INSTRUCTOR’S
SOLUTIONS MANUAL
MULTIVARIABLE
WILLIAM ARDIS
Collin County Community College
THOMAS’ CALCULUS
TWELFTH EDITION
BASED ON THE ORIGINAL WORK BY
George B. Thomas, Jr.
Massachusetts Institute of Technology
AS REVISED BY
Maurice D. Weir
Naval Postgraduate School
Joel Hass
University of California, Davis
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 Addison-Wesley from electronic files supplied by the author.
Copyright © 2010, 2005, 2001 Pearson Education, Inc.
Publishing as Pearson Addison-Wesley, 75 Arlington Street, Boston, MA 02116.
All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or
transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise,
without the prior written permission of the publisher. Printed in the United States of America.
ISBN-13: 978-0-321-60072-1
ISBN-10: 0-321-60072-X
1 2 3 4 5 6 BB 14 13 12 11 10
PREFACE TO THE INSTRUCTOR
This Instructor's Solutions Manual contains the solutions to every exercise in the 12th Edition of THOMAS' CALCULUS
by Maurice Weir and Joel Hass, including the Computer Algebra System (CAS) exercises. The corresponding Student's
Solutions Manual omits the solutions to the even-numbered exercises as well as the solutions to the CAS exercises (because
the CAS command templates would give them all away).
In addition to including the solutions to all of the new exercises in this edition of Thomas, we have carefully revised or
rewritten every solution which appeared in previous solutions manuals to ensure that each solution
ì conforms exactly to the methods, procedures and steps presented in the text
ì is mathematically correct
ì includes all of the steps necessary so a typical calculus student can follow the logical argument and algebra
ì includes a graph or figure whenever called for by the exercise, or if needed to help with the explanation
ì is formatted in an appropriate style to aid in its understanding
Every CAS exercise is solved in both the MAPLE and MATHEMATICA computer algebra systems. A template showing
an example of the CAS commands needed to execute the solution is provided for each exercise type. Similar exercises within
the text grouping require a change only in the input function or other numerical input parameters associated with the problem
(such as the interval endpoints or the number of iterations).
For more information about other resources available with Thomas' Calculus, visit http://pearsonhighered.com.
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
TABLE OF CONTENTS
10 Infinite Sequences and Series 569
10.1 Sequences 569
10.2 Infinite Series 577
10.3 The Integral Test 583
10.4 Comparison Tests 590
10.5 The Ratio and Root Tests 597
10.6 Alternating Series, Absolute and Conditional Convergence 602
10.7 Power Series 608
10.8 Taylor and Maclaurin Series 617
10.9 Convergence of Taylor Series 621
10.10 The Binomial Series and Applications of Taylor Series 627
Practice Exercises 634
Additional and Advanced Exercises 642
11 Parametric Equations and Polar Coordinates 647
11.1 Parametrizations of Plane Curves 647
11.2 Calculus with Parametric Curves 654
11.3 Polar Coordinates 662
11.4 Graphing in Polar Coordinates 667
11.5 Areas and Lengths in Polar Coordinates 674
11.6 Conic Sections 679
11.7 Conics in Polar Coordinates 689
Practice Exercises 699
Additional and Advanced Exercises 709
12 Vectors and the Geometry of Space 715
12.1
12.2
12.3
12.4
12.5
12.6
Three-Dimensional Coordinate Systems 715
Vectors 718
The Dot Product 723
The Cross Product 728
Lines and Planes in Space 734
Cylinders and Quadric Surfaces 741
Practice Exercises 746
Additional Exercises 754
13 Vector-Valued Functions and Motion in Space 759
13.1
13.2
13.3
13.4
13.5
13.6
Curves in Space and Their Tangents 759
Integrals of Vector Functions; Projectile Motion 764
Arc Length in Space 770
Curvature and Normal Vectors of a Curve 773
Tangential and Normal Components of Acceleration 778
Velocity and Acceleration in Polar Coordinates 784
Practice Exercises 785
Additional Exercises 791
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
14 Partial Derivatives 795
14.1
14.2
14.3
14.4
14.5
14.6
14.7
14.8
14.9
14.10
Functions of Several Variables 795
Limits and Continuity in Higher Dimensions 804
Partial Derivatives 810
The Chain Rule 816
Directional Derivatives and Gradient Vectors 824
Tangent Planes and Differentials 829
Extreme Values and Saddle Points 836
Lagrange Multipliers 849
Taylor's Formula for Two Variables 857
Partial Derivatives with Constrained Variables 859
Practice Exercises 862
Additional Exercises 876
15 Multiple Integrals 881
15.1
15.2
15.3
15.4
15.5
15.6
15.7
15.8
Double and Iterated Integrals over Rectangles 881
Double Integrals over General Regions 882
Area by Double Integration 896
Double Integrals in Polar Form 900
Triple Integrals in Rectangular Coordinates 904
Moments and Centers of Mass 909
Triple Integrals in Cylindrical and Spherical Coordinates 914
Substitutions in Multiple Integrals 922
Practice Exercises 927
Additional Exercises 933
16 Integration in Vector Fields 939
16.1
16.2
16.3
16.4
16.5
16.6
16.7
16.8
Line Integrals 939
Vector Fields and Line Integrals; Work, Circulation, and Flux 944
Path Independence, Potential Functions, and Conservative Fields 952
Green's Theorem in the Plane 957
Surfaces and Area 963
Surface Integrals 972
Stokes's Theorem 980
The Divergence Theorem and a Unified Theory 984
Practice Exercises 989
Additional Exercises 997
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
CHAPTER 10 INFINITE SEQUENCES AND SERIES
10.1 SEQUENCES
2
13
2
14
3
"
1. a" œ 11#1 œ 0, a# œ "
## œ 4 , a$ œ 3# œ 9 , a% œ 4# œ 16
1
2. a" œ 1!1 œ 1, a# œ #"! œ 2" , a$ œ 3!1 œ 61 , a% œ 4!1 œ 24
3.
#
$
%
&
)
(1)
(1)
"
"
"
a" œ (#1)1 œ 1, a# œ (4"
1 œ 3 , a$ œ 6 1 œ 5 , a% œ 8 1 œ 7
4. a" œ 2 (1)" œ 1, a# œ 2 (1)# œ 3, a$ œ 2 (1)$ œ 1, a% œ 2 (1)% œ 3
#
$
%
5. a" œ #2# œ #" , a# œ 22$ œ #" , a$ œ 2#% œ #" , a% œ 22& œ #"
#
$
%
15
6. a" œ 2 # " œ "# , a# œ 2 2# 1 œ 43 , a$ œ 2 2$ 1 œ 87 , a% œ 2 2% " œ 16
15
31
63
"
7. a" œ 1, a# œ 1 "# œ #3 , a$ œ #3 #"# œ 74 , a% œ 74 #"$ œ 15
8 , a& œ 8 #% œ 16 , a' œ 32 ,
255
511
1023
a( œ 127
64 , a) œ 128 , a* œ 256 , a"! œ 512
ˆ #" ‰
ˆ "6 ‰
"
3 œ 6 , a% œ 4
"
"
a* œ 362,880
, a"! œ 3,628,800
8. a" œ 1, a# œ "# , a$ œ
œ #"4 , a& œ
ˆ #"4 ‰
5
"
"
œ 1#"0 , a' œ 7#"0 , a( œ 5040
, a) œ 40,320
,
9. a" œ 2, a# œ (1)# (2) œ 1, a$ œ (1)2 (1) œ "# , a% œ
(1)% ˆ "# ‰
(1)& ˆ "4 ‰
œ 4" , a& œ
œ 8" ,
#
#
10. a" œ 2, a# œ 1†(#2) œ 1, a$ œ 2†(31) œ 32 , a% œ
3†ˆ 23 ‰
4†ˆ " ‰
œ "# , a& œ 5 # œ 52 , a' œ 3" ,
4
#
$
"
"
"
a' œ 16
, a( œ 3"# , a) œ 64
, a* œ 1#"8 , a"! œ 256
a( œ 27 , a) œ "4 , a* œ 29 , a"! œ "5
11. a" œ 1, a# œ 1, a$ œ 1 1 œ 2, a% œ 2 1 œ 3, a& œ 3 2 œ 5, a' œ 8, a( œ 13, a) œ 21, a* œ 34, a"! œ 55
12. a" œ 2, a# œ 1, a$ œ "# , a% œ
ˆ "# ‰
1
ˆ"‰
œ "# , a& œ ˆ# " ‰ œ 1, a' œ 2, a( œ 2, a) œ 1, a* œ "# , a"! œ "#
#
13. an œ (1)n1 , n œ 1, 2, á
14. an œ (1)n , n œ 1, 2, á
15. an œ (1)n1 n# , n œ 1, 2, á
16. an œ ("n)#
n1
n 1
, n œ 1, 2, á
17. an œ 3a2n 2b , n œ 1, 2, á
5
18. an œ n2n
an 1b , n œ 1, 2, á
19. an œ n# 1, n œ 1, 2, á
20. an œ n 4 , n œ 1, 2, á
21. an œ 4n 3, n œ 1, 2, á
22. an œ 4n 2 , n œ 1, 2, á
23. an œ 3nn! 2 , n œ 1, 2, á
24. an œ 5nn 1 , n œ 1, 2, á
3
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
570
Chapter 10 Infinite Sequences and Series
25. an œ 1 (#1)
n 1
, n œ 1, 2, á
26. an œ
27. n lim
2 (0.1)n œ 2 Ê converges
Ä_
n "# (1)n ˆ "# ‰
œ Ú n# Û, n œ 1, 2, á
#
(Theorem 5, #4)
n (")
28. n lim
œ n lim
1 (n1) œ 1 Ê converges
n
Ä_
Ä_
n
n
ˆ"‰ 2
" 2n
2
n
29. n lim
œ n lim
œ n lim
œ 1 Ê converges
Ä _ 1 #n
Ä _ ˆ"‰ 2
Ä_ #
n
2n "
30. n lim
œ n lim
Ä _ 1 3È n
Ä_
%
" 5n
31. n lim
œ n lim
Ä _ n% 8n$
Ä_
2Èn Š È"n ‹
Š È"n 3‹
Š n"% ‹ 5
1 ˆ 8n ‰
œ _ Ê diverges
œ 5 Ê converges
n3
n3
"
32. n lim
œ n lim
œ n lim
œ 0 Ê converges
Ä _ n# 5n 6
Ä _ (n 3)(n 2)
Ä _ n#
(n 1)(n 1)
n 2n 1
33. n lim
œ n lim
œ n lim
(n 1) œ _ Ê diverges
n1
n1
Ä_
Ä_
Ä_
#
Š "# ‹ n
$
"n
n
34 n lim
œ n lim
œ _ Ê diverges
Ä _ 70 4n#
Ä _ Š 70# ‹ 4
n
36. n lim
(1)n ˆ1 "n ‰ does not exist Ê diverges
Ä_
35. n lim
a1 (1)n b does not exist Ê diverges
Ä_
ˆ n #n " ‰ ˆ1 "n ‰ œ lim ˆ "# #"n ‰ ˆ1 n" ‰ œ #" Ê converges
37. n lim
Ä_
nÄ_
nb1
(")
œ 0 Ê converges
39. n lim
Ä _ #n 1
ˆ2 #"n ‰ ˆ3 #"n ‰ œ 6 Ê converges
38. n lim
Ä_
)
ˆ "# ‰ œ lim ("
œ 0 Ê converges
40. n lim
Ä_
n Ä _ #n
n
n
2n
É n 2n
41. n lim
œ Ên lim
Š 2 ‹ œ È2 Ê converges
1 œ É n lim
Ä_
Ä _ n1
Ä _ 1 "
n
"
ˆ "0 ‰ œ _ Ê diverges
42. n lim
œ n lim
Ä _ (0.9)n
Ä_ 9
n
ˆ 1 n" ‰‹ œ sin 1# œ 1 Ê converges
43. n lim
sin ˆ 1# n" ‰ œ sin Šn lim
Ä_
Ä_ #
44. n lim
n1 cos (n1) œ n lim
(n1)(1)n does not exist Ê diverges
Ä_
Ä_
sin n
45. n lim
œ 0 because n" Ÿ sinn n Ÿ n" Ê converges by the Sandwich Theorem for sequences
Ä_ n
#
#
sin n
46. n lim
œ 0 because 0 Ÿ sin#n n Ÿ #"n Ê converges by the Sandwich Theorem for sequences
Ä _ #n
n
"
^
47. n lim
œ n lim
œ 0 Ê converges (using l'Hopital's
rule)
Ä _ #n
Ä _ #n ln 2
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 10.1 Sequences
n
n
n
#
$
n
3 (ln 3)
3 (ln 3)
3
3 ln 3
^
48. n lim
œ n lim
œ n lim
œ n lim
œ _ Ê diverges (using l'Hopital's
rule)
6n
6
Ä _ n$
Ä _ 3n#
Ä_
Ä_
ln (n ")
49. n lim
œ n lim
Èn
Ä_
Ä_
ˆn " 1‰
" ‹
Š #È
n
2È n
œ n lim
œ n lim
Ä _ n1
Ä_
Š È2n ‹
1 Š n" ‹
œ 0 Ê converges
ˆ"‰
ln n
n
50. n lim
œ n lim
œ 1 Ê converges
Ä _ ln 2n
Ä_ ˆ2‰
2n
51. n lim
81În œ 1 Ê converges
Ä_
(Theorem 5, #3)
52. n lim
(0.03)1În œ 1 Ê converges
Ä_
(Theorem 5, #3)
ˆ1 7n ‰ œ e( Ê converges
53. n lim
Ä_
(Theorem 5, #5)
n
n
)
"
ˆ1 "n ‰ œ lim ’1 ("
Ê converges
54. n lim
n “ œe
Ä_
nÄ_
n
(Theorem 5, #5)
n
È
55. n lim
10n œ n lim
101În † n1În œ 1 † 1 œ 1 Ê converges
Ä_
Ä_
#
n
n
È
ˆÈ
56. n lim
n# œ n lim
n‰ œ 1# œ 1 Ê converges
Ä_
Ä_
ˆ3‰
57. n lim
Ä_ n
1 În
Ä_ 1În œ " œ 1 Ê converges
œ nlim
1
n
lim 31În
n
(Theorem 5, #3 and #2)
(Theorem 5, #2)
(Theorem 5, #3 and #2)
Ä_
58. n lim
(n 4)1ÎÐn4Ñ œ x lim
x1Îx œ 1 Ê converges; (let x œ n 4, then use Theorem 5, #2)
Ä_
Ä_
ln n
Ä_ 1În œ _ œ _ Ê diverges
59. n lim
œ nlim
1
n
Ä _ n1În
lim ln n
n
Ä_
(Theorem 5, #2)
n
60. n lim
cln n ln (n 1)d œ n lim
ln ˆ n n 1 ‰ œ ln Šn lim
‹ œ ln 1 œ 0 Ê converges
Ä_
Ä_
Ä _ n1
n
n
È
61. n lim
4n n œ n lim
4È
n œ 4 † 1 œ 4 Ê converges
Ä_
Ä_
(Theorem 5, #2)
n
È
62. n lim
32n1 œ n lim
32 a1Înb œ n lim
3# † 31În œ 9 † 1 œ 9 Ê converges
Ä_
Ä_
Ä_
"†2†3â(n 1)(n)
n!
ˆ " ‰ œ 0 and nn!n
63. n lim
œ n lim
Ÿ n lim
n†n†nân†n
Ä _ nn
Ä_
Ä_ n
(4)
64. n lim
œ 0 Ê converges
Ä _ n!
n
"
œ_
'n
Š (10n! ) ‹
n!
66. n lim
œ n lim
Ä _ 2n 3n
Ä_
"
œ_
ˆ 6n!n ‰
1ÎÐln nÑ
n!
0 Ê n lim
œ 0 Ê converges
Ä _ nn
(Theorem 5, #6)
n!
65. n lim
œ n lim
Ä _ 106n
Ä_
ˆ"‰
67. n lim
Ä_ n
(Theorem 5, #3)
Ê diverges
Ê diverges
(Theorem 5, #6)
(Theorem 5, #6)
œ n lim
exp ˆ ln"n ln ˆ n" ‰‰ œ n lim
exp ˆ ln 1lnnln n ‰ œ e" Ê converges
Ä_
Ä_
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
571
572
Chapter 10 Infinite Sequences and Series
n
ˆ1 n" ‰n ‹ œ ln e œ 1 Ê converges
68. n lim
ln ˆ1 "n ‰ œ ln Šn lim
Ä_
Ä_
(Theorem 5, #5)
" ‰‰
ˆ 3n " ‰ œ lim exp ˆn ln ˆ 3n
69. n lim
œ n lim
exp Š ln (3n 1) " ln (3n 1) ‹
3n 1
Ä _ 3n 1
nÄ_
Ä_
n
n
3
3
6n
#Î$
ˆ6‰
œ n lim
exp 3n 1 "3n 1 œ n lim
exp Š (3n 1)(3n
Ê converges
1) ‹ œ exp 9 œ e
Ä_
Ä_
Š ‹
#
n#
"
ˆ n ‰ œ lim exp ˆn ln ˆ n n 1 ‰‰ œ lim exp Š ln n ln" (n 1) ‹ œ lim exp n
70. n lim
ˆ ‰
Ä _ n1
nÄ_
nÄ_
nÄ_
n
n"1
Š n"# ‹ n
#
œ n lim
exp Š n(nn 1) ‹ œ e" Ê converges
Ä_
n
ˆ x ‰
71. n lim
Ä _ 2n 1
1 În
œ n lim
x ˆ #n " 1 ‰
Ä_
1În
1)
œ x n lim
exp ˆ "n ln ˆ #n " 1 ‰‰ œ x n lim
exp Š ln (2n
‹
n
Ä_
Ä_
œ x n lim
exp ˆ 2n21 ‰ œ xe! œ x, x 0 Ê converges
Ä_
n
ˆ1 n"# ‰ œ lim exp ˆn ln ˆ1 n"# ‰‰ œ lim exp 72. n lim
Ä_
nÄ_
nÄ_
ln Š1 n"# ‹
ˆ n" ‰
Š 2$ ‹‚Š1 n"# ‹
exp – n
œ n lim
Ä_
Š n"# ‹
—
œ n lim
exp ˆ n# 2n1 ‰ œ e! œ 1 Ê converges
Ä_
3 †6
36
73. n lim
œ n lim
œ 0 Ê converges
Ä _ 2cn †n!
Ä _ n!
n
n
n
ˆ 10 ‰n
(Theorem 5, #6)
ˆ 12 ‰n ˆ 10 ‰n
ˆ 120 ‰n
11
11
11
121
74. n lim
lim
lim
œ 0 Ê converges
n
n œ
n
n
n
n œ
n
Ä _ ˆ 9 ‰ ˆ 11 ‰
n Ä _ ˆ 12 ‰ ˆ 9 ‰ ˆ 12 ‰ ˆ 11 ‰
n Ä _ ˆ 108 ‰ 1
10
12
11
10
11
12
110
(Theorem 5, #4)
e e
e "
2e
75. n lim
tanh n œ n lim
œ n lim
œ n lim
œ n lim
" œ 1 Ê converges
Ä_
Ä _ en e n
Ä _ e2n 1
Ä _ 2e2n
Ä_
n
n
e
76. n lim
sinh (ln n) œ n lim
Ä_
Ä_
77. n lim
Ä_
2n
ln n
2n
n ˆ n" ‰
e ln n
œ n lim
œ_
2
#
Ä_
Ê diverges
ˆcos ˆ "n ‰‰ Š n"# ‹
cos ˆ n" ‰
sin ˆ "n ‰
n# sin ˆ n" ‰
œ
lim
œ
lim
œ "#
2n 1 œ n lim
Ä _ Š2 " ‹
nÄ_
n Ä _ # ˆ 2n ‰
Š 2# 2$ ‹
#
n
n
78. n lim
n ˆ1 cos "n ‰ œ n lim
Ä_
Ä_
Èn sinŠ È1 ‹ œ lim
79. n lim
n
Ä_
nÄ_
n
Ê converges
n
sin ˆ "n ‰‘ Š "# ‹
ˆ" cos "n ‰
n
œ
œ n lim
lim
sin ˆ "n ‰ œ 0 Ê converges
ˆ "n ‰
nÄ_
Ä_
Š"‹
n#
sinŠ È1n ‹
Èn
1
œ n lim
Ä_
cos Š È1n ‹Š
1
2n3Î2
1
‹
2n3Î2
œ n lim
cos Š È1n ‹ œ cos 0 œ 1 Ê converges
Ä_
80. n lim
a3n 5n b1În œ n lim
exp’lna3n 5n b1În “ œ n lim
exp’ lna3 n 5 b “ œ n lim
exp–
Ä_
Ä_
Ä_
Ä_
n
n
œ n lim
exp’
Ä_
Š 35n ‹ln 3 ln 5
ˆ 35nn ‰ 1
n
3n ln 3 b 5n ln 5
3n b 5n
1
—
ˆ 3 ‰n ln 3 ln 5
exp’ 5 ˆ 3 ‰n 1 “ œ expaln 5b œ 5
“ œ n lim
Ä_
81. n lim
tan" n œ 1# Ê converges
Ä_
5
"
82. n lim
tan" n œ 0 † 1# œ 0 Ê converges
Ä _ Èn
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 10.1 Sequences
573
n
ˆ " ‰n È" n œ lim Šˆ 3" ‰n Š È" ‹ ‹ œ 0 Ê converges
83. n lim
Ä_ 3
nÄ_
2
(Theorem 5, #4)
2
#
n
1‰
!
È
84. n lim
n# n œ n lim
exp ’ ln ann nb “ œ n lim
exp ˆ 2n
n# n œ e œ 1 Ê converges
Ä_
Ä_
Ä_
#!!
"**
(ln n)
85. n lim
n
Ä_
200 (ln n)
œ n lim
n
Ä_
&
Š 5(lnnn) ‹
"*)
200†199 (ln n)
œ n lim
n
Ä_
200!
œ á œ n lim
œ 0 Ê converges
Ä_ n
%
(ln n)
86. n lim
œ n lim
Ä _ Èn
Ä_ –
"
Š #Èn
%
$
10(ln n)
80(ln n)
3840
œ n lim
œ n lim
œ á œ n lim
œ 0 Ê converges
Èn
Èn
Ä_
Ä_
Ä _ Èn
‹ —
È #
87. n lim
Šn Èn# n‹ œ n lim
Šn Èn# n‹ Š n Èn# n ‹ œ n lim
Ä_
Ä_
Ä_
n
n n
n
"
œ n lim
Ä _ 1 É1 "
n È n# n
n
œ "# Ê converges
È #
È n# 1 È n# n
È #
"
"
œ n lim
Š
‹ Š Èn# 1 Èn# n ‹ œ n lim
1 n
Ä _ È n# 1 È n# n
Ä_
n 1 È n# n
n 1 n n
88. n lim
Ä_ È #
œ n lim
Ä_
É1 n"# É1 "n
ˆ n" 1‰
œ 2 Ê converges
" ' "
ln n
"
89. n lim
dx œ n lim
œ n lim
œ 0 Ê converges
Ä_ n 1 x
Ä_ n
Ä_ n
n
(Theorem 5, #1)
" ˆ "
' " dx œ n lim
90. n lim
1‰ œ p" 1 if p 1 Ê converges
’ " " “ œ n lim
Ä _ 1 xp
Ä _ 1p xpc1 1
Ä _ 1p npc1
n
n
72
2
91. Since an converges Ê n lim
a œ L Ê n lim
a
œ n lim
Ê L œ 1 72
L Ê La1 Lb œ 72 Ê L L 72 œ 0
Ä_ n
Ä _ n1
Ä _ 1 an
Ê L œ 9 or L œ 8; since an 0 for n
1ÊLœ8
an 6
6
2
92. Since an converges Ê n lim
a œ L Ê n lim
a
œ n lim
Ê L œ LL 2 Ê LaL 2b œ L 6 Ê L L 6 œ 0
Ä_ n
Ä _ n1
Ä _ an 2
Ê L œ 3 or L œ 2; since an 0 for n
2ÊLœ2
È8 2an Ê L œ È8 2L Ê L2 2L 8 œ 0 Ê L œ 2
93. Since an converges Ê n lim
a œ L Ê n lim
a
œ n lim
Ä_ n
Ä _ n1
Ä_
or L œ 4; since an 0 for n
3ÊLœ4
È8 2an Ê L œ È8 2L Ê L2 2L 8 œ 0 Ê L œ 2
94. Since an converges Ê n lim
a œ L Ê n lim
a
œ n lim
Ä_ n
Ä _ n1
Ä_
or L œ 4; since an 0 for n
2ÊLœ4
È5an Ê L œ È5L Ê L2 5L œ 0 Ê L œ 0 or L œ 5; since
95. Since an converges Ê n lim
a œ L Ê n lim
a
œ n lim
Ä_ n
Ä _ n1
Ä_
an 0 for n
1ÊLœ5
ˆ12 Èan ‰ Ê L œ Š12 ÈL‹ Ê L2 25L 144 œ 0
96. Since an converges Ê n lim
a œ L Ê n lim
a
œ n lim
Ä_ n
Ä _ n1
Ä_
Ê L œ 9 or L œ 16; since 12 Èan 12 for n
97. an 1 œ 2 a1n , n
1ÊLœ9
1, a1 œ 2. Since an converges Ê n lim
a œ L Ê n lim
a
œ n lim
Š2 a1n ‹ Ê L œ 2 L1
Ä_ n
Ä _ n1
Ä_
Ê L2 2L 1 œ 0 Ê L œ 1 „ È2; since an 0 for n
1 Ê L œ 1 È2
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
574
Chapter 10 Infinite Sequences and Series
È1 an Ê L œ È1 L
1, a1 œ È1. Since an converges Ê n lim
a œ L Ê n lim
a
œ n lim
Ä_ n
Ä _ n1
Ä_
98. an 1 œ È1 an , n
È
È
Ê L2 L 1 œ 0 Ê L œ 1 „2 5 ; since an 0 for n
1 Ê L œ 1 2 5
99. 1, 1, 2, 4, 8, 16, 32, á œ 1, 2! , 2" , 2# , 2$ , 2% , 2& , á Ê x" œ 1 and xn œ 2nc2 for n
2
100. (a) 1# 2(1)# œ 1, 3# 2(2)# œ 1; let f(aß b) œ (a 2b)# 2(a b)# œ a# 4ab 4b# 2a# 4ab 2b#
œ 2b# a# ; a# 2b# œ 1 Ê f(aß b) œ 2b# a# œ 1; a# 2b# œ 1 Ê f(aß b) œ 2b# a# œ 1
#
#
#
#
"
2a 4ab 2b
‰ 2 œ a 4ab 4b(a (b) r#n 2 œ ˆ aa2b
œ a(aa b)2b# b œ „"
b)#
b
y# Ê rn œ Ê 2 „ Š yn ‹
#
#
#
#
n
n. Let ba represent the (n 1)th fraction where ba
In the first and second fractions, yn
for n a positive integer
lim rn œ È2.
3.
Now the nth fraction is aa2b
b and a b
2b
2n 2
1 and b
n Ê yn
n1
n. Thus,
nÄ_
101. (a) f(x) œ x# 2; the sequence converges to 1.414213562 ¸ È2
(b) f(x) œ tan (x) 1; the sequence converges to 0.7853981635 ¸ 14
(c) f(x) œ ex ; the sequence 1, 0, 1, 2, 3, 4, 5, á diverges
102. (a) n lim
nf ˆ n" ‰ œ
Ä_
f(?x)
?x œ
lim
?x Ä !b
lim
?x Ä !b
f(0?x) f(0)
œ f w (0), where ?x œ n"
?x
(b) n lim
n tan" ˆ "n ‰ œ f w (0) œ 1 " 0# œ 1, f(x) œ tan" x
Ä_
(c) n lim
n ae1În 1b œ f w (0) œ e! œ 1, f(x) œ ex 1
Ä_
(d) n lim
n ln ˆ1 2n ‰ œ f w (0) œ 1 22(0) œ 2, f(x) œ ln (1 2x)
Ä_
#
#
#
103. (a) If a œ 2n 1, then b œ Ú a# Û œ Ú 4n #4n 1 Û œ Ú2n# 2n "# Û œ 2n# 2n, c œ Ü a# Ý œ Ü2n# 2n "# Ý
#
œ 2n# 2n 1 and a# b# œ (2n 1)# a2n# 2nb œ 4n# 4n 1 4n% 8n$ 4n#
#
œ 4n% 8n$ 8n# 4n 1 œ a2n# 2n 1b œ c# .
(b) a lim
Ä_
#
Ú a# Û
#
Ü a# Ý
#
2n 2n
œ a lim
œ 1 or a lim
Ä _ 2n# 2n 1
Ä_
#
Ú a# Û
#
Ü a# Ý
œ a lim
sin ) œ
Ä_
lim
) Ä 1 Î2
sin ) œ 1
Š 21 ‹
2n1 ‰
104. (a) n lim
(2n1)1Î a2nb œ n lim
exp ˆ ln2n
œ n lim
exp 2n#1 œ n lim
exp ˆ #"n ‰ œ e! œ 1;
Ä_
Ä_
Ä_
Ä_
n
n
n! ¸ ˆ ne ‰ È
2n1 , Stirlings approximation Ê È
n! ¸ ˆ ne ‰ (2n1)1Î a2nb ¸ ne for large values of n
(b)
n
40
50
60
n
È
n!
15.76852702
19.48325423
23.19189561
n
e
14.71517765
18.39397206
22.07276647
ˆ"‰
ln n
"
n
105. (a) n lim
œ n lim
œ n lim
œ0
Ä _ nc
Ä _ cncc1
Ä _ cnc
(b) For all % 0, there exists an N œ eÐln %ÑÎc such that n eÐln %ÑÎc Ê ln n lnc % Ê ln nc ln ˆ "% ‰
Ê nc "% Ê n"c % Ê ¸ n"c 0¸ % Ê lim n"c œ 0
nÄ_
106. Let {an } and {bn } be sequences both converging to L. Define {cn } by c2n œ bn and c2nc1 œ an , where
n œ 1, 2, 3, á . For all % 0 there exists N" such that when n N" then kan Lk % and there exists N#
such that when n N# then kbn Lk %. If n 1 2max{N" ß N# }, then kcn Lk %, so {cn } converges to L.
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 10.1 Sequences
575
107. n lim
n1În œ n lim
exp ˆ "n ln n‰ œ n lim
exp ˆ n" ‰ œ e! œ 1
Ä_
Ä_
Ä_
108. n lim
x1În œ n lim
exp ˆ "n ln x‰ œ e! œ 1, because x remains fixed while n gets large
Ä_
Ä_
109. Assume the hypotheses of the theorem and let % be a positive number. For all % there exists a N" such that
when n N" then kan Lk % Ê % an L % Ê L % an , and there exists a N# such that when
n N# then kcn Lk % Ê % cn L % Ê cn L %. If n max{N" ß N# }, then
L % an Ÿ bn Ÿ cn L % Ê kbn Lk % Ê n lim
b œ L.
Ä_ n
110. Let % !. We have f continuous at L Ê there exists $ so that kx Lk $ Ê kf(x) f(L)k %. Also, an Ä L Ê there
exists N so that for n N kan Lk $ . Thus for n N, kf(an ) f(L)k % Ê f(an ) Ä f(L).
1) 1
3n 1
3n 4
3n 1
#
#
an Ê 3(n
(n 1) 1 n 1 Ê n # n 1 Ê 3n 3n 4n 4 3n 6n n 2
111. an1
"
Ê 4 2; the steps are reversible so the sequence is nondecreasing; 3n
n 1 3 Ê 3n 1 3n 3
Ê 1 3; the steps are reversible so the sequence is bounded above by 3
1) 3)!
(2n 3)!
(2n 5)!
(2n 3)!
(2n 5)!
(n 2)!
an Ê (2(n
((n 1) 1)! (n 1)! Ê (n 2)! (n 1)! Ê (2n 3)! (n 1)!
112. an1
Ê (2n 5)(2n 4) n 2; the steps are reversible so the sequence is nondecreasing; the sequence is not
3)!
bounded since (2n
(n 1)! œ (2n 3)(2n 2)â(n 2) can become as large as we please
nb1 nb1
n n
nb1 nb1
113. an1 Ÿ an Ê 2(n 31)! Ÿ 2n!3 Ê 2 2n 33n
Ÿ (n n!1)! Ê 2 † 3 Ÿ n 1 which is true for n
5; the steps are
reversible so the sequence is decreasing after a& , but it is not nondecreasing for all its terms; a" œ 6, a# œ 18,
a$ œ 36, a% œ 54, a& œ 324
5 œ 64.8 Ê the sequence is bounded from above by 64.8
an Ê 2 n2 1 #n"b1
114. an1
2 n2 #"n Ê 2n n2 1
"
"
#nb1 #n
Ê n(n 2 1)
#n"b1 ; the steps are
reversible so the sequence is nondecreasing; 2 2n #"n Ÿ 2 Ê the sequence is bounded from above
115. an œ 1 "n converges because n" Ä 0 by Example 1; also it is a nondecreasing sequence bounded above by 1
116. an œ n "n diverges because n Ä _ and n" Ä 0 by Example 1, so the sequence is unbounded
117. an œ 2 2n 1 œ 1 #"n and 0 #"n "n ; since "n Ä 0 (by Example 1) Ê #"n Ä 0, the sequence converges; also it is
n
a nondecreasing sequence bounded above by 1
n
118. an œ 2 3n 1 œ ˆ 23 ‰ 3"n ; the sequence converges to ! by Theorem 5, #4
n
119. an œ a(1)n 1b ˆ nn 1 ‰ diverges because an œ 0 for n odd, while for n even an œ 2 ˆ1 n" ‰ converges to 2; it
diverges by definition of divergence
120. xn œ max {cos 1ß cos 2ß cos 3ß á ß cos n} and xn1 œ max {cos 1ß cos 2ß cos 3ß á ß cos (n 1)}
so the sequence is nondecreasing and bounded above by 1 Ê the sequence converges.
121. an
È
an1 Í 1 Èn 2n
È
and 1 Èn 2n
" È2(n 1)
Èn 1
Í Èn 1 È2n# 2n
xn with xn Ÿ 1
Èn È2n# 2n Í Èn 1
È2 ; thus the sequence is nonincreasing and bounded below by È2 Ê it converges
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Èn
576
Chapter 10 Infinite Sequences and Series
122. an
an1 Í n n 1
(n 1) "
n1
Í n# 2n 1
n# 2n Í 1
0 and n n 1
1; thus the sequence is
nonincreasing and bounded below by 1 Ê it converges
123.
n
n
n"
n
4nb1 3n
ˆ 34 ‰n1 Í 1 43 and
œ 4 ˆ 34 ‰ so an an1 Í 4 ˆ 34 ‰
4 ˆ 34 ‰
Í ˆ 34 ‰
4n
n
4 ˆ 34 ‰
4; thus the sequence is nonincreasing and bounded below by 4 Ê it converges
124. a" œ 1, a# œ 2 3, a$ œ 2(2 3) 3 œ 2# a22 "b † 3, a% œ 2 a2# a22 "b † 3b 3 œ 2$ a2$ 1b 3,
a& œ 2 c2$ a2$ 1b 3d 3 œ 2% a2% 1b 3, á , an œ 2n" a2n" 1b 3 œ 2n" 3 † 2n1 3
œ 2n1 (1 3) 3 œ 2n 3; an an1 Í 2n 3 2n1 3 Í 2n 2n1 Í 1 Ÿ 2
so the sequence is nonincreasing but not bounded below and therefore diverges
125. Let 0 M 1 and let N be an integer greater than 1 MM . Then n N Ê n 1 MM Ê n nM M
Ê n M nM Ê n M(n 1) Ê n n 1 M.
126. Since M" is a least upper bound and M# is an upper bound, M" Ÿ M# . Since M# is a least upper bound and M"
is an upper bound, M# Ÿ M" . We conclude that M" œ M# so the least upper bound is unique.
127. The sequence an œ 1 ("#) is the sequence #" , #3 , #" , #3 , á . This sequence is bounded above by #3 ,
n
but it clearly does not converge, by definition of convergence.
128. Let L be the limit of the convergent sequence {an }. Then by definition of convergence, for #% there
corresponds an N such that for all m and n, m N Ê kam Lk #% and n N Ê kan Lk #% . Now
kam an k œ kam L L an k Ÿ kam Lk kL an k #% #% œ % whenever m N and n N.
129. Given an % 0, by definition of convergence there corresponds an N such that for all n N,
kL" an k % and kL# an k %. Now kL# L" k œ kL# an an L" k Ÿ kL# an k kan L" k % % œ 2%.
kL# L" k 2% says that the difference between two fixed values is smaller than any positive number 2%.
The only nonnegative number smaller than every positive number is 0, so kL" L# k œ 0 or L" œ L# .
130. Let k(n) and i(n) be two order-preserving functions whose domains are the set of positive integers and whose
ranges are a subset of the positive integers. Consider the two subsequences akÐnÑ and aiÐnÑ , where akÐnÑ Ä L" ,
aiÐnÑ Ä L# and L" Á L# . Thus ¸akÐnÑ aiÐnÑ ¸ Ä kL" L# k 0. So there does not exist N such that for all m, n N
Ê kam an k %. So by Exercise 128, the sequence Öan × is not convergent and hence diverges.
131. a2k Ä L Í given an % 0 there corresponds an N" such that c2k N" Ê ka2k Lk %d . Similarly,
a2k1 Ä L Í c2k 1 N# Ê ka2k1 Lk %d . Let N œ max{N" ß N# }. Then n N Ê kan Lk % whether
n is even or odd, and hence an Ä L.
132. Assume an Ä 0. This implies that given an % 0 there corresponds an N such that n N Ê kan 0k %
Ê kan k % Ê kkan kk % Ê kkan k 0k % Ê kan k Ä 0. On the other hand, assume kan k Ä 0. This implies that
given an % 0 there corresponds an N such that for n N, kkan k 0k % Ê kkan kk % Ê kan k %
Ê kan 0k % Ê an Ä 0.
#
#
#
#
133. (a) f(x) œ x# a Ê f w (x) œ 2x Ê xn1 œ xn xn#xn a Ê xn1 œ 2xn 2xaxnn ab œ xn2xn a œ
ˆxn xa ‰
n
#
(b) x" œ 2, x# œ 1.75, x$ œ 1.732142857, x% œ 1.73205081, x& œ 1.732050808; we are finding the positive
number where x# 3 œ 0; that is, where x# œ 3, x 0, or where x œ È3 .
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 10.2 Infinite Series
577
134. x" œ 1, x# œ 1 cos (1) œ 1.540302306, x$ œ 1.540302306 cos (1 cos (1)) œ 1.570791601,
x% œ 1.570791601 cos (1.570791601) œ 1.570796327 œ 1# to 9 decimal places. After a few steps, the
arc axnc1 b and line segment cos axnc1 b are nearly the same as the quarter circle.
135-146. Example CAS Commands:
Mathematica: (sequence functions may vary):
Clear[a, n]
a[n_]; = n1 / n
first25= Table[N[a[n]],{n, 1, 25}]
Limit[a[n], n Ä 8]
Mathematica: (sequence functions may vary):
Clear[a, n]
a[n_]; = n1 / n
first25= Table[N[a[n]],{n, 1, 25}]
Limit[a[n], n Ä 8]
The last command (Limit) will not always work in Mathematica. You could also explore the limit by enlarging your table
to more than the first 25 values.
If you know the limit (1 in the above example), to determine how far to go to have all further terms within 0.01 of the
limit, do the following.
Clear[minN, lim]
lim= 1
Do[{diff=Abs[a[n] lim], If[diff < .01, {minN= n, Abort[]}]}, {n, 2, 1000}]
minN
For sequences that are given recursively, the following code is suggested. The portion of the command a[n_]:=a[n] stores
the elements of the sequence and helps to streamline computation.
Clear[a, n]
a[1]= 1;
a[n_]; = a[n]= a[n 1] (1/5)(n1)
first25= Table[N[a[n]], {n, 1, 25}]
The limit command does not work in this case, but the limit can be observed as 1.25.
Clear[minN, lim]
lim= 1.25
Do[{diff=Abs[a[n] lim], If[diff < .01, {minN= n, Abort[]}]}, {n, 2, 1000}]
minN
10.2 INFINITE SERIES
n
a1 r b
1. sn œ a (1
r) œ
2 ˆ1 ˆ "3 ‰ ‰
1 ˆ "3 ‰
a1 r b
2. sn œ a (1
r) œ
" ‰n ‰
9 ‰ˆ
ˆ 100
1 ˆ 100
"
1 ˆ 100 ‰
n
n
1 ˆ " ‰
Ê n lim
s œ 1 2ˆ " ‰ œ 3
Ä_ n
3
ˆ 9 ‰
"
Ê n lim
s œ 1 100
ˆ " ‰ œ 11
Ä_ n
100
n
a1 r b
#
3. sn œ a (1
s œ ˆ "3 ‰ œ 32
r) œ 1 ˆ "# ‰ Ê n lim
Ä_ n
#
n
4. sn œ 11 ((2)2) , a geometric series where krk 1 Ê divergence
n
5.
"
"
"
(n 1)(n #) œ n 1 n #
Ê sn œ ˆ #" 3" ‰ ˆ 3" 4" ‰ á ˆ n " 1 n " # ‰ œ #" n " # Ê n lim
s œ #"
Ä_ n
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
578
6.
Chapter 10 Infinite Sequences and Series
5
5
5
n(n 1) œ n n 1
Ê sn œ ˆ5 25 ‰ ˆ 25 35 ‰ ˆ 35 45 ‰ á ˆ n 5 1 n5 ‰ ˆ n5 n 5 1 ‰ œ 5 n 5 1
Ê n lim
s œ5
Ä_ n
"
"
7. 1 "4 16
64
á , the sum of this geometric series is 1 ˆ" " ‰ œ 1 "ˆ " ‰ œ 45
4
4
8.
" ‰
ˆ 16
"
"
"
"
16 64 256 á , the sum of this geometric series is 1 ˆ "4 ‰ œ 1#
9.
ˆ 74 ‰
7
7
7
7
4 16 64 á , the sum of this geometric series is 1 ˆ "4 ‰ œ 3
5
5
10. 5 54 16
64
á , the sum of this geometric series is 1 ˆ5 " ‰ œ 4
4
11. (5 1) ˆ 5# "3 ‰ ˆ 54 9" ‰ ˆ 58 #"7 ‰ á , is the sum of two geometric series; the sum is
5
1 "ˆ " ‰ œ 10 #3 œ 23
#
1 ˆ "# ‰
3
12. (5 1) ˆ 5# "3 ‰ ˆ 54 9" ‰ ˆ 58 #"7 ‰ á , is the difference of two geometric series; the sum is
5
1 "ˆ " ‰ œ 10 #3 œ 17
#
1 ˆ "# ‰
3
" ‰
13. (1 1) ˆ 1# "5 ‰ ˆ 14 25
ˆ 18 1#"5 ‰ á , is the sum of two geometric series; the sum is
1
1 "ˆ " ‰ œ 2 56 œ 17
6
1 ˆ "# ‰
5
8
16
4
8
14. 2 45 25
125
á œ 2 ˆ1 25 25
125
á ‰ ; the sum of this geometric series is 2 Š 1 "ˆ 2 ‰ ‹ œ 10
3
5
15. Series is geometric with r œ 25 Ê ¹ 25 ¹ 1 Ê Converges to 1 1 2 œ 53
5
16. Series is geometric with r œ 3 Ê ¹3¹ 1 Ê Diverges
1
17. Series is geometric with r œ 18 Ê ¹ 18 ¹ 1 Ê Converges to 1 8 1 œ 17
8
2
18. Series is geometric with r œ 23 Ê ¹ 23 ¹ 1 Ê Converges to 1 ˆ3 2 ‰ œ 25
3
_
n
_
Š 23 ‹
23 ˆ " ‰
19. 0.23 œ ! 100
œ 1 100
œ 23
10#
99
ˆ " ‰
n œ0
n œ0
100
_
n
7 ˆ " ‰
21. 0.7 œ ! 10
œ
10
nœ0
_
7
Š 10
‹
"
1 Š 10
‹
n
1 ‰ˆ 6 ‰ˆ " ‰
23. 0.06 œ ! ˆ 10
œ
10
10
n œ0
_
n
nœ0
6
Š 100
‹
"
1 Š 10
‹
414 ˆ " ‰
24. 1.414 œ 1 ! 1000
œ1
10$
n œ0
_
n
d ˆ " ‰
22. 0.d œ ! 10
œ
10
œ 79
234
Š 1000
‹
n
234 ˆ " ‰
20. 0.234 œ ! 1000
œ
10$
"
1 Š 1000
‹
d
Š 10
‹
"
1 Š 10
‹
œ d9
6
"
œ 90
œ 15
414
Š 1000
‹
"
1 Š 1000
‹
"413
œ 1 414
999 œ 999
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
œ 234
999
Section 10.2 Infinite Series
_
Š 123& ‹
n
! 123& ˆ " $ ‰ œ 124 25. 1.24123 œ 124
100 10
10
100
10
1 Š "$ ‹
n œ0
_
Š 142,857
' ‹
n
10
1 Š "' ‹
n œ0
28.
lim n œ lim 11 œ 1 Á 0 Ê diverges
nÄ_ n 10
nÄ_
lim
n an 1 b
n
2n 1
2
œ lim n2 n 5n
6 œ lim 2n 5 œ lim 2 œ 1 Á 0 Ê diverges
2
nÄ_ an 2ban 3b
nÄ_
nÄ_
lim 1 œ 0 Ê test inconclusive
nÄ_ n 4
30.
1
lim 2 n œ lim 2n
œ 0 Ê test inconclusive
nÄ_ n 3
nÄ_
32.
33.
34.
3,142,854
116,402
œ 3 142,857
10' 1 œ 999,999 œ 37,037
10
29.
31.
123,999
41,333
124
124
123
œ 100
10&123
10# œ 100 99,900 œ 99,900 œ 33,300
10
ˆ 10" ' ‰ œ 3 26. 3.142857 œ 3 ! 142,857
10'
27.
579
nÄ_
lim cos 1n œ cos 0 œ 1 Á 0 Ê diverges
nÄ_
n
lim ne œ
nÄ_ e n
n
lim n e
nÄ_ e 1
n
œ lim een œ lim 11 œ 1 Á 0 Ê diverges
nÄ_
nÄ_
lim ln 1n œ _ Á 0 Ê diverges
nÄ_
lim cos n 1 œ does not exist Ê diverges
nÄ_
35. sk œ ˆ1 "2 ‰ ˆ "2 "3 ‰ ˆ "3 "4 ‰ á ˆ k " 1 k" ‰ ˆ k" k " 1 ‰ œ 1 k " 1 Ê
œ lim ˆ1 k " 1 ‰ œ 1, series converges to 1
lim sk
kÄ_
kÄ_
3 ‰
36. sk œ ˆ 31 34 ‰ ˆ 34 39 ‰ ˆ 39 16
á Š ak 3 1b2 k32 ‹ Š k32 ak 3 1b2 ‹ œ 3 ak 3 1b2 Ê
lim sk
kÄ_
œ lim Š3 ak 3 1b2 ‹ œ 3, series converges to 3
kÄ_
37. sk œ ŠlnÈ2 lnÈ1‹ ŠlnÈ3 lnÈ2‹ ŠlnÈ4 lnÈ3‹ á ŠlnÈk lnÈk 1‹ ŠlnÈk 1 lnÈk‹
œ lnÈk 1 lnÈ1 œ lnÈk 1 Ê
lim sk œ lim lnÈk 1 œ _; series diverges
kÄ_
kÄ_
38. sk œ atan 1 tan 0b atan 2 tan 1b atan 3 tan 2b á atan k tan ak 1bb atan ak 1b tan kb
œ tan ak 1b tan 0 œ tan ak 1b Ê lim sk œ lim tan ak 1b œ does not exist; series diverges
kÄ_
kÄ_
39. sk œ ˆcos1 ˆ 12 ‰ cos1 ˆ 13 ‰‰ ˆcos1 ˆ 13 ‰ cos1 ˆ 14 ‰‰ ˆcos1 ˆ 14 ‰ cos1 ˆ 15 ‰‰ á
ˆcos1 ˆ 1k ‰ cos1 ˆ k 1 1 ‰‰ ˆcos1 ˆ k 1 1 ‰ cos1 ˆ k 1 # ‰‰ œ 13 cos1 ˆ k 1 # ‰
Ê
lim sk œ lim ’ 13 cos1 ˆ k 1 # ‰“ œ 13 12 œ 16 , series converges to 16
kÄ_
kÄ_
40. sk œ ŠÈ5 È4‹ ŠÈ6 È5‹ ŠÈ7 È6‹ á ŠÈk 3 Èk 2‹ ŠÈk 4 Èk 3‹
œ Èk 4 2 Ê
lim sk œ lim ’Èk 4 2“ œ _; series diverges
kÄ_
kÄ_
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
580
Chapter 10 Infinite Sequences and Series
4
"
"
"‰
" ‰
" ‰
ˆ
ˆ" "‰ ˆ"
ˆ "
41. (4n 3)(4n
1) œ 4n 3 4n 1 Ê sk œ 1 5 5 9 9 13 á 4k 7 4k 3
ˆ 4k " 3 4k " 1 ‰ œ 1 4k " 1 Ê
lim sk œ lim ˆ1 4k " 1 ‰ œ 1
kÄ_
kÄ_
A(2n 1) B(2n 1)
6
A
B
42. (2n 1)(2n
Ê A(2n 1) B(2n 1) œ 6 Ê (2A 2B)n (A B) œ 6
1) œ 2n 1 2n 1 œ
(2n 1)(2n 1)
Ê œ
k
k
2A 2B œ 0
ABœ0
6
ʜ
Ê 2A œ 6 Ê A œ 3 and B œ 3. Hence, ! (2n 1)(2n
œ 3 ! ˆ #n " 1 #n " 1 ‰
1)
A Bœ6
ABœ6
n œ1
nœ1
œ 3 Š "1 "3 "3 "5 "5 "7 á #(k "1) 1 2k " 1 #k " 1 ‹ œ 3 ˆ1 #k " 1 ‰ Ê the sum is
lim 3 ˆ1 #k " 1 ‰ œ 3
kÄ_
#
#
#
#
A(2n1)(2n1) B(2n1) C(2n1)(2n1) D(2n1)
A
B
C
D
43. (2n1)40n
# (2n1)# œ (2n1) (2n1)# (2n1) (2n1)# œ
(2n1)# (2n1)#
Ê A(2n 1)(2n 1)# B(2n 1)# C(2n 1)(2n 1)# D(2n 1)# œ 40n
Ê A a8n$ 4n# 2n 1b B a4n# 4n 1b C a8n$ 4n# 2n 1b œ D a4n# 4n 1b œ 40n
Ê (8A 8C)n$ (4A 4B 4C 4D)n# (2A 4B 2C 4D)n (A B C D) œ 40n
Ú
Ú
8A 8C œ 0
8A 8C œ 0
Ý
Ý
Ý
Ý
B Dœ 0
4A 4B 4C 4D œ 0
A BC Dœ 0
Ê œ
Ê 4B œ 20 Ê B œ 5
Ê Û
Ê Û
œ
œ
2D œ 20
2A
4B
2C
4D
40
A
2
B
C
2D
20
2B
Ý
Ý
Ý
Ý
Ü A B C D œ 0
Ü A B C D œ 0
and D œ 5 Ê œ
k
ACœ0
Ê C œ 0 and A œ 0. Hence, ! ’ (#n1)40n
# (2n1)# “
A 5 C 5 œ 0
n œ1
k
œ 5 ! ’ (#n" 1)# (#n" 1)# “ œ 5 Š 1" 9" 9" #"5 #"5 á (2(k1)" 1)# (#k" 1)# (#k" 1)# ‹
n œ1
œ 5 Š1 (2k" 1)# ‹ Ê the sum is n lim
5 Š1 (2k" 1)# ‹ œ 5
Ä_
1
"
"
"‰
" ‰
"
"
"
"
ˆ
ˆ" "‰ ˆ"
44. n#2n
(n 1)# œ n# (n 1)# Ê sk œ 1 4 4 9 9 16 á ’ (k 1)# k# “ ’ k# (k 1)# “
Ê
lim sk œ lim ’1 (k " 1)# “ œ 1
kÄ_
kÄ_
45. sk œ Š1 È"2 ‹ Š È"2 È"3 ‹ Š È"3 È"4 ‹ á Š È "
k1
Ê
È" ‹ Š È" Èk" 1 ‹ œ 1 Èk" 1
k
k
lim sk œ lim Š1 Èk" 1 ‹ œ 1
kÄ_
kÄ_
"
" ‰
"
" ‰
46. sk œ ˆ "# #""Î# ‰ ˆ #"Î#
#"Î$
ˆ #"Î$
#"Î%
á ˆ #1ÎÐ"k 1Ñ #1"Îk ‰ ˆ #1"Îk #1ÎÐ"k1Ñ ‰ œ #" #1ÎÐ"k1Ñ
Ê
lim sk œ "# "1 œ "#
kÄ_
47. sk œ ˆ ln"3 ln"# ‰ ˆ ln"4 ln"3 ‰ ˆ ln"5 ln"4 ‰ á Š ln (k" 1) ln"k ‹ Š ln (k" 2) ln (k" 1) ‹
œ ln"# ln (k" 2) Ê
lim sk œ ln"#
kÄ_
48. sk œ ctan" (1) tan" (2)d ctan" (2) tan" (3)d á ctan" (k 1) tan" (k)d
ctan" (k) tan" (k 1)d œ tan" (1) tan" (k 1) Ê lim sk œ tan" (1) 1# œ 14 1# œ 14
kÄ_
49. convergent geometric series with sum
È
"
œ È22 1 œ 2 È2
"
1 ŠÈ ‹
2
50. divergent geometric series with krk œ È2 1
51. convergent geometric series with sum
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Š 3# ‹
1 Š "# ‹
œ1
Section 10.2 Infinite Series
52. n lim
(1)n1 n Á 0 Ê diverges
Ä_
581
53. n lim
cos (n1) œ n lim
(1)n Á 0 Ê diverges
Ä_
Ä_
54. cos (n1) œ (1)n Ê convergent geometric series with sum
55. convergent geometric series with sum
"
œ 56
1 Š "5 ‹
#
"
œ e#e1
1 Š "# ‹
e
56. n lim
ln 3"n œ _ Á 0 Ê diverges
Ä_
57. convergent geometric series with sum
2
18
2
2 œ 20
"
9 9 œ 9
1 Š 10
‹
58. convergent geometric series with sum
"
œ x x 1
1 Š "x ‹
59. difference of two geometric series with sum
n
"
" " œ 3 3# œ 3#
1 Š 23 ‹
1 Š3‹
n
ˆ1 "n ‰ œ lim ˆ1 "
‰ œ e" Á 0 Ê diverges
60. n lim
n
Ä_
nÄ_
n
n †n â n
62. n lim
œ n lim
n lim
n œ _ Ê diverges
Ä _ n!
Ä _ 1†#ân
Ä_
n
n!
61. n lim
œ _ Á 0 Ê diverges
Ä _ 1000n
_
_
_
_
_
n
n
_
n
_
n
63. ! 2 4n 3 œ ! 42n ! 43n œ ! ˆ 21 ‰ ! ˆ 43 ‰ ; both œ ! ˆ 21 ‰ and ! ˆ 43 ‰ are geometric series, and both converge
n
n
n œ1
n
n œ1
n
nœ1
nœ1
nœ1
n œ1
n œ1
_
1
_
3
n
n
since r œ 12 Ê ¹ 12 ¹ 1 and r œ 34 Ê ¹ 34 ¹ 1, respectivley Ê ! ˆ 12 ‰ œ 1 2 1 œ 1 and ! ˆ 34 ‰ œ 1 4 3 œ 3 Ê
2
4
n œ1
_
nœ1
! 2 n 3 œ 1 3 œ 4 by Theorem 8, part (1)
n œ1
64.
n
n
4
4
lim 23n 4n œ
n
2n
n
ˆ 1 ‰n "
"
lim 43nn " œ lim ˆ 23 ‰n " œ 11 œ 1 Á 0 Ê diverges by nth term test for divergence
nÄ_
nÄ_ 4n
_
_
n œ1
n œ1
nÄ_
4
65. ! ln ˆ n n 1 ‰ œ ! cln (n) ln (n 1)d Ê sk œ cln (1) ln (2)d cln (2) ln (3)d cln (3) ln (4)d á
cln (k 1) ln (k)d cln (k) ln (k 1)d œ ln (k 1) Ê
lim sk œ _, Ê diverges
kÄ_
66. n lim
a œ n lim
ln ˆ 2n n 1 ‰ œ ln ˆ "# ‰ Á 0 Ê diverges
Ä_ n
Ä_
1
67. convergent geometric series with sum 1 "ˆ e ‰ œ 1 e
1
1
68. divergent geometric series with krk œ 1e e ¸ 23.141
22.459 1
_
_
n œ0
n œ0
_
_
n œ0
n œ0
69. ! (1)n xn œ ! (x)n ; a œ 1, r œ x; converges to 1 "(x) œ 1 " x for kxk 1
70. ! (1)n x2n œ ! ax# b ; a œ 1, r œ x# ; converges to 1 " x# for kxk 1
n
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
582
Chapter 10 Infinite Sequences and Series
71. a œ 3, r œ x # 1 ; converges to
_
_
72. ! (#1) ˆ 3 "sin x ‰ œ !
n
n
n œ0
n œ0
3
œ 3 6 x for 1 x"
# 1 or 1 x 3
1 Šx # "‹
ˆ "# ‰
" ˆ " ‰n
"
"
"
# 3 sin x ; a œ # , r œ 3 sin x ; converges to 1 Š
3 sin x ‹
3 sin x
3 sin x
"
"
"
ˆ
‰
œ 2(4
sin x) œ 8 2 sin x for all x since 4 Ÿ 3 sin x Ÿ # for all x
73. a œ 1, r œ 2x; converges to 1 " 2x for k2xk 1 or kxk #"
74. a œ 1, r œ x"# ; converges to
#
"
œ x#x 1 for ¸ x1# ¸ 1 or kxk 1.
1 Š #" ‹
x
75. a œ 1, r œ (x 1)n ; converges to 1 (x" 1) œ # " x for kx 1k 1 or 2 x 0
76. a œ 1, r œ 3 # x ; converges to
"
œ x 2 1 for ¸ 3 # x ¸ 1 or 1 x 5
1 Š3 # x‹
77. a œ 1, r œ sin x; converges to 1 "sin x for x Á (2k 1) 1# , k an integer
78. a œ 1, r œ ln x; converges to 1 "ln x for kln xk 1 or e" x e
_
79. (a) !
nœ2
_
80. (a) !
nœ1
_
"
(n 4)(n 5)
"
(b) ! (n 2)(n
3)
5
(n 2)(n 3)
5
(b) ! (n 2)(n
1)
n œ0
_
(c) !
n œ3
Š "# ‹
1 Š "# ‹
3
(b) one example is 3# 34 38 16
á œ
n 1
82. The series ! kˆ 12 ‰
Š 3# ‹
1 Š "# ‹
n œ0
_
_
nœ1
nœ1
nœ1
_
_
_
nœ1
nœ1
nœ1
5
(n 19)(n 18)
œ 3
Š "# ‹
1 Š "# ‹
is a geometric series whose sum is
_
nœ20
œ1
"
(c) one example is 1 "# 4" 8" 16
á œ 1
n
n œ5
_
"
81. (a) one example is "# 4" 8" 16
á œ
_
_
"
(c) ! (n 3)(n
#)
œ 0.
Š k# ‹
1 Š "# ‹
n
œ k where k can be any positive or negative number.
_
_
nœ1
nœ1
_
_
nœ1
nœ1
83. Let an œ bn œ ˆ "# ‰ . Then ! an œ ! bn œ ! ˆ "# ‰ œ 1, while ! Š bann ‹ œ ! (1) diverges.
n
n
n
84. Let an œ bn œ ˆ "# ‰ . Then ! an œ ! bn œ ! ˆ "# ‰ œ 1, while ! aan bn b œ ! ˆ 4" ‰ œ 3" Á AB.
n
n
_
_
_
_
n œ1
nœ1
nœ1
nœ1
n
85. Let an œ ˆ 4" ‰ and bn œ ˆ "# ‰ . Then A œ ! an œ 3" , B œ ! bn œ 1 and ! Š bann ‹ œ ! ˆ "# ‰ œ 1 Á AB .
86. Yes: ! Š a"n ‹ diverges. The reasoning: ! an converges Ê an Ä 0 Ê a"n Ä _ Ê ! Š a"n ‹ diverges by the
nth-Term Test.
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 10.3 The Integral Test
87. Since the sum of a finite number of terms is finite, adding or subtracting a finite number of terms from a series
that diverges does not change the divergence of the series.
88. Let An œ a" a# á an and n lim
A œ A. Assume ! aan bn b converges to S. Let
Ä_ n
Sn œ (a" b" ) (a# b# ) á (an bn ) Ê Sn œ (a" a# á an ) (b" b# á bn )
Ê b" b# á bn œ Sn An Ê n lim
ab" b# á bn b œ S A Ê ! bn converges. This
Ä_
contradicts the assumption that ! bn diverges; therefore, ! aan bn b diverges.
89. (a)
(b)
2
1r œ 5
Š 13
2 ‹
1r
#
Ê 52 œ 1 r Ê r œ 53 ; 2 2 ˆ 35 ‰ 2 ˆ 35 ‰ á
#
$
3
13
13 ˆ 3 ‰
13 ˆ 3 ‰
ˆ 3 ‰ á
œ 5 Ê 13
13
10 œ 1 r Ê r œ 10 ; 2 #
10 # 10
# 10
90. 1 eb e2b á œ 1"eb œ 9 Ê "9 œ 1 eb Ê eb œ 98 Ê b œ ln ˆ 89 ‰
91. sn œ 1 2r r# 2r$ r% 2r& á r2n 2r2n1 , n œ 0, 1, á
Ê sn œ a1 r# r% á r2n b a2r 2r$ 2r& á 2r2n1 b Ê n lim
s œ 1 " r# 1 2rr#
Ä_ n
2r
#
œ 11 r# , if kr k 1 or krk 1
92. L sn œ 1 a r a a11 rr b œ 1arr
n
n
#
#
93. area œ 2# ŠÈ2‹ (1)# Š È"2 ‹ á œ 4 2 1 #" á œ 1 4 " œ 8 m#
#
#
94. (a) L" œ 3, L# œ 3 ˆ 43 ‰ , L$ œ 3 ˆ 43 ‰ , á , Ln œ 3 ˆ 43 ‰
(b)
nc1
Ê n lim
L œ n lim
3 ˆ 43 ‰
Ä_ n
Ä_
nc1
œ_
È
È
Using the fact that the area of an equilateral triangle of side length s is 43 s2 , we see that A" œ 43 ,
È
È
È
È
È
È
È
2
2
A# œ A" 3Š 43 ‹ˆ "3 ‰ œ 43 1#3 , A$ œ A# 3a4bŠ 43 ‹ˆ 3"2 ‰ œ 43 123 #73 ,
È
È
2
2
A% œ A$ 3a4b2 Š 43 ‹ˆ 3"3 ‰ , A5 œ A4 3a4b3 Š 43 ‹ˆ 3"4 ‰ , . . . ,
An œ
n
n
n
È3
! 3a4bk2 Š È3 ‹ˆ "2 ‰k1 œ È3 ! 3È3a4bk$ ˆ " ‰k1 œ È3 3È3Œ! 4kkc$1 .
4 4
3
4
9
4
9
kœ2
kœ2
kœ2
n
1
È3
kc$
È
È
È
1 ‰
lim An œ n lim
3È3Œ! 49k 1 œ 43 3È3Œ 1 36 4 œ 43 3È3ˆ 20
œ 43 ˆ1 53 ‰
nÄ_
Ä_ Œ 4
9
kœ2
È
œ 43 ˆ 85 ‰ œ 85 A"
10.3 THE INTEGRAL TEST
_1
1; '1
1. faxb œ x12 is positive, continuous, and decreasing for x
œ lim
bÄ_
ˆ 1b 1‰ œ 1 Ê '
_1
1
x
_
bÄ_
ˆ 54 b0.8 54 ‰ œ _ Ê '
'1b x1 dx œ lim
2
bÄ_
b
’ 1x “
1
! 12 converges
2 dx converges Ê
n œ1
n
_ 1
1; '1
2. faxb œ x10.2 is positive, continuous, and decreasing for x
œ lim
x2 dx œ b lim
Ä_
_ 1
1
x0.2 dx diverges Ê
_
x0.2 dx œ b lim
Ä_
'1b x1 dx œ lim
0.2
bÄ_
! 10.2 diverges
n œ1
n
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
b
’ 54 x0.8 “
1
583
584
Chapter 10 Infinite Sequences and Series
_
1; '1
3. faxb œ x2 1 4 is positive, continuous, and decreasing for x
œ lim ˆ 12 tan1 b2 12 tan1 12 ‰ œ 14 12 tan1 12 Ê
bÄ_
'1b x 1 4 dx œ lim
’lnlx 4l“
'1b e2x dx œ lim
’ 12 e2x “
_
2
_ 1
1; '1
x 4 dx œ b lim
Ä_
_
b
bÄ_
1
2
n œ1
_
alnlb 4l ln 5b œ _ Ê '1 x 1 4 dx diverges Ê ! n 1 4 diverges
bÄ_
’ 21 tan1 x2 “
2
'1_ x 1 4 dx converges Ê ! n 1 4 converges
4. faxb œ x 1 4 is positive, continuous, and decreasing for x
œ lim
'1b x 1 4 dx œ lim
1
x2 4 dx œ b lim
Ä_
b
bÄ_
1
n œ1
_
1; '1 e2x dx œ lim
5. faxb œ e2x is positive, continuous, and decreasing for x
_ 2x
ˆ 2e12b 2e12 ‰ œ 2e12 Ê ' e
1
œ lim
bÄ_
_
ˆ ln1b ln12 ‰ œ ln12 Ê '
œ lim
bÄ_
_
2; '2
'3 x x 4 dx œ lim '3 x x 4 dx œ lim
2
_
_
n œ3
n œ1
'2b xaln1xb dx œ lim
2
b
’ ln1x “
bÄ_
1, f w axb œ ax42x4b2 0 for x 2, thus f is decreasing for x
2
b
b
bÄ_
1
dx œ lim
xaln xb2
bÄ_
2
1
dx converges Ê ! naln1nb2 converges
xaln xb2
n œ2
7. faxb œ x2 x 4 is positive and continuous for x
2
1
n œ1
_
2
_
b
bÄ_
dx converges Ê ! e2n converges
6. faxb œ xaln1xb2 is positive, continuous, and decreasing for x
_
bÄ_
bÄ_
’ 21 lnax2 4b“ œ lim
bÄ_
3
_
3;
ˆ 12 lnab2 4b 12 lna13b‰ œ _ Ê '
_
3
x
x2 4 dx
diverges Ê ! n2 n 4 diverges Ê ! n2 n 4 œ 15 28 ! n2 n 4 diverges
n œ3
2, f w axb œ 2 xln2 x 0 for x e, thus f is decreasing for x
2
2
8. faxb œ lnxx is positive and continuous for x
_ ln x2
'3
dx œ lim
x
bÄ_
_
'3 lnxx dx œ lim
b
_
2
_ ln x2
b
2
bÄ_
’2aln xb“ œ lim
_
2
bÄ_
3
a2aln bb 2aln 3bb œ _ Ê '3
x
3;
dx
2
diverges Ê ! lnnn diverges Ê ! lnnn œ ln24 ! lnnn diverges
n œ3
n œ2
n œ3
2
9. faxb œ exxÎ3 is positive and continuous for x
_ x2
'7 e
xÎ3
dx œ lim
bÄ_
2
xÎ3
bÄ_
18b
lim
Š 3a6b
‹ e327
7Î3 œ
ebÎ3
œ lim
bÄ_
_
'7 ex dx œ lim
b
bÄ_
ax 6 b
1, f w axb œ x3e
0 for x 6, thus f is decreasing for x
xÎ3
b
2
54
lim
’ e3xxÎ3 e18x
xÎ3 exÎ3 “ œ
7
327
'
ˆ b54
‰ 327
Î3
7Î3 œ 7Î3 Ê
e
e
e
_
2
bÄ_
_ x2
7
exÎ3
7;
54
e327
Š 3b eb18b
7Î3 ‹ œ
Î3
2
_
2
dx converges Ê ! ennÎ3 converges
n œ7
2
25
36
! nnÎ3 converges
Ê ! ennÎ3 œ e11Î3 e24Î3 e91 e16
4Î3 e5Î3 e2 e
n œ7
n œ1
10. faxb œ x2 x 2x4 1 œ axx14b2 is continuous for x
decreasing for x
œ lim
bÄ_
_
_ x4
8; '8
x1
3
1
3
”'8 ax 1b2 dx '8 ax 1b2 dx• œ b lim
”'8 x 1 dx '8 ax 1b2 dx•
bÄ_
Ä_
dx œ lim
ax 1b
2
b
’lnlx 1l x 3 1 “ œ lim
8
_
2, f is positive for x 4, and f w axb œ ax71xb3 0 for x 7, thus f is
bÄ_
b
b
b
_ x4
ˆlnlb 1l b 3 1 ln 7 37 ‰ œ _ Ê '
8
ax 1b2
b
dx diverges
_
1
2
3
Ê ! n2 n 2n4 1 diverges Ê ! n2 n 2n4 1 œ 2 14 0 16
25
36
! n2 n 2n4 1 diverges
nœ8
nœ2
"
11. converges; a geometric series with r œ 10
1
n œ8
12. converges; a geometric series with r œ "e 1
n
13. diverges; by the nth-Term Test for Divergence, n lim
œ1Á0
Ä _ n1
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 10.3 The Integral Test
_ 5
14. diverges by the Integral Test; '1 x 5 1 dx œ 5 ln (n 1) 5 ln 2 Ê '1
n
_
_
n œ1
nœ1
x 1 dx
Ä _
15. diverges; ! È3n œ 3 ! È"n , which is a divergent p-series (p œ #" )
_
_
n œ1
n œ1
"
16. converges; ! nÈ2n œ 2 ! n$Î#
, which is a convergent p-series (p œ 3# )
17. converges; a geometric series with r œ "8 1
_
_
_
_
n œ1
nœ1
nœ1
n œ1
18. diverges; ! n8 œ 8 ! 1n and since ! "n diverges, 8 ! 1n diverges
_ ln x
19. diverges by the Integral Test: '2 lnxx dx œ "# aln# n ln 2b Ê '2
n
x
dx Ä _
Ô t œ lndxx ×
_
b
dt œ x
Ä 'ln 2 tetÎ2 dt œ lim 2tetÎ2 4etÎ2 ‘ ln 2
bÄ_
Õ dx œ et dt Ø
œ lim 2ebÎ2 (b 2) 2eÐln 2ÑÎ2 (ln 2 2)‘ œ _
_ ln x
20. diverges by the Integral Test: '2
Èx dx;
bÄ_
21. converges; a geometric series with r œ 23 1
n
n
n
5
5 ln 5
ˆ ln 5 ‰ ˆ 54 ‰ œ _ Á 0
22. diverges; n lim
œ n lim
œ n lim
Ä _ 4n 3
Ä _ 4n ln 4
Ä _ ln 4
_
_
n œ0
n œ0
23. diverges; ! n21 œ 2 ! n " 1 , which diverges by the Integral Test
24. diverges by the Integral Test: '1 2xdx 1 œ "# ln (2n 1) Ä _ as n Ä _
n
n
n
2
2 ln 2
25. diverges; n lim
a œ n lim
œ n lim
œ_Á0
1
Ä_ n
Ä _ n1
Ä_
dx
26. diverges by the Integral Test: '1 Èx ˆÈ
;
x 1‰ –
n
Èn
27. diverges; n lim
œ n lim
Ä _ ln n
Ä_
"
Š 2È
‹
n
Š "n ‹
œ n lim
Ä_
Ènb1 du
u œ Èx "
'
ˆÈn 1‰ ln 2 Ä _ as n Ä _
Ä
dx
u œ ln
2
du œ Èx —
Èn
# œ_Á0
n
ˆ1 n" ‰ œ e Á 0
28. diverges; n lim
a œ n lim
Ä_ n
Ä_
29. diverges; a geometric series with r œ ln"# ¸ 1.44 1
30. converges; a geometric series with r œ ln"3 ¸ 0.91 1
_
31. converges by the Integral Test: '3
Š "x ‹
u œ ln x
dx; ”
(ln x) È(ln x)# 1
du œ "x dx •
_
Ä 'ln 3 È "#
u
u 1
du
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
585
586
Chapter 10 Infinite Sequences and Series
b
œ lim csec" kukd ln 3 œ lim csec" b sec" (ln 3)d œ lim cos" ˆ "b ‰ sec" (ln 3)‘
bÄ_
bÄ_
bÄ_
œ cos" (0) sec" (ln 3) œ 1# sec" (ln 3) ¸ 1.1439
_
32. converges by the Integral Test: '1
"
x a1 ln# xb dx œ
'1_ 1 Š(ln‹x) dx; ” u œ ln" x • Ä '0_ 1"u du
"
x
du œ x dx
#
#
œ lim ctan" ud 0 œ lim atan" b tan" 0b œ 1# 0 œ 1#
b
bÄ_
bÄ_
33. diverges by the nth-Term Test for divergence; n lim
n sin ˆ "n ‰ œ n lim
Ä_
Ä_
sin ˆ "n ‰
sin x
œ1Á0
ˆ "n ‰ œ xlim
Ä0 x
34. diverges by the nth-Term Test for divergence; n lim
n tan ˆ "n ‰ œ n lim
Ä_
Ä_
Š n"# ‹ sec# ˆ n" ‰
tan ˆ "n ‰
œ
lim
ˆ "n ‰
nÄ_
Š " ‹
n#
œ n lim
sec# ˆ "n ‰ œ sec# 0 œ 1 Á 0
Ä_
_
35. converges by the Integral Test: '1
uœe
ex
1 e2x dx; ” du œ ex dx •
x
Ä 'e
_
b
"
ctan" ud e
1 u# du œ n lim
Ä_
œ lim atan" b tan" eb œ 1# tan" e ¸ 0.35
bÄ_
_
36. converges by the Integral Test: '1
u œ ex ×
_
_
2
du œ ex dx Ä ' u(1 2 u) du œ ' ˆ 2u u 2 1 ‰ du
1 ex dx;
e
e
Õ dx œ " du Ø
u
Ô
b
œ lim 2 ln u u 1 ‘ e œ lim 2 ln ˆ b b 1 ‰ 2 ln ˆ e e 1 ‰ œ 2 ln 1 2 ln ˆ e e 1 ‰ œ 2 ln ˆ e e 1 ‰ ¸ 0.63
bÄ_
bÄ_
_ 8 tanc" x
37. converges by the Integral Test: '1
1 x#
_ x
38. diverges by the Integral Test: '1
dx; ”
u œ tan" x
'11ÎÎ42 8u du œ c4u# d 11ÎÎ24 œ 4 Š 14# 161# ‹ œ 341#
dx • Ä
du œ 1 x#
u œ x# 1
x# 1 dx; ” du œ 2x dx •
_
39. converges by the Integral Test: '1 sech x dx œ 2 lim
_ du
Ä #" '2
4
b
œ lim #" ln u‘ 2 œ lim
"
(ln b ln 2) œ _
bÄ_ #
bÄ_
'1b 1 eae b dx œ 2 lim ctan" ex d b1
x
x #
bÄ_
œ 2 lim atan" eb tan" eb œ 1 2 tan" e ¸ 0.71
bÄ_
bÄ_
_
40. converges by the Integral Test: '1 sech# x dx œ lim
bÄ_
œ 1 tanh 1 ¸ 0.76
_
'1b sech# x dx œ lim ctanh xd b1 œ lim (tanh b tanh 1)
bÄ_
bÄ_
41. '1 ˆ x a 2 x " 4 ‰ dx œ lim ca ln kx 2k ln kx 4kd 1 œ lim ln (bb2)4 ln ˆ 35 ‰ ;
a
b
bÄ_
a
_, a 1
lim (bb2)4 œ a lim (b 2)a1 œ œ
1, a œ 1
bÄ_
bÄ_
a
bÄ_
Ê the series converges to ln ˆ 53 ‰ if a œ 1 and diverges to _ if
a 1. If a 1, the terms of the series eventually become negative and the Integral Test does not apply. From
that point on, however, the series behaves like a negative multiple of the harmonic series, and so it diverges.
_
x1
b1
‰
ˆ2‰
42. '3 ˆ x " 1 x 2a
1 dx œ lim ’ln ¹ (x 1)2a ¹“ œ lim ln (b 1)2a ln 42a ; lim
b
bÄ_
1, a œ "#
"
œ
2a
c
1
_, a "
b Ä _ #a(b 1)
#
œ lim
3
bÄ_
b"
2a
b Ä _ (b 1)
Ê the series converges to ln ˆ #4 ‰ œ ln 2 if a œ #" and diverges to _ if
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 10.3 The Integral Test
587
if a "# . If a "# , the terms of the series eventually become negative and the Integral Test does not apply.
From that point on, however, the series behaves like a negative multiple of the harmonic series, and so it diverges.
43. (a)
(b) There are (13)(365)(24)(60)(60) a10* b seconds in 13 billion years; by part (a) sn Ÿ 1 ln n where
n œ (13)(365)(24)(60)(60) a10* b Ê sn Ÿ 1 ln a(13)(365)(24)(60)(60) a10* bb
œ 1 ln (13) ln (365) ln (24) 2 ln (60) 9 ln (10) ¸ 41.55
_
_
_
n œ1
n œ1
n œ1
"
44. No, because ! nx
œ x" ! n" and ! n" diverges
_
_
_
nœ1
nœ1
nœ1
45. Yes. If ! an is a divergent series of positive numbers, then ˆ "# ‰ ! an œ ! ˆ a#n ‰ also diverges and a#n an .
_
There is no “smallest" divergent series of positive numbers: for any divergent series ! an of positive numbers
n œ1
_
! ˆ an ‰ has smaller terms and still diverges.
n œ1
#
_
_
_
nœ1
nœ1
nœ1
46. No, if ! an is a convergent series of positive numbers, then 2 ! an œ ! 2an also converges, and 2an
an .
There is no “largest" convergent series of positive numbers.
47. (a) Both integrals can represent the area under the curve faxb œ Èx1 1 , and the sum s50 can be considered an
50
approximation of either integral using rectangles with ?x œ 1. The sum s50 œ ! Èn1 1 is an overestimate of the
n œ1
51
integral '1 Èx1 1 dx. The sum s50 represents a left-hand sum (that is, the we are choosing the left-hand endpoint of
each subinterval for ci ) and because f is a decreasing function, the value of f is a maximum at the left-hand endpoint of
each sub interval. The area of each rectangle overestimates the true area, thus '1
51
50
1
! È 1 . In a similar
Èx 1 dx n1
n œ1
50
manner, s50 underestimates the integral '0 Èx1 1 dx. In this case, the sum s50 represents a right-hand sum and because
f is a decreasing function, the value of f is aminimum at the right-hand endpoint of each subinterval. The area of each
rectangle underestimates the true area, thus ! Èn1 1 '0
50
50
n œ1
œ ’2Èx 1“ œ 2È52 2È2 ¸ 11.6 and '0
51
50
1
1
Èx 1 dx. Evaluating the integrals we find
50
1
È
Èx 1 dx œ ’2 x 1“
0
'151 Èx1 1 dx
œ 2È51 2È1 ¸ 12.3. Thus,
50
11.6 ! Èn1 1 12.3.
n œ1
nb1
(b) sn 1000 Ê '1
Ên
2
nb1
1
È
œ 2Èn 1 2È2 1000 Ê n Š500 2È2‹ Èx 1 dx œ ’2 x 1“
1
251415.
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
¸ 251414.2
588
Chapter 10 Infinite Sequences and Series
30
_
n œ1
n œ1
_
48. (a) Since we are using s30 œ ! n14 to estimate ! n14 , the error is given by ! n14 . We can consider this sum as an estimate
nœ31
of the area under the curve faxb œ x14 when x
30 using rectangles with ?x œ 1 and ci is the right-hand endpoint of
each subinterval. Since f is a decreasing function, the value of f is a minimum at the right-hand endpoint of each
1
subinterval, thus ! n14 '30 x14 dx œ lim '30 x14 dx œ lim ’ 3x1 3 “ œ lim Š 3b1 3 3a30
‹ ¸ 1.23 ‚ 105 .
b3
_
_
b
b
bÄ_
nœ31
bÄ_
bÄ_
30
5
Thus the error 1.23 ‚ 10 Þ
(b) We want S sn 0.000001 Ê 'n
_
œ
1
x4 dx 0.000001 Ê
4
_
1
x3 dx 0.01 Ê
œ 2n1 2 0.01 Ê n È50 ¸ 7.071 Ê n
4
bÄ_
3 1000000
lim ˆ 3b1 3 3n1 3 ‰ œ 3n1 3 0.000001 Ê n É
¸ 69.336 Ê n
3
bÄ_
49. We want S sn 0.01 Ê 'n
3
bÄ_
n
70.
'n_ x1 dx œ lim 'nb x1 dx œ lim ’ 2x1 “b œ lim ˆ 2b1 2n1 ‰
3
bÄ_
3
bÄ_
2
2
bÄ_
n
2
8
8 Ê S ¸ s8 œ ! n13 ¸ 1.195
n œ1
50. We want S sn 0.1 Ê 'n
'
_
œ
'n_ x1 dx œ lim 'nb x1 dx œ lim ’ 3x1 “b
b
b
1
1
1
1 ˆ x ‰
lim
tan
’
“
2 4 dx œ
x2 4 dx 0.1 Ê blim
x
2
2
Ä_ n
bÄ_
n
1
1
1
1
1
1 ˆ b ‰
1 ˆ n ‰‰
1 ˆ n ‰
ˆ
ˆ
‰
lim tan
œ 4 2 tan
2 2 tan
2
2 0.1 Ê n 2tan 2 0.2 ¸ 9.867 Ê n
bÄ_ 2
10 Ê S ¸ s10
10
œ ! n2 1 4 ¸ 0.57
n œ1
51. S sn 0.00001 Ê 'n
_
1
x1.1 dx 0.00001 Ê
'n_ x1 dx œ lim 'nb x1 dx œ lim ’ x10 “b œ lim ˆ b10 n10 ‰
1.1
bÄ_
1.1
bÄ_
0.1
bÄ_
n
0.1
0.1
10
œ n10
Ê n 1060
0.1 0.00001 Ê n 1000000
52. S sn 0.01 Ê 'n
_
œ
1
dx 0.01 Ê
xaln xb3
'n_ xaln1xb dx œ lim 'nb xaln1xb dx œ lim ’ 2aln1xb “b
3
bÄ_
3
È
lim Š 2aln1bb2 2aln1nb2 ‹ œ 2aln1nb2 0.01 Ê n e 50 ¸ 1177.405 Ê n
bÄ_
n
n
k œ1
k œ1
bÄ_
2
n
1178
53. Let An œ ! ak and Bn œ ! 2k aa2k b , where {ak } is a nonincreasing sequence of positive terms converging to
0. Note that {An } and {Bn } are nondecreasing sequences of positive terms. Now,
Bn œ 2a# 4a% 8a) á 2n aa2n b œ 2a# a2a% 2a% b a2a) 2a) 2a) 2a) b á
ˆ2aa2n b 2aa2n b á 2aa2n b ‰ Ÿ 2a" 2a# a2a$ 2a% b a2a& 2a' 2a( 2a) b á
ðóóóóóóóóóóóóóóñóóóóóóóóóóóóóóò
2n1 terms
_
ˆ2aa2nc1 b 2aa2nc1 1b á 2aa2n b ‰ œ 2Aa2n b Ÿ 2 ! ak . Therefore if ! ak converges,
k œ1
then {Bn } is bounded above Ê ! 2k aa2k b converges. Conversely,
_
An œ a" aa# a$ b aa% a& a' a( b á an a" 2a# 4a% á 2n aa2n b œ a" Bn a" ! 2k aa2k b .
k œ1
_
Therefore, if ! 2 aa2k b converges, then {An } is bounded above and hence converges.
k
k œ1
_
_
_
n œ2
n œ2
n œ2
"
" ! "
! 2 n a a2 n b œ ! 2 n n "
54. (a) aa2n b œ 2n ln"a2n b œ 2n †n(ln
# †n(ln 2) œ ln #
2) Ê
n , which diverges
_
Ê ! n ln" n diverges.
n œ2
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 10.3 The Integral Test
_
_
_
_
n œ1
nœ1
nœ1
nœ1
589
n
(b) aa2n b œ #"np Ê ! 2n aa2n b œ ! 2n † #"np œ ! a2n"bpc1 œ ! ˆ #p"c1 ‰ , a geometric series that
converges if #p"c1 1 or p 1, but diverges if p Ÿ 1.
_
55. (a) '2
'_
u œ ln x
cpb1
dx
ucp du œ lim ’ up 1 “ œ lim Š 1 " p ‹ cbp1 (ln 2)p1 d
x(ln x)p ; ” du œ dx • Ä
ln 2
b
Ä
_
bÄ_
ln 2
x
"
c
pb1
,p1
p 1 (ln 2)
Ê the improper integral converges if p 1 and diverges if p 1.
œœ
b
_, p "
_ dx
For p œ 1: '2
p œ 1.
cln (ln x)d 2 œ lim cln (ln b) ln (ln 2)d œ _, so the improper integral diverges if
x ln x œ b lim
Ä_
bÄ_
b
_
(b) Since the series and the integral converge or diverge together, ! n(ln" n)p converges if and only if p 1.
n œ2
56. (a) p œ 1 Ê the series diverges
(b) p œ 1.01 Ê the series converges
_
_
n œ2
n œ2
(c) ! n aln" n$ b œ 3" ! n(ln" n) ; p œ 1 Ê the series diverges
(d) p œ 3 Ê the series converges
nb1
57. (a) From Fig. 10.11(a) in the text with f(x) œ "x and ak œ k" , we have '1
"
"
"
"
x dx Ÿ 1 # 3 á n
Ÿ 1 '1 f(x) dx Ê ln (n 1) Ÿ 1 "# "3 á "n Ÿ 1 ln n Ê 0 Ÿ ln (n 1) ln n
n
Ÿ ˆ1 "# 3" á n" ‰ ln n Ÿ 1. Therefore the sequence ˜ˆ1 #" 3" á n" ‰ ln n™ is bounded above by
1 and below by 0.
nb1
(b) From the graph in Fig. 10.11(b) with f(x) œ "x , n " 1 'n
"
x dx œ ln (n 1) ln n
Ê 0 n " 1 cln (n 1) ln nd œ ˆ1 #" "3 á n" 1 ln (n 1)‰ ˆ1 #" 3" á n" ln n‰ .
If we define an œ 1 "# œ 3" "n ln n, then 0 an1 an Ê an1 an Ê {an } is a decreasing sequence of
nonnegative terms.
#
58. ex Ÿ ex for x
_
_
#
b
1, and '1 ecx dx œ lim cex d " œ lim ˆeb e1 ‰ œ ec1 Ê '1 ecx dx converges by
bÄ_
bÄ_
_
n#
the Comparison Test for improper integrals Ê ! e
n œ0
_
59. (a) s10 œ ! n"3 œ 1.97531986; '11 x13 dx œ lim
10
bÄ_
n œ1
_
#
œ 1 ! en converges by the Integral Test.
nœ1
1 ‰
1
'11b x3 dx œ lim ’ x2c “ b œ lim ˆ 2b1 242
œ 242
and
2
bÄ_
11
1 ‰
1
'10_ x1 dx œ lim '10b x3 dx œ lim ’ x2c “ b œ lim ˆ 2b1 200
œ 200
bÄ_
2
2
3
(b)
bÄ_
bÄ_
10
bÄ_
2
1
1
Ê 1.97531986 242
s 1.97531986 200
Ê 1.20166 s 1.20253
_
s œ ! n"3 ¸ 1.20166 2 1.20253 œ 1.202095; error Ÿ 1.20253 2 1.20166 œ 0.000435
n œ1
_
60. (a) s10 œ ! n"4 œ 1.082036583; '11 x14 dx œ lim
10
bÄ_
nœ1
1 ‰
1
'11b x4 dx œ lim ’ x3c “ b œ lim ˆ 3b1 3993
œ 3993
and
3
bÄ_
11
1 ‰
1
'10_ x1 dx œ lim '10b x4 dx œ lim ’ x3c “ b œ lim ˆ 3b1 3000
œ 3000
bÄ_
3
4
bÄ_
bÄ_
10
bÄ_
3
1
1
Ê 1.082036583 3993
s 1.082036583 3000
Ê 1.08229 s 1.08237
_
(b) s œ ! n"4 ¸ 1.08229 2 1.08237 œ 1.08233; error Ÿ 1.08237 2 1.08229 œ 0.00004
n œ1
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
3
590
Chapter 10 Infinite Sequences and Series
10.4 COMPARISON TESTS
_
1. Compare with ! n"2 , which is a convergent p-series, since p œ 2 1. Both series have nonnegative terms for n
1. For
n œ1
n
1, we have n2 Ÿ n2 30 Ê n12
1
n2 30 .
_
Then by Comparison Test, ! n2 1 30 converges.
n œ1
_
2. Compare with ! n"3 , which is a convergent p-series, since p œ 3 1. Both series have nonnegative terms for n
1. For
n œ1
n
1, we have n4 Ÿ n4 2 Ê n14
1
n
n 4 2 Ê n 4
n
1
n 4 2 Ê n 3
n
n 4 2
_
n1
n 4 2 .
Then by Comparison Test, ! nn412
n œ1
converges.
_
3. Compare with ! È"n , which is a divergent p-series, since p œ #" Ÿ 1. Both series have nonnegative terms for n
2. For
n œ2
n
2, we have Èn 1 Ÿ Èn Ê Èn1 1
1
Èn .
_
Then by Comparison Test, ! Èn1 1 diverges.
n œ2
_
4. Compare with ! "n , which is a divergent p-series, since p œ 1 Ÿ 1. Both series have nonnegative terms for n
2. For
n œ2
n
2, we have n2 n Ÿ n2 Ê n2 1 n
n
1
n2
n2 œ n Ê n2 n
1
n
n2 Ê n2 n
n
1
n.
n2 n
_
Thus ! nn22n diverges.
n œ2
_
5. Compare with ! n3"Î2 , which is a convergent p-series, since p œ 32 1. Both series have nonnegative terms for n
1.
n œ1
_
2
2
n
n
1, we have 0 Ÿ cos2 n Ÿ 1 Ê cos
Ÿ n31Î2 . Then by Comparison Test, ! cos
converges.
n3Î2
n3Î2
For n
n œ1
_
6. Compare with ! 3"n , which is a convergent geometric series, since lrl œ ¹ 13 ¹ 1. Both series have nonnegative terms for
n œ1
n
1. For n
_
1, we have n † 3n
3n Ê n†13n Ÿ 31n . Then by Comparison Test, ! n†13n converges.
_
_
_
nœ1
nœ1
È
n œ1
È
7. Compare with ! n3Î52 . The series ! n31Î2 is a convergent p-series, since p œ 32 1, and the series ! n3Î52
nœ1
_
œ È5 ! n31Î2 converges by Theorem 8 part 3. Both series have nonnegative terms for n
1. For n
1, we have
n œ1
n3 Ÿ n4 Ê 4n3 Ÿ 4n4 Ê n4 4n3 Ÿ n4 4n4 œ 5n4 Ê n4 4n3 Ÿ 5n4 20 œ 5an4 4b Ê nn44n4 Ÿ 5.
4
3
_
È
Ê n na4n44b Ÿ 5 Ê nn444 Ÿ n53 Ê É nn444 Ÿ É n53 œ n3Î52 Then by Comparison Test, ! É nn444 converges.
3
n œ1
_
8. Compare with ! È"n , which is a divergent p-series, since p œ #" Ÿ 1. Both series have nonnegative terms for n
n œ1
n
1, we have Èn
Ê n2 2 nÈn n
Èn 1
ÊÈ2
n 3
1
Èn .
1 Ê 2È n
n2 3 Ê
2 Ê 2È n 1
n ˆn 2 È n 1 ‰
n2 3
3 Ê nˆ2Èn 1‰
1Ê
_
Èn 1
Then by Comparison Test, ! È 2
n œ1
n 2È n 1
n2 3
n 3
3 Ê 2 nÈ n n
3n
ˆÈ n 1 ‰
1
n Ê
n2 3
2
ˆÈ
‰
1
Ê nn2 31
n Ê
diverges.
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
3
2
É 1n
1. For
Section 10.4 Comparison Tests
_
9. Compare with ! n"2 , which is a convergent p-series, since p œ 2 1. Both series have positive terms for n
nc2
n3 c n2 b 3
1 În 2
œ lim
_
!
n œ1
nÄ_
n œ1
591
1. lim bann
nÄ_
3n 4n
6n 4
6
œ lim n3nn22n 3 œ lim 3n
lim 6n
2 2n œ
2 œ lim 6 œ 1 0. Then by Limit Comparison Test,
3
2
2
nÄ_
nÄ_
nÄ_
nÄ_
n2
n3 n2 3 converges.
_
10. Compare with ! È"n , which is a divergent p-series, since p œ #" Ÿ 1. Both series have positive terms for n
É nn2bb12
n œ1
1. lim bann
nÄ_
n
n n
2n 1
2
È1 œ 1 0. Then by Limit Comparison
œ lim 1ÎÈn œ lim É nn2 2 œ É lim n2 2 œ É lim
2n œ É lim 2 œ
2
nÄ_
2
nÄ_
nÄ_
nÄ_
nÄ_
_
Test, ! É nn212 diverges.
n œ1
_
11. Compare with ! "n , which is a divergent p-series, since p œ 1 Ÿ 1. Both series have positive terms for n
nan b 1b
Šn2
œ lim
n œ2
b 1‹an c 1b
3
2
2
nÄ_
_
Test,
nÄ_
2n
2
6
œ lim n3 nn2+nn 1 œ lim 3n3n
lim 6n
2 2n 1 œ
6n 2 œ lim 6 œ 1 0. Then by Limit Comparison
1 În
nÄ_
2. lim bann
nÄ_
nÄ_
nÄ_
! 2 nan 1b diverges.
an 1ban 1b
n œ2
_
12. Compare with ! 2"n , which is a convergent geometric series, since lrl œ ¹ 12 ¹ 1. Both series have positive terms for
n œ1
2n
n
lim an œ lim 13 bÎ24nn
nÄ_ bn
nÄ_
1.
n
_
n
n
4
! 2 n converges.
œ lim 3 4 4n œ lim 44n ln
ln 4 œ 1 0. Then by Limit Comparison Test,
34
nÄ_
nÄ_
n œ1
_
13. Compare with ! È"n , which is a divergent p-series, since p œ 12 Ÿ 1. Both series have positive terms for n
n œ1
5n
n†4n
È
œ lim 1ÎÈn œ lim 54n œ
n
nÄ_
nÄ_
_
n
lim ˆ 5 ‰ œ
nÄ_ 4
_
1. lim bann
nÄ_
n
_. Then by Limit Comparison Test, ! È5n†4n diverges.
n œ1
n
14. Compare with ! ˆ 25 ‰ , which is a convergent geometric series, since lrl œ ¹ 25 ¹ 1. Both series have positive terms for
n œ1
ˆ 2n b 3 ‰
15 ‰n
15 ‰n
15 ‰
n 1. lim bann œ lim a5n2Îb54bn œ lim ˆ 10n
œ exp lim lnˆ 10n
œ exp lim n lnˆ 10n
10n 8
10n 8
nÄ_
nÄ_
nÄ_ 10n 8
nÄ_
nÄ_
b 15 ‰
10
lnˆ 10n
10b 8
70n2
70n2
10n b 8
œ exp lim
œ exp lim 10n b151În10n
œ exp lim a10n 15
2
2
1 În
ba10n 8b œ exp nlim
nÄ_
nÄ_
nÄ_
Ä_ 100n 230n 120
n
_
n
140
3‰
7Î10
œ exp lim 200n140n
0. Then by Limit Comparison Test, ! ˆ 2n
converges.
230 œ exp lim 200 œ e
5n 4
nÄ_
nÄ_
n œ1
_
15. Compare with ! "n , which is a divergent p-series, since p œ 1 Ÿ 1. Both series have positive terms for n
n œ2
2. lim bann
nÄ_
_
"
ln n
œ lim 1În œ lim lnnn œ lim 11În œ lim n œ _. Then by Limit Comparison Test, ! ln"n diverges.
nÄ_
nÄ_
nÄ_
nÄ_
n œ2
_
16. Compare with ! n"2 , which is a convergent p-series, since p œ 2 1. Both series have positive terms for n
n œ1
œ lim
nÄ_
lnŠ1 n"2 ‹
1În2
2
" Š n3 ‹
1
n2
Š n23 ‹
1
œ lim
nÄ_
_
1. lim bann
nÄ_
œ lim 1 1 " œ 1 0. Then by Limit Comparison Test, ! lnˆ1 n"2 ‰ converges.
nÄ_
n2
n œ1
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
592
Chapter 10 Infinite Sequences and Series
_
17. diverges by the Limit Comparison Test (part 1) when compared with ! È"n , a divergent p-series:
n œ1
"
lim
nÄ_
Œ #Èn È
$ n
Š " ‹
Èn
Èn
œ n lim
lim ˆ " ‰ œ #"
$ n œ
Ä _ 2È n È
n Ä _ # n 1Î6
18. diverges by the Direct Comparison Test since n n n n Èn 0 Ê n 3Èn "n , which is the nth
_
term of the divergent series ! "n or use Limit Comparison Test with bn œ n"
nœ1
#
19. converges by the Direct Comparison Test; sin2n n Ÿ #"n , which is the nth term of a convergent geometric series
n
20. converges by the Direct Comparison Test; 1 ncos
Ÿ n2# and the p-series ! n"# converges
#
2n
21. diverges since n lim
œ 23 Á 0
Ä _ 3n 1
"
22. converges by the Limit Comparison Test (part 1) with n$Î#
, the nth term of a convergent p-series:
lim
nÄ_
Š nn# È"n ‹
ˆ n n " ‰ œ 1
œ n lim
Ä_
" ‹
Š $Î#
n
23. converges by the Limit Comparison Test (part 1) with n"# , the nth term of a convergent p-series:
lim
nÄ_
Š n(n 10n1)(n" 2) ‹
Š n"# ‹
#
10n n
20n 1
20
œ n lim
œ n lim
œ n lim
œ 10
Ä _ n# 3n 2
Ä _ 2n 3
Ä_ 2
24. converges by the Limit Comparison Test (part 1) with n"# , the nth term of a convergent p-series:
lim
n# (n
5n$ 3n
2) Šn# 5‹ Š n"# ‹
nÄ_
$
#
5n 3n
15n 3
30n
œ n lim
œ n lim
œ n lim
œ5
Ä _ n$ 2n# 5n 10
Ä _ 3n# 4n 5
Ä _ 6n 4
n
n
n
n ‰
25. converges by the Direct Comparison Test; ˆ 3n n 1 ‰ ˆ 3n
œ ˆ "3 ‰ , the nth term of a convergent geometric series
"
26. converges by the Limit Comparison Test (part 1) with n$Î#
, the nth term of a convergent p-series:
lim
nÄ_
"
Š $Î#
‹
$
n
Š È $"
n
2
‹
É n n$ 2 œ lim É1 n2$ œ 1
œ n lim
Ä_
nÄ_
_
"
! " diverges
27. diverges by the Direct Comparison Test; n ln n Ê ln n ln ln n Ê "n ln"n ln (ln
n) and
n
n œ3
_
28. converges by the Limit Comparison Test (part 2) when compared with ! n"# , a convergent p-series:
n œ1
#
lim
nÄ_
’ (lnn$n) “
Š n"# ‹
#
(ln n)
œ n lim
œ n lim
n
Ä_
Ä_
2(ln n) Š n" ‹
1
ln n
œ 2 n lim
œ0
Ä_ n
29. diverges by the Limit Comparison Test (part 3) with "n , the nth term of the divergent harmonic series:
lim
nÄ_
’ Èn1ln n “
ˆ n" ‰
Èn
œ n lim
œ n lim
Ä _ ln n
Ä_
"
Š 2È
‹
n
ˆ n" ‰
œ n lim
Ä_
Èn
2 œ_
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 10.4 Comparison Tests
593
"
30. converges by the Limit Comparison Test (part 2) with n&Î%
, the nth term of a convergent p-series:
#
lim
nÄ_
n)
’ (ln$Î#
“
n
"
Š &Î%
‹
n
ˆ 2 lnn n ‰
#
(ln n)
œ n lim
œ n lim
Ä _ n"Î%
Ä_
"
Š $Î%
‹
4n
ln n
œ 8 n lim
œ 8 n lim
Ä _ n"Î%
Ä_
ˆ n" ‰
Š
"
‹
4n$Î%
"
œ 32 n lÄ
im_ n"Î%
œ 32 † 0 œ 0
31. diverges by the Limit Comparison Test (part 3) with "n , the nth term of the divergent harmonic series:
lim
nÄ_
ˆ 1 "ln n ‰
n
"
œ n lim
œ n lim
nœ_
ˆ "n ‰ œ n lim
Ä _ 1 ln n
Ä _ Š"‹
Ä_
n
_ ln (x 1)
32. diverges by the Integral Test: '2
x1
_
b
dx œ 'ln 3 u du œ lim "2 u# ‘ ln 3 œ lim
"
ab
bÄ_ #
bÄ_
#
ln# 3b œ _
"
33. converges by the Direct Comparison Test with n$Î#
, the nth term of a convergent p-series: n# 1 n for
"
"
n 2 Ê n# an# 1b n$ Ê nÈn# 1 n$Î# Ê $Î#
or use Limit Comparison Test with 1# .
nÈ n# 1
n
n
"
34. converges by the Direct Comparison Test with n$Î#
, the nth term of a convergent p-series: n# 1 n#
Èn
#
"
"
Ê n# 1 Ènn$Î# Ê nÈn1 n$Î# Ê n# 1 n$Î#
or use Limit Comparison Test with n$Î#
.
_
_
_
n œ1
n œ1
n œ1
35. converges because ! "n2nn œ ! n2" n ! "
#n which is the sum of two convergent series:
_
!
nœ1
_
"
"
"
! "n is a convergent geometric series
n2n converges by the Direct Comparison Test since n#n #n , and
2
nœ1
_
_
36. converges by the Direct Comparison Test: ! nn# 22n œ ! ˆ n2" n n"# ‰ and n2" n n"# Ÿ #"n n"# , the sum of
n
n œ1
nœ1
the nth terms of a convergent geometric series and a convergent p-series
37. converges by the Direct Comparison Test: 3nc1" 1 3n"c1 , which is the nth term of a convergent geometric series
nc1
ˆ " 3"n ‰ œ 3" Á 0
38. diverges; n lim
Š 3 3n" ‹ œ n lim
Ä_
Ä_ 3
_
n
39. converges by Limit Comparison Test: compare with ! ˆ 15 ‰ , which is a convergent geometric series with lrl œ 15 1,
n œ1
lim
nÄ_
1
1
Š n2n b
b 3n † 5n ‹
a 1 Î5 b n
n1
1
œ n lim
œ n lim
œ 0.
Ä _ n2 3n
Ä _ 2n 3
_
n
40. converges by Limit Comparison Test: compare with ! ˆ 34 ‰ , which is a convergent geometric series with lrl œ 15 1,
n œ1
3
Š 23n b
b 4n ‹
n
n
lim
n
n Ä _ a 3 Î4 b
8 ‰n
ˆ 12
1
8 12
œ n lim
lim
œ 11 œ 1 0.
n
n 12n œ
9
ˆ
Ä_
n Ä _ 129 ‰ 1
n
n
Š 2n 2cnn ‹
n
_
41. diverges by Limit Comparison Test: compare with ! 1n , which is a divergent p-series, n lim
Ä_
n œ1
†
1 În
2 n
œ n lim
Ä _ 2n
n
2n aln 2b2
2n ln 2 1
œ n lim
œ n lim
œ 1 0.
Ä _ 2n ln 2
Ä _ 2n aln 2b2
_
_
n œ1
n œ1
42. diverges by the definition of an infinite series: ! lnˆ n n 1 ‰ œ ! ln n ln an 1b‘, sk œ aln 1 ln 2b aln 2 ln 3b
Þ Þ Þ alnak 1b ln kb aln k ln ak 1bb œ ln ak 1b Ê lim sk œ _
kÄ_
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
594
Chapter 10 Infinite Sequences and Series
_
_
_
n œ2
n œ2
nœ2
43. converges by Comparison Test with ! nan 1 1b which converges since ! nan 1 1b œ ! ’ n 1 1 n1 “, and
sk œ ˆ1 12 ‰ ˆ 12 13 ‰ Þ Þ Þ ˆ k 1 2 k 1 1 ‰ ˆ k 1 1 k1 ‰ œ 1 k1 Ê lim sk œ 1; for n
Ê nan 1ban 2b!
nan 1b Ê n!
2, an 2b!
kÄ_
nan 1b Ê n!1 Ÿ nan11b
_
an
c 1bx
n 2x
44. converges by Limit Comparison Test: compare with ! n13 , which is a convergent p-series, n lim
Ä _ 1 În 3
a
b
n œ1
n a n 1 bx
n
2n
2
œ n lim
œ n lim
œ n lim
œ n lim
œ10
Ä _ an 2ban 1bnan 1bx
Ä _ n2 3n 2
Ä _ 2n 3
Ä_ 2
3
2
45. diverges by the Limit Comparison Test (part 1) with "n , the nth term of the divergent harmonic series:
lim
nÄ_
ˆsin "n ‰
sin x
œ1
ˆ "n ‰ œ xlim
Ä0 x
46. diverges by the Limit Comparison Test (part 1) with "n , the nth term of the divergent harmonic series:
lim
nÄ_
ˆtan "n ‰
ˆsin " ‰
Š " ‹ ˆ " ‰n œ lim ˆ cos" x ‰ ˆ sinx x ‰ œ 1 † 1 œ 1
ˆ "n ‰ œ n lim
Ä _ cos "
xÄ0
n
n
c"
1
_
_
1
47. converges by the Direct Comparison Test: tann1.1 n n#1.1 and ! n#1.1 œ 1# ! n"1.1 is the product of a
n œ1
nœ1
convergent p-series and a nonzero constant
c"
ˆ1‰
Þ
Þ
_
48. converges by the Direct Comparison Test: sec" n 1# Ê secn1 3 n n#1 3 and !
n œ1
_
ˆ 1# ‰
1 ! "
n1 3 œ #
n1 3 is the
Þ
Þ
n œ1
product of a convergent p-series and a nonzero constant
49. converges by the Limit Comparison Test (part 1) with n"# : n lim
Ä_
n
Š coth
‹
n#
Š n"# ‹
c2n
cn
e e
œ n lim
coth n œ n lim
Ä_
Ä _ en ecn
n
"e
œ n lim
œ1
Ä _ 1 ec2n
50. converges by the Limit Comparison Test (part 1) with n"# : n lim
Ä_
n
Š tanh
‹
n#
Š n"# ‹
c2n
e e
œ n lim
tanh n œ n lim
Ä_
Ä _ en e n
n
n
"e
œ n lim
œ1
Ä _ 1 ec2n
51. diverges by the Limit Comparison Test (part 1) with 1n : n lim
Ä_
52.
1
Š nÈ
n n‹
ˆ 1n ‰
converges by the Limit Comparison Test (part 1) with n"# : n lim
Ä_
1
œ n lim
n n œ 1.
Ä_ È
Èn n
Š n# ‹
Š n"# ‹
n
È
œ n lim
nœ1
Ä_
53. 1 2 3" á n œ ˆ n(n" 1) ‰ œ n(n 2 1) . The series converges by the Limit Comparison Test (part 1) with n"# :
#
lim
nÄ_
Š nan 2b 1b ‹
Š n"# ‹
#
2n
4n
4
œ n lim
œ n lim
œ n lim
œ 2.
Ä _ n# n
Ä _ 2n 1
Ä_ 2
"
6
6
54. 1 2# 3"# á n# œ n(n b 1)(2n
b 1) œ n(n 1)(2n 1) Ÿ n$ Ê the series converges by the Direct Comparison Test
6
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
1
Section 10.4 Comparison Tests
595
an
55. (a) If n lim
œ 0, then there exists an integer N such that for all n N, ¹ bann 0¹ 1 Ê 1 bann 1
Ä _ bn
Ê an bn . Thus, if ! bn converges, then ! an converges by the Direct Comparison Test.
an
(b) If n lim
œ _, then there exists an integer N such that for all n N, bann 1 Ê an bn . Thus, if
Ä _ bn
! bn diverges, then ! an diverges by the Direct Comparison Test.
_
56. Yes, ! ann converges by the Direct Comparison Test because ann an
n œ1
an
57. n lim
œ _ Ê there exists an integer N such that for all n N, bann 1 Ê an bn . If ! an converges,
Ä _ bn
then ! bn converges by the Direct Comparison Test
58. ! an converges Ê n lim
a œ 0 Ê there exists an integer N such that for all n N, 0 Ÿ an 1 Ê an# an
Ä_ n
Ê ! a#n converges by the Direct Comparison Test
59. Since an 0 and n lim
a œ _ Á 0, by nth term test for divergence, ! an diverges.
Ä_ n
an
60. Since an 0 and n lim
a n2 † an b œ 0, compare !an with ! n"# , which is a convergent p-series; n lim
Ä_
Ä _ 1 În 2
œ n lim
a n2 † an b œ 0 Ê !an converges by Limit Comparison Test
Ä_
_
q
_
61. Let _ q _ and p 1. If q œ 0, then ! alnnnp b œ ! n1p , which is a convergent p-series. If q Á 0, compare with
n œ2
_
!
n œ2
1
nr where 1 r p, then n lim
Ä_
aln nbq
np
1 În r
nœ2
aln nbq
aln nbq
œ n lim
, and p r 0. If q 0 Ê q 0 and n lim
Ä _ npcr
Ä _ npcr
qaln nbqc1 ˆ 1 ‰
q
qc1
qaln nb
aln nb
1
n
œ n lim
œ 0. If q 0, n lim
œ n lim
œ n lim
. If q 1 Ÿ 0 Ê 1 q
cq
Ä _ aln nb npcr
Ä _ npcr
Ä _ ap rbnpcrc1
Ä _ ap rbnpcr
lim
qaln nbqc1
n Ä _ ap rbnpcr
q
œ n lim
œ 0, otherwise, we apply L'Hopital's Rule again. n lim
Ä _ ap rbnpcr aln nb1cq
Ä_
qc2
qaq 1baln nb
œ n lim
. If q 2 Ÿ 0 Ê 2 q
Ä _ ap rb2 npcr
qc2
qaq 1baln nb
0 and n lim
Ä _ ap rb2 npcr
0 and
qaq 1baln nbqc2 ˆ 1n ‰
ap rb2 npcrc1
q aq 1 b
œ n lim
œ 0; otherwise, we
Ä _ ap rb2 npcr aln nb2cq
apply L'Hopital's Rule again. Since q is finite, there is a positive integer k such that q k Ÿ 0 Ê k q
0. Thus, after k
qaq 1bâaq k 1baln nbqck
q a q 1 bâa q k 1 b
applications of L'Hopital's Rule we obtain n lim
œ n lim
œ 0. Since the limit is
Ä_
Ä _ ap rbk npcr aln nbkcq
ap rbk npcr
_
q
0 in every case, by Limit Comparison Test, the series ! alnnnpb converges.
n œ1
_
q
_
62. Let _ q _ and p Ÿ 1. If q œ 0, then ! alnnnp b œ ! n1p , which is a divergent p-series. If q 0, compare with
n œ2
_
!
nœ2
1
np , which is a divergent p-series. Then n lim
Ä_
where 0 p r Ÿ 1. n lim
Ä_
lim
ar pbn
rcpc1
n Ä _ aqbaln nbcqc1 ˆ 1n ‰
aln nbq
np
1 În r
nœ2
aln nbq
np
1 În p
_
œ n lim
aln nbq œ _. If q 0 Ê q 0, compare with ! n1r ,
Ä_
nœ2
aln nbq
nrcp
œ n lim
œ n lim
cq since r p 0. Apply L'Hopital's to obtain
Ä _ npcr
Ä _ aln nb
rcp
ar pbn
œ n lim
. If q 1 Ÿ 0 Ê q 1
Ä _ aqbaln nbcqc1
rcp
ar pbn aln nb
0 and n lim
aqb
Ä_
2 rcpc1
qb1
œ _,
2 rcp
a r pb n
a r pb n
otherwise, we apply L'Hopital's Rule again to obtain n lim
œ n lim
. If
Ä _ aqbaq 1baln nbcqc2 ˆ 1 ‰
Ä _ aqbaq 1baln nbcqc2
n
q 2 Ÿ 0 Ê q 2
a r pb2 nrcp
a r pb2 nrcp aln nbqb2
0 and n lim
œ n lim
œ _, otherwise, we
aqbaq 1b
Ä _ aqbaq 1baln nbcqc2
Ä_
apply L'Hopital's Rule again. Since q is finite, there is a positive integer k such that q k Ÿ 0 Ê q k
k rcp
k rcp
0. Thus, after
qbk
a r pb n
a r pb n aln nb
k applications of L'Hopital's Rule we obtain n lim
œ n lim
œ _.
Ä _ aqbaq 1bâaq k 1baln nbcqck
Ä _ aqbaq 1bâaq k 1b
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
596
Chapter 10 Infinite Sequences and Series
_
q
Since the limit is _ if q 0 or if q 0 and p 1, by Limit comparison test, the series ! alnnpncbr diverges. Finally if q 0
_
q
_
_
q
nœ1
and p œ 1 then ! alnnnp b œ ! alnnnb . Compare with ! 1n , which is a divergent p-series. For n
nœ2
Ê aln nbq
_
nœ2
q
1 Ê alnnnb
1
n . Thus
3, ln n
1
nœ2
_
q
! aln nb diverges by Comparison Test. Thus, if _ q _ and p Ÿ 1,
n œ2
n
q
the series ! alnnpncbr diverges.
nœ1
63. Converges by Exercise 61 with q œ 3 and p œ 4.
64. Diverges by Exercise 62 with q œ 12 and p œ 12 .
65. Converges by Exercise 61 with q œ 1000 and p œ 1.001.
66. Diverges by Exercise 62 with q œ 15 and p œ 0.99.
67. Converges by Exercise 61 with q œ 3 and p œ 1.1.
68. Diverges by Exercise 62 with q œ 12 and p œ 12 .
69. Example CAS commands:
Maple:
a := n -> 1./n^3/sin(n)^2;
s := k -> sum( a(n), n=1..k );
# (a)]
limit( s(k), k=infinity );
pts := [seq( [k,s(k)], k=1..100 )]:
# (b)
plot( pts, style=point, title="#69(b) (Section 10.4)" );
pts := [seq( [k,s(k)], k=1..200 )]:
# (c)
plot( pts, style=point, title="#69(c) (Section 10.4)" );
pts := [seq( [k,s(k)], k=1..400 )]:
# (d)
plot( pts, style=point, title="#69(d) (Section 10.4)" );
evalf( 355/113 );
Mathematica:
Clear[a, n, s, k, p]
a[n_]:= 1 / ( n3 Sin[n]2 )
s[k_]= Sum[ a[n], {n, 1, k}]
points[p_]:= Table[{k, N[s[k]]}, {k, 1, p}]
points[100]
ListPlot[points[100]]
points[200]
ListPlot[points[200]
points[400]
ListPlot[points[400], PlotRange Ä All]
To investigate what is happening around k = 355, you could do the following.
N[355/113]
N[1 355/113]
Sin[355]//N
a[355]//N
N[s[354]]
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 10.5 The Ratio and Root Tests
597
N[s[355]]
N[s[356]]
_
_
70. (a) Let S œ ! n12 , which is a convergent p-series. By Example 5 in Section 10.2, ! nan 1 1b converges to 1. By Theorem 8,
n œ1
n œ1
_
_
_
_
n œ1
_
n œ1
n œ1
nœ1
S œ ! n12 œ ! nan 1 1b ! n12 ! nan 1 1b
_
_
nœ1
nœ1
œ ! nan 1 1b ! Š n12 nan 1 1b ‹ also converges.
_
_
n œ1
n œ1
(b) Since ! nan 1 1b converges to 1 (from Example 5 in Section 10.2), S œ 1 ! Š n12 nan 1 1b ‹ œ 1 ! n2 an1 1b
n œ1
_
(c)
_
Ä 0 faster than the terms of ! n12 .
The new series is comparible to ! n13 , so it will converge faster because its terms
n œ1
1000
n œ1
1000
(d) The series 1 ! n2 an1 1b gives a better approximation. Using Mathematica, 1 ! n2 an1 1b œ 1.644933568, while
nœ1
1000000
!
n œ1
nœ1
1
1
n2 œ 1.644933067. Note that 6
œ 1.644934067. The error is 4.99 ‚ 107 compared with 1 ‚ 106 .
2
10.5 THE RATIO AND ROOT TESTS
1.
2.
3.
2n
n! 0 for all n
2nb"
1; lim Œ n2n" ! œ
nÄ_
a
n
nÄ_
n!
an
b1b b 2
1; lim Œ 3nnbb21 œ
n2
3n 0 for all n
nÄ_
b1bc1b!
b2
b
b
anb1b2
_
nœ1
_
nÄ_
aan
a
n
n
3n
aa
nÄ_
3
3 ‰
n3 ‰
! n n 2 converges
ˆ1‰ 1
lim ˆ n3
œ lim ˆ 3n
n †3 † n 2
6 œ lim 3 œ 3 1 Ê
3
1; lim Œ nbnc1 1b!1 œ
nÄ_
an 1 b!
0 for all n
an 1b2
_
2 †2
n!
! 2 converges
ˆ 2 ‰
lim Š an"
b†n! † 2n ‹ œ lim n " œ 0 1 Ê
n!
b
nÄ_
nÄ_
2
nœ1
"b
n 2n n
3n 4n 1
lim Š na†nan21b2b! † aann 1b! ‹ œ lim Š n2 4n 4 ‹ œ lim Š 2n 4 ‹
3
nÄ_
2
2
nÄ_
nÄ_
1b!
œ lim ˆ 6n 2 4 ‰ œ _ 1 Ê ! aann diverges
1 b2
nÄ_
4.
n œ1
2nb1
n†3n 1 0 for all n
_
2an1b1
b an1b 1
1; lim n1 2†3n1
nÄ_
n 1
a
n†3
nb1
lim Š an21b†3†n2 1 †3 † n2†3n1 ‹ œ lim ˆ 3n2n 3 ‰ œ lim ˆ 23 ‰ œ 23 1
œ
n 1
nÄ_
nÄ_
nÄ_
nb1
Ê ! n2†3nc1 converges
nœ1
5.
n4
4n 0 for all n
b1b4
an
1; lim Œ 4nnb4 1 œ
nÄ_
n
4
4
lim Š an4n †14b † 4n4 ‹ œ lim Š n 4n 4n6n4 4n 1 ‹
n
nÄ_
4
3
2
nÄ_
_
4
œ lim ˆ 14 1n 2n3 2 n13 4n1 4 ‰ œ 14 1 Ê ! n4n converges
nÄ_
6.
n œ1
3nb2
ln n 0 for all n
3anb1bb2
2; lim Œ ln3nnbb21 œ
nÄ_
a
ln n
_
nb2
3
lim Š ln3an †31b † 3lnnbn2 ‹ œ lim Š ln 3anlnn1b ‹ œ lim Š n1 ‹ œ lim ˆ 3n n 3 ‰
b
nÄ_
nÄ_
nÄ_
nb1
nÄ_
nb2
œ lim ˆ 31 ‰ œ 3 1 Ê ! 3ln n diverges
nÄ_
7.
n œ2
n 2 a n 2 b!
0 for all n
nx32n
an
1; lim nÄ_
b 1b2aan b 1b b 2b!
b 1bx32an 1b
an
n2 an 2b!
nx32n
œ
2
3ban 2b!
lim Š an an1ba1nb
† n2 annx3 2b! ‹ œ lim Š n 9n5n3 9n7n2 3 ‹
†nx32n †32
2n
nÄ_
3
nÄ_
_
7
! n an 2n2b! converges
ˆ 6n 15 ‰
ˆ6‰ 1
œ lim Š 3n27n2 15n
18n ‹ œ lim 54n 18 œ lim 54 œ 9 1 Ê
n x3
2
nÄ_
nÄ_
nÄ_
2
n œ1
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
2
598
8.
Chapter 10 Infinite Sequences and Series
an
n †5 n
a2n 3b lnan 1b 0 for all n
a a
nÄ_
1b†a2n 3b
lim Š 5an † lnlnaann 21bb ‹ œ
na2n 5b
œ
b
lim Π2 n 1
1;
b 1b†5n 1
3b lnaan 1b
n†5n
3b lnan 1b
a2n
a2n 3b lnan 1b
n 1b†5 †5
lim Š a2na
‹
n †5 n
5b lnan 2b †
n
œ
1b
nÄ_
1
lnan 1b
15
25 ‰
lim Š 10n 2n2 25n
† lim Š n b1 1 ‹
‹ † lim Š lnan 2b ‹ œ lim ˆ 20n
5n
4n 5
2
nÄ_
nÄ_
nÄ_
_
nÄ_
nÄ_
nb2
n †5
!
‰
ˆ n2‰
ˆ 1‰
œ lim ˆ 20
4 † lim n 1 œ 5 † lim 1 œ 5 † 1 œ 5 1 Ê
a2n 3b lnan 1b diverges
nÄ_
9.
7
a2n 5bn
nÄ_
nÄ_
n
10. a3n4 bn
0 for all n
3‰
11. ˆ 4n
3n 5
n
n 1
_
nÄ_
n œ1
_
4
n
1; lim É
œ
a3n bn
nÄ_
nÄ_
n œ1
n
n
3‰
ˆ 4n
2; lim É
œ
3n 5
n 1
n
nÄ_
diverges
8
2n
3 1n ‰
n
1; lim É
ˆ
n
3‰
lim ˆ 43 ‰ œ 43 1 Ê ! ˆ 4n
diverges
3n 5
nÄ_
1; lim Ê’lnˆe2 1n ‰“
Ê ! ’lnˆe2 1n ‰“
_
3‰
lim ˆ 4n
3n 5 œ
nÄ_
0 for all n
n
4 ‰
lim ˆ 3n
œ 0 1 Ê ! a3n4 bn converges
n
n 1
_
n
È
lim Š 2n 7 5 ‹ œ 0 1 Ê ! a2n 7 5bn converges
nÄ_
0 for all n
12. ’lnˆe2 1n ‰“
n œ2
7
n
1; lim É
œ
a2n 5bn
0 for all n
n
nÄ_
n œ1
1 1 În
œ
lim ’lnˆe2 1n ‰“
nÄ_
œ lnae2 b œ 2 1
n œ1
13. ˆ
0 for all n
nÄ_
n
14. ’sinŠ È1n ‹“
n2
15. ˆ1 n1 ‰
16. n1"bn
8
2n œ
3 1n ‰
lim Œ ˆ
nÄ_
n
È
8
0 for all n
17. converges by the Ratio Test:
n
nÄ_
n œ1
_
n
n2
lim ˆ1 n1 ‰ œ e1 1 Ê ! ˆ1 n1 ‰ converges
nÄ_
"
n
2; lim É
n1bn œ
nÄ_
n2
lim Ɉ1 n1 ‰ œ
n
n
lim sinŠ È1n ‹ œ sina0b œ 0 1 Ê ! ’sinŠ È1n ‹“ converges
nÄ_
1;
8
n œ1
_
n
0 for all n
1
n
n
1; lim Ê
’sinŠ È1n ‹“ œ
0 for all n
_
œ 9 1 Ê ! ˆ3 1 ‰2n converges
3 1 ‰
2
nÄ_
nœ1
n
È
nÄ_
”
lim anb1 œ n lim
n Ä _ an
Ä_
_
n
È
1
! 1"bn converges
lim Š n È
n n‹ œ 0 1 Ê
n
lim Š n1În 1 1 ‹ œ
nÄ_
È
(n b 1) 2
2 n b1 •
È
n 2
n œ2
È
È
n
(n 1) 2
ˆ1 n" ‰ 2 ˆ "# ‰ œ "# 1
œ n lim
† 2È2 œ n lim
Ä _ #nb1
Ä_
n
” #n •
Š (nenbb1)1 ‹
2
18. converges by the Ratio Test:
lim anb1 œ n lim
n Ä _ an
Ä_
anb1
19. diverges by the Ratio Test: n lim
œ n lim
Ä _ an
Ä_
anb1
20. diverges by the Ratio Test: n lim
œ n lim
Ä _ an
Ä_
#
Š nen ‹
Š (nenbb1)!
1 ‹
ˆ en!n ‰
b 1)! ‹
Š (n
10nb1
ˆ 10n!n ‰
"!
21. converges by the Ratio Test:
lim anb1 œ n lim
n Ä _ an
Ä_
"!
2
n
(n ")! e
n"
œ n lim
† n! œ n lim
e œ_
Ä _ enb1
Ä_
n
(n ")! 10
n
œ n lim
† n! œ n lim
œ_
Ä _ 10nb1
Ä _ 10
Š (n10bn1)1 ‹
Š n10n ‹
#
(n 1)
œ n lim
† ne2 œ n Ä
lim ˆ1 n" ‰ ˆ "e ‰ œ "e 1
_
Ä _ enb1
n
"!
(n ")
"
ˆ1 n" ‰"! ˆ 1"0 ‰ œ 10
œ n lim
† 10
1
n"! œ n lim
Ä _ 10n 1
Ä_
n
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 10.5 The Ratio and Root Tests
n
599
n
ˆ nn 2 ‰ œ lim ˆ1 n2 ‰ œ e# Á 0
22. diverges; n lim
a œ n lim
Ä_ n
Ä_
nÄ_
(1)
ˆ 45 ‰ c2 (1)n d Ÿ ˆ 45 ‰ (3) which is the nth term of a convergent
23. converges by the Direct Comparison Test: 2(1.25)
n œ
n
n
n
geometric series
24. converges; a geometric series with krk œ ¸ 23 ¸ 1
n
n
ˆ1 3n ‰ œ lim ˆ1 n3 ‰ œ e$ ¸ 0.05 Á 0
25. diverges; n lim
a œ n lim
Ä_ n
Ä_
nÄ_
" ‰n
ˆ1 3n
26. diverges; n lim
a œ n lim
œ n lim
1
Ä_ n
Ä_
Ä_ Š "3 ‹
n
n
"Î$
œe
27. converges by the Direct Comparison Test: lnn$n nn$ œ n"# for n
¸ 0.72 Á 0
2, the nth term of a convergent p-series.
n 1În
n
a(ln n) b
n (ln n)
n
È
É
28. converges by the nth-Root Test: n lim
an œ n lim
n 1În
nn œ n lim
Ä_
Ä_
Ä_
an b
ln n
œ n lim
œ n lim
Ä_ n
Ä_
Š "n ‹
1
œ01
29. diverges by the Direct Comparison Test: "n n"# œ n n# 1 "# ˆ "n ‰ for n 2 or by the Limit Comparison Test (part 1)
with "n .
n 1În
n
n
n
ˆ n" n"# ‰ œ lim ˆˆ n" n"# ‰ ‰
È
É
30. converges by the nth-Root Test: n lim
an œ n lim
Ä_
Ä_
nÄ_
31. diverges by the Direct Comparison Test: lnnn n" for n
ˆ " n"# ‰ œ 0 1
œ n lim
Ä_ n
3
(n 1) ln (n 1)
anb1
32. converges by the Ratio Test: n lim
œ n lim
† n ln2 (n) œ "# 1
#nb1
Ä _ an
Ä_
n
(n 2)(n 3)
anb1
n!
33. converges by the Ratio Test: n lim
œ n lim
† (n 1)(n
(n 1)!
2) œ 0 1
Ä _ an
Ä_
$
(n 1)
anb1
34. converges by the Ratio Test: n lim
œ n lim
† ne$ œ "e 1
Ä _ an
Ä _ en 1
n
(n 4)!
anb1
n4
35. converges by the Ratio Test: n lim
œ n lim
† 3! n! 3 œ n lim
œ 3" 1
Ä _ an
Ä _ 3! (n 1)! 3nb1 (n 3)!
Ä _ 3(n 1)
n
nb1
(n 1)2 (n 2)!
anb1
2
2‰
ˆ n n 1 ‰ ˆ 32 ‰ ˆ nn 36. converges by the Ratio Test: n lim
œ n lim
† n2n3(n n!
3nb1 (n 1)!
1 œ 3 1
1)! œ n lim
Ä _ an
Ä_
Ä_
n
(n 1)!
1)!
anb1
n"
37. converges by the Ratio Test: n lim
œ n lim
† (2n n!
œ n lim
œ01
Ä _ an
Ä _ (2n 3)!
Ä _ (2n 3)(2n 2)
(n 1)!
anb1
"
ˆ n ‰ œ lim
38. converges by the Ratio Test: n lim
œ n lim
† n œ n lim
Ä _ an
Ä _ (n 1)nb1 n!
Ä _ n1
n Ä _ ˆ n b " ‰n
n
n
n
œ n lim
Ä_ ˆ
"
n
1 "n ‰
œ "e 1
n n
È
"
n
n
n
È
39. converges by the Root Test: n lim
an œ n lim
œ n lim
œ n lim
œ01
Ä_
Ä _ É (ln n)n
Ä _ ln n
Ä _ ln n
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
600
Chapter 10 Infinite Sequences and Series
n n
È
n
n
n
È
40. converges by the Root Test: n lim
an œ n lim
œ n lim
Ä_
Ä _ É (ln n)nÎ2
Ä_ È
ln n
Ä_
Ä_
n n
lim È
n
Èln n œ 0 1
lim
n
œ
n
È
n œ 1‹
Šn lim
Ä_
n! ln n
ln n
n
"
"
41. converges by the Direct Comparison Test: n(n
2)! œ n(n 1)(n 2) n(n 1)(n 2) œ (n 1)(n #) n#
which is the nth-term of a convergent p-series
$ n
n 1
3
an 1
3
n
ˆ 3 ‰ œ 3# 1
42. diverges by the Ratio Test: n lim
œ n lim
† n32n œ n lim
Ä _ an
Ä _ (n 1)$ 2n 1
Ä _ (n 1)3 #
an1bx‘2
2
an1b
anb1
n 2n 1
43. converges by the Ratio Test: n lim
œ n lim
† a2nbx œ n lim
œ n lim
œ 41 1
Ä _ an
Ä _ 2(n 1)‘x ‘2
Ä _ (2n 2)(2n 1)
Ä _ 4n2 6n 2
2
nx
anb1
44. converges by the Ratio Test: n lim
œ n lim
Ä _ an
Ä_
n
n
n
n
a2n 5bˆ2nb1 3‰
† a2n 33ba22n 3b œ n lim
’ 2n 5 † 2†6 4†2 3†3 6 “
3nb1 2
Ä _ 2n 3 3†6n 9†3n 2†2n 6
œ n lim
’ 2n 5 “ † n lim
’ 2†6 4†2 3†3 6 “ œ 1 † 23 œ 23 1
Ä _ 2n 3
Ä _ 3 † 6 n 9 † 3 n 2† 2 n 6
n
n
n
anb1
45. converges by the Ratio Test: n lim
œ n lim
Ä _ an
Ä_
anb1
46. converges by the Ratio Test: n lim
œ n lim
Ä _ an
Ä_
ˆ 1 b nsin n ‰ an
œ01
an
"n
Š 1 b tan
n
‹ an
an
approaches 1 1# while the denominator tends to _
anb1
47. diverges by the Ratio Test: n lim
œ n lim
Ä _ an
Ä_
" tan
œ n lim
n
Ä_
"n
œ 0 since the numerator
c1‰
ˆ 3n
3n 1
2n b 5 an
œ n lim
œ 3# 1
an
Ä _ 2n 5
2
‰
48. diverges; an1 œ n n 1 an Ê an1 œ ˆ n n 1 ‰ ˆ n n 1 an1 ‰ Ê an1 œ ˆ n n 1 ‰ ˆ n n 1 ‰ ˆ nn 1 an2
a
"
n
n
1
n
2
3
"
Ê an1 œ ˆ n 1 ‰ ˆ n ‰ ˆ n 1 ‰ â ˆ # ‰ a" Ê an1 œ n 1 Ê an1 œ n 1 , which is a constant times the
general term of the diverging harmonic series
anb1
49. converges by the Ratio Test: n lim
œ n lim
Ä _ an
Ä_
50. converges by the Ratio Test:
lim anb1 œ n lim
n Ä _ an
Ä_
anb1
51. converges by the Ratio Test: n lim
œ n lim
Ä _ an
Ä_
Š 2n ‹ an
an
2
œ n lim
œ01
Ä_ n
Èn n
Π# an
an
œ n lim
Ä_
Š 1 bnln n ‹ an
an
n n
È
n
œ "# 1
"ln n
"
œ n lim
œ n lim
œ01
n
Ä_
Ä_ n
52. nnln10n 0 and a" œ #" Ê an 0; ln n 10 for n e"! Ê n ln n n 10 Ê nnln10n 1
Ê an1 œ nnln10n an an ; thus an1 an
"
#
Ê n lim
a Á 0, so the series diverges by the nth-Term Test
Ä_ n
3
3
6 "
%! "
2 "
2 "
2 "
É
É
É
É
53. diverges by the nth-Term Test: a" œ "3 , a# œ É
3 , a$ œ Ê
3 œ
3 , a% œ ËÊ
3 œ
3 ,á ,
%
n! "
n! "
n "
É
an œ É
a œ 1 because šÉ
3 Ê n lim
3 › is a subsequence of š
3 › whose limit is 1 by Table 8.1
Ä_ n
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 10.5 The Ratio and Root Tests
# $
#
' %
'
#%
54. converges by the Direct Comparison Test: a" œ "# , a# œ ˆ "# ‰ , a$ œ Šˆ "# ‰ ‹ œ ˆ "# ‰ , a% œ Šˆ "# ‰ ‹ œ ˆ "# ‰ , á
n!
n
Ê an œ ˆ "# ‰ ˆ "# ‰ which is the nth-term of a convergent geometric series
2
anb1
55. converges by the Ratio Test: n lim
œ n lim
Ä _ an
Ä_
nb1
n"
œ n lim
œ #" 1
Ä _ 2n 1
(n 1)! (n 1)!
† 2(2n)!
lim 2(n 1)(n 1)
n n! n! œ
(2n 2)!
n Ä _ (2n #)(2n 1)
(3n 3)!
1)! (n 2)!
anb1
56. diverges by the Ratio Test: n lim
œ n lim
† n! (n (3n)!
Ä _ an
Ä _ (n 1)! (n 2)! (n 3)!
(3n 3)(3 2)(3n 1)
2 ‰ ˆ 3n 1 ‰
œ n lim
œ n lim
3 ˆ 3n
n#
n 3 œ 3 † 3 † 3 œ 27 1
Ä _ (n 1)(n 2)(n 3)
Ä_
n
n!
n (n!)
n
È
57. diverges by the Root Test: n lim
an ´ n lim
œ n lim
œ_1
Ä_
Ä _ É an n b#
Ä _ n#
n!
n (n!)
n (n!)
ˆ " ‰ ˆ 2n ‰ ˆ 3n ‰ â ˆ n n 1 ‰ ˆ nn ‰
É
58. converges by the Root Test: n lim
œ n lim
œ n lim
œ n lim
É
ann bn
Ä_
Ä_
Ä _ nn
Ä_ n
nn#
n
n
"
Ÿ n lim
œ01
Ä_ n
"
n
n n
n
È
59. converges by the Root Test: n lim
an œ n lim
œ n lim
œ n lim
œ01
Ä_
Ä _ É 2 n#
Ä _ #n
Ä _ #n ln 2
n
n
n
n
n
È
60. diverges by the Root Test: n lim
an œ n lim
œ n lim
œ_1
Ä_
Ä _ É a#n b#
Ä_ 4
n
1†3† â †(2n 1)(2n 1)
anb1
4 2 n!
2n "
61. converges by the Ratio Test: n lim
œ n lim
† 1 †3 † â
œ 4" 1
4nb1 2nb1 (n 1)!
†(2n 1) œ n lim
Ä _ an
Ä_
Ä _ (4†#)(n 1)
n
n
(2n 1)
1†2†3†4 â (2n 1)(2n)
(2n)!
62. converges by the Ratio Test: an œ (2†14†3ââ#n)
a3n 1b œ (2†4 â 2n)# a3n 1b œ a2n n!b# a3n 1b
#
a3 1 b
(2n ")(2n 2) a3 1b
(2n 2)!
Ê n lim
† a2 n!b(2n)!
œ n lim
2# (n 1)# a3n 1 1b
Ä _ c2nb1 (n 1)!d# a3nb1 1b
Ä_
n
n
n
cn
œ n lim
Š 4n 6n 2 ‹ a1 3 b œ 1 † 3" œ 3" 1
Ä _ 4n# 8n 4 a3 3cn b
#
anb1
"
ˆ n ‰ œ 1p œ 1 Ê no conclusion
63. Ratio: n lim
œ n lim
† n œ n lim
Ä _ an
Ä _ (n 1)p 1
Ä_ n1
p
p
"
"
n "
n
È
É
Root: n lim
an œ n lim
n n‰p œ (1)p œ 1 Ê no conclusion
np œ n lim
Ä_
Ä_
Ä _ ˆÈ
p
ˆ"‰
p
anb1
ln n
n"
"
n
64. Ratio: n lim
œ n lim
† (ln1n) œ ’n lim
œ Šn lim
“ œ ”n lim
‹
Ä _ an
Ä _ (ln (n 1))p
Ä _ ln (n 1)
Ä _ ˆ n b" 1 ‰ •
Ä_ n
p
œ (1)p œ 1 Ê no conclusion
"
n
n
È
Root: n lim
an œ n lim
É
(ln n)p œ
Ä_
Ä_
"
p
lim (ln n)1În ‹
ŠnÄ_
ln (ln n)
Ê n lim
ln f(n) œ n lim
œ n lim
n
Ä_
Ä_
Ä_
ˆ n ln" n ‰
1
n
È
œ n lim
eln fÐnÑ œ e! œ 1; therefore n lim
an œ
Ä_
Ä_
n)
; let f(n) œ (ln n)1În , then ln f(n) œ ln (ln
n
"
œ n lim
œ 0 Ê n lim
(ln n)1În
Ä _ n ln n
Ä_
"
p
lim (ln n)1În ‹
ŠnÄ_
_
œ (1)" p œ 1 Ê no conclusion
(n ") 2
65. an Ÿ 2nn for every n and the series ! #nn converges by the Ratio Test since n lim
† n œ "# 1
Ä _ 2nb1
_
n
nœ1
Ê ! an converges by the Direct Comparison Test
nœ1
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
p
601
602
Chapter 10 Infinite Sequences and Series
2
2anb1b
n2
n2 b2nb1
1; lim n2bn21 ! œ
nÄ_
66. 2n! 0 for all n
a
nÄ_
n!
_
2nb1
†4 ‰
ˆ 2†4 1ln 4 ‰
lim Š a2n1b†n! † 2n!n2 ‹ œ lim Š 2n1 ‹ œ lim ˆ n2
1 œ lim
b
n
nÄ_
n
nÄ_
nÄ_
n2
œ _ 1 Ê ! 2n! diverges
n œ1
10.6 ALTERNATING SERIES, ABSOLUTE AND CONDITIONAL CONVERGENCE
1.
converges by the Alternating Convergence Test since: un œ È1n 0 for all n
Ê Èn11 Ÿ È1n Ê un1 Ÿ un ;
lim un œ
nÄ_
1; n
1Ê n1
_
_
n œ1
n œ1
n Ê Èn 1
Èn
lim 1 œ 0.
nÄ_ Èn
"
2. converges absolutely Ê converges by the Alternating Convergence Test since ! kan k œ ! n$Î#
which is a
convergent p-series
3. converges Ê converges by Alternating Series Test since: un œ n31 n 0 for all n
Ê an 1b3n1
n 3n Ê an11b3nb1 Ÿ n13n Ê un1 Ÿ un ;
lim un œ
lim un œ
nÄ_
ln n Ê aln an 1bb2
2; n
3n
2Ên1
n
aln nb2 Ê aln an11bb2 Ÿ aln1nb2 Ê aln an41bb2 Ÿ aln4nb2 Ê un1 Ÿ un ;
lim 4 2 œ 0.
nÄ_ aln nb
5. converges Ê converges by Alternating Series Test since: un œ n2 n 1 0 for all n
Ê n3 2n2 2n
Ê n2n1
n Ê 3n 1
lim 1 n œ 0.
nÄ_ n 3
nÄ_
4. converges Ê converges by Alternating Series Test since: un œ aln4nb2 0 for all n
Ê ln an 1b
1Ên1
1; n
n3 n2 n 1 Ê nan2 2n 2b
n 1
Ê un1 Ÿ un ;
an1b2 1
lim un œ
nÄ_
1 Ê 2n2 2n
1; n
n3 n2 n 1 Ê nŠan 1b2 1‹
n2 n 1
an2 1ban 1b
lim 2 n œ 0.
nÄ_ n 1
6. diverges Ê diverges by nth Term Test for Divergence since:
2
lim n2 5 œ 1 Ê
nÄ_ n 4
7. diverges Ê diverges by nth Term Test for Divergence since:
lim 2n2 œ _ Ê
n
nÄ_
5
lim a1bn1 nn2 4 œ does not exist
2
nÄ_
lim a1bn1 2n2 œ does not exist
n
nÄ_
_
_
n œ1
n œ1
n
8. converges absolutely Ê converges by the Absolute Convergence Test since ! kan k œ ! an10
1bx , which converges by the
Ratio Test, since lim anabn 1 œ
nÄ_
lim n 10
2 œ 0 1
nÄ_
_
n
n
n
n ‰
ˆ n ‰ Á 0 Ê ! (1)n1 ˆ 10
9. diverges by the nth-Term Test since for n 10 Ê 10
1 Ê n lim
diverges
Ä _ 10
n œ1
10. converges by the Alternating Series Test because f(x) œ ln x is an increasing function of x Ê ln"x is decreasing
Ê un
un1 for n
1; also un
0 for n
"
1 and n lim
œ0
Ä _ ln n
11. converges by the Alternating Series Test since f(x) œ lnxx Ê f w (x) œ 1 x#ln x 0 when x e Ê f(x) is decreasing
Ê un
un1 ; also un
0 for n
ln n
1 and n lim
u œ n lim
œ n lim
Ä_ n
Ä_ n
Ä_
Š "n ‹
1
œ0
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 10.6 Alternating Series, Absolute and Conditional Convergence
603
12. converges by the Alternating Series Test since f(x) œ ln a1 x" b Ê f w (x) œ x(x"
1) 0 for x 0 Ê f(x) is decreasing
Ê un
un1 ; also un
0 for n
ˆ1 n" ‰‹ œ ln 1 œ 0
1 and n lim
u œ n lim
ln ˆ1 "n ‰ œ ln Šn lim
Ä_ n
Ä_
Ä_
13. converges by the Alternating Series Test since f(x) œ
Ê un
unb1 ; also un
0 for n
Èx "
x1
1 and n lim
u œ n lim
Ä_ n
Ä_
È
3 n1
14. diverges by the nth-Term Test since n lim
œ n lim
Ä _ Èn 1
Ä_
_
_
nœ1
nœ1
1 x 2È x
Ê f w (x) œ 2Èx (x 1)# 0 Ê f(x) is decreasing
Èn "
n1 œ 0
3É1 "n
1 Š È"n ‹
œ3Á0
n
" ‰
15. converges absolutely since ! kan k œ ! ˆ 10
a convergent geometric series
nb1
" ‰
16. converges absolutely by the Direct Comparison Test since ¹ (1) n (0.1) ¹ œ (10)" n n ˆ 10
which is the nth term
n
n
of a convergent geometric series
_
_
n œ1
n œ1
"
"
17. converges conditionally since È"n Èn" 1 0 and n lim
œ 0 Ê convergence; but ! kan k œ ! n"Î#
Ä _ Èn
is a divergent p-series
"
18. converges conditionally since 1 "Èn 1 È"n 1 0 and n lim
œ 0 Ê convergence; but
Ä _ 1 Èn
_
_
nœ1
nœ1
! kan k œ !
"
is a divergent series since 1"Èn
1 Èn
_
_
n œ1
nœ1
_
"
"
and ! n"Î#
is a divergent p-series
#È n
nœ1
19. converges absolutely since ! kan k œ ! n$n1 and n$n1 n"# which is the nth-term of a converging p-series
n!
20. diverges by the nth-Term Test since n lim
œ_
Ä _ #n
_
"
"
21. converges conditionally since n " 3 (n 1)
œ 0 Ê convergence; but ! kan k
3 0 and n lim
Ä _ n 3
n œ1
_
œ!
n œ1
"
"
n 3 diverges because n 3
_
"
! " is a divergent series
4n and
n
n œ1
_
22. converges absolutely because the series ! ¸ sinn# n ¸ converges by the Direct Comparison Test since ¸ sinn# n ¸ Ÿ n"#
n œ1
3n
23. diverges by the nth-Term Test since n lim
œ1Á0
Ä _ 5n
nb1
nb1
24. converges absolutely by the Direct Comparison Test since ¹ (n2)5n ¹ œ n25n 2 ˆ 25 ‰ which is the nth term
n
of a convergent geometric series
25. converges conditionally since f(x) œ x"# "x Ê f w (x) œ ˆ x2$ x"# ‰ 0 Ê f(x) is decreasing and hence
un unb1 0 for n
_
_
n œ1
nœ1
_
_
n œ1
n œ1
ˆ " "n ‰ œ 0 Ê convergence; but ! kan k œ ! 1n#n
1 and n lim
Ä _ n#
œ ! n"# ! "n is the sum of a convergent and divergent series, and hence diverges
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
604
Chapter 10 Infinite Sequences and Series
26. diverges by the nth-Term Test since n lim
a œ n lim
101În œ 1 Á 0
Ä_ n
Ä_
27. converges absolutely by the Ratio Test: n lim
Š uunbn 1 ‹ œ n lim
Ä_
Ä_ ”
(n")# ˆ 23 ‰
n
n# ˆ 23 ‰
n 1
•œ 3 1
2
1d
28. converges conditionally since f(x) œ x ln" x Ê f w (x) œ cln(x(x)
ln x)# 0 Ê f(x) is decreasing
"
2 and n lim
œ 0 Ê convergence; but by the Integral Test,
Ä _ n ln n
Ê un unb1 0 for n
'2_ x dxln x œ lim '2b Šln x‹ dx œ lim cln (ln x)d b2 œ lim cln (ln b) ln (ln 2)d œ _
"
x
bÄ_
_
_
n œ1
n œ1
bÄ_
bÄ_
Ê ! kan k œ ! n ln" n diverges
_
#
29. converges absolutely by the Integral Test since '1 atan" xb ˆ 1 " x# ‰ dx œ lim ’ atan # xb “
"
bÄ_
œ lim ’atan
bÄ_
"
#
"
bb atan
œ
(x ln x)#
œ n lim
Ä_
Š "n ‹
1 Š n" ‹
_
_
n œ1
nœ1
! kan k œ !
1
#
#
1#
1b “ œ "# ’ˆ 1# ‰ ˆ 14 ‰ “ œ 332
#
30. converges conditionally since f(x) œ x lnlnx x Ê f w (x) œ
1 Š lnxx ‹ ln x Š lnxx ‹
b
œ (x1lnln x)x # 0 Ê un
Š "x ‹ (x ln x) (ln x) Š1 x" ‹
(x ln x)#
ln n
un1 0 when n e and n lim
Ä _ n ln n
œ 0 Ê convergence; but n ln n n Ê n"ln n "n Ê nlnlnn n "n so that
ln n
n ln n diverges by the Direct Comparison Test
n
31. diverges by the nth-Term Test since n lim
œ1Á0
Ä _ n1
_
_
n œ1
nœ1
n
32. converges absolutely since ! kan k œ ! ˆ "5 ‰ is a convergent geometric series
nb1
("00)
n!
33. converges absolutely by the Ratio Test: n lim
† (100)
lim "00 œ 0 1
Š uunbn 1 ‹ œ n lim
n œ
Ä_
Ä _ (n1)!
n Ä _ n1
_
_
n œ1
n œ1
"
"
"
34. converges absolutely by the Direct Comparison Test since ! kan k œ ! n# 2n
1 and n# 2n 1 n# which is the
nth-term of a convergent p-series
_
_
n œ1
n œ1
_
"
35. converges absolutely since ! kan k œ ! ¹ (nÈ1)n ¹ œ ! n$Î#
is a convergent p-series
n
n œ1
_
_
n œ1
nœ1
36. converges conditionally since ! cosnn1 œ ! (n1) is the convergent alternating harmonic series, but
_
_
n œ1
n œ1
! kan k œ !
n
"
n diverges
1)
n
È
kan k œ n lim
37. converges absolutely by the Root Test: n lim
Š (n(2n)
n ‹
Ä_
Ä_
n
1 În
n"
œ n lim
œ "# 1
Ä _ #n
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 10.6 Alternating Series, Absolute and Conditional Convergence
#
605
#
a(n 1)!b
(n 1)
38. converges absolutely by the Ratio Test: n lim
† (2n)! œ n lim
œ "4 1
¹ anb1 ¹ œ n lim
Ä _ an
Ä _ ((2n 2)!) (n!)#
Ä _ (2n 2)(2n 1)
(2n)!
(n ")(n 2)â(2n)
39. diverges by the nth-Term Test since n lim
kan k œ n lim
œ n lim
2n n
Ä_
Ä _ 2n n! n
Ä_
(n 1)(n 2)â(n (n 1))
ˆ n # 1 ‰
œ n lim
n lim
#nc1
Ä_
Ä_
n 1
œ_Á0
(n 1)! (n 1)! 3
40. converges absolutely by the Ratio Test: n lim
¹ anabn 1 ¹ œ n lim
(2n 3)!
Ä_
Ä_
nb1
1)!
† (2n
n! n! 3n
#
(n 1) 3
œ n lim
œ 43 1
Ä _ (2n 2)(2n 3)
Èn 1 Èn Èn 1 Èn
† Èn 1 Èn œ Èn "1 Èn and š Èn 1" Èn › is a
1
41. converges conditionally since
_
(")
decreasing sequence of positive terms which converges to 0 Ê ! Èn converges; but
1 Èn
n
n œ1
_
_
n œ1
nœ1
"
Èn 1 Èn
! kan k œ !
diverges by the Limit Comparison Test (part 1) with È"n ; a divergent p-series:
"
lim
nÄ_ Èn 1 Èn
œ lim
"
Èn n Ä _
Èn
Èn 1 Èn œ n lim
Ä_
1
œ "#
É1 1n 1
È #
n n
42. diverges by the nth-Term Test since n lim
ŠÈn# n n‹ œ n lim
ŠÈn# n n‹ † Š Ènn# ‹
Ä_
Ä_
n n
n
œ n lim
œ n lim
Ä _ È n# n n
Ä_
"
œ "# Á 0
É1 "n 1
43. diverges by the nth-Term Test since n lim
ŠÉn Èn Èn‹ œ n lim
ŠÉ n È n È n ‹ Ä_
Ä_ –
œ n lim
Ä_
Èn
É n Èn Èn
œ n lim
Ä_
É n Èn Èn
Én Èn Èn —
"
œ #" Á 0
"
É 1 Èn 1
44. converges conditionally since š Èn "Èn 1 › is a decreasing sequence of positive terms converging to 0
_
(")
Ê ! Èn Èn 1 converges; but n lim
Ä_
n
n œ1
_
"
Èn 1 ‹
Èn
"
œ n lim
œ n lim
œ "#
Ä _ È n È n 1
Ä _ 1 É 1 "
Š È"n ‹
n
Š Èn
_
so that ! Èn "Èn 1 diverges by the Limit Comparison Test with ! È"n which is a divergent p-series
nœ1
nœ1
n
n
2
45. converges absolutely by the Direct Comparison Test since sech (n) œ en 2ecn œ e2n2e 1 2e
e2n œ en which is the
nth term of a convergent geometric series
_
_
n œ1
nœ1
46. converges absolutely by the Limit Comparison Test (part 1): ! kan k œ ! en 2ecn
Apply the Limit Comparison Test with e1n , the n-th term of a convergent geometric series:
2
cn
2e
2
e ce
lim
œ n lim
œ n lim
œ2
n Ä _ Œ 1n Ä _ en ecn
Ä _ 1 ec2n
n
n
e
_
nb1
)
1
1
1
1
47. 14 16 18 10
12
14
Þ Þ Þ œ ! 2("
an 1b ; converges by Alternating Series Test since: un œ 2an 1b 0 for all n
n œ1
n2
n 1 Ê 2 an 2 b
2an 1b Ê 2aan 11b 1b Ÿ 2an11b Ê un1 Ÿ un ;
lim un œ
nÄ_
lim 1 œ 0.
nÄ_ 2an1b
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
1;
606
Chapter 10 Infinite Sequences and Series
_
_
_
n œ1
n œ1
n œ1
1
1
1
1
1
48. 1 14 91 16
25
36
49
64
Þ Þ Þ œ ! an ; converges by the Absolute Convergence Test since ! kan k œ ! n"#
which is a convergent p-series
49. kerrork ¸(1)' ˆ "5 ‰¸ œ 0.2
50. kerrork ¸(1)' ˆ 10" & ‰¸ œ 0.00001
&
""
51. kerrork ¹(1)' (0.01)
5 ¹ œ 2 ‚ 10
52. kerrork k(1)% t% k œ t% 1
53. kerrork 0.001 Ê un1 0.001 Ê an 11b2 3 0.001 Ê an 1b2 3 1000 Ê n 1 È997 ¸ 30.5753 Ê n
54. kerrork 0.001 Ê un1 0.001 Ê an n 1b21 1 0.001 Ê an 1b2 1 1000an 1b Ê n ¸ 998.9999 Ê n
31
998È9982 4a998b
2
999
55. kerrork 0.001 Ê un1 0.001 Ê
1
È n 1‹
3 0.001 Ê Šan 1b 3
ˆan 1b 3Èn 1‰
3
1000
2
È
Ê ŠÈn 1‹ 3Èn 1 10 0 Ê Èn 1 œ 3 29 40 œ 2 Ê n œ 3 Ê n
4
56. kerrork 0.001 Ê un1 0.001 Ê lnalnan1 3bb 0.001 Ê lnalnan 3bb 1000 Ê n 3 ee
1000
¸ 5.297 ‚ 10323228467
which is the maximum arbitrary-precision number represented by Mathematica on the particular computer solving this
problem..
'
"
57. (2n)!
105 ' Ê (2n)! 105 œ 200,000 Ê n
'
58. n!" 105 ' Ê 105 n! Ê n
59. (a) an
9 Ê 1 1 #"! 3!" 4!" 5!" 6!" 7!" 8!" ¸ 0.367881944
an1 fails since 3" #"
_
_
nœ1
nœ1
5 Ê 1 #"! 4!" 6!" 8!" ¸ 0.54030
n
_
n
_
n
n
(b) Since ! kan k œ ! ˆ 3" ‰ ˆ "# ‰ ‘ œ ! ˆ "3 ‰ ! ˆ "# ‰ is the sum of two absolutely convergent
nœ1
nœ1
series, we can rearrange the terms of the original series to find its sum:
"
ˆ "3 "9 27
á ‰ ˆ "# "4 "8 á ‰ œ
ˆ "3 ‰
1 ˆ 3" ‰
ˆ"‰
1 #ˆ " ‰ œ #" 1 œ #"
#
"
"
60. s#! œ 1 "# 3" 4" á 19
20
¸ 0.6687714032 Ê s#! #" † #"1 ¸ 0.692580927
_
61. The unused terms are ! (1)j 1 aj œ (1)n 1 aan 1 an 2 b (1)n 3 aan 3 an 4 b á
jœn 1
œ (1)n 1 caan 1 an 2 b aan 3 an 4 b á d . Each grouped term is positive, so the remainder
has the same sign as (1)n 1 , which is the sign of the first unused term.
n
n
k œ1
k œ1
62. sn œ 1"†2 #"†3 3"†4 á n(n " 1) œ ! k(k " 1) œ ! ˆ k" k " 1 ‰
œ ˆ1 "# ‰ ˆ "# 3" ‰ ˆ 3" 4" ‰ ˆ 4" 5" ‰ á ˆ n" n " 1 ‰ which are the first 2n terms
of the first series, hence the two series are the same. Yes, for
n
sn œ ! ˆ k" k" 1 ‰ œ ˆ1 "# ‰ ˆ "# 3" ‰ ˆ 3" 4" ‰ ˆ 4" 5" ‰ á ˆ n " 1 n" ‰ ˆ n" n " 1 ‰ œ 1 n " 1
k œ1
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 10.6 Alternating Series, Absolute and Conditional Convergence
607
ˆ1 n " 1 ‰ œ 1 Ê both series converge to 1. The sum of the first 2n 1 terms of the first
Ê n lim
s œ n lim
Ä_ n
Ä_
ˆ1 n " 1 ‰ œ 1.
series is ˆ1 n " 1 ‰ n " 1 œ 1. Their sum is n lim
s œ n lim
Ä_ n
Ä_
_
_
_
_
n œ1
n œ1
n œ1
n œ1
63. Theorem 16 states that ! kan k converges Ê ! an converges. But this is equivalent to ! an diverges Ê ! kan k diverges
_
_
n œ1
n œ1
64. ka" a# á an k Ÿ ka" k ka# k á kan k for all n; then ! kan k converges Ê ! an converges and these imply that
_
_
n œ1
n œ1
º ! an º Ÿ ! kan k
_
65. (a) ! kan bn k converges by the Direct Comparison Test since kan bn k Ÿ kan k kbn k and hence
n œ1
_
! aan bn b converges absolutely
n œ1
_
_
_
(b) ! kbn k converges Ê ! bn converges absolutely; since ! an converges absolutely and
nœ1
_
nœ1
nœ1
_
_
! bn converges absolutely, we have ! can (bn )d œ ! aan bn b converges absolutely by part (a)
nœ1
_
_
_
nœ1
nœ1
_
n œ1
nœ1
nœ1
nœ1
(c) ! kan k converges Ê kkk ! kan k œ ! kkan k converges Ê ! kan converges absolutely
_
_
_
nœ1
nœ1
nœ1
66. If an œ bn œ (1)n È"n , then ! (1)n È"n converges, but ! an bn œ ! n" diverges
67. s" œ "# , s# œ "# 1 œ "# ,
"
"
"
"
s$ œ "# 1 "4 "6 8" 10
1"# 14
16
18
#"0 2"# ¸ 0.5099,
s% œ s$ 3" ¸ 0.1766,
"
"
"
"
"
"
"
s& œ s% #"4 #"6 #"8 30
3"# 34
36
38
40
42
44
¸ 0.512,
s' œ s& 5" ¸ 0.312,
"
"
"
"
"
"
"
"
"
"
"
s( œ s' 46
48
50
52
54
56
58
60
62
64
66
¸ 0.51106
N" 1
68. (a) Since ! kan k converges, say to M, for % 0 there is an integer N" such that º ! kan k Mº #%
nœ1
N" 1
N" 1
_
_
_
nœ1
nœ1
nœN"
nœN"
nœN"
Í » ! kan k ! kan k ! kan k » #% Í » ! kan k» #% Í ! kan k #% . Also, ! an
converges to L Í for % 0 there is an integer N# (which we can choose greater than or equal to N" ) such
_
that ksN# Lk #% . Therefore, ! kan k #% and ksN# Lk #% .
nœN"
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
608
Chapter 10 Infinite Sequences and Series
_
k
nœ1
nœ1
(b) The series ! kan k converges absolutely, say to M. Thus, there exists N" such that º ! kan k Mº %
whenever k N" . Now all of the terms in the sequence ekbn kf appear in ekan kf. Sum together all of the
N
terms in ekbn kf, in order, until you include all of the terms ekan kf nœ" 1 , and let N# be the largest index in the
N#
N#
_
n œ1
nœ1
n œ1
sum ! kbn k so obtained. Then º ! kbn k Mº % as well Ê ! kbn k converges to M.
10.7 POWER SERIES
_
nb1
1. n lim
¹ uunbn 1 ¹ 1 Ê n lim
¹ xxn ¹ 1 Ê kxk 1 Ê 1 x 1; when x œ 1 we have ! (1)n , a divergent
Ä_
Ä_
n œ1
_
series; when x œ 1 we have ! 1, a divergent series
n œ1
(a) the radius is 1; the interval of convergence is 1 x 1
(b) the interval of absolute convergence is 1 x 1
(c) there are no values for which the series converges conditionally
nb1
2. n lim
¹ uunbn 1 ¹ 1 Ê n lim
¹ (x(x5)5)n ¹ 1 Ê kx 5k 1 Ê 6 x 4; when x œ 6 we have
Ä_
Ä_
_
_
n œ1
nœ1
! (1)n , a divergent series; when x œ 4 we have ! 1, a divergent series
(a) the radius is 1; the interval of convergence is 6 x 4
(b) the interval of absolute convergence is 6 x 4
(c) there are no values for which the series converges conditionally
nb1
1)
"
"
3. n lim
¹ uunbn 1 ¹ 1 Ê n lim
¹ (4x
(4x 1)n ¹ 1 Ê k4x 1k 1 Ê 1 4x 1 1 Ê # x 0; when x œ # we
Ä_
Ä_
_
_
_
_
_
n œ1
n œ1
n œ1
n œ1
n œ1
have ! (1)n (1)n œ ! (1)2n œ ! 1n , a divergent series; when x œ 0 we have ! (1)n (1)n œ ! (1)n ,
a divergent series
(a) the radius is "4 ; the interval of convergence is "# x 0
(b) the interval of absolute convergence is "# x 0
(c) there are no values for which the series converges conditionally
nb1
4. n lim
¹ uunbn 1 ¹ 1 Ê n lim
¹ (3xn2)1
Ä_
Ä_
ˆ n ‰ 1 Ê k3x 2k 1
† (3x n 2)n ¹ 1 Ê k3x 2k n lim
Ä _ n1
_
Ê 1 3x 2 1 Ê "3 x 1; when x œ 3" we have ! ("n ) which is the alternating harmonic series and is
n
n œ1
_
conditionally convergent; when x œ 1 we have ! "n , the divergent harmonic series
n œ1
(a) the radius is "3 ; the interval of convergence is 3" Ÿ x 1
(b) the interval of absolute convergence is "3 x 1
(c) the series converges conditionally at x œ "3
nb1
kx 2 k
2)
5. n lim
† (x 10
¹ uunbn 1 ¹ 1 Ê n lim
¹ (x 10
nb1
10 1 Ê kx 2k 10 Ê 10 x 2 10
2)n ¹ 1 Ê
Ä_
Ä_
n
_
_
nœ1
nœ1
Ê 8 x 12; when x œ 8 we have ! (")n , a divergent series; when x œ 12 we have ! 1, a divergent series
(a) the radius is "0; the interval of convergence is 8 x 12
(b) the interval of absolute convergence is 8 x 12
(c) there are no values for which the series converges conditionally
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 10.7 Power Series
nb1
609
6. n lim
k2xk 1 Ê k2xk 1 Ê "# x "# ; when x œ "# we have
¹ uunbn 1 ¹ 1 Ê n lim
¹ (2x)
(2x)n ¹ 1 Ê n lim
Ä_
Ä_
Ä_
_
_
! (")n , a divergent series; when x œ " we have ! 1, a divergent series
#
n œ1
n œ1
(a) the radius is "# ; the interval of convergence is "# x "#
(b) the interval of absolute convergence is "# x "#
(c) there are no values for which the series converges conditionally
nb1
(n 2)
(n 1)(n 2)
1)x
7. n lim
1 Ê kxk 1
¹ uunbn 1 ¹ 1 Ê n lim
¹ (n (n
3) † nxn ¹ 1 Ê kxk n lim
Ä_
Ä_
Ä _ (n 3)(n)
_
Ê 1 x 1; when x œ 1 we have ! (")n n n # , a divergent series by the nth-term Test; when x œ " we
n œ1
_
have ! n n # , a divergent series
n œ1
(a) the radius is "; the interval of convergence is " x "
(b) the interval of absolute convergence is " x "
(c) there are no values for which the series converges conditionally
nb1
8. n lim
¹ uunbn 1 ¹ 1 Ê n lim
¹ (x n2)1
Ä_
Ä_
ˆ n ‰ 1 Ê kx 2k 1
† (x n 2)n ¹ 1 Ê kx 2k n lim
Ä _ n1
_
Ê 1 x 2 1 Ê 3 x 1; when x œ 3 we have ! "n , a divergent series; when x œ " we have
n œ1
_
!
n œ1
(1)n
n , a convergent series
(a) the radius is "; the interval of convergence is 3 x Ÿ "
(b) the interval of absolute convergence is 3 x "
(c) the series converges conditionally at x œ 1
nb1
x
9. n lim
†
¹ uunbn 1 ¹ 1 Ê n lim
¹
Ä_
Ä _ (n 1)Èn 1 3nb1
nÈ n 3n
xn ¹ 1
n
n
Ê kx3k Šn lim
‹ ŠÉn lim
‹1
Ä _ n1
Ä _ n1
_
)
Ê kx3k (1)(1) 1 Ê kxk 3 Ê 3 x 3; when x œ 3 we have ! ("
, an absolutely convergent series;
n$Î#
n
n œ1
_
1
when x œ 3 we have ! n$Î#
, a convergent p-series
n œ1
(a) the radius is 3; the interval of convergence is 3 Ÿ x Ÿ 3
(b) the interval of absolute convergence is 3 Ÿ x Ÿ 3
(c) there are no values for which the series converges conditionally
nb1
Èn
n
10. n lim
1 Ê kx 1k 1
¹ uunbn 1 ¹ 1 Ê n lim
¹ (xÈn1) 1 † (x 1)n ¹ 1 Ê kx 1k Én lim
Ä_
Ä_
Ä _ n1
_
)
Ê 1 x 1 1 Ê 0 x 2; when x œ 0 we have ! ("
, a conditionally convergent series; when x œ 2
n"Î#
n
n œ1
_
we have ! n"Î# , a divergent series
1
n œ1
(a) the radius is 1; the interval of convergence is 0 Ÿ x 2
(b) the interval of absolute convergence is 0 x 2
(c) the series converges conditionally at x œ 0
nb1
ˆ " ‰ 1 for all x
11. n lim
† n! ¹ 1 Ê kxk n lim
¹ uunbn 1 ¹ 1 Ê n lim
¹ x
Ä_
Ä _ (n 1)! xn
Ä _ n1
(a) the radius is _; the series converges for all x
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
610
Chapter 10 Infinite Sequences and Series
(b) the series converges absolutely for all x
(c) there are no values for which the series converges conditionally
nb1
nb1
ˆ " ‰ 1 for all x
12. n lim
¹ uunbn 1 ¹ 1 Ê n lim
¹ 3 x † 3nn!xn ¹ 1 Ê 3 kxk n lim
Ä_
Ä _ (n 1)!
Ä _ n1
(a) the radius is _; the series converges for all x
(b) the series converges absolutely for all x
(c) there are no values for which the series converges conditionally
nb1 2nb2
ˆ 4n ‰ œ 4x# 1 Ê x# 41
13. n lim
¹ uunbn 1 ¹ 1 Ê n lim
¹ 4 n x 1 † 4n nx2n ¹ 1 Ê x# n lim
Ä_
Ä_
Ä _ n1
_
n
Ê 12 x 12 ; when x œ 12 we have ! 4n ˆ 12 ‰
2n
n œ1
_
!
n œ1
_
œ ! n1 , a divergent p-series; when x œ 12 we have
nœ1
_
4 ˆ 1 ‰2n
œ ! n1 , a divergent p-series
n 2
n
n œ1
(a) the radius is 12 ; the interval of convergence is 12 x 12
(b) the interval of absolute convergence is 12 x 12
(c) there are no values for which the series converges conditionally
nb1
14. n lim
† n 3 ¹ 1 Ê lx 1l n lim
¹ uunbn 1 ¹ 1 Ê n lim
¹ (x 1)
Š n ‹ œ 13 lx 1l 1
Ä_
Ä _ an 1b2 3nb1 (x 1)n
Ä _ 3an 1b2
2
_
n
2
_
Ê 2 x 4; when x œ 2 we have ! (n2 3)3n œ ! (n1)
, an absolutely convergent series; when x œ 4 we have
2
n
n œ1
_
n
nœ1
_
n
! (3)
! 12 , an absolutely convergent series.
2 n œ
n œ1
n 3
n œ1
n
(a) the radius is 3; the interval of convergence is 2 Ÿ x Ÿ 4
(b) the interval of absolute convergence is 2 Ÿ x Ÿ 4
(c) there are no values for which the series converges conditionally
nb1
x
15. n lim
†
¹ uunbn 1 ¹ 1 Ê n lim
¹
Ä_
Ä _ È(n 1)# 3
È n# 3
¹1
xn
_
#
n 3
Ê kxk Én lim
" Ê kxk 1
Ä _ n# 2n 4
Ê 1 x 1; when x œ 1 we have ! È("# ) , a conditionally convergent series; when x œ 1 we have
n œ1
_
"
! È#
n œ1
n 3
n
n 3
, a divergent series
(a) the radius is 1; the interval of convergence is 1 Ÿ x 1
(b) the interval of absolute convergence is 1 x 1
(c) the series converges conditionally at x œ 1
n 1
x
16. n lim
†
¹ uun n 1 ¹ 1 Ê n lim
¹
Ä_
Ä _ È(n 1)# 3
È n# 3
¹1
xn
#
n 3
Ê kxk Én lim
" Ê kxk 1
Ä _ n# 2n 4
_
_
nœ1
nœ1
)
Ê 1 x 1; when x œ 1 we have ! Èn"# 3 , a divergent series; when x œ 1 we have ! È("
,
n# 3
a conditionally convergent series
(a) the radius is 1; the interval of convergence is 1 x Ÿ 1
(b) the interval of absolute convergence is 1 x 1
(c) the series converges conditionally at x œ 1
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
n
Section 10.7 Power Series
3)
17. n lim
¹ uunbn 1 ¹ 1 Ê n lim
¹ (n 1)(x
5nb1
Ä_
Ä_
nb1
611
ˆ n n " ‰ 1 Ê kx 5 3k 1
† n(x 5 3)n ¹ 1 Ê kx 5 3k n lim
Ä_
n
_
_
Ê kx 3k 5 Ê 5 x 3 5 Ê 8 x 2; when x œ 8 we have ! n(5n5) œ ! (1)n n, a divergent
n
n œ1
_
n œ1
_
n
series; when x œ 2 we have ! n5n œ ! n, a divergent series
n œ1
5
n œ1
(a) the radius is 5; the interval of convergence is 8 x 2
(b) the interval of absolute convergence is 8 x 2
(c) there are no values for which the series converges conditionally
nb1
#
#
(n 1) n
1
18. n lim
† 4 annxn 1b ¹ 1 Ê kx4k n lim
¹ uunbn 1 ¹ 1 Ê n lim
¹ (n 1)x
¹ a b ¹ 1 Ê kxk 4
Ä_
Ä _ 4nb1 an# 2n 2b
Ä _ n an# 2n 2b
n
_
_
Ê 4 x 4; when x œ 4 we have ! n(n#1)1 , a conditionally convergent series; when x œ 4 we have ! n# n 1 ,
n
n œ1
n œ1
a divergent series
(a) the radius is 4; the interval of convergence is 4 Ÿ x 4
(b) the interval of absolute convergence is 4 x 4
(c) the series converges conditionally at x œ 4
19. n lim
¹ uunbn 1 ¹ 1 Ê n lim
¹
Ä_
Ä_
Èn 1 xnb1
n
† È3n xn ¹ 1
3nb1
ˆ n n 1 ‰ 1 Ê k3xk 1 Ê kxk 3
Ê k3xk Én lim
Ä_
_
_
n œ1
n œ1
Ê 3 x 3; when x œ 3 we have ! (1)n Èn , a divergent series; when x œ 3 we have ! Èn, a divergent series
(a) the radius is 3; the interval of convergence is 3 x 3
(b) the interval of absolute convergence is 3 x 3
(c) there are no values for which the series converges conditionally
20. n lim
¹ uunbn 1 ¹ 1 Ê n lim
¹
Ä_
Ä_
nbÈ
1
n 1 (2x5)nb1
¹1
n n (2x5)n
È
Ê k2x 5k n lim
Š
Ä_
nbÈ
1
n1
‹1
n n
È
t
lim È
t
Ä_ n 1 Ê k2x 5k 1 Ê 1 2x 5 1 Ê 3 x 2; when x œ 3 we have
Ê k2x 5k Œ tlim
Èn
n
_
Ä_
_
n
n
n
! (1) È
È
n, a divergent series since n lim
n œ 1; when x œ 2 we have ! È
n, a divergent series
Ä_
n œ1
n œ1
(a) the radius is "# ; the interval of convergence is 3 x 2
(b) the interval of absolute convergence is 3 x 2
(c) there are no values for which the series converges conditionally
_
_
_
n œ1
n œ1
21. First, rewrite the series as ! a2 (1)n bax 1bn1 œ ! 2ax 1bn1 ! (1)n ax 1bn1 . For the series
n œ1
_
n
! 2ax 1bn1 : lim ¹ unb1 ¹ 1 Ê lim ¹ 2ax1nbc1 ¹ 1 Ê lx 1l lim 1 œ lx 1l 1 Ê 2 x 0; For the
un
nÄ_
n Ä _ 2 ax 1 b
nÄ_
n œ1
_
nb1
n
(1) ax1b
series ! (1)n ax 1bn1 : n lim
1 œ lx 1l 1
¹ uunbn 1 ¹ 1 Ê n lim
¹
¹ 1 Ê lx 1ln lim
Ä_
Ä _ (1)n ax1bnc1
Ä_
n œ1
_
Ê 2 x 0; when x œ 2 we have ! a2 (1)n ba1bn1 , a divergent series; when x œ 0 we have
n œ1
_
! a2 (1)n b, a divergent series
n œ1
(a) the radius is 1; the interval of convergence is 2 x 0
(b) the interval of absolute convergence is 2 x 0
(c) there are no values for which the series converges conditionally
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
612
Chapter 10 Infinite Sequences and Series
( 1)
22. n lim
¹ uunbn 1 ¹ 1 Ê n lim
¹
Ä_
Ä_
3
ax 2bnb1
9n
† (1)n 32n3nax 2bn ¹ 1 Ê lx 2ln lim
œ 9lx 2l 1
3an 1b
Ä _ n1
nb1 2nb2
_
_
19
17
! (1) 3 ˆ 1 ‰ œ ! 1 , a divergent series; when x œ 19 we have
Ê 17
9 x 9 ; when x œ 9 we have
3n
9
3n
9
n 2n
n œ1
_
_
n
nœ1
! (1) 3 ˆ 1 ‰n œ ! (1) , a conditionally convergent series.
n œ1
(a)
(b)
(c)
n 2n
3n
9
n
3n
n œ1
19
the radius is 19 ; the interval of convergence is 17
9 xŸ 9
19
the interval of absolute convergence is 17
9 x 9
the series converges conditionally at x œ 19
9
nb1
23.
lim ¹ uunbn 1 ¹ 1
nÄ_
Ê n lim
Ä_ »
Š1 n " 1 ‹
xnb1
Š1 "n ‹ xn
n
"
t
lim Š1 t ‹
e
Ä_
» 1 Ê kxk lim Š1 " ‹n 1 Ê kxk ˆ e ‰ 1 Ê kxk 1
n
nÄ_
t
_
n
Ê 1 x 1; when x œ 1 we have ! (1)n ˆ1 "n ‰ , a divergent series by the nth-Term Test since
n œ1
_
n
n
lim ˆ1 "n ‰ œ e Á 0; when x œ 1 we have ! ˆ1 n" ‰ , a divergent series
nÄ_
n œ1
(a) the radius is "; the interval of convergence is 1 x 1
(b) the interval of absolute convergence is 1 x 1
(c) there are no values for which the series converges conditionally
24. n lim
¹ uunbn 1 ¹ 1 Ê n lim
¹ ln (nxnln1)xn
Ä_
Ä_
nb1
ˆ " ‰
n 1
ˆ n ‰ 1 Ê kxk 1
1 Ê kxk n lim
¹ 1 Ê kxk n lim
Ä _ º ˆ"‰ º
Ä _ n1
n
_
Ê 1 x 1; when x œ 1 we have ! (1)n ln n, a divergent series by the nth-Term Test since n lim
ln n Á 0;
Ä_
n œ1
_
when x œ 1 we have ! ln n, a divergent series
n œ1
(a) the radius is 1; the interval of convergence is 1 x 1
(b) the interval of absolute convergence is 1 x 1
(c) there are no values for which the series converges conditionally
nb1 nb1
x
25. n lim
¹ uunbn 1 ¹ 1 Ê n lim
¹ (n 1)
nn xn
Ä_
Ä_
n
ˆ1 n" ‰ ‹ Š lim (n 1)‹ 1
¹ 1 Ê kxk Šn lim
Ä_
nÄ_
Ê e kxk n lim
(n 1) 1 Ê only x œ 0 satisfies this inequality
Ä_
(a) the radius is 0; the series converges only for x œ 0
(b) the series converges absolutely only for x œ 0
(c) there are no values for which the series converges conditionally
26. n lim
¹ uunbn 1 ¹ 1 Ê n lim
¹ (n n!1)!(x(x4)4)n
Ä_
Ä_
nb1
(n 1) 1 Ê only x œ 4 satisfies this inequality
¹ 1 Ê kx 4k n lim
Ä_
(a) the radius is 0; the series converges only for x œ 4
(b) the series converges absolutely only for x œ 4
(c) there are no values for which the series converges conditionally
nb1
ˆ n ‰ 1 Ê kx # 2k 1 Ê kx 2k 2
27. n lim
† n2 ¹ 1 Ê kx # 2k n lim
¹ uunbn 1 ¹ 1 Ê n lim
¹ (x 2)
Ä_
Ä _ (n 1) 2nb1 (x 2)n
Ä _ n1
n
_
_
n œ1
n œ1
! (1)
Ê 2 x 2 2 Ê 4 x 0; when x œ 4 we have ! "
n , a divergent series; when x œ 0 we have
n
the alternating harmonic series which converges conditionally
(a) the radius is 2; the interval of convergence is 4 x Ÿ 0
(b) the interval of absolute convergence is 4 x 0
(c) the series converges conditionally at x œ 0
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
nb1
,
Section 10.7 Power Series
nb1
613
nb1
(n 2)(x 1)
ˆ n 2 ‰ 1 Ê 2 kx 1k 1
28. n lim
¹ uunbn 1 ¹ 1 Ê n lim
¹ ((2)2)n (n
1)(x 1)n ¹ 1 Ê 2 kx 1k n lim
Ä_
Ä_
Ä _ n1
_
Ê kx 1k "# Ê "# x 1 "# Ê "# x 3# ; when x œ #" we have ! (n 1) , a divergent series; when x œ 3#
_
n œ1
we have ! (1) (n 1), a divergent series
n
n œ1
(a) the radius is "# ; the interval of convergence is "# x #3
(b) the interval of absolute convergence is "# x 3#
(c) there are no values for which the series converges conditionally
nb1
#
#
x
n
ln n
29. n lim
† n(lnxnn) ¹ 1 Ê kxk Šn lim
¹ uunbn 1 ¹ 1 Ê n lim
¹
‹ Šn lim
‹ 1
Ä_
Ä _ (n 1) aln (n 1)b#
Ä _ n1
Ä _ ln (n 1)
ˆ"‰
#
#
n1
n
Ê kxk (1) Œn lim
1 Ê kxk Šn lim
‹ 1 Ê kxk 1 Ê 1 x 1; when x œ 1 we have
Ä _ ˆ " ‰
Ä_ n
nb1
_
_
nœ1
nœ1
n
"
! (1) # which converges absolutely; when x œ 1 we have !
n(ln n)
n(ln n)# which converges
(a) the radius is "; the interval of convergence is 1 Ÿ x Ÿ 1
(b) the interval of absolute convergence is 1 Ÿ x Ÿ 1
(c) there are no values for which the series converges conditionally
nb1
ln (n)
x
n
30. n lim
† n lnxn(n) ¹ 1 Ê kxk Šn lim
¹ uunbn 1 ¹ 1 Ê n lim
¹
‹ Šn lim
‹1
Ä_
Ä _ (n 1) ln (n 1)
Ä _ n1
Ä _ ln (n 1)
_
Ê kxk (1)(1) 1 Ê kxk 1 Ê 1 x 1; when x œ 1 we have ! (nln1)n , a convergent alternating series;
n
n œ2
_
when x œ 1 we have ! n ln" n which diverges by Exercise 38, Section 9.3
n œ2
(a) the radius is "; the interval of convergence is 1 Ÿ x 1
(b) the interval of absolute convergence is 1 x 1
(c) the series converges conditionally at x œ 1
2nb3
$Î#
5)
n
31. n lim
† (4x n 5)2n 1 ¹ 1 Ê (4x 5)# Šn lim
¹ uunbn 1 ¹ 1 Ê n lim
¹ (4x
‹
(n 1)$Î#
Ä_
Ä_
Ä _ n1
_
$Î#
2nb1
Ê k4x 5k 1 Ê 1 4x 5 1 Ê 1 x 3# ; when x œ 1 we have ! (n1)$Î#
n œ1
_
1 Ê (4x 5)# 1
_
œ ! n"
$Î# which is
n œ1
2nb1
absolutely convergent; when x œ 3# we have ! ("n)$Î# , a convergent p-series
n œ1
(a) the radius is "4 ; the interval of convergence is 1 Ÿ x Ÿ 3#
(b) the interval of absolute convergence is 1 Ÿ x Ÿ 3#
(c) there are no values for which the series converges conditionally
nb2
32. n lim
¹ uunbn 1 ¹ 1 Ê n lim
¹ (3x2n1)4
Ä_
Ä_
ˆ 2n 2 ‰ 1 Ê k3x 1k 1
† (3x2n1)2nb1 ¹ 1 Ê k3x 1k n lim
Ä _ 2n 4
_
nb1
1)
Ê 1 3x 1 1 Ê 23 x 0; when x œ 23 we have ! (2n
1 , a conditionally convergent series;
n œ1
_
when x œ 0 we have !
n œ1
_
(")nb1
! " , a divergent series
2n 1 œ
#n 1
nœ1
(a) the radius is "3 ; the interval of convergence is 32 Ÿ x 0
(b) the interval of absolute convergence is 23 x 0
(c) the series converges conditionally at x œ 23
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
614
Chapter 10 Infinite Sequences and Series
nb1
x
ˆ 1 ‰ 1 for all x
33. n lim
† 2†4†6xân a2nb ¹ 1 Ê kxk n lim
¹ uunbn 1 ¹ 1 Ê n lim
¹
Ä_
Ä _ 2†4†6âa2nba2an 1bb
Ä _ 2n 2
(a) the radius is _; the series converges for all x
(b) the series converges absolutely for all x
(c) there are no values for which the series converges conditionally
3 5 7 2n 1 2 n 1
1x
34. n lim
¹ uunbn 1 ¹ 1 Ê n lim
¹ † † âa an ba1ba2 2nb1 b b
Ä_
Ä_
nb2
2n 3 n
† 3†5†7âan2n2 1bxnb1 ¹ 1 Ê kxk n lim
Š a2an 1bb2 ‹ 1 Ê only
Ä_
2
2 n
x œ 0 satisfies this inequality
(a) the radius is 0; the series converges only for x œ 0
(b) the series converges absolutely only for x œ 0
(c) there are no values for which the series converges conditionally
_
nan 1b
n n
35. For the series ! 121222â
and 12 22 â n2 œ nan 1ba6 2n 1b so that we can
â n2 x , recall 1 2 â n œ
2
n œ1
nan b 1b
_
_
nb1
rewrite the series as ! Œ n n b 1 2 2n b 1 xn œ ! ˆ 2n 3 1 ‰xn ; then n lim
† a2n3xn 1b ¹ 1
¹ uunbn 1 ¹ 1 Ê n lim
¹ 3x
Ä_
Ä _ a2 an 1 b 1 b
a
n œ1
Ê kx k
ba
b
nœ1
6
_
lim ¹ a2n 1b ¹ 1 Ê kxk 1 Ê 1 x 1; when x œ 1 we have ! ˆ 2n 3 1 ‰a1bn , a conditionally
n Ä _ a2n 3b
n œ1
_
convergent series; when x œ 1 we have ! ˆ
n œ1
3 ‰
2n 1 , a divergent series.
(a) the radius is 1; the interval of convergence is 1 Ÿ x 1
(b) the interval of absolute convergence is 1 x 1
(c) the series converges conditionally at x œ 1
_
36. For the series ! ŠÈn 1 Èn‹ax 3bn , note that Èn 1 Èn œ
n œ1
_
Èn 1 Èn Èn 1 Èn
† Èn 1 Èn œ Èn 11 Èn so that we
1
nb1
n
can rewrite the series as ! Ènax13b Èn ; then n lim
†
¹ uunbn 1 ¹ 1 Ê n lim
¹ ax 3 b
Ä_
Ä _ Èn 2 Èn 1
Èn 1 Èn
n œ1
ax 3bn
_
Èn 1 Èn
¹1
n
b
Ê lx 3ln lim
1 Ê lx 3l 1 Ê 2 x 4; when x œ 2 we have ! Èn a11
, a conditionally
Èn
Ä _ Èn 2 Èn 1
n œ1
_
convergent series; when x œ 4 we have ! Èn 11 Èn , a divergent series;
n œ1
(a) the radius is 1; the interval of convergence is 2 Ÿ x 4
(b) the interval of absolute convergence is 2 x 4
(c) the series converges conditionally at x œ 2
nb1
lx l
an 1bxx
an 1 b
37. n lim
† 3†6†n9xâxna3nb ¹ 1 Ê lxln lim
¹ uunbn 1 ¹ 1 Ê n lim
¹
¹
¹ 1 Ê 3 1 Ê lxl 3 Ê R œ 3
Ä_
Ä _ 3†6†9âa3nba3an 1bb
Ä _ 3 an 1 b
2 nb1
2
2
4 lx l
2 4 6 2n 2 n 1
x
38. n lim
† aa22†5†4†8†6ââaa3n2nbb12 xbbn ¹ 1 Ê lxln lim
¹ a2n 2b ¹ 1 Ê 9 1
¹ uunbn 1 ¹ 1 Ê n lim
¹ a † † âa ba a bbb
Ä_
Ä _ a2†5†8âa3n 1ba3an 1b 1bb2
Ä _ a3n 2b2
Ê lxl 94 Ê R œ 94
2 nb1
2
lx l
n 1 x
an 1 b
39. n lim
† 2 a2nbx ¹ 1 Ê lxln lim
¹ uunbn 1 ¹ 1 Ê n lim
¹ aa bxb
¹
¹ 1 Ê 8 1 Ê lxl 8 Ê R œ 8
Ä_
Ä _ 2nb1 a2an 1bbx anxb2 xn
Ä _ 2a2n 2ba2n 1b
n2
n
n
n
n ‰
n
Ɉ
ˆ n ‰ 1 Ê lxle1 1 Ê lxl e Ê R œ e
È
40. n lim
un 1 Ê n lim
xn 1 Ê lxl n lim
n1
Ä_
Ä_
Ä _ n1
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 10.7 Power Series
nb1
615
nb1
41. n lim
3 1 Ê lxl 31 Ê 31 x 31 ; at x œ 31 we have
¹ uunbn 1 ¹ 1 Ê n lim
¹ 3 3n xxn ¹ 1 Ê lxl n lim
Ä_
Ä_
Ä_
_
_
_
_
_
! 3n ˆ 1 ‰n œ ! a1bn , which diverges; at x œ 1 we have ! 3n ˆ 1 ‰n œ ! 1 , which diverges. The series ! 3n xn
3
n œ0
3
n œ0
3
n œ0
nœ0
_
œ ! a3xbn is a convergent geometric series when 1 x 1 and the sum is
3
n œ0
3
nœ0
1
1 3x .
nb1
e
4
42. n lim
1 1 Ê lex 4l 1 Ê 3 ex 5 Ê ln 3 x ln 5;
¹ uunbn 1 ¹ 1 Ê n lim
¹ a aex 4b bn ¹ 1 Ê lex 4l n lim
Ä_
Ä_
Ä_
x
_
n
_
_
nœ0
nœ0
_
n
at x œ ln 3 we have ! ˆeln 3 4‰ œ ! a1bn , which diverges; at x œ ln 5 we have ! ˆeln 5 4‰ œ ! 1, which
nœ0
_
nœ0
diverges. The series ! aex 4bn is a convergent geometric series when ln 3 x ln 5 and the sum is 1 a1ex 4b œ 5 1 ex .
n œ0
2nb2
43. n lim
¹ uunbn 1 ¹ 1 Ê n lim
¹ (x 4n1)1
Ä_
Ä_
n
† (x 4 1)2n ¹ 1 Ê
(x 1)#
lim
4
nÄ_
_
k1k 1 Ê (x 1)# 4 Ê kx 1k 2
_
_
Ê 2 x 1 2 Ê 1 x 3; at x œ 1 we have ! (42)n œ ! 44n œ ! 1, which diverges; at x œ 3
2n
n œ0
_
_
2n
n
nœ0
n œ0
_
n
we have ! 24n œ ! 44n œ ! 1, a divergent series; the interval of convergence is 1 x 3; the series
n œ0
_
!
n œ0
nœ0
nœ0
_
# n
(x ")2n
œ ! Šˆ x # 1 ‰ ‹ is a convergent geometric series when 1 x 3 and the sum is
4n
nœ0
"
œ
"
1 Šxc
# ‹
#
"
’
4
(x
4
")# “ œ 4 x# 2x 1 œ 3 2x x#
4
4
2nb2
44. n lim
¹ uunbn 1 ¹ 1 Ê n lim
¹ (x 9n1)1
Ä_
Ä_
n
† (x 9 1)2n ¹ 1 Ê
(x 1)#
lim k1k 1
9
nÄ_
_
Ê (x 1)# 9 Ê kx 1k 3
_
Ê 3 x 1 3 Ê 4 x 2; when x œ 4 we have ! (93)n œ ! 1 which diverges; at x œ 2 we have
n œ0
_
2n
n œ0
_
! 32nn œ ! " which also diverges; the interval of convergence is 4 x 2; the series
9
n œ0
_
nœ0
_
n
! (x n1) œ ! Šˆ x1 ‰# ‹ is a convergent geometric series when 4 x 2 and the sum is
9
3
2n
n œ0
n œ0
"
1Š
# œ
xb1
‹
3
"
’
9
(x 1)#
“
9
œ 9 x# 9 2x 1 œ 8 2x9 x#
45. n lim
¹ uunbn 1 ¹ 1 Ê n lim
Ä_
Ä_ º
n
ˆÈx 2‰nb1
† ˆÈx2 2‰n º 1
2nb1
Ê ¸Èx 2¸ 2 Ê 2 Èx 2 2 Ê 0 Èx 4
_
_
Ê 0 x 16; when x œ 0 we have ! (1)n , a divergent series; when x œ 16 we have ! (1)n , a divergent
nœ0
nœ0
_
Èx 2 n
series; the interval of convergence is 0 x 16; the series ! Š # ‹ is a convergent geometric series when
n œ0
0 x 16 and its sum is
"
1Œ
Èx c 2 œ
# Œ
2c
"
Èx 2
#
œ 4 2Èx
nb1
46. n lim
¹ uunbn 1 ¹ 1 Ê n lim
¹ (ln(lnx)x)n ¹ 1 Ê kln xk 1 Ê 1 ln x 1 Ê e" x e; when x œ e" or e we
Ä_
Ä_
_
_
_
nœ0
nœ0
nœ0
obtain the series ! 1n and ! (1)n which both diverge; the interval of convergence is e" x e; ! (ln x)n œ 1 "ln x
"
when e
xe
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
616
Chapter 10 Infinite Sequences and Series
#
47. n lim
¹ uunbn 1 ¹ 1 Ê n lim
Š x 3 1 ‹
Ä_
Ä_ º
n1
#
† ˆ x# 3 1 ‰ º 1 Ê ax 3 1b n lim
k1k 1 Ê x 3 " 1 Ê x# 2
Ä_
n
#
_
Ê kxk È2 Ê È2 x È2 ; at x œ „ È2 we have ! (1)n which diverges; the interval of convergence is
n œ0
_
#
È2 x È2 ; the series ! Š x 3 1 ‹
n
is a convergent geometric series when È2 x È2 and its sum is
n œ0
"
œ 3 c x"# c 1 œ # 3 x#
#
1 Š x 3b 1 ‹
Š
‹
3
#
ax 1 b
48. n lim
¹ uun n 1 ¹ 1 Ê n lim
¹ 2n 1
Ä_
Ä_
n 1
† ax# 2 1bn ¹ 1 Ê kx# 1k 2 Ê È3 x È3 ; when x œ „ È3 we
n
_
_
n
#
have ! 1n , a divergent series; the interval of convergence is È3 x È3 ; the series ! Š x 2 1 ‹ is a
n œ0
n œ0
convergent geometric series when È3 x È3 and its sum is
nb1
49. n lim
¹ (x #n3)
b1
Ä_
"
"
œ
#
1 Šx 2 1‹
2
œ 3 2 x#
Šx# 1 ‹
#
_
n
† (x 2 3)n ¹ 1 Ê kx 3k 2 Ê 1 x 5; when x œ 1 we have ! (1)n which diverges;
nœ1
_
when x œ 5 we have ! (1) which also diverges; the interval of convergence is 1 x 5; the sum of this
n
n œ1
"
œ x2 1 .
3
1 Šxc
# ‹
convergent geometric series is
n
If f(x) œ 1 #" (x 3) 4" (x 3)# á ˆ #" ‰ (x 3)n á
n
œ x 2 1 then f w (x) œ #" "# (x 3) á ˆ #" ‰ n(x 3)n1 á is convergent when 1 x 5, and diverges
2
2
when x œ 1 or 5. The sum for f w (x) is (x 1)# , the derivative of x 1 .
50. If f(x) œ 1 "# (x 3) 4" (x 3)# á ˆ "# ‰ (x 3)n á œ x 2 1 then ' f(x) dx
n
$
#
œ x (x 4 3) (x 123) á ˆ "# ‰
_
n (x 3)n 1
á .
n 1
_
At x œ 1 the series ! n21 diverges; at x œ 5
n œ1
2
the series ! (n1)
1 converges. Therefore the interval of convergence is 1 x Ÿ 5 and the sum is
n
n œ1
2 ln kx 1k (3 ln 4), since ' x 2 1 dx œ 2 ln kx 1k C, where C œ 3 ln 4 when x œ 3.
#
%
'
)
"!
5x
7x
9x
11x
51. (a) Differentiate the series for sin x to get cos x œ 1 3x
3! 5! 7! 9! 11! á
#
%
'
)
"!
x
œ 1 x#! x4! x6! x8! 10!
á . The series converges for all values of x since
lim
nÄ_
2nb2
"
¹ (2nx 2)! † a#xn#8b! ¹ œ x2 n lim
Š
‹ œ 0 1 for all x.
Ä _ a2n 1ba2n 2b
$ $
& &
( (
* *
"" ""
$
&
(
*
""
32x
128x
512x
2048x
(b) sin 2x œ 2x 23!x 25!x 27!x 29!x 2 11!x á œ 2x 8x
3! 5! 7! 9! 11! á
"
"‰ $
‰ # ˆ
(c) 2 sin x cos x œ 2 (0 † 1) (0 † 0 1 † 1)x ˆ0 † "
# 1 † 0 0 † 1 x 0 † 0 1 † # 0 † 0 1 † 3! x
ˆ0 † 4!" 1 † 0 0 † #" 0 † 3!" 0 † 1‰ x% ˆ0 † 0 1 † 4!" 0 † 0 #" † 3!" 0 † 0 1 † 5!" ‰ x&
$
&
16x
ˆ0 † 6!" 1 † 0 0 † 4!" 0 † 3!" 0 † #" 0 † 5!" 0 † 1‰ x' á ‘ œ 2 ’x 4x
3! 5! á “
$ $
& &
( (
* *
"" ""
œ 2x 23!x 25!x 27!x 29!x 2 11!x á
52. (a)
d
2x
3x#
4x$
5x%
x#
x$
x%
x
x
x
x
x ae b œ 1 2! 3! 4! 5! á œ 1 x #! 3! 4! á œ e ; thus the derivative of e is e itself
(b) ' ex dx œ ex C œ x x# x3! x4! x5! á C, which is the general antiderivative of ex
#
$
%
&
(c) ex œ 1 x x#! x3! x4! x5! á ; ecx † ex œ 1 † 1 (1 † 1 1 † 1)x ˆ1 † #"! 1 † 1 #"! † 1‰ x#
ˆ1 † 3!" 1 † #"! #"! † 1 3!" † 1‰ x$ ˆ1 † 4!" 1 † 3!" #"! † #"! 3!" † 1 4!" † 1‰ x%
#
$
%
&
ˆ1 † 5!" 1 † 4!" #"! † 3!" 3!" † #"! 4!" † 1 5!" † 1‰ x& á œ 1 0 0 0 0 0 á
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 10.8 Taylor and Maclaurin Series
617
17x
62x
53. (a) ln ksec xk C œ ' tan x dx œ ' Šx x3 2x
15 315 2835 á ‹ dx
$
#
%
'
)
&
(
*
"!
#
%
'
)
"!
x
17x
31x
x
x
17x
31x
œ x# 1x# 45
2520
14,175
á C; x œ 0 Ê C œ 0 Ê ln ksec xk œ x# 12
45
2520
14,175
á ,
converges when 1# x 1#
$
&
(
*
%
'
)
x)
d
17x
62x
2x
17x
62x
#
(b) sec# x œ d(tan
œ dx
Šx x3 2x
dx
15 315 2835 á ‹ œ 1 x 3 45 315 á , converges
when 1# x 1#
#
%
'
#
%
'
61x
x
5x
61x
(c) sec# x œ (sec x)(sec x) œ Š1 x# 5x
24 720 á ‹ Š1 # 24 720 á ‹
5
5 ‰ %
61
5
5
61 ‰ '
œ 1 ˆ "# "# ‰ x# ˆ 24
"4 24
x ˆ 720
48
48
720
x á
%
'
)
62x
1
1
œ 1 x# 2x3 17x
45 315 á , # x #
61x
54. (a) ln ksec x tan xk C œ ' sec x dx œ ' Š1 x2 5x
24 720 á ‹ dx
#
$
&
(
*
$
&
(
*
%
'
x
61x
277x
œ x x6 24
5040
72,576
á C; x œ 0 Ê C œ 0 Ê ln ksec x tan xk
x
61x
277x
œ x x6 24
5040
72,576
á , converges when 1# x 1#
#
%
'
$
&
(
x)
d
61x
5x
61x
277x
(b) sec x tan x œ d(sec
œ dx
Š1 x# 5x
dx
24 720 á ‹ œ x 6 120 1008 á , converges
when 1# x 1#
#
%
'
$
&
(
61x
x
2x
17x
(c) (sec x)(tan x) œ Š1 x# 5x
24 720 á ‹ Šx 3 15 315 á ‹
$
&
(
2
5 ‰ &
17
"
5
61 ‰ (
277x
œ x ˆ "3 #" ‰ x$ ˆ 15
"6 24
x ˆ 315
15
72
720
x á œ x 5x6 61x
120 1008 á ,
1# x 1#
_
_
n œ0
n œk
_
55. (a) If f(x) œ ! an xn , then f ÐkÑ (x) œ ! n(n 1)(n 2)â(n (k 1)) an xnk and f ÐkÑ (0) œ k!ak
Ê
ÐkÑ
ÐkÑ
ak œ f k!(0) ; likewise if f(x) œ ! bn xn , then bk œ f k!(0)
n œ0
Ê ak œ bk for every nonnegative integer k
_
(b) If f(x) œ ! an xn œ 0 for all x, then f ÐkÑ (x) œ 0 for all x Ê from part (a) that ak œ 0 for every nonnegative integer k
n œ0
10.8 TAYLOR AND MACLAURIN SERIES
1. f(x) œ e2x , f w (x) œ 2e2x , f ww (x) œ 4e2x , f www (x) œ 8e2x ; f(0) œ e2a0b œ ", f w (0) œ 2, f ww (0) œ 4, f www (0) œ 8 Ê P! (x) œ 1,
P" (x) œ 1 2x, P# (x) œ 1 x 2x# , P$ (x) œ 1 x 2x# 43 x3
2. f(x) œ sin x, f w (x) œ cos x , f ww (x) œ sin x , f www (x) œ cos x; f(0) œ sin 0 œ 0, f w (0) œ 1, f ww (0) œ 0, f www (0) œ 1
Ê P! (x) œ 0, P" (x) œ x, P# (x) œ x, P$ (x) œ x 16 x3
3. f(x) œ ln x, f w (x) œ "x , f ww (x) œ x"# , f www (x) œ x2$ ; f(1) œ ln 1 œ 0, f w (1) œ 1, f ww (1) œ 1, f www (1) œ 2 Ê P! (x) œ 0,
P" (x) œ (x 1), P# (x) œ (x 1) "# (x 1)# , P$ (x) œ (x 1) "# (x 1)# 3" (x 1)$
4. f(x) œ ln (1 x), f w (x) œ 1 " x œ (1 x)" , f ww (x) œ (1 x)# , f www (x) œ 2(1 x)$ ; f(0) œ ln 1 œ 0,
#
#
$
f w (0) œ 11 œ 1, f ww (0) œ (1)# œ 1, f www (0) œ 2(1)$ œ 2 Ê P! (x) œ 0, P" (x) œ x, P# (x) œ x x# , P$ (x) œ x x# x3
5. f(x) œ "x œ x" , f w (x) œ x# , f ww (x) œ 2x$ , f www (x) œ 6x% ; f(2) œ "# , f w (2) œ 4" , f ww (2) œ 4" , f www (x) œ 83
Ê P! (x) œ "# , P" (x) œ "# "4 (x 2), P# (x) œ "# "4 (x 2) "8 (x 2)# ,
"
P$ (x) œ "# "4 (x 2) "8 (x 2)# 16
(x 2)$
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
618
Chapter 10 Infinite Sequences and Series
6. f(x) œ (x 2)" , f w (x) œ (x 2)# , f ww (x) œ 2(x 2)$ , f www (x) œ 6(x 2)% ; f(0) œ (2)" œ "# , f w (0) œ (2)#
#
œ 4" , f ww (0) œ 2(2)$ œ 4" , f www (0) œ 6(2)% œ 38 Ê P! (x) œ #" , P" (x) œ #" x4 , P# (x) œ #" x4 x8 ,
#
$
x
P$ (x) œ "# x4 x8 16
È2
È2
1
w ˆ1‰
# ,f
4 œ cos 4 œ # ,
È
È
È
È
È
f ww ˆ 14 ‰ œ sin 14 œ #2 , f www ˆ 14 ‰ œ cos 14 œ #2 Ê P! œ #2 , P" (x) œ #2 #2 ˆx 14 ‰ ,
È
È
È
È
È
È
È
#
#
$
P# (x) œ #2 #2 ˆx 14 ‰ 42 ˆx 14 ‰ , P$ (x) œ #2 #2 ˆx 14 ‰ 42 ˆx 14 ‰ 1#2 ˆx 14 ‰
7. f(x) œ sin x, f w (x) œ cos x, f ww (x) œ sin x, f www (x) œ cos x; f ˆ 14 ‰ œ sin 14 œ
8. f(x) œ tan x, f w (x) œ sec2 x, f ww (x) œ 2sec2 x tan x, f www (x) œ 2sec4 x 4sec2 x tan2 x; f ˆ 14 ‰ œ tan 14 œ 1 ,
f w ˆ 14 ‰ œ sec2 ˆ 14 ‰ œ 2 , f ww ˆ 14 ‰ œ 2sec2 ˆ 14 ‰ tan ˆ 14 ‰ œ 4 , f www ˆ 14 ‰ œ 2sec4 ˆ 14 ‰ 4sec2 ˆ 14 ‰ tan2 ˆ 14 ‰ œ 16 Ê P! (x) œ 1 ,
2
2
P" (x) œ 1 2 ˆx 14 ‰ , P# (x) œ 1 2 ˆx 14 ‰ 2 ˆx 14 ‰ , P$ (x) œ 1 2 ˆx 14 ‰ 2 ˆx 14 ‰ 83 ˆx 14 ‰
3
9. f(x) œ Èx œ x"Î# , f w (x) œ ˆ "# ‰ x"Î# , f ww (x) œ ˆ 4" ‰ x$Î# , f www (x) œ ˆ 83 ‰ x&Î# ; f(4) œ È4 œ 2,
"
3
f w (4) œ ˆ "# ‰ 4"Î# œ "4 , f ww (4) œ ˆ "4 ‰ 4$Î# œ 32
,f www (4) œ ˆ 38 ‰ 4&Î# œ 256
Ê P! (x) œ 2, P" (x) œ 2 "4 (x 4),
"
"
P# (x) œ 2 4" (x 4) 64
(x 4)# , P$ (x) œ 2 "4 (x 4) 64
(x 4)# 51"# (x 4)$
10. f(x) œ (1 x)"Î# , f w (x) œ "# (1 x)"Î# , f ww (x) œ "4 (1 x)$Î# , f www (x) œ 38 (1 x)&Î# ; f(0) œ (1)"Î# œ 1,
f w (0) œ "# (1)"Î# œ "# , f ww (0) œ "4 (1)$Î# œ "4 , f www (0) œ 83 (1)&Î# œ 83 Ê P! (x) œ 1,
1 $
P" (x) œ 1 2" x, P# (x) œ 1 2" x 8" x# , P$ (x) œ 1 2" x 8" x# 16
x
11. f(x) œ ex , f w (x) œ ex , f ww (x) œ ex , f www (x) œ ex Ê á f ÐkÑ (x) œ a1bk ex ; f(0) œ ea0b œ ", f w (0) œ 1,
_
f ww (0) œ 1, f www (0) œ 1, á ß f ÐkÑ (0) œ (1)k Ê ex œ 1 x 12 x# 16 x3 á œ ! (n!1) xn
n
n œ0
12. f(x) œ x ex , f w (x) œ x ex ex , f ww (x) œ x ex 2ex , f www (x) œ x ex 3ex Ê á f ÐkÑ (x) œ x ex k ex ; f(0) œ a0bea0b œ 0,
_
f w (0) œ 1, f ww (0) œ 2, f www (0) œ 3, á ß f ÐkÑ (0) œ k Ê x x# 12 x3 á œ ! an 1 1b! xn
n œ0
13. f(x) œ (1 x)" Ê f w (x) œ (1 x)# , f ww (x) œ 2(1 x)$ , f www (x) œ 3!(1 x)% Ê á f ÐkÑ (x)
œ (1)k k!(1 x)k1 ; f(0) œ 1, f w (0) œ 1, f ww (0) œ 2, f www (0) œ 3!, á ß f ÐkÑ (0) œ (1)k k!
_
_
n œ0
nœ0
Ê 1 x x# x$ á œ ! (x)n œ ! (1)n xn
x
3
w
ww
$ www
%
14. f(x) œ 21 Ê á f ÐkÑ (x) œ 3ak!b(1 x) k 1 ; f(0) œ 2,
x Ê f (x) œ (1 x)# , f (x) œ 6(1 x) , f (x) œ 18(1 x)
_
f w (0) œ 3, f ww (0) œ 6, f www (0) œ 18, á ß f ÐkÑ (0) œ 3ak!b Ê 2 3x 3x# 3x$ á œ 2 ! 3xn
n œ1
_
n 2nb1
) x
15. sin x œ ! ("
(#n1)!
n œ0
_
n 2nb1
) x
16. sin x œ ! ("
(#n1)!
nœ0
_
2nb1
Ê sin 3x œ ! ("(#) n(3x)
1)!
n
n œ0
_
Ê sin x# œ !
nœ0
_
2n 1
(")n ˆ x# ‰
(#n1)!
_
n 2nb1 2nb1
œ ! (")(#3n1)!x
n œ0
_
n 2nb1
$ $
& &
œ 3x 33!x 35!x á
$
&
œ ! #(2n"1)(2nx 1)! œ #x 2$x†3! 2&x†5! á
nœ0
)x
7x
7x
7x
17. 7 cos (x) œ 7 cos x œ 7 ! ("
(2n)! œ 7 #! 4! 6! á , since the cosine is an even function
n 2n
#
%
'
n œ0
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 10.8 Taylor and Maclaurin Series
_
1) x
18. cos x œ ! ((2n)!
n 2n
n œ0
_
(1x)
Ê 5 cos 1x œ 5 ! (1)(#n)!
œ 5 512!x 514!x 516!x á
n
# #
2n
% %
' '
nœ0
cx
œ "# ’Š1 x# x#! x3! x4! á ‹ Š1 x x#! x3! x4! á ‹“ œ 1 x#! x4! x6! á
cx
œ "# ’Š1 x x#! x3! x4! á ‹ Š1 x x#! x3! x4! á ‹“ œ x x3! x5! x6! á
19. cosh x œ e #e
x
_
619
#
$
%
#
$
%
#
%
'
2n
x
œ ! (2n)!
n œ0
20. sinh x œ e #e
x
_
#
$
%
#
$
%
$
&
'
2n 1
œ ! (2nx 1)!
n œ0
21. f(x) œ x% 2x$ 5x 4 Ê f w (x) œ 4x$ 6x# 5, f ww (x) œ 12x# 12x, f www (x) œ 24x 12, f Ð4Ñ (x) œ 24
Ê f ÐnÑ (x) œ 0 if n 5; f(0) œ 4, f w (0) œ 5, f ww (0) œ 0, f www (0) œ 12, f Ð4Ñ (0) œ 24, f ÐnÑ (0) œ 0 if n 5
24 %
$
%
$
Ê x% 2x$ 5x 4 œ 4 5x 12
3! x 4! x œ x 2x 5x 4
n
x
6
22. f(x) œ x x 1 Ê f w (x) œ a2x
; f ww (x) œ ax 2 1b3 ; f www (x) œ ax Ê f ÐnÑ (x) œ axa11bbnnbx 1 ; f(0) œ 0, f w (0) œ 0, f ww (0) œ 2,
1 b4
x 1 b2
#
#
f www (0) œ 6, f ÐnÑ (0) œ a1bn nx if n
_
2 Ê x# x3 x4 x5 Þ Þ Þ œ ! a1bn xn
n œ2
23. f(x) œ x$ 2x 4 Ê f w (x) œ 3x# 2, f ww (x) œ 6x, f www (x) œ 6 Ê f ÐnÑ (x) œ 0 if n 4; f(2) œ 8, f w (2) œ 10,
6
#
$
f ww (2) œ 12, f www (2) œ 6, f ÐnÑ (2) œ 0 if n 4 Ê x$ 2x 4 œ 8 10(x 2) 12
2! (x 2) 3! (x 2)
œ 8 10(x 2) 6(x 2)# (x 2)$
24. f(x) œ 2x$ x# 3x 8 Ê f w (x) œ 6x# 2x 3, f ww (x) œ 12x 2, f www (x) œ 12 Ê f ÐnÑ (x) œ 0 if n
f w (1) œ 11, f ww (1) œ 14, f www (1) œ 12, f ÐnÑ (1) œ 0 if n 4 Ê 2x$ x# 3x 8
12
#
$
#
$
œ 2 11(x 1) 14
2! (x 1) 3! (x 1) œ 2 11(x 1) 7(x 1) 2(x 1)
4; f(1) œ 2,
25. f(x) œ x% x# 1 Ê f w (x) œ 4x$ 2x, f ww (x) œ 12x# 2, f www (x) œ 24x, f Ð4Ñ (x) œ 24, f ÐnÑ (x) œ 0 if n 5;
f(2) œ 21, f w (2) œ 36, f ww (2) œ 50, f www (2) œ 48, f Ð4Ñ (2) œ 24, f ÐnÑ (2) œ 0 if n 5 Ê x% x# 1
48
24
#
$
%
#
$
%
œ 21 36(x 2) 50
2! (x 2) 3! (x 2) 4! (x 2) œ 21 36(x 2) 25(x 2) 8(x 2) (x 2)
26. f(x) œ 3x& x% 2x$ x# 2 Ê f w (x) œ 15x% 4x$ 6x# 2x, f ww (x) œ 60x$ 12x# 12x 2,
f www (x) œ 180x# 24x 12, f Ð4Ñ (x) œ 360x 24, f Ð5Ñ (x) œ 360, f ÐnÑ (x) œ 0 if n 6; f(1) œ 7,
f w (1) œ 23, f ww (1) œ 82, f www (1) œ 216, f Ð4Ñ (1) œ 384, f Ð5Ñ (1) œ 360, f ÐnÑ (1) œ 0 if n 6
216
384
360
#
$
%
&
Ê 3x& x% 2x$ x# 2 œ 7 23(x 1) 82
2! (x 1) 3! (x 1) 4! (x 1) 5! (x 1)
œ 7 23(x 1) 41(x 1)# 36(x 1)$ 16(x 1)% 3(x 1)&
27. f(x) œ x# Ê f w (x) œ 2x$ , f ww (x) œ 3! x% , f www (x) œ 4! x& Ê f ÐnÑ (x) œ (1)n (n 1)! xn2 ;
f(1) œ 1, f w (1) œ 2, f ww (1) œ 3!, f www (1) œ 4!, f ÐnÑ (1) œ (1)n (n 1)! Ê x"#
_
œ 1 2(x 1) 3(x 1)# 4(x 1)$ á œ ! (1)n (n 1)(x 1)n
n œ0
28. f(x) œ a1 1 xb3 Ê f w (x) œ 3(1 x)4 , f ww (x) œ 12(1 x)5 , f www (x) œ 60 (1 x)6 Ê f ÐnÑ (x) œ an 2 2b! (1 x)n3 ;
fa0b œ 1, f w a0b œ 3, f ww a0b œ 12, f www a0b œ 60, á , f ÐnÑ a0b œ an 2 2b! Ê a1 1 xb3 œ 1 3x 6x# 10x3 á
_
œ ! an 2ba2 n 1b xn
n œ0
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
620
Chapter 10 Infinite Sequences and Series
29. f(x) œ ex Ê f w (x) œ ex , f ww (x) œ ex Ê f ÐnÑ (x) œ ex ; f(2) œ e# , f w (2) œ e# , á f ÐnÑ (2) œ e#
#
_
$
#
Ê ex œ e# e# (x 2) e# (x 2)# e3! (x 2)$ á œ ! en! (x 2)n
n œ0
30. f(x) œ 2x Ê f w (x) œ 2x ln 2, f ww (x) œ 2x (ln 2)# , f www (x) œ 2x (ln 2)3 Ê f ÐnÑ (x) œ 2x (ln 2)n ; f(1) œ 2, f w (1) œ 2 ln 2,
f ww (1) œ 2(ln 2)# , f www (1) œ 2(ln 2)$ , á , f ÐnÑ (1) œ 2(ln 2)n
#
_
Ê 2x œ 2 (2 ln 2)(x 1) 2(ln# 2) (x 1)# 2(ln3!2) (x 1)3 á œ ! 2(ln 2)n!(x1)
3
n
n
n œ0
31. f(x) œ cosˆ2x 12 ‰, f w (x) œ 2 sinˆ2x 12 ‰, f ww (x) œ 4 cosˆ2x 12 ‰, f www (x) œ 8 sinˆ2x 12 ‰,
f a4b axb œ 24 cosˆ2x 12 ‰ß f a5b axb œ 25 sinˆ2x 12 ‰ß . . ; fˆ 14 ‰ œ 1, f w ˆ 14 ‰ œ 0, f ww ˆ 14 ‰ œ 4, f www ˆ 14 ‰ œ 0, f a4b ˆ 14 ‰ œ 24 ,
2
4
f a5b ˆ 14 ‰ œ 0, . . ., f Ð2nÑ ˆ 14 ‰ œ a1bn 22n Ê cosˆ2x 12 ‰ œ 1 2ˆx 14 ‰ 32 ˆx 14 ‰ . . .
_
n 2n
œ ! aa12nb b2x ˆx 14 ‰
2n
n œ0
7 Î2
32. f(x) œ Èx 1, f w (x) œ 12 ax 1b1Î2 , f ww (x) œ 14 ax 1b3Î2 , f www (x) œ 38 ax 1b5Î2 , f a4b (x) œ 15
, . . .;
16 ax 1b
1
1
3
15
1
1
1
5
f(0) œ 1, f w (0) œ , f ww (0) œ , f www (0) œ , f a4b (0) œ , . . . Ê Èx 1 œ 1 x x2 x3 x4 Þ Þ Þ
2
4
8
16
_
2
8
16
128
n
33. The Maclaurin series generated by cos x is ! aa2n1bbx x2n which converges on a_, _b and the Maclaurin series generated
n œ0
_
by 1 2 x is 2 ! xn which converges on a1, 1b. Thus the Maclaurin series generated by faxb œ cos x 1 2 x is given by
n œ0
_
_
n œ0
nœ0
n
! a1b x2n 2 ! xn œ 1 2x 5 x2 Þ Þ Þ Þ which converges on the intersection of a_, _b and a1, 1b, so the
a2nbx
2
interval of convergence is a1, 1b.
_
n
34. The Maclaurin series generated by ex is ! xnx which converges on a_, _b. The Maclaurin series generated by
n œ0
_
n
faxb œ a1 x x be is given by a1 x x2 b ! xnx œ 1 12 x2 23 x3 Þ Þ Þ Þ which converges on a_, _bÞ
2
x
n œ0
_
n
a1b
2n1
35. The Maclaurin series generated by sin x is ! a2n
which converges on a_, _b and the Maclaurin series
1bx x
n œ0
_
generated by lna1 xb is ! a1nb
n œ1
nc1
xn which converges on a1, 1b. Thus the Maclaurin series genereated by
_
_
n
a1b
a 1 b
2n1
faxb œ sin x † lna1 xb is given by Œ ! a2n
Π! n
1bx x
n œ0
n œ1
nc1
xn œ x2 12 x3 61 x4 Þ Þ Þ Þ which converges on
the intersection of a_, _b and a1, 1b, so the interval of convergence is a1, 1b.
_
n
a1b
2n1
36. The Maclaurin series generated by sin x is ! a2n
which converges on a_, _b. The Maclaurin series
1bx x
n œ0
_
n
2
_
n
_
n
a 1 b
a 1 b
a 1 b
2n1
2n1
2n1
genereated by faxb œ x sin2 x is given by xŒ ! a2n
œ xŒ ! a2n 1bx x
Π! a2n 1bx x
1bx x
n œ0
nœ0
n œ0
2 7
œ x3 13 x5 45
x . . . which converges on a_, _bÞ
_
ÐnÑ
37. If ex œ ! f n!(a) (x a)n and f(x) œ ex , we have f ÐnÑ (a) œ ea f or all n œ 0, 1, 2, 3, á
n œ0
!
"
#
#
Ê e œ ea ’ (x 0!a) (x 1!a) (x 2!a) á “ œ ea ’1 (x a) (x 2!a) á “ at x œ a
x
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 10.9 Convergence of Taylor Series
621
38. f(x) œ ex Ê f ÐnÑ (x) œ ex for all n Ê f ÐnÑ (1) œ e for all n œ 0, 1, 2, á
#
$
Ê ex œ e e(x 1) #e! (x 1)# 3!e (x 1)$ á œ e ’1 (x 1) (x 2!1) (x 3!1) á “
ww
www
39. f(x) œ f(a) f w (a)(x a) f #(a) (x a)# f 3!(a) (x a)$ á Ê f w (x)
www
Ð4Ñ
œ f w (a) f ww (a)(x a) f 3!(a) 3(x a)# á Ê f ww (x) œ f ww (a) f www (a)(x a) f 4!(a) 4 † 3(x a)# á
Ê f ÐnÑ (x) œ f ÐnÑ (a) f Ðn1Ñ (a)(x a) f
Ðn 2Ñ (a)
Ê f(a) œ f(a) 0, f w (a) œ f w (a) 0, á , f
#
Ðn Ñ
(x a)# á
(a) œ f ÐnÑ (a) 0
40. E(x) œ f(x) b! b" (x a) b# (x a)# b$ (x a)$ á bn (x a)n
Ê 0 œ E(a) œ f(a) b! Ê b! œ f(a); from condition (b),
lim
xÄa
f(x) f(a) b" (x a) b# (x a)# b$ (x a)$ á bn (x a)n
œ0
(x a)n
w
#
f (x) b" 2b# (x a) 3b$ (x a) á nbn (x a)
Ê xlim
n(x a)n 1
Äa
n 1
œ0
f ww (x) 2b# 3! b$ (x a) á n(n ")bn (x a)n 2
Ê b" œ f (a) Ê xlim
œ0
n(n 1)(x a)n 2
Äa
n 3
www
f (x) 3! b$ á n(n 1)(n 2)bn (x a)
Ê b# œ "# f ww (a) Ê xlim
œ0
n(n 1)(n #)(x a)n 3
Äa
w
f
œ b$ œ 3!" f www (a) Ê xlim
Äa
ÐnÑ
(x) n! bn
œ0
n!
Ê bn œ n!" f ÐnÑ (a); therefore,
ww
ÐnÑ
g(x) œ f(a) f w (a)(x a) f 2!(a) (x a)# á f n!(a) (x a)n œ Pn (x)
#
41. f(x) œ ln (cos x) Ê f w (x) œ tan x and f ww (x) œ sec# x; f(0) œ 0, f w (0) œ 0, f ww (0) œ 1 Ê L(x) œ 0 and Q(x) œ x2
42. f(x) œ esin x Ê f w (x) œ (cos x)esin x and f ww (x) œ ( sin x)esin x (cos x)# esin x ; f(0) œ 1, f w (0) œ 1, f ww (0) œ 1
#
Ê L(x) œ 1 x and Q(x) œ 1 x x#
"Î#
43. f(x) œ a1 x# b
Ê f w (x) œ x a1 x# b
$Î#
and f ww (x) œ a1 x# b
$Î#
3x# a1 x# b
&Î#
; f(0) œ 1, f w (0) œ 0,
#
f ww (0) œ 1 Ê L(x) œ 1 and Q(x) œ 1 x#
#
44. f(x) œ cosh x Ê f w (x) œ sinh x and f ww (x) œ cosh x; f(0) œ 1, f w (0) œ 0, f ww (0) œ 1 Ê L(x) œ 1 and Q(x) œ 1 x#
45. f(x) œ sin x Ê f w (x) œ cos x and f ww (x) œ sin x; f(0) œ 0, f w (0) œ 1, f ww (0) œ 0 Ê L(x) œ x and Q(x) œ x
46. f(x) œ tan x Ê f w (x) œ sec# x and f ww (x) œ 2 sec# x tan x; f(0) œ 0, f w (0) œ 1, f ww œ 0 Ê L(x) œ x and Q(x) œ x
10.9 CONVERGENCE OF TAYLOR SERIES
_
#
_
1. ex œ 1 x x#! á œ ! xn! Ê e5x œ 1 (5x) (#5x)
á œ 1 5x 5#x! 53!x á œ ! (1)n!5 x
!
#
n
# #
$ $
nœ0
_
#
nœ0
2. ex œ 1 x x#! á œ ! xn! Ê exÎ2 œ 1 ˆ #x ‰ n
nœ0
&
_
3. sin x œ x x3! x5! á œ ! ((#1)nx1)!
$
n 2n 1
n œ0
&
_
4. sin x œ x x3! x5! á œ ! ((#1)nx1)!
$
n n n
nœ0
n 2n 1
_
n n
ˆ x# ‰#
#
$
á œ 1 x# 2x# #! 2x$ 3! á œ ! (21)n n!x
#!
nœ0
$
_
&
x
Ê 5 sin (x) œ 5 ’(x) (3!x) (5!x) á “ œ ! 5((1)
#n1)!
n 1 2n 1
n œ0
Ê sin 1#x œ 1#x _
n 2n 1 2n 1
ˆ 1#x ‰$
ˆ 1#x ‰&
ˆ 1#x ‰(
1
x
! (1)
3! 5! 7! á œ
22n 1 (#n1)!
nœ0
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
622
Chapter 10 Infinite Sequences and Series
_
1) x
5. cos x œ ! ((2n)!
n 2n
n œ0
_
Ê cos 5x2 œ !
n œ0
5 x
625x
15625x
œ ! (1)(2n)!
œ 1 25x
á
6!
#! 4! n 2n 4n
4
$
1bn x2n
cos x œ ! a(2n)!
Ê
n œ0
8
12
n œ0
2n
"Î#
n
_ a1b ŒŠ x# ‹
"Î#
$Î#
$
cos Š xÈ ‹ œ cos ŒŠ x# ‹ œ !
(#n)!
2
_
6.
_
2n
(1)n 5x2 ‘
(2n)!
_
(1)n x3n
2n (2n)!
œ!
nœ0
nœ0
$
'
*
œ 1 2x†2! 2#x†4! 2$x†6! á
_
a1bnc1 ˆx2 ‰
n
_
nc1 n
7. lna1 xb œ ! a1bn x Ê lna1 x2 b œ !
n œ1
nœ1
_
n 2nb1
1b x
8. tan1 x œ ! a2n
1
nœ0
9.
_
a1bn ˆ3x4 ‰
2n 1
Ê tan1 a3x4 b œ !
nœ0
_
n
nc1
œ ! a1b n x œ x2 x2 x3 x4 . . .
2n
4
6
8
nœ1
2nb1
_
2nb1 8nb4
n
œ ! a1b 3 n
x
nœ0
_
_
_
n œ0
nœ0
nœ0
2187 28
20
œ 3x4 9x12 243
5 x 7 x ...
1
! a1bn xn Ê 13 3 œ ! a1bn ˆ 3 x3 ‰n œ ! a1bn ˆ 3 ‰n x3n œ 1 3 x3 9 x6 27 x9 . . .
1x œ
4
4
4
16
64
1 4x
_
_
_
n
10. 1 1 x œ ! xn Ê 2 1 x œ #" 1 1 " x œ #" ! ˆ #" x‰ œ ! ˆ #" ‰
#
n œ0
_
nœ0
_
n
_
n œ0
_
n 2nb1
1) x
12. sin x œ ! ((2n
1)!
nœ0
_
1) x
13. cos x œ ! ((2n)!
n 2n
n œ0
1 3
x œ #" 14 x 18 x2 16
x ...
nœ0
xn
! xnb1 œ x x# x$ x% x& á
#!
n! œ
n!
3!
4!
11. ex œ ! xn! Ê xex œ x Œ !
n œ0
n 1 n
n œ0
_
_
n 2nb1
n 2nb3
1) x
x
x
x
$
Ê x# sin x œ x# Œ ! ((#1)nx1)! œ ! ((2n
1)! œ x 3! 5! 7! á
nœ0
&
(
*
nœ0
_
x
x
x
x
x
x
x
Ê x# 1 cos x œ x# 1 ! ((1)
#n)! œ # 1 1 2 4! 6! 8! 10! á
#
#
#
n 2n
#
%
'
)
"!
n œ0
_
x
x
œ x4! x6! x8! 10!
á œ ! ((1)
#n)!
%
'
)
"!
n 2n
n œ2
_
n 2nb1
1) x
14. sin x œ ! ((2n
1)!
n œ0
$
&
_
$
n 2nb1
$
Ê sin x x x3! œ Œ ! ((#1)nx1)! x x3!
n œ0
(
*
""
$
&
(
*
_
""
1) x
x
x
œ Šx x3! x5! x7! x9! 11!
á ‹ x x3! œ x5! x7! x9! 11!
á œ ! ((2n
1)!
n 2n 1
n œ2
_
1) x
15. cos x œ ! ((2n)!
n 2n
n œ0
_
1) x
16. cos x œ ! ((2n)!
n 2n
n œ0
_
_
n 2n 2nb1
(1x)
Ê x cos 1x œ x ! (1)(#n)!
œ ! (")(#1n)!x
n
2n
nœ0
nœ0
_
_
# 2n
ax b
x
Ê x# cos ax# b œ x# ! (1)(#n)!
œ ! ("(#) n)!
n
n œ0
_
n 4n 2
n œ0
% &
' (
'
"!
"%
œ x# x2! x4! x6! á
%
#
# $
œ x 12!x 14!x 16!x á
'
)
(2x)
(2x)
(2x)
(2x)
17. cos# x œ "# cos#2x œ "# "# ! (1)(2n)!
œ #" "# ’1 (2x)
2! 4! 6! 8! á “
n
2n
n œ0
_
_
n œ1
n œ1
#
n
2n
n 2n 1 2n
(2x)%
(2x)'
(2x))
! (1) (2x) œ 1 ! (1) 2 x
œ 1 (2x)
2†2! 2†4! 2†6! 2†8! á œ 1 2†(2n)!
(2n)!
#
%
'
#
%
'
(2x)
(2x)
(2x)
(2x)
(2x)
2x ‰
18. sin# x œ ˆ 1cos
œ "# "# cos 2x œ "# "# Š1 (2x)
#
#! 4! 6! á ‹ œ 2†2! 2†4! 2†6! á
_
nb1
_
(2x)
2
x
œ ! (1)#†(2n)!
œ ! (1) (2n)!
n œ1
2n
n
2n 1
2n
nœ1
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 10.9 Convergence of Taylor Series
_
_
n œ0
n œ0
#
19. 1x 2x œ x# ˆ 1"2x ‰ œ x# ! (2x)n œ ! 2n xn2 œ x# 2x$ 2# x% 2$ x& á
_
nc1
_
nc1 n n 1
20. x ln (1 2x) œ x ! (1) n (2x) œ ! (1) n2 x
n œ1
n
n œ1
# $
$ %
% &
œ 2x# 2 #x 2 4x 2 5x á
_
_
_
nœ0
nœ1
nœ0
"
d ˆ " ‰
#
! nxn1 œ ! (n 1)xn
21. 1 " x œ ! xn œ 1 x x# x$ á Ê dx
1x œ (1x)# œ 1 2x 3x á œ
_
#
d ˆ " ‰
d
"
d
#
#
! n(n 1)xn2
22. a12xb$ œ dx
#
1x œ dx Š (1x)# ‹ œ dx a1 2x 3x á b œ 2 6x 12x á œ
n œ2
_
œ ! (n 2)(n 1)xn
n œ0
3
5
7
23. tan1 x œ x 13 x3 15 x5 17 x7 Þ Þ Þ Ê x tan1 x2 œ xŠx2 13 ax2 b 15 ax2 b 17 ax2 b Þ Þ Þ ‹
_
n 4nc1
1b x
œ x3 13 x7 15 x11 17 x15 Þ Þ Þ œ ! a2n
1
n œ1
3
5
7
b
a2xb
a2xb
24. sin x œ x x3! x5! x7! á Ê sin x † cos x œ "# sin 2x œ "# Š2x a2x
3! 5! 7! á ‹
3
5
7
_
2nb1
4x
! (1) 2 x
œ x 43!x 165!x 647!x á œ x 23x 2x
15 315 á œ
(#n1)!
3
5
7
3
5
7
n 2n
n œ0
2
3
2
3
25. ex œ 1 x x2! x3! á and 1 1 x œ 1 x x2 x3 á Ê ex 1 1 x
_
4
! ˆ 1 a1bn ‰xn
œ Š1 x x2! x3! á ‹ a1 x x2 x3 á b œ 2 32 x2 56 x3 25
24 x á œ
n!
n œ0
3
5
7
2
4
6
26. sin x œ x x3! x5! x7! á and cos x œ 1 x2! x4! x6! á Ê cos x sin x
2
4
6
3
5
7
2
3
4
5
6
7
œ Š1 x2! x4! x6! á ‹ Šx x3! x5! x7! á ‹ œ 1 x x2! x3! x4! x5! x6! x7! á
_
n 2nb1
1) x
œ ! Š ((2n)!
((#1)nx1)! ‹
n 2n
n œ0
2
3
4
27. lna1 xb œ x 12 x2 13 x3 14 x4 á Ê x3 lna1 x2 b œ x3 Šx2 12 ax2 b 13 ax2 b 14 ax2 b á ‹
_
nc1
1 9
œ 13 x3 16 x5 19 x7 12
x á œ ! a13nb x2n1
n œ1
28. lna1 xb œ x 12 x2 13 x3 14 x4 á and lna1 xb œ x 12 x2 13 x3 14 x4 á Ê lna1 xb lna1 xb
_
œ ˆx 12 x2 13 x3 14 x4 á ‰ ˆx 12 x2 13 x3 14 x4 á ‰ œ 2x 23 x3 25 x5 á œ ! 2n 2 1 x2n1
n œ0
2
3
2
3
3
5
7
29. ex œ 1 x x2! x3! á and sin x œ x x3! x5! x7! á Ê ex † sin x
3
5
7
1 5
œ Š1 x x2! x3! á ‹Šx x3! x5! x7! á ‹ œ x x2 13 x3 30
x ÞÞÞÞ
30. lna1 xb œ x 12 x2 31 x3 41 x4 á and 1 " x œ 1 x x# x$ á Ê ln1a1xxb œ lna1 xb † 1 " x
7 4
œ ˆx 12 x2 13 x3 14 x4 á ‰a1 x x# x$ á b œ x 12 x2 56 x3 12
x ÞÞÞÞ
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
623
624
Chapter 10 Infinite Sequences and Series
2
31. tan1 x œ x 13 x3 15 x5 17 x7 Þ Þ Þ Ê atan1 xb œ atan1 xbatan1 xb
44 8
6
œ ˆx 13 x3 15 x5 17 x7 Þ Þ Þ‰ˆx 13 x3 15 x5 17 x7 Þ Þ Þ‰ œ x2 23 x4 23
45 x 105 x Þ Þ Þ Þ
3
5
7
2
4
6
32. sin x œ x x3! x5! x7! á and cos x œ 1 x2! x4! x6! á Ê cos2 x † sin x œ cos x † cos x † sin x
3
5
7
b
a2xb
a2xb
7 3
61 5
1247 7
œ cos x † "# sin 2x œ "# Š1 x2! x4! x6! á ‹Š2x a2x
3! 5! 7! á ‹ œ x 6 x 120 x 5040 x Þ Þ Þ
2
3
5
4
6
7
2
3
33. sin x œ x x3! x5! x7! á and ex œ 1 x x2! x3! á
3
5
7
3
5
2
7
3
5
3
7
Ê esin x œ 1 Šx x3! x5! x7! á ‹ 12 Šx x3! x5! x7! á ‹ 16 Šx x3! x5! x7! á ‹ á
œ 1 x 12 x2 18 x4 Þ Þ Þ Þ
34. sin x œ x x3! x5! x7! á and tan1 x œ x 13 x3 51 x5 71 x7 Þ Þ Þ Ê sinatan1 xb œ ˆx 13 x3 51 x5 71 x7 Þ Þ Þ‰
3
5
7
3
5
7
1 ˆ
1 ˆ
16 ˆx 31 x3 51 x5 71 x7 Þ Þ Þ‰ 120
x 13 x3 15 x5 17 x7 Þ Þ Þ‰ 5040
x 13 x3 15 x5 17 x7 Þ Þ Þ‰ á
5 7
œ x 12 x3 38 x5 16
x ÞÞÞ
35. Since n œ 3, then f a4b axb œ sin x, lf a4b axbl Ÿ M on Ò0, 0.1Ó Ê lsin xl Ÿ 1 on Ò0, 0.1Ó Ê M œ 1. Then lR3 a0.1bl Ÿ 1 l0.14x 0l
4
œ 4.2 ‚ 106 Ê error Ÿ 4.2 ‚ 106
36. Since n œ 4, then f a5b axb œ ex , lf a5b axbl Ÿ M on Ò0, 0.5Ó Ê lex l Ÿ Èe on Ò0, 0.5Ó Ê M œ 2.7. Then
lR4 a0.5bl Ÿ 2.7 l0.55x 0l œ 7.03 ‚ 104 Ê error Ÿ 7.03 ‚ 104
5
&
37. By the Alternating Series Estimation Theorem, the error is less than kx5!k Ê kxk& a5!b a5 ‚ 10% b Ê kxk& 600 ‚ 10%
5
Ê kxk È
6 ‚ 10# ¸ 0.56968
%
#
38. If cos x œ 1 x# and kxk 0.5, then the error is less than ¹ (.5)
24 ¹ œ 0.0026, by Alternating Series Estimation Theorem;
#
since the next term in the series is positive, the approximation 1 x# is too small, by the Alternating Series Estimation
Theorem
c$ $
39. If sin x œ x and kxk 10$ , then the error is less than a103! b ¸ 1.67 ‚ 1010 , by Alternating Series Estimation Theorem;
$
The Alternating Series Estimation Theorem says R# (x) has the same sign as x3! . Moreover, x sin x
Ê 0 sin x x œ R# (x) Ê x 0 Ê 10$ x 0.
#
$
#
#
x
40. È1 x œ 1 x# x8 16
á . By the Alternating Series Estimation Theorem the kerrork ¹ 8x ¹ (0.01)
8
œ 1.25 ‚ 10&
c $
Ð0Þ1Ñ
c $
$
41. kR# (x)k œ ¹ e3!x ¹ 3
(0.1)$
1.87 ‚ 104 , where c is between 0 and x
3!
%
42. kR# (x)k œ ¹ e3!x ¹ (0.1)
3! œ 1.67 ‚ 10 , where c is between 0 and x
#
%
'
#
$ %
& '
(2x)
(2x)
2x ‰
2x
2 x
2 x
43. sin# x œ ˆ 1 cos
œ "# "# cos 2x œ "# "# Š1 (2x)
#
2! 4! 6! á ‹ œ #! 4! 6! á
#
$ %
& '
$
&
(
(2x)
(2x)
(2x)
d
d
2 x
2 x
Ê dx
asin# xb œ dx
Š 2x
2! 4! 6! á ‹ œ 2x 3! 5! 7! á Ê 2 sin x cos x
$
&
(
(2x)
(2x)
œ 2x (2x)
3! 5! 7! á œ sin 2x, which checks
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 10.9 Convergence of Taylor Series
#
%
'
)
#
$ %
& '
625
( )
(2x)
(2x)
(2x)
2x
2 x
2 x
2 x
44. cos# x œ cos 2x sin# x œ Š1 (2x)
#! 4! 6! 8! á ‹ Š #! 4! 6! 8! á ‹
#
$ %
& '
" %
"
2 x
2 x
2 '
#
)
œ 1 2x
#! 4! 6! á œ 1 x 3 x 45 x 315 x á
45. A special case of Taylor's Theorem is f(b) œ f(a) f w (c)(b a), where c is between a and b Ê f(b) f(a) œ f w (c)(b a),
the Mean Value Theorem.
46. If f(x) is twice differentiable and at x œ a there is a point of inflection, then f ww (a) œ 0. Therefore,
L(x) œ Q(x) œ f(a) f w (a)(x a).
ww
47. (a) f ww Ÿ 0, f w (a) œ 0 and x œ a interior to the interval I Ê f(x) f(a) œ f (c# # ) (x a)# Ÿ 0 throughout I
Ê f(x) Ÿ f(a) throughout I Ê f has a local maximum at x œ a
ww
(b) similar reasoning gives f(x) f(a) œ f (c# # ) (x a)#
0 throughout I Ê f(x)
f(a) throughout I Ê f has a
local minimum at x œ a
48. f(x) œ (1 x)" Ê f w (x) œ (1 x)# Ê f ww (x) œ 2(1 x)$ Ê f Ð3Ñ (x) œ 6(1 x)%
10
"
"
ˆ 10 ‰
Ê f Ð4Ñ (x) œ 24(1 x)& ; therefore 1" x ¸ 1 x x# x$ . kxk 0.1 Ê 10
11 1x 9 Ê ¹ (1x)& ¹ 9
&
%
Ð4Ñ
&
%
&
Ð4Ñ
‰ Ê the error e$ Ÿ ¹ max f 4! (x) x ¹ (0.1)% ˆ 10
‰ œ 0.00016935 0.00017, since ¹ f 4!(x) ¹ œ ¹ (1"x)& ¹ .
Ê ¹ (1x x)& ¹ x% ˆ 10
9
9
49. (a) f(x) œ (1 x)k Ê f w (x) œ k(1 x)k1 Ê f ww (x) œ k(k 1)(1 x)k2 ; f(0) œ 1, f w (0) œ k, and f ww (0) œ k(k 1)
Ê Q(x) œ 1 kx k(k # ") x#
"
"
(b) kR# (x)k œ ¸ 3†3!2†" x$ ¸ 100
Ê kx$ k 100
Ê 0x
"
or 0 x .21544
100"Î$
50. (a) Let P œ x 1 Ê kxk œ kP 1k .5 ‚ 10n since P approximates 1 accurate to n decimals. Then,
P sin P œ (1 x) sin (1 x) œ (1 x) sin x œ 1 (x sin x) Ê k(P sin P) 1k
$
3n
œ ksin x xk Ÿ kx3!k 0.125
.5 ‚ 103n Ê P sin P gives an approximation to 1 correct to 3n decimals.
3! ‚ 10
_
_
n œ0
n œk
51. If f(x) œ ! an xn , then f ÐkÑ (x) œ ! n(n 1)(n 2)â(n k 1)an xnk and f ÐkÑ (0) œ k! ak
Ê
ÐkÑ
ak œ f k!(0) for k a nonnegative integer.
Therefore, the coefficients of f(x) are identical with the corresponding
coefficients in the Maclaurin series of f(x) and the statement follows.
52. Note: f even Ê f(x) œ f(x) Ê f w (x) œ f w (x) Ê f w (x) œ f w (x) Ê f w odd;
f odd Ê f(x) œ f(x) Ê f w (x) œ f w (x) Ê f w (x) œ f w (x) Ê f w even;
also, f odd Ê f(0) œ f(0) Ê 2f(0) œ 0 Ê f(0) œ 0
(a) If f(x) is even, then any odd-order derivative is odd and equal to 0 at x œ 0. Therefore,
a" œ a$ œ a& œ á œ 0; that is, the Maclaurin series for f contains only even powers.
(b) If f(x) is odd, then any even-order derivative is odd and equal to 0 at x œ 0. Therefore,
a! œ a# œ a% œ á œ 0; that is, the Maclaurin series for f contains only odd powers.
53-58. Example CAS commands:
Maple:
f := x -> 1/sqrt(1+x);
x0 := -3/4;
x1 := 3/4;
# Step 1:
plot( f(x), x=x0..x1, title="Step 1: #53 (Section 10.9)" );
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
626
Chapter 10 Infinite Sequences and Series
# Step 2:
P1 := unapply( TaylorApproximation(f(x), x = 0, order=1), x );
P2 := unapply( TaylorApproximation(f(x), x = 0, order=2), x );
P3 := unapply( TaylorApproximation(f(x), x = 0, order=3), x );
# Step 3:
D2f := D(D(f));
D3f := D(D(D(f)));
D4f := D(D(D(D(f))));
plot( [D2f(x),D3f(x),D4f(x)], x=x0..x1, thickness=[0,2,4], color=[red,blue,green], title="Step 3: #57 (Section 9.9)" );
c1 := x0;
M1 := abs( D2f(c1) );
c2 := x0;
M2 := abs( D3f(c2) );
c3 := x0;
M3 := abs( D4f(c3) );
# Step 4:
R1 := unapply( abs(M1/2!*(x-0)^2), x );
R2 := unapply( abs(M2/3!*(x-0)^3), x );
R3 := unapply( abs(M3/4!*(x-0)^4), x );
plot( [R1(x),R2(x),R3(x)], x=x0..x1, thickness=[0,2,4], color=[red,blue,green], title="Step 4: #53 (Section 10.9)" );
# Step 5:
E1 := unapply( abs(f(x)-P1(x)), x );
E2 := unapply( abs(f(x)-P2(x)), x );
E3 := unapply( abs(f(x)-P3(x)), x );
plot( [E1(x),E2(x),E3(x),R1(x),R2(x),R3(x)], x=x0..x1, thickness=[0,2,4], color=[red,blue,green],
linestyle=[1,1,1,3,3,3], title="Step 5: #53 (Section 10.9)" );
# Step 6:
TaylorApproximation( f(x), view=[x0..x1,DEFAULT], x=0, output=animation, order=1..3 );
L1 := fsolve( abs(f(x)-P1(x))=0.01, x=x0/2 );
# (a)
R1 := fsolve( abs(f(x)-P1(x))=0.01, x=x1/2 );
L2 := fsolve( abs(f(x)-P2(x))=0.01, x=x0/2 );
R2 := fsolve( abs(f(x)-P2(x))=0.01, x=x1/2 );
L3 := fsolve( abs(f(x)-P3(x))=0.01, x=x0/2 );
R3 := fsolve( abs(f(x)-P3(x))=0.01, x=x1/2 );
plot( [E1(x),E2(x),E3(x),0.01], x=min(L1,L2,L3)..max(R1,R2,R3), thickness=[0,2,4,0], linestyle=[0,0,0,2],
color=[red,blue,green,black], view=[DEFAULT,0..0.01], title="#53(a) (Section 10.9)" );
abs(`f(x)`-`P`[1](x) ) <= evalf( E1(x0) );
# (b)
abs(`f(x)`-`P`[2](x) ) <= evalf( E2(x0) );
abs(`f(x)`-`P`[3](x) ) <= evalf( E3(x0) );
Mathematica: (assigned function and values for a, b, c, and n may vary)
Clear[x, f, c]
f[x_]= (1 x)3/2
{a, b}= {1/2, 2};
pf=Plot[ f[x], {x, a, b}];
poly1[x_]=Series[f[x], {x,0,1}]//Normal
poly2[x_]=Series[f[x], {x,0,2}]//Normal
poly3[x_]=Series[f[x], {x,0,3}]//Normal
Plot[{f[x], poly1[x], poly2[x], poly3[x]}, {x, a, b},
PlotStyle Ä {RGBColor[1,0,0], RGBColor[0,1,0], RGBColor[0,0,1], RGBColor[0,.5,.5]}];
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 10.10 The Binomial Series
627
The above defines the approximations. The following analyzes the derivatives to determine their maximum values.
f''[c]
Plot[f''[x], {x, a, b}];
f'''[c]
Plot[f'''[x], {x, a, b}];
f''''[c]
Plot[f''''[x], {x, a, b}];
Noting the upper bound for each of the above derivatives occurs at x = a, the upper bounds m1, m2, and m3 can be defined
and bounds for remainders viewed as functions of x.
m1=f''[a]
m2=-f'''[a]
m3=f''''[a]
r1[x_]=m1 x2 /2!
Plot[r1[x], {x, a, b}];
r2[x_]=m2 x3 /3!
Plot[r2[x], {x, a, b}];
r3[x_]=m3 x4 /4!
Plot[r3[x], {x, a, b}];
A three dimensional look at the error functions, allowing both c and x to vary can also be viewed. Recall that c must be a
value between 0 and x, so some points on the surfaces where c is not in that interval are meaningless.
Plot3D[f''[c] x2 /2!, {x, a, b}, {c, a, b}, PlotRange Ä All]
Plot3D[f'''[c] x3 /3!, {x, a, b}, {c, a, b}, PlotRange Ä All]
Plot3D[f''''[c] x4 /4!, {x, a, b}, {c, a, b}, PlotRange Ä All]
10.10 THE BINOMIAL SERIES
1. (1 x)"Î# œ 1 "# x ˆ "# ‰ ˆ "# ‰ x#
2. (1 x)"Î$ œ 1 "3 x ˆ "3 ‰ ˆ 23 ‰ x#
#!
#!
3. (1 x)"Î# œ 1 "# (x) ˆ "# ‰ ˆ "# ‰ ˆ 3# ‰ x$
3!
ˆ 3" ‰ ˆ 32 ‰ ˆ 53 ‰ x$
#!
#!
ˆ x ‰#
x ‰#
ˆ
(4)(3) 3
4
6. ˆ1 x3 ‰ œ 1 4 ˆ x3 ‰ #!
"Î#
8. a1 x# b
"Î$
œ 1 "# x$ œ 1 "3 x# "Î#
9. ˆ1 1x ‰ œ 1 "# ˆ x1 ‰ ˆ "# ‰ ˆ "# ‰ (2x)#
(2)(3) #
#
5. ˆ1 x# ‰ œ 1 # ˆ x# ‰ #!
7. a1 x$ b
3!
ˆ "# ‰ ˆ 3# ‰ (x)#
4. (1 2x)"Î# œ 1 "# (2x) ˆ "# ‰ ˆ 3# ‰ ˆ 5# ‰ (x)$
3!
Š "# ‹ Š "# ‹ Š #3 ‹ (2x)$
3!
$
3!
(4)(3)(2) ˆ x3 ‰
#!
ˆ "3 ‰ ˆ 43 ‰ ax# b#
#!
ˆ "# ‰ ˆ "# ‰ ˆ 1x ‰#
5 $
á œ 1 3" x 9" x# 81
x á
(2)(3)(4) ˆ x# ‰
ˆ "# ‰ ˆ 3# ‰ ax$ b#
#!
" $
á œ 1 "# x "8 x# 16
x á
3!
$
á œ 1 x 12 x# 12 x$ á
á œ 1 x 34 x# "# x$
(4)(3)(2)(1) ˆ x3 ‰
4!
ˆ #" ‰ ˆ #3 ‰ ˆ #5 ‰ ax$ b$
3!
ˆ 3" ‰ ˆ 43 ‰ ˆ 73 ‰ ax# b$
3!
ˆ "# ‰ ˆ "# ‰ ˆ 3# ‰ ˆ 1x ‰$
3!
5 $
á œ 1 "# x 38 x# 16
x á
4
4 3
1 4
0 á œ 1 34 x 32 x2 27
x 81
x
5 *
á œ 1 "# x$ 38 x' 16
x á
'
á œ 1 "3 x# 29 x% 14
81 x á
"
á œ 1 #"x 8x1 # 16x
$ á
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
628
Chapter 10 Infinite Sequences and Series
ˆ "3 ‰ ˆ 43 ‰ x#
x
10. È
œ xa1 xb"Î3 œ xŒ1 ˆc "3 ‰x 3
1x
#
$
ˆ "3 ‰ ˆ 43 ‰ ˆ 73 ‰ x$
#!
14 4
á œ x 31 x# 92 x3 81
x á
3!
%
11. (1 x)% œ 1 4x (4)(3)x
(4)(3)(2)x
(4)(3)(2)x
œ 1 4x 6x# 4x$ x%
#!
3!
4!
# #
$
# $
ax b
12. a1 x# b œ 1 3x# (3)(2)#!ax b (3)(2)(1)
œ 1 3x# 3x% x'
3!
#
$
2x)
2x)
13. (1 2x)$ œ 1 3(2x) (3)(2)(#
(3)(2)(1)(
œ 1 6x 12x# 8x$
!
3!
%
14. ˆ1 #x ‰ œ 1 4 ˆ #x ‰ (4)(3) ˆ x# ‰
#
#!
(4)(3)(2) ˆ x# ‰
3!
$
(4)(3)(2)(1) ˆ x# ‰
15. '0 sin x# dx œ '0 Šx# x3! x5! á ‹ dx œ ’ x3 7x†3! á “
0Þ2
0 Þ2
'
"!
%
4!
$
!Þ#
(
!
(
" %
œ 1 2x 23 x# "# x$ 16
x
$
¸ ’ x3 “
!Þ#
!
¸ 0.00267 with error
kEk Ÿ (.2)
7†3! ¸ 0.0000003
16. '0
0Þ2
ecx "
dx œ
x
'00 2 x" Š1 x x#! x3! x4! á 1‹ dx œ '00 2 Š1 x# x6 24x á ‹ dx
Þ
#
#
$
x
œ ’x x4 18
á“
17. '0
0Þ1
!Þ#
!
$
Þ
%
%
'00 1 Š1 x2 3x8 á ‹ dx œ ’x 10x á “ !Þ" ¸ [x]!Þ"
! ¸ 0.1 with error
Þ
"
È1 x% dx œ
!Þ#&
$È
$
¸ 0.19044 with error kEk Ÿ (0.2)
96 ¸ 0.00002
%
)
&
!
&
kEk Ÿ (0.1)
10 œ 0.000001
18. '!
#
1 x# dx œ '0
0Þ25
#
%
$
&
x
á“
Š1 x3 x9 á ‹ dx œ ’x x9 45
&
!Þ#&
!
$
¸ ’x x9 “
!Þ#&
!
¸ 0.25174 with error
kEk Ÿ (0.25)
45 ¸ 0.0000217
19. '0
0Þ1
sin x
x dx œ
'00 1 Š1 x3! x5! x7! á ‹ dx œ ’x 3x†3! 5x†5! 7x†7! á “ !Þ" ¸ ’x 3x†3! 5x†5! “ !Þ"
Þ
#
%
'
$
&
(
$
&
!
!
7
12
¸ 0.0999444611, kEk Ÿ (0.1)
7†7! ¸ 2.8 ‚ 10
x
x
20. '0 exp ax# b dx œ '0 Š1 x# x2! x3! x4! á ‹ dx œ ’x x3 10
42
á“
0Þ1
0 Þ1
%
'
)
$
&
(
!Þ"
!
$
&
9
12
¸ 0.0996676643, kEk Ÿ (0.1)
216 ¸ 4.6 ‚ 10
21. a1 x% b
"Î#
œ (1)"Î# ˆ "# ‰ ˆ "# ‰ ˆ 3# ‰ ˆ 5# ‰
4!
Š "# ‹
1
(1)"Î# ax% b ˆ "# ‰ ˆ "# ‰
#!
%
%
#
(1)$Î# ax% b )
%
3!
(1)&Î# ax% b
)
"'
"#
"'
&
!Þ"
!
9
11
¸ 0.100001, kEk Ÿ (0.1)
72 ¸ 1.39 ‚ 10
x‰
x
22. '0 ˆ 1 xcos
dx œ '0 Š "# x4! x6! x8! 10!
á ‹ dx ¸ ’ x# 3x†4! 5x†6! 7x†8! 9†x10! “
#
1
1
¸ 0.4863853764,
#
%
'
)
$
&
1
(
kEk Ÿ 11†112! ¸ 1.9 ‚ 1010
t
t
23. '0 cos t# dt œ '0 Š1 t# 4!
t6! á ‹ dt œ ’t 10
9t†4! 13t †6! á “
1
$
(1)(Î# ax% b á œ 1 x# x8 x16 5x
128 á
x
Ê '0 Š1 x# x8 x16 5x
128 á ‹ dx ¸ ’x 10 “
0Þ1
"#
ˆ "# ‰ ˆ "# ‰ ˆ 3# ‰
%
)
"#
&
*
"$
"
!
*
(
x
x
¸ ’x x3 10
42
“
"
!
Ê kerrork 13"†6! ¸ .00011
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
!Þ"
!
Section 10.10 The Binomial Series
t
t
t
24. '0 cos Èt dt œ '0 Š1 #t 4!
6!
8!
á ‹ dt œ ’t t4 3t†4! 4t†6! 5t†8! á “
1
1
#
$
%
#
$
%
&
Ê kerrork 5†"8! ¸ 0.000004960
t
25. F(x) œ '0 Št# 3!
t5! t7! á ‹ dt œ ’ t3 7t†3! 11t †5! 15t †7! á “
x
'
"!
"%
$
(
""
"&
Ê kerrork 15"†7! ¸ 0.000013
x
$
"
!
(
""
¸ x3 7x†3! 11x †5!
!
t
t
26. F(x) œ '0 Št# t% 2!
3!
t4! t5! á ‹ dt œ ’ t3 t5 7t†2! 9t†3! 11t †4! 13t †5! á “
x
$
'
&
(
)
*
"!
"#
$
&
(
629
*
""
"$
""
x
!
¸ x3 x5 7x†2! 9x†3! 11x †4! Ê kerrork 13"†5! ¸ 0.00064
t
27. (a) F(x) œ '0 Št t3 t5 t7 á ‹ dt œ ’ t2 1t # 30
á “ ¸ x# 1x# Ê kerrork (0.5)
30 ¸ .00052
x
$
&
(
#
#
%
x
'
%
'
#
'
%
!
)
$#
(b) kerrork 33"†34 ¸ .00089 when F(x) ¸ x# 3x†4 5x†6 7x†8 á (1)"& 31x†32
28. (a) F(x) œ '0 Š1 2t t3 t4 á ‹ dt œ ’t 2t†2 3t†3 4t†4 5t†5 á “ ¸ x x## 3x# 4x# 5x#
x
#
$
#
$
%
x
&
#
$
%
&
!
'
Ê kerrork (0.5)
6# ¸ .00043
#
$
%
$"
x
(b) kerrork 32" # ¸ .00097 when F(x) ¸ x x## x3# x4# á (1)$" 31
#
#
$
#
29. x"# aex (1 x)b œ x"# ŠŠ1 x x# x3! á ‹ 1 x‹ œ #" 3!x x4! á Ê lim
xÄ0
ex (1 x)
x#
#
œ lim Š "# 3!x x4! á ‹ œ "#
xÄ0
$
#
%
#
$
%
$
&
(
2x
2x
30. x" aex ex b œ x" ’Š1 x x#! x3! x4! á ‹ Š1 x x#! x3! x4! á ‹“ œ x" Š2x 2x
3! 5! 7! á ‹
#
%
'
#
ex e x
2x%
2x'
œ x lim
Š2 2x
x
3! 5! 7! á ‹ œ 2
Ä_
2x
2x
œ 2 2x
3! 5! 7! á Ê lim
xÄ0
#
#
#
%
'
#
#
%
t
t
t
t
31. t"% Š1 cos t t# ‹ œ t"% ’1 t# Š1 t# 4!
6!
á ‹“ œ 4!" 6!
8!
á Ê lim
" cos t Š t# ‹
t%
tÄ0
#
%
t
t
"
œ lim Š 4!" 6!
8!
á ‹ œ 24
tÄ0
$
$
$
&
#
$
%
32. )"& Š) )6 sin )‹ œ )"& Š) )6 ) )3! )5! á ‹ œ 5!" )7! )9! á Ê lim
)Ä0
)%
#
sin ) ) Š )6 ‹
)&
œ lim Š 5!" )7! 9! á ‹ œ 1"#0
)Ä0
$
&
#
%
33. y"$ ay tan" yb œ y"$ ’y Šy y3 y5 á ‹“ œ "3 y5 y7 á Ê lim
yÄ0
#
%
y tan " y
œ lim Š 3" y5 y7 á ‹
y$
yÄ0
œ "3
34.
tanc" y sin y
œ
y$ cos y
Ê lim
yÄ0
y$
y&
y$
y&
Œy 3 5 á Œy 3! 5! á
y$ cos y
tanc" y sin y
œ lim
y$ cos y
yÄ0
23y#
"
Œ 6 5! á
cos y
y$
œ
Π6 23y&
5! á
y$ cos y
œ
"
Π6 23y#
5! á
cos y
œ 6"
#
#
35. x# Š1 e1Îx ‹ œ x# ˆ1 1 x"# #"x% 6x" ' á ‰ œ 1 #"x# 6x" % á Ê x lim
x# Še1Îx 1‹
Ä_
ˆ1 #x" # 6x" % á ‰ œ 1
œ x lim
Ä_
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
630
Chapter 10 Infinite Sequences and Series
36. (x 1) sin ˆ x " 1 ‰ œ (x 1) Š x " 1 3!(x " 1)$ 5!(x " 1)& á ‹ œ 1 3!(x " 1)# 5!(x " 1)% á
Ê x lim
(x 1) sin ˆ x " 1 ‰ œ x lim
Š1 3!(x " 1)# 5!(x " 1)% á ‹ œ 1
Ä_
Ä_
%
37.
'
#
%
#
%
x
x
x
x
x
x
#
Œx # 3 á
Œ1 # 3 á
Œ1 # 3 á
ln a1 x# b
ln a1 x# b
œ
œ
Ê
lim
œ
lim
œ 2! œ 2
#
#
#
%
1 cos x
x Ä 0 1 cos x
xÄ0
1 Š1 x#! x4! á‹
Š #"! x4! á‹
Š #"! x4! á‹
(x 2)(x 2)
#
38. lnx(x41) œ
’(x 2) (x c 2)#
#
x 2
œ lim
(x c 2)$
á“
3
#
x Ä 2 ’1 x c# 2 (x 32) á“
œ
x2
’1 x # 2 Ê lim
(x c 2)#
á“
3
x# 4
x Ä 2 ln (x 1)
œ4
81 10
4 6
39. sin 3x2 œ 3x2 92 x6 40
x . . . and 1 cos 2x œ 2x2 23 x4 45
x . . . Ê lim
sin 3x2
x Ä 0 1 cos 2x
10
8
3x2 92 x6 81
3 92 x4 81
40 x . . .
40 x . . .
œ 32
2 2
4 4
2 2 x4 4 x6 . . . œ lim
2x
2
x
x
. . .
xÄ0
xÄ0
3
45
3
45
œ lim
6
9
12
1 11
1
40. lna1 x3 b œ x3 x2 x3 x4 . . . and x sin x2 œ x3 16 x7 120
x 5040
x15 Þ Þ Þ Ê lim
xÄ0
œ lim
xÄ0
6
9
12
x3 x2 x3 x4 . . .
1 7
1 11
1
3
x 6 x 120 x 5040
x15 Þ Þ Þ
3
6
9
1 x2 x3 x4 . . .
1 4
1 8
1
1 6 x 120 x 5040
x12 Þ Þ Þ
œ lim
xÄ0
œ1
41. 1 1 21x 31x 41x Þ Þ Þ œ e1 œ e
3
4
5
3
2
1
1
1 4
1
42. ˆ 14 ‰ ˆ 14 ‰ ˆ 14 ‰ Þ Þ Þ œ ˆ 14 ‰ ”1 ˆ 14 ‰ ˆ 14 ‰ Þ Þ Þ • œ 64
1 1Î4 œ 64 3 œ 48
2
4
2
6
4
6
43. 1 432 2x 434 4x 436 6x Þ Þ Þ œ 1 21x ˆ 34 ‰ 41x ˆ 34 ‰ 61x ˆ 34 ‰ Þ Þ Þ œ cosˆ 34 ‰
2
3
4
5
7
5
7
44. 21 2†122 3†123 4†124 Þ Þ Þ œ ˆ 21 ‰ 21 ˆ #1 ‰ 31 ˆ #1 ‰ 41 ˆ #1 ‰ Þ Þ Þ œ lnˆ1 21 ‰ œ lnˆ 23 ‰
3
45. 13 313 3x 315 5x 317 7x Þ Þ Þ œ 13 31x ˆ 13 ‰ 51x ˆ 13 ‰ 71x ˆ 13 ‰ Þ Þ Þ œ sinˆ 13 ‰ œ
3
5
7
3
È3
2
46. 23 323 †3 325 †5 327 †7 Þ Þ Þ œ ˆ 32 ‰ 31 ˆ 32 ‰ 51 ˆ 32 ‰ 71 ˆ 32 ‰ Þ Þ Þ œ tan1 ˆ 23 ‰
3
5
7
3
47. x3 x4 x5 x6 Þ Þ Þ œ x3 a1 x x2 x3 Þ Þ Þ b œ x3 ˆ 1 1 x ‰ œ 1 x x
2 2
4 4
6 6
48. 1 32xx 34xx 36xx Þ Þ Þ œ 1 21x a3xb2 41x a3xb4 61x a3xb6 Þ Þ Þ œ cosa3xb
2
3
3
49. x3 x5 x7 x9 Þ Þ Þ œ x3 Š1 x2 ax2 b ax2 b Þ Þ Þ ‹ œ x3 ˆ 1 +1x2 ‰ œ 1 +x x2
2
3
4
50. x2 2x3 22xx 23xx 24xx Þ Þ Þ œ x2 Š1 2x a2x2xb a2x3xb a2x4xb Þ Þ Þ ‹ œ x2 e2x
2 4
3 5
4 6
d
d ˆ 1 ‰
1
51. 1 2x 3x2 4x3 5x4 Þ Þ Þ œ dx
a1 x x2 x3 x4 x5 Þ Þ Þ b œ dx
1 x œ a1 x b 2
52. 1 x2 x3 x4 x5 Þ Þ œ 1x Šx x2 x3 x4 x5 Þ Þ ‹ œ 1x lna1 xb œ lna1xxb
2
3
4
2
3
4
5
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
lnˆ1 x3 ‰
x sin x2
Section 10.10 The Binomial Series
$
#
%
#
$
%
$
631
&
x‰
x
x
x
x
x
x
x
x
53. ln ˆ 11 x œ ln (1 x) ln (1 x) œ Šx # 3 4 á ‹ Šx # 3 4 á ‹ œ 2 Šx 3 5 á ‹
#
$
%
"
54. ln (1 x) œ x x# x3 x4 á (1)n x á Ê kerrork œ ¹ (")n x ¹ œ n10
n when x œ 0.1;
"
"
n10n 10)
n 1 n
Ê n10n 10) when n
$
&
(
n 1 n
8 Ê 7 terms
*
) x
55. tan" x œ x x3 x5 x7 x9 á ("2n
1
"
"
#n1 10$
n 1 2n 1
á Ê kerrork œ ¹ (1)2nx1
n 1 2n 1
¹ œ #n"1 when x œ 1;
st
Ê n 1001
# œ 500.5 Ê the first term not used is the 501 Ê we must use 500 terms
&
(
n1 2n1
*
56. tan" x œ x x3 x5 x7 x9 á (1)2nx1
$
¸ 2n 1 ¸ œ x#
á and n lim
† 2n 1 ¹ œ x# n lim
¹x
Ä _ 2n 1 x2n1
Ä _ #n 1
2n 1
_
Ê tan" x converges for kxk 1; when x œ 1 we have ! (2n1)1 which is a convergent series; when x œ 1
n
n œ1
_
nc1
we have ! (2n1)1 which is a convergent series
Ê the series representing tan" x diverges for kxk 1
n œ1
$
&
(
*
57. tan" x œ x x3 x5 x7 x9 á (1)2n x1
n 1 2n 1
" ‰
á and when the series representing 48 tan" ˆ 18
has an
error less than "3 † 10' , then the series representing the sum
" ‰
" ‰
48 tan" ˆ 18
32 tan" ˆ 57
20 tan" ˆ #"39 ‰ also has an error of magnitude less than 10' ; thus
2nc1
Š"‹
"
kerrork œ 48 #18n 1 3†10
' Ê n
4 using a calculator Ê 4 terms
x
x
58. ln (sec x) œ '0 tan t dt œ '0 Št t3 2t15 á ‹ dt ¸ x# 12
45
á
x
59. (a) a1 x# b
lim
nÄ_
"Î#
x
#
$
%
&
#
'
%
$
'
&
(
5x
"
¸ 1 x# 3x8 5x
x ¸ x x6 3x
16 Ê sin
40 112 ; Using the Ratio Test:
2nb3
(2n 1)(2n1)
(2n 1)(2n 1)x
2†4†6â(2n)(2n ")
#
¹ 12††34††56â
¹
¹1
â(2n)(2n 2)(2n 3) † 1†3†5â(2n 1)x2nb1 ¹ 1 Ê x n lim
Ä _ (2n 2)(2n 3)
Ê kxk 1 Ê the radius of convergence is 1. See Exercise 69.
(b)
"Î#
d
"
xb œ a1 x# b
dx acos
60. (a) a1 t# b
"Î#
$
&
(
$
&
"‰ ˆ
¸ (1)"Î# ˆ "# ‰ (1)$Î# at# b #
ˆ
#3 ‰ (1) &Î# at# b
#!
#
ˆ #" ‰ ˆ #3 ‰ ˆ #5 ‰ (1) (Î# at# b$
3!
3†5t
t
3t
5t
x
3x
5x
"
œ 1 t# 23t
# †2! 2$ †3! Ê sinh x ¸ '0 Š1 # 8 16 ‹ dt œ x 6 40 112
#
%
'
x
#
%
'
$
&
(
"
3
(b) sinh" ˆ 4" ‰ ¸ 4" 384
40,960
œ 0.24746908; the error is less than the absolute value of the first unused
(
(
5x
1
x
3x
5x
Ê cos" x œ 1# sin" x ¸ 1# Šx x6 3x
40 112 ‹ ¸ # x 6 40 112
5x
term, 112
, evaluated at t œ 4" since the series is alternating Ê kerrork 5 ˆ 4" ‰
112
(
¸ 2.725 ‚ 10'
"
d ˆ 1 ‰
"
d
#
$
#
$
61. 1"
x œ 1 (x) œ 1 x x x á Ê dx 1 x œ 1 x# œ dx a1 x x x á b
œ 1 2x 3x# 4x$ á
d ˆ " ‰
2x
d
#
%
'
$
&
62. 1 " x# œ 1 x# x% x' á Ê dx
1 x# œ a1 x# b# œ dx a1 x x x á b œ 2x 4x 6x á
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
632
Chapter 10 Infinite Sequences and Series
8â(2n 2)†(2n)
63. Wallis' formula gives the approximation 1 ¸ 4 ’ 3†23††45††45††67††67†â
(2n 1)†(2n 1) “ to produce the table
n
µ1
10
3.221088998
20
3.181104886
30
3.167880758
80
3.151425420
90
3.150331383
93
3.150049112
94
3.149959030
95
3.149870848
100
3.149456425
At n œ 1929 we obtain the first approximation accurate to 3 decimals: 3.141999845. At n œ 30,000 we still do
not obtain accuracy to 4 decimals: 3.141617732, so the convergence to 1 is very slow. Here is a Maple CAS
procedure to produce these approximations:
pie :=
proc(n)
local i,j;
a(2) := evalf(8/9);
for i from 3 to n do a(i) := evalf(2*(2*i2)*i/(2*i1)^2*a(i1)) od;
[[j,4*a(j)] $ (j = n5 .. n)]
end
_
_
_
64. (a) faxb œ 1 !ˆ mk ‰xk Ê f w axb œ !ˆ mk ‰k xk1 Ê a1 xb † f w axb œ a1 xb!ˆ mk ‰k xk1
kœ1
kœ1
_
_
kœ1
_
_
_
_
kœ2
kœ1
œ !ˆ mk ‰k xk 1 x † !ˆ mk ‰k xk1 œ !ˆ mk ‰k xk 1 !ˆ mk ‰k xk œ ˆ m1 ‰a1b x0 !ˆ mk ‰k xk1 !ˆ mk ‰k xk
kœ1
kœ1
_
kœ1
_
œ m ! ˆ mk ‰k xk 1 ! ˆ mk ‰k xk
kœ2
kœ1
w
kœ1
_
_
kœ2
kœ1
_
_
kœ1
kœ1
k
‰
Note that: ! ˆ mk ‰k xk 1 œ ! ˆ k m
1 ak 1b x .
_
_
kœ2
kœ1
k
! ˆ m ‰k xk
‰
Thus, a1 xb † f axb œ m ! ˆ mk ‰k xk 1 ! ˆ mk ‰k xk œ m ! ˆ k m
1 ak 1b x k
_
_
kœ1
kœ1
k
! ’ˆˆ m ‰ak 1b ˆ m ‰k ‰xk “.
‰
ˆm‰ k
œ m ! ’ˆ k m
1 ak 1b x k k x “ œ m k1
k
m†am "bâam ak 1b 1b
m ‰
ˆm‰
Note that: ˆ k ak 1b m†am "bâk!am k 1b k
1 ak 1b k k œ
ak1b!
œ m†am "k!bâam kb m†am "bâk!am k 1b k œ m†am "bâk!am k 1b aam kb kb œ m m†am "bâk!am k 1b œ mˆ mk ‰.
_
_
_
kœ1
kœ1
kœ1
! ’ˆmˆ m ‰ ‰xk “ œ m m!ˆ m ‰xk
‰
ˆm‰ ‰ k
Thus, a1 xb † f w axb œ m ! ’ˆˆ k m
1 ak 1b k k x “ œ m k
k
_
†faxb
œ mŒ1 !ˆ mk ‰xk œ m † faxb Ê f w axb œ m
a1xb if " x 1.
kœ1
(b) Let gaxb œ a1 xbm faxb Ê gw axb œ ma1 xbm1 faxb a1 xbm f w axb
†faxb
m1
œ ma1 xbm1 faxb a1 xbm † m
faxb a1 xbm1 † m † faxb œ 0.
a1xb œ ma1 xb
_
(c) gw axb œ 0 Ê gaxb œ c Ê a1 xbm faxb œ c Ê faxb œ a1cxbcm œ ca1 xbm . Since faxb œ 1 !ˆ mk ‰xk
kœ1
_
Ê fa0b œ 1 !ˆ mk ‰a0bk œ 1 0 œ 1 Ê ca1 0bm œ 1 Ê c œ 1 Ê faxb œ a1 xbm .
kœ1
65. a1 x# b
"Î#
œ a1 ax# bb
"Î#
ˆ "# ‰ ˆ 3# ‰ ˆ 5# ‰ (1) (Î# ax# b$
3!
œ (1)"Î# ˆ "# ‰ (1)$Î# ax# b ˆ "# ‰ ˆ 3# ‰ (1)c&Î# ax# b#
_
#!
1†3†5x
! 1†3†5ân(2n1)x
á œ 1 x# 12†#3x
†#! 2$ †3! á œ 1 # †n!
#
%
'
2n
n œ1
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 10.10 The Binomial Series
Ê sin" x œ '0 a1 t# b
x
633
1†3†5â(2n 1)x
1)x
dt œ '0 Œ1 ! 1†3†5â#(2n
n †n!
dt œ x ! #†4â(2n)(2n 1) ,
_
x
"Î#
_
2n
n œ1
2nb1
nœ1
where kxk 1
_ dt
66. ctan" td x œ 1# tan" x œ 'x
_
_
1 t#
_
_ "
Š 1# ‹
œ 'x – t " — dt œ 'x
1Š ‹
t#
œ 'x ˆ t"# t"% t"' t") á ‰ dt œ lim
bÄ_
t#
ˆ1 t"# t"% t"' á ‰ dt
"t 3t"$ 5t"& 7t"( á ‘ b œ "x 3x" $ 5x" & 7x" ( á
x
Ê tan" x œ 1# x" 3x" $ 5x" & á , x 1; ctan" td c_ œ tan" x 1# œ ' _ 1 dt t#
x
x
œ
lim
b Ä _
"t 3t"$ 5t"& 7t"( á ‘ x œ "x 3x" $ 5x" & 7x" ( á Ê tan" x œ 1# x" 3x" $ 5x" & á ,
b
x 1
67. (a) ei1 œ cos (1) i sin (1) œ 1 i(0) œ 1
(b) ei1Î4 œ cos ˆ 14 ‰ i sin ˆ 14 ‰ œ È" Èi œ Š È" ‹ (1 i)
2
2
2
(c) ei1Î2 œ cos ˆ 1# ‰ i sin ˆ 1# ‰ œ 0 i(1) œ i
68. ei) œ cos ) i sin ) Ê ei) œ ei()) œ cos ()) i sin ()) œ cos ) i sin );
i)
ei) ei) œ cos ) i sin ) cos ) i sin ) œ 2 cos ) Ê cos ) œ e #e
i)
i)
e e
ci)
;
i)
ci)
sin ) œ e #ie
œ cos ) i sin ) (cos ) i sin )) œ 2i sin ) Ê
#
$
%
))
))
))
69. ex œ 1 x x#! x3! x4! á Ê ei) œ 1 i) (i2!
(i3!
(i4!
á and
#
$
%
#
$
%
#
$
%
))
))
ei) œ 1 i) (2!i)) (3!i)) (4!i)) á œ 1 i) (i#)!) (i3!
(i4!
á
#
$
%
#
$
%
))
))
(i4!
á‹
Š1 i) (i)) (i)) (i)) á‹ Š1 i) (i#)!) (i3!
i)
ci)
#!
3!
4!
Ê e #e œ
#
%
'
œ 1 )#! )4! )6! á œ cos );
#
$
#
%
#
$
%
))
))
(i4!
á‹
Š1 i) (i)) (i)) (i)) á‹ Š1 i) (i#)!) (i3!
#!
3!
4!
ei) eci)
œ
#i
$
&
(
œ ) )3! )5! )7! á œ sin )
#i
70. ei) œ cos ) i sin ) Ê ei) œ eiÐ)Ñ œ cos ()) i sin ()) œ cos ) i sin )
i)
(a) ei) ei) œ (cos ) i sin )) (cos ) i sin )) œ 2 cos ) Ê cos ) œ e #e
(b) ei) ei) œ (cos ) i sin )) (cos ) i sin )) œ 2i sin ) Ê i sin ) œ
#
$
%
$
&
ci)
œ cosh i)
e eci)
i)
2
œ sinh i)
(
71. ex sin x œ Š1 x x#! x3! x4! á ‹ Šx x3! x5! x7! á ‹
" &
œ (1)x (1)x# ˆ "6 #" ‰ x$ ˆ 6" "6 ‰ x% ˆ 1#"0 1"# #"4 ‰ x& á œ x x# "3 x$ 30
x á ;
ex † eix œ eÐ1iÑx œ ex (cos x i sin x) œ ex cos x i aex sin xb Ê ex sin x is the series of the imaginary part
_
#
$
%
of eÐ1iÑx which we calculate next; eÐ1iÑx œ ! (xn!ix) œ 1 (x ix) (x #!ix) (x 3!ix) (x 4!ix) á
n
n œ0
œ 1 x ix #"! a2ix# b 3!" a2ix$ 2x$ b 4!" a4x% b 5!" a4x& 4ix& b 6!" a8ix' b á Ê the imaginary part
" &
" '
of eÐ1iÑx is x #2! x# 3!2 x$ 5!4 x& 6!8 x' á œ x x# "3 x$ 30
x 90
x á in agreement with our
product calculation. The series for ex sin x converges for all values of x.
d ˆ ÐaibÑ ‰
d
ceax (cos bx i sin bx)d œ aeax (cos bx i sin bx) eax (b sin bx bi cos bx)
72. dx
e
œ dx
œ aeax (cos bx i sin bx) bieax (cos bx i sin bx) œ aeÐaibÑx ibeÐaibÑx œ (a ib)eÐaibÑx
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
634
Chapter 10 Infinite Sequences and Series
73. (a) ei)" ei)# œ (cos )" i sin )" )(cos )# i sin )# ) œ (cos )" cos )# sin )" sin )# ) i(sin )" cos )# sin )# cos )" )
œ cos()" )# ) i sin()" )# ) œ eiÐ)" )# Ñ
) i sin ) ‰
"
"
(b) ei) œ cos()) i sin()) œ cos ) i sin ) œ (cos ) i sin )) ˆ cos
cos ) i sin ) œ cos ) i sin ) œ ei)
74. aa# bib# eÐabiÑx C" iC# œ ˆ aa# bib# ‰ eax (cos bx i sin bx) C" iC#
ax
œ a# e b# (a cos bx ia sin bx ib cos bx b sin bx) C" iC#
ax
œ a# e b# [(a cos bx b sin bx) (a sin bx b cos bx)i] C" iC#
œ e (a cosa#bxb#b sin bx) C" ie (a sina#bxb#b cos bx) iC# ;
ax
ax
eÐabiÑx œ eax eibx œ eax (cos bx i sin bx) œ eax cos bx ieax sin bx, so that given
' eÐabiÑx dx œ aa bib eÐabiÑx C" iC# we conclude that ' eax cos bx dx œ e (a cosa bxb b sin bx) C"
and ' eax sin bx dx œ e (a sinabxbb cos bx) C#
ax
#
#
#
#
ax
#
#
CHAPTER 10 PRACTICE EXERCISES
1. converges to 1, since n lim
a œ n lim
Š1 (n1) ‹ œ 1
Ä_ n
Ä_
n
2
2. converges to 0, since 0 Ÿ an Ÿ È2n , n lim
0 œ 0, n lim
œ 0 using the Sandwich Theorem for Sequences
Ä_
Ä _ Èn
ˆ 1 2n2 ‰ œ lim ˆ #"n 1‰ œ 1
3. converges to 1, since n lim
a œ n lim
Ä_ n
Ä_
nÄ_
n
4. converges to 1, since n lim
a œ n lim
c1 (0.9)n d œ 1 0 œ 1
Ä_ n
Ä_
™ œ e0ß 1ß 0ß 1ß 0ß 1ß á f
5. diverges, since ˜sin n1
#
6. converges to 0, since {sin n1} œ {0ß 0ß 0ß á }
#
ln n
7. converges to 0, since n lim
a œ n lim
œ 2 n lim
Ä_ n
Ä_ n
Ä_
Š "n ‹
1
Š 2n 2b 1 ‹
ln (2n")
8. converges to 0, since n lim
a œ n lim
œ n lim
n
Ä_ n
Ä_
Ä_
1
1Š "n ‹
ˆ n nln n ‰ œ lim
9. converges to 1, since n lim
a œ n lim
Ä_ n
Ä_
nÄ_
$
œ0
1
ln a2n 1b
10. converges to 0, since n lim
a œ n lim
œ n lim
n
Ä_ n
Ä_
Ä_
Š
œ0
œ1
6n#
‹
2n$ 1
1
12n
2
œ n lim
œ n lim
œ0
Ä _ 6n#
Ä_ n
n
ˆ n n 5 ‰ œ lim Š1 (n5) ‹ œ ec5 by Theorem 5
11. converges to ec5 , since n lim
a œ n lim
Ä_ n
Ä_
nÄ_
n
ˆ1 "n ‰
12. converges to "e , since n lim
a œ n lim
Ä_ n
Ä_
n
"
"
œ n lim
n œ
e by Theorem 5
Ä _ ˆ1 " ‰
n
1 În
3
œ n lim
œ 13 œ 3 by Theorem 5
Ä _ n1În
1 În
3
œ n lim
œ 11 œ 1 by Theorem 5
Ä _ n1În
ˆ3 ‰
13. converges to 3, since n lim
a œ n lim
Ä_ n
Ä_ n
ˆ3‰
14. converges to 1, since n lim
a œ n lim
Ä_ n
Ä_ n
cn
1În
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Chapter 10 Practice Exercises
1În
2 1
15. converges to ln 2, since n lim
a œ n lim
n a21În 1b œ n lim
œ n lim
"
Ä_ n
Ä_
Ä_
Ä_
–
Š 21În ln 2‹
n#
Š n#" ‹
Šn‹
—
635
œ n lim
21În ln 2
Ä_
œ 2! † ln 2 œ ln 2
2
n
È
16. converges to 1, since n lim
a œ n lim
2n 1 œ n lim
exp Š ln (2nn 1) ‹ œ n lim
exp Œ 2n1b 1 œ e! œ 1
Ä_ n
Ä_
Ä_
Ä_
(n 1)!
17. diverges, since n lim
a œ n lim
œ n lim
(n 1) œ _
n!
Ä_ n
Ä_
Ä_
(4)
18. converges to 0, since n lim
a œ n lim
œ 0 by Theorem 5
Ä_ n
Ä _ n!
n
Š"‹
Š"‹
#
#
"
19. (2n 3)(2n
1) œ #n 3 2n 1 Ê sn œ –
Š "# ‹
3
Š "# ‹
5
—–
Š "# ‹
5
Š "# ‹
7
Š "# ‹
Š "# ‹
— á – #n 3 2n 1 — œ
Š "# ‹
3
Š "‹
#
2n 1
Š "‹
#
"
"
Ê n lim
s œ n lim
2n 1— œ 6
Ä_ n
Ä _ –6
20. n(n2 1) œ n2 n 2 1 Ê sn œ ˆ #2 32 ‰ ˆ 32 42 ‰ á ˆ n2 n 2 1 ‰ œ #2 n 2 1 Ê n lim
s
Ä_ n
ˆ1 n 2 1 ‰ œ 1
œ n lim
Ä_
9
3
3
3 ‰
3 ‰
ˆ3 3‰ ˆ3 3‰ ˆ3
ˆ 3
21. (3n 1)(3n
2) œ 3n 1 3n 2 Ê sn œ # 5 5 8 8 11 á 3n 1 3n 2
ˆ 3 3n 3 2 ‰ œ 3#
œ 3# 3n3# Ê n lim
s œ n lim
Ä_ n
Ä_ #
8
2
2
2 ‰
2 ‰
2 ‰
ˆ 92 13
22. (4n 3)(4n
ˆ 132 17
ˆ 172 21
á ˆ 4n2 3 4n 2 1 ‰
1) œ 4n 3 4n 1 Ê sn œ
ˆ 29 4n21 ‰ œ 29
œ 29 4n21 Ê n lim
s œ n lim
Ä_ n
Ä_
_
_
n œ0
nœ0
23. ! en œ ! e"n , a convergent geometric series with r œ "e and a œ 1 Ê the sum is
_
_
n œ1
n œ0
"
œ e e 1
1 Š "e ‹
‰ a convergent geometric series with r œ 4" and a œ 43 Ê the sum is
24. ! (1)n 43n œ ! ˆ 43 ‰ ˆ "
4
n
ˆ 34 ‰
œ 35
1ˆ c4" ‰
25. diverges, a p-series with p œ "#
_
_
n œ1
nœ1
26. ! n5 œ 5 ! "n , diverges since it is a nonzero multiple of the divergent harmonic series
"
27. Since f(x) œ x"Î#
Ê f w (x) œ #x"$Î# 0 Ê f(x) is decreasing Ê an1 an , and
_
_
1
lim a œ lim
œ 0, the
n Ä _ n n Ä _ Èn
)
! È" diverges, the given series converges conditionally.
series ! ("
Èn converges by the Alternating Series Test. Since
n
nœ1
n
nœ1
28. converges absolutely by the Direct Comparison Test since #n" $ n"$ for n
1, which is the nth term of a convergent
p-series
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
636
Chapter 10 Infinite Sequences and Series
29. The given series does not converge absolutely by the Direct Comparison Test since ln (n" 1) n " 1 , which is
"
the nth term of a divergent series. Since f(x) œ ln (x" 1) Ê f w (x) œ (ln (x 1))
# (x 1) 0 Ê f(x) is decreasing
"
Ê an1 an , and n lim
a œ n lim
œ 0, the given series converges conditionally by the Alternating
Ä_ n
Ä _ ln (n 1)
Series Test.
_
30. '2
"
x(ln x)# dx œ b lim
Ä_
'2b x(ln" x) dx œ lim c(ln x)" d b2 œ lim ˆ ln"b ln"2 ‰ œ ln"# Ê the series
#
bÄ_
bÄ_
converges absolutely by the Integral Test
31. converges absolutely by the Direct Comparison Test since lnn$n nn$ œ n"# , the nth term of a convergent p-series
n
n
32. diverges by the Direct Comparison Test for en n Ê ln ˆen ‰ ln n Ê nn ln n Ê ln nn ln (ln n)
Ê n ln n ln (ln n) Ê lnln(lnnn) "n , the nth term of the divergent harmonic series
33. n lim
Ä_
Š È "#
n
n
Š n"# ‹
1
‹
#
n
œ Én lim
œ È1 œ 1 Ê converges absolutely by the Limit Comparison Test
Ä _ n# 1
$
34. Since f(x) œ x$3x 1 Ê f w (x) œ 3xaxa$21xb# b 0 when x
#
#
2 Ê an1 an for n
3n
2 and n lim
œ 0, the
Ä _ n$ 1
series converges by the Alternating Series Test. The series does not converge absolutely: By the Limit
#
Comparison Test, n lim
Ä_
Š n$3n 1 ‹
ˆ n" ‰
$
3n
œ n lim
œ 3. Therefore the convergence is conditional.
Ä _ n$ 1
n2
35. converges absolutely by the Ratio Test since n lim
œ01
’ n2 † n! “ œ n lim
Ä _ (n 1)! n 1
Ä _ (n 1)#
#
(") an 1b
36. diverges since n lim
a œ n lim
does not exist
Ä_ n
Ä _ 2n# n 1
n
nb1
3
37. converges absolutely by the Ratio Test since n lim
† n! “ œ n lim
œ01
’ 3
Ä _ (n 1)! 3n
Ä _ n1
n 2 3
6
n
È
É
38. converges absolutely by the Root Test since n lim
an œ n lim
œ01
nn œ n lim
Ä_
Ä_
Ä_ n
n n
39. converges absolutely by the Limit Comparison Test since n lim
Ä_
40. converges absolutely by the Limit Comparison Test since n lim
Ä_
nb1
"
Š $Î#
‹
n
Š Èn(n "1)(n
Š n"# ‹
Š È "#
n n
‹
2)
n(n 1)(n 2)
œ Én lim
œ1
n$
Ä_
#
1
‹
#
n an 1 b
œ Én lim
œ1
n%
Ä_
kx 4 k
ˆ n ‰ 1 Ê kx 3 4k 1
41. n lim
¹ uunbn 1 ¹ 1 Ê n lim
¹ (x 4) † n3 ¹ 1 Ê 3 n lim
Ä_
Ä _ (n 1)3nb1 (x 4)n
Ä _ n1
n
_
_
1) 3
Ê kx 4k 3 Ê 3 x 4 3 Ê 7 x 1; at x œ 7 we have ! (n3
œ ! ("n ) , the alternating
n
_
n œ1
_
n n
n
n œ1
3
! " , the divergent harmonic series
harmonic series, which converges conditionally; at x œ 1 we have! n3
n œ
n
n œ1
n
n œ1
(a) the radius is 3; the interval of convergence is 7 Ÿ x 1
(b) the interval of absolute convergence is 7 x 1
(c) the series converges conditionally at x œ 7
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Chapter 10 Practice Exercises
"
42. n lim
œ 0 1, which holds for all x
¹ uunbn 1 ¹ 1 Ê n lim
¹ (x1) † (2n1)! ¹ 1 Ê (x 1)# n lim
Ä_
Ä _ (2n1)! (x1)2nc2
Ä _ (#n)(2n1)
2n
(a) the radius is _; the series converges for all x
(b) the series converges absolutely for all x
(c) there are no values for which the series converges conditionally
nb1
n
43. n lim
1 Ê k3x 1k 1
¹ uunbn 1 ¹ 1 Ê n lim
¹ (3x(n1)1)# † (3x n 1)n ¹ 1 Ê k3x 1k n lim
Ä_
Ä_
Ä _ (n 1)#
#
#
_
nc1
_
2nc1
Ê 1 3x 1 1 Ê 0 3x 2 Ê 0 x 23 ; at x œ 0 we have ! (1) n# (1) œ ! ("n)#
n
n œ1
n œ1
_
œ ! n"# , a nonzero constant multiple of a convergent p-series, which is absolutely convergent; at x œ 23 we
n œ1
_
_
have ! (1)n# (1) œ ! ("n)#
n 1
n
n œ1
n 1
, which converges absolutely
nœ1
(a) the radius is "3 ; the interval of convergence is 0 Ÿ x Ÿ 32
(b) the interval of absolute convergence is 0 Ÿ x Ÿ 23
(c) there are no values for which the series converges conditionally
44. n lim
¹ uunbn 1 ¹ 1 Ê n lim
¹ n 2 † (2x 2nb1)1
Ä_
Ä _ 2n 3
nb1
k2x 1k
2
1
† 2n
lim ¸ n 2 † 2n " ¸ 1
n 1 † (2x 1)n ¹ 1 Ê
2
n Ä _ 2n 3 n 1
n
Ê k2x # 1k (1) 1 Ê k2x 1k 2 Ê 2 2x 1 2 Ê 3 2x 1 Ê #3 x "# ; at x œ #3 we have
_
_
! n 1 † (2)
œ ! (") (n1) which diverges by the nth-Term Test for Divergence since
n
n œ1
n
2n 1
#
n œ1
n
2n 1
_
_
n1
2
! n " , which diverges by the nth-Term Test
lim ˆ n 1 ‰ œ "# Á 0; at x œ #" we have ! 2n
1 † #n œ
2n 1
n Ä _ 2n 1
n
n œ1
n œ1
(a) the radius is 1; the interval of convergence is 3# x "#
(b) the interval of absolute convergence is 3# x "#
(c) there are no values for which the series converges conditionally
nb1
¸ˆ n ‰ ˆ n " 1 ‰¸ 1 Ê kxek lim ˆ n " 1 ‰ 1
45. n lim
† n ¹ 1 Ê kxk n lim
¹ uunbn 1 ¹ 1 Ê n lim
¹ x
Ä_
Ä _ (n 1)nb1 xn
Ä _ n1
nÄ_
n
n
Ê kxek † 0 1, which holds for all x
(a) the radius is _; the series converges for all x
(b) the series converges absolutely for all x
(c) there are no values for which the series converges conditionally
nb1
Èn
n
46. n lim
† xn ¹ 1 Ê kxk n lim
1 Ê kxk 1; when x œ 1 we have
¹ uunbn 1 ¹ 1 Ê n lim
¹ x
Ä_
Ä _ Èn 1
Ä _ Én1
_
_
! (È1) , which converges by the Alternating Series Test; when x œ 1 we have ! È" , a divergent p-series
n œ1
n
n
nœ1
n
(a) the radius is 1; the interval of convergence is 1 Ÿ x 1
(b) the interval of absolute convergence is 1 x 1
(c) the series converges conditionally at x œ 1
2nb1
47. n lim
¹ uunbn 1 ¹ 1 Ê n lim
¹ (n 32)x
nb1
Ä_
Ä_
_
_
n œ1
n œ1
#
3
x
ˆ n 2 ‰ 1 Ê È3 x È3;
† (n 1)x
2n 1 ¹ 1 Ê
3 n lim
Ä _ n1
n
the series ! nÈ31 and ! nÈ31 , obtained with x œ „ È3, both diverge
(a) the radius is È3; the interval of convergence is È3 x È3
(b) the interval of absolute convergence is È3 x È3
(c) there are no values for which the series converges conditionally
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
637
638
Chapter 10 Infinite Sequences and Series
2nb3
48. n lim
¹ uunbn 1 ¹ 1 Ê n lim
¹ (x 2n1)x
3
Ä_
Ä_
1
#
ˆ 2n 1 ‰ 1 Ê (x 1)# (1) 1
† (x 2n
1)2nb1 ¹ 1 Ê (x 1) n lim
Ä _ 2n 3
_
Ê (x 1)# 1 Ê kx 1k 1 Ê 1 x 1 1 Ê 0 x 2; at x œ 0 we have ! (1)#n(1)1
n
2nb1
n œ1
_
œ ! (1)
n œ1
_
3nb1
2n 1
_
œ ! (1)
n œ1
nc1
2n 1
which converges conditionally by the Alternating Series Test and the fact
_
2nb1
that ! 2n " 1 diverges; at x œ 2 we have ! (1)2n (1)
1
n œ1
n
nœ1
_
(1)
œ ! 2n
1 , which also converges conditionally
n
nœ1
(a) the radius is 1; the interval of convergence is 0 Ÿ x Ÿ 2
(b) the interval of absolute convergence is 0 x 2
(c) the series converges conditionally at x œ 0 and x œ 2
(n 1)x
49. n lim
¹ uunbn 1 ¹ 1 Ê n lim
¹ cschcsch
(n)xn
Ä_
Ä_
c"
nb1
Š enb1 c ecnc1 ‹
1
¹ 1 Ê kxk n lim
Ä_ » ˆ 2 ‰ »
2
en c ecn
_
kx k
Ê kxk n lim
¹ e1 ee 2n 2 ¹ 1 Ê e 1 Ê e x e; the series ! a „ ebn csch n, obtained with x œ „ e,
Ä_
2n 1
n œ1
both diverge since n lim
a „ e) csch n Á 0
Ä_
n
(a) the radius is e; the interval of convergence is e x e
(b) the interval of absolute convergence is e x e
(c) there are no values for which the series converges conditionally
nb1
c2nc2
c2n
(n 1)
50. n lim
† 1 e ¹ 1 Ê kxk 1
¹ uunbn 1 ¹ 1 Ê n lim
¹ x xncoth
¹1e
coth (n) ¹ 1 Ê kxk n lim
Ä_
Ä_
Ä _ 1 ec2nc2 1 ec2n
_
Ê 1 x 1; the series ! a „ 1bn coth n, obtained with x œ „ 1, both diverge since n lim
a „ 1bn coth n Á 0
Ä_
n œ1
(a) the radius is 1; the interval of convergence is 1 x 1
(b) the interval of absolute convergence is 1 x 1
(c) there are no values for which the series converges conditionally
51. The given series has the form 1 x x# x$ á (x)n á œ 1 " x , where x œ 4" ; the sum is 1 "ˆ " ‰ œ 54
4
#
$
n
$
&
2n 1
#
%
2n
52. The given series has the form x x# x3 á (1)n1 xn á œ ln (1 x), where x œ 23 ; the sum is
ln ˆ 53 ‰ ¸ 0.510825624
53. The given series has the form x x3! x5! á (1)n (2nx 1)! á œ sin x, where x œ 1; the sum is sin 1 œ 0
x
54. The given series has the form 1 x2! x4! á (1)n (2n)!
á œ cos x, where x œ 13 ; the sum is cos 13 œ "#
#
#
55. The given series has the form 1 x x2! x3! á xn! á œ ex , where x œ ln 2; the sum is eln Ð2Ñ œ 2
$
n
&
x
"
56. The given series has the form x x3 x5 á (1)n (2n
x, where x œ È"3 ; the sum is
1) á œ tan
2n 1
tan" Š È"3 ‹ œ 16
57. Consider 1 " 2x as the sum of a convergent geometric series with a œ 1 and r œ 2x Ê 1 " 2x
_
_
n œ0
nœ0
œ 1 (2x) (2x)# (2x)$ á œ ! (2x)n œ ! 2n xn where k2xk 1 Ê kxk "#
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Chapter 10 Practice Exercises
639
58. Consider 1 " x$ as the sum of a convergent geometric series with a œ 1 and r œ x$ Ê 1 " x$ œ 1 a"x$ b
#
_
$
œ 1 ax$ b ax$ b ax$ b á œ ! (1)n x3n where kx$ k 1 Ê kx$ k 1 Ê kxk 1
n œ0
_
_
n 2nb1
2nb1
1) (1x)
59. sin x œ ! ((2n1)x 1)! Ê sin 1x œ ! ((2n
1)!
nœ0
n
nœ0
_
nœ0
2nb1
_
n 2n
n œ0
_
Ê cos ˆx5Î3 ‰ œ !
nœ0
(1)n ˆx5Î3 ‰
(2n)!
2n
_ (1)n Š x3 ‹
È
Ê cosŠ Èx 5 ‹ œ !
_
_ ˆ 1 x ‰n
n 2n
n œ0
3
5
(2n)!
nœ0
63. ex œ ! xn! Ê eÐ1xÎ2Ñ œ !
n
n œ0
n œ0
_
_
n!
œ!
# n
_
#
n œ0
n 2nb1 2nb1
nœ0
_
n 10nÎ3
x
œ ! (1)(#n)!
nœ0
2n
_
1) x
62. cos x œ ! ((2n)!
_
_
œ ! (32n1)b12(#n x 1)!
(2n 1)!
nœ0
n 2nb1 2nb1
nœ0
_ (1)n Š 2x ‹
3
n 2nb1
!
60. sin x œ ! ((2n1)x 1)! Ê sin 2x
3 œ
1) x
61. cos x œ ! ((2n)!
_
œ ! (1)(#1n 1)!x
_
x
œ ! (5n1)(#n)!
n 6n
nœ0
1 n xn
#n n!
64. ex œ ! xn! Ê ex œ ! an!x b œ ! (1)n! x
#
n
n œ0
nœ0
n œ0
"Î#
65. f(x) œ È3 x# œ a3 x# b
Ê f www (x) œ 3x$ a3 x# b
3
9
f www (1) œ 32
83 œ 32
n 2n
Ê f w (x) œ x a3 x# b
&Î#
"Î#
Ê f ww (x) œ x# a3 x# b
$Î#
a3 x# b
"Î#
$Î#
3x a3 x# b
; f(1) œ 2, f w (1) œ "# , f ww (1) œ 8" #" œ 83 ,
#
$
Ê È3 x# œ 2 (x 1) 3(x$ 1) 9(x& 1) á
2†1!
2 †2!
2 †3!
66. f(x) œ 1 " x œ (1 x)" Ê f w (x) œ (1 x)# Ê f ww (x) œ 2(1 x)$ Ê f www (x) œ 6(1 x)% ; f(2) œ 1, f w (2) œ 1,
f ww (2) œ 2, f www (2) œ 6 Ê 1" x œ 1 (x 2) (x 2)# (x 2)$ á
67. f(x) œ x " 1 œ (x 1)" Ê f w (x) œ (x 1)# Ê f ww (x) œ 2(x 1)$ Ê f www (x) œ 6(x 1)% ; f(3) œ 4" ,
f w (3) œ 4"# , f ww (3) œ 42$ , f www (2) œ 4%6 Ê x" 1 œ "4 4"# (x 3) 4"$ (x 3)# 4"% (x 3)$ á
68. f(x) œ x" œ x" Ê f w (x) œ x# Ê f ww (x) œ 2x$ Ê f www (x) œ 6x% ; f(a) œ "a , f w (a) œ a"# , f ww (a) œ a2$ ,
f www (a) œ a%6 Ê "x œ "a a"# (x a) a"$ (x a)# a"% (x a)$ á
69. '0 exp ax$ b dx œ '0 Š1 x$ x2! x3! x4! á ‹ dx œ ’x x4 7x†2! 10x †3! 13x †4! á “
1Î2
1Î2
'
*
"#
%
(
"!
"$
"
"
"
¸ "# #%"†4 #( †"7†2! 2"! †10
†3! 2"$ †13†4! 2"' †16†5! ¸ 0.484917143
"Î#
!
70. '0 x sin ax$ b dx œ '0 x Šx$ x3! x5! x7! x9! á ‹ dx œ '0 Šx% x3! x5! x7! x9! á ‹ dx
1
1
&
""
"(
*
#$
"&
#"
#(
1
"!
"'
##
#)
"
#*
œ ’ x5 11x †3! 17x †5! 23x †7! 29x †9! á “ ¸ 0.185330149
!
71. '1
1Î2
tanc" x
dx œ
x
"Î#
x
'11Î2 Š1 x3 x5 x7 x9 x11 á ‹ dx œ ’x x9 25x 49x 81x 121
á“
#
%
'
)
"!
$
&
(
*
""
!
¸ "# 9†"2$ 5#"†#& 7#"†#( 9#"†2* 11#"†2"" 13#"†2"$ 15#"†2"& 17#"†#"( 19#"†#"* 21#"†##" ¸ 0.4872223583
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
640
Chapter 10 Infinite Sequences and Series
72. '0
1Î64
tan " x
Èx dx œ
'01Î64 È"x Šx x3$ x5& x7( á ‹ dx œ '01Î64 ˆx"Î# "3 x&Î# "5 x*Î# 7" x"$Î# á ‰ dx
"Î'%
2 ""Î#
2
œ 23 x$Î# #21 x(Î# 55
x
105
x"&Î# á ‘ !
$
&
œ ˆ 3†28$ 212†8( 55†28"" 1052†8"& á ‰ ¸ 0.0013020379
#
7 Šx x x á ‹
%
7 Š1 x x á ‹
3!
5!
3!
5!
7 sin x
œ lim
œ lim
œ #7
2x
# #
$ $
#
$ #
x Ä 0 e 1
x Ä 0 Š2x 2 2!x 2 3!x á‹
x Ä 0 Š2 2#!x 2 3!x á‹
73. lim
74.
#
$
#
$
Š1 ) )#! )3! á‹ Š1 ) )#! )3! á‹ 2)
)
c)
lim e )e sin)2) œ lim
)Ä0
)Ä0
$
&
) Š) )3! )5! á‹
$
&
$
&
2 Š )3! )5! á‹
œ lim
)Ä0
Š )3! )5! á‹
#
œ lim
)Ä0
2 Š 3!" )5! á‹
œ2
#
Š 3!" )5! á‹
#
75.
#
2 cos t
lim ˆ "
t"# ‰ œ lim t 2t# (12 cos t) œ t lim
t Ä 0 # 2 cos t
tÄ0
Ä0
%
t
á
t# 2 2 Œ1 t# 4!
#
%
2t# Š1 1 t# t4! á‹
%
œ lim
tÄ0
'
2 Œ t4! t6! á
'
Št% 2t4! á‹
#
œ lim
tÄ0
76. lim
"
2 Š 4!
t6! á‹
#
Š1 2t4! á‹
Š sinh h ‹ cos h
hÄ0
œ lim
h#
œ 1"#
h#
œ lim
h%
h#
h#
hÄ0
h#
h#
h%
h%
h'
h'
Œ #! 3! 5! 4! 6! 7! á
h#
hÄ0
h%
Œ1 3! 5! á Œ1 #! 4! á
#
#
%
%
Š #"! 3!" h5! h4! h6! h7! á ‹ œ 3"
œ lim
hÄ0
%
%
1 Š1 z# z3 á‹
Šz# z3 á‹
" cos# z
œ
lim
œ
lim
#
$
$
&
#
$
%
z Ä 0 ln (1 z) sin z
z Ä 0 Šz z# z3 á‹ Šz z3! z5! á‹
z Ä 0 Š z# 2z3 z4 á‹
77. lim
#
œ lim
Š1 z3 á‹
#
z
z Ä 0 Š "# 2z
3 4 á‹
œ 2
y#
y#
y#
œ lim
œ lim
#
%
'
#
%
'
#
cos
y
cosh
y
y
y
y
y
y
y
2y
2y'
yÄ0
y Ä 0 Œ1 á Œ1 á
y Ä 0 Œ
4!
6!
4!
6!
#
#!
# 6! á
78. lim
œ lim
"
y Ä 0 Œ1 2y% á
6!
œ 1
$
79. lim ˆ sinx$3x xr# s‰ œ lim –
xÄ0
xÄ0
&
(3x)
Š3x (3x)
6 120 á‹
x$
#
r
xr# s— œ lim Š x3# #9 81x
40 á x# s‹ œ 0
xÄ0
Ê xr# x3# œ 0 and s #9 œ 0 Ê r œ 3 and s œ #9
80. The approximation sin x ¸ 6 6xx# is better than sin x ¸ x.
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Chapter 10 Practice Exercises
(3n 1)(3n 2)x
81. n lim
¹ 2†5†8#â
†4†6â(2n)(2n 2)
Ä_
nb1
â(2n)
¸ 3n 2 ¸ 1 Ê kxk 23
† #†52†8†4â†6(3n
1)xn ¹ 1 Ê kxk n lim
Ä _ 2n 2
Ê the radius of convergence is 23
nb1
(2n1)(2n3)(x1)
82. n lim
¹ 3†5†47†â
9†14â(5n1)(5n4)
Ä_
¸ 2n 3 ¸ 1 Ê kxk 52
† 34††59††714ââ(2n(5n1)x1)n ¹ 1 Ê kxk n lim
Ä _ 5n 4
Ê the radius of convergence is 52
n
n
n
kœ2
kœ2
kœ2
83. ! ln ˆ1 k"# ‰ œ ! ln ˆ1 "k ‰ ln ˆ1 "k ‰‘ œ ! cln (k 1) ln k ln (k 1) ln kd
œ cln 3 ln 2 ln 1 ln 2d cln 4 ln 3 ln 2 ln 3d cln 5 ln 4 ln 3 ln 4d cln 6 ln 5 ln 4 ln 5d
á cln (n 1) ln n ln (n 1) ln nd œ cln 1 ln 2d cln (n 1) ln nd
after cancellation
n
_
k œ2
k œ2
1 ‰
1‰
Ê ! ln ˆ1 k"# ‰ œ ln ˆ n2n
Ê ! ln ˆ1 k"# ‰ œ n lim
ln ˆ n 2n
œ ln "# is the sum
Ä_
n
n
k œ2
k œ2
84. ! k# " 1 - œ "# ! ˆ k " 1 k " 1 ‰ œ #" ˆ 11 3" ‰ ˆ #" 4" ‰ ˆ "3 5" ‰ ˆ 4" 6" ‰ á ˆ n " # n" ‰
#
2(n 1) 2n
ˆ n " 1 n " 1 ‰‘ œ #" ˆ 1" #" n" n " 1 ‰ œ #" ˆ #3 n" n " 1 ‰ œ #" ’ 3n(n 1)2n(n
“ œ 3n4n(n n 1)2
1)
_
"ˆ3
Ê ! k#"1 œ n lim
1n n 1 1 ‰ œ 43
Ä_ # 2
k œ2
1)x
85. (a) n lim
¹ 1†4†7â(3n(3n2)(3n
3)!
Ä_
3nb3
(3n ")
$
† 1†4†7â(3n)!
(3n 2)x3n ¹ 1 Ê kx k n lim
Ä _ (3n 1)(3n 2)(3n 3)
œ kx$ k † 0 1 Ê the radius of convergence is _
_
_
n œ1
_
n œ1
_
n œ1
nœ2
(3n 2) 3n
! 1†4†7â(3n 2) x3nc1
(b) y œ 1 ! 1†4†7â
x Ê dy
(3n)!
dx œ
(3n 1)!
Ê
d# y
! 1†4†7â(3n 2) x3nc2 œ x ! 1†4†7â(3n5) x3nc2
dx# œ
(3n 2)!
(3n3)!
_
(3n 2) 3n
œ x Œ1 ! 1†4†7â
x œ xy 0
(3n)!
n œ1
Ê a œ 1 and b œ 0
_
86. (a)
x#
x#
#
#
#
#
#
$
#
$
%
&
! (1)n xn which
1 x œ 1 (x) œ x x (x) x (x) x (x) á œ x x x x á œ
n œ2
converges absolutely for kxk 1
_
_
n œ2
nœ2
(b) x œ 1 Ê ! (1)n xn œ ! (1)n which diverges
_
_
87. Yes, the series ! an bn converges as we now show. Since ! an converges it follows that an Ä 0 Ê an 1
n œ1
nœ1
_
_
n œ1
n œ1
for n some index N Ê an bn bn for n N Ê ! an bn converges by the Direct Comparison Test with ! bn
_
88. No, the series ! an bn might diverge (as it would if an and bn both equaled n) or it might converge (as it would if
n œ1
an and bn both equaled "n ).
_
_
n œ1
k œ1
!(xk1 xk ) œ lim (xn1 x" ) œ lim (xn1 ) x" Ê both the series and
89. ! (xn1 xn ) œ n lim
Ä_
nÄ_
nÄ_
sequence must either converge or diverge.
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
641
642
Chapter 10 Infinite Sequences and Series
90. It converges by the Limit Comparison Test since n lim
Ä_
Š 1 banan ‹
an
_
"
œ n lim
œ 1 because ! an converges
Ä _ 1 a n
n œ1
and so an Ä 0.
_
a" ˆ #" ‰ a# ˆ "3 "4 ‰ a% ˆ "5 "6 "7 "8 ‰ a)
91. ! ann œ a" a## a3$ a4% á
n œ1
"
"
" ‰
ˆ 9" 10
11
á 16
a"' á
92. an œ ln"n for n
2 Ê a#
a$
"
# (a# a% a) a"' á ) which is a divergent series
á , and ln"# ln"4 ln"8 á œ ln"# # ln" 2 3 ln" 2 á
a%
_
œ ln"# ˆ1 "# "3 á ‰ which diverges so that 1 ! n ln" n diverges by the Integral Test.
n œ2
CHAPTER 10 ADDITIONAL AND ADVANCED EXERCISES
_
1. converges since (3n #")Ð2n 1ÑÎ2 (3n "2)$Î# and ! (3n "2)$Î# converges by the Limit Comparison Test:
n œ1
lim
nÄ_ Š
"
Š $Î#
‹
n
(3n
"
‹
2)$Î#
ˆ 3n n 2 ‰
œ n lim
Ä_
$Î#
œ 3$Î#
_
2. converges by the Integral Test: '1 atan" xb
$
$
#
dx
x # 1
"
$
"
b
$
$
1
œ lim ’ atan 3 xb “ œ lim ’ atan 3 bb 192
“
bÄ_
"
bÄ_
$
71
1
1
œ Š 24
192
‹ œ 192
c2n
e
3. diverges by the nth-Term Test since n lim
a œ n lim
(1)n tanh n œ lim (1)n Š 11 (1)n
ec2n ‹ œ n lim
Ä_ n
Ä_
Ä_
bÄ_
does not exist
4. converges by the Direct Comparison Test: n! nn Ê ln (n!) n ln (n) Ê lnln(n!)
(n) n
Ê logn (n!) n Ê lognn $(n!) n"# , which is the nth-term of a convergent p-series
12
1†2
12
ˆ 42††53 ‰ ˆ 31††42 ‰
5. converges by the Direct Comparison Test: a" œ 1 œ (1)(3)(2)
# , a# œ 3†4 œ (2)(4)(3)# , a$ œ
_
12
12
"2
!
ˆ 35††64 ‰ ˆ 24††53 ‰ ˆ 13††24 ‰ œ (4)(6)(5)
œ (3)(5)(4)
# , a% œ
# , á Ê 1 (n 1)(n 3)(n 2)# represents the
n œ1
given series and (n 1)(n 123)(n 2)# 12
, which is the nth-term of a convergent p-series
n%
anb1
n
6. converges by the Ratio Test: n lim
œ n lim
œ01
Ä _ an
Ä _ (n 1)(n 1)
7. diverges by the nth-Term Test since if an Ä L as n Ä _, then L œ 1 " L Ê L# L 1 œ 0 Ê L œ 1 „#
_
_
n œ1
n œ1
8. Split the given series into ! 32n"b1 and ! 32n2n ; the first subseries is a convergent geometric series and the
n 2n
É
second converges by the Root Test: n lim
32n œ n lim
Ä_
Ä_
n
n n
È
2È
È
9
œ "9†1 œ "9 1
9. f(x) œ cos x with a œ 13 Ê f ˆ 13 ‰ œ 0.5, f w ˆ 13 ‰ œ #3 , f ww ˆ 13 ‰ œ 0.5, f www ˆ 13 ‰ œ
È
#
È
È3
Ð4Ñ
ˆ 13 ‰ œ 0.5;
# ,f
$
cos x œ "# #3 ˆx 13 ‰ 4" ˆx 13 ‰ 1#3 ˆx 13 ‰ á
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
È5
Á0
Chapter 10 Additional and Advanced Exercises
643
10. f(x) œ sin x with a œ 21 Ê f(21) œ 0, f w (21) œ 1, f ww (21) œ 0, f www (21) œ 1, f Ð4Ñ (21) œ 0, f Ð5Ñ (21) œ 1,
$
&
(
f Ð6Ñ (21) œ 0, f Ð7Ñ (21) œ 1; sin x œ (x 21) (x 3!21) (x 5!21) (x 7!21) á
#
$
11. ex œ 1 x x#! x3! á with a œ 0
12. f(x) œ ln x with a œ 1 Ê f(1) œ 0, f w (1) œ 1, f ww (1) œ 1,f www (1) œ 2, f Ð4Ñ (1) œ 6;
#
$
%
ln x œ (x 1) (x # 1) (x 3 1) (x 4 1) á
13. f(x) œ cos x with a œ 221 Ê f(221) œ 1, f w (221) œ 0, f ww (221) œ 1, f www (221) œ 0, f Ð4Ñ (221) œ ",
f Ð5Ñ (221) œ 0, f Ð6Ñ (221) œ 1; cos x œ 1 "# (x 221)# 4!" (x 221)% 6!" (x 221)' á
14. f(x) œ tan" x with a œ 1 Ê f(1) œ 14 , f w (1) œ "# , f ww (1) œ "# , f www (1) œ "# ;
#
$
tan" x œ 14 (x 2 1) (x 4 1) (x 121) á
n
15. Yes, the sequence converges: cn œ aan bn b1În Ê cn œ b ˆˆ ba ‰ 1‰
ˆ a ‰n ln ˆ a ‰
œ ln b lim ˆb a ‰n 1b œ ln b nÄ_
b
1 În
n
lnˆˆ ba ‰ 1‰
n
nÄ_
Ê n lim
c œ ln b lim
Ä_ n
0†ln ˆ ba ‰
c œ eln b œ b.
0 1 œ ln b since 0 a b. Thus, n lim
Ä_ n
_
_
_
2
16. 1 10
103 # 107 $ 102 % 103 & 107 ' á œ 1 ! 103n2 2 ! 103n3 1 ! 1073n
_
_
_
n œ0
n œ0
n œ0
œ 1 ! 103n2 1 ! 103n3 2 ! 103n7 3 œ 1 n œ1
n œ1
2 ‰
ˆ 10
Š 103# ‹
"‰
1 ˆ 10
$ "‰
1 ˆ 10
n œ1
Š 107$ ‹
$ "‰
1 ˆ 10
$
30
7
999237
œ 1 200
œ 412
999 999 999 œ
999
333
kb1
17. sn œ ! 'k
n 1
k œ0
dx
1 x#
Ê sn œ '0 1 dxx# '1 1 dxx# á 'nc1 1 dxx# Ê sn œ '0 1 dxx#
1
2
n
n
Ê n lim
s œ n lim
atan" n tan" 0b œ 1#
Ä_ n
Ä_
nb1
#
1)
18. n lim
† (n 1)(2x
¹ uunbn 1 ¹ œ n lim
¹ (n 1)x
¹ œ n lim
¹ x † (n 1) ¹ œ ¸ 2x x 1 ¸ 1
nxn
Ä_
Ä _ (n 2)(2x 1)n 1
Ä _ 2x 1 n(n 2)
n
Ê kxk k2x 1k ; if x 0, kxk k2x 1k Ê x 2x 1 Ê x 1; if "# x 0, kxk k2x 1k
Ê x 2x 1 Ê 3x 1 Ê x "3 ; if x #" , kxk k2x 1k Ê x 2x 1 Ê x 1. Therefore,
the series converges absolutely for x 1 and x "3 .
19. (a) No, the limit does not appear to depend on the value of the constant a
(b) Yes, the limit depends on the value of b
(c)
cos ˆ a ‰ n
s œ Š1 n n ‹
œ n lim
Ä_
a
ˆa‰
ˆa‰
01
n sin n cos n
cos ˆ na ‰
10
1 n
a‰ n
ˆ
cos n
1Îb
lim Š1 nÄ_
Ê ln s œ
ln Œ1 œ
bn
cos ˆ na ‰
n
ˆ n" ‰
Ê n lim
ln s œ
Ä_
œ 1 Ê n lim
sœe
Ä_
"
1
"
cos ˆ na ‰ Œ
n
a
a
a
n sin ˆ n ‰ cos ˆ n ‰
n#
Š "# ‹
n
¸ 0.3678794412; similarly,
‹ œe
n 1În
_
an ‰
20. ! an converges Ê n lim
a œ 0; n lim
’ˆ 1 sin
“
#
Ä_ n
Ä_
n œ1
an ‰
ˆ 1sin
œ n lim
œ
#
Ä_
Ä_
1sin Šnlim an ‹
#
œ "# Ê the series converges by the nth-Root Test
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
0
œ 1sin
#
644
Chapter 10 Infinite Sequences and Series
nb1 nb1
21. n lim
¹ uunbn 1 ¹ 1 Ê n lim
¹ b x † ln n ¹ 1 Ê kbxk 1 Ê b" x b" œ 5 Ê b œ „ "5
Ä_
Ä _ ln (n 1) bn xn
22. A polynomial has only a finite number of nonzero terms in its Taylor series, but the functions sin x, ln x and
ex have infinitely many nonzero terms in their Taylor expansions.
$ $
$
Šax a 3!x á‹ Šx x3! á‹ x
$
&
sin (ax) sin xx
œ
lim
œ lim ’ a x# 2 a3! 3!" Š a5! 5!" ‹ x# á “
$
$
x
x
xÄ0
xÄ0
xÄ0
$
xx
is finite if a 2 œ 0 Ê a œ 2; lim sin 2x xsin
œ 23! 3!" œ 76
$
xÄ0
23. lim
# #
24.
lim cos#axx# b œ 1
xÄ0
Ê lim
Ê b œ 1 and a œ „ 2
25. (a)
(b)
#x#
xÄ0
(n 1)#
un
œ 1 2n n"#
unb1 œ
n#
un
n1
1
0
unb1 œ n œ 1 n n#
% %
a x
a x
Œ1 # 4! á b
#
xÄ0
_
Ê C œ 2 1 and ! n"# converges
n œ1
_
Ê C œ 1 Ÿ 1 and ! "n diverges
n œ1
6
26.
# #
œ 1 Ê lim Š "2x#b a4 a48x á ‹ œ 1
5n#
3
Š4‹
Š#‹
2n(2n 1)
un
4n# 2n
5
unb1 œ (2n 1)# œ 4n# 4n 1 œ 1 n 4n# 4n 1 œ 1 n #
Ê C œ 3# 1 and kf(n)k œ 4n# 5n4n 1 œ
5
Ÿ5
Š4 4n "# ‹
_
_
n œ1
n œ1
nœ1
n#
after long division
_
Ê ! un converges by Raabe's Test
n
_
– Š4n# c 4n b 1‹ —
n œ1
27. (a) ! an œ L Ê an# Ÿ an ! an œ an L Ê ! an# converges by the Direct Comparison Test
(b) converges by the Limit Comparison Test: n lim
Ä_
an
Š1c
an ‹
_
an
"
œ n lim
œ 1 since ! an converges and
Ä _ 1 an
#
$
n œ1
therefore x lim
a œ0
Ä_ n
28. If 0 an 1 then kln (1 an )k œ ln (1 an ) œ an a#n a3n á an an# an$ á œ 1 anan ,
a positive term of a convergent series, by the Limit Comparison Test and Exercise 27b
_
_
n œ1
nœ1
d
29. (1 x)" œ 1 ! xn where kxk 1 Ê (1"x)# œ dx
(1 x)" œ ! nxn1 and when x œ #" we have
#
$
n 1
4 œ 1 2 ˆ "# ‰ 3 ˆ "# ‰ 4 ˆ "# ‰ á n ˆ "# ‰
á
_
_
_
_
nœ1
nœ1
nœ1
#
2xx#
! n(n 1)xn1 œ 2 $ Ê ! n(n 1)xn œ 2x $
30. (a) ! xn1 œ 1xx Ê ! (n 1)xn œ (1
x)# Ê
(1x)
(1x)
nœ1
Ê
_
! n(n n 1) œ
x
n œ1
2
x
2x#
$ œ (x 1)$ , kxk 1
Š1 x" ‹
_
$
#
(b) x œ ! n(nxn ") Ê x œ (x 2x
1)$ Ê x 3x x 1 œ 0 Ê x œ 1 Š1 #
n œ1
"Î$
È57 "Î$
È
Š1 957 ‹
9 ‹
¸ 2.769292, using a CAS or calculator
_
31. (a)
"
d ˆ " ‰
d
#
$
#
$
! nxn1
(1x)# œ dx 1x œ dx a1 x x x á b œ 1 2x 3x 4x á œ
n œ1
_
(b)
2
n 1
from part (a) we have ! n ˆ 56 ‰ ˆ "6 ‰ œ ˆ "6 ‰ ’ 1 "ˆ 5 ‰ “ œ 6
6
n œ1
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Chapter 10 Additional and Advanced Exercises
_
(c) from part (a) we have ! npn1 q œ (1 q p)# œ qq# œ q"
n œ1
_
_
k œ1
k œ1
"
_
_
_
k œ1
kœ1
kœ1
ˆ ‰
32. (a) ! pk œ ! 2k œ 1#ˆ " ‰ œ 1 and E(x) œ ! kpk œ ! k2k œ "# ! k21k œ ˆ "# ‰ #
"
# œ 2
1ˆ "# ‰‘
by Exercise 31(a)
_
_
k œ1
k œ1
_
kc1
k
ˆ5‰
_
_
k œ1
kœ1
_
kc1
(b) ! pk œ ! 56k œ "5 ! ˆ 65 ‰ œ ˆ 5" ‰ ’ 1 6ˆ 5 ‰ “ œ 1 and E(x) œ ! kpk œ ! k 56k œ 6" ! k ˆ 65 ‰
k œ1
6
k1
k œ1
œ ˆ "6 ‰ "
# œ 6
1ˆ 56 ‰‘
_
_
_
kœ1
kœ1
(c) ! pk œ ! k(k"1) œ ! ˆ k" k " 1 ‰ œ
kœ1
_
œ!
k œ1
_
_
lim ˆ1 k " 1 ‰ œ 1 and E(x) œ ! kpk œ ! k Š k(k " 1) ‹
kÄ_
kœ1
kœ1
"
k 1 , a divergent series so that E(x) does not exist
C! e kt! ˆ1 e nkt! ‰
1e kt!
kt!
!
lim R œ 1C! ee kt! œ ektC! 1
nÄ_ n
e 1 a1 e n b
e " a1 e "! b
"
(b) Rn œ 1 e "
Ê R" œ e ¸ 0.36787944 and R"! œ 1 e " ¸ 0.58195028;
33. (a) Rn œ C! ekt! C! e2kt! á C! enkt! œ
Ê Rœ
R œ e" 1 ¸ 0.58197671; R R"! ¸ 0.00002643 Ê R RR"! 0.0001
Þ1n
e Þ1 ˆ1 e Þ1n ‰ R
, # œ #" ˆ eÞ1 " 1 ‰ ¸ 4.7541659; Rn R# Ê 1eÞ1 e 1 ˆ #" ‰ ˆ eÞ1 " 1 ‰
1 e Þ1
n
n
1 enÎ10 "# Ê enÎ10 "# Ê 10
ln ˆ "# ‰ Ê 10
ln ˆ "# ‰ Ê n 6.93
(c) Rn œ
Ê
Ê nœ7
!
34. (a) R œ ektC! Ê Rekt! œ R C! œ CH Ê ekt! œ CCHL Ê t! œ k" ln Š CCHL ‹
1
"
(b) t! œ 0.05
ln e œ 20 hrs
2 ‰
(c) Give an initial dose that produces a concentration of 2 mg/ml followed every t! œ 0.0" # ln ˆ 0.5
¸ 69.31 hrs
by a dose that raises the concentration by 1.5 mg/ml
0.1 ‰
"
‰
(d) t! œ 0.2
ln ˆ 0.03
œ 5 ln ˆ 10
3 ¸ 6 hrs
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
645
646
Chapter 10 Infinite Sequences and Series
NOTES:
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
CHAPTER 11 PARAMETRIC EQUATIONS AND
POLAR COORDINATES
11.1 PARAMETRIZATIONS OF PLANE CURVES
1. x œ 3t, y œ 9t# , _ t _ Ê y œ x#
2. x œ Èt , y œ t, t
0 Ê x œ È y
#
or y œ x , x Ÿ 0
3. x œ 2t 5, y œ 4t 7, _ t _
Ê x 5 œ 2t Ê 2(x 5) œ 4t
Ê y œ 2(x 5) 7 Ê y œ 2x 3
5. x œ cos 2t, y œ sin 2t, 0 Ÿ t Ÿ 1
Ê cos# 2t sin# 2t œ 1 Ê x# y# œ 1
4. x œ 3 3t, y œ 2t, 0 Ÿ t Ÿ 1 Ê y# œ t
Ê x œ 3 3 ˆ y# ‰ Ê 2x œ 6 3y
Ê y œ 2 23 x, ! Ÿ x Ÿ $
6. x œ cos (1 t), y œ sin (1 t), 0 Ÿ t Ÿ 1
Ê cos# (1 t) sin# (1 t) œ 1
Ê x# y# œ 1, y !
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
648
Chapter 11 Parametric Equations and Polar Coordinates
7. x œ 4 cos t, y œ 2 sin t, 0 Ÿ t Ÿ 21
Ê
#
16 cos# t
t
4 sin
œ1
16
4
Ê
y#
x#
16 4 œ 1
9. x œ sin t, y œ cos 2t, 12 Ÿ t Ÿ 12
Ê y œ cos 2t œ 1 2sin# t Ê y œ 1 2x2
11. x œ t2 , y œ t6 2t4 , _ t _
3
2
Ê y œ at2 b 2at2 b Ê y œ x3 2x2
13. x œ t, y œ È1 t# , 1 Ÿ t Ÿ 0
Ê y œ È1 x#
8. x œ 4 sin t, y œ 5 cos t, 0 Ÿ t Ÿ 21
#
#
#
#
sin t
cos t
x
Ê 16 16
25 25
œ 1 Ê 16
#y5 œ 1
10. x œ 1 sin t, y œ cos t 2, 0 Ÿ t Ÿ 1
Ê sin# t cos# t œ 1 Ê ax 1b# ay 2b# œ 1
2
12. x œ t t 1 , y œ tt 1 , 1 t 1
2x
Ê t œ x x 1 Ê y œ 2x
1
14. x œ Èt 1, y œ Èt, t 0
Ê y# œ t Ê x œ Èy# 1, y
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
0
Section 11.1 Parametrizations of Plane Curves
15. x œ sec# t 1, y œ tan t, 1# t 1#
16. x œ sec t, y œ tan t, 1# t 1#
17. x œ cosh t, y œ sinh t, _ 1 _
Ê cosh# t sinh# t œ 1 Ê x# y# œ 1
18. x œ 2 sinh t, y œ 2 cosh t, _ t _
Ê 4 cosh# t 4 sinh# t œ 4 Ê y# x# œ 4
19. (a) x œ a cos t, y œ a sin t, 0 Ÿ t Ÿ 21
20. (a) x œ a sin t, y œ b cos t, 1# Ÿ t Ÿ 5#1
Ê sec# t 1 œ tan# t Ê x œ y#
(b) x œ a cos t, y œ a sin t, 0 Ÿ t Ÿ 21
(c) x œ a cos t, y œ a sin t, 0 Ÿ t Ÿ 41
649
Ê sec# t tan# t œ 1 Ê x# y# œ 1
(d) x œ a cos t, y œ a sin t, 0 Ÿ t Ÿ 41
(b) x œ a cos t, y œ b sin t, 0 Ÿ t Ÿ 21
(c) x œ a sin t, y œ b cos t, 1# Ÿ t Ÿ 9#1
(d) x œ a cos t, y œ b sin t, 0 Ÿ t Ÿ 41
21. Using a"ß $b we create the parametric equations x œ " at and y œ $ bt, representing a line which goes
through a"ß $b at t œ !. We determine a and b so that the line goes through a%ß "b when t œ ".
Since % œ " a Ê a œ &. Since " œ $ b Ê b œ %. Therefore, one possible parameterization is x œ " &t,
y œ $ %t, 0 Ÿ t Ÿ ".
22. Using a"ß $b we create the parametric equations x œ " at and y œ $ bt, representing a line which goes through
a"ß $b at t œ !. We determine a and b so that the line goes through a$ß #b when t œ ". Since $ œ " a Ê a œ %.
Since # œ $ b Ê b œ &. Therefore, one possible parameterization is x œ " %t, y œ $ &t, 0 Ÿ t Ÿ ".
23. The lower half of the parabola is given by x œ y# " for y Ÿ !. Substituting t for y, we obtain one possible
parameterization x œ t# ", y œ t, t Ÿ 0Þ
24. The vertex of the parabola is at a"ß "b, so the left half of the parabola is given by y œ x# #x for x Ÿ ". Substituting
t for x, we obtain one possible parametrization: x œ t, y œ t# #t, t Ÿ ".
25. For simplicity, we assume that x and y are linear functions of t and that the pointax, yb starts at a#ß $b for t œ ! and passes
through a"ß "b at t œ ". Then x œ fatb, where fa!b œ # and fa"b œ ".
Since slope œ ??xt œ "#
"! œ $, x œ fatb œ $t # œ # $t. Also, y œ gatb, where ga!b œ $ and ga"b œ ".
3
Since slope œ ??yt œ "
"! œ 4. y œ gatb œ %t $ œ $ %t.
One possible parameterization is: x œ # $t, y œ $ %t, t
!.
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
650
Chapter 11 Parametric Equations and Polar Coordinates
26. For simplicity, we assume that x and y are linear functions of t and that the pointax, yb starts at a"ß #b for t œ ! and
passes through a!ß !b at t œ ". Then x œ fatb, where fa!b œ " and fa"b œ !.
a"b
Since slope œ ??xt œ ! "!
œ ", x œ fatb œ "t a"b œ " t. Also, y œ gatb, where ga!b œ # and ga"b œ !.
#
Since slope œ ??yt œ !"!
œ #. y œ gatb œ #t # œ # #t.
One possible parameterization is: x œ " t, y œ # #t, t
27. Since we only want the top half of a circle, y
!.
0, so let x œ 2cos t, y œ 2lsin tl, 0 Ÿ t Ÿ 41
28. Since we want x to stay between 3 and 3, let x œ 3 sin t, then y œ a3 sin tb2 œ 9sin# t, thus x œ 3 sin t, y œ 9sin# t,
0Ÿt_
dy
dy
x
x
29. x# y# œ a# Ê 2x 2y dy
dx œ 0 Ê dx œ y ; let t œ dx Ê y œ t Ê x œ yt. Substitution yields
y# t# y# œ a# Ê y œ È1a t# and x œ È1att , _ t _
30. In terms of ), parametric equations for the circle are x œ a cos ), y œ a sin ), 0 Ÿ ) 21. Since ) œ as , the arc
length parametrizations are: x œ a cos as , y œ a sin as , and 0 Ÿ as 21 Ê 0 Ÿ s Ÿ 21a is the interval for s.
31. Drop a vertical line from the point ax, yb to the x-axis, then ) is an angle in a right triangle, and from trigonometry we
know that tan ) œ yx Ê y œ x tan ). The equation of the line through a0, 2b and a4, 0b is given by y œ 12 x 2. Thus
4 tan )
1
x tan ) œ 12 x 2 Ê x œ 2 tan4) 1 and y œ 2 tan
) 1 where 0 Ÿ ) 2 .
32. Drop a vertical line from the point ax, yb to the x-axis, then ) is an angle in a right triangle, and from trigonometry we
know that tan ) œ yx Ê y œ x tan ). Since y œ Èx Ê y2 œ x Ê ax tan )b2 œ x Ê x œ cot2 ) Ê y œ cot ) where
0 ) Ÿ 12 .
33. The equation of the circle is given by ax 2b2 y2 œ 1. Drop a vertical line from the point ax, yb on the circle to the
x-axis, then ) is an angle in a right triangle. So that we can start at a1, 0b and rotate in a clockwise direction, let
x œ 2 cos ), y œ sin ), 0 Ÿ ) Ÿ 21.
34. Drop a vertical line from the point ax, yb to the x-axis, then ) is an angle in a right triangle, whose height is y and whose
base is x 2. By trigonometry we have tan ) œ x y 2 Ê y œ ax 2b tan ). The equation of the circle is given by
x2 y2 œ 1 Ê x2 aax 2btan )b2 œ 1 Ê x2 sec2 ) 4x tan2 ) 4tan2 ) 1 œ 0. Solving for x we obtain
xœ
2
4tan2 ) „ Éa4tan2 )b2 4 sec2 ) a4tan2 ) 1b
2È1 3tan2 )
œ 4tan ) „2 sec
œ 2sin2 ) „ cos )Ècos2 ) 3sin2 )
2)
2 sec2 )
œ 2 2cos2 ) „ cos )È4cos2 ) 3 and y œ Š2 2cos2 ) „ cos )È4cos2 ) 3 2‹ tan )
œ 2sin ) cos ) „ sin )È4cos2 ) 3. Since we only need to go from a1, 0b to a0, 1b, let
x œ 2 2cos2 ) cos )È4cos2 ) 3, y œ 2sin ) cos ) sin )È4cos2 ) 3, 0 Ÿ ) Ÿ tan1 ˆ 1 ‰.
2
To obtain the upper limit for ), note that x œ 0 and y œ 1, using y œ ax 2b tan ) Ê 1 œ 2 tan ) Ê ) œ tan1 ˆ 12 ‰.
35. Extend the vertical line through A to the x-axis and let C be the point of intersection. Then OC œ AQ œ x
2
2
and tan t œ OC
œ x2 Ê x œ tan2 t œ 2 cot t; sin t œ OA
Ê OA œ sin2 t ; and (AB)(OA) œ (AQ)# Ê AB ˆ sin2 t ‰ œ x#
#
2
2
2
sin
t
Ê AB ˆ sin t ‰ œ ˆ tan t ‰ Ê AB œ tan# t . Next y œ 2 AB sin t Ê y œ 2 ˆ 2tansin# tt ‰ sin t œ
#
sin t
#
#
#
2 2tan
# t œ 2 2 cos t œ 2 sin t. Therefore let x œ 2 cot t and y œ 2 sin t, 0 t 1 .
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 11.1 Parametrizations of Plane Curves
36. Arc PF œ Arc AF since each is the distance rolled and
Arc PF
œ nFCP Ê Arc PF œ b(nFCP); ArcaAF œ )
b
Ê Arc AF œ a) Ê a) œ b(nFCP) Ê nFCP œ ba );
nOCG œ 1# ); nOCG œ nOCP nPCE
œ nOCP ˆ 1# !‰ . Now nOCP œ 1 nFCP
œ 1 ba ). Thus nOCG œ 1 ba ) 1# ! Ê 1# )
œ 1 ba ) 1# ! Ê ! œ 1 ba ) ) œ 1 ˆ ab b )‰ .
Then x œ OG BG œ OG PE œ (a b) cos ) b cos ! œ (a b) cos ) b cos ˆ1 a b b )‰
œ (a b) cos ) b cos ˆ a b b )‰ . Also y œ EG œ CG CE œ (a b) sin ) b sin !
œ (a b) sin ) b sin ˆ1 a b b )‰ œ (a b) sin ) b sin ˆ a b b )‰ . Therefore
x œ (a b) cos ) b cos ˆ a b b )‰ and y œ (a b) sin ) b sin ˆ a b b )‰ .
aˆa‰
If b œ 4a , then x œ ˆa 4a ‰ cos ) 4a cos Š ˆ a ‰4 )‹
4
a
3a
a
œ 3a
4 cos ) 4 cos 3) œ 4 cos ) 4 (cos ) cos 2) sin ) sin 2))
a
#
#
œ 3a
4 cos ) 4 a(cos )) acos ) sin )b (sin ))(2 sin ) cos ))b
a
a
2a
$
#
#
œ 3a
4 cos ) 4 cos ) 4 cos ) sin ) 4 sin ) cos )
a
3a
$
#
$
œ 3a
4 cos ) 4 cos ) 4 (cos )) a1 cos )b œ a cos );
aˆa‰
a
3a
a
y œ ˆa 4a ‰ sin ) 4a sin Š ˆ a ‰4 )‹ œ 3a
4 sin ) 4 sin 3) œ 4 sin ) 4 (sin ) cos 2) cos ) sin 2))
4
a
#
#
œ 3a
4 sin ) 4 a(sin )) acos ) sin )b (cos ))(2 sin ) cos ))b
a
a
2a
#
$
#
œ 3a
4 sin ) 4 sin ) cos ) 4 sin ) 4 cos ) sin )
3a
a
#
$
œ 3a
4 sin ) 4 sin ) cos ) 4 sin )
3a
a
#
$
$
œ 3a
4 sin ) 4 (sin )) a1 sin )b 4 sin ) œ a sin ).
37. Draw line AM in the figure and note that nAMO is a right
angle since it is an inscribed angle which spans the diameter
of a circle. Then AN# œ MN# AM# . Now, OA œ a,
AN
AM
a œ tan t, and a œ sin t. Next MN œ OP
Ê OP# œ AN# AM# œ a# tan# t a# sin# t
Ê OP œ Èa# tan# t a# sin# t
#
sin t
œ (a sin t)Èsec# t 1 œ acos
t . In triangle BPO,
$
sin t
#
x œ OP sin t œ acos
t œ a sin t tan t and
y œ OP cos t œ a sin# t Ê x œ a sin# t tan t and y œ a sin# t.
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
651
652
Chapter 11 Parametric Equations and Polar Coordinates
38. Let the x-axis be the line the wheel rolls along with the y-axis through a low point of the trochoid
(see the accompanying figure).
Let ) denote the angle through which the wheel turns. Then h œ a) and k œ a. Next introduce xw yw -axes
parallel to the xy-axes and having their origin at the center C of the wheel. Then xw œ b cos ! and
yw œ b sin !, where ! œ 3#1 ). It follows that xw œ b cos ˆ 3#1 )‰ œ b sin ) and yw œ b sin ˆ 3#1 )‰
œ b cos ) Ê x œ h xw œ a) b sin ) and y œ k yw œ a b cos ) are parametric equations of the trochoid.
#
#
#
39. D œ É(x 2)# ˆy "# ‰ Ê D# œ (x 2)# ˆy "# ‰ œ (t 2)# ˆt# "# ‰ Ê D# œ t% 4t 17
4
#
Ê d adtD b œ 4t$ 4 œ 0 Ê t œ 1. The second derivative is always positive for t Á 0 Ê t œ 1 gives a local
minimum for D# (and hence D) which is an absolute minimum since it is the only extremum Ê the closest
point on the parabola is (1ß 1).
#
#
#
40. D œ Ɉ2 cos t 34 ‰ (sin t 0)# Ê D# œ ˆ2 cos t 34 ‰ sin# t Ê d adtD b
œ 2 ˆ2 cos t 34 ‰ (2 sin t) 2 sin t cos t œ (2 sin t) ˆ3 cos t 3# ‰ œ 0 Ê 2 sin t œ 0 or 3 cos t 3# œ 0
#
#
#
#
Ê t œ 0, 1 or t œ 13 , 531 . Now d dtaD# b œ 6 cos# t 3 cos t 6 sin# t so that d dtaD# b (0) œ 3 Ê relative
#
#
#
#
maximum, d dtaD# b (1) œ 9 Ê relative maximum, d dtaD# b ˆ 13 ‰ œ 92 Ê relative minimum, and
d # aD # b ˆ 5 1 ‰
œ 9#
dt#
3
Ê relative minimum. Therefore both t œ 13 and t œ 531 give points on the ellipse closest to
È
È
the point ˆ 34 ß !‰ Ê Š1ß #3 ‹ and Š1ß #3 ‹ are the desired points.
41. (a)
(b)
(c)
42. (a)
(b)
(c)
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 11.1 Parametrizations of Plane Curves
43.
44. (a)
(b)
(c)
45. (a)
(b)
46. (a)
(b)
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
653
654
Chapter 11 Parametric Equations and Polar Coordinates
47. (a)
(b)
(c)
48. (a)
(b)
(c)
(d)
11.2 CALCULUS WITH PARAMETRIC CURVES
dy
dy
dy/dt
2 cos t
1. t œ 14 Ê x œ 2 cos 14 œ È2, y œ 2 sin 14 œ È2; dx
dt œ 2 sin t, dt œ 2 cos t Ê dx œ dx/dt œ 2 sin t œ cot t
Ê dy
dx ¹
#
w
tœ 14
#
œ cot 14 œ 1; tangent line is y È2 œ 1 Šx È2‹ or y œ x 2È2 ; dy
dt œ csc t
w
#
#
/dt
d y
"
csc t
Ê ddxy# œ dy
dx/dt œ 2 sin t œ 2 sin$ t Ê dx# ¹
tœ 14
œ È 2
È
2. t œ 6" Ê x œ sin ˆ21 ˆ 6" ‰‰ œ sin ˆ 13 ‰ œ #3 , y œ cos ˆ21 ˆ 6" ‰‰ œ cos ˆ 13 ‰ œ #" ; dx
dt œ 21 cos 21t,
dy
dy
sin 21t
Ê dx
œ 2211cos
21t œ tan 21t Ê dx ¹
dy
dt œ 21 sin 21t
È
tœc 16
œ tan ˆ21 ˆ 6" ‰‰ œ tan ˆ 13 ‰ œ È3;
w
#
#
d y
21 sec 21t
#
tangent line is y "# œ È3 ’x Š #3 ‹“ or y œ È3x 2; dy
dt œ 21 sec 21t Ê dx# œ 21 cos 21t
#
œ cos$"21t Ê ddxy# ¹
tœc 16
œ 8
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 11.2 Calculus With Parametric Curves
dy
dy
dy/dt
2 sin t
3. t œ 14 Ê x œ 4 sin 14 œ 2È2, y œ 2 cos 14 œ È2; dx
dt œ 4 cos t, dt œ 2 sin t Ê dx œ dx/dt œ 4 cos t
œ "# tan t Ê dy
dx ¹
dyw
"
#
dt œ # sec t
tœ 14
œ "# tan 14 œ "# ; tangent line is y È2 œ "# Šx 2È2‹ or y œ "# x 2È2 ;
#
w
d y
dy /dt
Ê dx
# œ dx/dt œ
"# sec# t
"
4 cos t œ 8 cos$ t
#
d y
Ê dx
#¹
È
tœ 14
œ 42
È
È
3 sin t
dy
dy
È
È3
4. t œ 231 Ê x œ cos 231 œ "# , y œ È3 cos 231 œ #3 ; dx
dt œ sin t, dt œ 3 sin t Ê dx œ sin t œ
Ê dy
dx ¹
w
È
tœ 231
#
Ê ddxy# ¹
#
d y
0
œ È3 ; tangent line is y Š #3 ‹ œ È3 x ˆ #" ‰‘ or y œ È3 x; dy
dt œ 0 Ê dx# œ sin t œ 0
œ0
tœ 231
dy
dy
dy/dt
dy
"
1
5. t œ 14 Ê x œ 14 , y œ "# ; dx
dt œ 1, dt œ #Èt Ê dx œ dx/dt œ 2Èt Ê dx ¹
w
#
tœ 14
œ
"
#É "4
œ 1; tangent line is
w
#
/dt
" $Î#
" $Î#
y "# œ 1 † ˆx 4" ‰ or y œ x 4" ; dy
Ê ddxy# œ dy
Ê ddxy# ¹
dt œ 4 t
dx/dt œ 4 t
tœ 14
œ 2
dy
#
#
6. t œ 14 Ê x œ sec# ˆ 14 ‰ 1 œ 1, y œ tan ˆ 14 ‰ œ 1; dx
dt œ 2 sec t tan t, dt œ sec t
#
dy
"
"
sec t
Ê dy
dx œ 2 sec# t tan t œ 2 tan t œ # cot t Ê dx ¹
tœc 14
œ "# cot ˆ 14 ‰ œ "# ; tangent line is
w
" csc# t
#
d y
"
"
#
$
#
y (1) œ "# (x 1) or y œ "# x "# ; dy
dt œ # csc t Ê dx# œ 2 sec# t tan t œ 4 cot t
#
Ê ddxy# ¹
tœc 14
œ "4
dy
dy
dy/dt
#
7. t œ 16 Ê x œ sec 16 œ È23 , y œ tan 16 œ È"3 ; dx
dt œ sec t tan t, dt œ sec t Ê dx œ dx/dt
#
dy
œ secsect tant t œ csc t Ê dx
¹
dyw
dt œ csc t cot t
tœ 16
œ csc 16 œ 2; tangent line is y È"3 œ 2 Šx È23 ‹ or y œ 2x È3 ;
#
w
#
/dt
d y
csc t cot t
$
Ê ddxy# œ dy
dx/dt œ sec t tan t œ cot t Ê dx# ¹
tœ 16
œ 3È3
ˆ 3 ‰ (3t) "Î#
"
"Î# dy
#
8. t œ 3 Ê x œ È3 1 œ 2, y œ È3(3) œ 3; dx
, dt œ 3# (3t)"Î# Ê dy
dt œ # (t 1)
dx œ ˆ " ‰ (t1) "Î#
#
È
œ 3 Èt3t 1 œ dy
dx ¹
È
tœ3
31
œ 3È3(3)
œ 2; tangent line is y 3 œ 2[x (2)] or y œ 2x 1;
È3t 3 (t 1) "Î# ‘3Èt 1 3 (3t) "Î# ‘
#
#
œ 2tÈ3t3Èt1
dt œ
3t
dyw
#
Ê ddxy# ¹
tœ3
#
Ê ddxy# œ
Š 2tÈ3t3Èt 1 ‹
Š 2Èt1 1 ‹
œ tÈ33t
œ "3
$
dy
dy
dy/dt
dy
4t
$
#
9. t œ 1 Ê x œ 5, y œ 1; dx
dt œ 4t, dt œ 4t Ê dx œ dx/dt œ 4t œ t Ê dx ¹
w
#
w
#
d y
dy /dt
d y
"
2t
y 1 œ 1 † (x 5) or y œ x 4; dy
dt œ 2t Ê dx# œ dx/dt œ 4t œ # Ê dx# ¹
dy
" dy
"
10. t œ 1 Ê x œ 1, y œ 2; dx
dt œ t# , dt œ t Ê dx œ
w
#
tœc1
ˆ "t ‰
Š t"# ‹
d y
y (2) œ 1(x 1) or y œ x 1; dy
dt œ 1 Ê dx# œ
œ t Ê dy
dx ¹
1
œ t#
Š t"# ‹
tœc1
tœ1
œ (1)# œ 1; tangent line is
œ "#
œ 1; tangent line is
#
Ê ddxy# ¹
tœ1
œ1
È
dy
dy
dy/dt
11. t œ 13 Ê x œ 13 sin 13 œ 13 #3 , y œ 1 cos 13 œ 1 #" œ #" ; dx
dt œ 1 cos t, dt œ sin t Ê dx œ dx/dt
È
Š #3 ‹
È
sin ˆ 1 ‰
dy
œ 1 sincost t Ê dx
¹ 1 œ 1cos 3ˆ 1 ‰ œ ˆ " ‰ œ È3 ; tangent line is y "# œ È3 Šx 13 #3 ‹
tœ 3
3
#
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
655
656
Chapter 11 Parametric Equations and Polar Coordinates
w
È
#
w
1
ˆ
‰
(1 cos t)(cos t) (sin t)(sin t)
d y
dy /dt
1
1 cos t
Ê y œ È3x 1 3 3 2; dy
œ 1 cos
dt œ
(1cos t)#
t Ê dx# œ dx/dt œ 1 cos t
#
d y
1
œ (1 cos
t)# Ê dx# ¹
tœ 13
œ 4
dy
dy
cos t
12. t œ 12 Ê x œ cos 12 œ 0, y œ 1 sin 12 œ 2; dx
dt œ sin t, dt œ cos t Ê dx œ sin t œ cot t
dy
Ê dx
¹
w
tœ 2
1
#
#
#
d y
d y
csc t
#
$
œ cot 1# œ 0; tangent line is y œ 2; dy
dt œ csc t Ê dx# œ sin t œ csc t Ê dx# ¹
2
at 1b
dy
dy
dy
1
1
13. t œ 2 Ê x œ 2 1 1 œ 31 , y œ 2 2 1 œ 2; dx
dt œ at 1b2 , dt œ at 1b2 Ê dx œ at 1b2 Ê dx ¹
tœ 12
œ 1
2
tœ2
1b
œ aa22 œ 9;
1 b2
w
4 at 1 b
4 at 1 b3
4a 2 1b3
d# y
d# y
tangent line is y œ 9x 1; dy
dt œ at 1b3 Ê dx# œ at 1b3 Ê dx# ¹ tœ2 œ a2 1b3 œ 108
dy
dy
e
t dy
t
14. t œ 0 Ê x œ 0 e0 œ 1, y œ 1 e0 œ 0; dx
dt œ 1 e , dt œ e Ê dx œ 1 et Ê dx ¹
t
w
#
#
d y
d y
e
e
tangent line is y œ 12 x 12 ; dy
dt œ a1 et b2 Ê dx# œ a1 et b3 Ê dx# ¹
t
t
œ 1ee0 œ 12 ;
0
tœ0
e
œ a1 œ 18
e 0 b3
0
tœ0
dx
4t
2 dx
15. x3 2t# œ 9 Ê 3x2 dx
dt 4t œ 0 Ê 3x dt œ 4t Ê dt œ 3x2 ;
dy
dy
dy/dt
6t
t
2y$ 3t# œ 4 Ê 6y# dy
dt 6t œ 0 Ê dt œ 6y# œ y# ; thus dx œ dx/dt œ
#
Š yt# ‹
t(3x2 )
3x2
œ y# (4t) œ 4y# ; t œ 2
Š c4t ‹
3x2
Ê x 2(2) œ 9 Ê x 8 œ 9 Ê x œ 1 Ê x œ 1; t œ 2 Ê 2y 3(2)# œ 4
3
3
$
3
Ê 2y$ œ 16 Ê y$ œ 8 Ê y œ 2; therefore dy
dx ¹
" ˆ
Èt‰
16. x œ É5 Èt Ê dx
dt œ # 5 "Î#
2
tœ2
3
œ 34aa"#bb# œ 16
ˆ "# t"Î# ‰ œ "
4È t É 5 È t
" "Î#
; y(t 1) œ Èt Ê y (t 1) dy
dt œ # t
" #yÈt
"
dy
" #yÈt 4Èt É5 Èt
" #yÈt
dy
dy
#Èt y
#tÈt 2Èt
"
dt
Ê at 1b dy
œ È
†
"
dt œ #Èt y Ê dt œ at 1b œ #tÈt 2Èt ; thus dx œ dx œ
"
dt
œ
4
Èt É5 Èt
#
tat" b
#ˆ" #yÈt‰É& Èt
; t œ 4 Ê x œ É5 È4 œ È3; t œ 4 Ê y † 3 œ È4 Ê y œ 23
"t
therefore,
dy
dx ¹ tœ4 œ
2Š" 2a 23 bÈ4‹É& È4
"4
œ
10È3
9
dx
2t1
"Î# dx
"Î# ‰ dx
È
ˆ
È
17. x 2x$Î# œ t# t Ê dx
dt 3x
dt œ 2t 1 Ê 1 3x
dt œ 2t 1 Ê dt œ 13x"Î# ; y t 1 2t y œ 4
Èt 1 y ˆ " ‰ (t 1)"Î# 2Èy 2t ˆ " y"Î# ‰ dy œ 0 Ê dy Èt 1 y 2Èy Š t ‹ dy œ 0
Ê dy
Èy
#
#
dt
dt
dt
dt
2È t 1
y
dy
È
Ê ŠÈt 1 Èt y ‹ dy
dt œ 2Èt1 2 y Ê dt œ
cyÈy c 4yÈt b 1
dy
dy/dt
dx œ dx/dt œ
Œ 2Èy (t b 1) b 2tÈt b 1 Š
2t b 1
‹
1 b 3x"Î#
Š 2Èct yb 1 2Èy‹
ŠÈt 1 Èy ‹
t
œ
yÈ y 4yÈt 1
; thus
2È y (t 1) 2tÈt 1
; t œ 0 Ê x 2x$Î# œ 0 Ê x ˆ1 2x"Î# ‰ œ 0 Ê x œ 0; t œ 0
tœ0
œ
È4 4(4)È0 1
Œ 2È4(0 1) 2(0)È0 1 4
Ê yÈ0 1 2(0)Èy œ 4 Ê y œ 4; therefore dy
dx ¹
2(0) 1
Œ 1 3(0)"Î# œ 6
dx
dx
dx
1 x cos t
18. x sin t 2x œ t Ê dx
dt sin t x cos t 2 dt œ 1 Ê (sin t 2) dt œ 1 x cos t Ê dt œ sin t2 ;
dy
sin t t cos t 2
t sin t 2t œ y Ê sin t t cos t 2 œ dy
; t œ 1 Ê x sin 1 2x œ 1
dt ; thus dx œ
ˆ 1 c x cos t ‰
sin t b 2
Ê
sin 1 1 cos 1 2
x œ 1# ; therefore dy
œ 2411 8 œ 4
dx ¹ tœ1 œ
1 Š 1 ‹ cos 1
–
#
sin 1 2
—
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 11.2 Calculus With Parametric Curves
657
dy
dy
dx
2
2
2
2
19. x œ t3 t, y 2t3 œ 2x t2 Ê dx
dt œ 3t 1, dt 6t œ 2 dt 2t Ê dt œ 2a3t 1b 2t 6t œ 2t 2
dy
2t 2
Ê dy
dx œ 3t2 1 Ê dx ¹
tœ1
œ 32aa11bb221 œ 1
dy
1 ˆ dx
dx
dx
t
t
‰
20. t œ lnax tb, y œ t et Ê 1 œ x t dt 1 Ê x t œ dt 1 Ê dt œ x t 1, dt œ t e e ;
dy
te e
Ê dy
dx œ x t 1 ; t œ 0 Ê 0 œ lnax 0b Ê x œ 1 Ê dx ¹
t
21. A œ '0
21
œ a10be0e1 œ 12
0
t
tœ0
0
y dx œ '0 aa1 cos tbaa1 cos tbdt œ a2 '0 a1 cos tb2 dt œ a2 '0 a1 2cos t cos2 tbdt
21
21
21
2t ‰
œ a2 '0 ˆ1 2cos t 1 cos
dt œ a2 '0 ˆ 23 2cos t 12 cos 2t‰dt œ a2 ” 23 t 2sin t 14 sin 2t•
2
21
21
21
0
œ a2 a31 0 0b 0 œ 31 a2
22. A œ '0 x dy œ '0 at t2 baet bdt ”u œ t t2 Ê du œ a1 2tbdt; dv œ aet bdt Ê v œ et •
1
1
1
œ et at t2 bº '0 et a1 2tbdt
1
t
t
”u œ 1 2t Ê du œ 2dt; dv œ e dt Ê v œ e •
0
1
1
0
0
œ et at t2 bº ”et a1 2tbº '0 2et dt• œ ”et at t2 b et a1 2tb 2et •º
1
1
0
œ ae1 a0b e1 a1b 2e1 b ae0 a0b e0 a1b 2e0 b œ 1 3e1 œ 1 3e
23. A œ 2'1 y dx œ 2'1 ab sin tbaa sin tbdt œ 2ab'0 sin2 t dt œ 2ab'0
0
1
0
1
1
'
1
1 cos 2t
dt œ ab 0 a1 cos 2tb dt
2
œ ab’t 12 sin 2t“ œ abaa1 0b !b œ 1 ab
0
24. (a) x œ t2 , y œ t6 , 0 Ÿ t Ÿ 1 Ê A œ '0 y dx œ '0 at6 b2t dt œ '0 2t7 dt œ ’ 14 t8 “ œ 14 0 œ 14
1
1
1
1
(b) x œ t , y œ t , 0 Ÿ t Ÿ 1 Ê A œ '0 y dx œ '0 at b3t dt œ '0 3t
3
1
9
1
9
2
1
11
0
1
1 12
dt œ ’ 4 t “
0
œ 14 0 œ 14
#
#
dy
ˆ dx ‰ Š dy
Éasin tb# a1 cos tb# œ È2 2 cos t
25. dx
dt œ sin t and dt œ 1 cos t Ê Ê dt
dt ‹ œ
cos t ‰
È2 ' É sin t dt
Ê Length œ '0 È2 2 cos t dt œ È2 '0 Ɉ 11 cos t (1 cos t) dt œ
1 cos t
0
1
1
t
œ È2 '0 È1sin
dt (since sin t
cos t
1
1
#
0 on [0ß 1]); [u œ 1 cos t Ê du œ sin t dt; t œ 0 Ê u œ 0,
t œ 1 Ê u œ 2] Ä È2 '0 u"Î# du œ È2 2u"Î# ‘ ! œ 4
2
#
#
#
dy
#
ˆ dx ‰ Š dy
Éa3t# b# (3t)# œ È9t% 9t# œ 3tÈt# 1 Šsince t
26. dx
dt œ 3t and dt œ 3t Ê Ê dt
dt ‹ œ
È3
Ê Length œ '0
Ä '1
4
0 on ’0ß È3“‹
3tÈt# 1 dt; ’u œ t# 1 Ê 3# du œ 3t dt; t œ 0 Ê u œ 1, t œ È3 Ê u œ 4“
%
3 "Î#
du œ u$Î# ‘ " œ (8 1) œ 7
# u
#
dy
"Î#
Èt# a2t 1b œ Éat 1b# œ kt 1k œ t 1 since 0 Ÿ t Ÿ 4
‰# Š dy
27. dx
Ê Êˆ dx
dt œ t and dt œ (2t 1)
dt
dt ‹ œ
Ê Length œ '0 at 1b dt œ ’ t2 t“ œ a8 4b œ 12
4
#
%
!
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
658
Chapter 11 Parametric Equations and Polar Coordinates
#
"Î#
ˆ dx ‰# Š dy
Éa2t 3b a1 tb# œ Èt# 4t 4 œ kt 2k œ t 2
28. dx
and dy
dt œ a2t 3b
dt œ 1 t Ê Ê dt
dt ‹ œ
since 0 Ÿ t Ÿ 3 Ê Length œ '0 (t 2) dt œ ’ t2 2t“ œ 21
#
3
3
#
!
#
#
dy
ˆ dx ‰ Š dy
Éa8t cos tb# a8t sin tb# œ È64t# cos# t 64t# sin# t
29. dx
dt œ 8t cos t and dt œ 8t sin t Ê Ê dt
dt ‹ œ
œ k8tk œ 8t since 0 Ÿ t Ÿ 1# Ê Length œ '0
1Î2
1Î#
8t dt œ c4t# d !
œ 1#
#
#
dy
#
ˆ " ‰
ˆ dx ‰ Š dy
30. dx
dt œ sec ttan t asec t tan t sec tb cos t œ sec t cos t and dt œ sin t Ê Ê dt
dt ‹
œ Éasec t cos tb# asin tb# œ Èsec# t 1 œ Ètan# t œ ktan tk œ tan t since 0 Ÿ t Ÿ 13
Ê Length œ '0
1Î3
tan t dt œ '0
1Î3
1Î$
sin t
œ ln "# ln 1 œ ln 2
cos t dt œ c ln kcos tkd !
dy
ˆ dx ‰ Š dy
Éasin tb# acos tb# œ 1 Ê Area œ ' 21y ds
31. dx
dt œ sin t and dt œ cos t Ê Ê dt
dt ‹ œ
#
#
œ '0 21a2 sin tba1bdt œ 21 c2t cos td #!1 œ 21[a41 1b a0 1b] œ 81#
21
"Î#
"Î#
Èt t" œ É t# 1 Ê Area œ ' 21x ds
‰# Š dy
32. dx
and dy
Ê Êˆ dx
dt œ t
dt œ t
dt
dt ‹ œ
t
#
È3
œ '0
È3
21 ˆ 23 t$Î# ‰ É t t " dt œ 431 '0
#
tÈt# 1 dt; cu œ t# 1 Ê du œ 2t dt; t œ 0 Ê u œ 1,
’t œ È3 Ê u œ 4“ Ä '1 231 Èu du œ 491 u$Î# ‘ " œ 2891
4
È3
%
Note: '0 21 ˆ 23 t$Î# ‰ É t t 1 dt is an improper integral but limb fatb exists and is equal to 0, where
tÄ!
#
#
fatb œ 21 ˆ 23 t$Î# ‰ É t t " . Thus the discontinuity is removable: define Fatb œ fatb for t 0 and Fa0b œ 0
È3
Ê '0
Fatb dt œ 2891 .
#
#
dy
È2 Ê Êˆ dx ‰# Š dy ‹ œ Ê1# Št È2‹ œ Ét# 2È2 t 3 Ê Area œ ' 21x ds
33. dx
dt œ 1 and dt œ t dt
dt
È2
œ 'cÈ2 21 Št È2‹ Ét# 2È2 t 3 dt; ’u œ t# 2È2 t 3 Ê du œ Š2t 2È2‹ dt; t œ È2 Ê u œ 1,
’t œ È2 Ê u œ 9“ Ä '1 1Èu du œ 23 1u$Î# ‘ " œ 231 a27 1b œ 5231
9
*
' 21y ds œ '0
‰ Š dy
34. From Exercise 30, ʈ dx
dt
dt ‹ œ tan t Ê Area œ
#
1Î$
œ 21 c cos td !
#
1 Î3
21 cos t tan t dt œ 21 '0
1 Î3
sin t dt
œ 21 "# (1)‘ œ 1
dy
È2# 1# œ È5 Ê Area œ ' 21y ds œ ' 21at 1bÈ5 dt
ˆ dx ‰# Š dy
35. dx
dt œ 2 and dt œ 1 Ê Ê dt
dt ‹ œ
0
#
#
1
"
œ 21È5 ’ t2 t“ œ 31È5. Check: slant height is È5 Ê Area is 1a1 2bÈ5 œ 31È5 .
!
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 11.2 Calculus With Parametric Curves
659
dy
Èh# r# Ê Area œ ' 21y ds œ ' 21rtÈh# r# dt
ˆ dx ‰# Š dy
36. dx
dt œ h and dt œ r Ê Ê dt
dt ‹ œ
0
#
1
œ 21rÈh# r# '0 t dt œ 21rÈh# r# ’ t2 “ œ 1rÈh# r# . Check: slant height is Èh# r# Ê Area is
1
#
"
!
1rÈh# r# .
dy
37. Let the density be $ œ 1. Then x œ cos t t sin t Ê dx
dt œ t cos t, and y œ sin t t cos t Ê dt œ t sin t
#
#
È(t cos t)# (t sin t)# œ ktk dt œ t dt since 0 Ÿ t Ÿ 1 . The curve's mass is
‰ Š dy
Ê dm œ 1 † ds œ ʈ dx
dt
dt ‹ dt œ
#
M œ ' dm œ '0
1Î2
1Î2
1Î2
#
t dt œ 18 . Also Mx œ ' µ
y dm œ '0 asin t t cos tb t dt œ '0 t sin t dt '0 t# cos t dt
1Î2
1Î#
œ csin t t cos td !
#
x
yœ M
M œ
Š3 14 ‹
1 Î2
1Î#
œ ccos t t sin td !
y
xœ M
M œ
#
œ 3 14 , where we integrated by parts. Therefore,
' µx dm œ '0
œ 24
1# 2. Next, My œ
#
Š 18 ‹
ˆ 3#1 3‰
#
Š 18 ‹
1Î#
ct# sin t 2 sin t 2t cos td !
1Î#
ct# cos t 2 cos t 2t sin td !
acos t t sin tb t dt œ '0
1 Î2
t cos t dt '0
1 Î2
t# sin t dt
œ 3#1 3, again integrating by parts. Hence
24
ˆ 12 24 24
‰
œ 12
1 1# . Therefore axß yb œ 1 1# ß 1# 2 .
dy
t
t
t
t
t
38. Let the density be $ œ 1. Then x œ et cos t Ê dx
dt œ e cos t e sin t, and y œ e sin t Ê dt œ e sin t e cos t
#
‰# Š dy
Éaet cos t et sin tb# aet sin t et cos tb# dt œ È2e2t dt œ È2 et dt.
Ê dm œ 1 † ds œ ʈ dx
dt
dt ‹ dt œ
1
1
The curve's mass is M œ ' dm œ '0 È2 et dt œ È2 e1 È2 . Also Mx œ ' µ
y dm œ '0 aet sin tb ŠÈ2 et ‹ dt
È2 Š e21 " ‹
x
œ '0 È2 e2t sin t dt œ È2 ’ e5 (2 sin t cos t)“ œ È2 Š e5 5" ‹ Ê y œ M
M œ È
1
1
2t
21
!
5
5
2 e1 È 2
21
œ 5 eae1 "1b .
1
1
1
2t
21
Next My œ ' µ
x dm œ '0 aet cos tb ŠÈ2 et ‹ dt œ '0 È2 e2t cos t dt œ È2 ’ e5 a2 cos t sin tb“ œ È2 Š 2e5 52 ‹
!
21
y
Ê xœ M
M œ
È2 Š 2e5 25 ‹
È 2 e1 È 2
21
21
21
œ 52eae1 12b . Therefore axß yb œ Š 52eae1 12b ß 5 eae1 11b ‹.
dy
39. Let the density be $ œ 1. Then x œ cos t Ê dx
dt œ sin t, and y œ t sin t Ê dt œ 1 cos t
#
‰# Š dy
Éasin tb# a1 cos tb# dt œ È2 2 cos t dt. The curve's mass
Ê dm œ 1 † ds œ ʈ dx
dt
dt ‹ dt œ
is M œ ' dm œ '0 È2 2 cos t dt œ È2'0 È1 cos t dt œ È2 '0 É2 cos# ˆ #t ‰ dt œ 2 '0 ¸cos ˆ #t ‰¸ dt
1
1
1
1
œ 2 '0 cos ˆ #t ‰ dt ˆsince 0 Ÿ t Ÿ 1 Ê 0 Ÿ #t Ÿ 1# ‰ œ 2 2 sin ˆ 2t ‰‘ ! œ 4. Also Mx œ ' µ
y dm
1
1
œ '0 at sin tb ˆ2 cos #t ‰ dt œ '0 2t cos ˆ #t ‰ dt '0 2 sin t cos ˆ #t ‰ dt
1
1
1
1
1
Mx
œ 2 4 cos ˆ 2t ‰ 2t sin ˆ #t ‰‘ ! 2 "3 cos ˆ #3 t‰ cos ˆ "# t‰‘ ! œ 41 16
3 Ê yœ M œ
ˆ41 16
‰
3
œ 1 43 .
4
1
1
sin ˆ 3 t‰
cos t cos ˆ #t ‰ dt œ 2 ’sin ˆ 2t ‰ 3# “ œ 2 23
0
!
Next My œ ' µ
x dm œ '0 acos tbˆ2 cos #t ‰ dt œ '
1
ˆ 43 ‰
"
4 œ 3 .
y
œ 43 Ê x œ M
M œ
Therefore axß yb œ ˆ 3" ß 1 43 ‰.
#
dy
3t
#
40. Let the density be $ œ 1. Then x œ t$ Ê dx
dt œ 3t , and y œ # Ê dt œ 3t Ê dm œ 1 † ds
#
‰# Š dy
Éa3t# b# (3t)# dt œ 3 ktk Èt# 1 dt œ 3tÈt# 1 dt since 0 Ÿ t Ÿ È3. The curve's mass
œ ʈ dx
dt
dt ‹ dt œ
È3
is M œ ' dm œ '0
È3
œ #9 '0
$Î#
3tÈt# 1 dt œ ’at# 1b “
È3
!
È
3 #
œ 7. Also Mx œ ' µ
y dm œ '0 3t# Š3tÈt# 1‹ dt
Mx
17.4
' µx dm
t$ Èt# 1 dt œ 87
5 œ 17.4 (by computer) Ê y œ M œ 7 ¸ 2.49. Next My œ
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
660
Chapter 11 Parametric Equations and Polar Coordinates
È3
È3
œ '0 t$ † 3t Èat# 1b dt œ 3 '0 t% Èt# 1 dt ¸ 16.4849 (by computer) Ê x œ My œ 16.4849
¸ 2.35.
7
M
Therefore, axß yb ¸ a2.35ß 2.49b.
41. (a)
‰ Š dy
Éa2 sin 2tb# a2 cos 2tb# œ 2
Ê Êˆ dx
dt
dt ‹ œ
Ê Length œ '0
œ1
1Î2
(b)
#
#
dy
dx
dt œ 2 sin 2t and dt œ 2 cos 2t
1Î#
2 dt œ c2td !
dy
dx
ˆ dx ‰# Š dy
dt œ 1 cos 1t and dt œ 1 sin 1t Ê Ê dt
dt ‹
#
œ Éa1 cos 1tb# a1 sin 1tb# œ 1
Ê Length œ '1Î2 1 dt œ c1td "Î# œ 1
1 Î2
"Î#
dy
w
42. (a) x œ gayb has the parametrization x œ gayb and y œ y for c Ÿ y Ÿ d Ê dx
dy œ g ayb and dy œ 1; then
dx
dx
'
' È1 [gw ayb]# dy
Length œ 'c ÊŠ dy
dy ‹ Š dy ‹ dy œ c Ê1 Š dy ‹ dy œ c
d
#
#
#
d
3 1Î2
(b) x œ y3Î2 , 0 Ÿ y Ÿ 43 Ê dx
Ê L œ '0
dy œ 2 y
4 Î3
d
É1 ˆ 32 y1Î2 ‰# dy œ '
4Î3
0
É1 94 y dy œ ” 49 † 23 ˆ1 94 y‰
3Î2
4Î3
•
0
8
8
œ 27
a4b3Î2 27
a1b3Î2 œ 56
27
dx
(c) x œ 23 y2Î3 , 0 Ÿ y Ÿ 1 Ê dy
œ y1Î3 Ê L œ '0 É1 ay1Î3 b dy œ '0 É1 y21Î3 dy œ lim 'a É y y2Î3 1 dy
1
1
#
1
2Î3
a Ä0
œ lim 32 'a ˆy2Î3 1‰
1
1 Î2
a Ä0
1
ˆ 23 y1Î3 ‰ dy œ lim ” 32 † 23 ˆy2Î3 1‰3Î2 • œ lim Ša2b3Î2 ˆa2Î3 1‰3Î2 ‹ œ 2È2 1
a Ä0
a Ä0
a
dy
2
43. x œ a1 2 sin )bcos ), y œ a1 2 sin )bsin ) Ê dx
d) œ 2cos ) sin )a1 2 sin )b, d) œ 2cos ) sin ) cos )a1 2 sin )b
2cos ) sin ) cos )a1 2 sin )b
4cos ) sin ) cos )
2 sin 2) cos )
Ê dy
œ 2cos
2 ) 2sin2 ) sin ) œ 2 cos 2) sin )
dx œ
2cos2 ) sin )a1 2 sin )b
(a) x œ a1 2 sina0bbcosa0b œ 1, y œ a1 2 sina0bbsina0b œ 0; dy
dx º
)œ0
(b)
sina2a0bb cosa0b
01
1
œ 22 cos
a2a0bb sina0b œ 2 0 œ 2
2 sinˆ2ˆ 1 ‰‰ cosˆ 1 ‰
x œ ˆ1 2 sinˆ 1# ‰‰cosˆ 1# ‰ œ 0, y œ ˆ1 2 sinˆ 1# ‰‰sinˆ 1# ‰ œ 3; dy
œ 2 cosˆ2ˆ#1 ‰‰ sinˆ 1# ‰ œ 0201 œ 0
dx º
#
#
)œ1/2
(c) x œ ˆ1 2 sinˆ 431 ‰‰cosˆ 431 ‰ œ
È3 1
È
dy
, y œ ˆ1 2 sinˆ 431 ‰‰sinˆ 431 ‰ œ 3 2 3 ; dx
º
2
2 sinˆ2ˆ 41 ‰‰ cosˆ 41 ‰
)œ41/3
œ 2 cosˆ2ˆ 431 ‰‰ sinˆ 431 ‰
3
3
È3 1
2È 3 1
œ
È2 œ È3 2 œ Š4 3È3‹
1 3
2
2
dy
dy
d y
sin t
d dy
cos t
44. x œ t, y œ 1 cos t, 0 Ÿ t Ÿ 21 Ê dx
dt œ 1, dt œ sin t Ê dx œ 1 œ sin t Ê dt Š dx ‹ œ cos t Ê dx2 œ 1 œ cos t. The
2
d y
maximum and minimum slope will occur at points that maximize/minimize dy
dx , in other words, points where dx2 œ 0
2
d y
Ê cos t œ 0 Ê t œ 12 or t œ 321 Ê dx
± ± 2 œ 31Î2
1 Î2
(a) the maximum slope is dy
dx º
tœ1Î2
(a)
œ sinˆ 12 ‰ œ 1, which occurs at x œ 12 , y œ 1 cosˆ 12 ‰ œ 1
the minimum slope is dy
œ sinˆ 321 ‰ œ 1, which occurs at x œ 321 , y œ 1 cosˆ 321 ‰ œ 1
dx º
tœ31Î2
#
#
2 a2 cos t 1b
t 1b
dy
dy
dy/dt
dy
2 cos 2t
45. dx
; then dx
œ 0 Ê 2 a2 cos
œ0
dt œ cos t and dt œ 2 cos 2t Ê dx œ dx/dt œ cos t œ
cos t
cos t
Ê 2 cos# t 1 œ 0 Ê cos t œ „ È"2 Ê t œ 14 , 341 , 541 , 741 . In the 1st quadrant: t œ 14 Ê x œ sin 14 œ
È
È2
# and
y œ sin 2 ˆ 14 ‰ œ 1 Ê Š #2 ß 1‹ is the point where the tangent line is horizontal. At the origin: x œ 0 and y œ 0
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Section 11.2 Calculus With Parametric Curves
661
Ê sin t œ 0 Ê t œ 0 or t œ 1 and sin 2t œ 0 Ê t œ 0, 1# , 1, 3#1 ; thus t œ 0 and t œ 1 give the tangent lines at
the origin. Tangents at origin: dy
dx ¹
œ 2 Ê y œ 2x and dy
dx ¹
tœ0
tœ1
œ 2 Ê y œ 2x
dy
dy
dy/dt
3(cos 2t cos t sin 2t sin t)
3 cos 3t
46. dx
dt œ 2 cos 2t and dt œ 3 cos 3t Ê dx œ dx/dt œ 2 cos 2t œ
2 a2 cos# t1b
#
#
#
#
t) 2 sin t cos t sin td
t 1 2 sin tb
t) a4 cos t 3b
œ 3 ca2 cos t 12ba(cos
œ (3 cos t) 2a2a2cos
œ (3 cos
; then
2 cos# t 1b
cos# t 1b
2 a2 cos# t 1b
dy
dx œ 0
#
t) a4 cos t3b
Ê (3 cos
œ 0 Ê 3 cos t œ 0 or 4 cos# t 3 œ 0: 3 cos t œ 0 Ê t œ 1# , 3#1 and
2 a2 cos# t 1b
È
4 cos# t 3 œ 0 Ê cos t œ „ #3 Ê t œ 16 , 561 , 761 , 1161 . In the 1st quadrant: t œ 16 Ê x œ sin 2 ˆ 16 ‰ œ
È3
#
È
and y œ sin 3 ˆ 16 ‰ œ 1 Ê Š #3 ß 1‹ is the point where the graph has a horizontal tangent. At the origin: x œ 0
and y œ 0 Ê sin 2t œ 0 and sin 3t œ 0 Ê t œ 0, 1# , 1, 3#1 and t œ 0, 13 , 231 , 1, 431 , 531 Ê t œ 0 and t œ 1 give
the tangent lines at the origin. Tangents at the origin: dy
dx ¹
tœ0
(31)
3
3
œ 32 cos
cos (21) œ # Ê y œ # x
dy
cos 0
3
3
œ 23 cos
0 œ # Ê y œ # x, and dx ¹
tœ1
dy
47. (a) x œ aat sin tb, y œ aa1 cos tb, 0 Ÿ t Ÿ 21 Ê dx
dt œ aa1 cos tb, dt œ a sin t Ê Length
œ '0 Éaaa1 cos tbb# aa sin tb# dt œ '0 Èa# 2a# cos t a# cos# t a# sin# t dt
21
21
œ aÈ2'0 È1 cos t dt œ aÈ2'0 É2 sin2 ˆ 2t ‰ dt œ 2a'0 sinˆ 2t ‰ dt œ ’4a cosˆ 2t ‰“
21
21
21
21
0
œ 4a cos 1 4a cosa0b œ 8a
dy
(b) a œ 1 Ê x œ t sin t, y œ 1 cos t, 0 Ÿ t Ÿ 21 Ê dx
dt œ 1 cos t, dt œ sin t Ê Surface area œ
œ '0 21a1 cos tbÉa1 cos tb# asin tb# dt œ '0 21a1 cos tbÈ1 2 cos t cos# t sin# t dt
21
21
3Î2
œ 21'0 a1 cos tbÈ2 2 cos t dt œ 2È21'0 a1 cos tb3Î2 dt œ 2È21'0 ˆ1 cos ˆ2 † 2t ‰‰ dt
21
21
21
3 Î2
œ 2È21'0 ˆ2 sin2 ˆ 2t ‰‰ dt œ 81'0 sin3 ˆ 2t ‰ dt
21
21
’u œ 2t Ê du œ 21 dt Ê dt œ 2 du; t œ 0 Ê u œ 0, t œ 21 Ê u œ 1“
œ 161'0 sin3 u du œ 161'0 sin2 u sin u du œ 161'0 a1 cos2 u bsin u du œ 161'0 sin u du 161'0 cos2 u sin u du
1
1
1
1
1
œ ’161cos u 1631 cos3 u“ œ ˆ161 1631 ‰ ˆ161 1631 ‰ œ 6431
0
48. x œ t sin t, y œ 1 cos t, 0 Ÿ t Ÿ 21; Volume œ '0 1 y2 dx œ '0 1a1 cos tb2 a1 cos tbdt
21
21
2t ‰
œ 1'0 a1 3cos t 3cos2 t cos3 tbdt œ 1'0 ˆ1 3cos t 3ˆ 1 cos
cos2 t cos t‰dt
2
21
21
œ 1'0 ˆ 52 3cos t 32 cos 2t a1 sin2 tb cos t‰dt œ 1'0 ˆ 52 4cos t 32 cos 2t sin2 t cos t‰dt
21
21
21
œ 1’ 52 t 4sin t 34 sin 2t 31 sin3 t “
0
œ 1a51 0 0 0b 0 œ 512
47-50. Example CAS commands:
Maple:
with( plots );
with( student );
x := t -> t^3/3;
y := t -> t^2/2;
a := 0;
b := 1;
N := [2, 4, 8 ];
for n in N do
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1
662
Chapter 11 Parametric Equations and Polar Coordinates
tt := [seq( a+i*(b-a)/n, i=0..n )];
pts := [seq([x(t),y(t)],t=tt)];
L := simplify(add( student[distance](pts[i+1],pts[i]), i=1..n ));
# (b)
T := sprintf("#47(a) (Section 11.2)\nn=%3d L=%8.5f\n", n, L );
P[n] := plot( [[x(t),y(t),t=a..b],pts], title=T ):
# (a)
end do:
display( [seq(P[n],n=N)], insequence=true );
ds := t ->sqrt( simplify(D(x)(t)^2 + D(y)(t)^2) ):
# (c)
L := Int( ds(t), t=a..b ):
L = evalf(L);
11.3 POLAR COORDINATES
1. a, e; b, g; c, h; d, f
2. a, f; b, h; c, g; d, e
3. (a) ˆ2ß 1# 2n1‰ and ˆ2ß 1# (2n 1)1‰ , n an integer
(b) (#ß 2n1) and (#ß (2n 1)1), n an integer
(c) ˆ2ß 3#1 2n1‰ and ˆ2ß 3#1 (2n 1)1‰ , n an integer
(d) (#ß (2n 1)1) and (#ß 2n1), n an integer
4. (a) ˆ3ß 14 2n1‰ and ˆ3ß 541 2n1‰ , n an integer
(b) ˆ3ß 14 2n1‰ and ˆ3ß 541 2n1‰ , n an integer
(c) ˆ3ß 14 2n1‰ and ˆ3ß 341 2n1‰ , n an integer
(d) ˆ3ß 14 2n1‰ and ˆ3ß 341 2n1‰ , n an integer
5. (a) x œ r cos ) œ 3 cos 0 œ 3, y œ r sin ) œ 3 sin 0 œ 0 Ê Cartesian coordinates are ($ß 0)
(b) x œ r cos ) œ 3 cos 0 œ 3, y œ r sin ) œ 3 sin 0 œ 0 Ê Cartesian coordinates are ($ß 0)
(c) x œ r cos ) œ 2 cos 21 œ 1, y œ r sin ) œ 2 sin 21 œ È3 Ê Cartesian coordinates are Š1ß È3‹
3
(d)
3
x œ r cos ) œ 2 cos 731 œ 1, y œ r sin ) œ 2 sin 731 œ È3
Ê Cartesian coordinates are Š1ß È3‹
(e) x œ r cos ) œ 3 cos 1 œ 3, y œ r sin ) œ 3 sin 1 œ 0 Ê Cartesian coordinates are (3ß 0)
(f) x œ r cos ) œ 2 cos 1 œ 1, y œ r sin ) œ 2 sin 1 œ È3 Ê Cartesian coordinates are Š1ß È3‹
3
3
(g) x œ r cos ) œ 3 cos 21 œ 3, y œ r sin ) œ 3 sin 21 œ 0 Ê Cartesian coordinates are (3ß 0)
(h) x œ r cos ) œ 2 cos ˆ 1 ‰ œ 1, y œ r sin ) œ 2 sin ˆ 1 ‰ œ È3 Ê Cartesian coordinates are Š1ß È3‹
3
3
6. (a) x œ È2 cos 14 œ 1, y œ È2 sin 14 œ 1 Ê Cartesian coordinates are (1ß 1)
(b) x œ 1 cos 0 œ 1, y œ 1 sin 0 œ 0 Ê Cartesian coordinates are (1ß 0)
(c) x œ 0 cos 1# œ 0, y œ 0 sin 1# œ 0 Ê Cartesian coordinates are (!ß 0)
(d) x œ È2 cos ˆ 1 ‰ œ 1, y œ È2 sin ˆ 1 ‰ œ 1 Ê Cartesian coordinates are (1ß 1)
(e)
4
4
3È 3
51
51
x œ 3 cos 6 œ 2 , y œ 3 sin 6 œ 3#
È
Ê Cartesian coordinates are Š 3 # 3 ß 3# ‹
(f) x œ 5 cos ˆtan" 43 ‰ œ 3, y œ 5 sin ˆtan" 43 ‰ œ 4 Ê Cartesian coordinates are ($ß 4)
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Section 11.3 Polar Coordinates
663
(g) x œ 1 cos 71 œ 1, y œ 1 sin 71 œ 0 Ê Cartesian coordinates are (1ß 0)
(h) x œ 2È3 cos 21 œ È3, y œ 2È3 sin 21 œ 3 Ê Cartesian coordinates are ŠÈ3ß 3‹
3
3
7. (a) a1, 1b Ê r œ È12 12 œ È2, sin ) œ È12 and cos ) œ È12 Ê ) œ 14 Ê Polar coordinates are ŠÈ2, 14 ‹
(b) a3, 0b Ê r œ Éa3b2 02 œ 3, sin ) œ 0 and cos ) œ 1 Ê ) œ 1 Ê Polar coordinates are a3, 1b
2
(c) ŠÈ3, 1‹ Ê r œ ÊŠÈ3‹ a1b2 œ 2, sin ) œ 12 and cos ) œ
È3
111
ˆ 111 ‰
2 Ê ) œ 6 Ê Polar coordinates are 2, 6
(d) a3, 4b Ê r œ Éa3b2 42 œ 5, sin ) œ 45 and cos ) œ 35 Ê ) œ 1 arctanˆ 43 ‰ Ê Polar coordinates are
ˆ5, 1 arctanˆ 43 ‰‰
8. (a) a2, 2b Ê r œ Éa2b2 a2b2 œ 2È2, sin ) œ È12 and cos ) œ È12 Ê ) œ 341 Ê Polar coordinates are
Š2È2, 341 ‹
(b) a0, 3b Ê r œ È02 32 œ 3, sin ) œ 1 and cos ) œ 0 Ê ) œ 12 Ê Polar coordinates are ˆ3, 12 ‰
2
È
(c) ŠÈ3, 1‹ Ê r œ ÊŠÈ3‹ 12 œ 2, sin ) œ 12 and cos ) œ 23 Ê ) œ 561 Ê Polar coordinates are ˆ2, 561 ‰
5
ˆ 12 ‰
(d) a5, 12b Ê r œ É52 a12b2 œ 13, sin ) œ 12
13 and cos ) œ 12 Ê ) œ arctan 5 Ê Polar coordinates are
ˆ13, arctanˆ 12
‰‰
5
9. (a) a3, 3b Ê r œ È32 32 œ 3È2, sin ) œ È12 and cos ) œ È12 Ê ) œ 541 Ê Polar coordinates are
Š3È2, 541 ‹
(b) a1, 0b Ê r œ Éa1b2 02 œ 1, sin ) œ 0 and cos ) œ 1 Ê ) œ 0 Ê Polar coordinates are a1, 0b
2
È
(c) Š1, È3‹ Ê r œ Êa1b2 ŠÈ3‹ œ 2, sin ) œ 23 and cos ) œ 12 Ê ) œ 531 Ê Polar coordinates are
ˆ2, 531 ‰
(d) a4, 3b Ê r œ É42 a3b2 œ 5, sin ) œ 35 and cos ) œ 45 Ê ) œ 1 arctanˆ 34 ‰ Ê Polar coordinates are
ˆ5, 1 arctanˆ 43 ‰‰
10. (a) a2, 0b Ê r œ Éa2b2 02 œ 2, sin ) œ 0 and cos ) œ 1 Ê ) œ 0 Ê Polar coordinates are a2, 0b
(b) a1, 0b Ê r œ È12 02 œ 1, sin ) œ 0 and cos ) œ 1 Ê ) œ 1 or ) œ 1 Ê Polar coordinates are a1, 1b or
a1, 1b
(c) a0, 3b Ê r œ É02 a3b2 œ 3, sin ) œ 1 and cos ) œ 0 Ê ) œ 12 Ê Polar coordinates are ˆ3, 12 ‰
È
È
2
2
È
(d) Š 23 , 12 ‹ Ê r œ ÊŠ 23 ‹ ˆ 21 ‰ œ 1, sin ) œ 12 and cos ) œ 23 Ê ) œ 761 or ) œ 561 Ê Polar coordinates
are ˆ1, 761 ‰ or ˆ1, 561 ‰
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664
Chapter 11 Parametric Equations and Polar Coordinates
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
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Section 11.3 Polar Coordinates
665
26.
27. r cos ) œ 2 Ê x œ 2, vertical line through (#ß 0)
28. r sin ) œ 1 Ê y œ 1, horizontal line through (0ß 1)
29. r sin ) œ 0 Ê y œ 0, the x-axis
30. r cos ) œ 0 Ê x œ 0, the y-axis
31. r œ 4 csc ) Ê r œ sin4 ) Ê r sin ) œ 4 Ê y œ 4, a horizontal line through (0ß 4)
3
32. r œ 3 sec ) Ê r œ cos
) Ê r cos ) œ 3 Ê x œ 3, a vertical line through (3ß 0)
33. r cos ) r sin ) œ 1 Ê x y œ 1, line with slope m œ 1 and intercept b œ 1
34. r sin ) œ r cos ) Ê y œ x, line with slope m œ 1 and intercept b œ 0
35. r# œ 1 Ê x# y# œ 1, circle with center C œ (!ß 0) and radius 1
36. r# œ 4r sin ) Ê x# y# œ 4y Ê x# y# 4y 4 œ 4 Ê x# (y 2)# œ 4, circle with center C œ (0ß 2) and radius 2
37. r œ sin )52 cos ) Ê r sin ) 2r cos ) œ 5 Ê y 2x œ 5, line with slope m œ 2 and intercept b œ 5
38. r# sin 2) œ 2 Ê 2r# sin ) cos ) œ 2 Ê (r sin ))(r cos )) œ 1 Ê xy œ 1, hyperbola with focal axis y œ x
)‰ˆ " ‰
39. r œ cot ) csc ) œ ˆ cos
Ê r sin# ) œ cos ) Ê r# sin# ) œ r cos ) Ê y# œ x, parabola with vertex (0ß 0)
sin )
sin )
which opens to the right
sin ) ‰
40. r œ 4 tan ) sec ) Ê r œ 4 ˆ cos
Ê r cos# ) œ 4 sin ) Ê r# cos# ) œ 4r sin ) Ê x# œ 4y, parabola with
#)
vertex œ (!ß 0) which opens upward
41. r œ (csc )) er cos ) Ê r sin ) œ er cos ) Ê y œ ex , graph of the natural exponential function
42. r sin ) œ ln r ln cos ) œ ln (r cos )) Ê y œ ln x, graph of the natural logarithm function
43. r# 2r# cos ) sin ) œ 1 Ê x# y# 2xy œ 1 Ê x# 2xy y# œ 1 Ê (x y)# œ 1 Ê x y œ „ 1, two parallel
straight lines of slope 1 and y-intercepts b œ „ 1
44. cos# ) œ sin# ) Ê r# cos# ) œ r# sin# ) Ê x# œ y# Ê kxk œ kyk Ê „ x œ y, two perpendicular
lines through the origin with slopes 1 and 1, respectively.
45. r# œ 4r cos ) Ê x# y# œ 4x Ê x# 4x y# œ 0 Ê x# 4x 4 y# œ 4 Ê (x 2)# y# œ 4, a circle with
center C(2ß 0) and radius 2
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Chapter 11 Parametric Equations and Polar Coordinates
46. r# œ 6r sin ) Ê x# y# œ 6y Ê x# y# 6y œ 0 Ê x# y# 6y 9 œ 9 Ê x# (y 3)# œ 9, a circle with
center C(0ß 3) and radius 3
47. r œ 8 sin ) Ê r# œ 8r sin ) Ê x# y# œ 8y Ê x# y# 8y œ 0 Ê x# y# 8y 16 œ 16 Ê x# (y 4)# œ 16, a
circle with center C(0ß 4) and radius 4
48. r œ 3 cos ) Ê r# œ 3r cos ) Ê x# y# œ 3x Ê x# y# 3x œ 0 Ê x# 3x 94 y# œ 94
#
Ê ˆx 3# ‰ y# œ 49 , a circle with center C ˆ 3# ß !‰ and radius 3#
49. r œ 2 cos ) 2 sin ) Ê r# œ 2r cos ) 2r sin ) Ê x# y# œ 2x 2y Ê x# 2x y# 2y œ 0
Ê (x 1)# (y 1)# œ 2, a circle with center C(1ß 1) and radius È2
50. r œ 2 cos ) sin ) Ê r# œ 2r cos ) r sin ) Ê x# y# œ 2x y Ê x# 2x y# y œ 0
#
Ê (x 1)# ˆy "# ‰ œ 54 , a circle with center C ˆ1ß "# ‰ and radius
È5
#
È
51. r sin ˆ) 16 ‰ œ 2 Ê r ˆsin ) cos 16 cos ) sin 16 ‰ œ 2 Ê #3 r sin ) "# r cos ) œ 2 Ê
Ê È3 y x œ 4, line with slope m œ " and intercept b œ 4
È3
È3
"
# y # xœ 2
È3
È
52. r sin ˆ 231 )‰ œ 5 Ê r ˆsin 231 cos ) cos 231 sin )‰ œ 5 Ê #3 r cos ) "# r sin ) œ 5 Ê
Ê È3 x y œ 10, line with slope m œ È3 and intercept b œ 10
È3
"
# x # yœ 5
53. x œ 7 Ê r cos ) œ 7
54. y œ 1 Ê r sin ) œ 1
55. x œ y Ê r cos ) œ r sin ) Ê ) œ 14
56. x y œ 3 Ê r cos ) r sin ) œ 3
57. x# y# œ 4 Ê r# œ 4 Ê r œ 2 or r œ 2
58. x# y# œ 1 Ê r# cos# ) r# sin# ) œ 1 Ê r# acos# ) sin# )b œ 1 Ê r# cos 2) œ 1
#
#
59. x9 y4 œ 1 Ê 4x# 9y# œ 36 Ê 4r# cos# ) 9r# sin# ) œ 36
60. xy œ 2 Ê (r cos ))(r sin )) œ 2 Ê r# cos ) sin ) œ 2 Ê 2r# cos ) sin ) œ 4 Ê r# sin 2) œ 4
61. y# œ 4x Ê r# sin# ) œ 4r cos ) Ê r sin# ) œ 4 cos )
62. x# xy y# œ 1 Ê x# y# xy œ 1 Ê r# r# sin ) cos ) œ 1 Ê r# (1 sin ) cos )) œ 1
63. x# (y 2)# œ 4 Ê x# y# 4y 4 œ 4 Ê x# y# œ 4y Ê r# œ 4r sin ) Ê r œ 4 sin )
64. (x 5)# y# œ 25 Ê x# 10x 25 y# œ 25 Ê x# y# œ 10x Ê r# œ 10r cos ) Ê r œ 10 cos )
65. (x 3)# (y 1)# œ 4 Ê x# 6x 9 y# 2y 1 œ 4 Ê x# y# œ 6x 2y 6 Ê r# œ 6r cos ) 2r sin ) 6
66. (x 2)# (y 5)# œ 16 Ê x# 4x 4 y# 10y 25 œ 16 Ê x# y# œ 4x 10y 13
Ê r# œ 4r cos ) 10r sin ) 13
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Section 11.4 Graphing in Polar Coordinates
67. (!ß )) where ) is any angle
68. (a) x œ a Ê r cos ) œ a Ê r œ cosa ) Ê r œ a sec )
(b) y œ b Ê r sin ) œ b Ê r œ sinb ) Ê r œ b csc )
11.4 GRAPHING IN POLAR COORDINATES
1. 1 cos ()) œ 1 cos ) œ r Ê symmetric about the
x-axis; 1 cos ()) Á r and 1 cos (1 ))
œ 1 cos ) Á r Ê not symmetric about the y-axis;
therefore not symmetric about the origin
2. 2 2 cos ()) œ 2 2 cos ) œ r Ê symmetric about the
x-axis; 2 # cos ()) Á r and 2 2 cos (1 ))
œ 2 2 cos ) Á r Ê not symmetric about the y-axis;
therefore not symmetric about the origin
3. 1 sin ()) œ 1 sin ) Á r and 1 sin (1 ))
œ 1 sin ) Á r Ê not symmetric about the x-axis;
1 sin (1 )) œ 1 sin ) œ r Ê symmetric about
the y-axis; therefore not symmetric about the origin
4. 1 sin ()) œ 1 sin ) Á r and 1 sin (1 ))
œ 1 sin ) Á r Ê not symmetric about the x-axis;
1 sin (1 )) œ 1 sin ) œ r Ê symmetric about the
y-axis; therefore not symmetric about the origin
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667
668
Chapter 11 Parametric Equatins and Polar Coordinates
5. 2 sin ()) œ 2 sin ) Á r and 2 sin (1 ))
œ 2 sin ) Á r Ê not symmetric about the x-axis;
2 sin (1 )) œ 2 sin ) œ r Ê symmetric about the
y-axis; therefore not symmetric about the origin
6. 1 2 sin ()) œ 1 2 sin ) Á r and 1 2 sin (1 ))
œ 1 2 sin ) Á r Ê not symmetric about the x-axis;
1 2 sin (1 )) œ 1 2 sin ) œ r Ê symmetric about the
y-axis; therefore not symmetric about the origin
7. sin ˆ #) ‰ œ sin ˆ #) ‰ œ r Ê symmetric about the y-axis;
sin ˆ 21#) ‰ œ sin ˆ 2) ‰ , so the graph is symmetric about the
x-axis, and hence the origin.
8. cos ˆ #) ‰ œ cos ˆ #) ‰ œ r Ê symmetric about the x-axis;
cos ˆ 21#) ‰ œ cos ˆ 2) ‰ , so the graph is symmetric about the
y-axis, and hence the origin.
9. cos ()) œ cos ) œ r# Ê (rß )) and (rß )) are on the
graph when (rß )) is on the graph Ê symmetric about the
x-axis and the y-axis; therefore symmetric about the origin
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 11.4 Graphing in Polar Coordinates
10. sin (1 )) œ sin ) œ r# Ê (rß 1 )) and (rß 1 )) are on
the graph when (rß )) is on the graph Ê symmetric about
the y-axis and the x-axis; therefore symmetric about the
origin
11. sin (1 )) œ sin ) œ r# Ê (rß 1 )) and (rß 1 ))
are on the graph when (rß )) is on the graph Ê symmetric
about the y-axis and the x-axis; therefore symmetric about
the origin
12. cos ()) œ cos ) œ r# Ê (rß )) and (rß )) are on
the graph when (rß )) is on the graph Ê symmetric about
the x-axis and the y-axis; therefore symmetric about the
origin
13. Since a „ rß )b are on the graph when (rß )) is on the graph
ˆa „ rb# œ 4 cos 2( )) Ê r# œ 4 cos 2)‰ , the graph is
symmetric about the x-axis and the y-axis Ê the graph is
symmetric about the origin
14. Since (rß )) on the graph Ê (rß )) is on the graph
ˆa „ rb# œ 4 sin 2) Ê r# œ 4 sin 2)‰ , the graph is
symmetric about the origin. But 4 sin 2()) œ 4 sin 2)
Á r# and 4 sin 2(1 )) œ 4 sin (21 2)) œ 4 sin (2))
œ 4 sin 2) Á r# Ê the graph is not symmetric about
the x-axis; therefore the graph is not symmetric about
the y-axis
15. Since (rß )) on the graph Ê (rß )) is on the graph
ˆa „ rb# œ sin 2) Ê r# œ sin 2)‰ , the graph is
symmetric about the origin. But sin 2()) œ ( sin 2))
sin 2) Á r# and sin 2(1 )) œ sin (21 2))
œ sin (2)) œ ( sin 2)) œ sin 2) Á r# Ê the graph
is not symmetric about the x-axis; therefore the graph is
not symmetric about the y-axis
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
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Chapter 11 Parametric Equatins and Polar Coordinates
16. Sincea „ rß )b are on the graph when (rß )) is on the
graph ˆa „ rb# œ cos 2()) Ê r# œ cos 2)‰, the
graph is symmetric about the x-axis and the y-axis Ê the
graph is symmetric about the origin.
17. ) œ 1# Ê r œ 1 Ê ˆ1ß 1# ‰ , and ) œ 1# Ê r œ 1
w
)r cos )
Ê ˆ1ß 1# ‰ ; rw œ ddr) œ sin ); Slope œ rrw sin
cos )r sin )
#
sin )r cos )
1‰
ˆ
œ sin ) cos )r sin ) Ê Slope at 1ß # is
sin# ˆ 1# ‰(1) cos 1#
sin 1# cos 1# (1) sin 1#
œ 1; Slope at ˆ1ß 1# ‰ is
sin# ˆ 1# ‰(1) cos ˆ 1# ‰
œ1
sin ˆ 1# ‰ cos ˆ 1# ‰(1) sin ˆ 1# ‰
18. ) œ 0 Ê r œ 1 Ê ("ß 0), and ) œ 1 Ê r œ 1
dr
Ê ("ß 1); rw œ d)
œ cos );
w
)r cos )
cos ) sin )r cos )
Slope œ rrw sin
cos )r sin ) œ cos ) cos )r sin )
0 sin 0(1) cos 0
) sin )r cos )
œ coscos
Ê Slope at ("ß 0) is coscos
# )r sin )
# 0(1) sin 0
1 sin 1(1) cos 1
œ 1; Slope at ("ß 1) is coscos
œ1
# 1(1) sin 1
19. ) œ 14 Ê r œ 1 Ê ˆ"ß 14 ‰ ; ) œ 14 Ê r œ 1
Ê ˆ1ß 14 ‰ ; ) œ 341 Ê r œ 1 Ê ˆ"ß 341 ‰ ;
) œ 341 Ê r œ 1 Ê ˆ1ß 341 ‰ ;
dr
rw œ d)
œ 2 cos 2);
)r cos )
2 cos 2) sin )r cos )
Slope œ rr sin
cos )r sin ) œ 2 cos 2) cos )r sin )
w
w
2 cos ˆ 1 ‰ sin ˆ 1 ‰(1) cos ˆ 1 ‰
Ê Slope at ˆ1ß 14 ‰ is 2 cos ˆ 1# ‰ cos ˆ41 ‰(1) sin ˆ 14 ‰ œ 1;
#
4
4
2 cos ˆ 1 ‰ sin ˆ 1 ‰(1) cos ˆ 1 ‰
Slope at ˆ1ß 14 ‰ is 2 cos ˆ 1# ‰ cos ˆ 41 ‰(1) sin ˆ 41 ‰ œ 1;
#
Slope at ˆ1ß 341 ‰ is
Slope at ˆ1ß 341 ‰ is
4
2 cos Š 3#1 ‹ sin Š 341 ‹(1) cos Š 341 ‹
2 cos Š 3#1 ‹ cos Š 341 ‹(1) sin Š 341 ‹
4
œ 1;
2 cos Š 3#1 ‹ sin Š 341 ‹(1) cos Š 341 ‹
2 cos Š 3#1 ‹ cos Š 341 ‹(1) sin Š 341 ‹
œ 1
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 11.4 Graphing in Polar Coordinates
20. ) œ 0 Ê r œ 1 Ê (1ß 0); ) œ 12 Ê r œ 1 Ê ˆ1ß 12 ‰ ;
) œ 1# Ê r œ 1 Ê ˆ"ß 12 ‰ ; ) œ 1 Ê r œ 1
dr
Ê (1ß 1); rw œ d)
œ 2 sin 2);
)r cos )
2 sin 2) sin )r cos )
Slope œ rr sin
cos )r sin ) œ 2 sin 2) cos )r sin )
w
w
2 sin 0 sin 0cos 0
Ê Slope at (1ß 0) is 2 sin 0 cos 0sin 0 , which is undefined;
2 sin 2 ˆ 1 ‰ sin ˆ 1 ‰(1) cos ˆ 1 ‰
Slope at ˆ1ß 12 ‰ is 2 sin 2 ˆ 12 ‰ cos ˆ21 ‰(1) sin ˆ 21 ‰ œ 0;
2
2
2
2 sin 2 ˆ 1 ‰ sin ˆ 1 ‰(1) cos ˆ 1 ‰
Slope at ˆ1ß 12 ‰ is 2 sin 2 ˆ 1# ‰ cos ˆ #1 ‰(1) sin ˆ #1 ‰ œ 0;
#
#
#
2 sin 21 sin 1cos 1
Slope at ("ß 1) is 2 sin 21 cos 1sin 1 , which is undefined
21. (a)
(b)
22. (a)
(b)
23. (a)
(b)
24. (a)
(b)
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Chapter 11 Parametric Equatins and Polar Coordinates
25.
26. r œ 2 sec ) Ê r œ cos2 ) Ê r cos ) œ 2 Ê x œ 2
27.
28.
29. Note that (rß )) and (rß ) 1) describe the same point in the plane. Then r œ 1 cos ) Í 1 cos () 1)
œ 1 (cos ) cos 1 sin ) sin 1) œ 1 cos ) œ (1 cos )) œ r; therefore (rß )) is on the graph of
r œ 1 cos ) Í (rß ) 1) is on the graph of r œ 1 cos ) Ê the answer is (a).
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 11.4 Graphing in Polar Coordinates
30. Note that (rß )) and (rß ) 1) describe the same point in the plane. Then r œ cos 2) Í sin ˆ2() 1)) 1# ‰
œ sin ˆ2) 5#1 ‰ œ sin (2)) cos ˆ 5#1 ‰ cos (2)) sin ˆ 5#1 ‰ œ cos 2) œ r; therefore (rß )) is on the graph of
r œ sin ˆ2) 1# ‰ Ê the answer is (a).
31.
33. (a)
34. (a)
32.
(b)
(c)
(b)
(d)
(c)
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Chapter 11 Parametric Equatins and Polar Coordinates
(d)
(e)
11.5 AREA AND LENGTHS IN POLAR COORDINATES
1. A œ '0 "# )# d) œ 16 )3 ‘ ! œ 16
1
1
3
2)
2. A œ '1Î4 "# a2 sin )b# d) œ 2'1Î4 sin2 ) d) œ 2'1Î4 1 cos
d) œ '1Î4 a1 cos 2)bd) œ ) 12 sin 2)‘1Î4
2
1Î2
1Î2
1Î2
1Î2
1Î2
œ ˆ 12 0‰ ˆ 14 12 ‰ œ 14 12
3. A œ '0
21
"
#
# (4 2 cos )) d) œ
2 ) ‰‘
'021 "# a16 16 cos ) 4 cos# )b d) œ '021 8 8 cos ) 2 ˆ 1 cos
d)
#
œ '0 (9 8 cos ) cos 2)) d) œ 9) 8 sin ) "2 sin 2)‘ ! œ 181
21
#1
4. A œ '0
21
'
'
21
21
"
" #
" #
2) ‰
#
#
ˆ1 2 cos ) 1 cos
d)
# [a(1 cos ))] d) œ 0 # a a1 2 cos ) cos )b d) œ # a
#
0
21
ˆ #3 2 cos ) #" cos 2)‰ d) œ #" a# #3 ) 2 sin ) "4 sin 2)‘ #1 œ 3# 1a#
!
0
œ "# a# '
5. A œ 2 '0
1Î4
"
#
# cos 2) d) œ
1Î%
4)
'01Î4 1 cos
d) œ "# ) sin44) ‘ ! œ 18
#
6)
6. A œ '1Î6 "# acos 3)b2 d) œ "# ' 1Î6 cos2 3) d) œ "# '1Î6 1 cos
d) œ "4 '1Î6 a1 cos 6)b d)
2
1 Î6
1 Î6
1Î6
1 Î6
1 Î6
1
œ "4 ) 16 sin 6)‘ 1Î6 œ "4 ˆ 16 0‰ "4 ˆ 16 0‰ œ 12
7. A œ '0
1Î2
"
# (4 sin 2)) d) œ
8. A œ (6)(2)'0
1Î6
'01Î2 2 sin 2) d) œ c cos 2)d 1! Î# œ 2
"
# (2 sin 3)) d) œ 12
'01Î6 sin 3) d) œ 12 cos33) ‘ 1! Î' œ 4
9. r œ 2 cos ) and r œ 2 sin ) Ê 2 cos ) œ 2 sin )
Ê cos ) œ sin ) Ê ) œ 14 ; therefore
A œ 2 '0
1Î4
œ '0
1Î4
"
#
# (2 sin )) d) œ
2) ‰
4 ˆ 1 cos
d) œ '0
#
'01Î4 4 sin# ) d)
1Î4
(2 2 cos 2)) d)
1Î%
œ c2) sin 2)d ! œ 1# 1
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 11.5 Area and Lengths in Polar Coordinates
10. r œ 1 and r œ 2 sin ) Ê 2 sin ) œ 1 Ê sin ) œ "#
Ê ) œ 16 or 561 ; therefore
A œ 1(1)# '1Î6
51Î6
"
#
#
# c(2 sin )) 1 d d)
œ 1 '1Î6 ˆ2 sin# ) "# ‰ d)
51Î6
œ 1 '1Î6 ˆ1 cos 2) "# ‰ d)
51Î6
&1Î'
œ 1 '1Î6 ˆ "# cos 2)‰ d) œ 1 "2 ) sin#2) ‘ 1Î'
51Î6
1
œ 1 ˆ 511# "# sin 531 ‰ ˆ 12
"# sin 13 ‰ œ 41 63
È3
11. r œ 2 and r œ 2(1 cos )) Ê 2 œ 2(1 cos ))
Ê cos ) œ 0 Ê ) œ „ 1# ; therefore
A œ 2 '0
1Î2
œ '0
1Î2
œ '0
1Î2
œ '0
1Î2
"
"
#
# [2(1 cos ))] d) # area of the circle
4 a1 2 cos ) cos# )b d) ˆ "# 1‰ (2)#
2) ‰
4 ˆ1 2 cos ) 1 cos
d) 21
#
(4 8 cos ) 2 2 cos 2)) d) 21
1Î#
œ c6) 8 sin ) sin 2)d !
2 1 œ 51 8
12. r œ 2(1 cos )) and r œ 2(1 cos )) Ê 1 cos )
œ 1 cos ) Ê cos ) œ 0 Ê ) œ 1# or 3#1 ; the graph also
gives the point of intersection (0ß 0); therefore
A œ 2 '0
1Î2
"
#
# [2(1 cos ))] d) 2
œ '0 4a1 2cos ) cos# )bd)
'11Î2 "# [2(1 cos ))]# d)
1Î2
'1Î2 4 a1 2 cos ) cos# )bd)
1
œ '0
1Î2
œ '0
1Î2
2) ‰
2) ‰
4 ˆ1 2 cos ) 1 cos
d) '1Î2 4 ˆ1 2 cos ) 1 cos
d)
#
#
1
(6 8 cos ) 2 cos 2)) d) '1Î2 (6 8 cos ) 2 cos 2)) d)
1
1Î#
œ c6) 8 sin ) sin 2)d !
c6) 8 sin ) sin 2)d 11Î# œ 61 16
13. r œ È3 and r# œ 6 cos 2) Ê 3 œ 6 cos 2) Ê cos 2) œ "#
Ê ) œ 16 (in the 1st quadrant); we use symmetry of the
graph to find the area, so
A œ 4 '0 ” "# (6 cos 2)) "# ŠÈ3‹ • d)
1Î6
#
œ 2 '0 (6 cos 2) 3) d) œ 2 c3 sin 2) 3)d !
1Î6
1Î'
œ 3È 3 1
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Chapter 11 Parametric Equatins and Polar Coordinates
14. r œ 3a cos ) and r œ a(1 cos )) Ê 3a cos ) œ a(1 cos ))
Ê 3 cos ) œ 1 cos ) Ê cos ) œ "# Ê ) œ 13 or 13 ;
the graph also gives the point of intersection (0ß 0); therefore
A œ 2 '0
1Î3
"
#
#
#
# c(3a cos )) a (1 cos )) d d)
œ '0 a9a# cos# ) a# 2a# cos ) a# cos# )b d)
1Î3
œ '0
1Î3
a8a# cos# ) 2a# cos ) a# b d)
œ '0 c4a# (1 cos 2)) 2a# cos ) a# d d)
1Î3
œ '0 a3a# 4a# cos 2) 2a# cos )b d)
1Î3
1Î$
œ c3a# ) 2a# sin 2) 2a# sin )d !
È
œ 1a# 2a# ˆ "# ‰ 2a# Š #3 ‹ œ a# Š1 1 È3‹
15. r œ 1 and r œ 2 cos ) Ê 1 œ 2 cos ) Ê cos ) œ "#
Ê ) œ 231 in quadrant II; therefore
A œ 2'21Î3 "# c(2 cos ))# 1# d d) œ '21Î3 a4 cos# ) 1b d)
1
1
œ '21Î3 [2(1 cos 2)) 1] d) œ '21Î3 (1 2 cos 2)) d)
1
1
È
œ c) sin 2)d 1#1Î$ œ 13 #3
16. r œ 6 and r œ 3 csc ) Ê 6 sin ) œ 3 Ê sin ) œ "#
Ê ) œ 16 or 561 ; therefore A œ '1Î6
51Î6
"
#
#
# a6 9 csc )b d)
&1Î'
œ '1Î6 ˆ18 9# csc# )‰ d) œ 18) 9# cot )‘ 1Î'
51Î6
œ Š151 9# È3‹ Š31 9# È3‹ œ 121 9È3
17. r œ sec ) and r œ 4 cos ) Ê 4 cos ) œ sec ) Ê cos2 ) œ 14
Ê ) œ 13 , 231 , 431 , or 531 ; therefore
A œ 2'0
1Î3
œ '0
1Î3
"
#
#
# a16 cos ) sec )b d)
a8 8 cos 2) sec# )b d)
1Î3
œ c8) 4 sin 2) tan )d0
œ Š 831 2È3 È3‹ a0 0 0b œ 831 È3
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Section 11.5 Area and Lengths in Polar Coordinates
677
18. r œ 3 csc ) and r œ 4 sin ) Ê 4 sin ) œ 3 csc ) Ê sin2 ) œ 34
Ê ) œ 13 , 231 , 431 , or 531 ; therefore
A œ 41 2'1Î3 "# a16 sin# ) 9 csc# )b d)
1 Î2
œ 41 '1Î3 a8 8 cos 2) 9 csc# )b d)
1Î2
1Î2
œ 41 c8) 4 sin 2) 9 cot )d1Î3
œ 41 ’a41 0 0b Š 831 2È3 3È3‹“
œ 831 È3
È
È
19. (a) r œ tan ) and r œ Š #2 ‹ csc ) Ê tan ) œ Š #2 ‹ csc )
È
È
Ê sin# ) œ Š #2 ‹ cos ) Ê 1 cos# ) œ Š #2 ‹ cos )
È
Ê cos# ) Š #2 ‹cos ) 1 œ 0 Ê cos ) œ È2 or
È2
# (use the quadratic formula)
Ê ) œ 14 (the solution
in the first quadrant); therefore the area of R" is
A" œ '0
1Î4
"
"
#
# tan ) d) œ #
'01Î4 asec# ) 1b d) œ "# ctan ) )d 1! Î% œ "# ˆtan 14 14 ‰ œ "# 18 ; AO œ Š È#2 ‹ csc 1#
È2
È2
È2 #
È2
È2
1
" È2
"
#
# and OB œ Š # ‹ csc 4 œ 1 Ê AB œ Ê1 Š # ‹ œ # Ê the area of R# is A# œ # Š # ‹ Š # ‹ œ 4 ;
œ
therefore the area of the region shaded in the text is 2 ˆ "# 18 "4 ‰ œ #3 14 . Note: The area must be found this way
since no common interval generates the region. For example, the interval 0 Ÿ ) Ÿ 14 generates the arc OB of r œ tan )
È2
# csc ).
but does not generate the segment AB of the liner œ
È
_ on the line r œ #2 csc ).
(b)
lim
) Ä 1Î2
œ
lim
Instead the interval generates the half-line from B to
tan ) œ _ and the line x œ 1 is r œ sec ) in polar coordinates; then
) Ä 1Î2c
sin )
" ‰
ˆ cos
) cos ) œ
lim
) Ä 1 Î2 c
ˆ sincos) ) 1 ‰ œ
lim
) Ä 1Î2c
(tan ) sec ))
) ‰
ˆ cos
sin ) œ 0 Ê r œ tan ) approaches
lim
) Ä 1 Î2 c
c
r œ sec ) as ) Ä 1# Ê r œ sec ) (or x œ 1) is a vertical asymptote of r œ tan ). Similarly, r œ sec ) (or x œ 1)
is a vertical asymptote of r œ tan ).
20. It is not because the circle is generated twice from ) œ 0 to 21. The area of the cardioid is
2)
A œ 2 '0 "# (cos ) 1)# d) œ '0 acos# ) 2 cos ) 1b d) œ '0 ˆ 1 cos
2 cos ) 1‰ d)
#
1
1
1
1
#
œ 32) sin42) 2 sin )‘ ! œ 3#1 . The area of the circle is A œ 1 ˆ "# ‰ œ 14 Ê the area requested is actually 3#1 14 œ 541
È5
dr
21. r œ )# , 0 Ÿ ) Ÿ È5 Ê d)
œ 2); therefore Length œ '0
È5
œ '0 k)k È)# 4 d) œ (since )
È5
0) '0
Éa)# b# (2))# d) œ '
È5
0
È ) % 4) # d)
)È)# 4 d); u œ )# 4 Ê "# du œ ) d); ) œ 0 Ê u œ 4,
) œ È5 Ê u œ 9“ Ä '4 "# Èu du œ "# 23 u$Î# ‘ % œ 19
3
*
9
22. r œ Èe 2 , 0 Ÿ ) Ÿ 1 Ê ddr) œ Èe 2 ; therefore Length œ '0 ÊŠ Èe 2 ‹ Š Èe 2 ‹ d) œ '0 Ê2 Š e# ‹ d)
)
)
1
)
#
)
#
1
œ '0 e) d) œ e) ‘ ! œ e1 1
1
1
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
2)
678
Chapter 11 Parametric Equatins and Polar Coordinates
dr
23. r œ 1 cos ) Ê d)
œ sin ); therefore Length œ '0 È(1 cos ))# ( sin ))# d)
21
1
œ 2 '0 È2 2 cos ) d) œ 2'0 É 4(1 #cos )) d) œ 4 '0 É 1 #cos ) d) œ 4 '0 cos ˆ #) ‰ d) œ 4 2 sin 2) ‘ ! œ 8
1
1
1
1
24. r œ a sin# #) , 0 Ÿ ) Ÿ 1, a 0 Ê ddr) œ a sin #) cos #) ; therefore Length œ '0 Ɉa sin# #) ‰ ˆa sin #) cos #) ‰ d)
1
#
#
1
œ '0 Éa# sin% #) a# sin# #) cos# #) d) œ '0 a ¸sin #) ¸ Ésin# #) cos# #) d) œ (since 0 Ÿ ) Ÿ 1) a ' sin ˆ #) ‰ d)
1
1
0
1
œ 2a cos 2) ‘ ! œ 2a
6 sin )
'
25. r œ 1 6cos ) , 0 Ÿ ) Ÿ 1# Ê ddr) œ (1 cos ))# ; therefore Length œ 0
1Î2
œ '0
1Î2
36 sin )
'
É (1 36
cos ))# a1 cos )b% d) œ 6 0
1Î2
#
#
#
6 sin )
ʈ 1 6cos ) ‰ Š (1 cos ))# ‹ d)
"
sin# )
¸ 1cos
¸
) É 1 (1 cos ))# d)
cos# ) sin# )
œ ˆsince 1 "cos ) 0 on 0 Ÿ ) Ÿ 1# ‰ 6 '0 ˆ 1 "cos ) ‰ É 1 2 cos(1)cos
d)
) )#
1Î2
cos )
È '
œ 6 '0 ˆ 1 "cos ) ‰ É (12 2cos
) ) # d) œ 6 2 0
1Î2
1Î2
œ 3'0 sec$ #) d) œ 6'0
1Î2
1Î4
d)
œ 6È 2
(1 cos ))$Î#
'01Î2
1Î%
sec$ u du œ (use tables) 6 Œ sec u2tan u ‘ !
' 1Î2 ¸sec$ #) ¸ d)
d)
œ3 0
ˆ2 cos# #) ‰$Î#
"# '0
1Î4
sec u du
1Î%
œ 6 Š È"2 2" ln ksec u tan uk‘ ! ‹ œ 3 ’È2 ln Š1 È2‹“
sin )
sin )
' ˆ 2 ‰# Š (12cos
26. r œ 1 2cos ) , 1# Ÿ ) Ÿ 1 Ê ddr) œ (12cos
))# ; therefore Length œ 1Î2 Ê 1 cos )
) )# ‹ d )
#
1
)
4
sin )
' ¸ 2 ¸ É (1 (1cos )cos) ))sin
œ '1Î2 Ê (1 cos
d)
#
))# Š1 a1 cos )b# ‹ d) œ 1Î2 1 cos )
1
1
#
œ ˆsince 1 cos )
#
#
cos ) sin )
0 on 1# Ÿ ) Ÿ 1‰ 2 '1Î2 ˆ 1 "cos ) ‰ É 1 2 cos(1)cos
d)
))#
1
#
cos )
d)
È '
È '
œ 2 '1Î2 ˆ 1 "cos ) ‰ É (12 2cos
))# d) œ 2 2 1Î2 (1 cos ))$Î# œ 2 2 1Î2
1
1
œ '1Î2 csc$ ˆ #) ‰ d) œ ˆsince csc #)
1
1
#
d)
œ
ˆ2 sin# )# ‰$Î#
'11Î2 ¸csc$ #) ¸ d)
0 on 1# Ÿ ) Ÿ 1‰ 2 '1Î4 csc$ u du œ (use tables)
1Î2
2Œ csc u2cot u ‘ 1Î% #" '1Î4 csc u du œ 2 Š È" 2" ln kcsc u cot uk‘ 1Î% ‹ œ 2 ’ È" "# ln ŠÈ2 1‹“
2
2
1Î2
1Î#
1Î#
œ È2 ln Š1 È2‹
27. r œ cos$ 3) Ê ddr) œ sin 3) cos# 3) ; therefore Length œ '0
1Î4
œ '0
1Î4
œ '0
Écos' ˆ 3) ‰ sin# ˆ 3) ‰ cos% ˆ 3) ‰ d) œ '0
1Î4
1Î4 1cos ˆ 2) ‰
3
#
1Î%
d) œ "# ) 32 sin 23) ‘ !
Ɉcos$ 3) ‰# ˆ sin 3) cos# 3) ‰# d)
ˆcos# 3) ‰ Écos# ˆ 3) ‰ sin# ˆ 3) ‰ d) œ '
1Î4
0
cos# ˆ 3) ‰ d)
œ 18 38
dr
28. r œ È1 sin 2) , 0 Ÿ ) Ÿ 1È2 Ê d)
œ "# (1 sin 2))"Î# (2 cos 2)) œ (cos 2))(1 sin 2))"Î# ; therefore
Length œ '0
È
1 2
œ '0
È
1 2
2)
'
É(1 sin 2)) (1 cos
sin 2)) d) œ 0
#
sin 2)
'
É 212sin
2 ) d) œ 0
È
1 2
È2 d) œ ’È2 )“
È
1 2
1È#
!
#
#
sin 2) cos 2)
É 1 2 sin 2)1 d)
sin 2)
œ 21
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Section 11.6 Conic Sections
679
#
w
ˆ dx
‰ œ cf w ()) cos ) f()) sin )d#
29. Let r œ f()). Then x œ f()) cos ) Ê dx
d) œ f ()) cos ) f()) sin ) Ê
d)
w
œ cf w ())d# cos# ) 2f w ()) f()) sin ) cos ) [f())]# sin# ); y œ f()) sin ) Ê dy
d) œ f ()) sin ) f()) cos )
#
#
#
w
w
#
w
#
#
Ê Š dy
d) ‹ œ cf ()) sin ) f()) cos )d œ cf ())d sin ) 2f ())f()) sin ) cos ) [f())] cos ). Therefore
#
#
#
w
#
#
#
#
#
w
#
#
ˆ dx
‰# Š dy
ˆ dr ‰#
d)
d) ‹ œ cf ())d acos ) sin )b [f())] acos ) sin )b œ cf ())d [f())] œ r d) .
' Ér# ˆ ddr) ‰ d).
‰ Š dy
Thus, L œ '! ʈ dx
d)
d) ‹ d) œ !
"
#
"
#
#
dr
30. (a) r œ a Ê d)
œ 0; Length œ '0 Èa# 0# d) œ '0 kak d) œ ca)d #!1 œ 21a
21
21
dr
(b) r œ a cos ) Ê d)
œ a sin ); Length œ '0 È(a cos ))# (a sin ))# d) œ '0 Èa# acos# ) sin# )b d)
1
œ '0 kak d) œ ca)d 1! œ 1a
1
1
dr
(c) r œ a sin ) Ê d)
œ a cos ); Length œ '0 È(a cos ))# (a sin ))# d) œ '0 Èa# acos# ) sin# )b d)
1
1
œ '0 kak d) œ ca)d 1! œ 1a
1
31. (a) rav œ 21"0 '0 a(1 cos )) d) œ 2a1 c) sin )d #!1 œ a
21
(b) rav œ 21"0 '0 a d) œ #"1 ca)d #!1 œ a
21
(c) rav œ ˆ 1 ‰"ˆ 1 ‰ 'c1Î2 a cos ) d) œ 1" ca sin )d 1Î# œ 2a
1
1Î2
#
1Î#
#
32. r œ 2f()), ! Ÿ ) Ÿ " Ê ddr) œ 2f w ()) Ê r# ˆ ddr) ‰ œ [2f())]# c2f w ())d# Ê Length œ '! É4[f())]# 4 cf w ())d# d)
#
"
œ 2 '! É[f())]# cf w ())d# d) which is twice the length of the curve r œ f()) for ! Ÿ ) Ÿ " .
"
11.6 CONIC SECTIONS
#
1. x œ y8 Ê 4p œ 8 Ê p œ 2; focus is (2ß 0), directrix is x œ 2
#
2. x œ y4 Ê 4p œ 4 Ê p œ 1; focus is (1ß 0), directrix is x œ 1
#
3. y œ x6 Ê 4p œ 6 Ê p œ 3# ; focus is ˆ!ß 3# ‰ , directrix is y œ 3#
#
4. y œ x2 Ê 4p œ 2 Ê p œ 1# ; focus is ˆ!ß #1 ‰ , directrix is y œ 1#
5.
y#
x#
4 9 œ1
Ê c œ È4 9 œ È13 Ê foci are Š „ È13ß !‹ ; vertices are a „ 2ß 0b ; asymptotes are y œ „ 3# x
6.
y#
x#
4 9 œ1
Ê c œ È9 4 œ È5 Ê foci are Š0ß „ È5‹ ; vertices are a0ß „ 3b
7.
x#
#
2 y œ1
Ê c œ È2 1 œ 1 Ê foci are a „ 1ß 0b ; vertices are Š „ È2ß !‹
8.
y#
#
4 x œ1
Ê c œ È4 1 œ È5 Ê foci are Š0ß „ È5‹ ; vertices are a!ß „ 2b ; asymptotes are y œ „ 2x
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
680
Chapter 11 Parametric Equatins and Polar Coordinates
#
9. y# œ 12x Ê x œ 1y# Ê 4p œ 12 Ê p œ 3;
focus is ($ß !), directrix is x œ 3
#
11. x# œ 8y Ê y œ x 8 Ê 4p œ 8 Ê p œ 2;
focus is (!ß 2), directrix is y œ 2
#
"
13. y œ 4x# Ê y œ ˆx" ‰ Ê 4p œ "4 Ê p œ 16
;
4
" ‰
"
focus is ˆ!ß 16
, directrix is y œ 16
#
15. x œ 3y# Ê x œ ˆy" ‰ Ê 4p œ "3 Ê p œ 1"# ;
3
focus is ˆ 1"# ß !‰ , directrix is x œ 1"#
#
10. x# œ 6y Ê y œ x6 Ê 4p œ 6 Ê p œ 3# ;
focus is ˆ!ß 3# ‰ , directrix is y œ 3#
#
y
12. y# œ 2x Ê x œ #
Ê 4p œ 2 Ê p œ #" ;
focus is ˆ "# ß !‰ , directrix is x œ "#
#
"
14. y œ 8x# Ê y œ ˆx" ‰ Ê 4p œ 8" Ê p œ 32
;
8
" ‰
focus is ˆ!ß 32
, directrix is y œ 3"#
#
16. x œ 2y# Ê x œ ˆy" ‰ Ê 4p œ #" Ê p œ "8 ;
#
focus is ˆ "8 ß !‰ , directrix is x œ 8"
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 11.6 Conic Sections
#
#
y
17. 16x# 25y# œ 400 Ê #x5 16
œ1
Ê c œ Èa# b# œ È25 16 œ 3
#
19. 2x# y# œ 2 Ê x# y# œ 1
Ê c œ Èa# b# œ È2 1 œ 1
#
#
21. 3x# 2y# œ 6 Ê x# y3 œ 1
Ê c œ Èa# b# œ È3 2 œ 1
#
#
x
18. 7x# 16y# œ 112 Ê 16
y7 œ 1
Ê c œ Èa# b# œ È16 7 œ 3
#
#
20. 2x# y# œ 4 Ê x# y4 œ 1
Ê c œ Èa# b# œ È4 2 œ È2
#
#
x
22. 9x# 10y# œ 90 Ê 10
y9 œ 1
Ê c œ Èa# b# œ È10 9 œ 1
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
681
682
Chapter 11 Parametric Equations and Polar Coordinates
#
#
23. 6x# 9y# œ 54 Ê x9 y6 œ 1
Ê c œ Èa# b# œ È9 6 œ È3
#
#
y
x
24. 169x# 25y# œ 4225 Ê 25
169
œ1
Ê c œ Èa# b# œ È169 25 œ 12
#
25. Foci: Š „ È2ß !‹ , Vertices: a „ 2ß 0b Ê a œ 2, c œ È2 Ê b# œ a# c# œ 4 ŠÈ2‹ œ 2 Ê x4 y# œ 1
#
#
#
#
26. Foci: a!ß „ 4b , Vertices: a0ß „ 5b Ê a œ 5, c œ 4 Ê b# œ 25 16 œ 9 Ê x9 #y5 œ 1
27. x# y# œ 1 Ê c œ Èa# b# œ È1 1 œ È2 ;
asymptotes are y œ „ x
#
#
x
28. 9x# 16y# œ 144 Ê 16
y9 œ 1
Ê c œ Èa# b# œ È16 9 œ 5;
asymptotes are y œ „ 34 x
#
#
29. y# x# œ 8 Ê y8 x8 œ 1 Ê c œ Èa# b#
œ È8 8 œ 4; asymptotes are y œ „ x
#
#
30. y# x# œ 4 Ê y4 x4 œ 1 Ê c œ Èa# b#
œ È4 4 œ 2È2; asymptotes are y œ „ x
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 11.6 Conic Sections
683
#
#
31. 8x# 2y# œ 16 Ê x# y8 œ 1 Ê c œ Èa# b#
œ È2 8 œ È10 ; asymptotes are y œ „ 2x
32. y# 3x# œ 3 Ê y3 x# œ 1 Ê c œ Èa# b#
œ È3 1 œ 2; asymptotes are y œ „ È3x
#
#
33. 8y# 2x# œ 16 Ê y# x8 œ 1 Ê c œ Èa# b#
œ È2 8 œ È10 ; asymptotes are y œ „ x
y
x
34. 64x# 36y# œ 2304 Ê 36
64
œ 1 Ê c œ È a# b #
œ È36 64 œ 10; asymptotes are y œ „ 4
#
#
#
#
3
35. Foci: Š!ß „ È2‹ , Asymptotes: y œ „ x Ê c œ È2 and ba œ 1 Ê a œ b Ê c# œ a# b# œ 2a# Ê 2 œ 2a#
Ê a œ 1 Ê b œ 1 Ê y# x# œ 1
#
36. Foci: a „ 2ß !b , Asymptotes: y œ „ È"3 x Ê c œ 2 and ba œ È"3 Ê b œ Èa3 Ê c# œ a# b# œ a# a3 œ 4a3
Ê 4 œ 4a3 Ê a# œ 3 Ê a œ È3 Ê b œ 1 Ê x3 y# œ 1
#
#
#
#
y
37. Vertices: a „ 3ß 0b , Asymptotes: y œ „ 43 x Ê a œ 3 and ba œ 43 Ê b œ 43 (3) œ 4 Ê x9 16
œ1
#
#
x
38. Vertices: a!ß „ 2b , Asymptotes: y œ „ 12 x Ê a œ 2 and ba œ 12 Ê b œ 2(2) œ 4 Ê y4 16
œ1
39. (a) y# œ 8x Ê 4p œ 8 Ê p œ 2 Ê directrix is x œ 2,
focus is (#ß !), and vertex is (!ß 0); therefore the new
directrix is x œ 1, the new focus is (3ß 2), and the
new vertex is (1ß 2)
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
#
684
Chapter 11 Parametric Equations and Polar Coordinates
40. (a) x# œ 4y Ê 4p œ 4 Ê p œ 1 Ê directrix is y œ 1,
focus is (!ß 1), and vertex is (!ß 0); therefore the new
directrix is y œ 4, the new focus is (1ß 2), and the
new vertex is (1ß 3)
41. (a)
y#
x#
16 9 œ 1
Ê center is (!ß 0), vertices are (4ß 0)
(b)
(b)
and (%ß !); c œ Èa# b# œ È7 Ê foci are ŠÈ7ß 0‹
and ŠÈ7ß !‹ ; therefore the new center is (%ß $), the
new vertices are (!ß 3) and (8ß 3), and the new foci are
Š4 „ È7ß $‹
42. (a)
y#
x#
9 25 œ 1
Ê center is (!ß 0), vertices are (0ß 5)
and (0ß 5); c œ Èa# b# œ È16 œ 4 Ê foci are
(b)
(!ß 4) and (!ß 4) ; therefore the new center is (3ß 2),
the new vertices are (3ß 3) and (3ß 7), and the new
foci are (3ß 2) and (3ß 6)
43. (a)
y#
x#
16 9 œ 1
Ê center is (!ß 0), vertices are (4ß 0)
(b)
and (4ß 0), and the asymptotes are x4 œ „ y3 or
Èa# b# œ È25 œ 5 Ê foci are
y œ „ 3x
4 ;cœ
(5ß 0) and (5ß 0) ; therefore the new center is (2ß 0), the
new vertices are (2ß 0) and (6ß 0), the new foci
are (3ß 0) and (7ß 0), and the new asymptotes are
y œ „ 3(x 4 2)
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 11.6 Conic Sections
44. (a)
y#
x#
4 5 œ1
Ê center is (!ß 0), vertices are (0ß 2)
and (0ß 2), and the asymptotes are y2 œ
685
(b)
„ Èx5 or
2x
yœ „È
; c œ Èa# b# œ È9 œ 3 Ê foci are
5
(0ß 3) and (0ß 3) ; therefore the new center is (0ß 2),
the new vertices are (0ß 4) and (0ß 0), the new foci
are (0ß 1) and (0ß 5), and the new asymptotes are
2x
y2œ „ È
5
45. y# œ 4x Ê 4p œ 4 Ê p œ 1 Ê focus is ("ß 0), directrix is x œ 1, and vertex is (0ß 0); therefore the new
vertex is (2ß 3), the new focus is (1ß 3), and the new directrix is x œ 3; the new equation is
(y 3)# œ 4(x 2)
46. y# œ 12x Ê 4p œ 12 Ê p œ 3 Ê focus is (3ß 0), directrix is x œ 3, and vertex is (0ß 0); therefore the new
vertex is (4ß 3), the new focus is (1ß 3), and the new directrix is x œ 7; the new equation is (y 3)# œ 12(x 4)
47. x# œ 8y Ê 4p œ 8 Ê p œ 2 Ê focus is (0ß 2), directrix is y œ 2, and vertex is (0ß 0); therefore the new
vertex is (1ß 7), the new focus is (1ß 5), and the new directrix is y œ 9; the new equation is
(x 1)# œ 8(y 7)
48. x# œ 6y Ê 4p œ 6 Ê p œ #3 Ê focus is ˆ!ß #3 ‰ , directrix is y œ #3 , and vertex is (0ß 0); therefore the new
vertex is (3ß 2), the new focus is ˆ3ß "# ‰ , and the new directrix is y œ 7# ; the new equation is
(x 3)# œ 6(y 2)
49. x6 y9 œ 1 Ê center is (!ß 0), vertices are (0ß 3) and (!ß 3); c œ Èa# b# œ È9 6 œ È3 Ê foci are Š!ß È3‹
#
#
and Š!ß È3‹ ; therefore the new center is (#ß 1), the new vertices are (2ß 2) and (#ß 4), and the new foci
are Š#ß 1 „ È3‹ ; the new equation is (x 6 2) (y 9 1) œ 1
#
#
50. x2 y# œ 1 Ê center is (!ß 0), vertices are ŠÈ2ß !‹ and ŠÈ2ß !‹ ; c œ Èa# b# œ È2 1 œ 1 Ê foci are
#
(1ß 0) and ("ß !); therefore the new center is (3ß 4), the new vertices are Š3 „ È2ß 4‹ , and the new foci are (2ß 4)
#
and (4ß 4); the new equation is (x # 3) (y 4)# œ 1
51. x3 y# œ 1 Ê center is (!ß 0), vertices are ŠÈ3ß !‹ and ŠÈ3ß !‹ ; c œ Èa# b# œ È3 2 œ 1 Ê foci are
#
#
(1ß 0) and ("ß !); therefore the new center is (2ß 3), the new vertices are Š2 „ È3ß 3‹ , and the new foci are (1ß 3)
#
#
and (3ß 3); the new equation is (x 3 2) (y # 3) œ 1
x
52. 16
#y5 œ 1 Ê center is (!ß 0), vertices are (!ß &) and (!ß 5); c œ Èa# b# œ È25 16 œ 3 Ê foci are
#
#
(0ß 3) and (0ß 3); therefore the new center is (4ß 5), the new vertices are (4ß 0) and (4ß 10), and the new
#
#
foci are (4ß 2) and (4ß 8); the new equation is (x 164) (y #55) œ 1
53. x4 y5 œ 1 Ê center is (!ß 0), vertices are (2ß 0) and (2ß 0); c œ Èa# b# œ È4 5 œ 3 Ê foci are ($ß !) and
#
#
(3ß 0); the asymptotes are „ x# œ Èy5 Ê y œ „
È5x
# ; therefore the new center is (2ß 2), the new vertices are
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
686
Chapter 11 Parametric Equations and Polar Coordinates
(4ß 2) and (0ß 2), and the new foci are (5ß 2) and (1ß 2); the new asymptotes are y 2 œ „
#
#
È5 (x 2)
; the new
#
equation is (x 4 2) (y 5 2) œ 1
x
54. 16
y9 œ 1 Ê center is (!ß 0), vertices are (4ß 0) and (4ß 0); c œ Èa# b# œ È16 9 œ 5 Ê foci are (5ß !)
#
#
and (5ß 0); the asymptotes are „ x4 œ y3 Ê y œ „ 3x
4 ; therefore the new center is (5ß 1), the new vertices are
(1ß 1) and (9ß 1), and the new foci are (10ß 1) and (0ß 1); the new asymptotes are y 1 œ „ 3(x 4 5) ;
#
#
the new equation is (x 165) (y 9 1) œ 1
55. y# x# œ 1 Ê center is (!ß 0), vertices are (0ß 1) and (0ß 1); c œ Èa# b# œ È1 1 œ È2 Ê foci are
Š!ß „ È2‹ ; the asymptotes are y œ „ x; therefore the new center is (1ß 1), the new vertices are (1ß 0) and
(1ß 2), and the new foci are Š1ß 1 „ È2‹ ; the new asymptotes are y 1 œ „ (x 1); the new equation is
(y 1)# (x 1)# œ 1
56. y3 x# œ 1 Ê center is (!ß 0), vertices are Š0ß È3‹ and Š!ß È3‹ ; c œ Èa# b# œ È3 1 œ 2 Ê foci are (!ß #)
#
and (!ß 2); the asymptotes are „ x œ Èy3 Ê y œ „ È3x; therefore the new center is (1ß 3), the new vertices are
Š"ß $ „ È3‹ , and the new foci are ("ß &) and (1ß 1); the new asymptotes are y 3 œ „ È3 (x 1); the new equation is
(y 3)#
(x 1)# œ 1
3
57. x# 4x y# œ 12 Ê x# 4x 4 y# œ 12 4 Ê (x 2)# y# œ 16; this is a circle: center at C(2ß 0), a œ 4
58. 2x# 2y# 28x 12y 114 œ 0 Ê x# 14x 49 y# 6y 9 œ 57 49 9 Ê (x 7)# (y 3)# œ 1;
this is a circle: center at C(7ß 3), a œ 1
59. x# 2x 4y 3 œ 0 Ê x# 2x 1 œ 4y 3 1 Ê (x 1)# œ 4(y 1); this is a parabola: V(1ß 1), F(1ß 0)
60. y# 4y 8x 12 œ 0 Ê y# 4y 4 œ 8x 12 4 Ê (y 2)# œ 8(x 2); this is a parabola: V(#ß 2), F(!ß #)
#
61. x# 5y# 4x œ 1 Ê x# 4x 4 5y# œ 5 Ê (x 2)# 5y# œ 5 Ê (x 5 2) y# œ 1; this is an ellipse: the
center is (2ß 0), the vertices are Š2 „ È5ß 0‹ ; c œ Èa# b# œ È5 1 œ 2 Ê the foci are (4ß 0) and (!ß 0)
#
62. 9x# 6y# 36y œ 0 Ê 9x# 6 ay# 6y 9b œ 54 Ê 9x# 6(y 3)# œ 54 Ê x6 (y 9 3) œ 1; this is an ellipse:
#
the center is (0ß 3), the vertices are (!ß 0) and (!ß 6); c œ Èa# b# œ È9 6 œ È3 Ê the foci are Š0ß 3 „ È3‹
63. x# 2y# 2x 4y œ 1 Ê x# 2x 1 2 ay# 2y 1b œ 2 Ê (x 1)# 2(y 1)# œ 2
#
Ê (x1) (y 1)# œ 1; this is an ellipse: the center is (1ß 1), the vertices are Š" „ È2ß "‹ ;
2
c œ Èa# b# œ È2 1 œ 1 Ê the foci are (2ß 1) and (0ß 1)
64. 4x# y# 8x 2y œ 1 Ê 4 ax# 2x 1b y# 2y 1 œ 4 Ê 4(x 1)# (y 1)# œ 4
#
Ê (x 1)# (y41) œ 1; this is an ellipse: the center is (1ß 1), the vertices are (1ß 3) and
(1ß 1); c œ Èa# b# œ È4 1 œ È3 Ê the foci are Š1ß " „ È3‹
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 11.6 Conic Sections
65. x# y# 2x 4y œ 4 Ê x# 2x 1 ay# 4y 4b œ 1 Ê (x 1)# (y 2)# œ 1; this is a hyperbola:
the center is (1ß 2), the vertices are (2ß 2) and (!ß 2); c œ Èa# b# œ È1 1 œ È2 Ê the foci are Š1 „ È2ß #‹ ;
the asymptotes are y 2 œ „ (x 1)
66. x# y# 4x 6y œ 6 Ê x# 4x 4 ay# 6y 9b œ 1 Ê (x 2)# (y 3)# œ 1; this is a hyperbola:
the center is (2ß 3), the vertices are (1ß 3) and (3ß 3); c œ Èa# b# œ È1 1 œ È2 Ê the foci are
Š2 „ È2ß 3‹ ; the asymptotes are y 3 œ „ (x 2)
#
67. 2x# y# 6y œ 3 Ê 2x# ay# 6y 9b œ 6 Ê (y 6 3) x3 œ 1; this is a hyperbola: the center is (!ß $),
#
the vertices are Š!ß 3 „ È6‹ ; c œ Èa# b# œ È6 3 œ 3 Ê the foci are (0ß 6) and (!ß 0); the asymptotes are
y 3
È6 œ
„ Èx3 Ê y œ „ È2x 3
#
#
68. y# 4x# 16x œ 24 Ê y# 4 ax# 4x 4b œ 8 Ê y8 (x 2 2) œ 1; this is a hyperbola: the center is (2ß 0),
the vertices are Š2ß „ È8‹ ; c œ Èa# b# œ È8 2 œ È10 Ê the foci are Š2ß „ È10‹ ; the asymptotes are
y
È8 œ
„ xÈ2 Ê y œ „ 2(x 2)
2
#
69. (a) y# œ kx Ê x œ yk ; the volume of the solid formed by
Èkx
revolving R" about the y-axis is V" œ '0
Èkx
œ k1# '0
#
#
1 Š yk ‹ dy
#È
y% dy œ 1x 5 kx ; the volume of the right
circular cylinder formed by revolving PQ about the
y-axis is V# œ 1x# Èkx Ê the volume of the solid
formed by revolving R# about the y-axis is
#È
V$ œ V# V" œ 41x 5 kx . Therefore we can see the
ratio of V$ to V" is 4:1.
(b) The volume of the solid formed by revolving R# about the x-axis is V" œ '0 1 ŠÈkt‹ dt œ 1k'0 t dt
x
#
x
#
œ 1kx
# . The volume of the right circular cylinder formed by revolving PS about the x-axis is
#
V# œ 1 ŠÈkx‹ x œ 1kx# Ê the volume of the solid formed by revolving R" about the x-axis is
#
#
1kx
V$ œ V# V" œ 1kx# 1kx
# œ # . Therefore the ratio of V$ to V" is 1:1.
w(0)
w x
wx
wx
70. y œ ' w
H x dx œ H Š # ‹ C œ 2H C; y œ 0 when x œ 0 Ê 0 œ 2H C Ê C œ 0; therefore y œ 2H is the
#
#
#
#
equation of the cable's curve
71. x# œ 4py and y œ p Ê x# œ 4p# Ê x œ „ 2p. Therefore the line y œ p cuts the parabola at points (2pß p) and
(2pß p), and these points are È[2p (2p)]# (p p)# œ 4p units apart.
72. x lim
Š b x ba Èx# a# ‹ œ ba x lim
Šx Èx# a# ‹ œ ba x lim
Ä_ a
Ä_
Ä_ –
#
#
#
x
x a
œ ba x lim
’ x Èax # a #b “ œ ba x lim
’
Ä_
Ä_
Šx Èx# a# ‹ Šx Èx# a# ‹
x È x # a#
a#
“œ0
x È x# a#
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
—
687
688
Chapter 11 Parametric Equations and Polar Coordinates
#
73. Let y œ É1 x4 on the interval 0 Ÿ x Ÿ 2. The area of the inscribed rectangle is given by
#
#
A(x) œ 2x Š2É1 x4 ‹ œ 4xÉ1 x4 (since the length is 2x and the height is 2y)
#
Ê Aw (x) œ 4É1 x4 x#
.
É1 x4#
#
Thus Aw (x) œ 0 Ê 4É1 x4 x#
É1 x4#
#
œ 0 Ê 4 Š1 x4 ‹ x# œ 0 Ê x# œ 2
Ê x œ È2 (only the positive square root lies in the interval). Since A(0) œ A(2) œ 0 we have that A ŠÈ2‹ œ 4
is the maximum area when the length is 2È2 and the height is È2.
74. (a) Around the x-axis: 9x# 4y# œ 36 Ê y# œ 9 94 x# Ê y œ „ É9 94 x# and we use the positive root
#
#
Ê V œ 2 '0 1 ŠÉ9 94 x# ‹ dx œ 2 '0 1 ˆ9 94 x# ‰ dx œ 21 9x 34 x$ ‘ ! œ 241
2
2
(b) Around the y-axis: 9x# 4y# œ 36 Ê x# œ 4 49 y# Ê x œ „ É4 49 y# and we use the positive root
#
4 $‘
Ê V œ 2'0 1 ŠÉ4 49 y# ‹ dy œ 2 '0 1 ˆ4 49 y# ‰ dy œ 21 4y 27
y ! œ 161
3
3
$
75. 9x# 4y# œ 36 Ê y# œ 9x 4 36 Ê y œ „ 3# Èx# 4 on the interval 2 Ÿ x Ÿ 4 Ê V œ '2 1 Š #3 Èx# 4‹ dx
#
4
#
‰ ˆ8
‰‘ œ 941 ˆ 56
‰ 31
œ 941 '2 ax# 4b dx œ 941 ’ x3 4x“ œ 941 ˆ 64
3 16 3 8
3 8 œ 4 (56 24) œ 241
4
%
$
#
76. Let P" (pß y" ) be any point on x œ p, and let P(xß y) be a point where a tangent intersects y# œ 4px. Now
dy
2p
y y"
dy
2p
y# œ 4px Ê 2y dy
dx œ 4p Ê dx œ y ; then the slope of a tangent line from P" is x (p) œ dx œ y
#
#
y
y
Ê y# yy" œ 2px 2p# . Since x œ 4p
, we have y# yy" œ 2p Š 4p
‹ 2p# Ê y# yy" œ "# y# 2p#
È4y#" 16p#
œ y" „ Èy"# 4p# . Therefore the slopes of the two
#
#
and m# œ y È2p
Ê m" m# œ y# ay4p# 4p# b œ 1
y#" 4p#
"
"
"
Ê "# y# yy" 2p# œ 0 Ê y œ 2y" „
tangents from P" are m" œ y È2p
y# 4p#
"
"
Ê the lines are perpendicular
dy
x2
#
#
77. (x 2)# (y 1)# œ 5 Ê 2(x 2) 2(y 1) dy
dx œ 0 Ê dx œ y1 ; y œ 0 Ê (x 2) (0 1) œ 5
Ê (x 2)# œ 4 Ê x œ 4 or x œ 0 Ê the circle crosses the x-axis at (4ß 0) and (!ß 0); x œ 0
Ê (0 2)# (y 1)# œ 5 Ê (y 1)# œ 1 Ê y œ 2 or y œ 0 Ê the circle crosses the y-axis at (!ß 2) and (!ß !).
4 2
At (4ß 0): dy
dx œ 01 œ 2 Ê the tangent line is y œ 2(x 4) or y œ 2x 8
0 2
At (!ß !): dy
dx œ 01 œ 2 Ê the tangent line is y œ 2x
0 2
At (!ß #): dy
dx œ 21 œ 2 Ê the tangent line is y 2 œ 2x or y œ 2x 2
78. x# y# œ 1 Ê x œ „ È1 y# on the interval 3 Ÿ y Ÿ 3 Ê V œ 'c3 1 ˆÈ1 y# ‰ dy œ 2'0 1 ˆÈ1 y# ‰ dy
3
#
#
3
œ 21'0 a1 y# b dy œ 21 ’y y3 “ œ 241
3
$
$
!
#
79. Let y œ É16 16
9 x on the interval 3 Ÿ x Ÿ 3. Since the plate is symmetric about the y-axis, x œ 0. For a
vertical strip:
aµ
x ßµ
y b œ xß
#
É16 16
9 x
#
, length œ É16 9 x , width œ dx Ê area œ dA œ É16 9 x dx
16
#
#
Ê mass œ dm œ $ dA œ $É16 16
9 x dx. Moment of the strip about the x-axis:
É16 9 x
µ
8 #‰
#
ˆ
y dm œ
dx so the moment of the plate about the x-axis is
Š$ É16 16
#
9 x ‹ dx œ $ 8 9 x
16
#
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
16
#
Section 11.7 Conics in Polar Coordinates
689
8 $‘
Mx œ ' µ
y dm œ 'c3 $ ˆ8 89 x# ‰ dx œ $ 8x 27
x $ œ 32$ ; also the mass of the plate is
3
$
#
' É1 ˆ "3 x‰ dx œ 4$ 'c1 3È1 u# du where u œ x3 Ê 3 du œ dx; x œ 3
M œ 'c3 $ É16 16
9 x dx œ c3 4$
3
3
1
#
Ê u œ 1 and x œ 3 Ê u œ 1. Hence, 4$ 'c1 3È1 u# du œ 12$ 'c1 È1 u# du
1
œ 12$ ’ "2 ŠuÈ1 u# sin" u‹“
"
#
80. y œ Èx# 1 Ê dy
dx œ # ax 1b
È2
1
'
œ É 2x
x# 1 Ê S œ 0
#
–
"
"
1
œ 61$ Ê y œ MMx œ 6321$$ œ 3161 . Therefore the center of mass is ˆ!ß 3161 ‰ .
"Î#
#
(2x) œ È #x
x 1
#
#
#
dy
x
É1 x#x 1
Ê Š dy
dx ‹ œ x# 1 Ê Ê1 Š dx ‹ œ
È2
È2
1
È #
È #
'
'
É 2x
21yÊ1 Š dy
dx ‹ dx œ 0 21 x 1
x# 1 dx œ 0 21 2x 1 dx ;
#
#
#
2
u œ È2x
21 ' È #
21 "
Ä È
u 1 du œ È
’ 2 ŠuÈu# 1 ln Šu Èu# 1‹‹“ œ È12 ’2È5 ln Š2 È5‹“
—
0
2
2
!
du œ È2 dx
81. (a) tan " œ mL Ê tan " œ f w (x! ) where f(x) œ È4px ;
f w (x) œ "# (4px)"Î# (4p) œ È2p
Ê f w (x! ) œ È2p
4px
4px
!
2p
œ 2p
y! Ê tan " œ y! .
0
y!
(b) tan 9 œ mFP œ xy!! p œ x! p
Š x ! p c y2p ‹
y
!
9 tan "
(c) tan ! œ 1tan
tan 9 tan " œ
!
1 b Š x ! p ‹ Š y2p ‹
y
!
!
#
y# 2p(x p)
2p
! 2p
! p)
œ y!! (x! p ! 2p) œ 4px!y! (x2px
œ 2p(x
y! (x! p) œ y!
! p)
11.7 CONICS IN POLAR COORDINATES
#
y#
1. 16x# 25y# œ 400 Ê #x5 16
œ 1 Ê c œ È a# b #
œ È25 16 œ 3 Ê e œ c œ 3 ; F a „ 3ß 0b ;
a
5
directrices are x œ 0 „ ae œ „ ˆ 53 ‰ œ „ 25
3
5
#
x#
2. 7x# 16y# œ 112 Ê 16
y7 œ 1 Ê c œ Èa# b#
œ È16 7 œ 3 Ê e œ c œ 3 ; F a „ 3ß 0b ;
a
4
directrices are x œ 0 „ ae œ „ ˆ 43 ‰ œ „ 16
3
4
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
690
Chapter 11 Parametric Equations and Polar Coordinates
3. 2x# y# œ 2 Ê x# y2 œ 1 Ê c œ Èa# b#
œ È2 1 œ 1 Ê e œ c œ 1 ; F a0ß „ 1b ;
#
È2
a
directrices are y œ 0 „ ae œ „
È2
Š È12 ‹
œ „2
4. 2x# y# œ 4 Ê x# y4 œ 1 Ê c œ Èa# b#
#
#
œ È4 2 œ È2 Ê e œ ca œ
È2
È 2‹ ;
2 ; F Š0ß „
directrices are y œ 0 „ ae œ „ È22 œ „ 2È2
Š ‹
2
#
#
5. 3x# 2y# œ 6 Ê x# y3 œ 1 Ê c œ Èa# b#
œ È3 2 œ 1 Ê e œ c œ 1 ; F a0ß „ 1b ;
È3
a
directrices are y œ 0 „ ae œ
„
È3
Š È13 ‹
œ „3
x
6. 9x# 10y# œ 90 Ê 10
y9 œ 1 Ê c œ Èa# b#
œ È10 9 œ 1 Ê e œ c œ 1 ; F a „ 1ß 0b ;
#
#
a
directrices are x œ 0 „ ae œ „
È10
È10
Š È110 ‹
œ „ 10
7. 6x# 9y# œ 54 Ê x9 y6 œ 1 Ê c œ Èa# b#
#
#
œ È9 6 œ È3 Ê e œ ca œ
È3
È3ß 0‹ ;
3 ; FŠ „
directrices are x œ 0 „ ae œ „ È33 œ „ 3È3
Š ‹
3
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 11.7 Conics in Polar Coordinates
691
y#
x#
8. 169x# 25y# œ 4225 Ê 25
169
œ 1 Ê c œ Èa# b#
œ È169 25 œ 12 Ê e œ c œ 12 ; F a0ß „ 12b ;
a
13
169
directrices are y œ 0 „ ae œ „ ˆ13
12 ‰ œ „ 12
13
#
#
y
3
9. Foci: a0ß „ 3b , e œ 0.5 Ê c œ 3 and a œ ce œ 0.5
œ 6 Ê b# œ 36 9 œ 27 Ê #x7 36
œ1
#
#
y
x
10. Foci: a „ 8ß 0b , e œ 0.2 Ê c œ 8 and a œ ce œ 0.8# œ 40 Ê b# œ 1600 64 œ 1536 Ê 1600
1536
œ1
#
#
y
x
11. Vertices: a0ß „ 70b , e œ 0.1 Ê a œ 70 and c œ ae œ 70(0.1) œ 7 Ê b# œ 4900 49 œ 4851 Ê 4851
4900
œ1
#
#
y
x
12. Vertices: a „ 10ß 0b , e œ 0.24 Ê a œ 10 and c œ ae œ 10(0.24) œ 2.4 Ê b# œ 100 5.76 œ 94.24 Ê 100
94.24
œ1
9
13. Focus: ŠÈ5ß !‹ , Directrix: x œ È95 Ê c œ ae œ È5 and ea œ È95 Ê ae
e# œ È 5 Ê
Ê eœ
È5
3 .
Then PF œ
È5
3 PD
#
Ê ÊŠx È5‹ (y 0)# œ
È5
9
3 ¹x È5 ¹
È5
9
e# œ È 5
Ê e# œ 95
#
Ê Šx È5‹ y# œ 59 Šx È95 ‹
#
y
18
4 #
x
#
Ê x# 2È5 x 5 y# œ 59 Šx# È
x 81
5 ‹ Ê 9 x y œ 4 Ê 9 4 œ 1
5
#
#
a
16
ae
16
4
16
3
#
14. Focus: (%ß 0), Directrix: x œ 16
3 Ê c œ ae œ 4 and e œ 3 Ê e# œ 3 Ê e# œ 3 Ê e œ 4 Ê e œ
È3
È(x 4)# (y 0)# œ È3 ¸x 16 ¸ Ê (x 4)# y# œ 3 ˆx 16 ‰#
# PD Ê
#
3
4
3
y#
" #
3 ˆ #
32
256 ‰
16
x#
#
œ 4 x 3 x 9 Ê 4 x y œ 3 Ê ˆ 64 ‰ ˆ 16 ‰ œ 1
3
3
PF œ
È3
# . Then
Ê x# 8x 16 y#
"
4
1
#
15. Focus: (%ß 0), Directrix: x œ 16 Ê c œ ae œ 4 and ae œ 16 Ê ae
e# œ 16 Ê e# œ 16 Ê e œ 4 Ê e œ # . Then
PF œ 1 PD Ê È(x 4)# (y 0)# œ 1 kx 16k Ê (x 4)# y# œ 1 (x 16)# Ê x# 8x 16 y#
#
#
4
#
#
y
x
œ 14 ax# 32x 256b Ê 43 x# y# œ 48 Ê 64
48
œ1
È2
"
#
È
È
16. Focus: ŠÈ2ß !‹ , Directrix: x œ 2È2 Ê c œ ae œ È2 and ae œ 2È2 Ê ae
e# œ 2 2 Ê e# œ 2 2 Ê e œ #
#
#
Ê e œ È12 . Then PF œ È12 PD Ê ÊŠx È2‹ (y 0)# œ È12 ¹x 2È2¹ Ê Šx È2‹ y#
#
œ "# Šx 2È2‹ Ê x# 2È2 x 2 y# œ "# Šx# 4È2 x 8‹ Ê "# x# y# œ 2 Ê x4 y# œ 1
#
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
#
692
Chapter 11 Parametric Equations and Polar Coordinates
17. x# y# œ 1 Ê c œ Èa# b# œ È1 1 œ È2 Ê e œ ca
œ
È2
È2 ; asymptotes are y œ
1 œ
directrices are x œ 0 „ ae œ „ È"
„ x; F Š „ È2 ß !‹ ;
2
#
x#
18. 9x# 16y# œ 144 Ê 16
y9 œ 1 Ê c œ Èa# b#
œ È16 9 œ 5 Ê e œ c œ 5 ; asymptotes are
a
4
y œ „ 34 x; F a „ 5ß !b ; directrices are x œ 0 „ ae
œ „ "6
5
#
#
19. y# x# œ 8 Ê y8 x8 œ 1 Ê c œ Èa# b#
œ È8 8 œ 4 Ê e œ c œ 4 œ È2 ; asymptotes are
È8
a
y œ „ x; F a0ß „ 4b ; directrices are y œ 0 „ ae
È
œ „ È82 œ „ 2
20. y# x# œ 4 Ê y4 x4 œ 1 Ê c œ Èa# b#
#
#
È
œ È4 4 œ 2È2 Ê e œ ca œ 2 2 2 œ È2 ; asymptotes
are y œ „ x; F Š0ß „ 2È2‹ ; directrices are y œ 0 „ ae
œ „ È22 œ „ È2
21. 8x# 2y# œ 16 Ê x2 y8 œ 1 Ê c œ Èa# b#
#
#
È
œ È2 8 œ È10 Ê e œ ca œ È10
œ È5 ; asymptotes
2
are y œ „ 2x; F Š „ È10ß !‹ ; directrices are x œ 0 „ ae
È
œ „ È25 œ „ È210
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 11.7 Conics in Polar Coordinates
693
22. y# 3x# œ 3 Ê y3 x# œ 1 Ê c œ Èa# b#
œ È3 1 œ 2 Ê e œ c œ 2 ; asymptotes are
#
È3
a
y œ „ È3 x; F a0ß „ 2b ; directrices are y œ 0 „ ae
œ „
È3
Š È23 ‹
œ „ 3#
23. 8y# 2x# œ 16 Ê y2 x8 œ 1 Ê c œ Èa# b#
#
#
È
œ È2 8 œ È10 Ê e œ ca œ È10
œ È5 ; asymptotes
2
are y œ „ x# ; F Š0ß „ È10‹ ; directrices are y œ 0 „ ae
È
œ „ È25 œ „ È210
y
x
24. 64x# 36y# œ 2304 Ê 36
64
œ 1 Ê c œ È a# b #
œ È36 64 œ 10 Ê e œ c œ 10 œ 5 ; asymptotes are
#
#
a
6
3
y œ „ 43 x; F a „ 10ß !b ; directrices are x œ 0 „ ae
œ „ ˆ 65 ‰ œ „ 18
5
3
#
25. Vertices a!ß „ 1b and e œ 3 Ê a œ 1 and e œ ca œ 3 Ê c œ 3a œ 3 Ê b# œ c# a# œ 9 1 œ 8 Ê y# x8 œ 1
#
#
26. Vertices a „ 2ß !b and e œ 2 Ê a œ 2 and e œ ca œ 2 Ê c œ 2a œ 4 Ê b# œ c# a# œ 16 4 œ 12 Ê x4 1y# œ 1
#
27. Foci a „ 3ß !b and e œ 3 Ê c œ 3 and e œ ca œ 3 Ê c œ 3a Ê a œ 1 Ê b# œ c# a# œ 9 1 œ 8 Ê x# y8 œ 1
28. Foci a!ß „ 5b and e œ 1.25 Ê c œ 5 and e œ ca œ 1.25 œ 54 Ê c œ 54 a Ê 5 œ 54 a Ê a œ 4 Ê b# œ c# a#
#
#
y
œ 25 16 œ 9 Ê 16
x9 œ 1
2(1)
2
29. e œ 1, x œ 2 Ê k œ 2 Ê r œ 1 (1)
cos ) œ 1cos )
2(1)
2
30. e œ 1, y œ 2 Ê k œ 2 Ê r œ 1 (1)
sin ) œ 1sin )
30
31. e œ 5, y œ 6 Ê k œ 6 Ê r œ 1 6(5)
5 sin ) œ 15 sin )
8
32. e œ 2, x œ 4 Ê k œ 4 Ê r œ 1 4(2)
2 cos ) œ 12 cos )
ˆ " ‰ (1)
1
33. e œ "# , x œ 1 Ê k œ 1 Ê r œ 1 ˆ#" ‰ cos ) œ 2cos
)
#
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
694
Chapter 11 Parametric Equations and Polar Coordinates
ˆ " ‰ (2)
2
34. e œ "4 , x œ 2 Ê k œ 2 Ê r œ 1 ˆ4" ‰ cos ) œ 4cos
)
4
ˆ " ‰ (10)
35. e œ 5" , y œ 10 Ê k œ 10 Ê r œ 1 5ˆ " ‰ sin ) œ 510
sin )
5
ˆ " ‰ (6)
6
36. e œ "3 , y œ 6 Ê k œ 6 Ê r œ 1 ˆ3 " ‰ sin ) œ 3sin
)
3
37. r œ 1 "cos ) Ê e œ 1, k œ 1 Ê x œ 1
38. r œ 2 6cos ) œ 1 ˆ "3‰ cos ) Ê e œ "# , k œ 6 Ê x œ 6;
#
#
a a1 e b œ ke Ê a ’1 ˆ "# ‰ “ œ 3 Ê 34 a œ 3
#
Ê a œ 4 Ê ea œ 2
ˆ 25 ‰
ˆ5‰
10
#
10
#
39. r œ 10 25
5 cos ) Ê r œ 1 ˆ 5 ‰ cos ) œ 1 ˆ " ‰ cos )
Ê e œ "# , k œ 5 Ê x œ 5; a a1 e# b œ ke
#
5
Ê a ’1 ˆ "# ‰ “ œ 5# Ê 34 a œ 5# Ê a œ 10
3 Ê ea œ 3
2
40. r œ 224cos ) Ê r œ 1cos
) Ê e œ 1, k œ 2 Ê x œ 2
ˆ 400 ‰
25
16
41. r œ 16 400
8 sin ) Ê r œ 1 ˆ 8 ‰ sin ) Ê r œ 1 ˆ " ‰ sin )
16
e œ "# , k œ 50 Ê y œ 50; a a1 e# b œ ke
#
#
Ê a ’1 ˆ "# ‰ “ œ 25 Ê 34 a œ 25 Ê a œ 100
3
Ê ea œ 50
3
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 11.7 Conics in Polar Coordinates
42. r œ 3 312sin ) Ê r œ 1 4sin ) Ê e œ 1,
43. r œ 2 28sin ) Ê r œ 1 4sin ) Ê e œ 1,
kœ4 Ê yœ4
k œ 4 Ê y œ 4
44. r œ 2 4sin ) Ê r œ 1 ˆ "2‰ sin ) Ê e œ #" , k œ 4
#
#
Ê y œ 4; a a1 e# b œ ke Ê a ’1 ˆ "# ‰ “ œ 2
Ê 34 a œ 2 Ê a œ 83 Ê ea œ 43
45. r cos ˆ) 14 ‰ œ È2 Ê r ˆcos ) cos 14 sin ) sin 14 ‰
œ È2 Ê " r cos ) " r sin ) œ È2 Ê " x " y
È2
È2
È2
È2
œ È2 Ê x y œ 2 Ê y œ 2 x
46. r cos ˆ) 341 ‰ œ 1 Ê r ˆcos ) cos 341 sin ) sin 341 ‰ œ 1
È
È
Ê 22 r cos ) 22 r sin ) œ 1 Ê x y œ È2
Ê y œ x È 2
47. r cos ˆ) 231 ‰ œ 3 Ê r ˆcos ) cos 231 sin ) sin 231 ‰ œ 3
È
È
Ê 12 r cos ) 23 r sin ) œ 3 Ê "# x #3 y œ 3
Ê x È 3 y œ 6 Ê y œ
È3
È
3 x2 3
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
695
696
Chapter 11 Parametric Equations and Polar Coordinates
48. r cos ˆ) 13 ‰ œ 2 Ê r ˆcos ) cos 13 sin ) sin 13 ‰ œ 2
È
È
Ê 12 r cos ) 23 r sin ) œ 2 Ê "# x #3 y œ 2
Ê x È3 y œ 4 Ê y œ
È3
4È 3
3 x 3
È
È
49. È2 x È2 y œ 6 Ê È2 r cos ) È2 r sin ) œ 6 Ê r Š #2 cos ) #2 sin )‹ œ 3 Ê r ˆcos 14 cos ) sin 14 sin )‰
œ 3 Ê r cos ˆ) 14 ‰ œ 3
È
50. È3 x y œ 1 Ê È3 r cos ) r sin ) œ 1 Ê r Š #3 cos ) 1# sin )‹ œ #" Ê r ˆcos 16 cos ) sin 16 sin )‰
œ "# Ê r cos ˆ) 16 ‰ œ "#
51. y œ 5 Ê r sin ) œ 5 Ê r sin ) œ 5 Ê r sin ()) œ 5 Ê r cos ˆ 1# ())‰ œ 5 Ê r cos ˆ) 1# ‰ œ 5
52. x œ 4 Ê r cos ) œ 4 Ê r cos ) œ 4 Ê r cos () 1) œ 4
53.
54.
55.
56.
57. (x 6)# y# œ 36 Ê C œ (6ß 0), a œ 6
Ê r œ 12 cos ) is the polar equation
58. (x 2)# y# œ 4 Ê C œ (2ß 0), a œ 2
Ê r œ 4 cos ) is the polar equation
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 11.7 Conics in Polar Coordinates
59. x# (y 5)# œ 25 Ê C œ (!ß 5), a œ 5
Ê r œ 10 sin ) is the polar equation
60. x# (y 7)# œ 49 Ê C œ (!ß 7), a œ 7
Ê r œ 14 sin ) is the polar equation
61. x# 2x y# œ 0 Ê (x 1)# y# œ 1
Ê C œ (1ß 0), a œ 1 Ê r œ 2 cos ) is
the polar equation
62. x# 16x y# œ 0 Ê (x 8)# y# œ 64
Ê C œ (8ß 0), a œ 8 Ê r œ 16 cos ) is the
polar equation
#
63. x# y# y œ 0 Ê x# ˆy "# ‰ œ 4"
Ê C œ ˆ!ß "# ‰ , a œ "# Ê r œ sin ) is the
#
64. x# y# 34 y œ 0 Ê x# ˆy 32 ‰ œ 49
Ê C œ ˆ0ß 23 ‰ , a œ 23 Ê r œ 43 sin ) is the
polar equation
65.
polar equation
66.
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697
698
Chapter 11 Parametric Equations and Polar Coordinates
67.
68.
69.
70.
71.
72.
73.
74.
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Chapter 11 Practice Exercises
75. (a) Perihelion œ a ae œ a(1 e), Aphelion œ ea a œ a(1 e)
(b)
Planet
Perihelion
Aphelion
Mercury
0.3075 AU
0.4667 AU
Venus
0.7184 AU
0.7282 AU
Earth
0.9833 AU
1.0167 AU
Mars
1.3817 AU
1.6663 AU
Jupiter
4.9512 AU
5.4548 AU
Saturn
9.0210 AU
10.0570 AU
Uranus
18.2977 AU
20.0623 AU
Neptune
29.8135 AU
30.3065 AU
#
a1 0.2056 b
0.3707
76. Mercury: r œ (0.3871)
œ 1 0.2056
1 0.2056 cos )
cos )
#
a1 0.0068 b
0.7233
Venus: r œ (0.7233)
œ 1 0.0068
1 0.0068 cos )
cos )
#
0.0167 b
0.9997
Earth: r œ 11a10.0167
cos ) œ 1 0.0617 cos )
#
a1 0.0934 b
1.511
Mars: r œ (1.524)
œ 1 0.0934
1 0.0934 cos )
cos )
#
a1 0.0484 b
5.191
Jupiter: r œ (5.203)
œ 1 0.0484
1 0.0484 cos )
cos )
#
a1 0.0543 b
9.511
Saturn: r œ (9.539)
œ 1 0.0543
1 0.0543 cos )
cos )
#
a1 0.0460 b
19.14
Uranus: r œ (19.18)
œ 1 0.0460
1 0.0460 cos )
cos )
#
a1 0.0082 b
30.06
Neptune: r œ (30.06)
œ 1 0.0082
1 0.0082 cos )
cos )
CHAPTER 11 PRACTICE EXERCISES
1. x œ #t and y œ t 1 Ê 2x œ t Ê y œ 2x 1
2. x œ Èt and y œ 1 Èt Ê y œ 1 x
3. x œ "# tan t and y œ "# sec t Ê x# œ 4" tan# t
4. x œ 2 cos t and y œ 2 sin t Ê x# œ 4 cos# t and
and y# œ "4 sec# t Ê 4x# œ tan# t and
4y# œ sec# t Ê 4x# 1 œ 4y# Ê 4y# 4x# œ 1
y# œ 4 sin# t Ê x# y# œ 4
699
700
Chapter 11 Parametric Equations and Polar Coordinates
5. x œ cos t and y œ cos# t Ê y œ (x)# œ x#
6. x œ 4 cos t and y œ 9 sin t Ê x# œ 6 cos# t and
#
#
y
x
y# œ 81 sin# t Ê 16
81
œ1
#
#
y
7. 16x# 9y# œ 144 Ê x9 16
œ 1 Ê a œ 3 and b œ 4 Ê x œ 3 cos t and y œ 4 sin t, 0 Ÿ t Ÿ 21
8. x# y# œ 4 Ê x œ 2 cos t and y œ 2 sin t, 0 Ÿ t Ÿ 61
"
dy/dt
dy
tan t
#
9. x œ "# tan t, y œ "# sec t Ê dy
œ sec
"
dx œ dx/dt œ
t œ sin t Ê dx ¹
sec# t
sec t tan t
#
Ê x œ "# tan 13 œ
È3
"
1
# and y œ # sec 3 œ 1
tœ1Î3
œ sin 13 œ
È3
1
# ;tœ 3
È3
dyw /dt
" d# y
cos t
3
# x 4 ; dx# œ dx/dt œ "# sec# t œ 2 cos t
Ê yœ
#
Ê ddxy# ¹
tœ1Î3
œ 2 cos3 ˆ 13 ‰ œ "4
dy/dt
10. x œ " t"# , y œ " 3t Ê dy
dx œ dx/dt œ
Š t3# ‹
Š t2$ ‹
#
œ 32 t Ê dy
dx ¹
ˆ 3# ‰
w
/dt
y œ 1 #3 œ #" Ê y œ 3x "43 ; ddxy# œ dy
dx/dt œ
Èx
Š t2$ ‹
Èx
tœ2
œ 3# (2) œ 3; t œ 2 Ê x œ 1 #"# œ 54 and
#
œ 43 t$ Ê ddxy# ¹
3
tœ2
œ 43 (2)$ œ 6
3Î2
11. (a) x œ 4t2 , y œ t3 1 Ê t œ „ 2 Ê y œ Š „ 2 ‹ 1 œ „ x 8 1
(b) x œ cos t, y œ tan t Ê sec t œ 1x Ê tan2 t 1 œ sec2 t Ê y2 œ x12 1 œ 1 x2x Ê y œ „
2
È 1 x2
x
12. (a) The line through a1, 2b with slope 3 is y œ 3x 5 Ê x œ t, y œ 3t 5, _ t _
(b) ax 1b2 ay 2b2 œ 9 Ê x 1 œ 3 cos t, y 2 œ 3 sin t Ê x œ 1 3 cos t, y œ 2 3 sin t, 0 Ÿ t Ÿ 21
(c) y œ 4x2 x Ê x œ t, y œ 4t2 t, _ t _
2
2
(d) 9x2 4y2 œ 36 Ê x4 y9 œ 1 Ê x œ 2 cos t, y œ 3 sin t, 0 Ÿ t Ÿ 21
$Î#
13. y œ x"Î# x 3
" "Î#
" ˆ"
' É1 4" ˆ x" 2 x‰ dx
‰
Ê dy
#" x"Î# Ê Š dy
dx œ # x
dx ‹ œ 4 x 2 x Ê L œ 1
#
4
Ê L œ '1 É 4" ˆ x" 2 x‰ dx œ '1 É 4" ax"Î# x"Î# b dx œ '1 #" ˆx"Î# x"Î# ‰ dx œ #" 2x"Î# 23 x$Î# ‘ "
4
4
%
4
#
‰ 10
œ "# ˆ4 23 † 8‰ ˆ2 32 ‰‘ œ "# ˆ2 14
3 œ 3
#
#Î$
2 "Î$
4x
14. x œ y#Î$ Ê dx
Ê Š dx
dy œ 3 x
dy ‹ œ
9
œ '1
' É1 9x4#Î$ dy
Ê L œ '1 Ê1 Š dx
dy ‹ dy œ 1
8
#
8
'18 È9x#Î$ 4 ˆx"Î$ ‰ dx; u œ 9x#Î$ 4 Ê du œ 6y"Î$ dy; x œ 1 Ê u œ 13,
40
" '
" 2 $Î# ‘ %!
x œ 8 Ê u œ 40d Ä L œ 18
u"Î# du œ 18
œ #"7 40$Î# 13$Î# ‘ ¸ 7.634
3 u
"$
13
8È
9x#Î$ 4
dx œ 3"
3x"Î$
#
" "Î&
" ˆ #Î&
5 'Î&
15. y œ 12
x 58 x%Î& Ê dy
"# x"Î& Ê Š dy
2 x#Î& ‰
dx œ # x
dx ‹ œ 4 x
Ê L œ '1 É1 "4 ax#Î& 2 x#Î& b dx Ê L œ '1 É 4" ax#Î& 2 x#Î& b dx œ ' É "4 ax"Î& x"Î& b dx
32
32
32
1
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
#
Chapter 11 Practice Exercises
œ '1
$#
" ˆ "Î&
75 ‰
x"Î& ‰ dx œ "# 56 x'Î& 54 x%Î& ‘ " œ "# ˆ 56 † 2' 54 † 2% ‰ ˆ 56 54 ‰‘ œ "# ˆ 315
# x
6 4
"
285
œ 48
(1260 450) œ 1710
48 œ 8
32
" #
"
" %
"
"
" %
"
"
dx
'
16. x œ 1"# y$ "y Ê dx
dy œ 4 y y# Ê Š dy ‹ œ 16 y # y% Ê L œ 1 Ê1 Š 16 y # y% ‹ dy
#
2
" %
œ '1 É 16
y #" y"% dy œ '1 ÊŠ 4" y# y"# ‹ dy œ '1 Š 4" y# y"# ‹ dy œ ’ 1"# y$ y" “
2
#
2
2
#
"
8
œ ˆ 12
"# ‰ ˆ 1"# 1‰ œ 17# "# œ 13
12
#
#
dy
ˆ dx ‰ Š dy
17. dx
dt œ 5 sin t 5 sin 5t and dt œ 5 cos t 5 cos 5t Ê Ê dt
dt ‹
œ Éa5 sin t 5 sin 5tb# a5 cos t 5 cos 5tb#
œ 5Èsin# 5t #sin t sin 5t sin# t cos# t #cos t cos 5t cos# 5t œ &È# #asin t sin 5t cos t cos 5 tb
œ 5È#a" cos %tb œ 5É%ˆ "# ‰a" cos %tb œ "!Èsin# #t œ "!lsin #tl œ "!sin #t (since ! Ÿ t Ÿ 1# )
Ê Length œ '!
1 Î2
1Î#
"!sin #t dt œ c5 cos #td !
œ a&ba"b a&ba"b œ "!
#
#
dy
2
2
ˆ dx ‰ Š dy
Éa3t2 12tb# a3t2 12tb# œ È288t# "8t4
18. dx
dt œ 3t 12t and dt œ 3t 12t Ê Ê dt
dt ‹ œ
œ 3È2 ktkÈ16 t2 Ê Length œ '! 3È2 ktkÈ16 t2 dt œ 3È2'! t È16 t2 dt; ’u œ 16 t2 Ê du œ 2t dt
"
"
Ê "# du œ t dt; t œ 0 Ê u œ 16; t œ 1 Ê u œ 17“; 3 2 2 '16 Èu du œ 3 2 2 23 u3/2 ‘16 œ 3 2 2 Š 23 a17b3/2 23 a16b3/2 ‹
"7
È
È
È
17
È
œ 3 2 2 † 23 Ša17b3/2 64‹ œ È2Ša17b3/2 64‹ ¸ 8.617.
#
#
dy
ˆ dx ‰ Š dy
Éa$ sin )b# a$ cos )b# œ È$asin# ) cos# )b œ $
19. dx
d) œ $ sin ) and d) œ $ cos ) Ê Ê d)
d) ‹ œ
Ê Length œ '!
$1Î2
20. x œ t
œ
#
'
$ d) œ $'!
$1Î2
d) œ $ˆ $#1 !‰ œ *#1
$
dy
#
and y œ t3 t, È3 Ÿ t Ÿ È3 Ê dx
dt œ 2t and dt œ t " Ê Length œ
È3
È3
Èt% #t# " dt œ
'
È3
È 3
Èt% 2t# " dt œ
'
È3
È 3
Éat# "b# dt œ
'
È3
È 3
Éa2tb# at# "b# dt
È
'È33 at# "b dt œ ’ t3 t“ È3
3
È
3
œ 4È 3
È5
dy
'
21. x œ t# and y œ 2t, 0 Ÿ t Ÿ È5 Ê dx
dt œ t and dt œ 2 Ê Surface Area œ 0
#
*
21(2t)Èt# 4 dt œ '4 21u"Î# du
9
œ 21 23 u$Î# ‘ % œ 7631 , where u œ t# 4 Ê du œ 2t dt; t œ 0 Ê u œ 4, t œ È5 Ê u œ 9
dy
"
2
22. x œ t# 2t" and y œ 4Èt , È" Ÿ t Ÿ 1 Ê dx
dt œ 2t 2t# and dt œ Èt
2
Ê Surface Area œ '1ÎÈ2 21 ˆt# #"t ‰ ʈ2t 2t"# ‰ Š È2 t ‹ dt œ 21 '1ÎÈ2 ˆt# #"t ‰ Ɉ2t #"t# ‰ dt
1
#
#
1
#
œ 21 '1ÎÈ2 ˆt# 2t" ‰ ˆ2t 2t"# ‰ dt œ 21 '1ÎÈ2 ˆ2t$ 3# "4 t$ ‰ dt œ 21 "2 t% 3# t 8" t# ‘ "ÎÈ#
1
1
È
œ 21 Š2 3 4 2 ‹
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
"
701
702
Chapter 11 Parametric Equations and Polar Coordinates
23. r cos ˆ) 13 ‰ œ 2È3 Ê r ˆcos ) cos 13 sin ) sin 13 ‰
È
œ 2È3 Ê "# r cos ) #3 r sin ) œ 2È3
Ê r cos ) È3 r sin ) œ 4È3 Ê x È3 y œ 4È3
Ê yœ
È3
3 x4
24. r cos ˆ) 341 ‰ œ
œ
È2
#
È
È2
#
Ê r ˆcos ) cos 341 sin ) sin 341 ‰
È
Ê #2 r cos ) #2 r sin ) œ
Ê yœx1
È2
#
Ê x y œ 1
25. r œ 2 sec ) Ê r œ cos2 ) Ê r cos ) œ 2 Ê x œ 2
26. r œ È2 sec ) Ê r cos ) œ È2 Ê x œ È2
27. r œ 3# csc ) Ê r sin ) œ 3# Ê y œ 3#
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Chapter 11 Practice Exercises
28. r œ 3È3 csc ) Ê r sin ) œ 3È3 Ê y œ 3È3
29. r œ 4 sin ) Ê r# œ 4r sin ) Ê x# y# 4y œ 0
Ê x# (y 2)# œ 4; circle with center (!ß 2) and
radius 2.
30. r œ 3È3 sin ) Ê r# œ 3È3 r sin )
#
È
Ê x# y# 3È3 y œ 0 Ê x# Šy 3 # 3 ‹ œ 27
4 ;
È
È
circle with center Š!ß 3 # 3 ‹ and radius 3 # 3
31. r œ 2È2 cos ) Ê r# œ 2È2 r cos )
#
Ê x# y# 2È2 x œ 0 Ê Šx È2‹ y# œ 2;
circle with center ŠÈ2ß 0‹ and radius È2
32. r œ 6 cos ) Ê r# œ 6r cos ) Ê x# y# 6x œ 0
Ê (x 3)# y# œ 9; circle with center (3ß 0) and
radius 3
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
703
704
Chapter 11 Parametric Equations and Polar Coordinates
#
5‰
ˆ
33. x# y# 5y œ 0 Ê x# ˆy #5 ‰ œ 25
4 Ê C œ !ß #
and a œ 5# ; r# 5r sin ) œ 0 Ê r œ 5 sin )
34. x# y# 2y œ 0 Ê x# (y 1)# œ 1 Ê C œ (!ß 1) and
a œ 1; r# 2r sin ) œ 0 Ê r œ 2 sin )
#
35. x# y# 3x œ 0 Ê ˆx 3# ‰ y# œ 49 Ê C œ ˆ 3# ß !‰
and a œ 3# ; r# 3r cos ) œ 0 Ê r œ 3 cos )
36. x# y# 4x œ 0 Ê (x 2)# y# œ 4 Ê C œ (2ß 0)
and a œ 2; r# 4r cos ) œ 0 Ê r œ 4 cos )
37.
38.
39. d
40.
e
41.
l
42.
f
43. k
44.
h
45.
i
46.
j
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Chapter 11 Practice Exercises
2) ‰
47. A œ 2'0 "# r# d) œ '0 (2 cos ))# d) œ '0 a4 4 cos ) cos# )b d) œ '0 ˆ4 4 cos ) 1 cos
d)
#
1
1
1
1
œ '0 ˆ 9# 4 cos ) cos# 2) ‰ d) œ 92 ) 4 sin ) sin42) ‘ ! œ 9# 1
1
1
48. A œ '0
1Î3
"
#
# asin 3)b d) œ
1Î$
6) ‰
1
'01Î3 ˆ 1 cos
d) œ "4 ) "6 sin 6)‘ ! œ 12
#
49. r œ 1 cos 2) and r œ 1 Ê 1 œ 1 cos 2) Ê 0 œ cos 2) Ê 2) œ 1# Ê ) œ 14 ; therefore
A œ 4'0
1Î4
"
#
#
# c(1 cos 2)) 1 d d) œ 2
'01Î4 a1 2 cos 2) cos# 2) 1b d)
1Î%
œ 2'0 ˆ2 cos 2) "# cos# 4) ‰ d) œ 2 sin 2) "2 ) sin84) ‘ ! œ 2 ˆ1 18 0‰ œ 2 14
1Î4
50. The circle lies interior to the cardioid. Thus,
1Î2
A œ 2 ' 1Î2 "# [2(1 sin ))]# d) 1 (the integral is the area of the cardioid minus the area of the circle)
1Î2
1Î2
œ ' 1Î2 4 a1 2 sin ) sin# )b d) 1 œ ' 1Î2 (6 8 sin ) 2 cos 2)) d) 1 œ c6) 8 cos ) sin 2)d 1Î# 1
1Î#
œ c31 (31)d 1 œ 51
dr
51. r œ 1 cos ) Ê d)
œ sin ); Length œ '0 È(1 cos ))# ( sin ))# d) œ '0 È2 2 cos ) d)
21
21
#1
œ '0 É 4(1 #cos )) d) œ '0 2 sin #) d) œ 4 cos 2) ‘ ! œ (4)(1) (4)(1) œ 8
21
21
#
52. r œ 2 sin ) 2 cos ), 0 Ÿ ) Ÿ 1# Ê ddr) œ 2 cos ) 2 sin ); r# ˆ ddr) ‰ œ (2 sin ) 2 cos ))# (2 cos ) 2 sin ))#
œ 8 asin# ) cos# )b œ 8 Ê L œ '0 È8 d) œ ’2È2 )“
1Î2
1Î#
!
œ 2È2 ˆ 1# ‰ œ 1È2
#
#
#
53. r œ 8 sin$ ˆ 3) ‰ , 0 Ÿ ) Ÿ 14 Ê ddr) œ 8 sin# ˆ 3) ‰ cos ˆ 3) ‰ ; r# ˆ ddr) ‰ œ 8 sin$ ˆ 3) ‰‘ 8 sin# ˆ 3) ‰ cos ˆ 3) ‰‘
œ 64 sin% ˆ 3) ‰ Ê L œ '0 É64 sin% ˆ 3) ‰ d) œ '0
1Î4
1Î4
8 sin# ˆ 3) ‰ d) œ '0 8 ’
1Î4
1cos ˆ 23) ‰
“ d)
#
1Î%
œ '0 4 4 cos ˆ 23) ‰‘ d) œ 4) 6 sin ˆ 23) ‰‘ ! œ 4 ˆ 14 ‰ 6 sin ˆ 16 ‰ 0 œ 1 3
1Î4
#
#
sin 2)
54. r œ È1 cos 2) Ê ddr) œ "# (1 cos 2))"Î# (2 sin 2)) œ È1 sincos2)2) Ê ˆ ddr) ‰ œ 1 cos 2)
#
#
#
(1 cos 2)) sin 2)
sin 2)
cos 2) sin 2)
Ê r# ˆ ddr) ‰ œ 1 cos 2) 1 œ 1 2 cos 2)1cos
cos 2) œ
1 cos 2)
2)
#
#
#
1Î2
cos 2)
' 1Î2 È2 d) œ È2 1# ˆ 1# ‰‘ œ È2 1
œ 212cos
2) œ 2 Ê L œ
#
55. x# œ 4y Ê y œ x4 Ê 4p œ 4 Ê p œ 1;
therefore Focus is (0ß 1), Directrix is y œ 1
#
56. x# œ 2y Ê x# œ y Ê 4p œ 2 Ê p œ "# ;
therefore Focus is ˆ!ß "# ‰; Directrix is y œ "#
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
705
706
Chapter 11 Parametric Equations and Polar Coordinates
#
#
57. y# œ 3x Ê x œ y3 Ê 4p œ 3 Ê p œ 43 ;
58. y# œ 38 x Ê x œ ˆy8 ‰ Ê 4p œ 38 Ê p œ 32 ;
3
therefore Focus is ˆ 34 ß 0‰ , Directrix is x œ 34
#
therefore Focus is ˆ 23 ß !‰ , Directrix is x œ 23
#
#
#
y
59. 16x# 7y# œ 112 Ê x7 16
œ1
60. x# 2y# œ 4 Ê x4 y# œ 1 Ê c# œ 4 2 œ 2
Ê c# œ 16 7 œ 9 Ê c œ 3; e œ ca œ 34
Ê c œ È2 ; e œ ca œ
#
61. 3x# y# œ 3 Ê x# y3 œ 1 Ê c# œ 1 3 œ 4
Ê c œ 2; e œ ca œ 21 œ 2; the asymptotes are
#
È2
#
#
62. 5y# 4x# œ 20 Ê y4 x5 œ 1 Ê c# œ 4 5 œ 9
Ê c œ 3, e œ ca œ 3# ; the asymptotes are y œ „ È25 x
y œ „ È3 x
#
63. x# œ 12y Ê 1x# œ y Ê 4p œ 12 Ê p œ 3 Ê focus is (!ß 3), directrix is y œ 3, vertex is (0ß 0); therefore new
vertex is (2ß 3), new focus is (2ß 0), new directrix is y œ 6, and the new equation is (x 2)# œ 12(y 3)
#
y
64. y# œ 10x Ê 10
œ x Ê 4p œ 10 Ê p œ 5# Ê focus is ˆ 5# ß 0‰ , directrix is x œ 5# , vertex is (0ß 0); therefore new
vertex is ˆ "# ß 1‰ , new focus is (2ß 1), new directrix is x œ 3, and the new equation is (y 1)# œ 10 ˆx "# ‰
#
#
65. x9 #y5 œ 1 Ê a œ 5 and b œ 3 Ê c œ È25 9 œ 4 Ê foci are a!ß „ 4b , vertices are a!ß „ 5b , center is
(0ß 0); therefore the new center is ($ß 5), new foci are (3ß 1) and (3ß 9), new vertices are ($ß 10) and
#
#
($ß 0), and the new equation is (x 9 3) (y #55) œ 1
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Chapter 11 Practice Exercises
707
#
#
y
x
66. 169
144
œ 1 Ê a œ 13 and b œ 12 Ê c œ È169 144 œ 5 Ê foci are a „ 5ß 0b , vertices are a „ 13ß 0b , center
is (0ß 0); therefore the new center is (5ß 12), new foci are (10ß 12) and (0ß 12), new vertices are (18ß 12) and
#
#
5)
(8ß 12), and the new equation is (x169
(y 14412) œ 1
#
#
67. y8 x2 œ 1 Ê a œ 2È2 and b œ È2 Ê c œ È8 2 œ È10 Ê foci are Š0ß „ È10‹ , vertices are
Š0ß „ 2È2‹ , center is (0ß 0), and the asymptotes are y œ „ 2x; therefore the new center is Š2ß 2È2‹, new foci are
Š2ß 2È2 „ È10‹ , new vertices are Š2ß 4È2‹ and (#ß 0), the new asymptotes are y œ 2x 4 2È2 and
#
y 2È 2‹
Š
y œ 2x 4 2È2; the new equation is
#
8
#
(x # 2) œ 1
#
y
x
68. 36
64
œ 1 Ê a œ 6 and b œ 8 Ê c œ È36 64 œ 10 Ê foci are a „ 10ß 0b , vertices are a „ 6ß 0b , the center
is (0ß 0) and the asymptotes are y8 œ „ x6 or y œ „ 43 x; therefore the new center is (10ß 3), the new foci are
(20ß 3) and (0ß 3), the new vertices are (16ß 3) and (4ß 3), the new asymptotes are y œ 43 x 31
3 and
#
#
(x 10)
y œ 43 x 49
(y 643) œ 1
3 ; the new equation is
36
#
69. x# 4x 4y# œ 0 Ê x# 4x 4 4y# œ 4 Ê (x 2)# 4y# œ 4 Ê (x 4 2) y# œ 1, a hyperbola; a œ 2 and
b œ 1 Ê c œ È1 4 œ È5 ; the center is (2ß 0), the vertices are (!ß 0) and (4ß 0); the foci are Š2 „ È5 ß 0‹ and
the asymptotes are y œ „ x #2
#
70. 4x# y# 4y œ 8 Ê 4x# y# 4y 4 œ 4 Ê 4x# (y 2)# œ 4 Ê x# (y 4 2) œ 1, a hyperbola; a œ 1 and
b œ 2 Ê c œ È1 4 œ È5 ; the center is (!ß 2), the vertices are (1ß 2) and ("ß 2), the foci are Š „ È5ß 2‹ and
the asymptotes are y œ „ 2x 2
71. y# 2y 16x œ 49 Ê y# 2y 1 œ 16x 48 Ê (y 1)# œ 16(x 3), a parabola; the vertex is ($ß 1);
4p œ 16 Ê p œ 4 Ê the focus is (7ß 1) and the directrix is x œ 1
72. x# 2x 8y œ 17 Ê x# 2x 1 œ 8y 16 Ê (x 1)# œ 8(y 2), a parabola; the vertex is (1ß 2);
4p œ 8 Ê p œ 2 Ê the focus is (1ß 4) and the directrix is y œ 0
73. 9x# 16y# 54x 64y œ 1 Ê 9 ax# 6xb 16 ay# 4yb œ 1 Ê 9 ax# 6x 9b 16 ay# 4y 4b œ 144
#
#
Ê 9(x 3)# 16(y 2)# œ 144 Ê (x 163) (y 9 2) œ 1, an ellipse; the center is (3ß 2); a œ 4 and b œ 3
Ê c œ È16 9 œ È7 ; the foci are Š$ „ È7ß 2‹ ; the vertices are (1ß 2) and (7ß 2)
74. 25x# 9y# 100x 54y œ 44 Ê 25 ax# 4xb 9 ay# 6yb œ 44 Ê 25 ax# 4x 4b 9 ay# 6y 9b œ 225
#
#
Ê (x 2) (y 3) œ 1, an ellipse; the center is (2ß 3); a œ 5 and b œ 3 Ê c œ È25 9 œ 4; the foci are
9
25
(2ß 1) and (2ß 7); the vertices are (2ß 2) and (2ß 8)
75. x# y# 2x 2y œ 0 Ê x# 2x 1 y# 2y 1 œ 2 Ê (x 1)# (y 1)# œ 2, a circle with center (1ß 1) and
radius œ È2
76. x# y# 4x 2y œ 1 Ê x# 4x 4 y# 2y 1 œ 6 Ê (x 2)# (y 1)# œ 6, a circle with center (2ß 1)
and radius œ È6
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Chapter 11 Parametric Equations and Polar Coordinates
77. r œ 1 2cos ) Ê e œ 1 Ê parabola with vertex at (1ß 0)
78. r œ 2 8cos ) Ê r œ 1 ˆ "4‰ cos ) Ê e œ "# Ê ellipse;
#
ke œ 4 Ê "# k œ 4 Ê k œ 8; k œ ae ea Ê 8 œ ˆ a" ‰ "# a
#
ˆ " ‰ ˆ 16
‰ 8
Ê a œ 16
3 Ê ea œ #
3 œ 3 ; therefore the center is
ˆ 83 ß 1‰ ; vertices are ()ß 1) and ˆ 83 ß 0‰
79. r œ 1 26cos ) Ê e œ 2 Ê hyperbola; ke œ 6 Ê 2k œ 6
Ê k œ 3 Ê vertices are (2ß 1) and (6ß 1)
80. r œ 3 12sin ) Ê r œ 1 ˆ "4‰ sin ) Ê e œ "3 ; ke œ 4
3
Ê
"
3 kœ4
#
Ê k œ 12; a a1 e# b œ 4 Ê a ’1 ˆ 3" ‰ “
œ 4 Ê a œ 9# Ê ea œ ˆ "3 ‰ ˆ 9# ‰ œ 3# ; therefore the
center is ˆ 3# ß 3#1 ‰ ; vertices are ˆ3ß 1# ‰ and ˆ6ß 3#1 ‰
81. e œ 2 and r cos ) œ 2 Ê x œ 2 is directrix Ê k œ 2; the conic is a hyperbola; r œ 1 ekecos ) Ê r œ 1 (2)(2)
# cos )
Ê r œ 1 #4cos )
(4)(1)
82. e œ 1 and r cos ) œ 4 Ê x œ 4 is directrix Ê k œ 4; the conic is a parabola; r œ 1 ekecos ) Ê r œ 1 cos )
Ê r œ 1 4cos )
(2) ˆ " ‰
83. e œ "# and r sin ) œ 2 Ê y œ 2 is directrix Ê k œ 2; the conic is an ellipse; r œ 1 ekesin ) Ê r œ 1 ˆ " ‰#sin )
#
Ê r œ 2 2sin )
(6) ˆ " ‰
84. e œ "3 and r sin ) œ 6 Ê y œ 6 is directrix Ê k œ 6; the conic is an ellipse; r œ 1 ekesin ) Ê r œ 1 ˆ " ‰3sin )
3
Ê r œ 3 6sin )
85.
(a) Around the x-axis: 9x# 4y# œ 36 Ê y# œ 9 94 x# Ê y œ „ É9 94 x# and we use the positive root:
#
V œ 2 '0 1 ŠÉ9 94 x# ‹ dx œ 2 '0 1 ˆ9 94 x# ‰ dx œ 21 9x 34 x$ ‘ ! œ 241
2
2
#
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Chapter 11 Additional and Advanced Exercises
709
(b) Around the y-axis: 9x# 4y# œ 36 Ê x# œ 4 49 y# Ê x œ „ É4 49 y# and we use the positive root:
#
4 $‘ $
V œ 2 '0 1 ŠÉ4 49 y# ‹ dy œ 2'0 1 ˆ4 49 y# ‰ dy œ 21 4y 27
y ! œ 161
3
86.
3
9x# 4y# œ 36, x œ 4 Ê y# œ 9x 4 36 Ê y œ #3 Èx# 4 ; V œ '2 1 Š #3 Èx# 4‹ dx œ 941 '2 ax# 4b dx
4
#
#
4
%
$
24 ‰
31
‰ ˆ8
‰‘ œ 941 ˆ 56
œ 941 ’ x3 4x“ œ 941 ˆ 64
3 16 3 8
3 3 œ 4 (32) œ 241
#
87.
(a) r œ 1 ekcos ) Ê r er cos ) œ k Ê Èx# y# ex œ k Ê Èx# y# œ k ex Ê x# y#
œ k# 2kex e# x# Ê x# e# x# y# 2kex k# œ 0 Ê a1 e# b x# y# 2kex k# œ 0
(b) e œ 0 Ê x# y# k# œ 0 Ê x# y# œ k# Ê circle;
0 e 1 Ê e# 1 Ê e# 1 0 Ê B# 4AC œ 0# 4 a1 e# b (1) œ 4 ae# 1b 0 Ê ellipse;
e œ 1 Ê B# 4AC œ 0# 4(0)(1) œ 0 Ê parabola;
e 1 Ê e# 1 Ê B# 4AC œ 0# 4 a1 e# b (1) œ 4e# 4 0 Ê hyperbola
88.
Let (r" ß )" ) be a point on the graph where r" œ a)" . Let (r# ß )# ) be on the graph where r# œ a)# and
)# œ )" 21. Then r" and r# lie on the same ray on consecutive turns of the spiral and the distance between
the two points is r# r" œ a)# a)" œ a()# )" ) œ 21a, which is constant.
CHAPTER 11 ADDITIONAL AND ADVANCED EXERCISES
1. Directrix x œ 3 and focus (4ß 0) Ê vertex is ˆ 7# ß !‰
#
Ê p œ "# Ê the equation is x 7# œ y#
#
2. x# 6x 12y 9 œ 0 Ê x# 6x 9 œ 12y Ê (x123) œ y Ê vertex is (3ß 0) and p œ 3 Ê focus is (3ß 3) and the
directrix is y œ 3
3. x# œ 4y Ê vertex is (!ß 0) and p œ 1 Ê focus is (!ß 1); thus the distance from P(xß y) to the vertex is Èx# y#
and the distance from P to the focus is Èx# (y 1)# Ê Èx# y# œ 2Èx# (y 1)#
Ê x# y# œ 4 cx# (y 1)# d Ê x# y# œ 4x# 4y# 8y 4 Ê 3x# 3y# 8y 4 œ 0, which is a circle
4. Let the segment a b intersect the y-axis in point A and
intersect the x-axis in point B so that PB œ b and PA œ a
(see figure). Draw the horizontal line through P and let it
intersect the y-axis in point C. Let nPBO œ )
Ê nAPC œ ). Then sin ) œ yb and cos ) œ xa
#
#
Ê xa# by# œ cos# ) sin# ) œ 1.
5. Vertices are a!ß „ 2b Ê a œ 2; e œ ca Ê 0.5 œ #c Ê c œ 1 Ê foci are a0ß „ 1b
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710
Chapter 11 Parametric Equations and Polar Coordinates
6. Let the center of the ellipse be (xß 0); directrix x œ 2, focus (4ß 0), and e œ 23 Ê ae c œ 2 Ê ae œ 2 c
a
Ê a œ 32 (2 c). Also c œ ae œ 32 a Ê a œ 32 ˆ2 32 a‰ Ê a œ 34 94 a Ê 95 a œ 34 Ê a œ 12
5 ;x2œ e
28
28
8
#
#
#
‰ ˆ 3# ‰ œ 18
ˆ 28 ‰
Ê x 2 œ ˆ 12
5
5 Ê x œ 5 Ê the center is 5 ß 0 ; x 4 œ c Ê c œ 5 4 œ 5 so that c œ a b
#
#
80
‰ ˆ 58 ‰ œ 25
œ ˆ 12
; therefore the equation is
5
#
#
ˆx 28
‰#
‰#
25 ˆx 28
5
5
ˆy80 ‰ œ 1 or
5y
ˆ 144
‰
144
16 œ 1
25
25
7. Let the center of the hyperbola be (0ß y).
(a) Directrix y œ 1, focus (0ß 7) and e œ 2 Ê c ae œ 6 Ê ae œ c 6 Ê a œ 2c 12. Also c œ ae œ 2a
Ê a œ 2(2a) 12 Ê a œ 4 Ê c œ 8; y (1) œ ae œ #4 œ 2 Ê y œ 1 Ê the center is (0ß 1); c# œ a# b#
#
x
Ê b# œ c# a# œ 64 16 œ 48; therefore the equation is (y161) 48
œ1
#
(b) e œ 5 Ê c ae œ 6 Ê ae œ c 6 Ê a œ 5c 30. Also, c œ ae œ 5a Ê a œ 5(5a) 30 Ê 24a œ 30 Ê a œ 54
a
Ê c œ 25
4 ; y (1) œ e œ
ˆ 54 ‰
"
5 œ 4
Ê y œ 43 Ê the center is ˆ!ß 43 ‰ ; c# œ a# b# Ê b# œ c# a#
25
75
œ 625
16 16 œ # ; therefore the equation is
#
#
ˆy 34 ‰#
16 ˆy 34 ‰
x#
œ
1
or
2x
75
25
ˆ#‰
ˆ 16 ‰
25
75 œ 1
#
#
144
8. The center is (0ß 0) and c œ 2 Ê 4 œ a# b# Ê b# œ 4 a# . The equation is ya# bx# œ 1 Ê 49
a# b# œ 1
144
#
#
#
#
#
#
#
%
%
#
Ê 49
a# a4a# b œ 1 Ê 49 a4 a b 144a œ a a4 a b Ê 196 49a 144a œ 4a a Ê a 197a 196
œ 0 Ê aa# 196b aa# 1b œ 0 Ê a œ 14 or a œ 1; a œ 14 Ê b# œ 4 (14)# 0 which is impossible; a œ 1
#
Ê b# œ 4 1 œ 3; therefore the equation is y# x3 œ 1
#
#
b x"
b x
9. b# x# a# y# œ a# b# Ê dy
dx œ a# y ; at (x" ß y" ) the tangent line is y y" œ Š a# y" ‹ (x x" )
Ê a# yy" b# xx" œ b# x"# a# y"# œ a# b# Ê b# xx" a# yy" a# b# œ 0
#
#
b x"
b x
10. b# x# a# y# œ a# b# Ê dy
dx œ a# y ; at (x" ß y" ) the tangent line is y y" œ Š a# y" ‹ (x x" )
Ê b# xx" a# yy" œ b# x"# a# y"# œ a# b# Ê b# xx" a# yy" a# b# œ 0
11.
12.
13.
14.
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Chapter 11 Additional and Advanced Exercises
711
15. a9x# 4y# 36b a4x# 9y# 16b Ÿ 0
Ê 9x# 4y# 36 Ÿ 0 and 4x# 9y# 16 0
or 9x# 4y# 36 0 and 4x# 9y# 16 Ÿ 0
16. a9x# 4y# 36b a4x# 9y# 16b 0, which is the
complement of the set in Exercise 15
sin t
17. (a) x œ e2t cos t and y œ e2t sin t Ê x# y# œ e4t cos# t e4t sin# t œ e4t . Also yx œ ee2t cos
t œ tan t
" ˆ y ‰
#
#
% tan " ayÎxb
#
#
Ê t œ tan
Ê x y œe
is the Cartesian equation. Since r œ x y# and
x
2t
) œ tan" ˆ yx ‰ , the polar equation is r# œ e4) or r œ e2) for r 0
(b) ds# œ r# d)# dr# ; r œ e2) Ê dr œ 2e2) d)
#
#
Ê ds# œ r# d)# ˆ2e2) d)‰ œ ˆe2) ‰ d)# 4e4) d)#
œ 5e4) d)# Ê ds œ È5 e2) d) Ê L œ '0 È5 e2) d)
21
œ’
È5 e2) #1
È5
41
2 “ ! œ # ae 1b
#
18. r œ 2 sin$ ˆ 3) ‰ Ê dr œ 2 sin# ˆ 3) ‰ cos ˆ 3) ‰ d) Ê ds# œ r# d)# dr# œ 2 sin$ ˆ 3) ‰‘ d)# 2 sin# ˆ 3) ‰ cos ˆ 3) ‰ d)‘
œ 4 sin' ˆ 3) ‰ d)# 4 sin% ˆ 3) ‰ cos# ˆ 3) ‰ d)# œ 4 sin% ˆ 3) ‰‘ sin# ˆ 3) ‰ cos# ˆ 3) ‰‘ d)# œ 4 sin% ˆ 3) ‰ d)#
$1
Ê ds œ 2 sin# ˆ 3) ‰ d). Then L œ '0 2 sin# ˆ 3) ‰ d) œ '0 1 cos ˆ 23) ‰‘ d) œ ) 23 sin ˆ 23) ‰‘ ! œ 31
31
31
19. e œ 2 and r cos ) œ 2 Ê x œ 2 is the directrix Ê k œ 2; the conic is a hyperbola with r œ 1 ekecos )
4
Ê r œ 1 (2)(2)
2 cos ) œ 1 2 cos )
20. e œ 1 and r cos ) œ 4 Ê x œ 4 is the directrix Ê k œ 4; the conic is a parabola with r œ 1 ekecos )
(4)(1)
4
Ê r œ 1
cos ) œ 1 cos )
21. e œ "# and r sin ) œ 2 Ê y œ 2 is the directrix Ê k œ 2; the conic is an ellipse with r œ 1 ekesin )
2 ˆ"‰
Ê r œ 1 ˆ " ‰# sin ) œ 2 2sin )
#
22. e œ "3 and r sin ) œ 6 Ê y œ 6 is the directrix Ê k œ 6; the conic is an ellipse with r œ 1 ekesin )
6 ˆ"‰
Ê r œ 1 ˆ " ‰3 sin ) œ 3 6sin )
3
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#
712
Chapter 11 Parametric Equations and Polar Coordinates
23. Arc PF œ Arc AF since each is the distance rolled;
nPCF œ Arcb PF Ê Arc PF œ b(nPCF); ) œ ArcaAF
Ê Arc AF œ a) Ê a) œ b(nPCF) Ê nPCF œ ˆ ba ‰ );
nOCB œ 1# ) and nOCB œ nPCF nPCE
œ nPCF ˆ 1# !‰ œ ˆ ba ‰ ) ˆ 1# !‰ Ê 1# )
œ ˆ ba ‰ ) ˆ 1# !‰ Ê 1# ) œ ˆ ba ‰ ) 1# !
Ê ! œ 1 ) ˆ ba ‰ ) Ê ! œ 1 ˆ ab b ‰ ).
Now x œ OB BD œ OB EP œ (a b) cos ) b cos ! œ (a b) cos ) b cos ˆ1 ˆ a b b ‰ )‰
œ (a b) cos ) b cos 1 cos ˆˆ a b b ‰ )‰ b sin 1 sin ˆˆ a b b ‰ )‰ œ (a b) cos ) b cos ˆˆ a b b ‰ )‰ and
y œ PD œ CB CE œ (a b) sin ) b sin ! œ (a b) sin ) b sin ˆˆ a b b ‰ )‰
œ (a b) sin ) b sin 1 cos ˆˆ a b b ‰ )‰ b cos 1 sin ˆˆ a b b ‰ )‰ œ (a b) sin ) b sin ˆˆ a b b ‰ )‰ ;
therefore x œ (a b) cos ) b cos ˆˆ a b b ‰ )‰ and y œ (a b) sin ) b sin ˆˆ a b b ‰ )‰
ˆ dx ‰
24. x œ a(t sin t) Ê dx
dt œ a(1 cos t) and let $ œ 1 Ê dm œ dA œ y dx œ y dt dt
œ a(1 cos t) a (1 cos t) dt œ a# (1 cos t)# dt; then A œ '0 a# (1 cos t)# dt
21
œ a# '0 a1 2 cos t cos# tb dt œ a# '0 ˆ1 2 cos t "# "# cos 2t‰ dt œ a# 32 t 2 sin t sin4 2t ‘ !
21
21
#1
œ 31a# ; µ
x = x œ a(t sin t) and µ
y = "# y œ "# a(1 cos t) Ê Mx œ ' µ
y dm œ ' µ
y $ dA
œ '0
21
"
" $
#
#
# a(1 cos t) a (1 cos t) dt œ # a
'021 (1 cos t)$ dt œ a# '021 a1 3 cos t 3 cos# t cos$ tb dt
$
2t
œ a# '0 1 3 cos t #3 3 cos
a1 sin# tb (cos t)‘ dt œ a# ’ 25 t 3 sin t 3 sin4 2t sin t sin3 t “
#
21
$
$
œ 51#a .
$
51 a $
Š # ‹
Therefore y œ MMx œ 31a# œ 65 a.
$
#1
!
Also, My œ ' µ
x dm œ ' µ
x $ dA
œ '0 a(t sin t) a# (1 cos t)# dt œ a$ '0 at 2t cos t t cos# t sin t 2 sin t cos t sin t cos# tb dt
21
21
#
$
œ a$ ’ t2 2 cos t 2t sin t 4" t# 8" cos 2t 4t sin 2t cos t sin# t cos3 t “
# $
#1
!
œ 31# a$ . Thus
x œ My œ 3311aa# œ 1a Ê ˆ1aß 56 a‰ is the center of mass.
M
25.
<# tan <"
" œ <# <" Ê tan " œ tan (<# <" ) œ 1tan
tan <# tan <" ;
the curves will be orthogonal when tan " is undefined, or
when tan <# œ tan"<" Ê g (r)) œ "
r
w
’ f ()) “
w
#
w
w
Ê r œ f ( ) ) g ( ) )
26.
sin% ˆ ) ‰
4
ˆ)‰
r œ sin% ˆ 4) ‰ Ê ddr) œ sin$ ˆ 4) ‰ cos ˆ 4) ‰ Ê tan < œ sin$ ˆ ) ‰ cos
ˆ ) ‰ œ tan 4
4
27.
4
2a sin 3)
"
1
"
1
1
r œ 2a sin 3) Ê ddr) œ 6a cos 3) Ê tan < œ ˆ drr ‰ œ 6a
cos 3) œ 3 tan 3); when ) œ 6 , tan < œ 3 tan # Ê < œ #
d)
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Chapter 11 Additional and Advanced Exercises
28.
713
dr
(b) r) œ 1 Ê r œ )" Ê d)
œ )# Ê tan <k )œ1
(a)
"
œ )) # œ ) Ê
lim tan < œ _
)Ä_
Ê < Ä 1# from the right as the spiral winds in
around the origin.
29.
È
)
sin )
È3 at ) œ 1 ; since the product of
tan <" œ È33cos
œ cot ) is È"3 at ) œ 13 ; tan <# œ cos
) œ tan ) is
3
sin )
these slopes is 1, the tangents are perpendicular
30.
tan < œ ˆ drr ‰ œ a(1asincos) )) is 1 at ) œ 1# Ê < œ 14
d)
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Chapter 11 Parametric Equations and Polar Coordinates
NOTES:
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CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
12.1 THREE-DIMENSIONAL COORDINATE SYSTEMS
1. The line through the point a2ß 3ß 0b parallel to the z-axis
2. The line through the point a1ß 0ß 0b parallel to the y-axis
3. The x-axis
4. The line through the point a1ß 0ß 0b parallel to the z-axis
5. The circle x# y# œ 4 in the xy-plane
6. The circle x# y# œ 4 in the plane z = 2
7. The circle x# z# œ 4 in the xz-plane
8. The circle y# z# œ 1 in the yz-plane
9. The circle y# z# œ 1 in the yz-plane
10. The circle x# z# œ 9 in the plane y œ 4
11. The circle x# y# œ 16 in the xy-plane
12. The circle x# z# œ 3 in the xz-plane
13. The ellipse formed by the intersection of the cylinder x# y# œ 4 and the plane z œ y.
14. The circle formed by the intersection of the sphere x# y# z# œ 4 and the plane y œ x.
15. The parabola y œ x# in the the xy-plane.
16. The parabola z œ y# in the the plane x œ 1.
17. (a) The first quadrant of the xy-plane
(b) The fourth quadrant of the xy-plane
18. (a) The slab bounded by the planes x œ 0 and x œ 1
(b) The square column bounded by the planes x œ 0, x œ 1, y œ 0, y œ 1
(c) The unit cube in the first octant having one vertex at the origin
19. (a) The solid ball of radius 1 centered at the origin
(b) The exterior of the sphere of radius 1 centered at the origin
20. (a) The circumference and interior of the circle x# y# œ 1 in the xy-plane
(b) The circumference and interior of the circle x# y# œ 1 in the plane z œ 3
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716
Chapter 12 Vectors and the Geometry of Space
(c) A solid cylindrical column of radius 1 whose axis is the z-axis
21. (a) The solid enclosed between the sphere of radius 1 and radius 2 centered at the origin
(b) The solid upper hemisphere of radius 1 centered at the origin
22. (a) The line y œ x in the xy-plane
(b) The plane y œ x consisting of all points of the form (xß xß z)
23. (a) The region on or inside the parabola y œ x# in the xy-plane and all points above this region.
(b) The region on or to the left of the parabola x œ y# in the xy-plane and all points above it that are 2 units or less away
from the xy-plane.
24. (a) All the points the lie on the plane z œ 1 y.
(b) All points that lie on the curve z œ y3 in the plane x œ 2.
25. (a) x œ 3
(b) y œ 1
(c) z œ 2
26. (a) x œ 3
(b) y œ 1
(c) z œ 2
27. (a) z œ 1
(b) x œ 3
(c) y œ 1
28. (a) x# y# œ 4, z œ 0
(b) y# z# œ 4, x œ 0
(c) x# z# œ 4, y œ 0
29. (a) x# ay 2b# œ 4, z œ 0
(b) ay 2b# z# œ 4, x œ 0
(c) x# z# œ 4, y œ 2
30. (a) ax 3b# ay 4b# œ 1, z œ 1
(b) ay 4b# az 1b# œ 1, x œ 3
(c) ax 3b# az 1b# œ 1, y œ 4
31. (a) y œ 3, z œ 1
(b) x œ 1, z œ 1
(c) x œ 1, y œ 3
32. Èx# y# z# œ Éx# ay 2b# z# Ê x# y# z# œ x# ay 2b# z# Ê y# œ y# 4y 4 Ê y œ 1
33. x# y# z# œ 25, z œ 3 Ê x# y# œ 16 in the plane z œ 3
34. x# y# az 1b# œ 4 and x# y# az 1b# œ 4 Ê x# y# az 1b# œ x# y# az 1b# Ê z œ 0, x# y# œ 3
35. 0 Ÿ z Ÿ 1
36. 0 Ÿ x Ÿ 2, 0 Ÿ y Ÿ 2, 0 Ÿ z Ÿ 2
37. z Ÿ 0
38. z œ È1 x# y#
39. (a) ax 1b# ay 1b# az 1b# 1
(b) ax 1b# ay 1b# az 1b# 1
40. 1 Ÿ x# y# z# Ÿ 4
41. kP" P# k œ Éa3 1b# a3 1b# a0 1b# œ È9 œ 3
42. kP" P# k œ Éa2 1b# a5 1b# a0 5b# œ È50 œ 5È2
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Section 12.1 Three-Dimensional Coordinate Systems
717
43. kP" P# k œ Éa4 1b# a2 4b# a7 5b# œ È49 œ 7
44. kP" P# k œ Éa2 3b# a3 4b# a4 5b# œ È3
45. kP" P# k œ Éa2 0b# a2 0b# a2 0b# œ È3 † 4 œ 2È3
46. kP" P# k œ Éa0 5b# a0 3b# a0 2b# œ È38
47. center (2ß 0ß 2), radius 2È2
48. center ˆ1ß "# ß 3‰ , radius 5
49. center ŠÈ2ß È2ß È2‹ , radius È2
50. center ˆ!ß "3 ß 3" ‰ , radius 34
51. ax 1b# ay 2b# az 3b# œ 14
52. x# ay 1b# az 5b# œ 4
#
#
53. ax 1b# ˆy 12 ‰ ˆz 23 ‰ œ 16
81
54. x# ay 7b# z# œ 49
55. x# y# z# 4x 4z œ 0 Ê ax# 4x 4b y# az# 4z 4b œ 4 4
#
Ê ax 2b# ay 0b# az 2b# œ ŠÈ8‹ Ê the center is at a2ß 0ß 2b and the radius is È8
56. x# y# z# 6y 8z œ 0 Ê x# ay# 6y 9b az# 8z 16b œ 9 16 Ê ax 0b# ay 3b# az 4b# œ 5#
Ê the center is at a0ß 3ß 4b and the radius is 5
57. 2x# 2y# 2z# x y z œ 9 Ê x# "# x y# "# y z# "# z œ 9#
#
#
#
È
#
" ‰
" ‰
" ‰
3
Ê ˆx# "# x 16
ˆy# "# y 16
ˆz# "# z 16
œ 9# 16
Ê ˆx 4" ‰ ˆy 4" ‰ ˆz 4" ‰ œ Š 5 4 3 ‹
È
Ê the center is at ˆ "4 ß 4" ß 4" ‰ and the radius is 5 4 3
58. 3x# 3y# 3z# 2y 2z œ 9 Ê x# y# 23 y z# 23 z œ 3 Ê x# ˆy# 23 y "9 ‰ ˆz# 23 z "9 ‰ œ 3 29
#
#
È
#
Ê (x 0)# ˆy 3" ‰ ˆz 3" ‰ œ Š 329 ‹ Ê the center is at ˆ0ß 3" ß 3" ‰ and the radius is
59. (a) the distance between axß yß zb and axß 0ß 0b is Èy# z#
(b) the distance between axß yß zb and a0ß yß 0b is Èx# z#
(c) the distance between axß yß zb and a0ß 0ß zb is Èx# y#
60. (a) the distance between axß yß zb and axß yß 0b is z
(b) the distance between axß yß zb and a0ß yß zb is x
(c) the distance between axß yß zb and axß 0ß zb is y
61. kABk œ Éa1 a1bb# a1 2b# a3 1b# œ È4 9 4 œ È17
kBCk œ Éa3 1b# a4 a1bb# a5 3b# œ È4 25 4 œ È33
kCAk œ Éa1 3b# a2 4b# a1 5b# œ È16 4 16 œ È36 œ 6
Thus the perimeter of triangle ABC is È17 È33 6.
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
È29
3
718
Chapter 12 Vectors and the Geometry of Space
62. kPAk œ Éa2 3b# a1 1b# a3 2b# œ È1 4 1 œ È6
kPBk œ Éa4 3b# a3 1b# a1 2b# œ È1 4 1 œ È6
Thus P is equidistant from A and B.
63. Éax xb# ay a1bb# az zb# œ Éax xb# ay 3b# az zb# Ê ay 1b# œ ay 3b# Ê 2y 1 œ 6y 9
Êyœ1
64. Éax 0b# ay 0b# az 2b# œ Éax xb# ay yb# az 0b# Ê x# y2 az 2b# œ z#
#
2
Ê x# y2 4z 4 œ 0 Ê z œ x4 y4 1
65. (a) Since the entire sphere is below the xy-plane, the point on the sphere closest to the xy-plane is the point at the top of
the sphere, which occurs when x œ 0 and y œ 3 Ê 02 a3 3b2 az 5b2 œ 4 Ê z œ 5 „ 2 Ê z œ 3
Ê a0, 3, 3b.
(b) Both the center a0, 3, 5b and the the point a0, 7, 5b lie in the plane z œ 5, so the point on the sphere closest to
a0, 7, 5b should also be in the same plane. In fact it should lie on the line segment between a0, 3, 5b and a0, 7, 5b,
thus the point occurs when x œ 0 and z œ 5 Ê 02 ay 3b2 a5 5b2 œ 4 Ê y œ 3 „ 2 Ê y œ 5
Ê a0, 5, 5b.
66. Éax 0b# ay 0b# az 0b# œ Éax 0b# ay 4b# az 0b# œ Éax 3b# ay 0b# az 0b#
œ Éax 2b# ay 2b# az 3b#
Ê x# y2 z# œ x# y2 8y 16 z# œ x# 6x 9 y2 z# œ x# 4x y2 4y z# 6z 17
Solve: x# y2 z# œ x# y2 8y 16 z# Ê 0 œ 8y 16 Ê y œ 2
Solve: x# y2 z# œ x# 6x 9 y2 z# Ê 0 œ 6x 9 Ê x œ 32
Solve: x# y2 z# œ x# 4x y2 4y z# 6z 17 Ê 0 œ 4x 4y 6z 17 Ê 0 œ 4ˆ 32 ‰ 4a2b 6z 17
Ê z œ 12 Ê ˆ 32 , 2, 12 ‰
12.2 VECTORS
1. (a)
3a3b, 3a2b¡ œ 9, 6¡
(b) É92 a6b2 œ È117 œ 3È13
3. (a)
3 a2b, 2 5¡ œ 1, 3¡
(b) È12 32 œ È10
5. (a) 2u œ 2a3b, 2a2b¡ œ 6, 4¡
3v œ 3a2b, 3a5b¡ œ 6, 15¡
2u 3v œ 6 a 6b, 4 15¡ œ 12, 19¡
(b) É122 a19b2 œ È505
2. (a)
2a2b, 2a5b¡ œ 4, 10¡
(b) É42 a10b2 œ È116 œ 2È29
4. (a)
3 a2b, 2 5¡ œ 5, 7¡
(b) É52 a7b2 œ È74
6. (a) 2u œ 2a3b, 2a2b¡ œ 6, 4¡
5v œ 5a2b, 5a5b¡ œ 10, 25¡
2u 5v œ 6 a10b, 4 25¡ œ 16, 29¡
(b) Éa16b2 292 œ È1097
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 12.2 Vectors
7. (a)
3
3
3
9
6
5 u œ ¢ 5 a3b, 5 a2b£ œ ¢ 5 , 5 £
719
5
5
5
15 10
8. (a) 13
u œ ¢ 13
a3b, 13
a2b£ œ ¢ 13
, 13 £
4
4
4
8
5 v œ ¢ 5 a2b, 5 a5b£ œ ¢ 5 , 4£
12
12
12
24 60
13 v œ ¢ 13 a2b, 13 a5b£ œ ¢ 13 , 13 £
3
4
9
1 14
ˆ 8‰ 6
5 u 5 v œ ¢ 5 5 , 5 4£ œ ¢ 5 , 5 £
5
15
70
ˆ 24 ‰ 10 60
13
u 12
13 v œ ¢ 13 13 , 13 13 £ œ ¢3, 13 £
2
2
‰ œ
(b) Ɉ 15 ‰ ˆ 14
5
È"97
5
9.
2 1, 1 3¡ œ 1, 4¡
11.
0 2, 0 3¡ œ 2, 3¡
2
70 ‰
(b) Éa3b2 ˆ 13
œ
È6421
13
3
0£ œ 1, 1¡
10. ¢ 2a#4b 0, "
#
Ä
Ä
Ä
Ä
12. AB œ 2 1, 0 a1b¡ œ 1, 1¡, CD œ 2 a1b, 2 3¡ œ 1, 1¡, AB CD œ 0, 0¡
13. ¢cos 231 , sin 231 £ œ ¢ "# ,
È3
2 £
14. ¢cos ˆ 341 ‰, sin ˆ 341 ‰£ œ ¢ È"2 , È"2 £
15. This is the unit vector which makes an angle of 120‰ 90‰ œ 210‰ with the positive x-axis;
È
cos 210‰ , sin 210‰ ¡ œ ¢ 23 , "# £
16.
cos 135‰ , sin 135‰ ¡ œ ¢ È"2 , È"2 £
Ä
17. P" P# œ a2 5bi a9 7bj a2 a1bbk œ 3i 2j k
Ä
18. P" P# œ a3 1bi a0 2bj a5 0bk œ 4i 2j 5k
Ä
19. AB œ a10 a7bbi a8 a8bbj a1 1bk œ 3i 16j
Ä
20. AB œ a1 1bi a4 0bj a5 3bk œ 2i 4j 2k
21. 5u v œ 5 1, 1, 1¡ 2, 0, 3¡ œ 5, 5, 5¡ 2, 0, 3¡ œ 5 2, 5 0, 5 3¡ œ 3, 5, 8¡ œ 3i 5j 8k
22. 2u 3v œ 2 1, 0, 2¡ 3 1, 1, 1¡ œ 2, 0, 4¡ 3, 3, 3¡ œ 5, 3, 1¡ œ 5i 3j k
23. The vector v is horizontal and 1 in. long. The vectors u and w are "161 in. long. w is vertical and u makes a 45‰ angle with
the horizontal. All vectors must be drawn to scale.
(a)
(b)
(c)
(d)
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
720
Chapter 12 Vectors and the Geometry of Space
24. The angle between the vectors is 120‰ and vector u is horizontal. They are all 1 in. long. Draw to scale.
(a)
(b)
(c)
(d)
25. length œ k2i j 2kk œ È2# 1# (2)# œ 3, the direction is 23 i 3" j 32 k Ê 2i j 2k œ 3 ˆ 32 i 3" j 32 k‰
9
2
6
26. length œ k9i 2j 6kk œ È81 4 36 œ 11, the direction is 11
i 11
j 11
k Ê 9i 2 j 6k
9
2
6
œ 11 ˆ 11 i 11 j 11 k‰
27. length œ k5kk œ È25 œ 5, the direction is k Ê 5k œ 5(k)
9
16
28. length œ ¸ 35 i 45 k¸ œ É 25
25
œ 1, the direction is 35 i 45 k Ê 35 i 45 k œ 1 ˆ 35 i 45 k‰
#
29. length œ ¹ È16 i È16 j È"6 k¹ œ Ê3 Š È"6 ‹ œ É #" , the direction is È13 i È13 j È"3 k
Ê È16 i È16 j È"6 k œ É "# Š È13 i È13 j È"3 k‹
#
30. length œ ¹ È13 i È13 j È"3 k¹ œ Ê3 Š È"3 ‹ œ 1, the direction is È13 i È13 j È"3 k
Ê È13 i È13 j È"3 k œ 1 Š È13 i È13 j È"3 k‹
31. (a) 2i
(b) È3k
32. (a) 7j
(b) 3 5 2 i 4 5 2 k
È
È
(c)
3
2
10 j 5 k
(d) 6i 2j 3k
(c)
1
1
4 i 3 jk
(d)
a
a
a
È2 i È3 j È6 k
"
"
7
33. kvk œ È12# 5# œ È169 œ 13; kvvk œ 13
(12i 5k) Ê the desired vector is 13
(12i 5k)
v œ 13
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 12.2 Vectors
34. kvk œ É 4" 4" 4" œ
È3 v
1
1
1
# ; kv k œ È 3 i È 3 j È 3 k
721
Ê the desired vector is 3 Š È"3 i È"3 j È"3 k‹
œ È 3i È 3 j È 3k
3
4
3
4
35. (a) 3i 4j 5k œ 5È2 Š 5È
i 5È
j È"2 k‹ Ê the direction is 5È
i 5È
j È"2 k
2
2
2
2
(b) the midpoint is ˆ "# ß 3ß 5# ‰
36. (a) 3i 6j 2k œ 7 ˆ 37 i 67 j 27 k‰ Ê the direction is 37 i 67 j 27 k
(b) the midpoint is ˆ 5# ß 1ß 6‰
37. (a) i j k œ È3 Š È13 i È13 j È13 k‹ Ê the direction is È13 i È13 j È13 k
(b) the midpoint is ˆ 5# ß 7# ß 9# ‰
38. (a) 2i 2j 2k œ 2È3 Š È13 i È13 j È13 k‹ Ê the direction is È13 i È13 j È13 k
(b) the midpoint is ("ß "ß 1)
Ä
39. AB œ (5 a)i (1 b)j (3 c)k œ i 4j 2k Ê 5 a œ 1, 1 b œ 4, and 3 c œ 2 Ê a œ 4, b œ 3, and
c œ 5 Ê A is the point (4ß 3ß 5)
Ä
40. AB œ (a 2)i (b 3)j (c 6)k œ 7i 3j 8k Ê a 2 œ 7, b 3 œ 3, and c 6 œ 8 Ê a œ 9, b œ 0,
and c œ 14 Ê B is the point (9ß 0ß 14)
41. 2i j œ a(i j) b(i j) œ (a b)i (a b)j Ê a b œ 2 and a b œ 1 Ê 2a œ 3 Ê a œ 3# and
b œ a " œ "#
42. i 2j œ a(2i 3j) b(i j) œ (2a b)i (3a b)j Ê 2a b œ 1 and 3a b œ 2 Ê a œ 3 and
b œ 1 #a œ 7 Ê u" œ a(2i 3j) œ 6i 9j and u# œ b(i j) œ 7i 7j
43. 25‰ west of north is 90‰ 25‰ œ 115‰ north of east. 800 cos 115‰ , sin 115‰ ¡ ¸ 338.095, 725.046¡
44. Let u œ x, y¡ be represent the velocity of the plane alone, v œ 70 cos 60‰ , 70 sin 60‰ ¡ œ 35, 35È3¡, and let the
resultant u v œ 500, 0¡. Then x, y¡ 35, 35È3¡ œ 500, 0¡ Ê x 35, y 35È3¡ œ 500, 0¡
Ê x 35 œ 500 and y 35È3 œ 0 Ê x œ 465 and y œ 35È3 Ê u œ 465, 35È3¡
2
È
3
‰
‰
Ê lul œ Ê4652 Š35È3‹ ¸ 468.9 mph, and tan ) œ 35
465 Ê ) ¸ 7.4 Ê 7.4 south of east.
È
45. F1 œ ¢lF1 lcos 30‰ , lF1 lsin 30‰ £ œ ¢ 23 lF1 l, 12 lF1 l£, F2 œ ¢lF2 lcos 45‰ , lF2 lsin 45‰ £ œ ¢ È12 lF2 l, È12 lF2 l£, and
È
w œ 0, 100¡. Since F1 F2 œ 0, 100¡ Ê ¢ 23 lF1 l È12 lF2 l, 21 lF1 l È12 lF2 l£ œ 0, 100¡
È
Ê 23 lF1 l È12 lF2 l œ 0 and 21 lF1 l È12 lF2 l œ 100. Solving the first equation for lF2 l results in: lF2 l œ
È
È6
2 l F1 l .
Substituting this result into the second equation gives us: 12 lF1 l È12 Š 26 lF1 l‹ œ 100 Ê lF1 l œ 1 200È3 ¸ 73.205 N
È
6
Ê lF2 l œ 1100
¸ 89.658 N Ê F1 ¸ 63.397, 36.603¡ and F2 œ ¸ 63.397, 63.397¡
È3
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
722
Chapter 12 Vectors and the Geometry of Space
46. F1 œ 35 cos !, 35 sin !¡, F2 œ ¢lF2 lcos 60‰ , lF2 lsin 60‰ £ œ ¢ 12 lF2 l,
È3
2 lF2 l£, and w œ
0, 50¡. Since
È
F1 F2 œ 0, 50¡ Ê 35 cos ! 12 lF2 l, 35 sin ! 23 lF2 l¡ œ 0, 50¡ Ê 35 cos ! 21 lF2 l œ 0 and
È
35 sin ! 23 lF2 l œ 50. Solving the first equation for lF2 l results in: lF2 l œ 70 cos !. Substituting this result into the
second equation gives us: 35 sin ! 35È3 cos ! œ 50 Ê È3 cos ! œ 10 sin ! Ê 3cos2 ! œ 100 20 sin ! sin2 !
7
49
7
È
5„6 2
20
2
2
Ê 3a1 sin2 !b œ 100
. Since ! 0 Ê
49 7 sin ! sin ! Ê 196 sin ! 140 sin ! 47 œ 0 Ê sin ! œ
14
È
sin ! 0 Ê sin ! œ 5 146 2 Ê ! ¸ 74.42‰ , and lF2 l œ 70 cos ! ¸ 18.81 N.
47. F1 œ ¢lF1 lcos 40‰ , lF1 lsin 40‰ £, F2 œ 100 cos 35‰ , 100 sin 35‰ ¡, and w œ 0, w¡. Since F1 F2 œ 0, w¡
Ê ¢lF1 lcos 40‰ 100 cos 35‰ , lF1 lsin 40‰ 100 sin 35‰ £ œ 0, w¡ Ê lF1 lcos 40‰ 100 cos 35‰ œ 0 and
‰
lF1 lsin 40‰ 100 sin 35‰ œ w. Solving the first equation for lF1 l results in: lF1 l œ 100coscos4035
¸ 106.933 N. Substituting this
‰
result into the second equation gives us: w ¸ 126.093 N.
48. F1 œ lF1 l cos !, lF1 l sin !¡ œ 75 cos !, 75 sin !¡, F2 œ lF2 l cos " , lF2 l sin " ¡ œ 75 cos !, 75 sin !¡, and
w œ 0, 25¡. Since F1 F2 œ 0, 25¡ Ê 75 cos ! 75 cos !, 75 sin ! 75 sin !¡ œ 0, 25¡ Ê 150 sin ! œ 25
Ê ! ¸ 9.59‰ .
Ä
È
È
49. (a) The tree is located at the tip of the vector OP œ (5 cos 60°)i (5 sin 60°)j œ 5# i 5 # 3 j Ê P œ Š #5 ß 5 # 3 ‹
Ä
Ä
(b) The telephone pole is located at the point Q, which is the tip of the vector OP PQ
È
È
È
È
œ Š 52 i 5 # 3 j‹ (10 cos 315°)i (10 sin 315°)j œ Š 5# 10# 2 ‹ i Š 5 # 3 10# 2 ‹ j
Ê Q œ Š 5 10
#
È2
ß 5
È3 10È2
‹
#
50. Let t œ p q q and s œ p p q . Choose T on OP1 so that TQ is
parallel to OP2 , so that ˜TP1 Q is similar to ˜OP1 P2 . Then
Ä
Ä
kOTk
kOP1 k œ t Ê OT œ t OP1 so that T œ at x1 , t y1 , t z1 b.
Ä
Ä
TQk
Also, kkOP
œ s Ê TQ œ s OP2 œ s x2 , y2 , z2 ¡.
2k
Letting Q œ ax, y, zb, we have that
Ä
TQ œ x t x1 , y t y1 , z t z1 ¡ œ s x2 , y2 , z2 ¡
Thus x œ t x1 s x2 , y œ t y1 s y2 , z œ t z1 s z2 .
(Note that if Q is the midpoint, then qp œ 1 and t œ s œ "#
so that x œ "# x1 "# x2 œ x1 2 x2 , y œ y1 2 y2 , z œ z1 2 z2 so that this result agress with the midpoint formula.)
Ä
51. (a) the midpoint of AB is M ˆ 5# ß 5# ß 0‰ and CM œ ˆ 5# 1‰ i ˆ 5# 1‰ j (! 3)k œ 3# i 3# j 3k
Ä
(b) the desired vector is ˆ 23 ‰ CM œ 23 ˆ #3 i #3 j 3k‰ œ i j 2k
(c) the vector whose sum is the vector from the origin to C and the result of part (b) will terminate
at the center of mass Ê the terminal point of (i j 3k) (i j 2k) œ 2i 2j k is the point
(2ß 2ß 1), which is the location of the center of mass
Ä
52. The midpoint of AB is M ˆ #3 ß 0ß 5# ‰ and ˆ 23 ‰ CM œ 23 ˆ #3 1‰ i (0 2)j ˆ #5 1‰ k‘ œ 32 ˆ 5# i 2j 7# k‰
Ä
œ 53 i 43 j 73 k. The vector from the origin to the point of intersection of the medians is ˆ 53 i 43 j 73 k‰ OC
œ ˆ 53 i 43 j 73 k‰ (i 2j k) œ 23 i 23 j 43 k .
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 12.3 The Dot Product
723
53. Without loss of generality we identify the vertices of the quadrilateral such that A(0ß 0ß 0), B(xb ß 0ß 0),
C(xc ß yc ß 0) and D(xd ß yd ß zd ) Ê the midpoint of AB is MAB ˆ x#b ß 0ß 0‰ , the midpoint of BC is
MBC ˆ xb # xc ß y#c ß 0‰ , the midpoint of CD is MCD ˆ xc # xd ß yc # yd ß z#d ‰ and the midpoint of AD is
xb
MAD ˆ x#d ß y#d ß z#d ‰ Ê the midpoint of MAB MCD is Œ #
xb
xc
x
x
b c# d
#
ß yc 4 yd , z4d which is the same as the midpoint
xd
b
of MAD MBC œ Œ # # # ß yc 4 yd , z4d .
54. Let V" , V# , V$ , á , Vn be the vertices of a regular n-sided polygon and vi denote the vector from the center to
n
Vi for i œ 1, 2, 3, á , n. If S œ ! vi and the polygon is rotated through an angle of i(2n1) where i œ 1, 2, 3, á , n,
i œ1
then S would remain the same. Since the vector S does not change with these rotations we conclude that S œ 0.
55. Without loss of generality we can coordinatize the vertices of the triangle such that A(0ß 0), B(bß 0) and
Ä
C(xc ß yc ) Ê a is located at ˆ b # xc ß y#c ‰ , b is at ˆ x#c ß y#c ‰ and c is at ˆ b# ß 0‰ . Therefore, Aa œ ˆ b# x#c ‰ i ˆ y#c ‰ j ,
Ä
Ä
Ä
Ä
Ä
Bb œ ˆ x#c b‰ i ˆ y#c ‰ j , and Cc œ ˆ b# xc ‰ i (yc ) j Ê Aa Bb Cc œ 0 .
56. Let u be any unit vector in the plane. If u is positioned so that its initial point is at the origin and terminal point is at ax, yb,
then u makes an angle ) with i, measured in the counter-clockwise direction. Since kuk œ 1, we have that x œ cos ) and
y œ sin ). Thus u œ cos ) i sin ) j. Since u was assumed to be any unit vector in the plane, this holds for every unit
vector in the plane.
12.3 THE DOT PRODUCT
)
NOTE: In Exercises 1-8 below we calculate projv u as the vector Š kuk kcos
vk ‹ v , so the scalar multiplier of v is the number in
column 5 divided by the number in column 2.
v†u
kvk
kuk
cos )
kuk cos )
projv u
1. 25
5
5
1
5
2i 4 j È 5 k
2.
3
1
13
3
13
3
3 ˆ 35 i 45 k‰
3.
25
15
5
"
3
5
3
"
9 (10i 11j 2k)
4.
13
15
3
13
45
13
15
13
225 (2i 10j 11k)
5.
2
È34
È3
2
È3 È34
2
È34
"
17 (5j 3k)
6. È3 È2
È2
3
È3 È2
3È 2
È3 È2
È2
È3 È2
(i j)
#
7. 10 È17
È26
È21
10 È17
È546
10 È17
È26
10 È17
(5i j)
26
È30
6
È30
6
"
5
"
È30
"
"
"
5 ¢ È# , È3 £
8.
"
6
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
724
Chapter 12 Vectors and the Geometry of Space
(1)(2) (0)(1)
9. ) œ cos" Š kuuk†kvvk ‹ œ cos" Š È2#(2)(1)
‹ œ cos" Š È54È6 ‹ œ cos" Š È430 ‹ ¸ 0.75 rad
1# 0# È1# 2# ( 1)#
(2)(0) (1)(4)
10. ) œ cos" Š kuuk†kvvk ‹ œ cos" Š È2#(2)(3)
‹ œ cos" Š È910È25 ‹ œ cos" ˆ 23 ‰ ¸ 0.84 rad
( 2)# 1# È3# 0# 4#
11. ) œ cos" Š kuuk†kvvk ‹ œ cos"
Î
Ñ
ŠÈ3‹ ŠÈ3‹ (7)(1) (0)(2)
#
#
Ï ÊŠÈ3‹ (7)# 0# ÊŠÈ3‹ (1)# (2)# Ò
7
œ cos" Š È352È
‹
8
1
œ cos" Š È26
‹ ¸ 1.77 rad
12. ) œ cos" Š kuuk†kvvk ‹ œ cos"
Î
Ï
Ñ
(1)(1) ŠÈ2‹ (1) ŠÈ2‹ (1)
#
Ê(1)# ŠÈ2‹
#
ŠÈ2‹ È(1)# (1)# (1)# Ò
1
œ cos" Š È
‹
5 È3
1
œ cos" Š È15
‹ ¸ 1.83 rad
Ä
Ä
Ä
Ä
Ä
Ä
13. AB œ 3, 1¡, BC œ 1, 3¡, and AC œ 2, 2¡. BA œ 3, 1¡, CB œ 1, 3¡, CA œ 2, 2¡.
Ä
Ä
Ä
Ä
Ä
Ä
¹ AB ¹ œ ¹ BA ¹ œ È10, ¹ BC ¹ œ ¹ CB ¹ œ È10, ¹ AC ¹ œ ¹ CA ¹ œ 2È2,
Ä Ä
3a2b 1a2b
AB†AC
"
Angle at A œ cos" Ä
œ cos" Š È15 ‹ ¸ 63.435‰
Ä œ cos
È
È ¹AB¹ ¹AC¹
Š
10‹Š2
2‹
Ä Ä
a1ba3b a3ba1b
BC†BA
"
" 3
‰
Angle at B œ cos" Ä
Ä œ cos
œ cos ˆ 5 ‰ ¸ 53.130 , and
È
È
¹BC¹ ¹BA¹
Š
10‹Š
10‹
Ä Ä
1a2b 3a2b
CB†CA
"
Angle at C œ cos" Ä
œ cos" Š È15 ‹ ¸ 63.435‰
Ä œ cos
È
È ¹CB¹ ¹CA¹
Š
10‹Š2
2‹
Ä
Ä Ä
Ä
14. AC œ 2, 4¡ and BD œ 4, 2¡. AC † BD œ 2a4b 4a2b œ 0, so the angle measures are all 90‰ .
15. (a) cos ! œ kiik†kvvk œ kvak , cos " œ kjjk†kvvk œ kbvk , cos # œ kkkk†kvvk œ kvck and
#
#
#
cos# ! cos# " cos# # œ Š kvak ‹ Š kvbk ‹ Š kvck ‹ œ a kvbk kvk c œ kkvvkk kkvvkk œ 1
#
#
#
(b) kvk œ 1 Ê cos ! œ kvak œ a, cos " œ kvbk œ b and cos # œ kvck œ c are the direction cosines of v
16. u œ 10i 2k is parallel to the pipe in the north direction and v œ 10j k is parallel to the pipe in the east
direction. The angle between the two pipes is ) œ cos" Š kuuk†kvvk ‹ œ cos" Š È1042È101 ‹ ¸ 1.55 rad ¸ 88.88°.
17. The sum of two vectors of equal length is always orthogonal to their difference, as we can see from the equation
(v" v# ) † (v" v# ) œ v" † v" v# † v" v" † v# v# † v# œ kv" k # kv# k # œ 0
Ä Ä
18. CA † CB œ (v (u)) † (v u) œ v † v v † u u † v u † u œ kvk # kuk # œ 0 because kuk œ kvk since both equal
Ä
Ä
the radius of the circle. Therefore, CA and CB are orthogonal.
19. Let u and v be the sides of a rhombus Ê the diagonals are d" œ u v and d# œ u v
Ê d" † d# œ (u v) † (u v) œ u † u u † v v † u v † v œ kvk # kuk # œ 0 because kuk œ kvk , since a rhombus
has equal sides.
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 12.3 The Dot Product
725
20. Suppose the diagonals of a rectangle are perpendicular, and let u and v be the sides of a rectangle Ê the diagonals are
d" œ u v and d# œ u v. Since the diagonals are perpendicular we have d" † d# œ 0
Í (u v) † (u v) œ u † u u † v v † u v † v œ 0 Í kvk # kuk # œ 0 Í akvk kukb akvk kukb œ 0
Í akvk kukb œ 0 which is not possible, or akvk kukb œ 0 which is equivalent to kvk œ kuk Ê the rectangle is a square.
21. Clearly the diagonals of a rectangle are equal in length. What is not as obvious is the statement that equal diagonals
happen only in a rectangle. We show this is true by letting the adjacent sides of a parallelogram be the vectors
(v" i v# j) and (u" i u# j). The equal diagonals of the parallelogram are d" œ (v" i v# j) (u" i u# j) and
d# œ (v" i v# j) (u" i u# j). Hence kd" k œ kd# k œ k(v" i v# j) (u" i u# j)k œ k(v" i v# j) (u" i u# j)k
Ê k(v" u" )i (v# u# )jk œ k(v" u" )i (v# u# )jk Ê È(v" u" )# (v# u# )# œ È(v" u" )# (v# u# )#
Ê v#1 2v" u" u1# v## 2v# u# u## œ v1# 2v" u" u1# v## 2v# u# u## Ê 2(v" u" v# u# )
œ 2(v" u" v# u# ) Ê v" u" v# u# œ 0 Ê (v" i v# j) † (u" i u# j) œ 0 Ê the vectors (v" i v# j) and (u" i u# j)
are perpendicular and the parallelogram must be a rectangle.
22. If kuk œ kvk and u v is the indicated diagonal, then (u v) † u œ u † u v † u œ kuk # v † u œ u † v kvk #
œ u † v v † v œ (u v) † v Ê the angle cos" Š k(uuvvk )k†uuk ‹ between the diagonal and u and the angle
cos" Š k(uuvvk )k†vvk ‹ between the diagonal and v are equal because the inverse cosine function is one-to-one.
Therefore, the diagonal bisects the angle between u and v.
23. horizontal component: 1200 cosa8‰ b ¸ 1188 ft/s; vertical component: 1200 sina8‰ b ¸ 167 ft/s
2.5 lb
2.5 lb
‰
‰¡
24. kwkcosa33‰ 15‰ b œ 2.5 lb, so kwk œ cos
¸ 2.205, 1.432¡
18‰ . Then w œ cos 18‰ cos 33 , sin 33
25. (a) Since kcos )k Ÿ 1, we have ku † vk œ kuk kvk kcos )k Ÿ kuk kvk (1) œ kuk kvk .
(b) We have equality precisely when kcos )k œ 1 or when one or both of u and v is 0. In the case of nonzero
vectors, we have equality when ) œ 0 or 1, i.e., when the vectors are parallel.
26. (xi yj) † v œ kxi yjk kvk cos ) Ÿ 0 when 1# Ÿ ) Ÿ 1. This
means (xß y) has to be a point whose position vector makes
an angle with v that is a right angle or bigger.
27. v † u" œ (au" bu# ) † u" œ au" † u" bu# † u" œ a ku" k # b(u# † u" ) œ a(1)# b(0) œ a
28. No, v" need not equal v# . For example, i j Á i 2j but i † (i j) œ i † i i † j œ 1 0 œ 1 and
i † (i 2j) œ i † i 2i † j œ 1 2 † 0 œ 1.
2
29. projv u œ luv†lv2 v Ê Šu luv†lv2 v‹ † Š luv†lv2 v‹ œ u † Š luv†lv2 v‹ Š luv†lv2 v‹ † Š luv†lv2 v‹ œ Š luv†lv2 ‹au † vb Š luv†lv2 ‹ av † vb
2
2
œ aulv†vl2b aulv†vl4b lvl2 œ 0
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
726
Chapter 12 Vectors and the Geometry of Space
30. F œ 2i j 3k and v œ 3i j Ê projv F œ lFv†lv2 v œ
5
2
ŠÈ"!‹
a3i jb œ 23 i 21 j, is the vector parallel to v.
F projv F œ a2i j 3kb ˆ 32 i 12 j‰ œ 12 i 32 j 3k is the vector orthogonal to v.
31. P(x" ß y" ) œ P ˆx" , bc ba x" ‰ and Q(x# ß y# ) œ Q ˆx# ß bc ba x# ‰ are any two points P and Q on the line with b Á 0
Ä
Ä
Ê PQ œ (x# x" )i ba (x" x# )j Ê PQ † v œ (x# x" )i ba (x" x# )j‘ † (ai bj) œ a(x# x" ) b ˆ ba ‰ (x" x# )
Ä
œ 0 Ê v is perpendicular to PQ for b Á 0. If b œ 0, then v œ ai is perpendicular to the vertical line ax œ c.
Alternatively, the slope of v is ba and the slope of the line ax by œ c is ba , so the slopes are negative reciprocals
Ê the vector v and the line are perpendicular.
32. The slope of v is ba and the slope of bx ay œ c is ba , provided that a Á 0. If a œ 0, then v œ bj is parallel to
the vertical line bx œ c. In either case, the vector v is parallel to the line bx ay œ c.
33. v œ i 2j is perpendicular to the line x 2y œ c;
P(2ß 1) on the line Ê 2 2 œ c Ê x 2y œ 4
34. v œ 2i j is perpendicular to the line 2x y œ c;
P(1ß 2) on the line Ê (2)(1) 2 œ c
Ê 2x y œ 0
35. v œ 2i j is perpendicular to the line 2x y œ c;
P(2ß 7) on the line Ê (2)(2) 7 œ c
Ê 2x y œ 3
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 12.3 The Dot Product
36. v œ 2i 3j is perpendicular to the line 2x 3y œ c;
P(11ß 10) on the line Ê (2)(11) (3)(10) œ c
Ê 2x 3y œ 8
37. v œ i j is parallel to the line x y œ c;
P(2ß 1) on the line Ê a2b 1 œ c Ê x y œ 1
or x y œ 1.
38. v œ 2i 3j is parallel to the line 3x 2y œ c;
P(0ß 2) on the line Ê 0 2(2) œ c Ê 3x 2y œ 4
39. v œ i 2j is parallel to the line 2x y œ c;
P(1ß 2) on the line Ê 2(1) 2 œ c Ê 2x y œ 0
or 2x y œ 0.
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
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728
Chapter 12 Vectors and the Geometry of Space
40. v œ 3i 2j is parallel to the line 2x 3y œ c;
P(1ß 3) on the line Ê (2)(1) (3)(3) œ c
Ê 2x 3y œ 11 or 2x 3y œ 11
Ä
Ä
41. P(0ß 0), Q(1ß 1) and F œ 5j Ê PQ œ i j and W œ F † PQ œ (5j) † (i j) œ 5 N † m œ 5 J
42. W œ kFk (distance) cos ) œ (602,148 N)(605 km)(cos 0) œ 364,299,540 N † km œ (364,299,540)(1000) N † m
œ 3.6429954 ‚ 10"" J
Ä
43. W œ kFk ¹ PQ ¹ cos ) œ (200)(20)(cos 30°) œ 2000È3 œ 3464.10 N † m œ 3464.10 J
Ä
44. W œ kFk ¹ PQ ¹ cos ) œ (1000)(5280)(cos 60°) œ 2,640,000 ft † lb
In Exercises 45-50 we use the fact that n œ ai bj is normal to the line ax by œ c.
1
45. n" œ 3i j and n# œ 2i j Ê ) œ cos" Š knn""k†knn## k ‹ œ cos" Š È610È
‹ œ cos" Š È"2 ‹ œ 14
5
3 1
46. n" œ È3i j and n# œ È3i j Ê ) œ cos" Š knn""k†knn## k ‹ œ cos" Š È
‹ œ cos" ˆ "2 ‰ œ 231
4 È4
È
È
È
47. n" œ È3i j and n# œ i È3j Ê ) œ cos" Š knn""k†knn## k ‹ œ cos" Š È34È43 ‹ œ cos" Š 23 ‹ œ 16
48. n" œ i È3j and n# œ Š1 È3‹ i Š1 È3‹ j Ê ) œ cos" Š knn""k†knn## k ‹
œ cos" 1 È3 È3 3
4
œ cos" Š 2È
‹ œ cos" Š È"2 ‹ œ 14
8
È 1 3 É 1 2È 3 3 1 2È 3 3 4
7
49. n" œ 3i 4j and n# œ i j Ê ) œ cos" Š knn""k†knn## k ‹ œ cos" Š È325È
‹ œ cos" Š 5È
‹ ¸ 0.14 rad
2
2
10
"
50. n" œ 12i 5j and n# œ 2i 2j Ê ) œ cos" Š knn""k†knn## k ‹ œ cos" Š È24169
Š 2614
È8 ‹ œ cos
È2 ‹ ¸ 1.18 rad
12.4 THE CROSS PRODUCT
â
â
j
k â
âi
â
â
1. u ‚ v œ â 2 2 " â œ 3 ˆ 23 i 3" j 32 k‰ Ê length œ 3 and the direction is 23 i 3" j 32 k;
â
â
â " 0 " â
v ‚ u œ (u ‚ v) œ 3 ˆ 23 i "3 j 32 k‰ Ê length œ 3 and the direction is 32 i 3" j 32 k
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 12.4 The Cross Product
â
â i
â
2. u ‚ v œ â 2
â
â "
â
j kâ
â
3 0 â œ 5(k) Ê length œ 5 and the direction is k
â
1 0â
v ‚ u œ (u ‚ v) œ 5(k) Ê length œ 5 and the direction is k
â
â
j
k â
â i
â
â
3. u ‚ v œ â 2 2 4 â œ 0 Ê length œ 0 and has no direction
â
â
â " 1 2 â
v ‚ u œ (u ‚ v) œ 0 Ê length œ 0 and has no direction
â
âi
â
4. u ‚ v œ â 1
â
â0
â
j k â
â
1 1 â œ 0 Ê length œ 0 and has no direction
â
0 0 â
v ‚ u œ (u ‚ v) œ 0 Ê length œ 0 and has no direction
â
â
j kâ
âi
â
â
5. u ‚ v œ â 2 0 0 â œ 6(k) Ê length œ 6 and the direction is k
â
â
â 0 3 0 â
v ‚ u œ (u ‚ v) œ 6(k) Ê length œ 6 and the direction is k
â
âi j
â
6. u ‚ v œ (i ‚ j) ‚ (j ‚ k) œ k ‚ i œ â 0 0
â
â1 0
â
kâ
â
1 â œ j Ê length œ 1 and the direction is j
â
0â
v ‚ u œ (u ‚ v) œ j Ê length œ 1 and the direction is j
â
â
j
k â
â i
â
â
7. u ‚ v œ â 8 2 4 â œ 6i 12k Ê length œ 6È5 and the direction is È"5 i È25 k
â
â
2
1 â
â 2
v ‚ u œ (u ‚ v) œ (6i 12k) Ê length œ 6È5 and the direction is " i 2 k
È5
âi
â
â
8. u ‚ v œ â 3#
â
â1
È5
k ââ
â
1 â œ 2i 2j 2k Ê length œ 2È3 and the direction is È"3 i È13 j È"3 k
â
2â
v ‚ u œ (u ‚ v) œ (2i 2j 2k) Ê length œ 2È3 and the direction is " i 1 j " k
j
"#
1
È3
â
âi
â
9. u ‚ v œ â 1
â
â0
â
j kâ
â
0 0⠜ k
â
1 0â
â
âi
â
10. u ‚ v œ â 1
â
â0
È3
È3
â
j k â
â
0 1 ⠜ i k
â
1 0 â
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
729
730
Chapter 12 Vectors and the Geometry of Space
â
âi
â
11. u ‚ v œ â 1
â
â0
â
j k â
â
0 1 ⠜ i j k
â
1 1 â
â
j
âi
â
12. u ‚ v œ â 2 1
â
â1 2
â
âi
â
13. u ‚ v œ â 1
â
â1
â
j kâ
â
1 0 ⠜ 2k
â
1 0 â
â
âi
â
14. u ‚ v œ â 0
â
â1
â
â i
Ä
Ä
â
15. (a) PQ ‚ PR œ â 1
â
â 1
Ä
â
kâ
â
0 ⠜ 5k
â
0â
â
j kâ
â
1 2 ⠜ 2j k
â
0 0â
â
j k â
Ä
Ä
â
1 3 â œ 8i 4j 4k Ê Area œ "# ¹ PQ ‚ PR ¹ œ "# È64 16 16 œ 2È6
â
3 1 â
Ä
‚PR
"
(b) u œ PQ
Ä Ä œ È (2i j k)
6
¹PQ‚PR¹
â
âi
Ä
Ä
â
16. (a) PQ ‚ PR œ â 1
â
â2
Ä
j
0
2
Ä
â
kâ
Ä
Ä
â
2 â œ 4i 4j 2k Ê Area œ "# ¹ PQ ‚ PR ¹ œ "# È16 16 4 œ 3
â
0â
‚PR
"
(b) u œ PQ
Ä Ä œ 3 (2i 2j k)
¹PQ‚PR¹
â
âi
Ä
Ä
â
17. (a) PQ ‚ PR œ â 1
â
â1
Ä
j
1
1
Ä
â
kâ
Ä
Ä
È
â
1 â œ i j Ê Area œ "# ¹ PQ ‚ PR ¹ œ "# È1 1 œ #2
â
0â
‚PR
"
"
(b) u œ PQ
Ä Ä œ È (i j ) œ È ( i j )
2
2
¹PQ‚PR¹
â
âi
Ä
Ä
â
18. (a) PQ ‚ PR œ â 2
â
â1
Ä
Ä
j
1
0
â
k â
Ä
Ä
È
â
1 â œ 2i 3j k Ê Area œ "# ¹ PQ ‚ PR ¹ œ "# È4 9 1 œ #14
â
2 â
‚PR
"
(b) u œ PQ
(2i 3j k)
Ä Ä œ È
14
¹PQ‚PR¹
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 12.4 The Cross Product
â
â
â a" a# a$ â
â
â
19. If u œ a" i a# j a$ k, v œ b" i b# j b$ k, and w œ c" i c# j c$ k, then (u ‚ v) † w œ â b" b# b$ â ,
â
â
â c" c# c$ â
â
â
â
â
â b" b# b $ â
â c" c # c $ â
â
â
â
â
(v ‚ w) † u œ â c" c# c$ â and (w ‚ u) † v œ â a" a# a$ â which all have the same absolute value, since the
â
â
â
â
â a" a# a$ â
â b" b# b$ â
interchanging of two rows in a determinant does not change its absolute value Ê the volume is
â
â
â2 0 0â
â
â
k(u ‚ v) † wk œ abs â 0 2 0 â œ 8
â
â
â0 0 2â
â
â
â 1 1 1 â
â
â
1 2 ⠜ 4 (for details about verification, see Exercise 19)
20. k(u ‚ v) † wk œ abs â 2
â
â
â 1 2 1 â
â
â2 1
â
21. k(u ‚ v) † wk œ abs â 2 1
â
â1 0
â
0â
â
1 â œ k7k œ 7 (for details about verification, see Exercise 19)
â
2â
â
â
â 1 1 2 â
â
â
22. k(u ‚ v) † wk œ abs â 1 0 1 â œ 8 (for details about verification, see Exercise 19)
â
â
â 2 4 2 â
23. (a) u † v œ 6, u † w œ 81, v † w œ 18 Ê none are perpendicular
â
â
â
â
â
â
j
k â
j
k â
j k â
âi
â i
â i
â
â
â
â
â
â
1 1 â œ 0, v ‚ w œ â 0
1 5 â Á 0
(b) u ‚ v œ â 5 1 1 â Á 0, u ‚ w œ â 5
â
â
â
â
â
â
â 0 1 5 â
â 15 3 3 â
â 15 3 3 â
Ê u and w are parallel
24. (a) u † v œ 0, u ‚ w œ 0, u † r œ 31, v † w œ 0, v † r œ 0, w † r œ 0 Ê u ¼ v, u ¼ w, v ¼ w, v ¼ r
and w ¼ r
â
â
â
â
â
â
j
k â
j k â
â i
âi j k â
â i
â
â
â
â
â
â
2 1 ⠜ 0
(b) u ‚ v œ â 1 2 1 â Á 0, u ‚ w œ â 1 2 1 â Á 0, u ‚ r œ â 1
â
â
â
â
â 1 1 1 â
â 1 1 1 â
â1 0 1 â
â #
# â
â
â
â
â
â
â
j
kâ
j
kâ
j kâ
â i
â i
â i
â
â
â
â
â
â
1
1 â Á 0, w ‚ r œ â 1
0
1âÁ0
v ‚ w œ â 1 1 1 â Á 0, v ‚ r œ â 1
â
â
â 1 1 1 â
â 1 1 1 â
â 1 0 1â
â #
â #
# â
# â
Ê u and r are parallel
Ä
Ä
È
25. ¹ PQ ‚ F¹ œ ¹ PQ ¹ kFk sin (60°) œ 23 † 30 † #3 ft † lb œ 10È3 ft † lb
Ä
Ä
È
26. ¹ PQ ‚ F¹ œ ¹ PQ ¹ kFk sin (135°) œ 23 † 30 † #2 ft † lb œ 10È2 ft † lb
27. (a) true, kuk œ Èa#1 a## a3# œ Èu † u
(b) not always true, u † u œ kuk #
â
â
â
j kâ
â i
â i
â
â
â
(c) true, u ‚ 0 œ â u" u# u$ â œ 0i 0j 0k œ 0 and 0 ‚ u œ â 0
â
â
â
â0 0 0â
â u"
j
0
u#
â
kâ
â
0 â œ 0i 0j 0k œ 0
â
u$ â
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
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Chapter 12 Vectors and the Geometry of Space
â
â i
â
(d) true, u ‚ (u) œ â u"
â
â u"
j
u#
u#
â
k â
â
u$ â œ (u# u$ u# u$ )i (u" u$ u" u$ )j (u" u# u" u# )k œ 0
â
u$ â
(e) not always true, i ‚ j œ k Á k œ j ‚ i for example
(f) true, distributive property of the cross product
(g) true, (u ‚ v) † v œ u † (v ‚ v) œ u † 0 œ 0
(h) true, the volume of a parallelpiped with u, v, and w along the three edges is the same whether the plane containing u
and v or the plane containing v and w is used as the base plane, and the dot product is commutative.
28. (a) true, u † v œ u" v" u# v# u$ v$ œ v" u" v# u# v$ u$ œ v † u
â
â
â
â
j kâ
j kâ
â i
â i
â
â
â
â
(b) true, u ‚ v œ â u" u# u$ â œ â v" v# v$ â œ (v ‚ u)
â
â
â
â
â v" v# v $ â
â u" u # u $ â
â
â
â
â
j
k â
j kâ
â i
â i
â
â
â
â
(c) true, (u) ‚ v œ â u" u# u$ â œ â u" u# u$ â œ (u ‚ v)
â
â
â
â
v#
v$ â
â v"
â v" v# v$ â
(d) true, (cu) † v œ (cu" )v" (cu# )v# (cu$ )v$ œ u" (cv" ) u# (cv# ) u$ (cv$ ) œ u † (cv) œ c(u" v" u# v# u$ v$ )
œ c(u † v)
â
â
â
â
â
â
j kâ
j
k â
j
k â
â i
â i
â i
â
â
â
â
â
â
(e) true, c(u ‚ v) œ c â u" u# u$ â œ â cu" cu# cu$ â œ (cu) ‚ v œ â u" u# u$ â œ u ‚ (cv)
â
â
â
â
â
â
â v" v# v $ â
â v" v# v $ â
â cv" cv# cv$ â
#
(f) true, u † u œ u1# u## u3# œ ˆÈu1# u## u3# ‰ œ kuk #
(g) true, (u ‚ u) † u œ 0 † u œ 0
(h) true, u ‚ v ¼ u and u ‚ v ¼ v Ê (u ‚ v) † u œ v † (u ‚ v) œ 0
29. (a) projv u œ Š kvukk†vvk ‹ v
(e) (u ‚ v) ‚ (u ‚ w)
(b) (u ‚ v)
(f)
(c) a(u ‚ v) ‚ wb
(d) k(u ‚ v) † wk
kuk kvvk
30. ai ‚ jb ‚ j œ k ‚ j œ i; i ‚ aj ‚ jb œ i ‚ 0 œ 0. The cross product is not associative.
31. (a) yes, u ‚ v and w are both vectors
(c) yes, u and u ‚ w are both vectors
(b) no, u is a vector but v † w is a scalar
(d) no, u is a vector but v † w is a scalar
32. (u ‚ v) ‚ w is perpendicular to u ‚ v, and u ‚ v is perpendicular to both u and v Ê (u ‚ v) ‚ w is
parallel to a vector in the plane of u and v which means it lies in the plane determined by u and v.
The situation is degenerate if u and v are parallel so u ‚ v œ 0 and the vectors do not determine a plane.
Similar reasoning shows that u ‚ (v ‚ w) lies in the plane of v and w provided v and w are nonparallel.
33. No, v need not equal w. For example, i j Á i j, but i ‚ ai jb œ i ‚ i i ‚ j œ 0 k œ k and
i ‚ ai jb œ i ‚ aib i ‚ j œ 0 k œ k.
34. Yes. If u ‚ v œ u ‚ w and u † v œ u † w, then u ‚ (v w) œ 0 and u † (v w) œ 0. Suppose now that v Á w.
Then u ‚ (v w) œ 0 implies that v w œ ku for some real number k Á 0. This in turn implies that
u † (v w) œ u † (ku) œ k kuk # œ 0, which implies that u œ 0. Since u Á 0, it cannot be true that v Á w, so v œ w.
â
â
j kâ
â i
Ä
Ä
Ä
Ä
Ä
Ä
â
â
35. AB œ i j and AD œ i j Ê AB ‚ AD œ â 1 1 0 â œ 2k Ê area œ ¹ AB ‚ AD ¹ œ 2
â
â
â 1 1 0 â
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 12.4 The Cross Product
â
âi
Ä
Ä
Ä
Ä
â
36. AB œ 7i 3j and AD œ 2i 5j Ê AB ‚ AD œ â 7
â
â2
j
3
5
â
âi
Ä
Ä
Ä
Ä
â
37. AB œ 3i 2j and AD œ 5i j Ê AB ‚ AD œ â 3
â
â5
â
j kâ
Ä
Ä
â
2 0 â œ 13k Ê area œ ¹ AB ‚ AD ¹ œ 13
â
1 0â
â
âi
Ä
Ä
Ä
Ä
â
38. AB œ 7i 4j and AD œ 2i 5j Ê AB ‚ AD œ â 7
â
â2
â
j kâ
Ä
Ä
â
4 0 â œ 43k Ê area œ ¹ AB ‚ AD ¹ œ 43
â
5 0â
733
â
kâ
Ä
Ä
â
0 â œ 29k Ê area œ ¹ AB ‚ AD ¹ œ 29
â
0â
Ä
Ä
Ä
Ä Ä
Ä
Ä
39. AB œ 3i 2j 4k and DC œ 3i 2j 4k Ê AB is parallel to DC; BC œ 2i j and AD œ 2i j Ê BC is parallel to
â
â
j kâ
âi
Ä Ä
Ä
Ä
Ä
â
â
AD. AB ‚ BC œ â 3 2 4 â œ 4i 8j 7k Ê area œ lAB ‚ BC l œ È129
â
â
â 2 1 0 â
Ä
Ä
Ä
Ä
Ä
Ä Ä
40. AC œ i 4j and DB œ i 4j Ê AC is parallel to DB; AD œ i 3j 3k and CB œ i 3j 3k Ê AD is parallel
â
â
j kâ
â i
Ä Ä
Ä
Ä
Ä
â
â
to CB . AC ‚ AD œ â 1 4 0 â œ 12i 3j 7k Ê area œ lAC ‚ AD l œ È202
â
â
â 1 3 3 â
â
â i
Ä
Ä
Ä
Ä
â
41. AB œ 2i 3j and AC œ 3i j Ê AB ‚ AC œ â 2
â
â 3
â
âi
Ä
Ä
Ä
Ä
â
42. AB œ 4i 4j and AC œ 3i 2j Ê AB ‚ AC œ â 4
â
â3
j
3
1
â
kâ
Ä
Ä
â
0 â œ 11k Ê area œ "# ¹ AB ‚ AC ¹ œ 11
#
â
0â
â
j kâ
Ä
Ä
â
4 0 â œ 4k Ê area œ "# ¹ AB ‚ AC ¹ œ 2
â
2 0â
â
â i
Ä
Ä
Ä
Ä
â
43. AB œ 6i 5j and AC œ 11i 5j Ê AB ‚ AC œ â 6
â
â 11
â
j kâ
Ä
Ä
â
5 0 â œ 25k Ê area œ "# ¹ AB ‚ AC ¹ œ 25
#
â
5 0 â
â
â
j kâ
â i
Ä
Ä
Ä
Ä
Ä
Ä
â
â
44. AB œ 16i 5j and AC œ 4i 4j Ê AB ‚ AC œ â 16 5 0 â œ 84k Ê area œ "# ¹ AB ‚ AC ¹ œ 42
â
â
4 0â
â 4
â
â i
Ä
Ä
Ä
Ä
â
45. AB œ i 2j and AC œ i k Ê AB ‚ AC œ â 1
â
â 1
â
j k â
Ä
Ä
â
2 0 â œ 2i j 2k Ê area œ "# ¹ AB ‚ AC ¹ œ #3
â
0 1 â
â
â i
Ä
Ä
Ä
Ä
â
46. AB œ i j k and AC œ 3i 3k Ê AB ‚ AC œ â 1
â
â 3
â
j k â
Ä
Ä
È
â
1 1 â œ 3i 3k Ê area œ "# ¹ AB ‚ AC ¹ œ 3 # 2
â
0 3 â
â
â i
Ä
Ä
Ä
Ä
â
47. AB œ i 2j and AC œ j 2k Ê AB ‚ AC œ â 1
â
â 0
â
k â
Ä
Ä
È
â
0 â œ 4i 2j k Ê area œ "# ¹ AB ‚ AC ¹ œ #21
â
2 â
j
2
1
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
734
Chapter 12 Vectors and the Geometry of Space
â
â1
Ä
Ä
Ä
Ä
Ä
Ä
â
48. AB œ i 2j, AC œ 3j 2k and AD œ 3i 4j 5k Ê Š AB ‚ AC ‹ † AD œ â 0
â
â3
Ä
Ä
Ä
Ê volume œ ¹Š AB ‚ AC ‹ † AD ¹ œ 5
â
â i
â
49. If A œ a" i a# j and B œ b" i b# j, then A ‚ B œ â a"
â
â b"
"
# kA ‚ Bk œ
a
„ "# º "
b"
j
a#
b#
â
kâ
a
â
0â œ º "
b
â
"
0â
2
3
4
â
0â
â
2⠜ 5
â
5â
a#
k and the triangle's area is
b# º
a#
. The applicable sign is () if the acute angle from A to B runs counterclockwise
b# º
in the xy-plane, and () if it runs clockwise, because the area must be a nonnegative number.
Ä
Ä
50. If A œ a" i a# j, B œ b" i b# j, and C œ c" i c# j, then the area of the triangle is "# ¹ AB ‚ AC ¹ . Now,
â
â
i
j
kâ
â
Ä
Ä
Ä
Ä
b a"
b# a#
â
â
k Ê "# ¹ AB ‚ AC ¹
AB ‚ AC œ â b" a" b# a# 0 â œ º "
º
c" a"
c# a#
â
â
â c" a" c# a# 0 â
œ "# k(b" a" )(c# a# ) (c" a" )(b# a# )k œ "# ka" (b# c# ) a# (c" b" ) (b" c# c" b# )k
â
â
â a" a# 1 â
â
" â
œ „ # â b" b# 1 â . The applicable sign ensures the area formula gives a nonnegative number.
â
â
â c" c# 1 â
12.5 LINES AND PLANES IN SPACE
1. The direction i j k and P(3ß 4ß 1) Ê x œ 3 t, y œ 4 t, z œ 1 t
Ä
2. The direction PQ œ 2i 2j 2k and P(1ß 2ß 1) Ê x œ 1 2t, y œ 2 2t, z œ 1 2t
Ä
3. The direction PQ œ 5i 5j 5k and P(2ß 0ß 3) Ê x œ 2 5t, y œ 5t, z œ 3 5t
Ä
4. The direction PQ œ j k and P(1ß 2ß 0) Ê x œ 1, y œ 2 t, z œ t
5. The direction 2j k and P(!ß !ß !) Ê x œ 0, y œ 2t, z œ t
6. The direction 2i j 3k and P(3ß 2ß 1) Ê x œ 3 2t, y œ 2 t, z œ 1 3t
7. The direction k and P(1ß 1ß 1) Ê x œ 1, y œ 1, z œ 1 t
8. The direction 3i 7j 5k and P(2ß 4ß 5) Ê x œ 2 3t, y œ 4 7t, z œ 5 5t
9. The direction i 2j 2k and P(0ß 7ß 0) Ê x œ t, y œ 7 2t, z œ 2t
â
âi j
â
10. The direction is u ‚ v œ â 1 2
â
â3 4
â
kâ
â
3 â œ 2i 4j 2k and Pa2, 3, 0b Ê x œ 2 2t, y œ 3 4t, z œ 2t
â
5â
11. The direction i and P(0ß 0ß 0) Ê x œ t, y œ 0, z œ 0
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 12.5 Lines and Planes in Space
12. The direction k and P(0ß 0ß 0) Ê x œ 0, y œ 0, z œ t
Ä
13. The direction PQ œ i j 3# k and P(0ß 0ß 0) Ê x œ t,
y œ t, z œ 3# t, where 0 Ÿ t Ÿ 1
Ä
14. The direction PQ œ i and P(0ß 0ß 0) Ê x œ t, y œ 0, z œ 0,
where 0 Ÿ t Ÿ 1
Ä
15. The direction PQ œ j and P(1ß 1ß 0) Ê x œ 1, y œ 1 t,
z œ 0, where 1 Ÿ t Ÿ 0
Ä
16. The direction PQ œ k and P(1ß 1ß 0) Ê x œ 1, y œ 1, z œ t,
where 0 Ÿ t Ÿ 1
Ä
17. The direction PQ œ 2j and P(0ß 1ß 1) Ê x œ 0,
y œ 1 2t, z œ 1, where 0 Ÿ t Ÿ 1
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
735
736
Chapter 12 Vectors and the Geometry of Space
Ä
18. The direction PQ œ $i 2j and P(0ß 2ß 0) Ê x œ 3t,
y œ 2 2t, z œ 0, where 0 Ÿ t Ÿ 1
Ä
19. The direction PQ œ 2i 2j 2k and P(2ß 0ß 2)
Ê x œ 2 2t, y œ 2t, z œ 2 2t, where 0 Ÿ t Ÿ 1
Ä
20. The direction PQ œ i 3j k and P(1ß 0ß 1)
Ê x œ 1 t, y œ 3t, z œ 1 t, where 0 Ÿ t Ÿ 1
21. 3(x 0) (2)(y 2) (1)(z 1) œ 0 Ê 3x 2y z œ 3
22. 3(x 1) (1)(y 1) (1)(z 3) œ 0 Ê 3x y z œ 5
â
â
j kâ
â i
Ä
Ä
Ä
Ä
â
â
23. PQ œ i j 3k, PS œ i 3j 2k Ê PQ ‚ PS œ â 1 1 3 â œ 7i 5j 4k is normal to the plane
â
â
â 1 3 2 â
Ê 7(x 2) (5)(y 0) (4)(z 2) œ 0 Ê 7x 5y 4z œ 6
â
â
j kâ
â i
Ä
Ä
Ä
Ä
â
â
24. PQ œ i j 2k, PS œ 3i 2j 3k Ê PQ ‚ PS œ â 1 1 2 â œ i 3j k is normal to the plane
â
â
â 3 2 3 â
Ê (1)(x 1) (3)(y 5) (1)(z 7) œ 0 Ê x 3y z œ 9
25. n œ i 3j 4k, P(2ß 4ß 5) Ê (1)(x 2) (3)(y 4) (4)(z 5) œ 0 Ê x 3y 4z œ 34
26. n œ i 2j k, P(1ß 2ß 1) Ê (1)(x 1) (2)(y 2) (1)(z 1) œ 0 Ê x 2y z œ 6
27. œ
x œ 2t 1 œ s 2
2t s œ 1
4t 2s œ 2
Ê œ
Ê œ
Ê t œ 0 and s œ 1; then z œ 4t 3 œ 4s 1
y œ 3t 2 œ 2s 4
3t 2s œ 2
3t 2s œ 2
Ê 4(0) 3 œ (4)(1) 1 is satisfied Ê the lines intersect when t œ 0 and s œ 1 Ê the point of intersection is
x œ 1, y œ 2, and z œ 3 or P(1ß 2ß 3). A vector normal to the plane determined by these lines is
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 12.5 Lines and Planes in Space
737
â
â
âi j k â
â
â
n" ‚ n# œ â 2 3 4 â œ 20i 12j k, where n" and n# are directions of the lines Ê the plane
â
â
â 1 2 4 â
containing the lines is represented by(20)(x 1) (12)(y 2) (1)(z 3) œ 0 Ê 20x 12y z œ 7.
28. œ
œ 2s 2
xœ t
t 2s œ 2
Ê œ
Ê s œ 1 and t œ 0; then z œ t 1 œ 5s 6 Ê 0 1 œ 5(1) 6
t s œ 1
y œ t 2 œ s 3
is satisfied Ê the lines do intersect when s œ 1 and t œ 0 Ê the point of intersection is x œ 0, y œ 2 and z œ 1
â
â
j kâ
âi
â
â
or P(0ß 2ß 1). A vector normal to the plane determined by these lines is n" ‚ n# œ â 1 " 1 â œ 6i 3j 3k,
â
â
â2 1 5â
where n" and n# are directions of the lines Ê the plane containing the lines is represented by
(6)(x 0) (3)(y 2) (3)(z 1) œ 0 Ê 6x 3y 3z œ 3.
29. The cross product of i j k and 4i 2j 2k has the same direction as the normal to the plane
â
â
j k â
â i
â
â
Ê n œ â 1 " 1 â œ 6j 6k. Select a point on either line, such as P(1ß 2ß 1). Since the lines are given
â
â
â 4 2 2 â
to intersect, the desired plane is 0(x 1) 6(y 2) 6(z 1) œ 0 Ê 6y 6z œ 18 Ê y z œ 3.
30. The cross product of i 3j k and i j k has the same direction as the normal to the plane
â
â
j
k â
âi
â
â
n œ â 1 3 1 â œ 2i 2j 4k. Select a point on either line, such as P(0ß 3ß 2). Since the lines are
â
â
1 â
â1 1
given to intersect, the desired plane is (2)(x 0) (2)(y 3) (4)(z 2) œ 0 Ê 2x 2y 4z œ 14
Ê x y 2z œ 7.
â
âi j
â
31. n" ‚ n# œ â 2 1
â
â1 2
â
k â
â
1 ⠜ 3i 3j 3k is a vector in the direction of the line of intersection of the planes
â
1 â
Ê 3(x 2) (3)(y 1) 3(z 1) œ 0 Ê 3x 3y 3z œ 0 Ê x y z œ 0 is the desired plane containing
P! (2ß 1ß 1)
â
âi
Ä
â
32. A vector normal to the desired plane is P" P# ‚ n œ â 2
â
â4
â
j
k â
â
0 2 â œ 2i 12j 2k; choosing P" (1ß 2ß 3) as a point on
â
1 2 â
the plane Ê (2)(x 1) (12)(y 2) (2)(z 3) œ 0 Ê 2x 12y 2z œ 32 Ê x 6y z œ 16 is the
desired plane
â
âi
Ä
â
33. S(0ß 0ß 12), P(0ß 0ß 0) and v œ 4i 2j 2k Ê PS ‚ v œ â 0
â
â4
Ê dœ
Ä
¹PS‚v¹
kvk
È
j
0
2
â
kâ
â
12 â œ 24i 48j œ 24(i 2j)
â
2 â
È
5
È5 † 24 œ 2È30 is the distance from S to the line
œ È2416 1 44 4 œ 24
È24 œ
â
â
j
k â
â i
Ä
â
â
34. S(0ß 0ß 0), P(5ß 5ß 3) and v œ 3i 4j 5k Ê PS ‚ v œ â 5 5 3 â œ 13i 16j 5k
â
â
4 5 â
â 3
Ê dœ
Ä
¹PS‚v¹
kv k
œ
È169 256 25
È450
È9 œ 3 is the distance from S to the line
È9 16 25 œ È50 œ
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
738
Chapter 12 Vectors and the Geometry of Space
Ä
¹PS‚v¹
Ä
35. S(2ß 1ß 3), P(2ß 1ß 3) and v œ 2i 6j Ê PS ‚ v œ 0 Ê d œ kvk œ È040 œ 0 is the distance from S to the line
(i.e., the point S lies on the line)
â
âi
Ä
â
36. S(2ß 1ß 1), P(0ß 1ß 0) and v œ 2i 2j 2k Ê PS ‚ v œ â 2
â
â2
Ê dœ
Ä
¹PS‚v¹
kvk
œ
â
k â
â
1 ⠜ 2i 6j 4k
â
2 â
j
0
2
È4 36 16
È56
14
È4 4 4 œ È12 œ É 3 is the distance from S to the line
â
â
j kâ
â i
Ä
â
â
37. S(3ß 1ß 4), P(4ß 3ß 5) and v œ i 2j 3k Ê PS ‚ v œ â 1 4 9 â œ 30i 6j 6k
â
â
â 1 2 3 â
Ê dœ
Ä
¹PS‚v¹
kvk
œ
È900 36 36
È
È
È
È
œ È972
œ È486
œ È817†6 œ 9 742 is the distance from S to the line
È1 4 9
7
14
â
â i
Ä
â
38. S(1ß 4ß 3), P(10ß 3ß 0) and v œ 4i 4k Ê PS ‚ v œ â 11
â
â 4
Ê dœ
Ä
¹PS‚v¹
kv k
j
7
0
â
kâ
â
3 â œ 28i 56j 28k œ 28(i 2j k)
â
4â
È
œ 28 4È1 14 1 1 œ 7È3 is the distance from S to the line
Ä
39. S(2ß 3ß 4), x 2y 2z œ 13 and P(13ß 0ß 0) is on the plane Ê PS œ 11i 3j 4k and n œ i 2j 2k
Ä
9
Ê d œ ¹ PS † knnk ¹ œ ¹ È111 4648 ¹ œ ¹ È
¹œ3
9
Ä
40. S(0ß 0ß 0), 3x 2y 6z œ 6 and P(2ß 0ß 0) is on the plane Ê PS œ 2i and n œ 3i 2j 6k
Ä
Ê d œ ¹ PS † knnk ¹ œ ¹ È9 46 36 ¹ œ È649 œ 67
Ä
41. S(0ß 1ß 1), 4y 3z œ 12 and P(0ß 3ß 0) is on the plane Ê PS œ 4j k and n œ 4j 3k
Ä
Ê d œ ¹ PS † knnk ¹ œ ¹ È161639 ¹ œ 19
5
Ä
42. S(2ß 2ß 3), 2x y 2z œ 4 and P(2ß 0ß 0) is on the plane Ê PS œ 2j 3k and n œ 2i j 2k
Ä
Ê d œ ¹ PS † knnk ¹ œ ¹ È4216 4 ¹ œ 83
Ä
43. S(0ß 1ß 0), 2x y 2z œ 4 and P(2ß 0ß 0) is on the plane Ê PS œ 2i j and n œ 2i j 2k
Ä
4 1 0
Ê d œ ¹ PS † knnk ¹ œ ¹ È
¹ œ 53
414
Ä
44. S(1ß 0ß 1), 4x y z œ 4 and P(1ß 0ß 0) is on the plane Ê PS œ 2i k and n œ 4i j k
Ä
È
Ê d œ ¹ PS † knnk ¹ œ ¹ È16811 1 ¹ œ È918 œ 3 # 2
Ä
45. The point P(1ß 0ß 0) is on the first plane and S(10ß 0ß 0) is a point on the second plane Ê PS œ 9i, and
Ä
n œ i 2j 6k is normal to the first plane Ê the distance from S to the first plane is d œ ¹ PS † knnk ¹
œ ¹ È1 94 36 ¹ œ È9 , which is also the distance between the planes.
41
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 12.5 Lines and Planes in Space
739
46. The line is parallel to the plane since v † n œ ˆi j "# k‰ † (i 2j 6k) œ 1 2 3 œ 0. Also the point
Ä
S(1ß 0ß 0) when t œ 1 lies on the line, and the point P(10ß 0ß 0) lies on the plane Ê PS œ 9i. The distance from
Ä
S to the plane is d œ ¹ PS † knnk ¹ œ ¹ È1 49 36 ¹ œ È9 , which is also the distance from the line to the plane.
41
47. n" œ i j and n# œ 2i j 2k Ê ) œ cos" Š knn""k†knn## k ‹ œ cos" Š È22È1 9 ‹ œ cos" Š È"2 ‹ œ 14
523
48. n" œ 5i j k and n# œ i 2j 3k Ê ) œ cos" Š knn""k†knn## k ‹ œ cos" Š È
‹ œ cos" (0) œ 1#
27 È14
442
49. n" œ 2i 2j 2k and n# œ 2i 2j k Ê ) œ cos" Š knn""k†knn## k ‹ œ cos" Š È
‹ œ cos" Š 3"
È3 ‹ ¸ 1.76 rad
12 È9
50. n" œ i j k and n# œ k Ê ) œ cos" Š knn""k†knn## k ‹ œ cos" Š È31È1 ‹ ¸ 0.96 rad
5
51. n" œ 2i 2j k and n# œ i 2j k Ê ) œ cos" Š knn""k†knn## k ‹ œ cos" Š 2È94È61 ‹ œ cos" Š 3È
‹ ¸ 0.82 rad
6
18
26 ‰
52. n" œ 4j 3k and n# œ 3i 2j 6k Ê ) œ cos" Š knn""k†knn## k ‹ œ cos" Š È825È
¸ 0.73 rad
‹ œ cos" ˆ 35
49
53. 2x y 3z œ 6 Ê 2(1 t) (3t) 3(1 t) œ 6 Ê 2t 5 œ 6 Ê t œ "# Ê x œ #3 , y œ #3 and z œ "#
Ê ˆ 3# ß 3# ß #" ‰ is the point
41
54. 6x 3y 4z œ 12 Ê 6(2) 3(3 2t) 4(2 2t) œ 12 Ê 14t 29 œ 12 Ê t œ 41
14 Ê x œ 2, y œ 3 7 ,
20 27 ‰
ˆ
and z œ 2 41
7 Ê 2ß 7 ß 7 is the point
55. x y z œ 2 Ê (1 2t) (1 5t) (3t) œ 2 Ê 10t 2 œ 2 Ê t œ 0 Ê x œ 1, y œ 1 and z œ 0
Ê (1ß 1ß 0) is the point
56. 2x 3z œ 7 Ê 2(1 3t) 3(5t) œ 7 Ê 9t 2 œ 7 Ê t œ 1 Ê x œ 1 3, y œ 2 and z œ 5
Ê (4ß 2ß 5) is the point
â
âi
â
57. n" œ i j k and n# œ i j Ê n" ‚ n# œ â 1
â
â1
â
kâ
â
1 â œ i j, the direction of the desired line; (1ß 1ß 1)
â
0â
is on both planes Ê the desired line is x œ 1 t, y œ 1 t, z œ 1
j
1
1
â
â
j
k â
âi
â
â
58. n" œ 3i 6j 2k and n# œ 2i j 2k Ê n" ‚ n# œ â 3 6 2 â œ 14i 2j 15k, the direction of the
â
â
â 2 1 2 â
desired line; (1ß 0ß 0) is on both planes Ê the desired line is x œ 1 14t, y œ 2t, z œ 15t
â
â
j
k â
âi
â
â
59. n" œ i 2j 4k and n# œ i j 2k Ê n" ‚ n# œ â 1 2 4 â œ 6j 3k, the direction of the
â
â
â 1 1 2 â
desired line; (4ß 3ß 1) is on both planes Ê the desired line is x œ 4, y œ 3 6t, z œ 1 3t
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
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Chapter 12 Vectors and the Geometry of Space
â
â
j
k â
âi
â
â
60. n" œ 5i 2j and n# œ 4j 5k Ê n" ‚ n# œ â 5 2 0 â œ 10i 25j 20k, the direction of the
â
â
â 0 4 5 â
desired line; (1ß 3ß 1) is on both planes Ê the desired line is x œ 1 10t, y œ 3 25t, z œ 1 20t
61. L1 & L2: x œ 3 2t œ 1 4s and y œ 1 4t œ 1 2s Ê œ
2t 4s œ 2
2t 4s œ 2
Ê œ
4t 2s œ 2
2t s œ 1
Ê 3s œ 3 Ê s œ 1 and t œ 1 Ê on L1, z œ 1 and on L2, z œ 1 Ê L1 and L2 intersect at (5ß 3ß 1).
L2 & L3: The direction of L2 is "6 (4i 2j 4k) œ 3" (2i j 2k) which is the same as the direction
"
3 (2i j 2k) of L3; hence L2 and L3 are parallel.
L1 & L3: x œ 3 2t œ 3 2r and y œ 1 4t œ 2 r Ê œ
2t 2r œ 0
t rœ0
Ê œ
Ê 3t œ 3
4t r œ 3
4t r œ 3
Ê t œ 1 and r œ 1 Ê on L1, z œ 2 while on L3, z œ 0 Ê L1 and L2 do not intersect. The direction of L1
is È"21 (2i 4j k) while the direction of L3 is 3" (2i j 2k) and neither is a multiple of the other; hence
L1 and L3 are skew.
62. L1 & L2: x œ 1 2t œ 2 s and y œ 1 t œ 3s Ê œ
2t s œ 1
Ê 5s œ 3 Ê s œ 35 and t œ 45 Ê on L1,
t 3s œ 1
3
2
"
z œ 12
5 while on L2, z œ 1 5 œ 5 Ê L1 and L2 do not intersect. The direction of L1 is È (2i j 3k)
14
while the direction of L2 is È"11 (i 3j k) and neither is a multiple of the other; hence, L1 and L2 are
skew.
L2 & L3: x œ 2 s œ 5 2r and y œ 3s œ 1 r Ê œ
s 2r œ 3
Ê 5s œ 5 Ê s œ 1 and r œ 2 Ê on L2,
3s r œ 1
z œ 2 and on L3, z œ 2 Ê L2 and L3 intersect at (1ß 3ß 2).
L1 & L3: L1 and L3 have the same direction È"14 (2i j 3k); hence L1 and L3 are parallel.
63. x œ 2 2t, y œ 4 t, z œ 7 3t; x œ 2 t, y œ 2 "# t, z œ 1 3# t
64. 1(x 4) 2(y 1) 1(z 5) œ 0 Ê x 4 2y 2 z 5 œ 0 Ê x 2y z œ 7;
È2 (x 3) 2È2 (y 2) È2 (z 0) œ 0 Ê È2x 2È2y È2z œ 7È2
65. x œ 0 Ê t œ "# , y œ "# , z œ 3# Ê ˆ!ß "# ß 3# ‰ ; y œ 0 Ê t œ 1, x œ 1, z œ 3 Ê (1ß 0ß 3); z œ 0
Ê t œ 0, x œ 1, y œ 1 Ê (1ß 1ß 0)
66. The line contains (0ß 0ß 3) and ŠÈ3ß 1ß 3‹ because the projection of the line onto the xy-plane contains the origin
and intersects the positive x-axis at a 30° angle. The direction of the line is È3i j 0k Ê the line in question
is x œ È3t, y œ t, z œ 3.
67. With substitution of the line into the plane we have 2(1 2t) (2 5t) (3t) œ 8 Ê 2 4t 2 5t 3t œ 8
Ê 4t 4 œ 8 Ê t œ 1 Ê the point (1ß 7ß 3) is contained in both the line and plane, so they are not parallel.
68. The planes are parallel when either vector A" i B" j C" k or A# i B# j C# k is a multiple of the other or
when (A" i B" j C" k) ‚ aA# i B# j C# kb œ 0. The planes are perpendicular when their normals are
perpendicular, or(A" i B" j C" k) † (A# i B# j C# k) œ 0.
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 12.6 Cylinders and Quadric Surfaces
69. There are many possible answers. One is found as follows: eliminate t to get t œ x 1 œ 2 y œ z # 3
Ê x 1 œ 2 y and 2 y œ z # 3 Ê x y œ 3 and 2y z œ 7 are two such planes.
70. Since the plane passes through the origin, its general equation is of the form Ax By Cz œ 0. Since it meets
the plane M at a right angle, their normal vectors are perpendicular Ê 2A 3B C œ 0. One choice satisfying
this equation is A œ 1, B œ 1 and C œ 1 Ê x y z œ 0. Any plane Ax By Cz œ 0 with 2A 3B C œ 0
will pass through the origin and be perpendicular to M.
71. The points (aß 0ß 0), (0ß bß 0) and (0ß 0ß c) are the x, y, and z intercepts of the plane. Since a, b, and c are all
nonzero, the plane must intersect all three coordinate axes and cannot pass through the origin. Thus,
y
x
z
a b c œ 1 describes all planes except those through the origin or parallel to a coordinate axis.
72. Yes. If v" and v# are nonzero vectors parallel to the lines, then v" ‚ v# Á 0 is perpendicular to the lines.
Ä
Ä
73. (a) EP œ cEP" Ê x! i yj zk œ c c(x" x! )i y" j z" kd Ê x! œ c(x" x! ), y œ cy" and z œ cz" ,
where c is a positive real number
x !
(b) At x" œ 0 Ê c œ 1 Ê y œ y" and z œ z" ; at x" œ x! Ê x! œ 0, y œ 0, z œ 0; x lim
c œ x lim
Ä_
Ä _ x" x!
"
œ x lim
œ 1 Ê c Ä 1 so that y Ä y" and z Ä z"
Ä _ 1
!
!
!
74. The plane which contains the triangular plane is x y z œ 2. The line containing the endpoints of the line
segment is x œ 1 t, y œ 2t, z œ 2t. The plane and the line intersect at ˆ 23 ß 23 ß 23 ‰ . The visible section of the line
#
#
#
segment is Ɉ "3 ‰ ˆ 32 ‰ ˆ 32 ‰ œ 1 unit in length. The length of the line segment is È1# 2# 2# œ 3 Ê 23 of
the line segment is hidden from view.
12.6 CYLINDERS AND QUADRIC SURFACES
1. d, ellipsoid
2. i, hyperboloid
3. a, cylinder
4. g, cone
5. l, hyperbolic paraboloid
6. e, paraboloid
7. b, cylinder
8. j, hyperboloid
9. k, hyperbolic paraboloid
10. f, paraboloid
11. h, cone
12. c, ellipsoid
13. x# y# œ 4
14. z œ y# 1
15. x# 4z# œ 16
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Chapter 12 Vectors and the Geometry of Space
16. 4x# y# œ 36
17. 9x# y# z# œ 9
18. 4x# 4y# z# œ 16
19. 4x# 9y# 4z# œ 36
20. 9x# 4y# 36z# œ 36
21. x# 4y# œ z
22. z œ 8 x# y#
23. x œ 4 4y# z#
24. y œ 1 x# z#
25. x# y# œ z#
26. 4x# 9z# œ 9y#
27. x# y# z# œ 1
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 12.6 Cylinders and Quadric Surfaces
#
#
28. y# z# x# œ 1
29. z# x# y# œ 1
30. y4 x4 z# œ 1
31. y# x# œ z
32. x# y# œ z
33. z œ 1 y# x#
34. 4x# 4y# œ z#
35. y œ ax# z# b
36. 16x# 4y# œ 1
37. x# y# z# œ 4
38. x# z# œ y
39. x# z# œ 1
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
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Chapter 12 Vectors and the Geometry of Space
40. 16y# 9z# œ 4x#
41. z œ ax# y# b
43. 4y# z# 4x# œ 4
44. x# y# œ z
#
#
#
42. y# x# z# œ 1
#
45. (a) If x# y4 z9 œ 1 and z œ c, then x# y4 œ 99c Ê
œ 1Š
x#
Š
c#
9
9
‹
y#
’
4 ˆ9
c# ‰
9
“
œ 1 Ê A œ ab1
#
È 9 c#
È
#
‹ Š 2 93 c ‹ œ 21 a99 c b
3
(b) From part (a), each slice has the area 21 a99 z b , where 3 Ÿ z Ÿ 3. Thus V œ 2 '0 291 a9 z# b dz
3
#
œ 491 '0 a9 z# b dz œ 491 ’9z z3 “ œ 491 (27 9) œ 81
3
$
$
!
(c)
y#
x#
z#
a# b # c# œ 1
x#
Ê
a# Šc#
c#
–
z# ‹
y#
—
–
b# Šc#
c#
z# ‹
œ 1 Ê A œ 1 Ša
È c# z#
È #
#
‹ Š b cc z ‹
c
—
21ab
z
21ab ˆ 2 $ ‰
#
#
#
Ê V œ 2 '0 1cab
œ 413abc . Note that if r œ a œ b œ c,
# ac z b dz œ
c# ’c z 3 “ œ c#
3 c
c
$
$
then V œ 413r , which is the volume of a sphere.
#
#
c
!
#
46. The ellipsoid has the form Rx # Ry # zc# œ 1. To determine c# we note that the point (0ß rß h) lies on the surface
#
#
#
#
of the barrel. Thus, Rr # hc# œ 1 Ê c# œ Rh# Rr# . We calculate the volume by the disk method:
V œ 1 'ch y# dz. Now, Ry # cz# œ 1 Ê y# œ R# Š1 cz# ‹ œ R# ’1 z ahR# R# r b “ œ R# Š R h# r ‹ z#
h
#
#
#
#
Ê V œ 1 'ch ’R# Š R h# r ‹ z# “ dz œ 1 ’R# z "3 Š R h# r ‹ z$ “
h
#
#
#
#
h
ch
#
#
#
#
#
#
œ 21 R# h "3 aR# r# b h‘ œ 21 Š 2R3 h r3h ‹
œ 43 1R# h 23 1r# h, the volume of the barrel. If r œ R, then V œ 21R# h which is the volume of a cylinder of
radius R and height 2h. If r œ 0 and h œ R, then V œ 43 1R$ which is the volume of a sphere.
#
#
47. We calculate the volume by the slicing method, taking slices parallel to the xy-plane. For fixed z, xa# by# œ cz
gives the ellipse
x#
#
Š zac ‹
y#
#
Š zbc ‹
œ 1. The area of this ellipse is 1 ˆaÈ cz ‰ ˆbÈ cz ‰ œ 1abz
c (see Exercise 45a). Hence
1abz
1abh
the volume is given by V œ '0 1abz
c dz œ ’ 2c “ œ
c . Now the area of the elliptic base when z œ h is
h
A œ 1abh
c , as determined previously.
#
h
#
!
1abh#
‰ h œ "# (base)(altitude), as claimed.
Thus, V œ c œ "# ˆ 1abh
c
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 12.6 Cylinders and Quadric Surfaces
#
#
#
48. (a) For each fixed value of z, the hyperboloid xa# by# cz# œ 1 results in a cross-sectional ellipse
x#
–
a# Šc#
c#
z# ‹
y#
—
–
b# Šc#
z# ‹
c#
œ 1. The area of the cross-sectional ellipse (see Exercise 45a) is
—
#
#
A(z) œ 1 Š ac Èc# z# ‹ Š bc Èc# z# ‹ œ 1cab
# ac z b . The volume of the solid by the method of slices is
V œ '0 A(z) dz œ '0
h
(b)
h
" $‘ h
" $‰
1ab
1ab #
1ab ˆ #
#
#
#
#
œ 13cabh
# a3c h b
c# ac z b dz œ c# c z 3 z ! œ c# c h 3 h
1abh
#
#
#
#
A! œ A(0) œ 1ab and Ah œ A(h) œ 1cab
# ac h b , from part (a) Ê V œ 3c# a3c h b
#
#
#
h
1abh
c h
h 1ab
h
#
# ‘
œ 1abh
3 Š2 1 c # ‹ œ 3 Š 2 c# ‹ œ 3 21ab c# ac h b œ 3 (2A! Ah )
#
h
h
1ab
#
#
#
(c) Am œ A ˆ h# ‰ œ 1cab
# Šc 4 ‹ œ 4c# a4c h b Ê 6 (A! 4Am Ah )
1ab
1abh
1abh
#
#
#
# ‘
#
#
#
#
#
#
#
œ h6 1ab 1cab
# a4c h b c# ac h b œ 6c# ac 4c h c h b œ 6c# a6c 2h b
#
#
œ 1abh
3c# a3c h b œ V from part (a)
49. z œ y#
50. z œ 1 y#
51. z œ x# y#
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746
Chapter 12 Vectors and the Geometry of Space
52. z œ x# 2y#
(a)
(b)
(c)
(d)
53-58. Example CAS commands:
Maple:
with( plots );
eq := x^2/9 + y^2/36 = 1 - z^2/25;
implicitplot3d( eq, x=-3..3, y=-6..6, z=-5..5, scaling=constrained,
shading=zhue, axes=boxed, title="#89 (Section 11.6)" );
Mathematica: (functions and domains may vary):
In the following chapter, you will consider contours or level curves for surfaces in three dimensions. For the purposes of
plotting the functions of two variables expressed implicitly in this section, we will call upon the function ContourPlot3D.
To insert the stated function, write all terms on the same side of the equal sign and the default contour equating that
expression to zero will be plotted.
This built-in function requires the loading of a special graphics package.
<<Graphics`ContourPlot3D`
Clear[x, y, z]
ContourPlot3D[x2 /9 y2 /16 z2 /2 1, {x, 9, 9}, {y, 12, 12}, {z, 5, 5},
Axes Ä True, AxesLabel Ä {x, y, z}, Boxed Ä False,
PlotLabel Ä "Elliptic Hyperboloid of Two Sheets"]
Your identification of the plot may or may not be able to be done without considering the graph.
CHAPTER 12 PRACTICE EXERCISES
1. (a) 3 3, 4¡ 4 2, 5¡ œ 9 8, 12 20¡ œ 17, 32¡
(b) È172 322 œ È1313
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Chapter 12 Practice Exercises
2. (a)
3 2, 4 5¡ œ 1, 1¡
3. (a)
(b) Éa1b2 a1b2 œ È2
4. (a)
747
2a3b, 2a4b¡ œ 6, 8¡
(b) É62 a8b2 œ 10
5a2b, 5a5b¡ œ 10, 25¡
(b) É102 a25b2 œ È725 œ 5È29
5.
1
6 radians below the negative x-axis:
È
¢ #3 , "# £ [assuming counterclockwise].
È
6. ¢ #3 , "# £
7. 2Š È 21
4 12
‹a4i jb œ Š È817 i È217 j‹
8. 5
1
ˆ 3 i 54 j‰ œ 3i 4j
Ɉ 35 ‰2 ˆ 45 ‰2 5
9. length œ ¹È2i È2j¹ œ È2 2 œ 2, È2i È2j œ 2 Š È"2 i È"2 j‹ Ê the direction is È"2 i È"2 j
10. length œ ki jk œ È1 1 œ È2, i j œ È2 Š È"2 i È"2 j‹ Ê the direction is È"2 i È"2 j
11. t œ 12 Ê v œ (2 sin 12 )i ˆ2 cos 12 ‰ j œ 2i; length œ k2ik œ È4 0 œ 2; 2i œ 2aib Ê the direction is i
12. t œ ln 2 Ê v œ ˆeln 2 cosaln 2b eln 2 sinaln 2b‰ i ˆeln 2 sinaln 2b eln 2 cosaln 2b‰ j
œ a2 cosaln 2b 2 sinaln 2bb i a2 sinaln 2b 2 cosaln 2bb j œ 2c acosaln 2b sinaln 2bb i asinaln 2b cosaln 2bb j d
length œ k2c acosaln 2b sinaln 2bb i asinaln 2b cosaln 2bb j dk œ 2Éacosaln 2b sinaln 2bb2 acosaln 2b sinaln 2bb2
œ 2È2cos2 aln 2b 2sin2 aln 2b œ 2È2;
2c acosaln 2b sinaln 2bb i asinaln 2b cosaln 2bb j d œ 2È2Š acosaln 2b sinaln 2bb iÈ asinaln 2b cosaln 2bb j ‹
2
cosaln 2bb
Ê direction œ acosaln 2bÈ sinaln 2bb i asinaln 2bÈ
j
2
2
13. length œ k2i 3j 6kk œ È4 9 36 œ 7, 2i 3j 6k œ 7 ˆ 27 i 37 j 67 k‰ Ê the direction is 27 i 37 j 67 k
14. length œ ki 2j kk œ È1 4 1 œ È6, i 2j k œ È6 Š È16 i È26 j È16 k‹ Ê the direction is
1
2
1
È6 i È6 j È6 k
j 4k
4k
15. 2 kvvk œ 2 † È44# i(
œ 2 † 4i Èj33
œ È833 i È233 j È833 k
1)# 4#
16. 5 kvvk œ 5 †
ˆ 35 ‰ i ˆ 45 ‰ k
Ɉ 35 ‰# ˆ 45 ‰#
œ 5 †
ˆ 35 ‰ iˆ 45 ‰k
9
16
É 25
25
œ 3i 4k
â
âi
â
17. kvk œ È1 1 œ È2, kuk œ È4 1 4 œ 3, v † u œ 3, u † v œ 3, v ‚ u œ â 1
â
â2
j
1
1
â
k â
â
0 ⠜ 2i 2j k ,
â
2 â
u ‚ v œ (v ‚ u) œ 2i 2j k, kv ‚ uk œ È4 4 1 œ 3, ) œ cos" Š kvvk†kuuk ‹ œ cos" Š È" ‹ œ 14 ,
2
kuk cos ) œ È32 , projv u œ Š kvvkk†uvk ‹ v œ 32 (i j)
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Chapter 12 Vectors and the Geometry of Space
18. kvk œ È1# 1# 2# œ È6, kuk œ È(1)# (1)# œ È2, v † u œ (1)(1) (1)(0) (2)(1) œ 3,
â
â
j k â
â i
â
â
u † v œ 3, v ‚ u œ â 1 1 2 â œ i j k , u ‚ v œ (v ‚ u) œ i j k,
â
â
â 1 0 1 â
3
3
kv ‚ uk œ È(1)# (1)# 1# œ È3, ) œ cos" Š kvvk†kuuk ‹ œ cos" Š È
‹ œ cos" Š È12
‹
6 È2
È
È
È
œ cos" Š #3 ‹ œ 561 , kuk cos ) œ È2 † Š 2 3 ‹ œ 26 , projv u œ Š kvvkk†uvk ‹ v œ 63 (i j 2k) œ "# (i j k)
19. projv u œ Š kvvkk†uvk ‹ v œ 43 a2i j kb where v † u œ 8 and v † v œ 6
20. projv u œ Š kvvkk†uvk ‹ v œ 13 ai 2jb where v † u œ 1 and v † v œ 3
â
âi
â
21. u ‚ v œ â 1
â
â1
j
0
1
â
kâ
â
0⠜ k
â
0â
â
â
j kâ
âi
â
â
22. u ‚ v œ â 1 1 0 â œ 2k
â
â
â1 1 0â
23. Let v œ v" i v# j v$ k and w œ w" i w# j w$ k. Then kv 2wk # œ k(v" i v# j v$ k) 2(w" i w# j w$ k)k #
œ k(v" 2w" )i (v# 2w# )j (v$ 2w$ )kk # œ ˆÈ(v" 2w" )# (v# 2w# )# (v$ 2w$ )# ‰
#
œ av#" v## v3# b 4(v" w" v# w# v$ w$ ) 4 aw#1 w## w#3 b œ kvk # 4v † w 4 kwk #
œ kvk # 4 kvk kwk cos ) 4 kwk # œ 4 4(2)(3) ˆcos 13 ‰ 36 œ 40 24 ˆ "# ‰ œ 40 12 œ 28 Ê kv 2wk œ È28
œ 2È 7
â
j
â i
â
2
4
24. u and v are parallel when u ‚ v œ 0 Ê â
â
â 4 8
Ê 4a 40 œ 0 and 20 2a œ 0 Ê a œ 10
â
k â
â
5 â œ 0 Ê (4a 40)i (20 2a)j (0)k œ 0
â
a â
â
â
âi j k â
â
â
25. (a) area œ ku ‚ vk œ abs â 1 1 1 â œ k2i 3j kk œ È4 9 1 œ È14
â
â
â2 1 1 â
â
â
1 1 â
â 1
â
â
1
1 â œ 1a3 2b 1a6 a1bb 1a4 1b œ 1
(b) volume œ au ‚ vb † w œ â 2
â
â
â 1 2 3 â
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Chapter 12 Practice Exercises
â
â
â i j kâ
â
â
26. (a) area œ ku ‚ vk œ abs â 1 1 0 â œ kkk œ 1
â
â
â0 1 0â
â
â
â1 1 0â
â
â
(b) volume œ au ‚ vb † w œ â 0 1 0 â œ 1a1 0b 1a0 0b 0 œ 1
â
â
â1 1 1â
27. The desired vector is n ‚ v or v ‚ n since n ‚ v is perpendicular to both n and v and, therefore, also parallel to
the plane.
28. If a œ 0 and b Á 0, then the line by œ c and i are parallel. If a Á 0 and b œ 0, then the line ax œ c and j are
parallel. If a and b are both Á 0, then ax by œ c contains the points ˆ ca ß !‰ and ˆ0ß bc ‰ Ê the vector
ab ˆ ca i bc j‰ œ c(bi aj) and the line are parallel. Therefore, the vector bi aj is parallel to the line
ax by œ c in every case.
Ä
29. The line L passes through the point Pa0ß 0ß 1b parallel to v œ i j k. With PS œ 2i 2j k and
â
â
j kâ
â i
Ä
â
â
PS ‚ v œ â 2 2 1 â œ a2 1bi (2 1)j a2 2bk œ i 3j 4k, we find the distance
â
â
â 1 1 1 â
dœ
Ä
¹PS‚v¹
kv k
È
È
œ È119116
œ È26
œ
1
3
È78
3 .
Ä
30. The line L passes through the point Pa2ß 2ß 0b parallel to v œ i j k. With PS œ 2i 2j k and
â
â
j kâ
â i
Ä
â
â
PS ‚ v œ â 2 2 1 â œ a2 1bi a2 1bj a2 2bk œ i 3j 4k, we find the distance
â
â
â 1 1 1â
dœ
Ä
¹PS‚v¹
kv k
È
È
œ È119116
œ È26
œ
1
3
È78
3 .
31. Parametric equations for the line are x œ 1 3t, y œ 2, z œ 3 7t.
Ä
32. The line is parallel to PQ œ 0i j k and contains the point P(1ß 2ß 0) Ê parametric equations are
x œ 1, y œ 2 t, z œ t for 0 Ÿ t Ÿ 1.
Ä
33. The point P(4ß 0ß 0) lies on the plane x y œ 4, and PS œ (6 4)i 0j (6 0)k œ 2i 6k with n œ i j
Ê dœ
Ä
¹n†PS¹
knk
œ ¹ È210100 ¹ œ È22 œ È2.
Ä
34. The point P(0ß 0ß 2) lies on the plane 2x 3y z œ 2, and PS œ (3 0)i (0 0)j (10 2)k œ 3i 8k with
n œ 2i 3j k Ê d œ
Ä
¹n†PS¹
knk
œ ¹ È640981 ¹ œ È1414 œ È14.
35. P(3ß 2ß 1) and n œ 2i j k Ê (2)(x 3) (1)(y (2)) (1)(z 1) œ 0 Ê 2x y z œ 5
36. P(1ß 6ß 0) and n œ i 2j 3k Ê (1)(x (1)) (2)(y 6) (3)(z 0) œ 0 Ê x 2y 3z œ 13
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Chapter 12 Vectors and the Geometry of Space
Ä
Ä
Ä
Ä
37. P(1ß 1ß 2), Q(2ß 1ß 3) and R(1ß 2ß 1) Ê PQ œ i 2j k, PR œ 2i 3j 3k and PQ ‚ PR
â
â
j k â
â i
â
â
œ â 1 2 1 â œ 9i j 7k is normal to the plane Ê (9)(x 1) (1)(y 1) (7)(z 2) œ 0
â
â
â 2 3 3 â
Ê 9x y 7z œ 4
Ä
Ä
Ä
Ä
38. P(1ß 0ß 0), Q(0ß 1ß 0) and R(0ß 0ß 1) Ê PQ œ i j, PR œ i k and PQ ‚ PR
â
â
j kâ
â i
â
â
œ â 1 1 0 â œ i j k is normal to the plane Ê (1)(x 1) (1)(y 0) (1)(z 0) œ 0
â
â
â 1 0 1 â
Ê xyzœ1
39. ˆ0ß "# ß 3# ‰ , since t œ "# , y œ "# and z œ 3# when x œ 0; ("ß 0ß 3), since t œ 1, x œ 1 and z œ 3
when y œ 0; (1ß 1ß 0), since t œ 0, x œ 1 and y œ 1 when z œ 0
40. x œ 2t, y œ t, z œ t represents a line containing the origin and perpendicular to the plane 2x y z œ 4; this
line intersects the plane 3x 5y 2z œ 6 when t is the solution of 3(2t) 5(t) 2(t) œ 6
Ê t œ 23 Ê ˆ 43 ß 23 ß 23 ‰ is the point of intersection
41. n" œ i and n# œ i j È2k Ê the desired angle is cos" Š knn""k†knn## k ‹ œ cos" ˆ "# ‰ œ 13
42. n" œ i j and n# œ j k Ê the desired angle is cos" Š knn""k†knn## k ‹ œ cos" ˆ "# ‰ œ 13
â
âi
â
43. The direction of the line is n" ‚ n# œ â 1
â
â1
â
j kâ
â
2 1 â œ 5i j 3k. Since the point (5ß 3ß 0) is on
â
1 2 â
both planes, the desired line is x œ 5 5t, y œ 3 t, z œ 3t.
â
â
j
k â
âi
â
â
44. The direction of the intersection is n" ‚ n# œ â 1 2 2 â œ 6i 9j 12k œ 3(2i 3j 4k) and is the
â
â
â 5 2 1 â
same as the direction of the given line.
45. (a) The corresponding normals are n" œ 3i 'k and n# œ 2i 2j k and since n" † n#
œ (3)(2) (0)(2) (6)(1) œ 6 0 6 œ 0, we have that the planes are orthogonal
â
â
âi j k â
â
â
(b) The line of intersection is parallel to n" ‚ n# œ â 3 0 6 â œ 12i 15j 6k. Now to find a point in
â
â
â 2 2 1 â
the intersection, solve œ
3x 6z œ 1
3x 6z œ 1
Ê œ
Ê 15x 12y œ 19 Ê x œ 0 and y œ 19
12
2x 2y z œ 3
12x 12y 6z œ 18
"‰
19
"
Ê ˆ!ß 19
1# ß 6 is a point on the line we seek. Therefore, the line is x œ 12t, y œ 12 15t and z œ 6 6t.
â
âi
â
46. A vector in the direction of the plane's normal is n œ u ‚ v œ â 2
â
â1
â
j kâ
â
3 1 â œ 7i 3j 5k and P("ß 2ß 3) on
â
1 2 â
the plane Ê 7(x 1) 3(y 2) 5(z 3) œ 0 Ê 7x 3y 5z œ 14.
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Chapter 12 Practice Exercises
751
47. Yes; v † n œ (2i 4j k) † (2i j 0k) œ 2 † 2 4 † 1 1 † 0 œ 0 Ê the vector is orthogonal to the plane's normal
Ê v is parallel to the plane
Ä
48. n † PP! 0 represents the half-space of points lying on one side of the plane in the direction which the normal n points
â
âi
Ä
Ä
â
49. A normal to the plane is n œ AB ‚ AC œ â 2
â
â2
j
0
1
â
k â
Ä
â
1 â œ i 2j 2k Ê the distance is d œ ¹ APn†n ¹
â
0 â
j)†(i 2j #k)
œ ¹ (i 4È
¹ œ ¸ 1 38 0 ¸ œ 3
144
Ä
50. P(0ß 0ß 0) lies on the plane 2x 3y 5z œ 0, and PS œ 2i 2j 3k with n œ 2i 3j 5k Ê
Ä
4 6 15
25
d œ ¹ nk†PS
nk ¹ œ ¹ È4 9 25 ¹ œ È38
â
âi
â
51. n œ 2i j k is normal to the plane Ê n ‚ v œ â 2
â
â1
j
1
1
â
k â
â
1 â œ 0i 3j 3k œ 3j 3k is orthogonal
â
1 â
to v and parallel to the plane
52. The vector B ‚ C is normal to the plane of B and C Ê A ‚ (B ‚ C) is orthogonal to A and parallel to the plane of B
and C:
â
â
â
â
j
k â
âi j k â
â i
â
â
â
â
B ‚ C œ â 1 2 1 â œ 5i 3j k and A ‚ (B ‚ C) œ â 2 1 1 â œ 2i 3j k
â
â
â
â
â 1 1 2 â
â 5 3 1 â
Ê kA ‚ (B ‚ C)k œ È4 9 1 œ È14 and u œ " (2i 3j k) is the desired unit vector.
È14
â
âi
â
53. A vector parallel to the line of intersection is v œ n" ‚ n# œ â 1
â
â1
j
2
1
â
kâ
â
1 ⠜ 5i j 3 k
â
2â
Ê kvk œ È25 1 9 œ È35 Ê 2 Š kvvk ‹ œ È235 (5i j 3k) is the desired vector.
54. The line containing (0ß 0ß 0) normal to the plane is represented by x œ 2t, y œ t, and z œ t. This line
intersects the plane 3x 5y 2z œ 6 when 3(2t) 5(t) 2(t) œ 6 Ê t œ 23 Ê the point is ˆ 43 ß 23 ß 23 ‰ .
55. The line is represented by x œ 3 2t, y œ 2 t, and z œ 1 2t. It meets the plane 2x y 2z œ 2 when
26
7‰
2(3 2t) (2 t) 2(" 2t) œ 2 Ê t œ 89 Ê the point is ˆ 11
9 ß 9 ß 9 .
â
âi
â
56. The direction of the intersection is v œ n" ‚ n# œ â 2
â
â1
j
1
1
â
k â
â
1 â œ 3i 5j k Ê ) œ cos" Š kvvk†kiik ‹
â
2 â
œ cos" Š È335 ‹ ¸ 59.5°
57. The intersection occurs when (3 2t) 3(2t) t œ 4 Ê t œ 1 Ê the point is (1ß 2ß 1). The required line
â
â
âi j k â
â
â
must be perpendicular to both the given line and to the normal, and hence is parallel to â 2 2 1 â
â
â
â 1 3 1 â
œ 5i 3j 4k Ê the line is represented by x œ 1 5t, y œ 2 3t, and z œ 1 4t.
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Chapter 12 Vectors and the Geometry of Space
58. If P(aß bß c) is a point on the line of intersection, then P lies in both planes Ê a 2b c 3 œ 0 and
2a b c 1 œ 0 Ê (a 2b c 3) k(2a b c 1) œ 0 for all k.
â i
j
k ââ
â
Ä
Ä
â 3 2
4 ââ œ 26
i
j
k
59. The vector AB ‚ CD œ â
5 (2 7 2 ) is normal to the plane and A(2ß 0ß 3) lies on the
â 26
26 â
0
â 5
5 â
plane Ê 2(x 2) 7(y 0) 2(z (3)) œ 0 Ê 2x 7y 2z 10 œ 0 is an equation of the plane.
60. Yes; the line's direction vector is 2i 3j 5k which is parallel to the line and also parallel to the normal
4i 6j 10k to the plane Ê the line is orthogonal to the plane.
â
j
â i
Ä
Ä
â
61. The vector PQ ‚ PR œ â 2 1
â
â 3 0
â
kâ
â
3 ⠜ i 11j 3k is normal to the plane.
â
1â
Ä
Ä
(a) No, the plane is not orthogonal to PQ ‚ PR .
(b) No, these equations represent a line, not a plane.
Ä
Ä
(c) No, the plane (x 2) 11(y 1) 3z œ 0 has normal i 11j 3k which is not parallel to PQ ‚ PR .
(d) No, this vector equation is equivalent to the equations 3y 3z œ 3, 3x 2z œ 6, and 3x 2y œ 4
Ê x œ 43 23 t, y œ t, z œ 1 t, which represents a line, not a plane.
Ä
Ä
(e) Yes, this is a plane containing the point R(2ß 1ß 0) with normal PQ ‚ PR .
62. (a) The line through A and B is x œ 1 t, y œ t, z œ 1 5t; the line through C and D must be parallel and
is L" : x œ 1 t, y œ 2 t, z œ 3 5t. The line through B and C is x œ 1, y œ 2 2s, z œ 3 4s; the line
through A and D must be parallel and is L# : x œ 2, y œ 1 2s, z œ 4 4s. The lines L" and L# intersect
at D(2ß 1ß 8) where t œ 1 and s œ 1.
4k)†(i j 5k)
(b) cos ) œ (2j È
œ È315
20 È27
Ä Ä
Ä
Ä
Ä
18 Ä
9
†BC
(c) Š BA
Ä Ä ‹ BC œ 20 BC œ 5 (j 2k) where BA œ i j 5k and BC œ 2j 4k
BC†BC
(d) area œ k(2j 4k) ‚ (i j 5k)k œ k14i 4j 2kk œ 6È6
(e) From part (d), n œ 14i 4j 2k is normal to the plane Ê 14(x 1) 4(y 0) 2(z 1) œ 0
Ê 7x 2y z œ 8.
(f) From part (d), n œ 14i 4j 2k Ê the area of the projection on the yz-plane is kn † ik œ 14; the area of the
projection on the xy-plane is kn † jk œ 4; and the area of the projection on the xy-plane is kn † kk œ 2.
â
â i
Ä
Ä
Ä
â
63. AB œ 2i j k, CD œ i 4j k, and AC œ 2i j Ê n œ â 2
â
â 1
â
j k â
â
1 1 â œ 5i j 9k Ê the distance is
â
4 1 â
†(5i j 9k)
d œ ¹ (2i Èj)25
¹ œ È11
1 81
107
â
â i
Ä
Ä
Ä
â
64. AB œ 2i 4j k, CD œ i j 2k, and AC œ 3i 3j Ê n œ â 2
â
â 1
j
4
1
â
k â
â
1 â œ 7i 3j 2k Ê the distance
â
2 â
j)†(7i 3j 2k)
is d œ ¹ (3i È349
¹ œ È1262
94
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Chapter 12 Practice Exercises
65. x# y# z# œ 4
66. x# (y 1)# z# œ 1
67. 4x# 4y# z# œ 4
68. 36x# 9y# 4z# œ 36
69. z œ ax# y# b
70. y œ ax# z# b
71. x# y# œ z#
72. x# z# œ y#
73. x# y# z# œ 4
74. 4y# z# 4x# œ 4
75. y# x# z# œ 1
76. z# x# y# œ 1
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754
Chapter 12 Vectors and the Geometry of Space
CHAPTER 12 ADDITIONAL AND ADVANCED EXERCISES
1. Information from ship A indicates the submarine is now on the line L" : x œ 4 2t, y œ 3t, z œ "3 t; information from
ship B indicates the submarine is now on the line L# : x œ 18s, y œ 5 6s, z œ s. The current position of the sub is
ˆ6ß 3ß "3 ‰ and occurs when the lines intersect at t œ 1 and s œ 3" . The straight line path of the submarine contains both
points P ˆ2ß 1ß "3 ‰ and Q ˆ6ß 3ß 3" ‰ ; the line representing this path is L: x œ 2 4t, y œ 1 4t, z œ "3 . The
submarine traveled the distance between P and Q in 4 minutes Ê a speed of
minutes the submarine will move 20È2 thousand ft from Q along the line L
Ä
¹PQ¹
4
œ
È32
È2 thousand ft/min.
4 œ
In 20
Ê 20È2 œ È(2 4t 6)# (1 4t 3)# 0# Ê 800 œ 16(t 1)# 16(t 1)# œ 32(t 1)# Ê (t 1)# œ 800
32
œ 25 Ê t œ 6 Ê the submarine will be located at ˆ26ß 23ß "3 ‰ in 20 minutes.
2. H# stops its flight when 6 110t œ 446 Ê t œ 4 hours. After 6 hours, H" is at P(246ß 57ß 9) while H# is at (446ß 13ß 0).
The distance between P and Q is È(246 446)# (57 13)# (9 0)# ¸ 204.98 miles. At 150 mph, it would take
about 1.37 hours for H" to reach H# .
Ä
Ä
3. Torque œ ¹ PQ ‚ F¹ Ê 15 ft-lb œ ¹ PQ ¹ kFk sin 1# œ 34 ft † kFk Ê kFk œ 20 lb
4. Let a œ i j k be the vector from O to A and b œ i 3j 2k be the vector from O to B. The vector v orthogonal to a
and b Ê v is parallel to b ‚ a (since the rotation is clockwise). Now b ‚ a œ i j 2k; proja b œ ˆ aa††ba ‰a œ 2i 2j 2k
Ê a2, 2, 2b is the center of the circular path a1, 3, 2b takes Ê radius œ É12 a1b2 02 œ È2 Ê arc length per
a
second covered by the point is 3# È2 units/sec œ kvk (velocity is constant). A unit vector in the direction of v is kbb‚
‚a k
È3
È3
È 3k
2 i 2 j
a
3È
œ È"6 i È"6 j È26 k Ê v œ kvkŠ kbb‚
2Š È"6 i È"6 j È26 k‹ œ
‚a k ‹ œ #
4
3
5. (a) By the Law of Cosines we have cos ! œ 3 2a35ba
œ 53 and cos " œ 4 2a45ba
œ 54 Ê sin ! œ 45 and sin " œ 35
5b
5b
2
2
2
2
2
2
Ê F1 œ ¢lF1 lcos !, lF1 lsin !£ œ ¢ 35 lF1 l, 45 lF1 l£, F2 œ ¢lF2 lcos " , lF2 lsin " £ œ ¢ 45 lF2 l, 35 lF2 l£, and
w œ 0, 100¡. Since F1 F2 œ 0, 100¡ Ê ¢ 35 lF1 l 45 lF2 l, 45 lF1 l 35 lF2 l£ œ 0, 100¡ Ê 35 lF1 l 45 lF2 l œ 0
and 45 lF1 l 35 lF2 l œ 100. Solving the first equation for lF2 l results in: lF2 l œ 34 lF1 l. Substituting this result into the
9
second equation gives us: 45 lF1 l 20
lF1 l œ 100 Ê lF1 l œ 80 lb. Ê lF2 l œ 60 lb. Ê F1 œ 48, 64¡ and
F2 œ 48, 36¡ , and ! œ tan1 ˆ 43 ‰ and " œ tan1 ˆ 34 ‰
12
5
12
5
(b) By the Law of Cosines we have cos ! œ 5 2a13
œ 13
and cos " œ 122a1213ba13b 5 œ 13
Ê sin ! œ 12
5ba13b
13 and sin " œ 13
2
2
2
2
2
2
5
12
5
Ê F1 œ ¢lF1 lcos !, lF1 lsin !£ œ ¢ 13
lF1 l, 12
13 lF1 l£, F2 œ ¢lF2 lcos " , lF2 lsin " £ œ ¢ 13 lF2 l, 13 lF2 l£, and
5
12
5
¡
w œ 0, 200¡. Since F1 F2 œ 0, 200¡ Ê ¢ 13
lF1 l 13
lF2 l, 12
13 lF1 l 13 lF2 l£ œ 0, 200
5
12
5
5
Ê 13
lF1 l 13
lF2 l œ 0 and 12
13 lF1 l 13 lF2 l œ 200. Solving the first equation for lF2 l results in: lF2 l œ 12 lF1 l.
25
2400
Substituting this result into the second equation gives us: 12
13 lF1 l 156 lF1 l œ 200 Ê lF1 l œ 13 ¸ 184.615 lb.
12000 28800
12000 5000
¡
Ê lF2 l œ 1000
13 ¸ 76.923 lb. Ê F1 œ ¢ 1169 , 1169 £ ¸ 71.006, 170.414 and F2 œ ¢ 1169 , 1169 £
¸ 71.006, 29.586¡.
6. (a) T1 œ ¢lT1 lcos !, lT1 lsin !£, T2 œ ¢lT2 lcos " , lT2 lsin " £, and w œ 0, w¡. Since T1 T2 œ 0, w¡ Ê
¢lT1 lcos ! lT2 lcos " , lT1 lsin ! lT2 lsin " £ œ 0, w¡ Ê lT1 lcos ! lT2 lcos " œ 0 and
!
lT1 lsin ! lT2 lsin " œ w. Solving the first equation for lT2 l results in: lT2 l œ cos
cos " lT1 l. Substituting this result into
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Chapter 12 Additional and Advanced Exercises
755
w cos "
w cos "
! sin "
the second equation gives us: lT1 lsin ! coscos
" lT1 l œ w Ê lT1 l œ sin ! cos " cos ! sin " œ sin a! " b and
!
lT2 l œ sinwacos
! "b
(b)
w cos " cos a! " b d
w cos "
d
d
; d! alT1 lb œ 0 Ê w cos " cos a! " b œ 0 Ê cos a! " b œ 0
d! alT1 lb œ d! Š sin a! " b ‹ œ
sin2 a! " b
" cos a! " b
Ê ! " œ 12 Ê ! œ 12 " ; dd!2 alT1 lb œ dd! Š w cos
‹œ
sin2 a! " b
2
w cos " ˆcos2 a! " b 1‰
;
sin3 a! " b
d2
œ w cos " 0 Ê local minimum when ! œ 12 "
d!2 alT1 lbº
!œ 12 "
(c)
w cos ! cos a! " b d
d
d
w cos !
; d" alT2 lb œ 0 Ê w cos ! cos a! " b œ 0 Ê cos a! " b œ 0
d" alT2 lb œ d" Š sin a! " b ‹ œ
sin2 a! " b
! cos a! " b
Ê ! " œ 12 Ê " œ 12 !; dd"2 alT2 lb œ dd" Š w cos
‹œ
sin2 a! " b
2
w cos ! ˆcos2 a! " b 1‰
;
sin3 a! " b
d2
œ w cos ! 0 Ê local minimum when " œ 12 !
d!2 alT2 lbº
" œ 12 !
7. (a) If P(xß yß z) is a point in the plane determined by the three points P" (x" ß y" ß z" ), P# (x# ß y# ß z# ) and
Ä Ä
Ä
Ä
Ä
Ä
P$ (x$ ß y$ ß z$ ), then the vectors PP" , PP# and PP$ all lie in the plane. Thus PP" † (PP# ‚ PP$ ) œ 0
â
â
â x" x y" y z" z â
â
â
Ê â x# x y# y z# z â œ 0 by the determinant formula for the triple scalar product in Section 12.4.
â
â
â x$ x y$ y z$ z â
(b) Subtract row 1 from rows 2, 3, and 4 and evaluate the resulting determinant (which has the same value
as the given determinant) by cofactor expansion about column 4. This expansion is exactly the
determinant in part (a) so we have all points P(xß yß z) in the plane determined by P" (x" ß y" ß z" ),
P# (x# ß y# ß z# ), and P$ (x$ ß y$ ß z$ ).
8. Let L" : x œ a" s b" , y œ a# s b# , z œ a$ s b$ and L# : x œ c" t d" , y œ c# t d# , z œ c$ t d$ . If L" ² L# ,
â
â â
â
â a" c" b" d" â â kc" c" b" d" â
â
â â
â
then for some k, ai œ kci , i œ 1, 2, 3 and the determinant â a# c# b# d# â œ â kc# c# b# d# â œ 0,
â
â â
â
â a$ c$ b$ d$ â â kc$ c$ b$ d$ â
since the first column is a multiple of the second column. The lines L" and L# intersect if and only if the
Ú a s c t (b d ) œ 0
"
"
"
"
system Û a# s c# t (b# d# ) œ 0 has a nontrivial solution Í the determinant of the coefficients is zero.
Ü a$ s c$ t (b$ d$ ) œ 0
9. (a) Place the tetrahedron so that A is at a0, 0, 0b, the point P is on the y-axis, and ˜ABC lies in the xy-plane. Since
˜ABC is an equilateral triangle, all the angles in the triangle are 60‰ and since AP bisects BC Ê ˜ABP
is a 30‰ - 60‰ - 90‰ trinagle. Thus the coordinates of P are Š0, È3, 0‹, the coordinates of B are Š1, È3, 0‹, and the
coordinates of C are Š1, È3, 0‹. Let the coordinates of D be given by aa, b, cb. Since all of the faces are equilateral
Ä Ä
È
ab 3
†AB
1
trinagles Ê all the angles in each of the triangles are 60‰ Ê cosanDABb œ cosa60‰ b œ AD
Ä Ä œ a2ba2b œ 2
Ä Ä
lADllABl
È
a b 3
†AC
1
È
Ê a bÈ3 œ 2 and cosanDACb œ cosa60‰ b œ AD
Ä Ä œ a2ba2b œ 2 Ê a b 3 œ 2. Add the two equations
lADllACl
to obtain: 2bÈ3 œ 4 Ê b œ È23 . Substituting this value for b in the first equation gives us: a Š È23 ‹È3 œ 2
2
Ä
È
Ê a œ 0. Since lAD l œ Èa2 b2 c2 œ 2 Ê 02 Š È23 ‹ c2 œ 4 Ê c œ 2È32 . Thus the coordinates of D are
È
Ä Ä
†AP
2
1
1
‰
Š0, È23 , 2È32 ‹. cos ) œ cosanDAPb œ AD
Ä Ä œ È Ê ) œ cos Š È ‹ Ê 57.74 .
2 3
3
lADllAPl
(b) Since ˜ABC lies in the xy-plane Ê the normal to the face given by ˜ABC is n1 œ k. The face given by ˜BCD is an
Ä
Ä
È
È
adjacent face. The vectors DB œ i È13 j 2È32 k and DC œ i È13 j 2È32 k both lie in the plane containing
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
756
Chapter 12 Vectors and the Geometry of Space
â i
â
â
â
˜BCD. The normal to this plane is given by n2 œ â 1
â
â 1
â
j
1
È3
1
È3
adjacent faces is given by cos ) œ cosanDAPb œ lnn11ll†nn22 l œ
k
â
â
È â
2È32 ââ œ 4È2 j 2 k . The angle ) between two
È3
È3
â
2È 2 â
È3 â
2/È3
Ê ) œ cos1 ˆ 13 ‰ ¸ 70.53‰ .
a1bŠ6/È3‹
Ä
Ä
Ä
Ä
Ä
Ä
Ä
Ä
Ä
Ä
Ä
10. Extend CD to CG so that CD œ DG. Then CG œ t CF œ CB BG and t CF œ 3 CE CA, since ACBG is a
Ä
Ä
Ä
parallelogram. If t CF 3 CE CA œ 0, then t 3 1 œ 0 Ê t œ 4, since F, E, and A are collinear.
Ä
Ä
Ä
Ä
Therefore, CG œ 4 CF Ê CD œ 2 CF Ê F is the midpoint of CD.
Ä
11. If Q(xß y) is a point on the line ax by œ c, then P" Q œ (x x" )i (y y" )j , and n œ ai bj is normal to the
Ä
x" ) b(y y" )k
line. The distance is ¹projn P" Q¹ œ ¹ [(x x" )i È(y# y"#)j]†(ai bj) ¹ œ ka(x È
#
#
a b
a b
" by" ck
œ kaxÈ
, since c œ ax by.
a# b#
Ä
12. (a) Let Q(xß yß z) be any point on Ax By Cz D œ 0. Let QP" œ (x x" )i (y y" )j (z z" )k, and
Ä
B j Ck
B j Ck
n œ ÈAAi . The distance is ¹projn QP" ¹ œ ¹((x x" )i (y y" )j (z z" )k) † Š ÈAAi ‹¹
# B# C#
# B# C#
" (Ax By Cz)k
" By" Cz" Dk
œ kAx" By"È Cz
œ kAxÈ
.
#
#
#
#
#
#
A B C
A B C
(b) Since both tangent planes are parallel, one-half of the distance between them is equal to the radius of the
sphere, i.e., r œ " k3 9k œ È3 (see also Exercise 12a). Clearly, the points (1ß 2ß 3) and (1ß 2ß 3)
# È1 1 1
are on the line containing the sphere's center. Hence, the line containing the center is x œ 1 2t,
y œ 2 4t, z œ 3 6t. The distance from the plane x y z 3 œ 0 to the center is È3
(2 4t) (3 6t) 3k
Ê k(1 2t) È
œ È3 from part (a) Ê t œ 0 Ê the center is at (1ß 2ß 3). Therefore
111
an equation of the sphere is (x 1)# (y 2)# (z 3)# œ 3.
13. (a) If (x" ß y" ß z" ) is on the plane Ax By Cz œ D" , then the distance d between the planes is
D# k
" Cz" D# k
d œ kAx"È By
œ kAikD"Bj #
#
#
Ckk , since Ax" By" Cz" œ D" , by Exercise 12(a).
(b)
(c)
A B C
k12 6k
d œ È4 9 1 œ È614
k2(3) (1)(2) 2(1) 4k
2(1) Dk
œ k2(3) (1)(2)
È14
È14
Ê D œ 8 or 4 Ê the desired plane is
2x y 2x œ 8
Dk
(d) Choose the point (2ß 0ß 1) on the plane. Then k3È
œ 5 Ê D œ 3 „ 5È6 Ê the desired planes are
6
x 2y z œ 3 5È6 and x 2y z œ 3 5È6.
Ä
Ä
14. Let n œ AB ‚ BC and D(xß yß z) be any point in the plane determined by A, B and C. Then the point D lies in
Ä
Ä
Ä
Ä
this plane if and only if AD † n œ 0 Í AD † (AB ‚ BC ) œ 0.
â
âi
â
15. n œ i 2j 6k is normal to the plane x 2y 6z œ 6; v ‚ n œ â 1
â
â1
â
kâ
â
1 ⠜ 4i 5j k is parallel to the
â
6â
â
â
j kâ
âi
â
â
plane and perpendicular to the plane of v and n Ê w œ n ‚ (v ‚ n) œ â 1 2 6 â œ 32i 23j 13k is a
â
â
â 4 5 1 â
j
1
2
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Chapter 12 Additional and Advanced Exercises
757
vector parallel to the plane x 2y 6z œ 6 in the direction of the projection vector projP v. Therefore,
w
v †w
42
"
32
23
13
ˆ 32 23 13 ‰
projP v œ projw v œ Šv † kw
wk ‹ kwk œ Š kwk # ‹ w œ 32# 23# 13# w œ 1722 w œ 41 w œ 41 i 41 j 41 k
16. projz w œ projz v and w projz w œ v projz v Ê w œ (w projz w) projz w œ (v projz v) projz w
œ v 2 projz v œ v 2 Š kvz†kz# ‹ z
17. (a) u ‚ v œ 2i ‚ 2j œ 4k Ê (u ‚ v) ‚ w œ 0 ; (u † w)v (v † w)u œ 0v 0u œ 0; v ‚ w œ 4i Ê u ‚ (v ‚ w) œ 0;
(u † w)v (u † v)w œ 0v 0w œ 0
â
â
â
â
j
k â
j k â
âi
â i
â
â
â
â
(b) u ‚ v œ â 1 1 1 â œ i 4j 3k Ê (u ‚ v) ‚ w œ â 1 4 3 â œ 10i 2j 6k ;
â
â
â
â
â 2 1 2 â
â 1 2 1 â
(u † w)v (v † w)u œ 4(2i j 2k) 2(i j k) œ 10i 2j 6k;
â
â
â
â
j k â
j kâ
â i
âi
â
â
â
â
v ‚ w œ â 2 1 2 â œ 3i 4j 5k Ê u ‚ (v ‚ w) œ â 1 1 1 â œ 9i 2j 7k;
â
â
â
â
â 1 2 1 â
â3 4 5â
(u † w)v (u † v)w œ 4(2i j 2k) (1)(i 2j k) œ 9i 2j 7k
â
â
â
â
j kâ
j
k â
âi
âi
â
â
â
â
(c) u ‚ v œ â 2 1 0 â œ i 2j 4k Ê (u ‚ v) ‚ w œ â 1 2 4 â œ 4i 6j 2k ;
â
â
â
â
2 â
â 2 1 1 â
â1 0
(u † w)v (v † w)u œ 2(2i j k) 4(2i j) œ 4i 6j 2k;
â
â
â
â
j kâ
j kâ
âi
â i
â
â
â
â
1 0 ⠜ i 2 j 4k;
v ‚ w œ â 2 1 1 â œ 2 i 3 j k Ê u ‚ ( v ‚ w ) œ â 2
â
â
â
â
â1 0 2â
â 2 3 1 â
(u † w)v (u † v)w œ 2(2i j k) 3(i 2k) œ i 2j 4k
â
â
â
j k â
j
â i
â i
â
â
â
(d) u ‚ v œ â 1 1 2 â œ i 3j k Ê (u ‚ v) ‚ w œ â 1 3
â
â
â
â 1 0 1 â
â 2 4
â
k â
â
1 ⠜ 10i 10k ;
â
2 â
(u † w)v (v † w)u œ 10(i k) 0(i j 2k) œ 10i 10k;
â
â
â
â
j k â
j
k â
â i
âi
â
â
â
â
v ‚ w œ â 1 0 1 â œ 4i 4j 4k Ê u ‚ (v ‚ w) œ â 1 1 2 â œ 12i 4j 8k;
â
â
â
â
â 2 4 2 â
â 4 4 4 â
(u † w)v (u † v)w œ 10(i k) 1(2i 4j 2k) œ 12i 4j 8k
18. (a) u ‚ (v ‚ w) v ‚ (w ‚ u) w ‚ (u ‚ v) œ (u † w)v (u † v)w (v † u)w (v † w)u (w † v)u (w † u)v œ 0
(b) [u † (v ‚ i)]i [(u † (v ‚ j)]j [(u † (v ‚ k)]k œ [(u ‚ v) † i]i [(u ‚ v) † j]j [(u ‚ v) † k]k œ u ‚ v
u†w v†w
(c) (u ‚ v) † (w ‚ r) œ u † [v ‚ (w ‚ r)] œ u † [(v † r)w (v † w)r] œ (u † w)(v † r) (u † r)(v † w) œ º
u†r v†r º
19. The formula is always true; u ‚ [u ‚ (u ‚ v)] † w œ u ‚ [(u † v)u (u † u)v] † w
œ [(u † v)u ‚ u (u † u)u ‚ v] † w œ kuk # u ‚ v † w œ kuk # u † v ‚ w
20. If u œ (cos B)i (sin B)j and v œ (cos A)i (sin A)j , where A B, then u ‚ v œ ckuk kvk sin (A B)d k
â
â
j
kâ
â i
â
â
œ â cos B sin B 0 â œ (cos B sin A sin B cos A)k Ê sin (A B) œ cos B sin A sin B cos A, since
â
â
â cos A sin A 0 â
kuk œ 1 and kvk œ 1.
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758
Chapter 12 Vectors and the Geometry of Space
21. If u œ ai bj and v œ ci dj , then u † v œ kuk kvk cos ) Ê ac bd œ Èa# b# Èc# d# cos )
Ê (ac bd)# œ aa# b# b ac# d# b cos# ) Ê (ac bd)# Ÿ aa# b# b ac# d# b , since cos# ) Ÿ 1.
22. If u œ ai bj ck, then u † u œ a# b# c#
0 and u † u œ 0 iff a œ b œ c œ 0.
23. ku vk # œ (u v) † (u v) œ u † u 2u † v v † v Ÿ kuk # 2 kuk kvk kvk # œ akuk kvkb# Ê ku vk Ÿ kuk kvk
24. Let ! denote the angle between w and u, and " the angle between w and v. Let a œ kuk and b œ kvk . Then
#
cos ! œ kwwk†kuuk œ (avkwk bkuuk)†u œ (av†kuwk kubuk †u) œ (av†kuwk kubuk †u) œ aav†kuwk aba b œ v†ukwk ba , and likewise, cos " œ u†vkwk ba .
Since the angle between u and v is always Ÿ 1# and cos ! œ cos " , we have that ! œ " Ê w bisects the angle between
u and v .
25. (kukv kvku) † (kvku kukv) œ kukv † kvku kvku † kvku kukv † kukv kvku † kukv
œ kvku † kukv kvk# u † u kuk# v † v kvku † kukv œ kvk# kuk# kuk# kvk# œ 0
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
CHAPTER 13 VECTOR-VALUED FUNCTIONS
AND MOTION IN SPACE
13.1 CURVES IN SPACE AND THEIR TANGENTS
1. x œ t 1 and y œ t# 1 Ê y œ (x 1)# 1 œ x# 2x; v œ ddtr œ i 2tj Ê a œ ddtv œ 2j Ê v œ i 2j and a œ 2j
at t œ 1
1
2. x œ t t 1 and y œ 1t Ê x œ 1 y 1 œ 1 1 y Ê y œ x1 1; v œ ddtr œ at 11b2 i t12 j Ê a œ ddtv œ at 21b3 i t23 j
y
Ê v œ 4i 4j and a œ 16i 16j at t œ "#
3. x œ et and y œ 29 e2t Ê y œ 29 x# ; v œ ddtr œ et i 49 e2t j Ê a œ et i 89 e2t j Ê v œ 3i 4j and a œ 3i 8j at t œ ln 3
4. x œ cos 2t and y œ 3 sin 2t Ê x# "9 y# œ 1; v œ ddtr œ (2 sin 2t)i (6 cos 2t)j Ê a œ ddtv
œ (4 cos 2t)i (12 sin 2t)j Ê v œ 6j and a œ 4i at t œ 0
5. v œ ddtr œ (cos t)i (sin t)j and a œ ddtv œ (sin t)i (cos t)j
Ê for t œ 14 , v ˆ 14 ‰ œ
È
È2
È2
# i # j and
È
a ˆ 14 ‰ œ #2 i #2 j ; for t œ 1# , v ˆ 1# ‰ œ j and
a ˆ 1# ‰ œ i
6. v œ ddtr œ ˆ2 sin #t ‰ i ˆ2 cos #t ‰ j and a œ ddtv
œ ˆ cos #t ‰ i ˆ sin #t ‰ j Ê for t œ 1, v(1) œ 2i and
a(1) œ j ; for t œ 3#1 , v ˆ 3#1 ‰ œ È2 i È2 j and
a ˆ 3#1 ‰ œ
È2
È2
# i # j
7. v œ ddtr œ (1 cos t)i (sin t)j and a œ ddtv
œ (sin t)i (cos t)j Ê for t œ 1, v(1) œ 2i and a(1) œ j ;
for t œ 3#1 , v ˆ 3#1 ‰ œ i j and a ˆ 3#1 ‰ œ i
8. v œ ddtr œ i 2tj and a œ ddtv œ 2j Ê for t œ 1,
v(1) œ i 2j and a(1) œ 2j ; for t œ 0, v(0) œ i and
a(0) œ 2j ; for t œ 1, v(1) œ i 2j and a(1) œ 2j
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
760
Chapter 13 Vector-Valued Functions and Motion in Space
#
9. r œ (t 1)i at# 1b j 2tk Ê v œ ddtr œ i 2tj 2k Ê a œ ddt#r œ 2j ; Speed: kv(1)k œ È1# (2(1))# 2# œ 3;
i 2(1)j 2k
Direction: kvv(1)
œ 3" i 32 j 32 k Ê v(1) œ 3 ˆ 3" i 32 j 32 k‰
(1)k œ
3
#
$
#
10. r œ (1 t)i Èt 2 j t3 k Ê v œ ddtr œ i È2t2 j t# k Ê a œ ddt#r œ È22 j 2tk ; Speed: kv(1)k
#
v(1)
# #
œ Ê1# Š 2(1)
È2 ‹ (1 ) œ 2; Direction: kv(1)k œ
i È j (1# )k
2(1)
2
#
œ "# i È"2 j "# k Ê v(1)
œ 2 Š "# i È"2 j "# k‹
#
11. r œ (2 cos t)i (3 sin t)j 4tk Ê v œ ddtr œ (2 sin t)i (3 cos t)j 4k Ê a œ ddt#r œ (2 cos t)i (3 sin t)j ;
1
vˆ ‰
#
#
Speed: ¸v ˆ 1# ‰¸ œ Ɉ2 sin 1# ‰ ˆ3 cos 12 ‰ 4# œ 2È5; Direction: ¸v ˆ 1# ‰¸
#
2
3
4
œ Š #È
sin 1# ‹ i Š #È
cos 1# ‹ j #È
k œ È"5 i È25 k Ê v ˆ 1# ‰ œ 2È5 Š È"5 i È25 k‹
5
5
5
#
12. r œ (sec t)i (tan t)j 43 tk Ê v œ ddtr œ (sec t tan t)i asec# tb j 43 k Ê a œ ddt#r
#
#
#
œ asec t tan# t sec$ tb i a2 sec# t tan tb j ; Speed: ¸v ˆ 16 ‰¸ œ Ɉsec 16 tan 16 ‰ ˆsec# 16 ‰ ˆ 43 ‰ œ 2;
v ˆ1‰
Direction: ¸v ˆ 16 ‰¸ œ
6
ˆsec 1 tan 1 ‰ i ˆsec# 1 ‰ j 4 k
6
6
6
3
œ 3" i 32 j 32 k
#
Ê v ˆ 16 ‰ œ 2 ˆ 3" i 32 j 32 k‰
#
#
13. r œ (2 ln (t 1))i t# j t2 k Ê v œ ddtr œ ˆ t 2 1 ‰ i 2tj tk Ê a œ ddt#r œ ’ (t 21)# “ i 2j k ;
2
Š " 1 ‹ i 2(1)j (1)k
#
Speed: kv(1)k œ Ɉ 1 2 1 ‰ (2(1))# 1# œ È6; Direction: kvv(1)
È6
(1)k œ
œ È"6 i È26 j È16 k Ê v(1) œ È6 Š È"6 i È26 j È"6 k‹
#
14. r œ aet b i (2 cos 3t)j (2 sin 3t)k Ê v œ ddtr œ aet b i (6 sin 3t)j (6 cos 3t)k Ê a œ ddt#r
œ aet b i (18 cos 3t)j (18 sin 3t)k ; Speed: kv(0)k œ Éae! b# [6 sin 3(0)]# [6 cos 3(0)]# œ È37;
!
ae b i 6 sin 3(0)j 6 cos 3(0)k
Direction: kvv(0)
œ È"37 i È637 k Ê v(0) œ È37 Š È"37 i È637 k‹
È37
(0)k œ
#
15. v œ 3i È3 j 2tk and a œ 2k Ê v(0) œ 3i È3 j and a(0) œ 2k Ê kv(0)k œ Ê3# ŠÈ3‹ 0# œ È12 and
ka(0)k œ È2# œ 2; v(0) † a(0) œ 0 Ê cos ) œ 0 Ê ) œ 1#
16. v œ
È2
È2
# i Š # 32t‹ j and a œ 32j
Ê v(0) œ
È2
È2
# i # j and a(0) œ 32j
È
#
È
Ê kv(0)k œ ÊŠ #2 ‹ Š #2 ‹
È
È
#
È
16 2
œ 1 and ka(0)k œ È(32)# œ 32; v(0) † a(0) œ Š #2 ‹ (32) œ 16È2 Ê cos ) œ 1(32)
œ #2 Ê ) œ 341
1 ‰
#
17. v œ ˆ t# 2t 1 ‰ i ˆ t# 1 j tat 1b
"Î#
#
2t 2
k and a œ ’ “ i ’ at# 2t 1b# “ j ’ # " $Î# “ k Ê v(0) œ j and
at# 1b#
at 1b
a(0) œ 2i k Ê kv(0)k œ 1 and ka(0)k œ È2# 1# œ È5; v(0) † a(0) œ 0 Ê cos ) œ 0 Ê ) œ 12
18. v œ 23 (1 t)"Î# i 23 (1 t)"Î# j 3" k and a œ "3 (1 t)"Î# i 3" (1 t)"Î# j Ê v(0) œ 32 i 32 j 3" k and
#
#
#
#
#
a(0) œ "3 i 3" j Ê kv(0)k œ Ɉ 32 ‰ ˆ 32 ‰ ˆ 3" ‰ œ 1 and ka(0)k œ Ɉ "3 ‰ ˆ 3" ‰ œ
È2
2
2
3 ; v(0) † a(0) œ 9 9
œ 0 Ê cos ) œ 0 Ê ) œ 1#
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Section 13.1 Curves in Space and Their Tangents
761
19. r(t) œ (sin t)i at# cos tb j et k Ê v(t) œ (cos t)i (2t sin t)j et k ; t! œ 0 Ê v(t0 ) œ i k and
r(t0 ) œ P! œ (0ß 1ß 1) Ê x œ 0 t œ t, y œ 1, and z œ 1 t are parametric equations of the tangent line
20. r(t) œ t2 i a2t 1b j t3 k Ê v(t) œ 2t i 2 j 3t2 k ; t! œ 2 Ê va2b œ 4 i 2 j 12k and
r(t0 ) œ P! œ a4ß 3ß 8b Ê x œ 4 4t, y œ 3 2t, and z œ 8 12t are parametric equations of the tangent line
1
1
3
1
21. r(t) œ aln tbi tt 2 j at ln tbk Ê v(t) œ t i at 2b2 j aln t 1bk ; t! œ 1 Ê va1b œ i 3 j k and
r(t0 ) œ P! œ a0ß 0ß 0b Ê x œ 0 t œ t, y œ 0 13 t œ 13 t, and z œ 0 t œ t are parametric equations of the tangent line
22. r(t) œ (cos t)i asin tb j (sin 2t)k Ê v(t) œ ( sin t)i (cos t)j (2 cos 2t)k ; t! œ 1# Ê v(t0 ) œ i 2k and
r(t0 ) œ P! œ (0ß 1ß 0) Ê x œ 0 t œ t, y œ 1, and z œ 0 2t œ 2t are parametric equations of the tangent line
23. (a) v(t) œ (sin t)i (cos t)j Ê a(t) œ (cos t)i (sin t)j ;
(i) kv(t)k œ È(sin t)# (cos t)# œ 1 Ê constant speed;
(ii) v † a œ (sin t)(cos t) (cos t)(sin t) œ 0 Ê yes, orthogonal;
(iii) counterclockwise movement;
(iv) yes, r(0) œ i 0j
(b) v(t) œ (2 sin 2t)i (2 cos 2t)j Ê a(t) œ (4 cos 2t)i (4 sin 2t)j;
(i) kv(t)k œ È4 sin# 2t 4 cos# 2t œ 2 Ê constant speed;
(ii) v † a œ 8 sin 2t cos 2t 8 cos 2t sin 2t œ 0 Ê yes, orthogonal;
(iii) counterclockwise movement;
(iv) yes, r(0) œ i 0j
(c) v(t) œ sin ˆt 1# ‰ i cos ˆt 1# ‰ j Ê a(t) œ cos ˆt 1# ‰ i sin ˆt 1# ‰ j ;
(i)
kv(t)k œ Ésin# ˆt 1# ‰ cos# ˆt 1# ‰ œ 1 Ê constant speed;
(ii)
v † a œ sin ˆt 1# ‰ cos ˆt 1# ‰ cos ˆt 1# ‰ sin ˆt 1# ‰ œ 0 Ê yes, orthogonal;
(iii) counterclockwise movement;
(iv) no, r(0) œ 0i j instead of i 0j
(d) v(t) œ (sin t)i (cos t)j Ê a(t) œ (cos t)i (sin t)j ;
(i) kv(t)k œ È(sin t)# ( cos t)# œ 1 Ê constant speed;
(ii) v † a œ (sin t)(cos t) (cos t)(sin t) œ 0 Ê yes, orthogonal;
(iii) clockwise movement;
(iv) yes, r(0) œ i 0j
(e) v(t) œ (2t sin t)i (2t cos t)j Ê a(t) œ (2 sin t 2t cos t)i (2 cos t 2t sin t)j ;
(i)
kv(t)k œ Éc a2t sin tb d# (2t cos t)# œ È4t# asin# t cos# tb œ 2ktk œ 2t, t
0
Ê variable speed;
(ii) v † a œ 4 at sin# t t# sin t cos tb 4 at cos# t t# cos t sin tb œ 4t Á 0 in general Ê not orthogonal in general;
(iii) counterclockwise movement;
(iv) yes, r(0) œ i 0j
24. Let p œ 2i 2j k denote the position vector of the point a2, 2, 1b and let, u œ È"2 i È"2 j and v œ È"3 i È"3 j È"3 k.
Then r(t) œ p (cos t)u (sin t)v. Note that (2ß 2ß 1) is a point on the plane and n œ i j 2k is normal to
the plane. Moreover, u and v are orthogonal unit vectors with u † n œ v † n œ 0 Ê u and v are parallel to the
plane. Therefore, r(t) identifies a point that lies in the plane for each t. Also, for each t, (cos t)u (sin t)v
is a unit vector. Starting at the point Š2 È12 , 2 È12 , 1‹ the vector ratb traces out a circle of radius 1 and
center (2ß 2ß 1) in the plane x y 2z œ 2.
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
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Chapter 13 Vector-Valued Functions and Motion in Space
25. The velocity vector is tangent to the graph of y# œ 2x at the point (#ß #), has length 5, and a positive i
dy
component. Now, y# œ 2x Ê 2y dy
dx œ 2 Ê dx ¹
vector i "# j Ê the velocity vector is v œ
Ð#ß#Ñ
œ #2†2 œ "# Ê the tangent vector lies in the direction of the
5
ˆi "# j‰ œ È5 ˆi "# j‰ œ 2È5i È5j
5
É1 "4
Π#
26. (a)
(b) v œ (1 cos t)i (sin t)j and a œ (sin t)i (cos t)j ; kvk# œ (1 cos t)# sin# t œ 2 2 cos t Ê kvk# is at a max
when cos t œ 1 Ê t œ 1, 31, 51, etc., and at these values of t, kvk# œ 4 Ê max kvk œ È4 œ 2; kvk# is at a min
when cos t œ 1 Ê t œ 0, 21, 41, etc., and at these values of t, kvk# œ 0 Ê min kvk œ 0; kak# œ sin# t cos# t œ 1
for every t Ê max kak œ min kak œ È1 œ 1
27. dtd (r † r) œ r † ddtr ddtr † r œ 2r † ddtr œ 2 † 0 œ 0 Ê r † r is a constant Ê krk œ Èr † r is constant
28. (a)
(b)
d
du
d
du
dw ‰
ˆ dv
dt (u † v ‚ w) œ dt † (v ‚ w) u † dt (v ‚ w) œ dt † (v ‚ w) u † dt ‚ w v ‚ dt
œ ddtu † (v ‚ w) u † ddtv ‚ w u † v ‚ ddtw
d
dr
d# r
dr
dr
d# r
d# r
d# r
dr
d$ r
dr
d$ r
dt ’r † Š dt ‚ dt# ‹“ œ dt † Š dt ‚ dt# ‹ r † Š dt# ‚ dt# ‹ r † Š dt ‚ dt$ ‹ œ r † Š dt ‚ dt$ ‹ , since A † (A ‚ B) œ 0
and A † (B ‚ B) œ 0 for any vectors A and B
dg
dh
29. (a) u œ f(t)i g(t)j h(t)k Ê cu œ cf(t)i cg(t)j ch(t)k Ê dtd (cu) œ c df
dt i c dt j c dt k
dg
dh
du
œ c Š df
dt i dt j dt k‹ œ c dt
df
dg
df
dh
(b) f u œ f f(t)i f g(t)j f h(t)k Ê dtd (f u) œ ’ ddtf f(t) f df
dt “ i ’ dt g(t) f dt “ j ’ dt h(t) f dt “ k
dg
df
dh
du
œ ddtf [f(t)i g(t)j h(t)k] f ’ df
dt i dt j dt k“ œ dt u f dt
30. Let u œ f" (t)i f# (t)j f$ (t)k and v œ g" (t)i g# (t)j g$ (t)k. Then
u v œ [f" (t) g" (t)]i [f# (t) g# (t)]j [f$ (t) g$ (t)]k
Ê dtd (u v) œ [f"w (t) gw" (t)]i [f#w (t) gw# (t)]j [f$w (t) gw$ (t)]k
œ [f"w (t)i f#w (t)j f$w (t)k] [gw" (t)i gw# (t)j gw$ (t)k] œ ddtu ddtv ;
u v œ [f" (t) g" (t)]i [f# (t) g# (t)]j [f$ (t) g$ (t)]k
Ê dtd (u v) œ [f"w (t) gw" (t)]i [f#w (t) gw# (t)]j [f$w (t) gw$ (t)]k
œ [f"w (t)i f#w (t)j f$w (t)k] [gw" (t)i gw# (t)j gw$ (t)k] œ ddtu ddtv
31. Suppose r is continuous at t œ t! . Then lim r(t) œ r(t! ) Í lim [f(t)i g(t)j h(t)k]
t Ä t!
t Ä t!
œ f(t! )i g(t! )j h(t! )k Í lim f(t) œ f(t! ), lim g(t) œ g(t! ), and lim h(t) œ h(t! ) Í f, g, and h are
t Ä t!
t Ä t!
t Ä t!
continuous at t œ t! .
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 13.1 Curves in Space and Their Tangents
â
â i
â
32. lim [r" (t) ‚ r# (t)] œ lim â f" (t)
t Ä t!
t Ä t! â
â g" (t)
j
f# (t)
g# (t)
â â
i
k â ââ
â â lim f" (t)
f$ (t) â œ â t Ä t!
â
g$ (t) â ââ lim g" (t)
tÄt
!
j
lim f# (t)
t Ä t!
lim g# (t)
t Ä t!
â
â
lim f$ (t) ââ
t Ä t!
â
lim g$ (t) ââ
t Ä t!
k
œ lim r" (t) ‚ lim r# (t) œ A ‚ B
t Ä t!
t Ä t!
33. rw (t! ) exists Ê f w (t! )i gw (t! )j hw (t! )k exists Ê f w (t! ), gw (t! ), hw (t! ) all exist Ê f, g, and h are continuous at
t œ t! Ê r(t) is continuous at t œ t!
db
dc
34. u œ C œ ai bj ck with a, b, c real constants Ê ddtu œ da
dt i dt j dt k œ 0i 0j 0k œ 0
35-38. Example CAS commands:
Maple:
> with( plots );
r := t -> [sin(t)-t*cos(t),cos(t)+t*sin(t),t^2];
t0 := 3*Pi/2;
lo := 0;
hi := 6*Pi;
P1 := spacecurve( r(t), t=lo..hi, axes=boxed, thickness=3 ):
display( P1, title="#35(a) (Section 13.1)" );
Dr := unapply( diff(r(t),t), t );
# (b)
Dr(t0);
# (c)
q1 := expand( r(t0) + Dr(t0)*(t-t0) );
T := unapply( q1, t );
P2 := spacecurve( T(t), t=lo..hi, axes=boxed, thickness=3, color=black ):
display( [P1,P2], title="#35(d) (Section 13.1)" );
39-40. Example CAS commands:
Maple:
a := 'a'; b := 'b';
r := (a,b,t) -> [cos(a*t),sin(a*t),b*t];
Dr := unapply( diff(r(a,b,t),t), (a,b,t) );
t0 := 3*Pi/2;
q1 := expand( r(a,b,t0) + Dr(a,b,t0)*(t-t0) );
T := unapply( q1, (a,b,t) );
lo := 0;
hi := 4*Pi;
P := NULL:
for a in [ 1, 2, 4, 6 ] do
P1 := spacecurve( r(a,1,t), t=lo..hi, thickness=3 ):
P2 := spacecurve( T(a,1,t), t=lo..hi, thickness=3, color=black ):
P := P, display( [P1,P2], axes=boxed, title=sprintf("#39 (Section 13.1)\n a=%a",a) );
end do:
display( [P], insequence=true );
35-40. Example CAS commands:
Mathematica: (assigned functions, parameters, and intervals will vary)
The x-y-z components for the curve are entered as a list of functions of t. The unit vectors i, j, k are not inserted.
If a graph is too small, highlight it and drag out a corner or side to make it larger.
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
763
764
Chapter 13 Vector-Valued Functions and Motion in Space
Only the components of r[t] and values for t0, tmin, and tmax require alteration for each problem.
Clear[r, v, t, x, y, z]
r[t_]={ Sin[t] t Cos[t], Cos[t] t Sin[t], t^2}
t0= 31 / 2; tmin= 0; tmax= 61;
ParametricPlot3D[Evaluate[r[t]], {t, tmin, tmax}, AxesLabel Ä {x, y, z}];
v[t_]= r'[t]
tanline[t_]= v[t0] t r[t0]
ParametricPlot3D[Evaluate[{r[t], tanline[t]}], {t, tmin, tmax}, AxesLabel Ä {x, y, z}];
For 39 and 40, the curve can be defined as a function of t, a, and b. Leave a space between a and t and b and t.
Clear[r, v, t, x, y, z, a, b]
r[t_,a_,b_]:={Cos[a t], Sin[a t], b t}
t0= 31 / 2; tmin= 0; tmax= 41;
v[t_,a_,b_]= D[r[t, a, b], t]
tanline[t_,a_,b_]=v[t0, a, b] t r[t0, a, b]
pa1=ParametricPlot3D[Evaluate[{r[t, 1, 1], tanline[t, 1, 1]}], {t,tmin, tmax}, AxesLabel Ä {x, y, z}];
pa2=ParametricPlot3D[Evaluate[{r[t, 2, 1], tanline[t, 2, 1]}], {t,tmin, tmax}, AxesLabel Ä {x, y, z}];
pa4=ParametricPlot3D[Evaluate[{r[t, 4, 1], tanline[t, 4, 1]}], {t,tmin, tmax}, AxesLabel Ä {x, y, z}];
pa6=ParametricPlot3D[Evaluate[{r[t, 6, 1], tanline[t, 6, 1]}], {t,tmin, tmax}, AxesLabel Ä {x, y, z}];
Show[GraphicsRow[{pa1, pa2, pa4, pa6}]]
13.2 INTEGRALS OF VECTOR FUNCTIONS; PROJECTILE MOTION
1.
'01 ct$ i 7j (t 1)kd dt œ ’ t4 “ " i [7t] !" j ’ t2 t“ " k œ 4" i 7j 3# k
2.
'12 (6 6t)i 3Èt j ˆ t4 ‰ k‘ dt œ c6t 3t# d #" i 2t$Î# ‘ #" j c4t" d #" k œ 3i Š4È2 2‹ j 2k
3.
' 11Î%Î% c(sin t)i (1 cos t)j asec# tb kd dt œ c cos td 1Î%1Î% i ct sin td 1Î%1Î% j ctan td 1Î%1Î% k œ Š 1 #2È2 ‹ j 2k
4.
'01Î3 casec t tan tbi atan tbj a2 sin t cos tb kd dt œ '01Î3 [asec t tan tbi atan tbj asin 2tbk] dt
%
#
!
!
#
1Î$
1Î$
1Î$
œ csec td ! i c ln acos tbd ! j "# cos 2t‘ ! k œ i (ln 2)j 34 k
5.
'14 ˆ "t i 5 " t j #"t k‰ dt œ œ cln td %" i c ln (5 t)d %" j "# ln t‘ %" k œ (ln 4)i (ln 4)j (ln 2)k
6.
'01 Š È 2
7.
'01 Štet i et j k‹ dt œ ’ 12 et “ " i cet d "! j ctd !" k œ e 2 1 i e e 1 i k
8.
'1ln 3 atet i et j ln t kb dt œ ctet et d ln1 3 i cet d ln1 3 j ct ln t td 1ln 3 k
1 t#
2
"
È
"
È
i 1 3t# k‹ dt œ c2 sin" td ! i ’È3 tan" t“ k œ 1i 1 4 3 k
!
2
!
œ 3a ln 3 1bi a3 ebi aln 3alnaln 3b 1b 1b k
9.
'01Î2 cacos tbi asin 2tbj asin2 tb kd dt œ '01Î2 acos tbi asin 2tbj ˆ 12 12 cos 2t‰ k‘ dt œ
1 Î2
1 Î2
1 Î2
œ csin td ! i 12 cos t‘ ! j 12 t 14 sin 2t‘ ! k œ i j 14 k
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Section 13.2 Integrals of Vector Functions; Projectile Motion
1/4
765
1/4
10. '0 casec tbi atan2 tbj at sin tb kd dt œ '0 casec tbi asec2 t 1bj at sin tb kd dt
1
œ clnasec t tan tbd 1!/4 i ctan t td 1!/4 j ct cos t sin td !1/4 k œ lnŠ1 È2‹i ˆ1 14 ‰j Š È
È1 ‹k
4
2
2
11. r œ ' (ti tj tk) dt œ t# i t# j t# k C ; r(0) œ 0i 0j 0k C œ i 2j 3k Ê C œ i 2j 3k
#
#
#
#
#
#
Ê r œ Š t# 1‹ i Š t# 2‹ j Š t# 3‹ k
16
$‰
#
#
$‘
12. r œ ' c(180t)i a180t 16t# b jd dt œ 90t# i ˆ90t# 16
3 t j C ; r(0) œ 90(0) i 90(0) 3 (0) j C
$
‰
œ 100j Ê C œ 100j Ê r œ 90t# i ˆ90t# 16
3 t 100 j
13. r œ ' ˆ 3# (t 1)"Î# ‰ i e t j ˆ t " 1 ‰ k‘ dt œ (t 1)$Î# i e t j ln (t 1)k C ;
r(0) œ (0 1)$Î# i e ! j ln (0 1)k C œ k Ê C œ i j k
Ê r œ (t 1)$Î# 1‘ i a1 e t b j [1 ln (t 1)]k
14. r œ ' cat$ 4tb i tj 2t# kd dt œ Š t4 2t# ‹ i t2 j 2t3 k C ; r(0) œ ’ 04 2(0)# “ i 02 j 2(0)
3 kC
%
#
%
$
%
#
$
#
$
œ i j Ê C œ i j Ê r œ Š t4 2t# 1‹ i Š t# 1‹ j 2t3 k
15. ddtr œ ' (32k) dt œ 32tk C" ; ddtr (0) œ 8i 8j Ê 32(0)k C" œ 8i 8j Ê C" œ 8i 8j
Ê dr œ 8i 8j 32tk ; r œ ' (8i 8j 32tk) dt œ 8ti 8tj 16t# k C# ; r(0) œ 100k
dt
Ê 8(0)i 8(0)j 16(0)# k C# œ 100k Ê C# œ 100k Ê r œ 8ti 8tj a100 16t# b k
16. ddtr œ ' (i j k) dt œ (ti tj tk) C" ; ddtr (0) œ 0 Ê (0i 0j 0k) C" œ 0 Ê C" œ 0
#
#
#
Ê ddtr œ (ti tj tk) ; r œ ' (ti tj tk) dt œ Š t# i t# j t# k‹ C# ; r(0) œ 10i 10j 10k
#
#
#
Ê Š 0# i 0# j 0# k‹ C# œ 10i 10j 10k Ê C# œ 10i 10j 10k
#
#
#
Ê r œ Š t# 10‹ i Š t# 10‹ j Š t# 10‹ k
17. ddtv œ a œ 3i j k Ê v(t) œ 3ti tj tk C" ; the particle travels in the direction of the vector
(4 1)i (1 2)j (4 3)k œ 3i j k (since it travels in a straight line), and at time t œ 0 it has speed
2 Ê v(0) œ È9 21 1 (3i j k) œ C" Ê ddtr œ v(t) œ Š3t È6 ‹ i Št È2 ‹ j Št È2 ‹ k
11
Ê
11
11
r(t) œ Š 3# t# È611 t‹ i Š #" t# È211 t‹ j Š #" t# È211 t‹ k C# ; r(0) œ i 2j 3k œ C#
Ê r(t) œ Š 3# t# È611 t 1‹ i Š "# t# È211 t 2‹ j Š #" t# È211 t 3‹ k
œ Š "# t# È211 t‹ (3i j k) (i 2j 3k)
18. ddtv œ a œ 2i j k Ê v(t) œ 2ti tj tk C" ; the particle travels in the direction of the vector
(3 1)i (0 (1))j (3 2)k œ 2i j k (since it travels in a straight line), and at time t œ 0 it has speed 2
Ê v(0) œ È4 21 1 (2i j k) œ C" Ê ddtr œ v(t) œ Š2t È46 ‹ i Št È26 ‹ j Št È26 ‹ k
Ê r(t) œ Št# È46 t‹ i Š "# t# È26 t‹ j Š "# t# È26 t‹ k C# ; r(0) œ i j 2k œ C#
Ê r(t) œ Št# È46 t 1‹ i Š "# t# È26 t 1‹ j Š "# t# È26 t 2‹ k œ Š "# t# È26 t‹ (2i j k) (i j 2k)
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Chapter 13 Vector-Valued Functions and Motion in Space
m
m‰
19. x œ (v! cos !)t Ê (21 km)ˆ 1000
œ (840 m/s)(cos 60°)t Ê t œ (84021,000
1 km
m/s)(cos 60°) œ 50 seconds
v#
v#
!
20. R œ g! sin 2! and maximum R occurs when ! œ 45° Ê 24.5 km œ Š 9.8 m/s
# ‹ (sin 90°)
Ê v! œ È(9.8)(24,500) m# /s# œ 490 m/s
v#
#
m/s)(sin 45°)
m/s)
21. (a) t œ 2v! gsin ! œ 2(5009.8
¸ 72.2 seconds; R œ g! sin 2! œ (500
m/s#
9.8 m/s# (sin 90°) ¸ 25,510.2 m
5000 m
(b) x œ (v! cos !)t Ê 5000 m œ (500 m/s)(cos 45°)t Ê t œ (500 m/s)(cos
45°) ¸ 14.14 s; thus,
y œ (v! sin !)t "# gt# Ê y ¸ (500 m/s)(sin 45°)(14.14 s) "# a9.8 m/s# b (14.14 s)# ¸ 4020 m
#
#
!)
45°))
(c) ymax œ (v! sin
œ ((5002 m/s)(sin
¸ 6378 m
2g
a9.8 m/s# b
22. y œ y! (v! sin !)t "# gt# Ê y œ 32 ft (32 ft/sec)(sin 30°)t "# a32 ft/sec# b t# Ê y œ 32 16t 16t# ;
the ball hits the ground when y œ 0 Ê 0 œ 32 16t 16t# Ê t œ 1 or t œ 2 Ê t œ 2 sec since t 0; thus,
È
x œ (v! cos !) t Ê x œ (32 ft/sec)(cos 30°)t œ 32 Š #3 ‹ (2) ¸ 55.4 ft
v#
v#
#
# #
!
23. (a) R œ g! sin 2! Ê 10 m œ Š 9.8 m/s
Ê v! ¸ 9.9 m/s;
# ‹ (sin 90°) Ê v! œ 98 m s
#
m/s)
(b) 6m ¸ (9.9
9.8 m/s# (sin 2!) Ê sin 2! ¸ 0.59999 Ê 2! ¸ 36.87° or 143.12° Ê ! ¸ 18.4° or 71.6°
24. v! œ 5 ‚ 10' m/s and x œ 40 cm œ 0.4 m; thus x œ (v! cos !)t Ê 0.4m œ a5 ‚ 10' m/sb (cos 0°)t
Ê t œ 0.08 ‚ 10' s œ 8 ‚ 10) s; also, y œ y! (v! sin !)t "# gt#
#
Ê y œ a5 ‚ 10' m/sb (sin 0°) a8 ‚ 10) sb "# a9.8 m/s# b a8 ‚ 10) sb œ 3.136 ‚ 10"% m or
3.136 ‚ 10"# cm. Therefore, it drops 3.136 ‚ 10"# cm.
v#
#
m/s)
25. R œ g! sin 2! Ê 16,000 m œ (400
9.8 m/s# sin 2! Ê sin 2! œ 0.98 Ê 2! ¸ 78.5° or 2! ¸ 101.5° Ê ! ¸ 39.3°
or 50.7°
#
26. (a) R œ (2vg! ) sin 2! œ
4v#!
v#!
g sin 2! œ 4 Š g sin !‹ or 4 times the original range.
v#
(b) Now, let the initial range be R œ g! sin 2!. Then we want the factor p so that pv! will double the range
v#
#
Ê (pvg! ) sin 2! œ 2 Š g! sin 2!‹ Ê p# œ 2 Ê p œ È2 or about 141%. The same percentage will approximately
2
2
!b
!b
double the height: apv0 sin
œ 2av0 sin
Ê p# œ 2 Ê p œ È2.
2g
2g
27. The projectile reaches its maximum height when its vertical component of velocity is zero Ê dy
dt œ v0 sin ! gt œ 0
2
2
2
2
av0 sin !b
!b
!b
!
!
v0 sin !
"
Ê t œ v0 sin
Ê ymax œ av0 sin !bŠ v0 sin
av0 sin
œ av0 sin
. To find the flight time
g
g ‹ # gŠ g ‹ œ
g
2g
2g
!
we find the time when the projectile lands: av0 sin !bt "# g t2 œ 0 Ê tˆv0 sin ! "# g t‰ œ 0 Ê t œ 0 or t œ 2v0 sin
.
g
!
t œ 0 is the time when the projectile is fired, so t œ 2v0 sin
is the time when the projectile strikes the ground. The range is
g
v2
v2
!
!
the value of the horizontal component when t œ 2v0 sin
Ê R œ x œ av0 cos !bŠ 2v0 sin
‹ œ g0 a2 sin !cos !b œ g0 sin 2!.
g
g
The range is largest when sin 2! œ 1 Ê ! œ 45‰ .
R
28. When marble A is located R units downrange, we have x œ (v! cos !)t Ê R œ (v! cos !)t Ê t œ v! cos
! . At
R
R
"
that time the height of marble A is y œ y! (v! sin !)t "# gt# œ (v! sin !) Š v! cos
! ‹ # g Š v! cos ! ‹
#
R
R
Ê y œ R tan ! "# g Š v# cos
# ! ‹ . The height of marble B at the same time t œ v cos ! seconds is
!
!
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
#
Section 13.2 Integrals of Vector Functions; Projectile Motion
767
#
R
h œ R tan ! "# gt# œ R tan ! "# g Š v# cos
# ! ‹ . Since the heights are the same, the marbles collide regardless
!
of the initial velocity v! .
29. ddtr œ ' (gj) dt œ gtj C" and ddtr (0) œ (v! cos !)i (v! sin !)j Ê g(0)j C" œ (v! cos !)i (v! sin !)j
Ê C" œ (v! cos !)i (v! sin !)j Ê ddtr œ (v! cos !)i (v! sin ! gt)j ; r œ ' [(v! cos !)i (v! sin ! gt)j] dt
œ (v! t cos !)i ˆv! t sin ! "# gt# ‰ j C# and r(0) œ x! i y! j Ê [v! (0) cos !]i v! (0) sin ! "# g(0)# ‘ j C#
œ x! i y! j Ê C# œ x! i y! j Ê r œ (x! v! t cos !)i ˆy! v! t sin ! "# gt# ‰ j Ê x œ x! v! t cos ! and
y œ y! v! t sin ! "# gt#
v#
#
!)
30. The maximum height is y œ (v! sin
and this occurs for x œ #g! sin 2! œ
#g
v#! sin ! cos !
.
g
These equations describe
parametrically the points on a curve in the xy-plane associated with the maximum heights on the parabolic trajectories in
terms of the parameter (launch angle) !. Eliminating the parameter !, we have x# œ
œ
v%! sin# !
v% sin% !
v#
! g# œ g! (2y) (2y)#
g#
v#
#
v%
Ê x# 4 Šy 4g! ‹ œ 4g!# , where x
2v#
ˆv% sin# !‰ a1 sin# !b
v%! sin# ! cos# !
œ !
g#
g#
v#
v%
v%
Ê x# 4y# Š g ! ‹ y œ 0 Ê x# 4 ’y# Š 2g! ‹ y 16g! # “ œ 4g!#
0.
31. (a) At the time t when the projectile hits the line OR we
have tan " œ yx ; x œ [v! cos (! " )]t and
y œ [v! sin (! " )]t "# gt# 0 since R is
below level ground. Therefore let
kyk œ "# gt# [v! sin (! " )]t 0
"# gt# (v! sin (! " ))t‘
" gt v sin (! " )‘
œ # v! cos! (! ")
[v! cos (! " )]t
v! cos (! " ) tan " œ "# gt v! sin (! " )
t œ 2v! sin (! ") 2vg ! cos (! ") tan " , which is the time
so that tan " œ
Ê
Ê
when the projectile hits the downhill slope. Therefore,
x œ [v! cos (! " )] ’ 2v! sin (! ") 2vg ! cos (! ") tan " “ œ
2v#!
#
g ccos (! " ) tan " sin (! " ) cos (! " )d .
If x is
2v#
maximized, then OR is maximized: ddx! œ g ! [ sin 2(! " ) tan " cos 2(! " )] œ 0
Ê sin 2(! " ) tan " cos 2(! " ) œ 0 Ê tan " œ cot 2(! " ) Ê 2(! " ) œ 90° "
Ê ! " œ "# (90° " ) Ê ! œ "# (90° " ) œ "# of nAOR.
(b) At the time t when the projectile hits OR we have
tan " œ yx ; x œ [v! cos (! " )]t and
y œ [v! sin (! " )]t "# gt#
v sin (! " ) " gt‘
cv! sin (! " )d t "# gt#
œ ! v! cos (! ")#
[v! cos (! " )]t
v! cos (! " ) tan " œ v! sin (! " ) "# gt
t œ 2v! sin (! ") 2vg ! cos (! ") tan " , which is the time
Ê tan " œ
Ê
Ê
when the projectile hits the uphill slope. Therefore,
x œ [v! cos (! " )] ’ 2v! sin (! ") 2vg ! cos (! ") tan " “ œ
maximized, then OR is maximized: ddx! œ
2v#!
g
2v#!
#
g csin (! " ) cos (! " ) cos (! " ) tan " d .
If x is
[cos 2(! " ) sin 2(! " ) tan " ] œ 0
Ê cos 2(! " ) sin 2(! " ) tan " œ 0 Ê cot 2(! " ) tan " œ ! Ê cot 2(! " ) œ tan "
œ tan (" ) Ê 2(! " ) œ 90° (" ) œ 90° " Ê ! œ "# (90° " ) œ "# of nAOR. Therefore v! would bisect
nAOR for maximum range uphill.
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768
Chapter 13 Vector-Valued Functions and Motion in Space
32. v! œ 116 ft/sec, ! œ 45°, and x œ (v! cos !)t
Ê 369 œ (116 cos 45°)t Ê t ¸ 4.50 sec;
also y œ (v! sin !)t "# gt#
Ê y œ (116 sin 45°)(4.50) "# (32)(4.50)#
¸ 45.11 ft. It will take the ball 4.50 sec to travel
369 ft. At that time the ball will be 45.11 ft in
the air and will hit the green past the pin.
33. (a) (Assuming that "x" is zero at the point of impact:)
ratb œ axatbbi ayatbbj; where xatb œ a35 cos 27‰ bt and yatb œ 4 a35 sin 27‰ bt 16t2 .
‰ 2
2
!b
!
27
(b) ymax œ av0 sin
4 œ a35sin6427 b 4 ¸ 7.945 feet, which is reached at t œ v0 sin
œ 35sin
¸ 0.497 seconds.
2g
g
32
‰
(c) For the time, solve y œ 4 a35 sin 27‰ bt 16t2 œ 0 for t, using the quadratic formula
tœ
35 sin 27‰ Éa35 sin 27‰ b2 256
¸ 1.201 sec. Then the range is about xa1.201b œ a35 cos 27‰ ba1.201b ¸ 37.453 feet.
32
‰
2
(d) For the time, solve y œ 4 a35 sin 27 bt 16t œ 7 for t, using the quadratic formula
tœ
35 sin 27‰ Éa35 sin 27‰ b2 192
¸ 0.254 and 0.740 seconds. At those times the ball is about
32
‰
‰
xa0.254b œ a35 cos 27 ba0.254b ¸ 7.921 feet and xa0.740b œ a35 cos 27 ba0.740b ¸ 23.077 feet the impact point,
or about 37.453 7.921 ¸ 29.532 feet and 37.453 23.077 ¸ 14.376 feet from the landing spot.
(e) Yes. It changes things because the ball won't clear the net (ymax ¸ 7.945).
34. x œ x! (v! cos !)t œ 0 (v! cos 40°)t ¸ 0.766 v! t and y œ y! (v! sin !)t "# gt# œ 6.5 (v! sin 40°)t 16t#
¸ 6.5 0.643 v! t 16t# ; now the shot went 73.833 ft Ê 73.833 œ 0.766 v! t Ê t ¸ 96.383
sec; the shot lands when y œ 0
v!
#
148,635
Ê 0 œ 6.5 (0.643)(96.383) 16 Š 96.383
Ê v! ¸ É 148,635
v! ‹ Ê 0 ¸ 68.474 68.474 ¸ 46.6 ft/sec, the shot's initial
v#
!
speed
35. Flight time œ 1 sec and the measure of the angle of elevation is about 64° (using a protractor) so that t œ 2v! gsin !
#
v#
#
sin 64°)
64°
Ê 1 œ 2v! sin
Ê v! ¸ 17.80 ft/sec. Then ymax œ (17.802(32)
¸ 4.00 ft and R œ g! sin 2! Ê R œ (17.80)
sin 128°
32
32
¸ 7.80 ft Ê the engine traveled about 7.80 ft in 1 sec Ê the engine velocity was about 7.80 ft/sec
36. (a) ratb œ axatbbi ayatbbj; where xatb œ a145 cos 23‰ 14bt and yatb œ 2.5 a145 sin 23‰ bt 16t2 .
‰ 2
2
!b
23 b
!
23
(b) ymax œ av0 sin
2.5 œ a145sin
2.5 ¸ 52.655 feet, which is reached at t œ v0 sin
œ 145sin
¸ 1.771 seconds.
2g
64
g
32
‰
(c) For the time, solve y œ 2.5 a145 sin 23‰ bt 16t2 œ 0 for t, using the quadratic formula
tœ
145 sin 23‰ Éa145 sin 23‰ b2 160
¸ 3.585 sec. Then the range at t ¸ 3.585 is about x œ a145 cos 23‰ 14ba3.585b
32
¸ 428.311 feet.
(d) For the time, solve y œ 2.5 a145 sin 23‰ bt 16t2 œ 20 for t, using the quadratic formula
tœ
145 sin 23‰ Éa145 sin 23‰ b2 1120
¸ 0.342 and 3.199 seconds. At those times the ball is about
32
‰
xa0.342b œ a145 cos 23 14ba0.342b ¸ 40.860 feet from home plate and xa3.199b œ a145 cos 23‰ 14ba3.199b
¸ 382.195 feet from home plate.
(e) Yes. According to part (d), the ball is still 20 feet above the ground when it is 382 feet from home plate.
37. ddt2r k ddtr œ gj Ê Patb œ k and Qatb œ gj Ê ' Patb dt œ kt Ê vatb œ e' Patb dt œ ekt Ê ddtr œ va1tb ' vatb Qatb dt
2
œ gekt ' ekt j dt œ gekt ek j C1 ‘ œ gk j Cekt , where C œ gC1 ; apply the initial condition:
kt
g
dr
ˆg
‰
dt ¹tœ0 œ av0 cos !bi av0 sin !bj œ k j C Ê C œ av0 cos !bi k v0 sin ! j
Ê ddtr œ ˆv0 ekt cos !‰i ˆ gk ekt ˆ gk v0 sin !‰‰j, r œ ' c ˆv0 ekt cos !‰i ˆ gk ekt ˆ gk v0 sin !‰‰j ddt
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 13.2 Integrals of Vector Functions; Projectile Motion
769
ckt
gt
g
v
œ ˆ k0 ekt cos !‰i Š k e k ˆ k v0 sin !‰‹j C2 ; apply the initial condition:
!‰
!‰
j C2 Ê C2 œ ˆ vk0 cos !‰i ˆ kg2 v0 sin
j
ra0b œ 0 œ ˆ vk0 cos !‰i ˆ kg2 v0 sin
k
k
Ê ratb œ ˆ vk0 ˆ1 ekt ‰cos !‰i ˆ vk0 ˆ1 ekt ‰sin ! kg2 ˆ1 kt ekt ‰‰j
152 ‰
38. (a) ratb œ axatbbi ayatbbj; where xatb œ ˆ 0.12
a1 e0.12t bacos 20‰ b and
152 ‰
32 ‰
0.12t
yatb œ 3 ˆ 0.12
a1 e0.12t basin 20‰ b ˆ 0.12
b
2 a1 0.12t e
(b) Solve graphically using a calculator or CAS: At t ¸ 1.484 seconds the ball reaches a maximum height of about 40.435
feet.
(c) Use a graphing calculator or CAS to find that y œ 0 when the ball has traveled for ¸ 3.126 seconds. The range is
152 ‰ˆ
about xa3.126b œ ˆ 0.12
1 e0.12a3.126b ‰acos 20‰ b ¸ 372.311 feet.
(d) Use a graphing calculator or CAS to find that y œ 30 for t ¸ 0.689 and 2.305 seconds, at which times the ball is about
xa0.689b ¸ 94.454 feet and xa2.305b ¸ 287.621 feet from home plate.
(e) Yes, the batter has hit a home run since a graph of the trajectory shows that the ball is more than 14 feet above the
ground when it passes over the fence.
39. (a) 'a kr(t) dt œ 'a [kf(t)i kg(t)j kh(t)k] dt œ 'a [kf(t)] dt i 'a [kg(t)] dt j 'a [kh(t)] dt k
b
b
b
b
b
œ k Œ'a f(t) dt i 'a g(t) dt j 'a h(t) dt k œ k 'a r(t) dt
b
b
b
b
(b) 'a [r" (t) „ r# (t)] dt œ 'a acf" (t)i g" (t)j h" (t)kd „ cf# (t)i g# (t)j h# (t)kdb dt
b
b
œ 'a acf" (t) „ f# (t)d i [g" (t) „ g# (tb] j [h" (t) „ h# (t)] k) dt
b
œ 'a cf" (t) „ f# (t)d dt i 'a cg" (t) „ g# (t)d dt j 'a ch" (t) „ h# (t)d dt k
b
b
b
œ ”'a f" (t) dt i „ 'a f# (t) dt i• ”'a g" (t) dt j „ 'a g# (t) dt j • ”'a h" (t) dt k „ 'a h# (t) dt k•
b
b
b
b
b
b
œ 'a r" (t) dt „ 'a r# (t) dt
b
b
(c) Let C œ c" i c# j c$ k. Then 'a C † r(t) dt œ 'a cc" f(t) c# g(t) c$ h(t)d dt
b
b
œ c" 'a f(t) dt c# 'a g(t) dt c$ 'a h(t) dt = C † 'a r(t) dt;
b
b
b
b
'ab C ‚ r(t) dt œ 'ab cc# h(t) c$ g(t)d i cc$ f(t) c" h(t)d j cc" g(t) c# f(t)d k dt
œ ”c# 'a h(t) dt c$ 'a g(t) dt• i ”c$ 'a f(t) dt c" 'a h(t) dt• j ”c" 'a g(t) dt c# 'a f(t) dt• k
b
b
b
b
b
b
œ C ‚ 'a r(t) dt
b
40. (a) Let u and r be continuous on [aß b]. Then lim u(t)r(t) œ lim [u(t)f(t)i u(t)g(t)j u(t)h(t)k]
t Ä t!
t Ä t!
œ u(t! )f(t! )i u(t! )g(t! )j u(t! )h(t! )k œ u(t! )r(t! ) Ê ur is continuous for every t! in [aß b].
(b) Let u and r be differentiable. Then dtd (ur) œ dtd [u(t)f(t)i u(t)g(t)j u(t)h(t)k]
dg
df ‰
du
dh ‰
ˆ du
œ ˆ du
dt f(t) u(t) dt i Š dt g(t) u(t) dt ‹j dt h(t) u(t) dt k
dg
df
dh
du
dr
œ [f(t)i g(t)j h(t)k] du
dt u(t) Š dt i dt j dt k‹ œ r dt u dt
41. (a) If R" (t) and R# (t) have identical derivatives on I, then ddtR" œ dfdt" i dgdt" j dhdt" k œ dfdt# i dgdt# j dhdt# k
œ ddtR# Ê dfdt" œ dfdt# , dgdt" œ dgdt# , dhdt" œ dhdt# Ê f" (t) œ f# (t) c" , g" (t) œ g# (t) c# , h" (t) œ h# (t) c$
Ê f" (t)i g" (t)j h" (t)k œ [f# (t) c" ]i [g# (t) c# ]j [h# (t) c$ ]k Ê R" (t) œ R# (t) C, where
C œ c" i c# j c$ k.
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770
Chapter 13 Vector-Valued Functions and Motion in Space
(b) Let R(t) be an antiderivative of r(t) on I. Then Rw (t) œ r(t). If U(t) is an antiderivative of r(t) on I, then
Uw (t) œ r(t). Thus Uw (t) œ Rw (t) on I Ê U(t) œ R(t) C.
42. dtd 'a r(7 ) d7 œ dtd 'a [f(7 )i g(7 )j h(7 )k] d7 œ dtd 'a f(7 ) d7 i dtd 'a g(7 ) d7 j dtd 'a h(7 ) d7 k
t
t
t
t
t
œ f(t)i g(t)j h(t)k œ r(t). Since dtd 'a r(7 ) d7 œ r(t), we have that 'a r(7 ) d7 is an antiderivative of
t
t
r. If R is any antiderivative of r, then R(t) œ 'a r(7 ) d7 C by Exercise 41(b). Then R(a) œ 'a r(7 ) d7 C
t
a
œ 0 C Ê C œ R(a) Ê 'a r(7 ) d7 œ R(t) C œ R(t) R(a) Ê 'a r(7 ) d7 œ R(b) R(a).
t
b
1 ‰
43. (a) ratb œ axatbbi ayatbbj; where xatb œ ˆ 0.08
a1 e0.08t ba152 cos 20‰ 17.6b and
152 ‰
32 ‰
0.08t
yatb œ 3 ˆ 0.08
a1 e0.08t basin 20‰ b ˆ 0.08
b
2 a1 0.08t e
(b) Solve graphically using a calculator or CAS: At t ¸ 1.527 seconds the ball reaches a maximum height of about 41.893
feet.
(c) Use a graphing calculator or CAS to find that y œ 0 when the ball has traveled for ¸ 3.181 seconds. The range is
1 ‰ˆ
about xa3.181b œ ˆ 0.08
1 e0.08a3.181b ‰a152 cos 20‰ 17.6b ¸ 351.734 feet.
(d) Use a graphing calculator or CAS to find that y œ 35 for t ¸ 0.877 and 2.190 seconds, at which times the ball is about
xa0.877b ¸ 106.028 feet and xa2.190b ¸ 251.530 feet from home plate.
(e) No; the range is less than 380 feet. To find the wind needed for a home run, first use the method of part (d) to find that
1 ‰ˆ
y œ 20 at t ¸ 0.376 and 2.716 seconds. Then define xawb œ ˆ 0.08
1 e0.08a2.716b ‰a152 cos 20‰ wb, and solve
xawb œ 380 to find w ¸ 12.846 ft/sec.
#
#
#
sin !)
sin !)
!)
44. ymax œ (v! sin
Ê 34 ymax œ 3(v! 8g
and y œ (v! sin !)t "# gt# Ê 3(v! 8g
œ (v! sin !)t "# gt#
2g
Ê 3(v! sin !)# œ (8gv! sin !)t 4g# t# Ê 4g# t# (8gv! sin !)t 3(v! sin !)# œ 0 Ê 2gt 3v! sin ! œ 0 or
sin !
!
2gt v! sin ! œ 0 Ê t œ 3v!2gsin ! or t œ v! 2g
. Since the time it takes to reach ymax is tmax œ v! sin
,
g
sin !
then the time it takes the projectile to reach 34 of ymax is the shorter time t œ v! 2g
or half the time it takes
to reach the maximum height.
13.3 ARC LENGTH IN SPACE
1. r œ (2 cos t)i (2 sin t)j È5tk Ê v œ (2 sin t)i (2 cos t)j È5k
#
Ê kvk œ Ê(2 sin t)# (2 cos t)# ŠÈ5‹ œ È4 sin# t 4 cos# t 5 œ 3; T œ kvvk
1
1
œ ˆ 23 sin t‰ i ˆ 23 cos t‰ j 35 k and Length œ '0 kvk dt œ '0 3 dt œ c3td 1! œ 31
È
2. r œ (6 sin 2t)i (6 cos 2t)j 5tk Ê v œ (12 cos 2t)i (12 sin 2t)j 5k
Ê kvk œ È(12 cos 2t)# (12 sin 2t)# 5# œ È144 cos# 2t 144 sin# 2t 25 œ 13; T œ v
1
5
'
'
‰
ˆ 12
‰
œ ˆ 12
13 cos 2t i 13 sin 2t j 13 k and Length œ 0 kvk dt œ
1
kv k
13 dt œ c13td 1! œ 131
0
#
È
3. r œ ti 23 t$Î# k Ê v œ i t"Î# k Ê kvk œ É1# at"Î# b œ È1 t ; T œ kvvk œ È1" t i È1 t t k
)
and Length œ '0 È1 t dt œ 23 (1 t)$Î# ‘ ! œ 52
3
8
4. r œ (2 t)i (t 1)j tk Ê v œ i j k Ê kvk œ È1# (1)# 1# œ È3 ; T œ kvvk œ È"3 i È13 j È"3 k
$
and Length œ '0 È3 dt œ ’È3t“ œ 3È3
3
!
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 13.3 Arc Length in Space
771
5. r œ acos$ tb j asin$ tb k Ê v œ a3 cos# t sin tb j a3 sin# t cos tb k Ê kvk
œ Éa3 cos# t sin tb# a3 sin# t cos tb# œ Èa9 cos# t sin# tb acos# t sin# tb œ 3 kcos t sin tk ;
#
#
cos t sin t
3 sin t cos t
1
T œ kvvk œ 33kcos
t sin tk j 3 kcos t sin tk k œ ( cos t)j (sin t)k , if 0 Ÿ t Ÿ # , and
1Î2
Length œ '0
1Î2
1Î2
3 kcos t sin tk dt œ '0 3 cos t sin t dt œ '0
3
3
‘ 1Î# œ 3#
# sin 2t dt œ 4 cos 2t !
6. r œ 6t$ i 2t$ j 3t$ k Ê v œ 18t# i 6t# j 9t# k Ê kvk œ Éa18t# b# a6t# b# a9t# b# œ È441t% œ 21t# ;
"8t
6t
9t
6
2
3
'1 21t# dt œ c7t$ d " œ 49
T œ kvvk œ 21t
# i 21t# j 21t# k œ 7 i 7 j 7 k and Length œ
#
#
2
#
#
È
7. r œ (t cos t)i (t sin t)j 2 3 2 t$Î# k Ê v œ (cos t t sin t)i (sin t t cos t)j ŠÈ2 t"Î# ‹ k
#
Ê kvk œ Ê(cos t t sin t)# (sin t t cos t)# ŠÈ2 t‹ œ È1 t# 2t œ È(t 1)# œ kt 1k œ t 1, if t
1
1
0;
T œ kvvk œ ˆ cos tt t1sin t ‰ i ˆ sin ttt1cos t ‰ j Š t 2t1 ‹ k and Length œ '0 (t 1) dt œ ’ t2 t“ œ 12 1
È
"Î#
#
#
!
8. r œ (t sin t cos t)i (t cos t sin t)j Ê v œ (sin t t cos t sin t)i (cos t t sin t cos t)j
œ (t cos t)i (t sin t)j Ê kvk œ È(t cos t)# (t sin t)# œ Èt# œ ktk œ t if È2 Ÿ t Ÿ 2; T œ v
kvk
t‰
t‰
œ ˆ t cos
i ˆ t sin
j œ (cos t)i (sin t)j and Length œ 'È2 t dt œ ’ t2 “
t
t
2
#
#
È#
œ1
9. Let P(t! ) denote the point. Then v œ (5 cos t)i (5 sin t)j 12k and 261 œ '0 È25 cos# t 25 sin# t 144 dt
t!
œ '0 13 dt œ 13t! Ê t! œ 21, and the point is P(21) œ (5 sin 21ß 5 cos 21ß 241) œ (0ß 5ß 241)
t!
10. Let P(t! ) denote the point. Then v œ (12 cos t)i (12 sin t)j 5k and
131 œ '0 È144 cos# t 144 sin# t 25 dt œ '0 13 dt œ 13t! Ê t! œ 1, and the point is
t!
t!
P(1) œ (12 sin (1)ß 12 cos (1)ß 51) œ (0ß 12ß 51)
11. r œ (4 cos t)i (4 sin t)j 3tk Ê v œ (4 sin t)i (4 cos t)j 3k Ê kvk œ È(4 sin t)# (4 cos t)# 3#
œ È25 œ 5 Ê s(t) œ '0 5 d7 œ 5t Ê Length œ s ˆ 1# ‰ œ 5#1
t
12. r œ (cos t t sin t)i (sin t t cos t)j Ê v œ (sin t sin t t cos t)i (cos t cos t t sin t)j
t
#
œ (t cos t)i (t sin t)j Ê kvk œ È(t cos t)# (t cos t)# œ œ Èt# œ t, since 1 Ÿ t Ÿ 1 Ê s(t) œ ' 7 d7 œ t
#
Ê
#
0
ˆ 1 ‰#
#
Length œ s(1) s ˆ 1# ‰ œ 1# ## œ 381
#
13. r œ aet cos tb i aet sin tb j et k Ê v œ aet cos t et sin tb i aet sin t et cos tb j et k
Ê kvk œ Éaet cos t et sin tb# aet sin t et cos tb# aet b# œ œ È3e2t œ È3 et Ê s(t) œ '0 È3 e7 d7
t
È
œ È3 et È3 Ê Length œ s(0) s( ln 4) œ 0 ŠÈ3 e ln 4 È3‹ œ 3 4 3
14. r œ (1 2t)i (1 3t)j (6 6t)k Ê v œ 2i 3j 6k Ê kvk œ È2# 3# (6)# œ 7 Ê s(t) œ '0 7 d7 œ 7t
t
Ê Length œ s(0) s(1) œ 0 (7) œ 7
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Chapter 13 Vector-Valued Functions and Motion in Space
#
#
15. r œ ŠÈ2t‹ i ŠÈ2t‹ j a1 t# b k Ê v œ È2i È2j 2tk Ê kvk œ ÊŠÈ2‹ ŠÈ2‹ (2t)# œ È4 4t#
"
œ 2È1 t# Ê Length œ '0 2È1 t# dt œ ’2 Š 2t È1 t# "# ln Št È1 t# ‹‹“ œ È2 ln Š1 È2‹
1
!
16. Let the helix make one complete turn from t œ 0 to t œ 21.
Note that the radius of the cylinder is 1 Ê the
circumference of the base is 21. When t œ 21, the point P is
(cos 21ß sin 21ß 21) œ (1ß 0ß 21) Ê the cylinder is 21 units
high. Cut the cylinder along PQ and flatten. The resulting
rectangle has a width equal to the circumference of the
cylinder œ 21 and a height equal to 21, the height of the
cylinder. Therefore, the rectangle is a square and the portion
of the helix from t œ 0 to t œ 21 is its diagonal.
17. (a) r œ (cos t)i (sin t)j (" cos t)k, 0 Ÿ t Ÿ 21 Ê x œ cos t, y œ sin t, z œ 1 cos t Ê x# y#
œ cos# t sin# t œ 1, a right circular cylinder with the z-axis as the axis and radius œ 1. Therefore
P(cos tß sin tß 1 cos t) lies on the cylinder x# y# œ 1; t œ 0 Ê P(1ß 0ß 0) is on the curve; t œ 1# Ê Q(!ß 1ß 1)
Ä
Ä
is on the curve; t œ 1 Ê R(1ß 0ß 2) is on the curve. Then PQ œ i j k and PR œ 2i 2k
j k×
Ô i
Ä
Ä
Ê PQ ‚ PR œ 1 " " œ 2i 2k is a vector normal to the plane of P, Q, and R. Then the
Õ 2 0 2 Ø
plane containing P, Q, and R has an equation 2x 2z œ 2(1) 2(0) or x z œ 1. Any point on the curve
will satisfy this equation since x z œ cos t (1 cos t) œ 1. Therefore, any point on the curve lies on the
intersection of the cylinder x# y# œ 1 and the plane x z œ 1 Ê the curve is an ellipse.
(b) v œ ( sin t)i (cos t)j (sin t)k Ê kvk œ Èsin# t cos# t sin# t œ È1 sin# t Ê T œ v
kvk
(cos t)j (sin t)k
k
œ ( sin t)iÈ
Ê T(0) œ j , T ˆ 1# ‰ œ Èi2k , T(1) œ j , T ˆ 3#1 ‰ œ iÈ
1 sin# t
2
(c) a œ ( cos t)i (sin t)j (cos t)k ; n œ i k is
normal to the plane x z œ 1 Ê n † a œ cos t cos t
œ 0 Ê a is orthogonal to n Ê a is parallel to the
plane; a(0) œ i k , a ˆ 1# ‰ œ j , a a1b œ i k ,
‰œj
a ˆ 31
#
21
(d) kvk œ È1 sin# t (See part (b) Ê L œ '0 È1 sin# t dt
(e) L ¸ 7.64 (by Mathematica)
18. (a) r œ (cos 4t)i (sin 4t)j 4tk Ê v œ (4 sin 4t)i (4 cos 4t)j 4k Ê kvk œ È(4 sin 4t)# (4 cos 4t)# 4#
1Î2
œ È32 œ 4È2 Ê Length œ '0 4È2 dt œ ’4È2 t“
1Î#
!
œ 21È2
(b) r œ ˆcos #t ‰ i ˆsin #t ‰ j #t k Ê v œ ˆ #" sin #t ‰ i ˆ #" cos #t ‰ j #" k
#
#
#
Ê kvk œ Ɉ "# sin #t ‰ ˆ #" cos #t ‰ ˆ #" ‰ œ É 4" 4" œ
È2
#
41 È
Ê Length œ '0
#
2
È
dt œ ’ 22 t“
%1
!
œ 21 È 2
(c) r œ (cos t)i (sin t)j tk Ê v œ ( sin t)i (cos t)j k Ê kvk œ È( sin t)# ( cos t)# (1)# œ È1 1
œ È2 Ê Length œ 'c21 È2 dt œ ’È2 t“
0
!
#1
œ 21 È 2
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 13.4 Curvature and Normal Vectors of a Curve
773
19. nPQB œ nQOB œ t and PQ œ arc (AQ) œ t since
PQ œ length of the unwound string œ length of arc (AQ);
thus x œ OB BC œ OB DP œ cos t t sin t, and
y œ PC œ QB QD œ sin t t cos t
20. r œ acos t t sin tbi asin t t cos tbj Ê v œ asin t t cos t sin tbi acos t atasin tb cos tbbj
œ at cos tbi at sin tbj Ê kvk œ Éat cos tb2 at sin tb2 œ Èt2 œ ktk œ t, t
t
t sin t
0 Ê T œ kvvk œ t cos
t i t j
œ cos t i sin t j
21. v œ dtd ax0 t u1 bi dtd ay0 t u2 bj dtd az0 t u3 bk œ u1 i u2 j u3 k œ u, so satb œ '0 lvldt œ '0 luld7 œ '0 1 d7 œ t
t
t
t
22. ratb œ t i t2 j t3 k Ê vatb œ i 2t j 3t2 k Ê lvatbl œ Éa1b2 a2tb2 a3t2 b2 œ È1 4t2 9t4 . a0, 0, 0b Ê t œ 0
and a2, 4, 8b Ê t œ 2. Thus L œ '0 lvatbl dt œ '0 È1 4t2 9t4 dt. Using Simpson's rule with n œ 10 and
2
2
0
?x œ 2 10
œ 0.2 Ê L ¸ 0.2
3 Šlva0bl 4lva0.2bl 2lva0.4bl 4lva0.6bl 2lva0.8bl 4lva1bl 2lva1.2bl 4lva1.4bl
2lva1.6bl 4lva1.8bl lva2bl‹ ¸ 0.2
3 Š1 4a1.0837b 2a1.3676b 4a1.8991b 2a2.6919b 4a3.7417b
2a5.0421b 4a6.5890b 2a8.3800b 4a10.4134b 12.6886‹ œ 0.2
3 a143.5594b ¸ 9.5706
13.4 CURVATURE AND NORMAL VECTORS OF A CURVE
sin t ‰
È1# ( tan t)# œ Èsec# t œ ksec tk œ sec t, since
1. r œ ti ln (cos t)j Ê v œ i ˆ cos
t j œ i (tan t)j Ê kvk œ
tan t ‰
dT
1# t 1# Ê T œ kvvk œ ˆ sec" t ‰ i ˆ sec
t j œ (cos t)i (sin t)j ; dt œ ( sin t)i (cos t)j
ˆ dT ‰
Ê ¸ ddtT ¸ œ È( sin t)# ( cos t)# œ 1 Ê N œ ¸ ddtT ¸ œ ( sin t)i (cos t)j ; , œ k1vk † ¸ ddtT ¸ œ sec" t † 1 œ cos t.
dt
t tan t ‰
2. r œ ln (sec t)i tj Ê v œ ˆ secsec
i j œ (tan t)i j Ê kvk œ È( tan t)# 1# œ Èsec# t œ ksec tk œ sec t,
t
1
1
v
tan
since # t # Ê T œ kvk œ ˆ sec tt ‰ i ˆ sec1 t ‰ j œ (sin t)i (cos t)j ; ddtT œ (cos t)i (sin t)j
ˆ dT ‰
Ê ¸ ddtT ¸ œ È(cos t)# ( sin t)# œ 1 Ê N œ ¸ ddtT ¸ œ (cos t)i (sin t)j ; , œ k1vk † ¸ ddtT ¸ œ sec" t † 1 œ cos t.
dt
3. r œ (2t 3)i a5 t# b j Ê v œ 2i 2tj Ê kvk œ È2# (2t)# œ 2È1 t# Ê T œ kvvk œ È 2 # i È2t # j
2 1t
2 1t
Í
#
#
Í
Í
t
"
"
¸ ddtT ¸ œ t $ œ È " # i È t # j ; ddtT œ
$i $ j Ê
$
1 t
1 t
#
#
È
È
Š 1t ‹
Š 1t ‹
ŠÈ1 t# ‹
Ì ŠÈ1 t# ‹
ˆ dT ‰
œ É a1 "t# b# œ 1" t# Ê N œ ¸ ddtT ¸ œ È t # i È "
dt
1t
1 t#
j ; , œ k1vk † ¸ ddtT ¸ œ
"
† " # œ # a1 " t# b3/2
#È1 t# 1 t
4. r œ (cos t t sin t)i (sin t t cos t)j Ê v œ (t cos t)i (t sin t)j Ê kvk œ È( t cos t)# (t sin t)# œ Èt# œ ktk œ t, since
t 0 Ê T œ v œ (t cos t)i(t sin t)j œ (cos t)i (sin t)j ; dT œ ( sin t)i (cos t)j Ê ¸ dT ¸ œ È( sin t)# (cos t)#
kvk
t
dt
ˆ ddtT ‰
1 ¸ dT ¸
œ 1 Ê N œ ¸ dT ¸ œ ( sin t)i (cos t)j ; , œ kvk † dt œ "t † 1 œ "t
dt
dt
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
774
Chapter 13 Vector-Valued Functions and Motion in Space
5. (a) ,axb œ kva1xbk † ¹ dTdtaxb ¹. Now, v œ i f w axbj Ê kvaxbk œ É1 c f w axb d2 Ê T œ kvvk
1Î2
œ Š1 c f w axb d2 ‹
Í
Í
1Î2
i f w axbŠ1 c f w axb d2 ‹
2
f axbf axb
Í
xb
Ê ¹ dTa
” Š1 c f axb d2 ‹3 2 • dt ¹ œ
w
2
ww
f ax b
ww
Î
w
Ì
Thus ,axb œ
j. Thus ddtT axb œ
3Î2
2
Š1 c f axb d ‹
w
œË
f axbf axb
w
ww
i
3Î2
2
Š1 c f axb d ‹
w
c f axb d2 Š1 c f axb d2 ‹
ww
w
3
2
Š1 c f axb d ‹
w
œ
f ax b
ww
3Î2
2
Š1 c f axb d ‹
j
w
kf axbk
ww
2
¹1 c f axb d ¹
w
kf axbk
1
† k f a x bk 2 œ
3 2
2
a1 Òf axbÓ2 b1 2 k1 Òf axbÓ k
Š1 c f axb d ‹
ww
w
ww
w
Î
Î
w
#
d y
#
ˆ " ‰
(b) y œ ln (cos x) Ê dy
dx œ cos x ( sin x) œ tan x Ê dx# œ sec x Ê , œ
#
k sec# xk
x
œ ksec
sec$ xk
c1 ( tan x)# d$Î#
œ sec" x œ cos x, since 1# x 1#
(c) Note that f ww (x) œ 0 at an inflection point.
Þ
Þ
Þ
Þ
Þ
Þ
6. (a) r œ f(t)i g(t)j œ xi yj Ê v œ xi yj Ê kvk œ Èx# y# Ê T œ kvvk œ È Þ #x
y
Þ i ÈÞ# Þ# j
x y#
x y
Þ Þ ÞÞ Þ ÞÞ
Þ Þ ÞÞ Þ ÞÞ
Þ Þ ÞÞ Þ ÞÞ
Þ Þ ÞÞ Þ ÞÞ
Þ
Þ Þ ÞÞ Þ ÞÞ
yay x x yb 2
xax y y xb 2
yay x x yb
xax y y x b
ay# x# bay x x yb2
dT
dT ¸
¸
œ
i
j
Ê
œ
œ
Ê’
“
’
“
3/2
Þ
Þ
3/2
Þ
Þ
Ê
3/2
3/2
Þ
Þ 3
Þ
Þ
Þ
Þ
#
#
#
#
dt
dt
ax y b
ax y b
ax# y# b
ax# y# b
ax# y# b
Þ ÞÞ Þ ÞÞ
k y x x yk
Þ ÞÞ Þ ÞÞ
Þ ÞÞ Þ ÞÞ
k y x x yk
lxyyx l
Þ # † kxÞ # yÞ # k œ Þ # Þ # 3/2 .
x y
ax y b
œ kxÞ # yÞ # k ; , œ kv1k † ¸ ddtT ¸ œ È Þ #1
Þ
ÞÞ
Þ
ÞÞ
t
#
(b) r(t) œ ti ln (sin t)j , 0 t 1 Ê x œ t and y œ ln (sin t) Ê x œ 1, x œ 0; y œ cos
sin t œ cot t, y œ csc t
#
t
Ê , œ k csc #t 0$Î#k œ csc
csc$ t œ sin t
#
a1 cot t)b
"
Þ
t
"
(sinh t)i ln (cosh t)j Ê x œ tan" (sinh t) and y œ ln (cosh t) Ê x œ 1 cosh
sinh# t œ cosh t
$
#
ÞÞ
Þ
ÞÞ
ksech t sech t tanh tk
sinh t
#
œ sech t, x œ sech t tanh t; y œ cosh
œ ksech tk œ sech t
asech# t tanh# tb
t œ tanh t, y œ sech t Ê , œ
(c) r(t) œ tan
7. (a) r(t) œ f(t)i g(t)j Ê v œ f w (t)i gw (t)j is tangent to the curve at the point (f(t)ß g(t));
n † v œ c gw (t)i f w (t)jd † cf w (t)i gw (t)jd œ gw (t)f w (t) f w (t)gw (t) œ 0; n † v œ (n † v) œ 0; thus, n and n are
both normal to the curve at the point
(b) r(t) œ ti e2t j Ê v œ i 2e2t j Ê n œ 2e2t i j points toward the concave side of the curve; N œ knnk and
knk œ È4e4t 1 Ê N œ È2e 4t i È "
2t
1 4e
1 4e4t
j
(c) r(t) œ È4 t# i tj Ê v œ È t # i j Ê n œ i È t
4 t#
4t
#
j points toward the concave side of the curve;
Ê N œ "# ŠÈ4 t# i tj‹
N œ knnk and knk œ É1 4 t t# œ È 2
4 t#
8. (a) r(t) œ ti "3 t$ j Ê v œ i t# j Ê n œ t# i j points toward the concave side of the curve when t 0 and
n œ t# i j points toward the concave side when t 0 Ê N œ È "
1 t%
Nœ È"
1 t%
at# i jb for t 0
(b) From part (a), kvk œ È1 t% Ê T œ È "
1 t%
œ 12ktkt% ;
at# i jb for t 0 and
#
i Èt
1 t%
j Ê ddtT œ
2t$
2t
$Î# i $Î# j Ê
a1 t% b
a1 t% b
¸ ddtT ¸ œ É 4t6 %4t$2
a1 t b
ˆ dT ‰
%
$
t$
2t
N œ ¸ ddtT ¸ œ 12ktkt Š 2t% $Î# i i È t % j; t Á 0. N does not exist at t œ 0, where the
$Î# j‹ œ
ktkÈ1 t%
ktk 1 t
a1 t b
a1 t% b
dt
dt ¸
curve has a point of inflection; ddtT ¸ tœ0 œ 0 so the curvature , œ ¸ ddsT ¸ œ ¸ ddtT † ds
œ 0 at t œ 0 Ê N œ ," ddsT is
undefined. Since x œ t and y œ "3 t$ Ê y œ 3" x$ , the curve is the cubic power curve which is concave down for
x œ t 0 and concave up for x œ t 0.
9. r œ (3 sin t)i (3 cos t)j 4tk Ê v œ (3 cos t)i (3 sin t)j 4k Ê kvk œ È(3 cos t)# (3 sin t)# 4# œ È25
œ 5 Ê T œ kvvk œ ˆ 35 cos t‰ i ˆ 35 sin t‰ j 45 k Ê ddtT œ ˆ 53 sin t‰ i ˆ 53 cos t‰ j
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 13.4 Curvature and Normal Vectors of a Curve
#
#
775
ˆ dT ‰
3
Ê ¸ ddtT ¸ œ Ɉ 35 sin t‰ ˆ 35 cos t‰ œ 35 Ê N œ ¸ ddtT ¸ œ ( sin t)i (cos t)j ; , œ 15 † 35 œ 25
dt
10. r œ (cos t t sin t)i (sin t t cos t)j 3k Ê v œ (t cos t)i (t sin t)j Ê kvk œ È(t cos t)# (t sin t)# œ Èt#
œ ktk œ t, if t 0 Ê T œ kvvk œ (cos t)i (sin t)j , t 0 Ê ddtT œ ( sin t)i (cos t)j
ˆ dT ‰
Ê ¸ ddtT ¸ œ È( sin t)# (cos t)# œ 1 Ê N œ ¸ ddtT ¸ œ ( sin t)i (cos t)j ; , œ "t † 1 œ "t
dt
11. r œ aet cos tb i aet sin tb j 2k Ê v œ aet cos t et sin tb i aet sin t et cos tb j Ê
kvk œ Éaet cos t et sin tb# aet sin t et cos tb# œ È2e2t œ et È2 ;
cos t
t sin t
t sin t
T œ kvvk œ Š cos È
‹ i Š sin tÈ
‹ j Ê ddtT œ Š sinÈt 2 cos t ‹ i Š cos È
‹j
2
2
2
#
#
ˆ dT ‰
t sin t
Ê ¸ ddtT ¸ œ ÊŠ sinÈt cos t ‹ Š cos È
‹ œ 1 Ê N œ ¸ ddtT ¸ œ Š cosÈt sin t ‹ i Š sinÈt cos t ‹ j ;
2
2
2
dt
2
1
1
, œ kv1k † ¸ ddtT ¸ œ et È
† 1 œ et È
2
2
12. r œ (6 sin 2t)i (6 cos 2t)j 5tk Ê v œ (12 cos 2t)i (12 sin 2t)j 5k Ê kvk œ È(12 cos 2t)# (12 sin 2t)# 5#
5
dT
‰
ˆ 12
‰
ˆ 24
‰
ˆ 24
‰
œ È169 œ 13 Ê T œ kvvk œ ˆ 12
13 cos 2t i 13 sin 2t j 13 k Ê dt œ 13 sin 2t i 13 cos 2t j
#
#
ˆ dT ‰
24
‰ ˆ 24
‰ œ 13
Ê ¸ ddtT ¸ œ Ɉ 24
Ê N œ ¸ ddtT ¸ œ ( sin 2t)i (cos 2t)j ;
13 sin 2t
13 cos 2t
dt
1
24
, œ kv1k † ¸ ddtT ¸ œ 13
† 24
13 œ 169 .
$
#
13. r œ Š t3 ‹ i Š t# ‹ j , t 0 Ê v œ t# i tj Ê kvk œ Èt% t# œ tÈt# 1, since t 0 Ê T œ kvvk
œ Èt#t t i Èt#1 1 j Ê ddtT œ
1
i # t $Î# j
at# 1b$Î#
at 1 b
ˆ dT ‰
#
1
t
"
dt
œ É at1# œ t# 1 Ê N œ ¸ dT ¸ œ È #
1 b$
t 1
dt
i È #t
t 1
#
Ê ¸ ddtT ¸ œ ÊŠ # " $Î# ‹ Š # t $Î# ‹
at 1 b
j ; , œ kv1k † ¸ ddtT ¸ œ È 1#
t
#
at 1 b
t 1
1
† t# 1 œ
"
.
t at# 1b$Î#
14. r œ acos$ tb i asin$ tb j , 0 t 1# Ê v œ a3 cos# t sin tb i a3 sin# t cos tb j
Ê kvk œ Éa3 cos# t sin tb# a3 sin# t cos tb# œ È9 cos% t sin# t 9 sin% t cos# t œ 3 cos t sin t, since 0 t 1#
ˆ dT ‰
Ê T œ kvvk œ ( cos t)i (sin t)j Ê ddtT œ (sin t)i (cos t)j Ê ¸ ddtT ¸ œ Èsin# t cos# t œ 1 Ê N œ ¸ ddtT ¸
dt
œ (sin t)i (cos t)j; , œ kv1k † ¸ ddtT ¸ œ 3 cos1t sin t † 1 œ 3 cos1t sin t .
15. r œ ti ˆa cosh at ‰ j , a 0 Ê v œ i ˆsinh at ‰ j Ê kvk œ É1 sinh# ˆ at ‰ œ Écosh# ˆ at ‰ œ cosh at
Ê T œ kvvk œ ˆsech at ‰ i ˆtanh at ‰ j Ê ddtT œ ˆ "a sech at tanh at ‰ i ˆ "a sech# at ‰ j
ˆ dT ‰
Ê ¸ ddtT ¸ œ É a"# sech# ˆ at ‰ tanh# ˆ at ‰ a"# sech% ˆ at ‰ œ "a sech ˆ at ‰ Ê N œ ¸ ddtT ¸ œ ˆ tanh at ‰ i ˆsech at ‰ j ;
dt
1
"
t
"
# t
, œ k1vk † ¸ ddtT ¸ œ cosh
t † a sech ˆ a ‰ œ a sech ˆ a ‰.
a
16. r œ (cosh t)i (sinh t)j tk Ê v œ (sinh t)i (cosh t)j k Ê kvk œ Èsinh# t ( cosh t)# 1 œ È2 cosh t
Ê T œ kvvk œ Š È" tanh t‹ i È" j Š È" sech t‹ k Ê ddtT œ Š È" sech# t‹ i Š È" sech t tanh t‹ k
2
Ê
2
2
¸ ddtT ¸ œ É "# sech% t "# sech# t tanh# t œ È" sech t
2
, œ k1vk † ¸ ddtT ¸ œ È 1
2 cosh t
Ê
2
2
ˆ ddtT ‰
N œ ¸ dT ¸ œ (sech t)i (tanh t)k ;
dt
† È" sech t œ "# sech# t.
2
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
776
Chapter 13 Vector-Valued Functions and Motion in Space
$Î#
k2ak
œ k2ak a1 4a# x# b
a1 4a# x# b$Î#
17. y œ ax# Ê yw œ 2ax Ê yww œ 2a; from Exercise 5(a), ,(x) œ
Ê ,w (x) œ 3# k2ak a1 4a# x# b
&Î#
a8a# xb ; thus, ,w (x) œ 0 Ê x œ 0. Now, ,w (x) 0 for x 0 and ,w (x) 0 for
x 0 so that ,(x) has an absolute maximum at x œ 0 which is the vertex of the parabola. Since x œ 0 is the
only critical point for ,(x), the curvature has no minimum value.
18. r œ (a cos t)i (b sin t)j Ê v œ (a sin t)i (b cos t)j Ê a œ (a cos t)i (b sin t)j Ê v ‚ a
â
â
i
j
kâ
â
â
â
œ â a sin t b cos t 0 â œ abk Ê kv ‚ ak œ kabk œ ab, since a b 0; , (t) œ kvkv‚k$ak
â
â
â a cos t b sin t 0 â
$Î#
œ ab aa# sin# t b# cos# tb
&Î#
; ,w (t) œ #3 (ab) aa# sin# t b# cos# tb
œ 3# (ab) aa# b# b (sin 2t) aa# sin# t b# cos# tb
&Î#
a2a# sin t cos t 2b# sin t cos tb
; thus, ,w (t) œ 0 Ê sin 2t œ 0 Ê t œ 0, 1 identifying
points on the major axis, or t œ 1# , 3#1 identifying points on the minor axis. Furthermore, ,w (t) 0 for
0 t 1# and for 1 t 3#1 ; ,w (t) 0 for 1# t 1 and 3#1 t 21. Therefore, the points associated
with t œ 0 and t œ 1 on the major axis give absolute maximum curvature and the points associated with t œ 1#
and t œ 31
# on the minor axis give absolute minimum curvature.
#
#
19. , œ a# a b# Ê dda, œ aa#ab#bb# ; dda, œ 0 Ê a# b# œ 0 Ê a œ „ b Ê a œ b since a, b
0. Now, dda, 0 if
"
a b and dda, 0 if a b Ê , is at a maximum for a œ b and ,(b) œ b# b b# œ 2b
is the maximum value of ,.
20. (a) From Example 5, the curvature of the helix r(t) œ (a cos t)i (a sin t)j btk, a, b 0 is , œ a# a b# ; also
kvk œ Èa# b# . For the helix r(t) œ (3 cos t)i (3 sin t)j tk, 0 Ÿ t Ÿ 41, a œ 3 and b œ 1 Ê , œ # 3 # œ 3
41
and kvk œ È10 Ê K œ '0
3 1
%1
3 È
121
10 dt œ ’ È310 t“ œ È
10
10
!
10
(b) y œ x# Ê x œ t and y œ t# , _ t _ Ê r(t) œ ti t# j Ê v œ i 2tj Ê kvk œ È1 4t# ;
4
2
T œ È1 1 4t# i È12t 4t# j; ddtT œ a1 4t4t# b3/2 i a1 4t
j; ¸ ddtT ¸ œ É a16t
œ 1 24t# . Thus
# b3/2
1 4t# b3
2
, œ È1 1 4t# † 1 24t# œ
'
œaÄ
lim
_ a
0
_
2
3
ŠÈ1 4t# ‹
2
14t# dt b lim
Ä_
œaÄ
lim
a tan" 2ab lim
_
. Then K œ 'c_
2
_
$
ŠÈ1 4t# ‹
ŠÈ1 4t# ‹ dt œ ' _ 124t# dt
'0 1 24t dt œ a Älim_ ctan" 2td !a lim
bÄ_
b
#
a tan" 2bb œ 1# 1# œ 1
ctan" 2td 0
b
bÄ_
21. r œ ti (sin t)j Ê v œ i (cos t)j Ê kvk œ È1# (cos t)# œ È1 cos# t Ê ¸v ˆ 1# ‰¸ œ É1 cos# ˆ 1# ‰ œ 1; T œ kvvk
œ Èi cos t j2 Ê ddtT œ
1 cos t
sin t cos t
i sin2t 3/2 j Ê
a1 cos2 tb3/2
a1 cos tb
ksin tk
¸ ddtT ¸ œ 1 ¸ ddtT ¸
cos2 t ;
tœ 12
¸sin 1 ¸
œ 1 cos22ˆ 1 ‰ œ 11 œ 1. Thus ,ˆ 12 ‰ œ 11 † 1 œ 1
2
#
Ê 3 œ "1 œ 1 and the center is ˆ 1# ß 0‰ Ê ˆx 1# ‰ y# œ 1
2
1
22. r œ (2 ln t)i ˆt "t ‰ j Ê v œ ˆ 2t ‰ i ˆ1 t"# ‰ j Ê kvk œ É t42 ˆ1 t12 ‰ œ t t2 1 Ê T œ kvvk œ t2 2t 1 i tt2 1 j;
2
2
2 ˆt 1 ‰
4at2 1b 16t2
dT
4t
2
1 ¸ dT ¸
t2
2
2t2
2
¸ dT ¸
œ t2 dt œ at2 1b2 i at2 1b2 j Ê dt œ Ê
1 . Thus , œ kvk † dt œ t2 1 † t2 1 œ at2 1b2 Ê , a1b œ 22
at2 1b4
2
2
œ "# Ê 3 œ ," œ 2. The circle of curvature is tangent to the curve at P(0ß 2) Ê circle has same tangent as the curve
Ê v(1) œ 2i is tangent to the circle Ê the center lies on the y-axis. If t Á 1 (t 0), then (t 1)# 0
#
Ê t# 2t 1 0 Ê t# 1 2t Ê t t 1 2 since t 0 Ê t "t 2 Ê ˆt "t ‰ 2 Ê y 2 on both
sides of (0ß 2) Ê the curve is concave down Ê center of circle of curvature is (0ß 4) Ê x# (y 4)# œ 4
is an equation of the circle of curvature
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 13.4 Curvature and Normal Vectors of a Curve
23. y œ x# Ê f w (x) œ 2x and f ww (x) œ 2
Ê ,œ
k2 k
2
œ
a1 (2x)# b$Î#
a1 4x# b$Î#
%
24. y œ x4 Ê f w (x) œ x$ and f ww (x) œ 3x#
Ê ,œ
k3x# k
#
$Î#
Š1 ax$ b ‹
œ
3x#
a1 x' b$Î#
25. y œ sin x Ê f w (x) œ cos x and f ww (x) œ sin x
Ê ,œ
k sin xk
ksin xk
œ
a1 cos# xb$Î#
a1 cos# xb$Î#
26. y œ ex Ê f w (x) œ ex and f ww (x) œ ex
Ê ,œ
kex k
$Î#
#
Š1 aex b ‹
ex
$Î#
1 e2x ‰
œˆ
27-34. Example CAS commands:
Maple:
with( plots );
r := t -> [3*cos(t),5*sin(t)];
lo := 0;
hi := 2*Pi;
t0 := Pi/4;
P1 := plot( [r(t)[], t=lo..hi] ):
display( P1, scaling=constrained, title="#27(a) (Section 13.4)" );
CURVATURE := (x,y,t) ->simplify(abs(diff(x,t)*diff(y,t,t)-diff(y,t)*diff(x,t,t))/(diff(x,t)^2+diff(y,t)^2)^(3/2));
kappa := eval(CURVATURE(r(t)[],t),t=t0);
UnitNormal := (x,y,t) ->expand( [-diff(y,t),diff(x,t)]/sqrt(diff(x,t)^2+diff(y,t)^2) );
N := eval( UnitNormal(r(t)[],t), t=t0 );
C := expand( r(t0) + N/kappa );
OscCircle := (x-C[1])^2+(y-C[2])^2 = 1/kappa^2;
evalf( OscCircle );
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
777
778
Chapter 13 Vector-Valued Functions and Motion in Space
P2 := implicitplot( (x-C[1])^2+(y-C[2])^2 = 1/kappa^2, x=-7..4, y=-4..6, color=blue ):
display( [P1,P2], scaling=constrained, title="#27(e) (Section 13.4)" );
Mathematica: (assigned functions and parameters may vary)
In Mathematica, the dot product can be applied either with a period "." or with the word, "Dot".
Similarly, the cross product can be applied either with a very small "x" (in the palette next to the arrow) or with the word,
"Cross". However, the Cross command assumes the vectors are in three dimensions
For the purposes of applying the cross product command, we will define the position vector r as a three dimensional vector
with zero for its z-component. For graphing, we will use only the first two components.
Clear[r, t, x, y]
r[t_]={3 Cos[t], 5 Sin[t] }
t0= 1 /4; tmin= 0; tmax= 21;
r2[t_]= {r[t][[1]], r[t][[2]]}
pp=ParametricPlot[r2[t], {t, tmin, tmax}];
mag[v_]=Sqrt[v.v]
vel[t_]= r'[t]
speed[t_]=mag[vel[t]]
acc[t_]= vel'[t]
curv[t_]= mag[Cross[vel[t],acc[t]]]/speed[t]3 //Simplify
unittan[t_]= vel[t]/speed[t]//Simplify
unitnorm[t_]= unittan'[t] / mag[unittan'[t]]
ctr= r[t0] + (1 / curv[t0]) unitnorm[t0] //Simplify
{a,b}= {ctr[[1]], ctr[[2]]}
To plot the osculating circle, load a graphics package and then plot it, and show it together with the original curve.
<<Graphics`ImplicitPlot`
pc=ImplicitPlot[(x a)2 + (y b)2 == 1/curv[t0]2 , {x, 8, 8},{y, 8, 8}]
radius=Graphics[Line[{{a, b}, r2[t0]}]]
Show[pp, pc, radius, AspectRatio Ä 1]
13.5 TANGENTIAL AND NORMAL COMPONENTS OF ACCELERATION
1. r œ (a cos t)i (a sin t)j btk Ê v œ (a sin t)i (a cos t)j bk Ê kvk œ È(a sin t)# (a cos t)# b#
œ Èa# b# Ê aT œ d kvk œ 0; a œ (a cos t)i (a sin t)j Ê kak œ È(a cos t)# (a sin t)# œ Èa# œ kak
dt
#
Ê aN œ Ékak a#T œ Ékak# 0# œ kak œ kak Ê a œ (0)T kak N œ kak N
2. r œ (1 3t)i (t 2)j 3tk Ê v œ 3i j 3k Ê kvk œ È3# 1# (3)# œ È19 Ê aT œ dtd kvk œ 0; a œ 0
Ê aN œ Ékak# a#T œ 0 Ê a œ (0)T (0)N œ 0
3. r œ (t 1)i 2tj t# k Ê v œ i 2j 2tk Ê kvk œ È1# 2# (2t)# œ È5 4t# Ê aT œ "# a5 4t# b
# "Î#
œ 4t a5 4t b
È
#
"Î#
(8t)
#
Ê aT (1) œ È49 œ 34 ; a œ 2k Ê a(1) œ 2k Ê ka(1)k œ 2 Ê aN œ Ékak a#T œ É2# ˆ 34 ‰
È
2 5
œ É 20
Ê a(1) œ 43 T 2 3 5 N
9 œ 3
4. r œ (t cos t)i (t sin t)j t# k Ê v œ (cos t t sin t)i (sin t t cos t)j 2tk
"Î#
Ê kvk œ È(cos t t sin t)# (sin t t cos t)# (2t)# œ È5t# 1 Ê aT œ " a5t# 1b
(10t)
#
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 13.5 Tangential and Normal Components of Acceleration
œ È5t5t# 1 Ê aT (0) œ 0; a œ (2 sin t t cos t)i (2 cos t t sin t)j 2k Ê a(0) œ 2j 2k Ê ka(0)k
#
œ È2# 2# œ 2È2 Ê aN œ Ékak# a#T œ ÊŠ2È2‹ 0# œ 2È2 Ê a(0) œ (0)T 2È2N œ 2È2N
5. r œ t# i ˆt "3 t$ ‰ j ˆt "3 t$ ‰ k Ê v œ 2ti a1 t# b j a1 t# b k Ê kvk œ É(2t)# a1 t# b# a1 t# b#
œ È2 at% 2t# 1b œ È2 a1 t# b Ê aT œ 2tÈ2 Ê aT (0) œ 0; a œ 2i 2tj 2tk Ê a(0) œ 2i Ê ka(0)k œ 2
Ê aN œ Ékak# a#T œ È2# 0# œ 2 Ê a(0) œ (0)T 2N œ 2N
6. r œ aet cos tb i aet sin tb j È2et k Ê v œ aet cos t et sin tb i aet sin t et cos tb j È2et k
#
Ê kvk œ Êaet cos t et sin tb# aet sin t et cos tb# ŠÈ2et ‹ œ È4e2t œ 2et Ê aT œ 2et Ê aT (0) œ 2;
a œ aet cos t et sin t et sin t et cos tb i aet sin t et cos t et cos t et sin tb j È2et k
#
œ a2et sin tb i a2et cos tb j È2et k Ê a(0) œ 2j È2k Ê ka(0)k œ Ê2# ŠÈ2‹ œ È6
#
Ê aN œ Ékak# a#T œ ÊŠÈ6‹ 2# œ È2 Ê a(0) œ 2T È2N
7. r œ (cos t)i (sin t)j k Ê v œ ( sin t)i (cos t)j Ê kvk œ È( sin t)# (cos t)# œ 1 Ê T œ kvvk
È
È
œ ( sin t)i (cos t)j Ê T ˆ 14 ‰ œ #2 i #2 j ; ddtT œ ( cos t)i (sin t)j Ê ¸ ddtT ¸ œ È( cos t)# ( sin t)#
â
â
j
kâ
â i
ˆ ddtT ‰
È2
È2
â
â
1
œ 1 Ê N œ ¸ dT ¸ œ ( cos t)i (sin t)j Ê N ˆ 4 ‰ œ # i # j ; B œ T ‚ N œ â sin t cos t 0 â œ k
â
â
dt
â cos t sin t 0 â
Ê B ˆ 14 ‰ œ k , the normal to the osculating plane; r ˆ 14 ‰ œ
È
È2
È2
# i # jk
È
È
Ê P œ Š #2 ß #2 ß 1‹ lies on the
È
osculating plane Ê 0 Šx #2 ‹ 0 Šy #2 ‹ (z (1)) œ 0 Ê z œ 1 is the osculating plane; T is normal
È
È
È
È
È
È
to the normal plane Ê Š #2 ‹ Šx #2 ‹ Š #2 ‹ Šy #2 ‹ 0(z (1)) œ 0 Ê #2 x #2 y œ 0
Ê x y œ 0 is the normal plane; N is normal to the rectifying plane
È
È
È
È
È
È
Ê Š #2 ‹ Šx #2 ‹ Š #2 ‹ Šy #2 ‹ 0(z (1)) œ 0 Ê #2 x #2 y œ 1 Ê x y œ È2 is the
rectifying plane
8. r œ (cos t)i (sin t)j tk Ê v œ ( sin t)i (cos t)j k Ê kvk œ Èsin# t cos# t 1 œ È2 Ê T œ kvvk
œ Š È"2 sin t‹ i Š È"2 cos t‹ j È"2 k Ê ddtT œ Š È"2 cos t‹ i Š È"2 sin t‹ j Ê ¸ ddtT ¸
ˆ dT ‰
œ É "# cos# t "# sin# t œ È" Ê N œ ¸ ddtT ¸ œ ( cos t)i (sin t)j ; thus T(0) œ È" j È" k and N(0) œ i
2
2
2
dt
â
â
â i
j
k â
â
"
" ââ
"
"
Ê B(0) œ ââ 0
È2
È2 â œ È2 j È2 k , the normal to the osculating plane; r(0) œ i Ê P(1ß 0ß 0) lies on
â 1 0
0 ââ
â
the osculating plane Ê 0(x 1) È"2 (y 0) È"2 (z 0) œ 0 Ê y z œ 0 is the osculating plane; T is normal
to the normal plane Ê 0(x 1) È"2 (y 0) È"2 (z 0) œ 0 Ê y z œ 0 is the normal plane; N is normal to
the rectifying plane Ê 1(x 1) 0(y 0) 0(z 0) œ 0 Ê x œ 1 is the rectifying plane.
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
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780
Chapter 13 Vector-Valued Functions and Motion in Space
9. By Exercise 9 in Section 13.4, T œ ˆ 35 cos t‰ i ˆ 35 sin t‰ j 45 k and N œ ( sin t)i (cos t)j so that B œ T ‚ N
â i
j
k ââ
â
â3
â
œ â 5 cos t 35 sin t 45 â œ ˆ 45 cos t‰ i ˆ 45 sin t‰ j 35 k . Also v œ (3 cos t)i (3 sin t)j 4k
â
â
â sin t cos t 0 â
â
â
i
j
kâ
â
â
â
Ê a œ (3 sin t)i (3 cos t)j Ê ddta œ (3 cos t)i (3 sin t)j and v ‚ a œ â 3 cos t 3 sin t 4 â
â
â
â 3 sin t 3 cos t 0 â
œ (12 cos t)i (12 sin t)j 9k Ê kv ‚ ak2 œ a12 cos tb# a12 sin tb# a9b# œ 225. Thus
7œ
â
â 3 cos t
â
â 3 sin t
â
â 3 cos t
â
3 sin t 4 â
â
3 sin t 0 â
â
#
3 sin t 0 â
t9 cos# tb
36
4
œ 4†a9 sin225
œ 225
œ 25
225
10. By Exercise 10 in Section 13.4, T œ (cos t)i (sin t)j and N œ ( sin t)i (cos t)j ; thus B œ T ‚ N
â
â
j
kâ
â i
â
â
œ â cos t sin t 0 â œ acos# t sin# tb k œ k. Also v œ (t cos t)i (t sin t)j
â
â
â sin t cos t 0 â
Ê a œ atasin tb cos tbi at cos t sin tbj Ê ddta œ at cos t sin t sin tbi at sin t cos t cos tbj
â
â
i
j
kâ
â
â
â
t cos t
t sin t
0â
œ at cos t 2 sin tbi a2 cos t t sin tbj. Thus v ‚ a œ â
â
â
â at sin t cos tb at cos t sin tb 0 â
#
œ [(t cos t)(t cos t sin t) (t sin t)(t sin t cos t)]k œ t# k Ê kv ‚ ak2 œ at# b œ t4 . Thus
â
â
t cos t
t sin t
0â
â
â
â
sin t t cos t 0 â
â cos t t sin t
â
â
â 2 sin t t cos t 2 cos t t sin t 0 â
7œ
œ t04 œ 0
t4
t sin t
11. By Exercise 11 in Section 13.4, T œ Š cos È
‹ i Š sin tÈ2cos t ‹ j and N œ Š cosÈt 2 sin t ‹ i Š sinÈt 2 cos t ‹ j ; Thus
2
â
i
j
k ââ
â
â cos t sin t
sin t cos t
0 ââ œ ’Š cos# t 2 cos t sin t sin# t ‹ Š sin# t 2 sin t cos t cos# t ‹“ k
È2
È2
B œ T ‚ N œ ââ
2
2
â
â cos t sin t sin t cos t 0 â
â
â
È2
È2
a2tb
a2tb
œ ’Š 1 sin
‹ Š 1 sin
‹“ k œ k . Also, v œ aet cos t et sin tb i aet sin t et cos tb j
2
2
Ê a œ c et asin t cos tb et acos t sin tb d i c et acos t sin tb et asin t cos tb d j=a2et sin tb i a2et cos tb j
â
â
i
j
kâ
â
â
â
Ê ddta œ 2et acos t sin tb i 2et asin t cos tbj. Thus v ‚ a œ â et acos t sin tb et asin t cos tb 0 â œ 2e2t k
â
â
t
t
2e cos t
0â
â 2e sin t
#
Ê kv ‚ ak2 œ a2e2t b œ 4e4t . Thus 7 œ
â t
â e acos t sin tb
â
â
2et sin t
â
â 2et acos t sin tb
et asin t cos tb 0 ââ
â
2et cos t
0â
â
2et asin t cos tb 0 â
œ0
4e4t
5
‰
ˆ 12
‰
12. By Exercise 12 in Section 13.4, T œ ˆ 12
13 cos 2t i 13 sin 2t j 13 k and N œ ( sin 2t)i (cos 2t)j so
â
â
i
j
kâ
â
â ˆ 12
12
5 â
12
‰
ˆ
‰
ˆ5
‰
ˆ5
‰
B œ T ‚ N œ â 13 cos 2t
13 sin 2t
13 ââ œ 13 cos 2t i 13 sin 2t j 13 k . Also,
â
â a sin 2tb
a cos 2tb
0â
v œ (12 cos 2t)i (12 sin 2t)j 5k Ê a œ (24 sin 2t)i (24 cos 2t)j and ddta œ (48 cos 2t)i (48 sin 2t)j
â
â
i
j
kâ
â
â
â
v ‚ a œ â 12 cos 2t 12 sin 2t 5 â œ (120 cos 2t)i (120 sin 2t)j 288k Ê kv ‚ ak2
â
â
â 24 sin 2t 24 cos 2t 0 â
œ (120 cos 2t)# (120 sin 2t)# (288)# œ 120# acos# 2t sin# 2tb 288# œ 97344. Thus
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 13.5 Tangential and Normal Components of Acceleration
781
â
â
â 12 cos 2t 12 sin 2t 5 â
â
â
â 24 sin 2t 24 cos 2t 0 â
â
â
â 48 cos 2t 48 sin 2t 0 â
24†48b
10
7œ
œ 5†a97344
œ 169
97344
13. By Exercise 13 in Section 13.4, T œ at# t1b1/2 i at# 11b1/2 j and N œ Èt#1 1 i Èt#t 1 j so that B œ T ‚ N
â i
â #
â
j
k ââ
â
ât
t 0â
â t
â
1
â
â
0
a
d
#
â
â
œ â Èt# 1 Èt# 1
â œ k. Also, v œ t i tj Ê a œ 2ti j Ê dt œ 2i so that ââ 2t " 0 ââ œ 0 Ê 7 œ 0
t
â "
â
â 2 0 0â
â Èt# 1 Èt# 1 0 â
14. By Exercise 14 in Section 13.4, T œ ( cos t)i (sin t)j and N œ (sin t)i (cos t)j so that B œ T ‚ N
â
â
j
kâ
â i
â
â
œ â cos t sin t 0 â œ k . Also, v œ a3 cos# t sin tb i a3 sin# t cos tb j
â
â
cos t 0 â
â sin t
Ê a œ dtd a3 cos# t sin tb i dtd a3 sin# t cos tb j Ê ddta œ dtd ˆ dtd a3 cos# t sin tb‰ i dtd ˆ dtd a3 sin# t cos tb‰ j
â
â
â
3 cos# t sin t
3 sin# t cos t
0â
â d
â
d
#
0 ââ œ 0 Ê 7 œ 0
Ê ââ dt a3 cos# t sin tb
dt a3 sin t cos tb
â d ˆ d a3 cos# t sin tb‰ d ˆ d a3 sin# t cos tb‰ 0 â
â dt dt
â
dt dt
15. By Exercise 15 in Section 13.4, T œ kvvk œ ˆsech at ‰ i ˆtanh at ‰ j and N œ ˆ tanh at ‰ i ˆsech at ‰ j so that B œ T ‚ N
â
i
j
k ââ
â
â sech ˆ t ‰
t
tanh ˆ a ‰ 0 ââ œ k. Also, v œ i ˆsinh at ‰ j Ê a œ ˆ "a cosh at ‰ j Ê ddta œ a"# sinh ˆ at ‰ j so that
ω
a
â
â
â tanh ˆ at ‰ sech ˆ at ‰ 0 â
â
â
â1
sinh ˆ at ‰
0â
â
â
â 0 " cosh ˆ t ‰ 0 â œ 0 Ê 7 œ 0
a
a
â
â
â 0 " sinh ˆ t ‰ 0 â
â
â
a#
a
16. By Exercise 16 in Section 13.4, T œ Š È"2 tanh t‹ i È"2 j Š È"2 sech t‹ k and N œ (sech t)i (tanh t)k so that
â
â
â
â
i
j
k
â "
â
"
"
â
B œ T ‚ N œ â È2 tanh t È2 È2 sech t ââ œ Š È"2 tanh t‹ i È"2 j Š È"2 sech t‹ k. Also, v œ (sinh t)i (cosh t)j k
â sech t
0
tanh t ââ
â
â
â
j
kâ
â i
â
â
da
a œ (cosh t)i (sinh t)j Ê dt œ (sinh t)i (cosh t)j and v ‚ a œ â sinh t cosh t 1 â
â
â
â cosh t sinh t 0 â
œ (sinh t)i (cosh t)j acosh2 t sinh2 tbk œ (sinh t)i (cosh t)j k Ê kv ‚ ak# œ sinh# t cosh# t 1. Thus
â
â
â sinh t cosh t 1 â
â
â
â cosh t sinh t 0 â
â
â
sinh
t
cosh
t
0
â
â
"
7 œ sinh# t cosh# t1 œ sinh# t "
cosh# t 1 œ # cosh# t .
17. Yes. If the car is moving along a curved path, then , Á 0 and aN œ , kvk# Á 0 Ê a œ aT T aN N Á 0 .
18. kvk constant Ê aT œ dtd kvk œ 0 Ê a œ aN N is orthogonal to T Ê the acceleration is normal to the path
19. a ¼ v Ê a ¼ T Ê aT œ 0 Ê dtd kvk œ 0 Ê kvk is constant
20. a(t) œ aT T aN N , where aT œ dtd kvk œ dtd (10) œ 0 and aN œ , kvk# œ 100, Ê a œ 0T 100,N. Now, from
Exercise 5(a) Section 12.4, we find for y œ f(x) œ x# that , œ
kf ww (x)k
2
2
œ
; also,
$Î# œ
c1 (2x)# d$Î#
a1 4x# b$Î#
1 af w (x)b# ‘
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
782
Chapter 13 Vector-Valued Functions and Motion in Space
r(t) œ ti t# j is the position vector of the moving mass Ê v œ i 2tj Ê kvk œ È1 4t#
Ê T œ È1 " 4t# (i 2tj). At (0ß 0): T(0) œ i, N(0) œ j and ,(0) œ 2 Ê F œ ma œ m(100,)N œ 200m j ;
È
È
2
At ŠÈ2ß 2‹ : T ŠÈ2‹ œ "3 Ši 2È2j‹ œ 3" i 2 3 2 j , N ŠÈ2‹ œ 2 3 2 i 3" j , and , ŠÈ2‹ œ 27
Ê F œ ma
È
È
2 2
400 2
"
200
‰
œ m(100,)N œ ˆ 200
27 m Š 3 i 3 j‹ œ 81 m i 81 m j
2
2
3
2
d s
ds d s
‰
ˆ ds ‰
ˆ ds ‰
21. By a œ aT T aN N we have v ‚ a œ ˆ ds
dt T ‚ ’ dt2 T , dt N “ œ Š dt dt2 ‹aT ‚ Tb , dt aT ‚ Nb
3
lv‚al
ds
3
‰
œ ,ˆ ds
dt B. It follows that lv ‚ al œ , ¹ dt ¹ lBl œ , lvl Ê , œ lvl3
3
22. aN œ 0 Ê , kvk# œ 0 Ê , œ 0 (since the particle is moving, we cannot have zero speed) Ê the curvature is zero
so the particle is moving along a straight line
23. From Example 1, kvk œ t and aN œ t so that aN œ , kvk# Ê , œ kvaNk# œ tt# œ "t , t Á 0 Ê 3 œ ," œ t
24. r œ (x! At)i (y! Bt)j (z! Ct)k Ê v œ Ai Bj Ck Ê a œ 0 Ê v ‚ a œ 0 Ê , œ 0. Since the curve
is a plane curve, 7 œ 0.
25. If a plane curve is sufficiently differentiable the torsion is zero as the following argument shows:
r œ f(t)i g(t)j Ê v œ f w (t)i gw (t)j Ê a œ f ww (t)i gww (t)j Ê ddta œ f www (t)i gwww (t)j
Ê
â w
â
â f (t) gw (t) 0 â
â ww
â
â f (t) gww (t) 0 â
â www
â
â f (t) gwww (t) 0 â
7œ
œ0
#
kv ‚ a k
26. v œ aa sin tbi aa cos tbj b k and a œ aa cos tbi aa sin tbj
To find the torsion: 7 œ
#
â
â
â a sin t a cos t b â
â
â
â a cos t a sin t 0 â
â
â
â a sin t a cos t 0 â
2
Ša È a 2 b 2 ‹
œ
bˆa2 cos2 t a2 sin2 t‰
a2 bˆcos2 t sin2 t‰
œ a2 aa2 b2 b œ a2 b b2 Ê
a 2 aa2 b 2 b
#
#
7 w (b) œ aaa# bb# b# ;
#
7 w (b) œ 0 Ê aaa# bb# b# œ 0 Ê a# b# œ 0 Ê b œ „ a Ê b œ a since a, b 0. Also b a Ê 7 w 0 and b a
"
Ê 7 w 0 so 7max occurs when b œ a Ê 7max œ a# a a# œ 2a
27. r(t) œ f(t)i g(t)j h(t)k Ê v œ f w (t)i gw (t)j hw (t)k; v † k œ 0 Ê hw (t) œ 0 Ê h(t) œ C
Ê r(t) œ f(t)i g(t)j Ck and r(a) œ f(a)i g(a)j Ck œ 0 Ê f(a) œ 0, g(a) œ 0 and C œ 0 Ê h(t) œ 0.
28. From Exercise 26, v œ (a sin t)i (a cos t)j bk Ê kvk œ Èa# b# Ê T œ kvvk
ˆ dT ‰
œ È #" # c(a sin t)i (a cos t)j bkd ; ddtT œ È #" # c(a cos t)i (a sin t)jd Ê N œ ¸ ddtT ¸
a
b
a
b
dt
â
â
â
i
j
k â
â
â
a sin t
a cos t
b
œ (cos t)i (sin t)j ; B œ T ‚ N œ ââ Èa# b# Èa# b# Èa# b# ââ
â cos t
0 ââ
sin t
â
t
œ Èba#sin tb# i Èbacos
j Èa#a b# k Ê ddtB œ Èa#" b# c(b cos t)i (b sin t)jd Ê ddtB † N œ Èa#b b#
# b#
Ê 7 œ k"vk ˆ ddtB † N‰ œ Š È #" # ‹ Š È #b # ‹ œ a# b b# , which is consistent with the result in Exercise 26.
a
b
a
b
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 13.5 Tangential and Normal Components of Acceleration
29-32. Example CAS commands:
Maple:
with( LinearAlgebra );
r := < t*cos(t) | t*sin(t) | t >;
t0 := sqrt(3);
rr := eval( r, t=t0 );
v := map( diff, r, t );
vv := eval( v, t=t0 );
a := map( diff, v, t );
aa := eval( a, t=t0 );
s := simplify(Norm( v, 2 )) assuming t::real;
ss := eval( s, t=t0 );
T := v/s;
TT := vv/ss ;
q1 := map( diff, simplify(T), t ):
NN := simplify(eval( q1/Norm(q1,2), t=t0 ));
BB := CrossProduct( TT, NN );
kappa := Norm(CrossProduct(vv,aa),2)/ss^3;
tau := simplify( Determinant(< vv, aa, eval(map(diff,a,t),t=t0) >)/Norm(CrossProduct(vv,aa),2)^3 );
a_t := eval( diff( s, t ), t=t0 );
a_n := evalf[4]( kappa*ss^2 );
Mathematica: (assigned functions and value for t0 will vary)
Clear[t, v, a, t]
mag[vector_]:=Sqrt[vector.vector]
Print["The position vector is ", r[t_]={t Cos[t], t Sin[t], t}]
Print["The velocity vector is ", v[t_]= r'[t]]
Print["The acceleration vector is ", a[t_]= v'[t]]
Print["The speed is ", speed[t_]= mag[v[t]]//Simplify]
Print["The unit tangent vector is ", utan[t_]= v[t]/speed[t] //Simplify]
Print["The curvature is ", curv[t_]= mag[Cross[v[t],a[t]]] / speed[t]3 //Simplify]
Print["The torsion is ", torsion[t_]= Det[{v[t], a[t], a'[t]}] / mag[Cross[v[t],a[t]]]2 //Simplify]
Print["The unit normal vector is ", unorm[t_]= utan'[t] / mag[utan'[t]] //Simplify]
Print["The unit binormal vector is ", ubinorm[t_]= Cross[utan[t],unorm[t]] //Simplify]
Print["The tangential component of the acceleration is ", at[t_]=a[t].utan[t] //Simplify]
Print["The normal component of the acceleration is ", an[t_]=a[t].unorm[t] //Simplify]
You can evaluate any of these functions at a specified value of t.
t0= Sqrt[3]
{utan[t0], unorm[t0], ubinorm[t0]}
N[{utan[t0], unorm[t0], ubinorm[t0]}]
{curv[t0], torsion[t0]}
N[{curv[t0], torsion[t0]}]
{at[t0], an[t0]}
N[{at[t0], an[t0]}]
To verify that the tangential and normal components of the acceleration agree with the formulas in the book:
at[t]== speed'[t] //Simplify
an[t]==curv [t] speed[t]2 //Simplify
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
783
784
Chapter 13 Vector-Valued Functions and Motion in Space
13.6 VELOCITY AND ACCELERATION IN POLAR COORDINATES
1.
Þ
ÞÞ
.
..
d)
d)
d)
dt œ 3 œ ) Ê ) œ 0, r œ aa1 cos )b Ê r œ a sin ) dt œ 3a sin ) Ê r œ 3a cos ) dt œ 9a cos )
v œ a3a sin )bur aaa1 cos )bba3bu) œ a3a sin )bur 3aa1 cos )bu)
a œ Š9a cos ) aa1 cos )ba3b2 ‹ur aaa1 cos )b † 0 2a3a sin )ba3bbu)
œ a9a cos ) 9a 9a cos )bur a18a sin )bu) œ 9aa2 cos ) 1bur a18a sin )bu)
2.
Þ
ÞÞ
.
..
d)
d)
d) ‰
ˆ
dt œ 2t œ ) Ê ) œ 2, r œ a sin 2) Ê r œ a cos 2) † 2 dt œ 4ta cos 2) Ê r œ 4ta sin 2) † 2 dt 4a cos 2)
2
œ 16t a sin 2) 4a cos 2)
v œ a4ta cos 2)bur aa sin 2)ba2tbu) œ a4ta cos 2)bur a2ta sin 2)bu)
a œ ’a 16t2 a sin 2) 4a cos 2)b aa sin 2)ba2tb2 “ur aa sin 2)ba2b 2a4ta cos 2)ba2tb‘u)
œ ’16t2 a sin 2) 4a cos 2) 4t2 a sin 2)“ur 2a sin 2) 16t2 a cos 2)‘u)
œ ’20t2 a sin 2) 4a cos 2)“ur 2a sin 2) 16t2 a cos 2)‘u) œ 4aacos 2) 5t2 sin 2)bur 2aasin 2) 8t2 cos 2)bu)
3.
Þ
ÞÞ
.
..
d)
a)
Ê r œ ea ) † a ddt) œ 2a ea ) Ê r œ 2a ea ) † a ddt) œ 4a2 ea )
dt œ 2 œ ) Ê ) œ 0, r œ e
v œ ˆ2a ea ) ‰ur ˆea ) ‰a2bu) œ ˆ2a ea ) ‰ur ˆ2ea ) ‰u)
a œ ’ˆ 4a2 ea ) ‰ ˆea ) ‰a2b2 “ur ’ˆea ) ‰a0b 2ˆ2a ea ) ‰a2b“u) œ ’4a2 ea ) 4ea ) “ur ’0 8a ea ) “u)
œ 4ea ) aa2 1bur ˆ8a ea ) ‰u)
Þ
ÞÞ
.
..
4. ) œ 1 et Ê ) œ et Ê ) œ et , r œ aa1 sin tb Ê r œ a cos t Ê r œ a sin t
v œ aa cos tbur aaa1 sin tbbaet bu) œ aa cos tbur a et a1 sin tbu)
a œ ’a a sin tb aaa1 sin tbbaet b “ur ’aaa1 sin tbbaet b 2aa cos tbaet b“u)
2
œ ’a sin t a e2t a1 sin tb“ur ’a et a1 sin tb 2a et cos t“u)
œ aasin t e2t a1 sin tbbur a et aa1 sin tb 2cos tbu)
œ aasin t e2t a1 sin tbbur a et a2cos t 1 sin tbu)
Þ
ÞÞ
.
..
5. ) œ 2t Ê ) œ 2 Ê ) œ 0, r œ 2 cos 4t Ê r œ 8 sin 4t Ê r œ 32 cos 4t
v œ a8 sin 4tbur a2 cos 4tba2bu) œ 8asin 4tbur 4acos 4tbu)
a œ Ša32 cos 4tb a2 cos 4tba2b2 ‹ur aa2 cos 4tb † 0 2a8 sin 4tba2bbu)
œ a32 cos 4t 8 cos 4tbur a0 32sin 4tbu) œ 40acos 4tbur 32asin 4tbu)
r v#
! !
6. e œ GM
1 Ê v#! œ GM(er! 1) Ê v! œ É GM(er! 1) ;
Circle: e œ 0 Ê v! œ É GM
r!
2GM
Ellipse: 0 e 1 Ê É GM
r! v! É r!
Parabola: e œ 1 Ê v! œ É 2GM
r!
Hyperbola: e 1 Ê v! É 2GM
r!
GM
#
7. r œ GM
Ê v œ É GM
v# Ê v œ r
r which is constant since G, M, and r (the radius of orbit) are constant
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Chapter 13 Practice Exercises
?t)
r(t) r(t)
8. ?A œ "# kr(t ?t) ‚ r(t)k Ê ??At œ "# ¹ r(t ‚ r(t)¹ œ "# ¹ r(t ?t) ?
‚ r(t)¹
?t
t
" r(t ?t) r(t)
œ "# ¹ r(t ??t)t r(t) ‚ r(t) ?"t r(t) ‚ r(t)¹ œ #" ¹ r(t ??t)t r(t) ‚ r(t)¹ Ê dA
‚ r(t)¹
¹
dt œ ?tlim
?t
Ä0 #
Þ
œ "# ¸ ddtr ‚ r(t)¸ œ "# ¸r(t) ‚ ddtr ¸ œ "# kr ‚ rk
#
# %
#
r v#
# %
! !
9. T œ Š 2r!1va! ‹ È1 e# Ê T# œ Š 4r1# va# ‹ a1 e# b œ Š 4r1# va# ‹ ”1 Š GM
1‹ • (from Equation 5)
!
!
!
!
# %
ˆ41# a% ‰ a2GM r! v#! b
r# v %
r! v!#
2GMr! v!# r!# v!%
œ Š 4r1# va# ‹ ’ G!# M!# 2 Š GM
‹“ œ Š 4r1# va# ‹ ’
“œ
G# M#
r! G# M#
! !
! !
# %
œ a41# a% b Š
2GM r! v#!
# % ˆ " ‰ˆ 2 ‰
ˆ 2 ‰
2r! GM ‹ GM œ a41 a b 2a
GM (from Equation 10)
# $
#
#
1 a
41
Ê T# œ 4GM
Ê Ta$ œ GM
minutes
seconds
7
10. r œ 365.256 days œ 365.256 days ‚ 24 hours
day ‚ 60 hour ‚ 60 minute œ 31,558,118.4 seconds ¸ 3.16 ‚ 10 ,
41
G œ 6.6726 ‚ 1011 Nkg†m# , and the mass of the sun M œ 1.99 ‚ 1030 kg. Ta3 œ GM
Ê a3 œ T2 GM
41 2
2
Ê a3 œ a3.16 ‚ 107 b
2
2
2 ˆ6.6726‚10c11 ‰ˆ1.99‚1030 ‰
3
¸ 3.35863335 ‚ 1033 Ê a œ È
3.35863335 ‚ 1033
41 2
¸ 149757138111 m ¸ 149.757 billion km
CHAPTER 13 PRACTICE EXERCISES
1. r(t) œ (4 cos t)i ŠÈ2 sin t‹ j Ê x œ 4 cos t
#
#
x
and y œ È2 sin t Ê 16
y# œ 1;
v œ (4 sin t)i ŠÈ2 cos t‹ j and
a œ (4 cos t)i ŠÈ2 sin t‹ j ; r(0) œ 4 i , v(0) œ È2j ,
a(0) œ 4i ; r ˆ 14 ‰ œ 2È2i j , v ˆ 14 ‰ œ 2È2i j ,
a ˆ 1 ‰ œ 2È2i j ; kvk œ È16 sin# t 2 cos# t
4
Ê aT œ dtd kvk œ È 14 sin# tcos t # ; at t œ 0: aT œ 0, aN œ Ékak# 0 œ 4, a œ 0T 4N œ 4N, , œ kavNk# œ 42 œ 2;
16 sin t 2 cos t
È
È
È
4 2
4 2
4 2
aN
7
at t œ 14 : aT œ È871 œ 73 , aN œ É9 49
9 œ 3 , a œ 3 T 3 N, , œ kvk# œ 27
#
#
2. r(t) œ ŠÈ3 sec t‹ i ŠÈ3 tan t‹ j Ê x œ È3 sec t and y œ È3 tan t Ê x3 y3 œ sec# t tan# t œ 1;
Ê x# y# œ 3; v œ ŠÈ3 sec t tan t‹ i ŠÈ3 sec# t‹ j
and
a œ ŠÈ3 sec t tan# t È3 sec$ t‹ i Š2È3 sec# t tan t‹ j ;
r(0) œ È3i , v(0) œ È3j , a(0) œ È3i ;
kvk œ È3 sec# t tan# t 3 sec% t
#
$
%
t tan t 18 sec t tan t
Ê aT œ dtd kvk œ 6 sec
;
#
#
%
È
2
3 sec t tan t 3 sec t
at t œ 0: aT œ 0, aN œ Ékak# 0 œ È3,
a œ 0T È3N œ È3N, , œ kavNk# œ
È3
"
3 œ È3
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
785
786
Chapter 13 Vector-Valued Functions and Motion in Space
3. r œ È1" t# i È1t t# j Ê v œ t a1 t# b
$Î#
i a 1 t# b
$Î#
#
#
j Ê kvk œ Ê’t a1 t# b$Î# “ ’a1 t# b$Î# “
œ 1 " t# . We want to maximize kvk : ddtkvk œ a1 2tt# b# and ddtkvk œ 0 Ê a1 2tt# b# œ 0 Ê t œ 0. For t 0, a1 2tt# b# 0; for
t 0, a1 2tt# b# 0 Ê kvk max occurs when t œ 0 Ê kvk max œ 1
4. r œ aet cos tb i aet sin tb j Ê v œ aet cos t et sin tb i aet sin t et cos tb j
Ê a œ aet cos t et sin t et sin t et cos tb i aet sin t et cos t et cos t et sin tb j
œ a2et sin tb i a2et cos tb j . Let ) be the angle between r and a . Then ) œ cos" Š krrk†kaak ‹
œ cos" 2e2t sin t cos t2e2t sin t cos t
œ cos" Š 2e02t ‹ œ cos" 0 œ 1# for all t
Éaet cos tb# aet sin tb# Éa2et sin tb# a2et cos tb# â
â
â i j kâ
â
â
5. v œ 3i 4j and a œ 5i 15j Ê v ‚ a œ â 3 4 0 â œ 25k Ê kv ‚ ak œ 25; kvk œ È3# 4# œ 5
â
â
â 5 "5 0 â
"
Ê , œ kvkv‚k$ak œ 25
5$ œ 5
6. , œ
kyww k
x
2x $Î#
b
$Î# œ e a1 e
1 ayw b# ‘
œ ex a1 e2x b
$Î#
Ê ddx, œ ex a1 e2x b
3e3x a1 e2x b
&Î#
œ ex a1 e2x b
$Î#
ex ’ 3# a1 e2x b
&Î#
a2e2x b“
&Î#
&Î#
ca1 e2x b 3e2x d œ ex a1 e2x b
a1 2e2x b ;
d,
"
"
2x
2x
È2 Ê y œ " ; therefore , is at a
È
dx œ 0 Ê a1 2e b œ 0 Ê e œ # Ê 2x œ ln 2 Ê x œ # ln 2 œ ln
2
maximum at the point Š ln È2ß È"2 ‹
dy
dx
7. r œ xi yj Ê v œ dx
dt i dt j and v † i œ y Ê dt œ y. Since the particle moves around the unit circle
dy
dy
dy
dy
x dx
x
dx
x# y# œ 1, 2x dx
dt 2y dt œ 0 Ê dt œ y dt Ê dt œ y (y) œ x. Since dt œ y and dt œ x, we have
v œ yi xj Ê at (1ß 0), v œ j and the motion is clockwise.
dy
" # dx
# dx
8. 9y œ x$ Ê 9 dy
dt œ 3x dt Ê dt œ 3 x dt . If r œ xi yj , where x and y are differentiable functions of t,
dy
dy
dx
" # dx
"
#
then v œ dx
dt i dt j. Hence v † i œ 4 Ê dt œ 4 and v † j œ dt œ 3 x dt œ 3 (3) (4) œ 12 at (3ß 3). Also,
#
#
#
#
#
#
‰ ˆ 3" x# ‰ ddt#x . Hence a † i œ 2 Ê ddt#x œ 2 and
a œ ddt#x i ddt#y j and ddt#y œ ˆ 32 x‰ ˆ dx
dt
#
a † j œ ddt#y œ 23 (3)(4)# "3 (3)# (2) œ 26 at the point (xß y) œ (3ß 3).
9.
dr
dt orthogonal to r
Ê 0 œ ddtr † r œ "# ddtr † r "# r † ddtr œ "# dtd (r † r) Ê r † r œ K, a constant. If r œ xi yj , where
x and y are differentiable functions of t, then r † r œ x# y# Ê x# y# œ K, which is the equation of a circle
centered at the origin.
10. (a)
(b) v œ (1 1 cos 1t)i (1 sin 1t)j
Ê a œ a1# sin 1tb i a1# cos 1tb j ;
v(0) œ 0 and a(0) œ 1# j ;
v(1) œ 21i and a(1) œ 1# j ;
v(2) œ 0 and a(2) œ 1# j ;
v(3) œ 21i and a(3) œ 1# j
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Chapter 13 Practice Exercises
787
(c) Forward speed at the topmost point is kv(1)k œ kv(3)k œ 21 ft/sec; since the circle makes "# revolution per
second, the center moves 1 ft parallel to the x-axis each second Ê the forward speed of C is 1 ft/sec.
11. y œ y! (v! sin !)t "# gt# Ê y œ 6.5 (44 ft/sec)(sin 45°)(3 sec) "# a32 ft/sec# b (3 sec)# œ 6.5 66È2 144
¸ 44.16 ft Ê the shot put is on the ground. Now, y œ 0 Ê 6.5 22È2t 16t# œ 0 Ê t ¸ 2.13 sec (the
positive root) Ê x ¸ (44 ft/sec)(cos 45°)(2.13 sec) ¸ 66.27 ft or about 66 ft, 3 in. from the stopboard
#
#
45°)]
!)
12. ymax œ y! (v! sin
œ 7 ft [(80(2)ft/sec)(sin
¸ 57 ft
#g
a32 ft/sec# b
(v! sin !)t "# gt#
(v sin !) " gt
œ ! v! cos ! #
(v! cos !)t
t œ 2v! sin ! 2vg ! cos ! tan 9 , which is the time when the golf ball
13. x œ (v! cos !)t and y œ (v! sin !)t "# gt# Ê tan 9 œ yx œ
Ê v! cos ! tan 9 œ v! sin ! "# gt Ê
hits the upward slope. At this time x œ (v! cos !) Š 2v! sin ! 2vg ! cos ! tan 9 ‹ œ Š g2 ‹ av#! sin ! cos ! v#! cos# ! tan 9b .
v# sin ! cos ! v#! cos# ! tan 9
‹
cos 9
Now OR œ cosx 9 Ê OR œ Š g2 ‹ Š !
œŠ
2v#! cos !
cos ! tan 9
sin !
‹ Š cos
g
9 cos 9 ‹
œŠ
2v#! cos !
9 cos ! sin 9
‹ Š sin ! cos cos
‹
#9
g
2v# cos !
œ Š g !cos# 9 ‹ [sin (! 9)]. The distance OR is maximized
when x is maximized:
2v#!
dx
d! œ Š g ‹(cos 2! sin 2! tan 9) œ 0
Ê (cos 2! sin 2! tan 9) œ 0 Ê cot 2! tan 9 œ 0 Ê cot 2! œ tan (9) Ê 2! œ 1# 9 Ê ! œ 9# 14
5
14. (a) x œ v! (cos 40°)t and y œ 6.5 v! (sin 40°)t "# gt# œ 6.5 v! (sin 40°)t 16t# ; x œ 262 12
ft and y œ 0 ft
5
262.4167
262.4167
#
#
Ê 262 12
œ v! (cos 40°)t or v! œ (cos
40°)t and 0 œ 6.5 ’ (cos 40°)t “ (sin 40°)t 16t Ê t œ 14.1684
262.4167
Ê t ¸ 3.764 sec. Therefore, 262.4167 ¸ v! (cos 40°)(3.764 sec) Ê v! ¸ (cos 40°)(3.764
sec) Ê v! ¸ 91 ft/sec
#
2
40°)b
!)
(b) ymax œ y! (v! sin
¸ 6.5 a(91)(sin
¸ 60 ft
2g
(2)(32)
15. r œ (2 cos t)i (2 sin t)j t# k Ê v œ (2 sin t)i (2 cos t)j 2tk Ê kvk œ È(2 sin t)# (2 cos t)# (2t)#
œ 2È1 t# Ê Length œ '0 2È1 t# dt œ ’tÈ1 t# ln ¹t È1 t# ¹“
1Î4
1Î%
!
#
#
1
1
œ 14 É1 16
ln Š 14 É1 16
‹
16. r œ (3 cos t)i (3 sin t)j 2t$Î# k Ê v œ (3 sin t)i (3 cos t)j 3t"Î# k Ê kvk œ É(3 sin t)# (3 cos t)# a3t"Î# b
œ È9 9t œ 3È1 t Ê Length œ '0 3È1 t dt œ 2(1 t)$Î# ‘ ! œ 14
3
$
17. r œ 49 (1 t)$Î# i 49 (1 t)$Î# j "3 tk Ê v œ 32 (1 t)"Î# i 32 (1 t)"Î# j 3" k
#
#
#
Ê kvk œ É 23 (1 t)"Î# ‘ 23 (1 t)"Î# ‘ ˆ 3" ‰ œ 1 Ê T œ 32 (1 t)"Î# i 32 (1 t)"Î# j 3" k
Ê T(0) œ 32 i 32 j 3" k ; ddtT œ 3" (1 t)"Î# i 3" (1 t)"Î# j Ê ddtT (0) œ 3" i 3" j Ê ¸ ddtT (0)¸ œ
â
â
j
kâ
â i
â
2
2
" â
"
"
4
â
Ê N(0) œ È"2 i È"2 j ; B(0) œ T(0) ‚ N(0) œ ââ 3 3
3 â œ È i È j È k;
3 2
3 2
3 2
"
"
â
â
â È2 È2 0 â
a œ "3 (1 t)"Î# i "3 (1 t)"Î# j Ê a(0) œ "3 i "3 j and v(0) œ 23 i 23 j "3 k Ê v(0) ‚ a(0)
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
È2
3
#
788
Chapter 13 Vector-Valued Functions and Motion in Space
â i
â
â
œ â 23
â "
â 3
k ââ
È
Š 32 ‹
È2
È2
" â
k v ‚a k
"
"
4
i
j
k
k
v
a
k
œ
Ê
‚
œ
Ê
,
(0)
œ
œ
â
3
9
9
9
3
1$ œ 3 ;
kv k $
â
0â
j
23
"
3
Þ
Þ
a œ "6 (1 t)$Î# i "6 (1 t)$Î# j Ê a(0) œ "6 i "6 j Ê 7 (0) œ
â 2
â 3
â
â "
â 3
â "
â 6
23
"
3
"
6
kv‚ak
#
" ââ
3 â
0 ââ
0 ââ
2 ‰
ˆ 3" ‰ ˆ 18
œ
È
#
Š 32 ‹
œ 6"
18. r œ aet sin 2tb i aet cos 2tb j 2et k Ê v œ aet sin 2t 2et cos 2tb i aet cos 2t 2et sin 2tb j 2et k
Ê kvk œ Éaet sin 2t 2et cos 2tb# aet cos 2t 2et sin 2tb# a2et b# œ 3et Ê T œ kvvk
œ ˆ "3 sin 2t 23 cos 2t‰ i ˆ 3" cos 2t 23 sin 2t‰ j 23 k Ê T(0) œ 32 i 3" j 32 k ;
Ê ddtT (0) œ 32 i 34 j Ê ¸ ddtT (0)¸ œ 32 È5
â
â
j
kâ
â i
â
â
ˆ 23 i 43 j‰
2
1
2
4
2
5
â
Ê N(0) œ 2È5 œ È"5 i È25 j ; B(0) œ T(0) ‚ N(0) œ ââ 3
3
3 â œ È i È j È k;
3 5
3 5
3 5
Š 3 ‹
â " 2
â
0â
È5
â È5
dT
4
4
ˆ2
‰
ˆ 2
‰
dt œ 3 cos 2t 3 sin 2t i 3 sin 2t 3 cos 2t j
a œ a4et cos 2t 3et sin 2tb i a3et cos 2t 4et sin 2tb j 2et k Ê a(0) œ 4i 3j 2k and v(0) œ 2i j 2k
â
â
j kâ
âi
â
â
Ê v(0) ‚ a(0) œ â 2 " 2 â œ 8i 4j 10k Ê kv ‚ ak œ È64 16 100 œ 6È5 and kv(0)k œ 3
â
â
â 4 3 2 â
È
È
Ê ,(0) œ 6 3$ 5 œ 2 9 5 ;
Þ
a œ a4et cos 2t 8et sin 2t 3et sin 2t 6et cos 2tb i a3et cos 2t 6et sin 2t 4et sin 2t 8et cos 2tb j 2et k
Þ
œ a2et cos 2t 11et sin 2tb i a11 et cos 2t 2et sin 2tb j 2et k Ê a(0) œ 2i 11j 2k
Ê 7 (0) œ
â
â 2
â
â 4
â
â 2
1
3
11
kv‚ak#
â
2â
â
2â
â
2â
80
œ 180
œ 49
19. r œ ti "# e2t j Ê v œ i e2t j Ê kvk œ È1 e4t Ê T œ È " 4t i È e
2t
1e
dT
2e4t
2e2t
dt œ ˆ1 e4t ‰$Î# i ˆ1 e4t ‰$Î# j
1 e4t
j Ê T (ln 2) œ È"17 i È417 j ;
8
Ê ddtT (ln 2) œ 17È3217 i 17È
j Ê N (ln 2) œ È417 i È"17 j ;
17
â i
j
k ââ
â
â "
4
0 ââ œ k ; a œ 2e2t j Ê a(ln 2) œ 8j and v(ln 2) œ i 4j
È17
B (ln 2) œ T(ln 2) ‚ N(ln 2) œ ââ È17
â
"
â 4
â
â È17 È17 0 â
â
â
â i j kâ
â
â
Þ
8
Ê v(ln 2) ‚ a(ln 2) œ â " 4 0 â œ 8k Ê kv ‚ ak œ 8 and kv(ln 2)k œ È17 Ê ,(ln 2) œ 17È
; a œ 4e2t j
17
â
â
â0 8 0â
Þ
Ê a(ln 2) œ 16j Ê 7 (ln 2) œ
â
â1
â
â0
â
â0
4
8
16
k v ‚a k #
â
0â
â
0â
â
0â
œ0
20. r œ (3 cosh 2t)i (3 sinh 2t)j 6tk Ê v œ (6 sinh 2t)i (6 cosh 2t)j 6k
Ê kvk œ È36 sinh# 2t 36 cosh# 2t 36 œ 6È2 cosh 2t Ê T œ kvvk œ Š È"2 tanh 2t‹ i È"2 j Š È"2 sech 2t‹ k
"
8
dT
2
2
dT
#
Ê T(ln 2) œ 1715
È2 i È2 j 17È2 k ; dt œ Š È2 sech 2t‹ i Š È2 sech 2t tanh 2t‹ k Ê dt (ln 2)
#
#
#
È
8 2
8 ‰
8 ‰ ˆ 15 ‰
128
240
240
¸ ddtT (ln 2)¸ œ ÊŠ 128
œ Š È22 ‹ ˆ 17
i Š È22 ‹ˆ 17
‹ Š 289
È2 ‹ œ 17
17 k œ 289È2 i 289È2 k Ê
289È2
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Chapter 13 Practice Exercises
â
â
j
k â
â i
â 15
8 â
"
8
15
8
"
Ê N(ln 2) œ 17
i 17
k ; B(ln 2) œ T(ln 2) ‚ N(ln 2) œ ââ 17È2 È2 17È2 ââ œ 1715
È2 i È2 j 17È2 k ;
â 8
15 â
0 17 â
â 17
51
45
‰
ˆ 15 ‰
a œ (12 cosh 2t)i (12 sinh 2t)j Ê a(ln 2) œ 12 ˆ 17
8 i 12 8 j œ # i # j and
â i
j k ââ
â
â
45
51
45
51
‰
ˆ 17 ‰
6 ââ
v(ln 2) œ 6 ˆ 15
4
8 i 6 8 j 6k œ 4 i 4 j 6k Ê v(ln 2) ‚ a(ln 2) œ ââ 4
â
45
â 51
0â
2
#
È2 Ê ,(ln 2) œ
œ 135i 153j 72k Ê kv ‚ ak œ 153È2 and kv(ln 2)k œ 51
4
Þ
Þ
a œ (24 sinh 2t)i (24 cosh 2t)j Ê a(ln 2) œ 45i 51j Ê
153È2
$
È 2‹
Š 51
4
32
œ 867
;
â 45 51
â
6â
â 4
4
â 5" 45
â
â 2
â
0
2
â
â
â 45 51 0 â
32
7 (ln 2) œ kv‚ak# œ 867
21. r œ a2 3t 3t# b i a4t 4t# b j (6 cos t)k Ê v œ (3 6t)i (4 8t)j (6 sin t)k
Ê kvk œ È(3 6t)# (4 8t)# (6 sin t)# œ È25 100t 100t# 36 sin# t
"Î#
Ê ddtkvk œ "# a25 100t 100t# 36 sin# tb
(100 200t 72 sin t cos t) Ê aT (0) œ ddtkvk (0) œ 10;
a œ 6i 8j (6 cos t)k Ê kak œ È6# 8# (6 cos t)# œ È100 36 cos# t Ê ka(0)k œ È136
Ê aN œ Ékak# a#T œ È136 10# œ È36 œ 6 Ê a(0) œ 10T 6N
22. r œ (2 t)i at 2t# b j a1 t# b k Ê v œ i (1 4t)j 2tk Ê kvk œ È1# (1 4t)# (2t)#
"Î#
œ È2 8t 20t# Ê d kvk œ " a2 8t 20t# b
(8 40t) Ê aT œ d kvk (0) œ 2È2; a œ 4j 2k
dt
#
dt
#
Ê kak œ È4# 2# œ È20 Ê aN œ Ékak# a#T œ Ê20 Š2È2‹ œ È12 œ 2È3 Ê a(0) œ 2È2T 2È3N
23. r œ (sin t)i ŠÈ2 cos t‹ j (sin t)k Ê v œ (cos t)i ŠÈ2 sin t‹ j (cos t)k
#
Ê kvk œ Ê(cos t)# ŠÈ2 sin t‹ (cos t)# œ È2 Ê T œ kvvk œ Š È"2 cos t‹ i (sin t)j Š È"2 cos t‹ k ;
dT
"
"
dt œ Š È2 sin t‹ i (cos t)j Š È2 sin t‹ k
#
#
Ê ¸ ddtT ¸ œ ÊŠ È" sin t‹ ( cos t)# Š È" sin t‹ œ 1
2
2
â
â
i
j
k
â
â
â
â
d
T
"
"
ˆ dt ‰
cos
t
sin
t
cos
t
"
"
â
â
È
È
Ê N œ ¸ dT ¸ œ Š È sin t‹ i (cos t)j Š È sin t‹ k ; B œ T ‚ N œ â 2
2
â
2
2
dt
â " sin t cos t " sin t â
â È2
â
È2
â i
â
j
k â
â
â
â
"
"
È
È
œ È2 i È2 k ; a œ ( sin t)i Š 2 cos t‹ j (sin t)k Ê v ‚ a œ â cos t 2 sin t cos t â
â
â
â sin t È2 cos t sin t â
Þ
œ È2 i È2 k Ê kv ‚ ak œ È4 œ 2 Ê , œ kvkv‚k$ak œ 2 $ œ È"2 ; a œ ( cos t)i ŠÈ2 sin t‹ j (cos t)k
Ê 7œ
â
â cos t
â
â sin t
â
â
â cos t
ŠÈ2‹
â
È2 sin t
cos t ââ
È2 cos t sin t ââ
(cos t) ŠÈ2‹ ŠÈ2 sin t‹ (0) (cos t) ŠÈ2‹
È2 sin t cos t ââ
œ
œ0
#
4
k v ‚a k
24. r œ i (5 cos t)j (3 sin t)k Ê v œ (5 sin t)j (3 cos t)k Ê a œ (5 cos t)j (3 sin t)k
Ê v † a œ 25 sin t cos t 9 sin t cos t œ 16 sin t cos t; v † a œ 0 Ê 16 sin t cos t œ 0 Ê sin t œ 0 or cos t œ 0
Ê t œ 0, 1# or 1
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
789
790
Chapter 13 Vector-Valued Functions and Motion in Space
25. r œ 2i ˆ4 sin #t ‰ j ˆ3 1t ‰ k Ê 0 œ r † (i j) œ 2(1) ˆ4 sin #t ‰ (1) Ê 0 œ 2 4 sin #t Ê sin #t œ "# Ê #t œ 16
Ê t œ 13 (for the first time)
26. r(t) œ ti t# j t$ k Ê v œ i 2tj 3t# k Ê kvk œ È1 4t# 9t% Ê kv(1)k œ È14
Ê T(1) œ È"14 i È214 j È314 k , which is normal to the normal plane
Ê È"14 (x 1) È214 (y 1) È314 (z 1) œ 0 or x 2y 3z œ 6 is an equation of the normal plane. Next we
calculate N(1) which is normal to the rectifying plane. Now, a œ 2j 6tk Ê a(1) œ 2j 6k Ê v(1) ‚ a(1)
â
â
â i j kâ
#
È76
È19
â
â
œ â " 2 3 â œ 6i 6j 2k Ê kv(1) ‚ a(1)k œ È76 Ê ,(1) œ
œ kv(t)k Ê ddt#s ¹
; ds
$ œ
È
dt
7
14
È
â
â
tœ1
Š 14‹
â0 2 6â
œ "# a1 4t# 9t% b
"Î#
È
a8t 36t$ b¹
#
#
tœ1
#
‰ N Ê 2j 6k
œ È2214 , so a œ ddt#s T , ˆ ds
dt
È
2j3k
8
9 ‰
11
8
9
ˆ 11
œ È2214 Š iÈ
‹ 7È19
ŠÈ14‹ N Ê N œ 2È14
7 i 7 j 7 k Ê 7 (x 1) 7 (y 1) 7 (z 1)
19
14
14
œ 0 or 11x 8y 9z œ 10 is an equation of the rectifying plane. Finally, B(1) œ T(1) ‚ N(1)
â
â
j kâ
â i
È14
â
â
2 3 â œ È" (3i 3j k) Ê 3(x 1) 3(y 1) (z 1) œ 0 or 3x 3y z
œ Š 2È19 ‹ Š È" ‹ ˆ "7 ‰ â "
19
14
â
â
â 11 8 9 â
œ 1 is an equation of the osculating plane.
" ‰
27. r œ et i (sin t)j ln (1 t)k Ê v œ et i (cos t)j ˆ 1 t k Ê v(0) œ i j k ; r(0) œ i Ê (1ß 0ß 0) is on the line
Ê x œ 1 t, y œ t, and z œ t are parametric equations of the line
28. r œ ŠÈ2 cos t‹ i ŠÈ2 sin t‹ j tk Ê v œ ŠÈ2 sin t‹ i ŠÈ2 cos t‹ j k Ê v ˆ 14 ‰
œ ŠÈ2 sin 14 ‹ i ŠÈ2 cos 14 ‹ j k œ i j k is a vector tangent to the helix when t œ 14 Ê the tangent line
is parallel to v ˆ 14 ‰ ; also r ˆ 14 ‰ œ ŠÈ2 cos 14 ‹ i ŠÈ2 sin 14 ‹ j 14 k Ê the point ˆ1ß 1ß 14 ‰ is on the line
Ê x œ 1 t, y œ 1 t, and z œ 14 t are parametric equations of the line
#
#
29. x# œ av#! cos# !b t# and ˆy "# gt# ‰ œ av#! sin# !b t# Ê x# ˆy "# gt# ‰ œ v#! t#
Þ ÞÞ Þ ÞÞ
Þ ÞÞ Þ ÞÞ
ax x y yb#
ÞÞ
ÞÞ
Þ
Þ
ÞÞ
ÞÞ
ÞÞ
ÞÞ
x x y y
30. s œ dtd Èx# y# œ È Þ # Þ # Ê x# y# s # œ x# y# xÞ # yÞ #
x y
ÞÞ ÞÞ Þ
Þ
Þ ÞÞ
Þ ÞÞ Þ ÞÞ Þ ÞÞ
Þ ÞÞ Þ ÞÞ
Þ ÞÞ
Þ ÞÞ
Þ ÞÞ Þ ÞÞ
ax# y# b ax# y# b ax# x# 2x x y y y# y# b
ax y y x b#
x# y# y# x # 2x x y y
œ
œ
œ
Þ# Þ#
Þ# Þ#
Þ
Þ
x y
x y
x# y#
Þ ÞÞ Þ ÞÞ
Þ # Þ # $Î#
Þ
Þ
#
#
kx y y xk
ax y b
ÞÞ
ÞÞ
ÞÞ
x y
Ê È x# y# s # œ È Þ # Þ # Ê ÈÞÞ# ÞÞ# ÞÞ# œ kxÞ ÞÞy yÞ ÞÞxk œ ," œ 3
x y
x y s
31. s œ a) Ê ) œ as Ê 9 œ as 1# Ê dds9 œ "a Ê , œ ¸ "a ¸ œ "a since a 0
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Chapter 13 Additional and Advanced Exercises
y!
OT
6380
32. (1) ?SOT ¸ ?TOD Ê DO
OT œ SO Ê 6380 œ 6380437
#
Ê y! œ 6380
6817 Ê y! ¸ 5971 km;
(2) VA œ '5971 21x Ê1 Š dx
dy ‹ dy
#
6380
6380
œ 21'5971 È6380# y# Š È6380
# y# ‹ dy
6817
œ 21 '5971 6380 dy œ 21 c6380yd ')"(
&*("
6817
œ 16,395,469 km# ¸ 1.639 ‚ 10( km# ;
#
km
(3) percentage visible ¸ 16,395,469
41(6380 km)# ¸ 3.21%
CHAPTER 13 ADDITIONAL AND ADVANCED EXERCISES
1. (a) r()) œ (a cos ))i (a sin ))j b)k Ê ddtr œ [(a sin ))i (a cos ))j bk] ddt) ; kvk œ È2gz œ ¸ ddtr ¸
)
d) ¸
É a#2gb
É a#41gbb# œ 2É a#1gbb#
œ Èa# b# ddt) Ê ddt) œ É a#2gz
b# œ
b# Ê dt )œ#1 œ
(b)
)
d)
É a#2gb
b#
dt œ
Ê 2)
"Î#
d)
"Î#
Ê È
œ É a#2gb
œ É a#2gb
b# dt Ê 2)
b# t C; t œ 0 Ê ) œ 0 Ê C œ 0
)
œ É a#2gb
b# t
#
# #
gb t
Ê ) œ 2 aagbt
# b# b ; z œ b) Ê z œ 2 aa# b# b
(c) v(t) œ ddtr œ [(a sin ))i (a cos ))j bk] ddt) œ [(a sin ))i (a cos ))j bk] Š a# gbt
b# ‹ , from part (b)
i (a cos ))j bk
Ê v(t) œ ’ (a sin ))È
T;
“ Š È gbt
‹ œ È gbt
#
#
#
#
#
#
a b
a b
a b
#
d# r
ˆ d) ‰# [(a sin ))i (a cos ))j bk] ddt#)
dt# œ [(a cos ))i (a sin ))j] dt
#
gb
œ Š a# gbt
b# ‹ [(a cos ))i (a sin ))j] [(a sin ))i (a cos ))j bk] Š a# b# ‹
#
i (a cos ))j bk
œ ’ (a sin ))È
“ Š È #gb # ‹ a Š a# gbt
# #
b# ‹ [( cos ))i (sin ))j]
a
b
a
œ Èa#gb b# T a Š a# gbt
b# ‹
#
b
N (there is no component in the direction of B).
2. (a) r()) œ (a) cos ))i (a) sin ))j b)k Ê ddtr œ [(a cos ) a) sin ))i (a sin ) a) cos ))j bk] ddt) ;
kvk œ È2gz œ ¸ ddtr ¸ œ aa# a# )# b# b
(b) s œ '0 kvk dt œ '0 aa# a# )# b# b
t
t
"Î#
È2gb)
a a# ) # b#
ˆ ddt) ‰ Ê ddt) œ È #
"Î# d)
dt dt œ
'0t aa# a# )# b# b"Î# d) œ '0 aa# a# u# b# b"Î# du
)
œ '0 aÉ a a# b u# du œ a '0 Èc# u# du, where c œ
)
#
)
#
#
È a# b #
k ak
)
Ê s œ a ’ u# Èc# u# c# ln ¹u Èc# u# ¹“ œ #a Š)Èc# )# c# ln ¹) Èc# )# ¹ c# ln c‹
!
e)r!
(1 e)r! (e sin )) dr
(1 e)r! (e sin ))
dr
3. r œ 1(1e cos
) Ê d) œ (1 e cos ))# ; d) œ 0 Ê
(1 e cos ))# œ 0 Ê (1 e)r! (e sin )) œ 0
Ê sin ) œ 0 Ê ) œ 0 or 1. Note that ddr) 0 when sin ) 0 and ddr) 0 when sin ) 0. Since sin ) 0 on
e)r!
1 ) 0 and sin ) 0 on 0 ) 1, r is a minimum when ) œ 0 and r(0) œ 1(1e cos
0 œ r!
4. (a) f(x) œ x 1 "# sin x œ 0 Ê f(0) œ 1 and f(2) œ 2 1 "# sin 2
"
# since ksin 2k Ÿ 1; since f is continuous
on [0ß 2], the Intermediate Value Theorem implies there is a root between 0 and 2
(b) Root ¸ 1.4987011335179
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
791
792
Chapter 13 Vector-Valued Functions and Motion in Space
.
.
.
.
.
.
.
5. (a) v œ x i y j and v œ r ur r ) u) œ arb [(cos ))i (sin ))j] ˆr )‰ [( sin ))i (cos ))j] Ê v † i œ x and
.
.
.
.
.
. .
.
v † i œ r cos ) r ) sin ) Ê x œ r cos ) r ) sin ); v † j œ y and v † j œ r sin ) r ) cos )
.
. .
Ê y œ r sin ) r ) cos )
.
.
(b) ur œ (cos ))i (sin ))j Ê v † ur œ x cos ) y sin )
.
.
.
.
œ ˆr cos ) r ) sin )‰ (cos )) ˆr sin ) r ) cos )‰ (sin )) by part (a),
.
. .
.
Ê v † ur œ r ; therefore, r œ x cos ) y sin );
.
.
u) œ (sin ))i (cos ))j Ê v † u) œ x sin ) y cos )
.
.
.
.
.
œ ˆr cos ) r ) sin )‰ ( sin )) ˆr sin ) r ) cos )‰ (cos )) by part (a) Ê v † u) œ r ) ;
.
.
.
therefore, r ) œ x sin ) y cos )
#
#
#
d)
d r
d)
w
ww
ˆ d) ‰ f w ()) ddt#) ; v œ dr
6. r œ f()) Ê dr
dt œ f ()) dt Ê dt# œ f ()) dt
dt ur r dt u)
"Î#
"Î#
d) ‰
dr
d) ‰
ˆ
ˆ dr ‰# r# ˆ ddt) ‰# “ œ ’af w b# f # “
œ ˆcos ) dr
dt r sin ) dt i sin ) dt r cos ) dt j Ê kvk œ ’ dt
Þ ÞÞ Þ ÞÞ
d)
dr
kv ‚ ak œ kx y y xk , where x œ r cos ) and y œ r sin ). Then dx
dt œ (r sin )) dt (cos )) dt
#
#
#
ˆ ddt) ‰ ;
#
d)
dr
ˆ d) ‰ (r sin )) ddt#) (cos )) ddt#r ; dy
Ê ddt#x œ (2 sin )) ddt) dr
dt (r cos )) dt
dt œ (r cos )) dt (sin )) dt
#
#
#
#
ˆ d) ‰ (r cos )) ddt#) (sin )) ddt#r . Then kv ‚ ak
Ê ddt#y œ (2 cos )) ddt) dr
dt (r sin )) dt
$
#
#
#
$
d) d r
d) ˆ dr ‰
œ (after much algebra) r# ˆ ddt) ‰ r ddt#) dr
œ ˆ ddt) ‰ Šf 2 f † f ww 2af w b2 ‹
dt r dt dt# 2 dt dt
ww
w 2
fb
Ê , œ kvk‚vkak œ f wf†#f #2a$Î#
2
af b f ‘
#
#
d)
d r
d )
7. (a) Let r œ 2 t and ) œ 3t Ê dr
dt œ 1 and dt œ 3 Ê dt# œ dt# œ 0. The halfway point is (1ß 3) Ê t œ 1;
#
#
#
d)
d r
d)
ˆ d) ‰ “ ur ’r ddt#) 2 dr
v œ dr
dt ur r dt u) Ê v(1) œ ur 3u) ; a œ ’ dt# r dt
dt dt “ u) Ê a(1) œ 9ur 6u)
(b) It takes the beetle 2 min to crawl to the origin Ê the rod has revolved 6 radians
#
#
#
Ê L œ '0 É[f())]# cf w ())d# d) œ '0 Ɉ2 3) ‰ ˆ 3" ‰ d) œ '0 É4 43) )9 9" d)
6
6
6
œ '0 É 37 129 ) ) d) œ 3" '0 È() 6)# 1 d) œ 3" ’ ()#6) È() 6)# 1 #" ln ¸) 6 È() 6)# 1¸“
6
6
#
œ È37 " ln ŠÈ37 6‹ ¸ 6.5 in.
6
8. (a) x œ r cos ) Ê dx œ cos ) dr r sin ) d); y œ r sin ) Ê dy œ sin ) dr r cos ) d); thus
dx# œ cos# ) dr# 2r sin ) cos ) dr d) r# sin# ) d)# and
dy# œ sin# ) dr# 2r sin ) cos ) dr d) r# cos# ) d)# Ê ds2 œ dx# dy# dz# œ dr# r# d)# dz#
(b)
(c) r œ e) Ê dr œ e) d)
Ê L œ '0
ln 8
Èdr# r# d)# dz#
œ '0 Èe#) e#) e#) d)
ln 8
œ '0 È3e) d) œ ’È3 e) “
ln 8
ln 8
0
œ 8È3 È3 œ 7È3
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
'
!
Chapter 13 Additional and Advanced Exercises
â
â i
â
9. (a) ur ‚ u) œ â cos )
â
â sin )
(b)
j
sin )
cos )
793
â
kâ
â
0 â œ k Ê a right-handed frame of unit vectors
â
0â
du)
dur
d) œ ( sin ))i (cos ))j œ u) and d) œ ( cos ))i (sin ))j œ ur
Þ
ÞÞ
Þ Þ
Þ
ÞÞ
ÞÞ
ÞÞ
Þ
Þ
ÞÞ
(c) From Eq. (7), v œ rur r)u) zk Ê a œ v œ a r ur r ur b ˆr )u) r) u) r) u) ‰ z k
Þ#
ÞÞ
ÞÞ
ÞÞ
ÞÞ
œ Š r r) ‹ ur ˆr) 2r )‰ u) z k
#
10. L(t) œ r(t) ‚ mv(t) Ê ddtL œ ˆ ddtr ‚ mv‰ Šr ‚ m ddt#r ‹ Ê ddtL œ (v ‚ mv) (r ‚ ma) œ r ‚ ma ; F œ ma Ê krck$ r
œ ma Ê ddtL œ r ‚ ma œ r ‚ Š krck$ r‹ œ krck$ (r ‚ r) œ 0 Ê L œ constant vector
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
794
Chapter 13 Vector-Valued Functions and Motion in Space
NOTES:
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
CHAPTER 14 PARTIAL DERIVATIVES
14.1 FUNCTIONS OF SEVERAL VARIABLES
1. (a) fa0, 0b œ 0
(d) fa3, 2b œ 33
2. (a) fˆ2, 16 ‰ œ
È3
2
(b) fa1, 1b œ 0
(c) fa2, 3b œ 58
1‰
(b) fˆ3, 12
œ È12
(c) fˆ1, 14 ‰ œ È12
(b) fˆ1, 12 , 14 ‰ œ 85
(c) fˆ0, 13 , 0‰ œ 3
(b) fa2, 3, 6b œ 0
(c) fa1, 2, 3b œ È35
(d) fˆ 12 , 7‰ œ 1
3. (a) fa3, 1, 2b œ 45
(d) fa2, 2, 100b œ 0
4. (a) fa0, 0, 0b œ 7
(d) fŠ È42 , È52 , È62 ‹ œ É 21
2
5. Domain: all points axß yb on or above the line
yœx2
6. Domain: all points axß yb outside the circle
x2 y2 œ 4
7. Domain: all points axß yb not liying on the graph
of y œ x or y œ x3
8. Domain: all points axß yb not liying on the graph
of x2 y2 œ 25
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796
Chapter 14 Partial Derivatives
9. Domain: all points axß yb satisfying
x2 1 Ÿ y Ÿ x2 1
10. Domain: all points axß yb satisfying
ax 1bay 1b 0
11. Domain: all points axß yb satisfying
ax 2bax 2bay 3bay 3b 0
12. Domain: all points axß yb inside the circle
x2 y2 œ 4 such that x2 y2 Á 3
13.
14.
15.
16.
17. (a) Domain: all points in the xy-plane
(b) Range: all real numbers
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 14.1 Functions of Several Variables
(c) level curves are straight lines y x œ c parallel to the line y œ x
(d) no boundary points
(e) both open and closed
(f) unbounded
18. (a) Domain: set of all axß yb so that y x 0 Ê y x
(b) Range: z 0
(c) level curves are straight lines of the form y x œ c where c
(d) boundary is Èy x œ 0 Ê y œ x, a straight line
0
(e) closed
(f) unbounded
19. (a) Domain: all points in the xy-plane
(b) Range: z 0
(c) level curves: for f(xß y) œ 0, the origin; for faxß yb œ c 0, ellipses with center a0ß 0b and major and minor
axes along the x- and y-axes, respectively
(d) no boundary points
(e) both open and closed
(f) unbounded
20. (a) Domain: all points in the xy-plane
(b) Range: all real numbers
(c) level curves: for faxß yb œ 0, the union of the lines y œ „ x; for faxß yb œ c Á 0, hyperbolas centered at
a0ß 0b with foci on the x-axis if c 0 and on the y-axis if c 0
(d) no boundary points
(e) both open and closed
(f) unbounded
21. (a) Domain: all points in the xy-plane
(b) Range: all real numbers
(c) level curves are hyperbolas with the x- and y-axes as asymptotes when faxß yb Á 0, and the x- and y-axes
when f(xß y) œ 0
(d) no boundary points
(e) both open and closed
(f) unbounded
22. (a) Domain: all axß yb Á a0ß yb
(b) Range: all real numbers
(c) level curves: for faxß yb œ 0, the x-axis minus the origin; for faxß yb œ c Á 0, the parabolas y œ c x# minus the
origin
(d) boundary is the line x œ 0
(e) open
(f) unbounded
23. (a) Domain: all axß yb satisfying x# y# 16
(b) Range: z "4
(c) level curves are circles centered at the origin with radii r 4
(d) boundary is the circle x# y# œ 16
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797
798
Chapter 14 Partial Derivatives
(e) open
(f) bounded
24. (a) Domain: all axß yb satisfying x# y# Ÿ 9
(b) Range: 0 Ÿ z Ÿ 3
(c) level curves are circles centered at the origin with radii r Ÿ 3
(d) boundary is the circle x# y# œ 9
(e) closed
(f) bounded
25. (a) Domain: axß yb Á a0ß 0b
(b) Range: all real numbers
(c) level curves are circles with center a0ß 0b and radii r 0
(d) boundary is the single point a0ß 0b
(e) open
(f) unbounded
26. (a) Domain: all points in the xy-plane
(b) Range: 0 z Ÿ 1
(c) level curves are the origin itself and the circles with center a0ß 0b and radii r 0
(d) no boundary points
(e) both open and closed
(f) unbounded
27. (a) Domain: all axß yb satisfying 1 Ÿ y x Ÿ 1
(b) Range: 1# Ÿ z Ÿ 1#
(c) level curves are straight lines of the form y x œ c where 1 Ÿ c Ÿ 1
(d) boundary is the two straight lines y œ 1 x and y œ 1 x
(e) closed
(f) unbounded
28. (a) Domain: all axß yb, x Á 0
(b) Range: 1# z 1#
(c) level curves are the straight lines of the form y œ c x, c any real number and x Á 0
(d) boundary is the line x œ 0
(e) open
(f) unbounded
29. (a) Domain: all points axß yb outside the circle x# y# œ 1
(b) Range: all reals
(c) Circles centered ar the origin with radii r 1
(d) Boundary: the cricle x# y# œ 1
(e) open
(f) unbounded
30. (a) Domain: all points axß yb inside the circle x# y# œ 9
(b) Range: z ln 9
(c) Circles centered ar the origin with radii r 9
(d) Boundary: the cricle x# y# œ 9
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 14.1 Functions of Several Variables
(e) open
(f) bounded
31. f
32. e
33. a
34. c
35. d
36. b
37. (a)
(b)
38. (a)
(b)
39. (a)
(b)
40. (a)
(b)
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
799
800
Chapter 14 Partial Derivatives
41. (a)
(b)
42. (a)
(b)
43. (a)
(b)
44. (a)
(b)
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 14.1 Functions of Several Variables
45. (a)
(b)
46. (a)
(b)
47. (a)
(b)
48. (a)
(b)
#
#
801
49. faxß yb œ 16 x# y# and Š2È2ß È2‹ Ê z œ 16 Š2È2‹ ŠÈ2‹ œ 6 Ê 6 œ 16 x# y# Ê x# y# œ 10
50. faxß yb œ Èx# 1 and a1ß 0b Ê z œ È1# 1 œ 0 Ê x# 1 œ 0 Ê x œ 1 or x œ 1
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
802
Chapter 14 Partial Derivatives
51. faxß yb œ Èx y2 3 and a3, 1b Ê z œ É3 a1b2 3 œ 1 Ê x y2 3 œ 1 Ê x y2 œ 4
#a 1 b a 1 b
2y x
x
52. faxß yb œ x 2y
y 1 and a1ß 1b Ê z œ a1b 1 + 1 œ 3 Ê 3 œ x y 1 Ê y œ 4x 3
53.
54.
55.
56.
57.
58.
59.
60.
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Section 14.1 Functions of Several Variables
803
61. faxß yß zb œ Èx y ln z at a3ß 1ß 1b Ê w œ Èx y ln z; at a3ß 1ß 1b Ê w œ È3 a1b ln 1 œ 2
Ê Èx y ln z œ 2
62. faxß yß zb œ ln ax# y z# b at a"ß #ß "b Ê w œ ln ax# y z# b ; at a"ß #ß "b Ê w œ ln a1 2 1b œ ln 4
Ê ln 4 œ ln ax# y z# b Ê x# y z# œ 4
63. gaxß yß zb œ Èx# y2 z# at Š1ß 1ß È2‹ Ê w œ Èx# y2 z# ; at Š1ß 1ß È2‹ Ê w œ Ê1# a1b2 ŠÈ2‹
#
œ 2 Ê 2 œ Èx# y2 z# Ê x# y2 z# œ 4
1 0 a2b
xyz
xyz
xyz
1
1
64. gaxß yß zb œ 2x
y z at a1ß 0ß 2b Ê w œ 2x y z ; at a1ß 0ß 2b Ê w œ 2a1b 0 a2b œ 4 Ê 4 œ 2x y z
Ê 2x y z œ 0
_
n
65. faxß yb œ ! Š xy ‹ œ
n œ0
1
œ y y x for
1 Š xy ‹
¹ xy ¹ 1 Ê Domain: all points ax, yb satisfying lxl lyl;
at a1, 2b Ê since ¹ 12 ¹  1 Ê z œ 2 2 1 œ 2
Ê y y x œ 2 Ê y œ 2x
_
66. gaxß yß zb œ !
n œ0
(x b y)n
ÐxyÑÎz
Ê Domain: all points ax, y, zb satisfying z Á 0; at aln 4ß ln 9ß 2b
n! zn œ e
Ê w œ eÐln 4 ln 9ÑÎ2 œ eÐln 36ÑÎ2 œ eln 6 œ 6 Ê 6 œ eÐxyÑÎz Ê x z y œ ln 6
67. faxß yb œ 'x È d) # œ sin1 y sin1 x Ê Domain: all points
y
1)
ax, yb satisfying 1 Ÿ x Ÿ 1 and 1 Ÿ y Ÿ 1;
at a0, 1b Ê sin1 1 sin1 0 œ 12 Ê sin1 y sin1 x
œ 12 . Since 12 Ÿ sin1 y Ÿ 12 and 12 Ÿ sin1 x Ÿ 12 , in
order for sin1 y sin1 x to equal 12 , 0 Ÿ sin1 y Ÿ 12 and
12 Ÿ sin1 x Ÿ 0; that is 0 Ÿ y Ÿ 1 and 1 Ÿ x Ÿ 0. Thus
y œ sinˆ 1 sin1 x‰ œ È1 x2 , x Ÿ 0
2
68. gaxß yß zb œ 'x 1 dt t# '0 È d) # œ tan1 y tan1 x sin1 ˆ 2z ‰ Ê Domain: all points ax, y, zb satisfying 2 Ÿ z Ÿ 2;
y
z
4)
È
at Š0ß 1ß È3‹ Ê tan1 1 tan1 0 sin1 Š 3 ‹ œ 71 Ê tan1 y tan1 x sin1 ˆ z ‰ œ 71 . Since 1 Ÿ sin1 ˆ z ‰ Ÿ 1 ,
2
12
2
12
2
1
131
131
1
1
1
1 ‰ 1
1
1
ˆ 71
12 Ÿ tan y tan x Ÿ 12 Ê z œ 2 sin 12 tan y tan x , 12 Ÿ tan y tan x Ÿ 12
69-72. Example CAS commands:
Maple:
with( plots );
f := (x,y) -> x*sin(y/2) + y*sin(2*x);
xdomain := x=0..5*Pi;
ydomain := y=0..5*Pi;
x0,y0 := 3*Pi,3*Pi;
plot3d( f(x,y), xdomain, ydomain, axes=boxed, style=patch, shading=zhue, title="#69(a) (Section 14.1)" );
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2
2
804
Chapter 14 Partial Derivatives
plot3d( f(x,y), xdomain, ydomain, grid=[50,50], axes=boxed, shading=zhue, style=patchcontour, orientation=[-90,0],
title="#69(b) (Section 14.1)" );
# (b)
L := evalf( f(x0,y0) );
# (c)
plot3d( f(x,y), xdomain, ydomain, grid=[50,50], axes=boxed, shading=zhue, style=patchcontour, contours=[L],
orientation=[-90,0], title="#45(c) (Section 13.1)" );
73-76. Example CAS commands:
Maple:
eq := 4*ln(x^2+y^2+z^2)=1;
implicitplot3d( eq, x=-2..2, y=-2..2, z=-2..2, grid=[30,30,30], axes=boxed, title="#73 (Section 14.1)" );
77-80. Example CAS commands:
Maple:
x := (u,v) -> u*cos(v);
y := (u,v) ->u*sin(v);
z := (u,v) -> u;
plot3d( [x(u,v),y(u,v),z(u,v)], u=0..2, v=0..2*Pi, axes=boxed, style=patchcontour, contours=[($0..4)/2], shading=zhue,
title="#77 (Section 14.1)" );
69-60. Example CAS commands:
Mathematica: (assigned functions and bounds will vary)
For 69 - 72, the command ContourPlot draws 2-dimensional contours that are z-level curves of surfaces z = f(x,y).
Clear[x, y, f]
f[x_, y_]:= x Sin[y/2] y Sin[2x]
xmin= 0; xmax= 51; ymin= 0; ymax= 51; {x0, y0}={31, 31};
cp= ContourPlot[f[x,y], {x, xmin, xmax}, {y, ymin, ymax}, ContourShading Ä False];
cp0= ContourPlot[[f[x,y], {x, xmin, xmax}, {y, ymin, ymax}, Contours Ä {f[x0,y0]}, ContourShading Ä False,
PlotStyle Ä {RGBColor[1,0,0]}];
Show[cp, cp0]
For 73 - 76, the command ContourPlot3D will be used. Write the function f[x, y, z] so that when it is equated to zero, it
represents the level surface given.
For 73, the problem associated with Log[0] can be avoided by rewriting the function as x2 + y2 +z2 - e1/4
Clear[x, y, z, f]
f[x_, y_, z_]:= x2 y2 z2 Exp[1/4]
ContourPlot3D[f[x, y, z], {x, 5, 5}, {y, 5, 5}, {z, 5, 5}, PlotPoints Ä {7, 7}];
For 77 - 80, the command ParametricPlot3D will be used. To get the z-level curves here, we solve x and y in terms of z
and either u or v (v here), create a table of level curves, then plot that table.
Clear[x, y, z, u, v]
ParametricPlot3D[{u Cos[v], u Sin[v], u}, {u, 0, 2}, {v, 0, 2p}];
zlevel= Table[{z Cos[v], z sin[v]}, {z, 0, 2, .1}];
ParametricPlot[Evaluate[zlevel],{v, 0, 21}];
14.2 LIMITS AND CONTINUITY IN HIGHER DIMENSIONS
3x# y# 5
lim
#
#
Ðxß yÑ Ä Ð0ß 0Ñ x y 2
2.
lim
œ È œ0
4
Ðxß yÑ Ä Ð0ß 4Ñ Èy
x
#
#
0 5
5
œ 3(0)
0# 0# 2 œ #
1.
0
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 14.2 Limits and Continuity in Higher Dimensions
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
lim
Ðxß yÑ Ä Ð3ß 4Ñ
Èx# y# 1 œ È3# 4# 1 œ È24 œ 2È6
#
#
#
lim
"
Š x" y" ‹ œ #" ˆ "3 ‰‘ œ ˆ 6" ‰ œ 36
lim
sec x tan y œ (sec 0) ˆtan 14 ‰ œ (1)(1) œ 1
lim
cos Š xxy y 1 ‹ œ cos Š 000 0 1 ‹ œ cos 0 œ 1
Ðxß yÑ Ä Ð2ß 3Ñ
Ðxß yÑ Ä ˆ0ß 14 ‰
Ðxß yÑ Ä Ð0ß 0Ñ
lim
#
lim
#
exy œ e0 ln 2 œ eln ˆ 2 ‰ œ "#
ln k1 x# y# k œ ln k1 (1)# (1)# k œ ln 2
ey sin x
œ
lim
x
Ðxß yÑ Ä Ð0ß 0Ñ
Ðxß yÑ Ä Ð0ß 0Ñ
lim
lim
$
1
Ðxß yÑ Ä Ð0ß ln 2Ñ
Ðxß yÑ Ä Ð1ß 1Ñ
$
Ðxß yÑ Ä Ð1Î27ß 13 Ñ
aey b ˆ sinx x ‰ œ e! † lim ˆ sinx x ‰ œ 1 † 1 œ 1
xÄ0
3
1 ‰ 3
3 xy œ cos É
ˆ 27
cos È
1 œ cos ˆ 13 ‰ œ 21
1†sinˆ 16 ‰
1Î2
x sin y
1
#1 œ
x
1# 1 œ 2 œ 4
Ðxß yÑ Ä Ð1ß 1Î6Ñ
lim
lim
acos 0b "
cos y 1
11
y sin x œ 0 sin ˆ 1# ‰ œ 1 œ 2
lim
x# 2xy y#
(x y)#
œ
lim
œ
lim
xy
Ðx ß y Ñ Ä Ð 1 ß 1Ñ x y
Ðxß yÑ Ä Ð1ß 1Ñ
(x y) œ (" 1) œ 0
lim
x# y#
(x y)(x y)
œ
lim
x y œ Ðxß yÑlim
xy
Ä Ð1ß 1Ñ
Ðxß yÑ Ä Ð1ß 1Ñ
(x y) œ (1 1) œ 2
lim
xy y 2x 2
(x 1)(y 2)
œ
lim
œ
lim
x1
x1
Ðx ß y Ñ Ä Ð 1 ß 1Ñ
Ðxß yÑ Ä Ð1ß 1Ñ
Ðxß yÑ Ä ˆ 12 ß 0‰
Ðxß yÑ Ä Ð1ß 1Ñ
xÁy
Ðxß yÑ Ä Ð1ß 1Ñ
xÁy
Ðxß yÑ Ä Ð1ß 1Ñ
xÁ1
(y 2) œ (1 2) œ 1
xÁ1
y4
y4
"
lim
Ðxß yÑ Ä Ð0ß 0Ñ
xÁy
1
ˆÈ x È y ‰ ˆ È x È y 2 ‰
x y 2È x 2È y
œ
lim
œ
lim
Èx Èy
Èx Èy
Ðx ß y Ñ Ä Ð 0 ß 0Ñ
Ðxß yÑ Ä Ð0ß 0Ñ
ˆÈ x È y 2 ‰
xÁy
œ ŠÈ0 È0 2‹ œ 2
Note: (xß y) must approach (0ß 0) through the first quadrant only with x Á y.
18.
"
lim
œ
lim
œ
lim
œ #(2 1) œ #
#
#
Ðxß yÑ Ä Ð2ß 4Ñ x y xy 4x 4x
Ðxß yÑ Ä Ð2ß 4Ñ x(x 1)(y 4)
Ðxß yÑ Ä Ð2ß 4Ñ x(x 1)
y Á 4, x Á x#
y Á 4, x Á x#
x Á x#
xy4
lim
œ
lim
Ðxß yÑ Ä Ð2ß 2Ñ Èx y 2
Ðxß yÑ Ä Ð2ß 2Ñ
xyÁ4
xyÁ4
ˆÈx y 2‰ ˆÈx y 2‰
œ
Èx y 2
lim
Ðxß yÑ Ä Ð2ß 2Ñ
xyÁ4
ˆÈ x y 2 ‰
œ ŠÈ2 2 2‹ œ 2 2 œ 4
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
805
806
Chapter 14 Partial Derivatives
19.
lim
È2x y 2
È2x y 2
"
œ
lim
2x y 4 œ Ðxß yÑlim
Ä Ð2ß 0Ñ ˆÈ2x y 2‰ ˆÈ2x y 2‰
Ðxß yÑ Ä Ð2ß 0Ñ È2x y #
Ðxß yÑ Ä Ð2ß 0Ñ
2x y Á 4
2x y Á 4
œ È(2)(2)" 0 # œ 2 " 2 œ 4"
20.
È x È y 1
Èx Èy 1
"
œ
lim
x y 1 œ Ðxß yÑlim
Ä Ð4ß 3Ñ ˆÈx Èy 1‰ ˆÈx Èy 1‰
Ðxß yÑ Ä Ð4ß 3Ñ Èx Èy 1
lim
Ðxß yÑ Ä Ð4ß 3Ñ
xyÁ1
xyÁ1
"
œ È4 È
œ 2 " 2 œ 4"
31
21.
22.
lim
#
sinax# y# b
a r# b
œ lim sinra#r b œ lim 2r†cos
œ lim
x# y#
2r
rÄ0
rÄ0
rÄ0
lim
1 cosaxyb
œ lim 1 ucos u œ lim sin1 u œ 0
xy
uÄ0
uÄ0
Ðxß yÑ Ä Ð0ß 0Ñ
Ðxß yÑ Ä Ð0ß 0Ñ
cosar# b œ 1
23.
ax ybˆx2 xy y2 ‰
x3 y3
œ
lim
œ
lim
x
y
xy
Ðxß yÑ Ä Ð1ß "Ñ
Ðxß yÑ Ä Ð1ß "Ñ
Ðxß yÑ Ä Ð1ß "Ñ
24.
lim
lim
lim
4
4 œ
2
2 œ
2
2 œ a2 2ba22 22 b œ 32
Ðx ß y Ñ Ä Ð 2 ß 2 Ñ x y
Ðxß yÑ Ä Ð2ß 2Ñ ax ybax ybax y b
Ðxß yÑ Ä Ð2ß 2Ñ ax ybax y b
25.
26.
27.
28.
29.
30.
lim
xy
lim
T Ä Ð1ß 3ß 4Ñ
lim
lim
xy
1
1
2xy yz
2(1)(1) (1)(1)
œ 121" œ #"
x # z# œ
1# (1)#
asin# x cos# y sec# zb œ asin# 3 cos# 3b sec# 0 œ 1 1# œ 2
lim
T Ä ˆ 14 ß 12 ß 2‰
tan" (xyz) œ tan" ˆ "4 † 1# † 2‰ œ tan" ˆ 14 ‰
lim
ze 2y cos 2x œ 3e 2Ð0Ñ cos 21 œ (3)(1)(1) œ 3
lim
ln Èx# y# z# œ ln È2# (3)# 6# œ ln È49 œ ln 7
T Ä Ð1ß 0ß 3Ñ
1
Š "x y" "z ‹ œ 1" 3" 4" œ 12 124 3 œ 19
12
T Ä Ð 1 ß 1 ß 1Ñ
T Ä Ð3ß 3ß 0Ñ
ax2 xy y2 b œ Š12 a1ba1b a1b2 ‹ œ 3
T Ä Ð2 ß 3 ß 6 Ñ
31. (a) All axß yb
(b) All axß yb except a0ß 0b
32. (a) All axß yb so that x Á y
(b) All axß yb
33. (a) All axß yb except where x œ 0 or y œ 0
(b) All axß yb
34. (a) All axß yb so that x# 3x 2 Á 0 Ê ax 2bax 1b Á 0 Ê x Á 2 and x Á 1
(b) All axß yb so that y Á x#
35. (a) All axß yß zb
(b) All axß yß zb except the interior of the cylinder x# y# œ 1
36. (a) All axß yß zb so that xyz 0
(b) All axß yß zb
37. (a) All axß yß zb with z Á 0
(b) All axß yß zb with x# z# Á 1
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 14.2 Limits and Continuity in Higher Dimensions
38. (a) All axß yß zb except axß 0ß 0b
(b) All axß yß zb except a0ß yß 0b or axß 0ß 0b
39. (a) All axß yß zb such that z x2 y2 1
(b) All axß yß zb such that z Á Èx2 y2
807
40. (a) All axß yß zb such that x2 y2 z2 Ÿ 4
(b) All axß yß zb such that x2 y2 z2 9 except when x2 y2 z2 œ 25
41.
lim
Èx#x y# œ lim b Èx#x x# œ lim b È2x kxk œ lim b Èx2 x œ lim b È"2 œ È"2 ;
xÄ0
xÄ0
xÄ0
xÄ0
lim
Èx#x y# œ lim c È2x kxk œ lim c È2(xx) œ lim c
Ðxß yÑ Ä Ð0ß 0Ñ
along y œ x
x0
Ðxß yÑ Ä Ð0ß 0Ñ
along y œ x
x0
42.
lim
Ðxß yÑ Ä Ð0ß 0Ñ
along y œ 0
xÄ0
xÄ0
%
x%
œ lim x% x 0# œ 1;
lim
x% y#
xÄ0
Ðxß yÑ Ä Ð0ß 0Ñ
along y œ x#
%
"
x%
x%
œ lim x% xax# b# œ lim 2x
% œ #
x% y#
xÄ0
xÄ0
#
43.
44.
45.
46.
47.
48.
49.
50.
"
"
È2 œ È2
xÄ0
lim
%
%
akx# b
x% y#
k# x%
k#
œ lim xx% œ lim xx% œ 11 k#
x% y#
k# x%
akx# b#
xÄ0
xÄ0
lim
#
xy
x(kx)
k
œ lim kkx
; if k 0, the limit is 1; but if k 0, the limit is 1
# œ lim
kxyk œ xlim
Ä 0 kx(kx)k
x Ä 0 kx k
x Ä 0 kkk
lim
xy
x kx
k
œ 11 x y œ xlim
k
Ä 0 x kx
lim
x2 y
x2 kx
k
œ lim 1x œ 1kk
x y œ xlim
Ä 0 x kx
x Ä 0 k
lim
#
x# y
kx#
œ lim x kx
œ 1 k k
#
y
xÄ0
lim
x# y
kx4
k
4
2 4 œ 1 k2
x4 y2 œ xlim
Ä 0 x k x
lim
xy2 1
y2 1
œ lim ay 1b œ 2;
lim
y 1 œ ylim
Ä 1 y1
yÄ1
Ðx ß y Ñ Ä Ð 1 ß 1Ñ
Ðxß yÑ Ä Ð0ß 0Ñ
along y œ kx#
Ðxß yÑ Ä Ð0ß 0Ñ
along y œ kx
kÁ0
Ðx ß y Ñ Ä Ð 0 ß 0 Ñ
along y œ kx
k Á 1
Ðxß yÑ Ä Ð0ß 0Ñ
along y œ kx
kÁ1
Ðxß yÑ Ä Ð0ß 0Ñ
along y œ kx#
kÁ0
Ðxß yÑ Ä Ð0ß 0Ñ
along y œ kx#
Ðxß yÑ Ä Ð1ß 1Ñ
along x œ 1
lim
Ðxß yÑ Ä Ð1ß 1Ñ
along y œ 1
Ê different limits for different values of k
Ê different limits for different values of k, k Á 1
Ê different limits for different values of k, k Á 1
Ê different limits for different values of k, k Á 0
Ê different limits for different values of k
along y œ x
xy2 1
y3 1
œ lim ay2 y 1b œ 3
y 1 œ ylim
yÄ1
Ä 1 y1
xy 1
x 1
œ lim x11 œ 21 ;
lim
2
x2 y2 œ xlim
Ä 1 x 1
xÄ1
Ð x ß y Ñ Ä Ð 1 ß 1Ñ
along y œ x2
xy 1
x 3 1
x2 x 1
2 x4 œ lim
a
x
x
1bax2 1b
x2 y2 œ xlim
xÄ1
Ä1
œ 32
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
808
Chapter 14 Partial Derivatives
Ú 1 if y x%
51. fax, yb œ Û 1 if y Ÿ 0
Ü 0 otherwise
(a)
(b)
(c)
x%
lim
fax, yb œ 1 since any path through a0, 1b that is close to a0, 1b satisfies y
lim
fax, yb œ 0 since any path through a2, 3b that is close to a2, 3b does not satisfiy either y
lim
fax, yb œ 1 and
Ðxß yÑ Ä Ð0ß 1Ñ
Ðx ß y Ñ Ä Ð 0 ß 0 Ñ
along x œ 0
x2
52. fax, yb œ œ 3
x
(a)
lim
fax, yb œ 0 Ê
lim
Ðx ß y Ñ Ä Ð 0 ß 0Ñ
along y œ x2
fax, yb does not exist
lim
Ðxß yÑ Ä Ð0ß 0Ñ
if x 0
if x 0
fax, yb œ 32 œ 9 since any path through a3, 2b that is close to a3, 2b satisfies x
0
Ðxß yÑ Ä Ð3ß 2Ñ
(b)
(c)
x% or y Ÿ 0
Ðxß yÑ Ä Ð2ß 3Ñ
lim
fax, yb œ a2b3 œ 8 since any path through a2, 1b that is close to a2, 1b satisfies x 0
lim
fax, yb œ 0 since the limit is 0 along any path through a0, 0b with x 0 and the limit is also zero along
Ðxß yÑ Ä Ð2ß 1Ñ
Ðxß yÑ Ä Ð0ß 0Ñ
any path through a0, 0b with x
0
53. First consider the vertical line x œ 0 Ê
2
2 a0 b y
2x2 y
œ lim
4 y2 œ lim
4
x
y Ä 0 a0b y2
yÄ0
Ðxß yÑ Ä Ð0ß 0Ñ
lim
0 œ 0. Now consider any nonvertical
along x œ 0
through a0, 0b. The equation of any line through a0, 0b is of the form y œ mx Ê
lim
Ðxß yÑ Ä Ð0ß 0Ñ
along y œ mx
faxß yb œ
2x2 y
4 y2
x
Ðxß yÑ Ä Ð0ß 0Ñ
lim
along y œ mx
2
3
b
2mx3
2mx
œ lim x42x aamx
œ lim x4 2mx
lim
2 2 œ lim x2 ax2 m2 b œ lim ax2 m2 b œ 0. Thus
mxb2
xÄ0
x Ä 0 m x
xÄ0
xÄ0
Ðxß yÑ Ä Ð0ß 0Ñ
2x2 y
x4 y2 œ 0.
any line though a0, 0b
54. If f is continuous at (x! ß y! ), then
lim
Ðxß yÑ Ä Ðx! ß y! Ñ
f(xß y) must equal f(x! ß y! ) œ 3. If f is not continuous at
(x! ß y! ), the limit could have any value different from 3, and need not even exist.
55.
# #
lim
Ðxß yÑ Ä Ð0ß 0Ñ
Š1 x 3y ‹ œ 1 and
lim
Ðx ß y Ñ Ä Ð 0 ß 0Ñ
1œ1 Ê
# #
56. If xy 0,
lim
2 kxyk Š x 6y ‹
Ðx ß y Ñ Ä Ð 0 ß 0Ñ
kxyk
tan " xy
œ 1, by the Sandwich Theorem
xy
lim
Ðxß yÑ Ä Ð0ß 0Ñ
# #
œ
2xy Š x 6y ‹
lim
Ðxß yÑ Ä Ð0ß 0Ñ
xy
œ
lim
Ðxß yÑ Ä Ð0ß 0Ñ
ˆ2 xy
‰
6 œ 2 and
# #
2 kxyk
œ
lim
Ðxß yÑ Ä Ð0ß 0Ñ kxyk
Ðxß yÑ Ä Ð0ß 0Ñ
lim
œ
lim
Ðx ß y Ñ Ä Ð 0 ß 0 Ñ
ˆ2 xy
‰
6 œ 2 and
2 œ 2; if xy 0,
lim
Ðx ß y Ñ Ä Ð 0 ß 0Ñ
lim
2 kxyk Š x 6y ‹
kxyk
Ðxß yÑ Ä Ð0ß 0Ñ
2 kxyk
kxyk œ 2
Ê
lim
Ðxß yÑ Ä Ð0ß 0Ñ
# #
œ
lim
2xy Š x 6y ‹
xy
Ðxß yÑ Ä Ð0ß 0Ñ
4 4 cos Èkxyk
œ 2, by the Sandwich Theorem
kxyk
57. The limit is 0 since ¸sin ˆ "x ‰¸ Ÿ 1 Ê 1 Ÿ sin ˆ x" ‰ Ÿ 1 Ê y Ÿ y sin ˆ x" ‰ Ÿ y for y 0, and y y sin ˆ "x ‰
y Ÿ 0. Thus as (xß y) Ä (!ß !), both y and y approach 0 Ê y sin ˆ "x ‰ Ä 0, by the Sandwich Theorem.
58. The limit is 0 since ¹cos Š "y ‹¹ Ÿ 1 Ê 1 Ÿ cos Š y" ‹ Ÿ 1 Ê x Ÿ x cos Š y" ‹ Ÿ x for x
0, and x
x cos Š y" ‹
for x Ÿ 0. Thus as (xß y) Ä (!ß !), both x and x approach 0 Ê x cos Š "y ‹ Ä 0, by the Sandwich Theorem.
)
59. (a) f(xß y)k yœmx œ 1 2mm# œ 1 2tan
tan# ) œ sin 2). The value of f(xß y) œ sin 2) varies with ), which is the line's
angle of inclination.
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
y for
x
Section 14.2 Limits and Continuity in Higher Dimensions
(b) Since f(xß y)k yœmx œ sin 2) and since 1 Ÿ sin 2) Ÿ 1 for every ),
lim
Ðxß yÑ Ä Ð0ß 0Ñ
809
f(xß y) varies from 1 to 1
along y œ mx.
60. kxy ax# y# bk œ kxyk kx# y# k Ÿ kxk kyk kx# y# k œ Èx# Èy# kx# y# k Ÿ Èx# y# Èx# y# kx# y# k
#
#
#
#
# #
#
#
œ ax# y# b Ê ¹ xyxa#xy#y b ¹ Ÿ axx#yy#b œ x# y# Ê ax# y# b Ÿ xyxa#xy#y b Ÿ ax# y# b
Ê
61.
62.
63.
#
lim
Ðxß yÑ Ä Ð0ß 0Ñ
#
y
Šxy xx# y# ‹ œ 0 by the Sandwich Theorem, since
x$ xy#
œ lim
lim
#
#
Ðxß yÑ Ä Ð0ß 0Ñ x y
rÄ0
$
lim
Ðx ß y Ñ Ä Ð 0 ß 0Ñ
„ ax# y# b œ 0; thus, define fa0ß 0b œ 0
$
) sin# )b
r$ cos$ ) (r cos )) ar# sin# )b
œ lim r acos ) cos
œ0
r# cos# ) r# sin# )
1
rÄ0
$
$
$
lim
r acos ) sin )b
y
r cos ) r sin )
cos Š xx# “ œ cos 0 œ 1
y# ‹ œ lim cos Š r# cos# ) r# sin# ) ‹ œ lim cos ’
1
lim
y#
r# sin# )
œ lim
x# y# œ r lim
r#
Ä0
rÄ0
Ðxß yÑ Ä Ð0ß 0Ñ
Ðxß yÑ Ä Ð0ß 0Ñ
$
$
$
$
rÄ0
rÄ0
asin# )b œ sin# ); the limit does not exist since sin# ) is between
0 and 1 depending on )
64.
65.
2r cos )
2 cos )
2 cos )
lim
œ lim r cos ) œ cos ) ; the limit does not exist for cos ) œ 0
#
# œ lim
#
r Ä 0 r r cos )
rÄ0
Ðxß yÑ Ä Ð0ß 0Ñ x x y
2x
lim
Ðxß yÑ Ä Ð0ß 0Ñ
ky k
krk akcos )k ksin )kb
" kr cos )k kr sin )k
tan" ’ kxx#k ’
“ œ lim tan" ’
“;
y# “ œ lim tan
r#
r#
rÄ0
if r Ä 0 , then
rÄ0
lim b tan" ’ krk akcos )rk# ksin )kb “ œ
rÄ!
lim
r Ä !b
tan" ’ kcos )k r ksin )k “ œ 1# ; if r Ä 0 , then
lim tan" ’ krk akcos )rk# ksin )kb “ œ lim c tan" Š kcos )kr ksin )k ‹ œ 1# Ê the limit is 1#
rÄ!
r Ä !c
66.
x# y#
lim
#
# œ lim
rÄ0
Ðxß yÑ Ä Ð0ß 0Ñ x y
r# cos# ) r# sin# )
œ lim
r#
rÄ0
acos# ) sin# )b œ lim (cos 2)) which ranges between
rÄ0
1 and 1 depending on ) Ê the limit does not exist
67.
lim
Ðx ß y Ñ Ä Ð 0 ß 0Ñ
#
# #
#
ln Š 3x x#xy y# 3y ‹ œ lim ln Š 3r cos ) r cos r#) sin ) 3r sin ) ‹
#
#
%
#
#
#
#
rÄ0
œ lim ln a3 r# cos# ) sin# )b œ ln 3 Ê define f(0ß 0) œ ln 3
rÄ0
68.
lim
Ðxß yÑ Ä Ð0ß 0Ñ
(3r cos )) ar# sin# )b
3xy#
œ lim
x# y# œ r lim
r#
Ä0
rÄ0
3r cos ) sin# ) œ 0 Ê define f(0ß 0) œ 0
69. Let $ œ 0.1. Then Èx# y# $ Ê Èx# y# 0.1 Ê x# y# 0.01 Ê kx# y# 0k 0.01
Ê kf(xß y) f(!ß !)k 0.01 œ %.
70. Let $ œ 0.05. Then kxk $ and kyk $ Ê kfaxß yb fa0ß 0bk œ ¸ x# y 1 0¸ œ ¸ x# y 1 ¸ Ÿ kyk 0.05 œ %.
71. Let $ œ 0.005. Then kxk $ and kyk $ Ê kfaxß yb fa0ß 0bk œ ¸ xx#y1 0¸ œ ¸ xx#y1 ¸ Ÿ kx yk kxk kyk
0.005 0.005 œ 0.01 œ %.
kx yk
"
72. Let $ œ 0.01. Since 1 Ÿ cos x Ÿ 1 Ê 1 Ÿ 2 cos x Ÿ 3 Ê "3 Ÿ #cos
Ÿ ¸ 2 x cosy x ¸ Ÿ kx yk
x Ÿ 1 Ê
3
Ÿ kxk kyk . Then kxk $ and kyk $ Ê kfaxß yb fa0ß 0bk œ ¸ 2 x cosy x 0¸ œ ¸ 2 x cosy x ¸ Ÿ kxk kyk 0.01 0.01
œ 0.02 œ %.
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
810
Chapter 14 Partial Derivatives
73. Let $ œ 0.04. Since y2 Ÿ x2 y2 Ê x2 y y2 Ÿ 1 Ê xl2xlyy2 Ÿ lxl œ Èx2 Ÿ Èx2 y2 $ Ê kfaxß yb fa0ß 0bk
2
2
2
œ ¹ x2xy y2 0¹ 0.04 œ %.
74. Let $ œ 0.01. If lyl Ÿ 1, then y2 Ÿ lyl œ Èy2 Ÿ Èx2 y2 , so lxl œ Èx2 Ÿ Èx2 y2 Ê lxl y2 Ÿ 2Èx2 y2 . Since
y l
y
x
2
2
x2 Ÿ x2 y2 Ê x2 x y2 Ÿ 1 and y2 Ÿ x2 y2 Ê x2 y y2 Ÿ 1. Then lxx2 y2 Ÿ x2 y2 lxl x2 y2 y Ÿ lxl y 2$
3
2
2
4
2
2
y
Ê kfaxß yb fa0ß 0bk œ ¹ xx2 y2 0¹ 2a0.01b œ 0.002 œ % .
3
4
75. Let $ œ È0.015. Then Èx# y# z# $ Ê kf(xß yß z) f(!ß 0ß 0)k œ kx# y# z# 0k œ kx# y# z# k
#
#
œ ŠÈx# t# x# ‹ ŠÈ0.015‹ œ 0.015 œ %.
76. Let $ œ 0.2. Then kxk $ , kyk $ , and kzk $ Ê kf(xß yß z) f(!ß 0ß 0)k œ kxyz 0k œ kxyzk œ kxk kyk kzk (0.2)$
œ 0.008 œ %.
77. Let $ œ 0.005. Then kxk $ , kyk $ , and kzk $ Ê kf(xß yß z) f(!ß 0ß 0)k œ ¹ x# x y# yz#z 1 0¹
œ ¹ x# x y# yz#z 1 ¹ Ÿ kx y zk Ÿ kxk kyk kzk 0.005 0.005 0.005 œ 0.015 œ %.
78. Let $ œ tan" (0.1). Then kxk $ , kyk $ , and kzk $ Ê kf(xß yß z) f(!ß 0ß 0)k œ ktan# x tan# y tan# zk
Ÿ ktan# xk ktan# yk ktan# zk œ tan# x tan# y tan# z tan# $ tan# $ tan# $ œ 0.01 0.01 0.01 œ 0.03 œ %.
79.
lim
Ðxß yß zÑ Ä Ðx! ß y! ß z! Ñ
f(xß yß z) œ
lim
(x y z) œ x! y! z! œ f(x! ß y! ß z! ) Ê f is continuous at
lim
ax# y# z# b œ x!# y!# z!# œ f(x! ß y! ß z! ) Ê f is continuous at
Ðxß yß zÑ Ä Ðx! ß y! ß z! Ñ
every (x! ß y! ß z! )
80.
lim
Ðxß yß zÑ Ä Ðx! ß y! ß z! Ñ
f(xß yß z) œ
Ðxß yß zÑ Ä Ðx! ß y! ß z! Ñ
every point (x! ß y! ß z! )
14.3 PARTIAL DERIVATIVES
1.
`f
`f
` x œ 4x, ` y œ 3
2.
`f
`f
` x œ 2x y, ` y œ x 2y
3.
`f
`f
#
` x œ 2x(y 2), ` y œ x 1
4.
`f
`f
` x œ 5y 14x 3, ` y œ 5x 2y 6
5.
`f
`f
` x œ 2y(xy 1), ` y œ 2x(xy 1)
6.
`f
# `f
#
` x œ 6(2x 3y) , ` y œ 9(2x 3y)
7.
y
`f
x
`f
` x œ È x# y# , ` y œ È x# y#
8.
`f
2x#
, ` f œ $ $" y
`x œ É
$ $
x ˆ y ‰ `y
3É x ˆ ‰
9.
#
#
`f
"
`
"
`f
"
`
"
` x œ (x y)# † ` x (x y) œ (x y)# , ` y œ (x y)# † ` y (x y) œ (x y)#
#
#
#
#
#
#
10. `` xf œ ax ayx# b(1)y#b#x(2x) œ axy# yx# b# , `` yf œ ax ayx#b(0)y#b#x(2y) œ ax# 2xy
y # b#
#
(x y)(y)
y 1
(xy ")(1) (x y)(x)
`f
x "
11. `` xf œ œ (xy 1)(1)
œ (xy
œ (xy
(xy 1)#
1)# , ` y œ
(xy 1)#
1)#
#
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 14.3 Partial Derivatives
12. `` xf œ
y
y
"
` ˆy‰
`f
"
` ˆy‰
1
x
# † `x
x œ x# ’1 ˆ y ‰# “ œ x# y# , ` y œ 1 ˆ y ‰# † ` y x œ x ’1 ˆ y ‰# “ œ x# y#
1 ˆ xy ‰
x
x
x
13. `` xf œ eÐxy1Ñ † ``x (x y 1) œ eÐxy1Ñ , `` yf œ eÐxy1Ñ † ``y (x y 1) œ eÐxy1Ñ
14. `` xf œ ex sin (x y) ex cos (x y), `` yf œ ex cos (x y)
15. `` xf œ x " y † ``x (x y) œ x " y , `` yf œ x " y † ``y (x y) œ x " y
16. `` xf œ exy † ``x (xy) † ln y œ yexy ln y, `` yf œ exy † ``y (xy) † ln y exy † "y œ xexy ln y ey
xy
17. `` xf œ 2 sin (x 3y) † ``x sin (x 3y) œ 2 sin (x 3y) cos (x 3y) † ``x (x 3y) œ 2 sin (x 3y) cos (x 3y),
`f
`
`
` y œ 2 sin (x 3y) † ` y sin (x 3y) œ 2 sin (x 3y) cos (x 3y) † ` y (x 3y) œ 6 sin (x 3y) cos (x 3y)
18. `` xf œ 2 cos a3x y# b † ``x cos a3x y# b œ 2 cos a3x y# b sin a3x y# b † ``x a3x y# b
œ 6 cos a3x y# b sin a3x y# b ,
`f
`
`
#
#
#
#
#
` y œ 2 cos a3x y b † ` y cos a3x y b œ 2 cos a3x y b sin a3x y b † ` y a3x y b
œ 4y cos a3x y# b sin a3x y# b
19. `` xf œ yxyc1 , `` yf œ xy ln x
x
`f
"
`f
ln x
20. f(xß y) œ ln
ln y Ê ` x œ x ln y and ` y œ y(ln y)#
21. `` xf œ g(x), `` yf œ g(y)
_
22. f(xß y) œ ! (xy)n , kxyk 1 Ê f(xß y) œ 1 " xy Ê `` xf œ (1 "xy)# † ``x (1 xy) œ (1 yxy)# and
n œ0
`f
"
`
x
` y œ (1 xy)# † ` y (1 xy) œ (1 xy)#
23. fx œ y# , fy œ 2xy, fz œ 4z
24. fx œ y z, fy œ x z, fz œ y x
25. fx œ 1, fy œ Èy#y z# , fz œ Èy#z z#
26. fx œ x ax# y# z# b
$Î#
, fy œ y ax# y# z# b
$Î#
, fz œ z ax# y# z# b
$Î#
27. fx œ È1 yzx# y# z# , fy œ È1 xzx# y# z# , fz œ È1 xyx# y# z#
"
28. fx œ kx yzk È(x
, fy œ kx yzk È(xz yz)# 1 , fz œ kx yzk È(xy yz)# 1
yz)# 1
29. fx œ x 2y" 3z , fy œ x 2y2 3z , fz œ x 2y3 3z
yz
yz
"
`
`
30. fx œ yz † xy
† ``x (xy) œ (yz)(y)
xy œ x , fy œ z ln (xy) yz † ` y ln (xy) œ z ln (xy) xy † ` y (xy) œ z ln (xy) z,
fz œ y ln (xy) yz † ``z ln (xy) œ y ln (xy)
#
#
#
#
#
#
#
#
#
31. fx œ 2xe ax y z b , fy œ 2ye ax y z b , fz œ 2ze ax y z b
32. fx œ yzexyz , fy œ xzexyz , fz œ xyexyz
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
811
812
Chapter 14 Partial Derivatives
33. fx œ sech# (x 2y 3z), fy œ 2 sech# (x 2y 3z), fz œ 3 sech# (x 2y 3z)
34. fx œ y cosh axy z# b , fy œ x cosh axy z# b , fz œ 2z cosh axy z# b
35. ``ft œ 21 sin (21t !), ``!f œ sin (21t !)
Ð2uÎvÑ ` g
Ð2uÎvÑ
‰
‰
36. `` gu œ v# eÐ2uÎvÑ † ``u ˆ 2u
, ` v œ 2veÐ2uÎvÑ v# eÐ2uÎvÑ † ``v ˆ 2u
2ueÐ2uÎvÑ
v œ 2ve
v œ 2ve
37. `` 3h œ sin 9 cos ), ``9h œ 3 cos 9 cos ), `` h) œ 3 sin 9 sin )
38. ``gr œ 1 cos ), `` g) œ r sin ), `` gz œ 1
#
#
#
2V$ v
V$ v
V$ v
39. Wp œ V, Wv œ P $2gv , W$ œ Vv
2g , Wv œ 2g œ g , Wg œ #g#
A
h
40. ``Ac œ m, ``Ah œ #q , ``Ak œ mq , `` m
œ qk c, ``Aq œ km
q# #
#
#
#
#
41. `` xf œ 1 y, `` yf œ 1 x, `` xf# œ 0, `` yf# œ 0, ``y`fx œ ``x`fy œ 1
#
#
#
#
42. `` xf œ y cos xy, `` yf œ x cos xy, `` xf# œ y# sin xy, `` yf# œ x# sin xy, ``y`fx œ ``x`fy œ cos xy xy sin xy
#
#
#
#
43. `` gx œ 2xy y cos x, `` yg œ x# sin y sin x, `` xg# œ 2y y sin x, `` yg# œ cos y, ``y`gx œ ``x`gy œ 2x cos x
#
#
#
#
44. `` hx œ ey , `` hy œ xey 1, `` xh# œ 0, `` yh# œ xey , ``y`hx œ ``x`hy œ ey
#
#
#
#
` r
"
` r
` r
"
45. ``xr œ x" y , ``yr œ x" y , ``xr# œ (x"
y)# , ` y# œ (xy)# , ` y` x œ ` x` y œ (xy)#
46. ``xs œ ”
y
"
` ˆy‰
"
`s
"
` ˆy‰
"
x
ˆ y‰
ˆ1‰
# • † `x
x œ x # ” 1 ˆ y ‰# • œ x # y # , ` y œ ” 1 ˆ y ‰ # • † ` y x œ x ” 1 ˆ y ‰ # • œ x # y # ,
1 ˆ xy ‰
x
x
x
#
#
ax# y# b (1) y(2y)
y(2x)
2xy
2xy
x(2y)
` #s
` #s
` #s
` #s
œ axy# yx# b#
` x# œ ax# y# b# œ ax# y# b# , ` y# œ ax# y# b# œ ax# y# b# , ` y` x œ ` x` y œ
ax # y # b #
47. ``wx œ 2x tanaxyb x2 sec2 axyb † y œ 2x tanaxyb x2 y sec2 axyb, ``wy œ x2 sec2 axyb † x œ x3 sec2 axyb,
` #w
2
2
2
` x# œ 2tanaxyb 2x sec axyb † y 2xy sec axyb x y a2secaxybsecaxyb tanaxyb † yb
#
œ 2tanaxyb 4xy sec2 axyb 2x2 y2 sec2 axyb tanaxyb, `` yw# œ x3 a2secaxybsecaxyb tanaxyb † xb œ 2x4 sec2 axyb tanaxyb
` #w
` #w
2
2
3
2
2
3
2
` y` x œ ` x` y œ 3x sec axyb x a2secaxybsecaxyb tanaxyb † yb œ 3x sec axyb x y sec axyb tanaxyb
48. ``wx œ yex y † 2x œ 2xy ex y , ``wy œ a1bex y yex y † a1b œ ex y a1 yb ,
2
2
2
2
2
#
2
2
2
2
` #w
x2 y
2xyŠex y † 2x‹ œ 2yex y a1 2x2 b, `` yw# œ Šex y † a1b‹a1 yb ex y a1b
` x# œ 2y e
#
#
œ ex y ay 2b, ``y`wx œ ``x`wy œ Šex y † 2x‹a1 yb œ 2x ex y a1 yb
2
2
2
49. ``wx œ sinax2 yb x cosax2 yb † 2xy œ sinax2 yb 2x2 ycosax2 yb, ``wy œ x cosax2 yb † x2 œ x3 cosax2 yb,
` #w
2
2
2
2
2
3 2
2
` x# œ cosax yb † 2xy 4xy cosax yb 2x y sinax yb † 2xy œ 6xy cosax yb 4x y sinax yb,
` #w
` #w
` #w
3
2
2
5
2
2
2
3
2
2
2
4
2
` y# œ x sinax yb † x œ x sinax yb, ` y` x œ ` x` y œ 3x cosax yb x sinax yb † 2xy œ 3x cosax yb 2x y sinax yb
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 14.3 Partial Derivatives
50. ``wx œ
2
2
ˆx2 y‰ ax yba2xb
ˆx2 y‰a1b ax yb
y `w
œ xax2 2xy
, `y œ
œ ax2xybx2 ,
y b2
ax2 yb2
ax2 yb2
2
ˆx y‰ a2x 2yb ˆx 2xy y‰2ˆx y‰a2xb
2ˆx 3x y 3 xy y ‰
` #w
œ
,
2
` x# œ
2
ax 2 y b 3
’ax2 yb “
2
2
2
2
3
2
2
2
2
ˆx y‰ † 0 ˆx x‰2ˆx y‰† 1
#
#
ˆx y‰ a2x 1b ˆx 2xy y‰2ˆx y‰† 1
` #w
œ a2xx2 y2xb3 , ``y`wx œ ``x`wy œ
2
2
` y# œ
2
2
’ax2 yb “
’ax2 yb “
2
2
2
2
2
2
2xy y
œ 2x a3x
x2 yb3
3
2
#
#
6
` w
6
51. ``wx œ 2x 2 3y , ``wy œ 2x 3 3y , ``y`wx œ (2x 3y)# , and ` x` y œ (2x 3y)#
#
#
52. ``wx œ ex ln y yx , ``wy œ yx ln x, ``y`wx œ œ "y x" , and ``x`wy œ "y "x
#
#
53. ``wx œ y# 2xy$ 3x# y% , ``wy œ 2xy 3x# y# 4x$ y$ , ``y`wx œ 2y 6xy# 12x# y$ , and ``x`wy œ 2y 6xy# 12x# y$
#
#
54. ``wx œ sin y y cos x y, ``wy œ x cos y sin x x, ``y`wx œ cos y cos x 1, and ``x`wy œ cos y cos x 1
55. (a) x first
(b) y first
56. (a) y first three times
57. fx a1ß 2b œ lim
hÄ0
œ lim
hÄ0
hÄ0
(d) x first
(b) y first three times
(c) y first twice
(e) y first
(f) y first
(d) x first twice
#
#
f(1 hß 2) f(1ß 2)
œ lim c1 (1 h) 2 h6(1 h) d (2 6) œ lim h 6 a1 h2h h b 6
h
hÄ0
hÄ0
13h 6h#
œ lim
h
hÄ0
fy (1ß 2) œ lim
(c) x first
(13 6h) œ 13,
f(1ß 2 h) f(1ß 2)
(2 6)
œ lim c1 1 (2 h) h3(2 h)d (2 6) œ lim (2 6 2h)
h
h
hÄ0
hÄ0
œ lim (2) œ 2
hÄ0
58. fx a2ß 1b œ lim
hÄ0
œ lim
hÄ0
fa2 hß 1b fa2ß 1b
œ lim c4 2a2 hb 3 ha2 hbd a3 2b
h
hÄ0
a2h 1 hb 1
œ lim
h
hÄ0
1 œ 1,
4 4 3a1 hb 2a1 hb# ‘ a3 2b
fa2ß 1 hb fa2ß 1b
œ
lim
h
h
hÄ0
hÄ0
a3 3h 2 4h 2h# b 1
h 2h#
œ lim
œ lim
œ lim a1 2hb œ 1
h
h
hÄ0
hÄ0
hÄ0
fy a2ß 1b œ lim
59. fx a2ß 3b œ lim
hÄ0
È 2 a 2 h b 9 1 È 4 9 1
fa2 hß 3b fa2ß 3b
œ lim
h
h
hÄ0
È2h 4 2
œ lim
h
hÄ0
hÄ0
œ lim
fy a2ß 3b œ lim
hÄ0
œ lim
hÄ0
Š
È2h 4 2 È2h 4 2
2
œ 12 ,
È2h 4 2 ‹ œ hlim
h
Ä 0 È2h 4 2
È 4 3 a3 h b 1 È 4 9 1
fa2ß 3 hb fa2ß 3b
œ lim
h
h
hÄ0
È3h 4 2
œ lim
h
hÄ0
Š
È3h 4 2 È3h 4 2
3
œ 34
È3h 4 2 ‹ œ hlim
h
Ä 0 È2h 4 2
fa0 hß 0b fa0ß 0b
œ lim
h
hÄ0
hÄ0
60. fx a0ß 0b œ lim
fa0ß 0 hb fa0ß 0b
œ lim
h
hÄ0
hÄ0
fy a0ß 0b œ lim
sinŠh3 b 0‹
0
h2 b 0
h
sinŠ0 b h4 ‹
0
0 b h2
h
œ lim
sin h3
h3 œ 1
œ lim
4
sin h4
Šh † sinh4h ‹ œ 0 † 1 œ 0
h3 œ hlim
Ä0
hÄ0
hÄ0
61. (a) In the plane x œ 2 Ê fy axß yb œ 3 Ê fy a2ß 1b œ 3 Ê m œ 3
(b) In the plane y œ 1 Ê fx axß yb œ 2 Ê fy a2ß 1b œ 2 Ê m œ 2
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
813
814
Chapter 14 Partial Derivatives
62. (a) In the plane x œ 1 Ê fy axß yb œ 3y2 Ê fy a1ß 1b œ 3a1b2 œ 3 Ê m œ 3
(b) In the plane y œ 1 Ê fx axß yb œ 2x Ê fy a1ß 1b œ 2a1b œ 2 Ê m œ 2
63. fz ax! ß y! ß z! b œ lim
hÄ0
fz a1ß 2ß 3b œ lim
hÄ0
fax! ß y! ß z! hb fax! , y! ß z! b
;
h
#
#
fa1ß 2ß 3 hb fa1, 2ß 3b
œ lim 2a3 hbh 2a9b œ lim "2h h 2h œ lim a12 2hb œ 12
h
hÄ0
hÄ0
hÄ0
64. fy ax! ß y! ß z! b œ lim
hÄ0
fax! ß y! hß z! b fax! , y! ß z! b
;
h
#
fa1ß hß 3b fa1, 0ß 3b
œ lim a2h h9hb 0 œ lim
h
hÄ0
hÄ0
hÄ0
fy a1ß 0ß 3b œ lim
a2h 9b œ 9
65. y ˆ3z# `` xz ‰ x z$ 2y `` xz œ 0 Ê a3xz# 2yb `` xz œ y z$ Ê at (1ß 1ß 1) we have (3 2) `` xz œ 1 1 or `` xz œ 2
66. ˆ `` xz ‰ z x ˆ yx ‰ `` xz 2x `` xz œ 0 Ê ˆz xy 2x‰ `` xz œ x Ê at (1ß 1ß 3) we have (3 1 2) `` xz œ 1 or `` xz œ "6
`A
a
67. a# œ b# c# 2bc cos A Ê 2a œ a2bc sin Ab ``Aa Ê ``Aa œ bc sin
A ; also 0 œ 2b 2c cos A a2bc sin Ab ` b
cos A b
Ê 2c cos A 2b œ a2bc sin Ab ``Ab Ê ``Ab œ c bc
sin A
(sin A) ` a a cos A
cos A
`A
68. sina A œ sinb B Ê
œ 0 Ê asin Ab `` xa a cos A œ 0 Ê ``Aa œ asin
sin# A
A ; also
"
`
a
`
a
ˆ sin A ‰ ` B œ ba csc B cot Bb Ê ` B œ b csc B cot B sin A
69. Differentiating each equation implicitly gives 1 œ vx ln u ˆ vu ‰ ux and 0 œ ux ln v ˆ uv ‰ vx or
"
v
º 0 lnu v º
aln ub vx ˆ vu ‰ ux œ 1
Ê
œ
œ aln ubalnlnvvb 1
v
v
x
ln u
ˆ uv ‰ vx aln vb ux œ 0 Ÿ
u
º u
º
ln v
v
70. Differentiating each equation implicitly gives 1 œ a2xbxu a2ybyu and 0 œ a2xbxu yu or
"
2y
1 º
œ 2x1 4xy œ 2x 1 4xy
2y
º
2x 1
º0
a2xbxu a2ybyu œ 1
Ê xu œ 2x
a2xbxu yu œ 0 º
"
0º
yu œ #x 4xy œ 2x2x4xy œ 2x 2x4xy œ 1 1 2y ; next s œ x# y#
2x
º 2x
and
Ê ``us œ 2x `` ux 2y `` uy
2y
œ 2x Š 2x " 4xy ‹ 2y Š 1 " 2y ‹ œ 1 " #y 1 2y2y œ 11 2y
71. fx axß yb œ œ
0
0
if y 0
Ê fx axß yb œ 0 for all points ax, yb; at y œ 0, fy axß 0b œ lim fax, 0 hhb fax, 0b œ lim fax, hhb 0
if y 0
h Ä0
h Ä0
œ lim fax,h hb œ 0 because
h Ä0
lim fax,h hb œ
hÄ0c
3
lim h œ 0 and
h Ä0 c h
limb fax,h hb œ
h Ä0
2
limb hh œ 0 Ê fy axß yb œ œ
h Ä0
3y2
2y
if y 0
;
if y 0
fyx axß yb œ fxy axß yb œ 0 for all points ax, yb
72. At x œ 0, fx a0ß yb œ lim fa0 h, yhb fa0, yb œ lim fah, yhb 0 œ lim fah,h yb which does not exist because
h Ä0
œ
2
limc hh œ 0 and
hÄ0
fy axß yb œ œ
limb fah,h yb œ
h Ä0
h Ä0
È
limb hh œ
h Ä0
hÄ0
1
lim 1 œ _ Ê fx axß yb œ 2Èx
h Ä0 b È h
2x
if x 0
if x 0
lim fah,h yb
hÄ0c
;
0 if x 0
Ê fy axß yb œ 0 for all points ax, yb; fyx axß yb œ 0 for all points ax, yb, while fxy axß yb œ 0 for all
0 if x 0
points ax, yb such that x Á 0.
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 14.3 Partial Derivatives
#
#
#
#
#
815
#
73. `` xf œ 2x, `` yf œ 2y, `` zf œ 4z Ê `` xf# œ 2, `` yf# œ 2, `` zf# œ 4 Ê `` xf# `` yf# `` zf# œ 2 2 (4) œ 0
#
#
#
#
#
#
74. `` xf œ 6xz, `` yf œ 6yz, `` zf œ 6z# 3 ax# y# b , `` xf# œ 6z, `` yf# œ 6z, `` zf# œ 12z Ê `` xf# `` yf# `` zf#
œ 6z 6z 12z œ 0
75. `` xf œ 2ec2y sin 2x, `` yf œ 2ec2y cos 2x, `` xf# œ 4ec2y cos 2x, `` yf# œ 4ec2y cos 2x Ê `` xf# `` yf#
#
#
#
#
œ 4ec2y cos 2x 4ec2y cos 2x œ 0
#
#
#
#
#
#
#
#
76. `` xf œ x# x y# , `` yf œ x# y y# , `` xf# œ axy# yx# b# , `` yf# œ axx# yy# b# Ê `` xf# `` yf# œ axy# yx# b# axx# yy# b# œ 0
#
#
#
#
#
#
#
#
77. `` xf œ 3 , `` yf œ 2 , `` xf# œ 0 , `` yf# œ 0 Ê `` xf# `` yf# œ 0 0 œ 0
78. `` xf œ
1 Îy
#
1 Š xy ‹
œ y# y x# , `` yf œ
x Îy 2
#
#
#
#
œ y#xx# , `` xf# œ ay ayx# b†0x#b2y†2x œ ay#2xy
, ` f œ ay axy#b†0 x#ab2 xb†2y œ ay# 2xy
x# b2 ` y#
x # b2
#
#
1 Š xy ‹
#
Ê `` xf# `` yf# œ ay#2xy
ay# 2xy
œ0
x# b2
x# b2
#
#
79. `` xf œ "# ax# y# z# b
$Î#
$Î# ` f
$Î#
, ` y œ "# ax# y# z# b
a2yb
$Î#
$Î#
#
#
#
#
a2xb œ x ax# y# z# b
$Î# ` f
, ` z œ "# ax# y# z b
a2zb œ z ax y z b
;
#
&Î#
` f
#
#
# $Î#
#
#
#
# &Î# ` # f
#
#
# $Î#
3x ax y z b
, ` y # œ ax y z b
3y# ax# y# z# b
,
` x # œ ax y z b
#
#
#
#
$Î#
&Î#
` f
#
#
#
3z# ax# y# z# b
Ê `` xf# `` yf# `` zf#
` z # œ ax y z b
œ y ax# y# z# b
$Î#
3x# ax# y# z# b
$Î#
3z# ax# y# z# b
œ ’ ax# y# z# b
’ ax# y# z# b
&Î#
&Î#
$Î#
3y# ax# y# z# b
$Î#
a3x# 3y# 3z# b ax# y# z# b
“ ’ ax# y# z# b
“ œ 3 ax# y# z# b
&Î#
#
“
&Î#
œ0
#
80. `` xf œ 3e3x4y cos 5z, `` yf œ 4e3x4y cos 5z, `` zf œ 5e3x4y sin 5z; `` xf# œ 9e3x4y cos 5z, `` yf# œ 16e3x4y cos 5z,
` #f
3x4y
cos 5z
` z# œ 25e
#
#
#
Ê `` xf# `` yf# `` zf# œ 9e3x4y cos 5z 16e3x4y cos 5z 25e3x4y cos 5z œ 0
#
#
#
#
81. ``wx œ cos (x ct), ``wt œ c cos (x ct); `` xw# œ sin (x ct), `` tw# œ c# sin (x ct) Ê `` tw# œ c# [ sin (x ct)] œ c# `` xw#
#
#
82. ``wx œ 2 sin (2x 2ct), ``wt œ 2c sin (2x 2ct); `` xw# œ 4 cos (2x 2ct), `` tw# œ 4c# cos (2x 2ct)
#
#
Ê `` tw# œ c# [4 cos (2x 2ct)] œ c# `` xw#
83. ``wx œ cos (x ct) 2 sin (2x 2ct), ``wt œ c cos (x ct) 2c sin (2x 2ct);
` #w
` #w
#
#
` x# œ sin (x ct) 4 cos (2x 2ct), ` t# œ c sin (x ct) 4c cos (2x 2ct)
#
#
Ê `` tw# œ c# [ sin (x ct) 4 cos (2x 2ct)] œ c# `` xw#
#
#
#
#
#
1
` w
c
` w
"
#
# ` w
84. ``wx œ x " ct , ``wt œ x c ct ; `` xw# œ (x ct)# , ` t# œ (x ct)# Ê ` t# œ c ’ (x ct)# “ œ c ` x#
#
85. ``wx œ 2 sec# (2x 2ct), ``wt œ 2c sec# (2x 2ct); `` xw# œ 8 sec# (2x 2ct) tan (2x 2ct),
` #w
#
#
` t# œ 8c sec (2x 2ct) tan (2x 2ct)
#
#
Ê ux `` tw# œ c# [8 sec# (2x 2ct) tan (2x 2ct)] œ c# `` xw#
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
816
Chapter 14 Partial Derivatives
86. ``wx œ 15 sin (3x 3ct) exbct , ``wt œ 15c sin (3x 3ct) cexbct ; `` xw# œ 45 cos (3x 3ct) exbct ,
#
` #w
#
# xbct
` t# œ 45c cos (3x 3ct) c e
Ê `` tw# œ c# c45 cos (3x 3ct) exbct d œ c# `` xw#
#
#
#
#
#
#
#
87. ``wt œ `` uf ``ut œ `` uf (ac) Ê `` tw# œ (ac) Š `` uf# ‹ (ac) œ a# c# `` uf# ; ``wx œ `` uf `` ux œ `` uf † a Ê `` xw# œ Ša `` uf# ‹ † a
#
#
#
#
#
œ a# `` uf# Ê `` tw# œ a# c# `` uf# œ c# Ša# `` uf# ‹ œ c# `` xw#
88. If the first partial derivatives are continuous throughout an open region R, then by Theorem 3 in this section of the text,
f(xß y) œ f(x! ß y! ) fx (x! ß y! ) ?x fy (x! ß y! ) ?y %" ?x %# ?y, where %" , %# Ä 0 as ?x, ?y Ä 0. Then as
(xß y) Ä (x! ß y! ), ?x Ä 0 and ?y Ä 0 Ê
lim
f(xß y) œ f(x! ß y! ) Ê f is continuous at every point (x! ß y! ) in R.
Ðxß yÑ Ä Ðx! ß y! Ñ
89. Yes, since fxx , fyy , fxy , and fyx are all continuous on R, use the same reasoning as in Exercise 76 with
fx (xß y) œ fx (x! ß y! ) fxx (x! ß y! ) ?x fxy (x! ß y! ) ?y %" ?x %# ?y and
fy (xß y) œ fy (x! ß y! ) fyx (x! ß y! ) ?x fyy (x! ß y! ) ?y s%" ?x s%# ?y. Then
lim
fx (xß y) œ fx (x! ß y! )
Ðxß yÑ Ä Ðx! ß y! Ñ
and
lim
Ðxß yÑ Ä Ðx! ß y! Ñ
fy (xß y) œ fy (x! ß y! ).
90. To find ! and " so that ut œ uxx Ê ut œ " sina! xbe" t and ux œ ! cosa! xbe" t Ê uxx œ !2 sina! xbe" t ; then
ut œ uxx Ê " sina! xbe" t œ !2 sina! xbe" t , thus ut œ uxx only if " œ !2
h†02
91. fx a0, 0b œ lim fa0 hß 0hb fa0ß 0b œ lim h2 0h4
h Ä0
0
0†h2
œ lim 0h œ 0; fy a0, 0b œ lim fa0ß 0 hhb fa0ß 0b œ lim 02 hh4
hÄ0
hÄ0
hÄ0
ˆky2 ‰y2
ky4
k
lim
f ax, yb œ lim aky2 b2 y4 œ lim k2 y4 y4 œ lim k2 1 œ k2 k 1 Ê
yÄ0
yÄ0
yÄ0
Ðxß yÑ Ä Ð0ß 0Ñ
hÄ0
0
œ lim 0h œ 0;
hÄ0
different limits for different
along x œ ky2
values of k Ê
lim
Ðxß yÑ Ä Ð0ß 0Ñ
f ax, yb does not exist Ê f ax, yb is not continuous at a0, 0b Ê by Theorem 4, f ax, yb is not
differentiable at a0, 0b.
92. fx a0, 0b œ lim fa0 hß 0hb fa0ß 0b œ lim fahß 0hb 1 œ lim 1 h 1 œ 0; fy a0, 0b œ lim fa0ß 0 hhb fa0ß 0b œ lim fa0ß hhb 1 œ lim 1 h 1 œ 0;
h Ä0
lim
Ðxß yÑ Ä Ð0ß 0Ñ
along y œ x2
h Ä0
f ax, yb œ lim 0 œ 0 but
yÄ0
h Ä0
lim
Ðxß yÑ Ä Ð0ß 0Ñ
along y œ 1.5x2
h Ä0
f ax, yb œ lim 1 œ 1 Ê
yÄ0
h Ä0
lim
Ðxß yÑ Ä Ð0ß 0Ñ
h Ä0
f ax, yb does not exist
Ê f ax, yb is not continuous at a0, 0b Ê by Theorem 4, f ax, yb is not differentiable at a0, 0b.
14.4 THE CHAIN RULE
1. (a)
(b)
2. (a)
dy
`w
`w
dx
dw
` x œ 2x, ` y œ 2y, dt œ sin t, dt œ cos t Ê dt œ 2x sin t 2y cos t œ 2 cos t sin t 2 sin t cos t
œ 0; w œ x# y# œ cos# t sin# t œ 1 Ê dw
dt œ 0
dw
dt (1) œ 0
dy
`w
`w
dx
` x œ 2x, ` y œ 2y, dt œ sin t cos t, dt œ sin t cos t
Ê dw
dt
œ (2x)( sin t cos t) (2y)( sin t cos t)
œ 2(cos t sin t)(cos t sin t) 2(cos t sin t)(sin t cos t) œ a2 cos# t 2 sin# tb a2 cos# t 2 sin# tb
œ 0; w œ x# y# œ (cos t sin t)# (cos t sin t)# œ 2 cos# t 2 sin# t œ 2 Ê dw
dt œ 0
(b)
dw
dt (0) œ 0
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 14.4 The Chain Rule
3. (a)
(x y) dx
`w
" `w
" `w
dz
"
, dt œ 2 cos t sin t, dy
`x œ z , `y œ z , `z œ
z#
dt œ 2 sin t cos t, dt œ t#
x y
y
2
2
cos# t sin# t
x
cos# t
sin# t
Ê dw
dt œ z cos t sin t z sin t cos t z# t# œ Š " ‹ at# b œ 1; w œ z z œ Š " ‹ Š " ‹ œ t
t#
(b)
4. (a)
(b)
5. (a)
t
t
Ê dw
dt œ 1
dw
dt (3) œ 1
2y
dy
`w
2x
`w
`w
2z
dx
dz
"Î#
` x œ x# y# z# , ` y œ x# y# z# , ` z œ x# y# z# , dt œ sin t, dt œ cos t, dt œ 2t
2 cos t sin t 2 sin t cos t 4 ˆ4t"Î# ‰ t "Î#
2y cos t
2x sin t
4zt "Î#
Ê dw
dt œ x# y# z# x# y# z# x# y# z# œ
cos# t sin# t 16t
16
16
#
#
#
#
#
œ 116t ; w œ ln ax y z b œ ln acos t sin t 16tb œ ln (1 16t) Ê dw
dt œ 1 16t
dw
16
dt (3) œ 49
dy
`w
" dx
2t
"
dz
x `w
x `w
t
` x œ 2ye , ` y œ 2e , ` z œ z , dt œ t# 1 , dt œ t# 1 , dt œ e
"
#
#
x
x
t
4yte
2e
e
Ê dw
dt œ t# 1 t# 1 z
œ (4t) atant# tb1at 1b 2 at#t 11b eet œ 4t tan" t 1; w œ 2yex ln z œ a2 tan" tb at# 1b t
"
ˆ 2 ‰ #
Ê dw
tb (2t) 1 œ 4t tan" t 1
dt œ t# 1 at 1b a2 tan
(b)
6. (a)
(b)
7. (a)
t
dw
ˆ1‰
dt (1) œ (4)(1) 4 1 œ 1 1
dy
x cos xy
`w
`w
`w
dx
" dz
tc1
Ê dw
etc1
` x œ y cos xy, ` y œ x cos xy, ` z œ 1, dt œ 1, dt œ t , dt œ e
dt œ y cos xy t
œ (ln t)[cos (t ln t)] t cos (tt ln t) etc1 œ (ln t)[cos (t ln t)] cos (t ln t) etc1 ; w œ z sin xy
tc1
œ etc1 sin (t ln t) Ê dw
[cos (t ln t)] ln t t ˆ "t ‰‘ œ etc1 (1 ln t) cos (t ln t)
dt œ e
dw
dt (1) œ 1 (1 0)(1) œ 0
x
4ex ln y
`z
`z `x
`z `y
4ex
x
ˆ cos v ‰
4e ysin v
` u œ ` x ` u ` y ` u œ a4e ln yb u cos v Š y ‹ (sin v) œ
u
v)(sin v)
œ 4(u cos v)uln (u sin v) 4(u cos
œ (4 cos v) ln (u sin v) 4 cos v;
u sin v
`z
`z `x
`z `y
4ex
4ex u cos v
x
x
ˆ u sin v ‰
` v œ ` x ` v ` y ` v œ a4e ln yb u cos v Š y ‹ (u cos v) œ a4e ln yb (tan v) y
#
v)(u cos v)
cos v
œ [4(u cos v) ln (u sin v)](tan v) 4(u cosu sin
œ (4u sin v) ln (u sin v) 4usin
v
v ;
`z
sin
x
z œ 4e ln y œ 4(u cos v) ln (u sin v) Ê ` u œ (4 cos v) ln (u sin v) 4(u cos v) ˆ u sinvv ‰
v‰
œ (4 cos v) ln (u sin v) 4 cos v; also `` vz œ (4u sin v) ln (u sin v) 4(u cos v) ˆ uu cos
sin v
#
cos v
œ (4u sin v) ln (u sin v) 4usin
v
(b) At ˆ2ß 14 ‰ : `` uz œ 4 cos 14 ln ˆ2 sin 14 ‰ 4 cos 14 œ 2È2 ln È2 2È2 œ È2 (ln 2 2);
(4)(2) ˆcos# 14 ‰
`z
1
1‰
ˆ
œ 4È2 ln È2 4È2 œ 2È2 ln 2 4È2
ˆsin 1 ‰
` v œ (4)(2) sin 4 ln 2 sin 4 4
8. (a)
`z
`u œ –
`z
`v œ –
Š "y ‹
#
Š xy ‹
Š #x ‹
— cos v – Š x ‹# 1 — sin v œ x# y# x# y# œ
1
y cos v
y
x sin v
(u sin v)(cos v) (u cos v)(sin v)
œ 0;
u#
y
Š"‹
Š #x ‹
— (u sin v) – Š x ‹# 1 — u cos v œ x# y# x# y# œ
Šx‹ 1
y
y
#
y
yu sin v
xu cos v
(u sin v)(u sin v) (u cos v)(u cos v)
u#
y
"
#
‰
œ sin# v cos# v œ 1; z œ tan" Š xy ‹ œ tan" (cot v) Ê `` uz œ 0 and `` vz œ ˆ 1 cot
# v a csc vb
œ sin# v"
cos# v œ 1
(b) At ˆ1.3ß 16 ‰ : `` uz œ 0 and `` vz œ 1
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
817
818
9. (a)
Chapter 14 Partial Derivatives
`w
`w `x
`w `y
`w `z
` u œ ` x ` u ` y ` u ` z ` u œ (y z)(1) (x z)(1) (y x)(v) œ x y 2z v(y x)
œ (u v) (u v) 2uv v(2u) œ 2u 4uv; ``wv œ ``wx `` xv ``wy `` vy ``wz `` vz
#
œ (y z)(1) (x z)(1) (y x)(u) œ y x (y x)u œ 2v (2u)u œ 2v 2u ;
w œ xy yz xz œ au# v# b au# v uv# b au# v uv# b œ u# v# 2u# v Ê ``wu œ 2u 4uv and
(b)
10. (a)
`w
#
` v œ 2v 2u
#
At ˆ "# ß 1‰ : ``wu œ 2 ˆ "# ‰ 4 ˆ "# ‰ (1) œ 3 and ``wv œ 2(1) 2 ˆ "# ‰ œ #3
2y
`w
2x
2z
v
v
v
v
v
` u œ Š x# y# z# ‹ ae sin u ue cos ub Š x# y# z# ‹ ae cos u ue sin ub Š x# y# z# ‹ ae b
v
u
‰ aev sin u uev cos ub
œ ˆ u# e2v sin# u 2ueu# esin
2v cos# u u# e2v
v
cos u
v
‰ v
ˆ u# e2v sin# u 2ue
u# e2v cos# u u# e2v ae cos u ue sin ub
v
‰ aev b œ 2u ;
ˆ u# e2v sin# u u2ue
# e2v cos# u u# e2v
2y
`w
2x
2z
v
v
v
` v œ Š x# y# z# ‹ aue sin ub Š x# y# z# ‹ aue cos ub Š x# y# z# ‹ aue b
v
u
‰ auev sin ub
œ ˆ u# e2v sin# u 2ueu# esin
2v cos# u u# e2v
v
cos u
‰ v
ˆ u# e2v sin# u 2ue
u# e2v cos# u u# e2v aue cos ub
‰ auev b œ 2; w œ ln au# e2v sin# u u# e2v cos# u u# e2v b œ ln a2u# e2v b
ˆ u# e2v sin# u u2ue
# e2v cos# u u# e2v
v
œ ln 2 2 ln u 2v Ê ``wu œ 2u and ``wv œ 2
2
(b) At a2ß 0b: ``wu œ #
œ 1 and ``wv œ 2
11. (a)
rp
pq
qrrppq
`u
`u `p
`u `q
`u `r
"
œ 0;
` x œ ` p ` x ` q ` x ` r ` x œ q r (q r)# (q r)# œ
(q r)#
rp
pq
qrrppq
2p 2r
`u
`u `p
`u `q
`u `r
"
œ (q
` y œ ` p ` y ` q ` y ` r ` y œ q r (q r)# (q r)# œ
(q r)#
r)#
(2x 2y 2z) (2x 2y 2z)
z
`u
`u `p
`u `q
`u `r
œ
œ (z y)# ; ` z œ ` p ` z ` q ` z ` r ` z
(2z 2y)#
rp
pq
ppq
2p
4y
y
"
œ q r (q r)# (q r)# œ q r (qr œ 2q
r)#
(q r)# œ (2z 2y)# œ (z y)# ;
y
(z y) y(1)
y(1)
`u
`u
u œ pq qr œ 2z 2y
œ (z z y)# , and `` uz œ (z (zy)(0)
2y œ z y Ê ` x œ 0, ` y œ
(z y)#
y)#
œ (zyy)#
2
(b) At ŠÈ3ß 2ß 1‹ : `` ux œ 0, `` uy œ (1 " 2)# œ 1, and `` uz œ (1 2)# œ 2
12. (a)
z ln y
`u
eqr
eqr cos x
qr
"
pb (0) aqeqr sin" pb (0) œ È
œ Èe cos#x œ yz if 1# x 1# ;
` x œ È1 p# (cos x) are sin
1 p#
1 sin x
# "
#
"
qr
#
z ˆz‰y x
`u
eqr
qr
"
pb Š zy ‹ aqeqr sin" pb (0) œ z re ysin p œ
œ xzyz1 ;
` y œ È1 p# (0) are sin
y
z
qr
"p
`u
eqr
qr
"
pb (2z ln y) aqeqr sin" pb ˆ z"# ‰ œ a2zreqr sin" pb (ln y) qe sin
` z œ È1 p# (0) are sin
z#
#
z
œ (2z) ˆ "z ‰ ayz x ln yb az ln yz#b ay b x œ xyz ln y; u œ ez ln y sin" (sin x) œ xyz if 1# Ÿ x Ÿ 1# Ê `` ux œ yz ,
`u
z1
, and `` uz œ
` y œ xzy
œ xyz ln y from direct calculations
(b) At ˆ 14 ß "# ß "# ‰ : `` xu œ ˆ "# ‰
"Î#
œ È2, `` yu œ ˆ 14 ‰ ˆ "# ‰ ˆ "# ‰
Ð"Î#Ñ"
È
œ 1 4 2 , `` uz œ ˆ 14 ‰ ˆ "# ‰
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
"Î#
ln ˆ "# ‰ œ 1
È2 ln 2
4
Section 14.4 The Chain Rule
` z dx
` z dy
13. dz
dt œ ` x dt ` y dt
` z du
` z dv
` x dw
14. dz
dt œ ` u dt ` v dt ` w dt
15. ``wu œ ``wx `` xu ``wy `` yu ``wz `` uz
`w
`w `x
`w `y
`w `z
`v œ `x `v `y `v `z `v
16. ``wx œ ``wr ``xr ``ws ``xs ``wt ``xt
`w
`w `r
`w `s
`w `t
`y œ `r `y `s `y `t `y
17. ``wu œ ``wx `` xu ``wy `` yu
`w
`w `x
`w `y
`v œ `x `v `y `v
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
819
820
Chapter 14 Partial Derivatives
18. ``wx œ ``wu `` ux ``wv `` vx
`w
`w `u
`w `v
`y œ `u `y `v `y
19. ``zt œ `` xz ``xt `` yz ``yt
`z
`z `x
`z `y
`s œ `x `s `y `s
dy ` u
20. ``yr œ du
`r
`u
21. ``ws œ dw
du ` s
`w
dw ` u
` t œ du ` t
22. ``wp œ ``wx `` px ``wy `` py ``wz `` pz ``wv `` vp
dy
` w dy
` w dx
23. ``wr œ ``wx dx
dr ` y dr œ ` x dr since dr œ 0
`w
` w dx
` w dy
` w dy
dx
` s œ ` x ds ` y ds œ ` y ds since ds œ 0
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 14.4 The Chain Rule
821
25. Let F(xß y) œ x$ 2y# xy œ 0 Ê Jx (xß y) œ 3x# y
24. ``ws œ ``wx ``xs ``wy ``ys
#
3x y
Fx
and Fy (xß y) œ 4y x Ê dy
dx œ Fy œ (4y x)
4
Ê dy
dx (1ß 1) œ 3
y3
Fx
26. Let F(xß y) œ xy y# 3x 3 œ 0 Ê Fx (xß y) œ y 3 and Fy (xß y) œ x 2y Ê dy
dx œ Fy œ x 2y
Ê dy
dx (1ß 1) œ 2
2x y
Fx
27. Let F(xß y) œ x# xy y# 7 œ 0 Ê Fx (xß y) œ 2x y and Fy (xß y) œ x 2y Ê dy
dx œ Fy œ x 2y
4
Ê dy
dx (1ß 2) œ 5
28. Let F(xß y) œ xey sin xy y ln 2 œ 0 Ê Fx (xß y) œ ey y cos xy and Fy (xß y) œ xey x sin xy 1
e y cos xy
dy
Fx
Ê dy
dx œ Fy œ xey x sin xy 1 Ê dx (!ß ln 2) œ (2 ln 2)
y
29. Let F(xß yß z) œ z$ xy yz y$ 2 œ 0 Ê Fx (xß yß z) œ y, Fy (xß yß z) œ x z 3y# , Fz (xß yß z) œ 3z# y
#
Ê `` xz œ FFxz œ 3z# y y œ 3z#y y Ê `` xz (1ß 1ß 1) œ "4 ; `` yz œ Fyz œ x3z#zy3y œ x 3zz# 3y
y
F
#
Ê `` yz (1ß 1ß 1) œ 34
30. Let F(xß yß z) œ "x y" "z 1 œ 0 Ê Fx (xß yß z) œ x"# , Fy (xß yß z) œ y"# , Fz (xß yß z) œ z"#
Ê `` xz œ FFxz œ Š x"# ‹
Š z"# ‹
#
œ xz# Ê `` xz (2ß 3ß 6) œ 9; `` yz œ Fyz œ F
Š y"# ‹
Š z"# ‹
#
œ yz# Ê `` yz (2ß 3ß 6) œ 4
31. Let F(xß yß z) œ sin (x y) sin (y z) sin (x z) œ 0 Ê Fx (xß yß z) œ cos (x y) cos (x z),
Fy (xß yß z) œ cos (x y) cos (y z), Fz (xß yß z) œ cos (y z) cos (x z) Ê `` xz œ FFxz
(x y) cos (x z)
cos (x y) cos (y z)
y
`z
`z
`z
œ cos
cos (y z) cos (x z) Ê ` x (1ß 1ß 1) œ 1; ` y œ Fz œ cos (y z) cos (x z) Ê ` y (1ß 1ß 1) œ 1
F
32. Let F(xß yß z) œ xey yez 2 ln x 2 3 ln 2 œ 0 Ê Fx (xß yß z) œ ey 2x , Fy (xß yß z) œ xey ez , Fz (xß yß z) œ yez
Ê `` xz œ FFxz œ ˆey 2x ‰
yez
Ê `` xz (1ß ln 2ß ln 3) œ 3 ln4 2 ; `` yz œ Fyz œ xeyez e Ê `` yz (1ß ln 2ß ln 3) œ 3 ln5 2
F
y
z
33. ``wr œ ``wx ``xr ``wy ``yr ``wz ``zr œ 2(x y z)(1) 2(x y z)[ sin (r s)] 2(x y z)[cos (r s)]
œ 2(x y z)[1 sin (r s) cos (r s)] œ 2[r s cos (r s) sin (r s)][1 sin (r s) cos (r s)]
Ê ``wr ¸ rœ1ßsœ1 œ 2(3)(2) œ 12
`w ¸
‰
ˆ"‰
ˆ 2v ‰ v
ˆ4‰ ˆ4‰
34. ``wv œ ``wx `` xv ``wy `` vy ``wz `` vz œ y ˆ 2v
u x(1) z (0) œ (u v) u u Ê ` v uœ1ßvœ2 œ (1) 1 1 œ 8
#
"
35. ``wv œ ``wx `` xv ``wy `` yv œ ˆ2x xy# ‰ (2) ˆ "x ‰ (1) œ ’2(u 2v 1) (u2u2vv1)2 # “ (2) u 2v
1
Ê ``wv ¸ uœ0ßvœ0 œ 7
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
822
Chapter 14 Partial Derivatives
36. `` uz œ `` xz `` xu `` yz `` uy œ (y cos xy sin y)(2u) (x cos xy x cos y)(v)
œ cuv cos au$ v uv$ b sin uvd (2u) cau# v# b cos au$ v uv$ b au# v# b cos uvd (v)
Ê `` uz ¸ uœ0ßvœ1 œ 0 (cos 0 cos 0)(1) œ 2
dz ` x
5
`z ¸
5
u
ˆ 5 ‰ u
37. `` uz œ dx
` u œ 1 x# e œ ’ 1 aeu ln vb# “ e Ê ` u uœln 2ßvœ1 œ ’ 1 (2)# “ (2) œ 2;
`z
dz ` x
5
ˆ 5 ‰ ˆ "v ‰ œ ’
“ ˆ "v ‰
` v œ dx ` v œ 1 x#
1 aeu ln vb#
È
Ê `` vz ¸ uœln 2ßvœ1 œ ’ 1 5(2)# “ (1) œ 1
È
v3
v3
dz ` q
2
"
"
"
`z ¸
"
38. `` uz œ dq
` u œ Š q ‹ Š 1 u# ‹ œ Š Èv 3 tan " u ‹ Š 1 u# ‹ œ atan " ub a1 u# b Ê ` u uœ1ßvœ2 œ atan " 1b a1 1# b œ 1 ;
dz ` q
tan " u
tan " u
`z
"
"
"
` v œ dq ` v œ Š q ‹ Š 2Èv 3 ‹ œ Š Èv 3 tan " u ‹ Š 2Èv 3 ‹ œ #(v 3)
Ê `` vz ¸ uœ1ßvœ2 œ "#
`x
`x
2
2 s t ` w
s t
w
w
39. Let x œ s3 t2 Ê w œ fas3 t2 b œ faxb Ê ``ws œ dw
, ` t œ dw
dx ` s œ f axb † 3s œ 3s e
dx ` t œ f axb † 2t œ 2t e
3
2
3
40. Let x œ t s2 and y œ st Ê w œ fˆt s2 ß st ‰ œ faxß yb Ê ``ws œ ``wx ``xs ``wy ``ys œ fx ax, yb † 2t s fy ax, yb † 1t
œ at s2 bˆ st ‰ † 2t s ˆts2 ‰2 1
s4 t
5s4 t ` w
`w `x
`w `y
s
4
2
2 † t œ 2s t 2 œ 2 ; ` t œ ` x ` t ` y ` t œ fx ax, yb † s fy ax, yb † t2
ˆ 2 ‰2
ts
œ at s2 bˆ st ‰ † s2 2
5
5
† ˆ ts2 ‰ œ s5 s2 œ s2
` V dI
` V dR
dI
dR
41. V œ IR Ê ``VI œ R and `` VR œ I; dV
dt œ ` I dt ` R dt œ R dt I dt Ê 0.01 volts/sec
dI
œ (600 ohms) dI
dt (0.04 amps)(0.5 ohms/sec) Ê dt œ 0.00005 amps/sec
` V da
` V db
` V dc
da
db
dc
42. V œ abc Ê dV
dt œ ` a dt ` b dt ` c dt œ (bc) dt (ac) dt (ab) dt
¸
Ê dV
œ (2 m)(3 m)(1 m/sec) (1 m)(3 m)(1 m/sec) (1 m)(2 m)(3 m/sec) œ 3 m$ /sec
dt
aœ1ßbœ2ßcœ3
` S da
` S db
` S dc
and the volume is increasing; S œ 2ab 2ac 2bc Ê dS
dt œ ` a dt ` b dt ` c dt
db
dc
dS ¸
œ 2(b c) da
dt 2(a c) dt 2(a b) dt Ê dt
aœ1ßbœ2ßcœ3
œ 2(5 m)(1 m/sec) 2(4 m)(1 m/sec) 2(3 m)(3 m/sec) œ 0 m# /sec and the surface area is not changing;
` D da
` D db
` D dc
"
db
dc ‰
ˆa da
¸
D œ Èa# b# c# Ê dD
Ê dD
#
#
dt œ ` a dt ` b dt ` c dt œ È #
dt b dt c dt
dt
aœ1ßbœ2ßcœ3
a b c
"
œ Š È14
‹ [(1 m)(1 m/sec) (2 m)(1 m/sec) (3 m)(3 m/sec)] œ È614 m/sec 0
m
Ê the diagonals are
decreasing in length
43. `` xf œ `` uf `` xu `` vf `` xv ``wf ``wx œ `` uf (1) `` vf (0) ``wf (1) œ `` uf ``wf ,
`f
`f `u
`f `v
`f `w
`f
`f
`f
`f
`f
` y œ ` u ` y ` v ` y ` w ` y œ ` u (1) ` v (1) ` w (0) œ ` u ` v , and
`f
`f `u
`f `v
`f `w
`f
`f
`f
`f
`f
`f
`f
`f
` z œ ` u ` z ` v ` z ` w ` z œ ` u (0) ` v (1) ` w (1) œ ` v ` w Ê ` x ` y ` z œ 0
44. (a)
(b)
(c)
`y
`w
`x
`w
" `w
` r œ fx ` r fy ` r œ fx cos ) fy sin ) and ` ) œ fx (r sin )) fy (r cos )) Ê r ` ) œ fx sin ) fy cos )
`w
#
ˆ cosr ) ‰ ``w) œ fx sin ) cos ) fy cos# )
` r sin ) œ fx sin ) cos ) fy sin ) and
Ê fy œ (sin )) ``wr ˆ cosr ) ‰ ``w) ; then ``wr œ fx cos ) (sin )) ``wr ˆ cosr ) ‰ ``w) ‘ (sin )) Ê fx cos )
œ ``wr asin# )b ``wr ˆ sin ) rcos ) ‰ ``w) œ a1 sin# )b ``wr ˆ sin ) rcos ) ‰ ``w) Ê fx œ (cos )) ``wr ˆ sinr ) ‰ ``w)
#
#
#
afx b# œ acos# )b ˆ ``wr ‰ ˆ 2 sin )r cos ) ‰ ˆ ``wr ``w) ‰ Š sinr# ) ‹ ˆ ``w) ‰ and
#
#
#
afy b# œ asin# )b ˆ ``wr ‰ ˆ 2 sin )r cos ) ‰ ˆ ``wr ``w) ‰ Š cosr# ) ‹ ˆ ``w) ‰ Ê afx b# afy b# œ ˆ ``wr ‰ r"# ˆ ``w) ‰
#
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
#
2
Section 14.4 The Chain Rule
823
45. wx œ ``wx œ ``wu `` xu ``wv `` xv œ x ``wu y ``wv Ê wxx œ ``wu x ``x ˆ ``wu ‰ y ``x ˆ ``wv ‰
#
#
#
#
#
#
#
#
œ ``wu x Š `` uw# `` xu ``v`wu `` xv ‹ y Š ``u`wv `` xu `` vw# `` xv ‹ œ ``wu x Šx `` uw# y ``v`wu ‹ y Šx ``u`wv y `` vw# ‹
#
#
#
œ ``wu x# `` uw# 2xy ``v`wu y# `` vw# ; wy œ ``wy œ ``wu `` uy ``wv `` vy œ y ``wu x ``wv
#
#
#
#
Ê wyy œ ``wu y Š `` uw# `` uy ``v`wu `` yv ‹ x Š ``u`wv `` yu `` vw# `` yv ‹
#
#
#
#
#
#
#
œ ``wu y Šy `` uw# x ``v`wu ‹ x Šy ``u`wv x `` vw# ‹ œ ``wu y# `` uw# 2xy ``v`wu x# `` vw# ; thus
#
#
wxx wyy œ ax# y# b `` uw# ax# y# b `` vw# œ ax# y# b (wuu wvv ) œ 0, since wuu wvv œ 0
46. ``wx œ f w (u)(1) gw (v)(1) œ f w (u) gw (v) Ê wxx œ f ww (u)(1) gww (v)(1) œ f ww (u) gww (v);
`w
w
w
` y œ f (u)(i) g (v)(i)
Ê wyy œ f ww (u) ai# b gww (v) ai# b œ f ww (u) gww (v) Ê wxx wyy œ 0
` f dx
` f dy
` f dz
47. fx (xß yß z) œ cos t, fy (xß yß z) œ sin t, and fz (xß yß z) œ t# t 2 Ê df
dt œ ` x dt ` y dt ` z dt
#
œ (cos t)( sin t) (sin t)(cos t) at# t 2b(1) œ t# t 2; df
dt œ 0 Ê t t 2 œ 0 Ê t œ 2
or t œ 1; t œ 2 Ê x œ cos (2), y œ sin (2), z œ 2 for the point (cos (2)ß sin (2)ß 2); t œ 1 Ê x œ cos 1,
y œ sin 1, z œ 1 for the point (cos 1ß sin 1ß 1)
` w dx
` w dy
` w dz
2y
# 2y
# 2y
ˆ " ‰
48. dw
dt œ ` x dt ` y dt ` z dt œ a2xe cos 3zb ( sin t) a2x e cos 3zb t# a3x e sin 3zb (1)
# 2y
3z
œ 2xe2y cos 3z sin t 2x et cos
3x# e2y sin 3z; at the point on the curve z œ 0 Ê t œ z œ 0
#
#
2(1) (4)(1)
¸
Ê dw
0œ4
#
dt Ð1ßln 2ß0Ñ œ 0 49. (a)
`T
`T
` x œ 8x 4y and ` y œ 8y 4x
` T dx
` T dy
Ê dT
dt œ ` x dt ` y dt œ (8x 4y)( sin t) (8y 4x)(cos t)
#
œ (8 cos t 4 sin t)( sin t) (8 sin t 4 cos t)(cos t) œ 4 sin# t 4 cos# t Ê ddtT# œ 16 sin t cos t;
dT
dt œ 0
Ê 4 sin# t 4 cos# t œ 0 Ê sin# t œ cos# t Ê sin t œ cos t or sin t œ cos t Ê t œ 14 , 541 , 341 , 741 on
the interval 0 Ÿ t Ÿ 21;
È
d# T
1
1
dt# ¹ tœ 1 œ 16 sin 4 cos 4 0
È
Ê T has a minimum at (xß y) œ Š #2 ß #2 ‹ ;
4
È
È
d# T
31
31
dt# ¹ tœ 31 œ 16 sin 4 cos 4 0
Ê T has a maximum at (xß y) œ Š #2 ß #2 ‹ ;
d# T
51
51
dt# ¹ tœ 51 œ 16 sin 4 cos 4 0
Ê T has a minimum at (xß y) œ Š #2 ß #2 ‹ ;
d# T
71
71
dt# ¹ tœ 71 œ 16 sin 4 cos 4 0
Ê T has a maximum at (xß y) œ Š #2 ß #2 ‹
4
4
4
È
È
È
È
(b) T œ 4x# 4xy 4y# Ê ``Tx œ 8x 4y, and ``Ty œ 8y 4x so the extreme values occur at the four points
È
È
È
È
found in part (a): T Š #2 ß #2 ‹ œ T Š #2 ß #2 ‹ œ 4 ˆ "# ‰ 4 ˆ "# ‰ 4 ˆ "# ‰ œ 6, the maximum and
È
È
È
È
T Š #2 ß #2 ‹ œ T Š #2 ß #2 ‹ œ 4 ˆ #" ‰ 4 ˆ #" ‰ 4 ˆ #" ‰ œ 2, the minimum
50. (a)
`T
`T
` x œ y and ` y œ x
` T dx
` T dy
È
È
Ê dT
dt œ ` x dt ` y dt œ y Š2 2 sin t‹ x Š 2 cos t‹
œ ŠÈ2 sin t‹ Š2È2 sin t‹ Š2È2 cos t‹ ŠÈ2 cos t‹ œ 4 sin# t 4 cos# t œ 4 sin# t 4 a1 sin# tb
#
"
"
#
#
œ 4 8 sin# t Ê ddtT# œ 16 sin t cost t; dT
dt œ 0 Ê 4 8 sin t œ 0 Ê sin t œ # Ê sin t œ „ È
31 51 71
4 , 4 , 4 on the interval 0 Ÿ t Ÿ 21;
d# T
ˆ1‰
dt# ¹ tœ 1 œ 8 sin 2 4 œ 8
Ê T has a maximum at (xß y) œ (2ß 1);
d# T
ˆ 31 ‰ œ 8
dt# ¹ tœ 31 œ 8 sin 2 4
Ê T has a minimum at (xß y) œ (2ß 1);
4
4
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
2
Ê t œ 14 ,
824
Chapter 14 Partial Derivatives
d# T
ˆ 51 ‰ œ 8
dt# ¹ tœ 51 œ 8 sin 2 4
Ê T has a maximum at (xß y) œ (2ß 1);
4
d# T
ˆ 71 ‰ œ 8
dt# ¹ tœ 71 œ 8 sin 2 4
Ê T has a minimum at (xß y) œ (2ß 1)
4
(b) T œ xy 2 Ê ``Tx œ y and ``Ty œ x so the extreme values occur at the four points found in part (a):
T(2ß 1) œ T(2ß 1) œ 0, the maximum and T(2ß 1) œ T(2ß 1) œ 4, the minimum
` G du
` G dx
w
'
51. G(uß x) œ 'a g(tß x) dt where u œ f(x) Ê dG
dx œ ` u dx ` x dx œ g(uß x)f (x) a gx (tß x) dt; thus
u
u
F(x) œ '0 Èt% x$ dt Ê Fw (x) œ Éax# b% x$ (2x) '0
x#
x#
` È %
t x$ dt œ 2xÈx) x$ `x
52. Using the result in Exercise 51, F(x) œ 'x# Èt$ x# dt œ '1 Èt$ x# dt Ê Fw (x)
x#
1
œ ’ Éax# b$ x# x# '
x#
#
` È $
t x# dt “ œ x Èx' x# 1 `x
'x1 È x
#
t$ x#
dt
14.5 DIRECTIONAL DERIVATIVES AND GRADIENT VECTORS
1.
`f
`f
` x œ 1, ` y œ 1
Ê ™ f œ i j ; f(2ß 1) œ 1
Ê 1 œ y x is the level curve
2.
2y
`f
2x
`f
`f
` x œ x# y# Ê ` x ("ß ") œ 1; ` y œ x# y#
Ê `` yf ("ß ") œ 1 Ê ™ f œ i j ; f(1ß 1) œ ln 2
#
#
#
#
Ê ln 2
œ ln ax y b Ê 2 œ x y is the level curve
3.
`g
2
`x œ y
Ê `` gx a2ß 1b œ 1; `` gy œ 2x y Ê `` gx a2ß 1b œ 4;
Ê ™ g œ i 4j ; ga2ß 1b œ 2 Ê x œ y2# is the level
curve
4.
`g
`x œ x
Ê `` gx ŠÈ2ß "‹ œ È2; `` gy œ y
Ê `` gy ŠÈ2ß "‹ œ 1 Ê ™ g œ È2 i j ;
#
#
g ŠÈ2ß "‹ œ "# Ê "# œ x# y# or 1 œ x# y# is the level
curve
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
'0x È3x
#
2
#
t% x$
dt
Section 14.5 Directional Derivatives and Gradient Vectors
5.
`f
1
` x œ È2x 3y
Ê `` xf (1ß 2) œ 21 ; `` yf œ 2È2x3 3y
Ê `` xf (1ß 2) œ 43 ; Ê ™ f œ 12 i 34 j ; f(1ß 2) œ 2
Ê 4 œ 2x 3y is the level curve
6.
y
`f
` x œ 2y2 Èx 2x3Î2
1
Ê `` xf a4ß 2b œ 16
;
Èx
`f
`f
1
1
1
` y œ 2y2 x Ê ` y a4ß 2b œ 4 Ê ™ f œ 16 i 4 j ;
f a4ß 2b œ 14 Ê y œ Èx is the level curve
7.
`f
z
` x œ 2x x
Ê `` xf (1ß 1ß 1) œ 3; `` yf œ 2y Ê `` yf ("ß "ß ") œ 2; `` zf œ 4z ln x Ê `` zf ("ß "ß ") œ 4;
thus ™ f œ 3i 2j 4k
8.
9.
`f
z
`f
11 ` f
` x œ 6xz x# z# 1 Ê ` x (1ß "ß ") œ # ; ` y œ 6yz
"
Ê `` zf ("ß "ß ") œ "# ; thus ™ f œ 11
# i 6j # k
Ê `` yf ("ß "ß ") œ 6; `` zf œ 6z# 3 ax# y# b x# z#x 1
`f
x
"
` x œ ax# y# z# b$Î# x
`f
Ê `` xf (1ß 2ß 2) œ 26
27 ; ` y œ y
y" Ê `` yf (1ß 2ß 2) œ 23
54 ;
ax# y# z# b$Î#
`f
23
26
23
23
` z (1ß 2ß 2) œ 54 ; thus ™ f œ 27 i 54 j 54 k
`f
z
"
` z œ ax# y# z# b$Î# z
Ê
10. `` xf œ exy cos z Èy 1 # Ê `` xf ˆ!ß !ß 16 ‰ œ
1x
`f
x y
sin z
` z œ e
È3
`f
x y
cos z sin" x
# 1; ` y œ e
Ê `` zf ˆ!ß !ß 16 ‰ œ #" ; thus ™ f œ Š
Ê `` yf ˆ0ß 0ß 16 ‰ œ
È3
# ;
È 3 2
È3
"
# ‹i # j # k
4i 3j
4
3
11. u œ kA
Ak œ È4# 3# œ 5 i 5 j ; fx (xß y) œ 2y Ê fx (5ß 5) œ 10; fy (xß y) œ 2x 6y Ê fy (5ß 5) œ 20
Ê ™ f œ 10i 20j Ê (Du f)P! œ ™ f † u œ 10 ˆ 45 ‰ 20 ˆ 35 ‰ œ 4
3i 4j
3
4
12. u œ kA
Ak œ È3# (4)# œ 5 i 5 j ; fx (xß y) œ 4x Ê fx (1ß 1) œ 4; fy (xß y) œ 2y Ê fy (1ß 1) œ 2
8
Ê ™ f œ 4i 2j Ê (Du f)P! œ ™ f † u œ 12
5 5 œ 4
12i 5j
y 2
5
x 2
13. u œ kA
œ 12
#
Ak œ È #
13 i 13 j ; gx axß yb œ axy 2b2 Ê gx a1ß 1b œ 3; gy axß yb œ axy 2b2 Ê gy a1ß 1b œ 3
2
2
12 5
15
21
Ê ™ g œ 3i 3j Ê aDu gbP! œ ™ g † u œ 36
13 13 œ 13
Š #y ‹
3i 2j
x
3
2
14. u œ kA
Ak œ È3# (2)# œ È13 i È13 j ; hx (xß y) œ ˆ y ‰#
x
ˆ"‰
hy (xß y) œ ˆ y ‰#x
x
1
ˆ #x ‰ È3
x# y#
Ê1 Š 4 ‹
1
ˆ #y ‰ È3
x# y#
Ê1 Š 4 ‹
Ê hx (1ß 1) œ "# ;
Ê hy (1ß 1) œ #3 Ê ™ h œ "# i #3 j Ê (Du h)P! œ ™ h † u œ 2È313 2È613
œ 2È313
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
825
826
Chapter 14 Partial Derivatives
3 i 6 j #k
3
6
2
15. u œ kA
Ak œ È3# 6# (2)# œ 7 i 7 j 7 k ; fx (xß yß z) œ y z Ê fx (1ß 1ß 2) œ 1; fy (xß yß z) œ x z
Ê fy (1ß 1ß 2) œ 3; fz (xß yß z) œ y x Ê fz (1ß 1ß 2) œ 0 Ê ™ f œ i 3j Ê (Du f)P! œ ™ f † u œ 37 18
7 œ 3
ijk
1
1
1
16. u œ kA
Ak œ È1# 1# 1# œ È3 i È3 j È3 k ; fx (xß yß z) œ 2x Ê fx (1ß 1ß 1) œ 2; fy (xß yß z) œ 4y
Ê fy (1ß 1ß 1) œ 4; fz (xß yß z) œ 6z Ê fz (1ß 1ß 1) œ 6 Ê ™ f œ 2i 4j 6k Ê (Du f)P! œ ™ f † u
œ 2 Š È"3 ‹ 4 Š È"3 ‹ 6 Š È"3 ‹ œ 0
2i j 2k
2
1
2
x
x
17. u œ kA
Ak œ È2# 1# (2)# œ 3 i 3 j 3 k ; gx (xß yß z) œ 3e cos yz Ê gx (0ß 0ß 0) œ 3; gy (xß yß z) œ 3ze sin yz
Ê gy (0ß 0ß 0) œ 0; gz (xß yß z) œ 3yex sin yz Ê gz (0ß 0ß 0) œ 0 Ê ™ g œ 3i Ê (Du g)P! œ ™ g † u œ 2
i 2j 2k
1
2
2
"
"‰
ˆ
18. u œ kA
Ak œ È1# 2# 2# œ 3 i 3 j 3 k ; hx (xß yß z) œ y sin xy x Ê hx 1ß 0ß # œ 1;
hy (xß yß z) œ x sin xy zeyz Ê hy ˆ"ß !ß #" ‰ œ #" ; hz (xß yß z) œ yeyz "z Ê hz ˆ"ß !ß #" ‰ œ 2 Ê ™ h œ i #" j 2k
Ê (Du h)P! œ ™ h † u œ "3 "3 43 œ 2
f
i j
"
"
19. ™ f œ (2x y) i (x 2y) j Ê ™ f(1ß 1) œ i j Ê u œ k™
™f k œ È(1)# 1# œ È i È j ; f increases
2
2
most rapidly in the direction u œ È"2 i È"2 j and decreases most rapidly in the direction u œ È"2 i È"2 j ;
(Du f)P! œ ™ f † u œ k ™ f k œ È2 and (Du f)P! œ È2
f
20. ™ f œ a2xy yexy sin yb i ax# xexy sin y exy cos yb j Ê ™ f(1ß 0) œ 2j Ê u œ k™
™f k œ j ; f increases most
rapidly in the direction u œ j and decreases most rapidly in the direction u œ j ; (Du f)P! œ ™ f † u œ k ™ f k
œ 2 and (Du f)P! œ 2
f
i 5j k
5
"
"
21. ™ f œ "y i Š yx# z‹ j yk Ê ™ f(4ß "ß ") œ i 5j k Ê u œ k™
™f k œ È1# (5)# (1)# œ 3È3 i 3È3 j 3È3 k ;
"
5
"
f increases most rapidly in the direction of u œ 3È
i 3È
j 3È
k and decreases most rapidly in the direction
3
3
3
"
5
"
u œ 3È
i 3È
j 3È
k ; (Du f)P! œ ™ f † u œ k ™ f k œ 3È3 and (Du f)P! œ 3È3
3
3
3
g
2i 2j k
"
2
2
22. ™ g œ ey i xey j 2zk Ê ™ g ˆ1ß ln 2ß "# ‰ œ 2i 2j k Ê u œ k™
™gk œ È # # # œ 3 i 3 j 3 k ;
2
2
1
2
2
"
g increases most rapidly in the direction u œ 3 i 3 j 3 k and decreases most rapidly in the direction
u œ 23 i 23 j 3" k ; (Du g)P! œ ™ g † u œ k ™ gk œ 3 and (Du g)P! œ 3
f
"
"
"
23. ™ f œ ˆ "x x" ‰ i Š y" y" ‹ j ˆ "z "z ‰ k Ê ™ f("ß "ß ") œ 2i 2j 2k Ê u œ k™
™f k œ È3 i È3 j È3 k ;
f increases most rapidly in the direction u œ È"3 i È"3 j È"3 k and decreases most rapidly in the direction
u œ È"3 i È"3 j È"3 k; (Du f)P! œ ™ f † u œ k ™ f k œ 2È3 and (Du f)P! œ 2È3
™h
2y
2i 3 j 6k
24. ™ h œ Š x# 2x
y# 1 ‹ i Š x# y# 1 1‹ j 6k Ê ™ h("ß "ß 0) œ 2i 3j 6k Ê u œ k™hk œ È2# 3# 6#
œ 27 i 37 j 67 k ; h increases most rapidly in the direction u œ 27 i 37 j 67 k and decreases most rapidly in the
direction u œ 27 i 37 j 67 k ; (Du h)P! œ ™ h † u œ k ™ hk œ 7 and (Du h)P! œ 7
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 14.5 Directional Derivatives and Gradient Vectors
827
25. ™ f œ 2xi 2yj Ê ™ f ŠÈ2ß È2‹ œ 2È2 i 2È2 j
Ê Tangent line: 2È2 Šx È2‹ 2È2 Šy È2‹ œ 0
Ê È2x È2y œ 4
26. ™ f œ 2xi j Ê ™ f ŠÈ2ß 1‹ œ 2È2 i j
Ê Tangent line: 2È2 Šx È2‹ (y 1) œ 0
Ê y œ 2È2x 3
27. ™ f œ yi xj Ê ™ f(2ß 2) œ 2i 2j
Ê Tangent line: 2(x 2) 2(y 2) œ 0
Ê yœx4
28. ™ f œ (2x y)i (2y x)j Ê ™ f(1ß 2) œ 4i 5j
Ê Tangent line: 4(x 1) 5(y 2) œ 0
Ê 4x 5y 14 œ 0
29. ™ f œ a2x ybi ax 2y 1bj
(a) ™ fa1, 1b œ 3i 4j Ê l ™ fa1, 1bl œ 5 Ê Du fa1, 1b œ 5 in the direction of u œ 35 i 45 j
(b) ™ fa1, 1b œ 3i 4j Ê l ™ fa1, 1bl œ 5 Ê Du fa1, 1b œ 5 in the direction of u œ 35 i 45 j
(c) Du fa1, 1b œ 0 in the direction of u œ 45 i 35 j or u œ 45 i 35 j
(d) Let u œ u1 i u2 j Ê lul œ Èu12 u22 œ 1 Ê u12 u22 œ 1; Du fa1, 1b œ ™ fa1, 1b † u œ a3i 4jb † au1 i u2 jb
2
25 2
24
œ 3u1 4u2 œ 4 Ê u2 œ 43 u1 1 Ê u12 ˆ 43 u1 1‰ œ 1 Ê 16
u1 23 u1 œ 0 Ê u1 œ 0 or u1 œ 25
;
7
24
7
u1 œ 0 Ê u2 œ 1 Ê u œ j, or u1 œ 24
25 Ê u2 œ 25 Ê u œ 25 i 25 j
(e) Let u œ u1 i u2 j Ê lul œ Èu12 u22 œ 1 Ê u12 u22 œ 1; Du fa1, 1b œ ™ fa1, 1b † u œ a3i 4jb † au1 i u2 jb
2
8
24
2
œ 3u1 4u2 œ 3 Ê u1 œ 43 u2 1 Ê ˆ 43 u2 1‰ u22 œ 1 Ê 25
9 u2 3 u2 œ 0 Ê u2 œ 0 or u2 œ 25 ;
7
7
24
u2 œ 0 Ê u1 œ 1 Ê u œ i, or u2 œ 24
25 Ê u2 œ 25 Ê u œ 25 i 25 j
.
30. ™ f œ ax 2yyb2 i ax 2xyb2 j
(a) ™ fˆ 21 , 23 ‰ œ 3i j Ê l ™ fˆ 21 , 23 ‰l œ È10 Ê Du fˆ 21 , 23 ‰ œ È10 in the direction of u œ È310 i È110 j
(b) ™ fˆ 21 , 23 ‰ œ 3i j Ê l ™ fˆ 21 , 23 ‰l œ È10 Ê Du fa1, 1b œ È10 in the direction of u œ È310 i È110 j
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
828
Chapter 14 Partial Derivatives
(c) Du fˆ 12 , 23 ‰ œ 0 in the direction of u œ È110 i È310 j or u œ È110 i È310 j
(d) Let u œ u1 i u2 j Ê lul œ Èu12 u22 œ 1 Ê u12 u22 œ 1; Du fˆ 12 , 32 ‰ œ ™ fˆ 12 , 32 ‰ † u œ a3i jb † au1 i u2 jb
È
6
œ 3u1 u2 œ 2 Ê u2 œ 3u1 2 Ê u12 a3u1 2b2 œ 1 Ê 10u12 12u1 3 œ 0 Ê u1 œ 6 „
10
È
6
u1 œ 6 Ê u2 œ 2 103
10
È
6
Ê u œ 6 i 2 103
10
È6
È
6
Ê u œ 6 i 2 103
10
È6
È
6
j, or u1 œ 6 Ê u2 œ 2 103
10
È6
È6
j
(e) Let u œ u1 i u2 j Ê lul œ Èu12 u22 œ 1 Ê u12 u22 œ 1; Du fˆ 12 , 32 ‰ œ ™ fˆ 12 , 32 ‰ † u œ a3i jb † au1 i u2 jb
œ 3u1 u2 œ 1 Ê u2 œ 1 3u1 Ê u12 a1 3u1 b2 œ 1 Ê 10u12 6u1 œ 0 Ê u1 œ 0 or u1 œ 35 ;
u1 œ 0 Ê u2 œ 1 Ê u œ j, or u1 œ 35 Ê u2 œ 45 Ê u œ 35 i 45 j
31. ™ f œ yi (x 2y)j Ê ™ f(3ß 2) œ 2i 7j ; a vector orthogonal to ™ f is v œ 7i 2j Ê u œ kvvk œ È77#i(2j 2)#
œ È753 i È253 j and u œ È753 i È253 j are the directions where the derivative is zero
#
#
32. ™ f œ ax#4xy
i ax#4x yy# b# j Ê ™ f("ß ") œ i j ; a vector orthogonal to ™ f is v œ i j
y # b#
Ê u œ kvvk œ È1i #j 1# œ È12 i È12 j and u œ È12 i È12 j are the directions where the derivative is zero
33. ™ f œ (2x 3y)i (3x 8y)j Ê ™ f(1ß 2) œ 4i 13j Ê k ™ f(1ß 2)k œ È(4)# (13)# œ È185 ; no, the
maximum rate of change is È185 14
34. ™ T œ 2yi (2x z)j yk Ê ™ T(1ß 1ß 1) œ 2i j k Ê k ™ T(1ß 1ß 1)k œ È(2)# 1# 1# œ È6 ; no, the
minimum rate of change is È6 3
35. ™ f œ fx ("ß #)i fy ("ß #)j and u" œ È1i #j 1# œ È"2 i È"2 j Ê (Du" f)(1ß 2) œ fx (1ß 2) Š È"2 ‹ fy (1ß 2) Š È"2 ‹
œ 2È2 Ê fx (1ß 2) fy (1ß 2) œ 4; u# œ j Ê (Du# f)(1ß 2) œ fx (1ß 2)(0) fy (1ß 2)(1) œ 3 Ê fy (1ß 2) œ 3
Ê fy (1ß 2) œ 3; then fx (1ß 2) 3 œ 4 Ê fx (1ß 2) œ 1; thus ™ f(1ß 2) œ i 3j and u œ kvvk œ È(1)i#2(j2)#
œ È15 i È25 j Ê (Du f)P! œ ™ f † u œ È"5 È65 œ È75
f
36. (a) (Du f)P œ 2È3 Ê k ™ f k œ 2È3; u œ kvvk œ È1# i1j#k(1)# œ È13 i È13 j È"3 k; thus u œ k™
™f k
Ê ™ f œ k ™ f k u Ê ™ f œ 2È3 Š È"3 i È"3 j È"3 k‹ œ 2i 2j 2k
(b) v œ i j Ê u œ kvvk œ È1i #j 1# œ È"2 i È"2 j Ê (Du f)P! œ ™ f † u œ 2 Š È"2 ‹ 2 Š È"2 ‹ 2(0) œ 2È2
37. The directional derivative is the scalar component. With ™ f evaluated at P! , the scalar component of ™ f in the
direction of u is ™ f † u œ (Du f)P! .
38. Di f œ ™ f † i œ (fx i fy j fz k) † i œ fx ; similarly, Dj f œ ™ f † j œ fy and Dk f œ ™ f † k œ fz
39. If (xß y) is a point on the line, then T(xß y) œ (x x! )i (y y! )j is a vector parallel to the line Ê T † N œ 0
Ê A(x x! ) B(y y! ) œ 0, as claimed.
` (kf)
` (kf)
`f
`f
`f
`f
ˆ `f ‰
ˆ `f ‰
40. (a) ™ (kf) œ ``(kf)
x i ` y j ` z k œ k ` x i k Š ` y ‹ j k ` z k œ k Š ` x i ` y j ` z k‹ œ k ™ f
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 14.6 Tangent Planes and Differentials
829
(b) ™ (f g) œ ` (f`x g) i ` (f`y g) j ` (f`z g) k œ Š `` xf `` xg ‹ i Š `` yf `` gy ‹ j Š `` zf `` gz ‹ k
œ `` xf i `` gx i `` yf j `` gy j `` zf k `` gz k œ Š `` xf i `` yf j `` zf k‹ Š `` gx i `` gy j `` gz k‹ œ ™ f ™ g
(c) ™ (f g) œ ™ f ™ g (Substitute g for g in part (b) above)
` (fg)
` (fg)
`g
`g
`g
`f
`f
`f
(d) ™ (fg) œ ``(fg)
x i `y j `z k œ Š `x g `x f ‹ i Š `y g `y f ‹ j Š `z g `z f ‹ k
œ ˆ `` xf g‰ i Š `` xg f ‹ i Š `` yf g‹ j Š `` gy f ‹ j ˆ `` zf g‰ k Š `` gz f ‹ k
œ f Š `` gx i `` gy j `` gz k‹ g Š `` xf i `` yf j `` zf k‹ œ f ™ g g ™ f
(e) ™ Š gf ‹ œ
` Š gf ‹
`x
i
` Š gf ‹
`y
j
` Š gf ‹
`z
g `f f `g
g `f f `g
g `f f `g
k œ Š ` x g # ` x ‹ i Œ ` y g # ` y j Š ` z g# ` z ‹ k
`f
`f
`f
`g
`g
`g
g Š ` x i ` y j ` z k‹
f Š ` x i ` y j ` z k‹
g ` f i g ``yf j g `` fz k
f `` gx i f `` gy j f ``gz k
Œ
œ
g#
g#
g#
g#
œ Œ `x
f
f™g
g™f f™g
œ g™
g# g# œ
g#
14.6 TANGENT PLANES AND DIFFERENTIALS
1. (a) ™ f œ 2xi 2yj 2zk Ê ™ f(1ß 1ß 1) œ 2i 2j 2k Ê Tangent plane: 2(x 1) 2(y 1) 2(z 1) œ 0
Ê x y z œ 3;
(b) Normal line: x œ 1 2t, y œ 1 2t, z œ 1 2t
2. (a) ™ f œ 2xi 2yj 2zk Ê ™ f(3ß 5ß 4) œ 6i 10j 8k Ê Tangent plane: 6(x 3) 10(y 5) 8(z 4) œ 0
Ê 3x 5y 4z œ 18;
(b) Normal line: x œ 3 6t, y œ 5 10t, z œ 4 8t
3. (a) ™ f œ 2xi 2k Ê ™ f(2ß 0ß 2) œ 4i 2k Ê Tangent plane: 4(x 2) 2(z 2) œ 0
Ê 4x 2z 4 œ 0 Ê 2x z 2 œ 0;
(b) Normal line: x œ 2 4t, y œ 0, z œ 2 2t
4. (a) ™ f œ (2x 2y)i (2x 2y)j 2zk Ê ™ f(1ß 1ß 3) œ 4j 6k Ê Tangent plane: 4(y 1) 6(z 3) œ 0
Ê 2y 3z œ 7;
(b) Normal line: x œ 1, y œ 1 4t, z œ 3 6t
5. (a) ™ f œ a1 sin 1x 2xy zexz b i ax# zb j axexz yb k Ê ™ f(0ß 1ß 2) œ 2i 2j k Ê Tangent plane:
2(x 0) 2(y 1) 1(z 2) œ 0 Ê 2x 2y z 4 œ 0;
(b) Normal line: x œ 2t, y œ 1 2t, z œ 2 t
6. (a) ™ f œ (2x y)i (x 2y)j k Ê ™ f(1ß 1ß 1) œ i 3j k Ê Tangent plane:
1(x 1) 3(y 1) 1(z 1) œ 0 Ê x 3y z œ 1;
(b) Normal line: x œ 1 t, y œ 1 3t, z œ 1 t
7. (a) ™ f œ i j k for all points Ê ™ f(0ß 1ß 0) œ i j k Ê Tangent plane: 1(x 0) 1(y 1) 1(z 0) œ 0
Ê x y z 1 œ 0;
(b) Normal line: x œ t, y œ 1 t, z œ t
8. (a) ™ f œ (2x 2y 1)i (2y 2x 3)j k Ê ™ f(2ß 3ß 18) œ 9i 7j k Ê Tangent plane:
9(x 2) 7(y 3) 1(z 18) œ 0 Ê 9x 7y z œ 21;
(b) Normal line: x œ 2 9t, y œ 3 7t, z œ 18 t
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
830
Chapter 14 Partial Derivatives
2y
9. z œ f(xß y) œ ln ax# y# b Ê fx (xß y) œ x# 2x
y# and fy (xß y) œ x# y# Ê fx (1ß 0) œ 2 and fy (1ß 0) œ 0 Ê from
Eq. (4) the tangent plane at (1ß 0ß 0) is 2(x 1) z œ 0 or 2x z 2 œ 0
#
#
#
#
#
#
10. z œ f(xß y) œ e ax y b Ê fx (xß y) œ 2xe ax y b and fy (xß y) œ 2ye ax y b Ê fx (0ß 0) œ 0 and fy (!ß !) œ 0
Ê from Eq. (4) the tangent plane at (0ß 0ß 1) is z 1 œ 0 or z œ 1
11. z œ f(Bß y) œ Èy x Ê fx (xß y) œ "# (y x)"Î# and fy (xß y) œ "# (y x)"Î# Ê fx (1ß 2) œ "# and fy ("ß #) œ "#
Ê from Eq. (4) the tangent plane at (1ß 2ß 1) is "# (x 1) "# (y 2) (z 1) œ 0 Ê x y 2z 1 œ 0
12. z œ f(Bß y) œ 4x# y# Ê fx (xß y) œ 8x and fy (xß y) œ #y Ê fx (1ß 1) œ 8 and fy ("ß 1) œ # Ê from Eq. (4) the
tangent plane at (1ß 1ß 5) is 8(x 1) 2(y 1) (z 5) œ 0 or 8x 2y z 5 œ 0
13. ™ f œ i 2yj 2k Ê ™ f(1ß 1ß 1) œ i 2j 2k and ™ g œ i for all points; v œ ™ f ‚ ™ g
â
â
â i j kâ
â
â
Ê v œ â " 2 2 â œ 2j 2k Ê Tangent line: x œ 1, y œ 1 2t, z œ 1 2t
â
â
â" 0 0â
14. ™ f œ yzi xzj xyk Ê ™ f(1ß 1ß 1) œ i j k; ™ g œ 2xi 4yj 6zk Ê ™ g(1ß 1ß 1) œ 2i 4j 6k ;
â
â
â i j kâ
â
â
Ê v œ ™ f ‚ ™ g Ê â " 1 1 â œ 2i 4j 2k Ê Tangent line: x œ 1 2t, y œ 1 4t, z œ 1 2t
â
â
â2 4 6â
15. ™ f œ 2xi 2j 2k Ê ™ f ˆ1ß 1ß "# ‰ œ 2i 2j 2k and ™ g œ j for all points; v œ ™ f ‚ ™ g
â
â
â i j kâ
â
â
Ê v œ â 2 2 2 â œ 2i 2k Ê Tangent line: x œ 1 2t, y œ 1, z œ "# 2t
â
â
â0 1 0â
16. ™ f œ i 2yj k Ê ™ f ˆ "# ß 1ß "# ‰ œ i 2j k and ™ g œ j for all points; v œ ™ f ‚ ™ g
â
â
â i j kâ
â
â
Ê v œ â 1 2 1 â œ i k Ê Tangent line: x œ "# t, y œ 1, z œ "# t
â
â
â0 1 0â
17. ™ f œ a3x# 6xy# 4yb i a6x# y 3y# 4xb j 2zk Ê ™ f(1ß 1ß 3) œ 13i 13j 6k ; ™ g œ 2xi 2yj 2zk
â
â
j
k â
â i
â
â
Ê ™ g("ß "ß $) œ 2i 2j 6k ; v œ ™ f ‚ ™ g Ê v œ â "3 13 6 â œ 90i 90j Ê Tangent line:
â
â
2
6 â
â 2
x œ 1 90t, y œ 1 90t, z œ 3
18. ™ f œ 2xi 2yj Ê ™ f ŠÈ2ß È2ß 4‹ œ 2È2 i 2È2 j ; ™ g œ 2xi 2yj k Ê ™ g ŠÈ2ß È2ß 4‹
â i
j
k ââ
â
â
â
œ 2È2i 2È2j k ; v œ ™ f ‚ ™ g Ê v œ â 2È2 2È2 0 â œ 2È2 i 2È2 j Ê Tangent line:
â
â
â 2È2 2È2 1 â
x œ È2 2È2 t, y œ È2 2È2 t, z œ 4
3
4
12
19. ™ f œ Š x# yx# z# ‹ i Š x# yy# z# ‹ j Š x# yz# z# ‹ k Ê ™ f(3ß 4ß 12) œ 169
i 169
j 169
k;
2k
9
9 ‰
u œ kvvk œ È33#i 66#j œ 37 i 76 j 27 k Ê ™ f † u œ 1183
and df œ ( ™ f † u) ds œ ˆ 1183
(0.1) ¸ 0.0008
(2)#
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 14.6 Tangent Planes and Differentials
831
2k
20. ™ f œ aex cos yzb i azex sin yzb j ayex sin yzb k Ê ™ f(0ß 0ß 0) œ i ; u œ kvvk œ È22#i 22#j (2)#
œ È13 i È13 j È13 k Ê ™ f † u œ È13 and df œ ( ™ f † u) ds œ È13 (0.1) ¸ 0.0577
Ä
21. ™ g œ (1 cos z)i (1 sin z)j (x sin z y cos z)k Ê ™ g(2ß 1ß 0) œ 2i j k; A œ P! P" œ 2i 2j 2k
i 2 j 2k
1
1
1
Ê u œ kvvk œ È(22)
# 2# 2# œ È3 i È3 j È3 k Ê ™ g † u œ 0 and dg œ ( ™ g † u) ds œ (0)(0.2) œ 0
22. ™ h œ c1y sin (1xy) z# d i c1x sin (1xy)d j 2xzk Ê ™ h(1ß 1ß 1) œ (1 sin 1 1)i (1 sin 1)j 2k
Ä
k
œ i 2k ; v œ P! P" œ i j k where P" œ (!ß !ß !) Ê u œ kvvk œ È1i#j1# œ È13 i È13 j È13 k
1#
Ê ™ h † u œ È33 œ È3 and dh œ ( ™ h † u) ds œ È3(0.1) ¸ 0.1732
È
23. (a) The unit tangent vector at Š "# ß #3 ‹ in the direction of motion is u œ
È3
"
# i # j;
È
È
™ T œ (sin 2y)i (2x cos 2y)j Ê ™ T Š "# ß #3 ‹ œ Šsin È3‹ i Šcos È3‹ j Ê Du T Š "# ß #3 ‹ œ ™ T † u
œ
È3
È3 " cos È3 ¸ 0.935° C/ft
# sin
#
` T dx
` T dy
(b) r(t) œ (sin 2t)i (cos 2t)j Ê v(t) œ (2 cos 2t)i (2 sin 2t)j and kvk œ 2; dT
dt œ ` x dt ` y dt
È
œ ™ T † v œ Š ™ T † kvvk ‹ kvk œ (Du T) kvk , where u œ kvvk ; at Š "# ß #3 ‹ we have u œ
È3
"
# i # j from part (a)
È
3
È3 " cos È3‹ † 2 œ È3 sin È3 cos È3 ¸ 1.87° C/sec
Ê dT
dt œ Š # sin
#
24. (a) ™ T œ (4x yz)i xzj xyk Ê ™ T(8ß 6ß 4) œ 56i 32j 48k ; r(t) œ 2t# i 3tj t# k Ê the particle is
at the point P()ß 6ß 4) when t œ 2; v(t) œ 4ti 3j 2tk Ê v(2) œ 8i 3j 4k Ê u œ kvvk
736
œ È889 i È389 j È489 k Ê Du T(8ß 6ß 4) œ ™ T † u œ È"89 [56 † 8 32 † 3 48 † (4)] œ È
° C/m
89
(b)
dT
` T dx
` T dy
dt œ ` x dt ` y dt œ
736
È89 œ 736° C/sec
¸
™ T † v œ ( ™ T † u) kvk Ê at t œ 2, dT
dt œ Du T tœ2 v(2) œ Š È89 ‹
25. (a) f(!ß 0) œ 1, fx (xß y) œ 2x Ê fx (0ß 0) œ 0, fy (xß y) œ 2y Ê fy (0ß 0) œ 0 Ê L(xß y) œ 1 0(x 0) 0(y 0) œ 1
(b) f(1ß 1) œ 3, fx (1ß 1) œ 2, fy (1ß 1) œ 2 Ê L(xß y) œ 3 2(x 1) 2(y 1) œ 2x 2y 1
26. (a) f(!ß 0) œ 4, fx (xß y) œ 2(x y 2) Ê fx (0ß 0) œ 4, fy (xß y) œ 2(x y 2) Ê fy (0ß 0) œ 4
Ê L(xß y) œ 4 4(x 0) 4(y 0) œ 4x 4y 4
(b) f(1ß 2) œ 25, fx (1ß 2) œ 10, fy (1ß 2) œ 10 Ê L(xß y) œ 25 10(x 1) 10(y 2) œ 10x 10y 5
27. (a) f(0ß 0) œ 5, fx (xß y) œ 3 for all (xß y), fy (xß y) œ 4 for all (xß y) Ê L(xß y) œ 5 3(x 0) 4(y 0) œ 3x 4y 5
(b) f(1ß 1) œ 4, fx (1ß 1) œ 3, fy (1ß 1) œ 4 Ê L(xß y) œ 4 3(x 1) 4(y 1) œ 3x 4y 5
28. (a) f(1ß 1) œ 1, fx (xß y) œ 3x# y% Ê fx (1ß 1) œ 3, fy (xß y) œ 4x$ y$ Ê fy (1ß 1) œ 4
Ê L(xß y) œ 1 3(x 1) 4(y 1) œ 3x 4y 6
(b) f(0ß 0) œ 0, fx (!ß 0) œ 0, fy (0ß 0) œ 0 Ê L(xß y) œ 0
29. (a) f(0ß 0) œ 1, fx (xß y) œ ex cos y Ê fx (0ß 0) œ 1, fy (xß y) œ ex sin y Ê fy (0ß 0) œ 0
Ê L(xß y) œ 1 1(x 0) 0(y 0) œ x 1
(b) f ˆ0ß 1# ‰ œ 0, fx ˆ0ß 1# ‰ œ 0, fy ˆ0ß 1# ‰ œ 1 Ê L(xß y) œ 0 0(x 0) 1 ˆy 1# ‰ œ y 1#
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
832
Chapter 14 Partial Derivatives
30. (a) f(0ß 0) œ 1, fx (xß y) œ e2yx Ê fx (!ß !) œ 1, fy (xß y) œ 2e2yx Ê fy (0ß 0) œ 2
Ê L(xß y) œ 1 1(x 0) 2(y 0) œ x 2y 1
(b) f(1ß 2) œ e$ , fx (1ß 2) œ e$ , fy (1ß 2) œ 2e$ Ê L(xß y) œ e$ e$ (x 1) 2e$ (y 2) œ e$ x 2e$ y 2e$
31. (a) Wa20, 25b œ 11‰ F; Wa30, 10b œ 39‰ F; Wa15, 15b œ 0‰ F
(b) Wa10, 40b œ 65.5‰ F; Wa50, 40b œ 88‰ F; Wa60, 30b œ 10.2‰ F;
5.72
0.0684t
`W
`W
0.16
(c) Wa25, 5b œ 17.4088‰ F; ``W
V œ v0.84 v0.84 Ê ` V a25, 5b œ 0.36; ` T œ 0.6215 0.4275v
Ê ``W
T a25, 5b œ 1.3370 Ê LaV, Tb œ 17.4088 0.36aV 25b 1.337aT 5b œ 1.337T 0.36V 15.0938
(d) i) Wa24, 6b ¸ La24, 6b œ 15.7118 ¸ 15.7‰ F
ii) Wa27, 2b ¸ La27, 2b œ 22.1398 ¸ 22.1‰ F
ii) Wa5, 10b ¸ La5, 10b œ 30.2638 ¸ 30.2‰ F This value is very different because the point a5, 10b is not
close to the point a25, 5b.
5.72
0.0684t
`W
`W
0.16
32. Wa50, 20b œ 59.5298‰ F; ``W
V œ v0.84 v0.84 Ê ` V a50, 20b œ 0.2651; ` T œ 0.6215 0.4275v
Ê ``W
T a50, 20b œ 1.4209 Ê LaV, Tb œ 59.5298 0.2651aV 50b 1.4209aT 20b
œ 1.4209T 0.2651V 17.8568
(a) Wa49, 22b ¸ La49, 22b œ 62.1065 ¸ 62.1‰ F
(b) Wa53, 19b ¸ La53, 19b œ 58.9042 ¸ 58.9‰ F
(c) Wa60, 30b ¸ La60, 30b œ 76.3898 ¸ 76.4‰ F
33. f(2ß 1) œ 3, fx (xß y) œ 2x 3y Ê fx (2ß 1) œ 1, fy (xß y) œ 3x Ê fy (2ß 1) œ 6 Ê L(xß y) œ 3 1(x 2) 6(y 1)
œ 7 x 6y; fxx (xß y) œ 2, fyy (xß y) œ 0, fxy (xß y) œ 3 Ê M œ 3; thus kE(xß y)k Ÿ ˆ "# ‰ (3) akx 2k ky 1kb#
Ÿ ˆ 3# ‰ (0.1 0.1)# œ 0.06
34. f(2ß 2) œ 11, fx (xß y) œ x y 3 Ê fx (2ß 2) œ 7, fy (xß y) œ x y# 3 Ê fy (2ß 2) œ 0
Ê L(xß y) œ 11 7(x 2) 0(y 2) œ 7x 3; fxx (xß y) œ 1, fyy (xß y) œ "# , fxy (xß y) œ 1
Ê M œ 1; thus kE(xß y)k Ÿ ˆ "# ‰ (1) akx 2k ky 2kb# Ÿ ˆ #1 ‰ (0.1 0.1)# œ 0.02
35. f(0ß 0) œ 1, fx (xß y) œ cos y Ê fx (0ß 0) œ 1, fy (xß y) œ 1 x sin y Ê fy (0ß 0) œ 1
Ê L(xß y) œ 1 1(x 0) 1(y 0) œ x y 1; fxx (xß y) œ 0, fyy (xß y) œ x cos y, fxy (xß y) œ sin y Ê Q œ 1;
thus kE(xß y)k Ÿ ˆ "# ‰ (1) akxk kykb# Ÿ ˆ #1 ‰ (0.2 0.2)# œ 0.08
36. f("ß #) œ 6, fx (xß y) œ y# y sin (x 1) Ê fx (1ß 2) œ 4, fy (xß y) œ 2xy cos (x 1) Ê fy (1ß 2) œ 5
Ê L(xß y) œ 6 4(x 1) 5(y 2) œ 4x 5y 8; fxx (xß y) œ y cos (x 1), fyy (xß y) œ 2x,
fxy (xß y) œ 2y sin (x 1); kx 1k Ÿ 0.1 Ê 0.9 Ÿ x Ÿ 1.1 and ky 2k Ÿ 0.1 Ê 1.9 Ÿ y Ÿ 2.1; thus the max of
kfxx (xß y)k on R is 2.1, the max of kfyy (xß y)k on R is 2.2, and the max of kfxy (xß y)k on R is 2(2.1) sin (0.9 1)
Ÿ 4.3 Ê M œ 4.3; thus kE(xß y)k Ÿ ˆ "# ‰ (4.3) akx 1k ky 2kb# Ÿ (2.15)(0.1 0.1)# œ 0.086
37. f(0ß 0) œ 1, fx (xß y) œ ex cos y Ê fx (0ß 0) œ 1, fy (xß y) œ ex sin y Ê fy (0ß 0) œ 0
Ê L(xß y) œ 1 1(x 0) 0(y 0) œ 1 x; fxx (xß y) œ ex cos y, fyy (xß y) œ ex cos y, fxy (xß y) œ ex sin y;
kxk Ÿ 0.1 Ê 0.1 Ÿ x Ÿ 0.1 and kyk Ÿ 0.1 Ê 0.1 Ÿ y Ÿ 0.1; thus the max of kfxx (xß y)k on R is e0Þ1 cos (0.1)
Ÿ 1.11, the max of kfyy (xß y)k on R is e0Þ1 cos (0.1) Ÿ 1.11, and the max of kfxy (xß y)k on R is e0Þ1 sin (0.1)
Ÿ 0.12 Ê M œ 1.11; thus kE(xß y)k Ÿ ˆ "# ‰ (1.11) akxk kykb# Ÿ (0.555)(0.1 0.1)# œ 0.0222
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Section 14.6 Tangent Planes and Differentials
833
38. f(1ß 1) œ 0, fx (xß y) œ "x Ê fx (1ß 1) œ 1, fy (xß y) œ y" Ê fy (1ß 1) œ 1 Ê L(xß y) œ 0 1(x 1) 1(y 1)
œ x y 2; fxx (xß y) œ x"# , fyy (xß y) œ y"# , fxy (xß y) œ 0; kx 1k Ÿ 0.2 Ê 0.98 Ÿ x Ÿ 1.2 so the max of
"
kfxx (xß y)k on R is (0.98)
# Ÿ 1.04; ky 1k Ÿ 0.2 Ê 0.98 Ÿ y Ÿ 1.2 so the max of kfyy (xß y)k on R is
"
(0.98)# Ÿ 1.04
Ê M œ 1.04; thus kE(xß y)k Ÿ ˆ #" ‰ (1.04) akx 1k ky 1kb# Ÿ (0.52)(0.2 0.2)# œ 0.0832
39. (a) f("ß "ß ") œ 3, fx (1ß 1ß 1) œ y zkÐ1ß1ß1Ñ œ 2, fy (1ß 1ß 1) œ x zkÐ1ß1ß1Ñ œ 2, fz (1ß 1ß 1) œ y xkÐ1ß1ß1Ñ œ 2
Ê L(xß yß z) œ 3 2(x 1) 2(y 1) 2(z 1) œ 2x 2y 2z 3
(b) f(1ß 0ß 0) œ 0, fx (1ß 0ß 0) œ 0, fy (1ß 0ß 0) œ 1, fz (1ß 0ß 0) œ 1 Ê L(xß yß z) œ 0 0(x 1) (y 0) (z 0) œ y z
(c) f(0ß 0ß 0) œ 0, fx (0ß 0ß 0) œ 0, fy (0ß 0ß 0) œ 0, fz (0ß 0ß 0) œ 0 Ê L(xß yß z) œ 0
40. (a) f(1ß 1ß 1) œ 3, fx (1ß 1ß 1) œ 2xkÐ"ß"ß"Ñ œ 2, fy (1ß 1ß 1) œ 2ykÐ"ß"ß"Ñ œ 2, fz (1ß 1ß 1) œ 2zkÐ"ß"ß"Ñ œ 2
Ê L(xß yß z) œ 3 2(x 1) 2(y 1) 2(z 1) œ 2x 2y 2z 3
(b) f(0ß 1ß 0) œ 1, fx (0ß 1ß 0) œ 0, fy (!ß 1ß 0) œ 2, fz (0ß 1ß 0) œ 0 Ê L(xß yß z) œ 1 0(x 0) 2(y 1) 0(z 0)
œ 2y 1
(c) f(1ß 0ß 0) œ 1, fx (1ß 0ß 0) œ 2, fy (1ß 0ß 0) œ 0, fz (1ß 0ß 0) œ 0 Ê L(xß yß z) œ 1 2(x 1) 0(y 0) 0(z 0)
œ 2x 1
41. (a) f(1ß 0ß 0) œ 1, fx (1ß 0ß 0) œ Èx# xy# z# ¹
fz (1ß 0ß 0) œ Èx# zy# z# ¹
Ð1ß0ß0Ñ
Ð1ß0ß0Ñ
œ 1, fy (1ß 0ß 0) œ Èx# yy# z# ¹
Ð1 ß0 ß0 Ñ
œ 0,
œ 0 Ê L(xß yß z) œ 1 1(x 1) 0(y 0) 0(z 0) œ x
(b) f(1ß 1ß 0) œ È2, fx (1ß 1ß 0) œ È"2 , fy (1ß 1ß 0) œ È"2 , fz (1ß 1ß 0) œ 0
Ê L(xß yß z) œ È2 È"2 (x 1) È"2 (y 1) 0(z 0) œ È"2 x È"2 y
(c) f(1ß 2ß 2) œ 3, fx (1ß 2ß 2) œ "3 , fy (1ß 2ß 2) œ 23 , fz (1ß 2ß 2) œ 23 Ê L(xß yß z) œ 3 "3 (x 1) 23 (y 2) 23 (z 2)
œ "3 x 32 y 32 z
42. (a) f ˆ 12 ß 1ß 1‰ œ 1, fx ˆ 1# ß 1ß 1‰ œ y cosz xy ¸ ˆ 1 ß"ß"‰ œ 0, fy ˆ 1# ß 1ß 1‰ œ x cosz xy ¸ ˆ 1 ß"ß"‰ œ 0,
#
xy
fz ˆ 1# ß 1ß 1‰ œ sin
z# ¹ ˆ 1 ß"ß"‰ œ 1
Ê
#
#
L(xß yß z) œ 1 0 ˆx 1# ‰ 0(y 1) 1(z 1) œ 2 z
(b) f(2ß 0ß 1) œ 0, fx (2ß 0ß 1) œ 0, fy (2ß 0ß 1) œ 2, fz (2ß 0ß 1) œ 0 Ê L(xß yß z) œ 0 0(x 2) 2(y 0) 0(z 1) œ 2y
43. (a) f(0ß 0ß 0) œ 2, fx (0ß 0ß 0) œ ex k Ð!ß!ß!Ñ œ 1, fy (0ß 0ß 0) œ sin (y z)k Ð!ß!ß!Ñ œ 0,
fz (0ß 0ß 0) œ sin (y z)k Ð!ß!ß!Ñ œ 0 Ê L(xß yß z) œ 2 1(x 0) 0(y 0) 0(z 0) œ 2 x
(b) f ˆ0ß 1# ß 0‰ œ 1, fx ˆ0ß 1# ß 0‰ œ 1, fy ˆ0ß 1# ß 0‰ œ 1, fz ˆ0ß 1# ß 0‰ œ 1 Ê L(xß yß z)
œ 1 1(x 0) 1 ˆy 12 ‰ 1(z 0) œ x y z 1# 1
(c) f ˆ0ß 14 ß 14 ‰ œ 1, fx ˆ0ß 14 ß 14 ‰ œ 1, fy ˆ0ß 14 ß 14 ‰ œ 1, fz ˆ0ß 14 ß 14 ‰ œ 1 Ê L(xß yß z)
œ 1 1(x 0) 1 ˆy 14 ‰ 1 ˆz 14 ‰ œ x y z 1# 1
44. (a) f(1ß 0ß 0) œ 0, fx (1ß 0ß 0) œ (xyz)yz# 1 ¹
fz (1ß 0ß 0) œ (xyz)xy# 1 ¹
Ð"ß!ß!Ñ
Ð"ß!ß!Ñ
œ 0, fy (1ß 0ß 0) œ (xyz)xz# 1 ¹
Ð"ß!ß!Ñ
œ 0,
œ 0 Ê L(xß yß z) œ 0
(b) f(1ß 1ß 0) œ 0, fx (1ß 1ß 0) œ 0, fy (1ß 1ß 0) œ 0, fz (1ß 1ß 0) œ 1 Ê L(xß yß z) œ 0 0(x 1) 0(y 1) 1(z 0) œ z
(c) f(1ß 1ß 1) œ 14 , fx (1ß 1ß 1) œ "# , fy (1ß 1ß 1) œ "# , fz (1ß 1ß 1) œ "# Ê L(xß yß z) œ 14 "# (x 1) "# (y 1) "# (z 1)
œ "# x "# y "# z 14 #3
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834
Chapter 14 Partial Derivatives
45. f(xß yß z) œ xz 3yz 2 at P! (1ß 1ß 2) Ê f(1ß 1ß 2) œ 2; fx œ z, fy œ 3z, fz œ x 3y Ê L(xß yß z)
œ 2 2(x 1) 6(y 1) 2(z 2) œ 2x 6y 2z 6; fxx œ 0, fyy œ 0, fzz œ 0, fxy œ 0, fyz œ 3
Ê M œ 3; thus, kE(xß yß z)k Ÿ ˆ "# ‰ (3)(0.01 0.01 0.02)# œ 0.0024
46. f(xß yß z) œ x# xy yz "4 z# at P! (1ß 1ß 2) Ê f(1ß 1ß 2) œ 5; fx œ 2x y, fy œ x z, fz œ y "# z
Ê L(xß yß z) œ 5 3(x 1) 3(y 1) 2(z 2) œ 3x 3y 2z 5; fxx œ 2, fyy œ 0, fzz œ "# , fxy œ 1, fxz œ 0,
fyz œ 1 Ê M œ 2; thus kE(xß yß z)k Ÿ ˆ "# ‰ (2)(0.01 0.01 0.08)# œ 0.01
47. f(xß yß z) œ xy 2yz 3xz at P! (1ß 1ß 0) Ê f(1ß 1ß 0) œ 1; fx œ y 3z, fy œ x 2z, fz œ 2y 3x
Ê L(xß yß z) œ 1 (x 1) (y 1) (z 0) œ x y z 1; fxx œ 0, fyy œ 0, fzz œ 0, fxy œ 1, fxz œ 3,
fyz œ 2 Ê M œ 3; thus kE(xß yß z)k Ÿ ˆ "# ‰ (3)(0.01 0.01 0.01)# œ 0.00135
48. f(xß yß z) œ È2 cos x sin (y z) at P! ˆ0ß 0ß 14 ‰ Ê f ˆ0ß 0ß 14 ‰ œ 1; fx œ È2 sin x sin (y z),
fy œ È2 cos x cos (y z), fz œ È2 cos x cos (y z) Ê L(xß yß z) œ 1 0(x 0) (y 0) ˆz 14 ‰
œ y z 14 1; fxx œ È2 cos x sin (y z), fyy œ È2 cos x sin (y z), fzz œ È2 cos x sin (y z),
fxy œ È2 sin x cos (y z), fxz œ È2 sin x cos (y z), fyz œ È2 cos x sin (y z). The absolute value of
each of these second partial derivatives is bounded above by È2 Ê M œ È2; thus kE(xß yß z)k
Ÿ ˆ " ‰ ŠÈ2‹ (0.01 0.01 0.01)# œ 0.000636.
#
49. Tx (xß y) œ ey ey and Ty (xß y) œ x aey ey b Ê dT œ Tx (xß y) dx Ty (xß y) dy
œ aey ey b dx x aey ey b dy Ê dTkÐ#ßln 2Ñ œ 2.5 dx 3.0 dy. If kdxk Ÿ 0.1 and kdyk Ÿ 0.02, then the
maximum possible error in the computed value of T is (2.5)(0.1) (3.0)(0.02) œ 0.31 in magnitude.
#
21rh dr 1r dh
50. Vr œ 21rh and Vh œ 1r# Ê dV œ Vr dr Vh dh Ê dV
œ 2r dr "h dh; now ¸ drr † 100¸ Ÿ 1 and
1 r# h
V œ
¸ dh
¸
¸ dV
¸ ¸ˆ2 drr ‰ (100) ˆ dh
‰
¸
¸ dr
¸ ¸ dh
¸
h † 100 Ÿ 1 Ê
V † 100 Ÿ
h (100) Ÿ 2 r † 100 h † 100 Ÿ 2(1) 1 œ 3 Ê 3%
dy
51. dx
x Ÿ 0.02, y Ÿ 0.03
dy
2
(a) S œ 2x2 4xy Ê dS œ a4x 4ybdx 4x dy œ a4x2 4xyb dx
x 4xy y Ÿ a4x 4xyba0.02b a4xyba0.03b
œ 0.04a2x2 b 0.05a4xyb Ÿ 0.05a2x2 b 0.05a4xyb œ a0.05ba2x2 4xyb œ 0.05S
2 dy
2
2
2
(b) V œ x2 y Ê dV œ 2xy dx x2 dy œ 2x2 y dx
x x y y Ÿ a2x yba0.02b ax yba0.03b œ 0.07ax yb=0.07V
52. V œ 431 r3 1 r2 h Ê dV œ a41 r2 21 rhbdr 1 r2 dh; r œ 10, h œ 15, dr œ 12 and dh œ 0 Ê
dV œ Š41a10b2 21 a10ba15b‹ˆ 12 ‰ 1 a10b2 a0b œ 3501 cm3
53. Vr œ 21rh and Vh œ 1r# Ê dV œ Vr dr Vh dh Ê dV œ 21rh dr 1r# dh Ê dVkÐ5ß12Ñ œ 1201 dr 251 dh;
kdrk Ÿ 0.1 cm and kdhk Ÿ 0.1 cm Ê dV Ÿ (1201)(0.1) (251)(0.1) œ 14.51 cm$ ; V(5ß 12) œ 3001 cm$
1
Ê maximum percentage error is „ 14.5
3001 ‚ 100 œ „ 4.83%
54. (a)
#
"
"
"
R œ R" R#
#
Ê R"# dR œ R"# dR" R"# dR# Ê dR œ Š RR" ‹ dR" Š RR# ‹ dR#
"
#
"
"
(b) dR œ R# ’Š R"# ‹ dR" Š R"# ‹ dR# “ Ê dRk Ð100 400Ñ œ R# ’ (100)
# dR" (400)# dR# “ Ê R will be more
"
#
ß
"
"
sensitive to a variation in R" since (100)
# (400)#
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Section 14.6 Tangent Planes and Differentials
#
835
#
(c) From part (a), dR œ Š RR" ‹ dR" Š RR# ‹ dR# so that R" changing from 20 to 20.1 ohms Ê dR" œ 0.1 ohm
and R# changing from 25 to 24.9 ohms Ê dR# œ 0.1 ohms; R" œ R"" R"# Ê R œ 100
9 ohms
ˆ 100 ‰#
ˆ 100 ‰#
dR ¸
9
9
Ê dRk Ð20 25Ñ œ (20)
‚ 100
# (0.1) (25)# (0.1) ¸ 0.011 ohms Ê percentage change is R
Ð20 25Ñ
ß
ß
œ 0.011
ˆ 100 ‰ ‚ 100 ¸ 0.1%
9
55. A œ xy Ê dA œ x dy y dx; if x y then a 1-unit change in y gives a greater change in dA than a 1-unit change in x.
Thus, pay more attention to y which is the smaller of the two dimensions.
56. (a) fx (xß y) œ 2x(y 1) Ê fx (1ß 0) œ 2 and fy (xß y) œ x# Ê fy (1ß 0) œ 1 Ê df œ 2 dx 1 dy Ê df is more
sensitive to changes in x
dx
"
(b) df œ 0 Ê 2 dx dy œ 0 Ê 2 dx
dy 1 œ 0 Ê dy œ #
57. (a) r# œ x# y# Ê 2r dr œ 2x dx 2y dy Ê dr œ xr dx yr dy Ê dr|Ð$ß%Ñ œ ˆ 35 ‰ a „ 0.01b ˆ 45 ‰ a „ 0.01b
Š
y
#‹
x
1
x
¸ drr ‚ 100¸ œ ¸ „ 0.014
¸
œ „ 0.07
5 œ „ 0.014 Ê
5 ‚ 100 œ 0.28%; d) œ ˆ y ‰#
Š"‹
dx ˆ y ‰#x
0.04
3 ‰
œ y#yx# dx y# x x# dy Ê d)|Ð$ß%Ñ œ ˆ 254 ‰ a „ 0.01b ˆ 25
a „ 0.01b œ …25
„#0.03
5
x
1
dy
Ê maximum change in d) occurs when dx and dy have opposite signs (dx œ 0.01 and dy œ 0.01 or vice
„0.0028
" ˆ 4 ‰
¸ d)) ‚ 100¸ œ ¸ 0.927255218
versa) Ê d) œ „#0.07
‚ 100¸
5 ¸ „ 0.0028; ) œ tan
3 ¸ 0.927255218 Ê
¸ 0.30%
(b) the radius r is more sensitive to changes in y, and the angle ) is more sensitive to changes in x
58. (a) V œ 1r# h Ê dV œ 21rh dr 1r# dh Ê at r œ 1 and h œ 5 we have dV œ 101 dr 1 dh Ê the volume is
about 10 times more sensitive to a change in r
"
(b) dV œ 0 Ê 0 œ 21rh dr 1r# dh œ 2h dr r dh œ 10 dr dh Ê dr œ 10
dh; choose dh œ 1.5
Ê dr œ 0.15 Ê h œ 6.5 in. and r œ 0.85 in. is one solution for ?V ¸ dV œ 0
59. f(aß bß cß d) œ º
a b
œ ad bc Ê fa œ d, fb œ c, fc œ b, fd œ a Ê df œ d da c db b dc a dd; since
c dº
kak is much greater than kbk , kck , and kdk , the function f is most sensitive to a change in d.
60. ux œ ey , uy œ xey sin z, uz œ y cos z Ê du œ ey dx axey sin zb dy (y cos z) dz
Ê duk ˆ2ßln 3ß 12 ‰ œ 3 dx 7 dy 0 dz œ 3 dx 7 dy Ê magnitude of the maximum possible error
Ÿ 3(0.2) 7(0.6) œ 4.8
‰
61. QK œ "# ˆ 2KM
h
"Î#
" ˆ 2KM ‰"Î# ˆ 2KM ‰
ˆ 2M
‰ , QM œ "# ˆ 2KM
‰"Î# ˆ 2K
‰
h
h
h , and Qh œ #
h
h#
‰
Ê dQ œ "# ˆ 2KM
h
‰
œ "# ˆ 2KM
h
"Î#
ˆ 2M
‰ dK "# ˆ 2KM
‰"Î# ˆ 2K
‰ dM "# ˆ 2KM
‰"Î# ˆ 2KM
‰ dh
h
h
h
h
h#
2K
2KM
2M
‘
h dK h dM h# dh Ê dQk Ð2ß20ß0Þ0.05Ñ
"Î#
œ "# ’ (2)(2)(20)
0.05 “
"Î#
(2)(2)
(2)(2)(20)
’ (2)(20)
0.05 dK 0.05 dM (0.05)# dh“ œ (0.0125)(800 dK 80 dM 32,000 dh)
Ê Q is most sensitive to changes in h
62. A œ "# ab sin C Ê Aa œ "# b sin C, Ab œ "# a sin C, Ac œ "# ab cos C
Ê dA œ ˆ "# b sin C‰ da ˆ "# a sin C‰ db ˆ "# ab cos C‰ dC; dC œ k2°k œ k0.0349k radians, da œ k0.5k ft,
db œ k0.5k ft; at a œ 150 ft, b œ 200 ft, and C œ 60°, we see that the change is approximately
dA œ "# (200)(sin 60°) k0.5k "# (150)(sin 60°) k0.5k "# (200)(150)(cos 60°) k0.0349k œ „ 338 ft#
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836
Chapter 14 Partial Derivatives
63. z œ f(xß y) Ê g(xß yß z) œ f(xß y) z œ 0 Ê gx (xß yß z) œ fx (xß y), gy (xß yß z) œ fy (xß y) and gz (xß yß z) œ 1
Ê gx (x! ß y! ß f(x! ß y! )) œ fx (x! ß y! ), gy (x! ß y! ß f(x! ß y! )) œ fy (x! ß y! ) and gz (x! ß y! ß f(x! ß y! )) œ 1 Ê the tangent
plane at the point P! is fx (x! ß y! )(x x! ) fy (x! ß y! )(y y! ) [z f(x! ß y! )] œ 0 or
z œ fx (x! ß y! )(x x! ) fy (x! ß y! )(y y! ) f(x! ß y! )
64. ™ f œ 2xi 2yj œ 2(cos t t sin t)i 2(sin t t cos t)j and v œ (t cos t)i (t sin t)j Ê u œ kvvk
t)i (t sin t)j
œ È(t(tcos
œ (cos t)i (sin t)j since t 0 Ê (Du f)P! œ ™ f † u
cos t)# (t sin t)#
œ 2(cos t t sin t)(cos t) 2(sin t t cos t)(sin t) œ 2
65. ™ f œ 2xi 2yj 2zk œ (2 cos t)i (2 sin t)j 2tk and v œ ( sin t)i (cos t)j k Ê u œ kvvk
sin t)i (cos t)j k
t
"
œ È((sin
œ Š Èsin t ‹ i Š cos
È ‹ j È k Ê (Du f)P! œ ™ f † u
t)# (cos t)# 1#
2
2
2
t
"
2t
œ (2 cos t) Š Èsin2 t ‹ (2 sin t) Š cos
È2 ‹ (2t) Š È2 ‹ œ È2
(Du f) ˆ 14 ‰ œ
Ê (Du f) ˆ 41 ‰ œ 2È12 , (Du f)(0) œ 0 and
1
2È 2
66. r œ Èti Ètj 4" (t 3)k Ê v œ #" t"Î# i "# t"Î# j 4" k ; t œ 1 Ê x œ 1, y œ 1, z œ 1 Ê P! œ (1ß 1ß 1)
and v(1) œ "# i "# j "4 k ; f(xß yß z) œ x# y# z 3 œ 0 Ê ™ f œ 2xi 2yj k
Ê ™ f(1ß 1ß 1) œ 2i 2j k ; therefore v œ "4 ( ™ f) Ê the curve is normal to the surface
67. r œ Èti Ètj (2t 1)k Ê v œ "# t"Î# i "# t"Î# j 2k ; t œ 1 Ê x œ 1, y œ 1, z œ 1 Ê P! œ (1ß 1ß 1) and
v(1) œ "# i "# j 2k ; f(xß yß z) œ x# y# z 1 œ 0 Ê ™ f œ 2xi 2yj k Ê ™ f(1ß 1ß 1) œ 2i 2j k ;
now va1b † ™ fa1ß 1ß 1b œ 0, thus the curve is tangent to the surface when t œ 1
14.7 EXTREME VALUES AND SADDLE POINTS
1. fx (xß y) œ 2x y 3 œ 0 and fy (xß y) œ x 2y 3 œ 0 Ê x œ 3 and y œ 3 Ê critical point is (3ß 3);
#
œ 3 0 and fxx 0 Ê local minimum of
fxx (3ß 3) œ 2, fyy (3ß 3) œ 2, fxy (3ß 3) œ 1 Ê fxx fyy fxy
f(3ß 3) œ 5
2. fx (xß y) œ 2y 10x 4 œ 0 and fy (xß y) œ 2x 4y 4 œ 0 Ê x œ 23 and y œ 43 Ê critical point is ˆ 23 ß 43 ‰ ;
#
œ 36 0 and fxx 0 Ê local maximum of
fxx ˆ 23 ß 43 ‰ œ 10, fyy ˆ 23 ß 43 ‰ œ 4, fxy ˆ 23 ß 43 ‰ œ 2 Ê fxx fyy fxy
f ˆ 23 ß 43 ‰ œ 0
3. fx (xß y) œ 2x y 3 œ 0 and fy (xß y) œ x 2 œ 0 Ê x œ 2 and y œ 1 Ê critical point is (2ß 1);
#
œ 1 0 Ê saddle point
fxx (2ß 1) œ 2, fyy (2ß 1) œ 0, fxy (2ß 1) œ 1 Ê fxx fyy fxy
ˆ 6 69 ‰
4. fx (xß y) œ 5y 14x 3 œ 0 and fy (xß y) œ 5x 6 œ 0 Ê x œ 65 and y œ 69
#5 Ê critical point is 5 ß 25 ;
#
‰
ˆ 6 69 ‰
ˆ 6 69 ‰
fxx ˆ 65 ß 69
25 œ 14, fyy 5 ß 25 œ 0, fxy 5 ß 25 œ 5 Ê fxx fyy fxy œ 25 0 Ê saddle point
5. fx (xß y) œ 2y 2x 3 œ 0 and fy (xß y) œ 2x 4y œ 0 Ê x œ 3 and y œ 3# Ê critical point is ˆ3ß 32 ‰ ;
#
œ 4 0 and fxx 0 Ê local maximum of
fxx ˆ3ß 32 ‰ œ 2, fyy ˆ3ß 32 ‰ œ 4, fxy ˆ3ß 32 ‰ œ 2 Ê fxx fyy fxy
f ˆ3ß 3# ‰ œ 17
#
6. fx (xß y) œ 2x 4y œ 0 and fy (xß y) œ 4x 2y 6 œ 0 Ê x œ 2 and y œ 1 Ê critical point is (2ß 1);
#
œ 12 0 Ê saddle point
fxx (2ß 1) œ 2, fyy (2ß 1) œ 2, fxy (2ß 1) œ 4 Ê fxx fyy fxy
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Section 14.7 Extreme Values and Saddle Points
837
7. fx (xß y) œ 4x 3y 5 œ 0 and fy (xß y) œ 3x 8y 2 œ 0 Ê x œ 2 and y œ 1 Ê critical point is (2ß 1);
#
œ 23 0 and fxx 0 Ê local minimum of f(2ß 1) œ 6
fxx (2ß 1) œ 4, fyy (2ß 1) œ 8, fxy (2ß 1) œ 3 Ê fxx fyy fxy
8. fx (xß y) œ 2x 2y 2 œ 0 and fy (xß y) œ 2x 4y 2 œ 0 Ê x œ 1 and y œ 0 Ê critical point is (1ß 0);
#
œ 4 0 and fxx 0 Ê local minimum of f(1ß 0) œ 0
fxx (1ß 0) œ 2, fyy (1ß 0) œ 4, fxy (1ß 0) œ 2 Ê fxx fyy fxy
9. fx (xß y) œ 2x 2 œ 0 and fy (xß y) œ 2y 4 œ 0 Ê x œ 1 and y œ 2 Ê critical point is (1ß 2); fxx (1ß 2) œ 2,
#
œ 4 0 Ê saddle point
fyy (1ß 2) œ 2, fxy (1ß 2) œ 0 Ê fxx fyy fxy
10. fx (xß y) œ 2x 2y œ 0 and fy (xß y) œ 2x œ 0 Ê x œ 0 and y œ 0 Ê critical point is (0ß 0); fxx (0ß 0) œ 2,
#
œ 4 0 Ê saddle point
fyy (0ß 0) œ 0, fxy (0ß 0) œ 2 Ê fxx fyy fxy
8y
8x
‰
11. fx axß yb œ È56x2 112x
8 œ 0 and fy axß yb œ È56x2 8y
œ 0 Ê critical point is ˆ 16
2 16x 31
7 ß0 ;
8y2 16x 31
8
8
64
#
‰
ˆ 16 ‰
ˆ 16 ‰
fxx ˆ 16
7 ß 0 œ 15 , fyy 7 ß 0 œ 15 , fxy 7 ß 0 œ 0 Ê fxx fyy fxy œ 225 0 and fxx 0 Ê local maximum of
16
‰
fˆ 16
7 ß0 œ 7
12. fx axß yb œ
2x
œ 0 and fy axß yb œ 2 2y2 2Î3 œ 0 Ê there are no solutions to the system fx axß yb œ 0 and
3ax2 y2 b2Î3
3ax y b
fy axß yb œ 0, however, we must also consider where the partials are undefined, and this occurs when x œ 0 and y œ 0
Ê critical point is a0ß 0b. Note that the partial derivatives are defined at every other point other than a0ß 0b. We cannot use
the second derivative test, but this is the only possible local maximum, local minimum, or saddle point. faxß yb has a local
3
maximum of fa0ß 0b œ 1 at a0ß 0b since faxß yb œ 1 È
x2 y2 Ÿ 1 for all axß yb other than a0ß 0b.
13. fx (xß y) œ 3x# 2y œ 0 and fy (xß y) œ 3y# 2x œ 0 Ê x œ 0 and y œ 0, or x œ 23 and y œ 23 Ê critical points
are (0ß 0) and ˆ 23 ß 23 ‰ ; for (0ß 0): fxx (0ß 0) œ 6xk Ð0ß0Ñ œ 0, fyy (0ß 0) œ 6yk Ð0ß0Ñ œ 0, fxy (0ß 0) œ 2
#
Ê fxx fyy fxy
œ 4 0 Ê saddle point; for ˆ 32 ß 32 ‰ : fxx ˆ 32 ß 32 ‰ œ 4, fyy ˆ 32 ß 32 ‰ œ 4, fxy ˆ 32 ß 32 ‰ œ 2
#
Ê fxx fyy fxy
œ 12 0 and fxx 0 Ê local maximum of f ˆ 23 ß 32 ‰ œ 170
27
14. fx (xß y) œ 3x# 3y œ 0 and fy (xß y) œ 3x 3y# œ 0 Ê x œ 0 and y œ 0, or x œ 1 and y œ 1 Ê critical points
#
are (0ß 0) and (1ß 1); for (!ß !): fxx (0ß 0) œ 6xk Ð0ß0Ñ œ 0, fyy (0ß 0) œ 6yk Ð0ß0Ñ œ 0, fxy (0ß 0) œ 3 Ê fxx fyy fxy
#
œ 9 0 Ê saddle point; for (1ß 1): fxx (1ß 1) œ 6, fyy (1ß 1) œ 6, fxy (1ß 1) œ 3 Ê fxx fyy fxy
œ 27 0 and fxx 0 Ê local maximum of f(1ß 1) œ 1
15. fx (xß y) œ 12x 6x# 6y œ 0 and fy (xß y) œ 6y 6x œ 0 Ê x œ 0 and y œ 0, or x œ 1 and y œ 1 Ê critical
#
points are (0ß 0) and (1ß 1); for (!ß !): fxx (0ß 0) œ 12 12xk Ð0ß0Ñ œ 12, fyy (0ß 0) œ 6, fxy (0ß 0) œ 6 Ê fxx fyy fxy
œ 36 0 and fxx 0 Ê local minimum of f(0ß 0) œ 0; for (1ß 1): fxx (1ß 1) œ 0, fyy (1ß 1) œ 6,
#
fxy (1ß 1) œ 6 Ê fxx fyy fxy
œ 36 0 Ê saddle point
16. fx (xß y) œ 3x# 6x œ 0 Ê x œ 0 or x œ 2; fy (xß y) œ 3y# 6y œ 0 Ê y œ 0 or y œ 2 Ê the critical points are
(0ß 0), (0ß 2), (2ß 0), and (2ß 2); for (!ß !): fxx (0ß 0) œ 6x 6k Ð0ß0Ñ œ 6, fyy (0ß 0) œ 6y 6k Ð0ß0Ñ œ 6,
#
fxy (0ß 0) œ 0 Ê fxx fyy fxy
œ 36 0 Ê saddle point; for (0ß 2): fxx (0ß 2) œ 6, fyy (0ß 2) œ 6, fxy (0ß 2) œ 0
#
Ê fxx fyy fxy
œ 36 0 and fxx 0 Ê local minimum of f(0ß 2) œ 12; for (2ß 0): fxx (2ß 0) œ 6,
#
fyy (2ß 0) œ 6, fxy (2ß 0) œ 0 Ê fxx fyy fxy
œ 36 0 and fxx 0 Ê local maximum of f(2ß 0) œ 4;
#
for (2ß 2): fxx (2ß 2) œ 6, fyy (2ß 2) œ 6, fxy (2ß 2) œ 0 Ê fxx fyy fxy
œ 36 0 Ê saddle point
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Chapter 14 Partial Derivatives
17. fx axß yb œ 3x2 3y2 15 œ 0 and fy axß yb œ 6x y 3y2 15 œ 0 Ê critical points are a2ß 1b, a2ß 1b, Š0ß È5‹, and
Š0ß È5‹; for a2ß 1b: fxx a2ß 1b œ 6xk a2ß1b œ 12, fyy a2ß 1b œ a6x 6ybk a2ß1b œ 18, fxy a2ß 1b œ 6yk a2ß1b œ 6
#
Ê fxx fyy fxy
œ 180 0 and fxx 0 Ê local minimum of fa2ß 1b œ 30; for a2ß 1b: fxx a2ß 1b œ 6xk a2ß1b
#
œ 12, fyy a2ß 1b œ a6x 6ybk a2ß1b œ 18, fxy a2ß 1b œ 6yk a2ß1b œ 6 Ê fxx fyy fxy
œ 180 0 and
fxx 0 Ê local maximum of fa2ß 1b œ 30; for Š0ß È5‹: fxx Š0ß È5‹ œ 6x¹
œ a6x 6ybk Š0ßÈ5‹ œ 6È5, fxy Š0ß È5‹ œ 6y¹
for Š0ß È5‹: fxx Š0ß È5‹ œ 6x¹
fxy Š0ß È5‹ œ 6y¹
Š0ßÈ5‹
Š0ßÈ5‹
Š0ßÈ5‹
Š0ßÈ5‹
œ 0, fyy Š0ß È5‹
#
œ 6È5 Ê fxx fyy fxy
œ 180 0 Ê saddle pointà
œ 0, fyy Š0ß È5‹ œ a6x 6ybk Š0ßÈ5‹ œ 6È5,
#
œ 6È5 Ê fxx fyy fxy
œ 180 0 Ê saddle point.
18. fx (xß y) œ 6x# 18x œ 0 Ê 6x(x 3) œ 0 Ê x œ 0 or x œ 3; fy (xß y) œ 6y# 6y 12 œ 0 Ê 6(y 2)(y 1) œ 0
Ê y œ 2 or y œ 1 Ê the critical points are (0ß 2), (0ß 1), (3ß 2), and (3ß 1); fxx (xß y) œ 12x 18,
fyy (xß y) œ 12y 6, and fxy (xß y) œ 0; for (!ß 2): fxx (0ß 2) œ 18, fyy (0ß 2) œ 18, fxy (0ß 2) œ 0
#
Ê fxx fyy fxy
œ 324 0 and fxx 0 Ê local maximum of f(0ß 2) œ 20; for (0ß 1): fxx (0ß 1) œ 18,
#
fyy (0ß 1) œ 18, fxy (0ß 1) œ 0 Ê fxx fyy fxy
œ 324 0 Ê saddle point; for (3ß 2): fxx (3ß 2) œ 18,
#
fyy (3ß 2) œ 18, fxy (3ß 2) œ 0 Ê fxx fyy fxy
œ 324 0 Ê saddle point; for (3ß 1): fxx (3ß 1) œ 18,
#
fyy (3ß 1) œ 18, fxy (3ß 1) œ 0 Ê fxx fyy fxy
œ 324 0 and fxx 0 Ê local minimum of f(3ß 1) œ 34
19. fx (xß y) œ 4y 4x$ œ 0 and fy (xß y) œ 4x 4y$ œ 0 Ê x œ y Ê x a1 x# b œ 0 Ê x œ 0, 1, 1 Ê the critical
points are (0ß 0), (1ß 1), and (1ß 1); for (!ß !): fxx (0ß 0) œ 12x# k Ð0ß0Ñ œ 0, fyy (0ß 0) œ 12y# k Ð0ß0Ñ œ 0,
#
fxy (0ß 0) œ 4 Ê fxx fyy fxy
œ 16 0 Ê saddle point; for (1ß 1): fxx (1ß 1) œ 12, fyy (1ß 1) œ 12, fxy (1ß 1) œ 4
#
Ê fxx fyy fxy
œ 128 0 and fxx 0 Ê local maximum of f(1ß 1) œ 2; for (1ß 1): fxx (1ß 1) œ 12,
#
fyy (1ß 1) œ 12, fxy (1ß 1) œ 4 Ê fxx fyy fxy
œ 128 0 and fxx 0 Ê local maximum of f(1ß 1) œ 2
20. fx (xß y) œ 4x$ 4y œ 0 and fy (xß y) œ 4y$ 4x œ 0 Ê x œ y Ê x$ x œ 0 Ê x a1 x# b œ 0 Ê x œ 0, 1, 1
Ê the critical points are (0ß 0), (1ß 1), and (1ß 1); fxx (xß y) œ 12x# , fyy (xß y) œ 12y# , and fxy (xß y) œ 4;
#
for (!ß 0): fxx (0ß 0) œ 0, fyy (0ß 0) œ 0, fxy (0ß 0) œ 4 Ê fxx fyy fxy
œ 16 0 Ê saddle point; for (1ß 1):
#
fxx (1ß 1) œ 12, fyy (1ß 1) œ 12, fxy (1ß 1) œ 4 Ê fxx fyy fxy
œ 128 0 and fxx 0 Ê local minimum of
#
f("ß 1) œ 2; for (1ß 1): fxx (1ß 1) œ 12, fyy (1ß 1) œ 12, fxy (1ß 1) œ 4 Ê fxx fyy fxy
œ 128 0 and
fxx 0 Ê local minimum of f(1ß 1) œ 2
21. fx (xß y) œ ax# y2x
œ 0 and fy (xß y) œ ax# y2y
œ 0 Ê x œ 0 and y œ 0 Ê the critical point is (!ß 0);
# 1b#
# 1b#
#
#
#
#
fxx œ a4xx# y2y# 1b2$ , fyy œ ax2x# y#4y 1b$2 , fxy œ ax# 8xy
; fxx (!ß !) œ 2, fyy (0ß 0) œ 2, fxy (0ß 0) œ 0
y# 1b$
#
Ê fxx fyy fxy
œ 4 0 and fxx 0 Ê local maximum of f(0ß 0) œ 1
22. fx (xß y) œ x1# y œ 0 and fy (xß y) œ x y1# œ 0 Ê x œ 1 and y œ 1 Ê the critical point is (1ß 1); fxx œ x2$ , fyy œ y2$ ,
#
fxy œ 1; fxx (1ß 1) œ 2, fyy (1ß 1) œ 2, fxy (1ß 1) œ 1 Ê fxx fyy fxy
œ 3 0 and fxx 2 Ê local minimum of f(1ß 1) œ 3
23. fx (xß y) œ y cos x œ 0 and fy (xß y) œ sin x œ 0 Ê x œ n1, n an integer, and y œ 0 Ê the critical points are
(n1ß 0), n an integer (Note: cos x and sin x cannot both be 0 for the same x, so sin x must be 0 and y œ 0);
fxx œ y sin x, fyy œ 0, fxy œ cos x; fxx (n1ß 0) œ 0, fyy (n1ß 0) œ 0, fxy (n1ß 0) œ 1 if n is even and fxy (n1ß 0) œ 1
#
if n is odd Ê fxx fyy fxy
œ 1 0 Ê saddle point.
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Section 14.7 Extreme Values and Saddle Points
839
24. fx (xß y) œ 2e2x cos y œ 0 and fy (xß y) œ e2x sin y œ 0 Ê no solution since e2x Á 0 for any x and the functions
cos y and sin y cannot equal 0 for the same y Ê no critical points Ê no extrema and no saddle points
25. fx axß yb œ a2x 4bex y 4x œ 0 and fy axß yb œ 2yex y 4x œ 0 Ê critical point is a2ß 0b; fxx a2ß 0b œ e24 , fxy a2ß 0b œ 0,
2
2
2
2
#
fyy a2ß 0b œ e24 Ê fxx fyy fxy
œ e48 0 and fxx 0 Ê local mimimum of fa2ß 0b œ e14
26. fx axß yb œ yex œ 0 and fy axß yb œ ey ex œ 0 Ê critical point is a0ß 0b; fxx a2ß 0b œ 0, fxy a2ß 0b œ 1, fyy a2ß 0b œ 1
#
Ê fxx fyy fxy
œ 1 0 Ê saddle point
27. fx axß yb œ 2xey œ 0 and fy axß yb œ 2yey ey ax2 y2 b œ 0 Ê critical points are a0ß 0b and a0ß 2b; for a0ß 0b:
fxx a0ß 0b œ 2ey k a0ß0b œ 2, fyy a0ß 0b œ a2ey 4yey ey ax2 y2 bbk a0ß0b œ 2, fxy a0ß 0b œ 2xey k a0ß0b œ 0
#
Ê fxx fyy fxy
œ 4 0 and fxx 0 Ê local mimimum of fa0ß 0b œ 0; for a0ß 2b: fxx a0ß 2b œ 2ey k a0ß2b œ e22 ,
#
fyy a0ß 2b œ a2ey 4yey ey ax2 y2 bbk a0ß2b œ e22 , fxy a0ß 2b œ 2xey k a0ß2b œ 0 Ê fxx fyy fxy
œ e44 0
Ê saddle point
28. fx axß yb œ ex ax2 2x y2 b œ 0 and fy axß yb œ 2yex œ 0 Ê critical points are a0ß 0b and a2ß 0b; for a0ß 0b:
fxx a0ß 0b œ ex ax2 4x 2 y2 bk a0ß0b œ 2, fyy a0ß 0b œ 2ex k a0ß0b œ 2, fxy a0ß 0b œ 2yex k a0ß0b œ 0
#
Ê fxx fyy fxy
œ 4 0 and fxx 0 Ê saddle point; for a2ß 0b: fxx a2ß 0b œ ex ax2 4x 2 y2 bk a2ß0b œ e22 ,
#
fyy a2ß 0b œ 2ex k a2ß0b œ e22 , fxy a2ß 0b œ 2yex k a2ß0b œ 0 Ê fxx fyy fxy
œ e44 0 and fxx 0 Ê local maximum
of fa2ß 0b œ e42
29. fx axß yb œ 4 x2 œ 0 and fy axß yb œ 1 y1 œ 0 Ê critical point is ˆ 21 , 1‰ ; fxx ˆ 21 , 1‰ œ 8, fyy ˆ 12 , 1‰ œ 1,
#
œ 8 0 and fxx 0 Ê local maximum of fˆ 12 , 1‰ œ 3 2ln 2
fxy ˆ 12 , 1‰ œ 0 Ê fxx fyy fxy
30. fx axß yb œ 2x x 1 y œ 0 and fy axß yb œ 1 x 1 y œ 0 Ê critical point is ˆ 21 , 23 ‰ ; fxx ˆ 21 , 23 ‰ œ 1, fyy ˆ 12 , 32 ‰ œ 1,
#
œ 2 0 Ê saddle point
fxy ˆ 12 , 32 ‰ œ 1 Ê fxx fyy fxy
On OA, f(xß y) œ f(0ß y) œ y# 4y 1 on 0 Ÿ y Ÿ 2;
f w (0ß y) œ 2y 4 œ 0 Ê y œ 2;
f(0ß 0) œ 1 and f(!ß #) œ 3
(ii) On AB, f(xß y) œ f(xß 2) œ 2x# 4x 3 on 0 Ÿ x Ÿ 1;
f w (xß 2) œ 4x 4 œ 0 Ê x œ 1;
f(0ß 2) œ 3 and f(1ß #) œ 5
(iii) On OB, f(xß y) œ f(xß 2x) œ 6x# 12x 1 on
0 Ÿ x Ÿ 1; endpoint values have been found above;
f w (xß 2x) œ 12x 12 œ 0 Ê x œ 1 and y œ 2, but ("ß #) is not an interior point of OB
(iv) For interior points of the triangular region, fx (xß y) œ 4x 4 œ 0 and fy (xß y) œ 2y 4 œ 0
Ê x œ 1 and y œ 2, but (1ß 2) is not an interior point of the region. Therefore, the absolute maximum is
1 at (0ß 0) and the absolute minimum is 5 at ("ß #).
31. (i)
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840
Chapter 14 Partial Derivatives
On OA, D(xß y) œ D(0ß y) œ y# 1 on 0 Ÿ y Ÿ 4;
Dw (0ß y) œ 2y œ 0 Ê y œ 0; D(!ß !) œ 1 and
D(!ß %) œ 17
(ii) On AB, D(xß y) œ D(xß 4) œ x# 4x 17 on
0 Ÿ x Ÿ 4; Dw (xß 4) œ 2x 4 œ 0 Ê x œ 2 and (2ß 4)
is an interior point of AB; D(#ß %) œ 13 and
D(%ß %) œ D(!ß %) œ 17
(iii) On OB, D(xß y) œ D(xß x) œ x# 1 on 0 Ÿ x Ÿ 4;
Dw (xß x) œ 2x œ 0 Ê x œ 0 and y œ 0, which is not an interior point of OB; endpoint values have been found
above
(iv) For interior points of the triangular region, fx (xß y) œ 2x y œ 0 and fy (xß y) œ x 2y œ 0 Ê x œ 0 and y œ 0,
which is not an interior point of the region. Therefore, the absolute maximum is 17 at (!ß %) and (%ß %), and the
absolute minimum is 1 at (0ß 0).
32. (i)
On OA, f(xß y) œ f(!ß y) œ y# on 0 Ÿ y Ÿ 2;
f w (0ß y) œ 2y œ 0 Ê y œ 0 and x œ 0; f(0ß 0) œ 0 and
f(0ß #) œ 4
(ii) On OB, f(xß y) œ f(xß 0) œ x# on 0 Ÿ x Ÿ 1;
f w (xß 0) œ 2x œ 0 Ê x œ 0 and y œ 0; f(0ß 0) œ 0 and
f(1ß 0) œ 1
(iii) On AB, f(xß y) œ f(xß 2x 2) œ 5x# 8x 4 on
0 Ÿ x Ÿ 1; f w (xß 2x 2) œ 10x 8 œ 0 Ê x œ 45
and y œ 25 ; f ˆ 45 ß 25 ‰ œ 45 ; endpoint values have been found above.
33. (i)
(iv) For interior points of the triangular region, fx (xß y) œ 2x œ 0 and fy (xß y) œ 2y œ 0 Ê x œ 0 and y œ 0, but (!ß 0) is
not an interior point of the region. Therefore the absolute maximum is 4 at (0ß 2) and the absolute minimum is 0 at
(0ß 0).
34. (i)
(ii)
On AB, T(xß y) œ T(!ß y) œ y# on 3 Ÿ y Ÿ 3;
Tw (0ß y) œ 2y œ 0 Ê y œ 0 and x œ 0; T(0ß 0) œ 0,
T(!ß 3) œ 9, and T(!ß 3) œ 9
On BC, T(xß y) œ T(xß 3) œ x# 3x 9 on 0 Ÿ x Ÿ 5;
Tw (xß 3) œ 2x 3 œ 0 Ê x œ 3# and y œ 3;
T ˆ 3# ß 3‰ œ 27
4 and T(5ß 3) œ 19
(iii) On CD, T(xß y) œ T(5ß y) œ y# 5y 5 on
3 Ÿ y Ÿ 3;Tw (5ß y) œ 2y 5 œ 0 Ê y œ 5# and
x œ 5;T ˆ5ß 5# ‰ œ 45
4 , T(&ß 3) œ 11 and T(5ß 3) œ 19
(iv) On AD, T(xß y) œ T(xß 3) œ x# 9x 9 on 0 Ÿ x Ÿ 5; Tw (xß 3) œ 2x 9 œ 0 Ê x œ 9# and y œ 3;
T ˆ 9# ß 3‰ œ 45
4 , T(!ß 3) œ 9 and T(&ß 3) œ 11
(v)
For interior points of the rectangular region, Tx (xß y) œ 2x y 6 œ 0 and Ty (xß y) œ x 2y œ 0 Ê x œ 4
and y œ 2 Ê (4ß 2) is an interior critical point with T(4ß 2) œ 12. Therefore the absolute maximum
is 19 at (5ß 3) and the absolute minimum is 12 at (4ß 2).
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 14.7 Extreme Values and Saddle Points
35. (i)
(ii)
841
On OC, T(xß y) œ T(xß 0) œ x# 6x 2 on
0 Ÿ x Ÿ 5; Tw (xß 0) œ 2x 6 œ 0 Ê x œ 3 and
y œ 0; T(3ß 0) œ 7, T(0ß 0) œ 2, and T(5ß 0) œ 3
On CB, T(xß y) œ T(5ß y) œ y# 5y 3 on
3 Ÿ y Ÿ 0; Tw (5ß y) œ 2y 5 œ 0 Ê y œ 5# and
x œ 5; T ˆ5ß 5# ‰ œ 37
4 and T(5ß 3) œ 9
(iii) On AB, T(xß y) œ T(xß 3) œ x# 9x 11 on
0 Ÿ x Ÿ 5; Tw (xß 3) œ 2x 9 œ 0 Ê x œ 9# and
y œ 3; T ˆ 9# ß 3‰ œ 37
4 and T(!ß 3) œ 11
(iv) On AO, T(xß y) œ T(!ß y) œ y# 2 on 3 Ÿ y Ÿ 0; Tw (0ß y) œ 2y œ 0 Ê y œ 0 and x œ 0, but (0ß 0) is
not an interior point of AO
(v) For interior points of the rectangular region, Tx (xß y) œ 2x y 6 œ 0 and Ty (xß y) œ x 2y œ 0 Ê x œ 4
and y œ 2, an interior critical point with T(%ß 2) œ 10. Therefore the absolute maximum is 11 at
(!ß 3) and the absolute minimum is 10 at (4ß 2).
36. (i)
(ii)
On OA, f(xß y) œ f(!ß y) œ 24y# on 0 Ÿ y Ÿ 1;
f w (0ß y) œ 48y œ 0 Ê y œ 0 and x œ 0, but (0ß 0) is
not an interior point of OA; f(!ß 0) œ 0 and
f(!ß 1) œ 24
On AB, f(xß y) œ f(xß 1) œ 48x 32x$ 24 on
0 Ÿ x Ÿ 1; f w (xß 1) œ 48 96x# œ 0 Ê x œ È"2 and
y œ 1, or x œ È"2 and y œ 1, but Š È"2 ß 1‹ is not in
the interior of AB; f Š È"2 ß 1‹ œ 16È2 24 and f(1ß 1) œ 8
(iii) On BC, f(xß y) œ f("ß y) œ 48y 32 24y# on 0 Ÿ y Ÿ 1; f w ("ß y) œ 48 48y œ 0 Ê y œ 1 and x œ 1, but
("ß ") is not an interior point of BC; f("ß 0) œ 32 and f("ß ") œ 8
(iv) On OC, f(xß y) œ f(xß 0) œ 32x$ on 0 Ÿ x Ÿ 1; f w (xß 0) œ 96x# œ 0 Ê x œ 0 and y œ 0, but (0ß 0) is not an
interior point of OC; f(!ß 0) œ 0 and f("ß 0) œ 32
(v) For interior points of the rectangular region, fx (xß y) œ 48y 96x# œ 0 and fy (xß y) œ 48x 48y œ 0
Ê x œ 0 and y œ 0, or x œ "# and y œ "# , but (0ß 0) is not an interior point of the region; f ˆ "# ß "# ‰ œ 2.
Therefore the absolute maximum is 2 at ˆ "# ß "# ‰ and the absolute minimum is 32 at (1ß 0).
37. (i)
On AB, f(xß y) œ f(1ß y) œ 3 cos y on 14 Ÿ y Ÿ 14 ;
f w (1ß y) œ 3 sin y œ 0 Ê y œ 0 and x œ 1;
È
È
f("ß 0) œ 3, f ˆ1ß 14 ‰ œ 3 # 2 , and f ˆ1ß 14 ‰ œ 3 # 2
(ii)
On CD, f(xß y) œ f($ß y) œ 3 cos y on 14 Ÿ y Ÿ 14 ;
f w (3ß y) œ 3 sin y œ 0 Ê y œ 0 and x œ 3;
È
È
f(3ß 0) œ 3, f ˆ3ß 14 ‰ œ 3 # 2 and f ˆ3ß 14 ‰ œ 3 # 2
(iii) On BC, f(xß y) œ f ˆxß 14 ‰ œ
È2
#
# a4x x b on
È
1 Ÿ x Ÿ 3; f w ˆxß 14 ‰ œ È2(2 x) œ 0 Ê x œ 2 and y œ 14 ; f ˆ2ß 14 ‰ œ 2È2, f ˆ1ß 14 ‰ œ 3 # 2 , and
È
f ˆ3ß 14 ‰ œ 3 # 2
È2
1‰
#
wˆ
È2(2 x) œ 0
# a4x x b on 1 Ÿ x Ÿ 3; f xß 4 œ
È
È
f ˆ2ß 14 ‰ œ 2È2, f ˆ1ß 14 ‰ œ 3 # 2 , and f ˆ3ß 14 ‰ œ 3 # 2
(iv) On AD, f(xß y) œ f ˆxß 14 ‰ œ
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Ê x œ 2 and y œ 14 ;
842
Chapter 14 Partial Derivatives
(v)
For interior points of the region, fx (xß y) œ (4 2x) cos y œ 0 and fy (xß y) œ a4x x# b sin y œ 0 Ê x œ 2
and y œ 0, which is an interior critical point with f(2ß 0) œ 4. Therefore the absolute maximum is 4 at
È
(2ß 0) and the absolute minimum is 3 # 2 at ˆ3ß 14 ‰ , ˆ3ß 14 ‰ , ˆ1ß 14 ‰ , and ˆ1ß 14 ‰ .
On OA, f(xß y) œ f(!ß y) œ 2y 1 on 0 Ÿ y Ÿ 1;
f w (0ß y) œ 2 Ê no interior critical points; f(0ß 0) œ 1
and f(0ß 1) œ 3
(ii) On OB, f(xß y) œ f(xß 0) œ 4x 1 on 0 Ÿ x Ÿ 1;
f w (xß 0) œ 4 Ê no interior critical points; f(1ß 0) œ 5
(iii) On AB, f(xß y) œ f(xß x 1) œ 8x# 6x 3 on
0 Ÿ x Ÿ 1; f w (xß x 1) œ 16x 6 œ 0 Ê x œ 38
and y œ 58 ; f ˆ 38 ß 58 ‰ œ 15
8 , f(0ß 1) œ 3, and f("ß 0) œ 5
38. (i)
(iv) For interior points of the triangular region, fx (xß y) œ 4 8y œ 0 and fy (xß y) œ 8x 2 œ 0
Ê y œ "# and x œ 4" which is an interior critical point with f ˆ 4" ß #" ‰ œ 2. Therefore the absolute maximum is 5 at
(1ß 0) and the absolute minimum is 1 at (0ß 0).
39. Let F(aß b) œ 'a a6 x x# b dx where a Ÿ b. The boundary of the domain of F is the line a œ b in the ab-plane, and
b
F(aß a) œ 0, so F is identically 0 on the boundary of its domain. For interior critical points we have:
`F
`F
#
#
` a œ a6 a a b œ 0 Ê a œ 3, 2 and ` b œ a6 b b b œ 0 Ê b œ 3, 2. Since a Ÿ b, there is only one
interior critical point (3ß 2) and F(3ß 2) œ 'c3 a6 x x# b dx gives the area under the parabola y œ 6 x x# that is
2
above the x-axis. Therefore, a œ 3 and b œ 2.
40. Let F(aß b) œ 'a a24 2x x# b
b
"Î$
dx where a Ÿ b. The boundary of the domain of F is the line a œ b and on this line F is
identically 0. For interior critical points we have: ``Fa œ a24 2a a# b
`F
# "Î$
œ0
` b œ a24 2b b b
"Î$
œ 0 Ê a œ 4, 6 and
Ê b œ 4, 6. Since a Ÿ b, there is only one critical point (6ß 4) and
F(6ß 4) œ 'c6 a24 2x x# b dx gives the area under the curve y œ a24 2x x# b
4
"Î$
that is above the x-axis.
Therefore, a œ 6 and b œ 4.
41. Tx (xß y) œ 2x 1 œ 0 and Ty (xß y) œ 4y œ 0 Ê x œ "# and y œ 0 with T ˆ "# ß 0‰ œ 4" ; on the boundary
È
x# y# œ 1: T(xß y) œ x# x 2 for 1 Ÿ x Ÿ 1 Ê Tw (xß y) œ 2x 1 œ 0 Ê x œ "# and y œ „ #3 ;
È
È
È
T Š "# ß #3 ‹ œ 94 , T Š #" ß #3 ‹ œ 94 , T(1ß 0) œ 2, and T("ß 0) œ 0 Ê the hottest is 2 4" ° at Š #" ß #3 ‹ and
È
Š "# ß #3 ‹ ; the coldest is "4 ° at ˆ "# ß 0‰ .
42. fx (xß y) œ y 2 2x œ 0 and fy (xß y) œ x "y œ 0 Ê x œ "# and y œ 2; fxx ˆ "# ß 2‰ œ x2# ¸ ˆ 1 ß2‰ œ 8,
2
fyy ˆ #" ß 2‰ œ y"# ¹ 1 œ 4" , fxy ˆ "# ß 2‰ œ 1
ˆ ß2‰
2
Ê
#
fxx fyy fxy
œ 1 0 and fxx 0
Ê a local minimum of f ˆ "# ß 2‰
œ 2 ln "# œ 2 ln 2
43. (a) fx (xß y) œ 2x 4y œ 0 and fy (xß y) œ 2y 4x œ 0 Ê x œ 0 and y œ 0; fxx (0ß 0) œ 2, fyy (0ß 0) œ 2,
#
œ 12 0 Ê saddle point at (0ß 0)
fxy (0ß 0) œ 4 Ê fxx fyy fxy
(b) fx (xß y) œ 2x 2 œ 0 and fy (xß y) œ 2y 4 œ 0 Ê x œ 1 and y œ 2; fxx (1ß 2) œ 2, fyy (1ß 2) œ 2,
#
œ 4 0 and fxx 0 Ê local minimum at ("ß #)
fxy (1ß 2) œ 0 Ê fxx fyy fxy
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Section 14.7 Extreme Values and Saddle Points
843
(c) fx (xß y) œ 9x# 9 œ 0 and fy (xß y) œ 2y 4 œ 0 Ê x œ „ 1 and y œ 2; fxx (1ß 2) œ 18xk Ð1ß2Ñ œ 18,
#
œ 36 0 and fxx 0 Ê local minimum at ("ß #);
fyy (1ß 2) œ 2, fxy (1ß 2) œ 0 Ê fxx fyy fxy
#
fxx (1ß 2) œ 18, fyy ("ß 2) œ 2, fxy ("ß 2) œ 0 Ê fxx fyy fxy
œ 36 0 Ê saddle point at ("ß 2)
44. (a) Minimum at (0ß 0) since f(xß y) 0 for all other (xß y)
(b) Maximum of 1 at (!ß !) since f(xß y) 1 for all other (xß y)
(c) Neither since f(xß y) 0 for x 0 and f(xß y) 0 for x 0
(d) Neither since f(xß y) 0 for x 0 and f(xß y) 0 for x 0
(e) Neither since f(xß y) 0 for x 0 and y 0, but f(xß y) 0 for x 0 and y 0
(f) Minimum at (0ß 0) since f(xß y) 0 for all other (xß y)
45. If k œ 0, then f(xß y) œ x# y# Ê fx (xß y) œ 2x œ 0 and fy (xß y) œ 2y œ 0 Ê x œ 0 and y œ 0 Ê (0ß 0) is the only
critical point. If k Á 0, fx (xß y) œ 2x ky œ 0 Ê y œ 2k x; fy (xß y) œ kx 2y œ 0 Ê kx 2 ˆ 2k x‰ œ 0
4‰
ˆ
ˆ 2‰
Ê kx 4x
k œ 0 Ê k k x œ 0 Ê x œ 0 or k œ „ 2 Ê y œ k (0) œ 0 or y œ „ x; in any case (0ß 0) is a
critical point.
#
46. (See Exercise 45 above): fxx (xß y) œ 2, fyy (xß y) œ 2, and fxy (xß y) œ k Ê fxx fyy fxy
œ 4 k# ; f will have a saddle point
at (0ß 0) if 4 k# 0 Ê k 2 or k 2; f will have a local minimum at (0ß 0) if 4 k# 0 Ê 2 k 2; the test is
inconclusive if 4 k# œ 0 Ê k œ „ 2.
47. No; for example f(xß y) œ xy has a saddle point at (aß b) œ (0ß 0) where fx œ fy œ 0.
#
48. If fxx (aß b) and fyy (aß b) differ in sign, then fxx (aß b) fyy (aß b) 0 so fxx fyy fxy
0. The surface must therefore have a
saddle point at (aß b) by the second derivative test.
49. We want the point on z œ 10 x# y# where the tangent plane is parallel to the plane x 2y 3z œ 0. To find a normal
vector to z œ 10 x# y# let w œ z x# y# 10. Then ™ w œ 2xi 2yj k is normal to z œ 10 x# y# at
(xß y). The vector ™ w is parallel to i 2j 3k which is normal to the plane x 2y 3z œ 0 if
"
‰
6xi 6yj 3k œ i 2j 3k or x œ "6 and y œ "3 . Thus the point is ˆ "6 ß "3 ß 10 36
9" ‰ or ˆ 6" ß 3" ß 355
36 .
50. We want the point on z œ x# y# 10 where the tangent plane is parallel to the plane x 2y z œ 0. Let
w œ z x# y# 10, then ™ w œ 2xi 2yj k is normal to z œ x# y# 10 at (xß y). The vector ™ w is parallel
‰
to i 2j k which is normal to the plane if x œ "# and y œ 1. Thus the point ˆ "# ß 1ß 4" 1 10‰ or ˆ #" ß 1ß 45
4 is the point
on the surface z œ x# y# 10 nearest the plane x 2y z œ 0.
51. daxß yß zb œ Éax 0b2 ay 0b2 az 0b2 Ê we can minimize daxß yß zb by minimizing Daxß yß zb œ x2 y2 z2 ;
3x 2y z œ 6 Ê z œ 6 3x 2y Ê Daxß yb œ x2 y2 a6 3x 2yb2 Ê Dx axß yb œ 2x 6a6 3x 2yb œ 0
and Dy axß yb œ 2y 4a6 3x 2yb œ 0 Ê critical point is ˆ 97 , 67 ‰ Ê z œ 37 ; Dxx ˆ 97 , 67 ‰ œ 20, Dyy ˆ 12 , 1‰ œ 10,
È
Dxy ˆ 12 , 1‰ œ 12 Ê Dxx Dyy D#xy œ 56 0 and Dxx 0 Ê local minimum of dˆ 97 , 67 , 37 ‰ œ 3 714
52. daxß yß zb œ Éax 2b2 ay 1b2 az 1b2 Ê we can minimize daxß yß zb by minimizing
Daxß yß zb œ ax 2b2 ay 1b2 az 1b2 ; x y z œ 2 Ê z œ x y 2
Ê Daxß yb œ ax 2b2 ay 1b2 ax y 3b2 Ê Dx axß yb œ 2ax 2b 2ax y 3b œ 0
and Dy axß yb œ 2ay 1b 2ax y 3b œ 0 Ê critical point is ˆ 83 , 13 ‰ Ê z œ 13 ; Dxx ˆ 83 , 13 ‰ œ 4, Dyy ˆ 83 , 13 ‰ œ 4,
Dxy ˆ 83 , 13 ‰ œ 2 Ê Dxx Dyy D#xy œ 12 0 and Dxx 0 Ê local minimum of dˆ 83 , 13 , 13 ‰ œ È2
3
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844
Chapter 14 Partial Derivatives
53. saxß yß zb œ x2 y2 z2 ; x y z œ 9 Ê z œ 9 x y Ê saxß yb œ x2 y2 a9 x yb2
Ê sx axß yb œ 2x 2a9 x yb œ 0 and sy axß yb œ 2y 2a9 x yb œ 0 Ê critical point is a3, 3b Ê z œ 3;
sxx a3, 3b œ 4, syy a3, 3b œ 4, sxy a3, 3b œ 2 Ê sxx syy s#xy œ 12 0 and sxx 0 Ê local minimum of sa3, 3, 3b œ 27
54. paxß yß zb œ xyz; x y z œ 3 Ê z œ 3 x y Ê paxß yb œ x ya3 x yb œ 3x y x2 y x y2
Ê px axß yb œ 3y 2xy y2 œ 0 and py axß yb œ 3x x2 2xy œ 0 Ê critical points are a0, 0b, a0, 3b, a3, 0b, and
a1, 1b; for a0, 0b Ê z œ 3; pxx a0, 0b œ 0, pyy a0, 0b œ 0, pxy a0, 0b œ 3 Ê pxx pyy p#xy œ 9 0 Ê saddle point;
for a0, 3b Ê z œ 0; pxx a0, 3b œ 6, pyy a0, 3b œ 0, pxy a0, 3b œ 3 Ê pxx pyy p#xy œ 9 0 Ê saddle point;
for a3, 0b Ê z œ 0; pxx a3, 0b œ 0, pyy a3, 0b œ 6, pxy a3, 0b œ 3 Ê pxx pyy p#xy œ 9 0 Ê saddle point;
for a1, 1b Ê z œ 1; pxx a1, 1b œ 2, pyy a1, 1b œ 2, pxy a1, 1b œ 1 Ê pxx pyy p#xy œ 3 0 and pxx 0 Ê local
maximum of pa1, 1, 1b œ 1
55. saxß yß zb œ xy yz xz; x y z œ 6 Ê z œ 6 x y Ê saxß yb œ xy ya6 x yb xa6 x yb
œ 6x 6y xy x2 y2 Ê sx axß yb œ 6 2x y œ 0 and sy axß yb œ 6 x 2y œ 0 Ê critical point is a2, 2b
Ê z œ 2; sxx a2, 2b œ 2, syy a2, 2b œ 2, sxy a2, 2b œ 1 Ê sxx syy s#xy œ 3 0 and sxx 0 Ê local maximum of
sa2, 2, 2b œ 12
56. daxß yß zb œ Éax 6b2 ay 4b2 az 0b2 Ê we can minimize daxß yß zb by minimizing
Daxß yß zb œ ax 6b2 ay 4b2 z2 ; z œ Èx2 y2 Ê Daxß yb œ ax 6b2 ay 4b2 x2 y2
œ 2x2 2y2 12x 8y 52 Ê Dx axß yb œ 4x 12 œ 0 and Dy axß yb œ 4y 8 œ 0 Ê critical point is a3, 2b
Ê z œ È13; Dxx a3, 2b œ 4, Dyy a3, 2b œ 4, Dxy a3, 2b œ 0 Ê Dxx Dyy D# œ 16 0 and Dxx 0 Ê local
xy
minimum of dŠ3, 2, È13‹ œ È26
57. Vaxß yß zb œ a2xba2yba2zb œ 8xyz; x2 y2 z2 œ 4 Ê z œ È4 x2 y2 Ê Vaxß yb œ 8xyÈ4 x2 y2 ,
x
0 and y
y 8y
y 8x
0 Ê Vx axß yb œ 32yÈ4 16x
œ 0 and Vy axß yb œ 32xÈ416x
œ 0 Ê critical points are
x2 y2
x2 y2
2
3
2
3
a0, 0b, Š È#3 , È#3 ‹, Š È#3 , È#3 ‹, Š È#3 , È#3 ‹, and Š È#3 , È#3 ‹. Only a0, 0b and Š È#3 , È#3 ‹ satisfy x
0 and y
0
2
2
È
Va0ß 0b œ 0 and VŠ È#3 , È#3 ‹ œ 364
È3 ; On x œ 0, 0 Ÿ y Ÿ 2 Ê Va0ß yb œ 8a0by 4 0 y œ 0, no critical points,
Va0ß 0b œ 0, Va0ß 2b œ 0; On y œ 0, 0 Ÿ x Ÿ 2 Ê Vaxß 0b œ 8xa0bÈ4 x2 02 œ 0, no critical points, Va0ß 0b œ 0,
2
Va0ß 2b œ 0; On y œ È4 x2 , 0 Ÿ x Ÿ 2 Ê VŠxß È4 x2 ‹ œ 8xÈ4 x2 Ê4 x2 ŠÈ4 x2 ‹ œ 0
#
#
#
no critical points, Va0ß 2b œ 0, Va2ß 0b œ 0. Thus, there is a maximum volume of 364
È3 if the box is È3 ‚ È3 ‚ È3 .
27
27
27
54
58. Saxß yß zb œ 2xy 2yz 2xz; xyz œ 27 Ê z œ xy
Ê Saxß yß zb œ 2xy 2yŠ xy
‹ 2xŠ xy
‹ œ 2xy 54
x y , x 0,
54
y 0; Sx axß yb œ 2y 54
x2 œ 0 and Sy axß yb œ 2x y2 œ 0 Ê Critical point is a3, 3b Ê z œ 3; Sxx a3, 3b œ 4,
Syy a3, 3b œ 4, Dxy a3, 3b œ 2 Ê Dxx Dyy D#xy œ 12 0 and Dxx 0 Ê local minimum of Sa3ß 3ß 3b œ 54
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Section 14.7 Extreme Values and Saddle Points
845
59. Let x œ height of the box, y œ width, and z œ length, cut out
squares of length x from corner of the material See diagram
at right. Fold along the dashed lines to form the box. From
the diagram we see that the length of the material is 2x y
and the width is 2x z. Thus a2x yba2x zb œ 12
2ˆ6 2 x2 xy‰
. Since Vax, y, zb œ x y z
2x y
ˆ
2x y 6 2 x2 xy‰
Ê Vax, yb œ
, where x 0, y 0.
2x y
2
3
2 2
ˆ
4 3y 4x y 4x y xy3 ‰
Vx ax, yb œ
œ 0 and
a2x yb2
Êzœ
Vy ax, yb œ
2ˆ12x2 4 x4 4x3 y x2 y2 ‰
œ 0 Ê critical points are ŠÈ3, 0‹, ŠÈ3, 0‹, Š È13 , È43 ‹,
a2x yb2
and Š È13 , È43 ‹. Only ŠÈ3, 0‹ and Š È13 , È43 ‹ satisfy x 0 and y 0. For ŠÈ3, 0‹: z œ 0; Vxx ŠÈ3, 0‹ œ 0,
#
Vyy ŠÈ3, 0‹ œ 2È3, Vxy ŠÈ3, 0‹ œ 4È3 Ê Vxx Vyy Vxy
œ 48 0 Ê saddle point. For Š È13 , È43 ‹: z œ È43 ;
1
4
2
1
4
4
16
#
Vxx Š È13 , È43 ‹ œ 380
È3 , Vyy Š È3 , È3 ‹ œ 3È3 , Vxy Š È3 , È3 ‹ œ 3È3 Ê Vxx Vyy Vxy œ 3 0 and
16
Vxx 0 Ê local maximum of VŠ È13 , È43 , È43 ‹ œ 3È
3
60. (a) (i) On x œ 0, f(xß y) œ f(0ß y) œ y# y 1 for 0 Ÿ y Ÿ 1; f w (0ß y) œ 2y 1 œ 0 Ê y œ "# and x œ 0;
f ˆ0ß "# ‰ œ 34 , f(0ß 0) œ 1, and f(0ß 1) œ 1
(ii)
On y œ 1, f(xß y) œ f(xß 1) œ x# x 1 for 0 Ÿ x Ÿ 1; f w (xß 1) œ 2x 1 œ 0 Ê x œ "# and y œ 1, but
ˆ "# ß 1‰ is outside the domain; f(0ß 1) œ 1 and f("ß ") œ 3
(iii) On x œ 1, f(xß y) œ f("ß y) œ y# y 1 for 0 Ÿ y Ÿ 1; f w (1ß y) œ 2y 1 œ 0 Ê y œ "# and x œ 1, but
ˆ1ß "# ‰ is outside the domain; f(1ß 0) œ 1 and f("ß ") œ 3
(iv) On y œ 0, f(xß y) œ f(xß 0) œ x# x 1 for 0 Ÿ x Ÿ 1; f w (xß 0) œ 2x 1 œ 0 Ê x œ "# and y œ 0;
f ˆ "# ß 0‰ œ 34 ; f(0ß 0) œ 1, and f("ß 0) œ 1
(v)
On the interior of the square, fx (xß y) œ 2x 2y 1 œ 0 and fy (xß y) œ 2y 2x 1 œ 0 Ê 2x 2y œ 1
Ê (x y) œ "# . Then f(xß y) œ x# y# 2xy x y 1 œ (x y)# (x y) 1 œ 34 is the absolute
minimum value when 2x 2y œ 1.
(b) The absolute maximum is f("ß ") œ 3.
61. (a)
(i)
dy
df
` f dx
` f dy
dx
dt œ ` x dt ` y dt œ dt dt œ 2 sin t 2 cos t œ 0
On the semicircle x# y# œ 4, y
Ê cos t œ sin t Ê x œ y
0, we have t œ 14 and x œ y œ È2 Ê f ŠÈ2ß È2‹ œ 2È2. At the
endpoints, f(2ß 0) œ 2 and f(#ß !) œ 2. Therefore the absolute minimum is f(2ß 0) œ 2 when t œ 1;
the absolute maximum is f ŠÈ2ß È2‹ œ 2È2 when t œ 1 .
4
(ii)
On the quartercircle x# y# œ 4, x 0 and y 0, the endpoints give f(!ß 2) œ 2 and f(#ß 0) œ 2.
Therefore the absolute minimum is f(2ß 0) œ 2 and f(!ß 2) œ 2 when t œ 0, 1# respectively; the absolute
maximum is f ŠÈ2ß È2‹ œ 2È2 when t œ 14 .
(b)
(i)
dg
` g dx
` g dy
dy
dx
#
#
dt œ ` x dt ` y dt œ y dt x dt œ 4 sin t 4 cos t œ 0
On the semicircle x# y# œ 4, y
Ê cos t œ „ sin t Ê x œ „ y.
0, we obtain x œ y œ È2 at t œ 14 and x œ È2, y œ È2 at
t œ 341 . Then g ŠÈ2ß È2‹ œ 2 and g ŠÈ2ß È2‹ œ 2. At the endpoints, g(2ß 0) œ g(#ß 0) œ 0.
Therefore the absolute minimum is g ŠÈ2ß È2‹ œ 2 when t œ 341 ; the absolute maximum is
g ŠÈ2 ß È2‹ œ 2 when t œ 14 .
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846
Chapter 14 Partial Derivatives
(ii)
On the quartercircle x# y# œ 4, x 0 and y 0, the endpoints give g(!ß 2) œ 0 and g(#ß 0) œ 0.
Therefore the absolute minimum is g(2ß 0) œ 0 and g(!ß 2) œ 0 when t œ 0, 1# respectively; the absolute
maximum is g ŠÈ2ß È2‹ œ 2 when t œ 14 .
(c)
(i)
(ii)
dy
dh
` h dx
` h dy
dx
dt œ ` x dt ` y dt œ 4x dt 2y dt œ (8 cos t)(2 sin t) (4 sin t)(2 cos t) œ 8 cos t sin t œ 0
Ê t œ 0, 1# , 1 yielding the points (2ß 0), (0ß 2) for 0 Ÿ t Ÿ 1.
#
#
On the semicircle x y œ 4, y 0 we have h(2ß 0) œ 8, h(0ß 2) œ 4, and h(2ß 0) œ 8. Therefore,
the absolute minimum is h(!ß 2) œ 4 when t œ 1# ; the absolute maximum is h(2ß 0) œ 8 and h(2ß 0) œ 8
when t œ 0, 1 respectively.
On the quartercircle x# y# œ 4, x
0 and y
absolute maximum is h(2ß 0) œ 8 when t œ 0.
62. (a)
(i)
0 the absolute minimum is h(0ß 2) œ 4 when t œ 1# ; the
dy
df
` f dx
` f dy
dx
dt œ ` x dt ` y dt œ 2 dt 3 dt œ 6 sin t 6 cos t œ 0
È
#
#
Ê sin t œ cos t Ê t œ 14 for 0 Ÿ t Ÿ 1.
È
0, f(xß y) œ 2x 3y œ 6 cos t 6 sin t œ 6 Š #2 ‹ 6 Š #2 ‹ œ 6È2
On the semi-ellipse, x9 y4 œ 1, y
at t œ 14 . At the endpoints, f(3ß 0) œ 6 and f(3ß 0) œ 6. The absolute minimum is f(3ß 0) œ 6 when
È
t œ 1; the absolute maximum is f Š 3 # 2 ß È2‹ œ 6È2 when t œ 14 .
(ii)
On the quarter ellipse, at the endpoints f(0ß 2) œ 6 and f(3ß 0) œ 6. The absolute minimum is f(3ß 0) œ 6
È
and f(0ß 2) œ 6 when t œ 0, 1 respectively; the absolute maximum is f Š 3 2 ß È2‹ œ 6È2 when t œ 1 .
#
#
4
` g dx
` g dy
dy
dx
#
#
(b) dg
dt œ ` x dt ` y dt œ y dt x dt œ (2 sin t)(3 sin t) (3 cos t)(2 cos t) œ 6 acos t sin tb œ 6 cos 2t œ 0
Ê t œ 14 , 341 for 0 Ÿ t Ÿ 1.
È
(i) On the semi-ellipse, g(xß y) œ xy œ 6 sin t cos t. Then g Š 3 # 2 ß È2‹ œ 3 when t œ 14 , and
È
g Š 3 # 2 ß È2‹ œ 3 when t œ 341 . At the endpoints, g(3ß 0) œ g($ß 0) œ 0. The absolute minimum is
È
È
g Š 3 # 2 ß È2‹ œ 3 when t œ 341 ; the absolute maximum is g Š 3 # 2 ß È2‹ œ 3 when t œ 14 .
(ii)
On the quarter ellipse, at the endpoints g(!ß 2) œ 0 and g($ß 0) œ 0. The absolute minimum is g(3ß 0) œ 0
È
and g(0ß 2) œ 0 at t œ 0, 1 respectively; the absolute maximum is g Š 3 2 ß È2‹ œ 3 when t œ 1 .
#
(c)
(i)
(ii)
#
4
dy
dh
` h dx
` h dy
dx
dt œ ` x dt ` y dt œ 2x dt 6y dt œ (6 cos t)(3 sin t) (12 sin t)(2 cos t) œ 6 sin t cos t œ 0
Ê t œ 0, 1# , 1 for 0 Ÿ t Ÿ 1, yielding the points (3ß 0), (0ß 2), and (3ß 0).
On the semi-ellipse, y 0 so that h(3ß 0) œ 9, h(0ß 2) œ 12, and h(3ß 0) œ 9. The absolute minimum is
h(3ß 0) œ 9 and h(3ß 0) œ 9 when t œ 0, 1 respectively; the absolute maximum is h(!ß 2) œ 12 when t œ 1# .
On the quarter ellipse, the absolute minimum is h(3ß 0) œ 9 when t œ 0; the absolute maximum is
h(!ß 2) œ 12 when t œ 1# .
dy
` f dx
` f dy
dx
63. df
dt œ ` x dt ` y dt œ y dt x dt
(i)
(ii)
"
"
x œ 2t and y œ t 1 Ê df
dt œ (t 1)(2) (2t)(1) œ 4t 2 œ 0 Ê t œ # Ê x œ 1 and y œ # with
f ˆ1ß "# ‰ œ "# . The absolute minimum is f ˆ1ß "# ‰ œ "# when t œ "# ; there is no absolute maximum.
For the endpoints: t œ 1 Ê x œ 2 and y œ 0 with f(2ß 0) œ 0; t œ 0 Ê x œ 0 and y œ 1 with
f(!ß 1) œ 0. The absolute minimum is f ˆ1ß "# ‰ œ "# when t œ "# ; the absolute maximum is f(0ß 1) œ 0
and f(#ß 0) œ 0 when t œ 1, 0 respectively.
(iii) There are no interior critical points. For the endpoints: t œ 0 Ê x œ 0 and y œ 1 with f(0ß 1) œ 0;
t œ 1 Ê x œ 2 and y œ 2 with f(2ß 2) œ 4. The absolute minimum is f(0ß 1) œ 0 when t œ 0; the absolute
maximum is f(2ß 2) œ 4 when t œ 1.
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 14.7 Extreme Values and Saddle Points
dy
df
` f dx
` f dy
dx
dt œ ` x dt ` y dt œ 2x dt 2y dt
4
4
2
(i) x œ t and y œ 2 2t Ê df
dt œ (2t)(1) 2(2 2t)(2) œ 10t 8 œ 0 Ê t œ 5 Ê x œ 5 and y œ 5 with
4
f ˆ 45 ß 25 ‰ œ "#65 25
œ 45 . The absolute minimum is f ˆ 45 ß 25 ‰ œ 45 when t œ 45 ; there is no absolute
64. (a)
(ii)
maximum along the line.
For the endpoints: t œ 0 Ê x œ 0 and y œ 2 with f(0ß 2) œ 4; t œ 1 Ê x œ 1 and y œ 0 with f(1ß 0) œ 1.
The absolute minimum is f ˆ 45 ß 25 ‰ œ 45 at the interior critical point when t œ 45 ; the absolute maximum is
f(0ß 2) œ 4 at the endpoint when t œ 0.
(b)
(i)
dg
` g dx
` g dy
2y
dy
2x
dx
dt œ ` x dt ` y dt œ ’ ax# y# b# “ dt ’ ax# y# b# “ dt
#
#
x œ t and y œ 2 2t Ê x# y# œ 5t# 8t 4 Ê dg
dt œ a5t 8t 4b [(2t)(1) (2)(2 2t)(2)]
#
œ a5t# 8t 4b (10t 8) œ 0 Ê t œ 45 Ê x œ 45 and y œ 25 with g ˆ 45 ß 25 ‰ œ ˆ "4 ‰ œ 54 . The absolute
5
maximum is g ˆ 45 ß 25 ‰ œ 54 when t œ 45 ; there is no absolute minimum along the line since x and y can be
(ii)
as large as we please.
For the endpoints: t œ 0 Ê x œ 0 and y œ 2 with g(0ß 2) œ "4 ; t œ 1 Ê x œ 1 and y œ 0 with g(1ß 0) œ 1.
The absolute minimum is g(0ß 2) œ "4 when t œ 0; the absolute maximum is g ˆ 45 ß 52 ‰ œ 45 when t œ 54 .
65. w œ am x1 b y1 b2 am x2 b y2 b2 â am xn b yn b2
w
Ê `` m
œ 2am x1 b y1 bax1 b 2am x2 b y2 bax2 b â 2am xn b yn baxn b
Ê ``wb œ 2am x1 b y1 ba1b 2am x2 b y2 ba1b â 2am xn b yn ba1b
`w
‘
` m œ 0 Ê 2 am x1 b y1 bax1 b am x2 b y2 bax2 b â am xn b yn baxn b œ 0
Ê m x21 b x1 x1 y1 m x#2 b x2 x2 y2 â m xn2 b xn xn yn œ 0
Ê max21 x2# â xn2 b bax1 x2 â xn b ax1 y1 x2 y2 â xn yn b œ 0
n
n
n
Ê m! ax2k b b! xk ! axk yk b œ 0
k œ1
k œ1
k œ1
`w
‘
` b œ 0 Ê 2 am x1 b y1 b am x2 b y2 b â am xn b yn b œ 0
Ê m x1 b y1 m x2 b y2 â m xn b yn œ 0
Ê max1 x2 â xn b ab b â bb ay1 y2 â yn b œ 0
n
n
n
n
n
k œ1
k œ1
k œ1
k œ1
k œ1
n
n
kœ1
n
kœ 1
n
n
n
k œ1
k œ1
k œ1
Ê m ! xk b ! 1 ! yk œ 0 Ê m ! xk bn ! yk œ 0 Ê b œ 1n Œ ! yk m! xk .
n
w
Substituting for b in the equation obtained for `` m
we get m ! ax2k b 1n Œ ! yk m! xk ! xk ! axk yk b œ 0.
n
n
k œ1
n
n
k œ1
n
k œ1
k œ1
kœ1
Multiply both sides by n to obtain m n ! ax2k b Œ ! yk m! xk ! xk n ! axk yk b œ 0
k œ1
n
n
n
k œ1
k œ1
n
n
k œ1
2
n
n
Ê m n ! ax2k b Œ ! xk Œ ! yk mŒ ! xk n ! axk yk b œ 0
k œ1
2
k œ1
kœ1
n
n
n
k œ1
k œ1
k œ1
n
n
n
Ê m n ! ax2k b mŒ ! xk œ n ! axk yk b Œ ! xk Œ ! yk kœ1
kœ1
n
2
n
Ê m–n! ax2k b Œ ! xk — œ n ! axk yk b Œ ! xk Œ ! yk k œ1
k œ1
n
Êmœ
k œ1
n
n
n ! axk yk bŒ ! xk Œ ! yk kœ1
kœ1
n
n
kœ1
kœ1
kœ1
2
n! ax2k bŒ ! xk k œ1
n
œ
n
kœ1
n
Π! xk Π! yk n ! axk yk b
kœ1
kœ1
n
2
kœ1
n
Π! x k n ! ax k b
kœ1
2
kœ1
To show that these values for m and b minimize the sum of the squares of the distances, use second derivative test.
n
n
` 2w
2
2
2
! ax2k b; ` 2 w œ 2 x1 2 x2 â 2 xn œ 2! xk ; ` 2 w2 œ 2 2 â 2 œ 2 n
œ
2
x
2
x
â
2
x
œ
2
2
n
#
1
`m
`m `b
`b
k œ1
k œ1
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
847
848
Chapter 14 Partial Derivatives
2
n
n
kœ1
kœ1
#
n
n
kœ1
k œ1
2
The discriminant is: Š `` mw2 ‹Š `` bw2 ‹ Š ``m w` b ‹ œ ”2 ! ax2k b•a2 nb ”2 ! xk • œ 4–n ! ax2k b Œ ! xk —.
2
n
2
2
2
n
Now, n ! ax2k b Œ ! xk œ nax12 x#2 â x#n b ax1 x2 â xn bax1 x2 â xn b
k œ1
kœ1
2
2
œ n x1 n x# â n x#n x21 x1 x2 â x1 xn x2 x1 x2# â x2 xn xn x1 xn x2 â x#n
œ an 1b x21 an 1b x2# â an 1b x#n 2 x1 x2 2 x1 x3 â 2 x1 xn 2 x2 x3 â 2 x2 xn â 2 xn1 xn
œ a x21 2 x1 x2 x#2 b a x12 2 x1 x3 x23 b â ax21 2 x1 xn xn# b ax2# 2 x2 x3 x23 b â a x#2 2 x2 xn xn# b
â ax2n1 2 xn1 xn x#n b
œ ax1 x2 b2 ax1 x3 b2 â ax1 xn b2 ax2 x3 b2 â ax2 xn b2 â axn1 xn b2 0.
2
n
n
2
2
2
2
Thus we have : Š `` mw2 ‹Š `` bw2 ‹ Š ``m w` b ‹ œ 4–n ! ax2k b Œ ! xk — 4a0b œ 0. If x1 œ x2 œ â œ xn then
kœ1
k œ1
2
n
Š `` mw2 ‹Š `` bw2 ‹ Š ``m w` b ‹ œ 0. Also, `` mw2 œ 2 ! ax2k b
2
2
2
2
k œ1
0. If x1 œ x2 œ â œ xn œ 0, then `` mw2 œ 0.
2
2
Provided that at least one xi is nonzero and different from the rest of xj , j Á i, then Š `` mw2 ‹Š `` bw2 ‹ Š ``m w` b ‹ 0 and
2
` 2w
` m2 0
2
2
Ê the values given above for m and b minimize w.
3(6)
3
66. m œ (0)(5)
(0)# 3(8) œ 4 and
b œ "3 5 34 (0)‘ œ 53
Ê y œ 34 x 53 ; y¸ xœ4 œ 14
3
1) 3("4)
67. m œ (2)((2)
œ 20
# 3(10)
13 and
9
‰ ‘
b œ "3 1 ˆ 20
13 (2) œ 13
9
71
¸
Ê y œ 20
13 x 13 ; y xœ4 œ 13
3(8)
3
68. m œ (3)(5)
(3)# 3(5) œ 2 and
b œ "3 5 32 (3)‘ œ 16
Ê y œ 32 x 16 ; y¸ xœ4 œ 37
6
k
1
2
3
D
xk
2
0
2
0
yk
0
2
3
5
x#k
4
0
4
8
xk yk
0
0
6
6
k
1
2
3
D
xk
1
0
3
2
yk
2
1
4
1
x#k
1
0
9
10
xk yk
2
0
12
14
k
1
2
3
D
xk
0
1
2
3
yk
0
2
3
5
x#k
0
1
4
5
xk yk
0
2
6
8
69-74. Example CAS commands:
Maple:
f := (x,y) -> x^2+y^3-3*x*y;
x0,x1 := -5,5;
y0,y1 := -5,5;
plot3d( f(x,y), x=x0..x1, y=y0..y1, axes=boxed, shading=zhue, title="#69(a) (Section 14.7)" );
plot3d( f(x,y), x=x0..x1, y=y0..y1, grid=[40,40], axes=boxed, shading=zhue, style=patchcontour, title="#69(b)
(Section 14.7)" );
fx := D[1](f);
# (c)
fy := D[2](f);
crit_pts := solve( {fx(x,y)=0,fy(x,y)=0}, {x,y} );
fxx := D[1](fx);
# (d)
fxy := D[2](fx);
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 14.8 Lagrange Multipliers
849
fyy := D[2](fy);
discr := unapply( fxx(x,y)*fyy(x,y)-fxy(x,y)^2, (x,y) );
for CP in {crit_pts} do
# (e)
eval( [x,y,fxx(x,y),discr(x,y)], CP );
end do;
# (0,0) is a saddle point
# ( 9/4, 3/2) is a local minimum
Mathematica: (assigned functions and bounds will vary)
Clear[x,y,f]
f[x_,y_]:= x2 y3 3x y
xmin= 5; xmax= 5; ymin= 5; ymax= 5;
Plot3D[f[x,y], {x, xmin, xmax}, {y, ymin, ymax}, AxesLabel Ä {x, y, z}]
ContourPlot[f[x,y], {x, xmin, xmax}, {y, ymin, ymax}, ContourShading Ä False, Contours Ä 40]
fx= D[f[x,y], x];
fy= D[f[x,y], y];
critical=Solve[{fx==0, fy==0},{x, y}]
fxx= D[fx, x];
fxy= D[fx, y];
fyy= D[fy, y];
discriminant= fxx fyy fxy2
{{x, y}, f[x, y], discriminant, fxx} /.critical
14.8 LAGRANGE MULTIPLIERS
1.
™ f œ yi xj and ™ g œ 2xi 4yj so that ™ f œ - ™ g Ê yi xj œ -(2xi 4yj) Ê y œ 2x- and x œ 4yÈ
Ê x œ 8x-# Ê - œ „ 42 or x œ 0.
CASE 1: If x œ 0, then y œ 0. But (0ß 0) is not on the ellipse so x Á 0.
#
È
CASE 2: x Á 0 Ê - œ „ 42 Ê x œ „ È2y Ê Š „ È2y‹ 2y# œ 1 Ê y œ „ "# .
È
È
Therefore f takes on its extreme values at Š „ 22 ß "# ‹ and Š „ 22 ß "# ‹ . The extreme values of f on the ellipse
È
are „ #2 .
2.
™ f œ yi xj and ™ g œ 2xi 2yj so that ™ f œ - ™ g Ê yi xj œ -(2xi 2yj) Ê y œ 2x- and x œ 2yÊ x œ 4x-# Ê x œ 0 or - œ „ 12 .
CASE 1: If x œ 0, then y œ 0. But (0ß 0) is not on the circle x# y# 10 œ 0 so x Á 0.
CASE 2: x Á 0 Ê - œ „ 12 Ê y œ 2x ˆ „ "# ‰ œ „ x Ê x# a „ xb# 10 œ 0 Ê x œ „ È5 Ê y œ „ È5.
Therefore f takes on its extreme values at Š „ È5ß È5‹ and Š „ È5ß È5‹ . The extreme values of f on the
circle are 5 and 5.
3.
™ f œ 2xi 2yj and ™ g œ i 3j so that ™ f œ - ™ g Ê 2xi 2yj œ -(i 3j) Ê x œ -# and y œ 3#Ê ˆ -# ‰ 3 ˆ 3#- ‰ œ 10 Ê - œ 2 Ê x œ 1 and y œ 3 Ê f takes on its extreme value at (1ß 3) on the line.
The extreme value is f("ß $) œ 49 1 9 œ 39.
4.
™ f œ 2xyi x# j and ™ g œ i j so that ™ f œ - ™ g Ê 2xyi x# j œ -(i j) Ê 2xy œ - and x# œ Ê 2xy œ x# Ê x œ 0 or 2y œ x.
CASE 1: If x œ 0, then x y œ 3 Ê y œ 3.
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
850
Chapter 14 Partial Derivatives
CASE 2: If x Á 0, then 2y œ x so that x y œ 3 Ê 2y y œ 3 Ê y œ 1 Ê x œ 2.
Therefore f takes on its extreme values at (!ß 3) and (2ß "). The extreme values of f are f(0ß 3) œ 0 and f(2ß 1) œ 4.
5. We optimize f(xß y) œ x# y# , the square of the distance to the origin, subject to the constraint
g(xß y) œ xy# 54 œ 0. Thus ™ f œ 2xi 2yj and ™ g œ y# i 2xyj so that ™ f œ - ™ g Ê 2xi 2yj
œ - ay# i 2xyjb Ê 2x œ -y# and 2y œ 2-xy.
CASE 1: If y œ 0, then x œ 0. But (0ß 0) does not satisfy the constraint xy# œ 54 so y Á 0.
CASE 2: If y Á 0, then 2 œ 2-x Ê x œ -" Ê 2 ˆ -" ‰ œ -y# Ê y# œ -2# . Then xy# œ 54 Ê ˆ -" ‰ ˆ -2# ‰ œ 54
Ê -$ œ " Ê - œ " Ê x œ 3 and y# œ 18 Ê x œ 3 and y œ „ 3È2.
27
3
Therefore Š$ß „ 3È2‹ are the points on the curve xy# œ 54 nearest the origin (since xy# œ 54 has points increasingly
far away as y gets close to 0, no points are farthest away).
6. We optimize f(xß y) œ x# y# , the square of the distance to the origin subject to the constraint g(xß y) œ x# y 2 œ 0.
Thus ™ f œ 2xi 2yj and ™ g œ 2xyi x# j so that ™ f œ - ™ g Ê 2x œ 2xy- and 2y œ x# - Ê - œ 2y
x# , since
2y
x œ 0 Ê y œ 0 (but g(0ß 0) Á 0). Thus x Á 0 and 2x œ 2xy ˆ x# ‰ Ê x# œ 2y# Ê a2y# b y 2 œ 0 Ê y œ 1 (since
y 0) Ê x œ „ È2 . Therefore Š „ È2ß 1‹ are the points on the curve x# y œ 2 nearest the origin (since x# y œ 2 has
points increasingly far away as x gets close to 0, no points are farthest away).
7. (a) ™ f œ i j and ™ g œ yi xj so that ™ f œ - ™ g Ê i j œ -(yi xj) Ê 1 œ -y and 1 œ -x Ê y œ -" and
x œ -" Ê -"# œ 16 Ê - œ „ 4" . Use - œ "4 since x 0 and y 0. Then x œ 4 and y œ 4 Ê the minimum value is 8
at the point (4ß 4). Now, xy œ 16, x 0, y 0 is a branch of a hyperbola in the first quadrant with the x-and y-axes
as asymptotes. The equations x y œ c give a family of parallel lines with m œ 1. As these lines move away from
the origin, the number c increases. Thus the minimum value of c occurs where x y œ c is tangent to the hyperbola's
branch.
(b) ™ f œ yi xj and ™ g œ i j so that ™ f œ - ™ g Ê yi xj œ -(i j) Ê y œ - œ x y y œ 16 Ê y œ 8
Ê x œ 8 Ê f()ß )) œ 64 is the maximum value. The equations xy œ c (x 0 and y 0 or x 0 and y 0
to get a maximum value) give a family of hyperbolas in the first and third quadrants with the x- and y-axes as
asymptotes. The maximum value of c occurs where the hyperbola xy œ c is tangent to the line x y œ 16.
8. Let f(xß y) œ x# y# be the square of the distance from the origin. Then ™ f œ 2xi 2yj and
™ g œ (2x y)i (2y x)j so that ™ f œ - ™ g Ê 2x œ -(2x y) and 2y œ -(2y x) Ê 2y2yx œ Ê 2x œ Š 2y2yx ‹ (2x y) Ê x(2y x) œ y(2x y) Ê x# œ y# Ê y œ „ x.
CASE 1: y œ x Ê x# x(x) x# 1 œ 0 Ê x œ „ È"3 and y œ x.
CASE 2: y œ x Ê x# x(x) (x)# 1 œ 0 Ê x œ „ 1 and y œ x. Thus f Š È"3 ß È"3 ‹ œ 23
œ f Š È"3 ß È"3 ‹ and f(1ß 1) œ 2 œ f(1ß 1).
Therefore the points (1ß 1) and (1ß 1) are the farthest away; Š È"3 ß È"3 ‹ and Š È"3 ß È"3 ‹ are the closest
points to the origin.
9. V œ 1r# h Ê 161 œ 1r# h Ê 16 œ r# h Ê g(rß h) œ r# h 16; S œ 21rh 21r# Ê ™ S œ (21h 41r)i 21rj and
™ g œ 2rhi r# j so that ™ S œ - ™ g Ê (21rh 41r)i 21rj œ - a2rhi r# jb Ê 21rh 41r œ 2rh- and 21r œ -r#
Ê r œ 0 or - œ 2r1 . But r œ 0 gives no physical can, so r Á 0 Ê - œ 2r1 Ê 21h 41r œ 2rh ˆ 2r1 ‰ Ê 2r œ h
Ê 16 œ r# (2r) Ê r œ 2 Ê h œ 4; thus r œ 2 cm and h œ 4 cm give the only extreme surface area of 241 cm# . Since
r œ 4 cm and h œ 1 cm Ê V œ 161 cm$ and S œ 401 cm# , which is a larger surface area, then 241 cm# must be the
minimum surface area.
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 14.8 Lagrange Multipliers
851
10. For a cylinder of radius r and height h we want to maximize the surface area S œ 21rh subject to the constraint
#
g(rß h) œ r# ˆ h# ‰ a# œ 0. Thus ™ S œ 21hi 21rj and ™ g œ 2ri h# j so that ™ S œ - ™ g Ê 21h œ 2-r and
#
21r œ -#h Ê 1rh œ - and 21r œ ˆ 1rh ‰ ˆ #h ‰ Ê 4r# œ h# Ê h œ 2r Ê r# 4r4 œ a# Ê 2r# œ a# Ê r œ Èa
Ê h œ aÈ2 Ê
2
S œ 21 Š Èa2 ‹ ŠaÈ2‹ œ 21a# .
#
#
x
11. A œ (2x)(2y) œ 4xy subject to g(xß y) œ 16
y9 1 œ 0; ™ A œ 4yi 4xj and ™ g œ x8 i 2y
9 j so that ™ A
2y
2y
32y
x
x
‰ ˆ 32y
‰
œ - ™ g Ê 4yi 4xj œ - ˆ 8 i 9 j‰ Ê 4y œ ˆ 8 ‰ - and 4x œ ˆ 9 ‰ - Ê - œ x and 4x œ ˆ 2y
9
x
ˆ „43 x‰#
œ 1 Ê x# œ 8 Ê x œ „ 2È2 . We use x œ 2È2 since x represents distance.
9
È
Then y œ 34 Š2È2‹ œ 3 # 2 , so the length is 2x œ 4È2 and the width is 2y œ 3È2.
#
x
Ê y œ „ 34 x Ê 16
#
#
2y
12. P œ 4x 4y subject to g(xß y) œ xa# yb# 1 œ 0; ™ P œ 4i 4j and ™ g œ 2x
a# i b# j so that ™ P œ - ™ g
#
#
#
#
2a
b
x
‰
ˆ 2y ‰
ˆ 2y ‰ 2a
Ê 4 œ ˆ 2x
a# - and 4 œ b# - Ê - œ x and 4 œ b# Š x ‹ Ê y œ Š a# ‹ x Ê a# œ 1 Ê aa# b# b x# œ a% Ê x œ È #a
#
a b#
#
, since x 0 Ê y œ Š ba# ‹ x œ È #b
#
a b#
#
#
#
#
Š ba# ‹ x#
b#
#
# #
œ 1 Ê xa# bax%
Ê width œ 2x œ È 2a
#
#
a b#
#
4a 4b
and height œ 2y œ Èa2b
Ê perimeter is P œ 4x 4y œ È
œ 4Èa# b#
# b#
a# b#
13. ™ f œ 2xi 2yj and ™ g œ (2x 2)i (2y 4)j so that ™ f œ - ™ g œ 2xi 2yj œ -[(2x 2)i (2y 4)j]
2#
#
Ê 2x œ -(2x 2) and 2y œ -(2y 4) Ê x œ -
1 and y œ -1 , - Á 1 Ê y œ 2x Ê x 2x (2x) 4(2x) œ 0
Ê x œ 0 and y œ 0, or x œ 2 and y œ 4. Therefore f(0ß 0) œ 0 is the minimum value and f(2ß 4) œ 20 is the maximum
value. (Note that - œ 1 gives 2x œ 2x 2 or ! œ 2, which is impossible.)
3
3 ‰
14. ™ f œ 3i j and ™ g œ 2xi 2yj so that ™ f œ - ™ g Ê 3 œ 2-x and 1 œ 2-y Ê - œ 2x
and 1 œ 2 ˆ 2x
y
#
Ê y œ x3 Ê x# ˆ x3 ‰ œ 4 Ê 10x# œ 36 Ê x œ „ È610 Ê x œ È610 and y œ È210 , or x œ È610 and
y œ È210 . Therefore f Š È610 ß È210 ‹ œ È2010 6 œ 2È10 6 ¸ 12.325 is the maximum value, and f Š È610 ß È210 ‹
œ 2È10 6 ¸ 0.325 is the minimum value.
15. ™ T œ (8x 4y)i (4x 2y)j and g(xß y) œ x# y# 25 œ 0 Ê ™ g œ 2xi 2yj so that ™ T œ - ™ g
Ê (8x 4y)i (4x 2y)j œ -(2xi 2yj) Ê 8x 4y œ 2-x and 4x 2y œ 2-y Ê y œ -2x1 , - Á 1
Ê 8x 4 ˆ -2x1 ‰ œ 2-x Ê x œ 0, or - œ 0, or - œ 5.
CASE 1: x œ 0 Ê y œ 0; but (0ß 0) is not on x# y# œ 25 so x Á 0.
CASE 2: - œ 0 Ê y œ 2x Ê x# (2x)# œ 25 Ê x œ „ È5 and y œ 2x.
#
CASE 3: - œ 5 Ê y œ 42x œ #x Ê x# ˆ #x ‰ œ 25 Ê x œ „ 2È5 Ê x œ 2È5 and y œ È5, or x œ 2È5
and y œ È5 .
Therefore T ŠÈ5ß 2È5‹ œ 0° œ T ŠÈ5ß 2È5‹ is the minimum value and T Š2È5ß È5‹ œ 125°
œ T Š2È5ß È5‹ is the maximum value. (Note: - œ 1 Ê x œ 0 from the equation 4x 2y œ 2-y; but we
found x Á 0 in CASE 1.)
16. The surface area is given by S œ 41r# 21rh subject to the constraint V(rß h) œ 43 1r$ 1r# h œ 8000. Thus
™ S œ (81r 21h)i 21rj and ™ V œ a41r# 21rhb i 1r# j so that ™ S œ - ™ V œ (81r 21h)i 21rj
œ - ca41r# 21rhb i 1r# jd Ê 81r 21h œ - a41r# 21rhb and 21r œ -1r# Ê r œ 0 or 2 œ r-. But r Á 0
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852
Chapter 14 Partial Derivatives
so 2 œ r- Ê - œ 2r Ê 4r h œ 2r a2r# rhb Ê h œ 0 Ê the tank is a sphere (there is no cylindrical part) and
4
$
3 1r œ 8000
Ê r œ 10 ˆ 16 ‰
"Î$
.
17. Let f(xß yß z) œ (x 1)# (y 1)# (z 1)# be the square of the distance from (1ß 1ß 1). Then
™ f œ 2(x 1)i 2(y 1)j 2(z 1)k and ™ g œ i 2j 3k so that ™ f œ - ™ g
Ê 2(x 1)i 2(y 1)j 2(z 1)k œ -(i 2j 3k) Ê 2(x 1) œ -, 2(y 1) œ 2-, 2(z 1) œ 3Ê 2(y 1) œ 2[2(x 1)] and 2(z 1) œ 3[2(x 1)] Ê x œ y # 1 Ê z 2 œ 3 ˆ y # 1 ‰ or z œ 3y # 1 ; thus
y1
ˆ 3y # 1 ‰ 13 œ 0 Ê y œ 2 Ê x œ 3# and z œ #5 . Therefore the point ˆ #3 ß 2ß 5# ‰ is closest (since no
# 2y 3
point on the plane is farthest from the point (1ß 1ß 1)).
18. Let f(xß yß z) œ (x 1)# (y 1)# (z 1)# be the square of the distance from (1ß 1ß 1). Then
™ f œ 2(x 1)i 2(y 1)j 2(z 1)k and ™ g œ 2xi 2yj 2zk so that ™ f œ - ™ g Ê x 1 œ -x, y 1 œ -y
#
‰# ˆ 1 " - ‰# œ 4
and z 1 œ -z Ê x œ 1 " - , y œ 1 " - , and z œ 1" - for - Á 1 Ê ˆ 1 " - ‰ ˆ 1"
Ê " " - œ „ È23 Ê x œ È23 , y œ È23 , z œ È23 or x œ È23 , y œ È23 , z œ È23 . The largest value of f
occurs where x 0, y 0, and z 0 or at the point Š È23 ß È23 ß È23 ‹ on the sphere.
19. Let f(xß yß z) œ x# y# z# be the square of the distance from the origin. Then ™ f œ 2xi 2yj 2zk and
™ g œ 2xi 2yj 2zk so that ™ f œ - ™ g Ê 2xi 2yj 2zk œ -(2xi 2yj 2zk) Ê 2x œ 2x-, 2y œ 2y-,
and 2z œ 2z- Ê x œ 0 or - œ 1.
CASE 1: - œ 1 Ê 2y œ 2y Ê y œ 0; 2z œ 2z Ê z œ 0 Ê x# 1 œ 0 Ê x# 1 œ 0 Ê x œ „ 1 and y œ z œ 0.
CASE 2: x œ 0 Ê y# z# œ 1, which has no solution.
Therefore the points on the unit circle x# y# œ 1, are the points on the surface x# y# z# œ 1 closest to the originÞ
The minimum distance is 1.
20. Let f(xß yß z) œ x# y# z# be the square of the distance to the origin. Then ™ f œ 2xi 2yj 2zk and
™ g œ yi xj k so that ™ f œ - ™ g Ê 2xi 2yj 2zk œ -(yi xj k) Ê 2x œ -y, 2y œ -x, and 2z œ Ê x œ -#y Ê 2y œ - Š -#y ‹ Ê y œ 0 or - œ „ 2.
CASE 1: y œ 0 Ê x œ 0 Ê z 1 œ 0 Ê z œ 1.
CASE 2: - œ 2 Ê x œ y and z œ 1 Ê x# (1) 1 œ 0 Ê x# 2 œ 0, so no solution.
CASE 3: - œ 2 Ê x œ y and z œ 1 Ê (y)y 1 1 œ 0 Ê y œ 0, again.
Therefore (0ß 0ß 1) is the point on the surface closest to the origin since this point gives the only extreme value
and there is no maximum distance from the surface to the origin.
21. Let f(xß yß z) œ x# y# z# be the square of the distance to the origin. Then ™ f œ 2xi 2yj 2zk and
™ g œ yi xj 2zk so that ™ f œ - ™ g Ê 2xi 2yj 2zk œ -(yi xj 2zk) Ê 2x œ y-, 2y œ x-, and
2z œ 2z- Ê - œ 1 or z œ 0.
CASE 1: - œ 1 Ê 2x œ y and 2y œ x Ê y œ 0 and x œ 0 Ê z# 4 œ 0 Ê z œ „ 2 and x œ y œ 0.
#
#
CASE 2: z œ 0 Ê xy 4 œ 0 Ê y œ 4x . Then 2x œ 4x - Ê - œ x# , and x8 œ x- Ê x8 œ x Š x# ‹
Ê x% œ 16 Ê x œ „ 2. Thus, x œ 2 and y œ 2, or x = 2 and y œ 2.
Therefore we get four points: (#ß 2ß 0), (2ß 2ß 0), (0ß 0ß 2) and (!ß 0ß 2). But the points (!ß 0ß 2) and (!ß !ß 2)
are closest to the origin since they are 2 units away and the others are 2È2 units away.
22. Let f(xß yß z) œ x# y# z# be the square of the distance to the origin. Then ™ f œ 2xi 2yj 2zk and
™ g œ yzi xzj xyk so that ™ f œ - ™ g Ê 2x œ -yz, 2y œ -xz, and 2z œ -xy Ê 2x# œ -xyz and 2y# œ -yxz
Ê x# œ y# Ê y œ „ x Ê z œ „ x Ê x a „ xb a „ xb œ 1 Ê x œ „ 1 Ê the points are (1ß 1ß 1), ("ß 1ß 1),
("ß "ß "), and (1ß 1, 1).
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Section 14.8 Lagrange Multipliers
853
23. ™ f œ i 2j 5k and ™ g œ 2xi 2yj 2zk so that ™ f œ - ™ g Ê i 2j 5k œ -(2xi 2yj 2zk) Ê 1 œ 2x-,
2 œ 2y-, and 5 œ 2z- Ê x œ #"- , y œ -" œ 2x, and z œ #5- œ 5x Ê x# (2x)# (5x)# œ 30 Ê x œ „ 1.
Thus, x œ 1, y œ 2, z œ 5 or x œ 1, y œ 2, z œ 5. Therefore f(1ß 2ß 5) œ 30 is the maximum value and
f(1ß 2ß 5) œ 30 is the minimum value.
24. ™ f œ i 2j 3k and ™ g œ 2xi 2yj 2zk so that ™ f œ - ™ g Ê i 2j 3k œ -(2xi 2yj 2zk) Ê 1 œ 2x-,
2 œ 2y-, and 3 œ 2z- Ê x œ #"- , y œ -" œ 2x, and z œ #3- œ 3x Ê x# (2x)# (3x)# œ 25 Ê x œ „ È514 .
Thus, x œ È514 , y œ È1014 , z œ È1514 or x œ È514 , y œ È1014 , z œ È1514 . Therefore f Š È514 ß È1014 ß È1514 ‹
œ 5È14 is the maximum value and f Š È514 ß È1014 , È1514 ‹ œ 5È14 is the minimum value.
25. f(xß yß z) œ x# y# z# and g(xß yß z) œ x y z 9 œ 0 Ê ™ f œ 2xi 2yj 2zk and ™ g œ i j k so that
™ f œ - ™ g Ê 2xi 2yj 2zk œ -(i j k) Ê 2x œ -, 2y œ -, and 2z œ - Ê x œ y œ z Ê x x x 9 œ 0
Ê x œ 3, y œ 3, and z œ 3.
26. f(xß yß z) œ xyz and g(xß yß z) œ x y z# 16 œ 0 Ê ™ f œ yzi xzj xyk and ™ g œ i j 2zk so that
™ f œ - ™ g Ê yzi xzj xyk œ -(i j 2zk) Ê yz œ -, xz œ -, and xy œ 2z- Ê yz œ xz Ê z œ 0 or y œ x.
But z 0 so that y œ x Ê x# œ 2z- and xz œ -. Then x# œ 2z(xz) Ê x œ 0 or x œ 2z# . But x 0 so that
32
x œ 2z# Ê y œ 2z# Ê 2z# 2z# z# œ 16 Ê z œ „ È45 . We use z œ È45 since z 0. Then x œ 32
5 and y œ 5
32
4
4096
which yields f Š 32
5 ß 5 ß È5 ‹ œ 25È5 .
27. V œ xyz and g(xß yß z) œ x# y# z# 1 œ 0 Ê ™ V œ yzi xzj xyk and ™ g œ 2xi 2yj 2zk so that
™ V œ - ™ g Ê yz œ -x, xz œ -y, and xy œ -z Ê xyz œ -x# and xyz œ -y# Ê y œ „ x Ê z œ „ x
Ê x# x# x# œ 1 Ê x œ È"3 since x 0 Ê the dimensions of the box are È13 by È13 by È13 for maximum
volume. (Note that there is no minimum volume since the box could be made arbitrarily thin.)
28. V œ xyz with xß yß z all positive and xa yb cz œ 1; thus V œ xyz and g(xß yß z) œ bcx acy abz abc œ 0
Ê ™ V œ yzi xzj xyk and ™ g œ bci acj abk so that ™ V œ - ™ g Ê yz œ -bc, xz œ -ac, and xy œ -ab
Ê xyz œ -bcx, xyz œ -acy, and xyz œ -abz Ê - Á 0. Also, -bcx œ -acy œ -abz Ê bx œ ay, cy œ bz, and
a
cx œ az Ê y œ ba x and z œ ac x. Then xa by zc œ 1 Ê xa b" ˆ ba x‰ "c ˆ ca x‰ œ 1 Ê 3x
a œ 1 Ê xœ 3
Ê y œ ˆ ba ‰ ˆ 3a ‰ œ b3 and z œ ˆ ca ‰ ˆ 3a ‰ œ 3c Ê V œ xyz œ ˆ 3a ‰ ˆ b3 ‰ ˆ 3c ‰ œ abc
27 is the maximum volume. (Note that
there is no minimum volume since the box could be made arbitrarily thin.)
29. ™ T œ 16xi 4zj (4y 16)k and ™ g œ 8xi 2yj 8zk so that ™ T œ - ™ g Ê 16xi 4zj (4y 16)k
œ -(8xi 2yj 8zk) Ê 16x œ 8x-, 4z œ 2y-, and 4y 16 œ 8z- Ê - œ 2 or x œ 0.
CASE 1: - œ 2 Ê 4z œ 2y(2) Ê z œ y. Then 4z 16 œ 16z Ê z œ 43 Ê y œ 43 . Then
#
#
4x# ˆ 43 ‰ 4 ˆ 43 ‰ œ 16 Ê x œ „ 43 .
2z
#
#
#
#
#
CASE 2: x œ 0 Ê - œ 2z
y Ê 4y 16 œ 8z Š y ‹ Ê y 4y œ 4z Ê 4(0) y ay 4yb 16 œ 0
Ê y# 2y 8 œ 0 Ê (y 4)(y 2) œ 0 Ê y œ 4 or y œ 2. Now y œ 4 Ê 4z# œ 4# 4(4)
Ê z œ 0 and y œ 2 Ê 4z# œ (2)# 4(2) Ê z œ „ È3.
°
°
The temperatures are T ˆ „ 43 ß 43 ß 43 ‰ œ 642 23 , T(0ß 4ß 0) œ 600°, T Š0ß 2ß È3‹ œ Š600 24È3‹ , and
°
T Š0ß 2ß È3‹ œ Š600 24È3‹ ¸ 641.6°. Therefore ˆ „ 43 ß 43 ß 43 ‰ are the hottest points on the space probe.
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
854
Chapter 14 Partial Derivatives
30. ™ T œ 400yz# i 400xz# j 800xyzk and ™ g œ 2xi 2yj 2zk so that ™ T œ - ™ g
Ê 400yz# i 400xz# j 800xyzk œ -(2xi 2yj 2zk) Ê 400yz# œ 2x-, 400xz# œ 2y-, and 800xyz œ 2z-.
È
Solving this system yields the points a!ß „ 1ß 0b , a „ 1ß 0ß 0b , and Š „ "# ß „ "# ß „ #2 ‹ . The corresponding
È
temperatures are T a!ß „ 1ß 0b œ 0, T a „ 1ß 0ß 0b œ 0, and T Š „ "# ß „ "# ß „ #2 ‹ œ „ 50. Therefore 50 is the
È
È
maximum temperature at Š "# ß "# ß „ #2 ‹ and Š "# ß "# ß „ #2 ‹ ; 50 is the minimum temperature at
È
È
Š "# ß "# ß „ #2 ‹ and Š #" ß #" ß „ #2 ‹ .
31. ™ U œ (y 2)i xj and ™ g œ 2i j so that ™ U œ - ™ g Ê (y 2)i xj œ -(2i j) Ê y # œ 2- and
x œ - Ê y 2 œ 2x Ê y œ 2x 2 Ê 2x (2x 2) œ 30 Ê x œ 8 and y œ 14. Therefore U(8ß 14) œ $128
is the maximum value of U under the constraint.
32. ™ M œ (6 z)i 2yj xk and ™ g œ 2xi 2yj 2zk so that ™ M œ - ™ g Ê (6 z)i 2yj xk
œ -(2xi 2yj 2zk) Ê 6 z œ 2x-, 2y œ 2y-, x œ 2z- Ê - œ 1 or y œ 0.
CASE 1: - œ 1 Ê 6 z œ 2x and x œ 2z Ê 6 z œ 2(2z) Ê z œ 2 and x œ 4. Then
(4)# y# 2# 36 œ 0 Ê y œ „ 4.
x
x ‰
CASE 2: y œ 0, 6 z œ 2x-, and x œ 2z- Ê - œ 2z
Ê 6 z œ 2x ˆ 2z
Ê 6z z# œ x#
Ê a6z z# b 0# z# œ 36 Ê z œ 6 or z œ 3. Now z œ 6 Ê x# œ 0 Ê x œ 0; z œ 3
Ê x# œ 27 Ê x œ „ 3È3.
Therefore we have the points Š „ 3È3ß 0ß 3‹ , (0ß 0ß 6), and a4ß „ 4ß 2b . Then M Š3È3ß 0ß 3‹ œ 27È3 60
¸ 106.8, M Š3È3ß 0ß 3‹ œ 60 27È3 ¸ 13.2, M(0ß 0ß 6) œ 60, and M(4ß 4ß 2) œ 12 œ M(4ß 4ß 2). Therefore,
the weakest field is at a4ß „ 4ß 2b .
33. Let g" (xß yß z) œ 2x y œ 0 and g# (xß yß z) œ y z œ 0 Ê ™ g" œ 2i j , ™ g# œ j k , and ™ f œ 2xi 2j 2zk
so that ™ f œ - ™ g" . ™ g# Ê 2xi 2j 2zk œ -(2i j) .(j k) Ê 2xi 2j 2zk œ 2-i (. -)j .k
Ê 2x œ 2-, 2 œ . -, and 2z œ . Ê x œ -. Then 2 œ 2z x Ê x œ 2z 2 so that 2x y œ 0
Ê 2(2z 2) y œ 0 Ê 4z 4 y œ 0. This equation coupled with y z œ 0 implies z œ 43 and y œ 43 . Then
#
#
x œ 23 so that ˆ 23 ß 43 ß 43 ‰ is the point that gives the maximum value f ˆ 23 ß 43 ß 43 ‰ œ ˆ 23 ‰ 2 ˆ 43 ‰ ˆ 43 ‰ œ 43 .
34. Let g" (xß yß z) œ x 2y 3z 6 œ 0 and g# (xß yß z) œ x 3y 9z 9 œ 0 Ê ™ g" œ i 2j 3k ,
™ g# œ i 3j 9k , and ™ f œ 2xi 2yj 2zk so that ™ f œ - ™ g" . ™ g# Ê 2xi 2yj 2zk
œ -(i 2j 3k) .(i 3j 9k) Ê 2x œ - ., 2y œ 2- 3., and 2z œ 3- 9.. Then 0 œ x 2y 3z 6
‰
œ "# (- .) (2- 3.) ˆ 9# - 27
# . 6 Ê 7- 17. œ 6; 0 œ x 3y 9z 9
"
9
27
81
Ê # (- .) ˆ3- # .‰ ˆ # - # .‰ 9 Ê 34- 91. œ 18. Solving these two equations for - and . gives
-.
2- 3.
3- 9.
78
81
9
- œ 240
œ 123
œ 59
. The minimum value is
59 and . œ 59 Ê x œ # œ 59 , y œ
#
59 , and z œ
#
21,771
81
123
9
369
f ˆ 59 ß 59 ß 59 ‰ œ 59# œ 59 . (Note that there is no maximum value of f subject to the constraints because
at least one of the variables x, y, or z can be made arbitrary and assume a value as large as we please.)
35. Let f(xß yß z) œ x# y# z# be the square of the distance from the origin. We want to minimize f(xß yß z) subject to the
constraints g" (xß yß z) œ y 2z 12 œ 0 and g# (xß yß z) œ x y 6 œ 0. Thus ™ f œ 2xi 2yj 2zk , ™ g" œ j 2k,
and ™ g# œ i j so that ™ f œ - ™ g" . ™ g# Ê 2x œ ., 2y œ - ., and 2z œ 2-. Then 0 œ y 2z 12
œ ˆ -# .# ‰ 2- 12 Ê #5 - "# . œ 12 Ê 5- . œ 24; 0 œ x y 6 œ .# ˆ -# .# ‰ 6 Ê "# - . œ 6
Ê - #. œ 12. Solving these two equations for - and . gives - œ 4 and . œ 4 Ê x œ .# œ 2, y œ - # . œ 4, and
z œ - œ 4. The point (2ß 4ß 4) on the line of intersection is closest to the origin. (There is no maximum distance from the
origin since points on the line can be arbitrarily far away.)
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Section 14.8 Lagrange Multipliers
855
36. The maximum value is f ˆ 23 ß 43 ß 43 ‰ œ 43 from Exercise 33 above.
37. Let g" (xß yß z) œ z 1 œ 0 and g# (xß yß z) œ x# y# z# 10 œ 0 Ê ™ g" œ k , ™ g# œ 2xi 2yj 2zk , and
™ f œ 2xyzi x# zj x# yk so that ™ f œ - ™ g" . ™ g# Ê 2xyzi x# zj x# yk œ -(k) .(2xi 2yj 2zk)
Ê 2xyz œ 2x., x# z œ 2y., and x# y œ 2z. - Ê xyz œ x. Ê x œ 0 or yz œ . Ê . œ y since z œ 1.
CASE 1: x œ 0 and z œ 1 Ê y# 9 œ 0 (from g# ) Ê y œ „ 3 yielding the points a0ß „ 3ß 1b.
CASE 2: . œ y Ê x# z œ 2y# Ê x# œ 2y# (since z œ 1) Ê 2y# y# 1 10 œ 0 (from g# ) Ê 3y# 9 œ 0
#
Ê y œ „ È3 Ê x# œ 2 Š „ È3‹ Ê x œ „ È6 yielding the points Š „ È6ß „ È3ß "‹ .
Now f a!ß „ 3ß 1b œ 1 and f Š „ È6ß „ È3ß "‹ œ 6 Š „ È3‹ 1 œ 1 „ 6È3. Therefore the maximum of f is
1 6È3 at Š „ È6ß È3ß 1‹, and the minimum of f is 1 6È3 at Š „ È6ß È3ß "‹ .
38. (a) Let g" (xß yß z) œ x y z 40 œ 0 and g# (xß yß z) œ x y z œ 0 Ê ™ g" œ i j k , ™ g# œ i j k , and
™ w œ yzi xzj xyk so that ™ w œ - ™ g" . ™ g# Ê yzi xzj xyk œ -(i j k) .(i j k)
Ê yz œ - ., xz œ - ., and xy œ - . Ê yz œ xz Ê z œ 0 or y œ x.
CASE 1: z œ 0 Ê x y œ 40 and x y œ 0 Ê no solution.
CASE 2: x œ y Ê 2x z 40 œ 0 and 2x z œ 0 Ê z œ 20 Ê x œ 10 and y œ 10 Ê w œ (10)(10)(20)
œ 2000
â
â
âi j k â
â
â
" â œ 2i 2j is parallel to the line of intersection Ê the line is x œ 2t 10,
(b) n œ â " "
â
â
â " " " â
y œ 2t 10, z œ 20. Since z œ 20, we see that w œ xyz œ (2t 10)(2t 10)(20) œ a4t# 100b (20)
which has its maximum when t œ 0 Ê x œ 10, y œ 10, and z œ 20.
39. Let g" (Bß yß z) œ y x œ 0 and g# (xß yß z) œ x# y# z# 4 œ 0. Then ™ f œ yi xj 2zk , ™ g" œ i j , and
™ g# œ 2xi 2yj 2zk so that ™ f œ - ™ g" . ™ g# Ê yi xj 2zk œ -(i j) .(2xi 2yj 2zk)
Ê y œ - 2x., x œ - 2y., and 2z œ 2z. Ê z œ 0 or . œ 1.
CASE 1: z œ 0 Ê x# y# 4 œ 0 Ê 2x# 4 œ 0 (since x œ y) Ê x œ „ È2 and y œ „ È2 yielding the points
Š „ È2ß „ È2ß !‹ .
CASE 2: . œ 1 Ê y œ - 2x and x œ - 2y Ê x y œ 2(x y) Ê 2x œ 2(2x) since x œ y Ê x œ 0 Ê y œ 0
Ê z# 4 œ 0 Ê z œ „ 2 yielding the points a!ß !ß „ 2b .
Now, f a!ß !ß „ 2b œ 4 and f Š „ È2ß „ È2ß !‹ œ 2. Therefore the maximum value of f is 4 at a!ß !ß „ 2b and the
minimum value of f is 2 at Š „ È2ß „ È2ß !‹ .
40. Let f(xß yß z) œ x# y# z# be the square of the distance from the origin. We want to minimize f(xß yß z) subject
to the constraints g" (xß yß z) œ 2y 4z 5 œ 0 and g# (xß yß z) œ 4x# 4y# z# œ 0. Thus ™ f œ 2xi 2yj 2zk ,
™ g" œ 2j 4k , and ™ g# œ 8xi 8yj 2zk so that ™ f œ - ™ g" . ™ g# Ê 2xi 2yj 2zk
œ -(2j 4k) .(8xi 8yj 2zk) Ê 2x œ 8x., 2y œ 2- 8y., and 2z œ 4- 2z. Ê x œ 0 or . œ "4 .
CASE 1: x œ 0 Ê 4(0)# 4y# z# œ 0 Ê z œ „ 2y Ê 2y 4(2y) 5 œ 0 Ê y œ "# , or 2y 4(2y) 5 œ 0
Ê y œ 56 yielding the points ˆ!ß "# ß "‰ and ˆ!ß 56 ß 53 ‰ .
CASE 2: . œ "4 Ê y œ - y Ê - œ 0 Ê 2z œ 4(0) 2z ˆ 4" ‰ Ê z œ 0 Ê 2y 4(0) œ 5 Ê y œ 5# and
#
(0)# œ 4x# 4 ˆ #5 ‰ Ê no solution.
"
ˆ " ‰
Then f ˆ!ß "# ß 1‰ œ 54 and f ˆ!ß 56 ß 35 ‰ œ 25 ˆ 36
"9 ‰ œ 125
36 Ê the point !ß # ß 1 is closest to the origin.
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
856
Chapter 14 Partial Derivatives
41. ™ f œ i j and ™ g œ yi xj so that ™ f œ - ™ g Ê i j œ -(yi xj) Ê 1 œ y- and 1 œ x- Ê y œ x
Ê y# œ 16 Ê y œ „ 4 Ê (4ß 4) and (%ß 4) are candidates for the location of extreme values. But as x Ä _,
y Ä _ and f(xß y) Ä _; as x Ä _, y Ä 0 and f(xß y) Ä _. Therefore no maximum or minimum value
exists subject to the constraint.
4
42. Let f(Aß Bß C) œ ! (Axk Byk C zk )# œ C# (B C 1)# (A B C 1)# (A C 1)# . We want
k œ1
to minimize f. Then fA (Aß Bß C) œ 4A 2B 4C, fB (Aß Bß C) œ 2A 4B 4C 4, and
fC (Aß Bß C) œ 4A 4B 8C 2. Set each partial derivative equal to 0 and solve the system to get A œ "# ,
B œ 3# , and C œ "4 or the critical point of f is ˆ #" ß 3# ß "4 ‰ .
43. (a) Maximize f(aß bß c) œ a# b# c# subject to a# b# c# œ r# . Thus ™ f œ 2ab# c# i 2a# bc# j 2a# b# ck and
™ g œ 2ai 2bj 2ck so that ™ f œ - ™ g Ê 2ab# c# œ 2a-, 2a# bc# œ 2b-, and 2a# b# c œ 2cÊ 2a# b# c# œ 2a# - œ 2b# - œ 2c# - Ê - œ 0 or a# œ b# œ c# .
CASE 1: - œ 0 Ê a# b# c# œ 0.
#
$
CASE 2: a# œ b# œ c# Ê f(aß bß c) œ a# a# a# and 3a# œ r# Ê f(aß bß c) œ Š r3 ‹ is the maximum value.
(b) The point ŠÈaß Èbß Èc‹ is on the sphere if a b c œ r# . Moreover, by part (a), abc œ f ŠÈaß Èbß Èc‹
#
$
#
Ÿ Š r3 ‹ Ê (abc)"Î$ Ÿ r3 œ a b3 c , as claimed.
n
44. Let f(x" ß x# ß á ß xn ) œ ! ai xi œ a" x" a# x# á an xn and g(x" ß x# ß á ß xn ) œ x"# x## á xn# 1. Then we
i œ1
a#
a#
#
want ™ f œ - ™ g Ê a" œ -(2x" ), a# œ -(2x# ), á , an œ -(2xn ), - Á 0 Ê xi œ 2a-i Ê 4-"# 4-## á 4a-n# œ 1
n
n
iœ1
i œ1
"Î#
Ê 4-# œ ! a#i Ê 2- œ Œ! a#i n
n
n
n
i œ1
i œ1
i œ1
i œ1
"Î#
Ê f(x" ß x# ß á ß xn ) œ ! ai xi œ ! ai ˆ #a-i ‰ œ #"- ! a#i œ Œ! a#i the maximum value.
45-50. Example CAS commands:
Maple:
f := (x,y,z) -> x*y+y*z;
g1 := (x,y,z) -> x^2+y^2-2;
g2 := (x,y,z) -> x^2+z^2-2;
h := unapply( f(x,y,z)-lambda[1]*g1(x,y,z)-lambda[2]*g2(x,y,z), (x,y,z,lambda[1],lambda[2]) );
hx := diff( h(x,y,z,lambda[1],lambda[2]), x );
hy := diff( h(x,y,z,lambda[1],lambda[2]), y );
hz := diff( h(x,y,z,lambda[1],lambda[2]), z );
hl1 := diff( h(x,y,z,lambda[1],lambda[2]), lambda[1] );
hl2 := diff( h(x,y,z,lambda[1],lambda[2]), lambda[2] );
sys := { hx=0, hy=0, hz=0, hl1=0, hl2=0 };
q1 := solve( sys, {x,y,z,lambda[1],lambda[2]} );
q2 := map(allvalues,{q1});
for p in q2 do
eval( [x,y,z,f(x,y,z)], p );
``=evalf(eval( [x,y,z,f(x,y,z)], p ));
end do;
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
# (a)
#(b)
# (c)
# (d)
is
Section 14.9 Taylor's Formula for Two Variables
Mathematica: (assigned functions will vary)
Clear[x, y, z, lambda1, lambda2]
f[x_,y_,z_]:= x y y z
g1[x_,y_,z_]:= x2 y2 2
g2[x_,y_,z_]:= x2 z2 2
h = f[x, y, z] lambda1 g1[x, y, z] lambda2 g2[x, y, z];
hx= D[h, x]; hy= D[h, y]; hz= D[h,z]; hL1=D[h, lambda1]; hL2= D[h, lambda2];
critical=Solve[{hx==0, hy==0, hz==0, hL1==0, hL2==0, g1[x,y,z]==0, g2[x,y,z]==0},
{x, y, z, lambda1, lambda2}]//N
{{x, y, z}, f[x, y, z]}/.critical
14.9 TAYLOR'S FORMULA FOR TWO VARIABLES
1. f(xß y) œ xey Ê fx œ ey , fy œ xey , fxx œ 0, fxy œ ey , fyy œ xey
Ê f(xß y) ¸ f(0ß 0) xfx (0ß 0) yfy (0ß 0) "# cx# fxx (0ß 0) 2xyfxy (0ß 0) y# fyy (0ß 0)d
œ 0 x † 1 y † 0 "# ax# † 0 2xy † 1 y# † 0b œ x xy quadratic approximation;
fxxx œ 0, fxxy œ 0, fxyy œ ey , fyyy œ xey
Ê f(xß y) ¸ quadratic "6 cx$ fxxx (!ß !) 3x# yfxxy (0ß 0) 3xy# fxyy (!ß !) y$ fyyy (0ß 0)d
œ x xy "6 ax$ † 0 3x# y † 0 3xy# † 1 y$ † 0b œ x xy "# xy# , cubic approximation
2. f(xß y) œ ex cos y Ê fx œ ex cos y, fy œ ex sin y, fxx œ ex cos y, fxy œ ex sin y, fyy œ ex cos y
Ê f(xß y) ¸ f(0ß 0) xfx (0ß 0) yfy (!ß 0) "# cx# fxx (!ß !) 2xyfxy (!ß !) y# fyy (0ß 0)d
œ 1 x † 1 y † 0 "# cx# † 1 2xy † 0 y# † (1)d œ 1 x "# ax# y# b , quadratic approximation;
fxxx œ ex cos y, fxxy œ ex sin y, fxyy œ ex cos y, fyyy œ ex sin y
Ê f(xß y) ¸ quadratic "6 cx$ fxxx (0ß 0) 3x# yfxxy (!ß 0) 3xy# fxyy (0ß 0) y$ fyyy (0ß 0)d
œ 1 x "# ax# y# b 6" cx$ † 1 3x# y † 0 3xy# † (1) y$ † 0d
œ 1 x "# ax# y# b 6" ax$ 3xy# b , cubic approximation
3. f(xß y) œ y sin x Ê fx œ y cos x, fy œ sin x, fxx œ y sin x, fxy œ cos x, fyy œ 0
Ê f(xß y) ¸ f(0ß 0) xfx (0ß 0) yfy (!ß 0) "# cx# fxx (0ß 0) 2xyfxy (0ß 0) y# fyy (0ß 0)d
œ 0 x † 0 y † 0 "# ax# † 0 2xy † 1 y# † 0b œ xy, quadratic approximation;
fxxx œ y cos x, fxxy œ sin x, fxyy œ 0, fyyy œ 0
Ê f(xß y) ¸ quadratic "6 cx$ fxxx (0ß 0) 3x# yfxxy (!ß 0) 3xy# fxyy (0ß 0) y$ fyyy (0ß 0)d
œ xy "6 ax$ † 0 3x# y † 0 3xy# † 0 y$ † 0b œ xy, cubic approximation
4. f(xß y) œ sin x cos y Ê fx œ cos x cos y, fy œ sin x sin y, fxx œ sin x cos y, fxy œ cos x sin y,
fyy œ sin x cos y Ê f(xß y) ¸ f(0ß 0) xfx (0ß 0) yfy (0ß 0) "# cx# fxx (0ß 0) 2xyfxy (0ß 0) y# fyy (0ß 0)d
œ 0 x † 1 y † 0 "# ax# † 0 2xy † 0 y# † 0b œ x, quadratic approximation;
fxxx œ cos x cos y, fxxy œ sin x sin y, fxyy œ cos x cos y, fyyy œ sin x sin y
Ê f(xß y) ¸ quadratic "6 cx$ fxxx (0ß 0) 3x# yfxxy (!ß 0) 3xy# fxyy (0ß 0) y$ fyyy (0ß 0)d
œ x "6 cx$ † (1) 3x# y † 0 3xy# † (1) y$ † 0d œ x 6" ax$ 3xy# b, cubic approximation
x
x
x
5. f(xß y) œ ex ln (1 y) Ê fx œ ex ln (1 y), fy œ 1 e y , fxx œ ex ln (1 y), fxy œ 1 e y , fyy œ (1 e y)#
Ê f(xß y) ¸ f(0ß 0) xfx (0ß 0) yfy (0ß 0) "# cx# fxx (0ß 0) 2xyfxy (0ß 0) y# fyy (0ß 0)d
œ 0 x † 0 y † 1 "# cx# † 0 2xy † 1 y# † (1)d œ y "# a2xy y# b , quadratic approximation;
x
x
x
fxxx œ ex ln (1 y), fxxy œ 1 e y , fxyy œ (1 e y)# , fyyy œ (1 2e
y)$
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
857
858
Chapter 14 Partial Derivatives
Ê f(xß y) ¸ quadratic "6 cx$ fxxx (0ß 0) 3x# yfxxy (!ß 0) 3xy# fxyy (0ß 0) y$ fyyy (0ß 0)d
œ y "2 a2xy y# b 6" cx$ † 0 3x# y † 1 3xy# † (1) y$ † 2d
œ y "# a2xy y# b 6" a3x# y 3xy# 2y$ b , cubic approximation
6. f(xß y) œ ln (2x y 1) Ê fx œ 2x 2y 1 , fy œ #x "y 1 , fxx œ (2x y4 1)# , fxy œ (2x y2 1)# ,
"
#
#
fyy œ (2x "
y 1)# Ê f(xß y) ¸ f(0ß 0) xfx (0ß 0) yfy (0ß 0) # cx fxx (0ß 0) 2xyfxy (0ß 0) y fyy (0ß 0)d
œ 0 x † 2 y † 1 "# cx# † (4) 2xy † (2) y# † (1)d œ 2x y "# a4x# 4xy y# b
œ (2x y) "# (2x y)# , quadratic approximation;
fxxx œ (2x 16y 1)$ , fxxy œ (2x 8y 1)$ , fxyy œ (2x 4y 1)$ , fyyy œ (2x 2y 1)$
Ê f(xß y) ¸ quadratic "6 cx$ fxxx (0ß 0) 3x# yfxxy (!ß 0) 3xy# fxyy (0ß 0) y$ fyyy (0ß 0)d
œ (2x y) "# (2x y)# 6" ax$ † 16 3x# y † 8 3xy# † 4 y$ † 2b
œ (2x y) "# (2x y)# 3" a8x$ 12x# y 6xy# y# b
œ (2x y) "# (2x y)# 3" (2x y)$ , cubic approximation
7. f(xß y) œ sin ax# y# b Ê fx œ 2x cos ax# y# b , fy œ 2y cos ax# y# b , fxx œ 2 cos ax# y# b 4x# sin ax# y# b ,
fxy œ 4xy sin ax# y# b , fyy œ 2 cos ax# y# b 4y# sin ax# y# b
Ê f(xß y) ¸ f(0ß 0) xfx (0ß 0) yfy (0ß 0) "# cx# fxx (0ß 0) 2xyfxy (0ß 0) y# fyy (0ß 0)d
œ 0 x † 0 y † 0 "# ax# † 2 2xy † 0 y# † 2b œ x# y# , quadratic approximation;
fxxx œ 12x sin ax# y# b 8x$ cos ax# y# b , fxxy œ 4y sin ax# y# b 8x# y cos ax# y# b ,
fxyy œ 4x sin ax# y# b 8xy# cos ax# y# b , fyyy œ 12y sin ax# y# b 8y$ cos ax# y# b
Ê f(xß y) ¸ quadratic "6 cx$ fxxx (0ß 0) 3x# yfxxy (!ß 0) 3xy# fxyy (0ß 0) y$ fyyy (0ß 0)d
œ x# y# "6 ax$ † 0 3x# y † 0 3xy# † 0 y$ † 0b œ x# y# , cubic approximation
8. f(xß y) œ cos ax# y# b Ê fx œ 2x sin ax# y# b , fy œ 2y sin ax# y# b ,
fxx œ 2 sin ax# y# b 4x# cos ax# y# b , fxy œ 4xy cos ax# y# b , fyy œ 2 sin ax# y# b 4y# cos ax# y# b
Ê f(xß y) ¸ f(0ß 0) xfx (0ß 0) yfy (0ß 0) "# cx# fxx (0ß 0) 2xyfxy (0ß 0) y# fyy (0ß 0)d
œ 1 x † 0 y † 0 "# cx# † 0 2xy † 0 y# † 0d œ 1, quadratic approximation;
fxxx œ 12x cos ax# y# b 8x$ sin ax# y# b , fxxy œ 4y cos ax# y# b 8x# y sin ax# y# b ,
fxyy œ 4x cos ax# y# b 8xy# sin ax# y# b , fyyy œ 12y cos ax# y# b 8y$ sin ax# y# b
Ê f(xß y) ¸ quadratic "6 cx$ fxxx (0ß 0) 3x# yfxxy (!ß 0) 3xy# fxyy (0ß 0) y$ fyyy (0ß 0)d
œ 1 "6 ax$ † 0 3x# y † 0 3xy# † 0 y$ † 0b œ 1, cubic approximation
9. f(xß y) œ 1 x" y Ê fx œ (1 x" y)# œ fy , fxx œ (1 x2 y)$ œ fxy œ fyy
Ê f(xß y) ¸ f(0ß 0) xfx (0ß 0) yfy (0ß 0) "# cx# fxx (0ß 0) 2xyfxy (0ß 0) y# fyy (0ß 0)d
œ 1 x † 1 y † 1 "# ax# † 2 2xy † 2 y# † 2b œ 1 (x y) ax# 2xy y# b
œ 1 (x y) (x y)# , quadratic approximation; fxxx œ (1 x6 y)% œ fxxy œ fxyy œ fyyy
Ê f(xß y) ¸ quadratic "6 cx$ fxxx (0ß 0) 3x# yfxxy (!ß 0) 3xy# fxyy (0ß 0) y$ fyyy (0ß 0)d
œ 1 (x y) (x y)# "6 ax$ † 6 3x# y † 6 3xy# † 6 y$ † 6b
œ 1 (x y) (x y)# ax$ 3x# y 3xy# y$ b œ 1 (x y) (x y)# (x y)$ , cubic approximation
#
y)
10. f(xß y) œ 1 x " y xy Ê fx œ (1 x1yy xy)# , fy œ (" x1yx xy)# , fxx œ (1 2(1
x y xy)$ ,
#
fxy œ (" x 1y xy)# , fyy œ (1 2(x"yx) xy)$
Ê f(xß y) ¸ f(0ß 0) xfx (0ß 0) yfy (0ß 0) "# cx# fxx (0ß 0) 2xyfxy (0ß 0) y# fyy (0ß 0)d
œ 1 x † 1 y † 1 "# ax# † 2 2xy † 1 y# † 2b œ 1 x y x# xy y# , quadratic approximation;
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 14.10 Partial Derivatives with Constrained Variables
$
6(1 y)(1 x)](1 y)
y)
fxxx œ (1 6(1
, fxxy œ [4(1 x y (1 xy)x ,
x y xy)%
y xy)%
$
x)(1 y)](1 x)
x)
fxyy œ [4(1 x y (1 xy)x 6(1
, fyyy œ (1 6(1
y xy)%
x y xy)%
Ê f(xß y) ¸ quadratic "6 cx$ fxxx (0ß 0) 3x# yfxxy (!ß 0) 3xy# fxyy (0ß 0) y$ fyyy (0ß 0)d
œ 1 x y x# xy y# "6 ax$ † 6 3x# y † 2 3xy# † 2 y$ † 6b
œ 1 x y x# xy y# x$ x# y xy# y$ , cubic approximation
11. f(xß y) œ cos x cos y Ê fx œ sin x cos y, fy œ cos x sin y, fxx œ cos x cos y, fxy œ sin x sin y,
fyy œ cos x cos y Ê f(xß y) ¸ f(0ß 0) xfx (0ß 0) yfy (0ß 0) "# cx# fxx (0ß 0) 2xyfxy (0ß 0) y# fyy (0ß 0)d
#
#
œ 1 x † 0 y † 0 "# cx# † (1) 2xy † 0 y# † (1)d œ 1 x# y# , quadratic approximation. Since all partial
derivatives of f are products of sines and cosines, the absolute value of these derivatives is less than or equal
to 1 Ê E(xß y) Ÿ "6 c(0.1)$ 3(0.1)$ 3(0.1)$ 0.1)$ d Ÿ 0.00134.
12. f(xß y) œ ex sin y Ê fx œ ex sin y, fy œ ex cos y, fxx œ ex sin y, fxy œ ex cos y, fyy œ ex sin y
Ê f(xß y) ¸ f(0ß 0) xfx (0ß 0) yfy (0ß 0) "# cx# fxx (0ß 0) 2xyfxy (0ß 0) y# fyy (0ß 0)d
œ 0 x † 0 y † 1 "# ax# † 0 2xy † 1 y# † 0b œ y xy , quadratic approximation. Now, fxxx œ ex sin y,
fxxy œ ex cos y, fxyy œ ex sin y, and fyyy œ ex cos y. Since kxk Ÿ 0.1, kex sin yk Ÿ ke0Þ1 sin 0.1k ¸ 0.11 and
kex cos yk Ÿ ke0Þ1 cos 0.1k ¸ 1.11. Therefore,
E(xß y) Ÿ "6 c(0.11)(0.1)$ 3(1.11)(0.1)$ 3(0.11)(0.1)$ (1.11)(0.1)$ d Ÿ 0.000814.
14.10 PARTIAL DERIVATIVES WITH CONSTRAINED VARIABLES
1. w œ x# y# z# and z œ x# y# :
Î x œ x(yß z) Ñ
y
yœy
Ä w Ê Š ``wy ‹ œ ``wx `` xy ``wy `` yy ``wz `` yz ; `` yz œ 0 and `` yz œ 2x `` yx 2y `` yy
(a) Œ Ä
z
z
Ï zœz Ò
œ 2x `` xy 2y Ê 0 œ 2x `` xy 2y Ê `` xy œ xy Ê Š ``wy ‹ œ (2x) ˆ xy ‰ (2y)(1) (2z)(0) œ 2y 2y œ 0
z
xœx Ñ
x
y œ y(xß z)
Ä w Ê ˆ ``wz ‰x œ ``wx `` xz ``wy `` yz ``wz `` zz ; `` xz œ 0 and `` zz œ 2x `` xz 2y `` yz
(b) Œ Ä
z
Ï zœz Ò
Î
"
Ê 1 œ 2y `` yz Ê `` yz œ #"y Ê ˆ ``wz ‰x œ (2x)(0) (2y) Š 2y
‹ (2z)(1) œ 1 2z
Î x œ x(yß z) Ñ
y
yœy
Ä w Ê ˆ ``wz ‰y œ ``wx `` xz ``wy `` yz ``wz `` zz ; `` yz œ 0 and `` zz œ 2x `` xz 2y `` yz
(c) Œ Ä
z
Ï zœz Ò
1
Ê 1 œ 2x `` xz Ê `` xz œ 2x
Ê ˆ ``wz ‰ œ (2x) ˆ #"x ‰ (2y)(0) (2z)(1) œ 1 2z
y
2. w œ x# y z sin t and x y œ t:
Î xœx Ñ
ÎxÑ
Ð yœy Ó
y
Ä Ð
Ä w Ê Š ``wy ‹ œ ``wx `` xy ``wy `` yy ``wz `` yz ``wt ``yt ; `` xy œ 0, `` yz œ 0, and
(a)
Ó
œ
z
z
xz
ÏzÒ
Ït œ x yÒ
ß
`t
`y œ 1
Ê Š ``wy ‹ œ (2x)(0) (1)(1) (1)(0) (cos t)(1) œ 1 cos t œ 1 cos (x y)
xßt
Îx œ t yÑ
ÎyÑ
Ð yœy Ó
z
Ä Ð
Ä w Ê Š ``wy ‹ œ ``wx `` xy ``wy `` yy ``wz `` yz ``wt ``yt ; `` yz œ 0 and ``yt œ 0
(b)
Ó
z
z
œ
zt
ÏtÒ
Ï tœt Ò
ß
Ê `` xy œ ``yt `` yy œ 1 Ê Š ``wy ‹ œ (2x)(1) (1)(1) (1)(0) (cos t)(0) œ 1 2at yb œ 1 2y 2t
zßt
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859
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Chapter 14 Partial Derivatives
Î xœx Ñ
ÎxÑ
Ð yœy Ó
y
Ä Ð
Ä w Ê ˆ ``wz ‰x y œ ``wx `` xz ``wy `` yz ``wz `` zz ``wt ``zt ; `` xz œ 0 and `` yz œ 0
(c)
Ó
œ
z
z
ÏzÒ
Ït œ x yÒ
ß
Ê ˆ ``wz ‰x y œ (2x)(0) (1)(0) (1)(1) (cos t)(0) œ 1
ß
Îx œ t yÑ
ÎyÑ
Ð yœy Ó
z
Ä Ð
Ä w Ê ˆ ``wz ‰y t œ ``wx `` xz ``wy `` yz ``wz `` zz ``wt ``zt ; `` yz œ 0 and ``zt œ 0
(d)
Ó
zœz
ÏtÒ
Ï tœt Ò
`
w
Ê ˆ ` z ‰ œ (2x)(0) (1)(0) (1)(1) (cos t)(0) œ 1
ß
yßt
Î xœx Ñ
ÎxÑ
Ð y œ t xÓ
z
Ä Ð
Ä w Ê ˆ ``wt ‰x z œ ``wx ``xt ``wy ``yt ``wz ``zt ``wt `` tt ; ``xt œ 0 and ``zt œ 0
(e)
Ó
zœz
ÏtÒ
Ï tœt Ò
Ê ˆ ``wt ‰ œ (2x)(0) (1)(1) (1)(0) (cos t)(1) œ 1 cos t
ß
xßz
Îx œ t yÑ
ÎyÑ
Ð yœy Ó
z
Ä Ð
Ä w Ê ˆ ``wt ‰y z œ ``wx ``xt ``wy ``yt ``wz ``zt ``wt `` tt ; ``yt œ 0 and ``zt œ 0
(f)
Ó
œ
z
z
ÏtÒ
Ï tœt Ò
Ê ˆ ``wt ‰ œ (2x)(1) (1)(0) (1)(0) (cos t)(1) œ cos t 2x œ cos t 2(t y)
ß
yßz
3. U œ f(Pß Vß T) and PV œ nRT
Î PœP Ñ
P
`V
`U `T
`U
V ‰
ˆ `U ‰
ˆ ` U ‰ ˆ nR
VœV
Ä U Ê ˆ ``UP ‰V œ ``UP `` PP `` U
(a) Œ Ä
V ` P ` T ` P œ ` P ` V (0) ` T
V
PV
ÏT œ
Ò
nR
V ‰
œ ``UP ˆ ``UT ‰ ˆ nR
ÎP œ V Ñ
V
U `V
`U `T
`U
ˆ ` U ‰ ˆ nR
‰ ˆ `U ‰
Ä U Ê ˆ ``UT ‰V œ ``UP `` TP `` V
(b) Œ Ä
VœV
`T `T `T œ `P
V ` V (0) ` T
T
Ï TœT Ò
‰ `U
œ ˆ ``UP ‰ ˆ nR
V `T
nRT
4. w œ x# y# z# and y sin z z sin x œ 0
Î xœx Ñ
x
yœy
Ä w Ê ˆ ``wx ‰y œ ``wx `` xx ``wy `` yx ``wz `` xz ; `` yx œ 0 and
(a) Œ Ä
y
Ï z œ z(xß y) Ò
z cos x
`z
1
(y cos z) `` xz (sin x) `` xz z cos x œ 0 Ê `` xz œ y cos
z sin x . At (0ß 1ß 1), ` x œ 1 œ 1
Ê ˆ ``wx ‰yk
(0ß1ß1)
œ (2x)(1) (2y)(0) (2z)(1)k Ð0ß1ß1Ñ œ 21#
Î x œ x(yß z) Ñ
y
yœy
Ä w Ê ˆ ``wz ‰y œ ``wx `` xz ``wy `` yz ``wz `` zz œ (2x) `` xz (2y)(0) (2z)(1)
(b) Œ Ä
z
Ï zœz Ò
œ (2x) `` xz 2z. Now (sin z) `` yz y cos z sin x (z cos x) `` xz œ 0 and `` yz œ 0
z sin x
Ê y cos z sin x (z cos x) `` xz œ 0 Ê `` xz œ y cos
. At (!ß "ß 1), `` xz œ (11)(1)0 œ 1"
z cos x
Ê ˆ ``wz ‰Ck (!,"ß1Ñ œ 2(0) ˆ 1" ‰ 21 œ 21
5. w œ x# y# yz z$ and x# y# z# œ 6
Î xœx Ñ
x
yœy
(a) Œ Ä
Ä w Ê Š ``wy ‹ œ ``wx `` yx ``wy `` yy ``wz `` yz
y
x
Ï z œ z(xß y) Ò
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 14.10 Partial Derivatives with Constrained Variables
œ a2xy# b (0) a2x# y zb (1) ay 3z# b `` yz œ 2x# y z ay 3z# b `` yz . Now (2x) `` xy 2y (2z) `` yz œ 0 and
`x
`y œ 0
Ê 2y (2z) `` yz œ 0 Ê `` yz œ yz . At (wß xß yß z) œ (4ß 2ß 1ß 1), `` yz œ "1 œ 1 Ê Š ``wy ‹ ¹
x (4ß2ß1ßc1)
œ c(2)(2)# (1) (1)d c1 3(1)# d (1) œ 5
Î x œ x(yß z) Ñ
y
yœy
(b) Œ Ä
Ä w Ê Š ``wy ‹ œ ``wx `` xy ``wy `` yy ``wz `` yz
z
z
Ï zœz Ò
œ a2xy# b `` yx a2x# y zb (1) ay 3z# b (0) œ a2x# yb `` yx 2x# y z. Now (2x) `` yx 2y (2z) `` yz œ 0 and
`z
`y œ 0
Ê (2x) `` xy 2y œ 0 Ê `` xy œ yx . At (wß xß yß z) œ (4ß 2ß 1ß 1), `` yx œ "2 Ê Š ``wy ‹ ¹
œ (2)(2)(1)
z
"# ‰ (2)(2)# (1) (1) œ 5
(4ß2ß1ßc1)
#ˆ
6. y œ uv Ê 1 œ v `` uy u `` vy ; x œ u# v# and `` xy œ 0 Ê 0 œ 2u `` uy 2v `` vy Ê `` vy œ ˆ uv ‰ `` uy Ê 1
#
#
œ v `` uy u Š uv `` uy ‹ œ Š v v u ‹ `` yu Ê `` yu œ v# v u# . At (uß v) œ ŠÈ2ß 1‹ , `` uy œ
"
#
1# ŠÈ2‹
œ 1
Ê Š `` uy ‹ œ 1
x
r
x œ r cos )
7. Œ Ä Œ
Ê ˆ ``xr ‰) œ cos ); x# y# œ r# Ê 2x 2y `` xy œ 2r ``xr and `` xy œ 0 Ê 2x œ 2r ``xr
)
y œ r sin ) Ê ``xr œ xr Ê ˆ ``xr ‰ œ È #x #
y
x y
8. If x, y, and z are independent, then ˆ ``wx ‰y z œ ``wx `` xx ``wy `` xy ``wz `` xz ``wt ``xt
ß
œ (2x)(1) (2y)(0) (4)(0) (1) ˆ ``xt ‰ œ 2x ``xt . Thus x 2z t œ 25 Ê 1 0 ``xt œ 0 Ê ``xt œ 1
Ê ˆ ``wx ‰ œ 2x 1. On the other hand, if x, y, and t are independent, then ˆ ``wx ‰
yßz
yßt
œ ``wx `` xx ``wy `` yx ``wz `` xz ``wt ``xt œ (2x)(1) (2y)(0) 4 `` xz (1)(0) œ 2x 4 `` xz .
Ê 1 2 `` xz 0 œ 0 Ê `` xz œ #" Ê ˆ ``wx ‰yßt œ 2x 4 ˆ #" ‰ œ 2x 2.
Thus, x 2z t œ 25
9. If x is a differentiable function of y and z, then f(xß yß z) œ 0 Ê `` xf `` xx `` yf `` yx `` zf `` xz œ 0 Ê `` xf `` yf `` yx œ 0
Ê Š `` xy ‹ œ `` f/f/`` yz . Similarly, if y is a differentiable function of x and z, Š `` yz ‹ œ `` f/f/`` xz and if z is a
z
x
differentiable function of x and y, ˆ `` xz ‰ œ `` f/f/`` xy . Then Š `` xy ‹ Š `` yz ‹ ˆ `` xz ‰
y
z
x
y
œ Š `` f/f/`` yz ‹ ˆ `` f/f/`` xz ‰ Š `` f/f/`` xy ‹ œ 1.
df ` u
df
`z
df ` u
df
`z
`z
10. z œ z f(u) and u œ xy Ê `` xz œ 1 du
` x œ 1 y du ; also ` y œ 0 du ` y œ x du so that x ` x y ` y
df ‰
df ‰
œ x ˆ1 y du
y ˆx du
œx
11. If x and y are independent, then g(xß yß z) œ 0 Ê `` gx `` xy `` yg `` yy `` gz `` yz œ 0 and `` xy œ 0 Ê `` yg `` gz `` yz œ 0
`y
Ê Š `` yz ‹ œ `` g/
g/` z , as claimed.
x
12. Let x and y be independent. Then f(xß yß zß w) œ 0, g(xß yß zß w) œ 0 and `` yx œ 0
Ê `` xf `` xx `` yf `` yx `` zf `` xz ``wf ``wx œ `` xf `` zf `` xz ``wf ``wx œ 0 and
`g `x
`g `y
`g `z
`g `w
`g
`g `z
`g `w
` x ` x ` y ` x ` z ` x ` w ` x œ ` x ` z ` x ` w ` x œ 0 imply
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
861
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Chapter 14 Partial Derivatives
`f `z
`f `w
`f
`z `x `w `x œ `x
`g `z `g `w œ `g
`z `x
`w `x
`x
Ê ˆ `` xz ‰y œ
c ``xf
» c `g
`x
`f
`z
» `g
`z
`f
`w
`g »
`w
`f
`w
`g »
`w
œ
``xf ``wg `` gx ``wf
`g `f
`f `g
`z `w `z `w
`f `g
`f `g
`z `w
`w `z
œ ``xf ``wg ``wf ``gx , as claimed.
Likewise, f(xß yß zß w) œ 0, g(xß yß zß w) œ 0 and `` yx œ 0 Ê `` xf `` yx `` yf `` yy `` zf `` yz ``wf ``wy
œ `` yf `` zf `` yz ``wf ``wy œ 0 and (similarly) `` gy `` gz `` yz ``wg ``wy œ 0 imply
`f `z
`f `w
`f
`z `y `w `y œ `y
`g `z `g `w œ `g
`z `y
`w `y
`y
Ê Š ``wy ‹ œ
x
`f
`z
» `g
`z
`f
`z
» `g
`z
c ``yf
c `` gy »
`f
`w
`g »
`w
`f `g `g `f
`f `g
`f `g
`z `w
`z `w
`w `z
œ `f`z`g`y `g`z ``f y œ ``fz ``gy ``yf ``zg , as claimed.
`z `w
CHAPTER 14 PRACTICE EXERCISES
1. Domain: All points in the xy-plane
Range: z 0
Level curves are ellipses with major axis along the y-axis
and minor axis along the x-axis.
2. Domain: All points in the xy-plane
Range: 0 z _
Level curves are the straight lines x y œ ln z with
slope 1, and z 0.
3. Domain: All (xß y) such that x Á 0 and y Á 0
Range: z Á 0
Level curves are hyperbolas with the x- and y-axes
as asymptotes.
4. Domain: All (xß y) so that x# y
Range: z 0
0
Level curves are the parabolas y œ x# c, c
0.
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Chapter 14 Practice Exercises
5. Domain: All points (xß yß z) in space
Range: All real numbers
Level surfaces are paraboloids of revolution with
the z-axis as axis.
6. Domain: All points (xß yß z) in space
Range: Nonnegative real numbers
Level surfaces are ellipsoids with center (0ß 0ß 0).
7. Domain: All (xß yß z) such that (xß yß z) Á (0ß !ß 0)
Range: Positive real numbers
Level surfaces are spheres with center (0ß 0ß 0) and
radius r 0.
8. Domain: All points (xß yß z) in space
Range: (0ß 1]
Level surfaces are spheres with center (0ß 0ß 0) and
radius r 0.
9.
lim
Ðxß yÑ Ä Ð1ß ln 2Ñ
ey cos x œ eln 2 cos 1 œ (2)(1) œ 2
2y
20
10.
lim
œ 0 cos 0 œ 2
Ðxß yÑ Ä Ð0ß 0Ñ x cos y
11.
lim
lim
œ
lim
œ 11 œ #
#
# œ
Ðxß yÑ Ä Ð1ß 1Ñ x y
Ðxß yÑ Ä Ð1ß 1Ñ (x y)(x y)
Ðxß yÑ Ä Ð1ß 1Ñ x y
xÁ „y
xÁ „y
12.
13.
14.
xy
xy
1
(xy 1) ax# y# xy 1b
x$ y$ 1
œ
lim
œ
lim
xy
1
xy 1
Ðxß yÑ Ä Ð1ß 1Ñ
Ðx ß y Ñ Ä Ð 1 ß 1Ñ
Ðxß yÑ Ä Ð1ß 1Ñ
lim
"
"
ax# y# xy 1b œ 1# † 1# 1 † 1 1 œ 3
lim
ln kx y zk œ ln k1 (1) ek œ ln e œ 1
lim
tan" (x y z) œ tan" (1 (1) (1)) œ tan" (1) œ 14
P Ä Ð1ß 1ß eÑ
P Ä Ð1 ß 1 ß 1 Ñ
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Chapter 14 Partial Derivatives
15. Let y œ kx# , k Á 1. Then
kx#
lim
œ
lim
#
#
# œ 1 k# which gives different limits for
Ðxß yÑ Ä Ð0ß 0Ñ x y
axß kx# b Ä Ð0ß 0Ñ x kx
#
yÁx
y
k
different values of k Ê the limit does not exist.
16. Let y œ kx, k Á 0. Then
#
x# y#
x# (kx)#
œ
lim
œ 1 k k which gives different limits for
xy
x(kx)
Ðxß yÑ Ä Ð0ß 0Ñ
(xß kxÑ Ä Ð0ß 0Ñ
lim
xy Á 0
different values of k Ê the limit does not exist.
17. Let y œ kx. Then
x# y#
x# k# x#
1 k#
lim
#
# œ x# k# x# œ 1 k# which gives different limits for different values
Ðxß yÑ Ä Ð0ß 0Ñ x y
of k Ê the limit does not exist so f(0ß 0) cannot be defined in a way that makes f continuous at the origin.
18. Along the x-axis, y œ 0 and
lim
Ðxß yÑ Ä Ð0ß 0Ñ
1, x 0
sin (x y)
sin x
œœ
, so the
kxkkyk œ xlim
", x 0
Ä 0 k xk
limit fails to exist
Ê f is not continuous at (0ß 0).
19. ``gr œ cos ) sin ), `` g) œ r sin ) r cos )
20. `` xf œ #" Š x# 2x
y# ‹ y
‹
x#
y #
1 ˆx‰
Š
y
2y
`f
"
œ x# x y# x# y y# œ xx# y# , ` y œ # Š x # y # ‹ Š 1x ‹
y #
1 ˆx‰
y
œ x# y y# x# x y# œ xx# y#
21. ``Rf" œ R"# , ``Rf# œ R"# , ``Rf$ œ R"#
"
#
$
22. hx (xß yß z) œ 21 cos (21x y 3z), hy (xß yß z) œ cos (21x y 3z), hz (xß yß z) œ 3 cos (21x y 3z)
`P
nT ` P
nR ` P
nRT
23. `` Pn œ RT
V , ` R œ V , ` T œ V , ` V œ V#
24. fr (rß jß Tß w) œ 2r"# j É 1Tw , fj (rß jß Tß w) œ #r"j# É 1Tw , fT (rß jß Tß w) œ ˆ #"rj ‰ Š È"1w ‹ Š 2È" T ‹
œ 4r"j É T1"w œ 4r"jT É 1Tw , fw (rß jß Tß w) œ ˆ #"rj ‰ É T1 ˆ "# w$Î# ‰ œ 4r"jw É 1Tw
#
#
#
#
` g
` g
"
25. `` gx œ "y , `` yg œ 1 yx# Ê `` xg# œ 0, `` yg# œ 2x
y $ , ` y ` x œ ` x` y œ y #
26. gx (xß y) œ ex y cos x, gy (xß y) œ sin x Ê gxx (xß y) œ ex y sin x, gyy (xß y) œ 0, gxy (xß y) œ gyx (xß y) œ cos x
#
#
#
#
#
27. `` xf œ 1 y 15x# x#2x 1 , `` yf œ x Ê `` xf# œ 30x ax2#2x1b# , `` yf# œ 0, ``y`fx œ ``x`fy œ 1
28. fx (xß y) œ 3y, fy (xß y) œ 2y 3x sin y 7ey Ê fxx (xß y) œ 0, fyy (xß y) œ 2 cos y 7ey , fxy (xß y) œ fyx (xß y) œ 3
"
t dy
29. ``wx œ y cos (xy 1), ``wy œ x cos (xy 1), dx
dt œ e , dt œ t 1
t
ˆ " ‰
Ê dw
dt œ [y cos (xy 1)]e [x cos (xy 1)] t1 ; t œ 0 Ê x œ 1 and y œ 0
¸ œ 0 † 1 [1 † (1)] ˆ 0" 1 ‰ œ 1
Ê dw
dt
t œ0
"Î# dy
30. ``wx œ ey , ``wy œ xey sin z, ``wz œ y cos z sin z, dx
, dt œ 1 "t , dz
dt œ t
dt œ 1
y "Î#
Ê dw
axey sin zb ˆ1 "t ‰ (y cos z sin z)1; t œ 1 Ê x œ 2, y œ 0, and z œ 1
dt œ e t
¸ œ 1 † 1 (2 † 1 0)(2) (0 0)1 œ 5
Ê dw
dt
t œ1
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Chapter 14 Practice Exercises
31. ``wx œ 2 cos (2x y), ``wy œ cos (2x y), ``xr œ 1, ``xs œ cos s, ``yr œ s, ``ys œ r
Ê ``wr œ [2 cos (2x y)](1) [ cos (2x y)](s); r œ 1 and s œ 0 Ê x œ 1 and y œ 0
Ê ``wr ¸
œ (2 cos 21) (cos 21)(0) œ 2; ``ws œ [2 cos (2x y)](cos s) [ cos (2x y)](r)
Ð1ß0Ñ
Ê
`w ¸
` s Ð1ß0Ñ œ (2 cos 21)(cos 0) (cos 21)(1) œ 2 1
`x
" ‰
`w ¸
2
u
ˆ x
ˆ2 "‰
32. ``wu œ dw
dx ` u œ 1 x# x# 1 a2e cos vb ; u œ v œ 0 Ê x œ 2 Ê ` u Ð0ß0Ñ œ 5 5 (2) œ 5 ;
`w
dw ` x
" ‰
u
ˆ x
` v œ dx ` v œ 1 x# x# 1 a2e sin vb
Ê ``wv ¸ Ð0ß0Ñ œ ˆ 52 5" ‰ (0) œ 0
dy
dz
33. `` xf œ y z, `` yf œ x z, `` zf œ y x, dx
dt œ sin t, dt œ cos t, dt œ 2 sin 2t
Ê df
dt œ (y z)(sin t) (x z)(cos t) 2(y x)(sin 2t); t œ 1 Ê x œ cos 1, y œ sin 1, and z œ cos 2
df ¸
Ê dt
œ (sin 1 cos 2)(sin 1) (cos 1 cos 2)(cos 1) 2(sin 1 cos 1)(sin 2)
t œ1
`s
dw
`w
dw ` s
dw
dw
`w
`w
dw
dw
34. ``wx œ dw
ds ` x œ (5) ds and ` y œ ds ` y œ (1) ds œ ds Ê ` x 5 ` y œ 5 ds 5 ds œ 0
1 y cos xy
Fx
35. F(xß y) œ 1 x y# sin xy Ê Fx œ 1 y cos xy and Fy œ 2y x cos xy Ê dy
dx œ Fy œ 2y x cos xy
œ 12yy xcoscosxyxy Ê at (xß y) œ (!ß 1) we have dy
dx ¹
Ð0ß1Ñ
œ 12" œ 1
xby
2y e
Fx
36. F(xß y) œ 2xy exy 2 Ê Fx œ 2y exy and Fy œ 2x exy Ê dy
dx œ Fy œ 2x exby
Ê at (xß y) œ (!ß ln 2) we have dy
dx ¹
Ð0ßln 2Ñ
2
œ 2 ln0 2 2 œ (ln 2 1)
#
#
37. ™ f œ ( sin x cos y)i (cos x sin y)j Ê ™ f k ˆ 14 14 ‰ œ "# i "# j Ê k ™ f k œ Ɉ "# ‰ ˆ "# ‰ œ È" œ
ß
È
È
2
È
È2
# ;
È
f
2
2
2
2
u œ k™
™f k œ # i # j Ê f increases most rapidly in the direction u œ # i # j and decreases most
È2
È2
È2
È2
# i # j ; (Du f)P! œ k ™ f k œ # and (Dcu f)P! œ # ;
7
u" œ kvvk œ È33i #4j4# œ 53 i 54 j Ê (Du" f)P! œ ™ f † u" œ ˆ "# ‰ ˆ 35 ‰ ˆ "# ‰ ˆ 45 ‰ œ 10
rapidly in the direction u œ
f
1
1
38. ™ f œ 2xec2y i 2x# ec2y j Ê ™ f k Ð1ß0Ñ œ #i #j Ê k ™ f k œ È2# (2)# œ 2È2; u œ k™
™f k œ È2 i È2 j
Ê f increases most rapidly in the direction u œ È12 i È12 j and decreases most rapidly in the direction
u œ È12 i È12 j ; (Du f)P! œ k ™ f k œ 2È2 and (Dcu f)P! œ 2È2 ; u" œ kvvk œ È1i #j 1# œ È12 i È12 j
Ê (Du" f)P! œ ™ f † u" œ (2) Š È"2 ‹ (2) Š È"2 ‹ œ 0
2
3
6
39. ™ f œ Š 2x 3y
6z ‹ i Š 2x 3y 6z ‹ j Š 2x 3y 6z ‹ k Ê ™ f k Ð 1ß 1ß1Ñ œ 2i 3j 6k ;
f
2i 3j 6k
2
3
6
2
3
6
u œ k™
™f k œ È2# 3# 6# œ 7 i 7 j 7 k Ê f increases most rapidly in the direction u œ 7 i 7 j 7 k and
decreases most rapidly in the direction u œ 27 i 37 j 67 k ; (Du f)P! œ k ™ f k œ 7, (Du f)P! œ 7;
u" œ kvvk œ 27 i 37 j 67 k Ê (Du" f)P! œ (Du f)P! œ 7
f
2
"
40. ™ f œ (2x 3y)i (3x 2)j (1 2z)k Ê ™ f k Ð0ß0ß0Ñ œ 2j k ; u œ k™
™f k œ È5 j È5 k Ê f increases most
rapidly in the direction u œ È25 j È"5 k and decreases most rapidly in the direction u œ È25 j È"5 k ;
k
(Du f)P! œ k ™ f k œ È5 and (Du f)P! œ È5 ; u" œ kvvk œ È1i#j1# œ È"3 i È"3 j È"3 k
1#
Ê (Du" f)P! œ ™ f † u" œ (0) Š È"3 ‹ (2) Š È"3 ‹ (1) Š È"3 ‹ œ È33 œ È3
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
865
866
Chapter 14 Partial Derivatives
41. r œ (cos 3t)i (sin 3t)j 3tk Ê v(t) œ (3 sin 3t)i (3 cos 3t)j 3k Ê v ˆ 13 ‰ œ 3j 3k
Ê u œ È"2 j È"2 k ; f(xß yß z) œ xyz Ê ™ f œ yzi xzj xyk ; t œ 13 yields the point on the helix (1ß 0ß 1)
Ê ™ f k Ð 1ß0ß1Ñ œ 1j Ê ™ f † u œ (1j) † Š È"2 j È"2 k‹ œ È12
42. f(xß yß z) œ xyz Ê ™ f œ yzi xzj xyk ; at (1ß 1ß 1) we get ™ f œ i j k Ê the maximum value of
Du f k
œ k ™ f k œ È3
Ð1ß1ß1Ñ
43. (a) Let ™ f œ ai bj at (1ß 2). The direction toward (2ß 2) is determined by v" œ (2 1)i (2 2)j œ i œ u
so that ™ f † u œ 2 Ê a œ 2. The direction toward (1ß 1) is determined by v# œ (1 1)i (1 2)j œ j œ u
so that ™ f † u œ 2 Ê b œ 2 Ê b œ 2. Therefore ™ f œ 2i 2j ; fx a1, 2b œ fy a1, 2b œ 2.
(b) The direction toward (4ß 6) is determined by v$ œ (4 1)i (6 2)j œ 3i 4j Ê u œ 35 i 45 j
Ê ™ f † u œ 14
5 .
44. (a) True
(b) False
(c) True
(d) True
45. ™ f œ 2xi j 2zk Ê
™ f k Ð0ß 1ß 1Ñ œ j 2k ,
™ f k Ð0ß0ß0Ñ œ j ,
™ f k Ð0ß 1ß1Ñ œ j 2k
46. ™ f œ 2yj 2zk Ê
™ f k Ð2ß2ß0Ñ œ 4j ,
™ f k Ð2ß 2ß0Ñ œ 4j ,
™ f k Ð2ß0ß2Ñ œ 4k ,
™ f k Ð2ß0ß 2Ñ œ 4k
47. ™ f œ 2xi j 5k Ê ™ f k Ð2ß 1ß1Ñ œ 4i j 5k Ê Tangent Plane: 4(x 2) (y 1) 5(z 1) œ 0
Ê 4x y 5z œ 4; Normal Line: x œ 2 4t, y œ 1 t, z œ 1 5t
48. ™ f œ 2xi 2yj k Ê ™ f k Ð1ß1ß2Ñ œ 2i 2j k Ê Tangent Plane: 2(x 1) 2(y 1) (z 2) œ 0
Ê 2x 2y z 6 œ 0; Normal Line: x œ 1 2t, y œ 1 2t, z œ 2 t
2y
`z ¸
`z
`z
49. `` xz œ x# 2x
y# Ê ` x Ð0ß1ß0Ñ œ 0 and ` y œ x# y# Ê ` y ¹
Ð0ß1ß0Ñ
œ 2; thus the tangent plane is
2(y 1) (z 0) œ 0 or 2y z 2 œ 0
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Chapter 14 Practice Exercises
50. `` xz œ 2x ax# y# b
#
Ê `` xz ¸ ˆ1 1 1 ‰ œ #" and `` yz œ 2y ax# y# b
ß ß
#
2
Ê `` yz ¹
ˆ1ß1ß 1 ‰
2
867
œ "# ; thus the tangent
plane is "# (x 1) "# (y 1) ˆz "# ‰ œ 0 or x y 2z 3 œ 0
51. ™ f œ ( cos x)i j Ê ™ f k Ð1ß1Ñ œ i j Ê the tangent
line is (x 1) (y 1) œ 0 Ê x y œ 1 1; the
normal line is y 1 œ 1(x 1) Ê y œ x 1 1
52. ™ f œ xi yj Ê ™ f k Ð1ß2Ñ œ i 2j Ê the tangent
line is (x 1) 2(y 2) œ 0 Ê y œ "# x #3 ; the normal
line is y 2 œ 2(x 1) Ê y œ 2x 4
53. Let f(xß yß z) œ x# 2y 2z 4 and g(xß yß z) œ y 1. Then ™ f œ 2xi 2j 2kk a1 1 12 b œ 2i 2j 2k
â
â
â i j kâ
â
â
and ™ g œ j Ê ™ f ‚ ™ g œ â 2 2 2 â œ 2i 2k Ê the line is x œ 1 2t, y œ 1, z œ "# 2t
â
â
â0 " 0â
ß ß
54. Let f(xß yß z) œ x y# z 2 and g(xß yß z) œ y 1. Then ™ f œ i 2yj kk a 12 1 12 b œ i 2j k and
â
â
â i j kâ
â
â
™ g œ j Ê ™ f ‚ ™ g œ â 1 2 1 â œ i k Ê the line is x œ "# t, y œ 1, z œ "# t
â
â
â0 " 0â
ß ß
55. f ˆ 14 ß 14 ‰ œ "# , fx ˆ 14 ß 14 ‰ œ cos x cos yk Ð1Î4ß1Î4Ñ œ "# , fy ˆ 14 ß 14 ‰ œ sin x sin yk Ð1Î4ß1Î4Ñ œ "#
Ê L(xß y) œ "# "# ˆx 14 ‰ "# ˆy 14 ‰ œ "# "# x "# y; fxx (xß y) œ sin x cos y, fyy (xß y) œ sin x cos y, and
fxy (xß y) œ cos x sin y. Thus an upper bound for E depends on the bound M used for kfxx k , kfxy k , and kfyy k .
With M œ
È2
È
1¸
1 ¸ ‰#
" È2 ˆ¸
¸
Ÿ 42 (0.2)# Ÿ 0.0142;
# we have kE(xß y)k Ÿ # Š # ‹ x 4 y 4
#
with M œ 1, kE(xß y)k Ÿ "# (1) ˆ¸x 14 ¸ ¸y 14 ¸‰ œ "# (0.2)# œ 0.02.
56. f(1ß 1) œ 0, fx (1ß 1) œ yk Ð1ß1Ñ œ 1, fy (1ß 1) œ x 6yk Ð1ß1Ñ œ 5 Ê L(xß y) œ (x 1) 5(y 1) œ x 5y 4;
fxx (xß y) œ 0, fyy (xß y) œ 6, and fxy (xß y) œ 1 Ê maximum of kfxx k , kfyy k , and kfxy k is 6 Ê M œ 6
Ê kE(xß y)k Ÿ "# (6) akx 1k ky 1kb# œ "# (6)(0.1 0.2)# œ 0.27
57. f(1ß 0ß 0) œ 0, fx (1ß 0ß 0) œ y 3zk Ð1ß0ß0Ñ œ 0, fy (1ß 0ß 0) œ x 2zk Ð1ß0ß0Ñ œ 1, fz (1ß 0ß 0) œ 2y 3xk Ð1ß0ß0Ñ œ 3
Ê L(xß yß z) œ 0(x 1) (y 0) 3(z 0) œ y 3z; f(1ß 1ß 0) œ 1, fx (1ß 1ß 0) œ 1, fy (1ß 1ß 0) œ 1, fz ("ß "ß !) œ 1
Ê L(xß yß z) œ 1 (x 1) (y 1) 1(z 0) œ x y z 1
58. f ˆ0ß !ß 14 ‰ œ 1, fx ˆ!ß 0ß 14 ‰ œ È2 sin x sin (y z)¹
ˆ0ß0ß 1 ‰
œ 0, fy ˆ!ß 0ß 14 ‰ œ È2 cos x cos (y z)¹
4
fz ˆ!ß 0ß 14 ‰ œ È2 cos x cos (y z)¹
ˆ0ß0ß 1 ‰
œ 1,
4
œ 1 Ê L(xß yß z) œ 1 1(y 0) 1 ˆz 14 ‰ œ 1 y z 14 ;
4
È2
È2
È2
È2
ˆ1 1 ‰
ˆ1 1 ‰
ˆ1 1 ‰
# , fx 4 ß 4 ß 0 œ # , fy 4 ß 4 ß 0 œ # , fz 4 ß 4 ß 0 œ #
È
È
È
È
È
È
È
È
L(xß yß z) œ #2 #2 ˆx 14 ‰ #2 ˆy 14 ‰ #2 (z 0) œ #2 #2 x #2 y #2 z
f ˆ 14 ß 14 ß 0‰ œ
Ê
ˆ0ß0ß 1 ‰
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868
Chapter 14 Partial Derivatives
59. V œ 1r# h Ê dV œ 21rh dr 1r# dh Ê dVk Ð1Þ5ß5280Ñ œ 21(1.5)(5280) dr 1(1.5)# dh œ 15,8401 dr 2.251 dh.
You should be more careful with the diameter since it has a greater effect on dV.
60. df œ (2x y) dx (x 2y) dy Ê df k Ð1ß2Ñ œ 3 dy Ê f is more sensitive to changes in y; in fact, near the point
(1ß 2) a change in x does not change f.
"
24
¸
61. dI œ R" dV RV# dR Ê dI¸ Ð24ß100Ñ œ 100
dV 100
# dR Ê dI dVœ1ßdRœ20 œ 0.01 (480)(.0001) œ 0.038,
" ‰
20 ‰
or increases by 0.038 amps; % change in V œ (100) ˆ 24
¸ 4.17%; % change in R œ ˆ 100
(100) œ 20%;
24
I œ 100
œ 0.24 Ê estimated % change in I œ dII ‚ 100 œ 0.038
0.24 ‚ 100 ¸ 15.83% Ê more sensitive to voltage change.
62. A œ 1ab Ê dA œ 1b da 1a db Ê dAk Ð10ß16Ñ œ 161 da 101 db; da œ „ 0.1 and db œ „ 0.1
¸ ¸ 2.61
¸
Ê dA œ „ 261(0.1) œ „ 2.61 and A œ 1(10)(16) œ 1601 Ê ¸ dA
A ‚ 100 œ 1601 ‚ 100 ¸ 1.625%
63. (a) y œ uv Ê dy œ v du u dv; percentage change in u Ÿ 2% Ê kduk Ÿ 0.02, and percentage change in v Ÿ 3%
dy
v du u dv
dv
dv
¸ du
¸ ¸ du
¸ ¸ dv
¸
Ê kdvk Ÿ 0.03; dy
œ du
y œ
uv
u v Ê ¹ y ‚ 100¹ œ u ‚ 100 v ‚ 100 Ÿ u ‚ 100 v ‚ 100
Ÿ 2% 3% œ 5%
dv
du
dv
du
dv
(b) z œ u v Ê dzz œ duu v œ u v u v Ÿ u v (since u 0, v 0)
dy
dv
¸
Ê ¸ dzz ‚ 100¸ Ÿ ¸ du
u ‚ 100 v ‚ 100 œ ¹ y ‚ 100¹
(0.425)(7)
(0.725)(7)
64. C œ 71.84w07425 h0 725 Ê Cw œ 71.84w
1 425 h0 725 and Ch œ 71.84w0 425 h1 725
Þ
Ê
Þ
Þ
Þ
Þ
Þ
2.975
5.075
dC œ 71.84w
1 425 h0 725 dw 71.84w0 425 h1 725 dh; thus when w œ 70 and h œ 180 we have
Þ
Þ
Þ
Þ
dCk Ð70ß180Ñ ¸ (0.00000225) dw (0.00000149) dh Ê 1 kg error in weight has more effect
65. fx (xß y) œ 2x y 2 œ 0 and fy (xß y) œ x 2y 2 œ 0 Ê x œ 2 and y œ 2 Ê (2ß 2) is the critical point;
#
œ 3 0 and fxx 0 Ê local minimum value
fxx (2ß 2) œ 2, fyy (#ß 2) œ 2, fxy (#ß 2) œ 1 Ê fxx fyy fxy
of f(#ß 2) œ 8
66. fx (xß y) œ 10x 4y 4 œ 0 and fy (xß y) œ 4x 4y 4 œ 0 Ê x œ 0 and y œ 1 Ê (0ß 1) is the critical point;
#
œ 56 0 Ê saddle point with f(0ß 1) œ 2
fxx (0ß 1) œ 10, fyy (0ß 1) œ 4, fxy (0ß 1) œ 4 Ê fxx fyy fxy
67. fx (xß y) œ 6x# 3y œ 0 and fy (xß y) œ 3x 6y# œ 0 Ê y œ 2x# and 3x 6 a4x% b œ 0 Ê x a1 8x$ b œ 0
Ê x œ 0 and y œ 0, or x œ "# and y œ "# Ê the critical points are (0ß 0) and ˆ "# ß "# ‰ . For (!ß !):
#
fxx (!ß !) œ 12xk Ð0ß0Ñ œ 0, fyy (!ß !) œ 12yk Ð0ß0Ñ œ 0, fxy (!ß 0) œ 3 Ê fxx fyy fxy
œ 9 0 Ê saddle point with
#
f(0ß 0) œ 0. For ˆ "# ß "# ‰: fxx œ 6, fyy œ 6, fxy œ 3 Ê fxx fyy fxy
œ 27 0 and fxx 0 Ê local maximum
"
"
"
value of f ˆ # ß # ‰ œ 4
68. fx (xß y) œ 3x# 3y œ 0 and fy (xß y) œ 3y# 3x œ 0 Ê y œ x# and x% x œ 0 Ê x ax$ 1b œ 0 Ê the critical
points are (0ß 0) and (1ß 1) . For (!ß !): fxx (!ß !) œ 6xk Ð0ß0Ñ œ 0, fyy (!ß !) œ 6yk Ð0ß0Ñ œ 0, fxy (!ß 0) œ 3
#
Ê fxx fyy fxy
œ 9 0 Ê saddle point with f(0ß 0) œ 15. For (1ß 1): fxx (1ß 1) œ 6, fyy (1ß 1) œ 6, fxy (1ß 1) œ 3
#
Ê fxx fyy fxy
œ 27 0 and fxx 0 Ê local minimum value of f(1ß 1) œ 14
69. fx (xß y) œ 3x# 6x œ 0 and fy (xß y) œ 3y# 6y œ 0 Ê x(x 2) œ 0 and y(y 2) œ 0 Ê x œ 0 or x œ 2 and
y œ 0 or y œ 2 Ê the critical points are (0ß 0), (0ß 2), (2ß 0), and (2ß 2) . For (!ß !): fxx (!ß !) œ 6x 6k Ð0ß0Ñ
#
œ 6, fyy (!ß !) œ 6y 6k Ð0ß0Ñ œ 6, fxy (!ß 0) œ 0 Ê fxx fyy fxy
œ 36 0 Ê saddle point with f(0ß 0) œ 0. For
#
(0ß 2): fxx (!ß 2) œ 6, fyy (0ß #) œ 6, fxy (!ß 2) œ 0 Ê fxx fyy fxy
œ 36 0 and fxx 0 Ê local minimum value of
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Chapter 14 Practice Exercises
869
#
f(!ß 2) œ 4. For (#ß 0): fxx (2ß 0) œ 6, fyy (#ß 0) œ 6, fxy (2ß 0) œ 0 Ê fxx fyy fxy
œ 36 0 and fxx 0
Ê local maximum value of f(2ß 0) œ 4. For (2ß 2): fxx (2ß 2) œ 6, fyy (2ß 2) œ 6, fxy (2ß 2) œ 0
#
Ê fxx fyy fxy
œ 36 0 Ê saddle point with f(2ß 2) œ 0.
70. fx (xß y) œ 4x$ 16x œ 0 Ê 4x ax# 4b œ 0 Ê x œ 0, 2, 2; fy (xß y) œ 6y 6 œ 0 Ê y œ 1. Therefore the critical
points are (0ß 1), (2ß 1), and (2ß 1). For (!ß 1): fxx (!ß 1) œ 12x# 16k Ð0ß1Ñ œ 16, fyy (!ß 1) œ 6, fxy (!ß 1) œ 0
#
Ê fxx fyy fxy
œ 96 0 Ê saddle point with f(0ß 1) œ 3. For (2ß 1): fxx (2ß 1) œ 32, fyy (2ß 1) œ 6,
#
fxy (2ß 1) œ 0 Ê fxx fyy fxy
œ 192 0 and fxx 0 Ê local minimum value of f(2ß 1) œ 19. For (#ß 1):
#
fxx (2ß 1) œ 32, fyy (#ß 1) œ 6, fxy (2ß 1) œ 0 Ê fxx fyy fxy
œ 192 0 and fxx 0 Ê local minimum value of
f(2ß 1) œ 19.
71. (i)
On OA, f(xß y) œ f(0ß y) œ y# 3y for 0 Ÿ y Ÿ 4
Ê f w (!ß y) œ 2y 3 œ 0 Ê y œ 3# . But ˆ!ß 3# ‰
is not in the region.
Endpoints: f(0ß 0) œ 0 and f(0ß 4) œ 28.
(ii) On AB, f(xß y) œ f(xß x 4) œ x# 10x 28
for 0 Ÿ x Ÿ 4 Ê f w (xß x 4) œ 2x 10 œ 0
Ê x œ 5, y œ 1. But (5ß 1) is not in the region.
Endpoints: f(4ß 0) œ 4 and f(!ß 4) œ 28.
(iii) On OB, f(xß y) œ f(xß 0) œ x# 3x for 0 Ÿ x Ÿ 4 Ê f w (xß 0) œ 2x 3 Ê x œ 3# and y œ 0 Ê ˆ 3# ß 0‰ is a
critical point with f ˆ 3# ß !‰ œ 94 .
Endpoints: f(0ß 0) œ 0 and f(%ß 0) œ 4.
(iv) For the interior of the triangular region, fx (xß y) œ 2x y 3 œ 0 and fy (xß y) œ x 2y 3 œ 0 Ê x œ 3
and y œ 3. But (3ß 3) is not in the region. Therefore the absolute maximum is 28 at (0ß 4) and the
absolute minimum is 94 at ˆ 3# ß !‰ .
On OA, f(xß y) œ f(0ß y) œ y# 4y 1 for
0 Ÿ y Ÿ 2 Ê f w (!ß y) œ 2y 4 œ 0 Ê y œ 2 and
x œ 0. But (0ß 2) is not in the interior of OA.
Endpoints: f(0ß 0) œ 1 and f(0ß 2) œ 5.
(ii) On AB, f(xß y) œ f(xß 2) œ x# 2x 5 for 0 Ÿ x Ÿ 4
Ê f w (xß 2) œ 2x 2 œ 0 Ê x œ 1 and y œ 2
Ê (1ß 2) is an interior critical point of AB with
f(1ß 2) œ 4. Endpoints: f(4ß 2) œ 13 and f(!ß 2) œ 5.
(iii) On BC, f(xß y) œ f(4ß y) œ y# 4y 9 for 0 Ÿ y Ÿ 2 Ê f w (4ß y) œ 2y 4 œ 0 Ê y œ # and x œ 4. But
(4ß 2) is not in the interior of BC. Endpoints: f(4ß 0) œ 9 and f(%ß 2) œ 13.
(iv) On OC, f(xß y) œ f(xß 0) œ x# 2x 1 for 0 Ÿ x Ÿ 4 Ê f w (xß 0) œ 2x 2 œ 0 Ê x œ 1 and y œ 0 Ê (1ß 0)
is an interior critical point of OC with f(1ß 0) œ 0. Endpoints: f(0ß 0) œ 1 and f(4ß 0) œ 9.
(v) For the interior of the rectangular region, fx (xß y) œ 2x 2 œ 0 and fy (xß y) œ 2y 4 œ 0 Ê x œ 1 and
y œ 2. But (1ß 2) is not in the interior of the region. Therefore the absolute maximum is 13 at (4ß 2)
and the absolute minimum is 0 at (1ß 0).
72. (i)
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870
Chapter 14 Partial Derivatives
73. (i)
On AB, f(xß y) œ f(2ß y) œ y# y 4 for
2 Ÿ y Ÿ 2 Ê f w (2ß y) œ 2y 1 Ê y œ "# and
x œ 2 Ê ˆ2ß "# ‰ is an interior critical point in AB
with f ˆ2ß "# ‰ œ 17
4 . Endpoints: f(2ß 2) œ 2 and
f(2ß 2) œ 2.
On BC, f(xß y) œ f(xß 2) œ 2 for 2 Ÿ x Ÿ 2
Ê f w (xß 2) œ 0 Ê no critical points in the interior of
BC. Endpoints: f(2ß 2) œ 2 and f(2ß 2) œ 2.
(iii) On CD, f(xß y) œ f(2ß y) œ y# 5y 4 for
2 Ÿ y Ÿ 2 Ê f w (2ß y) œ 2y 5 œ 0 Ê y œ 5# and x œ 2. But ˆ#ß 5# ‰ is not in the region.
(ii)
Endpoints: f(2ß 2) œ 18 and f(2ß 2) œ 2.
(iv) On AD, f(xß y) œ f(xß 2) œ 4x 10 for 2 Ÿ x Ÿ 2 Ê f w (xß 2) œ 4 Ê no critical points in the interior
of AD. Endpoints: f(2ß 2) œ 2 and f(2ß 2) œ 18.
(v) For the interior of the square, fx (xß y) œ y 2 œ 0 and fy (xß y) œ 2y x 3 œ 0 Ê y œ 2 and x œ 1
Ê (1ß 2) is an interior critical point of the square with f(1ß 2) œ 2. Therefore the absolute maximum
"‰
ˆ
is 18 at (2ß 2) and the absolute minimum is 17
4 at #ß # .
On OA, f(xß y) œ f(0ß y) œ 2y y# for 0 Ÿ y Ÿ 2
Ê f w (!ß y) œ 2 2y œ 0 Ê y œ 1 and x œ 0 Ê
(!ß 1) is an interior critical point of OA with
f(0ß 1) œ 1. Endpoints: f(0ß 0) œ 0 and f(0ß 2) œ 0.
(ii) On AB, f(xß y) œ f(xß 2) œ 2x x# for 0 Ÿ x Ÿ 2
Ê f w (xß 2) œ 2 2x œ 0 Ê x œ 1 and y œ 2
Ê (1ß 2) is an interior critical point of AB with
f(1ß 2) œ 1. Endpoints: f(0ß 2) œ 0 and f(2ß 2) œ 0.
(iii) On BC, f(xß y) œ f(2ß y) œ 2y y# for 0 Ÿ y Ÿ 2
Ê f w (2ß y) œ 2 2y œ 0 Ê y œ 1 and x œ 2
Ê (2ß 1) is an interior critical point of BC with f(2ß 1) œ 1. Endpoints: f(2ß 0) œ 0 and f(2ß 2) œ 0.
(iv) On OC, f(xß y) œ f(xß 0) œ 2x x# for 0 Ÿ x Ÿ 2 Ê f w (xß 0) œ 2 2x œ 0 Ê x œ 1 and y œ 0 Ê (1ß 0)
is an interior critical point of OC with f(1ß 0) œ 1. Endpoints: f(0ß 0) œ 0 and f(0ß 2) œ 0.
(v) For the interior of the rectangular region, fx (xß y) œ 2 2x œ 0 and fy (xß y) œ 2 2y œ 0 Ê x œ 1 and
y œ 1 Ê (1ß 1) is an interior critical point of the square with f(1ß 1) œ 2. Therefore the absolute maximum
is 2 at (1ß 1) and the absolute minimum is 0 at the four corners (0ß 0), (0ß 2), (2ß 2), and (2ß 0).
74. (i)
On AB, f(xß y) œ f(xß x 2) œ 2x 4 for
2 Ÿ x Ÿ 2 Ê f w (xß x 2) œ 2 œ 0 Ê no critical
points in the interior of AB. Endpoints: f(2ß 0) œ 8
and f(2ß 4) œ 0.
(ii) On BC, f(xß y) œ f(2ß y) œ y# 4y for 0 Ÿ y Ÿ 4
Ê f w (2ß y) œ 2y 4 œ 0 Ê y œ 2 and x œ 2
Ê (2ß 2) is an interior critical point of BC with
f(2ß 2) œ 4. Endpoints: f(2ß 0) œ 0 and f(2ß 4) œ 0.
(iii) On AC, f(xß y) œ f(xß 0) œ x# 2x for 2 Ÿ x Ÿ 2
Ê f w (xß 0) œ 2x 2 Ê x œ 1 and y œ 0 Ê (1ß 0) is an interior critical point of AC with f(1ß 0) œ 1.
Endpoints: f(2ß 0) œ 8 and f(2ß 0) œ 0.
(iv) For the interior of the triangular region, fx (xß y) œ 2x 2 œ 0 and fy (xß y) œ 2y 4 œ 0 Ê x œ 1 and
y œ 2 Ê (1ß 2) is an interior critical point of the region with f(1ß 2) œ 3. Therefore the absolute maximum
is 8 at (2ß 0) and the absolute minimum is 1 at (1ß 0).
75. (i)
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Chapter 14 Practice Exercises
76. (i)
(ii)
871
On AB, faxß yb œ faxß xb œ 4x# 2x% 16 for
2 Ÿ x Ÿ 2 Ê f w axß xb œ 8x 8x$ œ 0 Ê x œ 0
and y œ 0, or x œ 1 and y œ 1, or x œ 1 and y œ 1
Ê a0ß 0b, a1ß 1b, a1ß 1b are all interior points of AB
with fa0ß 0b œ 16, fa1ß 1b œ 18, and fa1ß 1b œ 18.
Endpoints: fa2ß 2b œ 0 and fa2ß 2b œ 0.
On BC, faxß yb œ fa2ß yb œ 8y y% for 2 Ÿ y Ÿ 2
3
Ê f w a2ß yb œ 8 4y$ œ 0 Ê y œ È
2 and x œ 2
3
Ê Š2ß È
2‹ is an interior critical point of BC with
3
3
f Š2ß È
2‹ œ 6 È
2. Endpoints: fa2ß 2b œ 32 and fa2ß 2b œ 0.
3
(iii) On AC, faxß yb œ faxß 2b œ 8x x% for 2 Ÿ x Ÿ 2 Ê f w axß 2b œ 8 4x$ œ 0 Ê x œ È
2 and y œ 2
3
3
3
Ê ŠÈ
2ß 2‹ is an interior critical point of AC with f ŠÈ
2ß 2‹ œ 6 È
2. Endpoints:
fa2ß 2b œ 0 and fa2ß 2b œ 32.
(iv) For the interior of the triangular region, fx axß yb œ 4y 4x$ œ 0 and fy axß yb œ 4x 4y$ œ 0 Ê x œ 0 and
y œ 0, or x œ 1 and y œ 1 or x œ 1 and y œ 1. But neither of the points a0ß 0b and a1ß 1b, or a1ß 1b are
interior to the region. Therefore the absolute maximum is 18 at (1ß 1) and (1ß 1), and the absolute minimum is
32 at a2ß 2b.
On AB, f(xß y) œ f(1ß y) œ y$ 3y# 2 for
1 Ÿ y Ÿ 1 Ê f w (1ß y) œ 3y# 6y œ 0 Ê y œ 0
and x œ 1, or y œ 2 and x œ 1 Ê (1ß 0) is an
interior critical point of AB with f(1ß 0) œ 2; (1ß 2)
is outside the boundary. Endpoints: f(1ß 1) œ 2
and f(1ß 1) œ 0.
(ii) On BC, f(xß y) œ f(xß 1) œ x$ 3x# 2 for
1 Ÿ x Ÿ 1 Ê f w (xß 1) œ 3x# 6x œ 0 Ê x œ 0
and y œ 1, or x œ 2 and y œ 1 Ê (0ß 1) is an
interior critical point of BC with f(!ß 1) œ 2; (2ß 1) is outside the boundary. Endpoints: f("ß 1) œ 0 and
f("ß 1) œ 2.
(iii) On CD, f(xß y) œ f("ß y) œ y$ 3y# 4 for 1 Ÿ y Ÿ 1 Ê f w ("ß y) œ 3y# 6y œ 0 Ê y œ 0 and x œ 1, or
y œ 2 and x œ 1 Ê ("ß 0) is an interior critical point of CD with f("ß 0) œ 4; (1ß 2) is outside the boundary.
Endpoints: f(1ß 1) œ 2 and f("ß 1) œ 0.
(iv) On AD, f(xß y) œ f(xß 1) œ x$ 3x# 4 for 1 Ÿ x Ÿ 1 Ê f w (xß 1) œ 3x# 6x œ 0 Ê x œ 0 and y œ 1,
or x œ 2 and y œ 1 Ê (0ß 1) is an interior point of AD with f(0ß 1) œ 4; (#ß 1) is outside the
boundary. Endpoints: f(1ß 1) œ 2 and f("ß 1) œ 0.
(v) For the interior of the square, fx (xß y) œ 3x# 6x œ 0 and fy (xß y) œ 3y# 6y œ 0 Ê x œ 0 or x œ 2, and
y œ 0 or y œ 2 Ê (0ß 0) is an interior critical point of the square region with f(!ß 0) œ 0; the points (0ß 2),
(2ß 0), and (2ß 2) are outside the region. Therefore the absolute maximum is 4 at (1ß 0) and the
absolute minimum is 4 at (0ß 1).
77. (i)
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Chapter 14 Partial Derivatives
On AB, f(xß y) œ f(1ß y) œ y$ 3y for 1 Ÿ y Ÿ 1
Ê f w (1ß y) œ 3y# 3 œ 0 Ê y œ „ 1 and x œ 1
yielding the corner points (1ß 1) and (1ß 1) with
f(1ß 1) œ 2 and f(1ß 1) œ 2.
(ii) On BC, f(xß y) œ f(xß 1) œ x$ 3x 2 for
1 Ÿ x Ÿ 1 Ê f w (xß 1) œ 3x# 3 œ 0 Ê no
solution. Endpoints: f("ß 1) œ 2 and f("ß 1) œ 6.
(iii) On CD, f(xß y) œ f("ß y) œ y$ 3y 2 for
1 Ÿ y Ÿ 1 Ê f w ("ß y) œ 3y# 3 œ 0 Ê no
solution. Endpoints: f(1ß 1) œ 6 and f("ß 1) œ 2.
(iv) On AD, f(xß y) œ f(xß 1) œ x$ 3x for 1 Ÿ x Ÿ 1 Ê f w (xß 1) œ 3x# 3 œ 0 Ê x œ „ 1 and y œ 1
yielding the corner points (1ß 1) and (1ß 1) with f(1ß 1) œ 2 and f(1ß 1) œ 2
(v) For the interior of the square, fx (xß y) œ 3x# 3y œ 0 and fy (xß y) œ 3y# 3x œ 0 Ê y œ x# and
x% x œ 0 Ê x œ 0 or x œ 1 Ê y œ 0 or y œ 1 Ê (!ß 0) is an interior critical point of the square
region with f(0ß 0) œ 1; (1ß 1) is on the boundary. Therefore the absolute maximum is 6 at ("ß 1) and
the absolute minimum is 2 at (1ß 1) and (1ß 1).
78. (i)
79. ™ f œ 3x# i 2yj and ™ g œ 2xi 2yj so that ™ f œ - ™ g Ê 3x# i 2yj œ -(2xi 2yj) Ê 3x# œ 2x- and
2y œ 2y- Ê - œ 1 or y œ 0.
CASE 1: - œ 1 Ê 3x# œ 2x Ê x œ 0 or x œ 23 ; x œ 0 Ê y œ „ 1 yielding the points (0ß 1) and (!ß 1); x œ 23
È
È
È
Ê y œ „ 35 yielding the points Š 32 ß 35 ‹ and Š 32 ß 35 ‹ .
CASE 2: y œ 0 Ê x# 1 œ 0 Ê x œ „ 1 yielding the points (1ß 0) and (1ß 0).
È
Evaluations give f a!ß „ 1b œ 1, f Š 23 ß „ 35 ‹ œ 23
27 , f("ß 0) œ 1, and f("ß 0) œ 1. Therefore the absolute
maximum is 1 at a!ß „ 1b and (1ß 0), and the absolute minimum is 1 at ("ß !).
80. ™ f œ yi xj and ™ g œ 2xi 2yj so that ™ f œ - ™ g Ê yi xj œ -(2xi 2yj) Ê y œ 2-x and
xy œ 2-y Ê x œ 2-(2-x) œ 4-# x Ê x œ 0 or 4-# œ 1.
CASE 1: x œ 0 Ê y œ 0 but (0ß 0) does not lie on the circle, so no solution.
CASE 2: 4-# œ 1 Ê - œ "# or - œ "# . For - œ "# , y œ x Ê 1 œ x# y# œ 2x# Ê x œ C œ „ È"2 yielding the
points Š È"2 ß È"2 ‹ and Š È"2 , È"2 ‹ . For - œ #" , y œ x Ê 1 œ x# y# œ 2x# Ê x œ „ È"2 and
y œ x yielding the points Š È"2 ß È"2 ‹ and Š È"2 , È"2 ‹ .
Evaluations give the absolute maximum value f Š È"2 ß È"2 ‹ œ f Š È"2 ß È"2 ‹ œ #" and the absolute minimum
value f Š È"2 ß È"2 ‹ œ f Š È"2 ß È"2 ‹ œ #" .
81. (i) f(xß y) œ x# 3y# 2y on x# y# œ 1 Ê ™ f œ 2xi (6y 2)j and ™ g œ 2xi 2yj so that ™ f œ - ™ g
Ê 2xi (6y 2)j œ -(2xi 2yj) Ê 2x œ 2x- and 6y 2 œ 2y- Ê - œ 1 or x œ 0.
È
È
CASE 1: - œ 1 Ê 6y 2 œ 2y Ê y œ "# and x œ „ #3 yielding the points Š „ #3 ß #" ‹ .
CASE 2: x œ 0 Ê y# œ 1 Ê y œ „ 1 yielding the points a!ß „ 1b .
È
Evaluations give f Š „ #3 ß "# ‹ œ "# , f(0ß 1) œ 5, and f(!ß 1) œ 1. Therefore "# and 5 are the extreme
values on the boundary of the disk.
(ii) For the interior of the disk, fx (xß y) œ 2x œ 0 and fy (xß y) œ 6y 2 œ 0 Ê x œ 0 and y œ "3
Ê ˆ!ß 13 ‰ is an interior critical point with f ˆ!ß 3" ‰ œ 3" . Therefore the absolute maximum of f on the
disk is 5 at (0ß 1) and the absolute minimum of f on the disk is "3 at ˆ!ß 3" ‰ .
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Chapter 14 Practice Exercises
873
82. (i) f(xß y) œ x# y# 3x xy on x# y# œ 9 Ê ™ f œ (2x 3 y)i (2y x)j and ™ g œ 2xi 2yj so that
™ f œ - ™ g Ê (2x 3 y)i (2y x)j œ -(2xi 2yj) Ê 2x 3 y œ 2x- and 2y x œ 2yx
x
Ê 2x(" -) y œ 3 and x 2y(1 -) œ 0 Ê 1 - œ 2y
and (2x) Š 2y
‹ y œ 3 Ê x# y# œ 3y
Ê x# œ y# 3y. Thus, 9 œ x# y# œ y# 3y y# Ê 2y# 3y 9 œ 0 Ê (2y 3)(y 3) œ 0
Ê y œ 3, 3# . For y œ 3, x# y# œ 9 Ê x œ 0 yielding the point (0ß 3). For y œ 3# , x# y# œ 9
È
È
È
3 3
3 3 3
27 3
Ê x# 94 œ 9 Ê x# œ 27
4 Ê x œ „ # . Evaluations give f(0ß 3) œ 9, f Š # ß # ‹ œ 9 4
È
È
¸ 20.691, and f Š 3 # 3 , 3# ‹ œ 9 274 3 ¸ 2.691.
(ii) For the interior of the disk, fx (xß y) œ 2x 3 y œ 0 and fy (xß y) œ 2y x œ 0 Ê x œ 2 and y œ 1
Ê (2ß 1) is an interior critical point of the disk with f(2ß 1) œ 3. Therefore, the absolute maximum of f on
È
È
the disk is 9 274 3 at Š 3 # 3 ß 3# ‹ and the absolute minimum of f on the disk is 3 at (2ß 1).
83. ™ f œ i j k and ™ g œ 2xi 2yj 2zk so that ™ f œ - ™ g Ê i j k œ -(2xi 2yj 2zk) Ê 1 œ 2x-,
1 œ 2y-, 1 œ 2z- Ê x œ y œ z œ -" . Thus x# y# z# œ 1 Ê 3x# œ 1 Ê x œ „ È"3 yielding the points
Š È"3 ß È"3 , È"3 ‹ and Š È"3 , È"3 , È"3 ‹ . Evaluations give the absolute maximum value of
f Š È"3 ß È"3 ß È"3 ‹ œ È33 œ È3 and the absolute minimum value of f Š È"3 ß È"3 ß È"3 ‹ œ È3.
84. Let f(xß yß z) œ x# y# z# be the square of the distance to the origin and g(xß yß z) œ x# zy 4. Then
™ f œ 2xi 2yj 2zk and ™ g œ 2xi zj yk so that ™ f œ - ™ g Ê 2x œ 2-x, 2y œ -z, and
2z œ -y Ê x œ 0 or - œ 1.
CASE 1: x œ 0 Ê zy œ 4 Ê z œ 4y and y œ 4z Ê 2 Š y4 ‹ œ -y and 2 ˆ 4z ‰ œ -z Ê -8 œ y# and
8
#
- œz
Ê y# œ z# Ê y œ „ z. But y œ x Ê z# œ 4 leads to no solution, so y œ z Ê z# œ 4
Ê z œ „ 2 yielding the points (0ß 2ß 2) and (0ß 2ß 2).
CASE 2: - œ 1 Ê 2z œ y and 2y œ z Ê 2y œ ˆ y# ‰ Ê 4y œ y Ê y œ 0 Ê z œ 0 Ê x# 4 œ 0 Ê
x œ „ 2 yielding the points (2ß 0ß 0) and (2ß !ß 0).
Evaluations give f(0ß 2ß 2) œ f(0ß 2ß 2) œ 8 and f(2ß 0ß 0) œ f(2ß !ß 0) œ 4. Thus the points (2ß 0ß 0) and
(2ß !ß 0) on the surface are closest to the origin.
85. The cost is f(xß yß z) œ 2axy 2bxz 2cyz subject to the constraint xyz œ V. Then ™ f œ - ™ g
Ê 2ay 2bz œ -yz, 2ax 2cz œ -xz, and 2bx 2cy œ -xy Ê 2axy 2bxz œ -xyz, 2axy 2cyz œ -xyz, and
2bxz 2cyz œ -xyz Ê 2axy 2bxz œ 2axy 2cyz Ê y œ ˆ bc ‰ x. Also 2axy 2bxz œ 2bxz 2cyz Ê z œ ˆ ca ‰ x.
#
#
Then x ˆ bc x‰ ˆ ca x‰ œ V Ê x$ œ cabV Ê width œ x œ Š cabV ‹
#
Height œ z œ ˆ ac ‰ Š cabV ‹
"Î$
#
œ Š abcV ‹
"Î$
"Î$
#
, Depth œ y œ ˆ bc ‰ Š cabV ‹
"Î$
#
œ Š bacV ‹
"Î$
, and
.
86. The volume of the pyramid in the first octant formed by the plane is V(aß bß c) œ "3 ˆ "# ab‰ c œ "6 abc. The point
(2ß 1ß 2) on the plane Ê 2a "b 2c œ 1. We want to minimize V subject to the constraint 2bc ac 2ab œ abc.
ac
ab
Thus, ™ V œ bc
6 i 6 j 6 k and ™ g œ (c 2b bc)i (2c 2a ac)j (2b a ab)k so that ™ V œ - ™ g
ac
ab
abc
Ê bc
6 œ -(c 2b bc), 6 œ -(2c 2a ac), and 6 œ -(2b a ab) Ê 6 œ -(ac 2ab abc),
abc
abc
6 œ -(2bc 2ab abc), and 6 œ -(2bc ac abc)
Ê -ac œ 2-bc and 2-ab œ 2-bc. Now - Á 0 since
a Á 0, b Á 0, and c Á 0 Ê ac œ 2bc and ab œ bc Ê a œ 2b œ c. Substituting into the constraint equation gives
y
2
2
2
x
z
a a a œ 1 Ê a œ 6 Ê b œ 3 and c œ 6. Therefore the desired plane is 6 3 6 œ 1 or x 2y z œ 6.
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874
Chapter 14 Partial Derivatives
87. ™ f œ (y z)i xj xk , ™ g œ 2xi 2yj , and ™ h œ zi xk so that ™ f œ - ™ g . ™ h
Ê (y z)i xj xk œ -(2xi 2yj) .(zi xk) Ê y z œ 2-x .z, x œ 2-y, x œ .x Ê x œ 0 or . œ 1.
CASE 1: x œ 0 which is impossible since xz œ 1.
CASE 2: . œ 1 Ê y z œ 2-x z Ê y œ 2-x and x œ 2-y Ê y œ (2-)(2-y) Ê y œ 0 or
4-# œ 1. If y œ 0, then x# œ 1 Ê x œ „ 1 so with xz œ 1 we obtain the points (1ß 0ß 1)
and (1ß 0ß 1). If 4-# œ 1, then - œ „ "# . For - œ "# , y œ x so x# y# œ 1 Ê x# œ "#
Ê x œ „ È"2 with xz œ 1 Ê z œ „ È2, and we obtain the points Š È"2 ß È"2 ß È2‹ and
Š È"2 ß È"2 ß È2‹ . For - œ "# , y œ x Ê x# œ "# Ê x œ „ È"2 with xz œ 1 Ê z œ „ È2,
and we obtain the points Š È"2 ß È"2 , È2‹ and Š È"2 ß È"2 ß È2‹ .
Evaluations give f(1ß 0ß 1) œ 1, f(1ß 0ß 1) œ 1, f Š È"2 ß È"2 ß È2‹ œ "# , f Š È"2 ß È"2 , È2‹ œ "# ,
f Š È"2 ß È"2 ß È2‹ œ 3# , and f Š È"2 ß È"2 ß È2‹ œ 3# . Therefore the absolute maximum is 3# at
Š È"2 ß È"2 ß È2‹ and Š È"2 ß È"2 ß È2‹ , and the absolute minimum is "# at Š È"2 ß È"2 ß È2‹ and
Š È"2 ß È"2 ß È2‹ .
88. Let f(xß yß z) œ x# y# z# be the square of the distance to the origin. Then ™ f œ 2xi 2yj 2zk ,
™ g œ i j k , and ™ h œ 4xi 4yj 2zk so that ™ f œ - ™ g . ™ h Ê 2x œ - 4x., 2y œ - 4y.,
and 2z œ - 2z. Ê - œ 2x(1 2.) œ 2y(1 2.) œ 2z(1 2.) Ê x œ y or . œ "# .
CASE 1: x œ y Ê z# œ 4x# Ê z œ „ 2x so that x y z œ 1 Ê x x 2x œ 1 or x x 2x œ 1
(impossible) Ê x œ "4 Ê y œ "4 and z œ "# yielding the point ˆ "4 ß "4 ß "# ‰ .
CASE 2: . œ "# Ê - œ 0 Ê 0 œ 2z(1 1) Ê z œ 0 so that 2x# 2y# œ 0 Ê x œ y œ 0. But the origin
(!ß 0ß 0) fails to satisfy the first constraint x y z œ 1.
Therefore, the point ˆ "4 ß 4" ß "# ‰ on the curve of intersection is closest to the origin.
89. (a) y, z are independent with w œ x# eyz and z œ x# y# Ê ``wy œ ``wx `` xy ``wy `` yy ``wz `` yz
œ a2xeyz b `` xy azx# eyz b (1) ayx# eyz b (0); z œ x# y# Ê 0 œ 2x `` xy 2y Ê `` xy œ yx ; therefore,
Š ``wy ‹ œ a2xeyz b ˆ xy ‰ zx# eyz œ a2y zx# b eyz
z
(b) z, x are independent with w œ x# eyz and z œ x# y# Ê ``wz œ ``wx `` xz ``wy `` yz ``wz `` zz
œ a2xeyz b (0) azx# eyz b `` yz ayx# eyz b (1); z œ x# y# Ê 1 œ 0 2y `` yz Ê `` yz œ #"y ; therefore,
1
z
ˆ ``wz ‰ œ azx# eyz b Š 2y
‹ yx# eyz œ x# eyz Šy 2y
‹
x
(c) z, y are independent with w œ x# eyz and z œ x# y# Ê ``wz œ ``wx `` xz ``wy `` yz ``wz `` zz
œ a2xeyz b `` xz azx# eyz b (0) ayx# eyz b (1); z œ x# y# Ê 1 œ 2x `` xz 0 Ê `` xz œ #"x ; therefore,
1 ‰
ˆ ``wz ‰ œ a2xeyz b ˆ 2x
yx# eyz œ a1 x# yb eyz
y
`V
`U `T
90. (a) T, P are independent with U œ f(Pß Vß T) and PV œ nRT Ê ``UT œ ``UP `` TP `` U
V `T `T `T
‰ ˆ ``VT ‰ ˆ ``UT ‰ (1); PV œ nRT Ê P ``VT œ nR Ê ``VT œ nR
œ ˆ ``UP ‰ (0) ˆ `` U
V
P ; therefore,
ˆ ``UT ‰ œ ˆ `` U
‰ ˆ nR
‰ `U
V
P `T
P
`U `P
`U `V
`U `T
(b) V, T are independent with U œ f(Pß Vß T) and PV œ nRT Ê `` U
V œ `P `V `V `V `T `V
U‰
œ ˆ ``UP ‰ ˆ ``VP ‰ ˆ `` V
(1) ˆ ``UT ‰ (0); PV œ nRT Ê V ``VP P œ (nR) ˆ ``VT ‰ œ 0 Ê ``VP œ VP ; therefore,
U
ˆ `` U
‰ œ ˆ ``UP ‰ ˆ VP ‰ `` V
V
T
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Chapter 14 Practice Exercises
875
91. Note that x œ r cos ) and y œ r sin ) Ê r œ Èx# y# and ) œ tan" ˆ yx ‰ . Thus,
`w
`w `r
`w `)
x
ˆ `w ‰
ˆ ` w ‰ y
` x œ ` r ` x ` ) ` x œ ` r Š È x # y # ‹ ` ) Š x # y# ‹
œ (cos )) ``wr ˆ sinr ) ‰ ``w) ;
y
`w
`w `r
`w `)
x
ˆ `w ‰
ˆ `w ‰
` y œ ` r ` y ` ) ) y œ ` r Š È x # y# ‹ ` ) Š x# y# ‹
œ (sin )) ``wr ˆ cosr ) ‰ ``w)
92. zx œ fu `` ux fv `` vx œ afu afv , and zy œ fu `` uy fv `` vy œ bfu bfv
`u
dw
`w
dw ` u
dw
" `w
dw
" `w
dw
93. `` uy œ b and `` ux œ a Ê ``wx œ dw
du ` x œ a du and ` y œ du ` y œ b du Ê a ` x œ du and b ` y œ du
Ê "a ``wx œ b" ``wy Ê b ``wx œ a ``wy
2(r s)
2(r s)
2y
2(r s)
"
`w
rs
94. ``wx œ x# 2x
y# 2z œ (r s)# (r s)# 4rs - œ 2 ar# 2rs s# b œ r s , ` y œ x# y# 2z œ #(r s)# œ (r s)# ,
and ``wz œ x# y2# 2z œ (r " s)# Ê ``wr œ ``wx ``xr ``wy ``yr ``wz ``zr œ r " s (rrs)s # ’ (r " s)# “ (2s) œ (r2rs)2s#
œ r 2 s and ``ws œ ``wx ``xs ``wy ``ys ``wz `` zs œ r " s (rrs)s # ’ (r " s)# “ (2r) œ r 2 s
95. eu cos v x œ 0 Ê aeu cos vb `` ux aeu sin vb `` vx œ 1; eu sin v y œ 0 Ê aeu sin vb `` ux aeu cos vb `` vx œ 0.
Solving this system yields `` ux œ eu cos v and `` vx œ eu sin v. Similarly, eu cos v x œ 0
Ê aeu cos vb `` uy aeu sin vb `` vy œ 0 and eu sin v y œ 0 Ê aeu sin vb `` uy aeu cos vb `` vy œ 1. Solving this
second system yields `` uy œ eu sin v and `` vy œ eu cos v. Therefore Š `` ux i `` uy j‹ † Š `` vx i `` vy j‹
œ caeu cos vb i aeu sin vb jd † caeu sin vb i aeu cos vb jd œ 0 Ê the vectors are orthogonal Ê the angle
between the vectors is the constant 1# .
96. `` g) œ `` xf `` x) `` yf `` y) œ (r sin )) `` xf (r cos )) `` yf
#
Ê `` )g# œ (r sin )) Š `` xf# `` x) ``y`fx `` y) ‹ (r cos )) `` xf (r cos )) Š ``x`fy `` x) `` yf# `` y) ‹ (r sin )) `` yf
#
#
#
#
œ (r sin )) Š `` x) `` y) ‹ (r cos )) (r cos )) Š `` x) `` y) ‹ (r sin ))
œ (r sin ) r cos ))(r sin ) r cos )) (r cos ) r sin )) œ (2)(2) (0 2) œ 4 2 œ 2 at
(rß )) œ ˆ2ß 1# ‰ .
97. (y z)# (z x)# œ 16 Ê ™ f œ 2(z x)i 2(y z)j 2(y 2z x)k ; if the normal line is parallel to the
yz-plane, then x is constant Ê `` xf œ 0 Ê 2(z x) œ 0 Ê z œ x Ê (y z)# (z z)# œ 16 Ê y z œ „ 4.
Let x œ t Ê z œ t Ê y œ t „ 4. Therefore the points are (tß t „ 4ß t), t a real number.
98. Let f(xß yß z) œ xy yz zx x z# œ 0. If the tangent plane is to be parallel to the xy-plane, then ™ f is
perpendicular to the xy-plane Ê ™ f † i œ 0 and ™ f † j œ 0. Now ™ f œ (y z 1)i (x z)j (y x 2z)k
so that ™ f † i œ y z 1 œ 0 Ê y z œ 1 Ê y œ 1 z, and ™ f † j œ x z œ 0 Ê x œ z. Then
z(1 z) (" z)z z(z) (z) z# œ 0 Ê z 2z# œ 0 Ê z œ "# or z œ 0. Now z œ "# Ê x œ "# and y œ "#
Ê ˆ "# ß "# ß "# ‰ is one desired point; z œ 0 Ê x œ 0 and y œ 1 Ê (0ß 1ß 0) is a second desired point.
99. ™ f œ -(xi yj zk) Ê `` xf œ -x Ê f(xß yß z) œ "# -x# g(yß z) for some function g Ê -y œ `` yf œ `` yg
Ê g(yß z) œ "# -y# h(z) for some function h Ê -z œ `` zf œ `` gz œ hw (z) Ê h(z) œ #" -z# C for some arbitrary
constant C Ê g(yß z) œ "# -y# ˆ "# -z# C‰ Ê f(xß yß z) œ "# -x# "# -y# "# -z# C Ê f(0ß 0ß a) œ "# -a# C
and f(0ß 0ß a) œ "# -(a)# C Ê f(0ß 0ß a) œ f(0ß 0ß a) for any constant a, as claimed.
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
876
Chapter 14 Partial Derivatives
œ lim f(0 su" ß 0 su# ßs0 su$ )f(0ß 0ß 0) , s 0
sÄ0
‰
100. ˆ df
ds u (0 0 0)
ß
ß ß
És# u#" s# u## s# u#$ 0
œ lim
sÄ0
s
,s0
sÉu#" u## u#$
œ lim
œ lim kuk œ 1;
s
sÄ0
sÄ0
however, ™ f œ Èx# xy# z# i Èx# yy# z# j Èx# zy# z# k fails to exist at the origin (0ß 0ß 0)
101. Let f(xß yß z) œ xy z 2 Ê ™ f œ yi xj k . At (1ß 1ß 1), we have ™ f œ i j k Ê the normal line is
x œ 1 t, y œ 1 t, z œ 1 t, so at t œ 1 Ê x œ 0, y œ 0, z œ 0 and the normal line passes through the origin.
102. (b) f(xß yß z) œ x# y# z# œ 4
Ê ™ f œ 2xi 2yj 2zk Ê at (2ß 3ß 3)
the gradient is ™ f œ 4i 6j 6k which is
normal to the surface
(c) Tangent plane: 4x 6y 6z œ 8 or
2x 3y 3z œ 4
Normal line: x œ 2 4t, y œ 3 6t, z œ 3 6t
CHAPTER 14 ADDITIONAL AND ADVANCED EXERCISES
1. By definition, fxy (!ß 0) œ lim
hÄ0
fx (0ß h) fx (0ß 0)
so we need to calculate the first partial derivatives in the
h
numerator. For (xß y) Á (0ß 0) we calculate fx (xß y) by applying the differentiation rules to the formula for
#
#
$
#
#
#
#
$
# $
ax y b (2x)
f(xß y): fx (xß y) œ xx#yyy# (xy) ax y b (2x)
œ xx#yyy# ax#4x yy# b# Ê fx (0ß h) œ hh# œ h.
ax # y # b #
For (xß y) œ (0ß 0) we apply the definition: fx (!ß 0) œ lim
hÄ0
fxy (0ß 0) œ lim
hÄ0
$
h 0
œ 1.
h
Similarly, fyx (0ß 0) œ lim
$ #
$
#
hÄ0
$
f(hß 0) f(0ß0)
œ lim 0 h 0 œ 0.
h
hÄ0
Then by definition
fy (hß 0) fy (!ß 0)
, so for (xß y) Á (0ß 0) we have
h
4x y
h
fy (xß y) œ xx#xy
y# ax# y# b# Ê fy (hß 0) œ h# œ h; for (xß y) œ (0ß 0) we obtain fy (0ß 0) œ lim
œ lim
hÄ0
2.
00
h œ 0.
`w
x
` x œ 1 e cos y
#
Then by definition fyx (0ß 0) œ lim
hÄ0
hÄ0
h0
h œ 1.
f(0ß h) f(!ß 0)
h
Note that fxy (0ß 0) Á fyx (0ß 0) in this case.
Ê w œ x ex cos y g(y); ``wy œ ex sin y gw (y) œ 2y ex sin y Ê gw (y) œ 2y
Ê g(y) œ y C; w œ ln 2 when x œ ln 2 and y œ 0 Ê ln 2 œ ln 2 eln 2 cos 0 0# C Ê 0 œ 2 C
Ê C œ 2. Thus, w œ x ex cos y g(y) œ x ex cos y y# 2.
3. Substitution of u u(x) and v œ v(x) in g(uß v) gives g(u(x)ß v(x)) which is a function of the independent
` g du
` g dv
` '
du
` '
dv
variable x. Then, g(uß v) œ 'u f(t) dt Ê dg
dx œ ` u dx ` v dx œ Š ` u u f(t) dt‹ dx Š ` v u f(t) dt‹ dx
v
v
v
` '
dv
du
dv
dv
du
œ Š ``u 'v f(t) dt‹ du
dx Š ` v u f(t) dt‹ dx œ f(u(x)) dx f(v(x)) dx œ f(v(x)) dx f(u(x)) dx
u
v
#
#
#
#
#
#
`r
d f ˆ `r ‰
` r
d f
`r
df ` r
4. Applying the chain rules, fx œ df
df
dr ` x Ê fxx œ Š dr# ‹ ` x
dr ` x# . Similarly, fyy œ Š dr# ‹ Š ` y ‹ dr ` y# and
#
#
#
#
` r
`r
x
` r
fzz œ Š ddr#f ‹ ˆ ``zr ‰ df
dr ` z# . Moreover, ` x œ Èx# y# z# Ê ` x# œ
#
Ê ``yr# œ
x# z#
`r
z
3 ; and ` z œ È #
x y # z#
ˆÈx# y# z# ‰
#
Ê ``z#r œ
y # z#
y
`r
3 ; `y œ È #
x y # z#
ˆÈx# y# z# ‰
x# y#
3 .
ˆÈ x # y # z # ‰
Next, fxx fyy fzz œ 0
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Chapter 14 Additional and Advanced Exercises
#
y z
‰
Ê Š ddr#f ‹ Š x# xy# z# ‹ ˆ df
dr Œ ˆÈ #
#
#
#
#
#
x y
‰
Š ddr#f ‹ Š x# yz # z# ‹ ˆ df
dr Œ ˆÈ #
#
#
#
y#
d# f
x y z# ‰
#
3
3
x # z#
Š dr# ‹ Š x# y# z# ‹ ˆ dr ‰ Œ ˆÈx# y# z# ‰3 df
d# f
x y z# ‰
877
d# f
œ 0 Ê dr# Š Èx# y# z# ‹ dr œ 0 Ê dr# r dr œ 0
2
df
2 df
df
2 dr
Ê drd (f w ) œ ˆ 2r ‰ f w , where f w œ df
Ê ln f w œ 2 ln r ln C Ê f w œ Cr# , or
dr Ê f œ r
w
w
df
#
dr œ Cr
Ê f(r) œ Cr b œ ar b for some constants a and b (setting a œ C)
5. (a) Let u œ tx, v œ ty, and w œ f(uß v) œ f(u(tß x)ß v(tß y)) œ f(txß ty) œ tn f(xß y), where t, x, and y are
independent variables. Then ntnc1 f(xß y) œ ``wt œ ``wu ``ut ``wv ``vt œ x ``wu y ``wv . Now,
`w
`w `u
`w `v
`w
`w
ˆ `w ‰
ˆ `w ‰
ˆ " ‰ ˆ ``wx ‰ . Likewise,
` x œ ` u ` x ` v ` x œ ` u (t) ` v (0) œ t ` u Ê ` u œ t
`w
`w `u
`w `v
ˆ `w ‰
ˆ `w ‰
` y œ ` u ` y ` v ` y œ ` u (0) ` v (t)
Ê ``wv œ ˆ "t ‰ Š ``wy ‹ . Therefore,
ntnc1 f(xß y) œ x ``wu y ``wv œ ˆ xt ‰ ˆ ``wx ‰ ˆ yt ‰ Š ``wy ‹. When t œ 1, u œ x, v œ y, and w œ f(xß y)
Ê ``wx œ `` xf and ``wy œ `` xf Ê nf(xß y) œ x `` xf y `` yf , as claimed.
(b) From part (a), ntnc1 f(xß y) œ x ``wu y ``wv . Differentiating with respect to t again we obtain
n(n 1)tnc2 f(xß y) œ x `` uw# ``ut x ``v`wu ``vt y ``u`wv ``ut y `` vw# ``vt œ x# `` uw# 2xy ``u`wv y# `` vw# .
#
#
#
#
#
#
#
#
#
#
#
#
Also from part (a), `` xw# œ ``x ˆ ``wx ‰ œ ``x ˆt ``wu ‰ œ t `` uw# `` ux t ``v`wu `` vx œ t# `` uw# , `` yw# œ ``y Š ``wy ‹
#
#
#
#
#
#
œ ``y ˆt ``wv ‰ œ t ``u`wv `` yu t `` vw# `` yv œ t# `` vw# , and ``y`wx œ ``y ˆ ``wx ‰ œ ``y ˆt ``wu ‰ œ t `` uw# `` yu t ``v`wu `` yv
#
#
#
#
#
#
#
œ t# ``v`wu Ê ˆ t"# ‰ `` xw# œ `` uw# , ˆ t"# ‰ `` yw# œ `` vw# , and ˆ t"# ‰ ``y`wx œ ``v`wu
‰ Š ``y`wx ‹ Š yt# ‹ Š `` yw# ‹ for t Á 0. When t œ 1, w œ f(xß y) and
Ê n(n 1)tnc2 f(xß y) œ Š xt# ‹ Š `` xw# ‹ ˆ 2xy
t#
#
#
#
#
#
#
#
#
we have n(n 1)f(xß y) œ x# Š `` xf# ‹ 2xy Š ``x`fy ‹ y# Š `` yf# ‹ as claimed.
6. (a) lim
rÄ0
sin 6r
sin t
œ 1, where t œ 6r
6r œ t lim
Ä0 t
ˆ sin6h6h ‰ 1
f(0 hß 0) f(0ß 0)
6h
6h 6
œ lim
œ lim sin 6h
œ lim 6 cos12h
h
h
6h#
hÄ0
hÄ0
hÄ0
hÄ0
œ lim 3612sin 6h œ 0
(applying l'Hopital's
rule twice)
s
hÄ0
(b) fr (0ß 0) œ lim
ˆ sin6r6r ‰ ˆ sin6r6r ‰
f(rß ) h) f(rß ))
œ lim
œ lim h0 œ 0
h
h
hÄ0
hÄ0
hÄ0
(c) f) (rß )) œ lim
7. (a) r œ xi yj zk Ê r œ krk œ Èx# y# z# and ™ r œ Èx# xy# z# i Èx# yy# z# j Èx# zy# z# k œ rr
(b) rn œ ˆÈx# y# z# ‰
n
Ê ™ arn b œ nx ax# y# z# b
ÐnÎ2Ñ 1
ÐnÎ2Ñ 1
ÐnÎ2Ñ 1
i ny ax# y# z# b
j nz ax# y# z# b
k œ nrn 2 r
#
(c) Let n œ 2 in part (b). Then "# ™ ar# b œ r Ê ™ ˆ "# r# ‰ œ r Ê r# œ #" ax# y# z# b is the function.
(d) dr œ dxi dyj dzk Ê r † dr œ x dx y dy z dz, and dr œ rx dx ry dy rz dz œ xr dx yr dy zr dz
Ê r dr œ x dx y dy z dz œ r † dr
(e) A œ ai bj ck Ê A † r œ ax by cz Ê ™ (A † r) œ ai bj ck œ A
dy
dy
` f dx
` f dy
`f
`f
dx
dx
8. f(g(t)ß h(t)) œ c Ê 0 œ df
dt œ ` x dt ` y dt œ Š ` x i ` y j‹ † Š dt i dt j‹ , where dt i dt j is the tangent vector
Ê ™ f is orthogonal to the tangent vector
9. f(xß yß z) œ xz# yz cos xy 1 Ê ™ f œ az# y sin xyb i (z x sin xy)j (2xz y)k Ê ™ f(0ß 0ß 1) œ i j
Ê the tangent plane is x y œ 0; r œ (ln t)i (t ln t)j tk Ê rw œ ˆ "t ‰ i (ln t 1)j k ; x œ y œ 0, z œ 1
Ê t œ 1 Ê rw (1) œ i j k . Since (i j k) † (i j) œ rw (1) † ™ f œ 0, r is parallel to the plane, and
r(1) œ 0i 0j k Ê r is contained in the plane.
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
878
Chapter 14 Partial Derivatives
10. Let f(xß yß z) œ x$ y$ z$ xyz Ê ™ f œ a3x# yzb i a3y# xzb j a3z# xyb k Ê ™ f(0ß 1ß 1) œ i 3j 3k
$
Ê the tangent plane is x 3y 3z œ 0; r œ Š t4 2‹ i ˆ 4t 3‰ j (cos (t 2)) k
#
Ê rw œ Š 3t4 ‹ i ˆ t4# ‰ j (sin (t 2)) k ; x œ 0, y œ 1, z œ 1 Ê t œ 2 Ê rw (2) œ 3i j . Since
rw (2) † ™ f œ 0 Ê r is parallel to the plane, and r(2) œ i k Ê r is contained in the plane.
#
11. `` xz œ 3x# 9y œ 0 and `` yz œ 3y# 9x œ 0 Ê y œ "3 x# and 3 ˆ "3 x# ‰ 9x œ 0 Ê "3 x% 9x œ 0
Ê x ax$ 27b œ 0 Ê x œ 0 or x œ 3. Now x œ 0 Ê y œ 0 or (!ß 0) and x œ 3 Ê y œ 3 or (3ß 3). Next
` #z
` #z
` #z
` x# œ 6x, ` y# œ 6y, and ` x` y œ 9.
#
#
#
#
#
#
#
For (!ß 0), `` xz# `` yz# Š ``x`zy ‹ œ 81 Ê no extremum (a saddle point),
#
#
and for (3ß 3), `` xz# `` yz# Š ``x`zy ‹ œ 243 0 and `` xz# œ 18 0 Ê a local minimum.
12. f(xß y) œ 6xyeÐ2x3yÑ Ê fx (xß y) œ 6y(1 2x)eÐ2x3yÑ œ 0 and fy (xß y) œ 6x(1 3y)eÐ2x3yÑ œ 0 Ê x œ 0 and
y œ 0, or x œ "# and y œ 3" . The value f(0ß 0) œ 0 is on the boundary, and f ˆ "# ß "3 ‰ œ e"2 . On the positive y-axis,
f(0ß y) œ 0, and on the positive x-axis, f(xß 0) œ 0. As x Ä _ or y Ä _ we see that f(xß y) Ä 0. Thus the absolute
maximum of f in the closed first quadrant is e"2 at the point ˆ #" ß 3" ‰ .
#
#
#
2y
2z
13. Let f(xß yß z) œ xa# yb# cz# 1 Ê ™ f œ 2x
a# i b# j c# k Ê an equation of the plane tangent at the point
#
#
#
2x!
2y!
2z!
!‰
ˆ 2y! ‰
ˆ 2z! ‰
ˆ x! ‰
ˆ y! ‰
ˆ z! ‰
P! (x! ß y! ß y! ) is ˆ 2x
a# x b# y c# z œ a# b# c# œ 2 or a# x b# y c# z œ 1.
#
#
#
The intercepts of the plane are Š xa! ß 0ß 0‹ , Š0ß by! ß 0‹ and Š!ß !ß zc! ‹ . The volume of the tetrahedron formed by the
#
#
#
#
"
plane and the coordinate planes is V œ ˆ "3 ‰ ˆ #" ‰ Š xa! ‹ Š by! ‹ Š cz! ‹ Ê we need to maximize V(xß yß z) œ (abc)
6 (xyz)
#
#
#
#
#
(abc)
2y
"
2x
"
subject to the constraint f(xß yß z) œ xa# by# cz# œ 1. Thus, ’ (abc)
6 “ Š x# yz ‹ œ a# -, ’ 6 “ Š xy# z ‹ œ b# -,
#
"
2z
#
#
#
and ’ (abc)
6 “ Š xyz# ‹ œ c# -. Multiply the first equation by a yz, the second by b xz, and the third by c xy. Then equate
the first and second Ê a# y# œ b# x# Ê y œ ba x, x 0; equate the first and third Ê a# z# œ c# x# Ê z œ ca x, x 0;
substitute into f(xß yß z) œ 0 Ê x œ Èa3 Ê y œ Èb3 Ê z œ Èc3 Ê V œ
È3
# abc.
14. 2(x u) œ -, 2(y v) œ -, 2(x u) œ ., and 2(y v) œ 2.v Ê x u œ v y, x u œ .# , and
y v œ .v Ê x u œ .v œ .# Ê v œ "# or . œ 0.
CASE 1: . œ 0 Ê x œ u, y œ v, and - œ 0; then y œ x 1 Ê v œ u 1 and v# œ u Ê v œ v# 1
È
Ê v# v 1 œ 0 Ê v œ 1 „ #1 4 Ê no real solution.
CASE 2: v œ #" and u œ v# Ê u œ 4" ; x 4" œ #" y and y œ x 1 Ê x 4" œ x #" Ê 2x œ 4" Ê x œ 8"
#
#
#
Ê y œ 7 . Then f ˆ " ß 7 ß " ß " ‰ œ ˆ " " ‰ ˆ 7 " ‰ œ 2 ˆ 3 ‰ Ê the minimum distance is 3 È2.
8
8
8
4
#
8
4
8
#
8
8
(Notice that f has no maximum value.)
15. Let (x! ß y! ) be any point in R. We must show
lim
Ðhß kÑ Ä Ð0ß 0Ñ
lim
Ðxß yÑ Ä Ðx! ß y! Ñ
f(xß y) œ f(x! ß y! ) or, equivalently that
kf(x! hß y! k) f(x! ß y! )k œ 0. Consider f(x! hß y! k) f(x! ß y! )
œ [f(x! hß y! k) f(x! ß y! k)] [f(x! ß y! k) f(x! ß y! )]. Let F(x) œ f(xß y! k) and apply the Mean Value
Theorem: there exists 0 with x! 0 x! h such that Fw (0 )h œ F(x! h) F(x! ) Ê hfx (0ß y! k)
œ f(x! hß y! k) f(x! ß y! k). Similarly, k fy (x! ß () œ f(x! ß y! k) f(x! ß y! ) for some ( with
y! ( y! k. Then kf(x! hß y! k) f(x! ß y! )k Ÿ khfx (0ß y! k)k kkfy (x! ß ()k . If M, N are positive real
numbers such that kfx k Ÿ M and kfy k Ÿ N for all (xß y) in the xy-plane, then kf(x! hß y! k) f(x! ß y! )k
Ÿ M khk N kkk . As (hß k) Ä 0, kf(x! hß y! k) f(x! ß y! )k Ä 0 Ê
lim
kf(x! hß y! k) f(x! ß y! )k
Ðhß kÑ Ä Ð0ß 0Ñ
œ 0 Ê f is continuous at (x! ß y! ).
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Chapter 14 Additional and Advanced Exercises
879
dr
16. At extreme values, ™ f and v œ ddtr are orthogonal because df
dt œ ™ f † dt œ 0 by the First Derivative Theorem for
Local Extreme Values.
17. `` xf œ 0 Ê f(xß y) œ h(y) is a function of y only. Also, `` yg œ `` xf œ 0 Ê g(xß y) œ k(x) is a function of x only.
Moreover, `` yf œ `` xg Ê hw (y) œ kw (x) for all x and y. This can happen only if hw (y) œ kw (x) œ c is a constant.
Integration gives h(y) œ cy c" and k(x) œ cx c# , where c" and c# are constants. Therefore f(xß y) œ cy c"
and g(xß y) œ cx c# . Then f(1ß 2) œ g(1ß 2) œ 5 Ê 5 œ 2c c" œ c c# , and f(0ß 0) œ 4 Ê c" œ 4 Ê c œ "#
Ê c# œ 9# . Thus, f(xß y) œ "# y 4 and g(xß y) œ "# x #9 .
18. Let g(xß y) œ Du f(xß y) œ fx (xß y)a fy (xß y)b. Then Du g(xß y) œ gx (xß y)a gy (xß y)b
œ fxx (xß y)a# fyx (xß y)ab fxy (xß y)ba fyy (xß y)b# œ fxx (xß y)a# 2fxy (xß y)ab fyy (xß y)b# .
19. Since the particle is heat-seeking, at each point (xß y) it moves in the direction of maximal temperature
increase, that is in the direction of ™ T(xß y) œ aec2y sin xb i a2ec2y cos xb j . Since ™ T(xß y) is parallel to
c2y
cos x
the particle's velocity vector, it is tangent to the path y œ f(x) of the particle Ê f w (x) œ 2eec2y sin
x œ 2 cot x.
Integration gives f(x) œ 2 ln ksin xk C and f ˆ 14 ‰ œ 0 Ê 0 œ 2 ln ¸sin 14 ¸ C Ê C œ 2 ln
#
È2
2
# œ ln Š È2 ‹
œ ln 2. Therefore, the path of the particle is the graph of y œ 2 ln ksin xk ln 2.
20. The line of travel is x œ t, y œ t, z œ 30 5t, and the bullet hits the surface z œ 2x# 3y# when
30 5t œ 2t# 3t# Ê t# t 6 œ 0 Ê (t 3)(t 2) œ 0 Ê t œ 2 (since t 0). Thus the bullet hits the
surface at the point (2ß 2ß 20). Now, the vector 4xi 6yj k is normal to the surface at any (xß yß z), so that
n œ 8i 12j k is normal to the surface at (2ß 2ß 20). If v œ i j 5k , then the velocity of the particle
†25 ‰
600
50 ‰
after the ricochet is w œ v 2 projn v œ v Š 2knvk†#n ‹ n œ v ˆ 2209
n œ (i j 5k) ˆ 400
209 i 209 j 209 k
391
995
œ 191
209 i 209 j 209 k .
21. (a) k is a vector normal to z œ 10 x# y# at the point (!ß 0ß 10). So directions tangential to S at (!ß 0ß 10) will
be unit vectors u œ ai bj . Also, ™ T(xß yß z) œ (2xy 4) i ax# 2yz 14b j ay# 1b k
Ê ™ T(!ß 0ß 10) œ 4i 14j k . We seek the unit vector u œ ai bj such that Du T(0ß 0ß 10)
œ (4i 14j k) † (ai bj) œ (4i 14j) † (ai bj) is a maximum. The maximum will occur when ai bj
has the same direction as 4i 14j , or u œ È"53 (2i 7j).
(b) A vector normal to S at (1ß 1ß 8) is n œ 2i 2j k . Now, ™ T(1ß 1ß 8) œ 6i 31j 2k and we seek the unit
vector u such that Du T(1ß 1ß 8) œ ™ T † u has its largest value. Now write ™ T œ v w , where v is parallel
to ™ T and w is orthogonal to ™ T. Then Du T œ ™ T † u œ (v w) † u œ v † u w † u œ w † u. Thus
Du T(1ß 1ß 8) is a maximum when u has the same direction as w . Now, w œ ™ T Š ™knTk#†n ‹ n
62 2 ‰
‰ i ˆ31 152
‰ j ˆ2 76
‰
œ (6i 31j 2k) ˆ 124 (2i 2j k) œ ˆ6 152
41
9
9
9 k
127
58
w
"
œ 98
9 i 9 j 9 k Ê u œ kwk œ È29,097 (98i 127j 58k).
22. Suppose the surface (boundary) of the mineral deposit is the graph of z œ f(xß y) (where the z-axis points up into the air).
Then `` xf i `` yf j k is an outer normal to the mineral deposit at (xß y) and `` xf i `` yf j points in the direction of steepest
ascent of the mineral deposit. This is in the direction of the vector `` xf i `` yf j at (0ß 0) (the location of the 1st borehole)
that the geologists should drill their fourth borehole. To approximate this vector we use the fact that (0ß 0ß 1000),
(0ß 100ß 950), and (100ß !ß 1025) lie on the graph of z œ f(xß y). The plane containing these three points is a good
â
â
j
k â
â i
â
â
"00 50 â
approximation to the tangent plane to z œ f(xß y) at the point (0ß 0ß 0). A normal to this plane is â 0
â
â
25 â
â "00 0
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
880
Chapter 14 Partial Derivatives
œ 2500i 5000j 10,000k, or i 2j 4k. So at (0ß 0) the vector `` xf i `` yf j is approximately i 2j . Thus the
geologists should drill their fourth borehole in the direction of È"5 (i 2j) from the first borehole.
23. w œ ert sin 1x Ê wt œ rert sin 1x and wx œ 1ert cos 1x Ê wxx œ 1# ert sin 1x; wxx œ c"# wt , where c# is the
positive constant determined by the material of the rod Ê 1# ert sin 1x œ c"# arert sin 1xb
# #
Ê ar c# 1# b ert sin 1x œ 0 Ê r œ c# 1# Ê w œ ec 1 t sin 1x
24. w œ ert sin kx Ê wt œ rert sin kx and wx œ kert cos kx Ê wxx œ k# ert sin kx; wxx œ c"# wt
# #
Ê k# ert sin kx œ c"# arert sin kxb Ê ar c# k# b ert sin kx œ 0 Ê r œ c# k# Ê w œ ec k t sin kx.
# #
# # #
#
Now, w(Lß t) œ 0 Ê ec k t sin kL œ 0 Ê kL œ n1 for n an integer Ê k œ nL1 Ê w œ ec n 1 tÎL sin ˆ nL1 x‰ .
# # #
#
As t Ä _, w Ä 0 since ¸sin ˆ nL1 x‰¸ Ÿ 1 and ec n 1 tÎL Ä 0.
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
CHAPTER 15 MULTIPLE INTEGRALS
15.1 DOUBLE AND ITERATED INTEGRALS OVER RECTANGLES
1.
'12 '04 2xy dy dx œ '12 cx y# d 40 dx œ '12 16x dx œ c8 x# d 21 œ 24
2.
'02 'c11 ax yb dy dx œ '02 xy 12 y# ‘ "" dx œ '02 2x dx œ c x# d 20 œ 4
3.
!
'c01 'c11 (x y 1) dx dy œ 'c01 ’ x2 yx x“" dy œ 'c01 (2y 2) dy œ cy# 2yd "
œ1
4.
'01 '01 Š1 x 2 y ‹ dx dy œ '01 ’x x6 x2y “" dy œ '01 Š 65 y2 ‹dy œ ’ 56 y y6 “1 œ 23
#
"
#
#
#
3
#
3
0
0
5.
'03 '02 a4 y# b dy dx œ '03 ’4y y3 “ # dx œ '03 163 dx œ 163 x‘30 œ 16
6.
'03 'c02 ax# y 2xyb dy dx œ '03 ’ x 2y xy# “! dx œ '03 a4x 2x# b dx œ ’2x# 2x3 “3 œ 0
7.
'01 '01 1 yx y dx dy œ '01 clnl1 x yld"0 dy œ '01 lnl1 yldy œ cy lnl1 yl y lnl1 yld 10 œ 2 ln 2 1
8.
'14 '04 ˆ 2x Èy‰ dx dy œ '14 41 x2 xÈy‘ !4 dy œ '14 ˆ4 4 y1/2 ‰dy œ 4y 38 y3/2 ‘41 œ 923
9.
'0ln 2 '1ln 5 e2x y dy dx œ '0ln 2 ce2x y dln" 5 dx œ '0ln 2 a5e2x e2x 1 b dx œ 52 e2x "# e2x 1 ‘0ln 2 œ 32 a5 eb
$
!
# #
$
#
!
10. '0 '1 x y ex dy dx œ '0 "# x y2 ex ‘" dx œ '0 32 x ex dx œ 32 x ex 32 ex ‘0 œ 32
1
2
1
2
1Î2
1
1
11. 'c1 '0 y sin x dx dy œ 'c1 cy cos xd10 Î2 dy œ 'c1 y dy œ "# y2 ‘ 1 œ 32
2
2
2
2
12. '1 '0 asin x cos yb dx dy œ '1 ccos x x cos yd01 dy œ '1 a2 1 cos yb dy œ c2y 1 sin yd121 œ 21
21
1
21
21
13. ' ' a6 y# 2 xbdA œ '0 '0 a6 y# 2 xb dy dx œ '0 c2 y3 2 x yd0 dx œ '0 a16 4 xb dx œ c16 x 2 x2 d0 œ 14
1
2
1
2
1
1
R
14. ' '
R
Èx
y2 dA œ
'04 '12 Èy x dy dx œ '04 ’ Èyx “2 dx œ '04 "# x1Î2 dx œ 31 x3Î2 ‘40 œ 38
2
1
15. ' ' x y cos y dA œ 'c1 '0 x y cos y dy dx œ 'c1 cx y sin y x cos yd10 dx œ 'c1 a2xb dx œ cx2 dc1 œ 0
1
1
1
1
R
16. ' ' y sinax yb dA œ 'c1 '0 y sinax yb dy dx œ 'c1 cy cosax yb sinax ybd10 dx
0
1
0
R
œ 'c1 asinax 1b 1 cosax 1b sin xbdx œ ccosax 1b 1 sinax 1b cos xdc0 1 œ 4
0
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
1
882
Chapter 15 Multiple Integrals
17. ' ' ex y dA œ '0
ln 2
R
'0ln 2 ex y dy dx œ '0ln 2 cex y dln0 2 dx œ '0ln 2 aex ln 2 ex b dx œ cex ln 2 ex dln0 2 œ "#
18. ' ' x y ex y dA œ '0 '0 x y ex y dy dx œ '0 ’ "# ex y “ dx œ '0 ˆ "# ex "# ‰ dx œ "# ex "# x‘0 œ "# ae2 3b
2
2
1
2
2
2
1
2
2
0
R
19. ' ' xx2 y 1 dA œ '0 '0 xx2 y 1 dy dx œ '0 ’ 4axx2y 1b “ dx œ '0 x24x 1 dx œ c2 lnlx2 1ld0 œ 2 ln 2
3
1
2
1
3
2
4
1
1
0
R
1
1
20. ' ' x2 y2y 1 dA œ '0 '0 ax yby2 1 dx dy œ '0 ctan1 ax ybd0 dy œ '0 tan1 y dy œ y tan1 y "# lnl1 y2 l‘0 œ 14 "# ln 2
1
1
1
1
R
21. '1 '1
2
2
1
xy dy dx œ
'12 x" (ln 2 ln 1) dx œ (ln 2) '12 x" dx œ (ln 2)#
22. '0 '0 y cos xy dx dy œ '0 csin xyd 1! dy œ '0 sin 1y dy œ 1" cos 1y‘ ! œ 1" (1 1) œ 12
1
1
1
1
"
23. V œ ' ' fax, yb dA œ 'c1 'c1 ax2 y2 b dy dx œ 'c1 x2 y 31 y3 ‘ 1 dx œ 'c1 ˆ2 x2 32 ‰ dx œ 32 x3 32 x‘ 1 œ 38
1
1
1
1
"
"
R
2 3‘
2‰
88
24. V œ ' ' fax, yb dA œ '0 '0 a16 x2 y2 b dy dx œ '0 16 y x2 y 13 y3 ‘0 dx œ '0 ˆ 88
3 2 x dx œ 3 x 3 x 0
2
2
2
2
2
2
R
œ 160
3
25Þ V œ ' ' fax, yb dA œ '0 '0 a2 x yb dy dx œ '0 2 y x y "# y2 ‘ ! dx œ '0 ˆ 32 x‰ dx œ 32 x "# x2 ‘ ! œ 1
1
1
1
"
1
"
R
26Þ V œ ' ' fax, yb dA œ '0 '0 y2 dy dx œ '0 ’ y4 “ dx œ '0 1 dx œ cxd40 œ 4
4
2
4
2
2
4
0
R
27Þ V œ ' ' fax, yb dA œ '0
1Î2
R
'01Î4 2 sin x cos y dy dx œ '01Î2 c2 sin x sin yd01Î4 dx œ '01Î2 ŠÈ2 sin x‹ dx œ ’È2 cos x“1Î2
0
œ È2
16
‰
16 ‘
28. V œ ' ' fax, yb dA œ '0 '0 a4 y2 b dy dx œ '0 4 y 13 y3 ‘0 dx œ '0 ˆ 16
3 dx œ 3 x ! œ 3
1
2
1
2
1
"
R
15.2 DOUBLE INTEGRALS OVER GENERAL REGIONS
1.
2.
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 15.2 Double Integrals Over General Regions
3.
4.
5.
6.
7.
8.
9. (a) '! 'x3 dy dx
(b) '! '0
10. (a) '! '0 dy dx
(b) '! 'yÎ2 dx dy
11. (a) '! 'x2 dy dx
(b) '! 'yÎ3 dx dy
12. (a) '! '1 dy dx
(b) '1 'ln y dx dy
#
3
3
#
8
2x
3x
ex
6
9
e2
y1Î3
dx dy
3
Èy
2
Èx
13. (a) '! '0
9
8
dy dx
(b) '0 'y2 dx dy
3
9
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
883
884
Chapter 15 Multiple Integrals
14. (a) '!
1Î4
'tan1 x dy dx
tanc1 y
(b) '0 '0
1
15. (a) '!
ln 3
dx dy
'e1c dy dx
x
(b) '1Î3 'ln y dx dy
1
ln 3
16. (a) '! '0 dy dx '1 'ln x dy dx
1
1
e
1
(b) '0 '0 dx dy
1
ey
17. (a) '! 'x
1
3 2x
dy dx
(b) '0 '0 dx dy '1 '0
1
y
18. (a) '1 'x2
2
x2
Èy
3
a 3 y bÎ 2
dx dy
dy dx
Èy
(b) '0 'Èy dx dy '1 'y 2 dx dy
1
3
19. '0 '0 (x sin y) dy dx œ '0 c x cos yd x! dx
1
x
1
œ '0 (x x cos x) dx œ ’ x2 (cos x x sin x)“
1
#
#
1
!
œ 1# 2
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 15.2 Double Integrals Over General Regions
20. '0 '0
1
y dy dx œ '0 ’ y2 “
1
sin x
#
sin x
!
dx œ '0 "# sin# x dx
1
œ "4 '0 (1 cos 2x) dx œ "4 x "2 sin 2x‘ ! œ 14
1
21. '1
ln 8
1
'0ln yexby dx dy œ '1ln 8 cexbyd !ln y dy œ '1ln 8 ayey eyb dy
œ c(y 1)ey ey d 1ln 8 œ 8(ln 8 1) 8 e
œ 8 ln 8 16 e
22. '1 'y dx dy œ '1 ay# yb dy œ ’ y3 y# “
2
y#
2
$
#
œ ˆ 83 2‰ ˆ "3 "# ‰ œ 37 3# œ 56
#
"
23. '0 '0 3y$ exy dx dy œ '0 c3y# exy d 0 dy
1
y#
1
y#
œ '0 Š3y# ey 3y# ‹ dy œ ’ey y$ “ œ e 2
1
$
"
$
!
24.
È
'14 '0 x 3# eyÎÈx dy dx œ '14 32 Èx eyÎÈx ‘ 0Èx dx
4
%
œ 3# (e 1) '1 Èx dx œ 23 (e 1) ˆ 32 ‰ x$Î# ‘ " œ 7(e 1)
25. '1 'x
2
2x
1cx
26. '0 '0
1
x
y dy dx œ
'12 cx ln yd x2x dx œ (ln 2) '12 x dx œ 3# ln 2
ax# y# b dy dx œ '0 ’x# y y3 “
1
$
" x
0
dx œ '0 ’x# (1 x) (13x) “ dx œ '0 ’x# x$ (13x) “ dx
1
1
$
$
% "
$
%
œ ’ x3 x4 (11#x) “ œ ˆ "3 4" 0‰ ˆ0 0 1"# ‰ œ "6
!
1cu
27. '0 '0
1
ˆv Èu‰ dv du œ ' ’ v2 vÈu“
0
1
#
" u
0
du œ '0 ’ 12u# u Èu(1 u)“ du
1
#
"
œ '0 Š "# u u# u"Î# u$Î# ‹ du œ ’ u2 u# u6 32 u$Î# 25 u&Î# “ œ #" #" "6 32 25 œ #" 25 œ 10
1
#
#
$
"
!
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
885
886
Chapter 15 Multiple Integrals
28. '1 '0 es ln t ds dt œ '1 ces ln td 0ln t dt œ '1 (t ln t ln t) dt œ ’ t2 ln t t4 t ln t t“
2
ln t
2
2
œ (2 ln 2 1 2 ln 2 2) ˆ "4 1‰ œ 4"
cv
#
#
#
"
29. 'c2 'v 2 dp dv œ 2'c2 cpd vv dv œ 2'c2 2v dv
0
0
0
œ 2 cv# d c2 œ 8
0
È1cs
30. '0 '0
1
#
È1cs
8t dt ds œ '0 c4t# d 0
1
#
ds
œ '0 4 a1 s# b ds œ 4 ’s s3 “ œ 38
1
$
"
!
1Î3
31. 'c1Î3 '0
sec t
1Î3
1Î3
3 cos t du dt œ ' 1Î3 c(3 cos t)ud 0sec t
œ 'c1Î3 3 dt œ 21
3Î2
32. '0
'14 2u 4 v 2u dv du œ '03Î2 2u v 4 ‘ 14 2u du
#
œ '0
3Î2
Ð4 y)Î2
33. '2 '0
4
x2
34. ' 2 '0
0
$Î2
a3 2ub du œ c3u u# d ! œ 92
dx dy
dy dx
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 15.2 Double Integrals Over General Regions
35. '0 'x# dy dx
1
x
È1cy
36. '0 '1cy dx dy
1
37. '1 'ln y dx dy
e
1
38. '1 '0 dy dx
2
ln x
39. '0 '0
9
40.
1
2
È9cy
16x dx dy
È
'04 '0 4cx y dy dx
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
887
888
Chapter 15 Multiple Integrals
È1cx
41. 'c1 '0
1
È4cy
42. 'c2 '0
2
#
3y dy dx
#
6x dx dy
43. '0 'ey x y dx dy
1
44. '0
1 Î2
e
c1
'0sin y x y2 dx dy
45. '1 'ln x ax ybdy dx
e3
46. '0
1Î3
3
È
'tan 3x Èx y dy dx
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 15.2 Double Integrals Over General Regions
47. '0 'x siny y dy dx œ '0 '0 siny y dx dy œ '0 sin y dy œ 2
1
1
1
1
y
48. '0 'x 2y# sin xy dy dx œ '0 '0 2y# sin xy dx dy
2
2
2
y
œ '0 c2y cos xyd 0y dy œ '0 a2y cos y# 2yb dy
2
2
#
œ c sin y# y# d ! œ 4 sin 4
49. '0 'y x# exy dx dy œ '0 '0 x# exy dy dx œ '0 cxexy d 0x dx
1
1
1
x
1
œ '0 axex xb dx œ ’ "2 ex x# “ œ e# 2
1
"
#
#
#
!
4cx#
50. '0 '0
2
xe2y
4y dy dx œ
œ '0 ’ #x(4ey) “
4
51. '0
2
# 2y
Èln 3 Èln 3
'y/2
Èln 3
œ '0
È4cy
0
È
'04 '0 4cy 4xey dx dy
dy œ '0 e# dy œ ’ e4 “ œ e 4 "
4 2y
1
2y
%
)
!
Èln 3
ex dx dy œ '0
#
#
#
Èln 3
2xex dx œ cex d 0
52. '0 'ÈxÎ3 ey dy dx œ '0 '0
3
2y
1
$
3y#
'02x ex dy dx
#
œ eln 3 1 œ 2
$
ey dx dy
œ '0 3y# ey dy œ cey d ! œ e 1
1
53. '0
1Î16
$
$
"
'y1Î2 cos a161x& b dx dy œ '01Î2 '0x cos a161x& b dy dx
%
"Î%
161x b
œ '0 x% cos a161x& b dx œ ’ sin a80
“
1
1Î2
&
"Î#
!
œ 80"1
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
889
890
Chapter 15 Multiple Integrals
54. '0 'È$ x y%"1 dy dx œ '0 '0 y%"1 dx dy
8
2
y$
2
œ '0 y%y1 dy œ "4 cln ay% 1bd ! œ ln417
2
#
$
55. ' ' ay 2x# b dA
R
xb1
1cx
œ 'c1 'cxc1 ay 2x# b dy dx '0 'xc1 ay 2x# b dy dx
0
1
x "
1x
œ 'c1 "2 y# 2x# y‘ x1 dx '0 2" y# 2x# y‘ x1 dx
0
1
œ 'c1 "# (x 1)# 2x# (x 1) "# (x 1)# 2x# (x 1)‘dx
0
'0 "# (1 x)# 2x# (1 x) "# (x 1)# 2x# (x 1)‘ dx
1
œ 4 'c1 ax$ x# b dx 4 '0 ax$ x# b dx
0
1
%
$
œ 4 ’ x4 x3 “
56. ' ' xy dA œ '0
0
2Î3
R
%
c1
"
%
$
3
4 ‰
8
4 ’ x4 x3 “ œ 4 ’ (41) (31) “ 4 ˆ 4" 3" ‰ œ 8 ˆ 12
12
œ 12
œ 32
$
!
'x2x xy dy dx '21Î3 'x2 x xy dy dx
2Î3
2x
2 x
œ '0 "2 xy# ‘ x dx '2Î3 2" xy# ‘ x dx
1
œ '0 ˆ2x$ "# x$ ‰ dx '2Î3 "# x(2 x)# "# x$ ‘ dx
2Î3
œ '0
2Î3
1
3 $
# x dx '21Î3 a2x x# b dx
"
2Î3
2‰
8 ‰‘
6
‰ ˆ
4 ˆ 2 ‰ ˆ 27
ˆ 36 16 ‰ 13
œ 38 x% ‘ 0 x# 23 x$ ‘ #Î$ œ ˆ 38 ‰ ˆ 16
œ 81
27
81 1 3 9 3
81 81 81 œ 81
2cx
57. V œ '0 'x
1
ax# y# b dy dx œ '0 ’x# y y3 “
1
$
2cx
(2x)
dx œ '0 ’2x# 7x3 (23x) “ dx œ ’ 2x3 7x
12 12 “
1
x
$
$
$
%
7
" ‰
‰ 4
œ ˆ 23 12
12
ˆ0 0 16
12 œ 3
2cx#
58. V œ 'c2 'x
1
x# dy dx œ 'c2 cx# yd x
2cx#
1
%
"
!
dx œ 'c2 a2x# x% x$ b dx œ 23 x$ 15 x& 14 x% ‘ #
1
"
32
16 ‰
63
ˆ 40 12 15 ‰ ˆ 320 384 240 ‰ 189
œ ˆ 23 "5 4" ‰ ˆ 16
3 5 4 œ 60 60 60 60 60 60 œ 60 œ 20
4cx#
59. V œ 'c4 '3x
1
4cx
(x 4) dy dx œ 'c4 cxy 4yd 3x
dx œ 'c4 cx a4 x# b 4 a4 x# b 3x# 12xd dx
1
1
#
625
‰ 157 "
œ 'c4 ax$ 7x# 8x 16b dx œ 41 x% 37 x$ 4x# 16x‘ % œ ˆ 4" 37 12‰ ˆ 64
3 64 œ 3 4 œ 12
1
"
È 4 cx
60. V œ '0 '0
2
#
(3 y) dy dx œ '0 ’3y y2 “
2
#
$
È 4c x
0
#
dx œ '0 ’3È4 x# Š 4#x ‹“ dx
2
#
#
918
œ ’ 32 xÈ4 x# 6 sin" ˆ x# ‰ 2x x6 “ œ 6 ˆ 1# ‰ 4 86 œ 31 16
6 œ
3
!
61. V œ '0 '0 a4 y# b dx dy œ '0 c4x y# xd ! dy œ '0 a12 3y# b dy œ c12y y$ d ! œ 24 8 œ 16
2
3
2
$
2
#
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 15.2 Double Integrals Over General Regions
4cx#
62. V œ '0 '0
2
a4 x# yb dy dx œ '0 ’a4 x# b y y2 “
2
#
#
4cx#
!
dx œ '0 "# a4 x# b dx œ '0 Š8 4x# x# ‹ dx
2
2
#
%
" &‘
32
48032096
œ 8x 43 x$ 10
x ! œ 16 32
œ 128
3 10 œ
30
15
2cx
63. V œ '0 '0
2
a12 3y# b dy dx œ '0 c12y y$ d !
2
xb1
#x
dx œ '0 c24 12x (2 x)$ d dx œ ’24x 6x# (24x) “ œ 20
2
%
#
!
1cx
64. V œ 'c1 'cxc1 (3 3x) dy dx '0 'xc1 (3 3x) dy dx œ 6 'c1 a1 x# b dx 6 '0 (1 x)# dx œ 4 2 œ 6
0
1
1Îx
0
1
65. V œ '1 'c1Îx (x 1) dy dx œ '1 cxy yd 1Î1xÎx dx œ '1 1 "x ˆ1 x" ‰‘dx œ 2 '1 ˆ1 x" ‰ dx œ 2 cx ln xd #"
2
2
2
2
œ 2(1 ln 2)
66. V œ 4 '0
1Î3
'0sec x a1 y# b dy dx œ 4 '01Î3 ’y y3 “ sec x dx œ 4 '01Î3 Šsec x sec3 x ‹ dx
$
$
0
1Î$
œ 23 c7 ln ksec x tan xk sec x tan xd !
œ 23 ’7 ln Š2 È3‹ 2È3“
67.
68.
69.
'1_ 'ec1 x"y dy dx œ '1_ ’ lnx y “ " dx œ '1_ ˆ x x ‰ dx œ lim
x
$
$
$
ec x
1/ ˆ1cx ‰
È1cx
1
70. 'c1 'c1/È1cx (2y 1) dy dx œ 'c1 cy# ydº
1
1/
# 1Î#
bÄ_
ˆ "b 1‰ œ 1
_ _
'c_
' _ ax 1b"ay 1b -dx dy œ 2 '0_ Š y 21 ‹ Š lim
#
1
b
c1/ a1cx b
œ 4 lim c csin" b 0d œ 21
bÄ1
#
x" ‘ b œ lim
1
dx œ 'c1 È 2 # dx œ 4 lim c csin" xd !
1 x
bÄ1
# 1Î#
#
#
71.
bÄ_
#
bÄ_
tan" b tan" 0‹ dy œ 21 lim
bÄ_
'0b y "1 dy
#
œ 21 Š lim tan" b tan" 0‹ œ (21) ˆ 1# ‰ œ 1#
bÄ_
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
891
892
Chapter 15 Multiple Integrals
_ _
_
_
72. '0 '0 xecÐx 2yÑ dx dy œ '0 e2y lim
bÄ_
_
œ '0 ec2y dy œ "# lim
bÄ_
cxex ex d b0 dy œ '0 e2y lim
bÄ_
abeb eb 1b dy
aec2b 1b œ "#
3
73. ' ' f(xß y) dA ¸ "4 f ˆ "# ß 0‰ 8" f(0ß 0) 8" f ˆ "4 ß 0‰ œ "4 ˆ "# ‰ 8" ˆ0 "4 ‰ œ 32
R
"
128
‰
ˆ 9 11 ‰
ˆ 7 13 ‰
ˆ 9 13 ‰
74. ' ' f(xß y) dA ¸ "4 ’f ˆ 47 ß 11
4 f 4 ß 4 f 4 ß 4 f 4 ß 4 “ œ 16 (29 31 33 35) œ 16 œ 8
R
75. The ray ) œ 16 meets the circle x# y# œ 4 at the point ŠÈ3ß 1‹ Ê the ray is represented by the line y œ Èx3 . Thus,
È3 È4cx
È3
x# b
' ' f(xß y) dA œ ' ' È È4x# dy dx œ ' ’a4 x# b Èx3 È4 x# “ dx œ ”4x x3$ a4È
0
xÎ 3
0
3 3 •
#
$Î#
R
_
_
_
_
1)
' ˆ x#3x x#3x ‰ dx œ 6 '2
76. '2 '0 ax# xb "(y1)#Î$ dy dx œ '2 ’ 3(y
ax# xb “ dx œ 2
2
œ 6 lim
'2b ˆ x" 1 "x ‰ dx œ 6 lim
œ 6 ’ lim
ln ˆ1 "b ‰ ln 2“ œ 6 ln 2
bÄ_
bÄ_
2cx
77. V œ '0 'x
1
2
"Î$
bÄ_
0
cln (x 1) ln xd 2b œ 6 lim
bÄ_
ax# y# b dy dx œ '0 ’x# y y3 “
1
$
2cx
$
$
$
2
2
'
'
1x
dx
x(x1)
[ln (b 1) ln b ln 1 ln 2]
dx
%
7
‰ 4
œ ˆ 23 12
1"# ‰ ˆ0 0 16
12 œ 3
78. '0 atan" 1x tan" xb dx œ '0 'x
0
x
(2x)
œ '0 ’2x# 7x3 (23x) “ dx œ ’ 2x3 7x
12 12 “
1
È3
"
1y# dy dx œ
%
"
!
'02 'yyÎ1 1"y dx dy '221 'y2Î1 1"y dx dy
#
#
21 ˆ
21
1 " ‰ y
2 y ‰
# #
"
‰
œ 0 11y# dy 2 1y1# dy œ ˆ 12"
y #"1 ln a1 y# b‘ 2
1 cln a1 y bd ! 2 tan
œ ˆ 1211 ‰ ln 5 2 tan" 21 2"1 ln a1 41# b 2 tan" 2 #"1 ln 5
œ 2 tan" 21 2 tan" 2 2"1 ln a1 41# b ln#5
2ˆ
79. To maximize the integral, we want the domain to include all points where the integrand is positive and to
exclude all points where the integrand is negative. These criteria are met by the points (xß y) such that
4 x# 2y# 0 or x# 2y# Ÿ 4, which is the ellipse x# 2y# œ 4 together with its interior.
80. To minimize the integral, we want the domain to include all points where the integrand is negative and to
exclude all points where the integrand is positive. These criteria are met by the points (xß y) such that
x# y# 9 Ÿ 0 or x# y# Ÿ 9, which is the closed disk of radius 3 centered at the origin.
81. No, it is not possible. By Fubini's theorem, the two orders of integration must give the same result.
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
È
œ 209 3
Section 15.2 Double Integrals Over General Regions
82. One way would be to partition R into two triangles with the
line y œ 1. The integral of f over R could then be written
as a sum of integrals that could be evaluated by integrating
first with respect to x and then with respect to y:
' ' f(xß y) dA
R
2cÐyÎ2Ñ
œ '0 '2c2y
1
ÐyÎ2Ñ
f(xß y) dx dy '1 'yc1
2
2
f(xß y) dx dy.
Partitioning R with the line x œ 1 would let us write the
integral of f over R as a sum of iterated integrals with
order dy dx.
83. ' b ' b e x y dx dy œ ' b ' b e y e x dx dy œ ' b e y Œ' b e x dx dy œ Œ' b e x dx Œ' b e y dy
b
b
#
b
#
#
b
#
b
#
#
b
#
b
#
#
b
#
#
œ Œ'cb ecx dx œ Œ2 '0 ecx dx œ 4 Œ'0 ecx dx ; taking limits as b Ä _ gives the stated result.
b
#
b
#
b
#
84. '0 '0 (yx1)#Î$ dy dx œ '0 '0 (yx1)#Î$ dx dy œ '0 (y"1)#Î$ ’ x3 “ dy œ 3" '0 (ydy1)#Î$
1
3
3
#
1
3
#
$
"
3
!
œ "3 lim c '0 (ydy1)#Î$ "3 lim b 'b (ydy1)#Î$ œ lim c (y 1)"Î$ ‘ 0 lim b (y 1)"Î$ ‘ b
bÄ1
bÄ1
bÄ1
bÄ1
b
3
b
3
3
3
œ ’ lim c (b 1)"Î$ (1)"Î$ “ ’ lim b (b 1)"Î$ (2)"Î$ “ œ (0 1) Š0 È
2‹ œ 1 È
2
bÄ1
bÄ1
85-88. Example CAS commands:
Maple:
f := (x,y) -> 1/x/y;
q1 := Int( Int( f(x,y), y=1..x ), x=1..3 );
evalf( q1 );
value( q1 );
evalf( value(q1) );
89-94. Example CAS commands:
Maple:
f := (x,y) -> exp(x^2);
c,d := 0,1;
g1 := y ->2*y;
g2 := y -> 4;
q5 := Int( Int( f(x,y), x=g1(y)..g2(y) ), y=c..d );
value( q5 );
plot3d( 0, x=g1(y)..g2(y), y=c..d, color=pink, style=patchnogrid, axes=boxed, orientation=[-90,0],
scaling=constrained, title="#89 (Section 15.2)" );
r5 := Int( Int( f(x,y), y=0..x/2 ), x=0..2 ) + Int( Int( f(x,y), y=0..1 ), x=2..4 );
value( r5);
value( q5-r5 );
85-94. Example CAS commands:
Mathematica: (functions and bounds will vary)
You can integrate using the built-in integral signs or with the command Integrate. In the Integrate command, the
integration begins with the variable on the right. (In this case, y going from 1 to x).
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
893
894
Chapter 15 Multiple Integrals
Clear[x, y, f]
f[x_, y_]:= 1 / (x y)
Integrate[f[x, y], {x, 1, 3}, {y, 1, x}]
To reverse the order of integration, it is best to first plot the region over which the integration extends. This can be done
with ImplicitPlot and all bounds involving both x and y can be plotted. A graphics package must be loaded. Remember to
use the double equal sign for the equations of the bounding curves.
Clear[x, y, f]
<<Graphics`ImplicitPlot`
ImplicitPlot[{x==2y, x==4, y==0, y==1},{x, 0, 4.1}, {y, 0, 1.1}];
f[x_, y_]:=Exp[x2 ]
Integrate[f[x, y], {x, 0, 2}, {y, 0, x/2}] Integrate[f[x, y], {x, 2, 4}, {y, 0, 1}]
To get a numerical value for the result, use the numerical integrator, NIntegrate. Verify that this equals the original.
Integrate[f[x, y], {x, 0, 2}, {y, 0, x/2}] NIntegrate[f[x, y], {x, 2, 4}, {y, 0, 1}]
NIntegrate[f[x, y], {y, 0, 1},{x, 2y, 4}]
Another way to show a region is with the FilledPlot command. This assumes that functions are given as y = f(x).
Clear[x, y, f]
<<Graphics`FilledPlot`
FilledPlot[{x2 , 9},{x, 0,3}, AxesLabels Ä {x, y}];
f[x_, y_]:= x Cos[y2 ]
Integrate[f[x, y], {y, 0, 9}, {x, 0, Sqrt[y]}]
"
85. '1 '1 xy
dy dx ¸ 0.603
86. '0 '0 ec ˆx by ‰ dy dx ¸ 0.558
87. '0 '0 tan" xy dy dx ¸ 0.233
88.
3
1
x
1
1
'0 '2ye dx dy
4
#
È
#
'c11 '0 1cx 3È1 x# y# dy dx ¸ 3.142
#
The following graph was generated using
Mathematica.
89. Evaluate the integrals:
1
1
x#
œ '0 '0 ex dy dx '2 '0 ex dy dx
2
x/2
#
4
1
#
œ "4 4" ˆe4 2È1 erfia2b 2È1 erfia4b‰
¸ 1.1494 ‚ 106
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 15.2 Double Integrals Over General Regions
90. Evaluate the integrals:
Èy
'0 'x x cosay2 bdy dx œ '0 '0 x cosay2 bdx dy
3
9
9
The following graph was generated using
Mathematica.
2
œ sina481b ¸ 0.157472
91. Evaluate the integrals:
'0 'y
2
4
È2y
3
Èx
ax2 y xy2 bdx dy œ '0 'x2 /32 ax2 y xy2 bdy dx
8
3
The following graph was generated using
Mathematica.
œ 67,520
693 ¸ 97.4315
92. Evaluate the integrals:
4cy#
'0 '0
2
È4cx
exy dx dy œ '0 '0
4
exy dy dx
The following graph was generated using
Mathematica.
¸ 20.5648
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
895
896
Chapter 15 Multiple Integrals
93. Evaluate the integrals:
The following graph was generated using
Mathematica.
'1 '0 x1 y dy dx
1
2
4
2
œ '0 '1 x1 y dx dy '1 'Èy x1 y dx dy
2
x#
‰
1 lnˆ 27
4 ¸ 0.909543
94. Evaluate the integrals:
Èx
'1 'y Èx1y dx dy œ '1 '1
2
8
8
2
3
2
The following graph was generated using
Mathematica.
3
1
Èx2 y2 dy dx
¸ 0.866649
15.3 AREA BY DOUBLE INTEGRATION
1.
'02 '02cx dy dx œ '02 (2 x) dx œ ’2x x2 “ # œ 2,
#
2
2cy
2
or '0 '0 dx dy œ '0 (2 y) dy œ 2
2.
!
'02 '2x4 dy dx œ '02 (4 2x) dx œ c4x x# d 02 œ 4,
4
yÎ2
4
or '0 '0 dx dy œ '0 y# dy œ 4
'c12 'yccy2 dx dy œ 'c12 ay# y 2b dy
#
3.
$
#
œ ’ y3 y# 2y“
"
#
œ ˆ "3 "# 2‰ ˆ 83 2 4‰ œ 9#
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 15.3 Area by Double Integration
'02 'cyycy dx dy œ '02 a2y y# b dy œ ’y# y3 “ #
#
4.
$
!
œ 4 83 œ 43
5.
'0ln 2 '0e dy dx œ '0ln 2 ex dx œ cex d 0ln 2 œ 2 1 œ 1
6.
'1e 'ln2 xln x dy dx œ '1e ln x dx œ cx ln x xd 1e
x
œ (e e) (0 1) œ 1
'01 'y2ycy dx dy œ '01 a2y 2y# b dy œ y# 23 y$ ‘ "!
#
7.
#
œ "3
'c11 '2yy cc12 dx dy œ 'c11 ay# 1 2y# 2b dy
#
8.
#
œ 'c1 a1 y# b dy œ ’y y3 “
1
9.
$
"
"
œ 34
'02 'y3y 1 dx dy œ '02 cxd 3y
y dy
œ '0 a2yb dy œ cy2 d0 œ 4
2
2
y
10. '1 '1 y 1 dx dy œ '0 cxdln
1 y dy
2
ln y
2
œ '1 aln y 1 yb dy œ ’y ln y 2y y2 “
2
2
2
1
œ 2 ln 2 12
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
897
898
Chapter 15 Multiple Integrals
11. '0 'xÎ2 1 dy dx '1 'xÎ2 1 dy dx
1
2x
2
3x
' 3x
œ '0 cyd2x
xÎ2 dx 1 cydxÎ2 dx
1
2
œ '0 ˆ 32 x‰dx '1 ˆ3 32 x‰dx
1
2
1
2
œ 34 x2 ‘0 3x 34 x2 ‘1 œ 32
Èx
Èx
12. '0 'x 1 dy dx '1 'x 2 1 dy dx
1
4
œ '0 cydxx dx '1 cydx 2 dx
1
4
È
Èx
œ '0 ˆÈx x‰dx '1 ˆÈx x 2‰dx
1
4
1
4
œ 23 x3Î2 12 x2 ‘0 23 x3Î2 12 x2 2x‘1 œ 13
3
13. '0 'y#Î3 dx dy œ '0 Š2y y3 ‹ dy œ ’y# y9 “
6
2y
6
#
$
'
!
œ 36 216
9 œ 12
2xcx#
14. '0 'cx
3
dy dx œ '0 a3x x# b dx œ 32 x# 3" x$ ‘ !
3
$
9
œ 27
# 9œ #
15. '0
'sincosx x dy dx
1Î4
œ '0 (cos x sin x) dx œ csin x cos xd 01Î4
1Î4
È
È
œ Š #2 #2 ‹ (0 1) œ È2 1
yb2
16. 'c1 'y# dx dy œ 'c1 ay 2 y# b dy œ ’ y2 2y y3 “
2
2
#
œ ˆ2 4 83 ‰ ˆ "# 2 3" ‰ œ 5 "# œ 9#
$
2
c1
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 15.3 Area by Double Integration
17. 'c1 'c2x dy dx '0 'cxÎ2 dy dx
0
1 x
2
1 x
œ ' 1 (1 x) dx '0 ˆ1 x# ‰ dx
0
2
#
œ ’x x2 “
0
2
#
c1
’x x4 “ œ ˆ1 "# ‰ (2 1) œ 3#
0
Èx
18. '0 'x#c4 dy dx '0 '0 dy dx
2
0
4
œ '0 a4 x# b dx '0 x"Î# dx
2
4
$
2
32
œ ’4x x3 “ 32 x$Î# ‘ 0 œ ˆ8 38 ‰ 16
3 œ 3
4
0
19. (a) average œ 1"# '0 '0 sin (x y) dy dx œ 1"# '0 c cos (x y)d 01 dx œ 1"# '0 [ cos (x 1) cos x] dx
1
1
1
1
œ 1"# c sin (x 1) sin xd 01 œ 1"# [( sin 21 sin 1) ( sin 1 sin 0)] œ 0
"
(b) average œ
1#
Š#‹
'01 '01Î2 sin (x y) dy dx œ 12 '01 c cos (x y)d 1! Î# dx œ 12 '01 cos ˆx 1# ‰ cos x‘ dx
#
#
1
œ 12# sin ˆx 1# ‰ sin x‘ 0 œ 12# ˆ sin 3#1 sin 1‰ ˆ sin 1# sin 0‰‘ œ 14#
20. average value over the square œ '0 '0 xy dy dx œ '0 ’ xy2 “ dx œ '0 #x dx œ 4" œ 0.25;
1
1
1
È1cx
average value over the quarter circle œ ˆ 1 ‰ '0 '0
1
"
4
"
#
1
!
#
xy dy dx œ 14 '0 ’ xy2 “
1
#
È1cx
#
dx
0
œ 12 '0 ax x$ b dx œ 12 ’ x2 x4 “ œ #"1 ¸ 0.159. The average value over the square is larger.
1
#
4
"
!
8
21. average height œ "4 '0 '0 ax# y# b dy dx œ 4" '0 ’x# y y3 “ dx œ "4 '0 ˆ2x# 38 ‰ dx œ "# ’ x3 4x
3 “ œ 3
2
2
2
$
#
2
$
!
22. average œ (ln"2)# 'ln 2
2 ln 2
'
#
'
2 ln 2
23. 'c5 'c2 10,000e
dy dx œ 10,000 a1 e# b 'c5
x
0
!
'ln2 2ln 2 xy" dy dx œ (ln"2) 'ln2 2ln 2 ’ lnxy “ 2 ln 2 dx
ln 2
2 ln 2
œ (ln"2)# ln 2 "x (ln 2 ln ln 2 ln ln 2) dx œ ˆ ln"2 ‰ ln 2
œ ˆ ln"# ‰ (ln 2 ln ln 2 ln ln 2) œ 1
5
#
5
y
k k
1 #
2 ln 2
dx
ˆ " ‰
x œ ln # cln xd ln 2
dx
œ 10,000 a1 e# b ’
kxk
1 #
'c!5 1 dx '!5 1 dx “
x
#
x
#
!
&
œ 10,000 a1 ec2 b 2 ln ˆ1 x# ‰‘ & 10,000 a1 ec2 b 2 ln ˆ1 x# ‰‘ !
œ 10,000 a1 e# b 2 ln ˆ1 5# ‰‘ 10,000 a1 e# b 2 ln ˆ1 5# ‰‘ œ 40,000 a1 e# b ln ˆ 72 ‰ ¸ 43,329
2ycy#
24. '0 'y#
1
100(y 1) dx dy œ '0 c100(y 1)xd y2y# cy dy œ '0 100(y 1) a2y 2y# b dy œ 200 '0 ay y$ b dy
1
#
%
#
1
"
œ 200 ’ y2 y4 “ œ (200) ˆ "4 ‰ œ 50
!
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
1
899
900
Chapter 15 Multiple Integrals
25. Let (xi ß yi ) be the location of the weather station in county i for i œ 1ß á ß 254. The average temperature
254
! T(xi ßyi ) ?i A
in Texas at time t! is approximately iœ1
A
, where T(xi ß yi ) is the temperature at time t! at the
weather station in county i, ?i A is the area of county i, and A is the area of Texas.
26. Let y œ faxb be a nonnegative, continuous function on Òa, bÓ, then A œ ' ' dA œ 'a '0 dy dx œ 'a cydf0axb dx œ 'a faxb dx
b
faxb
b
R
15.4 DOUBLE INTEGRALS IN POLAR FORM
1. x2 y2 œ 92 Ê r œ 9 Ê 12 Ÿ ) Ÿ 21, 0 Ÿ r Ÿ 9
2. x2 y2 œ 12 Ê r œ 1, x2 y2 œ 42 Ê r œ 4 Ê 12 Ÿ ) Ÿ 12 , 1 Ÿ r Ÿ 4
3. y œ x Ê ) œ 14 , y œ x Ê ) œ 341 , y œ 1 Ê r œ csc ) Ê 14 Ÿ ) Ÿ 341 , 0 Ÿ r Ÿ csc )
4. x œ 1 Ê r œ sec ), y œ È3x Ê ) œ 13 Ê 0 Ÿ ) Ÿ 13 , 0 Ÿ r Ÿ sec )
5. x2 y2 œ 12 Ê r œ 1, x œ 2È3 Ê r œ 2È3 sec ), y œ 2 Ê r œ 2 csc ); 2È3 sec ) œ 2 csc ) Ê ) œ 16
Ê 0 Ÿ ) Ÿ 1 , 1 Ÿ r Ÿ 2È3 sec ); 1 Ÿ ) Ÿ 1 , 1 Ÿ r Ÿ 2È3 csc )
6
6
2
6. x2 y2 œ 22 Ê r œ 2, x œ 1 Ê r œ sec ); 2 œ sec ) Ê ) œ 13 or ) œ 13 Ê 13 Ÿ ) Ÿ 13 , sec ) Ÿ r Ÿ 2
7. x2 y2 œ 2x Ê r œ 2 cos ) Ê 12 Ÿ ) Ÿ 12 , 0 Ÿ r Ÿ 2 cos )
8. x2 y2 œ 2y Ê r œ 2 sin ) Ê 0 Ÿ ) Ÿ 1, 0 Ÿ r Ÿ 2 sin )
9.
È
'c11 '0 1cx dy dx œ '01 '01 r dr d) œ "# '01 d) œ 1#
#
È1cy
ax# y# b dx dy œ '0
'01 r$ dr d) œ 4" '01Î2 d) œ 18
È4cy
ax# y# b dx dy œ '0
'02 r$ dr d) œ 4 '01Î2 d) œ 21
10. '0 '0
1
11. '0 '0
2
1Î2
#
#
Èa cx
1Î2
12. 'ca 'cÈa#cx# dy dx œ '0 '0 r dr d) œ a2 '0 d) œ 1a#
#
a
21
#
a
#
13. '0 '0 x dx dy œ '1Î4 '0
6
1Î2
y
14. '0 '0 y dy dx œ '0
6 csc )
1Î4
21
r# cos ) dr d) œ 72 '1Î4 cot ) csc# ) d) œ 36 ccot# )d 1Î4 œ 36
1Î2
1Î2
'02 sec r# sin ) dr d) œ 83 '0 Î4 tan ) sec# ) d) œ 43
)
1
2
x
È3
'1x dy dx œ '1ÎÎ64 'csc 3 sec r dr d) œ ' ÎÎ64 ˆ 32 sec2 ) 12 csc2 )‰ d) œ 32 tan ) 12 cot )‘ ÎÎ64 œ 2 È3
15. '1
1
È
)
)
16. 'È2 'È4 y2 dy dx œ '1Î4 '2
2
y
1Î2
1
1
1
1
2 csc )
r dr d) œ '1Î6 a2csc2 ) 2b d) œ 2 cot ) 12 )‘ 1Î4 œ 2 12
1Î4
1Î2
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
b
Section 15.4 Double Integrals in Polar Form
17. 'c1 'cÈ1 c x# 1 È2x# y# dy dx œ '1
0
31Î2
0
'01 1 2r r dr d) œ 2 '131Î2 '01 ˆ1 1 " r ‰ dr d) œ 2 '131Î2 (1 ln 2) d)
œ (1 ln 2)1
È1 c x
18. 'c1 'cÈ1 c x# a1 x#2 y# b# dy dx œ 4 '0
1
19. '0
ln 2
1Î2
#
È
'01
' 1Î2 1 " r ‘ "! d) œ 2 '01Î2 d) œ 1
2r
dr d) œ 4 0
a1 r# b#
'0 (ln 2) c y eÈx y dx dy œ '01Î2 '0ln 2 rer dr d) œ '01Î2 a2 ln 2 1b d) œ 1# a2 ln 2 1b
#
#
#
#
È1 c y
20. 'c1 'cÈ1 c y# ln ax# y# 1b dx dy œ 4 '0
1Î2
#
1
È 2 x#
21. '0 'x
1
#
È2
ax 2yb dy dx œ '1Î4 '0
1 Î2
'01 ln ar# 1b r dr d) œ 2'01Î2 aln 4 1b d) œ 1aln 4 1b
1Î2
È
È
È
È
1 Î2
1 Î4
È2x x#
22. '1 '0
2
1
dy dx œ
ax # y # b #
3
3
0
œ '1Î4 Š 2 3 2 cos ) 4 3 2 sin )‹ d) œ ’ 2 3 2 sin ) 4 3 2 cos )“
1 Î2
œ
1
È 1 x#
È1 y#
'01 '0
1 Î4
)
'01Î4 'sec2cos) ) r1 r dr d) œ '01Î4 2r1 ‘2cos
d) œ œ '0 ˆ 21 cos2 ) 18 sec2 )‰ d)
sec )
4
2
x y dy dx or
x y dx dy
È3y
24. '1Î2 'È1 y# x dx dy or
1
È
È
'0 3Î2 'È1 1 x x dy dx 'È33Î2 'xÎ1 È3 x dy dx
#
25. '0 '0 y2 ax2 y2 b dy dx or
2
d)
2 ˆ1 È 2 ‰
3
1Î4
1
œ 14 ) 18 sin2 ) 18 tan )‘0 œ 16
23. '0 '0
È2
ar cos ) 2r sin )b r dr d) œ '1Î4 ’ r3 cos ) 2r3 sin )“
x
'02 'y2 y2 ax2 y2 b dx dy
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
901
902
Chapter 15 Multiple Integrals
26. '0 '0 ax2 y2 b dy dx or
3
4
3
'04 '03 ax2 y2 b3 dx dy
27. '0
1Î2
È
'02 2 sin 2 r dr d) œ 2 '0 Î2 (2 sin 2)) d) œ 2(1 1)
)
1
28. A œ 2 '0
'11 cos r dr d) œ '0 Î2 a2 cos ) cos# )b d) œ 8 4 1
29. A œ 2 '0
'012 cos 3 r dr d) œ 144 '0 Î6 cos# 3) d) œ 121
1Î2
1Î6
)
)
30. A œ '0 '0
21
4)Î3
31. A œ '0
1Î2
1
1
r dr d) œ 89 '0 )# d) œ 64271
21
$
'01 sin r dr d) œ 12 '0 Î2 ˆ 3# 2 sin ) cos#2) ‰ d) œ 381 1
)
32. A œ 4 '0
1Î2
1
'01 cos r dr d) œ 2 '0 Î2 ˆ 3# 2 cos ) cos#2) ‰ d) œ 321 4
)
1
33. average œ 14a# '0
'0a rÈa# r# dr d) œ 314a '01Î2 a$ d) œ 2a3
34. average œ 14a# '0
'0a r# dr d) œ 314a '01Î2 a$ d) œ 2a3
1Î2
#
1Î2
#
Èa cx
35. average œ 1"a# 'ca 'cÈa#cx# Èx# y# dy dx œ 1"a# '0 '0 r# dr d) œ 3a1 '0 d) œ 2a
3
#
a
21
#
21
a
36. average œ 1" ' ' c(1 x)# y# d dy dx œ 1" '0 '0 c(1 r cos ))# r# sin# )d r dr d)
21
1
R
œ 1"
'021 '01 ar$ 2r# cos ) rb dr d) œ 1" '021 ˆ 34 2 cos3 ) ‰ d) œ 1" 34 ) 2 sin3 ) ‘ 021 œ #3
Èe
Èe
37. '0 '1 Š lnrr ‹ r dr d) œ '0 '1 2 ln r dr d) œ 2'0 cr ln r rd1e d) œ 2'0 Èe ˆ "# 1‰ 1‘ d) œ 21ˆ2 Èe‰
21
21
#
21
21
"Î#
38. '0 '1 Š lnrr ‹ dr d) œ '0 '1 ˆ 2 lnr r ‰ dr d) œ '0 c(ln r)# d 1 d) œ '0 d) œ 21
21
e
21
#
39. V œ 2 '0
1Î2
21
e
21
e
'11 cos r# cos ) dr d) œ 23 '0 Î2 a3 cos# ) 3 cos$ ) cos% )b d)
)
1
1Î2
œ 23 "85) sin 2) 3 sin ) sin$ ) sin324) ‘ 0 œ 34 581
40. V œ 4 '0
1Î4
È
'0 2 cos 2 rÈ2 r# dr d) œ 43 '0 Î4 (2 2 cos 2))$Î# 2$Î# ‘ d)
)
1
'0 a1 cos# )b sin ) d) œ 213 2 323 ’ cos3 ) cos )“
œ 213 2 32
3
È
1Î4
È
$
1Î4
0
œ 61
È2 40È2 64
9
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 15.4 Double Integrals in Polar Form
_ _
41. (a) I# œ '0 '0 e ax y b dx dy œ '0
œ "# '0
#
1Î2
lim
bÄ_
t#
Î
1 2
#
'0_ Šer ‹ r dr d) œ '01Î2 ” lim '0b rer dr• d)
#
#
bÄ_
Šeb 1‹ d) œ "# '0
1Î2
#
_
903
d) œ 14 Ê I œ
È1
#
' 2e dt œ È21 '0 et dt œ Š È21 ‹ Š # ‹ œ 1, from part (a)
(b) x lim
Ä _ 0 È1
x
_ _
42. '0 '0
œ 14 lim
bÄ_
dx dy œ '0
Î
1 2
"
a1 x# y# b#
#
'0_
ˆ1 1 " b# ‰ œ 14
È1
r
dr d) œ 1# lim
a1 r# b#
bÄ_
È3Î2
43. Over the disk x# y# Ÿ 34 : ' ' 1 x"# y# dA œ '0 '0
21
R
21
21
œ '0 ˆ "# ln "4 ‰ d) œ (ln 2) '0 d) œ 1 ln 4
'0b
r
dr œ 14 lim
a1 r# b#
bÄ_
r
1 r# dr d) œ
1 " r# ‘ b
0
'021 2" ln a1 r# b‘ 0È3Î2 d)
Over the disk x# y# Ÿ 1: ' ' 1 x"# y# dA œ '0 '0 1 r r# dr d) œ '0 ’ lim c '0 1 r r# dr“ d)
œ '0
21
21
21
1
aÄ1
R
lim
a Ä 1c
x# y# Ÿ 1
a
"# ln a1 a# b‘ d) œ 21 † lim c
aÄ1
"# ln a1 a# b‘ œ 21 † _, so the integral does not exist over
44. The area in polar coordinates is given by A œ '! '0 r dr d) œ '! ’ r2 “
"
fÐ)Ñ
"
#
fÐ)Ñ
0
d) œ "# '! f # ()) d) œ '! "# r# d),
"
"
where r œ f())
45. average œ 1"a# '0 '0 c(r cos ) h)# r# sin# )d r dr d) œ 1"a# '0 '0 ar$ 2r# h cos ) rh# b dr d)
21
21
a
a
œ 1"a# '0 Š a4 2a h3cos ) a #h ‹ d) œ 1" '0 Š a4 2ah 3cos ) h# ‹ d) œ 1" ’ a4) 2ah 3sin ) h#) “
21
%
$
# #
21
#
#
#
#
21
0
œ "# aa# 2h# b
46. A œ '1Î4 'csc ) r dr d) œ "# '1Î4 a4 sin# ) csc# )b d)
31Î4
2 sin )
31Î4
œ "# c2) sin 2) cot )d 131Î4Î4 œ 1#
47-50. Example CAS commands:
Maple:
f := (x,y) -> y/(x^2+y^2);
a,b := 0,1;
f1 := x -> x;
f2 := x -> 1;
plot3d( f(x,y), y=f1(x)..f2(x), x=a..b, axes=boxed, style=patchnogrid, shading=zhue, orientation=[0,180], title="#47(a)
(Section 15.4)" );
# (a)
q1 := eval( x=a, [x=r*cos(theta),y=r*sin(theta)] );
# (b)
q2 := eval( x=b, [x=r*cos(theta),y=r*sin(theta)] );
q3 := eval( y=f1(x), [x=r*cos(theta),y=r*sin(theta)] );
q4 := eval( y=f2(x), [x=r*cos(theta),y=r*sin(theta)] );
theta1 := solve( q3, theta );
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
904
Chapter 15 Multiple Integrals
theta2 := solve( q1, theta );
r1 := 0;
r2 := solve( q4, r );
plot3d(0,r=r1..r2, theta=theta1..theta2, axes=boxed, style=patchnogrid, shading=zhue, orientation=[-90,0],
title="#47(c) (Section 15.4)" );
fP := simplify(eval( f(x,y), [x=r*cos(theta),y=r*sin(theta)] ));
# (d)
q5 := Int( Int( fP*r, r=r1..r2 ), theta=theta1..theta2 );
value( q5 );
Mathematica: (functions and bounds will vary)
For 47 and 48, begin by drawing the region of integration with the FilledPlot command.
Clear[x, y, r, t]
<<Graphics`FilledPlot`
FilledPlot[{x, 1}, {x, 0, 1}, AspectRatio Ä 1, AxesLabel Ä {x,y}];
The picture demonstrates that r goes from 0 to the line y=1 or r = 1/ Sin[t], while t goes from 1/4 to 1/2.
f:= y / (x2 y2 )
topolar={x Ä r Cos[t], y Ä r Sin[t]};
fp= f/.topolar //Simplify
Integrate[r fp, {t, 1/4, 1/2}, {r, 0, 1/Sin[t]}]
For 49 and 50, drawing the region of integration with the ImplicitPlot command.
Clear[x, y]
<<Graphics`ImplicitPlot`
ImplicitPlot[{x==y, x==2 y, y==0, y==1}, {x, 0, 2.1}, {y, 0, 1.1}];
The picture shows that as t goes from 0 to 1/4, r goes from 0 to the line x=2 y. Solve will find the bound for r.
bdr=Solve[r Cos[t]==2 r Sin[t], r]//Simplify
f:=Sqrt[x y]
topolar={x Ä r Cos[t], y Ä r Sin[t]};
fp= f/.topolar //Simplify
Integrate[r fp, {t, 0, 1/4}, {r, 0, bdr[[1, 1, 2]]}]
15.5 TRIPLE INTEGRALS IN RECTANGULAR COORDINATES
1.
'01 '01cx 'x1bz Fax, y, zb dy dz dx œ '01 '01cx 'x1bz dy dz dx œ '01 '01cx a1 x zb dz dx
œ '0 ’a1 xb xa1 xb a1 2 xb “dx œ '0
1
2.
3.
2
1
"
3
a1 xb2
dx œ ’ a1 6 xb “ œ 61
2
0
'01 '02 '03 dz dy dx œ '01 '02 3 dy dx œ '01 6 dx œ 6, '02 '01 '03 dz dx dy, '03 '02 '01 dx dy dz, '02 '03 '01 dx dz dy,
'03 '01 '02 dy dx dz, '01 '03 '02 dy dz dx
'01 '02c2x '03c3xc3yÎ2 dz dy dx
1
2 2x
œ '0 '0 ˆ3 3x 32 y‰ dy dx
1
œ '0 3(1 x) † 2(1 x) 34 † 4(1 x)# ‘ dx
1
"
œ 3 '0 (1 x)# dx œ c (1 x)$ d ! œ 1,
'02 '01cyÎ2 '03 3x 3yÎ2 dz dx dy, '01 '03 3x '02 2x 2zÎ3 dy dz dx,
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 15.5 Triple Integrals in Rectangular Coordinates
'03 '01 zÎ3 '02 2x 2zÎ3 dy dx dz, '02 '03 3yÎ2 '01 yÎ2 zÎ3 dx dz dy, '03 '02 2zÎ3 '01 yÎ2 zÎ3 dx dy dz
4.
È
'02 '03 '0 4 x dz dy dx œ '02 '03 È4 x# dy dx œ '02 3È4 x# dx œ 3# ’xÈ4 x# 4 sin" x# “ 2 œ 6 sin" 1 œ 31,
#
È4cx
'0 '0 '0
3
5.
2
È
È4cx
dz dx dy, '0 '0
#
2
#
È4cz
'0 dy dz dx, '0 '0
3
2
È
#
È4cz
'0 dy dx dz, '0 '0 '0
3
2
3
#
0
È4cz
dx dy dz, '0 '0 '0
3
2
#
dx dz dy
'c22 'cÈ44ccxx 'x8bcyx cy dz dy dx œ 4 '02 '0 4cx 'x8bcyx cy dz dy dx
#
#
#
#
È4cx
œ 4'0 '0
2
#
È4cx
œ 8 '0 '0
2
œ 8'0
1Î2
#
#
#
#
#
c8 2 ax# y# bd dy dx
#
a4 x# y# b dy dx
'02 a4 r# b r dr d) œ 8 '01Î2 ’2r# r4 “ # d)
%
!
1Î2
œ 32 '0 d) œ 32 ˆ 1# ‰ œ 161,
È
#
#
'c22 'cÈ44ccyy 'x8bcyx cy dz dx dy,
#
#
#
È
#
#
#
È
'c22 'y4 'cÈzzccyy dx dz dy 'c22 '48cy 'cÈ88cczzccyy dx dz dy,
#
#
#
#
#
#
È
È
È
È
È È
'0 'cÈzz 'cÈzzccyy dx dy dz '48 'cÈ88cczz 'cÈ88cczzccyy dx dy dz, 'c22 'x4 'cÈzzccxx dy dz dx 'c22 '48cx 'cÈ88cczzccxx dy dz dx,
È
È
È È
'04 'cÈzz 'cÈzzccxx dy dx dz '48 'cÈ88cczz 'cÈ88cczzccxx dy dx dz
4
#
#
#
#
#
#
#
#
#
#
#
#
#
#
6. The projection of D onto the xy-plane has the boundary
x# y# œ 2y Ê x# (y 1)# œ 1, which is a circle.
Therefore the two integrals are:
È
È
'02 'cÈ2y2yccyy 'x2yby dz dx dy and 'c11 '11cbÈ11ccxx 'x2yby dz dy dx
#
#
#
#
#
#
#
#
7.
'01 '01 '01 ax# y# z# b dz dy dx œ '01 '01 ˆx# y# 3" ‰ dy dx œ '01 ˆx# 23 ‰ dx œ 1
8.
'0 2 '03y'x8bc3yx cy dz dx dy œ '0 2 '03ya8 2x# 4y# b dx dy œ '0 2 8x 23 x$ 4xy# ‘ 03y dy
È
#
#
È2
È
#
È
#
È
%‘ 2
œ '0 a24y 18y$ 12y$ b dy œ 12y# "5
2 y 0 œ 24 30 œ 6
'1e '1e '1e xyz" dx dy dz œ '1e '1e ’ lnyzx “ e dy dz œ '1e '1e yz3 dy dz œ '1e ’ lnzy “ e dz œ '1e 6z dz œ 6
2
9.
3
2
3
1
3c3x
10. '0 '0
1
2
2
1
'03c3xcy dz dy dx œ '01 '03c3x (3 3x y) dy dx œ '01 (3 3x)# "# (3 3x)# ‘ dx œ #9 '01 (1 x)# dx
"
œ 3# c(1 x)$ d ! œ 3#
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
905
906
Chapter 15 Multiple Integrals
11. '0
1Î6
È
'01 ' 3# y sin z dx dy dz œ '01Î6 '01 5y sin z dy dz œ 5# '01Î6 sin z dz œ 5Š2 4 3‹
12. 'c1 '0 '0 ax y zb dy dx dz œ 'c1 '0 œ xy "2 y# zy‘0 dx dz œ 'c1 '0 a2x 2 2zb dx dz
1
1
2
1
1
1
2
1
œ 'c1 cx# 2x 2zxd 0 dz œ 'c1 a3 2zb dz œ c3z z# dc1 œ 6
1
È9 c x
13. '0 '0
3
È
1
È
'0 9 c x dz dy dx œ '03 '0 9 c x È9 x# dy dx œ '03 a9 x# b dx œ ’9x x3 “ $ œ 18
#
#
#
$
!
È4cy
2xby
14. '0 'cÈ4cy# '0
#
2
1
"
È4cy
È4cy
dz dx dy œ '0 'cÈ4cy# (2x y) dx dy œ '0 cx# xyd È4 y# dy œ '0 a4 y# b
#
2
#
2
c c
$Î# #
œ ’ 23 a4 y# b
2cx
15. '0 '0
1
2
"Î#
(2y) dy
“ œ 23 (4)$Î# œ 16
3
!
'02cxcydz dy dx œ '01 '02cx (2 x y) dy dx œ '01 (2 x)# "# (2 x)# ‘ dx œ "# '01 (2 x)# dx
"
œ "6 (2 x)$ ‘ ! œ 6" 86 œ 67
1cx#
'34cx cyx dz dy dx œ '01 '01cx x a1 x# yb dy dx œ '01 x ’a1 x# b# "# a1 x# b“ dx œ '01 "# x a1 x# b# dx
16. '0 '0
1
#
#
$ "
"
"
œ ’ 12
a1 x# b “ œ 12
!
17. '0 '0 '0 cos (u v w) du dv dw œ '0 '0 [sin (w v 1) sin (w v)] dv dw
1
1
1
1
1
œ '0 [( cos (w 21) cos (w 1)) (cos (w 1) cos w)] dw
1
œ c sin (w 21) sin (w 1) sin w sin (w 1)d 1! œ 0
Èe
18. '0 '1
1
œ
19. '0
2
2 Èe
6
1Î4
œ '0
È
lim
b Ä _
s
s
1
'01 s es ds œ 2 6Èe cs es es d 01 œ 2 6Èe
'0ln sec v 'c2t_ ex dx dt dv œ '01Î4 '0ln sec v
1Î4
ae2t eb b dt dv œ '0
1Î4
'0ln sec v e2t dt dv œ '01Î4 ˆ "# e2 ln sec v "# ‰ dv
#
1Î%
Š sec# v #" ‹ dv œ tan2 v 2v ‘ ! œ #" 18
È4cq
20. '0 '0 '0
7
È
'1e s es ln r alnttb dt dr ds œ '01 '1 e as es ln rb’ 31 aln tb3 “ e dr ds œ '01 '1 e s3e ln r dr ds œ '01 s3e cr ln r rd 1Èe ds
2
q
r 1 dp dq dr œ
1cx#
21. (a) 'c1 '0
1
#
!
'x1cz dy dz dx
1cy
1
cÈz
1
y#
1 cz
Èy
1cy
(e) '0 'cÈy '0
1
cÈz
(b) '0 '0 'c1
1
dy dz dx
(d) 'c1 '0 '0 dx dz dy
0
È1cz
(b) '0 'cÈ1cz 'x#
'cÈÈyy dx dz dy
22. (a) '0 '0 'c1
1
'07 '02 qÈr41 q dq dr œ '07 3(r " 1) ’ a4 q# b$Î# “ # dr œ 38 '07 r"1 dr œ 8 ln3 8 œ 8 ln 2
#
(d) '0 '0
1
#
1
0
1
y#
1
y#
1
1
cÈz
(c) '0 'c1
1
dy dx dz
23. V œ '0 'c1 '0 dz dy dx œ '0 'c1 y# dy dx œ 23 '0 dx œ 23
1
'cÈÈyy dx dy dz
dz dx dy
(e) 'c1 '0 '0 dz dx dy
1
1 cz
(c) '0 '0
1
dy dx dz
1
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
'01 dx dy dz
Section 15.5 Triple Integrals in Rectangular Coordinates
1cx
24. V œ '0 '0
1
'02c2z dy dz dx œ '01 '01cx (2 2z) dz dx œ '01 c2z z# d 01cx dx œ '01 a1 x# b dx œ ’x x3 “ " œ 32
$
!
È4cx
25. V œ '0 '0
4
È
'02cy dz dy dx œ '04 '0 4cx (2 y) dy dx œ '04 ’2È4 x ˆ 4#x ‰“ dx
%
20
œ 43 (4 x)$Î# "4 (4 x)# ‘ ! œ 43 (4)$Î# "4 (16) œ 32
3 4œ 3
cy
26. V œ 2 '0 'cÈ1cx# '0 dz dy dx œ 2 '0 'cÈ1cx# y dy dx œ '0 a1 x# b dx œ 23
1
0
2c2x
27. V œ '0 '0
1
1
0
1
'03c3xc3yÎ2 dz dy dx œ '01 '02 2x ˆ3 3x 3# y‰ dy dx œ '01 6(1 x)# 43 † 4(1 x)# ‘ dx
œ '0 3(1 x)# dx œ c (1 x)$ d ! œ 1
1
"
1cx
'0cos Ð1xÎ2Ñ dz dy dx œ '01 '01 x cos ˆ 1#x ‰ dy dx œ '01 ˆcos 1#x ‰ (1 x) dx
1
1
1Î2
"
œ '0 cos ˆ 1#x ‰ dx '0 x cos ˆ 1#x ‰ dx œ 12 sin 1#x ‘ ! 14 '0 u cos u du œ 12 14 ccos u u sin ud 01Î2
28. V œ '0 '0
1
#
#
œ 12 14# ˆ 1# 1‰ œ 14#
È1cx
29. V œ 8 '0 '0
1
4cx#
30. V œ '0 '0
2
#
È
È
'0 1cx dz dy dx œ 8 '01 '0 1cx È1 x# dy dx œ 8 '01 a1 x# b dx œ 163
#
#
'04cx cydz dy dx œ '02 '04cx a4 x# yb dy dx œ '02 ’a4 x# b# "# a4 x# b# “ dx
#
#
œ "# '0 a4 x# b dx œ '0 Š8 4x# x# ‹ dx œ 128
15
2
2
#
c Î
31. V œ '0 '0
ˆÈ16 y# ‰ 2
4
%
'04 ydx dz dy œ '04 '0 16 y Î2 (4 y) dz dy œ '04 È16#y (4 y) dy
ˆÈ
#‰
#
%
$Î#
œ '0 2È16 y# dy "# '0 yÈ16 y# dy œ yÈ16 y# 16 sin" y4 ‘ ! ’ 6" a16 y# b “
4
4
œ 16 ˆ 1# ‰ "6 (16)$Î# œ 81 32
3
È4cx
3cx
32. V œ 'c2 'cÈ4cx# '0
2
#
È4cx
dz dy dx œ 'c2 'cÈ4cx# (3 x) dy dx œ 2 'c2 (3 x)È4 x# dx
2
#
2
"
œ 12 sin
2cx
33. '0 '0
2
"
1 12 sin
(1) œ 12 ˆ 1# ‰ 12 ˆ 1# ‰ œ 121
#
#
’ 23 a4 x# b
$Î# #
'Ð24cx2xcyÑÎ2y2 dz dy dx œ '02 '02cx ˆ3 3x# 3y# ‰ dy dx
œ '0 3 ˆ1 x# ‰ (2 x) 34 (2 x)# ‘ dx
2
œ '0 ’6 6x 3x# 3(24 x) “ dx
2
#
#
$
#
œ ’6x 3x# x2 (24x) “ œ (12 12 4 0) 24 œ 2
$
!
2
œ 3 'c2 2È4 x# dx 2 'c2 xÈ4 x# dx œ 3 ’xÈ4 x# 4 sin" x2 “
2
%
$
!
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
“
#
907
908
Chapter 15 Multiple Integrals
34. V œ '0 'z 'z
4
8
8cz
dx dy dz œ '0 'z (8 2z) dy dz œ '0 (8 2z)() z) dz œ '0 a64 24z 2z# b dz
4
8
4
4
%
œ 64z 12z# 23 z$ ‘ ! œ 320
3
È4cx Î2
35. V œ 2 'c2 '0
2
#
È
'0x 2 dz dy dx œ 2 'c22 '0 4x Î2 (x 2) dy dx œ 'c22 (x 2)È4 x# dx
#
œ 'c2 2È4 x# dx 'c2 xÈ4 x# dx œ ’xÈ4 x# 4 sin" x# “
2
2
œ 4 ˆ 1# ‰ 4 ˆ 1# ‰ œ 41
1cy#
36. V œ 2 '0 '0
1
#
#
’ "3 a4 x# b
$Î# #
“
#
'0x by dz dx dy œ 2 '01 '01cy ax# y# b dx dy œ 2'01 ’ x3 xy# “ 1cy dy
#
#
#
#
$
0
1
1
1
#
œ 2 '0 a1 y# b ’ "3 a1 y# b y# “ dy œ 2 '0 a1 y# b ˆ 3" 3" y# 3" y% ‰ dy œ 23 '0 a1 y' b dy
"
(
œ 23 ’y y7 “ œ ˆ 23 ‰ ˆ 67 ‰ œ 47
!
37. average œ "8 '0 '0 '0 ax# 9b dz dy dx œ 8" '0 '0 a2x# 18b dy dx œ 8" '0 a4x# 36b dx œ 31
3
2
2
2
2
2
2
38. average œ "2 '0 '0 '0 (x y z) dz dy dx œ 2" '0 '0 (2x 2y 2) dy dx œ 2" '0 (2x 1) dx œ 0
1
1
2
1
1
1
39. average œ '0 '0 '0 ax# y# z# b dz dy dx œ '0 '0 ˆx# y# "3 ‰ dy dx œ '0 ˆx# 23 ‰ dx œ 1
1
1
1
1
1
1
40. average œ "8 '0 '0 '0 xyz dz dy dx œ 4" '0 '0 xy dy dx œ 2" '0 x dx œ 1
2
2
2
2
ax b
41. '0 '0 '2y 4 cos
dx dy dz œ '0 '0 '0
2È z
4
1
2
4
#
xÎ2
2
2
2
4 cos ax# b
dy dx dz œ
2È z
'04 '02 x cosÈzax b dx dz œ '04 ˆ sin# 4 ‰ z"Î# dz
#
%
œ (sin 4)z"Î# ‘ ! œ 2 sin 4
Èy
42. '0 '0 'x# 12xz ezy dy dx dz œ '0 '0 '0
1
1
1
#
1
1
12xz ezy dx dy dz œ '0 '0 6yz ezy dy dz œ '0 ’3ezy “ dz
1
#
1
#
1
#
!
1
œ 3'0 aez zb dz œ 3 cez 1d "! œ 3e 6
43. '0 'È$ z '0
1
1
ln 3
1e2x sin a1y# b
dx dy dz œ
y#
"
'01 'È1z 41 siny a1y b dy dz œ '01 '0y 41 siny a1y b dz dy
$
#
#
$
#
#
œ '0 41y sin a1y# b dy œ c2 cos a1y# bd ! œ 2(1) 2(1) œ 4
1
4cx#
44. '0 '0
2
"
È
2z
2z ‰
'0x sin
' 2 ' 4cx x4sin z2z dz dx œ '04 '0 4cz ˆ sin
' 4 ˆ sin 2z ‰ "
4 z dy dz dx œ 0 0
4 z x dx dz œ 0 4 z # (4 z) dz
#
%
%
#
œ "4 cos 2z‘ ! œ 4" "# sin# z‘ ! œ sin# 4
4cacx#
'a4cx cydz dy dx œ 154 Ê '01 '04cacx a4 x# y ab dy dx œ 154
1
1
1
#
#
#
4
4
Ê '0 ’a4 a x# b "# a4 a x# b “ dx œ 15
Ê "# '0 a4 a x# b dx œ 15
Ê '0 c(4 a)# 2x# (4 a) x% d dx
45. '0 '0
1
#
#
&
"
8
8
8
œ 15
Ê ’(4 a)# x 32 x$ (4 a) x5 “ œ 15
Ê (4 a)# 32 (4 a) "5 œ 15
Ê 15(4 a)# 10(4 a) 5 œ 0
!
Ê 3(4 a)# 2(4 a) 1 œ 0 Ê [3(4 a) 1][(4 a) 1] œ 0 Ê 4 a œ "3 or 4 a œ 1 Ê a œ 13
3 or a œ 3
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 15.6 Moments and Centers of Mass
#
1
1
46. The volume of the ellipsoid xa# by# zc# œ 1 is 4abc
so that 4(1)(2)(c)
œ 81 Ê c œ 3.
3
3
#
#
47. To minimize the integral, we want the domain to include all points where the integrand is negative and to exclude all
points where it is positive. These criteria are met by the points (xß yß z) such that 4x# 4y# z# 4 Ÿ 0 or
4x# 4y# z# Ÿ 4, which is a solid ellipsoid centered at the origin.
48. To maximize the integral, we want the domain to include all points where the integrand is positive and to exclude all
points where it is negative. These criteria are met by the points (xß yß z) such that 1 x# y# z# 0 or
x# y# z# Ÿ 1, which is a solid sphere of radius 1 centered at the origin.
49-52. Example CAS commands:
Maple:
F := (x,y,z) -> x^2*y^2*z;
q1 := Int( Int( Int( F(x,y,z), y=-sqrt(1-x^2)..sqrt(1-x^2) ), x=-1..1 ), z=0..1 );
value( q1 );
Mathematica: (functions and bounds will vary)
Clear[f, x, y, z];
f:= x2 y2 z
Integrate[f, {x,1,1}, {y,Sqrt[1 x2 ], Sqrt[1 x2 ]}, {z, 0, 1}]
N[%]
topolar={x Ä r Cos[t], y Ä r Sin[t]};
fp= f/.topolar //Simplify
Integrate[r fp, {t, 0, 21}, {r, 0, 1},{z, 0, 1}]
N[%]
15.6 MOMENTS AND CENTERS OF MASS
2cx#
1. M œ '0 'x
1
2cx#
3 dy dx œ 3'0 a2 x# xb dx œ 7# ; My œ '0 'x
1
1
2cx#
œ 3'0 a2x x$ x# b dx œ 45 ; Mx œ '0 'x
1
1
3x dy dx œ 3 '0 cxyd x2cx dx
1
3y dy dx œ 3# '0 cy# d x
2cx#
1
#
dx œ 3# '0 a4 5x# x% b dx œ 19
5
1
5
38
Ê x œ 14
and y œ 35
2. M œ $ '0 '0 dy dx œ $ '0 3 dx œ 9$ ; Ix œ $ '0 '0 y# dy dx œ $ '0 ’ y3 “ dx œ 27$ ;
3
3
3
Iy œ $ '0 '0 x dy dx œ $ '
3
3
3
3
$
cx yd ! dx œ $
0
#
#
3
3
$
0
'0 3x dx œ 27$
3
3
#
"'
#
' '
3. M œ '0 'y#Î2 dx dy œ '0 Š4 y y# ‹ dy œ 14
3 ; My œ 0 y# Î2 x dx dy œ # 0 cx d y# Î2 dy
2
4 y
2
2
#
4 y
2
4cy
4 y
' 'y#Î2 y dx dy œ '0 Š4y y# y# ‹ dy œ 103
œ "# '0 Š16 8y y# y4 ‹ dy œ 128
15 ; Mx œ 0
2
2
%
2
$
5
Ê x œ 64
35 and y œ 7
3cx
4. M œ '0 '0
3
3cx
dy dx œ '0 (3 x) dx œ 9# ; My œ '0 '0
3
3
x dy dx œ '0 cxyd 03cx dx œ '0 a3x x# b dx œ 9#
3
3
Ê x œ 1 and y œ 1, by symmetry
Èa cx
5. M œ '0 '0
a
#
#
Èa cx
dy dx œ 1a4 ; My œ '0 '0
#
a
#
#
È
x dy dx œ '0 cxyd 0 a cx dx œ '0 xÈa# x# dx œ a3
a
#
#
a
4a
Ê x œ y œ 31
, by symmetry
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
$
909
910
Chapter 15 Multiple Integrals
6. M œ '0 '0
1
dy dx œ '0 sin x dx œ 2; Mx œ '0 '0
1
sin x
1
sin x
œ "4 '0 (1 cos 2x) dx œ 14 Ê x œ 1# and y œ 18
1
È4cx
7. Ix œ 'c2 'cÈ4cx# y# dy dx œ 'c2 ’ y3 “
2
#
2
$
y dy dx œ "# '0 cy# d 0
1
sin x
dx œ "# '0 sin# x dx
1
È4cx
2
$Î#
dx œ 32 'c2 a4 x# b dx œ 41; Iy œ 41, by symmetry;
È
c 4cx
#
#
I o œ I x I y œ 81
8. Iy œ '1 '0
21
ˆsin# x‰Îx#
x# dy dx œ '1 asin# x 0b dx œ "# '1 (1 cos 2x) dx œ 1#
21
21
9. M œ 'c_ '0 dy dx œ ' _ ex dx œ
ex
0
œ
'b0 xex dx œ
lim
b Ä _
œ "#
'c_ e
0
_
b Ä _
'b e
0
dx œ "4
bÄ_
2
2
ycy#
#
Ix œ '0 'cy y (x y) dx dy œ '
2
È3Î2
È12 4y
12. M œ 'cÈ3Î2 '4y#
2cx
13. M œ '0 'x
1
1
ycy#
cy
2
È3Î2
5x dx dy œ 5 ' È3Î2 ’ x2 “
#
abeb eb b œ 1; Mx œ 'c_ '0 y dy dx
ex
b Ä _
x"#Î2 1‘ b œ 1
e
0
y
8
dy œ '0 Š y2 2y$ 2y# ‹ dy œ ’ 10
y# 2y3 “ œ 15
;
2
ycy#
È12 4y
#
4y#
%
&
%
$
#
!
dy œ '
2
'
64
;
Š y2 2y& 2y% ‹ dy œ 105
0
È3Î2
dy œ 5# ' È3Î2 a12 4y# 16y% b dy œ 23È3
2cx
(6x 3y 3) dy dx œ '0 6xy 3# y# 3y‘ x dx œ '0 a12 12x# b dx œ 8;
2cx
My œ '0 'x
#
0
Ê x œ 1 and y œ "4
# #
’ x 2y xy$ “
0
cy
#
b Ä _
ex
0
bÄ_
11. M œ '0 'cy (x y) dx dy œ '0 ’ x2 xy“
eb œ 1; My œ ' _ '0 x dy dx œ ' _ xex dx
0
lim
lim
'0b xe x#Î2 dx œ lim
x dy dx œ lim
ycy#
b Ä _
2x
#
e x Î2
10. My œ '0 '0
lim
cxex ex d b0 œ 1 lim
dx œ "# lim
b Ä _
2x
'b0 ex dx œ 1 0
1
1
2cx
x(6x 3y 3) dy dx œ '0 a12x 12x$ b dx œ 3; Mx œ '0 'x
1
1
y(6x 3y 3) dy dx
3
17
œ '0 a14 6x 6x# 2x$ b dx œ 17
# Ê x œ 8 and y œ 16
1
2ycy#
14. M œ '0 'y#
1
2ycy#
My œ '0 'y#
1
2ycy#
(y 1) dx dy œ '0 a2y 2y$ b dy œ "# ; Mx œ '0 'y#
1
1
4
y(y 1) dx dy œ '0 a2y# 2y% b dy œ 15
;
1
2ycy#
8
8
x(y 1) dx dy œ '0 a2y# 2y% b dy œ 154 Ê x œ 15
and y œ 15
; Ix œ '0 'y#
1
1
y# (y 1) dx dy
œ 2 '0 ay$ y& b dy œ "6
1
15. M œ '0 '0 (x y 1) dx dy œ '0 (6y 24) dy œ 27; Mx œ '0 '0 y(x y 1) dx dy œ '0 y(6y 24) dy œ 14;
1
6
1
1
6
1
14
' ' #
My œ '0 '0 x(x y 1) dx dy œ '0 (18y 90) dy œ 99 Ê x œ 11
3 and y œ 27 ; Iy œ 0 0 x (x y 1) dx dy
1
6
1
1
6
‰
œ 216 '0 ˆ y3 11
6 dy œ 432
1
32
16. M œ 'c1 'x# (y 1) dy dx œ 'c1 Š x# x# 3# ‹ dx œ 15
; Mx œ 'c1 'x# y(y 1) dy dx œ 'c1 Š 56 x3 x# ‹ dx
1
1
1
1
%
1
1
'
%
9
$
' '
' 3x x
' ' #
œ 48
35 ; My œ c1 x# x(y 1) dy dx œ c1 Š # # x ‹ dx œ 0 Ê x œ 0 and y œ 14 ; Iy œ c1 x# x (y 1) dy dx
1
1
1
&
16
œ 'c1 Š 3x2 x2 x% ‹ dx œ 35
1
#
'
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
1
1
Section 15.6 Moments and Centers of Mass
911
13
' '
' 7x x
17. M œ 'c1 '0 (7y 1) dy dx œ 'c1 Š 7x# x# ‹ dx œ 31
15 ; Mx œ c1 0 y(7y 1) dy dx œ c1 Š 3 2 ‹ dx œ 15 ;
x#
1
1
x#
1
%
1
'
%
' ' #
My œ 'c1 '0 x(7y 1) dy dx œ 'c1 Š 7x# x$ ‹ dx œ 0 Ê x œ 0 and y œ 13
31 ; Iy œ c1 0 x (7y 1) dy dx
x#
1
1
x#
1
&
œ 'c1 Š 7x# x% ‹ dx œ 75
1
'
x ‰
x ‰
x ‰
18. M œ '0 'c1 ˆ1 20
dy dx œ '0 ˆ2 10
dx œ 60; Mx œ '0 'c1 y ˆ1 20
dy dx œ '0 ’ˆ1 #x0 ‰ Š y# ‹“
20
1
20
20
1
20
#
"
"
dx œ 0;
x ‰
x
' 'c1 y# ˆ1 20x ‰ dy dx
My œ '0 'c1 x ˆ1 20
dy dx œ '0 Š2x 10
Ê x œ 100
‹ dx œ 2000
3
9 and y œ 0; Ix œ 0
20
1
20
20
#
1
x ‰
œ 23 '0 ˆ1 20
dx œ 20
20
19. M œ '0 'cy (y 1) dx dy œ '0 a2y# 2yb dy œ 53 ; Mx œ '0 'cy y(y 1) dx dy œ 2 ' ay$ y# b dy œ 76 ;
1
y
1
1
1
y
0
7
My œ '0 'cy x(y 1) dx dy œ '0 0 dy œ 0 Ê x œ 0 and y œ 10
; Ix œ '0 'cy y# (y 1) dx dy œ '0 a2y% 2y$ b dy
1
y
1
1
y
1
9
3
œ 10
; Iy œ '0 'cy x# (y 1) dx dy œ 3" '0 a2y% 2y$ b dy œ 10
Ê Io œ Ix Iy œ 56
1
y
1
20. M œ '0 'cy a3x# 1b dx dy œ '0 a2y$ 2yb dy œ 3# ; Mx œ '0 'cy y a3x# 1b dx dy œ '0 a2y% 2y# b dy œ 16
15 ;
1
y
1
1
y
1
5
&
$
' ' # #
'
My œ '0 'cy x a3x# 1b dx dy œ 0 Ê x œ 0 and y œ 32
45 ; Ix œ 0 cy y a3x 1b dx dy œ 0 a2y 2y b dy œ 6 ;
1
y
1
y
1
6
Iy œ '0 'cy x# a3x# 1b dx dy œ 2 '0 ˆ 35 y& 3" y$ ‰ dy œ 11
30 Ê Io œ Ix Iy œ 5
1
y
1
21. Ix œ '0 '0 '0 ay# z# b dz dy dx œ '0 '0 Šcy# c3 ‹ dy dx œ '0 Š cb3 c3b ‹ dx œ abc ab3 c b
a
b
c
a
b
a
$
$
#
$
#
œ M3 ab# c# b where M œ abc; Iy œ M3 aa# c# b and Iz œ M3 aa# b# b , by symmetry
Ð4 2yÑÎ3
22. The plane z œ 4 3 2y is the top of the wedge Ê Ix œ 'c3 'c2 'c4Î3
3
4
ay# z# b dz dy dx
Ð4 2yÑÎ3
y)
64
' ' '
œ 'c3 'c2 ’ 8y3 2y3 8(281
81
“ dy dx œ 'c3 104
3 dx œ 208; Iy œ c3 c2 c4Î3
3
4
#
$
3
$
3
4
ax# z# b dz dy dx
64
‰
œ 'c3 'c2 ’ (4 812y) x (4 3 2y) 4x3 81
“ dy dx œ 'c3 ˆ12x# 32
3 dx œ 280;
3
4
$
Ð4 2yÑÎ3
Iz œ 'c3 'c2 'c4Î3
3
4
#
3
#
' 3 ax# 2b dx œ 360
‰
ax# y# b dz dy dx œ ' 3 ' 2 ax# y# b ˆ 83 2y
3 dy dx œ 12
3
4
3
' ' '
23. M œ 4 '0 '0 '4y# dz dy dx œ 4 '0 '0 a4 4y# b dy dx œ 16 '0 23 dx œ 32
3 ; Mxy œ 4 0 0 4y# z dz dy dx
1
1
4
1
1
1
1
1
4
12
'0 dx œ 128
œ 2 '0 '0 a16 16y% b dy dx œ 128
5
5 Ê z œ 5 , and x œ y œ 0, by symmetry;
1
1
1
64y
7904
%
' 1976
‰
Ix œ 4 '0 '0 '4y# ay# z# b dz dy dx œ 4 '0 '0 ’ˆ4y# 64
3 Š4y 3 ‹“ dy dx œ 4 0 105 dx œ 105 ;
1
1
4
1
1
1
'
64y
# #
' ˆ 8 # 128 ‰ dx
‰
Iy œ 4 '0 '0 '4y# ax# z# b dz dy dx œ 4 '0 '0 ’ˆ4x# 64
3 Š4x y 3 ‹“ dy dx œ 4 0 3 x 7
1
1
4
1
1
1
'
' ' ' # #
' ' # ## # %
œ 4832
63 ; Iz œ 4 0 0 4y# ax y b dz dy dx œ 16 0 0 ax x y y y b dy dx
1
1
4
1
1
2
œ 16 '0 Š 2x3 15
‹ dx œ 256
45
1
#
ŠÈ4 x# ‹Î2
24. (a) M œ 'c2 'ŠcÈ4cx#‹Î2 '0
2
ŠÈ4 x# ‹Î2
2 x
Myz œ 'c2 'ŠcÈ4cx#‹Î2 '0
2
ŠÈ4 x# ‹Î2
dz dy dx œ ' 2 'Š È4 x#‹Î2 (2 x) dy dx œ ' 2 (2 x) ŠÈ4 x# ‹ dx œ 41;
2 x
2
ŠÈ4 x# ‹Î2
2
x dz dy dx œ ' 2 'Š È4 x#‹Î2 x(2 x) dy dx œ ' 2 x(2 x) ŠÈ4 x# ‹ dx œ 21;
2
2
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
912
Chapter 15 Multiple Integrals
ŠÈ4 x# ‹Î2
Mxz œ 'c2 'ŠcÈ4cx#‹Î2 '0
2
2 x
ŠÈ4 x# ‹Î2
y dz dy dx œ ' 2 'Š È4 x#‹Î2 y(2 x) dy dx
2
œ "# 'c2 (2 x) ’ 44x 44x “ dx œ 0 Ê x œ "# and y œ 0
2
#
#
ŠÈ4 x# ‹Î2
(b) Mxy œ 'c2 'ŠcÈ4cx#‹Î2 '0
2
2 x
ŠÈ4 x# ‹Î2
z dz dy dx œ "# ' 2 'Š È4 x# ‹Î2 (2 x)# dy dx œ "# ' 2 (2 x)# ŠÈ4 x# ‹ dx
2
2
œ 51 Ê z œ 54
È4cx
25. (a) M œ 4 '0 '0
2
#
'x4 y dz dy dx œ 4 '01Î2 '02 'r 4 r dz dr d) œ 4 '01Î2 '02 a4r r$ b dr d) œ 4 '01Î2 4 d) œ 81;
#
#
#
'0 d) œ 6431 Ê z œ 83 , and x œ y œ 0, by symmetry
Mxy œ '0 '0 'r# zr dz dr d) œ '0 '0 #r a16 r% b dr d) œ 32
3
21
2
21
4
Èc
(b) M œ 81 Ê 41 œ '0 '0
21
21
2
È
'r c r dz dr d) œ '021 '0 c acr r$ b dr d) œ '021 c4 d) œ c#1 Ê c# œ 8 Ê c œ 2È2,
#
#
#
since c 0
26. M œ 8; Mxy œ 'c1 '3 'c1 z dz dy dx œ 'c1 '3 ’ z2 “
1
5
1
1
5
#
dy dx œ 0; Myz œ 'c1 '3 'c1 x dz dy dx
"
1
"
5
1
œ 2 'c1 '3 x dy dx œ 4 'c1 x dx œ 0; Mxz œ 'c1 '3 'c1 y dz dy dx œ 2 'c1 '3 y dy dx œ 16 'c1 dx œ 32
1
5
1
1
5
1
1
5
1
Ê x œ 0, y œ 4, z œ 0; Ix œ 'c1 '3 'c1 ay# z# b dz dy dx œ 'c1 '3 ˆ2y# 23 ‰ dy dx œ 32 'c1 100 dx œ 400
3 ;
1
5
1
1
5
1
Iy œ 'c1 '3 'c1 ax# z# b dz dy dx œ 'c1 '3 ˆ2x# 23 ‰ dy dx œ 43 'c1 a3x# 1b dx œ 16
3 ;
1
5
1
1
5
1
400
‰
Iz œ 'c1 '3 'c1 ax# y# b dz dy dx œ 2 'c1 '3 ax# y# b dy dx œ 2 'c1 ˆ2x# 98
3 dx œ 3
1
5
1
1
5
1
Ð2 yÑÎ2
27. The plane y 2z œ 2 is the top of the wedge Ê IL œ 'c2 'c2 'c1
2
4
c(y 6)# z# d dz dy dx
49
#
œ 'c2 'c2 ’ (y 6)#(4 y) (2 24y) "3 “ dy dx; let t œ 2 y Ê IL œ 4 'c2 Š 13t
24 5t 16t 3 ‹ dt œ 1386;
2
4
#
4
$
$
M œ "# (3)(6)(4) œ 36
Ð2 yÑÎ2
28. The plane y 2z œ 2 is the top of the wedge Ê IL œ 'c2 'c2 'c1
2
4
c(x 4)# y# d dz dy dx
œ "# 'c2 'c2 ax# 8x 16 y# b (4 y) dy dx œ 'c2 a9x# 72x 162b dx œ 696; M œ "# (3)(6)(4) œ 36
2
4
2
2cx
'02cxcy 2x dz dy dx œ '02 '02cx a4x 2x# 2xyb dy dx œ '02 ax$ 4x# 4xb dx œ 43
2
2cx
2cxcy
2
2cx
2
8
8
(b) Mxy œ '0 '0 '0
2xz dz dy dx œ '0 '0 x(2 x y)# dy dx œ '0 x(23 x) dx œ 15
; Mxz œ 15
by
2
2cx
2cxcy
2
2cx
2
#
symmetry; Myz œ '0 '0 '0
2x# dz dy dx œ '0 '0 2x# (2 x y) dy dx œ '0 a2x x# b dx œ 16
15
29. (a) M œ '0 '0
2
$
Ê x œ 45 , and y œ z œ 25
Èx
30. (a) M œ '0 '0
2
2
#
Èx
(b) Myz œ '0 '0
È
'04cx kxy dz dy dx œ k'02 '0 x xy a4 x# b dy dx œ k# '02 a4x# x% b dx œ 32k
15
È
'04cx kx# y dz dy dx œ k '02 '0 x x# y a4 x# b dy dx œ k# '02 a4x$ x& b dx œ 8k3
#
Èx
Ê x œ 54 ; Mxz œ '0 '0
2
È
'04cx kxy# dz dy dx œ k'02 '0 x xy# a4 x# b dy dx œ k3 '02 ˆ4x&Î# x*Î# ‰ dx
#
Èx
œ 2562312k Ê y œ 4077 2 ; Mxy œ '0 '0
È
È
2
È
'04cx kxyz dz dy dx œ '02 '0 x xy a4 x# b# dy dx
#
8
œ k4 '0 a16x# 8x% x' b dx œ 256k
105 Ê z œ 7
2
31. (a) M œ '0 '0 '0 (x y z 1) dz dy dx œ '0 '0 ˆx y 3# ‰ dy dx œ '0 (x 2) dx œ 5#
1
1
1
1
1
1
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 15.6 Moments and Centers of Mass
4
‰
(b) Mxy œ '0 '0 '0 z(x y z 1) dz dy dx œ "# '0 '0 ˆx y 53 ‰ dy dx œ "# '0 ˆx 13
6 dx œ 3
1
1
1
1
1
1
8
Ê Mxy œ Myz œ Mxz œ 43 , by symmetry Ê x œ y œ z œ 15
(c) Iz œ '0 '0 '0 ax# y# b (x y z 1) dz dy dx œ '0 '0 ax# y# b ˆx y 3# ‰ dy dx
1
1
1
1
1
11
œ '0 ˆx$ 2x# "3 x 43 ‰ dx œ 11
6 Ê Ix œ Iy œ Iz œ 6 , by symmetry
1
32. The plane y 2z œ 2 is the top of the wedge.
Ð2 yÑÎ2
(a) M œ 'c1 'c2 'c1
1
4
4
1
Ð2 yÑÎ2
(b) Myz œ 'c1 'c2 'c1
1
(x 1) dz dy dx œ 'c1 'c2 (x 1) ˆ2 y# ‰ dy dx œ 18
Ð2 yÑÎ2
Mxz œ 'c1 'c2 'c1
1
4
Ð2 yÑÎ2
Mxy œ 'c1 'c2 'c1
1
4
Ð2 yÑÎ2
(c) Ix œ 'c1 'c2 'c1
1
4
Ð2 yÑÎ2
Iy œ 'c1 'c2 'c1
1
4
Ð2 yÑÎ2
Iz œ 'c1 'c2 'c1
1
4
1cz
Èz
4
x(x 1) dz dy dx œ 'c1 'c2 x(x 1) ˆ2 y# ‰ dy dx œ 6;
1
4
y(x 1) dz dy dx œ 'c1 'c2 y(x 1) ˆ2 y# ‰ dy dx œ 0;
1
4
z(x 1) dz dy dx œ "# 'c1 'c2 (x 1) Š y4 y‹ dy dx œ 0 Ê x œ 13 , and y œ z œ 0
1
4
#
(x 1) ay# z# b dz dy dx œ 'c1 'c2 (x 1) ’2y# "3 y# 3" ˆ" 2y ‰ “ dy dx œ 45;
1
4
$
$
(x 1) ax# z# b dz dy dx œ 'c1 'c2 (x 1) ’2x# "3 x#y 3" ˆ" y2 ‰ “ dy dx œ 15;
1
4
$
#
(x 1) ax# y# b dz dy dx œ 'c1 'c2 (x 1) ˆ2 y# ‰ ax# y# b dy dx œ 42
1
4
1cz
33. M œ '0 'zc1 '0 (2y 5) dy dx dz œ '0 'zc1 ˆz 5Èz‰ dx dz œ '0 2 ˆz 5Èz‰ (1 z) dz
1
1
1
$Î#
œ 2 '0 ˆ5z"Î# z 5z$Î# z# ‰ dz œ 2 10
"# z# 2z&Î# 3" z$ ‘ ! œ 2 ˆ 93 3# ‰ œ 3
3 z
"
1
È4cx
16c2 ˆx# by# ‰
34. M œ 'c2 'cÈ4cx# '2 ax#by# b
2
#
È
' 4cx Èx# y# c16 4 ax# y# bd dy dx
c2 cÈ4cx
Èx# y# dz dy dx œ '
2
#
#
œ 4 '0 '0 r a4 r# b r dr d) œ 4 '0 ’ 4r3 r5 “ d) œ 4 '0
21
21
2
$
&
#
21
!
64
5121
15 d) œ 15
35. (a) x œ Myz œ 0 Ê ' ' ' x$ (xß yß z) dx dy dz œ 0 Ê Myz œ 0
M
R
(b) IL œ ' ' ' kv hik# dm œ ' ' ' k(x h) i yjk# dm œ ' ' ' ax# 2xh h# y# b dm
D
D
D
œ ' ' ' ax# y# b dm 2h ' ' ' x dm h# ' ' ' dm œ Ix 0 h# m œ Ic m h# m
Þ
D
D
Þ
D
36. IL œ Ic m mh# œ 25 ma# ma# œ 75 ma#
Þ
Þ
#
#
#
37. (a) (xß yß z) œ ˆ #a ß #b ß #c ‰ Ê Iz œ Ic m abc ŠÉ a4 b4 ‹ Ê IcÞmÞ œ Iz abc aa4 b b
#
Þ
#
#
#
#
#
Þ
#
#
b
œ abc aa3 b b abc aa4 b b œ abc aa1# b b ; RcÞmÞ œ É IcMÞmÞ œ É a 12
#
#
#
#
#
(b) IL œ IcÞmÞ abc ŒÉ a4 ˆ b# 2b‰ œ abc aa12 b b abc aa 4 9b b œ abc a4a1# 28b b
#
#
IL
œ abc aa 3 7b b ; RL œ É M
œ É a 37b
Ð4 2yÑÎ3
38. M œ 'c3 'c2 'c4Î3
3
4
#
#
#
#
#
#
#
dz dy dx œ ' 3 ' 2 23 (4 y) dy dx œ ' 3 23 ’4y y2 “
3
4
#
3
#
#
%
#
dx œ 12 ' 3 dx œ 72;
3
x œ y œ z œ 0 from Exercise 22 Ê Ix œ IcÞmÞ 72 ŠÈ0# 0# ‹ œ IcÞmÞ Ê IL œ IcÞmÞ 72 ŠÉ16 16
9 ‹
‰ œ 1488
œ 208 72 ˆ 160
9
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
#
913
914
Chapter 15 Multiple Integrals
15.7 TRIPLE INTEGRALS IN CYLINDRICAL AND SPHERICAL COORDINATES
1.
È
'021 '01 'r 2cr dz r dr d) œ '021 '01 ’r a2 r# b"Î# r# “ dr d) œ '021 ’ "3 a2 r# b$Î# r3 “ " d)
#
$
!
œ '0 Š 2 3 23 ‹ d) œ
21
2.
È
#
$
%
!
91 Š8È2 7‹
2
'021 '0 Î2 '03 24r dz r dr d) œ '02 '0 Î2 a3r 24r$ b dr d) œ '02 32 r# 6r% ‘ !Î2 d) œ 3# '02 Š 4)1 164)1 ‹ d)
)
#
1
1
&
œ 3# ’ 12)1# 20)1% “
È
21
!
)
1
1
)
1
1
#
%
#
%
œ 1751
'01 '0 Î ' 3È44 rr z dz r dr d) œ '0 '0 Î "# c9 a4 r# b a4 r# bd r dr d) œ 4 '0 '0 Î a4r r$ b dr d)
) 1
#
1
) 1
1
) 1
#
œ 4 '0 ’2r# r4 “
1
%
'021 '01 'r 2cr
ˆ
5.
3
#
$
4.
41 ŠÈ2 "‹
'021 '03 'r Î318cr dz r dr d) œ '021 '03 ’r a18 r# b"Î# r3 “ dr d) œ '021 ’ "3 a18 r# b$Î# 12r “ $ d)
œ
3.
$Î#
#‰
"Î#
) Î1
!
œ 4 '0 Š 21)# 4)1% ‹ d) œ 37151
1
#
%
3 dz r dr d) œ 3 '0 '0 ’r a2 r# b
21
1
"Î#
r# “ dr d) œ 3 '0 ’ a2 r# b
21
"Î#
$
"
r3 “ d)
21
œ 3 '0 ŠÈ2 43 ‹ d) œ 1 Š6È2 8‹
!
6.
'021 '01 'c11ÎÎ22 ar# sin# ) z# b dz r dr d) œ '021 '01 ˆr$ sin# ) 12r ‰ dr d) œ '021 Š sin4 ) 24" ‹ d) œ 13
7.
21
z
3
'021 '03 '0zÎ3 r$ dr dz d) œ '021 '03 324
dz d) œ '0 20
d) œ 3101
8.
'c11 '021 '01bcos 4r dr d) dz œ 'c11 '02 2(1 cos ))# d) dz œ 'c11 61 d) œ 121
9.
'01 '0 z '021 ar# cos# ) z# b r d) dr dz œ '01 '0 z ’ r2) r sin4 2) z# )“ #1 r dr dz œ '01 '0 z a1r$ 21rz# b dr dz
#
%
)
1
È
È
œ '0 ’ 14r 1r# z# “
1
È4cr
10. '0 'rc2
2
%
#
Èz
#
È
#
dz œ '0 Š 14z 1z$ ‹ dz œ ’ 112z 14z “ œ 13
1
#
$
%
"
!
!
!
È
'021 (r sin ) 1) r d) dz dr œ '02 'rc24cr 21r dz dr œ 21'02 ’r a4 r# b"Î# r# 2r“ dr
$Î#
œ 21 ’ "3 a4 r# b
$
#
#
r3 r# “ œ 21 38 4 3" (4)$Î# ‘ œ 81
!
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 15.7 Triple Integrals in Cylindrical and Spherical Coordinates
È4cr
11. (a) '0 '0 '0
21
1
È3
#
dz r dr d)
È
(b) '0 '0
'01 r dr dz d) '021 'È23 '0 4cz r dr dz d)
(c) '0 '0
#
21
È4cr
1
12. (a) '0 '0 'r
21
1
'021 r d) dz dr
2cr#
dz r dr d)
È2cz
(b) '0 '0 '0 r dr dz d) '0 '1 '0
21
(c) '0 'r
1
z
2cr#
'0 r d) dz dr
1Î2
cos )
1Î2
1
1
#
Î
19. '0
1Î4
Î
Î
r$ dz dr d) œ ' 1Î2 '0 r% cos ) dr d) œ "5 ' 1Î2 cos ) d) œ 25
1 2
1 2
1
'04cr sin f(rß )ß z) dz r dr d)
1cos )
'04 f(rß )ß z) dz r dr d)
21. '0 '0 '0
1
1
2 sin 9
Î
3 cos )
1 2
Î
2 cos )
Î
csc )
18. ' 1Î2 'cos )
'0sec '02 r sin f(rß )ß z) dz r dr d)
)
1 2
16. ' 1Î2 '0
)
17. 'c1Î2 '1
1 2
r dr dz d)
21
r cos )
2 sin )
2
'03r f(rß )ß z) dz r dr d)
14. 'c1Î2 '0 '0
15. '0 '0
21
1
13. 'c1Î2 '0
#
20. '1Î4 '0
)
1 2
'05 r cos f(rß )ß z) dz r dr d)
)
'03 r sin f(rß )ß z) dz r dr d)
)
'02 r sin f(rß )ß z) dz r dr d)
)
3# sin 9 d3 d9 d) œ 83 '0 '0 sin% 9 d9 d) œ 83 '0 Š’ sin 94cos 9 “ 34 '0 sin# 9 d9‹ d)
1
1
1
1
$
!
1
1
1
1
1
œ 2 '0 '0 sin# 9 d9 d) œ '0 ) sin#2) ‘ ! d) œ '0 1 d) œ 1#
22. '0 '0
21
1Î4
'02 (3 cos 9) 3# sin 9 d3 d9 d) œ '021 '01Î4 4 cos 9 sin 9 d9 d) œ '021 c2 sin# 9d 1! Î% d) œ '021 d) œ 21
Ð1 cos 9ÑÎ2
23. '0 '0 '0
21
1
1
" ' '
" '
3# sin 9 d3 d9 d) œ 24
(1 cos 9)$ sin 9 d9 d) œ 96
c(1 cos 9)% d ! d)
0
0
0
21
1
21
1
" '
"
1
'
œ 96
a2% 0b d) œ 16
96 0 d) œ 6 (21) œ 3
0
21
24. '0
31Î2
œ 56
21
'01 '01 53$ sin$ 9 d3 d9 d) œ 54 '031Î2 '01 sin$ 9 d9 d) œ 54 '031Î2 Š’ sin 93cos 9 “ 1 23 '01 sin 9 d9‹ d)
#
'031Î2 c cos 9d 1! d) œ 53 '031Î2 d) œ 5#1
!
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
915
916
Chapter 15 Multiple Integrals
25. '0 '0
21
1Î3
'sec2 9 33# sin 9 d3 d9 d) œ '021 '01Î3 a8 sec$ 9b sin 9 d9 d) œ '021 8 cos 9 "2 sec# 9‘ !1Î$ d)
œ '0 (4 2) ˆ8 "# ‰‘ d) œ 5# '0 d) œ 51
21
21
26. '0 '0
21
1Î4
'0sec 9 3$ sin 9 cos 9 d3 d9 d) œ "4 '021 '01Î4 tan 9 sec# 9 d9 d) œ "4 '021 "2 tan# 9‘ !1Î% d) œ "8 '021 d) œ 14
1Î2
27. '0 'c1 '1Î4 3$ sin 29 d9 d) d3 œ '0 ' 1 3$ cos229 ‘ 1Î% d) d3 œ '0 ' 1 3# d) d3 œ '0 3#1 d3 œ ’ 138 “ œ 21
2
0
2
1Î#
0
2
0
2 $
$
%
#
!
28. '1Î6 'csc 9
'021 3# sin 9 d) d3 d9 œ 21 '11ÎÎ63 'csc2 csc9 9 3# sin 9 d3 d9 œ 231 '11ÎÎ63 c3$ sin 9d csc2 csc9 9 d9 œ 1431 '11ÎÎ63 csc# 9 d9 œ 28È1
29. '0 '0 '0
123 sin$ 9 d9 d) d3 œ '0 '0 Œ123 ’ sin 39 cos 9 “
1Î3
2 csc 9
1
1
3
1Î4
1
1
#
1Î%
!
83 '0 sin 9 d9 d) d3
1 Î4
23
3
103
53
#
'
œ '0 '0 Š È
83 ccos 9d ! ‹ d) d3 œ '0 '0 Š83 10
È2 ‹ d) d3 œ 1 0 Š83 È2 ‹ d3 œ 1 ’43 È2 “
2
1
1
œ
1
1
1Î%
1
#
"
!
Š4È2 5‹ 1
È2
1 Î2
1 Î2
1Î2
1Î2
1Î2
1 Î2
30. '1Î6 ' 1Î2 'csc 9 53% sin$ 9 d3 d) d9 œ '1Î6 ' 1Î2 a32 csc& 9b sin$ 9 d) d9 œ '1Î6 ' 1Î2 a32 sin$ 9 csc# 9b d) d9
2
œ 1 '1Î6 a32 sin$ 9 csc# 9b d9 œ 1 ’ 32 sin 39 cos 9 “
1Î2
#
È
1Î#
1Î'
6431 '1Î6 sin 9 d9 1 ccot 9d 1Î'
1Î2
1Î#
È3
È
È
ˆ 6431 ‰ Š #3 ‹ œ 3313 3 œ 111È3
3 1
1Î#
œ 1 Š 3224 3 ‹ 6431 ccos 9d 1Î' 1 ŠÈ3‹ œ
31. (a) x# y# œ 1 Ê 3# sin# 9 œ 1, and 3 sin 9 œ 1 Ê 3 œ csc 9; thus
'021 '01Î6 '02 3# sin 9 d3 d9 d) '021 '11ÎÎ62 '0csc 9 3# sin 9 d3 d9 d)
sin " Ð1Î3Ñ
(b) '0 '1 '1Î6
21
2
3# sin 9 d9 d3 d) '0 '0 '0
21
1Î6
2
3# sin 9 d9 d3 d)
32. (a) '0 '0
'0sec 9 3# sin 9 d3 d9 d)
21
1
1Î4
(b) '0 '0 '0 3# sin 9 d9 d3 d)
21
1Î4
È2
'0 '1
21
33. V œ '0 '0
21
1Î2
'cos1Î4" Ð"Î3Ñ 3# sin 9 d9 d3 d)
'cos2 9 3# sin 9 d3 d9 d) œ "3 '021 '01Î2 a8 cos$ 9b sin 9 d9 d)
œ 3" '0 ’8 cos 9 cos4 9 “
21
%
1Î#
!
311
‰
d) œ 3" '0 ˆ8 4" ‰ d) œ ˆ 31
12 (21) œ 6
21
'11 cos 9 3# sin 9 d3 d9 d) œ "3 '021 '01Î2 a3 cos 9 3 cos# 9 cos$ 9b sin 9 d9 d)
21
21
1Î#
111
' 21
ˆ 11 ‰
œ 3" '0 3# cos# 9 cos$ 9 14 cos% 9‘ ! d) œ 3" '0 ˆ 32 1 "4 ‰ d) œ 11
12 0 d) œ 12 (21) œ 6
34. V œ '0 '0
21
1Î2
1ccos 9
35. V œ '0 '0 '0
21
"
œ 12
(2)%
1
9)
3# sin 9 d3 d9 d) œ "3 '0 '0 (1 cos 9)$ sin 9 d9 d) œ 3" '0 ’ (" cos
“ d)
4
21
1
21
%
'021 d) œ 34 (21) œ 831
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
1
!
3
Section 15.7 Triple Integrals in Cylindrical and Spherical Coordinates
1Î#
9)
'01 cos 9 3# sin 9 d3 d9 d) œ "3 '021 '01Î2 (1 cos 9)$ sin 9 d9 d) œ 3" '021 ’ (" cos
d)
“
4
36. V œ '0 '0
21
"
œ 12
1Î2
%
!
'021 d) œ 12" (21) œ 16
37. V œ '0 '1Î4 '0
21
917
1Î2
3# sin 9 d3 d9 d) œ 83 '0 '1Î4 cos$ 9 sin 9 d9 d) œ 83 '0 ’ cos4 9 “
2 cos 9
21
1Î2
21
%
" ‰'
œ ˆ 83 ‰ ˆ 16
d) œ "6 (21) œ 13
0
1Î#
1Î%
d)
21
38. V œ '0 '1Î3 '0 3# sin 9 d3 d9 d) œ 83 '0 '1Î3 sin 9 d9 d) œ 83 '0 c cos 9d 1Î$ d) œ 43 '0 d) œ 831
21
1Î2
39. (a) 8'0
1Î2
È4cx
2
1Î2
(c) '0
1Î2
1Î2
21
'01Î2 '02 3# sin 9 d3 d9 d)
(c) 8 '0 '0
40. (a) '0
21
2
#
È
(b) 8'0
1Î2
'0 4cx cy dz dy dx
#
È
È
21
1Î#
È
'02 '0 4cr dz r dr d)
#
#
1Î2
'03Î 2 'r 9 r dz r dr d)
(b) '0
#
'01Î4 '03 3# sin 9 d3 d9 d)
È
'01Î4 '03 3# sin 9 d3 d9 d) œ 9 '01Î2 '01Î4 sin 9 d9 d) œ 9 '01Î2 Š È" 1‹ d) œ 91 Š2 4 2‹
2
41. (a) V œ '0 '0
21
1Î3
'sec2 9 3# sin 9 d3 d9 d)
È3cx
È3
È4cx cy
(c) V œ 'cÈ3 'cÈ3cx# '1
#
È3
#
(d) V œ '0 '0 ’r a4 r# b
21
21
È1cr
42. (a) Iz œ '0 '0 '0
21
1
#
'1
#
dz r dr d)
#
"Î#
dz dy dx
r“ dr d) œ '0 ” a4 3r b
21
# $Î#
#
r# •
'021 d) œ 531
œ 56
È3 È4cr
(b) V œ '0 '0
È$
!
d) œ '0 Š 3" #3 4 3 ‹ d)
21
$Î#
r# dz r dr d)
(b) Iz œ '0 '0
'01 a3# sin# 9b a3# sin 9b d3 d9 d), since r# œ x# y# œ 3# sin# 9 cos# ) 3# sin# 9 sin# ) œ 3# sin# 9
(c) Iz œ '0 '0
"
"
$
5 sin 9 d9 d) œ 5
21
1Î2
21
1Î2
'021 Œ’ sin 93cos 9 “ 1Î# 32 '01Î2 sin 9 d9 d) œ 152 '021 c cos 9d !1Î# d)
#
!
2
œ 15
(21) œ 4151
43. V œ 4 '0
'01 'r 4 14r dz r dr d) œ 4 '01Î2 '01 a5r 4r$ r& b dr d) œ 4 '01Î2 ˆ 5# 1 "6 ‰ d) œ 4 '01Î2 d) œ 831
44. V œ 4'0
'01 ' 1Èr1 r dz r dr d) œ 4 '01Î2 '01 Šr r# rÈ1r# ‹ dr d) œ 4 '01Î2 ’ r2 r3 "3 a1 r# b$Î# “ " d)
1Î2
1Î2
œ 4 '0
#
%
#
$
#
1Î2
'
1Î2
ˆ "# "3 "3 ‰ d) œ 2
0
45. V œ '31Î2 '0
21
3 cos )
!
d) œ 2 ˆ 1# ‰ œ 1
'0cr sin dz r dr d) œ '321Î2 '03 cos r# sin ) dr d) œ '321Î2 a9 cos$ )b (sin )) d) œ 94 cos% )‘ #$11Î#
)
1
)
1
œ 94 0 œ 94
c3 cos )
46. V œ 2 '1Î2 '0
1
#
'0r dz r dr d) œ 2 '1Î2 '0c3 cos r# dr d) œ 23 '1Î2 27 cos$ ) d)
œ 18 Œ’ cos )3 sin ) “
1
1
1Î#
)
1
32 '1Î2 cos ) d) œ 12 csin )d 11Î# œ 12
1
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
918
Chapter 15 Multiple Integrals
47. V œ '0
1Î2
È
'0sin '0 1 r dz r dr d) œ '0 Î2 '0sin rÈ1r# dr d) œ '0 Î2 ’ "3 a1 r# b$Î# “ sin d)
)
#
1
œ "3 '0 ’a1 sin# )b
1Î2
1Î#
1Î2
!
$Î#
1“ d) œ "3 '0 acos$ ) 1b d) œ "3 Œ’ cos )3 sin ) “
1 Î2
È
)
#
1
œ '0 ’ a1 cos )b
#
1Î#
œ 1# 32 ccos )d !
$Î#
)
1Î#
!
32 '0 cos ) d) 3) ‘ !
1 Î2
1Î#
21
21Î3
)
1
1“ d) œ '0 a1 sin
1Î2
$
!
'01Î2 sin ) d)
1Î#
#
)b d) œ ’) sin )3cos ) “
23
!
œ 1# 32 œ 316 4
49. V œ '0 '1Î3 '0 3# sin 9 d3 d9 d) œ '0 '1Î3
1Î6
#
'0cos '03 1 r dz r dr d) œ '0 Î2 '0cos 3rÈ1r# dr d) œ '0 Î2 ’ a1 r# b$Î# “ cos d)
1Î2
50. V œ '0
)
1
31
16 œ 418
œ 29 csin )d !
48. V œ '0
)
21
a
21Î3
a$
a$
3 sin 9 d9 d) œ 3
'021 c cos 9d #11Î$Î$ d) œ a3 '021 ˆ "# "# ‰ d) œ 213a
$
'01Î2 '0a 3# sin 9 d3 d9 d) œ a3 '01Î6 '01Î2 sin 9 d9 d) œ a3 '01Î6 d) œ a181
$
51. V œ '0 '0
21
1Î3
$
$
'sec2 9 3# sin 9 d3 d9 d)
œ "3 '0 '0 a8 sin 9 tan 9 sec# 9b d9 d)
21
1Î3
œ "3 '0 8 cos 9 "2 tan# 9‘ !
21
1Î$
d)
œ "3 '0 4 #" (3) 8‘ d) œ 3" '0
21
52. V œ 4 '0
1Î2
'0
œ 28
3
1Î2
21
5
5
51
# d) œ 6 (21) œ 3
'01Î4 'sec2 sec9 9 3# sin 9 d3 d9 d) œ 43 '01Î2 '01Î4 a8 sec$ 9 sec$ 9b sin 9 d9 d)
'01Î4 sec$ 9 sin 9 d9 d) œ 283 '01Î2 '01Î4 tan 9 sec# 9 d9 d) œ 283 '01Î2 2" tan# 9‘ !1Î% d) œ 143 '01Î2 d) œ 731
53. V œ 4 '0
'01 '0r dz r dr d) œ 4 '01Î2 '01 r$ dr d) œ '01Î2 d) œ 1#
54. V œ 4 '0
'01 'r r 1 dz r dr d) œ 4 '01Î2 '01 r dr d) œ 2 '01Î2 d) œ 1
55. V œ 8 '0
'1 2 '0r dz r dr d) œ 8 '01Î2 '1 2 r# dr d) œ 8 Š 2È32" ‹ '01Î2 d) œ 41 Š2 3 2"‹
56. V œ 8 '0
'1 2 '0 2 r dz r dr d) œ 8 '01Î2 '1 2 rÈ2 r# dr d) œ 8 '01Î2 ’ "3 a2 r# b$Î# “ # d) œ 83 '01Î2 d) œ 431
1Î2
1Î2
#
#
#
1Î2
1Î2
È
È
È
È
#
4cr sin )
2
È
dz r dr d) œ '0 '0 a4r r# sin )b dr d) œ 8 '0 ˆ1 sin3 ) ‰ d) œ 161
21
4cr cos )cr sin )
58. V œ '0 '0 '0
21
È
1
57. V œ '0 '0 '0
21
È
2
21
2
dz r dr d) œ '0 '0 c4r r# (cos ) sin ))d dr d) œ 83 '0 (3 cos ) sin )) d) œ 161
21
21
2
59. The paraboloids intersect when 4x# 4y# œ 5 x# y# Ê x# y# œ 1 and z œ 4
Ê V œ 4 '0
1Î2
'01 '4r5 r dz r dr d) œ 4 '01Î2 '01 a5r 5r$ b dr d) œ 20 '01Î2 ’ r2 r4 “ " d) œ 5'01Î2 d) œ 5#1
#
#
#
%
!
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
$
Section 15.7 Triple Integrals in Cylindrical and Spherical Coordinates
60. The paraboloid intersects the xy-plane when 9 x# y# œ 0 Ê x# y# œ 9 Ê V œ 4 '0
1Î2
œ 4 '0
1Î2
919
'13 '09cr dz r dr d)
#
'13 a9r r$ b dr d) œ 4 '01Î2 ’ 9r2 r4 “ $ d) œ 4 '01Î2 ˆ 814 174 ‰ d) œ 64 '01Î2 d) œ 321
#
%
"
È4cr
61. V œ 8 '0 '0 '0
21
1
#
dz r dr d) œ 8 '0 '0 r a4 r# b
21
œ 83 '0 ˆ3$Î# 8‰ d) œ
21
1
"Î#
dr d) œ 8 '0 ’ "3 a4 r# b
21
$Î# "
“ d)
!
41 Š8 3È3‹
3
62. The sphere and paraboloid intersect when x# y# z# œ 2 and z œ x# y# Ê z# z 2 œ 0
Ê (z 2)(z 1) œ 0 Ê z œ 1 or z œ 2 Ê z œ 1 since z 0. Thus, x# y# œ 1 and the volume is
given by the triple integral V œ 4 '0
1Î2
œ 4 '0 ’ "3 a2 r# b
1Î2
È
'01 'r 2 r dz r dr d) œ 4 '01Î2 '01 ’r a2 r# b"Î# r$ “ dr d)
#
#
1 Š8 È 2 7 ‹
7
r4 “ d) œ 4 '0 Š 2 3 2 12
‹ d) œ
$Î#
%
"
1Î2
È
6
!
63. average œ #"1 '0 '0 'c1 r# dz dr d) œ #"1 '0 '0 2r# dr d) œ 3"1 '0 d) œ 23
21
1
21
1
È1cr
21
1
3 ' '
64. average œ ˆ 4"1 ‰ '0 '0 'cÈ1cr# r# dz dr d) œ 41
2r# È1 r# dr d)
0
0
21
1
21
#
1
3
œ 231
'
"
’ "8 sin" r "8 rÈ1 r# a1 2r# b“ d) œ 1631
0
!
21
'021 ˆ 1# 0‰ d) œ 323 '021 d) œ ˆ 323 ‰ (21) œ 3161
65. average œ ˆ 4"1 ‰ '0 '0 '0 3$ sin 9 d3 d9 d) œ 1631 '0 '0 sin 9 d9 d) œ 831 '0 d) œ 43
21
1
21
1
1
21
3
66. average œ ˆ 2"1 ‰ '0 '0
21
1Î2
3
œ 1631
'01 3$ cos 9 sin 9 d3 d9 d) œ 831 '021 '01Î2 cos 9 sin 9 d9 d) œ 831 '021 ’ sin2 9 “ 1Î# d)
#
!
'021 d) œ ˆ 1631 ‰ (21) œ 38
67. M œ 4 '0
'01 '0r dz r dr d) œ 4 '01Î2 '01 r# dr d) œ 43 '01Î2 d) œ 231 ; Mxy œ '021 '01 '0r z dz r dr d)
21
1
21
œ "# '0 '0 r$ dr d) œ 18 '0 d) œ 14 Ê z œ MM œ ˆ 14 ‰ ˆ 231 ‰ œ 38 , and x œ y œ 0, by symmetry
1Î2
xy
68. M œ '0
'02 '0r dz r dr d) œ '01Î2 '02 r# dr d) œ 83 '01Î2 d) œ 431 ; Myz œ '01Î2 '02 '0r x dz r dr d)
2
2
r
2
1Î2
1Î2
1Î2
1Î2
œ '0 '0 r$ cos ) dr d) œ 4 '0 cos ) d) œ 4; Mxz œ '0 '0 '0 y dz r dr d) œ '0 '0 r$ sin ) dr d)
2
r
2
1Î2
1Î2
1Î2
1Î2
M
œ 4 '0 sin ) d) œ 4; Mxy œ '0 '0 '0 z dz r dr d) œ "# '0 '0 r$ dr d) œ 2 '0 d) œ 1 Ê x œ M œ 13 ,
1Î2
yz
M
y œ MMxz œ 13 , and z œ Mxy œ 34
69. M œ 831 ; Mxy œ '0 '1Î3 '0 z3# sin 9 d3 d9 d) œ '0 '1Î3 '0 3$ cos 9 sin 9 d3 d9 d) œ 4 '0 '1Î3 cos 9 sin 9 d9 d)
21
œ 4 '0 ’ sin2 9 “
21
#
70. M œ '0 '0
1Î#
1Î$
1Î2
21
2
1Î2
21
2
d) œ 4 '0 ˆ "# 38 ‰ d) œ "# '0 d) œ 1 Ê z œ Mxy œ (1) ˆ 831 ‰ œ 38 , and x œ y œ 0, by symmetry
21
21
M
È
'0a 3# sin 9 d3 d9 d) œ a3 '021 '01Î4 sin 9 d9 d) œ a3 '021 2 #È2 d) œ 1a Š23 2‹ ;
1Î4
1Î4
21
a
21
21
a '
Mxy œ '0 '0 '0 3$ sin 9 cos 9 d3 d9 d) œ a4 '0 '0 sin 9 cos 9 d9 d) œ 16
d) œ 18a
0
21
1Î4
1Î2
$
$
$
%
%
%
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
920
Chapter 15 Multiple Integrals
%
Ê z œ Mxy œ Š 18a ‹ –
M
Èr
71. M œ '0 '0 '0
21
4
3
— œ ˆ 8 ‰Š
3a
1a$ Š2È2‹
3 Š2È2‹ a
2 È2
, and x œ y œ 0, by symmetry
# ‹œ
16
Èr
'0 d) œ 1285 1 ; Mxy œ '0 '0 '0 z dz r dr d)
dz r dr d) œ '0 '0 r$Î# dr d) œ 64
5
21
21
4
21
4
'0 d) œ 6431 Ê z œ Mxy œ 65 , and x œ y œ 0, by symmetry
œ "# '0 '0 r# dr d) œ 32
3
21
21
4
1Î3
M
È1 r
1Î3
1Î3
72. M œ 'c1Î3 '0 ' È1 r# dz r dr d) œ ' 1Î3 '0 2rÈ1 r# dr d) œ ' 1Î3 ’ 23 a1 r# b
œ 23
'
#
1
1
$Î# "
“ d)
!
È
' 1Î3 '0 ' È11 rr r# cos ) dz dr d) œ 2 ' 11ÎÎ33 '01 r# È1 r# cos ) dr d)
1Î3
1Î3
d) œ ˆ 23 ‰ ˆ 231 ‰ œ 491 ; Myz œ
c1Î3
#
1
#
1 Î3
1 Î3
œ 2 'c1Î3 ’ 18 sin" r "8 rÈ1 r# a1 2r# b“ cos ) d) œ 18 ' 1Î3 cos ) d) œ 18 csin )d 1Î13Î3 œ ˆ 18 ‰ Š2 † #3 ‹ œ 1 8 3
Ê
"
È
È
!
Myz
9È 3
x œ M œ 32 , and y œ z œ 0, by symmetry
73. We orient the cone with its vertex at the origin and axis along the z-axis Ê 9 œ 14 . We use the the x-axis
which is through the vertex and parallel to the base of the cone Ê Ix œ '0 '0 'r ar# sin# ) z# b dz r dr d)
21
1
1
"
)
) ‘ #1
œ '0 '0 Šr$ sin# ) r% sin# ) 3r r3 ‹ dr d) œ '0 Š sin20 ) 10
sin802) 10
œ #10 15 œ 14
‹ d) œ 40
!
21
1
74. Iz œ '0 '0
21
1a
œ 815
21
%
Èa cr
#
a
#
'
cÈa cr
#
#
#
#
#
r$ dz dr d) œ '0 '0 2r$ Èa# r# dr d) œ 2 '0 ’Š r5 2a
15 ‹ aa r b
21
21
a
#
“ d) œ 2 '0
21
$Î# a
#
!
2 &
15 a d)
&
%
75. Iz œ '0 '0 'ˆ h ‰ r ax# y# b dz r dr d) œ '0 '0 ' ' hr r$ dz dr d) œ '0 '0 Šhr$ hra ‹ dr d) œ '0
21
a
21
h
a
hr
a
a
'0 d) œ 110ha
œ '0 h Š a4 5a
‹ d) œ ha
20
21
h
a
%
&
21
%
21
h
21
a
%
&
a
r
h ’ r4 5a
“ d)
!
a
%
" '
76. (a) M œ '0 '0 '0 z dz r dr d) œ '0 '0 "# r& dr d) œ 12
d) œ 16 ; Mxy œ '0 '0 '0 z# dz r dr d)
0
21
1
21
r#
21
1
21
r#
1
" '
1
œ 3" '0 '0 r( dr d) œ 24
d) œ 12
Ê z œ #" , and x œ y œ 0, by symmetry; Iz œ '0 '0 '0 zr$ dz dr d)
0
21
21
1
21
r#
1
" '
œ "# '0 '0 r( dr d) œ 16
d) œ 18
0
21
21
1
' '0 '0 zr# dz dr d)
(b) M œ '0 '0 '0 r# dz dr d) œ '0 '0 r% dr d) œ 5" '0 d) œ 21
5 ; Mxy œ 0
21
1
21
r#
21
1
21
1
r#
" '
5
œ 2" '0 '0 r' dr d) œ 14
d) œ 17 Ê z œ 14
, and x œ y œ 0, by symmetry; Iz œ '0 '0 '0 r% dz dr d)
0
21
21
1
21
r#
1
œ '0 '0 r' dr d) œ 7" '0 d) œ 21
7
21
21
1
77. (a) M œ '0 '0 'r z dz r dr d) œ "# '0 '0 ar r$ b dr d) œ 8" '0 d) œ 14 ; Mxy œ '0 '0 'r z# dz r dr d)
21
1
21
1
21
1
21
1
1
" '
œ 3" '0 '0 ar r% b dr d) œ 10
d) œ 15 Ê z œ 45 , and x œ y œ 0, by symmetry; Iz œ '0 '0 'r zr$ dz dr d)
0
21
21
1
1
" '
œ "# '0 '0 ar$ r& b dr d) œ 24
d) œ 12
0
21
21
1
1
21
1
(b) M œ '0 '0 'r z# dz r dr d) œ 15 from part (a); Mxy œ '0 '0 'r z$ dz r dr d) œ 4" '0 '0 ar r& b dr d)
21
1
1
21
1
21
1
1
" '
œ 12
d) œ 16 Ê z œ 56 , and x œ y œ 0, by symmetry; Iz œ '0 '0 'r z# r$ dz dr d) œ "3 '0 '0 ar$ r' b dr d)
0
21
21
1
1
" '
1
œ 28
d) œ 14
0
21
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
21
1
Section 15.7 Triple Integrals in Cylindrical and Spherical Coordinates
921
78. (a) M œ '0 '0 '0 3% sin 9 d3 d9 d) œ a5 '0 '0 sin 9 d9 d) œ 2a5 '0 d) œ 415a ; Iz œ '0 '0 '0 3' sin$ 9 d3 d9 d)
21
1
a
21
&
1
&
21
21
&
1
a
'0 d) œ 8a211
œ a7 '0 '0 a1 cos# 9b sin 9 d9 d) œ a7 '0 ’ cos 9 cos3 9 “ d) œ 4a
#1
21
(
1
21
(
1
$
(
21
(
!
21
a
21
21
1
1
29 )
(b) M œ '0 '0 '0 3$ sin# 9 d3 d9 d) œ a4 '0 '0 (1cos
d9 d) œ 18a '0 d) œ 1 4a ;
#
1
1
21
a
21
Iz œ '0 '0 '0 3& sin% 9 d3 d9 d) œ a6 '0 '0 sin% 9 d9 d)
1
21
21
21
1
1
œ a6 '0 Š’ sin 49 cos 9 “ 43 '0 sin# 9 d9‹ d) œ a8 '0 9# sin429 ‘ ! d) œ 116a '0 d) œ a 81
%
%
# %
'
$
'
'
'
' #
!
79. M œ '0 '0 '0
Èa cr
dz r dr d) œ '0 '0
21
a
21 $
a
d) œ 2ha3 1 ; Mxy œ '0 '0 '0
œ ha '0
3
h
a
#
#
21
21
#
a
a
'021 ’ 3" aa# r# b$Î# “ a d)
h È #
h
#
a r a r dr d) œ a
Èa cr
#
h
a
#
!
h ' '
z dz r dr d) œ 2a
aa# r r$ b dr d)
#
0
0
21
#
a
h '
œ 2a
Š a# a4 ‹ d) œ a h4 1 Ê z œ Š 1a4h ‹ ˆ 2ha3# 1 ‰ œ 83 h, and x œ y œ 0, by symmetry
#
0
21
#
%
%
# #
# #
80. Let the base radius of the cone be a and the height h, and place the cone's axis of symmetry along the z-axis
with the vertex at the origin. Then M œ 1a3 h and Mxy œ '0 '0 'ˆ h ‰ r z dz r dr d) œ "# '0
21
#
a
21
h
'0a Šh# r ha r$ ‹ dr d)
#
#
a
œ h# '0 ’ r2 4ar # “ d) œ h# '0 Š a# a4 ‹ d) œ h 8a '0 d) œ h a4 1 Ê z œ MMxy œ Š h a4 1 ‹ ˆ 1a3# h ‰ œ 43 h, and
21
#
#
a
%
#
21
#
#
# #
21
# #
# #
!
x œ y œ 0, by symmetry Ê the centroid is one fourth of the way from the base to the vertex
81. The density distribution function is linear so it has the form $ (3) œ k3 C, where 3 is the distance from the
center of the planet. Now, $ (R) œ 0 Ê kR C œ 0, and $ (3) œ k3 kR. It remains to determine the constant
k: M œ '0 '0 '0 (k3 kR) 3# sin 9 d3 d9 d) œ '0 '0 ’k 34 kR 33 “ sin 9 d9 d)
21
œ'
Ê
21
'
1
21
R
'
1
1
%
$
'
21
21
R
!
%
%
%
k Š R4 R3 ‹ sin 9 d9 d) œ 0 1k# R% c cos 9d 1! d) œ 0 6k R% d) œ k13R
0
0
‰ R œ 13M
$ (3) œ 13M
3 13M
R . At the center of the planet 3 œ 0 Ê $ (0) œ ˆ 13M
R$ .
R%
R%
R%
Ê k œ 13M
R%
82. The mass of the plant's atmosphere to an altitude h above the surface of the planet is the triple integral
M(h) œ '0 '0 'R .! ecÐ3RÑ 3# sin 9 d3 d9 d) œ 'R '0 '0 .! ecÐ3RÑ 3# sin 9 d9 d) d3
21
1
h
21
h
1
œ 'R '0 .! ecÐ3RÑ 3# ( cos 9)‘ ! d) d3 œ 2 'R '0 .! ecR ec3 3# d) d3 œ 41.! ecR 'R ec3 3# d3
h
21
1
œ 41.! ecR ’ 3 ec
c3
#
ch
#
œ 41.! ecR Š h ec
23ce#
c3
2hec#
ch
c3
2ec$ “
ch
h
21
h
h
R
(by parts)
#
2ec$ R ec
cR
2Rec#
cR
cR
2ec$ ‹ .
The mass of the planet's atmosphere is therefore M œ lim
hÄ_
#
2
M(h) œ 41.! Š Rc 2R
c# c$ ‹ .
83. (a) A plane perpendicular to the x-axis has the form x œ a in rectangular coordinates Ê r cos ) œ a Ê r œ cosa )
Ê r œ a sec ), in cylindrical coordinates.
(b) A plane perpendicular to the y-axis has the form y œ b in rectangular coordinates Ê r sin ) œ b Ê r œ sinb )
Ê r œ b csc ), in cylindrical coordinates.
84. ax by œ c Ê aar cos )b bar sin )b œ c Ê raa cos ) b sin )b œ c Ê r œ a cos ) c b sin ) .
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
922
Chapter 15 Multiple Integrals
85. The equation r œ fazb implies that the point ar, ), zb
œ afazb, ), zb will lie on the surface for all ). In particular
afazb, ) 1, zb lies on the surface whenever afazb, ), zb does
Ê the surface is symmetric with respect to the z-axis.
86. The equation 3 œ fa9b implies that the point a3, 9, )b œ afa9b, 9, )b lies on the surface for all ). In particular, if
afa9b, 9, )b lies on the surface, then afa9b, 9, ) 1b lies on the surface, so the surface is symmetric wiith respect to the
z-axis.
15.8 SUBSTITUTIONS IN MULTIPLE INTEGRALS
1. (a) x y œ u and 2x y œ v Ê 3x œ u v and y œ x u Ê x œ "3 (u v) and y œ 3" (2u v);
` (xßy)
` (ußv) œ »
"
3
23
"
"
2
"
3
" »œ 9 9 œ 3
3
(b) The line segment y œ x from (!ß 0) to (1ß 1) is x y œ 0
Ê u œ 0; the line segment y œ 2x from (0ß 0) to
(1ß 2) is 2x y œ 0 Ê v œ 0; the line segment x œ 1
from (1ß 1) to ("ß 2) is (x y) (2x y) œ 3
Ê u v œ 3. The transformed region is sketched at the
right.
2. (a) x 2y œ u and x y œ v Ê 3y œ u v and x œ v y Ê y œ "3 (u v) and x œ 3" (u 2v);
"
` (xßy)
3
œ
»1
` (ußv)
3
2
3
œ 9" 92 œ 3"
3" »
(b) The triangular region in the xy-plane has vertices (0ß 0),
(2ß 0), and ˆ 23 ß 23 ‰ . The line segment y œ x from (0ß 0)
to ˆ 23 ß 23 ‰ is x y œ 0 Ê v œ 0; the line segment
y œ 0 from (0ß 0) to (#ß 0) Ê u œ v; the line segment
x 2y œ 2 from ˆ 23 ß 23 ‰ to (2ß 0) Ê u œ 2. The
transformed region is sketched at the right.
"
3. (a) 3x 2y œ u and x 4y œ v Ê 5x œ 2u v and y œ "# (u 3x) Ê x œ 5" (2u v) and y œ 10
(3v u);
` (xßy)
` (ußv) œ »
2
5
1
10
15
6
1
"
3 » œ 50 50 œ 10
10
(b) The x-axis y œ 0 Ê u œ 3v; the y-axis x œ 0
Ê v œ 2u; the line x y œ 1
"
Ê "5 (2u v) 10
(3v u) œ 1
Ê 2(2u v) (3v u) œ 10 Ê 3u v œ 10. The
transformed region is sketched at the right.
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 15.8 Substitutions in Multiple Integrals
4. (a) 2x 3y œ u and x y œ v Ê x œ u 3v and y œ v x Ê x œ u 3v and y œ u 2v;
" 3
` (xßy)
` (ußv) œ º 1 2 º œ 2 3 œ 1
(b) The line x œ 3 Ê u 3v œ 3 or u 3v œ 3;
x œ 0 Ê u 3v œ 0; y œ x Ê v œ 0; y œ x 1
Ê v œ 1. The transformed region is the parallelogram
sketched at the right.
5.
'04 'yÐÎy2Î2Ñ 1 ˆx y# ‰ dx dy œ '04 ’ x2 xy# “ " dy œ "# '04 ’ˆ y# 1‰# ˆ y# ‰# ˆ y# 1‰ y ˆ y# ‰ y“ dy
#
y
2
y
2
œ "# '0 (y 1 y) dy œ "# '0 dy œ "# (4) œ 2
4
6.
4
' ' a2x# xy y# b dx dy œ ' ' (x y)(2x y) dx dy
R
R
ßy)
" ''
œ ' ' uv ¹ `` (x
uv du dv;
(ußv) ¹ du dv œ 3
G
G
We find the boundaries of G from the boundaries of R,
shown in the accompanying figure:
xy-equations for
Corresponding uv-equations
Simplified
for the boundary of G
uv-equations
the boundary of R
y œ 2x 4
"
2
3 (2u v) œ 3 (u v) 4
vœ4
y œ 2x 7
2
"
3 (2u v) œ 3 (u v) 7
vœ7
yœx2
1
"
3 (2u v) œ 3 (u v) 2
uœ2
yœx1
1
"
3 (2u v) œ 3 (u v) 1
u œ 1
'
ˆ 11 ‰ u
Ê "3 ' ' uv du dv œ "3 'c1 '4 uv dv du œ "3 'c1 u ’ v2 “ du œ 11
# c1 u du œ # ’ 2 “
2
7
2
#
2
%
G
7.
(
#
#
"
33
‰
œ ˆ 11
4 (4 1) œ 4
' ' a3x# 14xy 8y# b dx dy
R
œ ' ' (3x 2y)(x 4y) dx dy
R
ßy)
" ''
œ ' ' uv ¹ `` (x
uv du dv;
(ußv) ¹ du dv œ 10
G
G
We find the boundaries of G from the boundaries of R,
shown in the accompanying figure:
xy-equations for
the boundary of R
Corresponding uv-equations
Simplified
for the boundary of G
uv-equations
y œ 3# x 1
3
"
10 (3v u) œ 10 (2u v) 1
uœ2
y œ 3# x 3
"
3
10 (3v u) œ 10 (2u v) 3
uœ6
y œ 4" x
"
1
10 (3v u) œ 20 (2u v)
vœ0
y œ 4" x 1
"
1
10 (3v u) œ 20 (2u v) 1
vœ4
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
923
924
Chapter 15 Multiple Integrals
" ''
" ' '
" '
Ê 10
uv du dv œ 10
uv dv du œ 10
u ’ v2 “ du œ 45 '2 u du œ ˆ 54 ‰ ’ u2 “ œ ˆ 54 ‰ (18 2) œ 64
5
2
0
2
6
4
6
#
6
'
#
!
G
8.
%
#
' ' 2(x y) dx dy œ ' ' 2v ¹ `` (x(ußßy)v) ¹ du dv œ ' ' 2v du dv; the region G is sketched in Exercise 4
R
G
G
3c3v
Ê ' ' 2v du dv œ '0 'c3v 2v du dv œ '0 2v(3 3v 3v) dv œ '0 6v dv œ c3v# d ! œ 3
1
1
1
"
G
v" uv#
œ v" u v" u œ 2u
v ;
v
u º
y œ x Ê uv œ uv Ê v œ 1, and y œ 4x Ê v œ 2; xy œ 1 Ê u œ 1, and xy œ 9 Ê u œ 3; thus
(xßy)
9. x œ uv and y œ uv Ê yx œ v# and xy œ u# ; `` (u
ßv) œ J(uß v) œ º
' ' ŠÉ yx Èxy‹ dx dy œ ' ' (v u) ˆ 2uv ‰ dv du œ ' ' Š2u 2uv # ‹ dv du œ ' c2uv 2u# ln vd #" du
R
3
2
3
2
3
1
1
1
1
1
œ '1 a2u 2u# ln 2b du œ u# 23 u# ln 2‘ " œ 8 23 (26)(ln 2) œ 8 52
3 (ln 2)
3
10. (a)
$
"
` (xßy)
` (ußv) œ J(uß v) œ º v
0
œ u, and
uº
the region G is sketched at the right
(b) x œ 1 Ê u œ 1, and x œ 2 Ê u œ 2; y œ 1 Ê uv œ 1 Ê v œ "u , and y œ 2 Ê uv œ 2 Ê v œ 2u ; thus,
'12 '12 yx dy dx œ '12 '12ÎÎuu ˆ uvu ‰ u dv du œ '12 '12ÎÎuu uv dv du œ '12 u ’ v2 “ 2Îu du œ '12 u ˆ u2 2u" ‰ du
#
#
1Îu
#
#
3
3
œ #3 '1 u ˆ u"# ‰ du œ 3# cln ud #" œ 3# ln 2; '1 '1 yx dy dx œ '1 ’ x1 † y2 “ dx œ 3# '1 dx
x œ # cln xd " œ # ln 2
2
2
2
2
2
2
2
1
ßy)
11. x œ ar cos ) and y œ ar sin ) Ê ``(x
(rß)) œ J(rß ) ) œ º
a cos )
b sin )
ar sin )
œ abr cos# ) abr sin# ) œ abr;
br cos ) º
I! œ ' ' ax# y# b dA œ '0 '0 r# aa# cos# ) b# sin# )b kJ(rß ))k dr d) œ '0 '0 abr$ aa# cos# ) b# sin# )b dr d)
21
21
1
1
R
œ ab
4
'021 aa# cos# ) b# sin# )b d) œ ab4 ’ a2) a sin4 2) b2) b sin4 2) “ 21 œ ab1 aa4 b b
#
#
#
#
#
#
!
ßy)
12. `` (x
(ußv) œ J(uß v) œ º
È1cu#
1
a 0
œ ab; A œ ' ' dy dx œ ' ' ab du dv œ 'c1 'cÈ1cu# ab dv du
º
0 b
R
G
œ 2ab 'c1 È1u# du œ 2ab ’ u2 È1 u# "# sin" u“
1
"
"
œ ab csin" 1 sin" (1)d œ ab 1# ˆ 1# ‰‘ œ ab1
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 15.8 Substitutions in Multiple Integrals
13. The region of integration R in the xy-plane is
sketched in the figure at the right. The
boundaries of the image G are obtained as
follows, with G sketched at the right:
xy-equations for
Corresponding uv-equations
Simplified
for the boundary of G
uv-equations
the boundary of R
xœy
"
1
3 (u 2v) œ 3 (u v)
vœ0
x œ 2 2y
"
2
3 (u 2v) œ 2 3 (u v)
uœ2
yœ0
0 œ 13 (u v)
vœu
ßy)
"
'0
Also, from Exercise 2, `` (x
(ußv) œ J(uß v) œ 3 Ê
2Î3
'y2 2y (x 2y) eÐy xÑ dx dy œ '02 '0u ue v ¸ 3" ¸ dv du
œ "3 '0 u cecv d !u du œ 3" '0 u a1 ecu b du œ 3" ’u au ecu b u# ecu “ œ 3" c2 a2 ec2 b 2 ec2 1d
2
2
#
#
!
œ "3 a3ec2 1b ¸ 0.4687
14. x œ u v# and y œ v Ê 2x y œ (2u v) v œ 2u and
"
` (xßy)
` (ußv) œ J(uß v) œ º
"
#
v
º œ 1; next, u œ x #
0 "
œ x y# and v œ y, so the boundaries of the region of
integration R in the xy-plane are transformed to the
boundaries of G:
xy-equations for
Corresponding uv-equations
Simplified
for the boundary of G
uv-equations
the boundary of R
x œ y#
x œ #y 2
u #v œ #v
u #v œ #v 2
uœ2
yœ0
vœ0
vœ0
yœ2
vœ2
vœ2
ÐyÎ2Ñ 2
Ê '0 'yÎ2
2
uœ0
y$ (2x y) eÐ2xyÑ dx dy œ '0 '0 v$ (2u) e4u du dv œ '0 v$ ’ "4 e4u “ dv œ 4" '0 v$ ae16 1b dv
#
2
2
2
#
#
#
2
!
% #
œ "4 ae16 1b ’ v4 “ œ e16 1
!
v" uv#
œ v" u v" u œ 2u
v ;
v
u º
y œ x Ê uv œ uv Ê v œ 1, and y œ 4x Ê v œ 2; xy œ 1 Ê u œ 1, and xy œ 4 Ê u œ 2; thus
(xßy)
15. x œ uv and y œ uv Ê yx œ v# and xy œ u# ; `` (u
ßv) œ J(uß v) œ º
'12 '1yÎyax2 y2 b dx dy '24 'y4ÎÎ4yax2 y2 b dx dy œ '12 '12 Š uv u2 v2 ‹ ˆ 2uv ‰ du dv œ '12 '12 Š 2uv 2u3 v‹ du dv
2
3
2
3
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
925
926
Chapter 15 Multiple Integrals
u
1 4
15
15v ‰
15
15v
225
œ '1 ’ 2v
dv œ '1 ˆ 2v
dv œ ’ 4v
3 2 u v“
3 2
2 4 “ œ 16
2
#
4
2
2
"
2
"
ßy)
16. x œ u2 v2 and y œ 2uv; `` (x
(ußv) œ J(uß v) œ º
2u
2v
2v
œ 4u2 4v2 œ 4au2 v2 b ;
2u º
y œ 2È1 x Ê y2 œ 4a1 xb Ê a2uvb2 œ 4a1 au2 v2 bb Ê u œ „ 1; y œ 0 Ê 2uv œ 0 Ê u œ 0 or v œ 0;
x œ 0 Ê u2 v2 œ 0 Ê u œ v or u œ v; This gives us four triangular regions, but only the one in the quadrant where
both u, v are positive maps into the region R in the xy-plane.
È
'01 '02 1 x Èx2 y2 dx dy œ '01 '0u Éau2 v2 b2 a2uvb2 † 4au2 v2 b dv du œ 4'01 '0u au2 v2 b2 dv du
2
u
112 1 6 ‘ 2
56
'2 5
œ 4'1 u4 v 23 u2 v3 15 v5 ‘0 du œ 112
15 1 u du œ 15 6 u " œ 45
ßy)
17. (a) x œ u cos v and y œ u sin v Ê `` (x
(ußv) œ º
ßy)
(b) x œ u sin v and y œ u cos v Ê `` (x
(ußv) œ º
cos v u sin v
œ u cos# v u sin# v œ u
sin v
u cos v º
sin v
u cos v
œ u sin# v u cos# v œ u
cos v u sin v º
â
â cos v
â
z)
œ â sin v
18. (a) x œ u cos v, y œ u sin v, z œ w Ê ``(u(xßßvyßßw)
â
â 0
u sin v
u cos v
0
â
â2
â
z)
(b) x œ 2u 1, y œ 3v 4, z œ "# (w 4) Ê ``(u(xßßvyßßw)
œ â0
â
â0
â
â sin 9 cos )
â
19. â sin 9 sin )
â
â cos 9
œ (cos 9) º
3 cos 9 cos )
3 cos 9 sin )
3 sin 9
3 cos 9 cos )
3 cos 9 sin )
0
3
0
â
0â
â
0 â œ u cos# v u sin# v œ u
â
"â
â
0â
0 ââ œ (2)(3) ˆ #" ‰ œ 3
" â
# â
â
3 sin 9 sin ) â
â
3 sin 9 cos ) â
â
0
â
3 sin 9 sin )
sin 9 cos )
(3 sin 9) º
3 sin 9 cos ) º
sin 9 sin )
3 sin 9 sin )
3 sin 9 cos ) º
œ a3# cos 9b asin 9 cos 9 cos# ) sin 9 cos 9 sin# )b a3# sin 9b asin# 9 cos# ) sin# 9 sin# )b
œ 3# sin 9 cos# 9 3# sin$ 9 œ a3# sin 9b acos# 9 sin# 9b œ 3# sin 9
w
w
'
'
20. Let u œ gaxb Ê Jaxb œ du
dx œ g axb Ê a faub du œ gaab fagaxbbg axb dx in accordance with Theorem 7 in
gabb
b
Section 5.6. Note that gw axb represents the Jacobian of the transformation u œ gaxb or x œ g" aub.
21. '0 '0 'yÎ2
3
4
œ'
1
ÐyÎ2Ñ
xz
ˆ 2x # y 3z ‰ dx dy dz œ ' ' ’ x2 xy
# 3 “
%
#
#
’ (y 4 1) y4 yz
3 “ ! dz œ
0
3
'
3
4
0
0
#
3
"‰
ˆ 49 4z
3 4 dz œ
0
"ÐyÎ2Ñ
yÎ2
dy dz œ '0 '0 "# (y 1) y# 3z ‘ dy dz
3
4
'0 ˆ2 4z3 ‰ dz œ ’2z 2z3 “ $ œ 12
3
#
!
â
â
âa 0 0â
#
#
#
â
â
22. J(uß vß w) œ â 0 b 0 â œ abc; the transformation takes the ellipsoid region xa# by# cz# Ÿ 1 in xyz-space
â
â
â0 0 câ
into the spherical region u# v# w# Ÿ 1 in uvw-space ˆwhich has volume V œ 43 1‰
Ê V œ ' ' ' dx dy dz œ ' ' ' abc du dv dw œ 413abc
R
G
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Chapter 15 Practice Exercises
927
â
â
âa 0 0â
â
â
23. J(uß vß w) œ â 0 b 0 â œ abc; for R and G as in Exercise 22, ' ' ' kxyzk dx dy dz
â
â
R
â0 0 câ
œ ' ' ' a# b# c# uvw dw dv du œ 8a# b# c# '0
1Î2
G
# # #
œ 4a 3b c
'01Î2 '01 (3 sin 9 cos ))(3 sin 9 sin ))(3 cos 9) a3# sin 9b d3 d9 d)
'01Î2 '01Î2 sin ) cos ) sin$ 9 cos 9 d9 d) œ a b3 c '01Î2 sin ) cos ) d) œ a b6 c
# # #
# # #
â 1
â
â
24. u œ x, v œ xy, and w œ 3z Ê x œ u, y œ vu , and z œ "3 w Ê J(uß vß w) œ â uv#
â
â 0
0
"
u
0
0 ââ
"
0 ââ œ 3u
;
" â
â
3
' ' ' ax# y 3xyzb dx dy dz œ ' ' ' u# ˆ vu ‰ 3u ˆ vu ‰ ˆ w3 ‰‘ kJ(uß vß w)k du dv dw œ "3 ' ' ' ˆv vw
‰ du dv dw
u
0
0
1
3
D
œ "3
2
2
G
'0 '0 (v vw ln 2) dv dw œ 3" '03 (1 w ln 2) ’ v2 “ # dw œ 32 '03 (1 w ln 2) dw œ 32 ’w w2 ln 2“ $
3
2
#
#
!
!
œ 23 ˆ3 #9 ln 2‰ œ 2 3 ln 2 œ 2 ln 8
#
#
#
25. The first moment about the xy-coordinate plane for the semi-ellipsoid, xa# by# cz# œ 1 using the
transformation in Exercise 23 is, Mxy œ ' ' ' z dz dy dx œ ' ' ' cw kJ(uß vß w)k du dv dw
D
G
œ abc# ' ' ' w du dv dw œ aabc# b † aMxy of the hemisphere x# y# z# œ 1, z
G
#
0b œ abc4 1 ;
#
1
3 ‰
3
the mass of the semi-ellipsoid is 2abc
Ê z œ Š abc4 1 ‹ ˆ 2abc
3
1 œ 8 c
26. A solid of revolution is symmetric about the axis of revolution, therefore, the height of the solid is solely a function of r.
That is, y œ faxb œ farb. Using cylindrical coordinates with x œ r cos ), y œ y and z œ r sin ), we have
V œ ' ' ' r dy d) dr œ 'a '0 '0
farb
21
b
G
r dy d) dr œ 'a '0 c r y d0 d) dr œ 'a '0 r farb d) dr œ 'a c r)farb d201 dr
b
21
farb
b
21
b
'ab 21rfarbdr. In the last integral, r is a dummy or stand-in variable and as such it can be replaced by any variable name.
b
Choosing x instead of r we have V œ 'a 21xfaxbdx, which is the same result obtained using the shell method.
CHAPTER 15 PRACTICE EXERCISES
1.
'110 '01Îyyexy dx dy œ '110 cexy d !"Îy dy
10
œ '1 (e 1) dy œ 9e 9
'01 '0x eyÎx dy dx œ '01 x eyÎx ‘ !x dx
$
2.
$
œ '0 Šxex x‹ dx œ ’ "2 ex x# “ œ e # 2
1
#
#
#
"
!
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
928
3.
Chapter 15 Multiple Integrals
È
#
È9 4t
#
9
'03Î2 ' È99 4t4t t ds dt œ '03Î2 ctsd È
#
4t#
dt
œ '0 2tÈ9 4t# dt œ ’ "6 a9 4t# b
$Î# $Î#
3Î2
“
9
œ 6" ˆ0$Î# 9$Î# ‰ œ 27
6 œ #
4.
!
'01 'È2cy Èy xy dx dy œ '01 y ’ x2 “ 2cÈy dy
#
Èy
'0 y ˆ4 4Èy y y‰ dy
1
œ "#
œ '0 ˆ2y 2y$Î# ‰ dy œ ’y# 4y5 “ œ 5"
1
"
&Î#
!
'c02 '2x4 cb x4 dy dx œ 'c02 ax# 2xb dx
#
5.
$
œ ’ x3 x# “
!
#
œ ˆ 38 4‰ œ 43
'04 'c(Èy c4 c4)/2y dx dy œ '04 ˆ y c2 4 È4 y‰ dy
4
2
œ ’ y2 2y 32 a4 yb3/2 “ œ 4 8 23 † 43/2
0
4
œ 4 16
3 œ 3
6.
'01 'yÈy Èx dx dy œ '01 23 x$Î# ‘ yÈy dy
œ 32 '0 ˆy$Î% y$Î# ‰ dy œ 32 47 y(Î% 52 y&Î# ‘ !
"
1
4
œ 23 ˆ 47 25 ‰ œ 35
'01 'xx Èx dy dx œ '01 x1/2 ax x2 b dx œ '01 ˆx3/2 x5/2 ‰ dx
2
1
4
œ 25 x5/2 27 x7/2 ‘0 œ 25 27 œ 35
7.
È
È
'c33 '0Ð1Î2Ñ 9 x y dy dx œ 'c33 ’ y2 “ Ð1Î2Ñ 9 x dx
#
#
#
!
œ'
$
"
x$
a9 x# b dx œ ’ 9x
8 24 “ $
c3 8
3
27 ‰
9
ˆ 27 27 ‰ 27
œ ˆ 27
8 24 8 24 œ 6 œ #
È
'03Î2 'È99 4y4y y dx dy œ '03Î2 2yÈ94y# dy
#
#
3/2
œ "4 † 23 a94y# b3/2 º
0
9
œ 6" † 93/2 œ 27
6 œ #
'02 '04 x 2x dy dx œ '02 c2xyd 04 x dx
2
2
œ '0 a2xa4 x2 bb dx œ '0 a8x 2x3 b dx
2
8.
2
4
2
œ ’4x2 x2 “ œ 16 16
2 œ 8
È4 c y
'0 '0
4
!
È4 c y
2x dx dy œ '0 cx2 d 0
4
dy
œ '0 a4 yb dy œ ’ 4y y2 “ œ 16 16
2 œ 8
4
2
4
0
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Chapter 15 Practice Exercises
9.
'01 '2y2 4 cos ax# b dx dy œ '02 '0x/2 4 cos ax# b dy dx œ '02 2x cos ax# b dx œ csin ax# bd #! œ sin 4
10. '0 'yÎ2 ex dx dy œ '0 '0 ex dy dx œ '0 2xex dx œ cex d ! œ e 1
2
1
1
#
2x
1
#
11. '0 'È$ x y% " 1 dy dx œ '0 '0
8
2
y$
2
#
#
"
'02 y 4y 1 dy œ ln417
$
"
dx dy œ "4
y% 1
%
12. '0 'È$ y 21 sinx#a1x b dx dy œ '0 '0 21 sinx#a1x b dy dx œ '0 21x sin a1x# b dx œ c cos a1x# bd ! œ (1) (1) œ 2
1
1
x$
1
#
4 c x#
1
#
"
13. A œ 'c2 '2x b 4 dy dx œ 'c2 ax# 2xb dx œ 43
0
2cx
15. V œ '0 'x
1
4
ax# y# b dy dx œ '0 ’x# y y3 “
1
$
Èy
14. A œ '1 '2 c y dx dy œ '1 ˆÈy 2 y‰ dy œ 37
6
0
2cx
x
4
dx œ '0 ’2x# (23x) 7x3 “ dx œ ’ 2x3 (212x) 7x
12 “
1
$
$
%
$
"
7 ‰
2%
œ ˆ 23 12
12
12
œ 43
6 c x#
16. V œ 'c3 'x
2
x# dy dx œ 'c3 cx# yd x
"
!
dx œ 'c3 a6x# x% x$ b dx œ 125
4
6 c x#
2
%
2
17. average value œ '0 '0 xy dy dx œ '0 ’ xy2 “ dx œ '0 x2 dx œ 4"
1
1
1
"
#
1
!
È1 c x
18. average value œ ˆ "1 ‰ '0 '0
1
xy dy dx œ 14 '0 ’ xy2 “
#
1
4
È1 c x
#
È1 c x
#
!
dx œ 12 '0 ax x$ b dx œ #"1
1
"
19. 'c1 'cÈ1 c x# a1 x#2 y# b2 dy dx œ '0 '0 a1 2rr# b# dr d) œ '0 1 " r# ‘ ! d) œ #" '0 d) œ 1
1
21
#
21
1
È1 c y
21
20. 'c1 'cÈ1 c y# ln ax# y# 1b dx dy œ '0 '0 r ln ar# 1b dr d) œ '0 '1
1
21
#
21
1
2
"
"
# ln u du d) œ #
'021 cu ln u ud #" d)
œ "# '0 (2 ln 2 1) d) œ [ln (4) 1] 1
21
Ècos 2)
21. ax# y# b ax# y# b œ 0 Ê r% r# cos 2) œ 0 Ê r# œ cos 2) so the integral is 'c1Î4 '0
1Î4
#
1Î4
œ 'c1Î4 ’ 2 a1 " r# b “
1Î4
Ècos 2)
!
d) œ "#
1Î4
"
" '
"
' 11ÎÎ44 ˆ1 1 cos
‰
ˆ1 # cos
‰
) d)
2) d) œ #
1Î4
#
œ "# 'c1Î4 Š1 sec# ) ‹ d) œ "# ) tan2 ) ‘ 1Î4 œ 14 2
22. (a) ' '
R
1Î4
#
"
dx dy œ
a1 x # y # b #
'01Î3 '0sec
)
r
dr d) œ
a1 r# b#
œ '0 ’ "# # a1 "sec# )b “ d) œ #" '0
1Î3
1Î3
œ "# ’ È"2 tan" Èu2 “
(b) ' '
R
È$
!
"
dx dy œ
a1 x # y # b #
œ '0
1Î2
œ
r
dr d) œ
a1 r# b#
"
"
lim ’ " # a1 b# b “d) œ #
bÄ_ #
)
1
#
u œ tan )
sec# )
1 sec# ) d); ” du œ sec# ) d) •
È2
" É 3
#
4 tan
'01Î2 '0_
'0 Î3 ’ 2 a1 " r b “ sec d)
'0
1Î2
'01Î2 lim
bÄ_
!
È3
Ä #" '0
du
2 u #
b
’ 2 a1 " r# b “ d)
0
d) œ 14
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
r
dr d)
a1 r# b#
929
930
Chapter 15 Multiple Integrals
23. '0 '0 '0 cos (x y z) dx dy dz œ '0 '0 [sin (z y 1) sin (z y)] dy dz
1
1
1
1
1
œ '0 [ cos (z 21) cos (z 1) cos z cos (z 1)] dz œ 0
1
24. 'ln 6 '0
ln 7
ln 2
'lnln45 eÐxyzÑ dz dy dx œ 'lnln67 '0ln 2 eÐxyÑ dy dx œ 'lnln67 ex dx œ 1
xby
25. '0 '0 '0
1
x#
8
(2x y z) dz dy dx œ '0 '0 Š 3x# 3y# ‹ dy dx œ '0 Š 3x# x# ‹ dx œ 35
x#
1
1
#
#
%
'
e
' ' "
'
26. '1 '1 '0 2y
z$ dy dz dx œ 1 1 z dz dx œ 1 ln x dx œ cx ln x xd 1 œ 1
e
x
z
e
27. V œ 2 '0
1Î2
x
e
' 0cos y '0 2x dz dx dy œ 2 '01Î2 ' 0cos y 2x dx dy œ 2 '01Î2 cos# y dy œ 2 ’ y2 sin42y “ 1Î2 œ 1#
!
È4cx
28. V œ 4 '0 '0
2
œ ’x a4 x# b
#
È
'04cx dz dy dx œ 4 '02 '0 4cx a4 x# b dy dx œ 4 '02 a4 x# b$Î# dx
#
$Î#
#
#
6xÈ4 x# 24 sin" x2 “ œ 24 sin" 1 œ 121
!
29. average œ "3 '0 '0 '0 30xzÈx# y dz dy dx œ 3" '0 '0 15xÈx# y dy dx œ 3" '0 '0 15xÈx# y dx dy
1
3
œ "3 '0 ’5 ax# yb
3
1
1
3
3
1
“ dy œ "3 '0 5(1 y)$Î# 5y$Î# ‘ dy œ "3 2(1 y)&Î# 2y&Î# ‘ ! œ "3 2(4)&Î# 2(3)&Î# 2‘
$Î# "
$
3
!
œ "3 2 ˆ31 3&Î# ‰‘
30. average œ 413a$ '0 '0 '0 3$ sin 9 d3 d9 d) œ 163a1 '0 '0 sin 9 d9 d) œ 83a1 '0 d) œ 3a
4
21
È2
È2cy
1
È4cx cy
31. (a) 'cÈ2 'cÈ2cy# 'Èx#by#
#
21
a
#
3 dz dx dy
'02 33# sin 9 d3 d9 d)
(c) '0 '0
'r
1Î4
È2 È4cr
21
#
21
#
(b) '0 '0
21
1
È2
3 dz r dr d) œ 3 '0 '0 ’r a4 r# b
21
"Î#
r# “ dr d) œ 3 '0 ’ "3 a4 r# b
21
$Î#
œ '0 ˆ2$Î# 2$Î# 4$Î# ‰ d) œ Š8 4È2‹'0 d) œ 21 Š8 4È2‹
21
21
1Î2
1
1Î2
1
1Î2
$
r3 “
È#
!
d)
32. (a) 'c1Î2 '0 ' r# 21(r cos ))(r sin ))# dz r dr d) œ ' 1Î2 '0 ' r# 21r$ cos ) sin# ) dz r dr d)
r#
1Î2
(b) 'c1Î2 '0 ' r# 21r$ cos ) sin# ) dz r dr d) œ 84 '0
r#
1
r#
'01 r' sin# ) cos ) dr d) œ 12'01Î2 sin# ) cos ) d) œ 4
33. (a) '0 '0
'0sec 9 3# sin 9 d3 d9 d)
21
sec 9
21
21
21
1Î4
1Î4
1Î4
(b) '0 '0 '0 3# sin 9 d3 d9 d) œ 3" '0 '0 (sec 9)(sec 9 tan 9) d9 d) œ 3" '0 2" tan# 9‘ ! d) œ 6" '0 d) œ 13
21
1Î4
È1cx
È
1
r
1Î2
'0 x y (6 4y) dz dy dx
(b) '0 '0 '0 (6 4r sin )) dz r dr d)
csc 9
1Î2
1Î2
(c) '0 '1Î4 '0
(6 43 sin 9 sin )) a3# sin 9b d3 d9 d)
34. (a) '0 '0
1
#
#
#
(d) '0
'01 '0r (6 4r sin )) dz r dr d) œ '01Î2 '01 a6r# 4r$ sin )b dr d) œ '01Î2 c2r$ r% sin )d "! d)
1Î2
1Î#
œ '0 (2 sin )) d) œ c2) cos )d ! œ 1 1
1Î2
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Chapter 15 Practice Exercises
È3cx
1
È3 È3cx
È4cx cy
35. '0 'È1cx# '1
#
#
z# yx dz dy dx '1
#
'0
#
È
'1 4cx cy z# yx dz dy dx
#
#
36. (a) Bounded on the top and bottom by the sphere x# y# z# œ 4, on the right by the right circular
cylinder (x 1)# y# œ 1, on the left by the plane y œ 0
(b) '0
1Î2
È
'02 cos 'cÈ44ccrr dz r dr d)
)
#
#
È8cr
37. (a) V œ '0 '0 '2
21
œ'
2
#
dz r dr d) œ '0 '0 ŠrÈ8 r# 2r‹ dr d) œ '0 ’ "3 a8 r# b
21
21
"3 (4)$Î# 4 3" (8)$Î# ‘ d) œ
0
(b) V œ '0 '0
21
1Î4
21
2
È
'0
21
$Î#
1Î4
21
21
21
1Î3
21
È2
1Î%
!
d)
81 Š4È2 5‹
‹ d) œ
3
'02 (3 sin 9)# a3# sin 9b d3 d9 d) œ '021 '01Î3 '02 3% sin$ 9 d3 d9 d)
'0 '0 asin 9 cos# 9 sin 9b d9 d) œ 325 '0 ’ cos 9 cos3 9 “
œ 32
5
21
È
'0 d) œ 81 Š4 3 2 5‹
21
'2 sec8 9 3# sin 9 d3 d9 d) œ 83 '021 '01Î4 Š2È2 sin 9 sec$ 9 sin 9‹ d9 d)
œ 83 '0 Š2 "# 2È2‹ d) œ 83 '0 Š 5 #4
38. Iz œ '0 '0
!
4
4
È
È
3 Š2 3 2 8‹ d) œ 3 Š4 2 5‹
œ 83 '0 '0 Š2È2 sin 9 tan 9 sec# 9‹ d9 d) œ 83 '0 ’2È2 cos 9 "# tan# 9“
21
#
r # “ d)
1Î3
21
$
1Î$
!
d) œ 831
39. With the centers of the spheres at the origin, Iz œ '0 '0 'a $ (3 sin 9)# a3# sin 9b d3 d9 d)
21
1
b
œ $ ab 5 a b '0 '0 sin$ 9 d9 d) œ $ ab 5 a b '0 '0 asin 9 cos# 9 sin 9b d9 d)
&
21
&
1
&
21
&
1
œ $ ab 5 a b '0 ’ cos 9 cos3 9 “ d) œ 4$ ab15 a b '0 d) œ 81$ ab15 a b
&
21
&
$
1
&
&
21
&
&
!
1ccos )
40. Iz œ '0 '0 '0
21
1
1ccos )
(3 sin 9)# a3# sin 9b d3 d9 d) œ '0 '0 '0
21
1
3% sin$ 9 d3 d9 d)
œ "5 '0 '0 (1 cos 9)& sin$ 9 d9 d) œ '0 '0 (1 cos 9)' (1 cos 9) sin 9 d9 d);
21
1
21
1
#
21
2
21
21
u œ 1 cos 9
"
"
"
"
"
2u(
u)
'
)
” du œ sin 9 d9 • Ä 5 '0 '0 u (2 u) du d) œ 5 '0 ’ 7 8 “ d) œ 5 '0 ˆ 7 8 ‰ 2 d)
!
œ "5
'0
21 $ &
2 †2
56
d) œ 32
35
'0
21
d) œ 64351
41. M œ '1 '2Îx dy dx œ '1 ˆ2 2x ‰ dx œ 2 ln 4; My œ '1 '2Îx x dy dx œ '1 x ˆ2 2x ‰ dx œ 1;
2
2
2
2
2
2
Mx œ '1 '2Îx y dy dx œ '1 ˆ2 x2# ‰ dx œ 1 Ê x œ y œ # "ln 4
2
2
2
2y c y#
2y c y#
4y
y
64
' '
' # $
42. M œ '0 'c2y dx dy œ '0 a4y y# b dy œ 32
3 ; Mx œ 0 c2y y dx dy œ 0 a4y y b dy œ ’ 3 4 “ œ 3 ;
4
4
2y c y#
4
My œ '0 'c2y x dx dy œ '
4
4
#
’ a2y#y b
0
#
% %
y&
2y# “ dy œ ’ 10
y2 “ œ 128
5
!
4
$
4
2
!
Ê
$
44. (a) Io œ 'c2 'c1 ax# y# b dy dx œ 'c2 ˆ2x# 23 ‰ dx œ 40
3
2
1
%
Mx
y
12
xœ M
M œ 5 and y œ M œ 2
14x
43. Io œ '0 '2x ax# y# b (3) dy dx œ 3 '0 Š4x# 64
3 3 ‹ dx œ 104
2
%
2
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
931
932
Chapter 15 Multiple Integrals
4a b
' ' #
' 2a
(b) Ix œ 'ca 'cb y# dy dx œ 'ca 2b3 dx œ 4ab
Ê Io œ Ix Iy
3 ; Iy œ cb ca x dx dy œ cb 3 dy œ 3
a
b
a
#
$
b
$
a
b
$
$
#
4ab ab a b
4a b
œ 4ab
3 3 œ
3
$
$
45. M œ $ '0 '0
2xÎ3
3
' '
dy dx œ $ '0 2x
3 dx œ 3$ ; Ix œ $ 0 0
3
2xÎ3
3
8$ '
8$ ‰ 3
y# dy dx œ 81
x$ dx œ ˆ 81
Š 4 ‹ œ 2$
0
3
%
"3
46. M œ '0 'x# (x 1) dy dx œ '0 ax x$ b dx œ "4 ; Mx œ '0 'x# y(x 1) dy dx œ #" '0 ax$ x& x# x% b dx œ 120
;
1
x
1
1
x
1
2
8
13
My œ '0 'x# x(x 1) dy dx œ '0 ax# x% b dx œ 15
Ê x œ 15
and y œ 30
; Ix œ '0 'x# y# (x 1) dy dx
1
x
1
1
x
Ix
17
17
1
œ "3 '0 ax% x( x$ x' b dx œ 280
Ê Rx œ É M
œ É 70
; Iy œ '0 'x# x# (x 1) dy dx œ '0 ax$ x& b dx œ 12
1
1
x
1
47. M œ 'c1 'c1 ˆx# y# 3" ‰ dy dx œ 'c1 ˆ2x# 34 ‰ dx œ 4; Mx œ 'c1 'c1 y ˆx# y# "3 ‰ dy dx œ 'c1 0 dx œ 0;
1
1
1
1
1
1
My œ 'c1 'c1 x ˆx# y# "3 ‰ dy dx œ 'c1 ˆ2x$ 34 x‰ dx œ 0
1
1
1
48. Place the ?ABC with its vertices at A(0ß 0), B(bß 0) and C(aß h). The line through the points A and C is
Ða
y œ ha x; the line through the points C and B is y œ a h b (x b). Thus, M œ '0 'ayÎh
h
Ða bÑyÎh b
' '
œ b$ '0 ˆ1 yh ‰ dy œ $bh
# ; Ix œ 0 ayÎh
h
h
1Î3
1Î3
bÑyÎh b
y# $ dx dy œ b$ '0 Šy# yh ‹ dy œ $1bh#
h
1Î3
$
$ dx dy
$
1Î3
49. M œ ' 1Î3 '0 r dr d) œ 9# ' 1Î3 d) œ 31; My œ ' 1Î3 '0 r# cos ) dr d) œ 9 ' 1Î3 cos ) d) œ 9È3 Ê x œ 3 1 3 ,
3
3
È
and y œ 0 by symmetry
50. M œ '0
1Î2
13
'13 r dr d) œ 4 '01Î2 d) œ 21; My œ '01Î2 '13 r# cos ) dr d) œ 263 '01Î2 cos ) d) œ 263 Ê x œ 31
, and
13
y œ 31
by symmetry
51. (a) M œ 2 '0
1Î2
'11bcos r dr d)
)
(b)
2) ‰
œ '0 ˆ2 cos ) 1 cos
d) œ 8 4 1 ;
#
1Î2
1Î2
My œ 'c1Î2 '1
1 cos )
(r cos )) r dr d)
œ 'c1Î2 Šcos# ) cos$ ) cos3 ) ‹ d)
1Î2
%
1 32
œ 32 24151 Ê x œ 15
61 48 , and
y œ 0 by symmetry
)
52. (a) M œ 'c! '0 r dr d) œ 'c! a# d) œ a# !; My œ 'c! '0 (r cos )) r dr d) œ 'c! a cos
d) œ 2a 3sin !
3
!
!
a
!
#
!
a
$
!
2a sin !
Ê x œ 2a 3sin
œ0
! , and y œ 0 by symmetry; lim c x œ lim c
3!
2a
(b) x œ 51
and y œ 0
!Ä1
!Ä1
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
$
Chapter 15 Additional and Advanced Exercises
53. x œ u y and y œ v Ê x œ u v and y œ v
" "
Ê J(uß v) œ º
œ 1; the boundary of the
0 "º
image G is obtained from the boundary of R as
follows:
xy-equations for
Corresponding uv-equations
Simplified
the boundary of R
for the boundary of G
uv-equations
yœx
yœ0
vœuv
uœ0
vœ0
_
_ _ sÐuvÑ
Ê '0 '0 esx f(x yß y) dy dx œ '0 '0 e
x
vœ0
f(uß v) du dv
# s !t
$s "t
54. If s œ !x " y and t œ # x $ y where (!$ "# )# œ ac b# , then x œ !$
"# , y œ !$ "# ,
and J(sß t) œ (!$ " "#)# º
$
#
_ _
#
#
"
œ "
Ê 'c_ ' _ e as t b È " # ds dt
ac b
! º !$ "#
'021 '0_ rer dr d) œ È "
#
œÈ "
ac b#
#
ac b#
'021 d) œ È 1
ac b#
. Therefore, È 1
ac b#
œ 1 Ê ac b# œ 1# .
CHAPTER 15 ADDITIONAL AND ADVANCED EXERCISES
6cx#
1. (a) V œ 'c3 'x
2
6cx#
(c) V œ 'c3 'x
2
6cx#
(b) V œ 'c3 'x
2
x# dy dx
6cx#
x# dy dx œ 'c3 'x
2
'0x dz dy dx
#
&
%
a6x# x% x$ b dx œ ’2x$ x5 x4 “
#
$
œ 125
4
2. Place the sphere's center at the origin with the surface of the water at z œ 3. Then
9 œ 25 x# y# Ê x# y# œ 16 is the projection of the volume of water onto the xy-plane
c3
Ê V œ '0 '0 'cÈ25cr# dz r dr d) œ '0 '0 ŠrÈ25 r# 3r‹ dr d) œ '0 ’ "3 a25 r# b
21
4
21
21
4
$Î#
%
3# r# “ d)
!
21
21
521
œ '0 "3 (9)$Î# 24 3" (25)$Î# ‘ d) œ '0 26
3 d) œ 3
2crÐcos )
3. Using cylindrical coordinates, V œ '0 '0 '0
21
1
sin )Ñ
dz r dr d) œ '0 '0 a2r r# cos ) r# sin )b dr d)
21
1
œ '0 ˆ1 "3 cos ) "3 sin )‰ d) œ ) "3 sin ) 3" cos )‘ ! œ 21
21
#1
4. V œ 4 '0
1Î2
œ 4 '0
1Î2
È
'01 'r 2 r dz r dr d) œ 4 '01Î2 '01 ŠrÈ2 r# r$ ‹ dr d) œ 4'01Î2 ’ "3 a2 r# b$Î# r4 “ " d)
#
%
#
!
'
1Î2
È
È
Š "3 "4 2 3 2 ‹ d) œ Š 8 327 ‹ 0
d) œ
1 Š8È2 7‹
6
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
933
934
Chapter 15 Multiple Integrals
5. The surfaces intersect when 3 x# y# œ 2x# 2y# Ê x# y# œ 1. Thus the volume is
È1 c x
V œ 4 '0 '0
1
6. V œ 8 '0
1Î2
#
'2x3cx2ycy dz dy dx œ 4 '01Î2 '01 '2r3 r dz r dr d) œ 4 '01Î2 '01 a3r 3r$ b dr d) œ 3'01Î2 d) œ 31#
#
#
#
#
#
#
'01Î2 '02 sin 9 3# sin 9 d3 d9 d) œ 643 '01Î2 '01Î2 sin% 9 d9 d)
'0 ” sin 94cos 9 ¹
œ 64
3
1Î2
$
1Î#
!
43 '0 sin# 9 d9• d) œ 16 '0 92 sin429 ‘ !
1Î2
1Î2
1Î#
d) œ 41 '0 d) œ 21#
1Î2
7. (a) The radius of the hole is 1, and the
radius of the sphere is 2.
È3 È4cz
(b) V œ 2 '0 '0
21
8. V œ '0 '0
1
3 sin )
'1
È
#
È3
r dr dz d) œ '0 '0
21
a3 z# b dz d) œ 2È3 '0 d) œ 4È31
21
'0 9cr dz r dr d) œ '0 '03 sin rÈ9 r# dr d) œ '0 ’ "3 a9 r# b$Î# “ 3 sin d)
#
1
)
)
1
!
1
1
1
$Î#
$Î#
œ '0 ’ "3 a9 9 sin# )b 3" (9)$Î# “ d) œ 9'0 ’1 a1 sin# )b “ d) œ 9'0 a1 cos$ )b d)
œ '0 a1 cos ) sin# ) cos )b d) œ 9 ’) sin ) sin3 ) “ œ 91
1
$
1
!
#
#
9. The surfaces intersect when x# y# œ x y# 1 Ê x# y# œ 1. Thus the volume in cylindrical coordinates is
V œ 4 '0
1Î2
10. V œ '0
1Î2
'01 'r r 1 Î2 dz r dr d) œ 4 '01Î2 '01 Š #r r# ‹ dr d) œ 4'01Î2 ’ r4 r8 “ " d) œ "# '01Î2 d) œ 14
ˆ#
‰
$
%
!
'12 '0r sin cos dz r dr d) œ '0 Î2 '12 r$ sin ) cos ) dr d) œ '0 Î2 ’ r4 “ # sin ) cos ) d)
#
)
)
1
'0 sin ) cos ) d) œ 154 ’ sin2 ) “
œ 15
4
1Î2
#
_ ecax ecbx
11. '0
#
#
_
1Î#
!
1
"
œ 15
8
_
dx œ '0 'a exy dy dx œ 'a '0 exy dx dy œ 'a Š lim
b
b
b
tÄ_
x
%
'0t exy dx‹ dy
œ 'a lim ’ e y “ dy œ 'a lim Š "y e y ‹ dy œ 'a "y dy œ cln yd ab œ ln ˆ ba ‰
cxy
b
tÄ_
t
cyt
b
b
tÄ_
!
12. (a) The region of integration is sketched at the right
Ê '0
a sin "
È
'y cota "c y ln ax# y# b dx dy
#
#
œ '0 '0 r ln ar# b dr d);
"
a
a#
"
u œ r#
"
” du œ 2r dr • Ä # '0 '0 ln u du d)
œ "# '0 [u ln u u] !a d)
"
#
œ "# '0 ’2a# ln a a# lim t ln t“ d) œ a# '0 (2 ln a 1) d) œ a# " ˆln a "# ‰
"
"
#
tÄ0
È
a cos "
'0(tan ")x ln ax# y# b dy dx 'aacos " '0 a cx ln ax# y# b dy dx
(b) '0
#
#
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Chapter 15 Additional and Advanced Exercises
13. '0 '0 emÐxtÑ f(t) dt du œ '0 't emÐxtÑ f(t) du dt œ '0 (x t)emÐxtÑ f(t) dt; also
x
u
x
x
x
'0x '0v '0u emÐxtÑ f(t) dt du dv œ '0x 't x 't v emÐxtÑ f(t) du dv dt œ '0x 't x (v t)emÐxtÑ f(t) dv dt
x
x
x
œ '0 "2 (v t)# emÐxtÑ f(t)‘ t dt œ '0 (x # t) emÐxtÑ f(t) dt
#
14. '0 f(x) Š'0 g(xy)f(y) dy‹ dx œ '0 '0 g(xy)f(x)f(y) dy dx
1
x
1
x
œ '0 'y g(xy)f(x)f(y) dx dy œ '0 f(y) Œ'y g(xy)f(x) dx dy;
1
1
1
1
'01 '01 g akxykb f(x)f(y) dx dy œ '01 '0x g(xy)f(x)f(y) dy dx '01 'x1 g(yx)f(x)f(y) dy dx
1
1
1
1
œ '0 'y g(xy)f(x)f(y) dx dy '0 'x g(yx)f(x)f(y) dy dx
œ '0 'y g(xy)f(x)f(y) dx dy '0 'y g(xy)f(y)f(x) dx dy
ðóóóóóóóóóóóóñóóóóóóóóóóóóò
1
1
1
1
simply interchange x and y
variable names
œ 2'0 'y g(xy)f(x)f(y) dx dy, and the statement now follows.
1
1
15. Io (a) œ '0 '0
xÎa#
a
ax# y# b dy dx œ '0 ’x# y y3 “
a
$
#
œ a4 1"# a# ; Iow (a) œ #" a "6 a$ œ 0
%
Ê a
xÎa#
!
œ "3
x
x
x
dx œ '0 Š xa# 3a
' ‹ dx œ ’ 4a# 12a' “
a
Ê aœ
$
$
%
%
"
É "3 œ È
.
%
3
%
a
!
Since Io (a) œ "# #" a% 0, the
ww
value of a does provide a minimum for the polar moment of inertia Io (a).
64
16. Io œ '0 '2x ax# y# b (3) dy dx œ 3 '0 Š4x# 14x
3 3 ‹ dx œ 104
2
4
2
$
17. M œ 'c) 'b sec ) r dr d) œ 'c) Š a# b# sec# )‹ d)
)
)
a
#
#
œ a# ) b# tan ) œ a# cos" ˆ ba ‰ b# Š
È a# b#
‹
b
œ a# cos" ˆ ba ‰ bÈa# b# ; Io œ 'c) 'b sec ) r$ dr d)
)
a
œ "4 'c) aa% b% sec% )b d)
)
œ "4 'c) ca% b% a1 tan# )b asec# )bd d)
)
%
$
)
œ "4 ’a% ) b% tan ) b tan
“
3
%
%
%
$
)
)
)
)
œ a#) b tan
b tan
#
6
œ "# a% cos" ˆ ba ‰ "# b$ Èa# b# 6" b$ aa# b# b
2 ˆy# Î2‰
$Î#
y
18. M œ 'c2 '1cay#Î4b dx dy œ ' 2 Š1 y4 ‹ dy œ ’y 12
“
2
œ'
2
2c ˆy# Î2‰
#
’x “
c2 2 1
ay#
Î4b
2
dy œ '
#
$
2
3
3
a4 y# b dy œ 32
c2 32
#
#
2 ˆy# Î2‰
œ 38 ; My œ ' 2 '1 ay#Î4b x dx dy
2
'c2 a16 8y# y% b dy œ 163 ’16y 8y3 y5 “ #
2
$
&
!
My
3 ˆ
32 ‰
ˆ 3 ‰ ˆ 3215†8 ‰ œ 48
ˆ 48 ‰ ˆ 83 ‰ œ 56 , and y œ 0 by symmetry
œ 16
32 64
3 5 œ 16
15 Ê x œ M œ 15
19. '0 '0 emax ab x ßa y b dy dx œ '0 '0
a
b
# #
# #
a
bxÎa
eb x dy dx '0 '0
# #
b
ayÎb
# #
ea y dx dy
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
935
936
Chapter 15 Multiple Integrals
"
"
œ '0 ˆ ba x‰ eb x dx '0 ˆ ba y‰ ea y dy œ ’ 2ab
eb x “ ’ 2ba
ea y “ œ #"ab Šeb a 1‹ #"ab Šea b 1‹
a
b
# #
# #
# #
a
b
# #
!
# #
# #
!
# #
"
œ ab
Šea b 1‹
ßy)
' ` F(xßy)
' ` F(x"ßy) ` F(x` y!ßy) “ dx œ cF(x" ß y) F(x! ß y)d yy!"
20. 'y 'x ``F(x
x ` y dx dy œ y ’ ` y “ dy œ y ’ ` y
y"
x"
!
!
#
y"
x"
y"
!
x!
!
œ F(x" ß y" ) F(x! ß y" ) F(x" ß y! ) F(x! ß y! )
21. (a) (i)
(ii)
(iii)
(iv)
(b) '0
ln 2
Fubini's Theorem
Treating G(y) as a constant
Algebraic rearrangement
The definite integral is a constant number
'01Î2 ex cos y dy dx œ Œ'0ln 2 ex dx Œ'01Î2 cos y dy œ aeln 2 e0 b ˆsin 1# sin 0‰ œ (1)(1) œ 1
(c) '1 'c1 yx# dx dy œ Œ'1 y"# dy Œ'c1 x dx œ ’ y" “ ’ x2 “
2
1
2
#
1
#
"
"
"
œ ˆ "# 1‰ ˆ "# "# ‰ œ 0
22. (a) ™ f œ xi yj Ê Du f œ u" x u# y; the area of the region of integration is "#
1cx
Ê average œ 2'0 '0
1
(u" x u# y) dy dx œ 2 '0 u" x(1 x) "# u# (1 x)# ‘ dx
1
"
$
œ 2 ’u" Š x2 x3 ‹ ˆ "# u# ‰ (13x) “ œ 2 ˆ 6" u" 6" u# ‰ œ 3" (u" u# )
#
$
!
(b)
"
average œ area
_ _
23. (a) I# œ '0 '0
œ "# '0
1Î2
_
M
u" ' '
u# ' '
' ' (u" x u# y) dA œ area
x dA area
y dA œ u" Š My ‹ u# ˆ MMx ‰ œ u" x u# y
R
R
1Î2
e ˆx y ‰ dx dy œ '0
#
lim
bÄ_
#
R
'0_ ae r b r dr d) œ '01Î2 ” lim '0b re r dr• d)
#
#
bÄ_
aecb 1b d) œ "# '0 d) œ 14 Ê I œ
1Î2
#
_
(b) > ˆ "# ‰ œ '0 t"Î# et dt œ '0 ay# b
"Î# y#
e
_
È1
#
(2y) dy œ 2 '0 ey dy œ 2 Š # ‹ œ È1, where y œ Èt
È1
#
24. Q œ '0 '0 kr# (1 sin )) dr d) œ kR3 '0 (1 sin )) d) œ kR3 c) cos )d #!1 œ 21kR
3
21
R
$
21
$
Èh
Èh c x
$
25. For a height h in the bowl the volume of water is V œ 'cÈh 'cÈh c x# 'x#by# dz dy dx
Èh
Èh c x
Èh
#
h
œ 'cÈh 'cÈh c x# ah x# y# b dy dx œ '0 '0 ah r# b r dr d) œ '0 ’ hr2 r4 “
21
#
21
#
%
Èh
!
d) œ '0
21
h#
h# 1
4 d) œ # .
Since the top of the bowl has area 101, then we calibrate the bowl by comparing it to a right circular cylinder
#
whose cross sectional area is 101 from z œ 0 to z œ 10. If such a cylinder contains h#1 cubic inches of water
#
#
to a depth w then we have 101w œ h 1 Ê w œ h . So for 1 inch of rain, w œ 1 and h œ È20; for 3 inches of
#
20
rain, w œ 3 and h œ È60.
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Chapter 15 Additional and Advanced Exercises
26. (a) An equation for the satellite dish in standard position
is z œ "# x# "# y# . Since the axis is tilted 30°, a unit
vector v œ 0i aj bk normal to the plane of the
È3
#
È
"
v œ # j #3 k
water level satisfies b œ v † k œ cos ˆ 16 ‰ œ
Ê a œ È1 b# œ "# Ê
È
Ê "# (y 1) #3 ˆz "# ‰ œ 0
Ê z œ È"3 y Š "# È"3 ‹
is an equation of the plane of the water level. Therefore
È3 yb 2 c È3
the volume of water is V œ ' ' ' 1 x#b 1 y#
1
1
2
R
1
dz dy dx, where R is the interior of the ellipse
2
È 3 Ê 3 4 Š È 3 1‹
2
x# y# È23 y 1 È23 œ 0. When x œ 0, then y œ ! or y œ " , where ! œ
È
2
and " œ
È
Ê 43 4 Š 2 1‹
3
3
b1c 3 cy ‹
yb c È
' 'c b cÈ c ' È b
Ê Vœ
#
"Î#
"
Š
2
3 y
2
!
Š
È3 y# ‹
2
3 y
#
3
"Î#
2
1
1
1 #
2 x
1
2
4
2
#
1
3
1 dz dx dy
1 #
2 y
dz
dz
(b) x œ 0 Ê z œ "# y# and dy
œ y; y œ 1 Ê dy
œ 1 Ê the tangent line has slope 1 or a 45° slant
Ê at 45° and thereafter, the dish will not hold water.
27. The cylinder is given by x# y# œ 1 from z œ 1 to _ Ê ' ' ' z ar# z# b
_
œ '0 '0 '1
21
1
'021 '01 '1a
z
dz r dr d) œ a lim
Ä_
ar# z# b&Î#
' '0 ’ˆ "3 ‰
œ a lim
Ä_ 0
21
dV
D
rz
dz dr d)
ar# z# b&Î#
' 21 ' 1
a
1
&Î#
r
“ dr d) œ a lim
’ˆ 3" ‰ # r # $Î# ˆ "3 ‰ # r $Î# “ dr d)
Ä_ 0
0
ar# z# b$Î# 1
ar a b
ar 1 b
"
21
21
"Î#
"Î#
"Î#
"Î#
œ a lim
3" ar# 1b
3" ˆ2"Î# ‰ 3" aa# b
3" “ d)
’ 3" ar# a# b
“ d) œ a lim
’ 3" a1 a# b
Ä_ 0
Ä
_
0
!
È
È
"Î#
œ a lim
21 ’ 3" a1 a# b
3" Š #2 ‹ 3" ˆ "a ‰ 3" “ œ 21 ’ 3" ˆ 3" ‰ #2 “ .
Ä_
'
'
28. Let's see?
The length of the "unit" line segment is: L œ 2'0 dx œ 2.
1
È1 c x
The area of the unit circle is: A œ 4'0 '0
1
2
dy dx œ 1.
È1 c x
The volume of the unit sphere is: V œ 8'0 '0
1
2
È
'0 1 c x c y dz dy dx œ 43 1.
2
2
Therefore, the hypervolume of the unit 4-sphere should be:
È1cx
Vhyper œ 16'0 '0
1
2
È
È
'0 1cx cy '0 1cx cy cz dw dz dy dx.
2
2
2
2
2
Mathematica is able to handle this integral, but we'll use the brute force approach.
È1cx
Vhyper œ 16'0 '0
1
È
È
2
È
È
È
2
2
2
2
2
1
1cx
'0 1cx cy È1 x2 y2 É1 1 c xz c y dz dy dx œ –
œ 16'0 '0
2
2
È1cx
1
È1cx
œ 16'0 '0
1
2
2
2
2
2
2
2
z
È1 x2 y2 œ cos )
2
2
œ 16'0 '0
È
'0 1cx cy '0 1cx cy cz dw dz dy dx œ 16'01 '0 1cx '0 1cx cy È1 x2 y2 z2 dz dy dx
dz œ È1 x2 y2 sin ) d)
È1cx
a1 x2 y2 b'1/2 È1 cos2 ) sin ) d) dy dx œ 16'0 '0
0
1
2
—
a1 x2 y2 b'1/2 sin2 ) d) dy dx
0
'1
3/2
1
2
2
2È
È
2
1 x2 3" a1 x2 b ‹ dx
4 a1 x y b dy dx œ 41 0 Š 1 x x
œ 41'0 È1 x2 a1 x2 b 1 $ x ‘ dx œ 83 1'0 a1 x2 b
1
1
3
3/2
dx œ ”
0
x œ cos )
œ 83 1'1/2 sin4 ) d)
dx œ sin ) d) •
2) ‰
œ 83 1'1/2 ˆ 1 cos
d) œ 23 1'1/2 a1 2 cos 2) cos2 2)bd) œ 23 1'1/2 ˆ #3 2 cos 2) cos2 4) ‰d) œ 12
2
0
2
0
0
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
2
937
938
Chapter 15 Multiple Integrals
NOTES:
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
CHAPTER 16 INTEGRATION IN VECTOR FIELDS
16.1 LINE INTEGRALS
1. r œ ti a" tbj Ê x œ t and y œ 1 t Ê y œ 1 x Ê (c)
2. r œ i j tk Ê x œ 1, y œ 1, and z œ t Ê (e)
3. r œ a2 cos tbi a2 sin tbj Ê x œ 2 cos t and y œ 2 sin t Ê x# y# œ 4 Ê (g)
4. r œ ti Ê x œ t, y œ 0, and z œ 0 Ê (a)
5. r œ ti tj tk Ê x œ t, y œ t, and z œ t Ê (d)
6. r œ tj a2 2tbk Ê y œ t and z œ 2 2t Ê z œ 2 2y Ê (b)
#
7. r œ at# 1b j 2tk Ê y œ t# 1 and z œ 2t Ê y œ z4 1 Ê (f)
8. r œ a2 cos tbi a2 sin tbk Ê x œ 2 cos t and z œ 2 sin t Ê x# z# œ 4 Ê (h)
9. ratb œ ti a1 tbj , 0 Ÿ t Ÿ 1 Ê ddtr œ i j Ê ¸ ddtr ¸ œ È2 j ; x œ t and y œ 1 t Ê x y œ t (" t) œ 1
Ê 'C faxß yß zb ds œ '0 fatß 1 tß 0b ¸ ddtr ¸ dt œ '0 (1) ŠÈ2‹ dt œ ’È2 t“ œ È2
1
"
1
!
10. r(t) œ ti (1 t)j k , 0 Ÿ t Ÿ 1 Ê ddtr œ i j Ê ¸ ddtr ¸ œ È2; x œ t, y œ 1 t, and z œ 1 Ê x y z 2
œ t (1 t) 1 2 œ 2t 2 Ê 'C f(xß yß z) ds œ '0 (2t 2) È2 dt œ È2 ct# 2td ! œ È2
1
"
11. r(t) œ 2ti tj (2 2t)k , 0 Ÿ t Ÿ 1 Ê ddtr œ 2i j 2k Ê ¸ ddtr ¸ œ È4 1 4 œ 3; xy y z
œ (2t)t t (2 2t) Ê 'C f(xß yß z) ds œ '0 a2t# t 2b 3 dt œ 3 23 t$ "# t# 2t‘ ! œ 3 ˆ 23 "# 2‰ œ 13
#
1
"
12. r(t) œ (4 cos t)i (4 sin t)j 3tk , 21 Ÿ t Ÿ 21 Ê ddtr œ (4 sin t)i (4 cos t)j 3k
Ê ¸ ddtr ¸ œ È16 sin# t 16 cos# t 9 œ 5; Èx# y# œ È16 cos# t 16 sin# t œ 4 Ê 'C f(xß yß z) ds œ 'c21 (4)(5) dt
21
1
œ c20td ##
1 œ 801
13. r(t) œ (i 2j 3k) t(i 3j 2k) œ (1 t)i (2 3t)j (3 2t)k , 0 Ÿ t Ÿ 1 Ê ddtr œ i 3j 2k
Ê ¸ ddtr ¸ œ È1 9 4 œ È14 ; x y z œ (1 t) (2 3t) (3 2t) œ 6 6t Ê 'C f(xß yß z) ds
œ '0 (6 6t) È14 dt œ 6È14 ’t t2 “ œ Š6È14‹ ˆ "# ‰ œ 3È14
1
#
"
!
È
È
È
14. r(t) œ ti tj tk , 1 Ÿ t Ÿ _ Ê ddtr œ i j k Ê ¸ ddtr ¸ œ È3 ; x# y#3 z# œ t# t#3 t# œ 3t#3
_ È
_
Ê 'C f(xß yß z) ds œ '1 Š 3t#3 ‹ È3 dt œ 1t ‘ " œ lim ˆ b" 1‰ œ 1
bÄ_
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
940
Chapter 16 Integration in Vector Fields
15. C" : r(t) œ ti t# j , 0 Ÿ t Ÿ 1 Ê ddtr œ i 2tj Ê ¸ ddtr ¸ œ È1 4t# ; x Èy z# œ t Èt# 0 œ t ktk œ 2t
$Î#
0 Ê 'C f(xß yß z) ds œ '0 2tÈ1 4t# dt œ ’ "6 a" 4t# b “ œ 6" (5)$Î# "6 œ "6 Š5È5 1‹ ;
"
1
since t
!
"
C# : r(t) œ i j tk, 0 Ÿ t Ÿ 1 Ê
Ê ¸ ddtr ¸ œ 1; x Èy z# œ 1 È1 t# œ 2 t#
dr
dt œ k
"
Ê 'C f(xß yß z) ds œ '0 a2 t# b (1) dt œ 2t "3 t$ ‘ ! œ 2 3" œ 53 ; therefore 'C f(xß yß z) ds
1
#
œ 'C f(xß yß z) ds 'C f(xß yß z) ds œ 56 È5 3#
"
#
16. C" : r(t) œ tk , 0 Ÿ t Ÿ 1 Ê ddtr œ k Ê ¸ ddtr ¸ œ 1; x Èy z# œ 0 È0 t# œ t#
Ê 'C f(xß yß z) ds œ '0 at# b (1) dt œ ’ t3 “ œ 3" ;
1
$
"
!
"
C# : r(t) œ tj k, 0 Ÿ t Ÿ 1 Ê ddtr œ j Ê ¸ ddtr ¸ œ 1; x Èy z# œ 0 Èt 1 œ Èt 1
Ê 'C f(xß yß z) ds œ '0 ˆÈt 1‰ (1) dt œ 23 t$Î# t‘ ! œ 23 1 œ 3" ;
"
1
#
C$ : r(t) œ ti j k , 0 Ÿ t Ÿ 1 Ê ddtr œ i Ê ¸ ddtr ¸ œ 1; x Èy z# œ t È1 1 œ t
Ê 'C f(xß yß z) ds œ '0 (t)(1) dt œ ’ t2 “ œ "# Ê 'C f(xß yß z) ds œ 'C f ds 'C f ds 'C f ds œ 3" ˆ 3" ‰ "#
1
#
"
!
$
œ "6
"
#
$
17. r(t) œ ti tj tk , 0 a Ÿ t Ÿ b Ê ddtr œ i j k Ê ¸ ddtr ¸ œ È3 ; x#xyy#zz# œ t#ttt#tt# œ 1t
Ê 'C f(xß yß z) ds œ 'a ˆ 1t ‰ È3 dt œ ’È3 ln ktk “ œ È3 ln ˆ ba ‰ , since 0 a Ÿ b
b
b
a
18. r(t) œ aa cos tb j aa sin tb k , 0 Ÿ t Ÿ 21 Ê ddtr œ (a sin t) j (a cos t) k Ê ¸ ddtr ¸ œ Èa# sin# t a# cos# t œ kak ;
kak sin t, 0 Ÿ t Ÿ 1
Èx# z# œ È0 a# sin# t œ œ
Ê 'C f(xß yß z) ds œ '0 kak# sin t dt '1 kak# sin t dt
kak sin t, 1 Ÿ t Ÿ 21
1
1
21
#1
œ ca# cos td ! ca# cos td 1 œ ca# (1) a# d ca# a# (1)d œ 4a#
19. (a) ratb œ ti "# tj , 0 Ÿ t Ÿ 4 Ê ddtr œ i "# j Ê ¸ ddtr ¸ œ
dr
dt œ i 2tj
(b) ratb œ ti t j , 0 Ÿ t Ÿ 2 Ê
2
3Î2 2
1
œ ’ 12
a1 4t2 b
È5
2
Ê 'C x ds œ '0 t
4
È5
È5
2 dt œ 2
'04 t dt œ ’ È45 t2 “ 4 œ 4È5
Ê ¸ ddtr ¸ œ È1 4t2 Ê 'C x ds œ '0 t È1 4t2 dt
2
È
17 "
“ œ 17 12
!
20. (a) ratb œ ti 4tj , 0 Ÿ t Ÿ 1 Ê ddtr œ i 4j Ê ¸ ddtr ¸ œ È17 Ê 'C Èx 2y ds œ '0 Èt 2a4tb È17 dt
1
œ È17'0 È9t dt œ 3È17'0 Èt dt œ ’2È17 t2Î3 “ œ 2È17
1
1
1
!
(b) C" : ratb œ ti , 0 Ÿ t Ÿ 1 Ê ddtr œ i Ê ¸ ddtr ¸ œ 1; C2 : ratb œ i tj, 0 Ÿ t Ÿ 1 Ê ddtr œ j Ê ¸ ddtr ¸ œ 1
'C Èx 2y ds œ 'C Èx 2y ds 'C Èx 2y ds œ '01 Èt 2a0b dt '02 È1 2atb dt
1
2
2
1
2
È
È
1
œ '0 Èt dt '0 È1 2t dt œ 23 t2Î3 ‘ ! ’ 13 a1 2tb2Î3 “ œ 23 Š 5 3 5 31 ‹ œ 5 53 1
!
21. ratb œ 4ti 3tj , 1 Ÿ t Ÿ 2 Ê ddtr œ 4i 3j Ê ¸ ddtr ¸ œ 5 Ê 'C y ex ds œ 'c1 a3tb ea4tb † 5dt
2
16t
œ 15'c1 t e16t dt œ ’ 15
“
32 e
2
2
2
2
c1
2
2
15 16
15 16
64
64
œ 15
32 e 32 e œ 32 ae e b
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
!
Section 16.1 Line Integrals
941
22. ratb œ acos tbi asin tbj , 0 Ÿ t Ÿ 21 Ê ddtr œ asin tbi acos tbj Ê ¸ ddtr ¸ œ Èsin2 t cos2 t œ 1 Ê 'C ax y 3b ds
œ '0 acos t sin t 3b † 1 dt œ csin t cos t 3td 201 œ 61
21
23. ratb œ t2 i t3 j , 1 Ÿ t Ÿ 2 Ê ddtr œ 2ti 3t2 j Ê ¸ ddtr ¸ œ Éa2tb2 a3t2 b2 œ tÈ4 9t2 Ê 'C yx4Î3 ds
2
œ '1
2
ˆt2 ‰2
1
a4 9t2 b
† tÈ4 9t2 dt œ '1 t È4 9t2 dt œ ’ 27
3Î2 2
2
at3 b4Î3
“ œ 80
1
È10 13È13
27
24. ratb œ t3 i t4 j , 12 Ÿ t Ÿ 1 Ê ddtr œ 3t2 i 4t3 j Ê ¸ ddtr ¸ œ Éa3t2 b2 a4t3 b2 œ t2 È9 16t2 Ê 'C
1
a9 16t2 b
œ '1Î2 t3t † t2 È9 16t2 dt œ '1Î2 t È9 16t2 dt œ ’ 48
1
È4
1
3 Î2 1
“
1Î2
œ 125 4813
Èy
x ds
È13
25. C" : ratb œ ti t2 j , 0 Ÿ t Ÿ 1 Ê ddtr œ i 2tj Ê ¸ ddtr ¸ œ È1 4t2 ; C2 : ratb œ a1 tbi a1 tbj, 0 Ÿ t Ÿ 1
Ê ddtr œ i j Ê ¸ ddtr ¸ œ È2 Ê 'C ˆx Èy‰ds œ 'C ˆx Èy‰ds 'C ˆx Èy‰ds
1
2
œ '0 Št Èt2 ‹È1 4t2 dt '0 Ša1 tb È1 t‹ È2dt œ '0 2tÈ1 4t2 dt '0 Š1 t È1 t‹ È2dt
1
1
œ ’ 16 a1 4t2 b
1
3 Î2 1
1
0
0
Î
“ È2’t "# t2 23 a1 tb3 2 “ œ 5
1
È
È5 1
È
È
7 6 2 œ 5 5 76 2 1
6
26. C" : ratb œ ti , 0 Ÿ t Ÿ 1 Ê ddtr œ i Ê ¸ ddtr ¸ œ 1; C2 : ratb œ i tj, 0 Ÿ t Ÿ 1 Ê ddtr œ j Ê ¸ ddtr ¸ œ 1;
C3 : ratb œ a1 tbi j, 0 Ÿ t Ÿ 1 Ê ddtr œ i Ê ¸ ddtr ¸ œ 1; C4 : ratb œ a1 tbj, 0 Ÿ t Ÿ 1 Ê ddtr œ j Ê ¸ ddtr ¸ œ 1;
Ê 'C x2 y12 1 ds œ 'C x2 y12 1 ds 'C x2 y12 1 ds 'C x2 y12 1 ds 'C x2 y12 1 ds
œ'
1
dt
2
0 t 1
'
1
1
dt
2
0 t 2
'
2
1
dt
2
0 a1 tb 2
'
3
4
1
dt
2
0 a1 tb 1
1
1
0
0
1
œ ctan1 td 0 È12 ’tan1 Š Èt 2 ‹“ È12 ’tan1 Š tÈ
‹“ ctan1 a1 tbd 0 œ 12 È22 tan1 Š È12 ‹
2
1
1
#
#
dr
dr ¸
27. r(x) œ xi yj œ xi x# j , 0 Ÿ x Ÿ 2 Ê dx
œ i xj Ê ¸ dx
œ È1 x# ; f(xß y) œ f Šxß x# ‹ œ
œ '0 (2x)È1 x# dx œ ’ 23 a1 x# b
2
$Î# #
x$
#
Š x# ‹
È
“ œ 23 ˆ5$Î# 1‰ œ 10 35 2
!
28. r(t) œ a1 tbi #1 a1 tb2 j, 0 Ÿ t Ÿ 1 Ê ¸ ddtr ¸ œ É1 a1 tb# ; f(xß y) œ f Ša1 tbß #1 a1 tb2 ‹ œ
Ê 'C f ds œ '0
1
a1 tb 14 a1 tb4
É1 a1 tb
#
œ 2x Ê 'C f ds
a1 tb 14 a1 tb4
É1 a1 tb#
1
É1 a1 tb# dt œ ' Ša1 tb 14 a1 tb4 ‹ dt œ ’ "# a1 tb2 20
a1 tb5 “
1
"
0
!
œ 0 ˆ "# #"0 ‰ œ #110
29. r(t) œ (2 cos t) i (2 sin t) j , 0 Ÿ t Ÿ 1# Ê ddtr œ (2 sin t) i (2 cos t) j Ê ¸ ddtr ¸ œ 2; f(xß y) œ f(2 cos tß 2 sin t)
œ 2 cos t 2 sin t Ê 'C f ds œ '0 (2 cos t 2 sin t)(2) dt œ c4 sin t 4 cos td !
1Î2
1Î#
œ 4 (4) œ 8
30. r(t) œ (2 sin t) i (2 cos t) j , 0 Ÿ t Ÿ 14 Ê ddtr œ (2 cos t) i (2 sin t) j Ê ¸ ddtr ¸ œ 2; f(xß y) œ f(2 sin tß 2 cos t)
œ 4 sin# t 2 cos t Ê 'C f ds œ '0 a4 sin# t 2 cos t b (2) dt œ c4t 2 sin 2t 4 sin td 01Î% œ 1 2Š1 È2‹
1Î4
31. y œ x2 , 0 Ÿ x Ÿ 2 Ê ratb œ ti t2 j , 0 Ÿ t Ÿ 2 Ê ddtr œ i 2tj Ê ¸ ddtr ¸ œ È1 4t2 Ê A œ 'C fax, yb ds
È 1
3 Î2
œ 'C ˆx Èy‰ds œ '0 Št Èt2 ‹È1 4t2 dt œ '0 2tÈ1 4t2 dt œ ’ 16 a1 4t2 b “ œ 17 17
6
2
2
2
0
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
942
Chapter 16 Integration in Vector Fields
È13
3 Ê Aœ
32. 2x 3y œ 6, 0 Ÿ x Ÿ 6 Ê ratb œ ti ˆ2 23 t‰j , 0 Ÿ t Ÿ 6 Ê ddtr œ i 23 j Ê ¸ ddtr ¸ œ
œ 'C a4 3x 2ybds œ '0 ˆ4 3t 2ˆ2 23 t‰‰ 313 dt œ
È
6
È13
3
'C fax, yb ds
'06 ˆ8 35 t‰dt œ È313 8t 65 t2 ‘ 60 œ 26È13
33. r(t) œ at# 1b j 2tk , 0 Ÿ t Ÿ 1 Ê ddtr œ 2tj 2k Ê ¸ ddtr ¸ œ 2Èt# 1; M œ 'C $ (xß yß z) ds œ '0 $ (t) Š2Èt# 1‹ dt
1
œ '0 ˆ 3# t‰ Š2Èt# 1‹ dt œ ’at# 1b “ œ 2$Î# 1 œ 2È2 1
3/2 "
1
!
34. r(t) œ at# 1b j 2tk , 1 Ÿ t Ÿ 1 Ê ddtr œ 2tj 2k
Ê ¸ dr ¸ œ 2Èt# 1; M œ ' $ (xß yß z) ds
dt
C
œ 'c1
1
ˆ15Èat# 1b 2‰ Š2Èt# 1‹ dt
œ 'c1 30 at# 1b dt œ ’30 Š t3 t‹“
1
$
"
"
œ 60 ˆ 3" 1‰ œ 80;
Mxz œ 'C y$ (xß yß z) ds œ 'c1 at# 1b c30 at# 1bd dt
1
œ 'c1 30 at% 1b dt œ ’30 Š t5 t‹“
1
&
"
"
œ 60 ˆ 5" 1‰
48
œ 48 Ê y œ MMxz œ 80
œ 53 ; Myz œ 'C x$ (xß yß z) ds œ 'C 0 $ ds œ 0 Ê x œ 0; z œ 0 by symmetry (since $ is
independent of z) Ê (xß yß z) œ ˆ!ß 35 ß 0‰
35. r(t) œ È2t i È2t j a4 t# b k , 0 Ÿ t Ÿ 1 Ê ddtr œ È2i È2j 2tk Ê ¸ ddtr ¸ œ È2 2 4t# œ 2È1 t# ;
(a) M œ 'C $ ds œ '0 (3t) Š2È1 t# ‹ dt œ ’2 a1 t# b
1
$Î# "
“ œ 2 ˆ2$Î# 1‰ œ 4È2 2
!
(b) M œ 'C $ ds œ '0 a1b Š2È1 t# ‹ dt œ ’tÈ1 t# ln Št È1 t# ‹“ œ ’È2 ln Š1 È2‹“ a0 ln 1b
"
1
!
œ È2 ln Š1 È2‹
36. r(t) œ ti 2tj 23 t$Î# k , 0 Ÿ t Ÿ 2 Ê ddtr œ i 2j t"Î# k Ê ¸ ddtr ¸ œ È1 4 t œ È5 t;
M œ 'C $ ds œ '0 ˆ3È5 t‰ ˆÈ5 t‰ dt œ '0 3(5 t) dt œ 32 (5 t)# ‘ ! œ 3# a7# 5# b œ 3# (24) œ 36;
2
2
#
Myz œ 'C x$ ds œ '0 t[3(5 t)] dt œ '0 a15t 3t# b dt œ "25 t# t$ ‘ ! œ 30 8 œ 38;
2
2
#
Mxz œ 'C y$ ds œ '0 2t[3(5 t)] dt œ 2 '0 a15t 3t# b dt œ 76; Mxy œ 'C z$ ds œ '0 23 t$Î# [3(5 t)] dt
2
2
2
È2 œ 144 È2 Ê x œ Myz
œ '0 ˆ10t$Î# 2t&Î# ‰ dt œ 4t&Î# 47 t(Î# ‘ ! œ 4(2)&Î# 47 (2)(Î# œ 16È2 32
7
7
M
#
2
M
È
144 2
xy
Mxz
19
76
19
4 È
œ 38
2
36 œ 18 , y œ M œ 36 œ 9 , and z œ M œ 7†36 œ 7
dy
dz
37. Let x œ a cos t and y œ a sin t, 0 Ÿ t Ÿ 21. Then dx
dt œ a sin t, dt œ a cos t, dt œ 0
‰# Š dy
ˆ dz ‰# dt œ a dt; Iz œ ' ax# y# b $ ds œ ' aa# sin# t a# cos# tb a$ dt
Ê Êˆ dx
dt
dt ‹ dt
C
0
#
21
œ '0 a$ $ dt œ 21$ a$ .
21
38. r(t) œ tj (2 2t)k , 0 Ÿ t Ÿ 1 Ê ddtr œ j 2k Ê ¸ ddtr ¸ œ È5; M œ 'C $ ds œ '0 $ È5 dt œ $ È5;
1
Ix œ 'C ay# z# b $ ds œ '0 ct# (2 2t)# d $ È5 dt œ '0 a5t# 8t 4b $ È5 dt œ $ È5 53 t$ 4t# 4t‘ ! œ 53 $ È5 ;
1
1
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
"
Section 16.1 Line Integrals
943
Iy œ 'C ax# z# b $ ds œ '0 c0# (2 2t)# d $ È5 dt œ '0 a4t# 8t 4b $ È5 dt œ $ È5 43 t$ 4t# 4t‘ ! œ 43 $ È5 ;
1
1
"
Iz œ 'C ax# y# b $ ds œ '0 a0# t# b $ È5 dt œ $ È5 ’ t3 “ œ 3" $ È5
1
$
"
!
39. r(t) œ (cos t)i (sin t)j tk , 0 Ÿ t Ÿ 21 Ê ddtr œ ( sin t)i (cos t)j k Ê ¸ ddtr ¸ œ Èsin# t cos# t 1 œ È2;
(a) Iz œ 'C ax# y# b $ ds œ '0 acos# t sin# tb $ È2 dt œ 21$ È2
21
(b) Iz œ 'C ax# y# b $ ds œ '0 $ È2 dt œ 41$ È2
41
È
40. r(t) œ (t cos t)i (t sin t)j 2 3 2 t$Î# k , 0 Ÿ t Ÿ 1 Ê ddtr œ (cos t t sin t)i (sin t t cos t)j È2t k
Ê ¸ ddtr ¸ œ È(t 1)# œ t 1 for 0 Ÿ t Ÿ 1; M œ 'C $ ds œ '0 (t 1) dt œ "2 (t 1)# ‘ ! œ "# a2# 1# b œ #3 ;
1
"
Mxy œ 'C z$ ds œ '0 Š 2 3 2 t$Î# ‹ (t 1) dt œ 2 3 2 '0 ˆt&Î# t$Î# ‰ dt œ 2 3 2 27 t(Î# 25 t&Î# ‘ !
È
1
È
È
1
"
‰ 16 2 Ê z œ Mxy œ Š 1635 2 ‹ ˆ 23 ‰ œ 321052 ; Iz œ ' ax# y# b $ ds
œ 2 3 2 ˆ 27 52 ‰ œ 2 3 2 ˆ 24
35 œ 35
C
È
È
È
È
M
È
7
œ '0 at# cos# t t# sin# tb (t 1) dt œ '0 at$ t# b dt œ ’ t4 t3 “ œ "4 "3 œ 12
1
1
%
$
"
!
41. $ (xß yß z) œ 2 z and r(t) œ (cos t)j (sin t)k , 0 Ÿ t Ÿ 1 Ê M œ 21 2 as found in Example 3 of the text;
also ¸ ddtr ¸ œ 1; Ix œ 'C ay# z# b $ ds œ '0 acos# t sin# tb (2 sin t) dt œ '0 (2 sin t) dt œ 21 2
1
È
1
#
42. r(t) œ ti 2 3 2 t$Î# j t# k , 0 Ÿ t Ÿ 2 Ê ddtr œ i È2 t"Î# j tk Ê ¸ ddtr ¸ œ È1 2t t# œ È(1 t)# œ 1 t for
0 Ÿ t Ÿ 2; M œ 'C $ ds œ '0 ˆ t"1 ‰ (1 t) dt œ '0 dt œ 2; Myz œ 'C x$ ds œ '0 t ˆ t"1 ‰ (1 t) dt œ ’ t2 “ œ 2;
2
Mxz œ 'C y$ ds œ '
2
#
È
2È2 $Î#
32
dt œ ’ 4152 t&Î# “ œ 15
; Mxy œ
3 t
0
!
2
2
'C z$ ds œ '
#
$ #
t
dt œ ’ t6 “ œ 3%
0 #
!
2 #
#
!
Ê
M
x œ Myz œ 1,
xy
#
' # #
' ˆ 8 $ " % ‰ dt œ ’ 92 t% 20t “ œ 329 3220 œ 232
y œ MMxz œ 16
15 , and z œ M œ 3 ; Ix œ C ay z b $ ds œ 0 9 t 4 t
45 ;
2
M
Iy œ 'C ax z b $ ds œ '0 t
#
2
#
ˆ#
&
#
$
t&
64
4" t% ‰ dt œ ’ t3 20
“ œ 38 32
20 œ 15 ; Iz œ
!
#
!
'C ax y b $ ds
#
#
56
œ '0 ˆt# 89 t$ ‰ dt œ ’ t3 29 t% “ œ 83 32
9 œ 9
2
$
#
!
43-46. Example CAS commands:
Maple:
f := (x,y,z) -> sqrt(1+30*x^2+10*y);
g := t -> t;
h := t -> t^2;
k := t -> 3*t^2;
a,b := 0,2;
ds := ( D(g)^2 + D(h)^2 + D(k)^2 )^(1/2):
'ds' = ds(t)*'dt';
F := f(g,h,k):
'F(t)' = F(t);
Int( f, s=C..NULL ) = Int( simplify(F(t)*ds(t)), t=a..b );
`` = value(rhs(%));
# (a)
# (b)
# (c)
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
944
Chapter 16 Integration in Vector Fields
Mathematica: (functions and domains may vary)
Clear[x, y, z, r, t, f]
f[x_,y_,z_]:= Sqrt[1 30x2 10y]
{a,b}= {0, 2};
x[t_]:= t
y[t_]:= t2
z[t_]:= 3t2
r[t_]:= {x[t], y[t], z[t]}
v[t_]:= D[r[t], t]
mag[vector_]:=Sqrt[vector.vector]
Integrate[f[x[t],y[t],z[t]] mag[v[t]], {t, a, b}]
N[%]
16.2 VECTOR FIELDS, WORK, CIRCULATION, AND FLUX
1. f(xß yß z) œ ax# y# z# b
"Î#
Ê `` xf œ #" ax# y# z# b
$Î#
$Î#
`f
#
#
# $Î#
and `` zf œ z ax# y# z# b
` y œ y ax y z b
(2x) œ x ax# y# z# b
Ê
$Î#
; similarly,
™ f œ # xi #yj #zk$Î#
ax y z b
2. f(xß yß z) œ ln Èx# y# z# œ "# ln ax# y# z# b Ê `` xf œ "# Š x# y"# z# ‹ (2x) œ x# yx# z# ;
similarly, `` yf œ x# yy# z# and `` zf œ x# yz# z# Ê ™ f œ xx#i yyj#zzk#
`g
2y
`g
z
3. g(xß yß z) œ ez ln ax# y# b Ê `` gx œ x# 2x
y# , ` y œ x# y# and ` z œ e
z
Ê ™ g œ Š x#2xy# ‹ i Š x# 2y
y# ‹ j e k
4. g(xß yß z) œ xy yz xz Ê `` gx œ y z, `` gy œ x z, and `` gz œ y x Ê ™ g œ (y z)i (B z)j (x y)k
5. kFk inversely proportional to the square of the distance from (xß y) to the origin Ê È(M(xß y))# (N(xß y))#
x
œ x# k y# , k 0; F points toward the origin Ê F is in the direction of n œ Èx# i Èx#y y# j
y#
Ê F œ an , for some constant a 0. Then M(xß y) œ Èx#ax
and N(xß y) œ Èx#ay
y#
y#
Ê È(M(xß y))# (N(xß y))# œ a Ê a œ x# k y# Ê F œ
kx
i # ky# $Î# j , for any constant k 0
ax# y# b$Î#
ax y b
6. Given x# y# œ a# b# , let x œ Èa# b# cos t and y œ Èa# b# sin t. Then
r œ ŠÈa# b# cos t‹ i ŠÈa# b# sin t‹ j traces the circle in a clockwise direction as t goes from 0 to 21
Ê v œ ŠÈa# b# sin t‹ i ŠÈa# b# cos t‹ j is tangent to the circle in a clockwise direction. Thus, let
F œ v Ê F œ yi xj and F(0ß 0) œ 0 .
7. Substitute the parametric representations for r(t) œ x(t)i y(t)j z(t)k representing each path into the vector
field F , and calculate 'C F † ddtr .
(a) F œ 3ti 2tj 4tk and ddtr œ i j k Ê F † ddtr œ 9t Ê '0 9t dt œ 9#
1
(b) F œ 3t# i 2tj 4t% k and ddtr œ i 2tj 4t$ k Ê F † ddtr œ 7t# 16t( Ê '0 a7t# 16t( b dt œ 37 t$ 2t) ‘ !
1
œ 73 2 œ 13
3
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
"
Section 16.2 Vector Fields, Work, Circulation, and Flux
(c) r" œ ti tj and r# œ i j tk ; F" œ 3ti 2tj and ddtr" œ i j Ê F" † ddtr" œ 5t Ê '0 5t dt œ #5 ;
1
F# œ 3i 2j 4tk and ddtr# œ k Ê F# † ddtr# œ 4t Ê '0 4t dt œ 2 Ê #5 2 œ #9
1
8. Substitute the parametric representation for r(t) œ x(t)i y(t)j z(t)k representing each path into the vector
field F, and calculate 'C F † ddtr .
" ‰
dr
dr
"
"
' "
(a) F œ ˆ t# td ! œ 14
1 j and dt œ i j k Ê F † dt œ t# 1 Ê 0 t# 1 dt œ ctan
1
"
" ‰
dr
dr
2t
$
#
' 2t
(b) F œ ˆ t# 1 j and dt œ i 2tj 4t k Ê F † dt œ t# 1 Ê 0 t# 1 dt œ cln at 1bd ! œ ln 2
1
"
dr"
d r"
d r#
" ‰
"
"
(c) r" œ ti tj and r# œ i j tk ; F" œ ˆ t# 1 j and dt œ i j Ê F" † dt œ t# 1 ; F# œ # j and dt œ k
1
"
Ê F# † ddtr# œ 0 Ê '0 t# 1 dt œ 4
1
9. Substitute the parametric representation for r(t) œ x(t)i y(t)j z(t)k representing each path into the vector
field F, and calculate 'C F † ddtr .
"
(a) F œ Èti 2tj Ètk and ddtr œ i j k Ê F † ddtr œ 2Èt 2t Ê '0 ˆ2Èt 2t‰ dt œ 43 t$Î# t# ‘ ! œ "3
1
(b) F œ t# i 2tj tk and ddtr œ i 2tj 4t$ k Ê F † ddtr œ 4t% 3t# Ê '0 a4t% 3t# b dt œ 45 t& t$ ‘ ! œ "5
1
"
(c) r" œ ti tj and r# œ i j tk ; F" œ 2tj Èt k and ddtr" œ i j Ê F" † ddtr" œ 2t Ê '0 2t dt œ 1;
1
F# œ Èti 2j k and ddtr# œ k Ê F# † ddtr# œ 1 Ê '0 dt œ 1 Ê 1 1 œ 0
1
10. Substitute the parametric representation for r(t) œ x(t)i y(t)j z(t)k representing each path into the vector
field F, and calculate 'C F † ddtr .
(a) F œ t# i t# j t# k and ddtr œ i j k Ê F † ddtr œ 3t# Ê '0 3t# dt œ 1
1
(b) F œ t$ i t' j t& k and ddtr œ i 2tj 4t$ k Ê F † ddtr œ t$ 2t( 4t) Ê '0 at$ 2t( 4t) b dt
1
%
)
"
17
œ ’ t4 t4 94 t* “ œ 18
!
(c) r" œ ti tj and r# œ i j tk ; F" œ t# i and ddtr" œ i j Ê F" † ddtr" œ t# Ê '0 t# dt œ 3" ;
1
F# œ i tj tk and ddtr# œ k Ê F# † ddtr# œ t Ê '0 t dt œ "# Ê "3 #" œ 65
1
11. Substitute the parametric representation for r(t) œ x(t)i y(t)j z(t)k representing each path into the vector
field F, and calculate 'C F † ddtr .
(a) F œ a3t# 3tb i 3tj k and ddtr œ i j k Ê F † ddtr œ 3t# 1 Ê '0 a3t# 1b dt œ ct$ td ! œ 2
1
"
(b) F œ a3t# 3tb i 3t% j k and ddtr œ i 2tj 4t$ k Ê F † ddtr œ 6t& 4t$ 3t# 3t
"
Ê '0 a6t& 4t$ 3t# 3tb dt œ t' t% t$ 3# t# ‘ ! œ 3#
1
(c) r" œ ti tj and r# œ i j tk ; F" œ a3t# 3tb i k and ddtr" œ i j Ê F" † ddtr" œ 3t# 3t
"
Ê '0 a3t# 3tb dt œ t$ 32 t# ‘ ! œ "# ; F# œ 3tj k and ddtr# œ k Ê F# † ddtr# œ 1 Ê '0 dt œ 1
1
1
Ê "# 1 œ 12
12. Substitute the parametric representation for r(t) œ x(t)i y(t)j z(t)k representing each path into the vector
field F, and calculate 'C F † ddtr .
(a) F œ 2ti 2tj 2tk and ddtr œ i j k Ê F † ddtr œ 6t Ê '0 6t dt œ c3t# d ! œ 3
1
"
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
945
946
Chapter 16 Integration in Vector Fields
(b) F œ at# t% b i at% tb j at t# b k and ddtr œ i 2tj 4t$ k Ê F † ddtr œ 6t& 5t% 3t#
Ê '0 a6t& 5t% 3t# b dt œ ct' t& t$ d ! œ 3
1
"
(c) r" œ ti tj and r# œ i j tk ; F" œ ti tj 2tk and ddtr" œ i j Ê F" † ddtr" œ 2t Ê '0 2t dt œ ";
1
F# œ (1 t)i (t 1)j 2k and ddtr# œ k Ê F# † ddtr# œ 2 Ê '0 2 dt œ 2 Ê " 2 œ 3
1
3
13. x œ t, y œ 2t 1, 0 Ÿ t Ÿ 3 Ê dx œ dt Ê 'C ax yb dx œ '0 at a2t 1bb dt œ '0 at 1b dt œ "# t2 t‘ ! œ 15
2
3
3
14. x œ t, y œ t2 , 1 Ÿ t Ÿ 2 Ê dy œ 2t dt Ê 'C xy dy œ '1 tt2 a2tb dt œ '1 2 dt œ c2td21 œ 2
2
2
15. C1 : x œ t, y œ 0, 0 Ÿ t Ÿ 3 Ê dy œ 0; C2 : x œ 3, y œ t, 0 Ÿ t Ÿ 3 Ê dy œ dt Ê 'C ax2 y2 b dy
œ 'C ax2 y2 b dx 'C ax2 y2 b dx œ '0 at2 02 b † 0 '0 a32 t2 b dt œ '0 a9 t2 bdt œ 9t 13 t3 ‘ ! œ 36
3
1
3
3
3
2
16. C1 : x œ t, y œ 3t, 0 Ÿ t Ÿ 1 Ê dx œ dt; C2 : x œ 1 t, y œ 3, 0 Ÿ t Ÿ 1 Ê dx œ dt; C3 : x œ 0, y œ 3 t, 0 Ÿ t Ÿ 3
Ê dx œ 0 Ê 'C Èx y dx œ 'C Èx y dx 'C Èx y dx 'C Èx y dx
1
2
3
œ '0 Èt 3t dt '0 Èa1 tb 3 a1bdt '0 È0 a3 tb † 0 œ '0 2Èt dt '0 È4 t dt
1
1
3
1
1
1
1
È
œ 43 t2Î3 ‘ ! ’ 23 a4 tb2Î3 “ œ 43 Š2È3 16
3 ‹œ 2 34
!
17. ratb œ ti j t2 k , 0 Ÿ t Ÿ 1 Ê dx œ dt, dy œ 0, dz œ 2t dt
(a) 'C ax y zb dx œ '0 at 1 t2 b dt œ 12 t2 t 13 t3 ‘ ! œ 56
1
1
(b) 'C ax y zb dy œ '0 at 1 t2 b † 0 œ 0
1
(c) 'C ax y zb dz œ '0 at 1 t2 b 2t dt œ '0 a2t2 2t 2t3 b dt œ œ 23 t3 t2 12 t4 ‘ ! œ 56
1
1
1
18. ratb œ acos tbi asin tbj acos tbk , 0 Ÿ t Ÿ 1 Ê dx œ sin t dt, dy œ cos t dt, dz œ sin t dt
(a) 'C x z dx œ '0 acos tb acos tbasin tbdt œ '0 cos2 t sin tdt œ ’ 13 acos tb3 “ œ 23
1
1
1
!
1
1
1
1
(b) 'C x z dy œ '0 acos tb acos tbacos tbdt œ '0 cos3 t dt œ '0 a1 sin2 tb cos t dt œ ’ 13 asin tb3 sin t“ œ 0
(c) 'C x y z dz œ '0 acos tbasin tb acos tbasin tbdt œ '0 cos t sin
1
1
1
œ 18 '0 a1 cos 4tb dt œ 18 t 32
sin 4t‘ ! œ 18
1
2
2
t dt œ 14
'0 sin
1
2
2t dt œ 41
1
19. r œ ti t# j tk , 0 Ÿ t Ÿ 1, and F œ xyi yj yzk Ê F œ t$ i t# j t$ k and ddtr œ i 2tj k
Ê F † ddtr œ 2t$ Ê work œ '0 2t$ dt œ "#
1
20. r œ (cos t)i (sin t)j 6t k , 0 Ÿ t Ÿ 21, and F œ 2yi 3xj (x y)k
Ê F œ (2 sin t)i (3 cos t)j (cos t sin t)k and ddtr œ ( sin t)i (cos t)j 6" k Ê F † ddtr
œ 3 cos# t 2sin2 t "6 cos t 6" sin t Ê work œ '0 ˆ3 cos# t 2 sin2 t 6" cos t 6" sin t‰ dt
21
#1
œ 32 t 34 sin 2t t sin2 2t "6 sin t "6 cos t‘ ! œ 1
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
'
1
!
1 cos 4t
dt
2
0
Section 16.2 Vector Fields, Work, Circulation, and Flux
21. r œ (sin t)i (cos t)j tk , 0 Ÿ t Ÿ 21, and F œ zi xj yk Ê F œ ti (sin t)j (cos t)k and
dr
dt œ (cos t)i (sin t)j k
Ê F † ddtr œ t cos t sin# t cos t Ê work œ '0 at cos t sin# t cos tb dt
21
#1
œ cos t t sin t 2t sin4 2t sin t‘ ! œ 1
22. r œ (sin t)i (cos t)j 6t k , 0 Ÿ t Ÿ 21, and F œ 6zi y# j 12xk Ê F œ ti acos# tbj (12 sin t)k and
dr
"
dt œ (cos t)i (sin t)j 6 k
Ê F † ddtr œ t cos t sin t cos# t 2 sin t
Ê work œ '0 at cos t sin t cos# t 2 sin tb dt œ cos t t sin t 13 cos$ t 2 cos t‘ ! œ 0
21
#1
23. x œ t and y œ x# œ t# Ê r œ ti t# j , 1 Ÿ t Ÿ 2, and F œ xyi (x y)j Ê F œ t$ i at t# b j and
dr
dt œ i 2tj
Ê F † ddtr œ t$ a2t# 2t$ b œ 3t$ 2t# Ê 'C xy dx (x y) dy œ 'C F † ddtr dt œ 'c" a3t$ 2t# b dt
#
#
69
‰ ˆ 3 2 ‰ 45 18
œ 34 t% 32 t$ ‘ " œ ˆ12 16
3 4 3 œ 4 3 œ 4
24. Along (0ß 0) to (1ß 0): r œ ti , 0 Ÿ t Ÿ 1, and F œ (x y)i (x y)j Ê F œ ti tj and ddtr œ i Ê F † ddtr œ t;
Along (1ß 0) to (0ß 1): r œ (1 t)i tj , 0 Ÿ t Ÿ 1, and F œ (x y)i (x y)j Ê F œ (1 2t)i j and
dr
dr
dt œ i j Ê F † dt œ 2t;
Along (0ß 1) to (0ß 0): r œ (1 t)j , 0 Ÿ t Ÿ 1, and F œ (x y)i (x y)j Ê F œ (t 1)i (1 t)j and
dr
dr
dt œ j Ê F † dt œ t 1 Ê
"
œ c2t# td ! œ 2 1 œ 1
25. r œ xi yj œ y# i yj , 2
'C (x y) dx (x y) dy œ '01 t dt '01 2t dt '01 (t 1) dt œ '01 (4t 1) dt
dr
dr
1, and F œ x# i yj œ y% i yj Ê dy
œ 2yi j and F † dy
œ 2y& y
y
c1
c1
dr
4‰
3
63
39
Ê 'C F † T ds œ '2 F † dy
dy œ '2 a2y& yb dy œ 3" y' "# y# ‘ # œ ˆ 3" #" ‰ ˆ 64
3 # œ # 3 œ #
"
26. r œ (cos t)i (sin t)j , 0 Ÿ t Ÿ 1# , and F œ yi xj Ê F œ (sin t)i (cos t)j and ddtr œ ( sin t)i (cos t)j
Ê F † ddtr œ sin# t cos# t œ 1 Ê 'C F † dr œ '0
1Î2
(1) dt œ 1#
27. r œ (i j) t(i 2j) œ (1 t)i (1 2t)j , 0 Ÿ t Ÿ 1, and F œ xyi (y x)j Ê F œ a1 3t 2t# b i tj and
dr
dt œ i 2j
Ê F † ddtr œ 1 5t 2t# Ê work œ 'C F † ddtr dt œ '0 a1 5t 2t# b dt œ t 25 t# 23 t$ ‘ ! œ 25
6
1
"
28. r œ (2 cos t)i (2 sin t)j , 0 Ÿ t Ÿ 21, and F œ ™ f œ 2(x y)i 2(x y)j
Ê F œ 4(cos t sin t)i 4(cos t sin t)j and ddtr œ (2 sin t)i (2 cos t)j Ê F † ddtr
œ 8 asin t cos t sin# tb 8 acos# t cos t sin tb œ 8 acos# t sin# tb œ 8 cos 2t Ê work œ 'C ™ f † dr
œ 'C F † ddtr dt œ '0 8 cos 2t dt œ c4 sin 2td #!1 œ 0
21
29. (a) r œ (cos t)i (sin t)j , 0 Ÿ t Ÿ 21, F" œ xi yj , and F# œ yi xj Ê ddtr œ ( sin t)i (cos t)j ,
F" œ (cos t)i (sin t)j , and F# œ ( sin t)i (cos t)j Ê F" † ddtr œ 0 and F# † ddtr œ sin# t cos# t œ 1
Ê Circ" œ '0 0 dt œ 0 and Circ# œ '0 dt œ 21; n œ (cos t)i (sin t)j Ê F" † n œ cos# t sin# t œ 1 and
21
21
F# † n œ 0 Ê Flux" œ '0 dt œ 21 and Flux# œ '0 0 dt œ 0
21
21
(b) r œ (cos t)i (4 sin t)j , 0 Ÿ t Ÿ 21 Ê ddtr œ ( sin t)i (4 cos t)j , F" œ (cos t)i (4 sin t)j , and
F# œ (4 sin t)i (cos t)j Ê F" † ddtr œ 15 sin t cos t and F# † ddtr œ 4 Ê Circ" œ '0 15 sin t cos t dt
21
œ "25 sin# t‘ ! œ 0 and Circ# œ '0 4 dt œ 81; n œ Š È417 cos t‹ i Š È"17 sin t‹ j Ê F" † n
#1
21
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
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948
Chapter 16 Integration in Vector Fields
œ È417 cos# t È417 sin# t and F# † n œ È1517 sin t cos t Ê Flux" œ '0 (F" † n) kvk dt œ '0 Š È417 ‹ È17 dt
21
21
# ‘
œ 81 and Flux# œ '0 (F# † n) kvk dt œ '0 Š È1517 sin t cos t‹ È17 dt œ 15
2 sin t ! œ 0
21
21
#1
30. r œ (a cos t)i (a sin t)j , 0 Ÿ t Ÿ 21, F" œ 2xi 3yj , and F# œ 2xi (x y)j Ê ddtr œ (a sin t)i (a cos t)j ,
F" œ (2a cos t)i (3a sin t)j , and F# œ (2a cos t)i (a cos t a sin t)j Ê n kvk œ (a cos t)i (a sin t)j ,
F" † n kvk œ 2a# cos# t 3a# sin# t, and F# † n kvk œ 2a# cos# t a# sin t cos t a# sin# t
Ê Flux" œ '0 a2a# cos# t 3a# sin# tb dt œ 2a# 2t sin4 2t ‘ ! 3a# 2t sin4 2t ‘ ! œ 1a# , and
21
#1
#1
Flux# œ '0 a2a# cos# t a# sin t cos t a# sin# tb dt œ 2a# 2t sin4 2t ‘ ! a# csin# td ! a# 2t sin4 2t ‘ ! œ 1a#
21
#1
#1
#1
#
31. F" œ (a cos t)i (a sin t)j , ddtr" œ (a sin t)i (a cos t)j Ê F" † ddtr" œ 0 Ê Circ" œ 0; M" œ a cos t,
N" œ a sin t, dx œ a sin t dt, dy œ a cos t dt Ê Flux" œ 'C M" dy N" dx œ '0 aa# cos# t a# sin# tb dt
1
œ '0 a# dt œ a# 1;
1
F# œ ti , ddtr# œ i Ê F# † ddtr# œ t Ê Circ# œ 'ca t dt œ 0; M# œ t, N# œ 0, dx œ dt, dy œ 0 Ê Flux#
a
œ 'C M# dy N# dx œ 'ca 0 dt œ 0; therefore, Circ œ Circ" Circ# œ 0 and Flux œ Flux" Flux# œ a# 1
a
32. F" œ aa# cos# tb i aa# sin# tb j , ddtr" œ (a sin t)i (a cos t)j Ê F" † ddtr" œ a$ sin t cos# t a$ cos t sin# t
Ê Circ" œ '0 aa$ sin t cos# t a$ cos t sin# tb dt œ 2a3 ; M" œ a# cos# t, N" œ a# sin# t, dy œ a cos t dt,
1
$
dx œ a sin t dt Ê Flux" œ 'C M" dy N" dx œ '0 aa$ cos$ t a$ sin$ tb dt œ 43 a$ ;
1
F# œ t# i , ddtr# œ i Ê F# † ddtr# œ t# Ê Circ# œ 'ca t# dt œ 2a3 ; M# œ t# , N# œ 0, dy œ 0, dx œ dt
a
$
Ê Flux# œ 'C M# dy N# dx œ 0; therefore, Circ œ Circ" Circ# œ 0 and Flux œ Flux" Flux# œ 43 a$
33. F" œ (a sin t)i (a cos t)j , ddtr" œ (a sin t)i (a cos t)j Ê F" † ddtr" œ a# sin# t a# cos# t œ a#
Ê Circ" œ '0 a# dt œ a# 1 ; M" œ a sin t, N" œ a cos t, dx œ a sin t dt, dy œ a cos t dt
1
Ê Flux" œ 'C M" dy N" dx œ '0 aa# sin t cos t a# sin t cos tb dt œ 0; F# œ tj , ddtr# œ i Ê F# † ddtr# œ 0
1
Ê Circ# œ 0; M# œ 0, N# œ t, dx œ dt, dy œ 0 Ê Flux# œ 'C M# dy N# dx œ 'ca t dt œ 0; therefore,
a
Circ œ Circ" Circ# œ a# 1 and Flux œ Flux" Flux# œ 0
34. F" œ aa# sin# tb i aa# cos# tb j , ddtr" œ (a sin t)i (a cos t)j Ê F" † ddtr" œ a$ sin$ t a$ cos$ t
Ê Circ" œ '0 aa$ sin$ t a$ cos$ tb dt œ 43 a$ ; M" œ a# sin# t, N" œ a# cos# t, dy œ a cos t dt, dx œ a sin t dt
1
Ê Flux" œ 'C M" dy N" dx œ '0 aa$ cos t sin# t a$ sin t cos# tb dt œ 23 a$ ; F# œ t# j , ddtr# œ i Ê F# † ddtr# œ 0
1
Ê Circ# œ 0; M# œ 0, N# œ t# , dy œ 0, dx œ dt Ê Flux# œ 'C M# dy N# dx œ 'ca t# dt œ 23 a$ ; therefore,
a
Circ œ Circ" Circ# œ 43 a$ and Flux œ Flux" Flux# œ 0
35. (a) r œ (cos t)i (sin t)j , 0 Ÿ t Ÿ 1, and F œ (x y)i ax# y# b j Ê ddtr œ (sin t)i (cos t)j and
F œ (cos t sin t)i acos# t sin# tb j Ê F † ddtr œ sin t cos t sin# t cos t Ê 'C F † T ds
1
œ '0 a sin t cos t sin# t cos tb dt œ 2" sin# t #t sin4 2t sin t‘ ! œ 1#
1
(b) r œ (1 2t)i , 0 Ÿ t Ÿ 1, and F œ (x y)i ax# y# b j Ê ddtr œ 2i and F œ (1 2t)i (1 2t)# j Ê
F † ddtr œ 4t 2 Ê 'C F † T ds œ '0 (4t 2) dt œ c2t# 2td ! œ 0
1
"
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 16.2 Vector Fields, Work, Circulation, and Flux
949
(c) r" œ (1 t)i tj , 0 Ÿ t Ÿ 1, and F œ (x y)i ax# y# b j Ê ddtr" œ i j and F œ (1 2t)i a1 2t 2t# b j
Ê F † ddtr" œ (2t 1) a1 2t 2t# b œ 2t# Ê Flow" œ 'C F † ddtr" œ '0 2t# dt œ 32 ; r# œ ti (t 1)j ,
1
"
0 Ÿ t Ÿ 1, and F œ (x y)i ax# y# b j Ê ddtr# œ i j and F œ i at# t# 2t 1b j
œ i a2t# 2t 1b j Ê F † ddtr# œ 1 a2t# 2t 1b œ 2t 2t# Ê Flow# œ 'C F † ddtr# œ '0 a2t 2t# b dt
1
#
"
œ t# 23 t$ ‘ ! œ 3" Ê Flow œ Flow" Flow# œ 32 3" œ 1
36. From (1ß 0) to (0ß 1): r" œ (1 t)i tj , 0 Ÿ t Ÿ 1, and F œ (x y)i ax# y# b j Ê ddtr" œ i j ,
F œ i a1 2t 2t# b j , and n" kv" k œ i j Ê F † n" kv" k œ 2t 2t# Ê Flux" œ '0 a2t 2t# b dt
1
"
œ t# 23 t$ ‘ ! œ 3" ;
From (0ß 1) to (1ß 0): r# œ ti (1 t)j , 0 Ÿ t Ÿ 1, and F œ (x y)i ax# y# b j Ê ddtr# œ i j ,
F œ (1 2t)i a1 2t 2t# b j , and n# kv# k œ i j Ê F † n# kv# k œ (2t 1) a1 2t 2t# b œ 2 4t 2t#
"
Ê Flux# œ '0 a2 4t 2t# b dt œ 2t 2t# 23 t$ ‘ ! œ 23 ;
1
From (1ß 0) to (1ß 0): r$ œ (1 2t)i , 0 Ÿ t Ÿ 1, and F œ (x y)i ax# y# b j Ê ddtr$ œ 2i ,
F œ (1 2t)i a1 4t 4t# b j , and n$ kv$ k œ 2j Ê F † n$ kv$ k œ 2 a1 4t 4t# b
Ê Flux$ œ 2 '0 a1 4t 4t# b dt œ 2 t 2t# 43 t$ ‘ ! œ 32 Ê Flux œ Flux" Flux# Flux$ œ 3" 32 32 œ "3
1
"
37. (a) y œ 2x, 0 Ÿ x Ÿ 2 Ê ratb œ ti 2tj , 0 Ÿ t Ÿ 2 Ê ddtr œ i 2tj Ê F † ddtr œ Ša2tb2 i 2atba2tbj‹ † ai 2jb
œ 4t2 8t2 œ 12t2 Ê Flow œ 'C F † ddtr dt œ '0 12t2 dt œ c4t3 d ! œ 32
2
2
2
(b) y œ x2 , 0 Ÿ x Ÿ 2 Ê ratb œ ti t2 j , 0 Ÿ t Ÿ 2 Ê ddtr œ i 2tj Ê F † ddtr œ Šat2 b i 2atbat2 bj‹ † ai 2tjb
œ t4 4t4 œ 5t4 Ê Flow œ 'C F † ddtr dt œ '0 5t4 dt œ ct5 d ! œ 32
2
2
(c) answers will vary, one possible path is y œ 12 x3 , 0 Ÿ x Ÿ 2 Ê ratb œ ti "# t3 j , 0 Ÿ t Ÿ 2 Ê ddtr œ i 3t2 j
Ê F † ddtr œ Šˆ "# t3 ‰ i 2atbˆ "# t3 ‰j‹ † ai 3t2 jb œ 14 t6 32 t6 œ 74 t6 Ê Flow œ 'C F † ddtr dt œ '0 74 t6 dt œ 14 t7 ‘ !
2
2
œ 32
38. (a) C1 : ratb œ a1 tbi j , 0 Ÿ t Ÿ 2 Ê ddtr œ i Ê F † ddtr œ aa1bi aa1 tb 2a1bbjb † aib œ 1;
C2 : ratb œ i a1 tbj , 0 Ÿ t Ÿ 2 Ê ddtr œ j Ê F † ddtr œ aa1 tbi aa1b 2a1 tbbjb † ajb œ 2t 1;
C3 : ratb œ at 1bi j , 0 Ÿ t Ÿ 2 Ê ddtr œ i Ê F † ddtr œ aa1bi aat 1b 2a1bbjb † aib œ 1;
C4 : ratb œ i at 1bj , 0 Ÿ t Ÿ 2 Ê ddtr œ j Ê F † ddtr œ aat 1bi aa1b 2at 1bbjb † ajb œ 2t 1;
Ê Flow œ 'C F † ddtr dt œ 'C F † ddtr dt 'C F † ddtr dt 'C F † ddtr dt 'C F † ddtr dt
1
2
œ '0 a1b dt '0 a2t 1b dt '0 a1b dt '
2
2
2
3
4
2
2
2
a2t 1b dt œ ctd 2! ct2 td ! ctd !2 ct2 td !
0
œ 2 2 2 2 œ 0
(b) x2 y2 œ 4 Ê ratb œ a2cos tbi a2sin tbj , 0 Ÿ t Ÿ 21 Ê ddtr œ a2sin tbi a2cos tbj
Ê F † ddtr œ aa2sin tbi a2cos t 2a2sin tbbjb † aa2sin tbi a2cos tbjb œ 4sin2 t 4cos2 t 8sin t cos t
œ 4cos 2t 4sin 2t Ê Flow œ 'C F † ddtr dt œ '0 a4cos 2t 4sin 2tb dt œ c2sin 2t 2cos 2td 2!1 œ 0
21
(c) answers will vary, one possible path is:
C1 : ratb œ ti , 0 Ÿ t Ÿ 1 Ê ddtr œ i Ê F † ddtr œ aa0bi at 2a1bbjb † aib œ 0;
C2 : ratb œ a1 tbi tj , 0 Ÿ t Ÿ 1 Ê ddtr œ i j Ê F † ddtr œ ati aa1 tb 2tbjb † ai jb œ 1;
C3 : ratb œ a1 tbj , 0 Ÿ t Ÿ 1 Ê ddtr œ j Ê F † ddtr œ aa1 tbi a0 2a1 tbbjb † ajb œ 2t 1;
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
2
950
Chapter 16 Integration in Vector Fields
Ê Flow œ 'C F † ddtr dt œ 'C F † ddtr dt 'C F † ddtr dt 'C F † ddtr dt œ '0 a0b dt '0 a1b dt '0 a2t 1b dt
1
1
2
1
1
3
1
œ 0 ctd 1! ct2 td ! œ 1 a1b œ 0
39. F œ Èx#y y# i Èx#x y# j on x# y# œ 4;
at (2ß 0), F œ j ; at (0ß 2), F œ i ; at (2ß 0),
È
F œ j ; at (!ß 2), F œ i ; at ŠÈ2ß È2‹ , F œ 3 i " j ;
#
#
È
at ŠÈ2ß È2‹ , F œ #3 i #" j ; at ŠÈ2ß È2‹ ,
È
F œ #3 i #" j ; at ŠÈ2ß È2‹ , F œ
È3
"
# i # j
40. F œ xi yj on x# y# œ 1; at (1ß 0), F œ i ;
at (1ß 0), F œ i ; at (0ß 1), F œ j ; at (0ß 1),
È
È
F œ j ; at Š "# ß #3 ‹, F œ #" i #3 j ;
È
È
at Š "# ß #3 ‹, F œ "# i #3 j ;
È
È
at Š "# ß #3 ‹, F œ #" i #3 j ;
È
È
at Š "# ß #3 ‹, F œ "# i #3 j .
41. (a) G œ P(xß y)i Q(xß y)j is to have a magnitude Èa# b# and to be tangent to x# y# œ a# b# in a
counterclockwise direction. Thus x# y# œ a# b# Ê 2x 2yyw œ 0 Ê yw œ xy is the slope of the tangent
line at any point on the circle Ê yw œ ba at (aß b). Let v œ bi aj Ê kvk œ Èa# b# , with v in a
counterclockwise direction and tangent to the circle. Then let P(xß y) œ y and Q(xß y) œ x
Ê G œ yi xj Ê for (aß b) on x# y# œ a# b# we have G œ bi aj and kGk œ Èa# b# .
(b) G œ ˆÈx# y# ‰ F œ ŠÈa# b# ‹ F .
42. (a) From Exercise 41, part a, yi xj is a vector tangent to the circle and pointing in a counterclockwise
direction Ê yi xj is a vector tangent to the circle pointing in a clockwise direction Ê G œ Èyxi #xjy#
is a unit vector tangent to the circle and pointing in a clockwise direction.
(b) G œ F
43. The slope of the line through (xß y) and the origin is yx Ê v œ xi yj is a vector parallel to that line and
pointing away from the origin Ê F œ Èxxi #yjy# is the unit vector pointing toward the origin.
44. (a) From Exercise 43, Èxxi #yjy# is a unit vector through (xß y) pointing toward the origin and we want
kFk to have magnitude Èx# y# Ê F œ Èx# y# Š Èxxi #yjy# ‹ œ xi yj .
yj
(b) We want kFk œ Èx#C y# where C Á 0 is a constant Ê F œ Èx#C y# Š Èxxi #yjy# ‹ œ C Š xx#i y# ‹.
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 16.2 Vector Fields, Work, Circulation, and Flux
951
45. Yes. The work and area have the same numerical value because work œ 'C F † dr œ 'C yi † dr
‘
œ 'b [f(t)i] † i df
dt j dt
a
[On the path, y equals f(t)]
œ 'a f(t) dt œ Area under the curve
b
[because f(t) 0]
dr
46. r œ xi yj œ xi f(x)j Ê dx
œ i f w (x)j ; F œ Èx#k y# (xi yj) has constant magnitude k and points away
†f (x)
dr
d È #
from the origin Ê F † dx
œ Èx#kx y# Èk†xy#†f(x)y# œ kxÈx#k†f(x)
œ k dx
x [f(x)]# , by the chain rule
[f(x)]#
w
w
dr
d È #
Ê 'C F † T ds œ 'C F † dx
dx œ 'a k dx
x [f(x)]# dx œ k Èx# [f(x)]# ‘ a
b
b
œ k ˆÈb# [f(b)]# Èa# [f(a)]# ‰ , as claimed.
47. F œ 4t$ i 8t# j 2k and ddtr œ i 2tj Ê F † ddtr œ 12t$ Ê Flow œ '0 12t$ dt œ c3t% d ! œ 48
2
#
48. F œ 12t# j 9t# k and ddtr œ 3j 4k Ê F † ddtr œ 72t# Ê Flow œ '0 72t# dt œ c24t$ d ! œ 24
1
"
49. F œ (cos t sin t)i (cos t)k and ddtr œ ( sin t)i (cos t)k Ê F † ddtr œ sin t cos t 1
Ê Flow œ '0 ( sin t cos t 1) dt œ 2" cos# t t‘ ! œ ˆ #" 1‰ ˆ #" 0‰ œ 1
1
1
50. F œ (2 sin t)i (2 cos t)j 2k and ddtr œ (2 sin t)i (2 cos t)j 2k Ê F † ddtr œ 4 sin# t 4 cos# t 4 œ 0
Ê Flow œ 0
51. C" : r œ (cos t)i (sin t)j tk , 0 Ÿ t Ÿ 1# Ê F œ (2 cos t)i 2tj (2 sin t)k and ddtr œ ( sin t)i (cos t)j k
Ê F † ddtr œ 2 cos t sin t 2t cos t 2 sin t œ sin 2t 2t cos t 2 sin t
Ê Flow" œ '0
1Î2
1Î#
( sin 2t 2t cos t 2 sin t) dt œ 2" cos 2t 2t sin t 2 cos t 2 cos t‘ !
œ 1 1;
C# : r œ j 1# (1 t)k , 0 Ÿ t Ÿ 1 Ê F œ 1(1 t)j 2k and ddtr œ 1# k Ê F † ddtr œ 1
Ê Flow# œ '0 1 dt œ c1td "! œ 1;
1
C$ : r œ ti (1 t)j , 0 Ÿ t Ÿ 1 Ê F œ 2ti 2(1 t)k and ddtr œ i j Ê F † ddtr œ 2t
Ê Flow$ œ '0 2t dt œ ct# d ! œ 1 Ê Circulation œ (1 1) 1 1 œ 0
1
"
dy
dz
` f dx
` f dy
` f dz
"
dr
d
#
#
#
52. F † ddtr œ x dx
dt y dt z dt œ ` x dt ` y dt ` z dt , where f(xß yß z) œ # ax y x b Ê F † dt œ dt afaratbbb
by the chain rule Ê Circulation œ 'C F † ddtr dt œ 'a dtd afaratbbb dt œ farabbb faraabb. Since C is an entire ellipse,
b
rabb œ raab, thus the Circulation œ 0.
53. Let x œ t be the parameter Ê y œ x# œ t# and z œ x œ t Ê r œ ti t# j tk , 0 Ÿ t Ÿ 1 from (0ß 0ß 0) to (1ß 1ß 1)
Ê ddtr œ i 2tj k and F œ xyi yj yzk œ t$ i t# j t$ k Ê F † ddtr œ t$ 2t$ t$ œ 2t$ Ê Flow œ '0 2t$ dt
1
œ "#
` f dy
` z dz
df
dr
# $
)
54. (a) F œ ™ axy# z$ b Ê F † ddtr œ `` xf dx
dt ` y dt ` z dt œ dt , where f(xß yß z) œ xy z Ê C F † dt dt
œ 'a dtd afaratbbb dt œ farabbb faraabb œ 0 since C is an entire ellipse.
b
Ð2ß1ß 1Ñ
(b) 'C F † ddtr œ 'Ð1ß1ß1Ñ
d
# $
# $ Ð#ß"ß"Ñ
#
$
#
$
dt axy z b dt œ cxy z d Ð"ß"ß"Ñ œ (2)(1) (1) (1)(1) (1) œ 2 1 œ 3
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
952
Chapter 16 Integration in Vector Fields
55-60. Example CAS commands:
Maple:
with( LinearAlgebra );#55
F := r -> < r[1]*r[2]^6 | 3*r[1]*(r[1]*r[2]^5+2) >;
r := t -> < 2*cos(t) | sin(t) >;
a,b := 0,2*Pi;
dr := map(diff,r(t),t);
# (a)
F(r(t));
# (b)
q1 := simplify( F(r(t)) . dr ) assuming t::real;
# (c)
q2 := Int( q1, t=a..b );
value( q2 );
Mathematica: (functions and bounds will vary):
Exercises 55 and 56 use vectors in 2 dimensions
Clear[x, y, t, f, r, v]
f[x_, y_]:= {x y6 , 3x (x y5 2)}
{a, b}={0, 21};
x[t_]:= 2 Cos[t]
y[t_]:= Sin[t]
r[t_]:={x[t], y[t]}
v[t_]:= r'[t]
integrand= f[x[t], y[t]] . v[t] //Simplify
Integrate[integrand,{t, a, b}]
N[%]
If the integration takes too long or cannot be done, use NIntegrate to integrate numerically. This is suggested for exercises
57 - 60 that use vectors in 3 dimensions. Be certain to leave spaces between variables to be multiplied.
Clear[x, y, z, t, f, r, v]
f[x_, y_, z_]:= {y y z Cos[x y z], x2 x z Cos[x y z], z x y Cos[x y z]}
{a, b}={0, 21};
x[t_]:= 2 Cos[t]
y[t_]:= 3 Sin[t]
z[t_]:= 1
r[t_]:={x[t], y[t], z[t]}
v[t_]:= r'[t]
integrand= f[x[t], y[t],z[t]] . v[t] //Simplify
NIntegrate[integrand,{t, a, b}]
16.3 PATH INDEPENDENCE, POTENTIAL FUNCTIONS, AND CONSERVATIVE FIELDS
1.
`P
`N `M
`P `N
`M
`y œ x œ `z , `z œ y œ `x , `x œ z œ `y
2.
`P
`N `M
`P `N
`M
` y œ x cos z œ ` z , ` z œ y cos z œ ` x , ` x œ sin z œ ` y
3.
`P
`N
` y œ 1 Á 1 œ ` z
5.
`N
`M
`x œ 0 Á 1 œ `y
6.
`P
`N `M
`P `N
`M
x
` y œ 0 œ ` z , ` z œ 0 œ ` x , ` x œ e sin y œ ` y
Ê Conservative
Ê Not Conservative
4.
Ê Conservative
`N
`M
` x œ 1 Á 1 œ ` y
Ê Not Conservative
Ê Not Conservative
Ê Conservative
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 16.3 Path Independence, Potential Functions, and Conservative Fields
7.
`f
` x œ 2x
#
953
#
Ê f(xß yß z) œ x# g(yß z) Ê `` yf œ `` yg œ 3y Ê g(yß z) œ 3y# h(z) Ê f(xß yß z) œ x# 3y# h(z)
#
Ê `` zf œ hw (z) œ 4z Ê h(z) œ 2z# C Ê f(xß yß z) œ x# 3y# 2z# C
8.
`f
`x œ y z
Ê f(xß yß z) œ (y z)x g(yß z) Ê `` yf œ x `` yg œ x z Ê `` yg œ z Ê g(yß z) œ zy h(z)
Ê f(xß yß z) œ (y z)x zy h(z) Ê `` zf œ x y hw (z) œ x y Ê hw (z) œ 0 Ê h(z) œ C Ê f(xß yß z)
œ (y z)x zy C
9.
`f
y2z
`x œ e
y2z
œ xe
Ê f(xß yß z) œ xey2z g(yß z) Ê `` yf œ xey2z `` yg œ xey2z Ê `` yg œ 0 Ê f(xß yß z)
h(z) Ê `` zf œ 2xey2z hw (z) œ 2xey2z Ê hw (z) œ 0 Ê h(z) œ C Ê f(xß yß z) œ xey2z C
10. `` xf œ y sin z Ê f(xß yß z) œ xy sin z g(yß z) Ê `` yf œ x sin z `` yg œ x sin z Ê `` yg œ 0 Ê g(yß z) œ h(z)
Ê f(xß yß z) œ xy sin z h(z) Ê `` zf œ xy cos z hw (z) œ xy cos z Ê hw (z) œ 0 Ê h(z) œ C Ê f(xß yß z)
œ xy sin z C
11. `` zf œ y# z z# Ê f(xß yß z) œ #" ln ay# z# b g(xß y) Ê `` xf œ `` xg œ ln x sec# (x y) Ê g(xß y)
œ (x ln x x) tan (x y) h(y) Ê f(xß yß z) œ "# ln ay# z# b (x ln x x) tan (x y) h(y)
Ê `` yf œ y# y z# sec# (x y) hw (y) œ sec# (x y) y# y z# Ê hw (y) œ 0 Ê h(y) œ C Ê f(xß yß z)
œ "# ln ay# z# b (x ln x x) tan (x y) C
12. `` xf œ 1 yx# y# Ê f(xß yß z) œ tan" (xy) g(yß z) Ê `` yf œ 1 xx# y# `` yg œ 1 xx# y# È1 z y# z#
Ê `` gy œ È1 z y# z# Ê g(yß z) œ sin" (yz) h(z) Ê f(xß yß z) œ tan" (xy) sin" (yz) h(z)
Ê `` zf œ È1 y y# z# hw (z) œ È1 y y# z# "z Ê hw (z) œ "z Ê h(z) œ ln kzk C
Ê f(xß yß z) œ tan" (xy) sin" (yz) ln kzk C
13. Let F(xß yß z) œ 2xi 2yj 2zk Ê `` Py œ 0 œ ``Nz , ``Mz œ 0 œ `` Px , ``Nx œ 0 œ ``My Ê M dx N dy P dz is
exact; `` xf œ 2x Ê f(xß yß z) œ x# g(yß z) Ê `` yf œ `` yg œ 2y Ê g(yß z) œ y# h(z) Ê f(xß yß z) œ x# y# œ h(z)
Ð2ß3ß 6Ñ
Ê `` zf œ hw (z) œ 2z Ê h(z) œ z# C Ê f(xß yß z) œ x# y# z# C Ê 'Ð0ß0ß0Ñ 2x dx 2y dy 2z dz
œ f(2ß 3ß 6) f(!ß !ß !) œ 2# 3# (6)# œ 49
14. Let F(xß yß z) œ yzi xzj xyk Ê `` Py œ x œ ``Nz , ``Mz œ y œ `` Px , ``Nx œ z œ ``My Ê M dx N dy P dz is
exact; `` xf œ yz Ê f(xß yß z) œ xyz g(yß z) Ê `` yf œ xz `` yg œ xz Ê `` yg œ 0 Ê g(yß z) œ h(z) Ê f(xß yß z)
œ xyz h(z) Ê `` zf œ xy hw (z) œ xy Ê hw (z) œ 0 Ê h(z) œ C Ê f(xß yß z) œ xyz C
Ð3ß5ß0Ñ
Ê 'Ð1ß1ß2Ñ yz dx xz dy xy dz œ f(3ß 5ß 0) f(1ß 1ß 2) œ 0 2 œ 2
15. Let F(xß yß z) œ 2xyi ax# z# b j 2yzk Ê `` Py œ 2z œ ``Nz , ``Mz œ 0 œ `` Px , ``Nx œ 2x œ ``My
Ê M dx N dy P dz is exact; `` xf œ 2xy Ê f(xß yß z) œ x# y g(yß z) Ê `` yf œ x# `` gy œ x# z# Ê `` gy œ z#
Ê g(yß z) œ yz# h(z) Ê f(xß yß z) œ x# y yz# h(z) Ê `` zf œ 2yz hw (z) œ 2yz Ê hw (z) œ 0 Ê h(z) œ C
Ê f(xß yß z) œ x# y yz# C Ê 'Ð0ß0ß0Ñ 2xy dx ax# z# b dy 2yz dz œ f("ß #ß $) f(!ß !ß !) œ 2 2(3)# œ 16
Ð1ß2ß3Ñ
16. Let F(xß yß z) œ 2xi y# j ˆ 1 4 z# ‰ k Ê `` Py œ 0 œ ``Nz , ``Mz œ 0 œ `` Px , ``Nx œ 0 œ ``My
$
Ê M dx N dy P dz is exact; `` xf œ 2x Ê f(xß yß z) œ x# g(yß z) Ê `` yf œ `` yg œ y# Ê g(yß z) œ y3 h(z)
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
954
Chapter 16 Integration in Vector Fields
$
Ê f(xß yß z) œ x# y3 h(z) Ê `` zf œ hw (z) œ 1 4 z# Ê h(z) œ 4 tan" z C Ê f(xß yß z)
œ x# y3 4 tan" z C Ê 'Ð0ß0ß0Ñ 2x dx y# dy 1 4 z# dz œ f(3ß 3ß 1) f(!ß !ß !)
Ð3ß3ß1Ñ
$
1‰
œ ˆ9 27
3 4 † 4 (! ! 0) œ 1
17. Let F(xß yß z) œ (sin y cos x)i (cos y sin x)j k Ê `` Py œ 0 œ ``Nz , ``Mz œ 0 œ `` Px , ``Nx œ cos y cos x œ ``My
Ê M dx N dy P dz is exact; `` xf œ sin y cos x Ê f(xß yß z) œ sin y sin x g(yß z) Ê `` yf œ cos y sin x `` gy
œ cos y sin x Ê `` gy œ 0 Ê g(yß z) œ h(z) Ê f(xß yß z) œ sin y sin x h(z) Ê `` zf œ hw (z) œ 1 Ê h(z) œ z C
Ê f(xß yß z) œ sin y sin x z C Ê 'Ð1ß0ß0Ñ sin y cos x dx cos y sin x dy dz œ f(0ß 1ß 1) f(1ß !ß !)
Ð0ß1ß1Ñ
œ (0 1) (0 0) œ 1
18. Let F(xß yß z) œ (2 cos y)i Š "y 2x sin y‹ j ˆ "z ‰ k Ê `` Py œ 0 œ ``Nz , ``Mz œ 0 œ `` Px , ``Nx œ 2 sin y œ ``My
Ê M dx N dy P dz is exact; `` xf œ 2 cos y Ê f(xß yß z) œ 2x cos y g(yß z) Ê `` yf œ 2x sin y `` gy
œ y" 2x sin y Ê `` yg œ y" Ê g(yß z) œ ln kyk h(z) Ê f(xß yß z) œ 2x cos y ln kyk h(z) Ê `` zf œ hw (z) œ "z
Ê h(z) œ ln kzk C Ê f(xß yß z) œ 2x cos y ln kyk ln kzk C
Ê 'Ð0ß2ß1Ñ
Ð1ß1Î2ß2Ñ
2 cos y dx Š "y 2x sin y‹ dy "z dz œ f ˆ1ß 1# ß 2‰ f(!ß #ß ")
œ ˆ2 † 0 ln 1# ln 2‰ (0 † cos 2 ln 2 ln 1) œ ln 1#
#
`N `M
`P `N
`M
19. Let F(xß yß z) œ 3x# i Š zy ‹ j (2z ln y)k Ê `` Py œ 2z
y œ `z , `z œ 0 œ `x , `x œ 0 œ `y
Ê M dx N dy P dz is exact; `` xf œ 3x# Ê f(xß yß z) œ x$ g(yß z) Ê `` yf œ `` gy œ zy Ê g(yß z) œ z# ln y h(z)
#
Ê f(xß yß z) œ x$ z# ln y h(z) Ê `` zf œ 2z ln y hw (z) œ 2z ln y Ê hw (z) œ 0 Ê h(z) œ C Ê f(xß yß z)
œ x$ z# ln y C Ê 'Ð1ß1ß1Ñ 3x# dx zy dy 2z ln y dz œ f(1ß 2ß 3) f("ß "ß ")
Ð1ß2ß3Ñ
#
œ (1 9 ln 2 C) (1 0 C) œ 9 ln 2
#
`M
20. Let F(xß yß z) œ (2x ln y yz)i Š xy xz‹ j (xy)k Ê `` Py œ x œ ``Nz , ``Mz œ y œ `` Px , ``Nx œ 2x
y z œ `y
Ê M dx N dy P dz is exact; `` xf œ 2x ln y yz Ê f(xß yß z) œ x# ln y xyz g(yß z) Ê `` yf œ xy xz `` gy
#
œ xy xz Ê `` yg œ 0 Ê g(yß z) œ h(z) Ê f(xß yß z) œ x# ln y xyz h(z) Ê `` zf œ xy hw (z) œ xy Ê hw (z) œ 0
#
Ê h(z) œ C Ê f(xß yß z) œ x# ln y xyz C Ê 'Ð1ß2ß1Ñ (2x ln y yz) dx Š xy xz‹ dy xy dz
Ð2ß1ß1Ñ
#
œ f(2ß 1ß 1) f("ß 2ß 1) œ (4 ln 1 2 C) (ln 2 2 C) œ ln 2
21. Let F(xß yß z) œ Š "y ‹ i Š 1z yx# ‹ j ˆ zy# ‰ k Ê `` Py œ z"# œ ``Nz , ``Mz œ 0 œ `` Px , ``Nx œ y1# œ ``My
Ê M dx N dy P dz is exact; `` xf œ "y Ê f(xß yß z) œ xy g(yß z) Ê `` yf œ yx# `` gy œ "z yx#
Ê `` gy œ "z Ê g(yß z) œ yz h(z) Ê f(xß yß z) œ xy yz h(z) Ê `` zf œ zy# hw (z) œ zy# Ê hw (z) œ 0 Ê h(z) œ C
Ê f(xß yß z) œ yx yz C Ê 'Ð1ß1ß1Ñ "y dx Š 1z yx# ‹ dy zy# dz œ f(2ß 2ß 2) f("ß 1ß 1) œ ˆ 2# 2# C‰ ˆ "1 "1 C‰
Ð2ß2ß2Ñ
œ0
22. Let F(xß yß z) œ 2xxi#2yy#j z2z# k Šand let 3# œ x# y# z# Ê `` 3x œ 3x , `` 3y œ 3y , `` 3z œ 3z ‹
Ê `` Py œ 4yz
œ ``Nz , ``Mz œ 4xz
œ `` Px , ``Nx œ 4xy
œ ``My Ê M dx N dy P dz is exact;
3%
3%
3%
`f
2x
` x œ x # y # z#
`g
2y
Ê f(xß yß z) œ ln ax# y# z# b g(yß z) Ê `` yf œ x# 2y
y # z # ` y œ x # y # z#
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 16.3 Path Independence, Potential Functions, and Conservative Fields
955
w
Ê `` gy œ 0 Ê g(yß z) œ h(z) Ê f(xß yß z) œ ln ax# y# z# b h(z) Ê `` zf œ x# 2z
y# z# h (z)
w
#
#
#
œ x# 2z
y# z# Ê h (z) œ 0 Ê h(z) œ C Ê f(xß yß z) œ ln ax y z b C
Ð2ß2ß2Ñ
Ê 'Ð 1ß 1ß 1Ñ 2x dxx# 2yy#dyz#2z dz œ f(2ß 2ß 2) f("ß 1ß 1) œ ln 12 ln 3 œ ln 4
23. r œ (i j k) t(i 2j 2k) œ (1 t)i (1 2t)j (1 2t)k, 0 Ÿ t Ÿ 1 Ê dx œ dt, dy œ 2 dt, dz œ 2 dt
Ð2ß3ß 1Ñ
Ê 'Ð1ß1ß1Ñ y dx x dy 4 dz œ '0 (2t 1) dt (t 1)(2 dt) 4(2) dt œ '0 (4t 5) dt œ c2t# 5td ! œ 3
1
1
"
24. r œ t(3j 4k), 0 Ÿ t Ÿ 1 Ê dx œ 0, dy œ 3 dt, dz œ 4 dt Ê 'Ð0ß0ß0Ñ x# dx yz dy Š y# ‹ dz
Ð0ß3ß4Ñ
#
œ '0 a12t# b (3 dt) Š 9t# ‹ (4 dt) œ '0 54t# dt œ c18t# d ! œ 18
1
1
#
"
25. `` Py œ 0 œ ``Nz , ``Mz œ 2z œ `` Px , ``Nx œ 0 œ ``My Ê M dx N dy P dz is exact Ê F is conservative
Ê path independence
26. `` Py œ ˆÈ # yz#
x y z# ‰
$
œ ``Nz , ``Mz œ ˆÈ # xz#
x y z# ‰
$
œ `` Px , ``Nx œ ˆÈ # xy#
x y z# ‰
$
œ ``My
Ê M dx N dy P dz is exact Ê F is conservative Ê path independence
`M
27. `` Py œ 0 œ ``Nz , ``Mz œ 0 œ `` Px , ``Nx œ 2x
y# œ ` y Ê F is conservative Ê there exists an f so that F œ ™ f;
`f
2x
`x œ y
#
#
#
Ê f(xß y) œ xy g(y) Ê `` yf œ xy# gw (y) œ 1 y#x Ê gw (y) œ y"# Ê g(y) œ "y C
#
#
Ê f(xß y) œ xy y" C Ê F œ ™ Š x y 1 ‹
28. `` Py œ cos z œ ``Nz , ``Mz œ 0 œ `` Px , ``Nx œ ey œ ``My Ê F is conservative Ê there exists an f so that F œ ™ f;
x
`f
x
` x œ e ln y
Ê f(xß yß z) œ ex ln y g(yß z) Ê `` yf œ ey `` yg œ ey sin z Ê `` yg œ sin z Ê g(yß z)
x
x
œ y sin z h(z) Ê f(xß yß z) œ ex ln y y sin z h(z) Ê `` zf œ y cos z hw (z) œ y cos z Ê hw (z) œ 0
Ê h(z) œ C Ê f(xß yß z) œ ex ln y y sin z C Ê F œ ™ aex ln y y sin zb
29. `` Py œ 0 œ ``Nz , ``Mz œ 0 œ `` Px , ``Nx œ 1 œ ``My Ê F is conservative Ê there exists an f so that F œ ™ f;
`f
#
`x œ x y
Ê f(xß yß z) œ "3 x$ xy g(yß z) Ê `` yf œ x `` gy œ y# x Ê `` yg œ y# Ê g(yß z) œ "3 y$ h(z)
Ê f(xß yß z) œ "3 x$ xy 3" y$ h(z) Ê `` zf œ hw (z) œ zez Ê h(z) œ zez ez C Ê f(xß yß z)
œ "3 x$ xy 3" y$ zez ez C Ê F œ ™ ˆ "3 x$ xy 3" y$ zez ez ‰
(a) work œ 'A F † ddtr dt œ 'A F † dr œ 3" x$ xy 3" y$ zez ez ‘ Ð"ß!ß!Ñ œ ˆ 3" 0 0 e e‰ ˆ 3" 0 0 1‰
B
B
Ð"ß!ß"Ñ
œ1
(b) work œ 'A F † dr œ "3 x$ xy 3" y$ zez ez ‘ Ð"ß!ß!Ñ œ 1
B
Ð"ß!ß"Ñ
(c) work œ 'A F † dr œ "3 x$ xy 3" y$ zez ez ‘ Ð"ß!ß!Ñ œ 1
B
Ð"ß!ß"Ñ
Note: Since F is conservative, 'A F † dr is independent of the path from (1ß 0ß 0) to (1ß 0ß 1).
B
30. `` Py œ xeyz xyzeyz cos y œ ``Nz , ``Mz œ yeyz œ `` Px , ``Nx œ zeyz œ ``My Ê F is conservative Ê there exists an f so
that F œ ™ f; `` xf œ eyz Ê f(xß yß z) œ xeyz g(yß z) Ê `` yf œ xzeyz `` yg œ xzeyz z cos y Ê `` gy œ z cos y
Ê g(yß z) œ z sin y h(z) Ê f(xß yß z) œ xeyz z sin y h(z) Ê `` zf œ xyeyz sin y hw (z) œ xyeyz sin y
Ê hw (z) œ 0 Ê h(z) œ C Ê f(xß yß z) œ xeyz z sin y C Ê F œ ™ axeyz z sin yb
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
956
Chapter 16 Integration in Vector Fields
(a) work œ 'A F † dr œ cxeyz z sin yd Ð"ß!ß"Ñ
B
Ð"ß1Î#ß!Ñ
œ (1 0) (1 0) œ 0
(b) work œ 'A F † dr œ cxeyz z sin yd Ð"ß!ß"Ñ
B
Ð"ß1Î#ß!Ñ
(c) work œ 'A F † dr œ cxeyz z sin yd Ð"ß!ß"Ñ
B
Ð"ß1Î#ß!Ñ
œ0
œ0
Note: Since F is conservative, 'A F † dr is independent of the path from (1ß 0ß 1) to ˆ1ß 1# ß 0‰ .
B
31. (a) F œ ™ ax$ y# b Ê F œ 3x# y# i 2x$ yj ; let C" be the path from (1ß 1) to (0ß 0) Ê x œ t 1 and
y œ t 1, 0 Ÿ t Ÿ 1 Ê F œ 3(t 1)# (t 1)# i 2(t 1)$ (t 1)j œ 3(t 1)% i 2(t 1)% j
and r" œ (t 1)i (t 1)j Ê dr" œ dt i dt j Ê 'C F † dr" œ '0 c3(t 1)% 2(t 1)% d dt
1
"
1
"
œ '0 5(t 1)% dt œ c(t 1)& d ! œ 1; let C# be the path from (0ß 0) to (1ß 1) Ê x œ t and y œ t,
1
0 Ÿ t Ÿ 1 Ê F œ 3t% i 2t% j and r# œ ti tj Ê dr# œ dt i dt j Ê 'C F † dr# œ '0 a3t% 2t% b dt
1
œ '0 5t% dt œ 1 Ê 'C F † dr œ 'C F † dr" 'C F † dr# œ 2
"
#
#
Ð1ß1Ñ
(b) Since f(xß y) œ x$ y# is a potential function for F, 'Ð 1ß1Ñ F † dr œ f(1ß 1) f(1ß 1) œ 2
32. `` Py œ 0 œ ``Nz , ``Mz œ 0 œ `` Px , ``Nx œ 2x sin y œ ``My Ê F is conservative Ê there exists an f so that F œ ™ f;
`f
` x œ 2x cos y
Ê f(xß yß z) œ x# cos y g(yß z) Ê `` yf œ x# sin y `` yg œ x# sin y Ê `` yg œ 0 Ê g(yß z) œ h(z)
Ê f(xß yß z) œ x# cos y h(z) Ê `` zf œ hw (z) œ 0 Ê h(z) œ C Ê f(xß yß z) œ x# cos y C Ê F œ ™ ax# cos yb
(a) 'C 2x cos y dx x# sin y dy œ cx# cos yd Ð"ß!Ñ œ 0 1 œ 1
Ð!ß"Ñ
(b) 'C 2x cos y dx x# sin y dy œ cx# cos yd Ð"ß1Ñ œ 1 (1) œ 2
Ð"ß!Ñ
(c) 'C 2x cos y dx x# sin y dy œ cx# cos yd Ð"ß!Ñ œ 1 1 œ 0
Ð"ß!Ñ
(d) 'C 2x cos y dx x# sin y dy œ cx# cos yd Ð"ß!Ñ œ 1 1 œ 0
Ð"ß!Ñ
33. (a) If the differential form is exact, then `` Py œ ``Nz Ê 2ay œ cy for all y Ê 2a œ c, ``Mz œ `` Px Ê 2cx œ 2cx for
all x, and ``Nx œ ``My Ê by œ 2ay for all y Ê b œ 2a and c œ 2a
(b) F œ ™ f Ê the differential form with a œ 1 in part (a) is exact Ê b œ 2 and c œ 2
34. F œ ™ f Ê g(xß yß z) œ 'Ð0ß0ß0Ñ F † dr œ 'Ð0ß0ß0Ñ ™ f † dr œ f(xß yß z) f(0ß 0ß 0) Ê `` gx œ `` xf 0, `` yg œ `` yf 0, and
ÐxßyßzÑ
`g
`f
`z œ `z 0
ÐxßyßzÑ
Ê ™ g œ ™ f œ F, as claimed
35. The path will not matter; the work along any path will be the same because the field is conservative.
36. The field is not conservative, for otherwise the work would be the same along C" and C# .
37. Let the coordinates of points A and B be axA , yA , zA b and axB , yB , zB b, respectively. The force F œ ai bj ck is
conservative because all the partial derivatives of M, N, and P are zero. Therefore, the potential function is
fax, y, zb œ ax by cz C, and the work done by the force in moving a particle along any path from A to B is
faBb faAb œ f axB , yB , zB b faxA , yA , zA b œ aaxB byB czB Cb aaxA byA czA Cb
Ä
œ aaxB xA b bayB yA b cazB zA b œ F † BA
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 16.4 Green's Theorem in the Plane
38. (a) Let GmM œ C Ê F œ C ’
Ê
x
i # #y # $Î# j # #z # $Î# k“
ax# y# z# b$Î#
ax y z b
ax y z b
3yzC
3xyC
`P
`N `M
3xzC
`P `N
`M
` y œ ax# y# z# b&Î# œ ` z , ` z œ ax# y# z# b&Î# œ ` x , ` x œ ax# y# z# b&Î# œ ` y
Ê F œ ™ f for
yC
`g
`f
` y œ ax# y# z# b$Î# ` y
some f; `` xf œ
xC
Ê f(xß yß z) œ # #C # "Î# g(yß z) Ê
ax# y# z# b$Î#
ax y z b
`g
yC
œ # # # $Î# Ê ` y œ 0 Ê g(yß z) œ h(z) Ê `` zf œ # zC
hw (z) œ # zC
ax y z b
ax y# z# b$Î#
ax y# z# b$Î#
Ê h(z) œ C" Ê f(xß yß z) œ C
C" .
ax# y# z# b"Î#
Let C" œ 0 Ê f(xß yß z) œ
GmM
is a potential
ax# y# z# b"Î#
function for F.
(b) If s is the distance of (xß yß z) from the origin, then s œ Èx# y# z# . The work done by the gravitational field
F is work œ 'P F † dr œ ’ Èx#GmM
“
y # z#
P#
T#
"
T"
"
"
GmM
œ GmM
s# s" œ GmM Š s# s" ‹ , as claimed.
16.4 GREEN'S THEOREM IN THE PLANE
1. M œ y œ a sin t, N œ x œ a cos t, dx œ a sin t dt, dy œ a cos t dt Ê ``Mx œ 0, ``My œ 1, ``Nx œ 1, and
`N
` y œ 0;
Equation (3): )C M dy N dx œ '0 [(a sin t)(a cos t) (a cos t)(a sin t)] dt œ '0 0 dt œ 0;
21
21
' ' Š ``Mx ``Ny ‹ dx dy œ ' ' 0 dx dy œ 0, Flux
R
R
21
21
Equation (4): )C M dx N dy œ '0 [(a sin t)(a sin t) (a cos t)(a cos t)] dt œ '0 a# dt œ 21a# ;
' ' Š ``Nx ``My ‹ dx dy œ ' '
a
Èa c x
#
ca cc
R
#
2 dy dx œ 'ca 4Èa# x# dx œ 4 ’ x2 Èa# x# a# sin" xa “
a
#
a
ca
œ 2a# ˆ 1# 1# ‰ œ 2a# 1, Circulation
2. M œ y œ a sin t, N œ 0, dx œ a sin t dt, dy œ a cos t dt Ê ``Mx œ 0, ``My œ 1, ``Nx œ 0, and ``Ny œ 0;
#1
Equation (3): )C M dy N dx œ '0 a# sin t cos t dt œ a# 2" sin# t‘ ! œ 0; ' ' 0 dx dy œ 0, Flux
21
R
21
#1
Equation (4): )C M dx N dy œ '0 aa# sin# tb dt œ a# 2t sin4 2t ‘ ! œ 1a# ; ' ' Š ``Nx ``My ‹ dx dy
œ ' ' 1 dx dy œ '0 '0 r dr d) œ '
21
a
R
21
#
a# d) œ 1a# , Circulation
0
R
3. M œ 2x œ 2a cos t, N œ 3y œ 3a sin t, dx œ a sin t dt, dy œ a cos t dt Ê ``Mx œ 2, ``My œ 0, ``Nx œ 0, and
`N
` y œ 3;
Equation (3): )C M dy N dx œ '0 [(2a cos t)(a cos t) (3a sin t)(a sin t)] dt
21
œ '0 a2a# cos# t 3a# sin# tb dt œ 2a# 2t sin4 2t ‘ ! 3a# 2t sin4 2t ‘ ! œ 21a# 31a# œ 1a# ;
21
#1
#1
' ' Š ``Mx ``Ny ‹ œ ' ' 1 dx dy œ ' ' r dr d) œ ' a## d) œ 1a# , Flux
0
0
0
21
R
a
21
R
21
Equation (4): )C M dx N dy œ '0 [(2a cos t)(a sin t) (3a sin t)(a cos t)] dt
21
#1
œ '0 a2a# sin t cos t 3a# sin t cos tb dt œ 5a# 12 sin# t‘ ! œ 0; ' ' 0 dx dy œ 0, Circulation
R
4. M œ x# y œ a$ cos# t, N œ xy# œ a$ cos t sin# t, dx œ a sin t dt, dy œ a cos t dt
Ê ``Mx œ 2xy, ``My œ x2 , ``Nx œ y# , and ``Ny œ 2xy;
Equation (3): )C M dy N dx œ '0 aa% cos$ t sin t a% cos t sin$ tb œ ’ a4 cos% t a4 sin% t“
21
%
%
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
#1
!
œ 0;
957
958
Chapter 16 Integration in Vector Fields
' ' Š ``Mx ``Ny ‹ dx dy œ ' ' (2xy 2xy) dx dy œ 0, Flux
R
R
21
21
Equation (4): )C M dx N dy œ '0 aa% cos# t sin# t a% cos# t sin# tb dt œ '0 a2a% cos# t sin# tb dt
21
41
%1
œ '0 "# a% sin# 2t dt œ a4 '0 sin# u du œ a4 u2 sin42u ‘ ! œ 1#a ; ' ' Š ``Nx ``My ‹ dx dy œ ' ' ay# x# b dx dy
%
%
%
R
21
a
21
œ '0 '0 r# † r dr d) œ '0 a4 d) œ 1#a , Circulation
%
R
%
5. M œ x y, N œ y x Ê ``Mx œ 1, ``My œ 1, ``Nx œ 1, ``Ny œ 1 Ê Flux œ ' ' 2 dx dy œ '0 '0 2 dx dy œ 2;
1
1
R
Circ œ ' ' [1 (1)] dx dy œ 0
R
6. M œ x# 4y, N œ x y# Ê ``Mx œ 2x, ``My œ 4, ``Nx œ 1, ``Ny œ 2y Ê Flux œ ' ' (2x 2y) dx dy
R
1
1
1
1
"
"
œ '0 '0 (2x 2y) dx dy œ '0 cx# 2xyd ! dy œ '0 (1 2y) dy œ cy y# d ! œ 2; Circ œ ' ' (1 4) dx dy
R
œ '0 '0 3 dx dy œ 3
1
1
7. M œ y# x# , N œ x# y# Ê ``Mx œ 2x, ``My œ 2y, ``Nx œ 2x, ``Ny œ 2y Ê Flux œ ' ' (2x 2y) dx dy
œ '0 '0 (2x 2y) dy dx œ '0 a2x
3
x
3
R
#
$
x b dx œ "3 x$ ‘ ! œ 9; Circ œ
#
' ' (2x 2y) dx dy
3
x
3
œ '0 '0 (2x 2y) dy dx œ '0 x# dx œ 9
R
8. M œ x y, N œ ax# y# b Ê ``Mx œ 1, ``My œ 1, ``Nx œ 2x, ``Ny œ 2y Ê Flux œ ' ' (1 2y) dx dy
R
1
x
1
1
x
œ '0 '0 (1 2y) dy dx œ '0 ax x# b dx œ "6 ; Circ œ ' ' (2x 1) dx dy œ '0 '0 (2x 1) dy dx
œ '0 a2x
1
#
R
xb dx œ 76
9. M œ xy y2 , N œ x y Ê ``Mx œ y, ``My œ x 2y, ``Nx œ 1, ``Ny œ 1 Ê Flux œ ' ' ay a1bb dy dx
R
Èx
' ' a1 ax 2ybb dy dx
œ '0 'x2 ay 1b dy dx œ '0 ˆ "# x Èx "# x4 x# ‰ dx œ 11
60 ; Circ œ
1
1
R
Èx
7
œ '0 'x2 a1 x 2yb dy dx œ '0 ˆÈx x3Î2 x x# x3 x4 ‰ dx œ 60
1
1
10. M œ x 3y, N œ 2x y Ê ``Mx œ 1, ``My œ 3, ``Nx œ 2, ``Ny œ 1 Ê Flux œ ' ' a1 a1bb dy dx œ 0
R
È2
È2
2 x Î2
Circ œ ' ' a2 3b dy dx œ 'cÈ2 'È 2 c x Î2 a1b dy dx œ È22 'cÈ2 È2 x2 dx œ 1È2
Èa
R
2b
a
2b
11. M œ x3 y2 , N œ "# x4 y Ê ``Mx œ 3x2 y2 , ``My œ 2x3 y, ``Nx œ 2x3 y, ``Ny œ "# x4 Ê Flux œ ' ' ˆ3x2 y2 "# x4 ‰ dy dx
R
' ' a2x3 y 2x3 yb dy dx œ 0
œ '0 'x2 x ˆ3x2 y2 "# x4 ‰ dy dx œ '0 ˆ3x5 72 x6 3x7 x8 ‰ dx œ 64
9 ; Circ œ
2
x
2
R
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 16.4 Green's Theorem in the Plane
12. M œ 1 x y2 , N œ tan1 y Ê ``Mx œ 1 1 y2 , ``My œ a12xy2yb2 , ``Nx œ 0, ``Ny œ 1 1 y2 Ê Flux œ ' ' Š 1 1 y2 1 1 y2 ‹ dx dy
R
È1 y2
œ 'c1 'È1 y2
1
2
1 y2 dx dy œ
'
4È 1 y2
dx œ 41È2 41 ; Circ œ
c1 1 y2
1
È1 y2
' ' Š0 Š 2x y ‹‹ dy dx
a1 y 2 b 2
R
y
œ 'c1 'È1 y2 Š a1 2x
‹ dy dx œ 'c1 a0b dx œ 0
y 2 b2
1
1
13. M œ x ex sin y, N œ x ex cos y Ê ``Mx œ 1 ex sin y, ``My œ ex cos y, ``Nx œ 1 ex cos y, ``Ny œ ex sin y
Ècos 2)
1Î4
Ê Flux œ ' ' dx dy œ 'c1Î4 '0
R
Î
1Î%
r dr d) œ ' 1Î4 ˆ "# cos 2)‰ d) œ 4" sin 2)‘ 1Î% œ "# ;
1 4
Ècos 2)
1Î4
Circ œ ' ' a1 ex cos y ex cos yb dx dy œ ' ' dx dy œ 'c1Î4 '0
R
R
Î
r dr d) œ ' 1Î4 ˆ "# cos 2)‰ d) œ "#
1 4
2y
`N
14. M œ tan" yx , N œ ln ax# y# b Ê ``Mx œ x#yy# , ``My œ x# x y# , ``Nx œ x# 2x
y# , ` y œ x# y#
' ' ˆ r sinr# ) ‰ r dr d) œ '0 sin ) d) œ 2;
Ê Flux œ ' ' Š x#yy# x# 2y
y# ‹ dx dy œ 0 1
1
R
1
2
x
' ' ˆ r cosr# ) ‰ r dr d) œ '0 cos ) d) œ 0
Circ œ ' ' Š x# 2x
y# x# y# ‹ dx dy œ 0 1
1
1
2
R
15. M œ xy, N œ y# Ê ``Mx œ y, ``My œ x, ``Nx œ 0, ``Ny œ 2y Ê Flux œ ' ' (y 2y) dy dx œ '0 'x# 3y dy dx
1
x
R
1
1
x
1
œ '0 Š 3x# 3x# ‹ dx œ 5" ; Circ œ ' ' x dy dx œ '0 'x x dy dx œ '0 ax# x$ b dx œ 1"#
#
%
#
R
16. M œ sin y, N œ x cos y Ê ``Mx œ 0, ``My œ cos y, ``Nx œ cos y, ``Ny œ x sin y
Ê Flux œ ' ' (x sin y) dx dy œ '0
1Î2
R
'01Î2 (x sin y) dx dy œ '01Î2 Š 18 sin y‹ dy œ 18 ;
#
Circ œ ' ' [cos y ( cos y)] dx dy œ '0
1Î2
R
'01Î2 2 cos y dx dy œ '01Î2 1 cos y dy œ c1 sin yd 1Î#
œ1
!
17. M œ 3xy 1 x y# , N œ ex tan " y Ê ``Mx œ 3y 1 " y# , ``Ny œ 1 " y#
aÐ1
Ê Flux œ ' ' Š3y 1 " y# 1 " y# ‹ dx dy œ ' ' 3y dx dy œ '0 '0
21
R
R
œ '0 a$ (1 cos ))$ (sin )) d) œ ’ a4 (1 cos ))% “
21
#
$
#1
!
cos )Ñ
(3r sin )) r dr d)
œ 4a$ a4a$ b œ 0
18. M œ y ex ln y, N œ ey Ê ``My œ 1 ey , ``Nx œ ey Ê Circ œ ' ' ’ ey Š1 ey ‹“ dx dy œ ' ' (1) dx dy
x
x
x
x
x
R
R
1
3cx
1
1
œ 'c1 'x b 1 dy dx œ 'c1 ca3 x# b ax% 1bd dx œ 'c1 ax% x# 2b dx œ 44
15
#
%
19. M œ 2xy$ , N œ 4x# y# Ê ``My œ 6xy# , ``Nx œ 8xy# Ê work œ )C 2xy$ dx 4x# y# dy œ ' ' a8xy# 6xy# b dx dy
œ '0 '0 2xy dy dx œ '
1
x$
#
R
1
2 "!
2
x dx œ 33
0 3
20. M œ 4x 2y, N œ 2x 4y Ê ``My œ 2, ``Nx œ 2 Ê work œ )C (4x 2y) dx (2x 4y) dy
œ ' ' [2 (2)] dx dy œ 4 ' ' dx dy œ 4(Area of the circle) œ 4(1 † 4) œ 161
R
R
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
959
960
Chapter 16 Integration in Vector Fields
21. M œ y# , N œ x# Ê ``My œ 2y, ``Nx œ 2x Ê )C y# dx x# dy œ ' ' (2x 2y) dy dx
1cx
œ '0 '0
1
R
(2x 2y) dy dx œ '0 a3x 4x 1b dx œ cx 2x# xd ! œ 1 2 1 œ 0
1
#
"
$
22. M œ 3y, N œ 2x Ê ``My œ 3, ``Nx œ 2 Ê )C 3y dx 2x dy œ ' ' a2 3b dx dy œ '0 '0
1
sin x
a1bdy dx
R
œ '0 sin x dx œ 2
1
23. M œ 6y x, N œ y 2x Ê ``My œ 6, ``Nx œ 2 Ê )C (6y x) dx (y 2x) dy œ ' ' (2 6) dy dx
R
œ 4(Area of the circle) œ 161
24. M œ 2x y# , N œ 2xy 3y Ê ``My œ 2y, ``Nx œ 2y Ê )C a2x y# b dx (2xy 3y) dy œ ' ' (2y 2y) dx dy œ 0
R
25. M œ x œ a cos t, N œ y œ a sin t Ê dx œ a sin t dt, dy œ a cos t dt Ê Area œ "# )C x dy y dx
œ "# '0 aa# cos# t a# sin# tb dt œ "# '0 a# dt œ 1a#
21
21
26. M œ x œ a cos t, N œ y œ b sin t Ê dx œ a sin t dt, dy œ b cos t dt Ê Area œ "# )C x dy y dx
œ "# '0 aab cos# t ab sin# tb dt œ "# '0 ab dt œ 1ab
21
21
27. M œ x œ cos$ t, N œ y œ sin$ t Ê dx œ 3 cos# t sin t dt, dy œ 3 sin# t cos t dt Ê Area œ "# )C x dy y dx
3 '
œ "# '0 a3 sin# t cos# tb acos# t sin# tb dt œ "# '0 a3 sin# t cos# tb dt œ 38 '0 sin# 2t dt œ 16
sin# u du
0
21
21
21
41
%1
3 u
sin 2u ‘
œ 16
œ 38 1
2 4
!
28. C1 : M œ x œ t, N œ y œ 0 Ê dx œ dt, dy œ 0; C2 : M œ x œ a21 tb sina21 tb œ 21 t sin t, N œ y
œ 1 cosa21 tb œ 1 cos t Ê dx œ acos t 1b dt, dy œ sin t dt
Ê Area œ "# )C x dy y dx œ "# )C x dy y dx "# )C x dy y dx
"
œ "#
2
'021 a0bdt "# '021 ca21 t sin tbasin tb a1 cos tb acos t 1bd dt œ "# '021 a2 cos t t sin t 2 21 sin tb dt
œ 12 c3 sin t t cos t 2t 21 cos td20 1 œ 31
29. (a) M œ f(x), N œ g(y) Ê ``My œ 0, ``Nx œ 0 Ê )C f(x) dx g(y) dy œ ' ' Š ``Nx ``My ‹ dx dy œ ' ' 0 dx dy œ 0
R
(b) M œ ky, N œ hx Ê
`M
`N
` y œ k, ` x œ h
R
Ê )C ky dx hx dy œ ' ' Š ``Nx ``My ‹ dx dy
œ ' ' (h k) dx dy œ (h k)(Area of the region)
R
R
30. M œ xy# , N œ x# y 2x Ê ``My œ 2xy, ``Nx œ 2xy 2 Ê )C xy# dx ax# y 2xb dy œ ' ' Š ``Nx ``My ‹ dx dy
œ ' ' (2xy 2 2xy) dx dy œ 2 ' ' dx dy œ 2 times the area of the square
R
R
R
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 16.4 Green's Theorem in the Plane
961
31. The integral is 0 for any simple closed plane curve C. The reasoning: By the tangential form of Green's
Theorem, with M œ 4x$ y and N œ x% , )C 4x$ y dx x% dy œ ' ' ’ ``x ax% b ``y a4x$ yb“ dx dy
R
œ ' ' ðóóñóóò
a4x$ 4x$ b dx dy œ 0.
R
0
32. The integral is 0 for any simple closed curve C. The reasoning: By the normal form of Green's theorem, with
`
`
$
$
M œ x$ and N œ y$ , )C y$ dy x$ dx œ ' ' ”ðñò
` x ay b ï
` y ax b • dx dy œ 0.
R
0
0
33. Let M œ x and N œ 0 Ê ``Mx œ 1 and ``Ny œ 0 Ê )C M dy N dx œ ' ' Š ``Mx ``Ny ‹ dx dy Ê )C x dy
R
œ ' ' (1 0) dx dy Ê Area of R œ ' ' dx dy œ ) x dy; similarly, M œ y and N œ 0 Ê ``My œ 1 and
R
C
R
Ê )C M dx N dy œ ' ' Š ``Nx ``My ‹ dy dx Ê )C y dx œ ' ' (0 1) dy dx Ê )C y dx
`N
`x œ 0
R
œ ' ' dx dy œ Area of R
R
R
34. 'a f(x) dx œ Area of R œ )C y dx, from Exercise 33
b
' ' x $ (xßy) dA
' ' x dA
y
R
R
35. Let $ (xß y) œ 1 Ê x œ M
M œ ' ' $ (xßy) dA œ ' ' dA œ
R
' ' x dA
Ê Ax œ ' ' x dA œ ' ' (x 0) dx dy
R
A
R
R
R
R
R
œ )C x# dy, Ax œ ' ' x dA œ ' ' (0 x) dx dy œ ) xy dx, and Ax œ ' ' x dA œ ' ' ˆ 23 x "3 x‰ dx dy
#
R
œ)
R
" #
x dy 3" xy dx
C 3
Ê
"
#
C
)C x dy œ )
xy dx œ 3"
C
#
)C x dy xy dx œ Ax
#
36. If $ (xß y) œ 1, then Iy œ ' ' x# $ (xß y) dA œ ' ' x# dA œ ' ' ax# 0b dy dx œ "3 )C x$ dy,
R
R
R
' ' x# dA œ ' ' a0 x# b dy dx œ ) x# y dx, and ' ' x# dA œ ' ' ˆ 34 x# 4" x# ‰ dy dx
R
C
R
œ)
" $
x dy 4" x# y dx œ 4"
C 4
R
R
)C x dy x y dx Ê )C x dy œ )C x# y dx œ 4" )C x$ dy x# y dx œ Iy
$
"
3
#
$
37. M œ `` yf , N œ `` xf Ê ``My œ `` yf# , ``Nx œ `` xf# Ê )C `` yf dx `` xf dy œ ' ' Š `` xf# `` yf# ‹ dx dy œ 0 for such curves C
#
#
#
#
R
38. M œ "4 x# y "3 y$ , N œ x Ê ``My œ 14 x# y# , ``Nx œ 1 Ê Curl œ ``Nx ``My œ 1 ˆ "4 x# y# ‰ 0 in the interior of the
ellipse "4 x# y# œ 1 Ê work œ 'C F † dr œ ' ' ˆ1 4" x# y# ‰ dx dy will be maximized on the region
R
R œ {(xß y) | curl F}
0 or over the region enclosed by 1 œ "4 x# y#
2y
2y
2x
39. (a) ™ f œ Š x# 2x
y# ‹ i Š x# y# ‹ j Ê M œ x# y# , N œ x# y# ; since M, N are discontinuous at (0ß 0), we
compute 'C ™ f † n ds directly since Green's Theorem does not apply. Let x œ a cos t, y œ a sin t Ê dx œ a sin t dt,
dy œ a cos t dt, M œ 2a cos t, N œ 2a sin t, 0 Ÿ t Ÿ 21, so 'C ™ f † n ds œ 'C M dy N dx
œ '0 ˆ 2a cos t‰aa cos tb ˆ 2a sin t‰aa sin tb ‘dt œ '0 2acos2 t sin2 tbdt œ 41. Note that this holds for any
21
21
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
962
Chapter 16 Integration in Vector Fields
a 0, so 'C ™ f † n ds œ 41 for any circle C centered at a0, 0b traversed counterclockwise and 'C ™ f † n ds œ 41
if C is traversed clockwise.
(b) If K does not enclose the point (0ß 0) we may apply Green's Theorem: 'C ™ f † n ds œ 'C M dy N dx
œ ' ' Š ``Mx ``Ny ‹ dx dy œ ' ' Š ax2 y2 b2 ax2 y2 b2 ‹ dx dy œ ' ' 0 dx dy œ 0. If K does enclose the point
2 ˆy 2 x 2 ‰
R
2 ˆx 2 y 2 ‰
R
R
(0ß 0) we proceed as follows:
Choose a small enough so that the circle C centered at (0ß 0) of radius a lies entirely within K. Green's Theorem
applies to the region R that lies between K and C. Thus, as before, 0 œ ' ' Š ``Mx ``Ny ‹ dx dy
R
œ 'K M dy N dx 'C M dy N dx where K is traversed counterclockwise and C is traversed clockwise.
Hence by part (a) 0 œ ’ ' M dy N dx “ 41 Ê 41 œ ' M dy N dx œ 'K ™ f † n ds. We have shown:
K
K
'K ™ f † n ds œ œ 0
if (0ß 0) lies inside K
if (0ß 0) lies outside K
41
40. Assume a particle has a closed trajectory in R and let C" be the path Ê C" encloses a simply connected region
R" Ê C" is a simple closed curve. Then the flux over R" is )C F † n ds œ 0, since the velocity vectors F are
"
tangent to C" . But 0 œ )C F † n ds œ )C M dy N dx œ ' ' Š ``Mx ``Ny ‹ dx dy Ê Mx Ny œ 0, which is a
"
"
R"
contradiction. Therefore, C" cannot be a closed trajectory.
41. 'g ÐyÑ ``Nx dx dy œ N(g# (y)ß y) N(g" (y)ß y) Ê 'c 'g ÐyÑ ˆ ``Nx dx‰ dy œ 'c [N(g# (y)ß y) N(g" (y)ß y)] dy
g# ÐyÑ
d
"
g# ÐyÑ
d
"
d
d
d
c
œ 'c N(g# (y)ß y) dy 'c N(g" (y)ß y) dy œ 'c N(g# (y)ß y) dy 'd N(g" (y)ß y) dy œ 'C N dy 'C N dy
œ )C dy Ê )C N dy œ ' ' ``Nx dx dy
#
"
R
42. The curl of a conservative two-dimensional field is zero. The reasoning: A two-dimensional field F œ Mi Nj
can be considered to be the restriction to the xy-plane of a three-dimensional field whose k component is zero,
and whose i and j components are independent of z. For such a field to be conservative, we must have
`N
`M
`N
`M
` x œ ` y by the component test in Section 16.3 Ê curl F œ ` x ` y œ 0.
43-46. Example CAS commands:
Maple:
with( plots );#43
M := (x,y) -> 2*x-y;
N := (x,y) -> x+3*y;
C := x^2 + 4*y^2 = 4;
implicitplot( C, x=-2..2, y=-2..2, scaling=constrained, title="#43(a) (Section 16.4)" );
curlF_k := D[1](N) - D[2](M):
# (b)
'curlF_k' = curlF_k(x,y);
top,bot := solve( C, y );
# (c)
left,right := -2, 2;
q1 := Int( Int( curlF_k(x,y), y=bot..top ), x=left..right );
value( q1 );
Mathematica: (functions and bounds will vary)
The ImplicitPlot command will be useful for 43 and 44, but is not needed for 43 and 44. In 44, the equation of the line
from (0, 4) to (2, 0) must be determined first.
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 16.5 Surfaces and Area
963
Clear[x, y, f]
<<Graphics`ImplicitPlot`
f[x_, y_]:= {2x y, x 3y}
curve= x2 4y2 ==4
ImplicitPlot[curve, {x, 3, 3},{y, 2, 2}, AspectRatio Ä Automatic, AxesLabel Ä {x, y}];
ybounds= Solve[curve, y]
{y1, y2}=y/.ybounds;
integrand:=D[f[x,y][[2]], x] D[f[x,y][[1]], y]//Simplify
Integrate[integrand, {x, 2, 2}, {y, y1, y2}]
N[%]
Bounds for y are determined differently in 45 and 46. In 46, note equation of the line from (0, 4) to (2, 0).
Clear[x, y, f]
f[x_, y_]:= {x Exp[y], 4x2 Log[y]}
ybound = 4 2x
Plot[{0, ybound}, {x, 0,2. 1}, AspectRatio Ä Automatic, AxesLabel Ä {x, y}];
integrand:=D[f[x, y][[2]], x] D[f[x, y][[1]], y]//Simplify
Integrate[integrand, {x, 0, 2}, {y, 0, ybound}]
N[%]
16.5 SURFACES AND AREA
#
1. In cylindrical coordinates, let x œ r cos ), y œ r sin ), z œ ˆÈx# y# ‰ œ r# . Then r(rß )) œ (r cos ))i (r sin ))j r# k ,
0 Ÿ r Ÿ 2, 0 Ÿ ) Ÿ 21.
2. In cylindrical coordinates, let x œ r cos ), y œ r sin ), z œ 9 x# y# œ 9 r# . Then
r(rß )) œ (r cos ))i (r sin ))j a9 r# b k ; z 0 Ê 9 r# 0 Ê r# Ÿ 9 Ê 3 Ÿ r Ÿ 3, 0 Ÿ ) Ÿ 21. But
3 Ÿ r Ÿ 0 gives the same points as 0 Ÿ r Ÿ 3, so let 0 Ÿ r Ÿ 3.
È x# y#
Ê z œ #r . Then r(rß )) œ (r cos ))i (r sin ))j ˆ #r ‰ k .
#
r
For 0 Ÿ z Ÿ 3, 0 Ÿ # Ÿ 3 Ê 0 Ÿ r Ÿ 6; to get only the first octant, let 0 Ÿ ) Ÿ 1# .
3. In cylindrical coordinates, let x œ r cos ), y œ r sin ), z œ
4. In cylindrical coordinates, let x œ r cos ), y œ r sin ), z œ 2Èx# y# Ê z œ 2r. Then
r(rß )) œ (r cos ))i (r sin ))j 2rk . For 2 Ÿ z Ÿ 4, 2 Ÿ 2r Ÿ 4 Ê 1 Ÿ r Ÿ 2, and let 0 Ÿ ) Ÿ 21.
5. In cylindrical coordinates, let x œ r cos ), y œ r sin ); since x# y# œ r2 Ê z# œ 9 ax# y# b œ 9 r#
Ê z œ È9 r# , z 0. Then r(rß )) œ (r cos ))i (r sin ))j È9 r# k . Let 0 Ÿ ) Ÿ 21. For the domain
#
of r: z œ Èx# y# and x# y# z# œ 9 Ê x# y# ˆÈx# y# ‰ œ 9 Ê 2 ax# y# b œ 9 Ê 2r# œ 9
Ê r œ È32 Ê 0 Ÿ r Ÿ È32 .
6. In cylindrical coordinates, r(rß )) œ (r cos ))i (r sin ))j È4 r# k (see Exercise 5 above with x# y# z# œ 4,
instead of x# y# z# œ 9). For the first octant, let 0 Ÿ ) Ÿ 1# . For the domain of r: z œ Èx# y# and
#
x# y# z# œ 4 Ê x# y# ˆÈx# y# ‰ œ 4 Ê 2 ax# y# b œ 4 Ê 2r# œ 4 Ê r œ È2. Thus, let È2 Ÿ r Ÿ 2
(to get the portion of the sphere between the cone and the xy-plane).
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
964
Chapter 16 Integration in Vector Fields
7. In spherical coordinates, x œ 3 sin 9 cos ), y œ 3 sin 9 sin ), 3 œ Èx# y# z# Ê 3# œ 3 Ê 3 œ È3
Ê z œ È3 cos 9 for the sphere; z œ
È3
È3 cos 9
# œ
È
È
Ê cos 9 œ #" Ê 9 œ 13 ; z œ #3 Ê #3 œ È3 cos 9
Ê cos 9 œ "# Ê 9 œ 231 . Then r(9ß )) œ ŠÈ3 sin 9 cos )‹ i ŠÈ3 sin 9 sin )‹ j ŠÈ3 cos 9‹ k ,
1
21
3 Ÿ 9 Ÿ 3 and 0 Ÿ ) Ÿ 21.
8. In spherical coordinates, x œ 3 sin 9 cos ), y œ 3 sin 9 sin ), 3 œ Èx# y# z# Ê 3# œ 8 Ê 3 œ È8 œ 2È2
Ê x œ 2È2 sin 9 cos ), y œ 2È2 sin 9 sin ), and z œ 2È2 cos 9. Thus let
r(9ß )) œ Š2È2 sin 9 cos )‹ i Š2È2 sin 9 sin )‹ j Š2È2 cos 9‹ k ; z œ 2 Ê 2 œ 2È2 cos 9
Ê cos 9 œ È"2 Ê 9 œ 341 ; z œ 2È2 Ê 2È2 œ 2È2 cos 9 Ê cos 9 œ 1 Ê 9 œ 0. Thus 0 Ÿ 9 Ÿ 341 and
0 Ÿ ) Ÿ 21 .
9. Since z œ 4 y# , we can let r be a function of x and y Ê r(xß y) œ xi yj a4 y# b k . Then z œ 0
Ê 0 œ 4 y# Ê y œ „ 2. Thus, let 2 Ÿ y Ÿ 2 and 0 Ÿ x Ÿ 2.
10. Since y œ x# , we can let r be a function of x and z Ê r(xß z) œ xi x# j zk . Then y œ 2
Ê x# œ 2 Ê x œ „ È2. Thus, let È2 Ÿ x Ÿ È2 and 0 Ÿ z Ÿ 3.
11. When x œ 0, let y# z# œ 9 be the circular section in the yz-plane. Use polar coordinates in the yz-plane
Ê y œ 3 cos ) and z œ 3 sin ). Thus let x œ u and ) œ v Ê r(u,v) œ ui (3 cos v)j (3 sin v)k where
0 Ÿ u Ÿ 3, and 0 Ÿ v Ÿ 21.
12. When y œ 0, let x# z# œ 4 be the circular section in the xz-plane. Use polar coordinates in the xz-plane
Ê x œ 2 cos ) and z œ 2 sin ). Thus let y œ u and ) œ v Ê r(u,v) œ (2 cos v)i uj (3 sin v)k where
2 Ÿ u Ÿ 2, and 0 Ÿ v Ÿ 1 (since we want the portion above the xy-plane).
13. (a) x y z œ 1 Ê z œ 1 x y. In cylindrical coordinates, let x œ r cos ) and y œ r sin )
Ê z œ 1 r cos ) r sin ) Ê r(rß )) œ (r cos ))i (r sin ))j (1 r cos ) r sin ))k , 0 Ÿ ) Ÿ 21 and
0 Ÿ r Ÿ 3.
(b) In a fashion similar to cylindrical coordinates, but working in the yz-plane instead of the xy-plane, let
y œ u cos v, z œ u sin v where u œ Èy# z# and v is the angle formed by (xß yß z), (xß 0ß 0), and (xß yß 0)
with (xß 0ß 0) as vertex. Since x y z œ 1 Ê x œ 1 y z Ê x œ 1 u cos v u sin v, then r is a
function of u and v Ê r(uß v) œ (1 u cos v u sin v)i (u cos v)j (u sin v)k , 0 Ÿ u Ÿ 3 and 0 Ÿ v Ÿ 21.
14. (a) In a fashion similar to cylindrical coordinates, but working in the xz-plane instead of the xy-plane, let
x œ u cos v, z œ u sin v where u œ Èx# z# and v is the angle formed by (xß yß z), (yß 0ß 0), and (xß yß 0)
with vertex (yß 0ß 0). Since x y 2z œ 2 Ê y œ x 2z 2, then r(uß v)
œ (u cos v)i (u cos v 2u sin v 2)j (u sin v)k , 0 Ÿ u Ÿ È3 and 0 Ÿ v Ÿ 21.
(b) In a fashion similar to cylindrical coordinates, but working in the yz-plane instead of the xy-plane, let
y œ u cos v, z œ u sin v where u œ Èy# z# and v is the angle formed by (xß yß z), (xß 0ß 0), and (xß yß 0)
with vertex (xß 0ß 0). Since x y 2z œ 2 Ê x œ y 2z 2, then r(uß v)
œ (u cos v 2u sin v 2)i (u cos v)j (u sin v)k , 0 Ÿ u Ÿ È2 and 0 Ÿ v Ÿ 21.
15. Let x œ w cos v and z œ w sin v. Then (x 2)# z# œ 4 Ê x# 4x z# œ 0 Ê w# cos# v 4w cos v w# sin# v
œ 0 Ê w# 4w cos v œ 0 Ê w œ 0 or w 4 cos v œ 0 Ê w œ 0 or w œ 4 cos v. Now w œ 0 Ê x œ 0 and y œ 0,
which is a line not a cylinder. Therefore, let w œ 4 cos v Ê x œ (4 cos v)(cos v) œ 4 cos# v and z œ 4 cos v sin v.
Finally, let y œ u. Then r(uß v) œ a4 cos# vb i uj (4 cos v sin v)k , 1# Ÿ v Ÿ 1# and 0 Ÿ u Ÿ 3.
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 16.5 Surfaces and Area
16. Let y œ w cos v and z œ w sin v. Then y# (z 5)# œ 25 Ê y# z# 10z œ 0
Ê w# cos# v w# sin# v 10w sin v œ 0 Ê w# 10w sin v œ 0 Ê w(w 10 sin v) œ 0 Ê w œ 0 or
w œ 10 sin v. Now w œ 0 Ê y œ 0 and z œ 0, which is a line not a cylinder. Therefore, let w œ 10 sin v
Ê y œ 10 sin v cos v and z œ 10 sin# v. Finally, let x œ u. Then r(uß v) œ ui (10 sin v cos v)j a10 sin# vb k ,
0 Ÿ u Ÿ 10 and 0 Ÿ v Ÿ 1.
17. Let x œ r cos ) and y œ r sin ). Then r(rß )) œ (r cos ))i (r sin ))j ˆ 2 r#sin ) ‰ k , 0 Ÿ r Ÿ 1 and 0 Ÿ ) Ÿ 21
)‰
Ê rr œ (cos ))i (sin ))j ˆ sin# ) ‰ k and r) œ (r sin ))i (r cos ))j ˆ r cos
k
#
â
â
i
j
k
â
â
â
sin )
sin# ) ââ
Ê rr ‚ r) œ â cos )
â
) ââ
â r sin ) r cos ) r cos
#
œ Š r sin#) cos ) (sin ))(r# cos )) ‹ i Š r sin# ) r cos# ) ‹ j ar cos# ) r sin# )b k œ #r j rk
#
#
Ê krr ‚ r) k œ É r4 r# œ
È5 r
#
#
Ê A œ '0 '0
21
1
È5 r
# dr d) œ
'021 ’ È45 r “ " d) œ '021 d) œ 1È# 5
#
!
18. Let x œ r cos ) and y œ r sin ) Ê z œ x œ r cos ), 0 Ÿ r Ÿ 2 and 0 Ÿ ) Ÿ 21. Then
r(rß )) œ (r cos ))i (r sin ))j (r cos ))k Ê rr œ (cos ))i (sin ))j (cos ))k and
r) œ (r sin ))i (r cos ))j (r sin ))k
â
â
i
j
k â
â
â
â
sin ) cos ) â
Ê rr ‚ r) œ â cos )
â
â
â r sin ) r cos ) r sin ) â
œ ar sin# ) r cos# )b i (r sin ) cos ) r sin ) cos ))j ar cos# ) r sin# )b k œ ri rk
Ê krr ‚ r) k œ Èr# r# œ rÈ2 Ê A œ '0 '0 rÈ2 dr d) œ '0 ’ r 2 2 “ d) œ '0 2È2 d) œ 41È2
21
21
2
#È
#
21
!
19. Let x œ r cos ) and y œ r sin ) Ê z œ 2Èx# y# œ 2r, 1 Ÿ r Ÿ 3 and 0 Ÿ ) Ÿ 21. Then
r(rß )) œ (r cos ))i (r sin ))j 2rk Ê rr œ (cos ))i (sin ))j 2k and r) œ (r sin ))i (r cos ))j
â
â
i
j
kâ
â
â
â
sin ) 2 ⠜ (2r cos ))i (2r sin ))j ar cos# ) r sin# )b k
Ê rr ‚ r) œ â cos )
â
â
â r sin ) r cos ) 0 â
œ (2r cos ))i (2r sin ))j rk Ê krr ‚ r) k œ È4r# cos# ) 4r# sin# ) r# œ È5r# œ rÈ5
Ê A œ '0 '1 rÈ5 dr d) œ '0 ’ r 2 5 “ d) œ '0 4È5 d) œ 81È5
21
21
3
#È
$
21
"
È x# y#
20. Let x œ r cos ) and y œ r sin ) Ê z œ
œ 3r , 3 Ÿ r Ÿ 4 and 0 Ÿ ) Ÿ 21. Then
3
r(rß )) œ (r cos ))i (r sin ))j ˆ 3r ‰ k Ê rr œ (cos ))i (sin ))j ˆ 3" ‰ k and r) œ (r sin ))i (r cos ))j
â
â
i
j
kâ
â
â
" â
#
#
ˆ "
‰
ˆ"
‰
Ê rr ‚ r) œ â cos )
sin )
3 ââ œ 3 r cos ) i 3 r sin ) j ar cos ) r sin )b k
â
â r sin ) r cos ) 0 â
#
È
r 10
œ ˆ "3 r cos )‰ i ˆ 3" r sin )‰ j rk Ê krr ‚ r) k œ É 9" r# cos# ) 9" r# sin# ) r# œ É 10r
9 œ
3
Ê A œ '0 '3 r 310 dr d) œ '0 ’ r 6 10 “ d) œ '0
21
4
È
21
#È
%
$
21
È
7È10
d) œ 71 3 10
6
21. Let x œ r cos ) and y œ r sin ) Ê r# œ x# y# œ 1, 1 Ÿ z Ÿ 4 and 0 Ÿ ) Ÿ 21. Then
r(zß )) œ (cos ))i (sin ))j zk Ê rz œ k and r) œ ( sin ))i (cos ))j
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
965
966
Chapter 16 Integration in Vector Fields
â
â i
â
Ê r) ‚ rz œ â sin )
â
â 0
j
cos )
0
â
kâ
â
0 â œ (cos ))i (sin )) j Ê kr) ‚ rz k œ Ècos# ) sin# ) œ 1
â
1â
Ê A œ '0 '1 1 dr d) œ '0 3 d) œ 61
21
21
4
22. Let x œ u cos v and z œ u sin v Ê u# œ x# z# œ 10, 1 Ÿ y Ÿ 1, 0 Ÿ v Ÿ 21. Then
r(yß v) œ (u cos v)i yj (u sin v)k œ ŠÈ10 cos v‹ i yj ŠÈ10 sin v‹ k
â
i
â
â È
È
È
Ê rv œ Š 10 sin v‹ i Š 10 cos v‹ k and ry œ j Ê rv ‚ ry œ â 10 sin v
â
â
0
â
j
k
â
â
0 È10 cos v â
â
â
1
0
œ ŠÈ10 cos v‹ i ŠÈ10 sin v‹ k Ê krv ‚ ry k œ È10 Ê A œ '0 'c1 È10 du dv œ '0 ’È10u“
21
21
1
œ '0 2È10 dv œ 41È10
21
"
"
dv
23. z œ 2 x# y# and z œ Èx# y# Ê z œ 2 z# Ê z# z 2 œ 0 Ê z œ 2 or z œ 1. Since z œ Èx# y#
we get z œ 1 where the cone intersects the paraboloid. When x œ 0 and y œ 0, z œ 2 Ê the vertex of the
paraboloid is (0ß 0ß 2). Therefore, z ranges from 1 to 2 on the “cap" Ê r ranges from 1 (when x# y# œ 1) to 0
(when x œ 0 and y œ 0 at the vertex). Let x œ r cos ), y œ r sin ), and z œ 2 r# . Then
r(rß )) œ (r cos ))i (r sin ))j a2 r# b k , 0 Ÿ r Ÿ 1, 0 Ÿ ) Ÿ 21 Ê rr œ (cos ))i (sin ))j 2rk and
â
â
i
j
k â
â
â
â
sin ) 2r â
r) œ (r sin ))i (r cos ))j Ê rr ‚ r) œ â cos )
â
â
0 â
â r sin ) r cos )
œ a2r# cos )b i a2r# sin )b j rk Ê krr ‚ r) k œ È4r% cos# ) 4r% sin# ) r# œ rÈ4r# 1
"
a4r# 1b
Ê A œ '0 '0 rÈ4r# 1 dr d) œ '0 ’ 12
21
1
21
0,
“ d) œ '0 Š 5 15# " ‹ d) œ 16 Š5È5 1‹
$Î# "
21
È
!
24. Let x œ r cos ), y œ r sin ) and z œ x# y# œ r# . Then r(rß )) œ (r cos ))i (r sin ))j r# k , 1 Ÿ r Ÿ 2,
0 Ÿ ) Ÿ 21 Ê rr œ (cos ))i (sin ))j 2rk and r) œ (r sin ))i (r cos ))j
â
â
i
j
kâ
â
â
â
sin ) 2r â œ a2r# cos )b i a2r# sin )b j rk Ê krr ‚ r) k
Ê rr ‚ r) œ â cos )
â
â
â r sin ) r cos ) 0 â
"
œ È4r% cos# ) 4r% sin# ) r# œ rÈ4r# 1 Ê A œ '0 '1 rÈ4r# 1 dr d) œ '0 ’ 12
a4r# 1b
21
œ '0 Š 17
21
21
2
$Î# #
“ d)
"
È17 5È5
‹ d) œ 16 Š17È17 5È5‹
1#
25. Let x œ 3 sin 9 cos ), y œ 3 sin 9 sin ), and z œ 3 cos 9 Ê 3 œ Èx# y# z# œ È2 on the sphere. Next,
x# y# z# œ 2 and z œ Èx# y# Ê z# z# œ 2 Ê z# œ 1 Ê z œ 1 since z 0 Ê 9 œ 1 . For the lower
4
portion of the sphere cut by the cone, we get 9 œ 1. Then
r(9ß )) œ ŠÈ2 sin 9 cos )‹ i ŠÈ2 sin 9 sin )‹ j ŠÈ2 cos 9‹ k , 1 Ÿ 9 Ÿ 1, 0 Ÿ ) Ÿ 21
4
Ê r9 œ ŠÈ2 cos 9 cos )‹ i ŠÈ2 cos 9 sin )‹ j ŠÈ2 sin 9‹ k and r) œ ŠÈ2 sin 9 sin )‹ i ŠÈ2 sin 9 cos )‹ j
â
â
i
j
k
â
â
âÈ
â
È
È
2 cos 9 sin ) 2 sin 9 â
Ê r9 ‚ r) œ â 2 cos 9 cos )
â
â
â È2 sin 9 sin ) È2 sin 9 cos )
â
0
#
#
œ a2 sin 9 cos )b i a2 sin 9 sin )b j (2 sin 9 cos 9)k
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 16.5 Surfaces and Area
967
Ê kr9 ‚ r) k œ È4 sin% 9 cos# ) 4 sin% 9 sin# ) 4 sin# 9 cos# 9 œ È4 sin# 9 œ 2 ksin 9k œ 2 sin 9
Ê A œ '0 '1Î4 2 sin 9 d9 d) œ '0 Š2 È2‹ d) œ Š4 2È2‹ 1
21
1
21
26. Let x œ 3 sin 9 cos ), y œ 3 sin 9 sin ), and z œ 3 cos 9 Ê 3 œ Èx# y# z# œ 2 on the sphere. Next,
z œ 1 Ê 1 œ 2 cos 9 Ê cos 9 œ "# Ê 9 œ 231 ; z œ È3 Ê È3 œ 2 cos 9 Ê cos 9 œ
r(9ß )) œ (2 sin 9 cos ))i (2 sin 9 sin ))j (2 cos 9)k , 16 Ÿ 9 Ÿ 231 , 0 Ÿ ) Ÿ 21
È3
#
Ê 9 œ 16 . Then
Ê r9 œ (2 cos 9 cos ))i (2 cos 9 sin ))j (2 sin 9)k and
r) œ (2 sin 9 sin ))i (2 sin 9 cos )) j
â
â
i
j
k
â
â
â
â
Ê r9 ‚ r) œ â 2 cos 9 cos ) 2 cos 9 sin ) 2 sin 9 â
â
â
0
â 2 sin 9 sin ) 2 sin 9 cos )
â
œ a4 sin# 9 cos )b i a4 sin# 9 sin )b j (4 sin 9 cos 9)k
Ê kr9 ‚ r) k œ È16 sin% 9 cos# ) 16 sin% 9 sin# ) 16 sin# 9 cos# 9 œ È16 sin# 9 œ 4 ksin 9k œ 4 sin 9
Ê A œ '0 '1Î6 4 sin 9 d9 d) œ '0 Š2 2È3‹ d) œ Š4 4È3‹ 1
21
21Î3
21
27. The parametrization r(rß )) œ (r cos ))i (r sin ))j rk
at P! œ ŠÈ2ß È2ß 2‹ Ê ) œ 1 , r œ 2,
4
È
È
rr œ (cos ))i (sin ))j k œ #2 i #2 j k and
r) œ (r sin ))i (r cos ))j œ È2i È2j
â i
j
k ââ
â
âÈ
â
Ê rr ‚ r) œ â 2/2 È2/2 1 â
â
â
â È 2 È 2 0 â
œ È2i È2j 2k Ê the tangent plane is
0 œ ŠÈ2i È2j 2k‹ † ’Šx È2‹ i Šy È2‹ j (z 2)k“ Ê È2x È2y 2z œ 0, or x y È2z œ 0.
The parametrization r(rß )) Ê x œ r cos ), y œ r sin ) and z œ r Ê x# y# œ r# œ z# Ê the surface is z œ Èx# y# .
28. The parametrization r(9ß ))
œ (4 sin 9 cos ))i (4 sin 9 sin ))j (4 cos 9)k
at P! œ ŠÈ2ß È2ß 2È3‹ Ê 3 œ 4 and z œ 2È3
œ 4 cos 9 Ê 9 œ 16 ; also x œ È2 and y œ È2
Ê ) œ 14 . Then r9
œ (4 cos 9 cos ))i (4 cos 9 sin ))j (4 sin 9)k
œ È6i È6j 2k and
r) œ (4 sin 9 sin ))i (4 sin 9 cos ))j
â i
j
k ââ
â
â
â
œ È2i È2j at P! Ê r9 ‚ r) œ â È6 È6 2 â
â
â
â È2 È2 0 â
œ 2È2i 2È2j 4È3k Ê the tangent plane is
Š2È2i 2È2j 4È3k‹ † ’Šx È2‹ i Šy È2‹ j Šz 2È3‹ k“ œ 0 Ê È2x È2y 2È3z œ 16,
or x y È6z œ 8È2. The parametrization Ê x œ 4 sin 9 cos ), y œ 4 sin 9 sin ), z œ 4 cos 9
Ê the surface is x# y# z# œ 16, z 0.
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
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Chapter 16 Integration in Vector Fields
29. The parametrization r()ß z) œ (3 sin 2))i a6 sin# )b j zk
È
at P! œ Š 3 # 3 ß #9 ß 0‹ Ê ) œ 13 and z œ 0. Then
r) œ (6 cos 2))i (12 sin ) cos ))j
œ 3i 3È3j and rz œ k at P!
â i
j
k ââ
â
â
â
Ê r) ‚ rz œ â 3 3È3 0 â œ 3È3i 3j
â
â
â 0
0
1â
Ê the tangent plane is
È
Š3È3i 3j‹ † ’Šx 3 3 ‹ i ˆy 9 ‰ j (z 0)k“ œ 0
#
#
Ê È3x y œ 9. The parametrization Ê x œ 3 sin 2)
#
and y œ 6 sin# ) Ê x# y# œ 9 sin# 2) a6 sin# )b
œ 9 a4 sin# ) cos# )b 36 sin% ) œ 6 a6 sin# )b œ 6y Ê x# y# 6y 9 œ 9 Ê x# (y 3)# œ 9
30. The parametrization r(xß y) œ xi yj x# k at
P! œ (1ß 2ß 1) Ê rx œ i 2xk œ i 2k and ry œ j at P!
â
â
âi j k â
â
â
Ê rx ‚ ry œ â 1 0 2 â œ 2i k Ê the tangent plane
â
â
â0 " 0 â
is (2i k) † [(x 1)i (y 2)j (z 1)k] œ 0
Ê 2x z œ 1. The parametrization Ê x œ x, y œ y and
z œ x# Ê the surface is z œ x#
31. (a) An arbitrary point on the circle C is (xß z) œ (R r cos u, r sin u) Ê (xß yß z) is on the torus with
x œ (R r cos u) cos v, y œ (R r cos u) sin v, and z œ r sin u, 0 Ÿ u Ÿ 21, 0 Ÿ v Ÿ 21
(b) ru œ (r sin u cos v)i (r sin u sin v)j (r cos u)k and rv œ ((R r cos u) sin v)i ((R r cos u) cos v)j
â
â
i
j
k â
â
â
â
r sin u sin v
r cos u â
Ê ru ‚ rv œ â r sin u cos v
â
â
0 â
â (R r cos u) sin v (R r cos u) cos v
œ (R r cos u)(r cos v cos u)i (R r cos u)(r sin v cos u)j (r sin u)(R r cos u)k
Ê kru ‚ rv k# œ (R r cos u)# ar# cos# v cos# u r# sin# v cos# u r# sin# ub Ê kru ‚ rv k œ r(R r cos u)
Ê A œ '0 '0
21
21
arR r# cos ub du dv œ '0 21rR dv œ 41# rR
21
32. (a) The point (xß yß z) is on the surface for fixed x œ f(u) when y œ g(u) sin ˆ 1# v‰ and z œ g(u) cos ˆ 1# v‰
Ê x œ f(u), y œ g(u) cos v, and z œ g(u) sin v Ê r(uß v) œ f(u)i (g(u) cos v)j (g(u) sin v)k , 0 Ÿ v Ÿ 21,
aŸuŸb
(b) Let u œ y and x œ u# Ê f(u) œ u# and g(u) œ u Ê r(uß v) œ u# i (u cos v)j (u sin v)k , 0 Ÿ v Ÿ 21, 0 Ÿ u
#
#
#
33. (a) Let w# zc# œ 1 where w œ cos 9 and zc œ sin 9 Ê xa# yb# œ cos# 9 Ê xa œ cos 9 cos ) and yb œ cos 9 sin )
Ê x œ a cos ) cos 9, y œ b sin ) cos 9, and z œ c sin 9
Ê r()ß 9) œ (a cos ) cos 9)i (b sin ) cos 9)j (c sin 9)k
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 16.5 Surfaces and Area
(b) r) œ (a sin ) cos 9)i (b cos ) cos 9)j and r9 œ (a cos ) sin 9)i (b sin ) sin 9)j (c cos 9)k
â
â
i
j
k â
â
â
â
0 â
Ê r) ‚ r9 œ â a sin ) cos 9 b cos ) cos 9
â
â
â a cos ) sin 9 b sin ) sin 9 c cos 9 â
œ abc cos ) cos# 9b i aac sin ) cos# 9b j (ab sin 9 cos 9)k
Ê kr) ‚ r9 k# œ b# c# cos# ) cos% 9 a# c# sin# ) cos% 9 a# b# sin# 9 cos# 9, and the result follows.
A Ê '0 '0 kr) ‚ r9 k d9 d) œ '0 '0 c a# b# sin# 9 cos# 9 b# c# cos# ) cos% 9 a# c# sin# ) cos% 9 d d9 d)
21
1
21
1
1/2
34. (a) r()ß u) œ (cosh u cos ))i (cosh u sin ))j (sinh u)k
(b) r()ß u) œ (a cosh u cos ))i (b cosh u sin ))j (c sinh u)k
35. r()ß u) œ (5 cosh u cos ))i (5 cosh u sin ))j (5 sinh u)k Ê r) œ (5 cosh u sin ))i (5 cosh u cos ))j and
ru œ (5 sinh u cos ))i (5 sinh u sin ))j (5 cosh u)k
â
â
i
j
k
â
â
â
â
0
Ê r) ‚ ru œ â 5 cosh u sin ) 5 cosh u cos )
â
â
â
5 sinh u sin ) 5 cosh u â
â 5 sinh u cos )
œ a25 cosh# u cos )b i a25 cosh# u sin )b j (25 cosh u sinh u)k. At the point (x! ß y! ß 0), where x#0 y#0 œ 25
we have 5 sinh u œ 0 Ê u œ 0 and x! œ 25 cos ), y! œ 25 sin ) Ê the tangent plane is
5(x! i y! j) † [(x x! )i (y y! )j zk] œ 0 Ê x! x x0# y! y y0# œ 0 Ê x! x y! y œ 25
#
#
#
36. Let zc# w# œ 1 where zc œ cosh u and w œ sinh u Ê w# œ xa# by# Ê xa œ w cos ) and by œ w sin )
Ê x œ a sinh u cos ), y œ b sinh u sin ), and z œ c cosh u
Ê r()ß u) œ (a sinh u cos ))i (b sinh u sin ))j (c cosh u)k , 0 Ÿ ) Ÿ 21, _ u _
37. p œ k , ™ f œ 2xi 2yj k Ê k ™ f k œ È(2x)# (2y)# (1)# œ È4x# 4y# 1 and k ™ f † pk œ 1;
z œ 2 Ê x# y# œ 2; thus S œ ' ' k™f k dA œ ' ' È4x# 4y# 1 dx dy
R
k™f†pk
R
È2
$Î#
"
a4r# 1b “
œ ' ' È4r# cos# ) 4r# sin# ) 1 r dr d) œ '0 '0 È4r# 1 r dr d) œ '0 ’ 12
21
21
R
œ '0
21
È#
!
d)
13
13
6 d) œ 3 1
38. p œ k , ™ f œ 2xi 2yj k Ê k ™ f k œ È4x# 4y# 1 and k ™ f † pk œ 1; 2 Ÿ x# y# Ÿ 6
È6
k™f k
' ' È4x# 4y# 1 dx dy œ ' ' È4r# 1 r dr d) œ ' 'È È4r# 1 r dr d)
Ê S œ ' ' k™
f†pk dA œ
0
2
21
R
"
œ '0 ’ 12
a4r# 1b
21
R
$Î#
R
È'
“ È d) œ '0
21
#
49
49
6 d) œ 3 1
39. p œ k , ™ f œ i 2j 2k Ê k ™ f k œ 3 and k ™ f † pk œ 2; x œ y# and x œ 2 y# intersect at (1ß 1) and (1ß 1)
2 c y#
k™f k
' ' 3# dx dy œ ' ' #
Ê S œ ' ' k™
f†pk dA œ
c1 y
1
R
R
3
# dx dy œ
'c11 a3 3y# b dy œ 4
40. p œ k , ™ f œ 2xi 2k Ê k ™ f k œ È4x# 4 œ 2Èx# 1 and k ™ f † pk œ 2 Ê S œ ' '
œ''
R
2È x# 1
dx dy œ
#
È3
È3
'0 '0 Èx# 1 dy dx œ '0
x
R
k™f k
k™f†pk dA
È$
$Î#
xÈx# 1 dx œ ’ 3" ax# 1b “
œ 3" (4)$Î# 3" œ 37
!
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
969
970
Chapter 16 Integration in Vector Fields
41. p œ k , ™ f œ 2xi 2j 2k Ê k ™ f k œ È(2x)# (2)# (2)# œ È4x# 8 œ 2Èx# 2 and k ™ f † pk œ 2
Ê Sœ''
R
k™f k
k™f†pk dA œ
' ' 2Èx##2 dx dy œ ' ' Èx# 2 dy dx œ ' 3xÈx# 2 dx œ ’ax# 2b$Î# “
0
0
0
2
3x
2
R
#
!
œ 6È 6 2È 2
42. p œ k , ™ f œ 2xi 2yj 2zk Ê k ™ f k œ È4x# 4y# 4z# œ È8 œ 2È2 and k ™ f † pk œ 2z; x# y# z# œ 2 and
z œ Èx# y# Ê x# y# œ 1; thus, S œ ' '
œ È2 ' '
"
È2
È2 ax# y# b dA œ
R
'0 '
21
k™f k
k™f†pk dA œ
R
1
r dr d)
œ È2
0 È 2 r#
'0
2
È2 ' ' " dA
' ' 2È
#z dA œ
z
21
R
R
Š1 È2‹ d) œ 21 Š2 È2‹
43. p œ k , ™ f œ ci k Ê k ™ f k œ Èc# 1 and k ™ f † pk œ 1 Ê S œ ' '
R
21
1
21 È
c 1
œ '0 '0 Èc# 1 r dr d) œ '0
d) œ 1Èc# 1
#
k™f k
k™f†pk dA œ
' ' Èc# 1 dx dy
R
#
44. p œ k , ™ f œ 2xi 2zj Ê k ™ f k œ È(2x)# (2z)# œ 2 and k ™ f † pk œ 2z for the upper surface, z
Ê Sœ''
R
k™f k
k™f†pk dA œ
''
R
2
#z dA œ
''
R
"
È1 x# dy dx œ 2
1Î2
1Î2
'c1Î2 '0
1
È1 x# dy dx œ
'
1Î2
0
"
dx
1Î2 È1 x#
"Î#
œ csin" xd "Î# œ 16 ˆ 16 ‰ œ 13
45. p œ i , ™ f œ i 2yj 2zk Ê k ™ f k œ È1# (2y)# (2z)# œ È1 4y# 4z# and k ™ f † pk œ 1; 1 Ÿ y# z# Ÿ 4
k™f k
' ' È1 4y# 4z# dy dz œ ' ' È1 4r# cos# ) 4r# sin# ) r dr d)
Ê S œ ' ' k™
f†pk dA œ
0
1
21
R
R
"
a1 4r# b
œ '0 '1 È1 4r# r dr d) œ '0 ’ 12
21
21
2
2
“ d) œ '0
$Î# #
21
"
1
"
È
È
È
È
1# Š17 17 5 5‹ d) œ 6 Š17 17 5 5‹
46. p œ j , ™ f œ 2xi j 2zk Ê k ™ f k œ È4x# 4z# 1 and k ™ f † pk œ 1; y œ 0 and x# y z# œ 2 Ê x# z# œ 2;
k™f k
' ' È4x# 4z# 1 dx dz œ '
thus, S œ ' ' k™
f†pk dA œ
0
21
R
R
È
'0 2 È4r# 1 r dr d) œ '021 136 d) œ 133 1
#
#
#
47. p œ k , ™ f œ ˆ2x 2x ‰ i È15 j k Ê k ™ f k œ ʈ2x 2x ‰ ŠÈ15‹ (1)# œ É4x# 8 x4# œ Ɉ2x 2x ‰
k™f k
' ' a2x 2x" b dx dy
œ 2x 2x , on 1 Ÿ x Ÿ 2 and k ™ f † pk œ 1 Ê S œ ' ' k™
f†pk dA œ
R
R
œ '0 '1 a2x 2x" b dx dy œ '0 cx# 2 ln xd " dy œ '0 (3 2 ln 2) dy œ 3 2 ln 2
1
2
1
1
#
48. p œ k , ™ f œ 3Èx i 3Èy j 3k Ê k ™ f k œ È9x 9y 9 œ 3Èx y 1 and k ™ f † pk œ 3
k™f k
' ' Èx y 1 dx dy œ ' ' Èx y 1 dx dy œ ' 23 (x y 1)$Î# ‘ ! dy
Ê S œ ' ' k™
f†pk dA œ
0
0
0
1
R
1
1
"
R
1
"
4
4
4 œ '0 23 (y 2)$Î# 23 (y 1)$Î# ‘ dy œ 15
(y 2)&Î# 15
(y 1)&Î# ‘ ! œ 15
(3)&Î# (2)&Î# (2)&Î# 1‘
4
œ 15
Š9È3 8È2 1‹
49. fx (xß y) œ 2x, fy (xß y) œ 2y Ê Éfx# fy# 1 œ È4x# 4y# 1 Ê Area œ ' ' È4x# 4y# 1 dx dy
È3
œ '0 '0
21
R
È4r# 1 r dr d) œ 1 Š13È13 1‹
6
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 16.5 Surfaces and Area
971
50. fy (yß z) œ 2y, fz (yß z) œ 2z Ê Éfy# fz# 1 œ È4y# 4z# 1 Ê Area œ ' ' È4y# 4z# 1 dy dz
R
21
1
œ '0 '0 È4r# 1 r dr d) œ 16 Š5È5 1‹
51. fx (xß y) œ Èx#x y# , fy (xß y) œ Èx#y y# Ê Éfx# fy# 1 œ É x# x y# x# y y# 1 œ È2 Ê Area œ ' ' È2 dx dy
#
#
Rxy
œ È2aArea between the ellipse and the circleb œ È2a61 1b œ 51È2
52. Over Rxy : z œ 2 23 x 2y Ê fx (xß y) œ 23 , fy (xß y) œ 2 Ê Éfx# fy# 1 œ É 49 4 1 œ 73
Ê Area œ ' '
Rxy
7
7
ˆ 7 ‰ ˆ #3 ‰ œ #7 .
3 dA œ 3 (Area of the shadow triangle in the xy-plane) œ 3
Over Rxz : y œ 1 13 x "# z Ê fx (xß z) œ 13 , fz (xß z) œ "# Ê Èfx# fz# 1 œ É 19 "4 1 œ 76
Ê Area œ ' ' 76 dA œ 76 (Area of the shadow triangle in the xz-plane) œ ˆ 76 ‰ (3) œ 7# .
Rxz
Over Ryz : x œ 3 3y 3# z Ê fy (yß z) œ 3, fz (yß z) œ 3# Ê Éfy# fz# 1 œ É9 94 1 œ 7#
Ê Area œ ' ' 72 dA œ 72 (Area of the shadow triangle in the yz-plane) œ ˆ 72 ‰ (1) œ 7# .
Ryz
16
2 $Î#
53. y œ 23 z$Î# Ê fx (xß z) œ 0, fz (xß z) œ z"Î# Ê Èfx# fz# 1 œ Èz 1 ; y œ 16
Ê zœ4
3 Ê 3 œ 3 z
Ê Area œ '0 '0 Èz 1 dx dz œ '0 Èz 1 dz œ 23 Š5È5 1‹
4
1
4
4 c z#
54. y œ 4 z Ê fx (xß z) œ 0, fz (xß z) œ 1 Ê Èfx# fz# 1 œ È2 Ê Area œ ' ' È2 dA œ '0 '0
2
È2 dx dz
Rxz
œ È2 '0 a4 z# b dz œ 163 2
È
2
55. rax, yb œ x i y j fax, yb k Ê rx ax, yb œ i fx ax, yb k, ry ax, yb œ j fy ax, ybk
â
â
k â
âi j
â
â
Ê rx ‚ ry œ â 1 0 fx ax, yb â œ fx ax, yb i fy ax, yb j k
â
â
â 0 1 fy ax, yb â
Ê lrx ‚ ry l œ Éafx ax, ybb2 afy ax, ybb2 12 œ Éfx ax, yb2 fy ax, yb2 1
Ê d5 œ Éfx ax, yb2 fy ax, yb2 1 dA
56. S is obtained by rotating y œ faxb, a Ÿ x Ÿ b about the x-axis where faxb 0
(a) Let ax, y, zb be a point on S. Consider the cross section when x œ x‡ , the cross section is a circle with radius r œ fax‡ b.
The set of parametric equations for this circle are given by ya)b œ r cos ) œ fax‡ b cos ) and za)b œ r sin )
œ fax‡ b sin ) where 0 Ÿ ) Ÿ 21. Since x can take on any value between a and b we have xax, )b œ x, yax, )b
œ faxb cos ), zax, )b œ faxb sin ) where a Ÿ x Ÿ b and 0 Ÿ ) Ÿ 21. Thus rax, )b œ x i faxb cos ) j faxb sin ) k
(b) rx ax, )b œ i f w axb cos ) j f w axb sin ) k and r) ax, )b œ faxb sin ) j faxb cos ) k
â
â
j
k
âi
â
â
â
Ê rx ‚ r) œ â 1 f w axb cos ) f w axb sin ) â œ faxb † f w axb i faxb cos ) j faxb sin ) k
â
â
â 0 faxb sin ) faxb cos ) â
Ê lrx ‚ r) l œ Éafaxb † f w axbb2 afaxb cos )b2 afaxb sin )b2 œ faxbÉ1 af w axbb2
21
A œ 'a '0 faxbÉ1 af w axbb2 d) dx œ 'a ”ŒfaxbÉ1 af w axbb2 )• dx œ 'a 2 1 faxbÉ1 af w axbb2 dx
b
21
b
b
0
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
972
Chapter 16 Integration in Vector Fields
16.6 SURFACE INTEGRALS
â
âi
â
1. Let the parametrization be r(xß z) œ xi x# j zk Ê rx œ i 2xj and rz œ k Ê rx ‚ rz œ â 1
â
â0
j
2x
0
â
kâ
â
0â
â
"â
"
œ 2xi j Ê krx ‚ rz k œ È4x# 1 Ê ' ' G(xß yß z) d5 œ '0 '0 xÈ4x# 1 dx dz œ '0 ’ 12
a4x# 1b
3
œ'
2
3
S
$Î# #
“ dz
!
È "
"
Š17È17 1‹ dz œ 17 17
4
0 1#
3
2. Let the parametrization be r(xß y) œ xi yj È4 y# k , 2 Ÿ y Ÿ 2 Ê rx œ i and ry œ j È4y y# k
âi j
â
k
â
â
#
â1 0
â
0
Ê rx ‚ ry œ â
â œ y j k Ê krx ‚ ry k œ É 4 y y# 1 œ È42 y#
â 0 1 y â È 4 y#
â
È 4 y# â
Ê ' ' G(xß yß z) d5 œ '1 'c2 È4 y# Š È42 y# ‹ dy dx œ 24
4
2
S
3. Let the parametrization be r(9ß )) œ (sin 9 cos ))i (sin 9 sin ))j (cos 9)k (spherical coordinates with 3 œ 1
on the sphere), 0 Ÿ 9 Ÿ 1, 0 Ÿ ) Ÿ 21 Ê r9 œ (cos 9 cos ))i (cos 9 sin ))j (sin 9)k and
â
â
i
j
k â
â
â
â
r) œ ( sin 9 sin ))i (sin 9 cos ))j Ê r9 ‚ r) œ â cos 9 cos ) cos 9 sin ) sin 9 â
â
â
0 â
â sin 9 sin ) sin 9 cos )
œ asin# 9 cos )b i asin# 9 sin )b j (sin 9 cos 9)k Ê kr9 ‚ r) k œ Èsin% 9 cos# ) sin% 9 sin# ) sin# 9 cos# 9
œ sin 9; x œ sin 9 cos ) Ê G(xß yß z) œ cos# ) sin# 9 Ê ' ' G(xß yß z) d5 œ '0 '0 acos# ) sin# 9b (sin 9) d9 d)
21
1
S
21
c1
u œ cos 9
œ '0 '0 acos )b a1 cos 9b (sin 9) d9 d); ”
Ä '0 '1 acos# )b au# 1b du d)
•
du œ sin 9 d9
21
1
#
#
œ '0 acos# )b ’ u3 u“
21
$
"
"
d) œ 34 '0 cos# ) d) œ 34 2) sin42) ‘ ! œ 431
21
#1
4. Let the parametrization be r(9ß )) œ (a sin 9 cos ))i (a sin 9 sin ))j (a cos 9)k (spherical coordinates with 3 œ a,
a 0, on the sphere), 0 Ÿ 9 Ÿ 1# (since z 0), 0 Ÿ ) Ÿ 21 Ê r9 œ (a cos 9 cos ))i (a cos 9 sin ))j (a sin 9)k and
â
â
i
j
k
â
â
â
â
r) œ (a sin 9 sin ))i (a sin 9 cos ))j Ê r9 ‚ r) œ â a cos 9 cos ) a cos 9 sin ) a sin 9 â
â
â
0
â a sin 9 sin ) a sin 9 cos )
â
œ aa# sin# 9 cos )b i aa# sin# 9 sin )b j (a# sin 9 cos 9)k
Ê kr9 ‚ r) k œ Èa% sin% 9 cos# ) a% sin% 9 sin# ) a% sin# 9 cos# 9 œ a# sin 9; z œ a cos 9
Ê G(xß yß z) œ a# cos# 9 Ê ' ' G(xß yß z) d5 œ '0 '0
21
S
1Î2
aa# cos# 9b aa# sin 9b d9 d) œ 23 1a%
5. Let the parametrization be r(xß y) œ xi yj (4 x y)k Ê rx œ i k and ry œ j k
â
â
âi j k â
1
1
â
â
Ê rx ‚ ry œ â 1 0 1 â œ i j k Ê krx ‚ ry k œ È3 Ê ' ' F(xß yß z) d5 œ '0 '0 (4 x y) È3 dy dx
â
â
S
â 0 1 " â
œ '0 È3 ’4y xy y2 “ dx œ '0 È3 ˆ #7 x‰ dx œ È3 ’ 27 x x# “ œ 3È3
1
#
"
!
1
#
"
!
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 16.6 Surface Integrals
973
6. Let the parametrization be r(rß )) œ (r cos ))i (r sin ))j rk , 0 Ÿ r Ÿ 1 (since 0 Ÿ z Ÿ 1) and 0 Ÿ ) Ÿ 21
â
â
i
j
kâ
â
â
â
sin ) " â
Ê rr œ (cos ))i (sin ))j k and r) œ (r sin ))i (r cos ))j Ê rr ‚ r) œ â cos )
â
â
â r sin ) r cos ) 0 â
œ (r cos ))i (r sin ))j rk Ê krr ‚ r) k œ È(r cos ))# (r sin ))# r# œ rÈ2; z œ r and x œ r cos )
Ê F(xß yß z) œ r r cos ) Ê ' ' F(xß yß z) d5 œ '0 '0 (r r cos )) ŠrÈ2‹ dr d) œ È2 '0 '0 (1 cos )) r# dr d)
21
21
1
1
S
È
œ 213 2
7. Let the parametrization be r(rß )) œ (r cos ))i (r sin ))j a1 r# b k , 0 Ÿ r Ÿ 1 (since 0 Ÿ z Ÿ 1) and 0 Ÿ ) Ÿ 21
â
â
i
j
k â
â
â
â
sin ) 2r â
Ê rr œ (cos ))i (sin ))j 2rk and r) œ (r sin ))i (r cos ))j Ê rr ‚ r) œ â cos )
â
â
0 â
â r sin ) r cos )
œ a2r# cos )b i a2r# sin )b j rk Ê krr ‚ r) k œ Éa2r# cos )b# a2r# sin )b r# œ rÈ1 4r# ; z œ 1 r# and
x œ r cos ) Ê H(xß yß z) œ ar# cos# )b È1 4r# Ê ' ' H(xß yß z) d5
S
21
1
21
1
œ '0 '0 ar# cos# )b ŠÈ1 4r# ‹ ŠrÈ1 4r# ‹ dr d) œ '0 '0 r$ a1 4r# b cos# ) dr d) œ 11121
8. Let the parametrization be r(9ß )) œ (2 sin 9 cos ))i (2 sin 9 sin ))j (2 cos 9)k (spherical coordinates with
3 œ 2 on the sphere), 0 Ÿ 9 Ÿ 1 ; x# y# z# œ 4 and z œ Èx# y# Ê z# z# œ 4 Ê z# œ 2 Ê z œ È2 (since
4
È2
#
Ê 9 œ 14 , 0 Ÿ ) Ÿ 21; r9 œ (2 cos 9 cos ))i (2 cos 9 sin ))j (2 sin 9)k
â
â
i
j
k
â
â
â
â
and r) œ (2 sin 9 sin ))i (2 sin 9 cos ))j Ê r9 ‚ r) œ â 2 cos 9 cos ) 2 cos 9 sin ) 2 sin 9 â
â
â
0
â 2 sin 9 sin ) 2 sin 9 cos )
â
0) Ê 2 cos 9 œ È2 Ê cos 9 œ
z
œ a4 sin# 9 cos )b i a4 sin# 9 sin )b j (4 sin 9 cos 9)k
Ê kr9 ‚ r) k œ È16 sin% 9 cos# ) 16 sin% 9 sin# ) 16 sin# 9 cos# 9 œ 4 sin 9; y œ 2 sin 9 sin ) and
z œ 2 cos 9 Ê H(xß yß z) œ 4 cos 9 sin 9 sin ) Ê ' ' H(xß yß z) d5 œ '0 '0
21
œ '0 '0 16 sin# 9 cos 9 sin ) d9 d) œ 0
21
1Î4
1Î4
(4 cos 9 sin 9 sin ))(4 sin 9) d9 d)
S
9. The bottom face S of the cube is in the xy-plane Ê z œ 0 Ê G(xß yß 0) œ x y and f(xß yß z) œ z œ 0 Ê p œ k
and ™ f œ k Ê k ™ f k œ 1 and k ™ f † pk œ 1 Ê d5 œ dx dy Ê ' ' G d5 œ ' ' (x y) dx dy
œ '0 '0 (x y) dx dy œ '
a
a
S
a
#
Š a# ay‹ dy œ a$ .
0
R
Because of symmetry, we also get a$ over the face of the cube
in the xz-plane and a$ over the face of the cube in the yz-plane. Next, on the top of the cube, G(xß yß z)
œ G(xß yß a) œ x y a and f(xß yß z) œ z œ a Ê p œ k and ™ f œ k Ê k ™ f k œ 1 and k ™ f † pk œ 1 Ê d5 œ dx dy
' ' G d5 œ ' ' (x y a) dx dy œ ' ' (x y a) dx dy œ ' ' (x y) dx dy ' ' a dx dy œ 2a$ .
0
0
0
0
0
0
a
S
a
a
a
a
a
R
Because of symmetry, the integral is also 2a$ over each of the other two faces. Therefore,
' ' (x y z) d5 œ 3 aa$ 2a$ b œ 9a$ .
cube
10. On the face S in the xz-plane, we have y œ 0 Ê f(xß yß z) œ y œ 0 and G(xß yß z) œ G(xß 0ß z) œ z Ê p œ j and ™ f œ j
Ê k ™ f k œ 1 and k ™ f † pk œ 1 Ê d5 œ dx dz Ê ' ' G d5 œ ' ' (y z) d5 œ '0 '0 z dx dz œ '0 2z dz œ 1.
1
S
2
1
S
On the face in the xy-plane, we have z œ 0 Ê f(xß yß z) œ z œ 0 and G(xß yß z) œ G(xß yß 0) œ y Ê p œ k and ™ f œ k
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Chapter 16 Integration in Vector Fields
Ê k ™ f k œ 1 and k ™ f † pk œ 1 Ê d5 œ dx dy Ê ' ' G d5 œ ' ' y d5 œ '0 '0 y dx dy œ 1.
1
S
2
S
On the triangular face in the plane x œ 2 we have f(xß yß z) œ x œ 2 and G(xß yß z) œ G(2ß yß z) œ y z Ê p œ i and
1cy
™ f œ i Ê k ™ f k œ 1 and k ™ f † pk œ 1 Ê d5 œ dz dy Ê ' ' G d5 œ ' ' (y z) d5 œ '0 '0
1
S
œ '0 "# a1 y# b dy œ 3" .
(y z) dz dy
S
1
On the triangular face in the yz-plane, we have x œ 0 Ê f(xß yß z) œ x œ 0 and G(xß yß z) œ G(0ß yß z) œ y z
Ê p œ i and ™ f œ i Ê k ™ f k œ 1 and k ™ f † pk œ 1 Ê d5 œ dz dy Ê ' ' G d5 œ ' ' (y z) d5
S
1
1cy
œ '0 '0 (y z) dz dy œ "3 .
S
Finally, on the sloped face, we have y z œ 1 Ê f(xß yß z) œ y z œ 1 and G(xß yß z) œ y z œ 1 Ê p œ k and
™ f œ j k Ê k ™ f k œ È2 and k ™ f † pk œ 1 Ê d5 œ È2 dx dy Ê ' ' G d5 œ ' ' (y z) d5
œ '0 '0 È2 dx dy œ 2È2. Therefore, ' '
1
2
wedge
S
S
G(xß yß z) d5 œ 1 1 "3 3" 2È2 œ 38 2È2
11. On the faces in the coordinate planes, G(xß yß z) œ 0 Ê the integral over these faces is 0.
On the face x œ a, we have f(xß yß z) œ x œ a and G(xß yß z) œ G(aß yß z) œ ayz Ê p œ i and ™ f œ i Ê k ™ f k œ 1
and k ™ f † pk œ 1 Ê d5 œ dy dz Ê ' ' G d5 œ ' ' ayz d5 œ '0 '0 ayz dy dz œ ab4c .
c
S
b
# #
S
On the face y œ b, we have f(xß yß z) œ y œ b and G(xß yß z) œ G(xß bß z) œ bxz Ê p œ j and ™ f œ j Ê k ™ f k œ 1
and k ™ f † pk œ 1 Ê d5 œ dx dz Ê ' ' G d5 œ ' ' bxz d5 œ '0 '0 bxz dx dz œ a bc
4 .
c
S
a
#
#
S
On the face z œ c, we have f(xß yß z) œ z œ c and G(xß yß z) œ G(xß yß c) œ cxy Ê p œ k and ™ f œ k Ê k ™ f k œ 1
and k ™ f † pk œ 1 Ê d5 œ dy dx Ê ' ' G d5 œ ' ' cxy d5 œ '0 '0 cxy dx dy œ a 4b c . Therefore,
b
' ' G(xß yß z) d5 œ abc(ab 4ac bc) .
S
a
# #
S
S
12. On the face x œ a, we have f(xß yß z) œ x œ a and G(xß yß z) œ G(aß yß z) œ ayz Ê p œ i and ™ f œ i Ê k ™ f k œ 1
and k ™ f † pk œ 1 Ê d5 œ dz dy Ê ' ' G d5 œ ' ' ayz d5 œ 'cb 'cc ayz dz dy œ 0. Because of the symmetry
b
S
c
S
of G on all the other faces, all the integrals are 0, and ' ' G(xß yß z) d5 œ 0.
S
13. f(xß yß z) œ 2x 2y z œ 2 Ê ™ f œ 2i 2j k and G(xß yß z) œ x y (2 2x 2y) œ 2 x y Ê p œ k ,
k ™ f k œ 3 and k ™ f † pk œ 1 Ê d5 œ 3 dy dx; z œ 0 Ê 2x 2y œ 2 Ê y œ 1 x Ê ' ' G d5 œ ' ' (2 x y) d5
S
1
1cx
1
1
œ 3 '0 '0 (2 x y) dy dx œ 3 '0 (2 x)(1 x) "# (1 x)# ‘ dx œ 3 '0 Š #3 2x x# ‹ dx œ 2
S
#
14. f(xß yß z) œ y# 4z œ 16 Ê ™ f œ 2yj 4k Ê k ™ f k œ È4y# 16 œ 2Èy# 4 and p œ k Ê k ™ f † pk œ 4
Ê d5 œ
œ'
2È y# 4
dx dy
4
Èy 4
Ê ' ' G d5 œ 'c4 '0 ˆxÈy# 4‰ Š # ‹ dx dy œ 'c4 '0 x ay # 4b dx dy
4
1
#
4
1
#
S
%
"
" y$
#
‰ 56
a
y
4
b
dy
œ
4y
œ #" ˆ 64
’
“
#
4
3
3 16 œ 3
c4
!
4
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Section 16.6 Surface Integrals
975
15. fax, y, zb œ x y2 z œ 0 Ê ™ f œ i 2yj k Ê k ™ f k œ È4y# 2 œ È2È2y# 1 and p œ k Ê k ™ f † pk œ 1
È2È2y# 1
dx dy Ê
1
Ê d5 œ
œ È2'0 y
1
3È
' ' G d5 œ ' ' ax y2 xb È2È2y# 1 dx dy œ È2' ' y2 È2y# 1 dx dy
S
1
y
1
y
0
0
0
0
È
È
2y# 1 dy œ 6 630 2
16. fax, y, zb œ x2 y z œ 0 Ê ™ f œ 2xi j k Ê k ™ f k œ È4x# 2 œ È2È2x# 1 and p œ k Ê k ™ f † pk œ 1
È2È2x# 1
dx dy Ê
1
Ê d5 œ
È
È
œ 3 6 6 2
'
' ' G d5 œ ' ' x È2È2x# 1 dx dy œ È2' ' x È2x# 1 dx dy
S
1
1
1
1
c1
0
c1
0
È
È
dy œ 3 6 3 2
0
1
17. fax, y, zb œ 2x y z œ 2 Ê ™ f œ 2i j k Ê k ™ f k œ È6 and p œ k Ê k ™ f † pk œ 1 Ê d5 œ
2 c 2x
2 c 2x
Ê ' ' G d5 œ '0 '1 c x x ya2 2x ybÈ6 dy dx œ È6'0 '1 c x a2x y 2x2 y x y2 b dy dx
1
1
È6
1 dy dx
S
È
œ È6' ˆ 2 x 2x2 2x3 2 x4 ‰ dx œ 6
1
3
0
3
30
18. fax, y, zb œ x y œ " Ê ™ f œ i j Ê k ™ f k œ È2 and p œ j Ê k ™ f † pk œ 1
È2
1 dz dx Ê
Ê d5 œ
œ È2'
' ' G d5 œ ' ' ax a1 xb zb È2 dz dx œ È2' ' a2x z 1b dz dx
S
1
1
1
1
0
0
0
0
È
ˆ2x 32 ‰dx œ 22
0
1
19. Let the parametrization be r(xß y) œ xi yj a4 y# b k , 0 Ÿ x Ÿ 1, 2 Ÿ y Ÿ 2; z œ 0 Ê 0 œ 4 y#
â
â
k â
âi j
â
â
0 â œ 2yj k Ê F † n d5
Ê y œ „ 2; rx œ i and ry œ j 2yk Ê rx ‚ ry œ â 1 0
â
â
â 0 1 2y â
œ F † krxx ‚ryyk krx ‚ ry k dy dx œ (2xy 3z) dy dx œ c2xy 3a4 y# bd dy dx Ê ' ' F † n d5
r ‚r
œ '0 'c2 a2xy 3y 12b dy dx œ '0 cxy y
1
2
1
#
#
S
$
#
12yd # dx œ
'0 32 dx œ 32
1
20. Let the parametrization be r(xß y) œ xi x# j zk , 1 Ÿ x Ÿ 1, 0 Ÿ z Ÿ 2 Ê rx œ i 2xj and rz œ k
â
â
â i j kâ
â
â
rz
#
Ê rx ‚ rz œ â 1 2x 0 â œ 2xi j Ê F † n d5 œ F † krrxx ‚
‚rz k krx ‚ rz k dz dx œ x dz dx
â
â
â0 0 1â
Ê ' ' F † n d5 œ 'c1 '0 x# dz dx œ 43
1
2
S
21. Let the parametrization be r(9ß )) œ (a sin 9 cos ))i (a sin 9 sin ))j (a cos 9)k (spherical coordinates with
3 œ a, a 0, on the sphere), 0 Ÿ 9 Ÿ 1# (for the first octant) ß 0 Ÿ ) Ÿ 1# (for the first octant)
Ê r9 œ (a cos 9 cos ))i (a cos 9 sin ))j (a sin 9)k and r) œ (a sin 9 sin ))i (a sin 9 cos ))j
â
â
i
j
k
â
â
â
â
Ê r9 ‚ r) œ â a cos 9 cos ) a cos 9 sin ) a sin 9 â
â
â
0
â a sin 9 sin ) a sin 9 cos )
â
r ‚r
œ aa# sin# 9 cos )b i aa# sin# 9 sin )b j aa# sin 9 cos 9b k Ê F † n d5 œ F † kr99 ‚r)) k kr9 ‚ r) k d) d9
œ a$ cos# 9 sin 9 d) d9 since F œ zk œ (a cos 9)k Ê ' ' F † n d5 œ '0
1Î2
S
'01Î2 a$ cos# 9 sin 9 d9 d) œ 1a6
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$
976
Chapter 16 Integration in Vector Fields
22. Let the parametrization be r(9ß )) œ (a sin 9 cos ))i (a sin 9 sin ))j (a cos 9)k (spherical coordinates with
3 œ a, a 0, on the sphere), 0 Ÿ 9 Ÿ 1, 0 Ÿ ) Ÿ 21
Ê r9 œ (a cos 9 cos ))i (a cos 9 sin ))j (a sin 9)k and r) œ (a sin 9 sin ))i (a sin 9 cos ))j
â
â
i
j
k
â
â
â
â
Ê r9 ‚ r) œ â a cos 9 cos ) a cos 9 sin ) a sin 9 â
â
â
0
â a sin 9 sin ) a sin 9 cos )
â
r ‚r
œ aa# sin# 9 cos )b i aa# sin# 9 sin )b j aa# sin 9 cos 9b k Ê F † n d5 œ F † kr99 ‚r)) k kr9 ‚ r) k d) d9
œ aa$ sin$ 9 cos# 9 a$ sin$ 9 sin# ) a$ sin 9 cos# 9b d) d9 œ a$ sin 9 d) d9 since F œ xi yj zk
œ (a sin 9 cos ))i (a sin 9 sin ))j (a cos 9)k Ê ' ' F † n d5 œ '0 '0 a$ sin 9 d9 d) œ 41a$
21
1
S
23. Let the parametrization be r(xß y) œ xi yj (2a x y)k , 0 Ÿ x Ÿ a, 0 Ÿ y Ÿ a Ê rx œ i k and ry œ j k
â
â
âi j k â
â
â
r ‚r
Ê rx ‚ ry œ â 1 0 1 â œ i j k Ê F † n d5 œ F † krxx ‚ryyk krx ‚ ry k dy dx
â
â
â 0 1 1 â
œ [2xy 2y(2a x y) 2x(2a x y)] dy dx since F œ 2xyi 2yzj 2xzk
œ 2xyi 2y(2a x y)j 2x(2a x y)k Ê ' ' F † n d5
S
a
a
a
a
œ '0 '0 [2xy 2y(2a x y) 2x(2a x y)] dy dx œ '0 '0 a4ay 2y# 4ax 2x# 2xyb dy dx
a
œ '0 ˆ 43 a$ 3a# x 2ax# ‰ dx œ ˆ 43 3# 23 ‰ a% œ 13a
6
%
24. Let the parametrization be r()ß z) œ (cos ))i (sin ))j zk , 0 Ÿ z Ÿ a, 0 Ÿ ) Ÿ 21 (where r œ Èx# y# œ 1 on
â
â
j
kâ
â i
â
â
the cylinder) Ê r) œ ( sin ))i (cos ))j and rz œ k Ê r) ‚ rz œ â sin ) cos ) 0 â œ (cos ))i (sin ))j
â
â
0
1â
â 0
rz
#
#
Ê F † n d5 œ F † krr)) ‚
‚rz k kr) ‚ rz k dz d) œ acos ) sin )b dz d) œ dz d), since F œ (cos ))i (sin ))j zk
Ê ' ' F † n d5 œ '0
21
S
'0a 1 dz d) œ 21a
25. Let the parametrization be r(rß )) œ (r cos ))i (r sin ))j rk , 0 Ÿ r Ÿ 1 (since 0 Ÿ z Ÿ 1) and 0 Ÿ ) Ÿ 21
â
â
i
j
kâ
â
â
â
Ê rr œ (cos ))i (sin ))j k and r) œ (r sin ))i (r cos ))j Ê r) ‚ rr œ â r sin ) r cos ) 0 â
â
â
sin ) 1 â
â cos )
rr
$
#
#
œ (r cos ))i (r sin ))j rk Ê F † n d5 œ F † krr)) ‚
‚rr k kr) ‚ rr k d) dr œ ar sin ) cos ) r b d) dr since
F œ ar# sin ) cos )b i rk Ê ' ' F † n d5 œ '0 '0 ar$ sin ) cos# ) r# b dr d) œ '0 ˆ "4 sin ) cos# ) "3 ‰ d)
21
#1
"
œ 12
cos$ ) 3) ‘ ! œ 231
21
1
S
26. Let the parametrization be r(rß )) œ (r cos ))i (r sin ))j 2rk , 0 Ÿ r Ÿ 1 (since 0 Ÿ z Ÿ 2) and 0 Ÿ ) Ÿ 21
â
â
i
j
kâ
â
â
â
Ê rr œ (cos ))i (sin ))j 2k and r) œ (r sin ))i (r cos ))j Ê r) ‚ rr œ â r sin ) r cos ) 0 â
â
â
sin ) 2 â
â cos )
rr
œ (2r cos ))i (2r sin ))j rk Ê F † n d5 œ F † krr)) ‚
‚rr k kr) ‚ rr k d) dr
œ a2r$ sin# ) cos ) 4r$ cos ) sin ) rb d) dr since
F œ ar# sin# )b i a2r# cos )b j k Ê ' ' F † n d5 œ '0 '0 a2r$ sin# ) cos ) 4r$ cos ) sin ) rb dr d)
21
1
S
21
#1
œ '0 ˆ "2 sin# ) cos ) cos ) sin ) "2 ‰ d) œ "6 sin$ ) 12 sin# ) "# )‘ ! œ 1
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 16.6 Surface Integrals
27. Let the parametrization be r(rß )) œ (r cos ))i (r sin ))j rk , 1 Ÿ r Ÿ 2 (since 1 Ÿ z Ÿ 2) and 0 Ÿ ) Ÿ 21
â
â
i
j
kâ
â
â
â
Ê rr œ (cos ))i (sin ))j k and r) œ (r sin ))i (r cos ))j Ê r) ‚ rr œ â r sin ) r cos ) 0 â
â
â
sin ) 1 â
â cos )
rr
#
#
#
#
$
œ (r cos ))i (r sin ))j rk Ê F † n d5 œ F † krr)) ‚
‚rr k kr) ‚ rr k d) dr œ ar cos ) r sin ) r b d) dr
œ ar# r$ b d) dr since F œ (r cos ))i (r sin ))j r# k Ê ' ' F † n d5 œ '0 '1 ar# r$ b dr d) œ 7361
21
2
S
28. Let the parametrization be r(rß )) œ (r cos ))i (r sin ))j r# k , 0 Ÿ r Ÿ 1 (since 0 Ÿ z Ÿ 1) and 0 Ÿ ) Ÿ 21
â
â
i
j
kâ
â
â
â
Ê rr œ (cos ))i (sin ))j 2rk and r) œ (r sin ))i (r cos ))j Ê r) ‚ rr œ â r sin ) r cos ) 0 â
â
â
sin ) 2r â
â cos )
rr
$
#
$
#
œ a2r# cos )b i a2r# sin )b j rk Ê F † n d5 œ F † krr)) ‚
‚rr k kr) ‚ rr k d) dr œ a8r cos ) 8r sin ) 2rb d) dr
œ a8r$ 2rb d) dr since F œ (4r cos ))i (4r sin ))j 2k Ê ' ' F † n d5 œ '0 '0 a8r$ 2rb dr d) œ 21
21
1
S
29. g(xß yß z) œ z, p œ k Ê ™ g œ k Ê k ™ gk œ 1 and k ™ g † pk œ 1 Ê Flux œ ' ' F † n d5 œ ' ' (F † k) dA
R
S
œ '0 '0 3 dy dx œ 18
2
3
30. g(xß yß z) œ y, p œ j Ê ™ g œ j Ê k ™ gk œ 1 and k ™ g † pk œ 1 Ê Flux œ ' ' F † n d5 œ ' ' (F † j) dA
R
S
2
7
2
œ 'c1 '2 2 dz dx œ 'c1 2(7 2) dx œ 10(2 1) œ 30
2yj 2zk
31. ™ g œ 2xi 2yj 2zk Ê k ™ gk œ È4x# 4y# 4z# œ 2a; n œ #2xÈi œ xi yaj zk Ê F † n œ za ;
x # y# z#
#
' ' Š za ‹ ˆ az ‰ dA œ ' ' z dA œ ' ' Èa# ax# y# b dx dy
k ™ g † kk œ 2z Ê d5 œ 2a
2z dA Ê Flux œ
#
œ '0
1Î2
'0 Èa# r# r dr d) œ 1a6
a
R
R
R
$
2yj 2zk
32. ™ g œ 2xi 2yj 2zk Ê k ™ gk œ È4x# 4y# 4z# œ 2a; n œ #2xÈi œ xi yaj zk Ê F † n œ axy xy
a
x # y# z#
' ' F † n d5 œ ' ' 0 d 5 œ 0
œ 0; k ™ g † kk œ 2z Ê d5 œ 2a
2z dA Ê Flux œ
S
S
xy
z
z
' ' ˆ az ‰ ˆ za ‰ dA
33. From Exercise 31, n œ xi yaj zk and d5 œ za dA Ê F † n œ xy
a a a œ a Ê Flux œ
œ ' ' 1 dA œ 1a4
R
#
R
#
#
#
#
34. From Exercise 31, n œ xi yaj zk and d5 œ az dA Ê F † n œ zxa zya za œ z Š x ya z ‹ œ az
#
$
Ê Flux œ ' ' (za) ˆ az ‰ dx dy œ ' ' a# dx dy œ a# (Area of R) œ 4" 1a%
R
R
#
35. From Exercise 31, n œ xi yaj zk and d5 œ za dA Ê F † n œ xa ya za œ a Ê Flux
#
œ ' ' a ˆ az ‰ dA œ ' '
R
R
a#
z dA œ
' ' È # a# #
œ '0 a# ’Èa# r# “ d) œ 1a#
1Î2
a
R
a ax y# b
dA œ '0
1Î2
'0a È a
#
a # r#
#
r dr d)
$
!
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
977
978
Chapter 16 Integration in Vector Fields
y#
#
36.
From Exercise 31, n œ xi yaj zk and d5 œ az dA
Ê F†nœ
Ê Flux œ ' ' az dx dy œ ' ' Èa# aax# y# b dx dy œ '0
1Î2
R
R
#
Š xa ‹ Œ a Š za ‹
È x # y # z#
'0 È a
a
a # r#
#
Š aa ‹
œ
a
r dr d) œ 1#a
œ1
#
k
37. g(xß yß z) œ y# z œ 4 Ê ™ g œ 2yj k Ê k ™ gk œ È4y# 1 Ê n œ È2y4yj #1
2xy 3z
Ê F†nœ È
; p œ k Ê k ™ g † pk œ 1 Ê d5 œ È4y# 1 dA Ê Flux
4y# 1
2xy 3z
œ ' ' ŠÈ
‹ È4y# 1 dA œ ' ' (2xy 3z) dA; z œ 0 and z œ 4 y# Ê y# œ 4
4y# 1
R
R
Ê Flux œ ' ' c2xy 3 a4 y bd dA œ '0 'c2 a2xy 12 3y# b dy dx œ '0 cxy# 12y y$ d # dx
1
#
2
1
#
R
œ '0 32 dx œ 32
1
38. g(xß yß z) œ x# y# z œ 0 Ê ™ g œ 2xi 2yj k Ê k ™ gk œ È4x# 4y# 1 œ È4 ax# y# b 1
#
#
i 2yj k
8y 2
Ê n œ È2x
Ê F † n œ È8x
; p œ k Ê k ™ g † pk œ 1 Ê d5 œ È4 ax# y# b 1 dA
4 ax# y# b 1
4 ax# y# b 1
8y 2
Ê Flux œ ' ' Š È8x
‹ È4 ax# y# b 1 dA œ ' ' a8x# 8y# 2b dA; z œ 1 and x# y# œ z
4 ax # y # b 1
#
#
R
Ê x y œ 1 Ê Flux œ '0 '0 a8r 2b r dr d) œ 21
#
21
#
1
R
#
39. g(xß yß z) œ y ex œ 0 Ê ™ g œ ex i j Ê k ™ gk œ Èe2x 1 Ê n œ Èe 2xi j
x
e 1
Ê k ™ g † pk œ ex Ê d5 œ
Èe2x 1
dA
ex
Ê Flux œ ' ' Š È2e2x 2y ‹ Š
x
R
œ ' ' 4 dA œ ' ' 4 dy dz œ 4
R
1
2
0
1
e 1
Èe2x 1
‹ dA œ
ex
40. g(xß yß z) œ y ln x œ 0 Ê ™ g œ "x i j Ê k ™ gk œ É x"# 1 œ
Ê nœ
Š "x i j‹
Œ
Ê Flux œ ' ' Š È2xy
‹Š
#
1
R
x
È 1 x#
‹ dA œ
x
x
e 1
' ' 2exex 2ex dA
R
È 1 x#
since 1 Ÿ x Ÿ e
x
2xy
i x j
È1 x# œ È1 x# Ê F † n œ È1 x# ; p œ j Ê k ™ g † pk œ 1 Ê d5 œ
x
Ê F † n œ È2e2x 2y ; p œ i
È 1 x#
dA
x
'01 '1e 2y dx dz œ '1e '01 2 ln x dz dx œ '1e 2 ln x dx
œ 2 cx ln x xd " œ 2(e e) 2(0 1) œ 2
e
41. On the face z œ a: g(xß yß z) œ z Ê ™ g œ k Ê k ™ gk œ 1; n œ k Ê F † n œ 2xz œ 2ax since z œ a;
d5 œ dx dy Ê Flux œ ' ' 2ax dx dy œ '0 '0 2ax dx dy œ a% .
a
a
R
On the face z œ 0: g(xß yß z) œ z Ê ™ g œ k Ê k ™ gk œ 1; n œ k Ê F † n œ 2xz œ 0 since z œ 0;
d5 œ dx dy Ê Flux œ ' ' 0 dx dy œ 0.
R
On the face x œ a: g(xß yß z) œ x Ê ™ g œ i Ê k ™ gk œ 1; n œ i Ê F † n œ 2xy œ 2ay since x œ a;
d5 œ dy dz Ê Flux œ '0 '0 2ay dy dz œ a% .
a
a
On the face x œ 0: g(xß yß z) œ x Ê ™ g œ i Ê k ™ gk œ 1; n œ i Ê F † n œ 2xy œ 0 since x œ 0
Ê Flux œ 0.
On the face y œ a: g(xß yß z) œ y Ê ™ g œ j Ê k ™ gk œ 1; n œ j Ê F † n œ 2yz œ 2az since y œ a;
d5 œ dz dx Ê Flux œ '0 '0 2az dz dx œ a% .
a
a
On the face y œ 0: g(xß yß z) œ y Ê ™ g œ j Ê k ™ gk œ 1; n œ j Ê F † n œ 2yz œ 0 since y œ 0
Ê Flux œ 0. Therefore, Total Flux œ 3a% .
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 16.6 Surface Integrals
979
42. Across the cap: g(xß yß z) œ x# y# z# œ 25 Ê ™ g œ 2x i 2yj 2zk Ê k ™ gk œ È4x# 4y# 4z# œ 10
#
g
x i y j zk
Ê n œ k™
Ê F † n œ x5z y5z 5z ; p œ k Ê k ™ g † pk œ 2z since z
™gk œ
5
#
0 Ê d5 œ 10
2z dA
Ê Fluxcap œ ' ' F † n d5 œ ' ' Š x5z y5z 5z ‹ ˆ 5z ‰ dA œ ' ' ax# y# 1b dx dy œ '0 '0 ar# 1b r dr d)
cap
21
#
#
R
21
œ '0 72 d) œ 1441.
4
R
Across the bottom: g(xß yß z) œ z œ 3 Ê ™ g œ k Ê k ™ gk œ 1 Ê n œ k Ê F † n œ 1; p œ k Ê k ™ g † pk œ 1
Ê d5 œ dA Ê Fluxbottom œ ' ' F † n d5 œ ' ' 1 dA œ 1(Area of the circular region) œ 161. Therefore,
R
bottom
Flux œ Fluxcap Fluxbottom œ 1281
43. ™ f œ 2x i 2yj 2zk Ê k ™ f k œ È4x# 4y# 4z# œ 2a; p œ k Ê k ™ f † pk œ 2z since z
0 Ê d5 œ 2a
2z dA
œ az dA; M œ ' ' $ d5 œ 8$ (surface area of sphere) œ $1#a ; Mxy œ ' ' z$ d5 œ $ ' ' z ˆ za ‰ dA œ a$ ' ' dA
#
œ a$ '0
1Î2
'
S
$
$
M
r dr d) œ $14a Ê z œ Mxy œ Š $14a ‹ ˆ $12a# ‰ œ #a .
0
R
R
S
a
Because of symmetry, x œ y œ 2a Ê the centroid is
ˆ #a ß #a ß #a ‰ .
44. ™ f œ 2y j 2zk Ê k ™ f k œ È4y# 4z# œ È4 ay# z# b œ 6; p œ k Ê k ™ f † kk œ 2z since z
6
0 Ê d5 œ 2z
dA
œ 3z dA; M œ ' ' 1 d5 œ 'c3 '0 3z dx dy œ 'c3 '0 È93 y# dx dy œ 91; Mxy œ ' ' z d5 œ 'c3 '0 z ˆ 3z ‰ dx dy œ 54;
3
3
3
3
3
S
Mxz œ ' ' y d5 œ ' ' y ˆ 3z ‰ dx dy œ ' '
S
Therefore, x œ
3
3
3
c3
0
c3
Š 27
# 1‹
91
3
S
' ' x d5 œ ' '
3
3y
dx dy œ 0; Myz œ
0 È 9 y#
S
3
3
c3
0
3x
27
È9 y# dx dy œ # 1.
œ 3# , y œ 0, and z œ 9541 œ 16
45. Because of symmetry, x œ y œ 0; M œ ' ' $ d5 œ $ ' ' d5 œ (Area of S)$ œ 31È2 $ ; ™ f œ 2x i 2yj 2zk
S
S
2È x y z
Ê k ™ f k œ È4x# 4y# 4z# œ 2Èx# y# z# ; p œ k Ê k ™ f † pk œ 2z Ê d5 œ
dA
#
œ
È 2 È x# y#
Èx# y# ax# y# b
dA œ
dA Ê Mxy œ $
z
z
È
œ $ '0 '1 È2 r# dr d) œ 1413 2 $ Ê z œ
21
œ''
R
2
È 2 È x# y#
ax# y# b Š
‹ $ dA œ $ È2
z
È
#
' ' z Š È2 Èzx# y# ‹ dA œ $ ' ' È2 Èx# y# dA
141 2
Π3 $
31 È 2 $
#
#z
R
R
14 ‰
' ' ax# y# b $ d5
ˆ
œ 14
9 Ê axß yß zb œ 0ß 0ß 9 . Next, Iz œ
S
' ' ax# y# b dA œ $ È2 ' '
21
0
R
È
r$ dr d) œ 151# 2 $
1
2
Iz
Ê Rz œ É M
œ
È10
#
46. f(xß yß z) œ 4x# 4y# z# œ 0 Ê ™ f œ 8xi 8yj 2zk Ê k ™ f k œ È64x# 64y# 4z#
È
œ 2È16x# 16y# z# œ 2È4z# z# œ 2È5 z since z
0; p œ k Ê k ™ f † pk œ 2z Ê d5 œ 2 2z5 z dA œ È5 dA
Ê Iz œ ' ' ax# y# b $ d5 œ $ È5 ' ' ax# y# b dx dy œ $ È5 'c1Î2 '0
1Î2
2 cos )
R
S
È
r$ dr d) œ 3 #51$
47. (a) Let the diameter lie on the z-axis and let f(xß yß z) œ x# y# z# œ a# , z 0 be the upper hemisphere
Ê ™ f œ 2xi 2yj 2zk Ê k ™ f k œ È4x# 4y# 4z# œ 2a, a 0; p œ k Ê k ™ f † pk œ 2z since z
y
Ê d5 œ az dA Ê Iz œ ' ' $ ax# y# b ˆ az ‰ d5 œ a$ ' ' Èa# x dA œ a$ '0 '0 È #r
ax # y # b
#
S
œ a$ '0 ’r# Èa# r# 23 aa# r# b
21
“ d) œ a$ '0
$Î# a
!
#
R
21
2 $
41 %
3 a d) œ 3 a $
21
a
#
a r#
r dr d)
Ê the moment of inertia is 831 a% $ for
the whole sphere
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
0
980
Chapter 16 Integration in Vector Fields
(b) IL œ Ic.m. mh# , where m is the mass of the body and h is the distance between the parallel lines; now,
Ic.m. œ 831 a% $ (from part a) and m# œ ' ' $ d5 œ $ ' ' ˆ az ‰ dA œ a$ ' ' È # " # # dy dx
œ a$ '0 '0 È "#
21
a r#
R
S
a ax y b
R
r dr d) œ a$ '0 ’Èa# r# “ d) œ a$ '0 a d) œ 21a# $ and h œ a
21
a
21
a
!
Ê IL œ 831 a% $ 41a# $ a# œ 2031 a% $
48. Let z œ ha Èx# y# be the cone from z œ 0 to z œ h, h 0. Because of symmetry, x œ 0 and y œ 0;
#
#
#
2yh
z œ ha Èx# y# Ê f(xß yß z) œ ha# ax# y# b z# œ 0 Ê ™ f œ 2xh
a# i a# j 2zk
# %
# %
%
#
#
#
Ê k ™ f k œ É 4xa%h 4ya%h 4z# œ 2É ha% ax# y# b ha# ax# y# b œ 2Ɉ ha# ‰ ax# y# b ˆ ha# 1‰
#
#
œ 2Éz# ˆ h a# a ‰ œ ˆ 2za ‰ Èh# a# since z
œ
È h # a#
dA; M œ
a
0; p œ k Ê k ™ f † pk œ 2z Ê d5 œ
ˆ 2z
‰ È h # a#
a
dA
2z
' ' d5 œ ' ' Èh#a a# dA œ Èh#a a# a1a# b œ 1aÈh# a# ;
R
S
#
#
#
È #
È #
È #
Mxy œ ' ' z d5 œ ' ' z Š h a a ‹ dA œ h a a ' ' ha Èx# y# dx dy œ h ha# a '0 '0 r# dr d)
21
R
S
#
È #
œ 21ah 3h a
R
M
z œ Mxy œ 2h
3
Ê
a
‰
Ê the centroid is ˆ0ß 0ß 2h
3
16.7 STOKES' THEOREM
â i
â
â
1. curl F œ ™ ‚ F œ â ``x
â #
âx
j
`
`y
2x
k ââ
` â
` z â œ 0i 0j (2 0)k œ 2k and n œ k Ê curl F † n œ 2 Ê d5 œ dx dy
â
z# â
Ê )C F † dr œ ' ' 2 dA œ 2(Area of the ellipse) œ 41
R
â i
â
â
2. curl F œ ™ ‚ F œ â ``x
â
â 2y
j
`
`y
3x
k ââ
` â
` z â œ 0i 0j (3 2)k œ k and n œ k Ê curl F † n œ 1 Ê d5 œ dx dy
â
z# â
Ê )C F † dr œ ' ' dx dy œ Area of circle œ 91
R
â i
â
â
3. curl F œ ™ ‚ F œ â ``x
â
â y
j
`
`y
xz
k ââ
` â
ijk
Ê curl F † n
` z â œ xi 2xj (z 1)k and n œ
È3
â
#â
x
œ È"3 (x 2x z 1) Ê d5 œ
1cx
œ '0 '0
1
È3
1 dA
Ê )C F † dr œ ' '
1cx
[3x (1 x y) 1] dy dx œ '0 '0
1
œ '0 ˆ "# 3x #7 x# ‰ dx œ 56
R
"
È3 dA
È3 (3x z 1)
(4x y) dy dx œ '0 4x(1 x) "# (1 x)# ‘ dx
1
1
â
i
â
â
4. curl F œ ™ ‚ F œ â ``x
â #
â y z#
j
`
`y
x# z#
â
â
â
â œ (2y 2z)i (2z 2x)j (2x 2y)k and n œ i Èj 3 k
â
x# y# â
k
`
`z
Ê curl F † n œ È"3 (2y 2z 2z 2x 2x 2y) œ 0 Ê )C F † dr œ ' ' 0 d5 œ 0
S
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 16.7 Stoke's Theorem
â
i
â
â
5. curl F œ ™ ‚ F œ â ``x
â #
â y z#
j
981
â
â
â
â œ 2yi (2z 2x)j (2x 2y)k and n œ k
â
#
#â
x y
k
`
`y
x# y#
`
`z
Ê curl F † n œ 2x 2y Ê d5 œ dx dy Ê )C F † dr œ 'c1 'c1 (2x 2y) dx dy œ 'c1 cx# 2xyd " dy
1
1
1
"
œ 'c1 4y dy œ 0
1
â i
â
â
6. curl F œ ™ ‚ F œ â ``x
â # $
âx y
k ââ
` â
2xi 2yj 2zk
xi yj zk
# #
` z â œ 0i 0j 3x y k and n œ 2Èx# y# z# œ
4
â
z â
j
`
`y
1
Ê curl F † n œ 34 x# y# z; d5 œ 4z dA (Section 16.6, Example 6, with a œ 4)
œ 3 '0 '0 ar cos )b ar sin )b r dr d) œ 3'
21
2
#
#
#
Ê )C F † dr œ ' ' ˆ 34 x# y# z‰ ˆ 4z ‰ dA
' #
’ r6 “ (cos ) sin ))# d) œ 32
0
!
#
%1
21
'0
21
R
"
#
4 sin 2) d) œ 4
'041 sin# u du
œ 4 u2 sin42u ‘ ! œ 81
7. x œ 3 cos t and y œ 2 sin t Ê F œ (2 sin t)i a9 cos# tb j a9 cos# t 16 sin% tb sin eÈÐ6 sin t cos tÑÐ0Ñ k at the
base of the shell; r œ (3 cos t)i (2 sin t)j Ê dr œ (3 sin t)i (2 cos t)j Ê F † ddtr œ 6 sin# t 18 cos$ t
Ê ' ' ™ ‚ F † n d5 œ '0 a6 sin# t 18 cos$ tb dt œ 3t 32 sin 2t 6(sin t) acos# t 2b‘ ! œ 61
21
#1
S
â
i
â
â
`
8. curl F œ ™ ‚ F œ ââ
`x
â z "
â
#x
â
â
â
`
`
â œ 2j ; f(xß yß z) œ 4x# y z# Ê ™ f œ 8xi j 2zk
`y
`z
â
tan" y x 4 " z ââ
j
k
k™f k
f
"
2
Ê n œ k™
™f k and p œ j Ê k ™ f † pk œ 1 Ê d5 œ k™f†pk dA œ k ™ f k dA; ™ ‚ F † n œ k™f k (2j † ™ f) œ k™f k
Ê ™ ‚ F † n d5 œ 2 dA Ê ' ' ™ ‚ F † n d5 œ ' ' 2 dA œ 2(Area of R) œ 2(1 † 1 † 2) œ 41, where R
R
S
is the elliptic region in the xz-plane enclosed by 4x# z# œ 4.
9. Flux of ™ ‚ F œ ' ' ™ ‚ F † n d5 œ )C F † dr, so let C be parametrized by r œ (a cos t)i (a sin t)j ,
S
0 Ÿ t Ÿ 21 Ê ddtr œ (a sin t)i (a cos t)j Ê F † ddtr œ ay sin t ax cos t œ a# sin# t a# cos# t œ a#
Ê Flux of ™ ‚ F œ )C F † dr œ '0 a# dt œ 21a#
21
â i
â
â
10. ™ ‚ (yi) œ â ``x
â
â y
j
`
`y
0
k ââ
™f
2xi 2yj 2zk
` â
` z â œ k ; n œ k™f k œ 2Èx# y# z# œ xi yj zk
â
0 â
Ê ™ ‚ (yi) † n œ z; d5 œ "z dA (Section 16.6, Example 6, with a œ 1) Ê ' ' ™ ‚ (yi) † n d5
S
œ ' ' (z) ˆ "z dA‰ œ ' ' dA œ 1, where R is the disk x# y# Ÿ 1 in the xy-plane.
R
R
11. Let S" and S# be oriented surfaces that span C and that induce the same positive direction on C. Then
' ' ™ ‚ F † n" d5" œ ) F † dr œ ' ' ™ ‚ F † n# d5#
S"
C
S#
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
982
Chapter 16 Integration in Vector Fields
12. ' ' ™ ‚ F † n d5 œ ' ' ™ ‚ F † n d5 ' ' ™ ‚ F † n d5, and since S" and S# are joined by the simple
S
S"
S#
closed curve C, each of the above integrals will be equal to a circulation integral on C. But for one surface
the circulation will be counterclockwise, and for the other surface the circulation will be clockwise. Since the
integrands are the same, the sum will be 0 Ê ' ' ™ ‚ F † n d5 œ 0.
S
â i
j
k ââ
â
â `
â
`
13. ™ ‚ F œ â ` x ` y ``z â œ 5i 2j 3k ; rr œ (cos ))i (sin ))j 2rk and r) œ (r sin ))i (r cos ))j
â
â
â 2z 3x 5y â
â
â
i
j
k â
â
â
â
r)
sin ) 2r â œ a2r# cos )b i a2r# sin )b j rk ; n œ krrrr ‚
Ê rr ‚ r) œ â cos )
‚r) k and d5 œ krr ‚ r) k dr d)
â
â
0 â
â r sin ) r cos )
Ê ™ ‚ F † n d5 œ ( ™ ‚ F) † (rr ‚ r) ) dr d) œ a10r# cos ) 4r# sin ) 3rb dr d) Ê ' ' ™ ‚ F † n d5
œ '0 '0 a10r cos ) 4r sin ) 3rb dr d) œ '
21
2
#
S
21
"30 r$ cos ) 43 r$ sin ) 3# r# ‘ # d)
!
0
#
32
‰
œ '0 ˆ 80
3 cos ) 3 sin ) 6 d) œ 6(21) œ 121
21
â i
j
â
â
`
14. ™ ‚ F œ â ``x
`y
â
ây z z x
k ââ
â
`
#
#
` z â œ i 2j 2k ; rr ‚ r) œ a2r cos )b i a2r sin )b j rk and
â
x zâ
™ ‚ F † n d5 œ ( ™ ‚ F) † (rr ‚ r) ) dr d) (see Exercise 13 above) Ê ' ' ™ ‚ F † n d5
œ '0 '0 a2r cos ) 4r sin ) 2rb dr d) œ '
21
3
#
S
21
23 r$ cos ) 43 r$ sin ) r# ‘ $ d)
!
0
#
œ '0 a18 cos ) 36 sin ) 9b d) œ 9(21) œ 181
21
â i
â
â
j
k ââ
i
j
kâ
â
â
â `
â
â
`
` â
$
#
sin ) 1 â
15. ™ ‚ F œ â ` x
`y
` z â œ 2y i 0j x k ; rr ‚ r) œ â cos )
â #
â
â
â
â x y 2y$ z 3z â
â r sin ) r cos ) 0 â
œ (r cos ))i (r sin ))j rk and ™ ‚ F † n d5 œ ( ™ ‚ F) † (rr ‚ r) ) dr d) (see Exercise 13 above)
Ê ' ' ™ ‚ F † n d5 œ ' ' a2ry$ cos ) rx# b dr d) œ '0 '0 a2r% sin3 ) cos ) r$ cos# )b dr d)
21
œ'
1
R
S
21
#1
"
ˆ 25 sin3 ) cos ) 4" cos# )‰ d) œ 10
sin4 ) 4" ˆ #) sin42) ‰‘ ! œ 14
0
â i
â
â
16. ™ ‚ F œ â ``x
â
âx y
â
â
k ââ
i
j
k â
â
â
â
â
`
sin ) 1 â
` z â œ i j k ; rr ‚ r) œ â cos )
â
â
â
â r sin ) r cos ) 0 â
y z z xâ
œ (r cos ))i (r sin ))j rk and ™ ‚ F † n d5 œ ( ™ ‚ F) † (rr ‚ r) ) dr d) (see Exercise 13 above)
j
`
`y
‰
Ê ' ' ™ ‚ F † n d5 œ '0 '0 (r cos ) r sin ) r) dr d) œ '0 ’(cos ) sin ) 1) r# “ d) œ ˆ 25
# (21) œ 251
21
21
5
#
&
!
S
â i
â
â
i
j
k
j
k ââ
â
â
â
â `
â
âÈ
â
`
`
È
È
3
cos
cos
3
cos
sin
3
sin
9
)
9
)
9
i
j
k
r
r
œ
‚
œ
17. ™ ‚ F œ â ` x
0
0
5
;
â
â
â
9
)
`y
`z
â
â
â
â
#
È
È
â 3y 5 2x z 2 â
â 3 sin 9 sin )
â
3 sin 9 cos )
0
#
#
œ a3 sin 9 cos )b i a3 sin 9 sin )b j (3 sin 9 cos 9)k ; ™ ‚ F † n d5 œ ( ™ ‚ F) † (r9 ‚ r) ) d9 d) (see Exercise
# ‘
13 above) Ê ' ' ™ ‚ F † n d5 œ '0 '0 15 cos 9 sin 9 d9 d) œ '0 15
2 cos 9 !
21
S
1/2
21
1Î#
d) œ '0 15
# d) œ 151
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
21
Section 16.7 Stoke's Theorem
983
â
â
â
j
k ââ
i
j
k
â i
â
â
â `
â
â
`
` â
18. ™ ‚ F œ â ` x ` y ` z â œ 2zi j 2yk ; r9 ‚ r) œ â 2 cos 9 cos ) 2 cos 9 sin ) 2 sin 9 â
â #
â
â
â
0
ây
â 2 sin 9 sin ) 2 sin 9 cos )
â
z# x â
#
#
œ a4 sin 9 cos )b i a4 sin 9 sin )b j (4 sin 9 cos 9)k ; ™ ‚ F † n d5 œ ( ™ ‚ F) † (r9 ‚ r) ) d9 d) (see Exercise
13 above) Ê ' ' ™ ‚ F † n d5 œ ' ' a8z sin# 9 cos ) 4 sin# 9 sin ) 8y sin 9 cos )b d9 d)
R
S
21
1/2
œ '0 '0 a16 sin# 9 cos 9 cos ) 4 sin# 9 sin ) 16 sin# 9 sin ) cos )b d9 d)
21
1Î#
œ '0 "36 sin$ 9 cos ) 4 ˆ 9# sin429 ‰ (sin )) 16 ˆ 9# sin429 ‰ (sin ) cos ))‘ ! d)
21
# ‘ #1
‰
"6
œ '0 ˆ 16
3 cos ) 1 sin ) 41 sin ) cos ) d) œ 3 sin ) 1 cos ) 21 sin ) ! œ 0
19. (a) F œ 2xi 2yj 2zk Ê curl F œ 0 Ê )C F † dr œ ' ' ™ ‚ F † n d5 œ ' ' 0 d5 œ 0
S
S
(b) Let f(xß yß z) œ x y z Ê ™ ‚ F œ ™ ‚ ™ f œ 0 Ê curl F œ 0 Ê )C F † dr œ ' ' ™ ‚ F † n d5 œ ' ' 0 d5
# # $
S
S
œ0
(c) F œ ™ ‚ (xi yj zk) œ 0 Ê ™ ‚ F œ 0 Ê )C F † dr œ ' ' ™ ‚ F † n d5 œ ' ' 0 d5 œ 0
S
S
(d) F œ ™ f Ê ™ ‚ F œ ™ ‚ ™ f œ 0 Ê )C F † dr œ ' ' ™ ‚ F † n d5 œ ' ' 0 d5 œ 0
S
20. F œ ™ f œ "# ax# y# z# b
$Î#
# $Î#
(2x)i "# ax# y# z# b
$Î#
# $Î#
S
(2y)j "# ax# y# z# b
$Î#
(2z)k
# $Î#
i y ax# y# z b
j z ax# y# z b
k
œ x ax# y# z b
dr
(a) r œ (a cos t)i (a sin t)j , 0 Ÿ t Ÿ 21 Ê dt œ (a sin t)i (a cos t)j
Ê F † ddtr œ x ax# y# z# b
$Î#
(a sin t) y ax# y# z# b
$Î#
(a cos t)
t‰
t‰
œ ˆ a cos
(a sin t) ˆ a sin
(a cos t) œ 0 Ê )C F † dr œ 0
a$
a$
(b) )C F † dr œ ' ' ™ ‚ F † n d5 œ ' ' ™ ‚ ™ f † n d5 œ ' ' 0 † n d5 œ ' ' 0 d5 œ 0
S
S
S
â i
â
â
21. Let F œ 2yi 3zj xk Ê ™ ‚ F œ â ``x
â
â 2y
j
`
`y
3z
S
k ââ
2i 2j k
` â
` z â œ 3i j 2k ; n œ
3
â
x â
Ê ™ ‚ F † n œ 2 Ê )C 2y dx 3z dy x dz œ )C F † dr œ ' ' ™ ‚ F † n d5 œ ' ' 2 d5
S
S
œ 2 ' ' d5, where ' ' d5 is the area of the region enclosed by C on the plane S: 2x 2y z œ 2
S
S
â i
â
â
22. ™ ‚ F œ â ``x
â
â x
j
`
`y
y
k ââ
` â
`z ⠜ 0
â
z â
23. Suppose F œ Mi Nj Pk exists such that ™ ‚ F œ Š `` Py ``Nz ‹ i ˆ ``Mz `` Px ‰ j Š ``Nx ``My ‹ k
#
#
œ xi yj zk . Then ``x Š `` Py ``Nz ‹ œ ``x (x) Ê ``x`Py ``x`Nz œ 1. Likewise, ``y ˆ ``Mz `` Px ‰ œ ``y (y)
#
#
#
#
Ê ``y`Mz ``y`Px œ 1 and ``z Š ``Nx ``My ‹ œ ``z (z) Ê ``z`Nx ``z`My œ 1. Summing the calculated equations
#
#
#
#
#
#
Ê Š ``x`Py ``y`Px ‹ Š ``z`Nx ``x`Nz ‹ Š ``y`Mz ``z`My ‹ œ 3 or 0 œ 3 (assuming the second mixed partials are
equal). This result is a contradiction, so there is no field F such that curl F œ xi yj zk .
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
984
Chapter 16 Integration in Vector Fields
24. Yes: If ™ ‚ F œ 0 , then the circulation of F around the boundary C of any oriented surface S in the domain of
F is zero. The reason is this: By Stokes's theorem, circulation œ )C F † dr œ ' ' ™ ‚ F † n d5 œ ' ' 0 † n d5 œ 0.
S
S
#
25. r œ Èx# y# Ê r% œ ax# y# b Ê F œ ™ ar% b œ 4x ax# y# b i 4y ax# y# b j œ Mi Nj
Ê )C ™ ar% b † n ds œ )C F † n ds œ )C M dy N dx œ ' ' Š ``Mx ``Ny ‹ dx dy
R
œ ' ' c4 ax# y# b 8x# 4 ax# y# b 8y# d dA œ ' ' 16 ax# y# b dA œ 16 ' ' x# dA 16 ' ' y# dA
R
R
R
R
œ 16Iy 16Ix .
#
#
#
#
#
#
#
#
26. `` Py œ 0, ``Nz œ 0, ``Mz œ 0, `` Px œ 0, ``Nx œ axy# yx# b# , ``My œ axy# yx# b# Ê curl F œ ’ axy# yx# b# axy# yx# b# “ k œ 0 .
However, x# y# œ 1 Ê r œ (cos t)i (sin t)j Ê ddtr œ (sin t)i (cos t)j
#1
Ê F œ a sin tb i acos tb j Ê F † ddtr œ sin# t cos# t œ 1 Ê )C F † dr œ ) 1 dt œ 21 which is not zero.
!
16.8 THE DIVERGENCE THEOREM AND A UNIFIED THEORY
y i x j
1. F œ È
Ê div F œ
x# y#
xy xy
œ0
ax# y# b$Î#
2. F œ xi yj Ê div F œ 1 1 œ 2
#
# $Î#
#
#
#
# "Î#
#
ax y z b
i yj zk)
3. F œ GM(x
Ê div F œ GM ” ax y z abx# y3x
•
# z # b$
ax# y# z# b$Î#
#
# $Î#
#
#
#
# "Î#
#
#
# $Î#
#
#
#
#
# "Î#
ax y z b
ax y z b 3z ax y z b
GM ” ax y z abx# y3y
• GM ”
•
# z # b$
ax# y# z# b$
#
# #
#
#
#
#
#
#
#
z b ax y z b
œ GM ’ 3 ax y z b # 3 ax# y #(Î#
“œ0
ax y z b
4. z œ a# r# in cylindrical coordinates Ê z œ a# ax# y# b Ê v œ aa# x# y# b k Ê div v œ 0
5.
`
`
`
` x (y x) œ 1, ` y (z y) œ 1, ` z (y x) œ 0 Ê
6.
`
`
`
#
#
#
` x ax b œ 2x, ` y ay b œ 2y, ` x az b œ 2z
™ † F œ 2 Ê Flux œ 'c1 'c1 'c1 2 dx dy dz œ 2 a2$ b œ 16
1
1
1
Ê ™ † F œ 2x 2y 2z
(a) Flux œ '0 '0 '0 (2x 2y 2z) dx dy dz œ '0 '0 cx# 2x(y z)d ! dy dz œ '0 '0 (1 2y 2z) dy dz
1
1
1
1
1
1
"
1
œ '0 cy(1 2z) y# d ! dz œ '0 (2 2z) dz œ c2z z# d ! œ 3
1
1
"
"
(b) Flux œ 'c1 'c1 'c1 (2x 2y 2z) dx dy dz œ 'c1 'c1 cx# 2x(y z)d " dy dz œ 'c1 'c1 (4y 4z) dy dz
1
1
1
1
1
1
"
1
œ 'c1 c2y# 4yzd " dz œ 'c1 8z dz œ c4z# d " œ 0
1
"
1
"
(c) In cylindrical coordinates, Flux œ ' ' ' (2x 2y 2z) dx dy dz
D
#
œ '0 '0 '0 (2r cos ) 2r sin ) 2z) r dr d) dz œ '0 '0 23 r$ cos ) 23 r$ sin ) zr# ‘ ! d) dz
1
21
2
1
21
16
' "36 sin ) 163 cos ) 4z)‘ #!1 dz œ '0 81z dz œ c41z# d "! œ 41
‰
œ '0 '0 ˆ 16
3 cos ) 3 sin ) 4z d) dz œ 0
1
7.
21
1
`
`
`
` x (y) œ 0, ` y (xy) œ x, ` z (z) œ 1
1
Ê ™ † F œ x 1; z œ x# y# Ê z œ r# in cylindrical coordinates
Ê Flux œ ' ' ' (x 1) dz dy dx œ '0 '0 '0 (r cos ) 1) dz r dr d) œ '0 '0 ar$ cos ) r# b r dr d)
21
œ'
D
#
r&
r%
cos
)
d) œ
’
“
5
4
0
!
21
2
r#
21
2
'021 ˆ 325 cos ) 4‰ d) œ 325 sin ) 4)‘ #!1 œ 81
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 16.8 The Divergence Theorem and a Unified Theory
8.
985
Ê ™ † F œ 2x 3 Ê Flux œ ' ' ' (2x 3) dV
`
`
`
#
` x ax b œ 2x, ` y (xz) œ 0, ` z (3z) œ 3
D
#
21
2
21
1
1
œ '0 '0 '0 (23 sin 9 cos ) 3) a3# sin 9b d3 d9 d) œ '0 '0 ’ 32 sin 9 cos ) 3$ “ sin 9 d9 d)
%
!
21
21
21
1
1
œ '0 '0 (8 sin 9 cos ) 8) sin 9 d9 d) œ '0 8 ˆ 92 sin429 ‰ cos ) 8 cos 9‘ ! d) œ '0 (41 cos ) 16) d) œ 321
9.
Ê Flux œ ' ' ' 3x dx dy dz
`
`
`
#
` x ax b œ 2x, ` y (2xy) œ 2x, ` z (3xz) œ 3x
D
2
1Î2
1 Î2
1 Î2
1 Î2
1 Î2
œ '0 '0 '0 (33 sin 9 cos )) a3# sin 9b d3 d9 d) œ '0 '0 12 sin# 9 cos ) d9 d) œ '0 31 cos ) d) œ 31
10. ``x a6x# 2xyb œ 12x 2y, ``y a2y x# zb œ 2, ``z a4x# y$ b œ 0 Ê ™ † F œ 12x 2y 2
Ê Flux œ ' ' ' (12x 2y 2) dV œ '0 '0
3
D
1Î2
'02 (12r cos ) 2r sin ) 2) r dr d) dz
16 ‰
' ˆ
‰
œ '0 '0 ˆ32 cos ) 16
3 sin ) 4 d) dz œ 0 32 21 3 dz œ 112 61
3
1Î2
3
11. ``x (2xz) œ 2z, ``y (xy) œ x, ``z az# b œ 2z Ê ™ † F œ x Ê Flux œ ' ' ' x dV
È16 c 4x
œ '0 '0
2
#
4cy
'0
È16 c 4x
x dz dy dx œ '0 '0
2
D
#
(xy 4x) dy dx œ '
2
’ 12 x a16 4x# b 4xÈ16 4x# “ dx
0
$Î# #
œ ’4x# "2 x% "3 a16 4x# b
“ œ 40
3
!
12. ``x ax$ b œ 3x# , ``y ay$ b œ 3y# , ``z az$ b œ 3z# Ê ™ † F œ 3x# 3y# 3z# Ê Flux œ ' ' ' 3 ax# y# z# b dV
21
a
21
21
1
1
œ 3 '0 '0 '0 3# a3# sin 9b d3 d9 d) œ 3 '0 '0 a5 sin 9 d9 d) œ 3 '0 2a5 d) œ 1251a
&
&
D
&
13. Let 3 œ Èx# y# z# . Then `` 3x œ 3x , `` 3y œ 3y , `` 3z œ 3z Ê ``x (3x) œ Š `` 3x ‹ x 3 œ x3 3 , ``y (3y) œ Š `` 3y ‹ y 3
#
#
#
#
#
œ y3 3, ``z (3z) œ Š `` 3z ‹ z 3 œ z3 3 Ê ™ † F œ x y3 z 33 œ 43, since 3 œ Èx# y# z#
#
È2
Ê Flux œ ' ' ' 43 dV œ '0 '0 '1
21
1
D
(43) a3# sin 9b d3 d9 d) œ '0 '0 3 sin 9 d9 d) œ '0 6 d) œ 121
21
1
21
14. Let 3 œ Èx# y# z# . Then `` 3x œ 3x , `` 3y œ 3y , `` 3z œ 3z Ê ``x Š 3x ‹ œ 3" Š 3x# ‹ `` 3x œ 31 3x$ . Similarly,
#
y
y#
`
"
`
z
"
z#
` y Š 3 ‹ œ 3 3$ and ` z Š 3 ‹ œ 3 3$
Ê Flux œ ' ' '
D
2
3 dV œ
#
#
#
Ê ™ † F œ 33 x 3y$ z œ 32
'021 '01 '12 Š 32 ‹ a3# sin 9b d3 d9 d) œ '021 '01 3 sin 9 d9 d) œ '021 6 d) œ 121
15. ``x a5x$ 12xy# b œ 15x# 12y# , ``y ay$ ey sin zb œ 3y# ey sin z, ``z a5z$ ey cos zb œ 15z# ey sin z
È2
Ê ™ † F œ 15x# 15y# 15z# œ 153# Ê Flux œ ' ' ' 153# dV œ '0 '0 '1
21
1
a153# b a3# sin 9b d3 d9 d)
D
œ '0 '0 Š12È2 3‹ sin 9 d9 d) œ '0 Š24È2 6‹ d) œ Š48È2 12‹ 1
21
1
21
` ˆ
2z
" y ‰
ˆ 2z ‰
16. ``x cln ax# y# bd œ x# 2x
y# , ` y x tan
x œ x –
Š "x ‹
1 ˆx‰ —
y #
` ˆ È #
# ‰ œ Èx# y#
œ x# 2z
y# , ` z z x y
2z
Èx# y# Ê Flux œ ' ' ' Š # 2x # # 2z # Èx# y# ‹ dz dy dx
Ê ™ † F œ x# 2x
y# x# y# x y
x y
D
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
986
Chapter 16 Integration in Vector Fields
È2
œ '0 '1
21
È
)
'c21 ˆ 2r cos
' 21 '1 2 ˆ6 cos ) 3r 3r# ‰ dr d)
‰
2z
r
r r dz r dr d) œ 0
#
#
œ '0 ’6 ŠÈ2 1‹ cos ) 3 ln È2 2È2 1“ d) œ 21 Š 3# ln 2 2È2 1‹
21
17. (a) G œ Mi Nj Pk Ê ™ ‚ G œ curl G œ Š `` Py ``Nz ‹ i ˆ ``Mz `` Px ‰ k Š ``Nx ``My ‹ k Ê ™ † ™ ‚ G
œ div(curl G) œ ``x Š `` Py ``Nz ‹ ``y ˆ ``Mz `` Px ‰ ``z Š ``Nx ``My ‹
#
#
#
#
#
#
œ ``x`Py ``x`Nz ``y`Mz ``y`Px ``z`Nx ``z`My œ 0 if all first and second partial derivatives are continuous
(b) By the Divergence Theorem, the outward flux of ™ ‚ G across a closed surface is zero because
outward flux of ™ ‚ G œ ' ' ( ™ ‚ G) † n d5
œ ' ' ' ™ † ™ ‚ G dV
S
[Divergence Theorem with F œ ™ ‚ G]
D
œ ' ' ' (0) dV œ 0
[by part (a)]
D
18. (a) Let F" œ M" i N" j P" k and F# œ M# i N# j P# k Ê aF" bF#
œ (aM" bM# )i (aN" bN# )j (aP" bP# )k Ê ™ † (aF" bF# )
œ ˆa ``Mx" b ``Mx# ‰ Ša ``Ny" b ``Ny# ‹ ˆa ``Pz" b ``Pz# ‰
œ a Š ``Mx" ``Ny" ``Pz" ‹ b Š ``Mx# ``Ny# ``Pz# ‹ œ a( ™ † F" ) b( ™ † F# )
(b) Define F" and F# as in part a Ê ™ ‚ (aF" bF# )
œ ’Ša ``Py" b ``Py# ‹ ˆa ``Nz" b ``Nz# ‰“ i ˆa ``Mz" b ``Mz# ‰ ˆa ``Px" b ``Px# ‰‘ j
’ˆa ``Nx" b ``Nx# ‰ Ša ``My" b ``My# ‹“ k œ a ’Š ``Py" ``Nz" ‹ i ˆ ``Mz" ``Px" ‰ j Š ``Nx" ``My" ‹ k“
b ’Š ``Py# ``Nz# ‹ i ˆ ``Mz# ``Px# ‰ j Š ``Nx# ``My# ‹ k“ œ a ™ ‚ F" b ™ ‚ F#
â
â
j
kâ
â i
â
â
(c) F" ‚ F# œ â M" N" P" â œ (N" P# P" N# )i (M" P# P" M# )j (M" N# N" M# )k Ê ™ † (F" ‚ F# )
â
â
â M# N# P# â
œ ™ † [(N" P# P" N# )i (M" P# P" M# )j (M" N# N" M# )k]
œ ``x (N" P# P" N# ) ``y (M" P# P" M# ) ``z (M" N# N" M# ) œ ˆP# ``Nx" N" ``Px# N# ``Px" P" ``Nx# ‰
ŠM" ``Py# P# ``My" P" ``My# M# ``Py" ‹ ˆM" ``Nz# N# ``Mz" N" ``Mz# M# ``Nz" ‰
œ M# Š ``Py" ``Nz" ‹ N# ˆ ``Mz" ``Px" ‰ P# Š ``Nx" ``My" ‹ M" Š ``Nz# ``Py# ‹ N" ˆ ``Px# ``Mz# ‰
P" Š ``My# ``Nx# ‹ œ F# † ™ ‚ F" F" † ™ ‚ F#
19. (a) div(gF) œ ™ † gF œ ``x (gM) ``y (gN) ``z (gP) œ Šg ``Mx M `` xg ‹ Šg ``Ny N `` yg ‹ Šg ``Pz P `` gz ‹
œ ŠM `` gx N `` gy P `` gz ‹ g Š ``Mx ``Ny ``Pz ‹ œ g ™ † F ™ g † F
(b) ™ ‚ (gF) œ ’ ``y (gP) ``z (gN)“ i ``z (gM) ``x (gP)‘ j ’ ``x (gN) ``y (gM)“ k
œ ŠP `` gy g `` Py N `` gz g ``Nz ‹ i ŠM `` gz g ``Mz P `` xg g `` Px ‹ j ŠN `` xg g ``Nx M `` yg g ``My ‹ k
œ ŠP `` yg N `` gz ‹ i Šg `` Py g ``Nz ‹ i ŠM `` gz P `` xg ‹ j ˆg ``Mz g `` Px ‰ j ŠN `` xg M `` yg ‹ k
Šg ``Nx g ``My ‹ k œ g ™ ‚ F ™ g ‚ F
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Section 16.8 The Divergence Theorem and a Unified Theory
987
20. Let F" œ M" i N" j P" k and F# œ M# i N# j P# k .
(a) F" ‚ F# œ (N" P# P" N# )i (P" M# M" P# )j (M" N# N" M# )k Ê ™ ‚ (F" ‚ F# )
œ ’ ``y (M" N# N" M# ) ``z (P" M# M" P# )“ i ``z (N" P# P" N# ) ``x (M" N# N" M# )‘ j
’ ``x (P" M# M" P# ) ``y (N" P# P" N# )“ k
and consider the i-component only: ``y (M" N# N" M# ) ``z (P" M# M" P# )
œ N# ``My" M" ``Ny# M# ``Ny" N" ``My# M# ``Pz" P" ``Mz# P# ``Mz" M" ``Pz#
œ ŠN# ``My" P# ``Mz" ‹ ŠN" ``My# P" ``Mz# ‹ Š ``Ny# ``Pz# ‹ M" Š ``Ny" ``Pz" ‹ M#
œ ŠM# ``Mx" N# ``My" P# ``Mz" ‹ ŠM" ``Mx# N" ``My# P" ``Mz# ‹ Š ``Mx# ``Ny# ``Pz# ‹ M"
Š ``Mx" ``Ny" ``Pz" ‹ M# . Now, i-comp of (F# † ™ )F" œ ŠM# ``x N# ``y P# ``z ‹ M"
œ ŠM# ``Mx" N# ``My" P# ``Mz" ‹ ; likewise, i-comp of (F" † ™ )F# œ ŠM" ``Mx# N" ``My# P" ``Mz# ‹ ;
i-comp of ( ™ † F# )F" œ Š ``Mx# ``Ny# ``Pz# ‹ M" and i-comp of ( ™ † F" )F# œ Š ``Mx" ``Ny" ``Pz" ‹ M# .
Similar results hold for the j and k components of ™ ‚ (F" ‚ F# ). In summary, since the corresponding
components are equal, we have the result
™ ‚ ( F " ‚ F # ) œ ( F # † ™ ) F " ( F " † ™ ) F # ( ™ † F # ) F " ( ™ † F" ) F#
(b) Here again we consider only the i-component of each expression. Thus, the i-comp of ™ (F" † F# )
œ ``x (M" M# N" N# P" P# ) œ ˆM" ``Mx# M# ``Mx" N" ``Nx# N# ``Nx" P" ``Px# P# ``Px" ‰
i-comp of (F" † ™ )F# œ ŠM" ``Mx# N" ``My# P" ``Mz# ‹ ,
i-comp of (F# † ™ )F" œ ŠM# ``Mx" N# ``My" P# ``Mz" ‹ ,
i-comp of F" ‚ ( ™ ‚ F# ) œ N" Š ``Nx# ``My# ‹ P" ˆ ``Mz# ``Px# ‰ , and
i-comp of F# ‚ ( ™ ‚ F" ) œ N# Š ``Nx" ``My" ‹ P# ˆ ``Mz" ``Px" ‰ .
Since corresponding components are equal, we see that
™ (F" † F# ) œ (F" † ™ )F# (F# † ™ )F" F" ‚ ( ™ ‚ F# ) F# ‚ ( ™ ‚ F" ), as claimed.
21. The integral's value never exceeds the surface area of S. Since kFk Ÿ 1, we have kF † nk œ kFk knk Ÿ (1)(1) œ 1 and
' ' ' ™ † F d 5 œ ' ' F † n d5
D
[Divergence Theorem]
S
Ÿ ' ' kF † nk d5
[A property of integrals]
S
Ÿ ' ' (1) d5
ckF † nk Ÿ 1d
S
œ Area of S.
22. Yes, the outward flux through the top is 5. The reason is this: Since ™ † F œ ™ † (xi 2yj (z 3)k œ 1 2 1
œ 0, the outward flux across the closed cubelike surface is 0 by the Divergence Theorem. The flux across the top is
therefore the negative of the flux across the sides and base. Routine calculations show that the sum of these latter fluxes is
5. (The flux across the sides that lie in the xz-plane and the yz-plane are 0, while the flux across the xy-plane is 3.)
Therefore the flux across the top is 5.
23. (a)
`
`
`
` x (x) œ 1, ` y (y) œ 1, ` z (z) œ 1
Ê ™ † F œ 3 Ê Flux œ ' ' ' 3 dV œ 3 ' ' ' dV œ 3(Volume of the solid)
D
D
(b) If F is orthogonal to n at every point of S, then F † n œ 0 everywhere Ê Flux œ ' ' F † n d5 œ 0. But the flux is
S
3 (Volume of the solid) Á 0, so F is not orthogonal to n at every point.
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
988
Chapter 16 Integration in Vector Fields
24. ™ † F œ 2x 4y 6z 12 Ê Flux œ '0 '0 '0 (2x 4y 6z 12) dz dy dx œ '0 '0 (2x 4y 9) dy dx
a
b
1
a
b
œ '0 a2xb 2b# 9bb dx œ a# b 2ab# 9ab œ ab(a 2b 9) œ f(aß b); `` af œ 2ab 2b# 9b and
a
`f
`f
`f
#
` b œ a 4ab 9a so that ` a œ 0 and ` b œ 0
Ê b(2a 2b 9) œ 0 and a(a 4b 9) œ 0 Ê b œ 0 or
2a 2b 9 œ 0, and a œ 0 or a 4b 9 œ 0. Now b œ 0 or a œ 0 Ê Flux œ 0; 2a 2b 9 œ 0 and
a 4b 9 œ 0 Ê 3a 9 œ 0 Ê a œ 3 Ê b œ 3# so that f ˆ3ß 3# ‰ œ 27
# is the maximum flux.
25. ' ' F † n d5 œ ' ' ' ™ † F dV œ ' ' ' 3 dV Ê "3 ' ' F † n d5 œ ' ' ' dV œ Volume of D
D
S
D
D
S
26. F œ C Ê ™ † F œ 0 Ê Flux œ ' ' F † n d5 œ ' ' ' ™ † F dV œ ' ' ' 0 dV œ 0
D
S
D
27. (a) From the Divergence Theorem, ' ' ™ f † n d5 œ ' ' ' ™ † ™ f dV œ ' ' ' ™ # f dV œ ' ' ' 0 dV œ 0
S
D
S
D
D
D
(b) From the Divergence Theorem, ' ' f ™ f † n d5 œ ' ' ' ™ † f ™ f dV. Now,
f ™ f œ ˆf `` xf ‰ i Šf `` yf ‹ j ˆf `` zf ‰ k
#
#
#
#
#
#
Ê ™ † f ™ f œ ’f `` xf# ˆ `` xf ‰ “ ”f `` yf# Š `` yf ‹ • ’f `` zf# ˆ `` zf ‰ “
œ f ™ # f k ™ f k# œ 0 k ™ f k# since f is harmonic Ê ' ' f ™ f † n d5 œ ' ' ' k ™ f k# dV, as claimed.
D
S
28. From the Divergence Theorem, ' ' ™ f † n d5 œ ' ' ' ™ † ™ f dV œ ' ' ' Š `` xf# `` yf# `` zf# ‹ dV. Now,
#
D
S
#
#
D
f(xß yß z) œ ln Èx# y# z# œ "# ln ax# y# z# b Ê `` xf œ x# yx# z# , `` yf œ x# yy# z# , `` zf œ x# yz# z#
#
#
#
#
#
#
#
#
#
z
z
Ê `` xf# œ ax#xy#yz#zb# , `` yf# œ ax#xyy# , ` f œ ax#xyy# , Ê `` xf# `` yf# `` zf#
z# b# ` z#
z# b#
#
#
#
z
œ ax#xyy# œ x# y"# z# Ê ' ' ™ f † n d5 œ ' ' '
z # b#
#
œ '0
1Î2
#
#
D
S
#
dV
x # y # z#
#
œ '0
1Î2
#
'01Î2 '0a 3 3sin 9 d3 d9 d)
#
#
'01Î2 a sin 9 d9 d) œ '01Î2 ca cos 9d 1! Î# d) œ '01Î2 a d) œ 1#a
29. ' ' f ™ g † n d5 œ ' ' ' ™ † f ™ g dV œ ' ' ' ™ † Šf `` gx i f `` gy j f `` gz k‹ dV
D
S
D
#
#
#
œ ' ' ' Šf `` xg# `` xf `` gx f `` yg# `` yf `` gy f `` zg# `` zf `` gz ‹ dV
D
#
#
#
œ ' ' ' ’f Š `` xg# `` yg# `` zg# ‹ Š `` xf `` gx `` yf `` gy `` zf `` gz ‹“ dV œ ' ' ' af ™ # g ™ f † ™ gb dV
D
D
30. By Exercise 29, ' ' f ™ g † n d5 œ ' ' ' af ™ # g ™ f † ™ gb dV and by interchanging the roles of f and g,
D
S
' ' g ™ f † n d5 œ ' ' ' ag ™ # f ™ g † ™ f b dV. Subtracting the second equation from the first yields:
D
S
' ' a f ™ g g ™ f b † n d5 œ ' ' ' af ™ # g g ™ # f b dV since ™ f † ™ g œ ™ g † ™ fÞ
D
S
31. (a) The integral ' ' ' p(tß xß yß z) dV represents the mass of the fluid at any time t. The equation says that
D
the instantaneous rate of change of mass is flux of the fluid through the surface S enclosing the region D:
the mass decreases if the flux is outward (so the fluid flows out of D), and increases if the flow is inward
(interpreting n as the outward pointing unit normal to the surface).
(b) ' ' '
D
`p
d
` t dV œ dt
' ' ' p dV œ ' ' pv † n d5 œ ' ' ' ™ † pv dV Ê ``3t œ ™ † pv
D
S
D
Since the law is to hold for all regions D, ™ † pv ``pt œ 0, as claimed
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Chapter 16 Practice Exercises
32. (a) ™ T points in the direction of maximum change of the temperature, so if the solid is heating up at the
point the temperature is greater in a region surrounding the point Ê ™ T points away from the point
Ê ™ T points toward the point Ê ™ T points in the direction the heat flows.
(b) Assuming the Law of Conservation of Mass (Exercise 31) with k ™ T œ pv and c3T œ p, we have
d
dt
' ' ' c3T dV œ ' ' k ™ T † n d5 Ê the continuity equation, ™ † (k ™ T) ``t (c3T) œ 0
D
S
Ê c3 ``Tt œ ™ † (k ™ T) œ k ™ # T Ê ``Tt œ ck3 ™ # T œ K ™ # T, as claimed
CHAPTER 16 PRACTICE EXERCISES
dy
1. Path 1: r œ ti tj tk Ê x œ t, y œ t, z œ t, 0 Ÿ t Ÿ 1 Ê f(g(t)ß h(t)ß k(t)) œ 3 3t# and dx
dt œ 1, dt œ 1,
dz
dt œ 1
‰ Š dy
ˆ dz ‰ dt œ È3 dt Ê ' f(xß yß z) ds œ ' È3 a3 3t# b dt œ 2È3
Ê Êˆ dx
dt
dt ‹ dt
C
0
#
#
#
1
dy
Path 2: r" œ ti tj , 0 Ÿ t Ÿ 1 Ê x œ t, y œ t, z œ 0 Ê f(g(t)ß h(t)ß k(t)) œ 2t 3t# 3 and dx
dt œ 1, dt œ 1,
dz
dt œ 0
‰ Š dy
ˆ dz ‰ dt œ È2 dt Ê ' f(xß yß z) ds œ ' È2 a2t 3t# 3b dt œ 3È2 ;
Ê Êˆ dx
dt
dt ‹ dt
0
#
#
#
1
C"
dy
dz
r# œ i j tk Ê x œ 1, y œ 1, z œ t Ê f(g(t)ß h(t)ß k(t)) œ 2 2t and dx
dt œ 0, dt œ 0, dt œ 1
‰ Š dy
ˆ dz ‰ dt œ dt Ê ' f(xß yß z) ds œ ' (2 2t) dt œ 1
Ê Êˆ dx
dt
dt ‹ dt
C
0
#
#
#
1
#
Ê 'C f(xß yß z) ds œ 'C f(xß yß z) ds 'C f(xß yß z) œ 3È2 1
"
#
dy
dz
2. Path 1: r" œ ti Ê x œ t, y œ 0, z œ 0 Ê f(g(t)ß h(t)ß k(t)) œ t# and dx
dt œ 1, dt œ 0, dt œ 0
‰# Š dy
ˆ dz ‰# dt œ dt Ê ' f(xß yß z) ds œ ' t# dt œ "3 ;
Ê Êˆ dx
dt
dt ‹ dt
C
0
#
1
"
dy
dz
r# œ i tj Ê x œ 1, y œ t, z œ 0 Ê f(g(t)ß h(t)ß k(t)) œ 1 t and dx
dt œ 0, dt œ 1, dt œ 0
‰# Š dy
ˆ dz ‰# dt œ dt Ê ' f(xß yß z) ds œ ' (1 t) dt œ 32 ;
Ê Êˆ dx
dt
dt ‹ dt
C
0
#
1
#
dy
dz
r$ œ i j tk Ê x œ 1, y œ 1, z œ t Ê f(g(t)ß h(t)ß k(t)) œ 2 t and dx
dt œ 0, dt œ 0, dt œ 1
‰# Š dy
ˆ dz ‰# dt œ dt Ê ' f(xß yß z) ds œ ' (2 t) dt œ 32
Ê Êˆ dx
dt
dt ‹ dt
C
0
#
1
$
Ê 'Path 1 f(xß yß z) ds œ 'C f(xß yß z) ds 'C f(xß yß z) ds 'C f(xß yß z) ds œ 10
3
"
#
$
dy
dz
Path 2: r% œ ti tj Ê x œ t, y œ t, z œ 0 Ê f(g(t)ß h(t)ß k(t)) œ t# t and dx
dt œ 1, dt œ 1, dt œ 0
‰# Š dy
ˆ dz ‰# dt œ È2 dt Ê ' f(xß yß z) ds œ ' È2 at# tb dt œ 56 È2;
Ê Êˆ dx
dt
dt ‹ dt
C
0
#
1
%
r$ œ i j tk (see above) Ê 'C f(xß yß z) ds œ 32
$
È
Ê 'Path 2 f(xß yß z) ds œ 'C f(xß yß z) ds 'C f(xß yß z) ds œ 56 È2 3# œ 5 26 9
$
%
dy
dz
Path 3: r& œ tk Ê x œ 0, y œ 0, z œ t, 0 Ÿ t Ÿ 1 Ê f(g(t)ß h(t)ß k(t)) œ t and dx
dt œ 0, dt œ 0, dt œ 1
‰ Š dy
ˆ dz ‰ dt œ dt Ê ' f(xß yß z) ds œ ' t dt œ "2 ;
Ê Êˆ dx
dt
dt ‹ dt
C
0
#
#
#
1
&
dy
dz
r' œ tj k Ê x œ 0, y œ t, z œ 1, 0 Ÿ t Ÿ 1 Ê f(g(t)ß h(t)ß k(t)) œ t 1 and dx
dt œ 0, dt œ 1, dt œ 0
‰ Š dy
ˆ dz ‰ dt œ dt Ê ' f(xß yß z) ds œ ' (t 1) dt œ 12 ;
Ê Êˆ dx
dt
dt ‹ dt
C
0
#
#
#
1
'
dy
dz
r( œ ti j k Ê x œ t, y œ 1, z œ 1, 0 Ÿ t Ÿ 1 Ê f(g(t)ß h(t)ß k(t)) œ t# and dx
dt œ 1, dt œ 0, dt œ 0
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
989
990
Chapter 16 Integration in Vector Fields
‰# Š dy
ˆ dz ‰# dt œ dt Ê ' f(xß yß z) ds œ ' t# dt œ 13
Ê Êˆ dx
dt
dt ‹ dt
C
0
#
1
(
Ê 'Path 3 f(xß yß z) ds œ 'C f(xß yß z) ds 'C f(xß yß z) ds 'C f(xß yß z) ds œ "# "# "3 œ 32
&
'
(
3. r œ (a cos t)j (a sin t)k Ê x œ 0, y œ a cos t, z œ a sin t Ê fag(t)ß h(t)ß k(t)b œ Èa# sin# t œ a ksin tk and
dy
dx
dz
dt œ 0, dt œ a sin t, dt œ a cos t
#
#
#
‰ Š dy
ˆ dz ‰ dt œ a dt
Ê Êˆ dx
dt
dt ‹ dt
Ê 'C f(xß yß z) ds œ '0 a# ksin tk dt œ '0 a# sin t dt '1 a# sin t dt œ 4a#
21
1
21
4. r œ (cos t t sin t)i (sin t t cos t)j Ê x œ cos t t sin t, y œ sin t t cos t, z œ 0
Ê fag(t)ß h(t)ß k(t)b œ È(cos t t sin t)# (sin t t cos t)# œ È1 t# and dx œ sin t sin t t cos t
dt
dz
œ t cos t, dy
dt œ cos t cos t t sin t œ t sin t, dt œ 0
Ê
#
‰# Š dy
ˆ dz ‰# dt
ʈ dx
dt
dt ‹ dt
È3
œ Èt# cos# t t# sin# t dt œ ktk dt œ t dt since 0 Ÿ t Ÿ È3 Ê 'C f(xß yß z) ds œ '0
5.
tÈ1 t# dt œ 73
`P
"
$Î#
œ ``Nz , ``Mz œ "# (x y z)$Î# œ `` Px , ``Nx œ "# (x y z)$Î# œ ``My
` y œ # (x y z)
Ê M dx N dy P dz is exact; `` xf œ Èx "y z Ê f(xß yß z) œ 2Èx y z g(yß z) Ê `` yf œ Èx "y z `` gy
œ Èx "y z Ê `` gy œ 0 Ê g(yß z) œ h(z) Ê f(xß yß z) œ 2Èx y z h(z) Ê `` zf œ Èx "y z hw (z)
Ð4ß 3ß0Ñ dx dy dz
œ Èx "y z Ê hw (x) œ 0 Ê h(z) œ C Ê f(xß yß z) œ 2Èx y z C Ê 'Ð 1ß1ß1Ñ
Èx y z
œ f(4ß 3ß 0) f(1ß 1ß 1) œ 2È1 2È1 œ 0
6.
`P
"
`N `M
`P `N
`M
` y œ #Èyz œ ` z , ` z œ 0 œ ` x , ` x œ 0 œ ` y
Ê M dx N dy T dz is exact; `` xf œ 1 Ê f(xß yß z)
œ x g(yß z) Ê `` yf œ `` yg œ É yz Ê g(yß z) œ 2Èyz h(z) Ê f(xß yß z) œ x 2Èyz h(z)
Ê `` zf œ É yz hw (z) œ É yz Ê hw (z) œ 0 Ê h(z) œ C Ê f(xß yß z) œ x 2Èyz C
Ê 'Ð1ß1ß1Ñ dx É yz dy É yz dz œ f(10ß 3ß 3) f(1ß 1ß 1) œ (10 2 † 3) (1 2 † 1) œ 4 1 œ 5
Ð10ß3ß3Ñ
7.
`M
`P
` z œ y cos z Á y cos z œ ` x Ê F is not conservative; r œ a2 cos tbi a2 sin tbj k, 0 Ÿ t Ÿ 21
Ê dr œ a2 sin tbi a2 cos tbj Ê 'C F † dr œ '0 c a2 sin tbasina1bba2 sin tb a2 cos tbasina1bba2 cos tbd dt
21
œ 4 sina1b'0 asin2 t cos2 tbdt œ 81 sina1b
21
8.
`P
`N `M
`P `N
`M
#
` y œ 0 œ ` z , ` z œ 0 œ ` x , ` x œ 3x œ ` y
Ê F is conservative Ê 'C F † dr œ 0
9. Let M œ 8x sin y and N œ 8y cos x Ê ``My œ 8x cos y and ``Nx œ 8y sin x Ê 'C 8x sin y dx 8y cos x dy
œ ' ' (8y sin x 8x cos y) dy dx œ '0
1/2
R
'01/2 (8y sin x 8x cos y) dy dx œ '01/2 a1# sin x 8xb dx œ 1# 1# œ 0
10. Let M œ y# and N œ x# Ê ``My œ 2y and ``Nx œ 2x Ê 'C y# dx x# dy œ ' ' (2x 2y) dx dy
21
2
21
œ '0 '0 (2r cos ) 2r sin )) r dr d) œ '0 16
3 (cos ) sin )) d) œ 0
R
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Chapter 16 Practice Exercises
991
11. Let z œ 1 x y Ê fx (xß y) œ 1 and fy (xß y) œ 1 Ê Éfx# fy# 1 œ È3 Ê Surface Area œ ' ' È3 dx dy
R
œ È3(Area of the circular region in the xy-plane) œ 1È3
12. ™ f œ 3i 2yj 2zk , p œ i Ê k ™ f k œ È9 4y# 4z# and k ™ f † pk œ 3
Ê Surface Area œ ' '
R
È9 4y# 4z#
dy dz œ
3
È
'021 '0 3 È9 3 4r r dr d) œ "3 '021 Š 74 È21 94 ‹ d) œ 16 Š7È21 9‹
#
13. ™ f œ 2xi 2yj 2zk , p œ k Ê k ™ f k œ È4x# 4y# 4z# œ 2Èx# y# z# œ 2 and k ™ f † pk œ k2zk œ 2z since
z
0 Ê Surface Area œ ' '
R
œ '0 ’È1 r# “
21
"ÎÈ#
!
2
2z dA œ
' ' "z dA œ ' ' È
R
R
"
dx dy œ
1 x# y#
È
'021 '01Î 2 È "
1 r#
r dr d)
d) œ '0 Š1 È"2 ‹ d) œ 21 Š1 È"2 ‹
21
14. (a) ™ f œ 2xi 2yj 2zk , p œ k Ê k ™ f k œ È4x# 4y# 4z# œ 2Èx# y# z# œ 4 and k ™ f † pk œ 2z since
4
0 Ê Surface Area œ ' ' 2z
dA œ ' '
z
R
2
z dA œ 2
R
'01/2 '02 cos
)
2
È4 r# r dr d) œ 41 8
(b) r œ 2 cos ) Ê dr œ 2 sin ) d); ds# œ r# d)# dr# (Arc length in polar coordinates)
Ê ds# œ (2 cos ))# d)# dr# œ 4 cos# ) d)# 4 sin# ) d)# œ 4 d)# Ê ds œ 2 d); the height of the
cylinder is z œ È4 r# œ È4 4 cos# ) œ 2 ksin )k œ 2 sin ) if 0 Ÿ ) Ÿ 1# Ê Surface Area œ 'c1Î2 h ds
1/2
œ 2 '0 (2 sin ))(2 d)) œ 8
1/2
15. f(xß yß z) œ xa yb cz œ 1 Ê ™ f œ ˆ "a ‰ i ˆ b" ‰ j ˆ "c ‰ k Ê k ™ f k œ É a"# b"# c"# and p œ k Ê k ™ f † pk œ "c
since c 0 Ê Surface Area œ ' '
É "# "# "#
R
a
b
c
Š "c ‹
dA œ cÉ a"# b"# c"# ' ' dA œ #" abcÉ a"# b"# c"# ,
R
since the area of the triangular region R is "# ab. To check this result, let v œ ai ck and w œ ai bj; the area can be
found by computing "# kv ‚ wk.
16. (a) ™ f œ 2yj k , p œ k Ê k ™ f k œ È4y# 1 and k ™ f † pk œ 1 Ê d5 œ È4y# 1 dx dy
Ê ' ' g(xß yß z) d5 œ ' ' È4yyz# 1 È4y# 1 dx dy œ ' ' y ay# 1b dx dy œ 'c1 '0 ay$ yb dx dy
1
S
œ 'c1 3 ay
1
$
R
3
R
%
# "
yb dy œ 3 ’ y4 y# “ œ 0
"
(b) ' ' g(xß yß z) d5 œ ' ' È4yz# 1 È4y# 1 dx dy œ 'c1 '0 ay# 1b dx dy œ 'c1 3 ay# 1b dy
1
3
1
R
S
"
$
œ 3 ’ y3 y“ œ 4
"
17. ™ f œ 2yj 2zk , p œ k Ê k ™ f k œ È4y# 4z# œ 2Èy# z# œ 10 and k ™ f † pk œ 2z since z
5
' ' g(xß yß z) d5 œ ' ' ax% yb ay# z# b ˆ 5z ‰ dx dy
Ê d5 œ 10
2z dx dy œ z dx dy œ
S
œ''
R
%
ax yb (25) Š È255 y# ‹ dx dy œ
'0 '
4
R
1
125y
x% dx dy œ
0 È25 y#
0
'04 È2525y y dy œ 50
#
18. Define the coordinate system so that the origin is at the center of the earth, the z-axis is the earth's axis (north is the
positive z direction), and the xz-plane contains the earth's prime meridian. Let S denote the surface which is Wyoming so
"Î#
then S is part of the surface z œ aR# x# y# b
. Let Rxy be the projection of S onto the xy-plane. The surface area of
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
992
Chapter 16 Integration in Vector Fields
Wyoming is ' ' 1 d5 œ ' ' Ê1 ˆ `` xz ‰ Š `` yz ‹ dA œ ' ' É R# xx# y# R# yx# y# 1 dA œ ' '
#
#
S
Rxy
œ ') 'R sin 45° R aR# r# b
)#
R sin 49°
Rxy
"Î#
"
œ ') R aR# r# b
)#
"Î# R sin 49°
"
R sin 45°
¹
#
#
Rxy
R
dA
aR# x# y# b"Î#
r dr d) (where )" and )# are the radian equivalent to 104°3w and 111°3w , respectively)
œ ') R aR# R# sin# 45°b
)#
"Î#
"
R aR# R# sin# 49°b
"Î#
d)
71
71
œ ()# )" )R# (cos 45° cos 49°) œ 180
R# (cos 45° cos 49°) œ 180
(3959)# (cos 45° cos 49°) ¸ 97,751 sq. mi.
19. A possible parametrization is r(9ß )) œ (6 sin 9 cos ))i (6 sin 9 sin ))j (6 cos 9)k (spherical coordinates);
now 3 œ 6 and z œ 3 Ê 3 œ 6 cos 9 Ê cos 9 œ " Ê 9 œ 21 and z œ 3È3 Ê 3È3 œ 6 cos 9
Ê
È
cos 9 œ #3
#
Ê
9 œ 16
Ê
3
1
21
6 Ÿ 9 Ÿ 3 ; also 0 Ÿ ) Ÿ 21
#
20. A possible parametrization is r(rß )) œ (r cos ))i (r sin ))j Š r# ‹ k (cylindrical coordinates);
#
#
now r œ Èx# y# Ê z œ r# and 2 Ÿ z Ÿ 0 Ê 2 Ÿ r# Ÿ 0 Ê 4
r#
0 Ê 0 Ÿ r Ÿ 2 since r
0;
also 0 Ÿ ) Ÿ 21
21. A possible parametrization is r(rß )) œ (r cos ))i (r sin ))j (1 r)k (cylindrical coordinates);
now r œ Èx# y# Ê z œ 1 r and 1 Ÿ z Ÿ 3 Ê 1 Ÿ 1 r Ÿ 3 Ê 0 Ÿ r Ÿ 2; also 0 Ÿ ) Ÿ 21
22. A possible parametrization is r(xß y) œ xi yj ˆ3 x y# ‰ k for 0 Ÿ x Ÿ 2 and 0 Ÿ y Ÿ 2
23. Let x œ u cos v and z œ u sin v, where u œ Èx# z# and v is the angle in the xz-plane with the x-axis
Ê r(uß v) œ (u cos v)i 2u# j (u sin v)k is a possible parametrization; 0 Ÿ y Ÿ 2 Ê 2u# Ÿ 2 Ê u# Ÿ 1
Ê 0 Ÿ u Ÿ 1 since u 0; also, for just the upper half of the paraboloid, 0 Ÿ v Ÿ 1
24. A possible parametrization is ŠÈ10 sin 9 cos )‹ i ŠÈ10 sin 9 sin )‹ j ŠÈ10 cos 9) k , 0 Ÿ 9 Ÿ 12 and 0 Ÿ ) Ÿ 1#
â
âi
â
25. ru œ i j , rv œ i j k Ê ru ‚ rv œ â "
â
â"
â
j kâ
â
" 0 â œ i j 2k Ê kru ‚ rv k œ È6
â
" " â
Ê Surface Area œ ' ' kru ‚ rv k du dv œ '0 '0 È6 du dv œ È6
1
1
Ruv
26. ' ' axy z# b d5 œ '0 '0 c(u v)(u v) v# d È6 du dv œ È6'0 '0 au# 2v# b du dv
1
S
1
1
1
œ È6 '0 ’ u3 2uv# “ dv œ È6 '0 ˆ "3 2v# ‰ dv œ È6 "3 v 23 v$ ‘ ! œ 36 œ É 23
1
$
"
1
"
È
!
â
â
i
j
kâ
â
â
â
sin ) 0 â
27. rr œ (cos ))i (sin ))j , r) œ (r sin ))i (r cos ))j k Ê rr ‚ r) œ â cos )
â
â
â r sin ) r cos ) " â
œ (sin ))i (cos ))j rk Ê krr ‚ r) k œ Èsin# ) cos# ) r# œ È1 r# Ê Surface Area œ ' ' krr ‚ r) k dr d)
œ '0
21
Rr)
'0 È1 r# dr d) œ '0 ’ 2r È1 r# "# ln Šr È1 r# ‹“ d) œ '0 ’ "# È2 "# ln Š1 È2‹“ d)
21
1
"
21
!
œ 1 ’È2 ln Š1 È2‹“
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Chapter 16 Practice Exercises
993
28. ' ' Èx# y# 1 d5 œ '0 '0 Èr# cos# ) r# sin# ) 1 È1 r# dr d) œ '0 '0 a1 r# b dr d)
21
S
œ '0 ’r r3 “ d) œ '0
21
$
"
21
!
21
1
1
4
8
3 d) œ 3 1
29. `` Py œ 0 œ ``Nz , ``Mz œ 0 œ `` Px , ``Nx œ 0 œ ``My Ê Conservative
30. `` Py œ
3zy
œ ``Nz , ``Mz œ # # 3xz # &Î# œ `` Px , ``Nx œ # #3xy # &Î# œ ``My
ax# y# z# b &Î#
ax y z b
ax y z b
Ê Conservative
31. `` Py œ 0 Á yez œ ``Nz Ê Not Conservative
y
`P `N
z
`M
32. `` Py œ (x xyz)# œ ``Nz , ``Mz œ (x yz)
# œ ` x , ` x œ (x yz)# œ ` y Ê Conservative
33. `` xf œ 2 Ê f(xß yß z) œ 2x g(yß z) Ê `` yf œ `` yg œ 2y z Ê g(yß z) œ y# zy h(z)
Ê f(xß yß z) œ 2x y# zy h(z) Ê `` zf œ y hw (z) œ y 1 Ê hw (z) œ 1 Ê h(z) œ z C
Ê f(xß yß z) œ 2x y# zy z C
34. `` xf œ z cos xz Ê f(xß yß z) œ sin xz g(yß z) Ê `` yf œ `` yg œ ey Ê g(yß z) œ ey h(z)
Ê f(xß yß z) œ sin xz ey h(z) Ê `` zf œ x cos xz hw (z) œ x cos xz Ê hw (z) œ 0 Ê h(z) œ C
Ê f(xß yß z) œ sin xz ey C
35. Over Path 1: r œ ti tj tk , 0 Ÿ t Ÿ 1 Ê x œ t, y œ t, z œ t and dr œ (i j k) dt Ê F œ 2t# i j t# k
Ê F † dr œ a3t# 1b dt Ê Work œ '0 a3t# 1b dt œ 2;
1
Over Path 2: r" œ ti tj , 0 Ÿ t Ÿ 1 Ê x œ t, y œ t, z œ 0 and dr" œ (i j) dt Ê F" œ 2t# i j t# k
Ê F" † dr" œ a2t# 1b dt Ê Work" œ '0 a2t# 1b dt œ 53 ; r# œ i j tk , 0 Ÿ t Ÿ 1 Ê x œ 1, y œ 1, z œ t and
1
dr# œ k dt Ê F# œ 2i j k Ê F# † dr# œ dt Ê Work# œ '0 dt œ 1 Ê Work œ Work" Work# œ 53 1 œ 83
1
36. Over Path 1: r œ ti tj tk , 0 Ÿ t Ÿ 1 Ê x œ t, y œ t, z œ t and dr œ (i j k) dt Ê F œ 2t# i t# j k
Ê F † dr œ a3t# 1b dt Ê Work œ '0 a3t# 1b dt œ 2;
1
Over Path 2: Since f is conservative, )C F † dr œ 0 around any simple closed curve C. Thus consider
'curve F † dr œ 'C F † dr 'C F † dr , where C" is the path from (0ß 0ß 0) to (1ß 1ß 0) to ("ß "ß ") and C# is the path
from (1ß 1ß 1) to (!ß !ß !). Now, from Path 1 above, 'C F † dr œ 2 Ê 0 œ 'curve F † dr œ 'C F † dr (2)
Ê 'C F † dr œ 2
"
#
#
"
"
37. (a) r œ aet cos tb i aet sin tb j Ê x œ et cos t, y œ et sin t from (1ß 0) to ae21 ß 0b Ê 0 Ÿ t Ÿ 21
Ê ddtr œ aet cos t et sin tb i aet sin t et cos tb j and F œ
#
t
t
xi yj
œ ae2t cos#tb i 2tae sin# tb$Î#j
ax# y# b$Î#
ae cos t e sin tb
#
t‰
t‰
dr
cos t
sin t cos t
t
œ ˆ cos
i ˆ sin
sinet t sin tecos
‹ œ et
t
e2t
e2t j Ê F † dt œ Š et et
Ê Work œ '0 et dt œ 1 e21
21
(b) F œ
œ
xi yj
ax# y# b$Î#
y
ax# y# b$Î#
21
Ê `` xf œ
x
ax# y# b$Î#
Ê f(xß yß z) œ ax# y# b
Ê g(yß z) œ C Ê f(xß yß z) œ ax# y# b
"Î#
"Î#
g(yß z) Ê `` yf œ
y
`` gy
ax# y# b$Î#
is a potential function for F Ê 'C F † dr
œ f ae ß 0b f(1ß 0) œ 1 e21
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
994
Chapter 16 Integration in Vector Fields
38. (a) F œ ™ ax# zey b Ê F is conservative Ê )C F † dr œ 0 for any closed path C
(b) 'C F † dr œ 'Ð1ß0ß0Ñ
Ð1ß0ß21Ñ
â i
â
â
39. ™ ‚ F œ â ``x
â #
ây
™ ax# zey b † dr œ ax# zey bk Ð"ß!ß#1Ñ ax# zey bk Ð"ß!ß!Ñ œ 21 0 œ 21
k ââ
` â
2i 6 j 3k
2
6
3
` z â œ 2yk; unit normal to the plane is n œ È4 36 9 œ 7 i 7 j 7 k
â
y 3z# â
j
`
`y
k™f k
7
Ê ™ ‚ F † n œ 67 y; p œ k and f(xß yß z) œ 2x 6y 3z Ê k ™ f † pk œ 3 Ê d5 œ k™
f†pk dA œ 3 dA
Ê )C F † dr œ ' ' 76 y d5 œ ' ' ˆ 76 y‰ ˆ 37 dA‰ œ ' ' 2y dA œ '0 '0 2r sin ) r dr d) œ '0
21
R
â i
â
â
40. ™ ‚ F œ â ``x
â #
âx y
n œ È"2 j È"2 k
R
21
1
R
2
3 sin ) d) œ 0
â
â
â
â œ 8yi ; the circle lies in the plane f(xß yß z) œ y z œ 0 with unit normal
â
4y z â
j
k
`
`y
`
`z
#
xy
Ê ™ ‚ F † n œ 0 Ê )C F † dr œ ' ' ™ ‚ F † n d5 œ ' ' 0 d5 œ 0
R
R
È2, dy œ È2, dz œ 2t
41. (a) r œ È2ti È2tj a4 t# b k , 0 Ÿ t Ÿ 1 Ê x œ È2t, y œ È2t, z œ 4 t# Ê dx
dt œ
dt
dt
‰ Š dy
ˆ dz ‰ dt œ È4 4t# dt Ê M œ ' $ (xß yß z) ds œ ' 3tÈ4 4t# dt œ "4 (4 4t)$Î# ‘
Ê Êˆ dx
dt
dt ‹ dt
!
C
0
#
#
#
"
1
œ 4È2 2
(b) M œ 'C $ (xß yß z) ds œ '0 È4 4t# dt œ ’tÈ1 t# ln Št È1 t# ‹“ œ È2 ln Š1 È2‹
"
1
!
dy
dz
"Î#
42. r œ ti 2tj 32 t$Î# k , 0 Ÿ t Ÿ 2 Ê x œ t, y œ 2t, z œ 32 t$Î# Ê dx
dt œ 1, dt œ 2, dt œ t
‰# Š dy
ˆ dz ‰# dt œ Èt 5 dt Ê M œ ' $ (xß yß z) ds œ ' 3È5 t Èt 5 dt
Ê Êˆ dx
dt
dt ‹ dt
C
0
#
2
œ '0 3(t 5) dt œ 36; Myz œ 'C x$ ds œ '0 3t(t 5) dt œ 38; Mxz œ 'C y$ ds œ '0 6t(t 5) dt œ 76;
2
2
2
È2 Ê x œ yz œ 38 œ 19 , y œ Mxz œ 76 œ 19 , z œ xy œ
Mxy œ 'C z$ ds œ '0 2t$Î# (t 5) dt œ 144
7
M
36
18
M
36
9
M
2
M
M
È 2‹
Š 144
7
36
œ 47 È2
È
È
#
#
dy
È2 t"Î# , dz œ t
43. r œ ti Š 2 3 2 t$Î# ‹ j Š t# ‹ k , 0 Ÿ t Ÿ 2 Ê x œ t, y œ 2 3 2 t$Î# , z œ t# Ê dx
dt œ 1, dt œ
dt
#
#
#
‰ Š dy
ˆ dz ‰ dt œ È1 2t t# dt œ È(t 1)# dt œ kt 1k dt œ (t 1) dt on the domain given.
Ê Êˆ dx
dt
dt ‹ dt
Then M œ 'C $ ds œ '0 ˆ t " 1 ‰ (t 1) dt œ '0 dt œ 2; Myz œ 'C x$ ds œ '0 t ˆ t " 1 ‰ (t 1) dt œ '0 t dt œ 2;
2
2
2
2
32
Mxz œ 'C y$ ds œ '0 Š 2 3 2 t$Î# ‹ ˆ t " 1 ‰ (t 1) dt œ '0 2 3 2 t$Î# dt œ 15
; Mxy œ 'C z$ ds
2
È
2
È
œ '0 Š t# ‹ ˆ t"1 ‰ (t 1) dt œ '0 t# dt œ 43 Ê x œ Myz œ #2 œ 1; y œ MMxz œ
2
œ
2 #
#
ˆ 43 ‰
2
# œ 3 ; Ix œ
M
ˆ 32
‰
Mxy
16
15
# œ 15 ; z œ M
64
'C ay# z# b $ ds œ '02 Š 89 t$ t4 ‹ dt œ 232
' # #
'2 # t
45 ; Iy œ C ax z b $ ds œ 0 Št 4 ‹ dt œ 15 ;
%
%
Iz œ 'C ay# x# b $ ds œ '0 ˆt# 89 t$ ‰ dt œ 56
9
2
44. z œ 0 because the arch is in the xy-plane, and x œ 0 because the mass is distributed symmetrically with respect
#
#
#
‰ Š dy
ˆ dz ‰ dt
to the y-axis; r(t) œ (a cos t)i (a sin t)j , 0 Ÿ t Ÿ 1 Ê ds œ ʈ dx
dt
dt ‹ dt
œ È(a sin t)# (a cos t)# dt œ a dt, since a
0; M œ 'C $ ds œ 'C (2a y) ds œ '0 (2a a sin t) a dt
1
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Chapter 16 Practice Exercises
995
œ 2a2 1 2a# ; Mxz œ 'C y$ dt œ 'C y(2a y) ds œ '0 (a sin t)(2a a sin t) dt œ '0 a2a# sin t a# sin# tb dt
1
1
#
Š4a# a 1 ‹
1
#
81
ˆ 81 ‰
œ 2a# cos t a# ˆ 2t sin4 2t ‰ ‘ ! œ 4a# a#1 Ê y œ 2a# 1 2a
# œ 41 4 Ê axß yß zb œ 0ß 41 4 ß 0
#
t
t
45. r(t) œ aet cos tb i aet sin tb j et k , 0 Ÿ t Ÿ ln 2 Ê x œ et cos t, y œ et sin t, z œ et Ê dx
dt œ ae cos t e sin tb ,
dy
dz
t
t
t
dt œ ae sin t e cos tb, dt œ e
#
#
#
‰ Š dy
ˆ dz ‰ dt
Ê Êˆ dx
dt
dt ‹ dt
œ Éaet cos t et sin tb# aet sin t et cos tb# aet b# dt œ È3e2t dt œ È3 et dt; M œ 'C $ ds œ '0 È3 et dt
ln 2
œ È3; Mxy œ 'C z$ ds œ '0
ln 2
È
M
ŠÈ3 et ‹ aet b dt œ '0 È3 e2t dt œ 3 # 3 Ê z œ Mxy œ
ln 2
Œ
3
È3
#
È3
œ 3# ;
È
Iz œ 'C ax# y# b $ ds œ '0 ae2t cos# t e2t sin# tb ŠÈ3 et ‹ dt œ '0 È3 e3t dt œ 7 3 3
ln 2
ln 2
dy
46. r(t) œ (2 sin t)i (2 cos t)j 3tk , 0 Ÿ t Ÿ 21 Ê x œ 2 sin t, y œ 2 cos t, z œ 3t Ê dx
dt œ 2 cos t, dt œ 2 sin t,
dz
dt œ 3
‰# Š dy
ˆ dz ‰# dt œ È4 9 dt œ È13 dt; M œ ' $ ds œ ' $ È13 dt œ 21$ È13;
Ê Êˆ dx
dt
dt ‹ dt
C
0
#
21
Mxy œ 'C z$ ds œ '0 (3t) Š$ È13‹ dt œ 6$1# È13; Myz œ 'C x$ ds œ '0 (2 sin t) Š$ È13‹ dt œ 0;
21
21
13
Mxz œ 'C y$ ds œ '0 (2 cos t) Š$ È13‹ dt œ 0 Ê x œ y œ 0 and z œ Mxy œ 62$1
œ 31 Ê (!ß !ß 31) is the
$1È13
21
#È
M
center of mass
47. Because of symmetry x œ y œ 0. Let f(xß yß z) œ x# y# z# œ 25 Ê ™ f œ 2xi 2yj 2zk
Ê k ™ f k œ È4x# 4y# 4z# œ 10 and p œ k Ê k ™ f † pk œ 2z, since z 0 Ê M œ ' ' $ (xß yß z) d5
R
' ' 5 dA œ 5(Area of the circular region) œ 801; Mxy œ ' ' z$ d5 œ ' ' 5z dA
‰
œ ' ' z ˆ 10
#z dA œ
R
R
R
œ ' ' 5È25 x# y# dx dy œ '0 '0 Š5È25 r# ‹ r dr d) œ '0
21
21
4
R
R
490
980
3 d) œ 3 1
Ê zœ
Š 980
3 1‹
801
œ 49
12
' ' ax# y# b $ d5 œ ' ' 5 ax# y# b dx dy œ ' ' 5r$ dr d) œ ' 320 d) œ 6401
‰
Ê axß yß zb œ ˆ0ß !ß 49
12 ; Iz œ
0
0
0
21
R
4
21
R
48. On the face z œ 1: g(xß yß z) œ z œ 1 and p œ k Ê ™ g œ k Ê k ™ gk œ 1 and k ™ g † pk œ 1 Ê d5 œ dA
Ê I œ ' ' ax# y# b dA œ 2 '0
1/4
R
'0sec r$ dr d) œ 23 ; On the face z œ 0: g(xß yß z) œ z œ 0 Ê ™ g œ k and p œ k
)
Ê k ™ gk œ 1 Ê k ™ g † pk œ 1 Ê d5 œ dA Ê I œ ' ' ax# y# b dA œ 23 ; On the face y œ 0: g(xß yß z) œ y œ 0
R
Ê ™ g œ j and p œ j Ê k ™ gk œ 1 Ê k ™ g † pk œ 1 Ê d5 œ dA Ê I œ ' ' ax# 0b dA œ '0 '0 x# dx dz œ "3 ;
1
1
R
On the face y œ 1: g(xß yß z) œ y œ 1 Ê ™ g œ j and p œ j Ê k ™ gk œ 1 Ê k ™ g † pk œ 1 Ê d5 œ dA
Ê I œ ' ' ax# 1# b dA œ '0 '0 ax# 1b dx dz œ 43 ; On the face x œ 1: g(xß yß z) œ x œ 1 Ê ™ g œ i and p œ i
1
1
R
Ê k ™ gk œ 1 Ê k ™ g † pk œ 1 Ê d5 œ dA Ê I œ ' ' a1# y# b dA œ '0 '0 a1 y# b dy dz œ 43 ; On the face
1
1
R
x œ 0: g(xß yß z) œ x œ 0 Ê ™ g œ i and p œ i Ê k ™ gk œ 1 Ê k ™ g † pk œ 1 Ê d5 œ dA
Ê I œ ' ' a0# y# b dA œ '0 '0 y# dy dz œ 13 Ê Iz œ 23 23 "3 43 43 3" œ 14
3
1
1
R
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996
Chapter 16 Integration in Vector Fields
49. M œ 2xy x and N œ xy y Ê ``Mx œ 2y 1, ``My œ 2x, ``Nx œ y, ``Ny œ x 1 Ê Flux œ ' ' Š ``Mx ``Ny ‹ dx dy
1
1
œ ' ' (2y 1 x 1) dy dx œ '0 '0 (2y x) dy dx œ 3# ; Circ œ ' ' Š ``Nx ``My ‹ dx dy
R
œ ' ' (y 2x) dy dx œ '0 '
1
R
R
R
1
(y 2x) dy dx œ "#
0
50. M œ y 6x# and N œ x y# Ê ``Mx œ 12x, ``My œ 1, ``Nx œ 1, ``Ny œ 2y Ê Flux œ ' ' Š ``Mx ``Ny ‹ dx dy
R
œ ' ' (12x 2y) dx dy œ '0 'y (12x 2y) dx dy œ '0 a4y# 2y 6b dy œ 11
3 ;
1
1
1
R
Circ œ ' ' Š ``Nx ``My ‹ dx dy œ ' ' (1 1) dx dy œ 0
R
R
51. M œ cosx y and N œ ln x sin y Ê ``My œ sinx y and ``Nx œ sinx y Ê )C ln x sin y dy cosx y dx
œ ' ' Š ``Nx ``My ‹ dx dy œ ' ' Š sinx y sinx y ‹ dx dy œ 0
R
R
52. (a) Let M œ x and N œ y Ê ``Mx œ 1, ``My œ 0, ``Nx œ 0, ``Ny œ 1 Ê Flux œ ' ' Š ``Mx ``Ny ‹ dx dy
R
œ ' ' (1 1) dx dy œ 2 ' ' dx dy œ 2(Area of the region)
R
R
(b) Let C be a closed curve to which Green's Theorem applies and let n be the unit normal vector to C. Let
F œ xi yj and assume F is orthogonal to n at every point of C. Then the flux density of F at every point
of C is 0 since F † n œ 0 at every point of C Ê ``Mx ``Ny œ 0 at every point of C
Ê Flux œ ' ' Š ``Mx ``Ny ‹ dx dy œ ' ' 0 dx dy œ 0. But part (a) above states that the flux is
R
R
2(Area of the region) Ê the area of the region would be 0 Ê contradiction. Therefore, F cannot be
orthogonal to n at every point of C.
53. ``x (2xy) œ 2y, ``y (2yz) œ 2z, ``z (2xz) œ 2x Ê ™ † F œ 2y 2z 2x Ê Flux œ ' ' ' (2x 2y 2z) dV
D
1
1
1
1
1
1
œ '0 '0 '0 (2x 2y 2z) dx dy dz œ '0 '0 (1 2y 2z) dy dz œ '0 (2 2z) dz œ 3
54. ``x (xz) œ z, ``y (yz) œ z, ``z (1) œ 0 Ê ™ † F œ 2z Ê Flux œ ' ' ' 2z r dr d) dz
È25 c r
œ '0 '0 '3
21
4
D
#
2z dz r dr d) œ '0
21
'0 ra16 r# b dr d) œ '0 64 d) œ 1281
21
4
55. ``x (2x) œ 2, ``y (3y) œ 3, ``z (z) œ 1 Ê ™ † F œ 4; x# y# z# œ 2 and x# y# œ z Ê z œ 1
È2cr
Ê x# y# œ 1 Ê Flux œ ' ' ' 4 dV œ 4 '0 '0 'r#
21
D
1
#
dz r dr d) œ 4 '0 '0 ŠrÈ2 r# r$ ‹ dr d)
21
1
7
œ 4'0 Š 12
32 È2‹ d) œ 23 1 Š7 8È2‹
21
56. ``x (6x y) œ 6, ``y (x z) œ 0, ``z (4yz) œ 4y Ê ™ † F œ 6 4y; z œ Èx# y# œ r
Ê Flux œ ' ' ' (6 4y) dV œ '0
1/2
D
'01 '0r (6 4r sin )) dz r dr d) œ '01/2 '01 a6r# 4r$ sin )b dr d)
œ '0 (2 sin )) d) œ 1 1
1/2
Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.
Chapter 16 Additional and Advanced Exercises
997
57. F œ yi zj xk Ê ™ † F œ 0 Ê Flux œ ' ' F † n d5 œ ' ' ' ™ † F dV œ 0
D
S
58. F œ 3xz# i yj z$ k Ê ™ † F œ 3z# 1 3z# œ 1 Ê Flux œ ' ' F † n d5 œ ' ' ' ™ † F dV
È16cx Î2
œ '0 '0
4
#
D
S
'0
yÎ2
1 dz dy dx œ '
%
16x#
x$
dx
œ
x
œ 83
Š
‹
’
“
16
48
0
!
4
59. F œ xy# i x# yj yk Ê ™ † F œ y# x# 0 Ê Flux œ ' ' F † n d5 œ ' ' ' ™ † F dV
S
œ ' ' ' ax# y# b dV œ '0 '0 'c1 r# dz r dr d) œ '0 '0 2r$ dr d) œ '0
21
1
21
1
21
1
D
D
"
# d) œ 1
60. (a) F œ (3z 1)k Ê ™ † F œ 3 Ê Flux across the hemisphere œ ' ' F † n d5 œ ' ' ' ™ † F dV œ ' ' ' 3 dV
D
S
œ 3 ˆ "# ‰ ˆ 43 1a$ ‰ œ 21a$
D
(b) f(xß yß z) œ x# y# z# a# œ 0 Ê ™ f œ 2xi 2yj 2zk Ê k ™ f k œ È4x# 4y# 4z# œ È4a# œ 2a since
a 0 Ê n œ 2xi 2y#aj 2zk œ xi yaj zk Ê F † n œ (3z 1) ˆ za ‰ ; p œ k Ê ™ f † p œ ™ f † k œ 2z
Ê k ™ f † pk œ 2z since z
k™f k
2a
a
' ' F † n d5 œ ' ' (3z 1) ˆ za ‰ ˆ az ‰ dA
0 Ê d5 œ k™
f†pk œ 2z dA œ z dA Ê
S
Rxy
œ ' ' (3z 1) dx dy œ ' ' ˆ3Èa# x# y# 1‰ dx dy œ '0 '0 Š3Èa# r# 1‹ r dr d)
21
Rxy
a
Rxy
œ '0 Š a# a$ ‹ d) œ 1a# 21a$ , which is the flux across the hemisphere. Across the base we find
21
#
F œ [3(0) 1]k œ k since z œ 0 in the xy-plane Ê n œ k (outward normal) Ê F † n œ 1 Ê Flux across the
base œ ' ' F † n d5 œ ' ' 1 dx dy œ 1a# . Therefore, the total flux across the closed surface is
Rxy
S
#
$
#
a1a 21a b 1a œ 21a$ .
CHAPTER 16 ADDITIONAL AND ADVANCED EXERCISES
1. dx œ (2 sin t 2 sin 2t) dt and dy œ (2 cos t 2 cos 2t) dt; Area œ "# )C x dy y dx
œ "# '0 [(2 cos t cos 2t)(2 cos t 2 cos 2t) (2 sin t sin 2t)(2 sin t 2 sin 2t)] dt
21
œ "# '0 [6 (6 cos t cos 2t 6 sin t sin 2t)] dt œ "# '0 (6 6 cos t) dt œ 61
21
21
2. dx œ (2 sin t 2 sin 2t) dt and dy œ (2 cos t 2 cos 2t) dt; Area œ "# )C x dy y dx
œ "# '0 [(2 cos t cos 2t)(2 cos t 2 cos 2t) (2 sin t sin 2t)(2 sin t 2 sin 2t)] dt
21
œ "# '0 [2 2(cos t cos 2t sin t sin 2t)] dt œ "# '0 (2 2 cos 3t) dt œ "# 2t 23 sin 3t‘ ! œ 21
21
21
#1
3. dx œ cos 2t dt and dy œ cos t dt; Area œ "# )C x dy y dx œ "# '0 ˆ "# sin 2t cos t sin t cos 2t‰ dt
1
œ #" '0 csin t cos# t (sin t) a2 cos# t 1bd dt œ "# '0 a sin t cos# t sin tb dt œ "# "3 cos$ t cos t‘ ! œ 3" 1 œ 23
1
1
4. dx œ (2a sin t 2a cos 2t) dt and dy œ (b cos t) dt; Area œ "# )C x dy y dx
œ "# '0 ca2ab cos# t ab cos t sin 2tb a2ab sin# t 2ab sin t cos 2tbd dt
21
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998
Chapter 16 Integration in Vector Fields
œ "# '0 c2ab 2ab cos# t sin t 2ab(sin t) a2 cos# t 1bd dt œ "# '0 a2ab 2ab cos# t sin t 2ab sin tb dt
21
21
#1
œ "# 2abt 23 ab cos$ t 2ab cos t‘ ! œ 21ab
5. (a) F(xß yß z) œ zi xj yk is 0 only at the point (0ß 0ß 0), and curl F(xß yß z) œ i j k is never 0 .
(b) F(xß yß z) œ zi yk is 0 only on the line x œ t, y œ 0, z œ 0 and curl F(xß yß z) œ i j is never 0 .
(c) F(xß yß z) œ zi is 0 only when z œ 0 (the xy-plane) and curl F(xß yß z) œ j is never 0 .
y j zk
x i y j zk
cy
#
6. F œ yz# i xz# j 2xyzk and n œ Èxxi , so F is parallel to n when yz# œ cx
# y # z# œ
R
R , xz œ R ,
#
#
yz
xz
c
#
#
#
#
È2x. Also,
and 2xyz œ cz
R Ê x œ y œ 2xy Ê y œ x Ê y œ „ x and z œ „ R œ 2x Ê z œ „
x# y# z# œ R# Ê x# x# 2x# œ R# Ê 4x# œ R# Ê x œ „ R# . Thus the points are: Š R# ß R# ß
Š R# ß R# ß È2R
È2R
È2R
R
R È2R
R
R
R
R È2R
R
R
# ‹ , Š # ß # ß # ‹ , Š # ß # ß # ‹ , Š # ß # ß # ‹ , Š # ß # ß # ‹ ,
Š R# ß R# ß
È2R
È2R
R R
# ‹ , Š # ß # ß # ‹
È2R
# ‹,
7. Set up the coordinate system so that (aß bß c) œ (0ß Rß 0) Ê $ (xß yß z) œ Èx# (y R)# z#
œ Èx# y# z# 2Ry R# œ È2R# 2Ry ; let f(xß yß z) œ x# y# z# R# and p œ i
™f k
2R
Ê ™ f œ 2xi 2yj 2zk Ê k ™ f k œ 2Èx# y# z# œ 2R Ê d5 œ kk™
f†ik dz dy œ 2x dz dy
Ê Mass œ ' ' $ (xß yß z) d5 œ ' ' È2R# 2Ry ˆ Rx ‰ dz dy œ R ' '
Ryz
S
È c
R
œ 4R 'cR '0
R#
y#
Ryz
È2R# 2Ry
ÈR# y# z# dz dy œ 4R
È2R# 2Ry
ÈR# y# z# dz dy
'cRR É2R2 2Ry sin1 Š ÈR z y ‹»
2
ÈR cy
#
#
dy
2
0
3/2
1‰
2
œ 21R'cR É2R2 2Ry dy œ 21Rˆ 3R a2R 2Ryb »
R
R
cR
œ 1613R
$
â
â
i
j
kâ
â
â
â
sin ) 0 â
8. r(rß )) œ (r cos ))i (r sin ))j )k , 0 Ÿ r Ÿ 1, 0 Ÿ ) Ÿ 21 Ê rr ‚ r) œ â cos )
â
â
â r sin ) r cos ) " â
œ (sin ))i (cos ))j rk Ê krr ‚ r) k œ È1 r# ; $ œ 2Èx# y# œ 2Èr# cos# ) r# sin# ) œ 2r
Ê Mass œ ' ' $ (xß yß z) d5 œ '0
21
S
'01 2rÈ1 r# dr d) œ '021 ’ 23 a1 r# b$Î# “ " d) œ '021 23 Š2È2 1‹ d)
!
œ 431 Š2È2 1‹
9. M œ x# 4xy and N œ 6y Ê ``Mx œ 2x 4y and ``Nx œ 6 Ê Flux œ '0 '0 (2x 4y 6) dx dy
b
a
œ '0 aa# 4ay 6ab dy œ a# b 2ab# 6ab. We want to minimize f(aß b) œ a# b 2ab# 6ab œ ab(a 2b 6).
b
Thus, fa (aß b) œ 2ab 2b# 6b œ 0 and fb (aß b) œ a# 4ab 6a œ 0 Ê b(2a 2b 6) œ 0 Ê b œ 0 or b œ a 3.
Now b œ 0 Ê a# 6a œ 0 Ê a œ 0 or a œ 6 Ê (0ß 0) and (6ß 0) are critical points. On the other hand, b œ a 3
Ê a# 4a(a 3) 6a œ 0 Ê 3a# 6a œ 0 Ê a œ 0 or a œ 2 Ê (0ß 3) and (#ß ") are also critical points. The flux at
(0ß 0) œ 0, the flux at (6ß 0) œ 0, the flux at (0ß 3) œ 0 and the flux at (2ß 1) œ 4. Therefore, the flux is minimized at (2ß 1)
with value 4.
10. A plane through the origin has equation ax by cz œ 0. Consider first the case when c Á 0. Assume the plane is given
by z œ ax by and let f(xß yß z) œ x# y# z# œ 4. Let C denote the circle of intersection of the plane with the sphere.
By Stokes's Theorem, )C F † dr œ ' ' ™ ‚ F † n d5 , where n is a unit normal to the plane. Let
S
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Chapter 16 Additional and Advanced Exercises
999
â
â
â i j kâ
â
â
r(xß y) œ xi yj (ax by)k be a parametrization of the surface. Then rx ‚ ry œ â " 0 a â œ ai bj k
â
â
â0 " bâ
â i
j
k ââ
â
â
â
k
`
`
Ê d5 œ krx ‚ ry k dx dy œ Èa# b# 1 dx dy. Also, ™ ‚ F œ â ` x ` y ``z â œ i j k and n œ Èaai #bjb
#1
â
â
â z
x y â
Ê ' ' ™ ‚ F † n d5 œ ' ' a b 1 Èa# b# 1 dx dy œ ' ' (a b 1) dx dy œ (a b 1) ' ' dx dy. Now
Rxy
S
x# y# (ax by)# œ 4 Ê
È a# b# 1
Rxy
#
#
‰
Š a 4 1 ‹ x# Š b 4 1 ‹ y# ˆ ab
# xy œ 1
Rxy
Ê the region Rxy is the interior of the ellipse
#
#
b 1
Ax# Bxy Cy# œ 1 in the xy-plane, where A œ a 4 1 , B œ ab
# , and C œ
4 . The area of the ellipse is
21
41
È4AC B# œ Èa# b# 1
41(a b 1)
1)
Ê )C F † dr œ h(aß b) œ È
. Thus we optimize H(aß b) œ (aa#bb
#1 :
a# b # 1
#
#
2(a b 1) ab# 1 a abb
`H
œ 0 and ``Hb œ 2(a b aa#1)aab# 11b2 b abb œ 0
`a œ
aa# b# 1b2
Ê a b 1 œ 0, or b# 1 a ab œ 0
and a# 1 b ab œ 0 Ê a b 1 œ 0, or a# b# (b a) œ 0 Ê a b 1 œ 0, or (a b)(a b 1) œ 0
Ê a b 1 œ 0 or a œ b. The critical values a b 1 œ 0 give a saddle. If a œ b, then 0 œ b# 1 a ab
Ê a# 1 a a# œ 0 Ê a œ 1 Ê b œ 1. Thus, the point (aß b) œ (1ß 1) gives a local extremum for )C F † dr
Ê z œ x y Ê x y z œ 0 is the desired plane, if c Á 0.
Note: Since h(1ß 1) is negative, the circulation about n is clockwise, so n is the correct pointing normal for
the counterclockwise circulation. Thus ' ' ™ ‚ F † (n) d5 actually gives the maximum circulation.
S
If c œ 0, one can see that the corresponding problem is equivalent to the calculation above when b œ 0, which does not
lead to a local extreme.
11. (a) Partition the string into small pieces. Let ?i s be the length of the ith piece. Let (xi ß yi ) be a point in the
ith piece. The work done by gravity in moving the ith piece to the x-axis is approximately
Wi œ (gxi yi ?i s)yi where xi yi ?i s is approximately the mass of the ith piece. The total work done by
gravity in moving the string to the x-axis is D Wi œ D gxi y#i ?i s Ê Work œ 'C gxy# ds
i
i
1/2
1/2
(b) Work œ 'C gxy# ds œ '0 g(2 cos t) a4 sin# tbÈ4 sin# t 4 cos# t dt œ 16g '0 cos t sin# t dt
$
œ ’16g Š sin3 t ‹“
1Î#
!
œ 16
3 g
' x(xy) ds
' y(xy) ds
(c) x œ 'C
and y œ 'C
; the mass of the string is 'C xy ds and the weight of the string is
xy ds
xy ds
C
C
g 'C xy ds. Therefore, the work done in moving the point mass at axß yb to the x-axis is
W œ Šg 'C xy ds‹ y œ g 'C xy# ds œ 16
3 g.
12. (a) Partition the sheet into small pieces. Let ?i 5 be the area of the ith piece and select a point (xi ß yi ß zi ) in
the ith piece. The mass of the ith piece is approximately xi yi ?i 5 . The work done by gravity in moving the
ith piece to the xy-plane is approximately (gxi yi ?i 5 )zi œ gxi yi zi ?i 5 Ê Work œ ' ' gxyz d5.
(b) ' ' gxyz d5 œ g ' '
xy(1 x y)È1 (1)# (1)# dA œ È3g
Rxy
S
'
1
1cx
'0 '0
1
S
axy x# y xy# b dy dx
'
1
"x
œ È3g 0 "2 xy# "# x# y# "3 xy$ ‘ ! dx œ È3g 0 "6 x #" x# "# x$ "6 x% ‘ dx
È
" #
" &‘ "
" ‰
œ È3g 12
x 6" x$ 6" x% 30
x ! œ È3g ˆ 1"# 30
œ 203g
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1000 Chapter 16 Integration in Vector Fields
(c) The center of mass of the sheet is the point axß yß zb where z œ Mxy with Mxy œ ' ' xyz d5 and
M
S
M œ ' ' xy d5. The work done by gravity in moving the point mass at axß yß zb to the xy-plane is
S
gMz œ gM Š MMxy ‹ œ gMxy œ ' ' gxyz d5 œ
S
È3g
20 .
13. (a) Partition the sphere x# y# (z 2)# œ 1 into small pieces. Let ?i 5 be the surface area of the ith piece and let
(xi ß yi ß zi ) be a point on the ith piece. The force due to pressure on the ith piece is approximately w(4 zi )?i 5 . The
total force on S is approximately D w(4 zi )?i 5. This gives the actual force to be ' ' w(4 z) d5.
i
S
(b) The upward buoyant force is a result of the k-component of the force on the ball due to liquid pressure.
The force on the ball at (xß yß z) is w(4 z)(n) œ w(z 4)n , where n is the outer unit normal at (xß yß z).
Hence the k-component of this force is w(z 4)n † k œ w(z 4)k † n . The (magnitude of the) buoyant force
on the ball is obtained by adding up all these k-components to obtain ' ' w(z 4)k † n d5.
S
(c) The Divergence Theorem says ' ' w(z 4)k † n d5 œ ' ' ' div(w(z 4)k) dV œ ' ' ' w dV, where D
D
S
D
is x# y# (z 2)# Ÿ 1 Ê ' ' w(z 4)k † n d5 œ w ' ' ' 1 dV œ 43 1w, the weight of the fluid if it
D
S
were to occupy the region D.
14. The surface S is z œ Èx# y# from z œ 1 to z œ 2. Partition S into small pieces and let ?i 5 be the area of the
ith piece. Let (xi ß yi ß zi ) be a point on the ith piece. Then the magnitude of the force on the ith piece due to
liquid pressure is approximately Fi œ w(2 zi )?i 5 Ê the total force on S is approximately
#
D Fi œ D w(2 zi )?i 5 Ê the actual force is ' ' w(2 z) d5 œ ' ' w ˆ2 Èx# y# ‰ É1 x# x y# x# y y# dA
#
i
S
Rxy
#
œ ' ' È2 w ˆ2 Èx# y# ‰ dA œ '0 '1 È2w(2 r) r dr d) œ '0 È2w r# 3" r$ ‘ " d) œ '0
21
2
21
21
Rxy
È
2È2w
d) œ 4 321w
3
15. Assume that S is a surface to which Stokes's Theorem applies. Then )C E † dr œ ' ' ( ™ ‚ E) † n d5
S
œ ' ' ˆ ``Bt ‰ † n d5 œ ``t ' ' B † n d5. Thus the voltage around a loop equals the negative of the rate of
S
S
change of magnetic flux through the loop.
16. According to Gauss's Law, ' ' F † n d5 œ 41GmM for any surface enclosing the origin. But if F œ ™ ‚ H
S
then the integral over such a closed surface would have to be 0 by the Divergence Theorem since div F œ 0.
17. )C f ™ g † dr œ ' ' ™ ‚ (f ™ g) † n d5
(Stokes's Theorem)
S
œ ' ' (f ™ ‚ ™ g ™ f ‚ ™ g) † n d5
(Section 16.8, Exercise 19b)
S
œ ' ' [(f)(0) ™ f ‚ ™ g] † n d5
(Section 16.7, Equation 8)
S
œ ' ' ( ™ f ‚ ™ g) † n d5
S
18. ™ ‚ F" œ ™ ‚ F# Ê ™ ‚ (F# F" ) œ 0 Ê F# F" is conservative Ê F# F" œ ™ f; also, ™ † F" œ ™ † F#
Ê ™ † (F# F" ) œ 0 Ê ™ # f œ 0 (so f is harmonic). Finally, on the surface S, ™ f † n œ (F# F" ) † n
œ F# † n F" † n œ 0. Now, ™ † (f ™ f) œ ™ f † ™ f f ™ # f so the Divergence Theorem gives
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Chapter 16 Additional and Advanced Exercises 1001
' ' ' k ™ f k# dV ' ' ' f ™ # f dV œ ' ' ' ™ † (f ™ f) dV œ ' ' f ™ f † n d5 œ 0, and since ™ # f œ 0 we have
D
D
D
S
' ' ' k ™ f k# dV 0 œ 0 Ê ' ' ' kF# F" k # dV œ 0 Ê F# F" œ 0 Ê F# œ F" , as claimed.
D
D
19. False; let F œ yi xj Á 0 Ê
â i
â
â
™ † F œ ``x (y) ``y (x) œ 0 and ™ ‚ F œ â ``x
â
â x
j
`
`y
y
k ââ
` â
` z â œ 0i 0j 0k œ 0
â
0 â
20. kru ‚ rv k# œ kru k# krv k# sin# ) œ kru k# krv k# a1 cos# )b œ kru k# krv k# kru k# krv k# cos# ) œ kru k# krv k# (ru † rv )#
Ê kru ‚ rv k# œ EG F# Ê d5 œ kru ‚ rv k du dv œ ÈEG F# du dv
21. r œ xi yj zk Ê ™ † r œ 1 1 1 œ 3 Ê ' ' ' ™ † r dV œ 3 ' ' ' dV œ 3V Ê V œ "3 ' ' ' ™ † r dV
œ "3 ' ' r † n d5, by the Divergence Theorem
D
D
S
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D