h
h 0
h
lim
h 0
(cos x) lim
sin h
h 0 h
(cos x cos h sin x sin h ) cos x
h
lim sin x
(sin x) lim (cos x)(0) (sin x)
(cos x )(cos h 1) sin x sin h
h
h 0
lim
(e)
(sin x cos h cos x sin h ) sin x
h
lim cos x (sin x) lim
h 0
cos x
(sin x)(0) (cos x) 180 180
cos( x h ) cos x
d
(d) In degrees,
h 0
lim cos x
h 0
cos h 1
h
sin h
h
h 0
sin h
h 0 h
180
2
180
d 2 sin x 3 cos x;
180
dx 180
d2
dx 2
(sin x)
d2
dx 2
d sin x 2 cos x; d 3 (cos x ) d 2 cos x 3 sin x
(cos x) dx
180
dx 180
180
180
dx3
d cos x
dx 180
sin x;
d3
dx3
sin x
180
(sin x)
Copyright 2018 Pearson Education, Inc.
137
138
3.6
Chapter 3 Derivatives
THE CHAIN RULE
1. f (u ) 6u 9 f (u ) 6 f ( g ( x)) 6; g ( x) 12 x 4 g ( x) 2 x3 ;
therefore
dy
dx
f ( g ( x)) g ( x) 6 2 x3 12 x3
2. f (u ) 2u 3 f (u ) 6u 2 f ( g ( x)) 6(8 x 1) 2 ; g ( x) 8 x 1 g ( x) 8;
dy
therefore f ( g ( x)) g ( x) 6(8 x 1) 2 8 48(8 x 1)2
dx
3. f (u ) sin u f (u ) cos u f ( g ( x)) cos(3 x 1); g ( x) 3 x 1 g ( x) 3;
dy
therefore f ( g ( x)) g ( x ) (cos(3 x 1))(3) 3cos(3 x 1)
dx
4.
f (u ) cos u f (u ) sin u f ( g ( x )) sin( x /3); g ( x ) x /3 g ( x ) 1/3; therefore,
dy
dx
5.
f ( g ( x )) g ( x ) sin( x /3)( 1/3) (1/3)sin( x /3)
f ( u ) u f (u )
1
2 u
dy
dx
cos x
2 sin x
f ( g ( x )) g ( x )
f ( g ( x ))
1 ;
2 sin x
g ( x ) sin x g ( x ) cos x; therefore,
6. f (u ) sin u f (u ) cos u f ( g ( x)) cos( x cos x); g ( x) x cos x g ( x) 1 sin x;
dy
therefore f g ( x )) g ( x) (cos( x cos x))(1 sin x)
dx
7.
f (u ) tan u f (u ) sec 2 u f ( g ( x )) sec2 ( x 2 ); g ( x ) x 2 g ( x ) 2 x;
dy
dx
therefore
8.
f ( g ( x )) g ( x ) sec2 ( x 2 )(2 x ) 2 x sec2 ( x 2 )
f (u ) sec u f (u ) sec u tan u f ( g ( x )) sec 1x 7 x tan( 1x 7 x ); g ( x ) 1x 7 x
dy
g ( x ) 12 7; therefore, dx f ( g ( x )) g ( x ) 12 7 sec( 1x 7 x ) tan( 1x 7 x )
x
x
9. With u (2 x 1), y u 5 :
dy
dx
10. With u (4 3x ), y u 9 :
11. With u 1 7x , y u 7 :
12. With u
13. With u
dy
dx
dy
dx
1, y u 10:
x2
8
x 1x , y u 4 :
dy
dx
5u 4 2 10(2 x 1)4
dy du
du dx
x
2
dy du
du dx
dy du
du dx
dy du
du dx
dy
dx
9u8 ( 3) 27(4 3x )8
7u 8 17 1 7x
10u 11
dy du
du dx
4u 3
8
1
1
4 x
4 x
x
4
1
1
x2
4
x
2
x2
8
1
11
x 1x
dy
dy
14. With u 3 x 2 4 x 6, y u1/2 : dx du du
12 u 1/2 (6 x 4) 32x 2
dx
3x 4 x 6
Copyright 2018 Pearson Education, Inc.
3
x
4
1
1
x2
Section 3.6 The Chain Rule
139
dy
dy
(sec u tan u )(sec2 x) (sec(tan x) tan(tan x)) sec2 x
15. With u tan x, y sec u: dx du du
dx
dy
dy
( csc2 u ) 12 12 csc2 1x
16. With u 1x , y cot u: dx du du
dx
x
x
dy
dy
17. With u tan x, y u 3: dx du du
3u 2 sec2 x 3tan 2 x sec2 x
dx
18. With u cos x, y 5u 4 :
dy
du du
(20u 5 )( sin x) 20(cos 5 x)(sin x)
dx
dy
dx
d (3 t ) 1 (3 t ) 1/2 1
19. p 3 t (3 t )1/2 dt 12 (3 t ) 1/2 dt
2
2 3 t
dp
3
d (2r r 2 ) 1 (2r r 2 ) 2/3 (2 2r )
2 2r
20. q 2r r 2 (2r r 2 )1/3 dr 13 (2r r 2 )2/3 dr
3
3(2 r r 2 ) 2/3
dq
d (3t ) 4 ( sin 5t ) d (5t ) 4 cos 3t 4 sin 5t 4 (cos 3t sin 5t )
34 cos 3t dt
21. s 34 sin 3t 54 cos 5t ds
5
dt
dt
d 3 t sin 3 t d 3 t 3 cos 3 t 3 sin 3 t
cos 32 t dt
22. s sin 32 t cos 32 t ds
dt
2
2
dt 2
2
2
2
2
32 cos 32 t sin 32 t
2
csc (cot csc )
23. r (csc cot )1 ddr (csc cot )2 dd (csc cot ) csc cot csc2
csccsc
cot
(csc cot )
(csc cot ) 2
24. r 6(sec tan )3/2 ddr 6 32 (sec tan )1/2 dd (sec tan ) 9 sec tan (sec tan sec2 )
25. y x 2 sin 4 x x cos 2 x
d (sin 4 x ) sin 4 x d ( x 2 ) x d (cos 2 x ) cos 2 x d ( x )
x 2 dx
dx
dx
dx
d (sin x)) 2 x sin 4 x x( 2 cos 3 x d (cos x)) cos 2 x
x 2 (4sin 3 x dx
dx
dy
dx
x 2 (4sin 3 x cos x) 2 x sin 4 x x((2 cos 3 x) ( sin x)) cos 2 x
4 x 2 sin 3 x cos x 2 x sin 4 x 2 x sin x cos 3 x cos 2 x
26. y 1x sin 5 x 3x cos3 x y
27.
1 d (sin 5 x) sin 5 x d 1 x d (cos3 x ) cos3
3 dx
x dx
dx x
1 ( 5sin 6 x cos x) (sin 5 x) 1 x ((3cos 2 x)( sin x)) (cos3 x ) 1
3
3
x
x2
5x sin 6 x cos x 12 sin 5 x x cos2 x sin x 13 cos3 x
x
1 (3 x 2)6 4 1
y 18
2
2x
1
6(5 2 x)
4
2x 1
1
x2
2
x
1
3
dy
2x
4
6 (3 x 2)5 d (3 x 2) ( 1) 4 1
dx 18
dx
2 x2
6 (3 x 2)5 3 ( 1) 4 1
18
2
28. y (5 2 x) 3 18
(3x 2)
2
dy
dx
5
1
x3
1
x 3 (4
3(5 2 x)4 (2) 84
6
(5 2 x )4
2 1
x
3
1 )2
2 x2
2x 1
3
2
x2
x2
Copyright 2018 Pearson Education, Inc.
2
d ( x)
x dx
3
d 4 1
dx
2
2x
140
Chapter 3 Derivatives
29. y (4 x 3) 4 ( x 1) 3
dy
dx
d ( x 1) ( x 1) 3 (4)(4 x 3)3 d (4 x 3)
(4 x 3)4 (3)( x 1) 4 dx
dx
(4 x 3) 4 (3)( x 1)4 (1) ( x 1)3 (4)(4 x 3)3 (4) 3(4 x 3) 4 ( x 1) 4 16(4 x 3)3 ( x 1) 3
(4 x 3)3
( x 1)4
[3(4 x 3) 16( x 1)]
30. y (2 x 5) 1 ( x 2 5 x)6
6 ( x 2 5 x )5
2( x 2 5 x )6
(4 x 3)3 (4 x 7)
( x 1)4
(2 x 5)1 (6)( x 2 5 x)5 (2 x 5) ( x 2 5 x)6 (1)(2 x 5) 2 (2)
dy
dx
(2 x 5) 2
d (tan (2 x1/2 )) tan (2 x1/2 ) d ( x ) 0
31. h( x) x tan 2 x 7 h( x) x dx
dx
2
1/2
x sec (2 x
d (2 x1/2 ) tan(2 x1/2 )
) dx
x sec2 2 x 1 tan 2 x x sec 2 2 x tan 2 x
x
1x k ( x) x2 dxd sec 1x sec 1x dxd ( x2 ) x2 sec 1x tan 1x dxd 1x 2 x sec 1x
x 2 sec 1x tan 1x 1 2 x sec 1x 2 x sec 1x sec 1x tan 1x
x
32. k ( x) x 2 sec
2
33. f ( x) 7 x sec x f ( x) 12 (7 x sec x)1/2 ( x (sec x tan x) (sec x) 1)
34. g ( x)
tan 3 x
( x 7)4
35. f ( )
g ( x)
sin
1 cos
2
( x 7)4 (sec2 3 x3) (tan 3 x )4( x 7)3 .1
4 2
[( x 7) ]
f ( ) 2
(2sin )(cos cos 2 sin 2 )
(1 cos )3
36. g (t )
13sin2t3t
1
x sec x tan x sec x
2 7 x sec x
( x 7)3 (3( x 7) sec2 3 x 4 tan 3 x
8
( x 7)
(3( x 7) sec2 3 x 4 tan 3 x )
( x 7)5
) (sin )( sin )
1sincos dd 1sincos 12sincos (1cos )(cos(1 cos
)
2
(2 sin )(cos 1)
(1 cos )3
2t g (t )
13sin
3t
2 sin
(1 cos )2
(1sin 3t )( 2) (3 2t )(3cos 3t )
(1sin 3t ) 2
t 9 cos 3t 6t cos 3t
2 2sin 3(1
sin 3t )
37. r sin( 2 ) cos(2 ) ddr sin( 2 )( sin 2 ) dd (2 ) cos(2 ) (cos( 2 )) dd ( 2 )
sin( 2 )( sin 2 )(2) (cos 2 )(cos ( 2 ))(2 ) 2sin( 2 )sin(2 ) 2 cos(2 ) cos( 2 )
sec2 1 1 tan 1 (sec tan ) 2 1
tan tan sec
sec2 1 1 tan 1 sec tan sec
2
2
38. r sec tan 1 ddr sec
1
2
2
1
sec
cos
39. q sin
t
t 1
t
t 1
dq
cos
t
t 1
2(t 1) t
2(t 1)3/ 2
40. q cot( sint t )
dq
dt
d
dt
t 2
2(t 1)3/ 2
csc 2
t
t 1
cos
t
t 1
2
dt cos
2 1
2
t 1
d
t 1(1) t . dt
t 1
cos
1
t 1
t 1
t
t 1
sint t dtd sint t csc2 sint t t costt sin t
2
dy
d sin( t 2) 2sin( t 2) cos( t 2) d ( t 2)
41. y sin 2 ( t 2) dt 2sin( t 2) dt
dt
2 sin( t 2) cos( t 2)
Copyright 2018 Pearson Education, Inc.
t
2 t 1
t 1
Section 3.6 The Chain Rule
42. y sec 2 t
dy
dt
d (sec t ) (2sec t )(sec t tan t ) d ( t ) 2 sec2 t tan t
(2sec t ) dt
dt
43. y (1 cos 2t ) 4
d (1 cos 2t ) 4(1 cos 2t ) 5 ( sin 2t ) d (2t ) 8sin 2t
4(1 cos 2t ) 5 dt
dt
dy
dt
(1 cos 2t )5
2 1 cot 2t csc2 2t dtd 2t
3
2
3
dy
d 1 cot t
dt 2 1 cot 2t
dt
44. y 1 cot 2t
2
45. y (t tan t )10
141
dy
10 (t tan t )9
dt
2
9
10
2t
3
1cot 2t
csc2
t sec2 t 1 tan t 10t9 tan9 t (t sec2 t tan t )
10t10 tan 9 t sec t 10t tan t
46. y (t 3/4 sin t ) 4/3 t 1 (sin t ) 4/3 dt t 1
dy
(sin t )1/3 (4t cos t 3cos t )
43 (sin t )1/3 cos t t 2 (sin t )4/3 4(sin t3) t
1/3
cos t
(sin t ) 4/3
t2
3t 2
47. y
48. y
53tt 42
t2
t 4t
3
3
5
2
dy
dt 3 3t
t 4t
2
(t 3 4t )(2t ) t 2 (3t 2 4)
3
( t 4t )
2
3t 4 2t 4 8t 2 3t 4 4t 2
( t 4t ) 2
( t 3 4t ) 2
3
3t 4 ( t 4 4t 2 )
4
2
t ( t 4t )
4
3t 2 (t 2 4)
(t 2 4) 4
6
6 (5t 2)3(3t 4)5
dy
(5t 2)6
130(5t 2)4
dt 5 53tt 42
5 53tt 42 15t 615t2 20 5
26 2
2
6
(3t 4) (5t 2)
(3t 4)6
(5t 2)
(5t 2)
dy
d cos (2t 5) cos(cos (2t 5)) ( sin (2t 5)) d (2t 5)
49. y sin (cos (2t 5)) dt cos(cos (2t 5)) dt
dt
2 cos(cos(2t 5))(sin(2t 5))
dydt sin 5sin 3t dtd 5sin 3t sin 5sin 3t 5cos 3t dtd 3t
53 sin 5sin 3t cos 3t
50. y cos 5sin 3t
3
2
2
t dy 3 1 tan 4 t d 1 tan 4 t 3 1 tan 4 t 4 tan 3 t d tan t
51. y 1 tan 4 12
12 dt
12
12
12 dt
12
dt
2
2
4
4
3
2
3
2
t
t
t
t
t
t
1
12 1 tan 12 tan 12 sec 12 12 1 tan 12 tan 12 sec 12
3
2
dy
52. y 16 1 cos 2 (7t ) dt 63 [1 cos 2 (7t )]2 2 cos(7t )( sin(7t ))(7) 7 1 cos 2 (7t ) (cos(7t ) sin(7t ))
d (1 cos (t 2 )) 1 (1 cos (t 2 )) 1/2 sin(t 2 ) d (t 2 )
53. y (1 cos (t 2 ))1/2 dt 12 (1 cos (t 2 ))1/2 dt
dt
2
dy
2
12 (1 cos (t ))
54. y 4sin
1/2
1 t
2 cos 1 t
1 t 2 t
2
(sin (t )) 2t
dy
dt
4 cos
cos 1 t
t sin (t 2 )
1 cos (t 2 )
1 t 1 t 4 cos 1 t
d
dt
1
2 1 t
d 1 t
dt
t t t
dy
55. y tan 2 (sin 3 t ) dt 2 tan(sin 3 t ) sec 2 (sin 3 t ) (3sin 2 t (cos t )) 6 tan(sin 3 t ) sec 2 (sin 3 t ) sin 2 t cos t
Copyright 2018 Pearson Education, Inc.
142
56.
Chapter 3 Derivatives
dy
y cos 4 (sec 2 3t ) dt 4 cos3 sec 2 (3t )
sin(sec2 (3t ) 2 sec(3t ) sec(3t ) tan(3t ) 3
24 cos3 sec2 (3t ) sin sec2 (3t ) sec 2 (3t ) tan(3t )
57. y 3t (2t 2 5) 4
dy
dt
3t 4 (2t 2 5)3 (4t ) 3 (2t 2 5)4 3(2t 2 5)3 [16t 2 2t 2 5] 3(2t 2 5)3 (18t 2 5)
3
dy
1/2
58. y 3t 2 1 t dt 12 3t 2 1 t
1
1
1
1
3
2
1
t
2 2 1t
2 3t 2
2 3t 2 1t
59. y 1 1x
3
3
x2
1
x
2
12 1t
4 1t
1 t
1
x2
2
1
x2
6
x3
1
x
2
3
x2
1
x
6
x4
2
1
x
6
x3
1
x
1/2
1 (1 t ) 1/2 ( 1)
2
2 1t 1
2 1t
1 y
1 1 1
y 3 1 1x
2 1
1t
1
2
3
x2
2
1 1x 1 1x dxd x3
6 1 1 1 1 1 6 1 1 1 2
x x x x x x
x
2
d
dx
2
2
3
3
y 1 x 12 x1/2 12 1 x x1/2
2
3
y 12 1 x 12 x 3/2 x 1/2 (2) 1 x 12 x 1/2
2
3
3
1 3/2
1
1/2
1
1
1
1
2 2 x
1 x x 1 x 2 x 1 x 2 x 1 x 1
3
3
21x 1 x 1 12 1 21x 1 x 32 1
2 x
2 x
1
60. y 1 x
2
2
d csc(3 x 1))
61. y 19 cot (3x 1) y 19 csc 2 (3 x 1)(3) 13 csc2 (3x 1) y 23 (csc(3 x 1) dx
d (3 x 1)) 2 csc 2 (3 x 1) cot(3 x 1)
23 csc(3 x 1)( csc(3x 1) cot(3x 1) dx
62. y 9 tan
3x y 9 sec2 3x 13 3sec2 3x y 3 2sec 3x sec 3x tan 3x 13 2sec2 3x tan 3x
63. y x(2 x 1)4 y x 4(2 x 1)3 (2) 1 (2 x 1)4 (2 x 1)3 (8 x (2 x 1)) (2 x 1)3 (10 x 1)
y (2 x 1)3 (10) 3(2 x 1)2 (2)(10 x 1) 2(2 x 1)2 (5(2 x 1) 3(10 x 1)) 2(2 x 1) 2 (40 x 8)
16(2 x 1)2 (5 x 1)
64. y x 2 ( x3 1)5 y x 2 5( x3 1) 4 (3 x 2 ) 2 x( x3 1)5 x ( x3 1) 4 [15 x3 2 ( x3 1)] ( x3 1) 4 (17 x 4 2 x)
y ( x3 1) 4 (68 x3 2) 4 ( x3 1)3 (3 x 2 ) (17 x 4 2 x) 2 ( x3 1)3 [( x3 1) (34 x3 1) 6 x 2 (17 x 4 2 x)]
136 x6 47 x3 1
2 x3 1
65.
3
f ( x) x( x 4)3 f ( x) x 3( x 4)2 ( x 4)3 ( x 4)2 3 x ( x 4) ( x 4)2 4 x 4 0 x 4 or
x 1; and f ( x) ( x 4)2 (4) 2( x 4) 4 x 4 4( x 4) ( x 4) 2( x 1) 4( x 4) 3 x 6 0 x 4
or x 2.
66.
f ( x) sec2 x 2 tan x f ( x) 2sec x sec x tan x 2sec 2 x 2sec2 x tan x 1 0 tan x 1 x 4
or x 54 ; and f ( x) 2sec 2 x sec2 x 4sec x sec x tan x(tan x 1) 2sec2 x sec 2 x 2 tan 2 x 2 tan x
Copyright 2018 Pearson Education, Inc.
Section 3.6 The Chain Rule
143
2sec2 x 1 tan 2 x 2 tan 2 x 2 tan x 2sec 2 x 3tan 2 x 2 tan x 1 0 3tan 2 x 2 tan x 1 0
tan x
2 ( 2) 2 4(3)(1)
2(3)
67. g ( x) x g ( x )
2 6 8 (complex number) f ( x) 0 has no solutions
1
2 x
g (1) 1 and g (1) 12 ; f (u ) u 5 1 f (u ) 5u 4 f ( g (1)) f (1) 5;
therefore, ( f g )(1) f ( g (1)) g (1) 5 12
68. g ( x) (1 x )1 g ( x) (1 x) 2 (1)
5
2
1
(1 x ) 2
g (1)
1
2
and g (1) 14 ; f (u ) 1 u1 f (u )
1
u2
f ( g (1)) f 12 4; therefore, ( f g )(1) f ( g (1)) g (1) 4 14 1
69. g ( x) 5 x g ( x)
5
2 x
10 10 csc2 10u
g(1) 5 and g (1) 52 ; f (u ) cot 10u f (u ) csc 2 10u
csc 2 ; therefore, ( f g )(1) f ( g (1)) g (1) 5
f ( g (1)) f (5) 10
2
10
10 2
4
14 4 and g 14 ; f (u) u sec2 u f (u) 1 2 sec u sec u tan u
1 2sec2 u tan u f g 14 f 4 1 2sec2 4 tan 4 5; therefore, ( f g ) 14 f g 14 g 14 5
70. g ( x) x g ( x) g
71. g ( x) 10 x 2 x 1 g ( x) 20 x 1 g (0) 1 and g (0) 1; f (u )
2u
u 2 1
f (u )
2
22u 22 f ( g (0)) f (1) 0; therefore, ( f g )(0) f ( g (0)) g (0) 0 1 0
u 2 1(2)(2u )(2u )
2
u 2 1
(u 1)
1 1 g ( x ) 2 g ( 1) 0 and g ( 1) 2;
x2
x3
u 1 (u 1)(1) (u 1)(1) 2(u 1)(2) 4(u 1) f ( g ( 1))
u 1
(u 1) 2
(u 1)3
(u 1)3
72. g ( x)
2
2
d u 1
f (u ) uu 11 f (u ) 2 uu 11 du
u 1
f (0) 4; therefore,
( f g )(1) f ( g (1)) g (1) (4)(2) 8
73. y f ( g ( x)), f (3) 1, g (2) 5, g (2) 3 y f ( g ( x)) g ( x) y x 2 f ( g (2)) g (2) f (3) 5
(1) 5 5
74. r sin( f (t )), f (0) 3 , f (0) 4
dr
dt
cos( f (t )) f (t ) dr
dt
t 0
dy
2 f (2) 2 13 23
2 f ( x) dx
x2
dy
dy
f (3) g (3) 2 5
(b) y f ( x ) g ( x) dx f ( x) g ( x) dx
75. (a) y 2 f ( x)
dy
dx
x 3
dy
dy
dx
f ( x) g ( x) g ( x) f ( x) dx
f (3) g (3) g (3) f (3)
x 3
3 5 (4)(2 ) 15 8
(c) y f ( x) g ( x )
(d) y
f ( x)
g ( x)
g ( x ) f ( x ) f ( x ) g ( x )
dy
dx
(e) y f ( g ( x))
2
[ g ( x )]
dy
dx
(f ) y ( f ( x))1/2
dy
dx
1
g (2) f (2) f (2) g (2) (2) 3 (8)( 3) 37
dy
6
2
dx x 2
22
[ g (2)]
dy
f ( g (2)) g (2) f (2)(3) 13 (3) 1
dx x 2
f ( g ( x)) g ( x)
f ( x )
dy
12 ( f ( x)) 1/2 f ( x)
dx
2 f ( x)
x 2
cos( f (0)) f (0) cos 3 4 12 4 2
f (2)
2 f (2)
Copyright 2018 Pearson Education, Inc.
13
2 8
1 1
6 8
12 2
242
144
Chapter 3 Derivatives
5
2( g (3))3 g (3) 2(4) 3 5 32
2( g ( x))3 g ( x ) dx
x 3
dy
dy
(h) y (( f ( x )) 2 ( g ( x)) 2 )1/2 dx 12 (( f ( x)) 2 ( g ( x))2 ) 1/2 (2 f ( x) f ( x) 2 g ( x) g ( x)) dx
(g) y ( g ( x))2
dy
dx
dy
12 (( f (2)) 2 ( g (2)) 2 )1/2 (2 f (2) f (2) 2 g (2) g (2)) 12 (82 22 )1/2 (2 8 13 2 2 (3))
76. (a) y 5 f ( x) g ( x)
dy
dx
5 f ( x) g ( x) dx
5 f (1) g (1) 5 13 38 1
x 1
y f ( x)( g ( x))3
dy
dx
f ( x)(3( g ( x)) 2 g ( x)) ( g ( x))3 f ( x)
(b)
dy
3 f (0)( g (0))2 g (0) ( g (0))3 f (0) 3(1)(1) 2
(c) y
f ( x)
dy
dx
g ( x ) 1
(d) y f ( g ( x))
(e) y g ( f ( x ))
( g ( x ) 1) f ( x ) f ( x ) g ( x )
dy
dx
dy
dx
( g ( x ) 1)
f ( g ( x)) g ( x)
g ( f ( x)) f ( x)
(f ) y ( x11 f ( x))2
dy
dx
(g) y f ( x g ( x))
dy
dx
32
3
2
43
13 (1)3 (5) 6
dy
dx x 1
dy
dx x 0
dy
dx x 0
3
2( x11 f ( x))
2(1 3) 3 11 13
43 94
2
dy
dx x 0
( g (1) 1) f (1) f (1) g (1)
( g (1) 1)
2
1
3
dy
dx x 0
f (0 g (0))(1 g (0)) f (1) 1 13
ds
dt
dds ddt : s cos dds sin dds 3 sin 32 1 so that
2
78.
dy
dt
79. With y x, we should get
(a)
(b)
dy
dx
2x 7
dy
dx x 1
9 so that
dy
dt
ds
dt
dy dx
dx dt
9 13 3
1 for both (a) and (b):
(a) y u 3
dy
du
u
( x 1)
1
2 u
again as expected.
2
dy
dx
32 x1/2 for both (a) and (b):
dy
dy
3u 2 ; u x du
1 ; therefore, dx du du
3u 2 1 3( x )2 1 32 x,
dx
dx
2 x
2 x
2 x
as expected.
dy
(b) y u du
xx11
21 1 2
80. With y x3/2 , we should get
dy
dy
3 x 2 ; therefore, dx du du
; u x3 du
1 3 x 2 1 3 3x 2 32 x1/2 ,
dx
dx
2 u
2 x
0011
2
( x 1)1( x 1)1
( x 1)
4( x 1)
(1)2 1. y 2 xx 11
2 ( x 1) 2 2
( x 1) 2
( x 1)
( x 1)3
4(0 1)
y x 0
34 4 y 1 4( x 0) y 4 x 1
(0 1)3
1
81. y
dds ddt 1 5 5
dy
dy
dy
y 7 du 15 ; u 5 x 35 du
5; therefore, dx du du
15 5 1, as expected
dx
dx
dy
dy
dy
y 1 u1 du 12 ; u ( x 1)1 du
( x 1) 2 (1) 1 2 ; therefore dx du du
dx
dx
( x 1)
u
2
1
1
1
( x 1)
1, again as expected
2
( x 1)2
( x 1) 2
( x 1) 1
u
5
1
( 41)2
1 1
3
9
11x10 f ( x) dydx x1 2(1 f (1))3 (11 f (1))
77.
dy
dx
( 41) 13 (3) 83
13 13
g ( f (0)) f (0) g (1)(5) 83 (5) 40
3
f ( x g ( x))(1 g ( x))
y x2 7 x 5
f ( g (0)) g (0) f (1)
13
dy dx
:
dx dt
x2
5
3 17
and x 0 y
Copyright 2018 Pearson Education, Inc.
Section 3.6 The Chain Rule
82. y x 2 x 7 and x 2 y (2) 2 (2) 7 9 3. y
y x 2
2(2) 1
dy
dx x 1
x2 x 7
1/2
(2 x 1)
2 x 1
2 x2 x7
63 12 y 3 12 ( x 2) y 12 x 2
2 (2) 2 (2) 7
2 sec2 4x
dy
2sec 2 4x 4
dx
sec 2 ( ) slope
2
4
83. y 2 tan 4x
(a)
1
2
145
of tangent is ; thus, y (1) 2 tan 4 2 and y (1) tangent line is
given by y 2 ( x 1) y x 2
(b) y 2 sec 2 4x and the smallest value the secant function can have in 2 x 2 is 1 the minimum
value of y is 2 and that occurs when 2 2 sec2 4x 1 sec 2 4x 1 sec 4x x 0.
84. (a) y sin 2 x y 2 cos 2 x y (0) 2 cos(0) 2 tangent to y sin 2 x at the origin is y 2 x;
y sin 2x y 12 cos 2x y (0) 12 cos 0 12 tangent to y sin 2x at the origin is
y 12 x. The tangents are perpendicular to each other at the origin since the product of their slopes is 1.
(b) y sin(mx ) y m cos(mx ) y (0) m cos 0 m; y sin
mx y m1 cos mx
y (0) m1 cos(0) m1 . Since m m1 1, the tangent lines are perpendicular at the origin.
(c) y sin(mx ) y m cos(mx). The largest value cos(mx) can attain is 1 at x 0 the largest value y can
attain is | m | because y m cos (mx) m cos mx m 1 m . Also, y sin mx y m1 cos mx
1 cos x
m
m
y
1
m
cos
x
m
1
m
the largest value y can attain is
1
m
.
(d) y sin(mx ) y m cos( mx) y (0) m slope of curve at the origin is m. Also, sin(mx ) completes m
periods on [0, 2 ]. Therefore the slope of the curve y sin(mx) at the origin is the same as the number of
periods it completes on [0, 2 ]. In particular, for large m, we can think of “compressing” the graph of
y sin x horizontally which gives more periods completed on [0, 2 ], but also increases the slope of the
graph at the origin.
A sin(2 bt )(2 b) 2 bA sin(2 bt ). If we replace b with 2b to double the
85. s A cos(2 bt ) v ds
dt
frequency, the velocity formula gives v 4 bA sin(4 bt ) doubling the frequency causes the velocity to
double. Also v 2 bA sin(2 bt ) a dv
4 2b 2 A cos(2 bt ). If we replace b with 2b in the acceleration
dt
formula, we get a 16 2 b 2 A cos(4 bt ) doubling the frequency causes the acceleration to quadruple.
Finally, a 4 2b 2 A cos(2 bt ) j da
8 3b3 A sin(2 bt ). If we replace b with 2b in the jerk formula,
dt
we get j 64 3b3 A sin(4 bt ) doubling the frequency multiplies the jerk by a factor of 8.
2 ( x 101) 25 y 37 cos 2 ( x 101)
86. (a) y 37 sin 365
365
2 74 cos 2 ( x 101) . The temperature
365
365 365
2 ( x 101) is l and
is increasing the fastest when y is as large as possible. The largest value of cos 365
2 ( x 101) 0 x 101 on day 101 of the year ( April 11), the temperature is
occurs when 365
increasing the fastest.
cos 2 (101 101) 74 cos(0) 74 0.64 F/day
(b) y (101) 74
365
365
365
365
12 (1 4t ) 1/2 (4) 2(1 4t ) 1/2 v(6) 2(1 4 6)1/2 52 m/sec; v 2(1 4t )1/2
4 m/sec 2
12 2(1 4t ) 3/2 (4) 4(1 4t )3/2 a (6) 4(1 4 6) 3/2 125
87. s (1 4t )1/2 v
a dv
dt
ds
dt
88. We need to show a
k
2 s
2
dv
dt
is constant: a
dv
dt
d k s k a dv ds dv v
dv
dv and dv
ds
ds dt
ds
ds dt
ds
2 s
k s k2 which is a constant.
Copyright 2018 Pearson Education, Inc.
146
Chapter 3 Derivatives
89. v proportional to
2
k2
v k for some constant k dv
dv
ds dv
v k3/ 2 k
k3/ 2 Thus, a dv
dt
ds dt
ds
ds
s
s
2s
2s
acceleration is a constant times
1
s2
dx
dt
90. Let
1
s
1
s2
so a is inversely proportional to s 2.
dv f ( x )
f ( x). Then, a dv
dv
dx dx
dt
dx dt
L
g
91. T 2
f ( x) dxd ( f ( x)) f ( x) f ( x) f ( x), as required.
d dx
dx dt
dT
2 1 L g1 L Therefore, dT
dT
dL kL k L 12 2 k Lg kT
, as
dL
du
dL du
2
gL
gL
g
g g
2 g
required.
92. No. The chain rule says that when g is differentiable at 0 and f is differentiable at g (0), then f o 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.
sin 2( x h ) sin 2 x
93. As h 0, the graph of y
h
approaches the graph of y 2 cos 2 x because
sin 2( x h ) sin 2 x
d (sin 2 x ) 2 cos 2 x.
lim
dx
h
h 0
94. As h 0, the graph of y
cos[( x h ) 2 ]cos( x 2 )
h
2
approaches the graph of y 2 x sin ( x ) because
cos[( x h )2 ] cos( x 2 )
h
h 0
lim
d [cos ( x 2 )] 2 x sin ( x 2 ).
dx
dy
dx
1 x 3/4 ,
4
95. From the power rule, with y x1/4 , we get
dy
dx
1
2
x
d
dx
x 2 1 x 21x
96. From the power rule, with y x3/4 , we get
dy
dx
1
2 x x
d x x
dx
dy
dx
14 x 3/4 . From the chain rule, y
x
in agreement.
34 x 1/4 . From the chain rule, y x x
dy
dx 1 x 1 x 1 32 x 3 x 3 x 34 x 1/4 , in agreement.
2 x
2 x x
2 x x
4 x x
4 x
x
Copyright 2018 Pearson Education, Inc.
Section 3.6 The Chain Rule
97.
x sin
f ( x)
(a) 1 sin
1x ,
x0
0,
x0
1x 1
( x 0)
Theorem lim x sin
x 0
x x sin
1x 0 , i.e., xlim
0
x 0 lim
1x x and xlim
0
x 0
147
x, so by the Sandwich
f ( x) 0 ; and lim f ( x) lim 0 0; thus, lim f ( x) 0
x 0
x 0
x 0
f (0), and f is continuous at x 0.
(b) If x 0, then f ( x) 0.
If x 0, then f ( x) x cos
(c)
f (0) lim
h 0
f (0 h ) f (0)
h
f ( h)
h 0 h
If h 0, then lim
If h 0, then lim
h 0
2
cos
sin 1x x x .
1
f ( h)
lim h .
h 0
lim 0h lim 0 0.
h 0
h 0
f ( h)
h
lim
not differentiable at x 0.
98.
1x x 1 sin 1x
h sin
h 0
1h
h
lim sin 1h does not exist. Thus, f (0) does not exist and f is
h 0
x 2 cos 2 , x 0
x
f ( x)
0, x 0
x 2 0 lim x 2 , so by the Sandwich Theorem
2x 1 x2 x2 cos 2x x2 and xlim
0
x 0
2
2
lim x cos x 0 , i.e., lim f ( x) 0 f (0) , and f is continuous at x 0.
x 0
x 0
(a) 1 cos
(b) For x 0, f ( x) x 2 sin
(c)
(d)
2x x2 2 x cos 2x
2
h 2 cos
h2
2sin 2x 2 x cos 2x
1 cos h2 1 h h cos h2 h (h 0) or h h cos h2 h (h 0); in either case,
lim h 0 lim h so by the Sandwich Theorem lim h cos h2 0 , i.e., f (0) 0.
h 0
h 0
h 0
f (0 h ) f (0)
h
h 0
f (0) lim
f ( h)
lim h lim
h 0
h 0
h
lim h cos h2 ; we know
h 0
lim 2sin 2x lim 2 x cos 2x lim 2sin 2x 0 does not exist, i.e., lim f ( x) does not exist so
x 0
x 0
x 0
x0
lim 2 x cos 2x 0 , and lim 2sin 2x does not exist, so lim f ( x) lim 2sin 2x 2 x cos 2x
x0
x 0
x 0
x 0
not continuous at x 0.
99. (a)
(b)
f is even f ( x) f ( x) f ( x) f ( x) (1) f ( x), i.e., f is odd.
f is odd f ( x) f ( x) f ( x) f ( x) (1) f ( x), i.e., f is even.
Copyright 2018 Pearson Education, Inc.
f is
148
Chapter 3 Derivatives
100. (a)
(b)
df
dt
1.27324sin 2t 0.42444sin 6t 0.2546 sin10t 0.18186 sin14t
df
dg
(c) The curve of y dt approximates y dt
the best when t is not , 2 , 0, 2 , nor .
101. (a)
(b)
(c)
dh
dt
2.5464 cos(2t ) 2.5464 cos (6t ) 2.5465 cos (10t ) 2.54646 cos(14t ) 2.54646 cos (18t )
dh/dt
10
2
0
t
2
10
3.7
IMPLICIT DIFFERENTIATION
1. x 2 y xy 2 6 :
Step 1:
Step 2:
Step 3:
Step 4:
x
dy
2 dy
y 2 x x 2 y dx y 2
dx
dy
dy
x 2 dx 2 xy dx 2xy y 2
dy 2
( x 2 xy ) 2xy y 2
dx
dy
2 xy y 2
2
dx
x 2 xy
2. x3 y 3 18 xy 3 x 2 3 y 2
dy
dx
1 0
dy
dy
dy
6 y x2
18 y 18 x dx (3 y 2 18 x) dx 18 y 3 x 2 dx 2
y 6 x
Copyright 2018 Pearson Education, Inc.
Section 3.7 Implicit Differentiation
3. 2 xy y 2 x y :
Step 1:
Step 2:
Step 3:
Step 4:
2x
dy
dy
dy
2 y 2 y dx 1 dx
dx
dy
dy dy
2 x dx 2 y dx dx 1 2 y
dy
(2 x 2 y 1) 1 2 y
dx
dy
1 2 y
2 x 2 y 1
dx
dy
4. x3 xy y 3 1 3 x 2 y x dx 3 y 2
dy
dx
dy
dy
dx
0 (3 y 2 x) dx y 3x 2
5. x 2 ( x y )2 x 2 y 2 :
dy
dy
Step 1: x 2 2( x y ) 1 dx ( x y ) 2 (2 x) 2 x 2 y dx
dy
dy
Step 2: 2 x 2 ( x y ) dx 2 y dx 2 x 2 x 2 ( x y ) 2 x( x y )2
dy
Step 3: dx 2 x 2 ( x y ) 2 y 2 x [1 x( x y ) ( x y ) 2 ]
dy
dx
y 3 x 2
3 y2 x
Step 4:
149
2 x 1 x ( x y ) ( x y )2
2 x 2 ( x y ) 2 y
x 1 x ( x y ) ( x y ) 2
y x2 ( x y )
dy
dy
x 1 x 2 xy x 2 2 xy y 2
2
3
dy
dx
3 xy 2 7 y
dy
y (3 xy 7)
dy
3
2
2
3
2
x y x y
x yx y
6. (3xy 7)2 6 y 2(3 xy 7) 3x dx 3 y 6 dx 2(3 xy 7)(3x)
x2 x 3x y xy
dy
6 dx 6 y (3 xy 7)
dx [6 x(3xy 7) 6] 6 y (3xy 7) dx x (3 xy 7) 1
13 x 2 y 7 x
dy
( x 1) ( x 1)
7. y 2
x 1
x 1
8. x3
2 x y
x 4 3x3 y
x 3 y
2 4 x 3 9 x 2 y
y
9.
2 y dx
( x 1)2
2
( x1)2
dy
dx 1 2
y ( x1)
2x y 4 x3 9 x 2 y 3x3 y 2 y (3x3 1) y 2 4 x3 9 x 2 y
3 x3 1
dy
dy
x sec y 1 sec y tan y dx dx sec y1tan y
dy
dy
dy
dy
10. xy cot( xy ) x dx y csc2 ( xy ) x dx y x dx x csc 2 ( xy ) dx y csc2 ( xy ) y
dy
dx
x x csc2 ( xy ) y csc2 ( xy ) 1
dy
dy
dx
y csc2 ( xy ) 1
x 1 csc2 ( xy )
y
x
dy
11. x tan( xy ) 0 1 sec 2 ( xy ) y x dx 0 x sec 2 ( xy ) dx 1 y sec 2 ( xy )
1
x sec2 ( xy )
y
x
2
cos ( xy )
x
y
x
2
dy
dx
1 y sec2 ( xy )
x sec2 ( xy )
cos ( xy ) y
x
dy
dy
dy
12. x 4 sin y x3 y 2 4 x3 (cos y ) dx 3x 2 y 2 x3 2 y dx (cos y 2 x3 y ) dx 3 x 2 y 2 4 x3
13. y sin
1 xy y cos (1)
dy
1
y
dx
1y cos
1
y
y
sin x
1
y
1
y
1 dy
y 2 dx
y
y sin
3 x 2 y 2 4 x3
cos y 2 x3 y
dy
dy
dy
sin 1y dx x dx y dx 1y cos 1y sin 1y x y
2
cos xy
1
y
dy
dx
1
y
Copyright 2018 Pearson Education, Inc.
150
Chapter 3 Derivatives
14. x cos(2 x 3 y ) y sin x x sin(2 x 3 y )(2 3 y ) cos(2 x 3 y ) y cos x y sin x
2 x sin(2 x 3 y ) 3 xy sin(2 x 3 y ) cos(2 x 3 y ) y cos x y sin x
cos(2 x 3 y ) 2 x sin(2 x 3 y ) y cos x (sin x 3x sin(2 x 3 y )) y
cos(2 x 3 y ) 2 x sin(2 x 3 y ) y cos x
y
sin x 3 x sin(2 x 3 y )
15. 1/2 r1/2 1 12 1/2 12 r 1/2 ddr 0 ddr 1
2 r
1
2
ddr 2 r r
2
16. r 2 32 2/3 34 3/4 ddr 1/2 1/3 1/4 ddr 1/2 1/3 1/4
17. sin(r )
1
2
r cos( r )
cos(r ) r ddr 0 ddr [ cos(r )] r cos(r ) ddr cos( r ) r , cos(r ) 0
18. cos r cot r ( sin r ) ddr csc2 ddr
dy
dx
19. x 2 y 2 1 2 x 2 yy 0 2 yy 2 x
y
y ( 1) xy
y
2
y x
y
since y x d
xy
y
2
20. x 2/3 y 2/3 1 23 x 1/3 23 y 1/3
Differentiating again, y
d2y
dx 2
1 x 2/3 y 1/3
3
dy
dx
dr
d
1 sin r csc2
y
dx
2
y
21. y 2 x 2 2 x 2 yy 2 x 2 y
y 2 x2
y
3
d2y
dx 2
y
x 2/3
y1/3
3 x 4/3
2 x2
2y
d2y
dx 2
y
1
1
2 y 1/ 2 1
1/ 2
(y
y 3/ 2
1)
1/3
y
dx
2
y
2
2 y 2 xy
( x 2 y )3
( x 2 y ) y y (1 2 y)
( x2 y)
2
1 x 2/3
3
1
y 1
d 2 y y y 2 ( x 1)2
x 1
y
dx 2
y3
( y 1)1; then y ( y 1)2 y
1
( y 1)3
1
2 y 3/ 2 ( y 1/ 2 1)3
dy
dx
3/2
1
1
y 1/ 2 1
1/2
y
y y
2 1 y
( x 2 y ) ( x 2 y ) y 1 2
y
y
y
y
y 1
( x 2 y)
2
; we can differentiate the
1 y 0 y 1/2 1 y 12 [ y ]2 y 3/2
3
24. xy y 2 1 xy y 2 yy 0 xy 2 yy y y ( x 2 y ) y y
d2y
1
y3
1
3 y1/3 x 2/3
equation y y 1/2 1 1 again to find y : y 12 y
2
y ( x 1)
y ( x 1) y
x y1 ; then y
2
2
y3
x 2/3
23. 2 y x y y 1/2 y 1 y y y 1/2 1 1
d ( y ) d x
, dx
dx
y
y 2 (1 y 2 )
y1/3
x1/3 13 y 2/3 1/3
x
22. y 2 2 x 1 2 y 2 y y 2 2 y y (2 y 2) 2 y
( y 1) 2 ( y 1)1
dx 2
2
1csc
sin r
1/3
dy
dy
y 1/3
0 dx 23 y 1/3 23 x 1/3 y dx x 1/3 x
;
y
x1/3 ( 13 y 2/3 ) y y1/3 ( 13 x 2/3 )
1 y1/3 x 4/3
3
d2y
y xy ; now to find
2
dr
d
y
( x2 y)
y
;
( x2 y )
1 [ y ( x 2 y ) y ( x 2 y )2 y 2 ]
( x2 y)
2
( x2 y)
2 y( x y)
( x 2 y )3
Copyright 2018 Pearson Education, Inc.
2 y ( x 2 y )2 y 2
( x 2 y )3
Section 3.7 Implicit Differentiation
2
x
25. 3 sin y y x3 cos y y y 3 x 2 3x 2 y cos y y 3x 2 (1 cos y ) y y 13cos
;
y
(1cos y )
2
2
x
sin y13cos
y 1 cos y
1cos y
2
(1cos y )
6 x 6 x cos y 3 x
(1cos y )6 x 3 x 2 sin y y
y
2
6 x 6 x cos y 6 x cos y 6 x cos 2 y 9 x 4 sin y
(1cos y )3
6 x 12 x cos y 6 x cos 2 y 9 x 4 sin y
(1 cos y )3
26. sin y x cos y 2 cos y y x sin y y cos y cos y y x sin y y cos y
cos y x sin y y cos y y cos y x sin y
cos y
2
27. x3 y 3 16 3 x 2 3 y 2 y 0 3 y 2 y 3 x 2 y x 2 ; we differentiate y 2 y x 2 to find y :
y
2
2
2
y y y [2 y y ] 2x y y 2 x 2 y[ y ] y
2
d y
dx 2 (2,2)
3332
32
2
2 x 2 y x 2
y
y2
y (0,1) 12 we obtain y
(0, 1)
( 2)
12 (1)(0) 1
4
y
( x2 y)
y
y
2
2 xy 3 2 x 4
y5
dy
2 2 x
dy
dx
x 1
2 y3 y
y
( x 2 y )( y) ( y )(1 2 y)
( x 2 y )2
dy
dx ( 2, 1)
dy
dx
dy
2 2 y dx 4 y 3
dy
dx ( 2, 1)
1 and
dy
dy
dx
2 2 x
dy
1
30. ( x 2 y 2 )2 ( x y )2 at (1, 0) and (1, 1) 2 ( x 2 y 2 ) 2 x 2 y dx 2( x y ) 1 dx
dy
[2 y ( x 2
dx
and
dy
dx (1, 1)
; since
4
29. y 2 x 2 y 4 2 x at (2, 1) and (2, 1) 2 y dx 2 x 4 y 3
4
2 x 2 x3
2
28. xy y 2 1 xy y 2 yy 0 y ( x 2 y ) y y
dy
(2 y 4 y 3 )
dx
2
y 2 ) ( x y )] 2 x ( x 2 y 2 ) ( x y )
dy
dx
2 x ( x 2 y 2 ) ( x y )
2 y ( x2 y 2 )( x y )
dy
dx (1,0)
1
1
2 x y
;
2 y x
7
y 3 4 ( x 2)
31. x 2 xy y 2 1 2 x y xy 2 yy 0 ( x 2 y ) y 2 x y y
(a) the slope of the tangent line m y (2, 3)
7
4
the tangent line is
(b) the normal line is y 3 74 ( x 2) y 74 x 29
7
32. x 2 y 2 25 2 x 2 yy 0 y xy ;
(a) the slope of the tangent line m y
(3, 4)
xy
(3, 4)
3
4
the tangent line is y 4 34 ( x 3) y 34 x 25
4
(b) the normal line is y 4 43 ( x 3) y 43 x
Copyright 2018 Pearson Education, Inc.
y 74 x 12
151
152
Chapter 3 Derivatives
y
33. x 2 y 2 9 2 xy 2 2 x 2 yy 0 x 2 yy xy 2 y x ;
(a) the slope of the tangent line m y
y
x
( 1, 3)
( 1, 3)
13 x 83
(b) the normal line is y 3 13 ( x 1) y
3 the tangent line is y 3 3( x 1) y 3x 6
34. y 2 2 x 4 y 1 0 2 yy 2 4 y 0 2( y 2) y 2 y
1 ;
y 2
(a) the slope of the tangent line m y ( 2, 1) 1 the tangent line is y 1 1( x 2) y x 1
(b) the normal line is y 1 1( x 2) y x 3
35. 6 x 2 3xy 2 y 2 17 y 6 0 12 x 3 y 3xy 4 yy 17 y 0 y (3x 4 y 17) 12 x 3 y
12 x 3 y
y 3 x 4 y 17 ;
(a) the slope of the tangent line m y
12 x 3 y
3 x 4 y 17 ( 1, 0)
( 1, 0)
y 76 x 76
6
7
the tangent line is y 0 76 ( x 1)
(b) the normal line is y 0 76 ( x 1) y 76 x 76
36. x 2 3 xy 2 y 2 5 2 x 3xy 3 y 4 yy 0 y 4 y 3 x 3 y 2 x y
(a) the slope of the tangent line m y
( 3, 2)
3 y 2 x
4 y 3 x ( 3, 2)
3 y 2 x
;
4 y 3x
0 the tangent line is y 2
(b) the normal line is x 3
37. 2 xy sin y 2 2 xy 2 y (cos y ) y 0 y (2 x cos y ) 2 y y
(a) the slope of the tangent line m y
1,
2
2 y
2 x cos y
y 2 x
1, 2
2 y
;
2 x cos y
2 the tangent line is y 2 2 ( x 1)
(b) the normal line is y 2 2 ( x 1) y 2 x 2 2
38. x sin 2 y y cos 2 x x(cos 2 y )2 y sin 2 y 2 y sin 2 x y cos 2 x y (2 x cos 2 y cos 2 x)
sin 2 y 2 y sin 2 x
sin 2 y 2 y sin 2 x y cos 2 x 2 x cos 2 y ;
(a) the slope of the tangent line m y
y 2 2 x 4 y 2 x
4 2
,
sin 2 y 2 y sin 2 x
cos 2 x 2 x cos 2 y
4 2
,
2 the tangent line is
2
(b) the normal line is y 2 12 x 4 y 12 x 58
2 cos( x y )
39. y 2sin( x y ) y 2[cos( x y )] ( y ) y [1 2 cos( x y )] 2 cos ( x y ) y 1 2 cos( x y ) ;
(a) the slope of the tangent line m y
2 cos( x y )
(1, 0)
1 2 cos( x y )
y 2 x 2
(b) the normal line is y 0 21 ( x 1) y 2x 21
(1, 0)
2 the tangent line is y 0 2 ( x 1)
40. x 2 cos 2 y sin y 0 x 2 (2 cos y )( sin y ) y 2 x cos 2 y y cos y 0 y [ 2 x 2 cos y sin y cos y ]
2 x cos 2 y y
2 x cos 2 y
2
2 x cos y sin y cos y
;
Copyright 2018 Pearson Education, Inc.
Section 3.7 Implicit Differentiation
(a) the slope of the tangent line m y
(0, )
(b) the normal line is x 0
2 x cos 2 y
2
2 x cos y sin y cos y (0, )
0 the tangent line is y
41. Solving x 2 xy y 2 7 and y 0 x 2 7 x 7 7,0 and
2
2
153
7,0 are the points where the curve
2 x y
crosses the x-axis. Now x xy y 7 2 x y xy 2 yy 0 ( x 2 y ) y 2x y y x 2 y
2 x y
m x 2 y the slope at 7,0 is m 2 7 2 and the slope at
7
7, 0 is m 2 7 2. Since the
7
slope is 2 in each case, the corresponding tangents must be parallel.
dy
dy
dy
y2
42. xy 2 x y 0 x dx y 2 dx 0 dx 1 x ; the slope of the line 2 x y 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
y2
of the tangent is 12 . Therefore, 1 x 12 2 y 4 1 x x 3 2 y. Substituting in the original equation,
y ( 3 2 y ) 2( 3 2 y ) y 0 y 2 4 y 3 0 y 3 or y 1. If y 3, then x 3 and
y 3 2( x 3) y 2 x 3. If y 1, then x 1 and y 1 2( x 1) y 2 x 3.
43. y 4 y 2 x 2 4 y 3 y 2 yy 2 x 2(2 y 3 y ) y 2 x y
3
, 23
4
x
y 2 y3
is
3 1
,
4 2
x
y 2 y3
3
4
12
2 8
3
, 23
4
2 3
4 2
3
4
3 6 3
8
2
1
4
13
2 4
x ; the
y 2 y3
slope of the tangent line at
213 1; the slope of the tangent line at 43 , 12 is
3
44. y 2 (2 x ) x3 2 yy (2 x) y 2 (1) 3x 2 y
y 2 3 x 2
; the
2 y (2 x )
slope of the tangent line is m
y 2 3 x 2
2 y (2 x )
(1, 1)
42 2 the tangent line is y 1 2( x 1) y 2 x 1; the normal line is y 1 12 ( x 1) y 12 x 32
45. y 4 4 y 2 x 4 9 x 2 4 y 3 y 8 yy 4 x3 18 x y (4 y 3 8 y ) 4 x3 18 x y
x (2 x 2 9)
y (2 y 2 4)
m; (3, 2): m
( 3)(189)
2(8 4)
27
;(3, 2):
8
m
27 ;(3,
8
2): m
27 ; (3, 2):
8
46. x3 y 3 9 xy 0 3 x 2 3 y 2 y 9 xy 9 y 0 y (3 y 2 9 x) 9 y 3 x 2 y
(a) y (4, 2)
5
4
(b) y 0
3 y x2
and y (2, 4)
2
y 3 x
4;
5
2
2
0 3 y x 2 0 y x3 x3 x3
3
4 x3 18 x
4 y 3 8 y
m
9 y 3 x 2
3 y 2 9 x
3
2 x3 9 x
2 y 4 y
27
8
3 y x2
y 2 3 x
2
9 x x3 0 x 6 54 x3 0
x3 ( x3 54) 0 x 0 or x 3 54 33 2 there is a horizontal tangent at x 33 2. To find the
corresponding y -value, we will use part (c).
(c)
dx
dy
0
x3/2 x
y 2 3 x
3 y x2
3/2
0 y 2 3 x 0 y 3 x ; y 3 x x3
3x
3
9 x 3 x 0 x3 6 3 x3/2 0
6 3 0 or x3/2 0 or x3/2 6 3 x 0 or x 3 108 33 4. Since the equation
x3 y 3 9 xy 0 is symmetric in x and y, the graph is symmetric about the line y x. That is, if ( a, b) is
a point on the folium, then so is (b, a). Moreover, if y ( a, b) m, then y ( a , b ) m1 . Thus, if the folium has
a horizontal tangent at ( a, b), it has a vertical tangent at (b, a ) so one might expect that with a horizontal
Copyright 2018 Pearson Education, Inc.
154
Chapter 3 Derivatives
tangent at x 3 54 and a vertical tangent at x 33 4, the points of tangency are
3 54, 33 4 and
33 4, 3 54 , respectively. One can check that these points do satisfy the equation x3 y3 9 xy 0.
47. x 2 2 xy 3 y 2 0 2 x 2 xy 2 y 6 yy 0 y (2 x 6 y ) 2 x 2 y y
the tangent line m y
x y
(1, 1)
x y
3 y x
the slope of
3 yx
1 the equation of the normal line at (1, 1) is y 1 1( x 1)
(1, 1)
y x 2. To find where the normal line intersects the curve we substitute into its equation:
x 2 2 x(2 x) 3(2 x) 2 0 x 2 4 x 2 x 2 3(4 4 x x 2 ) 0 4 x 2 16 x 12 0 x 2 4 x 3 0
( x 3)( x 1) 0 x 3 and y x 2 1. Therefore, the normal to the curve at (1, 1) intersects the
curve at the point (3, 1). Note that it also intersects the curve at (1, 1).
q
48. Let p and q be integers with q 0 and suppose that y x p x p /q . Then y q x p Since p and q are integers
and assuming y is a differentiable function of x,
p 1
p
x
q ( x p /q ) q 1
p x p 1
q x p p/q
p
. x p 1( p p /q )
q
d ( yq )
dx
p 1
p p 1
d ( x p ) qy q 1 dy px p 1 dy px
dx
q x q 1
q 1
dx
dx
qy
y
p ( p /q ) 1
x
q
y 0
21y If a normal is drawn from (a, 0) to ( x1 , y1 ) on the curve its slope satisfies x1 a 2 y1
1
y1 2 y1 ( x1 a ) or a x1 12 Since x1 0 on the curve, we must have that a 12 . By symmetry, the two
x x
points on the parabola are x1 , x1 and x1 , x1 . For the normal to be perpendicular, x 1a a x1 1
1 1
2
x1
2
1
1
1
1
1 x1 ( a x1 ) x1 x1 2 x1 x1 4 and y1 2 Therefore, 4 , 12 and a 34 .
2
49. y 2 x
dy
dx
( a x1 )
50. 2 x 2 3 y 2 5 4 x 6 yy 0 y 32 yx y (1, 1) 32 yx
(1, 1)
23 and y (1,1) 32 yx
23 ; also,
(1, 1)
2
2
2
y 2 x3 2 yy 3x 2 y 32xy y (1, 1) 32xy
32 and y (1, 1) 32xy
32 . Therefore the
(1, 1)
(1, 1)
tangents to the curves are perpendicular at (1, 1) and (1, 1) (i.e., the curves are orthogonal at these two points of
intersection).
51. (a) x 2 y 2 4, x 2 3 y 2 (3 y 2 ) y 2 4 y 2 1 y 1. If y 1 x 2 (1) 2 4 x 2 3 x 3.
If y 1 x 2 (1)2 4 x 2 3 x 3.
dy
dy
dy
dy
x 2 y 2 4 2 x 2 y dx 0 m1 dx xy and x 2 3 y 2 2 x 6 y dx m2 dx 3xy
3, 1 : m1 dydx 13 3 and m2 dydx 3(1)3 33 m1 m2 3 33 1
dy
dy
At 3, 1 : m1 dx ( 1)3 3 and m2 dx 3( 31) 33 m1 m2 3 33 1
3
dy
dy
At 3, 1 : m1 dx 1 3 and m2 dx 3(1)3 33 m1 m2 3 33 1
3
3
dy
dy
At 3, 1 : m1 dx ( 1) 3 and m2 dx 3( 1) 33 m1 m2 3 33 1
At
(b) x 1 y 2 , x 13 y 2 ,
If y
1
2
3
3
2
13 y2 1 y2 y2 34 y 23 . If y
x 1
dy
3
2
2
dy
14 . x 1 y 2 1 2 y dx m1
3
2
2
3
2
x 1
dy
dx
21y and x 13 y 2
dy
y dx m2 dx 23y
Copyright 2018 Pearson Education, Inc.
14 .
Section 3.7 Implicit Differentiation
At
At
: m
, : m
1, 3
4 2
1
4
1
3
2
155
dy
dy
3 1
1 1 and m2 dx 3 3 m1 m2 1
dx
2( 3 /2)
3
3
3
2( 3 /2)
3
dy
dy
3
3
1
1
1
3
1 dx 2( 3 /2) 3 and m2 dx 2( 3/2) 3 m1 m2
3
3
1
32xy 1 x2 y x2 x3
(0)
x4 x3 x 4 4 x3 0 x3 ( x 4) 0 x 0 or x 4. If x 0 y 2 0 and 13 32xy 1 is
52. y 13 x b, y 2 x3
dy
dx
dy
2
dy
13 and 2 y dx 3x 2 dx 32xy 13
2
2
2
2
4
indeterminate at (0, 0). If x 4 y
53. xy 3 x 2 y 6 x 3 y 2
dy
dx
y x
3
(4)2
2
2 dy
dx
2
8. At (4, 8), y 13 x b 8 13 (4) b b
dy
2
dy
28 .
3
y 3 2 xy
y 3 2 xy
2 xy 0 dx 3xy 2 x 2 y 3 2 xy dx
2 2;
3 xy 2 x 2
3 xy x
2
2
dx x 2 y (2 x dx ) 0 dx ( y 3 2 xy ) 3xy 2 x 2 dx 3 xy x ;
also, xy 3 x 2 y 6 x (3 y 2 ) y 3 dy
3
dy
dy
dy
y 2 xy
dx appears to equal 1 The two different treatments view the graphs as functions symmetric across the
thus dy
dy
dx
line y x, so their slopes are reciprocals of one another at the corresponding points (a, b) and (b, a).
dy
dy
dy
dx
dy
2
3 x
(2 y 2 sin y cos y ) 3x 2 dx 2 y 2sin
y cos y
2
2 dx
3
2
2
x
dx 2 sin y cos y 2 y ; thus dx appears to
2 sin y3cos
;
also,
x
y
sin
y
3
x
2
y
2
sin
y
cos
y
2
y 2 y
dy
dy
dy
54. x3 y 2 sin 2 y 3 x 2 2 y dx (2sin y )(cos y ) dx
equal
1
dy
dx
3x
The two different treatments view the graphs as functions symmetric across the line y x so their
slopes are reciprocals of one another at the corresponding points ( a, b) and (b, a ).
55–62.
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 #55(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, /4};
eqn x Tan[y/x] 2;
ImplicitPlot[eqn,{x, x0 3, x0 3},{y, y0 3, y0 3}]
Copyright 2018 Pearson Education, Inc.
156
Chapter 3 Derivatives
eqn/.{x x0, y y0}
eqn/.{ y y[x]}
D[%, x]
Solve[%, y'[ x]]
slope y '[x]/.First[%]
m slope/.{x x0, y[x] y0}
tanline y y0 m (x x0)
ImplicitPlot[{eqn, tanline}, {x, x0 3, x0 3},{y, y0 3, y0 3}]
3.8
RELATED RATES
1. A r 2
dA
dt
2. S 4 r 2
3. y 5 x,
dx
dt
dx
dt
dS
dt
dy
dy
dy
dt
dy
2 2 dx
3 dt 0 2 dx
3(2) 0 dx
3
dt
dt
dt
dy
dy
3 dt 2 x dx
; when x 1 dt 2(1)(3) 6
dt
dy
dt
6. x y 3 y,
7. x 2 y 2 25,
dy dy
5 dx
3 y 2 dt dt ; when y 2 dx
3(2) 2 (5) (5) 55
dt
dt
dx
dt
2 2 x dx
2y
dt
9. L x 2 y 2 ,
dL
dt
dy
dx
dt
1,
(5)( 1) (12)(3)
(5)2 (12) 2
dy
dt
dy
0; when x 3 and y 4 2(3)(2) 2( 4) dt 0 dt 32
3 dL
dt
1
2 x2 y 2
4 y 1.
x 2 (2)2 y 3 27
3
dy
x dx y
dy
2 x dx
2 y dt dt2 dt2 ; when x 5 and y 12
dt
x y
31
13
10. r s 2 v3 12, dr
4,
dt
dy
dt
4 , dy 1 3 x 2 y 2 dy 2 xy 3 dx 0; when
dt
27 dt
2
dt
2
3
1 2(2) 1 dx 0 dx 9
3(2) 2 13
2
3 dt
2
dt
8. x 2 y3
Thus
8 r dr
dt
2 dt 5 dx
dt 5(2) 10
dt
4. 2 x 3 y 12,
5. y x 2 ,
2 r dr
dt
ds 3 dr
dt
dt
dv 1
4 2(1)(3) 3(2)2 dv
0
6
dt
dt
2 s ds
3v 2 dv
0; when r 3 and s 1 (3) (1) 2 v3 12 v 2
dt
dt
m dS 12 x dx ; when x 3 dS 12(3)( 5) 180 m 2
S 6 x 2 , dx
5 min
dt
min
dt
dt
dt
3
2
3 dx
2 dx
m
dV
dV
m
(b) V x , dt 5 min dt 3 x dt ; when x 3 dt 3(3) ( 5) 135 min
11. (a)
2
in ; V x3 dV 3 x 2 dx ; when x 3
72 in
dS
12 x dx
72 12(3) dx
dx
2 sec
sec
dt
dt
dt
dt
dt
dt
3
3(3) 2 (2) 54 in
sec
12. S 6 x 2 ,
dV
dt
dS
dt
Copyright 2018 Pearson Education, Inc.
Section 3.8 Related Rates
13. (a) V r 2 h
(c) V r 2 h
dV
dt
dV
dt
r 2 dh
dt
r2
dh
dt
(b) V r 2 h
2 rh dr
dt
13 r 2 dh
14. (a) V 13 r 2 h dV
dt
dt
1 r 2 dh 2 rh dr
(c) dV
3
dt
dt 3
dt
15. (a)
(c)
(d)
16. (a)
dV
dt
dV
dt
dR
dt
R
I dR
dt
1
12
1
2 1 2 3
2 rh dr
dt
(b) V 13 r 2 h dV
32 rh dr
dt
dt
1 volt/sec
dl
dt
dV
dt
157
dR
1I
dt
12 (3)
dl 1
dt
3
1 dV V dI
I dt
I dt
(b)
dV R dI dR
dt
dt
dt
3 ohms/sec, R is increasing
2
amp/sec
P RI 2 dP
I 2 dR
2 RI dI
dt
dt
dt
(b) P RI 2 0
dP
dt
2
dI I dI 2 P dI
I 2 dR
2 RI dI
dR
2 RI
dt
dt
dt
I 2 dt
I 2 dt
I 3 dt
P
17. (a) s x 2 y 2 ( x 2 y 2 )1/2
ds
dt
(b) s x 2 y 2 ( x 2 y 2 )1/2
ds
dt
x
dx
x 2 y 2 dt
x
dx
x 2 y 2 dt
(c) s x 2 y 2 s 2 x 2 y 2 2s ds
dt
y
dy
x y dt
dy
2 x dx
2 y dt 2 s 0
dt
2
2
dy
y dy
2 x dx
2 y dt dx
x dt
dt
dt
dy
18. (a) s x 2 y 2 z 2 s 2 x 2 y 2 z 2 2 s ds
2 x dx
2 y dt 2 z dz
dt
dt
dt
ds
dt
x
2
2
x y z
2
(b) From part (a) with
(c) From part (a) with
19. (a) A
(c) A
d
dt
y
2
2
2
1 ab sin dA 1 ab cos d
dt
dt
2
2
1 ab sin dA 1 ab cos d 1 b sin da
dt
dt
dt
2
2
2
20. Given A r 2 ,
21. Given
dy
2 z 2 2 dz
x y z dt
x y z dt
y
dy
dx 0 ds
2 z 2 2 dz
dt
dt
2
2
2 dt
x y z
x y z dt
ds 0 0 2 x dx 2 y dy 2 z dz dx y dy
dt
dt
dt
dt
dt
x dt
dx
dt
dr
dt
xz dz
0
dt
(b) A 12 ab sin
1 a sin db
dt
2
dA
dt
12 ab cos ddt 12 b sin da
dt
1 cm 2 /min.
0.01 cm/sec, and r 50 cm. Since dA
2 r dr
, then dA
2 (50) 100
dt
dt
dt r 50
2 cm/sec, dw
2 cm/sec, 12 cm and w 5 cm.
dt
(a) A w dA
dw
w ddt dA
12(2) 5(2) 14 cm 2 /sec, increasing
dt
dt
dt
2 ddt 2 dw
2(2) 2(2) 0 cm/sec, constant
(b) P 2 2 w dP
dt
dt
(c) D w2 2 ( w2 2 )1/2
14
cm/sec, decreasing
13
dD
dt
1
2
w2 2
2w dwdt 2 ddt dDdt w w
1/2
dy
dw
dt
2
d
dt
2
(5)(2) (12)( 2)
25144
yz dx
xz dt xy dz
dV
(3)(2)(1) (4)(2)(2) (4)(3)(1) 2 m3 /sec
dt
dt
dt (4, 3, 2)
dy
(b) S 2 xy 2 xz 2 yz dS
(2 y 2 z ) dx
(2 x 2 z ) dt (2 x 2 y ) dz
dt
dt
dt
22. (a) V xyz
dV
dt
dS
dt (4, 3, 2)
(10)(1) (12)( 2) (14)(1) 0 m 2 /sec
Copyright 2018 Pearson Education, Inc.
158
Chapter 3 Derivatives
(c) x 2 y 2 z 2 ( x 2 y 2 z 2 )1/2
ddt (4, 3, 2)
23. Given:
dx
dt
x
dx
x y z dt
2
2
2
y
dy
x y z dt
2
(1) (2) (1) 0 m/sec
3
29
4
29
2
2
dz
z
x 2 y 2 z 2 dt
2
29
5 ft/sec, the ladder is 13 ft long, and x 12, y 5 at the instant of time
125 (5) 12 ft/sec, the ladder is sliding down the wall
dy
The area of the triangle formed by the ladder and walls is A 12 xy dA
12 x dt y dx
. The area is
dt
dt
(a) Since x 2 y 2 169
(b)
d
dt
dy
dt
xy
dx
dt
changing at 12 [12(12) 5(5)] 119
59.5 ft 2 /sec.
2
x sin d 1 dx d 1 dx
(c) cos 13
dt
dt
13 dt
13sin dt
dy
24. s 2 y 2 x 2 2 s ds
2 x dx
2 y dt
dt
dt
ds
dt
dy
15 (5) 1 rad / sec
1s x dx
y dt ds
1 [5(442) 12(481)] 614 knots
dt
dt
169
25. Let s represent the distance between the girl and the kite and x represents the horizontal distance between the
400(25)
girl and kite s 2 (300) 2 x 2 ds
xs dx
500 20 ft/sec.
dt
dt
1 in/min. Also V 6 r 2 dV 12 r dr
3000
dt
dt
1
3
dV
12
(1.9)
0.0076
.
The
volume
is
changing
at
about
0.0239
in
/min.
dt
3000
26. When the diameter is 3.8 in., the radius is 1.9 in. and
dr
dt
2
3
h
27. V 13 r 2 h, h 83 (2r ) 34r r 43h V 13 43h h 1627
9
90 0.1119 m/sec 11.19 cm/sec
(a) dh
(10) 256
2
dt
(b) r
h4
16 4
43h dr
43 dh
dt
dt
2
75 h3
4
dy
dt
dh
dt
2
4( 50)
dV
2254 h dh
dh
dt h 5
dt
dt
225 (5)2
2258 154 0.0849 m/sec 8.49 cm/sec
dh dr
15
15
dt h 5
2
2 dt
dy
29. (a) V 3 y 2 (3R y ) dV
3 [2 y (3R y ) y 2 (1)] dt
dt
we have
2
dr
dt
169h
15 0.1492 m/sec 14.92 cm/sec
90
43 256
32
28. (a) V 13 r 2 h and r 152h V 13 152h h
8 0.0113 m/min 1.13 cm/min
225
(b) r 152h
dV
dt
dy
dt
1
3 (6 Ry 3 y 2 ) dV
at R 13 and y 8
dt
1 ( 6) 1 m/min
144
24
(b) The hemisphere is one the circle r 2 (13 y )2 169 r 26 y y 2 m
(c) r (26 y y 2 )1/2
dr
dt
y 8
dr
dt
dy
12 (26 y y 2 ) 1/2 (26 2 y ) dt
13 y dy
26 y y 2 dt
dr
dt
dr
dt
4k r 2 4 r 2 dr
dr
k , a constant.
dt
dt
2885 m/min
138
1
26864 24
30. If V 43 r 3 , S 4 r 2 , and dV
kS 4k r 2 , then dV
4 r 2
dt
dt
Therefore, the radius is increasing at a constant rate.
31. If V 43 r 3 , r 5, and
100 ft 3 /min, then dV
4 r 2 dr
dr
1 ft/min. Then S 4 r 2
dt
dt
dt
dS
8 r dr
8 (5)(1) 40 ft 2 /min, the rate at which the surface area is increasing.
dt
dt
dV
dt
Copyright 2018 Pearson Education, Inc.
Section 3.8 Related Rates
159
32. Let s represent the length of the rope and x the horizontal distance of the boat from the dock.
xs ds
2s ds
. Therefore, the boat is approaching the dock at
(a) We have s 2 x 2 36 dx
dt
dt
dt
s 36
dx
dt
s 10
10
102 36
(2) 2.5 ft/sec.
8
sin ddt 62 dr
ddt 2 6 dr
. Thus, r 10, x 8, and sin 10
r dt
r sin dt
3 rad/sec
ddt 26 8 (2) 20
10
(b) cos
6
r
10
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 2 h 2 x 2
1 [68(1) 51(17)] 11 ft/sec.
85
ds
1s h dh
x dx
ds
dt
dt
dt
dt
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
V 9 h dV
9 dh
the rate the coffee is rising is dh
91 dV
910 in/min.
dt
dt
dt
dt
3
(b) Let h the height of the coffee in the pot. From the figure, the radius of the filter r h2 V 13 r 2 h 12h ,
the volume of the filter. The rate the coffee is falling is dh
4 2 dV
254 (10) 58 in/min.
dt
dt
h
35. y QD 1
dy
dt
dQ
1 (0) 233 ( 2) 466 L/min increasing about 0.2772 L/min
D 1 dt QD 2 dD
41
dt
1681
(41)2
36. Let P ( x, y ) represent a point on the curve y x 2 and the angle of inclination of a line containing P and the
2
y
ddt cos 2 dx
. Since dx
10 m/sec and
origin. Consequently, tan x tan xx x sec2 ddt dx
dt
dt
dt
cos 2 x 3
x2
y 2 x2
2
23
9 32
1 , we have d
10
dt
x 3
1 rad/sec.
37. The distance from the origin is s x 2 y 2 and we wish to find
ds
dt (5, 12)
dy
(5)( 1) (12)( 5)
12 ( x 2 y 2 ) 1/2 2 x dx
2 y dt
5m/sec
dt
25144
(5, 12)
s
38. Let s distance of the car from the foot of perpendicular in the textbook diagram tan 132
1 ds d cos 2 ds ; ds 264 and 0 d 2 rad/sec. A half second later the car has
sec2 ddt 132
dt
dt
132 dt
dt
dt
264 (since s increases)
traveled 132 ft right of the perpendicular | | 4 , cos 2 12 , and ds
dt
1
(2)
ddt 132
(264) 1 rad/sec.
39. Let s 16t 2 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 50 and since the triangle LOQ and triangle
30 h h 50 16t 2
PRQ are similar we have I 50
h
and I
30(50 16t 2 )
50 (50 16t 2 )
dI
dt t 1
2
15002 30 dI
1500
3
dt
16t
8t
1500 ft/sec.
Copyright 2018 Pearson Education, Inc.
160
Chapter 3 Derivatives
sec2 ddt 802
40. When x represents the length of the shadow, then tan 80
x
We are given that ddt 0.27
3
2000
rad/ min . At x 60, cos 53 dx
dt
dx dx
dt
x dt
x 2 sec2 d
dt
80
2
sec 2 d .
x 80
dt
316 ft/min 0.589 ft/min 7.1 in./min.
3
and sec 53
ddt 2000
41. The volume of the ice is V 43 r 3 43 43 dV
4 r 2 dr
dr
5 in./min when dV
10 in 3 /min,
dt
dt
dt
dt r 6 72
the thickness of the ice is decreasing at 725 in/min. The surface area is S 4 r 2 ds
8 r dr
dS
dt
dt
dt
r 6
5 10 in 2 /min, the outer surface area of the ice is decreasing at 10 in 2 /min.
48 72
3
3
42. Let s represent the horizontal distance between the car and plane while r is the line-of-sight distance between
2r dr
ds
5 (160) 200 mph speed of plane speed
the car and plane 9 s 2 r 2 ds
dt
dt
dt
r 5
r 9
16
of car 200 mph 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
dx
dt
16 ft/sec
(a) s 2 x 2 8100 2 s ds
2 x dx
ds
xs dx
. When the player is 30 ft from first base, x 60 s 30 13
dt
dt
dt
dt
60 ( 16) 32 8.875 ft/sec
and ds
dt
30 13
90
s
(b) sin 1
13
d
d
cos 1 dt1 902 ds
dt1 2 90 ds
s90
ds . Therefore, x 60 and s 30 13
k dt
s dt
s cos 1 dt
d
8 rad/sec; cos 90 sin d 2 90 ds d 2
90
90 ds
dt1
32 65
2
2 dt
s
dt
s 2 sin 2 dt
s 2 dt
30 13 (60) 13
90 ds .
sk dt
(c)
d1
dt
90 ds
s 2 cos 1 dt
lim
x 0
d 2
dt
dx
dt
Therefore, x 60 and s 30 13
90
x 8100
2
90
dx
x 2 8100 dt
90
xs dx
902
2 x
dt
s
s
s
(15)
1 rad/sec; d 2
6
dt
d
90
30 13 (60)
13
x 908100
90 ds
s sin 2 dt
2
lim
8 rad / sec.
32 65
2
90
2 x
s s
d1
x 0 dt
dx
dt
x
s
dx
dt
90
s2
dx
dt
lim dt2 16 rad/sec
x 0
44. Let a represent the distance between point O and ship A, b the distance between point O and ship B, and D the
21D 2a da
2b db
distance between the ships. By the Law of Cosines, D 2 a 2 b 2 2ab cos120 dD
dt
dt
dt
2a da 2b db a db b da . When a 5, da 14, b 3, and db 21, then dD 413 where D 7. The
dt
dt
dt
dt
dt
dt
2D
dt
29.5
knots
apart.
ships are moving dD
dt
1
2D
45. The hour hand moves clockwise from 4 at 30/hr = 0.5/min. The minute hand, starting at 12, chases the hour
hand at 360°/hr = 6°/min. Thus, the angle between them is decreasing and is changing at
0.5/min − 6/min = −5.5/min.
46. The volume of the slick in cubic feet is V 43 a2 b2 , where a is the length of the major axis and b is the
a2 316 a dbdt b dadt . Convert all measurements to feet and
316 2(5280)(10) 43 (5280)(30) 316 (224,400) 132,183 ft 3 /hr
length of the minor axis.
substitute:
47.
d
dt
dx
dt
dV
dt
dV
dt
3
4
a d
2 dt
b
2
b d
2 dt
rad 6 rad , tan x sec 2 d dx ; x 1 km so
3 circles
2circle
dt
dt
4
min
min
km
sec2 4 (6 ) (2)(6 ) 12 min
Copyright 2018 Pearson Education, Inc.
3.9 Linearization and Differentials
3.9
161
LINEARIZATION AND DIFFERENTIALS
1. f ( x) x3 2 x 3 f ( x) 3x 2 2 L( x ) f (2)( x 2) f (2) 10( x 2) 7 L( x) 10 x 13 at x 2
2. f ( x) x 2 9 ( x 2 9)1/2 f ( x)
12 ( x2 9)1/2 (2 x)
54 ( x 4) 5 L( x) 54 x 95 at x 4
x
x 2 9
L( x) f (4)( x 4) f (4)
3. f ( x) x 1x f ( x) 1 x 2 L( x) f (1) f (1)( x 1) 2 0( x 1) 2
4. f ( x) x1/3 f ( x)
1
3 x 2/3
1 ( x 8) 2 L ( x) 1 x 4
L( x) f (8)( x (8)) f (8) 12
12
3
5. f ( x) tan x f ( x) sec2 x L( x) f ( ) f ( )( x ) 0 1( x ) x
6. (a) f ( x) sin x f ( x) cos x L( x ) f (0) f (0)( x 0) x L( x) x
(b) f ( x) cos x f ( x) sin x L( x) f (0) f (0)( x 0) 1 L( x) 1
(c) f ( x) tan x f ( x) sec2 x L( x) f (0) f (0)( x 0) x L( x) x
7. f ( x) x 2 2 x f ( x) 2 x 2 L( x) f (0)( x 0) f (0) 2( x 0) 0 L( x) 2 x at x 0
8. f ( x) x 1 f ( x) x 2 L( x) f (1)( x 1) f (1) (1)( x 1) 1 L( x) x 2 at x 1
9. f ( x) 2 x 2 4 x 3 f ( x) 4 x 4 L( x) f (1)( x 1) f (1) 0( x 1) (5) L( x) 5 at x 1
10.
f ( x) 1 x f ( x) 1 L( x) f (8)( x 8) f (8) 1( x 8) 9 L( x) x 1 at x 8
11.
1 ( x 8) 2 L( x) 1 x 4 at x 8
f ( x) 3 x x1/3 f ( x) 13 x 2/3 L( x) f (8)( x 8) f (8) 12
12
3
12.
f ( x)
x
x 1
f ( x)
(1)( x 1) (1)( x )
( x 1) 2
1
( x 1)2
L( x) f (1)( x 1) f (1) 14 ( x 1) 12 L( x) 14 x 14
at x 1
13. f ( x) k (1 x) k 1. We have f (0) 1 and f (0) k . L( x) f (0) f (0)( x 0) 1 k ( x 0) 1 kx
14. (a) f ( x) (1 x)6 [1 ( x)]6 1 6( x) 1 6 x
1
(b) f ( x) 12x 2 1 ( x) 2[1 (1)( x)] 2 2 x
(c) f ( x) 1 x
1/2
1 12 x 1 2x
(d) f ( x) 2 x 2 2 1
x2
2
1/2
(e) f ( x ) (4 3x )1 3 41 3 1 34x
(f)
x
f ( x ) 1
2
x
2/3
2 1 12
13
x2
2
2 1
x2
4
41 3 1 13 34x 41 3 1 4x
1 2x x
2/3
x
2x
1 23
1 6 3 x
x
2
15. (a) (1.0002)50 (1 0.0002)50 1 50(0.0002) 1 .01 1.01
(b) 3 1.009 (1 0.009)1/3 1 13 (0.009) 1 0.003 1.003
Copyright 2018 Pearson Education, Inc.
162
Chapter 3 Derivatives
12 ( x 1)1/2 cos x L f ( x) f (0)( x 0) f (0)
32 ( x 0) 1 L f ( x ) 32 x 1, the linearization of f ( x); g ( x) x 1 ( x 1)1/2 g ( x ) 12 ( x 1)1/2
16. f ( x) x 1 sin x ( x 1)1/2 sin x f ( x)
Lg ( x) g (0)( x 0) g (0) 12 ( x 0) 1 Lg ( x) 12 x 1, the linearization of g ( x); h( x) sin x
h( x) cos x Lh ( x) h(0)( x 0) h(0) (1)( x 0) 0 Lh ( x) x, the linearization of h( x).
L f ( x) Lg ( x) Lh ( x) implies that the linearization of a sum is equal to the sum of the linearizations.
17. y x3 3 x x3 3 x1/2 dy 3x 2 32 x 1/2 dx dy 3 x 2
18. y x 1 x 2 x (1 x 2 )1/2 dy (1) (1 x 2 )1/2 ( x)
1 x2
1/2
12 x dx
dx
12 (1 x2 )1/2 (2 x) dx
2
1 x 2
2
(2)(1 x 2 ) (2 x )(2 x )
dy
dx 2 22x 2 dx
2 2
(1 x )
(1 x )
19. y
2x
1 x 2
20. y
2 x
3 1 x
(1 x 2 ) x 2 dx
3
2 x
x 1/ 2 31 x1/ 2 2 x1/ 2 3 x 1/ 2
1/ 2
1 2
2
1
dx 3 x 33 dx dy
dx
2 x 1/ 2 dy
1/ 2 2
2
1/ 2
x
x )2
3
(1
9(1
x
)
31 x
91 x
21. 2 y 3/2 xy x 0 3 y1/2 dy y dx x dy dx 0 (3 y1/2 x) dy (1 y ) dx dy
1 y
dx
3 y x
22. xy 2 4 x3/2 y 0 y 2 dx 2 xy dy 6 x1/2 dx dy 0 (2 xy 1) dy (6 x1/2 y 2 )dx dy
23. y sin (5 x ) sin (5 x1/2 ) dy (cos (5 x1/2 ))
52 x1/2 dx dy
5cos 5 x
2 x
dx
24. y cos ( x 2 ) dy [ sin ( x 2 )](2 x)dx 2 x sin ( x 2 )dx
25. y 4 tan
dy 4 sec ( x ) dx dy 4x sec dx
x3
3
2 x3
3
2
2
2 x3
3
26. y sec x 2 1 dy [sec ( x 2 1) tan ( x 2 1)](2 x) dx 2 x [sec ( x 2 1) tan ( x 2 1)]dx
27. y 3csc (1 2 x ) 3csc (1 2 x1/2 ) dy 3( csc (1 2 x1/2 )) cot (1 2 x1/2 ) ( x 1/2 ) dx
dy 3 csc (1 2 x ) cot (1 2 x ) dx
x
28. y 2 cot
2 cot x dy 2 csc ( x
1
x
1/2
2
1/2
) 12 ( x 3/2 ) dx dy 1 csc2
x3
29. f ( x ) x 2 2 x, x0 1, dx 0.1 f ( x) 2 x 2
(a) f f ( x0 dx) f ( x0 ) f (1.1) f (1) 3.41 3 0.41
(b) df f ( x0 ) dx [2(1) 2](0.1) 0.4
(c) | f df | |0.41 0.4| 0.01
Copyright 2018 Pearson Education, Inc.
dx
1
x
6 x y2
2 xy 1
dx
3.9 Linearization and Differentials
163
30. f ( x ) 2 x 2 4 x 3, x0 1, dx 0.1 f ( x) 4 x 4
(a) f f ( x0 dx) f ( x0 ) f (.9) f (1) .02
(b) df f ( x0 ) dx [4(1) 4](.1) 0
(c) | f df | |.02 0| .02
31. f ( x ) x3 x, x0 1, dx 0.1 f ( x ) 3x 2 1
(a) f f ( x0 dx) f ( x0 ) f (1.1) f (1) .231
(b) df f ( x0 )dx [3(1) 2 1](.1) .2
(c) | f df | |.231 .2| .031
32. f ( x) x 4 , x0 1, dx 0.1 f ( x) 4 x3
(a) f f ( x0 dx) f ( x0 ) f (1.1) f (1) .4641
(b) df f ( x0 )dx 4(1)3 (.1) .4
(c) | f df | |.4641 .4| .0641
33. f ( x) x 1 , x0 0.5, dx 0.1 f ( x) x 2
(a) f f ( x0 dx) f ( x0 ) f (.6) f (.5) 13
(b) df f ( x0 )dx (4)
101 52
1
(c) | f df | | 13 52 | 15
34. f ( x) x3 2 x 3, x0 2, dx 0.1 f ( x) 3x 2 2
(a) f f ( x0 dx) f ( x0 ) f (2.1) f (2) 1.061
(b) df f ( x0 )dx (10)(0.10) 1
(c) | f df | |1.061 1| .061
35. V 43 r 3 dV 4 r02 dr
36. V x3 dV 3 x02 dx
37. S 6 x 2 dS 12 x0 dx
38. S r r 2 h 2 r (r 2 h 2 )1/2 , h constant
dS
2 r02 h 2
r02 h 2
dr , h constant
39. V r 2 h, height constant dV 2 r0 h dr
dS
dr
(r 2 h 2 )1/2 r r (r 2 h 2 )1/2 dS
dr
r 2 h 2 r 2
2
r h2
40. S 2 rh dS 2 r dh
41. Given r 2 m, dr .02 m
(a) A r 2 dA 2 r dr 2 (2)(.02) .08 m 2
(100%) 2%
(b) .08
4
42. C 2 r and dC 2 in. dC 2 dr dr 1 the diameter grew about 2 in.; A r 2
dA 2 r dr 2 (5) 1 10 in.2
43. The volume of a cylinder is V r 2 h. When h is held fixed, we have dV
2 rh, and so dV 2 rh dr .
dr
For h 30 in., r 6 in., and dr 0.5 in., the volume of the material in the shell is approximately
dV 2 rh dr 2 (6)(30)(0.5) 180 565.5 in 3 .
Copyright 2018 Pearson Education, Inc.
164
Chapter 3 Derivatives
44. Let angle of elevation and h height of building. Then h 30 tan , so dh 30 sec2 d . We want
| d | 0.04sin cos
| dh | 0.04h, which gives: |30sec2 d | 0.04 |30 tan | 12 | d | 0.04sin
cos
cos
| d | 0.04sin 512 cos 512 0.01 radian. The angle should be measured with an error of less than
0.01 radian (or approximately 0.57 degrees), which is a percentage error of approximately 0.76%.
45. The percentage error in the radius is
dr
2
dt
100
2 r
r
percentage error in calculating the circle’s circumference is
dCdt 100
2 r dr
. The percentage error in calculating the circle’s area is given by
dt
dAdt 100
dC
dt
(a) Since C 2 r
drdt 100 2%.
2
dr . The
dt
C
drdt 100 2%.
r
2
(b) Since A r
dA
dt
2 r drdt 100 2 drdt 100 2(2%) 4%.
A
r
r2
46. The percentage error in the edge of the cube is
2
dS
dt
(a) Since S 6 x
12 x dx
.
dt
dxdt 100 0.5%.
x
The percentage error in the cube’s surface area is
dSdt 100 12 x dxdt 100
S
6 x2
dx
dt
2 x 100 2(0.5%) 1%
(b) Since V x3 dV
3 x 2 dx
. The percentage error in the cube’s volume is Vdt 100
dt
dt
dV
dx
3 dt 100 3(0.5%) 1.5%
3x 100
2 dx
dt
3
x
x
47. V h3 dV 3 h 2 dh ; recall that ΔV dV. Then | V | (1%)(V )
|3 h 2 dh |
of h is 13 %.
(1)( h3 )
100
| dh |
1 h
300
(1)( h3 )
100
| dV |
(1)( h3 )
100
13 % h. Therefore the greatest tolerated error in the measurement
48. (a) Let Di represent the interior diameter. Then V r 2 h
h
Di 2
2
Di2 h
4
and h 10 V
5 Di2
2
2
2
1 5 Di Di
dV 5 Di dDi . Recall that V dV . We want | V | (1%)(V ) | dV | 100
2 40
Di2
dD
5 Di dDi 40 D i 200. The inside diameter must be measured to within 0.5%.
i
(b) Let De represent the exterior diameter, h the height and S the area of the painted surface. S De h
dS hdDe
dS
s
dDe
.
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 tank’s exterior diameter must be measured to
within 5%.
49. Given D 100 cm, dD 1 cm, V 43
26 102 %
106
104
106
2
106
6
D2
3
3
4
D
dV 2 D 2 dD 2 (100) 2 (1) 102 . Then
6
% 3% 26
10
6
104
Copyright 2018 Pearson Education, Inc.
dV
V
(100%)
3.9 Linearization and Differentials
165
D3 D3
3
D 2 dD; recall that V dV . Then V (3%)V 3
D
dV
200
100 6
6
2
3
2
3
D
D
D
D
dD
(1%)
D
dV 200 2 dD 200
the allowable percentage error in measuring the
100
diameter is 1%.
50. V 43 r 3 43
51. W a
b
g
D2
3
a bg
1
2
dW bg dg
b dg
g
2
dWmoon
dWearth
b dg
2
(5.2)
b dg
2
(32)
32 2 37.87, so a change of gravity
5.2
on the moon has about 38 times the effect that a change of the same magnitude has on Earth.
52. C (t )
4 8t 3
(1 t 3 ) 2
changes from
0.06, where t is measured in hours. When the time changes from 20 min to 30 min, t in hours
to 12 , so the differential estimate for the change in C is
1
3
C 13 12 13 16 C 13 0.584 mg/mL.
53. The relative change in V is estimated by
dV / dr r
V
4 kr3 r
kr4
4 r .
r
1.1r and r 0.1r. The approximate relative increase in V is thus
54. (a) T 2
L
g
1/2
If the radius increases by 10%, r changes to
4(0.1r )
r
0.4 or 40%.
dT 2 L 12 g 3/2 dg Lg 3/2 dg
(b) If g increases, then dg 0 dT 0. The period T decreases and the clock ticks more frequently. Both the
pendulum speed and clock speed increase.
(c) 0.001 100(9803/2 ) dg dg 0.977 cm/sec2 the new g 979 cm/sec2
55. (a) i. Q(a ) f (a ) implies that b0 f (a).
ii. Since Q ( x) b1 2b2 ( x a ), Q(a ) f (a ) implies that b1 f (a ).
f ( a )
iii. Since Q ( x) 2b2 , Q (a ) f (a ) implies that b2 2 .
In summary, b0 f (a ), b1 f ( a ), and b2
f ( a )
.
2
2
(b) f ( x ) (1 x) 1 ; f ( x) 1(1 x) 2 ( 1) (1 x) ; f ( x) 2(1 x) 3 ( 1) 2(1 x) 3 Since
f (0) 1, f (0) 1, and f (0) 2, the coefficients are b0 1, b1 1, b2 22 1. The quadratic
approximation is Q( x) 1 x x 2 .
(c)
As one zooms in, the two graphs quickly become
indistinguishable. They appear to be identical.
Copyright 2018 Pearson Education, Inc.
166
Chapter 3 Derivatives
(d) g ( x) x 1; g ( x) 1x 2 ; g ( x) 2 x 3
Since g (1) 1, g (1) 1, and g (1) 2, the coefficients are b0 1, b1 1, b2
2
approximation is Q( x) 1 ( x 1) ( x 1) .
2
2
1. The quadratic
As one zooms in, the two graphs quickly become
indistinguishable. They appear to be identical.
(e) h( x) (1 x)1/2 ; h( x) 12 (1 x) 1/2 ; h( x) 14 (1 x) 3/2
Since h(0) 1, h(0) 12 , and h(0) 14 , the coefficients are b0 1, b1 12 , b2
2
approximation is Q ( x) 1 2x x8 .
14
2
18 . The quadratic
As one zooms in, the two graphs quickly become
indistinguishable. They appear to be identical.
(f) The linearization of any differentiable function u ( x) at x a is L( x) u (a) u (a )( x a ) b0 b1 ( x a ),
where b0 and b1 are the coefficients of the constant and linear terms of the quadratic approximation. Thus,
the linearization for f ( x) at x 0 is 1 x; the linearization for g ( x) at x 1 is 1 ( x 1) or 2 x; and the
linearization for h( x) at x 0 is 1 2x .
56. E ( x) f ( x) g ( x) E ( x) f ( x) m( x a ) c. Then E (a ) 0 f (a ) m(a a) c 0 c f (a).
E ( x)
f ( x ) m( x a ) c
f ( x) f (a)
0 lim x a m 0 (since c f ( a ))
Next we calculate m: lim x a 0 lim
xa
x a
x a
x a
f (a ) m 0 m f ( a ). Therefore, g ( x) m( x a ) c f (a )( x a ) f ( a ) is the linear approximation,
as claimed.
5760.
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 (, 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_ ]: x 3 x 2 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]]
Copyright 2018 Pearson Education, Inc.
Chapter 3 Practice Exercises
167
Plot[err[x], {x, x1, x2}]
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 x5 0.125 x 2 0.25 x
dy
dx
2. y 3 0.7 x3 0.3 x7
dy
dx
5 x 4 0.25 x 0.25
2.1x 2 2.1x6
3. y x3 3( x 2 2 )
dy
dx
3x 2 3(2 x 0) 3 x 2 6 x 3x( x 2)
4. y x7 7 x 11
dy
dx
7 x6 7
5. y ( x 1) 2 ( x 2 2 x)
2
dy
dx
( x 1) 2 (2 x 2) ( x 2 2 x )(2( x 1)) 2( x 1)[( x 1) 2 x ( x 2)]
dy
dx
(2 x 5)( 1)(4 x) 2 ( 1) (4 x) 1 (2) (4 x) 2 [(2 x 5) 2(4 x)] 3(4 x) 2
2( x 1)(2 x 4 x 1)
6. y (2 x 5)(4 x)1
dy
7. y ( 2 sec 1)3 d 3( 2 sec 1)2 (2 sec tan )
2
8. y 1 csc2 4
9. s
10. s
t
1 t
ds
dt
1 ds
dt
t 1
2
1
sin 2 x
( t 1) (0) 1
t 1
dy
dx
2
1
2 t
2 t
csc cot
2
2
2 1 csc2 4 (csc cot )
ds
dt
1
t 1
2
(4 tan x )(sec 2 x) (2sec x)(sec x tan x) 2sec2 x tan x
sin2 x csc2 x 2 csc x
13. s cos 4 (1 2t )
14. s cot 3
2
1 t 21 t t 21 t 1 t t
1
2
2
2
t
t
t
2
1
2
1
1
t
t
11. y 2 tan 2 x sec 2 x
12. y
dy
d 2 1 csc2 4
dy
dx
(2 csc x)( csc x cot x) 2( csc x cot x) (2 csc x cot x)(1 csc x)
4 cos3 (1 2t )( sin(1 2t ))(2) 8cos3 (1 2t )sin(1 2t )
2t dsdt 3cot 2 2t csc2 2t t 2 t6 cot 2 2t csc2 2t
15. s (sec t tan t )5
2
ds
dt
2
5(sec t tan t ) 4 sec t tan t sec2 t 5(sec t )(sec t tan t )5
Copyright 2018 Pearson Education, Inc.
168
Chapter 3 Derivatives
16. s csc5 (1 t 3t 2 )
ds
dt
5csc4 (1 t 3t 2 ) ( csc (1 t 3t 2 ) cot (1 t 3t 2 )) (1 6t )
5(6t 1) csc5 (1 t 3t 2 ) cot (1 t 3t 2 )
17. r 2 sin (2 sin )1/2 ddr 12 (2 sin )1/2 (2 cos 2sin ) cos sin
2 sin
12 (cos )1/2 ( sin ) 2(cos )1/2 cossin 2
18. r 2 cos 2 (cos )1/2 ddr 2
20. r sin 1 ddr cos 1 1 1
2 1
dy
dx
21. y 12 x 2 csc 2x
12 x 2 csc 2x cot 2x
2 x cos x
23. y x 1/2 sec (2 x) 2
1/2
8x
or
dy
dx
2
x2 csc 2x 12 2 x csc 2x cot 2x x csc 2x
2
x sin
x
x
2
2
cos 1
x 1/2 sec (2 x) 2 tan(2 x)2 (2(2 x) 2) sec (2 x)2 12 x 3/2
sec (2 x) tan (2 x)
1
2 x3/ 2
2 1 1
2 1
2 1 x sin x 2 2 x cos
dy
dx
22. y 2 x sin x
1 x 3/2 sec (2 x ) 2
2
2
2
2 cos sin
cos
12 (2 )1/2 (2) cos 22
19. r sin 2 sin(2 )1/2 ddr cos(2 )1/2
cos
1 x1/2 sec (2 x ) 2
2
16 tan (2 x) x
2
2
sec (2 x) 16 x tan(2 x) 1
dy
x1/2 ( csc ( x 1)3 cot ( x 1)3 ) (3( x 1)2 ) csc ( x 1)3 12 x 1 2
dx
csc( x 1)3
3 x ( x 1)2 csc ( x 1)3 cot ( x 1)3
12 x csc ( x 1)3 1x 6( x 1) 2 cot ( x 1)3 or 1 csc ( x 1)3
2 x
2 x
3
2
3
or 1 csc ( x 1) 1 6 x( x 1) cot ( x 1)
2 x
24. y x csc( x 1)3 x1/2 csc ( x 1)3
25. y 5cot x 2
dy
dx
26. y x 2 cot 5 x
27.
5( csc2 x 2 )(2 x) 10 x csc2 ( x 2 )
dy
dx
x 2 ( csc 2 5 x)(5) (cot 5 x)(2 x) 5 x 2 csc2 5 x 2 x cot 5 x
y x 2 sin 2 (2 x 2 )
dy
dx
2
x 2 (2sin (2 x 2 )) (cos (2 x 2 ))(4 x) sin 2 (2 x 2 )(2 x)
8 x3 sin(2 x 2 ) cos(2 x ) 2 x sin 2 (2 x 2 )
28. y x 2 sin 2 ( x3 )
29. s
t4t1
30. s
1
15(15t 1)3
2
ds
dt
dy
dx
x 2 (2 sin ( x3 )) (cos ( x3 ))(3 x 2 ) sin 2 ( x3 )(2 x 3 ) 6 sin ( x3 ) cos ( x3 ) 2 x 3 sin 2 ( x3 )
2 t4t1
3 (t 1)(4) (4t )(1)
(t 1)2
2
t4t1
3
4
(t 1)2
1 (15t 1) 3 ds 1 ( 3)(15t 1) 4 (15)
15
15
dt
(t 1)
8t 3
3
(15t 1) 4
Copyright 2018 Pearson Education, Inc.
Chapter 3 Practice Exercises
31. y
32.
x
x 1
y
2
2 x
2 x 1
x2 x
x2
33. y
dy
dx
2
2
1 1x
x
x 1
2
dy
dx
1/2
( x 1)
x (1) ( x1)2 x
1
2 x
( x 1)2
( x 1)3
2 x
2 x
2 x 1
dy
dx
1
x
1
x
(2 x 1)
2
1/2
12 1 1x
12
34. y 4 x x x 4 x( x x1/2 )1/2
(2 x 1)
dy
dx
x
4x
1 x
( x 1)3
4 x
1
x
(2 x 1)3
4
(2 x 1)3
1
2 x 2 1 1x
12 ( x x1/2 )1/2 1 12 x1/2 ( x x1/2 )1/2 (4)
( x x ) 1 2 2 x 1 1 4( x x ) ( x x )1 2 (2 x x 4 x 4 x ) 6 x 5
2 x
x
169
x
x
sin (cos 1)(cos ) (sin )( sin ) 2 sin cos 2 cos sin 2
ddr 2 cos
1
cos 1
(cos 1)2
(cos 1) 2
(2sin )(1cos )
2sin
3
2
sin
35. r cos
1
2
(cos 1)
1
36. r 1sin
cos
2
(cos 1)
1 (1 cos )(cos ) (sin 1)(sin ) 2(sin 1) (cos cos 2 sin 2 sin )
ddr 2 1sin
(1cos )3
cos
(1cos ) 2
2(sin 1)(cos sin 1)
(1cos )3
37. y (2 x 1) 2 x 1 (2 x 1)3/2
dy
dx
32 (2 x 1)1/2 (2) 3 2 x 1
38. y 20(3 x 4)1/4 (3 x 4) 1/5 20(3 x 4)1/20
dy
dx
20
201 (3x 4)19/20 (3) (3x 34)
19/ 20
dy
dx
40. y (3 cos3 3 x) 1/3
2
13 (3 cos3 3 x) 4/3 (3cos 2 3x)( sin 3 x)(3) 3cos 33x sin 34/3x
dy
dx
3 32 (5 x 2 sin 2 x)5/2 [10 x (cos 2 x)(2)]
9(5 x cos 2 x )
39. y 3(5 x 2 sin 2 x)3/2
5 x2 sin 2 x
5/ 2
(3 cos 3 x )
y2
41. xy 2 x 3 y 1 ( xy y ) 2 3 y 0 xy 3 y 2 y y ( x 3) 2 y y x 3
dy
dy
dy
dy
42. x 2 xy y 2 5 x 2 2 x x dx y 2 y dx 5 0 x dx 2 y dx 5 2 x y
dy
dx
5 2 x y
x2 y
dy
43. x3 4 xy 3 y 4/3 2 x 3 x 2 4 x dx 4 y 4 y1/3
dy
2 4 x dx 4 y1/3
dy
dx
2 3x 2 4 y
2 3 x 2 4 y
dy
dy
dy
dx
dy
( x 2 y ) 5 2x
dx
dx (4 x 4 y1/3 ) 2 3x 2 4 y dx
4 x 4 y1/3
44. 5 x 4/5 10 y 6/5 15 4 x 1/5 12 y1/5
dy
dx
0 12 y1/5 dx 4 x 1/5 dx 13 x 1/5 y 1/5 1 1/5
3( xy )
dy
45. ( xy )1/2 1 12 ( xy ) 1/2 x dx y 0 x1/2 y 1/2
dy
dy
dx
x 1/2 y1/2
dy
dy
dx
x 1 y
Copyright 2018 Pearson Education, Inc.
dy
dx
y
x
y
170
Chapter 3 Derivatives
dy
dy
46. x 2 y 2 1 x 2 2 y dx y 2 (2 x) 0 2 x 2 y dx 2 xy 2
dy
47. y 2 xx1 2 y dx
48. y 2 11 xx
1/2
( x 1)(1) ( x )(1)
( x 1) 2
dy
dx
dy
dx
y 4 11 xx 4 y 3
49. p3 4 pq 3q 2 2 3 p 2
dp
dq
dp
dq
(1 x )2
dp
dy
dx
4 p q dq 6q 0 3 p 2
6q 4 p
y
x
1
2 y ( x 1) 2
(1 x )(1) (1 x )( 1)
dy
dx
dp
dq
1
2 y 3 (1 x ) 2
dp
4q dp 6q 4 p
5 p2 2 p
5/ 2
dp
dp
2 r sin 2 s sin 2 s
dr
cos 2 s
ds
(2 r 1)(sin 2 s )
cos 2 s
drds 2sin s cos s 0 drds (cos 2s) 2r sin 2s 2sin s cos s
(2r 1)(tan 2 s )
dr
1 2s 0
52. 2rs r s s 2 3 2 r s dr
ds
ds
dr
ds
(2 s 1) 1 2s 2r
dy
dr
ds
1 2 s 2r
2 s 1
y 2 ( 2 x ) ( x 2 ) 2 y dx
2
dy
dy
d2y
x
0 dx 2 2
dx
y
dx
y4
2
4
2 xy 2 (2 yx 2 ) x 2
2 xy 2 2 xy
2 xy 3 2 x 4
d2y
y
2
y4
y5
dx
y4
dy
dy
dy
dy
d2y
2
y 1 2x 2 y dx 22 dx 12 dx ( yx 2 )1 2 ( yx 2 )2 y (2 x) x 2 dx
dx
x
yx
2 1
2 xy x 2
2 xy 2 1
d2y
yx
53. (a) x3 y 3 1 3 x 2 3 y 2
dx 2
y2 x4
y3 x4
dy
dy
54. (a) x 2 y 2 1 2 x 2 y dx 0 2 y dx 2 x
(b)
dp
(10 p 2)
dq
3(5 p 1)
51. r cos 2 s sin 2 s r ( sin 2 s )(2) (cos 2s )
(b)
4 q ) 6q 4 p
3 p 2 4q
50. q (5 p 2 2 p )3/2 1 32 (5 p 2 2 p )5/2 10 p dq 2 dq 23 (5 p 2 2 p)5/2
dp
dq
dp
(3 p 2
dq
dy
dx
x
y
d2y
dx
2
dy
y (1) x dx
y
2
yx
y
y x
2
x
y
2
y
3
2
dy
dx
x
y
13 (since y 2 x 2 1)
y
55. (a) Let h( x) 6 f ( x) g ( x) h( x ) 6 f ( x) g ( x) h(1) 6 f (1) g (1) 6
12 (4) 7
(b) Let h( x) f ( x) g 2 ( x) h( x) f ( x) (2 g ( x)) g ( x) g 2 ( x) f ( x) h(0) 2 f (0) g (0) g (0) g 2 (0) f (0)
2(1)(1) 12 (1) 2 ( 3) 2
( g ( x ) 1) f ( x ) f ( x ) g ( x )
f ( x)
( g (1) 1) f (1) f (1) g (1)
(51)
1 3(4)
2
5
h(1)
12
(c) Let h( x) g ( x ) 1 h( x)
( g (1) 1) 2
( g ( x ) 1) 2
(51) 2
(d) Let h( x) f ( g ( x)) h( x) f ( g ( x)) g ( x) h(0) f ( g (0)) g (0) f (1) 12 12 12 14
(e) Let h( x) g ( f ( x)) h( x) g ( f ( x)) f ( x) h(0) g ( f (0)) f (0) g (1) f (0) (4)(3) 12
(f ) Let h( x) ( x f ( x))3/2 h( x) 32 ( x f ( x))1/2 (1 f ( x)) h(1) 32 (1 f (1))1/2 (1 f (1))
32 (1 3)1/2 1 12 92
Copyright 2018 Pearson Education, Inc.
Chapter 3 Practice Exercises
171
(g) Let h( x) f ( x g ( x)) h( x) f ( x g ( x))(1 g ( x)) h(0) f ( g (0))(1 g (0))
f (1) 1 12 12 23 43
56. (a) Let h( x) x f ( x) h( x) x f ( x) f ( x) 1 h(1) 1 f (1) f (1) 1 15 ( 3)
2 x
2 1
12 1013
(b) Let h( x) ( f ( x))1/2 h( x) 12 ( f ( x))1/2 ( f ( x)) h(0) 12 ( f (0))1/2 f (0) 12 (9)1/2 (2) 13
x h( x) f x 2 1 x h(1) f 1 21 1 15 12 101
(c) Let h( x) f
(d) Let h( x) f (1 5 tan x) h( x) f (1 5 tan x)(5sec2 x) h(0) f (1 5 tan 0)(5sec2 0)
f (1)(5) 15 (5) 1
f ( x)
(e) Let h( x) 2 cos x h( x)
(2 cos x ) f ( x ) f ( x )( sin x )
(2 cos x ) 2
(21) f (0) f (0)(0)
h(0)
(2 1) 2
3( 2)
9
2
h(1) 10 sin 2 (2 f (1) f (1)) f 2 (1) 10 cos 2 2 20( 3) 15 0 12
23
(f ) Let h( x) 10sin 2x f 2 ( x) h( x) 10sin 2x (2 f ( x) f ( x)) f 2 ( x) 10 cos 2x
57. x t 2
thus,
dy
dt
dx 2t ; y 3sin 2 x
dt
dy dx
6 cos (2t 2 ) 2t
dx dt
58. t (u 2 2u )1/3
2
1/3
dy
3(cos 2 x)(2) 6 cos 2 x
dx
dy
6 cos(0) 0 0
dt t 0
13 (u 2 2u )2/3 (2u 2) 23 (u 2 2u )2/3 (u 1); s t 2 5t
dt
du
2
ds ds dt
2(u 2u ) 5; thus du
dt du
2
1/3
ds
du
[2(2 2(2)) 5] 23
u 2
dw
ds
dw dr
dr ds
cos
2
1/3
[2(u 2u )
(2
59. r 8sin s 6
6 cos(2t 2 2 ) 6 cos (2t 2 );
2
2(2))
2/3
5]
8sin s 6
8sin s 6 2
60. 2 t 1 2 t 2 ddt
8cos s dw
6
2
2u )
1/3
(2 1) 2(2 8
8cos s 6 ; w sin ( r 2)
dr
ds
(u
2
3
ds
dw
dr
cos
cos
8sin
2
s 0
2/3
(u 1)
ds
dt
2t 5
5)(82/3 ) 2(2 2 5) 14 92
r 2
21r
8sin s 6
cos 8sin s 6 2
2
6 2 8cos 6 (cos 0)(8) 23
2 4
8sin 6
; thus,
3
ddt 0 ddt (2 t 1) 2 ddt 2t 1 ; r ( 2 7)1/3
2
ddr 13 ( 2 7)2/3 (2 ) 23 ( 2 7) 2/3 ; now t 0 and 2t 1 1 so that ddt
1 1
t 0, 1 1
and ddr
23 (1 7) 2/3 16 dr
ddr
ddr
16 (1) 16
dt
1
t 0
dy
dx
61. y 3 y 2 cos x 3 y 2
d2y
dx
2
dy
dx 2sin x
dy
(3 y 2 1)( 2 cos x ) ( 2sin x ) 6 y dx
2
(3 y 1)
2
d2y
dx
2
x
2/3
2 y 1/3 dy
dx
3
y
x
2/3 2
d
2
dy
dx
2/3
dy
dx
y
dy
t 0
(3 y 2 1) 2sin x
dx 2 (0, 1)
62. x1/3 y1/3 4 13 x 2/3 13 y 2/3
t 0
dy
dx
x
2sin
2
3 y 1
(31)( 2 cos 0)( 2sin 0)(60)
dy
d
2
y
(8, 8)
dx 2
8
2/3
2 81/3 ( 1)
3
4/3
2sin(0)
31
y 2/3
dy
0 dx 2/3 dx
1; dx 2/3
x
x
(8, 8)
2 x 1/3
3
12
(31)2
y 2/3
dy
dx (0, 1)
82/3
8
Copyright 2018 Pearson Education, Inc.
2 81/3
3
13 13 23 1
82/3
4
6
0;
172
Chapter 3 Derivatives
1
1
f (t h ) f ( t )
2t 1(2t 2 h 1)
2( t h ) 1 2 t 1
1 and f (t h)
1
(2t 2h 1)(2t 1) h
2t 1
2(t h ) 1
h
h
f ( t h ) f (t )
2
lim (2t 2h21)(2t 1) 2 2
f (t ) lim
h
(2t 2 h 1)(2t 1)
(2t 1)
h 0
h 0
63. f (t )
64. g ( x) 2 x 2 1 and g ( x h) 2( x h)2 1 2 x 2 4 xh 2h 2 1
4 xh 2 h 2
h
g ( x h) g ( x)
h
4 x 2h g ( x) lim
h 0
g ( x h) g ( x)
h
h
(2t 2h21)(2
t 1) h
(2 x 2 4 xh 2 h 2 1) (2 x 2 1)
h
lim (4 x 2h) 4 x
h 0
65. (a)
(b)
lim f ( x) lim x 2 0 and lim f ( x) lim x 2 0 lim f ( x) 0. Since lim f ( x) 0 f (0) it
x 0
x 0
x 0
x 0
x 0
x 0
x 0
x 0
follows that f is continuous at x 0.
(c) lim f ( x ) lim (2 x) 0 and lim f ( x) lim (2 x) 0 lim f ( x) 0. Since this limit exists, it
x 0
x 0
follows that f is differentiable at x 0.
x 0
66. (a)
(b)
lim f ( x) lim x 0 and lim f ( x ) lim tan x 0 lim f ( x) 0. Since lim f ( x) 0 f (0), it
x 0
x 0
x 0
x 0
x 0
x 0
x 0
x 0
x 0
x 0
x 0
follows that f is continuous at x 0.
(c) lim f ( x) lim 1 1 and lim f ( x) lim sec 2 x 1 lim f ( x) 1. Since this limit exists it follows
that f is differentiable at x 0.
67. (a)
(b)
lim f ( x) lim x 1 and lim f ( x ) lim (2 x) 1 lim f ( x) 1. Since lim f ( x ) 1 f (1), it
x 1
x 1
x 1
x 1
x 1
x 1
x 1
x 1
x 1
follows that f is continuous at x 1.
(c) lim f ( x ) lim 1 1 and lim f ( x) lim 1 1 lim f ( x ) lim f ( x), so lim f ( x) does not
exist f is not differentiable at x 1.
x 1
x 1
Copyright 2018 Pearson Education, Inc.
x 1
x 1
Chapter 3 Practice Exercises
68. (a)
173
lim f ( x ) lim sin 2 x 0 and lim f ( x) lim mx 0 lim f ( x) 0, independent of m; since
x 0
x 0
x 0
x 0
x 0
f (0) 0 lim f ( x ) it follows that f is continuous at x 0 for all values of m.
(b)
x 0
lim f ( x) lim (sin 2 x) lim 2 cos 2 x 2 and lim f ( x) lim (mx) lim m m f is
x 0
x 0
x 0
x 0
x 0
differentiable at x 0 provided that lim f ( x) lim f ( x) m 2.
x 0
69. y 2x 2 x1 4 12 x (2 x 4) 1
2 2(2 x 4) 2 1
(2 x 5)(2 x 3) 0 x
70.
x 0
x 0
12 2(2 x 4)2 ; the slope of the tangent is 32 32 12 2(2 x 4) 2
(2 x 4)2 1 4 x 2 16 x 16 1 4 x 2 16 x 15 0
dy
dx
1
(2 x 4) 2
52 or x
32 52 , 95 and 23 , 14 are points on the curve where the slope is 32 .
1 dy
1
1
;
, so the points where the slope is 2 are
1 2 . The derivative is equal to 2 when x
2 x dx
2
2x
1
1
, 0 and
, 0 .
2
2
y x
dy
dx
71. y 2 x3 3 x 2 12 x 20
dy
6 x 2 6 x 12; the tangent is parallel to the x-axis when dx 0
6 x 2 6 x 12 0 x 2 x 2 0 ( x 2)( x 1) 0 x 2 or x 1 (2, 0) and (1, 27) are points
on the curve where the tangent is parallel to the x-axis.
72. y x3
dy
dx
dy
3 x 2 dx
( 2, 8)
12; an equation of the tangent line at (2, 8) is y 8 12( x 2)
y 12 x 16; x-intercept: 0 12 x 16 x 43 43 , 0 ; y -intercept : y 12(0) 16 16 (0, 16)
dy
dx
73. y 2 x3 3 x 2 12 x 20
6 x 2 6 x 12
11 24; 6 x 2 6 x 12 24
24
2
2
x x 2 4 x x 6 0 ( x 3)( x 2) 0 x 2 or x 3 (2, 16) and (3, 11) are points
x .
where the tangent is perpendicular to y 1 24
x when
(a) The tangent is perpendicular to the line y 1 24
dy
dx
dy
(b) The tangent is parallel to the line y 2 12 x when dx 12 6 x 2 6 x 12 12 x 2 x 0
x( x 1) 0 x 0 or x 1 (0, 20) and (1, 7) are points where the tangent is parallel to y 2 12 x.
x
74. y sin
x
dy
dx
x ( cos x ) ( sin x )(1)
x
2
dy
2
m1 dx
2 1 and m2
x
the tangents intersect at right angles.
dy
75. y tan x, 2 x 2 dx sec 2 x; now the slope
of y 2x is 12 the normal line is parallel to
dy
y 2x when dx 2. Thus, sec 2 x 2 12 2
cos x
cos 2 x 12 cos x 1 x 4 and x 4
2
for 2 x 2 4 , 1 and 4 , 1 are points
where the normal is parallel to y 2x .
Copyright 2018 Pearson Education, Inc.
dy
2
dx x 2
1. Since m1 m1
2
174
Chapter 3 Derivatives
dy
dx
76. y 1 cos x
sin x
dy
dx
1
2 , 1
the tangent at 2 , 1 is the line y 1 x 2
y x 2 1; the normal at 2 , 1 is
y 1 (1) x y x 1
12
2
dy
2 x and y x dx 1; the parabola is tangent to y x when 2 x 1 x 12 y 12 ; thus,
2
C C 14
77. y x 2 C
1
2
2
78. y x3
dy
dx
dy
dx
dy
3 x 2 dx
3a 2 the tangent line at ( a, a3 ) is y a 3 3a 2 ( x a ). The tangent line
xa
intersects y x3 when x3 a3 3a 2 ( x a ) ( x a ) ( x 2 xa a 2 ) 3a 2 ( x a ) ( x a)( x 2 xa 2a 2 ) 0
dy
( x a ) 2 ( x 2a ) 0 x a or x 2a. Now dx
3( 2a ) 2 12a 2 4 (3a 2 ), so the slope at x 2a
x 2 a
is 4 times as large as the slope at (a, a3 ) where x a.
79. The line through (0, 3) and (5, 2) has slope m
1 the line through (0, 3) and (5, 2) is
dy
, so the curve is tangent to y x 3 dx 1 c 2
( x 1)
( x 1)2 c, x 1. Moreover, y xc1 intersects y x 3 xc1 x 3, x 1
y x 3; y
c dy
x 1
dx
3 ( 2)
0 5
c
( x 1)2
c ( x 1)( x 3), x 1. Thus c c ( x 1)2 ( x 1)( x 3) ( x 1)[ x 1 ( x 3)] 0,
x 1 ( x 1)(2 x 2) 0 x 1 (since x 1 ) c 4.
80. Let b, a 2 b 2 be a point on the circle x 2 y 2 a 2 . Then x 2 y 2 a 2 2 x 2 y dx 0 dx xy
2
2
dy
b
dx
normal line through b, a 2 b 2 has slope ab b normal line is
2
2
x a
a b
2
2
2
2
2
a
b2
y a 2 b 2 ab b ( x b) y a 2 b 2
x a 2 b 2 y a bb x which passes
b
through the origin.
dy
dy
81. x 2 2 y 2 9 2 x 4 y dx 0
dy
dx
dy
dy
2xy dx
14 the tangent line is y 2 14 ( x 1) 14 x 94
(1, 2)
and the normal line is y 2 4( x 1) 4 x 2.
82.
2
3 x 1
dy
dy
. At the point (0, 1),
0
2y
dx
dx
dy
3
3
3
, so the tangent line has the equation ( y 1) ( x 0) or y x 1 and the normal line has the
2
2
2
dx
2
2
equation ( y 1) ( x 0) or y x 1.
3
3
x 13 y 2 2
d
dx
x 13 y 2
2
0 3 x 1 2 y
dy y 2
dy
dx
2 the tangent
dx x 5
(3, 2)
line is y 2 2( x 3) 2 x 4 and the normal line is y 2 21 ( x 3) 12 x 72 .
dy
dy
83. xy 2 x 5 y 2 x dx y 2 5 dx 0
dy
( x 5)
dx
y 2
Copyright 2018 Pearson Education, Inc.
Chapter 3 Practice Exercises
84. ( y x) 2 2 x 4 2( y x)
dy
dx
dy
dy
1 y x
175
dy
1 2 ( y x) dx 1 ( y x) dx y x dx
34 the tangent
(6, 2)
line is y 2 34 ( x 6) 34 x 52 and the normal line is y 2 43 ( x 6) 43 x 10.
85. x xy 6 1
is y 1 54 ( x 4)
1
2 xy
54
x
dy
dx
dy
dy
2 xy y
x
dy
dx
45 the tangent line
(4, 1)
x 6 and the normal line is y 1 54 ( x 4) 54 x 11
.
5
y 0 x dx y 2 xy dx
1/ 2
dy
x1/ 2 dx
14 the tangent line is
2y
(1, 4)
y 4 14 ( x 1) 14 x 17
and the normal line is y 4 4( x 1) 4 x.
4
dy
86. x3/2 2 y 3/2 17 32 x1/2 3 y1/2. dx 0
dy
dx
dy
dy
dy
dy
dy dy
87. x3 y 3 y 2 x y x3 3 y 2 dx y 3 (3x 2 ) 2 y dx 1 dx 3 x3 y 2 dx 2 y dx dx 1 3 x 2 y 3
1 3 x2 y3
dy
dy
dy
dy
dx (3 x3 y 2 2 y 1) 1 3 x 2 y 3 dx 3 2
dx
24 , but dx
is undefined. Therefore, the
3 x y 2 y 1
curve has
slope 12
88. y sin( x sin x)
(1, 1)
(1, 1)
at (1, 1) but the slope is undefined at (1, 1).
dy
dx
[cos( x sin x)](1 cos x); y 0 sin( x sin x) 0 x sin x k , k 2, 1, 0, 1, 2
dy
(for our interval) cos( x sin x) cos( k ) 1. Therefore, dx 0 and y 0 when 1 cos x 0 and x k .
For 2 x 2 , these equations hold when k 2, 0, and 2(since cos( ) cos 1.) Thus the curve has
horizontal tangents at the x-axis for the x-values 2 , 0, and 2 (which are even integer multiples of ) the
curve has an infinite number of horizontal tangents.
89. B graph of f , A graph of f . Curve B cannot be the derivative of A because A has only negative slopes
while some of B’s values are positive.
90. A graph of f , B graph of f . 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
2
x 0 2 x x
95. lim
lim
x 0
3 x tan 7 x
2x
x 0
96. lim
sinx x (2 x11) (1) 11 1
lim cos17 x sin7 x7 x 1 32 1 1 72 2
3x sin 7 x 32 x
x0 2 x 2 x cos 7 x
0
lim
2
7
sin r lim sin r 2 r 1 1 (1) lim cos 2 r 1 (1) 1 1
97. lim tan
sin 2 r
2
1
2
tan 2 r 2
2
2 r r0 r
r0 2 r
r0
Copyright 2018 Pearson Education, Inc.
176
Chapter 3 Derivatives
sin(sin )
lim
sin(sin ) sin
sin
0
0
sin(sin )
x
sin
lim sin lim x 1
x 0
0
98. lim
99.
lim
2
lim 4 tan 2 tan 1
2
100.
tan 5
1 2 cot 2
lim
5cot 2 7 cot 8
0
sin (sin )
.
0 sin
4 1
tan
lim
2
1
tan 2
1 5
tan 2
1 2
2
cot
lim
Let x sin . Then x 0 as 0
(4 0 0)
(1 0) 4
(0 2)
0 5 cot7 82
cot
(500) 52
2x 2x sin x
2x
2x sin x (1)(1)(1) 1
x lim
x sin x lim
x sin x
101. lim 2x2sin
lim
lim
2 x
x
x
2
cos x x 0 2(1 cos x )
x
x 0
x0 2 2sin 2x x 0 sin 2
x0 sin 2 sin 2 x
102. lim
0
1 cos
2
lim
2 lim sin 2 sin 2 1 (1)(1) 1 1
2 2
2
0 2
2 2
2sin 2
0
tan(tan x )
103. lim tanx x lim cos1 x sinx x 1; let tan x 0 as x 0 lim g ( x ) lim tan x
x0
x0
x 0
x0
Therefore, to make g continuous at the origin, define g (0) 1.
lim tan 1.
0
tan(tan x )
tan(tan x )
sin x 1 1 lim sin x (using the result of # 98); let sin x
104. lim f ( x) lim sin(sin x ) lim tan x sin(sin
x ) cos x
x 0
x0
x0
x0 sin(sin x )
sin
x
0 as x 0 lim sin(sin x ) lim sin 1. Therefore, to make f continuous at the origin,
0
x0
define f (0) 1.
105. (a) S 2 r 2 2 rh and h constant dS
4 r dr
2 h dr
(4 r 2 h) dr
dt
dt
dt
dt
(b) S 2 r 2 2 rh and r constant dS
2 r dh
dt
dt
(c) S 2 r 2 2 rh dS
4 r dr
2 r dh
h dr
(4 r 2 h) dr
2 r dh
dt
dt
dt
dt
dt
dt
dr 2 r dh (2r h) dr r dh dr r dh
0
0
(4
r
2
h
)
(d) S constant dS
dt
dt
2 r h dt
dt
dt
dt
dt
106. S r r 2 h 2 dS
r
dt
(a) h constant dh
0
dt
(b) r constant
dr
dt
0
r drdt h dhdt
r 2 h2
2 dr
dS r dt
dt
r 2 h2
dS
dt
r 2 h 2 dr
;
dt
r 2 h2
rh dh
r 2 h 2 dt
dr
dt
2
r 2 h 2 2r 2 dr
r h dt
2
(c) In general, dS
r 2 h 2 2r 2 dr
2rh 2 dh
dt
dt
r h
r h dt
107. A r 2 dA
2 r dr
; so r 10 and dr
2 m /sec dA
(2 )(10) 2 40 m 2 /sec
dt
dt
dt
dt
3s 2 ds
ds
12 dV
; so s 20 and dV
1200 cm3 /min ds
1 2 (1200) 1 cm/min
108. V s3 dV
dt
dt
dt
dt
dt
30(20)
3s dt
Copyright 2018 Pearson Education, Inc.
Chapter 3 Practice Exercises
dR
dR
dR
177
dR
12 dt1 12 dt2 Also, R1 75 ohms
109. dt1 1 ohm/sec, dt2 0.5 ohm/sec; and R1 R1 R1 12 dR
R dt
1
2
R1
R2
1 1 R 30 ohms. Therefore, from the derivative equation,
and R2 50 ohms R1 75
50
1 dR
(30) 2 dt
1
(75)2
(1)
5000 5625
1 1
(0.5) 5625
dR
(900) 56255000
5000
dt
1
(50)2
9(625)
50(5625)
1 0.02 ohm/sec.
50
R dR X dX
110. dR
3 ohms/sec and dX
2 ohms/sec; Z R 2 X 2 dZ
dt 2 2dt so that R 10 ohms and
dt
dt
dt
R X
(10)(3) (20)( 2)
dZ
1
X 20 ohms dt
0.45 ohm/sec.
2
2
5
10 20
dy
111. Given dx
10 m/sec and dt 5 m/sec, let D be the distance from the origin D 2 x 2 y 2
dt
dy
dy
2 D dD
2 x dx
2 y dt D dD
x dx
y dt . When ( x, y ) (3, 4), D 32 (4) 2 5 and
dt
dt
dt
dt
5 dD
(3)(10) (4)(5) dD
10
2. Therefore, the particle is moving away from the origin at 2 m/sec
dt
dt
5
(because the distance D is increasing).
112. Let D be the distance from the origin. We are given that dD
11 units/sec. Then D 2 x 2 y 2 x 2 ( x3/2 ) 2
dt
x 2 x3 2 D dD
2 x dx
3 x 2 dx
x(2 3x) dx
; x 3 D 32 33 6 and substitution in the derivative
dt
dt
dt
dt
dx
4 units/sec.
equation gives (2)(6)(11) (3)(2 9) dx
dt
dt
4r r 52 h.
113. (a) From the diagram we have 10
h
2
3
(b) V 13 r 2 h 13 52 h h 475h
dV
dt
2
125 ft/min.
425h dh
, so dV
5 and h 6 dh
144
dt
dt
dt
114. From the sketch in the text, s r ds
r ddt dr
. Also r 1.2 is constant dr
0 ds
r ddt (1.2) ddt .
dt
dt
dt
dt
6 ft/sec and r 1.2 ft ddt 5 rad/sec
Therefore, ds
dt
115. (a) From the sketch in the text, ddt 0.6 rad/sec and x tan . Also x tan dx
sec2 ddt ; at point A,
dt
x 0 0 dx
(sec2 0)(0.6) 0.6. Therefore the speed of the light is 0.6 53 km/sec when it
dt
reaches point A.
(3/5) rad 1 rev 60sec
2 rad min
sec
(b)
18
revs/min
b a
b
116. From the figure, ar BC
. We are given
r
b2 r 2
that r is constant. Differentiation gives,
1 da
r dt
b2 r 2
db
dt
(b)
b2 r 2
b
b2 r 2
dbdt
. Then, b 2r and
2
2
(2r ) r ( 0.3r ) (2 r )
db 0.3r da r
dt
dt
(2 r ) 2 r 2
3r 2 ( 0.3r )
3r
2 r ( 0.3 r )
(2 r )2 r 2
4 r 2 (0.3 r )
3r 2
(3r 2 )( 0.3r ) (4 r 2 )(0.3r )
3 3r 2
0.3r r
3 3
10 3
m/sec. Since da
is positive, the distance OA is
dt
increasing when OB 2r , and B is moving toward O at the rate of 0.3r m/sec.
Copyright 2018 Pearson Education, Inc.
178
Chapter 3 Derivatives
117. (a) If f ( x ) tan x and x 4 , then f ( x) sec 2 x,
f 4 1 and f 4 2. The linearization of
f ( x ) is L( x) 2 x 4 ( 1) 2 x 2 2 .
(b) if f ( x) sec x and x 4 , then f ( x ) sec x tan x,
f 4 2 and f 4 2. The linearization of
f ( x) is L( x ) 2 x 4 2 2x
2(4 )
.
4
2
1
118. f ( x) 1 tan
f ( x) sec x 2 . The linearization at x 0 is L( x) f (0)( x 0) f (0) 1 x.
x
(1 tan x )
119. f ( x) x 1 sin x 0.5 ( x 1)1/2 sin x 0.5 f ( x) 12 ( x 1) 1/2 cos x
L( x) f (0)( x 0) f (0) 1.5( x 0) 0.5 L( x) 1.5 x 0.5 , the linearization of f ( x ) .
2
1 x 3.1 2(1 x) 1 ( x 1)1/2 3.1 f ( x) 2(1 x) 2 (1) 12 ( x 1) 1/2
1 x
2
1
L( x) f (0)( x 0) f (0) 2.5 x 0.1 , the linearization of f ( x) .
2 2 1 x
(1 x)
120. f ( x)
121. S r r 2 h 2 , r constant dS r 12 (r 2 h 2 ) 1/2 2h dh
h0 to h0 dh dS
r h0 ( dh )
rh
r 2 h2
dh. Height changes from
r 2 h02
2
r | dr | r . The measurement of the
122. (a) S 6r 2 dS 12r dr. We want | dS | (2%) S |12r dr | 12
100
100
edge r must have an error less than 1%.
(b) When V r 3 , then dV 3r 2 dr. The accuracy of the volume is
(dr )(100%) (100%) 3%
3
r
3
r
r
100
dVV (100%) 3rr dr (100%)
3
2
2
3
123. C 2 r r 2C , S 4 r 2 C , and V 43 r 3 C 2 . It also follows that dr 21 dC , dS 2C
dC and
6
2
dV C 2 dC. Recall that C 10 cm and dC 0.4 cm.
2
210 (100%) (.04)(100%) 4%
0.2
cm dr
(100%) 0.2
(a) dr 0.4
2
r
Copyright 2018 Pearson Education, Inc.
Chapter 3 Additional and Advanced Exercises
179
(100%) 8%
(0.4) 8 cm dS
(100%) 8 100
(b) dS 20
S
2
6 2 (100%) 12%
(c) dV 10 2 (0.4) 202 cm dV
(100%) 202 1000
V
2
15
h 14 ft. The same triangles imply that 20h a a6 h 120a 1 6
124. Similar triangles yield 35
h
6
1 120 1 2 .0444 ft 0.53 inches.
dh 120a 2 da 120
da 120
12
2
2
2
45
12
a
CHAPTER 3
a
15
ADDITIONAL AND ADVANCED EXERCISES
1. (a) sin 2 2sin cos dd (sin 2 ) dd (2sin cos ) 2 cos 2 2[(sin )( sin ) (cos )(cos )]
cos 2 cos 2 sin 2
(b) cos 2 cos 2 sin 2 dd (cos 2 ) dd (cos 2 sin 2 )
2 sin 2 (2 cos )( sin ) (2 sin )(cos )
sin 2 cos sin sin cos sin 2 2sin cos
2. The derivative of sin ( x a ) sin x cos a cos x sin a with respect to x is cos( x a ) cos x cos a sin x sin a,
which is also an identity. This principle does not apply to the equation x 2 2 x 8 0, since x 2 2 x 8 0 is
not an identity: it holds for 2 values of x (2 and 4), but not for all x.
3. (a) f ( x) cos x f ( x) sin x f ( x) cos x, and g ( x) a bx cx 2 g ( x) b 2cx g ( x) 2c;
also, f (0) g (0) cos(0) a a 1; f (0) g (0) sin(0) b b 0; f (0) g (0)
cos(0) 2c c 12 . Therefore, g ( x) 1 12 x 2 .
(b) f ( x ) sin( x a ) f ( x) cos( x a ), and g ( x) b sin x c cos x g ( x) b cos x c sin x; also
f (0) g (0) sin( a ) b sin(0) c cos(0) c sin a; f (0) g (0) cos( a ) b cos(0) c sin(0)
b cos a. Therefore, g ( x ) sin x cos a cos x sin a.
(c) When f ( x ) cos x, f ( x) sin x and f (4) ( x) cos x; when g ( x) 1 12 x 2 , g ( x) 0 and g (4) ( x) 0.
Thus f (0) 0 g (0) so the third derivatives agree at x 0 . However, the fourth derivatives do not
agree since f (4) (0) 1 but g (4) (0) 0. In case (b), when f ( x) sin( x a ) and
g ( x) sin x cos a cos x sin a , notice that f ( x) g ( x) for all x, not just x 0. Since this is an identity, we
have f ( n) ( x) g ( n ) ( x ) for any x and any positive integer n.
4. (a) y sin x y cos x y sin x y y sin x sin x 0; y cos x y sin x
y cos x y y cos x cos x 0; y a cos x b sin x y a sin x b cos x
y a cos x b sin x y y (a cos x b sin x) (a cos x b sin x) 0
(b) y sin(2 x) y 2 cos(2 x) y 4sin(2 x) y 4 y 4sin(2 x ) 4sin(2 x) 0. Similarly,
y cos(2 x) and y a cos(2 x) b sin(2 x) satisfy the differential equation y 4 y 0. In general,
y cos(mx), y sin(mx) and y a cos (mx) b sin (mx) satisfy the differential equation y m 2 y 0.
5. If the circle ( x h)2 ( y k ) 2 a 2 and y x 2 1 are tangent at (1, 2), then the slope of this tangent is
m 2 x (1, 2) 2 and the tangent line is y 2 x. The line containing (h, k) and (1, 2) is perpendicular to
y 2 x kh12 12 h 5 2k the location of the center is (5 2k , k ). Also, ( x h) 2 ( y k ) 2 a 2
1 ( y)2
x h ( y k ) y 0 1 ( y )2 ( y k ) y 0 y k y . At the point (1, 2) we know y 2 from the
tangent line and that y 2 from the parabola. Since the second derivatives are equal at (1, 2) we obtain
1 (2) 2
2 k 2 k
we have that a
9 . Then h
2
5 5
.
2
5 2k 4 the circle is ( x 4)2 y 92
2
Copyright 2018 Pearson Education, Inc.
a 2 . Since (1, 2) lies on the circle
180
Chapter 3 Derivatives
2
x
, where 0 x 60.
6. The total revenue is the number of people times the price of the fare: r ( x) xp x 3 40
2
dr 3 x
x
1 dr 3 x 3 x 2 x 3 3 x 1 x .
2 x 3 40
The marginal revenue is dx
40
dx
40
40
40
40
40
40
dr 0 x 40 (since x 120 does not belong to the domain). When 40 people are on the bus the
Then dx
x
marginal revenue is zero and the fare is p (40) 3 40
2
( x 40)
$4.00.
dy
7. (a) y uv dt du
v u dv
(0.04u )v u (0.05v) 0.09uv 0.09 y the rate of growth of the total
dt
dt
production is 9% per year.
dy
(b) If du
0.02u and dv
0.03v, then dt ( 0.02u )v (0.03v)u 0.01uv 0.01 y, increasing at 1% per year.
dt
dt
8. When x 2 y 2 225, then y xy . The tangent line
to the balloon at (12, 9) is y 9 43 ( x 12)
y 43 x 25. The top of the gondola is
15 8 23 ft below the center of the balloon. The
intersection of y 23 and y 43 x 25 is at the far
right edge of the gondola 23 43 x 25 x 32 .
Thus the gondola is 2 x 3 ft wide.
9. Answers will vary. Here is one possibility.
s (0) 10 cos 4 10
2
10. s (t ) 10 cos t 4 v(t )
(a)
ds
dt
2
10 sin t 4 a (t ) dv
d 2s 10 cos t 4
dt
dt
(b) Left : 10, Right:10
(c) Solving 10 cos t 4 10 cos t 4 1 t 34 when the particle is farthest to the left. Solving
10 cos t 4 10 cos t 4 1 t 4 , but t 0 t 2 4 74 when the particle is farthest to
0, v 0, a 34 10, and a 74 10.
Solving 10 cos t 4 0 t 4 v 4 10, v 4 10 and a 4 0.
the right. Thus, v
(d)
3
4
7
4
s (t ) 64t 16t 2 v(t ) ds
64 32t 32(2 t ). The maximum height is reached when v(t ) 0
dt
t 2 sec. The velocity when it leaves the hand is v(0) 64 ft/sec.
(b) s (t ) 64t 2.6t 2 v(t ) ds
64 5.2t. The maximum height is reached when v(t ) 0 t 12.31sec.
dt
The maximum height is about s (12.31) 393.85 ft.
11. (a)
12. s1 3t 3 12t 2 18t 5 and s2 t 3 9t 2 12t v1 9t 2 24t 18 and v2 3t 2 18t 12; v1 v2
9t 2 24t 18 3t 2 18t 12 2t 2 7t 5 0 (t 1)(2t 5) 0 t 1 sec and t 2.5 sec.
Copyright 2018 Pearson Education, Inc.
Chapter 3 Additional and Advanced Exercises
181
k 2 x dx
m dv
k 22vx dx
m dv
kx 1v dx
. Then substituting
13. m v 2 v02 k x02 x 2 m 2v dv
dt
dt
dt
dt
dt
dt
dx v m dv kx, as claimed.
dt
dt
t t
t t
14. (a) x At 2 Bt C on [t1 , t2 ] v dx
2 At B v 1 2 2 2 A 1 2 2 B A(t1 t2 ) B is
dt
the instantaneous velocity at the midpoint. The average velocity over the time interval is
At22 Bt2 C At12 Bt1 C t2 t1 [ At2 t1 B] A(t t ) B.
vav xt
2 1
t t
t t
2
2
1
1
(b) On the graph of the parabola x At 2 Bt C , the slope of the curve at the midpoint of the interval [t1 , t2 ]
is the same as the average slope of the curve over the interval.
15. (a) To be continuous at x requires that lim sin x lim (mx b) 0 m b m b ;
x
x
cos x, x
(b) If y
is differentiable at x , then lim cos x m m 1 and b .
m, x
x
f ( x ) f (0)
x 0
x 0
x 0 f (0). f (0) lim
16. f ( x) is continuous at 0 because lim 1cos
x
x 1 cos x
lim 1cos
1 cos x
x2
x 0
x 0
2
1
lim sinx x
1 cos x
x 0
lim
x 0
1 cos x 0
x
x
12 . Therefore f (0) exists with value 12 .
17. (a) For all a, b and for all x 2, f is differentiable at x. Next, f differentiable at x 2 f continuous at
x 2 lim f ( x) f (2) 2a 4a 2b 3 2a 2b 3 0. Also, f differentiable at x 2
x 2
a, x 2
f ( x)
. In order that f (2) exist we must have a 2a(2) b a 4a b 3a b.
2ax b, x 2
Then 2a 2b 3 0 and 3a b a 34 and b 94 .
(b) For x 2, the graph of f is a straight line having a slope of 34 and passing through the origin; for x 2, the
graph of f is a parabola. At x 2, the value of the y -coordinate on the parabola is 32 which matches the
y -coordinate of the point on the straight line at x 2. In addition, the slope of the parabola at the match up
point is 34 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 1, g is differentiable at x. Next, g differentiable at x 1 g continuous
at x 1 lim g ( x) g (1) a 1 2b a b b 1. Also, g differentiable at x 1
g ( x)
x 1
a , x 1
. In
3ax 2 1, x 1
order that g (1) exist we must have a 3a(1)2 1 a 3a 1 a 12 .
(b) For x 1, the graph of g is a straight line having a slope of 12 and a y -intercept of 1. For x 1, the
graph of g is a cubic. At x 1, the value of the y -coordinate on the cubic is 32 which matches the
y -coordinate of the point on the straight line at x 1. In addition, the slope of the cubic at the match up
point is 12 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.
d ( f ( x )) d ( f ( x )) f ( x)( 1) f ( x ) f ( x ) f ( x ) f is even.
19. f odd f ( x) f ( x) dx
dx
d ( f ( x)) d ( f ( x)) f ( x)(1) f ( x) f ( x) f ( x ) f is odd.
20. f even f ( x) f ( x) dx
dx
Copyright 2018 Pearson Education, Inc.
182
Chapter 3 Derivatives
h ( x ) h ( x0 )
f ( x ) g ( x ) f ( x0 ) g ( x0 )
lim
x x0
x x0
x x0
x x0
f ( x ) g ( x ) f ( x ) g ( x0 ) f ( x ) g ( x0 ) f ( x0 ) g ( x0 )
g ( x) g ( x )
f ( x) f ( x )
lim
lim f ( x) x x 0 lim g ( x0 ) x x 0
x
x
0
0
0
x x0
x x0
x x0
g
(
x
)
g
(
x
)
g
(
x
)
g
(
x
)
f ( x0 ) lim x x 0 g ( x0 ) f ( x0 ) 0 lim x x 0 g ( x0 ) f ( x0 ) g ( x0 ) f ( x0 ), if g is
0
0
x x
x x
21. Let h( x) ( fg )( x) f ( x) g ( x) h( x) lim
0
0
continuous at x0 . Therefore ( fg ) ( x) is differentiable at x0 if f ( x0 ) 0, and ( fg )( x0 ) g ( x0 ) f ( x0 ).
22. From Exercise 21 we have that fg is differentiable at 0 if f is differentiable at 0, f (0) 0 and g is continuous at 0.
(a) If f ( x) sin x and g ( x) | x |, then | x | sin x is differentiable because f (0) cos(0) 1, f (0) sin (0) 0
and g ( x) | x | is continuous at x 0.
(b) If f ( x) sin x and g ( x) x 2/3 , then x 2/3 sin x is differentiable because
f (0) cos (0) 1, f (0) sin (0) 0 and g ( x) x 2/3 is continuous at x 0.
(c) If f ( x) 1 cos x and g ( x) 3 x, then 3 x (1 cos x) is differentiable because f (0) sin (0) 0,
f (0) 1 cos (0) 0 and g ( x) x1/3 is continuous at x 0.
(d) If f ( x) x and g ( x) x sin 12 , then x 2 sin 1x is differentiable because f (0) 1, f (0) 0 and
sin
lim x sin 1x lim 1
x 0
x 0
1
x
x
lim sin t
x t
0 (so g is continuous at x 0 ).
23. If f ( x) x and g ( x) x sin 1x , then x 2 sin 1x is differentiable at x 0 because f (0) 1, f (0) 0 and
sin
lim x sin 1x lim 1
x 0
x 0
1
x
lim
t
x
sin t
t
0 (so g is continuous at x 0 ). In fact, from Exercise 21,
h(0) g (0) f (0) 0. However, for x 0, h( x) x 2 cos 1x 12 2 x sin 1x . But
x
lim h( x) lim cos 1x 2 x sin 1x does not exist because cos 1x has no limit as x 0. Therefore,
x 0
x 0
the derivative is not continuous at x 0 because it has no limit there.
24.
x 2 , x is rational
f ( x)
0, x is irrational
f (0 h ) f (0)
h
h 0
f (0) lim
f ( h)
f ( h)
lim h ; if h is irrational, then lim h lim 0h lim 0 0;
h 0
h 0
h 0
h 0
f ( h)
2
f ( h)
if h is rational, then lim h lim hh lim h 0; thus f (0) lim h 0 and f is differentiable at x 0.
h 0
h 0
h 0
h 0
25.
dx
dt
2cm sec, and by the Law of Cosines, 25 ( x 2 9) ((4 x) 2 36) 2 x 2 9 (4 x)2 36 cos
0 2 x dx
2(4 x) dx
2 12 ( x 2 9)
dt
dt
2 x 2 9 12 ((4 x) 36)
12
12
2 x dx
(4 x)2 36 cos
dt
(2(4 x) dx
)cos 2 x 2 9 (4 x )2 36 ( sin ) ddt ;
dt
Copyright 2018 Pearson Education, Inc.
Chapter 3 Additional and Advanced Exercises
183
Let x 4 25 25 36 2(5)(6)cos cos 53 sin 54 , then
0 2(4)(2)
2(5)(6) 54 ddt ddt 757 rad sec
2(4)
(2)(6) 53
5
26. From the given conditions we have f ( x h) f ( x) f (h), f (h) 1 hg (h) and lim g (h) 1. Therefore,
h 0
f ( h ) 1
lim f ( x) h f ( x) lim g (h) f ( x) 1 f ( x)
f ( x)
h 0
h 0
f ( x) f ( x) and f ( x) exists at every value of x.
f ( x h) f ( x )
lim
h
h 0
27. Step 1:
Step 2:
f ( x ) f ( h) f ( x )
lim
h
h 0
The formula holds for n 2 (a single product) since y u1u2
dy
dx
du
du
dx1 u2 u1 dx2
Assume the formula holds for n k :
du
du
du
dy
y u1u2 uk dx dx1 u2u3 uk u1 dx2 u3 uk ... u1u2 uk 1 dxk .
d (u u u )
du
2
k
uk 1 u1u2 uk dxk 1
If y u1u2 uk uk 1 u1u2 uk uk 1, then dx 1 dx
du
du
du
du
dx1 u2 u3 uk u1 dx2 u3 uk u1u2 uk 1 dxk uk 1 u1u2 uk dxk 1
dy
du
du1
du
u u uk 1 u1 dx2 u3 uk 1 u1u2 uk 1 dxk
dx 2 3
du
uk 1 u1u2 uk dxk 1 .
Thus the original formula holds for n (k 1) whenever it holds for n k .
m !( k 1) m !( m k )
m
m
m
m!
m!
m!
m!
28. Recall m
k k !( m k )! . Then 1 1!( m 1)! m and k k 1 k !( m k )! ( k 1)!( m k 1)! ( k 1)!( m k )!
m !( m 1)
( m 1)!
1
( k 1)!( m k )! ( k 1)!(( m 1) ( k 1))! m
k 1 . Now, we prove Leibniz’s rule by mathematical induction.
Step 1:
d (uv )
If n 1, then dx u dv
v du
. Assume that the statement is true for n k , that is:
dx
dx
k
k
k 1
k 2
2
d (uv )
k
d k 1v u d k v .
d ku v k d k u1 dv
2 d k u2 d 2v ... kk1 du
k
k
dx
dv k 1
dx
Step 2:
dx
dx
dx
dx
dx
dx
d d (uv ) d u v d u dv k d k u dv k d k 1u d 2 v
dx
dx k dx k 1
dx
dx k dx dx k dx
dx k 1 dx 2
k
1
2
k
2
3
2
k
1
k
d u v
2k d k u1 d 2v 2k d k u2 d 3v ... kk1 d u2 d k v1 kk1 du
dx dx k
dx
dx
dx
dx
dx dx
d k v u d k 1u d k 1u v ( k 1) d k u dv k k d k 1u d 2v
du
1
2 k 1
dx dx k
dx k 1 dx k 1
dx k dx
dx
dx 2
k 1
k 1
k
k
k du d k v
k 1 d k 1u d 2v
d
v
d
u
d
u
dv
k 1 k dx k u k 1 k 1 v (k 1) k dx 2
...
dx
dx
dk
dx
dx k 1 dx 2
k
k
1
d v u d v.
kk 1 du
k 1
dx k
If n k 1, then
d
k 1
k
(uv )
dx
k
k 1
k 1
dx
Therefore the formula (c) holds for n (k 1) whenever it holds for n k .
2
T 2g
29. (a) T 2 4g L L 2 L
(1sec2 )(32.2 ft/sec 2 )
4 2
4
2
(b) T 2 4g L T 2
g
L; dT 2
1
g 2 L
L 0.8156 ft
dL dL; dT
Lg
(0.8156ft)(32.2ft/sec2 )
(0.01 ft) 0.00613 sec.
(c) Since there are 86, 400 sec in a day, we have we have (0.00613 sec)(86, 400 sec/day) 529.6 sec/day, or
8.83 min/day; the clock will lose about 8.83 min/day.
3s 2 ds
k (6 s 2 ) ds
2k . If s0 the initial length of the cube’s side, then s1 s0 2k
30. v s3 dv
dt
dt
dt
s
s
( v )1/3
0
2k s0 s1. Let t the time it will take the ice cube to melt. Now, t 20k s 0 s
1/3
1/3
0 1
(v0 ) 34 v0
1 1/3 11 hr.
1
34
Copyright 2018 Pearson Education, Inc.
CHAPTER 4
4.1
APPLICATIONS OF DERIVATIVES
EXTREME VALUES OF FUNCTIONS
1. An absolute minimum at x c2 , an absolute maximum at x b. Theorem 1 guarantees the existence of such
extreme values because h is continuous on [a, b].
2. An absolute minimum at x b, an absolute maximum at x c. Theorem 1 guarantees the existence of such
extreme values because f is continuous on [a, b].
3. No absolute minimum. An absolute maximum at x 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 a and an absolute maximum at x c. Note that y 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 c and an absolute maximum at x a. Note that y 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 (1, 0), local maximum at (1, 0).
8. Minima at (2, 0) and (2, 0), maximum at (0, 2).
9. Maximum at (0, 5). Note that there is no minimum since the endpoint (2, 0) is excluded from the graph.
10. Local maximum at (3, 0), local minimum at (2, 0), maximum at (1, 2), minimum at (0, 1).
11. Graph (c), since this is 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 function 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 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 2018 Pearson Education, Inc.
185
186
Chapter 4 Applications of Derivatives
16. f has an absolute max at x 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 2 but does not have an
absolute min. Since the function is not continuous at
x 1, we do not meet the conditions of Theorem 1.
18. f has an absolute max at x 4 but does not have an
absolute min. Since the function is not continuous at
x 0, we do not meet the conditions of Theorem 1.
19. f has an absolute max at x 2 and an absolute min at
x 32 . Since the interval on which f is defined,
0 x 2 , is an open interval we do not meet the
conditions of Theorem 1.
20. f has an absolute max at x 0 and an absolute min
at x 2 and x 1 but does not have an absolute
y
(0, 1)
maximum. Since f is defined on a union of halfopen intervals, we do not meet the conditions of
Theorem 1.
y f ( x)
1
0
21. f ( x) 23 x 5 f ( x) 23 no critical points;
f (2) 19
, f (3) 3 the absolute maximum
3
is 3 at x 3 and the absolute minimum is 19
3
at x 2
Copyright 2018 Pearson Education, Inc.
2
x
Section 4.1 Extreme Values of Functions
22. f ( x) x 4 f ( x) 1 no critical points;
f ( 4) 0, f (1) 5 the absolute maximum is 0
at x 4 and the absolute minimum is 5 at x 1
23. f ( x ) x 2 1 f ( x) 2 x a critical point at
x 0; f (1) 0, f (0) 1, f (2) 3 the absolute
maximum is 3 at x 2 and the absolute minimum is
1 at x 0
24.
f ( x ) 4 x 3 f ( x ) 3x 2 a critical point at
x 0; f ( 2) 12, f (0) 4, f (1) 3 the absolute
maximum is 12 at x 2 and the absolute
minimum is 3 at x 1
y
(2, 12)
10
5
f ( x ) 4 x3
(1, 3)
2
25. F ( x)
1
x2
1
0
x 2 F ( x) 2 x 3 23 , however
x
x 0 is not a critical point since 0 is not in the domain;
F (0.5) 4, F (2) 0.25 the absolute maximum
is 0.25 at x 2 and the absolute minimum is 4 at
x 0.5
26. F ( x) 1x x 1 F ( x) x 2
1
x2
, however
x 0 is not a critical point since 0 is not in the
domain; F (2) 12 , F (1) 1 the absolute
maximum is 1 at x 1 and the absolute minimum
is 12 at x 2
Copyright 2018 Pearson Education, Inc.
1
x
187
188
Chapter 4 Applications of Derivatives
27. h( x) 3 x x1/3 h( x) 13 x 2/3 a critical point
at x 0; h(1) 1, h(0) 0, h(8) 2 the
absolute maximum is 2 at x 8 and the absolute
minimum is 1 at x 1
28. h( x) 3 x 2/3 h( x) 2 x 1/3 a critical point at
x 0; h(1) 3, h(0) 0, h(1) 3 the absolute
maximum is 0 at x 0 and the absolute minimum is
3 at x 1 and x 1
29. g ( x) 4 x 2 (4 x 2 )1/2
g ( x) 12 (4 x 2 )1/2 ( 2 x)
x
4 x 2
critical
points at x 2 and x 0, but not at x 2 because 2
is not in the domain;
g (2) 0, g (0) 2, g (1) 3 the absolute
maximum is 2 at x 0 and the absolute minimum is
0 at x 2
30.
g ( x) 5 x 2 (5 x 2 )1/2
g ( x) 12 (5 x 2 )1/2 (2 x)
x
5 x 2
critical points at x 5 and x 0, but not at
x 5 because 5 is not in the domain;
f 5 0, f (0) 5
the absolute maximum is 0 at x 5 and the
absolute minimum is 5 at x 0
31. f ( ) sin f ( ) cos 2 is a critical
point, but 2π is not a critical point because 2 is
not interior to the domain; f
2 1, f 2 1,
f 56 12 the absolute maximum is 1 at 2
and the absolute minimum is 1 at 2
32. f ( ) tan f ( ) sec2 f has no critical
points in 3 , 4 . The extreme values therefore
3
3 and f 4 1
the absolute maximum is 1 at 4 and
the absolute minimum is 3 at 3
occur at the endpoints: f
Copyright 2018 Pearson Education, Inc.
Section 4.1 Extreme Values of Functions
189
33. g ( x) csc x g ( x) (csc x)(cot x) a critical
point at x 2 ; g 3 2 , g π2 1, g 23 2
3
the absolute maximum is
2
3
3
at x and x 2 ,
3
3
and the absolute minimum is 1 at x 2
34. g ( x) sec x g ( x) (sec x)(tan x) a critical
point at x 0; g 3 2, g (0) 1, g 6 2 the
3
absolute maximum is 2 at x and the absolute
minimum is 1 at x 0
35.
3
f (t ) 2 | t | 2 t 2 2 (t 2 )1/2
f (t ) 12 (t 2 ) 1/2 (2t )
t
t2
|tt| a critical
point at t 0; f (1) 1, f (0) 2, f (3) 1 the
absolute maximum is 2 at t 0 and the absolute
minimum is 1 at t 3
36. f (t ) | t 5| (t 5) 2 ((t 5) 2 )1/2
f (t ) 12 ((t 5)2 ) 1/2 (2(t 5))
t 5
(t 5)2
5 a critical point at t 5; f (4) 1, f (5) 0,
| tt 5|
f (7) 2 the absolute maximum is 2 at t 7 and
the absolute minimum is 0 at t 5
37. f ( x) x 4/3 f ( x) 43 x1/3 a critical point at x 0; f (1) 1, f (0) 0, f (8) 16 the absolute
maximum is 16 at x 8 and the absolute minimum is 0 at x 0
38. f ( x) x5/3 f ( x) 53 x 2/3 a critical point at x 0; f (1) 1, f (0) 0, f (8) 32 the absolute
maximum is 32 at x 8 and the absolute minimum is 1 at x 1
39. g ( ) 3/5 g ( ) 53 2/5 a critical point at 0; g (32) 8, g (0) 0, g (1) 1 the absolute
maximum is 1 at 1 and the absolute minimum is 8 at 32
40. h( ) 3 2/3 h( ) 2 1/3 a critical point at 0; h(27) 27, h(0) 0, h(8) 12 the absolute
maximum is 27 at 27 and the absolute minimum is 0 at 0
41. y x 2 6 x 7 y 2 x 6 2 x 6 0 x 3. The critical point is x 3.
42. f ( x) 6 x 2 x3 f ( x) 12 x 3 x 2 12 x 3 x 2 0 3 x(4 x) 0 x 0 or x 4. The critical points are
x 0 and x 4.
Copyright 2018 Pearson Education, Inc.
190
Chapter 4 Applications of Derivatives
43. f ( x) x(4 x)3 f ( x) x[3(4 x) 2 (1)] (4 x)3 (4 x)2 [3 x (4 x )] (4 x)2 (4 4 x)
4(4 x) 2 (1 x) 4(4 x)2 (1 x) 0 x 1 or x 4. The critical points are x 1 and x 4.
44. g ( x) ( x 1)2 ( x 3)2 g ( x ) ( x 1)2 2( x 3)(1) 2( x 1)(1) ( x 3)2 2( x 3) ( x 1)[( x 1) ( x 3)]
4( x 3)( x 1) ( x 2) 4( x 3)( x 1)( x 2) 0 x 3 or x 1 or x 2. The critical points are x 1, x 2,
and x 3.
45. y x 2 2x y 2 x
2
x2
2 x3 2
x2
3
3
2 x 2 2 0 2 x3 2 0 x 1; 2 x 2 2 undefined x 2 0 x 0.
x
x
The domain of the function is (, 0) (0, ), thus x 0 is not the domain, so the only critical point is x 1.
46. f ( x )
x2
x2
2
f ( x)
( x 2)2 x x 2 (1)
( x 2)2
x2 4 x
( x 2)2
2
2
x 4 x2 0 x 2 4 x 0 x 0 or x 4; x 4 x2 undefined
( x 2)
( x 2)
( x 2) 0 x 2. The domain of the function is (, 2) (2, ), thus x 2 is not the domain, so the only
critical points are x 0 and x 4
47. y x 2 32 x y 2 x
16
x
2 x3/2 16
x
3/2
3/2
2 x 16 0 2 x3/2 16 0 x 4; 2 x 16 undefined
x
x
x 0 x 0. The critical points are x 4 and x 0.
48. g ( x) 2 x x 2 g ( x)
1 x
2 x x2
1 x
2 x x2
0 1 x 0 x 1;
1 x
2 x x2
undefined 2 x x 2 0
2 x x 2 0 x 0 or x 2. The critical points are x 0, x 1, and x 2.
49.
y x3 3 x 2 24 x 7 y 3 x 2 6 x 24 3 x 2 x 4 0 x 2 or x 4. The critical points are
x 2 and x 4.
50.
y x 3 x 2 3 y 1 3 23 x 1 3
x1 3 2
x1 3
0 x1 3 2 0 x 8;
The critical points are x 0 and x 8.
51. y x 2/3 (1) 23 x 1/3 ( x 2)
5x 4
33 x
crit. pt.
derivative
extremum
x 54
0
local max
12 101/3
25
undefined
local min
0
x0
x1 3 2
x1 3
value
1.034
Copyright 2018 Pearson Education, Inc.
undefined x1 3 0 x 0.
Section 4.1 Extreme Values of Functions
2
52. y x 2/3 (2 x) 23 x 1/3 ( x 2 4) 8 x3 8
3 x
crit. pt.
derivative
extremum
value
x 1
0
minimum
3
x0
undefined
local max
0
x 1
0
minimum
3
53. y x
1
2 4 x
(2 x ) (1) 4 x 2
x 2 (4 x 2 )
4 x
2
crit. pt.
derivative
extremum
value
x 2
undefined
local max
0
x 2
0
minimum
x 2
0
maximum
2
2
undefined
local min
0
x2
54. y x 2
2
1
2 3 x
5 x 2 12 x
2 3 x
(1) 2 x 3 x
4 2 x 2
4 x 2
x 2 (4 x )(3 x )
2 3 x
crit. pt.
derivative
extremum
value
x0
0
minimum
0
x 12
5
0
local max
144 151/2
125
x3
undefined
minimum
0
4.462
2, x 1
55. y
1, x 1
crit. pt.
derivative
extremum
value
x 1
undefined
minimum
2
1, x 0
56. y
2 2 x, x 0
crit. pt.
derivative
extremum
value
x0
undefined
local min
3
x 1
0
local mix
4
Copyright 2018 Pearson Education, Inc.
191
192
Chapter 4 Applications of Derivatives
2 x 2, x 1
57. y
2 x 6, x 1
crit. pt.
derivative
extremum
value
x 1
0
maximum
5
x 1
undefined
local min
1
x3
0
maximum
5
1 x 2 1 x 15 , x 1
2
4
58. We begin by determining whether f ( x) is defined at x 1, where f ( x) 4
x3 6 x 2 8 x,
x 1
Clearly, f ( x) 12 x 12 if x 1, and lim f (1 h) 1. Also, f ( x) 3 x 2 12 x 8 if x 1, and
h 0
lim f (1 h) 1. Since f is continuous at x 1, we have that f (1) 1.
h 0
12 x 12 , x 1
Thus, f ( x)
3 x 2 12 x 8, x 1
Note that 12 x 12 0 when x 1, and 3 x 2 12 x 8 0 when x
12 122 4(3)(8)
2(3)
126 48 2 2 3 3 .
But 2 2 3 3 0.845 1, so the critical points occur at x 1 and x 2 2 3 3 3.155.
crit. pt.
derivative
extremum
value
x 1
0
local max
4
x 3.155
0
local min
3.079
59. (a) No, since f ( x) 23 ( x 2) 1/3 , which is undefined at x 2.
(b) The derivative is defined and nonzero for all x 2. Also, f (2) 0 and f ( x ) 0 for all x 2.
(c) No, f ( x) need not have a global maximum because its domain is all real numbers. Any restriction of f to a
closed interval of the form [a, b] 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.
x3 9 x, x 3 or 0 x 3
3 x3 9, x 3 or 0 x 3
. Therefore, f ( x)
.
60. Note that f ( x)
3
3
x 9 x, 3 x 0 or x 3
3 x 9, 3 x 0 or x 3
(a) No, since the left- and right-hand derivatives at x 0, are 9 and 9, respectively.
(b) No, since the left- and right-hand derivatives at x 3, are 18 and 18, respectively.
(c) No, since the left- and right-hand derivatives at x 3, are 18 and 18, respectively.
Copyright 2018 Pearson Education, Inc.
Section 4.1 Extreme Values of Functions
193
(d) The critical points occur when f ( x) 0 (at x 3) and when f ( x) is undefined (at x 0 and x 3).
The minimum value is 0 at x 3, at x 0, and at x 3; local maxima occur at 3, 6 3 and 3, 6 3 .
61.
y x11 x3 x 5 y 11x10 3 x 2 1 0 for all x y is an increasing function. Thus y has no
extrema.
62.
y 3x tan x y 3 sec2 x 0 for all x y is an increasing function. Thus y has no extrema.
63. Yes, since f ( x) | x | x 2 ( x 2 )1/2 f ( x) 12 ( x 2 ) 1/2 (2 x)
x
( x 2 )1/ 2
| xx | is not defined at x 0. Thus it is
not required that f be zero at a local extreme point since f may be undefined there.
64. If f (c) is a local maximum value of f, then f ( x) f (c) for all x in some open interval (a, b) containing c. Since
f is even, f ( x) f ( x) f (c) f (c) for all x in the open interval (b, a ) containing c. That is, f assumes
a 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.
65. If g (c) is a local minimum value of g, then g ( x) g (c) for all x in some open interval (a, b) containing c.
Since g is odd, g ( x) g ( x) g (c) g (c) for all x in the open interval (b, 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.
66. 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) x for x . (Any other linear function f ( x) mx b with
m 0 will do as well.)
67. (a) V ( x) 160 x 52 x 2 4 x3
V ( x) 160 104 x 12 x 2 4( x 2)(3 x 20)
The only critical point in the interval (0, 5) is at x 2. The maximum value of V ( x) is 144 at x 2.
(b) The largest possible volume of the box is 144 cubic units, and it occurs when x 2 units.
68. (a) f ( x) 3ax 2 2bx c is a quadratic, so it can have 0, 1, or 2 zeros, which would be the critical points of f.
The function f ( x) x3 3 x has two critical points at x 1 and x 1. The function f ( x) x3 1 has one
critical point at x 0. The function f ( x) x3 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.)
69. s 12 gt 2 v0 t s0
Thus s
70.
g
v0
g
1
2
v0 2
g
ds
dt
gt v0 0 t
v0
s
v0
g
0
v0
. Now
g
gt
s (t ) s0 t 2 v0 0 t 0 or t
v2
20g s0 s0 is the maximum height over the interval 0 t
2v0
.
g
2v0
.
g
2sin t 2 cos t , solving dI
0 tan t 1 t 4 n where n is a nonnegative integer (in this exercise
dt
t is never negative) the peak current is 2 2 amps.
dI
dt
Copyright 2018 Pearson Education, Inc.
194
Chapter 4 Applications of Derivatives
71. Maximum value is 11 at x 5; minimum value is 5
on the interval [3, 2]; local maximum at (5, 9)
72. Maximum value is 4 on the interval [5, 7];
minimum value is 4 on the interval [2, 1].
73. Maximum value is 5 on the interval [3, );
minimum value is 5 on the interval (, 2].
74. Minimum value is 4 on the interval [1, 3]
75–80.
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 #75(a)" );
Df : D(f );
plot( Df(x), domain, color black, title "Section 4.1 #75(b)" )
StatPt : fsolve( Df(x) 0, domain )
SingPt : NULL;
EndPt : op(rhs(domain));
Copyright 2018 Pearson Education, Inc.
Section 4.2 The Mean Value Theorem
195
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 x1.86081 (singular point).
Mathematica: (functions may vary):
<<Miscellaneous `RealOnly`
Clear[f,x]
a 1; b 10/3;
f[x_ ] 2 2x 3 x 2/3
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
1.
THE MEAN VALUE THEOREM
When f ( x) x 2 2 x 1 for 0 x 1, then
2. When f ( x) x 2/3 for 0 x 1, then
f (1) f (0)
10
f (1) f (0)
1 0
8 .
f (c) 1 23 c 1/3 c 27
3. When f ( x) x 1x for 12 x 2, then
f (2) f (1/2)
2 1/2
4. When f ( x) x 1 for 1 x 3, then
f (3) f (1)
31
5. When f ( x ) x3 x 2 for 1 x 2, then
1 7
3
f (c) 3 2c 2 c 12 .
f (c) 0 1
f (c) 22
f (2) f ( 1)
2 ( 1)
1
c2
1
2 c 1
c 1.
c 32 .
f (c) 2 3c 2 2c c 13 7 .
1.22 and 13 7 0.549 are both in the interval 1 x 2.
x3 2 x 0
g (2) g ( 2)
, then 2( 2) g (c) 3 g (c). If 2 x 0, then g ( x) 3 x 2 3 g (c)
6. When g ( x)
2
x 0 x 2
2
3c 3 c 1. Only c 1 is in the interval. If 0 x 2, then g ( x) 2 x 3 g (c) 2c 3 c 32 .
7. Does not; f ( x) is not differentiable at x 0 in (1, 8).
8. Does; f ( x) is continuous for every point of [0, 1] and differentiable for every point in (0, 1).
9. Does; f ( x) is continuous for every point of [0, 1] and differentiable for every point in (0, 1).
10. Does not; f ( x) is not continuous at x 0 because lim f ( x) 1 0 f (0).
x 0
11. Does not; f is not differentiable at x 1 in (2, 0).
Copyright 2018 Pearson Education, Inc.
196
Chapter 4 Applications of Derivatives
12. Does; f ( x) is continuous for every point of [0, 3] and differentiable for every point in (0, 3).
13. Since f ( x) is not continuous on 0 x 1, Rolle’s Theorem does not apply: lim f ( x) lim x 1 0 f (1).
x 1
x 1
14. Since f ( x) must be continuous at x 0 and x 1 we have lim f ( x) a f (0) a 3 and
x 0
lim f ( x) lim f ( x) 1 3 a m b 5 m b. Since f ( x) must also be differentiable at x 1
x 1
x 1
we have lim f ( x) lim f ( x ) 2 x 3| x 1 m |x 1 1 m. Therefore, a 3, m 1 and b 4.
x 1
x 1
15. (a)
(b) Let r1 and r2 be zeros of the polynomial P( x) x n an 1 x n 1 a1 x a0 , then P(r1 ) P (r2 ) 0.
Since polynomials are everywhere continuous and differentiable, by Rolle’s Theorem P(r ) 0 for some r
between r1 and r2 , where P( x) nx n 1 (n 1)an 1 x n 2 a1.
16. With f both differentiable and continuous on [a, b] and f (r1 ) f (r2 ) f (r3 ) 0 where r1 , r2 and r3 are in [a, b],
then by Rolle’s Theorem there exists a c1 between r1 and r2 such that f (c1 ) 0 and a c2 between r2 and r3 such
that f (c2 ) 0. Since f is both differentiable and continuous on [a, b], Rolle’s Theorem again applies and we
have a c3 between c1 and c2 such that f (c3 ) 0. To generalize, if f has n 1 zeros in [a, b] and f ( n) is continuous
on [a, b], then f ( n) has at least one zero between a and b.
17. Since f exists throughout [a, b] the derivative function f is continuous there. If f has more than one zero
in [a, b], say f (r1 ) f (r2 ) 0 for r1 r2 , then by Rolle’s Theorem there is a c between r1 and r2 such that
f (c) 0, contrary to f 0 throughout [a, b]. Therefore f has at most one zero in [a, b]. The same argument
holds if f 0 throughout [a, b].
18. If f ( x ) is a cubic polynomial with four or more zeros, then by Rolle’s Theorem f ( x) has three or more zeros,
f ( x) has 2 or more zeros and f ( x) has at least one zero. This is a contradiction since f ( x) is a non-zero
constant when f ( x) is a cubic polynomial.
19. With f (2) 11 0 and f (1) 1 0 we conclude from the Intermediate Value Theorem that
f ( x) x 4 3 x 1 has at least one zero between 2 and 1. Then 2 x 1 8 x3 1 32 4 x3 4
29 4 x3 3 1 f ( x) 0 for 2 x 1 f ( x) is decreasing on [2, 1] f ( x) 0 has exactly one
solution in the interval (2, 1).
20. f ( x) x3
4
x2
7 f ( x) 3x 2
8
x3
0 on (, 0) f ( x ) is increasing on (, 0). Also, f ( x) 0 if x 2
and f ( x) 0 if 2 x 0 f ( x) has exactly one zero in (, 0).
21. g (t ) t t 1 4 g (t )
1
2 t
1
2 t 1
0 g (t ) is increasing for t in (0, ); g (3) 3 2 0 and
g (15) 15 0 g (t ) has exactly one zero in (0, ).
22. g (t ) 11 t 1 t 3.1 g (t )
1
(1t )2
1
2 1t
0 g (t ) is increasing for t in (1, 1); g (0.99) 2.5 and
g (0.99) 98.3 g (t ) has exactly one zero in (1, 1).
Copyright 2018 Pearson Education, Inc.
Section 4.2 The Mean Value Theorem
197
23. r ( ) sin 2 3 8 r ( ) 1 23 sin 3 cos 3 1 13 sin 23 0 on (, ) r ( ) is increasing on
(, ); r (0) 8 and r (8) sin 2
8
3
0 r ( ) has exactly one zero in (, ).
24. r ( ) 2 cos 2 2 r ( ) 2 2sin cos 2 sin 2 0 on (, ) r ( ) is increasing on
(, ); r (2 ) 4 cos(2 ) 2 4 1 2 0 and r (2 ) 4 1 2 0 r ( ) has exactly one
zero in (, ).
25. r ( ) sec
1
3
5 r ( ) (sec )(tan )
3
4
2
0 on 0, 2 r ( ) is increasing on 0, 2 ; r (0.1) 994
and r (1.57) 1260.5 r ( ) has exactly one zero in 0, .
2
26. r ( ) tan cot r ( ) sec 2 csc2 1 sec2 cot 2 0 on 0, 2 r ( ) is increasing on
2 4
0, ; r 0 and r (1.57) 1254.2 r ( ) has exactly one zero in 0, .
4
27. By Corollary 1, f ( x) 0 for all x f ( x) C , where C is a constant. Since f (1) 3 we have
C 3 f ( x) 3 for all x.
28. g ( x) 2 x 5 g ( x) 2 f ( x) for all x. By Corollary 2, f ( x) g ( x) C for some constant C. Then
f (0) g (0) C 5 5 C C 0 f ( x) g ( x) 2 x 5 for all x.
29. g ( x) x 2 g ( x) 2 x f ( x) for all x. By Corollary 2, f ( x) g ( x) C.
(a) f (0) 0 0 g (0) C 0 C C 0 f ( x) x 2 f (2) 4
(b) f (1) 0 0 g (1) C 1 C C 1 f ( x) x 2 1 f (2) 3
(c) f (2) 3 3 g (2) C 3 4 C C 1 f ( x) x 2 1 f (2) 3
30. g ( x) mx g ( x) m, a constant. If f ( x) m, then by Corollary 2, f ( x) g ( x ) b mx b where b is a
constant. Therefore all functions whose derivatives are constant can be graphed as straight lines y mx b.
31. (a) y
x2
2
C
(b) y
x3
3
C
(c) y
x4
4
C
32. (a) y x 2 C
(b) y x 2 x C
(c) y x3 x 2 x C
33. (a) y x 2 y 1x C
(b) y x 1x C
(c) y 5 x 1x C
34. (a) y 12 x 1/2 y x1/2 C y x C
(b) y 2 x C
35. (a) y 12 cos 2t C
(b) y 2sin 2t C
2
(c) y 2 x 2 x C
(c) y 12 cos 2t 2 sin 2t C
36. (a) y tan C
(b) y 1/2 y 23 3/2 C
(c)
37. f ( x) x 2 x C ; 0 f (0) 02 0 C C 0 f ( x) x 2 x
38. g ( x) 1x x 2 C ; 1 g (1) 11 (1)2 C C 1 g ( x) 1x x 2 1
Copyright 2018 Pearson Education, Inc.
y 23 3/2 tan C
198
Chapter 4 Applications of Derivatives
39. r ( ) 8 cot C ; 0 r 4 8 4 cot 4 C 0 2 1 C C 2 1
r ( ) 8 cot 2 1
40. r (t ) sec t t C ; 0 r (0) sec(0) 0 C C 1 r (t ) sec t t 1
9.8t 5 s 4.9t 2 5t C ; at s 10 and t 0 we have C 10 s 4.9t 2 5t 10
41. v ds
dt
32t 2 s 16t 2 2t C ; at s 4 and t
42. v ds
dt
1
2
we have C 1 s 16t 2 2t 1
1cos( t )
43. v
ds
dt
sin( t ) s 1 cos( t ) C ; at s 0 and t 0 we have C 1 s
44. v
ds
dt
2 cos 2t s sin 2t C ; at s 1 and t 2 we have C 1 s sin 2t 1
dv
32 v 32t C; at v = 20 and t = 0 we have C 20 v 32t 20
dt
ds
v
32t 20 s 16t 2 20t C; at s = 5 and t = 0 we have C 5 s 16t 2 20t 5
dt
45. a
46. a 9.8 v 9.8t C1; at v 3 and t 0 we have C1 3 v 9.8t 3 s 4.9t 2 3t C2 ; at s 0 and
t 0 we have C2 0 s 4.9t 2 3t
47. a 4sin(2t ) v 2 cos(2t ) C1; at v 2 and t 0 we have C1 0 v 2 cos(2t ) s sin(2t ) C2 ; at
s 3 and t 0 we have C2 3 s sin(2t ) 3
9
cos 3t v 3 sin 3t C1; at v 0 and t 0 we have C1 0 v 3 sin 3t s cos 3t C2 ; at
s 1 and t 0 we have C2 0 s cos 3t
48. a
2
49. If T (t ) is the temperature of the thermometer at time t, then T (0) 19 C and T (14) 100 C. From the Mean
T (14) T (0)
Value Theorem there exists a 0 t0 14 such that 140 8.5 C / sec T (t0 ), the rate at which the
temperature was changing at t t0 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 t 2 is
The Mean Value Theorem says that for some 0 t0 2, d ( t0 )
automobile at time t0 (which is read on the speedometer).
d (2) d (0)
. The
20
d (2) d (0)
.
20
value d ( t0 ) is the speed of the
54. a (t ) v (t ) 1.6 v(t ) 1.6t C ; at (0, 0) we have C 0 (t ) 1.6t. When t 30, then v(30) 48 m/sec.
Copyright 2018 Pearson Education, Inc.
Section 4.2 The Mean Value Theorem
55. The conclusion of the Mean Value Theorem yields
11
b a
ba
2
1
c2
c2
2
aabb a b c
56. The conclusion of the Mean Value Theorem yields bb aa 2c c
199
ab .
a b .
2
57. f ( x) [cos x sin( x 2) sin x cos( x 2)] 2sin( x 1) cos( x 1) sin( x x 2) sin 2( x 1)
sin(2 x 2) sin (2 x 2) 0. Therefore, the function has the constant value f (0) sin 2 1 0.7081
which explains why the graph is a horizontal line.
58. (a) f ( x) ( x 2)( x 1) x( x 1)( x 2) x5 5 x3 4x is one possibility.
(b) Graphing f ( x) x5 5 x3 4 x and f ( x) 5 x 4 15 x 2 4 on [3, 3] by [7, 7] we see that each
x-intercept of f ( x) lies between a pair of x-intercepts of f ( x), as expected by Rolle’s Theorem.
(c) Yes, since sin is continuous and differentiable on (, ).
59. f ( x) must be zero at least once between a and b by the Intermediate Value Theorem. Now suppose that f ( x) is
zero twice between a and b. Then by the Mean Value Theorem, f ( x) would have to be zero at least once
between the two zeros of f ( x), but this can’t be true since we are given that f ( x) 0 on this interval.
Therefore, f ( x) is zero once and only once between a and b.
60. Consider the function k ( x) f ( x) g ( x). k ( x) is
continuous and differentiable on [a, b], and since
k (a ) f (a ) g (a ) and k (b) f (b) g (b), by the
Mean Value Theorem, there must be a point c in
(a, b) where k (c) 0. But since k (c) f (c) g (c),
this means that f (c) g (c), 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 ( x) 1 for 1 x 4 f ( x) is differentiable on 1 x 4 f is continuous on 1 x 4 f satisfies the
f (4) f (1)
f (4) f (1)
f (c) for some c in 1 x 4 f (c) 1
1
conditions of the Mean Value Theorem
4 1
3
f (4) f (1) 3
Copyright 2018 Pearson Education, Inc.
200
Chapter 4 Applications of Derivatives
62. 0 f ( x)
1
2
for all x f ( x) exists for all x, thus f is differentiable on (1, 1) f is continuous on [1, 1]
f satisfies the conditions of the Mean Value Theorem
f (1) f ( 1)
f (1) f ( 1)
1 ( 1)
f (c) for some c in [1, 1]
0
12 0 f (1) f (1) 1. Since f (1) f (1) 1 f (1) 1 f (1) 2 f (1), and
2
since 0 f (1) f (1) we have f (1) f (1). Together we have f (1) f (1) 2 f (1).
63. Let f (t ) cos t and consider the interval [0, x] where x is a real number. f is continuous on [0, x] and f is
differentiable on (0, x) since f (t ) sin t f satisfies the conditions of the Mean Value Theorem
f ( x ) f (0)
x (0) f (c) for some c in [0, x] cosxx 1 sin c. Since 1 sin c 1 1 sin c 1
1 cosxx 1 1. If x 0, 1 cosxx 1 1 x cos x 1 x |cos x 1| x | x | . If x 0, 1 cosxx 1 1
x cos x 1 x x cos x 1 x ( x) cos x 1 x |cos x 1| x | x | . Thus, in both cases,
we have |cos x 1| | x | . If x 0, then |cos 0 1| |1 1| |0| |0|, thus |cos x 1| | x | is true for all x.
64. Let f ( x) sin x for a x b. From the Mean Value Theorem there exists a c between a and b such that
sin b sin a
sin b sin a
sin b sin a
cos c 1 b a 1 b a 1 |sin b sin a | | b a | .
ba
65. Yes. By Corollary 2 we have f ( x) g ( x) c since f ( x) g ( x). If the graphs start at the same point x a,
then f (a ) g (a ) c 0 f ( x) g ( x).
66. Assume f is differentiable and | f ( w) f ( x)| | w x | for all values of w and x. Since f is differentiable,
f ( w) f ( x )
f ( x) exists and f ( x) lim
using the alternative formula for the derivative. Let g ( x) x ,
w x
w x
f ( w) f ( x )
w x
w x
| f ( w) f ( x )|
1 as long
|w x|
which is continuous for all x. By Theorem 10 from Chapter 2, | f ( x)| lim
| f ( w) f ( x )|
.
|w x|
w x
lim
Since f ( w) f ( x) w x for allw and x
from Chapter 2, f ( x) lim
w x
| f ( w) f ( x )|
|w x|
lim
w x
f ( w) f ( x )
w x
as w x. By Theorem 5
lim 1 1 f ( x) 1 1 f ( x) 1.
w x
f (b ) f ( a )
f (c) for some point c between a and b. Since b a 0 and
67. By the Mean Value Theorem we have b a
f (b) f (a), we have f (b) f (a) 0 f (c) 0.
68. The condition is that f should be continuous over [a, b]. The Mean Value Theorem then guarantees the
f (b ) f ( a )
f (c). If f is continuous, then it has a minimum and
existence of a point c in (a, b) such that b a
maximum value on [a, b], and min f f (c) max f , as required.
69.
f ( x) (1 x 4 cos x) 1 f ( x) (1 x 4 cos x)2 (4 x3 cos x x 4 sin x)
x3 (1 x 4 cos x) 2 (4 cos x x sin x) 0 for 0 x 0.1 f ( x) is decreasing when 0 x 0.1
min f 0.9999 and max f 1. Now we have 0.9999
f (0.1) 1
0.1
1 0.09999 f (0.1) 1 0.1
1.09999 f (0.1) 1.1.
70. f ( x) (1 x 4 ) 1 f ( x) (1 x 4 ) 2 (4 x3 )
4 x3
(1 x 4 )3
0 for 0 x 0.1 f ( x) is increasing when
0 x 0.1 min f 1 and max f 1.0001. Now we have 1
0.1 f (0.1) 2 0.10001 2.1 f (0.1) 2.10001.
f (0.1) 2
0.1
1.0001
Copyright 2018 Pearson Education, Inc.
Section 4.3 Monotonic Functions and the First Derivative Test
71. (a) Suppose x 1, then by the Mean Value Theorem
(b)
f ( x ) f (1)
x 1
201
0 f ( x ) f (1). Suppose x 1, then by the
f ( x ) f (1)
Mean Value Theorem
0 f ( x) f (1). Therefore f ( x) 1 for all x since f (1) 1.
x 1
f ( x ) f (1)
f ( x ) f (1)
Yes. From part (a), lim
0 and lim
0. Since f (1) exists, these two one-sided
x 1
x 1
x 1
x 1
limits
are equal and have the value f (1) f (1) 0 and f (1) 0 f (1) 0.
72. From the Mean Value Theorem we have
has only one solution c
4.3
q
2p .
f (b ) f ( a )
ba
f (c) where c is between a and b. But f (c ) 2 pc q 0
(Note: p 0 since f is a quadratic function.)
MONOTONIC FUNCTIONS AND THE FIRST DERIVATIVE TEST
1. (a) f ( x) x( x 1) critical points at 0 and 1
(b) f | | increasing on ( , 0) and (1, ), decreasing on (0, 1)
0
1
(c) Local maximum at x 0 and a local minimum at x 1
2. (a) f ( x) ( x 1)( x 2) critical points at 2 and 1
(b) f | | increasing on (, 2) and (1, ), decreasing on (2, 1)
2
1
(c) Local maximum at x 2 and a local minimum at x 1
3. (a) f ( x) ( x 1) 2 ( x 2) critical points at 2 and 1
(b) f | | increasing on (2, 1) and (1, ), decreasing on (, 2)
2
1
(c) No local maximum and a local minimum at x 2
4. (a) f ( x) ( x 1) 2 ( x 2)2 critical points at 2 and 1
(b) f | | increasing on (, 2) (2, 1) (1, ), never decreasing
2
(c) No local extrema
5. (a)
(b)
1
f ( x ) ( x 1)( x 2)( x 3) critical points at 2, 1, 3
f | | | increasing on ( 2, 1) and (3, ), decreasing on ( , 2) and (1, 3)
2
1
3
(c) Local maximum at x 1, local minima at x 2 and x 3
6. (a) f ( x) ( x 7)( x 1)( x 5) critical points at 5, 1 and 7
(b) f | | | increasing on (5, 1) and (7, ), decreasing on (, 5) and (1, 7)
5
1
7
(c) Local maximum at x 1, local minima at x 5 and x 7
7. (a) f ( x)
x 2 ( x 1)
( x 2)
critical points at x 0, x 1 and x 2
(b) f )( | | increasing on (, 2) and (1, ), decreasing on (2, 0) and (0, 1)
2
0
(c) Local minimum at x 1
8. (a) f ( x)
( x 2)( x 4)
( x 1)( x 3)
1
critical points at x 2, x 4, x 1, and x 3
(b) f | )( | )( increasing on (, 4), (1, 2) and (3, ), decreasing on
4
1
2
3
(4, 1) and (2, 3)
(c) Local maximum at x 4 and x 2
Copyright 2018 Pearson Education, Inc.
202
Chapter 4 Applications of Derivatives
9. (a) f ( x) 1
4
x2
2
x 2 4 critical points at x 2, x 2 and x 0.
x
(b) f | )( | increasing on (, 2) and (2, ), decreasing on (2, 0) and (0, 2)
2
2
0
(c) Local maximum at x 2, local minimum at x 2
10. (a) f ( x) 3
6
x
3 x 6 critical points at x 4 and x 0
x
(b) f ( | increasing on (4, ), decreasing on (0, 4)
4
0
(c) Local minimum at x 4
11. (a) f ( x) x 1/3 ( x 2) critical points at x 2 and x 0
(b) f | )( increasing on (, 2) and (0, ), decreasing on (2, 0)
2
0
(c) Local maximum at x 2, local minimum at x 0
12. (a) f ( x) x 1/2 ( x 3) critical points at x 0 and x 3
(b) f ( | increasing on (3, ), decreasing on (0, 3)
3
0
(c) No local maximum and a local minimum at x 3
13. (a) f ( x) (sin x 1)(2 cos x 1), 0 x 2 critical points at x 2 , x 23 , and x 43
(b) f [ | | | ] increasing on
2
0
2
3
2
3
, 2 and 4 , 2
3
4
3
2
23 , 43 , decreasing on 0, 2 ,
(c) Local maximum at x 43 and x 0, local minimum at x
2π
3
and x 2
14. (a) f ( x) (sin x cos x)(sin x cos x), 0 x 2 critical points at x 4 , x 34 , x 54 , and x 74
(b) f [ | | | | ] increasing on 4 , 34 and
0
3 , 5
4
4
4
3
4
and 74 , 2
5
4
7
4
2
54 , 74 , decreasing on 0, 4 ,
(c) Local maximum at x 0, x 34 and x 74 , local minimum at x 4 , x 54 and x 2
15. (a) Increasing on (2, 0) and (2, 4), decreasing on (4, 2) and (0, 2)
(b) Absolute maximum at (4, 2), local maximum at (0, 1) and (4, 1); Absolute minimum at (2, 3), local
minimum at (2, 0)
16. (a) Increasing on (4, 3.25), (1.5, 1), and (2, 4), decreasing on (3.25, 1.5) and (1, 2)
(b) Absolute maximum at (4, 2), local maximum at (3.25, 1) and (1, 1); Absolute minimum at (1.5, 1), local
minimum at (4, 0) and (2, 0)
17. (a) Increasing on (4, 1), (0.5, 2), and (2, 4), decreasing on (1, 0.5)
(b) Absolute maximum at (4, 3), local maximum at (1, 2) and (2, 1); No absolute minimum, local minimum
at (4, 1) and (0.5, 1)
18. (a) Increasing on (4, 2.5), (1, 1), and (3, 4), decreasing on (2.5, 1) and (1, 3)
(b) No absolute maximum, local maximum at (2.5, 1), (1, 2) and (4, 2); No absolute minimum, local
minimum at (1, 0) and (3, 1)
Copyright 2018 Pearson Education, Inc.
Section 4.3 Monotonic Functions and the First Derivative Test
203
19. (a) g (t ) t 2 3t 3 g (t ) 2t 3 a critical point at t 32 ; g | , increasing on
3/2
(b)
, 32 , decreasing on 32 ,
at t 32 , absolute maximum is 21
at t 32
local maximum value of g 32 21
4
4
20. (a) g (t ) 3t 2 9t 5 g (t ) 6t 9 a critical point at t 32 ; g | , increasing on ,
decreasing on
3/2
32 ,
3
2
,
at t 32 , absolute maximum is 47
at t 32
(b) local maximum value of g 32 47
4
4
21. (a) h( x) x3 2 x 2 h( x) 3 x 2 4 x x(4 3 x) critical points at x 0, 43 h | | ,
0
4/3
increasing on 0, 43 , decreasing on (, 0) and 43 ,
at x 43 ; local minimum value of h(0) 0 at x 0, no absolute
(b) local maximum value of h 43 32
27
extrema
22. (a) h( x) 2 x3 18 x h( x) 6 x 2 18 6 x 3 x 3 critical points at x 3
h | | , increasing on , 3 and
3
3
(b) a local maximum is h 3 12 3 at x 3; local minimum is h
extrema
3, , decreasing on 3, 3
3 12
3 at x 3, no absolute
23. (a) f ( ) 3 2 4 3 f ( ) 6 12 2 6 (1 2 ) critical points at 0, 12
12 , decreasing on (, 0) and 12 ,
f | | , increasing on 0,
0
1/2
(b) a local maximum is f
12 14 at 12 , a local minimum is f (0) 0 at 0, no absolute extrema
2 critical points at 2
f | | , increasing on 2, 2 , decreasing on , 2 and 2,
2
2
a local maximum is f 2 4 2 at 2, a local minimum is f 2 4 2 at 2, no
24. (a) f ( ) 6 3 f ( ) 6 3 2 3
(b)
2
absolute extrema
25. (a) f (r ) 3r 3 16r f (r ) 9r 2 16 no critical points f , increasing on (, ), never
decreasing
(b) no local extrema, no absolute extrema
26. (a) h(r ) (r 7)3 h(r ) 3(r 7) 2 a critical point at r 7 h | , increasing on
(, 7) (7, ), never decreasing
(b) no local extrema, no absolute extrema
7
27. (a) f ( x) x 4 8 x 2 16 f ( x) 4 x3 16 x 4 x( x 2)( x 2) critical points at x 0 and x 2
f | | | , increasing on (2, 0) and (2, ), decreasing on (, 2) and (0, 2)
2
0
2
(b) a local maximum is f (0) 16 at x 0, local minima are f (2) 0 at x 2, no absolute maximum;
absolute minimum is 0 at x 2
Copyright 2018 Pearson Education, Inc.
204
Chapter 4 Applications of Derivatives
28. (a) g ( x) x 4 4 x3 4 x 2 g ( x) 4 x3 12 x 2 8 x 4 x ( x 2)( x 1) critical points at x 0, 1, 2
g | | | , increasing on (0, 1) and (2, ), decreasing on (, 0) and (1, 2)
0
1
2
(b) a local maximum is g (1) 1 at x 1, local minima are g (0) 0 at x 0 and g (2) 0 at x 2, no absolute
maximum; absolute minimum is 0 at x 0, 2
29. (a) H (t ) 32 t 4 t 6 H (t ) 6t 3 6t 5 6t 3 (1 t )(1 t ) critical points at t 0, 1
H | | | , increasing on (, 1) and (0, 1), decreasing on (1, 0) and (1, )
1
0
1
(b) the local maxima are H (1) 12 at t 1 and H (1) 12 at t 1, the local minimum is H (0) 0 at t 0,
absolute maximum is 12 at t 1; no absolute minimum
30. (a) K (t ) 15t 3 t 5 K (t ) 45t 2 5t 4 5t 2 (3 t )(3 t ) critical points at t 0, 3
K | | | , increasing on (3, 0) (0, 3), decreasing on (, 3) and (3, )
3
0
3
(b) a local maximum is K (3) 162 at t 3, a local minimum is K (3) 162 at t 3, no absolute extrema
3
x 1
31. (a) f ( x) x 6 x 1 f ( x) 1
x 1 3
x 1
critical points at x 1 and x 10 f ( | ,
10
1
increasing on (10, ), decreasing on (1, 10)
(b) a local minimum is f (10) 8, a local and absolute maximum is f (1) 1, absolute minimum of 8 at x 10
32. (a) g ( x) 4 x x 2 3 g ( x)
2
x
2 2 x3/ 2
x
2x
critical points at x 1 and x 0 g ( | ,
increasing on (0, 1), decreasing on (1, )
(b) a local minimum is f (0) 3, a local maximum is f (1) 6, absolute maximum of 6 at x 1
33. (a) g ( x) x 8 x 2 x(8 x 2 )1/2 g ( x) (8 x 2 )1/2 x
critical points at x 2, 2 2 g
on 2 2, 2 and 2, 2 2
1
0
12 (8 x2 )1/2 (2 x) 22(22 xx)(22 2x) x
( | | ) , increasing on (2, 2), decreasing
2
2 2
2
2 2
(b) local maxima are g (2) 4 at x 2 and g 2 2 0 at x 2 2, local minima are g (2) 4 at
x 2 and g 2 2 0 at x 2 2, absolute maximum is 4 at x 2; absolute minimum is 4 at x 2
34. (a) g ( x) x 2 5 x x 2 (5 x)1/2 g ( x) 2 x(5 x )1/2 x 2
12 (5 x)1/2 (1) 52x(45xx) critical points
at x 0, 4 and 5 g | | ), increasing on (0, 4), decreasing on (, 0) and (4, 5)
0
4
5
(b) a local maximum is g (4) 16 at x 4, a local minimum is 0 at x 0 and x 5, no absolute maximum;
absolute minimum is 0 at x 0, 5
35. (a) f ( x)
x 2 3
x2
f ( x)
2 x ( x 2) ( x 2 3)(1)
( x 2)
2
( x 3)( x 1)
( x 2)2
critical points at x 1, 3
f | )( | , increasing on (, 1) and (3, ), decreasing on (1, 2) and (2, 3),
1
2
3
discontinuous at x 2
(b) a local maximum is f (1) 2 at x 1, a local minimum is f (3) 6 at x 3, no absolute extrema
36. (a) f ( x )
x3
3x2 1
f ( x)
3 x 2 (3 x 2 1) x3 (6 x )
2
(3 x 1)
2
3 x 2 ( x 2 1)
(3 x 2 1)2
a critical point at x 0 f | ,
increasing on (, 0) (0, ), and never decreasing
(b) no local extrema, no absolute extrema
Copyright 2018 Pearson Education, Inc.
0
Section 4.3 Monotonic Functions and the First Derivative Test
37. (a) f ( x) x1/3 ( x 8) x 4/3 8 x1/3 f ( x) 43 x1/3 83 x 2/3
4( x 2)
3 x 2/3
205
critical points at x 0, 2
f | )( , increasing on (2, 0) (0, ), decreasing on (, 2)
2
0
(b) no local maximum, a local minimum is f (2) 6 3 2 7.56 at x 2, no absolute maximum; absolute
minimum is 6 3 2 at x 2
5( x 2)
33 x
38. (a) g ( x) x 2/3 ( x 5) x5/3 5 x 2/3 g ( x) 53 x 2/3 10
x 1/3
3
5( x 2)
33 x
critical points at
x 2 and x 0 g | )( , increasing on (, 2) and (0, ), decreasing on (2, 0)
2
0
3
(b) local maximum is g (2) 3 4 4.762 at x 2, a local minimum is g (0) 0 at x 0, no absolute
extrema
39. (a) h( x) x1/3 ( x 2 4) x 7/3 4 x1/3 h( x) 73 x 4/3 43 x 2/3
x 0, 2 h
7
, 0 and 0,
2
7
2
7
(b) local maximum is h
absolute extrema
|
2/ 7
2
7
7 x2
3 x2
)( | , increasing on ,
0
24 3 2
77/6
2/ 7
3.12 at x
2
7
7 x 2
3
2
7
and
critical points at
2
7
, the local minimum is h 2
40. (a) k ( x) x 2/3 ( x 2 4) x8/3 4 x 2/3 k ( x) 83 x5/3 83 x 1/3
7
8( x 1)( x 1)
33 x
, , decreasing on
24 3 2
77/ 6
3.12, no
critical points at x 0, 1
k | )( | , increasing on (1, 0) and (1, ), decreasing on (, 1) and (0, 1)
1
0
1
(b) local maximum is k (0) 0 at x 0, local minima are k ( 1) 3 at x 1, no absolute maximum;
absolute minimum is 3 at x 1
41. (a) f ( x) 2 x x 2 f ( x) 2 2 x a critical point at x 1 f | ] and f (1) 1 and f (2) 0
a local maximum is 1 at x 1, a local minimum is 0 at x 2.
(b) There is an absolute maximum of 1 at x 1; no absolute minimum.
(c)
1
2
42. (a) f ( x) ( x 1) 2 f ( x) 2( x 1) a critical point at x 1 f | ] and
1
0
f (1) 0, f (0) 1 a local maximum is 1 at x 0, a local minimum is 0 at x 1
(b) no absolute maximum; absolute minimum is 0 at x 1
(c)
Copyright 2018 Pearson Education, Inc.
206
Chapter 4 Applications of Derivatives
43. (a) g ( x ) x 2 4 x 4 g ( x) 2 x 4 2( x 2) a critical point at x 2 g [ | and
1
2
g (1) 1, g (2) 0 a local maximum is 1 at x 1, a local minimum is g (2) 0 at x 2
(b) no absolute maximum; absolute minimum is 0 at x 2
(c)
44. (a) g ( x) x 2 6 x 9 g ( x) 2 x 6 2( x 3) a critical point at x 3 g [ | and
4
3
g (4) 1, g (3) 0 a local maximum is 0 at x 3, a local minimum is 1 at x 4
(b) absolute maximum is 0 at x 3; no absolute minimum
(c)
45. (a) f (t ) 12t t 3 f (t ) 12 3t 2 3(2 t )(2 t ) critical points at t 2 f [ | |
3
2
2
and f (3) 9, f (2) 16, f (2) 16 local maxima are 9 at t 3 and 16 at t 2, a local minimum
is 16 at t 2
(b) absolute maximum is 16 at t 2; no absolute minimum
(c)
46. (a) f (t ) t 3 3t 2 f (t ) 3t 2 6t 3t (t 2) critical points at t 0 and t 2 f | | ]
0
2
3
and f (0) 0, f (2) 4, f (3) 0 a local maximum is 0 at t 0 and t 3, a local minimum is 4 at t 2
(b) absolute maximum is 0 at t 0, 3; no absolute minimum
(c)
Copyright 2018 Pearson Education, Inc.
Section 4.3 Monotonic Functions and the First Derivative Test
47. (a) h( x)
x3
3
207
2 x 2 4 x h( x) x 2 4 x 4 ( x 2) 2 a critical point at x 2 h [ | and
h(0) 0 no local maximum, a local minimum is 0 at x 0
(b) no absolute maximum; absolute minimum is 0 at x 0
(c)
0
2
48. (a) k ( x) x3 3 x 2 3 x 1 k ( x) 3 x 2 6 x 3 3( x 1) 2 a critical point at x 1 k | ]
1
and k (1) 0, k (0) 1 a local maximum is 1 at x 0, no local minimum
(b) absolute maximum is 1 at x 0; no absolute minimum
(c)
49. (a) f ( x) 25 x 2 f ( x)
x
25 x 2
0
critical points at x 0, x 5, and x 5 f ( | ),
5
0
f (5) 0, f (0) 5, f (5) 0 local maximum is 5 at x 0; local minimum of 0 at x 5 and x 5
(b) absolute maximum is 5 at x 0; absolute minimum of 0 at x 5 and x 5
(c)
50. (a) f ( x) x 2 2 x 3,3 x f ( x)
2 x 2
x 2 2 x 3
only critical point in 3 x is at x 3
f [ , f (3) 0 local minimum of 0 at x 3, no local maximum
3
(b) absolute minimum of 0 at x 3, no absolute maximum
(c)
Copyright 2018 Pearson Education, Inc.
5
208
Chapter 4 Applications of Derivatives
51. (a) g ( x)
x 2 , 0
x 2 1
2
x 1 g ( x) x 2 4 x21 only critical point in 0 x 1 is x 2 3 0.268
( x 1)
g [ | ), g 2 3
0
0.268
1
3
4 3 6
1.866 local minimum of
3
4 3 6
at x 2 3, local
maximum at x 0.
(b) absolute minimum of
(c)
52. (a) g ( x)
x2
4 x 2
3
4 3 6
at x 2 3, no absolute maximum
, 2 x 1 g ( x)
8x
(4 x 2 ) 2
only critical point in 2 x 1 is x 0
g ( | ], g (0) 0 local minimum of 0 at x 0, local maximum of 13 at x 1.
2
0
1
(b) absolute minimum of 0 at x 0, no absolute maximum
(c)
53. (a) f ( x) sin 2 x, 0 x f ( x) 2 cos 2 x, f ( x) 0 cos 2 x 0 critical points are x 4 and x 34
f [ | | ] , f (0) 0, f 4 1, f
0
3
4
4
34 1, f ( ) 0 local maxima are 1 at x 4
and 0 at x , and local minima are 1 at x 34 and 0 at x 0.
(b) The graph of f rises when f 0, falls when f 0, and has local
extreme values where f 0. The function f has a local minimum
value at x 0 and x 34 , where the values
of f change from negative to positive. The function f has a local
maximum value at x and x 4 , where the values of f change
from positive to negative.
54. (a) f ( x ) sin x cos x, 0 x 2 f ( x) cos x sin x, f ( x) 0 tan x 1 critical points are x 34
and x 74 f [ | | ] , f (0) 1, f
0
3
4
7
4
2
34
2, f 74 2, f (2 ) 1 local
maxima are 2 at x 34 and 1 at x 2 , and local minima are 2 at x 74 and 1 at x 0.
Copyright 2018 Pearson Education, Inc.
Section 4.3 Monotonic Functions and the First Derivative Test
209
(b) The graph of f rises when f 0, falls when
f 0, and has local extreme values where
f 0. The function f has a local minimum
value at x 0 and x 74 , where the values
of f change from negative to positive. The
function f has a local maximum value at x 2
and x 34 , where the values of f change from
positive to negative.
55. (a) f ( x) 3 cos x sin x, 0 x 2 f ( x) 3 sin x cos x, f ( x) 0 tan x
are x 6 and x 76 f [ | | ] , f (0) 3, f 6 2, f
0
7
6
6
2
1
3
critical points
76 2, f (2 )
local maxima are 2 at x 6 and 3 at x 2 , and local minima are 2 at x 76 and 3 at x 0.
(b) The graph of f rises when f 0, falls when
f 0, and has local extreme values where
f 0. The function f has a local minimum
value at x 0 and x 76 , where the values
of f change from negative to positive. The
function f has a local maximum value at
x 2 and x 6 , where the values of f
change from positive to negative.
56. (a) f ( x) 2 x tan x, 2 x 2 f ( x) 2 sec 2 x, f ( x ) 0 sec2 x 2 critical points are
x 4 and x 4 f ( | | ) , f 4 2 1, f 4 1 2 local maximum
2
4
4
2
is 2 1 at x 4 , and local minimum is 1 2 at x 4 .
(b) The graph of f rises when f 0, falls when
f 0, and has local extreme values where
f 0. The function f has a local minimum
value at x 4 , where the values of f change
from negative to positive. The function f has a
local maximum value at x 4 , where the
values of f change from positive to negative.
2x f ( x) 12 cos 2x , f ( x) 0 cos 2x 12 a critical point at x 23
f [ | ] and f (0) 0, f 23 3 3, f (2 ) local maxima are
0
2 /3
2
57. (a) f ( x) 2x 2sin
0 at x 0 and at x 2 , a local minimum is 3 3 at x 23
(b) The graph of f rises when f 0, falls when
f 0, and has a local minimum value at the
point where f changes from negative to
positive.
Copyright 2018 Pearson Education, Inc.
3
210
Chapter 4 Applications of Derivatives
58. (a) f ( x ) 2 cos x cos 2 x f ( x) 2 sin x 2 cos x sin x 2(sin x)(1 cos x) critical points at
x , 0, f [ | ] and f ( ) 1, f (0) 3, f ( ) 1 a local maximum is
0
1 at x , a local minimum is 3 at x 0
(b) The graph of f rises when f 0, falls when
f 0, and has local extreme values where
f 0. The function f has a local minimum
value at x 0, where the values of f change
from negative to positive.
59. (a) f ( x) csc 2 x 2 cot x f ( x) 2(csc x)( csc x)(cot x) 2( csc2 x) 2(csc2 x) (cot x 1) a critical
point at x 4 f ( | ) and f 4 0 no local maximum, a local minimum is 0 at x 4
0
/4
(b) The graph of f rises when f 0, falls when
f 0, and has a local minimum value at the
point where f 0 and the values of f change
from negative to positive. The graph of f
steepens as f ( x) .
60. (a) f ( x) sec2 x 2 tan x f ( x) 2(sec x)(sec x)(tan x) 2sec2 x (2sec2 x) (tan x 1) a critical point
at x 4 f ( | ) and f 4 0 no local maximum, a local minimum is 0 at x 4
/2
/4
/2
(b) The graph of f rises when f 0, falls when
f 0, and has a local minimum value where
f 0 and the values of f change from
negative to positive.
61.
local minimum at x 1, no local maximum.
62.
local minimum at x 2, local maximum at x 0
Copyright 2018 Pearson Education, Inc.
Section 4.3 Monotonic Functions and the First Derivative Test
211
63. h( ) 3cos 2 h( ) 32 sin 2 h [ ] , (0, 3) and (2 , 3) a local maximum is 3 at 0,
64. h( ) 5sin 2 h( )
5
2
minimum is 0 at 0
65. (a)
(b)
2
0
a local minimum is 3 at 2
cos 2 h [ ], (0, 0) and ( , 5) a local maximum is 5 at , a local
0
(c)
66. (a)
(b)
(c)
(d)
67. (a)
(b)
68. (a)
(b)
Copyright 2018 Pearson Education, Inc.
(d)
212
Chapter 4 Applications of Derivatives
69. The function f ( x) x sin
1x has an infinite number of local maxima and minima on its domain, which is
( , 0) (0, ). The function sin x has the following properties: a) it is continuous on (, ); b) it is
periodic; and c) its range is [1, 1]. Also, for a 0, the function
1 , 0
a
0,
1 .
a
1
x
1
x
has a range of (, a ] [a, )
1
x
1 when x is in [ 1, 0) (0, 1]. This means sin 1x
takes on the values of 1 and 1 infinitely many times on [ 1, 0) (0, 1], namely at 1x 2 , 32 , 52 ,.
on
In particular, if a 1, then
1 or
x 2 , 32 , 52 , . Thus sin 1x has infinitely many local maxima and minima in [ 1, 0) (0, 1]. On the
interval (0, 1], 1 sin 1x 1 and since x 0 we have x x sin 1x x. On the interval
1 and since x 0 we have x x sin x. Thus f ( x) is bounded by the lines
y x and y x. Since sin 1x oscillates between 1 and 1 infinitely many times on [ 1, 0) (0, 1] then f will
[ 1, 0), 1 sin
1
x
1
x
oscillate between y x and y 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 sin 1x 1 and lim x sin 1x 1. The
x
x
graph of f does not have any absolute maxima, but it does have two absolute minima.
vertex is at x 2ba . Thus when a 0,
increasing on , 2ab and decreasing
a 0, f
f is
on
| ; for a 0, f
b /2 a
b 2 c a x b 2 b 2 4 ac , a parabola whose
4a
2a
4a
b
increasing on 2a , and decreasing on , 2ab ; when a 0, f
b , . Also note that f ( x) 2ax b 2a x b for
2a
2a
70. f ( x) ax 2 bx c a x 2 ba x c a x 2 ba x
b2
4a 2
| .
is
b /2 a
71. f ( x) ax 2 bx f ( x) 2a x b, f (1) 2 a b 2, f (1) 0 2a b 0 a 2, b 4
f ( x) 2 x 2 4 x
72. f ( x) ax3 bx 2 cx d f ( x) 3ax 2 2bx c, f (0) 0 d 0, f (1) 1 a b c d 1,
f (0) 0 c 0, f (1) 0 3a 2b c 0 a 2, b 3, c 0, d 0 f ( x) 2 x3 3 x 2
4.4
CONCAVITY AND CURVE SKETCHING
1. y
x3
3
2
x2 2 x 13 y x 2 x 2 ( x 2)( x 1) y 2 x 1 2 x 12 . The graph is rising on
and concave down on , 12 . Consequently,
a local maximum is 32 at x 1, a local minimum is 3 at x 2, and 12 , 34 is a point of inflection.
(, 1) and (2, ), falling on (1, 2), concave up on
2. y
x4
4
1,
2
2 x 2 4 y x3 4 x x( x 2 4) x( x 2)( x 2) y 3 x 2 4
3x 2
is rising on (2, 0) and (2, ), falling on (, 2) and (0, 2), concave up on ,
2 , 16
3 9
and
2 , 2 . Consequently, a local
3
3
2 , 16 are points of inflection.
3 9
concave down on
3. y 34 ( x 2 1)2/3 y
2
3
3 x 2 . The graph
and
2 ,
3
and
maximum is 4 at x 0, local minima are 0 at x 2, and
34 23 ( x2 1)1/3 (2 x) x( x2 1)1/3 , y )1 ( 0| 1)( the
graph is rising on (1, 0) and (1, ), falling on (, 1) and (0, 1) a local maximum is
minima are 0 at x 1; y ( x 2 1)1/3 ( x) 13 ( x 2 1)4/3 (2 x)
2
x 3
3 3 ( x 2 1)4
Copyright 2018 Pearson Education, Inc.
,
3
4
at x 0, local
Section 4.4 Concavity and Curve Sketching
y | ) ( )( | the graph is concave up on , 3 and
3
1
1
3
3
down on 3, 3 points of inflection at 3, 3 44
213
3, , concave
9 x1/3 ( x 2 7) y 3 x 2/3 ( x 2 7) 9 x1/3 (2 x) 3 x 2/3 ( x 2 1), y | )( | the
4. y 14
14
14
2
graph is rising on (, 1) and (1, ), falling on (1, 1) a local maximum is
27
7
1
1
0
at x 1, a local minimum is
27
at x 1; y x 5/3 ( x 2 1) 3 x1/3 2x1/3 x 5/3 x 5/3 (2 x 2 1), y )( the graph is
7
concave up on (0, ), concave down on (, 0) a point of inflection at (0, 0).
0
5. y x sin 2 x y 1 2 cos 2 x, y [ | | ] the graph is rising on 3 , 3 ,
2 /3
/3
/3
2 /3
falling on 23 , 3 and 3 , 23 local maxima are 23
are 3
3
2
at x 3 and 23
3
2
3
2
at x 23 ; y 4sin 2 x, y
the graph is concave up on 2 , 0 and 2 , 23 , concave down
at 2 , 2 , (0, 0), and 2 , 2
at x 23 and 3
3
2
at x 3 , local minima
[ | | | ]
2 /3
on 23
/2
0
/2
2 /3
, 2 and 0, 2 points of inflection
6. y tan x 4 x y sec2 x 4, y ( | | ) the graph is rising on 2 , 3 and
/3
/2
π , π , falling on π , π a local maximum is 3 4 at x , a local minimum is
3 2
3 3
/3
/2
3
3
3 43 at x 3 ;
y 2(sec x)(sec x)(tan x) 2(sec2 x)(tan x), y ( | ) the graph is concave up on 0, 2 ,
/2
0
concave down on 2 , 0 a point of inflection at (0, 0)
/2
7. If x 0, sin x sin x and if x 0, sin x sin( x)
sin x. From the sketch the graph is rising on
32 , 2 , 0, 2 and 32 , 2 , falling on
3
3
2 , 2 , 2 , 0 and 2 , 2 ; local minima
are 1 at x 32 and 0 at x 0; local maxima are
1 at x 2 and 0 at x 2 ; concave up on
(2 , ) and ( , 2 ), and concave down on
( , 0) and (0, ) points of inflection are
( , 0) and ( , 0)
8. y 2 cos x 2 x y 2 sin x 2, y [
|
| | ] rising on
3 /4
4 , 54
/4
5 /4
3 /2
34 , 4 and 54 , 32 , falling on , 34 and
local maxima are 2 2 at x ,
2 4 2 at x 4 and 32 2 at x 32 , and local minima are 2 34 2 at x 34 and 2 54 2
at x 54 ; y 2 cos x, y [ | | ] concave up on , 2 and 2 , 32 ,
/2
/2
concave down on 2 , 2 points of inflection at
3 /2
2 , 22
and
, 2
2
Copyright 2018 Pearson Education, Inc.
2
214
Chapter 4 Applications of Derivatives
9. When y x 2 4 x 3, then y 2 x 4 2( x 2)
and y 2. The curve rises on (2, ) and falls on
(, 2). At x 2 there is a minimum. Since y 0,
the curve is concave up for all x.
10. When y 6 2 x x 2 , then y 2 2 x 2(1 x)
and y 2. The curve rises on (, 1) and falls on
(1, ). At x 1 there is a maximum. Since y 0,
the curve is concave down for all x.
11. When y x3 3 x 3, then y 3x 2 3
3( x 1)( x 1) and y 6 x. The curve rises on
(, 1) (1, ) and falls on (1, 1). At x 1 there is
a local maximum and at x 1 a local minimum. The
curve is concave down on (, 0) and concave up on
(0, ). There is a point on inflection at x 0.
12. When y x(6 2 x)2 , then
y 4 x(6 2 x) (6 2 x) 2 12(3 x)(1 x) and
y 12(3 x) 12(1 x) 24( x 2). The curve
rises on (, 1) (3, ) and falls on (1, 3). The curve
is concave down on (, 2) and concave up on
(2, ). At x 2 there is a point of inflection.
13. When y 2 x3 6 x 2 3, then y 6 x 2 12 x
6 x( x 2) and y 12 x 12 12( x 1). The
curve rises on (0, 2) and falls on (, 0) and (2, ).
At x 0 there is a local minimum and at x 2 a local
maximum. The curve is concave up on (, 1) and
concave down on (1, ). At x 1 there is a point of
inflection.
Copyright 2018 Pearson Education, Inc.
Section 4.4 Concavity and Curve Sketching
14. When y 1 9 x 6 x 2 x3 , then y 9 12 x 3 x 2
3( x 3)( x 1) and y 12 6 x 6( x 2). The
curve rises on (3, 1) and falls on (, 3) and
(1, ). At x 1 there is a local maximum and at
x 3 a local minimum. The curve is concave up on
(, 2) and concave down on (2, ). At x 2
there is a point of inflection.
15. When y ( x 2)3 1, then y 3( x 2)2 and
y 6( x 2). The curve never falls and there are no
local extrema. The curve is concave down on (, 2)
and concave up on (2, ). At x 2 there is a point of
inflection.
16. When y 1 ( x 1)3 , then y 3( x 1)2 and
y 6( x 1). The curve never rises and there are no
local extrema. The curve is concave up on (, 1)
and concave down on (1, ). At x 1 there is a
point of inflection.
17. When y x 4 2 x 2 , then y 4 x3 4 x
4 x( x 1)( x 1) and y 12 x 2 4
12 x
1
3
x . The curve rises on (1, 0)
1
3
and (1, ) and falls on (, 1) and (0, 1). At x 1
there are local minima and at x 0 a local maximum.
The curve is concave up on ,
and concave down on
points of inflection.
1 , 1
3
3
1
3
and
. At x
1
3
1 ,
3
there are
18. When y x 4 6 x 2 4, then y 4 x3 12 x
4 x x 3 x 3 and y 12 x 2 12
and 0, 3 , and falls on 3, 0 and
12( x 1)( x 1). The curve rises on , 3
3, . At
x 3 there are local maxima and at x 0 a local
minimum. The curve is concave up on (1,1) and
concave down on (, 1) and (1, ). At x 1 there
are points of inflection.
Copyright 2018 Pearson Education, Inc.
215
216
Chapter 4 Applications of Derivatives
19. When y 4 x3 x 4 , then
y 12 x 2 4 x3 4 x 2 (3 x) and y 24 x 12 x 2
12 x(2 x). The curve rises on (, 3) and falls on
(3, ). At x 3 there is a local maximum, but there is
no local minimum. The graph is concave up on (0, 2)
and concave down on (, 0) and (2, ). There are
inflection points at x 0 and x 2.
20. When y x 4 2 x3 , then y 4 x3 6 x 2 2 x 2 (2 x 3)
and y 12 x 2 12 x 12 x( x 1). The curve rises on
32 , and falls on , 32 . There is a local
32 , but
minimum at x
no local maximum. The
curve is concave up on (, 1) and (0, ), and
concave down on (1, 0). At x 1 and x 0 there
are points of inflection.
21. When y x5 5 x 4 , then
y 5 x 4 20 x3 5 x3 ( x 4) and y 20 x3 60 x 2
20 x 2 ( x 3). The curve rises on (, 0) and (4, ),
and falls on (0, 4). There is a local maximum at x 0,
and a local minimum at x 4. The curve is concave
down on (, 3) and concave up on (3, ). At x 3
there is a point of inflection.
22. When y x
2x 5
4
, then
x(4) 2x 5 12 2x 5 52x 5 ,
2
3
and y 3 2x 5 12 52x 5 2x 5 52
2
5 2x 5 ( x 4). The curve is rising on (, 2) and
y 2x 5
4
3
3
(10, ), and falling on (2, 10). There is a local
maximum at x 2 and a local minimum at x 10.
The curve is concave down on (, 4) and concave
up on (4, ). At x 4 there is a point of inflection.
23. When y x sin x, then y 1 cos x and y sin x.
The curve rises on (0, 2 ). At x 0 there is a local
and absolute minimum and at x 2 there is a local
and absolute maximum. The curve is concave down
on (0, ) and concave up on ( , 2 ). At x there is
a point of inflection.
Copyright 2018 Pearson Education, Inc.
Section 4.4 Concavity and Curve Sketching
24. When y x sin x, then y 1 cos x and y sin x.
The curve rises on (0, 2 ). At x 0 there is a local
and absolute minimum and at x 2 there is a local
and absolute maximum. The curve is concave up on
(0, ) and concave down on ( , 2 ). At x there is
a point of inflection.
25. When y 3x 2 cos x, then y 3 2sin x and
y 2 cos x. The curve is increasing on 0, 43 and
5
3
, 2 , and decreasing on
4
3
5
3
,
. At x 0 there
is a local and absolute minimum, at x 43 there is a
local maximum, at x 53 there is a local minimum,
and at x 2 there is a local and absolute maximum.
The curve is concave up on 0, 2 and 32 , 2 , and
2 2
is concave down on , 3 . At x and x 3
2
there are points of inflection.
2
26. When y 43 x tan x, then y 43 sec2 x and
y 2sec2 x tan x. The curve is increasing on
6 , 6 , and decreasing on 2 , 6 and 6 , 2 .
At x there is a local minimum, at x there is
6
6
a local maximum, there are no absolute maxima or
absolute minima. The curve is concave up on
2 , 0 , and is concave down on 0, 2 . At x 0
there is a point of inflection.
27. When y sin x cos x, then y sin 2 x cos2 x
cos 2x and y 2sin 2 x. The curve is increasing
on 0, 4 and 34 , , and decreasing on 4 , 34 . At
x 0 there is a local minimum, at x 4 there is
a local and absolute maximum, at x 3 there is a
4
local and absolute minimum, and at x there is
a local maximum. The curve is concave down on
0, 2 , and is concave up on 2 , . At x 2 there is
a point of inflection.
28. When y cos x 3 sin x, then y sin x 3 cos x
and y cos x 3 sin x. The curve is increasing on
0, 3 and 43 , 2 , and decreasing on 3 , 43 . At
x 0 there is a local minimum, at x 3 there is
a local and absolute maximum, at x 43 there is a
local and absolute minimum, and at x 2 there is
a local maximum. The curve is concave down on
Copyright 2018 Pearson Education, Inc.
217
218
Chapter 4 Applications of Derivatives
0, 56 and 116 , 2 , and is concave up on
56 , 116 . At x 56 and x 116 there are points
of inflection.
4 x 9/5 .
29. When y x1/5 , then y 15 x 4/5 and y 25
The curve rises on (, ) and there are no extrema.
The curve is concave up on (, 0) and concave
down on (0, ). At x 0 there is a point of inflection.
6 x 8/5 .
30. When y x 2/5 , then y 52 x 3/5 and y 25
The curve is rising on (0, ) and falling on (, 0).
At x 0 there is a local and absolute minimum.
There is no local or absolute maximum. The curve is
concave down on (, 0) and (0, ). There are no
points of inflection, but a cusp exists at x 0.
31. When y
y
x
x 2 1
, then y
3 x . The
( x 2 1)5/ 2
1
( x 2 1)3/ 2
and
curve is increasing on (, ).
There are no local or absolute extrema. The curve is
concave up on (, 0) and concave down on (0, ).
At x 0 there is a point of inflection.
32. When y
1 x 2
2 x 1
, then y
( x 2)
(2 x 1) 2 1 x 2
and
4 x3 12 x 2 7 . The curve is decreasing on
(2 x 1)3 (1 x 2 )3/ 2
1, 12 and 12 , 1 . There are no absolute extrema,
y
there is a local maximum at x 1 and a local
minimum at x 1. The curve is concave up on
(1, 0.92) and 12 , 0.69 , and concave down on
0.92, and (0.69, 1). At x 0.92 and x 0.69
12
there are points of inflection.
33. When y 2 x 3 x 2/3 , then y 2 2 x 1/3 and
y 23 x 4/3 . The curve is rising on (, 0) and (1, ),
and falling on (0, 1). There is a local maximum at
x 0 and a local minimum at x 1. The curve is
concave up on (, 0) and (0, ). There are no points
of inflection, but a cusp exists at x 0.
Copyright 2018 Pearson Education, Inc.
Section 4.4 Concavity and Curve Sketching
34. When y 5 x 2/5 2 x, then y 2 x 3/5 2
2 x 3/5 1 and y 65 x 8/5 . The curve is rising
on (0, 1) and falling on (, 0) and (1, ). There is
a local minimum at x 0 and a local maximum at
x 1. The curve is concave down on (, 0) and
(0, ). There are no points of inflection, but a cusp
exists at x 0.
35. When y x 2/3
52 x 52 x2/3 x5/3 , then
y 53 x 1/3 53 x 2/3 53 x 1/3 (1 x) and
y 95 x 4/3 10
x 1/3 95 x 4/3 (1 2 x). The curve
9
is rising on (0, 1) and falling on (, 0) and (1, ).
There is a local minimum at x 0 and a local
maximum at x 1. The curve is concave up on
, 12 and concave down on 12 , 0 and (0, ).
There is a point of inflection at x
at x 0.
12
and a cusp
36. When y x 2/3 ( x 5) x5/3 5 x 2/3 , then
y 53 x 2/3 10
x 1/3 53 x 1/3 ( x 2) and
3
y 10
x 1/3 10
x 4/3 10
x 4/3 ( x 1). The curve
9
9
9
is rising on (, 0) and (2, ), and falling on (0, 2).
There is a local minimum at x 2 and a local
maximum at x 0. The curve is concave up on
(1, 0) and (0, ), and concave down on (, 1).
There is a point of inflection at x 1 and a cusp
at x 0.
37. When y x 8 x 2 x(8 x 2 )1/2 , then
y (8 x 2 )1/2 ( x ) 12 (8 x 2 )1/2 (2 x)
2 1/2
(8 x )
(8 2 x 2 )
y 12 (8 x 2 )
2
2 x ( x 12)
(8 x 2 )3
32
2(2 x )(2 x )
2 2x 2 2x
and
(2 x)(8 2 x 2 ) (8 x 2 )
12
(4 x)
. The curve is rising on (2, 2), and falling
on 2 2, 2 and 2, 2 2 . There are local minima
x 2 and x 2 2, and local maxima at x 2 2
and x 2. The curve is concave up on 2 2, 0 and
concave down on 0, 2 2 . There is
a point of inflection at x 0.
Copyright 2018 Pearson Education, Inc.
219
220
Chapter 4 Applications of Derivatives
32 (2 x2 )1/2 (2 x)
3 x 2 x 2 3x 2 x 2 x and
y (3)(2 x 2 )1/2 (3x) 12 (2 x 2 )1/2 (2 x)
6(1 x )(1 x )
. The curve is rising on 2, 0 and
2 x 2 x
falling on 0, 2 . There is a local maximum at x 0,
38. When y (2 x 2 )3/2 , then y
and local minima at x 2. The curve is concave
down on (1, 1) and concave up on 2, 1 and
1, 2 . There are points of inflection at x 1.
x
39. When y 16 x 2 , then y
y
16
. The
(16 x 2 )3/ 2
16 x 2
and
curve is rising on (4, 0) and
falling on (0, 4). There is a local and absolute
maximum at x 0 and local and absolute minima at
x 4 and x 4. The curve is concave down on
(4, 4). There are no points of inflection.
40. When y x 2 2x , then y 2 x
3
2
x2
2 x3 2
x2
and
y 2 43 2 x 3 4 . The curve is falling on (, 0) and
x
x
(0, 1), and rising on (1, ). There is a local minimum at
x 1. There are no absolute maxima or absolute minima.
The curve is concave up on , 3 2 and (0, ), and
concave down on 3 2, 0 . There is a point of inflection
3
at x 2.
41. When y
and y
x 2 3 ,
x2
then y
2 x ( x 2) ( x 2 3)(1)
( x 2) 2
2
2
(2 x 4)( x 2) ( x 4 x 3)2( x 2)
( x 2) 4
( x 3)( x 1)
( x 2)2
2 . The
( x 2)3
curve
is rising on (, 1) and (3, ), and falling on (1, 2) and
(2, 3). There is a local maximum at x 1 and a local
minimum at x 3. The curve is concave down on
(, 2) and concave up on (2, ). There are no points
of inflection because x 2 is not in the domain.
3
42. When y x3 1, then y
x2
( x 1) 2/3
3
and y
2x
.
( x3 1)5/3
The curve is rising on (, 1), (1, 0), and (0, ). There
are no local or absolute extrema. The curve is concave up
on (, 1) and (0, ), and concave down on (1, 0).
There are points of inflection at x 1 and x 0.
Copyright 2018 Pearson Education, Inc.
Section 4.4 Concavity and Curve Sketching
43. When y
8 x , then
x2 4
8( x 2 4)
y
2
( x 4)
2
16 x ( x
and y
2
2
12)
( x 4)3
.
The curve is falling on (, 2) and (2, ), and is rising
on (2, 2). There is a local and absolute minimum at
x 2, and a local and absolute maximum at x 2. The
curve is concave down on , 2 3 and 0, 2 3 , and
concave up on 2 3, 0 and 2 3, . There are points
of inflection at x 2 3, x 0, and x 2 3. y 0 is a
horizontal asymptote.
44. When y
5 ,
x 5
4
then y
20 x3
( x 4 5)2
and y
100 x 2 ( x 4 3)
( x 4 5)3
.
The curve is rising on (, 0), and is falling on (0, ).
There is a local and absolute maximum at x 0, and there
is no local or absolute minimum. The curve is concave up
on , 4 3 and 4 3, , and concave down on 4 3, 0
4
4
and 0, 3 . There are points of inflection at x 3 and
4
x 3. There is a horizontal asymptote of y 0.
x 2 1, | x | 1
2 x, | x | 1
45. When y | x 2 1|
, then y
2
2 x, | x | 1
1 x , | x | 1
2, | x | 1
. The curve rises on (1, 0) and (1, )
and y
2, | x | 1
and falls on (, 1) and (0, 1). There is a local maximum
at x 0 and local minima at x 1. The curve is concave
up on (, 1) and (1, ), and concave down on (1, 1).
There are no points of inflection because y is not
differentiable at x 1 (so there is no tangent line at
those points).
x 2 2 x, x 0
46. When y | x 2 2 x | 2 x x 2 , 0 x 2,
2
x 2 x, x 2
2 x 2, x 0
2, x 0
then y 2 2 x, 0 x 2, and y 2, 0 x 2 .
2 x 2, x 2
2, x 2
The curve is rising on (0, 1) and (2, ), and falling on
(, 0) and (1, 2). There is a local maximum at x 1 and
local minima at x 0 and x 2. The curve is concave up
on (, 0) and (2, ), and concave down on (0, 2). There
are no points of inflection because y is not differentiable
at x 0 and x 2 (so there is no tangent at those points).
1 , x0
x , x 0
2 x
, then y
47. When y | x|
1
x , x 0
2 x , x 0
Copyright 2018 Pearson Education, Inc.
221
222
Chapter 4 Applications of Derivatives
x 3/ 2 ,
x0
4
and y
.
3/ 2
( x ) , x 0
4
Since lim y and lim y there is a cusp at
x 0
x 0
x 0. There is a local minimum at x 0, but no local
maximum. The curve is concave down on (, 0) and
(0, ). There are no points of inflection.
x 4, x 4
, then
48. When y | x 4 |
4 x, x 4
( x 4)3/ 2
1 ,x4
,x4
2 x4
4
and y
.
y
3/ 2
1
2 4 x , x 4
(4 x )
,
x
4
4
Since lim y and lim y there is a cusp at
x 4
x 4
x 4. There is a local minimum at x 4, but no local
maximum. The curve is concave down on (, 4) and
(4, ). There are no points of inflection.
49. When y
y
x
9 x 2
, then y
(9 x 2 )(1) x ( 2 x )
2 2
(9 x )
(9 x 2 )2 (2 x ) ( x 2 9)2(9 x 2 )( 2 x )
(9 x 2 )4
x 2 9
(9 x 2 ) 2
2 x ( x 2 27)
(9 x 2 )3
and
. The curve is
rising on (, 3), (3, 3), and (3, ). The curve is
concave down on (3, 0) and (3, ), and concave up on
(, 3) and (0, 3). There is a point of inflection at x 0.
2
50. When y 1x x , then y
y
2
2
(1 x )(2 x ) x 2 ( 1)
(1 x )2
(1 x ) (2 2 x ) (2 x x )2(1 x )( 1)
(1 x )4
2
2 x x2 and
(1 x )
2 .
(1 x )3
The curve is
rising on (0, 1) and (1, 2), and falling on (, 0) and
(2, ). There is a local minimum at x 0 and a local
maximum at x 2. The curve is concave up on (, 1), and
concave down on (1, ).
51. y 2 x x 2 (1 x)(2 x ), y | |
1
2
rising on ( 1, 2), falling on ( , 1) and (2, )
there is a local maximum at x 2 and a local
minimum at x 1; y 1 2 x, y |
1/2
concave up on , 12 , concave down on 12 ,
a point of inflection at x 12
Copyright 2018 Pearson Education, Inc.
Section 4.4 Concavity and Curve Sketching
52. y x 2 x 6 ( x 3)( x 2), y | |
2
3
rising on (, 2) and (3, ), falling on (2, 3)
there is a local maximum at x 2 and a local minimum at
x 3; y 2 x 1, y |
1/2
concave up on 12 , , concave down on , 12
a point of inflection at x 12
53. y x( x 3)2 , y | | rising on (0, ), falling
0
3
on (, 0) no local maximum, but there is a local minimum at
x 0; y ( x 3)2 x(2) ( x 3) 3( x 3)( x 1), y
| | concave up on (, 1) and (3, ), concave
1
3
down on (1, 3) points of inflection at x 1 and x 3
54. y x 2 (2 x), y | | rising on (, 2), falling
0
2
on (2, ) there is a local maximum at x 2, but no local
minimum; y 2 x(2 x) x 2 (1) x(4 3 x), y
43 , concave down on , 0
| | concave up on 0,
0
and
4/3
43 , points of inflection at x 0 and x 43
55. y x( x 2 12) x x 2 3 x 2 3 ,
y | | | rising on 2 3, 0 and
2
2 3
0
2 3
3, , falling on , 2 3 and 0, 2 3 a local
maximum at x 0, local minima at
x 2 3; y 1 ( x 2 12) x(2 x) 3( x 2)( x 2),
y | | concave up on (, 2) and (2, ),
2
2
concave down on (2, 2) points of inflection
at x 2
56. y ( x 1)2 (2 x 3), y | | rising on
3/2
1
32 , , falling on , 32 no local maximum,
a local minimum at x 32 ;
y 2( x 1)(2 x 3) ( x 1)2 (2) 2( x 1)(3 x 2),
y | | concave up on , 23 and (1, ),
2/3
1
concave down on 23 , 1 points of inflection at x 23 and
x 1
57. y (8 x 5 x 2 )(4 x) 2 x(8 5 x)(4 x) 2 ,
y | | | rising on 0, 85 , falling on
0
(, 0) and
8/5
4
85 , a local maximum at x 85 ,
Copyright 2018 Pearson Education, Inc.
223
224
Chapter 4 Applications of Derivatives
a local minimum at x 0;
y (8 10 x)(4 x) 2 (8 x 5 x 2 )(2)(4 x)(1)
4(4 x)(5 x 2 16 x 8), y |
8 2 6
5
concave
up on , 8 25
8 2 6 8 2 6
, 5
5
x4
6
and
8 2 6
5
| |
8 2 6
5
4
, 4 , concave down on
and (4, ) points of inflection at x
8 2 6
5
and
58. y ( x 2 2 x)( x 5)2 x( x 2)( x 5) 2 ,
y | | | rising on (, 0) and (2, ),
0
2
5
falling on (0, 2) a local maximum at x 0,
a local minimum at x 2;
y (2 x 2)( x 5)2 2( x 2 2 x)( x 5)
2( x 5)(2 x 2 8 x 5), y | | |
,
x5
4 6
2
5
4 6 4 6
, 2
and (5, ), concave down on
2
and 42 6 , 5 points of inflection at x 42 6
concave up on
4 6
2
4 6
2
and
59. y sec2 x, y ( ) rising on 2 , 2 , never falling
/2
/2
no local extrema;
y 2(sec x)(sec x)(tan x) 2 (sec 2 x) (tan x),
y ( | ) concave up on 0, 2 , concave down
/2
/2
0
on 2 , 0 , 0 is a point of inflection.
60. y tan x, y ( | ) rising on 0, 2 , falling on
/2
/2
0
2 , 0 no local maximum, a local minimum at
x 0; y sec2 x, y ( )
/2
/2
concave up on 2 , 2 no points of inflection
61. y cot 2 , y ( | ) rising on (0, ) , falling on
0
2
( , 2 ) a local maximum at , no
local minimum; y 12 csc2 2 , y ( ) never concave
0
2
up, concave down on (0, 2 ) no points of inflection
Copyright 2018 Pearson Education, Inc.
Section 4.4 Concavity and Curve Sketching
62. y csc2 2 , y ( ) rising on (0, 2 ) , never falling
2
0
no local extrema;
y 2 csc csc
2
2
cot 2 12
csc2 2 cot 2 , y ( | )
0
2
concave up on ( , 2 ), concave down on (0, )
a point of inflection at
63. y tan 2 1 (tan 1)(tan 1),
y ( | | ) rising on 2 , 4 and
/4
/4
/2
, , falling on , a local maximum at , a
/2
4 2
local minimum at
4
4
4
; y 2 tan sec2 ,
4
y ( | ) concave up on 0, 2 , concave down
/2
/2
0
on 2 , 0 a point of inflection at 0
64. y 1 cot 2 (1 cot )(1 cot ),
y ( | | ) rising on 4 , 34 , falling on
/4
3 /4
0, 4 and 34 , a local maximum
at 3 , a local minimum at ;
0
4
4
y 2(cot )( csc2 ), y ( | )
/2
0
concave up on 0, 2 , concave down on 2 ,
a point of infection at 2
65. y cos t , y [ | | ] rising on 0, 2 and
/2
3 /2
2
3 , 2 , falling on , 3 local maxima at t and t 2 ,
2
2
2 2
local minima at t 0 and t 3 ; y sin t , y [ | ]
0
2
0
2
concave up on ( , 2 ), concave down on (0, ) a point of
inflection at t
66. y sin t , y [ | ] rising on (0, ), falling on
0
2
( , 2 ) a local maximum at t ,
local minima at t 0 and t 2 ; y cos t ,
y [ | | ] concave up on 0, 2 and
0
/2
3 /2
2
32 , 2 , concave down on 2 , 32
points of inflection at t 2 and t 32
67. y ( x 1) 2/3 , y ) ( rising on (, ), never
1
falling no local extrema; y 23 ( x 1) 5/3 , y ) (
1
Copyright 2018 Pearson Education, Inc.
225
226
Chapter 4 Applications of Derivatives
concave up on (, 1), concave down on (1, ) a point of
inflection and vertical tangent at x 1
68. y ( x 2)1/3 , y )( rising on (2, ), falling on
2
(, 2) no local maximum, but a local minimum at
x 2; y 13 ( x 2)4/3 , y )( concave down on
2
(, 2) and (2, ) no points of inflection, but there is a cusp at
x2
69. y x 2/3 ( x 1), y )( | rising on (1, ),
0
1
falling on (, 1) no local maximum, but
a local minimum at x 1; y 13 x 2/3 23 x 5/3
13 x 5/3 ( x 2), y | )( concave up on
2
0
(, 2) and (0, ), concave down on (2, 0) points of
inflection at x 2 and x 0, and a vertical tangent at x 0
70. y x 4/5 ( x 1), y | ) ( rising on (1, 0) and
1
0
(0, ), falling on (, 1) no
local maximum, but a local minimum at x 1;
y 15 x 4/5 54 x 9/5 15 x 9/5 ( x 4), y )( |
0
4
concave up on (, 0) and (4, ), concave down on (0, 4)
points of inflection at x 0 and x 4, and a vertical tangent
at x 0
2 x, x 0
, y | rising on (, ) no
71. y
2 x, x 0
0
2, x 0
, y )( concave up
local extrema; y
2, x 0
0
on (0, ), concave down on (, 0) a point of inflection at
x0
72.
x 2 , x 0
y
, y | rising on (0, ), falling
2
0
x , x 0
on (, 0) no local maximum,
2 x, x 0
,
but a local minimum at x 0; y
2 x, x 0
y | concave up on (, )
0
no point of inflection
Copyright 2018 Pearson Education, Inc.
Section 4.4 Concavity and Curve Sketching
73. The graph of y f ( x) the graph of y f ( x) is concave up on
(0, ), concave down on (, 0) a point of inflection at x 0;
the graph of y f ( x) y | | the graph
y f ( x) has both a local maximum and a local minimum
74. The graph of y f ( x) y | the graph of
y f ( x) has a point of inflection, the graph of
y f ( x) y | | the graph of y f ( x) has
both a local maximum and a local minimum
75. The graph of y f ( x) y | |
the graph of y f ( x) has two points of inflection, the graph of
y f ( x) y | the graph of y f ( x) has a local
minimum
76. The graph of y f ( x) y | the graph of
y f ( x) has a point of inflection; the graph of
y f ( x) y | | the graph of y f ( x) has
both a local maximum and a local minimum
77. y
2 x 2 x 1
x 2 1
Since 1 and 1 are roots of the denominator, the domain is
( , 1) ( 1, 1) (1, ).
1
2
y
; y
( x 1)
2
( x 1)
( x 1)3
There are no critical points. The function is decreasing on its
domain. There are no inflection points. The function is concave
down on ( , 1) ( 1, 1) and concave up on (1, ). The
numerator and denominator share a factor of x 1. Dividing out
this common factor gives y 2xx11 ( x 1), which shows that
x 1 is a vertical asymptote. Now dividing numerator and
denominator by x gives y
2(1/ x )
,
1(1/ x )
which shows that y 2 is a
Copyright 2018 Pearson Education, Inc.
227
228
Chapter 4 Applications of Derivatives
horizontal asymptote. The graph will have a hole at x 1,
2( 1) 1
y 1( 1)1 23 . The x-intercept is
78.
1.
2
2
y 2x 49
x 5 x 14
Since 7 and 2 are roots of the denominator, the domain is
( , 7) ( 7, 2) (2, ).
5
10
y
; y
( x 7)
2
( x 2)
( x 1)3
There are no critical points. The function is increasing on its
domain. There are no inflection points. The function is concave
up on ( , 7) ( 7, 2) and concave down on (2, ). The
numerator and denominator share a factor of x 7. Dividing out
this common factor gives y xx 72 ( x 7), which shows that
x 1 is a vertical asymptote. Now dividing numerator and
1(7/ x )
denominator by x gives y 1(2/ x ) , which shows that y 1 is a
horizontal asymptote. The graph will have a hole at x 7,
y
79.
( 1) 7
( 7) 2
14
. The x-intercept is
9
7.
2
4
y x 21
x
Since 0 is a root of the denominator, the domain is
( , 0) (0, ).
y
2 x4 2
3
; y 2
6
x
x4
There are critical points at x 1. The function is increasing on
( 1, 0) (1, ) and decreasing on ( , 1) (0, 1). There are no
inflection points. The function is concave up on its domain. The
y-axis is a vertical asymptote. Dividing numerator and
denominator by x 2 gives y
x 2 1/ x 2 ,
1
which shows that there
are no horizontal asymptotes. For large x , the graph is close to
the graph of y x 2 .
80.
2
y x2x 4
Since 0 is a root of the denominator, the domain is
( , 0) (0, ).
2
y x 24 ; y
2x
4
x3
There are no critical points at x 2. The function is increasing
on ( , 2) (2, ) and decreasing on ( 2, 0) (0, 2). There
are no inflection points. The function is concave down on ( , 0)
and concave up on (0, ). The y-axis is a vertical asymptote.
Dividing numerator and denominator by x gives y
shows that the line y
x
2
x 4/ x ,
2
which
is an asymptote.
Copyright 2018 Pearson Education, Inc.
Section 4.4 Concavity and Curve Sketching
81. y
1
x 2 1
Since 1 and 1 are roots of the denominator, the domain is
( , 1) ( 1, 1) (1, ).
y
2x ;
( x 1)2
2
2
y 6 x2 23
( x 1)
There is a critical point at x 0, where the function has a local
maximum. The function is increasing on ( , 1) ( 1, 0) and
decreasing on (0, 1) (1, ). The function is concave up on
( , 1) (1, ) and concave down on ( 1, 1). The lines x 1
and x 1 are vertical asymptotes. The x-axis is a horizontal
asymptote.
82.
2
y x2
x 1
Since 1 and 1 are roots of the denominator, the domain is
( , 1) ( 1, 1) (1, ).
y
2x ;
( x 1)2
2
2
y 6 x2 23
( x 1)
There is a critical point at x 0, where the function has a local
maximum. The function is increasing on ( , 1) ( 1, 0) and
decreasing on (0, 1) (1, ). There are no inflection points. The
function is concave up on ( , 1) (1, ) and concave down on
( 1, 1). The lines x 1 and x 1 are vertical asymptotes.
Dividing numerator and denominator by x 2 gives y
1
1(1/ x 2 )
which shows that the line y 1 is a horizontal asymptote. The xintercept is 0 and the y-intercept is 0.
83.
2
y x 22
x 1
Since 1 and 1 are roots of the denominator, the domain is
( , 1) ( 1, 1) (1, ).
y
2x ;
( x 2 1)2
2
y 6 x2 23
( x 1)
There is a critical point at x 0, where the function has a local
maximum. The function is increasing on ( , 1) ( 1, 0) and
decreasing on (0, 1) (1, ). There are no inflection points. The
function is concave up on ( , 1) (1, ) and concave down on
( 1, 1). The lines x 1 and x 1 are vertical asymptotes.
Dividing numerator and denominator by x 2 gives y
1(2/ x 2 )
1(1/ x 2 )
which shows that the line y 1 is a horizontal asymptote. The
x-intercepts are 2 and the y-intercept is 2 .
Copyright 2018 Pearson Education, Inc.
229
230
84.
Chapter 4 Applications of Derivatives
2
y x 2 4
x 2
2 and 2 are roots of the denominator, the domain is
Since
, 2
y
4x ;
( x 2 2)2
2, 2
y
2, .
2
4(3 x 2)
( x 2 2)3
There is a critical point at x 0, where the function has a local
minimum. The function is increasing on 0, 2
2, and
decreasing on , 2 2, 0 . There are no inflection
points. The function is concave up on 2, 2 and concave
down on , 2
2, . The lines x 2 and x 2
are vertical asymptotes. Dividing numerator and denominator by
x 2 gives y
1(4/ x 2 )
1(2/ x 2 )
which shows that the line y 1 is a
horizontal asymptote. The x-intercepts are 2 and the
y-intercept is 2 .
85.
2
y xx1
Since 1 is a root of the denominator, the domain is
( , 1) ( 1, ).
2
y x 2 x2 ; y
( x 1)
2
( x 1)3
There is a critical point at x 0, where the function has a local
minimum, and a critical point at x 2 where the functions has a
local maximum. The function is increasing on ( , 2) (0, )
and decreasing on ( 2, 1) ( 1, 0). There are no inflection
points. The function is concave up on ( 1, ) and concave down
on ( , 1) . The line x 1 is a vertical asymptote. Dividing
numerator by denominator gives y x 1
1 ,
x 1
which shows
that the line y x 1 is an oblique asymptote. (See Section 2.6.)
The x-intercept is 0 and the y-intercept is 0.
86.
2
y xx 14
Since 1 is a root of the denominator, the domain is
( , 1) ( 1, ).
2
y x 2 x 2 4 ; y
( x 1)
6
( x 1)3
There are no critical points. The function is decreasing on its
domain. There are no inflection points. The function is concave up
on ( 1, ) and concave down on ( , 1) . The line x 1 is a
vertical asymptote. Dividing numerator by denominator gives
y 1 x x31 , which shows that the line y 1 x is an oblique
Copyright 2018 Pearson Education, Inc.
Section 4.4 Concavity and Curve Sketching
asymptote. (See Section 2.6.) The x-intercepts are 2 and the
y-intercept is 4.
87.
2
y x xx11
Since 1 is a root of the denominator, the domain is
( , 1) (1, ).
2
y x 2 x2 ; y 2 3
( x 1)
x 1
There is a critical point at x 0, where the function has a local
maximum, and a critical point at x 2 where the function has a
local minimum. The function is increasing on ( , 0) (2, )
and decreasing on (0, 1) (1, 2). There are no inflection points.
The function is concave up on (1, ) and concave down on
( , 1). The line x 1 is a vertical asymptote. Dividing
numerator by denominator gives y x
1
x 1
which shows that
the line y x is an oblique asymptote. (See Section 2.6.) The yintercept is 1.
88.
2
y x xx11
Since 1 is a root of the denominator, the domain is
( , 1) (1, ).
2
y 2 x x 2 ; y 2 3
( x 1)
x 1
There is a critical point at x 0, where the function has a local
minimum, and a critical point at x 2 where the function has a
local maximum. The function is increasing on (0, 1) (1, 2) and
decreasing on ( , 0) (2, ). There are no inflection points.
The function is concave up on ( , 1) and concave down on
(1, ). The line x 1 is a vertical asymptote. Dividing numerator
by denominator gives y x
1
x 1
which shows that the line
y x is an oblique asymptote. (See Section 2.6.) The yintercept is 1.
89.
3
3
2
( x 1)
y x 32x 3 x 1 ( x 1)( x 2)
x x 2
Since 1 and 2 are roots of the denominator, the domain is
( , 2) ( 2, 1) (1, ).
y
( x 1)( x 5)
,
( x 2)2
x 1; y
18 ,
( x 2)3
x 1
Since 1 is not in the domain, the only critical point is at x 5,
where the function has a local maximum. The function is
increasing on ( , 5) (1, ) and decreasing on
( 5, 2) ( 2, 1). There are no inflection points. The function is
concave up on ( 2, 1) (1, ) and concave down on ( , 2).
Copyright 2018 Pearson Education, Inc.
231
232
Chapter 4 Applications of Derivatives
The line x 2 is a vertical asymptote. Dividing numerator by
the denominator gives y x 4 x 9 2 which shows that the line
y x 4 is an oblique asymptote. (See Section 2.6.) The y-
intercept is
90.
1.
2
The graph has a hole at the point (1, 0).
3
( x 1)( x 2 x 2)
y x x 2 2 ( x 1)( x )
x x
Since 1 and 0 are roots of the denominator, the domain is
( , 0) (0, 1) (1, ).
y x
2
2 ,
x2
x 1; y
4
x2
, x 1
There is a critical point at x 2 where the function has a local
minimum, and a critical point at x 2 where the function has a
local maximum. The function is increasing on 2, 0 0, 2
and decreasing on , 2
2, . There are no inflection
points. The function is concave up on ( , 0) and concave down
on (0, 1) (1, ). The y-axis is a vertical asymptote. Dividing
numerator by denominator gives y x 1 2x which shows that
the line y x 1 is an oblique asymptote. (See Section 2.6.)
The graph has a hole at the point (1, 4).
91. y
x
x 2 1
Since 1 and 1 are roots of the denominator, the domain is
( , 1) ( 1, 1) (1, ).
2
3
y x2 1 2 ; y 2 x2 6 x3
( x 1)
( x 1)
There are no critical points. The function is decreasing on its
domain. There is an inflection point at x 0. The function is
concave up on ( 1, 0) (1, ) and concave down on
( , 1) (0, 1). The lines x 1 and x 1 are vertical
asymptotes. Dividing numerator and denominator by x 2 gives
y 1/ x 2 which show that the x-axis is a horizontal asymptote.
1(1/ x )
The x-intercept is 0 and the y-intercept is 0.
92. y
x 1
x 2 ( x 2)
Since 0 and 2 are roots of the denominator, the domain is
( , 0) (0, 2) (2, ).
2
3
2
y 2 x3 5 x 24 ; y 6 x 244 x 403x 24
x ( x 2)
x ( x 2)
There are no critical points. The function is increasing on ( , 0)
and decreasing on (0, 2) (2, ). There is an inflection point at
approximately x 1.223. The function is concave up on
Copyright 2018 Pearson Education, Inc.
Section 4.4 Concavity and Curve Sketching
( , 0) (0, 1.223) (2, ) and concave down on (1.223, 2).
The lines x 0 (the y-axis) and x 2 are vertical asymptotes.
Dividing numerator and denominator by x 3 gives
y
(1/ x 2 ) (1/ x 3 )
1 (2/ x )
which shows that the x-axis is a horizontal
asymptote. The x-intercept is 1.
93. y
8
x2 4
y
The domain is ( , ).
16 x ;
( x 2 4)2
y
16(3 x 2 4)
( x 2 4)3
There is a critical point at x 0, where the function has a local
maximum. The function is increasing on ( , 0) and decreasing
on (0, ). There are inflection points at x 2 / 3 and at
x 2 / 3. The function is concave up on
, 2 / 3 2 / 3, and concave down on
2 / 3, 2 / 3 . Dividing numerator and denominator by x 2
gives y
8/ x 2
1(4/ x 2 )
which shows that the x-axis is a horizontal
asymptote. The y-intercept is 2.
4x
The domain is ( , ).
x2 4
4( x 2 4)
8 x ( x 2 12)
y 2 2 ; y 2 3
( x 4)
( x 4)
94. y
There is a critical point at x 2, where the function has a local
minimum, and at x 2, where the function has a local maximum.
The function is increasing on ( 2, 2) and decreasing on
( , 2) (2, ). There are inflection points at
x 2 3, x 0, and x 2 3. The function is concave up on
2 3, 0 2 3, and concave down on
, 2 3 0, 2 3 . Dividing numerator and denominator by
x 2 gives y
4/ x
1(4/ x 2 )
which shows that the x-axis is a horizontal
asymptote. The x-intercept is 0 and the y-intercept is 0.
95.
Point
P
Q
R
S
T
y
y
0
0
Copyright 2018 Pearson Education, Inc.
233
234
96.
98.
Chapter 4 Applications of Derivatives
97.
99.
100.
101.
There are points of inflection at x 3, x 1, and x 2
102.
There are points of inflection at x 1, x 0, and x 2
103.
There are local maxima at x 1 and x 4. There is a local minimum at x 2. There are points of inflection
at x 0 and x 3.
Copyright 2018 Pearson Education, Inc.
Section 4.4 Concavity and Curve Sketching
235
104.
There is a local maximum at x 2. There are local minima at x 1 and x 4. There are points of inflection
at x 3, x 2, x 1, and x 3.
105. 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 |displacement| is increasing as t increases, 0 t 2 and
6 t 9.5; the body is moving toward the origin when |displacement| 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 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 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.
106. (a) The body is moving away from the origin when |displacement| 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 |displacement| 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 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 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.
107. The marginal cost is
dc
dx
which changes from decreasing to increasing when its derivative
d 2c
dx 2
is zero. This is a
point of inflection of the cost curve and occurs when the production level x is approximately 60 thousand units.
108. The marginal revenue is
dr
dt
and it is increasing when its derivative
d 2r
dt 2
is positive the curve is concave up
2
0 t 2 and 5 t 9; marginal revenue is decreasing when d 2r 0 the curve is concave down
dt
2 t 5 and 9 t 12.
109. When y ( x 1)2 ( x 2), then y 2( x 1)( x 2) ( x 1)2 . The curve falls on (, 2) and rises on (2, ).
At x 2 there is a local minimum. There is no local maximum. The curve is concave upward on (, 1) and
5 , , and concave downward on 1, 5 . At x 1 or x 5 there are inflection points.
3
3
3
110. When y ( x 1) 2 ( x 2)( x 4), then y 2( x 1)( x 2)( x 4) ( x 1) 2 ( x 4) ( x 1)2 ( x 2)
( x 1)[2( x 2 6 x 8) ( x 2 5 x 4) ( x 2 3x 2)] 2( x 1)(2 x 2 10 x 11). The curve rises on (, 2) and
(4, ) and falls on (2, 4). At x 2 there is a local maximum and at x 4 a local minimum. The curve is concave
downward on (, 1) and
5 3
2
5 3 5 3
, 2
2
and concave upward on 1, and
5 3
2
there are inflection points.
Copyright 2018 Pearson Education, Inc.
5 3
,
2
. At x 1,
5 3
2
and
236
Chapter 4 Applications of Derivatives
111. The graph must be concave down for x 0 because
f ( x) 12 0.
x
112. 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.
113. The curve will have a point of inflection at x 1 if 1 is a solution of y 0; y x3 bx 2 cx d
y 3 x 2 2bx c y 6 x 2b and 6(1) 2b 0 b 3.
114. (a) f ( x) ax 2 bx c a x 2 ba x c a x 2 ba x
b2
4a 2
b2
4a
c a x 2ba
2
2
b 4a4 ac a parabola whose
b 2 4ac
vertex is at x 2ba the coordinates of the vertex are 2ba , 4 a
(b) The second derivative, f ( x) 2a, describes concavity when a 0 the parabola is concave up and
when a 0 the parabola is concave down.
115. A quadratic curve never has an inflection point. If y ax 2 bx c where a 0, then y 2ax b and y 2a.
Since 2a is a constant, it is not possible for y to change signs.
116. A cubic curve always has exactly one inflection point. If y ax3 bx 2 cx d where a 0, then
y 3ax 2 2bx c and y 6ax 2b. Since 3ab is a solution of y 0, we have that y changes its sign at
x 3ba and y exists everywhere (so there is a tangent at x 3ba ). Thus the curve has an inflection point at
x b . There are no other inflection points because y changes sign only at this zero.
3a
117. y ( x 1)( x 2), when y 0 x 1 or x 2; y | | points of inflection at x 1
and x 2
1
2
118. y x 2 ( x 2)3 ( x 3), when y 0 x 3, x 0 or x 2; y | | | points of
inflection at x 3 and x 2
3
0
2
119. y a x3 bx 2 cx y 3a x 2 2bx c and y 6a x 2b; local maximum at x 3
3a (3) 2 2b(3) c 0 27a 6b c 0; local minimum at x 1 3a (1)2 2b(1) c 0
3a 2b c 0; point of inflection at (1, 11) a(1)3 b(1)2 c(1) 11 a b c 11 and
6a (1) 2b 0 6a 2b 0. Solving 27 a 6b c 0, 3a 2b c 0, a b c 11, and 6a 2b 0
a 1, b 3, and c 9 y x3 3x 2 9 x
Copyright 2018 Pearson Education, Inc.
Section 4.4 Concavity and Curve Sketching
120. y
x2 a
bx c
y
bx 2 2cx ab ; local
(bx c )2
minimum at (1, 2)
maximum at x 3
2
b ( 1) 2c ( 1) a b
(b ( 1) c )2
b (3)2 2c (3) ab
(b (3) c ) 2
0 b 2c a b 0 and
0 9b 6c ab 0; local
( 1) 2 a
b ( 1) c
2 a 2b 2c 1.
Solving 9b 6c ab 0, b 2c a b 0, and a 2b 2c 1 a 3, b 1, and c 1 y
121. If y x5 5 x 4 240, then y 5 x3 ( x 4) and
y 20 x 2 ( x 3). The zeros of y' are extrema, and
there is a point of inflection at x 3.
122. If y x3 12 x 2 then y 3 x( x 8) and y 6( x 4).
The zeros of y and y are extrema, and points of
inflection, respectively.
123. If y 54 x5 16 x 2 25, then y 4 x( x3 8) and
y 16( x3 2). The zeros of y and y are extrema,
and points of inflection, respectively.
124. If y
x4
4
3
x3 4 x 2 12 x 20, then
y x3 x 2 8 x 12 ( x 3)( x 2)2 . So y has a
local minimum at x 3 as its only extreme value.
Also y 3x 2 2x 8 (3 x 4)( x 2) and there
are inflection points at both zeros, 43 and 2, of y .
125. The graph of f falls where f 0, rises where f 0,
and has horizontal tangents where f 0. It has
local minima at points where f changes from
negative to positive and local maxima where f
changes from positive to negative. The graph of f is
concave down where f 0 and concave up where
f 0. It has an inflection point each time f
changes sign, provided a tangent line exists there.
Copyright 2018 Pearson Education, Inc.
x 2 3 .
x 1
237
238
Chapter 4 Applications of Derivatives
126. The graph f is concave down where f 0, and
concave up where f 0. It has an inflection point
each time f changes sign, provided a tangent line
exists there.
4.5
APPLIED OPTIMIZATION
1. Let and w represent the length and width of the rectangle, respectively. With an area of 16 in.2 , we have that
2( 2 16)
()( w) 16 w 16 1 the perimeter is P 2 2w 2 32 1 and P () 2 322
. Solving
2
P() 0
2( 4)( 4)
2
0 4, 4. Since 0 for the length of a rectangle, must be 4 and w 4 the
perimeter is 16 in., a minimum since P () 163 0.
2. Let x represent the length of the rectangle in meters (0 x 4). Then the width is 4 x and the area is
A( x) x(4 x) 4 x x 2 . Since A( x) 4 2 x, the critical point occurs at x 2. Since, A( x) 0 for 0 x 2
and A( x) 0 for 2 x 4, this critical point corresponds to the maximum area. The rectangle with the largest
area measures 2 m by 4 2 2 m, so it is a square.
Graphical Support:
3. (a) The line containing point P also contains the points (0, 1) and (1, 0) the line containing P is y 1 x
a general point on that line is ( x, 1 x).
(b) The area A( x) 2 x(1 x), where 0 x 1.
(c) When A( x) 2 x 2 x 2 , then A( x) 0 2 4 x 0 x 12 . Since A(0) 0 and A(1) 0, we conclude
that A
12 12 sq units is the largest area. The dimensions are 1 unit by 12 unit.
4. The area of the rectangle is A 2 xy 2 x (12 x 2 ),
where 0 x 12. Solving A( x) 0 24 6 x2 0
x 2 or 2. Now 2 is not in the domain, and
since A(0) 0 and A 12 0, we conclude that
A(2) 32 square units is the maximum area. The
dimensions are 4 units by 8 units.
Copyright 2018 Pearson Education, Inc.
Section 4.5 Applied Optimization
239
5. The volume of the box is V ( x) x(15 2 x)(8 2 x)
120 x 46 x 2 4 x3 , where 0 x 4. Solving
V ( x) 0 120 92 x 12 x 2 4(6 x)(5 3 x) 0
x 53 or 6, but 6 is not in the domain. Since
V (0) V (4) 0, V 53 2450
91 in3 must be the
27
maximum volume of the box with dimensions
14 35 5 inches.
3
3 3
6. The area of the triangle is A 12 ba
where 0 b 20. Then
dA
db
400 b 2 ,
b
2
400 b 2
1
2
2
b2
2 400 b 2
200b 2 0 the interior critical point is b 10 2.
400 b
When b 0 or 20, the area is zero A 10 2 is the
2
2
maximum area. When a b 400 and b 10 2,
the value of a is also 10 2 the maximum area
occurs when a b.
7. The area is A( x) x(800 2 x), where 0 x 400.
Solving A( x) 800 4 x 0 x 200. With
A(0) A(400) 0, the maximum area is
A(200) 80, 000 m 2 . The dimensions are 200 m by
400 m.
8. The area is 2 xy 216 y 108
. The amount of
x
fence needed is P 4 x 3 y 4 x 324 x 1 , where
0 x; dP
4 324
0 x 2 81 0 the critical
2
dx
x
points are 0 and 9, but 0 and 9 are not in the
domain. Then P (9) 0 at x 9 there is a
minimum the dimensions of the outer rectangle are
18 m by 12 m 72 meters of fence will be needed.
9. (a) We minimize the weight tS where S is the surface area, and t is the thickness of the steel walls of the
tank. The surface area is S x 2 4 xy where x is the length of a side of the square base of the tank, and y
. Therefore, the weight of the tank is
is its depth. The volume of the tank must be 500 ft 3 y 500
2
2
w( x) t x
x
2000
x
. The critical value is
. Treating the thickness as a constant gives w( x) t 2 x 2000
x
2
at x 10. Since w(10) t 2 4000
0, there is a minimum at x 10. Therefore, the optimum dimensions
3
10
of the tank are 10 ft on the base edges and 5 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 1125 ft 3 , we have that yx 2 1125 y 1125
. The cost of building the
2
2
tank is c( x) 5 x 30 x
, where 0 x. Then c( x)
1125
x2
x
10 x 33750
x2
0 the critical points are 0 and 15,
but 0 is not in the domain. Thus, c(15) 0 at x 15 we have a minimum. The values of x 15 ft and
y 5 ft will minimize the cost.
Copyright 2018 Pearson Education, Inc.
240
Chapter 4 Applications of Derivatives
(b) The cost function c 5( x 2 4 xy ) 10 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 ( x 2 4 xy ), it can be deduced that the unit cost to fabricate the tanks is $5/ft 2 . Normally,
excavation costs are per unit volume of excavated material. Consequently, the total excavation cost can be
taken as 10 xy
10x ( x2 y). This suggests that the unit cost of excavation is $10/ft
x
2
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
$10/ft 2
15 ft
$0.67
$183 . The total cost of the least expensive tank is $3375, which is the sum of $2625 for
3
ft
yd
fabrication and $750 for the excavation.
11. The area of the printing is ( y 4)( x 8) 50.
Consequently, y x50
4. The area of the paper is
8
A( x) x
50 4
x 8
, where 8 x. Then
x508 4 x ( x508)
4( x 8)2 400
( x 8)2 0
the critical points are 2 and 18, but 2 is not in the
domain. Thus A(18) 0 at x 18 we have a
minimum. Therefore the dimensions 18 by 9 inches
minimize the amount of paper.
A( x)
2
12. The volume of the cone is V 13 r 2 h, where r x 9 y 2 and h y 3 (from the figure in the text). Thus,
V ( y ) 3 (9 y 2 )( y 3) 3 (27 9 y 32 y 2 y 3 ) V ( y ) 3 (9 6 y 3 y 2 ) (1 y )(3 y ). The critical
points are 3 and 1, but 3 is not in the domain. Thus V (1) (6 6(1)) 0 at y 1 we have a maximum
volume of V (1) 3 (8)(4)
32
3
cubic units.
3
ab sin
, where
2
ab cos
0 2 .
2
13. The area of the triangle is A( )
0 . Solving A( ) 0
ab sin
Since A( ) 2 A 2 0, there is a
maximum at 2 .
. The amount
14. A volume V r 2 h 100 h 1000
2
r
of material is the surface area given by the sides and
bottom of the can S 2 rh r 2 2000
r2 ,
r
3
0 r. Then dS
2000
2 r 0 r 21000 0.
2
dr
r
The critical points are 0 and
domain. Since
d 2s
dr 2
4000
r3
10
3
r
, but 0 is not in the
2 0, we have a
minimum surface area when r
h
1000
r2
10
3
10
3
cm and
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.
Copyright 2018 Pearson Education, Inc.
Section 4.5 Applied Optimization
241
15. With a volume of 1000 cm3 and V r 2 h, then h 1000
. The amount of aluminum used per can is
2
r
0 8r
A 8r 2 2 rh 8r 2 2000
. Then A(r ) 16r 2000
2
r
r
3
1000
r2
0 the critical points are 0 and 5, but
r 0 results in no can. Since A(r ) 16 1000
and h:r 8: .
0 we have a minimum at r 5 h 40
r3
16. (a) The base measures 10 2x in. by 1522 x in., so the volume formula is V ( x)
x (10 2 x )(15 2 x )
2
2 x3 25 x 2 75 x.
(b) We require x 0, 2 x 10, and 2 x 15. Combining these requirements, the domain is the interval (0, 5).
(c) The maximum volume is approximately 66.02 in.3 when x 1.96 in.
(d) V ( x) 6 x 2 50 x 75. The critical point occurs when V ( x) 0, at x
50 ( 50) 2 4(6)(75)
2(6)
50 700
12
2565 7 , that is, x 1.96 or x 6.37. We discard the larger value because it is not in the domain. Since
V ( x ) 12 x 50, which is negative when x 1.96, the critical point corresponds to the maximum volume.
The maximum volume occurs when x
255 7
6
1.96, which confirms 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 V ( x) 2 x(24 2 x)(18 2x ) 8 x3 168 x 2 862 x.
(b) We require x 0, 2 x 18, and 2 x 12. Combining these requirements, the domain is the interval (0, 9).
(c) The maximum volume is approximately 1309.95 in.3 when x 3.39 in.
(d) V ( x) 24 x 2 336 x 864 24( x 2 14 x 36). The critical point is at x
14 ( 14)2 4(1)(36)
2(1)
142 52
7 13, that is, x 3.39 or x 10.61. We discard the larger value because it is not in the domain. Since
V ( x) 24(2 x 14) which is negative when x 3.39, the critical point corresponds to the maximum
volume. The maximum value occurs at x 7 13 3.39, which confirms the results in (c).
(e) 8 x3 168 x 2 862 x 1120 8( x3 21x 2 108 x 140) 0 8( x 2)( x 5)( x 14) 0. Since 14 is not in
the domain, the possible values of x are x 2 in. or x 5 in.
(f ) The dimensions of the resulting box are 2x in., (24 2 x) in., and (18 2 x ). Each of these measurements
must be positive, so that gives the domain of (0, 9).
18. If the upper right vertex of the rectangle is located at ( x, 4 cos 0.5 x) for 0 x , then the rectangle has width
2x and height 4 cos 0.5x, so the area is A( x) 8 x cos 0.5 x.. Solving A( x) 0 graphically for 0 x , we find
Copyright 2018 Pearson Education, Inc.
242
Chapter 4 Applications of Derivatives
that x 2.214. Evaluating 2x and 4 cos 0.5x for x 2.214, the dimensions of the rectangle are approximately
4.43 (width) by 1.79 (height), and the maximum area is approximately 7.923.
19. Let the radius of the cylinder be r cm, 0 r 10. Then the height is 2 100 r 2 and the volume is
2 r 3 4 r (100 r 2 )
1
(2r ) 2 100 r 2 (2r )
V (r ) 2 r 2 100 r 2 cm3 . Then, V (r ) 2 r 2
2
100 r 2
2 100 r
2 r (200 3r 2 )
100 r 2
. The critical point for 0 r 10 occurs at r
and V (r ) 0 for 10
r 10
2
3
2
3
200
3
10
2 . Since V ( r )
3
0 for 0 r 10
2
3
r 10, the critical point corresponds to the maximum volume. The dimensions are
8.16 cm and h
20
3
11.55 cm, and the volume is
4000
3 3
2418.40 cm3 .
20. (a) From the diagram we have 4 x 108 and
V x 2 . The volume of the box is
V ( x) x 2 (108 4 x), where 0 x 27. Then
V ( x) 216 x 12 x 2 12 x(18 x) 0 the
critical points are 0 and 18, but x 0 results in
no box. Since V ( x) 216 24 x 0 at x 18
we have a maximum. The dimensions of the
box are 18 18 36 in.
(b) In terms of length, V () x2
1084
2
. The
graph indicates that the maximum volume
occurs near 36, which is consistent with the
result of part (a).
21. (a) From the diagram we have 3h 2 w 108 and
V h 2 w V (h) h 2 54 32 h 54h 2 32 h3 .
Then V (h) 108h 92 h 2 92 h(24 h) 0
h 0 or h 24, but h 0 results in no box.
Since V (h) 108 9h 0 at h 24, we
have a maximum volume at h 24 and
w 54 32 h 18.
(b)
Copyright 2018 Pearson Education, Inc.
Section 4.5 Applied Optimization
243
22. From the diagram the perimeter is P 2r 2h r ,
where r is the radius of the semicircle and h is the
height of the rectangle. The amount of light
transmitted proportional to A 2rh 14 r 2
r ( P 2r r ) 14 r 2 rP 2r 2 43 r 2 . Then
dA
dr
P 4r 32 r 0 r 82 3P
(4 ) P
2h P 84 3P 823P 83 . Therefore,
2 r 8 gives the proportions that admit the most
h
4
light since
d2A
dr 2
4 32 0.
23. The fixed volume is V r 2 h 23 r 3 h
V
23r , where h is the height of the cylinder and r is the radius
r2
of the hemisphere. To minimize the cost we must minimize surface area of the cylinder added to twice the
surface area of the hemisphere. Thus, we minimize C 2 rh 4 r 2 2 r
dC
dr
Then
4V
1/3
1/3
32/3
the cost.
2V2 16
r 0 V 83 r 3 r
3
r
1/3
1/3
1/3
1/3
1/3
23 V
23 V1/3 3 24V 1/3
32
1/3
32
V
r2
23r 4 r 2 2rV 83 r 2 .
1/3
83V
. From the volume equation, h V 2 23r
r
3V 1/3 . Since d 2C 4V 16 0, these dimensions
3
dr 2
r3
do minimize
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( ) cos sin cos , 0 2 . Then A( ) sin cos 2 sin 2
(2sin 2 sin 1) (2sin 1)(sin 1) so A( ) 0 sin 12 or sin 1 6 because
sin 1 when 0 2 . Also, A( ) 0 for 0 6 and A( ) 0 for 6 2 . Therefore, at 6 there
is a maximum.
25. (a) From the diagram we have: AP x, RA L x 2 , PB 8.5 x,
CH DR 11 RA 11 L x 2 , QB x 2 (8.5 x)2 ,
HQ 11 CH QB 11 11 L x 2 x 2 (8.5 x) 2
2
2
L x 2 x 2 (8.5 x)2 , RQ RH HQ
2
2
2
2
2
(8.5) 2 L x 2 x 2 (8.5 x 2 ) . It follows that RP PQ RQ
2
L2 x 2 L2 x 2 x 2 ( x 8.5) 2 (8.5) 2
L2 x 2 L2 x 2 2 L2 x 2 17 x (8.5)2 17 x (8.5) 2 (8.5) 2
17 2 x 2 4( L2 x 2 )(17 x (8.5) 2 ) L2 x 2
17 x3
17 x (8.5) 2
(b) If f ( x)
17 x3
17 x
4 x3
4 x 17
172
2
3
3
44x x17 2 x2x8.5 .
17 2 x 2
4[17 x (8.5)2 ]
is minimized, then L2 is minimized. Now f ( x)
f ( x) 0 when x 51
. Thus L2 is minimized when x
8
51 .
8
4 x 2 (8 x 51)
(4 x 17) 2
Copyright 2018 Pearson Education, Inc.
f ( x) 0 when x 51
and
8
244
Chapter 4 Applications of Derivatives
(c) When x
51 ,
8
then L 11.0 in.
26. (a) From the figure in the text we have P 2 x 2 y y P2 x. If P 36, then y 18 x. When the
cylinder is formed, x 2 r r 2x and h y h 18 x. The volume of the cylinder is
2
3 x (12 x )
4
3
V r 2 h V ( x) 18 x4 x . Solving V ( x)
3
cylinder. Then V ( x) 3
x
2
0 x 0 or 12; but when x 0 there is no
V (12) 0 there is a maximum at x 12. The values of x 12 cm
and y 6 cm give the largest volume.
(b) In this case V ( x) x 2 (18 x). Solving V ( x) 3 x(12 x) 0 x 0 or 12; but x 0 would result in
no cylinder. Then V ( x) 6 (6 x) V (12) 0 there is a maximum at x 12. The values of
x 12 cm and y 6 cm give the largest volume.
27. Note that h 2 r 2 3 and so r 3 h 2 . Then the volume is given by V 3 r 2 h 3 (3 h 2 )h h 3 h3 for
0 h 3, and so
0 h 1, and
dV
dh
dV
dh
r 2 (1 r 2 ). The critical point (for h 0 ) occurs at h 1. Since
dV
dh
0 for
0 for 1 h 3, the critical point corresponds to the maximum volume. The cone of
greatest volume has radius 2 m, height 1 m, and volume 23 m3 .
y
28. Let d ( x 0) 2 ( y 0)2 x 2 y 2 and ax b 1 y ba x b. We can minimize d by minimizing
2
D x 2 y 2 x 2 ba x b
2
2
2
2
2 x b 2 x ba 0 x 2ab 2 is the critical point y ba
D
x
a
y
b
a
ab 2
a 2 b 2
2
2b 2
a2
2
2
D 2 x 2 ba x b ba 2 x 2b2 x 2ab . D 0
a
a b
ab 2
a b2
2
0 the critical point is a local minimum
b
ab 2
a 2 b2
,
a 2b . D 2 2b 2
a b2
a2
2
a b is the point on the
a 2 b 2
2
line
1 that is closest to the origin.
29. Let S ( x) x 1x , x 0 S ( x) 1
only consider x 1. S ( x)
2
x3
1
x2
S (1)
30. Let S ( x) 1x 4 x 2 , x 0 S ( x)
S ( x)
2
x3
8 S
12 (1/2)2
3
1
x2
x 2 1 . S ( x) 0 x 2 1 0 x 2 1 0
x2
x2
2 0 local minimum when x 1
13
3
3
x
x
1.
2
x 1. Since x 0, we
8 x 8 x 21 . S ( x) 0 8 x 21 0 8 x3 1 0 x 12 .
8 0 local minimum when x
31. The length of the wire b perimeter of the triangle circumference of the circle. Let x length of a side of the
equilateral triangle P 3 x, and let r radius of the circle C 2 r. Thus b 3 x 2 r r b23 x .
The area of the circle is r 2 and the area of an equilateral triangle whose sides are x is 12 ( x)
Thus, the total area is given by A
A
3
2
x 23 (b 3x)
3
2
3 2
x
4
2
b 3 x 2
r 2 43 x 2 b23 x 43 x 2 4
x 23b 29 x . A 0
3
2
x 23b 29 x 0 x 3b .
3 9
Copyright 2018 Pearson Education, Inc.
x
3
2
3 2
x .
4
Section 4.5 Applied Optimization
A
3
2
b23x b 3x b
9b
3 9
3b
9b m is the length of the
3 9
3 9
3 b
m is the length of the circular segment.
3 9
29 0 local minimum at the critical point. P 3
triangular segment and C 2
245
32. The length of the wire b perimeter of the triangle circumference of the circle. Let x length of a side of the
square P 4 x, and let r radius of the circle C 2 r. Thus b 4 x 2 r r b24 x . The area of the
circle is r 2 and the area of a square whose sides are x is x 2 . Thus, the total area is given by A x 2 r 2
x2
x
b
4
b4 x 2
2
x2
b 4 x 2
4
A 2 x 24 (b 4 x) 2 x 2b 8 x, A 0 2 x 2b 8 x 0
. A 2 8 0 local minimum at the critical point. P 4
b24 x b 4 x b 44b 4b m is the length of the circular segment.
square segment and C 2
4b 44b m is the length of the
33. Let ( x, y ) x, 43 x be the coordinates of the corner that intersects the line. Then base 3 x and height
y 43 x, thus the area of the rectangle is given by A (3 x) 43 x 4 x 43 x 2 , 0 x 3. A 4 83 x, A 0
x 32 . A 43 A 32 0 local maximum at the critical point. The base 3 32 32 and the height
43 32 2.
34. Let ( x, y ) x, 9 x 2 be the coordinates of the corner that intersects the semicircle. Then base 2 x and
2
height y 9 x , thus the area of the inscribed rectangle is given by A (2 x) 9 x 2 , 0 x 3. Then
A 2 9 x 2 (2 x)
x
9 x
2
2(9 x 2 ) 2 x 2
9 x
2
18 4 x2 , A 0 18 4 x 2 0 x 3 2 2 , only x
2
4 x
3 2
2
lies in
0 x 3. A is continuous on the closed interval 0 x 3 A has an absolute maxima and absolute minima.
A(0) 0, A(3) 0, and A
is 3 22 .
3 2 9 absolute maxima. Base of rectangle is 3
3 2
2
3 2
2
2 and height
35. (a) f ( x) x 2 ax f ( x) x 2 (2 x3 a ), so that f ( x ) 0 when x 2 implies a 16
(b) f ( x) x 2 ax f ( x) 2 x 3 ( x3 a), so that f ( x) 0 when x 1 implies a 1
36. If f ( x) x3 ax 2 bx, then f ( x) 3 x 2 2ax b and f ( x) 6 x 2a.
(a) A local maximum at x 1 and local minimum at x 3 f (1) 0 and f (3) 0 3 2a b 0 and
27 6a b 0 a 3 and b 9.
(b) A local minimum at x 4 and a point inflection at x 1 f (4) 0 and f (1) 0 48 8a b 0 and
6 2a 0 a 3 and b 24.
37. The height of the cone is h y 1, where
r 2 y 2 12 r 2 1 y 2 . The volume of the cone
is V 13 r 2 h 13 (1 y 2 )( y 1)
V 13 (1 y y 2 y 3 ) V 13 (1 2 y 3 y 2 )
13 (1 y )(1 3 y ) 0 critical points are 1 and
1,
3
but 1 is not in the domain. Thus V
13 0
at y 13 we have a maximum. Therefore r
and h
4
3
2 2
3
maximize the volume of the cone.
Copyright 2018 Pearson Education, Inc.
246
Chapter 4 Applications of Derivatives
38. Since y 20 x3 60 x 3 x5 5 x 4 , the slope equation is S y 60 x 2 60 15 x 4 20 x3 .
S 120 x 60 x3 60 x 2 60 x(2 x 2 x) 60 x( x 1)( x 2) 0 critical points are 0, 1, and 2. Thus
S (0) 0, S (1) 0, and S (2) 0 at x 1 and x 2 we have maxima. But S (1) 85 and
S (2) 220 the maximum slope of 220 occurs at x 2, y 264.
39. Since y 3x x 2 y 3 2 x the slope of the tangent line at x a is 3 2a and the equation of the
tangent line at x a is y (3a a 2 ) (3 2a )( x a ). If x 0 y a 2 . If y 0 x
2
area of the described triangle is A 12 a 2 2 aa 3
critical points are 0,
3,
2
and 2, but 0 and
3
2
a4
4 a 6
a2 .
2 a 3
Thus the
(4 a 6)4 a 3 a 4 (4) 12 a3 ( a 2)
0
A
2
2
(4 a 6)
(4 a 6)
are not in the domain. Thus
A(2) 0
at a 2 we have a
minimum. Therefore a 2 determines a minimum area of 8.
40. The circular base of the resulting
cone has circumference
2 (1) 2 2 r radius
r 22 .
Since r 2 h 2 12 the height of the cone is h
V 13 r 2 h
1
24 2
1
24 2
4 2
4 2
(2 ) 2 (4 2 )1 2 V
2(2 )(1) (4 2 )1 2
critical points are 0, 2 , 4 ,
V (0) 0 V (2 ). Thus V
determines a radius r
2,
3
6 2 6
3
6 2 6
3
1
24 2
3
1 (2 )
2
2 12
24 (4 )
, and
6 2 6
3
0 at
a height h
2 ,
3
12
(4 2 )1 2
2
volume of cone is
(2 ) 2 12 (4 2 ) 1 2 (4 2 )
2(2 )(4 2 )1 2
1
; but 4 and
623 6
6 2 6
3
0.367
(2 )(3 2 12 4 2 )
24 2 ( (4 ))1 2
0
are not in the domain, and
we have a maximum. Therefore
and a maximum volume of
V
4
9 3
6 2 6
3
0.806.
41. (a) s (t ) 16t 2 96t 112 v(t ) s (t ) 32t 96. At t 0, the velocity is v(0) 96 ft/sec.
(b) The maximum height occurs when v(t ) 0, when t 3. The maximum height is s (3) 256 ft and it occurs
at t 3 sec.
(c) Note that s (t ) 16t 2 96t 112 16(t 1)(t 7), so s 0 at t 1 or t 7. Choosing the positive value
of t, the velocity when s 0 is v(7) 128 ft/sec.
42.
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 4 x 2 mi at 2 mph and walk 6 x mi at 5 mph. The total amount of time to reach
the village is f ( x)
4 x 2
2
65 x hours (0 x 6). Then f ( x) 12
1
2 4 x 2
Copyright 2018 Pearson Education, Inc.
(2 x) 15
x
2 4 x 2
15 . Solving
Section 4.5 Applied Optimization
x
f ( x) 0. we have:
2 4 x 2
247
15 5 x 2 4 x 2 25 x 2 4 4 x 2 21x 2 16 x 4 . We discard
21
the negative value of x because it is not in the domain. Checking the endpoints and critical point, we have
f (0) 2.2, f
2.12, and f (6) 3.16. Jane should land her boat
4
21
from the point nearest her boat.
43.
8
x
4
21
0.87 miles down the shoreline
x h27 h 8 216
and L( x) h 2 ( x 27) 2
x
8 216x
2
( x 27)2 when x 0. Note that L( x)
is minimized when f ( x) 8 216
x
minimized. If f ( x) 0, then
2
( x 27)2 is
2 8 216
216
2( x 27) 0
2
x
x
( x 27) 1 1728
0 x 27 (not acceptable
3
x
since distance is never negative) or x 12 . Then
L(12) 2197 46.87 ft.
44. (a) s1 s2 sin t sin t 3 sin t sin t cos 3 sin 3 cos t sin t 12 sin t
t or 4
3
3
(b) The distance between the particles is s (t ) | s1 s2 | sin t sin t 3
s (t )
sin t
3 cos t cos t 3 sin t
2 sin t 3 cos t
since
0, 3 , 56 , 43 , 116 , 2 ; then s (0)
3
,
2
d
dx
sin t
3 cos t cos t 3 sin t
s 3 0, s 56 1, s
sin t 3 cos t
43 0, s 116 1, s(2 )
we can conclude that at t
3
2 sin t 3 cos t
tan t 3
| x | | xx | critical times and endpoints are
greatest distance between the particles is 1.
(c) Since s (t )
1
2
3
cos t
2
3
2
the
and 43 , s (t ) has cusps and the
distance between the particles is changing the fastest near these points.
k
d2
45. I
, let x distance the point is from the stronger light source 6 x distance the point is from the other
light source. The intensity of illumination at the point from the stronger light is I1
illumination at the point from the weaker light is I 2
the intensity of the second light k1 8k2 . I1
I
16 k2
x
3
x 4 m. I
2 k2
(6 x )
48k2
x4
3
16(6 x )3 k2 2 x3k2
3
6 k2
(6 x )4
x (6 x )
3
I (4)
k2
(6 x ) 2
8 k2
x2
and I 0
48k2
44
6 k2
(6 4)4
k1
x2
, and intensity of
. Since the intensity of the first light is eight times
. The total intensity is given by I I1 I 2
16(6 x )3 k2 2 x3k2
x3 (6 x )3
v02
g
2v 2
2v 2
4v02
g
k2
(6 x ) 2
0 local minimum. The point should be 4 m from the
2
4v 2
sin 2 ddR
g0 cos 2 and ddR
0 g0 cos 2 0 4 . d R2 g0 sin 2
d
d 2R
d 2
4
x2
0 16(6 x)3 k2 2 x3 k2 0
stronger light source.
46. R
8k2
2
4v
sin 2 4 g0 0 local maximum. Thus, the firing angle of 4 45
will maximize the range R.
Copyright 2018 Pearson Education, Inc.
248
Chapter 4 Applications of Derivatives
47. (a) From the diagram we have d 2 4r 2 w2 . The strength of the beam is S kwd 2 kw (4r 2 w2 ).
When r 6, then S 144kw kw3 . Also, S ( w) 144k 3kw2 3k (48 w2 ) so S ( w) 0 w 4 3;
S 4 3 0 and 4 3 is not acceptable. Therefore S 4 3 is the maximum strength. The dimensions of
the strongest beam are 4 3 by 4 6 inches.
(c)
(b)
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).
48. (a) From the situation we have w2 144 d 2 . The stiffness of the beam is S kwd 3 kd 3 (144 d 2 )1/2 ,
where 0 d 12. Also, S (d )
(b)
4 kd 2 (108 d 2 )
144 d 2
critical points at 0, 12, and 6 3. Both d 0 and d 12
cause S 0. The maximum occurs at d 6 3. The dimensions are 6 by 6 3 inches.
(c)
Both graphs indicate the same maximum value and are consistent with each other. The changing of k has
no effect.
49. (a) s 10 cos( t ) v 10 sin( t ) speed |10 sin( t )| 10 |sin( t ) | the maximum speed is
10 31.42 cm/sec since the maximum value of |sin ( t )| is 1; the cart is moving the fastest at t 0.5 sec,
1.5 sec, 2.5 sec and 3.5 sec when |sin ( t )| is 1. At these times the distance is s 10 cos 2 0 cm and
2
2
a 10 cos ( t ) | a | 10 |cos ( t )| | a | 0 cm/sec
2
(b) | a | 10 2 |cos ( t )| is greatest at t 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 | s | 10 cm from the rest position and the speed is 0 cm/sec.
50. (a) 2sin t sin 2t 2sin t 2sin t cos t 0 (2sin t )(1 cos t ) 0 t k where k is a positive integer
(b) The vertical distance between the masses is s(t ) | s1 s2 | ( s1 s2 ) 2
s (t )
s
1/2
((sin 2t 2sin t )2 )1/2
2t 2sin t )
12 ((sin 2t 2sin t )2 )1/2 (2)(sin 2t 2sin t )(2 cos 2t 2 cos t) 2(cos 2t |sin2 cos2t t)(sin
2sin t |
4(2 cos t 1)(cos t 1)(sin t )(cos t 1)
|sin 2t 2sin t |
critical times at 0, 23 , , 43 , 2 ; then s (0) 0,
23 sin 43 2sin 23 3 23 ,
43 3 23 , s(2 ) 0
208t 144
s ( ) 0, s 43 sin 83 2sin
the greatest distance is 3 23 at t 23 and 43
51. (a) s (12 12t ) 2 (8t ) 2 ((12 12t )2 64t 2 )1/2
(b) ds
12 ((12 12t )2 64t 2 ) 1/2 [2(12 12t )(12) 128t ]
dt
ds
dt t1
(12 12t 2 ) 64t 2
8 knots
Copyright 2018 Pearson Education, Inc.
ds
dt t 0
12 knots and
Section 4.5 Applied Optimization
(d) The graph supports the conclusions in parts (b)
and (c).
(c) The graph indicates that the ships did not see
each other because s (t ) 5 for all values of t.
ds
t dt
(e) lim
lim
249
208 144t
2
t 144 1 1 64
t
2
(208t 144)2
2
2
t 144(1t ) 64t
lim
2082
144 64
208 4 13 which equals the square
root of the sums of the squares of the individual speeds.
52. The distance OT TB is minimized when OB is a
straight line. Hence 1 2 .
53. If v kax kx 2 , then v ka 2kx and v 2k , so v 0 x a2 . At x
v
2k 0. The maximum value of v is
ka
4
a
2
2
a
2
there is a maximum since
.
54. (a) According to the graph, y (0) 0.
(b) According to the graph, y ( L) 0.
(c) y (0) 0, so d 0. Now y ( x) 3ax 2 2bx c, so y (0) 0 implies that c 0. Therefore, y ( x) ax3 bx 2
and y ( x) 3ax 2 2bx. then y ( L) aL3 bL2 H and y ( L) 3aL2 2bL 0, so we have two linear
. Substituting into the first
equations in two unknowns a and b. The second equation gives b 3aL
2
3
H , or
equation, we have aL3 3aL
2
y ( x)
2 H3 x3
L
3 H2
L
x , or y ( x) H 2
2
aL3
2
H , so a 2 H3 . Therefore, b 3 H2 and the equation for y is
x
L
3
3
x
L
2
L
L
.
55. The profit is p nx nc n( x c) [a( x c) 1 b(100 x)]( x c) a b(100 x)( x c)
a (bc 100b) x 100bc bx 2 . Then p ( x) bc 100b 2bx and p ( x) 2b. Solving
p ( x) 0 x 2c 50. At x 2c 50 there is a maximum profit since p ( x) 2b 0 for all x.
56. Let x represent the number of people over 50. The profit is p( x) (50 x)(200 2 x) 32(50 x) 6000
2 x 2 68 x 2400. Then p ( x) 4 x 68 and p 4. Solving p ( x) 0 x 17. At x 17 there is a
maximum since p (17) 0. It would take 67 people to maximize the profit.
57. (a) A(q ) kmq 1 cm h2 q, where q 0 A(q) kmq 2 h2
points are
2 km , 0, and
h
2 km , but
h
only
2km
h
hq 2 2 km
2q2
and A(q) 2kmq 3 . The critical
is in the domain. Then A
is a minimum average weekly cost.
Copyright 2018 Pearson Education, Inc.
2 km
h
0 at q
2km
h
there
250
Chapter 4 Applications of Derivatives
(b) A(q )
( k bq ) m
cm h2 q
q
3
A(q) 2kmq
kmq 1 bm cm h2 q, where q 0 A(q ) 0 at q
0 so the most economical quantity to order is still q
average weekly cost.
2km
h
2km
h
as in (a). Also
which minimizes the
c( x)
58. We start with c( x) the cost of producing x items, x 0, and x 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
d c ( x ) 0 xc( x ) c ( x ) 0 (by the quotient rule) xc( x) c( x) 0 (multiply both sides by x 2 )
2
dx
x
x
c( x)
c ( x) x where c( x) 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
minimum.)
59. The profit p ( x) r ( x) c( x) 6 x ( x3 6 x 2 15 x) x3 6 x 2 9 x, where x 0. Then p ( x) 3 x 2 12 x 9
3( x 3)( x 1) and p ( x) 6 x 12. The critical points are 1 and 3. Thus p (1) 6 0 at x 1 there is a
local minimum, and p (3) 6 0 at x 3 there is a local maximum. But p (3) 0 the best you can do is
break even.
c( x)
60. The average cost of producing x items is c ( x) x x 2 20 x 20, 000 c( x) 2 x 20 0 x 10, the
only critical value. The average cost is c (10) $19,900 per item is a minimum cost because c(10) 2 0.
61. Let x the length of a side of the square base of the box and h the height of the box. V x 2 h 48 h
The total cost is given by C 6 x 2 4(4 xh) 6 x 2 16 x
6x
48
x2
4; C 12 1536
x2
3
C 0 12 x 2 768 0 12 x3 768 0 x
x
48
x 4 h 2 3 and C (4) 6(4)2 768
288 the
4
4
2
768
, x 0 C 12 x 768
2
x
x
C (4)
12 1536
42
48 .
x2
3
12 x 768
x2
0 local minimum.
box is 4 ft 4 ft 3 ft, with a minimum cost of $288.
62. Let x the number of $10 increases in the charge per room, then price per room 50 10 x, and the number of
rooms filled each night 800 40 x the total revenue is R( x) (50 10 x)(800 40 x)
400 x 2 6000 x 40000, 0 x 20 R ( x) 800 x 6000; R ( x) 0 800 x 6000 0
x 15
; R( x) 800 R
2
63. We have
dR
dM
152 800 0 local maximum. The price per room is 50 10 152 $125.
CM M 2 . Solving
maximum.
d 2R
dM 2
C 2M 0 M
C . Also. d 3 R
2
dM 3
2 0 at M
C
2
there is a
64 . (a) If v cr0 r 2 cr 3 , then v 2cr0 r 3cr 2 cr 2r0 3r and v 2cr0 6cr 2c r0 3r . The solution of
2r
v 0 is r 0 or 30 , but 0 is not in the domain. Also, v 0 for r
there is a maximum.
(b) The graph confirms the findings in (a).
2 r0
3
Copyright 2018 Pearson Education, Inc.
and v 0 for r
2 r0
3
at r
2 r0
3
Section 4.5 Applied Optimization
2
65. If x 0, then x 1 0 x 2 1 2 x
then
16.
a 2 1
a
b 2 1
b
c 2 1
c
a x
2 1/ 2
2
x
66. (a) f ( x)
d 2 1
d
2
a x
2
f ( x)
function of x
d x
(b) g ( x)
b2 d x
b 2
2 3/ 2
of x (from part (b)):
dt
dx
x2 a2 x2
a2 x2
b 2 d x
(c) Since c1 , c2 0, the derivative
g ( x) 0
2. In particular if a, b, c and d are positive integers,
2 1/ 2
1/ 2
a2 x2 x2
a x
2 3/ 2
2
d x b2 d x
2
b2 d x
a2
2 1/ 2
2
0 f ( x) is an increasing
a x
2
2 3/ 2
b2 d x d x
2
2
b d x
2
2 3/ 2
0 g ( x) is a decreasing function of x
b d x
2
2
g ( x)
x 2 1
x
251
dt
dx
1
c1
dt
dx
is an increasing function of x (from part (a)) minus a decreasing function
f ( x) c1 g ( x)
2
is an increasing function of x.
d 2t
dx 2
1
c1
f ( x) c1 g ( x) 0 since f ( x) 0 and
2
67. At x c , the tangents to the curves are parallel. Justification: The vertical distance between the curves is
D ( x) f ( x) g ( x), so D ( x) f ( x) g ( x). The maximum value of D will occur at a point c where D 0. At
such a point, f (c) g (c) 0, or f (c ) g (c).
68. (a) f ( x ) 3 4 cos x cos 2 x is a periodic function with period 2
(b) No, f ( x) 3 4 cos x cos 2 x 3 4 cos x (2 cos 2 x 1) 2(1 2 cos x cos 2 x) 2(1 cos x)2 0
f ( x) is never negative.
69 . (a) If y cot x 2 csc x where 0 x , then y (csc x)
2 cot x csc x . Solving y 0 cos x 1
2
x . For 0 x we have y 0 and y 0 when x . Therefore, at x there is a maximum
4
value of y 1.
4
4
4
(b)
The graph confirms the findings in (a).
70. (a) If y tan x 3 cot x where 0 x 2x , then y sec 2 x 3csc 2 x. Solving y 0 tan x 3 x 3 ,
but 3 is not in the domain. Also, y 2sec 2 x tan x 6 csc 2 x cot x 0 for all 0 x 2 . Therefore at
x 3 there is a minimum value of y 2 3 .
Copyright 2018 Pearson Education, Inc.
252
Chapter 4 Applications of Derivatives
(b)
The graph confirms the findings in (a).
2
2
71. (a) The square of the distance is D( x) x 32 x 0 x 2 2 x 94 , so D( x) 2 x 2 and the critical
point occurs at x 1. Since D( x) 0 for x 1 and D( x) 0 for x 1, the critical point corresponds to the
minimum distance. The minimum distance is D(1)
(b)
The minimum distance is from the point
5
.
2
32 , 0 to the point (1, 1) on the graph of y
x, and this occurs at
the value x 1 where D( x), the distance squared, has its minimum value.
72. (a)
Calculus Method:
The square of the distance from the point 1, 3 to x, 16 x 2 is given by
2
D( x) ( x 1) 2 16 x 2 3 x 2 2 x 1 16 x 2 2 48 3 x 2 3 2 x 20 2 48 3 x 2 .
2
(
6
x) 2 6 x 2 . Solving D( x) 0 we have:
Then D( x) 2 12
2
483 x
2
483 x
6 x 2 48 3x 36 x 4(48 3 x ) 9 x 2 48 3 x 2 12 x 2 48 x 2 We discard x 2 as
an extraneous solution, leaving x 2. Since D ( x ) 0 for 4 x 2 and D ( x ) 0 for 2 x 4, the critical
point corresponds to the minimum distance. The minimum distance is D(2) 2 .
Geometry Method:
The semicircle is centered at the origin and has radius 4. The distance from the origin to 1, 3 is
12
3
2
2
2
2. The shortest distance from the point to the semicircle is the distance along the radius
containing the point 1, 3 . That distance is 4 2 2.
Copyright 2018 Pearson Education, Inc.
Section 4.6 Newton’s Method
253
(b)
The minimum distance is from the point 1, 3 to the point 2, 2 3 on the graph of y 16 x 2 , and
this occurs at the value x 2 where D ( x), the distance squared, has its minimum value.
4.6
1.
NEWTON’S METHOD
2
x x 1
y x 2 x 1 y 2 x 1 xn 1 xn n2 x n 1 ; x0 1 x1 1 12111 23 x2 23
n
6 9
x2 23 412
9
2
3
4 2 1
9 3
4 1
3
1 13 .61905; x 1 x 1 111 2 x 2 4 2 1 5 1.66667
21
2
0
1
3
21
4 1
2 1
2. y x3 3x 1 y 3 x 2 3 xn 1 xn
xn3 3 xn 1
3 xn2 3
; x0 0 x1 0 13 13 x2 13
1 11
27
1 3
3
1 29 0.32222
13 90
90
3. y x 4 x 3 y 4 x3 1 xn 1 xn
xn4 xn 3
4 xn3 1
; x0 1 x1 1 14113
6
5
x2 65
1296 6
3
625 5
864
1
125
750 1875 6 171 5763 1.16542; x 1 x 1 113 2 x 2 16 2 3
65 1296
0
1
2
4320 625
5 4945 4945
41
32 1
51
11
2 31 31 1.64516
2 x x2 1
4. y 2 x x 2 1 y 2 2 x xn 1 xn 2n 2 xn ; x0 0 x1 0 02001 12 x2 12
n
1 5 .41667; x 2 x 2 4 41
12 12
0
1
12
2 4
29 2.41667
12
5. y x 4 2 y 4 x3 xn 1 xn
113
54 2000
2500 113
2000
2387
2000
7.
625512
2000
54
113
2000
4 xn3
5 25
1
4
2 5
xn4 2
4 xn3
625
2
256
125
16
512
54 625
2000
; x0 1 x1 1 142 1 14 54 x2 54
625
2
256
125
16
1.1935
xn3 xn 3
3 xn2 1
2 xn3 3
3 xn2 1
; x0 1 x1 2313 54 1.25
54 3 221 1.214
2
182
3 54 1
2
4 5 1
52 2025
12
2 12
; x0 1 x1 1 142 54 x2 54
y x3 x 3 y 3x 2 1 xn1 xn
x2
x2 52
1.1935
6. From Exercise 5, xn 1 xn
54
xn4 2
5
2
3
Copyright 2018 Pearson Education, Inc.
1 14 1
2 1
254
Chapter 4 Applications of Derivatives
x0 0 x1 0 x2 x1 xn approaches as n .
x0 1 x1 is undefined since f (1) 0.
8. (a)
(b)
(c)
(d)
(e)
x0 2 2 x1 3 x1 x2 3 xn approaches 3 as n .
x0 4 2 x1 3 3 x2 x0 xn approaches 3 as n .
x0 5.5 x1 5.5 x2 x1 xn approaches as n .
f (x )
9. f ( x0 ) 0 and f ( x0 ) 0 xn 1 xn f ( xn ) gives x1 x0 x2 x0 xn x0 for all n 0. That is all, of
n
the approximations in Newton’s method will be the root of f ( x ) 0.
10. It does matter. If you start too far away from x 2 , the calculated values may approach some other root.
Starting with x0 0.5, for instance, leads to x 2 as the root, not x 2 .
f (x )
f (h)
11. If x0 h 0 x1 x0 f ( x0 ) h f ( h)
0
h
h
1
h
h 2 h h;
2 h
f ( h)
f (x )
if x0 h 0 x1 x0 f ( x0 ) h f ( h)
0
h
12.
h
1
2 h
h
h 2 h h.
f ( x) x1/3 f ( x)
xn 1 xn
x1/3
n
xn2/3
1
3
13 x2/3
2 xn ; x0 1
x1 2, x2 4, x3 8, and x4 16 and so
forth. Since xn 2 xn 1 we may conclude that
n xn .
13. i)
ii)
iii)
iv)
is equivalent to solving x3 3 x 1 0 .
is equivalent to solving x3 3 x 1 0 .
is equivalent to solving x3 3 x 1 0 .
is equivalent to solving x3 3 x 1 0 .
All four equations are equivalent.
14. f ( x) x 1 0.5sin x f ( x) 1 0.5cos x xn 1 xn
15. f ( x) tan x 2 x f ( x) sec2 x 2 xn 1 xn
xn 10.5sin xn
10.5cos xn
tan( xn ) 2 xn
sec2 xn
; if x0 1.5, then x1 1.49870
; x0 1 x1 1.2920445
x2 1.155327774 x16 x17 1.165561185
16. f ( x ) x 4 2 x3 x 2 2 x 2 f ( x) 4 x3 6 x 2 2 x 2 xn 1 xn
xn4 2 xn3 xn2 2 xn 2
x4 0.630115396; if x0 2.5, then x4 2.57327196
Copyright 2018 Pearson Education, Inc.
4 xn3 6 xn2 2 xn 2
; if x0 0.5, then
Section 4.6 Newton’s Method
255
17. (a) The graph of f ( x) sin 3 x 0.99 x 2 in the
window 2 x 2, 2 y 3 suggests three
roots. However, when you zoom in on the
x-axis near x 1.2, you can see that the graph
lies above the axis there. There are only two
roots, one near x 1, the other near x 0.4.
(b) f ( x) sin 3 x 0.99 x 2
f ( x) 3cos 3 x 2 x
xn 1 xn
sin 3 xn 0.99 xn2
3cos 3 xn 2 xn
and the solutions are approximately
0.35003501505249 and –1.0261731615301
18. (a) Yes, three times as indicted by the graphs
(b) f ( x) cos 3 x x f ( x) 3sin 3 x 1
cos(3 x ) x
xn 1 xn 3sin(3nx ) n1 ; at approximately
n
0.979367, 0.887726, and 0.39004 we have
cos 3x x
19. f ( x) 2 x 4 4 x 2 1 f ( x) 8 x3 8 x xn 1 xn
2 xn4 4 xn2 1
8 xn3 8 xn
; if x0 2, then x6 1.30656296; if
x0 0.5, then x3 0.5411961; the roots are approximately 0.5411961 and 1.30656296 because f ( x ) is an
even function.
20. f ( x) tan x f ( x) sec2 x xn 1 xn
approximate to be 3.14159.
tan( xn )
sec2 ( xn )
; x0 3 x1 3.13971 x2 3.14159 and we
21. From the graph we let x0 0.5 and f ( x) cos x 2 x
cos( x ) 2 x
xn 1 xn sin(nx ) 2n x1 .45063
n
x2 .45018 at x 0.45 we have cos x 2 x.
Copyright 2018 Pearson Education, Inc.
256
Chapter 4 Applications of Derivatives
22. From the graph we let x0 0.7 and
x cos( x )
f ( x) cos x x xn 1 xn 1nsin( x n)
n
x1 .73944 x2 .73908 at x 0.74
we have cos x x.
1 is the solution of x 2 ( x 1) 1
x
x
x3 x 2 1x f ( x) 3 x 2 2 x 12 . Let x0
x
23. The x-coordinate of the point of intersection of y x 2 ( x 1) and y
x3 x 2 1x 0 The x-coordinate is the root of f ( x)
xn 1 xn
xn3 xn2 x1
n
3 xn2 2 xn 12
xn
1
x1 0.83333 x2 0.81924 x3 0.81917 x7 0.81917 r 0.8192
24. The x-coordinate of the point of intersection of y x and y 3 x 2 is the solution of x 3 x 2
x 3 x 2 0 The x-coordinate is the root of f ( x ) x 3 x 2 f ( x ) 1 2 x . Let x0 1
xn 1 xn
xn 3 xn 2
1
2 xn
2 xn
2 x
x1 1.4 x2 1.35556 x3 1.35498 x7 1.35498 r 1.3550
25. If f ( x) x3 2 x 4, then f (1) 1 0 and f (2) 8 0 by the Intermediate Value Theorem the equation
x3 2 x 4 0 has a solution between 1 and 2. Consequently, f ( x) 3x 2 2 and xn 1 xn
xn3 2 xn 4
3 xn2 2
. Then
x0 1 x1 1.2 x2 1.17975 x3 1.179509 x4 1.1795090 the root is approximately 1.17951.
26. We wish to solve 8 x 4 14 x3 9 x 2 11x 1 0. Let f x 8 x 4 14 x3 9 x 2 11x 1, then
f ( x) 32 x3 42 x 2 18 x 11 xn 1 xn
x0
32 xn3 42 xn2 18 xn 11
.
approximation of corresponding root
–1.0
0.1
0.6
2.0
27.
8 xn4 14 xn3 9 xn2 11xn 1
–0.976823589
0.100363332
0.642746671
1.983713587
f (x )
x3 x
f ( x) 4 x 4 4 x 2 f ( x) 16 x3 8 x xi 1 xi f ( xi ) xi i 2 i . Iterations are performed using the
4 xi 2
i
procedure in this section.
(a) For x0 2 or x0 0.8, xi 1 as i gets large.
(b) For x0 0.5 or x0 0.25, xi 0 as i gets large.
(c) For x0 0.8 or x0 2, xi 1 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 x0
between x0
21
7
21
7
or x0
or x0
21
7
21
,
7
Newton’s method does not converge. The values of xi alternate
as i increases.
Copyright 2018 Pearson Education, Inc.
Section 4.7 Antiderivatives
257
28. (a) The distance can be represented by
D( x) ( x 2) 2 x 2 12
2
, where x 0. The distance
D( x) is minimized when f ( x) ( x 2) 2 x 2 12
minimized. If f ( x) ( x 2) 2 x 2 12
2
2
is
, then
f ( x) 4( x3 x 1) and f ( x) 4 (3 x 2 1) 0. Now
f ( x) 0 x3 x 1 0 x( x 2 1) 1 x 21 .
x 1
(b) Let g ( x)
1 x
x 2 1
2
( x 1)
1
2
x g ( x) ( x 1)
2
(2 x) 1
2 x
( x 2 1) 2
1 xn 1 xn
1
2 xn
xn 1
;
2
x
n
2 2
xn 1 1
x0 1 x4 0.68233 to five decimal places.
( xn 1)40
29. f ( x) ( x 1) 40 f ( x) 40( x 1)39 xn 1 xn
40( xn 1)39
39 xn 1
. With x0
40
2, our computer gave
x87 x88 x89 x200 1.11051, coming within 0.11051 of the root x 1.
30. Since s r 3 r 3r . Bisect the angle to obtain a right triangle with hypotenuse r and opposite side
of length 1. Then sin 2 1r sin 2r 1r sin
23r 1r sin 23r 1r 0. Thus the solution r is a root of
sin
cos 23r 1 ; r0 1 rn 1 rn
r1 1.00280
r
cos
3
f ( r ) sin 23r 1r f ( r ) 32
2r
3
2 rn
2
3
2 rn2
1
rn
3
2 rn
1
rn2
3
r2 1.00282 r3 1.00282 r 1.0028 1.00282
2.9916
4.7
ANTIDERIVATIVES
1. (a) x 2
(b)
x3
3
(c)
x3
3
x2 x
2. (a) 3x 2
(b)
x8
8
(c)
x8
8
3x 2 8 x
3. (a) x 3
(b)
x3
4. (a) x 2
(b)
x4 x3
(c)
3
2
3
(c) x3 x 2 3 x
3
x 2
2
2
x2 x
5. (a)
1
x
(b)
5
x
(c) 2 x 5x
6. (a)
1
x2
(b)
1
4x 2
(c)
x4
4
x3
(b)
x
(c)
2
3
8. (a) x 4/3
(b)
1 x 2/3
2
(c)
3 x 4/3
4
9. (a) x 2/3
(b)
x1/3
(c) x 1/3
7. (a)
Copyright 2018 Pearson Education, Inc.
12
2x
x3 2 x
32 x 2/3
258
Chapter 4 Applications of Derivatives
10. (a)
x1 2
(b)
x 1 2
(c)
11. (a) cos ( x )
(b)
3cos x
(c)
12. (a) sin ( x)
(b)
sin 2x
(c)
1 tan x
2
(b)
2 tan 3x
14. (a) cot x
(b)
cot 32x
(c) x 4 cot (2 x)
15. (a) csc x
(b)
1 csc(5 x )
5
(c) 2 csc 2x
16. (a) sec x
(b)
4 sec(3 x )
3
(c) 2 sec 2x
13. (a)
2
17.
( x 1) dx x2
19.
3t
21.
(2 x
23.
x12 x
24.
15 x23 2 x dx 15 2 x
25.
x
27.
28.
xC
18.
(5 6 x) dx 5 x 3x
t2 4t
3
5 x 7)dx 12 x 4 52 x 2 7 x C
22.
(1 x
1/3
13 dx
x
2
2
3
dt
t3
6
t4 C
3x5 )dx x 13 x3 12 x 6 C
2
2
2 x dx 15 x 2x2 22x C 5x 12 x 2 C
x
26.
x
5/4
dx
x 1/ 4
14
C 44 C
x
3/ 2
4/3
x 3 x dx x1/2 x1/3 dx x 3 x 4 C 23 x3/2 34 x 4/3 C
3
2
x
2
2 dx 12 x1/2 2 x 1/2 dx
x
29.
8 y y1/2 4 dy 8 y 2 y
30.
17 y5/1 4 dy 17 y
31.
2 x 1 x
32.
x
3
3
1/4
5/4
1 x3/ 2
2 3
dy 82y
2
2
1/ 4
( x 1) dx x 2 x 3 dx
x 1
1
2
x1/ 2
1/2
1 3/2
2 1 C 3 x 4x C
2
y3/ 4
2 3 C 4 y 2 83 y3/4 C
4
dy 17 y y
dx 2 x 2 x2 dx 22x
1
4
C
y
7
2 sin 2x sin x
C
2/3
dx x 2 C 32 x 2/3 C
3
2
2
1
3
3
x 2 13 dx x1 x3 13 x C 1x x3 3x C
3
cos (3 x)
32x
20.
2
2t dt t 3 t4 C
2
cos ( x )
(c) 23 tan
2
x 3 2
1/4 4 C
y
1
2 x1 C x 2 2x C
2
x2 C 1x 1 2 C
2x
Copyright 2018 Pearson Education, Inc.
Section 4.7 Antiderivatives
33.
t t t
t2
34.
4 t
t3
3/ 2
1/ 2
1/ 2
1/ 2
dt t 2 t 2 dt t 1/2 t 3/2 dt t 1 t 1 C 2 t 2 C
t
t
t
2
2
dt
4
t3
1/ 2
t 3 dt 4t 3 t 5/2 dt 4
t
35. 2 cos t dt 2sin t C
t 2
2
t 3/ 2
32
2
2
C t 2 3t 3/ 2 C
36. 5sin t dt 5 cos t C
37.
7 sin 3 d 21cos 3 C
39.
3csc
41.
csc2cot d 12 csc C
43.
(4sec x tan x 2sec
44.
12 (csc
45.
(sin 2 x csc
47.
1cos2 4t dt 12 12 cos 4t dt 12 t 12 sin44t C 2t sin84t C
48.
1cos2 6t dt 12 12 cos 6t dt 12 t 12 sin66t C 2t sin126t C
49.
3x
51.
(1 tan
52.
(2 tan
53.
cot
54.
(1 cot
55.
cos (tan sec ) d (sin 1) d cos C
56.
csccsc
sin
d csccsc
sin
57.
d (7 x 2)
dx 28
4(7 x 2)3 (7)
(7 x 2)3
C
28
3
2
2
2
x dx 3cot x C
2
38.
3cos 5 d 53 sin 5 C
40.
sec3 x dx tan3 x C
42.
52 sec tan d 52 sec C
46.
(2 cos 2 x 3sin 3x) dx sin 2 x cos 3x C
2
x) dx 4sec x 2 tan x C
x csc x cot x ) dx 12 cot x 12 csc x C
2
dx 3 x
2
x) dx 12 cos 2 x cot x C
3 1
3 1
C
50.
x
dx x
2 1
2
2
C
) d sec2 d tan C
2
) d (1 1 tan 2 ) d (1 sec2 ) d tan C
x dx (csc2 x 1) dx cot x x C
2
x) dx (1 (csc2 x 1)) dx (2 csc2 x) dx 2 x cot x C
4
2
sin d
sin
1sin1 d c o 1s d sec d tan C
2
2
Copyright 2018 Pearson Education, Inc.
259
260
Chapter 4 Applications of Derivatives
58.
d (3 x 5)
3
dx
59.
d 1 tan (5 x 1) C
dx 5
60.
d
dx
61.
d 1 C
dx x 1
1
(3 x 5)2 (3)
2
C
(3 x 5)
3
15 (sec2 (5x 1))(5) sec2 (5x 1)
3cot x31 C 3 csc2 x31 13 csc2 x31
(1)(1)( x 1)2 ( x11)
63. (a) Wrong:
d
x C
dx x 1
62.
2
d x 2 sin x C 2 x sin x x 2 cos x x sin x x 2
2
dx 2
2
2
d ( x cos x C ) cos x x sin x x sin x
dx
x (1)
1
( x(1)(1)
x 1)
( x 1)
2
2
cos x x sin x
(b) Wrong:
d ( x cos x sin x C ) cos x x sin x cos x x sin x
(c) Right: dx
(b) Right:
(c) Right:
sec3 C 3sec 2 (sec tan ) sec3 tan tan sec 2
3
3
d 1 tan 2 C 1 (2 tan ) sec 2 tan sec 2
2
d 2
d 1 sec 2 C 1 (2sec ) sec tan tan sec 2
d 2
2
64. (a) Wrong: dd
3(2 x 1)2 (2)
2(2 x 1) 2 (2 x 1)2
C
3
d ((2 x 1)3 C ) 3(2 x 1) 2 (2) 6(2 x 1) 2 3(2 x 1) 2
(b) Wrong: dx
65. (a) Wrong:
(c) Right:
3
d ((2 x 1)3
dx
66. (a) Wrong:
d ( x2
dx
C ) 6(2 x 1) 2
x C )1/2 12 ( x 2 x C )1/2 (2 x 1)
2 x 1
2x 1
2 x 2 x C
2 x 1 2 x 1
( x2 x)1/2 C 12 ( x2 x)1/2 (2x 1) 2 x x
3
d 1
3/2
1/2
3
d 1
Right: dx
3 2 x 1 C dx 3 (2 x 1) C 6 (2 x 1) (2)
(b) Wrong:
(c)
d (2 x 1)
3
dx
67. Right:
d
dx
d x 3
dx x 2
68. Wrong:
2
3
C 3
2
d sin( x )
dx
x
xx23
2 ( x 2)1 ( x 3)1
( x 2)
2
3
( x 3) 2
5
( x 2) ( x 2) 2
2
2x 1
15( x 3)2
( x 2) 4
xcos( x 2 )(2 x ) sin( x 2 )1 2 x 2 cos( x 2 ) sin( x 2 ) x cos( x 2 ) sin( x 2 )
C
x2
x2
x2
69. Graph (b), because
dy
dx
2 x y x 2 C. Then y (1) 4 C 3.
70. Graph (b), because
dy
dx
x y 12 x 2 C. Then y (1) 1 C 32 .
71.
dy
dx
2 x 7 y x 2 7 x C ; at x 2 and y 0 we have 0 22 7(2) C C 10 y x 2 7 x 10
72.
dy
dx
2
2
2
10 x y 10 x x2 C ; at x 0 and y 1 we have 1 10(0) 02 C C 1 y 10 x x2 1
Copyright 2018 Pearson Education, Inc.
Section 4.7 Antiderivatives
73.
dy
dx
2
2
12 x x 2 x y x 1 x2 C ; at x 2 and y 1 we have 1 21 22 C C 12
dy
dx
9 x 2 4 x 5 y 3 x3 2 x 2 5 x C ; at x 1 and y 0 we have 0 3(1)3 2(1) 2 5(1) C
dy
dx
3 x 2/3 y
261
x
2
2
y x 1 x2 12 or y 1x x2 12
74.
C 10 y 3 x3 2 x 2 5 x 10
75.
3 x1/3
1
3
C 9 y 9 x1/3 C ; at x 1 and y 5 we have 5 9(1)1/3 C C 4
y 9 x1/3 4
76.
dy
dx
1 12 x 1/2 y x1/2 C ; at x 4 and y 0 we have 0 41/2 C C 2 y x1/2 2
2 x
77.
ds
dt
1 cos t s t sin t C ; at t 0 and s 4 we have 4 0 sin 0 C C 4 s t sin t 4
78.
ds
dt
cos t sin t s sin t cos t C ; at t and s 1 we have 1 sin cos C C 0
s sin t cos t
79.
dr
d
sin r cos ( ) C ; at r 0 and 0 we have 0 cos ( 0) C C 1 r cos ( ) 1
80.
dr
d
cos r 1 sin ( ) C ; at r 1 and 0 we have 1 1 sin( 0) C C 1 r 1 sin ( ) 1
81.
dv
dt
12 sec t tan t v 12 sec t C ; at v 1 and t 0 we have 1 12 sec (0) C C 12 v 12 sec t 12
82.
dv
dt
8t csc2 t v 4t 2 cot t C ; at v 7 and t 2 we have 7 4 2
2
v 4t cot t 7
83.
d2y
dx 2
2
cot 2 C C 7 2
2
dy
dy
2 6 x dx 2 x 3 x 2 C1; at dx 4 and x 0 we have 4 2(0) 3(0)2 C1 C1 4
dy
dx
2 x 3 x 2 4 y x 2 x3 4 x C2 ; at y 1 and x 0 we have 1 02 03 4(0) C2 C2 1
y x 2 x3 4 x 1
d2y
dy
dy
dy
84.
0 dx C1; at dx 2 and x 0 we have C1 2 dx 2 y 2 x C2 ; at y 0 and x 0 we have
0 2(0) C2 C2 0 y 2 x
85.
d 2r
dt 2
23 2t 3 dr
t 2 C1; at dr
1 and t 1 we have 1 (1) 2 C1 C1 2
dt
dt
d 2s
dt 2
dx 2
t
dr
dt
t 2 2
r t 1 2t C2 ; at r 1 and t 1 we have 1 11 2(1) C2 C2 2 r t 1 2t 2 or r 1t 2t 2
86.
3t
8
ds
dt
3t 2
16
2
3(4)
t 2 s t 3 C ; at
C1; at ds
3 and t 4 we have 3 16 C1 C1 0 ds
316
2
dt
16
dt
3
3
4 C C 0 s t
s 4 and t 4 we have 4 16
2
2
16
Copyright 2018 Pearson Education, Inc.
262
87.
Chapter 4 Applications of Derivatives
d3y
dx
3
6
dy
d2y
dx
2
2
6 x C1; at
dx 3 x 8 x C2 ; at
dy
dx
d2y
dx
2
8 and x 0 we have 8 6(0) C1 C1 8
0 and x 0 we have 0 3(0) 2 8(0) C2 C2 0
d2y
6x 8
dx 2
dy
3x2 8 x
dx
3
2
y x3 4 x 2 C3 ; at y 5 and x 0 we have 5 03 4(0)2 C3 C3 5 y x 4 x 5
88.
d 3
dt 3
12
2
2
2
0 d 2 C1; at d 2 2 and t 0 we have d 2 2 ddt 2t C2 ; at ddt 12 and t 0 we have
dt
dt
dt
2(0) C2 C2 12 ddt 2t 12 t 2 12 t C3 ; at 2 and t 0 we have
2 02 12 (0) C3 C3 2 t 2 12 t 2
89. y (4) sin t cos t y cos t sin t C1; at y 7 and t 0 we have 7 cos (0) sin (0) C1 C1 6
y cos t sin t 6 y sin t cos t 6t C2 ; at y 1 and t 0 we have
1 sin (0) cos (0) 6(0) C2 C2 0 y sin t cos t 6t y cos t sin t 3t 2 C3 ; at
y 1 and t 0 we have 1 cos (0) sin (0) 3(0)2 C3 C3 0 y cos t sin t 3t 2
y sin t cos t t 3 C4 ; at y 0 and t 0 we have 0 sin (0) cos (0) 03 C4
C4 1 y sin t cos t t 3 1
90.
y (4) cos x 8sin(2 x) y sin x 4 cos (2 x) C1; at y 0 and x 0 we have
0 sin(0) 4 cos(2(0)) C1 C1 4 y sin x 4 cos(2 x) 4 y cos x 2sin(2 x) 4 x C2 ; at
y 1 and x 0 we have 1 cos(0) 2sin(2(0)) 4(0) C2 C2 0 y cos x 2sin(2 x) 4 x
y sin x cos(2 x ) 2 x 2 C3 ; at y 1 and x 0 we have 1 sin(0) cos(2(0)) 2(0) 2 C3 C3 0
y sin x cos(2 x) 2 x 2 y cos x 1 sin(2 x) 2 x3 C4 ; at y 3 and x 0 we have
2
3 cos(0)
1 sin(2(0)) 2 (0)3
2
3
3
C4 C4 4 y cos x 12 sin(2 x) 23 x3 4
91. m y 3 x 3 x1/2 y 2 x3/2 C ; at (9, 4) we have 4 2(9)3/2 C C 50 y 2 x3/2 50
92. (a)
d2y
dx 2
dy
dy
6 x dx 3 x 2 C1; at y 0 and x = 0 we have 0 3(0) 2 C1 C1 0 dx 3 x 2
y x3 C2 ; at y = 1 and x = 0 we have C2 1 y x3 1
(b) One, because any other possible function would differ from x3 1 by a constant that must be zero because
of the initial conditions
93.
Copyright 2018 Pearson Education, Inc.
Section 4.7 Antiderivatives
263
94.
95.
96.
97.
dy
dx
1 43 x1/3 y 1 43 x1/3 dx x x 4/3 C ; at (1, 0.5) on the curve we have
0.5 1 14/3 C C 0.5 x x 4/3 12
98.
99.
dy
dx
2
x 1 y ( x 1)dx x2 x C ; at (1, 1) on the curve we have
1
( 1)2
2
dy
dx
sin x cos x y (sin x cos x)dx cos x sin x C ; at (, 1) on the curve we have
dy
dx
1 sin x 12 x 1/2 sin x y 12 x 1/2 sin x dx x1/2 cos x C ; at (1, 2) on the curve
2
(1) C C 12 y x2 x 12
1 = cos() sin() + C C = 2 y = cos x sin x 2
100.
2 x
we have 2 11/2 cos (1) C C 0 y x cos x
101. (a)
ds
dt
9.8t 3 s 4.9t 2 3t C ; (i) at s = 5 and t = 0 we have C = 5 s 4.9t 2 3t 5;
displacement = s(3) s(1) = ((4.9)(9) 9 + 5) (4.9 3 + 5) = 33.2 units; (ii) at s = 2 and t = 0 we have
Copyright 2018 Pearson Education, Inc.
264
Chapter 4 Applications of Derivatives
C = 2 s 4.9t 2 3t 2; displacement = s(3) s(1) = ((4.9)(9) 9 2) (4.9 3 2) = 33.2 units;
(iii) at s s0 and t = 0 we have C s0 s 4.9t 2 3t s0 ;
displacement s (3) s (1) ((4.9)(9) 9 s0 ) (4.9 3 s0 ) 33.2 units
(b) True. Given an antiderivative f(t) of the velocity function, we know that the body’s position function is
s = f(t) + C for some constant C. Therefore, the displacement from t = a to t = b is
(f(b) + C) (f(a) + C) = 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.
102. a(t ) v (t ) 20 v(t) = 20t + C; at (0, 0) we have C = 0 v(t) = 20t. When t = 60, then
v(60) = 20(60) = 1200 m/sec.
103. Step 1:
d 2s
dt 2
ds
dt
k
kt C1 ; at ds
88 and t = 0 we have
dt
2
2
C1 88 ds
kt 88 s k t2 88t C2 ; at s = 0 and t = 0 we have C2 0 s kt2 88t
dt
Step 2:
ds
dt
0 0 kt 88 t 88
k
Step 3: 242
104.
d 2s
dt 2
k
88k
2
2
(88) 2 (88)2
(88)2
88 88
242
242
k 16
2k
2k
k
k
k ds
k dt kt C ; at ds
44 when t = 0 we have
dt
dt
44 = k(0) + C C = 44
ds
dt
2
kt 44 s kt2 44t C1 ; at s = 0 when t = 0 we have
2
2
k (0)
0 kt 44 0 t 44
and
0 2 44(0) C1 C1 0 s kt2 44t. Then ds
dt
k
s
44
k
k
44k
2
2
44 44
45 968
1936
45 k 968
21.5 ft 2 .
k
k
k
45
sec
105. (a) v a dt (15t1/2 3t 1/2 )dt 10t 3/2 6t1/2 C ;
ds (1)
dt
4 4 10(1)3/2 6(1)1/2 C C 0 v 10t 3/2 6t1/2
(b) s v dt (10t 3/2 6t1/2 )dt 4t 5/2 4t 3/2 C ;
s (1) 0 0 4(1)5/2 4(1)3/2 C C 0 s 4t 5/2 4t 3/2
106.
d 2s
dt 2
5.2 ds
5.2t C1 ; at ds
0 and t = 0 we have C1 0 ds
5.2t s 2.6t 2 C2 ; at s = 4
dt
dt
dt
and t = 0 we have C2 4 s 2.6t 2 4. Then s 0 0 2.6t 2 4 t
107.
d 2s
dt 2
4
2.6
1.24 sec, since t > 0
2
a ds
a dt at C ; ds
v0 when t = 0 C v0 ds
at v0 s at2 v0t C1; s s0
dt
dt
dt
when t = 0 s0
a (0)2
2
2
v0 (0) C1 C1 s0 s at2 v0t s0
108. The appropriate initial value problem is: Differential Equation:
d 2s
dt 2
g with Initial Conditions: ds
v0 and
dt
s s0 when t = 0. Thus ds
g dt gt C1; ds
(0) v0 v0 ( g )(0) C1 C1 v0 ds
gt v0 .
dt
dt
dt
Copyright 2018 Pearson Education, Inc.
Section 4.7 Antiderivatives
265
Thus s ( gt v0 )dt 12 gt 2 v0t C2 ; s (0) s0 12 ( g )(0)2 v0 (0) C2 C2 s0
Thus s 12 gt 2 v0t s0 .
109 (a)
(c)
(e)
(f)
(b) g ( x) dx x 2 C1 x C
f ( x) dx 1 x C1 x C
(d) g ( x) dx ( x 2) C1 x C
f ( x) dx 1 x C1 x C
[ f ( x) g ( x)]dx 1 x ( x 2) C1 x x C
[ f ( x) g ( x)]dx 1 x ( x 2) C1 x x C
110. 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) G ( x) C for all x. In particular, F ( x0 ) G ( x0 ) C , so C F ( x0 ) G ( x0 ) 0. Hence F ( x) G ( x)
for all x.
111114 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 4.6 #111" );
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 111–114.
Clear x, y, yprime
yprime[x_] Cos[x]2 Sin[x];
initxvalue π; 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 114 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).
Copyright 2018 Pearson Education, Inc.
266
Chapter 4 Applications of Derivatives
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 {RGBColor[1,0,0], RGBColor[0,0,1]}]
CHAPTER 4
PRACTICE EXERCISES
1. Minimum value is 1 at x 2.
2. To find the exact values, note that y 3x 2 2,
which is zero when x 23 . Local maximum at
at
2, 4 4 6
3
9
2, 4 4 6
3
9
(0.816, 5.089); local minimum
(0.816, 2.911)
3. To find the exact values, note that y 3 x 2 2 x 8
(3x 4)( x 2), which is zero when x 2 or
x 43 . Local maximum at (2, 17); local minimum at
43 , 2741
4. Note that y 5 x 2 ( x 5)( x 3), which is zero at
x 0, x 3, and x 5. Local maximum at (3, 108);
local minimum at (5, 0); (0, 0) is neither a maximum
nor a minimum.
Copyright 2018 Pearson Education, Inc.
Chapter 4 Practice Exercises
5. Minimum value is 0 when x 1 or x 1.
6. Note that y
x 2
,
x
which is zero at x 4 and is
undefined when x 0. Local maximum at (0, 0);
absolute minimum at (4, 4)
7. The actual graph of the function has asymptotes
at x 1, so there are no extrema near these values.
(This is an example of grapher failure.) There is
a local minimum at (0, 1).
8. Maximum value is 2 at x 1;
minimum value is 0 at x 1 and x 3.
9. Maximum value is 12 at x 1;
minimum value is 12 at x 1.
Copyright 2018 Pearson Education, Inc.
267
268
Chapter 4 Applications of Derivatives
10. Maximum value is 12 at x 0;
minimum value is 12 as x 2.
11. No, since f ( x) x3 2 x tan x f ( x) 3x 2 2 sec2 x 0 f ( x) is always increasing on its domain
12. No, since g ( x) csc x 2 cot x g ( x) csc x cot x 2 csc 2 x
g ( x) is always decreasing on its domain
cos x
sin 2 x
22
sin x
1 (cos x 2) 0
sin 2 x
13. No absolute minimum because lim (7 x)(11 3 x)1/3 . Next f ( x) (11 3x)1/3 (7 x)(11 3x) 2/3
(113 x ) (7 x )
(113 x )2/3
4(1 x )
(113 x ) 2/3
x
x 1 and x 11
are critical points. Since f 0 if x 1 and f 0
3
if x 1, f (1) 16 is the absolute maximum.
14. f ( x)
ax b
x 2 1
f ( x)
a ( x 2 1) 2 x ( ax b )
( x 2 1)2
require also that f (3) 1. Thus 1
f ( x)
2(3 x 1)( x 3)
( x 2 1) 2
3a b
8
( ax 2 2bx a )
( x 2 1)2
1 (9a 6b a ) 0 5a 3b 0. We
; f (3) 0 64
3a b 8. Solving both equations yields a 6 and b 10. Now,
so that f | | | | . Thus f changes sign at x 3 from
1
1/3
1
3
positive to negative so there is a local maximum at x 3 which has a value f (3) 1.
15. Yes, because at each point of [0, 1) except x 0, the function’s value is a local minimum value as well as a
local maximum value. At x 0 the function’s value, 0, is not a local minimum value because each open
interval around x 0 on the x-axis contains points to the left of 0 where f equals 1.
16. (a) The first derivative of the function f ( x) x3 is zero at x 0 even though f has no local extreme value at
x 0.
(b) Theorem 2 says only that if f is differentiable and f has a local extreme at x c then f (c) 0. It does not
assert the (false) reverse implication f (c) 0 f has a local extreme at x c.
17. 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 x b then the existence of absolute extrema is guaranteed on
that interval.
18. The absolute maximum is | 1| 1 and the absolute minimum is |0| 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 [ 1, 1), so there is nothing to contradict.
Copyright 2018 Pearson Education, Inc.
Chapter 4 Practice Exercises
269
19. (a) There appear to be local minima at x 1.75
and 1.8. Points of inflection are indicated at
approximately x 0 and x 1.
(b) f ( x) x 7 3x5 5 x 4 15 x 2 x 2 ( x 2 3)( x3 5). The pattern y | | | |
3
indicates a local maximum at x 3 5 and local minima at x 3.
3
0
5
3
(c)
20. (a) The graph does not indicate any local
extremum. Points of inflection are indicated
at approximately x 34 and x 1.
(b) f ( x) x 7 2 x 4 5 103 x 3 ( x3 2)( x 7 5). The pattern f )( | | indicates a
x
7
(c)
21.
3
0
7
5
3
2
local maximum at x 5 and a local minimum at x 2.
(a) g (t ) sin 2 t 3t g (t ) 2sin t cos t 3 sin(2t ) 3 g 0 g (t ) is always falling and hence must
decrease on every interval in its domain.
(b) One, since sin 2 t 3t 5 0 and sin 2 t 3t 5 have the same solutions: f (t ) sin 2 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 [3, 0].
Copyright 2018 Pearson Education, Inc.
270
Chapter 4 Applications of Derivatives
dy
22. (a) y tan d sec2 0 y tan is always rising on its domain y tan increases on every
interval in its domain
(b) The interval 4 , is not in the tangent’s domain because tan is undefined at 2 . Thus the tangent
need not increase on this interval.
23. (a) f ( x ) x 4 2 x 2 2 f ( x) 4 x3 4 x. Since f (0) 2 0, f (1) 1 0 and f ( x) 0 for 0 x 1, we
may conclude from the Intermediate Value Theorem that f ( x) has exactly one solution when 0 x 1.
(b) x 2
24. (a) y
2 4 8
2
x
x 1
0 x 2 3 1 and x 0 x .7320508076 .8555996772
y
1
( x 1)2
0, for all x in the domain of xx1 y xx1 is increasing in every interval in its
domain.
(b) y x3 2 x y 3 x 2 2 0 for all x the graph of y x3 2 x is always increasing and can never
have a local maximum or minimum
25. Let V (t ) represent the volume of the water in the reservoir at time t, in minutes, let V (0) a0 be the initial amount
and V (1440) a0 (1400)(43,560)(7.58) gallons be the amount of water contained in the reservoir after the rain,
where 24 hr 1440 min. Assume that V (t ) is continuous on [0, 1440] and differentiable on (0, 1440). The Mean
Value Theorem says that for some t0 in (0, 1440) we have V (t0 )
456,160,320 gal
1440 min
V (1400) V (0)
1440 0
a0 (1440)(43,560)(7.48) a0
1440
316, 778 gal/min. Therefore at t0 the reservoir’s volume was increasing at a rate in excess of
225,000 gal/min.
26. Yes, all differentiable functions g ( x) having 3 as a derivative differ by only a constant. Consequently, the
d (3 x ). Thus g ( x ) 3 x K , the same form as F ( x ).
difference 3 x g ( x) is a constant K because g ( x) 3 dx
x
x 1
d
x
dx x 1
27. No,
1 x11 xx1 differs from x11 by the constant 1. Both functions have the same derivative
( x 1) x (1)
d 1 .
1 2 dx
2
x 1
( x 1)
28. f ( x) g ( x)
2x
( x 2 1) 2
( x 1)
f ( x ) g ( x) C for some constant C the graphs differ by a vertical shift.
29. The global minimum value of 12 occurs at x 2.
30. (a) The function is increasing on the intervals [3, 2] and [1, 2].
(b) The function is decreasing on the intervals [2, 0) and (0, 1].
(c) The local maximum values occur only at x 2, and at x 2; local minimum values occur at x 3 and
at x 1 provided f is continuous at x 0.
31. (a) t 0, 6, 12
(b)
t 3, 9
(c)
6 t 12
(d)
0 t 6, 12 t 14
32. (a) t 4
(b)
at no time
(c)
0t 4
(d)
4t 8
Copyright 2018 Pearson Education, Inc.
Chapter 4 Practice Exercises
33.
34.
35.
36.
37.
38.
39.
40.
41.
42.
Copyright 2018 Pearson Education, Inc.
271
272
Chapter 4 Applications of Derivatives
43. (a) y 16 x 2 y | | the curve is rising on (4, 4), falling on (, 4) and (4, )
4
4
a local maximum at x 4 and a local minimum at x 4; y 2 x y | the curve is
0
concave up on (, 0), concave down on (0, ) a point of inflection at x 0
(b)
44. (a) y x 2 x 6 ( x 3)( x 2) y | | the curve is rising on (, 2) and (3, ),
2
3
falling on (2, 3) local maximum at x 2 and a local minimum at x 3; y 2 x 1 y |
concave up on
1 ,
2
, concave down on , a point of inflection at x
1
2
1/2
1
2
(b)
45. (a) y 6 x ( x 1)( x 2) 6 x3 6 x 2 12 x y | | | the graph is rising on ( 1, 0)
1
0
2
and (2, ), falling on (, 1) and (0, 2) a local maximum at x 0, local minima at x 1 and
x y | |
and , , concave down on , points of
x 2; y 18 x 2 12 x 12 6 (3 x 2 2 x 2) 6 x 13 7
the curve is concave up on , 13 7
inflection at x 13 7
1 7
3
1 7
3
1 7
3
1 7
3
1 7 1 7
3
3
(b)
46 . (a) y x 2 (6 4 x) 6 x 2 4 x3 y | | the curve is rising on ,
0
3/2
3
2
, falling on
32 , a local maximum at x 32 ; y 12 x 12 x2 12 x(1 x) y 0| 1| concave
up on (0, 1), concave down on ( , 0) and (1, ) points of inflection at x 0 and x 1
(b)
Copyright 2018 Pearson Education, Inc.
Chapter 4 Practice Exercises
273
47. (a) y x 4 2 x 2 x 2 ( x 2 2) y | | | the curve is rising on , 2 and
2
0
2
2, , falling on 2, 2 a local maximum at x 2 and a local minimum at x 2;
3
y 4 x 4 x 4 x( x 1)( x 1) y | | | concave up on (1, 0) and (1, ),
1
0
1
concave down on (, 1) and (0, 1) points of inflection at x 0 and x 1
(b)
48. (a) y 4 x 2 x 4 x 2 (4 x 2 ) y | | | the curve is rising on ( 2, 0) and (0, 2),
2
0
2
falling on ( , 2) and (2, ) a local maximum at x 2, a local minimum at x 2; y 8 x 4 x3
4 x (2 x 2 ) y | | | concave up on , 2 and 0, 2 , concave
down on 2, 0 and
(b)
2
0
2
2, points of inflection at x 0 and x 2
49. The values of the first derivative indicate that the curve is rising on (0, ) and falling on ( , 0). The slope of
the curve approaches as x 0 , and approaches as x 0 and x 1. The curve should therefore have a
cusp and local minimum at x 0, and a vertical tangent at x 1.
Copyright 2018 Pearson Education, Inc.
274
Chapter 4 Applications of Derivatives
12 and (1, ), and falling on (, 0) and
50. The values of the first derivative indicate that the curve is rising on 0,
12 , 1 . The derivative changes from positive to negative at x 12 , indicating a local maximum there. The slope
of the curve approaches as x 0 and x 1 , and approaches as x 0 and as x 1 , indicating
cusps and local minima at both x 0 and x 1.
51. The values of the first derivative indicate that the curve is always rising. The slope of the curve approaches
as x 0 and as x 1, indicating vertical tangents at both x 0 and x 1.
52. The graph of the first derivative indicates that the curve is rising on 0, 17 16 33 and
17 33
, , falling on
16
x 17 16 33 . The derivative
( , 0) and 17 16 33 , 17 16 33 a local maximum at x 17 16 33 , a local minimum at
approaches as x 0 and x 1, and approaches as x 0 , indicating a cusp and local minimum at
x 0 and a vertical tangent at x 1.
Copyright 2018 Pearson Education, Inc.
Chapter 4 Practice Exercises
53. y
x 1
x 3
55. y
x 2 1
x
57. y
59. y
54. y
2x
x 5
x 1x
56. y
x 2 x 1
x
x3 2
2x
2
x2 1x
58. y
x 4 1
x2
x 2 12
x2 4
x 2 3
1
60. y
x2
x 4
1
1 x 43
1
x 3
2
2
275
2 x10
5
x 1 1x
x
4
x2 4
61. (a) Maximize f ( x) x 36 x x1/2 (36 x)1/2 where 0 x 36
f ( x) 12 x 1/2 12 (36 x )1/2 (1)
36 x x
2 x 36 x
derivative fails to exist at 0 and 36; f (0) 6, and
f (36) 6 the numbers are 0 and 36
Copyright 2018 Pearson Education, Inc.
276
Chapter 4 Applications of Derivatives
(b) Maximize g ( x) x 36 x x1/2 (36 x)1/2 where 0 x 36 g ( x) 12 x 1/2 12 (36 x) 1/2 (1)
36 x x critical points at 0, 18 and 36; g (0) 6, g (18) 2 18 6 2 and g (36) 6 the numbers
2 x 36 x
are 18 and 18
62. (a) Maximize f ( x ) x (20 x) 20x1/2 x3/2 where 0 x 20 f ( x) 10 x 1/2 32 x1/2
20 3 x 0
2 x
x 0 and x 20
are critical points; f (0) f (20) 0 and f 20
20
20 20
40 20 the numbers
3
3
3
3
are
20
3
and
40 .
3
(b) Maximize g ( x) x 20 x x (20 x)1/2 where 0 x 20 g ( x)
2 20 x 1
2 20 x
3 3
0 20 x 12
x 79
. The critical points are x 79
and x 20. Since g 79
and g (20) 20, the numbers must be
81
4
4
4
4
79 and 1 .
4
4
63. A( x) 12 (2 x)(27 x 2 ) for 0 x 27
A( x) 3(3 x)(3 x) and A( x) 6 x. The
critical points are 3 and 3, but 3 is not in the
domain. Since A(3) 18 0 and A 27 0, the
maximum occurs at x 3 the largest area
is A(3) 54 sq units.
64. The volume is V x 2 h 32 h
area is S ( x) x 2 4 x
S ( x )
32
x2
2( x 4)( x 2 4 x 16)
x2
32 . The surface
x2
2 128
x x , where x 0
the critical points are 0
and 4, but 0 is not in the domain. Now
S (4) 2 256
0 at x 4 there is a minimum.
3
4
The dimensions 4 ft by 4 ft by 2 ft minimize the
surface area.
65. From the diagram we have
2
h2
2
r2
3
2
r 2 124h . The volume of the cylinder is
2
V r 2 h 124h h 4 (12h h3 ), where
0 h 2 3. Then V (h) 34 (2 h)(2 h) the
critical points are 2 and 2, but 2 is not in the
domain. At h 2 there is a maximum since
V (2) 3 0. The dimensions of the largest
cylinder are radius 2 and height 2.
66. From the diagram we have x radius and y height
12 2x and V ( x) 13 x 2 (12 2 x), where 0 x 6
V ( x) 2 x(4 x) and V (4) 8 . The critical
points are 0 and 4; V (0) V (6) 0 x 4 gives the
maximum. Thus the values of r 4 and h 4 yield
the largest volume for the smaller cone.
Copyright 2018 Pearson Education, Inc.
Chapter 4 Practice Exercises
277
40510x x , where p is the profit on grade B tires and 0 x 4. Thus
2p
P ( x)
( x 2 10 x 20) the critical points are 5 5 , 5, and 5 5 , but only 5 5 is in the
(5 x )
domain. Now P ( x) 0 for 0 x 5 5 and P ( x) 0 for 5 5 x 4 at x 5 5 there is a local
maximum. Also P(0) 8 p, P 5 5 4 p 5 5 11 p, and P(4) 8 p at x 5 5 there is an
absolute maximum. The maximum occurs when x 5 5 and y 2 5 5 , the units are hundreds of tires,
67. The profit P 2 px py 2 px p
2
i.e., x 276 tires and y 553 tires.
68. (a) The distance between the particles is | f (t )| where f (t ) cos t cos t 4 . Then,
f (t ) sin t sin t 4 . Solving f (t ) 0 graphically, we obtain t 1.178, t 4.320, and so on.
Alternatively, f (t ) 0 may be solved analytically as follows. f (t ) sin t 8 8 sin t 8 8
sin t 8 cos 8 cos t 8 sin 8 sin t 8 cos 8 cos t 8 sin 8 2sin 8 cos t 8 so the
critical points occur when cos t 8 0, or t 38 k . At each of these values, f (t ) cos 38 0.765
units, so the maximum distance between the particles is 0.765 units.
(b) Solving cos t cos t 4 graphically, we obtain t 2.749, t 5.890, and so on.
Alternatively, this problem can be solved analytically as follows.
cos t cos t 4
8 8
8 8
cos t 8 cos 8 sin t 8 sin 8 cos t 8 cos 8 sin t 8 sin 8
2 sin t 8 sin 8 0
sin t 8 0; t 78 k
cos t cos t
The particles collide when t 78 2.749. (Plus multiples of if they keep going.)
69. The dimensions will be x in. by 10 2x in. by 16 2x in., so V ( x) x(10 2 x)(16 2 x) 4 x3 52 x 2 160 x for
0 x 5. Then V ( x) 12 x 2 104 x 160 4( x 2)(3 x 20), so the critical point in the correct domain is x 2.
This critical point corresponds to the maximum possible volume because V ( x) 0 for 0 x 2 and V ( x) 0
for 2 x 5. 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.3
Copyright 2018 Pearson Education, Inc.
278
Chapter 4 Applications of Derivatives
Graphical support:
70. The length of the ladder is d1 d 2 8sec 6 csc .
We wish to maximize I ( ) 8sec 6 csc
I ( ) 8sec tan 6 csc cot . Then I ( ) 0
8sin 3 6 cos3 0 tan
3
6
2
d1 4 4 3 36 and d 2 3 36 4 3 36 the
length of the ladder is about 4 3 36
4 3 36
3/2
4 3 36
19.7 ft.
71. g ( x) 3x x3 4 g (2) 2 0 and g (3) 14 0 g ( x) 0 in the interval [2, 3] by the Intermediate
Value Theorem. Then g ( x) 3 3x 2 xn 1 xn
3 xn xn3 4
forth to x5 2.195823345.
33 xn2
; x0 2 x1 2.22 x2 2.196215, and so
72. g ( x) x 4 x3 75 g (3) 21 0 and g (4) 117 0 g ( x) 0 in the interval [3, 4] by the Intermediate
x 4 x3 75
Value Theorem. Then g ( x) 4 x3 3 x 2 xn 1 xn n 3 n 2 ; x0 3 x1 3.259259 x2 3.229050,
4 xn 3 xn
and so forth to x5 3.22857729.
3
73.
(x
74.
8t
75.
3
76.
4
2
5 x 7) dx x4 52x 7 x C
3
2
4
3
2
3
2
t2 t dt 84t t6 t2 C 2t 4 t6 t2 C
3/ 2
1
t 42 dt 3t1/2 4t 2 dt 3t 3 4t 1 C 2t 3/2 4t C
t
21 t t34 dt 12 t
1/2
2
3
1/ 2
3t 4 dt 12 t 1 (3t 3) C t 13 C
t
2
77. Our trial solution based on the chain rule is ( r 1 5) C. Differentiate the solution to check:
d
dr
1 C 1 . Thus
( r 5)
( r 5)2
(r dr5)
2
( r 1 5) C.
Copyright 2018 Pearson Education, Inc.
Chapter 4 Practice Exercises
78. Our trial solution based on the chain rule is
d
dr
6
3 2 C
. Thus
3
r 2
r 2
3
r 2
2
C. Differentiate the solution to check:
6 dr
r 2 3 r 3 2 2 C.
79. Our trial solution based on the chain rule is ( 2 1)3/2 C. Differentiate the solution to check:
d
d
( 2 1)3/2 C 3 2 1. Thus 3 2 1 d ( 2 1)3/2 C.
80. Our trial solution based on the chain rule is 7 2 C. Differentiate the solution to check:
d 7 2 C
. Thus
d 7 2 C .
d
2
2
7
7
81. Our trial solution based on the chain rule is 13 (1 x 4 )3/4 C. Differentiate the solution to check:
d 1 (1 x 4 )3/4
dx 3
C x3 (1 x 4 ) 1/4 . Thus
x
3
(1 x 4 ) 1/4 dx 13 (1 x 4 )3/4 C.
82. Our trial solution based on the chain rule is 85 (2 x)8/5 C. Differentiate the solution to check:
d 5 (2 x )8/5 C (2 x )3/5 . Thus (2 x )3/5 dx 5 (2 x )8/5 C .
dx 8
8
s C. Differentiate the solution to check:
83. Our trial solution based on the chain rule is 10 tan 10
d
ds
10 tan s C sec 2 s . Thus sec 2 s ds 10 tan s C.
10
10
10
10
84. Our trial solution based on the chain rule is 1 cot s C. Differentiate the solution to check:
d 1 cot s C csc 2 s. Thus csc 2 s ds 1 cot s C .
ds
85. Our trial solution based on the chain rule is 1 csc 2 C. Differentiate the solution to check:
2
d
d
1 csc 2 C csc 2 cot 2 . Thus
2
csc
2 cot 2 d 1 csc 2 C.
2
86. Our trial solution based on the chain rule is 3sec 3 C. Differentiate the solution to check:
d 3sec C sec tan . Thus sec tan 3sec C .
3 3
3
3
3
3
d
87. Our trial solution based on the chain rule is 2x sin 2x C. Differentiate the solution to check:
d x
dx 2
sin 2x C 12 12 cos 2x sin 2 4x . Thus sin 2 4x dx 2x sin 2x C.
88. Our trial solution based on the chain rule is 2x 12 sin x C. Differentiate the solution to check:
d x 1 sin x C 1 1 cos x cos 2 x . Thus cos 2 x dx x 1 sin x C.
2
dx 2 2
2
2 2
2 2
Copyright 2018 Pearson Education, Inc.
279
280
Chapter 4 Applications of Derivatives
2
89. y x 21 dx (1 x 2 ) dx x x 1 C x 1x C ; y 1 when x 1 1 11 C 1 C 1
x
y x 1x 1
2
90. y x 1x dx ( x 2 2 12 ) dx ( x 2 2 x 2 ) dx
x3
3
3
2 x x 1 C x3 2 x 1x C ;
x
3
y 1 when x 1 13 2 11 C 1 C 13 y x3 2 x 1x 13
91. dr
15 t 3 dt (15t1/2 3t 1/2 ) dt 10t 3/2 6t1/2 C ; dr
8 when t 1 10(1)3/2 6(1)1/2 C 8
dt
dt
t
C 8. Thus dr
10t 3/2 6t1/2 8 r (10t 3/2 6t1/2 8) dt 4t 5/2 4t 3/2 8t C ; r 0 when t 1
dt
4(1)5/2 4(1)3/2 8(1) C1 0 C1 0. Therefore, r 4t 5/2 4t 3/2 8t
92.
d 2r
dt 2
2
cos t dt sin t C ; r 0 when t 0 sin 0 C 0 C 0. Thus, d 2r sin t
dt
sin t dt cos t C1; r 0 when t 0 1 C1 0 C1 1. Then
dr
dt
dr
dt
cos t 1 r (cos t 1) dt sin t t C2 ; r 1 when t 0 0 0 C2 1 C2 1. Therefore,
r sin t t 1
CHAPTER 4
ADDITIONAL AND ADVANCED EXERCISES
1. If M and m are the maximum and minimum values, respectively, then m f ( x) M for all x I . If m M
then f is constant on I.
2. No, the function f ( x)
3 x 6, 2 x 0
9 x2 , 0 x 2
has an absolute minimum value of 0 at x 2 and an absolute
maximum value of 9 at x 0, but it is discontinuous at x 0.
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 0, f 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 | | | | indicates a local maximum at x 1 and a local minimum
1
at x 3.
2
3
4
5. (a) If y 6( x 1)( x 2) 2 , then y 0 for x 1 and y 0 for x 1. The sign pattern is
f | | f has a local minimum at x 1. Also y 6( x 2) 2 12( x 1)( x 2)
1
2
6( x 2)(3 x) y 0 for x 0 or x 2, while y 0 for 0 x 2. Therefore f has points of inflection at
x 0 and x 2. There is no local maximum.
(b) If y 6 x ( x 1)( x 2), then y 0 for x 1 and 0 x 2; y 0 for 1 x 0 and x 2. The sign pattern
is y | | | . Therefore f has a local maximum at x 0 and local minima at x 1 and
1
0
2
x 2. Also, y 18 x 13 7 x 13 7 , so y 0 for 13 7 x 13 7 and y 0 for all other x f
1 7
has points of inflection at x 3 .
f (6) f (0)
60
6. The Mean Value Theorem indicates that
indicates the most that f can increase is 12.
f (c) 2 for some c in (0, 6). Then f (6) f (0) 12
Copyright 2018 Pearson Education, Inc.
Chapter 4 Additional and Advanced Exercises
281
7. If f is continuous on [ a, c) and f ( x ) 0 on [ a, c), then by the Mean Value Theorem for all x [a, c ) we have
f (c) f ( x)
0 f (c) f ( x) 0 f ( x) f (c). Also if f is continuous on (c, b] and f ( x) 0 on (c, b], then
c x
for all x (c, b] we have
x [a, b].
f ( x ) f (c )
x c
0 f ( x) f (c) 0 f ( x) f (c). Therefore f ( x) f (c) for all
8. (a) For all x, ( x 1) 2 0 ( x 1) 2 (1 x 2 ) 2 x (1 x 2 ) 12
(b) There exists c (a, b) such that
| f (b) f (a)| 12 |b a | .
c
1 c 2
f (b ) f ( a )
f (b ) f ( a )
ba
ba
c2
1 c
x 1.
2
1 x 2
1
2 , from part (a)
9. No. Corollary 1 requires that f ( x) 0 for all x in some interval I, not f ( x ) 0 at a single point in I.
10. (a) h( x) f ( x) g ( x) h( x) f ( x) g ( x) f ( x) g ( x) which changes signs at x a since f ( x), g ( x) 0
when x a, f ( x), g ( x) 0 when x a and f ( x), g ( x) 0 for all x. Therefore h( x) does have a local
maximum at x a.
(b) No, let f ( x) g ( x ) x3 which have points of inflection at x 0, but h( x ) x 6 has no point of inflection
(it has a local minimum at x 0).
11. From (ii), f (1)
1 a
bc 2
0 a 1; from (iii), either 1 lim f ( x) or 1 lim f ( x). In either case,
x
1 1x
x 1
lim
2
2
x bx cx 2 x bx c x
1
1
1 x
1 x
lim f ( x) lim
x
c 0, then lim
2
x bx x
12.
dy
dx
lim
x
2
x
x
1 b 0 and c 1. For if b 1, then lim
1 1x
2
x x c x
0 and if
. Thus a 1, b 0, and c 1.
2
3 x 2 2kx 3 0 x 2k 64k 36 x has only one value when 4k 2 36 0 k 2 9 or k 3.
13. The area of the ∆ABC is A( x) 12 (2) 1 x 2
(1 x 2 )1/2 , where 0 x 1. Thus A( x) x
2
1 x
0 and 1 are critical points. Also A (1) 0 so
A(0) 1 is the maximum. When x 0 the ∆ABC is
isosceles since AC BC 2.
14.
f ( c h ) f ( c )
lim
f (c) for 12 | f (c) | 0 there exists a 0 such that 0 | h |
h
h 0
f ( c h ) f ( c )
f ( c h )
f (c) 12 | f (c) |
f (c) 12 | f (c) | . Then f (c ) 0 12 | f (c) |
h
h
f ( c h )
f (c) 12 | f (c) | h f (c) 12 | f (c) | . If f (c) 0, then | f (c) | f (c)
f ( c h )
f ( c h )
32 f (c) h 12 f (c) 0; likewise if f (c ) 0, then 0 12 f (c) h 32 f (c).
(a) If f (c) 0, then h 0 f (c h) 0 and 0 h f (c h) 0. Therefore, f (c) is a local
maximum.
(b) If f (c) 0, then h 0 f (c h) 0 and 0 h f (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
2y
, where
g
g is the acceleration of gravity.
The product of time and exit velocity (rate) yields the distance the water travels:
Copyright 2018 Pearson Education, Inc.
282
Chapter 4 Applications of Derivatives
64(h y ) 8 g2 (hy y 2 )1/2 , 0 y h D ( y ) 4 g2 (hy y 2 ) 1/2 (h 2 y ) 0, h2 and h
2y
g
D( y )
are critical points. Now D(0) 0, D
h2 8
2
g
h
h2 h2
2 1/2
4h
2
g
and D(h) 0 the best place to
drill the hole is at y h2 .
16. From the figure in the text, tan( )
ba
h
tan ah
1 ah tan
b a ;
n
tan tan
tan( ) 1 tan tan ; and tan ah . These equations give
h tan a
h a tan . Solving for tan gives tan
bh
h 2 a (b a )
or (h 2 a(b a )) tan bh.
d
Differentiating both sides with respect to h gives 2h tan (h 2 a (b a )) sec 2 dh b. Then
d
2
2
2
0 2h tan b 2h 2 bh
b 2bh bh ab(b a ) h a (b a ) h a(a b).
dh
h a (b a )
17. The surface area of the cylinder is S 2 r 2 2 rh.
From the diagram we have Rr HH h h RHR rH
and S (r ) 2 r (r h) 2 r r H r HR
2 1
H
R
r
2
2 Hr , where 0 r R.
Case 1: H R S (r ) is a quadratic equation containing the origin and concave upward S (r ) is maximum
at r R.
Case 2: H R S ( r ) is a linear equation containing the origin with a positive slope S ( r ) is maximum
at r R.
Case 3: H R S (r ) is a quadratic equation containing the origin and concave downward.
RH . For simplification
Then dS
4 1 HR r 2 H and dS
0 4 1 HR r 2 H 0 r 2( H
dr
dr
R)
we let r*
RH .
2( H R )
(a) If R H 2 R, then 0 H 2 R H 2( H R) r*
RH
2( H R )
R. Therefore, the maximum occurs at
the right endpoint R of the interval 0 r R because S ( r ) is an increasing function of r.
2
(b) If H 2 R, then r* 22RR R S (r ) is maximum at r R.
RH
R r* R. Therefore, S (r ) is
(c) If H 2 R, then 2 R H 2 H H 2( H R ) 2( HH R ) 1 2( H
R)
a maximum at r r*
RH .
2( H R )
Conclusion: If H (0, 2 R ], then the maximum surface area is at r R. If H (2 R, ), then the maximum is
RH .
at r r* 2( H
R)
18. f ( x) mx 1 1x f ( x) m
If f
0, then
1
m
1
x2
and f ( x)
2
x3
0 when x 0. Then f ( x) 0 x
1
m
yields a minimum.
m 1 m 2 m 1 0 m 14 . Thus the smallest acceptable value for m is 14 .
Copyright 2018 Pearson Education, Inc.
Chapter 4 Additional and Advanced Exercises
19. By similar triangles
10 y z
x
x
z
283
86 z 34 x, and
86 z 10 y 43 x
25 x; then area of
10 y 43 x 34 x y 10 12
25 x 10x 25 x 2
rectangle is A xy x 10 12
12
A 10 25
x 0 critical point is x 12
. Thus
5
6
A
125 0
x 12
and y 5 determine a
5
maximum area of 12.
20. The box has dimensions x by x by y , and L is a
diagonal of the square. We have L2 x 2 x 2
1 L2
2
x 2 , and by similar triangles
4
3
y
3 12 L
y 4 23 L. Then volume of box is V x 2 y
12 L2 (4 23 L) 2 L2 13 L3 V 4 L L2
L(4 L) 0 critical points are L 0 and L 4,
but V (0) 0. Thus V (4) 0 L 4 determines a
.
maximum volume of V 32
3
21. (a) The profit function is P( x) (c ex) x (a bx) ex 2 (c b) x a. P ( x) 2ex c b 0
x c2eb . P ( x) 2e 0 if e 0 so that the profit function is maximized at x c2eb .
(b) The price therefore that corresponds to a production level yielding a maximum profit is
p x c b c e c2eb c 2b dollars.
2e
(c) The weekly profit at this production level is P( x) e
c2eb
2
(c b )
c2eb a (c4be )
2
a.
2
(d) The tax increases cost to the new profit function is F ( x) (c ex) x (a bx tx) ex (c b t ) x a.
Now F ( x) 2ex c b t 0 when x t b2ec c 2bet . Since F ( x) 2e 0 if e 0, F is maximized
when x
c b t
2e
units per week. Thus the price per unit is p c e
increases the cost per unit by c 2b t c 2b
t
2
The x-intercept occurs when 1x 3 0
3 x 13 .
c2bet c2bt dollars. Thus, such a tax
dollars if units are priced to maximize profit.
22. (a)
(b) By Newton’s method, xn 1 xn
f ( xn )
.
f ( xn )
1
x
Here f ( xn ) xn2
1 . So x
n 1
xn2
xn xn 3 xn2 2 xn 3 xn2 xn (2 3 xn ).
Copyright 2018 Pearson Education, Inc.
xn
1 3
xn
1
xn2
xn
1
xn
3 xn2
284
Chapter 4 Applications of Derivatives
23. x1 x0
and
a
x0q 1
f ( x0 )
f ( x0 )
x0
x0q a
qx0q 1
with weights m0
In the case where x0
a
x0q 1
qx0q x0q a
qx0q 1
q 1
q
x0q ( q 1) a
qx0q 1
x0
so that x is a weighted average of x
q 1
q
a
1
x0q 1 q
1
0
and m1 1q .
we have x0q a and x1
q 1
a
x0q 1 q
a 1
x0q 1 q
a
x0q 1
dy
q 1 1
q
q
dy
24. We have that ( x h) 2 ( y h)2 r 2 and so 2( x h) 2( y h) dx 0 and 2 2 dx 2( y h)
dy
dy
dy
2 x 2 y dx 2h 2h dx , by the former. Solving for h, we obtain h
dy
equation yields 2 2 dx 2 y
d2y
dx 2
x y dx
dy
1 dx
a .
x0q 1
d2y
dx 2
0 hold. Thus
. Substituting this into the second
dy
x y dy
dy
d 2 y x y
2 dydx 0. Dividing by 2 results in 1 dx y 2 dydx 0.
dx
1 dx
1 dx
25. (a) a (t ) s (t ) k ( k 0) s (t ) kt C1 , where s (0) 88 C1 88 s (t ) kt 88. So
2
2
2
s (t ) kt2 88t C2 where s (0) 0 C2 0 so s (t ) kt2 88t. Now s (t ) 100 when kt2 88t 100.
Solving for t we obtain t 88
882 200 k
k
. At such t we want s (t ) 0, thus k 88
882 200 k
k
88 0 or
2
2
38.72 ft/sec 2 .
k 88 88k 200 k 88 0. In either case we obtain 882 200k 0 so that k 88
200
2
44t where k is as
(b) The initial condition that s (0) 44 ft/sec implies that s (t ) kt 44 and s (t ) kt
2
44
above. The car is stopped at a time t such that s (t ) kt 44 0 t . At this time the car has
traveled a distance s
k 44 2
2 k
44
k
44
44
k
k
442
2k
968
k
968
velocity quarters stopping distance.
25 feet. Thus halving the initial
200
882
26. h( x) f 2 ( x) g 2 ( x) h( x) 2 f ( x) f ( x) 2 g ( x) g ( x) 2[ f ( x) f ( x) g ( x) g ( x)]
2[ f ( x) g ( x) g ( x)( f ( x))] 2 0 0. Thus h( x) c, a constant. Since h(0) 5, h( x) 5 for all x in the
domain of h. Thus h(10) 5.
27. Yes. The curve y x satisfies all three conditions since
dy
dx
1 everywhere, when x 0, y 0, and
everywhere.
d2y
dx 2
0
28. y 3 x 2 2 for all x y x3 2 x C where 1 13 2 1 C C 4 y x3 2 x 4.
29. s (t ) a t 2 v s (t )
t 3
3
C. We seek v0 s (0) C. We know that s (t*) b for some t* and s is at a
maximum for this t*. Since s (t )
1/3
that t* (3C )
C
. So
[ (3C )1/3 ]4
12
(4b )3/ 4
. Thus v0
3
t 4
12
1/3
C (3C )
s (0)
t 4 Ct and also s (t*)
12
(3C )1/3 34C b 31/3 C 4/3
Ct k and s (0) 0 we have that s (t )
(4b )3/ 4
3
1/3
b (3C )
(C 312C )
b
2 2 3/4
b .
3
30. (a) s (t ) t1/2 t 1/2 v(t ) s (t ) 23 t 3/2 2t1/2 k where v(0) k
4 t 5/2 4 t 3/2 4 t k where s (0) k 4 . Thus s (t )
(b) s (t ) 15
2
2
15
3
3
4 v (t ) 2 t 3/2 2t1/2
3
3
4 t 5/2 4 t 3/2 4 t 4 .
15
3
3
15
Copyright 2018 Pearson Education, Inc.
43 .
0 so
4b
3
Chapter 4 Additional and Advanced Exercises
285
31. The graph of f ( x ) ax 2 2bx c with a 0 is a parabola opening upwards. Thus f ( x) 0 for all x if
f ( x ) 0 for at most one real value of x. The solutions to f ( x ) 0 are, by the quadratic equation
2b (2b ) 2 4 ac
. Thus
2a
we require (2b) 2 4ac 0 b 2 ac 0.
32. (a) Clearly f ( x) ( a1 x b1 ) 2 ( an x bn ) 2 0 for all x. Expanding we see
a12 a22 an2 x 2 2 a1b1 a2b2 an bn x b12 b22 bn2 0. Thus
a1b1 a2b2 anbn 2 a12 a22 an2 b12 b22 bn2 0 by Exercise 29. Thus
a1b1 a2b2 an bn 2 a12 a22 an2 b12 b22 bn2 .
f ( x) a12 x 2 2a1b1 x b12 an2 x 2 2an bn x bn2
(b) Referring to Exercise 31: It is clear that f ( x) 0 for some real x b 2 4ac 0, by quadratic formula.
Now notice that this implies that f ( x) (a1 x b1 )2 (an x bn ) 2
a12 a22 an2 x 2 2 a1b1 a2 b2 an bn x b12 b22 bn2 0
2
a1b1 a2b2 an bn a12 a22 an2 b12 b22 bn2 But now
a1b1 a2 b2 an bn a12 a22 an2 b12 b22 bn2 0
2
f ( x) 0 ai x bi 0
for all i 1, 2, , n ai x bi 0 for all i 1, 2, , n.
33. Let z be the length of AB and AC.
The coordinates of point C on the circle x 2 y 2 1 are (x, y) and the coordinates of point B are ( 1, z). Then
z 2 (1 x) 2 y 2 1 2 x x 2 (1 x 2 ) z 2 2 x 2 (1, z ) (1,
2 x 2) and ( x, y ) ( x,
The slope through points B and C equals the slope through points C and D
t x
( x 1) 1 x 2
1 x 2 2 x 2
x
( x 1) 1 x 1 x
1 x 1 x 2 1 x
x
( x 1) 1 x
1 x 2
1 x 2 2 x 2
x 1
1 x2 ) .
1 x 2 0
x t
.
1 2
( x 1) 12 (1 x )
( 1)
( x 1) 1 x
Then the limit of t as B approaches A is lim t lim x
1 lim
1 (1 x ) 1 2 ( 1)
1 x 2
x
x x
2
1 01 2 1 4 3 .
2 2
Copyright 2018 Pearson Education, Inc.
1 x
CHAPTER 5 INTEGRALS
5.1
AREA AND ESTIMATING WITH FINITE SUMS
1. f ( x) x 2
Since f is increasing on [0, 1], we use left endpoints to
obtain lower sums and right endpoints to obtain upper
sums.
1
2
i0
3
12 12 02
2
14 14 02
2i
12
2
(a) x
1 0
2
12 and xi ix 2i a lower sum is
(b) x
1 0
4
14 and xi ix 4i a lower sum is
1 0
2
1 0
4
2
2
2
2
30 15
14 and xi ix 4i an upper sum is 4i 14 14 14 12 34 12 14 16
32
i 1
(c) x
(d) x
1
2
4i
i 0
2
and xi ix 2i an upper sum is 2i
i 1
4
2. f ( x) x3
2
12
1
2
1
8
14 12 34
2
2
2
7
1 7
4 8 32
1 2 2 5
2 1 8
Since f is increasing on [0, 1], we use left endpoints to
obtain lower sums and right endpoints to obtain upper
sums.
1
3
i0
3
2i
12 12 03
12
3
14 14 03
14 12 34
3
(a) x
1 0
2
12 and xi ix 2i a lower sum is
(b) x
1 0
4
14 and xi i x 4i a lower sum is
(c) x
1 0
2
(d) x
1 0
4
4
3
3
3
3
14 and xi ix 4i an upper sum is 4i 14 14 14 12 34 13 100
25
256 64
i 1
1
2
and xi ix
i
2
4i
i0
2
1
16
3
3
3
3
3
9
an upper sum is 2i 12 12 12 13 12 98 16
i 1
Copyright 2018 Pearson Education, Inc.
287
36
9
256 64
288
Chapter 5 Integrals
3. f ( x) 1x
(a) x
(b) x
(c) x
(d) x
4.
Since f is decreasing on [1, 5], we use left endpoints to
obtain upper sums and right endpoints to obtain lower
sums.
51
2
2
(b) x
i 1
4
51 1 and x 1 i x 1 i a lower sum is
1 1 1 1 1 1 1 77
i
xi
4
60
2 3 4 5
i1
1
5 1
1 2 2 1 1 8
2 and xi 1 i x 1 2i an upper sum is
2
3
3
xi
i 0
3
5 1
1 1 1 1 1 1 1 25
1 and xi 1 ix 1 i an upper sum is
2 3 4
4
xi
12
i0
2 ( 2)
2
2 ( 2)
4
Since f is increasing on [2, 0] and decreasing on
[0, 2], we use left endpoints on [2, 0] and right
endpoints on [0, 2] to obtain lower sums and use right
endpoints on [2, 0] and left endpoints on [0, 2] to
obtain upper sums.
f ( x) 4 x 2
(a) x
2 and xi 1 i x 1 2i a lower sum is x1 2 2 13 15 16
15
i
2 and xi 2 ix 2 2i a lower sum is 2 (4 (2)2 ) 2 (4 22 ) 0
1
4
i 0
i 3
2
3
i 1
i 2
1 and xi 2 i x 2 i a lower sum is (4 ( xi ) 2 ) 1 (4 ( xi ) 2 ) 1
1((4 (2) 2 ) (4 (1)2 ) (4 12 ) (4 22 )) 6
2 ( 2)
(c) x 2 2 and xi 2 ix 2 2i an upper sum is 2 (4 (0) 2 ) 2 (4 02 ) 16
(d) x
2 ( 2)
4
1 and xi 2 ix 2 i an upper sum is (4 ( xi ) 2 ) 1 (4 ( xi )2 ) 1
1((4 (1)2 ) (4 02 ) (4 02 ) (4 12 )) 14
5. f ( x) x 2
1 0
2
2
1
12
1 0
4
14
Using 2 rectangles x
1
2
2
f 14 f 43 12 4 43 1032 165
Using 4 rectangles x
1
4
14
f 18 f 83 f 85 f 87
18 83 85 78
2
2
Copyright 2018 Pearson Education, Inc.
2
2
21
64
Section 5.1 Area and Estimating with Finite Sums
6. f ( x) x3
1 0
2
3
1
12
1 0
4
14
Using 2 rectangles x
289
3
12 4 43 22864 327
12 f 14 f 43
Using 4 rectangles x
14 f 18 f 83 f 85 f 87
3
3
3
3
31
14 1 3 35 7 4963 124
128
8
4 8
83
7. f ( x)
1
x
51
2
2 2( f (2) f (4))
51
4
1
Using 2 rectangles x
2 12 14 23
Using 4 rectangles Δx
32 f 52 f 72 f 92 1 23 52 72 92
1 f
496
3 1488
496 315
57 9 57 9
8. f ( x) 4 x 2
Using 2 rectangles x
2 ( 2)
2
2
2( f (1) f (1)) 2(3 3) 12
Using 4 rectangles x
2 ( 2)
1
4
1 f 3
2
2
1 f 32 f 12 f
1 4 32
16
2
1
4 2
2
4
94 2 14 2 16 102 11
12
4
2
23
9. (a) D (0)(1) (12)(1) (22)(1) (10)(1) (5)(1) (13)(1) (11)(1) (6)(1) (2)(1) (6)(1) 87 inches
(b) D (12)(1) (22)(1) (10)(1) (5)(1) (13)(1) (11)(1) (6)(1) (2)(1) (6)(1) (0)(1) 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) 5220 meters (NOTE: 5 minutes 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) 4920 meters (NOTE: 5 minutes 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) 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) 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 22.7 sec.
At 22.7 sec, the velocity was approximately 120 mi/hr.
Copyright 2018 Pearson Education, Inc.
2
290
Chapter 5 Integrals
13. (a) Because the acceleration is decreasing, an upper estimate is obtained using left endpoints in summing
acceleration t. Thus, t 1 and speed [32.00 19.41 11.77 7.14 4.33](1) 74.65 ft/sec
(b) Using right endpoints we obtain a lower estimate: speed [19.41 11.77 7.14 4.33 2.63](1)
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 3 seconds is s [32.00 51.41 63.18](1) 146.59 ft.
14. (a) The speed is a decreasing function of time right endpoints give a 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 32 from the speed at each time increment
t 1sec. Thus, the speed 240 ft/sec after 5 seconds.
(b) A lower estimate for height attained is h [368 336 304 272 240](1) 1520 ft.
15. Partition [0, 2] into the four subintervals [0, 0.5], [0.5, 1], [1, 1.5], and [1.5, 2]. The midpoints of these
subintervals are m1 0.25, m2 0.75, m3 1.25, and m4 1.75. The heights of the four approximating
1 , f ( m ) (0.75)3 27 , f ( m ) (1.25)3 125 , and f (m ) (1.75)3
rectangles are f (m1 ) (0.25)3 64
2
3
4
64
64
3
3
3
1 3 1
31
3
5
7
1
1
1
1
Notice that the average value is approximated by 2 4
4
4
4
2
2
2
2 16
approximate area under
length 1of [0,2]
. We use this observation in solving the next several exercises.
curve f ( x) x3
343
64
16. Partition [1,9] into the four subintervals [1, 3], [3, 5], [5, 7], and [7, 9]. The midpoints of these subintervals are
m1 2, m2 4, m3 6, and m4 8. The heights of the four approximating rectangles are f (m1 ) 12 ,
f (m2 ) 14 , f (m3 ) 16 , and f (m4 ) 18 . The width of each rectangle is x 2. Thus,
25 .
25 average value
area
Area 2 12 2 14 2 16 2 81 12
128 96
length of [1,9]
25
17. Partition [0, 2] into the four subintervals [0, 0.5], [0.5, 1], [1, 1.5], and [1.5, 2]. The midpoints of the
subintervals are m1 0.25, m2 0.75, m3 1.25, and m4 1.75. The heights of the four approximating
rectangles are f (m1 ) 12 sin 2 4 12 12 1, f (m2 ) 12 sin 2 34 12 12 1, f (m3 ) 12 sin 2 54
12
1
2
2
12 12 1, and f (m4 ) 12 sin 2 74 12
Thus, Area (1 1 1 1)
1
2
2
1. The width of each rectangle is x 12 .
12 2 average value lengthareaof [0, 2] 22 1.
18. Partition [0, 4] into the four subintervals [0, 1], [1, 2], [2, 3], and [3, 4]. The midpoints of the subintervals
are m1 12 , m2 32 , m3 52 , and m4 72 . The heights of the four approximating rectangles are
4
1
f (m1 ) 1 cos 42 1 cos 8
1 cos
3
8
4
4
3
0.27145 (to 5 decimal places), f (m2 ) 1 cos 42
4
5
0.97855, f (m3 ) 1 cos 42 1 cos 58
4
Copyright 2018 Pearson Education, Inc.
0.97855, and
4
Section 5.1 Area and Estimating with Finite Sums
4
7
f (m4 ) 1 cos 42 1 cos 78
4
291
0.27145. The width of each rectangle is x 1. Thus,
Area (0.27145)(1) (0.97855)(1) (0.97855)(1) (0.27145)(1) 2.5 average value
area
length of [0,4]
2.5
4
85
19. Since the leakage is increasing, an upper estimate uses right endpoints and a lower estimate uses left endpoints:
(a) upper estimate (70)(1) (97)(1) (136)(1) (190)(1) (265)(1) 758 gal,
lower estimate (50)(1) (70)(1) (97)(1) (136)(1) (190)(1) 543 gal.
(b) upper estimate (70 97 136 190 265 369 516 720) 2363 gal,
lower estimate (50 70 97 136 190 265 369 516) 1693 gal.
(c) worst case: 2363 720t 25, 000 t 31.4 hrs;
best case: 1693 720t 25, 000 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 (0.2)(30) (0.25)(30) (0.27)(30) (0.34)(30) (0.45)(30) (0.52)(30) 60.9 tons
lower estimate (0.05)(30) (0.2)(30) (0.25)(30) (0.27)(30) (0.34)(30) (0.45)(30) 46.8 tons
(b) Using the lower (best case) estimate: 46.8 (0.52)(30) (0.63)(30) (0.70)(30) (0.81)(30) 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 2. Area
2
2
2
(b) Think of the octagon as a collection of 16 right triangles with a hypotenuse of length 1 and an acute angle
measuring 216 8 .
Area 16
12 sin 8 cos 8 4 sin 4 2
2 2.828
(c) Think of the 16-gon as a collection of 32 right triangles with a hypotenuse of length 1 and an acute angle
.
measuring 232 16
Area /
(d) Each area is less than the area of the circle, . As n increase, the area approaches .
22. (a) Each of the isosceles triangles is made up of two right triangles having hypotenuse 1 and an acute angle
measuring 22n n The area of each isosceles triangle is AT 2 12 sin n cos n 12 sin 2n .
n sin 2
n
n 2
(b) The area of the polygon is AP nAT n2 sin 2n , so lim
lim
n
(c) Multiply each area by r 2 .
AT 12 r 2 sin 2n
AP n2 r 2 sin 2n
lim AP r 2
n
23–26. Example CAS commands:
Maple:
with( Student[Calculus 1] );
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)
Copyright 2018 Pearson Education, Inc.
sin
2
n
2n
292
Chapter 5 Integrals
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
Avg[n] : evalf(add(y,y Ylist)/nops(Ylist));
# (c)
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 π, , powers, roots, fractions, etc. are available in Palettes.
Never insert a space between the name of a function and its argument.
Clear[x]
f[x_] : x Sin[1/x]
{a, b}{π/4, π}
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.
k 1
3
2.
k 1
6k
k 1
6(1)
6(2)
11 21 62 12
7
3
k 1
k
11
2 1
3 1
1 2 3 0 12 23 76
4
3. cos k cos(1 ) cos(2 ) cos(3 ) cos(4 ) 1 1 1 1 0
k 1
Copyright 2018 Pearson Education, Inc.
Section 5.2 Sigma Notation and Limits of Finite Sums
4.
293
5
sin k sin(1 ) sin(2 ) sin(3 ) sin(4 ) sin(5 ) 0 0 0 0 0 0
k 1
3
5. (1)k 1 sin k (1)11 sin 1 (1) 21 sin 2 (1)31 sin 3 0 1
k 1
3
2
32 2
4
6. (1)k cos k (1)1 cos(1 ) (1) 2 cos(2 ) (1)3 cos(3 ) (1)4 cos(4 ) (1) 1 ( 1) 1 4
k 1
7. (a)
(b)
(c)
6
2k 1 211 221 231 241 251 261 1 2 4 8 16 32
k 1
5
2k 20 21 22 23 24 25 1 2 4 8 16 32
k 0
4
2k 1 211 201 211 221 231 241 1 2 4 8 16 32
k 1
All of them represent 1 2 4 8 16 32
8. (a)
(b)
(c)
6
(2)k 1 (2)11 (2)21 ( 2)31 ( 2) 41 ( 2)51 ( 2)6 1 1 2 4 8 16 32
k 1
5
(1)k 2k (1)0 20 (1)1 21 ( 1) 2 22 ( 1)3 23 ( 1) 4 24 ( 1)5 25 1 2 4 8 16 32
k 0
3
(1)k 1 2k 2 (1)21 22 2 (1) 11 21 2 (1)01 20 2 (1)11 21 2 (1)21 22 2 (1)31 23 2
k 2
1 2 4 8 16 32;
(a) and (b) represent 1 2 4 8 16 32; (c) is not equivalent to the other two
9. (a)
(b)
(c)
10. (a)
(b)
(c)
4
k 2
2
( 1) k 1
k 1
( 1)k
( 1) 2 1
( 1)31
( 1) 4 1
21 31 41 1 12 13
( 1)0
( 1)1
( 1)2
k 1 01 11 21 1 12 13
k 0
1
( 1) k
( 1) 1
( 1)0
( 1)1
k 2 1 2 0 2 1 2 1 12 13
k 1
(a) and (c) are equivalent; (b) is not equivalent to the other two.
4
(k 1)2 (1 1)2 (2 1)2 (3 1)2 (4 1)2 0 1 4 9
k 1
3
(k 1) 2 (1 1) 2 (0 1) 2 (1 1)2 (2 1)2 (3 1) 2 0 1 4 9 16
k 1
1
k 2 (3)2 (2) 2 (1) 2 9 4 1
k 3
(a) and (c) are equivalent to each other; (b) is not equivalent to the other two.
6
11. k
k 1
5
14. 2k
k 1
4
12. k 2
k 1
5
15. (1) k 1 1k
k 1
4
13.
1
k 1 2
5
k
16. (1) k
Copyright 2018 Pearson Education, Inc.
k 1
k
5
294
Chapter 5 Integrals
17. (a)
3ak 3 ak 3(5) 15
n
n
k 1
n b
(b)
(c)
k 1
n
k
6
1
6
k 1
n
bk 16 (6) 1
k 1
n
n
k 1
n
k 1
n
k 1
n
k 1
n
k 1
n
k 1
(ak bk ) ak bk 5 6 1
(d) (ak bk ) ak bk 5 6 11
(e)
18. (a)
(c)
19. (a)
(c)
20. (a)
n
(bk 2ak ) bk 2 ak 6 2(5) 16
k 1
k 1
n
n
k 1
n
k 1
n
k 1
8ak 8 ak 8(0) 0
n
(ak 1) ak 1 0 n n
k 1
k 1
10
k
k 1
10
10(10 1)
2
n
55
k
k 1
13
13(131)
2
91
n
k 1
k 1
k 1
n
k 1
10
10(101)(2(10) 1)
k2
385
6
(b)
k 1
10(10 1) 2
k 3 2 552 3025
k 1
13
k 1
n
(bk 1) bk 1 1 n
(d)
k 1
n
250bk 250 bk 250(1) 250
(b)
13
13(131)(2(13) 1)
k2
819
6
(b)
k 1
2
13(131)
(c) k 2 912 8281
k 1
3
7
7
k 1
k 1
6
6
6
k 1
k 1
k 1
6
6
6
k 1
k 1
k 1
21. 2k 2 k 2
7(7 1)
2
56
23. (3 k 2 ) 3 k 2 3(6)
24. (k 2 5) k 2 5
6(6 1)(2(6) 1)
6
6(6 1)(2(6) 1)
6
5
5
5
k 1
k 1
k 1
k 1
7
7
7
7
k 1
k 1
k 1
k 1
26. k (2k 1) (2k 2 k ) 2 k 2 k 2
27.
k 1
3
k
k 1
2
5
1
225
3
5
k k
k 1
k 1
5
3
2
k 1
k 1
7 3 7
7
7
28. k k4 k 14 k 3
k 1
k 1 k 1
k 1
5(51)
2
5(6) 61
5
k3
225
5
73
25. k (3k 5) (3k 2 5k ) 3 k 2 5 k 3
5
5
k
22. 15k 15
15
5(51)(2(5) 1)
6
7(7 1)(2(7) 1)
6
2
1 5(51)
225
2
7(7 1) 2
2
5
5(51)
2
7(7 1)
2
308
5(51) 3
3376
2
2
1 7(7 1)
4
2
240
588
Copyright 2018 Pearson Education, Inc.
Section 5.2 Sigma Notation and Limits of Finite Sums
29. (a)
7
295
500
3 3(7) 21
(b) 7 7(500) 3500
k 1
k 1
264
262
k 3
j1
(c) Let j k 2 k j 2; if k 3 j 1 and if k 264 j 262 10 10 10(262) 2620
36
28
28
28
k 9
j 1
j 1
j 1
17
15
k 3
j 1
30. (a) Let j k 8 k j 8; if k 9 j 1 and if k 36 j 28 k ( j 8) j 8
28(281)
2
8(28) 630
(b) Let j k 2 k j 2; if k 3 j 1 and if k 17 j 15 k 2 ( j 2) 2
15
15
2
2
15
15
j 1
j 1
( j 4 j 4) j 4 j 4
j 1
j 1
15(151)(2(15) 1)
6
15(151)
4 2
4(15) 1240 480 60 1780
71
(c) Let j k 17 k j 17; if k 18 j 1 and if k 71 j 54 k (k 1)
54
54
54
54
54
j 1
j 1
j 1
k 3
( j 17)(( j 17) 1) ( j 2 33 j 272) j 2 33 j 272
j 1
54(54 1)(2(54) 1)
6
n
(b)
k 1
n
n
n
k 1
k 1
k 1
(k 1) k 1
(c)
n ( n 1)
2
n
1n 2n 1n 2n n 1 2n2
k 1
n
k
k 1 n
2
1 n ( n 1)
2
n2
k 1
n n
2
n
(c)
n
c cn
2
32. (a)
34.
272(54) 53955 49005 14688 117648
4 4n
31. (a)
33.
j 1
54(54 1)
33 2
(b)
n
k 1
c
n
nc n c
n 1
2n
50
2
k 1 k 2 (22 12 ) (32 22 ) (42 32 ) (502 492 ) (512 502 ) 512 12 2600
k 1
20
sin k 1 sin k (sin1 sin 2) (sin 2 sin 3) (sin 3 sin 4)
k 2
(sin18 sin19) (sin19 sin 20) sin1 sin 20
35.
30
k 7
36.
40
k 1
3 4
26 26 27
4 5
3 27 3 3 3 2 3
25
1
k ( k 1)
40
1k k11
k 1
5 6
k 4 k 3
11 12 12 13 13 14
391 401 401 411 1 411 4041
Copyright 2018 Pearson Education, Inc.
296
Chapter 5 Integrals
37. (a)
(b)
(c)
38. (a)
(b)
(c)
39. (a)
(b)
(c)
40. (a)
(b)
(c)
41. | x1 x0 | |1.2 0| 1.2, | x2 x1 | | 1.5 1.2| 0.3, | x3 x2 | 2.3 1.5 0.8, | x4 x3 | 2.6 2.3 0.3,
and | x5 x4 | |3 2.6| 0.4; the largest is || P || 1.2.
42. | x1 x0 | | 1.6 (2)| 0.4,| x2 x1 | | 0.5 ( 1.6) | 1.1,| x3 x2 | | 0 (0.5) | 0.5,
| x4 x3 | |0.8 0| 0.8, and | x5 x4 | |1 0.8| 0.2; the largest is || P || 1.1.
Copyright 2018 Pearson Education, Inc.
Section 5.2 Sigma Notation and Limits of Finite Sums
43. f ( x) 1 x 2
Let x 1n0 1n and ci ix ni . The right-hand sum is
n
n
n
2
1 ci2 1n 1n 1 ni 13 n 2 i 2
n i 1
i 1
i 1
n3
n3
n
1 i2
3
n i 1
2 n3 12
2 n3 12
lim 1 6 n
n
Let x
n
n ( n 1)(2 n 1)
1
6n
3 0
n
3
n
n
1 2n
3
n
3
3n 2 n
6 n3
lim 1 ci2 1n
. Thus,
n
6
1
44. f ( x ) 2 x
297
n i 1
1 13
2
3
and ci ix
n
3i .
n
The right-hand sum
n3 i1 6ni n3 18n i1i 18n n(n21) 9n n9n .
i 1
n
Thus, lim 6ni n3 lim 9n 9n lim 9 9n 9.
n
n i 1
n n
is 2ci
2
2
2
2
2
2
45. f ( x ) x 2 1
Let x
i 1
30
n
3
n
n3 i1
n
is ci2 1
n
n
i 2 n3 n
27
n
i 1
18 27
n
3i 2
n
1
27 n ( n 1)(2 n 1)
6
n3
9
n
n2
3i . The right-hand
n
n
3 3 9i 2 1
2
n
n
i1 n
and ci ix
3
sum
9(2 n3 3n 2 n )
2 n3
3
n3
3. Thus, lim ci2 1
2
n i 1
18 27n 92
n
9 3 12.
lim
3
2
n
46. f ( x) 3x 2
Let x 1n0
n
3ci2
i 1
1
n
1n i13 ni 1n n3 i1i2 n3 n(n1)(26 n1)
n
2 3n 12
n
2
2n
lim
n
n
2
3
3
2
2 n 3n3 n
47. f ( x ) x x 2 x(1 x)
and ci ix ni . The right-hand sum is
2 3n 12
n
2
2
2
3
n
. Thus, lim 3ci2 1n
n i 1
1.
Let x 1n0 1n and ci ix ni . The right-hand sum is
n
n
n
n
2
ci ci2 1n ni ni 1n 12 i 13 i 2
n i 1
n i 1
i 1
i 1
2
3
2
n ( n 1)
n ( n 1)(2 n 1)
n
n
2
n
1
1
2
3
2 3n3 n
2
6
n
1 1n
2
2 n3 12
n
6
n
n
2n
. Thus, lim ci ci2
Copyright 2018 Pearson Education, Inc.
n i 1
1n
6n
298
Chapter 5 Integrals
1 1 2 3n 12
lim 2n 6 n
n
Let x 1n0
48. f ( x) 3 x 2 x 2
i 1
n
3ci 2ci2
12 62 56 .
and ci ix ni . The right-hand sum is
1
n
1n i1 3ni 2 ni
n
3 n ( n 1) 2 n ( n 1)(2 n 1)
2
6
n2
n3
3 1
n
3 3n 2 n n2
2
n i 1
Let x 1n0
n
2ci3
2 n 2 ( n 2 2 n 1)
4n4
32 23 13
.
6
3
n
2 i3
n 4 i 1
1 2n 12
1
n
2
n 22n 1
Let x
0 ( 1)
n
n
2
2 n ( n 1)
2
n4
n
1n and ci 1 ix 1 ni .
n
The right-hand sum is ci2 ci3
n
1 ni
i 1
. Thus, lim 2ci3 1n
n i 1
n
2
2n
1 2n 12
lim 2 n 12 .
n
50. f ( x ) x 2 x3
1n
and ci ix ni . The right-hand sum is
1
n
1 2 i
n
n
i 1
i 1
n
1
n
. Thus, lim 3ci 2ci2
3
3 3 2 n3 12
lim 2 n 3 n
n
49. f ( x) 2 x3
n
n
3 i 2 i2
2
3
n i 1
n i 1
3n 2 3n 2 n 2 3n 1
2n2
3n 2
2
i 1
1 ni
2
2
3
n
3
n
1n
2
3
5i 4i
i
1
n 2 n n 2 n3
i 1
n
n
1
n
n
n2 52i 4i3 i 4 n2 52 i 43 i 2 14 i3
n
n
n
n i 1
n i 1
n i 1
i1
i 1
n2 (n)
5
2
2
5
2 5n2n 5 4n 62n 2 n 22n 1 2 2 n
3n
5 5
lim 2 2 n
n
5.3
4 6n
3
4n
4 6n 22
n
3
1 n2 12
n
4
2
n2
1 2n
7.
2 52 34 14 12
4
1
n2
5 n ( n 1)
2
n2
4 n ( n 1)(2 n 1)
6
n3
n
. Thus, lim ci2 ci3 1n
n i 1
THE DEFINITE INTEGRAL
2 2
1.
0 x
4.
1 1x dx
4
dx
0
3
2.
1 2x dx
5.
2 11x dx
3
5
3.
7 ( x
6.
0
Copyright 2018 Pearson Education, Inc.
1
2
3 x) dx
4 x 2 dx
2
1 n ( n 1)
4
2
n
Section 5.3 The Definite Integral
7.
0
/4 (sec x) dx
9. (a)
(c)
8.
/4
0
(tan x) dx
2
2 g ( x) dx 0
2
2
1 3 f ( x) dx 31 f ( x) dx 3(4) 12
5
5
(b)
1
2
(d)
2 f ( x) dx 1
(e)
1 [ f ( x) g ( x)] dx 1 f ( x) dx 1 g ( x) dx 6 8 2
5
5
5
1 [4 f ( x) g ( x)] dx 41 f ( x) dx 1 g ( x) dx 4(6) 8 16
(f )
10. (a)
(b)
5
5 g ( x) dx 1 g ( x) dx 8
f ( x) dx f ( x) dx 6 (4) 10
1
5
5
9
5
9
1 2 f ( x) dx 2 1 f ( x) dx 2(1) 2
9
9
9
7 [ f ( x) h( x)] dx 7 f ( x) dx 7 h( x) dx 5 4 9
9
9
9
(c)
7 [2 f ( x) 3h( x)] dx 2 7
(d)
9 f ( x) dx 1 f ( x) dx (1) 1
7
9
9
1 f ( x) dx 1 f ( x) dx 7 f ( x) dx 1 5 6
7
9
9
9
9 [h( x) f ( x)] dx 7 [ f ( x) h( x)] dx 7 f ( x) dx 7 h( x) dx 5 4 1
(e)
(f )
11. (a)
(c)
12. (a)
(c)
13. (a)
(b)
14. (a)
(b)
1
f ( x) dx 3 h( x) dx 2(5) 3(4) 2
7
9
2
2
1 f (u) du 1 f ( x) dx 5
1
2
2 f (t ) dt 1 f (t ) dt 5
3
(d)
0
0 g (t ) dt 3 g (t ) dt 2
0
0
3[ g ( x)] dx 3 g ( x) dx
4
(b)
4
(b)
2
(d)
2
0
3
3
1
1 h(r ) dr 1 h(r ) dr 1 h(r ) dr 6 0 6
3
1
3
h(u ) du h(u ) du h(u ) du 6
1
3
1
Copyright 2018 Pearson Education, Inc.
3
0
3 g (u ) du 3 g (t ) dt 2
0 g (r )
0
3 2 dr 12 3 g (t ) dt
3 f ( z ) dz 0 f ( z ) dz 0 f ( z ) dz 7 3 4
3
4
4 f (t ) dt 3 f (t ) dt 4
3
2
1 3 f ( z ) dz 3 1 f ( z ) dz 5
2
2
1 [ f ( x)] dx 1 f ( x) dx 5
1
2
(
2) 1
299
300
Chapter 5 Integrals
15. The area of the trapezoid is A 12 ( B b)h
12 (5 2)(6) 21
2 2x 3 dx 21 square units
4
16. The area of the trapezoid is A 12 ( B b)h
12 (3 1)(1) 2
3/2
1/2
(2 x 4) dx 2 square units
17. The area of the semicircle is A 12 r 2 12 (3) 2
3
92
9 x 2 dx 92 square units
3
18. The graph of the quarter circle is A 14 r 2 14 (4) 2
4
0
4
16 x 2 dx 4 square units
19. The area of the triangle on the left is A 12 bh
12 (2)(2) 2. The area of the triangle on the right is
A 12 bh 12 (1)(1) 12 . Then, the total area is 2.5
1
2
| x| dx 2.5 square units
Copyright 2018 Pearson Education, Inc.
Section 5.3 The Definite Integral
20. The area of the triangle is A 12 bh 12 (2)(1) 1
1
(1 | x|) dx 1 square unit
1
21. The area of the triangular peak is A 12 bh 12 (2)(1) 1.
The area of the rectangular base is S w (2)(1) 2.
1
Then the total area is 3 (2 | x|) dx 3 square
1
units
22. y 1 1 x 2 y 1 1 x 2 ( y 1)2 1 x 2
x 2 ( y 1)2 1, a circle with center (0, 1) and radius
of 1 y 1 1 x 2 is the upper semicircle. The area
of this semicircle is A 12 r 2 12 (1) 2 2 . The area
of the rectangular base is A w (2)(1) 2. Then the
1
total area is 2 2 1 1 x 2 dx 2 2
1
square units
23.
b
2
0 2x dx 12 (b)( b2 ) b4
24.
b
0 4 x dx 12 b(4b) 2b
Copyright 2018 Pearson Education, Inc.
2
301
302
25.
Chapter 5 Integrals
b
a 2s ds 12 b(2b) 12 a(2a) b
2
2
27. (a)
1 x 2 dx
1
2
(b) 3x 1 x dx
1
0
29.
31.
33.
1
2
3
0
2
x dx
d
37.
a
0
1
2
3
12
1
13x dx 0 3x dx 1
3
(2 a )2
2
0
32.
34.
7
3
1
24
36.
2
a2
3 b
3a 2
2
3
41.
3 7 dx 7(1 3) 14
43.
0 (2t 3) dt 21 t dt 0 3 dt 2 22
44.
0 t 2 dt 0
3
b
3
1
1
2
2
t dt
2
2
0
2
a2 )
4 x 2 dx 14 [ (2)2 ]
2.5
0.5
x dx
5 2
(2.5) 2
2
5 2
2
2
0.3 2
(0.3)
0 s d s 3
r dr
d
3a
x dx
3
2
2
2
2
2
24
2
3
24
3
3
0.009
2
3a
(0.5) 2
2
2
3
/2 2
0
2
2
a2 a 2
38.
a
40.
0
42.
0 5 x dx 5 0 x dx 5 22
3b 2
x dx
2
(3b )3
3
2
9b3
2
2
02 3(2 0) 4 6 2
2 dt
2
1 x 2 dx 12 [(1)(3)] 12 [(1)(3)] 12 [ (1) 2 ] 2
2
0
x dx
2
(b)
30.
39.
2
a 3t d t 12 b(3b) 12 a(3a) 32 (b
1 x 2 dx 12 [(1)(3)] 14 [ (1)2 ] 4 23
1
2
2 32
3 7
0
b
26.
3
dt
x dx
b 2
(1)2
2
0
13x dx 1
3
x dx
2a
3
2
(2 )2
2
7 2
1/2 2
t
0
35.
2
2
a2
4 x 2 dx 12 [ (2)2 ] 2
1 3x
28. (a)
2
2
02 2 2 0 1 2 1
Copyright 2018 Pearson Education, Inc.
2
2
02 10
Section 5.3 The Definite Integral
32 74
45.
2 1 2z dz 21 dz 2 2z dz 21 dz 12 1 z dz 1[1 2] 12 22
46.
3 (2 z 3) dz 3 2 z dz 3 3 dz 2 0 z dz 3 3 dz 2 32 02 3[0 3] 9 9 0
47.
1 3u
1
1
0
1
0
2
1
2
0
2
3
0
2
12 1 12
2
2
3 1 3
7
1
1/2
u 2 du 24 u 2 du u 2 du 24 13 23 24 38 7
1/2
0
0
24u 2 du 24
1/2
49.
0 (3x
2
2
1
3
2
3
2
2
2
2
x 5) dx 3 x 2 dx x dx 5 dx 3 23 03 22 02 5[2 0] (8 2) 10 0
0
0
0
1
1
1
1
x 5) dx (3 x 2 x 5) dx 3 x 2 dx x dx 5 dx 3
0
0
0
0
32 5 72
0
1 (3x
2
51. Let x b n 0 bn and let x0 0, x1 x, x2 2x, ,
xn 1 (n 1)x, xn nx b. Let the ck 's be the right
endpoints of the subintervals c1 x1 , c2 x2 , and so on.
The rectangles defined have areas:
2
3
f c1 x f x x 3 x x 3 x
2
f c2 x f 2x x 3 2x x 3 2
f c3 x f 3x x 3 3x x 3 3
2
2
2
x 3
x 3
f cn x f nx x 3 nx x 3n 2 x
2
n
3
n
Then Sn f (ck )x 3k 2 (x)3
3
n
k 1
2
3(x) k 3
b3
2
3
3
3
3
3
3
2
2
1
du 3 u 2 du 3 u 2 du u 2 du 3 23 03 13 03 3 23 13 3 73 7
1
0
0
2
48.
50.
1
303
2
k 1
3
n
1
n2
b3
n3
b
0
k 1
n ( n 1)(2 n 1)
6
b3
n 2
3 x 2 dx lim
2
3
n
1
n2
b .
3
Copyright 2018 Pearson Education, Inc.
13
3
3
03
12
2
02 5(1 0)
2
304
Chapter 5 Integrals
52. Let x b n 0 bn and let x0 0, x1 x, x2 2x, . . . ,
xn 1 (n 1)x, xn nx b. Let the ck 's be the right
endpoints of the subintervals c1 x1 , c2 x2 , and so on.
The rectangles defined have areas:
2
3
f c1 x f x x x x x
2
f c2 x f 2x x 2x x 2
f c3 x f 3x x 3x x 3
2
2
2
x 3
x 3
f cn x f nx x nx x n 2 x
2
3
n
n
n
k 1
k 1
Then Sn f (ck )x k 2 (x)3 (x)3 k 2
3
b
n3
n ( n 1)(2 n 1)
6
b
n
3
n
6
3
b
x 2 dx lim 6b
0
k 1
2
2
3
3
n
1
n2
1
n2
b3 .
3
53. Let x b n 0 bn and let x0 0, x1 x, x2 2x, ,
xn 1 (n 1)x, xn nx b. Let the ck 's be the right
endpoints of the subintervals c1 x1 , c2 x2 , and so on.
The rectangles defined have areas:
f c1 x f x x 2 x x 2 x
2
f c2 x f 2x x 2 2x x 2 2 x
f c3 x f 3x x 2 3x x 2 3 x
2
2
f cn x f nx x 2 nx x 2 n x
n
n
n
k 1
k 1
2
Then Sn f (ck )x 2k (x)2 2(x)2 k 2
n ( n 1)
2
k 1
2
b 1
b2
n2
1
n
b
2 x dx lim b 2 1 1n b 2 .
0
54. Let x
n
b 0
n
b
n
and let x0 0, x1 x, x2 2x, ,
xn 1 (n 1)x, xn nx b. Let the ck 's be the right
endpoints of the subintervals c1 x1 , c2 x2 , and so on.
The rectangles defined have areas:
2x 1 x 12 x 2 x
2
f c2 x f 2x x 22 x 1 x 12 2 x x
2
f c3 x f 3x x 32 x 1 x 12 3 x x
f c1 x f x x
f cn x f nx x
n2x 1 x 12 n x 2 x
Copyright 2018 Pearson Education, Inc.
Section 5.3 The Definite Integral
n ( n 1)
1 k ( x ) 2 x 1 ( x ) 2 k x 1 1 b
b (n)
2 n 2 n
2
2
k 1
k 1
k 1
k 1
b
14 b 2 1 1n b 2x 1 dx lim 14 b 2 1 1n b 14 b 2 b.
0
n
n
n
n
Then Sn f (ck )x
55. av( f )
1
3
56. av( f )
3
3
0
1
3
( x 2 1) dx
3 2
3
1
x dx 1
1dx
3 0
3 0
3 0 1 1 0.
310 03 x2 dx 13 12 03 x2 dx
2
3.
2
110 01 (3x2 1) dx 3 01 x2 dx 011 dx
13
3
110 01 (3x2 3) dx 3 01 x2 dx 013 dx
3(1 0) 2.
13
3
310 03 (t 1)2 dt 13 03 t 2 dt 23 03 t dt 13 031 dt
33
3
2 32
3 2
2
2
2
(1 0) 2.
59. av( f )
13
33
3
58. av( f )
3
3
57. av( f )
3
1
3 0
3
16
n
02 13 (3 0) 1.
Copyright 2018 Pearson Education, Inc.
305
306
Chapter 5 Integrals
60. av( f )
1
1 ( 2)
1
2
2
2 2
0
13
3
1 ( 2)
3 3
61. (a) av( g )
3
1 1 t 2 dt 1 1
3 2
3 2
1 3
2 2.
1
1( 1)
1 0 ( x 1)
2 1
1
1
(| x| 1)dx
1
dx 12 ( x 1) dx
0
0
1 0 1 dx 1 1 x
2 1
2 0
12 x dx
1
1
dx 12 1 dx
2 ( 1)2
12 02 2 12 (0 (1)) 12
12 .
(b) av( g )
1 3x
2 1
(c) av( g )
0
12
2
1
31( 1)
3
1
(| x | 1) dx
1
0 ( 1)
x dx
1
2
2
3
dx 12 1 dx 12 32 12 12 (3 1) 1.
1
0
2
02 12 (1 0)
311 13 (| x | 1) dx 12 13 ( x 1) dx
1 1 (| x | 1) dx 1 3 (| x | 1) dx
4 1
4 1
1 ( 1 2) 1 (see parts (a) and (b)
4
4
62. (a) av(h)
t dt
2 ( 2)2
t dt 13 12 2
1
0
13 t 2 dt 13
13
t t dt
2
0
2
0
1
above).
0
| x | dx ( x) dx
( 1)2
2
1
12 .
Copyright 2018 Pearson Education, Inc.
Section 5.3 The Definite Integral
(b) av(h)
12
2
307
110 01 | x | dx 01 x dx
2
02 12 .
(c) av(h)
1
1( 1)
1
1
| x | dx
0
1
12 | x | dx | x | dx
0
1
1
1
1
1
2 2 2 2 (see parts (a) and (b) above).
b a and let c
k
n
n (b a )
a n
and
63. Consider the partition P that subdivides the interval [a, b] into n subintervals of width x
the right endpoint of each subinterval. So the partition is P a, a
ck a
k (b a )
. We
n
n
n
k 1
k 1
get the Riemann sum f (ck ) x c b n a
As n and P 0 this expression remains c(b a ). Thus,
ba ,
n
a
c (b a )
n
b
2(b a )
, ...,
n
n
1
k 1
c (b a )
n
n
c(b a ).
a c dx c(b a) .
64. Consider the partition P that subdivides the interval [0, 2] into n subintervals of width x
be the right endpoint of each subinterval. So the partition is P 0,
We get the Riemann sum
n
n
n
n
n
k 1
k 1
k 1
k 1
k 1
2,
n
2 n2 , . . . , n n2
f (ck )x 2 2nk 1 n2 n2 4nk 1 n82 k n2 1 n82
P 0 the expression
4( n 1)
n
2 has the value 4 2 6. Thus,
n ( n 1)
2
n2 n
4( n 1)
n
b a na 2
n
2 a (b a )2 n ( n 1)
2
n2
(b a )3 n ( n 1)(2 n 1)
6
n3
2
n
and let ck
2k .
n
2 . As n and
2
0 (2 x 1) dx 6.
2(b a )
the right endpoint of each subinterval. So the partition is P a, a b n a , a n , . . . , a
n
n
n
k (b a )
k (b a ) 2
ck a n . We get the Riemann sum
f (ck ) x ck2 b n a b n a
a n
k 1
k 1
k 1
n
n
n
n
2 a (b a )
(b a ) 2
2 2ak (b a ) k 2 (b a )2 b a
b n a
n a2 n
k 2
k2
a
2
n
n
n
k 1
k 1
k 1
k 1
20
n
2 and ck k n2
65. Consider the partition P that subdivides the interval [a, b] into n subintervals of width x
be
b a and let c
k
n
n (b a )
and
n
(b a )a 2 a(b a) 2 nn1
Copyright 2018 Pearson Education, Inc.
(b a )3 ( n 1)(2 n 1)
6
n2
be
308
Chapter 5 Integrals
(b a )a 2 a(b a) 2
1 1n
1
(b a)a 2 a (b a )2 1
b 2
a x
dx
b3
3
( b a )3
6
( b a )3
6
2 n3
1
n2
As n and P 0 this expression has value
1
2 ba 2 a3 ab 2 2a 2 b a3 13 (b3 3b 2 a 3ba 2 a3 )
b3
3
3
a3 . Thus,
3
a3 .
0 ( 1)
n
1 and
n
let ck be the right endpoint of each subinterval. So the partition is P 1, 1 1n , 1 2 1n , , 1 n 1n 0
n
n
k
k 2 1
and ck 1 k 1n 1 kn . We get the Riemann sum
f (ck )x
1 n 1 n n
k 1
k 1
n
n
n
n
2
n ( n 1)
n ( n 1)(2 n 1)
3
k
k
2k
2
1n
k 13 k 2 n2 n 32 2 13
1 n 1 n n n 1 n2
6
n
n
n
k 1
k 1
k 1
k 1
3( n 1) ( n 1)(2 n 1)
2 2 n
. As n and || P || 0 this expression has value 2 32 13 56 .
6n2
0
Thus,
( x x 2 ) dx 56 .
1
66. Consider the partition P that subdivides the interval [1, 0] into n subintervals of width x
2 ( 1)
n
67. Consider the partition P that subdivides the interval [1, 2] into n subintervals of width x
let ck be the right endpoint of each subinterval. So the partition is P 1, 1
n
and ck 1 k n3 1 3nk . We get the Riemann sum f (ck )x
3
n
3 18nk 27nk
n
n
n
2 6nk 1 18
1 722 k
n
2
2
n
k 1
k 1
k 1
36( n 1) 27( n 1)(2 n 1)
.
As
n
and
||
P
|| 0
n
2n2
2
2
18
Thus,
1 (3x
k 1
n
81
k2
n3
k 1
n
2 1
k 1
n
3
n
and
3 , 1 2 3 , , 1 n 3
n
n
n
2
3k
3k
3
3 1 n
18
n 722
n
n ( n 1)
2
813
n
n
1 n
2
n ( n 1)(2 n 1)
6
this expression has value 18 36 27 9.
2 x 1)dx 9.
68. Consider the partition P that subdivides the interval [1, 1] into n subintervals of width x
1 ( 1)
n
n2 and let
ck be the right endpoint of each subinterval. So the partition is P 1, 1 n2 , 1 2 n2 , , 1 n n2 1 and
n
ck 1 k n2 1 2nk . We get the Riemann sum f (ck )x
2
n
1 6nk 12nk
n
2
k 1
n2 n 122
n
2 6
Thus,
1
1 1n
1
1 x
3
n ( n 1)
2
4
2
3
8k3
n
k 1
n ( n 1)(2 n 1)
6
2 3n 12
n
1 n2 12
n
1
n
k 1
k 1
ck3 n2 n2 1
n
n
n
n
n2 1 n6 k 122 k 2 83 k 3
n
n
k 1
k 1
k 1
k 1
243
n
n
4
1
164
n
n ( n 1) 2
2
2 6 nn1 4
( n 1)(2 n 1)
n2
4
2k 3
n
( n 1)2
n2
1 1n
1
. As n and || P || 0 this expression has value 2 6 8 4 0.
dx 0.
b a and let c be
k
n
2(b a )
n (b a )
b
a
is P a, a n , a n , , a n b and
3
n
n
k (b a )
ck3 b n a b n a
a n
k 1
k 1
69. Consider the partition P that subdivides the interval [a, b] into n subintervals of width x
the right endpoint of each subinterval. So the partition
ck a
2 6
k (b a )
. We
n
n
get the Riemann sum f (ck )x
k 1
Copyright 2018 Pearson Education, Inc.
Section 5.3 The Definite Integral
ba
n
n
3a 2 k (b a ) 3ak 2 (b a )2 k 3 (b a )3
a3 n n2 n3
k 1
b a na3
n
2
3a ( b a )
(b a )a3
(b a )a3
2
n ( n 1)
2
3a ( b a )
3
value (b a) a3
3a 2 ( b a ) 2
2
a ( b a )3
2
2 n3
a (b a)3
1
n2
1
(b a ) 4
4
a3
(b a )
4
3a 2 ( b a )
n
n ( n 1) 2
2
n4
4
(b a ) ( n 1)2
2
4
n
(b a ) 4
4
b4
4
n
k 1
n ( n 1)(2 n 1)
6
n2
n3
2
2
3a (b a ) n 1 a (b a )3 ( n 1)(2 n 1)
n 2
2
n2
1
3a 2 (b a )2 1 n
2
1
ba
n
1 n2
n
k
3a ( b a ) 2
n
k 1
2
n
k2
(b a )3
n
k 1
3
309
n
k 1
k 3
1
n2
. As n and || P || 0 this expression has
1
4
a4 . Thus,
b 3
a x
dx
b4
4
4
a4 .
70. Consider the partition P that subdivides the interval [0, 1] into n subintervals of width x 1n0
1
n
and let ck be
the right endpoint of each subinterval. So the partition is P 0, 0 1n , 0 2 1n , , 0 n 1n 1 and
n
n
n
k
k 3
ck 0 k 1n kn . We get the Riemann sum
f (ck )x
3ck ck3 1n 1n
3 n n
k 1
k 1
k 1
2
1
n
1 n 2
n
1 1
n ( n 1)
n ( n 1) 2
( n 1) 2
1n n3
k 13
k 3 32 2 14 2
32 nn1 14 2 32 1 n 14 1 n . As n
n
n
n
n
k 1
k 1
and
1
|| P || 0 this expression has value 32 14 54 . Thus, (3 x x3 ) dx 54 .
0
71. To find where x x 2 0, let x x 2 0 x(1 x) 0 x 0 or x 1. If 0 x 1, then 0 x x 2 a 0 and
b 1 maximize the integral.
72. To find where x 4 2 x 2 0, let x 4 2 x 2 0 x 2 ( x 2 2) 0 x 0 or x 2. By the sign graph,
++++++ 0 0 0 +++++++, we can see that x 4 2 x 2 0 on 2, 2 a 2 and b 2
2
0
minimize the integral.
73.
f ( x)
1
1 x 2
2
is decreasing on [0, 1] maximum value of f occurs at 0 max f f (0) 1; minimum value of
f occurs at 1 min f f (1)
1
112
12 . Therefore, (1 0) min f
1 1
0 1 x 2
dx (1 0) max f
1
12 1 2 dx 1. That is, an upper bound 1 and a lower bound 12 .
0 1 x
74. See Exercise 73 above. On [0, 0.5], max f
(0.5 0) min f
and min f
Then
1
4
1
112
52
0.5
0
1, min f
f ( x) dx (0.5 0) max f
0.5. Therefore (1 0.5) min f
0.5 1
1 x 2
0
1
1 02
dx
1
1
0.5 1 x 2
dx 12 52
1
13
20
2
5
1
0.5 1
1 x 2
0
1
0.5 1 x 2
1
1 (0.5)2
dx 12 . On [0.5, 1], max f
dx (1 0.5) max f
1 1
0 1 x 2
1
4
1
1 (0.5)2
1
1
0.5 1 x 2
0.8
dx 52 .
9 .
dx 10
75. 1 sin x 2 1 for all x (1 0)(1) sin x 2 dx (1 0)(1) or
0
0.8. Therefore
1
0 sin x
equal 2.
Copyright 2018 Pearson Education, Inc.
2
1
dx 1 sin x 2 dx cannot
0
310
Chapter 5 Integrals
76. f ( x) x 8 is increasing on [0, 1] max f f (1) 1 8 3 and min f f (0) 0 8 2 2. Therefore,
(1 0) min f
1
x 8 dx (1 0) max f 2 2
0
1
0
x 8 dx 3.
b
77. If f ( x) 0 on [a, b], then min f 0 and max f 0 on [a, b]. Now, (b a ) min f f ( x) dx (b a ) max f .
a
b
Then b a b a 0 (b a) min f 0 f ( x) dx 0.
a
b
78. If f ( x) 0 on [ a, b], then min f 0 and max f 0. Now, (b a ) min f f ( x) dx (b a ) max f . Then
a
b
b a b a 0 (b a ) max f 0 f ( x) dx 0.
a
1
1
1
0
0
0
79. sin x x for x 0 sin x x 0 for x 0 (sin x x) dx 0 (see Exercise 78) sin x dx x dx 0
2
2
1
1
1
1
sin x dx x dx sin x dx 12 02 sin x dx 12 . Thus an upper bound is 12 .
0
0
0
0
0 on , sec x 1 dx 0 (see Exercise 77)
since [0, 1] is contained in , sec x dx 1 dx 0 sec x dx 1 dx
sec x dx 1 dx x dx sec x dx (1 0) sec x dx . Thus a lower bound is .
80. sec x 1
x2
2
x2
2
on 2 , 2 sec x 1
2
1
1
0
0
2
1 1 2
2 0
2
1
1
0
0
1
b
2
b
1
ba a
1
x2
2
0
x2
2
1 13
2 3
0
81. Yes, for the following reasons: av( f )
b
1
1
0
0
1
x2
2
7
6
7
6
0
f ( x) dx is a constant K. Thus
b
b
a av( f ) dx a K dx K (b a) a av( f ) dx (b a) K (b a) b1a a
b
f ( x) dx f ( x) dx.
a
82. All three rules hold. The reasons: On any interval [ a, b] on which f and g are integrable, we have:
b
b
b
b
b
(a) av( f g ) b 1 a [ f ( x) g ( x)]dx b 1 a f ( x) dx g ( x) dx b 1 a f ( x) dx b 1 a g ( x) dx
a
a
a
a
a
av( f ) av( g )
(b) av(kf )
(c) av( f )
b
1
kf ( x)
ba a
b
1
ba a
dx
f ( x) dx
1
ba
k b f ( x) dx k 1 b f ( x) dx k av( f )
a
b a a
b
1
g ( x)
ba a
dx since f ( x) g ( x) on [a, b], and
b
1
g ( x)
ba a
dx av( g ).
Therefore, av( f ) av( g ).
83. (a) U max1 x max 2 x max n x where max1 f ( x1 ), max 2 f ( x2 ) , , max n f ( xn ) since f is
increasing on [a, b]; L min1 x min 2 x min n x where min1 f ( x0 ), min 2 f ( x1 ) , ,
min n f ( xn 1 ) since f is increasing on [a, b]. Therefore
U L (max1 min1 )x (max 2 min 2 )x (max n min n )x
( f ( x1 ) f ( x0 ))x ( f ( x2 ) f ( x1 ))x ( f ( xn ) f ( xn 1 ))x ( f ( xn ) f ( x0 )) x
( f (b) f (a)) x.
(b) U max1 x1 max 2 x2 max n xn where max1 f ( x1 ), max 2 f ( x2 ) , , max n f ( xn ) since f
is increasing on [a, b]; L min1 x1 min 2 x2 ... min n xn where min1 f ( x0 ), min 2 f ( x1 ), ,
min n f ( xn 1 ) since f is increasing on [a, b]. Therefore
U L (max1 min1 ) x1 (max 2 min 2 )x2 (max n min n ) xn
Copyright 2018 Pearson Education, Inc.
Section 5.3 The Definite Integral
311
( f ( x1 ) f ( x0 )) x1 ( f ( x2 ) f ( x1 ))x2 ( f ( xn ) f ( xn 1 )) xn
( f ( x1 ) f ( x0 )) xmax ( f ( x2 ) f ( x1 ))xmax ( f ( xn ) f ( xn 1 )) xmax . Then
U L ( f ( xn ) f ( x0 )) xmax ( f (b) f (a)) xmax f (b) f (a) xmax since f (b) f (a). Thus
lim (U L) lim ( f (b) f (a )) xmax 0, since xmax P .
P 0
P 0
84. (a) U max1 x max 2 x max n x where
max1 f ( x0 ), max 2 f ( x1 ), , max n f ( xn 1 )
since f is decreasing on [a, b];
L min1 x min 2 x min n x where
min1 f ( x1 ), min 2 f ( x2 ), , min n f ( xn )
since f is decreasing on [ a, b]. Therefore
U L (max1 min1 ) x (max 2 min 2 ) x
... (max n min n ) x
( f ( x0 ) f ( x1 )) x ( f ( x1 ) f ( x2 )) x
... ( f ( xn 1 ) f ( xn )) x
( f ( x0 ) f ( xn )) x ( f (a) f (b)) x.
(b) U max1 x1 max 2 x 2 ... max n xn where max1 f ( x0 ), max 2 f ( x1 ), , max n f ( xn 1 )
since f is decreasing on [a, b]; L min1 x1 min 2 x2 min n xn where
min1 f ( x1 ), min 2 f ( x2 ), , min n f ( xn ) since f is decreasing on [ a, b]. Therefore
U L (max1 min1 )x1 (max 2 min 2 )x2 (max n min n )xn
( f ( x0 ) f ( x1 ))x1 ( f ( x1 ) f ( x2 ))x2 ( f ( xn 1 ) f ( xn ))xn ( f ( x0 ) f ( xn ))xmax
( f (a ) f (b)xmax f (b) f (a) xmax since f (b) f (a). Thus
lim (U L) lim f (b) f (a) xmax 0, since xmax P .
P 0
P 0
85. (a) Partition 0, 2 into n subintervals, each of length x 2n with points x0 0, x1 x,
x2 2x,... , xn nx 2 . Since sin x is increasing on 0, 2 , the upper sum U is the sum of the areas
of the circumscribed rectangles of areas f ( x1 )x (sin x)x, f ( x2 )x (sin 2x)x,... ,
f ( xn )x (sin nx)x.
cos x cos n 12 x
cos 4n cos n 12 2n
x
Then U (sin x sin 2x ... sin nx)x 2
2n
2sin 2x
2sin 4n
cos 4n cos
2 4n
4 n sin 4n
(b) The area is
/2
0
cos 4n cos
2 4n
sin
4n
4n
sin x dx lim
cos 4n cos
sin
n
2 4n 1cos 2
4n
4n
1
1.
n
86. (a) The area of the shaded region is xi mi which is equal to L.
i 1
n
(b) The area of the shaded region is xi M i which is equal to U.
i 1
(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.
Copyright 2018 Pearson Education, Inc.
312
Chapter 5 Integrals
n
n
i 1
i 1
87. By Exercise 86, U L xi M i xi mi where M i max { f ( x) on the ith subinterval} and
n
n
i 1
i 1
mi min { f ( x) on ith subinterval}. Thus U L ( M i mi )xi xi provided xi for each
n
n
i 1
i 1
i 1, , n. Since xi xi (b a) the result, U L (b a ) 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
mi/hr 37.5 mi/hr. In
8 hours, for an average value of 300
8
terms of average value of functions, the function whose
average value we seek is v(t )
average value is
89–94.
(30)(5) (50)(3)
8
30, 0 t 5
50, 5 t 8 ,
and the
37.5.
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
Avg[n] : evalf(add(y,y Ylist)/nops(Ylist));
# (c)
end do;
avg : FunctionAverage( f(x), x a..b, output value );
Copyright 2018 Pearson Education, Inc.
Section 5.4 The Fundamental Theorem of Calculus
313
evalf( avg );
FunctionAverage(f(x),x a..b, output plot);
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 );
95–98.
# (d)
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, π}; n 10; dx (b a)/n;
f Sin[x]2 ;
xvals Table[N[x],{x, a, b dx, dx}];
yvals f /.x xvals;
boxes MapThread[Line[{{#1, 0},{#1, #3},{#2, #3},{#2, 0}]&,{xvals, xvals dx, yvals}];
Plot[f, {x, a, b}, Epilog 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, π}; n 10; dx (b a)/n;
f Sin[x]2 ;
xvals Table[N[x], {x, a dx, b, dx}];
yvals f /.x xvals;
boxes MapThread[Line[{{#1, 0},{#1, #3},{#2, #3},{#2, 0}]&,{xvals, dx,xvals, yvals}];
Plot[f, {x, a, b}, Epilog 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, π}; 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 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 boxes];
Sum[yvals[[i]] dx, {i, 1, Length[yvals]}]//N
5.4
1.
THE FUNDAMENTAL THEOREM OF CALCULUS
2
2
0 x( x 3) dx 0 ( x
2
3
2 2
(2)3 3(2)2 (0)3 3(0)2
3x) dx x3 32x 3 2 3 2 10
3
0
Copyright 2018 Pearson Education, Inc.
314
2.
Chapter 5 Integrals
1 x
1
2
1
3
(1)3
( 1)3
2 x 3 dx x3 x 2 3 x 3 (1)2 3(1) 3 (1)2 3(1) 20
1
3
2
2
1
1
3
1
1
124
3.
dx
3 3 1
4
3
125
125
( x 3) 2 (5) (1)
2 ( x 3)
4.
1
1 x
299
x 300
dx
300
1
1
1
1
(1)300 ( 1)300
1 1 0
300
300
4
4
x3
x4
44
14
1 753
5. 3x 2 dx x 3 43 13 64 16 1
4
16
16
16
16
16
1
1
6.
4
x4
3
2
2 x 2 x 3 dx 4 x 3x
1
3
34
( 2)4
81
105
32 3(3)
( 2)2 3( 2) 6
4
4
4
4
7.
0 x
8.
1
9.
0
1
2
32 6/5
x
/3
1
x dx x3 32 x3/2
0
3
32
dx 5 x 1/5 52 (5)
1
0 (1 cos x) dx [ x sin x]0
11.
/4
3 /4
/3
0
(2 tan 0) 2
sin u
2
1 cos 2t
2
2
dt
4
cos u
13.
14.
/3 sin
2
t dt
/3
4
cos u 0
3 0 2 3
( sin ) (0 sin 0)
csc 4
csc cot d [ csc ]3/4/4 csc 34
du
/3
5
2
2sec2 x dx [2 tan x]0 /3 2 tan 3
10.
12.
0
13 32 0 1
2 2 0
4
4
4
(1/2) 1
1 1 cos 2t dt 1 t 1 sin 2t
2 4 2 12 (0) 14 sin 2(0) 12 2 14 sin 2 2 4
/2 2 2
0
0
Use the double angle formula cos 2t 1 2sin 2 t which implies that sin 2 t
Copyright 2018 Pearson Education, Inc.
1 cos(2t )
.
2
Section 5.4 The Fundamental Theorem of Calculus
/3
/3 sin
315
/3
/3 1 cos 2t
t sin 2t
t dt
dt
/3
2
4 /3
2
2
1 3 1
3
3
6 4 2 6 4
2
3
4
/4
15.
0
16.
0
/6
tan 2 x dx
/4
0
(sec2 x 1) dx [tan x x]0 /4 tan 4 4 (tan(0) 0) 1 4
(sec x tan x)2 dx
/6
0
/6
(sec2 x 2sec x tan x tan 2 x ) dx
0
(2sec2 x 2sec x tan x 1) dx
(2 tan 0 2sec 0 0) 2
[2 tan x 2sec x x]0 /6 2 tan 6 2sec 6 6
/8
17.
0
18.
sin 2 x dx 12 cos 2 x
4
4sec t dt
/8
0
1
19.
1
20.
23.
24.
4
4
2
1
3
3
3
4
4
u7
2 2
1
u5
y3
1/3
du
1
1
u8 1
u 5 du 16
4u 4
2
3
3
3
2( 3) 2 4( 3) 10 3
18
( 2 )8
16
1 4 16
4(1)
4
1
1
2
4
3
4
y3
( 1)3
( 3)3
( y 2 2 y 2 ) dy 3 2 y 1 3 ( 21) 3 ( 23)
3
3
1
2
(1 s 3 2 ) ds s
dx
x
2(8) 53 (8)5/3
u7
2 2
1
ds
1/3
2/3
8 x 1 2 x
4
3
(t 3 t 2 4t 4) dt t4 t3 2t 2 4t
3
dy
3
3 4
3
2( 3)2 4 3 4
3
3
3
2 s2 s
1
s2
1
(t 1)(t 2 4) dt
1 y 5 2 y
1
1
3
3
( 1)3
(r 2 2r 1) dr r3 r 2 r 3 (1)2 (1) 13 12 1 83
1
1
3
3
(r 1)2 dr
22.
12 cos 2(0) 24 2
12 cos 2
(4sec2 t t 2 ) dt 4 tan t t
3
/3
3
t
4 tan 4 4 tan 3 (4(1) 4) 4 3 3 4 3 3
4
3
2
21.
3 6 2
8
1/3
1 2 x
3(8)2/3
x 2 x 2/3
x1/3
3 (8) 4/3
4
2
s 1
2
2
2
2
1
2
1
22
3
2 23/4 1 2 4 8 1
3
8
dx (2 x 2/3 2 x 1/3 x1/3 ) dx 2 x 53 x5/3 3x 2/3 34 x 4/3
1
1
2(1) 53 (1)5/3 3(1)2/3 43 (1)4/3 13720
Copyright 2018 Pearson Education, Inc.
316
Chapter 5 Integrals
25.
sin 2 x dx
/2 2sin
/2 2sin2sinx cosx x dx /2 cos x dx sin x / 2 (sin( )) sin 2 1
x
26.
0
/3
(cos x sec x)2 dx
2
/3
0
(cos2 x 2 sec2 x) dx
/3 cos 2 x 1
0
2 sec2 x dx
2
/3
/3 1
cos 2 x 5 sec2 x dx 1 sin 2 x 5 x tan x
0
2
4
0
2
14 sin 2(0) 52 (0) tan(0) 56 9 83
14 sin 2 3 52 3 tan 3
4
0
4
0
4
0
4
cos x dx [sin x]0 /2
0 12 cos x cos x dx 0
/2 1
2
(cos x cos x ) dx
sin 2 sin 0 1
29.
0
1
31.
2
x
1 x 2
/3
0
33. (a)
x
x
5
1
5
2
2
0
x
(b)
d
dx
34. (a)
1
35. (a)
(b)
dx
0
2 1 cos 1 2 2 cos 1
5
/3
1 (cos x cos x )
1
1
sin sin 0
2
2
2
dx x(1 x 2 )1/2 dx 1 x 2
sin 2 x cos x dx
0
0
2
dx 2 cos x
26 5
(sin x)2 cos x dx 13 (sin x)3
/3
0
cos t dt [sin t ]0 x sin x sin 0 sin x
d (sin x ) cos x
dx
(b)
/2
1
x cos x 2 dx sin x 2
2
2 sin
30.
32.
/2
/2 2
12 x1/2 cos2 xx
x
d
0 cos t dt (cos x ) dx ( x ) (cos x )
13 sin 3 3 13 sin 3 (0)
d
dx
3
8
x
0 cos t dt
12 x1/2 cos2 xx
d sin x 3t 2 dt d (sin 3 x 1) 3sin 2 x cos x
3t 2 dt [t 3 ]1sin x sin 3 x 1 dx
1
dx
sin
x
d
d (sin x) 3sin 2 x cos x
3t 2 dt (3sin 2 x) dx
dx 1
sin x
t4
0
4
d t
dt
0
t4
t4
u du u1/2 du 23 u 3/2 23 (t 4 )3/2 0 23 t 6
0
0
u du t 4
4
d t
dt 0
dtd (t 4 ) t 2 (4t3 ) 4t5
Copyright 2018 Pearson Education, Inc.
/2
28.
2
( 4) 2
2
4 | x | dx 4 | x | dx 0 | x | dx 4 x dx 0 x dx x2 4 x2 0 02
2
42 02
2 2 16
2
27.
u du
4t 5
d 2 t6
dt 3
Section 5.4 The Fundamental Theorem of Calculus
36. (a)
tan
tan
sec2 y dy [tan y ]0tan tan (tan ) 0 tan (tan ) dd
sec2 y dy
0
dd (tan(tan )) (sec 2 (tan ))sec 2
0
x 3 2/3
t
dt
0
37. (a)
x3
3t1/3
3( x 0) 3 x
0
3
d x t 2/3 dt
dx 0
d (3 x )
dx
3
dxd x3 x 2 3x2 3
3
d x t 2/3 dt
dx 0
(b)
tan sec 2 y dy (sec 2 (tan )) d (tan ) (sec2 (tan )) sec2
0
d
d
d
(b)
2/3
x3
t
3 dx x5 3sin 1 x
15 t 5/2 3sin 1 t
x
5
2
1 x
0
t
d
d 1 t 5/2 3sin 1 t 1 5 t 3/2 3
1
1 t 1 2 12 t 3/2 23 1
x 4 3 2 dx dt
5
5 2
dt 0
2 2
t 1t
1 x
1 t
0
38. (a)
t 4
t 4
d
x
dt 0
(b)
39. y
x
0
41.
y
42.
y x
1 t 2 dt
0
x
dy
dx
sin t 2 dt
x2
2
sin t 3 dt
2 x 2 sin x6
x
t2
1 t 4
x2
2
x
0
dy
dx
1
t
2
d
dt
t t 2
1 t 1/2
2
12 t 3/2 32
x1
1 t
dt
dy
dx
3
1t
dy
dx
sin( x )2
1
t 1t
1x , x 0
dxd ( x ) (sin x) 12 x1/2 sin2 xx
2
2
d x sin t 3 dt 1 x sin t 3 dt x sin ( x 2 )3
x dx
2
2
2
d ( x 2 ) x sin t 3 dt
dx
2
sin t 3 dt
x t2
3 t 2 4
44.
x
y (t 3 1)10 dt
0
45.
y
sin x
dt
0
1t 2
46.
y
0
3
dt
dy
dx
, x 2
3
40. y
sin t 2 dt
y
tan x dt
1t 2
t
4
1 x2
43.
2
dt
dx
3
1 x 2
dy
dx
dy
dx
x2
x 4
2
x2
x 4
2
0
x
3 (t 3 1)10 dt
0
dy
dx
d x (t 3
dx 0
d (sin x)
1
1sin 2 x dx
1
1 tan 2 x
2
d (tan x)
dx
x
1)10 dt 3( x3 1)10 (t 3 1)10 dt
0
1
cos 2 x
(cos x)
cos x
cos x
sec1 x sec2 x 1
2
Copyright 2018 Pearson Education, Inc.
cos x
cos x
1 since x 2
2
317
318
Chapter 5 Integrals
47. x 2 2 x 0 x( x 2) 0 x 0 or x 2;
Area
2
3
( x 2 2 x)dx
0
2
( x 2 2 x)dx
2
( x 2 2 x)dx
0
2
0
2
3
3
3
x3 x 2 x3 x 2 x3 x 2
3
2
0
3
( 2)3
(
3)
3 (2) 2 3 (3)2
3
3
( 2)
03 02 3 (2) 2
23
3
2
2
03
3
02
28
3
48. 3 x 2 3 0 x 2 1 x 1;
because of symmetry about the y -axis,
1
2
Area 2 (3 x 2 3)dx (3 x 2 3)dx
1
0
2 [ x3 3 x]10 [ x3 3 x]12
2[((13 3(1)) (03 3(0))) ((23 3(2)) (13 3(1))] 2(6) 12
49. x3 3x 2 2 x 0 x( x 2 3 x 2) 0
x( x 2)( x 1) 0 x 0, 1, or 2;
1
2
0
1
Area ( x3 3 x 2 2 x)dx ( x3 3 x 2 2 x )dx
1
2
4
4
4
x4 x3 x 2 x4 x3 x 2 14 13 12
0
1
4
4
24 23 22 14 13 12 12
50.
04
4
03 02
x1/3 x 0 x1/3 1 x 2/3 0 x1/3 0 or 1 x 2/3 0 x 0 or
1 x
2/3
2
x 0 or 1 x x 0 or x 1;
0
1
8
1
0
1
Area ( x1/3 x) dx ( x1/3 x) dx ( x1/3 x) dx
0
1
8
2
2
2
34 x 4/3 x2 43 x 4/3 x2 43 x 4/3 x2
1
0
1
2
2
( 1)
34 (0)4/3 02 34 (1) 4/3 2
2
2
34 (1) 4/3 12 34 (0) 4/3 02
3 (8) 4/3
4
2
82
3 (1) 4/3
4
2
12
14 14 (20 43 12 ) 83
4
Copyright 2018 Pearson Education, Inc.
Section 5.4 The Fundamental Theorem of Calculus
319
51. The area of the rectangle bounded by the lines y 2, y 0, x , and x 0 is 2 . The area under the curve
y 1 cos x on [0, ] is
0 (1 cos x) dx [ x sin x]0
( sin ) (0 sin 0) . Therefore the area of the
shaded region is 2 .
52. The area of the rectangle bounded by the lines by the lines x 6 , x 56 , y sin 6
1
2
sin 56 , and y 0 is
6 3 . The area under the curve y sin x on 6 , 56 is 5/6/6 sin x dx [ cos x]5/6/6
cos 56 cos 6 23 23 3. Therefore the area of the shaded region is 3 3 .
1 5
2 6
53. On 4 , 0 : The area of the rectangle bounded by the lines y 2, y 0, 0, and 4 is 2 4
4 2 . The area between the curve y sec tan and y 0 is
0
sec tan d sec /4
0
/4
( sec 0) sec 4 2 1. Therefore the area of the shaded region on 4 , 0 is 4 2 ( 2 1).
On 0, 4 : The area of the rectangle bounded by 4 , 0, y 2, and y 0 is 2 4 4 2 . The area
under the curve y sec tan is
/4
0
/4
sec tan d sec 0
sec 4 sec 0 2 1. Therefore the area of
the shaded region on 0, 4 is 4 2 ( 2 1). Thus, the area of the total shaded region is
2
4
2 1 4 2 2 1 2 2 .
2 2 . The area
54. The area of the rectangle bounded by the lines y 2, y 0, t 4 , and t 1 is 2 1 4
under the curve y sec 2 t on 4 , 0 is
curve y 1 t 2 on [0, 1] is
1
0 (1 t
2
0
4 sec
1
2
t dt [tan t ]0 4 tan 0 tan 4 1. The area under the
2
3
3
) dt t t3 1 13 0 03 32 . Thus, the total area under the curves
0
on 4 , 1 is 1 23 53 . Therefore the area of the shaded region is 2 2 53 13 2 .
55.
x
dy
y 1t dt 3 dx 1x and y ( ) 1t dt 3 0 3 3 (d) is a solution to this problem.
56.
y sec t dt 4 dx sec x and y (1) sec t dt 4 0 4 4 (c) is a solution to this problem.
1
1
57.
y sec t dt 4 dx sec x and y (0) sec t dt 4 0 4 4 (b) is a solution to this problem.
0
0
58.
x
1
dy
y 1t dt 3 dx 1x and y (1) 1t dt 3 0 3 3 (a) is a solution to this problem.
1
1
59.
y sec t dt 3
x
dy
x
dy
x
2
1
0
60.
y
x
1
1 t 2 dt 2
Copyright 2018 Pearson Education, Inc.
320
Chapter 5 Integrals
61. Area
b /2
b /2
h x dx hx
4h
b2
2
b /2
4 hx3
2
3b b /2
2
2
4 h b
4 h b2
h b2 22 h b2
3b
3b 2
bh bh bh bh bh 2 bh
bh
2
6
2
6
3
3
62. k 0 one arch of y sin kx will occur over the interval 0, k the area
63.
1k cos
dc
dx
k
k
1k cos (0)
/k
0
sin kx dx k1 cos kx
2
k
x
1 12 x 1/2 c 12 t 1/2 dt [t1/2 ]0x x; c(100) c (1) 100 1 $9.00
0
2 x
3
3
64. r 2 2 2 dx 2 1 1 2 dx 2 x
0
0
( x 1)
( x 1)
1
1
2 3 4 1 2 2 4 4.5 or $4500
x11 0 2 3 (311) 0 (011)
3
65. (a) t 0 T 85 3 25 0 70F; t 16 T 85 3 25 16 76F;
t 25 T 85 3 25 25 85F
25
25
1 85t 2(25 t )3/2
(b) average temperature 2510 85 3 25 t dt 25
0
0
1 85(25) 2(25 25)3/2 1 85(0) 2(25 0)3/2 75F
25
25
66. (a) t 0 H 0 1 5(0)1/3 1 ft; t 4 H 4 1 5(4)1/3 5 53 4 10.17 ft;
t 8 H 8 1 5(8)1/3 13 ft
(b) average height 81 0
0
8
8
t 1 5t1/3 dt 81 32 (t 1)3/2 15
t 4/3
4
0
18 23 (8 1)3/2 15
(8)4/3 81 32 (0 1)3/2 15
(0) 4/3 29
9.67 ft
4
4
3
x
d x f (t ) dt d ( x 2 2 x 1) 2 x 2
f (t ) dt x 2 2 x 1 f ( x) dx
1
dx
x
f (t ) dt x cos x f ( x)
67.
1
68.
0
69.
f ( x) 2
x 1 9
2 1t
d x
dx 0
f (t ) dt cos x x sin x f (4) cos (4) (4) sin (4) 1
11
dt f ( x) 1 (9x 1) x92 f (1) 3; f (1) 2 19 t dt 2 0 2;
2
L( x) 3( x 1) f (1) 3( x 1) 2 3x 5
Copyright 2018 Pearson Education, Inc.
/k
0
Section 5.4 The Fundamental Theorem of Calculus
70.
g ( x) 3
x2
1
g (1) 3
321
sec(t 1) dt g ( x ) (sec( x 2 1))(2 x) 2 x sec( x 2 1) g ( 1) 2( 1) sec ((1)2 1) 2;
( 1) 2
1
1
sec(t 1) dt 3 sec(t 1) dt 3 0 3;
1
L( x) 2( x (1)) g (1) 2( x 1) 3 2 x 1
71. (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 g (1) f (1) 0.
False, since g (1) f (1) 0.
True, since g (1) 0 and g (1) f (1) 0.
False: g ( x) f ( x) 0, so g never changes sign.
True, since g (1) f (1) 0 and g ( x) f ( x) is an increasing function of x (because f ( x) 0).
72. Let a x0 x1 x2 xn b be any partition of [a, b] and left F be any antiderivative of f.
n
(a) [ F ( xi ) F ( xi 1 )]
i 1
[ F ( x1 ) F ( x0 )] [ F ( x2 ) F ( x1 )] [ F ( x3 ) F ( x2 )] [ F ( xn 1 ) F ( xn 2 )] [ F ( xn ) F ( xn 1 )]
F ( x0 ) F ( x1 ) F ( x1 ) F ( x2 ) F ( x2 ) F ( xn 1 ) F ( xn 1 ) F ( xn )
F ( xn ) F ( x0 ) F (b) F (a )
(b) Since F is any antiderivative of f on [a, b] F is differentiable of [a, b] F is continuous on [a, b].
Consider any subinterval [ xi 1, xi ] in [a, b], then by the Mean Value Theorem there is at least one
number ci in ( xi 1, xi ) such that [ F ( xi ) F ( xi 1 )] F (ci )( xi xi 1 ) f (ci )( xi xi 1 ) f (ci )xi .
n
n
Thus F (b) F (a) [ F ( xi ) F ( xi 1 )] f (ci )xi .
i 1
i 1
n
(c) Taking the limit of F (b) F (a) f (ci )xi we obtain lim ( F (b) F (a)) lim f (ci )xi
P
P
0
0
i 1
i 1
n
b
F (b) F (a) f ( x) dx
a
73. (a) v
ds
dt
(b) a
df
dt
3
d t f ( x ) dx f (t ) v(5) f (5) 2 m/sec
dt
0
is negative since the slope of the tangent line at t = 5 is negative
(c) s f ( x) dx 12 (3)(3)
0
9
2
m since the integral is the area of the triangle formed by y = f(x), the x-axis
and x = 3
(d) t = 6 since from t = 6 to t = 9, the region lies below the x-axis
(e) At t = 4 and t = 7, since there are horizontal tangents there
(f) Toward the origin between t = 6 and t = 9 since the velocity is negative on this interval. Away from the
origin between t = 0 and t = 6 since the velocity is positive there.
(g) Right or positive side, because the integral of f from 0 to 9 is positive, there being more area above the
x-axis than below it.
74. If the marginal cost is
x2
x
115, by the net change theorem the production cost is
1000 2
x
2
t
1 3 1 2
t
p( x )
x x 115 x. Thus the average cost per unit for 600 units is
115 dt
3000
4
0 1000 2
p(600)
85.
600
Copyright 2018 Pearson Education, Inc.
322
Chapter 5 Integrals
7578.
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 "#75(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 "75(b) (Section 5.4)" );
incr : solve( dF(x)>0, x );
decr : solve( dF(x)<0, x );
# (c)
df : D(f );
# (d)
p3 : plot( [df(x),F(x)], x a..b, legend ["y f '(x)","y F(x)"], title "#75(d) (Section 5.4)" ):
p3;
q2 : solve( df(x) 0, x );
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 "75(d) (Section 5.4)" );
79–82.
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);
cp : solve( dF(x) 0, x );
solve( dF(x)>0, x );
solve( dF(x)<0, x );
# (b)
d2F : D(dF);
solve( d2F(x) 0, x );
# (c)
plot( F(x), x -1..1, title "#79(d) (Section 5.4)" );
83.
Example CAS commands:
Maple:
f : `f `;
Copyright 2018 Pearson Education, Inc.
Section 5.5 Indefinite Integrals and the Substitution Method
q1: Diff( Int( f(t), t a..u(x) ), x );
d1: value( q1 );
84.
Example CAS commands:
Maple:
f : `f `;
q2 : Diff( Int( f(t), t a..u(x) ), x,x );
value( q2 );
75–84. 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, 2π}; 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 79 84.
Clear[x, f, F, u]
a 0; f[x_] x 2 2x 3
u[x_] 1 x 2
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}]
After determining an appropriate value for b, the following can be entered
b 4;
Plot[{F[x],{x, a, b}]
5.5
INDEFINITE INTEGRALS AND THE SUBSTITUTION METHOD
1. Let u 2 x 4 du 2 dx 12 du dx
5
2(2 x 4) dx 2u
51
du
2
u 5 du 16 u 6 C 16 (2 x 4)6 C
2. Let u 7 x 1 du 7 dx 17 du dx
7
7 x 1 dx 7(7 x 1)1/2 dx 7u1/2 17 du u1/2 du 23 u 3/2 C 23 (7 x 1)3/2 C
3. Let u x 2 5 du 2 x dx 12 du x dx
2 x( x
2
4
5) dx 2u 4 12 du u 4 du 13 u 3 C 13 ( x 2 5)3 C
4. Let u x 4 1 du 4 x3 dx 14 du x3 dx
3
( x44 x1)2 dx 4 x
3
( x 4 1)2 dx 4u 2 14 du u 2 du u 1 C 4 1 C
x 1
Copyright 2018 Pearson Education, Inc.
323
324
Chapter 5 Integrals
5. Let u 3 x 2 4 x du (6 x 4)dx 2(3x 2)dx 12 du (3x 2)dx
(3x 2)(3x
2
4 x) 4 dx u 4 12 du
6. Let u 1 x du
(1 x )1/3
x
1
2
u
4
1 u 5 C 1 (3 x 2 4 x)5 C
du 10
10
1 dx 2 du 1 dx
2 x
x
1/3
1/3 1
x)
dx u 2 du
x
dx (1
2 u1/3du 2 34 u 4/3 C 32 (1 x ) 4/3 C
7. Let u 3 x du 3 dx 13 du dx
sin 3x dx 13 sin u du 13 cos u C 13 cos 3x C
8. Let u 2 x 2 du 4 x dx 14 du x dx
x sin (2 x
2
) dx 14 sin u du 14 cos u C 14 cos 2 x 2 C
9. Let u 2t du 2 dt 12 du dt
sec 2t tan 2t dt 12 sec u tan u du 12 sec u C 12 sec 2t C
10. Let u 1 cos 2t du 12 sin 2t dt 2 du sin 2t dt
2
3
1 cos 2t sin 2t dt 2u du 32 u C 23 1 cos 2t
2
3
C
11. Let u 1 r 3 du 3r 2 dr 3du 9r 2 dr
2
1/2
1/2
3 1/2
9r dr3 3u du 3(2)u C 6(1 r ) C
1 r
12. Let u y 4 4 y 2 1 du (4 y 3 8 y ) dy 3 du 12 ( y 3 2 y ) dy 12( y 4 4 y 2 1) 2 ( y 3 2 y ) dy
3u 2 du u 3 C ( y 4 4 y 2 1)3 C
13. Let u x3/2 1 du 32 x1/2 dx 23 du x dx
14. Let u 1x du 12 dx
x sin 2 ( x3 2 1) dx 23 sin 2 u du 23 u2 14 sin 2u C 13 ( x3/2 1) 16 sin (2 x3/2 2) C
1
x2
x
15. (a) Let u cot 2 du 2 csc2 2 d 12 du csc 2 2 d
csc
2
2
2
2 cot 2 d 12 u du 12 u2 C u4 C 14 cot 2 2 C
(b) Let u csc 2 du 2 csc 2 cot 2 d 12 du csc 2 cot 2 d
csc
2
2
2
2 cot 2 d 12 u du 12 u2 C u4 C 14 csc 2 2 C
16. (a) Let u 5 x 8 du 5 dx 15 du dx
du
1 u 1/2 du 1 (2u1/2 ) C 2 u1/2 C
5
5
5
1/2
1
2
(5) dx 5 du dx
Let u 5 x 8 du 2 (5 x 8)
5 x 8
dx 2 du 2 u C 2 5 x 8 C
5
5
5
5 x 8
(b)
cos 2 1x dx cos 2 (u ) du cos 2 (u ) du u2 14 sin 2u C 21x 14 sin 2x C 21x 14 sin 2x C
dx
5 x 8
15 1
u
52 5 x 8 C
Copyright 2018 Pearson Education, Inc.
Section 5.5 Indefinite Integrals and the Substitution Method
17. Let u 3 2 s du 2 ds 12 du ds
32 u3/2 C 13 (3 2s)3/2 C
3 2s ds u 12 du 12 u1/2 du 12
18. Let u 5s 4 du 5 ds 15 du ds
1
5s 4
ds 1 15 du 15 u 1/2 du 15 (2u1/2 ) C 52 5s 4 C
u
19. Let u 1 2 du 2 d 12 du d
4
54 u5/4 C 52 (1 2 )5/4 C
1 2 d 4 u 12 du 12 u1/4 du 12
20. Let u 7 3 y 2 du 6 y dy 12 du 3 y dy
23 u3/2 C 13 (7 3 y 2 )3/2 C
7 3 y 2 dy u 12 du 12 u1/2 du 12
3y
21. Let u 1 x du 1 dx 2 du 1 dx
dx
1
x (1 x )2
2 du
u2
2 x
x
u2 C 2 C
1 x
22. Let u sin x du cos x dx
sin x 1 sin 2 x cos x dx u1/2 u5/2 du 23 u 3/2 72 u 7/2 C 23 sin 3/2 x 72 sin 7/2 x C
23. Let u 3 x 2 du 3dx 13 du dx
sec
2
(3x 2) dx (sec 2 u ) 13 du 13 sec 2u du 13 tan u C 13 tan(3x 2) C
24. Let u tan x du sec 2 x dx
2
2
2
3
3
tan x sec x dx u du 13 u C 13 tan x C
3x du 13 cos 3x dx 3 du cos 3x dx
5
5
6
6
sin 3x cos 3x dx u (3 du) 3 16 u C 12 sin 3x C
25. Let u sin
2x du 12 sec2 2x dx 2 du sec2 2x dx
7
2
7
8
8
tan 2x sec 2x dx u (2 du ) 2 18 u C 14 tan 2x C
26. Let u tan
3
2
r 1 du r dr 6 du r 2 dr
27. Let u 18
6
r
2 r3
18
5
6
6
r3 1 C
1 dr u 5 (6 du ) 6 u 5 du 6 u6 C 18
5
r du 1 r 4 dr 2 du r 4 dr
28. Let u 7 10
2
r
4
7 dr u (2 du) 2 u du 2 C 7
r5
10
3
3
3
u4
4
1
2
r5
10
4
C
29. Let u x3/2 1 du 32 x1/2 dx 23 du x1 2 dx
1/2
x
sin( x3/2 1) dx (sin u ) 23 du 23 sin u du 23 ( cos u ) C 23 cos( x3/2 1) C
Copyright 2018 Pearson Education, Inc.
325
326
Chapter 5 Integrals
v2 du 12 csc v2 cot v2 dv 2du csc v2 cot v2 dv
csc v2 cot v 2 dv 2du 2u C 2 csc v 2 C
30. Let u csc
31. Let u cos(2t 1) du 2sin(2t 1) dt 12 du sin(2t 1) dt
sin(2t 1)
cos2 (2t 1) dt 12 udu2 21u C 2 cos(21 t 1) C
32. Let u sec z du sec z tan z dz
1/2
1/2
sec z tan z dz 1 du u du 2u C 2 sec z C
sec z
u
33. Let u 1t 1 t 1 1 du t 2 dt du 12 dt
t
cos 1t 1 dt (cos u )(du ) cos u du sin u C sin 1t 1 C
1
t2
34. Let u t 3 t1/2 3 du 12 t 1/2 dt 2du 1 dt
t
1
t
cos( t 3) dt (cos u )(2 du ) 2 cos u du 2sin u C 2sin( t 3) C
35. Let u sin 1 du cos 1 12 d du 12 cos 1 d
1 sin 1 cos 1 d u du 1 u 2 C 1 sin 2 1 C
2
2
2
36. Let u csc du csc cot
cos
d
sin 2
x
1 x
u
x 1 dx
x5
1
x2
1
x2
1
x2
x 2 1 dx
x2
41. Let u 1
2
sin
C
1
x2
1
x2
1 1x dx u du u1/2 du 23 u 3/2 C
x 3 3
x11
3
x3
du
1
x3
2
x3
1
2
3
1 1x
3/2
dx 12 du
1
x2
1
x3
2
3
2 1x
3/2
C
dx
dx u 12 du
1
2
1/2
3/2
u du 13 u C 13 1
1
x2
3/2
C
du 94 dx 19 du 14 dx
dx
x
1
x4
x 3 3
x3
C
dx
2 1x dx u du u1/2 du 32 u 3/2 C
40. Let u 1
1
x3
cot csc d
dx
x 1 dx
x
39. Let u 2 1x du
x12
dx u 1 du u1/2 u 1/2 du 23 u 3/2 2u1/2 C 23 (1 x )3/2 2(1 x )1/2 C
38. Let u 1 1x du
1
cot csc d 2 du 2u C 2 csc C
1
37. Let u 1 x x u 1 dx du
2 1 d 2du
x
dx
1
x4
1
3
x3
dx u 91 du
1
9
1/2
u
du
2 u 3/2
27
Copyright 2018 Pearson Education, Inc.
C
2
27
1
3
x3
3/2
C
Section 5.5 Indefinite Integrals and the Substitution Method
327
42. Let u x3 1 du 3x 2 dx 13 du x 2 dx
x 4 dx
x3 1
x2
x3 1
dx
1 1
u 3
du 13 u 1/2 du 23 u1/2 C 23 ( x3 1)3/2 C
43. Let u x 1. Then du dx and x u 1. Thus x( x 1)10 dx (u 1) u10 du (u11 u10 )du
1 u12 1 u11 C 1 ( x 1)12 1 ( x 1)11 C
12
11
12
11
44. Let u 4 x. Then du 1 dx and (1)du dx and x 4 u. Thus x 4 xdx (4 u ) u (1) du
(4 u )(u1/2 ) du (u 3/2 4u1/2 ) du 52 u 5/2 83 u 3/2 C 52 (4 x)5/2 83 (4 x)3/2 C
45. Let u 1 x. Then du 1 dx and (1)du dx and x 1 u. Thus ( x 1)2 (1 x)5 dx
(2 u )2 u 5 (1) du (u 7 4u 6 4u 5 ) du 18 u8 74 u 7 23 u 6 C 18 (1 x)8 74 (1 x)7 23 (1 x)6 C
46. Let u x 5. Then du dx and x u 5. Thus ( x 5)( x 5)1/3 dx (u 10)u1/3 du (u 4/3 10u1/3 ) du
73 u 7/3 15
u 4/3 C 73 ( x 5)7/3 15
( x 5)4/3 C
2
2
47. Let u x 2 1. Then du 2 x dx and 12 du x dx and x 2 u 1. Thus x3 x 2 1 dx (u 1) 12 u du
12 (u 3/2 u1/2 ) du 12 52 u 5/2 23 u 3/2 C 15 u 5/2 13 u 3/2 C 15 ( x 2 1)5/2 13 ( x 2 1)3/2 C
48. Let u x3 1 du 3x 2 dx and x3 u 1. So 3x5 x3 1 dx (u 1) u du (u 3/2 u1/2 ) du
52 u 5/2 23 u 3/2 C 52 ( x3 1)5/2 23 ( x3 1)3/2 C
49. Let u x 2 4 du 2 x dx and 12 du x dx. Thus
14 u 2 C 14 ( x 2 4) 2 C
x
( x 2 4)3
dx ( x 2 4)3 x dx u 3 12 du
x
50. Let u 2 x 1 x 12 (u 1) dx 12 du. Thus
(2 x 1)2/3
1 3 u 4/3 3u1/3
4 4
1 ( u 1)
2
2/3
C 163 (2 x 1)4/3 43 (2 x 1)1/3 C
u
1
2
u
3
12 du 14 u1/3 u 2/3 du
51. (a) Let u tan x du sec2 x dx; v u 3 dv 3u 2 du 6dv 18u 2 du; w 2 v dw dv
2
2
18(2tan tanx secx)
3
6
2u 3
2
x dx
C
18u 2
(2u 3 ) 2
6
2 tan 3 x
du
6 dv
(2 v ) 2
6 dw
w2
6 w2 dw 6 w1 C 26 v C
C
(b) Let u tan 3 x du 3 tan 2 x sec 2 x dx 6 du 18 tan 2 x sec2 x dx; v 2 u dv du
2
2
18(2tan tanx secx)
3
2
x dx
6 du
(2 u ) 2
6 dv
v2
6v C 26 u C
6
2 tan 3 x
2
2
C
(c) Let u 2 tan 3 x du 3 tan 2 x sec2 x dx 6 du 18 tan x sec x dx
2
2
18(2tan tanx secx)
3
2
x dx
6 du
u2
u6 C
6
2 tan 3 x
C
Copyright 2018 Pearson Education, Inc.
du
328
Chapter 5 Integrals
52. (a) Let u x 1 du dx; v sin u dv cos u du; w 1 v 2 dw 2v dv 12 dw v dv
1 sin 2 ( x 1) sin( x 1) cos( x 1) dx 1 sin 2 u sin u cos u du v 1 v 2 dv
12 w dw 13 w3/2 C 13 (1 v 2 )3/2 C 13 (1 sin 2 u )3/2 C 13 (1 sin 2 ( x 1))3/2 C
(b) Let u sin( x 1) du cos( x 1) dx; v 1 u 2 dv 2u du 12 dv u du
1 sin 2 ( x 1) sin( x 1) cos( x 1) dx u 1 u 2 du 12 v dv 12 v1/2 dv
12 23 v3/2 C 13 v3/2 C 13 (1 u 2 )3/2 C 13 (1 sin 2 ( x 1))3/2 C
(c) Let u 1 sin 2 ( x 1) du 2sin( x 1) cos( x 1) dx 12 du sin( x 1) cos( x 1) dx
1 sin 2 ( x 1) sin( x 1) cos( x 1) dx 12 u du 12 u1/2 du
1 2 u 3/2
2 3
C 13 (1 sin 2 ( x 1))3/2 C
1 du (2r 1) dr ; v u dv
53. Let u 3(2r 1) 2 6 du 6(2r 1)(2) dr 12
(2 r 1) cos 3(2 r 1) 2 6
2
3(2 r 1) 6
dr cos u
u
1 du
12
(cos v) 16 dv 16 sin v C 16 sin
1
2 u
du 16 dv
u C
16 sin 3(2r 1)2 6 C
54. Let u cos du sin
sin
cos3
d
sin
cos3
2 1 d 2du sin d
d 23/du
2 u 3/2 du 2(2u 1/2 ) C 4 C
2
u
u
4
cos
C
55. Let u 3t 2 1 du 6t dt 2du 12t dt
s 12t (3t 2 1)3 dt u 3 (2 du ) 2 14 u 4 C 12 u 4 C 12 (3t 2 1)4 C ;
s 3 when t 1 3 12 (3 1)4 C 3 8 C C 5 s 12 (3t 2 1)4 5
56. Let u x 2 8 du 2 x dx 2 du 4 x dx
y 4 x( x 2 8) 1/3 dx u 1/3 (2du ) 2 32 u 2/3 C 3u 2/3 C 3( x 2 8)2/3 C ;
y 0 when x 0 0 3(8)2/3 C C 12 y 3( x 2 8) 2/3 12
du dt
57. Let u t 12
dt 8sin 2 u du 8 u 1 sin 2u C 4 t 2sin 2t C ;
s 8sin 2 t 12
2 4
12
6
2sin C C 8 1 9
s 8 when t 0 8 4 12
6
3
3
s4 t
12
2sin 2t 6 9 3 4t 2sin 2t 6 9
58. Let u 4 du d
r 3cos 2 4 d 3cos 2u du 3 u2 14 sin 2u C 23
r 8 when 0 8 38 34 sin 2 C C 2 34 r 32
r 32 34 sin 2 2 8 34 r 32 34 cos 2 8 34
59. Let u 2t 2 du 2 dt 2 du 4 dt
4 43 sin 2 2 C;
4 34 sin 2 2 2 34
4sin 2t 2 dt (sin u )(2 du ) 2 cos u C1 2 cos 2t 2 C1 ;
100 we have 100 2 cos 2 C1 C1 100 ds
at t 0 and ds
2 cos 2t 2 100
dt
dt
ds
dt
Copyright 2018 Pearson Education, Inc.
1
12 u
du
Section 5.6 Definite Integral Substitutions and the Area Between Curves
s 2 cos 2t 2 100 dt (cos u 50) du sin u 50u C2 sin 2t 2 50 2t 2 C2 ;
at t 0 and s 0 we have 0 sin 2 50 2 C2 C2 1 25
s sin 2t 100t 25 (1 25 ) s sin 2t 100t 1
2
2
60. Let u tan 2 x du 2sec2 2 x dx 2du 4sec 2 2 x dx; v 2 x dv 2dx 12 dv dx
dy
dx
4sec 2 2 x tan 2 x dx u (2 du ) u 2 C1 tan 2 2 x C1 ;
dy
dx
2
dy
4 we have 4 0 C1 C1 4 dx tan 2 2 x 4 (sec 2 2 x 1) 4 sec2 2 x 3
y (sec 2 x 3) dx (sec2 v 3) 12 dv 12 tan v 23 v C2 12 tan 2 x 3 x C2 ;
at x 0 and
at x 0 and y 1 we have 1
1 (0) 0 C
2
2
C2 1 y 12 tan 2 x 3x 1
61. Let u 2t du 2 dt 3 du 6dt
s 6 sin 2t dt (sin u )(3 du ) 3 cos u C 3 cos 2t C ;
at t 0 and s 0 we have 0 3cos 0 C C 3 s 3 3cos 2t s 2 3 3cos( ) 6 m
62. Let u t du dt du 2 dt
v 2 cos t dt (cos u )( du ) sin u C1 sin( t ) C1;
sin( t ) 8 s ( sin( t ) 8) dt
at t 0 and v 8 we have 8 (0) C1 C1 8 v ds
dt
sin u du 8t C2 cos( t ) 8t C2 ; at t 0 and s 0 we have 0 1 C2 C2 1
s 8t cos ( t ) 1 s (1) 8 cos 1 10 m
5.6
DEFINITE INTEGRAL SUBSTITUTIONS AND THE AREA BETWEEN CURVES
1. (a) Let u y 1 du dy; y 0 u 1, y 3 u 4
3
0
4
4
y 1 dy u1/2 du 23 u 3/2 23 (4)3/2 23 (1)3/2 23 (8) 23 (1) 14
3
1
1
(b) Use the same substitution for u as in part (a); y 1 u 0, y 0 u 1
0
1
1
1
y 1 dy u1/2 du 23 u 3/2
0
0
23 (1)3/2 0 23
2. (a) Let u 1 r 2 du 2r dr 12 du r dr ; r 0 u 1, r 1 u 0
1
0 r
0
0
1 r 2 dr 12 u du 13 u 3/2 0 13 (1)3/2 13
1
1
(b) Use the same substitution for u as in part (a); r 1 u 0, r 1 u 0
1
1 r
0
1 r 2 dr 12 u du 0
0
3. (a) Let u tan x du sec 2 x dx; x 0 u 0, x 4 u 1
/4
0
1
2
2
1
tan x sec2 x dx u du u2 12 0
0
0
1
2
(b) Use the same substitution as in part (a); x 4 u 1, x 0 u 0
0
/4 tan x sec
2
0
2
0
x dx u du u2 0 12 12
1
1
Copyright 2018 Pearson Education, Inc.
329
330
Chapter 5 Integrals
4. (a) Let u cos x du sin x dx du sin x dx; x 0 u 1, x u 1
0 3cos
2
1
x sin x dx
1
3u 2 du [u 3 ] 11 (1)3 ((1)3 ) 2
(b) Use the same substitution as in part (a); x 2 u 1, x 3 u 1
3
2 3cos
2
x sin x dx
1
1
3u 2 du 2
5. (a) u 1 t 4 du 4t 3 dt 14 du t 3 dt ; t 0 u 1, t 1 u 2
1 3
t (1 t 4 )3
0
2
2
u 4 24 14 15
dt 14 u 3 du 16
1
1 16 16 16
(b) Use the same substitution as in part (a); t 1 u 2, t 1 u 2
1 3
1t
2
(1 t 4 )3 dt 14 u 3 du 0
2
6. (a) Let u t 2 1 du 2t dt 12 du t dt ; t 0 u 1, t 7 u 8
0
7
8 1 1/3
u
1 2
t (t 2 1)1/3 dt
du
12 34 u 4/3 1 83 (8)4/3 83 (1)4/3 458
8
(b) Use the same substitution as in part (a); t 7 u 8, t 0 u 1
0
7
t (t 2 1)1/3 dt
1 1 1/3
u
82
8 1 1/3
u
1 2
du
du 45
8
7. (a) Let u 4 r 2 du 2r dr 12 du rdr ; r 1 u 5, r 1 u 5
1
5
1 (45rr2 )2 dr 55 12 u
2
du 0
(b) Use the same substitution as in part (a); r 0 u 4, r 1 u 5
1
5
0 (45rr2 )2 dr 5 4 12 u
2
5
8. (a) Let u 1 v3/2 du 32 v1/2 dv
1 10 v
0 (1 v3/ 2 )2
2 1 20
1 u2 3
dv
du 5 12 u 1 5 12 (5) 1 5 12 (4)1 18
4
du
20 du
3
20 2 u 2
3 1
10 v dv; v 0 u 1, v 1 u 2
1
du 20
3 u
2
1
1 1 10
20
3 2 1
3
(b) Use the same substitution as in part (a); v 1 u 2, v 4 u 1 43/2 9
70
1 (1v3/ 2 )2 dv 2 u12 203 du 203 u1 2 203 19 12 203 187 27
4 10 v
9
9
9. (a) Let u x 2 1 du 2 x dx 2 du 4 x dx; x 0 u 1, x 3 u 4
0
3
4x
x 2 1
dx
4 2
u
1
4
du 2u 1/2 du [4u1/2 ] 14 4(4)1/2 4(1)1/2 4
1
(b) Use the same substitution as in part (a); x 3 u 4, x 3 u 4
3
4x
3
2
x 1
dx
4 2
u
4
du 0
10. (a) Let u x 4 9 du 4 x3 dx 14 du x3 dx; x 0 u 9, x 1 u 10
1
0
x3
x 4 9
dx
10 1
9 4
10
u 1/2 du 14 (2)u1/2 12 (10)1/2 12 (9)1/2 102 3
9
Copyright 2018 Pearson Education, Inc.
Section 5.6 Definite Integral Substitutions and the Area Between Curves
(b) Use the same substitution as in part (a); x 1 u 10, x 0 u 9
0
1
9
10
dx 14 u 1/2 du 14 u 1/2 du 3 210
10
9
x 9
x3
4
1
1
11. (a) Let u 4 5t t (u 4), dt du; t 0 u 4, t 1 u 9.
5
5
1
0 t
4 5t dt
1 9
1 9
(u 4) u du u 3/2 4u1/2 du
25 4
25 4
9
1 2 5/2 8 3/2
1 2
8
8
2
506
u u (243) (27) (32) 8)
25 5
3
3
3
4 25 5
5
375
(b) Use the same substitution as in (a); t 1 u 9, t 9 u 49.
9
1
t 4 5t dt
49
1 49 3/2
1 2
8
u 4u1/2 du u5/2 u 3/2
25 9
25 5
3
9
1 2
8
8
2
86,744
(16,807) (343) (243) 27
25 5
3
5
3
375
12. (a) Let u 1 cos 3t du 3sin 3t dt 13 du sin 3t dt ; t 0 u 0, t 6 u 1 cos 2 1
/6
1
2
1
(1 cos 3t ) sin 3t dt 13 u du 13 u2 16 (1) 2 16 (0) 2 16
0
0
(b) Use the same substitution as in part (a); t 6 u 1, t 3 u 1 cos 2
0
/6 (1 cos 3t )sin 3t dt 1 13 u du 13 u2 1
/3
2
2
2
16 (2) 2 16 (1) 2 12
13. (a) Let u 4 3sin z du 3cos z dz 13 du cos z dz; z 0 u 4, z 2 u 4
2
0
cos z
4 3sin z
dz
4 1 1
du
u 3
4
0
(b) Use the same substitution as in part (a); z u 4 3sin( ) 4, z u 4
cos z
4 3sin z
dz
4 1 1
u 3
4
du 0
14. (a) Let u 2 tan 2t du 12 sec 2 2t dt 2 du sec 2 2t dt ; t 2 u 2 tan
/2 2 tan 2t sec
0
2 t
2
2
dt u (2 du ) [u 2 ] 12 22 12 3
1
(b) Use the same substitution as in part (a); t 2 u 1, t 2 u 3
/2
/2 (2 tan 2t ) sec
2 t
2
3
dt 2 u du [u 2 ] 13 32 12 8
1
15. Let u t 5 2t du (5t 4 2) dt ; t 0 u 0, t 1 u 3
1
0
3
3
t 5 2t (5t 4 2) dt u1/2 du 23 u 3/2 23 (3)3/2 23 (0)3/2 2 3
0
0
Copyright 2018 Pearson Education, Inc.
4 1, t 0 u 2
331
332
Chapter 5 Integrals
16. Let u 1 y du
4
1
dy
2 y (1 y ) 2
dy
2 y
; y 1 u 2, y 4 u 3
3
3
12 du u 2 du [u 1 ]32 13 12 16
2u
2
17. Let u cos 2 du 2sin 2 d 12 du sin 2 d ; 0 u 1, 6 u cos 2 6
/6
0
1/2 3
u
1
cos 3 2 sin 2 d
12 du 12 11/2 u 3 du 12 u2 1
2
1/2
1
12
4
18. Let u tan 6 du 16 sec2 6 d 6 du sec2 6 d ; u tan 6
3 /2
cot 5 6
d 1/
1
sec 2 6
3
u
5
2
1
2
1 2 34
4(1)
1 ,
3
32 u tan 4 1
1
1
4
3
3
u
(6 du ) 6 4
4 3
1/ 3 2u 4 1/ 3
2(1)
2 1
3
12
4
19. Let u 5 4 cos t du 4sin t dt 14 du sin t dt ; t 0 u 5 4 cos 0 1, t u 5 4 cos 9
1/4
0 5(5 4 cos t )
9
9
9
sin t dt 5u1/4 14 du 54 u1/4 du 54 54 u 5/4 95/4 1 35/2 1
1
1
1
20. Let u 1 sin 2t du 2 cos 2t dt 12 du cos 2t dt ; t 0 u 1, t 4 u 0
/4
0
0
0
(1 sin 2t )3/2 cos 2t dt 12 u 3/2 du 12 52 u 5/2 15 (0)5/2 15 (1)5/2 15
1
1
21. Let u 4 y y 2 4 y 3 1 du (4 2 y 12 y 2 ) dy; y 0 u 1, y 1 u 4(1) (1)2 4(1)3 1 8
1
0 (4 y y
2
8
4 y 3 1)2/3 (12 y 2 2 y 4) dy u 2/3 du [3u1/3 ] 18 3(8)1/3 3(1)1/3 3
1
22. Let u y 3 6 y 2 12 y 9 du (3 y 2 12 y 12) dy 13 du ( y 2 4 y 4) dy; y 0 u 9, y 1 u 4
1
0 ( y
3
4
4
6 y 2 12 y 9)1/2 ( y 2 4 y 4) dy 13 u 1/2 du 13 (2u1/2 ) 23 (4)1/2 23 (9)1/2 23 (2 3) 23
9
9
3
23. Let u 3/2 du 32 1/2 d 32 du d ; 0 u 0, 2 u
3
0
2
cos 2 ( 3/2 ) d cos 2 u
0
23 du 23 u2 14 sin 2u 0 32 2 14 sin 2 32 (0) 3
24. Let u 1 1t du t 2 dt ; t 1 u 0, t 12 u 1
1/2 2
1
t
sin 2 1 1t dt
1
0
sin 2 u du
1
u2 14 sin 2u 0
02 14 sin 0
12 14 sin(2)
12 14 sin 2
25. Let u 4 x 2 du 2 x dx 12 du x dx; x 2 u 0, x 0 u 4, x 2 u 0
A
0
2
4
0
4
4
x 4 x 2 dx x 4 x 2 dx 12 u1/2 du 12 u1/2 du 2 12 u1/2 du u1/2 du
0
0
4
0
0
2
4
23 u 3/2 23 (4)3/2 23 (0)3/2 16
3
0
Copyright 2018 Pearson Education, Inc.
Section 5.6 Definite Integral Substitutions and the Area Between Curves
333
26. Let u 1 cos x du sin x dx; x 0 u 0, x u 2
2
2
0 (1 cos x) sin x dx 0 u du u2 0 22
2
2
2
02 2
27. Let u 1 cos x du sin x dx du sin x dx; x u 1 cos ( ) 0, x 0 u 1 cos 0 2
A
0
2
2
2
3(sin x) 1 cos x dx 3u1/2 (du ) 3 u1/2 du 2u 3/2 2(2)3/2 2(0)3/2 25/2
0
0
0
28. Let u sin x du cos x dx 1 du cos x dx; x 2 u sin 2 0, x 0 u
Because of symmetry about x 2 , A 2
0
(cos x )(sin(
/2 2
sin x )) dx 2 2 (sin u ) 1 du
0
sin u du [ cos u ]0 ( cos ) ( cos 0) 2
0
29. For the sketch given, a 0, b ; f ( x) g ( x) 1 cos 2 x sin 2 x
A
(1 cos 2 x )
2
0
1 cos 2 x
;
2
sin 2 x
dx 12 (1 cos 2 x) dx 12 x 2 12 [( 0) (0 0)] 2
0
0
30. For the sketch given, a 3 , b 3 ; f (t ) g (t ) 12 sec2 t (4 sin 2 t ) 12 sec2 t 4sin 2 t ;
A
/3 1
sec 2 t 4sin 2 t
/3 2
dt 12 /3/3 sec2 t dt 4 sin 2t dt 12 /3/3 sec2t dt 4 /3/3 (1cos2 2t ) dt
/3
/3
sin 2t
12
sec2t dt 2
(1 cos 2t ) dt 12 [tan t ]/3/3 2 t 2
/3
/3
/3
/3
3 4 3 3 43
31. For the sketch given, a 2, b 2; f ( x) g ( x) 2 x 2 ( x 4 2 x 2 ) 4 x 2 x 4 ;
A
2
2
2
5
64 64 320192 128
(4 x 2 x 4 ) dx 43x x5 32
32
32
32
3
5
5
3
5
15
15
3
2
2
32. For the sketch given, c 0, d 1; f ( y ) g ( y ) y 2 y 3 ;
1
1
1
1
1
(1 0) (10)
y3 y 4
1
A ( y 2 y 3 ) dy y 2 dy y 3 dy 3 4 3 4 13 14 12
0
0
0
0 0
33. For the sketch given, c 0, d 1; f ( y ) g ( y ) (12 y 2 12 y 3 ) (2 y 2 2 y ) 10 y 2 12 y3 2 y;
1
1
1
1
1
1
1
A (10 y 2 12 y 3 2 y ) dy 10 y 2 dy 12 y 3 dy 2 y dy 10
y3 12 y 4 22 y 2
3 0 4
0
0
0
0
0
0
10
0 (3 0) (1 0) 43
3
34. For the sketch given, a 1, b 1; f ( x) g ( x) x 2 (2 x 4 ) x 2 2 x 4 ;
1
3
5
1
12 22
A ( x 2 2 x 4 ) dx x3 25x 13 52 13 52 32 54 1015
15
1
1
35. We want the area between the line y 1, 0 x 2, and the curve y
(formed by y x and y 1) with base 1 and height 1. Thus, A
8
2 12
1
2
2
3
1
2
2
2
0
x2
4
, minus the area of a triangle
1 dx
x2
4
5
6
Copyright 2018 Pearson Education, Inc.
1 (1)(1)
2
2
x 1
x 12
0 2
3
334
Chapter 5 Integrals
36. We want the area between the x-axis and the curve y x 2 , 0 x 1 plus the area of a triangle (formed by
1
3
1
x 1, x y 2, and the x-axis) with base 1 and height 1. Thus, A x 2 dx 12 (1)(1) x3 12 13 12
0
0
37. AREA A1 A2
A1: For the sketch given, a 3 and we find b by solving the equations y x 2 4 and y x 2 2 x
simultaneously for x: x 2 4 x 2 2 x 2 x 2 2 x 4 0 2( x 2)( x 1) x 2 or x 1 so
b 2: f ( x) g ( x) ( x 2 4) ( x 2 2 x) 2 x 2 2 x 4 A1
2
2
3
(2 x 2 2 x 4) dx
3
2
23x 22x 4 x 16
4 8 (18 9 12) 9 16
11
;
3
3
3
3
A2: For the sketch given, a 2 and b 1: f ( x) g ( x ) ( x 2 2 x) ( x 2 4) 2 x 2 2 x 4
A2
1
1
3
(2 x 2 2 x 4) dx 23x x 2 4 x 23 1 4 16
48
3
2
2
23 1 4 16
4 8 9;
3
Therefore, AREA A1 A2 11
9
3
38
3
38. AREA A1 A2
A1: For the sketch given, a 2 and b 0: f ( x) g ( x) (2 x3 x 2 5 x) ( x 2 3x ) 2 x3 8 x
A1
0
0
4
2
(2 x3 8 x) dx 24x 8 2x 0 (8 16) 8;
2
2
A2: For the sketch given, a 0 and b 2: f ( x) g ( x) ( x 2 3 x) (2 x3 x 2 5 x) 8 x 2 x3
2
2
4
2
A2 (8 x 2 x3 ) dx 8 2x 24x (16 8) 8;
0
0
Therefore, AREA A1 A2 16
39. AREA A1 A2 A3
A1: For the sketch given, a 2 and b 1: f ( x) g ( x ) ( x 2) (4 x 2 ) x 2 x 2
A1
1
1
3
2
( x 2 x 2) dx x3 x2 2 x 13 12 2 83 24 4 73 12 1463 11
;
6
2
2
A2: For the sketch given, a 1 and b 2: f ( x) g ( x ) (4 x 2 ) ( x 2) ( x 2 x 2)
2
2
A2 ( x 2 x 2) dx x3 x2 2 x
1
1
3
2
83 42 4 13 12 2 3 8 12 92 ;
A3: For the sketch given, a 2 and b 3: f ( x) g ( x) ( x 2) (4 x 2 ) x 2 x 2
3
3
2
3
A3 ( x 2 x 2) dx x3 x2 2 x 27
92 6 83 42 4 9 92 83 ;
3
2
2
Therefore, AREA A1 A2 A3 11
92 9 92 83 9 56
6
40. AREA A1 A2 A3
A1: For the sketch given, a 2 and b 0: f ( x) g ( x)
A1 13
0
0
x3
3
49
6
3
x 3x x3 43 x 13 ( x3 4 x)
( x3 4 x) dx 13 x4 2 x 2 0 13 (4 8) 43 ;
2
2
4
Copyright 2018 Pearson Education, Inc.
5
6
Section 5.6 Definite Integral Substitutions and the Area Between Curves
A2: For the sketch given, a 0 and we find b by solving the equations y
3
3
x and y 3x simultaneously
x 3x x3 43 x 0 3x ( x 2)( x 2) 0 x 2, x 0, or x 2 so b 2: f ( x) g ( x)
3
4 2
2
2
3x x3 x 13 ( x3 4 x) A2 13 ( x3 4 x) dx 13 (4 x x3 ) 13 2 x 2 x4 13 (8 4) 34 ;
0
0
0
for x:
x
3
x3
3
335
A3: For the sketch given, a 2 and b 3: f ( x) g ( x)
3
x3
3
x 3x 13 ( x3 4 x)
4
3
25 ;
A3 13 ( x3 4 x) dx 13 x4 2 x 2 13 81
2 9 16
8 13 81
14 12
4
4
2
4
2
25
Therefore, AREA A1 A2 A3 43 43 12
32 25
12
19
4
41. a 2, b 2;
f ( x ) g ( x) 2 ( x 2 2) 4 x 2
2
2
3
(4 x 2 ) dx 4 x x3 8 83 8 83
2
2
8
32
24
2 3 3 3
A
42. a 1, b 3;
f ( x) g ( x) (2 x x 2 ) (3) 2 x x 2 3
3
3
3
A (2 x x 2 3)dx x 2 x3 3x
1
1
9 27
9 1 13 3 11 13 32
3
3
43. a 0, b 2;
2
f ( x) g ( x) 8 x x 4 A (8 x x 4 )dx
0
2
2
5
8 2x x5 16 32
80532 48
5
5
0
44. Limits of integration: x 2 2 x x x 2 3x 0
x( x 3) 0 a 0 and b 3;
f ( x) g ( x) x ( x 2 2 x) 3x x 2
3
2
3
3
A (3x x 2 )dx 32x x3 27
9 27218 92
2
0
0
45. Limits of integration: x 2 x 2 4 x 2 x 2 4 x 0
2 x( x 2) 0 a 0 and b 2;
f ( x) g ( x) ( x 2 4 x) x 2 2 x 2 4 x
2
3
2
2
A (2 x 2 4 x)dx 23x 42x
0
0
16
16
32
48
8
3 2 6 3
Copyright 2018 Pearson Education, Inc.
336
Chapter 5 Integrals
46. Limits of integration: 7 2 x 2 x 2 4 3 x 2 3 0
3( x 1)( x 1) 0 a 1 and b 1;
f ( x) g ( x) (7 2 x 2 ) ( x 2 4) 3 3 x 2
1
3
1
A (3 3x 2 )dx 3 x x3 3 1 13 1 13
1
1
2
6 3 4
47. Limits of integration: x 4 4 x 2 4 x 2 x 4 5 x 2 4 0
( x 2 4)( x 2 1) 0 ( x 2)( x 2)( x 1)( x 1) 0
x 2, 1,1, 2; f ( x) g ( x ) ( x 4 4 x 2 4) x 2
x 4 5 x 2 4 and
g (x) f ( x) x 2 ( x 4 4 x 2 4) x 4 5 x 2 4
A
1
x 4 5 x 2 4 dx ( x 4 5 x 2 4) dx
2
1
1
2
( x 4 5 x 2 4) dx
1
1
1
2
5
3
5
3
5
3
x5 53x 4 x x5 53x 4 x 5x 53x 4 x
2
1
1
180 8
15 53 4 32
40
8 15 53 4 15 53 4 32
40
8 15 53 4 60
60
30015
5
3
5
3
5
3
48. Limits of integration: x a 2 x 2 0 x 0 or
a 2 x 2 0 x 0 or a 2 x 2 0 x a, 0, a;
A
0
a
a
x a 2 x 2 dx x a 2 x 2 dx
0
0
12 32 (a 2 x 2 )3/2 12 32 (a 2
a
2 3/2 1 2 3/2 2 a3
1
3 (a ) 3 (a )
3
a
x 2 )3/2
0
x, x 0
x
and
x, x 0
5 y x 6 or y 5x 65 ; for x 0; x 5x 65
5 x x 6 25( x) x 2 12 x 36
x 2 37 x 36 0 ( x 1)( x 36) 0
x 1, 36 (but x 36 is not a solution);
49. Limits of integration: y
for x 0 : 5 x x 6 25 x x 2 12 x 36
x 2 13x 36 0 ( x 4)( x 9) 0
x 4,9; there are three intersection points and
Copyright 2018 Pearson Education, Inc.
Section 5.6 Definite Integral Substitutions and the Area Between Curves
A
0
1
x5 6
x dx
0 x 5 6 x dx 4
4
9
0
x x 5 6 dx
4
9
3/2
( x 6)2
( x 6)2
( x 6)2
10 23 x 10 23 x3/2 23 x3/2 10
1
0
4
36 25 2 100 2 43/2 36 0 2 93/2 225 2 43/2 100 50 20 5
10
10
3
3
10 3
10
3
10
3
10
3
10
x 2 4, x 2 or x 2
50. Limits of integration: y | x 2 4|
2
4 x , 2 x 2
2
for x 2 and x 2 : x 2 4 x2 4
2 x 2 8 x 2 8 x 2 16 x 4; for 2 x 2 :
2
4 x 2 x2 4 8 2 x 2 x 2 8 x 2 0 x 0; by
symmetry of the graph,
2
2
A 2 x2 4 (4 x 2 ) dx
0
2
4
2 x2 4 x 2 4
2
2 82 0 2 32 64
16 86
6
dx 2 x2 0 2 8x x6 2
2
3
3
4
40 563 643
51. Limits of integration: c 0 and d 3;
f ( y) g ( y) 2 y 2 0 2 y 2
3
3
2 y3
A 2 y 2 dy 3 2 9 18
0
0
52. Limits of integration: y 2 y 2 ( y 1)( y 2) 0
c 1 and d 2; f ( y ) g ( y ) ( y 2) y 2
2
2
y3
y2
A ( y 2 y 2 ) dy 2 2 y 3
1
1
8
8
4
1
1
1
1
2 4 3 2 2 3 6 3 2 2 3 92
53. Limits of integration: 4 x y 2 4 and
4 x 16 y y 2 4 16 y y 2 y 20 0
( y 5)( y 4) 0 c 4 and d 5;
16 y
y 2 4 y 2 y 20
f ( y) g ( y) 4 4
4
5
5
y3 y 2
A 14 ( y 2 y 20) dy 14 3 2 20 y
4
4
125
1
1 64 16 80
4 3 25
100
2
4 3
2
14 189
92 180 243
3
8
Copyright 2018 Pearson Education, Inc.
337
338
Chapter 5 Integrals
54. Limits of integration: x y 2 and x 3 2 y 2
y 2 3 2 y 2 3 y 2 3 0 3( y 1)( y 1) 0
c 1 and d 1; f ( y ) g ( y ) (3 2 y 2 ) y 2
3 3 y 2 3(1 y 2 ) A 3
1
1
1
(1 y 2 ) dy
y3
3 y 3 3 1 13 3 1 13 3 2 1 13 4
1
55. Limits of integration: x y 2 and x 2 3 y 2
y
y2 2 3 y2 2 y2 2 0
2( y 1)( y 1) 0 c 1 and d 1;
x y 0
f ( y ) g ( y ) (2 3 y 2 ) ( y 2 ) 2 2 y 2 2(1 y 2 )
1
1
1
y3
A 2 (1 y 2 ) dy 2 y 3
1
1
2 1 13 2 1 13 4 23 83
56. Limits of integration: x y 2/3 and
x 2 y 4 y 2/3 2 y 4 c 1 and d 1;
1
f ( y ) g ( y ) (2 y 4 ) y 2/3 A (2 y 4 y 2/3 ) dy
1
1
57. Limits of integration: x y 2 1 and x | y | 1 y 2
y 2 1 | y | 1 y 2 y 4 2 y 2 1 y 2 (1 y 2 )
y4 2 y2 1 y2 y4 2 y4 3 y2 1 0
(2 y 2 1)( y 2 1) 0 2 y 2 1 0 or y 2 1 0 y 2 12
or y 2 1 y 22 or y 1.
Substitution shows that
2
2
are not solutions y 1;
for 1 y 0, f ( x) g ( x) y 1 y 2 ( y 2 1)
1 y 2 y (1 y 2 )1/2 , and by symmetry of the graph,
0
0
0
A 2 1 y 2 y (1 y 2 )1/2 dy 2 (1 y 2 ) dy 2 y (1 y 2 )1/2 dy
1
1
1
0
0
0
1
y5
2 y 5 53 y 5/3 2 15 53 2 15 53
1
2 2 15 53 12
5
x 3y2 2
1
2
y3
2(1 y 2 )3/ 2
2 y 3 2 12
2 (0 0) 1 13 23 0 2
3
1
1
Copyright 2018 Pearson Education, Inc.
1
2
x
Section 5.6 Definite Integral Substitutions and the Area Between Curves
58. AREA A1 A2
Limits of integration: x 2 y and
x y3 y 2 y3 y 2 2 y 0
y ( y 2 y 2) y ( y 1)( y 2) 0 y 1, 0, 2:
for 1 y 0, f ( y ) g ( y ) y 3 y 2 2 y
0
0
y 4 y3
A1 ( y3 y 2 2 y ) dy 4 3 y 2
1
1
5 ;
0 14 13 1 12
for 0 y 2, f ( y ) g ( y ) 2 y y 3 y 2
2
2
y4
y3
A2 (2 y y 3 y 2 ) dy y 2 4 3
0
0
5 8 37
4 16
83 0 83 ; Therefore, A1 A2 12
4
3 12
59. Limits of integration: y 4 x 2 4 and y x 4 1
x 4 1 4 x 2 4 x 4 4 x 2 5 0
( x 2 5)( x 1)( x 1) 0 a 1 and b 1;
f ( x ) g ( x ) 4 x 2 4 x 4 1 4 x 2 x 4 5
A
1
1
(4 x 2 x 4 5) dx 43x x5 5 x
1
1
3
5
43 15 5 43 15 5 2 43 15 5 104
15
60. Limits of integration: y x3 and y 3 x 2 4
x3 3x 2 4 0 ( x 2 x 2)( x 2) 0
( x 1)( x 2) 2 0 a 1 and b 2;
f ( x) g ( x) x3 (3x 2 4) x3 3 x 2 4
A
2
2
4
3
( x3 3x 2 4) dx x4 33x 4 x
1
1
16
24
8 14 1 4 27
4
3
4
61. Limits of integration: x 4 4 y 2 and x 1 y 4
4 4 y2 1 y4 y4 4 y2 3 0
y 3 y 3 ( y 1)( y 1) 0 c 1 and d 1
since x 0; f ( y ) g ( y ) (4 4 y 2 ) (1 y 4 )
1
3 4 y 2 y 4 A (3 4 y 2 y 4 ) dy
1
4 y3
3 y 3
1
y5
5
1
56
2 3 43 15 15
Copyright 2018 Pearson Education, Inc.
339
340
Chapter 5 Integrals
62. Limits of integration: x 3 y 2 and x
y2
3 y2
y2
4
3 y 2 4 4 3 0 34 ( y 2)( y 2) 0
y2
c 2 and d 2; f ( y ) g ( y ) (3 y 2 ) 4
2
2
y3
y2
y2
3 1 4 A 3 1 4 dy 3 y 12
2
2
8
8
16
3 2 12 2 12 3 4 12 12 4 8
63. a 0, b ; f ( x) g ( x ) 2sin x sin 2 x
A
0 (2sin x sin 2 x)dx 2 cos x cos22 x 0
2(1) 12 2 1 12 4
64. a 3 , b 3 ; f ( x) g ( x) 8cos x sec2 x
A
/3
/3 (8cos x sec
2
x)dx [8 sin x tan x]/3/3
8 23 3 8 23 3 6 3
65. a 1, b 1; f ( x ) g ( x) (1 x 2 ) cos 2x
x x3 2 sin 2x 1 13 2 1 13 2
1
2
2
4
4
2 3 3
1
A 1 x 2 cos 2x dx
1
1
3
66. A A1 A2
a1 1, b1 0 and a2 0, b2 1;
f1 ( x) g1 ( x) x sin 2x and f 2 ( x) g 2 ( x)
sin 2x x by symmetry about the origin,
1
A1 A2 2 A1 A 2 sin 2x x dx
0
2
1
0 2 2 0 12 2 1 0
4
2 2 cos 2x x2
2 2 12 2 42
Copyright 2018 Pearson Education, Inc.
Section 5.6 Definite Integral Substitutions and the Area Between Curves
67. a 4 , b 4 ; f ( x) g ( x) sec2 x tan 2 x
A
/4
/4
/4
/4
/4
/4
(sec2 x tan 2 x) dx
[sec 2 x (sec2 x 1)] dx
1 dx [ x]/4/4 4 4 2
68. c 4 , d 4 ; f ( y ) g ( y )
tan 2 y ( tan 2 y ) 2 tan 2 y 2(sec2 y 1)
A
/4
/4
2 sec2 y 1 dy 2[(tan y y )]/4/4
2 1 4 1 4 4 1 4 4
69. c 0, d 2 ; f ( y ) g ( y )
3 sin y cos y 0 3sin y cos y
/2
/2
sin y cos y dy 3 23 (cos y )3/2
0
0
2(0 1) 2
A 3
70. a 1, b 1; f ( x) g ( x) sec2 3x x1/3
A
1
x
sec2 x
3
1
1/3
dx
1
3 tan 3x 34 x 4/3
1
3 3 43 3 3 34 6 3
71. A A1 A2
Limits of integration: x y 3 and x y y y3
y 3 y 0 y ( y 1)( y 1) 0 c1 1, d1 0
and c2 0, d 2 1; f1 ( y ) g1 ( y ) y 3 y and
f 2 ( y ) g 2 ( y ) y y 3 by symmetry
about the origin, A1 A2 2 A2 A
1
1
y2 y4
2 ( y y 3 ) dy 2 2 4 2
0
0
12 14 12
Copyright 2018 Pearson Education, Inc.
341
342
Chapter 5 Integrals
72. A A1 A2
Limits of integration: y x3 and y x5 x3 x5
x5 x3 0 x3 ( x 1)( x 1) 0 a1 1, b1 0
and a2 0, b2 1; f1 ( x) g1 ( x) x3 x5 and
f 2 ( x) g 2 ( x) x5 x3 by symmetry about the
1
origin, A1 A2 2 A2 A 2 ( x3 x5 ) dx
0
1
2 x4 x6 2 14 16 16
0
4
6
73. A A1 A2
Limits of integration: y x and y
1
x2
x 12 , x 0
x3 1 x 1, f1 ( x) g1 ( x) x 0 x
1
2
1
A1 x dx x2
0
0
x 2 A2
2 2
1 x
1 ; f ( x) g ( x)
2
2 2
x
12 0
x
2
dx x1 12 1 12 ;
1
A A1 A2 12 12 1
74. Limits of integration: sin x cos x x 4 a 0 and
b 4 ; f ( x) g ( x) cos x sin x
A
2
2
/4
0
(cos x sin x) dx [sin x cos x ]0 /4
22 (0 1) 2 1
75. (a) The coordinates of the points of intersection of the
line and parabola are c x 2 x c and y c
(b) f ( y ) g ( y ) y y 2 y the area of
c
the lower section is, AL [ f ( y ) g ( y )] dy
0
c
y dy 2 23 y 3/2 43 c3/2 . The area of
0
0
the entire shaded region can be found by setting c 4 : A 43 43/2 438 32
. Since we want c to divide
3
the region into subsections of equal area we have A 2 AL 32
2 34 c3/2 c 42/3
3
2
c
c
c
c
3/ 2
3
(c x 2 ) dx cx x3
2 c3/2 c 3
c
c
c
3/2
32
4
3 c . Again, the area of the whole shaded region can be found by setting c 4 A 3 . From the
condition A 2 AL , we get 43 c3/2 32
c 42/3 as in part (b).
3
(c) f ( x) g ( x) c x 2 AL
[ f ( x) g ( x)] dx
Copyright 2018 Pearson Education, Inc.
Section 5.6 Definite Integral Substitutions and the Area Between Curves
76. (a) Limits of integration: y 3 x 2 and y 1
3 x 2 1 x 2 4 a 2 and b 2;
f ( x) g ( x) (3 x 2 ) (1) 4 x 2
2
2
(4 x 2 ) dx 4 x x3
2
2
8
8
16
32
8 3 8 3 16 3 3
A
3
(b) Limits of integration: let x 0 in y 3 x 2
y 3; f ( y ) g ( y ) 3 y 3 y
3
2(3 y )1/2 A 2 (3 y )1/2 dy 2
1
43 0 (3 1)3/2
3
43 (8) 323
77. Limits of integration: y 1 x and y
1/2
1 (3 y)
3
2(3 y )3/ 2
(1) dy (2)
3
1
2
x
1 x 2 , x 0 x x 2 x (2 x) 2
x
x 4 4 x x2 x2 5x 4 0
( x 4)( x 1) 0 x 1, 4 (but x 4 does not
satisfy the equation); y 2 and y 4x 2 4x
x
x
8 x x 64 x3 x 4. Therefore,
AREA A1 A2 : f1 ( x) g1 ( x) 1 x1/2 x4
4
4
A2 2 x 1/2 4x dx 4 x1/2 x8 4 2 16
4 18 4 15
17
; Therefore,
8
8
8
1
1
1
2
1
A1 1 x1/2 4x dx x 23 x3/2 x8 1 23 18 0 37
; f 2 ( x) g 2 ( x) 2 x 1/2 4x
24
0
0
2
51 88 11
AREA A1 A2 37
17
3724
24
8
24
3
78. Limits of integration: ( y 1) 2 3 y
y2 2 y 1 3 y y2 y 2 0
( y 2)( y 1) 0 y 2 since y 0; also,
2 y 3 y 4 y 9 6 y y 2 y 2 10 y 9 0
( y 9)( y 1) 0 y 1 since y 9 does not satisfy
the equation;
AREA A1 A2
1
1
2 y 3/ 2
f1 ( y ) g1 ( y ) 2 y 0 2 y1/2 A1 2 y1/2 dy 2 3 34 ;
0
0
2
2
f 2 ( y ) g 2 ( y ) (3 y ) ( y 1)2 A2 [3 y ( y 1)2 ] dy 3 y 12 y 2 13 ( y 1)3
1
1
6 2 13 3 12 0 1 13 12 76 . Therefore, A1 A2 43 76 15
25
6
Copyright 2018 Pearson Education, Inc.
343
344
Chapter 5 Integrals
79. Area between parabola and y a 2 : A 2
a
0 (a
2
3
Area of triangle AOC: 12 (2a)(a 2 ) a3 ; limit of ratio lim
a
3
a0 4 a
3
80.
b
b
A 2 f ( x) dx
a
a
f ( x) dx 2
b
a
f ( x) dx
a
3
3
x 2 ) dx 2 a 2 x 13 x3 2 a3 a3 0 43a ;
0
b
a
f ( x) dx
3
4
b
a
which is independent of a.
f ( x) dx 4
81. Neither one; they are both zero. Neither integral takes into account the changes in the formulas for the region’s
upper and lower bounding curves at x 0. The area of the shaded region is actually
A
0
1
0
1
x ( x ) dx 0 x ( x ) dx 1 2 x dx 0 2 x dx 2.
1
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 [a, b] will be less than the area between the curves (see Exercise 71).
83. Let u 2 x du 2 dx
1
2
du dx; x 1 u 2, x 3 u 6
1 sinx2 x dx 2 sin u2 u 12 du 2 sinu u du [ F (u )]2 F (6) F (2)
6
3
6
6
84. Let u 1 x du dx du dx; x 0 u 1, x 1 u 0
1
0
0 f (1 x) dx 1
0
1
1
1
0
0
f (u )(du ) f (u ) du f (u ) du f ( x) dx
85. (a) Let u x du dx; x 1 u 1, x 0 u 0
f odd f ( x) f ( x). Then
0
1 f ( x)dx
0
0
f (u )(du )
0
1 f (u ) (du ) 1
1
f (u ) du
1
f (u ) du 3
0
(b) Let u x du dx; x 1 u 1, x 0 u 0
f even f ( x) f ( x). Then
86. (a) Consider
/2
0
1
1
0 f (u ) du 3
0
0
0
0
a f ( x) dx a f (u ) du a
0
a
a f ( x) dx a f ( x) dx 0
(b)
0
1
f (u )(du ) f (u )du
a f ( x) dx when f is odd. Let u x du dx du dx and x a u a and
x 0 u 0. Thus
a
0
1 f ( x)dx
/2
/2 sin x dx [ cos x] /2
a
f ( x) dx f ( x) dx
0
a
a
0
0
f (u ) du f (u ) du f ( x) dx. Thus
a
0
f ( x) dx 0.
cos 2 cos 2 0 0 0.
Copyright 2018 Pearson Education, Inc.
Section 5.6 Definite Integral Substitutions and the Area Between Curves
87. Let u a x du dx; x 0 u a, x a u 0
I
a
f ( x ) dx
0 f ( x) f (a x)
0
a
(du )
a f ( a x ) dx
f ( x ) dx
f ( x) f (a x)
f
x
f
a
x
(
)
(
)
0
0
Therefore, 2 I a I a2 .
I I
88. Let u
xy 1
t
x
xy
t
a
f ( a u )
f ( a u ) f ( u )
du
xy
t2
1
a f ( x) f (a x)
0 f ( x) f (a x)
a
0
f ( a x ) dx
f ( x) f (a x)
a
dx dx [ x]0a a 0 a.
0
t du 1 dt 1 du 1 dt ; t x u y, t xy u 1. Therefore,
dt xy
t
u
t
dt u1 du
y
a f ( a u ) du
0 f (u ) f ( a u )
11
yu
y1
1 u
du
y1
1 t
du
dt
89. Let u x c du dx; x a c u a, x b c u b
b c
a c
b
b
a
a
f ( x c) dx f (u ) du f ( x) dx
90. (a)
(c)
(b)
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_] x 2 Cos[x]
g[x_] x 3 x
Plot[{f[x], g[x]}, {x, 2, 2}]
After examining the plots, the initial guesses for FindRoot can be determined.
Copyright 2018 Pearson Education, Inc.
345
346
Chapter 5 Integrals
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 0.4 sec. The distance traveled over each subinterval, using the
midpoint rule, is h 12 (vi vi 1 )t , where vi is the velocity at the left endpoint and vi 1 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 vi 1 at the right endpoint. Using this methodology we build the
following table based on the figure in the text:
t (sec) 0 0.4
v (fps) 0 10
h (ft) 0 2
t (sec)
v (fps)
h (ft)
6.4
50
643.2
0.8
25
9
1.2
55
25
6.8
37
660.6
1.6
100
56
7.2
25
672
2.0
190
114
7.6
12
679.4
2.4
180
188
2.8
165
257
3.2
150
320
3.6
140
378
4.0
130
432
4.4
115
481
4.8
105
525
5.2
90
564
5.6
76
592
6.0
65
620.2
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).
2. (a) Each time subinterval is of length t 1 sec. The distance traveled over each subinterval, using the
midpoint rule, is s 12 (vi vi 1 ) t , where vi is the velocity at the left, and vi 1 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 vi 1 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)
v (m/sec)
s (m)
0
0
0
1
0.5
0.25
2
1.2
1.1
3
2
2.7
4
3.4
5.4
5
4.5
9.35
6
4.8
14
7
4.5
18.65
8
3.5
22.65
(b) The graph shows the distance traveled by the
moving body as a function of time for 0 t 10.
Copyright 2018 Pearson Education, Inc.
9
2
25.4
10
0
26.4
Chapter 5 Practice Exercises
10
3. (a)
k 1
10
(b)
ak
4
1
4
10
ak 14 (2) 12
k 1
(bk 3ak )
(d)
4. (a)
(b)
k 1
10
10
k 1
k 1
k 1
10
10
k 1
k 1
k 1
10
52 bk 52 bk 52 (10) 25 0
20
20
k 1
20
k 1
k 1
3ak 3 ak 3(0) 0
20
20
(ak bk ) ak bk
12
k 1
20
(d)
k 1
10
(ak bk 1) ak bk 1 2 25 (1)(10) 13
k 1
20
(c)
10
bk 3 ak 25 3(2) 31
k 1
10
(c)
10
2bk
7
k 1
20
12
k 1
20
07 7
k 1
20
72 bk 12 (20) 72 (7) 8
k 1
20
(ak 2) ak 2 0 2(20) 40
k 1
k 1
k 1
5. Let u 2 x 1 du 2 dx 12 du dx; x 1 u 1, x 5 u 9
5
1 (2 x 1)
1/2
9 1/2 1
2
1 u
dx
6. Let u x 2 1 du 2 x dx
3
1 x( x
2
7. Let u
9
1
2
du x dx; x 1 u 0, x 3 u 8
1/3
4/3
0 u 12 du 83 u 0 83 (16 0) 6
8
8
1)1/3 dx
x
2
du u1/2 3 1 2
1
2 du dx; x u 2 , x 0 u 0
cos 2x dx /2 (cos u )(2 du ) [2sin u ] /2
0
0
0
2 sin 0 2sin 2 2(0 (1)) 2
8. Let u sin x du cos x dx; x 0 u 0, x 2 u 1
/2
0
9. (a)
(sin x)(cos x) dx
2
1
1
2
0 u du u2 0 12
2
2 f ( x) dx 13 2 3 f ( x) dx 13 (12) 4
5
5
2
(b)
2 f ( x) dx 2 f ( x) dx 2 f ( x) dx 6 4 2
(c)
5
(d)
2 ( g ( x)) dx 2 g ( x) dx (2) 2
(e)
2
g ( x) dx
5
2
g ( x) dx 2
5
2
5
5
f ( x) g ( x)
5
dx
1 5
5 2
f ( x) dx 15
5
2
g ( x) dx 15 (6) 15 (2) 85
Copyright 2018 Pearson Education, Inc.
347
348
Chapter 5 Integrals
10. (a)
0 g ( x) dx 17 0 7 g ( x) dx 17 (7) 1
2
2
2
2
1
(b)
1 g ( x) dx 0 g ( x) dx 0 g ( x) dx 1 2 1
(c)
2
0
2
f ( x) dx f ( x) dx
0
2
2
(d)
0
2 f ( x) dx 2 f ( x) dx 2( ) 2
(e)
0 [ g ( x) 3 f ( x)] dx 0 g ( x) dx 3 0
0
2
2
2
f ( x ) dx 1 3
11. x 2 4 x 3 0 ( x 3)( x 1) 0 x 3 or x 1;
1
3
0
1
Area ( x 2 4 x 3) dx ( x 2 4 x 3) dx
1
3
3
3
x3 2 x 2 3x x3 2 x 2 3x
0
1
3
2
1
3 2(1) 3(1) 0
3
3
33 2(32 ) 3(3) 13 2(1) 2 3(1)
8
1
1
3 1 0 3 1 3
2
12. 1 x4 0 4 x 2 0 x 2;
Area
2
2
2
1 dx 1 dx
3
x2
4
x2
4
2
3
x3 x x3
x 12
2 12 2
3
3
2 2 ( 2) 3 33 2 23
2 12
12
12
12
3
13
43 43 4 43 4
13. 5 5 x 2/3 0 1 x 2/3 0 x 1;
1
8
Area (5 5 x 2/3 ) dx (5 5 x 2/3 ) dx
1
1
1
8
5/3
5/3
5 x 3 x 5 x 3 x
1
1
5/3
5/3
5(1) 3(1)
5(1) 3(1)
5(8) 3(8)5/3 5(1) 3(1)5/3
[2 (2)] [(40 96) 2] 62
Copyright 2018 Pearson Education, Inc.
Chapter 5 Practice Exercises
14. 1 x 0 x 1;
1
4
Area (1 x ) dx (1 x ) dx
0
1
1
4
x 23 x3/2 x 23 x3/2
0
1
3/2
2
2
1 3 (1)
0 4 3 (4)3/2 1 23 (1)3/2
16
1
1
3 4 3 3 2
15.
f ( x ) x, g ( x )
1
x2
, a 1, b 2
b
A [ f ( x) g ( x)] dx
a
2
1
16.
x dx
1
x2
x2
2
2
1x
1
24 12 12 1 1
b
f ( x) x, g ( x) 1 , a 1, b 2 [ f ( x) g ( x)] dx
a
x
A
2
1
x dx
1
x
x2
2
2
2 x
1
42 2 2 12 2 7 42 2
b
1
17. f ( x) (1 x )2 , g ( x) 0, a 0, b 1 A [ f ( x) g ( x)] dx
0 (1
a
1
(1 2 x1/2 x) dx x 43 x3/2
0
1
x2
2 0
1 34 12 16 (6 8 3)
b
18. f ( x) (1 x3 )2 , g ( x) 0, a 0, b 1 A [ f ( x) g ( x)] dx
a
1
0 (1 x
1
19. f ( y ) 2 y 2 , g ( y ) 0, c 0, d 3
d
3
0
0
1
6
4
7
9
x x2 x7 1 12 17 14
0
c
1
x ) 2 dx (1 2 x x) dx
A [ f ( y ) g ( y )] dy (2 y 2 0) dy
3
2 y 2 dy 23 [ y 3 ]30 18
0
Copyright 2018 Pearson Education, Inc.
3 2
) dx
1
0 (1 2 x
3
x 6 ) dx
349
350
Chapter 5 Integrals
20. f ( y ) 4 y 2 , g ( y ) 0, c 2, d 2
d
2
c
2
A [ f ( y ) g ( y )] dy
4 y
y3
3
2
2
2 8 83
(4 y 2 ) dy
32
3
y2
y 2
21. Let us find the intersection points: 4 4
y 2 y 2 0 ( y 2)( y 1) 0 y 1 or
y2
y 2 c 1, d 2; f ( y ) 4 , g ( y )
d
2 y2 y2
A [ f ( y ) g ( y )] dy 4 4
1
c
y2
4
dy
2
2
y3
y2
14 ( y 2 y 2 ) dy 14 2 2 y 3
1
1
14 42 4 83 12 2 13 89
y 2 4
y 16
22. Let us find the intersection points: 4 4
y 2 y 20 0 ( y 5)( y 4) 0 y 4 or
y 2 4
y 16
y 5 c 4, d 5; f ( y ) 4 , g ( y ) 4
d
5 y 16 y 2 4
A [ f ( y ) g ( y )] dy 4 4 dy
4
c
2
3 5
5
y
y
2
14 ( y 20 y ) dy 14 2 20 y 3
4
4
125 16 80 64
1
4 25
100
3
2
3
2
9
9
1
1
4 2 180 63 4 2 117 81 (9 234) 243
8
23. f ( x) x, g ( x) sin x, a 0, b 4
b
/4
a
0
A [ f ( x) g ( x)] dx
/4
2
x2 cos x
0
( x sin x) dx
2
32 22 1
24. f ( x) 1, g ( x) |sin x |, a 2 , b 2
b
/2
a
/2
A [ f ( x) g ( x)] dx
0
/2
2
(1 sin x) dx
/2
0
/2
0
(1 |sin x |) dx
(1 sin x) dx
(1 sin x) dx 2[ x cos x]0 /2
2 2 1 2
Copyright 2018 Pearson Education, Inc.
Chapter 5 Practice Exercises
25. a 0, b , f ( x) g ( x) 2sin x sin 2 x
0
0
A (2sin x sin 2 x) dx 2 cos x cos22 x
2 (1) 12 2 1 12 4
26. a 3 , b 3 , f ( x) g ( x) 8cos x sec2 x
A
/3
/3
(8cos x sec2 x) dx [8sin x tan x]/3/3
8 23 3 8 23 3 6 3
27. f ( y )
y , g ( y ) 2 y, c 1, d 2
d
2
A [ f ( y ) g ( y )] dy [ y (2 y )] dy
c
1
1
2
y2
y 2 y dy 23 y 3/2 2 y 2
1
2
43 2 4 2 23 2 12 34 2 76 8 26 7
28. f ( y ) 6 y, g ( y ) y 2 , c 1, d 2
d
2
A [ f ( y ) g ( y )] dy (6 y y 2 ) dy
c
1
2
y2
y3
6 y 2 3 12 2 83 6 12 13
1
3 13
4 73 12 2414
6
6
29. f ( x) x3 3 x 2 x 2 ( x 3) f ( x) 3 x 2 6 x 3 x( x 2) f | | f (0) 0 is a
0
3
2
3
maximum and f (2) 4 is a minimum. A ( x3 3x 2 ) dx x4 x3
0
0
30.
a
a
A (a1/2 x1/2 )2 dx (a 2 ax1/2 x) dx ax 43 ax3/2
0
0
4
a
x2
2 0
2
2
a 2 1 43 12 a6 (6 8 3) a6
Copyright 2018 Pearson Education, Inc.
814 27 274
2
a 2 43 a a a a2
351
352
Chapter 5 Integrals
1
31. The area above the x-axis is A1 ( y 2/3 y ) dy
0
1
y
3y
1 ; the area below the x-axis is
5 2 10
0
0
0
3 y5/3 y 2
11 the
A2 ( y 2/3 y ) dy 5 2 10
1
1
total area is A1 A2 65
5/3
32.
A
2
/4
0
(cos x sin x) dx
3 /2
5 /4
5 /4
/4
(sin x cos x ) dx
(cos x sin x) dx
[sin x cos x]0 /4 [ cos x sin x]5/4/4
[sin x cos x]35 /2
/4
22 22 (0 1) 22 22 22 22
(1 0) 22 22 8 22 2 4 2 2
x
d2y
dy
1
33.
y x 2 1t dt dx 2 x 1x 2 2 12 ; y (1) 1 1t 1 and y (1) 2 1 3
1
1
dx
x
34.
y (1 2 sec t ) dt
x
0
dy
dx
1 2 sec x
d2y
dx 2
0
x 0 y (1 2 sec t ) dt 0 and x 0
0
x sin t
t
35.
y
5
36.
y
1
x
dt 3
dy
dx
2
dy
dx
12 (sec x)1/2 (sec x tan x)
sec x (tan x);
1 2 sec 0 3
5
sinx x ; x 5 y sint t dt 3 3
5
1
dy
2 sin 2 t dt 2 so that dx 2 sin 2 x; x 1 y
2 sin 2 t dt 2 2
1
37. Let u cos x du sin x dx du sin x dx
2(cos x)
1/2
1/ 2
sin x dx 2u 1/2 (du ) 2 u 1/2 du 2 u 1 C 4u1/2 C 4(cos x)1/2 C
2
38. Let u tan x du sec 2 x dx
3/2
2
3/2
(tan x) sec x dx u du
u 1/ 2
12
C 2u 1/2 C
39. Let u 2 1 du 2 d 12 du d
[2 1 2 cos (2 1)] d (u 2 cos u )
2
sin (2 1) C , where C C1
2
(tan x )1/ 2
C
12 du u4 sin u C1 (241)
1
4
2
is still an arbitrary constant
Copyright 2018 Pearson Education, Inc.
2
sin (2 1) C1
Chapter 5 Practice Exercises
40. Let u 2 du 2 d
1
2
du d
2sec2 (2 ) d 1 2sec 2 u
1
2
1/2
u
1 du
2
tan u C (2 )1/2 tan (2 ) C
u
41.
t 2t t 2t dt t
42.
(t 1)2 1
t
dt t
4
2
2t
t4
2
12 (u 1/2 2sec2 u) du 12 u
1/ 2
1
2
353
1
2 (2 tan u ) C
3
1
3
42 dt (t 2 4t 2 ) dt t3 4 t1 C t3 4t C
t
dt
1
t2
2
t3
dt (t
2
2t 3 ) dt
t 1
( 1)
2
C
t 2
2
1 1
t t2
C
43. Let u 2t 3/2 du 3 t dt 13 du t dt
t sin 2t 3/2 dt 13 sin u du 13 cos u C 13 cos (2t 3/2 ) C
44. Let u 1 sec du sec tan d sec tan 1 sec d u1/2 du 23 u 3/2 C 23 (1 sec )3/2 C
45.
cos 2
(sinsin22cos
2 )
1 du
2
d
1 1
2 u3
du , where u sin 2 cos 2 du (2cos 2 2sin 2 )d
(sin 2 cos 2 )d
46.
3
1 1 u 2
2 2
C
1
4(sin 2 cos 2 ) 2
C
cos sin(sin ) d sin u du, where u sin
du cos d
cos u C cos(sin ) C
1
2
4 x 7) dx [ x3 2 x 2 7 x]11 [13 2(1) 2 7(1)] [(1)3 2(1)2 7(1)] 6 (10) 16
47.
1 (3x
48.
0 (8s
49.
1 v4 dv 1 4v
50.
1
51.
1 tdtt 1 t dt 1 t
1
3
12s 2 5) ds [2 s 4 4 s3 5s ]10 [2(1)4 4(1)3 5(1)] 0 3
2
2
2
2
dv [4v 1 ]12
24 14 2
27 4/3
dx [3 x 1/3 ]127 3(27)1/3 (3(1)1/3 ) 3
4
4
x
3/ 2
4 3/2
dt [2t 1/2 ]14
2
4
52. Let x 1 u dx 12 u 1/2 du 2 dx
4 (1 u )1/ 2
u
1
3
du x1/2 (2 dx) 2
2
23 x
du ; u
u
3/2 3
( 2)
1
13 3(1) 2
1
1 x 2, u 4 x 3
4 (33/2 ) 4 (23/2 ) 4 3 8 2 4 (3 3 2 2)
3
3
3
2 3
53. Let u 2 x 1 du 2 dx 18 du 36 dx; x 0 u 1, x 1 u 3
3
1 36 dx
0 (2 x1)
3
3
2
3
18u3 du 18u2 92
1
1 u 1
8
9
32
9
12
Copyright 2018 Pearson Education, Inc.
354
Chapter 5 Integrals
54. Let u 7 5r du 5 dr 15 du dr ; r 0 u 7, r 1 u 2
1
0
dr
3
(7 5r )
2
1
2
0
7
(7 5r ) 2/3 du u 2/3 15 du
15 [3u1/3 ]72
3 37
5
55. Let u 1 x 2/3 du 23 x 1/3 dx 32 du x 1/3 dx; x 18 u 1
1
1/8
x 1/3 (1 x 2/3 )3/2 dx
34
5/2
53 (0)5/2 53
1
18
1625
1/2
2
1 du
36
x3 dx; x 0 u 1, x
25/16 3/2 1
u
36
1
5
0
25/16
1 u 1/ 2
du 36
1
2 1
(sin 2 u )
1
2
u 1 9
/4
cos 2 4t 4 dt
/3
0
60.
/4
25
16
1
90
15 du 15 u2 sin42u 0
5
0 sin 0
2 sin10
20
20
2
3 /4
/4 (cos
2
u)
3 /4
14 du 14 u2 sin42u /4 14 38
sin
32 1 sin 2
4
4
8
4
sec2 d [tan ]0 /3 tan 3 tan 0 3
3 /4
csc2 x dx [ cot x]3/4/4 cot 34 cot 4 2
x du
6
3
cot 2 6x dx
61. Let u
4
25/16
1 1
8 16
16
8
59.
12
1 u 1/2
18
1
58. Let u 4t 4 du 4 dt 14 du dt ; t 0 u 4 , t 4 u 34
0
43 , x 1 u 1 12/3 0
27 3
160
1 (1) 1/2
18
5r dr
2/3
0
57. Let u 5r du 5 dr 15 du dr ; r 0 u 0, r u 5
0 sin
18
0
5/ 2
u 3/2 32 du 32 u 5 53 u 5/2
3/4
3/4
2 3/4
0
56. Let u 1 9 x 4 du 36 x3 dx
1/2 3
x (1 9 x 4 ) 3/2 dx
0
32
1 dx 6 du dx; x
6
/2
6 cot 2 u du 6
/6
u 6 , x 3 u 2
/2
/6 (csc
2
u 1) du [6( cot u u )] /2
/6
6 cot 2 2 6 cot 6 6 6 3 2
62. Let u 3 du 13 d 3du d ; 0 u 0, u 3
0 tan
2
d
3
0 sec
2
3
1 d
/3
0
3 3
63.
/3 sec x tan x dx [sec x] /3 sec 0 sec 3 1 2 1
0
3(sec 2 u 1) du [3 tan u 3u ]0 /3 3 tan 3 3 3 (3 tan 0 0)
0
Copyright 2018 Pearson Education, Inc.
Chapter 5 Practice Exercises
64.
3 /4
/4
355
csc z cot z dz [ csc z ]3/4/4 csc 34 csc 4 2 2 0
65. Let u sin x du cos x dx; x 0 u 0, x 2 u 1
/2
0
1
5(sin x)3/2 cos x dx
0 5u
3/2
du 5
52 u5/2 0
1
[2u 5/2 ]10 2(1)5/2 2(0)5/2 2
66. Let u sin 3 x du 3cos 3 x dx 13 du cos 3 x dx; x 2 u sin 32 1, x 2 u sin 32 1
/2
/215 sin
4
1
15u 4
1
3 x cos 3x dx
13 du 1
1
5u 4 du [u 5 ]11 (1)5 (1)5 2
67. Let u 1 3sin 2 x du 6sin x cos x dx 12 du 3sin x cos x dx; x 0 u 1, x 2 u 1 3sin 2 2 4
/2 3sin x cos x
0
13sin 2 x
dx
4 1 1
1 u 2
du
4 1 1/2
u
1 2
4
1/ 2
du 12 u 1 [u1/2 ]14 41/2 11/2 1
2 1
68. Let u 1 7 tan x du 7 sec2 x dx 17 du sec2 x dx; x 0 u 1 7 tan 0 1, x 4 u 1 7 tan 4 8
/4
0
sec2 x
(1 7 tan x ) 2/3
dx
8 1 1
du
1 u 2/3 7
18 17 u 2/3du 17 u
1/3
1
3
8
8
1/3 3 1/3
3 1/3
3
7 u 1 7 (8) 7 (1)
1
3
7
1
2
2
mx 2 bx 1 m(1) b(1) m( 1) b(1) 1 (2b) b
2
2
2
2
2
1
k
2
2
2
k
m ( k )
m( k )
(b) av( f ) k (1 k ) (mx b) dx 21k mx2 bx 21k 2 b(k ) 2 b(k )
k
k
21k (2bk ) b
1
1 (mx b) dx 12
69. (a) av( f ) 1(11)
70. (a)
(b)
71.
1
30
1
a0
3
3
3x dx 13
a
ax dx 1a
yav
0
yav
0
0
a
0
3
3 x1/2 dx 33 23 x3/2
0
3 2
(3)3/2
3 3
23 (0)3/2
3
3
(2 3) 2
a
a x1/2 dx aa 23 x3/2 aa 23 (a )3/2 23 (0)3/2 aa 23 a a 23 a
0
b
1 f ( x) dx 1 [ f ( x )]ba 1 [ f (b) f (a )] f (b) f ( a ) so the average value of f over [ a, b] is
f av
ba a
ba
ba
ba
the slope of the secant line joining the points (a, f (a )) and (b, f (b)), which is the average rate of change
of f over [a, b].
b
72. Yes, because the average value of f on [ a, b] is b 1 a f ( x) dx. If the length of the interval is 2, then b a 2
a
and the average value of the function is
73. We want to evaluate
365
1
3650 0
1
f ( x) dx 365
365
0
1 b
2 a
f ( x) dx.
2 ( x 101) 25 dx 37
37 sin 365
365
365
0
2 ( x 101) dx 25
sin 365
365
365
0
dx
2 ( x 101) is 2 365 and that we are integrating this function over an
Notice that the period of y sin 365
2
365
interval of length 365. Thus the value of
37 365 sin 2 ( x 101)
365 0
365
25
dx 365
Copyright 2018 Pearson Education, Inc.
365
0
37 0 25 365 25.
dx is 365
365
356
74.
Chapter 5 Integrals
675
1
(8.27 105 (26T
675 20 20
1.87T 2 )) dT
1 8.27T
655
675
2
3
26T 5 1.87T5
210
310
20
2
3
2
3
1 8.27(675) 26(675) 1.87(675) 8.27(20) 26(20) 1.87(20)
655
5
5
5
5
210
310
210
310
1 (3724.44 165.40) 5.43 the average value of C on [20, 675]. To find the temperature T at
655
v
which Cv 5.43, solve 5.43 8.27 105 (26T 1.87T 2 ) for T. We obtain 1.87T 2 26T 284000 0
26 (26) 2 4(1.87)( 284000)
2(1.87)
2124996
26 3.74
. So T 382.82 or T 396.72. Only T 396.72 lies in the
interval [20, 675], so T 396.72C.
T
dy
dy
75. dx 2 cos3 x
d (7 x 2 ) 14 x 2 cos3 (7 x 2 )
76. dx 2 cos3 (7 x 2 ). dx
77.
dy
dx
d x 6 dt 6
dx
1 3t 4
3 x 4
78.
dy
dx
d 2
d sec x 1
d
sec x tan x
1
1
dx
sec x t 2 1 dt dx 2 t 2 1 dt sec2 x 1 dx (sec x) 1sec2 x
79. Yes. The function f, being differentiable on [a, b], is then continuous on [a, b]. The Fundamental Theorem of
Calculus says that every continuous function on [a, b] is the derivative of a function on [a, b].
80. The second part of the Fundamental Theorem of Calculus states that if F ( x) is an antiderivative of f ( x )
on [a, b], then
1
f ( x) dx F (b) F (a). In particular, if F ( x) is an antiderivative of 1 x 4 on [0, 1], then
1 x 4 dx F (1) F (0).
0
1
y
x
82.
y
0
1
cos x 1t 2
x
1 t 2 dt
81.
b
a
1
1cos 2 x
1
dt
1 t 2 dt
cos x 1
1t 2
0
d (cos x )
dx
dy
dx
d
dx
x 1 t 2 dt d x 1 t 2 dt 1 x 2
dx 1
1
dy
d cos x 1 dt d cos x 1 dt
dt dx dx
dx 0
0 1t 2
1t 2
sin1 x ( sin x) sin1 x csc x
2
83. 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 15 0236 362 54 54251 51249.5 49.5254 54264.4 64.42 67.5 67.52 42 5961 ft 2 . The cost is
Area ($2.10/ft 2 ) (5961 ft 2 )($2.10/ft 2 ) $12,518.10 the job cannot be done for $11,000.
84. (a) Before the chute opens for A, a 32 ft/sec2 . Since the helicopter is hovering v0 0 ft/sec
v 32 dt 32t v0 32t. Then s0 6400 ft s 32t dt 16t 2 s0 16t 2 6400.
At t 4sec, s 16(4)2 6400 6144 ft when A’s chute opens;
(b) For B, s0 7000 ft, v0 0, a 32 ft/sec 2 v 32 dt 32t v0 32t s
32t dt 16t
2
s0 16t 2 7000. At t 13 sec, s 16(13)2 7000 4296 ft when B’s chute opens;
(c) After the chutes open, v 16 ft/sec s 16 dt 16t s0 . For A, s0 6144 ft and for B, s0 4296 ft.
Therefore, for A, s 16t 6144 and for B, s 16t 4296. When they hit the ground, s 0 for A,
Copyright 2018 Pearson Education, Inc.
Chapter 5 Additional and Advanced Exercises
357
0 16t 6144 t 6144
384 seconds, and for B, 0 16t 4296 t 4296
268.5 seconds to hit the
16
16
ground after the chutes open, Since B’s chutes opens 58 seconds after A’s opens B hits the ground first.
CHAPTER 5
ADDITIONAL AND ADVANCED EXERCISES
1. (a) Yes, because
1
1
0 f ( x) dx 17 0 7 f ( x) dx 17 (7) 1
(b) No. For example,
2
5
2. (a) True:
1
0
1
3/ 2
3/2
4 2 3/2
4 2
0 8 x dx 2 2 x 32 3 1 0 3 4
0
1
8 x dx [4 x 2 ]10 4, but
5
f ( x) dx f ( x) dx 3
2
5
5
5
2
5
5
2 [ f ( x) g ( x)] dx 2 f ( x) dx 2 g ( x) dx 2 f ( x) dx 2 f ( x) dx 2 g ( x) dx
(b) True:
43 2 9
5
5
5
the other hand, f ( x) g ( x) [ g ( x) f ( x)] 0
3.
5
2 f ( x)dx 4 3 7 2 2 g ( x) dx 2 [ f ( x) g ( x)] dx 0 2 [ g ( x) f ( x)] dx 0. On
(c) False:
5
2
[ g ( x) f ( x)] dx 0 which is a contradiction.
x
x
x
y a1 f (t ) sin a ( x t ) dt a1 f (t ) sin ax cos at dt a1 f (t ) cos ax sin at dt
0
0
0
sin ax x
a
0
f (t ) cos at dt
cos ax x
a
0
dy
f (t ) sin at dt dx
x
x
sin ax d x
cos ax d x
cos ax f (t ) cos at dt a dx
f (t ) cos at dt sin ax f (t ) sin at dt a dx
f (t ) sin at dt
0
0
0
0
x
x
0
0
cos ax f (t ) cos at dt sinaax ( f ( x) cos ax) sin ax f (t ) sin at dt cosaax ( f ( x) sin ax)
x
dy
d2y
x
dx cos ax f (t ) cos at dt sin ax f (t ) sin at dt. Next, 2
0
0
dx
x
x
x
d
d x f (t ) sin at dt
a sin ax f (t ) cos at dt (cos ax) dx
f (t ) cos at dt a cos ax f (t ) sin at dt (sin ax) dx
0
0
0
0
x
x
0
0
a sin ax f (t ) cos at dt (cos ax) f ( x ) cos ax a cos ax f (t ) sin at dt (sin ax) f ( x) sin ax
x
x
0
0
a sin ax f (t ) cos at dt a cos ax f (t ) sin at dt f ( x). Therefore, y a 2 y
a cos ax f (t ) sin at dt a sin ax f (t ) cos at dt f ( x) a 2 sinaax f (t ) cos at dt
0
0
0
f ( x). Note also that y (0) y (0) 0.
x
4.
x
y
0
1
1
2
1
1 4t
2
1
1 4 y
2
1 4 y2
x
dt
d ( x)
dx
dy
dx
1/2
(8 y )
dy
dx
d y
1
dx 0 1 4t 2
x
dt
1 4 y 2 . Then
dy
dx
4y
dy
dx
1 4 y
2
4y
d
dy
d2y
dx 2
y
0
d
dx
1
1 4t 2
from the chain rule
1 4y
dt
1 4 y2
cos ax x
a
0
f (t ) sin at dt
dy
dx
d
dy
2
dy
dx
14 y 4 y. Thus d y 4 y, and the constant of proportionality
2
1 4 y
2
2
dx 2
is 4.
Copyright 2018 Pearson Education, Inc.
358
Chapter 5 Integrals
x2
0
5. (a)
2
d x
dx 0
f (t ) dt x cos x
f (t ) dt cos x x sin x f ( x 2 )(2 x) cos x x sin x
cos x x sin x . Thus, x 2 f (4) cos 2 2 sin 2 1
4
4
2x
3 f ( x)
3
3
3
t3
13 f ( x) 13 f ( x) x cos x f ( x)
0
f ( x2 )
f ( x) 2
0
(b)
t dt
3 x cos x f ( x) 3 3 x cos x
f (4) 3 3(4) cos 4 3 12
6.
a
0
f ( x) dx
a
a2
2
a2 sin a 2 cos a. Let F (a) f (t ) dt f (a ) F (a). Now F ( a )
0
a2
2
a2 sin a 2 cos a
f (a) F (a ) a 12 sin a a2 cos a 2 sin a f 2 2 12 sin 2 22 cos 2 2 sin 2 2 12 2
7.
b
1
f ( x) dx b 2 1 2 f (b)
d b
db 1
f ( x) dx
d
dx
8. The derivative of the left side of the equation is:
d x
dx 0
right side of the equation is:
d
dx
1
2
(b 2 1)1/2 (2b)
b
b 2 1
f ( x)
1
2
x
x 2 1
x
x u
0 0 f (t ) dt du 0 f (t ) dt ; the derivative of the
f (u )( x u ) du
d x
dx 0
x
d
f (u ) x du dx
u f (u ) du
0
x
x f (u ) du d
d
0
dx 0 u f (u ) du 0 f (u ) du x dx 0 f (u ) du x f ( x) 0 f (u ) du x f ( x) x f ( x)
x
x
x
x
x
f (u ) du. Since each side has the same derivative, they differ by a constant, and since both sides equal 0
0
x u
x
when x 0, the constant must be 0. Therefore, f (t ) dt du f (u )( x u ) du.
0 0
0
9.
dy
dx
3 x 2 2 y (3x 2 2) dx x3 2 x C. Then (1, 1) lies on the curve 13 2(1) C 1
C 4 y x3 2 x 4
10. The acceleration due to gravity downward is 32 ft/sec2 v 32 dt 32t v0 , where v0 is the initial
velocity v 32t 32 s (32t 32) dt 16t 2 32t C. If the release point, at t 0, is s 0, then
C 0 s 16t 2 32t. Then s 17 17 16t 2 32t 16t 2 32t 17 0. The discriminant of this
quadratic equation is 64 which says there is no real time when s 17 ft. You had better duck.
11.
3
0
8 f ( x) dx 8 x
2/3
3
dx 4 dx
0
0
53 x5/3 [4 x]30
8
3
0 5 (8)5/3 (4(3) 0)
12.
3
0
4 f ( x) dx 4
0
96
5
12
36
5
3
x dx ( x 2 4) dx
0
3
23 ( x)3/2 x3 4 x
4
0
3/2 33
16
2
0 3 (4)
3 4(3) 0 3 3
3
7
3
Copyright 2018 Pearson Education, Inc.
Chapter 5 Additional and Advanced Exercises
13.
2
1
359
2
0 g (t ) dt 0 t dt 1 sin t dt
1
2
t2 1 cos t
1
0
2
14.
12 0 1 cos 2 1 cos 12 2
2
1
2
1 z dz (7 z 6) 1/3 dz
0 h( z ) dz 0
1
1
2
3 (7 z 6) 2/3
23 (1 z )3/2 14
0
1
23 (1 1)3/2 23 (1 0)3/2
2/3
3
3
14 (7(2) 6) 14 (7(1) 6)2/3
6
3
55
2
3 7 14 42
15.
1
2
1
2 f ( x) dx 2 dx 1 (1 x
2
2
) dx 2 dx
1
1
[ x]12 x x3 [2 x]12
1
3
( 1)3
(1 (2)) 1 13 1 3 2(2) 2(1)
13
2
2
1 3 3 4 2 3
3
16.
2
0
1
1 h(r ) dr 1 r dr 0 (1 r
0
2
2
) dr dr
1
1
2
3
r2 r r3 [r ]12
1
0
2
3
( 1)
0 2 1 13 0 (2 1) 12 23 1
b
1
ba a
17. Ave. value
12
2
1
2
0
18. Ave. value
2
2
2
f ( x) dx
1
2
b
1
ba a
2
1
20 0
2
1
2
f ( x) dx
f ( x) dx
1
2
7
6
1 x dx 2 ( x 1) dx 1 x 2 1 1 x 2 x 2
1
0
2 2 0 2 2
1
1
2
3
1
3 0 0
1
2
3
f ( x) dx 13 dx 0 dx dx 13 [1 0 0 3 2]
0
1
2
2
3
19. Let f ( x) x5 on [0, 1]. Partition [0, 1] into n subintervals with x 1n0 1n . Then 1n , n2 , , nn are the right-hand
is the upper sum for f ( x) x
j 5 1
n
endpoints of the subintervals. Since f is increasing on [0, 1], U n
j
1 1 2
n
n 1n nlim
n n
n
[0, 1] lim
j 1
5
5
5
j 1
nn
5
1
15 25 n5 1 x5 dx x 6
nlim
n6
0
6 0
Copyright 2018 Pearson Education, Inc.
1
6
5
on
360
Chapter 5 Integrals
20. Let f ( x) x3 on [0, 1]. Partition [0, 1] into n subintervals with x 1n0 1n . Then 1n , n2 , , nn are the right-hand
is the upper sum for f ( x) x on
j 3 1
n
endpoints of the subintervals. Since f is increasing on [0, 1], U n
n
[0, 1] lim
j 1
lim
j 3 1
n
n
1
n n
1
n
3
2
n
3
j 1
3
1
1 2 n 1 x3 dx x 1
nlim
n
0
4 0 4
n 3
n
3
3
3
4
4
21. Let y f ( x) on [0, 1]. Partition [0, 1] into n subintervals with x 1n0 1n . Then 1n , n2 , , nn are the right-hand
is a Riemann sum of y f ( x) on
endpoints of the subintervals. Since f is continuous on [0, 1], f n
j
j 1
1
n
1 f 1 f 2 f n f ( x ) dx
n n
n 0
f n 1n nlim
n
n
[0, 1] lim
1
j
j 1
22. (a)
1 [2 4 6 2n] lim 1 2
2
n n
n n n
lim
f ( x) x
(c)
lim 1
n n
0
n2
215 n15 ] lim 1n 1n
n
on [0, 1]
1 [115
16
n n
15
(b) lim
1
n4 6n 2nn 2 x dx [ x 2 ]10 1, where f ( x) 2 x on [0, 1]
15
15
nn
15
1 15
0 x
16
1
1
1
1 , where
dx x16 16
0
sin sin 2 sin n sin n dx 1 cos x 1 cos 1 cos 0 2 , where
n
n 0
n
0
f ( x) sin x on [0, 1]
115 215 n15 lim 1 lim 1 [115 215 n15 ] lim 1 1 x15 dx 0
n n n n16
n n 0
(see part (b) above)
(e) lim 115 115 215 n15 lim n16 [115 215 n15 ]
n n
n n
1
lim n lim 116 [115 215 n15 ] lim n x15 dx (see part (b) above)
n n n
n 0
1
17
n n
(d) lim
161 0
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 2n . The area of each triangle is
2
2
An 12 r 2 sin n the area of the polygon is A nAn nr2 sin n nr2 sin 2n .
(b)
2
2
lim A lim nr2 sin 2n lim n2r sin 2n lim r 2
n
n
n
n
2n r 2
2n
sin
lim
2n r 2
2n
sin
2 / n
24. Partition [0, 1] into n subintervals, each of length x 1n with the points x0 0, x1 1n , x2 n2 , , xn nn 1.
The inscribed rectangles so determined have areas f ( x0 ) x (0)2 x, f ( x1 )x 1n
2
2
2
x,
f ( x2 ) x n2 x, , f ( xn 1 ) nn1 x. The sum of these areas is
2
2
2
2
2
2
( n 1) 2
( n 1) 2
2
Sn 02 1n n2 nn1 x 1 2 22 2 1n 1 3 23 3 . Then
n
n
n
n
n n
3
1 2
( n 1)2
12 22
lim Sn lim 3 3 3 x dx 13 13 .
0
n
n
n
n n
Copyright 2018 Pearson Education, Inc.
Chapter 5 Additional and Advanced Exercises
25. (a)
361
1
g (1) f (t ) dt 0
1
3
(b) g (3) f (t ) dt 12 (2)(1) 1
1
(c)
g (1)
1
1
f (t ) dt
1
1
f (t ) dt 14 ( 22 )
(d) g ( x) f ( x) 0 x 3, 1, 3 and the sign chart for g ( x) f ( x) is | | | . So g has a
3
relative maximum at x 1.
(e) g ( 1) f ( 1) 2 is the slope and g (1)
1
1
1
3
f (t )dt , by (c). Thus the equation is y 2( x 1)
y 2x 2 .
(f ) g ( x) f ( x) 0 at x 1 and g ( x) f ( x) is negative on ( 3, 1) and positive on ( 1, 1) so there is an
inflection point for g at x 1. We notice that g ( x) f ( x) 0 for x on ( 1, 2) and g ( x) f ( x) 0 for x
on (2, 4), even though g (2) does not exist, g has a tangent line at x 2, so there is an inflection point at
x 2.
(g) g is continuous on [3, 4] and so it attains its absolute maximum and minimum values on this interval. We
saw in (d) that g ( x) 0 x 3, 1, 3. We have that g (3)
3
1
1
3
g (1) f (t ) dt 0
1
3
2
f (t ) dt 22 2
4
g (3) f (t ) dt 1
1
f (t ) dt
g (4) f (t ) dt 1 12 1 1 12
1
1
Thus, the absolute minimum is 2 and the absolute maximum is 0. Thus, the range is [2 , 0].
26.
x
x
y sin x cos 2t dt 1 sin x cos 2t dt 1 y cos x cos(2 x); when x we have
y cos cos(2 ) 1 1 2. And y sin x 2sin(2 x); when x , y sin
x cos 2t dt 1 0 0 1 1.
1x x x1 1x 1x 2x
d 1
dt f ( x) 1x dx
11 dx
dx
x
x
x 1
1/ x t
27.
f ( x)
28.
f ( x)
sin x 1
cos x 1t 2
29.
g ( y)
2 y
30.
f ( x)
dt f ( x)
1
1sin 2 x
d (sin x)
dx
sin t 2 dt g ( y ) sin 2 y
y
x 3
x
2
1cos1 x dxd (cos x) coscos xx sinsin xx cos1 x sin1 x
2
2
2 y sin y y
2 d
dy
2
d
dy
2
sin 4 y
y
sin y
2 y
d ( x 3) x(5 x) dx ( x 3)(2 x) x(5 x )
t (5 t ) dt f ( x) ( x 3)(5 ( x 3)) dx
dx
6 x x 2 5 x x 2 6 6 x. Thus f ( x) 0 6 6 x 0 x 1. Also, f ( x) 6 0 x 1 gives
a maximum.
Copyright 2018 Pearson Education, Inc.
CHAPTER 6 APPLICATIONS OF DEFINITE INTEGRALS
6.1
1.
2.
VOLUMES USING CROSS-SECTIONS
A( x)
A( x )
(diagonal)2
2
x x
(diameter) 2
4
2
2
4
2 1 x 2
2
1 2 x 2 x 4 ; a 1, b 1;
4
V A( x) dx 1 2 x 2 x 4 dx x 23 x3
1
a
1
x5
5 1
2
3.
4
4
b
2 x; a 0, b 4; V A( x) dx 2 x dx x 2 16
0
0
a
2 x 2 x 2
1
b
2
2 1 23 15 16
15
2
A( x ) (edge)2 1 x 2 1 x 2 2 1 x 2 4 1 x 2 ; a 1, b 1;
1
3
b
1
V A( x) dx 4 1 x 2 dx 4 x x3 8 1 13 16
3
1
1
a
4.
A( x )
b
(diagonal) 2
2
2
2 1 x
2
2
2
2
2 1 x 2 ; a 1, b 1;
1 x 2 dx 2 x x3 4 1 13 83
1
1
V A( x) dx 2
a
1 x 2 1 x 2
2
1
1
3
5. (a) STEP 1) A( x ) 12 (side) (side) sin 3 12 2 sin x 2 sin x sin 3 3 sin x
STEP 2) a 0, b
b
STEP 3) V A( x) dx 3 sin x dx 3 cos x 3(1 1) 2 3
a
0
0
(b) STEP 1) A( x ) (side)2 2 sin x
2
sin x 4 sin x
STEP 2) a 0, b
b
a
0
STEP 3) V A( x) dx 4 sin x dx 4 cos x 0 8
6. (a) STEP 1) A( x )
(diameter) 2
4 (sec x tan x) 2 4 sec 2 x tan 2 x 2sec x tan x
4
4 sec 2 x sec2 x 1 2 sin2x
cos x
STEP 2) a 3 , b 3
b
a
/3
/3
2sec2 x 1 2sin2 x dx 4 2 tan x x 2 cos1 x
4
/3
/3
cos x
STEP 3) V A( x) dx
4 2 3 3 2 11 2 3 3 2 11 4 4 3 23
2
2
Copyright 2018 Pearson Education, Inc.
363
364
Chapter 6 Applications of Definite Integrals
(b) STEP 1) A( x) (edge)2 (sec x tan x) 2 2sec2 x 1 2 sin2x
STEP 2) a 3 , b 3
b
/3
a
/3
STEP 3) V A( x) dx
2sec
2
cos x
x 1 2 sin2 x dx 2 2 3 3 4 3 23
cos x
7. (a) STEP 1) A( x ) (length) (height) (6 3x ) (10) 60 30 x
STEP 2) a 0, b 2
2
2
b
STEP 3) V A( x) dx (60 30 x) dx 60 x 15 x 2 (120 60) 0 60
0
0
a
(b) STEP 1) A( x ) (length) (height) (6 3 x)
20 2(6 3 x )
2
(6 3x)(4 3x) 24 6x 9x
2
STEP 2) a 0, b 2
b
2
a
0
STEP 3) V A( x)dx
24 6 x 9 x2 dx 24 x 3x2 3x3 0 (48 12 24) 0 36
2
8. (a) STEP 1) A( x ) 12 (base) (height)
x 2x (6) 6 x 3x
STEP 2) a 0, b 4
6 x1/2 3 x dx 4 x3/2 32 x 2 (32 24) 0 8
0
0
b
STEP 3) V A( x) dx
a
(b) STEP 1) A( x) 12
4
4
diameter
2
2
2
x x3/ 2 14 x 2
xx
12 2 2 2
8 x x3/2 14 x 2
4
STEP 2) a 0, b 4
1 x3 8 64 16 (0)
x x3/2 14 x 2 dx 12 x 2 52 x5/2 12
5
3 8
15
0 8
0
b
STEP 3) V A( x) dx 8
a
9.
A( y ) 4 (diameter) 2 4
4
4
5 y2 0
2
54 y 4 ;
d
c 0, d 2; V A( y ) dy
c
2 5
0 4
2
y5
y 4 dy 54 5 4 25 0 8
0
2
10.
2
A( y ) 12 (leg)(leg) 12 1 y 2 1 y 2 12 2 1 y 2 2 1 y 2 ; c 1, d 1;
1
d
V A( y ) dy 2 1 y 2 dy 2 y
c
1
1
y3
3
1
4 1 13 83
Copyright 2018 Pearson Education, Inc.
Section 6.1 Volumes Using Cross-Sections
b
h
11. The slices perpendicular to the edge labeled 5 are triangles, and by similar triangles we have
4
3
365
h 34 b.
The equation of the line through (5, 0) and (0, 4) is y 54 x 4, thus the length of the base 54 x 4 and
the height 34 54 x 4 53 x 3. Thus A( x) 12 (base) (height) 12 54 x 4 53 x 3
6 x2
25
b
5 6 2
x
0 25
12
x 6 and V A( x) dx
5
a
5
2 x3 6 x 2 6 x (10 30 30) 0 10
12
x 6 dx 25
5
5
0
12. The slices parallel to the base are squares. The cross section of the pyramid is a triangle, and by similar
triangles we have bh 53 b 53 h. Thus A( y ) (base) 2
53 y
2
9
25
d
5 9
0 25
y 2 V A( y ) dy
c
y 2 dy
5
3 y 3 15 0 15
25
0
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 ) (sidelength) 2 s 2 ;
STEP 2) a 0, b h;
b
h
a
0
STEP 3) V A( x) dx s 2 dx s 2 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 V s 2 h
14. 1)
2)
The solid and the cone have the same altitude
of 12.
The cross sections of the solid are disks of
diameter x
2x 2x . If we place the vertex of
the cone at the origin of the coordinate system
and make its axis or symmetry coincide with
the x-axis then the cone’s cross sections will
be circular disks of diameter
3)
x
4
4x
x
2
(see accompanying figure).
The solid and the cone have equal altitudes and
identical parallel cross sections. From
Cavalier’s Principle we conclude that the solid
and the cone have the same volume.
15. Slices made parallel to the flat base of the solid at y are
squares of area
2
A( y ) (2 x) 2 2 1 y 2 4(1 y 2 )
1
1
V A( y ) dy 4(1 y 2 ) dy 4 y
0
0
y3
3
1
8
3
0
Copyright 2018 Pearson Education, Inc.
366
Chapter 6 Applications of Definite Integrals
16. Slices made parallel to the flat surface at y are
rectangles of area
A( y ) (2 x)(10) 20 16 y 2
2
2
4
4
V A( y ) dy 20 16 y 2 dy
2
0
2
17. R( x) y 1 2x V R( x) dx 1 2x
2
4
2
8
12
3y
2
2
0
2
2
dx 1 x
0
x2
4
dx x
x2
2
2
x
12
0
3
2
3
2
2 3y 2
dy
0 2
V R( y ) dy
2
29
0 4
2
y 2 dy 43 y 3 43 8 6
0
18.
R( y ) x
19.
R ( y ) tan 4 y ; u 4 y du 4 dy 4 du dy; y 0 u 0, y 1 u 4 ;
0
2
1
1
/4
/4
2
/4
1 sec2 u du 4 u tan u 0
V R ( y ) dy tan 4 y dy 4
tan 2 u du 4
0
0
0
0
4 4 1 0 4
20.
R ( x) sin x cos x; R( x) 0 a 0 and b 2 are the limits of integration;
V
/2
0
R( x) dx
2
/2
0
(sin x cos x)2 dx
0
x 0 u 0, x 2 u V
2
dx; u 2 x du 2 dx du
dx
;
8
4
8
2
4
0
R ( x) x 2 V R( x) dx
2
0
x2
0
22.
4
2
2
2
5
2
dx x 4 dx x5
0
0
2
32
5
R ( x) x3 V R( x) dx
2
0
x3
0
2
2
2
2
1
sin 2 u du u 1 sin 2u 0 0
0 8
21.
/2 (sin 2 x )2
2
7
2
dx x6 dx x7 128
7
0
0
Copyright 2018 Pearson Education, Inc.
8
16
Section 6.1 Volumes Using Cross-Sections
23.
3
R ( x) 9 x 2 V R( x) dx
2
3
2
3 9 x dx 9 x x3 3
3
3
3
2 π 18 36π
2 9(3) 27
3
24.
1
R ( x) x x 2 V R ( x) dx
2
0
0
1
2
1
x x 2 dx x 2 2 x3 x 4 dx
0
1
3
4
5
x3 24x x5 13 12 15
0
(10 15 6)
30
30
25.
R ( x ) cos x V
/2
0
26.
R ( x) dx
2
/2
cos x dx sin x 0
R ( x ) sec x V
/4
/4
(1 0)
R ( x) dx
2
/4
/4
/4
27.
/2
0
sec 2 x dx tan x /4 [1 (1)] 2
R ( x) 2 sec x tan x V
/4
0
/4
/4
0
0
R ( x) dx
2
2
2 sec x tan x dx
2 2
2 sec x tan x sec 2 x tan 2 x dx
/4
/4
2 dx 2 2 sec x tan x dx
0
0
/4
0
tan x 2 sec2 x dx
/4
3
/4
/4
2 x 0 2 2 sec x 0 tan3 x
0
2 0 2 2 2 1 13 (13 0)
2 2 2 11
3
Copyright 2018 Pearson Education, Inc.
367
368
28.
Chapter 6 Applications of Definite Integrals
R ( x ) 2 2sin x 2(1 sin x) V
/2
0
4(1 sin x) 2 dx 4
/2
0
4
/2
4
/2 3 cos 2 x
2sin x
2
0
2
4 32 x sin42 x 2 cos x
4
R( x) dx
2
1 sin 2 x 2sin x dx
1 1 (1 cos 2 x) 2sin x dx
2
0
29.
/2
0
3
4
/2
0
0 0 (0 0 2) (3 8)
1
R ( y ) 5 y 2 V R ( y ) dy
2
1
1
15 y
4
dy
1
y 5 [1 (1)] 2
1
30.
2
2
2
R ( y ) y 3/2 V R( y ) dy y 3 dy
0
0
2
y4
4 4
0
31.
R ( y ) 2sin 2 y V
/2
0
/2
0
R ( y ) dy
2
/2
2sin 2 y dy cos 2 y 0
[1 (1)] 2
32.
0
cos
2
33.
0
y
R ( y ) cos 4 V R ( y ) dy
2
dy 4 sin
y
4
2
y 0
4
4[0 ( 1)] 4
2
3
3
0
0
R( y ) 2 y 1 V R 2 dy 4 y 1 dy
0
3
4 12 y 2 y 4 92 3 30
Copyright 2018 Pearson Education, Inc.
Section 6.1 Volumes Using Cross-Sections
34.
1
2y
R( y )
2
y 1
1
V R( y ) dy
2 y y2 1
0
2
0
2
dy; [u y 2 1 du 2 y dy;
y 0 u 1, y 1 u 2]
2
2
1
1
V u 2 du u1 12 (1) 2
b
35. For the sketch given, a 2 , b 2 ; R( x) 1, r ( x) cos x; V
a
/2
/2
(1 cos x) dx 2
/2
0
/2
(1 cos x) dx 2 x sin x 0
d
/4
0
2
1
R( x)
2
1
r ( x)
2
R( y )
2
dx
3
1
1 x 2 dx x x3 1 13 0 23
0
0
1
38. r ( x) 2 x and R ( x) 2 V
0
R ( x)
1
2
r ( x)
2
dx
2
1
(4 4 x) dx 4 x x2 4 1 12 2
0
0
39. r ( x) x 2 1 and R( x) x 3
2
V
1
R( x)
2
r ( x)
2
dx
2
2
( x 3)2 x 2 1 dx
1
2
x 2 6 x 9 x 4 2 x 2 1 dx
1
x 4 x 2 6 x 8 dx x5 x3 62x
1
2
5
3
15 13 62 8
33
3 28 3 8 530533 1175
5
2
dx
r ( y)
2
2
0
2
dy
1 tan 2 y dy 0 /4 2 sec2 y dy 2 y tan y 0 /4 2 1 2
37. r ( x) x and R ( x) 1 V
r ( x)
2 2 1 2 2
36. For the sketch given, c 0, d 4 ; R( y ) 1, r ( y ) tan y; V
c
R ( x)
2
8x
1
32
8 24 16
5 3 2
Copyright 2018 Pearson Education, Inc.
369
370
Chapter 6 Applications of Definite Integrals
40. r ( x) 2 x and R ( x) 4 x 2
2
V
1
R( x)
2
r ( x)
2
dx
2
16 8 x 2 x 4 4 4 x x 2 dx
1
2
2
12 4 x 9 x 2 x 4 dx 12 x 2 x 2 3x3 x5
1
1
2
2
4 x 2 (2 x)2 dx
1
5
24 8 24 32
12 2 3 15 15 33
108
5
5
5
41. r ( x) sec x and R ( x ) 2
V
/4
/4
/4
/4
R ( x) r ( x)
2
2
dx
2 sec2 x dx 2 x tan x/4/4
2 1 2 1 ( 2)
42.
1
R ( x) sec x and r ( x) tan x V
0
1
1
R ( x)
2
r ( x)
2
dx
sec2 x tan 2 x dx 1 dx x 0
0
1
0
1
43. r ( y ) 1 and R ( y ) 1 y V
0
R( y )
2
2
dy
2
dy
r ( y)
1
1
(1 y ) 2 1 dy 1 2 y y 2 1 dy
0
0
1
1
y3
2 y y 2 dy y 2 3 1 13 43
0
0
44.
1
R ( y ) 1 and r ( y ) 1 y V
0
R ( y )
2
r ( y)
1
1
1 (1 y )2 dy 1 1 2 y y 2 dy
0
0
1
1
y3
2 y y 2 dy y 2 3 1 13 23
0
0
Copyright 2018 Pearson Education, Inc.
Section 6.1 Volumes Using Cross-Sections
45.
4
y V
R ( y ) 2 and r ( y )
0
R ( y )
2
dy
r ( y)
2
4
4
y2
(4 y ) dy 4 y 2 (16 8) 8
0
0
46.
R ( y ) 3 and r ( y ) 3 y 2
3
V
0
R ( y )
2
r ( y)
2
dy
3
3
3 3 y 2 dy 3 y 2 dy y 3
0
0
3 0
47.
3
1
R ( y ) 2 and r ( y ) 1 y V
0
R ( y )
dy
r ( y)
2
2
2
1
4 1 y dy 1 2 y y dy
0
1
3 43 12 18683 76
1
y2
3 2 y y dy 3 y 43 y 3/2 2
0
0
48.
1
R ( y ) 2 y1/3 and r ( y ) 1 V
0
R( y )
2
r ( y)
2
dy
2
1
1
2 y1/3 1 dy 4 4 y1/3 y 2/3 1 dy
0
0
1
1
3 y 5/3
3 4 y1/3 y 2/3 dy 3 y 3 y 4/3 5
0
0
3 3 53 35
49. (a) r ( x) x and R ( x) 2
4
V
0
R ( x)
2
r ( x)
2
dx
4
2
4
(4 x) dx 4 x x2 (16 8) 8
0
0
R( y) r( y) dy y dy
x V R( x) r ( x) dx 2 x dx
2
(b) r ( y ) 0 and R ( y ) y 2 V
2
2
0
(c) r ( x) 0 and R ( x) 2
4
0
4 4
4
2
2
0
x x dx 4 x 8 x3
3/ 2
2 4
0
y5
5
4
2
0
4
x2
2 0
16 64
16
83
3
2
Copyright 2018 Pearson Education, Inc.
2
0
32
5
371
372
Chapter 6 Applications of Definite Integrals
2
(d) r ( y ) 4 y 2 and R ( y ) 4 V
0
R( y )
r ( y)
2
2
dy
2
16 4 y 2
0
2
dy
2
2
2
y5
16 16 8 y 2 y 4 dy 8 y 2 y 4 dy 83 y 3 5 64
32
224
3
5
15
0
0
0
y
50. (a) r ( y ) 0 and R ( y ) 1 2
R( y) r ( y) dy
1 dy 1 y dy
2
2
V
2
0
y 2
2
2
0
y2
4
2
0
2
y2
y3
8 2
y 2 12 2 24 12
3
0
y
(b) r ( y ) 1 and R ( y ) 2 2
2
V
0
R ( y )
2
r ( y)
2
dy 2
2
y2
1 dy 4 2 y 4 1 dy
0
y 2
2
2
0
2
2
y2
y3
8 2 2 8
3 2 y 4 dy 3 y y 2 12 6 4 12
3
3
0
0
R( x)
1
2
51. (a) r ( x) 0 and R ( x) 1 x 2 V
1
r ( x)
2
dx
1
1 x 2 dx 1 2 x 2 x 4 dx
1
1
2
1
1
3
5
10 3 16
x 23x x5 2 1 23 15 2 1515
15
1
1
(b) r ( x) 1 and R ( x) 2 x 2 V
1
R ( x)
2
r ( x)
2
dx 2 x
2 2
1
1
1 dx
1
4 4 x 2 x 4 1 dx 3 4 x 2 x 4 dx 3x 43 x3 x5 2 3 43 15
1
1
1
1
5
215 (45 20 3) 56
15
1
(c) r ( x) 1 x 2 and R( x) 2 V
1
R ( x)
2
r ( x)
2
dx
1
2
4 1 x 2 dx
1
1
1
4 1 2 x 2 x 4 dx 3 2 x 2 x 4 dx 3x 23 x3 x5 2 3 23 15
1
1
1
1
5
215 (45 10 3) 64
15
52. (a) r ( x) 0 and R( x) bh x h
b
V
0
b
0
R ( x)
bh x h
2
2
r ( x)
2
dx
b h2
0 b2
dx
b
2
x 2 2bh x h 2 dx
3
2
2
h 2 x 2 xb x h 2 b3 b b h3 b
0
3b
Copyright 2018 Pearson Education, Inc.
1
Section 6.1 Volumes Using Cross-Sections
R( y )
h
y
2
(b) r ( y ) 0 and R ( y ) b 1 h V
0
r ( y)
2
h
dy b 1
y 2
h
2 h
0
373
dy
2
h
2 y y2
y2
y3
b 2 1 h 2 dy b 2 y h 2 b 2 h h h3 b3 h
0
3
h
h
0
53.
R ( y ) b a 2 y 2 and r ( y ) b a 2 y 2
a
V
a
R ( y )
2
r ( y)
2
2
b a y
a
a
a
a
2
dy
2
2
b a 2 y 2 dy
4b a 2 y 2 dy 4b
a
a 2 y 2 dy
a
2
4b area of semicircle of radius a 4b 2a 2a 2 b 2
54. (a) A cross section has radius r 2 y and area r 2 2 y. The volume is
(b) V (h) A(h)dh, so
dV
dh
A(h). Therefore
For h 4, the area is 2 (4) 8 , so
55. (a)
dh
dt
dV
dt
dV dh
dh dt
A(h) dh
, so
dt
3
dh
dt
5
0 2 ydy y
2 5
0
25 .
A(1h ) dV
.
dt
3
81 3 units
83 units
.
sec
sec
(ha)
y
a 2 y 2 dy a 2 y 3
a 2 h a3 3 a3 a3
a
a
h (3a h )
a 2 h 13 h3 3h 2 a 3ha 2 a3 a3 a 2 h h3 h 2 a ha 2
3
R( y ) a 2 y 2 V
ha
3
3
(b) Given
dV
dh
dV
dt
ha
3
2
3
. From part (a), V (h)
0.2 m3 /sec and a 5 m, find dh
dt h 4
10 h h 2
dV
dt
dV dh
dh dt
h(10 h)
dh
dt
3
dh
dt h 4
h 2 (15 h )
3
0.2
4 (10
4)
5 h 2 3h
1
(20 )(6)
3
1201 m/sec.
56. Suppose the solid is produced by revolving
y 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
b
a
n
d kˆ 2
2
R( y ) dy
2
k 1
y.
57. 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 A1 R 2 h 2 ( R 2 h 2 ). The cross section of the
2
R 2 h 2 . Therefore its area is A2 R 2 h 2 R 2 h 2 .
We can see that A1 A2 . The altitudes of both solids are R. Applying Cavalieri’s Principle we find
hemisphere is a disk of radius
Volume of Hemisphere (Volume of Cylinder) (Volume of Cone) R 2 R 13 R 2 R 23 R3 .
Copyright 2018 Pearson Education, Inc.
374
58.
Chapter 6 Applications of Definite Integrals
0
6
60536 365 cm3.
0
12 x3 x5 12 63 65 63 12 36 196
144
5 0 144
5
144
5
144
The plumb bob will weigh about W (8.5)
59.
x 36 x 2 V 6 R( x) 2 dx 6 x 2 36 x 2 dx 6 36 x 2 x 4 dx
R ( x) 12
144
144
R ( y ) 256 y 2 V
7
16
365 192 gm, to the nearest gram.
2
7
y
256 y 2 dy 256 y 3
16
16
R( y ) dy
0
7
3
3
3
3
3
(256)(7) 73 (256)(16) 163 73 256(16 7) 163 1053 cm3 3308 cm3
60. (a)
R( x)2 dx 0 (c sin x)2 dx 0 c2 2c sin x sin 2 x dx
0
R ( x) | c sin x |, so V
c2 2c sin x 1cos2 2 x dx 0 c2 12 2c sin x cos22 x dx c2 12 x 2c cos x sin42 x 0
c 2 2 2c 0 (0 2c 0) c 2 2 4c . Let V (c) c 2 2 4c . We find the
0
and V (1) 32 4 2 (4 ) .
(2c 4) 0 c 2 is a critical point, and V 2 4 2 8
extreme values of V (c) : dV
dc
2
2
2 4 2 4; Evaluate V at the endpoints: V (0) 2
2
2
Now we see that the function’s absolute minimum value is 2 4, taken on at the critical point c 2 .
(See also the accompanying graph.)
2
(b) From the discussion in part (a) we conclude that the function’s absolute maximum value is 2 , taken on
at the endpoint c 0.
(c) The graph of the solid’s volume as a function
of c for 0 c 1 is given at the right. As c
moves away from [0, 1] 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 [0, 1].
61. Volume of the solid generated by rotating the region bounded by the x-axis and y f ( x) from x a to
b
x b about the x-axis is V [ f ( x)]2 dx 4 , and the volume of the solid generated by rotating the same
a
b
region about the line y 1 is V [ f ( x) 1]2 dx 8 . Thus
b
a
f ( x )
2
a
2 f ( x) 1 f ( x)
2
b
b
b
a f ( x) 1 dx a f ( x)
2
dx 4 (2 f ( x) 1) dx 4 2
b
b
a
a
2
dx 8 4
b
f ( x) dx dx 4
a
b
f ( x) dx 12 (b a ) 2 f ( x) dx 4b2 a
a
a
62. Volume of the solid generated by rotating the region bounded by the x-axis and y f ( x) from x a to x b
b
about the x-axis is V f ( x) dx 6 , and the volume of the solid generated by rotating the same
2
a
Copyright 2018 Pearson Education, Inc.
Section 6.2 Volumes Using Cylindrical Shells
b
region about the line y 2 is V f ( x) 2 dx 10 . Thus
2
a
b
a f ( x) 2
2
b
dx f ( x) dx 10 6
2
a
b
a
f ( x )
2
4 f ( x) 4 f ( x)
2
b
b
b
b
b
a
a
a
a
a
dx 4
(4 f ( x) 4) dx 4 4 f ( x) dx 4 dx 4 f ( x) dx (b a) 1 f ( x) dx 1 b a
6.2
VOLUMES USING CYLINDRICAL SHELLS
1. For the sketch given, a 0, b 2;
b
shell
V 2 radius
a
shell
height
dx
2
2
3
2
2
x 4 2 4 16
2 x 1 x4 dx 2 x x4 dx 2 x2 16
2 16
0
0
0
2 3 6
2
2. For the sketch given, a 0, b 2;
b
shell
V 2 radius
a
shell
height
dx
2
0
2
2 x 2 x4 dx 2
2
0
2x dx 2 x
x3
4
2
2
x 2 4 1 6
16
0
4
3. For the sketch given, c 0, d 2;
d
shell
V 2 radius
c
shell
height
dy
2
0
2
y4
y dy 2 4 2
0
0
2 3
2 y y 2 dy 2
4. For the sketch given, c 0, d 3;
d
shell
V 2 radius
c
shell
height
3
3
y4
2 y 3 3 y 2 dy 2 y 3 dy 2 4 92
0
0
0
dy
3
5. For the sketch given, a 0, b 3;
b
V 2
a
shell
radius
shell
height
dx
3
0
2 x x 2 1 dx;
u x 2 1 du 2 x dx; x 0 u 1, x 3 u 4
4
4
V u1/2 du 23 u 3/2 23 43/2 1 23 (8 1) 143
1
1
6. For the sketch given, a 0, b 3;
b
shell
V 2 radius
a
shell
height
dx 2 x
3
9x
0
x3 9
dx;
[u x3 9 du 3 x 2 dx 3 du 9 x 2 dx; x 0 u 9, x 3 u 36]
36
36
3u 1/2 du 6 2u1/2 12
9
9
V 2
36 9 36
Copyright 2018 Pearson Education, Inc.
375
376
Chapter 6 Applications of Definite Integrals
7. a 0, b 2;
b
shell
V 2 radius
a
shell
height
2
dx
2
0
2 x x 2x dx
2
2
2 x 2 32 dx 3x 2 dx x3 8
0
0
0
8. a 0, b 1;
b
shell
V 2 radius
a
shell
height
dx 2 x 2 x dx
1
x
2
0
dx 3x dx x
1
2 32x
0
1
2
0
9. a 0, b 1;
b
shell
V 2 radius
a
3 1
0
2
shell
height
dx 2 x (2 x) x dx
1
2
0
1
3
4
1
2 2 x x 2 x3 dx 2 x 2 x3 x4
0
0
5
2 1 13 14 2 121243 10
12
6
10. a 0, b 1;
dx 2 x 2 x x dx
2 x 2 2 x dx 4 x x dx
b
V 2
a
shell
radius
1
1
shell
height
1
2
0
2
2
0
3
0
1
2
4
4 x2 x4 4 12 14
0
11. a 0, b 1;
b
shell
V 2 radius
a
shell
height
dx 2 x
1
0
x (2 x 1) dx
1
1
2 x3/2 2 x 2 x dx 2 52 x5/2 23 x3 12 x 2
0
0
20 15 7
2 52 23 12 2 1230
15
12. a 1, b 4;
b
shell
V 2 radius
a
shell
height
dx
4
1
4
2 x 32 x 1/2 dx
4
3 x1/2 dx 3 23 x3/2 2 43/2 1
1
1
2 (8 1) 14
Copyright 2018 Pearson Education, Inc.
Section 6.2 Volumes Using Cylindrical Shells
13. (a)
sin x
sin x, 0 x
x x , 0 x
x f ( x)
; since sin 0 0 we have
x f ( x)
x0
x0
0,
x,
sin x, 0 x
x f ( x)
x f ( x) sin x, 0 x
x0
sin x,
b
(b) V 2
a
shell
radius
shell
height
dx 2 x f ( x) dx and x f ( x) sin x, 0 x by part (a)
0
V 2 sin x dx 2 cos x 0 2 ( cos cos 0) 4
0
14. (a)
2
x tanx x , 0 x 4
x g ( x)
x0
x 0,
tan 2 x, 0 x /4
x g ( x)
; since tan 0 0 we have
x0
0,
tan 2 x, 0 x /4
x g ( x)
x g ( x) tan 2 x, 0 x /4
2
x0
tan x,
b
(b) V 2
a
shell
radius
V 2
/4
0
dx
shell
height
tan 2 x dx 2
/4
0
/4
0
2 x g ( x) dx and x g ( x) tan 2 x, 0 x /4 by part (a)
sec2 x 1 dx 2 tan x x0 /4 2 1 4 4 2
15. c 0, d 2;
d
dy
y y dy 2
V 2
c
2
2
0
2 52
shell
radius
3/2
2
5
d
V 2
c
23 2
8 2
5
shell
height
56 403
2
0
y4
4
2
2 y y ( y ) dy
y3
3
2
0
83 16
dy
dy 2
shell
radius
y3 y 2
0
16
2 y 5/ 2
5
3
16. c 0, d 2;
2
0
2
3 2 5
16
15
2
shell
height
2
5
13
2 y y 2 ( y ) dy
y3
3
2
16
0
24 13
17. c 0, d 2;
d
dy 2 y 2 y y dy
2 y y dy 2 2
V 2
c
2
2
0
32
shell
radius
2
2
shell
height
3
13 14 3212 83
2
0
2 y3
3
y4
4
2
0
16
3
16
4
Copyright 2018 Pearson Education, Inc.
2
377
378
Chapter 6 Applications of Definite Integrals
18. c 0, d 1;
d
V 2
c
1
shell
radius
2 y y y 2
0
1
y4
4
0
y3
2 3
d
c
shell
radius
2 2 y 2 dy
0
20. c 0, d 2;
d
c
shell
radius
2
0
1
2
y dy
3
0
13 14 6
shell
height
dy 2 y y ( y) dy
1
0
1
4 y 3
3 0
1
V 2
1
shell
height
2
19. c 0, d 1;
V 2
dy 2 y 2 y y
dy 2 y y dy
shell
height
43
dy
2
0
y
2 y y 2 dy
2
2 y2
0 2
dy 3 y 3 83
0
2
21. c 0, d 2;
d
V 2
c
shell
radius
shell
height
dy
2
0
2 y (2 y ) y 2 dy
2
y
y
2 y y 2 y 3 dy 2 y 2 3 4
0
0
2
3
2
4
2 4 83 16
6 (48 32 48) 163
4
22. c 0, d 1;
d
V 2
c
shell
radius
shell
height
dy 2 y (2 y) y dy
1
2
0
1
2 2 y y 2 y 3 dy 2 y 2
0
y3
3
1
y4
4
0
2 1 13 14 6 (12 4 3) 56
b
2
23. (a) V 2
a
(b) V
b
a
6
8 83
shell
radius
shell
radius
32
shell
height
shell
height
dx
dx
2
0
2
0
2
2
2 x (3x)dx 6 x 2 dx 2 x3 16
0
0
4 x x 2 dx 6 2 x 2 13 x3
0
0
2 (4 x) (3x)dx 6
2
Copyright 2018 Pearson Education, Inc.
2
Section 6.2 Volumes Using Cylindrical Shells
b
(c) V 2
a
6
shell
radius
shell
radius
shell
height
dx
shell
height
dy
83 2 28
d
(d) V 2
c
d
c
shell
radius
d
(f ) V 2
c
shell
radius
x2 x dx 6 13 x3 12 x2 0
2
6
0
6
2 y 13 y 2 dy 2 y 2 19 y3 0
6
6
2 (7 y ) 2 13 y dy 2 14 13
y 13 y 2 dy
3
0
2 (84 78 24) 60
dy
shell
height
2 y 2 13 y dy 2
0
6
1 y3
9
0
2
0
6
dy
shell
height
y2
2 14 y 13
6
2 ( x 1) (3x)dx 6
0
2 (36 24) 24
(e) V 2
2
0
379
6
2 ( y 2) 2 13 y dy 2
0
6
0
4 43 y 13 y 2 dy
6
2 4 y 23 y 2 19 y3 2 (24 24 24) 48
0
b
24. (a) V 2
a
shell
radius
2 16 32
5
b
(b) V 2
a
shell
radius
shell
height
96
5
shell
height
dx
2
0
dx
2
0
2
2 (3 x) 8 x3 dx 2
2
b
a
shell
radius
shell
height
dx
2
0
d
2
(e) V
d
c
2
d
c
2π
shell
radius
96 384
7
(f ) V 2
384
7
b
25. (a) V 2
a
shell
radius
2
0
12
shell
radius
8
8 4/3
1/3
0
0
8
shell
height
8
1/3
0
b
shell
radius
8
y 7/3 6 (128)
0 7
6
7
1/3
0
768
7
8
y 4/3 dy 2 6 y 4/3 73 y 7/3
0
576
7
dy 2 ( y 1) y
8
shell
height
1/3
0
y 4/3 y1/3 dy 2 73 y 7/3 43 y 4/3
0
0
dx 2
8
8
936π
7
shell
height
dx
2
1
2
a
336
5
2 (2 x) x 2 x 2 dx 2
2 4 x x3 14 x 4 2 (8 8 4) 2 4 1 14
1
(b) V 2
dy 2 y y dy 2 y dy
dy 2 (8 y) y dy 2 8 y
shell
height
shell
radius
24 8x 3x3 x4 dx
264
5
2
c
2
2 ( x 2) 8 x3 dx 2 16 8 x 2 x3 x 4 dx
2 16 x 4 x 2 12 x 4 15 x5 2 32 16 8 32
5
0
(d) V 2
2
0
2 24 x 4 x 2 34 x 4 15 x5 2 48 16 12 32
5
0
(c) V 2
2 x 8 x3 dx 2 8 x x 4 dx 2 4 x 2 15 x5
0
0
shell
height
2
dx
2
1
4 3x2 x3 dx
27
2
2 ( x 1) x 2 x 2 dx 2
2
1
2
1
2 2 x 32 x 2 14 x 4 2 (4 6 4) 2 2 23 14
1
2 3x x3 dx
27
2
Copyright 2018 Pearson Education, Inc.
380
Chapter 6 Applications of Definite Integrals
d
(c) V 2
c
shell
radius
1
4
0
1
4 y3/2 dy 2
85 (1) 2
d
(d) V 2
c
1
3/2
1
4
0
1
2
5/2 1
0
8
5
645 643 16 2 52 13 1 725
shell
radius
1
4
0
1
4
2
y ( y 2) dy
2 5/2
5
dy 2 (4 y) y y dy
dy 2 y y 6 y 4 y 8 dy
shell
height
4 4 y y3/2
0
dy 2 y y y dy 2 y
y y 2 y dy y 2 y
shell
height
4
13 y3 y 2
1
2 (4 y )
y ( y 2) dy
3/2
1
1
4
4 83 y3/2 52 y 5/2 2 13 y 3 52 y5/2 3 y 2 83 y 3/2 8 y
0
1
4
83 52 2 643 645 48 643 32 2 13 52 3 83 8 1085 .
b
26. (a) V 2
a
shell
radius
dx
shell
height
1
1
2 (1 x) 4 3x 2 x 4 dx 2
1
2 16 x6 15 x5 34 x 4 x3 2 x 2 4 x 2
1
d
(b) V 2
c
shell
radius
1
4 y5/4 dy 4
0
dy 2 y
1
shell
height
4
3 1
0
4
1
1
x5 x4 3x3 3x2 4 x 4 dx
16 15 34 1 2 4 2 16 15 34 1 2 4 565
4
y 4 y dy 2 y
1
4 y
3
4 y
3
dy
y 4 ydy [u 4 y y 4 u du du; y 1 u 3, y 4 u 0]
1
3
0
3
169 y 9/4 4 (4 u ) u du 169 (1) 4 4 u u 3/2 du 169 4 83 u 3/2 52 u 5/2
0
0
3 3
3 0
3
169 4 8 3 18
3 169 885 872
5
45
3
d
27. (a) V 2
c
shell
radius
shell
height
14 15 2420 65
24
d
(b) V 2
c
shell
radius
dy 2 (1 y) 12 y
1
shell
height
2
0
1
d
c
shell
radius
1 8 2
y
0 5
24
24
60
13
y3
5
(32 39 12)
d
(d) V 2
c
1
shell
radius
dy 2 y 12 y
y dy 24 y y
24
0
4
24
12
shell
height
24 y3 y 4 52 y 2
0
1
shell
height
8
5
8 3
15
13
20
1
y3 dy 24 (1 y ) y 2 y3 dy
0
1
y3 y 4 y5
24 y 2 2 y 3 y 4 dy 24 3 2 5 24
0
0
(c) V 2
1
1
1
y 4 y5
dy 2 y 12 y 2 y3 dy 24 y3 y 4 dy 24 4 5
0
0
0
13 12 15 24 301 45
2
1
y3 dy 24
0
4
1
y5
5
0
24
85 y y 2 y3 dy
158 1320 15
2
dy 2 y 12 y y dy 24 y y
y dy 24 y y y dy 24 y
1
0
2 3
5
2
2
5
1
3
0
1 2
0 5
2
3 3
5
4
152 203 15 2460 (8 9 12) 2412 2
Copyright 2018 Pearson Education, Inc.
2
5
2 3
15
2
y 3 dy
3
20
y4
1
y5
5
0
Section 6.2 Volumes Using Cylindrical Shells
d
28. (a) V 2
c
shell
radius
2
y4 y6
2 4 24 2
0
d
(b) V 2
c
shell
radius
dy
2
2
y2 y4 y2
y4
y5
2 y 2 4 2 dy 2 y y 2 4 dy 2 y 3 4 dy
0
0
0
shell
height
24
4
6
2
2 32
24
dy
14 244 32 14 16 32 242 83
y2 y4
2 (2 y ) 2 4
0
shell
height
2
y2
2
2
2 y4
dy 0 2 (2 y ) y 4 dy
2
2
y4
y5
2 y3 y5 y 4 y 6
2 2 y 2 2 y 3 4 dy 2 3 10 4 24 2
0
0
d
(c) V 2
c
shell
radius
dy
shell
height
2
0
163 1032 164 6424 85
2
y2 y4 y2
y4
2 (5 y ) 2 4 2 dy 2 (5 y ) y 2 4 dy
0
2
2
y5
5 y3 5 y5 y 4 y 6
2 5 y 2 54 y 4 y 3 4 dy 2 3 20 4 24 2
0
0
d
(d) V 2
c
2 3
y
0
2
shell
radius
dy 2 y
2
shell
height
0
d
0
shell
radius
1
shell
height
y y dy 2 y 3/2 y 2
0
1
2 52 y 5/2 13 y 3 2
0
b
About y -axis: V 2
1
a
2
y4
y4 y2
4 2 dy 2 y 85 y 2 4 dy
0
2
c
1
y2
2
4
6
3
5
5 y 4 dy 2 y y 5 y 5 y 2
85 y 2 32
4
24
24
160
0
y5
4
29. (a) About x-axis: V 2
2 y
5
8
16
64
8
403 160
20
4
24
160 4
164 6424 2440 160
dy
dy
52 13 215
shell
radius
1
shell
height
dx
2 x x x 2 dx 2 x 2 x3 dx
0
0
1
3
4
2 x3 x4 2
0
13 14 6
b
(b) About x-axis: R ( x) x and r ( x) x 2 V
a
1
3
5
x3 x5
0
1
b
30. (a) V
a
R( x)
2
2
r ( x)
2
d
y and r ( y ) y V
c
R( y )
2
1
0
r ( y)
12 13 6
r ( x)
2
dx 2
4
0
x
2
2
x 2 dx
3 x 2 2 x 4 dx x4 x 2 4 x
0 4
0
dx x
2
x 4 dx
13 15 215
About y -axis: R( y )
y 2 y3
2 3
0
R ( x )
4
3
4
16 16 16 16
Copyright 2018 Pearson Education, Inc.
2
dy y y dy
1
0
2
381
382
Chapter 6 Applications of Definite Integrals
b
(b) V 2
a
shell
radius
dx
shell
height
4
4
0
3
2 x 2 x6 2 16 64
6
0
b
(c) V 2
a
shell
radius
dx
shell
height
4
0
4
b
a
R ( x)
2
4 3 2
x
0 4
d
31. (a) V 2
c
2
2
1
2
0
2
4
4
2 (4 x) 2x 2 x dx 2 (4 x) 2 2x dx 2 8 4 x x2 dx
r ( x)
0
32
3
3
2 8 x 2 x 2 x6 2 32 32 64
6
0
(d) V
2
4
4
2 x 2x 2 x dx 2 x 2 2x dx 2 2 x x2 dx
dx
2
x
(8 x) 6 2
0
64
3
4
0
0
2
4
2
dx 0 64 16 x x 36 6 x
x2
4
4
3
10 x 28 dx x4 5 x 2 28 x [16 (5) (16) (7) (16)] (3) (16) 48
0
shell
radius
shell
height
dy
2
2 y ( y 1) dy
1
2
y3 y 2
y 2 y dy 2 3 2
1
83 42 13 12
2 73 2 12 3 (14 12 3) 53
2
b
(b) V 2
a
shell
radius
dx
shell
height
2
1
2 x(2 x) dx 2
2
1
2x x2 dx 2 x2 x3 1
2
3
1238 331 2 34 32 43
b
shell
shell
V 2 radius
height
dx 12 2 103 x (2 x) dx 2 12 203 163 x x2 dx
a
2
x 83 x 2 13 x3 2 40
2 20
32 8 20 8 1 2 33 2
3
1
3 3 3 3 3 3
2 4 83 1 13 2
(c)
d
(d) V 2
c
d
32. (a) V 2
c
shell
radius
shell
height
dy
shell
radius
shell
height
dy
1
2
b
a
shell
radius
4
shell
height
dx
2 x 2 x dx 2
0
4
2
4
0
8
24
4
2 x x3/2 dx
5
2 x 2 52 x5/2 2 16 252
0
2 16 64
25 (80 64)
5
b
(c) V 2
a
shell
radius
shell
height
2 8 x 83 x3/2 x 2
2 y y 2 0 dy
0
2
y4
2 y3 dy 2 4 2
0
0
(b) V 2
2
2
( y 1)3
2 ( y 1)( y 1) dy 2 ( y 1)2 2 3 23
1
1
2
32
5
dx
4
4
2 (4 x) 2 x dx 2 8 4 x1/2 2 x x3/2 dx
0
4
2 x5/2
5
0
0
2 32 64
16 64
215 (240 320 192) 215 (112)
3
5
Copyright 2018 Pearson Education, Inc.
224
15
dx
Section 6.2 Volumes Using Cylindrical Shells
d
(d) V 2
c
2
shell
radius
shell
height
dy
0
1
shell
radius
2
4
4
2
dy 2 y y y dy
2 y y dy 2 2
V 2
dy 2 (1 y) y y dy
2 y y y y dy 2
d
c
2
y
2 y 2 y 3 dy 2 23 y3 4
0
0
2 (2 y ) y 2 dy 2
163 164 3212 (4 3) 83
33. (a) V 2
1
shell
height
d
y3
3
shell
radius
c
1
y5
5
3
2
13 14
d
34. (a) V 2
c
15
shell
radius
shell
height
4
15
1
5
3
y2
2
4
2
60
1
3
0
0
0
1
2
1
1
shell
height
2
3
0
0
(b)
2
y3
3
y4
4
(30 20 15 12)
y5
5
1
0
7
30
dy 2 y 1 y y dy
1
3
0
1
1
y 2 y3 y5
2 y y 2 y 4 dy 2 2 3 5
0
0
2
12 13 15 230 (15 10 6) 1115
(b) Use the washer method:
d
V
c
R( y )
2
r ( y)
2
dy 1 y y dy 1 y
1
3 2
2
1
0
1
0
2
y 6 2 y 4 dy
y3 y7 2 y5
(105 35 15 42) 97
y 3 7 5 1 13 17 52 105
105
0
(c) Use the washer method:
2
d
1
1
2
2
V R( y ) r ( y ) dy 1 y y 3 0 dy 1 2 y y 3 y y 3
0
0
c
dy
2
1
1
y3
y7
y 4 2 y5
1 y 2 y 6 2 y 2 y 3 2 y 4 dy y 3 7 y 2 2 5
0
0
(70 30 105 2 42) 121
1 13 17 1 12 52 210
210
d
(d) V 2
c
0
1
shell
radius
2 1 y y 3
dy 2 (1 y) 1 y y dy 2 (1 y) 1 y y dy
y y y dy 2 1 2 y y y y dy 2 y y
1
shell
height
1
3
0
2
4
1
2
3
4
0
3
0
2 1 1 13 14 15 260 (20 15 12)
23
30
Copyright 2018 Pearson Education, Inc.
2
y3
3
y4
4
1
y5
5
0
383
384
Chapter 6 Applications of Definite Integrals
d
35. (a) V 2
c
2
0
4
2
2 4
2
dy 2 y
225
5
b
36. (a) V 2
a
1
shell
radius
4
shell
height
4 2
32
423
5
444
b
2 x
2
x x8 dx 2
7
9
a
4
3/2
1
48
5
2
2
3
shell
radius
13 14 6
shell
height
dx 2 1 x 2 x x x dx 2 1 x x x dx
1
2
0
R( x)
2
1
dx x
1
r ( x)
2
1/2
1/16
1
1
12 23 14 212 (6 8 3) 6
1 dx
1
2 x1/2 x
(2 1) 2 14 16
1/16
7 9
1 16
16
d
(b) V 2
c
2
2
1
shell
radius
shell
height
dy 2 y
1
0
1
y4
1 dy
16
2
y
y2
y 3 16 dy 2 12 y 2 32
1
1 2
2 18 12 32
321
2 (8 1) 9
d
38. (a) V R ( y ) r ( y )
c
2
2
2
2
0
4
3
x4
x8 dx 2 52 x5/2 32
0
0
1
b
x
2 2 (32 20) 2 3 2 3
32
160
160
5
2
4
1
2 x 2 x 2 x3 dx 2 x2 23 x3 x4 2
0
0
37. (a) V
4
0
0
1
a
8
4
0
shell
height
3
4
2 x3 x4 2
0
(b) V 2
26
5
dx
dx 2 x 2 x x x dx
dx 2 x x dx
shell
radius
2 x x x 2
0
8 y y 2 dy
2
85 1 85 (8 5) 245
b
1
0
4
24 2
5
5
a
shell
height
2
(b) V 2
2
y4
2 y 3/2 y 3 dy 2 4 5 2 y 5/2 4
0
2
2
shell
radius
dy
2
1
1
y4
1 dy
16
y
1 1 1 1
13 y 3 16 24
8
3 16
1
( 2 6 16 3) 11
48
48
Copyright 2018 Pearson Education, Inc.
Section 6.2 Volumes Using Cylindrical Shells
a 2 radius height dx 1/4 2 ( x)
b
(b) V
2
shell
2
3
1
2
1
shell
21 1
3 8 32
4
3
1 16
1 dx 2
1
x
1
16
1
x
1/2
1/4
385
1
2
x dx 2 23 x 3/2 x2
1/4
(4 16 48 8 3) 11
48
48
39. (a) Disk: V V1 V2
b
b
V1 1 R1 ( x) dx and V2 2 R2 ( x) dx with R1 ( x)
2
a1
2
a2
x2
3
and R2 ( x) x ,
a1 2, b1 1; a2 0, b2 1 two integrals are required
(b) Washer: V V1 V2
b
V1 1
a1
R ( x )
R ( x)
b
V2 2
2
2
a2
r1 ( x)
2
1
2
r2 ( x)
2
dx with R ( x)
dx with R ( x)
d
c
shell
radius
x2
3
2
two integrals are required
(c) Shell: V 2
x2
3
1
shell
height
dy
d
c
2 y
shell
height
and r1 ( x) 0; a1 2 and b1 0;
and r2 ( x) x ; a2 0 and b2 1
dy where shell height y 3 y 2 2 2 y ;
2
2
2
c 0 and d 1. Only one integral is required. It is, therefore preferable to use the shell method.
However, whichever method you use, you will get V .
40. (a) Disk: V V1 V2 V3
d
Vi i Ri ( y ) dy, i 1, 2, 3 with R1 ( y ) 1 and c1 1, d1 1; R2 ( y )
2
ci
y and c2 0 and d 2 1;
R3 ( y ) ( y )1/4 and c3 1, d3 0 three integrals are required
(b) Washer: V V1 V2
d
Vi i
ci
R ( y )
2
i
ri ( y )
2
dy, i 1, 2 with R ( y) 1, r ( y)
1
y , c1 0 and d1 1;
1
R2 ( y ) 1, r2 ( y ) ( y )1/4 , c2 1 and d 2 0 two integrals are required
b
(c) Shell: V 2
a
shell
radius
shell
height
dx
b
a
2 x
shell
height
dx , where shell height x x x
2
4
2
x4 , a 0
and b 1 only one integral is required. It is, therefore preferable to use the shell method.
However, whichever method you use, you will get V 56 .
b
41. (a) V
a
R ( x)
2
r ( x)
4
16 x 1 x3
3
4
2
dx
4
b
a
shell
radius
shell
height
2
dx
1
500
3
1
4
4
25 x 2 (3)2 dx
64 64
64 64
3
3
(b) Volume of sphere 43 (5)3
42. V 2
4
25 x2 9 dx 44 16 x2 dx
256
3
256 244
Volume of portion removed 500
3
3
3
2 x sin x 2 1 dx; [u x 2 1 du 2 x dx;
x 1 u 0, x 1 u ] sin u du cos u 0 (1 1) 2
0
Copyright 2018 Pearson Education, Inc.
386
Chapter 6 Applications of Definite Integrals
b
43. V 2
a
shell
radius
2
2
shell
height
dx 2 x
r
0
r
r
x h dx 2 hr x 2 h x dx 2 3hr x3 h2 x 2
0
0
h
r
2 r 3h r 2h 13 r 2 h
d
44. V 2
c
shell
radius
shell
height
dy
r
0
2 y r 2 y 2 r 2 y 2
r
2
2
dy 4 0 y r y dy
[u r 2 y 2 du 2 y dy; y 0 u r 2 , y r u 0] 2
0
r2
2
43 r 3
b
b
a
a
b
45. Using the Shell Method we have 2 2 x f ( x)dx x f ( x)dx 1; 10 2 ( x 2) f ( x)dx
b
b
b
a
a
a
a
5 x f ( x)dx 2 f ( x)dx 5 1 2 f ( x)dx area of R is
b
a
f ( x)dx 2
b
b
b
a
a
a
46. Using the Shell Method we have 10 2 ( x 3) f ( x)dx 5 x f ( x)dx 3 f ( x)dx, but
b
b
a
6.3
1.
f ( x)dx 1 5 x f ( x)dx 3 volume about y-axis is
a
b
a x
f ( x)dx 2
ARC LENGTH
dy
dx
13 32 x 2 2
L
3
0
3
0
1/2
2x
x2 2 x
1 x 2 2 x 2 dx
3
1 2 x 2 x 4 dx
0
1 x2 dx 03 1 x2 dx x x3 0
2
3
3
3 27
12
3
2.
dy
dx
4
32 x L 1 94 x dx;
0
u 1 9 x du 9 dx 4 du dx;
4
4
9
x 0 u 1; x 4 u 10]
10
10
8 10 10 1
L u1/2 94 du 94 23 u 3/2 27
1
1
3.
dx
dy
2
dx
y 2 1 2 dy
y 4 12 1 4
4y
16 y
L
3
1
3
1
1 y 4 12
1
16 y 4
dy
3
1
y 4 12
1
16 y 4
2
r
r
u du 2 u1/2 du 43 u 3 2
0
0
dy
2
3 2
2
1
1
y 4 y 2 dy 1 y 4 y 2 dy
Copyright 2018 Pearson Education, Inc.
Section 6.3 Arc Length
4.
y3
3
9
3
y 1
4
1
( 1 4 3)
12
9
dx
dy
1 y1/2
2
L
9
1
( 2)
12
53
6
y2
y 2 dy y 2 dy
12 y 1/2
dx
dy
1 14
1
y
y
91
1 2
273 121 13 14 9 121 13 14
1
y
2
dy
2
1
4
1
y
9
1
1 9
2 1
1
4
1
y
y1/2 y 1/2 dy
9
9 y 3/ 2
12 32 y 3/2 2 y1/2 3 y1/2
1
1
5.
dx
dy
33
3
3
y3
L
2
1
2
1
16
4
13 1 11 13 323
1
4 y3
dx
dy
2
1 y 6 12
3
y
y 3
4
y 6 12
1
16 y 6
dy
2
4
1
4
1
8
2
1
2 3
dy 1 y
1
(16)(2)
1
16 y 6
1
32
y 3
4
y 6 12
1
16 y 6
y4
dy 4
dy
2
y 2
8
1
14 18
18 4 123
12832
32
6.
y 4 2 y 4
3
L 1 14 y 4 2 y 4 dy
2
3
14 y 4 2 y 4 dy
2
2
3
3
12 y 2 y 2 dy 12 y 2 y 2 dy
2
2
dx
dy
y2
2
1
2 y2
dx
dy
3
2
1
4
y3
12 3 y 1 12 27
1 83 12
3 3
2
12 26
83 12 12 6 12 13
3
4
Copyright 2018 Pearson Education, Inc.
387
388
7.
Chapter 6 Applications of Definite Integrals
dy 2
dx
x1/3 14 x 1/3
dy
dx
8
2/3
x 2/3 12 x16
2/3
1 x 2/3 12 x16 dx
1
L
8
1
x1/3 14 x1/3 dx
2
8
2/3
x 2/3 12 x16 dx
1
x1/3 14 x1/3 dx 34 x4/3 83 x2/3 1
8
83 2 x 4/3 x 2/3 83 2 24 22 (2 1)
1
8
8
1
83 (32 4 3) 99
8
8.
dy
dx
x2 2 x 1
(1 x)2 14
L
2
2
0
1
(1 x )2
dy 2
dx
1 (1 x)4 12
0
x 2 2 x 1 14
4
(4 x 4)2
(1 x) 4 12
(1 x ) 4
16
1
(1 x )2
(1 x) 4 12
(1 x ) 4
16
1
16(1 x )4
dx
dx
2
2
2
2
2 (1 x )
2 (1 x )
(1
x
)
dx; [u 1 x du dx; x 0 u 1, x 2 u 3]
(1
x
)
dx
0
4
4
0
2
x2 12
x
3
3
3
1 1 1 1081 4 3 106 53
L u 2 14 u 2 du u3 14 u 1 9 12
3 4
12
12
6
1
1
9.
dy
dx
4x
L
3
1
3
1
1
3
10.
dy
dx
4
1 x
x 4 12
x2
1
4 x2
x4
1
4 x4
L
1
1/2
1
1/2
1
2
dx
1
16 x 4
3
1
12
4x
x
2
x
dy 2
dx
8
1 x
1
2
4
18
16 x
2
x 4 12 1 4
16 x
1/2
1
4 x2
1
4 x4
2
4
y
x8 12
x3 1
3 4x
4
dx
2
0
1 1 1 53
9 12
3 3
6
1
16 x8
1
2
3
y
x5
1
5 12 x 3
14
4x
2
dx
x
y
1
0.5
1
5
1 4 dx x5 1 3
4x
12 x 1/2
373
15 121 1601 23 480
10
6
2
dx
x
y
8
3
dx
3
dx x3 41x
1
16 x
4
2
14
16 x
1
x8 12 1 8 dx
1/2 x
1
dy 2
dx
0
0.5
Copyright 2018 Pearson Education, Inc.
1
1.5
x
Section 6.3 Arc Length
11.
dx
dy
dx
sec 4 y 1 dy
L
/4
2
sec4 y 1
1 sec 4 y 1 dy
/4
/4
/4
sec 2 y dy
/4
tan y /4 1 (1) 2
12.
dy
dx
dy 2
3x 4 1 dx
L
1
3x4 1
1 3x 4 1 dx
2
1
2
3 x 2 dx
1
3
3 x3 33 1 (2)3 33 (1 8) 7 3 3
2
13. (a)
dy
dx
dy 2
2
L
1
(c)
14. (a)
4x
1 dx
2
1
dy 2
dx
1 4 x 2 dx
L 6.13
dy
dx
dy 2
sec2 x dx
0
L
/3
(c)
15. (a)
sec4 x
(b)
1 sec 4 x dx
L 2.06
dx
dy
dx
cos y dy
L
0
(c)
(b)
2
2 x dx
2
cos 2 y
(b)
1 cos 2 y dy
L 3.82
Copyright 2018 Pearson Education, Inc.
389
390
Chapter 6 Applications of Definite Integrals
16. (a)
dx
dy
y
1 y 2
L
1/2
1/2
dx
dy
1
1 y2
1/2
(c)
1/2
3
dy
dx
1/2
1/2
1
1 y 2
dy
dy
dx
dy
2
(b)
( y 1)2
dy 2
(b)
x 2 sin 2 x
1 x 2 sin 2 x dx
L 4.70
dy
dx
dy 2
tan x dx
/6
0
(b)
tan 2 x
1 tan 2 x dx
/6
0
sin 2 x cos 2 x
cos 2 x
dx
/6 dx
/6
sec x dx
0
cos x
0
L 0.55
dx
dy
dx
sec2 y 1 dy
L
/4
/3
/4
/3
(c)
dy
L
20. (a)
1 y
cos x cos x x sin x dx
0
(c)
2
L 9.29
L
19. (a)
(b)
y2
1 ( y 1) 2 dy
1
(c)
1 y
2
L 1.05
L
18. (a)
y2
1/2
dx
17. (a) 2 y 2 2 dy
(c)
2
2
(b)
sec2 y 1
1 sec 2 y 1 dy
/4
/3
| sec y | dy
sec y dy
L 2.20
Copyright 2018 Pearson Education, Inc.
Section 6.3 Arc Length
dy 2
dx
21. (a)
corresponds to
1
4x
dy
dx
here, so take
as
1 .
2 x
391
Then y x C and since (1, 1) lies on the curve,
C 0. So y x from (1, 1) to (4, 2).
(b) Only one. We know the derivative of the function and the value of the function at one value of x.
dx
dy
22. (a)
2
corresponds to
1
y4
dy
dx
here, so take
as
1 .
y2
Then x 1y C and, since (0, 1) lies on the curve,
C 1. So y 11x .
(b) Only one. We know the derivative of the function and the value of the function at one value of x.
23.
y
x
0
/4
0
24.
cos 2t dt
2/3 3/2
2/3
2 /4
1
2
4
2/3
32 32
y 3 2 x, 0 x 2
1
2 /4
x
dy
dx
x 1
1 1 x2/3 dx
32 (1) 2/3 32
25.
,
2
4
/4
0
/4
1
cos 2 x L
2 cos x dx 2 sin x 0
y 1 x
dy
dx
dy
dx
2
1 cos 2 x dx
/4
1 cos 2 x dx
0
/4
0
2 cos 2 x dx
2 sin 4 2 sin(0) 1
3
2
1 x
1
x
2/3
2/3 1/2
1 dx
23 x 1/3
1
x
2/3
2/3
1
2
1/ 2
L
x1/3
dx
1
2 /4
1 x
1
1/3
2 /4 x
dx
1
1
2 /4
1 x 2/3 1/ 2
dx
1
x1/3
1
x 1/3 dx 32 x 2/3
2 /4
2 /4
12 43 total length 8 34 6
2 L
2
0
1 (2) 2 dx
2
0
2
5 dx 5 x 2 5.
0
d (2 0)2 (3 (1))2 2 5
26. Consider the circle x 2 y 2 r 2 , we will find the length of the portion in the first quadrant, and multiply our
result by 4.
y r 2 x2 , 0 x r
4
r
r
0
2
r x
2
dx
dy
( y 3)( y 1)
6x
( y 3) 2 ( y 1)2
4 y ( y 3)
2
d
dy
r
dx
r 2 x2
9 x 2
dx
x
0
dx 4r
27. 9 x 2 y ( y 3)2
dy
dx
r 2 x2
d
dy
2
2
r
r
r 2 dx
1 2 x 2 dx 4 1 2x 2 dx 4
2
0
0
0
x2
r
x
r
r x
L 4
r
y ( y 3) 2 18 x dx 2 y ( y 3) ( y 3)2 3( y 3)( y 1)
dy
( y 3)( y 1)
dy;
6x
2
( y 3)( y 1)
ds 2 dx 2 dy 2
dy dy 2
6x
( y 1)2
dy 2 dy 2 4 y 1 dy 2
y 2 2 y 1 4 y
4y
dy 2
( y 1)2
4y
dy 2
Copyright 2018 Pearson Education, Inc.
( y 3)2 ( y 1) 2
36 x 2
dy 2 dy 2
392
Chapter 6 Applications of Definite Integrals
28. 4 x 2 y 2 64
d
dx
4 x2 y 2
d
dx
dy
0 dx 4yx dy 4yx dx;
64 8 x 2 y dy
dx
2
2
2
ds 2 dx 2 dy 2 dx 2 4yx dx dx 2 16 x2 dx 2 1 16 x2 dx 2
y
y
20 x 2 64
y2
2x
dx 2
dy 2
x
y2
4 x 2 64 16 x 2
y2
dx 2
dx 2
(5 x 2 16) dx 2
4
y2
1 dt
0
29.
y 2 16 x 2
dt , x 0 2 1
dy 2
dx
dy
dx
1 y f ( x) x C where C is any real
number.
30. (a) From the accompanying figure and definition of the
differential (change along the tangent line) we see
that dy f ( xk 1 ) xk length of kth tangent fin is
xk 2 (dy )2
xk 2 f ( xk 1 ) xk 2 .
n
n
2
2
(length of kth tangent fin) lim xk f ( xk 1 ) xk
n
n
(b) Length of curve lim
k 1
lim
n
n k 1
31.
k 1
b
1 f ( xk 1 ) xk 1 f ( x) dx
a
2
2
x 2 y 2 1 y 1 x 2 ; P 0, 14 , 12 , 34 , 1 L
14 0
2
15
4
1
2
12
1 2
4
3
2
15
4
2
4
2
2
xi xi 1 yi yi 1
k 1
43 12
2
7
4
3
2
2
1 43
2
0
7
4
2
1.55225
32. Let ( x1 , y1 ) and ( x2 , y2 ), with x2 x1 , lie on y mx b, where m
L
33.
x2
x1
1 m 2 dx 1 m 2 x x2 1 m 2 x2 x1 1
x
1
x2 x1 2 y2 y1 2
x2 x1 2 y2 y1 2
x
x
x2 x1
2
1
x2 x1
x2 x1 2
x
dy
y 2 x3/2 dx 3 x1/2 ; L( x) 1 3t1/2
0
dt 0x
2
y2 y1
,
x2 x1
y2 y1 2
x2 x1
then
dy
dx
m
x2 x1
x2 x1 2 y2 y1 2 .
1 9t dt ;
[u 1 9t du 9dt ; t 0 u 1, t x u 1 9 x] 19
19 x
1
19 x
2 u 3/2
2 (1 9 x)3/2 2 ;
u du 27
27
27
1
2 (10)3 2 2 2(10 10 1)
L(1) 27
27
27
Copyright 2018 Pearson Education, Inc.
Section 6.3 Arc Length
34.
3
dy
y x3 x 2 x 4 x1 4 dx x 2 2 x 1
1
4( x 1)2
( x 1)2
2
1
4( x 1) 2
;
2
x
x
[4(t 1)4 1]2
4(t 1)4 1
1 (t 1)2 1 2 dt 1
dt 1
dt
2
0
0
0
4(t 1)
16(t 1)4
4(t 1)
L( x)
x
0
x
16(t 1)4 16(t 1)8 8(t 1) 4 1
16(t 1)
4
dt
x
16(t 1)8 8(t 1)4 1
16(t 1)
0
4
dt
x
0
[4(t 1)4 1]2
16(t 1)
4
x 4(t 1)4 1
0 4(t 1) 2
dt
dt
x
x 1
(t 1)2 1 2 dt ; [u t 1 du dt ; t 0 u 1, t x u x 1] u 2 14 u 2 du
0
1
4(t 1)
x 1
13 u 3 14 u 1
1
35–40.
1 ; L(1) 8 1 1 59
13 ( x 1)3 4( x11) 13 14 13 ( x 1)3 4( x11) 12
3 8 12
24
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("#35(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)
Mathematica: (assigned function and values for a, b, and n may vary)
Clear[x, f ]
{a, b} {1, 1}; f[x_ ] Sqrt[1 x 2 ]
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 2018 Pearson Education, Inc.
393
394
6.4
Chapter 6 Applications of Definite Integrals
AREAS OF SURFACES OF REVOLUTION
1. (a)
dy
dx
sec2 x
S 2
/4
0
(c)
2. (a)
dy 2
dx
(b)
sec4 x
(tan x) 1 sec4 x dx
S 3.84
dy
dx
2x
dy 2
dx
(b)
4 x2
2
S 2 x 2 1 4 x 2 dx
0
(c)
3. (a)
S 53.23
dx 1 dx
xy 1 x 1y dy
dy
y2
2
1
y4
(b)
2
S 2 1y 1 y 4 dy
1
(c)
4. (a)
S 5.02
dx
dy
cos y
dx
dy
2
cos 2 y
(b)
S 2 (sin y ) 1 cos 2 y dy
0
(c)
S 14.42
Copyright 2018 Pearson Education, Inc.
Section 6.4 Areas of Surfaces of Revolution
5. (a)
x1/2 y1/2 3 y 3 x1/2
(b)
2
12 x1/2
2
dy 2
dx 1 3x 1/2
2
2
4
S 2 3 x1/2 1 1 3x 1/2 dx
1
dy
dx 2 3 x1/2
(c)
6. (a)
S 63.37
dx
dy
S 2
2
1
(c)
7. (a)
1 y
2
1/2 2
y2 y
1 1 y 1/2
1 y 1/2
dx
dy
(b)
dx
2
S 51.33
dx
dy
tan y
dx
dy
2
(b)
tan 2 y
y
/3 y
2
0 tan t dt 1 tan y dy
/3 y
2 tan t dt sec y dy
0 0
(c) S 2.08
S 2
1
0
8. (a)
dy
dx
x2 1
S 2
1
2
5 x
1
(c) S 8.55
1
dy 2
dx
5 x
1
y
x tan t dt
0
0.5
0
0.4
0.6
x
0.8
(b)
x2 1
y
t 2 1 dt 1 x 2 1 dx
3
y
2
t 2 1 dt x dx
x
1
t 2 1 dt
1
0
9.
0.2
4
b
dy 2
dy
y 2x dx 12 ; S 2 y 1 dx dx S 2 2x
0
a
1
2
x
4
2
4
1 14 dx 2 5 x dx 2 5 x2 4 5;
0
0
Geometry formula: base circumference 2 (2), slant height 42 22 2 5
3
Lateral surface area 12 (4 ) 2 5 4 5 in agreement with the integral value
Copyright 2018 Pearson Education, Inc.
395
396
10.
Chapter 6 Applications of Definite Integrals
dy
2
dx 2; S d 2 x 1 dx
y 2x x 2 y dy
dy
c
2
2
2
2 2 y 1 22 dy 4 5 y dy 2 5 y 2
0
0
0
2 5 4 8 5; Geometry formula: base circumference 2 (4), slant height 42 22 2 5
Lateral surface area 12 (8 ) 2 5 8 5 in agreement with the integral value
11.
dx
dy
b
12 ; S 2 y 1
a
25
dx 2
dy 2
dx
3
1
( x 1)
2
92 3 12 1 2 5 (4 2) 3
1
12
2
3
2
3
dx 2 5 ( x 1) dx 2 5 x2 x
1
1
5; Geometry formula: r1 12 12 1, r2 23 12 2, slant height
(2 1)2 (3 1) 2 5 Frustum surface area r1 r2 slant height (1 2) 5 3 5 in
agreement with the integral value
12.
dy
dx 2; S d 2 x 1 dx
y 2x 12 x 2 y 1 dy
dy
c
2
2
1
2
2 (2 y 1) 1 4 dy 2 5 (2 y 1) dy
1
2
2 5 y 2 y 2 5 (4 2) (1 1) 4 5; Geometry formula: r1 1, r2 3,
1
slant height (2 1)2 (3 1) 2 5 Frustum surface area (1 3) 5 4 5 in agreement with
the integral value
13.
dy
dx
x2
3
dy 2
dx
x4
9
3
4
2
S 29x 1 x9 dx;
0
u 1 x 4 du 4 x3 dx 1 du x3 dx;
9
9
4
9
x 0 u 1, x 2 u
25/9
2 23 u 3/2
1
14.
dy
dx
12 x 1/2
S
15/4
3/4
15.
dy
dx
1 (2 2 x )
2 2 x x2
S
1.5
0.5
2
1.5
2
1.5
0.5
0.5
dy 2
dx
1
4x
3/2 15/4
15/4
28
3
dy 2
1 x
dx
2 x x
2
2 2 x x 2 1
2 x x2
x 14 dx
3/2
3/2
43 15
14
34 14
4
3/4
1 43 (8 1)
3
du
1 3 12527 27 98
125
27
81
25/9 1/2 1
u 4
1
3/4
42
S 2
2 x 1 41x dx 2
2 23 x 14
43
3
25
9
(1 x ) 2
2 x x2
2 x x 2 1 2 x x 2
2 x x2
(1 x ) 2
2 x x2
dx
dx
dx 2 x 0.5 2
1.5
Copyright 2018 Pearson Education, Inc.
Section 6.4 Areas of Surfaces of Revolution
16.
dy
dx
1
2 x 1
dy 2
dx
1
4( x 1)
5
5
S 2 x 1 1 4( x11) dx 2
1
1
x 54 dx 2 32 x 54
5
2
1
43 5 54
43
17.
dx
dy
53
23
3/2
3
1 54
3/2
33 6 (125 27)
2
y2
dx
dy
2
3/2 5
1
4
3
( x 1) 14 dx
98
6
254
3/2
94
3/2
49
3
1 2 y 3
0 3
y4 S
1 y 4 dy;
u 1 y 4 du 4 y3 dy 1 du y 3 dy;
4
2
y 0 u 1, y 1 u 2 S 2
1
13 u1/2 14 du
2
2
6 u1 2 du 6 23 u 3/2 9 ( 8 1)
1
1
18.
13 y3/2 y1/2 0, when 1 y 3. To get positive
area, we take x 13 y 3/2 y1/2
2
dx 1 y1/2 y 1/2 dx
dy
2
dy 14 y 2 y 1
3
S 2 13 y 3/2 y1/2 1 14 y 2 y 1 dy
1
3 1 3/2
2 3 y y1/2 14 y 2 y 1 dy
1
x
3 1 3/2
y
1 3
2
3 1 2
y
1 3
y1/2
y
1/ 2
y 1/ 2
2
2
3
dy y1/2
1
y3
23 y 1 dy 9
y3
3
3
y
1
13 y 1 y1/2 y1
1/ 2
3
dy 1
13 y 1 ( y 1) dy
279 93 3 19 13 1 3 19 13 1
9 (18 1 3) 169
19.
dx
dy
1
4 y
4
15/4
0
dx
dy
2
1
4 y
S
15/4
0
2 2 4 y 1 41 y dy 4
15/4
5 y dy 4 23 (5 y )3/2
0
83 5 5 5 8 5 83
40 5 5 5
8
15/4
0
83 5 15
4
3/2
(4 y ) 1 dy
53/2 83
35 5
3
Copyright 2018 Pearson Education, Inc.
54
3/2
53/2
397
398
20.
Chapter 6 Applications of Definite Integrals
dx
dy
1
2 y 1
dx
dy
2
1
2 y 1
1
2 2 23 y3/2 43
5/8
21.
22.
dy
dx
x
dy 2
dx
2
2
0
1
5/8
3/2
1
2 2 y 1 1 2 y11 dy 2
85
3/2
1
4 2
3
1
x 2 S 2 x 1 x 2 dx
0
2
1 x2
3
1
(2 y 1) 1 dy 2
5/8
4 2 82 2 5 5
3
82 2
5 5
8 8
3/2 1
16 2 5 5
12
2 y1/2 dy
2
dy x x 2 2 dx ds 1 2 x 2 x 4 dx S 2
x 1 2 x 2 x 4 dx
0
x2 1 dx 2 0 2 x x2 1 dx 2 0 2 x3 x dx 2 x4 x2 0
2
x
1
5/8
2
2 2 1
3
0
3/2
y 13 x 2 2
2
S
4
2
2
2
44 22 4
2
23. ds dx 2 dy 2 y3 1 3 1 dy y 6 12 1 6 1 dy y 6 12 1 6 dy
y
y
y
4
16
16
2
2
2
2
y 3 1 3 dy y3 1 3 dy; S 2 y ds 2 y y3 1 3 dy 2π y 4 14 y 2 dy
1
1
1
4y
4y
4y
2
y5
2 5 14 y 1 2
1
dy
dx
24.
y cos x
25.
y a2 x2
S 2
325 18 15 14 2 315 18 240 (8 31 5) 25320
dy 2
dx
sin x
a
dy
dx
1
2
a2 x2
a2 x2 1
a
sin 2 x S 2
/2
/2
1/2
x2
a2 x2
x
(2 x)
dx 2
2
a x
2
(cos x) 1 sin 2 x dx
dy 2
dx
x2
a x2
2
a2 x2 x2 dx 2 aa a dx 2 a xaa
a
a
2 a [a (a)] (2 a )(2a ) 4 a 2
26.
y
r
h
x
dy
dx
r
h
dy 2
dx
r2
h2
hr
x
0 h
S 2
h
2
22r h 2 r 2 x2 22r h 2 r 2
h
h
0
r
h2
2
d
h
dx
dy
2
7
16
y
2
16 y
2
dx
dy
2
y2
162 y 2
; S
7
16
x
h2 r 2
h2
dx 2h r
h2 r 2 h x
0
h2
dx
h2 r 2
27. The area of the surface of one wok is S 2 x 1
c
hr
0 h
2
1 r 2 dx 2
dy. Now, x
dx
dy
2
2 162 y 2 1
y2
162 y 2
2
y 2 162 x 162 y 2
dy
7
16
2
162 y 2 y 2 dy
16 dy 32 9 288 904.78 cm 2 . The enamel needed to cover one surface of one wok is
V S 0.5 mm S 0.05 cm (904.78)(0.05) cm3 45.24 cm3 . For 5000 woks, we need
5000 V 5000 45.24 cm3 (5)(45.24) L 226.2 L 226.2 liters of each color are needed.
Copyright 2018 Pearson Education, Inc.
Section 6.4 Areas of Surfaces of Revolution
28.
2
dy
dx
y R2 x2
2
2x
12
2
r x
2
x
2
r x
dx
dy
2
x2
r x2
2
; S 2
ah
r 2 x2 1
a
x2
r x2
2
dx
2x
12
2
R x
2
x
2
R x
2
dx
dy
2
x2
R2 x2
; S 2
ah
a
R2 x2 1
x2
R2 x2
dx
R2 x2 x2 dx 2 R aah dx 2 Rh
ah
a
x 2 y 2 452 x 452 y 2
30. (a)
2
r 2 x2 x2 dx 2 r aah dx 2 rh, which is independent of a.
ah
a
29.
dy
dx
y r 2 x2
399
S
45
22.5
2 452 y 2 1
y2
dx
dy
y
2
45 y
dy 2
452 y 2
2
dx
dy
2
y2
2
45 y 2
;
452 y 2 y2 dy 2 454522.5 dy
45
22.5
(2 )(45)(67.5) 6075 square feet
(b) 19,085 square feet
31. (a) An equation of the tangent line segment is
(see figure) y f (mk ) f (mk ) x mk . When
x xk 1 we have
r1 f (mk ) f (mk )( xk 1 mk )
f (m ) f (m )
f (mk ) f (mk )
xk
2
k
k
xk
2
;
xk
2
f (mk )
when x xk we have
r2 f (mk ) f (mk ) xk mk
xk
2
f (mk ) f (mk )
;
(b) L2k xk r2 r1 xk f (mk )
2
2
2
xk 2 f (mk )xk 2 ,
Lk
xk
2
2
x 2 f (m )x 2
k k
k
as claimed
(c) From geometry it is a fact that the lateral surface area of the frustum obtained by revolving the tangent line
segment about the x-axis is given by Sk r1 r2 Lk 2 f (mk )
xk 2 f (mk )xk 2
parts (a) and (b) above. Thus, Sk 2 f (mk ) 1 f (mk ) xk .
2
n
n
Sk nlim
2 f (mk )
n
(d) S lim
k 1
32.
y 1 x
2/3 3/2
k 1
1
dy
dx
S 2 2 1 x 2/3
0
3
2
3/2
1 x
2/3 1/2
1
1
x
2/3
23
2
x
1/3
1 x
1/ 2
2/3
1/3
x
1
1 dx 4 1 x 2/3
0
0
2
a
u 1 x 2/3 du 23 x 1/3dx 32 du x 1/3 dx;
0
b
1 f (mk ) xk 2 f ( x) 1 f ( x) dx
3/2
dy 2
dx
2/3
1 x2/3
x
1
1
x 2/3
1
x 2/3 dx 4 1 x 2/3
0
x 0 u 1, x 1 u 0
S 4 u 3/2 32 du 6 52 u 5/2 6 0 52 125
1
1
Copyright 2018 Pearson Education, Inc.
3/2 1/3
x
dx;
using
400
6.5
Chapter 6 Applications of Definite Integrals
WORK AND FLUID FORCES
1. Work is area beneath graph W 12 (3)(20) (5)(14) (2)(8) 116 J
2. Work is area beneath the graph; assume that each tic mark on both axes is 1 unit, where each square unit
represents 16 N m 16J W 16
14 (2)2 52 3 88 16 J.
3. The force required to stretch the spring from its natural length of 2 m to a length of 5 m is F ( x ) kx.
3
3
3
The work done by F is W F ( x) dx k x dx k2 x 2
0
0
0
9k .
2
This work is equal to 1800 J
92 k 1800 k 400 N/m
4. (a) We find the force constant from Hooke’s Law: F kx k
F
x
k 800
200 lb/in.
4
2
2
0
0
(b) The work done to stretch the spring 2 inches beyond its natural length is W kx dx 200 x dx
2
2
200 x2 200(2 0) 400 in-lb 33.3 ft-lb
0
(c) We substitute F 1600 into the equation F 200 x to find 1600 200 x x 8 in.
5. We find the force constant from Hooke’s law: F kx. A force of 2 N stretches the rubber band to 0.02 m
N . The force of 4 N will stretch the rubber band y m, where F ky y
2 k (0.02) k 100 m
y
4N
N
100 m
100
0.04
y 0.04 m 4 cm. The work done to stretch the rubber band 0.04 m is W
0.04
2
x dx 100 x2
0
0
(100)(0.04) 2
2
F
x
k
90
1
N . The work done to
k 90 m
5
2
5
5
stretch the spring 5 m beyond its natural length is W kx dx 90 x dx 90 x2 (90)
0
0
0
7. (a) We find the spring’s constant from Hooke’s law: F kx k
F
x
(b) The work done to compress the assembly the first half inch is W
21,714
85
0.5
2
0.5
1.0
0.5
(0.5)2
2
(7238)(0.25)
2
1.0
1.0
7238 1 (0.5) 2
2
F
x
8. First, we find the force constant from Hooke’s law: F kx k
compresses the scale x 18 in, he/she must weigh F kx 2, 400
scale this far is W
1/8
0
1/8
2
kx dx 2400 x2
0
21,714
3
252 1125 J
lb
k 7238 in
0.5
0
x dx
905 in-lb. The work done to compress the assembly the
2
x dx 7238 x2
0.5
0.5
kx dx 7238
kx dx 7238
0
7238 x2 (7238)
0
second half inch is:
kx dx
0.08 J
6. We find the force constant from Hooke’s law: F kx k
W
0.04
0
F
k
2400
264
(7238)(0.75)
2
150
161
2714 in-lb
16 150 2, 400
lb .
in
If someone
18 300 lb. The work done to compress the
18.75 lb in.
2.5 ft-lb
16
Copyright 2018 Pearson Education, Inc.
Section 6.5 Work and Fluid Forces
401
9. 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 ) 0.624 x. The work done is: W
50
0
F ( x) dx
50
0
0.624x dx
50
2
0.624 x2 780 J
0
10. The weight of sand decreases steadily by 72 lb over the 18 ft, at 4 lb/ft. So the weight of sand when the
b
18
a
0
bag is x ft off the ground is F ( x ) 144 4 x. The work done is: W F ( x)dx (144 4 x) dx
18
144 x 2 x 2
0
1944 ft-lb
11. The force required to lift the cable is equal to the weight of the cable paid out: F ( x) (4.5)(180 x)
where x is the position of the car off the first floor. The work done is: W
4.5 180 x
180
x2
2 0
4.5 1802 180
2
2
180
0
2
4.5180
2
F ( x) dx 4.5
180
0
(180 x) dx
72,900 ft-lb
12. Since the force is acting toward the origin, it acts opposite to the positive x-direction. Thus F ( x)
b
The work done is W
k
k2
a
b
dx k
a
1
x2
b
dx k 1x k
a
k .
x2
b1 1a k (aabb)
13. Let r 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 (20 x), 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 0.8(20 x ). So: W
20
0
0.8 (20 x )dx 0.8 20 x
20
x2
2 0
160 ft-lb.
14. Let r 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 (20 x), 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 2(20 x).
So: W
20
0
2(20 x ) dx 2 20 x
20
x2
2 0
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.
15. We will use the coordinate system given.
(a) The typical slab between the planes at y and y y
has a volume of V (10)(12)y 120y ft3. The
force F required to lift the slab is equal to its weight:
F 62.4 V 62.4 120y lb. The distance through
which F must act is about y ft, so the work done
lifting the slab is about W force distance
62.4 120 y y ft-lb The work it takes to lift all
the water is approximately
20
20
0
0
W W 62.4 120 y y ft-lb.
Copyright 2018 Pearson Education, Inc.
402
Chapter 6 Applications of Definite Integrals
This is a Riemann sum for the function 62.4 120 y over the interval 0 y 20. The work of pumping the
tank empty is the limit of these sums:
W
20
0
20
y2
62.4 120 y dy (62.4)(120) 2 (62.4)(120)
0
(b) The time t it takes to empty the full tank with
4002 (62.4)(120)(200) 1,497,600 ft-lb
115 hp motor is t 250W
ft-lb
sec
1,497,600 ftlb
250 ft-lb
sec
5990.4 sec
1.664 hr 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
10
10
y2
W 62.4 120 y dy (62.4)(120) 2 (62.4)(120)
0
0
1497.6 sec 0.416 hr 25 min
(d) In a location where water weighs 62.26
1002 374,400 ft-lb and the time is t 250W
ft-lb
sec
lb :
ft 3
a) W (62.26)(24,000) 1,494,240 ft-lb .
5976.96 sec 1.660 hr t 1 hr and 40 min
b) t 1,494,240
250
In a location where water weighs 62.59
lb
ft 3
a) W (62.59)(24,000) 1,502,160 ft-lb
6008.64 sec 1.669 hr t 1 hr and 40.1 min
b) t 1,502,160
250
16. We will use the coordinate system given.
(a) The typical slab between the planes at y and y y has
a volume of V (20)(12) y 240y ft 3 . The force F
required to lift the slab is equal to its weight:
F 62.4V 62.4 240y lb. The distance through
which F must act is about y ft, so the work done lifting
the slab is about W force distance
20
62.4 240 y y ft-lb. The work it takes to lift all the water is approximately W W
10
20
62.4 240 y y ft-lb. This is a Riemann sum for the function 62.4 240 y over the interval
10
10 y 20. The work it takes to empty the cistern is the limit of these sums:
W
20
10
(b) t
20
y2
62.4 240 y dy (62.4)(240) 2 (62.4)(240)(200 50) (62.4)(240)(150) 2,246,400 ft-lb
10
W
275 ft-lb
sec
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
15
15
y2
W 62.4 240 y dy (62.4)(240) 2 (62.4)(240)
10
10
Then the time is t
W
275 ft-lb
sec
936,000
275
2252 1002 (62.4)(240) 1252 936, 000 ft.
3403.64 sec 56.7 min
(d) In a location where water weighs 62.26
lb :
ft 3
a) W (62.26)(240)(150) 2,241,360 ft-lb.
Copyright 2018 Pearson Education, Inc.
Section 6.5 Work and Fluid Forces
b) t
2,241,360
275
403
8150.40 sec 2.264 hours 2 hr and 15.8 min
c) W (62.26)(240)
1252 933,900 ft-lb;
In a location where water weighs 62.59
t
933,900
275
3396 sec 0.94 hours 56.6 min
lb :
ft 3
a) W (62.59)(240)(150) 2,253,240 ft-lb.
b) t
2,253,240
275
8193.60 sec 2.276 hours 2 hr and 16.56 min
c) W (62.59)(240)
1252 938,850 ft-lb;
t
938,850
275
3414 sec 0.95 hours 56.9 min
, thickness y, and height below the top of the tank (10 y). So the
work to pump the oil in this slab, W , is 57 (10 y ) π . The work to pump all the oil to top of the tank is
17. The slab is a disk of area x 2
y 2
2
y 2
2
10 57
0 4
W
10 y 2 y3 dy 57π4 103y
3
10
y4
4
0
11,875 ft lb 37,306 ft-lb
18. Each slab of oil is to be pumped to a height of 14 ft. So the work to pump a slab is (14 y )( )
250
3
the tank is half full and the volume of the original cone is V 13 r 2 h 13 52 (10)
volume
250π
6
ft 3 , and with half the volume the cone is filled to a height y,
3
500 57
4
0
So W
14 y
2
y
3
dy
3
57π 14 y
4 3
y4
4
3
500
0
250
6
13
y2
4
y 2
2
and since
ft 3 , half the
y y 3 500 ft.
60,042 ft-lb.
19. The typical slab between the planes at y and y y has a volume of V (radius)2 (thickness)
2
20
y 100 y ft 3 . The force F required to lift the slab is equal to its weight:
2
F 51.2V 51.2 100 y lb F 5120 y lb The distance through which F must act is about
30
30
0
0
(30 y ) ft. The work it takes to lift all the kerosene is approximately W W 5120 (30 y ) y ft-lb
which is a Riemann sum. The work to pump the tank dry is the limit of these sums:
30
30
y2
W 5120 (30 y ) dy 5120 30 y 2 5120 900
(5120)(450 ) 7,238,229.48 ft-lb
2
0
0
lb
ft 3
20. (a) Follow all the steps of Example 5 but make the substitution of 64.5
8
for 57
3
4
8
(10 y ) y 2 dy 64.5 10 y y 64.5 1083 84 64.5
W 64.5
4 3
4
4
3
4
4
0 4
0
lb .
ft 3
Then,
83 103 2 64.53 8
3
21.5 83 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.
8 57
(13
0 4
Then W
8
83 133 2 57348 7
3
4
13 y3 y 4
y ) y 2 dy 574 3 4 574 1338 84 574
0
(19 )(82 )(7)(2) 53,482.5 ft-lb
Copyright 2018 Pearson Education, Inc.
3
404
Chapter 6 Applications of Definite Integrals
21. The typical slab between the planes at y and y y has a volume of about V (radius)2 (thickness)
y y ft3. The force F ( y) required to lift this slab is equal to its weight: F ( y) 73 V
2
73 y y 73 y y lb. The distance through which F ( y ) must act to lift the slab to the top of the
2
reservoir is about (4 y ) ft, so the work done is approximately W 73 y (4 y ) y ft-lb. The work done
lifting all the slabs from y 0 ft to y 4 ft is approximately W
n
73 yk 4 yk y ft-lb.
k 0
4
of these Riemann sums as n , we get W 73 y (4 y ) dy 73
0
0 4 y y
4
2
Taking the limit
dy
4
ft-lb 2446.25 ft-lb.
73 2 y 2 13 y 3 73 32 64
2336
3
3
0
22. The typical slab between the planes at y and y y has volume of about V (length)(width)(thickness)
2 25 y 2 (10)y ft 3 . The force F ( y ) required to lift this slab is equal to its weight:
F ( y ) 53 V 53 2 25 y 2 (10) y 1060 25 y 2 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 (20 y ) ft, so the work done is
approximately W 1060 25 y 2 (20 y ) y ft-lb. The work done lifting all the slabs from y 5 ft to
n
y 5 ft is approximately W 1060 25 yk2 20 yk y ft-lb. Taking the limit of these Riemann sums as
k 0
5
5
5
5
n , we get W 1060 25 y 2 (20 y )dy 1060 (20 y ) 25 y 2 dy
5
5
1060 20 25 y 2 dy y 25 y 2 dy . To evaluate the first integral, we use we can interpret
5
5
5
5
25 y 2 dy as the area of the semicircle whose radius is 5, thus
5
5 20
25 y 2 dy 20
5
5
25 y 2 dy
20 12 (5) 2 250 . To evaluate the second integral let u 25 y 2 du 2 y dy; y 5 u 0,
5
5
0
y 5 u 0, thus y 25 y 2 dy 12 u du 0. Thus, 1060 20 25 y 2 dy
5
0
5
5
5
1060 20 25 y 2 dy y 25 y 2 dy 1060(250 0) 265000 832522 ft-lb
5
5
23. The typical slab between the planes at y and y y has a volume of about V (radius)2 (thickness)
2
25 y 2 y m3 . The force F ( y ) required to lift this slab is equal to its weight:
2
F ( y ) 9800 V 9800 25 y 2 y 9800 25 y 2 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 9800 25 y 2 (4 y )y N m. The work done lifting all the slabs from y 5 m to
0
y 0 m is approximately W 9800 25 y 2 (4 y )y N m. Taking the limit of these Riemann sums,
5
Copyright 2018 Pearson Education, Inc.
Section 6.5 Work and Fluid Forces
0
we get W 9800 25 y 2 (4 y ) dy 9800
5
0
5
0
405
100 25 y 4 y 2 y3 dy
y4
9800 100 y 25
y 2 43 y 3 4 9800 500 25225 43 125 625
15, 073, 099.75 J
2
4
5
24. The typical slab between the planes at y and y y has a volume of about V (radius)2 (thickness)
2
56 lb
100 y 2 y 100 y 2 y ft 3 . The force is F ( y ) 3 V 56 100 y 2 y lb. The
ft
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
(12 y ) ft, so the work done is W 56 100 y 2 (12 y )y lb ft. The work done lifting all the slabs from
10
y 0 ft to y 10 ft is approximately W 56 100 y 2 (12 y )y lb ft. Taking the limit of these
0
10
Riemann sums, we get W 56 100 y 2 (12 y ) dy 56
0
100 y2 (12 y) dy
10
100 y 2 12 y 3
y4
1200 100 y 12 y 2 y 3 dy 56 1200 y 2 3 4
0
0
56
10
0
10
56 12,000 10,000
4 1000 10,000
(56 ) 12 5 4 25 (1000) 967,611 ft-lb. It would cost
2
4
(0.5)(967, 611) 483,805¢ = $4838.05. Yes, you can afford to hire the firm.
25.
F m dv
mv dv
by the chain rule W
dt
dx
1
2
x2
x1
mv dv
dx m
dx
x2
x1
v dvdx dx m 12 v2 ( x) x
x2
1
m v 2 ( x2 ) v 2 ( x1 ) 12 mv22 12 mv12 , as claimed.
2 lb; mass
26. weight 2 oz 16
27. 90 mph
weight
32
90 mi 1 hr
1 min 5280 ft
1 hr 60 min 60 sec 1 mi
2561 slugs (160 ft / sec)2 50 ft-lb
1
8
1 slugs; W 1
32
256
2
132 ft/sec; m
0.3125 lb
32 ft/ sec2
0.3125
slugs;
32
0.3125 lb
W 12
(132ft/sec)2 85.1 ft-lb
2
32 ft/sec
28. weight 1.6 oz 0.1 lb m
0.1 lb
32 ft/ sec2
3201 slugs (280ft/ sec)2 122.5ft-lb
1 slugs; W 1
320
2
ft , v 153 mph 224.4 ft ; 2 oz 0.125 lb m
29. v1 0 mph 0 sec
2
sec
W
x2
x1
0.125 lb
32 ft/ sec2
1 slugs;
256
1 (224.4) 2 1 1 (0) 2 98.35 ft-lb
F ( x ) dx 12 mv22 12 mv12 12 256
2 256
30. weight 6.5 oz
6.5 lb
16
6.5 slugs (132ft/sec)2 110.6ft-lb
(16)(32)
6.5
m (16)(32)
slugs; W 12
Copyright 2018 Pearson Education, Inc.
406
Chapter 6 Applications of Definite Integrals
31. We imagine the milkshake divided into thin slabs by planes perpendicular to the y -axis at the points of a
partition of the interval [0, 7]. The typical slab between the planes at y and y y has a volume of about
V (radius)2 (thickness)
weight: F ( y )
4
9
V 49
y 17.5 2
14
y 17.5 2
14
y in 3 . The force F ( y ) required to lift this slab is equal to its
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 done lifting the slab is about
( y1417.5)
W 49
2
2
(8 y ) y in oz. The work done lifting all the slabs from y 0 to y 7 is approximately
7
W 4 2 ( y 17.5)2 (8 y ) y in oz which is a Riemann sum. The work is the limit of these sums as the
0
9.14
norm of the partition goes to zero:
7 4
( y 17.5)2 (8
0 9142
W
7
y ) dy 4π 2 2450 26.25 y 27 y 2 y 3 dy
914
0
7
4
y4
2
4 2 7 9 73 26.25 72 2450 7 91.32 in-oz
4 2 4 9 y 3 26.25
y
2450
y
2
2
0 914 4
914
32. We fill the pipe and the tank. To find the work required to fill the tank note that radius = 10 ft, then
V 100y ft 3 . The force required will be F = 62.4 V = 62.4 100 y = 6240 y lb. The distance
through which F must act is y so the work done lifting the slab is about W1 6240 y y lb ft. The work it
takes to lift all the water into the tank is: W1
385
385
360
360
W1 6240 y y lb ft. Taking the limit we end up
385
y
[3852 3602 ] 182,557,949 ft-lb
6240 y dy 6240 2
6240
2
360
To find the work required to fill the pipe, do as above, but take the radius to be 42 in. 16 ft. Then
with W1
385
2
360
1 y ft 3 and F 62.4 V
V 36
360
360 62.4
y dy
36
0
W2 W2 W2
0
62.4
36
y. Also take different limits of summation and integration:
62.4
36
y2
2
360
0
62.436 3602 352,864 ft-lb
2
The total work is W W1 W2 182,557,949 352,864 182,910,813 ft-lb. The time it takes to fill the tank
W 182,910,813 110,855 sec 31 hr
and the pipe is Time 1650
1650
33. Work
35,780,000 1000 MG
6,370,000
(1000) 5.975 10
r
4
2
dr 1000 MG
6.672 10
11
35,780,000 dr
6,370,000 r
1
6,370,000
2
1000 MG 1r
1
35,780,000
35,780,000
6,370,000
5.144 10
10
J
34. (a) Let be the x-coordinate of the second electron. Then r 2 ( 1)2
0
0 (231029 )
d
1 ( 1) 2
W F ( ) d
1
0
29
(23 1029 ) 1 1 11.5 1029
2310
1 1
2
(b) W W1 W2 where W1 is the work done against the field of the first electron and W2 is the work done
against the field of the second electron. Let be the x-coordinate of the third electron. Then r12 ( 1)2
and r22 ( 1) 2
Copyright 2018 Pearson Education, Inc.
Section 6.5 Work and Fluid Forces
W1
5 231029
r12
3
W2
5 231029
r22
3
d
5 231029
3 ( 1)
d
23 1029.
12
Therefore W W1 W2
5
d 23 1029 11 (23 1029 )
3
5
5 231029
3 ( 1)
2
d 23 1029 11 (23 1029 )
3
2
407
14 12 234 1029 , and
16 14 231210
29
(3 2)
234 1029 1223 1029 233 1029 7.67 1029 J
35. 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 x 5. If we let x denote the width of the right-hand half of the triangle at depth y, then x 5 y and
the total width is L( y ) 2 x 2(5 y ). The depth of the strip is ( y ). The force exerted by the water against
one side of the plate is therefore F
2
5
124.8
2
5
(124.8)
w( y ) L( y ) dy
2
5
62.4 ( y ) 2(5 y ) dy
5 y y2 dy 124.8 52 y 2 13 y3 5 124.8 52 4 13 8 52 25 13 125
2
1052 1173 (124.8) 3156234 1684.8 lb
36. An equation for the line of the plate’s right-hand edge is y x 3 x y 3. Thus the total width is
L( y ) 2 x 2( y 3). The depth of the strip is (2 y ). The force exerted by the water is
0
0
0
3
3
3
F w(2 y ) L( y ) dy 62.4 (2 y ) 2(3 y ) dy 124.8
9
2
(124.8) 18 9 (124.8)
27
2
1684.8 lb
6 y y 2 dy 124.8 6 y
b
37. (a) The width of the strip is L( y ) 4, the depth of the strip is (10 y ) F w
a
3
3
62.4(10 y )(4)dy 249.6 (10 y ) dy 249.6 10 y
0
0
y2
2
3
0
b
4
y3
3
0
3
F ( y) dy
a
y2
2
strip
depth
249.6 30 92 6364.8 lb
(b) The width of the strip is L( y ) 3, the depth of the strip is (10 y ) F w
4
4
62.4(10 y )(3) dy 187.2 (10 y ) dy 187.2 10 y
0
0
y2
2
strip
depth
F ( y) dy
187.2(40 8) 5990.4 lb
0
b
38. The width of the strip is L( y ) 2 25 y 2 , the depth of the strip is (6 y ) F w
a
strip
depth
F ( y) dy
5
5
5
5
62.4 (6 y ) 2 25 y 2 dy 124.8 (6 y ) 25 y 2 dy 124.8 6 25 y 2 dy y 25 y 2 dy
0
0
0
0
To evaluate the first integral, we use we can interpret
radius is 5, thus
5
0 6
25 y 2 dy 6
5
0
5
0
25 y 2 dy as the area of a quarter circle whose
25 y 2 dy 6 14 (5) 2
u 25 y 2 du 2 y dy; y 0. u 25, y 5 u 0, thus
1 25 u1/2 du
2 0
75
2
5
. To evaluate the second integral let
0 y
25 y 2 dy 12
0
25
25
5
5
13 u 3/2 125
. Thus, 124.8 6 25 y 2 dy y 25 y 2 dy 124.8
3
0
0
0
9502.7 lb.
Copyright 2018 Pearson Education, Inc.
u du
752 1253
408
Chapter 6 Applications of Definite Integrals
39. Using the coordinate system of Exercise 32, we find the equation for the line of the plate’s right-hand edge to
be y 2 x 4 x
(a)
y4
2
and L( y ) 2 x y 4. The depth of the strip is (1 y ).
0
0
4
4
3y
4 3 y y 2 dy 62.4 4 y 2
4
F w(1 y ) L( y ) dy 62.4 (1 y )( y 4) dy 62.4
0
(62.4) (4)(4)
( 62.4)( 120 64)
3
(3)(16)
2
(62.4)(16 24 64 )
64
3
3
2
1164.8 lb
(3)(16)
( 64.0)( 120 64) 1194.7 lb
(b) F (64.0) (4)(4) 2 64
3
3
40. Using the coordinate system given, we find an equation for the
line of the plate’s right-hand edge to be y 2 x 4 x
4 y
2
and L( y ) 2 x 4 y. The depth of the strip is (1 y )
1
1
0
0
F w (1 y )(4 y ) dy 62.4 y 2 5 y 4 dy
1
y3 5 y 2
62.4 3 2 4 y (62.4)
0
(62.4)(11)
6
13 52 4 (62.4) 215624
114.4 lb
41 Using the coordinate system given in the accompanying figure,
we see that the total width is L( y ) 63 and the depth of the strip
is (33.5 y ) F
33
0
33 64
0 123
64
123
w(33.5 y ) L( y ) dy
(33.5 y ) 63 dy
(63) 33.5 y
(64)(63)(33)(67 33)
(2)(123 )
y2
2
33
0
(63) (33.5 y) dy
(33.5)(33)
33
64
123
0
332
2
6463
123
1309 lb
42. Using the coordinate system given in the accompanying figure,
we see that the right-hand edge is x 1 y 2 so the total width
is L( y ) 2 x 2 1 y 2 and the depth of the strip is ( y ). The
force exerted by the water is therefore
0
F w ( y ) 2 1 y 2 dy
1
62.4
0
1
(62.4)
43. (a)
1 y 2 (2 y ) dy 62.4 23 1 y 2
23 (1 0) 41.6 lb
F 62.4
lb
ft 3
3/2 0
1
(8 ft) 25 ft 12480 lb
2
b
(b) The width of the strip is L( y ) 5, the depth of the strip is (8 y ) F w
a
5
5
62.4(8 y )(5) dy 312 (8 y ) dy 312 8 y
0
0
y2
2
5
0
strip
depth
312 40 25
8580 lb
2
Copyright 2018 Pearson Education, Inc.
F ( y) dy
y3
3
0
4
Section 6.5 Work and Fluid Forces
(c) The width of the strip is L( y ) 5, the depth of the strip is (8 y ), the height of the strip is
b
F w
a
strip
depth
312 2 8 y
y2
2
F ( y) dy
0
5/ 2
312 2
0
44. The width of the strip is L( y )
3
4
0
b
F w
a
93.6 12 y
3
strip
depth
5/ 2
F ( y) dy
2
2 3
y3
3
3 3y2 y2 3
62.4 (8 y )(5) 2 dy 312 2
5/ 2
0
40
2
409
2 dy
(8 y ) dy
25
9722.3
4
2
3
3 y , the depth of the strip is (6 y ), the height of the strip is
23 dy 93.63 02 3 12
72 36 12
3 8 3 1571.04 lb
62.4(6 y ) 34 2 3 y
2 3
0
93.6
3
dy
3 6 y 2 y 3 y 2 dy
45. The coordinate system is given in the text. The right-hand edge is x
y and the total width is
L( y ) 2 x 2 y .
1
(a) The depth of the strip is (2 y ) so the force exerted by the liquid on the gate is F w(2 y ) L( y ) dy
0
1
1
1
1
50(2 y ) 2 y dy 100 (2 y ) y dy 100 2 y1/2 y 3/2 dy 100 43 y3/2 52 y 5/2
0
0
0
0
100
(20 6) 93.33 lb
43 52 100
15
1
(b) We need to solve 160 w( H y ) 2 y dy for h. 160 100
0
23H 52 H 3 ft.
46. 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 52 x x 52 y. The total width is L( y ) 2 x 54 y and the
h
depth of the typical horizontal strip at level y is (h y ). Then the force is F w(h y ) L( y ) dy Fmax ,
0
h
where Fmax 6667 lb. Hence, Fmax w (h y ) 54 y dy (62.4)
0
(62.4)
3
max
(Base) 52 h V
1
2
(Base)(Height) 30, where Height h and
52 h2 (30) 12h2 12(9.288)2 1035ft3.
47. The pressure at level y is p( y ) w y the average
pressure is p
b
y2
b1 w 2
0
level
b,
2
2
9.288 ft. The
54 h2 h3 (62.4) 54 16 h3 (10.4) 54 h3 h 3 54 F10.4 3 54 6667
10.4
3
volume of water which the tank can hold is V
1
2
54 0h hy y 2 dy (62.4) 54 hy2
1 b
b 0
p ( y ) dy
1 b w y
b 0
dy
wb b2 wb2 . This is the pressure at
2
which is the pressure at the middle of the
plate.
Copyright 2018 Pearson Education, Inc.
y3
3
h
0
410
Chapter 6 Applications of Definite Integrals
b
b
b
b
y2
48. The force exerted by the fluid is F w(depth)(length) dy w y a dy ( w a) y dy ( w a) 2
0
0
0
0
w
ab 2
2
wb
2
(ab) p Area, where
p is the average value of the pressure.
49. When the water reaches the top of the tank the force on the movable side is
0
0
2 (62.4) 2
4 y 2 ( y )dy
3/2
3/2
2
332.8 ft-lb. The force
( 2 y ) dy (62.4) 23 4 y 2
(62.4) 3 4
2
2
compressing the spring is F 100 x so when the tank is full we have 332.8 100 x x 3.33 ft. Therefore
the movable end does not reach the required 5 ft to allow drainage the tank will overflow.
(62.4)
0
4 y2
1/2
50. (a) Using the given coordinate system we see that the total
width L( y ) 3 and the depth of the strip is (3 y ).
3
3
0
0
Thus, F w(3 y )L( y ) dy (62.4)(3 y ) 3 dy
3
(62.4)(3) (3 y ) dy (62.4)(3) 3 y
0
(62.4)(3) 9
9
2
y2
2
(62.4)(3) 842.4 lb
3
0
9
2
(b) Find a new water level Y such that FY (0.75)(842.4 lb) 631.8 lb. The new depth of the strip is (Y y )
Y
Y
0
0
and Y is the new upper limit of integration. Thus, FY w(Y y )L( y ) dy 62.4 (Y y ) 3 dy
Y
2
2
Y
y2
(62.4)(3) (Y y ) dy (62.4)(3) Yy 2 (62.4)(3) Y 2 Y2 (62.4)(3) Y2 . Therefore,
0
0
2 FY
Y (62.4)(3)
1263.6
6.75 2.598 ft. So, Y 3 Y 3 2.598 0.402 ft 4.8 in
187.2
6.6
MOMENTS AND CENTERS OF MASS
1. A typical piece of length dx has mass ( x) dx x dx its moment about x 0 is
(distance)(mass) x x dx mass M
4
1
x
4
x
1
4
1
x dx
x dx
4
2 x5 2
5
1
14
3
3
14
4
x dx 23 x3 2 23 (4)3 2 23 (1)3 2 16
23 14
; then
3
3
1
52 (4)5 2 52 (1)5 2 143 645 52 3593
2. A typical piece of length dx has mass ( x) dx (1 3x 2 ) dx its moment about x 0 is
3
3
(distance)(mass) x(1 3x 2 ) dx ( x 3x3 ) dx mass M (1 3 x 2 ) dx x x3 30 (30) 60;
3
3
3
then x
3
( x 3 x3 ) dx
3
3
(13 x 2 ) dx
3
x2 3 x 4
2 4 3
60
0
Copyright 2018 Pearson Education, Inc.
Section 6.6 Moments and Centers of Mass
411
3. A typical piece of length dx has mass ( x) dx x 1 dx its moment about x 0 is
x 1 dx 12 x 2 x
0
(distance)(mass) x x 1 dx x 2 x dx mass M
x
3 2
0 x x dx
15
2
13 x3 12 x2 0
3
15
2
3
3
0
92 3 15
; then
2
2 27 9
9 92 15
15 5
4. A typical piece of length dx has mass ( x) dx
8
x3
dx its moment about x 0 is
2
2
(distance)(mass) x 83 dx 82 dx mass M 83 dx 42 1 (4) 3; then
1 x
x
x
x 1
x
2 8
x2
1
dx
3
2
8
x 1
3
13 (4 (8))
4
3
5. A typical piece of length dx has mass ( x) dx its moment about x 0 is (distance)(mass) x ( x) dx
3
2
3
0
0
2
mass M ( x) dx 4 dx 5 dx 4 x 0 5 x 2 8 5 13; then x
2
3
3
0 x ( x ) dx
3
0 ( x ) dx
2
3
0 4 x dx 2 5 x dx
13
452 202 2641
2
3
1 2x 2 5 x 2 1 8
13
2
0
2 13
6. A typical piece of length dx has mass ( x) dx its moment about x 0 is (distance)(mass) x ( x) dx
1
2
2
1
2
mass M ( x) dx (2 x) dx x dx 2 x 12 x 2 12 x 2 2 12 2 12 3; then
0
1
0
0
1
2
x
13
0 x ( x ) dx
2
( x ) dx
0
1
2
0 x (2 x ) dx 1 x x dx
3
23 83 13 1
3 1
3 2
1
2
13 (2 x x 2 ) dx x 2 dx 13 x 2 x3 x3
1
0 1
0
7. 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 0.
It remains to find y
Mx
M
. We model the distribution of mass
with vertical strips. The typical strip has center of mass:
2
( x , y ) x, x 2 4 , length: 4 x 2 width: dx,
area: dA 4 x 2 dx, mass: dm dA 4 x 2 dx
4 x dx 16 x dx. The moment of the
16 x dx 16 x 16 2 16 2
The moment of the strip about the x-axis is y dm
2
plate about the x-axis is M x y dm 2
2
x2 4
2
4
2
4
2
2
2
x5
5 2
25
5
2
2
25
5
. The mass of the plate is M (4 x 2 ) dx 4 x x3 2 8 8
22 32 32
128
5
5
3 2
3
Therefore y
Mx
M
1285 12
323 5
The plate’s center of mass is the point x , y 0, 12
.
5
Copyright 2018 Pearson Education, Inc.
32
3
.
412
Chapter 6 Applications of Definite Integrals
8. Applying the symmetry argument analogous to the one
in Exercise 7, we find x 0. To find y Mx
, we use the
M
vertical strips technique. The typical strip has center of
mass: ( x , y ) x,
25 x 2
2
, length: 25 x , width: dx,
2
area: dA 25 x 2 dx, mass: dm dA 25 x 2 dx.
The moment of the strip about the x-axis is
2
2
y dm 252 x 25 x 2 dx 2 25 x 2 dx.
5
The moment of the plate about the x-axis is M x y dm 2 25 x 2
5
5
2
dx 2
5
5
625 50 x2 x4 dx
5
5
2 625 x 50
x3 x5 2 2 625 5 50
53 55 625 5 10
1 625 83 . The mass of the plate
3
3
3
5
5
3
3
5
is M dm 25 x 2 dx 25 x x3 2 53 53 43 53. Therefore y
5
5
Mx
M
83 10.
53 43
54
The plate’s center of mass is the point ( x , y ) (0, 10).
9. Intersection points: x x 2 x 2 x x 2 0
x(2 x ) 0 x 0 or x 2. The typical vertical
x x 2 ( x )
strip has center of mass: ( x , y ) x,
2
2
x, x2 , length: x x 2 ( x) 2 x x 2 ,
width: dx, area: dA 2 x x 2 dx, mass: dm dA
2
2 x x 2 dx. The moment of the strip about the x-axis is y dm x2 2 x x 2 dx; about the y -axis
2 x x2 dx 2 02 2 x3 x4 dx
2
2
2 x2 x5 2 23 25 2 23 1 54 45 ; M y x dm x 2 x x 2 dx
0
0
2
2
2 4 ; M dm 2 2 x x 2 dx
2 x 2 x3 dx 23 x3 x4 23 23 24 12
0
3
0
0
2
2
M
M
2 x x 2 dx x 2 x3 4 83 43 . Therefore, x M 43 43 1 and y M
0
0
2
0 2
it is x dm x (2 x x 2 ) dx. Thus, M x y dm
4
5
x2
5
4
4
4
3
45
y
43 53 ( x , y ) 1, 53 is the center of mass.
10. Intersection points: x 2 3 2 x 2 3 x 2 3 0
3( x 1)( x 1) 0 x 1. or x 1 Applying the
symmetry argument analogous to the one in Exercise 7,
we find x 0 The typical vertical strip has center of mass:
2 x 2 x 2 3
2
( x , y ) x,
x, x2 3 , length: 2 x 2 x 2 3
2
Copyright 2018 Pearson Education, Inc.
x
Section 6.6 Moments and Centers of Mass
413
3 1 x 2 , width: dx, area: dA 3 1 x 2 dx,
mass: dm dA 3 1 x 2 dx. The moment of the strip about the x-axis is
y dm 32 x 2 3 1 x 2 dx 32 x 4 3 x 2 x 2 3 dx 32 x 4 2 x 2 3 dx;
45 32 ;
x 4 2 x 2 3 dx 32 x5 23x 3x 32 2 15 32 3 3 310
5
15
1
1
1
1
M
M dm 3 1 x 2 dx 3 x x3 3 2 1 13 4 . Therefore, y M 532
85
4
1
1
M x y dm 32
1
5
1
3
3
x
( x , y ) 0, 85 is the center of mass.
y y3
11. The typical horizontal strip has center of mass: ( x , y ) 2 ,
length: y y3 , width: dy, area: dA y y 3 dy,
mass: dm dA y y 3 dy. The moment of the strip about the
y y3
y -axis is x dm 2 y y3 dy 2 y y3
2
dy
1
1
1
y
y
M x y dm y 2 y 4 dy 3 5 13 15 215 ; M y x dm 2 y 2 2 y 4 y 6 dy
0
0
0
2 y 2 2 y 4 y 6 dy; the moment about the x-axis is y dm y y y3 dy y 2 y 4 dy. Thus,
3
y3 2 y5
2 3 5
1
y7
7
0
2
5
4 ;
13 52 17 2 353425715 105
4
16
4 105
12 14 4 . Therefore, x MM 105
y
1
y2
M dm ( y y )3 dy 2
0
and y
Mx
M
1
y4
4
0
16 , 8 is the
215 4 158 ( x , y ) 105
15
center of mass.
12. Intersection points: y y 2 y y 2 2 y 0
y ( y 2) 0 y 0 or y 2 The typical horizontal
y 2 y y
strip has center of mass: ( x , y )
,
2
y2
y 2 ,
y ,
area: dA 2 y y 2 dy, mass: dm dA 2 y y 2 dy.
The moment about the y -axis is x dm 2 y 2 2 y y 2 dy 2 2 y3 y 4 dy; the moment about the x-axis
2
2
y
2y
is y dm y 2 y y 2 dy 2 y 2 y3 dy. Thus, M x y dm 2 y 2 y3 dy 3 4
0
0
length: y y 2 y 2 y y 2 , width: dy,
3
163 164 1612 (4 3) 43 ;
2
y4
M y x dm 2 2 y 3 y 4 dy 2 2
0
Copyright 2018 Pearson Education, Inc.
y5
5
2
0
2 8 32
5
4
414
Chapter 6 Applications of Definite Integrals
2
40532 45 ;
x
My
M
2
M dm 2 y y 2 dy y 2
0
45 43 53
and y
Mx
M
y3
3
2
0
4 83 43 . Therefore,
43 43 1 ( x , y ) 53 , 1 is the center of mass.
13. Applying the symmetry argument analogous to the one used in
Exercise 7, we find x 0. The typical vertical strip has center
of mass: ( x , y ) x,
cos x
2
, length: cos x, width: dx,
area: dA cos x dx, mass: dm dA cos x dx. The moment
of the strip about the x-axis is y dm cos2 x cos x dx
2 cos 2 x dx 2
1 cos 2 x
2
dx
(1 cos 2 x) dx; thus,
4
/2
/2
x sin 2 x
4 2 0 2 64 ; M dm
(1
cos
2
x
)
dx
4
2 /2
/2 4
M x y dm
/2
/2
/2
M
8 ( x , y ) 0, 8 is the center of mass.
cos x dx sin x /2 2 . Therefore, y Mx 4
2
14. Applying the symmetry argument analogous to the one used in
Exercise 7, we find x 0. The typical vertical strip has center
sec2 x
of mass: ( x , y ) x, 2 , length: sec2 x , width: dx,
area: dA sec2 x dx, mass: dm dA sec2 x dx. The moment
2
about the x-axis is y dm sec2 x ( sec2 x) dx
2 sec4 x dx. M x
/4
/4
y dm 2
/4
/4
sec4 x dx 2
/4
/4
tan x 2 sec2 x dx 2 /4 sec2 x dx 2 (tan3x )
/4
2
/4
/4
3
tan 2 x 1sec2 x dx
/4
/4
2 tan x /4
/4
/4
/4
sec2 x dx tan x /4 1 (1) 2 .
2 13 13 2 1 (1) 3 43 ; M dm
/4
M
21 32 ( x , y ) 0, 32 is the center of mass.
Therefore, y Mx 43
15. (a) Since the plate is symmetric about the line x y and its
density is constant, the distribution of mass is symmetric about
this line. This means that x y The typical vertical strip has
2
center of mass: ( x , y ) x, 92 x
length:
,
9 x 2 width: dx, area: dA 9 x 2 dx,
mass: dm dA 9 x 2 dx. The moment about the x -axis
is y dm
9 x 2
2
2
2
9 x dx 2 9 x dx
Copyright 2018 Pearson Education, Inc.
Section 6.6 Moments and Centers of Mass
3
Thus, M x y dm 2 9 x 2 dx 2
0
415
3
9 x x3 (27 9) 9 ; M dm dA dA
3 0
2
4 4
(Area of a quarter of a circle of radius 3) 94 9
. Therefore, y Mx
(9 ) 9
4
M
( x , y ) 4 , 4 is the center of mass.
(b) Applying the symmetry argument analogous to the
one used in Exercise 7, we find that x 0. The
typical vertical strip has the same parameters as in
3
part (a). Thus, M x y dm 2 9 x 2 dx
3
3
2 2 9 x 2 dx 2(9 ) 18 ;
0
M dm dA
M
2 4,
dA (Area of a semi-circle of radius 3) 92 9
. Therefore, y Mx (18 ) 9
2
the same y as in part (a) ( x , y ) 0, 4 is the center of mass.
16. By symmetry, x 1.
y
21
(2 x x 2 )2
0 2
Mx
2
0
3 x4
2
(2 x 2 4 x )2 dx
6 x 3 6 x 2 dx
3 x5 3 x 4 2 x 3
10
2
2
0
0
8
5
1
2
0
3 x
2
4 x 2 dx
x3 2 x 2 2 x
y
1, 25
2
2
M (2 x x 2 ) (2 x 2 4 x ) dx
0
y 2x x2
1
2
0
4
8
Mx 5
2
M
4
5
16
17. M y x
1
x dx 116 x3 2 dx 25 x5 2 1
16
25 165 2 15 2 25 1024 1 2046
5
23 x3 2 1
1
23 163 2 13 2 23 64 1
M
16
x dx
16
42
M
x My
2046
5
42
341
35
Copyright 2018 Pearson Education, Inc.
2
x
y 2x2 4 x
416
Chapter 6 Applications of Definite Integrals
16 1
1 2
Mx
2
y
x
2
16
dx 2 x dx
1
12 x2 1
16
2
2562 12
255
4
Mx
M
255
4
42
85
56
18. Applying the symmetry argument analogous to the one
used in Exercise 7, we find that y 0. The typical
vertical strip has center of mass: ( x , y ) x,
area: dA
2
x3
1
x3
( x, 0), length:
a
2
1
x3
2
, width: dx ,
x
a 2
1 x2
( a 2 1)
2
x3
dx, mass: dm dA 23 dx. The moment about the y -axis is x dm x 23 dx 22 dx. Thus,
M y x dm
1
x3
1
x3
x
1
2 ( a 1)
a
a
2 ( a 1)
dx 2 1x 2 1a 1 a ; M dm
. Therefore, x
My
M
a 2
1 x3
a2
( a 2 1)
a2a1 ( x , y )
x
a
dx 12 12 1
a
x 1
x 2.
a2a1 , 0 . Also, alim
19. Intersection points: x5 x 4
x5 x 4 x 4 ( x 1) 0 x 0 or x 1.
The typical vertical strip has center of mass:
( x , y ), x,
,
x 4 x5
2
length: x 4 x5 , width: dx; area: dA ( x 4 x5 )dx,
mass: dm dA ( x 4 x5 )dx. The moment of the
strip about the x-axis is
y dm
x 4 x5
2
M x y dm
2
(x
1
4
0 2 ( x
19 111 99 .
x5 )dx 2 ( x8 x10 ) dx. Thus
8
1
1 x11
x10 ) dx 2 19 x9 11
0
The moment of the strip about the y-axis is x dm x( x 4 x5 )dx ( x5 x6 )dx. Thus
1
1
M y x dm ( x5 x 6 ) dx 16 x6 17 x7
0
0
15 x5
15 16 30 . Therefore, x
My
M
30
42
is the center of mass. Since ( x )4
75
4
1
1 x6
6
0
x , y 75 , 10
33
16 17 42 ;
5
7
1
M dm ( x 4 x5 ) dx
0
and y
Mx
M
30 10
99
33
0.260 y 10
0.303 the center of mass lies
33
outside the region.
Copyright 2018 Pearson Education, Inc.
Section 6.6 Moments and Centers of Mass
20. Intersection points:
1
2
1 x2
4
x x
x x2 4 x
x 2 4 x x( x 4) 0 x 0 or x 4. The typical
vertical strip has center of mass: ( x , y ) x,
x 12 x, width: dx; area: dA
length:
mass: dm dA
x 12 x
,
2
x 12 x dx,
x 12 x dx.
(a) The moment of the strip about the x-axis is y dm
x 12 x
2
x 12 x dx 2 x 14 x 2 dx. Thus
4
4
1 x3 8 16 4 .
M x y dm 2 x 14 x 2 dx 2 12 x 2 12
3
3
0 2
0
(b) The moment of the strip about the y-axis is x dm x
x 12 x dx x3 2 12 x 2 dx. Thus
645 646 1532 .
The moment of the strip about the line x 5 is (5 x ) dm (5 x) x 12 x dx
4
4
M y x dm x3 2 12 x 2 dx 52 x5 2 16 x3
0
0
(c)
4
5 x1 2 52 x x3 2 12 x 2 dx. Thus M x 5 (5 x ) dm 5 x1 2 52 x x3 2 12 x 2 dx
0
4
68 .
10
x3 2 54 x 2 52 x5 2 16 x3 80
20 64
64
3
5
6 15
3
0
(d) The moment of the strip about the line x 1 is ( x 1) dm ( x 1)
4
x 12 x dx
x1 2 12 x x3 2 12 x 2 dx. Thus M x 1 ( x 1) dm x1 2 12 x x3 2 12 x 2 dx
0
4
23 x3 2 14 x 2 52 x5 2 16 x3
0
163 4 645 646 1552 .
(e) The moment of the strip about the line y 2 is (2 y ) dm 2
4
x 12 x
2
x 12 x dx
2 x1 2 32 x 18 x 2 dx. Thus M y 2 (2 y ) dm 2 x1 2 32 x 18 x 2 dx
0
4
1 x3
43 x3 2 34 x 2 24
0
323 12 6424 34 .
(f) The moment of the strip about the line y 3 is ( y 3) dm
x 12 x
2
3
x 12 x dx
3 x1 2 x 18 x 2 dx. Thus
4
4
1 x3 16 8 64 16 .
M y 3 ( y 3) dm 3 x1 2 x 18 x 2 dx 2 x3 2 12 x 2 24
24
3
0
0
4
(g) M dm
0
(h) x
My
M
32
15
4
3
4
x 12 x dx 23 x3 2 14 x 2
0
85 and y
Mx
M
4
3
4
3
1
163 4 43
85 , 1 is the center of mass.
Copyright 2018 Pearson Education, Inc.
417
418
Chapter 6 Applications of Definite Integrals
2
2 x 2
1 2
21. M x y dm
x dx
2
22 dx 12
1
x
2
2
x2
x
2
2
2
22 dx 2 x 2 dx 2 x 1 2 12 (1)
1
1 x
1
2
2 12 1; M y x dm x
1
dx
2
x2
x2 dx 212 x dx 2 x2 1 2 2 12 4 1 3;
2
1
2
2
x x2
2
M dm
2
2
2 dx 2 x 1 2(2 1) 2. So x
2
1
My
M
3
2
and y
Mx
M
1
dx x dx
2 2
2
x2
1
2
x2
12 ( x , y ) 32 , 12 is the center of mass.
22. We use the vertical strip approach:
x x2 dx
1
1
12 x 2 x 4 12 x dx 6 x3 x5 dx
0
0
2
1 x x
2
0
M x y dm
1
4
6
6 x4 x6 6
0
14 16 64 1 12 ;
1
1
1
12 1. So
M dm x x 2 dx 12 x 2 x3 dx 12 x3 x4 12 13 14 12
0
0
0
1
4
5
1
1
1
M y x dm x x x 2 dx x 2 x3 12 x dx 12 x3 x 4 dx 12 x4 x5 12 14 15
0
0
0
0
12
3;
20 5
x
My
M
3
4
M
and y Mx 12 53 , 12 is the center of mass.
3
5
b
23. (a) We use the shell method: V 2 shell
radius
a
4 1/2
x dx
1
16
16
4
2 x3/2
3
1
4
4 x
4
4
shell
height dx 1 2 x x x dx 16 1 x dx
16 23 8 23 323 (8 1) 224
3
(b) Since the plate is symmetric about the x-axis and its density ( x) 1x is a function of x alone, the
distribution of its mass is symmetric about the x-axis. This means that y 0. We use the vertical strip
dx x dx 8 x dx 8 2x
dx 8 dx 8 x
dx 8 2 x
approach to find x : M y x dm x
1
4
4
8(2 2 2) 16; M dm 4
1 x
8 1 (2) 8. So x
My
M
4
x
4
x
4
4
x
1
4
1
1
x
4 1/2
1
8 1
x x
1
x
4 3/2
1
16
2 ( x , y ) (2, 0) is the center of mass.
8
Copyright 2018 Pearson Education, Inc.
1/2 4
1
1/2 4
1
Section 6.6 Moments and Centers of Mass
419
(c)
b
4
24. (a) We use the disk method: V R ( x) dx
2
a
1
[1 4] 3
dx 4
4 2
x dx
1
4
x2
(b) We model the distribution of mass with vertical strips: M x y dm
4
4
4 1x 4 41 (1)
2x
1 2
1
2x dx 14 x2
x dx
2
4
4
3/ 2
4
4
4
2 x 3/2 dx 2 2 2 1 (2) 2; M y x dm x 2x dx 2 x1/2 dx 2 2 x3
1
x 1
1
1
1
4
4
4
4
2 16
23 28
; M dm 2x dx 2 xx dx 2 x 1/2 dx 2 2 x1/2 2(4 2) 4.
3
3
1
1
1
1
So x
My
M
M
43 73 and y Mx 24 12 ( x , y ) 73 , 12 is the center of mass.
28
(c)
25. The mass of a horizontal strip is dm dA 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
L
b
h y
h
L bh (h y ). Thus,
h
3
3
h
h
hy 2 y3
M x y dm y bh (h y ) dy hb hy y 2 dy hb 2 3 hb h2 h3
0
0
0
bh 2
12 13 bh6
2
;
h
2
h
h
y2
M dm bh (h y ) dy hb (h y ) dy hb hy 2 hb h 2 h2 2bh . So
0
0
0
2
M
y Mx bh
6
2
bh
h
3
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.
Copyright 2018 Pearson Education, Inc.
420
Chapter 6 Applications of Definite Integrals
26. From the symmetry about the y -axis it follows that x 0. It also
follows that the line through the points (0, 0) and (0, 3) is a
median y 13 (3 0) 1 ( x , y ) (0, 1).
27. From the symmetry about the line x y it follows that x y . It
also follows that the line through the points (0, 0) and
median y x 23
12 0
12 , 12 is a
13 ( x , y ) 13 , 13 .
28. From the symmetry about the line x y it follows that x y . It
also follows that the line through the point (0, 0) and
median y x
2 a
3 2
0
a2 , a2 is a
13 a ( x , y ) a3 , a3 .
29. The point of intersection of the median from the vertex (0, b) to
a2 y (b 0) 13 b3 and
( x , y ) a3 , b3 .
the opposite side has coordinates 0,
x a2 0 23 a3
30. From the symmetry about the line x
it follows that x a2 . It
a
2
a2 , 0 and a2 , b is
( x , y ) a2 , b3 .
also follows that the line through the points
a median y 13 (b 0)
31.
b
3
y x1/2 dy 12 x 1/2 dx ds (dx)2 (dy )2
Mx
2
0
2
x 1 41x dx
0
1 41x dx;
3/2 2
x 14 dx 23 x 14
0
3/2
3/2 2 9 3/2
3/2 2 27 1
13
23 2 14
14
3 4
14
3 8 8 6
32.
y x3 dy 3 x 2 dx dx (dx)2 3 x 2 dx
2
1 9 x 4 dx;
1
M x x3 1 9 x 4 dx;
0
[u 1 9 x 4 du 36 x3 dx
1 du
36
x3 dx;
10 1 1/2
u du
1 36
x 0 u 1, x 1 u 10] M x
10
2 u 3/2 103/2 1
36
54
3
1
Copyright 2018 Pearson Education, Inc.
Section 6.6 Moments and Centers of Mass
0
0
33. From Example 4 we have M x a(a sin )(k sin )d a 2 k sin 2 d
2
2
a 2 k (1 cos 2 )
2 0
2
421
d
a2k sin22 a 2k ; M y a(a cos )(k sin ) d a 2 k sin cos d a2k sin 2 0;
0
0
0
0
M
M ak sin d ak cos 0 2ak . Therefore, x My 0 and
0
2
M
y Mx a 2k
1
2 ak
a4 0, a4 is the center of mass.
34. M x y dm (a sin ) a d
0
0
a2 sin 1 k cos d
a2
/2
a2
/2
0
0
(sin )(1 k cos ) d a 2
/
sin d a 2 k
/2
/2
sin cos d a 2 / 2 sin d a 2 k
/2
a 2 k sin2
0
2
a 2 0 (1) a 2 k
/2
0
2
12 0 a2 (1) 0 a2 k 0 12 a2 a2k a2 a2k 2a2 a2 k a2 (2 k );
2
0
0
(cos )(1 k cos ) d a 2
/2
2
a2 cos 1 k cos d
(cos )(1 k cos )d
/2 1 cos 2
1 cos 2
d a 2
cos d a 2 k
d
2
2
/2
/2
2
2
sin 2
sin 2
a 2 sin 0 a2k 2
a 2 sin a2k 2
0
a2
/2
sin cos d
a 2 cos /2 a 2 k sin2
/2
M y x dm (a cos ) a d
a2
/2
0
a 2 cos 0
(sin )(1 k cos ) d
0
cos d a 2 k
0
2
2
2
2
a 2 (1 0) a2k 2 0 (0 0) a 2 (0 1) a2k ( 0) 2 0 a 2 a 4k a 2 a 4k 0;
0
0
M a d a
/2
a k sin 0
/2
1 k cos d a 0
/2
(1 k cos ) d
a (2 k )
a (2 k )
is the center of mass.
a ( 2 k ) 2k 0, 2a2ka
k
a k sin /2 2 k 0 a ( 0) 2 k a2 ak a 2 k a 2ak
M
M
a ( 2k ). So x My 0 and y Mx
35.
(1 k cos ) d a
2
f ( x) x 6, g ( x) x 2 , f ( x) g ( x) x 6 x 2
x 2 x 6 0 x 3, x 2; 1
3
3
M ( x 6) x 2 dx 12 x 2 6 x 13 x3
2
2
92 18 9 2 12 83 125
6
3
3
2
6 3 x 2 6 x x3 dx 6 1 x3 3 x 2 1 x 4
1
x 125/6
2 x( x 6) x dx 125
2
125 3
4
2
6 9 27 81 6 8 12 4 1 ;
125
4
125
3
2
Copyright 2018 Pearson Education, Inc.
422
Chapter 6 Applications of Definite Integrals
3
3
2
2 2
3 3 x 2 12 x 36 x 4 dx 3 1 x3 6 x 2 36 x 1 x5
1
y 125/6
2 12 ( x 6) x dx 125
2
125 3
5
2
3 9 54 108 243 3 8 24 72 32 4 1 , 4 is the center of mass.
125
5
125
3
5
2
36.
f ( x) 2, g ( x) x 2 ( x 1), f ( x) g ( x ) 2 x 2 ( x 1)
x3 x 2 2 0 x 1; 1
1
1
M 2 x 2 ( x 1) dx 2 x3 x 2 dx
0
0
1
2 x 14 x 4 13 x3 2 14 13 0 17
;
12
0
1
2
1
12 1 2 x x 4 x3 dx
x 17/12
0 x 2 x ( x 1) dx 17
0
1
12 x 2 1 x5 1 x 4 12 1 1 1 0 33 ;
17
5
4
5 4
85
0 17
11 2
2
02
1
y 17/12
2
1
6 1 4 x 6 2 x5 x 4 dx 6 4 x 1 x 7 1 x 6 1 x5
x 2 ( x 1) dx 17
17
7
3
5
0
0
6 4 1 1 1 0 698 33 , 698 is the center of mass.
17
7 3 5
595
85 595
37.
f ( x) x 2 , g ( x) x 2 ( x 1), f ( x) g ( x)
x 2 x 2 ( x 1) x3 2 x 2 0 x 0, x 2; 1
2
2
M x 2 x 2 ( x 1) dx 2 x 2 x3 dx
0
0
2
23 x3 14 x 4 16
4 0 43 ;
3
0
1 2 x x 2 x 2 ( x 1) dx 3 2 2 x3 x 4 dx
x 4/3
0
4 0
2
34 12 x 4 15 x5 34 8 32
0 56 ;
5
0
2
2
2
2
1 2 1 x2
y 4/3
x 2 ( x 1) dx 83 2 x5 x6 dx 83 13 x6 17 x7 83 64
128
0 87 65 , 87
0 2
3
7
0
0
is the center of mass.
38.
f ( x) 2 sin x, g ( x) 0, x 0, x 2 ; 1;
M
2
0
2 sin x dx 2 x cos x 02
(4 1) (0 1) 4 ;
2
2
x 41 x 2 sin x 0 dx 41 2 x x sin x dx
0
0
2
2
2
2
41 2 x dx 41 x sin x dx 41 x 2 41 sin x x cos x 0
0
0
0
41 4 2 0 41 (0 2 ) 0 221 ;
2
y 41 12 (2 sin x)2 (0) 2 dx
0
Copyright 2018 Pearson Education, Inc.
Section 6.6 Moments and Centers of Mass
423
2
2
2
81 4 4sin x sin 2 x dx 81 4 4 sin x dx 81 sin 2 x dx
0
0
0
2
2 1cos 2 x
2 dx
0 4 4 sin x dx 81 0
81
2
2
2
81 4 x 4 cos x 0 161 dx 161 cos 2 x dx
0
0
2
4
2
[u 2 x du 2dx, x 0 u 0, x 2 u 4 ] 81 4 x 4 cos x 0 161 x 0 321 cos u du
0
2
2
4
81 4 x 4 cos x 0 161 x 0 321 sin u 0 81 (8 4) 81 (0 4) 161 (2 ) 0 0 98 221 , 89
is the center of mass.
39. 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 (dx)2 (dy )2 . This implies that M x y ds,
M y x ds and M ds. If is constant, then x
My
M
M
x ds
x ds
y ds
y ds
length and y Mx
length .
ds
ds
40. Applying the symmetry argument analogous to the one used in Exercise 7, we find that x 0. The typical
a x2
2
2
vertical strip has center of mass: ( x , y ) x, 24 p , length: a 4x p , width: dx, area : dA a 4x p dx,
2
2 pa 1
2 pa 2
mass: dm dA a 4x p dx. Thus, M x y dm
2 pa
4
5
2
a x 2 dx 2 a 2 x x 2
2 pa
16 p
80 p 2
2
2 pa
x2
4p
2 pa
pa
a a dx
5
2 2 a 2 x x 2
80 p 0
2
16 2a 2 pa 64 8a
2a 2 pa 1 16
2a 2 pa 8080
80
80
5
2 pa
ax 12x p
2
3
8a pa
.
3
3
pa
So y
2 pa
2 ax 12x p
0
Mx
M
8a 2
5
pa
3
8a
pa
x2
4p
25 p 2 a 2
2a 2 pa
80 p 2
; M dm
2 pa
2 pa
2
pa
a 4x p dx
23 pa pa
4 4a pa 12 4
2 2a pa 12 p 4a pa 1 12
12
pa
3
5 a, as claimed
41. The centroid of the square is located at (2, 2). The volume is V (2 )( y )( A) (2 )(2)(8) 32 and the
surface area is S (2 )( y )( L) (2 )(2) 4 8 32 2 (where
42. The midpoint of the hypotenuse of the triangle is
8 is the length of a side).
32 , 3 y 2 x
is an equation of the median the line y 2 x contains the
centroid. The point
32 , 3 is 3 25
units from the origin the
x-coordinate of the centroid solves the equation
x 32
2
(2 x 3) 2 25 x 2 3 x 94 4 x 2 12 x 9
5
4
5 x 2 15 x 9 1 x 2 3x 2 ( x 2)( x 1) 0 x 1 since the centroid must lie inside the triangle
y 2. By the Theorem of Pappus, the volume is V (distance traveled by the centroid)(area of the
region) 2 (5 x ) 12 (3)(6) (2 )(4)(9) 72
Copyright 2018 Pearson Education, Inc.
424
Chapter 6 Applications of Definite Integrals
43. The centroid is located at (2, 0) V (2 )( x )( A) (2 )(2)( ) 4 2
44. We create the cone by revolving the triangle with
vertices (0, 0), (h, r ) and (h, 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 2 yL
2
2r
h 2 r 2 r r 2 h 2 . 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
r
2h
x. The x-coordinate of the centroid solves the equation
( x h)2 2rh x 2r
2
2
13 h 2 r4
2 r 2 4 h 2
2
2
2
2
2
4h 2r x 2 4 h2h r x r4
0 x 23h or 43h x 23h , since the centroid must lie inside
9
4h
r
2h
the triangle y
x 3r . By the Theorem of Pappus, V 2 3r 12 hr 13 r 2 h.
45. S 2 y L 4 a 2 (2 y )( a ) y 2a , and by symmetry x 0
46. S 2 L 2 a 2a ( a ) 2 a 2 ( 2)
47. V 2 y A 43 ab 2 (2 y ) 2ab y 34b and by symmetry x 0
2
a
48. V 2 A V 2 a 34a 2a
3
(3 4)
3
49. V 2 A (2 ) (area of the region) (distance from the centroid to the line y x a). We must find the
to y x a. The line containing the centroid and perpendicular to y x a has slope
1 and contains the point 0, 34a . This line is y x 34a . The intersection of y x a and y x 34a is
the point 4a 63a , 4a63a . Thus, the distance from the centroid to the line y x a is
distance from 0, 34a
4a63a 34a 64a 36a
2
2
2(4 a 3a )
6
2(4 a 3a ) a 2
2 a3 (4 3 )
V (2 )
6
6
2
50. The line perpendicular to y x a and passing through the centroid 0, 2a has equation y x 2a . The
intersection of the two perpendicular lines occurs when x a x 2a x
y
2 a a
2
a (2 )
.
2
. Thus the distance from the centroid to the line y x a is
Therefore, by the Theorem of Pappus the surface area is S 2
2 a a
2
x 2 a2a
2a2 a 0 2a2 a 22a
2
2
a (2 )
( a )
2
2 a 2 (2 ).
51. If we revolve the region about the y -axis: r a, h b A 12 ab,V 13 a 2b, and x . By the Theorem
of Pappus:
1 a 2b
3
2 x 12 ab x a3 ; If we revolve the region about the x-axis:
Copyright 2018 Pearson Education, Inc.
Section Chapter 6 Practice Exercises
425
r b, h a A 12 ab, V 13 b 2 a, and y . By the Theorem of Pappus:
1 b2 a
3
2 y 12 ab y b3 a3 , b3
of mass.
is the center
52. Let O(0, 0), P(a, c), and Q(a, b) be the vertices of the given triangle. If we revolve the region about the
x-axis: Let R be the point R (a, 0). The volume is given by the volume of the outer cone, radius RP c,
minus the volume of the inner cone, radius RQ b, thus V 13 c 2 a 13 b 2 a 13 a c 2 b 2 , the area is
given by the area of triangle OPR minus area of triangle OQR, A
1 ac 1 ab
2
2
1 a (c b),
2
and y . By
a c 2 b 2 2 y 12 a (c b) y c 3b ; If we revolve the region about the
y -axis: Let S and T be the points S (0, c) and T (0, b), respectively. Then the volume is the volume of the
cylinder with radius OR a and height RP c, minus the sum of the volumes of the cone with radius
SP a and height OS c and the portion of the cylinder with height OT b and radius TQ a
the Theorem of Pappus:
1
3
with a cone of height OT b and radius TQ a removed. Thus
V a 2 c 13 a 2 c a 2b 13 a 2 b 23 a 2 c 23 a 2 b 23 a 2 (a b). The area of the triangle is the
same as before, A 12 ac 12 ab 12 a(c b), and x . By the Theorem of Pappus:
2
3
CHAPTER 6
1.
2 a ( a b )
2 a ( a b )
a 2 (a b) 2 x 12 a(c b) x 3(c b) 3(c b ) , c 2b is the center of mass.
PRACTICE EXERCISES
A( x ) 4 (diameter) 2 4
x x2
2
4 x 2 x x 2 x 4 ; a 0, b 1
b
1
V A( x)dx 4 x 2 x5/2 x 4 dx
a
0
1
2
5
4 x2 74 x 7/2 x5 4 12 74 15
0
(35 40 14) 9
470
280
2.
2
A( x) 12 (side)2 sin 3 43 2 x x
43 4 x 4 x x x 2 ; a 0, b 4
b
4
V A( x)dx 43 4 x 4 x3/2 x 2 dx
a
0
4
3
43 2 x 2 85 x5/2 x3 43 32 8532 64
3
0
32 3
4
1 85 23 8153 (15 24 10) 8153
Copyright 2018 Pearson Education, Inc.
426
3.
Chapter 6 Applications of Definite Integrals
A( x ) 4 (diameter)2 4 (2sin x 2 cos x)2
4 4 sin 2 x 2sin x cos x cos 2 x (1 sin 2 x);
b
a 4 , b 54 V A( x) dx
a
5 /4
/4
(1 sin 2 x) dx x cos22 x
5 /4
/4
cos 5 x
cos
54 2 2 4 2 2 2
4.
A( x ) (edge)2
6 x
36 24
0
b
A( x ) dx
a
6
2
2
0
6 x
4
36 24 6 x 36 x 4 6 x3/2 x 2 ; a 0, b 6 V
6
3
6 x 36 x 4 6 x3/2 x 2 dx 36 x 24 6 23 x3/2 18 x 2 4 6 52 x5/2 x3
0
3
216 16 6 6 6 18 62 85 6 6 62 63 216 576 648 1728
72 360 1728
180051728
5
5
5.
4
4
0
4x x
5/2
2
3
4
4 x
2
2
4
b
x ; a 0, b 4 V
4 4 x x5/2 16
A( x) dx
4
a
32
4
1 78 52
72
35
A( x) 12 (edge)2 sin 3
x 4 dx 2 x 2 2 x 7/2 x5 32 32 8 2 32
16
4
7
516 0
4
7 5
(35 40 14)
835
6.
A( x ) 4 (diameter)2 4 2 x x4
72
5
3
4
2 x 2 x
2
4 3x; a 0, b 1
b
1
1
V A( x) dx 4 3 x dx 2 3 x 2 2 3
0
a
0
7. (a) disk method:
b
1
2
V R ( x) dx 3x 4 dx
2
1
a
1
1
9 x8 dx x9 2
1
1
(b) shell method:
b
V 2
a
1
1 5
shell
4
x
shell
radius height dx 0 2 x 3 x dx 2 30 x dx 2 3 6
0
6
1
Note: The lower limit of integration is 0 rather than 1.
(c) shell method:
b
V 2
a
1
shell
4
3x
shell
radius height dx 2 1 (1 x ) 3 x dx 2 5
3
1
6
x2 2
1
(d) washer method:
Copyright 2018 Pearson Education, Inc.
53 12 53 12 125
Chapter 6 Practice Exercises
b
R ( x ) 3, r ( x) 3 3x 4 3 1 x 4 V
a
R ( x )
2
dx
r ( x)
2
1
1
9 9 1 x 4
dx
2
1
5
9
1
1
9 1 1 2 x 4 x8 dx 9 2 x 4 x8 dx 9 25x x9 18 52 19 2513
1
1
1
8. (a) washer method:
4
x3
R( x)
, r ( x)
b
V
1
2
a
R ( x)
2
r ( x)
2
dx dx
2
1
2
4
x3
1 2
2
1 1 16 1 ( 2 10 64 5)
516
1 16
14 10
5
2
5
4
20
32 2
(b) shell method:
2
V 2 x
1
4
x3
(c) shell method:
b shell
a radius
V 2
2
1
3
2
x2
4 1
57
20
54 52
3
2
2 (1 2 2 1) 4 4 1 14 32
2
7 2
2
2
4
x3
2
2
49
4
16 1 2x 3 x 6 dx
49
4
5
16 x x 2 x5
1
49
4
16 2 14 5132 1 1 15
1
2
49
4
16
14 1601 15 49 16 (40 1 32) 494 7110
103
20
9. (a) disk method:
1
5
2
2
5
x 1 dx ( x 1) dx x2 x
1
1
(b) washer method:
V
5
d
R ( y ) 5, r ( y ) y 2 1 V
c
2
2
2
4x
1
R( x) r( x) dx
4 dx
b
V
16 x 5
5
26
5
2
2 8
4
1
4 1 2x dx
shell
height dx 2 1 (2 x ) 2 dx 2 1
x
x
x
2 42 4x x
x
(d) washer method:
a
2
2
12 dx 2 4 x 1 x4 2 42 1 4 14 2
1
427
2
r ( y)
2
dy
2
2
2
25 y 2 1 dy
2
y5
24 y 4 2 y 2 dy 24 y 5 23 y 3
2
2
25 y 4 2 y 2 1 dy
R( y )
252 5 12 1 242 4 8
2
2 24 2 32
23 8 32 3 52 13
5
32
15
(45 6 5) 1088
15
Copyright 2018 Pearson Education, Inc.
428
Chapter 6 Applications of Definite Integrals
(c) disk method:
R( y ) 5 y 2 1 4 y 2
d
2
V R ( y ) dy 4 y 2
2
2
c
2
8y
16 8 y 2 y 4 dy 16 y 3
2
2
2 32
64
15
64
3
32
5
64 1
(15 10 3)
2
3
1
5
dy
3
y5
5
2
2
512
15
10. (a) shell method:
d
V 2
c
shell
shell
radius height dy
4
4
y2
y3
2 y y 4 dy 2 y 2 4 dy
0
0
4
y3 y 4
2 3 16 2
0
(b) shell method:
b
V 2
a
2
643 644 212 64 323
4
shell
shell
radius height dx 0 2 x 2
2 x3/2 x 2 dx 2 54 x5/2 x3
0
0
x x dx 2
54 32 643 12815
(c) shell method:
b
V 2
a
4
shell
shell
radius height dx 0 2 (4 x ) 2
4
3
x3/2 2 x 2 54 x5/2 x3 2
2 16
3
0
(d) shell method:
d
V 2
c
4
3
4
4
x x dx 2 8 x1/2 4 x 2 x3/2 x 2 dx
0
163 8 32 54 32 643 64 43 1 54 23 64 1 54 645
4
4
y
shell
2
2 y
shell
radius height dy 0 2 (4 y ) y 4 dy 2 0 4 y y y 4 dy
2
3
4
y
y
4 y 2 y 2 4 dy 2 2 y 2 23 y 3 16 2 32 32 64 16 32 2 83 1 323
0
0
2
3
4
4
11. disk method:
R( x) tan x, a 0, b 3 V
/3
0
12. disk method:
0
0
V (2 sin x)2 dx
tan 2 x dx
/3
0
sec2 x 1 dx tan x x0 /3
3 3
4 4sin x sin 2 x dx 0 4 4sin x 1cos2 2 x dx
4 x 4 cos x 2x sin42 x 4 4 2 0 (0 4 0 0)
0
92 8 2 (9 16)
13. (a) disk method:
x2 2 x
0
V
2
2
x 4 4 x3 4 x 2 dx x5 x 4 43 x3 32
16 32
5
3
0
0
dx
2
5
2
(6 15 10) 16
16
15
15
Copyright 2018 Pearson Education, Inc.
3
Chapter 6 Practice Exercises
429
(b) washer method:
2
2
2
2
2
x 15
V 12 x 2 2 x 1 dx dx ( x 1)4 dx 2 5 2 52 85
0
0
0
0
(c) shell method:
2
2
shell
2
2
shell
radius height dx 2 0 (2 x ) x 2 x dx 2 0 (2 x ) 2 x x dx
2
2
2
2 4 x 2 x 2 2 x 2 x3 dx 2 x3 4 x 2 4 x dx 2 x4 43 x3 2 x 2 2 4 32
8
3
0
0
0
b
V 2
a
4
23 (36 32) 83
(d) washer method:
2
2
4 4 x 2 8 x x 4 4 x3 4 x 2 dx 8 x 4 4 x3 8 x 4 dx 8
0
0
2
2
2
2
2
V 2 x 2 2 x dx 22 dx 4 4 x 2 2 x x 2 2 x dx 8
0
0
0
2
5
x5 x 4 4 x 2 4 x 8
0
14. disk method:
V 2
/4
0
4 tan 2 x dx 8
/4
0
325 16 16 8 8 5 (32 40) 8 725 405 325
sec2 x 1 dx 8 tan x x0 /4 2 (4 )
15. The material removed from the sphere consists of a cylinder
and two “caps.” From the diagram, the height of the cylinder
3 22 , i.e. h 1. Thus
2
Vcy1 (2h) 3 6 ft 3 . To get the volume of a cap,
is 2h, where h 2
2
2
use the disk method and x 2 y 2 22 : Vcap x 2 dy
1
2
2
y3
4 y 2 dy 4 y 3 8 83 4 13
1
1
53 ft 3 . Therefore, Vremoved Vcy1 2Vcap 6 103
28
3
ft 3 .
2
x
16. We rotate the region enclosed by the curve y 12 1 4121
and the x-axis around the x-axis. To find the
b
volume we use the disk method: V R ( x) dx
a
12
11/2
11/2
264
3
1 dx 12 x
4 x2
121
11/2
4 x3
363 11/2
2
11/2
2
2
11/2
2
x
4x
12 1 4121
dx 11/212 1 121 dx
11/2
24 11
2
3634 112 132 1 3634 114 132 1 13
3
88 276 in 3
Copyright 2018 Pearson Education, Inc.
2
430
17.
Chapter 6 Applications of Definite Integrals
h
R ( x) b h a x a V R ( x) dx
h
0
bha
b h a
18.
2
1
3
h ba
h
0
0
2
x a dx
x 2a b h a x a 2 dx
2 2
2 x3
3
h
2
2a b h a x2 a 2 x
0
b a 2 h a(b a)h a 2 h
( a 2 ab b 2 ) h
3
x 2 3 y 2 3 1 y 2 3 1 x 2 3 y (1 x 2 3 )3 2
R ( x) (1 x 2 3 )3 2
2
1
1
2
By symmetry V 2 R ( x) dx 2 (1 x 2 3 )3 2 dx
0
0
1
1
0
0
2 (1 x 2 3 )3 dx 2 (1 3x 2 3 3 x 4 3 x 2 ) dx
1
32
2 x 95 x5 3 79 x 7 3 13 x3 2 1 95 97 13 105
0
19.
3/ 2
dy 2
y x1/2 x 3 dx 12 x 1/2 12 x1/2 dx
dy
4
2 x dx 14 x 1/2 x1/2
1
1
1
2
2
1
2 4 3 8 2 3 2 2 14
10
3
3
L
20.
4
1 1
4 x
x y 2/3
13
8
1
dx
dy
2
3
dy 2
1 y3 1
x 12
y
2
1
dy 2
dx
32 1 1/5
x
1 2
1 dx
4 y 2/3
9
40
dx
32
22.
1 2 u 3/2
18
3
13
12 x1/5 12 x 1/5 1
1
2
L
41
1 2
dx
8
1
1
4
1 14 1x 2 x dx
1
x1/2 x1/2 dx 12 2 x1/2 32 x3/2 1
dx
dy
4
2
dy
8
1
1
4
9 y 2/3
dy
8 9 y 2/3 4
1
3 y1/3
40 1/2
u du
13
dy
dx
dx
dy
2
2 x L
9 y 2/3 4 y 1/3 dy; [u 9 y 2/3 4 du 6 y 1/3dy; y 1 u 13, y 8 u 40]
1
L 18
21.
y 1/3
1 1
4 x
1
16
dx
dy
y 4 12
1
y4
1
4
y2
dy
2
1
1
27
403/2 133/2 7.634
14 x 2/5 12 14 x 2/5
1 x1/5
2
5 x 6/5 5 x 4/5
12 x 1/5 dx 12
8
1
y2
dx
dy
2
1 y4 1
16
2
1
y4
12 x 1/5
32
1
L
2
1
2
285
8
1 4 1
1 16
y 2
2
2
2 1 2
1 2 1
1
1 3 1
4 y y 2 dy 1 4 y y 2 dy 12 y y
1
128 12 121 1 127 12 1213
Copyright 2018 Pearson Education, Inc.
1
y4
dy
dy
Chapter 6 Practice Exercises
b
23. S 2 y 1
a
2
3
2x 1
0
b
24. S 2 y 1
a
dy 2
dx
dx; dx
2 x2
2 x 1
dx 2 2
dy
1
2 x 1
dy 2
dx
dy
dx
dy 2
3
2 x11 S 2 2 x 1 1 2 x11 dx
0
3
3
x 1 dx 2 2 23 ( x 1)3/2 2 2 23 (8 1)
0
0
dx;
dx
dy 2
x 2 dx
431
1
3
x 4 S 2 x3 1 x 4 dx 6
0
28 2
3
1
1 x 4 4 x3 dx
0
1
2
4 3/2
1 x
2 2 1
6
3
9
0
d
25. S 2 x 1
c
dx
dy
2
2
1
d
c
4
4 y y2
dy;
2
dx
dy
dx
dy; dy
S 2 4 y y 2
26. S 2 x 1
dx
dy
12 (42 y )
4 y y
2 y
2
4 y y
2
dx
1 dy
2
4 y y 2 4 4 y y 2
4 y y2
4
4 y y2
2
dy 4 dx 4
1
1
2 y
dx
1 dy
6
4 23 (4 y 1)3/2 6 (125 27) 6 (98)
2
2
1 41y
4 y 1
4y
6
4 y 1
2
4y
S 2 y
dy
6
2
4 y 1 dy
49
3
27. The equipment alone: the force required to lift the equipment is equal to its weight F1 ( x) 100 N . The
b
40
a
0
work done is W1 F1 ( x) dx 100 dx 100 x 0 4000 J; the rope alone: the force required to lift the
40
rope is equal to the weight of the rope paid out at elevation x F2 ( x) 0.8(40 x). The work done is
b
40
40
2
2
0.8(40 x) dx 0.8 40 x x2 0.8 402 402
0
a
0
is W W1 W2 4000 640 4640 J
W2 F2 ( x) dx
(0.8)(1600)
2
640 J; the total work
28. 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
x (6400) 1 x
224750
9500 lb. The work done is
4750
b
4750
x
W F ( x) dx
6400 1 9500
dx
a
0
4750
x
6400 x 29500
6400 4750 44750
3 (6400)(4750) 22,800,000 ft-lb
4750 4
0
is F ( x) 8 800
2
2
29. Using a proportionality constant of 1, the work in lifting the weight of w lb from r a to r is
r a wt dt w t2 r a w2 r
r
2
r
2
( r a )2
w (2ar a 2 ).
2
Copyright 2018 Pearson Education, Inc.
432
Chapter 6 Applications of Definite Integrals
30. Force constant: F kx 200 k (0.8) k 250 N/m; the 300 N force stretches the spring
300
250
x Fk
1.2 m; the work required to stretch the spring that far is then W
1.2
0
F ( x) dx
1.2
0
250 x dx
1.2
1.2
250 x dx 125 x 2 125(1.2)2 180 J
0
0
31. We imagine the water divided into thin slabs by planes
perpendicular to the y -axis at the points of a partition
of the interval [0,8]. The typical slab between the planes at
y and y y has a volume of about
V (radius) 2 (thickness)
54 y
2
y
25
16
y 2 y ft 3 .
The force F ( y ) required to lift this slab is equal to its
weight: F ( y ) 62.4V
(62.4)(25)
y 2 y
16
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
(62.4)(25)
y 2 (14
16
y )y ft lb. The
work done lifting all the slabs from y 0 to y 8 to the level 6 ft above the top is approximately
8
W
0
(62.4)(25)
16
y 2 (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
8 (62.4)(25)
y 2 (14
(16)
0
(62.4)
25
16
8
y ) dy
(62.4)(25) 8
16
0
14 y 2 y3 dy
143 83 84 418,208.81 ft-lb
14 y 3 y 4 (62.4) 25
4
16
3
0
4
32. The same as in Exercise 31, 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:
5 (62.4)(25) 2
y (8
16
0
W
(62.4)
y ) dy (62.4)
5
2516 05 8 y 2 y3 dy (62.4) 2516 83 y3 y4 0
4
2516 83 53 54 54,241.56 ft-lb
4
y
5 y . A typical
33. The tank’s cross section looks like the figure in Exercise 31 with right edge given by x 10
2
horizontal slab has volume V (radius) 2 (thickness)
y 2
2
y 4 y 2 y. The force required to lift this
slab is its weight: F ( y ) 60 y 2 y. The distance through which F ( y ) must act is (2 10 y ) ft, so the
10
work to pump the liquid is W 60 (12 y )
0
to empty the tank is
22,500 ft-lb
275 ft-lb/sec
y2
4
10
dy 15 12 y 3 y 4 22,500 ft-lb; the time needed
4
3
0
257sec
Copyright 2018 Pearson Education, Inc.
Chapter 6 Practice Exercises
433
34. A typical horizontal slab has volume about V (20)(2 x)y (20) 2 16 y 2 y and the force required
to lift this slab is its weight F ( y ) (57)(20) 2 16 y 2 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
0
0
0
W 57 (10 y )(20) 2 16 y 2 dy 2880 10 16 y 2 dy 1140 16 y 2
4
4
4
22,800 (area of a quarter circle having radius 4)
23 (1140)
4
4
b
5
5
1/2
(2 y ) dy
0
2 3/2
16 y
335,153.25 ft-lb
b
(22,800)(4 ) 48, 640
4
35. (a) Work W F ( x) dx 10 x3 2 dx 4 x5 2 4(55 2 ) 128 ft-lb
0
0
a
(b) Work W F ( x) dx 10 x3 2 dx 4 x5 2 4(55 2 ) 4 219.6 ft-lb
1
1
a
36. (a) First find the spring constant k : F ( x) kx 5 x 2 3 k (2) 5 (2)2 k 12
b
11
02
(b) Work W F ( x) dx
a
1
x 5 x 2 dx 16 (5 x 2 )3 2 16 (63 2 ) 16 (53 2 ) 0.586 J
0
37. Intersection points: 3 x 2 2 x 2 3x 2 3 0
3( x 1)( x 1) 0 x 1 or x 1. Symmetry
suggests that x 0. The typical vertical strip has center of
2 x 2 3 x 2
x 2 3 ,
x
,
mass: ( x , y ) x,
2
2
length: 3 x 2 2 x 2 3 1 x 2 , width: dx,
1
y dm 32 x 2 31 x 2 dx 32 x 4 2 x 2 3 dx M x y dm 32 x 4 2 x 2 3 dx
1
1
1
32 x5 23x 3 x 3 15 32 3 315 (3 10 45) 325 ; M dm 3 1 x 2 dx
1
1
area: dA 3 1 x 2 dx, and mass: dm dA 3 1 x 2 dx the moment about the x-axis is
5
3
1
3
M
3 x x3 6 1 13 4 y Mx
1
32
54
38. Symmetry suggests that x 0. The typical vertical strip
, length: x , width: dx,
has center of mass: ( x , y ) x,
x2
2
2
area: dA x 2 dx, mass: dm dA x 2 dx the
moment about the x-axis is y dm 2 x 2 x 2 dx
2
85 . Therefore, the centroid is ( x , y ) 0, 85 .
2
x5
2 x 4 dx M x y dm 2 x 4 dx 10
2
2
Copyright 2018 Pearson Education, Inc.
434
Chapter 6 Applications of Definite Integrals
39. The typical vertical strip has: center of mass: ( x , y )
x,
2
, length: 4 x4 , width: dx,
2
4 x4
2
dx the moment about the x-axis is
2
area: dA 4 x4 dx, mass: dm dA
2
4 x4
4 x2
4
2
4 dx 16 dx; moment about: x dm 4 x dx 4 x dx. Thus,
64
M y dm 16 dx 16 x
; M x dm
4 x dx 2 x (32 16) 16 ; M dm 4 dx 4 x
y dm
x2
4
2
4
x
x4
16
2 0
4
x3
4
0
x4
16
4
x5
516 0
2
2
4
x4
16 0
2
64
5
128
5
x2
4
x3
4
x2
4
4
x3
12 0
y
4
0
64 32 x M y 16 3 3 and y M x 128 3 12 . Centroid is ( x , y ) 3 , 12 .
16 12
3
32
2
M
2 5
M
532
5
40. A typical horizontal strip has: center of mass:
y2 2 y
( x , y ) 2 , y , length: 2 y y 2 , width: dy,
area: dA 2 y y 2 dy, mass: dm dA
2 y y 2 dy; the moment about the x-axis
is y dm y 2 y y 2 dy 2 y 2 y 3 dy;
the moment about the y -axis is x dm
y2 2 y
2
2 y y 2 dy 2 4 y 2 y 4 dy M x y dm
2
y
16 4 ;
2 y 2 y 3 dy 23 y 3 4 23 8 16
16
16
12
4
3
4
3
0
0
4
2
2
y
;
4 y 2 y 4 dy 2 34 y 3 5 2 438 32
32
5
15
0
0
2
5
2
M y x dm
2
y
2 y y 2 dy y 2 3
0
0
M dm
2
3
M
323 8 and y M x 4 3 1. Therefore, the centroid is ( x , y ) 8 , 1 .
4 83 43 x My 15
4 5
5
34
M
41. A typical horizontal strip has: center of mass:
y2 2 y
( x , y ) 2 , y , length: 2 y y 2 , width: dy,
area: dA 2 y y 2 dy, mass: dm dA
(1 y ) 2 y y 2 dy the moment about the
x-axis is y dm y (1 y ) 2 y y 2 dy
y 2 y
x dm 2 (1 y ) 2 y y 2 dy 12 4 y 2 y 4 (1 y ) dy 12 4 y 2 4 y3 y 4 y 5 dy M x y dm
2 y 2 2 y 3 y 3 y 4 dy 2 y 2 y3 y 4 dy; the moment about the y -axis is
2
Copyright 2018 Pearson Education, Inc.
Chapter 6 Practice Exercises
435
2
y
y
4 (11) 44 ;
2 y 2 y 3 y4 dy 23 y 3 4 5 16
16
32
16 13 14 52 16
(20 15 24) 15
3
4
5
60
15
0
0
4
2
5
2
3
5
6
2
y5
y6
M y x dm 12 4 y 2 4 y 3 y 4 y5 dy 12 43 y 3 y 4 5 6 12 432 24 25 26
0
0
2
2
4 43 2 54 86 4 2 54 24
; M dm (1 y ) 2 y y 2 dy 2 y y 2 y3 dy
5
0
0
2
83 95 and y MM 1544 83 4440 1011 . Therefore,
M
y3 y 4
y 2 3 4 4 83 16
83 x My 24
4
5
0
the center of mass is ( x , y )
x
95 , 1011 .
42. A typical vertical strip has: center of mass: ( x , y ) x,
mass: dm dA
3
x3/ 2
, length:
3
x3/ 2
dx the moment about the x-axis is y dm
about the y -axis is x dm x
(a)
3
2 x3/ 2
3
x3/ 2
dx
3
x1/ 2
, width: dx, area: dA 3/3 2 dx,
x
3
2 x3/ 2
3/3 2 dx 93 dx; the moment
2x
x
dx.
9
9
2
9
9
M x 12 93 dx 92 x2 209 ; M y x 3/3 2 dx 3 2 x1/2 12 ;
1
1
1
x
x
1
9 3
1 x3/ 2
M
(b) M x
9
1
9
My
3 and y
dx 6 x 1/2 4 x M 12
4
1
Mx
M
209 5
4
9
9
9
9
9
9
x 9
dx 92 1x 4; M y x 2 3/3 2 dx 2 x3/2 52; M x 3/3 2 dx 6 x1/2
3
1
1
1
x
x
x
1
1
2
M
M
12 x My 13
and y Mx 13
3
43.
2
2
2
b
2 y3
2
F W strip
L
(
y
)
dy
F
2
(62.4)(2
y
)(2
y
)
dy
249.6
2
y
y
dy
249.6
0
0
depth
y 3
a
0
(249.6) 4 83 (249.6) 43 332.8 lb
44.
a
75
5/6 10
3
0
(75)
45.
b
5/6
5/6 5
F W strip
75 56 y (2 y 4) dy 75
y 10
2 y 2 4 y dy
depth L( y ) dy F
3
3
0
73 y 2 y 2 dy 75 10
y 76 y 2
3
0
5/6
2 y3
3
0
3625 32 125
216
50 7
(75) 18
6
(75)(3075)
250 75 (25 216 175 9 250 3) 9216 118.63 lb.
259 175
216 3216 9216
b
F W
a
strip
depth
L( y) dy F 62.4
4
0
(9 y ) 2
y
2
4
1/2
3/2
dy
dy 62.40 9 y 3 y
4
(62.4)(176)
62.4 6 y 3/2 52 y 5/2 (62.4) 6 8 52 32 62.4
(48 5 64)
2196.48 lb
5
5
0
h
46. Place the origin at the bottom of the tank. Then F W
0
strip
depth
L( y) dy,
h the height of the mercury
h
h
column, strip depth h y, L( y ) 1 F 849(h y ) 1 dy 849 (h y ) dy 849 h y
0
0
Copyright 2018 Pearson Education, Inc.
y2
2
h
0
436
Chapter 6 Applications of Definite Integrals
2
849 h 2 h2 849
h 2 . Now solve
2
849 h 2
2
40000 to get h 9.707 ft. The volume of the mercury is
s 2 h 12 9.707 9.707 ft 3 .
CHAPTER 6
1. V
b
a
ADDITIONAL AND ADVANCED EXERCISES
x
f ( x)2 dx b2 ab a f (t )2 dt x 2 ax
2
2 x a
f ( x)
a
x
f ( x)2 dx a 2 a a f (t )2 dt x 2 x
0
2. V
for all x a f ( x) 2 x a
3. s ( x) Cx
x
0
f ( x)
x
0
for all x a f ( x) 2 x 1 f ( x)
2
2 x 1
1 f (t ) dt Cx 1 f ( x) C f ( x) C 2 1 for C 1
2
2
C 2 1 dt k . Then f (0) a a 0 k f ( x)
x
0
C 2 1 dt a f ( x) x C 2 1 a,
where C 1.
4. (a) The graph of f ( x) sin x traces out a path from (0, 0) to ( , sin ) whose length is
L
0
1 cos 2 d . The line segment from (0, 0) to ( , sin ) has length
( 0)2 (sin 0) 2 α 2 sin 2 . Since the shortest distance between two points is the length of
the straight line segment joining them, we have immediately that
1 cos 2 d 2 sin 2 if 0 2 .
(b) In general, if y f ( x) is continuously differentiable and f (0) 0, then
0
2
1 f (t ) dt 2 f 2 ( ) for 0.
0
5. We can find the centroid and then use Pappus’ Theorem to calculate the volume. f ( x) x, g ( x) x 2 ,
1
1
f ( x) g ( x) x x 2 x 2 x 0 x 0, x 1; 1; M x x 2 dx 12 x 2 13 x3
0
0
1
1 1 x x x 2 dx 6 1 x 2 x3 dx 6 1 x3 1 x 4 6 1 1 0 1 ;
12 13 0 16 ; x 1/6
0
0
4
3 4
2
3
0
2
1
1 1 1 x 2 x 2 dx 3 1 x 2 x 4 dx 3 1 x3 1 x5 3 1 1 0 2 The centroid is 1 , 2 .
y 1/6
0 2
3
5
3 5
3
2 5
0
0
12 , 52 to the axis of rotation, y x. To calculate this distance we must find the point
on y x that also lies on the line perpendicular to y x that passes through 12 , 52 . The equation of this line
9 . The point of intersection of the lines x y 9 and y x is
is y 52 1 x 12 x y 10
209 , 209 .
10
9 1 2 9 2 2 1 . Thus V 2
Thus, 10
20 5 10 2
101 2 16 30π 2 .
2
is the distance from
Copyright 2018 Pearson Education, Inc.
Chapter 6 Additional and Advanced Exercises
437
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
1
2
7.
12
and height 1. Thus, V 12
y 2 x ds
1
x
2
(1) 8 .
3
3
1 dx A 2 x 1x 1 dx 34 (1 x)3/2 28
0 3
0
8. This surface is a triangle having a base of 2 a and a height of 2 ak . Therefore the surface area is
1 (2 a )(2 ak )
2
9.
2 2 a 2 k .
2
2
3
2
4
F ma t 2 d 2 a tm v dx
3tm C ; v 0 when t 0 C 0 dx
3tm x 12t m C1 ;
dt
dt
dt
4
x 0 when t 0 C1 0 x 12t m . Then x h t (12 mh)1/4 . The work done is
W F dx
(12 mh )1/ 4
0
F (t ) dx
dt
dt
(12 mh )1/ 4 2 t 3
t 3m dt
0
(12 mh )1/ 4
1 t6
3m 6 0
181m (12mh)6/4
(12 mh )3/ 2
18m
12mh18 m12 mh 23h 2 3mh 43h 3mh
2 lb 12 in
1 ft
10. Converting to pounds and feet, 2 lb/in 1 in
24 lb/ft. Thus, F 24 x W
1/2
0
24 x dx
1/2
1 lb
1
1
12 x 2 3 ft lb. Since W 12 mv02 12 mv12 , where W 3 ft lb, m 10
32 ft/sec2 320 slugs,
0
and v1 0 ft/sec, we have 3
12 3201 v02 v02 3 640. For the projectile height, s 16t 2 v0t
(since
v
s 0 at t 0 ) ds
v 32t v0 . At the top of the ball’s path, v 0 t 320 and the height is
dt
v
s 16 320
2
v
v2
640 30 ft.
v0 320 640 364
11. From the symmetry of y 1 x n , n even, about the y -axis for 1 x 1, we have x 0. To find y
n
Mx
M
,
we use the vertical strips technique. The typical strip has center of mass: ( x , y ) x, 12x , length: 1 x n ,
width: dx, area: dA 1 x n dx, mass: dm 1 dA 1 x n dx. The moment of the strip about the x-axis is
1 x
n 2
y dm
2
1 1 x
2
1
dx M x
( n 1)(2 n 1) 2(2 n 1) ( n 1)
( n 1)(2 n 1)
n 2
1
2n 1
n 1
1
dx 2 12 1 2 x n x 2n dx x 2nx 1 x2n 1 1 n 2 1 2 n1 1
0
0
2
1
1
1 4 n 2 n 1
2n 2
2n (3nn1)(2
. Also, M dA 1 x n dx
( n 1)(2 n 1)
n 1)
1
1
1
n 1
1
2 1 x n dx 2 x xn 1 2 1 n11
0
0
is the location of the centroid. As n , y
2n .
n 1
1
2
Therefore, y
Mx
M
2
( n 1)
2n
( n 1)(2
2nn1 0, 2 nn1
n 1) 2 n
12 .
so the limiting position of the centroid is 0,
Copyright 2018 Pearson Education, Inc.
438
Chapter 6 Applications of Definite Integrals
12. Align the telephone pole along the x-axis as shown
in the accompanying figure. The slope of the top
89
14.5
8
length of pole is
40
85.5
11 . Thus, y
40 8 80
1 1 (14.5 9)
8 40
9 11 x
8 8 80
11 x is an equation of the line
81 9 80
representing the top of the pole. Then
9 8011 x dx;
2
2
b
40
40
11 x dx 1
11 x dx. Thus, x M
M y 2 dx 81 9 80
9 80
M
64
a
0
0
b
40
a
0
M y x y 2 dx
2
11 x dx
x 81 9 80
2
40
1
x
64 0
y
129,700
5623.3
23.06 (using a
calculator to compute the integrals). By symmetry about the x-axis, y 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 ( x, y ). If the plate lies to the right of the line, then the
moment of this strip about the line x b is ( x b) dm ( x b) dA the plate’s first moment about
x b is the integral ( x b) dA x dA b dA M y b A.
(b) If the plate lies to the left of the line, the moment of a vertical strip about the line x b is (b x ) dm
(b x ) dA the plate’s first moment about x b is (b x) dA b dA x dA b A M y .
14. (a) By symmetry of the plate about the x-axis, y 0. A typical vertical strip has center of mass: ( x , y )
( x, 0), length: 4 ax , width: dx, area: 4 ax dx, mass: dm dA kx 4 ax dx, for some
a
proportionality constant k. The moment of the strip about the y -axis is M y x dm 4k x 2 ax dx
0
a
4
a
3
My
M
y2
a
4a
y 2 4a 2
, y 8a ,
a
a
8k a
4k a x5/2 dx 4k a 72 x7/2 4k a1/2 72 a 7/2 7 . Also, M dm 4k x ax dx
0
0
0
a
8k a
4k a x3/2 dx 4k a 52 x5/2 4k a1/2 52 a5/2 5 . Thus, x
0
0
8k a 4
7 5 3 75 a
8k a
( x , y ) 57a , 0 is the center of mass.
(b) A typical horizontal strip has center of mass: ( x , y )
2
y2
y , length: a 4a ,
y2
y2
width: dy, area: a 4 a dy, mass: dm dA y a 4a dy. Thus,
M x y dm
2a
2 a
0
2a
y2
y2
y2
y y a 4a dy y 2 a 4a dy y 2 a 4 a dy
2
0
a
0
2a
4
2a
y4
y4
y5
y5
a5 8a 4 32 a 5 0;
ay 2 4a dy ay 2 4a dy a3 y 3 20 a
a3 y 3 20a 8a3 32
a
20
3
20 a
0
2 a
2a
0
0
2a
y2
y 2 4a 2
y
a
dy 81a
y
4 a
2 a 8a
2 a
M y x dm
16a 4 y y 5 dy 1
2 a
32 a
0
12
32 a
2a
2a
2
0
4a 2 y 2
y 2 4a 2 4a dy
16a4 y y5 dy 321a
0
2
1
32 a 2
2a
2 a
6
4 2 y
12
8a y 6
2 a 32 a
Copyright 2018 Pearson Education, Inc.
y 16a 4 y 4 dy
2a
4 2 y6
8a y 6
0
Chapter 6 Additional and Advanced Exercises
439
6
6
6
1 2 8a 4 4a 2 646a 1 2 8a 4 4a 2 646a 1 2 32a 6 323a 1 2 32 32a 6 43 a 4 ;
32 a
16 a
32a
16a
2
2
2a
2
a
0
2a
4a y
2
3
2
3
1
M dm
y
dy 41a
y 4a 2 y 2 dy 4a
2a 4a y y dy 41a 0 4a y y dy
2 a 4 a
2 a
0
2a
4
y4
y4
41a 2a 2 y 2 4
41a 2a 2 y 2 4 2 41a 2a 2 4a 2 164a 21a 8a 4 4a 4 2a3 . Therefore,
2 a
0
x
My
M
43 a4 21a 23a
and y
3
Mx
M
0 is the center of mass.
15. (a) On [0, a ] a typical vertical strip has center of mass: ( x , y ) x,
b2 x2 a 2 x2
2
,
b 2 x 2 a 2 x 2 , width: dx, area: dA b 2 x 2 a 2 x 2 dx, mass: dm dA
length:
b 2 x 2 a 2 x 2
dx. On [a, b] a typical vertical strip has center of mass: ( x , y ) x, b 2 x 2
2
,
b 2 x 2 , width: dx, area: dA b 2 x 2 dx, mass: dm dA b 2 x 2 dx. Thus,
length:
a
M x y dm 12 b 2 x 2 a 2 x 2
0
b 2 x 2 a 2 x 2 dx b 1 b 2 x 2 b 2 x 2 dx
a 2
b
a
2 b 2 a 2 x 2 b 2 x x3 2 b 2 a 2 a 2 b3 b3 b 2 a a3
0
a
2 ab 2 a3 2 23 b3 ab 2 a3 3b 3a b 3 a ;
a
b
a
b
2 b 2 x 2 a 2 x 2 dx 2 b 2 x 2 dx 2 b 2 a 2 dx 2 b 2 x 2 dx
0
a
0
a
3
3
3
3
3
a
M y x dm x b 2 x 2 a 2 x 2
0
a
x b2 x2
0
1/2
a
0
a
2
1/2
dx x a 2 x 2
3
3
3
dx b x b 2 x 2 dx
a
b
dx x b 2 x 2
a
a
1/2
dx
b
2 b2 x 2 3/ 2
2 a 2 x 2 3/ 2
2 b2 x 2 3/ 2
3
2
3
2
3
0
0
a
b3 a 3
3/2
3/2
3/2
3/2
0 a 2
0 b2 a 2
b3 a 3
3 b 2 a 2
b2
Mx;
3
3
3
3
3
2
2
We calculate the mass geometrically: M A 4b 4a
b 2 a 2 . Thus, x
4
(b)
b3 a 3
3
lim 4
b a 3
2
4
b 2 a 2
a ab b 2
a b
34
4
3
b3 a 3
b2 a 2
34
a2 a2 a2
aa
(b a ) a 2 ab b 2
4
3
3a 2
2a
2
ab b 2
3 ( a b )
(b a )(b a )
4 a
2a
;
likewise y
Mx
M
My
M
4 a 2 ab b 2
3 ( a b )
.
( x , y ) 2a , 2a is the limiting position of the
centroid as b a. This is the centroid of a circle of radius a (and the two circles coincide when b a ).
Copyright 2018 Pearson Education, Inc.
440
Chapter 6 Applications of Definite Integrals
16. Since the area of the triangle is 36, the diagram may
be labeled as shown at the right. The centroid of
the triangle is
a3 , 24a . The shaded portion is
144 36 108. Write ( x y ) for the centroid of the
remaining region. The centroid of the whole square
is obviously (6, 6). 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: 6
6
36
24a 108( y )
and y
144
8( a 1)
.
a
36
a3 108( x )
144
and
which we solve to get x 8 a9
Set x 7 in. (Given). It follows that a 9, whence y
64
9
7 19 in. The distances of the
centroid ( x , y ) from the other sides are easily computed. (If we set y 7 in. above, we will find x 7 19 . )
17. The submerged triangular plate is depicted in the
figure at the right. The hypotenuse of the triangle
has slope 1 y (2) ( x 0) x ( y 2)
is an equation of the hypotenuse. Using a typical
horizontal strip, the fluid pressure is
F (62.4)
2
6
strip
depth
strip
length
dy
(62.4)( y ) ( y 2) dy
2
y
y 2 2 y dy 62.4 3 y 2
6
6
62.4
2
3
(62.4)(112)
(62.4) 208
32
2329.6 lb
3
3
(62.4) 83 4 216
36
3
18. Consider a rectangular plate of length and width
w. The length is parallel with the surface of the fluid
of weight density . The force on one side of the
0
2
y2
( y )() dy 2 2w .
w
w
The average force on one side of the plate is
plate is F
0
0
2
y2
( y ) dy w 2 2w . Therefore the force 2w 2w (w)
w
w
(the average pressure up and down ) ∙ (the area of the plate).
Fav w
0
Copyright 2018 Pearson Education, Inc.
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 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 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 1 intersects the graph twice.
11. Domain: 0 x 1, Range: 0 y
12. Domain: x 1, Range: y 0
13. Domain: 1 x 1, Range: 2 y 2
14. Domain: x , Range: 2 y 2
Copyright 2018 Pearson Education, Inc.
441
442
Chapter 7 Transcendental Functions
15. Domain: 0 x 6, Range: 0 y 3
16. Domain: 2 x 1, Range: 1 y 3
17. The graph is symmetric about y x.
(b)
y 1 x 2 y 2 1 x 2 x 2 1 y 2 x 1 y 2 y 1 x 2 f 1 ( x)
18. The graph is symmetric about y x.
y
19. Step 1:
Step 2:
20. Step 1:
Step 2:
21. Step 1:
Step 2:
22. Step 1:
1
x
x
1
y
y
f 1 ( x)
y x2 1 x2 y 1 x
y x 1 f
1
y 1
( x)
y x 2 x y , since x 0.
y x f 1 ( x)
y x3 1 x3 y 1 x ( y 1)1/3
y 3 x 1 f 1 ( x)
y x 2 2 x 1 y ( x 1)2
Step 2:
y 1 x f
23. Step 1:
y ( x 1) 2
Step 2:
1
x
y x 1 f
1
y x 1, since x 1 x 1 y
( x)
y x 1, since x 1 x
1
y 1
( x)
Copyright 2018 Pearson Education, Inc.
Section 7.1 Inverse Functions and Their Derivatives
24. Step 1:
y x 2/3 x y 3/2
Step 2:
y x3/2 f 1 ( x)
25. Step 1:
y x5 x y1/5
Step 2:
y 5 x f 1 ( x);
5
Domain and Range of f 1 : all reals; f f 1 ( x) x1/5
26. Step 1:
y x 4 x y1/4
Step 2:
y 4 x f 1 ( x);
x and f 1 f ( x) x5
Domain of f 1 : x 0, Range of f 1 : y 0; f f 1 ( x) x1/4
27. Step 1:
Step 2:
443
4
1/5
x
x and f 1 f ( x) x 4
1/4
x
y x3 1 x3 y 1 x ( y 1)1/3
y 3 x 1 f 1 ( x);
Domain and Range of f 1 : all reals;
f f 1 ( x) ( x 1)1/3
28. Step 1:
Step 2:
3
1 ( x 1) 1 x and f 1 f ( x)
x 1 1 x
1/3
3
3 1/3
x
y 12 x 72 12 x y 72 x 2 y 7
y 2 x 7 f 1 ( x);
Domain and Range of f 1 : all reals;
f f 1 ( x) 12 (2 x 7) 72 x 72 72 x and f 1 f ( x) 2 12 x 72 7 ( x 7) 7 x
29. Step 1:
Step 2:
y 12 x 2 1y x 1
y
x
y
f 1 ( x)
1
x
Domain of f 1 : x 0, Range of f 1 : y 0; f f 1 ( x)
since x 0
30. Step 1:
y
Step 2:
y
Domain of f
f 1 f ( x)
1
x3
2
1
1x
x and f 1 f ( x)
1
x3 1y x 1/3
y
1
x1/3
1
3
1
x
f 1 ( x);
: x 0, Range of f 1 : y 0; f f 1 ( x)
1
x3
1
1
x
1/3
1x
1
1
x 1/3
3
1
x 1
x and
x
Copyright 2018 Pearson Education, Inc.
1
1
x2
1
1x
x
444
Chapter 7 Transcendental Functions
31. Step 1:
y
x 3
x2
Step 2:
y
2 x 3
x 1
y ( x 2) x 3 xy 2 y x 3 xy x 2 y 3 x
f 1 ( x);
Domain of f 1 : x 1, Range of f 1 : y 2; f f 1 ( x)
y
x
x 3
Step 2:
y
x3x1
1
y
2
3y 2
y 1
f 1 ( x);
: (, 0] (1, ), Range of f
0
x 3 x y x 3y x y x x 3y x
x3x1
2
x3x1 3
2
3x
x 1
and
xx32 3 2( x 3)3( x 2) 5 x x
xx32 1 ( x 3)( x 2) 5
32. Step 1:
x0
2xx13 3 (2 x 3)3( x 1) 5 x x
2xx13 2 (2 x 3)2( x 1) 5
2
f 1 f ( x)
Domain of f
2 y 3
y 1
3x
x 1
3x
3
x 1
1
3x
3 x 3( x 1)
1
:[0, 9) (9, ); f f
3x
3
x and f
1
2
( x)
f ( x)
x3x1
;
2
x3x1 3
If x 1 or
9 x
1 x x 3
3
2
x
x 3
x
x 3
2
99x x
33. Step 1:
Step 2:
y x 2 2 x, x 1 y 1 ( x 1)2 , x 1 y 1 x 1, x 1 x 1 y 1
y 1 x 1 f 1 ( x);
Domain of f 1 : [1, ), Range of f 1 : (, 1];
f f 1 ( x) 1 x 1
f 1 f ( x ) 1
34. Step 1:
Step 2:
Domain of f
1
x5 1
2
1/5
2 1 x 1 1 2 x 1 x 1 2 2 x 1 x and
y 5 1
y 5 1
y 5 2 x3 1 y 5 1 2 x3 2 x3 x 3 2
: (, ), Range of f
1/5
( x 1)2 , x 1 1 | x 1| 1 (1 x) x
f 1 ( x);
x 1 1 x
5
2
x2 2 x 1, x 1 1
y 2 x3 1
y3
5 1/5
1
: (, ); f f
x and f 1 f ( x)
1
( x) 2 3
1/5
1
3
x5 1
2
2
5
3 1/5
3 2 x 1 1
2
3
(2 x3 1) 1
2
Copyright 2018 Pearson Education, Inc.
3
2 x3
2
x
x5 1
2
1
1/5
Section 7.1 Inverse Functions and Their Derivatives
35. (a)
y 2x 3 2x y 3
x
(c)
36. (a)
32 f 1 ( x) 2x 23
y
2
df
dx x 1
(b)
2,
df 1
dx
x 1
1
2
y 1xx2 y xy x 2
(b)
y 2 xy x x( y 1)
x
(c)
f
f 1 ( x )
12 5 dfdx x
df 1
dx
37. (a)
y 2
y 1
x 5
3
(1 x )2
1
2
x 5
x 2
x 1
3
(1 x ) 2
x 1
2
12,
1
12
y 5 4x 4x 5 y
(b)
y
x 54 4 f 1 ( x) 54 4x
(c)
38. (a)
df
dx x 1/2
4,
y 2 x2 x2
2
df
dx x 5
df 1
dx
39. (a)
(c)
1
2
x 3
14
(b)
y
y f 1 ( x)
x 1
(c)
df 1
dx
x
2
4 x x 5 20,
x 50
1 x 1/2
f ( g ( x))
2 2
3 x
3
x 50
1
20
3
x, g ( f ( x)) x3 x
(b)
f ( x ) 3 x 2 f (1) 3, f (1) 3;
g ( x) 13 x 2/3 g (1) 13 , g (1) 13
(d) The line y 0 is tangent to f ( x) x3 at
(0, 0); the line x 0 is tangent to g ( x) 3 x at
(0, 0)
Copyright 2018 Pearson Education, Inc.
445
446
Chapter 7 Transcendental Functions
4 x x,
k (h( x)) 4 x
40. (a) h(k ( x))
x3
4
(c) h( x)
k ( x)
(b)
1/3 3
1
4
1/3
3 x 2 h (2) 3, h( 2) 3;
4
4 (4 x ) 2/3 k (2) 1 , k ( 2)
3
3
x3
4
(d) The line y 0 is tangent to h( x)
13
at (0, 0);
the line x 0 is tangent to k ( x ) (4 x )1/3 at
(0, 0)
41.
43.
df
dx
3x2 6 x
df 1
dx
45. (a)
x 4
df 1
dx
df 1
dx
x f (2)
y mx x
1
m
x f (3)
1
df
dx
1
df
dx
x 2
1
1
9
42.
3
44.
x 3
1
13
y f 1 ( x )
(b) The graph of y f
46.
1
m
df
dx
2x 4
dg 1
dx
x 0
dg 1
dx
df 1
dx
x f (5)
x f (0)
1
df
dx
1
dg
dx
x 5
x 0
1
6
1
2
x
( x) is a line through the origin with slope m1 .
y
b f 1 ( x ) 1 x b ; the graph of f 1 ( x ) is a line with slope 1 and y -intercept
y mx b x m m
m
m
m
b.
m
47. (a)
y x 1 x y 1 f 1 ( x) x 1
(b) y x b x y b f 1 ( x) x b
(c) Their graphs will be parallel to one another and
lie on opposite sides of the line y x
equidistant from that line.
48. (a)
y x 1 x y 1 f 1 ( x) 1 x; the
lines intersect at a right angle
y x b x y b f 1 ( x) b x; the
lines intersect at a right angle
(c) Such a function is its own inverse.
(b)
49. Let x1 x2 be two numbers in the domain of an increasing function f. Then, either x1 x2 or x1 x2 which
implies f ( x1 ) f ( x2 ) or f ( x1 ) f ( x2 ), since f ( x ) is increasing. In either case, f ( x1 ) f ( x2 ) and f is oneto-one. Similar arguments hold if f is decreasing.
Copyright 2018 Pearson Education, Inc.
Section 7.1 Inverse Functions and Their Derivatives
50.
447
1
df
df
f ( x) is increasing since x2 x1 13 x2 56 13 x1 56 ; dx 13 dx 11 3
3
51.
f ( x) is increasing since x2 x1 27 x23 27 x13 ; y 27 x3 x 13 y1/3 f 1 ( x) 13 x1/3 ;
df
dx
52.
df 1
24 x 2 dx 1 2
1 2/3 16 (1 x)2/3
6(1 x )
24 x 1 (1 x )1/3
2
3
3
f ( x) is decreasing since x2 x1 1 x2 1 x1 ; y (1 x)3 x 1 y1/3 f 1 ( x) 1 x1/3 ;
df
dx
54.
df
f ( x) is decreasing since x2 x1 1 8 x23 1 8 x13 ; y 1 8 x3 x 12 (1 y )1/3 f 1 ( x) 12 (1 x )1/3 ;
df
dx
53.
1
81x 2 dx 1 2
12/3 91 x 2/3
81x 1 x1/3
9x
3
1
1
1 1 x 2/3
3(1 x) 2 dx
2/3
3
3x
3(1 x )2 1 x1/3
df
f ( x) is increasing since x2 x1 x25/3 x15/3 ; y x5/3 x y 3/5 f 1 ( x) x3/5 ;
df
dx
1
df
53 x 2/3 dx 5 12/3
x
3
x
3/5
3
5 x 2/5
53 x 2/5
55. The function g ( x) is also one-to-one. The reasoning: f ( x) is one-to-one means that if x1 x2 then
f ( x1 ) f ( x2 ), so f ( x1 ) f ( x2 ) and therefore g ( x1 ) g ( x2 ). 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 x1 x2 then
f ( x1 ) f ( x2 ), so f (1x ) f (1x ) , and therefore h( x1 ) h( x2 ).
1
2
57. The composite is one-to-one also. The reasoning: If x1 x2 then g ( x1 ) g ( x2 ) because g is one-to-one. Since
g ( x1 ) g ( x2 ), we also have f g ( x1 ) f g ( x2 ) because f is one-to-one. Thus, f g is one-to-one
because x1 x2 f g ( x1 ) f g ( x2 ) .
58. Yes, g must be one-to-one. If g were not one-to-one, there would exist numbers x1 x2 in the domain of g
with g ( x1 ) g ( x2 ). For these numbers we would also have f g ( x1 ) f g ( x2 ) , contradicting the
assumption that f g is one-to-one.
59. ( g f )( x) x g f ( x) x g f ( x) f ( x) 1
60. W (a )
f (a)
f (a)
f 1 ( y )
W (t ) f 1 f (t )
t
2
2
a
a 2 dy 0 2 x f (a ) f ( x) dx S (a );
a
a 2 f (t ) t 2 a 2 f (t ); also
t
t
S (t ) 2 f (t ) x dx 2 x f ( x) dx f (t )t 2 f (t )a 2 2 x f ( x) dx
a
a
a
Copyright 2018 Pearson Education, Inc.
448
Chapter 7 Transcendental Functions
S (t ) t 2 f (t ) 2 t f (t ) a 2 f (t ) 2 t f (t ) t 2 a 2 f (t ) W (t ) S (t ). Therefore, W (t ) S (t )
for all t [a, b].
61–66. Example CAS commands:
Maple:
with(plots); # 61
f : x - sqrt(3* x-2);
domain: 2/3..4;
x0: 3;
Df : D(f);
# (a)
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 "# 61(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.
<<Miscellaneous `RealOnly`
Clear [x, y]
{a,b} {2, 1}; x0 1/2;
f[x _ ] (3x 2) / (2 x 11)
Plot [{f[x], f'[ x]}, {x, a, b}]
sol x Solve[y f[x], x]
g[y _ ] x / . sol x[[1]]
y0 f[x0]
f tan[x _ ] y0 f'[x0] (x-x0)
Copyright 2018 Pearson Education, Inc.
Section 7.1 Inverse Functions and Their Derivatives
449
g tan[y _ ] x0 1/ f'[x0] (y y0)
Plot [{f[x], f tan[x], g[x], gtan[x], Identity[x]},{x, a, b},
Epilog Line[{{x0, y0},{y0, x0}}], PlotRange {{a,b},{a,b}}, AspectRatio Automatic]
67–68. 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 "# 67(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)
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 "# 67(e) Section 7.1 " );
Mathematica: (assigned function and values for a, b, and x0 may vary)
For problems 67 and 68, 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 / . sol x[[1]]
Copyright 2018 Pearson Education, Inc.
450
Chapter 7 Transcendental Functions
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], Idenity[x]},{x, a, b},
Epilog Line[{{x0, y0},{y0, x0}}], PlotRange {{a, b},{a, b}}, AspectRatio Automatic]
7.2
NATURAL LOGARITHMS
1. (a) ln 0.75 ln
(b) ln
(c) ln
ln 3 ln 4 ln 3 ln 22 ln 3 2 ln 2
3
4
4 ln 4 ln 9 ln 22
9
1 ln1 ln 2 ln 2
2
1/2
1
2
(d) ln 3 9 13 ln 9 13 ln 32
ln 13.5
1
2
ln
27
2
1
2
(e) ln 0.056 ln
3. (a) ln sin ln
2
7
125
3
2
1
2
49
5
(b) ln 9.8 ln
ln 7 2 ln 5 2 ln 7 ln 5
(d) ln 1225 ln 352 2 ln 35 2 ln 5 2 ln 7
ln 7
3
ln 7 ln 5 ln 7 3 ln 5
(f )
ln 35 ln
ln 25
1
7
ln 5 ln 7 ln 7
2 ln 5
1
2
sin5 ln sin ln 5
sin
5
31x ln 3x 3x9 x ln ( x 3)
2
(b) ln 3x 9 x ln
(c)
ln 3
ln 33 ln 2 12 (3 ln 3 ln 2)
1 ln 1 3 ln 5 3 ln 5
2. (a) ln 125
(c) ln 7 7 ln 73/2
2
3
ln 3 12 ln 2
(e) ln 3 2 ln 3 ln 2
(f ) ln 13.5
ln 32 2 ln 2 2 ln 3
ln 4t 4 ln 2 ln 4t 4 ln 2 ln 2t 2 ln 2 ln
ln t
2t 2
2
2
4. (a) ln sec ln cos ln (sec )(cos ) ln 1 0
8x44 ln (2 x 1)
1/3
3
(t 1)(t 1)
3 ln t 2 1 ln (t 1) 3 ln t 2 1 ln (t 1) 3 13 ln t 2 1 ln (t 1) ln (t 1) ln (t 1)
(b) ln (8 x 4) ln 22 ln (8 x 4) ln 4 ln
(c)
5. ln
tt 1 2
2
t t 1 e 2 t e2t e2 e2 e2t t e2 1 t t e2
e 1
6. ln t 2 ln 8 ln t ln t 2 ln t ln 8 ln t 2 t ln 8 t 2 2t 8
t 2 2t 8 t 4 t 2 0 t 4 or t 2 (Not in domain) t 4
7.
y ln 3x y
9.
y ln t 2
dy
dt
31x (3) 1x
(2t)
1
t2
2
t
8.
y ln kx y
10.
y ln t 3/2
kx1 (k ) 1x
dy
dt
Copyright 2018 Pearson Education, Inc.
1
t 3/ 2
3 t1/2
2
23t
Section 7.2 Natural Logarithms
3x
11.
y ln 3x ln 3x 1
dy
dx
12.
y ln 10
ln 10 x 1
x
dy
dx
13.
y ln ( 1) d 11 (1) 11
15.
y ln x3
17.
y t (ln t )2
18.
y t ln t t (ln t )1/2
19.
y
20.
y x 2 ln x
21.
y
ln t
t
22.
y
1 ln t
t
23.
y 1 ln x y
24.
y 1 ln x y
25.
y ln (ln x) y
26.
y ln ln (ln x) y
27.
dy
y sin (ln ) cos (ln ) d sin (ln ) cos (ln ) cos (ln ) 1 sin (ln ) 1
sin (ln ) cos (ln ) cos (ln ) sin (ln ) 2 cos (ln )
28.
y ln (sec tan ) d
x4
4
dy
dx
dy
dt
4
dy
dt
t
2
) 1x
3
x
y ln (2 2) d
16.
y (ln x)3
d (ln t ) (ln t ) 2
(ln t ) 2 2t (ln t ) dt
dy
dx
dy
dx
dy
dt
dy
14.
2t ln t
t
dy
dx
4 x 2 ln x
x4 1
4 x
d (ln x)
3(ln x) 2 dx
t2
t
t (ln t )1/ 2
2t
(ln t )1/2
3
x2 1x 2x ln x 4 x6 (ln x)3 ( x 2 x ln x ) 4 x7 (ln x)3 8x7 (ln x)4
3
1t (1ln t )(1) 11ln t ln t
t2
t2
t2
1x (ln x) 1x 1x lnx x lnx x
(1 ln x )2
(1 ln x )2
(1 ln x ) ln x x 1x ( x ln x )
(1 ln x )
1
ln x
1
x
2
1
x (1 ln x )2
1x (1ln x)2 ln x 1
(1 ln x )2
ln x
(1 ln x ) 2
1
x ln x
1
d
ln (ln x ) dx
dy
1
2(ln t )1/ 2
416x x3 ln x
t2
(1 ln x )
3(ln x ) 2
x
(ln t )2 2 ln t
d (ln t ) (ln t )1/2
(ln t )1/2 12 t(ln t ) 1/2 dt
x3 ln x
212 (2) 11
1t (ln t )(1) 1ln t
ln x
x ln x
2
1
x3
4
1
10 x 1
3x
x
ln x 16
(10x
1
x
dy
dy
dx
2
1
3 x 1
ln (ln x) ln (ln1 x) ln1x dxd (ln x ) x (ln x)1ln (ln x)
sec tan sec2
sec tan
sec (tan sec )
tan sec
sec
Copyright 2018 Pearson Education, Inc.
451
452
Chapter 7 Transcendental Functions
1
x x 1
ln x 12 ln ( x 1) y 1x 12
29.
y ln
30.
y
31.
ln t
y 11ln
t
32.
y ln t ln t1/2
ln 11 xx
1
2
dy
dt
1
2
x11 2(2 xx(x1)1)x 23x(xx21)
1 x 1
ln (1 x) ln (1 x) y 12 11x 11x (1) 12 (11 xx)(1
x ) 1 x 2
(1ln t )
1t (1ln t ) t1 1t lnt t 1t lnt t
(1ln t )2
2
t (1ln t ) 2
12 ln t1/2
33.
1
y ln sec (ln ) d sec (ln
d sec (ln )
) d
sec (ln ) tan (ln ) d
d
sec (ln )
34.
y ln 1 2 ln
ln t1/2
1
2
1/2
1/2
dy
dt
1 1 t 1/2
t1/ 2 2
1
2
ln t1/2
(1ln t ) 2
1/2
d ln t1/2
dt
1 d
t1/ 2 dt
t1/2
dy
12 (ln sin ln cos ) ln (1 2 ln ) d
1 cos
2 sin
(ln )
tan (ln )
2
sin
cos
1 2 ln
4
cot tan
(1 2 ln )
1
2
1
4t ln t
dy
sin cos
1/2
35.
x 2 15
5 ln x 2 1 1 ln (1 x) y 52 x 1 1 (1) 10 x 1
y ln
2
x 2 1 2 1 x
x 2 1 2(1 x )
1 x
36.
y ln
37.
y
38.
y
( x 1)5
( x 2) 20
x2
x 2 /2
3
x
x
ln 3 x
33 x2
ln t dt
2
41.
y
43.
0
25
5 ln ( x 1) 20 ln ( x 2) y 12 x51 x20 2 52 ( x(x 2)1)(4(x x2)1) 52 ( x 31)(x x2 2)
dy
dx
d x 2 ln
ln x 2 dx
dy
dx
ln
2
3
ln 2 ln 3 ln
3
d 3 x ln
x dx
x
dy ln y 2 25 C
sin t
2 cos t
x2
2
d x2
| x|
dx 2 2 x ln | x | x ln 2
x ln 3 x 13 x2/3 ln x 12 x1/2
d
x dx
ln x
2 x
3 1x dx ln
2y
1
2
ln t dt
39.
2
2
3
0
0
40.
1 3x32 dx ln 3x 2 1 ln 2 ln 5 ln 52
42.
4r82r5 dr ln
4r 2 5 C
dt ln |2 cos t |0 ln 3 ln1 ln 3; or let u 2 cos t du sin t dt with t 0 u 1 and
sin t
3
3
t u 3 2cos t dt u1 du ln | u |1 ln 3 ln 1 ln 3
0
1
Copyright 2018 Pearson Education, Inc.
Section 7.2 Natural Logarithms
44.
/3 4 sin
0
1 4 cos
d [ln |1 4 cos |]0 /3 ln |1 2| ln 3 ln 13 ; or let u 1 4 cos du 4 sin d with
0 u 3 and 3 u 1
/3 4 sin
1 4 cos
0
1 1
3 u
d
du [ln | u |]13 ln 3 ln 13
2 2 ln x
x
1
45. Let u ln x du
1
x
dx; x 1 u 0 and x 2 u ln 2;
46. Let u ln x du
1
x
dx; x 2 u ln 2 and x 4 u ln 4;
dx
ln 2
0
2
2
2u du [u 2 ]ln
0 (ln 2)
2 x dxln x ln 2 u1 du ln u ln 2 ln (ln 4) ln (ln 2) ln ln 2 ln ln 2 ln ln 2 ln 2
4
ln 4
ln 4
1
x
47. Let u ln x du
4
ln 4 2
16
dx
2 x ln x
ln 4
ln 2
ln14 ln12
1 ln 16 u 1/2
2 ln 2
/2
0
tan
x
2
1
2 ln 2
1
ln 4
ln 16
2
t dt du ln | u | C ln |6 3 tan t | C
63sec
u
3 tan t
sec y tan y
2 sec y
dy du
ln | u | C ln |2 sec y | C
u
du 12 sin 2x dx 2 du sin 2x dx; x 0 u 1 and x 2 u 1 ;
2
dx
/2 sin 2x
cos 2x
0
1/ 2 du
u
1
dx 2
52. Let u sin t du cos t dt ; t 4 u
/2
ln12 2 ln1 2 ln12
du u1/2
ln 16 ln 2 4 ln 2 ln 2 2 ln 2 ln 2 ln 2
ln 2
50. Let u 2 sec y du sec y tan y dy;
x
2
1
ln 22
dx; x 2 u ln 2 and x 16 u ln16;
49. Let u 6 3 tan t du 3 sec2 t dt ;
51. Let u cos
2 ln 2
dx; x 2 u ln 2 and x 4 u ln 4;
1
x
48. Let u ln x du
ln 22
ln 4
du u1
2 x(lndxx)2 ln 2 u
2
/2 cos t
1
/4 cot t dt /4 sin t dt 1/
2 ln | u |1
1/ 2
1
2
/2 2 cot 3 d /2
2 cos
sin 3
3
1
d 6
3/2 du
u
1/2
2 ln 2 ln 2
and t 2 u 1;
1
2
ln | u |1/
du
2 u
2 ln
2
1
2
ln
ln 2
53. Let u sin 3 du 13 cos 3 d 6 du 2 cos 3 d ; 2 u
6 ln | u |1/2
3/2
6 ln
1
2
3
2
and u
ln
1
2
6 ln
3
2
/12
55.
2
6 tan 3 x dx
/12 6 sin 3 x
cos 3 x
0
dx
x 2x
dx
2 x 1 x
dx 2
1/ 2 du
u
1
; let u 1 x du
2 ln | u |1
1/ 2
1
2 x
dx;
2 ln
dx
2 x 1 x
1
2
u
du
ln 1 x C ln 1 x C
Copyright 2018 Pearson Education, Inc.
;
3 ln 27
u
54. Let u cos 3 x du 3sin 3 x dx 2du 6 sin 3x dx; x 0 u 1 and x 12
0
453
1
2
;
ln 1 2 ln 2 ln 2
ln | u | C
454
Chapter 7 Transcendental Functions
56. Let u sec x tan x du sec x tan x sec2 x dx (sec x )(tan x sec x) dx sec x dx
57.
sec x dx
ln (sec x tan x )
y
x 2 1 ( x 1) 2
dy
2 x 1
2 x ( x 1)
2
2 x 2 x 1 | x 1|
2
x 2 1 ( x 1) 2 x 2 x x 1
x 2 1 ( x 1)
x 1( x 1)
1/2
3/ 2
1/2
12
x 1
x 2 1 x 1
ln y 12 ln t ln (t 1) 1y dydt 12 1t t 11
t 1 1 1
t 1
1
t 1 t t 1 2 t 1 t (t 1) 2 t (t 1)
y t (t11) t (t 1)
dy
dt
x ( x 1) (2 x 1)
2 x ( x 1)
2
dt 12
61.
x( x 1) 1x x11
y t t 1 t t 1
2 y
ln y 12 ln x( x 1) 2 ln y ln ( x) ln ( x 1) y 1x x11
x2 1 ( x 1)2 ln y 12 ln x2 1 2 ln ( x 1) yy 12 x2 x1 x21
y
60.
(ln u )1 2 u1 du 2(ln u )1 2 C 2 ln (sec x tan x) C
12
59.
du
u ln u
y x( x 1) x( x 1)
y 12
58.
du ;
u
1
t ( t 1)
dy
ln y 12 ln t ln(t 1) 1y dt 12 1t t 11
2t 1
t (t 1)
2
2t 1
t t
2
3/ 2
y 3 (sin ) ( 3)1/2 sin ln y 12 ln ( 3) ln (sin )
1 dy
y d
2(13) cos
sin
dy
d 3 (sin ) 2(13) cot
62.
dy
d (tan ) 2 1
63.
221
2
dy
1
y (tan ) 2 1 (tan )(2 1)1/2 ln y ln (tan ) 12 ln (2 1) 1y d sec
tan
2
sec2
tan
211 (sec2 ) 2 1
tan
2 1
dy
dy
y t (t 1)(t 2) ln y ln t ln (t 1) ln (t 2) 1y dt 1t t 11 t 12 dt t (t 1)(t 2) 1t t 11 t 12
(t 1)(t 2) t (t 2) t (t 1)
2
t (t 1)(t 2)
t (t 1)(t 2)
3t 6t 2
64.
dy
1
y t (t 1)(
ln y ln1 ln t ln (t 1) ln (t 2) 1y dt 1t t 11 t 12
t 2)
65.
dy
dt
1
t (t 1)(t 2)
1
1 1
t t 1 t 2
1
t (t 1)(t 2)
(t 1)(t 2) t (t 2) t (t 1) 3t 2 6t 2
2
t (t 1)(t 2)
t 3 3t 2 2t
5 ln y ln ( 5) ln ln (cos )
y cos
1 dy
y d
dy
sin
5
15 1 cos
cos
d
Copyright 2018 Pearson Education, Inc.
15 1 tan
Section 7.2 Natural Logarithms
66.
y sin ln y ln ln (sin ) 12 ln (sec )
sec
dy
d sin 1 cot 12 tan
sec
67.
y
x x 2 1
( x 1) 2/3
y
y
69.
y3
(2 x 1)5
x ( x 2)
y 13
70.
1
x
y
y
1x
y
y
(sec )(tan )
2 sec
x 2
x 2 1 3( x 1)
1
2
10 ln ( x 1) 5 ln (2 x 1)
5 5
x 1 2 x 1
x 1(2 x 3)
2
1
x
1
x 2
2x
x 2 1
(2 x 1)5
x51 2 x51
yy 13 1x x1 2 x2 x1
( x 1)10
y
ln y 13 ln x ln( x 2) ln x 2 1
3 x ( x 2)
x 2 1
y 13 3
1 cos
sin
x 2
x 2 1 3( x 1)
ln y
x ( x 1)( x 2)
y3
71. (a)
x x 2 1
( x 1) 2/3
x 2 1
ln y ln x 12 ln x 2 1 23 ln ( x 1)
( x 1)10
68.
1 dy
y d
455
2
ln y 13 ln x ln ( x 1) ln ( x 2) ln x 2 1 ln (2 x 3)
x 1(2 x 3) x
x ( x 1)( x 2)
1
2
1 1
x 1 x 2
2x 2
x 2 1 2 x 3
sin x tan x 0 x 0; f ( x ) 0 for x 0 and f ( x ) 0 for
f ( x) ln (cos x) f ( x) cos
4
x
0 x 3 there is a relative maximum at x 0 with f (0) ln (cos 0) ln 1 0; f 4
ln 12 12 ln 2 and f 3 ln cos 3 ln 12 ln 2. Therefore, the absolute
ln cos 4
minimum occurs at x 3 with f 3 ln 2 and the absolute maximum occurs at x 0 with f (0) 0.
(b)
f ( x) cos (ln x) f ( x)
sin (ln x )
x
0 x 1; f ( x) 0 for
1
2
x 1 and f ( x) 0 for 1 x 2
there is a relative maximum at x 1 with f (1) cos (ln 1) cos 0 1; f
12 cos ln 12
cos ( ln 2) cos (ln 2) and f (2) cos (ln 2). Therefore, the absolute minimum occurs at x
x 2 with f
72. (a)
(b)
12 f (2) cos (ln 2), and the absolute maximum occurs at x 1 with
f ( x) x ln x f ( x) 1 1x ; if x 1, then f ( x) 0 which means that f ( x) is increasing
f (1) 1 ln1 1 f ( x) x ln x 0, if x 1 by part (a) x ln x if x 1
5
5
5
73.
1 ln 2 x ln x dx 1 ln x ln 2 ln x dx (ln 2)1 dx (ln 2)(5 1) ln 2
74.
A
f (1) 1.
/3
0
tan x dx 0
/4
ln 1 ln
1
2
ln
1
2
0
sin x
/4 cos x
tan x dx
ln1 ln 2 ln 2
3
2
dx
/3 sin x
0
cos x
4
ln 16
/3
dx ln |cos x | /4 ln |cos x |0
ln 2
Copyright 2018 Pearson Education, Inc.
0
1
2
and
456
Chapter 7 Transcendental Functions
75. (a)
2
2
g ( x) x ln x g ( x) x 2 ln x 1x ln x ln x 2 ln x critical points at x 1, e 2
g | | ,
1
e2
increasing on 0, e 2 and 1, , decreasing on e2 , 1
(b) local maximum at g e2 4e 2 0.54, local minimum, absolute minimum at g 1 0, no absolute
maximum
76. (a)
g ( x) x 2 2 x 4 ln x g ( x) 2 x 2 4 1x
not in the domain
g | ,
x2
increasing on 2, , decreasing on 0, 2
2 x 2 x 1
x
critical points at x 2, 1 but x 1 is
(b) local minimum, absolute minimum at g 2 4 ln 2 2.77, no local, absolute maxima
2
3
3
3
77. V 2 dy 4 y11 dy 4 ln | y 1|0 4 ln 4 ln1 4 ln 4
0 y 1
0
78. V
2
6
79. V 2
2
1/2
2
2
x 12 dx 2 1x dx 2 ln | x |1 2 2 ln 2 ln 12 2 2 ln 2 ln 24 ln 16
12
x
2
2
3
3
80. V 93x dx 27 33x dx 27 ln x3 9
0 x 9
0 x 9
27 ln 4 54 ln 2
81. (a)
2 cos x
2
dx ln (sin x) 6 ln 1 ln 12 ln 2
6 sin x
cot x dx
0 27 ln 36 ln 9 27 ln 4 ln 9 ln 9
3
2
2
2
2
2
2
8
2
2
y x8 ln x 1 y 1 4x 1x 1 x 4x 4 x 4x 4 L 1 y dx
4
8 x2 4
4 4x
y 2
(b) x 4
L
dx
8
8 x
4 4
2
1x dx x8 ln | x | 8 ln 8 2 ln 4 6 ln 2
4
1 dy
y
dx y 2 1 dx
2 ln 4 dy
8
y
dy
12
4
dx
dy
2
12 y 2 16
8y
4
2
12 y
4 8
dy
dy
y
1 8 2y
2y
2
2
y 2 16
y 2 16
1 8y 8y
y2
16
2
12
2 ln y 9 2 ln 12 1 2 ln 4
4
8 2 ln 3 8 ln 9
82. L
2
1
dy
1 12 dx dx 1x y ln | x | C ln x C since x 0 0 ln 1 C C 0 y ln x
x
Copyright 2018 Pearson Education, Inc.
Section 7.2 Natural Logarithms
2
2
1x dx
457
(b)
83. (a) M y x 1x dx 1, M x 21x
1
1
2
2
2
12 12 dx 21x 14 , M 1x dx
1 x
1
1
M
ln | x |1 ln 2 x My ln12 1.44 and
2
y
Mx
M
14 0.36
ln 2
dx
16
16 1
16
16
84. (a) M y x 1 dx x1/2 dx 32 x3/2 42; M x
1
1 2 x
1
1
x
1
x
1 16 1
2 1 x
dx
16
M
16
M
16
12 ln | x |1 ln 4; M 1 dx 2 x1 2 6 x My 7 and y Mx ln64
1
x
1
dx 4 dx 60, M dx 2 x dx
3; M
dx 4 dx 4 ln | x | 4 ln 16 x
16
(b) M y x 1
1
x
16
85.
Mx
M
16
x
1
16
4 x1 2
1
y
16
4
x
1
1
x
1
1
2 x
1
x
16 1
1 x
4
x
16 3 2
4
x
1
My
M
16
1
15
ln 16
and
3
4 ln 16
2
f ( x) ln x3 1 , domain of f: (1, ) f ( x) 33x ; f ( x) 0 3x 2 0 x 0, not in the domain:
x 1
3
f ( x) undefined x 1 0 x 1, not a domain. On (1, ), f ( x) 0 f is increasing on (1, ) f
is one-to-one
86.
g ( x) x 2 ln x , domain of g: x 0.652919 g ( x)
2 x 1x
2
2 x ln x
2 x 2 1
2 x x 2 ln x
; g ( x) 0 2 x 2 1 0
no real solutions; g ( x) undefined 2 x x 2 ln x 0 x 0 or x 0.652919, neither in domain. On
x 0.652919, g ( x) 0 g is increasing for x 0.652919 g is one-to-one
87.
88.
dy
dx
1 1x at (1, 3) y x ln | x | C ; y 3 at x 1 C 2 y x ln | x | 2
d2y
dy
dy
sec 2 x dx tan x C and 1 tan 0 C dx tan x 1 y (tan x 1) dx ln |sec x | x C1 and
0 ln |sec 0| 0 C1 C1 0 y ln |sec x | x
dx 2
L( x) f (0) f (0) x, and f ( x) ln (1 x) f ( x) x 0 11x
1 L( x) ln1 1 x L( x) x
x 0
(b) Let f ( x) ln ( x 1). Since f ( x ) 1 2 0 on [0, 0.1], the graph of f is concave down on this
89. (a)
( x 1)
interval and the largest error in the linear approximation will occur when x 0.1. This error is
0.1 ln(1.1) 0.00469 to five decimal places.
Copyright 2018 Pearson Education, Inc.
458
Chapter 7 Transcendental Functions
(c) The approximation y x for ln (1 x) is best
for smaller positive values of x; in particular
for 0 x 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., | E ( x)| 0.095 for 0 x 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 x 0.1. This is
consistent with the estimate given in part (b)
above.
d ln a 1 a 1 and d ln a ln x 0 1 1 . Since in a and
90. For all positive values of x, dx
x
x
dx
x
x
x ax x 2
ln a ln x have the same derivative, then ln ax ln a ln x C for some constant C. Since this equation holds
for all positive values of x, it must be true for x 1 ln a1 ln a ln 1 C ln a 0 C ln a1 ln a C.
Thus ln a ln a C C 0 ln ax ln a ln x.
91. (a)
(b)
x . Since |sin x | and |cos x | are less than
y acos
sin x
or equal to 1, we have for a 1
1 y 1 for all x.
a 1
a 1
Thus, lim y 0 for all x the graph of y looks
a
more and more horizontal as a .
92. (a) The graph of y x ln x appears to be
concave upward for all x 0.
(b)
y x ln x y 1 1x y 13 2 12 12 4x 1 0 x 4 x 16. Thus y 0 if
2 x
4x
x
x
0 x 16 and y 0 if x 16 so a point of inflection exists at x 16. The graph of y x ln x
closely resembles a straight line x 10 and it is impossible to discuss the point of inflection visually from
the graph.
Copyright 2018 Pearson Education, Inc.
Section 7.3 Exponential Functions
7.3
EXPONENTIAL FUNCTIONS
1. (a) e 0.3t 27 ln e 0.3t ln 33 (0.3t ) ln e 3 ln 3 0.3t 3 ln 3 t 10 ln 3
(b) e kt 12 ln ekt ln 21 kt ln e ln 2 t lnk2
(c) e(ln 0.2)t 0.4 eln 0.2
t
0.4 0.2t 0.4 ln 0.2t ln 0.4 t ln 0.2 ln 0.4 t
ln 0.4
ln 0.2
2. (a) e 0.01t 1000 ln e0.01t ln 1000 (0.01t ) ln e ln1000 0.01t ln1000 t 100 ln 1000
1 ln e kt ln101 kt ln e ln10 kt ln 10 t ln10
(b) e kt 10
k
(c) e(ln 2)t
1
2
t
eln 2
21 2t 21 t 1
3. e t x 2 ln e t ln x 2 t 2 ln x t 4(ln x) 2
2
2
2
4. e x e2 x 1 et e x 2 x 1 et ln e x 2 x 1 ln et t x 2 2 x 1
5. e2t 3et 0 et
2
3et 0 et et 3 0 et 3 t ln 3
6. e 2t 6 5e t e t
2
5et 6 0 e t 3 e t 2 0 et 3 t ln 3 or e t 2
t ln 2
7.
d ( 5 x ) y 5e 5 x
y e 5 x y e 5 x dx
8.
d 2 x y 2 e2 x 3
y e 2 x 3 y e2 x 3 dx
3
3
9.
y e5 7 x y e5 7 x
4
x x2
d
dx
y e 4
(5 7 x) y 7e57 x
x x2
d
dx
4
x x2 y
2
x
4 x x2
10.
ye
11.
y xe x e x y e x xe x e x xe x
12.
y (1 2 x)e 2 x y 2e 2 x (1 2 x)e 2 x
13.
y x 2 2 x 2 e x y (2 x 2)e x x 2 2 x 2 e x x 2 e x
14.
d (3 x )
y 9 x 2 6 x 2 e3 x y (18 x 6)e3 x 9 x 2 6 x 2 e3 x dx
d
dx
(2 x) y 2e 2 x 2(1 2 x)e 2 x 4 xe 2 x
2x e
y (18 x 6)e3 x 3 9 x 2 6 x 2 e3 x 27 x 2 e3 x
Copyright 2018 Pearson Education, Inc.
459
460
Chapter 7 Transcendental Functions
15.
y e (sin cos ) y e (sin cos ) e (cos sin ) 2e cos
16.
y ln 3 e ln 3 ln ln e ln 3 ln d 1 1
17.
y cos e
18.
dy
y 3e2 cos 5 d 3 2 e 2 cos 5 3 cos 5 e 2 dd (2 ) 5(sin 5 ) 3e2
dy
2
dy
d
e sin e e
sin e
2
2
d
d
2
2
d
d
2
( 2 ) 2 e sin e
2
2 e2 3cos 5 2 cos 5 5 sin 5
19.
y ln 3te t ln 3 ln t ln e t ln 3 ln t t dt 1t 1 1t t
20.
y ln 2et sin t ln 2 ln et ln sin t ln 2 t ln sin t
21.
dy
y ln e ln e ln 1 e ln 1 e d 1 1 dd 1 e 1 e 1
1 e
1 e
1 e
1 e
22.
y ln
dy
1
ln ln 1 d
dy
1
1
2
1
1
1
1
1
2
2 1
1
t cos t sin t
sin1 t dtd (sin t ) 1 cos
sin t
sin t
d
d
11 dd 1
1
2 1
dy
dt
1
2 1 1/ 2
23.
dy
d (cos t ) (1 t sin t )ecos t
y e(cos t ln t ) ecos t eln t tecos t dt ecos t tecos t dt
24.
dy
y esin t ln t 2 1 dt esin t (cos t ) ln t 2 1 2t esin t esin t ln t 2 1 (cos t ) 2t
25.
0
26.
y
ln x
d (ln x) sin x
sin et dt y sin eln x dx
x
e2 x
e4
x
e4 x dxd 4 x
d e 2 x ln e 4 x d e 4 x (2 x ) 2e2 x 4 x
ln t dt y ln e2 x dx
dx
4 xe2 x 4 xe 4 x
27. ln y e y sin x
4xe
2
x
2x
8e4 x
y ye (sin x) e
1
y
y
y
cos x y 1y e y sin x e y cos x
ye y cos x
1 ye y sin x
y
y
e cos x y 1 ye y sin x
y
28. ln xy e x y ln x ln y e x y 1x
1 ye x y
y y
xe x y 1
x
y
y 1 y e
1
y
x1 ye
x y
y 1y e x y e x y 1x
y xe x y 1
x y
Copyright 2018 Pearson Education, Inc.
Section 7.3 Exponential Functions
29. e 2 x sin ( x 3 y ) 2e 2 x 1 3 y cos ( x 3 y ) 1 3 y
y
2e 2 x
cos( x 3 y )
2e 2 x
cos( x 3 y )
3 y
461
1
2e2 x cos ( x 3 y )
3 cos ( x 3 y )
30. tan y e x ln x sec2 y y e x 1x y
xe 1 cos
x
2
y
x
2
x
31. 3 sin y y x3 cos y y y 3 x 2 3x 2 y cos y y 3x 2 (1 cos y ) y y 13cos
;
y
(1cos y )6 x 3 x 2 sin y y
y
(1cos y )2
2
6 x 6 x cos y 6 x cos y 6 x cos 2 y 9 x 4 sin y
(1cos y )3
4
(1 cos y )3
y
6 x 12 x cos y 6 x cos y 9 x sin y
32. ln y xe y 2
2
x
6 x 6 x cos y 3 x 2 sin y13cos
y 1 cos y
1cos y
(1cos y ) 2
ye y
1 xye
y xe y y (1)e y y xye y y ye y y xye y y ye y (1 xye y ) y ye y
1
y
; y
y
(1 xye y )( ye y y ye y ) ye y ye y xye y xye y y
(1 xye y )2
ye y xy 2 e2 y y e y y xye2 y y y 2e2 y xye 2 y y xy 2 e2 y y
y
y 2
(1 xye )
y 2e2 y (e y ye y )
ye y
1 xye y
y 2
(1 xye )
33.
e
35.
ln 2 e
37.
8e
39.
ln 4 e
40.
0
3x
5e x dx
ln 3 x
e3 x
3
1 xye y
1 xye
y
5e x C
ln 9
dx 2e x 2
2 e(ln 9)
ln 4
ln16
dx 4e x /4
0
(1 xye )
(1 xye )
dx 8e( x 1) C
ln16 x 4
y 2
y 3
ln 3
ln 9 x 2
y 2e2 y e y y ye y y
y 2e2 y xy 3e3 y ye2 y y 2 e2 y
dx e x
eln 3 eln 2 3 2 1
ln 2
( x 1)
e
2
4 e(ln16)
y 2e2 y (e y ye y ) y
(1 xye y )2
ye2 y 2 y 2e2 y xy 3e3 y
(1 xye y )3
34.
2e
36.
ln 2 e
38.
2e
x
0
3e2 x dx 2e x 32 e2 x C
x
(2 x 1)
0
dx e x
e0 eln 2 1 2 1
ln 2
dx e(2 x 1) C
e(ln 4) 2 2 eln 3 eln 2 2(3 2) 2
4
e0 4 eln 2 1 4(2 1) 4
41. Let u r1 2 du 12 r 1 2 dr 2 du r 1 2 dr ;
r
e r
dr e r
12
r 1 2 dr 2 eu du 2eu C 2e r
12
C 2e
r
C
42. Let u r1 2 du 12 r 1 2 dr 2 du r 1 2 dr ;
e
r
r
dr e r
12
r 1 2 dr 2 eu du 2e r
12
C 2e
r
C
Copyright 2018 Pearson Education, Inc.
462
Chapter 7 Transcendental Functions
43. Let u t 2 du 2t dt du 2t dt ;
44. Let u t 4 du 4t 3 dt
1
x
45. Let u
du
1
x2
du t 3 dt ;
1
4
dx du
46. Let u x 2 du 2 x 3 dx
1 x 2
e x
3
2
dx e x x 3 dx
1
2
1
2
e
u
1
x2
dx;
2te
t
t 2
2
dt eu du eu C e t C
3 t4
e dt
1/ x
ex
2
1
4
e
u
4
du 14 et C
dx eu du eu C e1 x C
du x 3 dx;
du 12 eu C 12 e x
2
2
C 12 e 1/ x C
47. Let u tan du sec2 d ; 0 u 0, 4 u 1;
4
0
1 etan sec2 d 0 4 sec2 d 01eu du tan 0 4 eu 0 tan 4 tan(0) e1 e0
1
(1 0) (e 1) e
48. Let u cot du csc2 d ; 4 u 1, 2 u 0;
4 1 e
2
cot
csc2 d 42 csc2 d 10 eu du cot 24 eu 1 cot 2 cot 4 e0 e1
0
(0 1) (1 e) e
sec t tan t dt ;
49. Let u sec t du sec t tan t dt du
e
sec ( t )
u
sec ( t ) tan ( t ) dt 1 eu du e C
esec( t )
C
50. Let u csc ( t ) du csc ( t ) cot ( t ) dt ;
e
csc( t )
csc( t ) cot ( t ) dt eu du eu C ecsc ( t ) C
51. Let u ev du ev dv 2 du 2ev dv; v ln 6 u 6 , v ln 2 u 2 ;
ln ( 2)
ln ( 6) 2e
v
cos ev dv 2
2
6
2
2
cos u du 2 sin u
6
2 sin 2 sin 6 2 1 12 1
2
52. Let u e x du 2 xe x dx; x 0 u 1, x ln u eln ;
0
ln
dx
2
2 xe x cos e x
2
53. Let u 1 e r du e r dr ;
54.
x
1
cos u du sin u 1 sin ( ) sin (1) sin (1) 0.84147
1eer dr u1 du ln | u | C ln 1 e
11e x dx ee x 1 dx; let u e
r
x
C
1 du e x dx du e x dx;
ee x 1 dx u1 du ln | u | C ln e
x
r
x
1 C
Copyright 2018 Pearson Education, Inc.
Section 7.3 Exponential Functions
55.
dy
dt
et sin et 2 y et sin et 2 dt ;
let u et 2 du et dt y sin u du cos u C cos et 2 C ; y (ln 2) 0
cos eln 2 2 C 0 cos (2 2) C 0 C cos 0 1; thus, y 1 cos et 2
56.
dy
dt
e t sec 2 et y e t sec2 e t dt ;
let u et du e t dt 1 du e t dt y 1 sec 2 u du 1 tan u C
1 tan et C ; y (ln 4) 2 1 tan e ln 4 C 2 1 tan 1 C 2
1 (1) C 2 C 3 ; thus, y 3 1 tan e t
57.
d2y
dx 2
2e x
2e x C ; x 0 and
dy
dx
dy
dx
0 0 2e0 C C 2; thus
dy
dx
2e x 2
y 2e x 2 x C1 ; x 0 and y 1 1 2e0 C1 C1 1 y 2e x 2 x 1 2 e x x 1
58.
d2y
dy
1 e 2t dt t 12 e 2t C ; t 1 and
dt 2
dy
t 12 e 2t 12 e 2 1 y 12 t 2 14 e 2t
dt
dy
dt
0 0 1 12 e 2 C C 12 e 2 1; thus
12 e2 1 t C1; t 1 and y 1 1 12 14 e2 12 e2 1 C1
C1 12 14 e2 y 12 t 2 14 e 2t 12 e 2 1 t 12 14 e 2
y 3 x y 3 x (ln 3)(1) 3 x ln 3
59.
y 2 x y 2 x ln 2
61.
y5
dy
ds
5
62.
y 2s
dy
ds
2 s (ln 2)2 s ln 22
63.
y x y x ( 1)
65.
y (cos )
66.
y (ln ) d (ln )( 1) 1
67.
y 7sec ln 7 d 7sec ln 7 (ln 7)(sec tan ) 7sec (ln 7)2 (sec tan )
68.
y 3tan ln 3 d 3tan ln 3 (ln 3) sec2 3tan (ln 3)2 sec2
69.
y 2sin 3t
s
2
2
s
60.
(ln 5)
12 s1 2 2ln 5s 5 s
2
dy
dy
dy
dy
dt
2
(ln 4)s2
64.
d 2(cos )
dy
s 2s
s2
y t1e
dy
dt
(1 e)t e
(sin )
2 1
(ln )( 1)
2sin 3t ln 2 (cos 3t )(3) (3 cos 3t ) 2sin 3t (ln 2)
Copyright 2018 Pearson Education, Inc.
463
464
Chapter 7 Transcendental Functions
70.
y 5 cos 2t
71.
y log 2 5
72.
y log3 (1 ln 3)
73.
y
ln x
ln 4
74.
y
x ln e
ln 25
75.
y x3 log10 x x3
ln 5
ln 2
2
lnlnx4
1 x2
ln10
dy
d
ln x
ln 4
2lnlnx5
77.
y log3
xx11
y log5
1
2
ln(1 ln 3)
ln 3
ln13 11ln 3 (ln 3) 11ln 3
3
x ln 4
2 ln1 5 ( x ln x) y 2 ln1 5 1 1x 2 xxln15
1 x3 ln
ln 10
ln x
ln 10
x y
1
ln 10
x3 1x 3x2 ln x ln110 x2 3x2 lnln10x
1
(2 ln r ) 1 2 ln r
lnln 3r lnln 9r (lnln3)(lnr 9) dydr (ln 3)(ln
r
r (ln 3)(ln 9)
9)
2
ln 3
7x
3x2
ln
xx 11
log5 3 x7x 2
80.
y log 7
sin cos
e 2
cos
(sin )(ln 7)
ln 5
y sin log 7 sin
ln 3
ln 3
79.
dy
dy
d
x 3 ln x y
2 ln
ln 4
ln 4
dy
dx
ln 7 x 12 ln (3 x 2)
d
3 x 2 log10 x
y log3 r log9 r
ln12 51 (5) ln1 2
x ln x
2 ln 5 2 ln 5
76.
78.
5 cos 2t ln 5 (sin 2t )(2) (2sin 2t ) 5 cos 2t (ln 5)
dy
dt
(ln 3) ln
xx 11 ln
ln 3
(ln 5) 2
7
27 x
ln
2(33x 2)
ln(sin ) ln(cos ) ln e ln 2
ln 7
y log10 e x
82.
5 5 y
y 2log
l n
5
2
x
ln10
ln5
83.
y 3log 2 t 3(ln t )
84.
y 3log8 log 2 t
(ln 2)
(ln 5) 2
(3 x 2) 3 x
2 x (3 x 2)
ln 3 7x x 2
ln 5
ln25
1
2
ln
37xx2
1
x (3 x 2)
lnln7 ddy sin lnln7 cos lnln7 ln1 7 sin log7 ln17 cos log7
81.
3 x7x 2
ln 5
(cossin
1 ln 2
)(ln 7) ln 7 ln 7
ln e x
ln10
xx11 ln ( x 1) ln ( x 1) dydx x11 x11 ( x1)(2x1)
y
ln (sin ) ln(cos ) ln 2
ln 7
ln17 (cot tan 1 ln 2)
1
ln10
2 lnln5 5 ln 55 (1) 5 ln1 5 5 ln 5 2log5 ( ln 51)5
2
2
ln 5 2 log5
2 lnln5
dy
dt
3(ln t )
3ln log 2 t
ln 8
3ln
(ln 2)
(ln 3)
lnln 2t dy
ln 8
dt
t ln1 2 1t log2 3 3log t
2
3
1
1
1
ln38 (ln t )/(ln
2) t ln 2 t (ln t )(ln 8) t (ln t )(ln 2)
Copyright 2018 Pearson Education, Inc.
Section 7.3 Exponential Functions
85.
y log 2 8t ln 2
86.
y
87.
5
x
t ln eln 3
ln 3
dx
sin t
3 ln 2(ln 2)(ln t ) 3 ln t dy 1
ln 8 ln t ln 2
ln 2
t ln 3sin t
ln 3
dt
ln 2
89.
90.
0
ln 3
dt
5x C
ln 5
2
5
1 1
0 2
d
t
t (sin t )(ln 3) t sin t dy sin t t cos t
88. Let u 3 3x du 3x ln 3 dx ln13 du 3x dx;
1
2
0
d
0
1
1
d 2 1
ln 2
0
15
ln 15
1 d
2 5
1
2
ln
0
1
1
2
ln
1
2
1
2
ln
3x
3 3 x
12
dx ln13 u1 du ln13 ln | u | C
1
2(ln1ln 2)
ln 33x
ln 3
1
2 ln 2
15 1 (1 25) 24 24
1
ln1ln 5 ln 5
ln 5 ln 15
ln 15
2
2
1
91. Let u x 2 du 2 x dx 12 du x dx; x 1 u 1, x 2 u 2;
1
2
x dx
2
x2
1 12 2
2
u
du
1
2
1
x
x
dx
4 x1/2
2
1
2 ln1 2 22 21 ln12
2
2u
ln 2 1
92. Let u x1/2 du 12 x 1/2 dx 2 du
42
dx
x
; x 1 u 1, x 4 u 2;
2
2
( u 1)
x 1/2 dx 2 2u du 2ln 2
1
1
ln12 23 22 ln42
93. Let u cos t du sin t dt du sin t dt ; t 0 u 1, t 2 u 0;
2 cos t
0
7
0
0
sin t dt 7u du ln7 7
1
1
u
ln17 70 7 ln67
94. Let u tan t du sec2 t dt ; t 0 u 0, t 4 u 1;
0 3
0 3
95. Let u x 2 x ln u 2 x ln x
1
13
ln 13
1 1 u
4 1 tan t
sec2 t dt
du
1 du
u dx
8
u
0
13 13
ln13
2 ln x (2 x )
1
0
2
3 ln 3
1x dudx 2u (ln x 1) 12 du x2 x (1 ln x) dx;
x 2 u 24 16, x 4 u 4 65,536;
4 2x
2 x
465
(1 ln x) dx
1 65,536 du
2 16
1
2
65,536
12 (65,536 16) 65,520
32, 760
u 16
2
Copyright 2018 Pearson Education, Inc.
C
466
Chapter 7 Transcendental Functions
2
96. Let u 1 2 x du 2 x 2 (2 x) ln 2dx
x2
1x22
97.
3x
99.
0
101.
3
3
1
2 ln 2
dx
x2
dx 3 x
dx x
2 1
3
3
0
2 1
C
dx x
2 1
98.
x
100.
1 x
C
3 1
2
2 1 x
2
du 2 x x dx
2
ln 1 2 x
2 ln 2
u1 du 2 ln1 2 ln | u | C
3 1
1
2 ln 2
e (ln 2) 1
2
2
C
e
dx xln 2
1
ln 2
eln 2 1ln 2
ln 2
2 1
ln 2
ln x
ln10
1x dx; u ln x du 1x dx
x)
ln x
1
1
ln10
C
1x dx ln10
12 u 2 C (ln2 ln10
u du ln10
log10 x
x
dx
2
102.
1x dx; u ln x du 1x dx; x 1 u 0, x 4 u ln 4
ln 4
4
(ln 4)
(ln 4)
x 1 dx ln 4 1 u du 1
1 u2
ln
ln12 12 (ln 4)2 2 ln 2 ln 4
x
ln
2
ln
2
ln
2
2
1
0
0
4 log 2 x
x
1
dx
4 ln x
ln 2
1
2
103.
4 ln 2 log 2 x
x
1
104.
105.
2 log 2 ( x 2)
x2
0
107.
108.
ln12 4(ln22)
2
dx (ln x)2 1 (ln e)2 (ln 1)2 1
(ln( x 2))
ln ( x 2) x 1 2 dx ln12 2
1 2
ln 2 0
(ln 2)2
2
dx
3
2
2
2
0
ln12 (ln24)
2
(ln 2)2
2
ln 2
10
ln (10 x) 101x dx ln10
10 10
ln10 1 10
ln(10 x )
20
2
10
1 10
10 (ln 100)
ln10
20
2
(ln1) 2
2
10 4(ln 10) 2 ln 10
ln10
20
2
dx
9 2 log10 ( x 1)
x 1
0
dx
9
2
ln
ln10 0
3 2 log 2 ( x 1)
x 1
2
dx
2 3 ln( x 1) 1
x 1
ln 2 2
e
e (ln 10)(2 ln x ) 1
(ln 10)
x
1
dx
dx
10 log10 (10 x )
x
1 10
4
(2 ln 2) 2 2(ln 2) 2
e 2 ln10 log10 x
x
ln 4
lnln 2x dx 14 lnxx dx 12 (ln x)2 1 12 [(ln 4)2 (ln1)2 ] 12 (ln 4)2
4 ln 2
x
1
1
106.
1
2
dx
2
( x 1)
1
x 1
dx
2
ln 10
2
ln 2
9
ln( x 1) 2
2
0
3
ln ( x 1) 2
2
2
2
ln10
(ln 10)
2
ln22 (ln22)
Copyright 2018 Pearson Education, Inc.
2
2
(ln 1)2
2
(ln1) 2
2
ln10
ln 2
1
ln 2
Section 7.3 Exponential Functions
109.
x logdx
10
x
(ln 10)
dx (ln10) dx; u ln x du
ln 10
ln x
1
x
1
ln x
2
2 (ln x )
x
x logdx8 x 2 x lndx 2 (ln 8)
111.
1
x
ln 8
dt ln | t |1
ln x
dx (ln 8)2
dt ln | t |1 ln e x ln1 x ln e x
(ln x ) 1
1
C
(ln 8) 2
ln x
C
1/ x 1
t
dt ln | t |1
ex
113.
1
114.
1 x 1 dt
ln a 1 t
1/ x
ln
ln1 ln1 ln | x | ln1 ln x, x 0
1
x
x
ln1a ln | t |
1
ln x
ln a
ln1 log x, x 0
ln
a
a
115. y ( x 1) x ln y ln( x 1) x x ln( x 1)
y
y
ln ( x 1) x ( x11) y ( x 1) x xx1 ln ( x 1)
116. y x 2 x 2 x y x 2 x 2 x ln y x 2 ln x 2 x 2 x ln x
y 2 x y x 2 (2 2 ln x) y
t t1 2
118. y t
dx
ln |ln x | ln1 ln (ln x ), x 1
ex 1
1 t
112.
117. y
1
x
ln1x 1x dx (ln10) u1 du (ln10) ln | u | C (ln 10) ln |ln x | C
110.
ln x 1
t
1
x
467
t
t
t
t
x
t1/2 ln y ln t1/2
t ln y ln t t
1/ 2
1/ 2
2
1
y x2
y 2 x 2 x 1x 2 ln x 2 2 ln x
x 2 x x 2 (2 2 ln x) 2 x 2 x x 2 x x 2 x ln x
2t ln t 1y dydt 12 (ln t ) 2t 1t ln2t 12 dydt t ln2t 12
t
t1/2 (ln t ) 1y dydt 12 t 1/2 (ln t ) t1/2 1t ln2t t2 dydt ln2t t 2 t
119. y (sin x) x ln y ln (sin x) x x ln (sin x)
120. y xsin x ln y ln xsin x (sin x)(ln x)
y
y
y
y
ln (sin x) x
x y (sin x) x ln (sin x ) x cot x
cos
sin x
(cos x)(ln x) (sin x)
1x sin x x(lnx x)(cos x)
sin x x (ln x )(cos x )
y xsin x
x
121. y sin x x y cos x x
d
dx
x x ; if u x x ln u ln x x x ln x uu x 1x 1 ln x 1 ln x
u x x (1 ln x) y cos x x x x (1 ln x) x x cos x x (1 ln x)
122. y (ln x)ln x ln y (ln x) ln (ln x)
y
ln(ln x ) 1
x
(ln x)
y
y
t
1x ln (ln x) (ln x) ln1x dxd (ln x) ln(lnx x) 1x
ln x
Copyright 2018 Pearson Education, Inc.
468
Chapter 7 Transcendental Functions
123. y x x3 y x ln y 3ln x ln y x 1y y ln y 3 1x 1y y x 2 y xy ln y 3 y xy
x 2 y xy 3 y xy ln y ( x 2 x) y 3 y xy ln y y
3 y xy ln y
x2 x
124. xsin y ln y sin y ln x ln(ln y ) sin y 1x cos y y ln x ln1y 1y y
y ln y sin y xy ln y ln x cos y y xy y ln y sin y xy xy ln y ln x cos y y
y ln y sin y ( x xy ln y ln x cos y ) y y
125. x y xy ln x xy ln y
1
x
y ln y sin y
x xy ln y ln x cos y
y ln y xy ln y x y 1y y 1 xy ln y x 2 ln y y x 2 y
1 xy ln y ( x 2 ln y x 2 ) y y
1 xy ln y
x 2 ln y x 2
126. e y y ln x y ln x ln y y ln x 1y y 1x ln y xyy x ln x y y ln y xyy x ln xy y ln y
( xy x ln x) y y ln y y
127.
x
2
f (t ) dt x ln x
y ln y
xy x ln x
f ( x) x 1x ln x f ( x) 1 ln x
x
128. f ( x) e2 f (t ) dt f (1) e2 and f ( x) f ( x)
f ( x) e
1
x C
f ( x )
f ( x)
2
1 ln f ( x) x C
e x eC Ce x , f (1) e2 Ce e2 C e f ( x) e e x e1 x
129. f ( x) e x 2 x f ( x) e x 2; f ( x) 0 e x 2 x ln 2; f (0) 1, the absolute maximum;
f (ln 2) 2 2 ln 2 0.613706, the absolute minimum; f (1) e 2 0.71828, a relative or local maximum
since f ( x) e x is always positive.
130. The function f ( x) 2esin( x /2) has a maximum whenever sin 2x 1 and a minimum whenever sin 2x 1.
Therefore the maximums occur at x 2k (2 ) and the minimums occur at x 3 2k (2 ), where k is
any integer. The maximum is 2e 5.43656 and the minimum is
2
e
0.73576.
131. f ( x) xe x f ( x) xe x (1) e x e x xe x f ( x) e x xe x (1) e x xe x 2e x
(a)
f ( x) 0 e x xe x e x (1 x) 0 e x 0 or 1 x 0 x 1, f (1) (1)e1 1e ; using second
derivative test, f (1) (1)e1 2e1 1e 0 absolute maximum at 1, 1e
(b)
f ( x) 0 xe
x
2e
x
e
x
( x 2) 0 e
f (1) 0 and f (3) e3 (3 2)
1
e3
x
0 or x 2 0 x 2, f (2) (2)e2
0 point of inflection at 2,
Copyright 2018 Pearson Education, Inc.
2
e2
2
e2
; since
Section 7.3 Exponential Functions
132. f ( x)
1e e e 2e e e
f ( x)
1e
1e
2x
ex
1 e2 x
2x
x
1e
3x
x
2x 2
e x 16e2 x e4 x
x
2x 2
469
1e e 3e e e 21e 2e
f ( x)
1 e
2x 2
3x
x
x
3x
2x
2x
2
2x 2
2x 3
e0
1 e2(0)
f ( x) 0 e x e3 x 0 e x 1 e2 x 0 e2 x 1 x 0; f (0)
(a)
1 e2 x
f (0)
2
0 e2 x 1 no real solutions. Using the second derivative test,
e0 16e2(0) e4(0)
1e
2(0) 3
4 0
8
12
absolute maximum at 0,
f ( x) 0 e x 1 6e2 x e4 x e x 0 or 1 6e2 x e4 x 0 e2 x
(b)
12 ; f ( x) undefined
ln 3 2 2
2
or x
ln 3 2 2
2
f ln 3 2 2
3 2 2,
ln 3 2 2
3 2 2
and f
; since
2
4 2 2
ln 3 2 2 3 2 2
f (1) 0, f (0) 0, and f (1) 0 points of inflection at
,
and
2
4 2 2
ln 3 2 2 3 2 2
,
.
2
4 2 2
x
( 6) 36 4
2
2
3 2 2
4 2 2
133. f ( x) x 2 ln 1x f ( x) 2 x ln 1x x 2 11 x 2 2 x ln 1x x x(2 ln x 1); f ( x) 0 x 0 or
x
ln x 12 . Since x 0 is not in the domain of f, x e 1/2 1 . Also, f ( x) 0 for 0 x 1 and
f ( x) 0 for x 1 . Therefore, f
e
f assumed at x
1
e
e
1
e
e
ln e 1e ln e1/2 21e ln e 21e is the absolute maximum value of
1 .
e
134. f ( x) ( x 3)2 e x f ( x) 2( x 3)e x ( x 3) 2 e x
( x 3)e x (2 x 3) ( x 1)( x 3)e x ; thus
f ( x) 0 for x 1 or x 3, and f ( x) 0 for
1 x 3 f (1) 4e 10.87 is a local maximum
and f (3) 0 is a local minimum. Since f ( x) 0 for
all x, f (3) 0 is also an absolute minimum.
ln 3
135.
0
136.
0
e2 x e x dx e2
2x
2 ln 2
137. L
ln 3
ex
0
e 2 ln 3
2
eln 3
e x/2 e x/2 dx 2e x/2 2e x/2 0
1
0
2 ln 2
x
1 e4 dx
dy
dx
ex / 2
2
e0
2
3
e0
9
2
1
2
1 82 2 2
2eln 2 2e ln 2 2e0 2e0 (4 1) (2 2) 5 4 1
y e x /2 C ; y (0) 0 0 e0 C C 1 y e x /2 1
Copyright 2018 Pearson Education, Inc.
470
Chapter 7 Transcendental Functions
1 dy 2 1 e 2 e dy
dy 2 dy e 2 e dy
ln 2 e y e y
2
0
138. S 2
ln 2 e y e y
2
0
2
ln 2
ln 2 e y e y 2
2
0
2
1
e2 x
4
2 x
12 e 4 dx
e
1
1
2
0
ln 3
ln 3 e x e x
e4 x 2e2 x 1 4e2 x
e
ln 2
ln 2 e x e x
12 e2 ln 2 2 ln 2 12 e2 ln 2 12 0 12
2x
1
2
dy
dx
dx
x
e x
2
ex ex
e x 1 e x 1
ln 3
1
e4 x 2e2 x 1
e
ln 2
2x
1
2
2e x ;
e2 x 1
dx
4
ln
dy
dx
sin x
cos x
dy
dx
e 1
ln 2
e
e
2x
2x
2
1
1
2
dx
2
dx
ln 3
ln 2
ln 3 e2 x 1
ln 2 e2 x 1
1
dx
4e 2 x
ln 3
ln 2
e2 x 1
e2 x 1
ex
e2 x 1
ex
2
dx
dx
4
cos x cot x
csc x
ln | u |3 2 ln
83
2
1 tan x dx
83 ln 32 ln 169
4
1 tan 2 x dx
0
(0) ln
ln sec 4 tan 4
cot x; L
4
6
2
1 cot x dx
4
6
4
sec2 x dx
0
2 1
1 cot 2 x dx
ln csc 6 cot 6
4
6
csc 2 x dx
4
csc x dx ln |csc x cot x | 6 ln csc 4 cot 4
2 1 ln 2 3 ln 2 3
d ( x ln x x C )
dx
144. average value
(b)
x
1 22 xe
0
4
(b) average value
145. (a)
tan x; L
sec x dx ln |sec x tan x |0
142. y ln csc x
6
ln 3
ln 3
2
dx; let u e x e x du e x e x dx, x ln 2 u eln 2 e ln 2 2 12 32 ,
141. y ln cos x
4
L
ln 2
8 31
du
32u
0
2 x
2x
1 e4 12 e 4 dx
dx 01 12 e x e x dx 12 e x e x 0 12 e 1e 0 e2e1
x ln 3 u eln 3 e ln 3 3 13 83
143. (a)
2 y
2y
2
140. y ln e x 1 ln(e x 1)
ln 2
2 0
e x e x dydx 12 e x e x ; L 01 1 12 e x e x dx 01
1
2
0
2 y
2y
1
4
12 4 2 ln 2 12 14 2 2 18 2 ln 2 1615 ln 2
139. y
ln 2 e y e y
2
0
2
e y e y
2
2 12 e2 y 2 y 12 e2 y
0
2
2
e y e y
2
x 1x ln x 1 0 ln x
1 e ln x
e 1 1
1 2 1 dx
21 1 x
2 1
e
dx e11 x ln x x 1 e11 (e ln e e) (1ln1 1) e11 (e e 1) e11
ln | x |1 ln 2 ln1 ln 2
2
f ( x) e x f ( x) e x ; L( x) f (0) f (0)( x 0) L( x) 1 x
f (0) 1 and L(0) 1 error 0; f (0.2) e0.2 1.22140 and L(0.2) 1.2 error 0.02140
Copyright 2018 Pearson Education, Inc.
Section 7.3 Exponential Functions
471
(c) Since y e x 0, the tangent line approximation always lies below the curve y e x . Thus
L( x) x 1 never overestimates e x .
146. (a)
y e x y e x 0 for all x the graph of y e x is always concave upward
ln b x
e
ln a
ln b x
(b) area of the trapezoid ABCD
12 ( AB CD)(ln b ln a )
e(ln a ln b)
(c)
ln b x
ln a e
2
2
ln a
ln b
e dx e 2 e (ln b ln a). Now 12 ( AB CD) is the height of the
ln a
midpoint M e(ln a ln b)
dx area of the trapezoid AEFD
since the curve containing the points B and C is linear
(ln b ln a )
ln b x
ln a
ln b
e dx e 2 e
(ln b ln a )
ln a
ln b
dx e x
eln b eln a b a, so part (b) implies that
ln a
ln a
ln b
a a b
e(ln a ln b ) 2 (ln b ln a) b a e 2 e
(ln b ln a ) e(ln a ln b ) 2 ln bb ln
a
2
ba
ln b ln a
eln a 2 eln b
2
2 2x
2 1 x 2
2 2x
0 1 x 2
147. A
dx 2
51
du
1 u
A 2
1
148. A 2
(1 x )
1
a b
2
eln a eln b
ba
ln b ln a
a b
2
ab
ba
ln b ln a
a b
2
dx; [u 1 x 2 du 2 x dx; x 0 u 1, x 2 u 5]
2 ln | u |1 2(ln 5 ln1) 2 ln 5
5
1 x
1 2
dx 2
1
1
1 x
dx 2 2 1 ln22
ln 2
1
12 2 ln22 32 ln32
149. (a) The vertical distance at x is V 2 x 3 ln x V 2 1x
is not in the domain. Thus V
12 0 at x 12
2 x 1
x
0 critical points are 0 and
1,
2
but 0
we have a minimum. Therefore x 12 determines a
minimum vertical distance of V 4 ln 2.
(b) The horizontal distance at y is H e y 12 ( y 3) H e y 12 0 critical point is y ln 12 . Thus
H ln 12 0 at y ln 12 we have a minimum. Therefore y ln 12 determines a minimum horizontal
distance of H 12 (4 ln 2).
150. The area of the rectangle is A xy xe x A (1)e x xe x (1 x)e x 0
critical point is 1. Thus A(1) 0 at x 1 we have a maximum. Therefore the dimensions 1 and
maximize the area of the rectangle.
Copyright 2018 Pearson Education, Inc.
1
e
472
Chapter 7 Transcendental Functions
151. From zooming in on the graph at the right, we
estimate the third root to be x 0.76666
152. The functions f ( x) xln 2 and g ( x) 2ln x appear
to have identical graphs for x 0. This is no
accident, because x ln 2 eln 2ln x eln 2
153. (a)
ln x
2ln x.
and f (3)
ln 3
ln 3
f ( x) 2 x f ( x) 2 x ln 2; L( x) 20 ln 2 x 20 x ln 2 1 0.69 x 1
(b)
154. (a)
f ( x) log3 x f ( x)
1 ,
x ln 3
L( x)
1 ( x 3) ln 3
3ln 3
ln 3
x 1 1 0.30 x 0.09
3ln 3 ln 3
(b)
155. (a) The point of tangency is ( p, ln p) and mtangent
1
p
since
dy
dx
1x . The tangent line passes through (0, 0)
the equation of the tangent line is y 1p x. The tangent line also passes through ( p, ln p )
ln p
(b)
d2y
dx 2
1
p
1
x2
p 1 p e, and the tangent line equation is y 1e x.
for x 0 y ln x is concave downward over its domain. Therefore, y ln x lies below the
graph of y 1e x for all x 0, x e, and ln x
x
e
for x 0, x e.
(c) Multiplying by e, e ln x x or ln x e x.
Copyright 2018 Pearson Education, Inc.
Section 7.4 Exponential Change and Separable Differential Equations
473
e
(d) Exponentiating both sides of ln x e x, we have eln x e x , or x e e x for all positive x e.
(e) Let x to see that e e . Therefore, e is bigger.
156. Using Newton’s Method: f ( x) ln( x) 1 f ( x)
1
x
xn 1 xn
ln xn 1
1
xn
xn 1 xn 2 ln xn . Then,
x1 2, x2 2.61370564, x3 2.71624393, and x5 2.71828183. Many other methods may be used. For
example, graph y ln x 1 and determine the zero of y.
7.4
EXPONENTIAL CHANGE AND SEPARABLE DIFFERENTIAL EQUATIONS
1. (a)
(b)
(c)
2. (a)
3.
y e x y e x 2 y 3 y 2 e x 3e x e x
y e x Ce 3 x 2 y e x 32 Ce3 x 2 2 y 3 y 2 e x 32 Ce3 x 2 3 e x Ce3 x 2 e x
y e x e3 x 2 y e x 32 e3 x 2 2 y 3 y 2 e x 32 e3 x 2 3 e x e3 x 2 e x
2
y 1x y 12 1x y 2
x
(b)
y x 13 y
1
( x 3)2
(c)
y x 1C y
2
( x13) y 2
1
( x C ) 2
2
( x 1C ) y 2
ex x2 y 1x et
x t
x t
y 1x et dt y 12 et dt 1x
1
x 1
x
t
x t
dt e x x 1x et dt e x xy e x
1
x 2 y xy e x
4.
y
1 x
y
5.
x
1
4
1
1
1 t dt y 2
4
2 x3
1 x 4
x
1
1 x
4
1
4 x3
1 x
4
x 1 t 4 dt
3 1
1 t 4 dt 1 y
1
1 x
4
y 1 y
2 x3
1 x 4
4
1 x
2 x3
1 x 4
y 1
y e x tan 1 2e x y e x tan 1 2e x e x 1 2 2e x e x tan 1 2e x 2 2 x
1 4e
1 2e x
y y
2
1 4e2 x
y y
2
2
2
1 4e2 x
y ( x 2)e x y e x 2 xe x
7.
y
2
( x 2) y e
x sin x cos x y sin x
x
x2
cos(
2)
sin x; y 2 ( 2) 0
y
xy y
; y ( ln 2) e( ln 2) tan 1 2e ln 2 2 tan 1 1 2 4 2
6.
cos x
x
1x
x2
2
2 xy; y (2) (2 2)e2 0
cosx x y sinx x xy xy sin x y
Copyright 2018 Pearson Education, Inc.
474
8.
Chapter 7 Transcendental Functions
x
ln x
y
9. 2 xy
2
y
dy
dx
ln x x
1x y
(ln x )
2
1
ln x
1
(ln x )2
x2 y
x2
ln x
x2
(ln x ) 2
x 2 y xy y 2 ; y (e)
e
ln e
e.
1 2 x1/2 y1/2 dy dx 2 y1/2 dy x 1/2 dx 2 y1/2 dy x 1/2 dx
23 y3/2 2x1/2 C1 23 y3/2 x1/2 C, where C 12 C1
x3
3
10.
dy
dx
x 2 y dy x 2 y1/2 dx y 1/2 dy x 2 dx y 1/2 dy x 2 dx 2 y1/2
11.
dy
dx
e x y dy e x e y dx e y dy e x dx e y dy e x dx e y e x C e y e x C
12.
dy
dx
3x 2 e y dy 3 x 2 e y dx e y dy 3 x 2 dx e y dy 3x 2 dx e y x3 C e y x3 C
13.
dy
dx
y cos
y dy
hand side, substitute u
sec
14.
2
2 xy
2
15.
x
dy
dx
y cos 2
y du
1
2 y
y
y
dy dx
1
y
dy 2 du
sec2 y
y
dy dx. In the integral on the left-
dy, and we have
u du dx 2 tan u x C x 2 tan y C
dy
dx
y3 2
3
2
1
2 xy
1 dy
dy
e y
x
x1 2
1
2
dx 2 ydy
1
x
dx 2 y1 2 dy x 1 2 dx 2 y1 2 dy x 1 2 dx
C1 2 y 3 2 3 x 32 C1 2
dy
dx
e ye x
x
dy
e ye x
x
right-hand side, substitute u x du
e
sec2 y
y dx
C 2 y1/2 13 x3 C
dx e y dy
1
2 x
dy
dy
dx
3
e
dx 2 du
dy 2 eu du e y 2eu C1 e y 2e
16. (sec x) dx e y sin x
y
x
x
3 x C , where C 32 C1
dx e y dy e
x
1
x
x
x
dx. In the integral on the
dx, and we have
C , where C C1
e y sin x cos x dy e y esin x cos x dx e y dy esin x cos x dx
e y dy esin x cos x dx e y esin x C1 e y esin x C , where C C1
17.
dy
dx
2 x 1 y 2 dy 2 x 1 y 2 dx
| y | 1 y sin x 2 C
18.
dy
dx
e2 x y
ex y
2y
x
e
dy
e2 x y
ex y
dx dy
e2 x e y
e xe y
dy
1 y
2
dx
2 x dx
ex
e2 y
dy
1 y 2
2 x dx sin 1 y x 2 C since
dx e2 y dy e x dx e2 y dy e x dx
2e C where C 2C1
Copyright 2018 Pearson Education, Inc.
e2 y
2
e x C1
Section 7.4 Exponential Change and Separable Differential Equations
19.
y2
dy
dx
y2
3 x 2 y 3 6 x 2 y 2 dy 3 x 2 y 3 2 dx
3
y 2
dy 3x 2 dx
y2
3
y 2
475
dy 3 x 2 dx
13 ln y 3 2 x3 C
20.
dy
dx
xy 3x 2 y 6 ( y 3)( x 2)
1 dy
y 3
( x 2)dx
1 dy
y 3
( x 2)dx
ln | y 3| 12 x 2 2 x C
21.
1 dy
x dx
22.
dy
dx
2
1
y y 2
2
y 2 y y21
2
y
e
1 e y
2
y
dy xe x dx
2
dy xe x dx 2 ln
y 2 12 e x C 4 ln
y y0 ekt 0.99 y0 y0 e1000 k k
1 dy
e y 1
ln 0.99
1000
24. (a)
(b)
2
dy xe x dx
2
y 2 e x C 4 ln
2
y 2 ex C
e x 1 dx
y
0.00001
(b) 0.9 e( 0.00001)t (0.00001)t ln (0.9) t
(c)
1
y2 y
1 dy e x 1 dx
e 1
dy e x 1 dx ln 1 e y e x x C ln 1 e y e x x C
e x y e x e y 1 e y 1 e x 1
23. (a)
2
ye x 2 ye x e x
ln (0.9)
0.00001
10,536 years
y y0 e(20,000) k y0 e0.2 y0 (0.82) 82%
dp
dh
kp p p0 e kh where p0 1013; 90 1013e20k k
p 1013e 6.05 2.389 millibars
(c) 900 1013e( 0.121) h 0.121h ln
ln (90) ln(1013)
20
0.121
900 h ln(1013) ln(900) 0.9777 km
1013
0.121
0.6 y y y0 e0.6t ; y0 100 y 100e0.6t y 100e0.6 54.88 grams when t 1 hr
25.
dy
dt
26.
A A0 ekt 800 1000e10k k
ln (0.8)
10
A 1000e(ln (0.8) 10)t , where A represents the amount of sugar
that remains after time t. Thus after another 14 hrs, A 1000e(ln (0.8) 10)24 585.35 kg
27. L( x) L0 e kx
L0
2
L0 e 18k ln 12 18k k
one-tenth of the surface value,
28. V (t ) V0 et
29.
40
L0
10
0.1V0 V0 et
ln 2
18
0.0385 L( x) L0 e 0.0385 x ; when the intensity is
L0 e 0.0385 x ln10 0.0385 x x 59.8 ft
40
when the voltage is 10% of its original value t 40 ln (0.1) 92.1 sec
y y0 ekt and y0 1 y ekt at y 2 and t 0.5 we have 2 e0.5k ln 2 0.5k k
Therefore, y e(ln 4)t y e24 ln 4 424 2.81474978 1014 at the end of 24 hrs
Copyright 2018 Pearson Education, Inc.
ln 2
0.5
ln 4.
476
30.
Chapter 7 Transcendental Functions
y y0 ekt and y (3) 10, 000 10, 000 y0 e3k ; also y (5) 40, 000 y0 e5k . Therefore
y0 e5k 4 y0 e3k e5k 4e3k e2k 4 k ln 2. Thus, y y0 e(ln 2)t 10, 000 y0 e3ln 2 y0 eln 8
10, 000 8 y0 y0 10,000
1250
8
31. (a) 10, 000ek (1) 7500 ek 0.75 k ln 0.75 and y 10, 000e(ln 0.75)t . Now 1000 10, 000e(ln 0.75)t
ln 0.1 (ln 0.75)t t
ln 0.1
ln 0.75
8.00 years (to the nearest hundredth of a year)
(b) 1 10, 000e(ln 0.75)t ln 0.0001 (ln 0.75)t t
ln 0.0001
ln 0.75
32.02 years (to the nearest hundredth
of a year)
dz
dy
k
k ( r ky ) kz. The equation dz / dt kz has solution z ce kt , so
dt
dt
1
r ky ce kt and y r ce kt .
k
1
(a) Since y (0) y0 , we have y0 ( r c ) and thus c r ky0 . So
k
1
r
r
y r [ r ky0 ]e kt y0 e kt .
k
k
k
32. Let z r ky. Then
r
r r
(b) Since k 0, lim y0 e kt .
k
k k
t
y
y r/k
y y0
t
33. Let y ( t ) be the population at time t , so t (0) 1147 and we are interested in t (20). If the population
continues to decline at 39% per year, the population in 20 years would be 1147 (0.61)20 0.06 1, so the
species would be extinct.
34. (a) We will ignore leap years. There are (60)(60)(24)(365) 31,536,000 seconds in a year. Thus, assuming
exponential growth, P 314,419,198ekt , with t in years, and
31,536,000 314,419,199
ln
0.0083583.
12
314,419,198
(You don’t really need to compute that logarithm: it will be very nearly equal to 1 over the denominator
of the fraction.)
314,419,199 314,419,198e12 k /31,536,000 k
(b) In seven years, P 314,419,198e(0.0083583)(7) 333,664,000 . (We certainly can’t estimate this
population to better than six significant digits.)
35.
0.9 P0 P0 ek k ln 0.9; when the well’s output falls to one-fifth of its present value P 0.2 P0
0.2 P0 P0 e(ln 0.9)t 0.2 e(ln 0.9)t ln (0.2) (ln 0.9)t t
ln 0.2
ln 0.9
15.28 yr
Copyright 2018 Pearson Education, Inc.
Section 7.4 Exponential Change and Separable Differential Equations
dp
dx
36. (a)
1 p
100
dp
p
477
1 dx ln p 1 x C p e( 0.01x C ) eC e 0.01x C e 0.01x ;
100
1
100
p(100) 20.09 20.09 C1e( 0.01)(100) C1 20.09e 54.61 p( x) 54.61e 0.01x (in dollars)
(b) p(10) 54.61e( 0.01)(10) $49.41, and p(90) 54.61e( 0.01)(90) $22.20
(c) r ( x) xp ( x) r ( x) p ( x) xp ( x);
p ( x ) .5461e 0.01x
r ( x) (54.61 .5461x)e 0.01x . Thus,
r ( x) 0 54.61 .5461x x 100. Since
r 0 for any x 100 and r 0 for x 100,
then r ( x) must be a maximum at x 100.
37.
A A0 e kt and A0 10 A 10ekt , 5 10ek (24360) k
then 0.2(10) 10e0.000028454t t
38.
39.
A0 e139k
0.05 A0 A0 e
t
y y0 e kt y0 e( k )(3 k ) y0 e3
40. (a)
(b)
A A0 e kt
1
k
1
2
y0
e3
e2.645k k
y0
20
0.000028454 A 10e0.000028454t ,
56563 years
1 e139 k k ln(0.5)
2
139
ln 0.05 600 days
0.00499
A A0 e kt and
1A
2 0
0.00499t
ln 0.2
0.000028454
ln (0.5)
24360
0.00499; then
(0.05)( y0 ) after three mean lifetimes less than 5% remains
ln 2
2.645
0.262
3.816 years
ln 2 t ln 20 ln 2 t t
(c) (0.05) A A exp 2.645
2.645
2.645 ln 20
ln 2
11.431 years
41. T Ts T0 Ts e kt , T0 90C, Ts 20C, T 60C 60 20 70e 10k
k
ln
74 0.05596
4
7
e10k
10
(a) 35 20 70e 0.05596t t 27.5 min is the total time it will take 27.5 10 17.5 minutes longer to
reach 35C
(b) T Ts T0 Ts e kt , T0 90C, Ts 15C 35 15 105e0.05596t t 13.26 min
42. T 65 T0 65 e kt 35 65 T0 65 e 10 k and 50 65 T0 65 e20k . Solving
30 T0 65 e 10k and 15 T0 65 e 20k simultaneously T0 65 e 10k 2 T0 65 e 20 k
e10k 2 k
ln 2
10
and 30
T0 65
10 k
e
65 60 5
ln 2
30 e10 10 T0 65 T0 65 30 eln 2
43. T Ts To Ts e kt 39 Ts 46 Ts e 10k and 33 Ts 46 Ts e 20k
33Ts
46 Ts
e20k e10 k
2
33Ts
46 Ts
39 Ts 2
46 Ts
33 Ts 46 Ts 39 Ts
1518 79Ts Ts2 1521 78Ts Ts2 Ts 3 Ts 3C
Copyright 2018 Pearson Education, Inc.
2
39 Ts
46 Ts
e 10 k and
478
Chapter 7 Transcendental Functions
44. 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 t0 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
t0
temperature Ts 70 Ts 60 Ts x Ts y Ts 10
76 0.00771
1 ln
T Ts T0 Ts e kt 60 Ts Ts 70 Ts Ts e20k 60 70e 20 k k 20
(a) T Ts T0 Ts e 0.00771t Ts x Ts 70 Ts Ts e(0.00771)(35) x 70e0.26985 53.44C
(b) T Ts T0 Ts e 0.00771t Ts y Ts 70 Ts Ts e (0.00771)(140)
y 70e1.0794 23.79C
(c) T Ts T0 Ts e 0.00771t Ts 10 Ts 70 Ts Ts e (0.00771)t0 10 70e0.00771t0
ln
1
17 0.00771t0 t0 0.00771
ln 17 252.39 252.39 20 232 minutes from now the
silver will be 10°C above room temperature
45. From Example 4, the half-life of carbon-14 is 5700 yr 12 c0 c0 e k (5700) k
c c0 e0.0001216t (0.445)c0 c0 e0.0001216t t
ln(0.445)
0.0001216
ln 2
5700
0.0001216
6659 years
46. From Exercise 45, k 0.0001216 for carbon-14.
(a) c c0 e0.0001216t (0.17)c0 c0 e0.0001216t t 14,571.44 years 12,571 BC
(b) (0.18)c0 c0 e0.0001216t t 14,101.41 years 12,101 BC
(c) (0.16)c0 c0 e0.0001216t t 15, 069.98 years 13, 070 BC
47. From Exercise 45, k 0.0001216 for carbon- 14 y y0 e0.0001216t . When t 5000
y y0 e0.0001216(5000) 0.5444 y0
y
y0
0.5444 approximately 54.44% remains
48. From Exercise 45, k 0.0001216 for carbon-14. Thus, c c0 e0.0001216t (0.995)c0 c0 e0.0001216t
t
ln(0.995)
0.0001216
41 years old
49. e (ln 2/5730)t 0.15
ln 2
5730ln(0.15)
t ln(0.15) t
15,683 years
5730
ln 2
50. (a) e (ln 2/5730)(500) 0.94131, or about 94%.
(b) We’ll assume that the error could be 1% of the original amount. If the percentage of carbon-14 remaining
5730ln(0.93131)
were 0.93131, the Ice Maiden’s actual age would be
588 years.
ln 2
7.5
INDETERMINATE FORMS AND L’HÔPITAL’S RULE
x 2
2
x2 x 4
1. lHôpital: lim
1
2x
x2
1
4
x 2
2
x2 x 4
or lim
x 2
x2 ( x 2)( x 2)
lim
1
x2 x 2
lim
Copyright 2018 Pearson Education, Inc.
1
4
Section 7.5 Indeterminate Forms and L’Hôpital’s Rule
sin 5 x
x 0 x
5cos 5 x
1
5 x 2 3 x
2
x 7 x 1
lim
x3 1
3
x 1 4 x x 3
lim
2. lHôpital: lim
x 0
10 x 3
x 14 x
3. lHôpital: lim
x lim
5. lHôpital: lim 1cos
2
sin 2 x
2
x 0 x (1 cos x )
6. lHôpital:
lim x2 2
x 2 x 4
9.
t 3 4t 15
2
t 3 t t 12
1
x 2 2 x
lim
lim
12.
x 8 x 2
2
x 12 x 5 x
13.
lim sint t lim
cos x
2
lim 4
x 6 x
3( 3)2 4
2( 3) 1
23
7
30 x
x 42 x
lim
16
x 24
32
116 x
x 24 x 5
lim
lim
lim
sin x x
x3
lim
0 or
30
x 42
1
16
x 0 cos x
2
x 1 ( x 1) 4 x 4 x 3
sin x
x 0 6 x
lim
lim
x 1
2
lim x 25
x 5 x 5
14.
lim sin 5t
t 0 2t
lim
lim
x 0
cos x
6
x
11cos
cos x
9t 2
2
12
t 1 t 1
5
7
5t
lim 5cos
2
t 0
16
18.
3
lim 3 lim
3
3 sin 3
3 cos 3
19.
1
lim 1sin lim cos lim sin ( 4)(
14
1)
2 1 cos 2 2 2sin 2 2 4 cos 2
0
1
Copyright 2018 Pearson Education, Inc.
0
10
lim
x 2 x 1
2
x 1 4 x 4 x 3
2x
x 5 1
3t 3 3
3
4
t 1 t t 3
2
lim 2 lim
23 2
2 cos(2 ) 2 sin(2 ) sin 2
5
7
lim
1 1
x 2 x3
lim
10.
17.
2 3
x x2
16
16
1
lim
cos x 1
2
x 0 3 x
1
x2
( x 1) x 2 x 1
2
lim 2 3x 3 x
x
x 1
x
2
16 x
x 0 sin x
lim
x 7
x lim (1cos x )
or lim 1cos
2
x2
x 0 x
x 0
1
2
cos t (2t ) 0
8 x2
cos
x 1
x 0
lim
lim
t 0
x 0
x 0
8.
15 x 2 2
2
x 21x
t 0
lim
x3 1
3
x 1 4 x x 3
5 3x
lim
or lim
1
4
lim
2
5 x 2 3 x
2
x 7 x 1
5
7
3 or lim
11
lim
lim 4 x23
x 3 x 1
3t 2 4
t 3 2t 1
3
lim 5 x 3 2 x
x 7 x 3
lim
5 lim sin5 x5 x 5 1 5
5 x 0
sinx x sinx x 1cos1 x 12
lim
11.
16.
lim
x 0
2
lim 2 3x 3 x
x
x 1
x
7.
15.
sin x
x 0 2 x
x
lim
10
x 14
lim
3x2
2
x 1 12 x 1
4. lHôpital: lim
x 0
sin 5 x
x 0 x
5 or lim
5
2
9
11
3
11
479
480
Chapter 7 Transcendental Functions
20.
x 1
lim
x 1 ln x sin( x )
21.
x2
ln(sec
x)
x 0
22.
lim
x 0
ln(csc x )
2x
lim
lim
x 2 x
1
1
x 1 x cos( x )
lim
2
2x
x 0 tan x
lim
secsecx tanx x
23.
t (1 cos t )
lim
t 0 t sin t
24.
t lim sin t t cos t lim
lim 1tsin
cos t
sin t
25.
t 0
2
26.
lim
x
27.
2
x 2 sec x
0
12
28.
29.
lim
1
0
x
x 0 2 1
ln( x 1)
x log 2 x
32.
log 2 x
x log 3 ( x 3)
x 0
2 x
cot x
ln x 2 2 x
ln x
ln
(ln 2) 2 x
lim
lim
x
30 ln 3
20 ln 2
ln( x 1)
x
2
12
sin t (sin t t cos t )
sin t
ln x
ln 2
lim
x 0
2
2
2
t cos t t sin t 1110 3
lim cos t coscos
t
1
t 0
1 (10)
1
2
sin1 x 11 1
lim
x
1
csc2 x
lim sin 2 x 1
2
x
3 (ln 3)(1) ln 3
0
1
12 ln1 ln 2 ln 2
lim
x
(1) 2 x ( x ) (ln 2) 2 x
3x ln 3
x
x 0 2 ln 2
lim
1
2
1
lim
lim
1
2
x 0
31.
lim
x
2
ln
lim
3x 1
x
x 0 2 1
lim
lim
cos x
3sin (ln 3)(cos )
1
0
lim
t 0
x 2
lim
30.
33.
2
x
0
x 2x
lim
lim
2 x tan x
3sin 1
lim
lim
cos t (cos t t sin t )
cos t
t 0
t 0
lim
x
(1cos t ) t (sin t )
1cos t
t 0
csccscx cotx x lim cot x lim csc2 x 12 1
2
2
2 x 2
x 2 2 x 2 x 2 2
x 2
lim
2
2
x 0 sec x
lim
lim
2
11
120 0
(ln 2)20
1
ln 2
ln 3
ln 2
x11 (ln 2) lim x (ln 2) lim 1 ln 2
1
x x
x x 1
x 1
(ln 2) lim
lnln 2x
ln( x 3)
ln 3
ln 3 lim ln x ln 3 lim
ln 2 x ln( x 3) ln 2 x
2 x2
2
x 2 x
1
x
1
x
1
x 3
x 3 ln 3 lim 1 ln 3
lnln 23 xlim
ln 2 x 1 ln 2
x
2
lim 2 x2 2 x lim 42 xx 22 lim 22 1
x 2 x
x 0
x 0
x 0
Copyright 2018 Pearson Education, Inc.
Section 7.5 Indeterminate Forms and L’Hôpital’s Rule
34.
lim
x 0
ln e x 1
ln x
35.
5 y 25 5
y
y 0
36.
ay a 2 a
lim
y
y 0
37.
lim
ex
x
e 1
lim
x 0
1
x
x
x
x
lim xe
lim e xxe 110 1
e x 1
e
x 0
x 0
(5 y 25)1/ 2 5
y
y 0
lim
ay a
lim
2
1/ 2
a
y
y 0
lim
y 0
lim
12 (5 y 25)1/ 2 (5)
1
12 ay a2
1/ 2
(a)
1
y 0
5
lim
y 0 2 5 y 25
a
y 0 2 ay a 2
12 , a 0
lim
1
2
lim ln 2 x ln( x 1) lim ln x2x1 ln lim x2x1 ln lim 12 ln 2
x
x
x
x
38.
lim (ln x ln sin x) lim ln sinx x ln lim sinx x ln lim cos1 x ln1 0
x 0
x 0
x 0
x 0
39.
lim ln(sin x ) lim
x 0
x 0
40.
41.
(ln x ) 2
2(ln x )
1x
cos x
sin x
lim
x 0
2(ln x )(sin x )
x cos x
(3 x 1)(sin x ) x
lim 3 xx1 sin1 x lim
x sin x
x 0
x 0
33 (1)(0)
110 62 3
lim
2(ln x )
lim cos x sinx x 1
x 0
x 0
3sin x (3 x 1)(cos x ) 1
sin x x cos x
lim
x 0
1 1
ln x ( x 1)
x
lim x11 ln1x lim ( x 1)(ln x ) lim
lim ( x ln1x)x x 1
(ln x ) ( x 1) 1
x 1
x 1
x 1
x 1
x
3cos x 3cos x (3 x 1)( sin x )
cos x cos x x sin x
lim (ln x11) 1 (01)1 1 12
x 1
42.
x cos x lim
lim (csc x cot x cos x) lim sin1 x cos
sin x
x 0
x 0
x 0
(1cos x ) (sin x )(cos x )
sin x
2
x sin 2 x 0 1 0 1
lim sin x cos
cos
x
1
x 0
43.
1 lim sin lim cos 1
lim cos
0 e 1 0 e 1 0 e
44.
lim
45.
46.
481
h 0
eh (1 h )
h
2
h
h
lim e2h1 lim e2 12
h 0
h 0
t
2
t
t
t
lim e tt lim e t 2t lim e t 2 lim et 1
t e 1
t
e
t e
t e
2
lim x 2 e x lim xx lim 2 xx lim 2x 0
x
x e
x e
x e
Copyright 2018 Pearson Education, Inc.
482
47.
Chapter 7 Transcendental Functions
lim x sin x
x 0 x tan x
e x 1
lim
1 cos x
2
x 0 x sec x tan x
lim
2
sin x
2
2
x 0 2 x sec x tan x 2 sec x
lim
2 e x 1 e x
2x
x
0
2
0
2x
x
48.
e 2e lim
4e 2e
lim x cos x sin x lim x 2cos
22 1
x sin x x 0 x sin x 2 cos x
x 0 x sin x
x 0
x 0
49.
sin cos lim 1sin 2 cos 2 lim 2sin 2 lim 2 cos 2 2
lim tan
2
sec 2 1
0
0
0 tan
0
50.
3 x x 2 lim
3cos 3 x 3 2 x
9sin 3 x 2
lim sinsin3 xxsin
lim 3cos 3 x 3 2 x lim
2x
x 0
x 0 2sin x cos 2 x cos x sin 2 x x 0 sin x cos 2 x sin 3 x x 0 2sin x sin 2 x cos x cos 2 x 3cos 3 x
24 12
51. The limit leads to the indeterminate form 1. Let f ( x) x1/(1 x ) ln f ( x) ln x1/(1 x ) 1ln xx . Now
ln x
1 x
x 1
lim ln f ( x) lim
x 1
lim
x 1
1x 1.
1
Therefore lim x1/(1 x ) lim f ( x) lim eln f ( x ) e1 1e
x 1
x 1
x 1
52. The limit leads to the indeterminate form 1. Let f ( x) x1/( x 1) ln f ( x) ln x1/( x 1)
1
ln x .
x 1
Now
x lim x 1. Therefore lim x1/( x 1) lim f ( x ) lim eln f ( x ) e1 e
lim ln f ( x) lim ln
x 1
1
x 1
x 1
x 1
x 1
x 1
x 1
53. The limit leads to the indeterminate form 0 . Let f ( x) (ln x)1/ x ln f ( x) ln(ln x)1/ x
ln(ln x )
x
x
lim ln f ( x ) lim
x
lim
x
x ln1 x 0. Therefore
1
x e
ln(ln x )
x e
lim
x e
x ln1 x 1 . Therefore
e
1
x
x
ln(ln x )
x e
lim ln f ( x)
x e
lim (ln x)1/( x e) lim f ( x) lim eln f ( x ) e1/ e
xe
x e
x e
x 1. Therefore
55. The limit leads to the indeterminate form 00. Let f ( x) x 1/ln x ln f ( x) ln
ln x
lim x 1/ln x lim f ( x) lim eln f ( x ) e1 1e
x 0
x 0
x 0
56. The limit leads to the indeterminate form 0 . Let f ( x) x1/ln x ln f ( x)
ln x
ln x
1. Therefore
lim x1/ln x lim f ( x ) lim e1n f ( x ) e1 e
x
x
x
57. The limit leads to the indeterminate form 0 . Let f ( x) (1 2 x)1/(2 ln x ) ln f ( x)
lim ln f ( x) lim
x
x
1/2
ln(1 2 x )
2 ln x
Now
lim (ln x)1/ x lim f ( x) lim eln f ( x ) e0 1
x
54. The limit leads to the indeterminate form 1. Let f ( x) (ln x)1/( x e) ln f ( x)
lim
ln(ln x )
.
x
ln(1 2 x )
2 ln x
lim 1x2 x lim 12 12 . Therefore lim (1 2 x)1/(2 ln x ) lim f ( x)
x
x
x
x
lim eln f ( x ) e
x
Copyright 2018 Pearson Education, Inc.
Section 7.5 Indeterminate Forms and L’Hôpital’s Rule
58. The limit leads to the indeterminate form 1. Let f ( x) e x x
ln e x x
lim ln f ( x) lim
x 0
x
x 0
lim
e x 1
x
e
x 0 x
1/ x
x 0
ln e x x
ln f ( x)
2. Therefore lim e x x
1/ x
x
lim f ( x) lim eln f ( x ) e2
x 0
x 0
59. The limit leads to the indeterminate form 00. Let f ( x) x x ln f ( x) x ln x ln f ( x)
lim ln f ( x) lim ln1 x lim
x 0
x 0
x
1x
x 0 1
x2
483
ln x
1x
lim ( x) 0. Therefore lim x x lim f ( x) lim eln f ( x )
x 0
x 0
x 0
x 0
e0 1
60. The limit leads to the indeterminate form 0 . Let f ( x) 1 1x
x 2
1
lim 1 x2
x 0 x
1
1
x 0 1 x
lim
x
x
lim xx1 0. Therefore lim 1 1x
x 0
x 0
61. The limit leads to the indeterminate form 1. Let f ( x)
xx12
x
ln f ( x)
x
ln 1 x 1
x
x 1
lim ln f ( x)
x 0
lim f ( x) lim eln f ( x ) e0 1
x 0
x 0
ln f ( x ) ln
xx12
ln xx12
ln( x 2) ln( x 1)
lim
lim
1
1
x
x
x x
xx12 xlim
lim ln f ( x) lim x ln
x
x ln xx12
3
lim ( x 2)(1x 1)
x 2
x
1 1
x 2 x 1
12
x
x
x2
lim ( x 32)(
lim 26xx1 lim 62 3. Therefore, lim xx12 lim f ( x) lim eln f ( x ) e3
x 1)
x
x
x
x
x
x
62. The limit leads to the indeterminate form 0 . Let f ( x)
lim ln f ( x)
x
2
x 4 x 1
3
2
x x 2 x x 2
0
lim
2
lim 1x ln xx 21
x
2 x4
2
x 3 x 4 x 1
lim
ln xx 21
lim x
x
2
lim
x
x 2 1
x2
1/ x
ln f ( x) ln
ln x 2 1 ln( x 2)
x
lim 6 x2 4 0. Therefore,
x
2x
lim
x
1/ x
lim xx 21
x
2
x 2 1
x 2 1
x2
x 1 2
1
1/ x
1x ln
x 2 1
x2
x 2 4 x 1
2
1)( x 2)
x
(
x
lim
lim f ( x) lim eln f ( x )
x
x
e 1
63.
1
lim x 2 ln x lim ln1 x lim x2
x 0
x 0 x 2 x 0 x3
64.
2(ln x ) 1
(ln x )2
x
lim x(ln x)2 lim 1 lim
lim 2 ln1 x lim
1
x 0
x 0 x x 0 x2 x 0 x x 0
65.
1
x
lim x tan 2 x lim
lim 2 1
1
cot x
x
csc
2 1
x 0
x 0
2
x 0
3
2
lim 2x x lim 32x 0
x 0
x 0
2
x
1
x2
Copyright 2018 Pearson Education, Inc.
lim
x 0
lim 2x 0
2 x2
x
x 0
484
66.
Chapter 7 Transcendental Functions
9 x 1
x 1
67. lim
x
68.
69.
lim 9 x 1
x x 1
x
sin x
lim
x 0
x 0
2
2
x
lim 9
x 1
1
lim sinx x
x lim
lim sec
tan x
x
70.
1
2
ln x lim
x
lim sin x ln x lim csc
lim sin xxtan x lim sin x sec x1 cos x tan x 10 0
x
csc x cot x
x 0
x 0
x 0
x 0
x 0
9 3
11 1
x
cos1 x cos
sin x
lim sin1 x 1
2
x
x
cos
sin x
lim cos x 1
1
x 0 sin x
x 0
cot x lim
lim csc
x
x 0
23 1 0
x
x 1 4
3
x
71.
2 x 3 x
x
x
x 3 4
72.
2x 4x
x
x
x 5 2
73.
x
x x
x ( x 1)
e
lim e x lim e x lim e x lim
x xe
x
x
x
74.
lim
lim
lim
42 lim 1 2 x 10 1
x
x
x 5 1 x 5 1 0 1
2
2
1
lim
2
2
x
e1/ x
lim
1/ x
x 0 e
lim
x 0
1
x
x
lim
x 0
e1/ x
1
x2
1
x2
x ( x 1)
(2 x 1)
1
lim e1/ x
x 0
75. Part (b) is correct because part (a) is neither in the
2 x2
x 0 2 x cos x
76. Part (b) is correct; the step lim
0
0
2
x 0 2 sin x
lim
nor
form and so lHôpital’s rule may not be used.
2 x2
x 0 2 x cos x
in part (a) is false because lim
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 2, 0 so that
1
2c
f (0) f ( 2)
g (0) g ( 2) 00 42 12 . Since f (c) 1 and g (c) 2c we have that
12 c 1.
(b) We seek c in (a, b) so that
1
2c
f ( c )
g ( c )
f ( c )
g ( c )
f (b ) f ( a )
g (b) g ( a ) b2 a 2 b 1 a . Since f (c) 1 and g (c) 2c we have that
b a
f ( c )
g ( c )
g (3) g (0) 9300 13 . Since f (c) c 2 4 and g (c) 2c we have
b 1 a c b 2 a .
(c) We seek c in (0, 3) so that
that
c2 4
2c
f (3) f (0)
13 c 13 37 c 13 37 .
Copyright 2018 Pearson Education, Inc.
Section 7.5 Indeterminate Forms and L’Hôpital’s Rule
79. If f ( x ) is to be continuous at x 0, then lim f ( x ) f (0) c f (0) lim
x 0
80.
lim 27 sin 3 x
x 0 30 x
tan 2 x
x3
x 0
lim
a
x2
lim 81cos 3 x
x 0 30
9 x 3sin 3 x
5 x3
lim 99 cos2 3 x
x 0
15 x
27 .
10
tan 2 x ax x 2 sin bx
x3
x 0
sinxbx lim
x 0
lim
will be in
2 sec2 2 x a bx 2 cos bx 2 x sin bx
3 x2
x 0
0
0
2
2
bx 2 x sin bx
lim (2sec 2 2 x a bx 2 cos bx 2 x sin bx) a 2 0 a 2; lim 2sec 2 x 2bx cos
2
x 0
x 0
3x
form if
2
2 2
bx 4bx cos bx 2 sin bx lim 32sec2 2 x tan 2 2 x 16sec4 2 x b3 x 2 cos bx 6b 2 x sin bx 6b cos bx
lim 8sec 2 x tan 2 x b x sin
6x
6
x 0
16 6b
6
485
x 0
0 16 6b 0 b 83
81. (a)
(b) The limit leads to the indeterminate form :
x 2 x 2 x
2
x
lim x x 2 x lim x x 2 x x x 2 x lim
lim
x
x x x x x x 2 x x x x 2 x
x
1
x 1 1 1x
lim
82.
1
1 1 0
12
2
2
lim x 2 1 x lim x xx 1 xx lim x x 21 x2 lim x
x
x
x
x
x
x
1
1
x2
1x
83. The graph indicates a limit near 1. The limit leads
to the indeterminate form
2 x 2 3 x3/ 2 x1/ 2 2
x 1
x 1
lim
4 92 12
1
2
0 : lim 2 x (3 x 1) x 2
0 x 1
x 1
4 x 92 x1/ 2 12 x 1/ 2
1
x 1
lim
415 1
84. (a) The limit leads to the indeterminate form 1. Let f ( x) 1 1x
lim
x
lim ln1 x1
1x x x1
ln 1 1x
x 2
1
lim 1 x2
x x
1
1
x 1 x
lim
x
ln f ( x) x ln 1 1x lim ln f ( x)
110 1 lim 1 1x
x
lim eln f ( x ) e1 e
x
Copyright 2018 Pearson Education, Inc.
x
x
lim f ( x)
x
486
Chapter 7 Transcendental Functions
(b)
1 1x x
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.
1
x2
(c) Let f ( x) 1
x
ln f ( x ) x ln 1 x 2
lim ln f ( x) lim
x
85. Let f (k ) 1
r
k
k
x
x
Therefore lim 1
x
ln 1 x 2
1
x2
x
1
x
4x
2
x 3 x 1
x x x
lim 64x 0.
x
ln f (k )
x
ln 1 rk 1
y
y x1/ x ln y lnxx y
k 1
k
ln 1 rk 1
lim
k
k
rk 2
1
lim 1 rk2
k k
r
1
k 1 rk
lim
lim krk r
k
k
1x ( x )ln x y
x2
which indicates a maximum value of y e
2
y
y x1/ x ln y ln2x y
x
k 1
lim f (k ) lim eln f ( k ) e r .
1/e
(b)
2
lim 23x lim
lim f ( x) lim eln f ( x ) e0 1
lim 1r r. Therefore lim 1 kr
k
k
86. (a)
2 x3
2
lim 1 x2
x x
x 2 x ln x
x . The sign pattern is y | |
1ln x
x2
x
4
0
e
when x e
2
1
x
1/ x
y
1 2 ln x
x3
x . The sign pattern is
1/ x 2
y | | which indicates a maximum of y e1/(2e) when x e
0
(c)
e
n
y x1/ x ln y lnnx
1x xn (ln x ) nxn 1
x
x
2n
y
x n 1 (1 n ln x )
x
2n
n
x1/ x . The sign pattern is
y | | which indicates a maximum of y e1/( ne) when x n e
n
0
(d)
87. (a)
e
1/ x n
n
n
lim x1/ x lim eln x
lim e(ln x ) x exp lim lnnx exp lim 1n e0 1
x
x
x
x x
x nx
y x tan
, xlim
x tan
1
x
1
x
sec2 1 1
tan 1x
x x2
lim 1 lim
lim sec2
1
x x x
x
x2
sec2 1 1
tan 1x
x x2
lim 1 lim
lim sec2
x
1
x x x
x2
x and as x .
lim x tan 1x
1x 1; x
1x 1 the horizontal asymptote is y 1 as
Copyright 2018 Pearson Education, Inc.
Section 7.5 Indeterminate Forms and L’Hôpital’s Rule
(b)
2x
2x
2x
2x
y 3 x e 3 x , lim 3 x e 3 x lim 3 2e3 x lim 4e3 x lim
2 x e
x 2 x e
x 2 3e
x 9e
32 the horizontal asymptotes are y 0 as x
88.
4
x
x 9e
and y 32
1
2
2
f (0 h ) f (0)
e1/ h 0 lim e1/ h lim h
f (0) lim
lim
2
h
h
h 0
h 0
h 0 h
h0 e1/ h
lim h2 e 1/ h
h 0
2
0; lim
0
3 x e2 x
3x
x 2 x e
as x .
1
h
h2
lim 1/ h2 2 lim 1/ h2
0
0
h
h
2e
e 3
h
89. (a) We should assign the value 1 to
f ( x) (sin x) x to make it continuous at x 0.
ln(sin x )
(b) ln f ( x) x ln(sin x)
1
x
lim ln f ( x) lim
x 0
x 0
ln(sin x )
1
x
lim
x 0
sin1 x (cos x) lim
1
2
x
x2
x
tan
x 0
lim 22x 0 lim f ( x) e0 1
x 0 sec x
x 0
(c) The maximum value of f ( x) is close to 1 near the point x 1.55 (see the graph in part (a)).
(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 .
(b) Let f ( x) (sin x) tan x
ln f ( x ) (tan x ) ln(sin x) lim ln f ( x)
2
x
lim
x
2
lim
2
x
sin1 x (cos x)
ln(sin x )
cot x
lim
cos x
( csc x )
0 lim
2
x
csc2 x
x
2
f ( x) e0 1.
Copyright 2018 Pearson Education, Inc.
3 e 2 x
ex
x 2 3
lim
487
488
Chapter 7 Integrals and Transcendental Functions
Similarly,
f ( x) e0 1. Therefore, lim f ( x) 1.
lim
2
x
x 2
(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) 4
(b) 3
(c) 6
2. (a) 4
(b) 3
(c) 6
3. (a) 6
(b) 4
(c) 3
4. (a) 6
(b) 4
(c) 3
5. (a) 3
(b) 34
(c) 6
6. (a) 4
(b) 3
(c) 6
(b) 6
(c)
8. (a) 34
(b) 6
(c) 23
7. (a)
3
4
9. sin cos 1
2
2
sin
2
3
1
4
15.
17.
19.
12. cot sin 1
21.
lim sin 1 x 2
14.
lim tan 1 x 2
16.
lim sec1 x 2
18.
x 1
x
x
lim csc1 x lim sin 1
x
x
dy
23. y sin 1 2t
dy
dt
26. y sec1 5s
dy
ds
2x
1 x
2 2
2
25. y sec 1 (2 s 1)
27.
1x 0
y cos 1 x 2 dx
1
dy
ds
2t
2
2
|2 s 1| 4 s 2 4 s
x 1
lim sec 1 x lim cos 1
1x 2
lim csc1 x lim sin 1
1x 0
x
x
x
x
1x sec1 x dydx |x|
1
|2 s 1| s 2 s
1
x 1 1
2
2
2 x
x 1
2
3
lim tan 1 x 2
|s| 25 s 2 1
2x
2
3
x
24. y sin 1 (1 t )
2
1 2t 2
2
5
dy
dx
1 x
1
lim cos 1 x
22. y cos 1
2 x4
3
2
x 1
20.
|2 s 1| (2 s 1) 2 1
|5 s| (5 s )2 1
y csc1 x 2 1
cot
tan 6 13
11. tan sin 1 12
13.
10. sec cos 1 12 sec 3 2
2
x4 2 x2
Copyright 2018 Pearson Education, Inc.
dy
dt
1
1(1t )2
1
x 2 1
1
2t t 2
Section 7.6 Inverse Trigonometric Functions
12 1
2
2
2x 1 |x| x 4 4
28. y csc 1
2x dydx
29. y sec1
1t cos1 t dydt
30. y sin 1
csc
| 2x |
1 t 2
3
3
t2
1
dy
dt
32.
y cot 1 t 1 cot 1 (t 1)1/2
33.
y ln tan 1 x
dy
dx
34.
y tan 1 (ln x)
dy
dx
35.
y csc1 et
dy
dt
dy
dt
1
2
1 x
tan 1 x
1x
|e |
e t
y cos 1 et
37.
y s 1 s 2 cos 1 s s 1 s 2
39.
1 e t
1 s2
38.
dy
dt
s2
1 s
2
1
1 s
2
y tan
2
x 1 csc
1
2
1/2
x tan
6
t t 4 9
1
2 t (1t )
12 (t 1)1/ 2
1 (t 1)1/ 2
2
1
2 t 1(1t 1)
1
2t t 1
1
2
1
e2t 1
e t
1/2
1e2t
cos 1 s
s 2 1
1 s2
y s 2 1 sec1 s s 2 1
1
dy
dt
1/ 2 2
t 4 9
9
t2
1
x 1 (ln x ) 2
e 1
36.
1 t
1
2
t
12 t 1/ 2
2t
tan x 1 x
et
t
2
t 2 1
3
t2
3
1 (ln x )2
23t
y cot 1 t cot 1 t1/2
2
| x| x 2 4
1t 2
31.
1 s
2
1 s
sec 1 s
1
x 1
2
1/2
dy
dx
dy
ds
2
1 s2
s 2 1
1 s
csc
2
1/2
x
12 1 s2
1/2
(2 s )
1
1 s 2
2 s 2
1 s 2
12 s2 1
1
s
dy
dx
1/2
(2s )
12 x2 1
1
|s| s 1
1/ 2
(2 x )
1/ 2 2
1 x 2 1
2
s
2
s 1
1
| x| x 2 1
1
2
|s| s 1
1
x x 2 1
for x 1
40.
y cot 1
489
1x tan 1 x 2 tan 1 x1 tan 1 x dydx 0 1xx
2
1 2
1
1 x 2
Copyright 2018 Pearson Education, Inc.
1 1
x 2 1 1 x 2
0
s|s|1
|s| s 2 1
1
| x| x 2 1
0,
490
41.
Chapter 7 Integrals and Transcendental Functions
y x sin 1 x 1 x 2 x sin 1 x 1 x 2
sin 1 x
42.
x
1 x 2
1 x 2
y ln x 2 4 x tan 1
1/2
dy
dx
sin 1 x x
1
x 2 2
1 2x
1
1 x 2
12 1 x2
1/2
(2 x)
sin 1 x
x
x
2
dy
dx
2x
x2 4
tan 1
1
1 x 2
1
1 y 2
y 0
43. 3tan 1 x sin 1 y 4 3
x
2
y
1 y 2
2x
x2 4
tan 1
2x 42 xx
2
tan 1
2x
3 1 y 2
, and x 1, y 1
32 y
2
1 x
1 x
y 32 0 0
1 y
44. sin 1 ( x y ) cos 1 ( x y ) 56
1( x y )2
(1 y)
1( x y )2
0, and x 0, y 12
1 y
3
2
1 y
3
2
0
2 y 0 y 0
45.
y cos 1 ( xy ) 34 2 y
2 1
2
1
2
1
1( xy )2
( xy y ) y cos 1 ( xy ) 0, and x 12 , y 2
y 2 y 34 0 y 2 2 34 y 0 4 y 8 2 3 y 0 (4 3 ) y 8 2 y 4832
46. 16(tan 1 3 y ) 2 9(tan 1 2 x)2 2 2 32
3tan 1 (3 y ) y
1 (3 y )2
18
2 tan 1 (2 x )
1 (2 x )2
0, and x 23 , y 13
48 4 y 9 3 0; y 41
47.
48.
dx sin 1 3x C
1
9 x 2
1
1 4 x 2
dx 12
2
1(2 x ) 2
12 sin
1
dx 12 du 2 , where u 2 x and du 2dx
u C 12 sin (2 x) C
49.
171 x2 dx 17 12 x2 dx
50.
913x
51.
x
2
dx 13
dx
25 x 2 2
1
1u
1
2
3 x2
du
u u 2 2
dx
1
17
1
3 3
tan 1 x C
17
tan 1
C
x
3
3
9
tan 1
C
x
3
, where u 5 x and du 5 dx
1 sec1 u C 1 sec 1 5 x C
2
2
2
2
Copyright 2018 Pearson Education, Inc.
Section 7.6 Inverse Trigonometric Functions
52.
x
dx
5 x2 4
du
u u 2 4
, where u 5 x and du 5 dx
12 sec 1 u2 C 12 sec1 25 x C
0
54.
0
2
3 2 /4
3 2 /4 du
12
, where u 2s and du 2ds; s 0 u 0, s 3 42 u 3 2 2
0
9 4 s
9 u 2
ds
2
3 2 /2
12 sin 1 u3
0
55.
2
1 2 2 du
2 0
8 u 2
0 8dt2t 2
56.
2
2 4dt3t 2
57.
2 /2
1
2
2 2
u
8 0
y 4 y 1
2
2
du
u u 2 1
dy
2
y 9 y 1
2
2
du
2
u u 1
2
1 2 2
8
tan
tan 1 0
1
4
1
14 4 0 16
1 tan 1 0
du 3dt ; t 2 u 2 3, t 2 u 2 3
3 tan 1 3
1
2 3 3
sec1 2 sec 1 | 2 | 4 3 12
1u
32 sin 1 u C 32 sin 1 2(r 1) C
60.
61.
6 dr
4 ( r 1)2
2(dxx 1)
2
6 du 2 , where u r 1 and du dr
4 u
6sin 1 u2
C 6sin 1 r 21 C
du 2 , where u x 1 and du dx
2u
1
2
tan 1 u C 1 tan 1 x 1 C
2
2
2
u 2
sec1 2 sec 1 | 2 | 4 3 12
32 du 2 , where u 2(r 1) and du 2dr
3dr
1 4( r 1)2
3
3 3
, where u 3 y and du 3dy; y 23 u 2, y 32 u 2
sec 1 | u |
2
59.
, where u 2 y and du 2dy; y 1 u 2, y
2
2 /3
tan
2
2/3
1
4
1 2 3 du , where u 3t and
3 2 3 4 u 2
2 3
1 1 tan 1 u
1 tan 1
2
3 2
2 3
2 3
dy
1
tan 1
1
8
12 sin 1 22 sin 1 0 12 4 0 8
, where u 2t and du 2dt ; t 0 u 0, t 2 u 2 2
sec 1 | u |
2
58.
1
4sin 1 2s 4 sin 1 12 sin 1 0 4 6 0 23
0
4 s
1 4 ds
53.
2
2
Copyright 2018 Pearson Education, Inc.
491
492
62.
Chapter 7 Integrals and Transcendental Functions
1(3dxx1)
13 du2 , where u 3 x 1 and du 3dx
2
63.
(2 x 1)
1u
1 tan 1 u C
3
12
dx
(2 x 1)2 4
( x 3)
dx
( x 3)2 25
65.
/2
du
, where u 2 x 1 and du 2dx
u u 2 4
1 1 sec 1 u
2 2
2
64.
13 tan 1 (3x 1) C
du
u u 2 25
1 sec 1 u
5
5
C 14 sec 1 2 x21 C
, where u x 3 and du dx
C 15 sec1 x 53 C
1
d 2
/2 12cos
1 1duu2 , where u sin and du cos d ;
(sin )2
1
2 u 1, 2 u 1
2 tan 1 u 2 tan 1 1 tan 1 (1) 2 4 4
1
66.
/4
2
xdx
/6 1csc
(cot x ) 2
1
du ,
3 1u 2
where u cot x and du csc 2 x dx; x 6 u 3, x 4 u 1
1
tan 1 u tan 1 1 tan 1 3 4 3 12
3
67.
ln 3 e x dx
1 e2 x
0
3 du
, where u e x and du e x dx; x 0 u
1u 2
3
tan 1 u tan 1 3 tan 1 1
3 4 12
1
1
68.
e / 4
1
4 dt
t 1 ln 2 t
4
/4 du
, where u ln t and du 1 dt ; t 1 u 0, t e /4 u
2
t
1 u
0
/4
4 tan 1 u
0
69.
ydy
1 y 4
1
2
du
1u 2
1
sec2 y dy
1 tan 2 y
72.
dx
x 2 4 x 3
dx
2 x x2
, where u y 2 and du 2 y dy
du 2 , where u tan y and du sec2 y dy
sin
71.
4 tan 1 4 tan 1 0 4 tan 1 4
u C 12 sin 1 y 2 C
12 sin
70.
1, x ln 3 u 3
1u
1
u C sin 1 (tan y ) C
dx
1 x 2 4 x 4
dx
1 x 2 2 x 1
dx
1( x 2)2
dx
1( x 1) 2
sin 1 ( x 2) C
sin 1 ( x 1) C
Copyright 2018 Pearson Education, Inc.
4
Section 7.6 Inverse Trigonometric Functions
0
73.
1
74.
1/2
0
dt
1 4 t 2 2t 1
6
6 dt
3 2t t
1
2
1
2 dt
1/2 4 4t 2 4t 1
3
6 dt
3 4t 4t
0
dt
1 22 (t 1)2
6
2
0
6 sin 1 t 21 6 sin 1
1
1
2 dt
1/2 22 (2t 1) 2
3
3 sin 1
493
12 sin 1 0 6 6 0
2t21 1/2 3 sin 1 12 sin 1 0
1
3 6 0 2
dy
dy
dy
C
1 y 1
2
75.
y 2 2 y 5 4 y 2 2 y 1 22 ( y 1)2 12 tan
76.
y
77.
1
78.
2 x
79.
xx244dx x2x 4dx x24 4dx; x2x 4dx 12 u1du
2
dy
2
6 y 10
8dx
x 2 x 2
2
4
x 4 dx
x2 4
dy
1 y 2 6 y 9
8
2 dx
6 x 10
2
80.
2
dx
1 1 x 2 x 1
2
4
2
dx
2 1 x 6 x 9
2
tan 1 ( y 3) C
dy
1 ( y 3)2
8
2
dx
1 1 ( x 1) 2
2
4
dx
2 1 ( x 3)2
12 ln x 2 4 2 tan 1
2
8 tan 1 ( x 1) 8 tan 1 1 tan 1 0 8 4 0 2
1
4
2 tan 1 ( x 3) 2 tan 1 1 tan 1 (1) 2 4 4
2
where u x 2 4 du 2 xdx 12 du xdx
2x C
t t6t210dt (t t3)2 1dt Let w t 3 w 3 t dw dt ww11 dw w w1dw w 11dw;
2
w w1dw 12 u1du where u w 1 du 2w dw 12 du w dw w w1dw w 11dw
2
2
2
2
2
2
2
2
12 ln w2 1 tan 1 ( w) C 12 ln (t 3)2 1 tan 1 (t 3) C 12 ln t 2 6t 10 tan 1 (t 3) C
81.
x x 2x91 dx 1 2xx 109 dx dx x2x 9 dx 10 x 19 dx; x22x 9 dx u1 du
2
2
2
2
u x 2 9 du 2 xdx dx
82.
2 x dx 10
1 dx
x 2 9
x 2 9
where
x ln x 2 9 10
tan 1
3
3x C
t 2tt2 13t 4dt t 2 2t 2t 12 dt t 2 dt t 22t 1dt 2 t 211dt; t 22t 1 dt u1 du
3
2
u t 2 1 du 2t dt (t 2)dt
83.
2
( x 1) dxx2 2 x ( x 1)
dx
x 2 2 x 11
2t dt 2 1 dt
t 2 1
t 2 1
dx
( x 1) ( x 1)2 1
where
12 t 2 2t ln t 2 1 2 tan 1 (t ) C
du
u u 2 1
, where u x 1 and du dx
sec 1 | u | C sec 1 | x 1| C
84.
( x 2)
dx
x 2 4 x 3
dx
dx
( x 2) x 2 4 x 4 1
( x 2) ( x 2)2 1
1
1
sec
| u | C sec
1
u u 2 1
du , where u x 2 and du dx
| x 2| C
Copyright 2018 Pearson Education, Inc.
494
85.
Chapter 7 Integrals and Transcendental Functions
esin
1 x
1 x 2
dx eu du, where u sin 1 x and du
eu C esin
86.
cos 1 x
e 1 x
2
1
x
sin 1 x
u3
3
tan 1 x
1 x 2
sin x
C
C
3
C
dx u1/2 du , where u tan 1 x and du
1
tan 1 y 1 y 2
dy
2
3
dx
1 x 2
3
1
23 u 3/2 C
89.
x
dx u 2 du, where u sin 1 x and du
1 x 2
1
dx
1 x 2
2
88.
C
dx eu du , where u cos 1 x and du
eu C ecos
87.
dx
1 x 2
tan 1 x
3/2
1
1 y 2
dy
tan 1 y
C
2
3
dx
1 x 2
tan 1 x
3
C
u1 du , where u tan 1 y and du
dy
1 y 2
ln | u | C ln tan 1 y C
90.
dy
1
sin 1 y 1 y 2
1
2
1 y
dy
sin 1 y
u1 du, where u sin 1 y and du
dy
1 y 2
ln | u | C ln sin 1 y C
91.
2
1
2 sec sec x
2
x x 2 1
dx
/3
/4 sec
2
u du , where u sec1 x and du
dx
x x 2 1
; x 2 u 4 , x 2 u 3
/3
tan u /4 tan 3 tan 4 3 1
92.
cos sec1 x
2
2/
3
2
x x 1
dx
/3
/6 cos u du, where u sec
/3
sin u /6 sin 3 sin 6
93.
1
x ( x 1) tan 1 x
2
9
dx 2
1 du
u 2 9
23 tan 1
1
x and du
dx
x x 2 1
C
2
3
u 6 , x 2 u 3
3 1
2
where u tan 1 x du
tan 1 x
3
; x
1
1
x
2
Copyright 2018 Pearson Education, Inc.
1
2 x
dx 2du
1
(1 x ) x
dx
Section 7.6 Inverse Trigonometric Functions
94.
e
x
sin 1 e x
1e 2 x
dx u du where u sin 1 e x du
1
2
95.
1
sin 1 e x
4
1
x dx
0 tan
0
1 x 2
4
1
3 cos(tan 1 3 x )
1 3
dx 13
3
1
3
cos u du, where u tan 1 3x du
u 3
13 sin u
97.
98.
2
lim x 11
x 1 sec x
lim
5
1 25 x 2
1
xlim
99.
100.
2
1
lim 2 tan 23 x
7x
x 0
13 sin 3 13 sin
4
1/ 2
x 1 sec
lim x tan 1 2x
x
3
lim
12 x2 1
| x|
x 1
x
tan 1 2 x 1
x 1
12 x
9 x4
lim 114
x
x 0
e x tan 1 e x
2x
x e x
102. lim
lim
x
e tan
dx 13 du
4 13 23 13 22
(2 x )
1
x 2 1
3
1 9 x 2
dx;
3 2
6
6
4
x 0 7 19 x
lim x | x | 1
x 1
2 x2
2
lim 1 4 x2
x x
lim
1/ 2
2
2
x 1 4 x
lim
1 x
e
2x
e x tan 1 e x
e2 x
e2 x 1
2e 1
lim
x
2
6
7
2 3 x4 1
2
2x
1 x 4
1 2
4
tan
x
x
1
lim
101. lim
lim
1
x2 2
1 sin 1 x
x 0 x sin x x 0 x
x
0
1 x2 3/ 2
1 x 2
x
3
19 x 2
5
x 1
2
lim
1
x 0
dx; x 0 u 0, x 1 u 4
2
x 13 u 4 , x
1
lim sin x 5 x
x 0
1
1 x 2
32
4
19 x 2
e x dx
C
u du, where u tan 1 x du
12 u 2
0
96.
2
1
1e2 x
2(0 1)
12
0 2
(1 0)3/ 2
e2 x
e2 x 1
4e2 x
22 1
2 e2 x
2
e2 x 1
e x tan 1 e x
lim
x
1 x
e2 x 3 lim tan 1 e x 13e2 x 0 0 0
lim tan xe
2
x
2
x 4e
4 e2 x 1 x 4e
4 e x e x
Copyright 2018 Pearson Education, Inc.
e2 x e 2 x 3
4e 2 x
2
e2 x 1
495
496
103.
Chapter 7 Integrals and Transcendental Functions
tan 1 x
lim
x x 1
x 0
2
tan 1
lim
x
x
2 x 1
x 0
tan 1
1
x (1 x )
lim
x 1
x
x (1 x )
3x2
2 x 1
x 0
2 tan 1
lim
x 0 (3 x 2) x
1
x (1 x )
lim
2
12 x 13 x 2
x 1
x
0
2 x x 1
x
2
lim
22 1
2
12 x 13 x 2
x
1
x 0
104.
lim 1 4 lim x lim
1
lim
2
1
1
x0 sin 1 x 1 x 2 x0 sin 1 x
x 0 sin 1 x
x 0 2 sin x
1 2
1 2
2x
sin 1 x 2
x
x
x
1
1 x 2
x
1 x 2
2
2
lim 2 1 x 12 x 1 11 1
x 0 1 x x 1 x sin x
x tan 1 x
2
1
dx
105. If y ln x 12 ln 1 x 2 tanx x C , then dy 1x x 2 1 x 2
1 x
x
1
1x x 2 1 2 tan 2 x dx
1 x
x
x1 x
106. If y
x4
4
x4
cos 1 5 x 54
dy x3 cos 1 5 x
107. If y x sin 1 x
dy sin 1 x
formula
2
2
1 25 x 2
x4
4
2
110.
dy
dx
dx tan
1
x2
x dx,
which verifies the formula
dx, then
5
1 25 x 2
5
4
x4
1 25 x 2
3
1
dx x cos 5 x dx, which verifies the formula
2 x sin 1 x
1 x
2
2
2 x
1 x
2
x
a
sin 1 x 2 1 x 2
1
1 x
2
1
dx sin x
2
dx, which verifies the
2
C , then dy ln a 2 x 2 22 x 2 2 2 2 dx
a x
1 x 2
a
aa xx 2 dx ln a2 x2 dx, which verifies the formula
2
2 x 2 1 x 2 sin 1 x C , then
ln a 2 x 2 2
dy
dx
x 1 x
108. If y x ln a 2 x 2 2 x 2a tan 1
109.
x 1 x 2 x3 x tan 1 x 1 x 2
1
1 x 2
dy
1 1
x 2 1
1
dy
2
2
2
2
dx
1 x 2
y sin 1 x C ; x 0 and y 0 0 sin 1 0 C C 0 y sin 1 x
1
1 x 2
1 dx y tan 1 ( x) x C ; x 0 and y 1 1 tan 1 0 0 C C 1
y tan ( x) x 1
111.
dy
dx
1
2
x x 1
dy
dx
2
x x 1
y sec1 | x | C ; x 2 and y sec1 2 C C sec1 2
3 23 y sec1 ( x) 23 , x 1
Copyright 2018 Pearson Education, Inc.
Section 7.6 Inverse Trigonometric Functions
112.
497
dy 1 2 2 2 dx y tan 1 x 2sin 1 x C ; x 0 and y 2
1
x
1 x
1 x
1
1
2 tan 0 2sin 0 C C 2 y tan 1 x 2sin 1 x 2
dy
dx
1
1 x 2
2
2
113. (a) The angle is the large angle between the wall and the right end of the blackboard minus the small angle
x cot 1 x .
between the left end of the blackboard and the wall cot 1 15
3
d
dt
(b)
1
15
1
x 2
15
1
3
1
x 2
3
15
225 x 2
3
9 x 2
540 12 x 2
225 x 2 9 x 2
x 0, consider only x 3 5 3 5 cot 1
first derivative test, ddt
x 3 5 6.7082 ft.
114. V
/3
0
115. V
x 1
132
0 and ddt
565
x 10
;
d
dt
0 540 12 x 2 0 x 3 5. Since
cot 0.729728 41.8103. Using the
1 3 5
3
3 5
15
132 0 local maximum of 41.8103° when
7085
22 (sec y ) 2 dy 4 y tan y /3 4 3
0
3
13 r 2 h 13 (3sin )2 (3cos ) 9 cos cos3 , where 0 2
dV
9 (sin ) 1 3cos 2 0 sin 0 or cos 1 the critical points are: 0, cos 1 1 , and
d
3
3
1
cos 1 ; but cos 1 1 is not in the domain. When
3
3
1 1
cos
54.7, we have a maximum volume.
3
0, we have a minimum and when
116. 65 90 90 180 65 65 tan 1
5021 65 22.78 42.22
117. Take each square as a unit square. From the diagram we have the following: the smallest angle has a tangent
of 1 tan 1 1; the middle angle has a tangent of 2 tan 1 2; and the largest angle has a tangent of
3 tan 1 3. The sum of these three angles is tan 1 1 tan 1 2 tan 1 3 .
118. (a) From the symmetry of the diagram, we see that sec1 x is the vertical distance from the graph of
y sec1 x to the line y and this distance is the same as the height of y sec1 x above the x-axis at x;
i.e., sec1 x sec1 ( x).
(b) cos 1 ( x) cos 1 x, where 1 x 1 cos 1 1x cos 1 1x , where x 1 or x 1
1
sec ( x) sec
1
x
119. sin 1 (1) cos 1 (1) 2 0 2 ;sin 1 (0) cos 1 (0) 0 2 2 ; and sin 1 (1) cos 1 (1) 2 2 . If
x (1, 0) and x a, then sin 1 ( x) cos 1 ( x) sin 1 ( a) cos 1 ( a) sin 1 a cos 1 a
sin 1 a cos 1 a 2 2 from Equations (3) and (4) in the text.
Copyright 2018 Pearson Education, Inc.
498
Chapter 7 Integrals and Transcendental Functions
120.
121. csc1 u 2 sec1 u
d
dx
csc1 u dxd 2 sec1 u 0 |u| u 1 |u| u 1 , | u | 1
du
dx
2
122. y tan 1 x tan y x
sec2 y
dydx 1 dydx
d (tan
dx
1
sec2 y
du
dx
2
d ( x)
y ) dx
1
1 x
2
1 2 , as
2
1 x
indicated by the triangle
123. f ( x) sec x f ( x) sec x tan x
df 1
dx
x b
1
df
dx x f 1 ( b )
1
sec sec1 b tan sec1 b
of sec1 x is always positive, we obtain the right sign by writing
124. cot 1 u 2 tan 1 u
d
dx
cot 1 u dxd 2 tan 1 u 0 1u
d sec 1 x
dx
du
dx
2
125. The function f and g have the same derivative (for x 0), namely
1
b b 2 1
1
| x| x 2 1
. Since the slope
.
du
dx
1u 2
1
. The
x ( x1)
functions therefore differ by a
constant. To identify the constant we can set x equal to 0 in the equation f ( x) g ( x) C , obtaining
sin 1 (1) 2 tan 1 (0) C 2 0 C C 2 . For x 0, we have sin 1 xx 11 2 tan 1 x 2 .
126. The functions f and g have the same derivative for x 0, namely
1 . The
1 x 2
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) g ( x) C , obtaining
tan
sin 1 1
1
2
1 C 4 4 C C 0. For x 0, we have sin 1 12 tan 1 1x .
x 1
2
3
3
1
1 dx tan 1 x
dx
tan 1 3 tan 1 33
2
3/3
3/3 1 x 2
3/3 1 x
2
3 6 2
127. V
3
128. Consider y r 2 x 2
dy
dx
x
r 2 x2
; Since
dy
dx
is undefined at x r and x r , we will find the length from
2
r/ 2
x 0 to x r (in other words, the length of 18 of a circle) L
1 2 x 2 dx
0
2
r x
r/ 2
0
2
r/ 2
1 2x 2 dx
r x
0
r 2 dx
r x2
2
r/ 2
0
r/ 2
r sin 1 x
dx
r sin 1 r / r 2 r sin 1 (0)
r
2
2
0
r x
r
r sin 1 1 0 r 4 4r . The total circumference of the circle is C 8L 8 4r 2 r.
2
Copyright 2018 Pearson Education, Inc.
Section 7.6 Inverse Trigonometric Functions
2
129. (a)
1
b
1
A( x) 4 (diameter)2 4 1 2 1 2 2 V A( x) dx dx2 tan 1 x
a
1
x
x
1
1
1
1 x 1 x
2
( )(2) 4 2
2
(b)
1
b
1
A( x) (edge)2 1 2 1 2 4 2 V A( x) dx 4dx2 4 tan 1 x
a
1
x
x
1
1
1
1 x 1 x
1
1
4[tan (1) tan (1)] 4 4 ( 4 ) 2
2
130. (a)
A( x) 4 (diameter)2 4 4 2 2 0 4 4 2
1 x
1 x
2 /2
V b A( x) dx
dx
2
2 /2 1 x 2
a
1 x
2 /2
2
sin 1 x
sin 1 22 sin 1 22 4 4 2
2 /2
(b)
A( x)
(diagonal)2
2
2
2 /2
2 /2
b
2 dx 2 sin 1 x
12 4 2 2 0 2 2 V A( x) dx
2
a
2 /2 1 x
2 /2
1 x
1 x
2 4 2
1 0.84107
131. (a) sec1 1.5 cos 1 1.5
1 0.72973
(b) csc1 (1.5) sin 1 1.5
(c) cot 1 2 2 tan 1 2 0.46365
132. (a) sec 1 (3) cos 1 13 1.91063
(b) csc1 1.7 sin 1
1.71 0.62887
(c) cot 1 (2) 2 tan 1 (2) 2.67795
133. (a) Domain: all real numbers except those having
the form 2 k where k is an integer.
Range: 2 y 2
(b) Domain: x ; Range: y
The graph of y tan 1 (tan x ) is periodic, the
graph of y tan(tan 1 x) x for x .
134. (a) Domain: x ; Range: 2 y 2
Copyright 2018 Pearson Education, Inc.
499
500
Chapter 7 Integrals and Transcendental Functions
(b) Domain: 1 x 1; Range: 1 y 1
The graph of y sin 1 (sin x) is periodic; the
graph of y sin (sin 1 x) x for 1 x 1.
135. (a) Domain: x ; Range: 0 y
(b) Domain: 1 x 1; Range: 1 y 1
The graph of y cos 1 (cos x) is periodic; the
graph of y cos (cos 1 x) x for 1 x 1.
136. Since the domain of sec1 x is (, 1] [1, ), we
have sec (sec1 x) x for | x | 1. The graph of
y sec(sec1 x) is the line y x with the open line
segment from (1, 1) to (1, 1) removed.
137. The graphs are identical for y 2sin 2 tan 1 x
4 sin tan 1 x cos tan 1 x
4
x
x
2
1
1
x
2
1
4x
x 2 1
from the triangle
138. The graphs are identical for y cos 2sec1 x
2
2
cos 2 sec 1 x sin 2 sec 1 x 12 x 21 22x
x
x
x
from the triangle
Copyright 2018 Pearson Education, Inc.
Section 7.7 Hyperbolic Functions
139. The values of f increase over the interval [1, 1]
because f 0, and the graph of f steepens as the
values of f increase toward the ends of the interval.
The graph of f is concave down to the left of the
origin where f 0, and concave up to the right of
the origin where f 0, There is an inflection point
at x 0 where f 0 and f has a local minimum
value.
140. The values of f increase throughout the interval
(, ) because f 0, and they increase most
rapidly near the origin where the values of f are
relatively large. The graph of f is concave up to the
left of the origin where f 0, and concave down to
the right of the origin where f 0. There is an
inflection point at x 0 where f 0 and f has a
local maximum value.
7.7
HYPERBOLIC FUNCTIONS
1. sinh x 34 cosh x 1 sinh 2 x 1 34
coth x
2. sinh x
1
tanh x
4
3
53 ,
sech x
1
cosh x
4,
5
3
9 25 5 , tanh x sinh x 4 3 ,
1 16
cosh x
5
16
4
5
2
4
and csch x
cosh x 1 sinh 2 x 1 16
9
25
9
1
sinh x
43
53 , tanh x
sinh x
cosh x
43 4 , coth x 1 5 ,
tanh x
4
53 5
1 3 , and csch x 1 3
sech x cosh
sinh x
4
x
5
, x 0 sinh x cosh 2 x 1
3. cosh x 17
15
coth x
1
tanh x
17
, sech x
8
1
cosh x
1
tanh x
13 , sech x
12
5. 2 cosh (ln x) 2
6. sinh (2 ln x)
eln x e ln x
2
e2 ln x e2ln x
2
1
cosh x
e
lnx
1
15 , and csch x
17
, x 0 sinh x cosh 2 x 1
4. cosh x 13
5
coth x
17 2
15
169
25
1
5 , and csch x
13
289
225
1
sinh x
144
25
1
64
225
sinh x
12
, tanh x cosh
5
x
1
sinh x
5
12
e
2
2
x2 1
x2
2
sinh x
cosh x
15
8
ln1 x x 1x
ln x
ln x
e 2e
8 , tanh x
15
4
x 21
2x
Copyright 2018 Pearson Education, Inc.
125 12 ,
135 13
158 8 ,
1715 17
501
502
Chapter 7 Integrals and Transcendental Functions
7. cosh 5 x sinh 5 x
e 5 x e 5 x
2
9. (sinh x cosh x) 4
e x e x
2
e5 x e 5 x
2
x
x
e 2e
e5 x
8. cosh 3x sinh 3 x
e
4
x 4
e3 x e3 x
2
3x
3 x
e 2e e3 x
e4 x
10. ln(cosh x sinh x) ln(cosh x sinh x) ln cosh 2 x sinh 2 x ln1 0
11. (a) sinh 2 x sinh( x x) sinh x cosh x cosh x sinh x 2sinh x cosh x
(b) cosh 2 x cosh( x x) cosh x cosh x sinh x sin x cosh 2 x sinh 2 x
12. cosh 2 x sinh 2 x
1
4
e x e x
2
2
e x e x
2
2e x 2e x 14 4e0 14 (4) 1
dy
dx
6 cosh 3x
2
14 e x e x e x e x e x e x e x e x
13 2 cosh 3x
13.
y 6sinh 3x
14.
dy
y 12 sinh 2 x 1 dx 12 cosh(2 x 1) (2) cosh(2 x 1)
15.
dy
y 2 t tanh t 2t1/2 tanh t1/2 dt sech 2 t1/2
16.
dy
y t 2 tanh 1t t 2 tanh t 1 dt sech 2 t 1 t 2 t 2 2t tanh t 1 sech 2 1t 2t tanh 1t
12 t 1/2 2t1/2 tanh t1/2 t 1/2 sech 2
17. y ln(sinh z )
dy
dz
z coth z
cosh
sinh z
18. y ln(cosh z )
dy
dz
sinh z tanh z
cosh
z
19.
dy
y (sech )(1 ln sech ) d
sech tanh
sech
sech sech tanh 1 ln sech
sech tanh sech tanh (1 ln sech ) (sech tanh ) 1 1 ln sech
sech tanh ln sech
20.
dy
y csch 1 ln csch d csch
csch coth
csch
1 ln csch csch coth
csch coth 1 ln csch csch coth csch coth 1 1 ln csch
csch coth ln csch
21.
y ln cosh v 12 tanh 2 v
dy
dv
sinh v
cosh v
12 2 tanh v sech2v tanh v tanh v sech2v
(tanh v ) 1 sech 2 v (tanh v) tanh 2 v tanh 3 v
Copyright 2018 Pearson Education, Inc.
t tanh t
t
Section 7.7 Hyperbolic Functions
22.
dy
v 1 2 coth v csch 2 v coth v coth v csch 2 v
y ln sinh v 12 coth 2 v dv cosh
sinh v
2
(coth v ) 1 csch 2 v (coth v) coth 2 v coth 3 v
e
23.
y x 2 1 sech ln x x 2 1
24.
y 4 x 2 1 csch ln 2x 4 x 2 1
25.
y sinh 1 x sinh 1 x1/2 dx
ln x
e
ln 2 x
2
2
e ln 2 x
12 x1/ 2
dy
x 1
2
e ln x
1/ 2 2
x 1 2x
2
4x 1
2
1 x
2
x x 1
1
2 x 1 x
(2)
2
2 x (2 x )1
2x
x 2 1
12 ( x 1)1/ 2
27.
y (1 ) tanh 1 d (1 ) 1 2 (1) tanh 1 11 tanh 1
1
28.
dy
y ( 2 2 ) tanh 1 ( 1) d 2 2 1 2 (2 2) tanh 1 ( 1)
1( 1)
dy
2
1
2 x (1 x )
y cosh 1 2 x 1 cosh 1 2 x 1
dx
2
4 x4 x1 4 x dydx 4
4 x2 1
26.
1/2
dy
dx
2
2( x 1)1/ 2 1
1
x 1 4 x 3
1
4 x 2 7 x 3
dy
2
2 2 (2 2) tanh 1 ( 1) (2 2) tanh 1 ( 1) 1
2
29.
1 t 1/ 2
dy
(1) coth 1 t1/2 1 coth 1 t
y (1 t ) coth 1 t (1 t ) coth 1 t1/2 dt (1 t ) 2
2 t
1 t1/ 2 2
30.
y 1 t 2 coth 1 t dt 1 t 2
31.
dy
y cos 1 x x sech 1 x dx 1 2 x 1 2
1 x
x 1 x
32.
y ln x 1 x 2 sech 1 x ln x 1 x 2
dy
11t 2t coth 1 t 1 2t coth 1 t
2
dy
dx 1x 1 x 2
33.
1/2
1
1 1 sech 1 x sech 1 x
(1) sech x
1 x 2
1 x 2
sech 1 x
1/2
1 1 1 x 2
2 x sech 1 x 1x 1x x 2 sech 1 x x 2 sech 1 x
2
2
1
1 x
1 x
x
x
1/2
ln 1 1
dy
y csch 1 12 d 2 2
12
1
12
2
ln(1) ln(2)
1
1 2
2
ln 2
1
12
2
Copyright 2018 Pearson Education, Inc.
503
504
34.
Chapter 7 Integrals and Transcendental Functions
y csch 1 2 d
(ln 2)2
dy
ln 22
2
1 2
1 2
2
2
2
2
35.
x
y sinh 1 (tan x) dx sec x 2 sec 2x sec
|sec x|
1 (tan x )
sec x
36.
y cosh 1 (sec x) dx
dy
dy
(sec x )(tan x )
sec 2 x 1
37. (a) If y tan 1 (sinh x) C , then
1
(b) If y sin (tanh x) C , then
38. If y
x2
2
dy
dx
dy
dx
(sec x )(tan x )
tan 2 x
|sec x||sec x|
|sec x|
(sec x )(tan x )
|tan x|
| sec x |
sec x, 0 x 2
cosh 2x cosh2x sech x, which verifies the formula
1 sinh x
2
x sech x, which verifies the formula
sech
x
sech
1 tanh x
sech 1 x 12 1 x 2 C , then
cosh x
sech 2 x
2
dy
dx
x sech 1 x
x2
2
x
1
1 x 2
4
2x
1 x 2
x sech 1 x, which verifies the
formula
39. If y
x 2 1
2
2
dy
1
coth 1 x 2x C , then dx x coth 1 x x 21
12 x coth 1 x, which verifies the formula
1 x 2
40. If y x tanh 1 x 12 ln 1 x 2 C , then
dy
dx
tanh 1 x x
tanh
1
1 x 2
1 2 x
2 1 x 2
1
x, which verifies the
formula
41.
sinh 2 x dx 12 sinh u du,
42.
cosh u
2
C
where u 2 x and du 2 dx
cosh 2 x
2
C
sinh 5x dx 5 sinh u du, where u 5x
and du 15 dx
5cosh u C 5cosh 5x C
43.
6 cosh 2x ln 3 dx 12 cosh u du,
where u 2x ln 3 and du 12 dx
12 sinh u C 12sinh 2x ln 3 C
44.
4 cosh (3x ln 2) dx 43 cosh u du, where u 3x ln 2 and du 3 dx
43 sinh u C 43 sinh(3 x ln 2) C
45.
sinh u
tanh 7x dx 7 cosh u du,
where u
x
7
and du 17 dx
x/7
x /7
7 ln | cosh u | C1 7 ln cosh 7x C1 7 ln e 2e
C1 7 ln e x /7 e x /7 7 ln 2 C1
7 ln e x /7 e x /7 C
Copyright 2018 Pearson Education, Inc.
Section 7.7 Hyperbolic Functions
46.
u du ,
coth 3 d 3 cosh
sinh u
where u and du d
3
3
/ 3
/
3 ln sinh u C1 3 ln sinh C1 3 ln e 2e
3
3
C1
3 ln e / 3 e / 3 3 ln 2 C1 3 ln e / 3 e / 3 C
47.
2
2
sech x 12 dx sech u du, where u x 12
and du dx
tanh u C tanh x 12 C
48.
csch
2
(5 x)dx csch 2u du , where u (5 x) and du dx
( coth u ) C coth u C coth (5 x) C
49.
sech t tanh t
t
dt 2 sech u tanh u du , where u t t1/2 and du dt
2 t
2( sech u ) C 2 sech t C
50.
csch ( ln t ) coth (ln t )
t
dt csch u coth u du , where u ln t and du dtt
csch u C csch(ln t ) C
51.
ln 4
ln 4
15/8 1
du
u
x dx
ln 2 coth x dx ln 2 cosh
3/4
sinh x
where u sinh x, du cosh x dx;
2 1
4 1
ln 2
ln 4
ln 2
ln 4
x ln 2 u sinh(ln 2) e 2e
2 2 34 , x ln 4 u sinh(ln 4) e 2e
2 4 15
8
15/8
ln | u |3/4 ln 15
ln 34 ln 15
. 4 ln 52
8
8 3
52.
ln 2
0
tanh 2 x dx
ln 2 sinh 2 x
dx
cosh 2 x
0
17/8 1
12
du where u cosh 2 x, du 2sinh (2 x ) dx,
u
1
4 1
ln 4
ln 4
x 0 u cosh 0 1, x ln 2 u cosh (2 ln 2) cosh (ln 4) e 2e
2 4 17
8
17/8
17
17
1
1
1
2 ln | u |1 2 ln 8 ln1 2 ln 8
53.
ln2
ln4 2e
cosh d
ln 2
ln 4
ln 2
2
ln 2 2
2e e 2e d
e 1 d e 2
ln 4
ln 4
2 ln 2
2 ln 4
1 ln 4 3 ln 2 2 ln 2 3 ln 2
e 2 ln 2 e 2 ln 4 18 ln 2 32
32
32
54.
ln 2
0
4e sinh d
ln 2
0
2 ln 2 e 2
2 ln 2
ln 2
2
ln 2
4e e 2e d 2
1 e2 d 2 e 2
0
0
0 2 ln 2
e0
2
1
8
12 2 ln 2 14 1 ln 4 34
Copyright 2018 Pearson Education, Inc.
505
506
55.
Chapter 7 Integrals and Transcendental Functions
/4
/4 cosh(tan ) sec
2
1
d cosh u du where u tan , du sec2 d , x 4 u 1, x 4 u 1,
1
1
1
1 1
1
1
1
sinh u 1 sinh(1) sinh(1) e 2e e 2 e e e 2 e e e e 1
56.
/2
0
1
2sinh(sin ) cos d 2 sinh u du where u sin , du cos d , x 0 u 0, x 2 u 1
0
1
1
2 cosh u 0 2(cosh1 cosh 0) 2 e 2e 1 e e1 2
57.
2 cosh(ln t )
t
1
dt
ln 2
0
cosh u du where u ln t , du 1t dt , x 1 u 0, x 2 u ln 2
sinh u 0
ln 2
58.
4 8 cosh x
x
1
ln 2
ln 2
2 1
sinh(ln 2) sinh(0) e 2e
0 2 2 43
2
dx 16 cosh u du where u x x1/2 , du 12 x 1/2 dx
1
dx
2 x
, x 1 u 1, x 4 u 2
2
2
1
16 sinh u 12 16(sinh 2 sinh1) 16 e 2e e2e 8 e 2 e 2 e e 1
59.
0
ln 2 cosh
dx 0ln 2 cosh2x1 dx 12 0ln 2 (cosh x 1)dx 12 sinh x x0 ln 2
2 x
2
1 2
ln 2
ln 2
12 (sinh 0 0) (sinh( ln 2) ln 2) 12 (0 0) e 2e ln 2 12 2 2 ln 2 12 1 14 ln 2
83 12 ln 2 83 ln 2
60.
ln10
0
61. sinh 1
0
125 ln 125
63. tanh 1 12 12 ln
25
144
1(1/2)
1 (1/2)
0
ln10
1 2 ln10 9.9 2 ln10
e ln10 2 ln10 10 10
1 ln 23
ln 3
3
2 3
53 ln 53
64. coth 1
1 ln 9 ln 3
54 12 ln (9/4)
(1/4) 2
66. csch 1
1
3
2 3
sinh 1 2x
sinh 1 3 sinh 0 sinh 1 3
0
4 x
(b) sinh 1 3 ln 3 3 1 ln 3 2
0
1/3
0
dx
2
6 dx
1 9 x
2
1 dx
, where u 3x, du 3 dx,
0 a 2 u 2
1
2sinh 1 u 2 sinh 1 1 sinh 1 0
0
2
(b) 2sinh 1 1 2 ln 1 12 1 2 ln 1 2
ln 3
62. cosh 1
25 1
9
ln 3
(9/25)
53 ln 1 1(3/5)
ln 3
65. sech 1
68. (a)
ln10
ln10
ln10
4sinh 2 2x dx
4 cosh2 x 1 dx 2
(cosh x 1)dx 2 sinh x x 0
2 sinh(ln 10) ln 10 (sinh 0 0) e
67. (a)
a 1
2sinh 1 1
Copyright 2018 Pearson Education, Inc.
4/3
1/ 3
ln 3 2
Section 7.7 Hyperbolic Functions
69. (a)
2
5/4 11x2 dx coth
1
507
2
x
coth 1 2 coth 1 54
5/4
(b) coth 1 2 coth 1 54 12 ln 3 ln
1 ln 1
9/4
1/4 2
3
1/2
1/2 1
1 x 2
dx tanh 1 x tanh 1 12 tanh 1 0 tanh 1 12
0
1
1/2
1 ln 3
(b) tanh 1 12 12 ln
11/2 2
70. (a)
0
71. (a)
1/5
3/13
dx
x 116 x
2
12/13
4/5
du
u a 2 u 2
, u 4 x, du 4 dx, a 1
12/13
sech 1u
sech 1 12
sech 1 54
13
4/5
1
sech 1 54 ln
(b) sech 1 12
13
1(12/13)2
(12/13)
1
ln
1 (4/5) 2
(4/5)
ln
13 169 144
12
ln
5 2516
4
ln 54 3 ln 13125 ln 2 ln 32 ln 2 23 ln 43
2
72. (a)
1
(b)
1
2
73. (a)
2
1 csch1 x 1 csch 11 csch 1 1 1 csch1 1 csch 11
2 1
2
2
2
2
2
x 4 x 2
dx
5/4
csch 1 12 csch 11 12 ln 2 (1/2)
ln 1 2 12 ln 12 25
cos x
0
2
1sin x
0
1
0
1u 2
dx
du where u sin x, du cos x dx;
0
sinh 1 u sinh 1 0 sinh 1 0 0
0
(b) sinh 1 0 sinh 1 0 ln 0 0 1 ln 0 0 1 0
74. (a)
e
1 x
dx
1 (ln x ) 2
1 du
0 a 2 u 2
, where u ln x, du 1x dx, a 1
1
sinh 1 u sinh 1 1 sinh 1 0 sinh 1 1
0
(b) sinh 1 1 sinh 1 0 ln 1 12 1 ln 0 02 1 ln 1 2
f ( x) f ( x)
f ( x) f ( x)
f ( x) f ( x)
f ( x) f ( x)
2 f ( x)
and O( x)
. Then E ( x) O( x )
2 f ( x ).
2
2
2
2
f x f ( x )
f ( x ) f ( ( x ))
f ( x) f ( x)
E ( x) E ( x) is even, and O ( x)
Also, E x
2
2
2
f ( x) f ( x)
O ( x) O ( x) is odd. Consequently, f ( x) can be written as a sum of an even and an odd
2
f ( x) f ( x)
f ( x) f ( x)
f ( x) f ( x)
function. f ( x)
because
0 if f is even, and f ( x)
because
2
2
2
f ( x) f ( x)
2 f ( x)
2 f ( x)
0 if f is odd. Thus, if f is even f ( x) 2 0 and if f is odd, f ( x) 0 2
2
75. Let E (x)
Copyright 2018 Pearson Education, Inc.
508
76.
Chapter 7 Integrals and Transcendental Functions
y
y
y sinh 1 x x sinh y x e 2e 2 x e y 1y 2 xe y e 2 y 1 e2 y 2 xe y 1 0
e
2
e y 2 x 24 x 4 e y x x 2 1 sinh 1 x y ln x x 2 1 Since e y 0, we cannot choose
e y x x 2 1 because x x 2 1 0.
mg
k
77. (a) v
tanh
gk
m
t
dv
dt
2
sech
mg
k
gk
m
t
gk
m
2
g sech
gk
m
t . Thus
gk
gk
m dv
mg sech 2 m t mg 1 tanh 2 m t mg kv 2 . Also, since tanh x 0 when x 0, v 0
dt
when t 0.
mg
mg
mg
mg
kg
kg
(b) lim v lim k tanh m t k lim tanh m t k (1) k
t
t
t
160
0.005
(c)
78. (a)
160,000
400 80 5 178.89 ft/sec
5
5
2
s (t ) a cos kt b sin kt ds
ak sin kt bk cos kt d 2s ak 2 cos kt bk 2 sin kt
dt
dt
2
2
k (a cos kt b sin kt ) k s (t ) acceleration is proportional to s. The negative constant k 2
implies that the acceleration is directed toward the origin.
(b) s (t ) a cosh kt b sinh kt
2
ak sinh kt bk cosh kt d 2s ak 2 cosh kt bk 2 sinh kt
ds
dt
2
dt
k 2 (a cosh kt b sinh kt ) k s (t ) acceleration is proportional to s. The positive constant k2 implies
that the acceleration is directed away from the origin.
79. V
2
0
80. V 2
cosh 2 x sinh 2 x dx 021 dx 2
ln 3
0
81.
ln 3
y 12 cosh 2 x y sinh 2 x L
ln 5
12
82. (a)
(b)
(c)
(d)
e 2 x e 2 x
2
0
1
4
e x e x
x
x
x e e
lim
e x e x
x
x
x e e
e x e x
2
x
lim sinh x lim
x
x
ex 1
ex
x e x 1
ex
1
ex
e x 1x
e
1
e x ex
2
lim
cosh 2 x dx 12 sinh 2 x
x
e2
1
1
e2 x
1
x 1 2 x
e
lim
1100 1
e2 x 1
2x
x e 1
lim
0
0
ex
x 2
lim
ex
x 2
lim
1
ex
1
ex
ex 1
x
ex e
x
x e x 1 e
x
e
1
ex
e x 1x
e
x
x
lim
ex
lim
lim
e x e x
2
x
lim sinh x lim
ex
x
lim tanh x lim
x
ln 5
0
5 15 56
lim tanh x lim
x
3 1/ 3
1 (sinh 2 x)2 dx
0
ln 5
3 1/ 3
2
sech 2 x dx 2 tanh x 0
1
2ex
Copyright 2018 Pearson Education, Inc.
0 1
0 1
1
ln 5
0
Section 7.7 Hyperbolic Functions
(e)
(f )
(g)
(h)
(i)
83. (a)
2
x
x
x e e
2
x
x e
lim sech x lim
x
lim coth x
x
lim
x
x
lim e x e x
x e e
lim
x
e x e x
e x e x
x 0
lim coth x lim
x 0
lim coth x
x 0
x
x
lim e x e x
e e
x 0
x
y
H
w
cosh
x
1
ex
ex
1
ex
ex
x
1
lim
e2 x 1
2x
x 0 e 1
x
e2 x 1
2x
x 0 e 1
e x lim
e
x
1
ex
2e x
2x
x e 1
. e x lim
e
1
1
1100 1
e2 x
1
x 1 2 x
e
ex
1
ex
e
1
2
x e
x
100 0
e x lim
ex
ex 1
x 0
ex
lim
x 1
e x
e
lim
x e x 1
ex
1
ex
ex 1
x 0
ex
lim
2
ex
1
1
x
e2 x
e1 lim
ex
e x 1x
e
lim
2
x
x e e
lim csch x lim
ex
1
x
0
0 1
0
Hw x tan dydx Hw Hw sinh Hw x sinh Hw x
(b) The tension at P is given by T cos H T H sec H 1 tan 2 H 1 sinh Hw x
H cosh
Hw x w Hw cosh Hw x wy
84. s a1 sinh ax sinh ax as ax sinh 1 as x a1 sinh 1 as; y 1a cosh ax
1
a
509
sinh 2 ax 1
1
a
a2 s2 1 s2
85. To find the length of the curve: y
1
a
1
a
2
cosh 2 ax
1
a2
cosh ax y sinh ax L
b
0
b
b1
0 a
a1 sinh ax a1 sinh ab. The area under the curve is A
0
b
1 (sinh ax) 2 dx L cosh ax dx
0
b
cosh ax dx 12 sinh ax
a
0
1
a2
sinh ab
a1 sinh ab which is the area of the rectangle of height 1a and length L as claimed, and which is illustrated
1a
below.
86.
(a) Let the point located at (cosh u , 0) be called T. Then A(u ) area of the triangle OTP minus the area
under the curve y x 2 1 from A to T A(u )
(b)
1
2
cosh u sinh u
cosh u
1
cosh 2 u sinh 2 u
12 cosh 2 u 12 sinh 2 u sinh 2 u 12 cosh 2 u sinh 2 u 12 (1) 12
A(u ) 12 cosh u sinh u
(c) A(u )
cosh u
1
1
2
x 2 1 dx A(u )
1
2
x 2 1 dx.
cosh 2 u 1 sinh u
A(u ) u2 C , and from part (a) we have A(0) 0 C 0 A(u ) u2 u 2 A
Copyright 2018 Pearson Education, Inc.
510
7.8
Chapter 7 Integrals and Transcendental Functions
RELATIVE RATES OF GROWTH
x 3
x
x e
1
x
x e
lim
1. (a) slower, lim
x3 sin 2 x
ex
x
2
ex
x
(c) slower, lim
4x
x
x e
(e) slower, lim
x1/ 2
x
x e
x
lim 4e
x
32
x
ex/ 2
x
x e
(g) same,
(h) slower, lim
2. (a) slower, lim
x
(b) slower, lim
x
(d) slower, lim
1
x 2
lim
e
x
x e
x
e x
x
x e
xe x
x
x e
(f ) faster, lim
x
3
2e
10
x
x e
0 lim
0
1
x
x (ln10) e
40 x3 30
ex
ex
1
2x
x e
lim
x
lim
x
lim
120 x 2
ex
ex
4 x3
2x
x 2e
lim
0 since
5
2e
1
x
x (ln10) xe
lim
ln x 1 x
x
lim
5
x 2e
1
x
lim
x (ln x 1)
lim
x
(e) slower, lim
lim
1 x 4
2x
x e
240 x
x
x e
lim
1x
0
240
x
x e
lim
ln x 11
x
x e
lim
12 x 2
2x
x 4e
lim
lim e x
x e
ln x
x
x e
lim 242xx
x 8e
e1
ex
3. (a) same, lim
x
1
(d) same, lim
x
x2 4 x
x2
lim 1 x
x xe
0 0
0
1
( x 3)2
( x x 1)
2 x4
x 2 x
lim
x 4 x3
x2
x2
lim 242 x
x 16
1
x e
1e
2
x 2
1
lim
lim
lim x3 1
x
ecos x
ex
cos x
e
x
x e
x e
x5 x 2
2
x x
x
x
x e
1
, so by the Sandwich Theorem we conclude that lim
lim
(b) slower, lim
(c) same, lim
1x
x
e
x
x e
e x 1
x
x e
lim
lim x
0 lim
(h) same, lim
0
lim
(g) slower, since for all reals we have 1 cos x 1 e 1 ecos x e1
1
0 by the Sandwich
0
x
1 x 4
ex
6 4sin 2 x
ex
1
2
lim
ex
1
x
x 2 xe
lim
x
1
0 since
ln x
x
x (ln10) e
x ln x x
52
x/2
x
lim
10 x 4 30 x 1
ex
x
(c) slower, lim
x e
log10 x
x
4
e
1
lim
(f ) slower, lim
ex
2
lim x
x e
since
lim
2
x
x e
12 x1/ 2
ex
3
x 2e
6 x 2 cos 2 x
ex
e
x
lim
x
x e
x
10x for all reals, and lim
lim
(d) faster, lim
lim
ex
x
6 4sin 2 x
ex
lim
x
x e
3 x 2 2sin x cos x
lim
(b) slower, lim
Theorem because
0
x 4 x3
4
x x
lim
lim
x
2 x 3
2x
2
x 2
lim
lim 1 1x 1 1
x
1
Copyright 2018 Pearson Education, Inc.
0
e1
ex
and also
0
Section 7.8 Relative Rates of Growth
x ln x
(e) slower, lim
x x
ln x
x x
lim
2
lim
x 3e x
2
x x
8 x2
2
x x
x lim 1
x
x
x e x e
x2 x
2
x x
10 x 2
2
x x
x
1
x
x e
lim
log10 x 2
x
x3 . x 2
2
x x
1
10
2
(ln1.1)(1.1) x
2x
x
lnln 3x
x ln x
x1/ 2
x ln x
x
x ln x
lim
x
lim 5ln x
x ln x
1
x
ex
ln
x x
(ln1.1)2 (1.1) x
2
x
lim
lim 1
x ln 3
1
ln 3
lim 1
x 2
1
2
1
1x
12 x1/ 2
1
x x
lim
x
x 2 x
lim
x
x 2
lim
lim x
x
lim 5 5
x ln x
(h) faster, lim
0
lim
lim
(g) slower, lim
x ln x
x
x ln x
(e) faster, lim
6. (a) same,
x
lim
ln
x ln x
(d) faster, lim
1 lim 1
ln10 x x 2
lim 1 100
1
x
22x 1
1
x x
1 ln x
x
lim 2
(c) same, lim
1
2 lim x
ln10 x 2 x
0
lim
x 2 100 x
x2
x
ln 2 x
x ln x
x
1
x 2
x 10 x
2
(b) same, lim
1 lim 2 ln x
ln10 x x 2
lim ( x 1)
(h) same, lim
(f ) same,
ln x2
ln10
lim 2
x x
lim
(1.1) x
log3 x
x ln x
0
x
x x
5. (a) same, lim
2
(e) faster, lim
x x
1
1
x3/ 2
x
2 x
lim x e2
x x
x
lim 1
lim 10 10
(b) same, lim
(g) faster, lim
0
x
4. (a) same, lim
(f ) slower, lim
lim 8 8
(h) same, lim
(d) slower, lim
(ln 2) 2 2 x
2
x
lim
lim
(g) slower, lim
(c) slower,
x 1
(ln 2)2 x
2x
x
2x
2
x x
(f ) slower, lim
1x 0
lim
x
lim 1
x x ln x
lim
log x 2
lim ln2x
x
x
ex
1x
0
lim xe x
x
ln x 2
ln 2
lim ln x
x
1 lim ln x 2
ln 2 x ln x
1 lim 2 ln x
ln 2 x ln x
1 lim 2 2
ln 2 x
ln 2
Copyright 2018 Pearson Education, Inc.
511
512
Chapter 7 Integrals and Transcendental Functions
log10 10 x
x ln x
lim
(b) same, lim
x ln x
lim
1
x
(c) slower, lim
x ln x
1
2
x
x ln x
1
2
x x ln x
lim
x
x
xx
x
x (ln x )
x /2 x
8.
0
1
x
0
lim
x
x
lim
e
0
lim e x /2 e x grows faster then e x /2 ; since for x ee we have ln x e and
ln x
e
1010x 1 lim 1 1
1x ln10 x ln10
1/ln xx lim 1 0
1
x x
x ln x
2
ln(2 x 5)
lim ln x lim 2 x1 5 lim 2 2x x 5 lim 22 lim 1 1
x
x x
x
x
x
(h) same,
ex
x/2
x e
1 lim
ln10 x
lim x 1
x e ln x
ln(ln x )
x ln x
lim
lim
(g) slower, lim
7.
1 lim ln10 x
ln10 x ln x
x 2 lim 1 2 lim x 2
x 2 xlim
x ln x
x
ln x
x
x 2 ln x
x ln x
(f ) slower,
x (ln x)
lim
(e) faster, lim
e x
lim ln
x x
1
x
(d) slower, lim
x
ln10
ln10
ln x
x e
x
(ln x) x grows faster then e x ; since x ln x for all x 0 and
x
x ln x
x
x x grows faster then (ln x) x . Therefore, slowest to fastest are:
lim
lim
, e , (ln x ) x , x x so the order is d, a, c, b
lim
x
(ln 2) x
x2
2x
x
x2
x
x 2
x 2 ; lim
ln(ln 2) (ln 2) x
lim
lim
ln(ln 2) 2 (ln 2) x
2
x
2x
x
x (ln 2)2
x
2
2 x
x (ln 2) 2
lim
lim
ln(ln 2) 2
2
lim (ln 2) x 0 (ln 2) x grows slower than
x
0 x 2 grows slower than 2 x ; lim
x
2x
x
x e
x
2
x e
lim
x
0 2x
grows slower than e . Therefore, the slowest to the fastest is: ln 2 , x 2 , 2 x and e so the order is c, b, a, d
x
x x
9. (a) false; lim
1
x
x x 5
(b) false; lim
11 1
(c) true; x x 5
(d) true; x 2 x
ex
2x
e
x
(e) true; lim
(f ) true;
x ln x
x
1 if x 1 (or sufficiently large)
1
x
x e
lim
ln x
x ln 2 x
x 2 5
x
1 if x 1 (or sufficiently large)
0
1 lnxx 1
(g) false; lim
(h) true;
x
2x
x
x 5
x
x
1
1
x
2 if x 1 (or sufficiently large)
1x lim 1 1
2
x 2 x x
lim
( x 5)2
x
x 5
x
1 5x 6 if x 1 (or sufficiently large)
Copyright 2018 Pearson Education, Inc.
Section 7.8 Relative Rates of Growth
10. (a) true;
(b) true;
x 13 x 1 if
1x x 3
1 1
x 2
x
1x
(c) false; lim
1 1
x 2
x
1x
e x
ex
1
x
x
ex
lim 1 1x 1
2 cos x
2
(d) true; 2 cos x 3
(e) true;
x 1 (or sufficiently large)
1 1x 2 if x 1 (or sufficiently large)
x
x
and
x
ex
3
2
if x is sufficiently large
0 as x 1
1
x ln x lim ln x lim x 0
2
x x
x x
x 1
ln(ln x ) ln x
ln x 1 if x is sufficiently
ln x
(f ) true; lim
(g) true;
ln x
2
x ln( x 1)
(h) false; lim
lim
1x
x 2 x
x 2 1
x
ex
2 if x is sufficiently large
large
x 2 1
2
x 2 x
lim
1
x 2
lim
1
2 x2
1
2
f ( x)
g ( x)
f ( x)
L 0 lim f ( x ) L1 0. Then g ( x ) L 1 if
x g ( x )
x
f ( x)
L 1 if x is sufficiently large f O( g ). Similarly,
g ( x)
11. If f ( x ) and g ( x) grow at the same rate, then lim
sufficiently large L 1
g ( x)
f ( x)
1
L
513
f ( x)
g ( x)
L 1
x is
1 g O( f ).
f ( x)
x g ( x )
0.
f ( x)
x g ( x )
0 when the degree of f is smaller
12. When the degree of f is less than the degree of g since in that case lim
13. When the degree of f is less than or equal to the degree of g since lim
f ( x)
x g ( x )
than the degree of g, and lim
a
b
(the ratio of the leading coefficients) when the degrees are the same.
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.
x11 lim x lim 1 1 and
1
x x
x x 1 x 1
15.
lim
ln( x 1)
x ln x
lim
16.
ln( x a )
x ln x
lim
17.
lim
lim
x
10 x 1
x
ln( x 999)
ln x
x
lim
x 1 a lim x lim 1 1. Therefore, the relative rates are the same.
1
x x
x x a x 1
lim
x
10 x 1
x
10 and lim
x
x 1
x
lim x 1
x x
1 1. Since the growth rate is transitive, we
conclude that 10 x 1 and x 1 have the same growth rate (that of
18.
lim
x
x4 x
x2
x 1999 lim x 1
1
x x
x x 999
lim
x4 x
4
x x
4
lim
1 and lim
x
3
x 4 x3
x2
x 4 x3
4
x x
lim
x ).
1. Since the growth rate is transitive, we
conclude that x x and x 4 x have the same growth rate (that of x 2).
Copyright 2018 Pearson Education, Inc.
514
19.
Chapter 7 Integrals and Transcendental Functions
nx n 1
x
x e
xn
x
e
x
lim
lim
n!
x
x e
lim
0 x n o e x for any non-negative integer n
20. If p( x) an x n an 1 x n 1 a1 x a0 , then
lim
p( x)
x e
x
xn a
x n 1
n 1 lim ex
x
x e
x
p( x)
x
an lim
x
x e
x1/ n
x ln x
lim
lim
x (1 n )/ n
x
n
1
x
x e
a0 lim
where each limit is zero (from Exercise
0 e grows faster than any polynomial.
19). Therefore, lim
21. (a)
x
x
x e
a1 lim
1n
x1/ n ln x o x1/ n for any positive integer n
1n xlim
(b) ln e17,000,000 17, 000, 000 e1710
6
1/106
e17 24,154,952.75
(c) x 3.430631121 1015
(d) In the interval 3.41 1015 ,3.45 1015 we have
ln x 10 ln(ln x). The graphs cross at about
3.4306311 1015.
22.
ln x
n
n 1
x an x an 1 x a1 x a0
lim
lim ln x
x x n
lim an
x
a
a
an 1
n 11 n0
x
x
x
lim 1/nx1
nx
x
an
1
n
x an nx
lim
0 ln x grows slower
than any non-constant polynomial (n 1)
23. (a)
lim
n log 2 n
n n (log 2 n )
2
grows slower then n(log 2 n) 2 ;
lim
n
n log 2 n
n3/2
(b)
lim log1 n 0 n log 2 n
2
n
lim
lnln n2
1/ 2
n n
1 lim
ln 2 n
1n
12 n1/ 2
1 0 n log n grows slower
ln22 lim 1/2
2
n n
3/2
than n . Therefore, n log 2 n grows at the
slowest rate the algorithm that takes
O(n log 2 n) steps is the most efficient in the
long run.
24. (a)
lim
n
log 2 n 2
n
lim
n
2
lnln n2
n
2
lim
(ln n )2
2
n n (ln 2)
log 2 n grows slower then n; lim
lim
n
log 2 n 2
n n log 2 n
2(ln n )
1n
ln 2 2
lim
n
2
lim ln n
ln 2 2 n n
log 2 n
n
lim
ln n
ln 2
1/ 2
n n
Copyright 2018 Pearson Education, Inc.
1 0
2
lim n
(ln 2) 2 n 1
1 lim ln n
ln 2 n n1/ 2
Chapter 7 Practice Exercises
1n 2 lim 1 0
1/ 2
ln 2 n n1/ 2
x n
ln12 lim
(b)
1
2
2
log 2 n grows slower than n log 2 n.
2
Therefore log 2 n grows at the slowest rate
the algorithm that takes O log 2 n
2
steps
is the most efficient in the long run.
25. It could take one million steps for a sequential search, but at most 20 steps for a binary search because
219 524, 288 1, 000, 000 1, 048,576 220.
26. It could take 450,000 steps for a sequential search, but at most 19 steps for a binary search because
218 262,144 450, 000 524, 288 219.
CHAPTER 7
PRACTICE EXERCISES
2 2 e
1.
y 10e x /5 dx (10) 15 e x /5 2e x /5
3.
1 e 4 x dy 1 x 4e4 x e 4 x (1) 1 4e4 x xe4 x 1 e 4 x 1 e 4 x xe 4 x
y 14 xe4 x 16
dx
4
4
4
16
4.
y x 2 e2/ x x 2 e2 x
5.
y ln sin 2 d
dy
dy
dy
1
6. y ln sec2 d
2
log 2 x2
2
ln x2
ln 2
7.
y
8.
y log5 (3x 7)
2(sin )(cos )
2x
2e 2 x
cos 2 cot
2sin
sin 2
2(sec )(sec tan )
sec2
2 tan
dy
dx ln12 x2 2
x ln 2 x
2
ln(3 x 7)
ln 5
dy
3x37 (ln 5)(33 x7)
dx ln15
dy
dt
8t (ln 8)(1) 8t (ln 8)
10. y 92t
dy
dt
92t (ln 9)(2) 92t (2 ln 9)
11. y 5 x3.6
dy
dx
5(3.6) x 2.6 18 x 2.6
12. y 2 x
dy
dx
dy
y 2e 2 x dx
1
1
1
dy
dx x 2 2 x 2 e2 x e2 x 2 x 2 2 x e2 x 2e2/ x 1 x
9. y 8t
2
2.
2 2 x
2 x
2 1
2 1
Copyright 2018 Pearson Education, Inc.
515
516
13.
Chapter 7 Integrals and Transcendental Functions
y ( x 2) x 2 ln y ln( x 2) x 2 ( x 2) ln( x 2)
y
y
( x 2)
x1 2 (1) ln( x 2)
dx ( x 2) x 2 ln( x 2) 1
dy
14.
1
y
y 2(ln x) x /2 ln y ln 2(ln x) x /2 ln(2) 2x ln(ln x ) y 0 2x lnx x ln(ln x) 12
y 2 ln1 x 12 ln(ln x) 2(ln x) x /2 (ln x) x /2 ln(ln x) ln1x
15.
y sin
16.
1
u
u 1u 2
2
1 u sin
1
1u 2
1
1 u
2 1/2
sin v
1
1/2
v
x y
1
1 x 2
cos 1 x
dy
dv
1 v
18.
y z cos 1 z 1 z 2 z cos 1 z 1 z 2
1 z
z
1 z
2
1/ 2
1/2
y 1 t 2 cot 1 2t dx 2t cot 1 2t 1 t 2
21.
y z sec 1 z z 2 1 z sec 1 z z 2 1
22.
2
| z| z 1
z
2
z 1
1
1t 2
dy
1/ 2 2
u
1u 2 1 1u 2
1
2v3/ 2 1v 1
sec1 z
1 z
2
z 1
1
2v3/ 2 v v1
dy
dz
cos 1 z
1/2
1
2
z
1 z
1t tan 1 t 1tt
2
2
u
|u| 1u 2
21x
sec1 x
x 1
v
2v3/ 2 v 1
1
2v v 1
1/2
( 2 z )
21t
142t
2
1/2
dy
dz z 12 sec 1 z (1) 12 z 2 1
2z
| z| z 1
sec 1 z , z 1
sec1 x
2 x 1
12 1 z 2
x
x 11/2 sec1 x1/2 x 11/2 x x1
dy
y 2 x 1sec 1 x 2( x 1)1/2 sec 1 x1/2 dx 2 12
2
23.
2
20.
z
( 2u )
cos 1 z
dy
y t tan 1 t 12 ln t dt tan 1 t t
1 1u 2
19.
1/ 2
1 x 2 cos 1 x
2
2
1
y ln cos
z
1u
12 v 3/ 2
17.
cos 1 z
1
2
,0 u 1
y sin 1 1
1
dy
du
1x
tan 1, 0
y csc 1 (sec ) d sec tan2 |tan
2
|
|sec | sec 1
dy
Copyright 2018 Pearson Education, Inc.
1 1/ 2
2
Chapter 7 Practice Exercises
24.
1
1
1
1
tan 1 x
y 1 x 2 e tan x y 2 xe tan x 1 x 2 e 2 2 xe tan x e tan x
1 x
25.
y
2
cos 2 x
cos 2 x
y
26.
ln y ln 2 x 1 ln(2) ln
2 x 2 1
2 x tan 2 x
x 2 1
y
2 x 2 1
2sin 2 x
x2 1 12 ln(cos 2 x) yy 0 x2 x1 12 cos
2x
2
2 x tan 2 x
cos 2 x x 2 1
1 ln(3 x 4) ln(2 x 4) y 1
y 10 32 xx44 ln y ln 10 32 xx44 10
y 10 3 x3 4 2 x24
517
3x34 x1 2
3 1 y 10 3 x 4 1
1
y 10
3x4 x2
2 x 4 10
27.
5
5
(t 1)(t 1)
y (t 2)(t 3) ln y 5 ln(t 1) ln(t 1) ln(t 2) ln(t 3) 1y
(t 1)(t 1)
dy
dt 5 (t 2)(t 3)
28.
5
cot
ln(sin )
2
dy
d
y (ln x)1/ln x ln y ln1x ln(ln x) y ln1x
31.
e
cos
sin
1 1/2 ln(sin )
2
ln1x 1x ln(ln x) (lnx1) 1x y ln x 1/ln x 1x(lnln(lnx)x)
y
30.
2
sin e x dx sin u du , where u e x and du e x dx
cos u C cos e x C
32.
e
t
cos 3et 2 dt 13 cos u du , where u 3et 2 and du 3et dt
13 sin u C 13 sin 3et 2 C
33.
e
x
sec 2 e x 7 dx sec 2 u du, where u e x 7 and du e x dx
tan u C tan e x 7 C
34.
e
y
2
y (sin ) ln y ln y (sin ) 1y
x
t 13
dudy u1 ln 2 12 u2u1
u
y 2u22 ln y ln 2 ln u u ln 2 12 ln u 2 1 1y
u 1
dy
d (sin )
1 1 1
t 1 t 1 t 2
t 11 t 11 t 11 t 13
u
dy
du 2u22 u1 ln 2 2u
u 1
u 1
29.
dy
dt
csc e y 1 cot e y 1 dy csc u cot u du, where u e y 1 and du e y dy
csc u C csc e y 1 C
Copyright 2018 Pearson Education, Inc.
2
518
35.
Chapter 7 Integrals and Transcendental Functions
sec
2
x e tan x dx eu du, where u tan x and du sec2 x dx
eu C e tan x C
36.
csc
2
x ecot x dx eu du , where u cot x and du csc 2 x dx
eu C ecot x C
37.
1
1
1 3x14 dx 13 7 u1 du,
13 ln u
38.
e ln x
x
1
1
7
where u 3 x 4, du 3 dx; x 1 u 7, x 1 u 1
13 ln | 1| ln | 7 | 13 0 ln 7 ln37
1
dx u1/2 du , where u ln x, du 1x dx; x 1 u 0, x e u 1
0
1
23 u 3/2 23 13/2 23 03/2
0
39.
0
tan
3x dx 0 cos dx 311/2 u1 du, where u cos 3x , du 13 sin 3x dx;
sin
x
3
x
3
3 ln | u |1
1/2
40.
2
3
1/4
3 ln
1/4
1
2
x 0 u 1, x u 12
ln |1 | 3ln 12 ln 23 ln 8
1/ 2 1
u
x dx 2
1/6 2 cot x dx 21/6 cos
1/2
sin x
du , where u sin x, du cos x dx;
x 16 u 12 , x 14 u 1
2
41.
2
x
ln | u |1/1/2 2
9 1
u
4
0 t 2 2t25 dt 25
2 ln
1
2
ln
1
2
2 ln1 1 ln 2 ln1 ln 2 2 1 ln 2 ln 2
2
2
du , where u t 2 25, du 2t dt ; t 0 u 25, t 4 u 9
9
9
ln | u |25 ln | 9 | ln | 25 | ln 9 ln 25 ln 25
42.
/6
1/2 1
du,
u
where u 1 sin t , du cos t dt ; t 2 u 2, t 6 u
/2 1cossint t dt 2
ln | u |2 ln
1/2
43.
tan(ln v )
dv
v
1
2
ln 2 ln1 ln 2 ln 2 2 ln 2 ln 4
sin u
tan u du cos u du , u ln v and du 1v dv
ln | cos u | C ln | cos(ln v ) | C
44.
v ln1 v dv u1 du,
where u ln v and du 1v dv
ln | u | C ln | ln v | C
Copyright 2018 Pearson Education, Inc.
1
2
Chapter 7 Practice Exercises
45.
(ln x )3
x
dx u 3 du, where u ln x and du 1x dx
46.
ln( x 5)
dx
x 5
C 12 (ln x)2 C
u du , where u ln( x 5) and du
47.
u 2
2
u2
2
ln( x 5)2
C
2
1 dx
x 5
C
2
2
1r csc 1 ln r dr csc u du,
where u 1 ln r and du 1r dr
cot u C cot 1 ln r C
48.
49.
cos1ln v
dv
v
x3
x2
2
sin u C sin 1 ln v C
dx 12 3u du, where u x 2 and du 2 x dx
1
2 ln 3
50.
cos u du , where u 1 ln v and du 1v dv
3u C 2 ln1 3 3x
2
C
sec 2 x dx 2u du , where u tan x and du sec
