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INSTRUCTOR'S
SOLUTIONS
MANUAL
for
lOMoARcPSD|6566483
INSTRUCTOR’S SOLUTIONS
MANUAL
Robert A. Adams
University of British Columbia
Calculus
Ninth Edition
Robert A. Adams
University of British Columbia
Christopher Essex
University of Western Ontario
dumperina
ISBN: 978-0-13-452876-2
Copyright © 2018 Pearson Canada Inc., Toronto, Ontario. All rights reserved. This work is protected by Canadian
copyright laws and is provided solely for the use of instructors in teaching their courses and assessing student learning.
Dissemination or sale of any part of this work (including on the Internet) will destroy the integrity of the work and is
not permitted. The copyright holder grants permission to instructors who have adopted Calculus: A Complete Course,
Single-Variable Calculus, or Calculus of Several Variables by Adams and Essex, to post this material online only if the
use of the website is restricted by access codes to students in the instructor’s class that is using the textbook and
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FOREWORD
These solutions are provided for the benefit of instructors using the textbooks:
Calculus: A Complete Course (9th Edition),
Single-Variable Calculus (9th Edition), and
Calculus of Several Variables (9th Edition)
by R. A. Adams and Chris Essex, published by Pearson Canada. For the most part, the solutions are detailed, especially in exercises on core material and techniques. Occasionally some
details are omitted—for example, in exercises on applications of integration, the evaluation of
the integrals encountered is not always given with the same degree of detail as the evaluation of
integrals found in those exercises dealing specifically with techniques of integration.
Instructors may wish to make these solutions available to their students. However, students
should use such solutions with caution. It is always more beneficial for them to attempt exercises and problems on their own, before they look at solutions done by others. If they examine
solutions as “study material” prior to attempting the exercises, they can lose much of the benefit that follows from diligent attempts to develop their own analytical powers. When they have
tried unsuccessfully to solve a problem, then looking at a solution can give them a “hint” for
a second attempt. Separate Student Solutions Manuals for the books are available for students.
They contain the solutions to the even-numbered exercises only.
November, 2016.
R. A. Adams
adms@math.ubc.ca
Chris Essex
essex@uwo.ca
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CONTENTS
Solutions for Chapter P
1
Solutions for Chapter 1
23
Solutions for Chapter 2
39
Solutions for Chapter 3
81
Solutions for Chapter 4
108
Solutions for Chapter 5
177
Solutions for Chapter 6
213
Solutions for Chapter 7
267
Solutions for Chapter 8
316
Solutions for Chapter 9
351
Solutions for Chapter 10
392
Solutions for Chapter 11
420
Solutions for Chapter 12
448
Solutions for Chapter 13
491
Solutions for Chapter 14
538
Solutions for Chapter 15
579
Solutions for Chapter 16
610
Solutions for Chapter 17
637
Solutions for Chapter 18
644
Solutions for Chapter 18-cosv9
671
Solutions for Appendices
683
NOTE: “Solutions for Chapter 18-cosv9” is only needed by users of Calculus of Several Variables (9th Edition), which includes extra material in Sections 18.2 and 18.5 that is found in Calculus: a Complete Course and in Single-Variable Calculus in Sections 7.9 and 3.7 respectively.
Solutions for Chapter 18-cosv9 contains only the solutions for the two Sections 18.2 and 18.9 in
the Several Variables book. All other Sections are in “Solutions for Chapter 18.”
It should also be noted that some of the material in Chapter 18 is beyond the scope of most
students in single-variable calculus courses as it requires the use of multivariable functions and
partial derivatives.
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INSTRUCTOR’S SOLUTIONS MANUAL
CHAPTER P.
SECTION P.1 (PAGE 10)
PRELIMINARIES
19.
Given: 1=.2 " x/ < 3.
CASE I. If x < 2, then 1 < 3.2 " x/ D 6 " 3x, so 3x < 5
and x < 5=3. This case has solutions x < 5=3.
CASE II. If x > 2, then 1 > 3.2 " x/ D 6 " 3x, so 3x > 5
and x > 5=3. This case has solutions x > 2.
Solution: ."1; 5=3/ [ .2; 1/.
20.
Given: .x C 1/=x # 2.
CASE I. If x > 0, then x C 1 # 2x, so x $ 1.
CASE II. If x < 0, then x C 1 $ 2x, so x # 1. (not
possible)
Solution: .0; 1!.
21.
Given: x 2 " 2x $ 0. Then x.x " 2/ $ 0. This is only
possible if x # 0 and x $ 2. Solution: Œ0; 2!.
Section P.1 Real Numbers and the Real Line
(page 10)
1.
2
D 0:22222222 ! ! ! D 0:2
9
2.
1
D 0:09090909 ! ! ! D 0:09
11
3. If x D 0:121212 ! ! !, then 100x D 12:121212 ! ! ! D 12 C x.
Thus 99x D 12 and x D 12=99 D 4=33.
4. If x D 3:277777 ! ! !, then 10x " 32 D 0:77777 ! ! ! and
100x " 320 D 7 C .10x " 32/, or 90x D 295. Thus
x D 295=90 D 59=18.
5.
1=7 D 0:142857142857 ! ! ! D 0:142857
22.
Given 6x 2 " 5x $ "1, then .2x " 1/.3x " 1/ $ 0, so either
x $ 1=2 and x # 1=3, or x $ 1=3 and x # 1=2. The latter
combination is not possible. The solution set is Œ1=3; 1=2!.
23.
Given x 3 > 4x, we have x.x 2 " 4/ > 0. This is possible
if x < 0 and x 2 < 4, or if x > 0 and x 2 > 4. The
possibilities are, therefore, "2 < x < 0 or 2 < x < 1.
Solution: ."2; 0/ [ .2; 1/.
2=7 D 0:285714285714 ! ! ! D 0:285714
3=7 D 0:428571428571 ! ! ! D 0:428571
4=7 D 0:571428571428 ! ! ! D 0:571428
note the same cyclic order of the repeating digits
5=7 D 0:714285714285 ! ! ! D 0:714285
24.
6=7 D 0:857142857142 ! ! ! D 0:857142
6. Two different decimal expansions can represent the same
number. For instance, both 0:999999 ! ! ! D 0:9 and
1:000000 ! ! ! D 1:0 represent the number 1.
7.
25.
x # 0 and x $ 5 define the interval Œ0; 5!.
8. x < 2 and x # "3 define the interval Œ"3; 2/.
9. x > "5 or x < "6 defines the union ."1; "6/ [ ."5; 1/.
10. x $ "1 defines the interval ."1; "1!.
11.
27.
If jxj D 3 then x D ˙3.
28.
If jx " 3j D 7, then x " 3 D ˙7, so x D "4 or x D 10.
29.
If j2t C 5j D 4, then 2t C 5 D ˙4, so t D "9=2 or
t D "1=2.
30.
Ifj1 " t j D 1, then 1 " t D ˙1, so t D 0 or t D 2.
31.
If j8 " 3sj D 9, then 8 " 3s D ˙9, so 3s D "1 or 17, and
s D "1=3 or s D 17=3.
13. If "2x > 4, then x < "2. Solution: ."1; "2/
15.
If 5x " 3 $ 7 " 3x, then 8x $ 10 and x $ 5=4. Solution:
."1; 5=4!
6"x
3x " 4
16. If
#
, then 6 " x # 6x " 8. Thus 14 # 7x
4
2
and x $ 2. Solution: ."1; 2!
17.
If 3.2 " x/ < 2.3 C x/, then 0 < 5x and x > 0. Solution:
.0; 1/
2
18. If x < 9, then jxj < 3 and "3 < x < 3. Solution:
."3; 3/
2
3
<
.
x"1
xC1
CASE I. If x > 1 then .x " 1/.x C 1/ > 0, so that
3.x C 1/ < 2.x " 1/. Thus x < "5. There are no solutions
in this case.
CASE II. If "1 < x < 1, then .x " 1/.x C 1/ < 0, so
3.x C 1/ > 2.x " 1/. Thus x > "5. In this case all
numbers in ."1; 1/ are solutions.
CASE III. If x < "1, then .x " 1/.x C 1/ > 0, so that
3.x C 1/ < 2.x " 1/. Thus x < "5. All numbers x < "5
are solutions.
Solutions: ."1; "5/ [ ."1; 1/.
Given:
12. x < 4 or x # 2 defines the interval ."1; 1/, that is, the
whole real line.
If 3x C 5 $ 8, then 3x $ 8 " 5 " 3 and x $ 1. Solution:
."1; 1!
4
x
#1C .
2
x
CASE I. If x > 0, then x 2 # 2x C 8, so that
x 2 " 2x " 8 # 0, or .x " 4/.x C 2/ # 0. This is possible for x > 0 only if x # 4.
CASE II. If x < 0, then we must have .x " 4/.x C 2/ $ 0,
which is possible for x < 0 only if x # "2.
Solution: Œ"2; 0/ [ Œ4; 1/.
Given:
26.
x > "2 defines the interval ."2; 1/.
14.
Given x 2 "x $ 2, then x 2 "x "2 $ 0 so .x "2/.x C1/ $ 0.
This is possible if x $ 2 and x # "1 or if x # 2 and
x $ "1. The latter situation is not possible. The solution
set is Œ"1; 2!.
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SECTION P.1 (PAGE 10)
ADAMS and ESSEX: CALCULUS 9
ˇs
ˇ
s
ˇ
ˇ
32. If ˇ " 1ˇ D 1, then " 1 D ˙1, so s D 0 or s D 4.
2
2
33. If jxj < 2, then x is in ."2; 2/.
4.
From A.0:5; 3/ to B.2; 3/, "x D 2 " 0:5 D 1:5 and
"y D 3 " 3 D 0. jABj D 1:5.
5.
Starting point: ."2; 3/. Increments "x D 4, "y D "7.
New position is ."2 C 4; 3 C ."7//, that is, .2; "4/.
6.
Arrival point: ."2; "2/. Increments "x D "5, "y D 1.
Starting point was ."2 " ."5/; "2 " 1/, that is, .3; "3/.
34. If jxj $ 2, then x is in Œ"2; 2!.
35. If js " 1j $ 2, then 1 " 2 $ s $ 1 C 2, so s is in Œ"1; 3!.
36. If jt C 2j < 1, then "2 " 1 < t < "2 C 1, so t is in
."3; "1/.
7.
37. If j3x "7j < 2, then 7"2 < 3x < 7C2, so x is in .5=3; 3/.
38. If j2x C 5j < 1, then "5 " 1 < 2x < "5 C 1, so x is in
."3; "2/.
ˇ
ˇx
x
ˇ
ˇ
39. If ˇ " 1ˇ $ 1, then 1 " 1 $ $ 1 C 1, so x is in Œ0; 4!.
2
2
ˇ
x ˇˇ
1
ˇ
40. If ˇ2 " ˇ < , then x=2 lies between 2 " .1=2/ and
2
2
2 C .1=2/. Thus x is in .3; 5/.
41. The inequality jx C 1j > jx " 3j says that the distance
from x to "1 is greater than the distance from x to 3, so
x must be to the right of the point half-way between "1
and 3. Thus x > 1.
42. jx " 3j < 2jxj , x 2 " 6x C 9 D .x " 3/2 < 4x 2
, 3x 2 C 6x " 9 > 0 , 3.x C 3/.x " 1/ > 0. This
inequality holds if x < "3 or x > 1.
8.
9.
x 2 C y 2 D 1 represents a circle of radius 1 centred at the
origin.
p
x 2 C y 2 D 2 represents a circle of radius 2 centred at
the origin.
x 2 C y 2 $ 1 represents points inside and on the circle of
radius 1 centred at the origin.
10. x 2 C y 2 D 0 represents the origin.
11.
y # x 2 represents all points lying on or above the
parabola y D x 2 .
12. y < x 2 represents all points lying below the parabola
y D x2.
13.
14.
The vertical line through ."2; 5=3/ is x D "2; the horizontal line through that point is y D 5=3.
p
p
The vertical line through . 2; "1:3/ is x D 2; the
horizontal line through that point is y D "1:3.
43. jaj D a if and only if a # 0. It is false if a < 0.
15.
44. The equation jx " 1j D 1 " x holds if jx " 1j D ".x " 1/,
that is, if x " 1 $ 0, or, equivalently, if x $ 1.
Line through ."1; 1/ with slope m D 1 is y D 1C1.xC1/,
or y D x C 2.
16.
Line through ."2; 2/ with slope m D 1=2 is
y D 2 C .1=2/.x C 2/, or x " 2y D "6.
17.
Line through .0; b/ with slope m D 2 is y D b C 2x.
45. The triangle inequality jx C yj $ jxj C jyj implies that
jxj # jx C yj " jyj:
Apply this inequality with x D a " b and y D b to get
ja " bj # jaj " jbj:
ˇ
ˇ
ˇ
ˇ
Similarly, ja " bj D jb " aj # jbj " jaj. Since ˇjaj " jbjˇ is
equal to either jaj " jbj or jbj " jaj, depending on the sizes
of a and b, we have
ˇ
ˇ
ˇ
ˇ
ja " bj # ˇjaj " jbjˇ:
Section P.2 Cartesian Coordinates in the
Plane (page 16)
1.
From A.0; 3/ to B.4; 0/, "xp
D 4 " 0 D 4 and
"y D 0 " 3 D "3. jABj D 42 C ."3/2 D 5.
2. From A."1; 2/ to B.4; "10/, "xpD 4 " ."1/ D 5 and
"y D "10 " 2 D "12. jABj D 52 C ."12/2 D 13.
3. From A.3; 2/ to B."1; "2/, "x
and
p D "1 " 3 D "4 p
"y D "2 " 2 D "4. jABj D ."4/2 C ."4/2 D 4 2.
2
18. Line through .a; 0/ with slope m D "2 is y D 0"2.x "a/,
or y D 2a " 2x.
19.
At x D 2, the height of the line 2x C 3y D 6 is
y D .6 " 4/=3 D 2=3. Thus .2; 1/ lies above the line.
20.
At x D 3, the height of the line x " 4y D 7 is
y D .3 " 7/=4 D "1. Thus .3; "1/ lies on the line.
21.
The line through .0; 0/ and .2; 3/ has slope
m D .3 " 0/=.2 " 0/ D 3=2 and equation y D .3=2/x or
3x " 2y D 0.
22.
The line through ."2; 1/ and .2; "2/ has slope
m D ."2 " 1/=.2 C 2/ D "3=4 and equation
y D 1 " .3=4/.x C 2/ or 3x C 4y D "2.
23.
The line through .4; 1/ and ."2; 3/ has slope
1
m D .3"1/=."2"4/ D "1=3 and equation y D 1" .x"4/
3
or x C 3y D 7.
24.
25.
The line through ."2; 0/ and .0; 2/ has slope
m D .2 " 0/=.0 C 2/ D 1 and equation y D 2 C x.
p
If m D "2 and
p b D 2, then the line has equation
y D "2x C 2.
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION P.2 (PAGE 16)
y
26. If m D "1=2 and b D "3, then the line has equation
y D ".1=2/x " 3, or x C 2y D "6.
27.
1:5x " 2y D "3
3x C 4y D 12 has x-intercept a D 12=3 D 4 and yintercept b D 12=4 D 3. Its slope is "b=a D "3=4.
y
x
Fig. P.2-30
3x C 4y D 12
31.
line through .2; 1/ parallel to y D x C 2 is y D x " 1; line
perpendicular to y D x C 2 is y D "x C 3.
32.
line through ."2; 2/ parallel to 2xCy D 4 is 2xCy D "2;
line perpendicular to 2x C y D 4 is x " 2y D "6.
33.
We have
x
Fig. P.2-27
3x C 4y D "6
2x " 3y D 13
28. x C 2y D "4 has x-intercept a D "4 and y-intercept
b D "4=2 D "2. Its slope is "b=a D 2=."4/ D "1=2.
y
%
6x C 8y D "12
6x " 9y D 39:
Subtracting these equations gives 17y D "51, so y D "3
and x D .13 " 9/=2 D 2. The intersection point is .2; "3/.
34.
x
We have
2x C y D 8
5x " 7y D 1
x C 2y D "4
%
14x C 7y D 56
5x " 7y D 1:
Adding these equations gives 19x D 57, so x D 3 and
y D 8 " 2x D 2. The intersection point is .3; 2/.
35.
If a ¤ 0 and b ¤ 0, then .x=a/ C .y=b/ D 1 represents
a straight line that is neither horizontal nor vertical, and
does not pass through the origin. Putting y D 0 we get
x=a D 1, so the x-intercept of this line is x D a; putting
x D 0 gives y=b D 1, so the y-intercept is y D b.
36.
The line .x=2/ " .y=3/ D 1 has x-intercept a D 2, and
y-intercept b D "3.
y
Fig. P.2-28
29.
p
p
p
p
2x " 3y D 2 has x-intercept
a D 2= 2 D 2
p
and y-intercept
p b pD "2= 3. Its slope is
"b=a D 2= 6 D 2=3.
y
2
p
2x "
p
x
x
x y
" D1
2
3
3y D 2
"3
Fig. P.2-29
Fig. P.2-36
30. 1:5x " 2y D "3 has x-intercept a D "3=1:5 D "2 and
y-intercept b D "3=."2/ D 3=2. Its slope is "b=a D 3=4.
37. The line through .2; 1/ and .3; "1/ has slope
m D ."1 " 1/=.3 " 2/ D "2 and equation
y D 1 " 2.x " 2/ D 5 " 2x. Its y-intercept is 5.
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SECTION P.2 (PAGE 16)
ADAMS and ESSEX: CALCULUS 9
38. The line through ."2; 5/ and .k; 1/ has x-intercept 3, so
also passes through .3; 0/. Its slope m satisfies
43.
B D .1; 3/; C D ."3; 2/
p
p
jABj D .1 " 2/2 C .3 C 1/2 D 17
p
p
p p
jAC j D ."3 " 2/2 C .2 C 1/2 D 34 D 2 17
p
p
jBC j D ."3 " 1/2 C .2 " 3/2 D 17:
p
Since jABj D jBC j and jAC j D 2jABj, triangle ABC
is an isosceles right-angled triangle with right angle at
B. Thus ABCD is a square if D is displaced from C
by the same amount A is from B, that is, by increments
"x D 2 " 1 D 1 and "y D "1 " 3 D "4. Thus
D D ."3 C 1; 2 C ."4// D ."2; "2/.
44.
If M D .xm ; ym / is the midpoint of P1 P2 , then the displacement of M from P1 equals the displacement of P2
from M :
0"5
1"0
DmD
D "1:
k"3
3C2
Thus k " 3 D "1, and so k D 2.
39. C D Ax C B. If C D 5; 000 when x D 10; 000 and
C D 6; 000 when x D 15; 000, then
10; 000A C B D 5; 000
15; 000A C B D 6; 000
Subtracting these equations gives 5; 000A D 1; 000, so
A D 1=5. From the first equation, 2; 000 C B D 5; 000,
so B D 3; 000. The cost of printing 100,000 pamphlets is
$100; 000=5 C 3; 000 D $23; 000.
xm " x1 D x2 " xm ;
45.
C DF
-30
46.
10 20 30 40 50 60 70 80 F
C D
47.
Fig. P.2-40
42.
A D .2; 1/; B D .6; 4/; C D .5; "3/
p
p
jABj D .6 " 2/2 C .4 " 1/2 D 25 D 5
p
p
jAC j D .5 " 2/2 C ."3 " 1/2 D 25 D 5
p
p
p
jBC j D .6 " 5/2 C .4 C 3/2 D 50 D 5 2:
Since jABj D jAC j, triangle ABC is isosceles.
p
A D .0; 0/; B D .1; 3/; C D .2; 0/
q
p
p
jABj D .1 " 0/2 C . 3 " 0/2 D 4 D 2
p
p
jAC j D .2 " 0/2 C .0 " 0/2 D 4 D 2
q
p
p
jBC j D .2 " 1/2 C .0 " 3/2 D 4 D 2:
Since jABj D jAC j D jBC j, triangle ABC is equilateral.
4
Let the coordinates of P be .x; 0/ and those of Q be
.X; "2X/. If the midpoint of PQ is .2; 1/, then
.x C X/=2 D 2;
5
.F " 32/
9
-40
."40; "40/
-50
41.
yq " y1 D 2.y2 " yq /:
Thus xq D .x1 C 2x2 /=3 and yq D .y1 C 2y2 /=3.
10
-20
If Q D .xq ; yq / is the point on P1 P2 that is two thirds of
the way from P1 to P2 , then the displacement of Q from
P1 equals twice the displacement of P2 from Q:
xq " x1 D 2.x2 " xq /;
30
-50 -40 -30 -20 -10
-10
ym " y1 D y2 " ym :
Thus xm D .x1 C x2 /=2 and ym D .y1 C y2 /=2.
40. "40ı and "40ı is the same temperature on both the
Fahrenheit and Celsius scales.
C
40
20
A D .2; "1/;
48.
.0 " 2X/=2 D 1:
The second equation implies that X D "1, and the second
then implies that x D 5. Thus P is .5; 0/.
p
.x " 2/2 C y 2 D 4 says that the distance of .x; y/ from
.2; 0/ is 4, so the equation represents a circle of radius 4
centred at .2; 0/.
p
p
.x " 2/2 C y 2 D x 2 C .y " 2/2 says that .x; y/ is
equidistant from .2; 0/ and .0; 2/. Thus .x; y/ must lie on
the line that is the right bisector of the line from .2; 0/ to
.0; 2/. A simpler equation for this line is x D y.
49.
The line 2x C ky D 3 has slope m D "2=k. This line is
perpendicular to 4x C y D 1, which has slope "4, provided
m D 1=4, that is, provided k D "8. The line is parallel to
4x C y D 1 if m D "4, that is, if k D 1=2.
50.
For any value of k, the coordinates of the point of intersection of x C 2y D 3 and 2x " 3y D "1 will also satisfy
the equation
.x C 2y " 3/ C k.2x " 3y C 1/ D 0
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION P.3 (PAGE 22)
because they cause both expressions in parentheses to be
0. The equation above is linear in x and y, and so represents a straight line for any choice of k. This line will
pass through .1; 2/ provided 1 C 4 " 3 C k.2 " 6 C 1/ D 0,
that is, if k D 2=3. Therefore, the line through the point
of intersection of the two given lines and through the point
.1; 2/ has equation
2
x C 2y " 3 C .2x " 3y C 1/ D 0;
3
13.
Together, x 2 C y 2 > 1 and x 2 C y 2 < 4 represent annulus
(washer-shaped region) consisting of all points that are
outside the circle of radius 1 centred at the origin and
inside the circle of radius 2 centred at the origin.
14.
Together, x 2 C y 2 $ 4 and .x C 2/2 C y 2 $ 4 represent the
region consisting of all points that are inside or on both
the circle of radius 2 centred at the origin and the circle
of radius 2 centred at ."2; 0/.
15.
Together, x 2 Cy 2 < 2x and x 2 Cy 2 < 2y (or, equivalently,
.x " 1/2 C y 2 < 1 and x 2 C .y " 1/2 < 1) represent
the region consisting of all points that are inside both the
circle of radius 1 centred at .1; 0/ and the circle of radius
1 centred at .0; 1/.
16.
x 2 C y 2 " 4x C 2y > 4 can be rewritten
.x " 2/2 C .y C 1/2 > 9. This equation, taken together
with x C y > 1, represents all points that lie both outside
the circle of radius 3 centred at .2; "1/ and above the line
x C y D 1.
17.
The interior of the circle with centre ."1; 2/ and radius
p
6 is given by .x C 1/2 C .y " 2/2 < 6, or
2
x C y 2 C 2x " 4y < 1.
or, on simplification, x D 1.
Section P.3 Graphs of Quadratic Equations
(page 22)
1.
x 2 C y 2 D 16
2. x 2 C .y " 2/2 D 4, or x 2 C y 2 " 4y D 0
3. .x C 2/2 C y 2 D 9, or x 2 C y 2 C 4y D 5
4. .x " 3/2 C .y C 4/2 D 25, or x 2 C y 2 " 6x C 8y D 0.
5.
x 2 C y 2 " 2x D 3
x 2 " 2x C 1 C y 2 D 4
.x " 1/ C y D 4
centre: .1; 0/; radius 2.
18. The exterior of the circle with centre .2; "3/ and radius 4 is given by .x " 2/2 C .y C 3/2 > 16, or
x 2 C y 2 " 4x C 6y > 3.
x 2 C y 2 C 4y D 0
19.
x 2 C y 2 < 2;
x#1
20.
x 2 C y 2 > 4;
.x " 1/2 C .y " 3/2 < 10
21.
The parabola with focus .0; 4/ and directrix y D "4 has
equation x 2 D 16y.
22.
The parabola with focus .0; "1=2/ and directrix y D 1=2
has equation x 2 D "2y.
23.
The parabola with focus .2; 0/ and directrix x D "2 has
equation y 2 D 8x.
24.
The parabola with focus ."1; 0/ and directrix x D 1 has
equation y 2 D "4x.
25.
y D x 2 =2 has focus .0; 1=2/ and directrix y D "1=2.
2
6.
2
2
2
x C y C 4y C 4 D 4
x 2 C .y C 2/2 D 4
centre: .0; "2/; radius 2.
7.
x 2 C y 2 " 2x C 4y D 4
x 2 " 2x C 1 C y 2 C 4y C 4 D 9
.x " 1/2 C .y C 2/2 D 9
centre: .1; "2/; radius 3.
8.
x 2 C y 2 " 2x " y C 1 D 0
x 2 " 2x C 1 C y 2 " y C 41 D
!
"2
.x " 1/2 C y " 12 D 14
centre: .1; 1=2/; radius 1=2.
1
4
y
9. x 2 C y 2 > 1 represents all points lying outside the circle
of radius 1 centred at the origin.
2
yDx 2 =2
.0;1=2/
2
10. x C y < 4 represents the open disk consisting of all
points lying inside the circle of radius 2 centred at the
origin.
11.
x
yD!1=2
.x C 1/2 C y 2 $ 4 represents the closed disk consisting of
all points lying inside or on the circle of radius 2 centred
at the point ."1; 0/.
12. x 2 C .y " 2/2 $ 4 represents the closed disk consisting of
all points lying inside or on the circle of radius 2 centred
at the point .0; 2/.
Fig. P.3-25
26.
y D "x 2 has focus .0; "1=4/ and directrix y D 1=4.
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SECTION P.3 (PAGE 22)
ADAMS and ESSEX: CALCULUS 9
y
y
Version (c)
yD1=4
y D x2
x
.3; 3/
Version (b)
.0;!1=4/
yD!x 2
Fig. P.3-26
x
4
Version (d)
27.
Version (a)
"3
x D "y 2 =4 has focus ."1; 0/ and directrix x D 1.
y
.4; "2/
Fig. P.3-29
a) has equation y D x 2 " 3.
xD1
b) has equation y D .x " 4/2 or y D x 2 " 8x C 16.
.!1;0/
x
c) has equation y D .x " 3/2 C 3 or y D x 2 " 6x C 12.
d) has equation y D .x " 4/2 " 2, or y D x 2 " 8x C 14.
xD!y 2 =4
30.
Fig. P.3-27
28. x D y 2 =16 has focus .4; 0/ and directrix x D "4.
y
31.
.4;0/
32.
x
33.
xD!4
xDy 2 =16
34.
Fig. P.3-28
a) If y D mx is shifted to the right by amount x1 , the
equation y D m.x " x1 / results. If .a; b/ satisfies this
equation, then b D m.a " x1 /, and so x1 D a " .b=m/.
Thus the shifted equation is
y D m.x " a C .b=m// D m.x " a/ C b.
b) If y D mx is shifted vertically by amount y1 ,
the equation y D mx C y1 results. If .a; b/
satisfies this equation, then b D ma C y1 , and
so y1 D b " ma. Thus the shifted equation is
y D mx C b " ma D m.x " a/ C b, the same
equation obtained in part (a).
p
y D .x=3/ C 1
p
4y D x C 1
p
y D .3x=2/ C 1
p
.y=2/ D 4x C 1
35.
y D 1 " x 2 shifted down 1, left 1 gives y D ".x C 1/2 .
36.
x 2 C y 2 D 5 shifted up 2, left 4 gives
.x C 4/2 C .y " 2/2 D 5.
37. y D .x " 1/2 " 1 shifted down 1, right 1 gives
y D .x " 2/2 " 2.
p
p
38. y D x shifted down 2, left 4 gives y D x C 4 " 2.
29.
6
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION P.3 (PAGE 22)
y
39. y D x 2 C 3, y D 3x C 1. Subtracting these equations
gives
x 2 " 3x C 2 D 0, or .x " 1/.x " 2/ D 0. Thus x D 1 or
x D 2. The corresponding values of y are 4 and 7. The
intersection points are .1; 4/ and .2; 7/.
9x 2 C16y 2 D144
x
40. y D x 2 " 6, y D 4x " x 2 . Subtracting these equations
gives
2x 2 " 4x " 6 D 0, or 2.x " 3/.x C 1/ D 0. Thus x D 3
or x D "1. The corresponding values of y are 3 and "5.
The intersection points are .3; 3/ and ."1; "5/.
41. x 2 C y 2 D 25, 3x C 4y D 0. The second equation says
that y D "3x=4. Substituting this into the first equation
gives 25x 2 =16 D 25, so x D ˙4. If x D 4, then the
second equation gives y D "3; if x D "4, then y D 3.
The intersection points are .4; "3/ and ."4; 3/. Note that
having found values for x, we substituted them into the
linear equation rather than the quadratic equation to find
the corresponding values of y. Had we substituted into
the quadratic equation we would have got more solutions
(four points in all), but two of them would have failed to
satisfy 3x C 4y D 12. When solving systems of nonlinear
equations you should always verify that the solutions you
find do satisfy the given equations.
42. 2x 2 C 2y 2 D 5, xy D 1. The second equation says that
y D 1=x. Substituting this into the first equation gives
2x 2 C .2=x 2 / D 5, or 2x 4 " 5x 2 C 2 D 0. This equation
factors to p
.2x 2 " 1/.x 2 "p2/ D 0, so its solutions are
x D ˙1= 2 and x D ˙ 2. The corresponding values
of y are givenpby p
y D 1=x. p
Therefore,
p the
p intersection
p
points
are
.1=
2;
2/,
."1=
2;
"
2/,
.
2; 1= 2/, and
p
p
." 2; "1= 2/.
Fig. P.3-44
45.
.x " 3/2
.y C 2/2
C
D 1 is an ellipse with centre at
9
4
.3; "2/, major axis between .0; "2/ and .6; "2/ and minor
axis between .3; "4/ and .3; 0/.
y
x
.3;!2/
.x!3/2 .yC2/2
C
D1
9
4
Fig. P.3-45
46.
43. .x 2 =4/ C y 2 D 1 is an ellipse with major axis between
."2; 0/ and .2; 0/ and minor axis between .0; "1/ and
.0; 1/.
.y C 1/2
D 4 is an ellipse with centre at
4
.1; "1/, major axis between .1; "5/ and .1; 3/ and minor
axis between ."1; "1/ and .3; "1/.
.x " 1/2 C
y
.x!1/2 C
.yC1/2
D4
4
y
x2
2
4 Cy D1
x
.1;!1/
x
Fig. P.3-46
Fig. P.3-43
44. 9x 2 C 16y 2 D 144 is an ellipse with major axis between
."4; 0/ and .4; 0/ and minor axis between .0; "3/ and
.0; 3/.
47.
.x 2 =4/ " y 2 D 1 is a hyperbola with centre at the
origin and passing through .˙2; 0/. Its asymptotes are
y D ˙x=2.
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SECTION P.3 (PAGE 22)
ADAMS and ESSEX: CALCULUS 9
y
y
x2
!y 2 D1
4
yD!x=2
x
.x " 1/.y C 2/ D 1
y D "2
x
yDx=2
xD1
Fig. P.3-47
Fig. P.3-50
51.
48. x 2 " y 2 D "1 is a rectangular hyperbola with centre at
the origin and passing through .0; ˙1/. Its asymptotes are
y D ˙x.
y
x 2 !y 2 D!1
yDx
a) Replacing x with "x replaces a graph with its reflection across the y-axis.
b) Replacing y with "y replaces a graph with its reflection across the x-axis.
52.
Replacing x with "x and y with "y reflects the graph in
both axes. This is equivalent to rotating the graph 180ı
about the origin.
53.
jxj C jyj D 1.
In the first quadrant the equation is x C y D 1.
In the second quadrant the equation is "x C y D 1.
In the third quadrant the equation is "x " y D 1.
In the fourth quadrant the equation is x " y D 1.
x
yD!x
y
Fig. P.3-48
1
jxj C jyj D 1
1
49. xy D "4 is a rectangular hyperbola with centre at the
origin and passing through .2; "2/ and ."2; 2/. Its asymptotes are the coordinate axes.
x
"1
y
"1
Fig. P.3-53
x
Section P.4 Functions and Their Graphs
(page 32)
xyD!4
Fig. P.3-49
50. .x " 1/.y C 2/ D 1 is a rectangular hyperbola with centre
at .1; "2/ and passing through .2; "1/ and .0; "3/. Its
asymptotes are x D 1 and y D "2.
8
1.
f .x/ D 1 C x 2 ; domain R, range Œ1; 1/
2.
f .x/ D 1 "
p
p
x; domain Œ0; 1/, range ."1; 1!
8 " 2x; domain ."1; 4!, range Œ0; 1/
3.
G.x/ D
4.
F .x/ D 1=.x " 1/; domain ."1; 1/ [ .1; 1/, range
."1; 0/ [ .0; 1/
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INSTRUCTOR’S SOLUTIONS MANUAL
5. h.t / D p
t
SECTION P.4 (PAGE 32)
a) is the graph of x.1 " x/2 , which is positive for x > 0.
; domain ."1; 2/, range R. (The equation
2"t
y D h.t / can be squared and rewritten as
t 2 C y 2 t " 2y 2 D 0, a quadratic equation in t having real
solutions for every real value of y. Thus the range of h
contains all real numbers.)
1
p
; domain Œ2; 3/ [ .3; 1/, range
1" x"2
."1; 0/ [ .0; 1/. The equation y D g.x/ can be solved
for
x D 2 ".1 ".1=y//2 so has a real solution provided y ¤ 0.
b) is the graph of x 2 " x 3 D x 2 .1 " x/, which is positive
if x < 1.
c) is the graph of x " x 4 , which is positive if 0 < x < 1
and behaves like x near 0.
d) is the graph of x 3 "x 4 , which is positive if 0 < x < 1
and behaves like x 3 near 0.
6. g.x/ D
7.
y
9.
x
0
˙0:5
˙1
˙1:5
˙2
y
graph (i)
graph (ii)
f .x/ D x 4
0
0:0625
1
5:0625
16
y
x
y
y D x4
x
y
graph (iii)
graph (iv)
x
x
x
Fig. P.4-7
Graph (ii) is the graph of a function because vertical lines
can meet the graph only once. Graphs (i), (iii), and (iv)
do not have this property, so are not graphs of functions.
Fig. P.4-9
10.
8.
y
y
graph (a)
graph (b)
x
f .x/ D x 2=3
0
˙0:5
˙1
˙1:5
˙2
0
0:62996
1
1:3104
1:5874
y
x
y
x
y
graph (c)
y D x 2=3
graph (d)
x
Fig. P.4-10
11.
x
Fig. P.4-8
x
f .x/ D x 2 C 1 is even: f ."x/ D f .x/
12. f .x/ D x 3 C x is odd: f ."x/ D "f .x/
x
is odd: f ."x/ D "f .x/
13. f .x/ D 2
x "1
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SECTION P.4 (PAGE 32)
ADAMS and ESSEX: CALCULUS 9
14.
f .x/ D
1
is even: f ."x/ D f .x/
x2 " 1
15.
f .x/ D
1
is odd about .2; 0/: f .2 " x/ D "f .2 C x/
x"2
26.
y
1
is odd about ."4; 0/:
xC4
f ."4 " x/ D "f ."4 C x/
16. f .x/ D
17.
yD.x!1/2 C1
2
f .x/ D x " 6x is even about x D 3: f .3 " x/ D f .3 C x/
x
18. f .x/ D x 3 " 2 is odd about .0; "2/:
f ."x/ C 2 D ".f .x/ C 2/
3
27.
y
3
19. f .x/ D jx j D jxj is even: f ."x/ D f .x/
20. f .x/ D jx C 1j is even about x D "1:
f ."1 " x/ D f ."1 C x/
p
21. f .x/ D 2x has no symmetry.
p
22. f .x/ D .x " 1/2 is even about x D 1:
f .1 " x/ D f .1 C x/
23.
yD1!x 3
x
28.
y
y
yD!x 2
x
x
yD.xC2/3
24.
29.
y
y
yD1!x 2
p
yD xC1
x
x
25.
30.
y
y
yD.x!1/2
p
yD xC1
x
x
10
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION P.4 (PAGE 32)
31.
36.
y
y
x
xD2
x
yD!jxj
yD
32.
1
2!x
37.
y
y
x
yD
xC1
yDjxj!1
yD1
x
x
xD!1
33.
38.
y
y
xD1
yDjx!2j
x
yD!1
x
yD 1!x
x
34.
y
39.
yD1Cjx!2j
y
y
.1;3/
yDf .x/C2
2
.2;2/
.1;1/
yDf .x/
x
2
x
x
Fig. P.4.39(a)
35.
Fig. P.4.39(b)
y
40.
y
y
.1;3/
yDf .x/C2
xD!2
2
.2;2/
x
1
x
yD
!1
2
xC2
Fig. P.4.40(a)
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yDf .x/!1 x
.2;!1/
Fig. P.4.40(b)
11
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SECTION P.4 (PAGE 32)
ADAMS and ESSEX: CALCULUS 9
y
0.8
0.6
41.
y
y D 0:68 x C 2
yD 2
x C 2x C 3
0.4
0.2
.!1;1/
yDf .xC2/
x
!2
-5
-4
-3
-2
-1
-0.2
1y D2"0:18
3
4
x
-0.4
-0.6
42.
y
.2;1/
1
-0.8
-1.0
Fig. P.4-47
yDf .x!1/
3
x
48.
Range is approximately ."1; 0:1! [ Œ2:9; 1/.
y
43.
4
y
3
2
2
yD!f .x/
.1;!1/
1
x
-4
-3 -2 -1
1
-1
44.
y
-2
2
3 4 5
x"2
yD
x.x C 2/
6 x
-3
.!1;1/
yDf .!x/
-4
x
!2
-5
Fig. P.4-48
45.
y
49.
y
yDf .4!x/
5
.3;1/
4
4 x
2
3
2
46.
y
1
-5
.!1;1/
-4
-3
.1;1/
1
2
3
4
x
y D x 4 " 6x 3 C 9x 2 " 1
Fig. P.4-49
x
Range is approximately Œ"0:18; 0:68!.
12
-1
-1
yD1!f .1!x/
47.
-2
Apparent symmetry about x D 1:5.
This can be confirmed by calculating f .3"x/, which turns
out to be equal to f .x/.
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION P.5 (PAGE 38)
Apparent symmetry about ."2; 2/.
This can be confirmed by calculating shifting the graph
right by 2 (replace x with x"2) and then down 2 (subtract
2). The result is "5x=.1 C x 2 /, which is odd.
50.
y
2
yD
2
3 " 2x C x
2 " 2x C x 2
53.
1
-5
-4
-3
-2
-1
1
2
3
4
x
Section P.5 Combining Functions to Make
New Functions (page 38)
-1
Fig. P.4-50
Apparent symmetry about x D 1.
This can be confirmed by calculating f .2"x/, which turns
out to be equal to f .x/.
1.
51.
y
4
y Dx"1
x"1
yD
x"2
3
2
1
-3
-2
-1
1
2
3
4
-1
5
6
2.
Fig. P.4-51
Apparent symmetry about .2; 1/, and about the lines
y D x " 1 and y D 3 " x.
1
,
x"2
so the graph is that of 1=x shifted right 2 units and up
one.
These can be confirmed by noting that f .x/ D 1 C
yD
2x 2 C 3x
x 2 C 4x C 5
p
f .x/ D x, g.x/ D x " 1.
D.f / D R, D.g/ D Œ1; 1/.
D.f C g/ D D.f " g/ D D.fg/ D D.g=f / D Œ1; 1/,
D.f =g/ D .1; 1/. p
.f C g/.x/ D x C x " 1
p
.f " g/.x/ D x " x " 1
p
.fg/.x/ D x x " 1
p
.f =g/.x/ D x= x " 1
p
.g=f /.x/ D . 1 " x/=x
x
y D "x C 3
-2
52.
If f is both even and odd the f .x/ D f ."x/ D "f .x/,
so f .x/ D 0 identically.
y
p
p
f .x/ D 1 " x, g.x/ D 1 C x.
D.f / D ."1; 1!, D.g/ D Œ"1; 1/.
D.f C g/ D D.f " g/ D D.fg/ D Œ"1; 1!,
D.f =g/ D ."1;p1!, D.g=fp/ D Œ"1; 1/.
.f C g/.x/ D 1 " x C 1 C x
p
p
.f " g/.x/ D 1 " x " 1 C x
p
.fg/.x/ D 1 " x 2
p
.f =g/.x/ D .1 " x/=.1 C x/
p
.g=f /.x/ D .1 C x/=.1 " x/
3.
y
5
4
y D x " x2
3
x
2
yDx
1
-7
-6
-5
-4
-3
-2
-1
1
-1
-2
2
x
y D "x 2
Fig. P.4-52
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SECTION P.5 (PAGE 38)
ADAMS and ESSEX: CALCULUS 9
4.
8.
y
y D "x
1
-2
x
1
y D x3 " x
-1
f .x/ D 2=x, g.x/ D x=.1 " x/.
f ı f .x/ D 2=.2=x/ D xI D.f ı f / D fx W x ¤ 0g
f ı g.x/ D 2=.x=.1 " x// D 2.1 " x/=xI
D.f ı g/ D fx W x ¤ 0; 1g
g ı f .x/ D .2=x/=.1 " .2=x// D 2=.x " 2/I
D.g ı f / D fx W x ¤ 0; 2g
g ı g.x/ D .x=.1 " x//=.1 " .x=.1 " x/// D x=.1 " 2x/I
D.g ı g/ D fx W x ¤ 1=2; 1g
-1
y D x3
-2
9.
5.
y
y D x C jxj
y D jxj
p
f .x/ D 1=.1 " x/, g.x/ D x " 1.
f ı f .x/ D 1=.1 " .1=.1 " x/// D .x " 1/=xI
D.f ı f / D fx W x ¤ 0; 1g
p
f ı g.x/ D 1=.1 " x " 1/I
D.f ı g/ D fx W x # 1; x ¤ 2g
p
p
g ı f .x/ D .1=.1 " x// " 1 D x=.1 " x/I
D.g ı f / D Œ0; 1/
q
p
g ı g.x/ D
x " 1 " 1I D.g ı g/ D Œ2; 1/
y D x D jxj
x
yDx
6.
y
4
y D jxj C jx " 2j
3
y D jxj
2
y D jx " 2j
1
-2
-1
10. f .x/ D .x C 1/=.x " 1/ D 1 C 2=.x " 1/, g.x/ D sgn .x/.
f ı f .x/ D 1 C 2=.1 C .2=.x " 1/ " 1// D xI
D.f ı f / D fx W x ¤ 1g
sgn x C 1
f ı g.x/ D
D 0I D.f ı g/ D ."1; 0/
sgn x " 1
#
$ n
xC1
1
if x < "1 or x > 1
g ı f .x/ D sgn
D
I
"1 if "1 < x < 1
x"1
D.g ı f / D fx W x ¤ "1; 1g
g ı g.x/ D sgn .sgn .x// D sgn .x/I D.g ı g/ D fx W x ¤ 0g
1
2
3
4
5
11.
12.
13.
14.
15.
16.
x
-1
7.
f .x/
g.x/
f ı g.x/
x2
xp" 4
x
2x 3 C 3
.x C 1/=x
1=.x C 1/2
xC1
xC4
x2
x 1=3
1=.x " 1/
x"1
.x C 1/2
x
jxj
2x C 3
x
1=x 2
f .x/ D x C 5, g.x/ D x 2 " 3.
f ı g.0/ D f ."3/ D 2; g.f .0// D g.5/ D 22
f .g.x// D f .x 2 " 3/ D x 2 C 2
g ı f .x/ D g.f .x// D g.x C 5/ D .x C 5/2 " 3
f ı f ."5/ D f .0/ D 5; g.g.2// D g.1/ D "2
f .f .x// D f .x C 5/ D x C 10
g ı g.x/ D g.g.x// D .x 2 " 3/2 " 3
14
17.
p
y D x. p
y D 2 C px: previous graph is raised 2 units.
y D 2 C 3p
C x: previous graph is shiftend left 3 units.
y D 1=.2 C 3 C x/: previous graph turned upside down
and shrunk vertically.
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION P.5 (PAGE 38)
y
21.
y
y D2C
y D2C
p
xC3
y D 1=.2 C
p
p
.1=2;1/
yDf .2x/
x
yD
x C 3/
x
1
p
x
22.
y
x
yDf .x=3/
Fig. P.5-17
18.
yD
p
23.
y
1 " 2x
y
y D 2x
yD p
6 x
3
.!2;2/
y D 2x " 1
1
yD1Cf .!x=2/
1 " 2x
x
x
yD p
1
1 " 2x
24.
y
"1
yD2f ..x!1/=2/
y D 1 " 2x
Fig. P.5-18
1
5
x
19.
y
25.
y
.1;2/
y D f .x/
.1; 1/
yD2f .x/
82
x
2
x
26.
20.
-2y
y
y D g.x/
.1; 1/
2
x
yD!.1=2/f .x/
x
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SECTION P.5 (PAGE 38)
27.
ADAMS and ESSEX: CALCULUS 9
F .x/ D Ax C B
(a) F ı F .x/ D F .x/
) A.Ax C B/ C B D Ax C B
) AŒ.A " 1/x C B! D 0
Thus, either A D 0 or A D 1 and B D 0.
(b) F ı F .x/ D x
) A.Ax C B/ C B D x
) .A2 " 1/x C .A C 1/B D 0
Thus, either A D "1 or A D 1 and B D 0
35.
a)
Let E.x/ D 12 Œf .x/ C f ."x/!.
Then E."x/ D 21 Œf ."x/ C f .x/! D E.x/. Hence,
E.x/ is even.
Let O.x/ D 12 Œf .x/ " f ."x/!.
Then O."x/ D 12 Œf ."x/ " f .x/! D "O.x/ and O.x/
is odd.
E.x/ C O.x/
D 12 Œf .x/ C f ."x/! C 12 Œf .x/ " f ."x/!
28. bxc D 0 for 0 $ x < 1; dxe D 0 for "1 $ x < 0.
D f .x/:
29. bxc D dxe for all integers x.
Hence, f .x/ is the sum of an even function and an
odd function.
30. d"xe D "bxc is true for all real x; if x D n C y where n
is an integer and 0 $ y < 1, then "x D "n " y, so that
d"xe D "n and bxc D n.
31.
y
y D x " bxc
b) If f .x/ D E1 .x/ C O1 .x/ where E1 is even and O1
is odd, then
E1 .x/ C O1 .x/ D f .x/ D E.x/ C O.x/:
Thus E1 .x/ " E.x/ D O.x/ " O1 .x/. The left side
of this equation is an even function and the right side
is an odd function. Hence both sides are both even
and odd, and are therefore identically 0 by Exercise
36. Hence E1 D E and O1 D O. This shows that f
can be written in only one way as the sum of an even
function and an odd function.
x
32. f .x/ is called the integer part of x because jf .x/j
is the largest integer that does not exceed x; i.e.
jxj D jf .x/j C y, where 0 $ y < 1.
y
Section P.6 Polynomials and Rational
Functions (page 45)
x
y D f .x/
1.
x 2 " 7x C 10 D .x C 5/.x C 2/
The roots are "5 and "2.
2.
x 2 " 3x " 10 D .x " 5/.x C 2/
The roots are 5 and "2.
3.
Fig. P.5-32
33. If f is even and g is odd, then: f 2 , g 2 , f ı g, g ı f ,
and f ı f are all even. fg, f =g, g=f , and g ı g are odd,
and f C g is neither even nor odd. Here are two typical
verifications:
4.
34. f even , f ."x/ D f .x/
f odd , f ."x/ D "f .x/
f even and odd ) f .x/ D "f .x/ ) 2f .x/ D 0
) f .x/ D 0
16
Rather than use the quadratic formula this time, let us
complete the square.
x 2 " 6x C 13 D x 2 " 6x C 9 C 4
D .x " 3/2 C 22
D .x " 3 " 2i /.x " 3 C 2i /:
f ı g."x/ D f .g."x// D f ."g.x// D f .g.x// D f ı g.x/
.fg/."x/ D f ."x/g."x/ D f .x/Œ"g.x/!
D "f .x/g.x/ D ".fg/.x/:
The others are similar.
p
"2 ˙ 4 " 8
D "1 ˙ i .
If x 2 C 2x C 2 D 0, then x D
2
The roots are "1 C i and "1 " i .
x 2 C 2x C 2 D .x C 1 " i /.x C 1 C i /.
The roots are 3 C 2i and 3 " 2i .
5.
16x 4 " 8x 2 C 1 D .4x 2 " 1/2 D .2x " 1/2 .2x C 1/2 . There
are two double roots: 1=2 and "1=2.
6.
x 4 C 6x 3 C 9x 2 D x 2 .x 2 C 6x C 9/ D x 2 .x C 3/2 . There
are two double roots, 0 and "3.
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INSTRUCTOR’S SOLUTIONS MANUAL
7.
SECTION P.6 (PAGE 45)
x 3 C 1 D .x C 1/.x 2 " x C 1/. One root is "1. The other
two are the solutions of x 2 " x C 1 D 0, namely
xD
1˙
p
2
1"4
D
p
3
1
˙
i:
2
2
15.
16.
We have
p !
p !
3
1
3
1
i
x" C
i :
x C 1 D .x C 1/ x " "
2
2
2
2
3
17.
8. x 4 " 1 D .x 2 " 1/.x 2 C 1/ D .x " 1/.x C 1/.x " i /.x C i /.
The roots are 1, "1, i , and "i .
9. x 6 " 3x 4 C 3x 2 " 1 D .x 2 " 1/3 D .x " 1/3 .x C 1/3 . The
roots are 1 and "1, each with multiplicity 3.
10.
5
4
x " x " 16x C 16 D .x " 1/.x " 16/
D .x " 1/.x 2 " 4/.x 4 C 4/
D .x " 1/.x " 2/.x C 2/.x " 2i /.x C 2i /:
The roots are 1, 2, "2, 2i , and "2i .
11.
18.
4
19.
x 5 C x 3 C 8x 2 C 8 D .x 2 C 1/.x 3 C 8/
D .x C 2/.x " i /.x C i /.x 2 " 2x C 4/
Three of the five roots are "2, i and "i . The remain2
ing two are
p solutions of x " 2x C 4 D 0, namely
p
2 ˙ 4 " 16
xD
D 1 ˙ 3 i . We have
2
p
p
20.
x Cx C8x C8 D .xC2/.x"i /.xCi /.x"aC 3 i /.x"a" 3 i /:
5
12.
3
2
x 9 " 4x 7 " x 6 C 4x 4 D x 4 .x 5 " x 2 " 4x 3 C 4/
D x 4 .x 3 " 1/.x 2 " 4/
D x 4 .x " 1/.x " 2/.x C 2/.x 2 C x C 1/:
Seven of the nine roots are: 0 (with multiplicity 4),
1, 2, and "2. The other two roots are solutions of
x 2 C x C 1 D 0, namely
xD
"1 ˙
p
2
1"4
14.
The denominator is x 3 " x D x.x " 1/.x C 1/ which is zero
if x D 0, 1, or "1. Thus the rational function is defined
for all real numbers except 0, 1, and "1.
x2
x2
x 2 C 5x C 3 " 5x " 3
x2
D
C 5x C 3
x 2 C 5x C 3
"5x " 3
:
D1C 2
x C 5x C 3
x3
x 3 C 2x 2 C 3x " 2x 2 " 3x
D
C 2x C 3
x 2 C 2x C 3
2
x.x C 2x C 3/ " 2x 2 " 3x
D
x 2 C 2x C 3
2.x 2 C 2x C 3/ " 4x " 6 C 3x
Dx"
x 2 C 2x C 3
xC6
Dx"2C 2
:
x C 2x C 3
x4 C x2
x.x 3 C x 2 C 1/ " x 3 " x C x 2
D
x3 C x2 C 1
x3 C x2 C 1
".x 3 C x 2 C 1/ C x 2 C 1 " x C x 2
DxC
x3 C x2 C 1
2x 2 " x C 1
:
D x"1C 3
x C x2 C 1
22.
Following the method of Example 6, we calculate
.x 2 "bxCa2 /.x 2 CbxCa2 / D x 4 Ca4 C.2a2 "b 2 /x 2 D x 4 Cx 2 C1
provided a D 1 and b 2 D "1 C 2a2 D 1, so b D ˙1. Thus
P .x/ D .x 2 " x C 1/.x 2 C x C 1/.
The required factorization of x 9 " 4x 7 " x 6 C 4x 4 is
13. The denominator is x 2 C 2x C 2 D .x C 1/2 C 1 which is
never 0. Thus the rational function is defined for all real
numbers.
x3 " 1
x 3 " 2x C 2x " 1
D
2
x "2
x2 " 2
x.x 2 " 2/ C 2x " 1
D
x2 " 2
2x " 1
DxC 2
:
x "2
4
2
As in Example
p 6, we wantpa D 4, so a D 2
and a D
2, b D ˙ 2a D ˙2. Thus
P .x/ D .x 2 " 2x C 2/.x 2 C 2x C 2/.
3
1
D" ˙
i:
2
2
p !
p !
1
1
3
3
x .x"1/.x"2/.xC2/ x " C
i
x" "
i :
2
2
2
2
The denominator is x 2 Cx"1, which is a quadratic polynomial whose roots can p
be found with the quadratic formula.
They are x D ."1 ˙ 1 C 4/=2. Hence the given rational
p
function is p
defined for all real numbers except ."1" 5/=2
and ."1 C 5/=2.
21.
p
4
The denominator is x 3 C x 2 D x 2 .x C 1/ which is zero
only if x D 0 or x D "1. Thus the rational function is
defined for all real numbers except 0 and "1.
23.
24.
Let P .x/ D an x n C an!1 x n!1 C ! ! ! C a1 x C a0 ,
where n # 1. By the Factor Theorem, x " 1 is a factor
of P .x/ if and only if P .1/ D 0, that is, if and only if
an C an!1 C ! ! ! C a1 C a0 D 0.
Let P .x/ D an x n C an!1 x n!1 C ! ! ! C a1 x C a0 , where
n # 1. By the Factor Theorem, x C 1 is a factor of
P .x/ if and only if P ."1/ D 0, that is, if and only if
a0 " a1 C a2 " a3 C ! ! ! C ."1/n an D 0. This condition says
that the sum of the coefficients of even powers is equal to
the sum of coefficients of odd powers.
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SECTION P.6 (PAGE 45)
ADAMS and ESSEX: CALCULUS 9
25. Let P .x/ D an x n C an!1 x n!1 C ! ! ! C a1 x C a0 , where
the coefficients ak , 0 $ k $ n are all real numbers, so
that ak D ak . Using the facts about conjugates of sums
and products mentioned in the statement of the problem,
we see that if z D x C iy, where x and y are real, then
4.
sin
5.
cos
6.
sin
P .z/ D an z n C an!1 z n!1 C ! ! ! C a1 z C a0
D an z n C an!1 z n!1 C ! ! ! C a1 z C a0
D P .z/:
If z is a root of P , then P .z/ D P .z/ D 0 D 0, and z is
also a root of P .
26. By the previous exercise, z D u " iv is also a root of
P . Therefore P .x/ has two linear factors x " u " iv
and x " u C iv. The product of these factors is the real
quadratic factor .x " u/2 " i 2 v 2 D x 2 " 2ux C u2 C v 2 ,
which must also be a factor of P .x/.
27.
By the previous exercise
P .x/
P .x/
D
D Q1 .x/;
x 2 " 2ux C u2 C v 2
.x " u " iv/.x " u C iv/
where Q1 , being a quotient of two polynomials with real
coefficients, must also have real coefficients. If z D u C iv
is a root of P having multiplicity m > 1, then it must also
be a root of Q1 (of multiplicity m " 1), and so, therefore,
z must be a root of Q1 , as must be the real quadratic
x 2 " 2ux C u2 C v 2 . Thus
7.
P .x/
Q1 .x/
D 2
D Q2 .x/;
.x 2 " 2ux C u2 C v 2 /2
x " 2ux C u2 C v 2
where Q2 is a polynomial with real coefficients. We can
continue in this way until we get
P .x/
D Qm .x/;
.x 2 " 2ux C u2 C v 2 /m
#
7#
12
$
1.
cos
#
3#
4
$
%
#&
#
1
D cos # "
D " cos D " p
4
4
2
3#
"3#
D " tan
D "1
2. tan
4
4
#
11#
D sin
12
12
%#
#&
D sin
"
3
4
#
#
#
#
D sin cos " cos sin
3! 4
3
4
p #
$ # $#
$
3
1
1
1
"
D
p
p
2
2
2
2
p
3"1
D
p
2 2
%
&
cos.# C x/ D cos 2# " .# " x/
%
&
D cos ".# " x/
D cos.# " x/ D " cos x
8.
sin.2# " x/ D " sin x
9.
sin
#
3#
"x
2
$
10.
3#
Cx
cos
2
!
%
%
# &&
D sin # " x "
2
%
#&
D sin x "
%# 2 &
D " sin
"x
2
D " cos x
D cos
3#
3#
cos x " sin
sin x
2
2
D ."1/." sin x/ D sin x
11.
cos x
sin x
C
cos x
sin x
sin2 x C cos2 x
D
cos x sin x
1
D
cos x sin x
tan x C cot x D
p
%
2#
#&
#
3
3. sin
D sin # "
D sin D
3
3
3
2
18
C
#
$
2#
#
5#
D cos
"
12
3
4
2#
#
2#
#
D cos
cos C sin
sin
3
4
3 ! 4
p #
$
$
# $#
1
3
1
1
C
D"
p
p
2
2
2
2
p
3"1
D p
2 2
where Qm no longer has z (or z) as a root. Thus z and z
must have the same multiplicity as roots of P .
Section P.7 The Trigonometric Functions
(page 57)
%#
#&
4
3
#
#
#
#
D sin cos C cos sin
4
3 p
4
3
p
1 1
3
1
1C 3
D p
p
Cp
D
22
2 2
2 2
D sin
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INSTRUCTOR’S SOLUTIONS MANUAL
12.
SECTION P.7 (PAGE 57)
$
cos x
sin x
"
tan x " cot x
cos x
sin x
$
D #
cos x
sin x
tan x C cot x
C
cos x
sin x
!
sin2 x " cos2 x
cos x sin x
!
D
sin2 x C cos2 x
cos x sin x
#
y
!=2
!
13.
4
2
2
20.
2
2
sin
cos x " sin x D .cos x " sin x/.cos x C sin x/
x
has period 4#.
2
y
D cos2 x " sin2 x D cos.2x/
14.
15.
16.
17.
.1 " cos x/.1 C cos x/ D 1 " cos2 x D sin2 x implies
1 " cos x
sin x
D
. Now
sin x
1 C cos x % &
x
1 " cos 2
1 " cos x
% x &2
D
sin x
sin 2
% 2
% x &&
1 " 1 " 2 sin2
2
D
x
x
2 sin cos
2
2
x
sin
2 D tan x
D
x
2
cos
2
% &
2 x
2
sin
%x &
1 " cos x
% 2x & D tan2
D
1 C cos x
2
2 cos2
2
cos x " sin x
.cos x " sin x/2
D
cos x C sin x
.cos x C sin x/.cos x " sin x/
cos2 x " 2 sin x cos x C sin2 x
D
cos2 x " sin2 x
1 " sin.2x/
D
cos.2x/
D sec.2x/ " tan.2x/
sin 3x D sin.2x C x/
D sin 2x cos x C cos 2x sin x
2
1
!
2! x
!1
Fig. P.7-20
21.
sin #x has period 2.
y
y D sin.#x/
1
1
2
4 x
3
!1
Fig. P.7-21
22.
cos
#x
has period 4.
2
y
1
1
5
3
x
!1
Fig. P.7-22
2
D 2 sin x cos x C sin x.1 " 2 sin x/
D 2 sin x.1 " sin2 x/ C sin x " 2 sin3 x
23.
y
3
D 3 sin x " 4 sin x
18.
2! x
Fig. P.7-19
D sin2 x " cos2 x
4
y D cos.2x/
1
2
cos 3x D cos.2x C x/
D cos 2x cos x " sin 2x sin x
%
#&
y D 2 cos x "
3
1
D .2 cos2 x " 1/ cos x " 2 sin2 x cos x
"#
D 2 cos3 x " cos x " 2.1 " cos2 x/ cos x
D 4 cos3 x " 3 cos x
#
x
-1
-2
-3
19. cos 2x has period #.
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SECTION P.7 (PAGE 57)
ADAMS and ESSEX: CALCULUS 9
24.
%
#&
y D 1 C sin x C
4
y
2
29.
1
sin x D " ;
2p
cos x D "
# <x<
3
2
1
tan x D p
3
1
"#
p
! 3
x
#
!1
-1
25.
#
3
;
<x<#
5
2
4
3
cos x D " ; tan x D "
5
4
x
2
sin x D
Fig. P.7-29
3#
1
where x is in Œ#;
!. Then,
2
2
1
5
sec2 x D 1 C D . Hence,
4
p 4
5
2
sec x D "
; cos x D " p ;
2
5
1
sin x D tan x cos x D " p .
5
#
31. c D 2; B D
3
1
a D c cos B D 2 & D 1
2
p
p
3
b D c sin B D 2 &
D 3
2
#
32. b D 2; B D
3
p
2
2
D tan B D 3 ) a D p
c
a
3 B
p
4
2
3
D sin B D
)cD p
a
c
2
3
#
33. a D 5; B D
C
6
1
5
b D a tan B D 5 & p D p
3
3
r
p
10
25
c D a2 C b 2 D 25 C
D p
3
3
a
34. sin A D ) a D c sin A
c
a
35.
D tan A ) a D b tan A
b
a
36. cos B D ) a D c cos B
c
b
D tan B ) a D b cot B
37.
a
a
a
38. sin A D ) c D
c
sin A
b
D cos A ) c D b sec A
39.
c
30.
5
3
x
!4
Fig. P.7-25
#
!. Then
2
sec2 x D 1 C tan2 x D 1 C 4 D 5. Hence,
p
1
1
sec x D 5 and cos x D
D p ,
sec x
5
2
sin x D tan x cos x D p .
5
26. tan x D 2 where x is in Œ0;
27.
3#
2
1
#
; " <x<0
3p
2
8
2p
D"
2
sin x D "
3
3
p
p
8
D "2 2
tan x D "
1
cos x D
1
x
3
p
! 8
Fig. P.7-27
h# i
5
where x is in
; # . Hence,
13
2
r
p
12
25
D
,
sin x D 1 " cos2 x D 1 "
169
13
12
tan x D " .
5
28. cos x D "
20
tan x D
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b
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INSTRUCTOR’S SOLUTIONS MANUAL
40. sin A D
a
c
41. sin A D
a
D
c
42. sin A D
a
a
D p
c
a2 C b 2
SECTION P.7 (PAGE 57)
50.
p
c 2 " b2
c
#
43. a D 4, b D 3, A D
4
sin A
3 1
3
sin B D b
D p D p
a
4 2
4 2
44. Given that a D 2; b D 2; c D 3:
Since a2 D b 2 C c 2 " 2bc cos A,
a2 " b 2 " c 2
cos A D
"2bc
3
4"4"9
D .
D
"2.2/.3/
4
p
sin B
sin A
1
D
D .
b
a
2
p
2
1
#
3#
Thus sin B D
D p ,BD
or
, and
2
4
4
2
#
$
%#
3#
#&
7#
#
#
C D #"
C
D
or C D # "
C
D
.
4
6
12
4
6 p 12
%#
7#
#&
1" 3
p
or
Thus, cos C D cos
D cos
C
D
12
4
3 p
2 2
%#
#
#&
1C 3
cos C D cos
D cos
"
D
p .
12
3
4
2 2
If a D 1; b D
2; A D 30ı , then
Hence,
c 2 D a2 C b 2 " 2ab cos C
p
D 1 C 2 " 2 2 cos C
p
p
D 3 " .1 " 3/ or 3 " .1 C 3/
p
p
D 2 C 3 or 2 " 3:
A
c
b
B
45. a D 2, b D 3, c D 4
b 2 D a2 C c 2 " 2ac cos B
4 C 16 " 9
11
Thus cos B D
D
2
&
2
&
4
16
r
p
p
112
256 " 121
135
D
sin B D 1 " 2 D
16
16
16
a
C
#
:
4
12
#
c 2 D a2 Cb 2 "2ab cos C D 4C9"2.2/.3/ cos D 13" p .
4
2
s
12
Hence, c D 13 " p ' 2:12479.
2
Hence, c D
p
2C
p
3 or
p
p
2 " 3.
C
p
2
1
1
!=6
A
B0
B 00
46. Given that a D 2; b D 3; C D
47.
#
#
5#
c D 3, A D , B D
implies C D
4
3
12
c
1
a
3
# $
D
)aD p
5#
sin A
sin C
2
sin
12
3
1
$
#
aD p
7#
2
sin
12
p
3 2 2
D p
p (by #5)
21C 3
6
D
p
1C 3
Fig. P.7-50
51.
Let h be the height of the pole and x be the distance from
C to the base of the pole.
Then h D x tan 50ı and h D .x C 10/ tan 35ı
Thus x tan 50ı D x tan 35ı C 10 tan 35ı so
10 tan 35ı
tan 50ı " tan 35ı
10 tan 50ı tan 35ı
hD
' 16:98
tan 50ı " tan 35ı
xD
The pole is about 16.98 metres high.
52.
48. Given that a D 2; b D 3; C D 35ı . Then
c 2 D 4 C 9 " 2.2/.3/ cos 35ı , hence c ' 1:78050.
See the following diagram. Since tan 40ı D h=a, therefore
a D h= tan 40ı . Similarly, b D h= tan 70ı .
Since a C b D 2 km, therefore,
49. a D 4, B D 40ı , C D 70ı
Thus A D 70ı .
b
4
sin 40ı
D
so
b
D
4
D 2:736
sin 40ı
sin 70ı
sin 70ı
h
h
C
D2
ı
tan 40
tan 70ı
ı
2.tan 40 tan 70ı /
' 1:286 km:
hD
tan 70ı C tan 40ı
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SECTION P.7 (PAGE 57)
ADAMS and ESSEX: CALCULUS 9
54.
Balloon
sin B D
40ı
70ı
b
a
s
# 2
$2
a C c 2 " b2
1"
2ac
p
4
4
"a " b " c 4 C 2a2 b 2 C 2b 2 c 2 C 2a2 c 2
:
D
2ac
h
A
From Exercise 53, area D 12 ac sin B. By Cosine Law,
a2 C c 2 " b 2
. Thus,
cos B D
2ac
B
Fig. P.7-52
1
ah
ac sin B
ab sin C
jBC jh D
D
D
2
2
2
2
1
By symmetry, area 4ABC also D bc sin A
2
A
p
"a4 " b4 " c4 C 2a2 b2 C 2b2 c2 C 2a2 c2
4
square units. Since,
Hence, Area D
53. Area 4ABC D
P
s.s " a/.s " b/.s " c/
bCcCa bCc"a a"bCc aCb"c
D
2 $#
2
#2
$2
1
2
2
2
2
D
.b C c/ " a
a " .b " c/
16
# %
$
&
1
a2 .b C c/2 C .b " c/2 " a4 " .b 2 " c 2 /2
D
16
&
1% 2 2
D
2a b C 2a2 c 2 " a4 " b 4 " c 4 C 2b 2 c 2
16
Fig. P.7-53
Thus
c
h
b
C
B
22
p
s.s " a/.s " b/.s " c/ = Area of triangle.
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INSTRUCTOR’S SOLUTIONS MANUAL
CHAPTER 1.
SECTION 1.1 (PAGE 63)
LIMITS AND CONTINUITY
8.
3.t C k/2 " 12.t C k/ C 1 " Œ3.t " k/2 " 12.t " k/ C 1!
.t C k/ " .t " k/
1 % 2
D
3t C 6t k C 3k 2 " 12t " 12k C 1 " 3t 2 C 6t k " 3k 2
2k
&
C 12t " 12k C 1
Section 1.1 Examples of Velocity, Growth
Rate, and Area (page 63)
1.
Average velocity =
.t C h/2 " t 2
"x
D
m/s.
"t
h
D
2.
Avg. vel. over Œ2; 2 C h!
h
1
0:1
0:01
0:001
0:0001
Average velocity over Œt " k; t C k! is
12t k " 24k
D 6t " 12 m/s;
2k
which is the velocity at time t from Exercise 7.
5.0000
4.1000
4.0100
4.0010
4.0001
9.
y
2
3. Guess velocity is v D 4 m/s at t D 2 s.
y D2C
1
4. Average velocity on Œ2; 2 C h! is
.2 C h/2 " 4
4 C 4h C h2 " 4
4h C h2
D
D
D 4 C h:
.2 C h/ " 2
h
h
1
2
1
sin.# t /
#
3
Fig. 1.1-9
4
5
t
As h approaches 0 this average velocity approaches 4 m/s
5.
6.
x D 3t 2 " 12t C 1 m at time t s.
Average velocity over interval Œ1; 2! is
.3 & 22 " 12 & 2 C 1/ " .3 & 12 " 12 & 1 C 1/
D "3 m/s.
2"1
Average velocity over interval Œ2; 3! is
.3 & 32 " 12 & 3 C 1/ " .3 & 22 " 12 & 2 C 1/
D 3 m/s.
3"2
Average velocity over interval Œ1; 3! is
.3 & 32 " 12 & 3 C 1/ " .3 & 12 " 12 & 1 C 1/
D 0 m/s.
3"1
At t D 1 the height is y D 2 ft and the weight is
moving downward.
10.
#
$
1
1
sin #.1 C h/ " 2 C sin #
#
#
h
sin.# C #h/
sin # cos.#h/ C cos # sin.#h/
D
D
#h
#h
sin.#h/
D"
:
#h
2C
Average velocity over Œt; t C h! is
3.t C h/2 " 12.t C h/ C 1 " .3t 2 " 12t C 1/
.t C h/ " t
6t h C 3h2 " 12h
D
D 6t C 3h " 12 m/s:
h
This average velocity approaches 6t " 12 m/s as h approaches 0.
At t D 1 the velocity is 6 & 1 " 12 D "6 m/s.
At t D 2 the velocity is 6 & 2 " 12 D 0 m/s.
At t D 3 the velocity is 6 & 3 " 12 D 6 m/s.
7.
Average velocity over Œ1; 1 C h! is
At t D 1 the velocity is v D "6 < 0 so the particle is
moving to the left.
At t D 2 the velocity is v D 0 so the particle is stationary.
At t D 3 the velocity is v D 6 > 0 so the particle is
moving to the right.
h
1:0000
0:1000
0:0100
0:0010
11.
Avg. vel. on Œ1; 1 C h!
0
-0.983631643
-0.999835515
-0.999998355
The velocity at t D 1 is about v D "1 ft/s. The “"”
indicates that the weight is moving downward.
12. We sketched a tangent line to the graph on page 55 in
the text at t D 20. The line appeared to pass through
the points .10; 0/ and .50; 1/. On day 20 the biomass is
growing at about .1 " 0/=.50 " 10/ D 0:025 mm2 /d.
13.
The curve is steepest, and therefore the biomass is growing
most rapidly, at about day 45.
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SECTION 1.1 (PAGE 63)
14.
a)
ADAMS and ESSEX: CALCULUS 9
profit
175
150
125
100
75
50
25
3.
4.
lim g.x/ D 0
x!1C
5.
lim g.x/ D 0
x!3C
6.
2011 2012 2013 2014 2015
7.
8.
b) Average rate of increase in profits between 2010 and
2012 is
174 " 62
112
D
D 56 (thousand$/yr).
2012 " 2010
2
c) Drawing a tangent line to the graph in (a) at
t D 2010 and measuring its slope, we find that the
rate of increase of profits in 2010 is about 43 thousand$/year.
Section 1.2 Limits of Functions
(page 71)
9.
10.
x
11.
12.
lim f .x/ D 1;
x!!1
lim f .x/ D 0;
x!0
lim f .x/ D 1:
x!1
2. From inspecting the graph
y
y D g.x/
1
1
2
3
x
Fig. 1.2-2
we see that
lim
12 " 1
0
x2 " 1
D
D D0
x!1 x C 1
1C1
2
lim
x2 " 1
D lim .x " 1/ D "2
x!!1
x!!1 x C 1
lim
x 2 " 6x C 9
.x " 3/2
D lim
2
x!3
x!3 .x " 3/.x C 3/
x "9
x"3
0
D lim
D D0
x!3 x C 3
6
lim
x 2 C 2x
x
"2
1
D lim
D
D
x!!2 x 2 " 4
x!!2 x " 2
"4
2
15.
limh!2
lim
1
does not exist; denominator approaches 0
4 " h2
but numerator does not approach 0.
3 C 4h
3h C 4h2
D lim
does not exist; denomih2 " h3
h!0 h " h2
nator approaches 0 but numerator does not approach 0.
p
p
p
x"3
. x " 3/. x C 3/
D lim
17. lim
p
x!9 .x " 9/. x C 3/
x!9 x " 9
x"9
1
1
D lim
p
D lim p
D
x!9 .x " 9/. x C 3/
x!9
6
xC3
p
4Ch"2
18. lim
h
h!0
4Ch"4
D lim p
h!0 h. 4 C h C 2/
1
1
D lim p
D
4
h!0
4ChC2
16.
limh!0
19.
lim
x!!
lim g.x/ does not exist
20.
x!1
24
.x " #/2
02
D 2 D0
#x
#
lim jx " 2j D j " 4j D 4
x!!2
(left limit is 1, right limit is 0)
lim g.x/ D 1;
lim g.x/ D 0:
x!2
t2
."4/2
D
D2
4"t
4C4
14.
Fig. 1.2-1
we see that
xC3
3C3
2
D
D
xC6
3C6
3
t!!4
1
1
lim
x!3
13.
y
lim 3.1 " x/.2 " x/ D 3."1/.2 " 2/ D 0
x!2
From inspecting the graph
"1
lim .x 2 " 4x C 1/ D 42 " 4.4/ C 1 D 1
x!4
Fig. 1.1-14
y D f .x/
lim g.x/ D 0
x!3!
year
1.
lim g.x/ D 1
x!1!
x!3
21.
lim
x!0
jx " 2j
j " 2j
D
D "1
x"2
"2
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INSTRUCTOR’S SOLUTIONS MANUAL
22.
SECTION 1.2 (PAGE 71)
'
jx " 2j
1;
if x > 2
D lim
x!2 x " 2
x!2 "1; if x < 2.
jx " 2j
Hence, lim
does not exist.
x!2 x " 2
lim
33.
t2 " 1
" 2t C 1
.t " 1/.t C 1/
t C1
D lim
lim
does not exist
t!1
t!1 t " 1
.t " 1/2
(denominator ! 0, numerator ! 2.)
p
4 " 4x C x 2
24. lim
x!2
x"2
jx " 2j
does not exist.
D lim
x!2 x " 2
p
p
t
t. 4 C t C 4 " t/
25. lim p
p
D lim
t!0
t!0 .4 C t / " .4 " t /
4Ct " 4"t
p
p
4Ct C 4"t
D lim
D2
t!0
2
p
x2 " 1
.x " 1/.x C 1/. x C 3 C 2/
26. lim p
D lim
x!1 x C 3 " 2
x!1
.x C 3/ " 4
p
p
D lim .x C 1/. x C 3 C 2/ D .2/. 4 C 2/ D 8
23.
lim
t!1 t 2
34.
35.
$
4
1
" 2
x!2 x " 2
x "4
xC2"4
1
1
D lim
D lim
D
x!2 .x " 2/.x C 2/
x!2 x C 2
4
lim
#
$
1
1
" 2
x!2 x " 2
x "4
xC2"1
D lim
x!2 .x " 2/.x C 2/
xC1
does not exist.
D lim
x!2 .x " 2/.x C 2/
lim
#
p
2 C x2 " 2 " x2
lim
x!0
x2
.2 C x 2 / " .2 " x 2 /
p
D lim
p
x!0 x 2 . 2 C x 2 C
2 " x2/
2x 2
%p
&
D lim
p
x!0 2
x
2 C x2/ C 2 " x2
1
2
p D p
D p
2C 2
2
p
x!1
27.
36.
t 2 C 3t
lim
t!0 .t C 2/2 " .t " 2/2
t .t C 3/
D lim 2
t!0 t C 4t C 4 " .t 2 " 4t C 4/
t C3
3
D lim
D
t!0
8
8
4s
.s C 1/2 " .s " 1/2
D lim
D4
s!0 s
s!0
s
p
y "4 yC3
29. lim
y!1
y 2 "p1
p
. y " 1/. y " 3/
"2
"1
D lim p
D
D
p
y!1 . y " 1/. y C 1/.y C 1/
4
2
28.
30.
31.
32.
lim
x3 C 1
x!!1 x C 1
.x C 1/.x 2 " x C 1/
D3
D lim
x!!1
xC1
j3x " 1j " j3x C 1j
x
.3x " 1/2 " .3x C 1/2
D lim
x!0 x .j3x " 1j C j3x C 1j/
"12x
"12
D lim
D
D "6
x!0 x .j3x " 1j C j3x C 1j/
1C1
lim
x!0
f .x/ D x 2
37.
lim
h!0
f .x/ D x 3
38.
lim
x 4 " 16
x!2 x 3 " 8
.x " 2/.x C 2/.x 2 C 4/
D lim
x!2 .x " 2/.x 2 C 2x C 4/
8
.4/.8/
D
D
4C4C4
3
lim
h!0
lim
x 2=3 " 4
x!8 x 1=3 " 2
.x 1=3 " 2/.x 1=3 C 2/
D lim
x!8
.x 1=3 " 2/
1=3
D lim .x
C 2/ D 4
lim
x!8
f .x C h/ " f .x/
.x C h/2 " x 2
D lim
h
h
h!0
2hx C h2
D lim
D lim 2x C h D 2x
h
h!0
h!0
f .x C h/ " f .x/
.x C h/3 " x 3
D lim
h
h
h!0
3x 2 h C 3xh2 C h3
D lim
h
h!0
D lim 3x 2 C 3xh C h2 D 3x 2
h!0
39.
f .x/ D 1=x
1
1
"
f .x C h/ " f .x/
x
C
h
x
lim
D lim
h
h
h!0
h!0
x " .x C h/
D lim
h!0 h.x C h/x
1
1
D lim "
D" 2
x
h!0 .x C h/x
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SECTION 1.2 (PAGE 71)
ADAMS and ESSEX: CALCULUS 9
f .x/ D 1=x 2
40.
48.
f .x/ D
41.
p
x
p
p
f .x/ D 1= x
lim
f .x C h/ " f .x/
D lim
h
h!0
D lim
h!0
D lim
h!0
D lim
h!0
D
"1
2x 3=2
p
lim
p
x!2!
50.
51.
44.
52.
45.
lim
p
x!0
3
1
54.
46.
55.
56.
.x " x < 0 if 0 < x < 1/
p
x3 " x D 0
lim
lim
p
lim
p
x!0C
57.
58.
61.
sin x D sin 2#=3 D
p
x 3 " x does not exist. (See # 9.)
x2 " x4 D 0
jx " aj
x 2 " a2
jx " aj
1
D"
D lim
x!a! .x " a/.x C a/
2a
lim
x!a!
lim cos x D cos #=3 D 1=2
lim
x 3 " x does not exist.
x!0C
60.
x!2!=3
2"x D2
x!0!
p
lim cos x D cos #=4 D 1= 2
x!!=3
p
lim
x!!2C
59.
x!!=4
2 " x does not exist.
p
lim
2"x D2
x!!2!
lim sin x D sin #=2 D 1
x!!=2
3=2
lim
0
x2 " 4
D D0
x!2! jx C 2j
4
lim
x2 " 4
0
D D0
x!2C jx C 2j
4
8
if x $ "1
<x " 1
if "1 < x $ 0
f .x/ D x 2 C 1
:
.x C #/2 if x > 0
lim f .x/ D lim x " 1 D "1 " 1 D "2
lim
.sin x/=x
˙1:0
˙0:1
˙0:01
˙0:001
0:0001
0:84147098
0:99833417
0:99998333
0:99999983
1:00000000
sin x
It appears that lim
D 1.
x!0 x
26
63.
x!0C
lim f .x/ D lim x 2 C 1 D 1
x!0!
65.
lim x 2 C 1 D 1 C 1 D 2
x!!1C
lim f .x/ D lim .x C #/2 D # 2
x!0C
64.
x!!1!
lim f .x/ D
x!!1C
47.
.a ¤ 0/
jx " aj
1
x"a
D lim 2
D
2
2
2
x!aC
x "a
x "a
2a
x!!1!
62.
x
1 " cos x
1
D .
x2
2
2"x D0
x!2C
53.
1
"p
x
xCh
h
p
p
x" xCh
p p
h x xCh
x " .x C h/
p
p p
p
h x x C h. x C x C h/
"1
p
p p
p
x x C h. x C x C h/
p
lim
x!aC
43.
0:45969769
0:49958347
0:49999583
0:49999996
0:50000000
It appears that lim
49.
p
xCh" x
h
xCh"x
D lim p
p
h!0 h. x C h C
x/
1
1
D lim p
p D p
h!0
2 x
xChC x
h!0
˙1:0
˙0:1
˙0:01
˙0:001
0:0001
x!0
f .x C h/ " f .x/
lim
D lim
h
h!0
h!0
42.
.1 " cos x/=x 2
x
1
1
"
f .x C h/ " f .x/
.x C h/2 x 2
lim
D lim
h
h
h!0
h!0
x 2 " .x 2 C 2xh C h2 /
D lim
h.x C h/2 x 2
h!0
2x C h
2x
2
D lim "
D" 4 D" 3
2
2
x
x
h!0 .x C h/ x
x!0!
If lim f .x/ D 2 and lim g.x/ D "3, then
x!4
x!4
%
&
a) lim g.x/ C 3 D "3 C 3 D 0
x!4
b) lim xf .x/ D 4 & 2 D 8
x!4
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 1.2 (PAGE 71)
%
&2
c) lim g.x/ D ."3/2 D 9
70.
y
x!4
g.x/
"3
d) lim
D
D "3
x!4 f .x/ " 1
2"1
0.8
0.6
yD
0.4
66. If lim x ! af .x/ D 4 and lim g.x/ D "2, then
0.2
x!a
%
&
a) lim f .x/ C g.x/ D 4 C ."2/ D 2
-0.08
x!a
-0.04
x!a
-0.4
c) lim 4g.x/ D 4."2/ D "8
x!a
Fig. 1.2-70
4
f .x/
D
D "2
d) lim
x!a g.x/
"2
If lim
x!2
limx!0 sin.2#x/= sin.3#x/ D 2=3
71.
f .x/ " 5
D 3, then
x"2
y
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
%
&
f .x/ " 5
.x " 2/ D 3.2 " 2/ D 0:
lim f .x/ " 5 D lim
x!2
x!2 x " 2
Thus limx!2 f .x/ D 5.
68. If lim
x!0
f .x/
D "2 then
x2
f .x/
D 0 & ."2/ D 0,
x2
f .x/
f .x/
D lim x 2 D 0&."2/ D 0.
x!0
x
x
limx!0 f .x/ D limx!0 x 2
and similarly, limx!0
0.08 x
0.04
-0.2
b) lim f .x/ ! g.x/ D 4 & ."2/ D "8
67.
sin.2#x/
sin.3#x/
-0.1
p
sin 1 " x
yD p
1 " x2
0.2
0.4
0.6
0.8
1.0
x
Fig. 1.2-71
p
sin 1 " x
p
' 0:7071
x!1!
1 " x2
lim
69.
72.
y
1.2
1.0
0.8
y
yD
sin x
x
0.2
-0.2
-0.4
1
2
x
-1.0
-1.2
Fig. 1.2-69
lim
x!0
sin x
D1
x
x
p
x" x
yD p
sin x
-0.6
-0.8
-1
0.8
-0.4
0.2
-2
0.6
-0.2
0.6
0.4
-3
0.4
Fig. 1.2-72
lim
x!0C
p
x" x
p
D "1
sin x
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SECTION 1.2 (PAGE 71)
ADAMS and ESSEX: CALCULUS 9
73.
y
y D "x
f .x/ D s sin
79.
jf .x/j $ g.x/ ) "g.x/ $ f .x/ $ g.x/
Since lim g.x/ D 0, therefore 0 $ lim f .x/ $ 0.
yDx
y D x sin.1=x/
0.1
-0.2
-0.1
x
0.1
1
is defined for all x ¤ 0; its domain is
x
."1; 0/ [ .0; 1/. Since j sin t j $ 1 for all t , we have
jf .x/j $ jxj and "jxj $ f .x/ $ jxj for all x ¤ 0.
Since limx!0 D ."jxj/ D 0 D limx!0 jxj, we have
limx!0 f .x/ D 0 by the squeeze theorem.
78.
x!a
x!a
Hence, lim f .x/ D 0.
x!a
-0.1
If lim g.x/ D 3, then either "3 $ lim f .x/ $ 3 or
x!a
x!a
limx!a f .x/ does not exist.
-0.2
Fig. 1.2-73
f .x/ D x sin.1=x/ oscillates infinitely often as x approaches 0, but the amplitude of the oscillations decreases
and, in fact, limx!0 f .x/ D 0. This is predictable because
jx sin.1=x/j $ jxj. (See Exercise 95 below.)
p
p
74. Since p
5 " 2x 2 $ f .x/ $ 5p" x 2 for "1p$ x $ 1, and
2 D lim
5 " x 2 D 5, we have
limx!0 5 " 2xp
x!0
limx!0 f .x/ D 5 by the squeeze theorem.
75. Since 2 " x 2 $ g.x/ $ 2 cos x for all x, and since
limx!0 .2 " x 2 / D limx!0 2 cos x D 2, we have
limx!0 g.x/ D 2 by the squeeze theorem.
76.
Section 1.3 Limits at Infinity and Infinite
Limits (page 78)
1.
2.
3.
y
yDx
4.
4
1
y D x2
.1; 1/
5.
-2
-1
x
1
Fig. 1.2-76
2
b) Since the graph of f lies between those of x and
x 4 , and since these latter graphs come together at
.˙1; 1/ and at .0; 0/, we have limx!˙1 f .x/ D 1 and
limx!0 f .x/ D 0 by the squeeze theorem.
77.
x 1=3 < x 3 on ."1; 0/ and .1; 1/. x 1=3 > x 3 on
."1; "1/ and .0; 1/. The graphs of x 1=3 and x 3 intersect at ."1; "1/, .0; 0/, and .1; 1/. If the graph of h.x/
lies between those of x 1=3 and x 3 , then we can determine
limx!a h.x/ for a D "1, a D 0, and a D 1 by the
squeeze theorem. In fact
lim h.x/ D "1;
x!!1
28
lim h.x/ D 0;
x!0
lim
x
1=x
0
D lim
D D0
x!1 1 " .4=x 2 /
x2 " 4
1
3x 3 " 5x 2 C 7
x!1 8 C 2x " 5x 3
5
7
3" C 3
3
x
x
D"
D lim
2
x!1 8
5
C 2 "5
x3
x
lim
x2 " 2
x " x2
2
1" 2
x D 1 D "1
D lim
x!!1 1
"1
"1
x
lim
x!!1
2
."1; 1/
1
1
x
D lim
D
x!1
2x " 3
2 " .3=x/
2
x!1
a)
3
lim
x!1
lim h.x/ D 1:
x!1
6.
7.
1
3
C 3
x2 C 3
x D0
D lim x
lim
2
x!!1
x!!1 x 3 C 2
1C 3
x
sin x
1C 2
1
x 2 C sin x
x
lim
D lim
cos x D 1 D 1
x!1 x 2 C cos x
x!1
1C 2
x
sin x
We have used the fact that limx!1 2 D 0 (and simx
ilarly for cosine) because the numerator is bounded while
the denominator grows large.
p
3x C 2 x
x!1
1"x
2
3C p
x
D lim
D "3
x!1 1
"1
x
lim
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INSTRUCTOR’S SOLUTIONS MANUAL
8.
lim p
x!1
2x " 1
3x 2
SECTION 1.3 (PAGE 78)
C#x C 1 $
1
x 2"
x
D lim
r
(but jxj D x as x ! 1/
x!1
1
1
jxj 3 C C 2
x
x
1
2"
2
x
D lim r
D p
x!1
1
1
3
3C C 2
x
x
2x " 1
9.
lim p
x!!1
3x 2 C x C 1
1
2"
2
x
D lim
r
D "p ,
x!!1
1
1
3
" 3C C 2
x
x
p
because x ! "1 implies that x < 0 and so x 2 D "x.
10.
lim
x!!1
2x " 5
2x " 5
2
D lim
D"
x!!1 ".3x C 2/
j3x C 2j
3
1
does not exist.
x!3 3 " x
1
D1
12. lim
x!3 .3 " x/2
11.
13.
25.
26.
27.
28.
lim
29.
1
D1
3"x
1
D "1
lim
x!3C 3 " x
lim
0
2x C 5
D
D0
"25
5x C 2
C2
2
2x C 5
does not exist.
16.
lim
x!!2=5 5x C 2
15.
lim
x!!5=2
17.
lim
x!!.2=5/!
30.
x 2 C 2x " x 2 C 2x
p
x!1
x 2 C 2x C x 2 " 2x
4x
D lim r
r
x!1
2
2
x 1C Cx 1"
x
x
4
4
D D2
D lim r
r
x!1
2
2
2
1C C 1"
x
x
D lim p
2x C 5
D "1
5x C 2
lim
x!!2=5C
22.
lim
x!1!
23.
31.
lim p
x!1
1
x 2 " 2x " x
p
x 2 " 2x C x
p
p
2
. x " 2x C x/. x 2 " 2x " x/
p
x 2 " 2x C x
D lim
x!1 x 2 " 2x " x 2
p
x. 1 " .2=x/ C 1/
2
D
D "1
D lim
x!1
"2x
"2
D lim
x!1
1
D1
jx " 1j
x"3
x"3
D "1
D lim
x!2 .x " 2/2
x 2 " 4x C 4
p
x2 " x
"1
lim
D lim p
D "1
x!1C x " x 2
x!1C
x2 " x
lim
x!2
24.
.x 2 C 2x/ " .x 2 " 2x/
p
D lim p
x!!1
x 2 C 2x C x 2 " 2x
4x
!
D lim
r
r
x!!1
2
2
."x/
1C C 1"
x
x
4
D "2
D"
1C1
#p
$
p
lim
x 2 C 2x " x 2 " 2x
x!1
2x C 5
D1
5x C 2
x
19.
lim
D "1
x!2C .2 " x/3
x
20. lim p
D1
x!1!
1 " x2
1
D1
21.
lim
x!1C jx " 1j
18.
3
xC 2
x3 C 3
x
lim
D1
D lim
2
x!1 x 2 C 2
x!1
1C 2
x
!
p
"
p
x x C 1 1 " 2x C 3
lim
2
x!1
7"
!
!
r
r6x C 4x
1
1
3
2
1C
x
p " 2C
x
x
x
#
$
D lim
x!1
7
6
x2
" C4
x2 x
p
1." 2/
1p
2
D
D"
4
4
# 2
$
x
x2
"2x 2
lim
"
D lim 2
D "2
x!1 x C 1
x!1 x " 1
x"1
%p
&
p
lim
x 2 C 2x " x 2 " 2x
x!!1
x!3!
14.
x C x3 C x5
x!1 1 C x 2 C x 3
1
C 1 C x2
2
x
D lim
D1
1
x!1 1
C
C
1
x3
x
lim
32.
lim p
x!!1
1
x2
C 2x " x
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D lim
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1
p
D0
jxj. 1 C .2=x/ C 1
29
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SECTION 1.3 (PAGE 78)
ADAMS and ESSEX: CALCULUS 9
33. By Exercise 35, y D "1 is a horizontal asymptote (at the
1
. Since
right) of y D p
x 2 " 2x " x
lim p
x!!1
1
x2
" 2x " x
D lim
x!!1
1
p
D 0;
jxj. 1 " .2=x/ C 1
y D 0pis also a horizontal asymptote (at the left).
Now x 2 " 2x " x D 0 if and only if x 2 " 2x D x 2 , that
is, if and only if x D 0. The given function is undefined
at x D 0, and where x 2 " 2x < 0, that is, on the interval Œ0; 2!. Its only vertical asymptote is at x D 0, where
1
D 1.
limx!0! p
2
x " 2x " x
2x " 5
2
2x " 5
2
34. Since lim
D
and lim
D " ,
x!1 j3x C 2j
x!!1 j3x C 2j
3
3
y D ˙.2=3/ are horizontal asymptotes of
y D .2x " 5/=j3x C 2j. The only vertical asymptote is
x D "2=3, which makes the denominator zero.
35.
45.
46.
47.
horizontal: y D 1; vertical: x D 1, x D 3.
lim bxc D 3
x!3C
48.
lim bxc D 2
x!3!
49.
lim bxc does not exist
x!3
50.
lim bxc D 2
x!2:5
51.
lim b2 " xc D lim bxc D 1
x!0C
52.
x!2!
lim bxc D "4
x!!3!
53.
lim C.t / D C.t0 / except at integers t0
t!t0
lim C.t / D C.t0 / everywhere
t!t0 !
lim f .x/ D 1
lim C.t / D C.t0 / if t0 ¤ an integer
x!0C
36.
lim f .x/ D 1
x!1
t!t0 C
lim f .x/ D 1
lim C.t / D C.t0 / C 1:5 if t0 is an integer
x!1
t!t0 C
37.
y
6:00
y
3
y D f .x/
4:50
2
y D C.t /
3:00
1:50
1
1
1
2
3
4
-1
5
6
x
2
3
4
x
Fig. 1.3-53
54.
lim f .x/ D L
x!0C
(a) If f is even, then f ."x/ D f .x/.
Hence, lim f .x/ D L.
Fig. 1.3-37
x!0!
(b) If f is odd, then f ."x/ D "f .x/.
Therefore, lim f .x/ D "L.
limx!2C f .x/ D 1
38.
lim f .x/ D 2
x!2!
39.
lim f .x/ D "1
x!0!
55.
lim f .x/ D A;
x!0C
lim f .x/ D B
x!0!
x!3!
40.
41.
a)
lim f .x/ D 2
b) lim f .x 3 " x/ D A (because x 3 " x > 0 if
x!4C
42.
lim f .x/ D 0
x!4!
43.
lim f .x/ D "1
x!5!
44.
lim f .x 3 " x/ D B (since x 3 " x < 0 if 0 < x < 1)
lim f .x/ D 1
x!3C
lim f .x/ D 0
x!5C
30
x!0C
x!0!
"1 < x < 0)
c) lim f .x 2 " x 4 / D A
x!0!
d)
lim f .x 2 " x 4 / D A (since x 2 " x 4 > 0 for
x!0C
0 < jxj < 1)
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INSTRUCTOR’S SOLUTIONS MANUAL
Section 1.4 Continuity
1.
SECTION 1.4 (PAGE 87)
(page 87)
8.
g is continuous at x D "2, discontinuous at x D "1; 0; 1,
and 2. It is left continuous at x D 0 and right continuous
at x D 1.
y
.1; 2/
2
y D g.x/
."1; 1/
-2
1
lim f .x/ D
x!!1!
1
2
x
Fig. 1.4-1
2. g has removable discontinuities at x D "1 and x D 2.
Redefine g."1/ D 1 and g.2/ D 0 to make g continuous
at those points.
D f ."1/ D
4. Function f is discontinuous at x D 1; 2; 3; 4, and 5. f
is left continuous at x D 4 and right continuous at x D 2
and x D 5.
y
12. C.t / is discontinuous only at the integers. It is continuous
on the left at the integers, but not on the right.
13.
1
15.
4
5
6
x
16.
-1
Fig. 1.4-4
5. f cannot be redefined at x D 1 to become continuous
there because limx!1 f .x/ .D 1/ does not exist. (1 is
not a real number.)
6. sgn x is not defined at x D 0, so cannot be either continuous or discontinuous there. (Functions can be continuous
or discontinuous only at points in their domains!)
'
x
if x < 0
7. f .x/ D
is continuous everywhere on the
x 2 if x # 0
real line, even at x D 0 where its left and right limits are
both 0, which is f .0/.
The least integer function dxe is continuous everywhere on
right continuous.
2
3
lim f .x/:
x!!1C
R except at the integers, where it is left continuous but not
14.
y D f .x/
2
lim x 2 D
x!!1C
'
2
if x ¤ 0 is continuous everywhere except
f .x/ D 1=x
0
if x D 0
at x D 0, where it is neither left nor right continuous since
it does not have a real limit there.
'
2
if x $ 1 is continuous everywhere ex10. f .x/ D x
0:987 if x > 1
cept at x D 1, where it is left continuous but not right
continuous because 0:987 ¤ 1. Close, as they say, but no
cigar.
11.
3. g has no absolute maximum value on Œ"2; 2!. It takes on
every positive real value less than 2, but does not take the
value 2. It has absolute minimum value 0 on that interval,
assuming this value at the three points x D "2, x D "1,
and x D 1.
1
lim x D "1 ¤ 1
x!!1!
9.
-1
3
'
x
if x < "1
is continuous everywhere on the
x 2 if x # "1
real line except at x D "1 where it is right continuous, but
not left continuous.
f .x/ D
17.
x2 " 4
Since
D x C 2 for x ¤ 2, we can define the
x"2
function to be 2 C 2 D 4 at x D 2 to make it continuous
there. The continuous extension is x C 2.
1 C t3
.1 C t /.1 " t C t 2 /
1 " t C t2
D
D
for
2
1"t
.1 C t /.1 " t /
1"t
t ¤ "1, we can define the function to be 3=2 at t D "1
to make it continuous there. The continuous extension is
1 " t C t2
.
1"t
Since
.t " 2/.t " 3/
t "2
t 2 " 5t C 6
D
D
for t ¤ 3,
t2 " t " 6
.t C 2/.t " 3/
t C2
we can define the function to be 1=5 at t D 3 to make it
t "2
continuous there. The continuous extension is
.
t C2
Since
p
p
p
xC 2
x2 " 2
.x " 2/.x C 2/
p
p
D
p
D
2
x4 " 4
.x C 2/.x 2 C 2/
p.x " 2/.x C 2/.x C 2/
for x p
¤ 2, we can define the function to be 1=4 at
there. The continuous
x D 2 to make it continuous
p
xC 2
extension is
p
. (Note: cancelling the
.x C 2/.x 2 C 2/
p
2 factors provides a further continuous extension to
xC p
x D " 2.
Since
limx!2C f .x/ D k " 4 and limx!2! f .x/ D 4 D f .2/.
Thus f will be continuous at x D 2 if k " 4 D 4, that is,
if k D 8.
18. limx!3! g.x/
D
3 " m
and limx!3C g.x/ D 1 " 3m D g.3/. Thus g will be
continuous at x D 3 if 3 " m D 1 " 3m, that is, if m D "1.
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SECTION 1.4 (PAGE 87)
ADAMS and ESSEX: CALCULUS 9
19. x 2 has no maximum value on "1 < x < 1; it takes all
positive real values less than 1, but it does not take the
value 1. It does have a minimum value, namely 0 taken on
at x D 0.
20. The Max-Min Theorem says that a continuous function
defined on a closed, finite interval must have maximum
and minimum values. It does not say that other functions
cannot have such values. The Heaviside function is not
continuous on Œ"1; 1! (because it is discontinuous at
x D 0), but it still has maximum and minimum values. Do
not confuse a theorem with its converse.
21.
Let the numbers be x and y, where x # 0, y # 0, and
x C y D 8. If P is the product of the numbers, then
f .x/ D
29.
f .x/ D x 3 C x " 1, f .0/ D "1, f .1/ D 1.
Since f is continuous and changes sign between 0 and 1,
it must be zero at some point between 0 and 1 by IVT.
30.
f .x/ D x 3 " 15x C 1 is continuous everywhere.
f ."4/ D "3; f ."3/ D 19; f .1/ D "13; f .4/ D 5:
Because of the sign changes f has a zero between "4 and
"3, another zero between "3 and 1, and another between
1 and 4.
31.
F .x/ D .x " a/2 .x " b/2 C x. Without loss of generality,
we can assume that a < b. Being a polynomial, F is
continuous on Œa; b!. Also F .a/ D a and F .b/ D b.
Since a < 21 .a C b/ < b, the Intermediate-Value Theorem
guarantees that there is an x in .a; b/ such that
F .x/ D .a C b/=2.
32.
Let g.x/ D f .x/ " x. Since 0 $ f .x/ $ 1 if 0 $ x $ 1,
therefore, g.0/ # 0 and g.1/ $ 0. If g.0/ D 0 let c D 0,
or if g.1/ D 0 let c D 1. (In either case f .c/ D c.)
Otherwise, g.0/ > 0 and g.1/ < 0, and, by IVT, there
exists c in .0; 1/ such that g.c/ D 0, i.e., f .c/ D c.
33.
The domain of an even function is symmetric about the
y-axis. Since f is continuous on the right at x D 0,
therefore it must be defined on an interval Œ0; h! for some
h > 0. Being even, f must therefore be defined on
Œ"h; h!. If x D "y, then
P D xy D x.8 " x/ D 8x " x 2 D 16 " .x " 4/2 :
Therefore P $ 16, so P is bounded. Clearly P D 16 if
x D y D 4, so the largest value of P is 16.
22. Let the numbers be x and y, where x # 0, y # 0, and
x C y D 8. If S is the sum of their squares then
S D x 2 C y 2 D x 2 C .8 " x/2
D 2x 2 " 16x C 64 D 2.x " 4/2 C 32:
Since 0 $ x $ 8, the maximum value of S occurs at
x D 0 or x D 8, and is 64. The minimum value occurs at
x D 4 and is 32.
23. Since T D 100 " 30x C 3x 2 D 3.x " 5/2 C 25, T will
be minimum when x D 5. Five programmers should be
assigned, and the project will be completed in 25 days.
24.
lim f .x/ D lim f ."y/ D lim f .y/ D f .0/:
x!0!
If x desks are shipped, the shipping cost per desk is
C D
245x " 30x 2 C x 3
D x 2 " 30x C 245
x
D .x " 15/2 C 20:
.x C 2/.x " 1/
x2 C x " 2
D
x3
x3
f .x/ > 0 on ."2; 0/ and .1; 1/
f .x/ < 0 on ."1; "2/ and .0; 1/.
28.
y!0C
y!0C
Thus, f is continuous on the left at x D 0. Being continuous on both sides, it is therefore continuous.
34.
This cost is minimized if x D 15. The manufacturer
should send 15 desks in each shipment, and the shipping
cost will then be $20 per desk.
f odd , f ."x/ D "f .x/
f continuous on the right , lim f .x/ D f .0/
x!0C
Therefore, letting t D "x, we obtain
lim f .x/ D lim f ."t / D lim "f .t /
x2 " 1
.x " 1/.x C 1/
D
x
x
f D 0 at x D ˙1. f is not defined at 0.
f .x/ > 0 on ."1; 0/ and .1; 1/.
f .x/ < 0 on ."1; "1/ and .0; 1/.
x!0!
25. f .x/ D
t!0C
t!0C
D "f .0/ D f ."0/ D f .0/:
Therefore f is continuous at 0 and f .0/ D 0.
26. f .x/ D x 2 C 4x C 3 D .x C 1/.x C 3/
f .x/ > 0 on ."1; "3/ and ."1; 1/
f .x/ < 0 on ."3; "1/.
35.
max 1:593 at "0:831, min "0:756 at 0:629
36.
max 0:133 at x D 1:437; min "0:232 at x D "1:805
27.
38.
max 1:510 at x D 0:465; min 0 at x D 0 and x D 1
39.
root x D 0:682
40.
root x D 0:739
x2 " 1
.x " 1/.x C 1/
f .x/ D 2
D
x "4
.x " 2/.x C 2/
f D 0 at x D ˙1.
f is not defined at x D ˙2.
f .x/ > 0 on ."1; "2/, ."1; 1/, and .2; 1/.
f .x/ < 0 on ."2; "1/ and .1; 2/.
32
37. max 10:333 at x D 3; min 4:762 at x D 1:260
41. roots x D "0:637 and x D 1:410
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 1.5 (PAGE 92)
42. roots x D "0:7244919590 and x D 1:220744085
43. fsolve gives an approximation to the single real root to 10
significant figures; solve gives the three roots (including a
complex conjugate pair) in exact form involving the quan%
p &1=3
; evalf(solve) gives approximations
tity 108 C 12 69
to the three roots using 10 significant figures for the real
and imaginary parts.
Section 1.5 The Formal Definition of Limit
(page 92)
1.
We require 39:9 $ L $ 40:1. Thus
10. We need 1 " 0:05 $ 1=.x C 1/ $ 1 C 0:05,
or 1:0526 # x C 1 # 0:9524. This will occur if
"0:0476 $ x $ 0:0526. In this case we can take
ı D 0:0476.
11.
x!1
12. To be proved: lim .5 " 2x/ D 1.
x!2
Proof: Let $ > 0 be given. Then j.5 " 2x/ " 1j < $ holds
if j2x " 4j < $, and so if jx " 2j < ı D $=2. This confirms
the limit.
13.
39:9 $ 39:6 C 0:025T $ 40:1
0:3 $ 0:025T $ 0:5
12 $ T $ 20:
x!0
14.
4.
5.
6.
x"2
D 0.
1 C x2
Proof: Let $ > 0 be given. Then
ˇ
ˇ
ˇ
ˇ x"2
jx " 2j
ˇ
ˇ
ˇ 1 C x 2 " 0ˇ D 1 C x 2 $ jx " 2j < $
To be proved: lim
x!2
2. Since 1.2% of 8,000 is 96, we require the edge length x
of the cube to satisfy 7904 $ x 3 $ 8096. It is sufficient
that 19:920 $ x $ 20:079. The edge of the cube must be
within 0.079 cm of 20 cm.
3 " 0:02 $ 2x " 1 $ 3 C 0:02
3:98 $ 2x $ 4:02
1:99 $ x $ 2:01
To be proved: lim x 2 D 0.
Let $ > 0 be given. p
Then jx 2 " 0j < $ holds if
jx " 0j D jxj < ı D $.
The temperature should be kept between 12 ı C and 20 ı C.
3.
To be proved: lim .3x C 1/ D 4.
Proof: Let $ > 0 be given. Then j.3x C 1/ " 4j < $ holds
if 3jx " 1j < $, and so if jx " 1j < ı D $=3. This confirms
the limit.
provided jx " 2j < ı D $.
15.
1 " 4x 2
D 2.
x!1=2 1 " 2x
Proof: Let $ > 0 be given. Then if x ¤ 1=2 we have
To be proved:
lim
ˇ
ˇ
ˇ
ˇ 1 " 4x 2
ˇ
ˇ
ˇ
ˇ D j.1 C 2x/ " 2j D j2x " 1j D 2 ˇx "
"
2
ˇ 1 " 2x
ˇ
ˇ
4 " 0:1 $ x 2 $ 4 C 0:1
1:9749 $ x $ 2:0024
p
1 " 0:1 $ x $ 1:1
0:81 $ x $ 1:21
provided jx " 12 j < ı D $=2.
16.
1
"2 " 0:01 $ $ "2 C 0:01
x
1
1
"
#x#"
2:01
1:99
"0:5025 $ x $ "0:4975
We need "0:03 $ .3x C 1/ " 7 $ 0:03, which is equivalent
to "0:01 $ x " 2 $ 0:01 Thus ı D 0:01 will do.
p
8. We need "0:01 $ 2x C 3 " 3 $ 0:01. Thus
Here ı D 0:02995 will do.
9. We need 8 " 0:2 $ x 3 $ 8:2, or 1:9832 $ x $ 2:0165.
Thus, we need "0:0168 $ x " 2 $ 0:0165. Here
ı D 0:0165 will do.
x 2 C 2x
D "2.
x!!2 x C 2
Proof: Let $ > 0 be given. For x ¤ "2 we have
To be proved:
lim
ˇ 2
ˇ
ˇ x C 2x
ˇ
ˇ
ˇ
ˇ x C 2 " ."2/ˇ D jx C 2j < $
7.
p
2:99 $ 2x C 3 $ 3:01
8:9401 $ 2x C 3 $ 9:0601
2:97005 $ x $ 3:03005
3 " 0:02995 $ x " 3 $ 0:03005:
ˇ
1 ˇˇ
<$
2ˇ
provided jx C 2j < ı D $. This completes the proof.
17.
1
1
D .
xC1
2
Proof: Let $ > 0 be given. We have
ˇ
ˇ ˇ
ˇ
ˇ 1
1 ˇˇ ˇˇ 1 " x ˇˇ
jx " 1j
ˇ
"
D
ˇ x C 1 2 ˇ ˇ 2.x C 1/ ˇ D 2jx C 1j :
To be proved: lim
x!1
If jx " 1j < 1, then 0 < x < 2 and 1 < x C 1 < 3, so that
jx C 1j > 1. Let ı D min.1; 2$/. If jx " 1j < ı, then
ˇ
ˇ
ˇ 1
1 ˇˇ
jx " 1j
2$
ˇ
"
ˇ x C 1 2 ˇ D 2jx C 1j < 2 D $:
This establishes the required limit.
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SECTION 1.5 (PAGE 92)
ADAMS and ESSEX: CALCULUS 9
xC1
1
D" .
x2 " 1
2
Proof: Let $ > 0 be given. If x ¤ "1, we have
ˇ
# $ˇ ˇ
# $ˇ
ˇ xC1
1 ˇˇ ˇˇ 1
1 ˇ
jx C 1j
ˇ
"
"
D
"
" ˇˇ D
:
ˇ x2 " 1
ˇ
ˇ
2
x"1
2
2jx " 1j
18. To be proved:
lim
25.
x!!1
If jx C 1j < 1, then "2 < x < 0, so "3 < x " 1 < "1 and
jx " 1j > 1. Ler ı D min.1; 2$/. If 0 < jx " ."1/j < ı
then jx " 1j > 1 and jx C 1j < 2$. Thus
ˇ
# $ˇ
ˇ xC1
1 ˇˇ
jx C 1j
2$
ˇ
ˇ x 2 " 1 " " 2 ˇ D 2jx " 1j < 2 D $:
This completes the required proof.
p
19. To be proved: lim x D 1.
a <x <aCı
26.
Proof: Let $ > 0 be given. We have
ˇ
ˇ
ˇ x"1 ˇ
p
ˇ $ jx " 1j < $
j x " 1j D ˇˇ p
x C 1ˇ
provided jx " 1j < ı D $. This completes the proof.
28.
20. To be proved: lim x 3 D 8.
x!2
Proof: Let $ > 0 be given. We have
jx 3 "8j D jx"2jjx 2 C2xC4j. If jx"2j < 1, then 1 < x < 3
and x 2 < 9. Therefore jx 2 C 2x C 4j $ 9 C 2 & 3 C 4 D 19.
If jx " 2j < ı D min.1; $=19/, then
$
& 19 D $:
jx " 8j D jx " 2jjx C 2x C 4j <
19
3
29.
2
1
To be proved: limx!1!
D "1. Proof: Let B > 0
x"1
1
< "B if 0 > x " 1 > "1=B,
be given. We have
x"1
that is, if 1 " ı < x < 1, where ı D 1=B:. This completes
the proof.
To be proved: limx!1 p
be given. We have
implies jf .x/ " Lj < $:
22. We say that limx!!1 f .x/ D L if the following condition
holds: for every number $ > 0 there exists a number
R > 0, depending on $, such that
x < "R
implies jf .x/ " Lj < $:
23. We say that limx!a f .x/ D "1 if the following condition holds: for every number B > 0 there exists a number
ı > 0, depending on B, such that
0 < jx " aj < ı
24.
implies f .x/ < "B:
We say that limx!1 f .x/ D 1 if the following condition
holds: for every number B > 0 there exists a number
R > 0, depending on B, such that
x>R
34
implies f .x/ > B:
1
x2 C 1
D 0. Proof: Let $ > 0
ˇ
ˇ
ˇ
ˇ
1
ˇp 1
ˇD p 1
< <$
ˇ
ˇ
x
x2 C 1
x2 C 1
We say that limx!a! f .x/ D L if the following condition
holds: for every number $ > 0 there exists a number
ı > 0, depending on $, such that
a"ı <x <a
implies f .x/ > B:
1
To be proved: limx!1C
D 1. Proof: Let B > 0
x"1
1
> B if 0 < x " 1 < 1=B, that
be given. We have
x"1
is, if 1 < x < 1 C ı, where ı D 1=B. This completes the
proof.
This completes the proof.
21.
implies f .x/ < "B:
We say that limx!a! f .x/ D 1 if the following condition
holds: for every number B > 0 there exists a number
ı > 0, depending on B, such that
a"ı <x <a
27.
x!1
We say that limx!aC f .x/ D "1 if the following condition holds: for every number B > 0 there exists a number
ı > 0, depending on R, such that
provided x > R, where R D 1=$. This completes the
proof.
30.
31.
p
To be proved: lim
px!1 x D 1. Proof: Let B >2 0 be
given. We have x > B if x > R where R D B . This
completes the proof.
To be proved: if lim f .x/ D L and lim f .x/ D M , then
x!a
x!a
L D M.
Proof: Suppose L ¤ M . Let $ D jL " M j=3. Then
$ > 0. Since lim f .x/ D L, there exists ı1 > 0 such that
x!a
jf .x/"Lj < $ if jx "aj < ı1 . Since lim f .x/ D M , there
x!a
exists ı2 > 0 such that jf .x/ " M j < $ if jx " aj < ı2 .
Let ı D min.ı1 ; ı2 /. If jx " aj < ı, then
3$ D jL " M j D j.f .x/ " M / C .L " f .x/j
$ jf .x/ " M j C jf .x/ " Lj < $ C $ D 2$:
This implies that 3 < 2, a contradiction. Thus the original
assumption that L ¤ M must be incorrect. Therefore
L D M.
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INSTRUCTOR’S SOLUTIONS MANUAL
REVIEW EXERCISES 1 (PAGE 93)
32. To be proved: if lim g.x/ D M , then there exists ı > 0
This completes the proof.
x!a
such that if 0 < jx " aj < ı, then jg.x/j < 1 C jM j.
Proof: Taking $ D 1 in the definition of limit, we obtain
a number ı > 0 such that if 0 < jx " aj < ı, then
jg.x/ " M j < 1. It follows from this latter inequality that
36.
To be proved: if lim f .x/ D L and lim f .x/ D M ¤ 0,
x!a
x!a
L
f .x/
D
.
then lim
x!a g.x/
M
Proof: By Exercises 33 and 35 we have
jg.x/j D j.g.x/"M /CM j $ jG.x/"M jCjM j < 1CjM j:
lim
x!a
f .x/
1
1
L
D lim f .x/ &
DL&
D
:
x!a
g.x/
g.x/
M
M
33. To be proved: if lim f .x/ D L and lim g.x/ D M , then
x!a
x!a
lim f .x/g.x/ D LM .
x!a
37. To be proved: if f is continuous at L and lim g.x/ D L,
Proof: Let $ > 0 be given. Since lim f .x/ D L, there
x!c
x!a
then lim f .g.x// D f .L/.
exists ı1 > 0 such that jf .x/ " Lj < $=.2.1 C jM j//
if 0 < jx " aj < ı1 . Since lim g.x/ D M , there ex-
x!c
Proof: Let $ > 0 be given. Since f is continuous at L,
there exists a number % > 0 such that if jy " Lj < % , then
jf .y/ " f .L/j < $. Since limx!c g.x/ D L, there exists
ı > 0 such that if 0 < jx " cj < ı, then jg.x/ " Lj < % .
Taking y D g.x/, it follows that if 0 < jx " cj < ı, then
jf .g.x// " f .L/j < $, so that limx!c f .g.x// D f .L/.
x!a
ists ı2 > 0 such that jg.x/ " M j < $=.2.1 C jLj// if
0 < jx " aj < ı2 . By Exercise 32, there exists ı3 > 0
such that jg.x/j < 1 C jM j if 0 < jx " aj < ı3 . Let
ı D min.ı1 ; ı2 ; ı3 /. If jx " aj < ı, then
jf .x/g.x/ " LM D jf .x/g.x/ " Lg.x/ C Lg.x/ " LM j
D j.f .x/ " L/g.x/ C L.g.x/ " M /j
$ j.f .x/ " L/g.x/j C jL.g.x/ " M /j
D jf .x/ " Ljjg.x/j C jLjjg.x/ " M j
$
$
<
.1 C jM j/ C jLj
2.1 C jM j/
2.1 C jLj/
$
$
$ C D $:
2
2
38.
Thus lim f .x/g.x/ D LM .
x!a
34. To be proved: if lim g.x/ D M where M ¤ 0, then
jg.x/ " Lj D jg.x/ " f .x/ C f .x/ " Lj
$ jg.x/ " f .x/j C jf .x/ " Lj
$ jh.x/ " f .x/j C jf .x/ " Lj
D jh.x/ " L C L " f .x/j C jf .x/ " Lj
$ jh.x/ " Lj C jf .x/ " Lj C jf .x/ " Lj
$
$
$
< C C D $:
3
3
3
x!a
there exists ı > 0 such that if 0 < jx " aj < ı, then
jg.x/j > jM j=2.
Proof: By the definition of limit, there exists ı > 0 such
that if 0 < jx " aj < ı, then jg.x/ " M j < jM j=2 (since
jM j=2 is a positive number). This latter inequality implies
that
jM j D jg.x/C.M "g.x//j $ jg.x/jCjg.x/"M j < jg.x/jC
jM j
:
2
Thus limx!a g.x/ D L.
It follows that jg.x/j > jM j " .jM j=2/ D jM j=2, as
required.
35. To be proved: if lim g.x/ D M where M ¤ 0, then
x!a
1
1
lim
D
.
x!a g.x/
M
Proof: Let $ > 0 be given. Since lim g.x/ D M ¤ 0,
To be proved: if f .x/ $ g.x/ $ h.x/ in an open interval
containing x D a (say, for a " ı1 < x < a C ı1 , where
ı1 > 0), and if limx!a f .x/ D limx!a h.x/ D L, then
also limx!a g.x/ D L.
Proof: Let $ > 0 be given. Since limx!a f .x/ D L,
there exists ı2 > 0 such that if 0 < jx " aj < ı2 , then
jf .x/ " Lj < $=3. Since limx!a h.x/ D L, there exists
ı3 > 0 such that if 0 < jx "aj < ı3 , then jh.x/"Lj < $=3.
Let ı D min.ı1 ; ı2 ; ı3 /. If 0 < jx " aj < ı, then
Review Exercises 1 (page 93)
1.
The average rate of change of x 3 over Œ1; 3! is
33 " 13
26
D
D 13:
3"1
2
x!a
there exists ı1 > 0 such that jg.x/ " M j < $jM j2 =2 if
0 < jx " aj < ı1 . By Exercise 34, there exists ı2 > 0
such that jg.x/j > jM j=2 if 0 < jx " aj < ı3 . Let
ı D min.ı1 ; ı2 /. If 0 < jx " aj < ı, then
ˇ
ˇ
ˇ 1
1 ˇˇ
jM " g.x/j
$jM j2 2
ˇ
"
ˇ g.x/ M ˇ D jM jjg.x/j < 2 jM j2 D $:
2.
The average rate of change of 1=x over Œ"2; "1! is
.1=."1// " .1=."2//
"1=2
1
D
D" :
"1 " ."2/
1
2
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REVIEW EXERCISES 1 (PAGE 93)
ADAMS and ESSEX: CALCULUS 9
3. The rate of change of x 3 at x D 2 is
17.
h!0
4. The rate of change of 1=x at x D "3=2 is
#
1
"3=2
$
lim .x 2 " 4x C 7/ D 1 " 4 C 7 D 4
19.
21.
22.
23.
25.
22
4
x2
D
D"
2
x!2 1 " x
1 " 22
3
7.
x2
does not exist. The denominator approaches 0
x!1 1 " x 2
(from both sides) while the numerator does not.
8.
lim
9.
10.
lim
x2 " 4
.x " 2/.x C 2/
xC2
D lim
D lim
x!2 x 2 " 4x C 4
x!2
x!2 x " 2
.x " 2/2
does not exist. The denominator approaches 0 (from both
sides) while the numerator does not.
x!2! x 2
x2 " 4
xC2
D lim
D "1
x!2! x " 2
" 4x C 4
27.
15.
lim
x!!2C x 2
29.
p
p
h. x C 3h C x/
lim p
p D lim
.x C 3h/ " x
h!0
h!0
x C 3h " x
p
p
p
2 x
x C 3h C x
D
D lim
3
3
h!0
p
lim
x " x2 D 0
x!0C
16.
lim
x!0
x " x 2 does not exist because
fined for x < 0.
36
x " .1=x 2 /
x3 " 1
D
lim
D "1
x!!1 1 C .4=x 2 /
x!!1 x 2 C 4
lim
x4
x2
D
lim
D1
x!1 x 2 " 4
x!1 1 " .4=x 2 /
lim
1
D1
x " x2
1
1
D p
lim p
D2
2
x!1=2
1=4
x"x
lim p
lim sin x does not exist; sin x takes the values "1 and 1
p
x " x 2 is not de-
in any interval .R; 1/, and limits, if they exist, must be
unique.
cos x
D 0 by the squeeze theorem, since
lim
x!1 x
1
cos x
1
" $
$
for all x > 0
x
x
x
and limx!1 ."1=x/ D limx!1 .1=x/ D 0.
1
D 0 by the squeeze theorem, since
x
1
"jxj $ x sin $ jxj for all x ¤ 0
x
and limx!0 ."jxj/ D limx!0 jxj D 0.
lim x sin
1
does not exist; sin.1=x 2 / takes the values "1
x2
and 1 in any interval ."ı; ı/, where ı > 0, and limits, if
they exist, must be unique.
p
lim Œx C x 2 " 4x C 1!
lim sin
x!!1
x 2 " .x 2 " 4x C 1/
p
x!!1 x "
x 2 " 4x C 1
4x " 1
p
D lim
x!!1 x " jxj 1 " .4=x/ C .1=x 2 /
xŒ4 " .1=x/!
D lim
p
x!!1 x C x 1 " .4=x/ C .1=x 2 /
4 " .1=x/
D lim
p
D 2:
x!!1 1 C
1 " .4=x/ C .1=x 2 /
Note how we have used jxj D "x (in the second last line),
because x ! "1.
p
lim Œx C x 2 " 4x C 1! D 1 C 1 D 1
D lim
h
p
2x C 100
.2=x/ C .100=x 2 /
D lim
D0
2
x!!1 x C 3
x!!1
1 C .3=x 2 /
lim
x!0
x!3
14.
1 " x2
.1=x 2 / " 1
1
D lim
D"
x!1 3 " .1=x/ " .1=x 2 /
"x"1
3
x!1 3x 2
x!0
28.
x"2
x2 " 4
D lim
D "1
x!!2C x C 2
C 4x C 4
p
2" x
4"x
1
12. lim
D lim
p
D"
x!4 x " 4
x!4 .2 C
4
x/.x " 4/
p
p
x2 " 9
.x " 3/.x C 3/. x C 3/
13. lim p
p D lim
x!3
x!3
x"3
x" 3
p
p
p
D lim .x C 3/. x C 3/ D 12 3
11.
26.
x2 " 4
.x " 2/.x C 2/
xC2
D lim
D lim
D "4
x!2 .x " 2/.x " 3/
x!2 x " 3
" 5x C 6
lim
lim
lim
x!1
lim
x!2 x 2
x " x 2 is not de-
x!0C
x!1
6.
p
x!1!
20.
2
2
C
3
D lim 2h " 3
h
h!0
2.3 C 2h " 3/
D lim
h!0 3.2h " 3/h
4
4
D lim
D" :
9
h!0 3.2h " 3/
x " x 2 does not exist because
fined for x > 1.
p
18. lim
x " x2 D 0
24.
5.
p
x!1
.2 C h/3 " 23
8 C 12h C 6h2 C h3 " 8
lim
D lim
h
h
h!0
h!0
D lim .12 C 6h C h2 / D 12:
1
"
".3=2/ C h
lim
h
h!0
lim
30.
x!1
31.
f .x/ D x 3 " 4x 2 C 1 is continuous on the whole real line
and so is discontinuous nowhere.
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INSTRUCTOR’S SOLUTIONS MANUAL
CHALLENGING PROBLEMS 1 (PAGE 94)
x
is continuous everywhere on its domain,
xC1
which consists of all real numbers except x D "1. It is
discontinuous nowhere.
'
x 2 if x > 2
is defined everywhere and disf .x/ D
x
if x $ 2
continuous at x D 2, where it is, however, left continuous
since limx!2! f .x/ D 2 D f .2/.
'
x 2 if x > 1
is defined and continuous evf .x/ D
x
if x $ 1
erywhere, and so discontinuous nowhere. Observe that
limx!1! f .x/ D 1 D limx!1C f .x/.
n
1 if x # 1
f .x/ D H.x " 1/ D
is defined everywhere
0 if x < 1
and discontinuous at x D 1 where it is, however, right
continuous.
n
1 if "3 $ x $ 3
is defined
f .x/ D H.9 " x 2 / D
0 if x < "3 or x > 3
everywhere and discontinuous at x D ˙3. It is right
continuous at "3 and left continuous at 3.
32. f .x/ D
33.
34.
35.
36.
37. f .x/ D jxj C jx C 1j is defined and continuous everywhere.
It is discontinuous nowhere.
n
jxj=jx C 1j if x ¤ "1
38. f .x/ D
is defined everywhere
1
if x D "1
and discontinuous at x D "1 where it is neither left nor
right continuous since limx!!1 f .x/ D 1, while
f ."1/ D 1.
Challenging Problems 1
1.
3
The instantaneous rate of change of x at x D c is
.c C h/3 " c 3
3c 2 h C 3ch2 C h3
D lim
D 3c 2 :
h
h
h!0
h!0
lim
2
2
2
If c D .a2 C ab C b 2 /=3, then 3c D a C ab C b , so
the average rate ofpchange over Œa; b! is the instantaneous
rate of change
at .a2 C ab C b 2 /=3.
p
2
Claim: .a C ab C b 2 /=3 > .a C b/=2.
Proof: Since a2 " 2ab C b 2 D .a " b/2 > 0, we have
4a2 C 4ab C 4b 2 > 3a2 C 6ab C 3b 2
s
a2 C ab C b 2
a2 C 2ab C b 2
>
D
3
4
a2 C ab C b 2
aCb
>
:
3
2
#
aCb
2
$2
For x near 0 we have jx " 1j D 1 " x and jx C 1j D x C 1.
Thus
lim
x!0
3.
x
x
1
D lim
D" :
x!0 .1 " x/ " .x C 1/
jx " 1j " jx C 1j
2
For x near 3 we have j5 " 2xj D 2x " 5, jx " 2j D x " 2,
jx " 5j D 5 " x, and j3x " 7j D 3x " 7. Thus
lim
x!3
4.
2x " 5 " .x " 2/
j5 " 2xj " jx " 2j
D lim
x!3
jx " 5j " j3x " 7j
5 " x " .3x " 7/
x"3
1
D lim
D" :
x!3 4.3 " x/
4
Let y D x 1=6 . Then we have
y2 " 4
x 1=3 " 4
D
lim
y!2 y 3 " 8
x!64 x 1=2 " 8
.y " 2/.y C 2/
D lim
y!2 .y " 2/.y 2 C 2y C 4/
4
1
y C2
D
D :
D lim 2
y!2 y C 2y C 4
12
3
lim
5.
Use a " b D
have
a2
a3 " b 3
to handle the denominator. We
C ab C b 2
p
3Cx"2
lim p
x!1 3 7 C x " 2
3Cx"4
.7 C x/2=3 C 2.7 C x/1=3 C 4
D lim p
&
x!1
.7 C x/ " 8
3CxC2
.7 C x/2=3 C 2.7 C x/1=3 C 4
4C4C4
D 3:
D lim
p
D
x!1
2C2
3CxC2
(page 94)
Let 0 < a < b. The average rate of change of x 3 over
Œa; b! is
b 3 " a3
D b 2 C ab C a2 :
b"a
p
2.
6.
"1 C
p
1Ca
, r! .a/ D
"1 "
p
1Ca
.
a
a
a) lima!0 r! .a/ does not exist. Observe that the right
limit is "1 and the left limit is 1.
rC .a/ D
b) From the following table it appears that
lima!0 rC .a/ D 1=2, the solution of the linear equation 2x " 1 D 0 which results from setting a D 0 in
the quadratic equation ax 2 C 2x " 1 D 0.
a
rC .a/
1
0:1
"0:1
0:01
"0:01
0:001
"0:001
0:41421
0:48810
0:51317
0:49876
0:50126
0:49988
0:50013
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CHALLENGING PROBLEMS 1 (PAGE 94)
p
1Ca"1
a!0
a
.1 C a/ " 1
D lim p
a!0 a. 1 C a C 1/
1
1
D :
D lim p
a!0
2
1CaC1
8.
c) lim rC .a/ D lim
a!0
7.
ADAMS and ESSEX: CALCULUS 9
TRUE or FALSE
a) If limx!a f .x/ exists
and limx!a
%
& g.x/ does not exist, then limx!a f .x/ C g.x/ does not exist.
%
&
TRUE, because if limx!a f .x/ C g.x/ were to exist
then
%
&
lim g.x/ D lim f .x/ C g.x/ " f .x/
x!a
x!a
%
&
D lim f .x/ C g.x/ " lim f .x/
x!a
9.
x!a
would also exist.
b) If neither
% limx!a f .x/
& nor limx!a g.x/ exists, then
limx!a f .x/ C g.x/ does not exist.
FALSE. Neither
limx!0 1=x &nor limx!0 ."1=x/ exist,
%
but limx!0 .1=x/ C ."1=x/ D limx!0 0 D 0 exists.
c) If f is continuous at a, then so is jf j.
TRUE. For any two real numbers u and v we have
ˇ
ˇ
ˇ
ˇ
ˇjuj " jvjˇ $ ju " vj:
a) To be proved: if f is a continuous function defined
on a closed interval Œa; b!, then the range of f is a
closed interval.
Proof: By the Max-Min Theorem there exist numbers
u and v in Œa; b! such that f .u/ $ f .x/ $ f .v/ for
all x in Œa; b!. By the Intermediate-Value Theorem,
f .x/ takes on all values between f .u/ and f .v/ at
values of x between u and v, and hence at points of
Œa; b!. Thus the range of f is Œf .u/; f .v/!, a closed
interval.
b) If the domain of the continuous function f is an
open interval, the range of f can be any interval
(open, closed, half open, finite, or infinite).
n
x2 " 1
"1 if "1 < x < 1
D
.
f .x/ D 2
1
if x < "1 or x > 1
jx " 1j
f is continuous wherever it is defined, that is at all points
except x D ˙1. f has left and right limits "1 and 1,
respectively, at x D 1, and has left and right limits 1 and
"1, respectively, at x D "1. It is not, however, discontinuous at any point, since "1 and 1 are not in its domain.
1
1
1
" D
D 1 !1
!
"2 .
2
1
x " x2
" x " 12
4 " 4 "xCx
4
Observe that f .x/ # f .1=2/ D 4 for all x in .0; 1/.
10. f .x/ D
11.
Suppose f is continuous on Œ0; 1! and f .0/ D f .1/.
a) To be proved: f .a/ D f .a C 12 / for some a in Œ0; 21 !.
Proof: If f .1=2/ D f .0/ we can take a D 0 and be
done. If not, let
This follows from
g.x/ D f .x C 21 / " f .x/:
juj D ju " v C vj $ ju " vj C jvj; and
jvj D jv " u C uj $ jv " uj C juj D ju " vj C juj:
Now we have
ˇ
ˇ
ˇ
ˇ
ˇjf .x/j " jf .a/jˇ $ jf .x/ " f .a/j
so the left side approaches zero whenever the right
side does. This happens when x ! a by the continuity of f at a.
d) If jf j is continuous at a, then so
' is f .
"1 if x < 0
FALSE. The function f .x/ D
is
1
if x # 0
discontinuous at x D 0, but jf .x/j D 1 everywhere,
and so is continuous at x D 0.
e) If f .x/ < g.x/ in an interval around a and if
limx!a f .x/ D L and limx!a g.x/ D M both
exist, then L < M .
'
2
if x ¤ 0 and let
FALSE. Let g.x/ D x
1
if x D 0
f .x/ D "g.x/. Then f .x/ < g.x/ for all x, but
limx!0 f .x/ D 0 D limx!0 g.x/. (Note: under the
given conditions, it is TRUE that L $ M , but not
necessarily true that L < M .)
38
Then g.0/ ¤ 0 and
g.1=2/ D f .1/ " f .1=2/ D f .0/ " f .1=2/ D "g.0/:
Since g is continuous and has opposite signs at x D 0
and x D 1=2, the Intermediate-Value Theorem assures
us that there exists a between 0 and 1/2 such that
g.a/ D 0, that is, f .a/ D f .a C 12 /.
b) To be proved: if n > 2 is an integer, then
f .a/ D f .a C n1 / for some a in Œ0; 1 " n1 !.
Proof: Let g.x/ D f .x C n1 / " f .x/. Consider
the numbers x D 0, x D 1=n, x D 2=n, : : : ,
x D .n " 1/=n. If g.x/ D 0 for any of these numbers, then we can let a be that number. Otherwise,
g.x/ ¤ 0 at any of these numbers. Suppose that the
values of g at all these numbers has the same sign
(say positive). Then we have
2
f .1/ > f . n!1
n / > !!! > f.n/ >
1
n
> f .0/;
which is a contradiction, since f .0/ D f .1/. Therefore there exists j in the set f0; 1; 2; : : : ; n " 1g such
that g.j=n/ and g..j C 1/=n/ have opposite sign. By
the Intermediate-Value Theorem, g.a/ D 0 for some a
between j=n and .j C 1/=n, which we had to prove.
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INSTRUCTOR’S SOLUTIONS MANUAL
CHAPTER 2.
SECTION 2.1 (PAGE 100)
DIFFERENTIATION
7.
Slope of y D
x C 1 at x D 3 is
p
p
4Ch"2
4ChC2
!p
m D lim
h
h!0
4ChC2
4Ch"4
D lim !p
"
h!0 h
hChC2
1
1
D :
D lim p
4
h!0
4ChC2
Section 2.1 Tangent Lines and Their Slopes
(page 100)
1.
Slope of y D 3x " 1 at .1; 2/ is
m D lim
h!0
3.1 C h/ " 1 " .3 & 1 " 1/
3h
D lim
D 3:
h
h!0 h
The tangent line is y " 2 D 3.x " 1/, or y D 3x " 1. (The
tangent to a straight line at any point on it is the same
straight line.)
Tangent line is y " 2 D
8.
!
1
1
1
m D lim
"
p
h!0 h
9Ch 3
p
p
3" 9Ch 3C 9Ch
D lim
p
!
p
h!0 3h 9 C h
3C 9Ch
9"9"h
D lim
p
p
h!0 3h 9 C h.3 C
9 C h/
1
1
D"
D" :
3.3/.6/
54
3. Slope of y D 2x 2 " 5 at .2; 3/ is
2.2 C h/2 " 5 " .2.22 / " 5/
h
h!0
8 C 8h C 2h2 " 8
D lim
h
h!0
D lim .8 C 2h/ D 8
m D lim
The tangent line at .9; 13 / is y D
1
y D 12 " 54
x.
h!0
9.
Slope of y D
6 " ."2 C h/ " ."2 C h/2 " 4
h
h!0
3h " h2
D lim
D lim .3 " h/ D 3:
h
h!0
h!0
m D lim
5. Slope of y D x 3 C 8 at x D "2 is
."2 C h/3 C 8 " ."8 C 8/
h
h!0
"8 C 12h " 6h2 C h3 C 8 " 0
D lim
h
h!0
!
"
D lim 12 " 6h C h2 D 12
m D lim
h!0
Tangent line is y " 0 D 12.x C 2/ or y D 12x C 24.
6. The slope of y D
m D lim
h!0
x2
1
"h
1
" 1 D lim 2
D 0:
2
h h C1
h!0 h C 1
The tangent line at .0; 1/ is y D 1.
"
1
.x
54
" 9/, or
2x
at x D 2 is
xC2
1
Tangent line is y " 1 D .x " 2/,
4
or x " 4y D "2.
p
10. The slope of y D 5 " x 2 at x D 1 is
p
5 " .1 C h/2 " 2
m D lim
h
h!0
5 " .1 C h/2 " 4
&
D lim %p
h!0 h
5 " .1 C h/2 C 2
"2 " h
1
D"
D lim p
2
2
h!0
5 " .1 C h/ C 2
1
at .0; 1/ is
C1
!
1
3
2.2 C h/
"1
2ChC2
m D lim
h
h!0
4 C 2h " 2 " h " 2
D lim
h.2 C h C 2/
h!0
1
h
D :
D lim
4
h!0 h.4 C h/
4. The slope of y D 6 " x " x 2 at x D "2 is
The tangent line at ."2; 4/ is y D 3x C 10.
1
.x " 3/, or x " 4y D "5.
4
1
The slope of y D p at x D 9 is
x
2. Since y D x=2 is a straight line, its tangent at any point
.a; a=2/ on it is the same line y D x=2.
Tangent line is y " 3 D 8.x " 2/ or y D 8x " 13.
p
The tangent line at .1; 2/ is y D 2 " 21 .x " 1/, or
y D 52 " 12 x.
11.
Slope of y D x 2 at x D x0 is
m D lim
h!0
.x0 C h/2 " x02
2x0 h C h2
D lim
D 2x0 :
h
h
h!0
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SECTION 2.1 (PAGE 100)
ADAMS and ESSEX: CALCULUS 9
Tangent line is y " x02 D 2x0 .x " x0 /,
or y D 2x0 x " x02 .
19.
1
at .a; a1 / is
12. The slope of y D
x
!
1
1
a"a"h
1
1
C
D lim
D " 2.
m D lim
a
a
h!0 h.a C h/.a/
h!0 h a C h
1
1
1
The tangent line at .a; / is y D " 2 .x " a/, or
a
a a
x
2
y D " 2.
a a
p
j0 C hj " 0
1
D lim
does not
13. Since limh!0
h
h!0 jhjsgn .h/
p
exist (and is not 1 or "1), the graph of f .x/ D jxj
has no tangent at x D 0.
14.
a)
Slope of y D x 3 at x D a is
.a C h/3 " a3
h
h!0
a3 C 3a2 h C 3ah2 C h3 " a3
D lim
h
h!0
2
D lim .3a C 3ah C h2 / D 3a2
m D lim
h!0
b) We have m D 3 if 3a2 D 3, i.e., if a D ˙1.
Lines of slope 3 tangent to y D x 3 are
y D 1 C 3.x " 1/ and y D "1 C 3.x C 1/, or
y D 3x " 2 and y D 3x C 2.
20.
The slope of y D x 3 " 3x at x D a is
i
1h
.a C h/3 " 3.a C h/ " .a3 " 3a/
h!0 h
i
1h 3
a C 3a2 h C 3ah2 C h3 " 3a " 3h " a3 C 3a
D lim
h!0 h
D lim Œ3a2 C 3ah C h2 " 3! D 3a2 " 3:
m D lim
The slope of f .x/ D .x " 1/4=3 at x D 1 is
.1 C h " 1/4=3 " 0
D lim h1=3 D 0:
h
h!0
h!0
m D lim
h!0
The graph of f has a tangent line with slope 0 at x D 1.
Since f .1/ D 0, the tangent has equation y D 0
15.
At points where the tangent line is parallel to the x-axis,
the slope is zero, so such points must satisfy 3a2 " 3 D 0.
Thus, a D ˙1. Hence, the tangent line is parallel to the
x-axis at the points .1; "2/ and ."1; 2/.
The slope of f .x/ D .x C 2/3=5 at x D "2 is
m D lim
h!0
."2 C h C 2/3=5 " 0
D lim h!2=5 D 1:
h
h!0
21.
.a C h/3 " .a C h/ C 1 " .a3 " a C 1/
h
h!0
3a2 h C 3ah2 C a3 " h
D lim
h
h!0
D lim .3a2 C 3ah C h2 " 1/ D 3a2 " 1:
m D lim
The graph of f has vertical tangent x D "2 at x D "2.
16. The slope of f .x/ D jx 2 " 1j at x D 1 is
j.1 C h/2 " 1j " j1 " 1j
j2h C h2 j
m D limh!0
D lim
,
h
h
h!0
which does not exist, and is not "1 or 1. The graph of
f has no tangent at x D 1.
'p
x
if x # 0
p
17. If f .x/ D
, then
" "x if x < 0
p
h
f .0 C h/ " f .0/
D lim
D1
lim
h
h!0C h
h!0C
p
f .0 C h/ " f .0/
" "h
lim
D lim
D1
h
h
h!0!
h!0!
h!0
The tangent at x D a is parallel to the line y D 2x C 5 if
3a2 " 1 D 2, that is, if a D ˙1. The corresponding points
on the curve are ."1; 1/ and .1; 1/.
22.
The tangent at x D a is perpendicular to the line
y D 4x " 3 if "1=a2 D "1=4, that is, if a D ˙2.
The corresponding points on the curve are ."2; "1=2/ and
.2; 1=2/.
18. The slope of y D x 2 " 1 at x D x0 is
Œ.x0 C h/2 " 1! " .x02 " 1/
h
h!0
2x0 h C h2
D lim
D 2x0 :
h
h!0
If m D "3, then x0 D " 32 . The tangent line with slope
m D "3 at ." 32 ; 54 / is y D 54 " 3.x C 23 /, that is,
y D "3x " 13
4 .
40
The slope of the curve y D 1=x at x D a is
1
1
"
a " .a C h/
1
D " 2:
m D lim a C h a D lim
h
a
h!0 ah.a C h/
h!0
Thus the graph of f has a vertical tangent x D 0.
m D lim
The slope of the curve y D x 3 " x C 1 at x D a is
23.
The slope of the curve y D x 2 at x D a is
m D lim
h!0
.a C h/2 " a2
D lim .2a C h/ D 2a:
h
h!0
The normal at x D a has slope "1=.2a/, and has equation
y " a2 D "
1
.x " a/;
2a
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x
1
C y D C a2 :
2a
2
lOMoARcPSD|6566483
INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 2.1 (PAGE 100)
y
This is the line x C y D k if 2a D 1, and so
k D .1=2/ C .1=2/2 D 3=4.
24.
2
The curves y D kx 2 and y D k.x " 2/2 intersect at .1; k/.
The slope of y D kx 2 at x D 1 is
m1 D lim
h!0
1
-3
k.1 C h/2 " k
D lim .2 C h/k D 2k:
h
h!0
-2
-1
-2
The slope of y D k.x " 2/ at x D 1 is
h!0
x
y D jx 2 " 1j " x
-3
Fig. 2.1-27
k.2 " .1 C h//2 " k
D lim ."2 C h/k D "2k:
h
h!0
The two curves intersect at right angles if
2k D "1=."2k/, that is, if 4k 2 D 1, which is satisfied
if k D ˙1=2.
2
-1
2
m2 D lim
1
28.
25. Horizontal tangents at .0; 0/, .3; 108/, and .5; 0/.
y
.3; 108/
Horizontal tangent at .a; 2/ and ."a; "2/ for all a > 1.
No tangents at .1; 2/ and ."1; "2/.
y
y D jx C 1j " jx " 1j
2
1
100
80
-3
-2
-1
60
1
2
x
-1
40
-2
y D x 3 .5 " x/2
20
-1
1
2
3
4
5
-3
Fig. 2.1-28
x
-20
Fig. 2.1-25
29.
26. Horizontal tangent at ."1; 8/ and .2; "19/.
y
Horizontal tangent at .0; "1/. The tangents at .˙1; 0/ are
vertical.
y
y D .x 2 " 1/1=3 2
20
."1; 8/ 10
-2
-1
y D 2x 3 " 3x 2 " 12x C 1
1
2
3
1
-3
x
-2
-1
1
2
x
-1
-10
-2
-20
.2; "19/
-3
Fig. 2.1-29
-30
Fig. 2.1-26
27.
Horizontal tangent at ."1=2; 5=4/. No tangents at ."1; 1/
and .1; "1/.
30.
Horizontal tangent at .0; 1/. No tangents at ."1; 0/ and
.1; 0/.
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SECTION 2.1 (PAGE 100)
ADAMS and ESSEX: CALCULUS 9
y
2.
y
y D ..x 2 " 1/2 /1=3
2
1
y D g 0 .x/
-1
-2
1
2
x
x
Fig. 2.1-30
31. The graph of the function f .x/ D x 2=3 (see Figure 2.1.7
in the text) has a cusp at the origin O, so does not have
a tangent line there. However, the angle between OP and
the positive y-axis does ! 0 as P approaches 0 along the
graph. Thus the answer is NO.
3.
y
y D h0 .x/
x
32. The slope of P .x/ at x D a is
m D lim
h!0
P .a C h/ " P .a/
:
h
Since P .a C h/ D a0 C a1 h C a2 h2 C ! ! ! C an hn and
P .a/ D a0 , the slope is
a0 C a1 h C a2 h2 C ! ! ! C an hn " a0
h
h!0
D lim a1 C a2 h C ! ! ! C an hn!1 D a1 :
m D lim
4.
y
h!0
Thus the line y D `.x/ D m.x " a/ C b is tangent to
y D P .x/ at x D a if and only if m D a1 and b D a0 ,
that is, if and only if
x
P .x/"`.x/ D a2 .x " a/2 C a3 .x " a/3 C ! ! ! C an .x " a/n
h
i
D .x " a/2 a2 C a3 .x " a/ C ! ! ! C an .x " a/n!2
y D k 0 .x/
D .x " a/2 Q.x/
where Q is a polynomial.
Section 2.2 The Derivative
(page 107)
5.
Assuming the tick marks are spaced 1 unit apart, the function f is differentiable on the intervals ."2; "1/, ."1; 1/,
and .1; 2/.
6.
Assuming the tick marks are spaced 1 unit apart, the function g is differentiable on the intervals ."2; "1/, ."1; 0/,
.0; 1/, and .1; 2/.
7.
y D f .x/ has its minimum at x D 3=2 where f 0 .x/ D 0
1.
y
y D f 0 .x/
x
42
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 2.2 (PAGE 107)
y
y
y D f .x/ D jx 3 " 1j
y D f .x/ D 3x " x 2 " 1
x
x
y
y D f 0 .x/
y
y D f 0 .x/
x
x
Fig. 2.2-7
Fig. 2.2-9
8. y D f .x/ has horizontal tangents at the points near 1=2
and 3=2 where f 0 .x/ D 0
y
10. y D f .x/ is constant on the intervals ."1; "2/, ."1; 1/,
and .2; 1/. It is not differentiable at x D ˙2 and
x D ˙1.
y
y D f .x/ D jx 2 " 1j " jx 2 " 4j
x
x
3
2
y D f .x/ D x " 3x C 2x C 1
y
y D f 0 .x/
y
x
x
y D f 0 .x/
Fig. 2.2-10
Fig. 2.2-8
11.
y D x 2 " 3x
.x C h/2 " 3.x C h/ " .x 2 " 3x/
h
h!0
2xh C h2 " 3h
D lim
D 2x " 3
h
h!0
dy D .2x " 3/ dx
y 0 D lim
9. y D f .x/ fails to be differentiable at x D "1, x D 0,
and x D 1. It has horizontal tangents at two points, one
between "1 and 0 and the other between 0 and 1.
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SECTION 2.2 (PAGE 107)
12.
f .x/ D 1 C 4x " 5x 2
ADAMS and ESSEX: CALCULUS 9
17.
2
2
1 C 4.x C h/ " 5.x C h/ " .1 C 4x " 5x /
h
4h " 10xh " 5h2
D lim
D 4 " 10x
h
h!0
df .x/ D .4 " 10x/ dx
f 0 .x/ D lim
h!0
13.
2
D lim p
p
h!0
2.t C h/ C 1 C 2t C 1
1
D p
2t C 1
1
dF .t / D p
dt
2t C 1
f .x/ D x 3
.x C h/3 " x 3
h
h!0
3x 2 h C 3xh2 C h3
D lim
D 3x 2
h
h!0
df .x/ D 3x 2 dx
f 0 .x/ D lim
14.
15.
16.
18.
1
3 C 4t (
)
ds
1
1
1
D lim
"
dt
3 C 4t
h!0 h 3 C 4.t C h/
4
3 C 4t " 3 " 4t " 4h
D"
D lim
.3 C 4t /2
h!0 h.3 C 4t /Œ3 C .4t C h/!
4
ds D "
dt
.3 C 4t /2
sD
2"x
g.x/ D
2Cx
2 " .x C h/ 2 " x
"
2CxCh
2Cx
0
g .x/ D lim
h
h!0
.2 " x " h/.2 C x/ " .2 C x C h/.2 " x/
D lim
h.2 C x C h/.2 C x/
h!0
4
D"
.2 C x/2
4
dg.x/ D "
dx
.2 C x/2
y D 31 x 3 " x
i
1h1
3
1 3
y 0 D lim
3 .x C h/ " .x C h/ " . 3 x " x/
h!0 h
&
1% 2
x h C xh2 C 31 h3 " h
D lim
h!0 h
D lim .x 2 C xh C 31 h2 " 1/ D x 2 " 1
h!0
dy D .x 2 " 1/ dx
44
p
2t C 1
p
p
2.t C h/ C 1 " 2t C 1
0
F .t / D lim
h
h!0
2t C 2h C 1 " 2t " 1
&
D lim %p
p
h!0 h
2.t C h/ C 1 C 2t C 1
F .t/ D
f .x/ D
3
4
p
2"x
p
p
3
2 " .x C h/ " 43 2 " x
0
4
f .x/ D lim
h
h!0
"
#
3
2"x"h"2Cx
D lim
p
p
h!0 4 h. 2 " .x C h/ C
2 " x/
3
D" p
8 2"x
3
df .x/ D " p
dx
8 2"x
19.
yDxC
1
x
1
1
"x"
x
C
h
x
y D lim
h
h!0
$
#
x"x"h
D lim 1 C
h.x C h/x
h!0
"1
1
D 1 C lim
D1" 2
x
h!0 .x C h/x
$
#
1
dy D 1 " 2 dx
x
0
20.
xChC
s
1Cs (
)
dz
1
sCh
s
D lim
"
ds
1Cs
h!0 h 1 C s C h
.s C h/.1 C s/ " s.1 C s C h/
1
D lim
D
h.1 C s/.1 C s C h/
.1 C s/2
h!0
1
dz D
ds
.1 C s/2
zD
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INSTRUCTOR’S SOLUTIONS MANUAL
F .x/ D p
21.
SECTION 2.2 (PAGE 107)
1
25.
1 C x2
1
1
p
"p
2
1 C x2
1 C .x C h/
F 0 .x/ D lim
h
h!0
p
26.
p
1 C x 2 " 1 C .x C h/2
D lim p
p
h!0 h 1 C .x C h/2 1 C x 2
1 C x 2 " 1 " x 2 " 2hx " h2
%p
& 27.
D lim p
p
p
h!0 h 1 C .x C h/2 1 C x 2
1 C x 2 C 1 C .x C h/2
"2x
x
D"
2.1 C x 2 /3=2
.1 C x 2 /3=2
x
dF .x/ D "
dx
.1 C x 2 /3=2
D
22.
yD
1
x2
(
28.
23.
yD p
1
"p
p
1Cx
1CxCh
y 0 .x/ D lim
h
h!0
p
p
1Cx" 1CxCh
D lim p
p
h!0 h 1 C x C h 1 C x
1Cx"1"x"h
p
D lim p
!p
"
p
h!0 h 1 C x C h 1 C x
1CxC 1CxCh
1
p
D lim " p
!p
"
p
h!0
1CxCh 1Cx
1CxC 1CxCh
1
D"
2.1 C x/3=2
1
dy D "
dx
2.1 C x/3=2
24.
y D x 3 " 2x
f .x/ " f .1/
x"1
0:71000
0:97010
0:99700
0:99970
0:9
0:99
0:999
0:9999
ˇ
ˇ
d 3
ˇ
.x " 2x/ˇ
ˇ
dx
1
1Cx
h.x/ D jx 2 C 3x C 2j fails to be differentiable where
x 2 C 3x C 2 D 0, that is, at x D "2 and x D "1. Note:
both of these are single zeros of x 2 C 3x C 2. If they
were higher order zeros (i.e. if .x C 2/n or .x C 1/n were
a factor of x 2 C 3x C 2 for some integer n # 2) then h
would be differentiable at the corresponding point.
x
)
1
1
1
" 2
x
h!0 h .x C h/2
x 2 " .x C h/2
2
D lim
D" 3
x
h!0 hx 2 .x C h/2
2
dy D " 3 dx
x
y 0 D lim
Since f .x/ D x sgn x D jxj, for x ¤ 0, f will become
continuous at x D 0 if we define f .0/ D 0. However,
f will still not be differentiable at x D 0 since jxj is not
differentiable at x D 0.
'
2
Since g.x/ D x 2 sgn x D xjxj D x 2 if x > 0 , g will
"x
if x < 0
become continuous and differentiable at x D 0 if we define
g.0/ D 0.
1
t2 " 3
t2 C 3
#
$
t2 " 3
1 .t C h/2 " 3
"
f 0 .t / D lim
.t C h/2 C 3 t 2 C 3
h!0 h
Œ.t C h/2 " 3!.t 2 C 3/ " .t 2 " 3/Œ.t C h/2 C 3!
D lim
h.t 2 C 3/Œ.t C h/2 C 3!
h!0
12t h C 6h2
12t
D lim
D 2
2
.t C 3/2
h!0 h.t C 3/Œ.t C h/2 C 3!
12t
df .t / D 2
dt
.t C 3/2
xD1
x
1:1
1:01
1:001
1:0001
D lim
h!0
f .x/ " f .1/
x"1
1:31000
1:03010
1:00300
1:00030
.1 C h/3 " 2.1 C h/ " ."1/
h
h C 3h2 C h3
h
h!0
D lim 1 C 3h C h2 D 1
D lim
h!0
29.
f .x/ D 1=x
f .x/ " f .2/
x
x"2
"0:26316
"0:25126
"0:25013
"0:25001
1:9
1:99
1:999
1:9999
x
2:1
2:01
2:001
2:0001
f .x/ " f .2/
x"2
"0:23810
"0:24876
"0:24988
"0:24999
1
"2
2 " .2 C h/
2Ch
f .2/ D lim
D lim
h
h!0
h!0 h.2 C h/2
1
1
D lim "
D"
4
h!0 .2 C h/2
0
f .t / D
30.
The slope of y D 5 C 4x " x 2 at x D 2 is
ˇ
5 C 4.2 C h/ " .2 C h/2 " 9
dy ˇˇ
D lim
ˇ
dx ˇ
h
h!0
xD2
D lim
h!0
"h2
D 0:
h
Thus, the tangent line at x D 2 has the equation y D 9.
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SECTION 2.2 (PAGE 107)
31. y D
ADAMS and ESSEX: CALCULUS 9
p
x C 6. Slope at .3; 3/ is
p
1
9Ch"3
9Ch"9
m D lim
D lim !p
" D :
h
6
h!0
h!0 h
9ChC3
Tangent line is y " 3 D
44.
yD
2
t2 C t
45.
xDa
46.
39.
40.
41.
42.
43.
1
dy
D x !2=3 for x ¤ 0
dx
3
dy
1
D " x !4=3 for x ¤ 0
dx
3
d !2:25
t
D "2:25t !3:25 for t > 0
dt
d 119=4
119 115=4
s
D
s
for s > 0
ds
4
ˇ
ˇ
1
d p ˇˇ
1 ˇˇ
D :
sˇ
D p ˇ
ˇ
ds
6
2 sˇ
sD9
sD9
# $
1
1
1
D "16
F .x/ D ; F 0 .x/ D " 2 ; F 0
x
x
4
ˇ
2 !5=3 ˇˇ
1
0
f .8/ D " x
D"
ˇ
ˇ
3
48
xD8
ˇ
ˇ
ˇ
dy ˇˇ
1
1
ˇ
D t !3=4 ˇ
D p
ˇ
ˇ
dt ˇ
4
8 2
tD4
46
1
.
a2
tD4
The intersection points of y D x 2 and x C 4y D 18 satisfy
4x 2 C x " 18 D 0
.4x C 9/.x " 2/ D 0:
Therefore x D " 94 or x D 2.
dy
The slope of y D x 2 is m1 D
D 2x.
dx
9
9
At x D " , m1 D " . At x D 2, m1 D 4.
4
2
The slope of x C 4y D 18, i.e. y D " 14 x C 18
4 , is
m2 D " 14 .
Thus, at x D 2, the product of these slopes is
.4/." 14 / D "1. So, the curve and line intersect at right
angles at that point.
35. g 0 .t / D 22t 21 for all t
38.
D
1
Normal has slope a2 , and equation y " D a2 .x " a/,
a
1
or y D a2 x " a3 C
a
34. f 0 .x/ D "17x !18 for x ¤ 0
37.
ˇ
1 ˇˇ
1
at x D a is " 2 ˇ
Slope of y D
x
x ˇ
Slope at t D a is
2
2
"
.a C h/2 C .a C h/ a2 C a
m D lim
h
h!0
2.a2 C a " a2 " 2ah " h2 " a " h/
D lim
hŒ.a C h/2 C a C h!.a2 C a/
h!0
"4a " 2h " 2
D lim
h!0 Œ.a C h/2 C a C h!.a2 C a/
4a C 2
D" 2
.a C a/2
2.2a C 1/
2
"
.t " a/
Tangent line is y D 2
a C a .a2 C a/2
36.
1
D p :
2 x0
Thus, the equation of the tangent line is
x C x0
1
p
y D x0 C p .x " x0 /, that is, y D p .
2 x0
2 x0
Thus, the tangent line has the equation
y D "1 " 23 .t C 2/, that is, y D " 32 t " 4.
33.
x at x D x0 is
xDx0
tD!2
3
"2 C h C Œ."2 C h/2 " 2!
D" :
D lim
hŒ."2 C h/2 " 2!
2
h!0
p
ˇ
dy ˇˇ
ˇ
dx ˇ
1
.x " 3/, or x " 6y D "15.
6
t
32. The slope of y D 2
at t D "2 and y D "1 is
t "2
ˇ
(
)
dy ˇˇ
"2 C h
1
"
."1/
D
lim
ˇ
dt ˇ
h!0 h ."2 C h/2 " 2
The slope of y D
47.
Let theˇ point of tangency be .a; a2 /. Slope of tangent is
d 2 ˇˇ
x ˇ
D 2a
dx ˇ
xDa
This is the slope from .a; a2 / to .1; "3/, so
a2 C 3
D 2a, and
a"1
a2 C 3 D 2a2 " 2a
a2 " 2a " 3 D 0
a D 3 or " 1
The two tangent lines are
(for a D 3): y " 9 D 6.x " 3/ or 6x " 9
(for a D "1): y " 1 D "2.x C 1/ or y D "2x " 1
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 2.2 (PAGE 107)
p
p
4a2 " 4b
D a ˙ a2 " b.
Hence t D
2
p
If b < a2 , i.e. a2 " b > 0, then t D a ˙ a2 " b
has two real solutions. Therefore, there will be two distinct tangent %lines passing through
.a; b/ with equations
&
p
y D b C 2 a ˙ a2 " b .x " a/. If b D a2 , then
t D a. There will be only one tangent line with slope 2a
and equation y D b C 2a.x " a/.
If b > a2 , then a2 " b < 0. There will be no real solution
for t . Thus, there will be no tangent line.
y
2a ˙
.a;a2 /
y D x2
x
.1;!3/
Fig. 2.2-47
48. The slope of y D
1
at x D a is
x
ˇ
dy ˇˇ
ˇ
dx ˇ
xDa
D"
51.
Suppose f is odd: f ."x/ D "f .x/. Then
f ."x C h/ " f ."x/
f 0 ."x/ D lim
h
h!0
f .x " h/ " f .x/
D lim "
h
h!0
.let h D "k/
f .x C k/ " f .x/
D lim
D f 0 .x/
k
k!0
Thus f 0 is even.
Now suppose f is even: f ."x/ D f .x/. Then
f ."x C h/ " f ."x/
f 0 ."x/ D lim
h
h!0
f .x " h/ " f .x/
D lim
h
h!0
f .x C k/ " f .x/
D lim
"k
k!0
D "f 0 .x/
so f 0 is odd.
52.
Let f .x/ D x !n . Then
1
:
a2
1
1
D "2, or a D ˙ p . Therea2
2
fore, the equations
of the
are
#
$ two straight lines #
$
p
p
1
1
and y D " 2 " 2 x C p ,
y D 2"2 x" p
2
p 2
or y D "2x ˙ 2 2.
If the slope is "2, then "
p
49. Let the point of tangency be
ˇ .a; a/
d p ˇˇ
1
Slope of tangent is
xˇ
D p
ˇ
dx
2 a
xDa
p
a"0
1
, so a C 2 D 2a, and a D 2.
Thus p D
aC2
2 a
1
The required slope is p .
2 2
y
p
.a; a/
p
yD x
x
!2
Fig. 2.2-49
50. If aˇ line is tangent to y D x 2 at .t; t 2 /, then its slope is
dy ˇˇ
D 2t . If this line also passes through .a; b/, then
ˇ
dx ˇ
xDt
its slope satisfies
t2 " b
D 2t;
t "a
that is t 2 " 2at C b D 0:
.x C h/!n " x !n
h
h!0
$
#
1
1
1
"
D lim
.x C h/n x n
h!0 h
x n " .x C h/n
D lim
h!0 hx n .x C h/n
x " .x C h/
&
D lim
h!0 hx n ..x C h/n
#
$
x n!1 C x n!2 .x C h/ C ! ! ! C .x C h/n!1
f 0 .x/ D lim
D"
1
& nx n!1 D "nx !.nC1/ :
x 2n
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SECTION 2.2 (PAGE 107)
53.
ADAMS and ESSEX: CALCULUS 9
f .x/ D x 1=3
If f 0 .aC/ is finite, call the half-line with equation
y D f .a/ C f 0 .aC/.x " a/, (x # a), the right tangent line to the graph of f at x D a. Similarly, if f 0 .a"/
is finite, call the half-line y D f .a/ C f 0 .a"/.x " a/,
(x $ a), the left tangent line. If f 0 .aC/ D 1 (or "1),
the right tangent line is the half-line x D a, y # f .a/ (or
x D a, y $ f .a/). If f 0 .a"/ D 1 (or "1), the right
tangent line is the half-line x D a, y $ f .a/ (or x D a,
y # f .a/).
The graph has a tangent line at x D a if and only if
f 0 .aC/ D f 0 .a"/. (This includes the possibility that
both quantities may be C1 or both may be "1.) In this
case the right and left tangents are two opposite halves of
the same straight line. For f .x/ D x 2=3 , f 0 .x/ D 32 x !1=3 :
At .0; 0/, we have f 0 .0C/ D C1 and f 0 .0"/ D "1.
In this case both left and right tangents are the positive
y-axis, and the curve does not have a tangent line at the
origin.
For f .x/ D jxj, we have
.x C h/1=3 " x 1=3
h
h!0
.x C h/1=3 " x 1=3
D lim
h
h!0
.x C h/2=3 C .x C h/1=3 x 1=3 C x 2=3
&
.x C h/2=3 C .x C h/1=3 x 1=3 C x 2=3
xCh"x
D lim
2=3
h!0 hŒ.x C h/
C .x C h/1=3 x 1=3 C x 2=3 !
1
D lim
2=3
h!0 .x C h/
C .x C h/1=3 x 1=3 C x 2=3
1
1 !2=3
D
D x
3
3x 2=3
f 0 .x/ D lim
54. Let f .x/ D x 1=n . Then
.x C h/1=n " x 1=n
f .x/ D lim
h
h!0
a"b
D lim n
a!b a " b n
0
D lim
a!b an!1
an!2 b
f 0 .x/ D sgn .x/ D
(let x C h D an , x D b n )
1
C an!3 b 2 C ! ! ! C b n!1
d n
.x C h/n " x n
x D lim
dx
h
h!0
(
n.n " 1/ n!2 2
1 n n n!1
x C x
hC
x
h
D lim
1
1&2
h!0 h
)
n.n " 1/.n " 2/ n!3 3
C
x
h C ! ! ! C hn " x n
1&2&3
(
n.n " 1/ n!2
x
h
D lim nx n!1 C h
1&2
h!0
)!
n.n " 1/.n " 2/ n!3 2
n!1
C
x
h C !!! C h
1&2&3
D nx n!1
1.
y D 3x 2 " 5x " 7;
2.
y D 4x 1=2 "
3.
f .x/ D Ax 2 C Bx C C;
4.
f .x/ D
5.
zD
6.
y D x 45 " x !45
7.
8.
9.
f 0 .aC/ D lim
h!0C
48
if x > 0
if x < 0.
Section 2.3 Differentiation Rules
(page 115)
56. Let
f .a C h/ " f .a/
h
f .a C h/ " f .a/
0
f .a"/ D lim
h
h!0!
1
"1
At .0; 0/, f 0 .0C/ D 1, and f 0 .0"/ D "1. In this case
the right tangent is y D x, .x # 0/, and the left tangent is
y D "x, .x $ 0/. There is no tangent line.
C
1
1
D x .1=n/!1 :
D
nb n!1
n
55.
n
10.
5
;
x
y 0 D 6x " 5:
y 0 D 2x !1=2 C 5x !2
2
6
C 2 " 2;
3
x
x
s5 " s3
;
15
f 0 .x/ D 2Ax C B:
f 0 .x/ D "
18
4
" 3
4
x
x
dz
1
1
D s4 " s2:
dx
3
5
y 0 D 45x 44 C 45x !46
g.t / D t 1=3 C 2t 1=4 C 3t 1=5
1
3
1
g 0 .t / D t !2=3 C t !3=4 C t !4=5
3
2
5
p
2
3
y D 3 t 2 " p D 3t 2=3 " 2t !3=2
t3
dy
D 2t !1=3 C 3t !5=2
dt
3
5
u D x 5=3 " x !3=5
5
3
du
D x 2=3 C x !8=5
dx
F .x/ D .3x " 2/.1 " 5x/
F 0 .x/ D 3.1 " 5x/ C .3x " 2/."5/ D 13 " 30x
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INSTRUCTOR’S SOLUTIONS MANUAL
11.
#
$
p
x2
1
x 5"x"
D 5 x " x 3=2 " x 5=2
3
3
5
5
3p
y0 D p "
x " x 3=2
6
2 x 2
yD
p
12. g.t / D
13.
14.
15.
18.
19.
20.
21.
22.
1
;
2t " 3
g 0 .t / D "
1
x 2 C 5x
1
2x C 5
y0 D " 2
.2x C 5/ D " 2
.x C 5x/2
.x C 5x/2
yD
4
;
3"x
y0 D
24.
4
.3 " x/2
#
2 " #t
#
#2
f 0 .t / D "
."#/
D
.2 " # t /2
.2 " # t /2
f .t / D
2
;
1 " y2
23.
2
.2t " 3/2
yD
16. g.y/ D
17.
SECTION 2.3 (PAGE 115)
g 0 .y/ D
25.
4y
.1 " y 2 /2
1 " 4x 2
4
D x !3 "
x3
x
4x 2 " 3
f 0 .x/ D "3x !4 C 4x !2 D
x4
p
u u"3
D u!1=2 " 3u!2
g.u/ D
u2
p
1
12 " u u
g 0 .u/ D " u!3=2 C 6u!3 D
2
2u3
26.
f .x/ D
27.
p
2 C t C t2
D 2t !1=2 C t C t 3=2
p
t
1
3p
dy
3t 2 C t " 2
D "t !3=2 C p C
p
tD
dt
2
2 t
2t t
yD
x"1
z D 2=3 D x 1=3 " x !2=3
x
dz
1 !2=3 2 !5=3
xC2
D x
C x
D
dx
3
3
3x 5=3
x3 " 4
xC1
.x C 1/.3x 2 / " .x 3 " 4/.1/
0
f .x/ D
.x C 1/2
3
2x C 3x 2 C 4
D
.x C 1/2
f .x/ D
ax C b
cx C d
.cx C d /a " .ax C b/c
f 0 .x/ D
.cx C d /2
ad " bc
D
.cx C d /2
f .x/ D
t 2 C 7t " 8
t2 " t C 1
.t 2 " t C 1/.2t C 7/ " .t 2 C 7t " 8/.2t " 1/
F 0 .t / D
.t 2 " t C 1/2
2
"8t C 18t " 1
D
.t 2 " t C 1/2
F .t/ D
f .x/ D .1 C x/.1 C 2x/.1 C 3x/.1 C 4x/
f 0 .x/ D .1 C 2x/.1 C 3x/.1 C 4x/ C 2.1 C x/.1 C 3x/.1 C 4x/
C 3.1 C x/.1 C 2x/.1 C 4x/ C 4.1 C x/.1 C 2x/.1 C 3x/
OR
f .x/ D Œ.1 C x/.1 C 4x/! Œ.1 C 2x/.1 C 3x/!
D .1 C 5x C 4x 2 /.1 C 5x C 6x 2 /
D 1 C 10x C 25x 2 C 10x 2 .1 C 5x/ C 24x 4
D 1 C 10x C 35x 2 C 50x 3 C 24x 4
f 0 .x/ D 10 C 70x C 150x 2 C 96x 3
28.
3 " 4x
3 C 4x
.3 C 4x/."4/ " .3 " 4x/.4/
f 0 .x/ D
.3 C 4x/2
24
D"
.3 C 4x/2
t 2 C 2t
t2 " 1
2
.t
" 1/.2t C 2/ " .t 2 C 2t /.2t /
z0 D
.t 2 " 1/2
2
2.t C t C 1/
D"
.t 2 " 1/2
f .r/ D .r !2 C r !3 " 4/.r 2 C r 3 C 1/
f 0 .r/ D ."2r !3 " 3r !4 /.r 2 C r 3 C 1/
f .x/ D
zD
p
1C t
p
1" t
p
p
1
1
.1 " t / p " .1 C t /." p /
ds
2 t
2 t
D
p
dt
.1 " t /2
1
p
D p
t .1 " t /2
sD
or
C .r !2 C r !3 " 4/.2r C 3r 2 /
f .r/ D "2 C r !1 C r !2 C r !3 C r " 4r 2 " 4r 3
29.
f 0 .r/ D "r !2 " 2r !3 " 3r !4 C 1 " 8r " 12r 2
p
y D .x 2 C 4/. x C 1/.5x 2=3 " 2/
p
y 0 D 2x. x C 1/.5x 2=3 " 2/
1
C p .x 2 C 4/.5x 2=3 " 2/
2 x
p
10
C x !1=3 .x 2 C 4/. x C 1/
3
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49
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SECTION 2.3 (PAGE 115)
30.
yD
D
y0 D
D
D
31.
32.
yD
ADAMS and ESSEX: CALCULUS 9
.x 2 C 1/.x 3 C 2/
.x 2 C 2/.x 3 C 1/
x 5 C x 3 C 2x 2 C 2
x 5 C 2x 3 C x 2 C 2
.x 5 C 2x 3 C x 2 C 2/.5x 4 C 3x 2 C 4x/
.x 5 C 2x 3 C x 2 C 2/2
.x 5 C x 3 C 2x 2 C 2/.5x 4 C 6x 2 C 2x/
"
.x 5 C 2x 3 C x 2 C 2/2
7
6
2x " 3x " 3x 4 " 6x 2 C 4x
.x 5 C 2x 3 C x 2 C 2/2
7
2x " 3x 6 " 3x 4 " 6x 2 C 4x
.x 2 C 2/2 .x 3 C 1/2
x
D
3x 2 C x
C 2x C 1
1
3x C 1
.6x 2 C 2x C 1/.6x C 1/ " .3x 2 C x/.12x C 2/
y0 D
.6x 2 C 2x C 1/2
6x C 1
D
.6x 2 C 2x C 1/2
2x C
36.
37.
38.
39.
&
33.
34.
35.
d
dx
d
dx
#
#
x2
f .x/
f .x/
x2
$ˇˇ
ˇ
ˇ
ˇ
$ˇˇ
ˇ
ˇ
ˇ
xD2
ˇ
f .x/.2x/ " x 2 f 0 .x/ ˇˇ
D
ˇ
ˇ
Œf .x/!2
D
xD2
$ˇˇ
ˇ
ˇ
ˇ
xD2
ˇ
.x 2 C f .x//f 0 .x/ " f .x/.2x C f 0 .x// ˇˇ
D
ˇ
ˇ
.x 2 C f .x//2
xD2
18 " 14
1
.4 C f .2//f 0 .2/ " f .2/.4 C f 0 .2//
D
D
D
.4 C f .2//2
62
9
# 2
$
#
$ˇˇ
d
x "4
d
8
ˇ
jxD!2 D
1" 2
ˇ
dx x 2 C 4
dx
x C4 ˇ
xD!2
ˇ
ˇ
8
ˇ
D 2
.2x/ˇ
ˇ
.x C 4/2
D"
p #ˇ
t .1 C t / ˇˇ
ˇ
ˇ
5"t
ˇ
"
#tD4
3=2 ˇ
d t Ct
ˇ
D
ˇ
ˇ
dt
5"t
d
dt
"
1
32
D"
64
2
tD4
.1/.4/ " .12/."1/
D
D 16
.1/2
p
x
f .x/ D
xC1
p
1
.x C 1/ p " x.1/
2 x
f 0 .x/ D
.x C 1/2
p
3
p " 2
1
2 2
f 0 .2/ D
D" p
9
18 2
ˇ
ˇ
d
Œ.1 C t /.1 C 2t /.1 C 3t /.1 C 4t /!ˇˇ
dt
tD0
D .1/.1 C 2t /.1 C 3t /.1 C 4t / C .1 C t /.2/.1 C 3t /.1 C 4t /C
ˇ
ˇ
.1 C t /.1 C 2t /.3/.1 C 4t / C .1 C t /.1 C 2t /.1 C 3t /.4/ˇˇ
tD0
D 1 C 2 C 3 C 4 D 10
41. y D
2
2
p , y0 D " %
p &2
3"4 x
3"4 x
#
$
4
" p
2 x
8
D4
."1/2 2
Tangent line has the equation y D "2 C 4.x " 1/ or
y D 4x " 6
Slope of tangent at .1; "2/ is m D
xD2
4
1
4f 0 .2/ " 4f .2/
D
D
D
16
16
4
&ˇˇ
%
&ˇˇ
d % 2
2 0
ˇ
x f .x/ ˇ
D 2xf .x/ C x f .x/ ˇˇ
dx
xD2
xD2
D 4f .2/ C 4f 0 .2/ D 20
50
40.
xD2
4
4f .2/ " 4f 0 .2/
D " D "1
Œf .2/!2
4
ˇ
x 2 f 0 .x/ " 2xf .x/ ˇˇ
D
ˇ
ˇ
x4
f .x/
x 2 C f .x/
tD4
ˇ
.5 " t /.1 C 23 t 1=2 / " .t C t 3=2 /."1/ ˇˇ
D
ˇ
ˇ
.5 " t /2
f .x/ D
.3 C 2x/."1 " 4x C 3x 2 / " .2 " x " 2x 2 C x 3 /.2/
.3 C 2x/2
2
.2 " x/.1 " x /
D
2x 3=2 .3 C 2x/
!
1 4x 3 C 5x 2 " 12x " 7
C 1" p
.3 C 2x/2
x
#
xD!2
6x 2
p
. x " 1/.2 " x/.1 " x 2 /
p
x.3 C 2x/
!
1
2 " x " 2x 2 C x 3
D 1" p
!
3 C 2x
x
!
!
1 !3=2 2 " x " 2x 2 C x 3
1
0
f .x/ D
x
C 1" p
2
3 C 2x
x
d
dx
42.
For y D
xC1
we calculate
x"1
y0 D
.x " 1/.1/ " .x C 1/.1/
2
D"
:
.x " 1/2
.x " 1/2
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 2.3 (PAGE 115)
y
At x D 2 we have y D 3 and y 0 D "2. Thus, the equation
of the tangent line is y D 3 " 2.x " 2/, or y D "2x C 7.
The normal line is y D 3 C 12 .x " 2/, or y D 12 x C 2.
%
b
a;
1
1 0
, y D1" 2
x
x
1
For horizontal tangent: 0 D y 0 D 1 " 2 so x 2 D 1 and
x
x D ˙1
The tangent is horizontal at .1; 2/ and at ."1; "2/
43. y D x C
2
3
2
48.
The slope of a horizontal line must be zero, so
p
4x.2 " x 2 / D 0, which impliesp
that x D 0 or x D ˙ 2.
At x D 0; y D 0 and at x D ˙ 2; y D 4.
Hence, there are two horizontal lines that are tangent to
the curve. Their equations are y D 0 and y D 4.
For
horizontal tangent we want 0 D y 0 D "
1
2
The tangent is horizontal only at
46. If y D
xD1
2x C 1
.
.x 2 C x C 1/2
49.
˙ 12 ,
xD1
1
D" :
2
!
The tangent to y D x 3 at .a; a3 / has equation
y D a3 C 3a2 .x " a/, or y D 3a2 x " 2a3 . This line
passes through .2; 8/ if 8 D 6a2 " 2a3 or, equivalently, if
a3 " 3a2 C 4 D 0. Since .2; 8/ lies on y D x 3 , a D 2 must
be a solution of this equation. In fact it must be a double
root; .a " 2/2 must be a factor of a3 " 3a2 C 4. Dividing
by this factor, we find that the other factor is a C 1, that is,
a3 " 3a2 C 4 D .a " 2/2 .a C 1/:
The two tangent lines to y D x 3 passing through .2; 8/
correspond to a D 2 and a D "1, so their equations are
y D 12x " 16 and y D 3x C 2.
1
.x C 2/.1/ " .x C 1/.1/
D
:
.x C 2/2
.x C 2/2
1
D 4;
.x C 2/2
ˇ
1 !3=2 ˇˇ
D" x
ˇ
ˇ
2
"
The product of the slopes is .2/ " 12 D "1. Hence, the
two curves intersect at right angles.
#
$
1 4
" ;
.
2 3
In order to be parallel to y D 4x, the tangent line must
have slope equal to 4, i.e.,
or .x C 2/2 D 41 :
50.
The tangent to y D x 2 =.x " 1/ at .a; a2 =.a " 1// has slope
ˇ
.x " 1/2x " x 2 .1/ ˇˇ
a2 " 2a
:
mD
D
ˇ
2
ˇ
.x " 1/
.a " 1/2
xDa
The equation of the tangent is
" 32
" 52 .
" 23 ,
Hence x C 2 D
and x D
or
At x D
y D "1, and at x D " 52 , y D 3.
the
is" parallel to y D 4x at the points
" tangent
!
!Hence,
" 32 ; "1 and " 25 ; 3 .
47.
ˇ
dy ˇˇ
ˇ
dx ˇ
xC1
, then
xC2
y0 D
1
Since p D y D x 2 ) x 5=2 D 1, therefore x D 1 at
x
theˇ intersection point. The slope of y D x 2 at x D 1 is
ˇ
1
2x ˇˇ
D 2. The slope of y D p at x D 1 is
x
xD1
1
2x C 1
, y0 D " 2
x2 C x C 1
.x C x C 1/2
Thus 2x C 1 D 0 and x D "
&
Fig. 2.3-47
2
y D 2x.4 " x / C x ."2x/ D 8x " 4x D 4x.2 " x /:
45. y D
1
a
x
44. If y D x 2 .4 " x 2 /, then
0
1
x
yD
Let the point of tangency be .a; a1 /. The slope of the tanb " a1
1
2
gent is " 2 D
. Thus b " a1 D a1 and a D .
a
0"a
b
b2
b2
so has equation y D b "
x.
Tangent has slope "
4
4
y"
a2 " 2a
a2
D
.x " a/:
a"1
.a " 1/2
This line passes through .2; 0/ provided
0"
a2 " 2a
a2
D
.2 " a/;
a"1
.a " 1/2
or, upon simplification, 3a2 " 4a D 0. Thus we can have
either a D 0 or a D 4=3. There are two tangents through
.2; 0/. Their equations are y D 0 and y D "8x C 16.
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51
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SECTION 2.3 (PAGE 115)
51.
ADAMS and ESSEX: CALCULUS 9
p
p
f .x C h/ " f .x/
h
f .x C h/ " f .x/
1
p
D lim
p
h
h!0
f .x C h/ C f .x/
d p
f .x/ D lim
dx
h!0
Proof: The case n D 2 is just the Product Rule. Assume
the formula holds for n D k for some integer k > 2.
Using the Product Rule and this hypothesis we calculate
.f1 f2 ! ! ! fk fkC1 /0
D Œ.f1 f2 ! ! ! fk /fkC1 !0
0
D .f1 f2 ! ! ! fk /0 fkC1 C .f1 f2 ! ! ! fk /fkC1
f 0 .x/
D p
2 f .x/
2x
x
d p 2
x C1D p
D p
dx
2 x2 C 1
x2 C 1
'
3
52. f .x/ D jx 3 j D x 3 if x # 0 . Therefore f is differen"x
if x < 0
tiable everywhere except possibly at x D 0, However,
D .f10 f2 ! ! ! fk C f1 f20 ! ! ! fk C ! ! ! C f1 f2 ! ! ! fk0 /fkC1
0
C .f1 f2 ! ! ! fk /fkC1
D f10 f2 ! ! ! fk fkC1 C f1 f20 ! ! ! fk fkC1 C ! ! !
0
C f1 f2 ! ! ! fk0 fkC1 C f1 f2 ! ! ! fk fkC1
so the formula is also true for n D k C 1. The formula is
therefore for all integers n # 2 by induction.
f .0 C h/ " f .0/
D lim h2 D 0
h
h!0C
h!0C
f .0 C h/ " f .0/
lim
D lim ."h2 / D 0:
h
h!0!
h!0!
lim
Thus f 0 .0/ exists and equals 0. We have
'
2
f 0 .x/ D 3x 2
"3x
Section 2.4 The Chain Rule
1.
2.
if x # 0
if x < 0.
n
d n=2
D x .n=2/!1 for n D 1, 2, 3, : : : .
x
53. To be proved:
dx
2
Proof: It is already known that the case n D 1 is true: the
derivative of x 1=2 is .1=2/x !1=2 .
Assume that the formula is valid for n D k for some
positive integer k:
3.
4.
5.
d k=2
k
x
D x .k=2/!1 :
dx
2
Then, by the Product Rule and this hypothesis,
6.
d .kC1/=2
d 1=2 k=2
x
D
x x
dx
dx
1
k
k C 1 .kC1/=2!1
D x !1=2 x k=2 C x 1=2 x .k=2/!1 D
x
:
2
2
2
Thus the formula is also true for n D k C 1. Therefore it
is true for all positive integers n by induction.
For negative n D "m (where m > 0) we have
d n=2
d 1
x
D
dx
dx x m=2
"1 m
D m x .m=2/!1
x 2
n
m
D " x !.m=2/!1 D x .n=2/!1 :
2
2
.f1 f2 ! ! ! fn /0
D f10 f2 ! ! ! fn C f1 f20 ! ! ! fn C ! ! ! C f1 f2 ! ! ! fn0
52
y D .2x C 3/6 ; y 0 D 6.2x C 3/5 2 D 12.2x C 3/5
%
x &99
y D 1"
3
# $
%
%
x &98
1
x &98
0
D "33 1 "
"
y D 99 1 "
3
3
3
f .x/ D .4 " x 2 /10
f 0 .x/ D 10.4 " x 2 /9 ."2x/ D "20x.4 " x 2 /9
d p
3x
dy
"6x
D
D "p
1 " 3x 2 D p
dx
dx
2 1 " 3x 2
1 " 3x 2
#
$
3 !10
F .t/ D 2 C
t
$
#
$
#
30
3 !11
3 !11 "3
D
2
C
F 0 .t / D "10 2 C
t
t2
t2
t
z D .1 C x 2=3 /3=2
z 0 D 32 .1 C x 2=3 /1=2 . 23 x !1=3 / D x !1=3 .1 C x 2=3 /1=2
7.
8.
3
5 " 4x
3
12
y0 D "
."4/ D
.5 " 4x/2
.5 " 4x/2
yD
y D .1 " 2t 2 /!3=2
y 0 D " 32 .1 " 2t 2 /!5=2 ."4t / D 6t .1 " 2t 2 /!5=2
9.
10.
y D j1 " x 2 j;
y 0 D "2xsgn .1 " x 2 / D
f .t / D j2 C t 3 j
f 0 .t / D Œsgn .2 C t 3 /!.3t 2 / D
11.
54. To be proved:
(page 120)
y D 4x C j4x " 1j
y 0 D 4 C 4.sgn .4x " 1//
'
8 if x > 14
D
0 if x < 14
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2x 3 " 2x
j1 " x 2 j
3t 2 .2 C t 3 /
j2 C t 3 j
lOMoARcPSD|6566483
INSTRUCTOR’S SOLUTIONS MANUAL
12.
SECTION 2.4 (PAGE 120)
18.
y D .2 C jxj3 /1=3
y
y 0 D 31 .2 C jxj3 /!2=3 .3jxj2 /sgn .x/
# $
x
D jxj2 .2 C jxj3 /!2=3
D xjxj.2 C jxj3 /!2=3
jxj
13.
15.
slope 0
%
!4
x"2
f .x/ D 1 C
3
!3
!# $
r
r
x"2
1
3
1
0
f .x/ D 4 1 C
3
2 x"2
3
!3
r
r
3
x"2
2
D
1C
3 x"2
3
r
#
z D uC
1
u"1
$!5=3
#
$!8=3 #
$
5
1
1
dz
D"
uC
1"
du
3
u"1
.u " 1/2
#
$#
$!8=3
5
1
1
D"
1"
uC
3
.u " 1/2
u"1
16.
17.
yD4xCj4x!1j
1
yD
p
2 C 3x C 4
#
$
3
1
0
p
y D "%
&
2
p
2 3x C 4
2 C 3x C 4
3
D" p
%
&2
p
2 3x C 4 2 C 3x C 4
14.
slope 8
p
x5 3 C x6
yD
.4 C x 2 /3
#
#
$)
(
p
3x 5
1
5
0
2 3
4
6
y D
.4 C x / 5x 3 C x C x
p
.4 C x 2 /6
3 C x6
!
h
i
p
" x 5 3 C x 6 3.4 C x 2 /2 .2x/
h
i
.4 C x 2 / 5x 4 .3 C x 6 / C 3x 10 " x 5 .3 C x 6 /.6x/
p
D
.4 C x 2 /4 3 C x 6
60x 4 " 3x 6 C 32x 10 C 2x 12
D
p
.4 C x 2 /4 3 C x 6
!21=3
d 1=4
d
x
D
dx
dx
q
20.
d 3=4
d
x
D
dx
dx
q
21.
1
3
d 3=2
d p 3
x D p .3x 2 / D x 1=2
x
D
3
dx
dx
2
2 x
22.
d
f .2t C 3/ D 2f 0 .2t C 3/
dt
23.
d
f .5x " x 2 / D .5 " 2x/f 0 .5x " x 2 /
dx
24.
25.
26.
27.
28.
29.
t
x
19.
y
yDj2Ct 3 j
&
1
4 ;1
p
1
1
1
x D pp & p D x !3=4
4
2
x 2 x
p
1
x xD p p
2 x x
#
p
x
xC p
2 x
$
D
3 !1=4
x
4
( # $)3
( # $)2 # $ # $
d
2
2
2
"2
f
D3 f
f0
dx
x
x
x
x2
# $ ( # $)2
6
2
2
D " 2f0
f
x
x
x
2f 0 .x/
d p
f 0 .x/
3 C 2f .x/ D p
D p
dx
2 3 C 2f .x/
3 C 2f .x/
p
p
d
2
f . 3 C 2t / D f 0 . 3 C 2t/ p
dt
2 3 C 2t
p
1
D p
f 0 . 3 C 2t /
3 C 2t
p
p
d
1
f .3 C 2 x/ D p f 0 .3 C 2 x/
dx
x
# %
&$
d
f 2f 3f .x/
dt
# %
&$
%
&
D f 0 2f 3f .x/ ! 2f 0 3f .x/ ! 3f 0 .x/
%
& # %
&$
D 6f 0 .x/f 0 3f .x/ f 0 2f 3f .x/
&
d %
f 2 " 3f .4 " 5t /
dx %
&%
&
D f 0 2 " 3f .4 " 5t / "3f 0 .4 " 5t / ."5/
%
&
D 15f 0 .4 " 5t /f 0 2 " 3f .4 " 5t /
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53
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SECTION 2.4 (PAGE 120)
30.
!ˇ
x 2 " 1 ˇˇ
ˇ
x2 C 1 ˇ
xD!2
p
x
" x 2 " 1.2x/ ˇˇ
.x 2 C 1/ p
2
ˇ
x "1
D
ˇ
ˇ
.x 2 C 1/2
xD!2
#
$
p
2
.5/ " p
" 3."4/
2
3
D p
D
25
25 3
p
d
dx
32.
ˇ
ˇ
d p
ˇ
3t " 7ˇ
ˇ
dt
33.
ˇ
ˇ
1
ˇ
f .4/ D "
ˇ
.2x C 1/3=2 ˇ
31.
37. Slope of y D .1 C x 2=3 /3=2ˇ at x D "1 is
#
$
p
2 !1=3 ˇˇ
3
.1 C x 2=3 /1=2
x
D" 2
ˇ
ˇ
2
3
xD!1
The tangent p
line at ."1; 23=2 / has equation
3=2
yD2
" 2.x C 1/.
38.
ˇ
ˇ
3
ˇ
D p
D p
ˇ
2 3t " 7 ˇ tD3
2 2
tD3
1
f .x/ D p
2x C 1
xD4
D"
xD!2
xD!2
D
y D .2b/8 D 256b 8 is
3
#
$
b
256b 8 C 1024ab 7 x "
, or
a
y D 210 ab 7 x " 3 & 28 b 8 .
39.
Slope of y D 1=.x 2 " x Cˇ3/3=2 at x D "2 is
ˇ
3
5
3
D " .9!5=2 /."5/ D
" .x 2 "xC3/!5=2 .2x"1/ˇˇ
2
2
162
xD!2
1
The tangent line at ."2; / has equation
27
5
1
C
.x C 2/.
yD
27
162
40.
Given that f .x/ D .x " a/m .x " b/n then
17
.12/ D 102
2
f 0 .x/ D m.x " a/m!1 .x " b/n C n.x " a/m .x " b/n!1
4
2.1 C x/.2 C x/.3 C x/ .4 C x/ C
D .x " a/m!1 .x " b/n!1 .mx " mb C nx " na/:
3.1 C x/.2 C x/2 .3 C x/2 .4 C x/4 C
0
If x ¤ a and x ¤ b, then f 0 .x/ D 0 if and only if
4.1 C x/.2 C x/2 .3 C x/3 .4 C x/3
2
3
4
3
4
F .0/ D .2 /.3 /.4 / C 2.1/.2/.3 /.4 /C
mx " mb C nx " na D 0;
3.1/.22 /.32 /.44 / C 4.1/.22 /.33 /.43 /
D 4.22 ! 33 ! 44 / D 110; 592
35.
which is equivalent to
#
%
&!1=2 &!6
y D x C .3x/5 " 2
#
%
&!1=2 $!7
y 0 D "6 x C .3x/5 " 2
#
&!3=2 %
&$
1%
.3x/5 " 2
5.3x/4 3
& 1"
2
#
%
&!3=2 $
15
D "6 1 " .3x/4 .3x/5 " 2
2
#
%
&!1=2 $!7
& x C .3x/5 " 2
36. The slope of y D
ˇ
dy ˇˇ
ˇ
dx ˇ
p
ˇ
ˇ
ˇ
D p
ˇ
2 1 C 2x 2 ˇ
4x
xD2
D
n
m
aC
b:
mCn
mCn
This point lies lies between a and b.
41. x.x 4 C 2x 2 " 2/=.x 2 C 1/5=2
42.
4.7x 4 " 49x 2 C 54/=x 7
43.
857; 592
44.
5=8
45.
The Chain Rule does not enable you to calculate the
derivatives of jxj2 and jx 2 j at x D 0 directly as a composition of two functions, one of which is jxj, because jxj
is not differentiable at x D 0. However, jxj2 D x 2 and
jx 2 j D x 2 , so both functions are differentiable at x D 0
and have derivative 0 there.
46.
It may happen that k D g.x C h/ " g.x/ D 0 for values
of h arbitrarily close to 0 so that the division by k in the
“proof” is not justified.
4
:
3
Thus, the equation of the tangent line at .2; 3/ is
y D 3 C 43 .x " 2/, or y D 43 x C 31 :
54
xD
1 C 2x 2 at x D 2 is
xD2
b
and
a
D
y
F .x/ D .1 C x/.2 C x/2 .3 C x/3 .4 C x/4
F 0 .x/ D .2 C x/2 .3 C x/3 .4 C x/4 C
xDb=a
The equation of the tangent line at x D
1
27
y D .x 3 C 9/17=2
ˇ
ˇ
ˇ
ˇ
17 3
0ˇ
15=2
2ˇ
yˇ
D
.x C 9/
3x ˇ
ˇ
ˇ
2
b
The slope of y D .ax C b/8 at x D is
a
ˇ
ˇ
ˇ
dy ˇˇ
ˇ
D 8a.ax C b/7 ˇ
D 1024ab 7 :
ˇ
ˇ
dx ˇ
xDb=a
3
0
34.
ADAMS and ESSEX: CALCULUS 9
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 2.5 (PAGE 126)
Section 2.5 Derivatives of Trigonometric
Functions (page 126)
23.
d
d 1
cos x
csc x D
D " 2 D " csc x cot x
dx
dx sin x
sin x
24.
1.
d
d cos x
" cos2 x " sin2 x
D "csc 2 x
cot x D
D
dx
dx sin x
sin2 x
25.
2.
y 0 D "3 sin 3x
3. y D cos 3x;
4. y D sin
x
;
5
y0 D
5. y D tan #x;
6. y D sec ax;
7.
1
x
cos :
5
5
27.
y 0 D # sec2 #x
28.
y 0 D a sec ax tan ax:
29.
y 0 D 3 csc2 .4 " 3x/
y D cot.4 " 3x/;
26.
# "x
1
# "x
d
sin
D " cos
dx
3
3
3
9. f .x/ D cos.s " rx/; f 0 .x/ D r sin.s " rx/
8.
30.
y 0 D A cos.Ax C B/
10. y D sin.Ax C B/;
d
sin.#x 2 / D 2#x cos.#x 2 /
dx
p
p
d
1
12.
cos. x/ D " p sin. x/
dx
2 x
p
" sin x
13. y D 1 C cos x; y 0 D p
2 1 C cos x
11.
14.
15.
16.
31.
32.
d
sin.2 cos x/ D cos.2 cos x/."2 sin x/
dx
D "2 sin x cos.2 cos x/
f .x/ D cos.x C sin x/
f 0 .x/ D ".1 C cos x/ sin.x C sin x/
33.
g.& / D tan.& sin & /
34.
g 0 .& / D .sin & C & cos & / sec2 .& sin & /
17.
u D sin3 .#x=2/;
18. y D sec.1=x/;
19.
20.
21.
22.
u0 D
3#
cos.#x=2/ sin2 .#x=2/
2
y 0 D ".1=x 2 / sec.1=x/ tan.1=x/
1
sin 2at /
2
0
F .t / D a cos at cos at " a sin at sin at
. D a cos 2at /
F .t/ D sin at cos at
.D
35.
36.
sin a&
cos b&
a cos b& cos a& C b sin a& sin b&
0
G .& / D
:
cos2 b&
&
%
d
sin.2x/ " cos.2x/ D 2 cos.2x/ C 2 sin.2x/
dx
d
d
.cos2 x " sin2 x/ D
cos.2x/
dx
dx
D "2 sin.2x/ D "4 sin x cos x
G.& / D
d
.tan x C cot x/ D sec2 x " csc2 x
dx
d
.sec x " csc x/ D sec x tan x C csc x cot x
dx
d
.tan x " x/ D sec2 x " 1 D tan2 x
dx
d
d
tan.3x/ cot.3x/ D
.1/ D 0
dx
dx
d
.t cos t " sin t / D cos t " t sin t " cos t D "t sin t
dt
d
.t sin t C cos t / D sin t C t cos t " sin t D t cos t
dt
d
sin x
.1 C cos x/.cos x/ " sin.x/." sin x/
D
dx 1 C cos x
.1 C cos x/2
cos x C 1
1
D
D
.1 C cos x/2
1 C cos x
d cos x
.1 C sin x/." sin x/ " cos.x/.cos x/
D
dx 1 C sin x
.1 C sin x/2
" sin x " 1
"1
D
D
2
.1 C sin x/
1 C sin x
d 2
x cos.3x/ D 2x cos.3x/ " 3x 2 sin.3x/
dx
p
g.t / D .sin t /=t
t cos t " sin t
1
&
g 0 .t / D p
t2
2 .sin t /=t
t cos t " sin t
D
p
2t 3=2 sin t
v D sec.x 2 / tan.x 2 /
v 0 D 2x sec.x 2 / tan2 .x 2 / C 2x sec3 .x 2 /
p
sin x
zD
p
1 C cos x
p
p
p
p
p
p
.1 C cos x/.cos x=2 x/ " .sin x/." sin x=2 x/
z0 D
p 2
.1 C cos x/
p
1
1 C cos x
p 2 D p
p
D p
2 x.1 C cos x/
2 x.1 C cos x/
d
sin.cos.tan t // D ".sec2 t /.sin.tan t // cos.cos.tan t //
dt
f .s/ D cos.s C cos.s C cos s//
f 0 .s/ D "Œsin.s C cos.s C cos s//!
& Œ1 " .sin.s C cos s//.1 " sin s/!
37. Differentiate both sides of sin.2x/ D 2 sin x cos x and
divide by 2 to get cos.2x/ D cos2 x " sin2 x.
38.
Differentiate both sides of cos.2x/ D cos2 x " sin2 x and
divide by "2 to get sin.2x/ D 2 sin x cos x.
39.
Slope of y D sin x at .#; 0/ is cos # D "1. Therefore
the tangent and normal lines to y D sin x at .#; 0/ have
equations y D ".x " #/ and y D x " #, respectively.
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SECTION 2.5 (PAGE 126)
ADAMS and ESSEX: CALCULUS 9
40. The slope of y D tan.2x/ at .0; 0/ is 2 sec2 .0/ D 2.
Therefore the tangent and normal lines to y D tan.2x/ at
.0; 0/ have equations y D 2x and y D "x=2, respectively.
p
41. Thepslope of y D
2 cos.x=4/ at .#; 1/ is
". 2=4/ sin.#=4/ Dp"1=4. Therefore the tangent and
normal lines to y D 2 cos.x=4/ at .#; 1/ have equations
y D 1 " .x " #/=4 and y D 1 C 4.x " #/, respectively.
42. The slope of y Dpcos2 x at .#=3; 1=4/ is
" sin.2#=3/ D " 3=2. Therefore the tangent and normal
lines to y D tan.2x/
at .0; 0/ have equations
p
y D .1=4/ " . 3=2/.x
" .#=3// and
p
y D .1=4/ C .2= 3/.x " .#=3//, respectively.
% #x &
% #x &
#
is y 0 D
cos
.
43. Slope of y D sin.x ı / D sin
180
180
180
At x D 45 the tangent line has equation
1
#
yD p C
p .x " 45/.
2
180 2
% x# &
44. For y D sec .x ı / D sec
we have
180
% x# &
% x# &
#
dy
D
sec
tan
:
dx
180
180
180
p
# p
# 3
.2 3/ D
:
180
90
90
Thus, the normal line has slope " p and has equation
# 3
90
y D 2 " p .x " 60/.
# 3
49.
y D x C sin x has a horizontal tangent at x D # because
dy=dx D 1 C cos x D 0 there.
50.
y D 2x C sin x has no horizontal tangents because
dy=dx D 2 C cos x # 1 everywhere.
51.
y D x C 2 sin x has horizontal tangents at x D 2#=3 and
x D 4#=3 because dy=dx D 1 C 2 cos x D 0 at those
points.
52.
y D x C 2 cos x has horizontal tangents at x D #=6 and
x D 5#=6 because dy=dx D 1 " 2 sin x D 0 at those
points.
53.
54.
2
sin.2x/
t an.2x/
D lim
D1&2D2
x!0
x
2x cos.2x/
lim sec.1 C cos x/ D sec.1 " 1/ D sec 0 D 1
x!!
55.
56.
57.
58.
At x D 60 the slope is
45. The slope of y D tan x at x D a is sec2 a. The tangent there is parallel
to y D 2x if sec2 a D 2, or
p
cos a D ˙1= 2. The only solutions in ."#=2; #=2/
are a D ˙#=4. The corresponding points on the graph are
.#=4; 1/ and ."#=4; 1/.
lim
x!0
% x &2
cos x D 12 & 1 D 1
x!0
x!0 sin x
#
$
#
$
# " # cos2 x
sin x 2
lim cos
D cos # D "1
D
lim
cos
#
x!0
x!0
x2
x
#
$
1 " cos h
2 sin2 .h=2/
1 sin.h=2/ 2
1
D
lim
D
lim
D
lim
h2
h2
h=2
2
h!0
h!0
h!0 2
lim x 2 csc x cot x D lim
f will be differentiable at x D 0 if
2 sin 0 C 3 cos 0 D b;
and
ˇ
ˇ
d
D a:
.2 sin x C 3 cos x/ˇˇ
dx
xD0
Thus we need b D 3 and a D 2.
59.
There are infinitely many lines through the origin that are
tangent to y D cos x. The two with largest slope are
shown in the figure.
y
46. The slope of y D tan.2x/ at x D a is 2 sec2 .2a/. The
tangent there is normal to y D "x=8 if 2 sec2 .2a/ D 8,
or cos.2a/ D ˙1=2. The only solutions in ."#=4; #=4/
are a D
points on the graph are
p ˙#=6. The corresponding
p
.#=6; 3/ and ."#=6; " 3/.
47.
48.
!!
56
2! x
y D cos x
d
sin x D cos x D 0 at odd multiples of #=2.
dx
d
cos x D " sin x D 0 at multiples of #.
dx
d
sec x D sec x tan x D 0 at multiples of #.
dx
d
csc x D " csc x cot x D 0 at odd multiples of #=2.
dx
Thus each of these functions has horizontal tangents at
infinitely many points on its graph.
d
tan x D sec2 x D 0 nowhere.
dx
d
cot x D " csc2 x D 0 nowhere.
dx
Thus neither of these functions has a horizontal tangent.
!
Fig. 2.5-59
The tangent to y D cos x at x D a has equation
y D cos a " .sin a/.x " a/. This line passes through
the origin if cos a D "a sin a. We use a calculator with
a “solve” function to find solutions of this equation near
a D "# and a D 2# as suggested in the figure. The solutions are a ' "2:798386 and a ' 6:121250. The slopes
of the corresponding tangents are given by " sin a, so they
are 0:336508 and 0:161228 to six decimal places.
60.
1
61.
p
" 2# C 3.2# 3=2 " 4# C 3/=#
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INSTRUCTOR’S SOLUTIONS MANUAL
62.
SECTION 2.6 (PAGE 131)
a) As suggested by the figure in the problem, the square
of the length of chord AP is .1"cos & /2 C.0"sin & /2 ,
and the square of the length of arc AP is & 2 . Hence
5.
y D x 1=3 " x !1=3
1
1
y 0 D x !2=3 C x !4=3
3
3
2 !5=3 4 !7=3
00
y D" x
" x
9
9
10 !8=3 28 !10=3
000
x
C x
y D
27
27
6.
y D x 10 C 2x 8
.1 C cos & /2 C sin2 & < & 2 ;
and, since squares cannot be negative, each term in
the sum on the left is less than & 2 . Therefore
0 $ j1 " cos & j < j& j;
0
y D 10x C 16x
0 $ j sin & j < j& j:
7.
Since lim"!0 j& j D 0, the squeeze theorem implies
that
lim 1 " cos & D 0; lim sin & D 0:
"!0
"!0
From the first of these, lim"!0 cos & D 1.
b) Using the result of (a) and the addition formulas for
cosine and sine we obtain
8.
lim cos.&0 C h/ D lim .cos &0 cos h " sin &0 sin h/ D cos &0
h!0
h!0
lim sin.&0 C h/ D lim .sin &0 cos h C cos &0 sin h/ D sin &0 :
h!0
h!0
9.
This says that cosine and sine are continuous at any
point &0 .
1.
y D .3 " 2x/7
12.
y 00 D 168.3 " 2x/5
y 000 D "1680.3 " 2x/4
3.
1
x
1
y 0 D 2x C 2
x
y D x2 "
y 00 D 2 "
y 000 D
2
x3
6
x4
13.
6
D 6.x " 1/!2
.x " 1/2
y 0 D "12.x " 1/!3
yD
p
ax C b
a
0
y D p
2 ax C b
2
y 000
y 00 D 2 sec2 x tan x
y 000 D 2 sec4 x C 4 sec2 x tan2 x
y D sec x
y 0 D sec x tan x
y 00 D sec x tan2 x C sec3 x
y 000 D sec x tan3 x C 5 sec3 x tan x
y D cos.x 2 /
a2
4.ax C b/3=2
3a3
D
8.ax C b/5=2
y 00 D "
y 000
y 00 D "2 sin.x 2 / " 4x 2 cos.x 2 /
y 000 D "12x cos.x 2 / C 8x 3 sin.x 2 /
sin x
x
cos x sin x
0
y D
" 2
x
x
2
.2
"
x
/
sin
x 2 cos x
y 00 D
"
x3
x2
2
3.x 2 " 2/ sin x
.6 " x / cos x
C
y 000 D
3
x
x4
yD
1
D x !1
x
f 0 .x/ D "x !2
f .x/ D
f 000 .x/ D "3Šx !4
y 000 D "144.x " 1/!5
yD
y D tan x
4
.x C 1/3
12
D
.x C 1/4
y 00 D "
f 00 .x/ D 2x !3
y 00 D 36.x " 1/!4
4.
x"1
xC1
2
y0 D
.x C 1/2
y 0 D "2x sin.x 2 /
y 0 D "14.3 " 2x/6
2.
y 000 D 720x 7 C 672x 5
yD
y D sec x
11.
y 00 D 90x 8 C 112x 6
7
p
y D .x 2 C 3/ x D x 5=2 C 3x 1=2
5
3
y 0 D x 3=2 C x !1=2
2
2
15 1=2 3 !3=2
00
y D
x
" x
4
4
15 !1=2 9 !5=2
000
x
C x
y D
8
8
0
10.
Section 2.6 Higher-Order Derivatives
(page 131)
9
f .4/ .x/ D 4Šx !5
Guess: f .n/ .x/ D ."1/n nŠx !.nC1/ .(/
Proof: (*) is valid for n D 1 (and 2, 3, 4).
!.kC1/
Assume f .k/ .x/ D ."1/k kŠx
for
%
& some k # 1
.kC1/
k
Then f
.x/ D ."1/ kŠ ".k C 1/ x !.kC1/!1
D ."1/kC1 .k C 1/Šx !..kC1/C1/ which is (*) for n D k C 1.
Therefore, (*) holds for n D 1; 2; 3; : : : by induction.
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SECTION 2.6 (PAGE 131)
14.
ADAMS and ESSEX: CALCULUS 9
1
D x !2
x2
f 0 .x/ D "2x !3
f 00 .x/ D "2."3/x !4 D 3Šx !4
f .3/ .x/ D "2."3/."4/x !5 D "4Šx !5
Conjecture:
Thus, the formula is also true for n D k C 1. Hence, it is
true for n # 2 by induction.
f .x/ D
f .n/ .x/ D ."1/n .n C 1/Šx !.nC2/
17.
f 00 .x/ D 2b 2 .a C bx/!3
for n D 1; 2; 3; : : :
f 000 .x/ D "3Šb 3 .a C bx/!4
Guess: f .n/ .x/ D ."1/n nŠb n .a C bx/!.nC1/ .(/
Proof: (*) holds for n D 1; 2; 3
Assume (*) holds for n D k:
f .k/ .x/ D ."1/k kŠb k .a C bx/!.kC1/
Then
%
&
f .kC1/ .x/ D ."1/k kŠb k ".k C 1/ .a C bx/!.kC1/!1 .b/
Proof: Evidently, the above formula holds for n D 1; 2
and 3. Assume it holds for n D k,
i.e., f .k/ .x/ D ."1/k .k C 1/Šx !.kC2/ : Then
d .k/
f .x/
dx
D ."1/k .k C 1/ŠŒ."1/.k C 2/!x !.kC2/!1
f .kC1/ .x/ D
D ."1/kC1 .k C 1/Šb kC1 .a C bx/..kC1/C1/
So (*) holds for n D k C 1 if it holds for n D k.
Therefore, (*) holds for n D 1; 2; 3; 4; : : : by induction.
D ."1/kC1 .k C 2/Šx !Œ.kC1/C2# :
Thus, the formula is also true for n D k C 1. Hence it is
true for n D 1; 2; 3; : : : by induction.
15.
1
D .2 " x/!1
f .x/ D
2"x
f 0 .x/ D C.2 " x/!2
f 00 .x/ D 2.2 " x/!3
f 000 .x/ D C3Š.2 " x/!4
Guess: f .n/ .x/ D nŠ.2 " x/!.nC1/ .(/
Proof: (*) holds for n D 1; 2; 3.
Assume f .k/ .x/ D kŠ.2 " x/!.kC1/ (i.e., (*) holds for
n D k)
%
&
Then f .kC1/ .x/ D kŠ ".k C 1/.2 " x/!.kC1/!1 ."1/
18. f .x/ D x 2=3
f 0 .x/ D 32 x !1=3
f 00 .x/ D 32 ." 13 /x !4=3
f 000 .x/ D 32 ." 13 /." 43 /x !7=3
Conjecture:
1 ! 4 ! 7 ! ! ! ! .3n " 5/ !.3n!2/=3
f .n/ .x/ D 2."1/n!1
x
for
3n
n # 2.
Proof: Evidently, the above formula holds for n D 2 and
3. Assume that it holds for n D k, i.e.
f .k/ .x/ D 2."1/k!1
!..kC1/C1/
D .k C 1/Š.2 " x/
:
Thus (*) holds for n D k C 1 if it holds for k.
Therefore, (*) holds for n D 1; 2; 3; : : : by induction.
p
16. f .x/ D x D x 1=2
0
f .x/ D 21 x !1=2
f 00 .x/ D 12 ." 12 /x !3=2
f 000 .x/ D 21 ." 12 /." 32 /x !5=2
f .4/ .x/ D 21 ." 12 /." 32 /." 52 /x !7=2
Conjecture:
f .n/ .x/ D ."1/n!1
1 ! 3 ! 5 ! ! ! .2n " 3/ !.2n!1/=2
x
2n
f .k/ .x/ D ."1/k!1
1 ! 3 ! 5 ! ! ! .2k " 3/ !.2k!1/=2
x
:
2k
Then
d .k/
f .kC1/ .x/ D
f .x/
dx
(
)
".2k " 1/ !Œ.2k!1/=2#!1
k!1 1 ! 3 ! 5 ! ! ! .2k " 3/
!
D ."1/
x
2
2k
1 ! 3 ! 5 ! ! ! .2k " 3/Œ2.k C 1/ " 3! !Œ2.kC1/!1#=2
x
:
D ."1/.kC1/!1
2kC1
58
1 ! 4 ! 7 ! ! ! ! .3k " 5/ !.3k!2/=3
x
:
3k
Then,
d .k/
f .x/
dx
(
)
1 ! 4 ! 7 ! ! ! ! .3k " 5/ ".3k " 2/ !Œ.3k!2/=3#!1
!
D 2."1/k!1
x
3
3k
1 ! 4 ! 7 ! ! ! ! .3k " 5/Œ3.k C 1/ " 5! !Œ3.kC1/!2#=3
x
:
D 2."1/.kC1/!1
3. k C 1/
f .kC1/ .x/ D
Thus, the formula is also true for n D k C 1. Hence, it is
true for n # 2 by induction.
.n # 2/:
Proof: Evidently, the above formula holds for n D 2; 3 and
4. Assume that it holds for n D k, i.e.
1
D .a C bx/!1
a C bx
f 0 .x/ D "b.a C bx/!2
f .x/ D
f .x/ D cos.ax/
f 0 .x/ D "a sin.ax/
19.
f 00 .x/ D "a2 cos.ax/
f 000 .x/ D a3 sin.ax/
f .4/ .x/ D a4 cos.ax/ D a4 f .x/
It follows that f .n/ .x/ D a4 f .n!4/ .x/ for n # 4, and
8̂ n
a cos.ax/
< n
"a sin.ax/
.n/
f .x/ D
n
:̂ "a cos.ax/
n
a sin.ax/
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if
if
if
if
n D 4k
n D 4k C 1
.k D 0; 1; 2; : : :/
n D 4k C 2
n D 4k C 3
lOMoARcPSD|6566483
INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 2.6 (PAGE 131)
Differentiating any of these four formulas produces the one
for the next higher value of n, so induction confirms the
overall formula.
20.
f .x/ D x cos x
f 0 .x/ D cos x " x sin x
f 00 .x/ D "2 sin x " x cos x
f 000 .x/ D "3 cos x C x sin x
The pattern suggests that
'
"nŠjxj!.nC1/ sgn x
f .n/ .x/ D
nŠjxj!.nC1/
23.
f .4/ .x/ D 4 sin x C x cos x
This suggests the formula (for k D 0; 1; 2; : : :)
8̂
n sin x C x cos x
<
n cos x " x sin x
.n/
f .x/ D
:̂ "n sin x " x cos x
"n cos x C x sin x
if
if
if
if
n D 4k
n D 4k C 1
n D 4k C 2
n D 4k C 3
Differentiating any of these four formulas produces the one
for the next higher value of n, so induction confirms the
overall formula.
21.
f .x/ D x sin.ax/
f 0 .x/ D sin.ax/ C ax cos.ax/
f 00 .x/ D 2a cos.ax/ " a2 x sin.ax/
f 000 .x/ D "3a2 sin.ax/ " a3 x cos.ax/
f 4/ .x/ D "4a3 cos.ax/ C a4 x sin.ax/
This suggests the formula
8̂
"nan!1 cos.ax/ C an x sin.ax/
ˆ
< n!1
na
sin.ax/ C an x cos.ax/
.n/
f .x/ D
n!1
ˆ na
cos.ax/ " an x sin.ax/
:̂
n!1
"na
sin.ax/ " an x cos.ax/
if
if
if
if
n D 4k
n D 4k C 1
n D 4k C 2
n D 4k C 3
for k D 0; 1; 2; : : :. Differentiating any of these four
formulas produces the one for the next higher value of n,
so induction confirms the overall formula.
22. f .x/ D
24.
25.
If y D sec.kx/, then y 0 D k sec.kx/ tan.kx/ and
y 00 D k 2 .sec2 .kx/ tan2 .kx/ C sec3 .kx//
.sgn x/ D 1:
f 00 .x/ D 2jxj!3 .sgn x/2 D 2jxj!3
f .3/ .x/ D "3Šjxj!4 sgn x
D k 2 y.2 sec2 .kx/ " 1/ D k 2 y.2y 2 " 1/:
2
Thus we can calculate successive derivatives of f using
the product rule where necessary, but will get only one
nonzero term in each case:
f .4/ .x/ D 4Šjxj!5 :
If y D tan.kx/, then y 0 D k sec2 .kx/ and
D 2k 2 .1 C tan2 .kx// tan.kx/ D 2k 2 y.1 C y 2 /:
If x ¤ 0 we have
d
sgn x D 0 and
dx
Differentiating this formula leads to the same formula with
n replaced by n C 1 so the formula is valid for all n # 1
by induction.
p
f .x/ D 1 " 3x D .1 " 3x/1=2
1
f 0 .x/ D ."3/.1 " 3x/!1=2
2# $
1
1
f 00 .x/ D
"
."3/2 .1 " 3x/!3=2
2
2
# $# $
1
3
1
"
"
."3/3 .1 " 3x/!5=2
f 000 .x/ D
2
2
2
# $# $# $
1
1
3
5
f .4/ .x/ D
"
"
"
."3/4 .1 " 3x/7=2
2
2
2
2
1 & 3 & 5 & ! ! ! & .2n " 3/ n
3
Guess: f .n/ .x/ D "
2n
.1 " 3x/!.2n!1/=2 .(/
Proof: (*) is valid for n D 2; 3; 4; (but not n D 1)
Assume (*) holds for n D k for some integer k # 2
1 & 3 & 5 & : : : & .2k " 3/ k
i.e., f .k/ .x/ D "
3
2k
!.2k!1/=2
.1 " 3x/
1 & 3 & 5 & ! ! ! & .2k " 3/ k
.kC1/
Then f
.x/ D "
3
2k
#
$
2.k " 1/
"
.1 " 3x/!.2k!1/=2!1 ."3/
2
%
&
1 & 3 & 5 & ! ! ! 2.k C 1/ " 1
D"
3kC1
2kC1
.1 " 3x/!.2.kC1/!1/=2
Thus (*) holds for n D k C 1 if it holds for n D k.
Therefore, (*) holds for n D 2; 3; 4; : : : by induction.
y 00 D 2k 2 sec 2 .kx/t an.kx/
d
1
D jxj!1 . Recall that
jxj D sgn x, so
jxj
dx
f 0 .x/ D "jxj!2 sgn x:
if n is odd
if n is even
26.
To be proved: if f .x/ D sin.ax C b/, then
'
."1/k an sin.ax C b/ if n D 2k
.n/
f .x/ D
."1/k an cos.ax C b/ if n D 2k C 1
for k D 0; 1; 2; : : : Proof: The formula works for k D 0
(n D 2 & 0 D 0 and n D 2 & 0 C 1 D 1):
'
f .0/ .x/ D f .x/ D ."1/0 a0 sin.ax C b/ D sin.ax C b/
f .1/ .x/ D f 0 .x/ D ."1/0 a1 cos.ax C b/ D a cos.ax C b/
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59
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SECTION 2.6 (PAGE 131)
ADAMS and ESSEX: CALCULUS 9
Section 2.7 Using Differentials and
Derivatives (page 137)
Now assume the formula holds for some k # 0.
If n D 2.k C 1/, then
f .n/ .x/ D
d .n!1/
d .2kC1/
f
.x/ D
f
.x/
dx
dx #
$
d
D
."1/k a2kC1 cos.ax C b/
dx
D ."1/kC1 a2kC2 sin.ax C b/
and if n D 2.k C 1/ C 1 D 2k C 3, then
#
d
."1/kC1 a2kC2 sin.ax C b/
f .n/ .x/ D
dx
1.
2.
3.
D ."1/kC1 a2kC3 cos.ax C b/:
27.
0:01
1
"y ' dy D " 2 dx D " 2 D "0:0025.
x
2
If x D 2:01, then y ' 0:5 " 0:0025 D 0:4975.
3 dx
3
"f .x/ ' df .x/ D p
D .0:08/ D 0:06
4
2 3x C 1
f .1:08/ ' f .1/ C 0:06 D 2:06.
#
#t
#
1
1
"h.t / ' dh.t / D " sin
dt " .1/
D" .
4
4
4
10#
40
$
#
1
1
1
' h.2/ "
D" :
h 2C
10#
40
40
%s&
1
1
"u ' du D sec2
ds D .2/."0:04/ D "0:04.
4
4
4
If s D # " 0:06, then u ' 1 " 0:04 ' 0:96.
Thus the formula also holds for k C 1. Therefore it holds
for all positive integers k by induction.
4.
If y D tan x, then
5.
If y D x 2 , then "y ' dy D 2x dx. If dx D .2=100/x,
then "y ' .4=100/x 2 D .4=100/y, so y increases by
about 4%.
6.
If y D 1=x, then "y ' dy D ."1=x 2 / dx. If
dx D .2=100/x, then "y ' ."2=100/=x D ."2=100/y, so
y decreases by about 2%.
7.
If y D 1=x 2 , then "y ' dy D ."2=x 3 / dx. If
dx D .2=100/x, then "y ' ."4=100/=x 2 D ."4=100/y,
so y decreases by about 4%.
8.
If y D x 3 , then "y ' dy D 3x 2 dx. If dx D .2=100/x,
then "y ' .6=100/x 3 D .6=100/y, so y increases by
about 6%.
p
p
x, then dy ' dy D .1=2
If y D
p x/ dx. If
"x D .2=100/x, then "y ' .1=100/ x D .1=100/y,
so y increases by about 1%.
y 0 D sec2 x D 1 C tan2 x D 1 C y 2 D P2 .y/;
where P2 is a polynomial of degree 2. Assume that
y .n/ D PnC1 .y/ where PnC1 is a polynomial of degree
n C 1. The derivative of any polynomial is a polynomial
of one lower degree, so
y .nC1/ D
dy
d
PnC1 .y/ D Pn .y/
D Pn .y/.1Cy 2 / D PnC2 .y/;
dx
dx
a polynomial of degree n C 2. By induction,
.d=dx/n tan x D PnC1 .tan x/, a polynomial of degree
n C 1 in tan x.
28.
.fg/00 D .f 0 g C fg 0 / D f 00 g C f 0 g 0 C f 0 g 0 C fg 00
D f 00 g C 2f 0 g 0 C fg 00
29.
.fg/.3/ D
d
.fg/00
dx
d
D
Œf 00 g C 2f 0 g 0 C fg 00 !
dx
D f .3/ g C f 00 g 0 C 2f 00 g 0 C 2f 0 g 00 C f 0 g 00 C fg .3/
.3/
.fg/.4/
00 0
0 00
C 3f 0 g .3/ C f 0 g .3/ C fg .4/
D f .4/ g C 4f .3/ g 0 C 6f 00 g 00 C 4f 0 g .3/ C fg .4/ :
nŠ
f .n!2/ g 00
.fg/.n/ D f .n/ g C nf .n " 1/g 0 C
2Š.n " 2/Š
nŠ
f .n!3/ g .3/ C ! ! ! C nf 0 g .n!1/ C fg .n/
C
3Š.n " 3/Š
n
X
nŠ
f .n!k/ g .k/ :
D
kŠ.n " k/Š
60
10. If y D x !2=3 , then "y ' dy D ."2=3/x !5=3 dx. If
dx D .2=100/x, then "y ' ."4=300/x 2=3 D ."4=300/y,
so y decreases by about 1.33%.
11.
.3/
D f g C 3f g C 3f g C fg :
d
D
.fg/.3/
dx
d
D
Œf .3/ g C 3f 00 g 0 C 3f 0 g 00 C fg .3/ !
dx
D f .4/ g C f .3/ g 0 C 3f .3/ g 0 C 3f 00 g 00 C 3f 00 g 00
kD0
9.
If V D 34 # r 3 , then "V ' d V D 4# r 2 dr. If r increases by 2%, then dr D 2r=100 and "V ' 8# r 3 =100.
Therefore "V =V ' 6=100. The volume increases by about
6%.
12. If V is the volume and x is the edge length of the cube
then V D x 3 . Thus "V ' d V D 3x 2 "x. If
"V D ".6=100/V , then "6x 3 =100 ' 3x 2 dx, so
dx ' ".2=100/x. The edge of the cube decreases by
about 2%.
13.
14.
Rate change of Area A with respect to side s, where
dA
A D s 2 , is
D 2s: When s D 4 ft, the area is changing
ds
2
at rate 8 ft /ft.
p
p
If A D s 2 , then s D A and ds=dA D 1=.2 A/.
If A D 16 m2 , then the side is changing at rate
ds=dA D 1=8 m/m2 .
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INSTRUCTOR’S SOLUTIONS MANUAL
15.
SECTION 2.7 (PAGE 137)
The diameter
D and area A of a circle are related by
p
D D 2 A=#. The rate of change of diameter with respect
p
to area is dD=dA D 1=.#A/ units per square unit.
The flow rate will increase by 10% if the radius is increased by about 2.5%.
23.
F D k=r 2 implies that dF=dr D "2k=r 3 . Since
dF=dr D 1 pound/mi when r D 4; 000 mi, we have
2k D 4; 0003 . If r D 8; 000, we have
dF=dr D ".4; 000=8; 000/3 D "1=8. At r D 8; 000 mi F
decreases with respect to r at a rate of 1/8 pounds/mi.
24.
If price = $p, then revenue is $R D 4; 000p " 10p 2 .
2
16. Since A D #D =4, the rate of change of area with respect
to diameter is dA=dD D #D=2 square units per unit.
17.
Rate of change of V D
4 3
# r with respect to radius r is
3
dV
D 4# r 2 . When r D 2 m, this rate of change is 16#
dr
m3 /m.
a) Sensitivity of R to p is dR=dp D 4; 000 " 20p. If
p D 100, 200, and 300, this sensitivity is 2,000 $/$, 0
$/$, and "2; 000 $/$ respectively.
18. Let A be the area of a square, s be its side length and L
be its diagonal. Then, L2 D s 2 C s 2 D 2s 2 and
dA
A D s 2 D 12 L2 , so
D L. Thus, the rate of change of
dL
the area of a square with respect to its diagonal L is L.
b) The distributor should charge $200. This maximizes
the revenue.
25.
19. If the radius of the circle is r then C D 2# r and
A D # r 2.
r
p p
A
Thus C D 2#
D 2 # A.
#
Rate of p
change of C with respect to A is
#
dC
1
D p D .
dA
r
A
20. Let s be the side length and V be the volume of a cube.
ds
D 31 V !2=3 . Hence,
Then V D s 3 ) s D V 1=3 and
dV
the rate of change of the side length of a cube with respect to its volume V is 31 V !2=3 .
21.
a) Marginal cost if x D 100 is
C 0 .100/ D 400 " 100 D $300.
b) C.101/ " C.100/ D 43; 299:50 " 43; 000 D $299:50
which is approximately C 0 .100/.
26.
b) To maximize daily profit, production should be 800
sheets/day.
a) At t D 5, water volume is changing at rate
ˇ
ˇ
ˇ
ˇ
ˇ
tD5
ˇ
ˇ
ˇ
D "700.20 " t /ˇ
ˇ
tD5
D "10; 500:
27.
Water is draining out at 10,500 L/min at that time.
At t D 15, water volume is changing at rate
dV
dt
ˇ
ˇ
ˇ
ˇ
ˇ
tD15
ˇ
ˇ
ˇ
D "700.20 " t /ˇ
ˇ
tD15
D "3; 500:
350 & .25 " 225/
V .15/ " V .5/
D
D "7; 000:
15 " 5
10
The average rate of draining is 7,000 L/min over that
interval.
22. Flow rate F D kr 4 , so "F ' 4kr 3 "r. If "F D F=10,
then
F
kr 4
"r '
D
D 0:025r:
40kr 3
40kr 3
80; 000
n2
C 4n C
n
100
80; 000
n
dC
D"
C
4
C
:
2
dn
n
50
dC
(a) n D 100;
D "2. Thus, the marginal cost of
dn
production is "$2.
C D
82
dC
D
' 9:11. Thus, the marginal cost
dn
9
of production is approximately $9.11.
(b) n D 300;
Water is draining out at 3,500 L/min at that time.
b) Average rate of change between t D 5 and t D 15 is
Daily profit if production is x sheets per day is $P .x/
where
P .x/ D 8x " 0:005x 2 " 1; 000:
a) Marginal profit P 0 .x/ D 8 " 0:01x. This is positive if
x < 800 and negative if x > 800.
Volume in tank is V .t / D 350.20 " t /2 L at t min.
dV
dt
Cost is $C.x/ D 8; 000 C 400x " 0:5x 2 if x units are
manufactured.
28.
Daily profit P D 13x " C x D 13x " 10x " 20 "
x2
1000
x2
D 3x " 20 "
1000
Graph of P is a parabola opening downward. P will be
maximum where the slope is zero:
0D
2x
dP
D3"
so x D 1500
dx
1000
Should extract 1500 tonnes of ore per day to maximize
profit.
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SECTION 2.7 (PAGE 137)
ADAMS and ESSEX: CALCULUS 9
29. One of the components comprising C.x/ is usually a fixed
cost, $S, for setting up the manufacturing operation. On a
per item basis, this fixed cost $S=x, decreases as the number x of items produced increases, especially when x is
small. However, for large x other components of the total
cost may increase on a per unit basis, for instance labour
costs when overtime is required or maintenance costs for
machinery when it is over used.
C.x/
Let the average cost be A.x/ D
. The minimal avx
erage cost occurs at point where the graph of A.x/ has a
horizontal tangent:
0D
4.
If f .x/ D cos x C .x 2 =2/, then f 0 .x/ D x " sin x > 0
for x > 0. By the MVT, if x > 0, then
f .x/ " f .0/ D f 0 .c/.x " 0/ for some c > 0, so
f .x/ > f .0/ D 1. Thus cos x C .x 2 =2/ > 1 and
cos x > 1 " .x 2 =2/ for x > 0. Since both sides of the
inequality are even functions, it must hold for x < 0 as
well.
5.
Let f .x/ D tan x. If 0 < x < #=2, then by the MVT
f .x/ " f .0/ D f 0 .c/.x " 0/ for some c in .0; #=2/.
Thus tan x D x sec2 c > x, since secc > 1.
6.
dA
xC 0 .x/ " C.x/
D
:
dx
x2
C.x/
D A.x/.
x
Thus the marginal cost C 0 .x/ equals the average cost at
the minimizing value of x.
Hence, xC 0 .x/ " C.x/ D 0 ) C 0 .x/ D
7.
Let f .x/ D .1 C x/r " 1 " rx where r > 1.
Then f 0 .x/ D r.1 C x/r!1 " r.
If "1 $ x < 0 then f 0 .x/ < 0; if x > 0, then f 0 .x/ > 0.
Thus f .x/ > f .0/ D 0 if "1 $ x < 0 or x > 0.
Thus .1 C x/r > 1 C rx if "1 $ x < 0 or x > 0.
Let f .x/ D .1 C x/r where 0 < r < 1. Thus,
f 0 .x/ D r.1 C x/r!1 . By the Mean-Value Theorem, for
x # "1, and x ¤ 0,
f .x/ " f .0/
D f 0 .c/
x"0
.1 " x/r " 1
D r.1 C c/r!1
)
x
30. If y D Cp !r , then the elasticity of y is
"
p dy
p
D " !r ."r/Cp !r!1 D r:
y dp
Cp
for some c between 0 and x. Thus,
.1 C x/r D 1 C rx.1 C c/r!1 .
If "1 $ x < 0, then c < 0 and 0 < 1 C c < 1. Hence
Section 2.8 The Mean-Value Theorem
(page 144)
1.
f .x/ D x 2 ;
2
.1 C c/r!1 > 1
r!1
rx.1 C c/
Hence, .1 C x/ < 1 C rx.
If x > 0, then
f 0 .x/ D 2x
2
f .b/ " f .a/
b "a
D
b"a
b"a
bCa
0
D f .c/ D 2c ) c D
2
c>0
1Cc >1
.1 C c/r!1 < 1
rx.1 C c/r!1 < rx:
1
1
2. If f .x/ D , and f 0 .x/ D " 2 then
x
x
f .2/ " f .1/
1
1
1
D " 1 D " D " 2 D f 0 .c/
2"1
2
2
c
p
Hence, .1 C x/r < 1 C rx in this case also.
Hence, .1 C x/r < 1 C rx for either "1 $ x < 0 or x > 0.
8.
If f .x/ D x 3 " 12x C 1, then f 0 .x/ D 3.x 2 " 4/.
The critical points of f are x D ˙2. f is increasing on
."1; "2/ and .2; 1/ where f 0 .x/ > 0, and is decreasing
on ."2; 2/ where f 0 .x/ < 0.
9.
If f .x/ D x 2 " 4, then f 0 .x/ D 2x. The critical point of
f is x D 0. f is increasing on .0; 1/ and decreasing on
."1; 0/.
2 lies between 1 and 2.
3. f .x/ D x 3 " 3x C 1, f 0 .x/ D 3x 2 " 3, a D "2, b D 2
f .b/ " f .a/
f .2/ " f ."2/
D
b"a
4
8 " 6 C 1 " ."8 C 6 C 1/
D
4
4
D D1
4
0
2
f .c/ D 3c " 3
2
3c 2 " 3 D 1 ) 3c 2 D 4 ) c D ˙ p
3
(Both points will be in ."2; 2/.)
62
.since x < 0/:
r
bCa D
where c D
< rx
.since r " 1 < 0/;
10. If y D 1 " x " x 5 , then y 0 D "1 " 5x 4 < 0 for all
x. Thus y has no critical points and is decreasing on the
whole real line.
11.
If y D x 3 C 6x 2 , then y 0 D 3x 2 C 12x D 3x.x C 4/. The
critical points of y are x D 0 and x D "4. y is increasing
on ."1; "4/ and .0; 1/ where y 0 > 0, and is decreasing
on ."4; 0/ where y 0 < 0.
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 2.8 (PAGE 144)
12. If f .x/ D x 2 C 2x C 2 then f 0 .x/ D 2x C 2 D 2.x C 1/.
Evidently, f 0 .x/ > 0 if x > "1 and f 0 .x/ < 0 if x < "1.
Therefore, f is increasing on ."1; 1/ and decreasing on
."1; "1/.
22.
There is no guarantee that the MVT applications for f
and g yield the same c.
23.
CPs x D 0:535898 and x D 7:464102
13. f .x/ D x 3 " 4x C 1
f 0 .x/ D 3x 2 " 4
24.
CPs x D "1:366025 and x D 0:366025
25.
CPs x D "0:518784 and x D 0
26.
CP x D 0:521350
27.
If x1 < x2 < : : : < xn belong to I , and f .xi / D 0,
.1 $ i $ n/, then there exists yi in .xi ; xiC1 / such that
f 0 .yi / D 0, .1 $ i $ n " 1/ by MVT.
28.
For x ¤ 0, we have f 0 .x/ D 2x sin.1=x/ " cos.1=x/
which has no limit as x ! 0. However,
f 0 .0/ D limh!0 f .h/= h D limh!0 h sin.1= h/ D 0
does exist even though f 0 cannot be continuous at 0.
29.
If f 0 exists on Œa; b! and f 0 .a/ ¤ f 0 .b/, let us assume,
without loss of generality, that f 0 .a/ > k > f 0 .b/. If
g.x/ D f .x/ " kx on Œa; b!, then g is continuous on Œa; b!
because f , having a derivative, must be continuous there.
By the Max-Min Theorem, g must have a maximum value
(and a minimum value) on that interval. Suppose the maximum value occurs at c. Since g 0 .a/ > 0 we must have
c > a; since g 0 .b/ < 0 we must have c < b. By Theorem
14, we must have g 0 .c/ D 0 and so f 0 .c/ D k. Thus f 0
takes on the (arbitrary) intermediate value k.
'
2
f .x/ D x C 2x sin.1=x/ if x ¤ 0
0
if x D 0.
2
f 0 .x/ > 0 if jxj > p
3
2
0
f .x/ < 0 if jxj < p
3
2
2
f is increasing on ."1; " p / and . p ; 1/:
3
3
2
2
f is decreasing on ." p ; p /:
3
3
14.
If f .x/ D x 3 C 4x C 1, then f 0 .x/ D 3x 2 C 4. Since
f 0 .x/ > 0 for all real x, hence f .x/ is increasing on the
whole real line, i.e., on ."1; 1/.
15.
f .x/ D .x 2 " 4/2
f 0 .x/ D 2x2.x 2 " 4/ D 4x.x " 2/.x C 2/
f 0 .x/ > 0 if x > 2 or "2 < x < 0
f 0 .x/ < 0 if x < "2 or 0 < x < 2
f is increasing on ."2; 0/ and .2; 1/.
f is decreasing on ."1; "2/ and .0; 2/.
"2x
1
then f 0 .x/ D
. Evidently,
x2 C 1
.x 2 C 1/2
0
0
f .x/ > 0 if x < 0 and f .x/ < 0 if x > 0. Therefore, f
is increasing on ."1; 0/ and decreasing on .0; 1/.
16. If f .x/ D
17.
f .x/ D x 3 .5 " x/2
0
2
30.
2
3
f .x/ D 3x .5 " x/ C 2x .5 " x/."1/
f .0 C h/ " f .0/
h
h C 2h2 sin.1= h/
D lim
h
h!0
D lim .1 C 2h sin.1= h/ D 1;
a) f 0 .0/ D lim
D x 2 .5 " x/.15 " 5x/
h!0
D 5x 2 .5 " x/.3 " x/
f .x/ > 0 if x < 0, 0 < x < 3, or x > 5
f 0 .x/ < 0 if 3 < x < 5
f is increasing on ."1; 3/ and .5; 1/.
f is decreasing on .3; 5/.
0
h!0
because j2h sin.1= h/j $ 2jhj ! 0 as h ! 0.
b) For x ¤ 0, we have
18. If f .x/ D x " 2 sin x, then f 0 .x/ D 1 " 2 cos x D 0 at
x D ˙#=3 C 2n#for n D 0; ˙1; ˙2; : : :.
f is decreasing on ."#=3 C 2n#; # C 2n#/.
f is increasing on .#=3 C 2n#; "#=3 C 2.n C 1/#/ for
integers n.
f 0 .x/ D 1 C 4x sin.1=x/ " 2 cos.1=x/:
There are numbers x arbitrarily close to 0 where
f 0 .x/ D "1; namely, the numbers x D ˙1=.2n#/,
where n D 1, 2, 3, : : : . Since f 0 .x/ is continuous at
every x ¤ 0, it is negative in a small interval about
every such number. Thus f cannot be increasing on
any interval containing x D 0.
0
19. If f .x/ D x C sin x, then f .x/ D 1 C cos x # 0
f 0 .x/ D 0 only at isolated points x D ˙#; ˙3#; :::.
Hence f is increasing everywhere.
20. If f .x/ D x C 2 sin x, then f 0 .x/ D 1 C 2 cos x > 0
if cos x > "1=2. Thus f is increasing on the intervals
.".4#=3/ C 2n#; .4#=3/ C 2n#/ where n is any integer.
21.
f .x/ D x 3 is increasing on ."1; 0/ and .0; 1/ because
f 0 .x/ D 3x 2 > 0 there. But f .x1 / < f .0/ D 0 < f .x2 /
whenever x1 < 0 < x2 , so f is also increasing on intervals containing the origin.
31.
Let a, b, and c be three points in I where f vanishes;
that is, f .a/ D f .b/ D f .c/ D 0. Suppose a < b < c. By
the Mean-Value Theorem, there exist points r in .a; b/ and
s in .b; c/ such that f 0 .r/ D f 0 .s/ D 0. By the MeanValue Theorem applied to f 0 on Œr; s!, there is some point
t in .r; s/ (and therefore in I ) such that f 00 .t / D 0.
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SECTION 2.8 (PAGE 144)
ADAMS and ESSEX: CALCULUS 9
32. If f .n/ exists on interval I and f vanishes at n C 1 distinct points of I , then f .n/ vanishes at at least one point
of I .
Proof: True for n D 2 by Exercise 8.
Assume true for n D k. (Induction hypothesis)
Suppose n D k C 1, i.e., f vanishes at k C 2 points of I
and f .kC1/ exists.
By Exercise 7, f 0 vanishes at k C 1 points of I .
By the induction hypothesis, f .kC1/ D .f 0 /.k/ vanishes at
a point of I so the statement is true for n D k C 1.
Therefore the statement is true for all n # 2 by induction.
(case n D 1 is just MVT.)
3.
x 2 C xy D y 3
Differentiate with respect to x:
2x C y C xy 0 D 3y 2 y 0
2x C y
y0 D 2
3y " x
4.
x 3 y C xy 5 D 2
3x 2 y C x 3 y 0 C y 5 C 5xy 4 y 0 D 0
"3x 2 y " y 5
y0 D 3
x C 5xy 4
5.
x 2 y 3 D 2x " y
2xy 3 C 3x 2 y 2 y 0 D 2 " y 0
2 " 2xy 3
y0 D
3x 2 y 2 C 1
6.
x 2 C 4.y " 1/2 D 4
2x C 8.y " 1/y 0 D 0, so y 0 D
33. Given that f .0/ D f .1/ D 0 and f .2/ D 1:
a) By MVT,
f 0 .a/ D
1"0
1
f .2/ " f .0/
D
D
2"0
2"0
2
for some a in .0; 2/.
b) By MVT, for some r in .0; 1/,
f 0 .r/ D
7.
f .1/ " f .0/
0"0
D
D 0:
1"0
1"0
Also, for some s in .1; 2/,
f 0 .s/ D
f .2/ " f .1/
1"0
D
D 1:
2"1
2"1
8.
Then, by MVT applied to f 0 on the interval Œr; s!, for
some b in .r; s/,
1"0
f 0 .s/ " f 0 .r/
D
s"r
s"r
1
1
D
>
s"r
2
f 00 .b/ D
9.
Section 2.9 Implicit Differentiation
(page 149)
1.
xy " x C 2y D 1
Differentiate with respect to x:
y C xy 0 " 1 C 2y 0 D 0
1"y
Thus y 0 D
2Cx
2. x 3 C y 3 D 1
3x 2 C 3y 2 y 0 D 0, so y 0 D "
64
x2
.
y2
x2
x2 C y
x"y
D
C1D
xCy
y
y
Thus xy "y 2 D x 3 Cx 2 y Cxy Cy 2 , or x 3 Cx 2 y C2y 2 D 0
Differentiate with respect to x:
3x 2 C 2xy C x 2 y 0 C 4yy 0 D 0
3x 2 C 2xy
y0 D " 2
x C 4y
p
x x C y D 8 " xy
1
p
xCyCx p
.1 C y 0 / D "y " xy 0
2 xCy
p
2.x C y/ C x.1 C y 0 /p
D "2 x C y.y C xy 0 /
3x C 2y C 2y x C y
y0 D "
p
x C 2x x C y
2x 2 C 3y 2 D 5
4x C 6yy 0 D 0
2
At .1; 1/: 4 C 6y 0 D 0, y 0 D "
3
2
Tangent line: y " 1 D " .x " 1/ or 2x C 3y D 5
3
since s " r < 2.
c) Since f 00 .x/ exists on Œ0; 2!, therefore f 0 .x/ is continuous there. Since f 0 .r/ D 0 and f 0 .s/ D 1, and
since 0 < 71 < 1, the Intermediate-Value Theorem
assures us that f 0 .c/ D 71 for some c between r and
s.
x
4.1 " y/
10. x 2 y 3 " x 3 y 2 D 12
2xy 3 C 3x 2 y 2 y 0 " 3x 2 y 2 " 2x 3 yy 0 D 0
At ."1; 2/: "16 C 12y 0 " 12 C 4y 0 D 0, so the slope is
28
7
12 C 16
D
D :
y0 D
12 C 4
16
4
Thus, the equation of the tangent line is
y D 2 C 74 .x C 1/, or 7x " 4y C 15 D 0:
11.
x % y &3
C
D2
y
x
4
4
x C y D 2x 3 y
4x 3 C 4y 3 y 0 D 6x 2 y C 2x 3 y 0
at ."1; "1/: "4 " 4y 0 D "6 " 2y 0
2y 0 D 2, y 0 D 1
Tangent line: y C 1 D 1.x C 1/ or y D x.
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 2.9 (PAGE 149)
y2
x"1
.x " 1/2yy 0 " y 2 .1/
0
1 C 2y D
.x " 1/2
At .2; "1/ we have 1 C 2y 0 D "2y 0 " 1 so y 0 D " 12 .
Thus, the equation of the tangent is
y D "1 " 21 .x " 2/, or x C 2y D 0:
p
13. 2x C y "p 2 sin.xy/ D #=2
2 C y 0 " 2 cos.xy/.y C xy 0 / D 0
At .#=4; 1/: 2 C y 0 " .1 C .#=4/y 0 / D 0, so
y 0 D "4=.4 " #/. The tangent has equation
12. x C 2y C 1 D
y D1"
14.
20.
y " x2
y2 " x
"2x C 2y 0 " 2y.y 0 /2
y 00 D
y2 " x
"
#
$
#
$2 #
2
y " x2
y " x2
D 2
"x C
"y
y "x
y2 " x
y2 " x
)
(
2
4xy
"2xy
D 2
D
:
y " x .y 2 " x/2
.x " y 2 /3
21.
x 2 C y 2 D a2
x
2x C 2yy 0 D 0 so x C yy 0 D 0 and y 0 D "
y
1 C y 0 y 0 C yy 00 D 0 so
x2
1C 2
0 2
1 C .y /
y
D"
y 00 D "
y
y
a2
y2 C x2
D"
D" 3
3
y
y
22.
Ax 2 C By 2 D C
2Ax C 2Byy 0 D 0 ) y 0 D "
y 00 D
y "1
y C xy 0 D 1 C y 0 ) y 0 D
1"x
y 0 C y 0 C xy 00 D y 00
2.y " 1/
2y 0
D
Therefore, y 00 D
1"x
.1 " x/2
2 C 8.y 0 /2 C 8yy 00 D 0.
y 00 D
2
2
24.
Maple gives the slope as
26.
#
Ax
By
$2
By
"A.By 2 C Ax 2 /
AC
D
D " 2 3:
B 2y3
B y
Maple gives 0 for the value.
2
"4y " x
1
"2 " 8.y /
1
x
D" "
D
D " 3:
8y
4y 16y 3
16y 3
4y
"A " B.y 0 /2
D
By
"A " B
23.
25.
0 2
Ax
By
2A C 2B.y 0 /2 C 2Byy 00 D 0.
Thus,
xy D x C y
18. x 2 C 4y 2 D 4; 2x C 8yy 0 D 0;
"x
Thus, y 0 D
and
4y
x 3 " 3xy C y 3 D 1
3x 2 " 3y " 3xy 0 C 3y 2 y 0 D 0
6x " 3y 0 " 3y 0 " 3xy 00 C 6y.y 0 /2 C 3y 2 y 00 D 0
Thus
y0 D
# "2
1
C
.x C #/:
2
4#.# " 1/
% #y &
x 2 17
16. cos
D
"
x
y
2
h
% #y &i #.xy 0 " y/
2xy " x 2 y 0
" sin
D
:
x p
x2
y2
0
3 #.3y " 1/
D 6 " 9y 0 ,
At .3; 1/: "
2 p 9
p
so y 0 D .108 " 3#/=.162 " 3 3#/. The tangent has
equation
p
108 " 3#
p .x " 3/:
y D1C
162 " 3 3#
17.
1 " 3x 2
3y 2 " 2y
0 2
00
0 2
6x " 2.y / " 2yy C 6y.y / C 3y 2 y 00 D 0
.1 " 3x 2 /2
" 6x
.2 " 6y/
0 2
.2
"
6y/.y
/
"
6x
.3y 2 " 2y/2
y 00 D
D
3y 2 " 2y
3y 2 " 2y
2 2
.2 " 6y/.1 " 3x /
6x
D
" 2
.3y 2 " 2y/3
3y " 2y
#&
4 %
x"
:
4"#
4
x sin.xy " y 2 / D x 2 " 1
sin.xy " y 2 / C x.cos.xy " y 2 //.y C xy 0 " 2yy 0 / D 2x.
At .1; 1/: 0 C .1/.1/.1 " y 0 / D 2, so y 0 D "1. The tangent
has equation y D 1 " .x " 1/, or y D 2 " x.
x3 " y2 C y3 D x
3x 2 " 2yy 0 C 3y 2 y 0 D 1 ) y 0 D
tan.xy 2 / D .2=#/xy
.sec2 .xy 2 //.y 2 C 2xyy 0 / D .2=#/.y C xy 0 /.
At ."#; 1=2/: 2..1=4/ " #y 0 / D .1=#/ " 2y 0 , so
y 0 D .# " 2/=.4#.# " 1//. The tangent has equation
yD
15.
19.
206
.
55
Maple gives the value "26.
Maple gives the value "
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.
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SECTION 2.9 (PAGE 149)
27.
ADAMS and ESSEX: CALCULUS 9
Ellipse: x 2 C 2y 2 D 2
2x C 4yy 0 D 0
30.
x
Slope of ellipse:
D"
2y
Hyperbola: 2x 2 " 2y 2 D 1
4x " 4yy 0 D 0
x
0
Slope of hyperbola: yH
D
y
' 2
x C 2y 2 D 2
At intersection points
2x 2 " 2y 2 D 1
1
2
2
2
3x D 3 so x D 1, y D
2
x x
x2
0 0
Thus yE
yH D "
D " 2 D "1
2y y
2y
Therefore the curves intersect at right angles.
0
yE
2x C 4yy 0 C y C xy 0 D 0
i.e. y 0 D "
Similarly, the slope of the hyperbola
.x; y/ satisfies
2y 0
2x
"
y D 0;
A2 B 2
2
2
Section 2.10 Antiderivatives and Initial-Value
Problems (page 155)
1.
2
B x
:
A2 y
2.
2 2
b CB
A a
x
D
! 2
.
y2
B 2 b2
a " A2
2
2
2
2
Since a " b D A C B , therefore B 2 C b 2 D a2 " A2 ,
x2
A2 a2
and 2 D 2 2 . Thus, the product of the slope of the
y
B b
two curves at .x; y/ is
3.
4.
5.
Thus,
b 2 B 2 A2 a2
b2 x B 2 x
" 2 ! 2 D " 2 2 ! 2 2 D "1:
a y A y
a A B b
6.
7.
8.
Therefore, the curves intersect at right angles.
9.
29. If z D tan.x=2/, then
1 D sec2 .x=2/
2
2
1 C tan .x=2/ dx
1 C z dx
1 dx
D
D
:
2 dz
2
dz
2
dz
2
D
2
sec2 .x=2/
1 " z2
2
"1D
2
1Cz
1 C z2
sin x D 2 sin.x=2/ cos.x=2/ D
66
10.
11.
Thus dx=dz D 2=.1 C z /. Also
cos x D 2 cos2 .x=2/ " 1 D
2x C y
:
4y C x
the only solution is x D 0, y D 0, and these values do not
satisfy the original equation. There are no points on the
given curve.
b2 x
:
a2 y
If the point .x; y/ is an intersection of the two curves,
then
y2
x2
y2
x2
C 2 D 2 " 2
2
b$
A # B
# a
$
1
1
1
1
2
2
x
"
Dy
C 2 :
A2 a2
B2
b
2
y0 D "
1
7
y
7
x C xy C y 2 C y 2 D 0; or .x C /2 C y 2 D 0;
4
4
2
4
x2
y2
" 2 D 1 at
2
A
B
or y 0 D
)
However, since x 2 C 2y 2 C xy D 0 can be written
x2
y2
28. The slope of the ellipse 2 C 2 D 1 is found from
a
b
2x
2y
C 2 y 0 D 0;
2
a
b
x
x"y
D C 1 , xy " y 2 D x 2 C xy C xy C y 2
xCy
y
, x 2 C 2y 2 C xy D 0
Differentiate with respect to x:
"1
12.
2 tan.x=2/
2z
D
:
1 C tan2 .x=2/
1 C z2
13.
Z
Z
Z
Z
Z
5 dx D 5x C C
x 2 dx D 31 x 3 C C
p
x dx D
x 12 dx D
x 3 dx D
2 3=2
x
CC
3
1 13
13 x
CC
1 4
x CC
4
Z
.x C cos x/ dx D
Z
1 C cos3 x
dx D
cos2 x
x2
C sin x C C
2
Z
Z
tan x cos x dx D sin x dx D " cos x C C
Z
Z
Z
Z
.sec2 xCcos x/ dx D tan xCsin xCC
1
.a2 " x 2 / dx D a2 x " x 3 C C
3
.A C Bx C C x 2 / dx D Ax C
.2x 1=2 C 3x 1=3 dx D
Z
B 2 C 3
x C x CK
2
3
4 3=2 9 4=3
x
C x
CC
3
4
Z
6.x " 1/
dx
D
.6x !1=3 " 6x !4=3 / dx
x 4=3
D 9x 2=3 C 18x !1=3 C C
$
Z # 3
x2
1 4 1 3 1 2
x
"
C x " 1 dx D
x " x C x "x CC
3
2
12
6
2
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INSTRUCTOR’S SOLUTIONS MANUAL
14.
105
Z
SECTION 2.10
.1 C t 2 C t 4 C t 6 / dt
29.
'
30.
Given that
D 105.t C 13 t 3 C 15 t 5 C 17 t 7 / C C
D 105t C 35t 3 C 21t 5 C 15t 7 C C
Z
1
15.
cos.2x/ dx D sin.2x/ C C
2
Z
%x &
%x &
16.
sin
dx D "2 cos
CC
2
2
Z
1
dx
D"
CC
17.
.1 C x/2
1Cx
Z
18.
sec.1 " x/ tan.1 " x/ dx D " sec.1 " x/ C C
19.
Z
p
20. Since
21.
Z
p
p
4
dx D 8 x C 1 C C:
xC1
23.
24.
25.
Z
Z
Z
Z
2
tan x dx D
p
Z
2x
x2 C 1
Z
Z
33.
1 " cos.2x/
x sin.2x/
26.
sin2 x dx D
dx D "
CC
2
2
4
(
1
y 0 D x " 2 ) y D x 2 " 2x C C
27.
2
y.0/ D 3
) 3 D 0 C C therefore C D 3
1
Thus y D x 2 " 2x C 3 for all x.
2
28. Given that
' 0
y D x !2 " x !3
y."1/ D 0;
Z
then y D .x !2 " x !3 / dx D "x !1 C 12 x !2 C C
and 0 D y."1/ D "."1/!1 C 12 ."1/!2 C C so C D " 32 .
1
1
3
Hence, y.x/ D " C
" which is valid on the
x
2x 2
2
interval ."1; 0/.
Z
'
y 0 D x !9=7
y.1/ D "4;
x !9=7 dx D " 72 x !2=7 C C .
Also, "4 D y.1/ D " 72 C C , so C D " 12 . Hence,
y D " 72 x !2=7 " 21 , which is valid in the interval .0; 1/.
' 0
y D cos x
For
, we have
y.#=6/ D 2
yD
Z
cos x dx D sin x C C
#
1
CC D CC %
6
2
3
(for all x):
y D sin x C
2
2 D sin
1
1
sin.2x/ dx D " cos.2x/ C C
2
4
x
sin.2x/
1 C cos.2x/
dx D C
CC
2
2
4
C 5 which is valid on the whole real
Given that
.sec x " 1/ dx D tan x " x C C
Z
C C and 5 D y.0/ D C .
32.
2
sin x cos x dx D
cos2 x dx D
p
dx D 2 x 2 C 1 C C:
3 4=3
4x
3 4=3
4x
Since y 0 D Ax 2 C Bx C C we have
A
B
y D x 3 C x 2 C C x C D. Since y.1/ D 1, therefore
3
2
A B
A B
1 D y.1/ D C C C C D. Thus D D 1 " " " C ,
3
2
3
2
and
A 3
B 2
y D .x " 1/ C .x " 1/ C C.x " 1/ C 1 for all x
3
2
2x sin.x 2 / dx D " cos.x 2 / C C
Z
x 1=3 dx D
y 0 D x 1=3
y.0/ D 5;
31.
then y D
d p 2
x
22. Since
x C1D p
, therefore
dx
x2 C 1
Z
'
Hence, y.x/ D
line.
d p
1
xC1D p
, therefore
dx
2 xC1
Z
p
y 0 D 3 x ) y D 2x 3=2 C C
y.4/ D 1 )
1 D 16 C C so C D "15
Thus y D 2x 3=2 " 15 for x > 0.
then y D
1
.2x C 3/3=2 C C
3
2x C 3 dx D
(PAGE 155)
34.
For
'
C D
3
2
y 0 D sin.2x/
, we have
y.#=2/ D 1
Z
1
sin.2x/ dx D " cos.2x/ C C
2
1
1
1
1 D " cos # C C D C C % C D
2
2
2
&
1%
yD
1 " cos.2x/
(for all x):
2
yD
35.
'
y 0 D sec2 x
; we have
y.0/ D 1
Z
y D sec2 x dx D tan x C C
For
1 D tan 0 C C D C % C D 1
y D tan x C 1
(for "#=2 < x < #=2/.
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SECTION 2.10 (PAGE 155)
36. For
'
ADAMS and ESSEX: CALCULUS 9
8 00
< y D cos x
41. For y.0/ D 0 we have
: 0
y .0/ D 1
y 0 D sec2 x
, we have
y.#/ D 1
yD
Z
sec2 x dx D tan x C C
y0 D
1 D tan # C C D C % C D 1
y D tan x C 1
(for #=2 < x < 3#=2/:
then y 0 D
Z
0 D " cos 0 C 0 C C2
y D 1 C x " cos x:
42.
8
< y 00 D x !4
y 0 .1/ D 2
:
y.1/ D 1;
x !4 dx D " 13 x !3 C C .
Z %
&
" 13 x !3 C 73 dx D 61 x !2 C 73 x C D;
and 1 D y.1/ D 16 C 73 C D, so that D D " 23 . Hence,
y.x/ D 16 x !2 C 37 x " 32 , which is valid in the interval
.0; 1/.
1 4
x " x C C1 .
4
0
Since y .0/ D 0, therefore 0 D 0 " 0 C C1 , and
1
y 0 D x 4 " x.
4
1 5 1 2
x " x C C2 .
Thus y D
20
2
Since y.0/ D 8, we have 8 D 0 " 0 C C2 .
1 5 1 2
Hence y D
x " x C 8 for all x.
20
2
40. Given that
8
< y 00 D 5x 2 " 3x !1=2
y 0 .1/ D 2
:
y.1/ D 0;
Z
we have y 0 D
5x 2 " 3x !1=2 dx D 53 x 3 " 6x 1=2 C C .
43.
39. Since y 00 D x 3 " 1, therefore y 0 D
5
3 "
1=2
6x
C 19
3 ,
Also, 2 D y 0 .1/ D
y 0 D 53 x 3 "
yD
Z %
6 C C so that C D
19
. Thus,
3
and
5 3
x " 6x 1=2 C 19
3
3
&
dx D
C2 D 1
8 00
< y D x C sin x
For y.0/ D 2
we have
: 0
y .0/ D 0
y0 D
Z
yD
x3
" sin x C x C 2:
6
.x C sin x/ dx D
B
B
2B
. Then y 0 D A " 2 , and y 00 D 3 .
x
x
x
Thus, for all x ¤ 0,
Let y D Ax C
x 2 y 00 C xy 0 " y D
B
B
2B
C Ax "
" Ax "
D 0:
x
x
x
We will also have y.1/ D 2 and y 0 .1/ D 4 provided
A C B D 2;
and A " B D 4:
These equations have solution A D 3, B D "1, so the
initial value problem has solution y D 3x " .1=x/.
44.
Let r1 and r2 be distinct rational roots of the equation
ar.r " 1/ C br C c D 0
Let y D Ax r1 C Bx r2
.x > 0/
Then y 0 D Ar1 x r1 !1 C Br2 x r2 !1 ,
and y 00 D Ar1 .r1 " 1/x r1 !2 C Br2 .r2 " 1/x r2 !2 . Thus
ax 2 y 00 C bxy 0 C cy
D ax 2 .Ar1 .r1 " 1/x r1 !2 C Br2 .r2 " 1/x r2 !2
5 4
x " 4x 3=2 C 19
x C D:
12
3
11
5
" 4 C 19
Finally, 0 D y.1/ D 12
3 C D so that D D " 4 .
5 4
19
11
3=2
Hence, y.x/ D 12 x " 4x
C 3 x" 4.
68
%
x2
" cos x C C1
2
0 D 0 " cos 0 C C1 % C1 D 1
$
Z # 2
x
x3
yD
" cos x C 1 dx D
" sin x C x C C2
2
6
2 D 0 " sin 0 C 0 C C2 % C2 D 2
Since 2 D y 0 .1/ D " 31 C C , therefore C D 37 ,
and y 0 D " 13 x !3 C 73 . Thus
yD
cos x dx D sin x C C1
1 D sin 0 C C1 % C1 D 1
Z
y D .sin x C 1/ dx D " cos x C x C C2
37. Since y 00 D 2, therefore y 0 D 2x C C1 .
Since y 0 .0/ D 5, therefore 5 D 0 C C1 , and y 0 D 2x C 5.
Thus y D x 2 C 5x C C2 .
Since y.0/ D "3, therefore "3 D 0C0CC2 , and C2 D "3.
Finally, y D x 2 C 5x " 3, for all x.
38. Given that
Z
C bx.Ar1 x r1 !1 C Br2 x r2 !1 / C c.Ax r1 C Bx r2 /
%
&
D A ar1 .r1 " 1/ C br1 C c x r1
&
C B.ar2 .r2 " 1/ C br2 C c x r2
D 0x r1 C 0x r2 ) 0 .x > 0/
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INSTRUCTOR’S SOLUTIONS MANUAL
8
< 4x 2 y 00 C 4xy 0 " y D 0
45.
y.4/ D 2
: 0
y .4/ D "2
.(/
SECTION 2.11
) a D 4; b D 4; c D "1
d) never accelerating to the left
e) particle is speeding up for t > 2
Auxilary Equation: 4r.r " 1/ C 4r " 1 D 0
4r 2 " 1 D 0
1
r D˙
2
By #31, y D Ax 1=2 C Bx !1=2 solves .(/ for x > 0.
A
B
Now y 0 D x !1=2 " x !3=2
2
2
Substitute the initial conditions:
B
2
A
B
"2 D "
4
16
2 D 2A C
)1 D A C
f) slowing down for t < 2
g) the acceleration is 2 at all times
h) average velocity over 0 $ t $ 4 is
16 " 16 C 3 " 3
x.4/ " x.0/
D
D0
4"0
4
B
4
B
:
4
)"8 DA"
2.
x D 4 C 5t " t 2 , v D 5 " 2t , a D "2.
a) The point is moving to the right if v > 0, i.e., when
t < 25 .
B
7
, so B D 18, A D " .
2
2
7
Thus y D " x 1=2 C 18x !1=2 (for x > 0).
2
46. Consider
8
< x 2 y 00 " 6y D 0
y.1/ D 1
: 0
y .1/ D 1:
Hence 9 D
b) The point is moving to the left if v < 0, i.e., when
t > 25 .
c) The point is accelerating to the right if a > 0, but
a D "2 at all t ; hence, the point never accelerates to
the right.
d) The point is accelerating to the left if a < 0, i.e., for
all t .
Let y D x r ; y 0 D rx r!1 ; y 00 D r.r " 1/x r!2 . Substituting
these expressions into the differential equation we obtain
e) The particle is speeding up if v and a have the same
sign, i.e., for t > 52 .
x 2 Œr.r " 1/x r!2 ! " 6x r D 0
Œr.r " 1/ " 6!x r D 0:
f) The particle is slowing down if v and a have opposite
sign, i.e., for t < 52 .
Since this equation must hold for all x > 0, we must have
g) Since a D "2 at all t , a D "2 at t D
r.r " 1/ " 6 D 0
r2 " r " 6 D 0
.r " 3/.r C 2/ D 0:
There are two roots: r1 D "2, and r2 D 3. Thus the
differential equation has solutions of the form
y D Ax !2 C Bx 3 . Then y 0 D "2Ax !3 C 3Bx 2 . Since
1 D y.1/ D A C B and 1 D y 0 .1/ D "2A C 3B, therefore
A D 52 and B D 35 . Hence, y D 25 x !2 C 53 x 3 .
Section 2.11 Velocity and Acceleration
(page 162)
1.
(PAGE 162)
dx
dv
x D t " 4t C 3, v D
D 2t " 4, a D
D2
dt
dt
a) particle is moving: to the right for t > 2
2
b) to the left for t < 2
c) particle is always accelerating to the right
3.
5
2
when v D 0.
h) The average velocity over Œ0; 4! is
8"4
x.4/ " x.0/
D
D 1.
4
4
dx
dv
x D t 3 " 4t C 1, v D
D 3t 2 " 4, a D
D 6t
dt
dt
p
a) particlepmoving: to the right for t < "2= 3 or
t > 2= 3,
p
p
b) to the left for "2= 3 < t < 2= 3
c) particle is accelerating: to the right for t > 0
d) to the left for t < 0
p
e) particle
p is speeding up for t > 2= 3 or for
"2= 3 < t < 0
p
f) particle is slowing
down for t < "2= 3 or for
p
0 < t < 2= 3
p
g) velocity is zeropat t D ˙2= 3. Acceleration at these
times is ˙12= 3.
h) average velocity on Œ0; 4! is
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D 12
4"0
69
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SECTION 2.11 (PAGE 162)
4.
ADAMS and ESSEX: CALCULUS 9
.t 2 C 1/.1/ " .t /.2t /
1 " t2
t
; vD
D 2
;
2
2
C1
.t C 1/
.t C 1/2
.t 2 C 1/2 ."2t / " .1 " t 2 /.2/.t 2 C 1/.2t /
2t .t 2 " 3/
aD
D
:
.t 2 C 1/4
.t 2 C 1/3
xD
7.
D D t 2 , D in metres, t in seconds
dD
D 2t
velocity v D
dt
Aircraft becomes airborne if
500
200; 000
D
m/s:
v D 200 km/h D
3600
9
250
Time for aircraft to become airborne is t D
s, that is,
9
about 27:8 s.
Distance travelled during takeoff run is t 2 ' 771:6 metres.
8.
Let y.t / be the height of the projectile t seconds after it is
fired upward from ground level with initial speed v0 . Then
t2
a) The point is moving to the right if v > 0, i.e., when
1 " t 2 > 0, or "1 < t < 1.
b) The point is moving to the left if v < 0, i.e., when
t < "1 or t > 1.
c) The point is accelerating to the right if a > 0, i.e.,
whenp2t .t 2 "p3/ > 0, that is, when
t > 3 or " 3 < t < 0.
y 00 .t / D "9:8; y 0 .0/ D v0 ; y.0/ D 0:
d) The point
p to the left if a < 0, i.e., for
p is accelerating
t < " 3 or 0 < t < 3.
Two antidifferentiations give
y D "4:9t 2 C v0 t D t .v0 " 4:9t /:
e) The particle is speeding
p up if v and a have the same
sign, i.e.,pfor t < " 3, or "1 < t < 0 or
1 < t < 3.
Since the projectile returns to the ground at t D 10 s,
we have y.10/ D 0, so v0 D 49 m/s. On Mars, the
acceleration of gravity is 3.72 m/s2 rather than 9.8 m/s2 ,
so the height of the projectile would be
f) The particle is p
slowing down if v and a have opposite
sign,pi.e., for " 3 < t < "1, or 0 < t < 1 or
t > 3.
"2."2/
1
g) v D 0 at t D ˙1. At t D "1, a D
D .
.2/3
2
2."2/
1
At t D 1, a D
D" .
.2/3
2
y D "1:86t 2 C v0 t D t .49 " 1:86t /:
The time taken to fall back to ground level on Mars would
be t D 49=1:86 ' 26:3 s.
9.
h) The average velocity over Œ0; 4! is
4
"0
x.4/ " x.0/
1
D 17
D
.
4
4
17
5. y D 9:8t " 4:9t 2 metres (t in seconds)
dy
D 9:8 " 9:8t
velocity v D
dt
dv
acceleration a D
D "9:8
dt
The acceleration is 9:8 m/s2 downward at all times.
Ball is at maximum height ˇwhen v D 0, i.e., at t D 1.
ˇ
Thus maximum height is y ˇ
D 9:8 " 4:9 D 4:9 metres.
tD1
Ball strikes the ground when y D 0, .t > 0/, i.e.,
0 D t .9:8 " 4:9t / so t D 2.
Velocity at t D 2 is 9:8 " 9:8.2/ D "9:8 m/s.
Ball strikes the ground travelling at 9.8 m/s (downward).
hD"
6. Given that y D 100 " 2t " 4:9t , the time t at which the
ball reaches the ground is the positive root of the equation
y D 0, i.e., 100 " 2t " 4:9t 2 D 0, namely,
tD
"2 C
p
4 C 4.4:9/.100/
' 4:318 s:
9:8
"100
The average velocity of the ball is
D "23:16 m=s.
4:318
Since "23:159 D v D "2 " 9:8t , then t ' 2:159 s.
70
v2
v0
g v02
! 2 C v0 !
D 0:
2 g
g
2g
An initial speed of 2v0 means the maximum height will
be 4v02 =2g D 4h. To get a maximum height of 2h an
p
initial speed of 2v0 is required.
10. To get to 3h metres above Mars, the ball would have to be
thrown upward with speed
q
p
p
vM D 6gM h D 6gM v02 =.2g/ D v0 3gM =g:
Since gM D 3:72 and g D 9:80, we have vM ' 1:067v0
m/s.
11.
2
The height of the ball after t seconds is
y.t / D ".g=2/t 2 C v0 t m if its initial speed was v0
m/s. Maximum height h occurs when dy=dt D 0, that is,
at t D v0 =g. Hence
If the cliff is h ft high, then the height of the rock t seconds after it falls is y D h "p
16t 2 ft. The rock hits the
ground (y D 0) at time p
t D h=16 s. Its speedpat that
time is v D "32t D "8 h D "160 ft/s. Thus h D 20,
and the cliff is h D 400 ft high.
12. If the cliff is h ft high, then the height of the rock t seconds after it is thrown down is y D h " 32t " 16t 2 ft. The
rock hits the ground (y D 0) at time
p
1p
"32 C 322 C 64h
D "1 C
tD
16 C h s:
32
4
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 2.11
v
Its speed at that time is
(PAGE 162)
.4; 96/
p
v D "32 " 32t D "8 16 C h D "160 ft/s:
Solving this equation for h gives the height of the cliff as
384 ft.
t
13. Let x.t / be the distance travelled by the train in the t seconds after the brakes are applied. Since d 2 x=dt 2 D "1=6
m/s2 and since the initial speed is v0 D 60 km/h D 100=6
m/s, we have
1
100
x.t / D " t 2 C
t:
12
6
The speed
of the train at time t is v.t / D ".t =6/ C .100=6/ m/s,
so it takes the train 100 s to come to a stop. In that time
it travels x.100/ D "1002 =12 C 1002 =6 D 1002 =12 ' 833
metres.
14.
15.
x D At 2 C Bt C C; v D 2At C B.
The average velocity over Œt1 ; t2 ! is
x.t2 / " x.t1 /
t2 " t1
At22 C Bt1 C C " At12 " Bt1 " C
D
t2 " t1
A.t22 " t12 / C B.t2 " t1 /
D
.t2 " t1 /
A.t2 C t1 /.t2 " t1 / C B.t2 " t1 /
D
.t2 " t1 /
D A.t2 C t1 / C B.
The
velocity
at$the midpoint of Œt1 ; t2 ! is
$
#
# instantaneous
t2 C t1
t 2 C t1
D 2A
C B D A.t2 C t1 / C B.
v
2
2
Hence, the average velocity over the interval is equal to
the instantaneous velocity at the midpoint.
8
< t2
0$t $2
s D 4t " 4
2<t <8
:
"68 C 20t " t 2 8 $ t $ 10
Note: s is continuous at 2 and 8 since 22 D 4.2/ " 4 and
4.8/ " 4 D "68 C 160
( " 64
2t
if 0 < t < 2
ds
D 4
velocity v D
if 2 < t < 8
dt
20 " 2t if 8 < t < 10
Since 2t ! 4 as t ! 2", therefore, v is continuous at 2
(.v.2/ D 4).
Since 20 " 2t ! 4 as t ! 8C, therefore v is continuous
at 8 .v.8/ D 4/. Hence the velocity is continuous for
0 < t < 10
(
2
if 0 < t < 2
dv
acceleration a D
D 0
if 2 < t < 8
dt
"2 if 8 < t < 10
is discontinuous at t D 2 and t D 8
Maximum velocity is 4 and is attained on the interval
2 $ t $ 8.
.14; "224/
Fig. 2.11-16
The rocket’s acceleration while its fuel lasted is the slope
of the first part of the graph, namely 96=4 D 24 ft/s.
17.
The rocket was rising until the velocity became zero, that
is, for the first 7 seconds.
18. As suggested in Example 1 on page 154 of the text, the
distance travelled by the rocket while it was falling from
its maximum height to the ground is the area between the
velocity graph and the part of the t -axis where v < 0. The
area of this triangle is .1=2/.14 " 7/.224/ D 784 ft. This
is the maximum height the rocket achieved.
19.
The distance travelled upward by the rocket while it was
rising is the area between the velocity graph and the part
of the t -axis where v > 0, namely .1=2/.7/.96/ D 336 ft.
Thus the height of the tower from which the rocket was
fired is 784 " 336 D 448 ft.
20.
Let s.t / be the distance the car travels in the t seconds
after the brakes are applied. Then s 00 .t / D "t and the
velocity at time t is given by
s 0 .t / D
Z
."t / dt D "
t2
C C1 ;
2
where C1 D 20 m/s (that is, 72km/h) as determined in
Example 6. Thus
s.t / D
$
Z #
t3
t2
dt D 20t "
C C2 ;
20 "
2
6
where C2 D 0 because s.0/ D 0. The time taken
p to come
to a stop is given by s 0 .t / D 0, so it is t D 40 s. The
distance travelled is
16. This exercise and the next three refer to the following figure depicting the velocity of a rocket fired from a tower as
a function of time since firing.
p
1
s D 20 40 " 403=2 ' 84:3 m.
6
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SECTION 2.11 (PAGE 162)
ADAMS and ESSEX: CALCULUS 9
Review Exercises 2 (page 163)
1.
8.
y D .3x C 1/2
.3x C 3h C 1/2 " .3x C 1/2
dy
D lim
dx
h
h!0
9x 2 C 18xh C 9h2 C 6x C 6h C 1 " .9x 2 C 6x C 1/
D lim
h
h!0
D lim .18x C 9h C 6/ D 18x C 6
d !4
d 1 C x C x2 C x3
D
.x C x !3 C x !2 C x !1 /
dx
x4
dx
D "4x !5 " 3x !4 " 2x !3 " x !2
9.
d
.4 " x 2=5 /!5=2
dx
D x !3=5 .4 " x 2=5 /!7=2
h!0
2.
3.
p
p
1 " .x C h/2 " 1 " x 2
d p
1 " x 2 D lim
dx
h
h!0
1 " .x C h/2 " .1 " x 2 /
p
D lim p
h!0 h. 1 " .x C h/2 C
1 " x2/
"2x " h
x
D lim p
p
D "p
h!0
1 " .x C h/2 C 1 " x 2
1 " x2
f .x/ D 4=x 2
10.
11.
12.
4
"1
.2 C h/2
0
f .2/ D lim
h
h!0
"4 " h
4 " .4 C 4h C h2 /
D lim
D "1
D lim
h.2 C h/2
h!0
h!0 .2 C h/2
4.
g.t / D
t "5
p
1C t
4Ch
p
"1
1C 9Ch
0
g .9/ D lim
h
h!0
p
p
.3 C h " 9 C h/.3 C h C 9 C h/
D lim
p
p
h!0 h.1 C
9 C h/.3 C h C 9 C h/
9 C 6h C h2 " .9 C h/
D lim
p
p
h!0 h.1 C
9 C h/.3 C h C 9 C h/
5Ch
p
p
D lim
h!0 .1 C
9 C h/.3 C h C 9 C h/
5
D
24
5. The tangent to y D cos.#x/ at x D 1=6 has slope
Its equation is
ˇ
dy ˇˇ
ˇ
dx ˇ
xD1=6
yD
D "# sin
p
3 #
"
2
2
#
#
D" :
6
2
#
$
1
x"
:
6
1
1 " cos x
d
D"
dx x " sin x
.x " sin x/2
72
13.
14.
d p
"2 cos x sin x
" sin x cos x
2 C cos2 x D p
D p
dx
2 2 C cos2 x
2 C cos2 x
d
.tan & " & sec2 & / D sec2 & " sec2 & " 2& sec2 & tan &
d&
D "2& sec2 & tan &
p
d
1 C t2 " 1
p
dt 1 C t 2 C 1
p
p
t
t
. 1 C t 2 C 1/ p
" . 1 C t 2 " 1/ p
1 C t2
1 C t2
p
D
. 1 C t 2 C 1/2
2t
D p
p
2
1 C t . 1 C t 2 C 1/2
.x C h/20 " x 20
d 20
D
x D 20x 19
h
dx
h!0
p
p
4x C 1 " 3
9 C 4h " 3
D lim 4
lim
x!2
x"2
h!0
ˇ 4h
d p ˇˇ
2
4
D
D p D
4 xˇ
ˇ
dx
3
2 9
lim
xD9
15.
lim
x!!=6
16.
cos..#=3/ C 2h/ " cos.#=3/
cos.2x/ " .1=2/
D lim 2
x " #=6
2h
h!0
ˇ
ˇ
d
ˇ
D2
cos x ˇ
ˇ
dx
xD!=3
p
D "2 sin.#=3/ D " 3
1
1
"
2
.1=x 2 / " .1=a2 /
."a C h/
."a/2
lim
D lim
x!!a
xCa
h
h!0
ˇ
d 1 ˇˇ
2
D
D 3
ˇ
dx x 2 ˇ
a
xD!a
6. At x D # the curve y D tan.x=4/ has slope
.sec2 .#=4//=4 D 1=2. The normal to the curve there
has equation y D 1 " 2.x " #/.
7.
4 C 3x C 2x 2 C x 3
x5
#
$
5
2
D " .4 " x 2=5 /!7=2 " x !3=5
2
5
D"
17.
d
f .3 " x 2 / D "2xf 0 .3 " x 2 /
dx
18.
p
p
p
p
p
1
d
f . x/f 0 . x/
p
Œf . x/!2 D 2f . x/f 0 . x/ p D
dx
2 x
x
19.
p
p
f .2x/g 0 .x=2/
d
f .2x/ g.x=2/ D 2f 0 .2x/ g.x=2/ C
p
dx
4 g.x=2/
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INSTRUCTOR’S SOLUTIONS MANUAL
20.
d f .x/ " g.x/
dx f .x/ C g.x/
(
1
D
f .x/ C g.x//.f 0 .x/ " g 0 .x//
.f .x/ C g.x//2
)
" .f .x/ " g.x//.f 0 .x/ C g 0 .x/
D
21.
22.
23.
24.
REVIEW EXERCISES 2 (PAGE 163)
2.f 0 .x/g.x/ " f .x/g 0 .x//
.f .x/ C g.x//2
32.
g.x/ D "3 cos.x=3/ C 6 sin.x=6/ C C:
If .#; 2/ lies on y D g.x/, then ".3=2/ C 3 C C D 2, so
C D 1=2 and g.x/ D "3 cos.x=3/ C 6 sin.x=6/ C .1=2/:
33.
d
f .x C .g.x//2 / D .1 C 2g.x/g 0 .x//f 0 .x C .g.x//2 /
dx
#
$
#
$
g.x 2 /
2x 2 g 0 .x 2 / " g.x 2 / 0 g.x 2 /
d
f
D
f
dx
x
x2
x
d
f .sin x/g.cos x/
dx
D .cos x/f 0 .sin x/g.cos x/ " .sin x/f .sin x/g 0 .cos x/
s
cos f .x/
d
dx sin g.x/
s
1 sin g.x/
D
2 cos f .x/
&
d
.x sin x C cos x/ D sin x C x cos x " sin x D x cos x
dx
d
.x cos x " sin x/ D cos x " x sin x " cos x D "x sin x
dx
Z
Z
34.
x cos x dx D x sin x C cos x C C
x sin x dx D "x cos x C sin x C C
If f 0 .x/ D f .x/ and g.x/ D x f .x/, then
g 0 .x/ D f .x/ C xf 0 .x/ D .1 C x/f .x/
g 00 .x/ D f .x/ C .1 C x/f 0 .x/ D .2 C x/f .x/
g 000 .x/ D f .x/ C .2 C x/f 0 .x/ D .3 C x/f .x/
Conjecture: g .n/ .x/ D .n C x/f .x/ for n D 1, 2, 3, : : :
Proof: The formula is true for n D 1, 2, and 3 as shown
above. Suppose it is true for n D k; that is, suppose
g .k/ .x/ D .k C x/f .x/. Then
"f 0 .x/ sin f .x/ sin g.x/ " g 0 .x/ cos f .x/ cos g.x/
.sin g.x//2
25. If x 3 yC2xy 3 D 12, then 3x 2 yCx 3 y 0 C2y 3 C6xy 2 y 0 D 0.
At .2; 1/: 12 C 8y 0 C 2 C 12y 0 D 0, so the slope there is
y 0 D "7=10. The tangent line has equation
7
y D 1 " 10
.x " 2/ or 7x C 10y D 24.
p
26. 3p2x sin.#y/ C 8ypcos.#x/ D 2
3 2 sin.#y/ C 3# 2x cos.#y/y 0 C 8y 0 cos.#x/
"8#y sin.#x/ D 0
p
At .1=3; 1=4/: 3pC #y 0 C 4y 0 " # 3 D 0, so the slope
# 3"3
.
there is y 0 D
# C4
$
Z
Z #
1 C x4
1
1
x3
2
27.
dx
D
C
x
dx
D
"
C
CC
x2
x2
x
3
Z
Z
p
2
1Cx
28.
p dx D .x !1=2 C x 1=2 / dx D 2 x C x 3=2 C C
3
x
Z
Z
2 C 3 sin x
29.
dx D .2 sec2 x C 3 sec x tan x/ dx
cos2 x
D 2 tan x C 3 sec x C C
Z
Z
30.
.2x C 1/4 dx D .16x 4 C 32x 3 C 24x 2 C 8x C 1/ dx
If g 0 .x/ D sin.x=3/ C cos.x=6/, then
&
d %
.k C x/f .x/
dx
D f .x/ C .k C x/f 0 .x/ D ..k C 1/ C x/f .x/:
g .kC1/ .x/ D
Thus the formula is also true for n D k C 1. It is therefore
true for all positive integers n by induction.
35.
36.
The tangent to y D x 3 C 2 at x D a has equation
y D a3 C 2 C 3a2 .x " a/, or y D 3a2 x " 2a3 C 2. This
line passes through the origin if 0 D "2a3 C 2, that is, if
a D 1. The line then has equation y D 3x.
p
2 C x 2 at x D a has slope
The
ptangent to y D
a= 2 C a2 and equation
yD
This line passes through .0; 1/ provided
5
D
16x
C 8x 4 C 8x 3 C 4x 2 C x C C
5
p
a2
1 D 2 C a2 " p
2 C a2
p
2
2 C a2 D 2 C a " a2 D 2
or, equivalently,
Z
.2x C 1/5
.2x C 1/4 dx D
CC
10
31. If f 0 .x/ D 12x 2 C 12x 3 , then f .x/ D 4x 3 C 3x 4 C C .
If f .1/ D 0, then 4 C 3 C C D 0, so C D "7 and
f .x/ D 4x 3 C 3x 4 " 7.
p
a
2 C a2 C p
.x " a/:
2 C a2
2 C a2 D 4
p
The possibilities are a D ˙ 2, and the equations
of the
p
corrresponding tangent lines are y D 1 ˙ .x= 2/.
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73
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REVIEW EXERCISES 2 (PAGE 163)
37.
ADAMS and ESSEX: CALCULUS 9
&
d % n
sin x sin.nx/
dx
D n sinn!1 x cos x sin.nx/ C n sinn x cos.nx/
D n sinn!1 xŒcos x sin.nx/ C sin x cos.nx/!
Observe that this rate is half the rate at which F decreases when r increases from R.
42.
D n sinn!1 x sin..n C 1/x/
y D sinn x sin.nx/ has a horizontal tangent at
x D m#=.n C 1/, for any integer m.
&
d % n
38.
sin x cos.nx/
dx
D n sinn!1 x cos x cos.nx/ " n sinn x sin.nx/
D n sinn!1 xŒcos x cos.nx/ " sin x sin.nx/!
D n sinn!1 x cos..n C 1/x/
&
d % n
cos x sin.nx/
dx
D "n cosn!1 x sin x sin.nx/ C n cosn x cos.nx/
Thus the isothermal compressibility of the gas is
1
1 dV
D
V dP
V
43.
D n cos
x cos..n C 1/x/
&
d % n
cos x cos.nx/
dx
D "n cosn!1 x sin x cos.nx/ " n cosn x sin.nx/
39. Q D .0; 1/. If P D .a; a2 / on the curve y D x 2 , then
the slope of y D x 2 at P is 2a, and the slope of PQ is
.a2 " 1/=a. PQ is normal to y D x 2 if a D 0 or
Œ.a2 " 1/=a!.2a/ D "1, that is, ifpa D 0 or a2 D 1=2.
The points P are .0; 0/ and .˙1= p2; 1=2/. The distances
from these points to Q are 1 and 3=2, respectively. The
distance from Q to the curve
y D x 2 is the shortest of
p
these distances, namely 3=2 units.
y2 D
"
74
ˇ
mg
d
ˇ
D "mk D "
.mkr/ˇ
:
rDR
dr
R
100
100 4:9 & 100
"
DhC
:
9:8
.9:8/2
19:6
400 4:9 & 400
400
D
"
:
9:8
.9:8/2
19:6
These two maximum heights are equal, so
hC
100
400
D
;
19:6
19:6
which gives h D 300=19:6 ' 15:3 m as the height of the
building.
44.
The first ball has initial height 60 m and initial velocity 0,
so its height at time t is
y1 D 60 " 4:9t 2 m:
The second ball has initial height 0 and initial velocity v0 ,
so its height at time t is
2mgR2
2mg
D"
:
3
R
R
As r decreases from R, F changes at rate
1
:
P
It reaches maximum height when dy2 =dt D 20 " 9:8t D 0,
that is, at t D 20=9:8 s. The maximum height of the
second ball is
a) For continuity of F .r/ at r D R we require
mg D mkR, so k D g=R.
rDR
D"
The height of the second ball at time t during its motion
is
y2 D 20t " 4:9t 2 :
D "n cosn!1 x sin..n C 1/x/
D"
$
Let the building be h m high. The height of the first ball
at time t during its motion is
y1 D h C
D "n cosn!1 xŒsin x cos.nx/ C cos x sin.nx/!
ˇ
d mgR2 ˇˇ
ˇ
dr r 2 ˇ
V
"
P
It reaches maximum height when dy1 =dt D 10 " 9:8t D 0,
that is, at t D 10=9:8 s. The maximum height of the first
ball is
n!1
b) As r increases from R, F changes at rate
#
y1 D h C 10t " 4:9t 2 :
D n cosn!1 xŒcos x cos.nx/ " sin x sin.nx/!
40. The average profit per tonne if x tonnes are exported is
P .x/=x, that is the slope of the line joining .x; P .x// to
the origin. This slope is maximum if the line is tangent
to the graph of P .x/. In this case the slope of the line is
P 0 .x/, the marginal profit.
(
mgR2
if r # R
41. F .r/ D
r2
mkr
if 0 $ r < R
P V D kT . Differentiate with respect to P holding T
constant to get
dV
D0
V CP
dP
y2 D v0 t " 4:9t 2 m:
The two balls collide at a height of 30 m (at time T , say).
Thus
30 D 60 " 4:9T 2
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30 D v0 T " 4:9T 2 :
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INSTRUCTOR’S SOLUTIONS MANUAL
CHALLENGING PROBLEMS 2 (PAGE 164)
Thus v0 T D 60 and T 2 D 30=4:9. The initial upward
speed of the second ball is
60
v0 D
D 60
T
r
4:9
' 24:25 m/s:
30
Challenging Problems 2 (page 164)
1.
At time T , the velocity of the first ball is
ˇ
dy1 ˇˇ
ˇ
dt ˇ
tDT
x 2 " mx C ma " a2 D 0
D "9:8T ' "24:25 m/s:
In order that this quadratic have only one solution x D a,
the left side must be .x " a/2 , so that m D 2a. The
tangent has slope 2a.
This won’t work for more general curves whose tangents
can intersect them at more than one point.
At time T , the velocity of the second ball is
ˇ
dy2 ˇˇ
ˇ
dt ˇ
The line through .a; a2 / with slope m has equation
y D a2 C m.x " a/. It intersects y D x 2 at points x
that satisfy
x 2 D a2 C mx " ma; or
tDT
D v0 " 9:8T D 0 m/s:
45. Let the car’s initial speed be v0 . The car decelerates at 20
ft/s2 starting at t D 0, and travels distance s in time t ,
where d 2 s=dt 2 D "20. Thus
2.
f .9 C 4h C h2 / " f .9/
f .x 2 C 5/ " f .9/
D lim
x!2
x"2
h
h!0
f .9 C 4h C h2 / " f .9/ 4h C h2
&
D lim
4h C h2
h
h!0
f .9 C k/ " f .9/
D lim
& lim .4 C h/
k
k!0
h!0
4
D f 0 .9/ & 4 D
9
p
p
f .x/ " 3
f .2 C h/ " 3
D lim
b) lim
x!2
x"2
h
h!0
f .2 C h/ " 9
1
D lim
&p
h
h!0
f .2 C h/ C 3
1
1
0
D f .2/ & D
:
6
12
a) lim
ds
D v0 " 20t
dt
x D v0 t " 10t 2 :
The car stops at time t D v0 =20. The stopping distance is
s D 160 ft, so
160 D
f 0 .x/ D 1=x, f .2/ D 9.
v02 v02
v2
"
D 0:
20 40
40
The car’s
p initial speed cannot exceed
v0 D 160 & 40 D 80 ft/s.
p
46. P D 2# L=g D 2#L1=2 g !1=2 .
a) If L remains constant, then
3.
dP
"P '
"g D "#L1=2 g !3=2 "g
dg
1 "g
"P
"#L1=2 g !3=2
'
:
"g D "
P
2 g
2#L1=2 g !1=2
If g increases by 1%, then "g=g D 1=100, and
"P =P D "1=200. Thus P decreases by 0:5%.
f 0 .4/ D 3, g 0 .4/ D 7, g.4/ D 4, g.x/ ¤ 4 if x ¤ 4.
%
&
f .x/ " f .4/
.x " 4/
a) lim f .x/ " f .4/ D lim
x!4
x!4
x"4
D f 0 .4/.4 " 4/ D 0
b) lim
x!4
b) If g remains constant, then
dP
"L D #L!1=2 g !1=2 "L
"P '
dL
1 "L
"P
#L!1=2 g !1=2
"L D
'
:
1=2
!1=2
P
2 L
2#L g
If L increases by 2%, then "L=L D 2=100, and
"P =P D 1=100. Thus P increases by 1%.
c) lim
x!4
d) lim
x!4
f .x/ " f .4/
f .x/ " f .4/
1
D lim
&
2
x!4
x " 16
x"4
xC4
3
1
0
D f .4/ & D
8
8
p
f .x/ " f .4/
f .x/ " f .4/
p
& . x C 2/
D lim
x!4
x"4
x"2
D f 0 .4/ & 4 D 12
x"4
f .x/ " f .4/
f .x/ " f .4/
D lim
&
1 1
x!4
x"4
.4 " x/=4x
"
x 4
D f 0 .4/ & ."16/ D "48
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CHALLENGING PROBLEMS 2 (PAGE 164)
ADAMS and ESSEX: CALCULUS 9
f .x/ " f .4/
f .x/ " f .4/
x"4
e) lim
D lim
x!4 g.x/ " 4
x!4 g.x/ " g.4/
x"4
f 0 .4/
3
D 0
D
g .4/
7
f) lim
x!4
n
7.
a) g.0/ D g.0 C 0/ D g.0/ C g.0/. Thus g.0/ D 0.
g.x C h/ " g.x/
h
g.x/ C g.h/ " g.x/
g.h/ " g.0/
D lim
D lim
h
h
h!0
h!0
D g 0 .0/ D k:
b) g 0 .x/ D lim
h!0
f .g.x// " f .4/
x"4
f .g.x// " f .4/ g.x/ " g.4/
D lim
&
x!4
g.x/ " 4
x"4
D f 0 .g.4// & g 0 .4/ D f 0 .4/ & g 0 .4/ D 3 & 7 D 21
x
if x D 1; 1=2; 1=3; : : :
.
x 2 otherwise
a) f is continuous except at 1=2, 1=3, 1=4, : : : . It is
continuous at x D 1 and x D 0 (and everywhere else).
Note that
4. f .x/ D
c) If h.x/ D g.x/ " kx, then h0 .x/ D g 0 .x/ " k D 0
for all x. Thus h.x/ is constant for all x. Since
h.0/ D g.0/ " 0 D 0, we have h.x/ D 0 for all
x, and g.x/ D kx.
8.
lim x 2 D 1 D f .1/;
lim x 2 D lim x D 0 D f .0/
x!0
b) If a D 1=2 and b D 1=3, then
f .a/ C f .b/
1
D
2
2
#
1
1
C
2
3
$
D
f .x C k/ " f .x/
(let k D "h)
k
f .x/ " f .x " h/
f .x " h/ " f .x/
D lim
:
D lim
"h
h
h!0
h!0
%
&
1 0
f .x/ C f 0 .x/
f 0 .x/ D
2#
1
f .x C h/ " f .x/
D
lim
2 h!0
h
$
f .x/ " f .x " h/
C lim
h
h!0
f .x C h/ " f .x " h/
D lim
:
2h
h!0
b) The change of variables used in the first part of (a)
shows that
a) f 0 .x/ D lim
k!0
x!1
x!0
Given that g 0 .0/ D k and g.x C y/ D g.x/ C g.y/, then
5
:
12
If 1=3 < x < 1=2, then f .x/ D x 2 < 1=4 < 5=12.
Thus the statement is FALSE.
lim
h!0
c) By (a) f cannot be differentiable at x D 1=2, 1=2,
: : :. It is not differentiable at x D 0 either, since
f .x C h/ " f .x/
h
lim
h!0
5. If h ¤ 0, then
9.
p
ˇ
ˇ
ˇ f .h/ " f .0/ ˇ
ˇ
ˇ D jf .h/j > jhj ! 1
ˇ
ˇ
h
jhj
jhj
jhj " jhj
0
f .0 C h/ " f .0 " h/
D lim
D lim D 0:
2h
h
h!0
h!0 h
The tangent to y D x 3 at x D 3a=2 has equation
yD
as h ! 0. Therefore f 0 .0/ does not exist.
6. Given that f .0/ D k, f .0/ ¤ 0, and
f .x C y/ D f .x/f .y/, we have
27
27a3
C 2
8
4a
f .0/ D 1:
Thus f .0/ D 1.
f .x C h/ " f .x/
f 0 .x/ D lim
h
h!0
f .x/f .h/ " f .x/
D f .x/f 0 .0/ D kf .x/:
D lim
h
h!0
76
27a3
27
C 2
8
4a
#
$
3a
:
x"
2
This line passes through .a; 0/ because
0
f .0/ D 0 or
f .x/ " f .x " h/
h
c) If f .x/ D jxj, then f 0 .0/ does not exist, but
f is differentiable elsewhere, including at x D 1
where its derivative is 2.
%
lim
h!0
are always equal if either exists.
h2 " 0
:
lim h " 0h D 1 ¤ 0 D lim
h
h!0
h!0
f .0/ D f .0C0/ D f .0/f .0/
and
#
$
3a
a"
D 0:
2
If a ¤ 0, the x-axis is another tangent to y D x 3 that
passes through .a; 0/.
The number of tangents to y D x 3 that pass through
.x0 ; y0 / is
three, if x0 ¤ 0 and y0 is between 0 and x03 ;
two, if x0 ¤ 0 and either y0 D 0 or y0 D x03 ;
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INSTRUCTOR’S SOLUTIONS MANUAL
CHALLENGING PROBLEMS 2 (PAGE 164)
The curve y D Ax 2 C Bx C C has slope m D 2Aa C B
at .a; Aa2 C Ba C C /. Thus a D .m " B/=.2A/, and the
tangent has equation
one, otherwise.
This is the number of distinct real solutions b of the cubic equation 2b 3 " 3b 2 x0 C y0 D 0, which states that the
tangent to y D x 3 at .b; b 3 / passes through .x0 ; y0 /.
y D Aa2 C Ba C C C m.x " a/
.m " B/2
B.m " B/
m.m " B/
C
CC "
4A
2A
2A
.m " B/2 .m " B/2
D mx C C C
"
4A
2A
D mx C f .m/;
10. By symmetry, any line tangent to both curves must pass
through the origin.
y
y D x 2 C 4x C 1
D mx C
where f .m/ D C " .m " B/2 =.4A/.
14.
x
Parabola y D x 2 has tangent y D 2ax " a2 at .a; a2 /.
Parabola y D Ax 2 C Bx C C has tangent
y D .2Ab C B/x " Ab 2 C C
y D "x 2 C 4x " 1
at .b; Ab 2 C Bb C C /. These two tangents coincide if
2Ab C B D 2a
Fig. C-2-10
The tangent to y D x 2 C 4x C 1 at x D a has equation
y D a2 C 4a C 1 C .2a C 4/.x " a/
The two curves have one (or more) common tangents if
.(/ has real solutions for a and b. Eliminating a between
the two equations leads to
D .2a C 4/x " .a2 " 1/;
which passes through the origin if a D ˙1. The two
common tangents are y D 6x and y D 2x.
11.
The slope of y D x 2 at x D a is 2a.
The slope of the line from .0; b/ to .a; a2 / is .a2 " b/=a.
This line is normal to y D x 2 if either a D 0 or
2a..a2 " b/=a/ D "1, that is, if a D 0 or 2a2 D 2b " 1.
There are three real solutions for a if b > 1=2 and only
one (a D 0) if b $ 1=2.
2
2
12. The point Q D .a; a / on y D x that is closest to
P D .3; 0/ is such that PQ is normal to y D x 2 at Q.
Since PQ has slope a2 =.a " 3/ and y D x 2 has slope 2a
at Q, we require
a2
1
D" ;
a"3
2a
which simplifies to 2a3 C a " 3 D 0. Observe that a D 1
is a solution of this cubic equation. Since the slope of
y D 2x 3 C x " 3 is 6x 2 C 1, which is always positive,
the cubic equation can have only one real solution. Thus
Q D .1; 1/ is the point on y D x 2 that is closest
p to P .
The distance from P to the curve is jPQj D 5 units.
13. The curve y D x 2 has slope m D 2a at .a; a2 /. The
tangent there has equation
y D a2 C m.x " a/ D mx "
m2
:
4
.(/
Ab 2 " C D a2 :
.2Ab C B/2 D 4Ab 2 " 4C;
or, on simplification,
4A.A " 1/b 2 C 4ABb C .B 2 C 4C / D 0:
This quadratic equation in b has discriminant
D D 16A2 B 2 "16A.A"1/.B 2 C4C / D 16A.B 2 "4.A"1/C /:
There are five cases to consider:
CASE I. If A D 1, B ¤ 0, then .(/ gives
bD"
B 2 C 4C
;
4B
aD
B 2 " 4C
:
4B
There is a single common tangent in this case.
CASE II. If A D 1, B D 0, then .(/ forces C D 0, which
is not allowed. There is no common tangent in this case.
CASE III. If A ¤ 1 but B 2 D 4.A " 1/C , then
bD
"B
D a:
2.A " 1/
There is a single common tangent, and since the points of
tangency on the two curves coincide, the two curves are
tangent to each other.
CASE IV. If A ¤ 1 and B 2 " 4.A " 1/C < 0, there are no
real solutions for b, so there can be no common tangents.
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CHALLENGING PROBLEMS 2 (PAGE 164)
ADAMS and ESSEX: CALCULUS 9
b) The tangent to y D x 4 " 2x 2 at x D a has equation
CASE V. If A ¤ 1 and B 2 " 4.A " 1/C > 0, there are
two distinct real solutions for b, and hence two common
tangent lines.
y
y
y D a4 " 2a2 C .4a3 " 4a/.x " a/
D 4a.a2 " 1/x " 3a4 C 2a2 :
Similarly, the tangent at x D b has equation
x
y D 4b.b 2 " 1/x " 3b 4 C 2b 2 :
x
one common
two common
tangent
tangents
tangent curves
y
These tangents are the same line (and hence a double
tangent) if
4a.a2 " 1/ D 4b.b 2 " 1/
y
" 3a4 C 2a2 D "3b 4 C 2b 2 :
x
The second equation says that either a2 D b 2 or
3.a2 C b 2 / D 2; the first equation says that
a3 " b 3 D a " b, or, equivalently, a2 C ab C b 2 D 1.
If a2 D b 2 , then a D "b (a D b is not allowed).
Thus a2 D b 2 D 1 and the two points are .˙1; "1/
as discovered in part (a).
If a2 Cb 2 D 2=3, then ab D 1=3. This is not possible
since it implies that
x
no common
tangent
Fig. C-2-14
15.
a) The tangent to y D x 3 at .a; a3 / has equation
0 D a2 C b 2 " 2ab D .a " b/2 > 0:
y D 3a2 x " 2a3 :
Thus y D "1 is the only double tangent to
y D x 4 " 2x 2 .
For intersections of this line with y D x 3 we solve
3
2
c) If y D Ax C B is a double tangent to
y D x 4 " 2x 2 C x, then y D .A " 1/x C B is a
double tangent to
y D x 4 " 2x 2 . By (b) we must have A " 1 D 0
and B D "1. Thus the only double tangent to
y D x 4 " 2x 2 C x is y D x " 1.
3
x " 3a x C 2a D 0
.x " a/2 .x C 2a/ D 0:
The tangent also intersects y D x 3 at .b; b 3 /, where
b D "2a.
b) The slope of y D x 3 at x D "2a is 3."2a/2 D 12a2 ,
which is four times the slope at x D a.
17.
a) The tangent to
y D f .x/ D ax 4 C bx 3 C cx 2 C dx C e
3
c) If the tangent to y D x at x D a were also tangent
at x D b, then the slope at b would be four times that
at a and the slope at a would be four times that at b.
This is clearly impossible.
d) No line can be tangent to the graph of a cubic polynomial P .x/ at two distinct points a and b, because
if there was such a double tangent y D L.x/, then
.x " a/2 .x " b/2 would be a factor of the cubic polynomial P .x/ " L.x/, and cubic polynomials do not
have factors that are 4th degree polynomials.
16.
a) y D x 4 " 2x 2 has horizontal tangents at points x
satisfying 4x 3 " 4x D 0, that is, at x D 0 and
x D ˙1. The horizontal tangents are y D 0 and
y D "1. Note that y D "1 is a double tangent; it is
tangent at the two points .˙1; "1/.
78
at x D p has equation
y D .4ap 3 C3bp 2 C2cpCd /x"3ap 4 "2bp 3 "cp 2 Ce:
This line meets y D f .x/ at x D p (a double root),
and
xD
"2ap " b ˙
p
b 2 " 4ac " 4abp " 8a2 p 2
:
2a
These two latter roots are equal (and hence correspond to a double tangent) if the expression under the
square root is 0, that is, if
8a2 p 2 C 4abp C 4ac " b 2 D 0:
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INSTRUCTOR’S SOLUTIONS MANUAL
CHALLENGING PROBLEMS 2 (PAGE 164)
This quadratic has two real solutions for p provided
its discriminant is positive, that is, provided
16a2 b 2 " 4.8a2 /.4ac " b 2 / > 0:
This condition simplifies to
3b 2 > 8ac:
For example, for y D x 4 " 2x 2 C x " 1, we have
a D 1, b D 0, and c D "2, so 3b 2 D 0 > "16 D 8ac,
and the curve has a double tangent.
b) From the discussion above, the second point of tangency is
so the formula above is true for n D 1. Assume it is
true for n D k, where k is a positive integer. Then
(
#
$)
d kC1
d
k#
k
cos.ax/
D
a
cos
ax
C
dx
2
dx kC1
$)
(
#
k#
k
D a "a sin ax C
2
#
$
.k
C 1/#
D akC1 cos ax C
:
2
Thus the formula holds for n D 1; 2; 3; : : : by induction.
b) Claim:
qD
"2ap " b
b
D "p "
:
2a
2a
The slope of PQ is
f .q/ " f .p/
b 3 " 4abc C 8a2 d
:
D
q"p
8a2
Calculating f 0 ..p C q/=2/ leads to the same expression, so the double tangent PQ is parallel to the
tangent at the point horizontally midway between P
and Q.
c) The inflection points are the real zeros of
f 00 .x/ D 2.6ax 2 C 3bx C c/:
This equation has distinct real roots provided
9b 2 > 24ac, that is, 3b 2 > 8ac. The roots are
p
"3b " 9b 2 " 24ac
rD
p12a
"3b C 9b 2 " 24ac
sD
:
12a
The slope of the line joining these inflection points is
b 3 " 4abc C 8a2 d
f .s/ " f .r/
D
;
s"r
8a2
so this line is also parallel to the double tangent.
18.
%
dn
n# &
cos.ax/ D an cos ax C
.
n
dx
2
Proof: For n D 1 we have
%
n# &
dn
sin.ax/ D an sin ax C
.
n
dx
2
Proof: For n D 1 we have
%
#&
d
sin.ax/ D a cos.ax/ D a sin ax C
;
dx
2
so the formula above is true for n D 1. Assume it is
true for n D k, where k is a positive integer. Then
(
#
$)
d kC1
k#
d
k
a
sin
ax
C
sin.ax/
D
dx
2
dx kC1
(
#
$)
k#
k
D a a cos ax C
2
#
$
.k
C 1/#
D akC1 sin ax C
:
2
Thus the formula holds for n D 1; 2; 3; : : : by induction.
c) Note that
d
.cos4 x C sin4 x/ D "4 cos3 x sin x C 4 sin3 x cos x
dx
D "4 sin x cos x.cos2 " sin2 x/
D "2 sin.2x/ cos.2x/
%
#&
D " sin.4x/ D cos 4x C
:
2
It now follows from part (a) that
a) Claim:
%
#&
d
cos.ax/ D "a sin.ax/ D a cos ax C
;
dx
2
%
n# &
dn
.cos4 x C sin4 x/ D 4n!1 cos 4x C
:
n
dx
2
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CHALLENGING PROBLEMS 2 (PAGE 164)
19.
ADAMS and ESSEX: CALCULUS 9
v (m/s)
d) The upward acceleration in Œ0; 3! was 39:2=3 ' 13:07
m/s2 .
.3; 39:2/
40
e) The maximum height achieved by the rocket is the
distance it fell from t D 7 to t D 15. This is the
area under the t -axis and above the graph of v on
that interval, that is,
30
20
10
t (s)
-10
2
4
6
8
10
12
14
49 C 1
12 " 7
.49/ C
.15 " 12/ D 197:5 m:
2
2
.15; "1/
-20
-30
f) During the time interval Œ0; 7!, the rocket rose a distance equal to the area under the velocity graph and
above the t -axis, that is,
-40
Fig. C-2-19
.12; "49/
1
.7 " 0/.39:2/ D 137:2 m:
2
a) The fuel lasted for 3 seconds.
b) Maximum height was reached at t D 7 s.
c) The parachute was deployed at t D 12 s.
80
Therefore the height of the tower was
197:5 " 137:2 D 60:3 m.
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INSTRUCTOR’S SOLUTIONS MANUAL
CHAPTER 3.
FUNCTIONS
TRANSCENDENTAL
Section 3.1 Inverse Functions
1.
SECTION 3.1 (PAGE 171)
f .x1 / D f .x2 / , x12 D x22 ; .x1 $ 0; x2 $ 0/
, x1 D x2
Thus f is one-to-one. Let y D f !1 .x/.
Then x D f .y/ p
D y 2 .y $ 0/.
p
therefore y D " x and f !1 .x/ D " x.
!1
D.f / D ."1; 0! D R.f /,
D.f !1 / D Œ0; 1/ D R.f /.
(page 171)
f .x/ D x " 1
f .x1 / D f .x2 / ) x1 " 1 D x2 " 1 ) x1 D x2 .
Thus f is one-to-one. Let y D f !1 .x/.
Then x D f .y/ D y " 1 and y D x C 1. Thus
f !1 .x/
D
x C 1.
D.f / D D.f !1 / D R D R.f / D R.f !1 /.
2. f .x/ D 2x " 1. If f .x1 / D f .x2 /, then 2x1 " 1 D 2x2 " 1.
Thus 2.x1 " x2 / D 0 and x1 D x2 . Hence, f is one-toone.
Let y D f !1 .x/. Thus x D f .y/ D 2y " 1, so
y D 21 .x C 1/. Thus f !1 .x/ D 12 .x C 1/.
D.f / D R.f !1 / D ."1; 1/.
R.f / D D.f !1 / D ."1; 1/.
8.
f .x/ D .1 " 2x/3 . If f .x1 / D f .x2 /, then
.1 " 2x1 /3 D .1 " 2x2 /3 and x1 D x2 . Thus, f is one-toone.
3
Let y D f !1
.x/. Then x D f .y/ D .1 " 2y/
so
p
p
1
1
!1
3
3
y D 2 .1 " x/. Thus, f .x/ D 2 .1 " x/.
D.f / D R.f !1 / D ."1; 1/.
R.f / D D.f !1 / D ."1; 1/.
9.
f .x/ D
p
3. f .x/ D x " 1
p
p
f .x1 / D f .x2 / , x1 " 1 D x2 " 1; .x1 ; x2 # 1/
, x1 " 1 D x2 " 1 D 0
, x1 D x2
Thus f is one-to-one.pLet y D f !1 .x/.
Then x D f .y/ D y " 1, and y D 1 C x 2 . Thus
f !1 .x/ D 1 C x 2 , .x # 0/.
D.f / D R.f !1 / D Œ1; 1/, R.f / D D.f !1 / D Œ0; 1/.
p
4. f .x/ D " x " 1 for x # 1.
p
p
If f .x1 / D f .x2 /, then " x1 " 1 D " x2 " 1 and
x1 " 1 D x2 " 1. Thus x1 D x2 and f is one-to-one.
p
Let y D f !1 .x/. Then x D f .y/ D " y " 1 so
x 2 D y " 1 and y D x 2 C 1. Thus, f !1 .x/ D x 2 C 1.
D.f / D R.f !1 / D Œ1; 1/. R.f / D D.f !1 / D ."1; 0!.
5. f .x/ D x 3
f .x1 / D f .x2 / , x13 D x23
) .x1 " x2 /.x12 C x1 x2 C x22 / D 0
) x1 D x2
Thus f is one-to-one. Let y D f !1 .x/.
Then x D f .y/ D y 3 so y D x 1=3 .
Thus f !1 .x/ D x 1=3 .
D.f / D D.f !1 / D R D R.f / D R.f !1 /.
p
6. f .x/ pD 1 C 3 x.
p If f .x1 / D f .x2 /, then
3
1 C x 1 D 1 C 3 x 2 so x1 D x2 . Thus, f is oneto-one.
p
Let y D f !1 .x/ so that x D f .y/ D 1 C 3 y. Thus
y D .x " 1/3 and f !1 .x/ D .x " 1/3 .
D.f / D R.f !1 / D ."1; 1/.
R.f / D D.f !1 / D ."1; 1/.
f .x/ D x 2 ; .x $ 0/
7.
1
: D.f / D fx W x ¤ "1g D R.f !1 /.
xC1
1
1
D
f .x1 / D f .x2 / ,
x1 C 1
x2 C 1
, x2 C 1 D x1 C 1
, x2 D x1
Thus f is one-to-one; Let y D f !1 .x/.
1
Then x D f .y/ D
y C1
1
1
so y C 1 D
and y D f !1 .x/ D " 1.
x
x
D.f !1 / D fx W x ¤ 0g D R.f /.
x
. If f .x1 / D f .x2 /, then
10. f .x/ D
1Cx
x1
x2
D
. Hence x1 .1 C x2 / D x2 .1 C x1 /
1 C x1
1 C x2
and, on simplification, x1 D x2 . Thus, f is one-to-one.
y
Let y D f !1 .x/. Then x D f .y/ D
and
1Cy
x
D f !1 .x/.
x.1 C y/ D y. Thus y D
1"x
D.f / D R.f !1 / D ."1; "1/ [ ."1; 1/.
R.f / D D.f !1 / D ."1; 1/ [ .1; 1/.
11.
1 " 2x
: D.f / D fx W x ¤ "1g D R.f !1 /
1Cx
1 " 2x1
1 " 2x2
f .x1 / D f .x2 / ,
D
1 C x1
1 C x2
, 1 C x2 " 2x1 " 2x1 x2 D 1 C x1 " 2x2 " 2x1 x2
, 3x2 D 3x1 , x1 D x2
Thus f is one-to-one. Let y D f !1 .x/.
1 " 2y
Then x D f .y/ D
1Cy
so x C xy D 1 " 2y
1"x
and f !1 .x/ D y D
.
2Cx
!1
D.f / D fx W x ¤ "2g D R.f /.
f .x/ D
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SECTION 3.1 (PAGE 171)
ADAMS and ESSEX: CALCULUS 9
x
. If f .x1 / D f .x2 /, then
x2 C 1
x1
x2
q
D q
:
.(/
2
x1 C 1
x22 C 1
12. f .x/ D p
20.
Thus x12 .x22 C 1/ D x22 .x12 C 1/ and x12 D x22 .
From (*), x1 and x2 must have the same sign. Hence,
x1 D x2 and f is one-to-one.
y
, and
Let y D f !1 .x/. Then x D f .y/ D p
y2 C 1
x2
x 2 .y 2 C 1/ D y 2 . Hence y 2 D
. Since f .y/ and y
1 " x2
x
, so
have the same sign, we must have y D p
1 " x2
x
.
f !1 .x/ D p
1 " x2
D.f / D R.f !1 / D ."1; 1/.
R.f / D D.f !1 / D ."1; 1/.
21.
13. g.x/ D f .x/ " 2
Let y D g !1 .x/. Then x D g.y/ D f .y/ " 2, so
f .y/ D x C 2 and g !1 .x/ D y D f !1 .x C 2/.
14.
15.
f .x/ D x 2 C 1 if x # 0, and f .x/ D x C 1 if x < 0.
If f .x1 / D f .x2 / then if x1 # 0 and x2 # 0 then
x12 C 1 D x22 C 1 so x1 D x2 ;
if x1 # 0 and x2 < 0 then x12 C 1 D x2 C 1 so x2 D x12
(not possible);
if x1 < 0 and x2 # 0 then x1 D x22 (not possible);
if x1 < 0 and x2 < 0 then x1 C 1 D x2 C 1 so x1 D x2 .
Therefore f is
Let y D f !1 .x/. Then
' one-to-one.
2
y C 1 if y # 0
x D f .y/ D
y C 1 ' if y < 0.
p
!1
x " 1 if x # 1
Thus f .x/ D y D
x"1
if x < 1.
y
h.x/ D f .2x/. Let y D h!1 .x/. Then x D h.y/ D f .2y/
and 2y D f !1 .x/. Thus h!1 .x/ D y D 21 f !1 .x/.
1
y D f .x/
k.x/ D "3f .x/. Let y D k !1 .x/. Then
x
x D k.y/ D "3f .y/, so f .y/ D " and
3
% x&
k !1 .x/ D y D f !1 " .
3
16. m.x/ D f .x " 2/. Let y D m!1 .x/. Then
x D m.y/ D f .y " 2/, and y " 2 D f !1 .x/.
Hence m!1 .x/ D y D f !1 .x/ C 2.
17.
1 C f .x/
. Let y D s !1 .x/.
1 " f .x/
1 C f .y/
Then x D s.y/ D
. Solving for f .y/ we obtain
1 " f .y/
#
$
x"1
x"1
. Hence s !1 .x/ D y D f !1
.
f .y/ D
xC1
xC1
s.x/ D
x
Fig. 3.1-21
22.
g.x/ D x 3 if x # 0, and g.x/ D x 1=3 if x < 0.
Suppose f .x1 / D f .x2 /. If x1 # 0 and x2 # 0 then
x13 D x23 so x1 D x2 .
Similarly, x1 D x2 if both are negative. If x1 and x2 have
opposite sign, then so do g.x1 / and g.x2 /.
Therefore g is
Let y D g !1 .x/. Then
' one-to-one.
y3
if y # 0
x D g.y/ D
y 1=3 'if y < 0.
1=3
if x # 0
Thus g !1 .x/ D y D x 3
x
if x < 0.
23.
If x1 and x2 are both positive or both negative, and
h.x1 / D h.x2 /, then x12 D x22 so x1 D x2 . If x1 and
x2 have opposite sign, then h.x1 / and h.x2 / are on opposite sides of 1, so cannot be equal.' Hence h is one-to-one.
y2 C 1
if y # 0
. If
If y D h!1 .x/, then x D h.y/ D
2
"y
C
1
if y < 0
p
p
y # 0, then y D' px " 1. If y < 0, then y D 1 " x.
Thus h!1 .x/ D px " 1 if x # 1
1 " x if x < 1
24.
y D f !1 .x/ , x D f .y/ D y 3 C y. To find y D f !1 .2/
we solve y 3 C y D 2 for y. Evidently y D 1 is the only
solution, so f !1 .2/ D 1.
1
. Let y D p !1 .x/.
p.x/ D
1 C f .x/
1
1
Then x D p.y/ D
so f .y/ D " 1,
1 C f#
.y/
x
$
1
"1 .
and p !1 .x/ D y D f !1
x
f .x/ " 3
Let y D q !1 .x/. Then
2
f .y/ " 3
and f .y/ D 2x C 3. Hence
x D q.y/ D
2
!1
!1
q .x/ D y D f .2x C 3/.
18. q.x/ D
19. r.x/ D 1 " 2f .3 " 4x/
Let y D r !1 .x/. Then x D r.y/ D 1 " 2f .3 " 4y/.
f .3 " 4y/ D
1"x
2 #
3 " 4y D f !1
and r
82
!1
1"x
2
$
#
#
$$
1
1"x
.x/ D y D
3 " f !1
.
4
2
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 3.1 (PAGE 171)
25. g.x/ D 1 if x 3 C x D 10, that is, if x D 2. Thus
g !1 .1/ D 2.
26. h.x/ D "3 if xjxj D "4, that is, if x D "2. Thus
h!1 ."3/ D "2.
27.
If y D f
31.
32.
!1
.x/ then x D f .y/.
dy
dy
1
1
Thus 1 D f 0 .y/
so
D 0
D
Dy
1
dx
dx
f .y/
y
(since f 0 .x/ D 1=x).
g.x/ D 2x C sin x ) g 0 .x/ D 2 C cos x # 1 for all x.
Therefore g is increasing, and so one-to-one and invertible
on the whole real line.
y D g !1 .x/ , x D g.y/ D 2y C sin y. For y D g !1 .2/,
we need to solve 2y C sin y " 2 D 0. The root is between 0
and 1; to five decimal places g !1 .2/ D y ' 0:68404. Also
28. f .x/ D 1 C 2x 3
Let y D f !1 .x/.
Thus x D f .y/ D 1 C 2y 3 .
dy
1
1
dy
D
so .f !1 /0 .x/ D
D
1 D 6y 2
2
!1
dx
dx
6y
6Œf .x/!2
29. If f .x/ D
p
y D f !1 .2/ , 2 D f .y/ D y 2 =.1 C y/. We must solve
p
2
2 C 2 y D y for y. There is a root between 2 and 3:
f !1 .2/ ' 2:23362 to 5 decimal places.
dx
dy
D .2 C cos y/
dx
dx
ˇ
dy ˇˇ
1
!1 0
.g / .2/ D
' 0:36036:
D
ˇ
dx ˇ
2 C cos y
1D
xD2
4x 3
, then
2
x C1
.x 2 C 1/.12x 2 / " 4x 3 .2x/
4x 2 .x 2 C 3/
f .x/ D
D
:
2
2
.x C 1/
.x 2 C 1/2
33.
0
Since f 0 .x/ > 0 for all x, except x D 0, f must be oneto-one and so it has an inverse.
4y 3
, and
If y D f !1 .x/, then x D f .y/ D 2
y C1
If f .x/ D x sec x, then f 0 .x/ D sec x C x sec x tan x # 1
for x in ."#=2; #=2/. Thus f is increasing, and so oneto-one on that interval. Moreover,
limx!!.!=2/C f .x/ D "1 and limx!.!=2/C f .x/ D 1,
so, being continuous, f has range ."1; 1/, and so f !1
has domain ."1; 1/.
Since f .0/ D 0, we have f !1 .0/ D 0, and
.f !1 /0 .0/ D
.y 2 C 1/.12y 2 y 0 / " 4y 3 .2yy 0 /
1 D f .y/ D
:
.y 2 C 1/2
0
.y 2 C 1/2
. Since f .1/ D 2, therefore
4y 4 C 12y 2
f !1 .2/ D 1 and
Thus y 0 D
ˇ
! !1 "0
.y 2 C 1/2 ˇˇ
f
.2/ D 4
ˇ
4y C 12y 2 ˇ
yD1
D
1
:
4
p
30. If f .x/ D x 3pC x 2 and y D f !1 .x/, then
x D f .y/ D y 3 C y 2 , so,
p
2yy 0
1 D y 0 3 C y2 C y p
2 3 C y2
)
y0 D
35.
p
3 C y2
:
3 C 2y 2
Since f ."1/ D "2 implies that f !1 ."2/ D "1, we have
!
f
"
!1 0
ˇ
p
3 C y 2 ˇˇ
."2/ D
ˇ
3 C 2y 2 ˇ
yD!1
D
34.
2
:
5
p
Note: f .x/ D x 3 C x 2 D "2 ) x 2 .3 C x 2 / D 4
) x 4 C 3x 2 " 4 D 0 ) .x 2 C 4/.x 2 " 1/ D 0.
Since .x 2 C 4/ D 0 has no real solution, therefore
x 2 " 1 D 0 and x D 1 or "1. Since it is given that
f .x/ D "2, therefore x must be "1.
1
1
D 0
D 1:
f 0 .f !1 .0/
f .0/
If y D .f ı g/!1 .x/, then x D f ı g.y/ D f .g.y//. Thus
g.y/ D f !1 .x/ and y D g !1 .f !1 .x// D g !1 ı f !1 .x/.
That is, .f ı g/!1 D g !1 ı f !1 .
x"a
bx " c
y "a
Let y D f !1 .x/. Then x D f .y/ D
and
by " c
cx " a
bxy " cx D y " a so y D
. We have
bx " 1
cx
"
a
x
"
a
D
, which simplifies to
f !1 .x/ D f .x/ if
bx " c
bx " 1
f .x/ D
b.1 " c/x 2 C .c 2 " 1/x C a..1 " c/ ) 0:
This equation is satisfied for (almost) all values of x provided that c D 1 a and b arbitrary, or c D "1 and
a D b D 0. In the latter case f .x/ ) x, which is certainly
self-inverse. For the former case, f will be self-inverse
provided f is one-to-one and so has an inverse. Since, for
c D 1,
ab " 1
f 0 .x/ D
.bx " 1/2
is positive (or negative) for all x ¤ Ê1=b provided ab > 1
(or ab < 1), we have that f is self-inverse in c D 1, and
a and b are arbitrary except that ab ¤ 1.
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SECTION 3.1 (PAGE 171)
ADAMS and ESSEX: CALCULUS 9
36. Let f .x/ be an even function. Then f .x/ D f ."x/.
Hence, f is not one-to-one and it is not invertible. Therefore, it cannot be self-inverse.
An odd function g.x/ may be self-inverse if its graph is
symmetric about the line x D y. Examples are g.x/ D x
and g.x/ D 1=x.
37. No. A function that is one-to-one on a single interval need
not be either increasing or decreasing. For example, consider the function defined on Œ0; 2! by
f .x/ D
'
x
"x
if 0 $ x $ 1
if 1 < x $ 2.
# $!2x
1
D log1=3
D "2x
3
7.
log1=3 3
8.
43=2 D 8
9.
10! log10 .1=x/ D
log4 8 D
)
3
2
)
p
2log4 8 D 23=2 D 2 2
1
Dx
1=x
!
"
10. Since loga x 1=.loga x/ D
x 1=.loga x/ D a1 D a.
1
loga x D 1, therefore
loga x
.loga b/.logb a/ D loga a D 1
%
&
12. logx x.logy y 2 / D logx .2x/ D logx x C logx 2
11.
It is one-to-one but neither increasing nor decreasing on
all of Œ0; 2!.
38. First we consider the case where the domain of f is a
closed interval. Suppose that f is one-to-one and continuous on Œa; b!, and that f .a/ < f .b/. We show that
f must be increasing on Œa; b!. Suppose not. Then there
are numbers x1 and x2 with a $ x1 < x2 $ b and
f .x1 / > f .x2 /. If f .x1 / > f .a/, let u be a number
such that u < f .x1 /, f .x2 / < u, and f .a/ < u. By
the Intermediate-Value Theorem there exist numbers c1 in
.a; x1 / and c2 in .x1 ; x2 / such that f .c1 / D u D f .c2 /,
contradicting the one-to-oneness of f . A similar contradiction arises if f .x1 / $ f .a/ because, in this case,
f .x2 / < f .b/ and we can find c1 in .x1 ; x2 / and c2 in
.x2 ; b/ such that f .c1 / D f .c2 /. Thus f must be increasing on Œa; b!.
D 1 C logx 2 D 1 C
1
log2 x
1
D1
2
13.
.log4 16/.log4 2/ D 2 &
14.
log15 75 C log15 3 D log15 225 D 2
.since 152 D 225/
15.
16.
17.
A similar argument shows that if f .a/ > f .b/, then
f must be decreasing on Œa; b!.
1.
2x
log6 9 C log6 4 D log6 36 D 2
# 2 2$
4 !3
2 log3 12 " 4 log3 6 D log3
24 ! 34
D log3 .3!2 / D "2
loga .x 4 C 3x 2 C 2/ C loga .x 4 C 5x 2 C 6/
p
" 4 loga x 2 C 2
%
&
%
&
D loga .x 2 C 2/.x 2 C 1/ C loga .x 2 C 2/.x 2 C 3/
" 2 log1 .x 2 C 2/
Finally, if the interval I where f is defined is not
necessarily closed, the same argument shows that if Œa; b!
is a subinterval of I on which f is increasing (or decreasing), then f must also be increasing (or decreasing)
on any intervals of either of the forms Œx1 ; b! or Œa; x2 !,
where x1 and x2 are in I and x1 $ a < b $ x2 . So f
must be increasing (or decreasing) on the whole of I .
18.
Section 3.2 Exponential and Logarithmic
Functions (page 175)
19.
p
33
p D 33!5=2 D 31=2 D 3
35
20.
log3 5 D .log10 5/=.log10 3 ' 1:46497
21.
22x D 5xC1 , 2x log10 2 D .x C 1/ log10 5,
x D .log10 5/=.2 log10 2 " log10 5/ ' "7:21257
p
p
x D log10 3,
x 2 D 3, 2 log
p 10
.log10 3/= 2
x D 10
' 2:17458
2. 2
1=2 1=2
3. .x
8
!3 !2
/
D2
Dx
4. . 21 /x 4x=2 D
1=2 3=2
2
6
D loga .x 2 C 1/ C loga .x 2 C 3/
D loga .x 4 C 4x 2 C 3/
2
D2 D4
22.
x
2
D1
2x
23.
logx 3 D 5, .log10 3/=.log10 x/ D 5,
log10 x D .log10 3/=5, x D 10.log10 3/=5 ' 1:24573
24.
log3 x D 5, .log10 x/=.log10 3/ D 5,
log10 x D 5 log10 3, x D 105 log10 3 D 35 D 243
5. log5 125 D log5 53 D 3
6. If log4 . 18 / D y then 4y D 18 , or 22y D 2!3 . Thus
2y D "3 and log4 . 18 / D y D " 32 .
84
log! .1 " cos x/ C log! .1 C cos x/ " 2 log! sin x
)
(
.1 " cos x/.1 C cos x/
sin2 x
D log!
D log!
2
sin x
sin2 x
D log! 1 D 0
p
p
2
yD3 p
, log10 y D 2 log10 3,
y D 10 2 log10 3 ' 4:72880
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 3.3 (PAGE 183)
# $
1
1
then au D D x !1 . Hence, a!u D x
x
x
and u D " #
loga$x.
1
D " loga x.
Thus, loga
x
25. Let u D loga
26. Let loga x D u, loga y D v.
Then x D au , y D av .
x
au
Thus
D v D au!v
y # a$
x
and loga
D u " v D loga x " loga y.
y
27.
y
u
y
u=y
Let u D loga .x /, then a D x and a
u
D loga x, or u D y loga x.
Therefore
y
Thus, loga .x y / D y loga x.
35.
h!0
log4 .x C 4/ " 2 log4 .x C 1/ D
axCh " ax
h
h!0
ax ah " ax
D lim
h
h!0
h
a
"1
D ax f 0 .0/ D ax k D kf .x/:
D ax lim
h
h!0
36.
D x.
Section 3.3 The Natural Logarithm and
Exponential Functions (page 183)
1
2
xC4
1
D
.x C 1/2
2
xC4
D 41=2 D 2
.x C 1/2
2x 2 C 3x " 2 D 0 but we need x C 1 > 0, so x D 1=2.
30. First observe that log9 x D log3 x= log3 9 D
2 log3 x C log9 x D 10
1
2
log3 x. Now
log3 x 2 C log3 x 1=2 D 10
log3 x
x
5=2
y D f !1 .x/ ) x D f .y/ D ay
dx
dy
)1D
D kay
dx
dx
dy
1
1
)
D
D
:
dx
kay
kx
Thus .f !1 /0 .x/ D 1=.kx/.
log4
5=2
ah " 1
D k. Thus
h
f 0 .x/ D lim
28. Let logb x D u, logb a D v.
Thus b u D x and b v D a.
Therefore x D b u D b v.u=v/ D au=v
u
logb x
and loga x D D
.
v
logb a
29.
f .x/ D ax and f 0 .0/ D lim
1.
p
e3
p D e 3!5=2 D e 1=2 D e
5
e
2.
ln.e 1=2 e 2=3 / D
3.
e 5 ln x D x 5
4.
e .3 ln 9/=2 D 93=2 D 27
5.
ln
6.
%
&2
e 2 ln cos x C ln e sin x D cos2 x C sin2 x D 1
7.
3 ln 4 " 4 ln 3 D ln
8.
4 ln
9.
%
&
2 ln x C 5 ln.x " 2/ D ln x 2 .x " 2/5
D 10
10
D 3 ; so x D .310 /2=5 D 34 D 81
31. Note that logx 2 D 1= log2 x.
Since limx!1 log2 x D 1, therefore limx!1 logx 2 D 0.
32. Note that logx .1=2/ D " logx 2 D "1= log2 x.
Since limx!0C log2 x D "1, therefore
limx!0C logx .1=2/ D 0.
33. Note that logx 2 D 1= log2 x.
Since limx!1C log2 x D 0C, therefore
limx!1C logx 2 D 1.
34. Note that logx 2 D 1= log2 x.
Since limx!1! log2 x D 0", therefore
limx!1! logx 2 D "1.
1
2
C
2
3
D
7
6
1
D ln e !3x D "3x
e 3x
p
64
43
D ln
4
3
81
x C 6 ln.x 1=3 / D 2 ln x C 2 ln x D 4 ln x
10. ln.x 2 C 6x C 9/ D lnŒ.x C 3/2 ! D 2 ln.x C 3/
11.
2xC1 D 3x
.x C 1/ ln 2 D x ln 3
ln 2
ln 2
xD
D
ln 3 " ln 2
ln.3=2/
12.
3x D 91!x ) 3x D 32.1!x/
)
x D 2.1 " x/
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2
3
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SECTION 3.3 (PAGE 183)
13.
14.
ADAMS and ESSEX: CALCULUS 9
1
5
D xC3
2x
8
"x ln 2 D ln 5 " .x C 3/ ln 8
D ln 5 " .3x C 9/ ln 2
2x ln 2 D ln 5 " 9 ln 2
ln 5 " 9 ln 2
xD
2 ln 2
2x
2 !3
2
y D ln ln x
34.
y D x ln x " x
# $
1
" 1 D ln x
y 0 D ln x C x
x
35.
D 4x D 22x ) x 2 " 3 D 2x
x " 2x " 3 D 0 ) .x " 3/.x C 1/ D 0
Hence; x D "1 or 3:
36.
ln.x=.2 " x// is defined if x=.2 " x/ > 0, that is, if
0 < x < 2. The domain is the interval .0; 2/.
37.
16. ln.x 2 " x " 2/ D lnŒ.x " 2/.x C 1/! is defined if
.x " 2/.x C 1/ > 0, that is, if x < "1 or x > 2. The
domain is the union ."1; "1/ [ .2; 1/.
38.
15.
17.
ln.2x "5/ > ln.7 "2x/ holds if 2x "5 > 0, 7 "2x > 0, and
2x " 5 > 7 " 2x, that is, if x > 5=2, x < 7=2, and 4x > 12
(i.e., x > 3). The solution set is the interval .3; 7=2/.
2
2
18. ln.x 2 " 2/ $ ln x holds
p if x > 2, x > 0, and x " 2 $ x.
Thus we need x > 2 and x 2 " x " 2 $ 0. This latter
inequality says that .x " 2/.x C 1/ $ 0, so it holds for
"1
p $ x $ 2. The solution set of the given inequality is
. 2; 2!.
39.
40.
41.
D .1 " 2x/e
22. y D x 2 e x=2 ;
42.
24.
y D ln j3x " 2j;
25. y D ln.1 C e x /
2
26. f .x/ D e x ;
27.
yD
e x C e !x
;
2
28. x D e
3t
x
30. y D
f 0 .x/ D .2x/e x
y0 D
44.
2
e x " e !x
2
y 0 D e x e .e
x/
D e xCe
1
ex
D 1"
;
1 C ex
1 C ex
31. y D e x sin x;
32. y D e !x cos x;
86
y0 D
45.
dx
1
D 3e 3t ln t C e 3t
dt
t
ln t;
29. y D e .e / ;
3
3x " 2
3
y0 D
3x " 2
ex
y0 D
1 C ex
ex
.1 C e x /2
y 0 D e x .sin x C cos x/
2 !3xC8/
g.x/ D t x x t ;
h.t / D t x " x t ;
y 0 D .2x " 3/.ln 2/2.x
;
y 0 D "e !x cos x " e !x sin x
46.
2 !3xC8/
g 0 .x/ D t x x t ln t C t xC1 x t!1
h0 .t / D xt x!1 " x t ln x
f .s/ D loga .bs C c/ D
ln.bs C c/
ln a
b
.bs C c/ ln a
ln.2x C 3/
g.x/ D logx .2x C 3/ D
ln x
#
$
# $
2
1
ln x
" Œln.2x C 3/!
2x
C
3
x
g 0 .x/ D
.ln x/2
2x ln x " .2x C 3/ ln.2x C 3/
D
x.2x C 3/.ln x/2
yDx
p
x
p
De
#
x ln x
p $
ln x
x
p C
x
2 x
#
$$
#
p
1
1
x
p
Dx
ln x C 1
x 2
# $ln x
1
Given that y D
, let u D ln x. Then x D e u and
x
# $u
1
2
yD
D .e !u /u D e !u . Hence,
eu
# $
# $
1
dy du
2 ln x 1 ln x
dy
2
:
D
!
D ."2ue !u /
D"
dx
du dx
x
x
x
p
x ln x
y D ln j sec x C tan xj
sec x tan x C sec2 x
sec x C tan x
D sec x
p
y D ln jx C x 2 " a2 j
2x
1C p
2p x 2 " a2 D p 1
y0 D
x C x 2 " a2
x 2 " a2
y0 D
x
y0 D
y D 2.x
y0 D e
y 0 D 2xe x=2 C 12 x 2 e x=2
23. y D ln.3x " 2/
y D 52xC1
f 0 .s/ D
43.
!2x
x2
2
2x
x2
"
D 2x ln x
y 0 D 2x ln x C
x
2
cos x
D cot x
y D ln j sin xj;
y0 D
sin x
y D x 2 ln x "
y 0 D 2.52xC1 / ln 5 D .2 ln 5/52xC1
19. y D e 5x ; y 0 D 5e 5x
20. y D xe x " x;
y 0 D e x C xe x " 1
x
21. y D 2x D xe !2x
e
y 0 D e !2x " 2xe !2x
1
x ln x
y0 D
33.
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INSTRUCTOR’S SOLUTIONS MANUAL
47.
48.
SECTION 3.3 (PAGE 183)
p
y D ln. x 2 C a2 " x/
x
"1
p
2 C a2
x
y0 D p
x 2 C a2 " x
1
D "p
x 2 C a2
53.
1
" sin x ln x C cos x
x
f 0 D xx
2 C1
x
.2 ln x C 1/
!
f .x/ D xe ax
f 0 .x/ D e ax .1 C ax/
49.
2/
g.x/ D x x
ln g D x x ln x
xx
1 0
g D x x .1 C ln x/ ln x C
0
g
x
#
$
1
0
xx x
g Dx x
C ln x C .ln x/2
x
Evidently g grows more rapidly than does f as x grows
large.
b)
D .cos x/x .ln cos x " x tan x/
"x
f .x/ D .x x /x D x .x
ln f .x/ D x 2 ln x
1 0
f D 2x ln x C x
f
y D .cos x/x " x cos x D e x ln cos x " e .cos x/.ln x/
#
"
$
#
1
0
x ln cos x
." sin x/
ln cos x C x
y De
cos x
"
#
1
.cos x/.ln x/
"e
" sin x ln x C cos x
x
cos x
a)
54.
f 00 .x/ D e ax .2a C a2 x/
Given that x x
f 000 .x/ D e ax .3a2 C a3 x/
::
:
::
x:
D a where a > 0, then
ln a D x x
f .n/ .x/ D e ax .nan!1 C an x/
Thus ln x D
50. Since
::
x:
ln x D a ln x:
1
ln a D ln a1=a , so x D a1=a .
a
d
.ax 2 C bx C c/e x D .2ax C b/e x C .ax 2 C bx C c/e x
dx
D Œax 2 C .2a C b/x C .b C c/!e x
D ŒAx 2 C Bx C C !e x :
55.
f .x/ D .x " 1/.x " 2/.x " 3/.x " 4/
ln f .x/ D ln.x " 1/ C ln.x " 2/ C ln.x " 3/ C ln.x " 4/
1
1
1
1
1
f 0 .x/ D
C
C
C
f .x/
x"1
x"2
x"3
x"4
#
$
1
1
1
1
f 0 .x/ D f .x/
C
C
C
x"1
x"2
x"3
x"4
56.
F .x/ D
Thus, differentiating .ax 2 C bx C c/e x produces another
function of the same type with different constants. Any
number of differentiations will do likewise.
y D ex
51.
2
y 0 D 2xe x
y 00 D 2e
x2
2
2
C 4x 2 e x D 2.1 C 2x 2 /e x
2
2
2
y 000 D 2.4x/e x C 2.1 C 2x 2 /2xe x D 4.3x C 2x 3 /e x
2
y .4/ D 4.3 C 6x 2 /e x C 4.3x C 2x 3 /2xe x
D 4.3 C 12x 2 C 4x 4 /e x
52.
f
.4/
4
!4
.x/ D ".3Š/2 .2x C 1/
2
2
f .x/ D ln.2x C 1/
f 00 .x/ D ."1/22 .2x C 1/!2
2
f 0 .x/ D 2.2x C 1/!1
f 000 .x/ D .2/23 .2x C 1/!3
Thus, if n D 1; 2; 3; : : : we have
f .n/ .x/ D ."1/n!1 .n " 1/Š2n .2x C 1/!n .
p
1 C x.1 " x/1=3
.1 C 5x/4=5
1
ln F .x/ D 2 ln.1 C x/ C 13 ln.1 " x/ " 54 ln.1 C 5x/
F 0 .x/
1
1
4
D
"
"
F .x/
2.1 C x/ 3.1 " x/ .1 C 5x/
"
#
(
)
1 1 4
1 1
23
0
F .0/ D F .0/
" "
D .1/
" "4 D"
2 3 1
2 3
6
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SECTION 3.3 (PAGE 183)
57.
ADAMS and ESSEX: CALCULUS 9
.x 2 " 1/.x 2 " 2/.x 2 " 3/
.x 2 C 1/.x 2 C 2/.x 2 C 3/
3&2&1
1
f .2/ D
D
;
f .1/ D 0
5&6&7
35
ln f .x/ D ln.x 2 " 1/ C ln.x 2 " 2/ C ln.x 2 " 3/
61.
f .x/ D
xDa
" ln.x 2 C 1/ " ln.x 2 C 2/ " ln.x 2 C 3/
1
2x
2x
2x
f 0 .x/ D 2
C 2
C 2
f .x/
x "1
x "2
x "3
2x
2x
2x
"
"
" 2
x C 1 # x2 C 2 x2 C 3
1
1
1
C 2
C 2
f 0 .x/ D 2xf .x/ 2
x "1
x "2
x "3
$
1
1
1
" 2
" 2
" 2
x C1 x C2 x C3
#
$
4 1
1
1 1 1 1
f 0 .2/ D
C C " " "
35 3
2
1 5 6 7
139
556
4
&
D
D
35 105
3675
2
Since f .x/ D .x " 1/g.x/ where g.1/ ¤ 0, then
f 0 .x/ D 2xg.x/ C .x 2 " 1/g 0 .x/ and
1
."1/."2/
D .
f 0 .1/ D 2g.1/ C 0 D 2 &
2&3&4
6
Therefore, a D 1 and line has slope e.
The line has equation y D ex.
y
.a;e a /
y D ex
x
Fig. 3.3-61
62.
2
58. Since y D x 2 e !x , then
2
2
Let the point of tangency be .a; e a /.
Tangent line has slope
ˇ
d x ˇˇ
ea " 0
D
e ˇ
D ea :
a"0
dx ˇ
ˇ
1 ˇˇ
1
The slope of y D ln x at x D a is y D ˇ
D : The
xˇ
a
xDa
line from .0; 0/ to .a; ln a/ is tangent to y D ln x if
0
ln a " 0
1
D
a"0
a
2
y 0 D 2xe !x " 2x 3 e !x D 2x.1 " x/.1 C x/e !x :
The tangent is horizontal at .0; 0/ and
i.e., if ln a D 1, or a D e. Thus, the line is y D
#
$
1
.
˙1;
e
y
x
.
e
.a; ln a/
59. f .x/ D xe !x
f 0 .x/ D e !x .1 " x/,
C.P. x D 1, f .1/ D
f 0 .x/ > 0 if x < 1 (f increasing)
f 0 .x/ < 0 if x > 1 (f decreasing)
x
1
e
y D ln x
Fig. 3.3-62
y
.1;1=e/ y
D x e !x
63.
x
Let the point of tangency be .a; 2a /. Slope of the tangent
is
ˇ
d x ˇˇ
2a " 0
D
2 ˇ
D 2a ln 2:
a"1
dx ˇ
xDa
1
1
; a D1C
.
Thus a " 1 D
ln 2
ln 2
a
1C.1= ln 2/
So the slope is 2 ln 2 D 2
ln 2 D 2e ln 2.
1
(Note: ln 21= ln 2 D
ln 2 D 1 ) 21= ln 2 D e)
ln 2
The tangent line has equation y D 2e ln 2.x " 1/.
Fig. 3.3-59
64.
1
D 4 then x D 14 and
x
y D ln 14 D " ln 4. The tangent line of slope 4 is
y D " ln 4 C 4.x " 14 /, i.e., y D 4x " 1 " ln 4.
60. Since y D ln x and y 0 D
88
The tangent line to y D ax which passes through the
origin is tangent at the point .b; ab / where
ˇ
d x ˇˇ
ab " 0
D
a ˇ
D ab ln a:
b"0
dx ˇ
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 3.3 (PAGE 183)
1
D ln a, so ab D a1= ln a D e. The line y D x
b
will intersect y D ax provided the slope of this tangent
e
line does not exceed 1, i.e., provided $ 1, or e ln a $ 1.
b
Thus we need a $ e 1=e .
Thus
70.
y
(a) If Aa CBb D 1 and Ba "Ab D 0, then A D
.b; ab /
and B D
Z
y D ax
x
1
x
DxC
y
y
#
$
y y " xy 0
1
x
D 1 " 2 y0
e xy .y C xy 0 / ln C e xy
2
y
x
y
y
#
$
1
At e;
we have
e $
#
1
1
C ey 0 2 C e 2 .e " e 3 y 0 / D 1 " e 2 y 0
e
e
e
2 C 2e 2 y 0 C 1 " e 2 y 0 D 1 " e 2 y 0 .
1
Thus the slope is y 0 D " 2 .
e
"b
(b) If Aa CBb D 0 and Ba "Ab D 1, then A D 2
a C b2
a
and B D 2
. Thus
a C b2
65. e xy ln
y
y
0
0
66. xe C y " 2x D ln 2 ) e C xe y C y " 2 D 0:
At .1; ln 2/, 2 C 2y 0 C y 0 " 2 D 0 ) y 0 D 0.
Therefore, the tangent line is y D ln 2.
67.
68.
f .x/ D Ax cos ln x C Bx sin ln x
f 0 .x/ D A cos ln x " A sin ln x C B sin ln x C B cos ln x
D .A C B/ cos ln x C .B " A/ sin ln x
1
If A D B D then f 0 .x/ D cos ln x.
Z 2
1
1
Therefore
cos ln x dx D x cos ln x C x sin ln x C C .
2
2
1
1
If B D , A D " then f 0 .x/ D sin ln x.
2 Z
2
1
1
Therefore
sin ln x dx D x sin ln x " x cos ln x C C .
2
2
FA;B .x/ D Ae x cos x C Be x sin x
d
FA;B .x/
dx
D Ae x cos x " Ae x sin x C Be x sin x C Be x cos x
D .A C B/e x cos x C .B " A/e x sin x D FACB;B!A .x/
b
. Thus
a2 C b 2
a
a2 C b 2
e ax cos bx dx
%
&
1
ax
ax
ae
cos
bx
C
be
sin
bx
C C:
D 2
a C b2
Fig. 3.3-64
y
d
.Ae ax cos bx C Be ax sin bx/
dx
D Aae ax cos bx " Abe ax sin bx C Bae ax sin bx
C Bbe ax cos bx
D .Aa C Bb/e ax cos bx C .Ba " Ab/e ax sin bx:
Z
e ax sin bx dx
%
&
1
ax
ax
ae
sin
bx
"
be
cos
bx
C C:
D 2
a C b2
71.
)
(
# $
1
"1
1
1
d
1
1
ln C ln x D
C D " C D 0:
dx
x
1=x x 2
x
x
x
1
C ln x D C (constant). Taking x D 1, we
x
1
get C D ln 1 C ln 1 D 0. Thus ln D " ln x.
x
#
$
x
1
1
ln D ln x
D ln x C ln D ln x " ln y:
y
y
y
Therefore ln
72.
73.
rx r!1
r
r
r
d
Œln.x r / " r ln x! D
" D " D 0.
dx
xr
x
x x
Therefore ln.x r /"r ln x D C (constant). Taking x D 1, we
get C D ln 1 " r ln 1 D 0 " 0 D 0. Thus ln.x r / D r ln x.
74. Let x > 0, and F .x/ be the area bounded by y D t 2 , the
t -axis, t D 0 and t D x. For h > 0, F .x C h/ " F .x/ is
the shaded area in the following figure.
y
d
FA;B .x/ D FACB;B!A .x/ we have
69. Since
dx
2
d
d
a)
FA;B .x/ D
FACB;B!A .x/ D F2B;!2A .x/
dx 2
dx
d3 x
d3
d
b)
e cos x D
F1;0 .x/ D
F0;!2 .x/
3
dx
dx 3
dx
D F!2;!2 .x/ D "2e x cos x " 2e x sin x
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y D t2
x
xCh
t
Fig. 3.3-74
89
lOMoARcPSD|6566483
SECTION 3.3 (PAGE 183)
ADAMS and ESSEX: CALCULUS 9
c) The tangent to y D 1=t at t D 2 has slope "1=4. Its
equation is
Comparing this area with that of the two rectangles, we
see that
hx 2 < F .x C h/ " F .x/ < h.x C h/2 :
Hence, the Newton quotient for F .x/ satisfies
x2 <
yD
Letting h approach 0 from the right (by the Squeeze Theorem applied to one-sided limits)
h!0C
1 1
2 x
" .x " 3/ or y D " :
3 9
3 9
If h < 0 and 0 < x C h < x, then
A1 D
F .x C h/ " F .x/
< x2;
h
F .x C h/ " F .x/
lim
D x2:
h
h!0!
A2 D
Combining these two limits, we obtain
1
t
1
A2
2
D
5
:
8
1
2
#
4
1
C
9
3
$
D
7
:
18
2.
x3
D 0 (exponential wins)
ex
lim x !3 e x D lim
ex
D1
x3
x!1
x!1
x!1
3.
2e x " 3
2 " 3e !x
2"0
D
D
lim
D2
x!1 e x C 5
x!1 1 C 5e !x
1C0
4.
1 " 2=.xe x /
1"0
x " 2e !x
D
D1
D
lim
x!1 x C 3e !x
x!1 1 C 3=.xe x /
1C0
lim
lim
lim x ln x D 0 .power wins/
x!0C
t
6.
b) If f .t / D 1=t , then f 0 .t / D "1=t 2 and
f 00 .t / D 2=t 3 > 0 for t > 0. Thus f 0 .t / is an
increasing function of t for t > 0, and so the graph
of f .t / bends upward away from any of its tangent
lines. (This kind of argument will be explored further
in Chapter 5.)
(page 191)
lim x 3 e !x D lim
x!1
5.
3
Fig. 3.3-75
90
1.
yD1=t
2
$
Section 3.4 Growth and Decay
a) The shaded area A in part (i) of the figure is less
than the area of the rectangle (actually a square) with
base from t D 1 to t D 2 and height 1=1 D 1. Since
ln 2 D A < 1, we have 2 < e 1 D e; i.e., e > 2.
(i)
(ii)
y
y
A1
3
1
C
4
2
7
73
5
C
D
> 1.
8
18
72
1
Thus 3 > e D e. Combining this with the result of
(a) we conclude that 2 < e < 3.
F .0/ D C D 0, therefore F .x/ D 31 x 3 . For x D 2,
the area of the region is F .2/ D 83 square units.
A
#
e) ln 3 > A1 C A2 D
F .x C h/ " F .x/
d
F .x/ D lim
D x2:
dx
h
h!0
Z
Therefore F .x/ D
x 2 dx D 31 x 3 C C . Since
yD1=t
1
2
The trapezoid bounded by x D 2, x D 3, y D 0, and
y D .2=3/ " .x=9/ has area
so similarly,
75.
x
:
4
d) The trapezoid bounded by x D 1, x D 2, y D 0, and
y D 1 " .x=4/ has area
F .x C h/ " F .x/
D x2:
h
.x C h/2 <
or y D 1 "
The tangent to y D 1=t at t D 3 has slope "1=9. Its
equation is
F .x C h/ " F .x/
< .x C h/2 :
h
lim
1 1
" .x " 2/
2 4
yD
lim
x!0C
7.
ln x
D "1
x
lim x.ln jxj/2 D 0
x!0
8.
.ln x/3
p
D0
x!1
x
lim
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.power wins/
lOMoARcPSD|6566483
INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 3.4 (PAGE 191)
9. Let N.t / be the number of bacteria present after t hours.
Then N.0/ D 100; N.1/ D 200:
dN
D kN we have N.t / D N.0/e k t D 100e k t .
Since
dt
k
Thus 200 D
and k D ln 2.
# 100e
$
5
Finally, N
D 100e .5=2/ ln 2 ' 565:685.
2
There will be approximately 566 bacteria present after
another 1 21 hours.
10. Let y.t / be the number of kg undissolved after t hours.
Thus, y.0/ D 50 and y.5/ D 20. Since y 0 .t / D ky.t /,
therefore y.t / D y.0/e k t D 50e k t . Then
20 D y.5/ D 50e
5k
)kD
1
5
ln
14.
3N.0/ D N.3/ D N.0/e 3k ) k D
7
10 D N.7/ D N.0/e
15.
16.
17.
1
ln.1:1/.
14
Since
$P invested at 4% compounded continuously grows to
$P .e 0:04 /7 D $P e 0:28 in 7 years. This will be $10,000 if
$P D $10; 000e !0:28 D $7; 557:84.
18. Let y.t / be the value of the investment after t years. Thus
y.0/ D 1000 and y.5/ D 1500. Since y.t / D 1000e k t and
1500 D y.5/ D 1000e 5k , therefore, k D 15 ln 32 .
a) Let t be the time such that y.t / D 2000, i.e.,
1000e k t D 2000
5 ln 2
1
D 8:55:
) t D ln 2 D
k
ln. 32 /
Hence, the doubling time for the investment is about
8.55 years.
1
1
ln ' 0:0004101:
1690 2
b) Let r% be the effective annual rate of interest; then
r
1000.1 C
/ D y.1/ D 1000e k
100
)r D 100.e k " 1/ D 100Œexp . 51 ln 32 / " 1!
a) P .100/ D 100e 100k ' 95:98, i.e., about 95.98%
remains after 100 years.
D 8:447:
b) P .1000/ D 100e 1000k ' 66:36, i.e., about 66.36%
remains after 1000 years.
The effective annual rate of interest is about 8.45%.
19.
13. Let P .t / be the percentage of the initial amount remaining
after t years.
Then P .t / D 100e k t and 99:57 D P .1/ D 100e k .
Thus k D ln.0:9957/:
The half-life T satisfies 50 D P .T / D 100e kT ,
1
ln.0:5/
so T D ln.0:5/ D
' 160:85.
k
ln.0:995/
The half-life is about 160.85 years.
Let W .t / be the weight t days after birth.
Thus W .0/ D 4000 and W .t / D 4000e k t :
I 0 .t / D kI.t / ) I.t/ D I.0/e k t D 40e k t ;
15
3
1
ln
D 100 ln ;
15 D I.0:01/ D 40e 0:01k ) k D
0:01 40
8
thus,
#
$
# $100t
3
3
I.t/ D 40 exp 100t ln
:
D 40
8
8
12. Let P .t / be the percentage remaining after t years. Thus
P 0 .t / D kP .t / and P .t / D P .0/e k t D 100e k t . Then,
50 D P .1690/ D 100e 1690k ) k D
ln 3:
) N.0/ D 107 e !.7=3/ ln 3 ' 770400:
Five days after birth, the baby weighs
W .5/ D 4000e .5=14/ ln.1:1/ ' 4138:50 ' 4139 grams.
Hence, 90% of the sugar will dissolved in about 12.56
hours.
Let P .t / be the percentage undecayed after t years.
Thus P .0/ D 100;
P .15/ D 70.
dP
D kP , we have P .t/ D P .0/e k t D 100e k t :
Since
dt
1
Thus 70 D P .15/ D 100e 15k so k D
ln.0:7/.
15
The half-life T satisfies if 50 D P .T / D 100e kT , so
15 ln.0:5/
1
' 29:15.
T D ln.0:5/ D
k
ln.0:7/
The half-life is about 29.15 years.
1
3
Also 4400 D W .14/ D 4000e 14k , is k D
1
1
5 ln.0:1/
T D ln
D
' 12:56:
k 10
ln.0:4/
11.
7k
There were approximately 770,000 bacteria in the culture
initially. (Note that we are approximating a discrete quantity (number of bacteria) by a continuous quantity N.t / in
this exercise.)
2
5:
If 90% of the sugar is dissolved at time T then
5 D y.T / D 50e kT , so
Let N.t / be the number of bacteria in the culture t days
after the culture was set up. Thus N.3/ D 3N.0/ and
N.7/ D 10 & 106 . Since N.t / D N.0/e k t , we have
Let the purchasing power of the dollar be P .t / cents after
t years.
Then P .0/ D 100 and P .t/ D 100e k t .
Now 91 D P .1/ D 100e k so k D ln.0:91/.
If 25 D P .t/ D 100k t then
ln.0:25/
1
' 14:7.
t D ln.0:25/ D
k
ln.0:91/
The purchasing power will decrease to $0.25 in about 14.7
years.
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SECTION 3.4 (PAGE 191)
ADAMS and ESSEX: CALCULUS 9
20. Let i % be the effective rate, then
! an original investment of
i
in one year. Let r% be
$A will grow to $A 1 C
100
the nominal rate per annum compounded n times per year,
then an original investment of $A will grow to
r
$A 1 C
100n
a) If u.x/ D a C bf .x/, then
u0 .x/ D bf 0 .x/ D bŒa C bf .x/! D bu.x/. This equation for u is the equation of exponential growth/decay.
Thus
u.x/ D C1 e bx ;
&
a
1%
C1 e bx " a D C e bx " :
f .x/ D
b
b
!n
dy
D a C by and y.0/ D y0 , then, from part (a),
dx
a
a
y D C e bx " ; y0 D C e 0 " :
b
b
Thus C D y0 C .a=b/, and
%
a & bx a
y D y0 C
e " :
b
b
b) If
in one year, if compounding is performed n times per year.
For i D 9:5 and n D 12, we have
!
!12
9:5
r
D $A 1 C
100
1200
&
% p
12
1:095 " 1 D 9:1098:
)r D 1200
$A 1 C
24.
The nominal rate of interest is about 9.1098%.
21.
Let x.t / be the number of rabbits on the island t years
after they were introduced. Thus x.0/ D 1;000,
x.3/ D 3;500, and x.7/ D 3;000. For t < 5 we have
dx=dt D k1 x, so
b) The differential equation for x.t / resembles that of
Exercise 21(b), except that y.x/ is replaced by x.t /,
and b is replaced by "b. Using the result of Exercise
21(b), we obtain, since x.0/ D 0,
%
a
a & !bt
e
C
x.t / D x.0/ "
b
b
&
a%
D
1 " e !bt :
b
x.t / D x.0/e k1 t D 1;000e k1 t
% e 2k1 D 3:5
%
&5=2
D 1;000 e 2k1
D 1;000.3:5/5=2
x.2/ D 1;000e 2k1 D 3;500
x.5/ D 1;000e 5k1
' 22;918:
c) We will have x.t / D 21 .a=b/ if 1 " e !bt D 12 , that
is, if e !bt D 12 , or "bt D ln.1=2/ D " ln 2. The
time required to attain half the limiting concentration
is t D .ln 2/=b.
For t > 5 we have dx=dt D k2 x, so that
x.t / D x.5/e k2 .t!5/
x.7/ D x.5/e 2k2 D 3;000
%
%
&5=2
x.10/ D x.5/35k2 D x.5/ e 2k2
' 142:
3;000
22;918
#
$
3;000 5=2
' 22;918
22;918
25.
e 2k2 '
so there are approximately 142 rabbits left after 10 years.
22. Let N.t / be the number of rats on the island t months
after the initial population was released and before the
first cull. Thus N.0/ D R and N.3/ D 2R. Since
N.t / D Re k t , we have e 3k D 2, so e k D 21=3 . Hence
N.5/ D Re 5k D 25=3 R. After the first 1,000 rats are
killed the number remaining is 25=3 R " 1;000. If this number is less than R, the number at the end of succeeding
5-year periods will decline. The minimum value of R for
which this won’t happen must satisfy 25=3 R " 1;000 D R,
that is, R D 1;000=.25=3 " 1/ ' 459:8. Thus R D 460 rats
should be brought to the island initially.
dx
D a " bx.t /.
dt
This says that x.t / is increasing if it is less than a=b
and decreasing if it is greater than a=b. Thus, the
limiting concentration is a=b.
a) The concentration x.t / satisfies
26.
Let T .t / be the reading t minutes after the Thermometer
is moved outdoors. Thus T .0/ D 72, T .1/ D 48.
dT
D k.T " 20/.
By Newton’s law of cooling,
dt
dV
If V .t / D T .t / " 20, then
D kV , so
dt
V .t / D V .0/e k t D 52e k t .
Also 28 D V .1/ D 52e k , so k D ln.7=13/.
Thus V .5/ D 52e 5 ln.7=13/ ' 2:354. At t D 5 the thermometer reads about T .5/ D 20 C 2:354 D 22:35 ı C.
Let T .t / be the temperature of the object t minutes after
its temperature was 45 ı C. Thus T .0/ D 45 and
dT
D k.T C 5/. Let
T .40/ D 20. Also
dt
u.t / D T .t / C 5, so u.0/ D 50, u.40/ D 25, and
dT
du
D
D k.T C 5/ D ku. Thus,
dt
dt
u.t / D 50e k t ;
23. f 0 .x/ D a C bf .x/.
92
25 D u.40/ D 50e 40k ;
1
25
1
1
)k D
ln
D
ln :
40 50
40 2
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 3.5 (PAGE 199)
We wish to know t such that T .t / D 0, i.e., u.t / D 5,
hence
5 D u.t / D 50e k t
# $
5
40 ln
50
# $ D 132:88 min:
tD
1
ln
2
30.
Hence, it will take about .132:88 " 40/ D 92:88 minutes
more to cool to 0 ı C.
27.
Let T .t / be the temperature of the body t minutes after it
was 5 ı .
Thus T .0/ D 5, T .4/ D 10. Room temperature = 20ı .
dT
By Newton’s law of cooling (warming)
D k.T " 20/.
dt
dV
If V .t / D T .t / " 20 then
D kV ,
dt
kt
kt
so V .t / D V .0/e D "15e .
# $
2
1
.
Also "10 D V .4/ D "15e 4k , so k D ln
4
3
ı
kt
If T .t / D 15 , then "5 D
# V$.t / D "15e
1
# $
ln
1
1
3
D 4 # $ ' 10:838.
so t D ln
2
k
3
ln
3
It will take a further 6.84 minutes to warm to 15 ı C.
31.
Ly0
;
y0 C .L " y0 /e !k
y2 D
.L " y1 /y0
y1 .L " y0 /
$2
D
Assuming k and L are positive, but y0 is negative, we
have t " > 0. The solution is therefore valid on ."1; t " /.
The solution approaches "1 as t ! t " ".
32.
Ly0
y0 C .L " y0 /e !2k
.L " y2 /y0
y2 .L " y0 /
t!1
33.
#
$
y
dy
D ky 1 "
:
dt
L
Since
#
$
dy
k
L 2 kL
C
D"
y"
;
dt
L
2
4
L
10; 000
D
' 7671 cases
1 C 49.9=49/3
1 C Me !3k
LkMe !3k
' 3; 028 cases/week:
y 0 .3/ D
.1 C Me !3k /2
y.3/ D
Section 3.5 The Inverse Trigonometric
Functions (page 199)
Assuming L ¤ 0, L D
29. The rate of growth of y in the logistic equation is
L
1 C Me !k t
L
200 D y.0/ D
1CM
L
1; 000 D y.1/ D
1 C Me !k
10; 000 D lim y.t / D L
y.t / D
Thus 200.1 C M / D L D 10; 000, so M D 49. Also
1; 000.1 C 49e !k / D L D 10; 000, so e !k D 9=49 and
k D ln.49=9/ ' 1:695.
Now simplify: y0 y2 .L " y1 /2 D y12 .L " y0 /.L " y2 /
y0 y2 L2 "2y1 y0 y2 LCy0 y12 y2 D y12 L2 "y12 .y0 Cy2 /LCy0 y12 y2
y12 .y0 C y2 / " 2y0 y1 y2
.
y12 " y0 y2
If y0 D 3, y1 D 5, y2 D 6, then
25.9/ " 180
45
LD
D
' 6:429.
25 " 18
7
Ly0
y0 C .L " y0 /e !k t
of the logistic equation is valid on any interval containing
t D 0 and not containing any point where the denominator
is zero. The denominator is zero if y0 D .y0 " L/e !k t ,
that is, if
#
$
y0
1
:
t D t " D " ln
k
y0 " L
Thus y1 .L " y0 /e !k D .L " y1 /y0 , and
y2 .L " y0 /e !2k D .L " y2 /y0 .
Square the first equation and thus eliminate e !k :
#
The solution
yD
28. By the solution given for the logistic equation, we have
y1 D
dy
L
is greatest when y D .
dt
2
Ly0
The solution y D
is valid on the
y0 C .L " y0 /e !k t
largest interval containing t D 0 on which the denominator
does not vanish.
If y0 > L then y0 C .L " y0 /e !k t D 0 if
y0
1
.
t D t " D " ln
k y0 " L
Then the solution is valid on .t " ; 1/.
limt!t ! C y.t / D 1.
thus
p
3
#
D
2
3
# $
2#
1
D
"
2
3
1.
sin!1
2.
cos!1
3.
tan!1 ."1/ D "
4.
sec!1
5.
sin.sin!1 0:7/ D 0:7
p
2D
#
4
#
4
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SECTION 3.5 (PAGE 199)
ADAMS and ESSEX: CALCULUS 9
q
6.
cos.sin!1 0:7/ D
7.
#
$
p
#
2#
tan!1 tan
D tan!1 ." 3/ D "
3
3
1 " sin2 . arcsin 0:7/
p
p
D 1 " 0:49 D 0:51
19.
8. sin!1 .cos 40ı / D 90ı " cos!1 .cos 40ı / D 50ı
9.
10.
11.
%
&
%
&
#
cos!1 sin."0:2/ D " sin!1 sin."0:2/
2
#
D C 0:2
2
20.
%
& q
sin cos!1 ." 13 / D 1 " cos2 . arccos ." 13 /
p
p
q
8
2 2
D
D 1 " 19 D
3
3
#
$
1
1
#
$
D
cos tan!1
1
2
sec tan!1
2
2
1
D s
$ D p5
#
1
1 C tan2 tan!1
2
21.
13.
sin.cos
!1
23.
15.
cos.tan!1 x/ D
17.
18.
1
1
D p
sec.tan!1 x/
1 C x2
tan. arctan x/ D x ) sec. arctan x/ D
25.
p
1 C x2
1
) cos. arctan x/ D p
1 C x2
x
) sin. arctan x/ D p
1 C x2
26.
94
f .t / D t tan!1 t
27.
r
x
1 " x2
:
t
1 C t2
u D z 2 sec!1 .1 C z 2 /
F .x/ D .1 C x 2 / tan!1 x
y D sin!1
y0 D r
sin.cos!1 x/
cos.cos!1 x
p
1 " x2
.by # 13/
D
x
p
x2
1
"1
1" 2 D
x
jxj
p
) tan.sec!1 x/ D x 2 " 1 sgn x
'p
x2 " 1
if x # 1
p
D
" x 2 " 1 if x $ "1
1
) sin.sec!1 x/ D
x
f .x/ D x sin!1 x
F 0 .x/ D 2x tan!1 x C 1
tan.cos!1 x/ D
cos.sec!1 x/ D
1
a
.x " b/
1"
a2
"1
(assuming) a > 0/:
D p
a2 " .x " b/2
2
du
z 2 .2z/
D 2z sec!1 .1 C z 2 / C
p
dz
.1 C z 2 / .1 C z 2 /2 " 1
2z 2 sgn .z/
p
D 2z sec!1 .1 C z 2 / C
.1 C z 2 / z 2 C 2
q
!
" p
x/ D 1 " sin2 sin!1 x D 1 " x 2
cos.sin
16.
24.
a
:
1 C .ax C b/2
x"b
a
1
f 0 .t / D tan!1 t C
p
x/ D 1 " cos2 .cos!1 x/
p
D 1 " x2
14.
y D cos!1
y0 D
f 0 .x/ D sin!1 x C p
12. tan.tan!1 200/ D 200
!1
y D tan!1 .ax C b/;
y0 D " r
22.
#
$
2x " 1
3
2
1
0
y D s
$2 3
#
2x " 1
1"
3
2
D p
9 " .4x 2 " 4x C 1/
1
D p
2 C x " x2
y D sin!1
%a&
x
1
% a &2
1"
x
G.x/ D
h
.jxj > jaj/
a i
a
" 2 D" p
x
jxj x 2 " a2
sin!1 x
sin!1 .2x/
2
1
" sin!1 x p
2
1
"
x
1
"
4x 2
G .x/ D
%
&2
sin!1 .2x/
p
p
1 " 4x 2 sin!1 .2x/ " 2 1 " x 2 sin!1 x
D
&2
%
p
p
1 " x 2 1 " 4x 2 sin!1 .2x/
0
sin!1 .2x/ p
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INSTRUCTOR’S SOLUTIONS MANUAL
28.
H.t / D
sin!1 t
sin #
t
sin t
0
H .t / D
D
29.
30.
p
SECTION 3.5 (PAGE 199)
35.
1
$
" sin!1 t cos t
1 " t2
sin2 t
1
p
" csc t cot t sin!1 t
.sin t / 1 " t 2
f .x/ D .sin!1 x 2 /1=2
2x
1
f 0 .x/ D .sin!1 x 2 /!1=2 p
2
1 " x4
x
D p
p
1 " x 4 sin!1 x 2
#
$
a
y D cos!1 p
a2 C x 2
!!1=2 "
#
a2
a 2
2 !3=2
y0 D " 1 " 2
"
.a
C
x
/
.2x/
a C x2
2
asgn .x/
a2 C x 2
p
x
31. y D a2 " x 2 C a sin!1
a
x
a
1
0
y D "p
Cr
2
2
2
a
a "x
x
1" 2
a
r
a"x
a"x
D
D p
.a > 0/
aCx
a2 " x 2
$ p
#
x
" 2ax " x 2
32. y D a cos!1 1 "
.a > 0/
a
#
"
$ !1=2 #
$
#
1
2a " 2x
x 2
"
" p
y 0 D "a 1 " 1 "
a
a
2 2ax " x 2
x
D p
2ax " x 2
# $
#x
!1 2x
33. tan
D 2
y
y
1
2y " 2xy 0
y 2 " 2xyy 0
D#
2
2
y
y4
4x
1C 2
y
4 " 4y 0
1 4 " 2y 0
D#
At .1; 2/
2
4
16
# "2
8 " 4y 0 D 4# " 4#y 0 ) y 0 D
# "1
# "2
At .1; 2/ the slope is
# "1
36.
37.
D
34. If y D sin!1 x, then y 0 D p
1
. If the slope is 2
1 "p
x2
1
3
then p
D 2 so that x D ˙
: Thus the equations
2
2
1"x
of the two tangentplines are
p $
#
$
#
3
#
3
#
and y D " C 2 x C
.
y D C2 x"
3
2
3
2
d
1
sin!1 x D p
> 0 on ."1; 1/.
dx
1 " x2
!1
Therefore, sin is increasing.
d
1
tan!1 x D
> 0 on ."1; 1/:
dx
1 C x2
!1
Therefore tan is increasing.
d
1
< 0 on ."1; 1/.
cos!1 x D " p
dx
1 " x2
Therefore cos!1 is decreasing.
Since the domain of sec!1 consists of two disjoint intervals ."1; "1! and Œ1; 1/, the fact that the derivative of
sec!1 is positive wherever defined does not imply that
sec!1 is increasing over its whole domain, only that it is
increasing on each of those intervals taken independently.
In fact, sec!1 ."1/ D # > 0 D sec!1 .1/ even though
"1 < 1.
d
d
1
csc!1 x D
sin!1
dx
dx
x#
$
1
1
D r
" 2
x
1
1" 2
x
1
D" p
jxj x 2 " 1
y
.1;!=2/
x
!1
y D csc
x
.!1;!!=2/
Fig. 3.5-37
38.
cot!1 x D arctan .1=x/I
d
1
1
"1
cot!1 x D
D"
2
1
dx
x
1 C x2
1C 2
x
y
!=2
y D cot!1 x
x
!!=2
Fig. 3.5-38
Remark: the domain of cot!1 can be extended to include
0 by defining, say, cot!1 0 D #=2. This will make cot!1
right-continuous (but not continuous) at x D 0. It is also
possible to define cot!1 in such a way that it is continuous
on the whole real line, but we would then lose the identity
cot!1 x D tan!1 .1=x/, which we prefer to maintain for
calculation purposes.
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SECTION 3.5 (PAGE 199)
39.
ADAMS and ESSEX: CALCULUS 9
#
$
d
d
1
.tan!1 x C cot!1 x/ D
tan!1 x C tan!1
dx
dx
x
$
#
1
1
1
D
C
" 2 D 0 if x ¤ 0
1
1 C x2
x
1C 2
x
Thus tan!1 x C cot!1 x D C1 (const. for x > 0)
#
#
At x D 1 we have
C D C1
4
4
#
for x > 0.
Thus tan!1 x C cot!1 x D
2
!1
!1
Also tan x C cot x D C2 for .x < 0/.
#
#
At x D "1, we get " " D C2 .
4
4
#
Thus tan!1 x C cot!1 x D " for x < 0.
2
42.
1
d
sin!1 .cos x/ D p
." sin x/
dx
1 " cos2 x
n
"1 if sin x > 0
D
1
if sin x < 0
sin!1 .cos x/ is continuous everywhere and differentiable everywhere except at x D n# for integers n.
y
y D sin!1 .cos x/
#=2
"#
40. If g.x/ D tan.tan!1 x/ then
2
x
Fig. 3.5-42
!1
sec .tan x/
1 C x2
1 C x2
1 C Œtan.tan!1 x/!2
D
D 1:
D
1 C x2
1 C x2
g 0 .x/ D
#
43.
If h.x/ D tan!1 .tan x/ then h is periodic with period #,
and
sec2 x
h0 .x/ D
D1
1 C tan2 x
tan!1 .tan x/ is continuous and differentiable everywhere except at x D .2n C 1/#=2 for integers n. It is not
defined at those points.
y
y D tan!1 .tan x/
#=2
provided that x ¤ .k C 21 /# where k is an integer. h.x/ is
#
not defined at odd multiples of .
2
y
d
1
tan!1 .tan x/ D
.sec2 x/ D 1 except at odd
dx
1 C tan2 x
multiples of #=2.
y
"#
.!=2;!=2/
#
x
yDtan.tan"1 x/
!!
x
!
x
Fig. 3.5-43
yDtan"1 .tan x/
Fig. 3.5.40(a)
41.
Fig. 3.5.40(b)
"1
d
cos!1 .cos x/ D p
." sin x/
dx
1
"
cos2 x
n
1
if sin x > 0
D
"1 if sin x < 0
44.
!1
cos .cos x/ is continuous everywhere and differentiable everywhere except at x D n# for integers n.
y
y D cos!1 .cos x/
#
1
d
tan!1 .cot x/ D
." csc2 x/ D "1 except at
dx
1 C cot2 x
integer multiples of #.
tan!1 .cot x/ is continuous and differentiable everywhere except at x D n# for integers n. It is not defined at
those points.
y
y D tan!1 .cot x/
#=2
"#
"#
#
x
Fig. 3.5-41
96
#
Fig. 3.5-44
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INSTRUCTOR’S SOLUTIONS MANUAL
45. If jxj < 1 and y D tan!1 p
and
tan y D p
x
x
1 " x2
SECTION 3.5 (PAGE 199)
because both x and sec y are negative. Thus y D sec!1 x
in this case also.
, then y > 0 , x > 0
1 " x2
1
x2
D
sec2 y D 1 C
1 " x2
1 " x2
sin2 y D 1 " cos2 y D 1 " .1 " x 2 / D x 2
sin y D x:
x
Thus y D sin!1 x and sin!1 x D tan!1 p
.
1 " x2
An alternative method of proof involves showing that the
derivative of the left side minus the right side is 0, and
both sides are 0 at x D 0.
p
p
46. If x # 1 and y D tan!1 x 2 " 1, then tan y D x 2 " 1
!1
and sec y D x, so that y D secp
x.
If x $ "1 and y D # " tan!1 x 2 " 1, then !2 < y < 3!
2 ,
so sec y < 0. Therefore
p
p
tan y D tan.# " tan!1 x 2 " 1/ D " x 2 " 1
49.
f 0 .x/ ) 0 on ."1;#"1/ $
x"1
" tan!1 x D C on ."1; "1/:
Thus f .x/ D tan!1
xC1
Evaluate the limit as x ! "1:
% #&
3#
lim f .x/ D tan!1 1 " "
D
x!!1
2
4
#
$
x"1
3#
Thus tan!1
" tan!1 x D
on ."1; "1/:
xC1
4
50.
Since f .x/ D x " tan!1 .tan x/ then
sec2 y D 1 C .x 2 " 1/ D x 2
sec y D x;
f 0 .x/ D 1 "
because both x and sec y are negative. Thus y D sec!1 x
in this case also.
x
, then y > 0 , x > 0 and
47. If y D sin!1 p
1 C x2
x
sin y D p
1 C x2
1
x2
D
cos2 y D 1 " sin2 y D 1 "
1 C x2
1 C x2
tan2 y D sec2 y " 1 D 1 C x 2 " 1 D x 2
tan y D x:
x
Thus y D tan!1 x and tan!1 x D sin!1 p
.
1 C x2
p
x2 " 1
48. If x # 1 and y D sin!1
, then 0 $ y < !2 and
x
p
x2 " 1
sin y D
x
1
x2 " 1
2
D 2
cos y D 1 "
x2
x
sec2 y D x 2 :
Thus sec y D x and y D sec!1p
x.
x2 " 1
If x $ "1 and y D # C sin!1
, then !2 $ y <
x
and sec y < 0. Therefore
!
p
p
x2 " 1
x2 " 1
!1
sin y D sin # C sin
D
x
x
cos2 y D 1 "
sec2 x
D1"1D0
1 C tan2 x
1
2 /#
where k is an integer. Thus, f is
#
constant on intervals not containing odd multiples of .
2
f .0/ D 0 but f .#/
%#D
& # "0 D #. There is no contradiction
here because f 0
is not defined, so f is not constant
2
on the interval containing 0 and #.
if x ¤ ".k C
51.
f .x/ D x " sin!1 .sin x/ ."# $ x $ #/
1
f 0 .x/ D 1 " p
cos x
1 " sin2 x
cos x
D1"
j cos xj
8
#
#
< 0 if " < x <
2
2
D
: 2 if "# < x < " # or # < x < #
2
2
#
Note: f is not differentiable at ˙ .
2
y
.!;!/
3!
2
x2 " 1
1
D 2
2
x
x
!!=2
!=2
x
y D f .x/
.!!;!!/
sec2 y D x 2
sec y D x;
Fig. 3.5-51
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SECTION 3.5 (PAGE 199)
52.
ADAMS and ESSEX: CALCULUS 9
1
) y D tan!1 x C C
1 C x2
y.0/ D C D 1
y0 D
Thus; y D tan!1 x C 1:
8̂
1
1
x
< y0 D
) y D tan!1 C C
2
9
C
x
3
3
53.
1
#
:̂ y.3/ D 2
2 D tan!1 1 C C
C D2"
3
12
1
x
#
Thus y D tan!1 C 2 " .
3
3
12
1
54. y 0 D p
) y D sin!1 x C C
1 " x2
y. 12 / D sin!1 . 12 / C C D 1
#
#
) CC D 1)C D 1" :
6
6
#
!1
Thus; y D sin x C 1 " :
6
8
x
4
< 0
) y D 4sin!1 C C
y D p
55.
2
5
25
"
x
:
y.0/ D 0
0D0CC )C D0
x
Thus y D 4sin!1 .
5
sinh.x ˙ y/
cosh.x ˙ y/
sinh x cosh y ˙ cosh x sinh y
D
cosh x cosh y ˙ sinh x sinh y
tanh x ˙ tanh y
D
1 ˙ tanh x tanh y
3.
tanh.x ˙ y/ D
4.
y D coth x D
y
e x C e !x
e x " e !x
y D sech x D
y
y D coth x
1
1
y D sech x
x
!1
Fig. 3.6.4(a)
y D csch x D
2
e x C e !x
x
Fig. 3.6.4(b)
2
e x " e !x
y
Section 3.6 Hyperbolic Functions
(page 205)
1.
2.
d
1
d
sech x D
dx
dx cosh x
1
D"
sinh x D " sech x tanh x
cosh2 x
d
d
1
csch x D
dx
dx sinh x
1
D"
cosh x D " csch x coth x
sinh2 x
d cosh
d
coth x D
dx
dx sinh x
1
sinh2 x " cosh2 x
D"
D " csch 2 x
D
sinh2 x
sinh2 x
cosh x cosh y C sinh x sinh y
D 41 Œ.e x C e !x /.e y C e !y / C .e x " e !x /.e y " e !y /!
D 14 .2e xCy C 2e !x!y / D 12 .e xCy C e !.xCy/ /
D cosh.x C y/:
sinh x cosh y C cosh x sinh y
D 41 Œ.e x " e !x /.e y C e !y / C .e x C e !x /.e y " e !y /!
D 12 .e xCy " e !.xCy/ / D sinh.x C y/:
cosh.x " y/ D coshŒx C ."y/!
D cosh x cosh."y/ C sinh x sinh."y/
D cosh x cosh y " sinh x sinh y:
sinh.x " y/ D sinhŒx C ."y/!
D sinh x cosh."y/ C cosh x sinh."y/
D sinh x cosh y " cosh x sinh y:
98
y D csch x
x
Fig. 3.6-4
5.
x
1C p
p
2C1
d
d
x
!1
p
sinh x D
ln.x C x 2 C 1/ D
dx
dx
x C x2 C 1
1
D p
2
x C1
x
1C p
p
2
d
d
cosh!1 x D
ln.x C x 2 " 1/ D
px " 1
dx
dx
x C x2 " 1
1
D p
x2 " 1
#
$
1Cx
d
d 1
tanh!1 x D
ln
dx
dx 2
1"x
1 1 " x 1 " x " .1 C x/."1/
1
D
D
21Cx
.1 " x/2
1 " x2
R
dx
D sinh!1 x C C
p
x2 C 1
R
dx
p
D cosh!1 x C C .x > 1/
x2 " 1
R dx
D tanh!1 x C C ."1 < x < 1/
1 " x2
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 3.6 (PAGE 205)
y
6. Let
y D sinh!1
Thus,
%x &
a
dy
, x D a sinh y ) 1 D a.cosh y/ .
dx
y D coth!1 x
%x&
d
1
sinh!1
D
dx
a
a cosh y
1
1
D q
D p
2 C x2
2
a
a 1 C sinh y
Z
x
dx
p
D sinh!1 C C:
.a > 0/
a
a2 C x 2
x
, x D a Cosh y D a cosh y
a
dy
for y # 0, x # a. We have 1 D a.sinh y/ . Thus,
dx
Let y D cosh!1
!1
Fig. 3.6-8
9.
x
1
d
cosh!1
D
dx
a
a sinh y
1
1
D q
D p
2
x " a2
a cosh2 y " 1
Z
x
dx
D cosh!1 C C:
.a > 0; x # a/
p
2
2
a
x "a
Since sech !1 x D cosh!1 .1=x/ is defined in terms of the
restricted function Cosh, its domain consists of the reciprocals of numbers in Œ1; 1/, and is therefore the interval
.0; 1!. The range of sech !1 is the domain of Cosh, that
is, Œ0; 1/. Also,
d
d
1
sech !1 x D
cosh!1
dx
dx
x# $
"1
1
"1
D p
:
D s# $
2
x
x 1 " x2
1 2
"1
x
x
dy
Let y D tanh!1
, x D a tanh y ) 1 D a.sech2 y/ .
a
dx
Thus,
x
1
d
tanh!1
D
dx
a
a sech2 y
a
a
D
D 2
2 " a2 tanh2 x
a
"
x2
a
Z
1
x
dx
D tanh!1 C C:
a2 " x 2
a
a
#
$
1 ln x
1
1
x2 " 1
.e " e ! ln x / D
x"
D
2
2
x
2x
#
$
1
1
x2 C 1
1
xC
D
b) cosh ln x D .e ln x C e ! ln x / D
2
2
x
2x
sinh ln x
x2 " 1
c) tanh ln x D
D 2
cosh ln x
x C1
x 2 C 1 C .x 2 " 1/
cosh ln x C sinh ln x
D 2
D x2
d)
cosh ln x " sinh ln x
.x C 1/ " .x 2 " 1/
#
$
1
xC1
8. The domain of coth!1 x D ln
consists of all x
2
x"1
xC1
> 0. Since
satisfying jxj > 1. For such x, we have
x"1
this fraction takes very large values for x close to 1 and
values close to 0 for x close to "1, the range of coth!1 x
consists of all real numbers except 0.
7.
a) sinh ln x D
d
d
coth!1 x D
tanh!1
dx
dx
1
D
1 " .1=x/2
x
1
y
y D Sech!1 x
1
x
Fig. 3.6-9
10.
csch !1 has domain and range consisting of all real numbers x except x D 0. We have
d
d
1
csch!1 x D
sinh!1
dx
dx
x # $
1
"1
"1
D s
# $2 x 2 D jxjpx 2 C 1 :
1
1C
x
1
x
"1
"1
D 2
:
x2
x "1
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y
y D csch!1 x
x
Fig. 3.6-10
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SECTION 3.6 (PAGE 205)
11.
ADAMS and ESSEX: CALCULUS 9
fA;B .x/ D Ae kx C Be !kx
2.
0
fA;B
.x/ D kAe kx " kBe !kx
auxiliary eqn
3.
auxiliary eqn
gC;D .x/ D C cosh kx C D sinh kx
0
gC;D
.x/ D kC cosh kx C kD sinh kx
00
gC;D
.x/ D k 2 C cosh kx C k 2 D sinh kx
00
Thus gC;D " k 2 gC;D D 0
cosh kx C sinh kx D e kx
cosh kx " sinh kx D e !kx
Thus fA;B .x/ D .A C B/ cosh kx C .A " B/ sinh kx, that
is,
fA;B .x/ D gACB;A!B .x/, and
c
D
gC;D .x/ D .e kx C e !kx / C .e kx " e !kx /,
2
2
that is gC;D .x/ D f.C CD/=2;.C !D/=2 .x/.
12. Since
4.
5.
auxiliary eqn
y 00 C 7y 0 C 10y D 0
auxiliary eqn r 2 C 7r C 10 D 0
.r C 5/.r C 2/ D 0
y D Ae
100
C Be
!2t
) r D "5; "2
!4t
C Bt e
) r D "4; "4
!4t
r 2 " 2r C 1 D 0 ) .r " 1/2 D 0
Thus; r D 1; 1; and y D Ae t C Bt e t :
auxiliary eqn
y 00 " 6y 0 C 10y D 0
r 2 " 6r C 10 D 0
y D Ae
8.
3t
) r D3˙i
cos t C Be 3t sin t
9y 00 C 6y 0 C y D 0
9r 2 C 6r C 1 D 0 ) .3r C 1/2 D 0
Thus; r D " 13 ; " 13 ; and y D Ae !.1=3/t C Bt e !.1=3/t :
auxiliary eqn
where A D 12 e !ka .L C M / and B D 12 e ka .L " M /.
Section 3.7 Second-Order Linear DEs with
Constant Coefficients (page 212)
r 2 C 8r C 16 D 0
y 00 " 2y 0 C y D 0
9.
y 00 " k 2 y D 0 ) y D hL;M .x/
D L cosh k.x " a/ C M sinh k.x " a/
y.a/ D y0 ) y0 D L C 0 ) L D y0 ,
v0
y 0 .a/ D v0 ) v0 D 0 C M k ) M D
k
Therefore y D hy0 ;v0 =k .x/
D y0 cosh k.x " a/ C .v0 =k/ sinh k.x " a/.
y 00 C 8y 0 C 16y D 0
y D Ae
hence, hL;M .x/ is a solution of y 00 " k 2 y D 0 and
!5t
4y 00 " 4y 0 " 3y D 0
7.
D Ae kx C Be !kx D fA;B .x/
) r D 0; "2
Thus; r1 D " 21 ; r2 D 23 ; and y D Ae !.1=2/t C Be .3=2/t :
00
hL;M
.x/ D Lk 2 cosh k.x " a/ C M k 2 sinh k.x " a/
hL;M .x/
& M%
&
L % kx!ka
e
C e !kxCka C
e kx!ka " e !kxCka
D
2
2
!
!
L !ka M !ka kx
L ka M ka !kx
D
e
C
e
e C
e "
e
e
2
2
2
2
r 2 C 2r D 0
4r 2 " 4r " 3 D 0 ) .2r C 1/.2r " 3/ D 0
6.
D k 2 hL;M .x/
y 00 C 2y 0 D 0
) r D "1; r D 3
y D A C Be !2t
hL;M .x/ D L cosh k.x " a/ C M sinh k.x " a/
1.
r 2 " 2r " 3 D 0
y D Ae !t C Be 3t
00
fA;B
.x/ D k 2 Ae kx C k 2 Be !kx
00
Thus fA;B
" k 2 fA;B D 0
13.
y 00 " 2y 0 " 3y D 0
y 00 C 2y 0 C 5y D 0
r 2 C 2r C 5 D 0 ) r D "1 ˙ 2i
y D Ae !t cos 2t C Be !t sin 2t
10. For y 00 " 4y 0 C 5y D 0 the auxiliary equation is
r 2 " 4r C 5 D 0, which has roots r D 2 ˙ i . Thus, the
general solution of the DE is y D Ae 2t cos t C Be 2t sin t .
11.
For y 00 C 2y 0 C 3y D 0 the auxiliary equation isp
r 2 C 2r C 3 D 0, which has solutions r D "1 ˙ 2i . Thus
the general solution
of the givenpequation is
p
y D Ae !t cos. 2t / C Be !t sin. 2t /.
12. Given that y 00 C y 0 C y D 0, hence r 2 C r C 1 D 0. Since
a D 1, b D 1 and c D 1, the discriminant is
D D b 2 " 4ac D "3 < 0 and ".b=2a/ D " 21 and
p
! D 3=2. Thus, the
general solution is p
#
$
#p $
3
3
!.1=2/t
t C Be !.1=2/t sin
t .
y D Ae
cos
2
2
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 3.7 (PAGE 212)
8 00
< 2y C 5y 0 " 3y D 0
13.
y.0/ D 1
: 0
y .0/ D 0
The DE has auxiliary equation 2r 2 C 5y " 3 D 0, with
roots r D 12 and r D "3. Thus y D Ae t=2 C Be !3t .
A
Now 1 D y.0/ D A C B, and 0 D y 0 .0/ D " 3B.
2
Thus B D 1=7 and A D 6=7. The solution is
6
1
y D e t=2 C e !3t .
7
7
14.
Given that y 00 C 10y 0 C 25y D 0, hence
r 2 C 10r C 25 D 0 ) .r C 5/2 D 0 ) r D "5. Thus,
y D Ae
!5t
C Bt e
!5t
y 0 D "5e !5t .A C Bt / C Be !5t :
Since
0 D y.1/ D Ae !5 C Be !5
2 D y 0 .1/ D "5e !5 .A C B/ C Be !5 ;
we have A D "2e 5 and B D 2e 5 .
Thus, y D "2e 5 e !5t C 2t e 5 e !5t D 2.t " 1/e !5.t!1/ .
15.
8 00
< y C 4y 0 C 5y D 0
y.0/ D 2
: 0
y .0/ D 0
The auxiliary equation for the DE is r 2 C 4r C 5 D 0,
which has roots r D "2 ˙ i . Thus
y D Ae !2t cos t C Be !2t sin t
y 0 D ."2Ae !2t C Be !2t / cos t " .Ae !2t C 2Be !2t / sin t:
Now 2 D y.0/ D A ) A D 2, and
2 D y 0 .0/ D "2A C B ) B D 6.
Therefore y D e !2t .2 cos t C 6 sin t /.
17.
Given that a > 0, b > 0 and c > 0:
Case 1: If D D b 2 " 4ac > 0 then the two roots are
p
"b ˙ b 2 " 4ac
:
r1;2 D
2a
Since
b 2 " 4ac < b 2
p
˙ b 2 " 4ac < b
p
"b ˙ b 2 " 4ac < 0
therefore r1 and r2 are negative. The general solution is
y.t / D Ae r1 t C Be r2 t :
If t ! 1, then e r1 t ! 0 and e r2 t ! 0.
Thus, lim y.t / D 0.
t!1
Case 2: If D D b 2 " 4ac D 0 then the two equal roots
r1 D r2 D "b=.2a/ are negative. The general solution is
y.t / D Ae r1 t C Bt e r2 t :
If t ! 1, then e r1 t ! 0 and e r2 t ! 0 at a faster rate
than Bt ! 1. Thus, lim y.t / D 0.
t!1
Case 3: If D D b 2 " 4ac < 0 then the general solution is
y D Ae !.b=2a/t cos.!t / C Be !.b=2a/t sin.!t /
p
4ac " b 2
where ! D
. If t ! 1, then the amplitude of
2a
!.b=2a/t
both terms Ae
! 0 and Be !.b=2a/t ! 0. Thus,
lim y.t / D 0.
t!1
18. The auxiliary equation ar 2 C br C c D 0 has roots
r1 D
16. The auxiliary equation r 2 " .2 C $/r C .1 C $/ factors to
.r " 1 " $/.r " 1/ D 0 and so has roots r D 1 C $ and
r D 1. Thus the DE y 00 " .2 C $/y 0 C .1 C $/y D 0
has general solution y D Ae .1C$/t C Be t . The function
e .1C$/t " e t
is of this form with A D "B D 1=$.
y$ .t / D
$
We have, substituting $ D h=t ,
e .1C$/t " e t
lim y$ .t / D lim
$!0
$!0
$
e tCh " e t
D t lim
h
h!0
#
$
d t
Dt
e D t et
dt
which is, along with e t , a solution of the CASE II DE
y 00 " 2y 0 C y D 0.
p
"b " D
;
2a
r2 D
p
"b C D
;
2a
where D D p b 2 " 4ac. Note that
a.r2 " r1 / D D D ".2ar1 C b/. If y D e r1 t u, then
y 0 D e r1 t .u0 Cr1 u/, and y 00 D e r1 t .u00 C2r1 u0 Cr12 u/. Substituting these expressions into the DE ay 00 C by 0 C cy D 0,
and simplifying, we obtain
e r1 t .au00 C 2ar1 u0 C bu0 / D 0;
or, more simply, u00 " .r2 " r1 /u0 D 0. Putting v D u0
reduces this equation to first order:
v 0 D .r2 " r1 /v;
which has general solution v D C e .r2 !r1 /t : Hence
Z
u D C e .r2 !r1 /t dt D Be .r2 !r1 /t C A;
and y D e r1 t u D Ae r1 t C Be r2 t .
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SECTION 3.7 (PAGE 212)
ADAMS and ESSEX: CALCULUS 9
by the previous problem. Therefore ay 00 C by 0 C cy D 0
has general solution
19. If y D A cos !t C B sin !t then
y 00 C ! 2 y D "A! 2 cos !t " B! 2 sin !t
y D Ae k t cos.!t / C Be k t sin.!t /:
C ! 2 .A cos !t C B sin !t / D 0
for all t . So y is a solution of (†).
20. If f .t / is any solution of .†/ then f 00 .t / D "! 2 f .t / for
all t . Thus,
24.
&2 %
&2 i
d h 2%
! f .t / C f 0 .t /
dt
D 2! 2 f .t /f 0 .t / C 2f 0 .t /f 00 .t /
2
0
2
y 0 .0/ D "5 ) B D " 52 :
Thus, y D 2 cos 2t " 25 sin 2t .
circular frequency = ! D 2, frequency =
1
!
D
' 0:318
2#
#
2#
period =
D # ' 3:14
!q
amplitude = .2/2 C ." 52 /2 ' 3:20
0
D 2! f .t /f .t / " 2! f .t /f .t / D 0
%
&2 %
&2
for all t . Thus, ! 2 f .t / C f 0 .t / is constant. (This
can be interpreted as a conservation of energy statement.)
21.
Because y 00 C 4y D 0, therefore y D A cos 2t C B sin 2t .
Now
y.0/ D 2 ) A D 2;
If g.t / satisfies .†/ and also g.0/ D g 0 .0/ D 0, then by
Exercise 20,
%
&2 %
&2
! 2 g.t / C g 0 .t /
%
&2 %
&2
D ! 2 g.0/ C g 0 .0/ D 0:
25.
Since a sum of squares cannot vanish unless each term
vanishes, g.t / D 0 for all t .
22. If f .t / is any solution of .†/, let
g.t / D f .t / " A cos !t " B sin !t where A D f .0/
and B! D f 0 .0/. Then g is also solution of .†/. Also
g.0/ D f .0/ " A D 0 and g 0 .0/ D f 0 .0/ " B! D 0.
Thus, g.t / D 0 for all t by Exercise 24, and therefore
f .x/ D A cos !t C B sin !t . Thus, it is proved that every
solution of .†/ is of this form.
26.
(easy to%calculate y 00 C ! 2 y D 0)
&
y D A cos.!t / cos.!c/ C sin.!t / sin.!c/
%
&
C B sin.!t / cos.!c/ " cos.!t / sin.!c/
%
&
D A cos.!c/ " B sin.!c/ cos !t
%
&
C A sin.!c/ C B cos.!c/ sin !t
4ac " b 2
b
and ! 2 D
which is
2a
4a2
kt
positive for Case III. If y D e u, then
23. We are given that k D "
%
&
y 0 D e k t u0 C ku
%
&
y 00 D e k t u00 C 2ku0 C k 2 u :
8 00
< y C 100y D 0
y.0/ D 0
: 0
y .0/ D 3
y D A cos.10t / C B sin.10t /
A D y.0/ D 0; 10B D y 0 .0/ D 3
3
yD
sin.10t /
10
%
&
%
&
y D A cos !.t " c/ C B sin !.t " c/
D A cos !t C B sin !t
where A D A cos.!c/ " B sin.!c/ and
B D A sin.!c/ C B cos.!c/
27.
For y 00 C y D 0, we have y D A sin t C B cos t . Since,
y.2/ D 3 D A sin 2 C B cos 2
y 0 .2/ D "4 D A cos 2 " B sin 2;
Substituting into ay 00 C by 0 C cy D 0 leads to
%
&
0 D e k t au00 C .2ka C b/u0 C .ak 2 C bk C c/u
%
&
D e k t au00 C 0 C ..b 2 =.4a/ " .b 2 =.2a/ C c/u
%
&
D a e k t u00 C ! 2 u :
00
A D 3 sin 2 " 4 cos 2
B D 4 sin 2 C 3 cos 2:
Thus,
2
Thus u satisfies u C ! u D 0, which has general solution
u D A cos.!t / C B sin.!t /
102
therefore
y D .3 sin 2 " 4 cos 2/ sin t C .4 sin 2 C 3 cos 2/ cos t
D 3 cos.t " 2/ " 4 sin.t " 2/:
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 3.7 (PAGE 212)
8
< y 00 C ! 2 y D 0
28.
y.a/ D A
: 0
y .a/ D B
%
& B
%
&
y D A cos !.t " a/ C sin !.t " a/
!
29. From Example 9, the spring constant is
k D 9 & 104 gm=sec2 . For a frequency of 10 Hz (i.e., a
circular
frequency ! D 20# rad=sec.), a mass m satisfying
p
k=m D 20# should be used. So,
32.
y D e k t ŒA cosh !.t " t0 /B sinh !.t " t0 /!
(
A !.t!t0 / A !!.t!t0 /
e
C e
D ek t
2
2
)
B !.t!t0 / B !!.t!t0 /
" e
C e
2
2
D A1 e .kC!/t C B1 e .k!!/t
9 & 104
k
D
D 22:8 gm:
mD
400# 2
400# 2
where A1 D .A=2/e !!t0 C .B=2/e !!t0 and
B1 D .A=2/e !t0 " .B=2/e !t0 . Under the conditions of this
problem we know that Rr D k ˙ ! are the two real roots
of the auxiliary equation ar 2 C br C c D 0, so e .k˙!/t
are independent solutions of ay 00 C by 0 C cy D 0, and our
function y must also be a solution. Since it involves two
arbitrary constants, it is a general solution.
The motion is determined by
8
< y 00 C 400# 2 y D 0
y.0/ D "1
: 0
y .0/ D 2
therefore, y D A cos 20# t C B sin 20# t and
y.0/ D "1 ) A D "1
2
1
y 0 .0/ D 2 ) B D
D
:
20#
10#
33.
D e k t ŒA cos !t cos !t0 C A sin !t sin !t0
C B sin !t cos !t0 " B cos !t sin !t0 !
D e k t ŒA1 cos !t C B1 sin !t !;
where A1 D A cos !t0 " B sin !t0 and
B1 D A sin !t0 C B cos !t0 . Under the conditions of
this problem we know that e k t cos !t and e k t sin !t are
independent solutions of ay 00 C by 0 C cy D 0, so our function y must also be a solution, and, since it involves two
arbitrary constants, it is a general solution.
8 00
< y C 2y 0 C 5y D 0
y.3/ D 2
: 0
y .3/ D 0
The DE has auxiliary equation r 2 C 2r C 5 D 0 with
roots r D "1 ˙ 2i . By the second previous problem, a general solution can be expressed in the form
y D e !t ŒA cos 2.t " 3/ C B sin 2.t " 3/! for which
1
Thus, y D " cos 20# t C
sin 20# t , with y in cm and t
10#
in r
second, gives the displacement at time t . The amplitude
1 2
is ."1/2 C .
/ ' 1:0005 cm.
10#
k
!
, !2 D
(k = spring const, m = mass)
30. Frequency D
2#
m
Since the spring does not change, ! 2 m D k (constant)
For m D 400 gm, ! D 2#.24/ (frequency = 24 Hz)
4# 2 .24/2 .400/
If m D 900 gm, then ! 2 D
900
2# & 24 & 2
D 32#.
so ! D
3
32#
Thus frequency =
D 16 Hz
2#
4# 2 .24/2 400
For m D 100 gm, ! D
100
!
so ! D 96# and frequency =
D 48 Hz.
2#
31. Using the addition identities for cosine and sine,
y D e k t ŒA cos !.t " t0 /B sin !.t " t0 /!
Expanding the hyperbolic functions in terms of exponentials,
y 0 D "e !t ŒA cos 2.t " 3/ C B sin 2.t " 3/!
C e !t Œ"2A sin 2.t " 3/ C 2B cos 2.t " 3/!:
The initial conditions give
2 D y.3/ D e !3 A
0 D y 0 .3/ D "e !3 .A C 2B/
Thus A D 2e 3 and B D "A=2 D "e 3 . The IVP has
solution
y D e 3!t Œ2 cos 2.t " 3/ " sin 2.t " 3/!:
34.
8 00
< y C 4y 0 C 3y D 0
y.3/ D 1
: 0
y .3/ D 0
The DE has auxiliary equation r 2 C 4r C 3 D 0 with
roots r D "2 C 1 D "1 and r D "2 " 1 D "3 (i.e.
k ˙ !, where k D "2 and ! D 1). By the second previous
problem, a general solution can be expressed in the form
y D e !2t ŒA cosh.t " 3/ C B sinh.t " 3/! for which
y 0 D "2e !2t ŒA cosh.t " 3/ C B sinh.t " 3/!
C e !2t ŒA sinh.t " 3/ C B cosh.t " 3/!:
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SECTION 3.7 (PAGE 212)
ADAMS and ESSEX: CALCULUS 9
The initial conditions give
The conditions for stopping the motion are met at t D 2#;
the mass remains at rest thereafter. Thus
84
1
< 5 cos t C 5 if 0 $ t $ #
2
1
x.t / D 5 cos t " 5 if # < t $ 2#
:1
if t > 2#
5
1 D y.3/ D e !6 A
0 D y 0 .3/ D "e !6 ."2A C B/
Thus A D e 6 and B D 2A D 2e 6 . The IVP has solution
y D e 6!2t Œcosh.t " 3/ C 2 sinh.t " 3/!:
Review Exercises 3 (page 213)
1.
35. Let u.x/ D c " k 2 y.x/. Then u.0/ D c " k 2 a.
Also u0 .x/ D "k 2 y 0 .x/, so u0 .0/ D "k 2 b. We have
%
&
u00 .x/ D "k 2 y 00 .x/ D "k 2 c " k 2 y.x/ D "k 2 u.x/
This IVP for the equation of simple harmonic motion has
solution
xD0
2.
u.x/ D .c " k 2 a/ cos.kx/ " kb sin.kx/
so that
&
1 %
y.x/ D 2 c " u.x/
k %
&
c
D 2 c " .c " k 2 a/ cos.kx/ C kb sin.kx/
k
b
c
D 2 .1 " cos.kx/ C a cos.kx/ C sin.kx/:
k
k
36. Since x 0 .0/ D 0 and x.0/ D 1 > 1=5, the motion will be
governed by x 00 D "x C .1=5/ until such time t > 0 when
x 0 .t / D 0 again.
Let u D x " .1=5/. Then u00 D x 00 D ".x " 1=5/ D "u,
u.0/ D 4=5, and u0 .0/ D x 0 .0/ D 0. This simple harmonic motion initial-value problem has solution
u.t / D .4=5/ cos t . Thus x.t / D .4=5/ cos t C .1=4/ and
x 0 .t / D u0 .t / D ".4=5/ sin t . These formulas remain
valid until t D # when x 0 .t / becomes 0 again. Note that
x.#/ D ".4=5/ C .1=5/ D ".3=5/.
104
f .x/ D sec2 x tan x ) f 0 .x/ D 2 sec2 x tan2 x C sec4 x > 0
for x in ."#=2; #=2/, so f is increasing and therefore
one-to-one and invertible there. The domain of f !1 is
."1; 1/, the range of f . Since f .#=4/ D 2, therefore
f !1 .2/ D #=4, and
.f !1 /0 .2/ D
3.
lim f .x/ D
x!˙1
4.
5.
6.
7.
Since x.#/ < ".1=5/, the motion for t > # will be
governed by x 00 D "x " .1=5/ until such time t > # when
x 0 .t / D 0 again.
Let v D x C .1=5/. Then v 00 D x 00 D ".x C 1=5/ D "v,
v.#/ D ".3=5/ C .1=5/ D ".2=5/, and
v 0 .#/ D x 0 .#/ D 0. Thius initial-value problem has
solution v.t / D ".2=5/ cos.t " #/ D .2=5/ cos t , so that
x.t / D .2=5/ cos t " .1=5/ and x 0 .t / D ".2=5/ sin t . These
formulas remain valid for t # # until t D 2# when x 0
becomes 0 again. We have x.2#/ D .2=5/ " .1=5/ D 1=5
and x 0 .2#/ D 0.
f .x/ D 3x C x 3 ) f 0 .x/ D 3.1 C x 2 / > 0 for all x, so f
is increasing and therefore one-to-one and invertible. Since
f .0/ D 0, therefore f !1 .0/ D 0, and
ˇ
ˇ
1
1
1
d
ˇ
.f !1 /.x/ˇ
D 0 !1
D 0
D :
ˇ
dx
f .f .0//
f .0/
3
lim
1
0 .f !1 .2//
f
x!˙1
x
ex2
D
1
f
0 .#=4/
D
1
:
8
D 0:
2
Observe f 0 .x/ D e !x .1 " 2x 2 / is positive if x 2 < 1=2
2
and ispnegative
p if x > 1=2. Thus f is increasing
p on
2; 1= 2/ and is decreasing on ."1; "1= 2/ and
."1= p
on .1= 2; 1/.
p
p
The maxpand min values p
of f are 1= 2e (at x D 1= 2)
and "1= 2e (at x D "1= 2).
y D e !x sin x, .0 $ x $ 2#/ has a horizontal tangent
where
dy
0D
D e !x .cos x " sin x/:
dx
This occurs if tan x D 1,
p The
pso x D #=4 or x D 5#=4.
points are .#=4; e !!=4 = 2/ and .5#=4; "e !5!=4 = 2/.
If f 0 .x/ D x for all x, then
d f .x/
f 0 .x/ " xf .x/
D
D 0:
2 =2
2
x
dx e
e x =2
2
Thus f .x/=e x =2 D C (constant) for all x.
Since f .2/ D 3, we have C D 3=e 2 and
2
2
f .x/ D .3=e 2 /e x =2 D 3e .x =2/!2 .
8.
Let the length, radius, and volume of the clay cylinder at
time t be `, r, and V , respectively. Then V D # r 2 `, and
dr
d`
dV
D 2# r`
C #r2 :
dt
dt
dt
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INSTRUCTOR’S SOLUTIONS MANUAL
REVIEW EXERCISES 3 (PAGE 213)
Since d V =dt D 0 and d `=dt D k` for some constant
k > 0, we have
2# r`
dr
D "k# r 2 `;
dt
)
13.
dr
kr
D" :
dt
2
y D aa C aa .1 C ln a/.x " a/:
This line passes through the origin if
0 D aa Œ1 " a.1 C ln a/!, that is, if .1 C ln a/a D 1. Observe
that a D 1 solves this equation. Therefore the slope of the
line is 11 .1 C ln 1/ D 1, and the line is y D x.
That is, r is decreasing at a rate proportional to itself.
9.
a) An investment of $P at r% compounded continuously
grows to $P e rT =100 in T years. This will be $2P
provided e rT =100 D 2, that is, rT D 100 ln 2. If
T D 5, then r D 20 ln 2 ' 13:86%.
14.
b) Since the doubling time is T D 100 ln 2=r, we have
"T '
100 ln 2
."0:5/ ' 0:1803 years:
13:8632
The doubling time will increase by about 66 days.
ˇ
ah " 1
a0Ch " a0
d x ˇˇ
10.
a) lim
D lim
D
a ˇ
D ln a.
h
h
dx ˇ
h!0
h!0
xD0
%
&
Putting h D 1=n, we get lim n a1=n " 1 D ln a.
n!1
11.
ln 2
ln x
D
is satisfied if x D 2 or x D 4 (because
x
2
ln 4 D 2 ln 2).
1
ln b
DmD
:
b
b
Thus ln b D 1, and b D e.
15.
b) Using the technique described in the exercise, we
calculate
%
&
10
210 21=2 " 1 ' 0:69338183
%
&
11
211 21=2 " 1 ' 0:69326449
Thus ln 2 ' 0:693.
&2 %
&2
d %
f .x/ D f 0 .x/
dx
%
&2
) 2f .x/f 0 .x/ D f 0 .x/
a)
b) The line y D mx through the origin intersects the
curve y D ln x at .b; ln b/ if m D .ln b/=b. The same
line intersects y D ln x at a different point .x; ln x/
if .ln x/=x D m D .ln b/=b. This equation will have
only one solution x D b if the line y D mx intersects
the curve y D ln x only once, at x D b, that is, if the
line is tangent to the curve at x D b. In this case m
is the slope of y D ln x at x D b, so
dT
100 ln 2
"r D "
"r:
dr
r2
If r D 13:863% and "r D "0:5%, then
"T ' "
y D x x D e x ln x ) y 0 D x x .1 C ln x/. The tangent to
y D x x at x D a has equation
Let the rate be r%. The interest paid by account A is
1; 000.r=100/ D 10r.
The interest paid by account B is 1; 000.e r=100 " 1/. This
is $10 more than account A pays, so
1; 000.e r=100 " 1/ D 10r C 10:
A TI-85 solve routine gives r ' 13:8165%.
16.
If y D cos!1 x, then x D cos y and 0 $ y $ #. Thus
tan y D sgn x
and # e < e ! follows because ln is increasing.
1
"1D
x2
p
1 " x2
:
x
1
#
1
D " cos!1
x p
2
x
1 " .1=x/2
#
!1
D " tan
2
1=x
p
#
D " sgn xtan!1 x 2 " 1:
2
#
!1
cos x D " sin!1 x.
2
csc!1 x D sin!1
17.
ln.# e / D e ln # < # D # ln e D ln e ! ;
y " 1 D sgn x
r
Since cot x D 1= tan x, cot!1 x D tan!1 .1=x/.
12. If f .x/ D .ln x/=x, then f 0 .x/ D .1 " ln x/=x 2 . Thus
f 0 .x/ > 0 if ln x < 1 (i.e., x < e) and f 0 .x/ < 0 if
ln x > 1 (i.e., x > e). Since f is increasing to the left
of e and decreasing to the right, it has a maximum value
f .e/ D 1=e at x D e. Thus, if x > 0 and x ¤ e, then
Putting x D # we obtain .ln #/=# < 1=e. Thus
sec2
p
Thus cos!1 x D tan!1 .. 1 " x 2 /=x/.
) f 0 .x/ D 0 or f 0 .x/ D 2f .x/:
Since f .x/ is given to be nonconstant, we have
f 0 .x/ D 2f .x/. Thus f .x/ D f .0/e 2x D e 2x .
ln x
1
< :
x
e
p
If y D cot!1 x, then x D cot y and 0 < y < #=2. Thus
p
p
csc y D sgn x 1 C cot2 y D sgn x 1 C x 2
sgn x
sin y D p
:
1 C x2
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REVIEW EXERCISES 3 (PAGE 213)
ADAMS and ESSEX: CALCULUS 9
1
sgn x
D sgn xsin!1 p
:
Thus cot!1 x D sin!1 p
2
1Cx
1 C x2
1
csc!1 x D sin!1 .
x
18. Let T .t / be the temperature of the milk t minutes after it
is removed from the refrigerator. Let U.t / D T .t / " 20.
By Newton’s law,
U 0 .t / D kU.t /
)
21.
Suppose that for some positive integer k, the inequality
ex > 1 C x C
holds for all x > 0. This is certainly true for k D 1, as
shown in the previous exercise. Apply the MVT to
g.t / D e t " 1 " t "
U.t / D U.0/e k t :
Now T .0/ D 5 ) U.0/ D "15 and
T .12/ D 12 ) U.12/ D "8. Thus
kD
1
12
g.x/
g.x/ " g.0/
D
D g 0 .c/
x
x"0
ln.8=15/:
for some c in .0; x/. Since x and g 0 .c/ are both positive,
so is g.x/. This completes the induction and shows the
desired inequality holds for x > 0 for all positive integers
k.
If T .s/ D 18, then U.s/ D "2, so "2 D "15e sk . Thus
sk D ln.2=15/, and
sD
ln.2=15/
ln.2=15/
D 12
' 38:46:
k
ln.8=15/
It will take another 38:46 " 12 D 26:46 min for the milk to
warm up to 18ı .
Challenging Problems 3 (page 214)
1.
)
a) .d=dx/x x D x x .1 C ln x/ > 0 if ln x > "1, that is, if
x > e !1 . Thus x x is increasing on Œe !1 ; 1/.
b) Being increasing on Œe !1 ; 1/, f .x/ D x x is invertible on that interval. Let g D f !1 . If y D x x , then
x D g.y/. Note that y ! 1 if and only if x ! 1.
We have
19. Let R be the temperature of the room, Let T .t / be the
temperature of the water t minutes after it is brought into
the room. Let U.t / D T .t / " R. Then
U 0 .t / D kU.t /
U.t / D U.0/e k t :
ln y D x ln x
ln.ln y/ D ln x C ln.ln x/
g.y/ ln.ln y/
x.ln x C ln.ln x//
D lim
lim
x!1
y!1
ln y
x ln x
$
#
ln.ln x/
:
D lim 1 C
x!1
ln x
We have
T .0/ D 96 ) U.0/ D 96 " R
T .10/ D 60 ) U.10/ D 60 " R ) 60 " R D .96 " R/e 10k
T .20/ D 40 ) U.20/ D 40 " R ) 40 " R D .96 " R/e 20k :
p
Now ln x <p x for sufficiently large x, so
ln.ln x/ < ln x for sufficiently large x.
ln.ln x/
1
Therefore, 0 <
< p
! 0 as x ! 1, and
ln x
ln x
so
g.y/ ln.ln y/
lim
D 1 C 0 D 1:
y!1
ln y
Thus
$
40 " R
60 " R 2
D e 20k D
96 " R
96 " R
.60 " R/2 D .96 " R/.40 " R/
#
3600 " 120R C R2 D 3840 " 136R C R2
16R D 240
R D 15:
Room temperature is 15ı .
20. Let f .x/ D e x " 1 " x. Then f .0/ D 0 and by the MVT,
f .x/ " f .0/
f .x/
D
D f 0 .c/ D e c " 1
x
x"0
for some c between 0 and x. If x > 0, then c > 0, and
f 0 .c/ > 0. If x < 0, then c < 0, and f 0 .c/ < 0. In either
case f .x/ D xf 0 .c/ > 0, which is what we were asked to
show.
106
t kC1
t2
" !!! "
2Š
.k C 1/Š
on the interval .0; x/ (where x > 0) to obtain
" 8 D U.12/ D U.0/e 12k D "15e 12k
e 12k D 8=15;
x2
xk
C !!! C
2Š
kŠ
2.
dv
D "g " kv.
dt
a) Let u.t / D "g " kv.t /. Then
and
dv
du
D "k
D "ku,
dt
dt
u.t / D u.0/e !k t D ".g C kv0 /e !k t
&
&
1%
1%
v.t / D " g C u.t / D " g " .g C kv0 /e !k t :
k
k
b) lim t!1 v.t / D "g=k
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INSTRUCTOR’S SOLUTIONS MANUAL
c)
3.
CHALLENGING PROBLEMS 3 (PAGE 214)
dy
g
g C kv0 !k t
D v.t / D " C
e ; y.0/ D y0
dt
k
k
gt
g C kv0 !k t
y.t / D " "
e
CC
k
k2
g C kv0
g C kv0
y0 D "0 "
C C ) C D y0 C
2
k
k2
&
gt
g C kv0 %
y.t / D y0 "
1 " e !k t
C
k
k2
dv
D "g C kv 2 (k > 0)
dt
r
p
g 1 " eu
a) Let u D 2t gk. If v.t / D
, then
k 1 C eu
r
dv
g .1 C e u /."e u / " .1 " e u /e u p
D
2 gk
dt
k
.1 C e u /2
u
"4ge
D
.1 C e u /2
#
$
.1 " e u /2
kv 2 " g D g
"
1
.1 C e u /2
"4ge u
dv
D
D
:
u
2
.1 C e /
dt
p
g 1 " e 2t gk
p .
Thus v.t / D
k 1 C e 2t gk
p
r
r
g e !2t gk " 1
g
b) lim v.t / D lim
p
D"
t!1
t!1
k e !2t gk C 1
k
p
r
g
1 1 C e 2t gk
c) If y.t / D y0 C
t " ln
, then
k
k
2
y.0/ D y0 and
p
p
r
1 2 gke 2t gk
dy
g
D
"
p
dt
k k 1 C e 2t gk
p
r
g 1 " e 2t gk
p D v.t /:
D
k 1 C e 2t gk
4.
#
$
dy
dp
D e !bt
" by .
If p D e !bt y, then
$ dt
# dt
dp
p
therefore transforms to
The DE
D kp 1 " !bt
dt
e M
$
#
dy
p
D by C kpe bt 1 " !bt
dt
e M
%
ky 2
y&
D .b C k/y "
D Ky 1 "
;
M
L
bCk
M . This is a standard
k
Logistic equation with solution (as obtained in Section 3.4)
given by
Ly0
yD
;
y0 C .L " y0 /e !Kt
where K D b C k and L D
where y0 D y.0/ D p.0/ D p0 . Converting this solution
back in terms of the function p.t /, we obtain
Lp0 e !bt
p0 C .L " p0 /e !.bCk/t
.b C k/Mp0
%
&
:
D
bt
p0 ke C .b C k/M " kp0 e !k t
p.t / D
r
Thus y.t / gives the height of the object at time t
during its fall.
Since p represents a percentage, we must have
.b C k/M=k < 100.
If k D 10, b D 1, M D 90, and p0 D 1, then
bCk
M D 99 < 100. The numerator of the final exk
pression for p.t / given above is a constant. Therefore p.t /
will be largest when the derivative of the denominator,
%
&
f .t / D p0 ke bt C .b Ck/M "kp0 e !k t D 10e t C980e !10t
is zero. Since f 0 .t / D 10e t " 9; 800e !10t , this will happen
at t D ln.980/=11. The value of p at this t is approximately 48.1. Thus the maximum percentage of potential
clients who will adopt the technology is about 48.1%.
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SECTION 4.1 (PAGE 220)
ADAMS and ESSEX: CALCULUS 9
CHAPTER 4. MORE APPLICATIONS OF
DIFFERENTIATION
7.
V D
dV
dr
4 3
# r , so
D 4# r 2 .
3
dt
dt
When r D 30 cm and d V =dt D 20 cm3 /s, we have
Section 4.1 Related Rates
1.
(page 220)
20 D 4#.30/2
If the side and area of the square at time t are x and A,
respectively, then A D x 2 , so
dA
dx
D 2x
:
dt
dt
20
1
dr
D
D
:
dt
3600#
180#
The radius is increasing at 1=.180#/ cm/s.
8.
The volume V of the ball is given by
If x D 8 cm and dx=dt D 2 cm/min, then the area is
increasing at rate dA=dt D 32 cm2 /min.
V D
2. As in Exercise 1, dA=dt D 2x dx=dt . If dA=dt D "2
ft2 /s and x D 8 ft, then dx=dt D "2=.16/. The side
length is decreasing at 1/8 ft/s.
D
D
)
dr
D
dt
9.
1
p
2 A#
$
dA
:
dt
dA
dr
1
D "2, and A D 100,
D " p . The
dt
dt
10 #
1
radius is decreasing at the rate
p cm/min when the
10 #
area is 100 cm2 .
5. For A D # r 2 , we have dA=dt D 2# r dr=dt. If
dA=dt D 1=3 km2 /h,
p then (a) dr=dt
p D 1=.6# r/ km/h, or
(b) dr=dt D 1=.6# A=#/ D 1=.6 #A/ km/h
6. Let the length, width, and area be l, w, and A at time t .
Thus A D lw.
dw
dl
dA
Dl
Cw
dt
dt
dt
108
10. Let V , r and h denote the volume, radius and height of
the cylinder at time t . Thus, V D # r 2 h and
dV
dr
dh
D 2# rh
C #r2 :
dt
dt
dt
If V D 60,
dV
dr
D 2, r D 5,
D 1, then
dt
dt
60
12
V
D
D
# r 2 # 25#
5# $
dh
1
dV
dr
D
" 2# rh
dt
# r 2 dt
dt
#
$
1
12
22
D
2 " 10#
D"
:
25#
5#
25#
hD
dl
dl
48
)
D"
D "4
dt
dt
12
The length is decreasing at 4 m/s.
dS
dx
D 12x
:
dt
dt
If V D 64 cm3 and d V =dt D 2 cm3 /s, then x D 4
cm and dx=dt D 2=.3 & 16/ D 1=24 cm/s. Therefore,
dS=dt D 12.4/.1=24/ D 2. The surface area is increasing
at 2 cm2 /s.
dA
dw
D 3,
D 0, we have
dt
dt
0 D 16 & 3 C 12
# 3
D ;
6
The volume V , surface area S, and edge length x of a
cube are related by V D x 3 and S D 6x 2 , so that
dx
dV
D 3x 2
;
dt
dt
When
When l D 16, w D 12,
D
The volume is decreasing at about 28.3 cm3 /h.
4. Let A and r denote the area and radius of the circle. Then
A
##
$3
dV
#
D .36/."0:5/ D "9# ' "28:3:
dt
2
4 cm/s, then
r
D
2
When D D 6 cm, dD=dt D ":5 cm/h. At that time
dA
D 40#.4/ D 160#.
dt
The area is increasing at 160# cm2 /s.
A D #r2 ) r D
#
dV
#
dD
D D2
:
dt
2
dt
dr
dA
D 2# r :
dt
dt
If r
4 3
4#
#r D
3
3
where D D 2r is the diameter of the ball. We have
3. Let the radius and area of the ripple t seconds after impact be r and A respectively. Then A D # r 2 . We have
dr
20 cm and
dt
dr
dt
The height is decreasing at the rate
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25#
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INSTRUCTOR’S SOLUTIONS MANUAL
11.
SECTION 4.1 (PAGE 220)
Let the length, width, depth, and volume at time t be l,
w, h and V respectively. Thus V D lwh, and
dl
dw
dh
dV
D
wh C lh
C lw :
dt
dt
dt
dt
If l D 6 cm, w =5cm, h= 4cm,
dw
D "2cm/s, then
dt
The
pdistance from the origin is increasing at a rate of
2= 5.
16.
From the figure, x 2 C k 2 D s 2 . Thus
dl
dh
D
D 1m/s, and
dt
dt
x
p
When angle P CA D 45ı , x D k and s D 2k. The radar
gun indicates p
that ds=dt D 100 km/h. Thus
dx=dt D 100 2k=k ' 141. The car is travelling at about
141 km/h.
dV
D 20 " 48 C 30 D 2:
dt
The volume is increasing at a rate of 2 cm3 /s.
A
12. Let the length, width and area at time t be x, y and A
respectively. Thus A D xy and
C
x
dA
dy
dx
Dx
Cy
:
dt
dt
dt
If
ds
dx
Ds
:
dt
dt
k
dA
dx
D 5,
D 10, x D 20, y D 16, then
dt
dt
s
P
dy
dy
31
5 D 20
C 16.10/ )
D" :
dt
dt
4
Thus, the width is decreasing at
13. y D x 2 . Thus
then
14.
Fig. 4.1-16
31
m/s.
4
dy
dx
dx
D 2x
. If x D "2 and
D "3,
dt
dt
dt
dy
D "4."3/ D 12. y is increasing at rate 12.
dt
Since x 2 y 3 D 72, then
2xy 3
dx
dy
dy
2y dx
C 3x 2 y 2
D0)
D"
:
dt
dt
dt
3x dt
dx
dy
8
D 2, then
D " . Hence, the
dt
dt
9
8
vertical velocity is " units/s.
9
We have
dx
dy
Cy
D1
xy D t ) x
dt
dt
dy
dx
y D tx 2 )
D x 2 C 2xt
dt
dt
If x D 3, y D 2,
15.
17.
We continue the notation of Exercise 16. If dx=dt
p D 90
km/h, and p
angle P CA D 30ı , p
then s D 2k, x D 3k, and
ds=dt D . 3k=2k/.90/ D 45 3 D 77:94. The radar gun
will read about 78 km/h.
18. Let the distances x and y be as shown at time t . Thus
dx
dy
x 2 C y 2 D 25 and 2x
C 2y
D 0.
dt
dt
dx
1
4
dy
If
D and y D 3, then x D 4 and C 3
D 0 so
dt
3
3
dt
dy
4
D" .
dt
9
4
m/s.
The top of the ladder is slipping down at a rate of
9
At t D 2 we have xy D 2, y D 2x 2 ) 2x 3 D 2 ) x D 1,
y D 2.
dy
dx
dx
dy
Thus
C2
D 1, and 1 C 4
D
.
dt
dt
dt
dt
dx
dy
dx
D1)
D0)
D 1 ).
So 1 C 6
dt
dt
dt p
Distance D from origin satisfies D D x 2 C y 2 . So
#
$
1
dx
dy
dD
D p
2x
C 2y
dt
dt
dt
2 x2 C y2
%
&
1
2
D p 1.0/ C 2.1/ D p :
5
5
5 m
y
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x
1/3 m/s
Fig. 4.1-18
109
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SECTION 4.1 (PAGE 220)
ADAMS and ESSEX: CALCULUS 9
We are given that dx=dt D 2 ft/s, so dy=dt D 24=13 ft/s
when x D 12 ft. Now the similar triangles in the figure
show that
s
sCy
D
;
6
15
so that s D 2y=3. Hence ds=dt D 48=39. The woman’s
shadow is changing at rate 48/39 ft/s when she is 12 ft
from the point on the path nearest the lamppost.
19. Let x and y be the distances shown in the following figure. From similar triangles:
x
xCy
2y
dx
2 dy
D
)xD
)
D
:
2
5
3
dt
3 dt
Since
1
dy
D " , then
dt
2
dx
1
d
1 1
5
D " and
.x C y/ D " " D " :
dt
3
dt
2 3
6
21.
Hence, the man’s shadow is decreasing at 31 m/s and the
shadow of his head is moving towards the lamppost at a
rate of 65 m/s.
If dC =dt D 600 when x D 12; 000, then dx=dt D 100.
The production is increasing at a rate of 100 tonnes per
day.
22.
5 m
2 m
y
x2
C D 10; 000 C 3x C
8; 000
#
$
x
dx
dC
D 3C
:
dt
4; 000 dt
Let x, y be distances travelled by A and B from their positions at 1:00 pm in t hours.
dx
dy
Thus
D 16 km/h,
D 20 km/h.
dt
dt
Let s be the distance between A and B at time t .
Thus s 2 D x 2 C .25 C y/2
x
2s
Fig. 4.1-19
ds
dx
dy
D 2x
C 2.25 C y/
dt
dt
dt
"
!
At 1:30 t D 21 we have x D 8, y D 10,
p
p
s D 82 C 352 D 1289 so
20.
p
1289
ds
D 8 & 16 C 35 & 20 D 828
dt
ds
828
D p
' 23:06. At 1:30, the ships are sepadt
1289
rating at about 23.06 km/h.
15
and
6
y
5
s
x
A
16 km/h
pos. of
A
at 1:00 p.m.
B
at 1:00 p.m.
x
25 km
s
pos. of
Fig. 4.1-20
y
Refer to the figure. s, y, and x are, respectively, the
length of the woman’s shadow, the distances from the
woman to the lamppost, and the distances from the woman
to the point on the path nearest the lamppost. From one of
triangles in the figure we have
If x D 12, then y D 13. Moreover,
2y
110
B
Fig. 4.1-22
23.
y 2 D x 2 C 25:
dy
dx
D 2x
:
dt
dt
20 km/h
Let & and ! be the angles that the minute hand and hour
hand made with the vertical t minutes after 3 o’clock.
Then
#
d&
D
rad=min
dt
30
d!
#
D
rad=min:
dt
360
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INSTRUCTOR’S SOLUTIONS MANUAL
Since & D 0 and ! D
&D
#
t
30
SECTION 4.1 (PAGE 220)
#
at t D 0, therefore
2
and ! D
25.
#
#
tC :
360
2
V D 13 # r 2 h D 31 #h3
dV
dh
dh
1 dV
D #h2
)
D
:
dt
dt
dt
#h2 dt
At the first time after 3 o’clock when the hands of the
clock are together, i.e., & D !,
1
dV
1
dV
D
and h D 3, then
D
. Hence, the
dt
2
dt
18#
1
m/min.
height of the pile is increasing at
18#
Let r, h, and V be the top radius, depth, and volume of
10
r
and
the water in the tank at time t . Then D
h
8
1 2
# 25 3
V D #r h D
h . We have
3
3 16
If
#
#
180
#
tD
tC )t D
:
)
30
360
2
11
4
Thus, the hands will be together at 16 11
minutes after 3
o’clock.
Let V , r and h be the volume, radius and height of the
cone. Since h D r, therefore
26.
12
"
!
9
1
# 25 2 dh
dh
16
:
D
3h
)
D
10
3 16
dt
dt
250#h2
3
1
dh
D
.
dt
250#
1
m/min when
The water level is rising at a rate of
250#
depth is 4 m.
When h D 4 m, we have
6
Fig. 4.1-23
24.
Let y be the height of balloon t seconds after release.
Then y D 5t m.
Let & be angle of elevation at B of balloon at time t .
Then tan & D y=100. Thus
10 m
r
1 dy
5
1
d&
D
D
D
sec &
dt
100 dt
100
20
!
" d&
1
2
1 C tan &
D
dt
20
(
% y &2 ) d&
1
D
:
1C
100
dt
20
2
d&
1
d&
1
When y D 200 we have 5
.
D
so
D
dt
20
dt
100
The angle of elevation of balloon at B is increasing at a
1
rad/s.
rate of
100
8 m
h
Fig. 4.1-26
27.
Let r and h be the radius and height of the water in the
tank at time t . By similar triangles,
r
10
5
D
) r D h:
h
8
4
The volume of water in the tank at time t is
V D
1 2
25# 3
#r h D
h :
3
48
Thus,
y
dV
25# 2 dh
dh
16 d V
D
h
)
D
:
dt
16
dt
dt
25#h2 dt
If
"
B
100 m
Fig. 4.1-24
A
dV
1
h3
D
"
and h D 4, then
dt
10 1000
#
$
16
1
43
9
dh
D
"
D
:
dt
.25#/.4/2 10 1000
6250#
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SECTION 4.1 (PAGE 220)
ADAMS and ESSEX: CALCULUS 9
9
Hence, the depth of water is increasing at
m/min
6250#
when the water is 4 m deep. The maximum depth occurs
dh
when
D 0, i.e.,
dt
16
25#h2
#
1
h3
"
10 1000
$
x
10 m/min
30 m
1
h3
"
D0
10 1000
p
3
) h D 100:
D0)
Thus, the maximum depth the water in the tank can get is
p
3
100 ' 4:64 m.
Fig. 4.1-29
30.
28. Let r, h, and V be the top radius, depth, and volume of
the water in the tank at time t . Then
Let P , x, and y be your position, height above centre, and
horizontal distance from centre at time t . Let & be the
angle shown. Then y D 10 sin & , and x D 10 cos & . We
have
dy
d&
D 10 cos &
;
dt
dt
3
1
r
D D
h
9
3
1 2
# 3
V D #r h D
h
3
27
dV
# dh
D h2 :
dt
9 dt
If
s
d&
D 1 rpm D 2# rad/min:
dt
6
dy
6
When x D 6, then cos & D
, so
D 10 &
& 12#.
10
dt
10
You are rising or falling at a rate of 12# m/min at the
time in question.
2
dh
D 20 cm/h D
m/h when h D 6 m, then
dt
10
P
dV
#
2
4#
D & 36 &
D
' 2:51 m3 /h.
dt
9
10
5
10 m
y
"
Since water is coming in at a rate of 10 m3 /h, it must be
leaking out at a rate of 10 " 2:51 ' 7:49 m3 /h.
C
x
3 m
r
Fig. 4.1-30
9 m
31.
h
Let x and y denote the distances of the two aircraft east
and north of the airport respectively at time t as shown in
the following diagram. Also let the distance between the
two aircraft be s, then s 2 D x 2 C y 2 . Thus,
2s
Fig. 4.1-28
29. Let x and s be the distance as shown. Then s 2 D x 2 C302
and
dx
ds
x dx
ds
D 2x
)
D
:
2s
dt
dt
dt
s dt
When x D 40,
p
dx
D 10, s D 402 C 302 D 50, then
dt
ds
40
D
.10/ D 8. Hence, one must let out line at 8
dt
50
m/min.
112
dx
dy
ds
D 2x
C 2y :
dt
dt
dt
dx
dy
D "200 and
D 150 when x D 144 and
dt
p dt
y D 60, we have s D 1442 C 602 D 156, and
Since
ds
1
D
Œ144."200/ C 60.150/! ' "126:9:
dt
156
Thus, the distance between the aircraft is decreasing at
about 126.9 km/h.
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 4.1 (PAGE 220)
150 km/h
34.
s
Let x and y be the distances travelled from the intersection point by the boat and car respectively in t minutes.
Then
y
1000
1000
dx
D 20 &
D
m/min
dt
60
3
dy
1000
4000
D 80 &
D
m/min
dt
60
3
The distance s between the boat and car satisfy
x
airport
200 km/h
Fig. 4.1-31
s 2 D x 2 C y 2 C 202 ;
32.
1 0:6 0:4
x y
3
dP
0:6 !0:4 0:4 dx
0:4 0:6 !0:6 dy
D
x
y
C
x y
:
dt
3
dt
3
dt
P D
After one minute, x D
Thus
1374:5
If dP =dt D 0, x D 40, dx=dt D 1, and y D 10; 000, then
Hence
dy
6y 0:4 y 0:6 dx
6y dx
D " 0:4
D"
D "375:
dt
x
4x 0:6 dt
4x dt
s
ds
dx
dy
Dx
Cy :
dt
dt
dt
4000
1000
,yD
so s ' 1374: m.
3
3
1000 1000
4000 4000
ds
D
C
' 1; 888; 889:
dt
3
3
3
3
ds
' 1374:2 m/min ' 82:45 km/h after 1 minute.
dt
The daily expenses are decreasing at $375 per day.
y
20 m
33. Let the position of the ant be .x; y/ and the position of its
shadow be .0; s/. By similar triangles,
Car
x
s
s"y
y
3y
D
)sD
:
x
3"x
3"x
Boat
Then,
dy
dx
3.3 " x/
C 3y
ds
dt
dt :
D
dt
.3 " x/2
dx
1 dy
1
If the ant is at .1; 2/ and
D ,
D " , then
dt
3 dt
4
3.2/." 14 / C 3.2/. 31 /
ds
1
D
D :
dt
4
8
Hence, the ant’s shadow is moving at
along the y-axis.
1
8
units/s upwards
y
S
Fig. 4.1-34
35.
Let h and b (measured in metres) be the depth and the
surface width of the water in the trough at time t . We
have
p
2
h
D tan 60ı D 3 ) b D p h:
. 12 b/
3
Thus, the volume of the water is
$
#
10
1
hb .10/ D p h2 ;
V D
2
3
and
y
ant
p
dV
20 dh
dh
3 dV
D p h
)
D
:
dt
dt
20h dt
3 dt
dV
1
D and h D 0:2 metres, then
dt
4
p
p
# $
3
3
1
dh
D
D
:
dt
20.0:2/ 4
16
p
3
m/min.
Hence, the water level is rising at
16
If
x
Fig. 4.1-33
3
lamp
x
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113
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SECTION 4.1 (PAGE 220)
ADAMS and ESSEX: CALCULUS 9
ı
60
b=2
10 m
b=2
30 cm
h
p
y
ı
30
3 m
32 Cx 2
x
s
Fig. 4.1-37
Fig. 4.1-35
a) By similar triangles,
Thus,
36. Let V and h be the volume and depth of water in the pool
at time t . If h $ 2, then
x
20
D
D 10;
h
2
so V D
1
xh8 D 40h2 :
2
y
3
30
D p
)yD p
:
2
2
10
3 Cx
9 C x2
dy
dx
dy dx
"30x
D
D
:
dt
dx dt
.9 C x 2 /3=2 dt
1
dx
D , then
dt
5
If x D 4 and
dy
"30.4/
D
dt
.9 C 16/3=2
If 2 $ h $ 3, then V D 160 C 160.h " 2/.
Hence, the free top end of the ladder is moving vertically downward at 24/125 m/s.
dV
dh
D 160 .
dt
dt
1
So surface of water is dropping at a rate of
160
m/min.
a) If h D 2:5m, then "1 D
b) By similar triangles,
p
x
32
Then,
dh
dh
dV
D 80h
D 80 .
dt
dt
dt
1
m/min.
So surface of water is dropping at a rate of
80
b) If h D 1m, then "1 D
C
x2
D
s
10x
:
)sD p
10
9 C x2
ds
ds dx
D
dt
dx dt
#
$
p
2x
. 9 C x 2 /.10/ " .10x/ p
2 9 C x 2 dx
D
.9 C x 2 /
dt
90
dx
D
:
.9 C x 2 /3=2 dt
20
8
3
1
# $
1
24
D"
:
5
125
If x D 4 and
x
h
1
dx
D , then
dt
5
ds
90
D
dt
.9 C 16/3=2
# $
1
18
D
:
5
125
This is the rate of change of the length of the horizontal projection of the ladder. The free top end of
the ladder is moving horizontally to the right at rate
1
18
7
dx ds
"
D "
D
m/s:
dt
dt
5 125
125
Fig. 4.1-36
38.
37. Let the various distances be as shown in the figure.
114
Let x, y, and s be distances shown at time t . Then
s 2 D x 2 C 16;
ds
dx
s
Dx
;
dt
dt
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.15 " s/2 D y 2 C 16
ds
dy
" .15 " s/
Dy :
dt
dt
lOMoARcPSD|6566483
INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 4.1 (PAGE 220)
1
dx
D , then s D 5 and
When x D 3 and
dt
2
p
p
2 D
y D 102 " 4#
$ 84.
3 1
3
ds
D
D
so
Also
dt
5 2
10
40.
dy
10 3
3
D "p
D " p ' 0:327:
dt
10
84
84
Crate B is moving toward Q at a rate of 0.327 m/s.
Let y be height of ball t seconds after it drops.
dy
d 2y
D "9:8,
j tD0 D 0, yj tD0 D 20, and
Thus
dt 2
dt
dy
D "9:8t:
dt
y D "4:9t 2 C 20;
Let s be distance of shadow of ball from base of pole.
s " 10
s
By similar triangles,
D
.
y
20
200
20s " 200 D sy, s D
20 " y
ds
ds
dy
20
Dy
Cs .
dt
dt
dt
P
15!s
B
4
y
a) At t D 1, we have
s
x
A
Q
Fig. 4.1-38
39. Let & be the angle of elevation, and x and y the horizontal and vertical distances from the launch site. We have
y
tan & D
x
dx
dy
"y
x
d&
dt
dt :
sec &
D
dt
x2
ds
200
4:9
D
."9:8/.
dt
4:9
The shadow is moving at a rate of 81.63 m/s after
one second.
b) As the r
ball hits the ground, y Dr0, s D 10,
20
20
dy
t D
, and
D "9:8
, so
4:9
dt
4:9
dy
ds
D 0 C 10 .
20
dt
dt
r
20
Now y D 0 implies that t D
. Thus
4:9
r
ds
1
20
D " .9:8/
' "9:90:
dt
2
4:9
2
)
At the instant in question
The shadow is moving at about 9.90 m/s when the
ball hits the ground.
10 m
p
dx
dy
D 4 cos 30ı D 2 3;
D 4 sin 30ı D 2;
dt
dt
x D 50 km;
y D 100 km:
Thus tan & D
100
D 2, sec2 & D 1 C tan2 & D 5, and
50
dy
D "9:8, y D 15:1,
dt
20!y
20 m
p
p
d&
1"2 3
1 50.2/ " 100.2 3/
D
D
' "0:0197:
dt
5
.50/2
125
y
10
s!10
s
Therefore, the angle of elevation is decreasing at about
0.0197 rad/s.
Fig. 4.1-40
4 km/s
ı
30
y
41. Let y.t / be the height of the rocket t seconds after it
blasts off. We have
"
d 2y
D 10;
dt 2
x
Fig. 4.1-39
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dy
DyD0
dt
115
lOMoARcPSD|6566483
SECTION 4.1 (PAGE 220)
ADAMS and ESSEX: CALCULUS 9
at t D 0. Hence y D 5t 2 , (y in metres, t in seconds).
Now
y
d&
dy=dt
tan & D
, so sec2 &
D
, and
2000
dt
2000
%
% y &2 & d&
10t
t
1C
D
D
2000
dt
2000
200
d&
t
1
D
!
dt
200
25t 4
1C
20002
t
800t
1
D
D
:
!
200
4002 C t 4
t4
1C
4002
At t D 10, we have
d&
8000
D
' 0:047 rad/s.
dt
4002 C 1002
Starting with x0 D 1:5, get x4 D x5 D 1:73205080757.
9.
g.x/ D x "
10. f .x/ D x 3 C 2x 2 " 2, f 0 .x/ D 3x 2 C 4x.
Newton’s formula xnC1 D g.xn /, where
g.x/ D x "
11.
f .x/ D x 4 " 8x 2 " x C 16, f 0 .x/ D 4x 3 " 16x " 1.
Newton’s formula xnC1 D g.xn /, where
g.x/ D x "
x 4 " 8x 2 " x C 16
3x 4 " 8x 2 " 16
D
:
3
4x " 16x " 1
4x 3 " 16x " 1
Starting with x0 D 1:5, get x4 D x5 D 1:64809536561.
Starting with x0 D 2:5, get x5 D x6 D 2:35239264766.
2 km
Fig. 4.1-41
Section 4.2 Finding Roots of Equations
(page 230)
12. f .x/ D x 3 C 3x 2 " 1, f 0 .x/ D 3x 2 C 6x.
Newton’s formula xnC1 D g.xn /, where
g.x/ D x "
1 !xn
e
starting with x C 0 D 0:3. Both x10
2
round to 0.35173.
Iterate xnC1 D
3. To solve cos.x=3/ D x, start with x0 D 0:9 and iterate
xnC1 D cos.xn =3/. x4 and x5 round to 0.95025.
4. To solve .x C 9/1=3 D x, start with x0 D 2 and iterate
xnC1 D .xn C 9/1=3 . x4 and x5 round to 2.24004.
13.
f .x/ D sin x " 1 C x, f 0 .x/ D cos x C 1.
Newton’s formula is xnC1 D g.xn /, where
5. To solve 1=.2 C x 2 / D x, start with x0 D 0:5 and iterate
xnC1 D 1=.2 C xn2 /. x6 and x7 round to 0.45340.
6. To solve x 3 C 10x " 10 D 0, start with x0 D 1 and iterate
1 3
xn . x7 and x8 round to 0.92170.
xnC1 D 1 " 10
f .x/ D x 2 " 2, f 0 .x/ D 2x.
Newton’s formula xnC1 D g.xn /, where
2x 3 C 3x 2 C 1
x 3 C 3x 2 " 1
D
:
3x 2 C 6x
3x 2 C 6x
Because f ."3/ D "1, f ."2/ D 3, f ."1/ D 1,
f .0/ D "1, f .1/ D 3, there are roots between "3 and
"2, between "1 and 0, and between 0 and 1.
Starting with x0 D "2:5, get x5 D x6 D "2:87938524157.
Starting with x0
D
"0:5, get
x4 D x5 D "0:652703644666.
Starting with x0 D 0:5, get x4 D x5 D 0:532088886328.
2. To solve 1 C 14 sin x D x, start with x0 D 1 and iterate
xnC1 D 1 C 41 sin xn . x5 and x6 round to 1.23613.
7.
x 3 C 2x 2 " 2
2x 3 C 2x 2 C 2
D
:
2
3x C 4x
3x 2 C 4x
Starting with x0 D 1:5, get x5 D x6 D 0:839286755214.
"
and x11
x 3 C 2x " 1
2x 3 C 1
D
:
2
3x C 2
3x 2 C 2
Starting with x0 D 0:5, get x3 D x4 D 0:45339765152.
y
1.
f .x/ D x 3 C 2x " 1, f 0 .x/ D 3x 2 C 2.
Newton’s formula xnC1 D g.xn /, where
g.x/ D x "
The graphs of sin x and 1 " x suggest a root near x D 0:5.
Starting with x0 D 0:5, get
x3 D x4 D 0:510973429389.
y
x2 " 2
x2 C 2
g.x/ D x "
D
:
2x
2x
y D1"x
y D sin x
Starting with x0 D 1:5, get x3 D x4 D 1:41421356237.
8. f .x/ D x 2 " 3, f 0 .x/ D 2x.
Newton’s formula xnC1 D g.xn /, where
2
g.x/ D x "
116
sin x " 1 C x
:
cos x C 1
0.5
1.0
2
x "3
x C3
D
:
2x
2x
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Fig. 4.2-13
1.5x
lOMoARcPSD|6566483
INSTRUCTOR’S SOLUTIONS MANUAL
14.
SECTION 4.2 (PAGE 230)
f .x/ D x 2 " cos x, f 0 .x/ D 2x C sin x.
Newton’s formula is xnC1 D g.xn /, where
g.x/ D x "
has a root near x D 0:6. Use of a solve routine or Newton’s Method gives x D 0:56984029099806.
x 2 " cos x
:
2x C sin x
sin x
. Since jf .x/j $ 1=.1 C x 2 / ! 0
1 C x2
as x ! ˙1 and f .0/ D 0, the maximum and minimum
values of f will occur at the two critical points of f that
are closest to the origin on the right and left, respectively.
For CP:
18. Let f .x/ D
The graphs of cos x and x 2 , suggest a root near
x D ˙0:8. Starting with x0 D 0:8, get
x3 D x4 D 0:824132312303. The other root is the negative of this one, because cos x and x 2 are both even functions.
y
.1 C x 2 / cos x " 2x sin x
.1 C x 2 /2
2
0 D .1 C x / cos x " 2x sin x
0 D f 0 .x/ D
y D x2
y D cos x
19.
-1.5 -1.0 -0.5
0.5 1.0 1.5x
Fig. 4.2-14
15.
Since tan x takes all real values between any two consecutive odd multiples of #=2, its graph intersects y D x
infinitely often. Thus, tan x D x has infinitely many solutions. The one between #=2 and 3#=2 is close to 3#=2,
so start with x0 D 4:5. Newton’s formula here is
xnC1 D xn "
with 0 < x < # for the maximum and "# < x < 0 for
the minimum. Solving this equation using a solve routine
or Newton’s Method starting, say, with x0 D 1:5, we get
x D ˙0:79801699184239: The corresponding max and
min values of f are ˙0:437414158279.
cos x
Let f .x/ D
. Note that f is an even function, and
1 C x2
that f has maximum value 1 at x D 0. (Clearly f .0/ D 1
and jf .x/j < 1 if x ¤ 0.) The minimum value will occur
at the critical points closest to but not equal to 0. For CP:
.1 C x 2 /." sin x/ " 2x cos x
.1 C x 2 /2
0 D .1 C x 2 / sin x C 2x cos x:
0 D f 0 .x/ D
The first CP to the right of zero is between #=2
and 3#=2, so start with x D 2:5, say, and get
x D 2:5437321475261. The minimum value is
f .x/ D "0:110639672192.
tan xn " xn
:
sec2 xn " 1
We get x3 D x4 D 4:49340945791.
y
20.
For x 2 D 0 we have xnC1 D xn " .xn2 =.2xn // D xn =2.
If x0 D 1, then x1 D 1=2, x2 D 1=4, x3 D 1=8.
a) xn D 1=2n , by induction.
b) xn approximates the root x D 0 to within 0.0001
provided 2n > 10; 000. We need n # 14 to ensure
this.
yDx
c) To ensure that xn2 is within 0.0001 of 0 we need
.1=2n /2 < 0:0001, that is, 22n > 10; 000. We need
n # 7.
!
x
y D tan x
21.
Fig. 4.2-15
d) Convergence of Newton approximations to the root
x D 0 of x 2 D 0 is slower than usual because the
derivative 2x of x 2 is zero at the root.
'p
x
if x # 0
,
f .x/ D p
' "x p if x < 0
x/
if x > 0
1=.2 p
f 0 .x/ D
.
"1=.2 "x/ if x < 0
The Newton’s Method formula says that
16. A graphing calculator shows that the equation
p
.1 C x 2 / x " 1 D 0
xnC1 D xn "
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f .xn /
D xn " 2xn D "xn :
f 0 .xn /
117
lOMoARcPSD|6566483
SECTION 4.2 (PAGE 230)
ADAMS and ESSEX: CALCULUS 9
If x0 D a, then x1 D "a, x2 D a, and, in general,
xn D ."1/n a. The approximations oscillate back and forth
between two numbers.
If one observed that successive approximations were oscillating back and forth between two values a and b, one
should try their average, .aCb/=2, as a new starting guess.
It may even turn out to be the root!
22. Newton’s Method formula for f .x/ D x 1=3 is
Hence
dxn D "
Since the values of xn are assumed to neither converge nor diverge, the exponential factor 2n will dominate for large n
26.
1=3
xn
xnC1 D xn "
!2=3
.1=3/xn
D xn " 3xn D "2xn :
If x0 D 1, then x1 D "2, x2 D 4, x3 D "8, x4 D 16, and,
in general, xn D ."2/n . The successive “approximations”
oscillate ever more widely, diverging from the root at
x D 0.
23. Newton’s Method formula for f .x/ D x 2=3 is
xnC1 D xn "
!1=3
.2=3/xn
D xn "
3
2 xn
D
" 12 xn :
27.
If x0 D 1, then x1 D "1=2, x2 D 1=4, x3 D "1=8,
x4 D 1=16, and, in general, xn D ."1=2/n . The successive approximations oscillate around the root x D 0, but
still converge to it (though more slowly than is usual for
Newton’s Method).
24.
Since xnC1 D
#
xn2 " 1
2xn
$2
D
#
xn2 C 1
2xn
$2
holds whenever u and v are in Œa; b!. Pick any x0 in
Œa; b!, and let x1 D f .x0 /, x2 D f .x1 /, and, in general, xnC1 D f .xn /. Let r be the fixed point of f in
Œa; b! found in Exercise 24. Thus f .r/ D r. We have
jx1 " rj D jf .x0 / " f .r/j $ Kjx0 " rj
jx2 " rj D jf .x1 / " f .r/j $ Kjx1 " rj $ K 2 jx0 " rj;
:
and, in general, by induction
It follows that
$2
2xn
xn2 C 1
1 " yn
D 4yn2
D 4yn .1 " yn /:
yn
ynC1 D
#
jxn " rj $ K n jx0 " rj:
Since K < 1, limn!1 K n D 0, so limn!1 xn D r. The
iterates converge to the fixed point as claimed in Theorem
6.
25. Let yj D sin2 .uj /.
a) Since sin2 .unC1 / D 4 sin2 .un /.1 " sin2 .un //
D 4 sin2 .un / cos2 .un / D sin2 .2un /; we have
unC1 D 2un . Thus unC1 D 2n u0 . It follows that
Section 4.3 Indeterminate Forms
(page 235)
1.
dyn D 2 sin.un / cos.un / 2n du0 :
b) Since yn D
1
, we have
1 C xn2
dyn D "
118
We are given that there is a constant K satisfying
0 < K < 1, such that
jf .u/ " f .v/j $ Kju " vj
xn2 " 1
, we have
2xn
2
1 C xnC1
D1C
Let g.x/ D f .x/ " x for a $ x $ b. g is continuous
(because f is), and since a $ f .x/ $ b whenever
a $ x $ b (by condition (i)), we know that g.a/ # 0 and
g.b/ $ 0. By the Intermediate-Value Theorem there exists
r in Œa; b! such that g.r/ D 0, that is, such that f .r/ D r.
The fixed point r is unique because if there were two such
fixed points, say r1 and r2 , then condition (ii) would imply
that
jr1 " r2 j D jf .r1 / " f .r2 /j $ Kjr1 " r2 j;
which is impossible if r1 ¤ r2 and K < 1.
2=3
xn
.1 C xn2 /2
2 sin.un / cos.un / 2n du0 :
2xn
2.
2xn
dxn :
.1 C xn2 /2
( )
0
0
3
3
D lim
D
x!0 4 sec2 4x
4
( )
0
ln.2x " 3/
lim
x!2 x 2 " 4
0
#
$
2
1
2x " 3
D
D :
2x
2
3x
x!0 tan 4x
lim
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INSTRUCTOR’S SOLUTIONS MANUAL
( )
sin ax
0
x!0 sin bx
0
a cos ax
a
D lim
D
x!0 b cos bx
b
( )
1 " cos ax
0
4.
lim
x!0 1 " cos bx
0
( )
0
a sin ax
D lim
x! 0 b sin bx
0
2
a cos ax
a2
D lim 2
D 2:
x!0 b cos bx
b
(
)
sin!1 x
0
5. lim
!1
x!0 tan
x
0
1 C x2
D lim p
D1
x!0
1 " x2
( )
x 1=3 " 1
0
lim 2=3
6.
x!1 x
0
"1
. 1 /x !2=3
1
D lim 32
D :
x!1 . /x !1=3
2
3
3.
7.
lim
lim x cot x
Œ0 & 1!
% x &
cos x
D lim
x!0 sin x
( )
0
x
D 1 & lim
x!0 sin x
0
1
D1
D lim
x!0 cos x
( )
1 " cos x
0
lim
x!0 ln.1 C x 2 /
0
sin x
$
D lim #
2x
x!0
1 C x2
sin x
D lim .1 C x 2 / lim
x!0
x!0 2x
1
cos x
D :
D lim
x!0
2
2
(
)
sin2 t
0
lim
t!! t " #
0
2 sin t cos t
D lim
D0
t!!
1
( )
10x " e x
0
lim
x!0
x
0
10x ln 10 " e x
D lim
D ln 10 " 1:
x!0
1
( )
0
cos 3x
lim
0
x!!=2 # " 2x
"3 sin 3x
3
3
D lim
D ."1/ D "
"2
2
2
x!!=2
SECTION 4.3 (PAGE 235)
12.
13.
9.
10.
11.
lim x sin
x!1
1
x
sin
D lim
Œ1 & 0!
1
x
( )
0
0
1
x
1
1
" 2 cos
1
x
x
D lim
D lim cos D 1:
1
x!1
x!1
x
" 2
x
x!1
14.
x!0
8.
( )
0
ln.ex/ " 1
x!1
sin #x
0
1
1
x
D lim
D" :
x!1 # cos.#x/
#
lim
15.
16.
17.
( )
x " sin x
0
x!0
x3
0
( )
0
1 " cos x
D lim
x!0
3x 2
0
( )
sin x
0
D lim
x!0 6x
0
1
cos x
D :
D lim
x!0
6
6
lim
( )
x " sin x
0
x!0 x " tan x
0
( )
1 " cos x
0
D lim
x!0 1 " sec2 x
0
1 " cos x
D lim .cos2 x/ 2
x!0
cos x " 1
cos x " 1
D "1 & lim
x!0 .cos x " 1/.cos x C 1/
1
D"
2
lim
( )
2 " x 2 " 2 cos x
0
x!0
x4
0
( )
0
"2x C 2 sin x
D lim
x!0
4x 3
0
1
x " sin x
D " lim
3
2 x!0
# $ x
1
1 1
D"
(by Exercise 14):
D"
2 6
12
lim
( )
sin2 x
0
x!0C tan x " x
0
( )
2 sin x cos x
0
D lim
x!0C sec2 x " 1
0
cos x
D 2 & 1 & lim
D1
x!0C 2 sec2 x tan x
lim
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SECTION 4.3 (PAGE 235)
18.
19.
( )
ln sin r
0
0
r!!=2 cos r
% cos r &
sin r D 0:
D lim
r!!=2 " sin r
21.
25.
lim
sin t
2
D
t
#
lim
t!!=2
20.
ADAMS and ESSEX: CALCULUS 9
( )
0
cos!1 x
x!1! x " 1
0
#
$
1
" p
1 " x 2 D "1:
D lim
x!1!
1
lim
lim x.2 tan!1 x " #/
x!1
!1
x"#
1
x
1
2 .
" 2
D lim
x!1 1 C x 2
x
2x 2
D "2
D lim "
x!1 1 C x 2
D lim
2 tan
x!1
lim
22.
.sec t " tan t /
t!.!=2/!
Œ0 & 1!
( )
0
0
26.
Œ1 " 1!
( )
0
0
1 " sin t
cos t
" cos t
D 0:
D lim
t!.!=2/! " sin t
D
23.
24.
2
Let y D .csc x/sin x .
Then ln y D sin2 x ln.csc x/
h1i
ln.csc x/
lim ln y D lim
x!0C
x!0C csc2 x
1
csc x cot x
"
csc x
D lim
x!0C "2 csc2 x cot x
1
D lim
D 0:
x!0C 2 csc2 x
sin2 x
Thus limx!0C .csc x/
D e 0 D 1.
lim
t!.!=2/!
$
1
1
" at
.1 " 1/
t!0 t
te
(
)
e at " 1
0
D lim
t!0 t e at
0
ae at
D lim at
Da
t!0 e
C at e at
lim
#
#
Since
hence
p
ln x
x ln x D lim !1=2
x!0C x
x!0C
# $
1
x
D 0;
D lim # $
1
x!0C
"
x !3=2
2
lim
lim x
p
x
x!0C
D lim e
x!0C
120
$
1
x
"
Œ1 " 1!
x!1C x " 1
ln x
( )
x ln x " x C 1
0
D lim
x!1C .x " 1/.ln x/
0
( )
0
ln x
D lim
1
x!1C
0
ln x C 1 "
x
1
x
D lim
1
x!1C 1
C 2
x
x
x
1
D lim D
D :
x!1C
xC1
2
lim
p
x ln x
D e 0 D 1:
( )
0
0
27.
( )
0
3 sin t " sin 3t
t!0 3 tan t " tan 3t
0
( )
3.cos t " cos 3t /
0
D lim
t!0 3.sec2 t " sec2 3t /
0
cos t " cos 3t
D lim
t!0 cos2 3t " cos2 t
cos2 t cos2 3t
cos 3t " cos t
D " lim
t!0 cos2 3t " cos2 t
1
1
D"
D " lim
t!0 cos 3t C cos t
2
lim
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INSTRUCTOR’S SOLUTIONS MANUAL
28.
Let y D
#
sin x
x
$1=x 2
SECTION 4.3 (PAGE 235)
32.
:
$
sin x
( )
0
x
lim ln y D lim
2
x!0
x!0
x
0
% x & # x cos x " sin x $
sin x
x2
D lim
x!0
2x ( )
x cos x " sin x
0
D lim
x!0
2x 2 sin x
0
"x sin x
D lim
x!0 4x sin x C 2x 2 cos x
( )
0
" sin x
D lim
x!0 4 sin x C 2x cos x
0
" cos x
1
D lim
D" :
x!0 6 cos x " 2x sin x
6
#
$ 2
sin x 1=x
Thus; lim
D e !1=6 :
x!0
x
ln
#
Let y D .1 C tan x/1=x :
( )
0
ln.1 C tan x/
lim ln y D lim
x!0
x!0
x
0
sec2 x
D lim
D 1:
x!0 1 C tan x
Thus; lim .1 C tan x/1=x D e:
x!0
33.
34.
( )
0
f .x C h/ " 2f .x/ C f .x " h/
h2
0
h!0
( )
f 0 .x C h/ " f 0 .x " h/
0
D lim
2h
0
h!0
f 00 .x C h/ C f 00 .x " h/
D lim
2
h!0
2f .x/
00
D f .x/
D
2
lim
f .x C 3h/ " 3f .x C h/ C 3f .x " h/ " f .x " 3h/
h3
0
0
3f .x C 3h/ " 3f .x C h/ " 3f 0 .x " h/ C 3f 0 .x " 3h/
D lim
3h2
h!0
3f 00 .x C 3h/ " f 00 .x C h/ C f 00 .x " h/ " 3f 00 .x " 3h/
D lim
2h
h!0
9f 000 .x C 3h/ " f 000 .x C h/ " f 000 .x " h/ C 9f 000 .x " 3h/
D lim
2
h!0
D8f 000 .x/:
lim
h!0
2
29. Let y D .cos 2t /1=t .
ln.cos 2t /
. We have
Then ln y D
t2
( )
ln.cos 2t /
0
2
t!0
t
0
( )
0
"2 tan 2t
D lim
t!0
2t
0
2 sec2 2t
D " lim
D "2:
t!0
1
lim ln y D lim
t!0
2
Therefore lim t!0 .cos 2t /1=t D e !2 .
30.
31.
csc x h 1 i
"
lim
x!0C ln x
1
" csc x cot x h 1 i
"
D lim
1
x!0C
1
x
( )
"x cos x
0
D lim
x!0C sin2 x
0
#
$
1
D " lim cos x lim
x!0C
x!0C 2 sin x cos x
D "1:
h1i
ln sin #x
lim
x!1! csc #x
1
# cos #x
sin #x
D lim
x!1! "# csc #x cot #x
"#
D
lim tan #x D 0
# x!1!
35.
Suppose that f and g are continuous on Œa; b! and differentiable on .a; b/ and g.x/ 6D 0 there. Let a < x < t < b,
and apply the Generalized Mean-Value Theorem; there
exists c in .x; t / such that
)
)
)
)
)
f .x/ " f .t /
f 0 .c/
D 0
g.x/ " g.t /
g .c/
)(
)
(
f .x/ " f .t /
g.x/
f 0 .c/
D 0
g.x/
g.x/ " g.t /
g .c/
(
)
f .x/ f .t /
f 0 .c/ g.x/ " g.t /
"
D 0
g.x/
g.x/
g .c/
g.x/
f .x/
f 0 .c/
g.t / f 0 .c/
f .t /
D 0
"
C
g.x/
g .c/
g.x/ g 0 .c/
g.x/
"
#
f .x/
f 0 .c/
1
f 0 .c/
D 0
C
f .t / " g.t / 0
g.x/
g .c/
g.x/
g .c/
"
#
f .x/
f 0 .c/
1
f 0 .c/
"LD 0
"LC
f .t / " g.t / 0
:
g.x/
g .c/
g.x/
g .c/
Since jm C nj $ jmj C jnj, therefore,
"
ˇ ˇ 0
ˇ
ˇ 0 ˇ#
ˇ
ˇ ˇ f .c/
ˇ
ˇ f .c/ ˇ
ˇ f .x/
1
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ g.x/ "Lˇ $ ˇ g 0 .c/ "Lˇ C jg.x/j jf .t /jCjg.t /jˇ g 0 .c/ ˇ :
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SECTION 4.3 (PAGE 235)
ADAMS and ESSEX: CALCULUS 9
Now suppose that $ is an arbitrary small positive number.
Since limc!aC f 0 .c/=g 0 .c/ D L, and since a < x < c < t,
we can choose t sufficiently close to a to ensure that
In particular,
ˇ 0
ˇ
ˇ f .c/
ˇ $
ˇ
ˇ< :
"
L
ˇ g 0 .c/
ˇ 2
ˇ 0 ˇ
ˇ f .c/ ˇ
$
ˇ
ˇ
ˇ g 0 .c/ ˇ < jLj C 2 :
f .x/ D x 5 C x 3 C 2x on .a; b!
f 0 .x/ D 5x 4 C 3x 2 C 2 > 0 for all x.
f has no min value, but has abs max value b 5 C b 3 C 2b
at x D b.
"1
1
10. f .x/ D
. Since f 0 .x/ D
< 0 for all x in
x"1
.x " 1/2
the domain of f , therefore f has no max or min values.
9.
11.
Since limx!aC jg.x/j D 1, we can choose x between a
and t sufficiently close to a to ensure that
%
1 h
$ &i $
jf .t /j C jg.t /j jLj C
< :
jg.x/j
2
2
It follows that
Thus limx!aC
ˇ
ˇ
ˇ $
ˇ f .x/
$
ˇ
ˇ
ˇ g.x/ " Lˇ < 2 C 2 D $:
1.
1
on Œ2; 3!
x"1
1
abs min 2 at x D 3, abs max 1 at x D 2.
12. f .x/ D
13.
Let f .x/ D jx " 1j on Œ"2; 2!: f ."2/ D 3, f .2/ D 1.
f 0 .x/ D sgn .x " 1/. No CP; SP x D 1, f .1/ D 0.
Max value of f is 3 at x D "2; min value is 0 at
x D 1.
14.
Let f .x/ D jx 2 " x " 2j D j.x " 2/.x C 1/j on Œ"3; 3!:
f ."3/ D 10, f .3/ D 4.
f 0 .x/ D .2x " 1/sgn .x 2 " x " 2/.
CP x D 1=2; SP x D "1, and x D 2. f .1=2/ D 9=4,
f ."1/ D 0, f .2/ D 0.
Max value of f is 10 at x D "3; min value is 0 at
x D "1 or x D 2.
1
2x
f .x/ D 2
, f 0 .x/ D " 2
x C1
.x C 1/2
f has abs max value 1 at x D 0; f has no min values.
f .x/
D L:
g.x/
Section 4.4 Extreme Values
(page 242)
f .x/ D x C 2 on Œ"1; 1!
f 0 .x/ D 1 so f is increasing.
f has absolute minimum 1 at x D "1 and absolute maximum 3 at x D 1.
2. f .x/ D x C 2 on ."1; 0!
abs max 2 at x D 0, no min.
3. f .x/ D x C 2 on Œ"1; 1/
f has absolute minimum 1 at x D "1 and has no absolute
maximum.
1
on .0; 1/
x"1
1
< 0 on .0; 1/
f 0 .x/ D "
.x " 1/2
f has no max or min values.
f .x/ D
15.
16.
f .x/ D .x C 2/.2=3/
no max, abs min 0 at x D "2.
17.
f .x/ D .x " 2/1=3 , f 0 .x/ D
f has no max or min values.
1
.x " 2/!2=3 > 0
3
y
4. f .x/ D x 2 " 1
no max, abs min "1 at x D 0.
5. f .x/ D x 2 " 1 on Œ"2; 3!
f has abs min "1 at x D 0, abs max 8 at x D 3, and
local max 3 at x D "2.
2
y D .x " 2/1=3
2
6. f .x/ D x " 1 on .2; 3/
no max or min values.
7.
Fig. 4.4-17
f .x/ D x 3 C x " 4 on Œa; b!
f 0 .x/ D 3x 2 C 1 > 0 for all x.
Therefore f has abs min a3 C a " 4 at x D a and abs max
b 3 C b " 4 at x D b.
18. f .x/ D x 2 C 2x, f 0 .x/ D 2x C 2 D 2.x C 1/
Critical point: x D "1.
f .x/ ! 1 as x ! ˙1.
8. f .x/ D x 3 C x " 4 on .a; b/
Since f 0 .x/ D 3x 2 C 1 > 0 for all x, therefore f is
increasing. Since .a; b/ is open, f has no max or min
values.
122
CP
f0
"
"1
C
"""""""""""""""
j """"""""""!x
abs
f
&
%
min
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 4.4 (PAGE 242)
y
Hence, f .x/ has no max value, and the abs min is "1 at
x D "1.
2
y D .x " 4/2
16
y
y D x 2 C 2x
x
!2
2
x
Fig. 4.4-20
.!1;!1/
21.
Fig. 4.4-18
f .x/ D x 3 .x " 1/2
f 0 .x/ D 3x 2 .x " 1/2 C 2x 3 .x " 1/
D x 2 .x " 1/.5x " 3/
3
CP x D 0; ; 1
5
3
19. f .x/ D x " 3x " 2
f 0 .x/ D 3x 2 " 3 D 3.x " 1/.x C 1/
CP
CP
CP
3
"
1
C
f0 C
0
C
5
"""""""""""
j""""""""""j"""""""""""j"""""""!x
loc
loc
f
%
%
max &
min %
CP
CP
f0
C
"1
"
1
C
""""""""""""""
j"""""""""""""
j"""""""""!x
loc
loc
f
%
&
%
max
min
f has no absolute extrema.
f has no absolute extrema.
y
y
!1
%
x
3 108
5 ; 55
&
1
y D x 3 " 3x " 2
x
y D x 3 .x " 1/2
.1;!4/
Fig. 4.4-19
Fig. 4.4-21
22.
20. f .x/ D .x 2 " 4/2 , f 0 .x/ D 4x.x 2 " 4/ D 4x.x C 2/.x " 2/
Critical points: x D 0; ˙2.
f .x/ ! 1 as x ! ˙1.
CP
CP
CP
f0 "
"2
C
0
"
C2
C
""""""""""""
j """""""""""
j """""""""""
j """""""!x
abs
abs
loc
f
&
max &
min %
min %
Hence, f .x/ has abs min 0 at x D ˙2 and loc max 16 at
x D 0.
f .x/ D x 2 .x " 1/2 ,
f 0 .x/ D 2x.x " 1/2 C 2x 2 .x " 1/ D 2x.2x " 1/.x " 1/
Critical points: x D 0; 21 and 1.
f .x/ ! 1 as x ! ˙1.
CP
CP
CP
1
f0
"
0
C
"
1
C
2
""""""""""""
j """""""""""
j """""""""""j"""""""!x
abs
abs
loc
f
&
max &
min %
min %
Hence, f .x/ has loc max
x D 0 and x D 1.
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16
at x D
1
2
and abs min 0 at
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SECTION 4.4 (PAGE 242)
ADAMS and ESSEX: CALCULUS 9
y
y
.1;0:5/
yD
.!1;!0:5/
y D x 2 .x " 1/2
x
x2 C 1
x
Fig. 4.4-24
2
25.
!1
1
2 ; 16
"
1
x
D1" 2
<1
x2 C 1
x C1
2x
f 0 .x/ D 2
.x C 1/2
f .x/ D
x
1
CP
f0
"
0
C
""""""""""""j""""""""!x
abs
f
&
min %
Fig. 4.4-22
23.
f .x/ D x.x 2 " 1/2
f 0 .x/ D .x 2 " 1/2 C 2x.x 2 " 1/2x
2
2
y
2
yD1
D .x " 1/.x " 1 C 4x /
2
2
D .x " 1/.5x " 1/
p
p
D .x " 1/.x C 1/. 5x " 1/. 5x C 1/
yD
CP
CP
CP
CP
"
1 C
f 0 C "1 " " p1 C p1
5
5
""""""""""
j """""""""
j """""""""
j """"""""j"""""!x
loc & loc % loc & loc %
f % max
max
min
min
x
Fig. 4.4-25
26.
p
p
f .˙1/ D 0, f .˙1= 5/ D ˙16=25 5
2
p
1= 5
x
x4
, f 0 .x/ D
1
x
Hence, f has abs max
x D "1.
2
y D x.x " 1/
p1
2
at x D 1 and abs min " p1 at
2
y
Fig. 4.4-23
24.
1 " x2
x
, f 0 .x/ D 2
2
x C1
.x C 1/2
Critical point: x D ˙1.
f .x/ ! 0 as x ! ˙1.
%
f .x/ D
27.
CP
CP
f0
"
"1
C
C1
"
""""""""""""""
j """""""""""""
j """""""""!x
abs
abs
f
&
&
%
max
min
Hence, f has abs max
x D "1.
124
1
2
at x D 1 and abs min " 12 at
1 " x4
C 1/3=2
.x 4
CP
CP
f0
"
"1
C
C1
"
""""""""""""""
j """""""""""""
j """""""""!x
abs
abs
f
&
%
&
max
min
p
!1= 5
f .x/ D p
C1
Critical points: x D ˙1.
f .x/ ! 0 as x ! ˙1.
y
!1
x2
C1
x2
1
!1;! p
2
&
%
1
1; p
2
&
yD p
x
x
x4 C 1
Fig. 4.4-26
p
f .x/ D x 2 " x 2 .jxj $ 2/
2
p
x
2.1 " x 2 /
f 0 .x/ D 2 " x 2 " p
D p
2 " x2
2 " x2
p
SP
CP
CP
SP
p
p
1
"
2
f 0 " 2 " "1 C
""""""j""""""""""
j """""""""j""""""""""
j "!x
loc & abs % abs & loc
f max
max
min
min
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INSTRUCTOR’S SOLUTIONS MANUAL
y
SECTION 4.4 (PAGE 242)
.1;1/
30.
f .x/ D x " 2 tan!1 x, f 0 .x/ D 1 "
Critical points: x D ˙1.
f .x/ ! ˙1 as x ! ˙1.
p
! 2
p
2
x2 " 1
2
D 2
2
1Cx
x C1
x
CP
CP
f0
C
"1
"
C1
C
""""""""""""""""
j """"""""""""""j"""""""""""!x
loc
loc
f
%
&
%
max
min
p
y D x 2 " x2
.!1;!1/
Fig. 4.4-27
#
Hence, f has loc max "1 C
at x D "1 and loc min
2
#
1"
at x D 1.
2
28. f .x/ D x C sin x, f 0 .x/ D 1 C cos x # 0
f 0 .x/ D 0 at x D ˙#; ˙3#; :::
f .x/ ! ˙1 as x ! ˙1.
Hence, f has no max or min values.
y
y
.2!;2!/
!
!"
!1;!1C 2
.!;!/
x
x
y D x C sin x
!
1;1!
y D x " 2 tan!1 x
"
!
2
Fig. 4.4-30
Fig. 4.4-28
29.
f .x/ D x " 2 sin x
f 0 .x/ D 1 " 2 cos x
#
CP: x D ˙ C 2n#
3
n D 0; ˙1; ˙2; ! ! !
alternating local maxima and minima
31.
y
yDx
!
3
y D x " 2 sin x
x
f .x/ D 2x " sin!1 x
."1 $ x $ 1/
1
0
f .x/ D 2 " p
2
p 1"x
2 1 " x2 " 1
D
p
1 " x2
3 " 4x 2
D p
p
1 " x 2 .2 1 " x 2 C 1/
p
3
;
SP: (EP:) x D ˙1
CP: x D ˙
2 !
p
%p
#&
3
3"
f ˙
D˙
2
3
SP
Fig. 4.4-29
CP
p
CP
p
SP
3
"
1
f 0 "1 " " 23 C
2
""""""j""""""""""
j """""""""j""""""""""
j "!x
loc & abs % abs & loc
f max
max
min
min
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SECTION 4.4 (PAGE 242)
ADAMS and ESSEX: CALCULUS 9
y
34.
p
3
!
2
!1
p 1
3
2
!1
y D 2x " sin
2
2
f .x/ D x 2 e !x , f 0 .x/ D 2xe !x .1 " x 2 /
Critical points: x D 0; ˙1.
f .x/ ! 0 as x ! ˙1.
CP
CP
CP
f0 C
"1
"
0
C
1
"
""""""""""""
j """""""""""
j """""""""""
j"""""""!x
abs
abs
abs
f
%
max &
max &
min %
x
x
Hence, f has abs max 1=e at x D ˙1 and abs min 0 at
x D 0.
y
.!1;1=e/
Fig. 4.4-31
2
32. f .x/ D e !x =2 , f 0 .x/ D "xe !x
Critical point: x D 0.
f .x/ ! 0 as x ! ˙1.
y D x 2 e !x
2 =2
35.
Hence,f has abs max 1 at x D 0 and no min value.
y
y D e !x
2 =2
x
Fig. 4.4-32
!x
f .x/ D x2
f 0 .x/ D 2!x C x."2!x ln 2/
D 2!x .1 " x ln 2/
y
ln x
.x > 0/
x
x
" ln x
1 " ln x
f 0 .x/ D x 2
D
x
x2
f .x/ ! "1 as x ! 0C (vertical asymptote),
f .x/ ! 0 as x ! 1 (horizontal asymptote).
ASY
CP
f0 0
C
e
"
""""""
j"""""""""
j"""""!x
abs &
f
% max
% &
1
e; e
x
CP
f 0 C 1= ln 2 "
""""""""""""j""""""""!x
abs
f %
&
max
%
&
yD
ln x
x
Fig. 4.4-35
1
1
ln 2 ; e ln 2
36.
Since f .x/ D jx C 1j,
x
y Dx2
!x
Fig. 4.4-33
126
x
f .x/ D
y
33.
2
Fig. 4.4-34
CP
f0
C
0
"
""""""""""""j""""""""!x
abs
f
%
max &
1
.1;1=e/
f 0 .x/ D sgn .x C 1/ D
'
1;
"1;
if x > "1;
if x < "1.
"1 is a singular point; f has no max but has abs min 0 at
x D "1.
f .x/ ! 1 as x ! ˙1.
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 4.4 (PAGE 242)
y
y
1
!!
!
x
y D jx C 1j
y D sin jxj
!1
Fig. 4.4-38
x
39.
Fig. 4.4-36
37.
f .x/ D jx 2 " 1j
f .x/ D j sin xj
.2n C 1/#
CP: x D ˙
, SP D ˙n#
2
f has abs max 1 at all CP.
f has abs min 0 at all SP.
f 0 .x/ D 2xsgn .x 2 " 1/
CP: x D 0
SP: x D ˙1
y D j sin xj
y
!!
SP
CP
SP
f
" "1 C
0
"
1
C
"""""""""""
j """""""""j""""""""""j""""""!x
abs % loc & abs %
f & min
max
min
0
!
x
2!
Fig. 4.4-39
40.
y
f .x/ D .x " 1/2=3 " .x C 1/2=3
f 0 .x/ D 32 .x " 1/!1=3 " 32 .x C 1/!1=3
Singular point at x D ˙1. For critical points:
.x " 1/!1=3 D .x C 1/!1=3 ) x " 1 D x C 1 ) 2 D 0, so
there are no critical points.
SP
SP
f0
C
"1
"
C1
C
""""""""""""""
j"""""""""""""
j"""""""""!x
abs
abs
f
%
&
%
max
min
y D jx 2 " 1j
1
Hence, f has abs max 22=3 at x D "1 and abs min "22=3
at x D 1.
.!1;22=3 /
!1
1
y
y D .x " 1/2=3 " .x C 1/2=3
x
x
.1;!22=3 /
Fig. 4.4-37
Fig. 4.4-40
38. f .x/ D sin jxj
3#
5#
#
f 0 .x/ D sgn .x/ cos jxj D 0 at x D ˙ ; ˙ ; ˙ ; :::
2
2
2
0 is a singular point. Since f .x/ is an even function, its
graph is symmetric about the origin.
CP
CP
SP
CP
CP
3#
#
#
3#
f " "
C "
"
0
C
"
C
2
2
2
2
""""""""""j""""""""""
j """""""""
j """""""""
j """""""""
j """""!x
abs % abs & loc % abs & abs %
f & min
max
max
min
min
p
41. f .x/ D x= x 2 C 1. Since
p
f 0 .x/ D
0
#
Hence, f has abs max 1 at x D ˙.4k C 1/ and abs min
2
#
"1 at x D ˙.4k C 3/ where k D 0, 1, 2, : : : and loc
2
min 0 at x D 0.
42.
2x
x2 C 1 " x p
1
2 x2 C 1
D 2
> 0;
x2 C 1
.x C 1/3=2
for all x, f cannot have any maximum or minimum value.
p
f .x/ D x= x 4 C 1. f is continuous on R, and
limx!˙1 f .x/ D 0. Since f .1/ > 0 and f ."1/ < 0,
f must have both maximum and minimum values.
p
f 0 .x/ D
4x 3
x4 C 1 " x p
1 " x4
2 x4 C 1
D 4
:
4
x C1
.x C 1/3=2
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SECTION 4.4 (PAGE 242)
ADAMS and ESSEX: CALCULUS 9
p
p
CP x D ˙1. f .˙1/
p D ˙1= 2. f has max value 1= 2
and min value "1= 2.
%
&
y
1
1; p
2
%
1
!1;! p
2
&
yD p
x
x4
48.
49.
x
C1
Fig. 4.4-42
p
43. f .x/ D x 4 " x 2 is continuous on Œ"2; 2!, and
f .˙2/ D 0.
f 0 .x/ D
p
2.2 " x 2 /
"2x
D p
:
4 " x2 C x p
2 4 " x2
4 " x2
p
p
maximum value 2
CP x Dp˙ 2. f .˙ 2/ D ˙2. f hasp
at x D 2 and min value "2 at x D " 2.
p
44. f .x/ D x 2 = 4 " x 2 is continuous on ."2; 2/, and
limx!!2C f .x/ D limx!2! f .x/ D 1. Thus f can
have no maximum value, but will have a minimum value.
p
"2x
2x 4 " x 2 " x 2 p
3
2 4 " x 2 D 8x " x :
f 0 .x/ D
4 " x2
.4 " x 2 /3=2
p
p
CP x D 0, x D ˙ 8. f .0/ D 0, and ˙ 8 is not in the
domain of f . f has minimum value 0 at x D 0.
45. f .x/ D 1=Œx sin x! is continuous on .0; #/, and
limx!0C f .x/ D 1 D limx!!! f .x/. Thus f can
have no maximum value, but will have a minimum value.
Since f is differentiable on .0; #/, the minimum value
must occur at a CP in that interval.
46. f .x/ D .sin x/=x is continuous and differentiable on R
except at x D 0 where it is undefined.
Since limx!0 f .x/ D 1 (Theorem 8 of Section 2.5), and
jf .x/j < 1 for all x ¤ 0 (because j sin xj < jxj), f cannot
have a maximum value.
Since limx!˙1 f .x/ D 0 and since f .x/ < 0 at some
points, f must have a minimum value occurring at a critical point. In fact, since jf .x/j $ 1=jxj for x ¤ 0 and f
is even, the minimum value will occur at the two critical
points closest to x D 0. (See Figure 2.20 In Section 2.5 of
the text.)
47.
If it exists, an absolute max value is the maximum of the
set of all the local max values. Hence, if a function has
an absolute max value, it must have one or more local
max values. On the other hand, if a function has a local max value, it may or may not have an absolute max
value. Since a local max value, say f .x0 / at the point
x0 , is defined such that it is the max within some interval
jx " x0 j < h where h > 0, the function may have greater
values, and may even approach 1 outside this interval.
There is no absolute max value in this latter case.
128
No. f .x/ D "x 2 has abs max value 0, but
g.x/ D jf .x/j D x 2 has no abs max value.
(
1
if x > 0
f .x/ D x sin x
0
if x < 0
jf .x/j $ jxj if x > 0 so limx!0C f .x/ D 0 D f .0/.
1
Therefore f is continuous at x D 0. Clearly x sin is
x
continuous at x > 0. Therefore f is continuous on Œ0; 1/.
Given any h > 0 there exists x1 in .0; h/ and x2 in .0; h/
such that f .x1 / > 0 D f .0/ and f .x2 / < 0 D f .0/.
Therefore f cannot be a local max or min value at 0.
1
Specifically, let positive integer n satisfy 2n# >
h
1
1
.
and let x1 D
,
x
D
2
#
3#
2n# C
2n# C
2
2
Then f .x1 / D x1 > 0 and f .x2 / < 0.
Section 4.5 Concavity and Inflections
(page 246)
1.
2.
3.
4.
p
1
1
x, f 0 .x/ D p , f 00 .x/ D " x !3=2
4
2 x
f 00 .x/ < 0 for all x > 0. f is concave down on .0; 1/.
f .x/ D
f .x/ D 2x " x 2 , f 0 .x/ D 2 " 2x, f 00 .x/ D "2 < 0.
Thus, f is concave down on ."1; 1/.
f .x/ D x 2 C 2x C 3, f 0 .x/ D 2x C 2, f 00 .x/ D 2 > 0.
f is concave up on ."1; 1/.
f .x/ D x " x 3 , f 0 .x/ D 1 " 3x 2 ,
f 00 .x/ D "6x.
f 00
C
0
"
""""""""""""""j"""""""""!x
f
^
infl
_
5.
f .x/ D 10x 3 " 3x 5 ;
f 0 .x/ D 30x 2 " 15x 4 ;
f 00 .x/ D 60.x " x 3 / D 60x.1 " x/.1 C x/:
f 00 C "1 "
0
C
1
"
"""""""""""j""""""""""j""""""""""
j """"""!x
f
^ infl _ infl ^ infl _
6.
f .x/ D 10x 3 C 3x 5 , f 0 .x/ D 30x 2 C 15x 4 ,
f 00 .x/ D 60x C 60x 3 D 60x.1 C x 2 /.
f 00
"
0
C
""""""""""""""j"""""""""!x
f
_
infl
^
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INSTRUCTOR’S SOLUTIONS MANUAL
7.
f .x/ D .3 " x 2 /2 ;
0
SECTION 4.5 (PAGE 246)
13.
2
3
f .x/ D "4x.3 " x / D "12x C 4x ;
f 00 .x/ D "12 C 12x 2 D 12.x " 1/.x C 1/:
$
#
.2n " 1/#
; n# , and conf is concave up on intervals
2 $
#
.2n C 1/#
n#
cave down on intervals n#;
. Points
are
2
2
inflection points.
f 00
C
"1
"
1
C
"""""""""""""j""""""""""""
j """"""""!x
f
^
infl
_
infl
^
14.
8.
f .x/ D .2 C 2x " x 2 /2 ;
00
2
f 0 .x/ D 2.2 C 2x " x 2 /.2 " 2x/;
f .x/ D 2.2 " 2x/ C 2.2 C 2x " x 2 /."2/
D 12x.x " 2/:
15.
00
f
C
0
"
2
C
"""""""""""""j""""""""""""
j """"""""!x
f
^
infl
_
infl
^
9.
f .x/ D x C sin 2x;
f 0 .x/ D 1 C 2 cos 2x;
f 00 .x/ D "4 sin 2x:
f .x/ D x " 2 sin x, f 0 .x/ D 1 " 2 cos x, f 00 .x/ D 2 sin x.
Inflection points: x D n#
% for n D 0; ˙1; ˙2;
& :::.
f is concave down on .2nC1/#; .2nC2/# and concave
%
&
up on .2n/#; .2n C 1/# .
f .x/ D tan!1 x; f 0 .x/ D
f 00 .x/ D
"2x
.
.1 C x 2 /2
f 00
C
0
"
""""""""""""""j"""""""""!x
f
^
infl
_
f .x/ D .x 2 " 4/3 ;
f 0 .x/ D 6x.x 2 " 4/2 ;
f 00 .x/ D 6.x 2 " 4/2 C 24x 2 .x 2 " 4/
16.
D 6.x 2 " 4/.5x 2 " 4/:
f .x/ D xe x , f 0 .x/ D e x .1 C x/,
f 00 .x/ D e x .2 C x/.
f 00
"
"2
C
""""""""""""""j"""""""""!x
f
_
infl
^
f 00 C "2 " " p2 C p2 " 2 C
5
5
""""""""""j"""""""""
j"""""""""
j""""""""j"""""!x
f ^ infl _ infl ^ infl _ infl ^
17.
2
x
3"x
,
, f 0 .x/ D 2
x2 C 3
.x C 3/2
2
2x.x " 9/
f 00 .x/ D
.
.x 2 C 3/3
10. f .x/ D
2
2
f .x/ D e !x ; f 0 .x/ D "2xe !x ;
2
f 00 .x/ D e !x .4x 2 " 2/.
f 00 C " p1
" p1
C
2
2
""""""""""""
j """"""""""
j""""""!x
f
^
infl _ infl ^
f 00 " "3 C
0
"
3
C
"""""""""""j""""""""""
j """""""""
j""""""!x
f
_ infl ^ infl _ infl ^
11.
1
;
1 C x2
f .x/ D sin x; f 0 .x/ D cos x; f 00 .x/ D " sin x.
f is concave down on intervals .2n#; .2n C 1/#/ and
concave up on intervals ..2n " 1/#; 2n#/, where n ranges
over the integers. Points x D n# are inflection points.
12. f .x/ D cos 3x, f 0 .x/ D "3 sin 3x, f 00 .x/ D "9 cos 3x.
!
"#
Inflection points: x D n C 12
for n D 0; ˙1; ˙2; :::.
3
#
$
4n C 1 4n C 3
f is concave up on
#;
# and concave
6
$ 6
#
4n C 3 4n C 5
#;
# .
down on
6
6
2 " ln.x 2 /
ln.x 2 /
, f 0 .x/ D
,
x
x2
2
"6
C
2
ln.x
/
f 00 .x/ D
.
x3
f has inflection point at x D ˙e 3=2 and f is undefined at
x D 0. f is concave up on ."e 3=2 ; 0/ and .e 3=2 ; 1/; and
concave down on ."1; "e 3=2 / and .0; e 3=2 /.
18. f .x/ D
19.
2x
;
1 C x2
2.1 " x 2 /
.1 C x 2 /.2/ " 2x.2x/
D
:
f 00 .x/ D
2
2
.1 C x /
.1 C x 2 /2
f .x/ D ln.1 C x 2 /;
f 0 .x/ D
f 00 " "1 C
1
"
""""""""""""
j """""""""j"""""""!x
f
_ infl ^ infl _
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SECTION 4.5 (PAGE 246)
20. f .x/ D .ln x/2 , f 0 .x/ D
f 00 .x/ D
ADAMS and ESSEX: CALCULUS 9
2
ln x,
x
2.1 " ln x/
for all x > 0.
x2
f 00
0
C
e
"
"""""""
j """""""""""
j """"""""!x
f
^
infl
_
21.
2
f 00 .x/ D 6x C 3 :
x
$
$
#
#
"1
1
00
>
0;
f
< 0:
p
f 00 p
4
4
3
3
1
"1
Therefore f has a loc min at p
and a loc max at p
.
4
4
3
3
25
x3
" 4x 2 C 12x " ;
3
3
f 0 .x/ D x 2 " 8x C 12;
f 00 .x/ D 2x " 8 D 2.x " 4/:
28.
f .x/ D
29.
f 00
"
4
C
""""""""""""""j"""""""""!x
f
_
infl
^
22. f .x/ D .x " 1/1=3 C .x C 1/1=3 ,
f 0 .x/ D 31 Œ.x " 1/!2=3 C .x C 1/!2=3 !,
f 00 .x/ D " 29 Œ.x " 1/!5=3 C .x C 1/!5=3 !.
f .x/ D 0 , x " 1 D ".x C 1/ , x D 0.
Thus, f has inflection point at x D 0. f 00 .x/ is undefined
at x D ˙1. f is defined at ˙1 and x D ˙1 are also inflection points. f is concave up on ."1; "1/ and .0; 1/;
and down on ."1; 0/ and .1; 1/.
23. According to Definition 4.3.1 and the subsequent discussion, f .x/ D ax C b has no concavity and therefore no
inflections.
24.
25.
f .x/ D 3x 3 " 36x " 3, f 0 .x/ D 9.x 2 " 4/, f 00 .x/ D 18x.
The critical points are
x D 2; f 00 .2/ > 0 ) local min;
x D "2; f 00 ."2/ < 0 ) local max:
f .x/ D x.x " 2/2 C 1 D x 3 " 4x 2 C 4x C 1
f 0 .x/ D 3x 2 " 8x C 4 D .x " 2/.3x " 2/
2
CP: x D 2, x D
3
# $
00
00
00 2
D "4 < 0.
f .x/ D 6x " 8; f .2/ D 4 > 0; f
3
Therefore, f has a loc min at x D 2 and a loc max at
2
xD .
3
4
4
26. f .x/ D x C , f 0 .x/ D 1 " 2 , f 00 .x/ D 8x !3 .
x
x
The critical points are
x D 2; f 00 .2/ > 0 ) local min;
x D "2; f 00 ."2/ < 0 ) local max:
27.
f .x/ D x 3 C
1
x
f 0 .x/ D 3x 2 "
130
1
3x 4 " 1
D
;
2
x
x2
1
CP: x D ˙ p
:
4
3
30.
31.
x
1 " x ln 2
, f 0 .x/ D
,
x
2
2x
ln
2.x
ln
2
"
2/
f 00 .x/ D
.
2x
The critical point
is
#
$
1
1
; f 00
< 0 ) local max:
xD
ln 2
ln 2
x
f .x/ D
1 C x2
.1 C x 2 / " x2x
1 " x2
f 0 .x/ D
D
.1 C x 2 /2
.1 C x 2 /2
CP: x D ˙1
.1 C x/2 ."2x/ " .1 " x 2 /2.1 C x 2 /2x
f 00 .x/ D
.1 C x 2 /4
3
"2x " 2x " 4x C 4x 3
"6x C 2x 3
D
D
.1 C x 2 /3
.1 C x 2 /3
1
1
00
00
f .1/ D " , f ."1/ D .
2
2
f has a loc max at 1 and a loc min at "1.
f .x/ D
f .x/ D xe x , f 0 .x/ D e x .1 C x/, f 00 .x/ D e x .2 C x/.
The critical point is x D "1.
f 00 ."1/ > 0; ) local min:
f .x/ D x ln x;
f 0 .x/ D 1 C ln x;
CP: x D
1
e
1
1
;
f 00 . / D e > 0:
x
e
1
f has a loc min at .
e
f 00 .x/ D
32.
33.
f .x/ D .x 2 "4/2 , f 0 .x/ D 4x 3 "16x, f 00 .x/ D 12x 2 "16.
The critical points are
x D 0; f 00 .0/ < 0 ) local max;
x D 2; f 00 .2/ > 0 ) local min;
x D "2; f 00 ."2/ > 0 ) local min:
f .x/ D .x 2 " 4/3
f 0 .x/ D 6x.x 2 " 4/2
CP: x D 0, x D ˙2
f 00 .x/ D 6.x 2 " 4/2 C 24x 2 .x 2 " 4/
D 6.x 2 " 4/.5x 2 " 4/
f 00 .0/ > 0; f 00 .˙2/ D 0:
f has a loc min at x D 0. Second derivative test yields
no direct information about ˙2. However, since f 00 has
opposite signs on opposite sides of the points 2 and "2,
each of these points is an inflection point of f , and therefore f cannot have a local maximum or minimum value at
either.
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INSTRUCTOR’S SOLUTIONS MANUAL
34.
f .x/ D .x 2 " 3/e x ;
0
2
SECTION 4.5 (PAGE 246)
38.
x
x
f .x/ D .x C 2x " 3/e D .x C 3/.x " 1/e ;
f 00 .x/ D .x 2 C 4x " 1/e x :
The critical points are
x D "3; f 00 ."3/ < 0 ) local max;
x D 1; f 00 .1/ > 0 ) local min.
35.
Suppose that f has an inflection point at x0 . To be
specific, suppose that f 00 .x/ < 0 on .a; x0 / and
f 00 .x/ > 0 on .x0 ; b/ for some numbers a and b satisfying a < x0 < b.
If the graph of f has a non-vertical tangent line at x0 ,
then f 0 .x0 / exists. Let
F .x/ D f .x/ " f .x0 / " f 0 .x0 /.x " x0 /:
2 !2x 2
f .x/ D x e
2
F .x/ represents the signed vertical distance between the
graph of f and its tangent line at x0 . To show that the
graph of f crosses its tangent line at x0 , it is sufficient
to show that F .x/ has opposite signs on opposite sides of
x0 .
Observe that F .x0 / D 0, and F 0 .x/ D f 0 .x/ " f 0 .x0 /,
so that F 0 .x0 / D 0 also. Since F 00 .x/ D f 00 .x/, the
assumptions above show that F 0 has a local minimum
value at x0 (by the First Derivative Test). Hence F .x/ > 0
if a < x < x0 or x0 < x < b. It follows (by Theorem
6) that F .x/ < 0 if a < x < x0 , and F .x/ > 0 if
x0 < x < b. This completes the proof for the case of a
nonvertical tangent.
If f has a vertical tangent at x0 , then its graph necessarily
crosses the tangent (the line x D x0 ) at x0 , since the
graph of a function must cross any vertical line through a
point of its domain that is not an endpoint.
2
f 0 .x/ D e !2x .2x " 4x 3 / D 2.x " 2x 3 /e !2x
1
CP: x D 0, x D ˙ p
2
2
f 00 .x/ D e !2x .2 " 20x 2 C 16x 4 /
#
$
1
4
f 00 .0/ > 0; f 00 ˙ p
D " < 0:
e
2
Therefore, f has a loc (and abs) min value at 0, and loc
1
(and abs) max values at ˙ p .
2
36. Since
f .x/ D
'
x2
"x 2
if x # 0
if x < 0,
we have
n
2x
if x # 0
D 2jxj
n"2x if x < 0
2
if x > 0
f 00 .x/ D
D 2sgn x:
"2 if x < 0
f 0 .x/ D
39.
fn0 .x/
37. Suppose f is concave up (i.e., f 00 .x/ > 0) on an open
interval containing x0 .
Let h.x/ D f .x/ " f .x0 / " f 0 .x0 /.x " x0 /.
Since h0 .x/ D f 0 .x/ " f 0 .x0 / D 0 at x D x0 , x D x0 is a
CP of h.
Now h00 .x/ D f 00 .x/. Since h00 .x0 / > 0, therefore h has a
min value at x0 , so h.x/ # h.x0 / D 0 for x near x0 .
Since h.x/ measures the distance y D f .x/ lies above the
tangent line y D f .x0 / C f 0 .x0 /.x " x0 / at x, therefore
y D f .x/ lies above that tangent line near x0 .
Note: we must have h.x/ > 0 for x near x0 , x ¤ x0 , for
otherwise there would exist x1 ¤ x0 , x1 near x0 , such that
h.x1 / D 0 D h.x0 /. If x1 > x0 , there would therefore
exist x2 such that x0 < x2 < x1 and f 0 .x2 / D f 0 .x0 /.
Therefore there would exist x3 such that x0 < x3 < x2 and
f 0 .x3 / D 0, a contradiction.
The same contradiction can be obtained if x1 < x0 .
n!1
n D 2; 3; 4; : : :
D nx
D 0 at x D 0
If n is even, fn has a loc min, gn has a loc max at x D 0.
If n is odd, fn has an inflection at x D 0, and so does gn .
0
f .x/ D 0 if x D 0. Thus, x D 0 is a critical point
of f . It is also an inflection point since the conditions of
Definition 3 are satisfied. f 00 .0/ does not exist. If a the
graph of a function has a tangent line, vertical or not, at
x0 , and has opposite concavity on opposite sides of x0 ,
the x0 is an inflection point of f , whether or not f 00 .x0 /
even exists.
f .x/ D x n
g.x/ D "x n D "f .x/;
40.
Let there be a function f such that
f 0 .x0 / D f 00 .x0 / D ::: D f .k!1/ .x0 / D 0;
f .k/ .x0 / ¤ 0
for some k # 2:
If k is even, then f has a local min value at x D x0 when
f .k/ .x0 / > 0, and f has a local max value at x D x0
when f .k/ .x0 / < 0.
If k is odd, then f has an inflection point at x D x0 .
'
!1=x 2
if x ¤ 0
41. f .x/ D e
0
if x D 0
a)
e !1=x
x!0C
xn
lim x !n f .x/ D lim
x!0C
D lim y n e
y!1
!y 2
2
.put y D 1=x/
D 0 by Theorem 5 of Sec. 4.4
Similarly, limx!0! x !n f .x/ D 0, and
limx!0 x !n f .x/ D 0.
P
b) If P .x/ D jnD0 aj x j then by (a)
lim P
x!0
# $
n
X
1
f .x/ D
aj lim x !j f .x/ D 0:
x!0
x
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SECTION 4.5 (PAGE 246)
ADAMS and ESSEX: CALCULUS 9
c) If x ¤ 0 and P1 .t / D 2t 3 , then
f 0 .x/ D
2 !1=x 2
e
D P1
x3
If x D 0, then
# $
1
f .x/:
x
f .x/ D lim
# $
1
Assume that f .k/ .x/ D Pk
f .x/ for some
x
k # 1, where Pk is a polynomial. Then
1
"0
h
D 0:
h
h2 sin
0
h!0
Thus 0 is a critical point of f . There are points x arbitrarily close to 0 where f .x/ > 0, for example
2
x D
, and other such points where f .x/ < 0,
.4n C 1/#
2
for example x D
. Therefore f does not have
.4n C 3/#
a local max or min at x D 0. Also, there are points
arbitrarily close to 0 where f 00 .x/ > 0, for example
1
, and other such points where f 00 .x/ < 0,
xD
.2n C 1/#
1
for instance x D
. Therefore f does not have con2n#
stant concavity on any interval .0; a/ where a > 0, so 0 is
not an inflection point of f either.
# $
# $ # $
1
1 0 1
1
P
f
.x/
C
P
P1
f .x/
k
x2 k x
x
x
# $
1
D PkC1
f .x/;
x
f .kC1/ .x/ D "
where PkC1 .t / D t 2 Pk0 .t / C P1 .t /Pk .t / is a polynomial.
# $
1
.n/
f .x/ for n ¤ 0, where
By induction, f
D Pn
n
Pn is a polynomial.
f .h/ " f .0/
D lim h!1 f .h/ D 0 by
h
h!0
(a). Suppose that f .k/ .0/ D 0 for some k # 1. Then
d) f 0 .0/ D limh!0
f .k/ .h/ " f .k/ .0/
h
h!0
D lim h!1 f .k/ .h/
h!0
# $
1
f .h/ D 0
D lim h!1 Pk
h
h!0
Section 4.6 Sketching the Graph of
a Function (page 255)
f .kC1/ .0/ D lim
1.
by (b).
Thus f .n/ .0/ D 0 for n D 1; 2; : : : by induction.
e) Since f 0 .x/ < 0 if x < 0 and f 0 .x/ > 0 if x > 0,
therefore f has a local min value at 0 and "f has a
loc max value there.
Function (d) appears to be the derivative of function (c),
and function (b) appears to be the derivative of function
(d). Thus graph (c) is the graph of f , (d) is the graph of
f 0 , (b) is the graph of f 00 , and (a) must be the graph of
the other function g.
y
y
(a)
(b)
4
3
2
1
f) If g.x/ D xf .x/ then g 0 .x/ D f .x/ C xf 0 .x/,
g 00 .x/ D 2f 0 .x/ C xf 00 .x/.
In general, g .n/ .x/ D nf .n!1/ .x/ C xf .n/ .x/ (by
induction).
Then g .n/ .0/ D 0 for all n (by (d)).
Since g.x/ < 0 if x < 0 and g.x/ > 0 if x > 0, g
cannot have a max or min value at 0. It must have an
inflection point there.
!5 !4 !3 !2 !1
!1
!2
!3
!4
!5
(c)
42. We are given that
f .x/ D
(
1
x 2 sin ; if x ¤ 0;
x
0;
if x D 0.
1 2 3
x
!5 !4 !3 !2 !1
!1
!2
!3
!4
!5
(d)
4
3
2
1
1 2 3
!2
!3
!4
!5
f 0 .x/ D 2x sin
132
4
y
!5 !4 !3 !2 !1
!1
If x ¤ 0, then
1
1
" cos
x
x
2
1
1
1
1
f 00 .x/ D 2 sin " cos " 2 sin :
x x
x x
x
4
3
2
1
2.
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4
x
3 4
x
1 2
3 4
x
y
4
3
2
1
!5 !4 !3 !2 !1
!1
!2
!3
!4
!5
Fig. 4.6-1
1 2
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INSTRUCTOR’S SOLUTIONS MANUAL
(a)
y
!5 !4 !3 !2 !1
!1
!2
!3
!4
!5
(c)
(b)
4
3
2
1
1 2 3
4
x
y
!5 !4 !3 !2 !1
!1
!2
!3
!4
!5
1 2 3
4
x
y
has inflections at 0 and ˙1:7 (approximately).
4
3
2
1
!5 !4 !3 !2 !1
!1
!2
!3
!4
!5
(d)
4
3
2
1
SECTION 4.6 (PAGE 255)
3.
1 2
3 4
x
1 2
3 4
x
y
4
3
2
1
!5 !4 !3 !2 !1
!1
!2
!3
!4
!5
f .x/ D x=.1 " x 2 / has slope 1 at the origin, so its graph
must be (c).
g.x/ D x 3 =.1 " x 4 / has slope 0 at the origin, but has the
same sign at all points as does f .x/, so its graph must be
(b).
p
h.x/ D .x 3 " x/= 1 C x 6 has no vertical asymptotes, so
its graph must
p be (d).
k.x/ D x 3 = jx 4 " 1j is positive for all positive x ¤ 1, so
its graph must be (a).
4.
(a)
Fig. 4.6-2
(b)
y
3
2
2
1
The function graphed in Fig. 4.2(a):
is odd, is asymptotic to y D 0 at ˙1,
is increasing on ."1; "1/ and .1; 1/,
is decreasing on ."1; 1/,
has CPs at x D "1 (max) and 1 (min),
is concave up on ."1; "2/ and .0; 2/ (approximately),
is concave down on ."2; 0/ and .2; 1/ (approximately),
has inflections at x D ˙2 (approximately).
The function graphed in Fig. 4.2(b):
is even, is asymptotic to y D 0 at ˙1,
is increasing on ."1:7; 0/ and .1:7; 1/ (approximately),
is decreasing on ."1; "1:7/ and .0; 1:7/ (approximately),
has CPs at x D 0 (max) and ˙1:7 (min) (approximately),
is concave up on ."2:5; "1/ and .1; 2:5/ (approximately),
is concave down on ."1; "2:5/, ."1; 1/, and .2:5; 1/
(approximately),
has inflections at ˙2:5 and ˙1 (approximately).
The function graphed in Fig. 4.2(c):
is even, is asymptotic to y D 2 at ˙1,
is increasing on .0; 1/,
is decreasing on ."1; 0/,
has a CP at x D 0 (min),
is concave up on ."1; 1/ (approximately),
is concave down on ."1; "1/ and .1; 1/ (approximately),
has inflections at x D ˙1 (approximately).
The function graphed in Fig. 4.2(d):
is odd, is asymptotic to y D 0 at ˙1,
is increasing on ."1; 1/,
is decreasing on ."1; "1/ and .1; 1/,
has CPs at x D "1 (min) and 1 (max),
is concave down on ."1; "1:7/ and .0; 1:7/ (approximately),
is concave up on ."1:7; 0/ and .1:7; 1/ (approximately),
!5 !4 !3 !2 !1
!1
1
1 2 3
4
x
!2
!5 !4 !3 !2 !1
!1
3 4
x
1 2
3 4
x
!3
!4
!4
y
(d)
3
y
3
2
2
1
1
!5 !4 !3 !2 !1
!1
1 2
!2
!3
(c)
y
3
1 2 3
!2
4
x
!5 !4 !3 !2 !1
!1
!2
!3
!4
!3
Fig. 4.6-4
!4
The function graphed in Fig. 4.4(a):
is odd, is asymptotic to x D ˙1 and y D x,
is increasing on ."1; "1:5/, ."1; 1/, and .1:5; 1/ (approximately),
is decreasing on ."1:5; "1/ and .1; 1:5/ (approximately),
has CPs at x D "1:5, x D 0, and x D 1:5,
is concave up on .0; 1/ and .1; 1/,
is concave down on ."1; "1/ and ."1; 0/,
has an inflection at x D 0.
The function graphed in Fig. 4.4(b):
is odd, is asymptotic to x D ˙1 and y D 0,
is increasing on ."1; "1/, ."1; 1/, and .1; 1/,
has a CP at x D 0,
is concave up on ."1; "1/ and .0; 1/,
is concave down on ."1; 0/ and .1; 1/,
has an inflection at x D 0.
The function graphed in Fig. 4.4(c):
is odd, is asymptotic to x D ˙1 and y D 0,
is increasing on ."1; "1/, ."1; 1/, and .1; 1/,
has no CP,
is concave up on ."1; "1/ and .0; 1/,
is concave down on ."1; 0/ and .1; 1/,
has an inflection at x D 0.
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SECTION 4.6 (PAGE 255)
ADAMS and ESSEX: CALCULUS 9
y
The function graphed in Fig. 4.4(d):
is odd, is asymptotic to y D ˙2,
is increasing on ."1; "0:7/ and .0:7; 1/ (approximately),
is decreasing on ."0:7; 0:7/ (approximately),
has CPs at x D ˙0:7 (approximately),
is concave up on ."1; "1/ and .0; 1/ (approximately),
is concave down on ."1; 0/ and .1; 1/ (approximately),
has an inflection at x D 0 and x D ˙1 (approximately).
y D f .x/
2
.1;1/
.3;1/
!1
2
x
5. f .0/ D 1 f .˙1/ D 0 f .2/ D 1
limx!1 f .x/ D 2, limx!!1 f .x/ D "1
SP
CP
f0
C
0
"
1
C
"""""""""""""j""""""""""""
j""""""""!x
loc
loc
f
%
&
%
max
min
yDx!1
Fig. 4.6-6
f 00
C
0
C
2
"
"""""""""""""j"""""""""""j"""""""""!x
f
^
^
infl
_
0 must be a SP because f 00 > 0 on both sides and it is a
loc max. 1 must be a CP because f 00 is defined there so
f 0 must be too.
y
yD2
y D f .x/
1
.2;1/
!1
7.
x
1
y D .x 2 " 1/3
y 0 D 6x.x 2 " 1/2
yD!1
00
D 6x.x " 1/2 .x C 1/2
y D 6Œ.x 2 " 1/2 C 4x 2 .x 2 " 1/!
Fig. 4.6-5
6. According to the given properties:
Oblique asymptote: y D x " 1.
Critical points: x D 0; 2. Singular point: x D "1.
Local max 2 at x D 0; local min 0 at
x D 2.
D 6.x 2 " 1/.5x 2 " 1/
p
p
D 6.x " 1/.x C 1/. 5x " 1/. 5x C 1/
From y: Asymptotes: none. Symmetry: even. Intercepts:
x D ˙1.
From y 0 : CP: x D 0, x D ˙1. SP: none.
SP
CP
CP
f 0 C "1 C
0
"
2
C
"""""""""""j""""""""""j"""""""""""j"""""""!x
loc
loc
f
%
%
max &
min %
Inflection points: x D "1; 1; 3.
f0
C
"1
"
1
C
3
"
""""""""""""
j """"""""""
j """"""""""j"""""""!x
f
^
infl _
infl ^
infl _
Since
lim
x!˙1
%
&
f .x/ C 1 " x D 0, the line y D x " 1 is an
oblique asymptote.
134
CP
CP
CP
y 0 " "1 " 0 C 1 C
""""""""j""""""""j""""""""
j""""!x
abs
y &
& min %
%
1
From y 00 : y 00 D 0 at x D ˙1, x D ˙ p .
5
y 00 C "1 " " p1 C p1 " 1 C
5
5
"""""""""j"""""""""
j """""""""
j """""""j"""""!x
y ^ infl _ infl ^ infl _ infl ^
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 4.6 (PAGE 255)
y
9.
y D .x 2 " 1/3
p
!1= 5
2"x
2
2
4
yD
D " 1; y 0 D " 2 ; y 00 D 3 :
x
x
x
x
From y: Asymptotes: x D 0, y D "1.
Symmetry: none obvious.
Intercept: .2; 0/. Points: ."1; "3/.
From y 0 : CP: none. SP: none.
ASY
y0
"
0
"
"""""""""""""
j """"""""!x
y
&
&
p
1= 5
!1
x
1
From y 00 : y 00 D 0 nowhere.
!1
Fig. 4.6-7
ASY
y 00
"
0
C
"""""""""""""
j """"""""!x
y
_
^
8. y D x.x 2 " 1/2 , y 0 D .x 2 " 1/.5x 2 " 1/, y 00 D 4x.5x 2 " 3/.
From y: Intercepts: .0; 0/, .1; 0/. Symmetry: odd (i.e.,
about the origin).
1
From y 0 : Critical point: x D ˙1; ˙ p .
5
y
CP
CP
CP
1
1
C p
"
1
C
y C "1 " " p
5
5
""""""""""
j """"""""""
j """""""""j""""""""""
j """""!x
loc &
loc
loc
loc
y % max
min % max & min %
yD
2"x
x
CP
0
.2;0/
x
!1
From y 00 : q
Inflection points at
x D 0; ˙
3
.
5
.!1;!3/
q
q
3
y 00 " " 35 C
0
"
C
5
""""""""""""j"""""""""""j""""""""""j"""""""!x
y _
infl
^ infl _
infl ^
Fig. 4.6-9
2
2
"4
x"1
D1"
, y0 D
, y 00 D
.
2
xC1
xC1
.x C 1/
.x C 1/3
From y: Intercepts: .0; "1/, .1; 0/. Asymptotes: y D 1
(horizontal), x D "1 (vertical). No obvious symmetry.
Other points: ."2; 3/.
From y 0 : No critical point.
10. y D
y
p1
5
q
3 1
5
ASY
y0
C
"1
C
""""""""""""""""j""""""""""""!x
y
%
%
x
From y 00 : No inflection point.
y D x.x 2 " 1/2
ASY
y 00
C
"1
"
""""""""""""""""j""""""""""""!x
y
^
_
Fig. 4.6-8
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135
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SECTION 4.6 (PAGE 255)
ADAMS and ESSEX: CALCULUS 9
y
y
yD
x"1
yD
xC1
%
3 27
!2; 4
.!2;3/
yD1
&
xD!1
x
1
xD!1
x3
1Cx
x
!1
Fig. 4.6-11
1
"2x
6x 2 " 8
, y0 D
, y 00 D
.
2
2
2
4Cx
.4 C x /
.4 C x 2 /3
From y: Intercept: .0; 14 /. Asymptotes: y D 0 (horizontal). Symmetry: even (about y-axis).
From y 0 : Critical point: x D 0.
12. y D
Fig. 4.6-10
11.
CP
y0
C
0
"
""""""""""""""""j""""""""""""!x
abs
y
%
&
max
2
From y 00 : y 00 D 0 at x D ˙ p .
3
x3
1Cx
.1 C x/3x 2 " x 3
3x 2 C 2x 3
y0 D
D
2
.1 C x/
.1 C x/2
2
2
.1 C x/ .6x C 6x / " .3x 2 C 2x 3 /2.1 C x/
y 00 D
.1 C x/4
2
2
6x.1 C x/ " 6x " 4x 3
6x C 6x 2 C 2x 3
D
D
3
.1 C x/
.1 C x/3
2
2x.3 C 3x C x /
D
.1 C x/3
From y:
Asymptotes: x D "1. Symmetry: none.
Intercepts .0; 0/. Points ."3=2; 27=4/.
3
From y 0 CP: x D 0, x D " .
2
yD
CP
ASY
CP
y 0 " " 32
C
"1
C
0
C
"""""""""""
j """"""""""
j """""""""j""""""!x
loc %
y & min
%
%
From y 00 : y 00 D 0 only at x D 0.
ASY
y
C
"1
"
0
C
""""""""""""""j"""""""""""""j"""""""""!x
y
^
_
infl
^
00
136
2
2
p
"p
"
C
3
3
"""""""""""""j""""""""""""j""""""""!x
y
^
infl
_
infl
^
y 00
C
y
1=4
yD
!2
p
3
1
4 C x2
p2
3
Fig. 4.6-12
13.
2x
1
; y0 D
2 " x2
.2 " x 2 /2
2
8x 2
4 C 6x 2
y 00 D
C
D
2
2
2
3
.2 " x /
.2 " x /
.2 "px 2 /3
From y: Asymptotes: y D 0, x D ˙ 2.
Symmetry: even.
Intercepts .0; 12 /. Points .˙2; " 21 /.
From y 0 : CP x D 0.
yD
ASY
CP
ASY
p
p
0
C
2 C
y 00 " " 2 "
"""""""""""
j """""""""j""""""""""j""""""!x
loc %
y &
& min
%
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x
lOMoARcPSD|6566483
INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 4.6 (PAGE 255)
y
y 00 W y 00 D 0 nowhere.
yD
ASY
ASY
p
p
y 00 " " 2 C
2
"
"""""""""""j"""""""""""
j """"""!x
y _
^
_
%
xD!1
2
2; 3
%
2
!2;! 3
y
p
xD! 2
x2
&
x
"1
&
x
xD1
p
xD 2
1=2
x
%
1
!2;! 2
%
&
1
2;! 2
yD
&
1
2 " x2
Fig. 4.6-13
Fig. 4.6-14
15.
x2
1
D1C 2
x2 " 1
x "1
"2x
y0 D 2
.x " 1/2
2.3x 2 C 1/
.x 2 " 1/2 " x2.x 2 " 1/2x
D
y 00 D "2
2
4
.x " 1/
.x 2 " 1/3
From y: Asymptotes: y #D 1, x $
D ˙1. Symmetry: even.
4
Intercepts .0; 0/. Points ˙2;
.
3
0
From y : CP x D 0.
yD
ASY
CP
ASY
y 0 C "1 C
0
"
1
"
""""""""""j""""""""""j""""""""""j""""""!x
loc &
y %
% max
&
14.
x
2x.x 2 C 3/
x2 C 1
, y 00 D
.
, y0 D " 2
2
"1
.x " 1/
.x 2 " 1/3
From y: Intercept: .0; 0/. Asymptotes: y D 0 (horizontal), x D ˙1 (vertical). Symmetry: odd. Other points:
.2; 23 /, ."2; " 23 /.
From y 0 : No critical or singular points.
yD
x2
From y 00 : y 00 D 0 nowhere.
ASY
ASY
y 00 C
"1
"
1
C
"""""""""""j"""""""""""
j """"""!x
y ^
_
^
y
ASY
ASY
y0
"
"1
"
1
"
""""""""""""""
j """""""""""""
j """""""""!x
y
&
&
&
%
4
!2; 3
&
yD
%
x2
"1
x2
4
2; 3
&
yD1
From y 00 : y 00 D 0 at x D 0.
x
xD!1
ASY
ASY
y 00 "
"1 C
0
"
1
C
"""""""""""
j"""""""""j""""""""""
j """"""!x
y _
^ infl _
^
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xD1
Fig. 4.6-15
137
lOMoARcPSD|6566483
SECTION 4.6 (PAGE 255)
ADAMS and ESSEX: CALCULUS 9
x 2 .x 2 " 3/ 00
2x.x 2 C 3/
x3
, y0 D
,y D
.
2
2
"1
.x " 1/
.x 2 " 1/3
From y: Intercept: .0; 0/. Asymptotes: x D ˙1 (vertical), y D p
x (oblique). Symmetry: odd. Other points:
$
#
p
3 3
.
˙ 3; ˙
2
p
From y 0 : Critical point: x D 0; ˙ 3.
16. y D
x2
CP
ASY
CP
ASY
CP
p
p
y 0 C " 3 " "1 " 0 "
1
"
3 C
"""""""""
j """""""""
j """""""j""""""""j"""""""""
j """!x
loc &
loc %
y % max
&
&
& min
From y 0 : CP: x D 0.
CP
y0
C
0
C
"""""""""""""
j """"""""!x
y
%
%
p
From y 00 : y 00 D 0 at x D 0, x D ˙ 3.
p
p
3 "
y 00 C " 3 " 0 C
"""""""""""
j """""""""
j """"""""
j """""!x
y ^ infl _ infl ^ infl _
From y 00 : y 00 D 0 at x D 0.
y
ASY
ASY
y 00
"
"1
C
0
"
1
C
"""""""""""""
j"""""""""""j""""""""""""
j """"""""!x
y
_
^
infl _
^
yD
%p
3;
y
%
p &
p
3 3
! 3;! 4
yDx
x3
C1
x2
p &
3 3
4
x
xD!1
yDx
p
! 3
p
3
x
Fig. 4.6-17
xD1
x2
2x
2.1 " 3x 2 /
0
00
,
y
D
,
y
D
.
x2 C 1
.x 2 C 1/2
.x 2 C 1/3
From y: Intercept: .0; 0/. Asymptotes: y D 1 (horizontal). Symmetry: even.
From y 0 : Critical point: x D 0.
x3
yD 2
x "1
18. y D
Fig. 4.6-16
17.
x3
x3 C x " x
x
D
Dx" 2
C1
x2 C 1
x C1
x 4 C 3x 2
x 2 .x 2 C 3/
.x 2 C 1/3x 2 " x 3 2x
D 2
D
y0 D
2
2
2
.x C 1/
.x C 1/
.x 2 C 1/2
2
2
3
4
2
2
.x C 1/ .4x C 6x/ " .x C 3x /2.x C 1/2x
y 00 D
.x 2 C 1/4
4x 5 C 10x 3 C 6x " 4x 5 " 12x 3
D
.x 2 C 1/3
2
2x.3 " x /
D
.x 2 C 1/3
From y: Asymptotes: y D x (oblique). Symmetry: odd.
Intercepts p
.0; 0/. p
Points .˙ 3; ˙ 34 3/.
yD
138
CP
y0
"
0
C
""""""""""""""""j""""""""""""!x
abs
y
&
%
min
x2
1
From y 00 : y 00 D 0 at x D ˙ p .
3
1
1
p
"p
C
"
3
3
"""""""""""""j""""""""""""j""""""""!x
y
_
infl
^
infl
_
y 00
"
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lOMoARcPSD|6566483
INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 4.6 (PAGE 255)
y
20.
yD1
x2
x2 C 1
yD
!1
p
3
p1
3
x
x2 " 2 0
2x
"2.3x 2 C 1/
,y D 2
, y 00 D
.
2
2
x "1
.x " 1/ p
.x 2 " 1/3
From y: Intercept: .0; 2/, .˙ 2; 0/. Asymptotes: y D 1
(horizontal), x D ˙1 (vertical). Symmetry: even.
From y 0 : Critical point: x D 0.
yD
ASY
CP
ASY
f0 "
"1
"
0
C
1
C
"""""""""""
j """""""""j""""""""""j""""""!x
loc %
f &
& min
%
Fig. 4.6-18
From y 00 : y 00 D 0 nowhere.
2
19.
x "4
3
Dx"1"
xC1
xC1
.x C 1/2 C 3
3
D
y0 D 1 C
.x C 1/2
.x C 1/2
6
y 00 D "
.x C 1/3
From y: Asymptotes: y D x " 1 (oblique), x D "1.
Symmetry: none.
Intercepts .0; "4/, .˙2; 0/.
From y 0 : CP: none.
yD
ASY
ASY
y0
"
"1
C
1
"
"""""""""""""
j """"""""""""
j """"""""!x
y
_
^
_
y
xD1
xD!1
ASY
y0
C
"1
C
"""""""""""""
j """"""""!x
y
%
%
2
p
! 2
yD1
p
yD
From y 00 : y 00 D 0 nowhere.
x
2
x2 " 2
x2 " 1
ASY
y 00
C
"1
"
"""""""""""""
j""""""""!x
y
^
_
Fig. 4.6-20
y
21.
!2
!1
2
yD
yDx!1
Fig. 4.6-19
!4
x2 " 4
xC1
x
x 3 " 4x
x.x " 2/.x C 2/
D
2
x "1
x2 " 1
2
2
.x
"
1/.3x
"
4/
" .x 3 " 4x/2x
y0 D
.x 2 " 1/2
3x 4 " 7x 2 C 4 " 2x 4 C 8x 2
D
.x 2 " 1/2
4
2
x Cx C4
D
.x 2 " 1/2
.x 2 " 1/2 .4x 3 C 2x/ " .x 4 C x 2 C 4/2.x 2 " 1/2x
y 00 D
.x 2 " 1/4
4x 5 " 2x 3 " 2x " 4x 5 " 4x 3 " 16x
D
.x 2 " 1/3
3
"6x " 18x
x2 C 3
D
D "6x 2
2
3
.x " 1/
.x " 1/3
From y: Asymptotes: y D x (oblique), x D ˙1.
Symmetry: odd. Intercepts .0; 0/, .˙2; 0/.
yD
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139
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SECTION 4.6 (PAGE 255)
ADAMS and ESSEX: CALCULUS 9
y
From y 0 : CP: none.
yD1
ASY
ASY
y0 C
"1
C
1
C
"""""""""""
j""""""""""j"""""""!x
y %
%
%
!1
1
yD
From y 00 : y 00 D 0 at x D 0.
x
x2 " 1
x2
ASY
ASY
y 00 C "1 " 0 C
1
"
""""""""""j"""""""""
j"""""""""
j """""!x
y ^
_ infl ^
_
Fig. 4.6-22
y
xD1
xD!1
!2
x
2
23.
yDx
yD
x 3 " 4x
x2 " 1
Fig. 4.6-21
x2 " 1
1
2
6
D 1 " 2 , y 0 D 3 , y 00 D " 4 .
x2
x
x
x
From y: Intercepts: .˙1; 0/. Asymptotes: y D 1 (horizontal), x D 0 (vertical). Symmetry: even.
From y 0 : No critical points.
22. y D
ASY
y0
"
0
C
""""""""""""""""j""""""""""""!x
y
&
%
From y 00 : y 00 is negative for all x.
140
x5
2x 3 " x
DxC 2
2
" 1/
.x " 1/2
2
2
4
5
.x " 1/ 5x " x 2.x 2 " 1/2x
y0 D
.x 2 " 1/4
5x 6 " 5x 4 " 4x 6
x 4 .x 2 " 5/
D
D
2
3
.x " 1/
.x 2 " 1/3
2
3
5
3
.x " 1/ .6x " 20x / " .x 6 " 5x 4 /3.x 2 " 1/2 2x
y 00 D
.x 2 " 1/6
7
5
3
6x " 26x C 20x " 6x 7 C 30x 5
D
.x 2 " 1/4
4x 3 .x 2 C 5/
D
.x 2 " 1/4
From y: Asymptotes: y #D x, x D ˙1. Symmetry:
odd.
$
p
25 p
5 .
Intercepts .0; 0/. Points ˙ 5; ˙
16
p
0
From y : CP x D 0, x D ˙ 5.
yD
.x 2
CP
ASY
CP
ASY
CP
p
p
1
"
5 C
y 0 C " 5 " "1 C 0 C
"""""""""j"""""""""j""""""""j""""""""
j""""""""j""""""""""!x
loc
loc &
y % max
%
%
&
min %
From y 00 : y 00 D 0 if x D 0.
ASY
ASY
y 00 "
"1
"
0
C
1
C
"""""""""""
j"""""""""j""""""""""
j """"""!x
y _
_ infl ^
^
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 4.6 (PAGE 255)
y
1
1
D
x 3 " 4x
x.x " 2/.x C 2/
3x 2 " 4
3x 2 " 4
D" 2 2
y0 D " 3
2
.x " 4x/
x .x " 4/2
3
2
.x " 4x/ .6x/ " .3x 2 " 4/2.x 3 " 4x/.3x 2 " 4/
y 00 D "
.x 3 " 4x/4
4
2
4
6x " 24x " 18x C 48x 2 " 32
D"
.x 3 " 4x/3
2
2
12.x " 1/ C 20
D
x 3 .x 2 " 4/3
From y: Asymptotes: y D 0, x D 0; "2; 2.
Symmetry:
# odd. No intercepts.
$ #
$
16
1
2
Points: ˙ p ; ˙ p , ˙3; ˙
15
3
3 3
2
0
From y : CP: x D ˙ p .
3
ASY
CP
ASY
CP
ASY
y 0 " "2 " " p2 C
0
C p2
"
2
"
3
3
""""""""""
j """""""""j"""""""""j"""""""""j""""""""""
j """""!x
loc %
loc &
y &
& min
% max
&
00
00
From y : y D 0 nowhere.
25.
yDx
p
! 5
!1
1
p
5
yD
.x 2
x
x5
" 1/2
Fig. 4.6-23
24.
.2 " x/2 0
.x " 2/.x " 6/
, y D"
,
x3
x4
p
p
2
2.x " 12x C 24/
2.x " 6 C 2 3/.x " 6 " 2 3/
00
y D
D
.
x5
x5
yD
From y: Intercept: .2; 0/. Asymptotes: y D 0 (horizontal), x D 0 (vertical). Symmetry: none obvious. Other
points: ."2; "2/, ."10; "0:144/.
From y 0 : Critical points: x D 2; 6.
yD
ASY
ASY
ASY
y 00 " "2 C 0
"
2 C
"""""""""j"""""""""
j """"""""j"""""!x
y _
^
_
^
ASY
CP
CP
y0 "
0
"
2
C
6
"
"""""""""""j"""""""""""
j """"""""""j"""""""!x
loc %
loc
y &
& min
max &
p
From y 00 : y 00 D 0 at x D 6 ˙ 2 3.
y
yD
p
p
y 00 " 0 C 6 C 2 3 " 6 " 2 3 C
"""""""""
j""""""""""
j """""""""""""
j """""""!x
y _
^
infl
_
infl
^
1
x 3 " 4x
!3
2
!p
3
xD!2
xD2
p2
3
3
x
y
yD
.2 " x/2
x3
.6;2=27/
2
.!10;!0:144/
Fig. 4.6-24
p
6C2 3
p
6!2 3
Fig. 4.6-25
x
26.
x
x
D
,
x2 C x " 2
.2 C x/.x " 1/
".x 2 C 2/
2.x 3 C 6x C 2/
y0 D
, y 00 D
.
2
2
.x C 2/ .x " 1/
.x C 2/3 .x " 1/3
From y: Intercepts: .0; 0/. Asymptotes: y D 0 (horizontal), x D 1, x D "2 (vertical). Other points: ."3; " 34 /,
.2; 12 /.
From y 0 : No critical point.
yD
ASY
ASY
y0
"
"2
"
1
"
""""""""""""""
j """""""""""""
j """""""""!x
y
&
&
&
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141
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SECTION 4.6 (PAGE 255)
ADAMS and ESSEX: CALCULUS 9
y
From y 00 : y 00 D 0 if f .x/ D x 3 C 6x C 2 D 0. Since
f 0 .x/ D 3x 2 C 6 # 6, f is increasing and can only have
one root. Since f .0/ D 2 and f ."1/ D "5, that root must
be between "1 and 0. Let the root be r.
yD
.!1;3/
ASY
ASY
y 00 "
"2 C
r
"
1
C
"""""""""""
j"""""""""j""""""""""
j """"""!x
y _
^ infl _
^
x 3 " 3x 2 C 1
x3
yD1
x
.1;!1/
y
yD
xD!2
Fig. 4.6-27
x
2
x Cx"2
28.
.2;1=2/
x
r
.!3;!3=4/
y D x C sin x, y 0 D 1 C cos x, y 00 D " sin x.
From y: Intercept: .0; 0/. Other points: .k#; k#/, where
k is an integer. Symmetry: odd.
From y 0 : Critical point: x D .2k C 1/#, where k is an
integer.
CP
CP
CP
f0
C
"#
C
#
"
3#
C
""""""""""""j"""""""""""
j """"""""""
j """""""!x
f
%
%
%
%
xD1
Fig. 4.6-26
From y 00 : y 00 D 0 at x D k#, where k is an integer.
27.
x 3 " 3x 2 C 1
1
3
D1" C 3
x3
x
x
3
3
3.x 2 " 1/
y0 D 2 " 4 D
x
x
x4
12
2 " x2
6
y 00 D " 3 C 5 D 6
x
x
x5
From y W Asymptotes: y D 1, x D 0. Symmetry: none.
Intercepts: since limx!0C y D 1, and limx!0! y D "1,
there are intercepts between "1 and 0, between 0 and 1,
and between 2 and 3.
1
Points: ."1; 3/, .1; "1/, .2; " 38 /, .3; /.
27
From y 0 : CP: x D ˙1.
yD
y 00 C "2# " "# C 0 " # C 2# "
"""""""""j""""""""j"""""""j""""""""
j """""""
j """""!x
y ^ infl _ infl ^ infl _ infl ^ infl _
y
2!
ASY
p
p
y 00 C " 2 "
0
C
2 "
""""""""""j""""""""""
j """"""""j"""""!x
y ^ infl _
^ infl _
142
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y D x C sin x
!
CP
ASY
CP
y 0 C "1 "
0
"
1
C
""""""""""
j """""""""
j """""""""
j """""!x
loc %
loc &
y % max
& min
p
From y 00 : y 00 D 0 at x D ˙ 2.
!
Fig. 4.6-28
2!
x
lOMoARcPSD|6566483
INSTRUCTOR’S SOLUTIONS MANUAL
29. y D x C 2 sin x;
y D 0 if x D 0
y 0 D 1 C 2 cos x;
SECTION 4.6 (PAGE 255)
1
From y 00 : y 00 D 0 at x D ˙ p .
2
y 00 D "2 sin x:
1
2#
y 0 D 0 if x D " , i.e., x D ˙
˙ 2n#
2
3
y 00 D 0 if x D ˙n#
From y:#Asymptotes: (none).
odd.
$ Symmetry:
#
$
p
p
2#
8#
2#
8#
Points: ˙ ; ˙
C 3 , ˙ ;˙
C 3 ,
3
3$
3
3
#
4# p
4#
" 3 .
˙ ;˙
3
3
2#
˙ 2n#.
From y 0 : CP: x D ˙
3
CP
CP
CP
CP
CP
4!
2!
2!
4!
y 0 " " 8!
C
"
C
"
C
"
C
3
3
3
3
3
""""""""""
j """""""""j"""""""""j""""""""""
j """""""""
j """""!x
loc % loc & loc % loc & loc %
y & min
max
max
min
min
00
1
1
"
C
"p
p
2
2
"""""""""""""j""""""""""""j""""""""!x
y
^
infl
_
infl
^
y 00
C
y
1
y D e !x
2
p1
2
!1
p
2
x
Fig. 4.6-30
31.
00
From y : y D 0 at x D ˙n#.
y 00 C "2# " "# C 0 " # C 2# "
"""""""""j"""""""""
j """""""j"""""""j""""""""
j """"!x
y ^ infl _ infl ^ infl _ infl ^ infl _
y D xe x ; y 0 D e x .1 C x/; y 00 D e x .2 C x/:
From y: Asymptotes: y D 0 (at x D "1).
Symmetry:
.0;$0/.
# none.$Intercept
#
1
2
Points: "1; " , "2; " 2 ,
e
e
From y 0 : CP: x D "1.
CP
y0
"
"1
C
"""""""""""""
j """"""""!x
abs
y
&
%
min
00
00
From y : y D 0 at x D "2.
y
y 00
"
"2
C
"""""""""""""
j """"""""!x
y
_
infl
^
1!
3
yDx
!
4!
3
2!
x
y
y D x ex
y D x C 2 sin x
x
%
!2;!
Fig. 4.6-29
2
2
30. y D e !x , y 0 D "2xe !x , y 00 D .4x 2 " 2/e !x .
From y: Intercept: .0; 1/. Asymptotes: y D 0 (horizontal). Symmetry: even.
From y 0 : Critical point: x D 0.
CP
y
C
0
"
""""""""""""""""j""""""""""""!x
abs
y
%
&
max
0
&%
!1;!
Fig. 4.6-31
32.
2
2
e2
1
e
&
y D e !x sin x
.x # 0/,
y 0 D e !x .cos x " sin x/, y 00 D "2e !x cos x.
From y: Intercept: .k#; 0/, where k is an integer.
Asymptotes: y D 0 as x ! 1.
#
C k#, where k is an
From y 0 : Critical points: x D
4
integer.
CP
CP
CP
5#
9#
#
"
C
"
y 0 C
4
4
4
"""""
j """"""""j""""""""""j""""""""""
j""""""!x
abs & abs %
loc
y
% max
max &
min
0
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143
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SECTION 4.6 (PAGE 255)
ADAMS and ESSEX: CALCULUS 9
From y 00 : y 00 D 0 at x D .k C
integer.
1
2 /#,
y
where k is an
.!1;1=e/
#
3#
5#
y 00 0 "
C
"
C
2
2
2
""""""
j """"""""j""""""""""
j """"""""""
j """"""!x
y
_ infl ^ infl _ infl ^
y
!a
34.
p "
"!=4 = 2
4 ;e
3!
2
!
2
y D x 2 e !x
x
x
p
p
y 00 C "2 " 2 " "2 C 2 C
"""""""""""""j""""""""""""""j"""""""""!x
y ^
infl
_
infl
^
y
2
2
y 0 D e !x .2x " 2x 3 / D 2x.1 " x 2 /e !x
.!2;4e "2 /
2
2 x
yDx e
2
y 00 D e !x .2 " 6x 2 " 2x.2x " 2x 3 //
2
D .2 " 10x 2 C 4x 4 /e !x
From y: Asymptotes: y D 0.
Intercept:
$ Symmetry: even.
# .0; 0/.
1
Points ˙1;
e
From y 0 : CP x D 0, x D ˙1.
p
!2! 2
35.
From y 00 : y 00 D 0 if
2x 4 " 5x 2 C 1 D 0
p
5 ˙ 25 " 8
2
x D
p4
5 ˙ 17
D
:
4
s
s
p
p
5 C 17
5 " 17
so x D ˙a D ˙
, x D ˙b D ˙
.
4
4
1 " ln x
y0 D
x2
$
1
x2 "
" .1 " ln x/2x
2 ln x " 3
x
00
y D
D
x4
x3
From y: Asymptotes: x D 0, y D 0.
Symmetry:
$ 0/.
# none.
$ # Intercept: .1;
1
3
3=2
Points: e;
, e ; 3=2 .
e
2e
0
From y : CP: x D e.
yD
ln x
;
x#
ASY
CP
y0
0
C
e
"
"""""""j"""""""""""""
j """"""""!x
abs
y
%
&
max
From y 00 : y 00 D 0 at x D e 3=2 .
y 00 C "a " "b C b " a C
""""""""""
j """"""""
j """"""""
j""""""""j"""""!x
y ^ infl _ infl ^ infl _ infl ^
144
p
!2C 2
Fig. 4.6-34
CP
CP
CP
y 0 C "1 "
0
C
1
"
""""""""""
j """""""""
j """""""""
j """""!x
abs & abs % abs &
y % max
max
min
2
y D x 2 e x , y 0 D .2x C x 2 /e x D x.2pC x/e x ,
p
y 00 D .x 2 C 4x C 2/e x D .x C 2 " 2/.x C 2 C 2/e x .
From y: Intercept: .0; 0/.
Asymptotes: y D 0 as x ! "1.
From y 0 : Critical point: x D 0, x D "2.
Fig. 4.6-32
33.
a
CP
CP
y0
C
"2
"
0
C
""""""""""""""
j """""""""""""
j """""""""!x
abs
loc
y
%
&
%
max
min
p
From y 00 : y 00 D 0 at x D "2 ˙ 2.
y D e !x sin x
5!
4
b
y D x 2 e !x
Fig. 4.6-33
!!
!
!b
.1;1=e/
ASY
y 00
0
"
e 3=2
C
""""""""
j """"""""""""
j """"""""!x
y
_
infl
^
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x
lOMoARcPSD|6566483
INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 4.6 (PAGE 255)
y
37.
.e;1=e/
e 3=2
1
yD
x
ln x
x
1
D .4 " x 2 /!1=2
4 " x2
x
1
y 0 D " .4 " x 2 /!3=2 ."2x/ D
2
.4 " x 2 /3=2
3
.4 " x 2 /3=2 " x .4 " x 2 /1=2 ."2x/
2
y 00 D
.4 " x 2 /3
4 C 2x 2
D
.4 " x 2 /5=2
From y: Asymptotes: x D ˙2. Domain "2 < x < 2.
Symmetry: even. Intercept: .0; 21 /.
From y 0 : CP: x D 0.
yD p
ASY
CP
ASY
y0
"2
"
0
C
2
"""""""j""""""""""""j""""""""""""j"""!x
abs
y
&
%
min
00
00
00
From y : y D 0 nowhere, y > 0 on ."2; 2/.
Therefore, y is concave up.
Fig. 4.6-35
ln x
.x > 0/,
x2
6 ln x " 5
1 " 2 ln x 00
,y D
.
y0 D
x3
x4
From y: Intercepts: .1; 0/. Asymptotes: y D 0, since
ln x
ln x
D 0, and x D 0, since lim
D "1.
lim
x!1 x 2
x!0C x 2
From y 0 : Critical point: x D e 1=2 .
36. y D
y
yD p
1
4 " x2
CP
p
y0
0
C
e
"
""""""j""""""""""""j""""""""""!x
abs
y
%
&
max
1=2
!2
From y 00 : y 00 D 0 at x D e 5=6 .
2
x
Fig. 4.6-37
38.
y 00
0
"
e 5=6
C
"""""""
j """"""""""""
j """""""""!x
y
_
infl
^
y
x
, y 0 D .x 2 C 1/!3=2 , y 00 D "3x.x 2 C 1/!5=2 .
x2 C 1
From y: Intercept: .0; 0/. Asymptotes: y D 1 as x ! 1,
and y D "1 as x ! "1. Symmetry: odd.
From y 0 : No critical point. y 0 > 0 and y is increasing for
all x.
From y 00 : y 00 D 0 at x D 0.
yD p
p
. e;.2e/"1 /
e 5=6
1
x
y 00
C
0
"
""""""""""""j""""""""!x
y
^
infl _
y
yD
yD1
ln x
x2
yD!1
Fig. 4.6-36
yD p
x
x2
x
C1
Fig. 4.6-38
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145
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SECTION 4.6 (PAGE 255)
39.
ADAMS and ESSEX: CALCULUS 9
y D .x 2 " 1/1=3
2
y 0 D x.x 2 " 1/!2=3
3
2
2
y 00 D Œ.x 2 " 1/!2=3 " x.x 2 " 1/!5=3 2x!
3
$
#3
x2
2 2
!5=3
1C
D " .x " 1/
3
3
From y: Asymptotes: none.
Symmetry: even. Intercepts: .˙1; 0/, .0; "1/.
From y 0 : CP: x D 0. SP: x D ˙1.
From f : Intercept: .0; 0/, .˙1; 0/. Asymptotes: none.
Symmetry: odd.
1
From f 0 : CP: x D ˙ . SP: x D 0.
e
CP
SP
CP
1
1
f
C
"
"
0
"
C
e
e
""""""""""""j""""""""""j"""""""""""
j """""""!x
loc
loc
f
%
&
max &
min %
0
SP
CP
SP
y 0 " "1 "
0
C
1 C
""""""""""
j """""""""j""""""""""
j """""!x
abs
y &
& min %
%
From f 00 : f 00 is undefined at x D 0.
f 00
"
0
C
"""""""""""""""""
j """""""""""!x
f
_
infl
^
From y 00 : y 00 D 0 nowhere.
y
y 00
"
"1 C
1
"
"""""""""""j""""""""""j"""""""!x
y _ infl ^ infl _
! !1
y
e
y D .x 2 " 1/1=3
; 1e
"
!1
e;
!1
!1
e
"
x
y D x ln jxj
1
x
Fig. 4.6-40
!1
sin x
.
1 C x2
Curve crosses asymptote at infinitely many points: x D n#
.n D 0; ˙1; ˙2; : : :/.
y
sin x
yD
1 C x2
41. y D 0 is an asymptote of y D
Fig. 4.6-39
40. According to Theorem 5 of Section 4.4,
yD
1
1Cx 2
lim x ln x D 0:
x!0C
Thus,
x
lim x ln jxj D lim x ln x D 0:
x!0
x!0C
If f .x/ D x ln jxj for x ¤ 0, we may define f .0/ such
that f .0/ D lim x ln jxj D 0. Then f is continuous on
x!0
yD!
the whole real line and
f 0 .x/ D ln jxj C 1;
146
f 00 .x/ D
1
sgn .x/:
jxj
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Fig. 4.6-41
1
1Cx 2
lOMoARcPSD|6566483
INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 4.8 (PAGE 267)
Section 4.7 Graphing with Computers
(page 261)
1.
> plot([exp(x)*ln(1+1/exp(x)),
> 0.71*exp(x-35.7)], x=33..38,
> y =0..2, style=point, symbol=[circle,
point],
> color = [red, black], numpoints = 1500);
(b) The problem is with g.x/. When x grows large
enough, the argument of the square root is evaluated as 22x , as the computer discards the 1. When
this happens, the argument of the logarithm vanishes
and the computer could be expected to return "1.
However we now arrive at the case of the previous
exercise. The computer will return a variety of complex numbers, infinite values, and finite real values,
produced by the computer evaluation of the logarithm
of the expression in question 2. The computer only
plots the finite real values, but all of them are completely spurious.
will produce the curve as shown in the given figure (in
red), and also the exponential curve (black) conforming to
the rightmost stripe as shown in this figure:
(c) The argument of the square root is 22x .1"2!2x /. The
computer will begin to encounter serious difficulties
for values of x beginning where 2!2x ' $, that is,
"2x D "52 or x D 26. This is evident in the figure.
The longest (rightmost) of the exponential stripes in the
given figure seems to begin at about .0:71; 35:7/. Accordingly, the plot command
2
1.8
1.6
1.4
1.2
y 1
0.8
0.6
0.4
0.2
0 33
4.
Since there are 64 " 52 " 2 D 10 bits left to represent the
exponent the absolute value of the exponent, the smallest
possible exponent is
!
"
"1111111111 D " 1 C 2 C 22 C ! ! ! C 29 ;
that is, "1023 in base 10. The smallest positive mantissa
is 0:000 ! ! ! 001 D 2!52 , so the smallest positive binary
floating-point number is
34
35
x
36
37
2!52 & 2!1023 D 2!52!1023
D 10!1075 log10 2 ' 10!324 :
38
5.
Fig. 4.7-1
(The red curves appear gray here.) Other exponential
stripes can be handled similarly
As in the previous exercise, there are 10 bits available
for the exponent, so the largest possible exponent is
1111111111 in base 2, or 1023 in base 10. The largest
possible mantissa is
0:111 ! ! ! 111
(52 digits)
1
1
1
D C 2 C ! ! ! C 52
2
2
2
1
D 1 " 52 ' 1:
2
2. The square of any number will tend to require an increased number of digits to represent it—especially for
squares of numbers with large numbers of digits to begin with. However on a computer the number of digits is
fixed so the least significant digits are discarded. The resulting number is different from the square of the original
number. Consequently the square root will not be quite
the same and the expression cannot be expected to vanish
exactly.
Thus the largest positive floating-point number is approximately 21023 D 101023 log10 2 ' 10308 .
Section 4.8 Extreme-Value Problems
(page 267)
3. (a) We have
&
%
p
g.x/ D ln 2x " 22x " 1
$
#
1
p
D " ln
2x " 22x " 1
!
p
2x C 22x " 1
p
p
D " ln
.2x " 22x " 1/.2x C 22x " 1/
%
&
p
D " ln 2x C 22x " 1 D f .x/:
1.
Let the numbers be x and 7 " x. Then 0 $ x $ 7. The
product is P .x/ D x.7 " x/ D 7x " x 2 .
P .0/ D P .7/ D 0 and P .x/ > 0 if 0 < x < 7. Thus
maximum P occurs at a CP:
0D
7
dP
D 7 " 2x ) x D :
dx
2
The maximum product is P .7=2/ D 49=4.
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SECTION 4.8 (PAGE 267)
ADAMS and ESSEX: CALCULUS 9
8
2. Let the numbers be x and
where x > 0. Their sum is
x
8
S D x C . Since S ! 1 as x ! 1 or x ! 0C, the
x
minimum sum must occur at a critical point:
7.
Let the dimensions of a rectangle be x and y. Then the
area is A D xy and the perimeter is P D 2x C 2y.
Given A we can express
P D P .x/ D 2x C
p
dS
8
0D
D 1 " 2 ) x D 2 2:
dx
x
0D
3. Let the numbers be x and 60 " x. Then 0 $ x $ 60.
Let P .x/ D x 2 .60 " x/ D 60x 2 " x 3 .
Clearly, P .0/ D P .60/ D 0 amd P .x/ > 0 if 0 < x < 60.
Thus maximum P occurs at a CP:
8.
Let the width and the length of a rectangle of given
perimeter 2P be x and P " x. Then the area of the rectangle is
A.x/ D x.P " x/ D P x " x 2 :
0D
5
4. Let the numbers be x and 16"x. Let P .x/ D x .16"x/ .
Since P .x/ ! "1 as x ! ˙1, so the maximum must
occur at a critical point:
The critical points are 0, 6 and 16. Clearly,
P .0/ D P .16/ D 0, and P .6/ D 216 & 105 . Thus, P .x/ is
maximum if the numbers are 6 and 10.
9.
Let the dimensions of the isosceles triangle be as shown.
Then 2x C 2y D P (given constant). The area is
p
A D xh D x y 2 " x 2 D x
5. Let the numbers be x and 10 " x. We want to minimize
0 $ x $ 10:
S.0/ D 100 and S.10/ D 1; 000. For CP:
0 D S 0 .x/ D 3x 2 " 2.10 " x/ D 3x 2 C 2x " 20:
p
The only positive CP is x D ."2 C 4 C 240/=6 ' 2:270.
Since S.2:270/ ' 71:450, the minimum value of S is
about 71.45.
6. If the numbers are x and n " x, then 0 $ x $ n and the
sum of their squares is
dA
P
D P " 2x ) x D
dx
2
P
Hence, the width and the length are
and
2
P
P
. Since the width equals the length, it
.P " / D
2
2
is a square.
0 D P 0 .x/ D 3x 2 .16 " x/5 " 5x 3 .16 " x/4
D x 2 .16 " x/4 .48 " 8x/:
x 2 D A D xy ) x D y:
)
Since A.x/ ! "1 as x ! ˙1 the maximum must occur
at a critical point:
Therefore, x D 0 or 40.
Max must correspond to x D 40. The numbers are 40 and
20.
3
dP
2A
D2" 2
dx
x
Thus min P occurs for x D y, i.e., for a square.
dP
D 120x " 3x 2 D 3x.40 " x/:
dx
S.x/ D x 3 C .10 " x/2 ;
.0 < x < 1/:
Evidently, minimum P occurs at a CP. For CP:
p
8
Thus, the smallest possible sum is 2 2 C p D 4 2.
2 2
p
0D
2A
;
x
s
#
$2
" x2:
Evidently, y # x so 0 $ x $ P =4. If x D 0 or x D P =4,
then A D 0. Thus the maximum of A must occur at a CP.
For max A:
s
Px
dA
P2
D
" Px " r
0D
;
dx
4
P2
" Px
2
4
P
P2
" 2P x " P x D 0, or x D . Thus y D P =3 and
2
6
the triangle is equilateral since all three sides are P =3.
i.e.,
S.x/ D x 2 C .n " x/2 :
Observe that S.0/ D S.n/ D n2 . For critical points:
y
y
h
0 D S 0 .x/ D 2x " 2.n " x/ D 2.2x " n/ ) x D n=2:
Since S.n=2/ D n2 =2, this is the smallest value of the
sum of squares.
148
P
"x
2
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x
x
Fig. 4.8-9
lOMoARcPSD|6566483
INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 4.8 (PAGE 267)
10. Let the various dimensions be as shown in the figure.
Since h D 10 sin & and b D 20 cos & , the area of the
triangle is
A.& / D 21 bh D 100 sin & cos &
D 50 sin 2&
for 0 < & <
12. Let x be as shown in the figure. The perimeter of the
rectangle is
p
P .x/ D 4x C 2 R2 " x 2
#
:
2
For critical points:
dP
"2x
D4C p
dx
R2 " x 2
p
2R
)2 R2 " x 2 D x ) x D p :
5
0D
#
Since A.& / ! 0 as & ! 0 and & ! , the maximum
2
must be at a critial point:
0 D A0 .& / D 100 cos 2& ) 2& D
.0 $ x $ R/:
#
#
)& D :
2
4
Since
d 2P
"2R2
D
<0
2
2
dx
.R " x 2 /3=2
Hence, the largest possible area is
( % &)
#
D 50 m2 :
A.#=4/ D 50 sin 2
4
therefore P .x/ is concave down on Œ0; R!, so it must have
2R
an absolute maximum value at x D p . The largest
5
perimeter is therefore
(Remark: alternatively, we may simply observe that the
largest value of sin 2& is 1; therefore the largest possible
area is 50.1/ D 50 m2 .)
P
#
2R
p
5
$
#
2R
D4 p
5
$
C
s
R2 "
10R
4R2
D p units:
5
5
p
.x; R2 !x 2 /
10
10
h
R
x
"
"
b=2
b=2
Fig. 4.8-10
11.
Let the corners of the rectangle be as shown.
p
The area of the rectangle is A D 2xy D 2x R2 " x 2 (for
0 $ x $ R).
If x D 0 or x D R then A D 0; otherwise A > 0.
Thus maximum
) point:
(pA must occur at a2 critical
x
dA
R2 " x 2 " p
) R2 " 2x 2 D 0.
0D
D2
dx
R2 " x 2
R
Thus x D p and the maximum area is
2
s
2
R
R
D R2 square units.
2p
R2 "
2
2
y
.x;y/
R
x
Fig. 4.8-11
x
Fig. 4.8-12
13.
Let the upper right
r corner be .x; y/ as shown. Then
x2
x # 0 and y D b 1 " 2 , so x $ a.
a
The area of the rectangle is
A.x/ D 4xy D 4bx
s
1"
x2
;
a2
.0 $ x $ a/:
Clearly, A D 0 if x D 0 or x D a, so maximum A must
occur at a critical
0 point:
1
s
2x 2
2
B
C
dA
x
a2
C
D 4b B
0D
@ 1 " a2 " r
2 A
dx
x
2 1" 2
a
2
2
b
x
a
x
Thus 1 " 2 " 2 D 0 and x D p . Thus y D p .
a
a
2
2
a b
The largest area is 4 p p D 2ab square units.
2 2
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SECTION 4.8 (PAGE 267)
ADAMS and ESSEX: CALCULUS 9
y
x2 y2
C
D1
a2 b 2
15.
.x;y/
NEED FIGURE If the sides of the triangle are 10 cm,
10 cm, andp2x cm, then the area of the triangle is
A.x/ D x 100 " x 2 cm2 , where 0 $ x $ 10. Evidently A.0/ D A.10/ D 0 and A.x/ > 0 for 0 < x < 10.
Thus A will be maximum at a critical point. For a critical
point
x
0 D A0 .x/ D
D
100 " x 2 " x
100 " x 2 " x 2
:
p
100 " x 2
#
$
1
."2x/
p
2 100 " x 2
2
Thus the
p critical point is given by 2x D 100, so
50. The maximum area of the triangle is
x p
D
A. 50/ D 50 cm2 .
Fig. 4.8-13
14.
p
See the diagrams below.
a) The area of the rectangle is A D xy. Since
y
b
b.a " x/
D )yD
:
a"x
a
a
16.
Thus, the area is
A D A.x/ D
bx
.a " x/
a
.0 $ x $ a/:
A.& / D
For critical points:
0 D A0 .x/ D
NEED FIGURE If the equal sides of the isosceles triangle
are 10 cm long and the angles opposite these sides are & ,
then the area of the triangle is
b
a
.a " 2x/ ) x D :
a
2
2b
Since A00 .x/ D "
< 0, A must have a maximum
a
a
value at x D . Thus, the largest area for the rectan2
gle is
#
$
a
ab
b %a&
a"
D
square units;
a 2
2
4
that is, half the area of the triangle ABC .
A
C
1
.10/.10 sin & / D 50 sin & cm2 ;
2
which is evidently has maximum value 50 cm2 when
& D #=2, that is, when the triangle is right-angled. This
solution requires no calculus, and so is easier than the one
given for the previous problem.
17.
Let the width and the height of the billboard be w and
h m respectively. The area of the board is A D wh. The
printed area is .w " 8/.h " 4/ D 100.
100w
100
and A D 4w C
; .w > 8/.
Thus h D 4 C
w"8
w"8
Clearly, A ! 1 if w ! 1 or w ! 8C. Thus minimum
A occurs at a critical point:
b
x
C
a!x
a
Fig. 4.8.14(a)
B
A
D
B
Fig. 4.8.14(b)
(b) This part has the same answer as part (a). To see
this, let CD ? AB, and solve separate problems for
the largest rectangles in triangles ACD and BCD
as shown. By part (a), both maximizing rectangles
have the same height, namely half the length of CD.
Thus, their union is a rectangle of area half of that of
triangle ABC .
150
dA
100
100w
D4C
"
dw
w " 8 .w " 8/2
100w D 4.w 2 " 16w C 64/ C 100w " 800
0D
y
w 2 " 16w " 136 D 0
p
p
16 ˙ 800
D 8 ˙ 10 2:
wD
2
p
Since w > 0 we must have w D 8 C 10 2.
p
100
Thus h D 4 C p D 4 C 5 2.
10 2
p
p
The billboard should be 8 C 10 2 m wide and 4 C 5 2 m
high.
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 4.8 (PAGE 267)
For maximum profit:
w
2
0D
d 2P
D "8 < 0 any critical point gives a local
dx 2
max.) The manufacturer should offer a rebate of $250 to
maximize profit.
4
h!4
(Since
h
4
w!8
20.
2
Fig. 4.8-17
18. Let x be the side of the cut-out squares. Then the volume
of the box is
V .x/ D x.70 " 2x/.150 " 2x/
If the manager charges $.40 C x/ per room, then .80 " 2x/
rooms will be rented.
The total income will be $.80 " 2x/.40 C x/ and the total
cost will be $.80 " 2x/.10/ C .2x/.2/. Therefore, the profit
is
P .x/ D .80 " 2x/.40 C x/ " Œ.80 " 2x/.10/ C .2x/.2/!
D 2400 C 16x " 2x 2
.0 $ x $ 35/:
0 D V 0 .x/ D 4.2625 " 220x C 3x 2 /
D 4.3x " 175/.x " 15/
175
:
) x D 15 or
3
21.
Head for point C on road x km east of A. Travel time is
T D
The only critical point in Œ0; 35! is x D 15. Thus, the
largest possible volume for the box is
V .15/ D 15.70 " 30/.150 " 30/ D 72; 000 cm3 :
150
x
p
122 C x 2
10 " x
C
:
15
39
10
12
C
D 1:0564 hrs
15
39
p
244
D 1:0414 hrs
T .10/ D
15
For critical points:
We have
T .0/ D
150!2x
70!2x
for x > 0:
If P 0 .x/ D 16 " 4x D 0, then x D 4. Since
P 00 .x/ D "4 < 0, P must have a maximum value at
x D 4. Therefore, the manager should charge $44 per
room.
Since V .0/ D V .35/ D 0, the maximum value will occur
at a critical point:
x
dP
D 4.500 " 2x/ ) x D 250:
dx
0D
70
dT
1
1
x
p
"
D
2
2
dx
15 12 C x
39
p
) 13x D 5 122 C x 2
) .132 " 52 /x 2 D 52 & 122 ) x D 5
Fig. 4.8-18
'
13
5
T .0/
C
D 0:9949 <
T .10/:
15
39
(Or note that
T .5/ D
19. Let the rebate be $x. Then number of cars sold per month
is
%x&
2000 C 200
D 2000 C 4x:
50
2
1
d T
D
dt 2
15
The profit per car is 1000 " x, so the total monthly profit
is
P D .2000 C 4x/.1000 " x/ D 4.500 C x/.1000 " x/
D 4.500; 000 C 500x " x 2 /:
D
p
122 C x 2 " p
x2
122 C x 2
122 C x 2
122
>0
15.122 C x 2 /3=2
so any critical point is a local minimum.)
To minimize travel time, head for point 5 km east of A.
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SECTION 4.8 (PAGE 267)
A
C
x
ADAMS and ESSEX: CALCULUS 9
(b) Since E t0 D a`w 2 .2w C 3u/=.w C u/2 > 0 for all
w > 0, E t is an increasing function of w. Thus its
smallest value occurs for the smallest value of w that
permits flight, namely w D s. The minimum energy
for the tailwind trip is E t D k`s 3 =.s C u/.
B
10!x
39 km/h
12 15 km/h
p
(c) The minimum energy for the tailwind part of the
round trip in part (b) is independent of u. However,
for the headwind part, the minimum Eh D v D 3u=2
only applies as long as v # s. Otherwise the plane
cannot fly. If u > 2s=3, then v D 3u=2 and the least
total energy for the trip is k`.s 3 =.s C u/ C 27u2 =4/.
If u < 2s=3, then the minimum value for Eh at
v D 3u=2 implies a speed too slow to stay airborne.
As Eh0 > 0 when v > 3u=2 the the least value for Eh
that is admissible happens for v D s. Thus the total
energy becomes 2k`s 4 =.s 2 " u2 /.
122 Cx 2
P
Fig. 4.8-21
22. This problem is similar to the previous one except that the
10 in the numerator of the second fraction in the expression for T is replaced with a 4. This has no effect on the
critical point of T , namely x D 5, which now lies outside
the appropriate interval 0 $ x $ 4. Minimum T must
occur at an endpoint. Note that
25.
4
12
C
D 0:9026
15
39
p
1
T .4/ D
122 C 42 D 0:8433:
15
T .0/ D
1
1
, A.1/ D
> A.0/. For CP:
16
4#
#
$
1
x 1"x
1
1
#
dA
D
"
)x
C
D )xD
:
0D
dx
2#
8
2#
8
8
4C#
Now A.0/ D
The minimum travel time corresponds to x D 4, that is, to
driving in a straight line to B.
Since
23. The time for the trip is 3000=.v " 100/, so the total energy
needed for the trip is
E D 3000k
Use x m for the circle and 1 " x m for square. The sum
of areas is
#
$
1"x 2
#x 2
C
A D # r 2 C s2 D
4# 2
4
2
x
.1 " x/2
D
.0 $ x $ 1/
C
4#
42
for A.
v3
;
v " 100
d 2A
1
1
D
C > 0, the CP gives local minimum
dx 2
2#
8
a) For max total area use none of wire for the square,
i.e., x D 1.
4
#
D
m
b) For minimum total area use 1 "
4C#
4C#
for square.
where k is a constant. Evidently we must have v > 100
or the trip would be impossible. The minimum value of E
will occur at a critical point where
1 metre
3
0D
2
1!x
x
dE
2v " 300v
D
:
dv
.v " 100/2
s
The minimum thus occurs at v D 150 knots, and the time
for the flight at this speed would be 3000=50 D 60 hours,
or 2.5 days. This is much slower than commercial airliners
can travel; the time for the flight at that speed is not much
shorter than a fast ship could cross the Atlantic ocean.
r
s
xDC D2! r
1!xDP D4s
Fig. 4.8-25
24.
(a) As shown in the previous exercise, the energy
for a flight with airspeed v into the headwind is
Eh D k`v 3 =.v " u/. Similarly, the energy for a flight
of the same distance with airspeed w and a tailwind
of speed u is E t D k`w 3 =.w C u/.
152
26.
Let the dimensions of the rectangle be as shown in the
figure. Clearly,
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 4.8 (PAGE 267)
y
Therefore, the area is
A.& / D xy
D .a sin & C b cos & /.a cos & C b sin & /
D ab C .a2 C b 2 / sin & cos &
1
D ab C .a2 C b 2 / sin 2&
2
for 0 $ & $
Y
#
:
2
#
If A0 .& / D .a2 C b 2 / cos 2& D 0, then & D . Since
4
#
A00 .& / D "2.a2 C b 2 / sin 2& < 0 when 0 $ & $ ,
2
#
therefore A.& / must have a maximum value at & D .
4
Hence, the area of the largest rectangle is
A
%# &
4
%# &
1
D ab C .a2 C b 2 / sin
2
2
1 2
1
2
D ab C .a C b / D .a C b/2
2
2
sq. units.
b
a
(Note: x D y D p C p indicates that the rectangle
2
2
containing the given rectangle with sides a and b, has
largest area when it is a square.)
p
.9; 3/
"
9
p
3
"
X
x
Fig. 4.8-27
28.
The longest beam will have length equal to the minimum
of L D x C y, where x and y are as shown in the figure
below:
a
b
xD
; yD
:
cos &
sin &
Thus,
#
$
a
b
#
L D L.& / D
C
0<& <
:
cos &
sin &
2
x
a
x
"
a
"
&
y
y
b
b
Fig. 4.8-28
Fig. 4.8-26
27.
Let the line have intercepts x, y as shown. Let & be angle
shown. The length of line is
p
9
3
LD
C
cos &
sin &
.0 < & <
#
/:
2
#
Clearly, L ! 1 if & ! 0C or & ! ".
2
Thus the minimum length occurs at a critical point.
For CP:
p
#
$
3 cos &
dL
9 sin &
1 3
3
p
0D
D
"
)
tan
&
D
d&
cos2 &
sin2 &
3
#
)& D
6
Shortest line segment has length
LD p
9
3=2
C
p
If L0 .& / D 0, then
a sin &
b cos &
D0
"
cos2 &
sin2 &
a sin3 & " b cos3 &
,
D0
cos2 & sin2 &
,
a sin3 & " b cos3 & D 0
b
,
tan3 & D
a
b 1=3
,
tan & D 1=3 :
a
#
Clearly, L.& / ! 1 as & ! 0C or & ! ". Thus, the
2!
1=3
b
. Using the
minimum must occur at & D tan!1
a1=3
b 1=3
, it follows that
a1=3
a1=3
b 1=3
cos & D p
; sin & D p
:
a2=3 C b 2=3
a2=3 C b 2=3
triangle above for tan & D
p
3
D 8 3 units:
1=2
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SECTION 4.8 (PAGE 267)
ADAMS and ESSEX: CALCULUS 9
Hence, the minimum is
a
L.& / D
a1=3
b
!C
b 1=3
p
p
a2=3 C b 2=3
a2=3 C b 2=3
%
&3=2
D a2=3 C b 2=3
units.
Thus, the maximum height of the fence is
!
p
p
32=3 " 1
"
2
32=3 " 1
h.& / D 6
31=3
!
D 2.32=3 " 1/3=2 ' 2:24 m:
31.
29. If the largest beam that can be carried horizontally around the corner is l m long (by Exercise 26,
l D .a2=3 C b 2=3 /2=3 m), then at the point of maximum
clearance, one end of the beam will be on the floor at the
outer wall of one hall, and the other will be on the ceiling
at the outer wall of the second hall. Thus the horizontal
projection of the beam will be l. So the beam will have
length
dS
"4
D 5 C 2y ) y 6 D 2 ) y D ˙21=6
dy
y
1
) x D ˙ 1=3
2
Thus the shortest distance from origin to curve is
r
r
1
3
31=2
1=3
SD
C
2
D
D 1=3 units.
2=3
2=3
2
2
2
0D
p
l 2 C c 2 D Œ.a2=3 C b 2=3 /3 C c 2 !1=2 units.
32.
30. Let & be the angle of inclination of the ladder. The height
of the fence is
#
h.& / D 6 sin & " 2 tan &
Let .x; y/ be a point on x 2 y 4 D 1. Then x 2 y 4 D 1
and the square of distance from .x; y/ to .0; 0/ is
1
S D x 2 C y 2 D 4 C y 2 , (y ¤ 0)
y
Clearly, S ! 1 as y ! 0 or y ! ˙1, so minimum S
must occur at a critical point. For CP:
The square of the distance from .8; 1/ to the curve
y D 1 C x 3=2 is
S D .x " 8/2 C .y " 1/2
$
#
0<& <
:
2
D .x " 8/2 C .1 C x 3=2 " 1/2
D x 3 C x 2 " 16x C 64:
Note that y, and therefore also S, is only defined for
x # 0. If x D 0 then S D 64. Also, S ! 1 if x ! 1.
For critical points:
dS
D 3x 2 C 2x " 16 D .3x C 8/.x " 2/
dx
) x D " 83 or 2:
0D
6 m
Only x D 2 is feasible. At x D 2 we
p have SpD 44 < 64.
Therefore the minimum distance is 44 D 2 11 units:
h
"
2 m
33.
Fig. 4.8-30
For critical points:
0 D h0 .& / D 6 cos & " 2 sec2 &
)3 cos & D sec2 & ) 3 cos3 & D 1
! "1=3
) cos & D 31
:
#
Since h00 .& / D "6 sin & "4 sec2 & tan & < 0 for 0 < & < ,
! "1=3 2
therefore h.& / must be maximum at & D cos!1 13
.
Then
sin & D
154
p
32=3 " 1
;
31=3
tan & D
p
32=3 " 1:
Let the cylinder have radius r and height h. By symmetry,
the centre of the cylinder is at the centre of the sphere.
Thus
h2
r2 C
D R2 :
4
The volume of cylinder is
$
#
h2
; .0 $ h $ 2R/:
V D # r 2 h D #h R2 "
4
Clearly, V D 0 if h D 0 or h D 2R, so maximum V
occurs at a critical point. For CP:
(
)
dV
h2 2h2
0D
D # R2 "
"
dh
4
4
2R
4 2
2
)hD p
)h D R
3
3
r
2
)r D
R:
3
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 4.8 (PAGE 267)
2R
The largest cylinder has height p units and radius
3
units.
r
2
R
3
35.
Let the box have base dimensions x m and height y m.
Then x 2 y D volume D 4.
Most economical box has minimum surface area (bottom
and sides). This area is
r
S D x 2 C 4xy D x 2 C 4x
h=2
R
D x2 C
#
4
x2
16
;
x
$
.0 < x < 1/:
h
Clearly, S ! 1 if x ! 1 or x ! 0C. Thus minimum S
occurs at a critical point. For CP:
0D
16
dS
D 2x " 2 ) x 3 D 8 ) x D 2 ) y D 1:
dx
x
Most economical box has base 2 m & 2 m and
height 1 m.
Fig. 4.8-33
34. Let the radius and the height of the circular cylinder be r
and h. By similar triangles,
h
H
H.R " r/
D
)hD
:
R"r
R
R
y
Hence, the volume of the circular cylinder is
# r 2 H.R " r/
V .r/ D # r 2 h D
R
#
3$
r
for 0 $ r $ R:
D #H r 2 "
R
Since V .0/ D V .R/ D 0, the maximum
# value2 of
$ V must
dV
3r
be at a critical point. If
D #H 2r "
D 0, then
dr
R
2R
. Therefore the cylinder has maximum volume if
r D
3
2R
its radius is r D
units, and its height is
3
x
x
Fig. 4.8-35
36.
#
$
2R
H R"
H
3
hD
D
units.
R
3
x
s
x
x
2 ft
r
H
h
R
2 ft
Fig. 4.8-34
Fig. 4.8-36
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SECTION 4.8 (PAGE 267)
ADAMS and ESSEX: CALCULUS 9
From the figure, if the side of the square base of the pyramid is 2x, then p
the slant height of triangular walls of the
pyramid is s D 2 " x. The vertical height of the pyramid
is
q
q
p
p
p
p
2
2
2
2
h D s " x D 2 " 2 2x C x " x D 2 1 " 2x:
38.
V D # r 2 h C 43 # r 3 :
h
Thus the volume of the pyramid is
p
q
p
4 2 2
V D
1 " 2x;
x
3
p
for 0 $ x $ 1= 2. V D 0 at both endpoints, so the
maximum will occur at an interior critical point. For CP:
p " q
p 2 #
p
4 2
2x
dV
D
2x 1 " 2x " p
0D
p
dx
3
2 1 " 2x
p 2
p
4x.1 " 2x/ D 2x
p
p
4x D 5 2x 2 ; x D 4=.5 2/:
p
p
p
V .4=.5 2// D p
32 2=.75
p 5/.3 The largest volume of such
a pyramid is 32 2=.75 5/ ft .
r
Fig. 4.8-38
If the cylindrical wall costs $k per unit area and the hemispherical wall $2k per unit area, then the total cost of the
tank wall is
C D 2# rhk C 8# r 2 k
37. Let the dimensions be as shown. The perimeter is
x
# C x C 2y D 10. Therefore,
2
%
#&
1C
x C 2y D 10; or .2 C #/x C 4y D 20:
2
The area of the window is
#
$
1 % x &2
x2
.2 C #/x
A D xy C #
D#
Cx 5"
:
2
2
8
4
V " 34 # r 3
C 8# r 2 k
#r2
2V k
16
D
C # r 2k
.0 < r < 1/:
r
3
D 2# rk
Since C ! 1 as r ! 0C or r ! 1, the minimum cost
must occur at a critical point. For critical points,
To maximize light admitted, maximize the area A. For
CP:
0D
0D
#x
2C#
2C#
20
dA
D
C5"
x"
x)xD
dx
4
4
4
4C#
10
)yD
:
4C#
dC
32
D "2V kr !2 C # rk
dr
3
,
rD
#
r3 D
#
$
4
1
3
# r2h C # r3 ) r D h
16#
3
4
) h D 4r D 4
#
3V
16#
y
39.
$1=3
Let D 0 be chosen so that mirror AB is the right bisector
of DD 0 . Let CD 0 meet AB at X. Therefore, the travel
time along CXD is
x
156
$1=3
:
Hence, in order to minimize the cost, the radius and length
#
$
3V 1=3
of the cylindrical part of the tank should be
16#
#
$
3V 1=3
and 4
units respectively.
16#
x=2
Fig. 4.8-37
3V
16#
Since V D # r 2 h C 34 # r 3 ,
20
To admit greatest amount of light, let width D
m
4C#
10
m.
and height (of the rectangular part) be
4C#
y
Let h and r be the length and radius of the cylindrical
part of the tank. The volume of the tank is
TX D
CX C XD
CX C XD 0
CD 0
D
D
:
speed
speed
speed
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 4.8 (PAGE 267)
p
The stiffness is S D wh3 D h3 4R2 " h2 for
.0 $ h $ 2R/. We have S D 0 if h D 0 or h D 2R.
For maximum stiffness:
If Y is any other point on AB, travel time along C YD is
TY D
C Y C YD
C Y C YD 0
CD 0
D
>
:
speed
speed
speed
0D
(The sum of two sides of a triangle is greater than the
third side.) Therefore, X minimizes travel time. Clearly,
XN bisects †CXD.
p
dS
h4
D 3h2 4R2 " h2 " p
:
dh
4R2 " h2
p
Thus 3.4R2 " h2 / D h2 so h D 3R, andpw D R.
The stiffest beam has width R and depth 3R.
D
N
C
w=2
A
"
h=2
B
"
R
Y
X
h
D0
w
Fig. 4.8-39
Fig. 4.8-41
40. If the path of the light ray is as shown in the figure then
the time of travel from A to B is
T D T .x/ D
p
a2 C x 2
C
v1
p
b 2 C .c " x/2
:
v2
42.
c
A
a
i
x
c!x
r
43.
b
B
Fig. 4.8-40
The curve y D 1 C 2x " x 3 has slope m D y 0 D 2 " 3x 2 .
Evidently m is greatest for x D 0, in which case y D 1
and m D 2. Thus the tangent line with maximal slope has
equation y D 1 C 2x.
dQ
D kQ3 .L " Q/5
.k; L > 0/
dt
Q grows at the greatest rate when f .Q/ D Q3 .L " Q/5 is
maximum, i.e., when
0 D f 0 .Q/ D 3Q2 .L " Q/5 " 5Q3 .L " Q/4
To minimize T , we look for a critical point:
0D
D Q2 .L " Q/4 .3L " 8Q/
1
x
1
c"x
dT
D
"
p
p
dx
v1 a2 C x 2 v2 b 2 C .c " x/2
1
1
D
sin i "
sin r:
v1
v2
Since f .0/ D f .L/ D 0 and f
most rapidly when Q D
3L
.
8
#
3L
8
)
$
Q D 0; L;
3L
:
8
> 0, Q is growing
Thus,
v1
sin i
D
:
sin r
v2
44.
Let h and r be the height and base radius of the cone and
R be the radius of the sphere. From similar triangles,
41. Let the width be w, and the depth be h. Therefore
# $2 % &
w 2
h
C
D R2 :
2
2
p
r
R
h"R
2r 2 R
hD 2
r " R2
h2 C r 2
)
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157
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SECTION 4.8 (PAGE 267)
ADAMS and ESSEX: CALCULUS 9
which simplifies to
t 2 C t C 1 D t 2 et :
h
h!R
p
Both sides of this equation are increasing functions but the
left side has smaller slope than the right side for t > 0.
Since the left side is 1 while the right side is 0 at t D 0,
there will exist a unique solution in t > 0. Using a graphing calculator or computer program we determine that the
critical point is approximately t D 1:05032. For this value
of t we have x ' 750:15, so the movement of cars will be
optimized by loading 750 cars for each sailing.
h2 Cr 2
R
R
r
46.
Fig. 4.8-44
Then the volume of the cone is
V D
1 2
2
r4
# r h D #R 2
3
3
r " R2
.R < r < 1/:
Clearly V ! 1 if r ! 1 or r ! RC. Therefore to
minimize V , we look for a critical point:
( 2
)
dV
2
.r " R2 /.4r 3 / " r 4 .2r/
D #R
dr
3
.r 2 " R2 /2
5
3 2
5
4r " 4r R " 2r D 0
p
r D 2R:
0D
,
,
2
Let distances and angles be as shown. Then tan ˛ D ,
x
12
tan.& C ˛/ D
x
2
tan & C
12
tan & C tan ˛
x
D
D
2
x
1 " tan & tan ˛
1 " tan &
x
12 24
2
tan & D tan & C
"
x #x 2
x
$
10
10x
24
; so tan & D 2
D f .x/:
tan & 1 C 2 D
x
x
x C 24
To maximize & (i.e., to get the best view of the mural), we
can maximize tan & D f .x/.
Since f .0/ D 0 and f .x/ ! 0 as x ! 1, we look for a
critical point.
( 2
)
x C 24 " 2x 2
0 D f 0 .x/ D 10
) x 2 D 24
.x 2 C 24/2
p
)xD2 6
p
Stand back 2 6 ft (' 4:9 ft) to see the mural best.
Hence, the smallest possible volume of a right circular
cone which can contain sphere of radius R is
#
$
4R4
2
8
V D #R
D #R3 cubic units.
3
2R2 " R2
3
45. If x cars are loaded, the total time for the trip is
10
x
T Dt C1C
1;000
1;000 t
where x D f .t / D !t
:
e Ct
2
We can minimize the average time per car (or, equivalently, maximize the number of cars per hour). The average time (in hours) per car is
AD
T
e !t C t
e !t C t
1
D
C
C
x
1;000
1;000t
1;000
(
#
$
)
! !t
"
1
1
D
e Ct 1C
C1 :
1;000
t
This expression approaches 1 as t ! 0C or t ! 1. For
a minimum we should look for a positive critical point.
Thus we want
Fig. 4.8-46
47.
Let r be the radius of the circular arc and & be the angle
shown in the left diagram below. Thus,
(
#
$
)
! !t
"
!
" 1
1
1
"e C 1 1 C
" e !t C t 2 ;
0D
1;000
t
t
158
"
˛
x
2r& D 100
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rD
50
:
&
lOMoARcPSD|6566483
INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 4.8 (PAGE 267)
r !
8
2#R3
p
#
D
3
9 3
y
" "
r
yDtan x
The largest cone has volume V
yDx
wall
!=2
cu. units.
!
"
fence
2! r
r
R
Fig. 4.8.47(a)
R
Fig. 4.8.47(b)
h
R
"
The area of the enclosure is
2&
# r 2 " .r cos & /.r sin & /
AD
2#
502 502 sin 2&
D
"
& # &2 2 $
1 sin 2&
D 502
"
&
2& 2
for 0 < & $ #. Note that A ! 1 as & ! 0C, and
for & D # we are surrounding the entire enclosure with
fence (a circle) and not using the wall at all. Evidently
this would not produce the greatest enclosure area, so the
maximum area must correspond to a critical point of A:
#
$
dA
2& 2 .2 cos 2& / " sin 2&.4& /
1
0D
D 502 " 2 "
d&
&
4& 4
1
cos 2&
sin 2&
,
C
D
&2
&2
&3
2
,
2& cos & D 2 sin & cos &
,
cos & D 0 or tan & D &:
Observe that tan & D & has no solutions in .0; #!. (The
graphs of y D tan & and y D & cross at & D 0 but
nowhere else between 0 and #.) Thus, the greatest enclo#
sure area must correspond to cos & D 0, that is, to & D .
2
The largest enclosure is thus semicircular, and has area
5000 2
2
.50/2 D
m .
#
#
48. Let the cone have radius r and height h.
Let sector of angle & from disk be used.
R
Then 2# r D R& so r D
&.
r2#
p
Rp 2
R2 & 2
Also h D R2 " r 2 D R2 "
D
4# " & 2
4# 2
2#
The cone has volume
# r2h
# R2 2 R p 2
4# " &
&
V D
D
3
3 4# 2 2#
p
R3
D
f .& / where f .& / D & 2 4# 2 " & 2 .0 $ & $ 2#/
24# 2
V .0/ D V .2#/ D 0 so maximum V must occur at a
critical point. For CP:
p
&3
0 D f 0 .& / D 2& 4# 2 " & 2 " p
4# 2 " & 2
8
) 2.4# 2 " & 2 / D & 2
) & 2 D # 2:
3
R"
Fig. 4.8-48
49.
Let the various distances be as labelled in the diagram.
h
a!x
y
a
x
L
y!h
h
Fig. 4.8-49
From the geometry of the various triangles in the diagram
we have
x 2 D h2 C .a " x/2 ) h2 D 2ax " a2
y 2 D a2 C .y " h/2 ) h2 D 2hy " a2
hence hy D ax. Then
a2 x 2
h2
2ax 3
a2 x 2
2
D
Dx C
2
2ax " a
2ax " a2
L2 D x 2 C y 2 D x 2 C
a
a
< x $ a. Clearly, L ! 1 as x ! C, and
2 p
2
L.a/ D 2a. For critical points of L2 :
for
0D
d.L2 /
.2ax " a2 /.6ax 2 / " .2ax 3 /.2a/
D
dx
.2ax " a2 /2
2 2
2a x .4x " 3a/
D
:
.2ax " a2 /2
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SECTION 4.8 (PAGE 267)
ADAMS and ESSEX: CALCULUS 9
#
)
a
3a
The only critical point in
; a is x D
. Since
2
4
p
# $
3 3a
3a
D
< L.a/, therefore the least possible
L
4
4
p
3 3a
cm.
length for the fold is
4
Section 4.9 Linear Approximations
(page 274)
1.
2
0
so that "x D "1=100. The edge length must decrease by
about 0:01 cm in to decrease the volume by 12 cm3 .
13.
The circumference C and radius r of the orbit are linked
by C D 2# r. Thus "C D 2# "r. If "r D "10 mi then
"C ' 2# "r D 20#. The circumference of the orbit will
decrease by about 20# ' 62:8 mi if the radius decreases
by 10 mi. Note that the answer does not depend on the
actual radius of the orbit.
14.
a D gŒR=.R C h/!2 implies that
0
f .x/ D x , f .x/ D 2x, f .3/ D 9, f .3/ D 6.
Linearization at x D 3: L.x/ D 9 C 6.x " 3/.
2. f .x/ D x !3 , f 0 .x/ D "3x !4 , f .2/ D 1=8,
f 0 .2/ D "3=16.
3
.x " 2/.
Linearization at x D 2: L.x/ D 81 " 16
p
p
3. f .x/ D 4 " x, f 0 .x/ D "1=.2 4 " x/, f .0/ D 2,
f 0 .0/ D "1=4.
Linearization at x D 0: L.x/ D 2 " 41 x.
p
p
4. f .x/ D 3 C x 2 , f 0 .x/ D x= 3 C x 2 , f .1/ D 2,
0
f .1/ D 1=2.
Linearization at x D 1: L.x/ D 2 C 21 .x " 1/.
"a '
If h D 0 and "h D 10 mi, then
"a ' "
15.
5. f .x/ D .1 C x/!2 , f 0 .x/ D "2.1 C x/!3 , f .2/ D 1=9,
f 0 .2/ D "2=27.
2
.x " 2/.
Linearization at x D 2: L.x/ D 91 " 27
6. f .x/ D x !1=2 , f 0 .x/ D ."1=2/x !3=2 , f .4/ D 1=2,
f 0 .4/ D "1=16.
1
Linearization at x D 4: L.x/ D 21 " 16
.x " 4/.
7.
f .x/ D sin x, f 0 .x/ D cos x, f .#/ D 0, f 0 .#/ D "1.
Linearization at x D #: L.x/ D ".x " #/.
0
8. f .x/ D cos.2x/,
p f .x/ D "2 sin.2x/, f .#=3/ D "1=2,
0
f .#=3/ D " 3.
p !
"
Linearization at x D #=3: L.x/ D " 12 " 3 x " !3 .
2
9. f .x/ D sinp
x, f 0 .x/ D 2 sin x cos x, f .#=6/ D 1=4,
0
f .#=6/ D 3=2.
p
!
"
Linearization at x D #=6: L.x/ D 14 C . 3=2/ x " !6 .
10. f .x/ D tan x, f 0 .x/ D sec2 x, f .#=4/ D!1, f 0 .#=4/
D 2.
"
Linearization at x D #=4: L.x/ D 1 C 2 x " !4 .
11.
If A and x are the area and side length of the square, then
A D x 2 . If x D 10 cm and "x D 0:4 cm, then
dA
"x D 2x "x D 20.0:4/ D 8:
"A '
dx
The area increases by about 8 cm2 .
12. If V and x are the volume and side length of the cube,
then V D x 3 . If x D 20 cm and "V D "12 cm3 , then
"12 D "V '
160
da
"2
"h:
"h D gR2
dh
.R C h/3
20 & 32
20g
D"
' 0:16 ft/s2 :
R
3960
1
1
f .x/ D x 1=2 f 0 .x/ D x !1=2 f 00 .x/ D " x !3=2
2
4
p
50 D f .50/ ' f .49/ C f 0 .49/.50 " 49/
99
1
D
' 7:071:
D7C
14
14
p
99
f 00 .x/ < 0 on Œ49; 50!, so error is negative: 50 <
14
1
1
1
jf 00 .x/j <
D
D
'
0:00073
D
k on
4 & 73
1372
4 & 493=2
.49; 50/.
1
k
D 0:00036. We have
Thus jerrorj $ .50 " 49/2 D
2
2744
p
99
99
1
"
$ 50 $
;
14 2744
14
16.
p
i.e., 7:071064 $ 50 $ 7:071429
p
Let f .x/ D x, then f 0 .x/ D 21 x !1=2 and
f 00 .x/ D " 41 x !3=2 . Hence,
p
47 D f .47/ ' f .49/ C f 0 .49/.47 " 49/
# $
1
48
D7C
."2/ D
' 6:8571429:
14
7
Clearly, if x # 36, then
jf 00 .x/j $
1
1
D
D K:
4 & 63
864
Since f 00 .x/ < 0, f is concave down. Therefore, the error
p
48
< 0 and
E D 47 "
7
dV
"x D 3x 2 "x D 1; 200 "x;
dx
jEj <
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1
.47 " 49/2 D
:
2
432
lOMoARcPSD|6566483
INSTRUCTOR’S SOLUTIONS MANUAL
Thus,
SECTION 4.9 (PAGE 274)
f 00 .0/ < 0 on Œ45ı ; 46ı ! so
p
48
1
48
"
< 47 <
7
432 p
7
6:8548 < 47 < 6:8572:
17.
1 % # &2
' 0:0001:
jErrorj < p
2 2 180
3
1
f .x/ D x 1=4 ; f 0 .x/ D x !3=4 ; f 00 .x/ D " x !7=4
4
16
p
4
85 D f .85/ ' f .81/ C f 0 .81/.85 " 81/
1
82
4
D3C
D
' 3:037:
D 3C
4 & 27
27
27
p
82
f 00 .x/ < 0 on Œ81; 85! so error is negative: 4 85 <
.
27
3
1
jf 00 .x/j <
D
D k on Œ81; 85!.
16 & 37
11; 664
k
Thus jErrorj $ .85 " 81/2 D 0:00069.
2
We have
#
$
#
# &
1
#2
1 %
"
< cos 46ı < p 1 "
p 1"
2
180 2 & 180
180
2
2
i.e., 0:694658 $ cos 46ı < 0:694765.
20.
1
#
, then jf 00 .x/j $ p . Since f 00 .x/ < 0 on
4
2
0 < x $ 90ı , f is concave down. Therefore, the error E
is negative and
p
1
82
82
4
"
< 85 <
;
27 1458
27
or 3:036351 $
p
4
If x $
85 $ 3:037037
1 % # &2
D 0:0038772:
jEj < p
2 2 30
1
1
2
18. Let f .x/ D , then f 0 .x/ D " 2 and f 00 .x/ D 3 .
x
x
x
Hence,
Thus,
1
D f .2:003/ ' f .2/ C f 0 .2/.0:003/
2:003
# $
1
1
.0:003/ D 0:49925:
D C "
2
4
If x # 2, then jf 00 .x/j $ 82 D 14 . Since f 00 .x/ > 0 for
x > 0, f is concave up. Therefore, the error
0:5906900 " 0:0038772 < sin
0:5868128 < sin
21.
jEj <
5&
%#
5
< 0:5906900
< 0:5906900:
Let f .x/ D sin x, then f 0 .x/ D cos x and
f 00 .x/ D " sin x. The linearization at x D # gives:
Since f 00 .x/ < 0 between 3:14 and #, the error E in the
above approximation is negative: sin.3:14/ < 0:001592654.
For 3:14 $ t $ #, we have
1
.0:003/2 D 0:000001125:
8
jf 00 .t /j D sin t $ sin.3:14/ < 0:001592654:
Thus,
Thus the error satisfies
1
< 0:49925 C 0:000001125
2:003
1
0:49925 <
< 0:499251125:
2:003
jEj $
0:49925 <
f .x/ D cos x; f 0 .x/ D " sin x; f 00 .x/ D " cos x
%#
# &
C
cos 46ı D cos
4
180
%# & % # &
#
' cos " sin
4
4
180
1 %
# &
D p 1"
' 0:694765:
180
2
0:001592654
.3:14 " #/2 < 0:000000002:
2
Therefore 0:001592652 < sin.3:14/ < 0:001592654.
22.
19.
%# &
sin.3:14/ ' sin #Ccos #.3:14"#/ D #"3:14 ' 0:001592654:
1
ED
" 0:49925 > 0
2:003
and
Let f .x/ D sin x, then f 0 .x/ D cos x and
f 00 .x/ D " sin x. Hence,
%#
%# &
%# &% # &
%# &
#&
Df
C
'f
Cf0
sin
5
6p 30
6
6
30
3%# &
1
' 0:5906900:
D C
2
2 30
Let f .x/ D sin x, then f 0 .x/ D cos x and
f 00 .x/ D " sin x. The linearization at x D 30ı D #=6
gives
!
"
!
sin.33ı / D sin !6 C 60
# %# &
#
' sin C cos
6p
6 60
3%# &
1
' 0:545345:
D C
2
2 60
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SECTION 4.9 (PAGE 274)
ADAMS and ESSEX: CALCULUS 9
Since f 00 .x/ < 0 between 30ı and 33ı , the error E in the
above approximation is negative: sin.33ı / < 0:545345.
For 30ı $ t $ 33ı , we have
for 47 $ x $ 49. Thus, on that interval, M $ f 00 .x/ $ N ,
where M D "0:000776 and N D "0:000729. By Corollary C,
jf 00 .t /j D sin t $ sin.33ı / < 0:545345:
N
M
.47 " 49/2 $ f .47/ $ L.47/ C .47 " 49/2
2p
2
6:855591 $ 47 $ 6:855685:
L.47/ C
Thus the error satisfies
jEj $
0:545345 % # &2
< 0:000747:
2
60
Using
p the midpoint of this interval as a new approximation
for 47 ensures that the error is no greater than half the
length of the interval:
Therefore
p
ı
0:545345 " 0:000747 < sin.33 / < 0:545345
0:544598 < sin.33ı / < 0:545345:
25.
23. From the solution to Exercise 15, the linearization to
f .x/ D x 1=2 at x D 49 has value at x D 50 given
by
"3
"3
"3
$
$ f 00 .x/ $
16.3/7
16.3:037037/7
16.851=4 /7
"1
"1
"1
$ f 00 .x/ $ p
$
3
3
4.7/
4.7:071429/3
4. 50/
for 81 $ x $ 85. Thus, on that interval, M $ f 00 .x/ $ N ,
where M D "0:000086 and N D "0:000079. By Corollary C,
for 49 $ x $ 50. Thus, on that interval, M $ f 00 .x/ $ N ,
where M D "0:000729 and N D "0:000707. By Corollary C,
M
N
.85 " 81/2 $ f .85/ $ L.85/ C .85 " 81/2
2
2
3:036351 $ 851=4 $ 3:036405:
M
N
.50 " 49/2 $ f .50/ $ L.50/ C .50 " 49/2
2p
2
7:071064 $ 50 $ 7:071075:
L.85/ C
L.50/ C
Using the midpoint of this interval as a new approximation
for 851=4 ensures that the error is no greater than half the
length of the interval:
Using
p the midpoint of this interval as a new approximation
for 50 ensures that the error is no greater than half the
length of the interval:
24.
L.47/ D f .49/ C f 0 .49/.47 " 49/ ' 6:8571429:
p
Also, 6:8548 $ p 47 $ 6:8572, and, since
f 00 .x/ D "1=.4. x/3 /,
"1
"1
"1
$ p
$ f 00 .x/ $
3
3
4.6:8548/
4.7/3
4. 47/
162
851=4 ' 3:036378;
jerrorj $ 0:000006:
From the solution to Exercise 16, the linearization to
f .x/ D x 1=2 at x D 49 has value at x D 47 given
by
From the solution to Exercise 17, the linearization to
f .x/ D x 1=4 at x D 81 has value at x D 85 given
by
Also, 3:036351 $ 851=4 $ 3:037037, and, since
f 00 .x/ D "3=.16.x 1=4 /7 /,
p
Also, 7:071064 $
p 3 50 $ 7:071429, and, since
00
f .x/ D "1=.4. x/ /,
50 ' 7:071070;
jerrorj $ 0:000047:
L.85/ D f .81/ C f 0 .81/.85 " 81/ ' 3:037037:
L.50/ D f .49/ C f 0 .49/.50 " 49/ ' 7:071429:
p
47 ' 6:855638;
26.
jerrorj $ 0:000028:
From the solution to Exercise 22, the linearization to
f .x/ D sin x at x D 30ı D #=6 has value at
x D 33ı D #=6 C #=60 given by
L.33ı / D f .#=6/ C f 0 .#=6/.#=60/ ' 0:545345:
Also, 0:544597 $ sin.33ı / $ 5:545345, and, since
f 00 .x/ D " sin x,
" sin.33ı / $ f 00 .x/ $ " sin.30ı /
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 4.10
If j& j $ 17ı D 17#=180, then
for 30ı $ x $ 33ı . Thus, on that interval,
M $ f 00 .x/ $ N , where M D "0:545345 and N D "0:5.
By Corollary C,
M
N
L.33ı / C
.#=60/2 $ sin.33ı / $ L.33ı / C .#=60/2
2
2
0:544597 $ sin.33ı / $ 0:544660:
jE.& /j
1
$
j& j
2
27.
f .2/ D 4,
f 0 .2/ D "1,
31.
jerrorj $ 0:000031:
0 $ f 00 .x/ $
1
if x > 0.
x
1.
2.
Thus L.3/ D 3. Also, since 1=.2x/ $ f 00 .x/ $ 1=x for
x > 0, we have for 2 $ x $ 3, .1=6/ $ f 00 .x/ $ .1=2/.
Thus
3.
29. The linearization of g.x/ at x D 2 is
L.x/ D g.2/ C g 0 .2/.x " 2/ D 1 C 2.x " 2/:
Thus L.1:8/ D 0:6.
If jg 00 .x/j $ 1 C .x " 2/2 for x > 0, then
jg 00 .x/j < 1 C ."0:2/2 D 1:04 for 1:8 $ x $ 2. Hence
1
g.1:8/ ' 0:6 with jerrorj < .1:04/.1:8 " 2/2 D 0:0208:
2
30. If f .& / D sin & , then f 0 .& / D cos & and f 00 .& / D " sin & .
Since f .0/ D 0 and f 0 .0/ D 1, the linearization of f at
& D 0 is L.& / D 0 C 1.& " 0/ D & .
If 0 $ t $ & , then f 00 .t / $ 0, so 0 $ sin & $ & .
If 0 # t # & , then f 00 .t / # 0, so 0 # sin & # & .
In either case, j sin t j $ j sin & j $ j& j if t is between 0 and
& . Thus the error E.& / in the approximation sin & ' &
satisfies
j& j 2
j& j3
jE.& / $
j& j D
:
2
2
' 0:044:
If f .x/ D e !x , then f .k/ .x/ D ."1/k e !x , so
f .k/ .0/ D ."1/k . Thus
x2 x3
x4
"
C
:
2Š
3Š
4Š
If f .x/ D cos x, then f 0 .x/
f 00 .x/ D " cos x, and f 000 .x/
ticular, f .#=4/ D f 000 .#=4/
p
f 0 .#=4/ D f 00 .#=4/ D "1= 2.
D " sin x,
D sinpx. In parD 1= 2 and
Thus
(
)
%
# &2 1 %
# &3
#& 1 %
1
"
x"
C
x"
:
P3 .x/ D p 1 " x "
4
2
4
6
4
2
# $
# $
1
1 1
2
.3 " 2/ $ f .3/ $ 3 C
.3 " 2/2 :
6
2 2
The best approximation for f .3/ is the midpoint of this
interval: f .3/ ' 3 61 .
$2
V D 34 # r 3 ) "V ' 4# r 2 "r
If r D 20:00 and "r D 0:20, then
"V ' 4#.20:00/2 .0:20/ ' 1005.
The volume has increased by about 1005 cm2 .
P4 .x/ D 1 " x C
28. The linearization of f .x/ at x D 2 is
1
3C
2
17#
180
Section 4.10 Taylor Polynomials
(page 283)
f .3/ ' f .2/ C f 0 .2/.3 " 2/ D 4 " 1 D 3.
f 00 .x/ # 0 ) error # 0 ) f .3/ # 3.
1
1
1
jf 00 .x/j $ $ if 2 $ x $ 3, so jErrorj $ .3 " 2/2 .
x
2
4
Thus 3 $ f .3/ $ 3 41
L.x/ D f .2/ C f 0 .2/.x " 2/ D 4 " .x " 2/:
#
Thus the percentage error is less than 5%.
Using the midpoint of this interval as a new approximation
for sin.33ı / ensures that the error is no greater than half
the length of the interval:
sin.33ı / ' 0:544629;
(PAGE 283)
f .x/ D ln x
1
f 0 .x/ D
x
"1
f 00 .x/ D 2
x
2
000
f .x/ D 3
x
"6
f .4/ .x/ D 4
x
Thus
f .2/ D ln 2
1
f 0 .2/ D
2
"1
00
f .2/ D
4
2
000
f .2/ D
8
"6
.4/
f .2/ D
16
1
1
1
1
P4 .x/ D ln 2C .x"2/" .x"2/2 C .x"2/3 " .x"2/4 :
2
8
24
64
4.
f .x/ D sec x
f 0 .x/ D sec x tan x
f 00 .x/ D 2 sec3 x " sec x
f 000 .x/ D .6 sec2 x " 1/ sec x tan x
f .0/ D 1
f 0 .0/ D 0
f 00 .0/ D 1
f 000 .0/ D 0
Thus P3 .x/ D 1 C .x 2 =2/.
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SECTION 4.10 (PAGE 283)
5.
f .x/ D x 1=2
1
f 0 .x/ D x !1=2
2
"1 !3=2
00
f .x/ D
x
4
3
f 000 .x/ D x !5=2
8
Thus
ADAMS and ESSEX: CALCULUS 9
Evidently f .2n/ .#=2/ D 0 and
f .2n!1/ .#=2/ D ."1/n 22n!1 . Thus
f .4/ D 2
1
f 0 .4/ D
4
"1
f 00 .4/ D
32
3
000
f .4/ D
256
%
# &3 25 %
# &5
# & 23 %
C
x"
"
x"
P2n!1 .x/ D "2 x "
2
3Š
2
5Š
2
&2n!1
2n!1 %
2
#
C ! ! ! C ."1/n
x"
:
.2n " 1/Š
2
1
1
1
.x " 4/3 :
P3 .x/ D 2 C .x " 4/ " .x " 4/2 C
4
64
512
9.
f .x/ D .1 " x/!1
6.
f .0/ D 1
f 0 .0/ D 1
f 00 .0/ D 2
f 000 .0/ D 3Š
::
:
f 0 .x/ D .1 " x/!2
f 00 .x/ D 2.1 " x/!3
000
!4
f .x/ D 3Š.1 " x/
::
:
f .n/ .x/ D nŠ.1 " x/!.nC1/
1
2Cx
"1
f 0 .x/ D
.2 C x/2
2Š
f 00 .x/ D
.2 C x/3
"3Š
f 000 .x/ D
.2 C x/4
::
:
."1/n nŠ
.n/
f .x/ D
.2 C x/nC1
f .n/ .0/ D nŠ
0 < Error $
1
3
"1
0
f .1/ D
9
2Š
00
f .1/ D
27
"3Š
000
f .1/ D 4
3
::
:
."1/n nŠ
f .n/ .1/ D
3nC1
f .1/ D
p
10. Since f .x/ D x, then f 0 .x/ D 12 x !1=2 ,
f 00 .x/ D " 41 x !3=2 and f 000 .x/ D 38 x !5=2 . Hence,
p
1 1
1
."1/n
" .x " 1/ C .x " 1/2 " ! ! ! C nC1 .x " 1/n :
3 9
27
3
f .x/ D sin.2x/
f 0 .x/ D 2 cos.2x/
8.
00
f .x/ D "2 sin.2x/
f 000 .x/ D "23 cos.2x/
f .4/ .x/ D 24 sin.2x/ D 24 f .x/
f
.5/
164
4
0
.x/ D 2 f .x/
::
:
f
.5/
.#=2/ D "2
::
:
f 000 .c/
.61 " 64/3 for
3Š
some c between 61 and 64. Clearly R2 < 0. If t # 49,
and in particular 61 $ t $ 64, then
The error is R2 D R2 .f I 64; 61/ D
Hence,
jR2 j $
f 000 .#=2/ D 23
f .4/ .#=2/ D 0
1
61 ' f .64/ C f 0 .64/.61 " 64/ C f 00 .64/.61 " 64/2
2
#
$
1
1
1
D 8 C ."3/ "
."3/2 ' 7:8103027:
16
2 2048
jf 000 .t /j $ 38 .49/!5=2 D 0:0000223 D K:
f .#=2/ D 0
f 0 .#=2/ D "2
f 00 .#=2/ D 0
2
5
< 0:000241:
81 & 256
Thus 2:07986 < 91=3 < 2:08010.
Thus
Pn .x/ D
f 00 .8/
.x " 8/2
2
1
1
D 2 C .x " 8/ "
.x " 8/2
12
9 & 32
1
1
"
' 2:07986
91=2 ' 2 C
12 288
f 000 .c/
10
1
for some c in
Error D
.9 " 8/3 D
3Š
27 & 6 X 8=3
Œ8; 9!.
For 8 $ c $ 9 we have c 8=3 # 88=3 D 28 D 256 so
Pn .x/ D 1 C x C x 2 C x 3 C ! ! ! C x n :
f .x/ D
f .x/ ' f .8/ C f 0 .8/.x " 8/ C
aD8W
Thus
7.
1
f .x/ D x 1=3 , f 0 .x/ D x !2=3 ,
3
10 !8=3
2
x
.
f 00 .x/ D " x !5=3 , f 000 .x/ D
9
27
K
j61 " 64j3 D 0:0001004:
3Š
Since R2 < 0, therefore,
5
7:8103027 " 0:0001004 <
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p
p
61 < 7:8103027
61 < 7:8103027:
lOMoARcPSD|6566483
INSTRUCTOR’S SOLUTIONS MANUAL
11.
SECTION 4.10
1
1
, f 0 .x/ D " 2 ,
x
x
"6
2
f 00 .x/ D 3 , f 000 .x/ D 4 .
x
x
2
a D 1 W f .x/ ' 1 " .x " 1/ C .x " 1/2
2
1
' 1 " .0:02/ C .0:02/2 D 0:9804.
1:02
f 000 .c/
1
Error D
.0:02/3 D " 4 .0:02/3 where
3Š
X
1 $ c $ 1:02.
1
" 0:9804 < 0,
Therefore, ".0:02/3 $
1:02
1
i.e., 0:980392 $
< 0:980400.
1:02
14.
f .x/ D
Since f .x/ D sin x, then f 0 .x/ D cos x, f 00 .x/ D " sin x
and f 000 .x/ D " cos x. Hence,
%#
#&
C
sin.47ı / D f
4
90
%# &
% # & % # &2
%# &% # & 1
'f
Cf0
C f 00
4
4
90
2
4
90
1 %# &
1 % # &2
1
" p
D p Cp
2
2 90
2 2 90
' 0:7313587:
f 000 .c/ % # &3
for some c between 45ı
3Š
90
ı
and 47 . Observe that R2 < 0. If 45ı $ t $ 47ı , then
The error is R2 D
12. Since f .x/ D tan!1 x, then
f 0 .x/ D
(PAGE 283)
1
jf 000 .t /j $ j " cos 45ı j D p D K:
2
2
"2x
"2 C 6x
1
; f 00 .x/ D
; f 000 .x/ D
:
1 C x2
.1 C x 2 /2
.1 C x 2 /3
Hence,
K % # &3
< 0:0000051:
3Š 90
Since R2 < 0, therefore
jR2 j $
Hence,
tan!1 .0:97/ ' f .1/ C f 0 .1/.0:97 " 1/ C 21 f 00 .1/.0:97 " 1/2
# $
#
1
1
D C ."0:03/ C "
."0:03/2
4
2
4
D 0:7701731:
f 000 .c/
."0:03/3 for some c between
The error is R2 D
3Š
0.97 and 1. Note that R2 < 0. If 0:97 $ t $ 1, then
jf 000 .t /j $ f 000 .1/ D
Hence,
jR2 j $
0:7313587 " 0:0000051 < sin.47ı / < 0:7313587
0:7313536 < sin.47ı / < 0:7313587:
15.
"2 C 6
< 0:5232 D K:
.1:97/3
f .x/ D sin x
f 0 .x/ D cos x
f 00 .x/ D " sin x
f 000 .x/ D " cos x
f .4/ .x/ D sin x
a D 0I n D 7:
x3
x5
x7
C0C
"0"
C R7 ;
3Š
5Š
7Š
5
7
3
x
x
x
C
"
C R7 .x/
Dx"
3Š
5Š
7Š
K
j0:97 " 1j3 < 0:0000024:
3Š
sin x D 0 C x " 0 "
Since R2 < 0,
0:7701731 " 0:0000024 < tan!1 .0:97/ < 0:7701731
16.
13. f .x/ D e x ,
f .k/ .x/ D e x for k D 1; 2; 3 : : :
x2
aD0W
f .x/ ' 1 C x C
2
.0:5/2
e !0:5 ' 1 " 0:5 C
D 0:625
2
f 000 .c/
ec
Error D
.0:5/3 D
."0:05/3 for some c between
6
6
"0:5 and 0. Thus
jErrorj <
.0:5/3
< 0:020834;
6
and "0:020833 < e !0:5 " 0:625 < 0, or
0:604 < e !0:5 < 0:625.
sin c 8
x for some c between 0 and x.
8Š
For f .x/ D cos x we have
where R7 .x/ D
0:7701707 < tan!1 .0:97/ < 0:7701731:
f
f 0 .x/ D " sin x
.4/
.x/ D cos x
f
f 00 .x/ D " cos x
.5/
.x/ D " sin x
f 000 .x/ D sin x
f .6/ .x/ D " cos x:
The Taylor’s Formula for f with a D 0 and n D 6 is
cos x D 1 "
x4 x6
x2
C
"
C R6 .f I 0; x/
2Š
4Š
6Š
where the Lagrange remainder R6 is given by
R6 D R6 .f I 0; x/ D
f .7/ .c/ 7
sin c 7
x D
x ;
7Š
7Š
for some c between 0 and x.
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SECTION 4.10 (PAGE 283)
ADAMS and ESSEX: CALCULUS 9
#
f .x/ D sin x
aD ;
nD4
4 &
%
1
#
1 1 %
# &2
1
"p
x"
sin x D p C p x "
4
4
2
2
2 2Š&
1 1 %
# &3
1 1 %
# 4
"p
x"
Cp
x"
C R4 .x/
4
4
2 3Š
2%4Š
&
5
#
1
where R4 .x/ D .cos c/ x "
5Š
4
#
and x.
for some c between
4
1
18. Given that f .x/ D
, then
1"x
17.
f 0 .x/ D
20.
Given that f .x/ D tan x, then
f 0 .x/ D sec2 x
f 00 .x/ D 2 sec2 x tan x
f .3/ .x/ D 6 sec4 x " 4 sec2 x
f .4/ .x/ D 8 tan x.3 sec4 x " sec2 x/:
Given that a D 0 and n D 3, the Taylor’s Formula is
tan x D f .0/ C f 0 .0/x C
2
1
; f 00 .x/ D
:
.1 " x/2
.1 " x/3
2 3
x C R3 .f I 0; x/
3Š
1
2
D x C x3 C x5:
3
15
DxC
In general,
f .n/ .x/ D
nŠ
:
.1 " x/.nC1/
The Lagrange remainder is
Since a D 0, f .n/ .0/ D nŠ. Hence, for n D 6, the Taylor’s
Formula is
1
D f .0/ C
1"x
6
X
nD1
f
R3 .f I 0; x/ D
.n/
.0/ n
x C R6 .f I 0; x/
nŠ
D 1 C x C x 2 C x 3 C x 4 C x 5 C x 6 C R6 .f I 0; x/:
f .7/ .c/ 7
x7
x D
7Š
.1 " c/8
21.
e 3x D e 3.xC1/ e !3
(
)
9
9
P3 .x/ D e !3 1 C 3.x C 1/ C .x C 1/2 C .x C 1/3 :
2
2
22.
For e u , P4 .u/ D 1 C u C
for some c between 0 and x.
19.
f .x/ D ln x
1
f 0 .x/ D
x
"1
f 00 .x/ D " 2
x
2Š
000
f .x/ D 3
x
"3Š
.4/
f .x/ D 4
x
4Š
.5/
f .x/ D 5
x
"5Š
.6/
f .x/ D 6
x
6Š
.7/
f
D 7
x
a D 1; n D 6
2
Then for e !x :
u3
u4
u2
C
C
. Let u D "x 2 .
2Š
3Š
4Š
P8 .x/ D 1 " x 2 C
23.
For sin2 x D
x8
x4 x6
"
C
:
2Š
3Š
4Š
&
1%
1 " cos.2x/ at x D 0, we have
2
(
#
$)
1
.2x/2
.2x/4
x4
P4 .x/ D
1" 1"
C
D x2 "
:
2
2Š
4Š
3
24.
%
&
sin x D sin # C .x " #/ D " sin.x " #/
P5 .x/ D ".x " #/ C
2Š
1
ln x D 0 C 1.x " 1/ " .x " 1/2 C .x " 1/3
2Š
3Š
3Š
4Š
5Š
" .x " 1/4 C .x " 1/5 " .x " 1/6 C R6 .x/
4Š
5Š
6Š
.x " 1/3 .x " 1/4
.x " 1/2
C
"
D .x " 1/ "
2
3
4
5
6
.x " 1/
.x " 1/
C
"
C R6 .x/
5
6
1
where R6 .x/ D 7 .x " 1/7 for some c between 1 and x.
7c
166
tan c.3 sec4 X " sec2 C / 4
f .4/ .c/ 4
x D
x
4Š
3
for some c between 0 and x.
The Langrange remainder is
R6 .f I 0; x/ D
f 00 .0/ 2 f 000 .0/ 3
x C
x C R3 .f I 0; x/
2Š
3Š
25.
.x " #/3 .x " #/5
"
3Š
5Š
1
at u D 0, P3 .u/ D 1 C u C u2 C u3 . Let
1"u
1
u D "2x 2 . Then for
at x D 0,
1 C 2x 2
For
P6 .x/ D 1 " 2x 2 C 4x 4 " 8x 6 :
26.
cos.3x " #/ D " cos.3x/
P8 .x/ D "1 C
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36 x 6 38 x 8
32 x 2 34 x 4
"
C
"
:
2Š
4Š
6Š
8Š
lOMoARcPSD|6566483
INSTRUCTOR’S SOLUTIONS MANUAL
27.
SECTION 4.10
Since x 3 D 0 C 0x C 0x 2 C x 3 C 0x 4 C ! ! ! we have
Pn .x/ D 0 if 0 $ n $ 2; Pn .x/ D x 3 if n # 3
32.
28.
29.
R2nC2 .f I 0; x/ D ."1/nC1
For tanh!1 x D
sin.1/ ' 1 "
1
1
1
C
"
' 0:84147
3Š
5Š 7Š
correct to five decimal places.
33.
x2 x3
x 2nC1
"
" !!! "
:
2
3
2n C 1
1
1
ln.1 C x/ " ln.1 " x/,
2
2
P2nC1 .x/ D x C
cos c
x 2nC3
.2n C 3/Š
for some c between 0 and x.
In order to use the formula to approximate sin.1/ correctly
to 5 decimal places, we need jR2nC2 .f I 0; 1/j < 0:000005.
Since j cos cj $ 1, it is sufficient to have
1=.2n C 3/Š < 0:000005. n D 3 will do since
1=9Š ' 0:000003. Thus
x3
x 2nC1
x2
C
" !!! C
:
P2nC1 .x/ D x "
2
3
2n C 1
P2nC1 .x/ D "x "
x5
x 2nC1
x3
C
" ! ! ! C ."1/n
C R2nC2 ;
3Š
5Š
.2n C 1/Š
where
30. For ln.1 C x/ at x D 0 we have
For ln.1 " x/ at x D 0 we have
In Taylor’s Formulas for f .x/ D sin x with a D 0, only
odd powers of x have nonzero coefficients. Accordingly
we can take terms up to order x 2nC1 but use the remainder after the next term 0x 2nC2 . The formula is
sin x D x "
1
sinh x D .e x " e !x /
#2
$
x2
x 2nC1
1
1CxC
C !!! C
P2nC1 .x/ D
2
2Š
.2n C 1/Š
#
$
1
x2
x 2nC1
"
1"xC
C !!! "
2
2Š
.2n C 1/Š
x3
x5
x 2nC1
Dx C
C
C !!! C
:
3Š
5Š
.2n C 1/Š
(PAGE 283)
f .x/ D .x " 1/2 ,
f 0 .x/ D 2.x " 1/, f 00 .x/ D 2.
2 2
f .x/ ' 1 " 2x C x D 1 " 2x C x 2
2
Error = 0
3
2
g.x/ D x C 2x C 3x C 4
Quadratic approx.: g.x/ ' 4 C 3x C 2x 2
Error D x 3
g 000 .c/ 3
Since g 000 .c/ D 6 D 3Š, error D
x
3Š
1
in the error formula for the quadratic
so that constant
3Š
approximation cannot be improved.
x5
x 2nC1
x3
C
C !!! C
:
3
5
2n C 1
34.
31. f .x/ D e !xn
e !x
if n is even
"e !x if n is odd
x2 x3
x5
e !x D 1 " x C
"
C ! ! ! C ."1/n
C Rn .x/
2Š
3Š
nŠ
nC1
X
where Rn .x/ D ."1/nC1
for some X between 0
.n C 1/Š
and x.
For x
D
1, we have
1
1
1
1
D1"1C
"
C ! ! ! C ."1/n C Rn .1/
e
2Š 3Š
nŠ
e !X x nC1
where Rn .1/ D ."1/nC1
for some X between
.n C 1/Š
"1 and 0.
1
. We want
Therefore, jRn .1/j <
.n C 1/Š
jRn .1/j < 0:000005 for 5 decimal places.
1
Choose n so that
< 0:000005. n D 8 will do
.n C 1/Š
since 1=9Š ' 0:0000027.
1
1
1
1
1
1
1
1
'
"
C
"
C
"
C
Thus
e
2Š 3Š
4Š 5Š
6Š 7Š
8Š
' 0:36788
(to 5 decimal places).
1 " x nC1 D .1 " x/.1 C x C x 2 C x 3 C ! ! ! C x n /. Thus
x nC1
1
D 1 C x C x2 C x3 C ! ! ! C xn C
:
1"x
1"x
f .n/ .x/ D
If jxj $ K < 1, then j1 " xj # 1 " K > 0, so
ˇ nC1 ˇ
ˇ
ˇx
1
nC1
ˇ
ˇ
j D O.x nC1 /
ˇ 1 " x ˇ $ 1 " K jx
as x ! 0. By Theorem 11, the nth-order
Maclaurin polynomial for 1=.1 " x/ must be
Pn .x/ D 1 C x C x 2 C x 3 C ! ! ! C x n .
35.
Differentiating
1
x nC1
D 1 C x C x2 C x3 C ! ! ! C xn C
1"x
1"x
with respect to x gives
1
n C 1 " nx n
D 1 C 2x C 3x 2 C ! ! ! C nx n!1 C
x :
.1 " x/2
.1 " x/2
Then replacing n with n C 1 gives
n C 2 " .n C 1/x nC1
1
D 1C2xC3x 2 C! ! !C.nC1/x n C
x
:
.1 " x/2
.1 " x/2
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SECTION 4.10 (PAGE 283)
ADAMS and ESSEX: CALCULUS 9
because jxj # 2p!1 . Clearly 2!t is the smallest value
when added to 1 that will not be discarded thus
If jxj $ K < 1, then j1 " xj # 1 " K > 0, and so
ˇ
ˇ
ˇ n C 2 " .n C 1/x nC1 ˇ
ˇ
ˇ $ n C 2 jx nC1 j D O.x nC1 /
x
ˇ
ˇ .1 " K/2
.1 " x/2
as x ! 0. By Theorem 11XXX, the nth-order
Maclaurin polynomial for 1=.1 " x/2 must be
Pn .x/ DD 1 C 2x C 3x 2 C ! ! ! C .n C 1/x n .
jF .x/ " xj $ $jxj:
4.
a) The result of question 3 suggests that computer produces a number C.1 C ˛/ for one expression and
C.1 C ˇ/ for the other. ˛ and ˇ will both be discarded positive numbers that are both less than $
by assumption. Thus for each expression, individually the computer returns C , however the difference
between these two expressions is .˛ " ˇ/C , where
j.˛ " ˇ/j < $. Thus the computer returns a value for
the difference which is less than $C but not necessarily equal to 0.
Section 4.11 Roundoff Error, Truncation
Error, and Computers (page 287)
1.
Since the normal rules of algebra (commutativity, associativity, distributivity, etc.) don’t apply to floating-point
calculations, we should not expect the plots of mathematically equivalent expressions to be the same in all cases.
b) Yes. In certain cases, internal errors may grow sufficiently that the computer returns a value different
from C , meaning that either or both of ˛ and ˇ may
not be entirely discarded. The algorithm evaluating
such expression may be in need of improvement in
such a case.
2. Since f .x/ " P4 .x/ D O.jxj5 / as x approaches 0, on
this very small interval centred at zero we would expect
the graph to be the horizontal line through the origin. Instead, there is a band of points having a peculiar structure.
The plot can vary between different implementations of
Maple on different operating systems, but some of the four
horizontal lines (actually envelop curves) proposed in the
exercise seem to provide natural boundaries for most of
the points.
c) Yes. In certain cases internal errors can be negligible
or zero, or they can cancel one another.
Consider
p
for example the expression. 1 - 12 , which will not
challenge machine precision at all.
3. As noted in section 4.7, a real number x can be represented in binary form as,
x D ˙0:d1 d2 : : : d t d tC1 d tC2 : : : & 2p
Review Exercises 4 (page 287)
1.
Since dr=dt D 2r=100 and V D .4=3/# r 3 , we have
4# 2 dr
2
6V
dV
D
3r
D 3V
D
:
dt
3
dt
100
100
where each of the base two digits di is either 0 or 1,
but d1 D 1, and p is the appropriate power of 2.
Consider the floating point number, y (having the
same sign as x) given by
Hence The volume is increasing at 6%/min.
2.
a) Since F must be continuous at r D R, we have
mgR2
D mkR;
R2
y D ˙0:d1 d2 : : : d t & 10p ;
which has only t significant binary digits. The distance
between x and y thus satisfies
jx " yj D 0:d tC1 d tC2 ! ! ! & 2
p!t
It follows that
jx " F .x/j $ 2p!t!1 $ 2!t jxj
168
g
:
R
b) The rate of change of F as r decreases from R is
$ˇˇ
#
d
ˇ
" .mkr/ ˇ
ˇ
dr
:
Since d tC1 may or may not be 1, this distance is less than
or equal to 2p!t . Thus F .x/, the nearest floating point
number to x, can be no further than half that distance
away, or
1
jx " F .x/j $ 2p!t D 2p!t!1 :
2
or k D
rDR
D "mk D "
mg
:
R
The rate of change of F as r increases from R is
#
$ˇ
d mgR2 ˇˇ
"
ˇ
ˇ
dr r 2
rDR
D"
2mgR2
mg
D "2
:
R3
R
Thus F decreases as r increases from R at twice the
rate at which it decreases as r decreases from R.
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INSTRUCTOR’S SOLUTIONS MANUAL
REVIEW EXERCISES 4 (PAGE 287)
Evidently P $ 0 if x $ 75, and P ! 0 as x ! 1. P
will therefore have a maximum value at a critical point in
.75; 1/. For CP:
3. 1=R D 1=R1 C 1=R2 . If R1 D 250 ohms and R2 D 1; 000
ohms, then 1=R D .1=250/ C .1=1; 000/ D 1=200,
so R D 200 ohms. If dR1 =dt D 100 ohms/min, then
1
" 2
R
1
2002
dR
1 dR1
1 dR2
D" 2
" 2
dt
R1 dt
R2 dt
dR
1
1
dR2
D
.100/ C
:
2
2
dt
250
1; 000 dt
a) If R remains constant, then dR=dt D 0, so
0D
x 2 " .x " 75/2x
dP
;
D 4:5 & 106
dx
x4
from which we obtain x D 150. She should charge $150
per bicycle and order N.150/ D 200 of them from the
manufacturer.
7.
dR2
1; 0002 & 100
D "1; 600:
D"
dt
2502
R
h
R2 is decreasing at 1,600 ohms/min.
R
b) If R is increasing at 10 ohms/min, then then
dR=dt D 10, and
dR2
D 1; 0002
dt
#
10
100
"
2002 2502
$
h!R
r
D "1; 350:
R2 is decreasing at 1,350 ohms/min.
Fig. R-4-7
Let r, h and V denote the radius, height, and volume of
the cone respectively. The volume of a cone is one-third
the base area times the height, so
4. If pV D 5:0T , then
dV
dT
dp
V Cp
D 5:0
:
dt
dt
dt
V D
1
# r 2 h:
3
From the small right-angled triangle in the figure,
a) If T D 400 K, d T =dt D 4 K/min, and V D 2:0 m3 ,
then d V =dt D 0, so dp=dt D 5:0.4/=2:0 D 10. The
pressure is increasing at 10 kPa/min.
b) If T D 400 K, d T =dt D 0, V D 2 m3 , and
d V =dt D 0:05 m3 /min, then p D 5:0.400/=2 D 1; 000
kPa, and 2 dp=dt C 1; 000.0:05/ D 0, so
dp=dt D "25. The pressure is decreasing at 25
kPa/min.
5. If x copies of the book are printed, the cost of printing
each book is
C D
10; 000
C 8 C 6:25 & 10!7 x 2 :
x
.h " R/2 C r 2 D R2 :
Thus r 2 D R2 " .h " R/2 and
&
&
#%
# %
2Rh2 " h3 :
V D V .h/ D h R2 " .h " R/2 D
3
3
The height of any inscribed cone cannot exceed the diameter of the sphere, so 0 $ h $ 2R. Being continuous, V .h/
must have a maximum value on this interval. Since V D 0
when h D 0 or h D 2R, and V > 0 if 0 < h < 2R, the
maximum value of V must occur at a critical point. (V
has no singular points.) For a critical point,
#
#
.4Rh " 3h2 / D h.4R " 3h/;
3
3
4R
:
h D 0 or h D
3
0 D V 0 .h/ D
Since C ! 1 as x ! 0C or x ! 1, C will be minimum at a critical point. For CP:
dC
10; 000
0D
C 12:5 & 10!7 x;
D"
dx
x2
so x 3 D 8 & 109 and x D 2 & 103 . 2,000 books should be
printed.
6. If she charges $x per bicycle, her total profit is $P , where
V 0 .h/ > 0 if 0 < h < 4R=3 and V 0 .h/ < 0 if
4R=3 < h < 2R. Hence h D 4R=3 does indeed give
the maximum value for V . The volume of the largest cone
can be inscribed in a sphere of radius R is
#
$
#
$
$ !
#
4R 3
4R
#
4R 2
"
D
2R
V
3
3
3
3
x " 75
:
P D .x " 75/N.x/ D 4:5 & 10
x2
6
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D
32
#R3 cubic units:
81
169
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REVIEW EXERCISES 4 (PAGE 287)
ADAMS and ESSEX: CALCULUS 9
Thus x D 10 or x D 100=3. The latter CP is not in the
interval Œ0; 25!, so the maximum occurs at x D 10. The
maximum volume of the box is V .10/ D 9; 000 cm3 .
8.
C
10. If x more trees are planted, the yield of apples will be
.x; C.x//
slope =
C.x/
= average cost
x
Y D .60 C x/.800 " 10x/
D 10.60 C x/.80 " x/
x
D 10.4; 800 C 20x " x 2 /:
Fig. R-4-8
This is a quadratic expression with graph opening downward; its maximum occurs at a CP:
a) For minimum C.x/=x, we need
0D
xC 0 .x/ " C.x/
d C.x/
D
;
dx x
x2
0D
so C 0 .x/ D C.x/=x; the marginal cost equals the
average cost.
b) The line from .0; 0/ to .x; C.x// has smallest slope at
a value of x which makes it tangent to the graph of
C.x/. Thus C 0 .x/ D C.x/=x, the slope of the line.
c) The line from .0; 0/ to .x; C.x// can be tangent to
the graph of C.x/ at more than one point. Not all
such points will provide a minimum value for the
average cost. (In the figure, one such line will make
the average cost maximum.)
dY
D 10.20 " 2x/ D 20.10 " x/:
dx
Thus 10 more trees should be planted to maximize the
yield.
11.
9.
y
side
flap
"
2 km
side
bottom
side
top
50 cm
Fig. R-4-11
side
flap
80 cm
Fig. R-4-9
It was shown in the solution to Exercise 41 in Section 3.2
that at time t s after launch, the tracking antenna rotates
upward at rate
d&
800t
D
D f .t /; say:
dt
4002 C t 4
If the edge of the cutout squares is x cm, then the volume
of the folded box is
Observe that f .0/ D 0 and f .t / ! 0 as t ! 1. For
critical points,
V .x/ D x.50 " 2x/.40 " x/
D 2x 3 " 130x 2 C 2; 000x;
and is valid for 0 $ x $ 25. Since V .0/ D V .25/ D 0, and
V .x/ > 0 if 0 < x < 25, the maximum will occur at a CP:
)
0 D V 0 .x/ D 6x 2 " 260x C 2; 000
D 2.3x 2 " 130x C 1; 000/
D 2.3x " 100/.x " 10/:
170
)
.4002 C t 4 / " 4t 4
.4002 C t 4 /2
4
3t D 4002 ;
or t ' 15:197:
0 D f 0 .t / D 800
(
The maximum rate at which the antenna must turn is
f .15:197/ ' 0:057 rad/s.
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INSTRUCTOR’S SOLUTIONS MANUAL
REVIEW EXERCISES 4 (PAGE 287)
At t D 0, we have x D 0 and y D 1; 000. Thus the
position of the ball at time t is given by
12. The narrowest hallway in which the table can be turned
horizontally through 180ı has width equal to twice the
greatest distance from the origin (the centre of the table)
to the curve x 2 C y 4 D 1=8 (the edge of the table). We
maximize the square of this distance, which we express as
a function of y:
S.y/ D x 2 C y 2 D y 2 C
1
" y4;
8
We can obtain the Cartesian equation for the path of the
cannon ball by solving the first equation for t and substituting into the second equation:
.0 $ y $ .1=8/1=4 /:
p
Note that S.0/ D 1=8 and S..1=8/1=4 / D 1= 8 > S.0/.
For CP:
0D
y D "16
dS
D 2y " 4y 3 D 2y.1 " 2y 2 /:
dy
13. Let the ball have radius r cm. Its weight is proportional
to the volume of metal it contains, so the condition of the
problem states that
"16
2x 2
1; 000
C x C 1; 000 D
:
2002
1 C .x=500/2
Graphing both sides of this equation suggests a solution
near x D 1; 900. Newton’s Method or a solve routine then
gives x ' 1; 873. The horizontal range is about 1,873 ft.
15.
1 4# 3
4# 3 4#
r "
.r " 2/3 D
r
3
3
2 3
r 3 " 12r 2 C 24r " 16 D 0:
The percentage error in the approximation
".g=L/ sin & ' ".g:L/& is
ˇ
ˇ
#
$
ˇ sin & " & ˇ
&
ˇ
ˇ
100 ˇ
D 100
"1 :
sin & ˇ
sin &
Graphing the left side of this latter equation with a graphics calculator shows a root between 9 and 10. A “solve
routine” or Newton’s Method then refines an initial guess
of, say, r D 9:5 to give r D 9:69464420373 cm for the
radius of the ball.
Since lim"!0 &=.sin & / D 1, the percentage error ! 0
as & ! 0. Also, &= sin & grows steadily larger as j& j increases from 0 towards #=2. Thus the maximum percentage error for j& j $ 20ı D #=9 will occur at & D #=9.
This maximum percentage error is
14.
y
100
trajectory
1;000
1C.x=500/2
dy
200
D "32t C p :
dt
2
$
#=9
" 1 ' 2:06%:
sin.#=9/
&
1%
1 " cos.2x/
2(
#
$)
1
22 x 2
24 x 4
26 x 6
D
1" 1"
C
"
C O.x 8 /
2
2Š
4Š
6Š
x4
2x 6
2
8
Dx "
C
C O.x /
3
45
3 sin2 x " 3x 2 C x 4
lim
x!0
x6
2
3x 2 " x 4 C x 6 C O.x 8 / " 3x 2 " x 4
15
D lim
x!0
x6
2
2
C O.x 2 / D
:
D lim
x!0 15
15
sin2 x D
17.
f .x/ D tan!1 x, f 0 .x/ D
Fig. R-4-14
If the origin is at sea level under the launch point, and
x.t / and y.t / are the horizontal and vertical coordinates
of the cannon ball’s position at time t s after it is fired,
then
d 2x
d 2y
D 0;
D "32:
2
dt
dt 2
p
At t D 0, we have dx=dt D dy=dt D 200= 2, so
#
16.
x
dx
200
D p ;
dt
2
2x 2
C x C 1; 000:
2002
The cannon ball strikes the ground when
The CPs are given by y D 0 (already considered),
p and
y 2 D 1=2, where S.y/ D 3=8. Since 3=8 > 1= 8, this is
the maximum
p value of S. The hallway must therefore be
at least 2 3=8 ' 1:225 m wide.
yD
200t
y D "16t 2 C p C 1; 000:
2
200t
xD p ;
2
f 000 .x/ D
6x 2 " 2
.
.1 C x 2 /3
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1
"2x
, f 00 .x/ D
,
1 C x2
.1 C x 2 /2
171
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REVIEW EXERCISES 4 (PAGE 287)
ADAMS and ESSEX: CALCULUS 9
x " 1 .x " 1/2
#
C
"
.
4
2
4
#
1
1
Thus tan!1 .1:1/ ' C
"
' 0:832898: On Œ1; 1:1!,
4
20 400
we have
This line passes through .1; 1/ if 1 D e a .2 " a/. A solve
routine gives a ' "1:1461932. The corresponding value
of e a is about 0.3178444. The car is at .a; e a /.
About x D 1, P2 .x/ D
jf 000 .x/j $
6.1:1/2 " 2
D 0:6575:
.1 C 1/3
Thus the error does not exceed
in absolute value.
Challenging Problems 4 (page 289)
0:6575
.1:1 " 1/3 ' :00011
3Š
1.
18. The second approximation x1 is the x-intercept of the
tangent to y D f .x/ at x D x0 D 2; it is the x-intercept
of the line 2y D 10x " 19. Thus x1 D 19=10 D 1:9.
dV
D kx 2 .V0 " V /.
dt
a) If V D x 3 , then 3x 2
k
dx
D .V0 " x 3 /:
dt
3
19.
y
y D cos x
b) The rate of growth of the edge is .k=3/.V0 " x 3 /,
1=3
which is positive if 0 $ x < x0 D V0 . The time
derivative of this rate is
y D .x " 1/2
"kx 2
x
1
y D cos x and y D .x " 1/2 intersect at x D 0 and at a
point x between x D 1 and x D #=2 ' 1:57. Starting
with an initial guess x0 D 1:3, and iterating the Newton’s
Method formula
20. The square of the distance from .2; 0/ to .x; ln x/ is
S.x/ D .x " 2/2 C .ln x/2 , for x > 0. Since S.x/ ! 1
as x ! 1 or x ! 0C, the minimum value of S.x/ will
occur at a critical point. For CP:
$
#
ln x
:
0 D S 0 .x/ D 2 x " 2 C
x
We solve this equation using a TI-85 solve routine;
x ' 1:6895797.
The minimum distance from the origin to
p
y D e x is S.x/ ' 0:6094586.
If the car is at .a; e a /, then its headlight beam lies along
the tangent line to y D e x there, namely
y D e a C e a .x " a/ D e a .1 C x " a/:
172
c) Initially, x grows at rate kV0 =3. The rate of growth
of x will be half of this if
k
kV0
.V0 " x 3 / D
;
3
6
.xn " 1/2 " cos xn
;
2.xn " 1/ C sin xn
we get x4 D x5 D 1:40556363276. To 10 decimal places
the two roots of the equation are x D 0 (exact), and
x D 1:4055636328.
21.
dx
k2
D " x 2 .V0 " x 3 / < 0
dt
3
for 0 < x < x0 . Thus the edge length is increasing at
a decreasing rate.
Fig. R-4-19
xnC1 D xn "
dx
dV
D
D kx 2 .V0 " x 3 /, so
dt
dt
that is, if x D .V0 =2/1=3 . Then V D V0 =2.
2.
dy
D ky.
dt
kt
Thus, y D C e . Given that at t D 0, y D 4, then
4 D y.0/ D C . Also given that at t D 10, y D 2, thus,
Let the speed of the tank be v where v D
1
ln 2:
2 D y.10/ D 4e 10k ) k D " 10
1
dy
1
D ." ln 2/y: The
dt
10
dy
1
slope of the curve xy D 1 is m D
D " . Thus, the
dx # x 2 $
1
equation of the tangent line at the point
; y0 is
y0
Hence, y D 4e .! 10 ln 2/t and v D
y D y0 " #
#
$
1
;
$2 x "
y0
1
y0
1
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INSTRUCTOR’S SOLUTIONS MANUAL
CHALLENGING PROBLEMS 4 (PAGE 289)
y
3.
yD
1
x
&
.1=y0 ;y0 /
y
x
x
b) Starting with x0 D 20, we iterate xnC1 D f .xn /. The
first three iterations give
Fig. C-4-2
x1 ' 11:03;
2
and the y-intercept is
y0
y D 2y0 . Let & be the angle between the gun and the
y-axis. We have
Hence, the x-intercept is x D
x
y
d&
sec2 &
dt
tan & D
)
#
$
2
1
4
y0
D
D 2 D 2
2y0
y
y0
"8 dy
D 3
:
y dt
a) If q D 0:99, the number of tests required is
T D N ..1=x/ C 1 " 0:99x /. T is a decreasing function for small values of x because the term 1=x dominates. It is increasing for large x because "0:99x
dominates. Thus T will have a minimum value at a
critical point, provided N is sufficiently large that the
CP is in .0; N /. For CP:
#
$
1
dT
x
D N " 2 " 0:99 ln.0:99/
0D
dx
x
!x
.0:99/
x2 D
" ln.0:99/
.0:99/!x=2
xD p
D f .x/; say:
" ln.0:99/
4.
x3 ' 10:51:
This suggests the CP is near 10.5. Since x must
be an integer, we test x D 10 and x D 11:
T .10/ ' 0:19562 and T .11/ ' 0:19557. The minimum cost should arise by using groups of 11 individuals.
p
P D 2# L=g D 2#L1=2 g !1=2 .
a) If L remains constant, then
dP
"g D "#L1=2 g !3=2 "g
dg
"P
"#L1=2 g !3=2
1 "g
'
"g D "
:
P
2 g
2#L1=2 g !1=2
"P '
Now
sec2 & D 1 C tan2 & D 1 C
x2 ' 10:54;
If g increases by 1%, then "g=g D 1=100, and
"P =P D "1=200. Thus P decreases by 0:5%.
y 4 C 16
16
D
;
4
y
y4
b) If g remains constant, then
so
dP
"L D #L!1=2 g !1=2 "L
dL
1 "L
"P
#L!1=2 g !1=2
"L D
'
:
P
2 L
2#L1=2 g !1=2
d&
8y
dy
8ky 2
D" 4
D" 4
:
dt
y C 16 dt
y C 16
The maximum value of
y2
occurs at a critical point:
y 4 C 16
.y 4 C 16/2y " y 2 .4y 3 /
.y 4 C 16/2
5
2y D 32y;
0D
,
or y D 2. Therefore the maximum rate of rotation of the
gun turret must be
"8k
"P '
22
1
D "k D
ln 2 ' 0:0693 rad/m,
24 C 16
10
and occurs when your tank is 2 km from the origin.
If L increases by 2%, then "L=L D 2=100, and
"P =P D 1=100. Thus P increases by 1%.
5.
dV
p
D "k y; V D Ay.
dt
dy
dV
dy
kp
p
a) A
D
D "k y, so
D"
y.
dt
dt
dt
A
#
$2
kt
p
y0 "
, then y.0/ D y0 , and
b) If y.t / D
2A
#
$#
$
k
dy
kt
p
D2
"
y0 "
dt
2A
2A
kp
D"
y.t /:
A
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CHALLENGING PROBLEMS 4 (PAGE 289)
ADAMS and ESSEX: CALCULUS 9
A D 0 if x D 0 or x D P =2 and A > 0 between these
values of x. The maximum area will therefore occur at a
critical point.
Thus the given expression does solve the initial-value
problem for y.
kT
p
p
c) If y.T / D 0, then
D y0 , so k D 2A y0 =T .
2A
Thus
y.t / D
#
p
y0 "
dA
P .P " x/.P " 4x/ " x.P " 2x/."1/
D
dx
4
.P " x/2
2
2
0 D P " 5P x C 4x C P x " 2x 2
0D
$
#
$
p
2A y0 t 2
t 2
D y0 1 "
:
2AT
T
2x 2 " 4P x C P 2 D 0:
This quadratic has two roots, but the only one in Œ0; P =2!
is
p
$
#
4P " 16P 2 " 8P 2
1
xD
DP 1" p :
4
2
&2
%
This value of x gives A.x/ D 21 P 2 1 " p1
un2 for
2
the maximum area of the triangle. (Note that the maximal
triangle is isosceles, as we might have guessed.)
d) Half the liquid drains out in time t1 , where
$
#
y0
t1 2
D
:
y0 1 "
T
2
p
Thus t1 D T .1 " .1= 2//.
6. If the depth of liquid in the tank at time t is y.t /, then
the surface of the liquid has radius r.t / D Ry.t /=H , and
the volume of liquid in the tank at that time is
#
V .t / D
3
#
Ry.t /
H
$2
y.t / D
8.
The slope of y D x 3 C ax 2 C bx C c is
y 0 D 3x 2 C 2ax C b;
&3
#R2 %
y.t
/
:
3H 2
which ! 1 as x ! ˙1. The quadratic expression y 0
takes each of its values at two different points except its
minimum value, which is achieved only at one point given
by y 00 D 6x C 2a D 0. Thus the tangent to the cubic
at x D "a=3 is not parallel to any other tangent. This
tangent has equation
p
By Torricelli’s law, d V =dt D "k y. Thus
dV
#R2 2 dy
p
3y
D
D "k y;
3H 2
dt
dt
a3
a3
ab
C
"
Cc
27
9
3 $
# 2
%
2
a
2a
a&
C
"
Cb xC
3
3
3
#
$
a3
a2
D"
CcC b"
x:
27
3
yD"
or, dy=dt D "k1 y !3=2 , where k1 D kH 2 =.#R2 /.
$
#
t 2=5
, then y.0/ D y0 , y.T / D 0, and
If y.t / D y0 1 "
T
#
$
#
$
2
t !3=5
1
dy
D y0 1 "
"
D "k1 y !3=2 ;
dt
5
T
T
where k1 D 2y0 =.5T /. Thus this function y.t / satisfies
the conditions of the problem.
7.
9.
P
A
If
p the triangle has legs x and y and hypotenuse
x 2 C y 2 , then
P DxCyC
p
x2 C y2
.P " x " y/2 D x 2 C y 2
Fig. C-4-9
a) The total resistance of path AP C is
kjAP j
kjP C j
C
r12
r22
$
#
h csc &
L " h cot &
C
:
Dk
r12
r22
RD
y.2P " 2x/ D P 2 " 2P x
P .P " 2x/
:
yD
2.P " x/
The area of the triangle is
h
C
P 2 C x 2 C y 2 C 2xy " 2P x " 2P y D x 2 C y 2
We have
$
# 2
dR
csc & cot &
csc &
"
;
D kh
d&
r12
r22
xy
P P x " 2x 2
AD
D
:
2
4 P "x
174
B
&
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INSTRUCTOR’S SOLUTIONS MANUAL
CHALLENGING PROBLEMS 4 (PAGE 289)
r2
csc &
D 12 , that
cot &
r2
is, cos & D .r2 =r1 /2 or & D cos!1 ..r2 =r1 /2 /.
This CP will give the minimum resistance if it
is in the interval of possible values of & , namely
Œtan!1 .h=L/; #=2!; otherwise the minimum will occur
for P D A. Thus, for large L, P should be chosen to
make cos & D .r2 =r1 /2 .
For a given depth y, R will be maximum if h.y " h/
is maximum. This occurs at the critical point
h D y=2 of the quadratic Q.h/ D h.y " h/.
so the CP of R is given by
c) By the result of part (c) of Problem 3 (with y replaced by y " h, the height of the surface of the water
above the drain in the current problem), we have
$
#
t 2
; for 0 $ t $ T .
y.t / " h D .y0 " h/ 1 "
T
b) This is the same problem as that in (a) except that
r1 and r2 are replaced with r12 and r22 , respectively.
Thus the minimum resistance corresponds to choosing
P so that cos & D .r2 =r1 /4 . This puts P closer to
B than it was in part (a), which is reasonable since
the resistance ratio between the thin and thick pipes is
greater than for the wires in part (a).
As shown above, the range of the spurt at time t is
s r
%
&
2
R.t / D k
h y.t / " h :
g
Since R D R0 when y D y0 , we have
R0
:
p
2
h.y0 " h/
g
r %
&
#
$
h y.t / " h
t
D R0 1 "
.
Therefore R.t / D R0 p
T
h.y0 " h/
kD s
10.
11.
y
x
h
25!x
25!2x
25!2x
R
x
25 cm
x
y
Fig. C-4-10
25 cm
a) Let the origin be at the point on the table directly under the hole. If a water particle leaves the tank with
horizontal velocity v, then its position .X.t /; Y.t //, t
seconds later, is given by
d 2X
D0
dt 2
dX
Dv
dt
X D vt
d 2Y
D "g
dt 2
dY
D "gt
dt
1
Y D " gt 2 C h:
2
The range R of the particle (i.e., of the spurt)pis the
value of X when Y D 0, that is, at time t D 2h=g.
p
Thus R D v 2h=g.
p
b) Since v D k y " h, the range R is a function of y,
the depth of water in the tank.
RDk
s
2p
h.y " h/:
g
Fig. C-4-11
Note that the vertical back wall of the dustpan is perpendicular to the plane of the top of the pan, not the bottom.
The volume of the pan is made up of three parts:
a triangular prism (the centre part) having height
x, width 25 " 2x, and depth y (all distances
2
2
2
in cm),
p where y C xp D .25 " x/ , and so
y D 625 " 50x D 5 25 " 2x, and
two triangular pyramids (one on each side) each
having height x and a right-triangular top with
dimensions x and y.
The volume of the pan is, therefore,
# $#
$
1
1
1
V D xy.25 " 2x/ C 2
xy x
2
3
2
#
$
1
2
D xy 25 " 2x C x
2
3
5 p
D x 25 " 2x.75 " 4x/ D V .x/:
6
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CHALLENGING PROBLEMS 4 (PAGE 289)
The appropriate values for x are 0 $ x $ 25=2. Note that
V .0/ D V .25=2/ D 0 and V .x/ > 0 in .0; 25=2/. The
maximum volume will therefore occur at a critical point:
0D
176
dV
25 4x 2 " 85x C 375
D"
p
dx
6
25 " 2x
ADAMS and ESSEX: CALCULUS 9
(after simplification). The quadratic in the numerator factors to .x " 15/.4x " 25/, so the CPs are x D 15 and
x D 25=4. Only x D 25=4 is in the required interval. The
maximum volume of the dustpan is V .25=4/ ' 921 cm3 .
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INSTRUCTOR’S SOLUTIONS MANUAL
CHAPTER 5.
SECTION 5.1 (PAGE 295)
INTEGRATION
16.
m
X
kD!5
Section 5.1 Sums and Sigma Notation
(page 295)
17.
n
X
iD1
1.
4
X
i 3 D 13 C 23 C 33 C 43
100
X
j
1
2
3
100
D C C C !!! C
j C1
2
3
4
101
iD1
2.
j D1
3.
n!1
X
."1/i
1
1
."1/n!1
D 1 " C " !!! C
i C1
2
3
n
i
2
3
n
20.
21.
n
X
23
24 25
."1/n 2n
."2/j
D " 2 C 2 " 2 C !!! C
2
.j " 2/
1
2
3
.n " 2/2
n
X
j2
1
4
9
n2
D
C
C
C
!
!
!
C
n3
n3
n3
n3
n3
5C6C7C8C9D
9
X
24.
iD5
9. 22 " 32 C 42 " 52 C ! ! ! " 992 D
99
X
25.
."1/i i 2
26.
13. 1 "
14.
15.
1
."1/
1
C " !!! C
4
9
n2
D
j D0
sin j D
100
X
iD1
ln m D ln 1 C ln 2 C ! ! ! C ln n D ln.nŠ/
e .nC1/=n " 1
e 1=n " 1
e i=n D
2 C 2 C ! ! ! C 2 (200 terms) equals 400
(
1 " x nC1
if x ¤ 1
1"x
nC1
if x D 1
(
1 C x 2nC1
if x ¤ "1
1 " x C x 2 " x 3 C ! ! ! C x 2n D
1Cx
2n C 1
if x D "1
2
n
1 C x C x C !!! C x D
Let f .x/ D 1 C x C x 2 C ! ! ! C x 100 D
Then
ix i!1
x 101 " 1
if x ¤ 1.
x"1
f 0 .x/ D 1 C 2x C 3x 2 C ! ! ! C 100x 99
iD1
D
xi
2n
X
."1/i x i
27.
iD0
n
X
iD1
1
2
3
4
n
C C C
C !!! C n D
2
4
8
16
2
99
X
100
X
#.# n " 1/
" 3n
# "1
.2i " i 2 / D 2nC1 " 2 " 61 n.n C 1/.2n C 1/
n
X
n
X
2.1; 000/.1; 001/
C 3; 000 D 1; 004; 000
2
d x 101 " 1
100x 101 " 101x 100 C 1
D
:
dx x " 1
.x " 1/2
iD0
12. 1 " x C x 2 " x 3 C ! ! ! C x 2n D
n!1
iD1
iD2
n
X
1 C x C x2 C x3 C ! ! ! C xn D
200
X
2
10. 1 C 2x C 3x 2 C 4x 3 C ! ! ! C 100x 99 D
11.
22.
n.n C 1/
n.n C 1/.2n C 7/
n.n C 1/.2n C 1/
C2
D
6
2
6
.2j C 3/ D
.# k " 3/ D
mD1
23.
i
8. 2 C 2 C 2 C ! ! ! C 2 (200 terms) equals
n
X
iD0
j D1
7.
n
X
iD1
j D3
6.
1;000
X
kD1
3 D 3 C 3 C 3 C !!!C 3
iD0
5.
19.
iD1
.i 2 C2i / D
j D1
n
X
iD1
4.
18.
mC6
X
1
1
D
2
k C1
..i " 6/2 C 1
D
i!1
."1/
i2
n
X
iD1
i
2i
22 " 32 C 42 " 52 C ! ! ! C 982 " 992
49
X
kD1
49
X
D"
28.
Œ.2k/2 " .2k C 1/2 ! D
49
X
Œ4k 2 " 4k 2 " 4k " 1!
kD1
49 & 50
Œ4k C 1! D "4
" 49 D "4; 949
2
kD1
Let s D
2
3
n
1
C C C ! ! ! C n : Then
2
4
8
2
sin.i " 1/
s
1
2
3
n
D C C
C ! ! ! C nC1 :
2
4
8
16
2
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SECTION 5.1 (PAGE 295)
ADAMS and ESSEX: CALCULUS 9
Subtracting these two sums, we get
1
1
1
1
n
s
D C C C ! ! ! C n " nC1
2
2
4
8
2
2
1 1 " .1=2n /
n
D
" nC1
2 1 " .1=2/
2
nC2
D 1 " nC1 :
2
Thus s D 2 C .n C 2/=2n .
29.
n %
n
n
&
X
X
X
f .i C 1/ " f .i / D
f .i C 1/ "
f .i /
iDm
D
iDm
nC1
X
j DmC1
Fig. 5.1-34
iDm
f .j / "
n
X
f .i /
35.
n
X
D f .n C 1/ " f .m/;
because each sum has only one term that is not cancelled
by a term in the other sum. It is called “telescoping” because the sum “folds up” to a sum involving only part of
the first and last terms.
30.
To show that
iDm
iD1
.k C 1/2 " k 2 D 2k C 1;
one for each k from 1 to n:
10
X
.n " .n " 1/ D 10 " 0 D 10; 000
m
X
.2j " 2j !1 / D 2m " 20 D 2m " 1
j D1
n.n C 1/
;
2
we write n copies of the identity
4
4
4
4
22 " 12 D 2.1/ C 1
32 " 22 D 2.2/ C 1
42 " 32 D 2.3/ C 1
::
:
.n C 1/2 " n2 D 2.n/ C 1:
nD1
31.
iD
Adding the left and right sides of these formulas we get
32.
2m #
X
1
1
"
i
i C1
iDm
33.
m
X
j D1
$
1
1
mC1
D
"
D
m 2m C 1
m.2m C 1/
m
X
1
D
j.j C 1/
j D1
#
1
1
"
j
j C1
$
D1"
.n C 1/2 " 12 D 2
1
n
D
nC1
nC1
Hence,
36.
34. The number of small shaded squares is 1 C 2 C
P! ! ! C n.
Since each has area 1, the total area shaded is niD1 i .
But this area consists of a large right-angled triangle of
area n2 =2 (below the diagonal), and n small triangles
(above the diagonal) each of area 1/2. Equating these areas, we get
iD1
178
iD1
iD
iD1
i C n:
1 2
n.n C 1/
.n C 2n C 1 " 1 " n/ D
.
2
2
P
The formula niD1 i D n.n C 1/=2 holds for n D 1, since
it says 1 D 1 in this case. Now assume that it holds for
P
n D some number k # 1; that is, kiD1 i D k.k C 1/=2.
Then for n D k C 1, we have
kC1
X
iD1
n
X
Pn
n
X
iD
k
X
iD1
i C.kC1/ D
k.k C 1/
.k C 1/.k C 2/
C.kC1/ D
:
2
2
2
iD
n
1
n.n C 1/
Cn D
:
2
2
2
Thus the formula also holds for n D k C 1. By induction,
it holds for all positive integers n.
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 5.1 (PAGE 295)
P
37. The formula niD1 i 2 D n.n C 1/.2n C 1/=6 holds for
n D 1, since it says 1 D 1 in this case. Now assume that
it
for n D some number k # 1; that is,
Pholds
k
2
iD1 i D k.k C 1/.2k C 1/=6. Then for n D k C 1, we
have
kC1
X
iD1
i2 D
D
D
D
D
k
X
iD1
iD1
r i!1 D
i 2 C .k C 1/2
40.
k.k C 1/.2k C 1/
C .k C 1/2
6
kC1 2
Œ2k C k C 6k C 6!
6
kC1
.k C 2/.2k C 3/
6
.k C 1/..k C 1/ C 1/.2.k C 1/ C 1/
:
6
k
X
iD1
n
X
iD1
Thus the formula also holds for n D k C 1. By induction,
it holds for all positive integers n.
P
38. The formula niD1 r i!1 D .r n " 1/=.r " 1/ (for r ¤ 1)
holds for n D 1, since it says 1 D 1 in this case. Now
assume
that it holds for n D some number k # 1; that is,
Pk
i!1
D .r k " 1/=.r " 1/. Then for n D k C 1, we
iD1 r
have
kC1
X
that L-shaped region has area i 3 . The sum of the areas of
the n L-shaped regions is the area of the large square of
side n.n C 1/=2, so
r i!1 C r k D
rk " 1
r kC1 " 1
C rk D
:
r "1
r "1
Thus the formula also holds for n D k C 1. By induction,
it holds for all positive integers n.
i3 D
#
n.n C 1/
2
$2
:
To show that
n
X
j D1
j 3 D 13 C 23 C 33 C ! ! ! C n3 D
n2 .n C 1/2
;
4
we write n copies of the identity
.k C 1/4 " k 4 D 4k 3 C 6k 2 C 4k C 1;
one for each k from 1 to n:
24 " 14 D 4.1/3 C 6.1/2 C 4.1/ C 1
34 " 24 D 4.2/3 C 6.2/2 C 4.2/ C 1
44 " 34 D 4.3/3 C 6.3/2 C 4.3/ C 1
::
:
4
4
.n C 1/ " n D 4.n/3 C 6.n/2 C 4.n/ C 1:
Adding the left and right sides of these formulas we get
.n C 1/4 " 14 D 4
D4
39.
n
X
j D1
n
X
j D1
j3 C 6
j3 C
n
X
j D1
j2 C 4
n
X
j D1
j Cn
4n.n C 1/
6n.n C 1/.2n C 1/
C
C n:
6
2
Hence,
n
4
n
X
j D1
::
:
so
3
1 2
!!!
3
Fig. 5.1-39
n
The L-shaped region with short side i is a square of side
i.i C 1/=2 with a square of side .i " 1/i=2 cut out. Since
i.i C 1/
2
D n2 .n C 1/2
n
X
j D1
2
1
#
j 3 D .n C 1/4 " 1 " n.n C 1/.2n C 1/ " 2n.n C 1/ " n
$2
#
$
.i " 1/i 2
2
i 4 C 2i 3 C i 2 " .i 4 " 2i 3 C i 2 /
D i 3;
D
4
"
j3 D
n2 .n C 1/2
:
4
P
41. The formula niD1 i 3 D n2 .n C 1/2 =4 holds for n D 1,
since it says 1 D 1 in this case. Now assume that it holds
for
Pkn D3 some2 number2k # 1; that is,
iD1 i D k .k C 1/ =4. Then for n D k C 1, we have
kC1
X
iD1
i3 D
k
X
iD1
2
i 3 C .k C 1/3
k .k C 1/2
.k C 1/2 2
C .k C 1/3 D
Œk C 4.k C 1/!
4
4
2
.k C 1/
D
.k C 2/2 :
4
D
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SECTION 5.1 (PAGE 295)
ADAMS and ESSEX: CALCULUS 9
Thus the formula also holds for n D k C 1. By induction,
it holds for all positive integers n.
P
42. To find jnD1 j 4 D 14 C 24 C 34 C ! ! ! C n4 , we write n
copies of the identity
(
)
3&2
3&3
3n
1 3
C
C
C !!! C
n!1 n n
n
n
n
3
D lim 2 .1 C 2 C 3 C ! ! ! C n/
n!1 n
3 n.n C 1/
3
D lim 2 !
D sq. units:
n!1 n
2
2
A D lim
.k C 1/5 " k 5 D 5k 4 C 10k 3 C 10k 2 C 5k C 1;
one for each k from 1 to n:
y
25 " 15 D 5.1/4 C 10.1/3 C 10.1/2 C 5.1/ C 1
35 " 25 D 5.2/4 C 10.2/3 C 10.2/2 C 5.2/ C 1
45 " 35 D 5.3/4 C 10.3/3 C 10.3/2 C 5.3/ C 1
::
:
.n C 1/5 " n5 D 5.n/4 C 10.n/3 C 10.n/2 C 5.n/ C 1:
.1;3/
yD3x
Adding the left and right sides of these formulas we get
5
5
.n C 1/ " 1 D 5
n
X
j D1
4
j C 10
n
X
j D1
3
j C 10
n
X
j D1
2
j C5
n
X
j D1
1
n
j C n:
Substituting
the known formulas for all the sums except
Pn
4
j D1 j , and solving for this quantity, gives
n
X
j D1
j4 D
n.n C 1/.2n C 1/.3n2 C 3n " 1/
:
30
n
X
iD1
n
X
iD1
n
X
iD1
n
X
iD1
i5 D
2.
7
7
1
1 8 1 7
n C n C n6 " n4 C n2
8
2
12
24
12
i8 D
1 9 1 8
n C n C !!!
9
2
1 11 1 10
n C n C !!!:
11
2
iD1
18 n.n C 1/
D lim 2
C 3 D 9 C 3 D 12sq. units:
n!1 n
2
3.
We would guess (correctly) that
This is similar to #1; the rectangles have width
.3 " 1/=n D 2=n and the i th has height the value of 2x " 1
at x D 1 C .2i=n/. The area is
#
$
n
X
2i
2
2C2 "1
n!1
n
n
A D lim
The area is the limit of the sum of the areas of the rectangles shown in the figure. It is
180
x
n
3
18 X
iC n
2
n!1 n
n
Section 5.2 Areas as Limits of Sums
(page 301)
1.
D1
iD1
D lim
i7 D
iD1
n!1 n
n n
This is similar to #1; the rectangles now have width
3=n and the i th has height 2.3i=n/C1, the value of 2xC1
at x D 3i=n. The area is
A D lim
1
1
1
1
1
i 6 D n7 C n6 C n5 " n3 C n
7
2
2
6
42
i 10 D
4
n
#
$
n
X
3
3i
2 C1
n!1
n
n
1 6 1 5
5
1
n C n C n4 " n2
6
2
12
12
n
X
3
n
Fig. 5.2-1
Of course we got Maple to do the donkey work!
43.
2
n
iD1
n
8 X
2
iC n
2
n!1 n
n
D lim
iD1
8 n.n C 1/
C 2 D 4 C 2 D 6sq. units:
D lim 2
n!1 n
2
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 5.2 (PAGE 301)
% a &3 n.n C 1/.2n C 1/
a
C .n/
n
6
n
a3 .n C 1/.2n C 1/
D
C a:
6
n2
3
a
Area D lim Sn D
C asq. units:
n!1
3
4. This is similar to #1; the rectangles have width
.2 " ."1//=n D 3=n and the i th has height the value of
3x C 4 at x D "1 C .3i=n/. The area is
D
#
$
n
X
3
3i
"3 C 3 C 4
n!1
n
n
A D lim
iD1
n
27 X
3
iC n
n!1 n2
n
D lim
y
iD1
yDx 2 C1
27 n.n C 1/
27
33
D lim 2
C3D
C3D
sq. units:
n!1 n
2
2
2
5. The area is the limit of the sum of the areas of the rectangles shown in the figure. It is
2
A D lim
n!1 n
"#
"
2
1C
n
$2
#
#
$
$ #
4 2
2n 2
C 1C
C !!! C 1 C
n
n
2
4
4
8
16
1C C 2 C1C C 2
n
n
n
n
n
#
2
4n
4n
C !!! C 1 C
C 2
n
n
#
$
8 n.n C 1/
8 n.n C 1/.2n C 1/
D lim 2 C 2 !
C 3 !
n!1
n
2
n
6
8
26
D2C4C D
sq. units:
3
3
D lim
n!1
y
yDx 2
x1 x2
a
x
Fig. 5.2-6
7.
The required area is (see the figure)
"#
$
#
$
3
3 2
3
A D lim
"1 C
C 2 "1 C
C3
n!1 n
n
n
#
#
$
$
6 2
6
C "1 C
C 2 "1 C
C3
n
n
#
#
$
$
#
3n
3n 2
C 2 "1 C
C3
C ! ! ! C "1 C
n
n
"#
$
3
6
32
6
D lim
1" C 2 "2C C3
n!1 n
n
n
n
$
#
2
6
12
12
C 2 "2C
C3
C 1"
n
n
n
#
$#
6n
6n
9n2
C !!! C 1 "
C 2 "2C
C3
n
n
n
#
$
27 n.n C 1/.2n C 1/
D lim 6 C 3 !
n!1
n
6
D 6 C 9 D 15sq. units:
y
1 1C 2
n
2n
1C n D3
x
yDx 2 C2xC3
Fig. 5.2-5
a
6. Divide Œ0; a! into n equal subintervals of length "x D
n
ia
by points xi D
, .0 $ i $ n/. Then
n
"
#
$
n % & #
X
ia 2
a
Sn D
C1
n
n
iD1
D
n
% a &3 X
n
iD1
i2 C
n
aX
.1/
n
!1
iD1
!1C
(Use Theorem 1(a) and 1(c).)
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3
n
!1C
x
3n
D2
n
Fig. 5.2-7
181
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SECTION 5.2 (PAGE 301)
ADAMS and ESSEX: CALCULUS 9
8.
y
"1
1
x
A
10.
y
2
y Dx "1
Fig. 5.2-8
x
2
The region in question lies between x D "1 and x D 1
and is symmetric about the y-axis. We can therefore double the area between x D 0 and x D 1. If we divide this
interval into n equal subintervals of width 1=n and use
the distance 0 " .x 2 " 1/ D 1 " x 2 between y D 0 and
y D x 2 " 1 for the heights of rectangles, we find that the
required area is
A
y D x 2 " 2x
Fig. 5.2-10
The height of the region at position x is
0 " .x 2 " 2x/ D 2x " x 2 . The “base” is an interval of
length 2, so we approximate using n rectangles of width
2=n. The shaded area is
#
$
n
X
1
i2
A D 2 lim
1" 2
n!1
n
n
iD1
$
#
n
X 1
i2
" 3
D 2 lim
n!1
n n
iD1
$
#
4
4
n n.n C 1/.2n C 1/
"
D 2 " D sq. units:
D 2 lim
3
n!1 n
6n
6
3
#
$
n
X
2
2i
4i 2
2 " 2
n!1
n
n
n
iD1
$
#
n
X 8i
8i 2
" 3
D lim
2
n!1
n
n
iD1
#
$
8 n.n C 1/
8 n.n C 1/.2n C 1/
D lim
"
n!1 n2
2
n3
6
4
8
D 4 " D sq. units:
3
3
A D lim
9.
y
2
4
x
A
y D1"x
11.
y
Fig. 5.2-9
The height of the region at position x is 0".1"x/ D x"1.
The “base” is an interval of length 2, so we approximate
using n rectangles of width 2=n. The shaded area is
#
$
n
X
2i
2
2C
"1
A D lim
n!1
n
n
iD1
$
n #
X
2
4i
D lim
C 2
n!1
n
n
iD1
$
#
n.n C 1/
2n
C4
D 2 C 2 D 4 sq. units:
D lim
n!1
n
2n2
182
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y D 4x " x 2 C 1
A
4
Fig. 5.2-11
x
lOMoARcPSD|6566483
INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 5.2 (PAGE 301)
The height of the region at position x is
4x " x 2 C 1 " 1 D 4x " x 2 . The “base” is an interval of
length 4, so we approximate using n rectangles of width
4=n. The shaded area is
Now we can use l’H^opital’s rule to evaluate
lim n.22=n " 1/ D lim
n!1
#
$
n
X
4
4i
16i 2
A D lim
4 " 2
n!1
n
n
n
iD1
$
#
n
X 64i
64i 2
"
D lim
n!1
n2
n3
iD1
#
$
64 n.n C 1/ 64 n.n C 1/.2n C 1/
D lim
"
n!1 n2
2
n3
6
32
64
D
sq. units:
D 32 "
3
3
12. Divide Œ0; b! into n equal subintervals of length "x D
by points xi D
Sn D
ib
, .0 $ i $ n/. Then
n
22=n ln 2
D lim
Thus the area is
3
square units.
2 ln 2
b
n
yD2x
!1
(Use Thm. 6.1.2(d).)
iD1
.b=n/n
14.
!1C
2
4
6
!1C !1C
n
n
n
!1C
x
2n
D1
n
Fig. 5.2-13
"# $
# $3
# $3 #
2b
nb
b
b 3
C
C !!! C
Area D lim
n!1 n
n
n
n
b4 3
.1 C 23 C 33 C ! ! ! C n3 /
n!1 n4
b4
b 4 n2 .n C 1/2
D
sq. units:
D lim 4 !
n!1 n
4
4
D lim
r
Area D lim Sn D .e " 1/ lim e lim r
n!1
r!0C
r!0C e " 1
1
D .e b " 1/.1/ lim r D e b " 1sq. units:
r!0C e
b
$
y
n
b
e
"1
D e .b=n/ .b=n/
n
e
"1
b
eb " 1
D e .b=n/ .b=n/
:
n
e
"1
b
Let r D :
n
"2
n2
n!1
iD1
&i!1
X%
b
D e .b=n/
e .b=n/
n
( )
0
0
"1
n2
D lim 2.2=n/C1 ln 2 D 2 ln 2:
n!1
#
$
n
n
X
b .ib=n/
b X% .b=n/ &i
e
D
e
n
n
iD1
n!1
22=n " 1
1
n
#
r
( )
0
0
y
yDx 3
13. The required area is
i
2 h !1C.2=n/
2
C 2!1C.4=n/ C ! ! ! C 2!1C.2n=n/
n!1 n
(
%
& %
&2
%
&n!1 )
22=n
1 C 22=n C 22=n C ! ! ! C 22=n
D lim
n!1 n
!
"n
22=n 22=n " 1
D lim
!
n!1 n
22=n " 1
1
2=n
D lim 2
&3&
n!1
n.22=n " 1/
1
D 3 lim
:
n!1 n.22=n " 1/
A D lim
b
n
2b
n
3b
n
.n!1/b nb
Db
n
n
x
Fig. 5.2-14
15.
Let t D
# $1=n
b
and let
a
x0 D a; x1 D at; x2 D at 2 ; : : : ; xn D at n D b:
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SECTION 5.2 (PAGE 301)
ADAMS and ESSEX: CALCULUS 9
The i th subinterval Œxi!1 ; xi ! has length "xi D
1
at i!1 .t " 1/. Since f .xi!1 / D i!1 , we form the sum
at
#
$
n
X
1
i!1
at .t " 1/
Sn D
at i!1
iD1
(# $1=n
)
b
D n.t " 1/ D n
"1 :
a
17.
y
A1
b
1
and c D . The area under the curve is
n
a
( )
cr " 1
0
A D lim Sn D lim
n!1
r!0C
r
0
# $
b
c r ln c
D ln c D ln
square units.
D lim
r!0C
1
a
2
x
Fig. 5.2-17
#
$
n
X
2
2i
1"
represents a sum of areas of n rectn
n
iD1
angles each of width 2=n and having heights equal to the
height to the graph y D 1"x at the points x D 2i=n. Half
of these rectangles have negative height, and limn!1 Sn
is the difference A1 " A2 of the areas of the two triangles in the figure above. It has the value 0 since the two
triangles have the same area.
y
sn D
This is not surprising because it follows from the definition
of ln.
y
1
x
1
A2
Let r D
yD
y D1"x
18.
y D 2 C 3x
A
a
x1
x2
b
x
1
x
Fig. 5.2-18
Fig. 5.2-15
#
$
n
n
X
X
1
3i
2n C 3i
D
2
C
represents a sum
n2
n
n
iD1
iD1
of areas of n rectangles each of width 1=n and having
heights equal to the height to the graph y D 2 C 3x at
the points x D i=n. Thus limn!1 Sn is the area of the
trapezoid in the figure above, and has the value
1.2 C 5/=2 D 7=2.
16.
y
sn D
2
y D 2.1 " x/
A
1
x
Fig. 5.2-16
19.
#
$
n
X
i
2
1"
represents a sum of areas of n rectsn D
n
n
iD1
angles each of width 1=n and having heights equal to the
height to the graph y D 2.1 " x/ at the points
x D i=n. Thus limn!1 Sn is the area A of the triangle in
the figure above, and therefore has the value 1.
184
n
X
1
Sn D
n
j D1
s
1"
# $2
j
n
D sum of areas of rectangles in the figure.
Thus the limit of Sn is the area of a quarter circle of unit
radius:
#
lim Sn D :
n!1
4
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 5.3 (PAGE 307)
y
5.
f .x/ D sin x on Œ0; #!; n D 6:
*
'
# # # 2# 5#
;
;#
P6 D 0; ; ; ;
6 3 2 3 6
"
#
p
p
#
1
3
3
1
L.f; P6 / D
0C C
C
C C0
6
2
2
2
2
p
#
D .1 C 3/ ' 1:43;
6 "
#
p
p
# 1
1
3
3
U.f; P6 / D
C
C1C1C
C
6 2
2
2
2
p
#
D .3 C 3/ ' 2:48:
6
6.
f .x/ D cos x on Œ0; 2#!; n D 4:
# $(
)
2#
#
3#
L.f; P4 / D
cos C cos # C cos # C cos
D "#:
4
2
2
# $(
)
2#
#
3#
U.f; P4 / D
cos 0 C cos C cos
C cos 2# D #:
4
2
2
p
yD 1!x 2
1
n
2
n
x
n!1 n D1
n
n
Fig. 5.2-19
Section 5.3 The Definite Integral
(page 307)
1.
y
yDcos x
f .x/ D x on Œ0; 2!; n D 8:
'
*
1 1 3
5 3 7
P8 D 0; ; ; ; 1; ; ; ; 1
4 2 4
4 2 4
(
)
2"0
1
1
3
5
3
7
L.f; P8 / D
0C C C C1C C C
D
8
4
2
4
4
2
4
(
)
1
3
5
3
7
2"0 1
C C C1C C C C2 D
U.f; P8 / D
8
4
2
4
4
2
4
!=2
3.
f .x/ D x 2 on Œ0; 4!; n D 4:
#
$
4"0
L.f; P4 / D
Œ0 C .1/2 C .2/2 C .3/2 ! D 14:
4
#
$
4"0
U.f; P4 / D
Œ.1/2 C .2/2 C .3/2 C .4/2 ! D 30:
4
f .x/ D e x on Œ"2; 2!; n D 4:
e4 " 1
' 4:22
" 1/
e4 " 1
U.f; P4 / D 1.e !1 C e 0 C e 1 C e 2 / D
' 11:48:
e.e " 1/
L.f; P4 / D 1.e !2 C e !1 C e 0 C e 1 / D
e 2 .e
7
4
9
4
Fig. 5.3-6
˚
f .x/ D x on Œ0; 1!. Pn D 0; n1 ; n2 ; : : : ;
f .x/ D ln x on Œ1; 2!; n D 5:
$(
)
#
6
7
8
9
2"1
ln 1 C ln C ln C ln C ln
L.f; P5 / D
5
5
5
5
5
' 0:3153168:
#
$(
)
2"1
6
7
8
9
U.f; P5 / D
ln C ln C ln C ln C ln 2
5
5
5
5
5
' 0:4539462:
n!1 n
;n
n
+
: We have
#
$
1
2
n"1
1
0 C C C !!! C
n
n
n
n
1 .n " 1/n
n"1
D 2 !
D
;
n#
2
2n
$
1 1
2
3
n
U.f; Pn / D
C C C !!! C
n n
n
n
n
1 n.n C 1/n
nC1
D 2 !
D
:
n
2
2n
L.f; Pn / D
Thus limn!1 L.f; Pn / D limn!1 U.f; Pn / D 1=2.
If P is any partition of Œ0; 1!, then
L.f; P / $ U.f; Pn / D
4.
2! x
3!=2
7.
2.
!
nC1
2n
for every n, so L.f; P / $ limn!1 U.f; Pn / D 1=2.
Similarly, U.f; P / # 1=2. If there exists any number I
such that L.f; P / $ I $ U.f; P / for all P , then I cannot be less than 1/2 (or there would exist a Pn such that
L.f; Pn / > I ), and, similarly, I cannot be greater than 1/2
(or there would exist a Pn such that U.f; Pn / < I ). Thus
R1
I D 1=2 and 0 x dx D 1=2.
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SECTION 5.3 (PAGE 307)
ADAMS and ESSEX: CALCULUS 9
˚
8. f .x/ D 1 " x on Œ0; 2!. Pn D 0; n2 ; n4 ; : : : ;
have
2n!2 2n
n ; n
+
By l’H^opital’s Rule,
: We
##
$ #
$
#
$$
2
4
2n
1"
C 1"
C !!! C 1 "
n
n
n
n
X
4
2
i
D n" 2
n
n
n!1
iD1
2
4 n.n C 1/
D " ! 0 as n ! 1;
D2" 2
n
2
n
##
$ #
$
#
$$
2
0
2
2n " 2
U.f; Pn / D
1"
C 1"
C !!! C 1 "
n
n
n
n
n!1
4 X
2
i
D n" 2
n
n
Thus
lim L.f; Pn / D lim U.f; Pn / D e 3 " 1 D
n!1
11.
iD0
D2"
Thus
Z
4 .n " 1/n
2
D ! 0 as n ! 1:
2
n
2
n
n!1
n
X
1
lim
n!1
n
iD1
12.
n
X
1
n!1
n
lim
iD1
2
.1 " x/ dx D 0.
0
e 3=n " 1
n!1
1=n
e 3=n ."3=n2 /
3e 3=n
D lim
D
lim
D 3:
n!1
n!1
"1=n2
1
lim n.e 3=n " 1/ D lim
2
L.f; Pn / D
n
13.
˚
+
n
9. f .x/ D x 3 on Œ0; 1!. Pn D 0; n1 ; n2 ; : : : ; n!1
n ; n : We
have (using the result of Exercise 51 (or 52) of Section
6.1)
L.f; Pn / D
1
n
#
# $3 # $3
1
n"1
0
C
C !!! C
n
n
n
$3 !
i
D
n
r
i "1
D
n
Z
1
p
3
e x dx:
0
x dx
0
Z
1
p
x dx
0
Z !
n
X
#
#i
sin x dx
sin
D
n!1
n
n
0
lim
iD1
14.
#
$ Z 2
n
X
2
2i
lim
ln 1 C
D
ln.1 C x/ dx
n!1
n
n
0
iD1
15.
n!1
1 X 3
1 .n " 1/2 n2
i
D
n4
n4
4
iD0
#
$2
1 n"1
1
! as n ! 1;
D
4
n
4
!
# $3 # $3
% n &3
2
1
1
C
C !!! C
U.f; Pn / D
n
n
n
n
r
Z
D
16.
#
$ Z 1
n
X
1 !1 2i " 1
tan
D
tan!1 x dx
lim
n!1
n
2n
0
iD1
(
)
2i " 1
i "1 i
Note that
is the midpoint of
;
.
2n
n n
Z 1
n
n
X
X
1
1
dx
n
D
lim
D
lim
n!1
n!1
n2 C i 2
n 1 C .i=n/2
1
C
x2
0
iD1
iD1
17.
b"a
and xi D a C i "x where 1 $ i $ n " 1.
Let "x D
n
Since f is continuous and nondecreasing,
n
1 X 3
1 n2 .n C 1/2
i D 4
4
n
n
4
iD1
#
$2
1
1 nC1
! as n ! 1:
D
4
n
4
D
Thus
Z
1
0
iD1
U.f; Pn / D f .x1 /"x C f .x2 /"x C ! ! ! C
f .xn!1 /"x C f .b/"x
)
(n!1
b"a X
f .xi / C f .b/ :
D
n
1
x dx D .
4
3
˚
+
3n
10. f .x/ D e x on Œ0; 3!. Pn D 0; n3 ; n6 ; : : : ; 3n!3
n ; n : We
have (using the result of Exercise 51 (or 52) of Section
6.1)
&
3 % 0=n
L.f; Pn / D
e
C e 3=n C e 6=n C ! ! ! C e 3.n!1/=n
n
3.e 3 " 1/
3 e 3n=n " 1
D
;
D
n e 3=n " 1
n.e 3=n " 1/
%
&
3 3=n
U.f; Pn / D
e
C e 6=n C e 9=n C ! ! ! C e 3n=n D e 3=n L.f; Pn /:
n
186
L.f; Pn / D f .a/"x C f .x1 /"xC
f .x2 /"x C ! ! ! C f .xn!1 /"x
(
)
n!1
X
b"a
f .a/ C
f .xi / ;
D
n
iD1
Thus,
U.f; Pn / " L.f; Pn /
(n!1
)
n!1
X
b"a X
D
f .xi / C f .b/ " f .a/ "
f .xi /
n
iD1
.b " a/.f .b/ " f .a//
D
:
n
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 5.4 (PAGE 312)
Since
3.
lim ŒU.f; Pn / " L.f; P " n/! D 0;
n!1
R2
!2 .x
C 2/ dx D
1
.4/.4/ D 8
2
y
.2;4/
therefore f must be integrable on Œa; b!.
18. P D fx0 < x1 < ! ! ! < xn g,
P 0 D fx0 < x1 < ! ! ! < xj !1 < x 0 < xj < ! ! ! < xn g.
Let mi and Mi be, respectively, the minimum and maximum values of f .x/ on the interval Œxi!1 ; xi !, for
1 $ i $ n. Then
L.f; P / D
U.f; P / D
n
X
iD1
n
X
iD1
!2
2
4.
R2
0
.3x C 1/ dx D shaded area D 12 .1 C 7/.2/ D 8
y
yD3xC1
If mj0 and Mj0 are the minimum and maximum values of
f .x/ on Œxj !1 ; x 0 !, and if mj00 and Mj00 are the corresponding values for Œx 0 ; xj !, then
mj0 # mj ;
mj00 # mj ;
Mj0 $ Mj ;
x
Fig. 5.4-3
mi .xi " xi!1 /;
Mi .xi " xi!1 /:
2
yDxC2
Mj00 $ Mj :
Therefore we have
Fig. 5.4-4
Mj .xj " xj !1 / # Mj0 .x 0 " xj !1 / C Mj00 .xj " x 0 /:
Hence L.f; P / $ L.f; P 0 / and U.f; P / # U.f; P 0 /.
00
5.
If P is any refinement of P we can add the new points
in P 00 to those in P one at a time, and thus obtain
L.f; P / $ L.f; P 00 /;
Rb
a
x dx D
b2
a2
"
2
2
y
U.f; P 00 / $ U.f; P /:
yDx
Section 5.4 Properties of the Definite
Integral (page 312)
1.
Z
b
f .x/ dx C
a
D
2.
Z
Z
c
Zb c
a
Z
2
f .x/ dx C
f .x/ dx "
3
Z
a
x
b
Fig. 5.4-5
a
f .x/ dx
Zc c
a
Z
f .x/ dx D 0
6.
R2
!1 .1
" 2x/ dx D A1 " A2 D 0
y
3
3f .x/ dx C
3f .x/ dx "
2f .x/ dx
0
1
0
Z 2
"
3f .x/ dx
1
Z 1
Z 2
D
.3 " 2/f .x/ dx C
.3 C 3 " 2 " 3/f .x/ dx
0
1
Z 3
C
.3 " 2/f .x/ dx
2
Z 3
D
f .x/ dx
0
x
2
mj .xj " xj !1 / $ mj0 .x 0 " xj !1 / C mj00 .xj " x 0 /;
A1
1
2
!1
2
x
A2
y D 1 " 2x
Fig. 5.4-6
7.
R p2 p
p
! 2
p
2 " t 2 dt D 12 #. 2/2 D #
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SECTION 5.4 (PAGE 312)
ADAMS and ESSEX: CALCULUS 9
p
yD 2!t 2
y
15.
p
! 2
Z
p
1
0
4 " x 2 dx D area A1 in figure below
1
area of circle " area A2
4
(see #14 below)
p !
1
2#
3
2
D .#2 / "
"
4
3
2
p
#
3
D C
3
2
D
t
p
2
Fig. 5.4-7
8.
9.
Z
Z
0
p
! 2
!
p
2 " x 2 dx D quarter disk D
1 p 2
#
#. 2/ D
4
2
y
sin.x 3 / dx D 0. (The integrand is an odd function
!!
and the interval of integration is symmetric about x D 0.)
Ra
10. !a .a " jsj/ ds D shaded area D 2. 21 a2 / D a2
P
yD
p
4 " x2
y
a
yDa!jsj
A1
A2
a
!a
s
11.
R1
5
!1 .u
" 3u3 C #/ du D #
R1
y
!1
O
Z
p
2
1
D area sector POQ " area triangle POR
1
1 p
D .#22 / " .1/ 3
6
p 2
2#
3
D
"
3
2
u
Fig. 5.4-11
p
Let y D 2x " x 2 ) y 2 C .x " 1/2 D 1:
Z 2p
#
1
2x " x 2 dx D shaded area D #.1/2 D :
2
2
0
y
17.
Z
18.
Z
2
6x 2 dx D 6
0
p
yD 2x!x 2
3
2
Z
2
0
.x 2 " 4/ dx D
D
2
x
19.
Z
2
2
!2
14.
Z
4
!4
Z
.e x " e !x / dx D 0 (odd function, symmetric interval)
3
!3
3
p
Z
3 p
9 " t 2 dt C
t 9 " t 2 dt
!3
#!3
$
1 2
D2
#3 C 0 D 9#
2
p
.2 C t / 9 " t 2 dt D 2
188
Z
20.
Z
2
0
21.
Z
0
.x 2 C
p
3
0
3
33
D 16
3
x 2 dx "
Z
2
0
x 2 dx " 4.3 " 2/
3
23
7
"
"4D
3
3
3
Z
2
.4 " t 2 / dt
$
#
32
23
D
D 2 2.4/ "
3
3
.v 2 " v/ dv D
1
x 2 dx D 6
Z
.4 " t / dt D 2
Fig. 5.4-12
13.
x
4 " x 2 dx D area A2 in figure above
1
1
2
Fig. 5.4-15
16.
12.
1
du D 2#
!1
Q
R
Fig. 5.4-10
0
2
23 22
"
D
3
2
3
13
1
C .#12 /
3
4
#
1
D C
3
4
1 " x 2 / dx D
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INSTRUCTOR’S SOLUTIONS MANUAL
22.
Z
6
x 2 .2 C sin x/ dx D
!6
Z
D4
23.
24.
Z
2
1
Z
26.
Z
4
1
1=3
Z
3
1=4
27.
28.
29.
Z
Z
4
Z
0
6
Z
6
!6
x 2 dx C 0 D
y
p
yD 4!x 2
x 2 sin x dx
4 3
.6 / D 288
3
1
dt D "
t
Fig. 5.4-31
32.
1=3
1
1
1
dt D " ln D ln 3
t
3
Z 1=4
1
1
ds "
ds
s
1 s
1
1
D ln 3 " ln D ln 3 C ln 4 D ln 12
4
1
ds D
s
Z
Z
3
33.
Z 2
1
1
ds
2 " .1=2/ 1=2 s
#
$
2
1
4
D
ln 2 " ln
D ln 2
3
2
3
Average value D
R2
!1
sgn x dx D 2 " 1 D 1
y
!1
2
Fig. 5.4-33
b
1
.x C 2/ dx
b"a a
(
)
1 2
1
.b " a2 / C 2.b " a/
D
b"a 2
1
4CaCb
D .b C a/ C 2 D
2
2
34.
Let
'
f .x/ D
1Cx
2
Z
2
!3
f .x/ dx D area.1/ C area.2/ " area.3/
D .2 & 2/ C 21 .1/.1/ " 12 .2/.2/ D 2 12 :
y
(1)
1 33
D3
x 2 dx D
3 3
Z 2
1
.4 " x 2 /1=2 dx
2"0 0
1
D .shaded area/
2#
$
#
1 1
#.2/2 D
D
2 4
2
yD2
(2)
!3
(3)
0
if x < 0
if x # 0.
Then
Z !
1
Average D
.1 C sin t / dt
# " ."#/ !!
(Z !
)
Z !
1
1 dt C
sin t dt
D
2# !!
!!
1
Œ2# C 0! D 1
D
2#
3
x
!1
Z
Z
yDsgn x
1
Z 4
1
.x C 2/ dx
Average D
4"0 0
(
)
1 1 2
D
.4 / C 2.4/ D 4
4 2
Average D
x
2
2
1
1
1
dt D
dt "
dt
t
t
1
1 t
D ln 4 " ln 2 D ln.4=2/ D ln 2
1
30. Average D
3"0
31.
!6
2x 2 dx C
1
dx D ln 2
x
2
25.
6
SECTION 5.4 (PAGE 312)
!1
2
x
yDxC1
.!3;!2/
Fig. 5.4-34
Average value D
35.
Z
2
0
g.x/ dx D
D
Z
1
0
3
x 2 dx C
Z
2
x dx
1
22 " 12
11
1
C
D
3
2
6
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SECTION 5.4 (PAGE 312)
36.
Z
Z 3
.2 " x/ dx C
.x " 2/ dx
0
2
ˇ
ˇ
$ˇ2 # 2
$ˇ3
#
ˇ
x 2 ˇˇ
x
" 2x ˇˇ
D 2x "
C
ˇ
2 ˇ
2
ˇ
3
0
j2 " xj dx D
Z
ADAMS and ESSEX: CALCULUS 9
2
I D
Z
2
0
p
.jx C 1j " jx " 1j C jx C 2j/ dx
D area A1 " area A2
1 5
5C8
1C2
1C2
1 1
41
D
.5/ C
.3/ "
.1/ "
.1/ "
.1/ D
2 3
2
2
2
2 3
2
9
5
"6"2C4D
2
2
D4"2"0C
4
!3
2
0
37.
Z
40.
y
4 " x 2 sgn.x " 1/ dx
yD
D area A1 " area A2 :
p
p
Area A1 D 16 #22 " 12 .1/. 3/ D 32 # " 12 3.
p
Area A2 D 41 #22 " area A1 D 31 # C 21 3.
p
Therefore I D .#=3/ " 3.
x2 " x
jx " 1j
A1
y
p
yD 4!x 2
p
2
3
A2
3
A1
!=3
1
2
x
A2
Fig. 5.4-40
Z
38.
R 3:5
0
Œx! dx D shaded area D 1 C 2 C 1:5 D 4:5:
y
Z
1 2
jx C 1jsgn x dx
4 !2
#Z 2
$
Z 0
1
.x C 1/ dx "
jx C 1j dx
D
4
!2
# 0
$
1 1C3
1
D
&2"2& &1&1
4
2
2
3
1
D1" D :
4
4
yDbxc
41.
1
2
Average D
x
3:5
3
x2 " x
dx
0 jx " 1j
D area A1 " area A2
1C3
1
7
D
.2/ " .1/.1/ D
2
2
2
Fig. 5.4-37
Fig. 5.4-38
y
39.
y
.2;3/
yDxC1
yD!.xC1/
1
A1
!2
A2 !1 A3
A1
.!2;!1/
y D jx C 1j " jx " 1j C jx C 2j
x
A2
42.
Z b%
a
2
x
!1
Fig. 5.4-41
Z
Z b
&
b
f .x/ dx "
f dx
f .x/ " f dx D
a
a
Z b
D .b " a/f " f
dx
a
Fig. 5.4-39
190
D .b " a/f " .b " a/f D 0
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INSTRUCTOR’S SOLUTIONS MANUAL
43.
Z b%
a
D
D
f .x/ " k
Z b%
a
Z b%
a
&2
SECTION 5.5 (PAGE 318)
dx
10.
Z
&2
f .x/ dx " 2k
b
f .x/ dx C k
a
2
Z
D .b " a/.k " f /2 C
a
This is minimum if k D f .
dx
a
&2
2
f .x/ dx " .b " a/f
Section 5.5 The Fundamental Theorem of
Calculus (page 318)
1.
2.
3.
Z
Z
Z
2
0
ˇ2
x 4 ˇˇ
16 " 0
x dx D
D4
ˇ D
4 ˇ
4
p
0
6.
1=2
7.
2
1
Z
1=2
2
9.
3$
x
2
"
x3
2
!=3
Z
0
13.
14.
Z
0
!!
2
Z
e
0
16.
D "1 " ."2/ D 1
17.
0
Z
1
18.
Z
19.
.x 4 C 6x 2 C 9/ dx
$ˇˇ2
# 5
x
ˇ
3
C 2x C 9x ˇ
D2
ˇ
5
#
$ 0
32
404
D2
C 16 C 18 D
5
5
ˇ
!
9
Z 9
p ˇˇ
p
1
2
x"p
dx D x 3=2 " 2 x ˇ
ˇ
3
x
4
4
# "
#
"
p
p
2 3=2
32
2 3=2
D
.9/
"2 9 "
.4/
"2 4 D
3
3
3
!1
p
0
Z
ˇ1
ˇ
dx
#
!1 ˇ
D tan x ˇ D
2
ˇ
1Cx
2
1=2
1
!1
p
dx
1 " x2
dx
4 " x2
!1
D sin
D sin!1
0
20.
Z
0
!2
21.
ˇ1=2
ˇ
#
ˇ
xˇ
D
ˇ
6
0
ˇ1
x ˇˇ
ˇ
2ˇ
!1
# $
1
1
" sin!1 "
D sin
2
2
# % #&
#
D " "
D
6
6
3
ˇ0
ˇ
1
dx
#
1
!1 x ˇ
D
tan
ˇ D 0 " tan!1 ."1/ D
2
ˇ
4Cx
2
2
2
8
!1
1
!!=4
x
ˇ1
Z 1
2x ˇˇ
1
3
2
x
2 dx D
"
D
ˇ D
ˇ
ln
2
ln
2
2
ln
2
2
ln
2
!1
2
ˇ!!=6
Z !!=6
ˇ
ˇ
cos x dx D sin x ˇ
ˇ
!!=4
ˇe
ae " 1
ax ˇˇ
a dx D
ˇ D
ln a ˇ
ln a
!1
#
$ˇ2
x 4 ˇˇ
1
dx D " 2 "
ˇ D "9=8
x
8 ˇ
.x 2 C 3/2 dx D 2
.e x " e !x / dx D 0 (odd function, symmetric interval)
!1
!1
Z
ˇ2!
ˇ
ˇ
.1 C sin u/ du D .u " cos u/ˇ D 2#
ˇ
ˇ!
ˇ
!
x
xˇ
e dx D e ˇ D e ! " e !!
ˇ
!!
Z
15.
ˇ!=3
p
ˇ
2"1
ˇ
sin & d& D " cos & ˇ
D
ˇ
2
!=4
2!
!2
ˇ4
2 3=2 ˇˇ
16
x dx D x ˇ D
ˇ
3
3
ˇ1
1
"1 ˇˇ
dx D
ˇ
x2
x ˇ
#
!2
8.
12.
0
1
Z
ˇ!=3
ˇ
p
#
ˇ
sec & d& D tan & ˇ
D tan D 3
ˇ
3
2
0
!=4
3
$
#
$ ˇˇ!1
Z !1 #
1
1
1
1
ˇ
"
dx D " C 2 ˇ
4.
ˇ
x2 x3
x
2x
!2
!2
#
$
1
1
1
7
D1C "
C
D
2
2
8
8
ˇ2
Z 2
ˇ
ˇ
2
3
2
5.
.3x " 4x C 2/ dx D .x " 2x C 2x/ˇ D 9
ˇ
!1
Z
11.
0
4
!=3
0
b
&2
f .x/ dx " 2k.b " a/f C k 2 .b " a/
Z b%
Z
Area R D
!2
R1
0
ˇ1
1
x 5 ˇˇ
x dx D
ˇ D sq. units:
5 ˇ
5
4
0
y
.1;1/
yDx 4
R
1
x
p
1
2" 2
1
D" Cp D p
2
2
2 2
Fig. 5.5-21
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SECTION 5.5 (PAGE 318)
22.
Area D
Z
e2
e
ADAMS and ESSEX: CALCULUS 9
ˇe 2
ˇ
1
ˇ
dx D ln x ˇ
ˇ
x
25.
For intersection of y D x 2 " 3x C 3 and y D 1, we have
x 2 " 3x C 3 D 1
e
D ln e 2 " ln e D 2 " 1 D 1 sq. units.
x 2 " 3x C 2 D 0
.x " 2/.x " 1/ D 0:
y
Thus x D 1 or x D 2. The indicated region has area
1
yD x
Area
e2 x
e
Fig. 5.5-22
23.
Z
4
.x 2 " 4x/ dx
# 3
$ˇˇ4
x
2 ˇ
D"
" 2x ˇ
ˇ
3
$ 0
#
32
64
" 32 D
sq. units:
D"
3
3
Area R D "
Z
2
.x 2 " 3x C 3/ dx
$ˇˇ2
# 3
3x 2
x
ˇ
"
C 3x ˇ
D1"
ˇ
3
2
1
(
)$
#
1 3
8
1
"6C6"
" C3
D1"
D sq. units:
3
3 2
6
Area R D 1 "
1
0
y
yDx 2 !3xC3
yD1
R
y
1
4 x
R
yDx 2 !4x
p
p
x
x
Since y D x and y D
intersect where x D , that
2
2
is, at x D 0 and x D 4, thus,
Since y D 5 " 2x " 3x 2 D .5 C 3x/.1 " x/, therefore y D 0
at x D " 53 and 1, and y > 0 if " 53 < x < 1. Thus, the
area is
1
!1
.5 " 2x " 3x 2 / dx D 2
Z
Z
Z 4
x
x dx "
dx
0
0 2
ˇ4
ˇ4
ˇ
2
x 2 ˇˇ
ˇ
D x 3=2 ˇ "
ˇ
ˇ
3
4 ˇ
Area D
Fig. 5.5-23
Z
4
p
0
0
4
16 16
"
D sq. units.
D
3
4
3
1
.5 " 3x 2 / dx
ˇ1
ˇ
3 ˇ
D 2.5x " x /ˇ
ˇ
y
0
p
yD x
0
D 2.5 " 1/ D 8 sq. units.
A
.4;2/
yDx=2
y
x
yD5!2x!3x 2
Fig. 5.5-26
27.
Area
1
!1
Fig. 5.5-24
192
x
Fig. 5.5-25
26.
24.
2
x
Area R D 2 & shaded area
#
$
Z 1
1
D2
"
x 2 dx
2
0
#
$
1 1
1
D2
"
D sq. units:
2 3
3
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 5.5 (PAGE 318)
y
y
.1;1/
xDy 2
y D e !x
A
R
x
!a
yDx 2
x
Fig. 5.5-30
Fig. 5.5-27
31.
Area R D
Z
3
0
0
y
3
.12 " x 2 / dx " 2
x dx
0
0
ˇ
$3
$ˇ3
#
#
1 2 ˇˇ
1 3 ˇˇ
D 2 12x " x ˇ " 2 x ˇ
ˇ
ˇ
3
2
Area D 2
2!
.1 " cos x/ dx
ˇ2!
ˇ
D .x " sin x/ˇ D 2#sq. units:
28. The two graphs intersect at .˙3; 3/, thus
Z
Z
0
yD1!cos x
0
R
D 2.36 " 9/ " 9 D 45 sq. units.
2!x
y
yD12!x 2
Fig. 5.5-31
Area
yDjxj
.3;3/
32.
Area D
Fig. 5.5-28
x
!1=3
1
ˇ27
3 2=3 ˇˇ
dx D x ˇ
ˇ
2
1
y
Z 1
x 1=3 dx "
x 1=2 dx
0
0
ˇ1
ˇ1
ˇ
ˇ
2
3 2
1
3
ˇ
ˇ
D x 4=3 ˇ " x 3=2 ˇ D " D
sq. units:
ˇ
ˇ
4
3
4 3
12
Area R D
Z
27
3
3
D .27/2=3 " D 12 sq. units.
2
2
x
29.
Z
1
0
yDx "1=3
Area
0
1
y
27 x
Fig. 5.5-32
.1;1/
yDx 1=3
33.
R
yDx 1=2
Z
3!=2
j cos xj dx D
0
Z
!=2
cos x dx "
0
Z
3!=2
cos x dx
!=2
ˇ!=2
ˇ3!=2
ˇ
ˇ
ˇ
ˇ
" sin x ˇ
D sin x ˇ
ˇ
ˇ
0
x
!=2
D1C1C1D3
Fig. 5.5-29
34.
ˇ0
ˇ
R0
ˇ
30. Area D !a e !x dx D "e !x ˇ D e a " 1 sq. units.
ˇ
Z
3
1
Z 3
dx
dx
C
2
2
1 x
2 x
ˇ2
ˇ3
1 ˇˇ
1 ˇˇ
1
D ˇ " ˇ D"
xˇ
xˇ
3
sgn .x " 2/
dx D "
x2
!a
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1
2
2
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SECTION 5.5 (PAGE 318)
35.
36.
ADAMS and ESSEX: CALCULUS 9
Average value
Z
1 2
D
.1 C x C x 2 C x 3 / dx
2 0
#
$ˇ2
x2
x3
x 4 ˇˇ
1
xC
C
C
D
ˇ
2
2
3
4 ˇ
0
#
$
8
16
1
2C2C C4 D
:
D
2
3
3
Z 2
1
Average value D
e 3x dx
2 " ."2/ !2
#
$ˇ2
1 1 3x ˇˇ
e
D
ˇ
ˇ
4 3
44.
45.
1= ln 2
0
38. Since
g.t / D
ˇ1= ln 2
2x ˇˇ
2 dx D .ln 2/
De"1
ˇ
ln 2 ˇ
x
46.
0
'
"Z
0
1
.0/ dt C
Z
1
3
39.
40.
41.
42.
d
dx
d
dt
d
dx
Z
x
2
Z
3
t
Z
x2
d 2
x
dx
D 2x
D 2x
D 2x
43.
d
dt
Z
194
Z
Z
Z
Z
t
!!
x2
0
x2
0
x2
0
x2
0
sin u
du
u
Z x2
sin u
sin u
d
du C x 2
du
u
dx 0
u
(
)
2x sin x 2
sin u
du C x 2
u
x2
sin u
du C 2x sin.x 2 /
u
47.
a
Z
sin "
a
cos &
" sin &
D
"
1 " cos2 &
1 " sin2 &
"1
1
D
"
D " csc & " sec &
sin &
cos &
Z t
F .t/ D
cos.x 2 / dx
Z
0
p
x
1
dx
1 " x2
#
cos.u2 / du
0
p
d
cos x
1
F . x/ D cos x p D p
dx
2 x
2 x
Z x2 p
H.x/ D 3x
e ! t dt
Z
Z
4
x2
p
t
e!
4
4
p
e!
4
t
dt C 3x.2xe !jxj /
dt C 3.2/.4e !2 /
24
e2
f .x/ D # C #
Z
x
f .t / dt
1
f 0 .x/ D #f .x/ % f .x/ D C e !x
# D f .1/ D C e ! % C D #e !!
2
1
.3 " 1/ D :
3
3
Z x2
sin t
d
sin t
dt D "
dt
t
dx 0
t
sin x 2
sin x 2
D "2x 2 D "2
x
x
1
dx "
1 " x2
D 3.0/ C 24e !2 D
1
sin t
sin x
dt D
t
x
" Z
#
t
sin x
d
sin x
sin t
dx D
"
dx D "
x
dt
x
t
3
0
cos "
H 0 .2/ D 3
ˇ3 #
"
ˇ
1
ˇ
0 C tˇ
1 dt D
ˇ
3
#
D
"Z
1
dx
1 " x2
H .x/ D 3
the average value of g.t / over [0,3] is
1
3
sin "
0
if 0 $ t $ 1,
if 1 < t $ 3,
0;
1;
cos "
p
F . x/ D
!2
Z
Z
d
D
d&
1 6
D
.e " e !6 /:
12
1
37. Avg. D
1= ln 2
d
d&
48.
f .x/ D # e !.x!1/ :
Z x
f .x/ D 1 "
f .t / dt
0
f 0 .x/ D "f .x/
1 D f .0/ D C
f .x/ D e !x :
49.
50.
51.
%
f .x/ D C e !x
The function 1=x 2 is not defined (and therefore not continuous) at x D 0, so the Fundamental Theorem of Calculus cannot be applied to it on the interval Œ"1; 1!. Since
Z 1
dx
1=x 2 > 0 wherever it is defined, we would expect
2
!1 x
to be positive if it exists at all (which it doesn’t).
Z x
sin x
sin t
dt , then F 0 .x/ D
and
If F .x/ D
2
1
C
t
1
C x2
17
F .17/ D 0.
#
$
R 2x!x 2
1
F .x/ D 0
cos
dt .
1 C t2
Note that 0 <
cos y
cos t
dy D
1 C y2
1 C t2
1
$ 1 for all t , and hence
1 C t2
0 < cos.1/ $ cos
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1
1 C t2
$
$ 1:
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 5.6 (PAGE 326)
y
The integrand is continuous for all t , so F .x/ is defined
and differentiable for all x. Since
limx!˙1 .2x " x 2 / D "1, therefore
limx!˙1 F .x/ D "1. Now
F 0 .x/ D .2 " 2x/ cos
#
1
1 C .2x " x 2 /2
$
yD
D0
only at x D 1. Therefore F must have a maximum value
at x D 1, and no minimum value.
52.
lim
n!1
1
n
"#
1C
1
n
$5
1
n
#
$
#
$ #
2 5
n 5
C 1C
C !!! C 1 C
n
n
D area below y D x 5 , above y D 0,
between x D 1 and x D 2
ˇ2
Z 2
21
1
1 6 ˇˇ
5
D
x dx D x ˇ D .26 " 1/ D
6 ˇ
6
2
1
2.
0
y
yDsin x
3.
::::::
2! 3!
n n
.n!1/!
n
!
x
4.
Fig. 5.5-53
54.
#
$
n
n
n
n
C
C
C
!
!
!
C
n!1 n2 C 1
n2 C 4
n2 C 9
2n2
#
$
2
2
2
n
n
n
n2
1
C
C
C
!
!
!
C
D lim
n!1 n
n2 C 1
n2 C 4
n2 C 9
2n2
1
0
lim
C
1
1
1
C
C
C !!! C
C
$
$
%
#
#
&
2
2
2
n
n!1
A
1
2
1C
1C
1C
n
n
n
1
D area below y D
, above y D 0,
1 C x2
between x D 0 and x D 1
ˇ1
Z 1
ˇ
#
1
!1 ˇ
dx
D
tan
x
D
ˇ D
2
ˇ
4
0 1Cx
D lim
1
n
B
B
B
@
0
1
x
Fig. 5.5-54
1.
#
$
#
#
2#
n#
lim
sin C sin
C ! ! ! C sin
n!1 n
n
n
n
D lim sum of areas of rectangles shown in figure
n!1
ˇ!
Z !
ˇ
ˇ
D
sin x dx D " cos x ˇ D 2
ˇ
0
!
n
2
n
Section 5.6 The Method of Substitution
(page 326)
1
53.
1
1Cx 2
5.
6.
Z
e 5!2x dx
Let u D 5 " 2x
du D "2 dx
Z
1
1
1
u
e du D " e u C C D " e 5!2x C C:
D"
2
2
2
Z
cos.ax C b/ dx Let u D ax C b
du D a dx
Z
1
1
D
cos u du D sin u C C
a
a
1
D sin.ax C b/ C C:
a
Z
p
3x C 4 dx Let u D 3x C 4
du D 3 dx
Z
1
2
2
1=2
D
u du D u3=2 C C D .3x C 4/3=2 C C:
3
9
9
Z
e 2x sin.e 2x / dx Let u D e 2x
du D 2e 2x dx
Z
1
1
sin u du D " cos u C C
D
2
2
1
D " cos.e 2x / C C:
2
Z
x dx
Let u D 4x 2 C 1
.4x 2 C 1/5
du D 8x dx
Z
1
"1
1
!5
u du D " u!4 C C D
C C:
D
8
32
32.4x 2 C 1/4
p
Z
p
sin x
dx Let u D x
p
x
dx
du D p
2 x
Z
D 2 sin u du D "2 cos u C C
p
D "2 cos x C C:
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SECTION 5.6 (PAGE 326)
7.
8.
Z
ADAMS and ESSEX: CALCULUS 9
2
xe x dx
Let u D x 2
du D 2x dx
Z
1
1
1 2
u
D
e du D e u C C D e x C C:
2
2
2
Z
3
x 2 2x C1 dx Let u D x 3 C 1
du D 3x 2 dx
Z
1 2u
1
u
2 du D
CC
D
3
3 ln 2
15.
16.
9.
10.
11.
12.
13.
14.
2x C1
C C:
3 ln 2
Z
cos x
dx Let u D sin x
4 C sin2 x
du D cos x dx
Z
du
D
4 C u2
#
$
1
u
1
1
D tan!1 C C D tan!1
sin x C C:
2
2
2
2
Z
sec2 x
p
dx Let u D tan x
1 " tan2 x
du D sec2 x dx
du
D p
1 " u2
D sin!1 u C C
D sin!1 .tan x/ C C:
Z x
e C1
dx
ex " 1
Z x=2
e
C e !x=2
D
dx Let u D e x=2 " e !x=2
!
"
e x=2 " e x=2
du D 12 e x=2 C e !x=2 dx
Z
du
D2
D 2 ln juj C C
ˇ
ˇ
ˇ
ˇ u
ˇ
ˇ
ˇ
ˇ
D 2 lnˇe x=2 " e !x=2 ˇ C C D lnˇe x C e !x " 2ˇ C C:
Z
17.
18.
19.
20.
ln t
dt
t
Let u D ln t
dt
du D
t
Z
1 2
1
D u du D u C C D .ln t /2 C C:
2
2
Z
ds
Let u D 4 " 5s
p
4 " 5s
du
D "5 ds
Z
1
du
D"
p
5
u
2p
2 1=2
CC D"
4 " 5s C C:
D" u
5
5
Z
xC1
dx Let u D x 2 C 2x C 3
p
2
x C 2x C 3
du D 2.x C 1/ dx
Z
p
p
1
1
p du D u C C D x 2 C 2x C 3 C C
D
2
u
196
21.
22.
t
dt Let u D t 2
4 " t4
du D 2t dt
Z
du
p
4 " u2
# 2$
u
1
t
sin!1 C C D sin!1
C C:
2
2
2
p
1
2
1
D
2
Z
D
3
D
Z
x2
dx Let u D x 3
2 C x6
du D 3x 2 dx #
$
Z
du
1
u
1
!1
D
p
tan
CC
D
p
3
2 C u2
3 2
2
# 3$
1
x
D p tan!1 p
C C:
3 2
2
Z
Z !x
dx
e dx
Let u D 1 C e !x
D
ex C 1
1 C e !x
du D "e !x dx
Z
du
D " ln juj C C D " ln.1 C e !x / C C:
D"
u
Z
Z
dx
e x dx
D
Let u D e x
e x C e !x
e 2x C 1
du D e x dx
Z
du
D tan!1 u C C
D
u2 C 1
D tan!1 e x C C:
Z
tan x ln cos x dx Let u D ln cos x
du D " tan x dx
Z
&2
1%
1 2
D " u du D " u C C D " ln cos x C C:
2
2
Z
xC1
p
dx
2
Z
Z 1"x
dx
x dx
C
Let u D 1 " x 2
D
p
p
2
1"x
1 " x2
du D "2x dx
in the first integral only
Z
p
1
du
D"
p C sin!1 x D " u C sin!1 x C C
2
u
p
D " 1 " x 2 C sin!1 x C C:
Z
Z
dx
dx
D
Let u D x C 3
x 2 C 6x C 13
.x C 3/2 C 4
du
D dx
Z
1
du
!1 u
D tan
CC
D
u2 C 4
2
2
1
xC3
D tan!1
C C:
2
2
Z
dx
dx
p
D p
Let u D 1 " x
2
4 C 2x " x
5 " .1 " x/2
#
$ du D "dx
Z
u
du
!1
D " sin
CC
D" p
p
5#
5#" u2 $
$
1"x
x"1
D " sin!1 p
C C D sin!1 p
C C:
5
5
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INSTRUCTOR’S SOLUTIONS MANUAL
23.
Z
sin3 x cos5 x dx
Z
D sin x.cos5 x " cos7 x/ dx
D
D
24.
SECTION 5.6 (PAGE 326)
Z
.u7 " u5 / du
28.
Let u D cos x
du D " sin x dx
u8 u6
cos8 x cos6 x
"
CC D
"
C C:
8
6
8
6
Z
29.
sin4 t cos5 t dt
Z
D sin4 t .1 " sin2 t /2 cos t dt
Let u D sin t
du D cos t dt
Z
u5 2u7
u9
4
6
8
D .u " 2u C u / du D
"
C
CC
5
7
9
1
2
1
D sin5 t " sin7 t C sin9 t C C:
5
7
9
25.
Z
sin ax cos2 ax dx
Let u D cos ax
du D "a sin ax dx
Z
1
u2 du
a
u3
1
D"
C C D " cos3 ax C C:
3a
3a
31.
D"
26.
30.
Z #
$
sin 2x 2
sin x cos x dx D
dx
2
Z
1 " cos 4x
x sin 4x
1
dx D "
C C:
D
4
2
8
32
Z
2
2
Z
Œ1 C cos.2x/!2
cos4 x dx D
dx
4
Z
1
D
Œ1 C 2 cos.2x/ C cos2 .2x/! dx
4
Z
x
sin.2x/
1
D C
C
1 C cos.4x/ dx
4
4
8
x
sin.2x/
x
sin.4x/
D C
C C
CC
4
4
8
32
sin.2x/
sin.4x/
3x
C
C
C C:
D
8
4
32
Z
sec5 x tan x dx Let u D sec x
du D sec x tan x dx
Z
u5
sec5 x
4
D u du D
CC D
C C:
5
5
Z
sec6 x tan2 x dx
Z
D sec2 x tan2 x.1 C tan2 x/2 dx Let u D tan x
du D sec2 x dx
Z
1 3 2 5 1 7
2
4
6
D .u C 2u C u / du D u C u C u C C
3
5
7
2
1
1
3
5
7
D tan x C tan x C tan x C C:
3
5
7
Z
p
tan x sec4 x dx
Z
p
tan x.1 C tan2 x/ sec2 x dx Let u D tan x
D
du D sec2 x dx
Z %
&
D
u1=2 C u5=2 du
2u3=2
2u7=2
C
CC
3
7
2
2
D .tan x/3=2 C .tan x/7=2 C C:
3
7
Z
sin!2=3 x cos3 x dx Let u D sin x
du D cos x dx
Z
3
1 " u2
du D 3u1=3 " u7=3 C C
D
7
u2=3
3 7=3
1=3
D 3 sin x " sin x C C:
7
Z
cos x sin4 .sin x/ dx Let u D sin x
du D cos x dx
$
Z
Z #
1 " cos 2u 2
4
D sin u du D
du
2
$
Z #
1 C cos 4u
1
1 " 2 cos 2u C
du
D
4
2
3u sin 2u
sin 4u
D
"
C
CC
8
4
32
1
1
3
sin.4 sin x/ C C:
D sin x " sin.2 sin x/ C
8
4
32
D
32.
27.
Z
$
Z #
1 " cos 2x 3
sin6 x dx D
dx
2
Z
1
D
.1 " 3 cos 2x C 3 cos2 2x " cos3 2x/ dx
8
Z
3
x 3 sin 2x
.1 C cos 4x/ dx
C
D "
8
16 Z
16
1
"
cos 2x.1 " sin2 2x/ dx Let u D sin 2x
8
du D 2 cos 2x dx
Z
5x 3 sin 2x
3 sin 4x
1
D
"
C
"
.1 " u2 / du
16
16
64
16
3 sin 4x sin 2x
sin3 2x
5x 3 sin 2x
"
C
"
C
CC
D
16
16
64
16
48
3
3 sin 4x
sin 2x
5x sin 2x
"
C
C
C C:
D
16
4
64
48
Z
33.
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SECTION 5.6 (PAGE 326)
34.
35.
36.
Z
sin3 .ln x/ cos3 .ln x/
dx
x
Let u D sin.ln x/
cos.ln x/
dx
du D
x
Z
1
1
D u3 .1 " u2 / du D u4 " u6 C C
4
6
1
1
D sin4 .ln x/ " sin6 .ln x/ C C:
4
6
Z
sin2 x
dx
4
Zcos x
D tan2 x sec2 x dx Let u D tan x
du D sec2 x dx
Z
3
u
1
D u2 du D
C C D tan3 x C C:
3
3
Z
Z
3
sin x
dx D tan3 x sec x dx
4x
cos
Z
D .sec2 x " 1/ sec x tan x dx Let u D sec x
du D sec x tan x dx
Z
D
Z
40.
1
3
sec3 x " sec x C C:
D"
Z
39.
Let u D # ln x
#
du D dx
x
ˇ!=2
Z
ˇ
1 !=2
1
ˇ
D
sin u du D " cos uˇ
ˇ
# 0
#
41.
Z
!=2
0
sin4 x dx D
Z
!=2
0
#
1 " cos 2x
2
$2
$
#
1 C cos 4x
1 " 2 cos 2x C
dx
2
0
ˇ!=2
ˇ!=2
ˇ!=2
3x ˇˇ
sin 2x ˇˇ
sin 4x ˇˇ
3#
:
D
"
C
D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
8
4
32
16
1
D
4
Z
!=2
0
Z
!
0
0
sin5 x dx
!=4
D
Z
!
!=4
.1 " cos2 x/2 sin x dx
Let u D cos x
du D " sin x dx
ˇ1=p2
ˇ
1
2
ˇ
D " p .1 " 2u2 C u4 / du D u " u3 C u5 ˇ
ˇ
3
5
1= 2
!1
!
1
1
1
2 1
43
8
D p " p C p " "1 C "
D p C :
3 5
15
2 3 2
20 2
60 2
Z
Let u D csc x
du D " csc x cot x dx
.u8 " 2u6 C u4 / du
u9
2u7 u5
C
"
CC
9
7
5
1
2
1
D " csc9 x C csc7 x " csc5 x C C:
9
7
5
Z
Z
4
cos x
dx D cot4 x csc4 x dx
8
sin
x
Z
D cot4 x.1 C cot2 x/ csc2 x dx Let u D cot x
du D " csc2 x dx
Z
5
7
u
u
D " u4 .1 C u2 / du D "
"
CC
5
7
1
1
D " cot5 x " cot7 x C C:
5
7
Z 4
x 3 .x 2 C 1/!1=2 dx Let u D x 2 C 1; x 2 D u " 1
0
du D 2x dx
Z
1 17
!1=2
.u " 1/u
du
D
2 1
#
$ˇˇ17
1 2 3=2
ˇ
D
u
" 2u1=2 ˇ
ˇ
2 3
1
p
p
p
2
17 17 " 1
14 17
" . 17 " 1/ D
C :
D
3
3
3
198
sin.# ln x/
dx
x
0
D"
38.
p
e
1
1
D " .0 " 1/ D :
#
#
42.
csc5 x cot5 x dx
Z
D csc x cot x csc4 x.csc2 x " 1/2 dx
Z
1
.u2 " 1/ du D 13 u3 " u C C
D
37.
ADAMS and ESSEX: CALCULUS 9
43.
Z
e2
e
D
44.
Z
Z
!1
dt
t ln t
2
1
! 2 =9
Let u D ln t
dt
du D
t
ˇ2
du
ˇ
D ln uˇ D ln 2 " ln 1 D ln 2:
1
u
2sin
p
x
p
p
Let u D sin x
p
cos x
du D p dx
2 x
ˇp3=2
Z p3=2
u ˇ
2.2 / ˇ
D 2 p 2u du D
ˇ
ln 2 ˇ p
1= 2
! 2 =16
cos
p
x
x
dx
1= 2
p
2 p3=2
D
.2
" 21= 2 /:
ln 2
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INSTRUCTOR’S SOLUTIONS MANUAL
45.
Z
Z
r
x
2 cos2 dx
2
0
ˇ!=2
Z
!=2
p
p
x
x ˇˇ
D 2
cos dx D 2 2 sin ˇ
D 2:
2
2ˇ
0
!=2
0
0
Z
1 C cos x dx D
p
Z
!=2
49.
We start with the addition formulas
cos.x C y/ D cos x cos y " sin x sin y
cos.x " y/ D cos x cos y C sin x sin y
0
Z
0
q
1 " cos
p
!=2
!!
2
"
" x dx
!=2
r
&
1%
cos.x C y/ C cos.x " y/
2
&
1%
cos.x " y/ " cos.x C y/ :
sin x sin y D
2
Let u D !2 " x
du D "dx
cos x cos y D
1 " cos u du
ˇ!=2
p %
u
u &ˇˇ
du D 2 "2 cos
D
2 sin
ˇ
2
2 ˇ
0
0
p
p
D "2 C 2 2 D 2. 2 " 1/:
Z 2
x
Area D
dx Let u D x 2 C 16
2
0 x C 16
du D 2x dx
ˇ20
Z 20
ˇ
du
1
1
ˇ
D
D ln uˇ
ˇ
2 16 u
2
16
# $
5
1
1
sq: units:
D .ln 20 " ln 16/ D ln
2
2
4
Z 2
x dx
Area R D
Let u D x 2
4 C 16
x
0
du D 2x dx
ˇ4
Z
ˇ
1 4 du
1
#
!1 u ˇ
D
D
tan
sq. units:
ˇ D
2
ˇ
2 0 u C 16
8
4
32
Z
and take half their sum and half their difference to obtain
1 " sin x dx
!=2
D"
47.
p
0
!=2
D
46.
SECTION 5.6 (PAGE 326)
Similarly, taking half the sum of the formulas
2
0
y
yD
x
x 4 C16
R
x
Fig. 5.6-47
48. The area bounded by the ellipse .x 2 =a2 / C .y 2 =b 2 / D 1 is
s
Z a
x2
4
b 1 " 2 dx Let x D au
a
0
dx D adu
Z 1p
2
D 4ab
1 " u du:
0
The integral is the area of a quarter circle of radius 1.
Hence
$
#
#.1/2
D #ab sq: units:
Area D 4ab
4
sin.x C y/ D sin x cos y C cos x sin y
sin.x " y/ D sin x cos y " cos x sin y;
we obtain
sin x cos y D
50.
&
1%
sin.x C y/ C sin.x " y/ :
2
We have
Z
cos ax cos bx dx
Z
1
D
Œcos.ax " bx/ C cos.ax C bx/! dx
2Z
Z
1
1
D
cosŒ.a " b/x! dx C
cosŒ.a C b/x! dx
2
2
Let u D .a " b/x, du D .a " b/ dx in the first integral;
let v D .a C b/x, dv D .a C b/ dx in the second integral.
Z
Z
1
1
cos u du C
cos v dv
D
2.a " b/
2.a C b/
(
)
1 sinŒ.a " b/x!
sinŒ.a C b/x!
D
C
C C:
2
.a " b/
.a C b/
Z
sin ax sin bx dx
Z
1
Œcos.ax " bx/ " cos.ax C bx/! dx
D
2
(
)
1 sinŒ.a " b/x! sinŒ.a C b/x!
D
"
C C:
2
.a " b/
.a C b/
Z
sin ax cos bx dx
Z
1
D
Œsin.ax C bx/ C sin.ax " bx/! dx
2 Z
Z
1
D Œ sinŒ.a C b/x! dx C sinŒ.a " b/x! dx!
2
(
)
cosŒ.a " b/x!
1 cosŒ.a C b/x!
C
C C:
D"
2
.a C b/
.a " b/
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SECTION 5.6 (PAGE 326)
51.
ADAMS and ESSEX: CALCULUS 9
For m D 0 we have
Z !
Z
f .x/ cos mx dx D
If m and n are integers, and m ¤ n, then
)
Z !(
cos mx cos nx
dx
!! sin mx sin nx
Z
&
1 !%
D
cos.m " n/x ˙ cos.m C n/x dx
2 !!
#
$ˇ!
sin.m C n/x ˇˇ
1 sin.m " n/x
˙
D
ˇ
ˇ
2
m"n
mCn
f .x/ dx
Z !
a0
dx
D
2 !!
k
X
C
.an cos.nx/ C bn sin.nx// dx
!!
!!
nD1
!!
D 0 ˙ 0 D 0:
Z !
sin mx cos nx dx
!!
Z
&
1 !%
D
sin.m C n/x C sin.m " n/x dx
2 !!
#
$ˇ!
cos.m " n/x ˇˇ
1 cos.m C n/x
C
D"
ˇ
ˇ
2
mCn
m"n
!!
D 0 (by periodicity).
!
D
a0
.2#/ C 0 C 0 D a0 #;
2
so the formula for am holds for m D 0 also.
Section 5.7 Areas of Plane Regions
(page 331)
1.
Z
1
.x " x 2 / dx
$ˇ1
# 2
x
x 3 ˇˇ
1
1 1
D
"
ˇ D " D sq. units.
2
3 ˇ
2 3
6
Area of R D
If m D n ¤ 0 then
Z !
sin mx cos mx dx
!!
Z
1 !
sin 2mx dx
D
2 !!
ˇ!
ˇ
1
ˇ
D"
cos 2mx ˇ D 0 (by periodicity).
ˇ
4m
0
0
y
.1;1/
yDx
R
yDx 2
!!
52. If 1 $ m $ k, we have
Z
Z !
a0 !
cos mx dx
f .x/ cos mx dx D
2 !!
!!
Z !
k
X
C
an
cos nx cos mx dx
nD1
C
k
X
nD1
bn
Z
x
Fig. 5.7-1
2.
!!
!
sin nx cos mx dx:
Z
1
p
. x " x 2 / dx
0
#
$ˇ1
2 3=2 1 3 ˇˇ
1
2 1
D
x
" x ˇ D " D sq. units.
ˇ
3
3
3 3
3
Area of R D
0
y
!!
By the previous exercise, all the integrals on the right side
are zero except the one in the first sum having n D m.
Thus the whole right side reduces to
Z !
Z !
1 C cos.2mx/
dx
am
cos2 .mx/ dx D am
2
!!
!!
am
D
.2# C 0/ D #am :
2
Thus
am D
1
#
Z
R
yDx 2
Fig. 5.7-2
f .x/ cos mx dx:
!!
.1;1/
x
!
A similar argument shows that
Z
1 !
f .x/ sin mx dx:
bm D
# !!
200
p
yD x
3.
Z
2
.8 " 2x 2 / dx
$ˇˇ2
#
4
64
ˇ
D 16x " x 3 ˇ D
sq. units.
ˇ
3
3
Area of R D 2
0
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INSTRUCTOR’S SOLUTIONS MANUAL
y
SECTION 5.7 (PAGE 331)
y
yD3!x 2
2yD4x!x 2
!2
2
6
x
R
1
x
R
2yC3xD6
yDx 2 !5
Fig. 5.7-5
Fig. 5.7-3
4.
6.
For intersections:
2
2
x " 2x D 6x " x ) 2x " 8x D 0
i:e:; x D 0 or 4:
Z 4h
i
Area of R D
6x " x 2 " .x 2 " 2x/ dx
0
Z 4
D
.8x " 2x 2 / dx
0
$ˇ4
#
2 3 ˇˇ
64
2
sq. units.
D 4x " x ˇ D
ˇ
3
3
0
y
For intersections:
7 C y D 2y 2 " y C 3 ) 2y 2 " 2y " 4 D 0
2.y " 2/.y C 1/ D 0 ) i:e:; y D "1 or 2:
Z 2
Area of R D
Œ.7 C y/ " .2y 2 " y C 3/! dy
!1
Z 2
D2
.2 C y " y 2 / dy
!1
$ˇ2
#
1 2 1 3 ˇˇ
D 2 2y C y " y ˇ D 9 sq. units.
ˇ
2
3
2
!1
y
yD6x!x 2
.4;8/
.9;2/
xD2y 2 !yC3
R
x!yD7
R
yDx 2 !2x
x
.6;!1/
x
Fig. 5.7-6
Fig. 5.7-4
5. For intersections:
7.
2y D 4x " x 2
2y C 3x D 6
*
)
4x " x 2 D 6 " 3x
x 2 " 7x C 6 D 0
.x " 1/.x " 6/ D 0
Thus intersections of the curves occur at x D 1 and x D 6.
We have
Z
6
#
3x
x2
Area of R D
"3C
2x "
2
2
1
ˇ6
$
# 2
ˇ
x3
7x
ˇ
"
" 3x ˇ
D
ˇ
4
6
$
Z
1
.x " x 3 / dx
# 2
$ˇ1
x
x 4 ˇˇ
1
D2
"
ˇ D sq. units.
2
4 ˇ
2
Area of R D 2
0
0
y
.1;1/
yDx
R
dx
R
yDx 3
x
.!1;!1/
1
245
1
125
D
" 36 C " 15 D
sq. units.
4
6
12
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Fig. 5.7-7
201
lOMoARcPSD|6566483
SECTION 5.7 (PAGE 331)
8.
Shaded area D
D
#
Z
1
0
ADAMS and ESSEX: CALCULUS 9
y
.x 2 " x 3 / dx
$ˇ1
1
1 3 1 4 ˇˇ
x " x ˇ D
sq. units.
ˇ
3
4
12
.4;2/
xD2y 2 !y!2
xDy 2
A
0
y
x
.1;!1/
.1;1/
yDx 2
Fig. 5.7-10
yDx 3
x
11.
Fig. 5.7-8
1
5 " 2x
DyD
.
x
2
2
Thus 2x " 5x C 2 D 0, i.e., .2x " 1/.x " 2/ D 0. The
graphs intersect at x D 1=2 and x D 2. Thus
For intersections:
$
1
5 " 2x
"
dx
2
x
1=2
#
$ˇˇ2
5x x 2
ˇ
D
"
" ln x ˇ
ˇ
2
2
Area of R D
9.
Z
#
2
1=2
1
p
. x " x 3 / dx
0
$ˇ1
#
5
2 3=2 x 4 ˇˇ
x
"
sq. units.
D
ˇ D
3
4 ˇ
12
Area of R D
Z
15
" 2 ln 2 sq. units.
D
8
y
0
y
%
&
1
;2
2
2xC2yD5
xDy 2
p
yD x
R
%
R
.1;1/
yD1=x
1
2; 2
&
yDx 3
x
x
Fig. 5.7-11
Fig. 5.7-9
12.
Z
1
Œ.1 " x 2 / " .x 2 " 1/2 ! dx
$ˇˇ1
#
Z 1
1
4
1
ˇ
sq. units.
D2
.x 2 " x 4 / dx D 2 x 3 " x 5 ˇ D
ˇ
3
5
15
0
Area of shaded region D 2
0
0
y
10.
For intersections:
2
2
yD.x 2 !1/2
2
yD1!x 2
y D 2y " y " 2 ) y " y " 2 D 0
.y " 2/.y C 1/ D 0 ) i:e:; y D "1 or 2:
Z 2
Area of R D
Œy 2 " .2y 2 " y " 2/! dy
A
!1
$ˇ2
#
1 2 1 3 ˇˇ
D
Œ2 C y " y ! dy D 2y C y " y ˇ
ˇ
2
3
!1
Z
D
2
9
sq. units.
2
202
A
2
x
!1
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INSTRUCTOR’S SOLUTIONS MANUAL
13. The curves y D
Thus
SECTION 5.7 (PAGE 331)
1
x2
and y D
intersect at x D ˙1.
2
1 C x2
y
1
#
R
0
!2
yD
1
1
x
2
Fig. 5.7-15
16.
2
Z
!
.sin y " .y 2 " # 2 // dy
$ˇ!
#
4# 3
y 3 ˇˇ
2
sq. units.
D " cos y C # y "
ˇ D
3 ˇ
3
Area A D
R
!1
R
!1
y
1
yD 2
x C1
x2
4
x2
yD5!x 2
$
1
x2
Area of R D 2
"
dx
1 C x2
2
0
ˇ
$1
#
#
1
x 3 ˇˇ
D 2 tan!1 x "
ˇ D " sq. units.
6 ˇ
2
3
Z
yD
!!
y
x
!!
xDsin y
Fig. 5.7-13
A
14.
For intersections:
4x
D 1 ) x 2 " 4x C 3 D 0
3 C x2
i:e:; x D 1 or 3:
#
Z 3"
4x
Shaded area D
" 1 dx
3 C x2
1
ˇ3
ˇ
ˇ
2
D Œ2 ln.3 C x / " x!ˇ D 2 ln 3 " 2 sq. units.
ˇ
x
xDy 2 !! 2
Fig. 5.7-16
17.
1
Area of R D
4x
3Cx 2
yD1
.1;1/
5!=4
!=4
.sin x " cos x/ dx
ˇ5!=4
ˇ
D ".cos x C sin x/ˇˇ
y
yD
Z
D
.3;1/
p
2C
y
p
p
!=4
2 D 2 2 sq. units.
yDsin x
x
5!=4
R
x
!=4
yDcos x
Fig. 5.7-14
15.
4
and y D 5 " x 2 intersect where
x2
x 4 " 5x 2 C 4 D 0, i.e., where .x 2 " 4/.x 2 " 1/ D 0. Thus
the intersections are at x D ˙1 and x D ˙2. We have
Fig. 5.7-17
The curves y D
#
$
4
5 " x 2 " 2 dx
x
1
$ˇ2
#
3
4 ˇˇ
4
x
C
D 2 5x "
ˇ D sq. units.
3
x ˇ
3
Area of R D 2
Z
2
1
18.
Area D
Z
!=2
.1 " sin2 x/ dx
!!=2
Z !=2
1 C cos.2x/
dx
2
$ˇ!=2
#
#
sin.2x/ ˇˇ
D
sq. units.
D xC
ˇ
ˇ
2
2
D2
0
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SECTION 5.7 (PAGE 331)
ADAMS and ESSEX: CALCULUS 9
y
y
yDtan x
yD
yD1
A
!! "
4 ;1
4x
!
yDsin2 x
x
! x
2
Fig. 5.7-21
Fig. 5.7-18
19.
Area A D
Z
22.
!=2
2
.sin x " sin x/ dx
0
For intersections: x 1=3 D tan.#x=4/. Thus x D ˙1.
Z
%
#x &
x 1=3 " tan
dx
4
0
#
$ˇ1
3 4=3 4 ˇˇ
#x ˇˇ ˇˇ
D2
x
" ln ˇsec
ˇ ˇ
ˇ
4
#
4
0
3
3
8 p
4
D " ln 2 D " ln 2 sq. units.
2 #
2 #
#
$ˇ!=2
sin x cos x " x ˇˇ
#
D " cos x C
D1"
sq. units.
ˇ
ˇ
2
4
Area A D 2
0
y
1
yDsin x
y
yDsin2 x
A
A
!x
2
yDtan.!x=4/
x
1
A
Fig. 5.7-19
yDx 1=3
20.
Area A D 2
D2
Z
!=4
0
Z
0
!=4
.cos2 x " sin2 x/ dx
ˇ!=4
ˇ
ˇ
cos.2x/ dx D sin.2x/ˇ
D 1 sq. units.
ˇ
0
Fig. 5.7-22
23.
For intersections: sec x D 2. Thus x D ˙#=3.
y
Area A D 2
A
.2 " sec x/ dx
yD2
Fig. 5.7-20
A
#
4x
D tan x ) x D 0 or :
For intersections:
#
4
$
Z !=4 #
4x
Area D
" tan x dx
#
0
$ˇˇ!=4
#
#
1
2 2
ˇ
D " ln 2 sq. units.
x " ln j sec xj ˇ
D
ˇ
#
8
2
0
204
0
y
! x
4
21.
!=3
ˇ!=3
ˇ
ˇ
D .4x " 2 ln j sec x C tan xj/ˇ
ˇ
0
p
4#
D
" 2 ln.2 C 3/ sq. units.
3
yDcos2 x
yDsin2 x
Z
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yDsec x
!
!3
!
3
Fig. 5.7-23
x
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INSTRUCTOR’S SOLUTIONS MANUAL
24.
p
For intersections: jxj D
Area A D 2
D
Z
SECTION 5.7 (PAGE 331)
y
2 cos.#x=4/. Thus x D ˙1.
&
#x
" x dx
4
0
!ˇ1
p
ˇ
8 2
#x
2 ˇ
sin
"x ˇ
ˇ
#
4
1
%p
2 cos
yDxC2
A
x1
0
x2
8
D " 1 sq. units.
#
Area of R D 4
p
!x
yD 2 cos
4
x
Z
D4
Z
D2
Z
A
yDjxj
1
x
Fig. 5.7-26
27.
y
yDe x
1
p
0
1
0
x 2 " x 4 dx
p
x 1 " x 2 dx
1
1=2
u
0
Let u D 1 " x 2
du D "2x dx
ˇ1
4 3=2 ˇˇ
4
du D u ˇ D sq. units.
ˇ
3
3
0
y
Fig. 5.7-24
y 2 Dx 2 !x 4
R
R
x
25. For intersections: x D sin.#x=2/. Thus x D ˙1.
Z
% #x
&
sin
" x dx
2
0
#
$ˇˇ1
4
#x
2 ˇ
D " cos
"x ˇ
ˇ
#
2
Area A D 2
1
0
Fig. 5.7-27
28.
4
D " 1 sq. units.
#
Loop area D 2
D2
y
Z
p
2
0
Z
0
!2
p
x 2 2 C x dx
.u2 " 2/2 u.2u/ du D 4
#
1
4
4
D 4 u7 " u5 C u3
7
5
3
A
1
Z
0
!x
yDsin 2
yDx
0
!2 A
x
y 2 Dx 4 .2Cx/
x
x2
.x C 2 " e x / dx
# 2
$ˇˇx2
x
x ˇ
D
C 2x " e ˇ
ˇ
2
Area A D
x1
.u6 " 4u4 C 4u2 / du
p
p
$ˇˇ 2
256 2
ˇ
sq. units:
ˇ D
ˇ
105
Fig. 5.7-25
Z
2
y
x
26. For intersections: e D x C2. There are two roots, both of
which must be found numerically. We used a TI-85 solve
routine to get x1 ' "1:841406 and x2 ' 1:146193. Thus
Let u2 D 2 C x
2u
du D dx
p
Fig. 5.7-28
29.
The tangent line to y D e x at x D 1 is y " e D e.x " 1/,
or y D ex. Thus
x1
' 1:949091 sq. units.
Z
1
.e x " ex/ dx
#
$ˇ1
ex 2 ˇˇ
e
D ex "
ˇ D " 1 sq. units.
2 ˇ
2
Area of R D
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SECTION 5.7 (PAGE 331)
ADAMS and ESSEX: CALCULUS 9
y
yDe x
3.
Z
.1;e/
R
xi D 1 C .2i=n/, .i D 0; 1; 2; : : : ; n/, "xi D 2=n.
yDex
3
f .x/ dx D lim
n!1
1
30. The tangent line to y D x 3 at .1; 1/ is y " 1 D 3.x " 1/, or
y D 3x " 2. The intersections of y D x 3 and this tangent
line occur where x 3 " 3x C 2 D 0. Of course x D 1
is a (double) root of this cubic equation, which therefore
factors to .x " 1/2 .x C 2/ D 0. The other intersection is at
x D "2. Thus
4.
1
.x 3 " 3x C 2/ dx
# 4
$ˇˇ1
x
3x 2
ˇ
D
"
C 2x ˇ
ˇ
4
2
p
Pn
Rn D p
iD1 .1=n/ 1 C .i=n/ is a Riemann sum for
f .x/ D 1 C x on the interval Œ0; 1!. Thus
!2
lim Rn D
!2
y
R
1
p
0
0
5.
Z
!
.2 " sin x/ dx D 2.2#/ "
!!
p
.1;1/
x
Z
1 C x dx
ˇ1
p
ˇ
4 2"2
2
3=2 ˇ
:
D .1 C x/ ˇ D
ˇ
3
3
n!1
27
15 3
sq. units.
D" " C6C2C4D
4
2
4
yDx 3
2
n
$ #
$
)
n (#
4i 2
2X
4i
4i
1C
C 2 " 2C
C3
n!1 n
n
n
n
iD1
)
n (
X
4
2
2 C 2 i2
D lim
n!1 n
n
# iD1
$
4
8 n.n C 1/.2n C 1/
D lim
nC 3
n!1 n
n
6
20
8
D4C D
3
3
Fig. 5.7-29
Z
iD1
.xi2 " 2xi C 3/
D lim
x
Area of R D
n
X
6.
yD3x!2
Z
5
0
p
!!
sin x dx D 4# " 0 D 4#
1 p 2
5#
#. 5/ D
4
4
R3%
x&
dx D area A1 " area A2 D 0
1 1"
2
y
.!2;!8/
!
5 " x 2 dx D 1/4 of the area of a circle of radius
D
7.
Z
x
yD1! 2
A1
Fig. 5.7-30
1
3
A2
x
Review Exercises 5 (page 332)
Fig. R-5-7
1.
1
1
j 2 C 2j C 1 " j 2
2j C 1
"
D
D 2
2
2
j
.j C 1/
j 2 .j C 1/2
j .j C 1/2
$
n #
n
X
X
1
1
2j C 1
D
"
j 2 .j C 1/2
j 2 .j C 1/2
j D1
j D1
8.
R!
0
cos x dx D area A1 " area A2 D 0
y
1
n C 2n
1
D 2 "
D
1
.n C 1/2
.n C 1/2
iD1
206
i.i C 10/ D
.30/.31/.61/
.30/.31/
C 10
D 14;105:
6
2
x
Fig. R-5-8
Z
!
1
1
.2 " sin.x 3 // dx D
Œ2.2#/ " 0! D 2
2# !!
2#
Z
1 5
5
1 3
jx " 2j dx D
D (via #9)
10. h D
3 0
3 2
6
9.
40 & 30 C 39 & 29 C ! ! ! C 12 & 2 C 11 & 1
30
X
!
A2
2
2. The number of balls is
D
yDcos x
A1
f D
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INSTRUCTOR’S SOLUTIONS MANUAL
11.
Z
f .t / D
12. f .x/ D
13. g.s/ D
14.
t
sin.x 2 / dx;
13
Z
sin x
!13
Z
1
p
4s
f 0 .x/ D
p
1 C sin2 x.cos x/
e cos "
ln x dx
e sin "
cos "
g 0 .& / D .ln.e
//e cos " ." sin & / " .ln.e sin " //e sin " cos &
D " sin & cos &.e cos " C e sin " /
15.
2f .x/ C 1 D 3
0
16.
Z
f .t / dt
x
21.
y D sin x and y D cos.2x/ intersect at x D #=6, but
nowhere else in the interval Œ0; #=6!. The area between
the curves in that interval is
ˇ!=6
Z !=6
!1
"ˇˇ
.cos.2x/ " sin x/ dx D 2 sin.2x/ C cos x ˇ
ˇ
0
0
p
p
p
3
3
3 3
D
C
"1D
" 1 sq. units.:
4
2
4
22.
y D 5 " x 2 and y D 4=x 2 meet where 5 " x 2 D 4=x 2 , that
is, where
x 4 " 5x 2 C 4 D 0
!3x=2
.x 2 " 1/.x 2 " 4/ D 0:
There are four intersections: x D ˙1 and x D ˙2. By
symmetry (see the figure) the total area bounded by the
curves is
$
#
$ˇ2
Z 2#
4
4 ˇˇ
4
x3
2
2
5 " x " 2 dx D 2 5x "
C
ˇ D sq. units.
ˇ
x
3
x
3
1
0
Now, solving for I , we get
17.
!
0
xf .sin x/ dx D I D
#
2
Z
y D 4x " x 2 and y D 3 meet where x 2 " 4x C 3 D 0, that
is, at x D 1 and x D 3. Since 4x " x 2 # 3 on Œ1; 3!, the
required area is
#
$ˇˇ3
Z 3
x3
4
ˇ
2
2
.4x " x " 3/ dx D 2x "
" 3x ˇ D sq. units.
ˇ
3
3
1
1
1
2f .x/ D "3f .x/ % f .x/ D C e
2f .1/ C 1 D 0
1
1
" D f .1/ D C e !3=2 % C D " e 3=2
2
2
1
f .x/ D " e .3=2/.1!x/ :
2
Z !
I D
xf .sin x/ dx Let x D # " u
0
dx D "du
Z 0
D"
.# " u/f .sin.# " u// du (but sin.# " u/ D sin u)
Z!!
Z !
D#
f .sin u/ du "
uf .sin u/ du
0
Z0 !
D#
f .sin x/ dx " I:
Z
x D y " y 4 and x D 0 intersect where y " y 4 D 0, that
is, at y D 0 and y D 1. Since y " y 4 # 0 on Œ0; 1!, the
required area is
# 2
$ˇ1
Z 1
y
y 5 ˇˇ
3
4
.y " y " 0/ dy D
"
sq. units.
ˇ D
ˇ
2
5
10
0
0
g 0 .s/ D "4e sin.4s/
20.
Z
g.& / D
19.
f 0 .t / D sin.t 2 /
1 C t 2 dt;
e sin u du;
REVIEW EXERCISES 5 (PAGE 332)
!
1
f .sin x/ dx:
y
0
y D 2 C x " x 2 and y D 0 intersect where 2 C x " x 2 D 0,
that is, where .2 " x/.1 C x/ D 0, namely at x D "1 and
x D 2. Since 2 C x " x 2 # 0 on Œ"1; 2!, the required area
is
Z
y D 5 " x2
A
$ˇ2
#
2
9
x 2 x 3 ˇˇ
2
"
.2 Cx "x / dx D 2x C
ˇ D sq. units.:
2
3 ˇ
2
!1
A
!1
1
18. The area bounded by y D .x " 1/2 , y D 0, and x D 0 is
Z
1
0
ˇ1
.x " 1/3 ˇˇ
1
.x " 1/ dx D
ˇ D sq. units.:
ˇ
3
3
2
0
4
yD 2
x
2
x
Fig. R-5-22
23.
Z
x 2 cos.2x 3 C 1/ dx
Let u D 2x 3 C 1
du D 6x 2 dx
Z
1
sin u
sin.2x 3 C 1/
D
cos u du D
CC D
CC
6
6
6
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REVIEW EXERCISES 5 (PAGE 332)
24.
Z
e
ADAMS and ESSEX: CALCULUS 9
ln x
dx
x
Since 1=.1 C t 2 / > 0 for all t , F .x/ will be minimum
when
x 2 " 2x D .x " 1/2 " 1
Let u D ln x
1
du D dx=x
ˇ1
Z 1
u2 ˇˇ
1
D
u du D
ˇ D
ˇ
2
2
0
is minimum, that is, when x D 1. The minimum value is
0
25.
Z
p
9t 2 C t 4 dt
0
Z 4 p
D
t 9 C t 2 dt
4
F .1/ D
Let u D 9 C t 2
du D 2t dt
ˇ25
Z
1 25 p
1 3=2 ˇˇ
98
D
u du D u ˇ D
ˇ
2 9
3
3
9
Z
sin3 .#x/ dx
Z
%
&
D sin.#x/ 1 " cos2 .#x/ dx Let u D cos.#x/
du D "# sin.#x/ dx
Z
1
2
.1 " u / du
D"
#
# 3
$
1 u
1
1
D
"u CC D
cos3 .#x/ " cos.#x/ C C
# 3
3#
#
Z ln 2
eu
du Let v D e u
4 C e 2u
0
dv D e u du
Z 2
dv
D
2
1 4Cv
ˇ2
1
v ˇˇ
1
1
#
D tan!1 ˇ D " tan!1
2
2ˇ
8
2
2
27.
!1
0
ˇ!1
ˇ
dt
#
!1 ˇ
D
tan
t
ˇ D" :
ˇ
1 C t2
4
0
F has no maximum value; F .x/ < #=2 for all x, but
F .x/ ! #=2 if x 2 " 2x ! 1, which happens as
x ! ˙1.
0
26.
Z
32.
f .x/ D 4x " x 2 # 0 if 0 $ x $ 4, and f .x/ < 0
Rb
otherwise. If a < b, then a f .x/ dx will be maximum if
Œa; b! D Œ0; 4!; extending the interval to the left of 0 or to
the right of 4 will introduce negative contributions to the
integral. The maximum value is
Z
4
0
33.
0
The average value of v.t / D dx=dt over Œt0 ; t1 ! is
1
t1 " t0
34.
#
$ˇ4
x 3 ˇˇ
32
.4x " x 2 / dx D 2x 2 "
:
ˇ D
3 ˇ
3
Z
t1
t0
ˇt1
ˇ
x.t1 / " x.t0 /
dx
1
ˇ
dt D
x.t /ˇ D
D vav :
ˇ
dt
t1 " t0
t1 " t0
t0
If y.t / is the distance the object falls in t seconds from its
release time, then
1
28.
Z
p
e
4
y 00 .t / D g;
tan2 .# ln x/
dx
x
Let u D # ln x
1
du D .#=x/ dx
Z
Z
1 !=4
1 !=4 2
tan u du D
.sec2 u " 1/ du
D
# 0
# 0
ˇ!=4
ˇ
1
1
1
ˇ
D "
D .tan u " u/ˇ
ˇ
#
#
4
29.
The average height during the time interval Œ0; T ! is
Z
Z
t
t
1
2t
30.
cos
sin2 dt D
sin2
dt
5
5
4
5
$
Z #
1
4t
D
1 " cos
dt
8
5
#
$
1
5
4t
D
t " sin
CC
8
4
5
Z x 2 !2x
1
dt .
31. F .x/ D
1 C t2
0
208
1
T
p
p
sin 2s C 1
p
ds Let u D 2s C 1
p
2s C 1
du D ds= 2s C 1
Z
p
D sin u du D " cos u C C D " cos 2s C 1 C C
2
and y 0 .0/ D 0:
Antidifferentiating twice and using the initial conditions
leads to
1
y.t / D gt 2 :
2
0
Z
y.0/ D 0;
35.
Z
T
0
1 2
g T3
gT 2
gt dt D
D
Dy
2
2T 3
6
#
$
T
p :
3
Let f .x/ D ax 3 C bx 2 C cx C d so that
Z
1
a
b
c
C C C d:
4
3
2
%
&
We want this integral to be f .x1 / C f .x2 / =2 for all
choices of a, b, c, and d . Thus we require that
0
f .x/ dx D
a.x13 C x23 / C b.x12 C x22 / C c.x1 C x 2 / C 2d
Z 1
2b
a
C c C 2d:
D2
f .x/ dx D C
2
3
0
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INSTRUCTOR’S SOLUTIONS MANUAL
CHALLENGING PROBLEMS 5 (PAGE 332)
It follows that x1 and x2 must satisfy
2.
1
C
D
2
2
x12 C x22 D
3
x2 C x2 D 1:
x13
x23
D cos.jt / cos. 12 t / " sin.jt / sin. 12 t /
" cos.jt / cos. 12 t / " sin.jt / sin. 12 t /
.1/
D "2 sin.jt / sin. 12 t /:
Therefore, we obtain a telescoping sum:
.2/
.3/
n
X
At first glance this system may seem overdetermined; there
are three equations in only two unknowns. However, they
do admit a solution as we now show. Squaring equation
(3) and subtracting equation (2) we get 2x1 x2 D 1=3.
Subtracting this latter equation from equation (2)
p then
gives .x2 " x1 /2 D 1=3, so that x2 " x1 D 1= 3 (the positive square root since we want x1 < x2 ). Adding and subtracting this equation
and p
equation (3) thenpproduces the
p
p
values x2 D . 3 C 1/=.2 3/ and x1 D . 3 " 1/=.2 3/.
These values also satisfy equation (1) since
x13 C x23 D .x2 C x2 /.x12 " x1 x2 C x22 / D 1 &
Challenging Problems 5
&
%
&
%
a) cos .j C 12 /t " cos .j " 12 /t
#
2 1
"
3 6
$
D
sin.jt /
j D1
2 sin. 12 t / j D1
1
D"
D
1
:
2
2 sin. 12 t /
Z
(page 332)
xi D 2i=n , 0 $ i $ n, f .x/ D 1=x on Œ1; 2!. Since f
is decreasing, f is largest at the left endpoint and smallest
at the right endpoint of any interval Œ2.i!1/=n ; 2i=n ! of the
partition. Thus
n
X
iD1
1
2.i!1/=n
!=2
sin x dx
0
iD1
n
X
1
L.f; Pn / D
.2i=n " 2.i!1/=n /
i=n
2
iD1
D
n
X
iD1
3.
.1 " 2!1=n / D n.1 " 2!1=n / D
U.f; Pn /
:
21=n
Now, by l’H^opital’s rule,
( )
21=x " 1
0
lim n.21=n " 1/ D lim
n!1
x!1
1=x
0
21=x ln 2."1=x 2 /
D lim
D ln 2:
x!1
"1=x 2
Thus limn!1 U.f; Pn / D limn!1 L.f; Pn / D ln s.
n
X
sin
j D1
#
j#
2n
$
#
2n
#
$
#
1
#
.2n C 1/#
D lim
cos
" cos
n!1 2n 2 sin.#=.4n//
4n
4n
#
$
#
.2n C 1/#
#=.4n/
lim cos
" cos
D lim
n!1 sin.#=.4n// n!1
4n
4n
%
#&
D 1 & cos 0 " cos
D 1:
2
.2i=n " 2.i!1/=n /
n
X
D
.21=n " 1/ D n.21=n " 1/
%
&
i
cos .n C 21 /t " cos. 12 t /
! 2! 3!
b) Let Pn D f0; 2n
; 2n ; 2n ; : : : n!
2n g be the partition of
Œ0; #=2! into n subintervals of equal length
"x D #=2n. Using t D #=2n in the formula obtained in part (a), we get
n!1
U.f; Pn / D
h
%
&
%
&i
cos .j C 12 /t " cos .j " 21 /t
h
%
&i
1
cos. 12 t / " cos .n C 12 /t :
1
2 sin. 2 t /
D lim
1.
n h
X
1
D"
&
%
&
%
a) sin .j C 21 /t " sin .j " 21 /t
D sin.jt / cos. 12 t / C cos.jt / sin. 12 t /
" sin.jt / cos. 12 t / C cos.jt / sin. 12 t /
D 2 cos.jt / sin. 12 t /:
Therefore, we obtain a telescoping sum:
n
X
cos.jt /
j D1
D
D
1
n h
%
&
%
&i
X
sin .j C 21 /t " sin .j " 21 /t
2 sin. 12 t / j D1
1
2 sin. 12 t /
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&
i
sin .n C 12 /t " sin. 12 t / :
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CHALLENGING PROBLEMS 5 (PAGE 332)
ADAMS and ESSEX: CALCULUS 9
! 2! 3!
b) Let Pn D f0; 3n
; 3n ; 3n ; : : : n!
3n g be the partition of
Œ0; #=3! into n subintervals of equal length
"x D #=3n. Using t D #=3n in the formula obtained in part (a), we get
Z
(obtained by the Binomial Theorem) for j D 1; 2; : : : ; n.
The left side telescopes, and we get
.n C 1/mC2 " 1mC2 D .m C 2/
!=2
cos x dx
0
D lim
n!1
n
X
j D1
#
j#
cos
3n
$
C
#
3n
C
n
X
n
X
1
.xi " xi!1 /
xi!1 xi
iD1
$
n #
X
1
1
"
(telescoping)
D
xi!1
xi
Thus
2
1
1
1
1
1
"
D1" D :
x0 xn
2
2
j mC1
Z
0
iD1
a
j D1
j D1
1
n!1 nkC1
nk
nkC1
C
C Pk!1 .n/;
kC1
2
D akC1 lim
n!1
kC1
j D
n.n C 1/
n1C1
n
D
C C P0 .n/;
2
1C1
2
where P0 .n/ D 0 certainly has degree $ 0. Now assume
that the formula above holds for k D 1; 2; 3; : : : ; m. We
will show that it also holds for k D m C 1. To this end,
sum the the formula
.j C1/mC2 "j mC2 D .mC2/j mC1 C
210
n!1
D akC1 lim
D
where Pk!1 is a polynomial of degree at most k " 1.
First check the case k D 1:
n
X
n
a X % a &j
n
n
x k dx D lim
5. We want to prove that for each positive integer k,
j D1
.m C 2/.m C 1/ nmC1
C !!!;
2
mC1
b) Using the technique of Example 2 in Section 6.2 and
the result above,
n
jk D
j mC1
j D1
so that the formula is also correct for k D m C 1. Hence it
is true for all positive integers k by induction.
X
dx
1
D lim
f .ci / "xi D .
n!1
x2
2
n
X
n
X
#
$
1
m C 2 mC1
mC2
mC1
D
n
C .m C 2/n
C !!! "
n
" !!!
mC2
2
mC1
mC2
n
n
D
C
C !!!
mC2
2
iD1
Z
1:
j D1
j D1
so xi!1 < ci < xi . We have
D
n
X
where the ! ! ! represent terms of degree m or lower in the
variable n. Solving for the remaining sum, we get
2
xi!1
< xiD1 xi D ci2 < xi2 ;
iD1
j D1
jm C !!! C
nmC2 C .m C 2/nmC1 C ! ! ! D .m C 2/
4. f .x/ D 1=x 2 , 1 D x0 < x1 < x2 < ! ! ! < xn D 2. If
p
ci D xi!1 xi , then
f .ci / "xi D
j D1
n
X
j mC1
Expanding the binomial power on the left and using the
induction hypothesis on the other terms we get
#
$
#
1
.2n C 1/#
#
sin
D lim
" sin
n!1 3n 2 sin.#=.6n//
6n
6n
#
$
#=.6n/
.2n C 1/#
#
D lim
lim sin
" sin
n!1 sin.#=.6n// n!1
6n
6n
p
% #
&
3
D 1 & sin " sin 0 D
:
3
2
n
X
.m C 2/.m C 1/
2
n
X
.m C 2/.m C 1/ m
j C! ! !C1
2
6.
a
:
kC1
#
n
X
jk
j D1
1
1
Pk!1 .n/
C
C
kC1
2n
nkC1
$
Let f .x/ D ax 3 C bx 2 C cx C d . We used Maple to
calculate the following:
The tangent to y D f .x/ at P D .p; f .p// has equation
y D g.x/ D ap 3 Cbp 2 Ccp Cd C.3ap 2 C2bp Cc/.x "p/:
This line intersects y D f .x/ at x D p (double root) and
at x D q, where
2ap C b
qD"
:
a
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INSTRUCTOR’S SOLUTIONS MANUAL
CHALLENGING PROBLEMS 5 (PAGE 332)
Similarly, the tangent to y D f .x/ at x D q has equation
8.
y D h.x/ D aq 3 C bq 2 C cq C d C .3aq 2 C 2bq C c/.x " q/;
y D g.x/ D ap 4 Cbp 3 Ccp 2 CdpCeC.4ap 3 C3bp 2 C2cpCd /.x"p/;
and intersects y D f .x/ at x D q (double root) and
x D r, where
r D"
and intersects y D f .x/ at x D p (double root) and at the
two points
2aq C b
4ap C b
D
:
a
a
xD
The area between y D f .x/ and the tangent line at P is
the absolute value of
Z q
.f .x/ " g.x/ dx
p
1
D"
12
#
Let f .x/ D ax 4 C bx 3 C cx 2 C dx C e. The tangent to
y D f .x/ at P D .p; f .p// has equation
81a4 p 4 C 108a3 bp 3 C 54a2 b 2 p 2 C 12ab 3 p C b 4
a3
$
:
which has two solutions, which we take to be p and q:
#
81a4 p 4 C 108a3 bp 3 C 54a2 b 2 p 2 C 12ab 3 p C b 4
a3
$
which is 16 times the area between y D f .x/ and the
tangent at P .
7.
r D"
;
# 3
$#
$
3b 4
b 2 c bd
b
bc
b
C
"
"
CeC
C
d
x
C
256a3 16a2 4a
8a2 2a
4a
and it intersects y D f .x/ at the points U and V with
x-coordinates
p p
2 3b 2 " 8ac
;
uD
4a
p p
"b C 2 3b 2 " 8ac
:
vD
4a
b
:
3a
"b "
Since
p"r D
"b C
(Both roots exist and are distinct provided 3b 2 > 8ac.)
The point T corresponds to x D t D .p C q/=2 D "b=4a.
The tangent to y D f .x/ at x D t has equation
y D h.x/ D "
We continue with the calculations begun in the previous
problem. P and Q are as they were in that problem, but
R D .r; f .r// is now the inflection point of y D f .x/,
given by f 00 .r/ D 0. Maple gives
p
3b 2 " 8ac
4a
p
b
"b " 3b 2 " 8ac
D "p "
:
qD
4a
2a
pD
.f .x/ " h.x/ dx
4
D"
3
2.b C 3ap/
b C 3ap
and r " q D
a
a
have the same sign, R must lie between Q and P on the
curve y D f .x/. The line QR has a rather complicated
equation y D k.x/, which we won’t reproduce here, but
the area between this Rline and the curve y D f .x/ is
q
the absolute value of r .f .x/ " k.x// dx, which Maple
evaluates to be
"
4
81
#
b 2 " 4ac " 4abp " 8a2 p 2
:
2a
8a2 p 2 C 4abp C 4ac " b 2 D 0;
r
q
p
If these latter two points coincide, then the tangent is a
“double tangent.” This happens if
The area between y D f .x/ and the tangent line at
Q D .q; f .q// is the absolute value of
Z
"2ap " b ˙
81a4 p 4 C 108a3 bp 3 C 54a2 b 2 p 2 C 12ab 3 p C b
a3
4$
V
S
T
B
Q
;
which is 16=27 of the area between the curve and its
tangent at P . This leaves 11=27 of that area to lie between the curve, QR, and the tangent, so QR divides the
area between y D f .x/ and its tangent at P in the ratio
16/11.
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A
U
R
P
Fig. C-5-8
211
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CHALLENGING PROBLEMS 5 (PAGE 332)
a) The areas between the curve y D f .x/ and the lines
PQ and U V are, respectively, the absolute values of
Z q
Z v
A1 D
.f .x/"g.x// dx and A2 D
.h.x/"f .x// dx:
p
ADAMS and ESSEX: CALCULUS 9
The region bounded by RS and the curve y D f .x/
is divided into three parts by A and B. The areas of
these three regions are the absolute values of
u
Maple calculates theseptwo integrals and simplifies the
ratio A1 =A2 to be 1= 2.
A1 D
b) The two inflection points A and B of f have xcoordinates shown by Maple to be
p
"3b " 3.3b 2 " 8ac/
and
˛D
p12a
"3b C 3.3b 2 " 8ac/
ˇD
:
12a
It then determines the four points of intersection of
the line y D k.x/ through these inflection points and
the curve. The other two points have x-coordinates
p
"3b " 15.3b 2 " 8ac/
and
rD
p 12a
"3b C 15.3b 2 " 8ac/
sD
:
12a
212
A2 D
A3 D
Z
˛
.k.x/ " f .x// dx
r
Z
ˇ
Z˛s
ˇ
.f .x/ " k.x// dx
.k.x/ " f .x// dx:
The expressions calculated by Maple for k.x/ and
for these three areas are very complicated, but Maple
simplifies the rations A3 =A1 and A2 =A1 to 1 and 2
respectively, as was to be shown.
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 6.1 (PAGE 339)
CHAPTER 6. TECHNIQUES OF
INTEGRATION
5.
Z
x cos x dx
U Dx
d V D cos x dx
d U D dx
V D sin x
Z
D x sin x " sin x dx
x 3 ln x dx
U D ln x d V D x 3 dx
dx
x4
dU D
V D
x Z
4
1 4
1
3
D x ln x "
x dx
4
4
1
1
D x 4 ln x " x 4 C C:
4
16
Section 6.1 Integration by Parts
(page 339)
1.
Z
6.
Z
x.ln x/3 dx D I3 where
Z
In D x.ln x/n dx
D x sin x C cos x C C:
2.
3.
Z
.x C 3/e 2x dx
U DxC3
d U D dx
1
D .x C 3/e 2x "
2
1
D .x C 3/e 2x "
2
Z
x 2 cos #x dx
U D x2
d U D 2x dx
4.
D
d V D e 2x dx
V D 21 e 2x
Z
1
e 2x dx
2
1 2x
e C C:
4
D
I3 D
D
D
D
d V D cos #x dx
sin #x
V D
#
Z
2
x 2 sin #x
x sin #x dx
"
D
#
#
U Dx
d V D sin #x dx
cos #x
d U D dx
V D"
#
#
$
Z
x 2 sin #x
2
x cos #x
1
D
"
"
C
cos #x dx
#
#
#
#
2
2
1 2
D x sin #x C 2 x cos #x " 3 sin #x C C:
#
#
#
Z
.x 2 " 2x/e kx dx
U D x 2 " 2x
d U D .2x " 2/ dx
d V D e kx
1
V D e kx
k
Z
D
7.
tan!1 x dx
U D tan!1 x
dx
dU D
1 CZx 2
D x tan!1 x "
D x tan!1 x "
8.
1
1
D .x 2 " 2x/e kx "
.2x " 2/e kx dx
k
k
U D x " 1 d V D e kx dx
1
d U D dx
V D e kx
k
(
)
Z
1
2 1
1
D .x 2 " 2x/e kx "
.x " 1/e kx "
e kx dx
k
k k
k
2
2
1 2
D .x " 2x/e kx " 2 .x " 1/e kx C 3 e kx C C:
k
k
k
Z
d V D x dx
U D .ln x/n
n
1
n!1
dx
d U D .ln x/
V D x2
x
2
Z
1 2
n
n
n!1
x .ln x/ "
x.ln x/
dx
2
2
n
1 2
x .ln x/n " In!1
2
2
1 2
3
3
x .ln x/ " I2
2
2(
)
1 2
3 1 2
2
3
x .ln x/ "
x .ln x/2 " I1
2
2 2
2
(
)
3
3
1 2
1
1 2
x .ln x/3 " x 2 .ln x/2 C
x .ln x/ " I0
2
4
2 2
2
Z
1 2
3
3
3
x dx
x .ln x/3 " x 2 .ln x/2 C x 2 .ln x/ "
2
4
4
4
(
)
3
3
3
x2
.ln x/3 " .ln x/2 C .ln x/ "
C C:
2
2
2
4
d V D dx
V Dx
x dx
1 C x2
1
ln.1 C x 2 / C C:
2
Z
x 2 tan!1 x dx
D
x2
1
x3
tan!1 x "
C ln.1 C x 2 / C C:
3
6
6
U D tan!1 x d V D x 2 dx
dx
x3
dU D
V D
2
1Cx
3
Z
x3
1
x3
!1
tan x "
dx
D
3
3
1 C x2
!
Z
x3
1
x
!1
D
dx
tan x "
x"
3
3
1 C x2
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SECTION 6.1 (PAGE 339)
9.
Z
ADAMS and ESSEX: CALCULUS 9
For n D 5 we have
x sin!1 x dx
d V D x dx
U D sin!1 x
dx
x2
dU D p
V
D
2
1 " x2
Z
x 2 dx
1 2 !1
1
D x sin x "
Let x D sin &
p
2
2
1 " x2
dx
D cos & d&
Z
1
1 2 !1
2
sin & d&
D x sin x "
2
2
1
1
D x 2 sin!1 x " .& " sin & cos & / C C
2
#
$ 4
1 2 1
1 p
D
x "
sin!1 x C x 1 " x 2 C C:
2
4
4
10.
Z
Z
0
12.
2
x 5 e !x dx D I2 where
Z
2
In D x .2nC1/ e !x dx
!=4
1
2
13.
I D
Z
sec x tan x "
1
2
ln j sec x C tan xj C C:
e 2x sin 3x dx
U D e 2x
d V D sin 3x dx
d U D 2e 2x dx
V D " 13 cos 3x
Z
1
2
D " e 2x cos 3x C
e 2x cos 3x dx
3
3
U D e 2x
d V D cos 3x dx
2x
d U D 2e dx
V D 13 sin 3x
#
$
2 1 2x
2
1
e sin 3x " I
D " e 2x cos 3x C
3
3 3
3
1 2x
2 2x
13
I D " e cos 3x C e sin 3x C C1
9
3
9
1 2x
I D
e .2 sin 3x " 3 cos 3x/ C C:
13
secn x dx
0
U D secn!2 x
d V D sec2 x dx
d U D .n " 2/ secn!2 x tan x dx
V D tan x
ˇ!=4
Z
ˇ
!=4
ˇ
D tan x secn!2 x ˇ
" .n " 2/
secn!2 x tan2 x dx
ˇ
0
0
p
D . 2/n!2 " .n " 2/.In " In!2 /:
p
.n " 1/In D . 2/n!2 C .n " 2/In!2 :
Therefore
p
. 2/n!2
n"2
In D
C
In!2 ;
n"1
n"1
214
tan2 x sec x dx
Thus; I D
1
2
D " e !x .x 4 C 2x 2 C 2/ C C:
2
Z
Z
D sec x tan x " ln j sec x C tan xj " I
2
In D
I D
p
2 2
3
sec5 x dx D I5 D
C I3
4
4
!
p
p
2
2
3
1
C
C I1
D
2
4
2
2
p
ˇ!=4
7 2
3
ˇ
D
C ln j sec x C tan xjˇ
0
8
8
p
p
3
7 2
C ln.1 C 2/:
D
8
8
U D tan x
d V D sec x tan x dx
d U D sec2 xZ dx
V D sec x
3
D sec x tan x " sec x dx
Z
D sec x tan x " .1 C tan2 x/ sec x dx
U D x 2n
d V D xe !x dx
2
.2n!1/
d U D 2nx
dx
V D " 12 e !x
Z
1
2
2
D " x 2n e !x C n x .2n!1/ e !x dx
2
1
2
D " x 2n e !x C nIn!1
2
"
#
Z
1 2 !x 2
1 4 !x 2
!x 2
C2 " x e
C xe
dx
I2 D " x e
2
2
11.
!=4
.n # 2/:
14.
I D
Z
xe
p
x
dx
Let x D w 2
dx D 2w dw
Z
D 2 w 3 e w dw D 2I3 where
Z
In D w n e w dw
U D wn
d U D nw n!1 dw
D w n e w " nIn!1 :
d V D e w dw
V D ew
I D 2I3 D 2w 3 e w " 6Œw 2 e w " 2.we w " I0 /!
p
p
p
D e x .2x x " 6x C 12 x " 12/ C C:
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INSTRUCTOR’S SOLUTIONS MANUAL
15.
16.
Z
1
1=2
SECTION 6.1 (PAGE 339)
sin!1 x
dx
x2
19.
dx
dV D 2
U D sin!1 x
x
dx
1
dU D p
V D"
1 " x2
x
ˇ1
Z 1
ˇ
dx
1
ˇ
C
D " sin!1 x ˇ
p
Let x D sin &
ˇ
x
1=2 x 1 " x 2
1=2
dx D cos & d&
Z !=2
#
#
csc & d&
D" C C
2
3
!=6
ˇ!=2
#
ˇ
D " " ln j csc & C cot & jˇ
!=6
6
p
p
#
#
D " " ln 1 C ln.2 C 3/ D ln.2 C 3/ " :
6
6
Z 1
p
p
x sin.# x/ dx Let x D w 2
0
dx D 2w dw
Z 1
2
D2
w sin.#w/ dw
20.
1
U D cos.ln x/
d V D dx
sin.ln x/
V Dx
dU D "
dx
x
ˇe
"
#
ˇ
ˇ
D e sin.1/ " x cos.ln x/ˇ C I
ˇ
0
1
21.
1
x2 x
" sin 2x " cos 2x C C:
4
4
8
1
Thus; I D Œe sin.1/ " e cos.1/ C 1!:
2
Z
ln.ln x/
dx Let u D ln x
x
dx
du D
x
Z
D
ln u du
U D ln u
du
dU D
Zu
U Dx
d V D sec2 x dx
d U D dx
V D tan x
Z
D x tan x " tan x dx
D
sin.ln x/ dx
d V D dx
U D sin.ln x/
cos.ln x/
V Dx
dx
dU D
x
ˇe Z
ˇ
e
ˇ
D x sin.ln x/ˇ "
cos.ln x/ dx
ˇ
1
U Dw
d U D dw
D u ln u "
D x tan x " ln j sec xj C C:
Z
Z
1
.x " x cos 2x/ dx
x sin2 x dx D
2
Z
x2 1
D
x cos 2x dx
"
4
2
U Dx
d V D cos 2x dx
d U D dx
V D 21 sin 2x
"
#
Z
x2 1 1
1
D
"
x sin 2x "
sin 2x dx
4
2 2
2
d V D dx
V Dx
1
d V D sin.#w/ dw
cos.#w/
V D"
#
ˇ1
Z 1
ˇ
2 2
4
ˇ
D " w cos.#w/ˇ C
w cos.#w/ dw
ˇ
#
# 0
18.
U D cos.ln x/
sin.ln x/
dU D "
dx
Zx
U D sin.ln x/
d V D dx
cos.ln x/
V Dx
dU D
dx
x
D x cos.ln x/ C x sin.ln x/ " I
&
1%
I D
x cos.ln x/ C x sin.ln x/ C C:
2
Z e
I D
sin.ln x/ dx
U D w2
d U D 2w dw
17.
cos.ln x/ dx
D x cos.ln x/ C
0
d V D cos.#w/ dw
sin.#w/
V D
#
"
#ˇ1
Z 1
ˇ
2
4 w
4
ˇ
D C
sin.#w/ ˇ " 2
sin.#w/ dw
ˇ
#
# #
# 0
0
ˇ1
ˇ
2
4
2
4
2
8
ˇ
D C 3 cos.#w/ˇ D C 3 ."2/ D " 3 :
ˇ
#
#
#
#
#
#
0
Z
2
x sec x dx
Z
I D
d V D du
V Du
du D u ln u " u C C
D .ln x/.ln.ln x// " ln x C C:
22.
Z
4
p
xe
p
x
dx
0
D2
Z
2
0
Let x D w 2
dx D 2w dw
w 2 e w dw D 2I2
See solution #16 for the formula
R
In D w n e w dw D w n e w " nIn!1 .
ˇ2
ˇ2
%
&
%
&
ˇ
ˇ
D 2 w 2 e w ˇˇ " 2I1 D 8e 2 " 4 we w ˇˇ " I0
0
0
Z 2
2
2
w
2
D 8e " 8e C 4
e dw D 4.e " 1/:
0
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SECTION 6.1 (PAGE 339)
23.
Z
ADAMS and ESSEX: CALCULUS 9
cos!1 x dx
Z
27.
d V D dx
U D cos!1 x
dx
V Dx
dU D "p
2
Z1 " x
x dx
D x cos!1 x C
p
2
p 1"x
!1
2
D x cos x " 1 " x C C:
24.
Z
D
D
x sec!1 x dx
U D sec!1 x
d V D x dx
1
dx
V D x2
dU D
p
2
jxj x 2 " 1
Z
1 2
1
jxj
D x sec!1 x "
p
dx
2
2
x2 " 1
p
1
1
D x 2 sec!1 x " sgn .x/ x 2 " 1 C C:
2
2
25.
Z
D
D
D
1
U D cos!1
D
d V D dx
V Dx
& 2 cos & d&
D
U D &2
d U D 2& d&
Z
D & 2 sin & " 2
d V D cos & d&
V D sin &
& sin & d&
U D&
d U D d&
d V D sin & d&
V D " cos
Z&
2
D & sin & " 2."& cos & C cos & d& /
2
28.
By the procedure used in Example 4 of Section 7.1,
Z
Z
e x cos x dx D 21 e x .sin x C cos x/ C C I
e x sin x dx D 21 e x .sin x " cos x/ C C:
Now
Z
xe x cos x dx
d V D e x cos x dx
V D 21 e x .sin x C cos x/
Z
D 12 xe x .sin C cos x/ " 21 e x .sin x C cos x/ dx
U Dx
d U D dx
D 12 xe x .sin C cos x/
D & sin & C 2& cos & " 2 sin & C C
p
D x.sin!1 x/2 C 2 1 " x 2 .sin!1 x/ " 2x C C:
216
d V D x dx
U D .tan!1 x/2
x2
2 tan!1 x dx
V D
dU D
2
1 C x2
Z 2 !1
2
x tan x
x
.tan!1 x/2 "
dx Let u D tan!1 x
2
1 C x2
dx
du D
1 C x2
Z
x2
.tan!1 x/2 " u tan2 u du
2
Z
x2
.tan!1 x/2 C .u " u sec2 u/ du
2
Z
x2
u2
.tan!1 x/2 C
" u sec2 u du
2
2
U Du
d V D sec2 u du
d U D du
V D tan u
Z
1 2
!1
.x C 1/.tan x/2 " u tan u C tan u du
2
1 2
.x C 1/.tan!1 x/2 " x tan!1 x C ln j sec uj C C
2
1
1 2
.x C 1/.tan!1 x/2 " x tan!1 x C ln.1 C x 2 / C C
2
2
:
1
x
1
$
#
1
" 2 dx
x
1
1" 2
x
ˇ2 Z
ˇ
2
dx
!1 1 ˇ
Let x D sec &
D x cos
p
ˇ "
xˇ
x2 " 1
1
1
dx D sec & tan & d&
Z !=3
2#
D
"0"
sec & d&
3
0
ˇ!=3
2#
ˇ
D
" ln j sec & C tan & jˇ
0
3
p
2#
D
" ln.2 C 3/:
3
Z
.sin!1 x/2 dx Let x D sin &
dx D cos & d&
Z
dU D "r
26.
D
2
sec!1 x dx
Z 2
1
D
cos!1
x
1
x.tan!1 x/2 dx
D
" 41 e x .sin x " cos x C sin x C
x
1 x
1
2 xe .sin x C cos x/ " 2 e sin x
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C C:
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INSTRUCTOR’S SOLUTIONS MANUAL
29.
Area D A D
Z
!
SECTION 6.1 (PAGE 339)
e !x sin x dx
32.
0
d V D sin x dx
U D e !x
d U D "eˇ!x dx
V D " cos x
Z !
ˇ!
!x
D "e cos x ˇˇ "
e !x cos x dx
U D e !x
d V D cos x dx
!x
d U D "e#
dx
Vˇ D sin$x
ˇ!
D e !! C 1 " e !x sin x ˇˇ C A
30. The tangent line to y D ln x at x D 1 is y D x " 1, Hence,
1
.1/.1/ C .1/.e " 2/ "
2
ˇe
ˇ
3
ˇ
D e " " .x ln x " x/ˇ
ˇ
2
Z
x n sin x dx
0
0
1 C e !!
units2
2
Shaded area D
!=2
U D x n!1
d V D cos x dx
n!2
d U D .n " 1/x
dx
V D sin x
ˇ!=2
"
#
Z !=2
ˇ
ˇ
D n x n!1 sin x ˇ
" .n " 1/
x n!2 sin x dx
ˇ
0
0
% # &n!1
" n.n " 1/In!2 ;
.n # 2/:
Dn
2
ˇ
Z !=2
ˇ!=2
ˇ
I0 D
sin x dx D " cos x ˇ
D 1:
ˇ
0
0
(
( % &
))
% # &5
% # &3
#
I6 D 6
" 6.5/ 4
" 4.3/ 2
" 2.1/I0
2
2
2
0
Thus Area D A D
Z
U D xn
d V D sin x dx
d U D nx n!1 dx
V D " cos x
ˇ!=2
Z !=2
ˇ
ˇ
D "x n cos x ˇ
Cn
x n!1 cos x dx
ˇ
0
0
0
In D
e
ln x dx
1
D
3 5
# " 15# 3 C 360# " 720:
16
1
3
5
D e " " e C e C 0 " 1 D e " sq. units:
2
2
33.
y
yDx!1
yD1
.e;1/
yDln x
1
x
In D
Z
sinn x dx
.n # 2/
U D sinn!1 x
d V D sin x dx
d U D .n " 1/ sinn!2 x cos x dx
V D " cos x
Z
D " sinn!1 x cos x C .n " 1/ sinn!2 x cos2 x dx
D " sinn!1 x cos x C .n " 1/.In!2 " In /
nIn D " sinn!1 x cos x C .n " 1/In!2
1
n"1
In D " sinn!1 x cos x C
In!2 :
n
n
Fig. 6.1-30
31.
In D
Z
Note: I0 D x C C , I1 D " cos x C C . Hence
.ln x/n dx
n
U D .ln x/
n!1 dx
d U D n.ln x/
x
d V D dx
V Dx
In D x.ln x/n " nIn!1 :
I4 D x.ln x/4 " 4I3
%
&
D x.ln x/4 " 4 x.ln x/3 " 3I2
%
&
D x.ln x/4 " 4x.ln x/3 C 12 x.ln x/2 " 2I1
D x.ln x/4 " 4x.ln x/3 C 12x.ln x/2
%
&
" 24 x ln x " x C C
%
&
D x .ln x/4 " 4.ln x/3 C 12.ln x/2 " 24 ln x C 24 C C:
1
5
I6 D " sin5 x cos x C I4
6
6#
$
5
1
3
1 5
" sin3 x cos x C I2
D " sin x cos x C
6
6
4
4
5
1 5
3
sin x cos x
D " sin x cos x "
6
24
#
$
5
1
1
C
" sin x cos x C I0
8
2
2
5
5
1 5
sin3 x cos x "
sin x cos x
D " sin x cos x "
6
24
16
5
C xCC
16
!
sin5 x
5 sin3 x
5 sin x
5x
" cos x
C
C
C C:
D
16
6
24
16
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SECTION 6.1 (PAGE 339)
ADAMS and ESSEX: CALCULUS 9
1
6
I7 D " sin6 x cos x C I5
7
7#
$
1 6
6
1
4
D " sin x cos x C
" sin4 x cos x C I3
7
7
5
5
6
1 6
4
sin x cos x
D " sin x cos x "
7
35
#
$
1
2
24
" sin2 x cos x C I1
C
35
3
3
1 6
6
8
4
D " sin x cos x "
sin x cos x "
sin2 x cos x
7
35
35
16
cos x C C
"
35
!
sin6 x
6 sin4 x
8 sin2 x
16
D " cos x
C
C
C
C C:
7
35
35
35
35.
Z
Z 2
dx
1
x C a2 " x 2
D
dx
2
2
n
2
.x C a /
a
.x 2 C a2 /n
Z
Z
1
dx
1
x2
D 2
"
dx
a
.x 2 C a2 /n!1 a2
.x 2 C a2 /n
In D
U Dx
d U D dx
D
x dx
C a2 /n
"1
V D
2.n " 1/.x 2 C a2 /n!1
dV D
.x 2
"x
2.n " 1/.x 2 C a2 /n!1
!
Z
1
dx
C
:
2.n " 1/
.x 2 C a2 /n!1
1
1
In!1 " 2
a2
a
2n " 3
x
C
In!1 :
2.n " 1/a2 .x 2 C a2 /n!1
2.n " 1/a2
1
x
Now I1 D tan!1 , so
a
a
3
x
C 2 I2
I3 D
4a2 .x 2 C a2 /2
4a
$
#
x
3
x
1
D
C
I
C
1
4a2 .x 2 C a2 /2
4a2 2a2 .x 2 C a2 /
2a2
3x
3
x
x
C 4 2
C 5 tan!1 C C:
D
4a2 .x 2 C a2 /2
8a .x C a2 /
8a
a
In D
34. We have
In D
Z
secn x dx
.n # 3/
U D secn!2 x
d V D sec2 x dx
n!2
d U D .n " 2/ sec
x tan x dx
V D tan x
Z
n!2
n!2
D sec
x tan x " .n " 2/ sec
x tan2 x dx
Z
D secn!2 x tan x " .n " 2/ secn!2 x.sec2 x " 1/ dx
D secn!2 x tan x " .n " 2/In C .n " 2/In!2 C C
n"2
1
.secn!2 x tan x/ C
In!2 C C:
In D
n
"
1
n
"1
Z
I1 D
Z
sec x dx D ln j sec x C tan xj C C I
sec2 x dx D tan x C C:
#
$
1
4 1
2
I6 D .sec4 x tan x/ C
sec2 x tan x C I2 C C
5
5 3
3
4
8
1
sec2 x tan x C
tan x C C:
D sec4 x tan x C
5
15 "
15
5 1
1
sec3 x tan xC
I7 D .sec5 x tan x/ C
6
6 4
#
$#
3 1
1
CC
sec x tan x C I1
4 2
2
I2 D
5
15
1
sec3 x tan x C
sec x tan xC
D sec5 x tan x C
6
24
48
15
ln j sec x C tan xj C C:
48
36.
Given that f .a/ D f .b/ D 0.
Z b
.x " a/.b " x/f 00 .x/ dx
a
U D .x " a/.b " x/
d V D f 00 .x/ dx
d U D .b C a " 2x/ dx
V D f 0 .x/
ˇb Z
ˇ
b
ˇ
D .x " a/.b " x/f 0 .x/ˇ "
.b C a " 2x/f 0 .x/ dx
ˇ
a
a
d V D f 0 .x/ dx
V D f .x/
ˇb
#
Z b
ˇ
ˇ
D 0 " .b C a " 2x/f .x/ˇ C 2
f .x/ dx
ˇ
a
a
Z b
D "2
f .x/ dx:
U D b C a " 2x
d U D "2dx
"
a
37. Given: f 00 and g 00 are continuous on Œa; b!, and
f .a/ D g.a/ D f .b/ D g.b/ D 0. We have
Z
b
f .x/g 00 .x/ dx
a
U D f .x/
d V D g 00 .x/ dx
0
d U D f .x/ dx
V D g 0 .x/
ˇb Z
ˇ
b
ˇ
D f .x/g 0 .x/ˇ "
f 0 .x/g 0 .x/ dx:
ˇ
a
a
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 6.2 (PAGE 348)
Multiplying the expression for I2nC1 by #=2 and
dividing by the expression for I2n , we obtain, by part
(b),
Similarly,
Z
b
a
ˇb Z
ˇ
b
ˇ
f .x/g.x/ dx D f .x/g.x/ˇ "
f 0 .x/g 0 .x/ dx:
ˇ
a
00
0
2n " 2
4 2 #
2n
!
!!! ! !
5 3 2 D # &1 D #;
lim 2n C 1 2n " 1
3 1 #
n!1 2n " 1 2n " 3
2
2
!
!!! ! !
2n
2n " 2
4 2 2
a
Thus we have
Z
b
a
f .x/g 00 .x/ dx "
Z
b
f 00 .x/g.x/ dx
or, rearranging the factors on the left,
a
ˇ
& ˇb
ˇ
0
0
D f .x/g .x/ " f .x/g.x/ ˇ D 0
ˇ
%
lim
a
n!1
2 2 4 4
2n
2n
#
! ! ! !!!
!
D :
1 3 3 5
2n " 1 2n C 1
2
by the assumptions on f and g. Thus
Z
a
b
f .x/g 00 .x/ dx D
Z
b
f 00 .x/g.x/ dx:
a
Section 6.2 Integrals of Rational Functions
(page 348)
This equation is also valid for any (sufficiently smooth)
functions f and g for which
f .b/g 0 .b/ " f 0 .b/g.b/ D f .a/g 0 .a/ " f 0 .a/g.a/:
Examples are functions which are periodic with period
b " a, or if f .a/ D f .b/ D f 0 .a/ D f 0 .b/ D 0, or if
instead g satisfies such conditions. Other combinations of
conditions on f and g will also do.
38. In D
Z
!=2
1.
Z
2 dx
D ln j2x " 3j C C:
2x " 3
2.
Z
1
dx
D " ln j5 " 4xj C C:
5 " 4x
4
cosn x dx.
0
3.
Z
4.
Z
a) For 0 $ x $ #=2 we have 0 $ cos x $ 1, and so
0 $ cos2nC2 x $ cos2nC1 x $ cos2n x. Therefore
0 $ I2nC2 $ I2nC1 $ I2n .
n"1
2n C 1
In!2 , we have I2nC2 D
I2n .
n
2n C 2
Combining this with part (a), we get
b) Since In D
2n C 1
I2nC2
I2nC1
D
$
$ 1:
2n C 2
I2n
I2n
The left side approaches 1 as n ! 1, so, by the
Squeeze Theorem,
lim
n!1
I2nC1
D 1:
I2n
c) By Example 6 we have, since 2n C 1 is odd and 2n is
even,
2n
2n " 2
4 2
!
!!! !
2n C 1 2n " 1
5 3
3 1 #
2n " 1 2n " 3
!
!!! ! ! :
D
2n
2n " 2
4 2 2
I2nC1 D
I2n
Z
x dx
#x C 2 " 2
1
D
dx
#x C 2
#
#x C 2
x
2
D " 2 ln j#x C 2j C C:
#
#
x2
dx D
x"4
D
5.
Z
!
16
xC4C
dx
x"4
x2
C 4x C 16 ln jx " 4j C C:
2
A
B
1
D
C
x2 " 9
x"3
xC3
Ax C 3A C Bx " 3B
D
x2 " 9
'
1
1
ACB
=0
)AD ; BD" :
)
3.A " B/ =1
6
6
Z
Z
Z
dx
dx
1
1
dx
D
"
x2 " 9
6
x"3 6
xC3
%
&
1
D
ln jx " 3j " ln jx C 3j C C
6 ˇ
ˇ
1 ˇ x " 3 ˇˇ
D ln ˇˇ
C C:
6
x C 3ˇ
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SECTION 6.2 (PAGE 348)
6.
7.
8.
9.
ADAMS and ESSEX: CALCULUS 9
1
A
B
D p
Cp
5 " x2
5"x
5Cx
p
.A C B/ 5 C .A " B/x
D
5 " x2
(
1
1
) A C B D p5 ) A D B D p :
2 5
A " B ZD #
0
$
Z
1
1
1
1
p
p
dx
D
p
C
dx
5 " x2
2 5
5"x
5Cx
%
&
p
p
1
D p " ln j 5 " xj C ln j 5 C xj C C
2 5 ˇ
ˇ
ˇ p5 C x ˇ
1
ˇ
ˇ
D p ln ˇ p
ˇ C C:
2 5 ˇ 5"xˇ
1
A
B
D
C
a2 " x 2
a"x
aCx
Aa C Ax C Ba " Bx
D
a2 " x 2
n
1
Aa C Ba D 1
)
)ADBD
:
A"B D0
2a
Thus
Z
Z
Z
dx
dx
1
1
dx
D
C
a2 " x 2
2a
a"x
2a
aCx
&
1 %
D
" ln ja " xj C ln ja C xj C C
2a ˇ
ˇ
ˇa C x ˇ
1
ˇ
ˇ C C:
D
ln
2a ˇ a " x ˇ
A
B
1
D
C
b 2 " a2 x 2
b " ax
b C ax
.A C B/b C .A " B/ax
D
b 2 " a2 x 2
1
)A D B D
$
Z #2b
Z
1
1
1
dx
D
C
dx
b 2 " a2 x 2
2b
b " ax
b C ax
#
$
1 " ln jb " axj
ln jb C axj
D
C
CC
2b
a
a
ˇ
ˇ
ˇ b C ax ˇ
1
ˇ C C:
D
ln ˇ
2ab ˇ b " ax ˇ
$
Z #
x"2
1" 2
dx
x Cx"2
Z
x"2
Dx"
dx:
x2 C x " 2
A
B
Ax " A C Bx C 2B
x"2
D
C
D
,
If 2
x Cx"2
xC2
x"1
x2 C x " 2
then A C B D 1 and "A C 2B D "2, so that A D 4=3 and
B D "1=3. Thus
Z
10.
11.
12.
13.
14.
x 2 dx
D
2
x Cx"2
Z
220
x2
Z
Z
x 2 dx
dx
dx
4
1
Dx"
C
Cx"2
3
xC2
3
x"1
1
4
D x " ln jx C 2j C ln jx " 1j C C:
3
3
15.
x
A
B
D
C
3x 2 C 8x " 3
3x " 1
xC3
.A C 3B/x C .3A " B/
D
3x 2 C 8x " 3
n
3
1
A C 3B D 1
; BD
:
)
)AD
3A
"
B
D
0
10
10
$
Z
Z #
x dx
1
1
3
D
C
dx
3x 2 C 8x " 3
10
3x " 1
xC3
3
1
ln j3x " 1j C
ln jx C 3j C C:
D
30
10
x"2
A
B
Ax C A C Bx
D C
D
2
x Cx
x
xC1
x2 C x
n
ACB D1
)
) A D "2; B D 3:
AD
Z
Z "2
Z
x"2
dx
dx
dx
D
3
"
2
x2 C x
xC1
x
D 3 ln jx C 1j " 2 ln jxj C C:
1
A
Bx C C
D C 2
x 3 C 9x
x
x C9
Ax 2 C 9A C Bx 2 C C x
D
x 3 C 9x
(
ACB D0
1
1
) C D0
) A D ; B D " ; C D 0:
9
9
9A ZD #
1
$
Z
1
1
x
dx
D
"
dx
x 3 C 9x
9
x x2 C 9
1
1
D ln jxj "
ln.x 2 C 9/ C K:
9
18
Z
dx
D
1 " 6x C 9x 2
Z
1
dx
D
C C:
.1 " 3x/2
3.1 " 3x/
Z
Z
x
x
dx
D
dx Let u D 3x C 1
2 C 6x C 9x 2
.3x C 1/2 C 1
du D 3 dx
Z
Z
Z
u"1
u
1
1
1
1
du
D
du
"
du
9
u2 C 1
9
u2 C 1
9
u2 C 1
1
1
D
ln.u2 C 1/ " tan!1 u C C
18
9
1
1
ln.2 C 6x C 9x 2 / " tan!1 .3x C 1/ C C:
D
18
9
Z
Z
9x 2 " 6x C 6x C 9
dx
2
Z6x " 9x
x
2x C 3
1
dx:
D" C
9
9
x.2 " 3x/
x2 C 1
1
dx D
6x " 9x 2
9
Now
2x C 3
A
B
2A " 3Ax C Bx
D C
D
x.2 " 3x/
x
2 " 3x
x.2 " 3x/
)2A D 3;
"3A C B D 2
3
13
)A D ;
BD
:
2
2
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 6.2 (PAGE 348)
Z
Therefore we have
Z
x2 C 1
dx
6x " 9x 2
Z
Z
x
1
dx
13
dx
D" C
C
9
6
x
18
2 " 3x
x
1
13
D " C ln jxj "
ln j2 " 3xj C C:
9
6
54
16. First divide to obtain
37x C 85
x3 C 1
Dx"7C
x 2 C 7x C 12
.x C 4/.x C 3/
A
B
37x C 85
D
C
.x C 4/.x C 3/
xC4
xC3
.A C B/x C 3A C 4B
D
x 2 C 7x C 12
n
A C B D 37
)
) A D 63; B D "26:
3A C 4B D 85
Now we have
!
Z
Z
x3 C 1
63
26
dx D
x"7C
"
dx
12 C 7x C x 2
xC4 xC3
D
17.
19.
A
Bx C C
a3
D
C 2
3
"a
x"a
x C ax C a2
Ax 2 C Aax C Aa2 C Bx 2 " Bax C C x " C a
D
x 38" a3
(
< A D a=3
ACB D0
) Aa " Ba C C D 0 ) B D "a=3
:
C D "2a2 =3:
Aa2 " C a D a3
Therefore we have
$
Z #
x3
a3
D
1C 3
dx
x 3 " a3
x " a3
Z
Z
a
dx
x C 2a
a
DxC
"
3
x " a 3 Z x 2 C ax C a2
2x C a C 3a
a
a
D x C ln jx " aj "
3
6
x 2 C ax C a2
a
a
D x C ln jx " aj " ln.x 2 C ax C a2 /
3
6
Z
dx
a2
"
%
a &2 3 2
2
C a
xC
2
4
a
a
D x C ln jx " aj " ln.x 2 C ax C a2 /
3
6
a2 2
!1 x C .a=2/
"
p tan
p
CK
2 3a
. 3a/=2
a
a
D x C ln jx " aj " ln.x 2 C ax C a2 /
3
6
a
!1 2x C a
" p tan
C K:
p
3
3a
Z
x2
" 7x C 63 ln jx C 4j " 26 ln jx C 3j C C:
2
1
A
B
C
D C
C
x.x 2 " a2 /
x
x"a
xCa
Ax 2 " Aa2 C Bx 2 C Bax C C x 2 " C ax
D
x.x 2 " a2 /
(
'
ACB CC D0
A D "1=a2
) B "C D0
)
B D C D 1=.2a2 /:
"Aa2 D 1
Thus we have
Z
dx
x.x 2 " a2 /
# Z
$
Z
Z
dx
dx
dx
1
"2
C
C
D
2a2
x
x"a
xCa
1
D
."2 ln jxj C ln jx " aj C ln jx C aj/ C K
2a2
jx 2 " a2 j
1
ln
C K:
D
2a2
x2
18. The partial fraction decomposition is
1
A
B
Cx C D
D
C
C 2
4
4
x "a
x"a
xCa
x C a2
A.x 3 C ax 2 C a2 x C a3 / C B.x 3 " ax 2 C a2 x " a3 /
D
x 4 " a4
C.x 3 " a2 x/ C D.x 2 " a2 /
C
x 4 " a4
8̂
ACB CC D0
<
aA " aB C D D 0
)
2
2
2
:̂ a A C a B " a C D 0
3
3
2
a A"a B "a D D1
1
1
1
)A D
; B D " 3 ; C D 0; D D " 2 :
4a3
4a
2a
x3
$
Z #
dx
1
1
1
2a
D
"
"
dx
x 4 " a4
4a3
x " a x C a x 2 C a2
ˇ
ˇ
% &
ˇx "aˇ
1
ˇ " 1 tan!1 x C K:
D
ln ˇˇ
ˇ
3
3
4a
xCa
2a
a
20.
Here the expansion is
1
A
Bx C C
D C 2
x 3 C 2x 2 C 2x
x
x C 2x C 2
A.x 2 C 2x C 2/ C Bx 2 C C x
D
x 3 C 2x 2 C 2
(
ACB D0
1
) 2A C C D 0 ) A D "B D ; C D "1;
2
2A D 1
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SECTION 6.2 (PAGE 348)
ADAMS and ESSEX: CALCULUS 9
so we have
Z
23.
Z
Z
dx
1
dx 1
xC2
D
"
dx
x 3 C 2x 2 C 2x
2
x
2
x 2 C 2x C 2
Let u D x C 1
du D dxZ
1
1
uC1
D ln jxj "
du
2
2
u2 C 1
1
1
1
D ln jxj " ln.u2 C 1/ " tan!1 u C K
2
4
2
1
1
1
D ln jxj " ln.x 2 C 2x C 2/ " tan!1 .x C 1/ C K:
2
4
2
21.
1
A
B
C
D
D
C
C
C
.x 2 " 1/2
x"1
.x " 1/2
xC1
.x C 1/2
%
1
D 2
A.x " 1/.x C 1/2 C B.x C 1/2
.x " 1/2
&
C C.x C 1/.x " 1/2 C D.x " 1/2
8̂
8̂
ACC D0
1
<A D "
<
ACB "C CD D0
4
)
)
:̂ "A C 2B " C " 2D D 0
:̂ B D C D D D 1 :
"A C B C C C D D 1
4
Thus
Z
dx
.x 2 " 1/2
Z
Z
dx
dx
1
"
C
D
4
x"1
.x " 1/2
!
Z
Z
dx
dx
C
C
xC1
.x C 1/2
#
$
1
1
1
D
ln jx C 1j " ln jx " 1j "
"
CK
4
x"1 xC1
ˇ
ˇ
1 ˇ x C 1 ˇˇ
x
D ln ˇˇ
"
C K:
4
x " 1 ˇ 2.x 2 " 1/
1
A
B
C
D C
C
3
2
x " 4x C 3x
x
x"1
x"3
A.x 2 " 4x C 3/ C B.x 2 " 3x/ C C.x 2 " x/
D
x 3 " 4x 2 C 3x
(
ACB CC D0
) "4A " 3B " C D 0
3A D 1
1
1
1
)AD ; BD" ; C D :
3
2
6
Therefore we have
Z
dx
3
x " 4x 2 C 3x
Z
Z
Z
dx 1
dx
dx
1
1
D
"
C
3
x
2
x"1
6
x"3
1
1
1
D ln jxj " ln jx " 1j C ln jx " 3j C K:
3
2
6
24.
22. Here the expansion is
The expansion is
A
B
C
D
x2
D
C
C
C
" 1/.x 2 " 4/
x"1
xC1
x"2
xC2
x2
1
1
D
D"
A D lim
x!1 .x C 1/.x 2 " 4/
2."3/
6
x2
1
1
B D lim
D
D
x!!1 .x " 1/.x 2 " 4/
"2."3/
6
4
1
x2
D
D
C D lim 2
x!2 .x " 1/.x C 2/
3.4/
3
x2
4
1
D D lim
D
D" :
x!!2 .x 2 " 1/.x " 2/
3."4/
3
A
Bx C C
x2 C 1
D
C 2
x3 C 8
xC2
x " 2x C 4
A.x 2 " 2x C 4/ C B.x 2 C 2x/ C C.x C 2/
D
x3 C 8
(
ACB D1
5
7
1
) "2A C 2B C C D 0 ) A D
BD
; C D" ;
12
12
3
4A C 2C D 1
.x 2
so we have
Z
Z
Z
x2 C 1
dx
7x " 4
5
1
dx
D
C
dx
x3 C 8
12
xC2
12
.x " 1/2 C 3
Let u D x " 1
du D dx Z
5
1
7u C 3
D
ln jx C 2j C
du
12
12
u2 C 3
7
5
ln jx C 2j C
ln.x 2 " 2x C 4/
D
12
24
1
x"1
C p tan!1 p C K:
4 3
3
222
Therefore
Z
x2
1
1
dx D " ln jx " 1j C ln jx C 1jC
.x 2 " 1/.x 2 " 4/
6
6
1
1
ln jx " 2j " ln jx C 2j C K:
3
3
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INSTRUCTOR’S SOLUTIONS MANUAL
25.
SECTION 6.2 (PAGE 348)
1
1
D 3
x 4 " 3x 3
x .x " 3/
A
B
C
D
D C 2 C 3 C
x
x
x
x"3
A.x 3 " 3x 2 / C B.x 2 " 3x/ C C.x " 3/ C Dx 3
D
x 3 .x " 3/
8̂
ACD D0
<
"3A C B D 0
)
:̂ "3B C C D 0
"3C D 1
8̂
A D "1=27
<
B D "1=9
)
:̂ C D "1=3
D D 1=27:
27.
28.
26. We have
Z
dt
.t " 1/.t 2 " 1/2
Z
dt
Let u D t " 1
D
.t " 1/3 .t C 1/2
du D dt
Z
du
D
u3 .u C 2/2
B
E
1
A
C
D
C
D C 2 C 3 C
u3 .u C 2/2
u
u
u
uC2
.u C 2/2
A.u4 C 4u3 C 4u2 / C B.u3 C 4u2 C 4u/
D
u3 .u C 2/2
2
C.u C 4u C 4/ C D.u4 C 2u3 / C Eu3
u3 .u C 2/2
8̂
ACD D0
ˆ
ˆ
< 4A C B C 2D C E D 0
) 4A C 4B C C D 0
ˆ
ˆ 4B C 4C D 0
:̂
4C D 1
3
1
1
3
1
)AD
; BD" ; C D ; DD" ; ED" :
16
4
4
16
8
Z
du
u3 .u C 2/2
Z
Z
Z
3
1
du 1
du
du
D
C
"
16
u
4
u2
4
u3
Z
Z
3
du
1
du
"
"
16
uC2 8
.u C 2/2
3
1
1
D
ln jt " 1j C
"
"
16
4.t " 1/ 8.t " 1/2
3
1
ln jt C 1j C
C K:
16
8.t C 1/
29.
e 2x
Z
dx
D
" 4e x C 4
Z
dx
.e x " 2/2
Let u D e x
du D e x dx
du
u.u " 2/2
1
A
B
C
D C
C
u.u " 2/2
u
u"2
.u " 2/2
A.u2 " 4u C 4/ C B.u2 " 2u/ C CC u
D
u.u " 2/2
(
ACB D0
1
1
1
) "4A " 2B C C D 0 ) A D ; B D " ; C D :
4
4
2
4A D 1
Z
Z
Z
Z
du 1
du
du
1
du
1
"
C
D
u.u " 2/2
4
u
4
u"2
2
.u " 2/2
1
1
1
1
D ln juj " ln ju " 2j "
CK
4
4
2 .u " 2/
1
x 1
C K:
D " ln je x " 2j "
4 4
2.e x " 2/
Z
d&
Let u D sin &
cos &.1 C sin & /
du D
Z cos & d&
Z
du
du
D
D
.1 " u2 /.1 C u/
.1 " u/.1 C u/2
1
A
B
C
D
C
C
.1 " u/.1 C u/2
1"u
1Cu
.1 C u/2
A.1 C 2u C u2 / C B.1 " u2 / C C.1 " u/
D
.1 " u/.1 C u/2
(
A"B D0
1
1
1
) 2A " C D 0
)AD ; B D ; C D :
4
4
2
ACB CC D1
Z
du
.1 " u/.1 C u/2
Z
Z
Z
du
du
du
1
1
1
C
C
D
4
1"u
4
1Cu
2
.1 C u/2
ˇ
ˇ
1 ˇ 1 C sin & ˇˇ
1
D ln ˇˇ
"
C C:
4
1 " sin & ˇ 2.1 C sin & /
D
Therefore
Z
dx
x 4 " 3x 3
Z
Z
Z
Z
1
1
dx 1
dx 1
dx
dx
D"
"
"
C
27 ˇ x ˇ 9
x2
3
x3
27
x"3
1
1
1 ˇˇ x " 3 ˇˇ
ln
C
C 2 C K:
D
27 ˇ x ˇ 9x
6x
Z
Since .x "1/.x 2 "1/.x 3 "1/ D .x "1/3 .x C1/.x 2 Cx C1/,
and the numerator has degree less than the denominator,
we have
A
B
3
D
Ex C F
x5 C x3 C 1
D
C
C
C
C
:
.x " 1/.x 2 " 1/.x 3 " 1/
x " 1 .x " 1/2 .x " 1/3 x C 1 x 2 C x C 1
30.
Since .x 4 " 16/2 D .x " 2/2 .x C 2/2 .x 2 C 4/2 , and the
numerator has degree less than the denominator, we have
123 " x 7
A
B
C
D
D
C
C
C
.x 4 " 16/2
x"2
.x " 2/2
xC2
.x C 2/2
Ex C F
Gx C H
C 2
C 2
:
x C4
.x C 4/2
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SECTION 6.2 (PAGE 348)
ADAMS and ESSEX: CALCULUS 9
31. First expand the denominator .x 2 " 4/.x C 2/2 D x 4 C 4x 3 " 16x " 16. Now
divide this into x 5 to obtain the quotient polynomial x " 4
and a remainder of degree less than 4. Since the denominator factors to .x C 2/3 .x " 2/, the required decomposition
is
.x 2
Section 6.3 Inverse Substitutions
(page 355)
1.
Z
2.
Z
A
B
C
D
x5
D x"4C
C
C
C
:
" 4/.x C 2/2
x C 2 .x C 2/2 .x C 2/3 x " 2
32. We have
.x 2 CkxC4/.x 2 "kxC4/ D x 4 C.8"k 2 /x 2 C16 D x 4 C4x 2 C16
provided 8 " k 2 D 4. Since k > 0, we have k D 2. Accordingly,
D
x4
x 4 C 4x C 16 " 4x " 16
D
4
2
x C 4x C 16
x 4 C 4x 2 C 16
4x 2 C 16
D1" 2
.x C 2x C 4/.x 2 " 2x C 4/
Ax C B
Cx C D
D1C 2
C 2
:
x C 2x C 4
x " 2x C 4
D
D
D
33. Since Q.x/ D .x " a1 /.x " a2 / ! ! ! .x " an /, we have
ln Q.x/ D ln.x " a1 / C ln.x " a2 / C ! ! ! C ln.x " an /;
3.
Z
4.
Z
5.
Z
and, differentiating both sides,
d
1
1
1
Q 0 .x/
D
Œln Q.x/! D
C
C !!! C
Q.x/
dx
x " a1
x " a2
x " an
(
)
1
1
1
1
1
D 0
C
C !!! C
:
Q.x/
Q .x/ x " a1
x " a2
x " an
Since
P .x/
A1
A2
An
D
C
C !!! C
;
Q.x/
x " a1
x " a2
x " an
we have
(
)
P .x/
1
1
1
C
C
!
!
!
C
Q 0 .x/ x " a1
x " a2
x " an
A1
A2
An
D
C
C !!! C
:
x " a1
x " a2
x " an
Multiply both sides by x " a1 and get
(
)
x " a1
x " a1
P .x/
1C
C !!! C
Q 0 .x/
x " a2
x " an
A2 .x " a1 /
An .x " a1 /
D A1 C
C !!! C
:
x " a2
x " an
Now let x D a1 and obtain
Similarly, Aj D
224
P .a1 /
D A1 .
Q 0 .a1 /
P .aj /
for 1 $ j $ n.
Q 0 .aj /
dx
Let u D 2x
1 " 4x 2
du D 2 dx
Z
1
du
1
1
p
D
D sin!1 u C C D sin!1 .2x/ C C:
2
2
2
1 " u2
p
x 2 dx
p
Let 2x D sin u
1 " 4x 2
2 dx D cos u du
Z
sin2 u cos u du
1
8 Z
cos u
u
sin 2u
1
.1 " cos 2u/ du D
"
CC
16
16
32
1
1
sin!1 2x "
sin u cos u C C
16
16
1 p
1
sin!1 2x " x 1 " 4x 2 C C:
16
8
x 2 dx
p
Let x D 3 sin &
9 " x2
dx D 3 cos & d&
Z
9 sin2 & 3 cos & d&
D
3 cos &
9
D .& " sin & cos & / C C
2
x 1 p
9
D sin!1 " x 9 " x 2 C C:
2
3 2
dx
p
Let x D 12 sin &
x 1 " 4x 2
dx D 12 cos & d&
Z
Z
cos & d&
p
D csc & d&
D
sin & 1 " sin2 &
ˇ
ˇ
p
ˇ 1
1 " 4x 2 ˇˇ
ˇ
"
D ln j csc & " cot & j C C D ln ˇ
ˇCC
ˇ
ˇ 2x
2x
ˇ
ˇ
ˇ 1 " p1 " 4x 2 ˇ
ˇ
ˇ
D ln ˇ
ˇ C C1 :
ˇ
ˇ
x
dx
p
Let x D 3 sin &
x2 9 " x2
dx D 3 cos & d&
Z
3 cos & d&
D
2
Z9 sin & 3 cos &
1
csc2 & d&
D
9
p
1
1 9 " x2
D " cot & C C D "
C C:
9
9
x
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INSTRUCTOR’S SOLUTIONS MANUAL
6.
7.
8.
Z
p
SECTION 6.3 (PAGE 355)
dx
Let x D 3 sin &
x 9 " x2
dx D 3 cos
Z
Z & d&
3 cos & d&
1
D
D
csc & d&
3 sin & 3 cos &
3
ˇ
ˇ
p
9 " x 2 ˇˇ
1
1 ˇˇ 3
D ln j csc & " cot & j C C D ln ˇ "
ˇCC
ˇ
3
3 ˇx
x
ˇ
ˇ
p
1 ˇˇ 3 " 9 " x 2 ˇˇ
D ln ˇ
ˇ C C:
ˇ
3 ˇ
x
Z
Z
xC1
x dx
dx D
C
p
p
9 " x2
9 " x2
p
x
D " 9 " x 2 C sin!1 C C:
3
Z
p
Z
p
11.
a
x
dx
"
9 " x2
Let x D 3 tan &
dx D 3 sec2 & d&
Z
Z
3 sec2 & d&
D sec & d&
D
3 sec &
a2 !x 2
9 C x2
9Cx 2
p
9 C x 2 / C C1 :
12.
x
"
3
Fig. 6.3-8
Z
dx
Let x D a tan &
.a2 C x 2 /3=2
dx D a sec2 & d&
Z
Z
a sec2 & d&
a sec2 & d&
D
D
a3 sec3 &
.a2 C a2 tan2 & /3=2
Z
1
1
x
D 2
C C:
cos & d& D 2 sin & C C D p
2
a
a
a a2 C x 2
p
Z
x 3 dx
p
Let u D 9 C x 2
9 C x2
du D 2x Zdx
Z
.u " 9/ du
1
1
D
.u1=2 " 9u!1=2 / du
D
p
2
2
u
1
D u3=2 " 9u1=2 C C
3
p
1
D .9 C x 2 /3=2 " 9 9 C x 2 C C:
3
Z p
p
dx
p
10.
dx
" x 2 /3=2
Let x D a sin &
dx D
Z a cos & d&
Z
a cos & d&
1
sec2 & d&
D
D 2
a3 cos3 &
a
x
1
1
C C:
D 2 tan & C C D 2 p
2
a
a
a " x2
.a2
Fig. 6.3-11
D ln j sec & C tan & j C C D ln.x C
9.
Z
x
"
a
Fig. 6.3-12
9 C x2
dx
x4
Let x D 3 tan &
dx D 3 sec2 & d&
Z
.3 sec & /.3 sec2 & / d&
D
81 tan4 &
Z
Z
3
1
sec &
1
cos &
d& Let u D sin &
D
d&
D
9
tan4 &
9
sin4 &
du D cos & d&
Z
1
du
1
1
D
D"
CC D"
CC
9
u4
27u3
27 sin3 &
.9 C x 2 /3=2
C C:
D"
27x 3
a2 Cx 2
13.
Z
x 2 dx
" x 2 /3=2
Let x D a sin &
dx D a cos & d&
Z 2 2
a sin & a cos & d&
D
a3 cos3Z&
Z
2
D tan & d& D .sec2 & " 1/ d&
.a2
D tan & " & C C
(see Fig. s6-5-17)
x
!1 x
D p
" sin
C C:
a
a2 " x 2
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SECTION 6.3 (PAGE 355)
14.
Z
ADAMS and ESSEX: CALCULUS 9
dx
.1 C 2x 2 /5=2
1
Let x D p tan &
2
1
dx D p sec2 & d&
2
Z
Z
1
1
sec2 & d&
D
p
D p
cos3 & d&
2 Z .1 C tan2 & /5=2
2
1
.1 " sin2 & / cos & d& Let u D sin &
D p
2
# du D cos
$ & d&
Z
1
1 3
1
2
.1 " u / du D p u " u C C
D p
3
2
2
1
1
D p sin & " p sin3 & C C
2
3 2
!3
p
p
2x
1
2x
D p p
" p
p
CC
2 1 C 2x 2 3 2
1 C 2x 2
D
17.
Z
dx
D
2
x C 2x C 10
18.
Z
dx
D
x2 C x C 1
19.
Z
4x 3 C 3x
C C:
3.1 C 2x 2 /3=2
p
1C2x 2
p
2x
"
1
Fig. 6.3-14
15.
Z
dx
1
xC1
D tan!1
C C:
2
.x C 1/ C 9
3
3
Z
dx
$2 # p $2 Let u D x C
#
1
3
du D dx
C
xC
2
2
#
$
Z
2
du
2
!1
p u CC
D
# p $2 D p tan
3
3
3
u2 C
2
#
$
2
2x C 1
D p tan!1
C C:
p
3
3
dx
.4x 2 C 4x C 5/2
Z
dx
D
%
&2 Let 2x C 1 D 2 tan &
.2x C 1/2 C 4
2 dx D 2 sec2 & d&
Z
Z
1
sec2 & d&
cos2 & d&
D
D
16 sec4 &
16
&
1%
D
& C sin & cos &
32
2x C 1
1
2x C 1
1
tan!1
C
C C:
D
32
2
16 4x 2 C 4x C 5
Z
dx
p
Let x D 2 sec & .x > 2/
x x2 " 4
dx D 2 sec & tan & d&
Z
2 sec & tan & d&
D
2 sec & 2 tan &
Z
&
1
1
x
d& D C C D sec!1 C C:
D
2
2
2
2
p
4x 2 C4xC5
2xC1
"
2
Fig. 6.3-19
p
x 2 !4
x
"
20.
Z
21.
Z
2
Fig. 6.3-15
16.
Z
dx
p
x 2 " a2
Let x D a sec & .a > 0/
dx D a sec & tan & d&
Z
a sec & tan & d&
D
aZ2 sec2 & a tan &
1
1
D 2
cos & d& D 2 sin & C C
a
a
p
1 x 2 " a2
C C:
D 2
a
x
x2
p
x 2 !a2
x
"
a
Fig. 6.3-16
226
Z
x dx
.x " 1/ C 1
D
dx Let u D x " 1
x 2 " 2x C 3
.x " 1/2 C 2
du D dx
Z
Z
du
u du
C
D
u2 C 2
u2 C 2
#
$
1
1
u
D ln.u2 C 2/ C p tan!1 p
CC
2
2
2#
$
1
x"1
1
C C:
D ln.x 2 " 2x C 3/ C p tan!1 p
2
2
2
x dx
p
2
Z 2ax " x
x dx
Let x " a D a sin &
D
p
a2 " .x " a/2
dx D a cos & d&
Z
.a C a sin & /a cos & d&
D
a cos &
D a.& " cos & / C C
x"a p
" 2ax " x 2 C C:
D a sin!1
a
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 6.3 (PAGE 355)
24.
a
x!a
"
p
2ax!x 2
Fig. 6.3-21
22.
Z
p
dx
.4x " x 2 /3=2
Z
dx
Let 2 " x D 2 sin u
D
Œ4 " .2 " x/2 !3=2
"dx
D 2 cos u du
Z
Z
2 cos u du
1
D"
sec2 u du
D"
8 cos3 u
4
1
1 x"2
D " tan u C C D p
C C:
4
4 4x " x 2
2
u
Z
x dx
.3 " 2x " x 2 /3=2
Z
x dx
D
%
&3=2 Let x C 1 D 2 sin &
4 " .x C 1/2
dx D 2 cos & d&
Z
.2 sin & " 1/2 cos & d&
D
8 cos3 &
Z
Z
1
1
sec & tan & d& "
sec2 & d&
D
2
4
1
1
D sec & " tan & C C
2
4
1
xC1
1
" p
CC
D p
4 3 " 2x " x 2
3 " 2x " x 2
3"x
1
D !p
C C:
4
3 " 2x " x 2
xC1
1
Fig. 6.3-24
25.
p
4x!x 2
x 2 C2xC2
u
2!x
Fig. 6.3-22
23.
Z
dx
dx
D
Let x C 1 D tan u
2
2
.x C 2x C 2/
Œ.x C 1/2 C 1!2
dx D sec2 u du
Z
Z
2
sec u du
D cos2 u du
D
sec4 u
Z
1
u
sin 2u
D
.1 C cos 2u/ du D C
CC
2
2
4
1
1
D tan!1 .x C 1/ C sin u cos u C C
2
2
1
xC1
1
C C:
D tan!1 .x C 1/ C
2
2 x 2 C 2x C 2
Z
Z
dx
Let x D tan &
.1 C x 2 /3
dx D sec2 d&
Z
Z
sec2 &
D
d&
D
cos4 & d&
sec6 &
$
Z #
1 C cos 2& 2
D
d&
2
$
Z #
1
1 C cos 4&
D
1 C 2 cos 2& C
d&
4
2
3&
sin 2&
sin 4&
D
C
C
CC
8
4
32
sin & cos &
sin 2& cos 2&
3&
C
C
CC
D
8
2
16
3&
sin & cos &
1
D
C
C sin & cos &.2 cos2 & " 1/ C C
8
2
8
#
$
3
1
x
1
x
2
!1
D tan x C !
C
!
"
1
CC
8
2 1 C x2
8 1 C x2 1 C x2
3
1
3
x
x
D tan!1 x C !
C !
CC
8
8 1 C x2
4 .1 C x 2 /2
3x 3 C 5x
3
C C:
D tan!1 x C
8
8.1 C x 2 /2
p
2
1Cx 2
xC1
"
p
x
"
1
3!2x!x 2
Fig. 6.3-23
Fig. 6.3-25
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SECTION 6.3 (PAGE 355)
26.
ADAMS and ESSEX: CALCULUS 9
Z
x 2 dx
Let x D tan u
.1 C x 2 /2
dx D sec2 u du
Z
Z
tan2 u sec2 u du
tan2 u du
D
D
sec4 u Z
sec2 u
Z
1
.1 " cos 2u/ du
D sin2 u du D
2
u sin u cos u
D "
CC
2
2
1
1 x
D tan!1 x "
C C:
2
2 1 C x2
p
1Cx 2
28.
x
Fig. 6.3-26
29.
30.
Z p
1 " x2
dx
Let x D sin &
x3
dx D cos & d&
Z
cos2 &
d& D I; where
D
3
Z sin &
I D cot2 & csc & d&
D " csc & cot & "
D " csc & cot & "
3
d V D cot & csc & d&
V D " csc &
csc & d&
Z
csc & d& " I:
Therefore
1
1
I D " csc & cot & C ln j csc & C cot & j C C
2p
2 ˇ
ˇ
p
1 ˇˇ 1
1 1 " x2
1 " x 2 ˇˇ
C ln ˇ C
D"
ˇCC
ˇ
2
x2
2 ˇx
x
p
p
1
1 1 " x2
1
D ln.1 C 1 " x 2 / " ln jxj "
C C:
2
2
2
x2
228
Z
Let x D 3 tan &
dx D 3 sec2 & d&
U D sec &
d V D sec2 & d&
d U D sec & tanZ& d&
V D tan &
D 9 sec & tan & " 9 sec & tan2 & d&
Z
D 9 sec & tan & " 9 sec &.sec2 & " 1/ d&
Z
Z
D 9 sec & tan & C 9 sec & d& " 9 sec3 & d&
1
U D cot &
d U D " csc2 Z& d&
Z p
9 C x 2 dx
3 sec & 3 sec2 & d&
Z
D 9 sec3 & d&
D
u
27.
I D
31.
D 9 sec & tan & C 9 ln j sec & C tan & j " I
ˇp
ˇ
" p
!
#
9 C x2 % x &
9 ˇˇ 9 C x 2
x ˇˇ
9
C ln ˇ
C ˇCC
I D
2
3
3
2 ˇ
3
3ˇ
&
%
p
p
9
1
D x 9 C x 2 C ln 9 C x 2 C x C C1 :
2
2
9
(where C1 D C " ln 3)
2
Z
dx
p
Let x D u2
2C x
dx
# 2u du
$
Z D
Z
2
2u du
D2
1"
du
D
2Cu
uC2
p
p
D 2u " 4 ln ju C 2j C C D 2 x " 4 ln.2 C x/ C C:
Z
dx
Let x D u3
1 C x 1=3
dx D 3u2 du
Z 2
u du
D3
Let v D 1 C u
1Cu
dv D du
$
Z #
Z 2
1
v " 2v C 1
dv D 3
v"2C
dv
D3
v
v
$
# 2
v
D3
" 2v C ln jvj C C
2
3
D .1 C x 1=3 /2 " 6.1 C x 1=3 / C 3 ln j1 C x 1=3 j C C:
2
Z
1 C x 1=2
.
dx Let x D u6
I D
1 C x 1=3
dx D 6u5 du
Z 8
Z
u C u5
1 C u3 5
6u
du
D
6
du:
D
2
1Cu
1 C u2
Division is required to render the last integrand as a polynomial with a remainder fraction of simpler form: observe
that
u8 D u8 C u6 " u6 " u4 C u4 C u2 " u2 " 1 C 1
5
D .u2 C 1/.u6 " u4 C u2 " 1/ C 1
u D u5 C u3 " u3 " u C u
D .u2 C 1/.u3 " u/ C u:
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 6.3 (PAGE 355)
Thus
35.
uC1
u8 C u5
D u6 " u4 C u3 C u2 " u " 1 C 2
:
u2 C 1
u C1
D
32.
36.
6 7=6 6 5=6 3 2=3
x
" x
C x
C 2x 1=2 " 3x 1=3 " 6x 1=6
7
5
2
C 3 ln.1 C x 1=3 / C 6 tan!1 x 1=6 C C:
p
x 2 " x2
p
dx Let u2 D x 2 C 1
x2 C 1
2u du D 2x dx
Z p
u 3 " u2 du
D
u
Z p
p
D
3 " u2 du Let u D 3 sin v
p
du D 3 cosZv dv
Z p
p
D . 3 cos v/ 3 cos v dv D 3 cos2 v dv
Z
3
.v C sin v cos v/ C C
2
p
$
#
3 u 3 " u2
3 !1 u
C
p
CC
D sin
2
2
3
3
0s
1
x2 C 1
3
1p 2
AC
.x C 1/.2 " x 2 / C C:
D sin!1 @
2
3
2
34.
Z
0
p
e x 1 " e 2x dx
Let e x D sin &
e x dx D cos & d&
Z !=2
ˇ!=2
1
ˇ
D
cos2 & d& D .& C sin & cos & /ˇ
!=6
2
!=6
p !
p
#
1 #
3
3
"
D "
:
D
2 3
4
6
8
Z !=2
cos x
p
dx Let u D sin x
0
1 C sin2 x
du D cos x dx
Z 1
du
p
Let u D tan w
D
1 C u2
0
du D sec2 w dw
Z !=4
Z !=4
2
sec w dw
D
D
sec w dw
sec w
0
0
ˇ!=4
ˇ
ˇ
D ln j sec w C tan wjˇ
ˇ
0
p
p
D ln j 2 C 1j " ln j1 C 0j D ln. 2 C 1/:
! ln 2
dx
x 2 C 2x C 2
p
3!1
D
Z
p
3
D
Z
Z
!1
0
2
dx
.x C 1/2 C 1
Let u D x C 1
du D dx
ˇp3
ˇ
#
du
ˇ
D tan!1 uˇ D :
ˇ
u2 C 1
3
0
dx
p
x2 9 " x2
Let x D 3 sin u
dx D 3 cos u du
Z xD2
Z
3 cos u du
1 xD2 2
D
D
csc u du
2
9 xD1
xD1 9 sin u.3 cos u/
ˇxD2
ˇxD2
!
p
ˇ
1
1
9 " x 2 ˇˇ
ˇ
D"
D ." cot u/ˇ
ˇ
ˇ
ˇ
9
9
x
xD1
xD1
p
p !
p
p
5
8
5
1
2 2
"
D
"
:
D"
9
2
1
9
18
1
D
33.
p
3!1
!1
Therefore
$
Z #
uC1
I D6
u6 " u4 C u3 C u2 " u " 1 C 2
du
u C1
u7 u5
u4
u3 u2
D6
"
C
C
"
"u
7
5
4
3
2
!
1
C ln.u2 C 1/ C tan!1 u C C
2
Z
3
u
x
p
9!x 2
Fig. 6.3-36
37.
A
Bt C C
Dt C E
t
D
C 2
C 2
.t C 1/.t 2 C 1/2
t C1
t C1
.t C 1/2
%
1
D
A.t 4 C 2t 2 C 1/ C B.t 4 C t 3 C t 2 C t /
2
2
.t C 1/.t C 1/
&
C C.t 3 C t 2 C t C 1/ C D.t 2 C t / C E.t C 1/
8̂
8̂
ACB D0
B D "A
ˆ
ˆ
ˆ
ˆ
<B C C D 0
<C D A
) 2A C B C C C D D 0 ) D D "2A
ˆ
ˆ
ˆB C C C D C E D 1
ˆ 2D D 1
:̂
:̂
ACC CE D0
E D "2A:
Thus A D "1=4 D C , B D 1=4, D D E D 1=2. We have
Z
t dt
.t C 1/.t 2 C 1/2
Z
Z
Z
1
1
1
dt
.t " 1/ dt
.t C 1/ dt
D"
C
C
4
t C1
4
t2 C 1
2
.t 2 C 1/2
1
1
1
D " ln jt C 1j C ln.t 2 C 1/ " tan!1 t
4
8 Z
4
dt
1
1
"
C
Let t D tan &
4.t 2 C 1/
2
.t 2 C 1/2
dt D sec2 & d&
1
1
1
2
D " ln jt C 1j C ln.t C 1/ " tan!1 t
4
8 Z
4
1
1
2
"
cos & d&
C
4.t 2 C 1/
2
1
1
1
D " ln jt C 1j C ln.t 2 C 1/ " tan!1 t
4
8
4
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SECTION 6.3 (PAGE 355)
ADAMS and ESSEX: CALCULUS 9
1
1
C .& C sin & cos & / C K
4.t 2 C 1/
4
1
1
1
D " ln jt C 1j C ln.t 2 C 1/ " tan!1 t
4
8
4
1
1
1 t
!1
"
C tan t C
CK
4.t 2 C 1/
4
4 t2 C 1
1 t "1
1
1
D
" ln jt C 1j C ln.t 2 C 1/ C K:
4 t2 C 1 4
8
"
1
.1 " u2 /.4 " u2 /
A
B
C
D
D
C
C
C
1"u
1 C u ˇ2 " u
2Cu
ˇ
1
1
ˇ
AD
D
ˇ
2
ˇ
.1 C u/.4 " u /
6
ˇ uD1
ˇ
1
1
ˇ
D
BD
ˇ
.1 " u/.4 " u2 / ˇ
6
ˇuD!1
ˇ
1
1
ˇ
C D
D"
ˇ
.1 " u2 /.2 C u/ ˇ
12
ˇ uD2
ˇ
1
1
ˇ
DD
D" :
ˇ
.1 " u2 /.2 " u/ ˇ
12
38. We have
Z
D
x dx
D
.x 2 " x C 1/2
Z
u du
.u2
C
3 2
4/
p
C
Z
1
2
Z
uD!2
h
x dx
.x " 12 /2 C
du
.u2
3
4
i2
Let u D x "
du D dx
40.
p
2.x " 12 / 3
"1
2
2
!1 2x " 1
C
p
tan
C
p
p
p
CC
2.x 2 " x C 1/
3 3
3
3 3 .2 x 2 " x C 1/2
2x " 1
x"2
2
C
C C:
D p tan!1 p
2 " x C 1/
3.x
3 3
3
D
v
Z
du
du
1
C
1"u
6
1Cu
!
Z
Z
du
du
1
1
"
"
12
2 " u 12
2Cu
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1
1"uˇ
1 ˇ 2 C u ˇˇ
ln
D ln ˇˇ
C
CK
6
1 C u ˇ 12 ˇ 2 " u ˇ
ˇ
ˇ
ˇ
ˇ
p
p
1 ˇˇ 2 C 1 " x 2 ˇˇ
1 ˇˇ 1 " 1 " x 2 ˇˇ
p
ln ˇ
p
D ln ˇ
ˇC
ˇCK
6 ˇ 1 C 1 " x 2 ˇ 12 ˇ 2 " 1 " x 2 ˇ
p
p
1 .1 " 1 " x 2 /2
1
.2 C 1 " x 2 /2
D ln
C
C K:
ln
6
x2
12
3 C x2
I D"
C 43 /2
3
Let u D
tan v,
p 2
3
sec2 v dv in the second integral.
du D
2
p
3
!
Z
sec2 v dv
1
1
1
2
D"
C
9
2 u2 C 43
2
sec4 v
16
Z
"1
4
D
cos2 v dv
C
p
2.x 2 " x C 1/
3 3
"1
2
D
C p .v C sin v cos v/ C C
2.x 2 " x C 1/
3 3
p
Thus
1
2
4u2 C3
p
2u
3
Fig. 6.3-38
1
6
Z
Z
dx
Let x D sec &
x 2 .x 2 " 1/3=2
dx D sec & tan & d&
Z
Z
sec & tan & d&
cos3 & d&
D
D
2
3
sec & tan &
sin2 &
Z
2
1 " sin &
D
cos & d& Let u D sin &
sin2 &
du D cos & d&
Z
1
1 " u2
du D " " u C C
D
u2
u
$
#
1
C sin & C C
D"
sin &
!
p
x
x2 " 1
C
C C:
D" p
x
x2 " 1
p
x
39.
I
Z
dx
p
Let 1 " x 2 D u2
x.3 C x 2 / 1 " x 2
"2x dx D 2u du
Z
du
:
D"
.1 " u2 /.4 " u2 /
230
"
1
Fig. 6.3-40
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INSTRUCTOR’S SOLUTIONS MANUAL
41.
I D
Z
SECTION 6.3 (PAGE 355)
Z
Z
Z
du
du
du
du 1
1
D
"
"
"
2
2
2
.1 " u /u
u
2
1"u 2
1Cu
1
1
1
D C ln j1 " uj " ln j1 C uj C K
u
2
ˇ 2 p
ˇ
1 ˇˇ 1 " 1 " x 2 ˇˇ
1
C ln ˇ
p
D p
ˇCK
2 ˇ 1 C 1 " x2 ˇ
1 " x2
&
%
p
1
D p
C ln 1 " 1 " x 2 " ln jxj C K:
1 " x2
dx
x.1 C x 2 /3=2
I D"
Let x D tan &
dx D sec2 & d&
Z
Z
2
cos2 & d&
sec & d&
D
D
3
tan & sec &
sin &
Z
cos2 & sin & d&
D
Let u D cos &
sin2 &
du D " sin & d&
Z
Z 2
du
u du
DuC
:
D"
1 " u2
u2 " 1
We have
1
1
D
2
u "1
2
#
$
1
1
:
"
u"1 uC1
Thus
43.
ˇ
ˇ
1 ˇˇ u " 1 ˇˇ
I D u C ln ˇ
CC
2
u C 1ˇ
ˇ
ˇ
1 ˇˇ cos & " 1 ˇˇ
D cos & C ln ˇ
CC
2
cos & C 1 ˇ
ˇ
ˇ
1
ˇ
ˇ
ˇp
"1ˇ
ˇ
1
1 ˇˇ 1 C x 2
ˇCC
D p
C ln ˇ
ˇ
2
1
2
1Cx
ˇp
C 1 ˇˇ
ˇ
1 C x2
!
p
1
1 C x2 " 1
1
C ln p
C C:
D p
2
1 C x2
1 C x2 C 1
p
Let x D tan.&=2/;
2 dx
2x
; d& D
sin & D
1 C x2
1 C x2
2 dx
Z
dx
1
C x2 D
D
2x
1 C x C x2
2C
2
1
C
x
Z
2x C 1
dx
2
D
CC
D p tan!1 p
#
$
3
3
1 2 3
xC
C
2
4
#
$
2
!1 2 tan.&=2/ C 1
D p tan
p
C C:
3
3
44.
1Cx 2
"
1
Fig. 6.3-41
42. We have
Z
dx
Let u2 D 1 " x 2
x.1 " x 2 /3=2
2u du
Z D "2x dx
Z
du
u du
D"
D"
.1 " u2 /u3
.1 " u2 /u2
1
A
B
C
D
D C 2 C
C
2
2
u .1 " u /
u
u
1"u
1Cu
A.u " u3 / C B.1 " u2 / C C.u2 C u3 / C D.u2 " u3 /
D
u2 .1 " u2 /
8̂
"A C C " D D 0
<
"B C C C D D 0
)
:̂ A D 0
BD1
1
1
) A D 0; B D 1; C D ; D D :
2
2
d&
2 C sin &
Z
x
I D
Z
Z
Z
!=2
&
2
, d& D
dx,
2
1
C
x2
0
2
1"x
2x
cos & D
, sin & D
.
2
1
C
x
1
C
x2
$
#
2
Z 1
dx
1 C x2
D
#
$ #
$
1 " x2
2x
0
1C
C
1 C x2
1 C x2
Z 1
Z 1
dx
dx
D2
D
0 2 C 2x
0 1Cx
ˇ1
ˇ
ˇ
D ln j1 C xjˇ D ln 2:
ˇ
d&
1 C cos & C sin &
Let x D tan
0
45.
Z
d&
3 C 2 cos &
Let x D tan.&=2/;
1 " x2
2 dx
cos & D
; d& D
1 C x2
1 C x2
2 dx
Z
2 dx
1 C x2
D
D
5 C x2
2 " 2x 2
3C
1 C x2
#
$
x
2
tan.&=2/
2
p
D p tan!1 p C C D p tan!1
C C:
5
5
5
5
Z
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SECTION 6.3 (PAGE 355)
46.
Area D
Z
Z
1
D
2x " x 2
Let u D x " 1
du D dx
1=2
0
dx
p
Z
ADAMS and ESSEX: CALCULUS 9
1
1=2
p
dx
1 " .x " 1/2
!1=2
p
Average value D
D
ˇ0
ˇ
ˇ
uˇ
ˇ
du
D sin!1
1 " u2
% #&
#
D0" "
D
sq: units:
6
6
D
48.
!1=2
D
D
47.
For intersection of y D
x4
9
and y D 1 we have
C 4x 2 C 4
D
Z
1 4
dx
4 0 .x 2 " 4x C 8/3=2
Z
dx
1 4
4 0 Œ.x " 2/2 C 4!3=2
Let x " 2 D 2 tan u
dx D 2 sec2 u du
Z !=4
1
2 sec2 u du
4 !!=4 8 sec3 u
ˇ!=4
Z !=4
ˇ
1
1
ˇ
cos u du D
sin uˇ
ˇ
16 !!=4
16
!!=4
p
#
$
2
1
1
1
p Cp
:
D
16
16
2
2
x 4 C 4x 2 C 4 D 9
x 4 C 4x 2 " 5 D 0
.x 2 C 5/.x 2 " 1/ D 0;
so the intersections are at x D ˙1. The required area is
#
$
9 dx
"
1
dx
x 4 C 4x 2 C 4
0
Z 1
p
dx
D 18
" 2 Let x D 2 tan &
2 C 2/2
p
.x
0
dx D 2 sec2 &
Z xD1 p
2
2 sec & d&
"2
D 18
4 sec4 &
xD0
Z xD1
9
D p
cos2 & d& " 2
2 xD0
ˇ
&ˇxD1
9 %
ˇ
"2
D p & C sin & cos & ˇ
ˇ
2 2
xD0
!ˇ1
p
9
2x ˇˇ
!1 x
D p tan p C 2
ˇ "2
x C2 ˇ
2 2
2
0
#
$
1
9
!1 1
2
" units
D p tan p
2
2 2
2
AD2
Z
1
y
yD
9
x 4 C4x 2 C4
49.
Area of R
Z pa2 !b2 %p
&
a2 " x 2 " b dx
D2
0
D2
Z
p
xD a2 !b 2
xD0
Let x D a sin &
dx D a cos & d&
p
a2 cos2 & d& " 2b a2 " b 2
ˇxDpa2 !b2
p
ˇ
" 2b a2 " b 2
D a .& C sin & cos & /ˇˇ
xD0
ˇpa2 !b2
ˇ
%
&
p
p
x
ˇ
D a2 sin!1 C x a2 " x 2 ˇ
" 2b a2 " b 2
ˇ
a
0
s
2
p
p
b
D a2 sin!1 1 " 2 C b a2 " b 2 " 2b a2 " b 2
a
p
b
D a2 cos!1 " b a2 " b 2 units2
a
2
y
p
. a2 !b 2 ;b/
R
R
yDb
b
yD1
a
x
x 2 Cy 2 Da2
!1
1
x
Fig. 6.3-49
Fig. 6.3-47
232
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INSTRUCTOR’S SOLUTIONS MANUAL
1
4,
50. The circles intersect at x D
A1 C A2 where
A1 D 2
D2
Z
Z
1
p
1=4
1 " x 2 dx
xD1
cos2 u du
SECTION 6.3 (PAGE 355)
so the common area is
51.
Let x D sin u
dx D cos u du
xD1=4
ˇxD1
ˇ
ˇ
D .u C sin u cos u/ˇ
ˇ
xD1=4
ˇxD1
ˇ
ˇ
D .sin x C x 1 " x 2 /ˇ
ˇ
xD1=4
p
1
15
#
sq. units.
D " sin!1 "
2
4
16
Z 1=4 p
A2 D 2
4 " .x " 2/2 dx Let x " 2 D 2 sin v
0
dx D 2 cos v dv
Z xD1=4
D8
cos2 v dv
p
!1
$
Z 4 #p
12
Required area D
25 " x 2 "
dx
x
3
Z 4p
Z 4
12
D
25 " x 2 dx "
dx
3
3 x
Let x D 5 sin u, dx D 5 cos u du in the first integral.
ˇ4
Z xD4
ˇ
ˇ
2
D
25 cos u du " 12 ln x ˇ
ˇ
xD3
3
ˇxD4
ˇ
25
4
ˇ
.u C sin u cos u/ˇ
D
" 12 ln
ˇ
2
3
xD3
ˇ4
#
$
ˇ
25
1 p
4
ˇ
!1 x
D
sin
C x 25 " x 2 ˇ " 12 ln
ˇ
2
5
2
3
3
#
$
25
4
!1 4
!1 3
D
sin
" sin
" 12 ln sq. units.
2
5
5
3
y
xyD12
.3;4/
.4;3/
xD0
ˇxD1=4
ˇ
ˇ
D 4.v C sin v cos v/ˇ
ˇ
xD0
"
#ˇxD1=4
#
$ #
$p
x
"
2
x"2
4x " x 2 ˇˇ
!1
D 4 sin
C
ˇ
ˇ
2
2
2
xD0
"
#
p
# $
#
7
15
7
D 4 sin!1 "
C
"
8
64
2
p
# $
7
7 15
D "4 sin!1
"
C 2# sq. units.
8
16
s 2 Cy 2 D25
x
Fig. 6.3-51
52.
Shaded area D 2
A1 C A2 D
p
15
2# $
# $
1
7
" sin!1
" 4 sin!1
sq. units:
4
8
5#
"
2
xDa
c
a
r
b 1"
% x &2
a
dx
Let x D a sin u
dx D a cos u du
cos2 u du
ˇx!a
ˇ
ˇ
D ab.u C sin u cos u/ˇ
ˇ
xDc
#
$ˇˇa
p
b
ˇ
!1 x
D ab sin
C x a2 " x 2 ˇ
ˇ
a
a
c
$
#
p
cb
#
!1 c
2
" sin
"
a " c 2 sq. units.
D ab
2
a
a
D 2ab
Hence, the common area is
Z
Z
xDc
y
xDc
.x!2/2 Cy 2 D4
y
1
xD 4
A
x 2 Cy 2 D1
x
x
x2 y2
C
D1
a2 b 2
Fig. 6.3-50
Fig. 6.3-52
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SECTION 6.3 (PAGE 355)
53.
Area of R
Z
p
ADAMS and ESSEX: CALCULUS 9
Section 6.4 Other Methods for Evaluating
Integrals (page 362)
p
Yp
1CY2 "
x 2 " 1 dx
2
1
Let x D sec &
dx D sec & tan & d&
Z tan"1 Y
Yp
D
1CY2 "
sec & tan2 & d&
2
0
Z tan"1 Y
Yp
2
1CY "
sec3 & d&
D
2
0
Z tan"1 Y
C
sec & d&
0
#
Yp
1
D
1 C Y 2 C " sec & tan &
2
2
D
1CY 2
1.
We try
I D
I 0 D 3Ae 3x si n.4x/ C 4Ae 3x cos.4x/
De
Z
2.
.
1
p
1CY 2 ;Y /
D
x e !x sin x dx D .a0 Ca1 x/e !x sin xC.b0 Cb1 x/e !x cos x:
a1 " a0 " b0 D 0
"a1 " b1 D 1
Z
dx
a sinh u du
p
D
2
2
2
a cosh2 u a sinh u
Zx " a
1
1
sech2 u du D 2 tanh u C C
a2
a
%
&
1
!1 x
cosh
tanh
CC
a2
a
s
x
1
x2
C
"1"
s
a
a2
x
x2
"1
"
1
a
a2
!
CC
s
a2
x
x2
1
s
C
"1C
a
a2
x
x2
"
"1
a
a2
p
x 2 " a2
C C1 :
a2 x
234
Z
provided that
x
Fig. 6.3-53
D
3 3x
4
e sin.4x/ " e 3x cos.4x/:
25
25
dI
dx
D a1 e !x sin x " .a0 C a1 x/e !x sin x C .a0 C a1 x/e !x cos x
C b1 e !x cos x " .b0 C b1 x/e !x cos x " .b0 C b1 x/e !x sin x
D xe !x sin x
R
D
e 3x sin.4x/ dx D
Differentiating this formula gives
x 2 !y 2 D1
D
sin.4x/
Here we try
I D
y
x2
C 3Be 3x cos.4x/ " 4Be 3x si n.4x/
3x
provided 3A " 4B D 1 and 4A C 3B D 0. This pair of
equations has solution a D 3=25, b D "4=25, so we have
1
t
1
Area D ln.sinh t C cosh t / D ln e t D units2
2
2
2
54.
e 3x sin.4x/ dx D Ae 3x sin.4x/CBe 3x cos.4x/CC
for some constants A, B and C . Differentiation gives
$ˇˇtan"1 Y
1
ˇ
" ln j sec & C tan & j C ln j sec & C tan & j ˇ
ˇ
2
0
p
p
p
Y
Y
1
2
2
D
1CY "
1 C Y C ln.Y C 1 C Y 2 /
2
2
2
p
1
2
2
D ln.Y C 1 C Y / units
2
If Y D sinh t , then we have
Z
Z
a0 C b1 " b0 D 0
a1 " b1 D 0:
This system has solution a0 D 0, a1 D b0 D b1 D "1=2.
Therefore,
Z
3.
1
1
x e !x sin x dx D " xe !x sin x " .1Cx/e !x cos x CC:
2
2
It is useful to make the change of variable u D "x 2 ,
du D "2x dx before guessing the form of the integral.
Z
2
x 5 e !x dx
Z
1
D"
u2 e u du
2
D .a2 u2 C a1 u C a0 /e u C C:
I D
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 6.4 (PAGE 362)
We will have
dI
D .2a2 u C a1 C a2 u2 C a1 u C a0 /e u
du
1
D " u2 e u
2
One suspects it has forgotten to use absolute values in
some of the logarithms.
7.
I D
provided a2 D "1=2, 2a2 C a1 D 0, and a1 C a0 D 0.
Thus a1 D 1 and a0 D "1, so
$
#
1
I D " u2 C u " 1 e u C C
2
#
$
Z
1 4
2
5 !x 2
x e
dx D "
x C x 2 C 1 e !x C C
2
4. As an alternative to the direct method of Example 3, we
begin with the change of variable u D ln x, or, equivalently, x D e u , so that dx D e u du.
I D
Z
x 2 .ln x/4 dx D
Z
Neither the author’s version of Maple nor his version of
Mathematics would do
Z
p
t5
3 " 2t 4
dt
as presented. Both did an integration by parts and left
an unevaluated integral. Both managed to evaluate the
integral after the substitution u D t 2 was made. (See
Exercise 4.) However, Derive had no trouble doing the
integral in its original form, giving as the answer
p
p
p
3 2 !1 6t 2 t 2 3 " 2t 4
sin
"
:
16
3
8
8.
Maple, Mathematica, and Derive readily gave
u4 e 3u du
Z
!
"
D a4 u4 C a3 u3 C a2 u2 C a1 u C a0 e 3u C C:
1
0
3#
1
1
dx D
C :
.x 2 C 1/3
32
4
We will have
!
"
dI
D 4a4 u3 C 3a3 u2 C 2a2 u C a1 e 3u
du
!
"
C 3 a4 u4 C a3 u3 C a2 u2 C a1 u C a0 e 3u
9.
Z
4 3u
Du e
provided 3a4 D 1, 3a3 C 4a4 D 0, 3a2 C 3a3 D 0,
3a1 C 2a2 D 0, and 3a0 C a1 D 0. Thus a4 D 1=3,
a3 D "4=9, a2 D 4=9, a1 D "8=27, and a0 D 8=81. We
now have
#
$
8
8
1 4 4 3 4 2
u " u C u " uC
e 3u C C
I D
3
9
9
27
81
Z
x 2 .ln x/4 dx
$
#
4.ln x/2 8 ln x
8
.ln x/4 4.ln x/3
"
C
"
C
CC
D x3
3
9
9
27
81
6. According to Maple
Z
1 C x C x2
dx
" 1/.x 4 " 16/2
ln.x " 1/ ln.x C 1/
7
D
"
"
300
900
15;360.x " 2/
613
1
79
"
ln.x " 2/ "
C
ln.x C 2/
460;800
5;120.x C 2/
153;600
47
ln.x 2 C 1/
C
ln.x 2 C 4/
"
900
115;200
23
6x C 8
"
tan!1 .x=2/ "
25;600
15;360.x 2 C 4/
.x 4
Use the 6th integral in the list involving
11.
x 2 ˙ a2 .
p
xp 2
x 2 dx
D
p
x " 2 C ln jx C x 2 " 2j C C
2
x2 " 2
10. Use the last integral in the list involving
Z p
p
.x 2 C 4/3 dx D
x 2 ˙ a2 .
p
p
x 2
.x C10/ x 2 C 4C6 ln jxC x 2 C 4jCC
4
Use the 8th integral in the list involving
p
making the change of variable x D 3t .
Z
p
p
x 2 ˙ a2 after
p
dt
p
Let x D 3t
2
p
3t C 5
dx D 3 dt
Z
3
dx
D p
p
3
x2 x2 C 5
p
p
p
x2 C 5
3t 2 C 5
D" 3
CC D"
CC
5x
5t
t2
12. Use the 8th integral in the miscellaneous algebraic set.
Z
dt
2
p
D p tan!1
t 3t " 5
5
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r
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CC
5
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SECTION 6.4 (PAGE 362)
ADAMS and ESSEX: CALCULUS 9
13. The 5th and 4th integrals in the exponential/logarithmic set
give
17.
Use the last integral in the miscellaneous algebraic set.
Z
dx
x"2
1
p
p
D
CC
2 /3
x 5 .ln x/4 4
4
.
4x
"
x
4x
" x2
4
3
x .ln x/ dx D
x .ln x/ dx
"
5
5
#
$
Z
x 5 .ln x/4 4 x 5 .ln x/3 3
D
"
"
x 4 .ln x/2 dx
5
5
5
2
18. Use the last integral in the miscellaneous algebraic set.
#
3$
4
Then complete the square, change variables, and use the
.ln
x/
4.ln
x/
D x5
"
second last integral in the elementary list.
5
25
$
# 5
Z
2
Z
2
12 x .ln x/
dx
x 5 ln x dx
"
C
p
25
5
5
#
$
. 4x " x 2 /4
.ln x/4 4.ln x/3
12.ln x/2 24 ln x
24
Z
dx
x"2 p
1
D x5
"
C
"
C
C C:
D
. 4x " x 2 /!2 C
5
25
125
625
3;125
8
8
4x " x 2
Z
dx
x"2
1
D
C
Let u D x " 2
8.4x " x 2 /
8
4 " .x " 2/2
14. We make a change of variable and then use the first two
du D dx
Z
integrals in the exponential/logarithmic set.
x"2
1
du
D
C
8.4x " x 2 /
8
4 " u2
Z
ˇ
ˇ
2
7 x
2
x"2
1 ˇˇ u C 2 ˇˇ
x e dx Let u D x
D
C
ln
CC
8.4x " x 2 /
32 ˇ u " 2 ˇ
du D 2x dx
Z
ˇ
ˇ
1
x"2
1 ˇ x ˇ
D
u3 e u du
C
ln ˇ
D
ˇCC
2
2
8.4x " x /
32
x"4
#
$
Z
1
D
u3 e u " 3 u2 e u du
2
Z
#
$
Z
1 !
u3 e u 3
19.
Since
J
.x/
D
cos.x sin t / dt we have, differentiat2 u
u
0
"
u e " 2 ue du
D
# 0
2
2
ing with respect to x through the integral,
$
# 3
u
3u2
u
D
"
C 3.u " 1/ e C C
Z
2
2
1 !h
00
0
# 6
$
"x sin2 t cos.x sin t / " sin t sin.x sin t /
xy
C
y
C
xy
D
x
3x 4
# 0
2
x2
i
D
"
C 3x " 3 e C C
2
2
C x cos.x sin t / dt
Z
i
1 !h
D
x cos2 t cos.x sin t / " sin t sin.x sin t / dt
# 0
15. Use integrals 14 and 12 in the miscellaneous algebraic set.
Z
i
1 ! d h
Z p
D
cos t sin.x sin t / dt
# 0 dt
x 2x " x 2 dx
iˇˇ!
1h
ˇ D0
Z
D
cos
t
sin.x
sin
t
/
ˇ
3 p
.2x " x 2 /3=2
#
0
2
C
D"
2x " x dx
3
3
.
x " 1p
.2x " x 2 /3=2
1
C
D"
2x " x 2 C sin!1 .x " 1/ C C
20.
3
2
2
d
2
2
(a) Since
erf.x/ D p e !x by the Fundamental
dx
#
16. Use integrals 17 and 16 in the miscellaneous algebraic set.
Theorem of Calculus, we have
p
Z p
Z
2x " x 2
#
!x 2
dx
e
dx
D
erf.x/ C C:
x2
2
p
Z
.2x " x 2 /3=2 1
2x " x 2
"
dx
D"
p
Z 1
x2
1
x
2
#
2
!x 2
(b)
lim
erf.x/
D
p
e
p
dx
D
D 1.
x!1
.2x " x 2 /3=2 p
2
#
#
!1
2
0
D"
"
2x
"
x
"
sin
.x
"
1/
C
C
2
x2
Since e !x is an even function, erf.x/ must be an
odd function, so limx!!1 erf.x/ D "1
Z
236
4
4
Z
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 6.5 (PAGE 370)
(c) If P .x/ is a polynomial of degree m > 0, then P 0 .x/
has degree m " 1 and
2.
Z
3.
Z
%
&
d
2
2
P .x/ e !x D P 0 .x/ " 2xP .x/ e !x ;
dx
which cannot equal erf.x/ because P 0 .x/ " 2xP .x/
must have degree m C 1 and so cannot equal
p
2
.2= #/P .x/ which has degree m. Thus P .x/e !x
cannot be an antiderivative of erf.x/.
Z
2
erf.x/ dx D P .x/erf.x/ C Q.x/ e !x C C;
where P and Q are polynomials to be determined.
Then
4.
dJ
D P 0 .x/ erf.x/
dx
#
$
2
2
C p P .x/ C Q 0 .x/ " 2xQ.x/ e !x
#
D erf.x/:
Hence we must have P 0 .x/ D 1 and
2
p P .x/ C Q 0 .x/ " 2xQ.x/ D 0. The first of
#
these DEs says that P .x/ D x C k; without loss of
generality we can take the constant k to be zero. The
second DE says that
5.
6.
2x
Q 0 .x/ " 2xQ.x/ D " p :
#
Z
1
2
erf.x/ dx D xerf.x/ C p e !x C C:
#
Section 6.5 Improper Integrals
ˇR
e !2x ˇˇ
e
dx D lim
ˇ
R!1 "2 ˇ
0
0
$
#
1
1
" R
D lim
R!1 2
2e
1
D :
This integral converges.
2
Z !1
Z !1
dx
dx
D
lim
2C1
2C1
R!!1
x
x
!1
R
(
)
!1
D lim tan ."1/ " tan!1 .R/
R!!1
#
# % #&
D :
D" " "
4
2
4
This integral converges.
Z
1
1
!1
!2x
ˇ1
ˇ
dx
1=3 ˇ
D
lim
3.x
C
1/
ˇ
c!!1C
.x C 1/2=3
c
%
&
D lim 3 21=3 " .1 C c/1=3
c!!1C
p
3
D 3 2:
This integral converges.
Z C
dx
dx
D
lim
2
2
C !a! 0 a2 " x 2
0 a "x
ˇ
ˇ ˇC
ˇ a C x ˇ ˇˇ
1
ˇ
ˇˇ
D lim
ln ˇ
C !a! 2a
a " x ˇˇ
Z
a
0
The right side has degree 1 and so must the left side.
0
Thus Q must have
p degree zero. Hence Q .x/ D 0
and Q.x/ D 1= #. Therefore
J D
1
dx Let u D 2x " 1
.2x " 1/2=3
3
du D 2 dx
Z
Z R
1 1 du
1
lim
u!2=3 du
D
D
2 5 u2=3
2 R!1 5
ˇR
ˇ
1
1=3 ˇ
D
lim 3u ˇ D 1 (diverges)
ˇ
2 R!1
5
(d) Let us try the form
J D
1
1
aCC
ln
D 1:
C !a! 2a
a"C
The integral diverges to infinity.
D lim
7.
Z
1
Let u D 1 " x
du D "dx
Z 1
Z 1
du
du
D
D
lim
1=3
c!0C c u1=3
0 u
ˇ1
3
3 2=3 ˇˇ
D lim u ˇ D
ˇ
c!0C 2
2
0
(page 370)
dx
.1 " x/1=3
c
1.
Z
1
1
dx
.x " 1/3
Let u D x " 1
2
du D dx
Z R
Z 1
du
du
D
lim
D
R!1 1 u3
u3
1
ˇR
#
$
"1 ˇˇ
1
1
1
D lim
D
lim
"
D
ˇ
R!1 2u2 ˇ
R!1 2
2R2
2
1
8.
Z
1
dx
p
x 1"x
Let u2 D 1 " x
2u du D "dx
Z 1
Z c
2u du
du
D
D 2 lim
2 /u
c!1!
.1
"
u
1
"
u2
0
0
ˇ
ˇˇˇc
1 ˇˇ u C 1 ˇˇˇ
D 2 lim
ln ˇ
ˇ D 1 (diverges)
c!1! 2
u " 1 ˇˇ
0
0
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SECTION 6.5 (PAGE 370)
9.
Z
!=2
cos x dx
.1 " sin x/2=3
Let u D 1 " sin x
0
du D " cos x dx
ˇ1
Z 1
ˇ
!2=3
D
u
du D lim 3u1=3 ˇˇ D 3:
c!0C
0
10.
ADAMS and ESSEX: CALCULUS 9
15.
0
D
c
The integral converges.
Z 1
xe !x dx
0
Z R
D lim
xe !x dx
16.
R!1 0
d V D e !x dx
V D "e !x
1
ˇR Z
ˇ
R
ˇ
D lim @"xe !x ˇ C
e !x dx A
ˇ
R!1
0
0
$
#
1
R
D lim " R " R C 1 D 1:
R!1
e
e
The integral converges.
Z 1=2
Z 1
dx
dx
p
D2
s
#
$
x.1 " x/
0
0
1
1 2
" x"
4
2
Z 1=2
dx
s
D 2 lim
#
$
c!0C c
1 2
1
" x"
4
2
ˇ1=2
ˇ
D 2 lim sin!1 .2x " 1/ˇˇ
D #:
U Dx
dU D
0 dx
11.
c!0C
12.
13.
D
c
This integral diverges to infinity.
238
21.
0
C !.!=2/!
0
This integral diverges to infinity.
Z 1
dx
Let u D ln x
x.ln
x/
e
dx
du D
x
ˇln R
Z ln R
ˇ
du
ˇ
D lim
D lim ln jujˇ
ˇ
R!1 1
R!1
u
0
ˇC
ˇ
ˇ
sec x dx D
lim ln j sec x C tan xjˇ
ˇ
C !.!=2/!
ln j sec C C tan C j D 1:
c!.!=2/!
c!.!=2/!
ln sec c D 1:
ˇc
ˇ
ln j sec xjˇˇ
R!1
The integral converges.
lim
lim
lim
This integral diverges to infinity.
Z e
dx
17.
p
Let u D ln x
1 x ln x
du D dx=x
ˇ
Z 1
p ˇ1
du
D
p D lim 2 uˇˇ D 2:
c!0C
u
0
c
This integral converges.
Z 1
dx
18.
Let u D ln x
x.ln x/2
e
dx
du D
x
#
$
Z ln R
du
1
D lim
D
lim
"
C
1
D 1:
R!1
R!1 1
u2
ln R
The integral converges.
Z 1
Z 0
Z 1
x dx
19.
I D
D
C
D I1 C I2
1 C x2
!1
0
Z!1
1
x dx
I2 D
Let u D 1 C x 2
1
C x2
0
du D 2x dx
Z
1 R du
D lim
D 1 (diverges)
R!1 2 1
u
Z 0
Z 1
Z 1
x dx
20.
I D
D
C
D I1 C I2
1 C x4
!1
0
Z!1
1
x dx
Let u D x 2
I2 D
1 C x4
0
du D 2x dx
ˇR
Z 1
ˇ
1
du
1
#
!1 ˇ
D
D
lim
tan
u
ˇ D
ˇ
2 0 1 C u2
2 R!1
4
This integral diverges to infinity.
Z 1
x dx
Let u D 1 C 2x 2
.1 C 2x 2 /3=2
0
du D 4x dx
$ˇR
#
Z 1
1
du
2 ˇˇ
1
D
lim " p ˇ
D
4 1 u3=2
4 R!1
u ˇ
0
tan x dx D
1
1
14.
!=2
D lim ln.ln R/ " ln 1 D 1:
The integral converges.
Z R
Z 1
x
x
dx
D
lim
dx
R!1 0 1 C 2x 2
1 C 2x 2
0
ˇR
ˇ
1
ˇ
D lim
ln.1 C 2x 2 /ˇ
ˇ
R!1 4
0
#
"
1
1
2
ln.1 C 2R / " ln 1 D 1:
D lim
R!1 4
4
1
D :
2
Z !=2
Z
#
Similarly, I1 D " . Therefore, I D 0.
4
Z 1
Z 0
Z 1
2
I D
xe !x dx D
C
D I1 C I2
!1
0
Z!1
1
2
xe !x dx Let u D x 2
I2 D
0
du D 2x dx
ˇR
Z 1
ˇ
1
1
1
ˇ
e !u du D
lim "e !u ˇ D
D
ˇ
2 0
2 R!1
2
0
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 6.5 (PAGE 370)
1
Similarly, I1 D " . Therefore, I D 0.
2
22.
I D
I2 D
Z
1
Z!1
1
0
e !jxj dx D
Z
0
!1
e x dx C
Z
0
26.
1
e !x dx D I1 C I2
0
Area of R D "
Z
1
ˇ1
ˇ
ln x dx D " lim .x ln x " x/ˇˇ
c!0C
0
x !2 e !1=x dx
1
C !0C C
C !0C !1=C
1
D lim .e !1 " e !1=C / D :
C !0C
e
Z R
Z
I2 D lim
x !2 e !1=x dx D lim
0
D ".0 " 1/ C lim .c ln c " c/
c!0C
2
D 1 " 0 D 1 units
R!1 1
!1=R
R!1 !1
e u du
1
D lim .e !1=R " e !1 / D 1 " :
R!1
e
Hence, the total area is I1 C I2 D 1 square unit.
y
1
1
D I1 C I2 :
1
Then let u D " and du D x !2 dx in both I1 and I2 :
x
Z 1
Z !1
I1 D lim
x !2 e !1=x dx D lim
e u du
e !x dx D 1
Similarly, I1 D 1. Therefore, I D 2.
23.
The required area is
Z 1
Area D
x !2 e !1=x dx
0
Z 1
Z
D
x !2 e !1=x dx C
x
R
27.
yDln x
First assume that p ¤ 1. Since a > 0 we have
ˇR
Z 1
x !pC1 ˇˇ
!p
x dx D lim
ˇ
R!1 "p C 1 ˇ
a
a
"a!pC1
1
D
C lim
R!1 .1 " p/Rp!1
1"p
(
1
if p > 1
D .p " 1/ap!1
1
if p < 1
ˇa
Z a
!pC1 ˇ
x
ˇ
x !p dx D lim
ˇ
c!0C "p C 1 ˇ
0
Fig. 6.5-23
24.
Area of shaded region D
Z
0
1
.e !x " e !2x / dx
c
#
$ˇR
1 !2x ˇˇ
!x
D lim "e C e
ˇ
ˇ
R!1
2
0
$
#
1
1
1
!2R
!R
D lim "e
C1"
D sq. units.
C e
R!1
2
2
2
c 1!p
a!pC1
C lim
D
c!0C p " 1
1"p
8
< a1!p
D 1 " p if p < 1
:
1
if p > 1.
y
1
yDe "x
28.
yDe "2x
x
Fig. 6.5-24
25.
#
$
4
2
"
dx
2x C 1 x C 2
1
%
&ˇˇR
D lim 2 ln.2x C 1/ " ln.x C 2/ ˇˇ
R!1
1
$
#
2R C 1
" 0 D 2 ln 2 sq. units.
D lim 2 ln
R!1
RC2
Area D
Z
1
29.
If p D 1 both integrals diverge as shown in Examples 2
and 6(a).
Z 1
Z 0
Z 1
x sgn x
"x
x
dx D
dx C
dx
!1 x C 2
!1 x C 2
0 xC2
!
!
Z 1
Z 0
2
2
dx C
1"
dx
D
"1 C
xC2
xC2
0
!1
ˇ0
ˇ1
ˇ
ˇ
16
ˇ
ˇ
D ."x C 2 ln jx C 2j/ˇ C .x " 2 ln jx C 2j/ˇ D ln :
ˇ
ˇ
9
!1
0
Z 2
x 2 sgn .x " 1/ dx
0
Z 1
Z 2
D
"x 2 dx C
x 2 dx
0
1
ˇ1
ˇ2
x 3 ˇˇ
x 3 ˇˇ
1
8 1
D" ˇ C
ˇ D " C " D 2:
3 ˇ
3 ˇ
3
3 3
0
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1
239
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SECTION 6.5 (PAGE 370)
30. Since
ADAMS and ESSEX: CALCULUS 9
1
x2
$ 3 for all x # 0, therefore
C1
x
35.
x5
Z
1
x2
dx
C1
0
Z 1
Z 1
x2
x2
dx
C
dx
D
5
5
x C1
1
0 x C1
Z 1
Z
1
x2
dx
$
dx C
5
x3
0 x C1
1
D I1 C I2 :
I D
x5
Since I1 is a proper integral (finite) and I2 is a convergent improper integral, (see Theorem 2), therefore I converges.
31.
The given integral diverges to infinity.
36.
sin x
$ 1. Then
x
Z !
sin x
dx $
.1/ dx D #:
x
0
Since sin x $ x for all x # 0, thus
I D
Z
!
0
sin x
dx D lim
$!0C
x
Z
$
!
Hence, I converges.
2x
on Œ0; #=2!, we have
#
Z 1
Z !=2
j sin xj
sin x
dx
#
dx
2
x
x2
0
0
Z !=2
dx
2
D 1:
#
# 0
x
37. Since sin x #
1
1
p # p on Œ1; 1/.
1 C Zx
2 x
Z 1
1
dx
dx
p .
Since
p diverges to infinity, so must
1C x
1
1 Z x
1
dx
Therefore
p also diverges to infinity.
1
C
x
0
32. Since
e !1
ex
#
on Œ"1; 1!. Thus
xC1
xC1
Z 1
Z
ex
1 1 dx
dx #
D 1:
e !1 x C 1
!1 x C 1
The given integral diverges to infinity.
y
p
x x
1
# p for all x > 1, therefore
x2 " 1
x
I D
Z
1
2
p
x x
dx #
x2 " 1
Z
1
2
yDsin x
dx
p D I1 D 1:
x
2x
yD !
Since I1 is a divergent improper integral, I diverges.
33.
Z
1
#Z
3
1
Z
1$
!
2
3
C
e !x dx.
e !x dx D
0
1
Z 1
3
Now
e !x dx is a proper integral, and is therefore
0
0
finite. Since x 3 # x on Œ1; 1/, we have
Z
Thus
Z
1
3
e !x dx $
1
Z
1
1
e !x dx D
1
:
e
39.
3
e !x dx converges.
0
34. On [0,1], p
Thus,
1
1
1
1
1
$ p . On Œ1; 1/, p
$ 2.
2
2
x
xCx
x
xCx
Z
Z
1
0
1
1
1
dx
dx
p
$
2
xCx
x
0
Z 1
dx
dx
p
:
$
x2
x C x2
1
p
Fig. 6.5-37
#p $
# p $2
p
x
x
x
$2
D ,
Since 0 $ 1 " cos x D 2 sin2
2
2
2
Z !2
Z !2
dx
dx
for x # 0, therefore
, which
p #2
x
1 " cos x
0
0
diverges to infinity.
On .0; #=2!, sin x < x, and so csc x # 1=x. Thus
Z !=2
Z !=2
dx
csc x dx >
D 1:
x
0
0
Z !=2
Therefore
csc x dx must diverge. (It is of the form
1 " 1.)
Z
40.
Since bothZof these integrals are convergent, therefore so is
1
dx
their sum
p
.
x C x2
0
240
38.
x
!!=2
Since ln x grows more slowly than any positive power of
x, therefore we have ln x $ kx 1=4 for some constant k
1
1
#
for x # 2
and every x # 2. Thus, p
3=4
x
ln
x
kx
Z 1
dx
p
diverges to infinity by comparison with
and
x
ln x
Z 12
dx
1
.
k 2 x 3=4
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INSTRUCTOR’S SOLUTIONS MANUAL
41.
Z
1
0
dx
D
xe x
#Z
1
C
0
Z
1
0
Z
1
1$
SECTION 6.5 (PAGE 370)
dx
. But
xe x
dx
1
#
xe x
e
Z
1
0
44.
E.$/ D
dx
D 1:
x
Z
1
0
2
e !x dx D
1
2
p
#.
1
0
2
x 2 e !x dx D lim
Z
R!1 0
R
2
x 2 e !x dx
1
0
2
x 4 e !x dx D lim
Z
R
R!1 0
45.
2
0
$K
Z
ˇ
ˇ
f .x/ dx ˇˇ
$
0
x p dx D
so E.$/ D O.$ pC1 / as $ ! 0C.
K
$ pC1 ;
pC1
Z
dx
x C x2
1
1=$
p
dx
D I1 .$/ C I2 .$/
x C x2
Since f is continuous on Œa; b!, there exists a positive
constant K such that jf .x/j $ K for a $ x $ b. If
a < c < b, then
ˇ
ˇZ
Z b
ˇ
ˇ b
ˇ
ˇ
f .x/ dx "
f .x/ dx ˇ
ˇ
ˇ
ˇ c
a
ˇZ a
ˇ
ˇ
ˇ
D ˇˇ
f .x/ dx ˇˇ $ K.c " a/ ! 0 as c ! a C :
Z
Thus lim
b
f .x/ dx D
c!aC c
Z
b
f .x/ dx.
a
Similarly
ˇ
ˇZ
Z b
ˇ
ˇ c
ˇ
ˇ
f .x/ dx "
f .x/ dx ˇ
ˇ
ˇ
ˇ a
a
ˇZ c
ˇ
ˇ
ˇ
D ˇˇ
f .x/ dx ˇˇ $ K.b " c/ ! 0 as c ! b " :
b
Z
Thus lim
$
p
c
2
U D x3
d V D xe !x dx
2
d U D 3x 2 dx
V D " 12 e !x
ˇR
"
#
Z
1 3 !x 2 ˇˇ
3 R 2 !x 2
D lim " x e
x e
dx
ˇ C
ˇ
R!1
2
2 0
0
Z 1
1
3
2
2
D " lim R3 e !R C
x 2 e !x dx
2 R!1
2 0
$
#
3 1p
3p
D0C
# D
#:
2 4
8
ˇZ
ˇ
jE.$/ D ˇˇ
!
I1 .$/
> $ .1=2/!k ! 1 as $ ! 0C, and I1 .$/ is not
$k
O.$ k /.
x 4 e !x dx
43. Since f .x/ D O.x p / as x ! 0C, we have jf .x/ $ Kx p
for all x in the interval .0; a/ for some constants K and
a $ 1. If 0 < $ < a it follows that
$
1=$
Thus
b) Similarly,
Z
0
$
Z
"
Z
Now we must show that no value of k larger than 1/2
will work. If k > 1=2, then
Z $
Z $
p
dx
dx
>
p D $:
I1 .$/ D
p
2
x
C
x
2
x
0
0
2
U Dx
d V D xe !x dx
2
d U D dx
V D " 12 e !x
ˇR
#
"
Z
1 R !x 2
1 !x 2 ˇˇ
e
dx
D lim " xe
ˇ C
ˇ
R!1
2
2 0
0
Z
1
1 1 !x 2
2
D " lim Re !R C
e
dx
2 R!1
2 0
1p
1p
#D
#:
D0C
4
4
1
Thus I1 .$/ D O.$ 1=2 / as $ ! 0C, and I2 .$/ D O.$/ as
$ ! 0C. Since $ < $ 1=2 (for 0 < $ < 1), both I1 .$/ and
I2 .$/ are O.$ 1=2 / as $ ! 0C, and so, therefore, is E.$/.
a) First we calculate
Z
Z
dx
C
D
p
x C x2
0
Z $
p
dx
jI1 .$/j <
p D2 $
x
Z0 1
dx
jI2 .$/j <
D $:
2
1=$ x
Thus the given integral must diverge to infinity.
42. We are given that
We have
46.
c
c!b! a
Z 1
(.x/ D
f .x/ dx D
Z
b
f .x/ dx.
a
t x!1 e !t dt .
0
a) Since lim t!1 t x!1 e !t=2 D 0, there exists T > 0 such
that t x!1 e !t=2 $ 1 if t # T . Thus
Z 1
Z 1
0$
t x!1 e !t dt $
e !t dt D 2e !T =2
T
and
Z
1
T
t x!1 e !t dt converges by the comparison
T
theorem.
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SECTION 6.5 (PAGE 370)
ADAMS and ESSEX: CALCULUS 9
The approximations are
If x > 0, then
Z
0$
T
t x!1 e !t dt <
0
Z
T
(
#
$
#
$
)
1 1
1
9
5
C 1C
C .1 C 1/ C 1 C
C
D 4:75
2 2
4
4
2
(#
$ #
$ #
$ #
$)
1
1
9
25
49
M4 D
1C
C 1C
C 1C
C 1C
2
16
16
16
16
D 4:625
1
T8 D .T4 C M4 / D 4:6875
2"
#
$ #
$ #
$
1
1
9
25
M8 D
1C
C 1C
C 1C
4
64
64
64
#
$ #
$ #
$
49
81
121
C 1C
C 1C
C 1C
64
64
64
#
$ #
$#
169
225
C 1C
C 1C
D 4:65625
64
64
T4 D
t x!1 dt
0
converges by Theorem 2(b). Thus the integral defining (.x/ converges.
Z 1
b) (.x C 1/ D
t x e !t dt
0
Z R
t x e !t dt
D lim
c!0C
R!1
c
U D tx
d V D e !t dt
x!1
dU D
dx
V D "e !t
0 xt
1
ˇR
Z R
ˇ
ˇ
D lim @"t x e !t ˇ C x
t x!1 e !t dt A
c!0C
ˇ
c
R!1
c
Z 1
x!1 !t
D0Cx
t
e dt D x(.x/:
T16 D
0
c) (.1/ D
Z
1
e
!t
0
1
.T8 C M8 / D 4:671875:
2
The exact errors are
dt D 1 D 0Š.
I " T4 D "0:0833333I
I " T8 D "0:0208333I
I " T16 D "0:0052083:
By (b), (.2/ D 1(.1/ D 1 & 1 D 1 D 1Š.
In general, if (.k C 1/ D kŠ for some positive integer
k, then
(.k C 2/ D .k C 1/(.k C 1/ D .k C 1/kŠ D .k C 1/Š.
Hence (.n C 1/ D nŠ for all integers n # 0, by
induction.
# $ Z 1
1
D
t !1=2 e !t dt Let t D x 2
d) (
2
0
dt DZ2x dx
Z 1
1
p
1 !x 2
2
e
2x dx D 2
e !x dx D #
D
x
0
0
# $
# $
1
1
1p
3
D (
D
#:
(
2
2
2
2
I " M4 D 0:0416667I
I " M8 D 0:0104167I
If f .x/ D 1 C x 2 , then f 00 .x/ D 2 D K, and
K.2 " 0/
1
D . Therefore, the error bounds are
12
3
# $2
1
' 0:0833333I
2
# $2
1 1
' 0:0208333I
" T8 j $
3 4
# $2
1 1
' 0:0052083:
" T16 j $
3 8
# $
1 1 2
" M4 j $
' 0:0416667I
6 2
# $2
1 1
' 0:0104167:
" M8 j $
6 4
Trapezoid W jI " T4 j $
jI
jI
Midpoint W jI
jI
Section 6.6 The Trapezoid and Midpoint
Rules (page 377)
1
3
Note that the actual errors are equal to these estimates
since f is a quadratic function.
1.
The exact value of I is
I D
Z
0
2
2.
The exact value of I is
$ˇ2
#
x 3 ˇˇ
.1 C x / dx D x C
ˇ
3 ˇ
2
0
I D
8
D 2 C ' 4:6666667:
3
242
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Z
0
1
e
!x
dx D "e
ˇ1
ˇ
ˇ
ˇ
!x ˇ
1
D 1 " ' 0:6321206:
e
0
lOMoARcPSD|6566483
INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 6.6 (PAGE 377)
The approximations are
The approximations are
#
$
#
#
#
3#
1
0 C sin C sin C sin
C
' 0:9871158
8
8
4
8
2
#
$
#
3#
5#
7#
#
sin
C sin
C sin
C sin
' 1:0064545
M4 D
8
16
16
16
16
1
T8 D .T4 C M4 / ' 0:9967852
2
#
3#
5#
#
sin
C sin
C sin
M8 D
16
32
32
32
T4 D 41 . 12 e 0 C e !1=4 C e !1=2 C e !3=4 C 21 e !1 /
T4 D
' 0:6354094
M4 D 41 .e !1=8 C e !3=8 C e !5=8 C e !7=8 /
' 0:6304774
T8 D 12 .T4 C M4 / ' 0:6329434
M8 D 18 .e !1=16 C e !3=16 C e !5=16 C e !7=16 C
e !9=16 C e !11=16 C e !13=16 C e !15=16 /
' 0:6317092
T16 D
1
2 .T8
7#
9#
11#
C sin
C sin
32
32 ! 32
13#
15#
C sin
C sin
' 1:0016082
32
32
C sin
C M8 / ' 0:6323263:
The exact errors are
I " T4 D "0:0032888I
I " T8 D "0:0008228I
I " T16 D "0:0002057:
T16 D
I " M4 D 0:0016432I
I " M8 D 0:0004114I
The actual errors are
I " T4 ' 0:0128842I
I " T8 ' 0:0032148I
I " T16 ' 0:0008033:
If f .x/ D e !x , then f .2/ .x/ D e !x . On [0,1],
jf .2/ .x/j $ 1. Therefore, the error bounds are:
Trapezoid W jI
jI
jI
jI
Midpoint W jI
jI
jI
1
.T8 C M8 / ' 0:9991967:
2
# $
1 1 2
" Tn j $
12 n
# $
1
1
" T4 j $
' 0:0052083I
12 16
# $
1
1
" T8 j $
' 0:001302I
12 64
#
$
1
1
' 0:0003255:
" T16 j $
12 256
# $2
1 1
" Mn j $
24 n
# $
1
1
' 0:0026041I
" M4 j $
24 16
# $
1
1
" M8 j $
' 0:000651:
24 64
I " M4 ' "0:0064545I
I " M8 ' "0:0016082I
If f .x/ D sin x, then f 00 .x/ D " sin x, and
jf 00 .x/j $ 1 D K. Therefore, the error bounds are:
& % # &2
1 %#
"0
' 0:020186I
12 2
8
%
&
%
&
# 2
1 #
"0
' 0:005047I
" T8 j $
12 2
16
& % # &2
1 %#
" T16 j $
"0
' 0:001262:
12 2
32
%
&
%
&
1 #
# 2
" M4 j $
"0
' 0:010093I
24 2
8
%
&
%
&
# 2
1 #
"0
' 0:002523:
" M8 j $
24 2
16
Trapezoid W jI " T4 j $
jI
jI
Midpoint W jI
jI
Note that the actual errors satisfy these bounds.
Note that the actual errors satisfy these bounds.
4.
3. The exact value of I is
I D
Z
0
!=2
sin x dx D 1:
The exact value of I is
I D
Z
1
0
dx
D tan!1
1 C x2
Copyright © 2018 Pearson Canada Inc.
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ˇ1
ˇ
#
ˇ
xˇ D
' 0:7853982:
ˇ
4
0
243
lOMoARcPSD|6566483
SECTION 6.6 (PAGE 377)
ADAMS and ESSEX: CALCULUS 9
The approximations are
(
# $)
1 1
16
4
16
1 1
T4 D
.1/ C
C C
C
4 2
17
5
25
2 2
' 0:7827941
(
)
64
64
64
1 64
C
C
C
M4 D
4 65
73
89
113
' 0:7867001
1
T8 D .T4 C M4 / ' 0:7847471
2"
1 256
256
256
256
M8 D
C
C
C
C
8 257
265
281
305
#
256
256
256
256
C
C
C
337
377
425
481
T16
' 0:7857237
1
D .T8 C M8 / ' 0:7852354:
2
The exact errors are
I " T4 D 0:0026041I I " M4 D "0:0013019I
I " T8 D 0:0006511I I " M8 D "0:0003255I
I " T16 D 0:0001628:
"2x
1
, then f 0 .x/ D
and
1 C x2
.1 C x 2 /2
6x 2 " 2
f 00 .x/ D
. On [0,1], jf 00 .x/j $ 4. Therefore,
.1 C x 2 /3
the error bounds are
Since f .x/ D
Trapezoid W jI
jI
jI
jI
Midpoint W jI
jI
jI
# $
4 1 2
" Tn j $
12 n
# $
4
1
" T4 j $
' 0:0208333I
12 16
# $
1
4
' 0:0052083I
" T8 j $
12 64
#
$
4
1
" T16 j $
' 0:001302:
12 256
# $
4 1 2
" Mn j $
24 n
# $
4
1
" M4 j $
' 0:0104167I
24 16
# $
4
1
" M8 j $
' 0:0026042:
24 64
The exact errors are much smaller than these bounds.
In part, this is due to very crude estimates made for
jf 00 .x/j.
5.
2
Œ3 C 2.5 C 8 C 7/ C 3! D 46
2
1
T8 D Œ3 C 2.3:8 C 5 C 6:7 C 8 C 8 C 7 C 5:2/ C 3! D 46:7
2
T4 D
244
6.
7.
M4 D 2.3:8 C 6:7 C 8 C 5:2/ D 47:4
2
T4 D 100 & Œ0 C 2.5:5 C 5 C 4:5/ C 0! D 3; 000 km2
2
1
T8 D 100 & Œ0 C 2.4 C 5:5 C 5:5 C 5 C 5:5 C 4:5 C 4/ C 0!
2
D 3; 400 km2
8.
M4 D 100 & 2.4 C 5:5 C 5:5 C 4/ D 3; 800 km2
9.
We have
T4 D 0:4
%
1
.1:4142/ C
2
1:3860 C 1:3026 C 1:1772
&
C 21 .0:9853/ ' 2:02622
M4 D .0:4/.1:4071 C 1:3510 C 1:2411 C 1:0817/ ' 2:03236
T8 D .T4 C M4 /=2 ' 2:02929
M8 D .0:2/.1:4124 C 1:3983 C 1:3702 C 1:3285
C 1:2734 C 1:2057 C 1:1258 C 1:0348/ ' 2:02982
T16 D .T8 C M8 /=2 ' 2:029555:
10. The approximations for I D
Z
1
2
e !x dx are
0
#
1 !1=256
e
C e !9=256 C e !25=256 C e !49=256 C
8
$
e !81=256 C e !121=256 C e !169=256 C e !225=256
M8 D
T16
' 0:7473
(
1 1
.1/ C e !1=256 C e !1=64 C e !9=256 C e !1=16 C
D
16 2
e !25=256 C e !9=64 C e !49=256 C e !1=4 C e !81=256 C
e !25=64 C e !121=256 C e !9=16 C e !169=256 C e !49=64 C
)
1
e !225=256 C e !1
2
' 0:74658:
2
2
Since f .x/ D e !x , we have f 0 .x/ D "2xe !x ,
2
2
f 00 .x/ D 2.2x 2 " 1/e !x , and f 000 .x/ D 4x.3 " 2x 2 /e !x .
000
Since f .x/ ¤ 0 on (0,1), therefore the maximum value
of jf 00 .x/j on Œ0; 1! must occur at an endpoint of that
interval. We have f 00 .0/ D "2 and f 00 .1/ D 2=e, so
jf 00 .x/j $ 2 on Œ0; 1!. The error bounds are
# $2
# $
2
1
1
) jI " M8 j $
n
24 64
' 0:00130:
# $
#
$
2
1
2 1 2
) jI " T16 j $
jI " Tn j $
12 n
12 256
' 0:000651:
jI " Mn j $
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24
lOMoARcPSD|6566483
INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 6.6 (PAGE 377)
Thus the constant in the error estimate for the Midpoint
Rule cannot be improved; no smaller constant will work
for f .x/ D x 2 .
According to the error bounds,
Z
11.
1
0
2
e !x dx D 0:747;
14.
accurate to two decimal places, with error no greater than
1 in the third decimal place.
Z !=2
sin x
sin x
dx. Note that lim
D 1.
I D
x!0 x
x
0
"
# 1
16
#
8
#
16
3#
4
#
C
sin
C sin C
sin
C sin
16 2
#
16
#
8
3#
16
#
4
# $#
16
5#
8
3#
16
7#
1 2
C
sin
C
sin
C
sin
C
5#
16
3#
8
7#
16
2 #
T8 D
' 1:3694596
"
#
32
3#
32
5#
32
7#
# 32
sin
C
sin
C
sin
C
sin
M8 D
16 #
32
3#
32
5#
32
7#
32
C
32
9#
32
11#
32
13#
sin
C
sin
C
sin
9#
32
11##
32
13#
32
32
15#
C
sin
' 1:3714136
15#
32
T16 D .T8 C M8 /=2 ' 1:3704366;
I ' 1:370:
12. The exact value of I is
I D
Z
The approximation is
1
0
ˇ1
1
x 3 ˇˇ
x dx D
ˇ D :
ˇ
3
3
2
0
(
)
1
1
1
T1 D .1/ .0/2 C .1/2 D :
2
2
2
The actual error is I " T1 D " 16 . However, since
f .x/ D x 2 , then f 00 .x/ D 2 on [0,1], so the error estimate
here gives
1
2
.1/2 D :
jI " T1 j $
12
6
Since this is the actual size of the error in this case, the
constant “12” in the error estimate cannot be improved
(i.e., cannot be made larger).
# $2
Z 1
1
1
1
.1/ D . The actual
13. I D
x 2 dx D . M1 D
3
2
4
0
1
1 1
.
error is I " M1 D " D
3 4
12
Since the second derivative of x 2 is 2, the error estimate
is
2
1
jI " M1 j $
.1 " 0/2 .12 / D
:
24
12
Let y D f .x/. We are given that m1 is the midpoint of
Œx0 ; x1 ! where x1 " x0 D h. By tangent line approximate
in the subinterval Œx0 ; x1 !,
f .x/ ' f .m1 / C f 0 .m1 /.x " m1 /:
The error in this approximation is
E.x/ D f .x/ " f .m1 / " f 0 .m1 /.x " m1 /:
If f 00 .t / exists for all t in Œx0 ; x1 ! and jf 00 .t /j $ K for
some constant K, then by Theorem 11 of Section 4.9,
jE.x/j $
K
.x " m1 /2 :
2
Hence,
jf .x/ " f .m1 / " f 0 .m1 /.x " m1 /j $
K
.x " m1 /2 :
2
We integrate both sides of this inequlity. Noting that
x1 " m1 D m1 " x0 D 21 h, we obtain for the left side
ˇZ
Z x1
ˇ x1
ˇ
f .x/ dx "
f .m1 / dx
ˇ
ˇ x0
x0
ˇ
Z x1
ˇ
ˇ
0
"
f .m1 /.x " m1 / dx ˇ
ˇ
x0
ˇZ
ˇx1 ˇ
ˇ x1
.x " m1 /2 ˇˇ ˇˇ
ˇ
0
Dˇ
f .x/ dx " f .m1 /h " f .m1 /
ˇ ˇ
ˇ x0
ˇ ˇ
2
x0
ˇ Z x1
ˇ
ˇ
ˇ
D ˇˇ
f .x/ dx " f .m1 /hˇˇ :
x0
Integrating the right-hand side, we get
ˇx1
K
K .x " m1 /3 ˇˇ
2
.x " m1 / dx D
ˇ
ˇ
2
3
x0 2
x0
# 3
3$
K h
h
K 3
D
C
D
h :
6
8
8
24
Z
x1
Hence,
ˇZ
ˇ
ˇ
ˇ
x1
x0
ˇZ
ˇ
D ˇˇ
ˇ
ˇ
f .x/ dx " f .m1 /hˇˇ
x1
x0
ˇ
ˇ
Œf .x/ " f .m1 / " f .m1 /.x " m1 /! dx ˇˇ
0
K 3
h :
$
24
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SECTION 6.6 (PAGE 377)
ADAMS and ESSEX: CALCULUS 9
A similar estimate holds on each subinterval Œxj !1 ; xj ! for
1 $ j $ n. Therefore,
ˇZ
ˇ
ˇ
ˇ
b
a
ˇ ˇ n #Z
ˇ ˇX
f .x/ dx " Mn ˇˇ D ˇˇ
$
$
xj
xj "1
j D1
ˇZ x
n
Xˇ j
j D1
n
X
j D1
ˇ
ˇ
xj "1
$ˇ
ˇ
f .x/ dx " f .mj /h ˇˇ
ˇ
ˇ
f .x/ dx " f .mj /hˇˇ
These errors are evidently much smaller than the corresponding errors for the corresponding Trapezoid Rule approximations.
3.
K 3
K 3
K.b " a/ 2
h D
nh D
h
24
24
24
Z
2
0
.1 C x 2 / dx D
(page 382)
#
#
3#
#
#
0 C 4 sin
C 2 sin C 4 sin
C 2 sin
48
16
8
16
4
!
3#
7#
#
5#
C 2 sin
C 4 sin
C sin
C 4 sin
16
8
16
2
Errors: I " S4 ' "0:0001346; I " S8 ' "0:0000083.
4.
1
e
0
D1"
!x
dx D "e
The exact value of I is
I D
Z
1
0
dx
D tan!1
1 C x2
The approximations are
ˇ1
ˇ
#
ˇ
xˇ D
' 0:7853982:
ˇ
4
0
"
#
# $
# $
# $
16
4
16
1
1
1C4
C2
C4
C
S4 D
12
17
5
25
2
' 0:7853922
"
# $
# $
# $
1
64
16
64
S8 D
1C4
C2
C4
C
24
65
17
73
#
# $
# $
#
$
# $
64
16
64
1
4
C4
C2
C4
C
2
5
89
25
113
2
2. The exact value of I is
ˇ1
ˇ
!x ˇ
' 0:7853981:
ˇ
ˇ
0
The actual errors are
1
' 0:6321206:
e
I " S4 D 0:0000060I I " S8 D 0:0000001;
The approximations are
1 0
.e C 4e !1=4 C 2e !1=2 C 4e !3=4 C e !1 /
12
' 0:6321342
1 0
S8 D
.e C 4e !1=8 C 2e !1=4 C 4e !3=8 C
24
2e !1=2 C 4e !5=8 C 2e !3=4 C 4e !7=8 C e !1 /
' 0:6321214:
accurate to 7 decimal places. These errors are evidently
much smaller than the corresponding errors for the corresponding Trapezoid Rule approximation.
S4 D
The actual errors are
5.
6.
I " S4 D "0:0000136I I " S8 D "0:0000008:
246
#
$
#
#
#
3#
#
0 C 4 sin C 2 sin C 4 sin
C sin
24
8
4
8
2
' 1:0001346
' 1:0000083:
The errors are zero because Simpson approximations are
exact for polynomials of degree up to three.
Z
sin x dx D 1.
0
14
' 4:6666667
3
"
#
$
#
$
1
9
1
1C4 1C
C 2.1 C 1/ C 4 1 C
S4 D
6
4
4
#
14
C .1 C 4/ D
3
"
#
$
#
$
#
$
1
1
1
9
S8 D
1C4 1C
C2 1C
C4 1C
12
16
4
16
#
$
#
$
25
9
C 2 .1 C 1/ C 4 1 C
C2 1C
16
4
#
#
$
49
14
C4 1C
C .1 C 4/ D
16
3
I D
!=2
S8 D
Section 6.7 Simpson’s Rule
I D
Z
S4 D
because nh D b " a.
1.
I D
1
Œ3 C 4.3:8 C 6:7 C 8 C 5:2/ C 2.5 C 8 C 7/ C 3!
3
' 46:93
S8 D
1
S8 D 100 & Œ0 C 4.4 C 5:5 C 5:5 C 4/ C 2.5:5 C 5 C 4:5/ C 0!
3
' 3; 533 km2
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lOMoARcPSD|6566483
INSTRUCTOR’S SOLUTIONS MANUAL
7.
SECTION 6.7 (PAGE 382)
If f .x/ D e !x , then f .4/ .x/ D e !x , and jf .4/ .x/j $ 1 on
Œ0; 1!. Thus
# $
1.1 " 0/ 1 4
' 0:000022
jI " S4 j $
180
4
# $
1.1 " 0/ 1 4
jI " S8 j $
' 0:0000014:
180
8
If f .x/ D sin x, then f .4/ .x/ D sin x, and jf .4/ .x/j $ 1
on Œ0; #=2!. Thus
1..#=2/ " 0/ % # &4
jI " S4 j $
' 0:00021
180
8
%
&
1..#=2/ " 0/ # 4
' 0:000013:
jI " S8 j $
180
16
Z b
8. Let I D
f .x/ dx, and the interval Œa; b! be subdivided
a
into 2n subintervals of equal length h D .b " a/=2n. Let
yj D f .xj / and xj D a C j h for 0 $ j $ 2n, then
#
$(
1 b"a
S2n D
y0 C 4y1 C 2y2 C ! ! !
3
2n
)
C 2y2n!2 C 4y2n!1 C y2n
D
1
3
#
b"a
2n
and
T2n D
1
2
Tn D
1
2
Since T2n D
#
#
1
2 .Tn
$(
)
2n!1
n!1
X
X
y0 C 4
yj " 2
y2j C y2n
j D1
j D1
b"a
2n
b"a
n
$#
$
2n!1
X
y0 C 2
yj C y2n
j D1
$#
$
n!1
X
y0 C 2
y2j C y2n :
D
1
3
9.
j D1
j D1
j D1
Tn C 2Mn
2T2n C Mn
4T2n " Tn
D
D
:
3
3
3
We use the results of Exercise 9 of Section 7.6 and Exercise 8 of this section.
I D
Z
1:6
f .x/ dx
0
0:4
.1:4142 C 4.1:3860/ C 2.1:3026/ C 4.1:1772/
3
C 0:9853/ ' 2:0343333
S8 D .T4 C 2M4 /=3 ' 2:0303133
S16 D .T8 C 2M8 /=3 ' 2:0296433:
S4 D
10. The approximations for I D
S8 D
C Mn / ) Mn D 2T2n " Tn , then
$(
)
2n!1
n!1
X
X
b"a
y0 C 4
yj " 2
y2j C y2n
2n
D S2n :
S2n D
j D1
Tn C 2.2T2n " Tn /
4T2n " Tn
Tn C 2Mn
D
D
3
3
3
2T2n C 2T2n " Tn
4T2n " Tn
2T2n C Mn
D
D
:
3
3
3
Hence,
Tn C 2Mn
2T2n C Mn
4T2n " Tn
D
D
:
3
3
3
Using the formulas of T2n and Tn obtained above,
4T2n " Tn
3"
#
$#
$
2n!1
X
1 4 b"a
y0 C 2
yj C y2n
D
3 2
2n
j D1
#
$#
$#
n!1
X
1 b"a
"
y0 C 2
y2j C y2n
2
n
#
Hence,
S16
Z
1
2
e !x dx are
0
# $(
#
1
1 C 4 e !1=64 C e !9=64 C e !25=64 C
8
$
#
$
)
!49=64
!1=16
!1=4
!9=16
!1
e
C2 e
Ce
Ce
Ce
1
3
' 0:7468261
# $(
#
1 1
D
1 C 4 e !1=256 C e !9=256 C e !25=256 C
3 16
e !49=256 C e !81=256 C e !121=256 C e !169=256 C
$
#
e !225=256 C 2 e !1=64 C e !1=16 C e !9=64 C e !1=4 C
$
)
e !25=64 C e !9=16 C e !49=64 C e !1
' 0:7468243:
2
2
If f .x/ D e !x , then f .4/ .x/ D 4e !x .4x 4 " 12x 2 C 3/.
On [0,1], jf .4/ .x/j $ 12, and the error bounds are
# $
12.1/ 1 4
180 n
# $
12 1 4
' 0:0000163
jI " S8 j $
180 8
# $4
1
12
' 0:0000010:
jI " S16 j $
180 16
jI " Sn j $
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SECTION 6.7 (PAGE 382)
ADAMS and ESSEX: CALCULUS 9
Comparing the two approximations,
I D
Z
1
0
Another possibility:
Z 1 x
Z 0
e dx
e x dx
C
D I1 C I2 :
I D
p
p
2
1"x
0
1 " x2
!1
2
e !x dx D 0:7468;
In I1 put 1 C x D u2 ; in I2 put 1 " x D u2 :
accurate to 4 decimal places.
11.
Z
D
I
1
I1 D
1
.
D
# 5
4
x dx
" 0 # $
5
1 4
1 4
C 14 D
S2 D
0 C4
.
6
2
24
I2 D
If f .x/ D x 4 , then f .4/ .x/ D 24.
# $
1
24.1 " 0/ 1 4
.
D
Error estimate: jI " S2 j $
180ˇ
2
120
ˇ
ˇ1
5ˇ
1
Actual error: jI " S2 j D ˇˇ " ˇˇ D
.
5 24
120
Thus the error estimate cannot be improved.
so I D 2
4.
12. The exact value of I is
I D
Z
The approximation is
1
S2 D
3
1
0
ˇ1
x 4 ˇˇ
1
x dx D
ˇ D :
4 ˇ
4
3
0
#
# $"
# $3
1
1
1
3
3
C1 D :
0 C4
2
2
4
5.
The actual error is zero. Hence, Simpson’s Rule is exact
for the cubic function f .x/ D x 3 . Since it is evidently
exact for quadratic functions f .x/ D Bx 2 C C x C D, it
must also be exact for arbitrary cubics
f .x/ D Ax 3 C Bx 2 C C x C D.
Section 6.8 Other Aspects of Approximate
Integration (page 388)
1.
2.
Z
Z
1
dx
Let x D u3
1=3 .1 C x/
x
0
Z 1
Z 1
u2 du
u du
:
D
3
D3
3
3
0 1Cu
0 u.1 C u /
1
0
p
D"
Z
ex
0
1
I D
248
!1
e x dx
p
D
1 " x2
Z
!=2
1
0
2 !1
eu
p
2 !1
eu
p
du
2 " u2
e
du
p
2 " u2
1!u2
2
C e 1!u
2 " u2
1
2
t
1
2 dt
dx D " 3
t
$
#
Z 0
1
2 dt
D
" 3
# $2 r
t
1
1
1
C1
C
2
2
t
t
Z 1
t dt
D2
:
4
3
0 t Ct C1
Z !=2
dx
Let sin x D u2
p
p
sin x
0
2u du D cos x dx D 1 " u4 dx
Z 1
u du
D2
p
0 u 1 " u4
Z 1
du
D2
Let 1 " u D v 2
p
0
.1 " u/.1 C u/.1 C y 2
"du D 2v dv
Z 1
v dv
p
D4
0 v .1 C 1 " v 2 /.1 C .1 " v 2 /2 /
Z 1
dv
p
D4
:
.2 " v 2 /.2 " 2v 2 C v 4 /
0
Let
Z
x2 C
1
0
dx
p
xC1
du.
dx
D
x4 C 1
Z
Let x D
1
0
dx
C
x4 C 1
Z
1
1
dx
D I1 C I2 :
x4 C 1
Hence,
Z
3. One possibility: let x D sin & and get
1
0
0
Z
Z 1
2
2e u !1 u du
p
D2
u 2 " u2
0
Z 1
2
2e 1!u u du
p
D2
u 2 " u2
0
1
dt
and dx D " 2 in I2 , then
t
t
#
$ Z 1
Z 0
dt
t2
1
dt:
" 2 D
I2 D
# $4
4
t
1
0 1Ct
1
C1
t
dx
Z
1
1
1
Let x D
Let t 2 D 1 " x
2t dt D "dx
Z 1
2
e 1!t
2
e 1!t dt:
2t dt D 2
t
0
1"x
6.
Z
Z
Z
1
0
dx
D
x4 C 1
e sin " d&:
!!=2
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D
Z
Z
1
0
1
0
#
1
x2
C
4
x C1
1 C x4
2
x C1
dx:
x4 C 1
$
dx
lOMoARcPSD|6566483
INSTRUCTOR’S SOLUTIONS MANUAL
7.
I D
Z
1
0
p
x dx D
SECTION 6.8 (PAGE 388)
2
' 0:666667:
3
Hence,
# $(
)
1 1
0 C 4.4e !4 / C e !1
3 2
' 0:1101549
# $"
1 1
0 C 4.16e !16 / C 2.4e !4 /
S4 D
3 4
#
$
#
16 !16=9
!1
e
Ce
C4
9
S2 D
!
1
1
C
' 0:603553
2
2
r
r !
1
3
1
2T2 C
C
' 0:643283
T4 D
4
4
4
r
r
r
r !
1
3
5
7
1
4T4 C
C
C
C
' 0:658130
T8 D
8
8
8
8
8
r
r
r
r
1
3
5
7
1
T16 D
8T8 C
C
C
C
16
16
16
16
16
r
r
r !
r
9
11
13
15
C
C
C
' 0:663581:
C
16
16
16
16
T2 D
1
0C
2
r
' 0:1430237
# $"
#
1 1
64
64
S8 D
0 C 4 64e !64 C e !64=9 C e !64=25 C
3 8
9
25
#
$
#
$
64 !64=49
16
e
C 2 16e !16 C 4e !4 C e !16=9 C e !1
49
9
' 0:1393877:
The errors are
I " T2
I " T4
I " T8
I " T16
Observe that, although these errors are decreasing, they are
not decreasing like 1=n2 ; that is,
p
This is because the second derivative of f .x/ D x is
f 00 .x/ D "1=.4x 3=2 /, which is not bounded on Œ0; 1!.
8. Let
1
1
I D
e
dx Let x D
t
1
dt
dx D " 2
t
#
$
Z 0
Z 1 !1=t 2
1
e
!.1=t/2
D
e
" 2 dt D
dt:
t
t2
1
0
!x 2
Observe that
h1i
t !2
e !1=t
D lim
2
2
1=t
t!0C e
t
1
"2t !3
D lim
t!0C e 1=t 2 ."2t !3 /
1
D lim
D 0:
t!0C e 1=t 2
2
lim
t!0C
9.
Referring to Example 5, we have
ex D 1 C x C
1
jI " T2n j >> jI " Tn j:
4
Z
Hence, I ' 0:14, accurate to 2 decimal places. These
approximations do not converge very quickly, because the
2
fourth derivative of e !1=t has very large values for some
values of t near 0. In fact, higher and higher derivatives
behave more and more badly near 0, so higher order methods cannot be expected to work well either.
' 0:0631
' 0:0234
' 0:0085
' 0:0031:
where Rn .f I 0; x/ D
x. Now
x2
xn
C !!! C
C Rn .f I 0; x/;
2Š
nŠ
e X x nC1
, for some X between 0 and
.n C 1/Š
jRn .f I 0; "x 2 /j $
x 2nC2
.n C 1/Š
if 0 $ x $ 1 for any x, since "x 2 $ X $ 0. Therefore
ˇZ 1
ˇ
Z 1
ˇ
ˇ
1
2
ˇ
ˇ
x 2nC2 dx
R
.f
I
0;
"x
/
dx
$
n
ˇ
ˇ .n C 1/Š
0
0
1
:
D
.2n C 3/.n C 1/Š
This error will
be less than 10!4 if .2n C 3/.n C 1/Š > 10; 000. Since
15 & 7Š > 10; 000, n D 6 will do. Thus we use seven terms
of the series (0 $ n $ 6):
Z 1
2
e !x dx
0
$
Z 1#
x4 x6
x 8 x 10
x 12
'
1 " x2 C
"
C
"
C
dx
2Š
3Š
4Š
5Š
6Š
0
1
1
1
1
1
1
"
C
"
C
D1" C
3
5 & 2Š 7 & 3Š
9 & 4Š 11 & 5Š
13 & 6Š
' 0:74684 with error less than 10!4 .
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249
lOMoARcPSD|6566483
SECTION 6.8 (PAGE 388)
10. We are given that
previous exercise
Z
Z
1
0
1
0
Z
1
e
!x 2
1
11.
ADAMS and ESSEX: CALCULUS 9
2
e !x dx D
1
2
p
2
Dividing
p the first two equations gives u D 3=5, so
u D 3=5. Then 3A=5 D 1=3, so A D 5=9, and finally,
B D 8=9.
# and from the
2
e !x dx D 0:74684. Therefore,
dx D
Z
1
0
e
!x 2
dx "
Z
14.
1
e
!x 2
dx
Z
0
1p
D
# " 0:74684
2
D 0:139
(to 3 decimal places).
1
!1
f .x/ dx D 2
Z
1
0
.bx 2 C d / dx D 2
Af ."u/ C Af .u/ D 2A.bu2 C d /:
#
b
Cd
3
2 r !
r !6 3
6
5
3
3
5 C 0 D 0:24000
x dx ' 4 "
C
9
5
5
!1
Z 1
Error D
x 6 dx " 0:24000 ' 0:04571
!1
"
r !
r !#
Z 1
3
3
5
8
cos x dx '
cos "
C cos
C
9
5
5
9
!1
Z
$
1
!1
p
Z
p
f .x/ dx ' f ."1= 3/ C f .1= 3/:
We have
15.
Z
1
#
d
b
C Cf
.bx C dx C f / dx D 2
F .x/ dx D 2
5
3
0
!1
AF ."u/ C BF .0/ C AF .u/ D 2A.bu4 C du2 C f / C Bf:
4
2
These two expressions are identical provided
1
Au4 D ;
5
250
1
!1
cos x dx " 1:68300 ' 0:00006
p
p
e x dx ' e ! 3=5 C e 3=5 ' 2:35034
!1
Z 1
Error D
e x dx " 2:35034 ' 0:00006:
1
1
Au2 D ;
3
B
AC
D 1:
2
I D
Z
1
2
e !x dx
0
#
$
1 0 1 !1
T00 D T1 D R0 D .1/
e C e
' 0:6839397
2
2
$
#
1
1 1 0
e C e !1=4 C e !1 ' 0:7313703
T10 D T2 D
2 2
2
&
1%
T20 D T4 D
2T2 C e !1=16 C e !9=16 ' 0:7429841
4
&
1%
0
4T4 C e !1=64 C e !9=64 C e !25=64 C e !49=64
T3 D T8 D
8
' 0:7458656
4T10 " T00
' 0:7471805
3
0
0
4T " T1
D S4 D 2
' 0:7468554
3
0
0
4T " T2
D S8 D 3
' 0:7468261
3
1
1
16T2 " T1
D R2 D
' 0:7468337
15
16T31 " T21
' 0:7468242
D
15
64T32 " T22
D R3 D
' 0:7468241
63
' 0:746824 to 6 decimal places.
T11 D S2 D R1 D
13. If F .x/ D ax 5 C bx 4 C cx 3 C dx 2 C ex C f , then, by
symmetry,
1
' 1:68300
1
!1
Z
Z
!1
#
#
$
$
1 4
1 4
2
x 4 dx ' " p
C p
D
9
3
3
!1
Z 1
2
2 2
Error D
x 4 dx " D " ' 0:17778
9
5 9
!1
#
$
#
$
Z 1
1
1
cos x dx ' cos " p
C cos p
' 1:67582
3
3
!1
Z 1
Error D
cos x dx " 1:67582 ' 0:00712
!1
Z 1
p
p
e x dx ' e !1= 3 C e 1= 3 ' 2:34270
!1
Z 1
Error D
e x dx " 2:34270 ' 0:00771:
Z
6
Error D
12. For any function f we use the approximation
1
f .x/ dx '
We have
These two expressionspare identical provided A D 1 and
u2 D 1=3, so u D 1= 3.
Z
i 8
p
p
5h
f ." 3=5/ C f . 3=5/ C f .0/:
9
9
1
!1
If f .x/ D ax 3 C bx 2 C cx C d , then, by symmetry,
Z
For any function f we use the approximation
T21
$
T31
T22
T32
T33
I
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lOMoARcPSD|6566483
INSTRUCTOR’S SOLUTIONS MANUAL
16. From Exercise 9 in Section 7.6, for I D
Z
SECTION 6.8 (PAGE 388)
1:6
The#transformation
is# not
$
$ suitable because the derivative of
1
1
1
sin
is " 2 cos
, which has very large values at
t
t
t
some points close to 0.
In order to approximate the integral I to an desired degree
of accuracy, say with error less than $ in absolute value,
we have to divide the integral into two parts:
Z 1
sin x
I D
dx
1
C x2
Z 1
Z!t
sin x
sin x
dx
C
dx
D
2
1
C
x
1
C x2
t
!
D I1 C I2 :
f .x/ dx,
0
T00 D T1 D 1:9196
T10 D T2 D 2:00188
T20 D T4 D 2:02622
T30 D T8 D 2:02929:
Hence,
R1 D T11 D
T21
D
R2 D T22 D
T31 D
T32 D
R3 D T33 D
17.
4T10 " T00
D 2:0346684
3
0
0
4T2 " T1
D 2:0343333 D S4
3
1
1
16T2 " T1
D 2:0346684
15
4T30 " T20
D 2:0303133 D S8
3
16T31 " T21
D 2:0300453
15
64T32 " T22
D 2:0299719:
63
&
2h %
y0 C 4y2 C y4
3
&
h%
1
T2 D S4 D
y0 C 4y1 C 2y2 C 4y3 C y4
3
16T21 " T11
2
R2 D T2 D
15
2h
16h
.y
C
4y
0
1 C 2y2 C 4y3 C y4 / " 3 .y0 C 4y2 C y4 /
D 3
15
&
h%
D
14y0 C 64y1 C 24y2 C 64y3 C 14y4
45
&
2h %
7y0 C 32y1 C 12y2 C 32y3 C 7y4
D
45
If t # tan
t
Now let ZA be a numerical approximation to the proper
t
sin x
integral
dx, having error less than $=2 in abso1
C x2
!
lute value. Then
jI " Aj D jI1 C I2 " Aj
$ jI1 " Aj C jI2 j
$
$
$ C D $:
2
2
T11 D S2 D
18. Let
I D
Z
# "$
, then
2
Z 1
Z 1
sin x
dx
dx
<
1 C x2
1 C x2
t
t
ˇ1
ˇ
$
#
ˇ
!1
D tan .x/ˇ D " tan!1 .t / $ :
ˇ
2
2
Hence, A is an approximation to the integral I with the
desired accuracy.
19.
sin x
x cos c " sin x
;
;
f 0 .x/ D
x
x2
2
x .cos x " x sin x " cos x/ " .x cos x " sin x/2x
f 00 .x/ D
x4
"x 2 sin x " 2x cos x C 2 sin x
D
:
x3
Now use l’H^opital’s Rule to get
f .x/ D
lim f 00 .x/
x!0
"2x sin x " x 2 cos x " 2 cos x C 2x sin x C 2 cos x
x!0
3x 2
cos x
1
D lim "
D" :
x!0
3
3
D lim
1
!
sin x
dx
1 C x2
1
t
dt
dx D " 2
t
Let x D
# $
1
#
$
Z 0 sin
1
t
# $ " 2 dt
D
1
t
1=!
1C 2
t
# $
1
Z 1=! sin
t
dt:
D
1 C t2
0
20.
If t is time and E is error, then for the trapezoid rule,
t 2 E is approximately constant. Since E D 6 & 10!16
when t D 175:777 seconds, the time we would expect our
computer to achieve an error of 10!32 is about
s
10!16
175:7772 & 6 & !32 seconds;
10
which is about 1,365 years.
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251
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SECTION 6.8 (PAGE 388)
21.
ADAMS and ESSEX: CALCULUS 9
If t is time and E is error, then t 4 E is approximately constant for the Simpson’s Rule case. Since E D 3:15 & 10!30
when t D 175:777 seconds, we would expect our computer
to achieve quadruple precision in time
#
$1=4
10!30
175:7774 & 3:15 & !32
10
5.
seconds;
3
A
B
D
C
4x 2 " 1
2x " 1
2x C 1
2Ax C A C 2Bx " B
D
4x 2 " 1
n
3
2A C 2B D 0
)
) A D "B D
A"B D3
2
#Z
$
Z
dx
3
dx
3 dx
D
"
4x 2 " 1
2
2x " 1
2x C 1
ˇ
ˇ
3 ˇˇ 2x " 1 ˇˇ
D ln ˇ
C C:
4
2x C 1 ˇ
Z
or about 12 minutes if it were using Simpson’s Rule.
Since our computer did the calculation more than 5,000
times faster than that, it must have been using an even
higher-order method.
Review Exercises on Techniques
of Integration (page 390)
1.
6.
Z
.x 2 C x " 2/ sin 3x dx
U D x2 C x " 2
d U D .2x C 1/ dx
d V D sin 3x
V D " 13 cos 3x
Z
D " 13 .x 2 C x " 2/ cos 3x C 31 .2x C 1/ cos 3x dx
A
B
x
D
C
2x 2 C 5x C 2
2x C 1
xC2
Ax C 2A C 2Bx C B
D
2x 2 C 5x C 2
n
A C 2B D 1
)
2A C B D 0
U D 2x C 1 d V D cos 3x dx
d U D 2 dx
V D 31 sin 3x
Thus A D "1=3 and B D 2=3. We have
Z
Z
Z
x dx
1
dx
2
dx
D
"
C
2
2x C 5x C 2
3
2x C 1
3
xC2
1
2
D ln jx C 2j " ln j2x C 1j C C:
3
6
2.
3.
4.
D " 13 .x 2 C x " 2/ cos 3x C 91 .2x C 1/ sin 3x
Z
" 92 sin 3x dx
D " 13 .x 2 C x " 2/ cos 3x C 91 .2x C 1/ sin 3x
2
C
cos 3x C C:
27
Z
x
dx Let u D x " 1
.x " 1/3
du
# dx
$
Z
Z D
uC1
1
1
D
du
D
C
du
u3
u2
u3
1
1
1
1
"
C C:
D" " 2 CC D"
u 2u
x " 1 2.x " 1/2
Z
sin3 x cos3 x dx
Z
D sin3 x.1 " sin2 x/ cos x dx Let u D sin x
du D cos x dx
Z
u4 u6
3
5
"
CC
D .u " u / du D
4
6
1
1
D sin4 x " sin6 x C C:
4
6
p 1=3
Z
p
.1 C x/
dx Let u D 1 C x
p
x
dx
du D p
2 x
Z
D 2 u1=3 du D 2. 34 /u4=3 C C
p
D 32 .1 C x/4=3 C C:
252
7.
Z p
D
Z
1 " x2
dx
x4
cos2 &
Let x D sin &
dx D cos & d&
d&
4
Z sin &
D csc2 & cot2 & d&
Z
Let v D cot &
dv D " csc2 & d&
v3
CC
3
!3
p
cot3 &
1
1 " x2
D"
CC D"
C C:
3
3
x
D"
v 2 dv D "
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1
x
"
p
1!x 2
Fig. RT-7
lOMoARcPSD|6566483
INSTRUCTOR’S SOLUTIONS MANUAL
8.
Z
x 3 cos.x 2 / dx
D
1
2
Z
REVIEW EXERCISES ON TECHNIQUES OF INTEGRATION (PAGE 390)
Let w D x 2
dw D 2x dx
13.
9.
10.
11.
cos.x 2 / C C:
p
1C4x
Z
x 2 dx
Let u D 5x 3 " 2
.5x 3 " 2/2=3
du D 15x 2 dx
Z
1
1
D
u!2=3 du D u1=3 C C
15
5
1
3
1=3
D .5x " 2/ C C:
5
1
Fig. RT-13
1
A
B
.A C B/x C .5A " 3B/
D
C
D
C 2x " 15
x"3
xC5
x 2 C 2x " 15
n
1
1
ACB D0
)
)AD ; B D" :
5A " 3B D 1
8
Z
Z 8
Z
dx
dx
1
1
dx
D
"
x 2 Cˇ 2x " 15
8
x
"
3
8
x
C5
ˇ
1 ˇˇ x " 3 ˇˇ
D ln ˇ
C C:
8
x C 5ˇ
14.
Z
15.
Z
dx
Let x D 2 tan &
.4 C x 2 /2
dx D 2 sec2 & d&
Z
Z
2
1
2 sec & d&
D
cos2 & d&
D
16 sec4 &
8
1
D
.& C sin & cos & / C C
16
#
$
1
1
x
!1 x
D
tan
C
C C:
16
2
8 4 C x2
cos x
dx Let u D sin x
1 C sin2 x
du D cos x dx
Z
du
D
D tan!1 u C C
1 C u2
D tan!1 .sin x/ C C:
D
Z
D
D
16.
sin3 x
dx D
7
Zcos x
Z
D
"
D
Fig. RT-11
.sin x C cos x/2 dx D
Z
D
.1 C sin 2x/ dx
Dx"
1
2
cos 2x C C:
.u3 C u5 / du D
Let u D tan x
du D sec2 x dx
u6
u4
C
CC
4
6
We have
D
2
tan3 x sec4 x dx
1
1
tan4 x C tan6 x C C:
4
6
Z
4Cx 2
Z
tan3 x.1 C tan2 x/ sec2 x dx
x
Z
2x
"
x2
p
12.
Let 2x D tan &
2x ln 2 dx D sec2 & d&
D
U Dw
d V D cos w dw
d U D dw Z V D sin w
D 21 w sin w " 12 sin w dw
1
2
p
2x 1 C 4x dx
Z
1
sec3 & d&
ln 2
&
1 %
sec & tan & C ln j sec & C tan & j C C
D
2 ln 2
&
p
1 % xp
D
2 1 C 4x C ln.2x C 1 C 4x / C C:
2 ln 2
w cos w dw
D 21 x 2 sin.x 2 / C
Z
D
q
Let x D 35 tan u
q
dx D 35 sec2 u du
q
Z . 3 tan2 u/. 3 sec2 u/ du
5
5
x 2 dx
.3 C 5x 2 /3=2
.3/3=2 sec3 u
Z
1
p
.sec u " cos u/ du
5 5
1
p .ln j sec u C tan uj " sin u/ C C
5 5
ˇp
!
p ˇˇ
p
ˇ 5x 2 C 3
5x ˇ
5x
1
ˇ
p
CC
p ln ˇ
C p ˇ" p
ˇ
5 5
3
3 ˇ
5x 2 C 3
%p
p &
1
x
C C0 ;
p ln 5x 2 C 3 C 5x " p
5 5
5 5x 2 C 3
p
1
where C0 D C " p ln 3:
5 5
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REVIEW EXERCISES ON TECHNIQUES OF INTEGRATION (PAGE 390)
p
20.
p
5x
5x 2 C3
u
p
3
Fig. RT-16
17.
I D
Z
e !x sin 2x dx
U D e !x
d U D "e !x dx
1
1
D " e !x cos 2x "
2
2
d V D sin 2x dx
1
V D " cos 2x
2
Z
21.
e !x cos 2x dx
d V D cos 2x dx
1
V D sin 2x
2
!
1 !x
1 1 !x
1
D " e cos 2x "
e sin 2x C I
2
2 2
2
18.
19.
Z
2x 2 C 4x " 3
dx D
x 2 C 5x
#
"
Z
6x C 3
dx
D
2"
x.x C 5/
I D
Z
U D cos.3 ln x/
3 sin.3 ln x/ dx
dU D "
Zx
D x cos.3 ln x/ C 3
d V D dx
V Dx
Z
x ln.1 C x 2 /
dx
1 C x2
22.
Z
23.
Z
sin2 x cos4 x dx
Z
1
D
.1 " cos 2x/Π12 .1 C cos 2x/!2 dx
2
Z
1
D
.1 C cos 2x " cos2 2x " cos3 2x/ dx
8
Z
1
1
1
.1 C cos 4x/ dx
sin 2x "
D xC
8 Z 16
16
1
"
.1 " sin2 2x/ cos 2x dx
8
1
x
1
1
x
sin 2x "
"
sin 4x "
sin 2x
D C
8
16
16 64
16
1
C
sin3 2x C C
48
sin 4x
sin3 2x
x
"
C
C C:
D
16
64
48
p
x 2 dx
p
Let x D 2 sin &
p
2
2"x
dx D 2 cos & d&
Z
2
D 2 sin & d& D & " sin & cos & C C
p
x 2 " x2
x
C C:
D sin!1 p "
2
2
sin.3 ln x/ dx
p
d V D dx
U D sin.3 ln x/
3 cos.3 ln x/ dx
V Dx
dU D
x%
&
D x cos.3 ln x/ C 3 x sin.3 ln x/ " 3I
I D
254
Let u D ln.1 C x 2 /
2x dx
du D
1 C x2
D
2x 2 C 10x " 6x " 3
dx
x 2 C 5x
6x C 3
A
B
.A C B/x C 5A
D C
D
x.x C 5/
x
xC5
x.x C 5/
n
27
3
ACB D6
:
)
)AD ; B D
5A
D
3
5
Z
Z
Z 5
3
dx 27
dx
I D 2 dx "
"
5
x
5
xC5
3
27
D 2x " ln jxj "
ln jx C 5j C C:
5
5
Z
I D cos.3 ln x/ dx
1
A
Bx C C
D C
4x 3 C x
x
4x 2 C 1
A.4x 2 C 1/ C Bx 2 C C x
D
4x 3 C x
'
4A C B D 0
)
) B D "4:
C D 0; A D 1
Z
Z
Z
1
dx
x dx
dx
D
"
4
4x 3 C x
x
4x 2 C 1
1
D ln jxj " ln.4x 2 C 1/ C C:
2
Z
u2
1
u du D
CC
2
4
&2
1%
ln.1 C x 2 / C C:
D
4
U D e !x
d U D "e !x dx
1
1
1
D " e !x cos 2x " e !x sin 2x " I
2 #
4
$ 4
1
!x 2
I D "e
cos 2x C sin 2x C C:
5
5
ADAMS and ESSEX: CALCULUS 9
2
x
"
1
3
x cos.3 ln x/ C x sin.3 ln x/ C C:
10
10
p
2!x 2
Fig. RT-23
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INSTRUCTOR’S SOLUTIONS MANUAL
24.
We have
Z
I D tan4 x sec x dx
27.
U D tan3 x
d U D 3 tan2 x Zsec2 x dx
D tan3 x sec x " 3
D tan3 x sec x " 3
REVIEW EXERCISES ON TECHNIQUES OF INTEGRATION (PAGE 390)
d V D tan x sec x dx
V D sec x
tan2 x.tan2 x C 1/ sec x dx
tan2 x sec x dx
U D tan x
d V D tan x sec x dx
d U D sec2 xZ dx
V D sec x
3
D tan x sec x " sec x dx
Z
D tan x sec x " .tan2 x C 1/ sec x dx
28.
We have
25.
1
4
C
tan3 x sec x "
3
8
3
8
tan x sec x
ln j sec x C tan xj C C:
Z
x 2 dx
Let u D 4x C 1
.4x C 1/10
du D 4 dx
$
Z #
u"1 2 1
1
D
du
4
4
u10
Z
1
.u!8 " 2u!9 C u!10 / du
D
64
1 !7
1 !8
1 !9
D"
u C
u "
u CC
448
256
576
#
$
1
1
1
1
D
"
C
"
C C:
64
7.4x C 1/7
4.4x C 1/8 9.4x C 1/9
26. We have
Z
%x &
dx
x sin!1
2
%x&
U D sin!1
d V D x dx
2
x2
dx
V D
dU D p
2
4 " x2
Z
%
&
2
2
x dx
x
1
x
sin!1
"
Let x D 2 sin u
D
p
2
2
2
4 " x2
dx D 2 cos u du
Z
% &
x2
!1 x
2
sin
" 2 sin u du
D
2
2
%x& Z
x2
D
sin!1
" .1 " cos 2u/ du
2
2
%x&
x2
D
sin!1
" u C sin u cos u C C
$2
#2 2
%x & 1 p
x
" 1 sin!1
C x 4 " x 2 C C:
D
2
2
4
Z
Z
dx
x dx
D
Let u D x 2
x 5 " 2x 3 C x
x 6 " 2x 4 C x 2
du D 2x dx
Z
Z
1
du
1
du
D
D
2
u3 " 2u2 C u
2
u.u " 1/2
1
A
B
C
D C
C
u.u " 1/2
u
u"1
.u " 1/2
A.u2 " 2u C 1/ C B.u2 " u/ C C u
D
u3 " 2u2 C u
(
ACB D0
) "2A " B C C D 0 ) A D 1; B D "1; C D 1:
Z AD1
Z
Z
du
du 1
du
1
1
D
"
2
u3 " 2u2 C u
2
u
2
u"1
Z
du
1
C
2
.u " 1/2
1
1
1 1
D ln juj " ln ju " 1j "
CK
2
2
2u"1
1
x2
1
D ln 2
"
C K:
2 jx " 1j 2.x 2 " 1/
I D
D tan x sec x " J " ln j sec x C tan xj C C
J D 21 tan x sec x " 12 ln j sec x C tan xj C C:
I D
Let u D cos 4x
du D "4 sin 4x dx
D"
3
D tan
Z x sec x " 3I " 3J where
J D
sin5 .4x/ dx
Z
D .1 " cos2 4x/2 sin 4x dx
Z
1
.1 " 2u2 C u4 / du
4
#
$
2 3 1 5
1
u" u C u CC
D"
4
3
5
1
1
1
D " cos 4x C cos3 4x "
cos5 4x C C:
4
6
20
tan2 x sec3 x dx
Z
Z
29.
Z
30.
Let
dx
2 C ex
Z
e !x dx
D
Let u D 2e !x C 1
2e !x C 1
du D "2e !x dx
Z
du
1
1
D " ln.2e !x C 1/ C C:
D"
2
u
2
In D
Z
x n 3x dx
U D xn
d U D nx n!1 dx
d V D 3x dx
3x
V D
ln 3
x n 3x
n
"
In!1 :
ln
3
ln
3
Z
3x
I0 D 3x dx D
C C:
ln 3
D
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REVIEW EXERCISES ON TECHNIQUES OF INTEGRATION (PAGE 390)
Hence,
I3 D
34.
Z
We have
Z
x 3 3x dx
"
!#
3 x 2 3x
2 x3x
1
x 3 3x
C C1
"
"
"
I0
D
ln 3
ln 3 ln 3
ln 3 ln 3
ln 3
"
#
x3
3x 2
6x
6
D 3x
C C1 :
"
C
"
2
3
ln 3 .ln 3/
.ln 3/
.ln 3/4
31.
35.
32. We have
!
Z
x2 C 1
2x C 1
dx D
1" 2
dx
x 2 C 2x C 2
x C 2x C 2
Z
2x C 1
Dx"
dx Let u D x C 1
.x C 1/2 C 1
du D dx
Z
2u " 1
Dx"
du
u2 C 1
2
D x " ln ju C 1j C tan!1 u C C
D x " ln.x 2 C 2x C 2/ C tan!1 .x C 1/ C C:
Z
33.
Z
dx
p
Let x D sin &
x2 1 " x2
dx D
Z cos & d&
Z
cos & d&
D csc2 & d&
D
sin2 & cos & p
1 " x2
D " cot & C C D "
C C:
x
Fig. RT-33
256
2x
p
1!4x 2
Fig. RT-35
Z
e 1=x
dx
x2
D"
x
p
1!x 2
x 3 dx
p
Let 2x D sin &
1 " 4x 2
2 dx D cos & d&
Z
Z
sin3 & cos & d&
1
1
D
.1 " cos2 & / sin & d&
D
16
cos &
16
#
$
1
1
D
" cos & C cos3 & C C
16
3
1p
1
1 " 4x 2 C C:
.1 " 4x 2 /3=2 "
D
48
16
"
37.
"
Z
1
36.
1
x 3 .ln x/2 dx
d V D x 3 dx
U D .ln x/2
1
2
V D x4
d U D ln x dx
4
x
Z
1
1
D x 4 .ln x/2 "
x 3 ln x dx
4
2
U D ln x
d V D x 3 dx
1
1
d U D dx
V D x4
x
4 Z
1
1
1
D x 4 .ln x/2 " x 4 ln x C
x 3 dx
4
8
8
"
#
1
1
x4
2
.ln x/ " ln x C
C C:
D
4
2
8
Z
sin2 x cos x
dx Let u D sin x
2 " sin x
du D cos x dx
Z 2
u du
D
Let 2 " u D v
2"u
du D "dv
$
Z
Z #
4 " 4v C v 2
4
D"
dv D
" C 4 " v dv
v
v
v2
D "4 ln jvj C 4v "
CC
2
1
D "4 ln j2 " uj C 4.2 " u/ " .2 " u/2 C C
2
1
D "4 ln.2 " sin x/ " 2 sin x " sin2 x C C1 :
2
ADAMS and ESSEX: CALCULUS 9
Z
Z
1
x
1
du D " 2 dx
x
Let u D
e u du D "e u C C D "e 1=x C C:
xC1
p
dx
x2 C 1 Z
p
dx
D x2 C 1 C
p
Let x D tan &
x2 C 1
dx D sec2 & d&
Z
p
D x 2 C 1 C sec & d&
p
D x 2 C 1 C ln j sec & C tan & j C C
p
p
D x 2 C 1 C ln.x C x 2 C 1/ C C:
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INSTRUCTOR’S SOLUTIONS MANUAL
REVIEW EXERCISES ON TECHNIQUES OF INTEGRATION (PAGE 390)
42.
p
1Cx 2
x
"
1
Fig. RT-37
38.
Z
e .x
D3
1=3 /
Z
Let x D u3
dx D 3u2 du
u2 e u du D 3I2
43.
See solution
to #16 of Section 6.6 for
Z
n u
In D u e dx D un e u " nIn!1 .
D 3Œu2 e u " 2.ue u " e u /! C C
De
39.
I D
.x 1=3 /
2=3
B
1
A
p C
p
D
u2 " 2
u" 2
uC 2
p
p
Au C 2A C Bu " 2B
D
u2 " 2
'
ACB D0
) p
2.A " B/ D 1
1
) A D "B D p :
2 2
1=3
C 6/ C C:
$
Z #
9x " 3
x "3
dx D
1C 3
dx:
x 3 " 9x
x " 9x
Z
.3x
3
" 6x
Assume that x # 1 and let x D sec u and
dx D sec u tan u du. Then
Z
x 2 dx
p
x2 " 1
Z
Z
sec3 u tan u du
D
D sec3 u du
tan u
1
1
D sec u tan u C ln j sec u C tan uj C C
2
2
p
1 p
1
D x x 2 " 1 C ln jx C x 2 " 1j C C:
2
2
Differentiation shows that this solution is valid for x $ "1
also.
Z
Z
.x C 1 " 1/ dx
x dx
D
Let u D x C 1
I D
x 2 C 2x " 1
.x C 1/2 " 2
du D dx
Z
Z
du
u"1
1
2
D
du D ln ju " 2j "
:
u2 " 2
2
u2 " 2
9x " 3
A
B
C
D C
C
x 3 " 9x
x
x"3
xC3
Ax 2 " 9A C Bx 2 C 3Bx C C x 2 " 3C x
D
x 3 " 9x 8
(
< A D 1=3
ACB CC D0
)
) B D 4=3
3B " 3C D 9
:
C D "5=3:
"9A D "3
Thus we have
Thus we have
Z
Z
Z
dx
4
dx
5
dx
1
C
"
3
x
3
x"3 3
xC3
1
4
5
D x C ln jxj C ln jx " 3j " ln jx C 3j C K:
3
3
3
I DxC
40.
41.
Z
p
10 xC2 dx
p
xC2
p
xC2
dx
du D p
2 xC2
Z
p
2
2
10u C C D
10 xC2 C C:
D 2 10u du D
ln 10
ln 10
Z
sin5 x cos9 x dx
Z
D .1 " cos2 x/2 cos9 x sin x dx Let u D cos x
du D " sin x dx
Z
D"
Let u D
.1 " 2u2 C u4 /u9 du
u12 u14
u10
C
"
CC
10
6
14
12
10
cos x cos x cos14 x
D
"
"
C C:
6
10
14
D"
44.
45.
Z
ˇ
ˇ
ˇ u " p2 ˇ
1
1
ˇ
ˇ
2
I D ln ju " 2j " p ln ˇ
p ˇCK
ˇ
2
2 2
u C 2ˇ
ˇ
ˇ
ˇ x C 1 " p2 ˇ
1
1
ˇ
ˇ
2
D ln jx C 2x " 1j " p ln ˇ
p ˇ C K:
2
2 2 ˇx C 1 C 2ˇ
2x " 3
Let u D 4 " 3x C x 2
du D ."3 C 2x/ dx
Z
p
p
du
D
p D 2 u C C D 2 4 " 3x C x 2 C C:
u
Z
x 2 sin!1 2x dx
D
D
D
D
D
p
4 " 3x C x 2
dx
U D sin!1 2x
d V D x 2 dx
2 dx
x3
dU D p
V
D
3
1 " 4x 2
Z
3
x3
2
x
dx
sin!1 2x "
Let v D 1 " 4x 2
p
3
3
1 " 4x 2
dv$D "8x dx
#
Z
1"v
2
1
x3
!1
sin 2x "
" dv
3
3
8
4v 1=2
Z %
&
x3
1
sin!1 2x C
v !1=2 " v 1=2 dv
3
48
1p
x3
1
sin!1 2x C
v " v 3=2 C C
3
24
72
x3
1p
1
sin!1 2x C
1 " 4x 2 " .1 " 4x 2 /3=2 C C:
3
24
72
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REVIEW EXERCISES ON TECHNIQUES OF INTEGRATION (PAGE 390)
46. Let
p
3x D sec u and
p
3 dx D sec u tan u du. Then
50.
Z p
3x 2 " 1
dx
x
1
Z tan u p sec u tan u du
3
D
1
p sec u
3 Z
Z
D tan2 u du D .sec2 u " 1/ du
p
p
D tan u " u C C D 3x 2 " 1 " sec!1 . 3x/ C C
#
$
p
1
D 3x 2 " 1 C sin!1 p
C C1 :
3x
47.
48.
1
4
Let x "
dx D
1
2
1
2
D
1
2
sin u
p
1
x! 2
x!x 2
Fig. RT-48
dx
p
Let x D u2
.4 C x/ x
dx DZ 2u du
Z
2u du
du
D
D2
.4 C u2 /u
4 C u2
p
x
2
!1 u
!1
D tan
C C D tan
C C:
2
2
2
52.
d V D x dx
x2
V D
2
3
3
2
Z
x2
dx
9 C x2
!
9
1"
dx
9 C x2
%x &
3x
9
C tan!1
C C:
2
2
3
3
2
Z
x2
2
Z
x4 " 1
dx
I D
x 3 C 2x 2
Z 4
x C 2x 3 " 2x 3 " 4x 2 C 4x 2 " 1
D
dx
x 3 C 2x 2
$
Z #
4x 2 " 1
D
x"2C 3
dx:
x C 2x 2
A
B
C
4x 2 " 1
D C 2 C
x 3 C 2x 2
x
x
xC2
Ax 2 C 2Ax C Bx C 2B C C x 2
D
x 3 C 2x82
(
< A D 1=4
ACC D4
) 2A C B D 0 ) B D "1=2
:
C D 15=4:
2B D "1
x2
" 2x C
2
x2
D
" 2x C
2
cos u du
1
8
1
2
258
dx
%x &
3 dx
9 C x2
%x&
tan!1
"
3
%x&
tan!1
"
3
%x&
tan!1
"
3
Thus
sin u cos u C C
p
D 81 sin!1 .2x " 1/ C 14 .2x " 1/ x " x 2 C C:
Z
x2
2
D
51.
3
I D
cos2 u du D 18 u C
u
%x&
U D tan!1
x2
D
2
Z p
D
49.
D
Z
1
sin4 2x dx
cos4 x sin4 x dx D
16
Z
1
D
.1 " cos 4x/2 dx
64
$
Z #
1
1 C cos 8x
D
1 " 2 cos 4x C
dx
64
2
#
$
sin 8x
1 3x sin 4x
"
C
CC
D
64 2
2
16
#
$
1
sin 8x
D
3x " sin 4x C
C C:
128
8
Z
x tan!1
dU D
Z
x " x 2 dx
Z q
1
1 2
D
4 " .x " 2 / dx
Z
ADAMS and ESSEX: CALCULUS 9
Z
Z
Z
15
1
dx 1
dx
dx
"
C
2
4
x
2
x
4
xC2
1
1
15
ln jxj C
C
ln jx C 2j C K:
4
2x
4
Let u D x 2 and du D 2x dx; then we have
Z
Z
Z
1
du
dx
x dx
D
D
:
I D
x.x 2 C 4/2
x 2 .x 2 C 4/2
2
u.u C 4/2
Since
1
A
B
C
D C
C
u.u C 4/2
u
uC4
.u C 4/2
A.u2 C 8u C 16/ C B.u2 C 4u/ C C u
D
u.u C 4/2
(
ACB D0
1
1
1
; BD" ; C D" ;
) 8A C 4B C C D 0 ) A D
16
16
4
16A D 1
therefore
Z
Z
Z
1
du
1
du
1
du
I D
"
"
32
u
32
uC4 8
.u C 4/2
ˇ
ˇ
1 ˇˇ u ˇˇ 1 1
D
ln ˇ
C
CC
32
u C 4ˇ 8 u C 4
ˇ
ˇ
1
1 ˇˇ x 2 ˇˇ
ln
C
C C:
D
32 ˇ x 2 C 4 ˇ 8.x 2 C 4/
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INSTRUCTOR’S SOLUTIONS MANUAL
53.
Z
REVIEW EXERCISES ON TECHNIQUES OF INTEGRATION (PAGE 390)
sin.2 ln x/
dx
x
therefore
Let u D 2 ln x
2
du D dx
x
Z
1
1
D
sin u du D " cos u C C
2
2
1
D " cos.2 ln x/ C C:
2
Z
sin.ln x/
dx
x2
57.
dx
dV D 2
U D sin.ln x/
x
cos.ln x/
1
dx
dU D
V D"
x
x
Z
sin.ln x/
cos.ln x/
D"
C
dx
x
x2
dx
U D cos.ln x/
dV D 2
x
sin.ln x/
"1
dU D "
dx
V
D
x
x
sin.ln x/ cos.ln x/
D"
"
" I;
x
x
58.
Z
i
1 h
sin.ln x/ C cos.ln x/ C C:
2x
"1
e 2 tan x
dx
1 C x2
Let u D 2 tan!1 x
2 dx
du D
1
C x2
Z
1 u
1
1
"1
u
e du D e C C D e 2 tan x C C:
D
2
2
2
61.
I D
x3 C x " 2
dx D
x2 " 7
Z
ln.3 C x 2 /
x dx
3 C x2
Z
x 3 " 7x C 8x " 2
dx
x2 " 7
!
Z
8x " 2
dx:
D
xC 2
x "7
Since
p
B
.A C B/x C .B " A/ 7
8x " 2
A
p
C
p
D
D
x2 " 7
x2 " 7
xC 7
x" 7
(
ACB D8
1
1
) B " A D " p2 ) A D 4 C p ; B D 4 " p ;
7
7
7
Let u D ln.3 C x 2 /
2x dx
du D
3 C x2
Z
&2
u2
1
1%
u du D
D
CC D
ln.3 C x 2 / C C:
2
4
4
Z
Z
cos7 x dx D .1 " sin2 x/3 cos x dx Let u D sin x
du D cos x dx
Z
Z
.1 " u2 /3 du D
.1 " 3u2 C 3u4 " u6 / du
D sin x " sin3 x C 53 sin5 x " 71 sin7 x C C:
Z
sin!1 .x=2/
dx Let u D sin!1 .x=2/
.4 " x 2 /1=2
dx
dx
du D p
D p
2 1 " .x 2 =4/
4 " x2
Z
&2
u2
1 % !1
D u du D
CC D
sin .x=2/ C C:
2
2
We have
Z
Z
tan4 .#x/ dx D tan2 .#x/Œsec2 .#x/ " 1! dx
Z
Z
2
2
D tan .#x/ sec .#x/ dx " Œsec2 .#x/ " 1! dx
D
56. We have
Z
x2
2
D u " u3 C 53 u5 " 17 u7 C C
60.
55.
D
D
59.
therefore
I D"
x2
2
I D
54. Since
I D
D
!
8x " 2
xC 2
dx
x "7
!Z
!Z
1
dx
1
dx
C 4C p
p C 4" p
p
7
xC 7
7
x" 7
!
!
p
p
1
1
C 4 C p ln jx C 7j C 4 " p ln jx " 7j C C:
7
7
Z
1
1
tan3 .#x/ " tan.#x/ C x C C:
3#
#
Z
.x C 1/ dx
p
2 C 6x C 10
x
Z
.x C 3 " 2/ dx
D
p
Let u D x C 3
.x C 3/2 C 1
du D dx
Z
.u " 2/ du
p
D
u2 C 1 Z
p
du
D u2 C 1 " 2 p
Let u D tan &
u2 C 1
du
D sec2 & d&
Z
p
D x 2 C 6x C 10 " 2 sec & d&
p
D x 2 C 6x C 10 " 2 ln j sec & C tan & j C C
%
&
p
p
D x 2 C 6x C 10 " 2 ln x C 3 C x 2 C 6x C 10 C C:
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REVIEW EXERCISES ON TECHNIQUES OF INTEGRATION (PAGE 390)
p
ADAMS and ESSEX: CALCULUS 9
p
x 2 C6xC10
x
"
"
1
62.
63.
Z
65.
Z
66.
x 1=2
dx Let x D u6
1 C x 1=3
dx D 6u5 du
Z
8
u
D6
du
u2 C 1
Z 8
u C u6 " u6 " u4 C u4 C u2 " u2 " 1 C 1
du
D6
u2 C 1
$
Z #
1
D6
du
u6 " u4 C u2 " 1 C 2
u C1
$
# 7
u5
u3
u
"
C
" u C tan!1 u C C
D6
7
5
3
p
6 7=6 6 5=6
D x
" x
C 2 x " 6x 1=6 C 6 tan!1 x 1=6 C C:
7
5
We have
Z
.x 2
dx
x.x 2 C x C 1/1=2
Z
dx
D
xŒ.x C 12 /2 C 43 !1=2
2
1
"
C C:
5.2 C x 2 /5=2 3.2 C x 2 /3=2
3
sec2 & d&
2
D
#p
$# p
$
3
3
1
tan & "
sec &
2
2
2
Z
Z
d&
2 sec & d&
D2 p
p
D
3 tan & " 1
3 sin & " cos &
Z p
3 sin & C cos &
d&
D2
3Zsin2 & " cos2 &
Z
p
cos & d&
sin & d&
D2 3
C
2
2
2
2 &
3
sin
&
"
cos
3
sin
& " cos2 &
Z
p Z
sin & d&
cos & d&
D2 3
C2
3 " 4 cos2 &
4 sin2 & " 1
Let u D cos & , du D " sin & d& in the first integral;
p
x 3 dx
Let x D 2 tan &
7=2
p
C 2/
dx D 2 sec2 & d&
p
Z p
2 2 tan3 & 2 sec2 & d&
D
p
7
Z 8 2 sec &
1
D p
sin3 & cos2 & d&
2 2Z
1
D p
.1 " cos2 & / cos2 & sin & d& Let u D cos &
2 2
$ " sin & d&
# 5 du3 D
Z
1
u
1
u
4
2
"
CC
D p
.u " u / du D p
3
2 20
2 2 5
!5
!3 1
p
p
1 @1
1
2
2
ACC
D p
"
p
p
3
2 2 5
2 C x2
2 C x2
D
2
!
Z
3
1
x2
1
C
dx
D
dx
2x 2 " 3
2
2x 2 " 3
!
p Z
1
x
3
1
p "p
p dx
D C
p
2
4
2x " 3
2x C 3
ˇ
ˇ
p
p
p
x
3 ˇˇ 2x " 3 ˇˇ
D C p ln ˇ p
p ˇ C C:
2
4 2 ˇ 2x C 3 ˇ
64.
e x .1 " e 2x /5=2 dx
Let e x D sin u
e x dx D cos u du
# $3 Z
Z
1
D cos6 u du D
.1 C cos 2u/3 du
2
Z
1
D
.1 C 3 cos 2u C 3 cos2 2u C cos3 2u/ du
8
Z
3
3
u
sin 2u C
.1 C cos 4u/ duC
D C
8
16
16
Z
1
.1 " sin2 2u/ cos 2u du
8
5u
3
3
sin 2u
D
C
sin 2u C
sin 4u C
16
16
64
16
1
3
sin 2u C C
"
48
5
1
D
sin!1 .e x / C sinŒ2 sin!1 .e x /!C
16
4
1
3
!1 x
sinŒ4 sin .e /! "
sin3 Œ2 sin!1 .e x /! C C
64
48
5
1 p
D
sin!1 .e x / C e x 1 " e 2x
16
2
%
&
3 xp
C e 1 " e 2x 1 " 2e 2x
16
%
&3=2
1
" e 3x 1 " e 2x
C C:
6
Z
p
Fig. RT-63
Fig. RT-61
Z
2Cx 2
xC3
260
Z
p
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p
1
3
Let x C D
tan &
2
2
p
3
sec2 & d&
dx D
2
lOMoARcPSD|6566483
INSTRUCTOR’S SOLUTIONS MANUAL
REVIEW EXERCISES ON TECHNIQUES OF INTEGRATION (PAGE 390)
let v D sin & , dv D cos & d& in the second integral.
Z
p Z
du
dv
C
2
D "2 3
3 " 4u2
4v 2 " 1
p Z
Z
3
du
1
du
"
D"
3
1
2
2
2
"u
" v2
4
p ˇ
ˇ 4
ˇ
ˇ
p # $#
$ ˇ cos & C 3 ˇ
ˇ
ˇ
3 1
2
p ln ˇˇ
p2 ˇˇ
D"
2
2
3
3
ˇ
ˇ
ˇ cos & "
ˇ
2
ˇ
ˇ
1ˇ
ˇ
# $
ˇ sin & C ˇ
1 1
ˇ
2 ˇˇ C C
.2/ ln ˇ
"
1ˇ
ˇ
2 2
ˇ sin & " ˇ
2
ˇ#
p $#
$ˇ
ˇ
1 ˇˇ
3
ˇ cos & "
sin & "
ˇ
1 ˇˇ
2
2 ˇ
p $#
D ln ˇ #
$ ˇ C C:
ˇ
2 ˇ
ˇ cos & C 3 sin & C 1 ˇ
ˇ
2
2 ˇ
69.
70.
p
2x C 1
3
and cos & D p
,
Since sin & D p
2
2
2 x CxC1
2 x CxC1
therefore
Z
67.
68.
ˇ
ˇ
p
1 ˇˇ .x C 2/ " 2 x 2 C x C 1 ˇˇ
dx
D ln ˇ
p
ˇCC:
2 ˇ .x C 2/ C 2 x 2 C x C 1 ˇ
x.x 2 C x C 1/1=2
71.
Z
1Cx
p dx Let x D u2
1C x
dx D 2u du
Z
u.1 C u2 /
du
D2
1Cu
Z 3
2
u C u " u2 " u C 2u C 2 " 2
D2
du
1Cu
$
Z #
2
D2
u2 " u C 2 "
du
1Cu
$
# 3
u2
u
"
C 2u " 2 ln j1 C uj C C
D2
3
2
p
p
2 3=2
D x
" x C 4 x " 4 ln.1 C x/ C C:
3
Z
D
D
D
D
x dx
Let u D x 2
4x 4 C 4x 2 C 5
du D 2x dx
Z
1
du
2
4u2 C 4u C 5
Z
1
du
Let w D 2u C 1
2
.2u C 1/2 C 4
dw D 2du
Z
% &
dw
1
1
!1 w
D tan
CC
2
4
w C4
8
2
#
$
1
1
tan!1 x 2 C
C C:
8
2
72.
Z
x dx
Let u D x 2 " 4
" 4/2
du D 2x dx
Z
1
du
1
D
D"
CC
2
u2
2u
1
1
D"
CC D" 2
C C:
2.x 2 " 4/
2x " 8
Use the partial fraction decomposition
1
A
Bx C C
D C 2
3
2
x Cx Cx
x
x CxC1
A.x 2 C x C 1/ C Bx 2 C C x
D
x3 C x2 C x
(
ACB D0
) A C C D 0 ) A D 1; B D "1; C D "1:
AD1
Therefore,
Z
dx
x3 C x2 C x
Z
Z
dx
xC1
D
"
dx Let u D x C 21
2
x
x CxC1
du D dx
Z
u C 12
D ln jxj "
du
u2 C 43
#
$
&
2x C 1
1
1 %
C C:
p
D ln jxj " ln x 2 C x C 1 " p tan!1
2
3
3
Z
x 2 tan!1 x dx
U D tan!1 x d V D x 2 dx
dx
x3
dU D
V D
2
1Cx
3
Z
x3
1
x 3 dx
!1
D
tan x "
3
3
1 C x2
Z 3
3
x Cx"x
1
x
tan!1 x "
dx
D
3
3
x2 C 1
x3
1
1
D
tan!1 x " x 2 C ln.1 C x 2 / C C:
3
6
6
Z
e x sec.e x / dx Let u D e x
du D e x dx
Z
D
73.
.x 2
sec u du D ln j sec u C tan uj C C
D ln j sec.e x / C tan.e x /j C C:
Z
x
2 dz
dx
Let z D tan ; dx D
I D
4 sin x " 3 cos x
2
1 C z2
1 " z2
2z
cos x D
; sin x D
1 C z2
1 C z2
2 dz
Z
1
C z2
D
8z
3 " 3z 2
"
2
1 C z2 Z
Z1Cz
dz
dz
D2
D2
:
3z 2 C 8z " 3
.3z " 1/.z C 3/
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REVIEW EXERCISES ON TECHNIQUES OF INTEGRATION (PAGE 390)
1
A
B
D
C
.3z " 1/.z C 3/
3z " 1
zC3
Az C 3A C 3Bz " B
D
.3z " 1/.z C 3/
'
n
A D 3=10
A C 3B D 0
)
)
B D "1=10:
3A " B D 1
Thus
Z
Z
3
dz
dz
1
"
5
3z " 1 5
zC3
1
1
D ln j3z " 1j " ln jz C 3j C C
5 ˇ
5
ˇ
1 ˇˇ 3 tan!1 .x=2/ " 1 ˇˇ
C C:
D ln ˇ
5
tan!1 .x=2/ C 3 ˇ
76.
I D
74.
75.
Z
dx
x 1=3 " 1
D
3 1=3
.x
" 1/2 C 6.x 1=3 " 1/ C 3 ln jx 1=3 " 1j C C:
2
Let x D .u C 1/3
dx D 3.u C 1/2 du
!
Z
Z
1
.u C 1/2
uC2C
D3
du D 3
du
u
u
!
u2
C 2u C ln juj C C
D3
2
Z
dx
tan x C sin x
Z
cos x dx
D
sin x.1 C cos x/
Let z D tan.x=2/;
1 " z2
;
cos x D
1 C z2
78.
2 dz
1 C z2
2z
sin x D
1 C z2
dx D
1 " z 2 2 dz
1 C z2 1 C z2
D
#
$
2z
1 " z2
1
C
1 C z2
1 C z2
Z
Z
.1 " z 2 / dz
1 " z2
1
D
D
dz
2
2
z.1 C z C 1 " z /
2
z
1
z2
D ln jzj "
CC
2
4
ˇ
ˇ
xˇ 1 %
x &2
1 ˇ
tan
C C:
D ln ˇtan ˇ "
2
2
4
2
Remark: Since
Z
x
sin2
x
2 D 1 " cos x ;
tan
D
x
2
1 C cos x
cos2
2
2
the answer can also be written as
ˇ
ˇ
1 ˇˇ 1 " cos x ˇˇ 1 1 " cos x
ln ˇ
" !
C C:
4
1 C cos x ˇ 4 1 C cos x
262
77.
79.
Z
ADAMS and ESSEX: CALCULUS 9
x dx
p
Z
x dx
p
4 " .2x C 1/2
Let u D 2x C 1
3 " 4x "
du D 2 dx
Z
u"1
1
D
du
p
4
4 " u2
%u&
1
1p
4 " u2 " sin!1
D"
CC
4
4
2
!
1 !1
1p
1
2
3 " 4x " 4x " sin
D"
xC
C C:
4
4
2
Z
4x 2
D
p
x
dx Let x D u2
1Cx
dx D 2u#du
$
Z
Z
1
u2 du
D
2
du
1
"
D2
1 C u2
1 C u2
!
"
p
p
D 2 u " tan!1 u C C D 2 x " 2 tan!1 x C C:
Z
p
1 C e x dx Let u2 D 1 C e x
2u du D e x dx!
Z
Z
2
2
2u du
D
2C 2
du
D
u2 " 1
u "1
!
Z
1
1
D
2C
"
du
u"1 uC1
ˇ
ˇ
ˇu"1ˇ
ˇCC
D 2u C ln ˇˇ
u C 1ˇ
ˇp
ˇ
ˇ 1 C ex " 1 ˇ
p
ˇ
ˇ
x
D 2 1 C e C ln ˇ p
ˇ C C:
ˇ 1 C ex C 1 ˇ
I D
Z
x 4 dx
D
x3 " 8
Z #
xC
8x
x3 " 8
$
dx:
8x
A
Bx C C
D
C 2
"8
x"2
x C 2x C 4
Ax 2 C 2Ax C 4A C Bx 2 " 2Bx C C x " 2C
D
x 3 " 8(
(
ACB D0
B D "A
) 2A " 2B C C D 8 ) C D 2A
4A " 2C D 0
6A D 8
x3
Thus A D 4=3, B D "4=3, C D 8=3. We have
Z
Z
dx
x"2
4
4
x2
C
"
dx
2
3
x"2 3
x 2 C 2x C 4
Z
x2
4
4
xC1"3
D
C ln jx " 2j "
dx
2
3
3
.x C 1/2 C 3
x2
4
2
D
C ln jx " 2j " ln.x 2 C 2x C 4/
2
3
3
4
xC1
C p tan!1 p C K:
3
3
I D
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INSTRUCTOR’S SOLUTIONS MANUAL
OTHER REVIEW EXERCISES 6 (PAGE 391)
80. By the procedure used in Example 4 of Section 7.1,
Z
e x cos x dx D 21 e x .sin x C cos x/ C C;
Z
e x sin x dx D 12 e x .sin x " cos x/ C C:
Now
Z
3.
Z
ˇ!=2
ˇ
ˇ
csc x dx D lim " ln j csc x C cot xjˇ
ˇ
c!0C
!=2
0
c
D lim ln j csc c C cot cj D 1 (diverges)
c!0C
4.
xe x cos x dx
Z
1
dx
D lim
R!1
x C x3
1
d V D e cos x dx
V D 21 e x .sin x C cos x/
Z
D 12 xe x .sin C cos x/ " 12 e x .sin x C cos x/ dx
D 12 xe x .sin C cos x/
D
" 14 e x .sin x " cos x C sin x C
1
xe x .sin x C cos x/ " 12 e x sin x
2
cos x/ C C
C C:
5.
Z
Other Review Exercises 6 (page 391)
1.
h
2.
d x
e .ax C b/ cos x C .cx C d / sin x
dx h
D e x .ax C b/ cos x C .cx C d / sin x C a cos x C c sin x
i
" .ax C b/ sin x C .cx C d / cos x
h%
&
D e x .a C c/x C b C a C d cos x
%
&
i
C .c " a/x C d C c " b sin x
Z
c!0C
$
dx
Let x D u2
dx D 2u du
x ln x dx
Z
1
u.2 ln u/2u du
0
Z
1
u2 ln u du
0
c
6.
Z
1
p
dx
x 1"
Z
Therefore
0
x2
7.
c
8.
>
1
!1
x r e !x dx
Z R
D lim
x r e !x dx
Z
1
0
dx
D 1 (diverges)
x
dx
diverges:
p
x 1 " x2
Z 1 Z 1
dx
C
D I1 C I2
p x D
xe
0
1
0
Z 1
Z 1
dx
dx
I1 D
p x <
p D2
xe
x
Z0 1
Z0 1
dx
1
e !x dx D
I2 D
p x <
e
xe
1
1
Thus I converges, and I < 2 C .1=e/.
I D
Z
1
R 60
Volume =
0
T6 D
0
because limR!1 Rr e !R D 0 for any r. In order to ensure
that limc!0C c r e !c D 0 we must have
limc!0C c r D 0, so we need r > 0.
1
x
"
x 1 C x2
4
4
4
D " lim c 3 ln c " .1 " c 3 / D "
3 c!0C
9
9
0
d V D e !x dx
U D xr
r!1
d U D rx
dr
V D "e !x
ˇR
Z
ˇ
1
ˇ
D lim "x r e !x ˇ C r
x r!1 e !x dx
c!0C
ˇ
0
R!1
c
Z 1
r !c
D lim c e C r
x r!1 e !x dx
1
#
U D ln u d V D u2 du
du
u3
dU D
V D
u
3
0
1
ˇ1
Z
ˇ
3
u
1 1 2
ˇ
D 4 lim @
ln uˇ "
u duA
ˇ
c!0C
3
3 c
1
c!0C
R!1
p
D4
i
If a C c D 1, b C a C d D 0, c " a D 0, and d C c " b D 0,
then a D c D "d D 1=2 and b D 0. Thus
Z
i
ex h
x cos x C .x " 1/ sin x C C:
I D xe x cos x dx D
2
If a C c D 0, b C a C d D 0, c " a D 1, and d C c " b D 0,
then b D c D "a D 1=2 and d D 0. Thus
Z
i
ex h
J D xe x sin x dx D
x sin x " .x " 1/ cos x C C:
2
1
0
D
R
#
$ˇˇR
1
ˇ
2
D lim ln jxj " ln.1 C x / ˇ
ˇ
R!1
2
$ 1
#
2
ln 2
R
1
C ln 2 D
ln
D lim
R!1 2
1 C R2
2
x
U Dx
d U D dx
Z
A.x/ dx. The approximation is
10 h
10; 200 C 2.9; 200 C 8; 000 C 7; 100
2
i
C 4; 500 C 2; 400/ C 100
' 364; 000 m3 .
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OTHER REVIEW EXERCISES 6 (PAGE 391)
9.
10.
S6 D
ADAMS and ESSEX: CALCULUS 9
10 h
10; 200 C 4.9; 200 C 7; 100 C 2; 400/
3
i
C 2.8; 000 C 4; 500/ C 100
Integrating both sides over Œ0; 1! leads at once to
Z
' 367; 000 m3
Z 1p
2 C sin.#x/ dx
I D
0
h
p
p
1 p
2 C 2. 2 C sin.#=4/ C 2 C sin.#=2/
T4 D
8
p
p i
C 2 C sin.3#=4/ C 2
0
x 4 .1 " x/4
22
> 0 on .0; 1/,
" # > 0, and so
x2 C 1
7
22
.
#<
7
Z 1
b) If I D
x 4 .1 " x/4 dx, then since 1 < x 2 C 1 < 2
0
on .0; 1/, we have
I >
' 1:626765
I ' 1:6
12.
13.
1
.T4 C M4 / ' 1:617996
2
1
S8 D .T4 C 2M4 / ' 1:62092
3
I ' 1:62
Z 1
x2
dx Let x D 1=t
I D
5
3
1=2 x C x C 1
dx D ".1=t 2 / dt
Z 2
Z 2
4
.1=t / dt
t dt
D
D
5
3
5
2
0 .1=t / C .1=t / C 1
0 t Ct C1
T4 ' 0:4444
M4 ' 0:4799
T8 ' 0:4622
M8 ' 0:4708
S8 ' 0:4681
S16 ' 0:4680
I ' 0:468 to 3 decimal places
#
$
0:730
2:198
a) T4 D 1
C 1:001 C 1:332 C 1:729 C
2
2
D 5:526
#
$
1
S4 D
0:730 C 2:198 C 4.1:001 C 1:729/ C 2.1:332/
3
D 5:504:
x 4 .1 " x/4
22
22
dx D
" 4tan!1 1 D
" #:
x2 C 1
7
7
Since
' 1:609230
p
1 hp
M4 D
2 C sin.#=8/ C 2 C sin.3#=8/
4
i
p
p
2 C sin.5#=8/ C 2 C sin.7#=8/
11.
1
4T8 " T4
D 5:504.
3
c) Yes, S4 D S8 suggests that Sn may be independent
of n, which is consistent with a polynomial of degree
not exceeding 3.
1
0
x 4 .1 " x/4
I
dx > :
x2 C 1
2
Thus I > .22=7/ " # > I =2, or
22
22 I
"I <# <
" :
7
7
2
T8 D
b) If T8 D 5:5095, then S8 D
Z
c) I D
R1
0
.x 8 " 4x 7 C 6x 6 " 4x 5 C x 4 / dx D
1
. Thus
630
22
1
22
1
"
<# <
"
:
7
630
7
1260
2.
a) In D
Z
.1 " x 2 /n dx
d V D dx
U D .1 " x 2 /n
2 n!1
d U D "2nx.1 " x /
dx
V Dx
Z
2 n
2
2 n!1
D x.1 " x / C 2n x .1 " x /
dx
Z
D x.1 " x 2 /n " 2n .1 " x 2 " 1/.1 " x 2 /n!1 dx
D x.1 " x 2 /n " 2nIn C 2nIn!1 ; so
2n
1
x.1 " x 2 /n C
In!1 :
In D
2n C 1
2n C 1
Z 1
b) Let Jn D
.1 " x 2 /n dx. Observe that J0 D 1. By
0
(a), if n > 0, then we have
ˇ1
x.1 " x 2 /n ˇˇ
2n
2n
Jn D
Jn!1 D
Jn!1 :
ˇ C
2n C 1 ˇ
2n C 1
2n C 1
0
Challenging Problems 6
(page 391)
Therefore,
1.
2
a) Long division of x C 1 into
x 4 .1 " x/4 D x 8 " 4x 7 C 6x 6 " 4x 5 C x 4 yields
4
x 4 .1 " x/4
D x 6 " 4x 5 C 5x 4 " 4x 2 C 4 " 2
:
x2 C 1
x C1
264
2n
2n " 2
4 2
!
! ! ! ! J0
2n C 1 2n " 1
5 3
Œ.2n/.2n " 2/ ! ! ! .4/.2/!2
22n .nŠ/2
D
D
:
.2n C 1/Š
.2n C 1/Š
Jn D
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INSTRUCTOR’S SOLUTIONS MANUAL
CHALLENGING PROBLEMS 6 (PAGE 391)
c) From (a):
5.
In!1 D
2n C 1
1
In "
x.1 " x 2 /n :
2n
2n
Thus
Z
.1 " x 2 /!3=2 dx D I!3=2
2."1=2/ C 1
1
I!1=2 "
x.1 " x 2 /!1=2
"1
"1
x
:
D p
1 " x2
D
3.
Z
Z
1
x m .ln x/n dx
0
1
0
D nIn!1 "
1
0
x n dx D
1
e
if n # 1
c) The formula
0
1
n
1X 1
A
In D nŠ @1 "
e
jŠ
j D0
holds for n D 0 by part (b). Assume that it holds for
some integer n D k # 0. Then by (b),
0
1
k
X
1
1
1
A" 1
IkC1 D .k C 1/Ik " D .k C 1/kŠ @1 "
e
e
jŠ
e
j D0
0
1
k
X
1
1
1
A
D .k C 1/Š @1 "
"
e
j Š e.k C 1/Š
j D0
1
0
kC1
X 1
1
A:
D .k C 1/Š @1 "
e
jŠ
j D0
Thus the formula holds for all n # 0, by induction.
d) Since limn!1 In D 0, we must have
0
1
n
1X 1
A D 0:
lim @1 "
n!1
e
jŠ
j D0
6.
Z 1
."1/
un e !u du
.m C 1/n 0
."1/n
(.n C 1/ (see #50 in Section 7.5)
D
.m C 1/n
."1/n nŠ
D
:
.m C 1/n
n
0
!x
n
X
1
.
n!1
jŠ
Thus e D lim
Let u D .m C 1/t
du D .m C 1/ dt
Z
0
Let x D e !t
dx D "e !t dt
e !mt ."t /n e !t dt
0
Z 1
t n e !.mC1/t dt
D ."1/n
D
x n e !x dx <
d V D e !x dx
U D xn
d U D nx n!1 dx
V D "e !x
ˇ1
Z 1
ˇ
ˇ
D "x n e !x ˇ C n
x n!1 e !x dx
ˇ
0
a) x 4 Cx 2 C1 D .x 2 C1/2 "x 2 D .x 2 "xC1/.x 2 CxC1/.
Thus
Z
Z
x2 C 1
x2 C 1
D
dx
x4 C x2 C 1
.x 2 " x C 1/.x 2 C x C 1/
$
Z #
1
1
1
D
C 2
dx
2
2
x "xC1
x CxC1
!
Z
1
1
1
C!
dx
D
!
"2
"2
2
x " 21 C 34
x C 21 C 34
#
$
1
2x C 1
2x " 1
C tan!1 p
D p tan!1 p
C C:
3
3
3
Im;n D
1
0
p
p
b) x 4 C1 D .x 2 C1/2 "2x 2 D .x 2 " 2xC1/.x 2 C 2xC1/.
Thus
Z 2
Z
x C1
x2 C 1
p
p
dx
D
x4 C 1
.x 2 " 2x C 1/.x 2 C 2x C 1/
$
Z #
1
1
1
p
C
p
D
dx
2
2
2
x C 2x C 1
1
0 x " 2x C 1
Z
1 B
1
1
C
D
C%
A dx
@%
&2
&2
2
1
1
3
3
x" p
xCp
C4
C4
2
2
p
p !
2
2
1
!1 2x "
!1 2x C
p
p
D p tan
C tan
C C:
3
3
3
4.
Z
1
,
nC1
because 0 < e < 1 on .0; 1/. Thus limn!1 In D 0
by the Squeeze Theorem.
ˇ1
Z 1
ˇ
1
!x
!x ˇ
b) I0 D
e dx D "e ˇ D 1 "
ˇ
e
0
0
Z 1
In D
x n e !x dx
a) 0 < In D
I D
Z
1
e
!Kx
0
j D0
ˇ1
$
#
e !Kx ˇˇ
1
1
dx D
1" K .
ˇ D
"K ˇ
K
e
0
For very large K, the value of I is very small (I < 1=K).
However,
D
1
1
.1 C ! ! !/ >
100
100
1
1
D
.1 C ! ! !/ >
300
300
1
1 !K=200
.e
C ! ! !/ <
:
D
100
100
T100 D
S100
M100
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CHALLENGING PROBLEMS 6 (PAGE 391)
In each case the ! ! ! represent terms much less than the
first term (shown) in the sum. Evidently M100 is smallest
if k is much greater than 100, and is therefore the best
approximation. T100 appears to be the worst.
7.
5
4
3
ADAMS and ESSEX: CALCULUS 9
8.
a) f 0 .x/ < 0 on Œ1; 1/, and limx!1 f .x/ D 0. Therefore
Z
2
a) Let f .x/ D Ax C Bx C C x C Dx C Ex C F .
Then
# 5
$
Z h
Bh
Dh3
f .x/ dx D 2
C
C Fh :
5
3
!h
1
2a C
2b
1
D ;
16
5
2a C
2b
1
D ;
4
3
2a C 2b C c D 1:
Solving these equations, we get a D 7=90, b D 16=45,
and c D 2=15. The approximation for the integral of
any function f on Œm " h; m C h! is
"
Z mCh
7
16
f .x/ dx ' 2h
f .m " h/ C f .m " 21 h/
90
45
m!h
#
16
7
2
C f .m/ C f .m C 21 h/ C f .m C h/ :
15
45
90
b) If m D h D 1=2, we obtain
"
Z 1
7 0 16 !1=4
2
e !x dx ' 1
e C e
C e !1=2
90
45
15
0
#
7 !1
16 !3=4
C e
C e
45
90
' 0:63212087501:
16 !3=8
e
C
45
16
C e !5=8 C
45
1
f 0 .x/ dx
Z R
D " lim
f 0 .x/ dx
1
R!1 1
R!1
Thus
ˇZ
ˇ
ˇ
ˇ
1
R
ˇ Z
ˇ
f 0 .x/ cos x dx ˇˇ $
Z
Thus lim
R
R!1 1
b)
Z
1
1
R
jf 0 .x/j dx ! 0 as R ! 1:
f 0 .x/ cos x dx exists.
f .x/ sin x dx
1
U D f .x/
d V D sin x dx
d U D f 0 .x/ dx
V D " cos x
ˇR Z
ˇ
1
ˇ
D lim f .x/ cos x ˇ C
f 0 .x/ cos x dx
ˇ
R!1
1
Z1 1
D "f .1/ cos.1/ C
f 0 .x/ cos x dxI
1
the integral converges.
c) f .x/ D 1=x satisfies the conditions of part (a), so
Z
1
1
sin x
dx
x
converges
by part (b). Similarly, it can be shown that
Z
1
1
7 !1=2
e
45
#
2 !3=4 16 !7=8
7 !1
e
C e
C e
15
45
90
Z
cos.2x/
dx
x
converges:
1
1
j sin xj
dx #
x
Z
1
1
1 " cos.2x/
:
2x
R1
The latter integral diverges
R 1 because 1 .1=x/ dx diverges to infinity while 1 .cos.2x//=.2x/ dx converges. Therefore
' 0:63212055883:
266
Z
But since j sin xj # sin2 x D 12 .1 " cos.2x//, we have
With two intervals having h D 1=4 and m D 1=4 and
m D 3=4, we get
"
Z 1
1 7 0 16 !1=8
2
!x
e dx '
e C e
C e !1=4
2
90
45
15
0
C
jf 0 .x/j dx D "
D lim .f .1/ " f .R// D f .1/:
Also
h
i
2h af ."h/ C bf ."h=2/ C cf .0/ C bf .h=2/ C af .h/
, !
"
D 2 a 2Bh5 C 2Dh3 C 2F
#
$
)
2Bh5
2Dh3
Cb
C
C 2F C cF h :
16
4
These expressions will be identical if the coefficients
of like powers of h on the two sides are identical.
Thus
1
Z
1
1
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j sin xj
dx
x
diverges to infinity.
lOMoARcPSD|6566483
INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 7.1 (PAGE 401)
CHAPTER 7. APPLICATIONS OF
INTEGRATION
3.
By slicing:
Section 7.1 Volumes by Slicing—Solids of
Revolution (page 401)
1.
Z
1
.x " x 4 / dx
$ˇ1
# 2
x 5 ˇˇ
3#
x
"
cu. units.
D#
ˇ D
ˇ
2
5
10
V D#
0
0
By slicing:
V D#
Z
1
0
x 4 dx D
By shells:
#
cu. units.
5
V D 2#
By shells:
V D 2#
Z
1
y.1 "
Z
1
D 2#
p
y/ dy
0
!ˇ1
y 2 2y 5=2 ˇˇ
#
D 2#
"
cu. units.
ˇ D
ˇ
2
5
5
p
y. y " y 2 / dy
!ˇ1
2y 5=2 y 4 ˇˇ
3#
"
cu. units.
ˇ D
5
4 ˇ
10
0
0
y
p
yD x
0
y
.1;1/
yDx 2
yDx 2
x
x
x
x
Fig. 7.1-3
4.
Fig. 7.1-1
Slicing:
2. Slicing:
V D#
Z
1
.1 " y/ dy
#
$ˇ1
1 2 ˇˇ
#
D# y" y ˇ D
cu: units:
ˇ
2
2
0
0
Shells:
V D 2#
Z
1
3
x dx
0
# 4 $ ˇˇ1
x
#
ˇ
D 2#
cu: units:
ˇ D
4 ˇ
2
0
y
Z
1
.y " y 4 / dy
#
$ˇ1
1 2 1 5 ˇˇ
3#
D#
y " y ˇ D
cu: units:
ˇ
2
5
10
V D#
0
0
Shells:
Z
1
x.x 1=2 " x 2 / dx
$ˇ1
#
3#
2 5=2 1 4 ˇˇ
x
" x ˇ D
cu: units:
D 2#
ˇ
5
4
10
V D 2#
0
0
y
.1;1/
p
yD x yDx 2
yDx 2
1 x
x
Fig. 7.1-4
Fig. 7.1-2
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SECTION 7.1 (PAGE 401)
ADAMS and ESSEX: CALCULUS 9
y
5. a) About the x-axis:
.1;1/
V D#
Z
2
0
Z
x 2 .2 " x/2 dx
yDx
yDx 2
2
.4x 2 " 4x 3 C x 4 / dx
0
# 3
$ˇ2
4x
x 5 ˇˇ
16#
4
D#
"x C
cu. units.
ˇ D
3
5 ˇ
15
D#
x
Fig. 7.1-6
0
b) About the y-axis:
7.
Z
x 2 .2 " x/ dy
# 3
$ˇ2
2x
x 4 ˇˇ
8#
D 2#
"
cu. units.
ˇ D
3
4 ˇ
3
V D 2#
a) About the x-axis:
2
V D 2#
0
(a)
Z
3
y.4y " y 2 " y/ dy
$ˇ3
#
27#
y 4 ˇˇ
3
cu. units.
D 2# y "
ˇ D
4 ˇ
2
0
0
0
y
b) About the y-axis:
yD2x!x 2
Z 3h
i
.4y " y 2 /2 " y 2 dy
0
Z 3
D#
.15y 2 " 8y 3 C y 4 / dy
0
#
$ˇ3
y 5 ˇˇ
108#
3
4
D # 5y " 2y C
cu. units.
ˇ D
5 ˇ
5
V D#
2
x
0
y
y
(b)
yD2x!x 2
.3;3/
xDy
2
x
xD4y!y 2
Fig. 7.1-5
x
6. Rotate about
Fig. 7.1-7
a) the x-axis
Z
1
.x 2 " x 4 / dx
#
$ˇ1
1 3 1 5 ˇˇ
2#
D#
x " x ˇ D
cu: units:
ˇ
3
5
15
V D#
0
8.
Rotate about
a) the x-axis
0
b) the y-axis
Z
D#
1
x.x " x 2 / dx
0
#
$ˇ1
1 3 1 4 ˇˇ
#
D 2#
x " x ˇ D
cu: units:
ˇ
3
4
6
V D 2#
0
268
V D#
Z
Z
!
0
!
0
Œ.1 C sin x/2 " 1! dx
.2 sin x C sin2 x/ dx
$ˇˇ!
#
#
#
ˇ
D "2# cos x C x " sin 2x ˇ
ˇ
2
4
0
1
D 4# C # 2 cu: units:
2
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 7.1 (PAGE 401)
y
b) the y-axis
.1=3;3/
V D 2#
Z
!
x sin x dx
3xC3yD10
0
U Dx
d V D sin x dx
d U D dx ˇ V D " cos x
(
)
ˇ! Z !
ˇ
D 2# "x cos x ˇ C
cos x dx
ˇ
0
1
yD x
x
0
Fig. 7.1-10
D 2# 2 cu: units:
9. a) About the x-axis:
Z 1#
V D#
4"
0
D 4# " #
D 4# " #
Z
Z
11.
1
.1 C x 2 /2
!=4
0
!=4
$
Let x D tan &
dx D sec2 & d&
dx
sec2 &
d&
sec4 &
.3;1=3/
Z 1
V D 2 & 2#
.2 " x/.1 " x/ dx
0
Z 1
D 4#
.2 " 3x C x 2 / dx
0
$ˇ1
#
3x 2
x 3 ˇˇ
10#
D 4# 2x "
C
cu. units.
ˇ D
2
3 ˇ
3
0
cos2 & d&
y
0
ˇ!=4
ˇ
#
D 4# " .& C sin & cos & /ˇˇ
2
0
#2 #
15#
#2
D 4# "
" D
"
cu. units.
8
4
4
8
b) About the y-axis:
$
Z 1 #
1
V D 2#
x 2"
dx
1 C x2
0
#
$ˇˇ1
1
ˇ
2
2
D 2# x " ln.1 C x / ˇ
ˇ
2
0
#
$
1
D 2# 1 " ln 2 D 2# " # ln 2 cu. units.
2
y
xCyD1
xD2
x
x
Fig. 7.1-11
12.
y
Z
1
Œ.1/2 " .x 2 /2 ! dx
#
$ˇ1
1 5 ˇˇ
D# x" x ˇ
ˇ
5
V D#
!1
8#
cu: units:
D
5
yD2
!1
y
yD
1
1Cx 2
yD1
x2
yD1!x 2
x
1
dx
x
x
Fig. 7.1-9
10. By symmetry, rotation about the x-axis gives the same
volume as rotation about the y-axis, namely
$
Z 3 #
1
10
"x"
dx
V D 2#
x
3
x
1=3
#
$ˇˇ3
5 2 1 3
ˇ
D 2#
x " x "x ˇ
ˇ
3
3
Fig. 7.1-12
13.
The volume remaining is
V D 2 & 2#
1=3
512#
cu: units:
D
81
x
D 2#
Z
3
Z
2
1
p
0
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p
x 4 " x 2 dx
u du D
ˇ3
4# 3=2 ˇˇ
u ˇ
ˇ
3
0
Let u D 4 " x 2
du D "2x dx
p
D 4# 3 cu. units.
269
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SECTION 7.1 (PAGE 401)
ADAMS and ESSEX: CALCULUS 9
4 3
32#
#2 D
cu. units.,
3
3
p
32#
" 4# 3 cu. units.
therefore the volume removed is
3
The percentage removed is
Since the volume of the ball is
15.
p
32#
p !
" 4# 3
3 3
3
& 100 D 100 1 "
' 35:
32#
8
3
About 35% of the volume is removed.
y
The volume remaining is
Z b
%
x&
dx
V D 2#
xh 1 "
b
a
# 2
$ˇb
x
x 3 ˇˇ
D 2#h
"
ˇ
2
3b ˇ
a
#
$
a3
2
D #h.b 2 " a2 / " #h b 2 "
3
b
#
3$
1
2a
D #h b 2 " 3a2 C
cu. units.
3
b
y
p
yD 4!x 2
h
dx
1
2
x
x
x y
C D1
b h
xDa
dx b
x
Fig. 7.1-13
14.
The radius of the hole is
the remaining volume is
V D#
Z
L=2
!L=2
"#
q
Fig. 7.1-15
16.
R2 " 14 L2 . Thus, by slicing,
2
R "x
2
$
$#
#
L2
2
dx
" R "
4
!ˇL=2
L2
1 3 ˇˇ
D 2#
x" x ˇ
ˇ
4
3
0
# 3
D L cu. units (independent of R).
6
y
x
Let a circular disk with radius a have centre at point
.a; 0/. Then the disk is rotated about the y-axis which
is one of its tangent lines. The volume is:
Z 2a p
V D 2 & 2#
x a2 " .x " a/2 dx Let u D x " a
0
du D dx
Z a
p
D 4#
.u C a/ a2 " u2 du
Z!a
Z ap
a p
D 4#
u a2 " u2 du C 4#a
a2 " u2 du
!a
!a
$
#
1
D 0 C 4#a #a2 D 2# 2 a3 cu: units:
2
(Note that the first integral is zero because the integrand is
odd and the interval is symmetric about zero; the second
integral is the area of a semicircle.)
p
yD R2 !x 2
y
R
q
L
2
L2
R2 ! 4
.x!a/2 Cy 2 Da2
x
2a
a
L
Fig. 7.1-16
Fig. 7.1-14
270
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INSTRUCTOR’S SOLUTIONS MANUAL
17.
SECTION 7.1 (PAGE 401)
Volume of the smaller piece:
Z a
V D#
.a2 " x 2 / dx
b
#
$ˇa
x 3 ˇˇ
2
D# a x"
ˇ
3 ˇ
b
#
$
a3 " b 3
2
D # a .a " b/ "
3
#
2
2
D .a " b/Œ3a " .a C ab C b 2 /!
3
#
D .a " b/2 .2a C b/ cu. units.
3
19.
The volume of the ellipsoid is
$
#
x2
b 2 1 " 2 dx
a
0
ˇa
#
3 $ˇ
x
4
ˇ
2
D 2#b x " 2 ˇ D #ab 2 cu. units.
3a ˇ
3
Z
V D 2#
a
0
y
yDb
b
q
1!
x2
a2
y
p
yD a2 !x 2
dx
b
x
a
dx
x
a
x
x
Fig. 7.1-19
20.
Fig. 7.1-17
18. Let the centre of the bowl be at (0, 30). Then the volume
of the water in the bowl is
Z 20 h
i
V D#
302 " .y " 30/2 dy
0
Z 20
D#
60y " y 2 dy
0
(
)ˇ20
1 3 ˇˇ
2
D # 30y " y ˇ
ˇ
3
V D#
Z
D 2#
ah
Z!aa
0
i
p
p
a2 " y 2 /2 " .b " a2 " y 2 /2 dy
.b C
p
4b a2 " y 2 dy
#a2
D 8#b
D 2# 2 a2 b cu. units.:
4
We used the area of a quarter-circle of radius a to evaluate
the last integral.
0
' 29322 cm3 :
y
The cross-section at height
p y is an annulus (ring)
having
inner
radius
b
"
a2 " y 2 and outer radius
p
2
2
b C a " y . Thus the volume of the torus is
21.
a) Volume of revolution about the x-axis is
Z
1
e !2x dx
ˇR
e !2x ˇˇ
#
cu. units.
D # lim
ˇ D
R!1 "2 ˇ
2
V D#
30
20
0
0
b) Volume of revolution about the y-axis is
x 2 C.y!30/2 D302
x
V D 2#
Fig. 7.1-18
Z
1
xe !x dx
0
ˇR
ˇ
D 2# lim ."xe !x " e !x /ˇˇ D 2# cu. units.
R!1
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SECTION 7.1 (PAGE 401)
ADAMS and ESSEX: CALCULUS 9
y
25.
1
yDe "x
dx
x
x
The volume os S is still the constant cross-sectional
area a2 =2 times the height b, that is, V D a2 b=2 cm3 .
Fig. 7.1-21
26.
22. The volume is
V D#
Z
1
x
!2k
1
D # lim
R!1
ˇR
x 1!2k ˇˇ
dx D # lim
ˇ
R!1 1 " 2k ˇ
#
R1!2k
C
:
1 " 2k
2k " 1
Z
8
0
D 3072#
In order for the solid to have finite volume we need
D 3072#
1
k> :
2
that is;
The region is symmetric about x D y so has the same
volume of revolution about the two coordinate axes. The
volume of revolution about the y-axis is
V D 2#
1
1 " 2k < 0;
Since all isosceles right-angled triangles having leg length
a cm are congruent, S does satisfy the condition for being
a prism given in early editions. It does not satisfy the
condition in this edition because one of the line segments
joining vertices of the triangular cross-sections, namely the
x-axis, is not parallel to the line joining the vertices of the
other end of the hypotenuses of the two bases.
D 3072#
Z
Z
Z
Z
!=2
y
27.
1
!=2
0
1
0
.1 " v 2 /2 v 4 dv
Let v D cos u
dv D " sin u du
0
4
#R3 . Expressing this volume
3
as the “sum” (i.e., integral) of volume elements that are
concentric spherical shells having thickness dr and varying radius r, and therefore having surface area kr 2 and
volume kr 2 dr, we obtain
The volume of the ball is
4
#R3 D
3
Z
R
0
kr 2 dr D
k 3
R :
3
Thus k D 4#.
A solid consisting of points on parallel line segments between parallel planes will certainly have congruent crosssections in planes parallel to and lying between the two
base planes, any solid satisfying the new definition will
certainly satisfy the old one. But not vice versa; congruent
cross-sections does not imply a family of parallel line segments giving all the points in a solid. For a counterexample, see the next exercise. Thus the earlier, incorrect definition defines a larger class of solids than does the current
definition. However, the formula V D Ah for the volume
of such a solid is still valid, as all congruent cross-sections
still have the same area, A, as the base region.
272
.1 " cos2 u/2 cos4 u sin u du
1
x
Fig. 7.1-23
24.
sin5 u cos4 u du
0
dx
x
Let x D 8 sin3 u
dx D 24 sin2 u cos u du
.v 4 " 2v 6 C v 8 / dv
$
#
1
8192#
1 2
" C
D
cu. units.
D 3072#
5 7
9
105
D 3072#
R1
23. The volume is V D 2# 1 x 1!k dx. This improper integral converges if 1 " k < "1, i.e., if k > 2. The solid has
finite volume only if k > 2.
yDx "k
x.4 " x 2=3 /3=2 dx
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dr
r
Fig. 7.1-27
lOMoARcPSD|6566483
INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 7.1 (PAGE 401)
R
D sin ˛, so R D .x C h/ sin ˛.
xCh
Using the result of Exercise #17, the volume of liquid displaced by the ball is
28. Using heights f .x/ estimated from the given graph, we
obtain
Z 9%
&2
V D#
f .x/ dx
1
#h 2
3 C 4.3:8/2 C 2.5/2 C 4.6:7/2 C 2.8/2
'
3
i
C 4.8/2 C 2.7/2 C 4.5:2/2 C 32 ' 938 cu. units.
Note that
V D
#
.R " x/2 .2R C x/:
3
We would like to consider V as a function of x for
"2R $ x $ R since V D 0 at each end of this interval, and V > 0 inside the interval. However, the actual interval of values of x for which the above formulation makes physical sense is smaller: x must satisfy
"R $ x $ h tan2 ˛. (The left inequality signifies nonsubmersion of the ball; the right inequality signifies that
the ball is tangent to the glass somewhere below the rim.)
We look for a critical point of V , considered as a function
of x. (As noted above, R is a function of x.) We have
29. Using heights f .x/ estimated from the given graph, we
obtain
Z 9
V D 2#
xf .x/ dx
1
h
2#
'
1.3/ C 4.2/.3:8/ C 2.3/.5/ C 4.4/.6:7/ C 2.5/.8/
3
i
C 4.6/.8/ C 2.7/.7/ C 4.8/.5:2/ C 9.3/ ' 1537 cu. units.
0D
30. Using heights f .x/ estimated from the given graph, we
obtain
Z 9
V D 2#
.x C 1/f .x/ dx
1
2# h
2.3/ C 4.3/.3:8/ C 2.4/.5/ C 4.5/.6:7/ C 2.6/.8/
'
3
i
C 4.7/.8/ C 2.8/.7/ C 4.9/.5:2/ C 10.3/ ' 1832 cu. units.
"
#
$
dV
#
dR
D
2.R " x/
" 1 .2R C x/
dx
3
dx
$#
#
dR
2
C1
C .R " x/ 2
dx
dR
.4R C 2x C 2R " 2x/ D 4R C 2x " .R " x/:
dx
Thus
#
$
R
6R sin ˛ D 3.R C x/ D 3 R C
"h
sin ˛
2R sin2 ˛ D R sin ˛ C R " h sin ˛
h sin ˛
h sin ˛
RD
:
D
2
cos 2˛ C sin ˛
1 " 2 sin ˛ C sin ˛
31. Let the ball have radius R, and suppose its centre is x
units above the top of the conical glass, as shown in the
figure. (Clearly the ball which maximizes liquid overflow
from the glass must be tangent to the cone along some
circle below the top of the cone — larger balls will have
reduced displacement within the cone. Also, the ball will
not be completely submerged.)
This value of R yields a positive value of V , and corresponds to x D R.2 sin ˛ " 1/. Since sin ˛ # sin2 ˛,
"R $ x D
h sin ˛.2 sin ˛ " 1/
h sin2 ˛
D h tan2 ˛:
$
cos2 ˛
1 C sin ˛ " 2 sin2 ˛
Therefore it gives the maximum volume of liquid displaced.
x
R
32.
h
h sec ˛
.hCx/ cos ˛
˛
Let P be the point .t; 52 " t /. The line through P perpendicular to AB has equation y D x C 52 " 2t , and meets the
curve xy D 1 at point Q with x-coordinate s equal to the
positive root of s 2 C . 52 " 2t /s D 1. Thus,
s
#
"
#
$2
5
5
1
2t " C
" 2t C 4 :
sD
2
2
2
Fig. 7.1-31
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SECTION 7.1 (PAGE 401)
ADAMS and ESSEX: CALCULUS 9
y
A.1=2;2/
yD
Section 7.2 More Volumes by Slicing
(page 405)
p
2 dt
1
x
P
xCyD
Q
5
2
1.
Z
V D
B.2;1=2/
2.
s
t
x
The volume element at P has radius
p
PQ D 2.t " s/
3
2
s
$2
#
p
5 1
5
" 2t C 45
D 24 "
4 2
2
3.
and thickness 2 dt . Hence, the volume of the solid is
s
#
$2
$#2
Z 2 "p #
p
5 1
5
V D#
2
"
" 2t C 4
2 dt
4 2
2
1=2
2
0s
1
#
$2
p Z 2 25 5
5
4
@
"
" 2t C 4A C
D 2 2#
16 4
2
1=2
(#
$2
)#
1 5
" 2t C 4 dt Let u D 2t " 52
4 2
du D 2!dt
p Z 3=2 41 5 p
u2
"
du
D 2#
u2 C 4 C
4
4
!3=2 16
!ˇ3=2
p
41
1 3 ˇˇ
D 2#
uC u ˇ
"
ˇ
16
12
!3=2
p
Z
5 2# 3=2 p 2
u C 4 du Let u D 2 tan v
4
!3=2
du D 2 sec2 v dv
p
Z tan"1 .3=4/
p
33 2#
" 5 2#
D
sec3 v dv
4
tan"1 .!3=4/
p
p Z tan"1 .3=4/ 3
33 2#
" 10 2#
sec v dv
D
4
0
p
p %
33 2#
" 5 2# sec v tan vC
D
4
ˇ "1
&ˇtan .3=4/
ˇ
ln j sec v C tan vj ˇ
ˇ
0
"
#
$#
p
33
15
D 2#
"5
C ln 2 " 0 " ln 1
4
16
$
#
p
57
" 5 ln 2 cu: units:
D 2#
16
0
A horizontal slice of thickness dz at height a has volume
d V D z.h " z/ dz. Thus the volume of the solid is
Z
h
.z.h " z/ dz D
0
#
hz 2 z 3
"
2
3
ˇ
$ˇh
3
ˇ
ˇ D h units3 :
ˇ
6
ˇ
0
A horizontal
p slice of thickness dz at height a has volume
d V D #z 1 " z 2 dz. Thus the volume of the solid is
Z
1
p
z 1 " z 2 dz
let u " 1 " z 2
ˇ1
ˇ
Z 1
p
#
# 2 3=2 ˇˇ
#
D
u du D
u ˇ D
units3 :
2 0
2 3
3
ˇ
V D
p
274
0
V D
Fig. 7.1-32
ˇ2
3 2 ˇˇ
3x dx D x ˇ D 6 m3
2 ˇ
2
0
0
4.
5.
Z
V D
3
1
ˇ3
x 3 ˇˇ
26
cu. units
ˇ D
3 ˇ
3
1
Z 6
.2 C z/.8 " z/ dz D
.16 C 6z " z 2 / dz
0
0
$ˇ6
#
z 3 ˇˇ
2
D 16z C 3z "
ˇ D 132 ft3
3 ˇ
V D
Z
x 2 dx D
6
0
6.
p
The area of an%equilateral
triangle
of edge x is
p p &
p
p
A.x/ D 21 x 23 x D 43 x sq. units. The volume of
the solid is
V D
7.
Z
4
1
ˇ4
p
p
3
3 2 ˇˇ
15 3
x dx D
x ˇ D
cu. units.
ˇ
4
8
8
p
1
The area of cross-section at height y is
A.y/ D
%
y&
2#.1 " .y= h//
.#a2 / D #a2 1 "
sq. units.
2#
h
The volume of the solid is
V D
Z
0
h
%
#a2 h
y&
dy D
cu. units.
#a2 1 "
h
2
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 7.2 (PAGE 405)
8. Since V D 4, we have
4D
Z
2
0
Thus k D 1.
ˇ2
x 4 ˇˇ
kx dx D k
ˇ D 4k:
4 ˇ
3
y
0
p
3y
x
2y
3
9. The volume between height 0 and height z is z . Thus
z3 D
Z
x 2 Cy 2 Dr 2
z
A.t / dt;
0
Fig. 7.2-12
where A.t / is the cross-sectional area at height t . Differentiating the above equation with respect to z, we get
3z 2 D A.z/. The cross-sectional area at height z is
3z 2 sq. units.
10. This is similar to Exercise 7. We have 4z D
Z
13.
The cross-section at distance y from the vertex of the partial cone is a semicircle of radius y=2 cm, and hence area
#y 2 =8 cm2 . The volume of the solid is
V D
z
A.t / dt ,
0
Z
12
0
# 2
#123
y dy D
D 72# cm3 .
8
24
so A.z/ D 4. Thus the square cross-section at height z has
side 2 units.
11.
V D2
D8
Z
z
Z r% p
&2
2 r 2 " y 2 dy
0
y
0
r
#
$ˇr
y 3 ˇˇ
16r 3
2
.r " y / dy D 8 r y "
cu. units.
ˇ D
3 ˇ
3
2
r
12
y
2
x
0
z
p
2
.12; 12; 0/
Fig. 7.2-13
r 2 !y 2
14.
y
p
x
xD
r 2 !y 2
The volume of a solid of given height h and given crosssectional area A.z/ at height z above the base is given
by
Z h
V D
A.z/ dz:
0
If two solids have the same height h and the same area
function A.z/, then they must necessarily have the same
volume.
Fig. 7.2-11
12. The area
triangle of base 2y is
p of an equilateral
p 2
1
.2y/.
3y/
D
3y
.
Hence,
the solid has volume
2
Z
r
p
3.r 2 " x 2 / dx
#
$ˇˇr
p
1
ˇ
D 2 3 r 2x " x3 ˇ
ˇ
3
V D2
15.
Let the x-axis be along the diameter shown in the figure,
with the origin at the centre of the base. The cross-section
perpendicular to the x-axis at x is a rectangle having base
p
aCb
a"b
2 r 2 " x 2 and height h D
C
x: Thus the
2
2
volume of the truncated cylinder is
0
0
4
D p r 3 cu: units:
3
#
$
p
aCb
a"b
C
x dx
.2 r 2 " x 2 /
2
2r
!r
Z r
2
p
#
r
.a C b/
D
cu. units.
.a C b/ r 2 " x 2 dx D
2
!r
V D
Z
r
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SECTION 7.2 (PAGE 405)
ADAMS and ESSEX: CALCULUS 9
17.
Cross-sections of the wedge removed perpendicular to the
x-axis are isosceles, right triangles. The volume of the
wedge removed from the log is
Z
1 p
. 400 " x 2 /2 dx
2
0
$ˇ20
#
16; 000
x 3 ˇˇ
cm3 :
D 400x "
ˇ D
3 ˇ
3
V D2
h
20
0
x
y
yD
p
r2
Fig. 7.2-15
x
r
" x2
z
16. The plane z D k meets the ellipsoid in the ellipse
# $2
k
a
b
c
y2
x2
"
# $2 # C "
# $2 # D 1
k
k
a2 1 "
b2 1 "
c
c
% x &2
that is,
C
% y &2
D1"
45ı
20
x
y
x
Fig. 7.2-17
which has area
# $2 #
k
:
A.k/ D #ab 1 "
c
"
The volume of the ellipsoid is found by summing volume
elements of thickness d k:
18. The solution is similar to that of Exercise 15 except that
the legs of the right-triangular cross-sections
are y " 10
p
p
instead of y, and x goes from "10 3 to 10 3 instead of
"20 to 20. The volume of the notch is
p
4; 000#
D 3; 000 3 "
' 1; 007 cm3 :
3
19.
The hole has the shape of two copies of the truncated cylinder of Exercise
15, placed base to base,
p
with a C b D 3 2 in and r D 2 in. Thus the
volume of wood
p volume of the hole) is
p removed (the
V D 2.#22 /.3 2=2/ D 12 2# in3 .
20.
One eighth of the region lying inside both cylinders is
shown in the figure. If the region is sliced by a horizontal
plane at height z, then the intersection is a rectangle with
area
p
p
A.z/ D b 2 " z 2 a2 " z 2 :
z
c
x2
a2
C
y2
b2
C
z2
D1
c2
(one-eighth of the
solid is shown)
k
A.k/
b
a
p
10 3
0
!c
4
D #abc cu: units:
3
Z
1 p
. 400 " x 2 " 10/2 dx
2
0
Z 10p3 %
&
p
D
500 " x 2 " 20 400 " x 2 dx
V D2
# $2 #
Z c
k
V D
dk
#ab 1 "
c
!c
#ˇc
"
ˇ
1
ˇ
D #ab k " 2 k 3 ˇ
ˇ
3c
"
y
The volume of the whole region is
x
V D8
Fig. 7.2-16
276
p
yD 400!x 2
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0
b
p
p
b 2 " z 2 a2 " z 2 dz:
lOMoARcPSD|6566483
INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 7.3 (PAGE 412)
z
5.
b
z
p
b 2 !z 2
A.z/p
a2 !z 2
a
y
6.
x
Fig. 7.2-20
21.
By the result given in Exercise 18 with a D 4 cm and
b D 2 cm, the volume of wood removed is
Z
V D8
2
p
0
4 " z2
p
16 " z 2 dz ' 97:28 cm3 :
(We used the numerical integration routine in Maple to
evaluate the integral.)
7.
Section 7.3 Arc Length and Surface Area
(page 412)
2 !1=3
x
;
3
p
4 !2=3
9x 2=3 C 4
dx D
ds D 1 C x
dx
9
3jxj1=3
p
Z 1
9x 2=3 C 4
dx Let u D 9x 2=3 C 4
LD2
3x 1=3
0
du D 6x !1=3 dx
Z
1 13 p
2.133=2 / " 16
units.
D
u du D
9 4
27
y D x 2=3 ;
r
y0 D
q
2.x C 1/3 D 3.y " 1/2 ; y D 1 C 23 .x C 1/3=2
q
y 0 D 32 .x C 1/1=2 ;
r
r
3x C 5
3x C 3
ds D 1 C
dx D
dx
2
2
ˇ0
p
Z 0
ˇ
p
2
1
ˇ
3x C 5 dx D
.3x C 5/3=2 ˇ
LD p
ˇ
9
2 !1
!1
p %
&
2 3=2
5
" 23=2 units.
D
9
x3
1
x2
1
C ; y0 D
" 2
12
x
4
x
s
# 2
$2
# 2
$
x
1
x
1
ds D 1 C
" 2
dx D
C 2 dx
4
x
4
x
$
# 3
$ˇˇ4
Z 4# 2
1
x
1 ˇ
x
C 2 dx D
"
LD
ˇ D 6 units.
ˇ
4
x
12
x
1
yD
1
1.
y D 2x " 1; y 0 D 2; ds D
Z 3p
p
LD
5 dx D 2 5 units.
p
1 C 22 dx
8.
1
2. y D ax C b, A $ x $ B, y 0 D a. The length is
LD
3.
Z
B
A
p
1 C a2 dx D
p
1 C a2 .B " A/ units.
p
p
y D 32 x 3=2 ; y 0 D x; ds D 1 C x dx
ˇ8
Z 8
ˇ
p
52
2
3=2 ˇ
LD
units.
1 C x dx D .1 C x/ ˇ D
ˇ
3
3
0
0
4.
y 2 D .x " 1/3 ;
Z 2r
y D .x " 1/3=2 ;
y0 D
3p
x"1
2
Z
1 2p
9
1 C .x " 1/ dx D
9x " 5 dx
4
2 1
1
ˇ2
ˇ
133=2 " 8
1
ˇ
.9x " 5/3=2 ˇ D
units.
D
ˇ
27
27
LD
x3
1
1
C
; y 0 D x2 " 2
3
4x
4x
s
#
$
#
$
1 2
1
dx D x 2 C 2 dx
ds D 1 C x 2 " 2
4x
4x
$
# 3
$ˇ2
Z 2#
x
1 ˇˇ
59
1
2
LD
"
units.
x C 2 dx D
ˇ D
ˇ
4x
3
4x
24
1
yD
1
9.
1
x
ln x x 2
"
; y0 D
"
2
4
2x
2
s
#
#
$2
$
1
1
x
x
ds D 1 C
dx D
"
C
dx
2x
2
2x
2
ˇ
$
#
$e
Z e#
x
ln x
x 2 ˇˇ
1
C
dx D
C
LD
ˇ
2x
2
2
4 ˇ
1
yD
1
1
e2 " 1
e2 C 1
D C
D
units.
2
4
4
10. If y D x 2 "
1
ln x
1
then y 0 D 2x "
and
8
8x
$
#
1 2
:
1 C .y 0 /2 D 2x C
8x
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SECTION 7.3 (PAGE 412)
ADAMS and ESSEX: CALCULUS 9
Thus the arc length is given by
s
#
$
Z 2
1 2
1 C 2x "
sD
dx
8x
1
$
Z 2#
1
D
2x C
dx
8x
1
ˇ
$ˇ2
#
1
1
ˇ
D x 2 C ln x ˇ D 3 C ln 2 units:
ˇ
8
8
15.
1
11.
Z
Z a
1 C sinh2 x dx D
cosh x dx
0
0
ˇa
ˇ
e a " e !a
ˇ
units:
D sinh x ˇ D sinh a D
ˇ
2
sD
a
p
ex " 1
;
2$x$4
ex C 1
x
x
e C 1 .e C 1/e x " .e x " 1/e x
y0 D x
e "1
.e x C 1/2
x
2e
D 2x
:
e "1
The length of the curve is
y D ln
LD
length D
Z
Z
1=2
!1=2
s
1C
1=2
4x 2
dx
.1 " x 2 /2
!1=2
13.
sD
Z
!=4 p
!=6
14.
0
Z
xD2
Let 2x D tan &
2 dx D sec2 & d&
1
sec3 &
2 xD0
&ˇˇxD2
1%
D
sec & tan & C ln j sec & C tan & j ˇˇ
4
xD0
&ˇˇ2
p
1% p
2x 1 C 4x 2 C ln.2x C 1 C 4x 2 / ˇˇ
D
4
0
p &
1% p
D
4 17 C ln.4 C 17/
4
p
p
1
D 17 C ln.4 C 17/ units.
4
D
278
16.
1C
2
4
4e 2x
dx
.e 2x " 1/2
We have
sD
!=4
y D x 2 ; 0 $ x $ 2; y 0 D 2x.
Z 2p
length D
1 C 4x 2 dx
s
Z
e
r
1
1 C 2 dx
x
1
Z ep 2
x C1
D
dx Let x D tan u, dx D sec2 u du
x
1
Z xDe
Z xDe
du
sec3 u
du D
D
2
tan u
xD1 cos u sin u
xD1
Z xDe
sin u du
D
v D cos u, dv D " sin u du
2
2
xD1 cos u sin u
Z xDe
dv
D"
:
2
2
xD1 v .1 " v /
1 C tan2 x dx
ˇ!=4
ˇ
ˇ
D
sec x dx D ln j sec x C tan xjˇ
ˇ
!=6
!=6
#
$
p
2
1
D ln. 2 C 1/ " ln p C p
3
3
p
2C1
units:
D ln p
3
Z
D
"2x
.
1 " x2
1 C x2
dx
2
!1=2 1 " x
$
Z 1=2 #
2
dx
D
"1 C
1 " x2
!1=2
ˇ
#
$ 1=2
1 C x ˇˇ
D "x C ln
D 2 ln 3 " 1 units
ˇ
1"x ˇ
D
Z
4
e 2x C 1
dx
2x " 1
2 e
ˇ4
Z 4 x
ˇ
e C e !x
x
!x ˇ
D
dx D ln je " e jˇ
x " e !x
e
2
2
#
$
#
$
1
1
4
2
D ln e " 4 " ln e " 2
e
e
$
# 8
e4 C 1
e " 1 e2
D ln
units.
D ln
4
4
e
e "1
e2
0
12. y D ln.1 " x 2 /, " 12 $ x $ 21 , y 0 D
Z
Since
1
A
B
C
D
D C 2C
C
2
"v /
v
v
1"v
1Cv
A.v " v 3 / C B.1 " v 2 / C C.v 2 C v 3 / C D.v 2 " v 3 /
D
v 2 .1 " v 2 /
8̂
"A C C " D D 0
<
"B C C C D D 0
)
:̂ A D 0
BD1
1
) A D 0; B D 1; C D D D ;
2
v 2 .1
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 7.3 (PAGE 412)
therefore,
19.
#
$#
1
1
1
1
C
C
dv
sD"
v2
2 1"v
1Cv
xD1
ˇ
ˇ
ˇ xDe
1 1 ˇ 1 C v ˇˇ ˇˇ
1
D " ln ˇˇ
/
.but v D cos u D p
ˇ
2 C1
v 2
1 " v ˇˇ
x
xD1
#ˇxDe
"
p
1 .1 C v/2 ˇˇ
2
x C 1 " ln
D
ˇ
2
j1 " v 2 j ˇ
xD1
3
2
20.
2
1
ˇe
ˇ
1
C
p
C
6p
ˇ
1
x2 C 1 7
x2 C 1
2
7ˇ
D6
4 x C 1 " 2 ln
5ˇ
1
ˇ
1" 2
1
x C1
#ˇe
"
p
p
1 2 C x 2 C 2 1 C x 2 ˇˇ
x 2 C 1 " ln
D
ˇ
ˇ
2
x2
1
p
2
p
2
p
p
1
2
C
e
C
2
1
C
e
1
" 2 C ln.3 C 2 2/
D e 2 C 1 " ln
2
e2
2
p
p
p
1
.3 C 2 2/e 2
2
D e C 1 " 2 C ln
p
units:
2 2 C e2 C 2 1 C e2
Z
17.
xDe
"
x 2=3 C y 2=3 D x 2=3 . By symmetry, the curve has congruent arcs in the four quadrants. For the first quadrant arc
we have
21.
%
&3=2
y D a2=3 " x 2=3
$
&1=2 # 2
3 % 2=3
!1=3
2=3
0
a
"x
" x
:
y D
2
3
y D x 1=3 ; 1 $ x $ 2; y 0 D
Length D
have
LD4
D 4a
a
0
1=3
D 4a1=3
s
Z
1C
a
2=3
"x
x 2=3
Z
1
0
1
0
Using a calculator we calculate some Simpson’s Rule approximations as described in Section 7.2:
S2 ' 1:59921
S8 ' 1:60025
S4 ' 1:60110
S16 ' 1:60023:
s
ds D
9x 2
dx D
1C
3 " 3x 2
s
3 C 6x 2
dx:
3 " 3x 2
The circumference of the ellipse is
4
Z
s
1
0
3 C 6x 2
dx ' 8:73775 units
3 " 3x 2
(with a little help from Maple’s numerical integration
routine.)
For the ellipse x 2 C 2y 2 D 2, we have 2x C 4yy 0 D 0, so
y 0 D "x=.2y/. Thus
Z
dx
p
1 C 16x 6 dx:
M4 D 1:03363
M8 D 1:03374
M16 D 1:00376:
s
x2
dx D
1C
4 " 2x 2
s
4 " x2
dx
4 " 2x 2
p
The length of the short arc from .0; 1/ to .1; 1= 2/ is
a
Z
p
1 C .4x 3 /2 dx D
1
. We
9x 4=3
For the ellipse 3x 2 C y 2 D 3, we have 6x C 2yy 0 D 0, so
y 0 D "3x=y. Thus
ds D
22.
18. The required length is
LD
1C
Thus the length is approximately 1.0338 units.
2=3
x !1=3 dx
0
ˇ
3 2=3 ˇˇa
x ˇ D 6a units.
2
0
f .x/ dx, where f .x/ D
1
T4 D 1:03406
T8 D 1:03385
T16 D 1:03378
Thus the length of the whole curve is
Z
R2
1 !2=3
x
.
3
r
23.
1
0
s
4 " x2
dx ' 1:05810 units
4 " 2x 2
(with a little help from Maple’s numerical integration
routine).
Z 2 p
S D 2#
jxj 1 C 4x 2 dx Let u D 1 C 4x 2
0
du D 8x dx
$ˇ17
Z 17
%
p
# 2 3=2 ˇˇ
#
u du D
D
u
ˇ
ˇ
4 1
4 3
1
p
#
D .17 17 " 1/ sq: units:
6
p
y D x 3 , 0 $ x $ 1. ds D 1 C 9x 4 dx.
The area of the surface of rotation about the x-axis is
S D 2#
D
To four decimal places the length is 1.6002 units.
#
18
Z
1
0
Z
p
x 3 1 C 9x 4 dx
10
1
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p
u du D
Let u D 1 C 9x 4
du D 36x 3 dx
#
.103=2 " 1/ sq. units.
27
279
lOMoARcPSD|6566483
SECTION 7.3 (PAGE 412)
24.
ADAMS and ESSEX: CALCULUS 9
q
y D x 3=2 , 0 $ x $ 1. ds D 1 C 94 x dx.
The area of the surface of rotation about the x-axis is
S D 2#
Z
1
x
0
128#
243
D
3=2
128#
D
243
128#
D
243
Z
3=2
0
r
9x
1C
dx
4
p
u4 1 C u2 du
Z tan!1 .3=2/
25.
2
Let 9x D 4u
9 dx D 8u du
Let u D tan v
du D sec2 v dv
D
0
5
3
At this stage it is convenient to use the reduction formula
26.
Z
secn v dv D
n"2
1
secn!2 v tan v C
n"1
n"1
Z
secn!2 v dv
(see Exercise 36 of Section 7.1) to reduce the powers of
secant down to 3, and then use
Z
a
0
sec3 v dv D
1
.sec a tan a C ln j sec a C tan aj:
2
D
D
D
D
Z
Z
32#
81
Let u D 1 C
du D
13=4
1
#
9x
dx
4
9x
4
9
dx
4
p
.u " 1/ u du
2 5=2 2 3=2
u
" u
5
3
$ˇˇ13=4
ˇ
ˇ
ˇ
1
.13=4/5=2 " 1 .13=4/3=2 " 1
"
5
3
!
sq. units.
We have
Z
1
ex
p
Let e x D tan &
e x dx D sec2 & d&
Z xD1
Z xD1 p
D 2#
1 C tan2 & sec2 & d& D 2#
sec3 & d&
xD0
xD0
(
)ˇˇxD1
ˇ
D # sec & tan & C ln j sec & C tan & j ˇ
:
ˇ
S D 2#
0
1 C e 2x dx
xD0
Since
p
1 C e2 ;
p
x D 0 ) tan & D 1; sec & D 2;
a
.sec7 v " 2 sec5 v C sec3 v/ dv
ˇa #
$Z a
Z a
sec5 v tan v ˇˇ
5
"2
sec5 v dv C
sec3 v dv
ˇ C
ˇ
6
6
0
0
0
ˇa
#
"
Z
sec5 a tan a 7 sec3 v tan v ˇˇ
3 a
3
"
sec v dv
ˇ C
ˇ
6
6
4
4 0
0
Z a
sec3 v dv
C
0
27.
Z
1 a
sec5 a tan a 7 sec3 a tan a
3
"
C
sec v dv
6
24
8 0
sec a tan a C ln j sec a C tan aj
sec5 a tan a 7 sec3 a tan a
"
C
:
6
24
16
0
Substituting a D arct an.3=2/ now gives the following
value for the surface area:
p
p !
28 13#
8#
3 C 13
SD
C
ln
sq. units.
81
243
2
280
0
x D 1 ) tan & D e; sec & D
We have
I D
x 1C
64#
D
81
.sec v " 2 sec v C sec v/ dv:
r
1
32#
D
81
tan4 v sec3 v dv
7
Z
S D 2#
0
Z tan!1 .3=2/
If y D x 3=2 , 0 $ x $ 1, is rotated about the y-axis, the
surface area generated is
therefore
)
( p
p
p
p
S D # e 1 C e 2 C ln j 1 C e 2 C ej " 2 " ln j 2 C 1j
"
#
p
p
p
1 C e2 C e
2
D # e 1 C e " 2 C ln p
sq: units:
2C1
If y D sin x; 0 $ x $ #, is rotated about the x-axis, the
surface area generated is
S D 2#
D 2#
D 2#
Z
!
0
Z
1
!1
Z
p
sin x 1 C cos2 dx
Let u D cos x
du D " sin x dx
p
1 C u2 du
!=4
!!=4
Let u D tan &
du D sec2 & d&
Z !=4
sec3 & d& D 4#
sec3 & d&
0
&ˇˇ!=4
D 2# sec & tan & C ln j sec & C tan & j ˇˇ
0
%p
p &
2 C ln.1 C 2/ sq. units.
D 2#
%
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INSTRUCTOR’S SOLUTIONS MANUAL
28.
SECTION 7.3 (PAGE 412)
# 2
$2 # 2
$2
x
x
1
1
D
1 C .y 0 /2 D 1 C
" 2
C 2
4
x
4
x
$# 2
$
Z 4# 3
1
x
1
x
C
C 2 dx
S D 2#
12
x
4
x
1
$
Z 4# 5
x
x
1
C C 3 dx
D 2#
48
3
x
1
$ ˇˇ4
# 6
2
x
1
x
ˇ
C
" 2 ˇ
D 2#
ˇ
288
6
2x
32.
dy
"x
D p
, and
dx
2 4 " x2
S D 2 & 2#
1
D
The top half of x 2 C 4y 2 D 4 is y D
D#
Z
D#
Z
275
# sq: units:
8
1
x3
C , 1 $ x $ 4, we have
29. For y D
x$
# 212
1
x
C 2 dx.
ds D
4
x
The surface generated by rotating the curve about the yaxis has area
$
Z 4 # 2
x
1
S D 2#
x
C 2 dx
4
x
1
$ˇˇ4
# 4
x
ˇ
C ln jxj ˇ
D 2#
ˇ
16
$ 1
#
255
C ln 4 sq. units.
D 2#
16
30. The area of the cone obtained by rotating the line
y D .h=r/x, 0 $ x $ r, about the y-axis is
ˇr
p
Z r p
2 C h2 x 2 ˇ
r
ˇ
S D 2#
x 1 C .h=r/2 dx D 2#
ˇ
r
2 ˇ
0
0
p
D # r r 2 C h2 sq. units.
Thus s
ds D
1C
dy
D0
dx
)
dy
x"b
D"
:
dx
y
.x " b/2
a
a
dx D dx D p
dx
y2
y
a2 " .x " b/2
(if y > 0).
The surface area of the torus obtained by rotating the circle about the line x D 0 is
Z bCa
a
dx Let u D x " b
S D 2 & 2#
xp
a2 " .x " b/2
b!a
du D dx
Z a
uCb
p
D 4#a
du
2
2
!a
Z a a "u
du
p
by symmetry
D 8#ab
a2ˇ " u2
0
a
u ˇˇ
D 8#ab sin!1 ˇ D 4# 2 ab sq. units.
aˇ
0
p
0
!=3
0
2
p
0
4 " x2
2
s
#
x
$2
p
dx
2 4 " x2
r
16
Let x D
sin &
3
r
16
dx D
cos & d&
3
16 " 3x 2 dx
1C
4
.4 cos & / p cos & d&
3
Z
16# !=3
cos2 & d&
D p
3 0
#
$ˇˇ!=3
8#
ˇ
D p & C sin & cos & ˇ
ˇ
3
0
p
2#.4# C 3 3/
p
sq: units:
D
3 3
33.
For the ellipse x 2 C 4y 2 D 4 we have
2x
dx
C 8y D 0
dy
)
dx
y
D "4 :
dy
x
The arc length element on the ellipse is given by
ds D
31. For the circle .x " b/2 C y 2 D a2 we have
2.x " b/ C 2y
2
Z
1p
4 " x 2 , so
2
D
s
1C
s
1C
#
dx
dy
$2
dy
16y 2
1p
dy D
4 C 12y 2 dy:
2
x
x
If the ellipse is rotated about the y-axis, the resulting surface has area
Z 1
1p
S D 2 & 2#
x
4 C 12y 2 dy
x
0
Z 1p
p
1 C 3y 2 dy
Let 3y D tan &
D 8#
p
0
3dy D sec2 & d&
Z !=3
8#
D p
sec3 & d&
3 0
&ˇˇ!=3
8# %
D p sec & tan & C ln j sec & C tan & j ˇˇ
2 3
0
p &
8# % p
D p 2 3 C ln.2 C 3/
2 3
p !
ln.2 C 3/
p
sq. units.
D 8# 1 C
2 3
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SECTION 7.3 (PAGE 412)
ADAMS and ESSEX: CALCULUS 9
Thus, the surface area depends only on the radius R of
the sphere, and the distance .b " a/ between the parellel
planes.
34. As in Example 4, the arc length element for the ellipse is
y
v
u
2
2
s
u 2 a "b 2
# $2
x
ta "
2
dy
a
ds D 1 C
dx D
dx:
2
2
dx
a "x
To get the area of the ellipsoid, we must rotate both the
upper and lower semi-ellipses (see the figure for Exercise
20 of Section 8.1):
S D 2 & 2#
Z a"
r
c"b 1"
0
cCb
r
1"
% x &2
a
!#
% x &2
a
a
!
C
Fig. 7.3-36
ds
v
u
2
2
u 2 a "b 2
x
a ta "
2
a
D 8#c
dx
a2 " x 2
(0
)
1
D 8#c
of the circumference of the ellipse
4
D 8#caE."/
37. If the curve y D x k , 0 < x $ 1, is rotated about the
y-axis, it generates a surface of area
S D 2#
D 2#
p
R !=2 p
a2 " b 2
and E."/ D 0
1 " "2 sin t dt as
a
defined in Example 4.
Z
s
!=2
0
Z
s
!=2
1C
#2
cos2 t dt
4
S D 2#
S D 2#
D 2#
282
a
Z
D 2#R
p
b
p
a
Z
b
a
R2
"
x2
s
1C
R2 " x 2 p
#
dy
dx
R
R2
" x2
$2
1
0
p
x 2 C k 2 x 2k dx:
Z
1
x k dx, which is infinite.
0
Z
1
0
Z
1
p
x 1 C k 2 x 2.k!1/ dx
p
0
0
36. Let the equation of the sphere be x 2 C y 2 D R2 . Then the
surface area between planes x D a and x D b
."R $ a < b $ R/ is
b
0
Z
p
x 1 C k 2 x 2.k!1/ dx
x 2!2k C k 2 x k dx
Z 1
p
x k dx < 1:
< 2# 1 C k 2
D 2#
38.
Z
1
If k # 0, the surface area S is finite, since x k is bounded
on .0; 1! in that case.
Hence we need only consider the case "1 < k < 0. In this
case 2 < 2 " 2k < 4, and
#2 #2
1C
"
sin2 t dt
4
4
0
s
Z !=2
#2
5p
2
4C#
1"
D
sin2 t dt
#
4 C #2
0
$
#
#
5p
:
D
4 C # 2E p
#
4 C #2
10
D
#
Z
If k $ "1, we have S # 2#k
35. From Example 3, the length is
10
sD
#
x
x 2 Cy 2 DR2
Z
where " D
b
Thus the area is finite if and only if k > "1.
Z 1 r
1
S D 2#
jxj 1 C 2 dx
x
0
Z 1p
D 2#
x 2 C 1 dx Let x D tan &
0
dx D sec2 & d&
Z !=4
D 2#
sec3 & d&
0
ˇ
&ˇ!=4
ˇ
D # sec & tan & C ln j sec & C tan & j ˇ
ˇ
0
p
p
D #Π2 C ln. 2 C 1/! sq: units:
%
dx
dx
dx D 2#R.b " a/ sq: units:
39.
a) Volume V D #
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R 1 dx
D # cu. units.
1
x2
lOMoARcPSD|6566483
INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 7.4 (PAGE 419)
b) The surface area is
S D 2#
> 2#
Z
1
Z 11
1
r
1
1
1 C 4 dx
x
x
dx
D 1:
x
3.
c) Covering a surface with paint requires applying a
layer of paint of constant thickness to the surface. Far
to the right, the horn is thinner than any prescribed
constant, so it can contain less paint than would be
required to cover its surface.
The mass of the plate is m D )0 & area D
The moment about x D 0 is
MxD0 D
a
0
)0
D
2
p
x)0 a2 " x 2 dx
Z
a2
p
u du
0
0
The mass of the wire is
Z L
Z L
#s
ds
mD
ı.s/ ds D
sin
L
0
0
ˇL
#s ˇˇ
L
2L
:
D " cos
ˇ D
#
Lˇ
#
MxD0
)0 a3
4
4a
D
D
. By symmetry,
m
3 #)0 a2
3#
x. $Thus the centre of mass of the plate is
4a
.
3#
Thus x D
y
# D
4a
;
3#
y
0
p
yD a2 !x 2
Since ı.s/ is symmetric about s D L=2 (that is,
ı..L=2/ " s/ D ı..L=2/ C s/), the centre of mass is at the
midpoint of the wire: s D L=2.
2. A slice of the wire of width dx at x has volume
d V D #.a C bx/2 dx. Therefore the mass of the whole
wire is
Z L
mD
ı0 #.a C bx/2 dx
0
Z L
D ı0 #
.a2 C 2abx C b 2 x 2 / dx
0
#
$
1
D ı0 # a2 L C abL2 C b 2 L3 :
3
Its moment about x D 0 is
Z L
MxD0 D
xı0 #.a C bx/2 dx
0
Z L
D ı0 #
.a2 x C 2abx 2 C b 2 x 3 / dx
#0
$
1 2 2 2
1
D ı0 #
a L C abL3 C b 2 L4 :
2
3
4
Thus, the centre of mass is
$
#
1
1 2 2 2
a L C abL3 C b 2 L4
ı0 #
2
3
4
$
#
xD
1
ı0 # a2 L C abL2 C b 2 L3
3
#
$
1 2 2
1 2 2
L a C abL C b L
2
3
4
D
:
1
a2 C abL C b 2 L2
3
Let u D a2 " x 2
du D "2x dx
ˇa 2
)0 2 3=2 ˇˇ
)0 a3
D
u ˇ D
:
ˇ
2 3
3
Section 7.4 Mass, Moments, and Centre of
Mass (page 419)
1.
Z
#)0 a2
.
4
dx
x
a
x
Fig. 7.4-3
4.
p
A vertical strip has area dA D a2 " x 2 dx. Therefore,
the mass of the quarter-circular plate is
Z
a
p
.)0 x/ a2 " x 2 dx
Let u D a2 " x 2
du D "2x dx
#
$ˇa2
Z a2
p
2 3=2 ˇˇ
1
1
1
u
D )0
u du D )0
ˇ D )0 a3 :
ˇ
2
2
3
3
0
mD
0
0
The moment about x D 0 is
MxD0 D
Z
a
0
p
)0 x 2 a2 " x 2 dx
D )0 a4
Z
!=2
Let x D a sin &
dx D a cos & d&
sin2 & cos2 & d&
0
Z
)0 a4 !=2 2
sin 2& d&
4
0
Z
#)0 a4
)0 a4 !=2
.1 " cos 4& / d& D
:
D
8
16
0
D
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SECTION 7.4 (PAGE 419)
ADAMS and ESSEX: CALCULUS 9
The moment about y D 0 is
MyD0 D
1
)0
2
D
1
)0
2
Z
The moment about x D 0 is
$
Z 3#
h3
dh
MxD0 D 10
h2 "
3
0
ˇ
# 3
$3
h
h4 ˇˇ
45
D 10
"
kg-m.
ˇ D
3
12 ˇ
2
a
x.a2 " x 2 / dx
!ˇa
a2 x 2
x 4 ˇˇ
1
"
ˇ D a4 )0 :
2
4 ˇ
8
0
0
0
The moment about y D 0 is
$#
$
Z 3 #
1
2
1
MyD0 D 10
2 " h h " h2 dh
3
3
0 2
$
Z 3#
2 2 1 3
D 10
h " h C h dh
3
9
0
ˇ3
# 2
3
4 $ˇ
h
2h
h ˇ
15
D 10
"
C
kg-m:
ˇ D
2
9
36 ˇ
2
3
3
#a and y D a. Hence, the centre of mass
16
8
3
3
is located at . #a; a/.
16
8
5. The mass of the plate is
Thus, x D
mD2
Z
4
0
D 2k
D 2k
Z
p
ky 4 " y dy
4
0
#
1=2
.4 " u/u
Let u D 4 " y
du D "dy
0
45
3
2
Thus, x D
D and y D
15
2
of mass is located at . 32 ; 21 /.
du
8 3=2 2 5=2
u
" u
3
5
$
#
$ˇˇ4
256k
ˇ
:
ˇ D
ˇ
15
0
MyD0 D 2
D 2k
4
0
Z
p
ky 2 4 " y dy
4
1=2
.16u
3=2
" 8u
4096k
15
16
!
D
. The centre of mass of the
105
256k
7
plate is .0; 16=7/.
y
p
xD 4!y
x
3
Fig. 7.4-6
7.
The mass of the plate is
Z a
ka3
:
mD
kx a dx D
2
0
By symmetry, y D a=2.
Z
MxD0 D
4
a
0
kx 2 a dx D
ka4
:
3
4
2
2a
ka
Thus x D
!
D
. The centre of mass of the
3
# 3 $ ka3
2a a
plate is
;
.
3 2
x
density ky
!2
2
y
a
Fig. 7.4-5
6. A vertical strip at h has area dA D .2 " 23 h/ dh. Thus, the
mass of the plate is
#
$
$
Z 3#
2
h2
.5h/ 2 " h dh D 10
dh
h"
3
3
0
0
$ˇ3
# 2
h3 ˇˇ
h
"
D 10
ˇ D 15 kg:
2
9 ˇ
mD
Z
3
0
284
1
. The centre
2
dh
h
Cu
Thus y D
D
2
yD2! 3 x
5=2
0
$
2
Let u D 4 " y
du D "dy
/ du
#
$ˇ4
32 3=2 16 5=2 2 7=2 ˇˇ
4096k
D 2k
u
" u
C u
:
ˇ D
ˇ
3
5
7
105
0
15
2
15
y
By symmetry, MxD0 D 0, so x D 0.
Z
#
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density
kx
a
Fig. 7.4-7
x
lOMoARcPSD|6566483
INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 7.4 (PAGE 419)
#
$
a
8. A vertical strip has area dA D 2 p " r dr. Thus, the
2
mass is
$)
Z a=p2 ( #
a
mD2
kr 2 p " r dr
2
0
$
Z a=p2 #
a
k
D 4k
p r " r 2 dr D p a3 g:
2
3 2
0
50
.8000/k
40
Thus, x D 3
D
. Since the density is inde10000k
3
5
pendent of y and z, y D and z D 5. Hence, the centre
2
of mass is located on the 20 cm long central axis of the
brick, two-thirds of the way from the least dense 10 & 5
face to the most dense such face.
y
5
Since the mass is symmetric about the y-axis, and the
plate is symmetric about both the x- and y-axis, therefore
the centre of mass must be located at the centre of the
square.
y
a
p
2
dx
a
yD p !x
2
a
p
dr
2
r
x
x
20
10
z
Fig. 7.4-10
x
Fig. 7.4-8
9.
mD
MxD0 D
Z
b
a
Z
b
a
Z
%
&
).x/ g.x/ " f .x/ dx
%
&
x).x/ g.x/ " f .x/ dx
11.
%
&
1
MyD0 D
x).x/ .g.x//2 " .f .x//2 dx
2 a
#
$
MxD0 MyD0
;
:
Centre of mass:
m
m
b
y
Choose axes through the centre of the ball as shown in the
following figure. The mass of the ball is
Z
R
.y C 2R/#.R2 " y 2 / dy
$ˇR
#
8
y 3 ˇˇ
2
D 4#R R y "
ˇ D #R4 kg:
3 ˇ
3
mD
yDg.x/
!R
0
density %.x/
By symmetry, the centre of mass lies along the y-axis; we
need only calculate y.
yDf .x/
a
b
x
MyD0 D
Fig. 7.4-9
MxD0 D 50k
Z
0
20
50 3 ˇˇ20
kx ˇ
x dx D
0
3
2
50
.8000/k g-cm:
D
3
!R
D 2#
Z
y.y C 2R/#.R2 " y 2 / dy
R
0
0
0
The moment about x D 0, i.e., the yz-plane, is
R
y 2 .R2 " y 2 / dy
#
$ˇR
3
y 5 ˇˇ
4
2y
D 2# R
"
#R5 :
ˇ D
3
5 ˇ
15
10. The slice of the brick shown in the figure has volume
d V D 50 dx. Thus, the mass of the brick is
Z 20
ˇ20
ˇ
mD
kx50 dx D 25kx 2 ˇ D 10000k g:
0
Z
4#R5
R
3
Thus y D
D
!
: The centre of mass is on
4
15
8#R
10
the line through the centre of the ball perpendicular to the
plane mentioned in the problem, at a distance R=10 from
the centre of the ball on the side opposite to the plane.
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SECTION 7.4 (PAGE 419)
ADAMS and ESSEX: CALCULUS 9
y
z
y
a
x
z
yC2R
!R
p
a2 !z 2
!a
a
y
a
!2R
x
Fig. 7.4-13
Fig. 7.4-11
12. A slice at height z has volume d V D #y 2 dz and density
kz g=cm3 . Thus, the mass of the cone is
Z
d m D *0 z dz
b
kz#y 2 dz
$
Z b #
z 2
dz
D #ka2
z 1"
b
0
$ˇb
# 2
2z 3
z 4 ˇˇ
z
"
C 2 ˇ
D #ka2
2
3b
4b ˇ
mD
A horizontal slice of the solid
p at height z with thickness
dz is a half-disk
of radius a2 " z 2 with centre of mass
p
4 a2 " z 2
at x D
, by Exercise 3 above. Its mass is
3#
0
and its moment about x D 0 is
dMxD0
0
1
D
#ka2 b 2 g:
12
The moment about z D 0 is
MzD0 D #ka2
Z
0
b
#
$
1
z 2
z2 1 "
dz D
#ka2 b 3 g-cm:
b
30
2b
. Hence, the centre of mass is on the axis
5
of the cone at height 2b=5 cm above the base.
Thus, z D
# 2
.a " z 2 /;
2
p
2
2
#*0
2
2 4 a "z
D dm x D
z.a " z /
2
3#
2*0
2
2 3=2
z.a " z / :
D
3
Thus the mass of the solid is
Z
#*0 a 2
.a z " z 3 / dz
mD
2 0
#
$ˇa
#*0 a2 z 2 z 4 ˇˇ
#*0 a4
D
"
:
ˇ D
ˇ
2
2
4
8
0
Also,
MzD0
z
b
#*0
D
2
Z
a
.a2 z 2 " z 4 / dz
#
$ˇa
#*0 a2 z 3 z 5 ˇˇ
#*0 a5
D
"
;
ˇ D
2
3
5 ˇ
15
0
0
dz
z
!
yDa 1!
z
b
"
y
a
Fig. 7.4-12
13. By symmetry, y D 0.
286
and z D
Finally,
#*0 a5
8a
8
D
!
.
15
#*0 a4
15
Z
2*0 a
z.a2 " z 2 /3=2 dz
3 0
Z 2
*0 a 3=2
u du
D
3 0
#
$ˇa2
*0 2 5=2 ˇˇ
2*0 a5
D
u
;
ˇ D
ˇ
3 5
15
MxD0 D
0
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Let u D a2 " z 2
du D "2z dz
lOMoARcPSD|6566483
INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 7.4 (PAGE 419)
2*0 a5
8
16a
!
D
.
15
#*0 a4# 15
$
8a
16a
; 0;
:
The centre of mass is
15
15
ka4
3
3a
!
D
. By symmetry, x D 0.
2
k#a3
2#
#
$
3a
Thus, the centre of mass of the plate is 0;
.
2#
Therefore, y D
so x D
14.
Assume the cone has its base in the xy-plane and its
vertex at height b on the z-axis. By symmetry, the centre of mass lies on the z-axis. A cylindrical shell of
thickness dx and radius x about the z-axis has height
z D b.1 " .x=a//. Since it’s density is constant kx, its
mass is
%
x&
dx:
d m D 2#bkx 2 1 "
a
Also its centre of mass is at half its height,
y shell D
16.
y
b%
x&
1"
:
2
a
ds
Thus its moment about z D 0 is
Hence
%
x &2
dx:
dMzD0 D y shell d m D #bkx 2 1 "
a
MzD0
Z
%
x&
#kba3
2#bkx 2 1 "
dx D
a
6
0
Z a
%
x &2
#kb 2 a3
2
D
#bkx 1 "
dx D
a
30
0
mD
s
L
!
"
x
a
and z D MzD0 =m D b=5. The centre of mass is on the
axis of the cone at height b=5 cm above the base.
15.
y
x 2 Cy 2 Da2
ds
" d"
!a
Fig. 7.4-16
L
The radius of the semicircle is . Let s measure the
#
distance along the wire from the point where it leaves
the positive% x-axis.
Thus, the density at position s is
#s &
ı.s/ D sin
g=cm. The mass of the wire is
L
mD
s
xa
Fig. 7.4-15
Consider the area element which is the thin half-ring
shown in the figure. We have
d m D ks #s ds D k# s ds:
k# 3
a .
3
Regard this area element as itself composed of smaller
elements at positions given by the angle & as shown. Then
$
#Z !
.s sin & /s d& ks ds
dMyD0 D
MyD0
0
ˇL
#s
L
#s ˇˇ
2L
sin
ds D " cos
g:
ˇ D
L
#
Lˇ
#
0
Since an arc element ds at position s is at height
L
#s
L
sin & D
sin
, the moment of the wire about
y D
#
#
L
y D 0 is
Thus, m D
0
L
MyD0 D
2
D 2ks 3 ds;
Z a
ka4
D 2k
:
s 3 ds D
2
0
Z
Z
L
0
L 2 #s
sin
ds
#
L
Let & D #s=L
d& D #ds=L
$ Z
L 2 ! 2
sin & d&
#
0
&ˇˇ!
2
L2 %
ˇ D L g-cm:
D
&
"
sin
&
cos
&
ˇ
2# 2
2#
0
D
#
Since the wire and the density function are both symmetric about the y-axis, we have MxD0 D 0.
$
#
L
.
Hence, the centre of mass is located at 0;
4
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SECTION 7.4 (PAGE 419)
17.
Z
1
ADAMS and ESSEX: CALCULUS 9
y
2
C e !kr .4# r 2 / dr
0
Z 1
p
2
D 4# C
r 2 e !kr dr Let u D k r
p
0
du D k dr
Z 1
4# C
2 !u2
u e
du
D 3=2
k
0
mD
dx
2
d V D ue !u du
2
V D " 12 e !u
0
1
ˇR
Z R
!u2 ˇ
4# C
"ue
1
2
ˇ
D 3=2 lim @
e !u duA
ˇ C
ˇ
R!1
2
2 0
k
0
#
$
Z
4# C
1 1 !u2
D 3=2 0 C
e
du
2 0
k
p
% # &3=2
5:57C
4# C #
DC
' 3=2 :
D 3=2
4
k
k
k
U Du
d U D du
18.
rD
D
D
D
D
1
m
Z
x
2.
r
MyD0 D 2
D4
p
Z
Z
9
0
3
0
ˇ9
# $
ˇ
2
3=2 ˇ
.9 " y/ ˇ D 36
9 " y dy D "2
ˇ
3
0
p
y 9 " y dy
ˇ3
ˇ
648
:
.9u2 " u4 / du D 4.3u3 " 15 u5 /ˇˇ D
5
0
y
9
yD9!x 2
dy
!3
y
3
x
Fig. 7.5-2
The area and moments of the region are
Z
D
Z
MxD0 D
Z
p
x r 2 " x 2 dx
4
4r
r3
!
D
D y by symmetry:
3 ##r 2
3#$
4r 4r
;
.
The centroid is
3# 3#
Let u2 D 9 " y
2u du D "dy
648
18
D
. Hence, the centroid is at
Thus, y D
5 & 36
5
#
$
18
0;
.
5
1
p
0
!=4
dx
1 C x2
Let x D tan &
dx D sec2 & d&
sec & d&
0
ˇ!=4
p
ˇ
D ln j sec & C tan & jˇˇ
D ln.1 C 2/
0
288
0
AD
Let u D r 2 " x 2
0
du D "2x dx
ˇr 2
Z r2
1
u3=2 ˇˇ
r3
1=2
D
u du D
ˇ D
ˇ
2 0
3
3
xD
9
(page 424)
#r2
AD
Z4
MxD0 D
Z
2
3.
1.
x D 0. A horizontal strip at y has
By symmetry,
p
p mass
d m D 2 9 " y dy and moment dMyD0 D 2y 9 " y dy
about y D 0. Thus,
mD2
rC e !kr .4# r 2 / dr
0
Z 1
4# C
2
r 3 e !kr dr Let u D kr 2
C # 3=2 k !3=2 0
du D 2kr dr
Z 1
4k 3=2 1
!u
ue du
p
# 2k 2 0
U Du
d V D e !u du
d U D du 0 V D "e !u
1
ˇR Z
ˇ
R
2
ˇ
lim @"ue !u ˇ C
e !u duA
p
ˇ
#k R!1
0
0
&
2
2 %
0
0 C lim .e " e !R D p :
p
R!1
#k
#k
Section 7.5 Centroids
x
r
Fig. 7.5-1
and
1
p
yD r 2 !x 2
r
MyD0 D
0
1
0
1
2
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Z
p
1
0
x dx
ˇ1
p
p
ˇ
D 1 C x 2 ˇˇ D 2 " 1
1 C x2
0
ˇ1
ˇ
dx
1
#
!1 ˇ
D tan x ˇ D :
1 C x2
2
8
0
lOMoARcPSD|6566483
INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 7.5 (PAGE 424)
p
2"1
#
p , and y D
p . The
ln.1 C 2/
8 ln.1
C 2/
!
p
2"1
#
centroid is
p ;
p
.
ln.1 C 2/ 8 ln.1 C 2/
Thus x D
y
yD p
5.
By symmetry, x D 0. We have
AD2
1
p
0
D2 4
1Cx 2
1
Z
x
MyD0
Fig. 7.5-3
3 %p
Z
&
4 " x 2 " 1 dx
!=3
0
cos2 & d& "
p
3
Let x D 2 sin &
! dx D 2 cos & d&
ˇ!=3
p
ˇ
D 4.& C sin & cos & /ˇˇ
"2 3
0
p !
p
#
4# p
3
D4
C
"2 3D
" 3
3
4
3
Z p3 %p
&2
1
4 " x 2 " 1 dx
D 2&
2 0
Z p3 %
&
p
D
5 " x 2 " 2 4 " x 2 dx
0
p
Z 3p
p
p
4 " x 2 dx
D5 3" 3"2
0
p !
p
p
#
3
4#
D4 3"4
C
D3 3"
:
3
4
3
p
p
9 3 " 4#
3
9 3 " 4#
!
p D
p . The
3
4#! " 3 3
4# " 3 3
p
9 3 " 4#
centroid is 0;
p .
4# " 3 3
Thus y D
4. The area of the sector is A D
x D 0 is
Z
p
r= 2
1
2
8 #r .
Its moment about
y
Z
p
x r 2 " x 2 dx
0
ˇr
ˇ
1 2
r3
r3
2 3=2 ˇ
D p " .r " x / ˇ
D p :
ˇ p
6 2 3
3 2
r= 2
MxD0 D
x 2 dx C
r
1
p
r= 2
r3
8r
8
D p . By symmetry, the
p &
2
#
r
3 2
3 2##
$
p
#
centroid must lie on the line y D x tan
D x. 2 " 1/.
8
p
8r. 2 " 1/
Thus, y D
p
.
3 2#
p
! 3
Thus, x D
y
yDx
p x
3
Fig. 7.5-5
6.
By symmetry, x D 0. The area is A D
moment about y D 0 is
MyD0
p
yD r 2 !x 2
p
yD 4!x 2 !1
1
2 #ab.
The
"
# $2 #
Z a
x
x2
b 1"
dx D b 2
1 " 2 dx
a
a
!a
0
ˇa
#
$
x 3 ˇˇ
2
D b 2 x " 2 ˇ D ab 2 :
3a ˇ
3
1
D
2
Z
a
2
0
pr
Fig. 7.5-4
r
2
x
Thus, y D
2
4b
2ab 2
&
D
.
3
#ab
3#
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SECTION 7.5 (PAGE 424)
ADAMS and ESSEX: CALCULUS 9
y
8.
yDb
q
1!
x2
a2
dx
!a
x
ax
Fig. 7.5-6
7.
The quadrilateral consists of two triangles, T1 and T2 , as
shown in the figure. The area and centroid of T1 are given
by
4&1
D 2;
2
0C3C4
7
x1 D
D ;
3
3
The region is the union of a half-disk and a#triangle.
$ The
4
and
centroid of the half-disk is known to be at 1;
3#
$
#
2 2
; " . The area of the semithat of the triangle is at
3 3
#
circle is
and the triangle is 2. Hence,
2
# $
%# &
2
3# C 8
MxD0 D
.1/ C .2/
D
I
2
3
6
# $
%# &# 4 $
2
2
MyD0 D
C .2/ "
D" :
2
3#
3
3
#
C 2, then
Since the area of the whole region is
2
3# C 8
4
xD
and y D "
.
3.# C 4/
3.# C 4/
A1 D
y
0C1C0
1
y1 D
D :
3
3
The area and centroid of T2 are given by
4&2
D 4;
2
0C2C4
D 2;
x2 D
3
y2 D
M1;yD0
M2;xD0 D 2 & 4 D 8
2
8
M2;yD0 D " & 4 D " :
3
3
Since areas and moments are additive, we have for the
whole quadrilateral
A D 2 C 4 D 6;
14
38
MxD0 D
C8D
;
3
3
2
x
Fig. 7.5-8
9.
14
3
2
3
1
!2
2
0"2C0
D" :
3
3
It follows that
7
&2D
3
1
D &2D
3
1!.x!1/2
yDx!2
A2 D
M1;xD0 D
p
yD
MyD0 D
2 8
" D "2:
3 3
38
19
"2
1
D
, and y D
D " . The centroid
Thus x D
3&6
9#
3
$ 6
19 1
of the quadrilateral is
;" .
9
3
A circular strip of the surface between heights y and
y C dy has area
dS D 2#x
r
dy
D 2#x dy D 2# r dy:
cos &
x
The total surface area is
S D 2# r
Z
0
r
dy D 2# r 2 :
The moment about y D 0 is
ˇr
Z r
ˇ
MyD0 D 2# r
y dy D # r.y 2 /ˇˇ D # r 3 :
0
0
3
#r
r
D . By symmetry, the centroid of the
Thus y D
2# r 2
2
hemispherical surface is on the axis of symmetry of the
hemisphere. It is halfway between the centre of the base
circle and the vertex.
y
y
.3;1/
T1
"
4
x
.x;y/
dS
T2
r
y
"
x
.2;!2/
Fig. 7.5-9
Fig. 7.5-7
290
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x
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 7.5 (PAGE 424)
z
h
10. By symmetry, x D y D 0. The volume is V D 32 # r 3 . A
thin slice of the solid at height z will have volume
d V D #y 2 dz D #.r 2 " z 2 / dz. Thus, the moment about
z D 0 is
Z
dz
z
r
!
z#.r 2 " z 2 / dz
$ ˇr
# 2 2
z 4 ˇˇ
#r4
r z
"
:
D#
ˇ D
2
4 ˇ
4
MzD0 D
yDr 1!
0
y
#r4
3r
3
D
&
. Hence, the centroid is on
4
2# r 3
8
the axis of the hemisphere at distance 3r=8 from the base.
Thus, z D
z
p
yD r 2 !z 2
Fig. 7.5-11
12. A band
$ z with vertical width dz has radius
# at height
z
, and has actual (slant) width
y Dr 1"
h
s
s
# $2
dy
r2
dz D 1 C 2 dz:
ds D 1 C
dz
h
Its area is
dz
y
%
z&
dA D 2# r 1 "
h
r
x
%
z &2
r2
dV D #r2 1 "
dz D # 2 .h " z/2 dz:
h
h
D
#r2
h2
#r2
D 2
h
h
0
Z
h
0
#
z.h " z/2 dz
Let u D h " z
du D "dz
.h " u/u2 du
hu3 u4
"
3
4
r2
dz:
h2
0
We have
Z
1C
The moment about z D 0 is
s
#
$
Z
z
r2 h
z 1"
dz
MzD0 D 2# r 1 C 2
h 0
h
s
ˇ
#
$h
p
r 2 z2
z 3 ˇˇ
1
D 2# r 1 C 2
"
ˇ D # rh r 2 C h2 :
h
2
3h ˇ
3
The cone has volume V D 31 # r 2 h. (See the following
figure.) The disk-shaped
slice with vertical width dz has
%
z&
, and therefore has volume
radius y D r 1 "
h
#r2
h2
s
Thus the area of the conical surface is
s
Z
p
r2 h %
z&
1"
dz D # r r 2 C h2 :
A D 2# r 1 C 2
h 0
h
Fig. 7.5-10
MzD0 D
"
r
0
11.
z
h
$ˇˇh
# r 2 h2
ˇ
:
ˇ D
ˇ
12
0
# r 2 h2
3
h
Therefore z D
!
D . The centroid of the
12
# r 2h
4
solid cone is on the axis of the cone, at a distance above
the base equal to one quarter of the height of the cone.
13.
p
# rh r 2 C h2
h
1
p
D . By
Thus, z D
&
3
3
# r r 2 C h2
symmetry, x D y D 0. Hence, the centroid is on the axis
of the conical surface, at distance h=3 from the base.
#
By symmetry, x D . The area and y-moment of the
2
region are given by
Z !
AD
sin x dx D 2
0
Z !
1
MyD0 D
sin2 x dx
2 0
ˇ!
ˇ
1
#
D .x " sin x cos x/ˇˇ D :
4
4
0
%
&
# #
#
;
.
Thus y D , and the centroid is
8
2 8
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SECTION 7.5 (PAGE 424)
ADAMS and ESSEX: CALCULUS 9
y
Hence x D r 2 !
2r
2
D
, and the centroid is
#r
#
y
#
$
2r 2r
;
.
# #
yDsin x
x 2 Cy 2 Dr 2
r
ds
!
!=2
x
r
Fig. 7.5-13
14.
"
The area of the region is
x xCdx
ˇ!=2
Z !=2
ˇ
ˇ
AD
cos x dx D sin x ˇ
D 1:
ˇ
0
r
x
Fig. 7.5-15
0
The moment about x D 0 is
Z !=2
MxD0 D
x cos x dx
16.
0
U Dx
d V D cos x dx
d U D dx
V D sin x
ˇ!=2 Z
ˇ
!=2
#
ˇ
"
sin x dx D " 1:
D x sin x ˇ
ˇ
2
0
By symmetry, the centroid of the solid lies on its vertical
axis of symmetry; let us continue to call this the y-axis.
We need only determine y S . Since D lies between y D 1
and y D 2, its centroid satisfies yD D 3=2. Also, by Exercise 11, the centroid of the solid cone satisfies y C D 3=4.
Thus C and D have moments about y D 0:
0
#
Thus, x D " 1. The moment about y D 0 is
2
Z
1 !=2
MyD0 D
cos2 x dx
2 0
#
$ˇˇ!=2
1
#
1
ˇ
x C sin 2x ˇ
D :
D
ˇ
4
2
8
MC;yD0 D
y
#
1
yDcos x
#
. The centroid is
8
$
#
#
" 1;
.
2
8
17.
!
2
x
#r
. By symmetry, x D y. An
2
element of the arc between x and x C dx has length
The arc has length L D
ds D
r dx
r dx
dx
D
D p
:
sin &
y
r 2 " x2
Thus
MxD0 D
292
Z
r
0
$# $
3
D #;
4
MD;yD0
# $
3
D .4#/
D 6#:
2
# $
2
8
3x D 2.1/ C 1
D
3
3
# $
# $
3
2
11
C1
D
:
3y D 2
2
3
3
Fig. 7.5-14
15.
4#
3
The region in figure (a) is the union of a rectangle of area
2 and centroid .1; 3=2/ and a triangle of area 1 and centroid .2=3; 2=3/. Therefore its area is 3 and its centroid is
.x; y/, where
dx
x
#
Thus MS;yD0 D # C 6# D 7#, and
z S D 7#=.16#=3/ D 21=16. The centroid of the solid
S is on its vertical axis of symmetry at height 21/16 above
the vertex of the conical part.
0
Thus, y D
The solid S in question consists of a solid cone C with
vertex at the origin, height 1, and top a circular disk of
radius 2, and a solid cylinder D of radius 2 and height
1 sitting on top of the cone. These solids have volumes
VC D 4#=3, VD D 4#, and VS D VC C VD D 16#=3.
Therefore, the centroid is .8=9; 11=9/.
18. The
p region in figure (b) is the union of a square of area
. 2/2 D 2 and centroid .0; 0/ and a triangle of area 1/2
and centroid .2=3; 2=3/. Therefore its area is 5/2 and its
centroid is .x; y/, where
ˇr
p
ˇ
xr dx
p
D "r r 2 " x 2 ˇˇ D r 2 :
2
2
r "x
0
5
1
x D 2.0/ C
2
2
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2
1
D :
3
3
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 7.5 (PAGE 424)
Therefore, x D y D 2=15, and the centroid is
.2=15; 2=15/.
23.
19. The region in figure (c) is the union of a half-disk of area
#=2 and centroid .0; 4=.3#// (by Example 1) and a triangle of area 1 and centroid .0; "1=3/. Therefore its area is
.#=2/ C 1 and its centroid is .x; y/, where x D 0 and
# C2
#
yD
2
2
#
4
3#
$
C1
#
"1
3
$
D
y
1
:
3
1
T
Therefore, the centroid is .0; 2=Œ3.# C 2/!/.
20. The region in figure (d) is the union of three half-disks,
one with area #=2 and centroid .0; 4=.3#//, and two
with areas #=8 and centroids ."1=2; "2=.3#// and
.1=2; "2=.3#//. Therefore its area is 3#=4 and its centroid is .x; y/, where
3#
.x/ D
4
3#
.y/ D
4
# $
# $
#
# "1
# 1
.0/ C
C
D0
2
8
2
8 2
# $
# $
# $
4
# "2
# "2
1
#
C
C
D :
2 3#
8 3#
8 3#
2
xD2
24.
By symmetry the centroid is .1; "2/.
y
Fig. 7.5-23
p
s 3
. Its centroid is at
The altitude h of the triangle is
2
h
s
height
D p above the base side. Thus, by Pappus’s
3
2 3
Theorem, the volume of revolution is
p !
#
$
s
3s
#s 3
s
V D 2#
p
&
D
cu: units:
2
2
4
2 3
p
h
s 3
The centroid of one side is
D
above the base.
2
4
Thus, the surface area of revolution is
p !
p
3s
.s/ D s 2 # 3 sq: units:
S D 2 & 2#
4
.1;1/
yD2x!x 2
x
.1;!2/
x
1
Therefore, the centroid is .0; 2=.3#//.
21.
"
!
The triangle T has centroid 31 ; 13 and area 21 . By Pappus’s Theorem the volume of revolution about x D 2 is
#
$
1
5#
1
D
cu. units.
V D & 2# 2 "
2
3
3
yD!2
s
h
s
Fig. 7.5-21
22. The line segment from .1; 0/ to .0; 1/ has centroid . 12 ; 12 /
p
and length 2. By Pappus’s Theorem, the surface area of
revolution about x D 2 is
$
p
1 p
A D 2# 2 "
2 D 3# 2 sq: units:
2
#
y
1
1
25.
For the purpose of evaluating the integrals in this problem
and the next, the definite integral routine in the TI-85 calculator
p was used. For the region bounded by y D 0 and
y D x cos x between x D 0 and x D #=2, we have
AD
r
1
2
Fig. 7.5-24
2
3
x
Fig. 7.5-22
Z
0
!=2
p
x cos x dx ' 0:704038
Z
1 !=2 3=2
x
cos x dx ' 0:71377
A 0
Z !=2
1
x cos2 x dx ' 0:26053:
yD
2A 0
xD
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SECTION 7.5 (PAGE 424)
ADAMS and ESSEX: CALCULUS 9
29.
26. The region bounded by y D 0 and y D ln.sin x/ between
x D 0 and x D #=2 lies below the x-axis, so
AD"
Z
!=2
ln.sin x/ dx ' 1:088793
0
Z
c
"1 !=2
x ln.sin x/ dx ' 0:30239
A 0
Z
&2
"1 !=2 %
yD
ln.sin x/ dx ' "0:93986:
2A 0
xD
27.
MxD0
MyD0
MxD0 D
Z
1
0
Z
1
0
Z
1
30.
ˇR
ˇ
"1
1
dx
ˇ
D
lim
ˇ D
R!1 2.1 C x/2 ˇ
.1 C x/3
2
0
x dx
.1 C x/3
Let u D x C 1
du D dx
u"1
du
u3
1
#
$ˇR
1
1 ˇˇ
1
1
D lim " C 2 ˇ D 1 " D
ˇ
R!1
u
2u
2
2
D
MyD0
Z
0
1
Z
c
d
d h%
&2 %
&2 i
g.y/ " f .y/
dy
%
&
y g.y/ " f .y/ dy:
Let us take L to be the y-axis and suppose that a plane
curve C lies between x D a and x D b where 0 < a < b.
Thus, r D x, the x-coordinate of the centroid of C. Let
ds denote an arc length element of C at position x. This
arc length element generates, on rotation about L, a circular band of surface area dS D 2#x ds, so the surface area
of the surface of revolution is
Z xDb
S D 2#
x ds D 2#MxD0 D 2#rs:
xDa
1
1
D
2
1
D
2
Z
D
c
The area and moments of the region are
AD
By analogy with the formulas for the region a $ x $ b,
f .x/ $ y $ g.y/, the region c $ y $ d ,
f .y/ $ x $ g.y/ will have centroid .MxD0 =A; MyD0 =A/,
where
Z d%
&
AD
g.y/ " f .y/ dy
ˇR
ˇ
dx
"1
1
ˇ
D
lim
:
ˇ D
6
5
R!1 10.1 C x/ ˇ
.1 C x/
10
31.
y
0
"
!
The centroid is 1; 15 .
y
1
yD
1
.x C 1/3
L
1
t
.#=4/ " t
t
1
p
2
x
p
P
x
2
.#=4/ " t
N
Fig. 7.5-27
28. The surface
Z 1area ispgiven by
2
2
S D 2#
e !x 1 C 4x 2 e !2x dx. Since
!1
M
2
lim 1 C 4x 2 e !2x D 1, this expression must be bounded
Fig. 7.5-31
x!˙1
2
for all x, that is, 1 $ 1 C Z4x 2 e !2x $ K 2 for some con1
p
2
stant K. Thus, S $ 2#K
e !x dx D 2K# #. The
!1
integral converges and the surface area is finite. Since the
2
whole curve y D e !x lies above the x-axis, its centroid
would have to satisfy y > 0. However, Pappus’s Theorem
would then imply that the surface of revolution would have
infinite area: S D 2#y & .length of curve/ D 1. The
curve cannot, therefore, have any centroid.
294
We need to find the x-coordinate xLMNP of the centre of
buoyancy, that is, of the centroid of quadrilateral LMNP .
From various triangles in the figure we can determine the
x-coordinates of the four points:
xL D " sec t;
xP D sec t;
xM D " sec t C .1 C tan t / sin t
xN D sec t C .1 " tan t / sin t
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 7.6 (PAGE 431)
Triangle LMN has area 1 C tan t , and the x-coordinate of
its centroid is
xLMN
" sec t " sec t C .1 C tan t / sin t C sec t C .1 " tan t / sin t
D
3
2 sin t " sec t
:
D
3
h
dh
6 m
Triangle LNP has area 1 " tan t , and the x-coordinate of
its centroid is
" sec t C sec t C sec t C .1 " tan t / sin t
3
sec t C .1 " tan t / sin t
D
:
3
2 m
2 m
Fig. 7.6-1
xLNP D
2.
Therefore,
xLMNP D
D
D
D
D
1h
.2 sin t " sec t /.1 C tan t /
6
i
C .sec t C sin t " sin t tan t /.1 " tan t /
i
1h
3 sin t " 2 sec t tan t C sin t tan2 t
6
"
#
2
sin2 t
sin t
3"
C
6
cos2 t
cos2 t
h
i
sin t
3 cos2 t C sin2 t " 2
2
6 cos t
i
i
sin t h
sin t h
2
2
cos
t
"
1
D
cos.2t
/
6 cos2 t
6 cos2 t
A vertical slice of water at position y with thickness dy is
in contact withpthe botttom over an area
8 sec & dy D 54 101 dy m2 , which is at depth
1
x D 10
y C 1 m. The force exerted on this area is then
p
1
y C 1/ 45 101 dy. Hence, the total force
dF D *g. 10
exerted on the bottom is
20 #
$
1
y C 1 dy
10
0
# 2
$ˇˇ20
p
y
4
ˇ
101 .1000/.9:8/
Cy ˇ
D
ˇ
5
20
F D
4p
101 *g
5
Z
0
' 3:1516 & 106 N:
20
which is positive provided 0 < t < #=4. Thus the beam
will rotate counterclockwise until an edge is on top.
y
1
y
dy
3
y
xD 10 C1
Section 7.6 Other Physical Applications
(page 431)
1.
a) The pressure at the bottom is p D 9; 800 & 6 N/m2 .
The force on the bottom is 4 & p D 235; 200 N.
b) The pressure at depth h metres is 9; 800h N/m2 .
The force on a strip between depths h and h C dh on
one wall of the tank is
dF D 9; 800h & 2 dh D 19; 600 h dh N.
Thus, the total force on one wall is
F D 19; 600
Z
Fig. 7.6-2
3.
A strip along the slant wall of the dam between depths h
and h C dh has area
dA D
26
200 dh
D 200 &
dh:
cos &
24
The force on this strip is
dF D 9; 800 h dA ' 2:12 & 106 h dh N.
Thus the total force on the dam is
6
0
"
x
h dh D 19; 600 & 18 D 352; 800 N.
F D 2:12 & 106
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Z
24
0
h dh ' 6:12 & 108 N.
295
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SECTION 7.6 (PAGE 431)
ADAMS and ESSEX: CALCULUS 9
5.
The unbalanced force is
F D 9; 800 & 5
20
h dh
# 2 $ˇˇ20
h ˇ
D 9; 800 & 5
ˇ ' 8:92 & 106 N.
2 ˇ
"
h
hCdh
Z
6
6
26
24
Fig. 7.6-3
5 m
p
4. The height of each triangular
p face is 2 3 m and the
height of the pyramid is 2 2 m. Let the angle between
r
2
the triangular face and the base be & , then sin & D
3
1
and cos & D p .
3
20 m
6 m
p
2 2
p
2 3
Fig. 7.6-5
2 "
4
6.
4
Fig. 7.6-4
The spring force is F .x/ D kx, where x is the amount of
compression. The work done to compress the spring 3 cm
is
100 N!cm D W D
Z
0
y
front view of
dy
p
10!2 2
p
dy sec "D 3dy
p
p
xD 2yC10!2 2
"
x
W D
ı
60
2
7.
Fig. 7.6-4
A vertical slice of water with thickness dy at a distance
y from the vertex of the pyramid exerts a force on the
shaded
strip shown in the front view,pwhich has areap
p
2 3y dy m2 and which is at depth 2y C 10 " 2 2
m. Hence, the force exerted on the triangular face is
Z 2 p
p p
F D *g
. 2y C 10 " 2 2/2 3y dy
0
(p
)ˇ2
p 2 ˇˇ
p
2 3
y C .5 " 2/y ˇ
D 2 3.9800/
ˇ
3
0
' 6:1495 & 105 N:
296
Z
4
3
kx dx D
4
side view of one face
ˇ3
1 2 ˇˇ
9
kx dx D kx ˇ D k:
ˇ
2
2
0
200
Hence, k D
N/cm. The work necessary to compress
9
the spring a further 1 cm is
one face
10
3
200
9
$
ˇ4
1 2 ˇˇ
700
x ˇ D
N!cm:
2 ˇ
9
3
A layer of water in the tank between depths h and h C dh
has weight dF D *g d V D 4*g dh. The work done
to raise the water in this layer to the top of the tank is
d W D h dF D 4*gh dh. Thus the total work done to
pump all the water out over the top of the tank is
W D 4*g
8.
#
Z
6
0
h dh D 4 & 9; 800 & 18 ' 7:056 & 105 N!m.
The horizontal cross-sectional area of the pool at depth h
is
'
160;
if 0 $ h $ 1;
A.h/ D
240 " 80h; if 1 < h $ 3.
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 7.6 (PAGE 431)
The work done to empty the pool is
Z 3
W D *g
hA.h/ dh
0
(Z 1
)
Z 3
D *g
160h dh C
240h " 80h2 dh
0
1
ˇ1 #
"
$ˇ3 #
ˇ
80 3 ˇˇ
2ˇ
2
D 9800 80h ˇ C 120h " h ˇ
ˇ
ˇ
3
0
1
10. When the water surface is y m above the centre of the
piston face ("R $ y $ R), a horizontal strip on the
piston face at height z m above thepcentre of the piston
face, having height dz has width 2 .R2 " z 2 / and so its
p
area is dA D 2 R2 " z 2 dz. The strip is y " z m below
the surface of the water, and the pressure at that depth is
*g.y " z/ D 9;800.y " z/ N/m2 . Thus, the
p force of the
water on that strip is dF D 19;600.y " z/ R2 " z 2 dz
Newtons. The total force of the water on the piston is
D 3:3973 & 106 N!m:
F D
!R
p
19;600.y " z/ R2 " z 2 dz
3
A.h/
11.
Fig. 7.6-8
9. A layer of water between depths y and y C dy
has volume d V D #.a2 " y 2 / dy and weight
dF D 9; 800#.a2 " y 2 / dy N. The work done to raise
this water to height h m above the top of the bowl is
d W D .h C y/ dF D 9; 800#.h C y/.a2 " y 2 / dy N!m.
Thus the total work done to pump all the water in the
bowl to that height is
Z a
W D 9; 800#
.ha2 C a2 y " hy 2 " y 3 / dy
0
)ˇa
(
y 4 ˇˇ
a2 y 2 hy 3
2
"
"
D 9; 800# ha y C
ˇ
2
3
4 ˇ
0
( 3
)
2a h
a4
D 9; 800#
C
3
4
#
$
8h
3a
C
8h
D 2450#a3 a C
N!m.
D 9; 800#a3
12
3
a
sin"1 .y=R/
!!=2
.y " R sin & / R2 cos2 & d&
(
h
1
Z
(let z D R sin & )
)ˇsin"1 .y=R/
ˇ
R2 y
R3
3
D 19;600
.& C sin & cos & / C
cos & ˇˇ
2
3
!!=2
( 2
)
"
R
#yR2
1! 2
!1 y
2 3=2
y sin
C
C
R "y
D 19;600
N:
2
R
4
3
8
y
y
D 19;600
20
dy
Z
Initially the water occupies the bottom half of a cylinder of length X. Half of this water (say, a bottom halfcylinder of length X=2, must be moved to fill the top
half of a cylinder of length X=2. By symmetry, we can
accomplish this by moving a thin horizontal slice of
water of thickness dy at distance y below the central
axis of the cylinder to height y above that axis, that is,
p
X
dy of wawe move a volume d V D 2 R2 " y 2
2
ter up a distance
2y.
The
work
required
to
do this is
p
d W D 2*gXy R2 " y 2 dy N!m. Thus the work to raise
the half-cylinder of water is
W D 2*gX
Z
D 9;800X
R
0
Z
p
y R2 " y 2 dy
R2
p
0
u du D
let u D R2 " y 2
19;600
XR3 N!m:
3
Since no work is lost to friction, the work to push the
piston in to distance X=2 is equal to this work needed to
raise the water.
12. Let the time required to raise the bucket to height h m be
h
t minutes. Given that the velocity is 2 m/min, then t D .
2
The weight of the bucket at time t is
h
16 kg " .1 kg/min/.t min/ D 16 "
kg. Therefore,
2
the work done required to move the bucket to a height of
10 m is
$
h
dh
2
0
$ˇ10
#
h2 ˇˇ
D 9:8 16h "
ˇ D 1323 N!m:
4 ˇ
W Dg
Fig. 7.6-9
Z
10 #
16 "
0
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SECTION 7.6 (PAGE 431)
ADAMS and ESSEX: CALCULUS 9
8.
Section 7.7 Applications in Business,
Finance, and Ecology (page 435)
1.
Cost D $4; 000 C
D $11; 000:
Z
1;000
0
1; 000
Z
52
0
that is, about 96,580.
x
0
4
p dt
1C t
p
x
let t D u2
$
Z px #
u
1
D8
du D 8
du
1"
1Cu
1Cu
0
%0p
&
p
D$8 x " ln.1 C x/ :
Z
9.
t e !t=10 dt D 100; 000 " 620; 000e !26=5
3. The monthly charge is
Z
V D
#
$
2x
6x 2
6 " 3 C 6 dx
10
10
2. The number of chips sold in the first year was
1
0
11.
V D
1;000e
!0:02t
0
ˇ10
1;000 !0:02t ˇˇ
e
dt D
ˇ D $9;063:46:
ˇ
"0:02
V D
7.
Z
10
0
V D
298
Z
2
1;000e
!0:08t
ˇ12
1;000 !0:08t ˇˇ
e
dt D
ˇ D $5;865:64:
ˇ
"0:08
2
Z
1
0
1;000e !0:02t dt D
1; 000
D $50; 000:
"0:02
After t years, money is flowing at $.1;000 C 100t / per
year. The present value of 10 years of payments discounted at 5% is
Z
10
0
.10 C t /e !0:05t dt
0
12. After t years, money is flowing at $1;000.1:1/t per year.
The present value of 10 years of payments discounted at
5% is
Z
10
e t ln.1:1/ e !0:05t dt
ˇ10
ˇ
1;000
t.ln.1:1/!0:05 ˇ
D
e
ˇ D $12; 650:23:
ˇ
ln.1:1/ " 0:05
V D 1;000
0
12
The present value of continuous payments of $1,000 per
year for all future time at a discount rate of 2% is
d V D e !0:05t dt
e !0:05t
V D
"0:05
ˇ10
Z
100 10 !0:05t
e !0:05t ˇˇ
e
dt
D 100.10 C t /
ˇ C
"0:05 ˇ
0:05 0
0
ˇ10
ˇ
100
!0:05t ˇ
D "4261:23 C
e
ˇ D $11; 477:54:
ˇ
".0:05/2
ˇ10
1;000 !0:05t ˇˇ
!0:05t
1;000e
dt D
e
ˇ D $7;869:39:
ˇ
"0:05
The present value of continuous payments of $1,000 per
year for 10 years beginning 2 years from now at a discount
rate of 8% is
10
U D 10 C t
d U D dt
0
6. The present value of continuous payments of $1,000 per
year for 10 years at a discount rate of 5% is
1;000e
ˇ35
1;000 !0:05t ˇˇ
e
dt D
ˇ D $8;655:13:
ˇ
"0:05
!0:05t
10
V D 100
400.10 C 5t /
dt ' $4; 750:37:
1 C 0:1t
10
35
10. The present value of continuous payments of $1,000 per
year beginning 10 years from now and continuing for all
future time at a discount rate of 5% is
Z 1
1;000 !0:5
V D
1;000e !0:05t dt D
e
D $12;130:61:
"0:05
10
5. The present value of continuous payments of $1,000 per
year for 10 years at a discount rate of 2% is
Z
Z
V D
4. The price per kg at time t (years) is $10 C 5t . Thus the
revenue per year at time t is 400.10C5t /=.1C0:1t / $/year.
The total revenue over the year is
Z
The present value of continuous payments of $1,000 per
year for 25 years beginning 10 years from now at a discount rate of 5% is
0
0
13.
The amount after 10 years is
A D 5; 000
Z
10
e
0
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0:05t
ˇ10
5;000 0:05t ˇˇ
dt D
e
ˇ D $64;872:13:
ˇ
0:05
0
lOMoARcPSD|6566483
INSTRUCTOR’S SOLUTIONS MANUAL
14.
SECTION 7.7 (PAGE 435)
Let T be the time required for the account balance to
reach $1;000;000. The $5; 000.1:1/t dt deposited in the
time interval Œt; t C dt ! grows for T " t years, so the balance after T years is
Z
For realistic growth functions, the maximum will occur
where Q 0 .x/ D 0, that is, where F 0 .x/ D ı.
17.
T
5; 000.1:1/t .1:06/T !t dt D 1; 000; 000
$
Z T#
1; 000; 000
1:1 t
dt D
D 200
.1:06/T
1:06
5; 000
0
#
"#
$
1:1 T
.1:06/T
" 1 D 200
ln.1:1=1:06/
1:06
0
.1:1/T " .1:06/T D 200 ln
x D .k " ı/
Let P .+/ be the value at time + < t that will grow to
$P D P .t / at time t . If the discount rate at time + is
ı.+/, then
d
P .+/ D ı.+/P .+/;
d+
or, equivalently,
dP .+/
D ı.+/ d +:
P .+/
#
$
23; 333:33
6.0:12/.23; 333:33/ 1 "
D $11; 900:
80; 000
18. We are given that k D 0:02, L D 150; 000, p D $10; 000.
The growth rate at population level x is
#
$
dx
x
D 0:02x 1 "
:
dt
150; 000
a) The maximum sustainable annual harvest is
ˇ
dx ˇˇ
ˇ
dt ˇ
Integrating this from 0 to t , we get
ln P .t/ " ln P .0/ D
Z
t
0
ı.+/ d + D ,.t /;
75; 000.$10; 000/.0:02/ D $15; 000; 000:
The present value of a stream of payments due at a rate
P .t / at time t from t D 0 to t D T is
0
P .t /e !&.t/ dt;
where
,.t / D
Z
d) At 5%, the interest would be
.5=2/.$15; 000/ D $37; 500; 000.
e) The total present value of all future harvesting revenue if the population level is maintained at 75,000
and ı D 0:05 is
t
ı.+/ d +:
0
Z
16. The analysis carried out in the text for the logistic growth
model showed that the total present value of future harvests could be maximized by holding the population size x
at a value that maximizes the quadratic expression
%
x&
" ıx:
Q.x/ D kx 1 "
L
0
19.
If the logistic model dx=dt D kx.1 " .x=L// is replaced
with a more general growth model dx=dt D F .x/, exactly
the same analysis leads us to maximize
Q.x/ D F .x/ " ıx:
D 0:02.75; 000/.0:5/ D 750 whales:
c) If the whole population of 75,000 is harvested and the
proceeds invested at 2%, the annual interest will be
P .0/ D P .t /e !&.t/ D P e !&.t/ :
T
xDL=2
b) The resulting annual revenue is $750p D $7; 500; 000.
and, taking exponentials of both sides and solving for
P .0/, we get
Z
0:07
L
D
.80; 000/ D 23; 333:33
2k
0:24
The annual revenue from harvesting to keep the population
at this level (given a price of $6 per fish) is
1:1
:
1:06
This equation can be solved by Newton’s method or using
a calculator “solve” routine. The solution is T ' 26:05
years.
15.
We are given L D 80; 000, k D 0:12, and ı D 0:05.
According to the analysis in the text, the present value of
future harvests will be maximized if the population level is
maintained at
1
e !0:05t 7; 500; 000 dt D
7; 500; 000
D $150; 000; 000:
0:05
If we assume that the cost of harvesting 1 unit of population is $C.x/ when the population size is x, then the
effective income from 1 unit harvested is $.p " C.x//.
Using this expression in place of the constant p in the
analysis given in the text, we are led to choose x to maximize
i
%
&h %
x&
" ıx :
Q.x/ D p " C.x/ kx 1 "
L
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SECTION 7.7 (PAGE 435)
ADAMS and ESSEX: CALCULUS 9
A reasonable cost function C.x/ will increase as x decreases (the whales are harder to find), and will exceed
p if x $ x0 , for some positive population level x0 . The
value of x that maximizes Q.x/ must exceed x0 , so the
model no longer predicts extinction, even for large discount rates ı. However, the optimizing population x may
be so low that other factors not accounted for in the simple logistic growth model may still bring about extinction
whether it is economically indicated or not.
Section 7.8 Probability
1.
6.
9
9
&
' 0:0225
60 60
9
f .3/ D 2 &
& 16 D 0:0500
60
1
9
& 16 C
D 0:0778
f .4/ D 2 &
60
36
9
2
f .5/ D 2 &
& 16 C
D 0:1056
60
36
3
9
& 16 C
D 0:1333
f .6/ D 2 &
60
36
9
4
f .7/ D 2 &
& 1160 C
D 0:1661
60
36
11
3
f .8/ D 2 &
& 16 C
D 0:1444
60
36
2
11
& 16 C
D 0:1167
f .9/ D 2 &
60
36
11
1
f .10/ D 2 &
& 16 C
D 0:0889
60
36
11
& 16 D 0:0611
f .11/ D 2 &
60
11
f .12/ D
& 1160 D 0:0336:
60
f .2/ D
(page 449)
The expected winnings on a toss of the coin are
$1 & 0:49 C $2 & 0:49 C $50 & 0:02 D $2:47:
If you pay this much to play one game, in the long term
you can expect to break even.
P
2. (a) We need 6nD1 Kn D 1. Thus 21K D 1, and
K D 1=21.
(b) Pr.X $ 3/ D .1=21/.1 C 2 C 3/ D 2=7:
3. From the second previous Exercise, the mean winings is
- D $2:47. Now
) 2 D 1 & 0:49 C 4 & 0:49 C 2;500 & 0:02 " -2
' 52:45 " 6:10 D 46:35:
The standard deviation is thus ) ' $6:81.
4. Since Pr.X D n/ D n=21, we have
-D
)2 D
6
X
nD1
6
X
nD1
nPr.X D n/ D
1 & 1 C 2 & 2 C !!! C 6 & 6
13
D
' 4:33
21
3
n2 Pr.X D n/ " -2 D
nD2
Similarly,
E.X 2 / D
12
X
nD2
n2 f .n/ ' 57:1783;
so the standard deviation of X is
)D
5. The mean of X is
The expectation of X 2 is
9
1
C.22 C32 C42 C52 /& C62 &1160 ' 15:7500:
60
6
Hence the standard deviation of X is
p
15:75 " 3:58332 ' 1:7059.
9
2
29
Also Pr.X $ 3/ D
C D
' 0:4833.
60
6
60
p
E.X 2 / " -2 ' 2:4124:
The mean is somewhat larger than the value (7) obtained for the unweighted dice, because the weighting
favours more 6s than 1s showing if the roll is repeated many times. The standard deviation is just a
tiny bit smaller than that found for the unweighted
dice (2.4152); the distribution of probability is just
slightly more concentrated around the mean here.
1
9
C .2 C 3 C 4 C 5/ & C 6 & 1160 ' 3:5833:
-D1&
60
6
300
(b) Multiplying each value f .n/ by n and summing, we
get
12
X
-D
nf .n/ ' 7:1665:
12 C 23 C ! ! ! C 63
" -2
21
20
169
D 21 "
D
' 2:22
9
9
p
20
)D
' 1:49:
3
E.X 2 / D 12 &
(a) Calculating as we did to construct the probability function in Example 2, but using the different values for the
probabilities of “1” and “6”, we obtain
7.
(a) The sample space consists of the eight triples
.H; H; H /, .H; H; T /, .H; T; H /, .T; H; H /,
.H; T; T /, .T; H; T /, .T; T; H /, and .T; T; T /.
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 7.8 (PAGE 449)
(b) We have
9.
We have f .x/ D C x on Œ0; 3!.
a) C is given by
Pr.H; H; H / D .0:55/3 D 0:166375
Pr.H; H; T / D Pr.H; T; H / D Pr.T; H; H / D .0:55/2 .0:45/
D 0:136125
Pr.H; T; T / D Pr.T; H; T / D Pr.T; T; H / D .0:55/.0:45/2
D 0:111375
Pr.T; T; T / D .0:45/3 D 0:091125:
(c) The probability function f for X is given by
1D
Hence, C D
Z
3
0
ˇ3
C 2 ˇˇ
9
C x dx D x ˇ D C:
2 ˇ
2
0
2
.
9
b) The mean is
f .0/ D .0:45/3 D 0:911125
Z
2
- D E.X/ D
9
f .1/ D 3 & .0:55/.0:45/2 D 0:334125
2
f .2/ D 3 & .0:55/ .0:45/ D 0:408375
f .3/ D .0:55/3 D 0:166375:
(d) Pr.X # 1/ D 1 " Pr.X D 0/ D 0:908875.
(e) E.X/ D 0&f .0/C1&f .1/C2&f .2/C3&f .3/ D 1:6500.
8. The number of red balls in the sack must be 0:6&20 D 12.
Thus there are 8 blue balls.
(a) The probability of pulling out one blue ball is 8=20.
If you got a blue ball, then there would be only 7
blue balls left among the 19 balls remaining in the
sack, so the probability of pulling out a second blue
ball is 7=19. Thus the probability of pulling out two
8
7
14
blue balls is
&
D
.
20 19
95
(b) The sample space for the three ball selection consists
of all eight triples of the form .x; y; z/, where each
of x; y; z is either R(ed) or B(lue). Let X be the
number of red balls among the three balls pulled out.
Arguing in the same way as in (a), we calculate
Pr.X D 0/ D Pr.B; B; B/ D
Thus the expected value of X is
D
0
0
ˇ3
9
2 4 ˇˇ
x ˇ D , the
x dx D
36 ˇ
2
3
0
9
1
"4D ;
2
2
p
and the standard deviation is ) D 1= 2.
c) We have
Pr.- " ) $ X $ - C )/ D
D
2
9
Z
'C(
x dx
'!(
4-)
.- C )/2 " .- " )/2
D
' 0:6285:
9
9
10. We have f .x/ D C x on Œ1; 2!.
a) To find C , we have
1D
Hence, C D
Z
2
1
ˇ2
C 2 ˇˇ
3
C x dx D x ˇ D C:
2 ˇ
2
1
2
.
3
b) The mean is
2
- D E.X/ D
3
Z
2
Since E.X / D
3
variance is
2
14
28
44
11
C1&
C2&
C3&
285
95
95
57
3
0
ˇ3
2 3 ˇˇ
x dx D
x ˇ D 2:
27 ˇ
2
) 2 D E.X 2 / " -2 D
8
7
6
14
&
&
D
20 19 18
285
' 0:0491
Pr.X D 1/ D Pr.R; B; B/ C Pr.B; R; B/ C Pr.B; B; R/
12
8
7
28
D3&
&
&
D
' 0:2947
20 19 18
95
Pr.X D 2/ D Pr.R; R; B/ C Pr.R; B; R/ C Pr.B; R; R/
8
44
12 11
&
&
D
' 0:4632
D3&
20 19 18
95
12 11 10
11
Pr.X D 3/ D Pr.R; R; R/ D
&
&
D
20 19 18
57
' 0:1930
E.X/ D 0 &
Z
2
Since E.X / D
9
variance is
2
3
9
D 1:8:
5
2
1
Z
ˇ2
2 3 ˇˇ
14
x dx D x ˇ D
' 1:556:
9 ˇ
9
2
1
2
1
ˇ2
1 4 ˇˇ
5
x dx D x ˇ D , the
6 ˇ
2
3
) 2 D E.X 2 / " -2 D
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13
5 196
"
D
2
81
162
301
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SECTION 7.8 (PAGE 449)
ADAMS and ESSEX: CALCULUS 9
Since
and the standard deviation is
r
13
)D
' 0:283:
162
E.X 2 / D
D
11.
2
3
Z
'C(
x dx
'!(
a) C is given by
0
1
D .# 2 " 4/:
2
ˇ1
Z 1
C 3 ˇˇ
C
2
1D
C x dx D x ˇ D :
3 ˇ
3
0
Hence, the variance is
0
Hence, C D 3.
) 2 D E.X 2 / " -2 D
b) The mean, variance, and standard deviation are
Z 1
3
- D E.X/ D 3
x 3 dx D
4
0
Z 1
9
3
9
3
) 2 D E.X 2 / " -2 D 3
x 4 dx "
D "
D
16
5 16
80
0
p
) D 3=80:
Pr.- " ) $ X $ - C )/ D 3
3
Z
'C(
)D
1D
Hence, C D
0
1
.
2
1
- D E.X/ D
2
Z
sin x dx
'!(
a) C is given by
1D
Z
0
1
2
C.x " x / dx D C
#
x2 x3
"
2
3
$ ˇˇ1
C
ˇ
ˇ D :
ˇ
6
0
b) The mean, variance, and standard deviation are
!
x sin x dx
0
U Dx
d V D sin x dx
d U D dx ˇ V D " cos x
(
)
ˇ! Z !
1
ˇ
D
"x cos x ˇ C
cos x dx
ˇ
2
0
0
#
D 1:571:
D
2
302
'C(
We have f .x/ D C.x " x 2 / on Œ0; 1!.
Hence, C D 6.
b) The mean is
Z
D"
13.
0
1
2
i
1h
cos.- C )/ " cos.- " )/
2
D sin - sin ) D sin ) ' 0:632:
D .- C )/ " .- " )/
r !3
r !3
3
3
3
3
D
C
"
"
' 0:668:
4
80
4
80
ˇ!
ˇ
ˇ
C sin x dx D "C cos x ˇ D 2C:
ˇ
#2 " 8
' 0:684:
4
Pr.- " ) $ X $ - C )/ D
'!(
a) To find C , we calculate
s
c) Then
x 2 dx
12. We have f .x/ D C sin x on Œ0; #!.
#2 " 8
#2 " 4 #2
"
D
' 0:467
2
4
4
and the standard deviation is
c) We have
!
x 2 sin x dx
0
U Dx
d V D cos x dx
d"U D dx
V D
ˇ! sin Zx
#
$#
ˇ
!
1 2
ˇ
D
# C 2 x sin x ˇ "
sin x dx
ˇ
2
0
We have f .x/ D C x 2 on Œ0; 1!.
Z
!
0
4-)
.- C )/2 " .- " )/2
D
' 0:5875:
3
3
3
Z
d V D sin x dx
U D x2
d U D 2x dx
ˇ! VZD " cos x )
(
ˇ
!
1
ˇ
"x 2 cos x ˇ C 2
D
x cos x dx
ˇ
2
0
c) We have
Pr.- " ) $ X $ - C )/ D
1
2
Z
1
1
.x 2 " x 3 / dx D
2
0
Z 1
1
2
2
2
) D E.X / " - D 6
.x 3 " x 4 / dx "
4
0
3
1
1
D
" D
10 4
20
p
) D 1=20:
- D E.X/ D 6
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 7.8 (PAGE 449)
c) Finally,
c) We have
Z
.1=2/C(
Pr.- " ) $ X $ - C )/
Z 'C(
D k2
xe !kx dx Let u D kx
'!(
du D k dx
Z k.'C( /
D
ue !u du
2
Pr.- " ) $ X $ - C )/ D 6
.x " x / dx
.1=2/!(
"
#
#
$
Z .1=2/C(
1
1 2
D6
dx
" x"
4
2
.1=2/!(
k.'!( /
Let u D x " 21
du D dx
)
(
)
Z ((
)
)3
1
2
D 12
" u du D 12
"
4
4
3
0
)
(
12 1
1
' 0:626:
D p
"
20 4 60
14.
ˇk.'C( / Z
ˇ
k.'C( /
ˇ
D "ue !u ˇ
C
e !u du
ˇ
k.'!( /
k.'!( /
p
p
p
p
D ".2 C 2/e !.2C 2/ C .2 " 2/e !.2! 2/
p
" e !.2C
' 0:738:
2/
p
C e !.2!
2/
It was shown in Section 6.1 (p. 349) that
Z
If In D
n !x
x e
Z
1
n !x
dx D "x e
Cn
Z
15.
x
n!1 !x
e
0
Since I0 D
Z
1
0
if n # 1:
e !x dx D 1, therefore In D nŠ for n # 1.
Let u D kx; then
0
x n e !kx dx D
1
k nC1
Z
1
0
1
nŠ
In D nC1 :
k nC1
k
un e !u du D
Now let f .x/ D C xe !kx on Œ0; 1/.
Z
1
0
xe !kx dx D
C
:
k2
b) The mean is
- D E.X/ D k
Since
E.X 2 / D k 2
Z
0
1
Z
1
0
2
e !x dx D
C
2
Z
1
!1
2
e !x dx D
C
p
2
#
:
b) The mean, variance, and standard deviation are
ˇ
Z 1
2 1
2
e !x ˇˇ
1
!x 2
-D p
dx D " p ˇ D p
xe
ˇ
# 0
#
#
0
Z 1
1
2
2
)2 D " C p
x 2 e !x dx
#
# 0
U Dx
d U D dx
D"
1
2
Cp
#
#
2
d V D xe !x dx
2
V D " 12 e !x
ˇ1
!
Z
x !x 2 ˇˇ
1 1 !x 2
" e
e
dx
ˇ C
ˇ
2
2 0
0
p $
#
#
1
1
1
D "
0C !
2 2
2 #
c) We have
Hence, C D k 2 .
2
1
1
2
D" Cp
#
#
r
1
1
)D
" ' 0:426:
2 #
a) To find C , observe that
1DC
Z
p
Thus C D 2= #.
R!1
1
1DC
x n e !x dx, then
In D lim "Rn e !R C nIn!1 D nIn!1
Z
dx:
a) We have
2 !kx
x e
0
dx D k
x 3 e !kx dx D k 2
then the variance is
) 2 D E.X 2 / " -2 D
#
2
#
6
k4
2
k3
$
$
D
4
2
6
"
D 2
k2 k2
k
p
2
and the standard deviation is ) D
.
k
D
2
:
k
6
;
k2
Z 'C(
2
2
e !x dx
Pr.- " ) $ X $ - C )/ D p
# '!(
p
Let x D z= 2
p
dx D dz= 2
r Z p
2.'C( /
2
2
D
e !z =2 dz:
p
#
2.'!( /
p
p
But 2.- " )/ ' 0:195 and 2.- C )/ ' 1:40.
Thus, if Z is a standard normal random variable, we
obtain by interpolation in the table on page 386 in the
text,
Pr.- " ) $ X $ - C )/ D 2Pr.0:195 $ Z $ 1:400/
' 2.0:919 " 0:577/ ' 0:68:
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SECTION 7.8 (PAGE 449)
16. No. The identity
constant C .
17.
Z
1
!1
ADAMS and ESSEX: CALCULUS 9
b) For f .x/ D ke !kx on Œ0; 1/, we know that
1
- D ) D
(Example 6). Thus - " 2) < 0 and
k
3
- C 2) D . We have
k
#
$
3
Pr.jX " -j # 2)/ D Pr X #
k
Z 1
Dk
e !kx dx
3=k
ˇ1
ˇ
D e !3 ' 0:050:
D "e !kx ˇˇ
C dx D 1 is not satisfied for any
1
2
2
p e !.x!'/ =2(
) 2#
Z 1
1
2
2
mean D p
xe !.x!'/ =2( dx
) 2# !1
f';( .x/ D
Let z D
dz D
x")
1
dx
)
Z 1
1
2
D p
.- C )z/e !z =2 dz
2# !1
Z 1
2
e !z =2 dz D D p
%2# !1 &
variance D E .x " -/2
Z 1
1
2
2
D p
.x " -/2 e !.x!'/ =2( dx
) 2# !1
Z 1
1
2
D p
) 2 z 2 e !z =2 dz D )Var.Z/ D )
) 2# !1
18. Since f .x/ D
2
#
Z
0
1
3=k
1
2
2
p e .x!'/ =2( , which has mean ) 2#
and standard deviation ), we have
c) For f';( .x/ D
Pr.jX " -j # 2)/ D 2Pr.X $ - " 2)/
Z '!2(
1
2
2
D2
p e !.x!'/ =2( dx
)
2#
!1
x"Let z D
)
1
dz D dx
)
Z !2
2
!z 2
e
dz
D p
2# !1
D 2Pr.Z $ "2/ ' 2 & 0:023 D 0:046
2
> 0 on Œ0; 1/ and
#.1 C x 2 /
dx
2
2 %# &
D lim
tan!1 .R/ D
D 1;
2
R!1 #
1Cx
# 2
therefore f .x/ is a probability density function on Œ0; 1/.
The expectation of X is
- D E.X/ D
2
#
Z
1
0
from the table in this section.
20.
Z 12
1
e !t=20 dt D 1 "
e !t=20 dt
20 0
12
ˇ12
ˇ
!t=20 ˇ
D1Ce
ˇ D e !12=20 ' 0:549:
ˇ
x dx
1 C x2
Pr.T # 12/ D
1
D lim
ln.1 C R2 / D 1:
R!1 #
-D
bCa
;
2
)D
b"a
p :
2 3
bCa
b"a
C p > b, and similarly,
2
3
- " 2) < a, therefore Pr.jX " -j # 2)/ D 0.
Since - C 2) D
304
Z
1
The probability that the system will last at least 12 hours
is about 0.549.
21.
a) The density function for the uniform distribution on
Œa; b! is given by f .x/ D 1=.b " a/, for a $ x $ b.
By Example 5, the mean and standard deviation are
given by
1
20
0
No matter what the cost per game, you should be willing
to play (if you have an adequate bankroll). Your expected
winnings per game in the long term is infinite.
19.
The density function for T is f .t / D ke !k t on Œ0; 1/,
1
1
D
(see Example 6). Then
where k D
20
If X is distributed normally, with mean - D 5; 000, and
standard deviation ) D 200, then
Pr.X # 5500/
Z 1
1
2
2
p
e !.x!5000/ =.2#200 / dx
D
200 2# 5500
x " 5000
Let z D
200
dx
dz D
200
Z 1
1
2
D p
e !z =2 dz
2# 5=2
D Pr.Z # 5=2/ D Pr.Z $ "5=2/ ' 0:006
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 7.9 (PAGE 458)
from the table in this section.
2.
22. If X is the random variable giving the spinner’s value,
then Pr.X D 1=4/ D 1=2 and the density function for the
other values of X is f .x/ D 1=2. Thus the mean of X is
#
$ Z 1
1
1 1
3
1
Pr X D
C
x f .x/ dx D C D :
4
4
8 4
8
0
- D E.X/ D
Also,
#
$ Z 1
1
1
1
19
1
Pr X D
C
C D
x 2 f .x/ dx D
16
4
32
6
96
0
9
11
19
2
2
2
"
D
:
) D E.X / " - D
96 64
192
E.X 2 / D
3.
p
11=192.
Thus ) D
23.
4.
(a) The integral exists for p D 0 as a degenerate case.
For p > 0,
Z
1
x p S˛ .x/dx
!1
D 2 lim
y!1
D 2 lim
y!1
#
(Z
%
%
&& )
x c˛ x !.1C˛/ C O x !.1C2˛/ dx
p
$
!
"
c˛
x p!˛ C O x p!2˛
p"˛
jxDy
!
5.
This limit will be finite if p < ˛. The case p D ˛ is
logarithmically divergent.
(b) The mean will not exist for ˛ $ 1. The variance will
not exist for ˛ < 2.
24.
1 " FS .x/ D Pr.X > x/ D
D
Z
D"
1
x
Z
1
˛
˛ !˛
.
!
"i
C O . !2˛
j#Dx
7.
c˛ ˛
x ;
˛
where higher order terms were discarded in the last step.
Section 7.9 First-Order Differential
Equations (page 458)
1.
y D 0 is a constant solution. Otherwise
y
dy
D
dx
2x
dy
dx
2
D
y
x
2 ln y D ln x C C1
)
y2 D C x
Y D 0 is a constant solution. Otherwise
dY
dY
D tY
)
D t dt
dt
Y
t2
2
ln Y D
C C1 ;
or Y D C e t =2
2
dx
D e x sin t
dt
Z
Z
sin t dt
"e !x D " cos t " C
) x D " ln.cos t C C /:
S˛ ../d .
*
x2
dy
D 2
)
y 2 dy D x 2 dx
dx
y
y3
x3
D
C C1 ;
or x 3 " y 3 D C
3
3
y D 0 is a constant solution. Otherwise,
dy
D x2y2
dx
Z
Z
dy
D
x 2 dx
y2
1
1
1
" D x3 C C
y
3
3
3
:
) yD" 3
x CC
e !x dx D
x
n
%
&o
c˛ . !.1C˛/ C O . !.1C2˛/ d .
h "c
6.
y D 1=3 is a constant solution. Otherwise
3y " 1
dy
D
dx
x Z
Z
dy
dx
D
3y " 1
x
1
1
ln j3y " 1j D ln jxj C ln C
3
3
3y " 1
DC
x3
1
) y D .1 C C x 3 /:
3
y D 1 and y D "1 are constant solutions, Otherwise
dy
dy
D dx
D 1 " y2
)
dx
1 " y2
#
$
1
1
1
C
dy D dx
2 1Cy
1"y
ˇ
ˇ
1 ˇˇ 1 C y ˇˇ
ln
D x C C1
2 ˇ1"yˇ
1Cy
D C e 2x
1"y
8.
dy
dx
Z
dy
1 C y2
tan!1 y
) y
or
yD
C e 2x " 1
C e 2x C 1
D 1 C y2
Z
D
dx
DxCC
D tan.x C C /:
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SECTION 7.9 (PAGE 458)
9.
ADAMS and ESSEX: CALCULUS 9
dy
dy
D 2 C ey
)
D dt
2 C ey
Zdt !y
Z
e dy
D
dt
2e !y C 1
1
" ln.2e !y C 1/ D t C C1
2
$
#
1
!y
2e C 1 D C2 e !2t ;
or y D " ln C e !2t "
2
10. y D 0 and y D 1 are constant solutions. For the other
solutions we have
dy
D y 2 .1 " y/
dx
Z
Z
dy
D
dx D x C K:
y 2 .1 " y/
Expand the left side in partial fractions:
1
A
B
C
D C 2 C
y 2 .1 " y/
y
y
1"y
A.y " y 2 / C B.1 " y/ C Cy 2
D
y 2 .1 " y/
(
"A C C D 0I
) A " B D 0I ) A D B D C D 1:
B D 1:
Hence,
$
Z #
Z
1
1
1
dy
D
C
C
dy
y 2 .1 " y/
y
y2
1"y
1
D ln jyj " " ln j1 " yj:
y
Therefore,
14.
yD
1
C
C 2:
x
x
We have
R
dy
Cy D e x . Let - D dx D x, then e ' D e x ,
dx
#
$
d x
x dy
x
x dy
.e y/ D e
Ce y De
C y D e 2x
dx
dx
dx
Z
1
) e x y D e 2x dx D e 2x C C:
2
Hence, y D
15.
1 x
e C C e !x .
2
$
#Z
dy
1 dx D e x
Cy Dx
- D exp
dx
d x
.e y/ D e x .y 0 C y/ D xe x
dx
Z
ex y D
dy
2
" y D x2
(linear)
dx x
#Z
$
2
1
" dx D 2
- D exp
x
x
1 dy
2
yD1
"
x 2 dx x 3
d y
D1
dx x 2
y
D x C C;
so y D x 3 C C x 2
x2
2y
1
dy
C
D
. Let
12. We have
x
x2
Z dx
2
-D
dx D 2 ln x D ln x 2 , then e ' D x 2 , and
x
d 2
dy
.x y/ D x 2
C 2xy
dx #
dx$
# $
2y
1
dy
C
D x2
D1
D x2
dx
x
x2
Z
) x2y D
dx D x C C
)
$
#Z
dy
2 dx D e 2x
C 2y D 3
- D exp
dx
d 2x
.e y/ D e 2x .y 0 C 2y/ D 3e 2x
dx
3
3
e 2x y D e 2x C C ) y D C C e !2x
2
2
and
ˇ
ˇ
ˇ y ˇ 1
ˇ " D x C K:
ln ˇˇ
1"yˇ y
11.
306
13.
xe x dx D xe x " e x C C
y D x " 1 C C e !x
16.
R
dy
We have
C 2e x y D e x . Let - D 2e x dx D 2e x ,
dx
then
d % 2ex &
x dy
x
e y D e 2e
C 2e x e 2e y
dx
dx
#
$
dy
x
x
D e 2e
C 2e x y D e 2e e x :
dx
Therefore,
x
e 2e y D
Hence, y D
Z
Let u D 2e x
du D 2e x dx
Z
1
1
x
e u du D e 2e C C:
D
2
2
1
x
C C e !2e .
2
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INSTRUCTOR’S SOLUTIONS MANUAL
17.
18.
!1"
dy
C 10y D 1;
y 10
D
dt Z
- D 10 dt D 10t
2
10
22.
d 10t
dy
.e y/ D e 10t
C 10e 10t y D e 10t
dt
dt
1 10t
e 10t y.t / D
e
CC
10
!1"
e
e
2e
2
D
CC ) C D
y 10 D 10
)
10
10
10
1
1
yD
C e 1!10t :
10
10
dy
C 3x 2 y D x 2 ;
y.0/ D 1
dx Z
-D
x 2 y 0 C y D x 2 e 1=x ;
y.1/ D 3e
1
y 0 C 2 y D e 1=x
Zx
1
1
-D
dx D "
x2
x
$
#
d % !1=x &
1
e
y D e !1=x y 0 C 2 y D 1
dx
x
Z
!1=x
e
y D 1 dx D x C C
23.
24.
22 D 02 C C %
p
y D 4 C x2:
x
1
y.0/ D 1
y.t /
dt
t .t C 1/
%
y.1/ D 1
0
dy
D e !y ;
dx
ey D x C C
i.e. e y dy D dx
%
3 D y.0/ D ln C
3
y D ln.x C e /:
25.
y D ln.x C C /
C D e3
%
Since a > b > 0 and k > 0,
lim x.t / D lim
t!1
26.
%
&
ab e .b!a/k t " 1
be .b!a/k t " a
ab.0 " 1/
D
D b:
0"a
t!1
Since b > a > 0 and k > 0,
lim x.t / D lim
t!1
y.#/ D 0 ) 0 D # 2 C C ) C D "# 2
21.
%
y
dy
D
;
for x > 0
dx
x.x C 1/
dx
dx
dx
dy
D
D
"
y
x.x C 1/
x
xC1
x
ln y D ln
C ln C
xC1
Cx
; % 1 D C =2
yD
xC1
2x
yD
:
xC1
Z x
y.x/ D 3 C
e !y dt % y.0/ D 3
2x dx D x 2 C C
y D .x 2 " # 2 /e ! sin x :
Z x
t
y.x/ D 2 C
dt %
0 y.t /
dy
x
D ;
i.e. y dy D x dx
dx
y
y2 D x2 C C
0
.y.t //2
dt
1 C t2
y.#/ D 0
d sin x
.e
y/ D e sin x .y 0 C .cos x/y/ D 2x
dx
Z
e sin x y D
Z
y.x/ D 1 C
y.1/ D 3e ) 3 D 1 C C ) C D 2
y 0 C .cos x/y D 2xe ! sin x ;
Z
- D cos x dx D sin x
x
y D 1=.1 " tan!1 x/:
y D .x C 2/e 1=x :
20.
Z
y.x/ D 1 C
y2
dy
D
;
i.e. dy=y 2 D dx=.1 C x 2 /
dx
1 C x2
1
" D tan!1 x C C
y
" 1 D 0 C C % C D "1
3x 2 dx D x 3
d x3
3 dy
3
3
.e y/ D e x
C 3x 2 e x y D x 2 e x
dx
dx
Z
1 3
3
3
e x y D x 2 e x dx D e x C C
3
1
2
y.0/ D 1 ) 1 D C C ) C D
3
3
2
1
3
y D C e !x :
3
3
19.
SECTION 7.9 (PAGE 458)
t!1
D lim
%
&
ab e .b!a/k t " 1
.b!a/k t " a
be
%
&
ab 1 " e .a!b/k t
b " ae .a!b/k t
ab.1 " 0/
D
D a:
b"0
t!1
y.0/ D 2
27.
The solution given, namely
C D4
xD
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!
"
ab e .b!a/k t " 1
be .b!a/k t " a
;
307
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SECTION 7.9 (PAGE 458)
ADAMS and ESSEX: CALCULUS 9
Let a2 D mg=k, where a > 0. Thus, we have
is indeterminate (0/0) if a D b.
If a D b the original differential equation becomes
Z
dx
D k.a " x/2 ;
dt
which is separable and yields the solution
Z
Z
dx
1
D
D
k
dt D k t C C:
a"x
.a " x/2
1
1
1
Since x.0/ D 0, we have C D , so
D kt C .
a
a"x
a
Solving for x, we obtain
kt
dv
D
CC
a2 ˇ" v 2 ˇ m
ˇa C vˇ
1
ˇ D kt C C
ln ˇˇ
2a
a"vˇ
m
r
ˇ
ˇ
ˇa C vˇ
kg
2ak
t
ˇ
ˇ
ln ˇ
D
C C1 D 2
t C C1
a"vˇ
m
m
p
aCv
D C2 e 2t kg=m :
a"v
a2 k t
:
1 C ak t
This solution also results from evaluating the limit of solution obtained for the case a ¤ b as b approaches a (using
l’H^opital’s Rule, say).
xD
dv
D mg " kv has constant solution
dt
v D mg=k. However this solution does not satisfy the
initial condition v.0/ D 0. For other solutions we separate
variables:
Z
Z
dv
D
dt
k
g" v
ˇm
ˇ
k ˇˇ
m ˇˇ
" lnˇg " v ˇ D t C C:
k
m
m
k
Since v.0/ D 0, therefore C D " ln g. Also, g " v
k
m
remains positive for all t > 0, so
g
m
ln
Dt
k
k
g" v
m
k
g" v
m D e !k t=m
g
&
mg %
) v D v.t / D
1 " e !k t=m :
k
mg
Note that lim v.t / D
, the constant solution of the
t!1
k
differential equation noted earlier. This limiting velocity
can be obtained directly from the differential equation by
dv
setting
D 0.
dt
p
29. y D mg=k is a constant solution of the equation. For
other solutions we proceed by separation of variables:
Assuming v.0/ D 0, we get C2 D 1. Thus
p
a C v D e 2t kg=m .a " v/
p
%
&
% p
&
v 1 C e 2t kg=m D a e 2t kg=m " 1
r
&
mg % 2t pkg=m
D
e
"1
k
p
r
2t kg=m
mg e
"1
p
vD
k e 2t kg=m C 1
28. The equationt m
dv
D mg " kv 2
dt
dv
k
D g " v2
dt
m
dv
D dt
k
g " v2
Z m
Z
k
dv
kt
D
dt D
C C:
mg
2
m
m
"v
k
m
308
Clearly v !
setting
30.
r
mg
as t ! 1. This also follows from
k
dv
D 0 in the given differential equation.
dt
The balance in the account after t years is y.t / and
y.0/ D 1000. The balance must satisfy
dy
y2
D 0:1y "
dt
1; 000; 000
dy
105 y " y 2
D
dt
106
Z
Z
dy
dt
D
105 y " y 2
106
$
Z #
1
1
t
C
1
C
dy D 6 " 5
105
y
105 " y
10
10
t
ln jyj " ln j105 " yj D
"C
10
5
10 " y
D e C !.t=10/
y
105
y D C !.t=10/
:
e
C1
Since y.0/ D 1000, we have
1000 D y.0/ D
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105
C1
eC
)
C D ln 99;
lOMoARcPSD|6566483
INSTRUCTOR’S SOLUTIONS MANUAL
and
Hence,
105
yD
99e !t=10 C 1
The balance after 1 year is
105
yD
99e !1=10 C 1
REVIEW EXERCISES 7 (PAGE 459)
:
.500 C t /5 x D 0:12
Z
.500 C t /5 dt D 0:02.500 C t /6 C C
) x D 0:02.500 C t / C C.500 C t /!5 :
' $1; 104:01:
Since x.0/ D 50, we have C D 1:25 & 1015 and
As t ! 1, the balance can grow to
x D 0:02.500 C t / C .1:25 & 1015 /.500 C t /!5 :
105
105
lim y.t / D lim .4:60!0:1t/
D
D $100; 000:
t!1
t!1 e
0C1
C1
After 40 min, there will be
For the account to grow to $50,000, t must satisfy
x D 0:02.540/ C .1:25 & 1015 /.540/!5 D 38:023 kg
100; 000
99e !t=10 C 1
!t=10
99e
C1D2
t D 10 ln 99 ' 46 years:
50; 000 D y.t / D
)
)
of salt in the tank.
Review Exercises 7 (page 459)
31. The hyperbolas xy D C satisfy the differential equation
y Cx
dy
D 0;
dx
or
dy
y
D" :
dx
x
1.
3 cm
Curves that intersect these hyperbolas at right angles must
x
dy
D , or x dx D y dy, a separated
therefore satisfy
dx
y
equation with solutions x 2 " y 2 D C , which is also a
family of rectangular hyperbolas. (Both families are degenerate at the origin for C D 0.)
1 cm
5 cm
32. Let x.t / be the number of kg of salt in the solution in the
tank after t minutes. Thus, x.0/ D 50. Salt is coming into
the tank at a rate of 10 g=L & 12 L=min D 0:12 kg=min.
Since the contents flow out at a rate of 10 L=min, the
volume of the solution is increasing at 2 L=min and thus,
at any time t , the volume of the solution is 1000 C 2t L.
x.t /
L. Hence,
Therefore the concentration of salt is
1000 C 2t
salt is being removed at a rate
Fig. R-7-1
The volume of thread that can be wound on the left spool
is #.32 " 12 /.5/ D 40# cm3 .
The height of the winding region of the right spool at
distance r from the central axis of the spool is of the form
h D A C Br. Since h D 3 if r D 1, and h D 5 if r D 3,
we have A D 2 and B D 1, so h D 2 C r. The volume of
thread that can be wound on the right spool is
Therefore,
5x
dx
D 0:12 "
dt
500 C t
dx
5
C
x D 0:12:
dt
500 C t
2#
5
Let - D
dt D 5 ln j500 C t j D ln.500 C t /5 for
500 C t
t > 0. Then e ' D .500 C t /5 , and
i
dx
d h
.500 C t /5 x D .500 C t /5
C 5.500 C t /4 x
dt
dy
#
$
dx
5x
D .500 C t /5
C
dy
500 C t
5
D 0:12.500 C t / :
1 cm 5 cm
3 cm
1 cm
5x.t /
x.t /
kg=L & 10 L=min D
kg=min:
1000 C 2t
500 C t
Z
3 cm
Z
3
1
$ˇ3
#
100#
r 3 ˇˇ
2
cm3 :
r.2 C r/ dr D 2# r C
ˇ D
3 ˇ
3
1
100
.1; 000/ D 833:33 m of
The right spool will hold
3 & 40
thread.
2.
Let A.y/ be the cross-sectional area of the bowl at height
y above the bottom. When the depth of water in the bowl
is Y , then the volume of water in the bowl is
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V .Y / D
Z
Y
A.y/ dy:
0
309
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REVIEW EXERCISES 7 (PAGE 459)
ADAMS and ESSEX: CALCULUS 9
The water evaporates at a rate proportional to exposed
surface area. Thus
60ı
dV
D kA.Y /
dt
dV dY
D kA.Y /
d Y dt
dY
A.Y /
D kA.Y /:
dt
10 cm
Hence d Y =dt D k; the depth decreases at a constant
rate.
3. The barrel is generated by revolving x D a " by 2 ,
."2 $ y $ 2/, about the y-axis. Since the top and bottom
disks have radius 1 ft, we have a " 4b D 1. The volume
of the barrel is
V D2
Z
0
2
Fig. R-7-4
5.
#.a " by 2 /2 dy
#
$ˇ2
2aby 3
b 2 y 5 ˇˇ
2
D 2# a y "
C
ˇ
ˇ
3
5
#
$0
16
32
DD 2# 2a2 " ab C b 2 :
3
5
1
cosh.ax/ from x D 0 to x D 1 is
a
Z 1q
Z 1
sD
1 C sinh2 .ax/ dx D
cosh.ax/ dx
0
0
ˇ1
ˇ
1
1
ˇ
D sinh.ax/ˇ D sinh a:
ˇ
a
a
The arc length of y D
0
1
sinh a D 2, that is, sinh a D 2a. Solving this
a
by Newton’s Method or a calculator solve function, we get
a ' 2:1773.
p
The area of revolution of y D x, .0 $ x $ 6/, about the
x-axis is
s
# $2
Z 6
dy
S D 2#
y 1C
dx
dx
0
r
Z 6
p
1
x 1C
dx
D 2#
4x
0
r
Z 6
1
D 2#
x C dx
4
0
#
$ ˇ6
(
)
4#
1 3=2 ˇˇ
4# 125 1
62#
D
xC
"
D
sq. units.
ˇ D
ˇ
3
4
3
8
8
3
We want
6.
Since V D 16 and a D 1 C 4b, we have
x
#
$
16
32
2# 2.1 C 4b/2 " b.1 C 4b/ C b 2 D 16
3
5
60
2
D 0:
128b C 80b C 15 "
#
Solving this quadratic gives two solutions, b ' 0:0476
and b ' "0:6426. Since the second of these leads to an
unacceptable negative value for a, we must have
b ' 0:0476, and so a D 1 C 4b ' 1:1904.
0
4. A vertical slice parallel to the top ridge of the solid at
distance
x to the right of the centre
p
p is a rectangle of base
2 100 " x 2 cm�and height 3.10 " x/ cm. Thus the
solid has volume
Z
p
3.10 " x/2 100 " x 2 dx
0
p Z 10 p
p Z
D 40 3
100 " x 2 dx " 4 3
V D2
10
p
0
0
Let u D 100 " x 2
du D "2x dx
p 100#
p Z 100 p
u du
D 40 3
"2 3
4
$ 0
#
p
4
D 1; 000 3 # "
cm3 :
3
310
10
p
x 100 " x 2 dx
7.
The region is a quarter-elliptic disk with semi-axes a D 2
and b D 1. The area of the region is A D #ab=4 D #=2.
The moments about the coordinate axes are
s
Z 2
x2
x2
dx Let u D 1 "
MxD0 D
x 1"
4
4
0
x
du D " dx
2
Z 1
p
4
D2
u du D
3
0
Z 2#
2$
1
x
MyD0 D
1"
dx
2 0
4
#
$ˇ2
x 3 ˇˇ
2
1
x"
D
ˇ D :
2
12 ˇ
3
0
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lOMoARcPSD|6566483
INSTRUCTOR’S SOLUTIONS MANUAL
CHALLENGING PROBLEMS 7 (PAGE 459)
Thus x D MxD0 =A D 8=.3#/ and
y D MyD0 =A& D 4=.3#/. The centroid is
%
or, equivalently, f .a/ C g.a/ D 2a. Thus f and g must
satisfy
8=.3#/; 4=.3#/ .
f .x/ C g.x/ D 2x
for every x > 0.
8.
y
11.
1
Z
dy
3y
dy
dx
D
)
D3
dx
x"1
y
x"1
) ln jyj D ln jx " 1j3 C ln jC j
) y D C.x " 1/3 :
Since y D 4 when x D 2, we have 4 D C.2 " 1/3 D C , so
the equation of the curve is y D 4.x " 1/3 .
3
x
12. The ellipses 3x 2 C 4y 2 D C all satisfy the differential
equation
Fig. R-7-8
6x C 8y
Let the disk have centre (and therefore centroid) at .0; 0/.
Its area is 9#. Let the hole have centre (and therefore
centroid) at .1; 0/. Its area is #. The remaining part has
area 8# and centroid at .x; 0/, where
13.
5
20; 000
20
dy D 20; 000 ln
' 27; 726 N!cm:
y
5
10. We are told that for any a > 0,
#
Z a h%
Z
&2 %
&2 i
f .x/ " g.x/
dx D 2#
0
a
0
h
i
x f .x/"g.x/ dx:
Differentiating both sides of this equation with respect to
a, we get
%
&2 %
&2
h
i
f .a/ " g.a/ D 2a f .a/ " g.a/ ;
Z
2
0
sin.2# t /e 0:04.2!t/ dt ' $8; 798:85:
(We omit the details of evaluation of the integral, which is
done by the method of Example 4 of Section 7.1.)
We are told that F D 1; 000 N when y D 20 cm. Thus
k D 20; 000 N!cm. The work done by the piston as it
descends to 5 cm is
20
The original $8,000 grows to $8; 000e 0:08 in two years.
Between t and t C dt , an amount $10; 000 sin.2# t / dt
comes in, and this grows to $10; 000 sin.2# t /e 0:04.2!t/ dt
by the end of two years. Thus the amount in the account
after 2 years is
8; 000e 0:08 C10; 000
k
kA
D :
F .y/ D P .y/A D
Ay
y
Z
dy
3x
D" :
dx
4y
The family is given by y 3 D C x 4 .
9. Let the area of cross-section of the cylinder be A. When
the piston is y cm above the base, the volume of gas in
the cylinder is V D Ay, and its pressure P .y/ satisfies
P .y/V D k (constant). The force exerted by the piston is
W D
or
A family of curves that intersect these ellipses at right
4y
dy
D
.
angles must therefore have slopes given by
dx
3x
Thus
Z
Z
dy
dx
3
D4
y
x
3 ln jyj D 4 ln jxj C ln jC j:
.9#/.0/ D .8#/x C .#/.1/:
Thus x D "1=8. The centroid of the remaining part is
1=8 ft from the centre of the disk on the side opposite the
hole.
dy
D 0;
dx
Challenging Problems 7 (page 459)
1.
a) The nth bead extends from x D .n " 1/# to x D n#,
and has volume
Z n!
Vn D #
e !2kx sin2 x dx
.n!1/!
Z
# n!
D
e !2kx .1 " cos.2x// dx
2 .n!1/!
Let x D u C .n " 1/#
Zdx D du
i
# ! !2ku !2k.n!1/! h
1 " cos.2u C 2.n " 1/#/ du
e
e
D
2 0
Z
# !2k.n!1/! ! !2ku
D e
e
.1 " cos.2u// du
2
0
D e !2k.n!1/! V1 :
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CHALLENGING PROBLEMS 7 (PAGE 459)
ADAMS and ESSEX: CALCULUS 9
e !2k n! V1
VnC1
D !2k.n!1/!
D e !2k! ; which deVn
e
V1
pends on k but not n.
or, equivalently, a.100 " k 2 /2 D 4. The volume of the pool
is
Thus
b) VnC1 =Vn D 1=2 if "2k# D ln.1=2/ D " ln 2, that is,
if k D .ln 2/=.2#/.
c) Using the result of Example 4 in Section 7.1, we calculate the volume of the first bead:
Z
# ! !2kx
V1 D
e
.1 " cos.2x// dx
2 0
ˇ!
ˇ!
# e !2kx .2 sin.2x/ " 2k cos.2x// ˇˇ
#e !2kx ˇˇ
D
ˇ "
ˇ
ˇ
"4k ˇ
2
4.1 C k 2 /
0
0
#
#
!2k!
.1 " e !2k! / "
.k
"
ke
/
D
4k
4.1 C k 2 /
#
!2k!
D
.1 " e
/:
4k.1 C k 2 /
#
.1 " e !2k! /
4k.1 C k 2 /
h
%
&2
%
&n!1 i
& 1 C e !2k! C e !2k! C ! ! ! C e !2k!
10
The volume of the hill is
VH D 2#a
Z
k
0
#
$
1
r.r 2 "100/.r 2 "k 2 / dr D 2#a 25k 4 " k 6 :
12
These two volumes must be equal, so k 2 D 100=3 and
k ' 5:77 m. Thus a D 4=.100 " k 2 /2 D 0:0009. The
volume of earth to be moved is VH with these values of a
and k, namely
"
By part (a) and Theorem 1(d) of Section 6.1, the sum
of the volumes of the first n beads is
Sn D
Z
r.100 " r 2 /.r 2 " k 2 / dr
$
#k
1 6
250; 000
2
4
D 2#a
" 2; 500k C 25k " k :
3
12
VP D 2#a
#
100
2#.0:0009/ 25
3
3.
$2
1
"
12
#
100
3
$4 #
' 140 m3 :
y D ax C bx 2 C cx 3
.h; r/
y
!2k n!
#
!2k! 1 " e
.1
"
e
/
4k.1 C k 2 /
1 " e !2k!
#
!2k n!
D
.1 " e
/:
4k.1 C k 2 /
D
x
Thus the total volume of all the beads is
V D lim Sn D
n!1
#
cu. units.:
4k.1 C k 2 /
Fig. C-7-3
f .x/ D ax C bx 2 C cx 3 must satisfy f .h/ D r, f 0 .h/ D 0,
and f 0 .x/ > 0 for 0 < x < h. The first two conditions
require that
ah C bh2 C ch3 D r
2.
10 m
a C 2bh C 3ch2 D 0;
1m
from which we obtain by solving for b and c,
bD
3r " 2ah
;
h2
cD
ah " 2r
:
h3
The volume of the nose cone is then
Fig. C-7-2
h.r/ D a.r 2 " 100/.r 2 " k 2 /;
0
2
2
where 0 < k < 10
2
h .r/ D 2ar.r " k / C 2ar.r " 100/ D 2ar.2r 2 " 100 " k 2 /:
The deepest point occurs where 2r 2 D 100 C k 2 , i.e.,
r 2 D 50 C .k 2 =2/. Since this depth must be 1 m, we
require
$#
$
# 2
k2
k
" 50
50 "
D "1;
a
2
2
312
V .a/ D #
Z h%
&2
#h
f .x/ dx D
.13ahr C 78r 2 C 2a2 h2 /:
210
0
Solving d V =da D 0 gives only one critical point,
a D "13r=.4h/. This is unacceptable, because the condition f 0 .x/ > 0 on .0; h/ forces us to require a # 0. In
fact
f 0 .x/ D a C
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3.ah " 2r/ 2
2.3r " 2ah/
xC
x
h2
h3
lOMoARcPSD|6566483
INSTRUCTOR’S SOLUTIONS MANUAL
CHALLENGING PROBLEMS 7 (PAGE 459)
To minimize this expression for a > 0 we should take
k D 0. This gives f .x/ D ax 2 .1 " x/. To minimize
V .a; k/ for a < 0 we should take k D "1. This gives
f .x/ D "ax.1 " x/2 . Since we want the maximum value
of f to be 2 in either case, we calculate the critical points
of these two possible functions. For a > 0 the CP is
x D 2=3 and f .2=3/ D 2 gives a D 27=2. The volume in
this case is V .27=2; 0/ D .27#=60/.78 " 0/. For a < 0 the
CP is x D 1=3 and f .1=3/ D 2 gives a D "27=2. The
volume in this case is
V ."27=2; "1/ D ".27#=60/.78 " 155/ D .27#=60/.77/.
Thus the minimum volume occurs for
f .x/ D .27=2/x.1 " x/2 , i.e. b D "a D 27=2.
is clearly positive for small x if a > 0. Its two roots are
x1 D h and x2 D h2 a=.3ah " 6r/. a must be restricted
so that x2 is not in the interval .0; h/. If a < 2r= h, then
x2 < 0. If 2r= h < a < 3r= h, then x2 > h. If a > 3r= h,
then 0 < x2 < h. Hence the interval of acceptable values
of a is 0 $ a $ 3r= h. We have
V .0/ D
13# r 2 h
;
35
V
#
3r
h
$
D
9# r 2 h
:
14
The largest volume corresponds to a D 3r= h, which is
the largest allowed value for a and so corresponds to the
bluntest possible nose. The corresponding cubic f .x/ is
f .x/ D
4.
r
.3h2 x " 3hx 2 C x 3 /:
h3
'
a C bx C cx 2 for 0 $ x $ 1
a) If f .x/ D
, then
2
for 1 $ x $ 3
' p C qx C rx
b C 2cx for 0 < x < 1
f 0 .x/ D
. We require that
q C 2rx for 1 < x < 3
aD1
aCbCc D2
b C 2c D m
p C 3q C 9r D 0
pCqCr D2
q C 2r D m:
The solutions of these systems are a D 1, b D 2 " m,
c D m " 1, p D 32 .1 " m/, q D 2m C 1, and
r D " 12 .1 C m/. f .x; m/ is f .x/ with these values
of the six constants.
b) The length of the spline is
L.m/ D
Z
1
0
Z
p
1 C .b C 2cx/2 dx C
1
3
p
1 C .q C 2rx/2 dx
with the values of b, c, q, and r determined above.
A plot of the graph of L.m/ reveals a minimum
value in the neighbourhood of m D "0:3. The derivative of L.m/ is a horrible expression, but Mathematica determined its zero to be about m D "0:281326,
and the corresponding minimum value of L is
about
p 4:41748. The polygonal line ABC has length
3 2 ' 4:24264, which is only slightly shorter.
5. Let b D ka so that the cross-sectional curve is given by
y D f .x/ D ax.1 " x/.x C k/:
The requirement that f .x/ # 0 for 0 $ x $ 1 is satisfied
provided either a > 0 and k # 0 or a < 0 and k $ "1.
The volume of the wall is
V .a; k/ D
Z
0
1
2#.15 C x/f .x/ dx D
#a
.78 C 155k/:
30
6.
Starting with V1 .r/ D 2r, and using repeatedly the formula
Vn .r/ D
Z
r
!r
p
Vn!1 . r 2 " x 2 / dx;
Maple gave the following results:
V1 .r/ D 2r
4
V3 .r/ D # r 3
3
8 2 5
# r
V5 .r/ D
15
16 3 7
V7 .r/ D
# r
105
32 4 9
# r
V9 .r/ D
945
V2 .r/ D # r 2
1
V4 .r/ D # 2 r 4
2
1
V6 .r/ D # 3 r 6
6
1 4 8
V8 .r/ D
# r
24
1 5 10
# r
V10 .r/ D
120
It appears that
1 n 2n
# r ;
and
nŠ
n
2
V2n!1 .r/ D
# n!1 r 2n!1
1 ! 3 ! 5 ! ! ! .2n " 1/
22n!1 .n " 1/Š n!1 2n!1
#
r
:
D
.2n " 1/Š
V2n .r/ D
These formulas predict that
V11 .r/ D
211 5Š 5 11
# r
11Š
and
V12 .r/ D
1 6 12
# r ;
6Š
both of which Maple is happy to confirm.
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CHALLENGING PROBLEMS 7 (PAGE 459)
7.
ADAMS and ESSEX: CALCULUS 9
With y and & as defined in the statement of the problem,
we have
8.
y
Q
y D f .x/
0 $ y $ 10 and
L
0 $ & < #:
P .x; y/
x
.L; 0/
The needle crosses a line if y < 5 sin & . The probability of
this happening is the ratio of the area under the curve to
the area of the rectangle in the figure, that is,
Fig. C-7-8
If Q D .0; Y /, then the slope of PQ is
y"Y
dy
D f 0 .x/ D
:
x"0
dx
Pr D
1
10#
Z
!
0
5 sin & d& D
2
Since jPQj D L, we have .y " Y /2 D L2 "
px . Since the
slope dy=dx is negative at P , dy=dx D " L2 " x 2 =x.
Thus
1
:
#
yD"
!
p
Z p 2
p
L " x2
L C L2 " x 2
" L2 " x 2 CC:
dx D L ln
x
x
Since y D 0 when x D L, we have C D 0 and the
equation of the tractrix is
LC
y D L ln
!
p
p
L2 " x 2
" L2 " x 2 :
x
Note that the first term can be written in an alternate way:
y
y D L ln
y D 10
#
L"
p
x
L2
"
x2
$
"
p
L2 " x 2 :
y D 5 sin x
#
Fig. C-7-7
314
9.
&
a) S.a; a; c/ is the area of the surface obtained by rotating the ellipse .x 2 =a2 / C .y 2 =c 2 / D 1p(where a > c)
about the y-axis. Since y 0 D "cx=.a a2 " x 2 /, we
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INSTRUCTOR’S SOLUTIONS MANUAL
CHALLENGING PROBLEMS 7 (PAGE 459)
have
S.a; a; c/ D 2 & 2#
Z
a
x
0
s
c) Since b D
1C
c2x2
dx
a2 .a2 " x 2 /
p
a4 " .a2 " c 2 /x 2
dx
p
x
a2 " x 2
0
Let x D a sin u
dx D a cos u du
Z
q
4# !=2
D
a sin u a4 " .a2 " c 2 /a2 sin2 u du
a 0
Z !=2
p
D 4#a
sin u a2 " .a2 " c 2 /.1 " cos2 u/ du
4#
D
a
Z
a
#
S.a; b; c/ '
b"c
a"c
#
$
b"c
a"c
aC
$
#
a"b
a"c
$
S.a; a; c/ C
c, we use
#
a"b
a"c
$
S.a; c; c/:
0
Let v D cos u
dv D " sin u du
Z 1p
D 4#a
c 2 C .a2 " c 2 /v 2 dv:
0
d) We cannot evaluate S.3; 2; 1/ even numerically at
this stage. The double integral necessary to calculate
it is not treated until a later chapter. (The value is
approximately 48.882 sq. units.) However, using the
formulas obtained above,
This integral can now be handled using tables or
computer algebra. It evaluates to
aC
2#ac 2
ln
S.a; a; c/ D 2#a C p
2
2
a "c
2
p
a2 " c 2
c
!
:
b) S.a; c; c/ is the area of the surface obtained by rotating the ellipse
p of part (a) about the y-axis. Since
y 0 D "cx=.a a2 " x 2 /, we have
S.a; c; c/ D 2 & 2#
Z
a
y
0
s
1C
S.3; 3; 1/ C S.3; 1; 1/
S.3; 2; 1/ '
2
$
#
p
1
6#
18#
D
18# C p ln.3 C 8/ C 2# C p cos!1 .1=3/
2
8
8
' 49:595 sq. units.
c2x2
dx
" x2/
a2 .a2
p
p
4#c
a4 " .a2 " c 2 /x 2
2 " x2
dx
a
p
2
a
a2 " x 2
Z0 a p
4#c
D 2
a4 " .a2 " c 2 /x 2 dx
a
0
s
Z a
a2 " c 2 2
D 4#c
1"
x dx
a4
0
c
2#a2 c
cos!1 :
D 2#c 2 C p
a
a2 " c 2
D
Z
a
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315
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SECTION 8.1 (PAGE 472)
ADAMS and ESSEX: CALCULUS 9
CHAPTER 8. CONICS, PARAMETRIC
CURVES, AND POLAR CURVES
Section 8.1 Conics
1.
8.
If x 2 C 4y 2 " 4y D 0, then
$
#
1
D 1;
x2 C 4 y2 " y C
4
(page 472)
or
#
$
1
0;
, semi2 p
#
$
3 1
1
major axis 1, semi-minor axis , and foci at ˙
;
.
2
2 2
This represents an ellipse with centre at
The ellipse with foci .0; ˙2/ has major axis along the yaxis and c D 2. If a D 3, then b 2 D 9 " 4 D 5. The
ellipse has equation
y
y2
x2
C
D 1:
5
9
x 2 C4y 2 !4yD0
1
1
2
2. The ellipse with foci .0; 1/ and .4; 1/ has c D 2, centre
.2; 1/, and major axis along y D 1. If $ D 1=2, then
a D c=$ D 4 and b 2 D 16 " 4 D 12. The ellipse has
equation
.x " 2/2
.y " 1/2
C
D 1:
16
12
3. A parabola with focus .2; 3/ and vertex .2; 4/ has a D "1
and principal axis x D 2. Its equation is
.x " 2/2 D "4.y " 4/ D 16 " 4y.
4. A parabola with focus at .0; "1/ and principal axis
along y D "1 will have vertex at a point of the
form .v; "1/. Its equation will then be of the form
.y C 1/2 D ˙4v.x " v/. The origin lies on this curve
if
1 D ˙4."v 2 /. Only the " sign is possible, and in this
case v D ˙1=2. The possible equations for the parabola
are .y C 1/2 D 1 ˙ 2x.
1
x
Fig. 8.1-8
9.
If 4x 2 C y 2 " 4y D 0, then
4x 2 C y 2 " 4y C 4 D 4
4x 2 C .y " 2/2 D 4
x2 C
.y " 2/2
D1
4
This is an ellipse with semi-axes 1 and 2, centred at .0; 2/.
y
4
5. The hyperbola with semi-transverse axis a D 1 and foci
.0; ˙2/ has transverse axis along the y-axis, c D 2, and
b 2 D c 2 " a2 D 3. The equation is
y2 "
.y " 12 /2
x2
C
D 1:
1
1
4
.!1;2/
4x 2 Cy 2 !4yD0
.1;2/
2
x2
D 1:
3
x
6. The hyperbola with foci at .˙5; 1/ and asymptotes
x D ˙.y " 1/ is rectangular, has centre at .0; 1/ and
has transverse axis along the line y D 1. Since c D 5
and a D b (because the asymptotes are perpendicular to
each other) we have a2 D b 2 D 25=2. The equation of the
hyperbola is
25
x 2 " .y " 1/2 D
:
2
7.
If x 2 C y 2 C 2x D "1, then .x C 1/2 C y 2 D 0. This
represents the single point ."1; 0/.
316
Fig. 8.1-9
10. If 4x 2 " y 2 " 4y D 0, then
4x 2 " .y 2 C 4y C 4/ D "4;
or
x 2 .y C 2/2
"
D "1:
1
4
This represents a hyperbola with centre at .0; "2/, semitransversepaxis 2, semi-conjugate axis 1, and foci at
.0; "2 ˙ 5/. The asymptotes are y D ˙2x " 2.
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 8.1 (PAGE 472)
!
"
This is a hyperbola with centre " 32 ; 1 , and asymptotes
p
the straight lines 2x C 3 D ˙2 2.y " 1/.
y
y
x
!2
4x 2 !y 2 !4yD0
3
.! ;1/
2
1
.!3;1/
x 2 !2y 2 C3xC4yD2
x
Fig. 8.1-10
11.
If x 2 C 2x " y D 3, then .x C 1/2 " y D 4.
Thus y D .x C 1/2 " 4. This is a parabola with vertex
."1; "4/, opening upward.
y
Fig. 8.1-13
14.
If 9x 2 C 4y 2 " 18x C 8y D "13, then
9.x 2 " 2x C 1/ C 4.y 2 C 2y C 1/ D 0
x
,9.x " 1/2 C 4.y C 1/2 D 0:
This represents the single point .1; "1/.
x 2 C2x!yD3
15.
.!1;!4/
9.x 2 " 2x C 1/ C 4.y 2 C 2y C 1/ D 23 C 9 C 4 D 36
Fig. 8.1-11
9.x " 1/2 C 4.y C 1/2 D 36
12. If x C 2y C 2y 2 D 1, then
,
If 9x 2 C 4y 2 " 18x C 8y D 23, then
.x " 1/2
.y C 1/2
C
D 1:
4
9
$
#
3
1
D "x
2 y2 C y C
4
2
#
$
3
1 2
x D "2 yC
:
2
2
This is an ellipse with centre .1; "1/, and semi-axes 2 and
3.
y
.1;2/
This represents a parabola with vertex at . 32 ; " 12 /, focus at
; " 12 / and directrix x D 13
.
. 11
8
8
y
.!1;!1/
x
.3;!1/
9x 2 C4y 2 !18xC8yD23
xC2yC2y 2 D1
x
%
13. If x 2 " 2y 2 C 3x C 4y D 2, then
#
$
9
3 2
" 2.y " 1/2 D
xC
2
4
!
"
3 2
2
xC 2
.y " 1/
"
D1
9
9
8
3 1
;!
2 2
&
.1;!4/
Fig. 8.1-15
16.
Fig. 8.1-12
4
.1;!1/
The equation .x " y/2 " .x C y/2 D 1 simplifies to
4xy D "1 and hence represents a rectangular hyperbola
with centre at the origin, asymptotes along the coordinate
axes, transverse axis! along"y D "x,
conjugate
axis along
!
"
y D x, vertices at 12 ; " 12 and " 12 ; 21 , semi-transverse
p
and semi-conjugate
q axes equal to 1= 2, semi-focal sepa-
ration equal to 12 C 21 D 1, and hence foci at the points
%
&
%
&
p
p1 ; " p1
and " p1 ; p1 . The eccentricity is 2.
2
2
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2
317
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SECTION 8.1 (PAGE 472)
ADAMS and ESSEX: CALCULUS 9
p
This is a rectangular hyperbola
with centre .0;
p " 2/,
p
The
semi-axes a D b D 2, and eccentricity 2. p
semi-focal separation is 2; the
foci
are
at
.˙2;
"
2/. The
p
asymptotes are u D ˙.v C 2/.
In terms of thep
original coordinates,
the centre is .1; "1/,
p
the foci are .˙ 2 C 1; ˙ 2 " 1/, and the asymptotes are
x D 1 and y D "1.
y
x
%
1 1
2;2
&
y
.x!y/2 !.xCy/2 D1
xyCx!yD2
Fig. 8.1-16
.1;!1/
17.
x
The parabola has focus at .3; 4/ and principal axis along
y D 4. The vertex must be at a point of the form .v; 4/,
in which case a D ˙.3 " v/ and the equation of the
parabola must be of the form
.y " 4/2 D ˙4.3 " v/.x " v/:
This curve passes through the origin if 16 D ˙4.v 2 " 3v/.
We have two possible equations for v: v 2 " 3v " 4 D 0
and v 2 " 3v C 4 D 0. The first of these has solutions
v D "1 or v D 4. The second has no real solutions. The
two possible equations for the parabola are
.y " 4/2 D 4.4/.x C 1/
2
.y " 4/ D 4."1/.x " 4/
or
or
y 2 " 8y D 16x
y 2 " 8y D "4x
18. The foci of the ellipse are .0; 0/ and .3; 0/, so the centre
is .3=2; 0/ and c D 3=2. The semi-axes a and b must
satisfy a2 " b 2 D 9=4. Thus the possible equations of the
ellipse are
y2
.x " .3=2//2
C
D 1:
.9=4/ C b 2
b2
19. For xy C x " y D 2 we have A D C D 0, B D 1. We
therefore rotate the coordinate axes (see text pages 407–
408) through angle & D #=4.
(Thus cot 2& D 0 D .A " C /=B.) The transformation is
1
x D p .u " v/;
2
Fig. 8.1-19
20.
We have x 2 C 2xy C y 2 D 4x " 4y C 4 and
A D 1, B D 2, C D 1, D D "4, E D 4 and
F D "4. We rotate the axes through angle & satisfying
#
tan 2& D B=.A " C / D 1 ) & D . Then A0 D 2,
4
p
B 0 D 0, C 0 D 0, D 0 D 0, E 0 D 4 2 and the transformed
equation is
p
2u2 C 4 2v " 4 D 0
)
$
#
p
1
u2 D "2 2 v " p
2
which represents
a& parabola with vertex at
%
1
p
.u; v/ D 0;
and principal axis along u D 0.
2
The distance
a
from
thepfocus to the vertex is given by
p
4a D 2 2, so apD 1= 2 and the focus is at .0; 0/. The
directrix is v D 2.
1
1
Since x D p .u " v/ and y D p .u C v/, the ver2
2
tex of the parabola in terms of xy-coordinates is ." 12 ; 12 /,
and the focus is .0; 0/. The directrix is x " y D 2. The
principal axis is y D "x.
y
yD!x
1
y D p .u C v/:
2
x 2 C2xyCy 2 D4x!4yC4
.!1=2;1=2/
The given equation becomes
x
1 2
1
1
.u " v 2 / C p .u " v/ " p .u C v/ D 2
2
2
2
p
u2 " v 2 " 2 2v D 4
%
p &2
u2 " v C 2 D 2
p
u2 .v C 2/2
"
D 1:
2
2
318
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Fig. 8.1-20
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INSTRUCTOR’S SOLUTIONS MANUAL
21.
SECTION 8.1 (PAGE 472)
For 8x 2 C 12xy C 17y 2 D 20, we have A D 8, B D 12,
C D 17, F D "20. Rotate the axes through angle & where
tan 2& D
Then A0 D 0, B 0 D 0, C 0 D 5, D 0 D
transformed equation is
B
12
4
D"
D" :
A"C
9
3
5v 2 C
Thus cos 2& D 3=5, sin 2& D "4=5, and
2 cos2 & " 1 D cos 2& D
3
5
)
cos2 & D
4
:
5
p
5u D 0
5, E 0 D 0 and the
1
v2 D " p u
5
which represents
a parabola
with vertex at .u; v/ D .0; 0/,
$
#
1
1
focus at " p ; 0 . The directrix is u D p and the
4 5
4 5
2
1
principal axis is v D 0. Since x D p u " p v and
5
5
1
2
y D p u C p v, in terms of the xy-coordinates, the ver5
5
#
$
1
1
tex is at .0; 0/, the focus at " ; "
. The directrix is
10 20
1
2x C y D 4 and the principal axis is 2y " x D 0.
2
1
We may therefore take cos & D p , and sin & D " p .
5
5
The transformation is therefore
2
1
2
1
x D p uC p v
uD p x" p y
5
5
5
5
1
2
1
2
y D "p u C p v
vD p xCp y
5
5
5
5
The coefficients of the transformed equation are
# $
# $
# $
4
2
1
A0 D 8
C 12 "
C 17
D5
5
5
5
B0 D 0
# $
# $
# $
2
4
1
" 12 "
C 17
D 20:
C0 D 8
5
5
5
y
x
x 2 !4xyC4y 2 C2xCyD0
The transformed equation is
5u2 C 20v 2 D 20;
)
p
xD2y
or
u2
C v 2 D 1:
4
This is an ellipse with centre
p .0; 0/, semi-axes a D 2 and
b D 1, and foci at u D ˙ 3, v D 0.
In terms of the original coordinates,
the centre is .0; 0/,
p
p !
2 3
3
the foci are ˙ p ; " p .
5
5
Fig. 8.1-22
y
23.
8x 2 C12xyC17y 2 D20
p
The distance from P to F is x 2 C y 2 .
The distance from P to D is x C p. Thus
p
x2 C y2
D$
xCp
x 2 C y 2 D $ 2 .x 2 C 2px C p 2 /
x
.1 " $ 2 /x 2 C y 2 " 2p$ 2 x D $ 2 p 2 :
Fig. 8.1-21
2
y
2
22. We have x "4xy C4y C2x Cy D 0 and A D 1, B D "4,
C D 4, D D 2, E D 1 and F D 0. We rotate the axes
through angle & satisfying tan 2& D B=.A " C / D 43 . Then
p
5
3
) cos 2& D
sec 2& D 1 C tan2 2& D
3 r
5
r
8̂
1 C cos 2&
4
2
ˆ
D
D p I
< cos & D
2
5
r
r 5
)
ˆ
1 " cos 2&
1
1
:̂ sin & D
D
D p :
2
5
5
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P D.x;y/
xD!p
D
F
x
Fig. 8.1-23
319
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SECTION 8.1 (PAGE 472)
24.
ADAMS and ESSEX: CALCULUS 9
y
Let the equation of the parabola be y 2 D 4ax. The focus
F is at .a; 0/ and vertex at .0; 0/. Then the distance from
the vertex
p to the focus is a. At x D a,
y D 4a.a/ D ˙2a. Hence, ` D 2a, which is twice the
distance from the vertex to the focus.
`
a c
x
x2 y2
!
D1
a2 b 2
y
y 2 D4ax
`
Fig. 8.1-26
x
.a;0/
27.
S2
Fig. 8.1-24
C2
`2
c2
25. We have 2 C 2 D 1. Thus
a
b
B
F2
#
$
c2
`2 D b 2 1 " 2
but c 2 D a2 " b 2
a
#
$
b2
a2 " b 2
D b2 2 :
D b2 1 "
2
a
a
P
F1
S1
C1
Therefore ` D b 2 =a.
y
b
A
x2 y2
C
D1
a2 b 2
`
c
a
x
V
Fig. 8.1-27
Let the spheres S1 and S2 intersect the cone in the circles
C1 and C2 , and be tangent to the plane of the ellipse at
the points F1 and F2 , as shown in the figure.
Let P be any point on the ellipse, and let the straight line
through P and the vertex of the cone meet C1 and C2
at A and B respectively. Then PF1 D PA, since both
segments are tangents to the sphere S1 from P . Similarly,
PF2 D PB.
Thus PF1 C PF2 D PA C PB D AB D constant
(distance from C1 to C2 along all generators of the cone
is the same.) Thus F1 and F2 are the foci of the ellipse.
Fig. 8.1-25
y2
x2
"
D 1. The
a2
b2
vertices
are
at
.˙a;
0/
and
the
foci
are
at
.˙c;
0/ where
p
p
c D a2 C b 2 . At x D a2 C b 2 ,
26. Suppose the hyperbola has equation
a2 C b 2 y 2
" 2 D1
a2
b
.a2 C b 2 /b 2 " a2 y 2 D a2 b 2
yD˙
Hence, ` D
320
b2
.
a
b2
:
a
28.
Let F1 and F2 be the points where the plane is tangent to
the spheres. Let P be an arbitrary point P on the hyperbola in which the plane intersects the cone. The spheres
are tangent to the cone along two circles as shown in the
figure. Let PAVB be a generator of the cone (a straight
line lying on the cone) intersecting these two circles at
A and B as shown. (V is the vertex of the cone.) We
have PF1 D PA because two tangents to a sphere from
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 8.2 (PAGE 478)
a point outside the sphere have equal lengths. Similarly,
PF2 D PB. Therefore
PF2 " PF1 D PB " PA D AB D constant,
F
P
Y
since the distance between the two circles in which the
spheres intersect the cone, measured along the generators
of the cone, is the same for all generators. Hence, F1 and
F2 are the foci of the hyperbola.
C
V
X
Q
P
F1
A
A
Fig. 8.1-29
V
B
Section 8.2 Parametric Curves
1.
F2
(page 478)
If x D t , y D 1 " t , .0 $ t $ 1/ then
x C y D 1. This is a straight line segment.
y
Fig. 8.1-28
1
xDt
yD1!t
.0$t$1/
x
1
Fig. 8.2-1
2.
29. Let the plane in which the sphere is tangent to the cone
meet AV at X. Let the plane through F perpendicular to
the axis of the cone meet AV at Y . Then VF D V X,
and, if C is the centre of the sphere, F C D XC . Therefore V C is perpendicular to the axis of the cone. Hence
YF is parallel to V C , and we have Y V D V X D VF .
If P is on the parabola, FP ? VF , and the line from P
to the vertex A of the cone meets the circle of tangency of
the sphere and the cone at Q, then
If x D 2 " t and y D t C 1 for 0 $ t < 1, then
y D 2 " x C 1 D 3 " x for "1 < x $ 2, which is a half
line.
y
xD2!t
yDtC1
.2;1/
FP D PQ D YX D 2V X D 2VF:
x
Fig. 8.2-2
Since FP D 2VF , FP is the semi-latus rectum of the
parabola. (See Exercise 18.) Therefore F is the focus of
the parabola.
3.
1
If x D 1=t , y D t " 1, .0 < t < 4/, then y D " 1. This
x
is part of a hyperbola.
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SECTION 8.2 (PAGE 478)
ADAMS and ESSEX: CALCULUS 9
y
y
&
%
1
;3
4
tD4
bxDay
1
yD x ! 1
tD0
a
x
tD1
x
yD!1
bxD!ay
Fig. 8.2-3
4. If x D
Fig. 8.2-6
1
t
and y D
for "1 < t < 1, then
1 C t2
1 C t2
1
1 C t2
D
Dx
.1 C t 2 /2
1 C t2
#
$
1 2
1
,
x"
C y2 D :
2
4
7.
x2 C y2 D
If x D 3 sin # t , y D 4 cos # t , ."1 $ t $ 1/, then
y2
x2
C
D 1. This is an ellipse.
9
16
y
tD0
x2 y2
9 C 16 D1
This curve consists of all points of the circle with centre
at . 21 ; 0/ and radius 12 except the origin .0; 0/.
y
x
tD1
tD!1 tD1
tD0
tD!1
x
Fig. 8.2-7
xD1=.1Ct 2 /
yDt=.1Ct 2 /
8.
Fig. 8.2-4
5. If x D 3 sin 2t , y D 3 cos 2t , .0 $ t $ #=3/, then
x 2 C y 2 D 9. This is part of a circle.
y
tD0
If x D cos sin s and y D sin sin s for "1 < s < 1, then
x 2 C y 2 D 1. The curve consists of the arc of this circle extending from .a; "b/ through .1; 0/ to .a; b/ where
a D cos.1/ and b D sin.1/, traversed infinitely often back
and forth.
y
xDcos sin s
yDsin sin s
x 2 Cy 2 D9
x
1 rad
!
tD
3
x
Fig. 8.2-5
6. If x D a sec t and y D b tan t for "
#
#
< t < , then
2
2
Fig. 8.2-8
x2 y2
" 2 D sec2 t " tan2 t D 1:
a2
b
9.
The curve is one arch of this hyperbola.
322
If x D cos3 t , y D sin3 t , .0 $ t $ 2#/, then
x 2=3 C y 2=3 D 1. This is an astroid.
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INSTRUCTOR’S SOLUTIONS MANUAL
y
SECTION 8.2 (PAGE 478)
and
tD!=2
1 D .cos2 t Csin2 t /2 D cos4 t Csin4 t C2 cos2 t sin2 t:
x 2=3 Cy 2=3 D1
Hence,
tD0
tD!
tD2!
1 C .x " y/2 D 2.cos4 t C sin4 t / D 2.x C y/:
(b) If x D sec4 t and y D tan4 t , then
tD3!=2
.x " y/2 D .sec4 t " tan4 t /2
Fig. 8.2-9
10. If x D 1 "
p
D .sec2 t C tan2 t /2
D sec4 t C tan4 t C 2 sec2 t tan2 t
4 " t 2 and y D 2 C t for "2 $ t $ 2 then
and
.x " 1/2 D 4 " t 2 D 4 " .y " 2/2 :
1 D .sec2 t "tan2 t /2 D sec4 t Ctan4 t "2 sec2 t tan2 t:
The parametric curve is the left half of the circle of radius 4 centred at .1; 2/, and is traced in the direction of
increasing y.
Hence,
1 C .x " y/2 D 2.sec4 t C tan4 t / D 2.x C y/:
y
p
xD1! 4!t 2
yD2Ct
!2$t$2
(c) Similarly, if x D tan4 t and y D sec4 t , then
1 C .x " y/2 D 1 C .y " x/2
D .sec2 t " tan2 t /2 C .sec4 t " tan4 t /2
.1;2/
D 2.tan4 t C sec4 t /
D 2.x C y/:
These three parametric curves above correspond to
different parts of the parabola 1C.x "y/2 D 2.x Cy/,
as shown in the following diagram.
x
Fig. 8.2-10
11.
y
x D cosh t , y D sinh t represents the right half (branch) of
the rectangular hyperbola x 2 " y 2 D 1.
xDtan4 t
yDsec4 t
The parabola
12. x D 2 " 3 cosh t , y D "1 C 2 sinh t represents the left half
(branch) of the hyperbola
2.xCy/D1C.x!y/2
1
xDcos4 t
yDsin4 t
.x " 2/2 .y C 1/2
"
D 1:
9
4
xDsec4 t
yDtan4 t
1
x
Fig. 8.2-14
13. x D t cos t , y D t sin t , .0 $ t $ 4#/ represents two
revolutions of a spiral curve winding outwards from the
origin in a counterclockwise direction. The point on the
curve corresponding to parameter value t is t units distant
from the origin in a direction making angle t with the
positive x-axis.
14.
(a) If x D cos4 t and y D sin4 t , then
15.
16.
The slope of y D x 2 at x is m D 2x. Hence the parabola
can be parametrized x D m=2, y D m2 =4,
."1 < m < 1/.
If .x; y/ is any point on the circle x 2 C y 2 D R2 other
than .R; 0/, then the line from .x; y/ to .R; 0/ has slope
y
. Thus y D m.x " R/, and
mD
x"R
x 2 C m2 .x " R/2 D R2
.x " y/2 D .cos4 t " sin4 t /2
h
i2
D .cos2 t C sin2 t /.cos2 t " sin2 t /
D .cos2 t " sin2 t /2
4
4
2
2
D cos t C sin t " 2 cos t sin t
.m2 C 1/x 2 " 2xRm2 C .m2 " 1/R2 D 0
h
i
.m2 C 1/x " .m2 " 1/R .x " R/ D 0
) xD
.m2 " 1/R
or x D R:
m2 C 1
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SECTION 8.2 (PAGE 478)
ADAMS and ESSEX: CALCULUS 9
y
The parametrization of the circle in terms of m is given by
Y
.m2 " 1/R
xD
m2 C 1
"
#
.m2 " 1/R
2Rm
yDm
"
R
D" 2
m2 C 1
m C1
b
P D .x; y/
T
a
t
X
x
where "1 < m < 1. This parametrization gives every
point on the circle except .R; 0/.
y
.x;y/
slope m
x
.R;0/
Fig. 8.2-18
x 2 Cy 2 DR2
19.
Fig. 8.2-16
If x D
3t
3t 2
,yD
, .t ¤ "1/, then
3
1Ct
1 C t3
x3 C y3 D
17.
y
As t ! "1, we see that jxj ! 1 and jyj ! 1, but
T
P D .x; y/
a
t
27t 3
27t 3
3
.1
C
t
/
D
D 3xy:
.1 C t 3 /3
.1 C t 3 /2
xCy D
X
x
3t .1 C t /
3t
D
! "1:
3
1Ct
1 " t C t2
Thus x C y D "1 is an asymptote of the curve.
y
tD1
Fig. 8.2-17
t!1
Using triangles in the figure, we see that the coordinates
of P satisfy
x D a sec t;
tD0
y D a sin t:
x
folium of Descartes
The Cartesian equation of the curve is
t!!1!
y2
a2
C 2 D 1:
2
a
x
The curve has two branches extending to infinity to the
left and right of the circle as shown in the figure.
Fig. 8.2-19
20.
18. The coordinates of P satisfy
x D a sec t;
The Cartesian equation is
324
y D b sin t:
Let C0 and P0 be the original positions of the centre of
the wheel and a point at the bottom of the flange whose
path is to be traced. The wheel is also shown in a subsequent position in which it makes contact with the rail at
R. Since the wheel has been rotated by an angle & ,
a2
y2
C
D 1.
b2
x2
OR D arc SR D a&:
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 8.2 (PAGE 478)
y
Thus, the new position of the centre is C D .a&; a/. Let
P D .x; y/ be the new position of the point; then
Tt
x D OR " PQ D a& " b sin.# " & / D a& " b sin &;
y D RC C CQ D a C b cos.# " & / D a " b cos &:
These are the parametric equations of the prolate cycloid.
y
P
Q
P t D.x;y/
b
"
"t
b
S
C0
t
Ct
t
A
a
C
x
a
O
x
R
P0
Fig. 8.2-21
Fig. 8.2-20
If a D 2 and b D 1, then x D 2 cos t , y D 0. This is a
straight line segment.
If a D 4 and b D 1, then
y
x D 3 cos t C cos 3t
D 3 cos t C .cos 2t cos t " sin 2t sin t /
%
&
D 3 cos t C .2 cos2 t " 1/ cos t " 2 sin2 t cos t
xDa"!b sin "
yDa!b cos "
x
2!a
D 2 cos t C 2 cos3 t " 2 cos t .1 " sin2 t / D 4 cos3 t
y D 3 sin t C sin 3t
D 3 sin t " sin 2t cos t " .cos 2t sin t /
%
&
D 3 sin t " 2 sin t cos2 t " .1 " 2 sin2 t / sin t
Fig. 8.2-20
D 2 sin t " 2 sin t C 2 sin3 t C 2 sin3 t D 4 sin3 t
This is an astroid, similar to that of Exercise 11.
22.
21.
Let t and & t be the angles shown in the figure below.
Then arc AT t D arc T t P t , that% is, at D b& t . The centre &C t
of the rolling circle is C t D .a " b/ cos t; .a " b/ sin t .
Thus
a)
From triangles in the figure,
x D jTXj D jOT j tan t D tan t
!
"
y D jOY j D si n !2 " t D jOY j cos t
D jOT j cos t cos t D cos2 t:
y
Since & t " t D
a"b
a
t "t D
t , therefore
b
b
yD1
T
x " .a " b/ cos t D b cos.& t " t /
y " .a " b/ sin t D "b sin.& t " t /:
Y
1
2
#
$
.a " b/t
x D .a " b/ cos t C b cos
b
#
$
.a " b/t
y D .a " b/ sin t " b sin
:
b
X
P D .x; y/
t
O
x
Fig. 8.2-22
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SECTION 8.2 (PAGE 478)
b)
ADAMS and ESSEX: CALCULUS 9
y
1
1
D sec2 t D 1Ctan2 t D 1Cx 2 . Thus y D
.
y
1 C x2
23. x D sin t;
y D sin.2t /
y
x
x
Fig. 8.2-23
28.
24.
x D sin t;
y D sin.3t /
y
x
tangent to x 2 C y 2 D 1. If n # 2 is an integer, the curve
closes after one revolution and has n cusps. The figure
shows the curve for n D 7. If n is a rational number but
not an integer, the curve will wind around the circle more
than once before it closes.
y
Fig. 8.2-24
25. x D sin.2t /;
y D sin.3t /
Fig. 8.2-27
$
#
1
1
cos t C cos..n " 1/t /
x D 1C
n
n
#
$
1
1
y D 1C
sin t " sin..n " 1/t /
n
n
represents a cycloid-like curve that is wound around the
%
&2
inside circle x 2 C y 2 D 1 C .2=n/ and is externally
y
x
x
Fig. 8.2-25
26. x D sin.2t /;
y D sin.5t /
Fig. 8.2-28
y
Section 8.3 Smooth Parametric Curves and
Their Slopes (page 483)
x
27.
Fig. 8.2-26
#
$
1
1
x D 1C
cos t " cos.nt /
n
n
#
$
1
1
y D 1C
sin t " sin.nt /
n
n
represents a cycloid-like curve that is wound around the
circle x 2 C y 2 D 1 instead of extending along the x-axis.
If n # 2 is an integer, the curve closes after one revolution
and has n " 1 cusps. The left figure below shows the curve
for n D 7. If n is a rational number, the curve will wind
around the circle more than once before it closes.
326
1.
2.
3.
y D 2t " 4
x D t2 C 1
dy
dx
D 2t
D2
dt
dt
No horizontal tangents. Vertical tangent at t D 0, i.e., at
.1; "4/.
x D t 2 " 2t
y D t 2 C 2t
dx
dy
D 2t " 2
D 2t C 2
dt
dt
Horizontal tangent at t D "1, i.e., at .3; "1/.
Vertical tangent at t D 1, i.e., at ."1; 3/.
x D t 2 " 2t
y D t 3 " 12t
dx
dy
D 2.t " 1/
D 3.t 2 " 4/
dt
dt
Horizontal tangent at t D ˙2, i.e., at .0; "16/ and .8; 16/.
Vertical tangent at t D 1, i.e., at ."1; "11/.
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INSTRUCTOR’S SOLUTIONS MANUAL
4.
5.
6.
7.
SECTION 8.3 (PAGE 483)
y D 2t 3 C 3t 2
x D t 3 " 3t
dy
dx
D 3.t 2 " 1/
D 6t .t C 1/
dt
dt
Horizontal tangent at t D 0, i.e., at .0; 0/.
Vertical tangent at t D 1, i.e., at ."2; 5/.
At t D "1 (i.e., at .2; 1/) both dx=dt and dy=dt change
sign, so the curve is not smooth there. (It has a cusp.)
2
10.
11.
y D t e 2t
dy
D e 2t .1 C 2t /
dt
dy
e !4 .1 " 4/
3
At t D "2;
D
D" :
dx
2e !4
2
13.
x D t 3 " 2t D "1
y D t C t 3 D 2 at t D 1
dx
dy
D 3t 2 " 2 D 1
D 1 C 3t 2 D 4 at t D 1
dt
dt
Tangent line: x D "1 C t , y D 2 C 4t . This line is at
."1; 2/ at t D 0. If you want to be at that point at t D 1
instead, use
2
x D t e !t =2
y D e !t
dx
dy
2
2
D .1 " t 2 /e !t =2
D "2t e !t
dt
dt
Horizontal tangent at t D 0, i.e., at .0; 1/.
Vertical tangent at t D ˙1, i.e. at .˙e !1=2 ; e !1 /.
y
x D sin t
dy
dx
D cos t
dt
dt
Horizontal tangent
integers n).
Vertical tangent at
."1; "1/.
x D sin.2t /
dx
D 2 cos.2t /
dt
Horizontal tangent
Vertical tangent at
p
.˙1; "1= 2/.
D t sin t
14.
at t D n#, i.e., at .0; "."1/n n#/ (for
t D .n C
1
2 /#,
x D t3 C t
dx
D 3t 2 C 1
dt
dy
At t D 1;
D
dx
x D e 2t
dx
D 2e 2t
dt
x D "1 C .t " 1/ D t " 2;
D sin t " t cos t
i.e. at .1; 1/ and
y D sin t
dy
D cos t
dt
at t D .n C 12 /#, i.e., at .0; ˙1/.
p
t D 12 .n C 21 /#, i.e., at .˙1; 1= 2/ and
15.
3t
3t 2
8.
xD
y
D
1 C t3
1 C t3
dx
dy
3.1 " 2t 3 /
3t .2 " t 3 /
D
D
dt
.1 C t 3 /2
dt
.1 C t 3 /2
Horizontal tangent at t D 0 and t D 21=3 , i.e., at .0; 0/ and
.21=3 ; 22=3 /.
Vertical tangent at t D 2!1=3 , i.e., at .22=3 ; 21=3 /. The
curve also approaches .0; 0/ vertically as t ! ˙1.
9.
12.
y D 2 C 4.t " 1/ D 4t " 2:
#
1
"p
4
2
dx
1
D 1 C sin t D 1 C p
dt
2
#
1
at t D
y D 1 " sin t D 1 " p
4
2
#
dy
1
at t D
D " cos t D " p
dt
2
#4
$
1
1
#
Tangent line: x D " p C 1 C p t ,
4
2
2
1
t
y D1" p " p .
2
2
x D t " cos t D
x D t 3 " t , y D t 2 is at .0; 1/ at t D "1 and t D 1. Since
dy
2t
˙2
D 2
D
D ˙1;
dx
3t " 1
2
the tangents at .0; 1/ at t D ˙1 have slopes ˙1.
16.
x D sin t , y D sin.2t / is at .0; 0/ at t D 0 and t D #.
Since
'
dy
2 cos.2t /
2
if t D 0
D
D
"2 if t D #,
dx
cos t
the tangents at .0; 0/ at t D 0 and t D # have slopes 2
and "2, respectively.
y D 1 " t3
dy
D "3t 2
dt
"3.1/2
3
D" :
3.1/2 C 1
4
x D t4 " t2
y D t 3 C 2t
dx
dy
D 4t 3 " 2t
D 3t 2 C 2
dt
dt
dy
3."1/2 C 2
5
At t D "1;
D
D" :
dx
4."1/3 " 2."1/
2
x D cos.2t /
y D sin t
dx
dy
D "2 sin.2t /
D cos t
dt
dt
# dy
cos.#=6/
1
At t D ;
D
D" :
6 dx
"2 sin.#=3/
2
17.
y D t2
x D t3
dy
dx
D 3t 2
D 2t both vanish at t D 0.
dt
dt
dy
2
dx
3t
D
has no limit as t ! 0.
D
! 0 as t ! 0,
dx
3t
dy
2
but dy=dt changes sign at t D 0. Thus the curve is not
smooth at t D 0. (In this solution, and in the next five, we
are using the Remark following Example 2 in the text.)
18.
x D .t " 1/4
y D .t " 1/3
dx
dy
D 4.t " 1/3
D 3.t " 1/2 both vanish at t D 1.
dt
dt
dx
4.t " 1/
Since
D
! 0 as t ! 1, and dy=dt does not
dy
3
change sign at t D 1, the curve is smooth at t D 1 and
therefore everywhere.
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SECTION 8.3 (PAGE 483)
19.
20.
21.
ADAMS and ESSEX: CALCULUS 9
y D t3
x D t sin t
dy
dx
D sin t C t cos t
D 3t 2 both vanish at t D 0.
dt
dt
dy
3t 2
6t
lim
D lim
D lim
D 0,
t!0 dx
t!0 sin t C t cos t
t!0 2 cos t " t sin t
but dx=dt changes sign at t D 0. dx=dy has no limit at
t D 0. Thus the curve is not smooth at t D 0.
22.
If x D f .t / D t 3 and y D g.t / D 3t 2 " 1, then
f 0 .t / D 3t 2 ; f 00 .t / D 6t I
g 0 .t / D 6t; g 00 .t / D 6:
Both f 0 .t / and g 0 .t / vanish at t D 0. Observe that
6t
2
dy
D 2 D :
dx
3t
t
x D t3
y D t " sin t
dx
dy
D 3t 2
D 1 " cos t both vanish at t D 0.
dt
dt
3t 2
6t
dx
D lim
D lim
D 6 and dy=dt
lim
t!0 1 " cos t
t!0 sin t
t!0 dy
does not change sign at t D 0. Thus the curve is smooth
at t D 0, and hence everywhere.
Thus,
lim
t!0C
dy
D 1;
dx
.3t 2 /.6/ " .6t /.6t /
2
d 2y
D
D" 4 <0
2
dx
.3t 2 /3
3t
dx
dy
D 2.t " 1/;
D 2.t " 2/
dt
dt
2
2
d y
d x
D
D2
dt 2
dt 2
2
d y
1
d dy
D
dx 2
dx=dt dt dx
1
d t "2
1
D
:
D
2.t " 1/ dt t " 1
2.t " 1/3
for all t , the curve is concave down everywhere.
y
xDt 3
yD3t 2 !1
x
!1
Directional information is as follows:
Fig. 8.3-22
23.
1
2
""""""""""""""""
j """""""""j""""""""!t
dx=dt
"
C
C
dy=dt
"
"
C
x
!
!
y
#
#
"
curve
.
&
%
The tangent is horizontal at t D 2, (i.e., .0; "4/), and
is vertical at t D 1 (i.e., at ."1; "3/. Observe that
d 2 y=dx 2 > 0, and the curve is concave up, if t > 1.
Similarly, d 2 y=dx 2 < 0 and the curve is concave down if
t < 1.
y
x
Fig. 8.3-21
328
x D t 3 " 3t , y D 2=.1 C t 2 /. Observe that y ! 0,
x ! ˙1 as t ! ˙1.
dx
dy
4t
D 3.t 2 " 1/;
D"
dt
dt
.1 C t 2 /2
dy
4t
D" 2
dx
3.t " 1/.1 C t 2 /2
d 2x
d 2y
4.3t 2 " 1/
D 6t;
D
2
2
dt
dt
.1 C t 2 /3
2
4.3t " 1/
4t .6t /
3.t 2 " 1/
"
d 2y
.1 C t 2 /3
.1 C t 2 /2
D
dx 2
Œ3.t 2 " 1/!3
4
60t C 48t 2 C 12
D
27.t 2 " 1/3 .1 C t 2 /3
Directional information:
xDt 2 !2t
yDt 2 !4t
tD2
dy
D "1
dx
and the curve has a cusp at t D 0, i.e., at .0; "1/. Since
If x D t 2 " 2t and y D t 2 " 4t , then
tD1
lim
t!0!
"1
0
1
""""""""""""""j"""""""""
j """"""""
j """"""""!t
dx=dt C
"
"
C
dy=dt C
C
"
"
x
!
!
y
"
"
#
#
curve %
.
&
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 8.4 (PAGE 487)
y
xDt 3 !3t!2
The tangent is horizontal at t D 0, i.e., .0; 2/, and vertical
at t D ˙1, i.e., .˙2; 1/.
yDt 2 !2t!2
tD!1; 2 x
"1
1
"""""""""""""""""j"""""""""""""
j """""""""!t
d 2y
C
"
C
dx 2
curve
^
_
^
tD1
Fig. 8.3-24
25.
y
yD
2
1 C t2
tD1
x D cos t C t sin t;
y D sin t " t cos t;
dy
dx
D t cos t;
D t sin t;
dt
dt
2
d x
D cos t " t sin t
dt 2
2
d y
D sin t C t cos t
dt 2
xDt 3 !3t
tD0
1
tD 2
tD!1
.t # 0/:
dy
D tan t
dt
dx d 2 y
dy d 2 x
"
d y
2
dt 2
D dt dt# $dt
dx 2
dx 3
dt
1
D
t cos3 t
"
!
Tangents are vertical at t D n C 12 #,
and horizontal at t D n# (n D 0; 1; 2; : : :).
x
2
Fig. 8.3-23
24.
If x D f .t / D t 3 " 3t " 2 and y D g.t / D t 2 " t " 2, then
f 0 .t / D 3t 2 " 3; f 00 .t / D 6t I
g 0 .t / D 2t " 1; g 00 .t / D 2:
#
$
1
27 9
, i.e., at " ; " .
2
8
4
The tangent is vertical at t D ˙1, i.e., ."4; "2/ and .0; 0/.
Directional information is as follows:
y
The tangent is horizontal at t D
tD!
tD3!=2 tD0
1
1
t
"1
2
""""""""""""""j""""""""""j""""""""""
j """"""""""!
f 0 .t /
C
"
"
C
g 0 .t /
"
"
C
C
x
!
!
y
#
#
"
"
curve &
.
%
tD2!
Fig. 8.3-25
Section 8.4 Arc Lengths and Areas for
Parametric Curves (page 487)
For concavity,
3.t 2 " 1/.2/ " .2t " 1/.6t /
2.t 2 " t C 1/
d 2y
D
D"
2
2
3
dx
Œ3.t " 1/!
9.t 2 " 1/3
which is undefined at t D ˙1, therefore
t
"1
1
"""""""""""""""""""
j """"""""""""""
j"""""""""""!
d 2y
"
C
"
dx 2
curve
_
^
_
x
1.
y D 2t 3 .0 $ t $ 1/
dy
D 6t 2
dt
Z 1p
Length D
.
.6t /2 C .6t 2 /2 dt
0
Z 1 p
D6
t 1 C t 2 dt Let u D 1 C t 2
0
du D 2t dt
ˇ2
Z 2
ˇ
p
p
ˇ
u du D 2u3=2 ˇ D 4 2 " 2 units
D3
ˇ
1
x D 3t 2
dx
D 6t
dt
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SECTION 8.4 (PAGE 487)
ADAMS and ESSEX: CALCULUS 9
2. If x D 1 C t 3 and y D 1 " t 2 for "1 $ t $ 2, then the arc
length is
Z 2p
.3t 2 /2 C ."2t /2 dt
sD
!1
Z 2 p
D
jt j 9t 2 C 4 dt
!1
#Z 1 Z 2 $ p
D
C
t 9t 2 C 4 dt Let u D 9t 2 C 4
0
0
du D 18t dt
#Z 13 Z 40 $
p
1
D
C
u du
18 4
4
%
&
p
p
1
13 13 C 40 40 " 16 units:
D
27
The length of the curve is
Z
6.
0
Z
7.
0
4. If x D ln.1 C t 2 / and y D 2 tan!1 t for 0 $ t $ 1, then
dx
dy
2t
2
I
:
D
D
dt
1 C t2
dt
1 C t2
The arc length is
s
Z 1
4t 2 C 4
dt
sD
.1 C t 2 /2
0
Z 1
dt
D2
p
Let t D tan &
1 C t2
0
dt D sec2 & d&
Z !=4
D2
sec & d&
0
ˇ!=4
ˇ
p
ˇ
D 2 ln j sec & C tan & jˇ
D 2 ln.1 C 2/ units.
ˇ
8.
D t 2 .4 C t 2 /:
330
4 C t 2 dt
y D sin t " t cos t .0 $ t $ 2#/
x D cos t C t sin t
dy
dx
D t cos t
D t sin t
dt
dt
Z 2! p
Length D
t 2 cos2 t C t 2 sin2 t dt
0
ˇ2!
Z 2!
t 2 ˇˇ
D
t dt D ˇ D 2# 2 units:
2ˇ
0
y D cos t .0 $ t $ #/
x D t C sin t
dy
dx
D 1 C cos t
D " sin t
dt
dt
Z !p
1 C 2 cos t C cos2 t C sin2 t dt
Length D
0
Z !p
Z !
t
D
4 cos2 .t =2/ dt D 2
cos dt
2
0
0
ˇ!
t ˇˇ
D 4 sin ˇ D 4 units:
2ˇ
x D sin2 t
dx
D 2 sin t cos t
dt
y D 2 cos t .0 $ t $ #=2/
dy
D "2 sin t
dt
Length
Z !=2 p
D
4 sin2 t cos2 t C 4 sin2 t dt
0
Z !=2
p
D2
sin t 1 C cos2 t dt Let cos t D tan u
0
" sin t dt D sec2 u du
Z !=4
D2
sec3 u du
5. x D t 2 sin t , y D t 2 cos t , .0 $ t $ 2#/.
C 4 cos2 t " 4t sin t cos t C t 2 sin2 t
0
0
0
dx
D 2t sin t C t 2 cos t
dt
dy
D 2t cos t " t 2 sin t
dt
# $2
(
ds
2
D t 4 sin2 t C 4t sin t cos t C t 2 cos2 t
dt
t
0
!=2
1
sin 2t dt
2
0
#
$ˇ!=2
cos 2t ˇˇ
D6a "
D 6a units:
ˇ
ˇ
2
p
Let u D 4 C t 2
du D 2t dt
ˇ4C4! 2
Z 4C4! 2
1 3=2 ˇˇ
1
1=2
u du D u ˇ
D
ˇ
2 4
3
4
#
$
8
D .1 C # 2 /3=2 " 1 units:
3
3. x D a cos3 t , y D a sin3 t , .0 $ t $ 2#/: The length is
Z 2! p
9a2 cos4 t sin2 t C 9a2 sin4 t cos2 t dt
0
Z 2!
D3a
j sin t cos t j dt
D12a
2!
0
#
$ˇˇ!=4
ˇ
D sec u tan u C ln.sec u C tan u/ ˇ
ˇ
0
p
p
D 2 C ln.1 C 2/ units:
)
9.
x D a.t " sin t /
dx
D a.1 " cos t /
dt
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y D a.1 " cos t / .0 $ t $ 2#/
dy
D a sin t
dt
lOMoARcPSD|6566483
INSTRUCTOR’S SOLUTIONS MANUAL
Z
SECTION 8.4 (PAGE 487)
q
a2 .1 " 2 cos t C cos2 t C sin2 t / dt
0
Z 2! p
Z 2! r
t
Da
2 " 2 cos t dt D a
sin2 dt
2
0
0
ˇ!
Z !
t
t ˇˇ
D 2a
sin dt D "4a cos ˇ D 4a units:
2
2ˇ
0
Length D
2!
the y-axis is
Z 2!
Sy D 2#
jxj ds
0
# $
Z 2!
t
D 2#
.at " a sin t /2a sin
dt
2
0
"
# $#
# $
# $
Z 2!
t
t
t
2
cos
sin
dt
D 4#a
t " 2 sin
2
2
2
0
# $
Z 2!
t
D 4#a2
t sin
dt
2
0
# $
# $
Z 2!
t
t
2
2
" 8#a
sin
cos
dt
2
2
0
ˇ
"
2!
# $ˇ
# $ #
Z 2!
t ˇ
t
2
dt " 0
D 4#a "2t cos
cos
ˇ C2
2 ˇ
2
0
0
10. If x D at " a sin t and y D a " a cos t for 0 $ t $ 2#,
then
dy
dx
D a " a cos t;
D a sin t I
dt
dt
p
ds D .a " a cos t /2 C .a sin t /2 dt
s
# $
p p
p
t
D a 2 1 " cos t dt D a 2 2 sin2
dt
2
# $
t
dt:
D 2a sin
2
a) The surface area generated by rotating the arch about
the x-axis is
Sx D 2#
Z
11.
x D e t cos t
dx
D e t .cos t " sin t /
dt
Arc length element:
jyj ds
# $
t
D 4#
.a " a cos t /2a sin
dt
2
0
# $
Z !
t
D 16#a2
sin3
dt
2
0
# $#
# $
Z !"
t
t
2
2
D 16#a
1 " cos
sin
dt
2
2
0
# $
t
Let u D cos
2
# $
t
1
dt
du D " sin
2
2
Z 0
D "32#a2
.1 " u2 / du
1
"
#ˇ1
1 3 ˇˇ
2
D 32#a u " u ˇ
ˇ
3
p
e 2t .cos t " sin t /2 C e 2t .sin t C cos t /2 dt
p t
D 2e dt:
ds D
Z
tD!=2
tD0
p Z
2#y ds D 2 2#
!=2
e 2t sin t dt
0
ˇ!=2
ˇ
e 2t
ˇ
.2 sin t " cos t /ˇ
D 2 2#
ˇ
5
0
p
2 2#
.2e ! C 1/ sq. units.
D
5
p
12. The area of revolution of the curve in Exercise 11 about
the y-axis is
Z
tD!=2
tD0
p Z
2#x ds D 2 2#
0
D
y D e t sin t .0 $ t $ #=2/
dy
D e t .sin t C cos t /
dt
The area of revolution about the x-axis is
2!
Z0 !
0
D 4#a2 Œ4# C 0! D 16# 2 a2 sq. units.
64 3
#a sq. units.
3
!=2
e 2t cos t dt
0
ˇ!=2
ˇ
p e 2t
ˇ
.2 cos t C sin t /ˇ
D 2 2#
ˇ
5
0
p
2 2# !
.e " 2/ sq. units.
D
5
b) The surface area generated by rotating the arch about
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SECTION 8.4 (PAGE 487)
13.
ADAMS and ESSEX: CALCULUS 9
y D 2t 3 .0 $ t $ 1/
x D 3t 2
dy
dx
D 6t
D 6t 2
dt
dt
Arc length
p element:
p
ds D 36.t 2 C t 4 / dt D 6t 1 C t 2 dt .
The area of revolution about the y-axis is
Z
tD1
tD0
2#x ds D 36#
D 18#
Z
Z
1
0
2
1
p
t 3 1 C t 2 dt
p
.u " 1/ u du
16.
Area of R D 4 &
Z
0
D "12a2
D 12a
Let u D 1 C t 2
du D 2t dt
.a sin3 t /."3a sin t cos2 t / dt
!=2
2
"
Z
0
sin4 t cos2 t dt
!=2
t
sin.4t / sin3 .2t /
"
"
16
64
48
D
3 2
#a sq. units.
8
y
$ˇ2
2 5=2 2 3=2 ˇˇ
u
" u
D 18#
ˇ
ˇ
5
3
1
p
72#
D
.1 C 2/ sq. units.
15
tD1
2#y ds D 24#
tD0
D 24#
Z
Z
Z
1
0
p
t 4 1 C t 2 dt
!=4
R
!a
Let t D tan u
dt D sec2 u du
tan4 u sec3 u du
17.
Fig. 8.4-16
%
#&
x D sin4 t , y D cos4 t , 0 $ t $
.
2
Z !=2
Area D
.cos4 t /.4 sin3 t cos t / dt
!=4
D4
Z
!=2
cos5 t .1 " cos2 t / sin t dt
Let u D cos t
du D " sin t dt
$
#
Z 1
1
1 1
5
7
"
D sq. units.
D4
.u " u / du D 6
6 8
6
0
0
y
xDsin4 t
yDcos4 t
2
x D t " 4t , y D t , ."2 $ t $ 2/.
Area D
Z
2
!2
Z
0$t$!=2
t 2 .3t 2 " 4/ dt
A
2
.3t 4 " 4t 2 / dt
0
$ˇ2
# 5
4t 3 ˇˇ
256
3t
"
sq. units.
D2
ˇ D
ˇ
5
3
15
D2
0
y
xDt 3 !4t
yDt 2
x
Fig. 8.4-17
1
2
18. If x D cos s sin s D sin 2s and y D sin2 s D
for 0 $ s $ 21 #, then
1
2
" 12 cos 2s
#
$
1 2
1
1
1
x2 C y "
D sin2 2s C cos2 2s D
2
4
4
4
A
x
which is the right half of the circle with radius
centre at .0; 12 /. Hence, the area of R is
" # $ #
1
1 2
#
#
D
sq. units.
2
2
8
Fig. 8.4-15
332
x
!a
We have omitted the details of evaluation of the final integral. See Exercise 24 of Section 8.3 for a similar evaluation.
15.
a
0
0
.sec7 u " 2 sec5 u C sec3 u/ du
D 24#
0
p &
#% p
D
7 2 C 3 ln.1 C 2/ sq. units.
2
3
a
xDa cos3 t
yDa sin3 t
The area of revolution of the curve of Exercise 13 about
the x-axis is
Z
0
(See Exercise 34 of Section 6.4.)
#
14.
# ˇ!=2
ˇ
ˇ
ˇ
ˇ
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2
and
lOMoARcPSD|6566483
INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 8.4 (PAGE 487)
y
upper half of the hyperbola from that of a right triangle:
xDcos s sin s
yDsin2 s
1
Shaded area D Area 4ABC " Area sector ABC
Z t0
1
D sec t0 tan t0 "
tan t .sec t tan t / dt
2
Z0t0
1
D sec t0 tan t0 "
.sec3 t " sec t / dt
2
0
(
1
1
D sec t0 tan t0 "
sec t tan t C
2
2
)ˇˇt0
1
ˇ
ln j sec t C tan t j " ln j sec t C tan t j ˇ
ˇ
2
1
2
x
Fig. 8.4-18
0
1
D ln j sec t0 C tan t0 j sq. units.
2
y
19. x D .2 C sin t / cos t , y D .2 C sin t / sin t , .0 $ t $ 2#/.
This is just the polar curve r D 2 C sin & .
Area D "
Z
Z
2!
.2 C sin t / sin t
0
2!
&
d %
.2 C sin t / cos t dt
dt
D"
.2 sin t C sin2 t /.cos2 t " 2 sin t " sin2 t / dt
0
Z 2! h
D
4 sin2 t C 4 sin3 t C sin4 t
0
i
" 2 sin t cos2 t " sin2 t cos2 t dt
Z 2! h
i
1 " cos 2t
." cos 2t / dt
D
2.1 " cos 2t / C
2
0
Z 2!
h
i
C
sin t 4 " 6 cos2 t dt
0
D 4# C
tDt0
#
9#
C0D
sq. units.
2
2
R tD0
x
xDsec t
yDtan t
Fig. 8.4-20
21.
See the figure below. The area is the area of a triangle
less the area under the hyperbola:
xD.2Csin t/ cos t
y
Z t0
1
sinh t sinh t dt
cosh t0 sinh t0 "
2
Z t0 0
1
cosh 2t " 1
D sinh 2t0 "
dt
4
2
0
1
1
1
D sinh 2t0 " sinh 2t0 C t0
4
4
2
t0
D
sq. units.
2
AD
yD.2Csin t/ sin t
0$t$2!
A
x
y
Fig. 8.4-19
.cosh t0 ;sinh t0 /
A
x
20. To find the shaded area we subtract the area under the
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Fig. 8.4-21
333
lOMoARcPSD|6566483
SECTION 8.4 (PAGE 487)
ADAMS and ESSEX: CALCULUS 9
22. If x D f .t / D at " a sin t and y D g.t / D a " a cos t ,
then the volume of the solid obtained by rotating about the
x-axis is
V D
Z
tD2!
tD0
Z 2!
D#
0
D #a3
D #a3
#y 2 dx D #
tD2!
Œg.t /!2 f 0 .t / dt
r D 5=.3 sin & " 4 cos & /
3r sin & " 4r cos & D 5
3y " 4x D 5
straight line.
4.
r D sin & C cos &
tD0
r 2 D r sin & C r cos &
2
x2 C y2 D y C x
#
$
$
#
1 2
1
1 2
C y"
D
x"
2
2
2
#
$
1 1
1
a circle with centre
;
and radius p .
2 2
2
.a " a cos t / .a " a cos t / dt
Z
2!
0
Z
Z
3.
.1 " cos t /3 dt
2!
.1 " 3 cos t C 3 cos2 t " cos3 t / dt
#
Z
3 2!
3
D #a 2# " 0 C
.1 C cos 2t / dt " 0
2 0
"
#
3
3
D #a 2# C .2#/ D 5# 2 a3 cu. units.
2
"0
y
5.
r 2 sin 2& D 1
2r 2 sin & cos & D 1
2xy D 1
a rectangular hyperbola.
6.
dx
x
tD0
23. Half of the volume corresponds to rotating x D a cos3 t ,
y D a sin3 t .0 $ t $ #=2/ about the x-axis. The whole
volume is
V D2
8.
rD p
D 2#
0
D 6#a3
D 6#a3
Z
Z
9.
0
.1 " cos2 t /3 cos2 t sin t dt
1
Let u D cos t
du D " sin t dt
.1 " 3u2 C 3u4 " u6 /u2 du
0
$
#
3 1
32#a3
1 3
" C "
cu. units.
D
D 6#a3
3 5
7 9
105
2.
r D 3 sec &
r cos & D 3
xD3
vertical straight line.
r D "2 csc & ) r sin & D "2
, y D "2
a horizontal line:
334
2
an ellipse.
1
1 " cos &
r "x D1
rD
r 2 D .1 C x/2
x 2 C y 2 D 1 C 2x C x 2
y 2 D 1 C 2x
10.
a parabola.
2
2 " cos &
2r " r cos & D 2
rD
4r 2 D .2 C x/2
Section 8.5 Polar Coordinates and Polar
Curves (page 493)
1.
cos2
x 2 C 4y 2 D 4
a2 sin6 t .3a cos2 t sin t / dt
!=2
2
& C 4 sin2 &
r cos & C 4r 2 sin2 & D 4
2
#y 2 ."dx/
!=2
a parabola.
r D sec &.1 C tan & /
r cos & D 1 C tan &
y
x D1C
x
x2 " x " y D 0
a parabola.
0
Z
r sin &
r cos &
7.
tD2!
Fig. 8.4-22
!=2
r D sec & tan & ) r cos & D
x2 D y
xDat!a sin t
yDa!a cos t
Z
r 2 D csc 2&
4x 2 C 4y 2 D 4 C 4x C x 2
3x 2 C 4y 2 " 4x D 4
11.
an ellipse.
2
1 " 2 sin &
r " 2y D 2
rD
x 2 C y 2 D r 2 D 4.1 C y/2 D 4 C 8y C 4y 2
x 2 " 3y 2 " 8y D 4
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a hyperbola.
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INSTRUCTOR’S SOLUTIONS MANUAL
12.
SECTION 8.5 (PAGE 493)
y
2
1 C sin &
r C r sin & D 2
5!
"D 6
rD
"D
!
6
r 2 D .2 " y/2
x
x 2 C y 2 D 4 " 4y C y 2
x 2 D 4 " 4y
a parabola.
13. r D 1 C sin & (cardioid)
rD1!2 sin "
y
Fig. 8.5-16
2
17.
r D 2 C cos &
y
x
!1
3
x
Fig. 8.5-13
14.
#
$
#
#
7#
If r D 1 " cos & C
, then r D 0 at & D " and
.
4
4
4
This is a cardioid.
Fig. 8.5-17
y
18. If r D 2 sin 2& , then r D 0 at & D 0; ˙
#
and #.
2
y
p p
. 2; 2/
!
"D! 4
rD1!cos."C
x
!
/
4
x
Fig. 8.5-14
rD2 sin 2"
15.
r D 1 C 2 cos &
r D 0 if & D ˙2#=3.
Fig. 8.5-18
y
2!=3
19.
r D cos 3& (three leaf rosette)
r D 0 at & D ˙#=6, ˙#=2, ˙5#=6.
y
1
3
x
!=6
x
!2!=3
Fig. 8.5-15
16. If r D 1 " 2 sin & , then r D 0 at & D
5#
#
and
.
6
6
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Fig. 8.5-19
335
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SECTION 8.5 (PAGE 493)
ADAMS and ESSEX: CALCULUS 9
y
#
3#
5#
20. If r D 2 cos 4& , then r D 0 at & D ˙ ; ˙ ; ˙
and
8
8
8
7#
˙ . (an eight leaf rosette)
8
!=3
y
3!
"D 8
x
!
"D 8
x
Fig. 8.5-23
rD2 cos 4"
24.
Fig. 8.5-20
21.
If r D ln & , then r D 0 at & D 1. Note that
y D r sin & D ln & sin & D .& ln & /
#
sin &
&
$
!0
as & ! 0C. Therefore, the (negative) x-axis is an asymptote of the curve.
p
r 2 D 4 sin 2& . Thus r D ˙2 sin 2& . This is a lemniscate.
r D 0 at & D 0, & D ˙#=2, and & D #.
y
y
x
x
rDln "
Fig. 8.5-24
Fig. 8.5-21
#
#
22. If r 2 D 4 cos 3& , then r D 0 at & D ˙ ; ˙ and
6
2
5#
˙ . This equation defines two functions of r, namely
6 p
r D ˙2 cos 3& . Each contributes 3 leaves to the graph.
25.
r D 3 cos & , and r D sin & both pass through the origin,
and so intersect
there. Also
p
p
sin & D 3 cos & ) tan & D p
3 ) & D #=3; 4#=3.
3=2; #=3!.
Both of these give the same point
Œ
p
Intersections: the origin and Π3=2; #=3!.
26.
r 2 D 2 cos.2& /, r D 1.
cos.2& / D 1=2 ) & D ˙#=6 or & D ˙5#=6.
Intersections: Œ1; ˙#=6! and Œ1; ˙5#=6!.
27.
r D 1 C cos & , r D 3 cos & . Both curves pass through the
origin, so intersect there. Also
3 cos & D 1Ccos & ) cos & D 1=2 ) & D ˙#=3.
Intersections: the origin and Œ3=2; ˙#=3!.
28.
Let r1 .& / D & and r2 .& / D & C #. Although the equation
r1 .& / D r2 .& / has no solutions, the curves r D r1 .& /
and r D r2 .& / can still intersect if r1 .&1 / D "r2 .&2 / for
two angles &1 and &2 having the opposite directions in the
polar plane. Observe that &1 D "n# and &2 D .n " 1/#
are two such angles provided n is any integer. Since
y
!
"D 6
x
r 2 D4 cos 3"
Fig. 8.5-22
p
r1 .&1 / D "n# D "r2 ..n " 1/#/;
p
23. r 2 D sin 3& . Thus r D ˙ sin 3& . This is a lemniscate.
r D 0 at & D 0, ˙#=3, ˙2#=3, #.
336
the curves intersect at any point of the form Œn#; 0! or
Œn#; #!.
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 8.6 (PAGE 497)
29. If r D 1=& for & > 0, then
lim y D lim
"!0C
"!0C
37. r D C C cos & sin.3& /
For C < 1 there appear to be 6 petals of 3 different sizes.
For C # 1 there are only 4 of 2 sizes, and these coalesce
as C increases.
sin &
D 1:
&
Thus y D 1 is a horizontal asymptote.
38.
y
y
yD1
rD1="
r D ln.& /
x
x
Fig. 8.5-29
30. The graph of r D cos n& has 2n leaves if n is an even
integer and n leaves if n is an odd integer. The situation
for r 2 D cos n& is reversed. The graph has 2n leaves
if n is an odd integer (provided negative values of r are
allowed), and it has n leaves if n is even.
Fig. 8.5-38
We will have Œln &1 ; &1 ! D Œln &2 ; &2 ! if
31. If r D f .& /, then
&2 D &1 C #
x D r cos & D f .& / cos &
y D r sin & D f .& / sin &:
32. r D cos & cos.m& /
For odd m this flower has 2m petals, 2 large ones and 4
each of .m " 1/=2 smaller sizes.
For even m the flower has m C 1 petals, one large and 2
each of m=2 smaller sizes.
and
ln &1 D " ln &2 ;
that is, if ln &1 C ln.&1 C #/ D 0. This equation has solution &1 ' 0:29129956. The corresponding intersection point has Cartesian coordinates
.ln &1 cos &1 ; ln &1 sin &1 / ' ."1:181442; "0:354230/.
39.
y
33. r D 1 C cos & cos.m& /
These are similar to the ones in Exercise 32, but the curve
does not approach the origin except for & D # in the case
of even m. The petals are joined, and less distinct. The
smaller ones cannot be distinguished.
r D 1=&
x
34. r D sin.2& / sin.m& /
For odd m there are m C 1 petals, 2 each of .m C 1/=2
different sizes.
For even m there are always 2m petals. They are of n
different sizes if m D 4n " 2 or m D 4n.
35. r D 1 C sin.2& / sin.m& /
These are similar to the ones in Exercise 34, but the petals
are joined, and less distinct. The smaller ones cannot be
distinguished. There appear to be m C 2 petals in both the
even and odd cases.
36. r D C C cos & cos.2& /
The curve always has 3 bulges, one larger than the other
two. For C D 0 these are 3 distinct petals. For 0 < C < 1
there is a fourth supplementary petal inside the large one.
For C D 1 the curve has a cusp at the origin. For C > 1
the curve does not approach the origin, and the petals become less distinct as C increases.
r D ln.& /
Fig. 8.5-39
The two intersections of r D ln & and r D 1=& for
0 < & $ 2# correspond to solutions &1 and &2 of
ln &1 D
1
;
&1
ln &2 D "
1
:
&2 C #
The first equation has solution &1 ' 1:7632228, giving the point ."0:108461; 0:556676/, and the second
equation has solution &2 ' 0:7746477, giving the point
."0:182488; "0:178606/.
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SECTION 8.6 (PAGE 497)
ADAMS and ESSEX: CALCULUS 9
y
Section 8.6 Slopes, Areas, and Arc Lengths
for Polar Curves (page 497)
1.
Area D
1
2
Z
2!
& d& D
0
"D
!
3
A
.2#/2
D # 2:
4
x
y
rDsin 3"
p
rD "
Fig. 8.6-4
"D0
"D2!x
5.
Fig. 8.6-1
Z
1 !=8
Total area D 16 &
cos2 4& d&
2 0
Z !=8
D4
.1 C cos 8& / d&
0
$ˇ!=8
#
sin 8& ˇˇ
#
sq. units.
D4 &C
D
ˇ
ˇ
8
2
ˇ2!
1 2! 2
& 3 ˇˇ
4
2. Area D
& d& D
ˇ D # 3 sq. units.
2 0
6 ˇ
3
Z
0
0
y
y
rDcos 4"
!=8
x
x
A
rD"
Fig. 8.6-5
Fig. 8.6-2
3. Area D 4 &
1
2
Z
!=4
a2 cos 2& d&
6.
0
ˇ!=4
ˇ
2 sin 2& ˇ
D 2a
D a2 sq. units.
ˇ
2 ˇ
The circles r D a and r D 2a cos & intersect
at & D ˙#=3. By symmetry, the common area is
4 & .area of sector " area of right triangle/ (see the figure), i.e.,
0
y
r 2 Da2 cos 2"
4&
"#
p $#
p
$ #
1 a 3a
4# " 3 3 2
1 2
#a "
D
a sq. units.
6
22 2
6
x
y
rD2a cos "
rDa
Fig. 8.6-3
A
4.
Area D
D
338
1
2
Z
!=3
0
sin2 3& d& D
1
4
Z
!=3
0
.1 " cos 6& / d&
#
$ˇˇ!=3
#
1
1
ˇ
D
& " sin 6& ˇ
sq. units.
ˇ
4
6
12
0
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Fig. 8.6-6
x
lOMoARcPSD|6566483
INSTRUCTOR’S SOLUTIONS MANUAL
7.
SECTION 8.6 (PAGE 497)
Z
#
1 !
.1 " cos & /2 d& "
Area D 2 &
2 !=2
2
$
Z ! #
#
1 C cos 2&
D
d& "
1 " 2 cos & C
2
2
!=2
#
$ˇˇ!
%
&
3
#
#
sin 2& ˇ
D
#"
" 2 sin & "
"
ˇ
ˇ
2
2
4
2
!=2
D
and & D ˙#=3. The shaded area is given by
1
2&
2
"Z
D
!
2
.1 C cos & / d& " 9
!=3
Z !
#
!=3
#
C 2 sq. units.
4
1 C 2 cos & C
Z
"
9
2
#
2#
3
Z
!=2
2
cos & d&
!=3
1 C cos 2&
2
!=2
!=3
.1 C cos 2& / d&
$
#
$
#
d&
$ˇ!
sin 2& ˇˇ
C 2 sin & C
ˇ
ˇ
4
!=3
ˇ
#
$ˇ!=2
sin 2& ˇ
9
&C
"
ˇ
ˇ
2
2
!=3
p
p !
p
#
3
3
9
#
D " 3"
C
D
sq. units.
4
8
4
2
4
y
3
D
2
rD1!cos "
rD1
x
y
Fig. 8.6-7
!=3
rD3 cos "
rD1Ccos "
x
8.
Z
1 2
1 !=2 2
#a C 2 &
a .1 " sin & /2 d&
2
2 0
$
Z !=2 #
#a2
1 " cos 2&
2
D
Ca
d&
1 " 2 sin & C
2
2
0
ˇ
#
$ˇ!=2
3
1
#a2
ˇ
C a2 & C 2 cos & " sin 2& ˇ
D
ˇ
2
2
4
0
#
$
5#
2
D
" 2 a sq. units.
4
Area D
Fig. 8.6-9
5#
#
and ˙ ,
6
6
the area inside the lemniscate and outside the circle is
10. Since r 2 D 2 cos 2& meets r D 1 at & D ˙
y
4&
rDa
x
A
1
2
Z
0
!=6 h
i
2 cos 2& " 12 d&
ˇ!=6
ˇ
p
ˇ
#
#
" D 3"
sq. units.
D 2 sin 2& ˇˇ
3
3
ˇ
0
y
rDa.1!sin "/
rD1
Fig. 8.6-8
A
A
x
r 2 D2 cos 2"
9. For intersections: 1 C cos & D 3 cos & . Thus 2 cos & D 1
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Fig. 8.6-10
339
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SECTION 8.6 (PAGE 497)
11.
ADAMS and ESSEX: CALCULUS 9
r D 0 at & D ˙2#=3. The shaded area is
1
2&
2
Z
!
14.
Z
2!
0
Da
2
.1 C 2 cos & / d&
2!=3
Z !
%
&
1 C 4 cos & C 2.1 C cos 2& / d&
2!=3
ˇ!
ˇ!
ˇ
ˇ
%# &
ˇ
ˇ
D3
C 4 sin & ˇ
C sin 2& ˇ
ˇ
ˇ
3
2!=3
2!=3
p
p
p
3
3 3
D#"
sq. units.
D# "2 3C
2
2
D
Da
rD1C2 cos "
15.
1
Z
Z
p
a2 C a2 & 2 d&
2!
0
p
1 C & 2 d&
"D2!
Let & D tan u
d& D sec2 u d&
sec3 u du
"D0
ˇ
&ˇ"D2!
a%
ˇ
sec u tan u C ln j sec u C tan uj ˇ
D
ˇ
2
"D0
ˇ"D2!
ˇ
i
h
p
p
a
ˇ
D
& 1 C & 2 C ln j 1 C & 2 C & j ˇ
ˇ
2
"D0
i
p
ah p
D
2# 1 C 4# 2 C ln.2# C 1 C 4# 2 / units:
2
y
2!=3
sD
3
x
r 2 D cos 2&
dr
2r
D "2 sin 2&
d& s
dr
sin 2&
D"
d&
r
)
p
sin2 2&
d& D sec 2& d&
cos 2&
Z !=4 p
sec 2& d&:
Length D 4
ds D
!2!=3
cos 2& C
0
y
Fig. 8.6-11
r 2 Dcos 2"
x
12.
Z
sD
!
0
Z
!
s
#
dr
d&
$2
C r 2 d& D
p
4 C & 2 d&
Z
!
0
p
4& 2 C & 4 d&
Let u D 4 C & 2
0
du D 2& d&
ˇ4C! 2
Z 4C! 2
p
1
1 3=2 ˇˇ
D
u du D u ˇ
ˇ
2 4
3
4
h
i
1
2 3=2
D
.4 C # /
" 8 units:
3
D
&
Fig. 8.6-15
16.
If r 2 D cos 2& , then
2r
dr
sin 2&
dr
D "2 sin 2& )
D "p
d&
d&
cos 2&
and
ds D
dr
13. r D e a" , ."# $ & $ #/.
D ae a" .
d&p
p
ds D e 2a" C a2 e 2a" d& D 1 C a2 e a" d& . The length
of the curve is
Z
!
!!
p
1 C a2 e a" d& D
p
s
cos 2& C
d&
sin2 2&
d& D p
:
cos 2&
cos 2&
a) Area of the surface generated by rotation about the xaxis is
Sx D 2#
1 C a2 a!
.e " e !a! / units.
a
D 2#
Z
!=4
r sin & ds
0
Z
!=4 p
cos 2& sin & p
cos 2&
ˇ!=4
ˇ
p
ˇ
D "2# cos & ˇ
D .2 " 2/# sq. units:
ˇ
0
0
340
d&
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 8.6 (PAGE 497)
y
b) Area of the surface generated by rotation about the yaxis is
Z !=4
Sy D 2#
r cos & ds
D 4#
!!=4
!=4 p
Z
0
cos 2& cos & p
d&
x
cos 2&
ˇ!=4
ˇ
p
ˇ
D 4# sin & ˇ
D 2 2# sq. units.
ˇ
0
17.
For r D 1 C sin & ,
Fig. 8.6-18
19.
r
1 C sin &
D
:
dr=d&
cos &
p
If & D #=4, then tan D 2 Cp1 and D 3#=8.
If & D 5#=4, then tan D 1 " 2 and D "#=8.
The line y D x meets the cardioid r D 1 C sin & at
the origin at an angle of 45ı , and also at first and third
quadrant points at angles of 67:5ı and "22:5ı as shown in
the figure.
tan
rD2 cos "
r 2 D2 sin 2"
D
y
The curves r D 1 " cos & and r D 1 " sin & intersect on the
rays & D #=4 and & D 5#=4, as well as at the origin. At
the origin their cusps clearly intersect at right angles.
For r D 1 " cos & , tan p1 D .1 " cos & /= sin & .
At & D #=4, tan 1 D 2 p
" 1, so 1 D #=8.
At & D 5#=4, tan 1 D ". 2 C 1/, so 1 D "3#=8.
For r D 1 " sin & , tan 2 Dp.1 " sin & /=." cos & /.
At & D #=4, tan 2 D 1p
" 2, so 2 D "#=8.
At & D 5#=4, tan 2 D 2 C 1, so 2 D 3#=8.
At #=4 the curves intersect at angle #=8 " ."#=8/ D #=4.
At 5#=4 the curves intersect at angle 3#=8 " ."3#=8/
D 3#=4 (or #=4 if you use the supplementary angle).
y
rD1Csin "
rD1!cos "
& D #=4
rD1!sin "
x
x
Fig. 8.6-17
18. The two curves r 2 D 2 sin 2& and r D 2 cos & intersect
where
2 sin 2& D 4 cos2 &
4 sin & cos & D 4 cos2 &
.sin & " cos & / cos & D 0
, sin & D cos & or cos & D 0;
)
(
p #
and P2 D .0; 0/.
i.e., at P1 D
2;
4
dr
For r 2 D 2 sin 2& we have 2r
D 4 cos 2& . At P1 we
d&
p
have r D 2 and dr=d& D 0. Thus the angle
between
the curve and the radial line & D #=4 is D #=2.
For r D 2 cos & we have dr=d& D "2 sin & , so the angle
between this curve
ˇ and the radial line & D #=4 satisfies
r ˇˇ
tan D
D "1, and
D 3#=4. The two
dr=d& ˇ
Fig. 8.6-19
20.
We have r D cos & C sin & . For horizontal tangents:
,
3#
#
: The tangents are horizontal at
Thus & D " or
8 # $8
" # $
#
#
#
#
cos
" sin
;"
and
8
8
8
" # $
#
# $
3#
3# 3#
cos
C sin
;
.
8
8
8
For vertical tangent:
"D!=4
3#
#
#
curves intersect at P1 at angle
" D .
4
2
4
The Figure shows that at the origin, P2 , the circle meets
the lemniscate twice, at angles 0 and #=2.
&
d %
dy
D
cos & sin & C sin2 &
d&
d&
D cos2 & " sin2 & C 2 sin & cos &
cos 2& D " sin 2& , tan 2& D "1:
0D
&
dx
d % 2
D
cos & C cos & sin &
d&
d&
D "2 cos & sin & C cos2 & " sin2 &
sin 2& D cos 2& , tan 2& D 1:
0D
,
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SECTION 8.6 (PAGE 497)
ADAMS and ESSEX: CALCULUS 9
Thus
There are vertical tangents at
" #& D$ #=8 of# 5#=8.
$ #
# #
#
C sin
;
and
cos
8
8
8
" # $
#
# $
5#
5# 5#
cos
C sin
;
.
8
8
8
There are no points on the curve where cos & D 0. Therefore, horizontal tangents occur only where
)
(
#
1
tan2 & D 1=3: There are horizontal tangents at p ; ˙
6
2
(
)
1
5#
and p ; ˙
.
6
2
For vertical tangents:
y
,
#
$
sin 2&
d
r cos & D "r sin & C cos & "
d&
r
cos 2& sin & D " sin 2& cos &
,
sin & D 0 or
rDcos "Csin "
0D
, .cos2 & " sin2 & / sin & D "2 sin & cos2 &
x
Fig. 8.6-20
3 cos2 & D sin2 &:
There are no points on the curve where tan2 & D 3, so the
only vertical tangents occur where sin & D 0, that is, at the
points with polar coordinates Œ1; 0! and Œ1; #!.
y
21.
r 2 Dcos 2"
r
D " cot & .
r D 2 cos & . tan D
dr=d&
For horizontal tangents we want tan D " tan & . Thus we
want " tan & D " cot & , and so p
& D ˙#=4 or ˙3#=4.
The tangents are horizontal at Œ 2; ˙#=4!.
For vertical tangents we want tan D cot & . Thus we
want " cot & D cot & , and so & D 0, ˙#=2, or #. There
are vertical tangents at the origin and at Œ2; 0!.
x
Fig. 8.6-22
y
"D!=4
23.
rD2 cos "
sin 2&
D
2 cos 2&
For horizontal tangents:
r D sin 2& . tan
D
1
2
tan 2& .
2
x
tan 2& D "2 tan &
2 tan &
D "2 tan &
2
1 " tan
% &
&
tan & 1 C .1 " tan2 & / D 0
"D!!=4
tan &.2 " tan2 & / D 0:
Fig. 8.6-21
p
p
Thus & D 0, #, ˙ tan!1 2, # ˙ tan!1 2.
There are horizontal tangents at the origin and the points
dr
22. We have r D cos 2& , and 2r
D "2 sin 2& . For horid&
zontal tangents:
2
#
$
sin 2&
d
r sin & D r cos & C sin & "
0D
d&
r
, cos 2& cos & D sin 2& sin &
, .cos2 & " sin2 & / cos & D 2 sin2 & cos &
,
342
cos & D 0 or
cos2 & D 3 sin2 &:
" p
#
p
2 2
!1
; ˙ tan
2
3
and
" p
#
p
2 2
!1
; # ˙ tan
2 :
3
Since the rosette r D sin 2& is symmetric about x D y,
there must be vertical tangents at the origin and at the
points
#
" p
2 2
!1 1
; ˙ tan p
3
2
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#
2 2
!1 1
; # ˙ tan p :
3
2
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INSTRUCTOR’S SOLUTIONS MANUAL
REVIEW EXERCISES 8 (PAGE 498)
y
y
rD2.1!sin "/
x
x
rDsin 2"
Fig. 8.6-25
Fig. 8.6-23
24.
We have r D e " and
dr
D e " . For horizontal tangents:
d&
d
r sin & D e " cos & C e " sin &
d&
#
tan & D "1 , & D " C k#;
4
0D
,
where k D 0; ˙1; ˙2; : : :. At the points
Œe k!!!=4 ; k# " #=4! the tangents are horizontal.
For vertical tangents:
d
r cos & D e " cos & " e " sin &
d&
#
tan & D 1 $ & D C k#:
4
0D
,
At the points Œe k!C!=4 ; k# C #=4! the tangents are vertical.
26.
dx
dy
D f 0 .& / cos & " f .& / sin &;
D f 0 .& / sin & C f .& / cos &
d& r
d&
%
&2 %
&2
f 0 .& / cos & " f .& / sin & C f 0 .& / sin & C f .& / cos & d&
ds D
(%
&2
%
&2
D f 0 .& / cos2 & " 2f 0 .& /f .& / cos & sin & C f .& / sin2 &
)1=2
%
&2
%
&2
d&
C f 0 .& / sin2 & C 2f 0 .& /f .& / sin & cos & C f .& / cos2 &
r
%
&2 %
&2
D
f 0 .& / C f .& / d&:
Review Exercises 8 (page 498)
1.
1 " sin &
.
cos &
For horizontal tangents tan D " cot & , so
25. r D 2.1 " sin & /, tan
D"
1 " sin &
sin &
D"
cos &
cos &
cos & D 0; or 2 sin & D 1:
"
The solutions are & D ˙#=2, ˙#=6, and ˙5#=6.
& D #=2 corresponds to the origin where the cardioid
has a cusp, and therefore no tangent. There are horizontal
tangents at Œ4; "#=2!, Œ1; #=6!, and Œ1; 5#=6!.
For vertical tangents tan D cot & , so
1 " sin &
cos &
D
cos &
sin &
sin2 & " sin & D cos2 & D 1 " sin2 &
"
x D r cos & D f .& / cos & , y D r sin & D f .& / sin & .
2.
x2
C y2 D 1
2
p
Ellipse, semi-major axis a D 2, along the x-axis. Semiminor axis b D 1.
c 2 D a2 " b 2 D 1. Foci: .˙1; 0/.
x 2 C 2y 2 D 2
x2 y2
9x 2 " 4y 2 D 36 ,
"
D1
4
9
Hyperbola, transverse axis along the x-axis.
Semi-transverse axis a D 2, semi-conjugate
axis b D 3.
p
c 2 D a2 C b 2 D 13. Foci: .˙ 13; 0/.
Asymptotes: 3x ˙ 2y D 0.
3.
x C y 2 D 2y C 3 , .y " 1/2 D 4 " x
Parabola, vertex .4; 1/, opening to the left, principal axis
y D 1.
a D "1=4. Focus: .15=4; 1/.
4.
2x 2 C 8y 2 D 4x " 48y
2.x 2 " 2x C 1/ C 8.y 2 C 6y C 9/ D 74
2 sin2 & " sin & " 1 D 0
.sin & " 1/.2 sin & C 1/ D 0
The solutions here are & D #=2 (the origin again),
& D "#=6 and & D "5#=6. There are vertical tangents at
Œ3; "#=6! and Œ3; "5#=6!.
,
.y C 3/2
.x " 1/2
C
D 1:
37
37=4
Ellipse,
p "3/, major axis along y D "3.
p centre .1;
a D 37, bpD 37=2, c 2 D a2 " b 2 D 111=4.
Foci: .1 ˙ 111=2; "3/.
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REVIEW EXERCISES 8 (PAGE 498)
ADAMS and ESSEX: CALCULUS 9
y
5. x D t , y D 2 " t , .0 $ t $ 2/.
Straight line segment from .0; 2/ to .2; 0/.
.!2;4/
xDt 3 !3t
yDt 3 C3t
6. x D 2 sin.3t /, y D 2 cos.3t /, .0 $ t $ 2/
Part of a circle of radius 2 centred at the origin from the
point .0; 2/ clockwise to .2 sin 6; 2 cos 6/.
x
.2;!4/
7.
x D cosh t , y D sinh2 t .
Parabola x 2 " y D 1, or y D x 2 " 1, traversed left to
right.
Fig. R-8-12
13.
8. x D e t , y D e !2t , ."1 $ t $ 1/.
Part of the curve x 2 y D 1 from .1=e; e 2 / to .e; 1=e 2 /.
9. x D cos.t =2/, y D 4 sin.t =2/, .0 $ t $ #/.
The first quadrant part of the ellipse 16x 2 C y 2 D 16,
traversed counterclockwise.
x D t 3 " 3t
dx
D 3.t 2 " 1/
dt
Horizontal tangent
Vertical tangent at
y D t3
dy
D 3t 2
dt
at t D 0, i.e., at .0; 0/.
t D ˙1, i.e., at .2; "1/ and ."2; 1/.
'
dy
t2
> 0 if jt j > 1
D 2
< 0 if jt j < 1
dx
t "1
Slope ! 1 as t ! ˙1.
y
Slope
10. x D cos t C sin t , y D cos t " sin t , .0 $ t $ 2#/
The circle x 2 C y 2 D 2, traversed clockwise, starting and
ending at .1; 1/.
tD1
11.
4
xD
y D t 3 " 3t
1 C t2
dy
dx
8t
D 3.t 2 " 1/
D"
2
2
dt
dt
.1 C t /
Horizontal tangent at t D ˙1, i.e., at .2; ˙2/.
Vertical tangent at t D 0, p
i.e., at .4; 0/.
Self-intersection at t D ˙ 3, i.e., at .1; 0/.
y
tD!1
p
tD˙ 3
tD0
x
Fig. R-8-13
14.
x D t 3 " 3t
dx
D 3.t 2 " 1/
dt
Horizontal tangent
."2; 16/.
Vertical tangent at
y D t 3 C 3t
x D t 3 " 3t
dy
dx
D 3.t 2 " 1/
D 3.t 2 C 1/
dt
dt
Horizontal tangent: none.
Vertical tangent at t D ˙1, i.e., at .2; "4/ and ."2; 4/.
'
dy
t2 C 1
> 0 if jt j > 1
D 2
< 0 if jt j < 1
dx
t "1
Slope ! 1 as t ! ˙1.
344
y D t 3 " 12t
dy
D 3.t 2 " 4/
dt
at t D ˙2, i.e., at .2; "16/ and
t D ˙1, i.e., at .2; 11/ and ."2; "11/.
Slope
Fig. R-8-11
Slope
x
'
t2 " 4
dy
> 0 if jt j > 2 or jt j < 1
D 2
< 0 if 1 < jt j < 2
dx
t "1
Slope ! 1 as t ! ˙1.
y
.!2;16/
tD1
12.
tD!1
.2;11/
.!2;!11/
xDt 3 !3t
yDt 3 !12t
.2;!16/
Fig. R-8-14
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INSTRUCTOR’S SOLUTIONS MANUAL
15.
REVIEW EXERCISES 8 (PAGE 498)
The curve x D t 3 " t , y D jt 3 j is symmetric about x D 0
since x is an odd function and y is an even function. Its
self-intersection occurs at a nonzero value of t that makes
x D 0, namely, t D ˙1. The area of the loop is
r D &;
! !3!
2
$& $
3!
2
"
y
rD&
Z 1
."x/ dy D "2
.t 3 " t /3t 2 dt
tD0
0
#
$ˇ1
3 4 ˇˇ
1
6
D "t C t ˇ D sq. units.
ˇ
2
2
AD2
Z
19.
tD1
x
0
y
Fig. R-8-19
x D t3 " t
y D jt 3 j
20.
r D j& j;
."2# $ & $ 2#/
y
rDj"j
tD˙1
x
tD0
x
Fig. R-8-15
16. The volume of revolution about the y-axis is
V D#
D#
Z
tD1
Fig. R-8-20
x 2 dy
tD0
Z 1
0
Z
21.
.t 6 " 2t 4 C t 2 /3t 2 dt
r D 1 C cos.2& /
y
1
.t 8 " 2t 6 C t 4 / dt
$
#
1
1 2
8#
" C
cu. units.
D 3#
D
9 7
5
105
D 3#
17.
rD1Ccos 2"
0
x
x D e t " t , y D 4e t=2 , .0 $ t $ 2/. Length is
Z
p
.e t " 1/2 C 4e t dt
0
Z 2p
Z 2
D
.e t C 1/2 dt D
.e t C 1/ dt
0
0
ˇ2
ˇ
D .e t C t /ˇˇ D e 2 C 1 units:
LD
2
Fig. R-8-21
22.
r D 2 C cos.2& /
y
rD2Ccos.2"/
0
18. Area of revolution about the x-axis is
Z
S D 2# 4e t=2 .e t C 1/ dt
$ˇˇ2
#
2 3t=2
t=2 ˇ
e
C 2e
D 8#
ˇ
ˇ
3
x
Fig. R-8-22
0
16# 3
D
.e C 3e " 4/ sq. units.
3
23.
r D 1 C 2 cos.2& /
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REVIEW EXERCISES 8 (PAGE 498)
ADAMS and ESSEX: CALCULUS 9
y
27.
rD1C2 cos 2"
p
the origin in the directions for
r D 1C 2 sin & approaches
p
which sin & D "1= 2, that is, & D "3#=4 and & D "#=4.
The smaller loop corresponds to values of & between these
two values. By symmetry, the area of the loop is
x
AD2&
D
r D 1 " sin.3& /
y
Z
!!=4
!!=2
!!=4
!!=2
p
.1 C 2 2 sin & C 2 sin2 & / d&
p
.2 C 2 2 sin & " cos.2& // d&
$ˇˇ!!=4
#
p
1
ˇ
D 2& " 2 2 cos & " sin.2& / ˇ
ˇ
2
Fig. R-8-23
24.
Z
1
2
#
1
# "3
D "2C D
sq. units.
2
2
2
rD1!sin.3"/
y
!!=2
p
rD1C 2 sin "
!=6
x
!3!=4
28.
25. Area of a large loop:
AD2&
D
Z
1
2
!=3
!=3
0
26. Area of a small loop:
AD2&
D
Z
1
2
!=2
!=3
!=3
346
p
Only . 2 " 1/=2 is between "1pand 1, so is a possible
2"1
value of cos & . Let &0 D cos!1
. Then
2
v
p
!2
u
p
p
u
2"1
1C2 2
t
sin &0 D 1 "
:
D
2
2
By symmetry, the area inside r D 1 C cos & to the left of
the line x D 1=4 is
.1 C 2 cos.2& //2 d&
Œ1 C 4 cos.2& / C 2.1 C cos.4& //! d&
$ˇˇ!=2
1
ˇ
D 3& C 2 sin.2& / C sin.4& / ˇ
ˇ
2
!=3
p
#
3 3
D "
sq. units.
2
4
#
1
4 cos &
4 cos2 & C 4 cos & " 1 D 0
p
p
"4 ˙ 16 C 16
˙ 2"1
D
:
cos & D
8
2
.1 C 2 cos.2& //2 d&
Œ1 C 4 cos.2& / C 2.1 C cos.4& //! d&
#
$ˇˇ!=3
1
ˇ
D 3& C 2 sin.2& / C sin.4& / ˇ
ˇ
2
0
p
3 3
sq. units.
D#C
4
!=2
r cos & D x D 1=4 and r D 1 C cos & intersect where
1 C cos & D
0
Z
x
Fig. R-8-27
Fig. R-8-24
Z
!!=4
#
$
1 C cos.2& /
1 C 2 cos & C
d& C cos &0 sin &0
2
"0
ˇ
#
$ˇ!
3
1
ˇ
D .# " &0 / C 2 sin & C sin.2& / ˇ
ˇ
2
4
"0
p
p
p
. 2 " 1/ 1 C 2 2
C
4
! q
!
p
p
p
3
2"1
2"9
!1
# " cos
C 1C2 2
sq. units.
D
2
2
8
AD2&
1
2
Z
!
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INSTRUCTOR’S SOLUTIONS MANUAL
y
xD1=4
CHALLENGING PROBLEMS 8 (PAGE 498)
rD1Ccos "
"0
C1
x
A1
S1
Fig. R-8-28
C
Challenging Problems 8
(page 498)
P
F1
F2
C2
A2
1.
The surface of the water is elliptical (see Problem 2 below) whose semi-minor axis is 4 cm, the radius of the
cylinder, and whose semi-major axis is 4 sec & cm because
of the tilt of the glass. The surface area is that of the ellipse
x D 4 sec & cos t; y D 4 sin t;
S2
Fig. C-8-2
Let P be any point on C . Let A1 A2 be the line through
P that lies on the cylinder, with A1 on C1 and A2 on C2 .
Then PF1 D PA1 because both lengths are of tangents
drawn to the sphere S1 from the same exterior point P .
Similarly, PF2 D PA2 . Hence
.0 $ t $ 2#/:
This area is
AD4
D4
Z
Z
tD!=2
PF1 C PF2 D PA1 C PA2 D A1 A2 ;
x dy
tD0
!=2
.4 sec & cos t /.4 cos t / dt
0
Z !=2
D 32 sec &
.1 C cos.2t // dt D 16# sec & cm2 :
which is constant, the distance between the centres of the
two spheres. Thus C must be an ellipse, with foci at F1
and F2 .
3.
0
"
4 cm
Given the foci F1 and F2 , and the point P on the ellipse,
construct N1 PN2 , the bisector of the angle F1 PF2 . Then
construct T1 P T2 perpendicular to N1 N2 at P . By the
reflection property of the ellipse, N1 N2 is normal to the
ellipse at P . Therefore T1 T2 is tangent there.
N2
T1
P
4 sec " cm
T2
&
&
N1
F1
F2
Fig. C-8-1
2. Let S1 and S2 be two spheres inscribed in the cylinder,
one on each side of the plane that intersects the cylinder
in the curve C that we are trying to show is an ellipse.
Let the spheres be tangent to the cylinder around the circles C1 and C2 , and suppose they are also tangent to the
plane at the points F1 and F2 , respectively, as shown in
the figure.
Fig. C-8-3
4.
Without loss of generality, choose the axes and axis scales
so that the parabola has equation y D x 2 . If P is the
point .x0 ; x02 / on it, then the tangent to the parabola at P
has equation
y D x02 C 2x0 .x " x0 /;
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CHALLENGING PROBLEMS 8 (PAGE 498)
ADAMS and ESSEX: CALCULUS 9
which intersects the principal axis x D 0 at .0; "x02 /.
Thus R D .0; "x02 / and Q D .0; x02 /. Evidently the vertex
V D .0; 0/ bisects RQ.
y
y
6.
P D Œr; & !
Œa; &0 !
r
Q
P D .x0 ; x02 /
a
&
L
&0
x
x
V
Fig. C-8-6
R
a) Let L be a line not passing through the origin, and
let Œa; &0 ! be the polar coordinates of the point on L
that is closest to the origin. If P D Œr; & ! is any point
on the line, then, from the triangle in the figure,
Fig. C-8-4
To construct the tangent at a given point P on a parabola
with given vertex V and principal axis L, drop a perpendicular from P to L, meeting L at Q. Then find R on L
on the side of V opposite Q and such that QV D VR.
Then PR is the desired tangent.
a
D cos.& " &0 /;
r
rD
a
:
cos.& " &0 /
b) As shown in part (a), any line not passing through the
origin has equation of the form
r D g.& / D
5.
y
b
or
a
D a sec.& " &0 /;
cos.& " &0 /
for some constants a and &0 . We have
c
g 0 .& / D a sec.& " &0 / tan.& " &0 /
2 ft
2 ft
a
x
g 00 .& / D a sec.& " &0 / tan2 .& " &0 /
%
&2
%
&2
C a sec3 .& " &0 / g.& / C 2 g 0 .& / " g.& /g 00 .& /
D a2 sec2 .& " &0 / C 2a2 sec2 .& " &0 / tan2 .& " &0 /
" a2 sec2 .& " &0 / tan2 .& " &0 / " a2 sec4 .& " &0 /
h
%
&
i
D a2 sec2 .& " &0 / 1 C tan2 .& " &0 / " sec4 .& " &0 /
D 0:
Fig. C-8-5
2
x
y2
C
D 1, with a D 2 and foci at
a2
b2
2
.0; ˙2/ so that c D 2 and b D a2 C c 2 D 8. The volume
of the barrel is
c) If r D g.& / is the polar equation of the tangent to
r D f .& / at & D ˛, then g.˛/ D f .˛/ and
g 0 .˛/ D f 0 .˛/. Suppose that
%
Let the ellipse be
$
2
2 #
y2
V D2
dy
#x 2 dy D 2#
4 1"
8
0
0
$ˇ2
#
40# 3
y 3 ˇˇ
ft :
D 8# y "
ˇ D
24 ˇ
3
Z
348
By part (b) we have
Z
0
&2
%
&2
f .˛/ C 2 f 0 .˛/ " f .˛/f 00 .˛/ > 0:
%
&2
%
&2
g.˛/ C 2 g 0 .˛/ " g.˛/g 00 .˛/ D 0:
Subtracting, and using g.˛/ D f .˛/ and
g 0 .˛/ D f 0 .˛/, we get f 00 .˛/ < g 00 .˛/. It follows
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INSTRUCTOR’S SOLUTIONS MANUAL
CHALLENGING PROBLEMS 8 (PAGE 498)
that f .& / < g.& / for values of & near ˛; that is, the
graph of r D f .& / is curving to the origin side of its
tangent at ˛. Similarly, if
%
&2
%
&2
f .˛/ C 2 f 0 .˛/ " f .˛/f 00 .˛/ < 0;
8.
Take the origin at station O as shown in the figure. Both
of the lines L1 and L2 pass at distance 100 cos $ from the
origin. Therefore, by Problem 6(a), their equations are
100 cos $
100 cos $
,
!
"- D
sin.& C $/
cos & " !2 " $
100 cos $
100 cos $
,
!!
"- D
:
rD
sin.& " $/
cos & " 2 C $
L1 W
then the graph is curving to the opposite side of the
tangent, away from the origin.
rD
L2 W
The search area A.$/ is, therefore,
7.
x.t /
0
B
A
x
&
R
$
1002 cos2 $
1002 cos2 $
d&
"
!
sin2 .& " $/ sin2 .& C $/
4 !$
Z ! C$ %
&
4
2
D 5; 000 cos $ !
csc2 .& " $/ " csc2 .& C $/ d&
A.$/ D
1
2
Z
!
4 C$
#
!$
,4 !
"
!
"D 5; 000 cos2 $ cot !4 C 2$ " 2 cot !4 C cot !4 " 2$
#
"
!!
"
"
!!
cos 4 C 2$
sin 4 C 2$
!!
" C
!
" "2
D 5; 000 cos2 $
sin 4 C 2$
cos !4 C 2$
, !
"
D 10; 000 cos2 $ csc !2 C 4$ " 1
r
D 10; 000 cos2 $.sec.4$/ " 1/ mi2 :
For $ D 3ı D #=60, we have A.$/ ' 222:8 square miles.
Also
A0 .$/ D "20; 000 cos $ sin $.sec.4$/ " 1/
C 40; 000 cos2 $ sec.4$/ tan.4$/
A .#=60/ ' 8645:
0
Fig. C-8-7
When the vehicle is at position x, as shown in the figure, the component of the gravitational force on it in the
direction of the tunnel is
ma.r/ cos & D "
When $ D 3ı , the search area increases at about
8645.#=180/ ' 151 square miles per degree increase
in $.
y
mg
mgr
cos & D "
x:
R
R
$
L2
L1
By Newton’s Law of Motion, this force produces an acceleration d 2 x=dt 2 along the tunnel given by
m
Area A.$/
100 mi
2
d x
mg
D"
x;
dt 2
R
$
#=4
that is
d 2x
C ! 2 x D 0;
dt 2
x
O
where
g
!2 D :
R
Fig. C-8-8
This is the equation ofpsimple harmonic motion, with period T D 2#=! D 2# R=g.
For R ' 3960 mi ' 2:09 & 107 ft, and g ' 32 ft/s2 , we
have T ' 5079 s ' 84:6 minutes. This is a rather short
time for a round trip between Atlanta and Baghdad, or any
other two points on the surface of the earth.
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CHALLENGING PROBLEMS 8 (PAGE 498)
9. The easiest way to determine which curve is which is
to calculate both their areas; the outer curve bounds the
larger area.
The curve C1 with parametric equations
x D sin t;
1
y D sin.2t /;
2
.0 $ t $ 2#/
ADAMS and ESSEX: CALCULUS 9
The curve C2 with polar equation r 2 D cos.2& / has area
4
A2 D
2
Z
0
!=4
ˇ!=4
ˇ
ˇ
cos.2& / d& D sin.2& /ˇ
D 1 sq. units.
ˇ
0
C1 is the outer curve, and the area between the curves is
1/3 sq. units.
y
has area
x
A1 D 4
D4
tD!=2
0
D4
D4
Z
Fig. C-8-9
y dx
tD0
Z !=2
Z
Let u D cos t
du D " sin t dt
350
Z
!=2
1
sin.2t / cos t dt
2
sin t cos2 t dt
0
0
1
u2 du D
4
sq. units.
3
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 9.1 (PAGE 507)
CHAPTER 9. SEQUENCES, SERIES,
AND POWER SERIES
Section 9.1 Sequences and Convergence
(page 507)
1.
2.
3.
4.
5.
6.
7.
8.
*
2
8 9
2n2
D
2
"
D
1;
;
;
:
:
:
is bounded,
n2 C 1
n2 C 1
5 5
positive, increasing, and converges to 2.
(
)
'
*
2n
4 3 8
D
1;
;
;
;
:
:
:
is bounded, positive, den2 C 1
5 5 17
creasing, and converges to 0.
'
*
'
*
."1/n
7 13
4"
D 5; ; ; : : : is bounded, positive, and
n
2 3
converges to 4.
(
)
*
# $
# $
'
1
1
1
D sin 1; sin
; sin
; : : : is bounded,
sin
n
2
3
positive, decreasing, and converges to 0.
' 2
*
'
*
'
*
n "1
1
3 8 15
D n"
D 0; ; ; ; : : : is bounded
n
n
2 3 4
below, positive, increasing, and diverges to infinity.
(
)
' n*
e
e % e &2 % e &3
D
;
;
; : : : is bounded, positive,
#n
# #
#
decreasing, and converges to 0, since e < #.
'#
' n *
$ *
p
e n
e
D
p
. Since e= # > 1, the sequence
#
# n=2
is bounded below, positive, increasing, and diverges to
infinity.
)
(
'
*
."1/n n
"1 2 "3
; : : : is bounded, alternating,
D
; ;
en
e e2 e3
and converges to 0.
'
*
'
*
14.
5
"2
5 " 2n
2
n
D lim
lim
D" :
7
3n " 7
3
3"
n
15.
4
n"
n2 " 4
n
lim
D lim
D 1:
5
nC5
1C
n
16.
n2
lim 3
D lim
n C1
'
9. f2n =nn g is bounded, positive, decreasing, and converges to
0.
# $n
.nŠ/2
1
2
3
n
1
10.
D
!!!
$
.
.2n/Š
nC1 nC2 nC3
2n
2
anC1
.n C 1/2
1
Also,
D
< . Thus the sequence
.2n C 2/.2n C 1/
2
'
*an
.nŠ/2
is positive, decreasing, bounded, and convergent
.2n/Š
to 0.
fn cos.n#=2/g D f0; "2; 0; 4; 0; "6; : : :g is divergent.
(
)
*
'
sin 2 sin 3
sin n
D sin 1;
;
; : : : is bounded and con12.
n
2
3
verges to 0.
11.
13. f1; 1; "2; 3; 3; "4; 5; 5; "6; : : :g is divergent.
17.
lim."1/n
1
n
1C
1
n3
D 0:
n
D 0:
n3 C 1
2
1
p
1" p C 2
n2 " 2 n C 1
1
n
n n
18. lim
D lim
D" :
1
1
1 " n " 3n2
3
" "3
n2 n
e n " e !n
1 " e !2n
D lim
D 1:
n
!n
e Ce
1 C e !2n
19.
lim
20.
lim n sin
21.
22.
23.
24.
25.
sin x
cos x
1
D lim
D lim
D 1:
x!0C x
x!0C
n
1
$
#
$
#
"3 n
n"3 n
D lim 1 C
D e !3 by l’H^opital’s
lim
n
n
Rule.
x
n
D lim
lim
x!1 ln.x C 1/
ln.n C 1/
1
$ D lim x C 1 D 1:
D lim #
x!1
x!1
1
xC1
p
p
nC1"n
lim. n C 1 " n/ D lim p
p D 0:
nC1C n
%
&
p
n2 " .n2 " 4n/
lim n " n2 " 4n D lim
p
n C n2 " 4n
4
4n
r
D 2:
p
D lim
D lim
4
n C n2 " 4n
1C 1"
n
p
p
lim. n2 C n " n2 " 1/
n2 C n " .n2 " 1/
D lim p
p
n2 C n C n2 " 1
nC1
!
D lim
r
r
1
1
n
1C C 1" 2
n
n
D lim r
1C
1C
1
C
n
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1
n
r
1"
1
n2
D
1
:
2
351
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SECTION 9.1 (PAGE 507)
26. If an D
#
n"1
nC1
$n
Thus, fan g is increasing by induction. Observe that
a1 < 5 and a2 < 5. If ak < 5 then
, then
$n
n
nC1
#
$ ,
#
$
1 n
1 n
D lim 1 "
lim 1 C
n
n
lim an D lim
D
27.
#
ADAMS and ESSEX: CALCULUS 9
n"1
n
$n #
e !1
D e !2
e
akC1 D
(by Theorem 6 of Section 3.4).
.nŠ/2
.1 ! 2 ! 3 ! ! ! n/.1 ! 2 ! 3 ! ! ! n/
D
.2n/Š
1 ! 2 ! 3 ! ! ! n ! .n C 1/ ! .n C 2/ ! ! ! 2n
# $n
1
2
3
n
1
D
!
!
!!!
$
:
nC1 nC2 nC3
nCn
2
Thus lim an D 0.
n
and lim
aD
32.
n2
D 0 since 2n grows much faster than n2
2n
p
!
4n
lim
nŠ
1 C 2akC1 >
p
!
D 0:
b) Since ln x $ x " 1,
#
$
#
$
1
1
ln ak D k ln 1 C
$k 1C "1 D1
k
k
1 C 2ak D akC1 :
p
p
p
p
1 C 2ak < 1 C 2.3/ D 7 < 9 D 3:
which implies that ak $ e for all k. Since fan g is
increasing, e is an upper bound for fan g.
33.
Therefore, an < 3 for all n, by induction. Since fan g
is increasing and bounded above, it converges. Let
lim an D a. Then
aD
p
1 C 2a ) a2 " 2a " 1 D 0 ) a D 1 ˙
p
2:
p
it is not appropriate. Hence, we
Since a D 1 " 2 < 0, p
must have lim an D 1 C 2.
p
31. Let a1 D 3 and anC1pD 15 C 2an for n D 1; 2; 3; : : :.
Then we have a2 D 21 > 3 D a1 . If akC1 > ak for
some k, then
akC2 D
352
25 D 5:
Since f 0 .x/ > 0, f .x/ must be an increasing function. Thus, fan g D fe f .xn / g is increasing.
Thus, fan g is increasing by induction. Observe that
a1 < 3 and a2 < 3. If ak < 3 then
akC1 D
p
x
f 0 .x/ D ln.x C 1/ C
" ln x " 1
x
C
1
#
$
xC1
1
D ln
"
x
xC1
Z xC1
dt
1
D
"
t
x
C
1
x
Z xC1
1
1
dt "
>
xC1 x
xC1
1
1
"
D 0:
D
xC1 xC1
#n
) 0 < an < .#=4/n . Since #=4 < 1,
1 C 22n
n
therefore .#=4/ ! 0 as n ! 1. Thus lim an D 0.
p
30. Let a1 D 1 and anC1p D 1 C 2an for n D 1; 2; 3; : : :.
Then we have a2 D 3 > a1 . If akC1 > ak for some k,
then
p
15 C 2.5/ D
Since a > a1 , we must have lim an D 5.
#
$
$
#
1
1 n
so ln an D n ln 1 C
.
Let an D 1 C
n
n
#
$
1
a) If f .x/ D x ln 1 C
D x ln.x C 1/ " x ln x, then
x
29. an D
akC2 D
p
15 C 2a ) a2 " 2a " 15 D 0 ) a D "3; or a D 5:
4
D 0 by Theorem 3(b). Hence,
nŠ
n2 22n
n2
n2 2n
D lim n !
D lim n
lim
nŠ
2
nŠ
2
15 C 2ak <
Therefore, an < 5 for all n, by induction. Since fan g
is increasing and bounded above, it converges. Let
lim an D a. Then
an D
28. We have lim
p
p
p
15 C 2akC1 > 15 C 2ak D akC1 :
Suppose fan g is ultimately increasing, say anC1 # an if
n # N.
Case I. If there exists a real number K such that an $ K
for all n, then lim an D a exists by completeness.
Case II. Otherwise, for every integer K, there exists
n # N such that an > K, and hence aj > K for all
j # n. Thus lim an D 1.
If fan g is ultimately decreasing, then either it is bounded
below, and therefore converges, or else it is unbounded
below, and therefore diverges to negative infinity.
34.
If fjan jg is bounded then it is bounded above, and there
exists a constant K such that jan j $ K for all n. Therefore, "K $ an $ K for all n, and so fan g is bounded
above and below, and is therefore bounded.
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 9.2 (PAGE 514)
35. Suppose limn!1 jan j D 0. Given any $ > 0,
there exists an integer N D N.$/ such that if
n > N , then jjan j " 0j < $. In this case
jan " 0j D jan j D jjan j " 0j < $, so limn!1 an D 0.
36.
3.
nD5
b) “If lim an D 1 and lim bn D "1, then
lim.an C bn / D 0” is FALSE. Let an D 1 C n
and bn D "n; then lim an D 1 and lim bn D "1
but lim.an C bn / D 1.
1
.2 C #/10
1
D
.2 C #/10
1
D
.2 C #/10
4.
d) “If neither fan g nor fbn g converges, then fan bn g
does not converge” is FALSE. Let an D bn D ."1/n ;
then lim an and lim bn both diverge. But
an bn D ."1/2n D 1 and fan bn g does converge
(to 1).
e) “If fjan jg converges, then fan g converges”
is FALSE. Let an D ."1/n . Then
limn!1 jan j D limn!1 1 D 1, but limn!1 an does
not exist.
5.
6.
1
!
3
1
1
1"
3
D
5
1
1"
1000
7.
(page 514)
D
5000
:
999
1
X
."5/2
."5/3
."5/4
."5/n
D
C
C
C !!!
2n
4
6
8
8
8
88
nD2
)
(
52
25
5
C 2 " !!!
D 4 1"
8
64
64
25
1
25
25
D 4 !
D
D
:
5
8
64 & 69
4416
2 1 C 64
i
iii
# $2
1
X
1
1
1
D
1
C
C !!! D
C
en
e
e
nD0
9.
D
1
1
C
C
C !!!
.2 C #/12
.2 C #/14
)
(
1
1
C
C !!!
1C
.2 C #/2
.2 C #/4
1
1
h
i:
!
D
1
8
.2 C #/ .2 C #/2 " 1
1"
.2 C #/2
#
"
#
$2
1
5
1
C !!!
D5 1C
C
103n
1000
1000
D
8.
1.
1
X
nD0
c) “If lim an D 1 and lim bn D "1, then
lim an bn D "1” is TRUE. Let R be an arbitrary, large positive number. Since lim an D
p1
R
and lim bn D
p "1, we must have an #
and bn $ " R, for all sufficiently large n. Thus
an bn $ "R, and lim an bn D "1.
1
1
1
1
1
C C
C !!! D
1C C
3
9
27
3
3
1
.2 C #/2n
D
a) “If lim an D 1 and lim bn D L > 0, then
lim an bn D 1” is TRUE. Let R be an arbitrary,
large positive number. Since lim an D 1, and L > 0,
2R
it must be true that an #
for n sufficiently large.
L
L
for n
Since lim bn D L, it must also be that bn #
2
2R L
sufficiently large. Therefore an bn #
D R for n
L 2
sufficiently large. Since R is arbitrary, lim an bn D 1.
Section 9.2 Infinite Series
1
X
1
1
1"
e
D
e
:
e"1
1 # $k
1
X
X
2
8e 3
2kC3
8e 4
3
D
8e
D
D
:
k!3
2
e
e"2
e
kD0
1"
kD0
e
P1
P
# j=2 cos.j#/ D j1D2 ."1/j # j=2 diverges because
limj !1 ."1/j # j=2 does not exist.
j D1
P1
nD1
3 C 2n
diverges to 1 because
2nC2
3
C1
n
3 C 2n
1
2
lim
D
lim
D > 0:
n!1 2nC2
n!1
4
4
!
# $2
1
C !!!
3
1
:
2
10.
1 # $
1 # $
1
X
1X 1 n 1X 2 n
3 C 2n
C
D
3nC2
3
3
9
3
nD0
nD0
# $n!1
1
X
1
3
3
3
3
3 "
D
2. 3 " C
"
C!!! D
4 16 64
4
1
C
nD1
1
4
D
12
:
5
1
D !
3
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nD0
1
1
1
1
1
5
C !
D C D :
1
2
9
2
3
6
1"
1"
3
3
353
lOMoARcPSD|6566483
SECTION 9.2 (PAGE 514)
11.
1
1
D
n.n C 2/
2
Since
#
ADAMS and ESSEX: CALCULUS 9
$
1
1
"
, therefore
n nC2
14.
1
1
1
1
C
C
C !!! C
1&3
2&4
3&5
n.n C 2/
(
1 1 1
1 1
1 1
1 1
D
" C " C " C " C !!!
2 1 3
2 4
3 5
4 6
)
1
1
1
1
1
1
" C
"
C "
C
n"2 n
n"1 nC1
n nC2
(
)
1
1
1
1
D
1C "
"
:
2
2 nC1 nC2
sn D
Thus lim sn D
Hence,
1
X
1
3
3
, and
D .
4
n.n C 2/
4
nD1
1
1
1
1
D
C
C
C !!!:
.2n " 1/.2n C 1/
1&3
3&5
5&7
1
1
D
.2n " 1/.2n C 1/
2
partial sum is
Since
#
$
1
1
, the
"
2n " 1 2n C 1
#
$
#
$
1
1
1 1 1
sn D
1"
C
"
C !!!
2
3
2 3 5
#
$
#
$
1
1
1
1
1
1
"
"
C
C
2 2n " 3 2n " 1
2 2n " 1 2n C 1
#
$
1
1
D
1"
:
2
2n C 1
Hence,
1
X
nD1
1
1
D lim sn D :
.2n " 1/.2n C 1/
2
1
1
D
.3n " 2/.3n C 1/
3
13. Since
fore
#
$
1
1
"
, there3n " 2 3n C 1
1
1
1
1
C
C
C !!! C
1&4
4&7
7 & 10
.3n " 2/.3n C 1/
(
1 1
1
1
1 1 1
" C " C "
C !!!
D
3 1 4
4 7
7 10
)
1
1
1
1
C
"
C
"
3n " 5 3n " 2
3n " 2 3n C 1
#
$
1
1
1
D
1"
!
:
3
3n C 1
3
16.
17.
Thus
354
nD1
1
1
D .
.3n " 2/.3n C 1/
3
1
1
D lim sn D :
n.n C 1/.n C 2/
4
1
1 1
1
>
D ! , therefore the partial sums
2n " 1
2n
2 n
of the given series
P exceed half those of the divergent harmonic series .1=2n/. Hence the given series diverges to
infinity.
Since
1
X
nD1
n
n
diverges to infinity since lim
D 1 > 0.
nC2
nC2
1
1
Since n!1=2 D p # for n # 1, we have
n
n
n
X
kD1
k !1=2 #
n
X
1
! 1;
k
kD1
P !1=2
as n ! 1 (harmonic series). Thus
n
diverges to
infinity.
#
$
1
X
1
1
1
2
D 2
C C C ! ! ! diverges to infinity
18.
nC1
2
3
4
nD1
since it is just twice the harmonic series with the first
term omitted.
n
"1 if n is odd
19. sn D "1 C 1 " 1 C ! ! ! C ."1/n D
.
P 0 n if n is even
Thus lim sn does not exist, and ."1/ diverges.
20.
sn D
P1
1
X
nD1
15.
nD1
(
)
1 1
2
1
1
D
"
C
;
n.n C 1/.n C 2/
2 n nC1
nC2
the partial sum is
#
$
#
$
1
2
1
1 1 2
1
sn D
1" C
C
" C
C !!!
2
2
3
2 2 3
4
#
$
#
$
1
2
1
1 1
2
1
1
" C
C
"
C
C
2 n"1 n
nC1
2 n nC1
nC2
#
$
1 1
1
1
D
"
C
:
2 2 nC1
nC2
12. Let
1
X
Since
21.
n.n C 1/
Since 1 C 2 C 3 C ! ! ! C n D
, the given series
2
P1
2
is nD1
which converges to 2 by the result of
n.n C 1/
Example 3 of this section.
The total distance is
"
#
# $2
3
3
C !!!
2C2 2& C2&
4
4
"
#
# $2
3
3
3
1C C
C !!!
D2C2&
2
4
4
D2 C
3
1"
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4
D 14 metres.
lOMoARcPSD|6566483
INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 9.3 (PAGE 524)
29.
“If an # c > 0 for all n, then
TRUE. We have
P
an diverges to infinity” is
sn D a1 C a2 C a3 C ! ! ! C an # c C c C c C ! ! ! C c D nc;
and nc ! 1 as n ! 1.
2 m
30.
Fig. 9.2-21
22. The balance at the end of 8 years is
31.
h
i
sn D 1000 .1:1/8 C .1:1/7 C ! ! ! C .1:1/2 C .1:1/
#
$
.1:1/8 " 1
D 1000.1:1/
' $12; 579:48:
1:1 " 1
23. For n > N let sn D
n
X
j D1
aj , and Sn D
Then sn D Sn C C , where C D
n
X
j DN
PN !1
j D1 aj .
P
an diverges and fbn g is bounded, then
an bn
1
1
and bn D
.
diverges” is FALSE. Let an D
n
n
C
1
P
Then
an D 1 and 0 $ bn $ 1=2. But
P
P
1
which converges by Example
an bn D
n.n C 1/
3.
“If
P
P
P 2
“If an > 0 and
an converges, then
an converges” is
TRUE.P
Since
an converges, therefore lim an D 0.
Thus there exists N such that 0 < an $ 1 for n # N .
Thus 0 < an2 $ an for n # N .
n
n
X
X
If Sn D
ak2 and sn D
ak , then fSn g is increasing
kDN
aj .
Sn $ sn $
We have
lim sn D lim Sn C C W
n!1
n!1
eitherPboth sides exist or neither does. Hence
and 1
nDN both converge or neither does.
P1
nD1
Thus
an
If fan g is ultimately positive, then the sequence fsn g of
partial sums of the series must be ultimately increasing.
By Theorem 2, if fsn g is ultimately increasing, then either it is bounded above, and therefore convergent, or else
it
Pis not boundedPabove and diverges to infinity. Since
an D lim sn ,
an must either converge when fsn g converges and lim sn D s exists, or diverge to infinity when
fsn g diverges to infinity.
P
25. If fan g is ultimately negative, then the series
an must
either converge (if its partial sums are bounded below), or
diverge to "1 (if its partial sums are not bounded below).
P
26. “If an D 0 forPevery n, then
an converge” is TRUE
n
because
s
D
0
D
0,
for
every
n, and so
n
kD0
P
an D lim sn D 0.
P
P
27. “If
an converges, then
1=a
Pn diverges to infinity” is
FALSE. A counterexample is ."1/n =2n .
P
P
28. “If
an and
bn both diverge, then so does
P
1
.an C bn /” is FALSE. Let an D
and
n
P
P
1
bn D " , then
an D 1 and
bn D "1 but
n
P
P
.an C bn / D .0/ D 0.
24.
kDN
and bounded above:
1
X
1
X
kD1
ak2 converges, and so
kDN
ak < 1:
1
X
ak2 converges.
kD1
Section 9.3 Convergence Tests for Positive
Series (page 524)
1.
2.
X 1
1
converges by comparison with
since
C1
n2
1
1
0< 2
< 2.
n C1
n
X
n2
1
X
1
X
n
1
since
converges
by
comparison
with
n4 " 2
n3
nD1
nD1
n &
lim # "$2 D 1;
1
n3
%
3.
X n2 C 1
n3
n4
and 0 < 1 < 1:
diverges to infinity by comparison with
C1
1
n2 C 1
> .
since 3
n C1
n
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n
355
lOMoARcPSD|6566483
SECTION 9.3 (PAGE 524)
4.
ADAMS and ESSEX: CALCULUS 9
p
1
X
1
n
converges
by
comparison
with
3=2
n2 C n C 1
n
nD1
nD1
since
p
$
#
n
n2 C n C 1
$
#
D 1; and 0 < 1 < 1:
lim
1
n3=2
1
X
1
X
1
p
ˇ
X ˇˇ
X 1
ˇ
ˇsin 1 ˇ converges by comparison with
so
.
ˇ n2 ˇ
n2
1
X
1
converges by comparison with the geometric
#n C 5
nD8
1 # $n
X
1
1
1
series
since 0 < n
< n.
#
# C5
#
3
14.
1
X
15.
1
a
dt
D
t ln t .ln ln t /2
X 1 " ."1/n
n4
since 0 $
16.
dt
p
D
t ln t ln ln t
X
series
10.
1
" n!
X 1
#n
#n
#n
D lim
" n!
1
D 1, the series
n!
#n
converges by comparison with the geometric
9. Since limn!1
#n
1"
nD0
11.
12.
X 1 C n4=3
diverges to infinity by comparison with the
2 C n5=3
X 1
divergent p-series
, since
n1=3
,
1
n1=3 C n5=3
1 C n4=3
D
lim
D 1:
lim
n!1 2 C n5=3
n1=3
2 C n5=3
1
X
Z
1
ln ln a
du
< 1 if ln ln a > 0:
u2
2
1 " ."1/n
$ 4.
n4
n
X 1
,
n4
1
X
2
2
2
1 C ."1/n
p
D 0 C p C 0 C p C 0 C p C !!!
n
2
4
6
nD1
D2
1
X
1
p X
1
1
p D 2
p
2k
k
kD1
kD1
diverges to infinity.
X
1
1
1
< n , the series
con2n .n C 1/
2
2n .n C 1/
X 1
.
verges by comparison with the geometric series
2n
1
X
n4
18.
converges by the ratio test since
nŠ
17.
Since
nD1
.n C 1/4
#
$
nC1 4 1
.n C 1/Š
D
lim
D 0:
lim
n
nC1
n4
nŠ
19.
X nŠ
diverges to infinity by the ratio test, since
n2 e n
* D lim
20.
.n C 1/Š
n2 e n
1
n2
!
D
lim
D 1:
.n C 1/2 e nC1
nŠ
e
nC1
1
X
.2n/Š6n
converges by the ratio test since
.3n/Š
nD1
n2
p diverges to infinity since
1Cn n
nD1
.2n C 2/Š6nC1
lim
.3n C 3/Š
lim
D lim
n2
p D 1.
1Cn n
356
ln ln 3
du
p D 1:
u
The series
.
1
X
1Cn
1Cn
diverges to infinity since lim
D 1 > 0.
2Cn
2Cn
1
converges by comparison with
7.
nD1
Z
1
converges by the integral test:
n ln n.ln ln n/2
nD2
nD8
X 1
Since .ln n/3 < n for large n,
diverges to
.ln n/3
X1
.
infinity by comparison with
n
1
X
1
8.
diverges to infinity by comparison with the
ln.3n/
nD1
1
X
1
1
1
since
>
for n # 1.
harmonic series
3n
ln.3n/
3n
diverges to infinity by the integral test,
n ln n ln ln n
since
Z 1
nD3
Z
5. Since sin x $ x for x # 0, we have
ˇ
ˇ
ˇ
ˇ
ˇsin 1 ˇ D sin 1 $ 1 ;
ˇ n2 ˇ
n2
n2
6.
13.
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.2n/Š6n
.3n/Š
.2n C 2/.2n C 1/6
D 0:
.3n C 3/.3n C 2/.3n C 1/
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INSTRUCTOR’S SOLUTIONS MANUAL
21.
SECTION 9.3 (PAGE 524)
p
1
X
n
converges by the ratio test, since
3n ln n
We use the approximation
nD2
s'
p
3n ln n
nC1
! p
nC1
3
ln.n C 1/
n
r
1
ln n
1
nC1
D lim
! lim
D < 1:
3
n
ln.n C 1/
3
* D lim
22.
23.
1
X
js "
,
.n C 1/100 2nC1
n100 2n
lim p
p
nŠ
.n C 1/Š
#
$100
nC1
1
p
D 0:
D lim 2
n
nC1
1
X
1 C nŠ
diverges by comparison with the harmonic
.1 C n/Š
nD1
1
X
1 C nŠ
nŠ
1
1
since
>
D
.
series
nC1
.1 C n/Š
.1 C n/Š
nC1
25.
2n
converges by the ratio test since
3n " n3
* D lim
D
26.
2
lim
3
3
3 "n
.n C 1/3
3n "
3
' 1:082
,
28.
1
is positive, continuous and decreasing
x3
on Œ1; 1/, for any n D 1; 2; 3; : : :, we have
Since f .x/ D
sn C AnC1 $ s $ sn C An
n3
2
2
3n
D < 1:
D lim
3
3
.n C 1/3
1"
3nC1
1"
where sn D
sn"
Z 1
n
X
dx
1
1
and
A
D
D
. If
n
k3
x3
2n2
n
kD1
1
D sn C .AnC1 C An /, then
2
(
)
An " AnC1
1 1
1
"
D
2
4 n2 .n C 1/2
1 2n C 1
< 0:001
D
4 n2 .n C 1/2
jsn " sn" j $
#
$
nn
e
1
1 n
D
D lim 1 C
< 1:
# n nŠ
#
n
#
f .x/ D 1=x 4 is positive, continuous, and decreasing on
Œ1; 1/. Let
An D
Z
n
1
#
$
1
1
1
"
$
2 3n3 3.n C 1/3
1 .n C 1/3 " n3
D
6 n3 .n C 1/3
1 3n2 C 3n C 1
7
D
<
:
6 n3 .n C 1/3
6n4
with error less than 0.001 in absolute value:
nD1
27.
sn" j
nD1
1
X
nn
converges by the ratio test since
# n nŠ
.n C 1/nC1
lim .nC1/
#
.n C 1/Š
:
#
$
1
X
1
1
1
1
1
1
1 1
1
"
'
s
D
1
C
C
C
C
C
C
C
6
n4
24
34
44
54
64
6 73
63
3n " n3
2nC1
!
3nC1 " .n C 1/3
2n
n
$
that is, if n4 > 7000=6. Since 64 D 1296 > 7000=6, n D 6
will do. Thus
nD1
X
1
1
C 3
3.n C 1/3
3n
We have used 3n2 C 3n C 1 $ 7n2 and n3 .n C 1/3 > n6
to obtain the last inequality. We will have js " sn" j < 0:001
provided
7
< 0:001;
6n4
X .2n/Š
converges by the ratio test, since
.nŠ/3
.2n C 2/Š .nŠ/3
.2n C 2/.2n C 1/
!
D lim
D 0 < 1:
..n C 1/Š/3 .2n/Š
.n C 1/3
#
The error satisfies
n100 2n
converges by the ratio test since
p
nŠ
nD0
* D lim
24.
1
D sn C
2
sn"
#
$ˇR
1 ˇˇ
dx
1
D
lim
"
ˇ D 3:
R!1
x4
3x 3 ˇ
3n
if n D 8. Thus, the error in the approximation s ' s8" is
less than 0.001.
29.
1
is positive, continuous and decreasing
x 3=2
on Œ1; 1/, for any n D 1; 2; 3; : : :, we have
Since f .x/ D
n
sn C AnC1 $ s $ sn C An
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357
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SECTION 9.3 (PAGE 524)
n
X
ADAMS and ESSEX: CALCULUS 9
Z
1
dx
2
D p . If
3=2
n
k 3=2
x
n
kD1
#
$
1
1
1
sn" D sn C .AnC1 C An / D sn C p C p
, then
2
n
nC1
where sn D
jsn " sn" j $
D
D
<
1
and An D
31.
kD1
An " AnC1
$
# 2
1
2
2
p "p
2
n
nC1
p
p
nC1" n
1
D p p
p
p
p p
n nC1
n n C 1. n C n C 1/
1
< 0:001
2n3=2
sn D
where a D tan!1
and b D tan!1
%n&
2
. Now
tan a " tan b
tan.a " b/ D
1#C tan a$tan b
%n&
nC1
"
2
2
#
$
D
n C 1 %n&
1C
2
2
2
D 2
n C n#C 4
$
2
, a " b D tan!1
:
n2 C n C 4
kD1
32.
1
!
1
D
2nC1 .n
D
nC2
< 0:001
2n .n C 1/Š.2n C 3/
C 1/Š
1"
1
2.n C 2/
1
1
1
1
if n D 4. Thus, s ' s4 D C 2 C 3 C 4 with
2
2 2Š
2 3Š
2 4Š
error less than 0.001.
1
X
1
We have s D
and
.2k " 1/Š
kD1
sn D
n
X
kD1
1
1
1
1
1
D
C
C
C !!! C
:
.2k " 1/Š
1Š
3Š
5Š
.2n " 1/Š
Then
1
1
1
C
C
C !!!
.2n C 1/Š
.2n C 3/Š
.2n C 5/Š
"
1
1
D
1C
C
.2n C 1/Š
.2n C 2/.2n C 3/
0 < s " sn D
We want error less than 0.001:
#
$
1
2
1
.a " b/ D tan!1
< 0:001
4
4
n2 C n C 4
2
,
< tan 0:004
2
n CnC4
, n2 C n > 2 cot.0:004/ " 4 ' 496:
358
1
1
1
1
C 2 C 3 C !!!C n :
2
2 2Š
2 3Š
2 nŠ
1
1
1
C nC2
C nC3
C !!!
2nC1 .n C 1/Š
2
.n C 2/Š
2
.n C 3/Š
(
)
1
1
1
D nC1
1C
C 2
C !!!
2
.n C 1/Š
2.n C 2/
2 .n C 2/.n C 3/
"
#
#
$2
1
1
1
1C
C
C !!!
< nC1
2
.n C 1/Š
2.n C 2/
2.n C 2/
1
If sn" D sn C .AnC1 C An /, then
2
$
D
D
n
nC1
2
1
2k kŠ
0 < s " sn
30. Again, we have sn C AnC1 $ s $ sn C An where
P
1
sn D nkD1 2
and
k C4
ˇ
Z 1
% & ˇ1
% &
1
dx
# 1
!1 x ˇ
!1 n
D
tan
"
tan
:
An D
D
ˇ
x2 C 4
2
2 ˇ
4 2
2
n
#
n
X
Then
"
if n # 63. Thus, the error in the approximation s ' s63
is
less than 0.001.
An " AnC1
jsn " sn" j $
2
"
#
$#
% & #
1 #
1
1
!1 n
!1 n C 1
D
" tan
" C tan
2 4
2
2
4
2
2
"
#
#
$
%n&
nC1
1
1
tan!1
" tan!1
D .a " b/;
D
4
2
2
4
"
n D 22 will do. The approximation s ' s22
has error less
than 0.001.
1
X
1
and
We have s D
2k kŠ
1
C !!!
.2n C 2/.2n C 3/.2n C 4/.2n C 5/
"
1
1
1C
C
<
.2n C 1/Š
.2n C 2/.2n C 3/
#
1
C !!!
Œ.2n C 2/.2n C 3/!2
2
3
6
1
6
.2n C 1/Š 4
7
7
5
1
1"
.2n C 2/.2n C 3/
4n2 C 10n C 6
1
< 0:001
D
.2n C 1/Š 4n2 C 10n C 5
D
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#
lOMoARcPSD|6566483
INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 9.3 (PAGE 524)
1
X
1
D
f .n/ converges by the integral test, and
1 C n2
nD1
nD1
its sum is less than #=2.
1
X
1
1
if n D 3. Thus, s ' s3 D 1 C
C
D 1:175 with error
3Š
5Š
less than 0.001.
33. We have s D
0 < s " sn
n
n!1
1
X 2k
X
2k
and sn D
. Thus
.2k/Š
.2k/Š
kD0
kD0
nC1
36.
Z
nC2
2
2
2
D
C
C
C !!!
.2n/Š
.2n C 2/Š
.2n C 4/Š
n (
2
2
1C
D
.2n/Š
.2n C 1/.2n C 2/
)
22
C
C !!!
.2n C 1/.2n C 2/.2n C 3/.2n C 4/
#
"
#
$2
2
2n
2
C !!!
<
1C
C
.2n/Š
.2n C 1/.2n C 2/
.2n C 1/.2n C 2/
D
2n
!
.2n/Š
Let u D ln ln t , du D
1
X
nDN
kD1
Thus
1
1
1
0 < s " sn D
C
C
C !!!
.n C 1/nC1
.n C 2/nC2
.n C 3/nC3
"
#
1
1
1
1C
<
C
C !!!
.n C 1/nC1
nC1
.n C 1/2
2
3
series
1
X
nDN
1
X
1
1
1
C 3 C 4 D 1:291 with
22
3
4
ˇR
ˇ
#
ˇ
f .x/ dx D lim tan!1 x ˇ D ;
ˇ
R!1
2
0
,n , and
1
X
an also converges.
nD1
1
1
and bn D 2 .
n
n
ln n
Since lim n1=n D 1 (because lim
D 0), we have
n
1=n
1=n
lim.an /
D 1 and
D 1. That is, ) D
1 for
X lim.bn /
X
both series. But
an diverges to infinity, while
bn
converges. Thus the case ) D 1 provides no information
on the convergence or divergence of a series.
38.
Let an D 2nC1 =nn . Then
lim
nD1
0
an converges by comparison with the geometric
CASE III. Let an D
1
< 0:001
n.n C 1/n
1
aN C2 $ ,N C2 ; : : : :
CASE II. Suppose ) > 1. Then .an /1=n # 1, and an # 1,
for all
Xsufficiently large values of n. Therefore lim an ¤ 0
and
an must diverge. Since an > 0 it diverges to
infinity.
1
error less than 0.001.
1
. Then f is decreasing on Œ1; 1/.
35. Let f .x/ D
1 C x2
1
X
Since
f .n/ is a right Riemann sum for
Z
aN C1 $ ,N C1 ;
nDN
7
1 5
1"
nC1
if n D 4. Thus, s ' s4 D 1 C
1
n.ln n/.ln ln n/ ! ! ! .lnj n/.lnj C1 n/p
aN $ ,N ;
n
X
1
1
1
1
1
D C 2 C 3 C !!! C n:
1
2
3
n
kk
Then
D
ln ln a
du
up
CASE I. Suppose ) < 1. Pick , such that ) < , < 1.
Then there exists N such that .an /1=n $ , for all n # N .
Therefore
kD1
1
6
4
.n C 1/nC1
1
37. Let an > 0 for all n. (Let’s forget the “ultimately” part.)
Let ) D lim.an /1=n .
1
X
1
and
34. We have s D
kk
D
a
Z
converges if and only if p > 1, where N is large enough
that lnj N > 1.
if n D 4. Thus, s ' s4 with error less than 0.001.
sn D
dt
D
t ln t .ln ln t /p
will converge if and only if p > 1. Thus,
1
X
1
will converge if and only if p > 1.
n ln n.ln ln n/p
nD3
Similarly,
1
2
1"
.2n C 1/.2n C 2/
2n
4n2 C 6n C 2
D
!
< 0:001
.2n/Š
4n2 C 6n
1
dt
and ln ln a > 0; then
t ln t
n!1
p
n
an D lim
n!1
Since this limit is less than 1,
root test.
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2 & 21=n
D 0:
n
P1
nD1
an converges by the
359
lOMoARcPSD|6566483
SECTION 9.3 (PAGE 524)
1 #
X
39.
nD1
since
n
nC1
$n2
"#
) D lim
n!1
40. Let an D
ADAMS and ESSEX: CALCULUS 9
converges by the root test of Exercise 31
n
nC1
$n2 #1=n
43.
b) Let sN
1
1
D lim #
$ D < 1:
n!1
e
1 n
1C
n
$
1 C k N C1
1
2
sn < !
1Ck
k
1"
2
!
#
$
2
1 C k N C1
2
D
1"
$
:
k.1 " k/
2
k.1 " k/
1"
1
X
n
2
$
.
2n
k.1 " k/
nD0
Since the maximum value of k.1 " k/ is 1=4 (at
k D 1=2), the best upper bound we get for s by this
method is s $ 8.
$
1 #
1
X
1Ck j
1 X
j
<
c) s " sn D
2j
k
2
j DnC1
j DnC1
#
$nC1
1 1Ck
1
D
!
1Ck
k
2
1"
2
.1 C k/nC1
G.k/
D
D n ;
k.1 " k/2n
2
.1 C k/nC1
where G.k/ D
. For minimum G.k/, look
k.1 " k/
for a critical point:
X 22n .nŠ/2
, we obtain
.2n/Š
22nC2 ..n C 1/Š/2
.2n/Š
4.n C 1/2
! 2n
D lim
D 1:
2
.2n C 2/Š
2 .nŠ/
.2n C 2/.2n C 1/
Thus the ratio test provides no information. However,
22n .nŠ/2
Œ2n.2n " 2/ ! ! ! 6 ! 4 ! 2!2
D
.2n/Š
2n.2n " 1/.2n " 2/ ! ! ! 3 ! 2 ! 1
2n
2n " 2
4 2
D
!
! ! ! ! ! ! > 1:
2n " 1 2n " 3
3 1
k.1 " k/.n C 1/.1 C k/n " .1 C k/nC1 .1 " 2k/
D0
k 2 .1 " k/2
.k " k 2 /.n C 1/ " .1 C k/.1 " 2k/ D 0
Since the terms exceed 1, the series diverges to infinity.
k 2 .n C 1/ " k.n C 1/ C 1 " k " 2k 2 D 0
.n " 1/k 2 " .n C 2/k C 1 D 0
p
.n C 2/ ˙ .n C 2/2 " 4.n " 1/
kD
2.n " 1/
p
n C 2 ˙ n2 C 8
D
:
2.n " 1/
42. We have
1 & 2 & 3 & 4 & ! ! ! & 2n
.2n/Š
D
22n .nŠ/2
.2 & 4 & 6 & 8 & ! ! ! & 2n/2
1 & 3 & 5 & ! ! ! & .2n " 1/
D
2 & 4 & 6 & ! ! ! & .2n " 2/ & 2n
3 5 7
2n " 1
1
1
D 1 & & & & !!! &
&
>
:
2 4 6
2n " 2 2n
2n
an D
1
X
.2n/Š
diverges to infinity by comparison
22n .nŠ/2
nD1
1
X
1
with the harmonic series
.
2n
For given n, the
p upper bound is minimal if
n C 2 " n2 C 8
kD
(for n # 2).
2.n " 1/
44.
If s D
1
X
kD1
nD1
360
#
Therefore, s D
P
Thus 1
nD1 an converges by the ratio test.
(Remark: the question contained a typo. It was intended
to ask that #33 be repeated, using the ratio test. That is a
little harder.)
Therefore
nD0
N
1 1 " r N C1
1 X n
r D !
;
D
k
k
1"r
nD0
where r D .1 C k/=n. Thus
anC1
2nC2
nn
D
! nC1
nC1
an
.n C 1/
2
2
1
2
D
!#
#
$n D
$
nC1
n
1 n
.n C 1/
1C
nC1
n
1
! 0 & D 0 as n ! 1:
e
* D lim
$
N #
N
X
1 X 1Ck n
n
<
D
2n
k
2
nD0
2nC1
. Then
nn
41. Trying to apply the ratio test to
a) If n is a positive integer and k > 0, then
1
.1 C k/n # 1 C nk > nk, so n < .1 C k/n .
k
ck D
1
X
kD1
1
, then we have
k 2 .k C 1/
sn C AnC1 $ s $ sn C An
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INSTRUCTOR’S SOLUTIONS MANUAL
where sn D
An D
Z
1
n
n
X
kD1
SECTION 9.4 (PAGE 531)
1
and
k 2 .k C 1/
dx
D
x 2 .x C 1/
Z
1
n
(b) Let Sn D
#
"1
1
1
C 2 C
x
x
xC1
ˇ1
ˇ
1
ˇ
D " ln x " C ln.x C 1/ˇ
ˇ
x
n
ˇ1
#
$
1
1 ˇˇ
D ln 1 C
" ˇ
x
xˇ
#
$n
1
1
D " ln 1 C
:
n
n
$
Pn
bi , where bn D
0 < bn D
dx
0<
1
X
iD1
$
1
1
C
C
!
!
!
4nC1
4
42
1
1
4
<
D f rac14nC1 & D
n
3
3&4
1;000
1
An " AnC1
" 2 #
$
$#
#
1
1 1
1
1
D
" ln 1 C
"
C ln 1 C
2 n
n
nC1
nC1
"
#
$#
2
n C 2n
1
1
C ln
D
2 n.n C 1/
n2 C 2n C 1
"
#
$#
n2 C 2n
1
1
C
"1
$
2 n.n C 1/
n2 C 2n C 1
1
X
1C
nD1
1
1
X
X
1
1
D
"
bn
2n C 1
2n
nD1
' 1"
nD1
5
X
bn
nD1
#
$ #
$ #
$
1 1
1 1
1 1
"
"
"
"
"
2 3
4 5
8 9
#
$ #
$
1
1
1
1
"
"
"
"
16 17
32 33
' 0:765 with error less than 0.001:
1
D
< 0:001
2n.n C 1/2
D 1"
if n D 8. Thus,
Section 9.4 Absolute and Conditional
Convergence (page 531)
1.
D 1:6450
with error less than 0.001.
P
n
45. s D 1
nD1 1=.2 C 1/.
(a) We have
2.
1
X
Since 210 D 1;024, s10 will approximate s to within
0:001.
#
provided 4n > 1;000=3. Thus n D 5 will
P do (but
n D 4 is insufficient). S5 approximates 1
nD1 bn to
within 0:001.
P
n
(c) Since 1
nD1 1=2 D 1, we have
jsn " sn" j $
1
1
1
1
D nC1 C nC2 C nC3 C ! ! !
2i C 1
2
2
2
iD1
#
$
1
1
1
D nC1 1 C C 2 C ! ! !
2
2
2
1
1
D n <
if 2n > 1;000:
2
1;000
2n C 1 " 2n
1
< n;
2n .2n C 1/
4
bi " Sn D bnC1 C bnC2 C bnC3 C ! ! !
<
1
X
1
1
D 1 C s8" D 1 C s8 C .A9 C A8 /
n2
2
nD1
)
(
1
1
1
1
C 2
C 2
C !!!C 2
C
D1C
2
2 .3/
3 .4/
8 .9/
"#
#
$ #
$
1
1
10
1
9
" ln
C
" ln
2
9
9
8
8
1
1
"
. Since
2n 2n C 1
we have
1
If sn" D sn C .AnC1 C An /, then
2
0 < s " sn D
iD1
X ."1/n
p
converges by the alternating series test (since
n
the terms alternate in sign, decrease in size, and approach
0). However, the convergence is only conditional, since
X 1
p diverges to infinity.
n
1
X
."1/n
converges absolutely since
C ln n
nD1
ˇ
ˇ
1
X
ˇ ."1/n ˇ
1
1
ˇ
ˇ
converges.
ˇ n2 C ln n ˇ $ n2 and
n2
n2
nD1
3.
X
X
cos.n#/
."1/n
D
converges
.n C 1/ ln.n C 1/
.n C 1/ ln.n C 1/
by the alternating series test, but only conditionally since
X
1
diverges to infinity (by the integral
.n C 1/ ln.n C 1/
test).
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SECTION 9.4 (PAGE 531)
4.
5.
6.
ADAMS and ESSEX: CALCULUS 9
1
1
X
X
1
."1/2n
D
is a positive, convergent geometric
2n
2n
nD1
nD1
series so must converge absolutely.
12.
X ."1/n .n2 " 1/
diverges since its terms do not apn2 C 1
proach zero.
1
X
."2/n
converges absolutely by the ratio test since
nŠ
nD1
13.
ˇ
ˇ
ˇ ."2/nC1
nŠ ˇˇ
1
lim ˇˇ
!
D 2 lim
D 0:
.n C 1/Š ."2/n ˇ
nC1
7.
X ."1/n
converges absolutely, since, for n # 1,
n# n
and
8.
1
X
10.
X
20n2 " n " 1
converges by the alternating series
n3 C n2 C 33
test (the terms are ultimately decreasing in size, and approach zero), but the convergence is only conditional since
X 20n2 " n " 1
diverges to infinity by comparison with
n3 C n2 C 33
X1
.
n
1
1
X
X
100."1/n
100 cos.n#/
D
converges by the alter2n C 3
2n C 3
nD1
nD1
nating series test but only conditionally since
1
X
and
nD1
11.
X
ˇ
ˇ
ˇ 100."1/n ˇ
100
ˇ
ˇ
ˇ 2n C 3 ˇ D 2n C 3
100
diverges to infinity.
2n C 3
kD1
n
X
k
k
,
and
s
D
."1/k!1 2
,
n
k2 C 1
k C1
kD1
14.
15.
."1/n
are alter.2n/Š
nating in sign and decreasing in size, the size of the error
in the approximation s ' sn does not exceed that of the
first omitted term:
1
js " sn j $
< 0:001
.2n C 2/Š
1
1
1
if n D 3. Hence s ' 1 "
C
" ; four terms will
2Š
4Š
6Š
approximate s with error less than 0.001 in absolute value.
P1
Since the terms of the series s D
If s D
1
X
."1/k!1
kD1
nD0
n
X
k
k
,
and
s
D
."1/k!1 k , then
n
k
2
2
kD1
nC1
js " sn j < nC1 < 0:001
2
if n D 13, because the series satisfies the conditions of the
alternating series test from the second term on.
16.
17.
Since the terms of the series s D
P1
n3
nD0 ."1/
n
are
nŠ
alternating in sign and ultimately decreasing in size (they
decrease after the third term), the size of the error in the
approximation s ' sn does not exceed that of the first
3nC1
omitted term (provided n # 3): js"sn j $
< 0:001
.n C 1/Š
if n D 12. Thus twelve terms will suffice to approximate s
with error less than 0.001 in absolute value.
X xn
Applying the ratio test to
p
, we obtain
nC1
ˇ
ˇ
r
p
ˇ x nC1
n C 1 ˇˇ
nC1
ˇ
* D lim ˇ p
!
D jxj:
ˇ D jxj lim
ˇ nC2
xn ˇ
nC2
Hence the series converges absolutely if jxj < 1, that is,
if "1 < x < 1. The series converges conditionally for
x D "1, but diverges for all other values of x.
.x " 2/n
. Apply the ratio test
n2 22n
ˇ
ˇ
ˇ .x " 2/nC1
n2 22n ˇˇ
jx " 2j
* D lim ˇˇ
&
D
<1
.n C 1/2 22nC2 .x " 2/n ˇ
4
18. Let an D
nŠ
nŠ
diverges since lim
D 1.
."100/n
100n
362
."1/k!1
nC1
< 0:001
.n C 1/2 C 1
if n D 999, because the series satisfies the conditions of
the alternating series test.
X 1
is a convergent geometric series.
#n
."1/n
1
X
js " sn j <
nD0
9.
If s D
then
ˇ
ˇ
ˇ ."1/n ˇ
1
ˇ
ˇ
ˇ n# n ˇ $ # n ;
"n
diverges to "1 since all terms are negative
n2 C 1
nD0
1
X
n
diverges to infinity by comparison with
and
n2 C 1
nD0
1
X
1
.
n
1
1
X
X
sin.n C 12 /#
."1/n
D
converges by the alterln ln n
ln ln n
nD10
nD10
1
X
1
nating series test but only conditionally since
ln ln n
nD10
1
X
1
.
diverges to infinity by comparison with
n
nD10
(ln ln n < n for n # 10.)
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 9.4 (PAGE 531)
if and only if jx " 2j < 4, that is "2 < x < 6. If x D "2,
1
1
X
X
."1/n
an D
then
, which converges absolutely. If
n2
nD1
nD1
1
1
X
X
1
x D 6, then
an D
, which also converges abson2
nD1
nD1
lutely. Thus, the series converges absolutely if "2 $ x $ 6
and diverges elsewhere.
19. Apply the ratio test to
X
."1/n
.x " 1/n
:
2n C 3
23.
ˇ
ˇ
ˇ .x " 1/nC1 2n C 3 ˇ
ˇ D jx " 1j:
* D lim ˇˇ
!
2n C 5
.x " 1/n ˇ
The series converges absolutely if jx " 1j < 1, that is,
if 0 < x < 2, and converges conditionally if x D 2. It
diverges for all other values of x.
#
$
1
3x C 2 n
20. Let an D
. Apply the ratio test
2n " 1
"5
ˇ
ˇ
ˇ 1 # 3x C 2 $nC1 2n " 1 # 3x C 2 $!n ˇ
ˇ
ˇ
&
* D lim ˇ
ˇ
ˇ
ˇ 2n C 1
"5
1
"5
ˇ
ˇ
ˇ 3x C 2 ˇ
ˇ<1
D ˇˇ
5 ˇ
24.
ˇ
2 ˇˇ
5
7
< , that is " < x < 1. If
3ˇ
3
3
1
1
X
X
7
1
x D " , then
an D
, which diverges.
3
2n " 1
nD1
nD1
1
1
X
X
."1/n
, which converges
If x D 1, then
an D
2n " 1
nD1
nD1
conditionally. Thus, the series converges absolutely if
7
" < x < 1, converges conditionally if x D 1 and
3
diverges elsewhere.
Apply the ratio test to
ˇ
ˇ
* D lim ˇˇ
X
xn
2n ln n
25.
:
ˇ
x nC1
2n ln n ˇˇ
jxj
ln n
jxj
!
D
lim
D
:
2nC1 ln.n C 1/
xn ˇ
2
ln.n C 1/
2
(The last limit can be evaluated by l’H^opital’s Rule.) The
given series converges absolutely if jxj < 2, that is, if
"2 < x < 2. By the alternating series test, it converges
conditionally if x D "2. It diverges for all other values of
x.
22. Let an D
.4x C 1/n
. Apply the ratio test
n3
ˇ
ˇ
ˇ
3ˇ
The series converges absolutely if ˇˇx C ˇˇ < 2, that is, if
2
1
7
" < x < . By the alternating series test it converges
2
2
7
conditionally at x D " . It diverges elsewhere.
2
#
$
1 n
1
. Apply the ratio test
1C
Let an D
n
x
ˇ
ˇ
$
#
$ ˇ ˇ
ˇ 1 #
1 nC1 n
1 !n ˇˇ ˇˇ
1ˇ
ˇ
* D lim ˇ
1C
&
1C
ˇ D ˇ1C ˇˇ < 1
ˇ
ˇn C 1
x
1
x
x
ˇ
ˇ
if and only if ˇˇx C
21.
1
1
< x < 0. If x D " , then
2
2
1
1
X
X
."1/n
an D
, which converges absolutely.
n3
nD1
nD1
1
1
X
X
1
, which also conIf x D 0, then
an D
n3
nD1
nD1
verges absolutely. Thus, the series converges absolutely
1
if " $ x $ 0 and diverges elsewhere.
2
X .2x C 3/n
Apply the ratio test to
:
n1=3 4n
ˇ
ˇ
ˇ
ˇ
ˇ .2x C 3/nC1
ˇx C 3 ˇ
n1=3 4n ˇˇ
j2x C 3j
ˇ
2
* D lim ˇ
!
D
:
ˇD
ˇ .n C 1/1=3 4nC1 .2x C 3/n ˇ
4
2
if and only if "
26.
if and only if jx C 1j < jxj, that is,
1
1
1
"2 <
< 0 ) x < " . If x D " , then
x
2
2
1
1
X
X
."1/n
an D
, which converges conditionally.
n
nD1
nD1
1
Thus, the series converges absolutely if x < " , con2
1
verges conditionally if x D " and diverges elsewhere. It
2
is undefined at x D 0.
1
X
1
1
1
sin.n#=2/
D 1 C 0 " C 0 C C 0 " C 0 C !!!
n
3
5
7
nD1
The alternating series test does not apply directly, but does
apply to the modified series with the zero terms deleted.
Since this latter series converges conditionally, the given
series also converges conditionally.
If
1
X
10
for
every
n
#
1.
Hence,
an converges
n2
nD1
1
X
10
absolutely by comparison with
.
n2
then jan j $
ˇ
ˇ
ˇ .4x C 1/nC1
ˇ
n3
ˇ D j4x C 1j < 1
* D lim ˇˇ
&
3
n
.n C 1/
.4x C 1/ ˇ
8̂
10
< ;
if n is even;
2
an D n"1
:̂
; if n is odd;
10n3
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363
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SECTION 9.4 (PAGE 531)
27.
ADAMS and ESSEX: CALCULUS 9
P
P
a) “ an converges implies ."1/n an converges” is
n
."1/
is a counterexample.
FALSE. an D
n
P
P
b) “P an converges and ."1/n an converges implies
an converges absolutely” is FALSE. The series of
Exercise 25 is a counterexample.
P
P
c) “ an converges absolutely implies ."1/n an converges absolutely” is TRUE, because
j."1/n an j D jan j.
29.
obtain
* D lim jxj
an D
1
D n ln n " n C 1:
y
It is evident that an decreases as n increases. To see
whether lim an D 0, take logarithms and use the inequality
ln.1 C x/ $ x:
yDln x
3
4
n!1 n
#
$
#
$
#
$
1
1
1
ln an D ln 1 "
C ln 1 "
C ! ! ! C ln 1 "
2
4
2n
1 1
1
$ " " " !!! "
2#4
2n $
1
1
1
D"
1C C !!! C
! "1 as n ! 1:
2
2
n
x
Fig. 9.4-28
n
b) Let an D
nŠx
. Apply the ratio test
nn
ˇ
ˇ
ˇ .n C 1/Šx nC1
nn ˇˇ
* D lim ˇˇ
&
.n C 1/nC1
nŠx n ˇ
jxj
jxj
D lim #
<1
$ D
e
1 n
1C
n
Thus lim an D 0, and the given series converges conditionally at x D "1 by the alternating series test.
30.
P
1
1
and qn D " . Then
pn diverges
2n " 1
2n
P
to 1 and
qn diverges to "1. Also, the alternating
harmonic series is the sum of all the pn s and qn s in a
specific order:
Let pn D
1
1
X
X
."1/n!1
D
.pn C qn /:
n
if and only if "e < x < e. If x D ˙e, then, by (a),
ˇ nˇ
ˇ nŠe ˇ
ln ˇˇ n ˇˇ D ln.nŠ/ C ln e n " ln nn
n
> .n ln n " n C 1/ C n " n ln n D 1:
ˇ nˇ
ˇ nŠe ˇ
) ˇˇ n ˇˇ > e:
n
Hence,
1
X
nD1
an converges absolutely if "e < x < e
and diverges elsewhere.
364
.2n C 2/.2n C 1/
D jxj:
4.n C 1/2
.2n/Š
1 & 2 & 3 & 4 & ! ! ! & 2n
D
22n .nŠ/2
.2 & 4 & 6 & 8 & ! ! ! & 2n/2
1 & 3 & 5 & ! ! ! & .2n " 1/
D
2 & 4 & 6 & ! ! ! & .2n " 2/ & 2n
1 3
2n " 3 2n " 1
D & & !!! &
&
2
4
$ # 2n$" 2 # 2n
$#
$
#
1
1
1
1
1"
!!! 1 "
1"
:
D 1"
2
4
2n " 2
2n
ln.nŠ/ D ln 1 C ln 2 C ln 3 C ! ! ! C ln n
D sum of area of the shaded rectangles
ˇn
Z n
ˇ
ˇ
ln t dt D .t ln t " t /ˇ
>
ˇ
1
2
X
X .2n/Šx n
D
an x n , we
2n
2
2 .nŠ/
P
Thus
an x n converges absolutely if "1 < x < 1, and diverges if x > 1 or x < "1. In Exercise 36 of Section 9.3
1
, so the given series definitely
it was shown that an #
2n
diverges at x D 1 and may at most converge conditionally
at x D "1. To see whether it does converge at "1, we
write, as in Exercise 36 of Section 9.3,
28. a) We have
1
Applying the ratio test to
nD1
nD1
a) Rearrange
the terms as follows: first add terms of
P
pn until the sumP
exceeds 2. Then add q1 . Then
add more terms of
pn until the sum exceeds 3.
Then add q2 . Continue
P in this way; at the nth stage,
add new terms from
pn until the sum exceeds
n C 1, and then add qn . All partial sums after the
nth stage exceed n, so the rearranged series diverges
to infinity.
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 9.5 (PAGE 541)
b) Rearrange the terms of the original alternating
P harmonic series as follows: first add terms of
qn until
the sum is less than "2. Then add p1 . The sum will
now be greater than
P "2. (Why?) Then resume adding
new terms from
qn until the sum is less than "2
again, and add p2 , which will raise the sum above "2
again. Continue in this way. After the nth stage, all
succeeding partial sums will differ from "2 by less
than 1=n, so the rearranged series will converge to
"2.
1
X
."1/n
, which converges.
n4
nD1
1
X
1
At x D "4, the series is
, which also converges.
n4
nD1
Hence, the interval of convergence is Œ"4; 4!.
At x D 4, the series is
5.
1
X
nD0
Section 9.5 Power Series
1.
1
X
2. We have
1
X
nD0
6.
3n.x C 1/n . The centre of convergence is
1
X
en
.4 " x/n . The centre of convergence is
n3
nD1
x D 4. The radius of convergence is
R D lim
3n
D 1:
3.n C 1/
1
e n .n C 1/3
!
D :
n3
e nC1
e
1
X
1
."1/n
, which converges.
, the series is
e
n3
nD1
1
X
1
1
, which also converges.
At x D 4 " , the series is
e
n3
nD1
#
"
1
1
.
Hence, the interval of convergence is 4 " ; 4 C
e
e
At x D 4 C
The series converges absolutely on ."2; 0/ and diverges
on ."1; "2/ and .0; 1/. At x D "2, the series is
1
X
3n."1/n , which diverges. At x D 0, the series is
nD0
1
X
nD0
!
"n
2n n3 x " 23 . Here
We have
x D "1. The radius of convergence is
R D lim
1
X
2n n3
1
R D lim nC1
D . The radius of convergence
2
.n C 1/3
2
is 1=2; the centre of convergence is 3=2; the interval of
convergence is .1; 2/.
(page 541)
ˇp
ˇ
ˇ n C 2ˇ
x 2n
ˇ
ˇ
For
p
we have R D lim ˇ p
ˇ D 1. The
ˇ n C 1ˇ
n
C
1
nD0
radius of convergence is 1; the centre of convergence is
0; the interval of convergence is ."1; 1/. (The series does
not converge at x D "1 or x D 1.)
n3 .2x " 3/n D
3n, which diverges to infinity. Hence, the interval of
nD0
convergence is ."2; 0/.
1
X
#
$n
For
8.
We have
2nC1 .n C 1/
1 xC2
we have R D lim
D 2.
n
2
2n n
nD1
The radius of convergence is 2; the centre of convergence
is "2. For x D "4 the series is an alternating harmonic
series, so converges. For x D 0, the series is a divergent
harmonic series. Therefore the interval of convergence is
Œ"4; 0/.
3. For
1
X
."1/n n
x . The centre of convergence is
n4 22n
nD1
x D 0. The radius of convergence is
4. We have
ˇ
ˇ
ˇ ."1/n .n C 1/4 22nC2 ˇ
ˇ
ˇ
R D lim ˇ 4 2n !
ˇ
n 2
."1/nC1
ˇ
ˇ#
$
ˇ nC1 4 ˇ
ˇ
ˇ
! 4ˇ D 4:
D limˇ
ˇ
ˇ
n
1 C 5n n
x we
nŠ
1 C 5n .n C 1/Š
have R D lim
D 1. The radius of
!
nŠ
1 C 5nC1
convergence is infinite; the centre of convergence is 0; the
interval of convergence is the whole real line ."1; 1/.
P1
7.
nD0
$
1
1 # $n #
X
X
4
.4x " 1/n
1 n
D
. The cenx
"
nn
n
4
nD1
nD1
tre of convergence is x D 41 . The radius of convergence
is
4n .n C 1/nC1
R D lim n !
n#
4nC1
$
1
nC1 n
D lim
.n C 1/ D 1:
4
n
Hence, the interval of convergence is ."1; 1/.
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SECTION 9.5 (PAGE 541)
ADAMS and ESSEX: CALCULUS 9
9. By Example 5(a),
C 3x 2
C
4x 3
x2
C
C 3x 2
C 2x 2
x2
C
2x
& 1
C
x
C
1
C
2x
x
1
1
C
Thus
3x
Thus
C 6x 2
C
!!! D
x3
C
!!! D
C
C
C
4x 3
3x 3
2x 3
x3
C
C
C
C
!!!
!!!
!!!
!!!
!!!
C
10x 3
C
!!! D
1
.1 " x/2
1
1"x
nD0
12.
13.
1
.1 " x/3
1
X
.n C 1/.n C 2/ n
1
D
x ;
3
.1 " x/
2
1
1
1
1
1 X % x &n
$ D
D #
x
2"x
2
2
2
nD0
1"
2
x
x2
x3
1
."2 < x < 2/:
D C 2 C 3 C 4 C !!!
2
2
2
2
1
1
1
D !
2"x
2 1" x
2
1
x
x2
x3
D C 2 C 3 C 4 !!!
2
2
2
2
for "2 < x < 2. Now differentiate to get
1
2x
3x 2
1
D 2 C 3 C 4 C !!!
2
.2 " x/
2
2
2
1
X
.n C 1/x n
;
."2 < x < 2/:
D
2nC2
nD0
for "1 < x < 1.
nD0
10. We have
1 C x C x2 C x3 C ! ! ! D
1
X
1
D
xn
1"x
14.
1
X
1
D
."2x/n
1 C 2x
nD0
nD0
D 1 " 2x C 22 x 2 " 23 x 3 C ! ! !
Z x
Z xX
1
dt
tn
D
dt
nC1
0 2"t
0 nD0 2
ˇx
ˇx
1
ˇ
ˇ
X
t nC1
ˇ
ˇ
" ln.2 " t /ˇ D
ˇ
nC1
ˇ
2
.n C 1/ ˇ
and
1 " x C x2 " x3 C ! ! ! D
15.
1
X
1
D
."1/n x n
1Cx
nD0
0
holds for "1 < x < 1. Since an D 1 and bn D ."1/n for
n D 0; 1; 2; : : :, we have
Cn D
n
X
j D0
."1/n!j D
'
4
1 C x C x C !!! D
1
X
x
" ln.2 " x/ C ln 2 D
0; if n is odd;
1; if n is even.
16.
2n
nD0
1 " 2x C x
366
0
1
X
xn
:
2n n
."2 $ x < 2/:
nD1
Let y D x " 1. Then x D 1 C y and
."1 < y < 1/
nD0
D
1 h
X
nD0
in
".x " 1/
D 1 " .x " 1/ C .x " 1/2 " .x " 1/3 C .x " 1/4 " ! ! !
.for 0 < x < 2/:
By long division:
2
nD0
." 21 < x < 12 /:
x nC1
nC1
2
.n C 1/
1
X
1
1
D
D
."y/n
x
1Cy
1
1
1
!
D
D
1"x 1Cx
1 " x2
for "1 < x < 1.
11.
nD0
1
X
ln.2 " x/ D ln 2 "
Then the Cauchy product is
2
1
X
1
D
.n C 1/x n , for "1 < x < 1.
.1 " x/2
1 C 2x
1
1 " 2x
2x
2x
C 3x 2 C 4x 3 C ! ! !
C x2
" x2
" 4x 2 C 2x 3
3x 2 " 2x 3
3x 2 " 6x 3
4x 3
C
C
C
C
17.
!!!
!!!
!!!
!!!
Let x C 2 D t , so x D t " 2. Then
1
X
1
1
.n C 1/t n
D
D
x2
.2 " t /2
2nC2
nD0
1
X
.n C 1/.x C 2/n
D
;
2nC2
nD0
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lOMoARcPSD|6566483
INSTRUCTOR’S SOLUTIONS MANUAL
18.
SECTION 9.5 (PAGE 541)
1"x
2
D
"1
1Cx
1Cx
D 2.1 " x C x 2 " x 3 C ! ! !/ " 1
1
X
D1C2
."x/n
."1 < x < 1/:
for "1 < x < 1. Therefore,
1
X
.n C 3/x n D
nD0
19. We have
nD0
D
1
X
n 2nC3
2 x
!
;
nD0
#
$
1
1
"p < x < p :
2
2
23.
20. Let y D x " 4. Then x D 4 C y and
1
1
1
D
D #
x
4Cy
4
1
1 (
1X
.x " 4/
D
"
4
4
nD0
D
y
4
)n
1C
$ D
1
1 X % y &n
"
4
4
nD0
24.
.x " 4/2 .x " 4/3
1 .x " 4/
"
C
"
C !!!
2
4
4
43
44
xn
nD0
1
x
x2
x3
C C
C
C !!!
3 #4
5
6
$
4
5
3
1 x
x
x
D 3
C
C
C !!!
x
3
4
5
#
$
x2
x3
x4
x2
1
C
C
C C!!! " x "
D 3 xC
x
2
3
4
2
(
2)
1
x
D 3 " ln.1 " x/ " x "
x
2
1
1
1
D " 3 ln.1 " x/ " 2 "
: ."1 $ x < 1; x ¤ 0/:
x
x
2x
We start with
1
1Cx
and differentiate to get
"1 C 2x " 3x 3 C 4x 3 " ! ! ! D "
Z 4
Z x
dt
dt
dt
D
C
t
t
t
1
4
1
#
Z x"
1 .t " 4/
.t " 4/2 .t " 4/3
D ln 4 C
"
C
"
C ! ! ! dt
4
42
43
44
4
Z
1
X
1 " x C x2 " x3 C x4 " ! ! ! D
for 0 < x < 8. Therefore,
ln x D
nD0
nx n C 3
3
x
C
D
.1 " x/2
1"x
3 " 2x
."1 < x < 1/:
D
.1 " x/2
nD1
1
X
x3
3
D
x
.2x 2 /n
1 " 2x 2
1
X
1
:
.1 C x/2
Now we multiply by "x 3 :
x
x 3 " 2x 4 C 3x 5 " 4x 6 C ! ! ! D
x3
:
.1 C x/2
Differentiating again we get
x " 4 .x " 4/2
.x " 4/3 .x " 4/4
C
"
C !!!
"
4
2 ! 42
3 ! 43
4 ! 44
.for 0 < x $ 8/:
D ln 4 C
3x 2 " 2 & 4x 3 C 3 & 5x 4 " 4 & 6x 5 C ! ! ! D
x 3 C 3x 2
:
.1 C x/3
Finally, we remove the factor x 2 :
21.
2
3
1 D 4x C 16x " 64x C ! ! !
3 " 2 & 4x C 3 & 5x 2 " 4 & 6x 3 C ! ! ! D
D1 C ."4x/ C ."4x/2 " ."4x/3 C ! ! !
! 1
"
1
1
D
D
;
" 4 < x < 14 :
1 " ."4x/
1 C 4x
All steps are valid for "1 < x < 1.
25.
22. We differentiate the series
1
X
nD0
xn D 1 C x C x2 C x3 C ! ! ! D
1
1"x
nD0
1
, for "1 < x < 1,
Since 1 C x 2 C x 4 C x 6 C ! ! ! D
1 " x2
we obtain by differentiation
2x C 4x 3 C 6x 5 C 8x 7 C ! ! ! D
2x
;
.1 " x 2 /2
or, on division by x,
and multiply by x to get
1
X
xC3
:
.1 C x/3
nx n D x C 2x 2 C 3x 3 C ! ! ! D
x
.1 " x/2
2 C 4x 2 C 6x 4 C 8x 6 C ! ! ! D
2
;
.1 " x 2 /2
for "1 < x < 1.
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SECTION 9.5 (PAGE 541)
ADAMS and ESSEX: CALCULUS 9
x2 x3 x4
C
"
C ! ! ! D ln.1 C x/ for "1 < x $ 1,
26. Since x "
2
3
4
therefore
x2 "
6
4
Now let x D "1=2:
1
X
8
x
x
x
C
"
C ! ! ! D ln.1 C x 2 /
2
3
4
nD1
27.
(
ln.1 C x 2 /
x2
1
1
X
if "1 $ x $ 1, x ¤ 0
if x D 0.
From Example 5(a),
1
X
nx n!1
31.
Since
1
X
."1/n!1
."1 < x < 1/:
n
1
9
D !
" D :
1 2
3n!1
4
1
"
nD1
3
#
$
1
X
1
3
."1/n!1
D
ln
1
C
D ln :
n2n
2
2
32.
1
X
1 9
3
n
D ! D .
3n
3 4
4
Putting x D 1=#, we get
nD1
1
X
1Cx
for "1 < x < 1.
.1 " x/3
30. From Example 5(a),
nD1
nx n!1 D
nD3
368
nD0
!
.for all x/:
nD0
n.n " 1/x n!2 D
2
;
.1 " x/3
."1 < x < 1/
.n C 1/nx n!1 D
2
;
.1 " x/3
."1 < x < 1/:
nD1
1
X
.3x/n
nŠ
1
X
e3n x n
D
nŠ
2.
nD2
1
X
e 3xC1 D e ! e 3x D e
."1 < x < 1/:
Differentiate with respect to x and then replace n by nC1:
1
X
1 1
5
1
D ln 2 " " D ln 2 " :
n2n
2 8
8
Section 9.6 Taylor and Maclaurin Series
(page 550)
1.
1
;
.1 " x/2
kD0
1
X
1
D ln 2
n2n
1
1
X
X
1 C !1
# 2 .# C 1/
.n C 1/2
k2
D
D
D
:
1
n
k!1
#
.# " 1/3
#
.1 " ! /3
nD0
kD1
1
X
# $kC1
#
$
1
X
."1/k
1
1
1
D
"
D
ln
1
"
D " ln 2:
n2n
kC1
2
2
Therefore
# $k!1
1
1
X
X
nC1
1
1
D
D !
k
" D 4:
1 2
2n
2
1
"
nD0
kD1
2
nD1
."1/
nD1
28. From Example 5(a) with x D 1=2,
n2 x n!1 D
In the series for ln.1 C x/ in Example 5(c), put x D "1=2
to get
1
X
nD1
29. From Example 7,
xn
D ln.1 C x/ for "1 < x $ 1,
n
nD1
1
X
1
X
n.n C 1/
8
D" :
2n
27
therefore
1
D
;
.1 " x/2
Putting x D 1=3, we get
Thus
."1/n
nD1
nD1
nD1
n.n C 1/
16
D
:
n!1
2
27
Finally, multiply by "1=2:
for "1 $ x $ 1, and, dividing by x 2 ,
x2 x4 x6
1" C " C! ! ! D
2
3
4
."1/n!1
.2x 3 /4 .2x 3 /6
.2x 3 /2
C
"
C !!!
2Š
4Š
6Š
22 x 6
24 x 12 26 x 18
D1"
C
"
C !!!
2Š
4Š
6Š
1
X ."1/n 4n
D
x 6n
.for all x/:
.2n/Š
cos.2x 3 / D 1 "
nD0
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INSTRUCTOR’S SOLUTIONS MANUAL
3.
SECTION 9.6 (PAGE 550)
%
#&
#
#
sin x "
D sin x cos " cos x sin
4
4
4
1
1
2nC1
X
1 X
x
1
x 2n
D p
."1/n
"p
."1/n
.2n C 1/Š
.2n/Š
2 nD0
2 nD0
(
)
1
x 2n
x 2nC1
1 X
C
.for all x/:
D p
."1/n "
.2n/Š
.2n C 1/Š
2
10.
nD0
4.
#
$
x2
ln.2 C x 2 / D ln 2 1 C
2
#
$
x2
D ln 2 C ln 1 C
2
"
#
# $2
# $3
2
1 x2
x
1 x2
"
D ln 2 C
C
" !!!
2
2 2
3 2
1
X
."1/n!1 x 2n
! n
n
2
nD1
p
p
.for " 2 $ x $ 2/:
D ln 2 C
cos.2x " #/ D " cos.2x/
22 x 2 24 x 4
26 x 6
"
C
" !!!
2Š
4Š
6Š
1
X ."1/n
D"
.2x/2n
.2n/Š
D "1 C
11.
nD0
1
X
."1/nC1 n 2n
4 .x/
D
.2n/Š
.for all x/:
1
X
."1/n x 2nC3
x
D
2nC1
3
3
.2n C 1/Š
D2
.for all x/:
nD0
6.
cos2
%x &
2
7.
12.
1
.1 C cos x/
2
!
x2
x4 x6
1
1C1"
C
"
C !!!
D
2
2Š
4Š
6Š
D
D1C
1Cx
D ln.1 C x/ " ln.1 " x/
1"x
1
1
X
xn
xn X
"
."1/n!1
D
n
n
nD1
1
X
nD0
5. x 2 sin
ln
1
1 X ."1/n 2n
x
2
.2n/Š
2
&
e 2x " 1
1 % 2
D 2 e 2x " 1
2
x
x
!
.2x 2 /2
.2x 2 /3
1
C
C !!! " 1
D 2 1 C 2x 2 C
x
2Š
3Š
22 x 2
23 x 4
24 x 6
C
C
C !!!
2Š
3Š
4Š
1
X
2nC1 2n
D
x
.for all x 6D 0/:
.n C 1/Š
.for all x/:
nD0
13.
cosh x " cos x D
D
nD0
#
3
9.
14.
i x 2nC1
.2n C 1/Š
nD0
# 2
$
6
x
x
x 10
D2
C
C
C !!!
2Š
6Š
10Š
1
4nC3
X
x
D2
.for all x/:
.4n C 3/Š
sinh x " sin x D
."1/n 52nC1 4nC2
x
.2n C 1/
$
1
1
for " p $ x $ p :
5
5
#
i x 2n
.2n/Š
nD0
tan!1 .5x 2 / D .5x 2 / "
nD0
1
X
nD0
1 " ."1/n
$
x2
x6
x 10
C
C
C !!!
2Š
6Š
10Š
1
4nC2
X
x
D2
.for all x/:
.4n C 2/Š
.for all x/:
.5x 2 /3
.5x 2 /5 .5x 2 /7
C
"
C !!!
3
5
7
1
X
."1/n
D
.5x 2 /2nC1
.2n C 1/
1 h
X
D2
nD0
8.
."1 < x < 1/:
D2C
nD1
1
sin x cos x D sin.2x/
2
1
X
22n x 2nC1
D
."1/n
.2n C 1/Š
nD1
nD1
x 2n!1
2n " 1
1 h
X
1 " ."1/n
nD0
15.
!
"
1Cx
D .1 C x 3 / 1 " x 2 C x 4 " x 6 C ! ! !
1 C x2
D 1 " x2 C x3 C x4 " x5 " x6 C x7 C x8 " ! ! !
1
%
&
X
D 1 " x2 C
."1/n x 2n!1 C x 2n
.jxj < 1/:
Let t D x C 1, so x D t " 1. We have
f .x/ D e !2x D e !2.t!1/
1
X
."2/n t n
D e2
nŠ
nD2
D e2
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nD0
1
X
nD0
."1/n 2n .x C 1/n
nŠ
.for all x/:
369
lOMoARcPSD|6566483
SECTION 9.6 (PAGE 550)
16. Let y D x "
ADAMS and ESSEX: CALCULUS 9
#
#
; then x D y C . Hence,
2
2
21.
$
#
#
D cos y
sin x D sin y C
2
y4
y2
C
" !!!
.for all y/
D1"
2Š
4Š
#
$2
#
$
1
1
#
# 4
D1"
C
" !!!
x"
x"
2Š
2
4Š
2
#
$
1
X
."1/n
# 2n
D
.for all x/:
x"
.2n/Š
2
nD0
1
p X
D 2
nD0
nD0
17.
Let t D x " #, so x D t C #. Then
f .x/ D cos x D cos.t C #/ D " cos t D "
1
X
."1/n
nD0
1
X
."1/nC1
.x " #/2n
D
.2n/Š
t 2n
.2n/Š
22.
Let y D x "
."1/n %
# &2nC1
x"
.2n C 1/Š
4
18. Let y D x " 3; then x D y C 3. Hence,
#
$
y
ln x D ln.y C 3/ D ln 3 C ln 1 C
3
1 % y &2 1 % y &3 1 % y &4
y
C
"
C !!!
D ln 3 C "
3 2 3
3 3
4 3
2
3
.x " 3/ .x " 3/
.x " 3/
.x " 3/4
D ln 3 C
"
C
"
C !!!
3
2 ! 32
3 ! 33
4 ! 34
1
X ."1/n!1
D ln 3 C
.x " 3/n
.0 < x $ 6/:
n ! 3n
nD1
$)
( #
x"2
ln.2 C x/ D lnŒ4 C .x " 2/! D ln 4 1 C
4
#
$
x"2
D ln 4 C ln 1 C
4
1
X
.x " 2/n
D ln 4 C
."1/n!1
."2 < x $ 6/:
n4n
nD1
20. Let t D x C 1. Then x D t " 1, and
e 2xC3 D e 2tC1 D e e 2t
1
X
2n t n
De
(for all t )
nŠ
#
#
; then x D y C . Thus,
8
8
nD0
1
X
e2n .x C 1/n
D
nŠ
(for all x).
nD0
23.
370
.for all x/:
$
#
#
cos2 x D cos2 y C
8
"
#
$#
1
#
D
1 C cos 2y C
2
4
"
#
1
1
1
1 C p cos.2y/ " p sin.2y/
D
2
2
2
"
#
2
1
.2y/
.2y/4
1
C
" !!!
D C p 1"
2
2Š
4Š
2 2
"
#
.2y/3
.2y/5
1
C
" !!!
" p 2y "
3Š
5Š
2 2
"
1
.2y/2
.2y/3
1
C
D C p 1 " 2y "
2
2Š
3Š
2 2
#
.2y/4 .2y/5
C
"
" !!!
4Š
5Š
"
#
$
#
$
#
1
1
22
# 2
D C p 1"2 x"
"
x"
2
8
2Š
8
2 2
#
#
$
#
$
#
$
3
4
#
24
#
25
# 5
23
C
"
" !!!
C
x"
x"
x"
3Š
8
4Š
8
5Š
8
"
#
$
1
1 X
22n!1
1
1
# 2n!1
."1/n
D C p C p
x"
2
.2n " 1/Š
8
2 2
2 2 nD1
#
#
$
2n
22n
#
C
x"
.for all x/:
.2n/Š
8
.for all x/:
nD0
19.
Let t D x " .#=4/, so x D t C .#=4/. Then
f .x/ D sin x " cos x
%
%
#&
#&
D sin t C
" cos t C
4
4
i
1 h
D p .sin t C cos t / " .cos t " sin t /
2
1
p
p X
t 2nC1
D 2 sin t D 2
."1/n
.2n C 1/Š
Let t D x C 2, so x D t " 2. We have
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INSTRUCTOR’S SOLUTIONS MANUAL
f .x/ D
1
1
D
D
x2
.t " 2/2
1
#
$
t 2
4 1"
2
1
1 X t n!1
D
n n!1
4
2
24.
D
1
4
D
1
4
nD1
1
X
nD1
1
X
nD0
SECTION 9.6 (PAGE 550)
In the first sum replace n by n " 1.
1
1
X
X
e !2 un
2e !2 un
"
.n " 1/Š
nŠ
nD1
nD0
#
1
X 1
2
1
D" 2 C
"
e
e 2 .n " 1/Š
nD1
#
1
X
2
1
1
D" 2 C
"
e
e 2 .n " 1/Š
xe x D
."2 $ t < 2/
n!1
n.x C 2/
2n!1
.n C 1/.x C 2/n
2n
."4 < x < 0/:
nD1
$
2
un
nŠ
$
2
.x C 2/n
nŠ
(for all x).
Let y D x " 1; then x D y C 1. Thus,
x
1Cy
1
D
D1" %
y&
1Cx
2Cy
2 1C
2
(
)
1
y % y &2 % y &3
D1"
1" C
"
C !!!
2
2
2
2
)
(
1
y y2
y3 y4
D
."1 < y < 1/
1 C " 2 C 3 " 4 C !!!
2
2
2
2
2
1
1
1
1
D C 2 .x " 1/ " 3 .x " 1/2 C 4 .x " 1/3 " ! ! !
2
2
2
2
1
1 X ."1/n!1
D C
.x " 1/n
.for 0 < x < 2/:
2
2nC1
27.
cos x D 1 "
x4
x2
C
" ! ! !.
2
24
1 C
1 "
x2
x4
C
" !!!
2
24
1
1 "
nD1
25. Let u D x " 1. Then x D 1 C u, and
x ln x D .1 C u/ ln.1 C u/
1
X
un
D .1 C u/
."1/n!1
n
."1 < u $ 1/
nD1
D
1
X
."1/n!1
nD1
1
X
un
unC1
C
."1/n!1
:
n
n
nD1
Replace n by n " 1 in the last sum.
1
X
n
n!1 u
1
X
28.
un
x ln x D
."1/
C
."1/
n
n"1
nD1
nD2
$
#
1
X
1
1
"
un
DuC
."1/n!1
n n"1
1
X
."1/n
D .x " 1/ C
.x " 1/n
n.n " 1/
x4
x6
x2
C
"
C !!!
2
24 720
sec x D 1 C
x2
5x 4
61x 6
C
C
C !!!:
2
24
720
Now we can differentiate and obtain
u!2
1
X
un
D .u " 2/e !2
nŠ
1
1
X
e !2 unC1 X 2e !2 un
"
:
nŠ
nŠ
nD0
sec x tan x D x C
(for all u)
nD0
nD0
If we divide the first four terms of the series
into 1 we obtain
26. Let u D x C 2. Then x D u " 2, and
xe D .u " 2/e
x2
5x 4
C
C ! ! !.
2
24
cos x D 1 "
.0 $ x $ 2/:
nD2
x
x4
x2
C
" !!!
2
24
x4
x2
"
C !!!
2
24
2
4
x
x
C !!!
"
4
2
5x 4
" !!!
24
n!2
nD2
D
Thus sec x D 1 C
5x 4
x2
C !!!
C
2
24
61x 5
5x 3
C
C !!!:
6
120
(Note: the same result can be obtained by multiplying
the first three nonzero terms of the series for sec x (from
Exercise 25) and tan x (from Example 6(b)).)
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SECTION 9.6 (PAGE 550)
ADAMS and ESSEX: CALCULUS 9
x3
x2
C
C !!!
2
6
.e x " 1/3
tan!1 .e x " 1/ D .e x " 1/ "
3
.e x " 1/5
C
" !!!
5
2
3
x
x
DxC
C
C !!!
2
6
#
$3
x2
x3
1
xC
C
C !!!
"
3
2
6
#
$5
x2
x3
1
xC
C
C !!! C !!!
C
5
2
6
3
2
!
"
x
1 3
x
C
"
DxC
x C !!! C !!!
2
6
3
x2 x3
DxC
"
C !!!
2
6
30. We have
"
#
x3
x5 x7
tan"1 x
e
" 1 D exp x "
C
"
C !!! " 1
3
5
7
#
$
#
$2
x5
1
x3
x3
C
" !!! C
x"
C !!!
D1C x"
3
5
2Š
3
1
3
C .x " ! ! !/ C ! ! ! " 1
3Š
x2
x3
x3
C
C
C higher degree terms
Dx"
3
2
6
x2 x3
DxC
"
C !!!:
2
6
ex " 1 D x C
29.
p
31. Let 1 C x D 1 C ax C bx 2 C ! ! !.
Then 1 C x D 1 C 2ax C .a2 C 2b/x 2 C ! ! !, so 2a D 1,
and a2 C 2b
p D 0. Thus a D 1=2 and b D "1=8.
Therefore 1 C x D 1 C .x=2/ " .x 2 =8/ C ! ! !.
35.
36.
x2
x4
x6
C
C
C !!!
3Š
5Š
7Š
1
e x " e !x
D sinh x D
x
2x
if x ¤ 0. The sum is 1 if x D 0.
1C
1
1
1
C
C
C !!!
2"& 2Š
4 & 3Š
8 & 4Š
#
# $2
# $3
1 1
1 1
1
C
C !!!
C
D2
2
2Š 2
3Š 2
%
&
D 2 e 1=2 " 1 :
1C
37. P .x/ D 1 C x C x 2 .
a) The Maclaurin series for P .x/ is 1 C x C x 2
(for all x).
b) Let t D x " 1, so x D t C 1. Then
P .x/ D P .t C 1/ D 1 C t C 1 C .t C 1/2 D 3 C 3t C t 2 :
The Taylor series for P .x/ about 1 is
3 C 3.x " 1/ C .x " 1/2 .
38.
If a ¤ 0 and jx " aj < jaj, then
1
1
1
1
D
D
x
a C .x " a/
a 1C x"a
a
)
(
x"a
.x " a/2 .x " a/3
1
C
"
C
!
!
!
:
1"
D
a
a
a2
a3
The radius of convergence of this series is jaj, and the
series converges to 1=x throughout its interval of convergence. Hence, 1=x is analytic at a.
39.
If a > 0 and t D x " a, then x D t C a and
#
$
t
ln x D ln.a C t / D ln a C ln 1 C
a
1
n
X
t
."a < t $ a/
D ln a C
."1/n!1 n
a
32. csc x does not have a Maclaurin series because
limx!0 csc x does not exist.
#
#
Let y D x " . Then x D y C
and sin x D cos y.
2
2
Therefore, using the result of Exercise 25,
5y 4
y2
C
C !!!
csc x D sec y D 1 C
2
24
%
&
%
1
# 2
5
# &4
D1C
x"
C
x"
C !!!:
2
2
24
2
33. 1 C x 2 C
34.
x4
x6
2
C
C ! ! ! D ex
2Š
3Š
.for all x/:
x 15
x 21
x 27
x9
C
"
C
" !!!
x3 "
3Š
&
4
5Š
&
16
7Š
&
64
9Š
& 256
#
"
#
$
#
$
3
5
1 x3
1 x3
x3
C
" !!!
"
D2
2
3Š 2
5Š 2
# 3$
x
D 2 sin
.for all x/:
2
372
D ln a C
nD1
1
X
."1/n!1
nD1
.x " a/n
an
.0 < x < 2a/:
Since the series converges to ln x on an interval of positive radius .a/, centred at a, ln is analytic at a.
40.
If
'
!1=x 2
; if x 6D 0;
f .x/ D e
0;
if x D 0;
then the Maclaurin series for f .x/ is the identically zero
series 0 C 0x C 0x 2 C ! ! ! since f .k/ .0/ D 0 for every k.
The series converges for every x, but converges to f .x/
only at x D 0, since f .x/ 6D 0 if x 6D 0. Hence, f cannot
be analytic at 0.
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INSTRUCTOR’S SOLUTIONS MANUAL
41.
x y
e e D
e xCy D
D
D
1
X
xn
nŠ
nD0
1
X
kD0
1
X
j D0
1
X
j D0
!
1
X
ym
mŠ
mD0
k
.x C y/
D
kŠ
!
1
X
kD0
x
jŠ
mD0
We have
1
kŠ
1
x j X y k!j
jŠ
.k " j /Š
kDj
1
j X
SECTION 9.6 (PAGE 550)
k
X
j D0
0
ˇtDx
1
f .kC1/ .t /.x " t /kC1 ˇˇ
f .x/ D Pk .x/ C @"
ˇ
ˇ
kŠ
kC1
tDc
1
Z x
.x " t /kC1 f .kC2/ .t / A
dt
C
kC1
c
kŠ
x j y k!j
j Š.k " j /Š
.let k " j D m/
f .kC1/ .c/
.x " c/kC1
.k C 1/Š
Z x
1
.x " t /kC1 f .kC2/ .t / dt
C
.k C 1/Š c
D PkC1 .x/ C EkC1 .x/:
ym
D ex ey :
mŠ
D Pk .x/ C
42. We want to prove that f .x/ D Pn .x/ C En .x/, where Pn
is the nth-order Taylor polynomial for f about c and
Z
1 x
En .x/ D
.x " t /n f .nC1/ .t / dt:
nŠ c
(a) The Fundamental Theorem of Calculus written in
the form
Z x
f .x/ D f .c/ C
f 0 .t / dt D P0 .x/ C E0 .x/
c
is the case n D 0 of the above formula. We now
apply integration by parts to the integral, setting
U D f 0 .t /;
d U D f 00 .t / dt;
d V D dt;
V D ".x " t /:
(We have broken our usual rule about not including a constant of integration with V . In this case
we have included the constant "x in V in order
to have V vanish when t D x.) We have
ˇtDx Z
ˇ
x
ˇ
0
f .x/ D f .c/ " f .t /.x " t /ˇ
C
.x " t /f 00 .t / dt
ˇ
c
tDc
Z x
0
D f .c/ C f .c/.x " c/ C
.x " t /f 00 .t / dt
Thus the formula is valid for n D k C 1 if it
is valid for n D k. Having been shown to be
valid for n D 0 (and n D 1), it must therefore
be valid for every positive integer n for which
En .x/ exists.
43.
If f .x/ D ln.1 C x/, then
1
"1
2
; f 000 .x/ D
;
; f 00 .x/ D
1Cx
.1 C x/2
.1 C x/3
"3Š
."1/n!1 .n " 1/Š
f .4/ .x/ D
; : : : ; f .n/ D
4
.1 C x/
.1 C x/n
f 0 .x/ D
and
f
f .0/ D 0; f 0 .0/ D 1; f 00 .0/ D "1; f 000 .0/ D 2;
.4/
Therefore, the Taylor Formula is
2
"3Š 4
"1 2
x C x3 C
x C !!!C
2Š
3Š
4Š
."1/n!1 .n " 1/Š n
x C En .x/
nŠ
f .x/ D x C
c
D P1 .x/ C E1 .x/:
We have now proved the case n D 1 of the
formula.
(b) We complete the proof for general n by mathematical induction. Suppose the formula holds for
some n D k:
.0/ D "3Š; : : : ; f .n/ .0/ D ."1/n!1 .n " 1/Š:
where
Z
1 x
.x " t /n f .nC1/ .t / dt
nŠ 0
Z
."1/n nŠ
1 x
.x " t /n
dt
D
nŠ 0
.1 C t /nC1
Z x
n
.x " t /
D ."1/n
dt:
.1
C t /nC1
0
En .x/ D
f .x/ D Pk .x/ C Ek .x/
Z
1 x
.x " t /k f .kC1/ .t / dt:
D Pk .x/ C
kŠ c
Again we integrate by parts. Let
U D f .kC1/ .t /;
d U D f .kC2/ .t / dt;
d V D .x " t /k dt;
"1
.x " t /kC1 :
V D
kC1
If 0 $ t $ x $ 1, then 1 C t # 1 and
jEn .x/j $
Z
x
0
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.x " t /n dt D
1
x nC1
$
!0
nC1
nC1
373
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SECTION 9.6 (PAGE 550)
ADAMS and ESSEX: CALCULUS 9
#
$
1Ct
t3
t5
D 2 tC
C
C ! ! ! for
1"t
3
5
"1 < t < 1. Thus
0
1
1
C
0 < cn " cnC1 D .2n C 1/@
2n C 1
3.2n C 1/3
1
1
C
C ! ! !A " 1
5.2n C 1/5
#
$
1
1
1
C
C
!
!
!
<
3 .2n C 1/2
.2n C 1/4
(geometric)
1
1
D
1
3.2n C 1/2
1"
.2n C 1/2
1
D
12.n2 C n/
#
$
1 1
1
D
"
:
12 n n C 1
as n ! 1.
If "1 < x $ t $ 0, then
(c) ln
ˇ
ˇ
ˇx " t ˇ
t "x
ˇ
ˇ
ˇ 1 C t ˇ D 1 C t $ jxj;
t "x
increases from 0 to "x D jxj as t increases
1Ct
from x to 0. Thus,
because
jEn .x/j <
1
1Cx
Z
0
jxj
jxjn dt D
jxjnC1
!0
1Cx
as n ! 1 since jxj < 1. Therefore,
f .x/ D x "
1
X
x2
x3 x4
xn
C
"
C !!! D
."1/n!1 ;
2
3
4
n
nD1
These inequalities
imply+ that fcn g is de˚
1
is increasing. Thus
creasing and cn " 12n
1
11
fcn g is bounded below by c1 " 12
D 12
and so limn!1 cn D c exists. Since
e cn D nŠn!.nC1=2/ e n , we have
nŠ
lim
D lim e cn D e c
n!1 nnC1=2 e !n
n!1
p
exists. It remains to show that e c D 2#.
for "1 < x $ 1.
44. We follow the steps outlined in the problem:
(d) The Wallis Product,
2n
2n
#
22446
!!!
D
lim
n!1 1 3 3 5 5
2n " 1 2n C 1
2
can be rewritten in the form
r
#
2n nŠ
;
lim
p
D
n!1 1 ! 3 ! 5 ! ! ! .2n " 1/ 2n C 1
2
Rj
(a) Note that ln.j " 1/ < j !1 ln x dx < ln j ,
j D 1; 2; : : :. For j D 0 the integral is improper
but convergent. We have
n ln n " n D
Z
n
ln x dx < ln.nŠ/ <
0
Z
nC1
ln x dx
or, equivalently,
1
D .n C 1/ ln.n C 1/ " n " 1 < .n C 1/ ln.n C 1/ " n:
lim
n!1
22n n2nC1 e !2n e 2cn
lim
#
:
2
p
D
e 2c
ec
D
:
2e c
2
22nC1=2 n2nC1=2 e !2n e c2n 2n
p
p
Thus e c =2 D #2, and e c D 2#, which
completes the proof of Stirling’s Formula.
n!1
cn " cnC1 D ln
374
r
Substituting nŠ D nnC1=2 e ! ne cn and a similar
expression for .2n/Š, we obtain
"
!
(b) If cn D ln.nŠ/ " n C 12 ln n C n, then
!
"
nŠ
" n C 12 ln n
.n C 1/Š
"
!
C n C 23 ln.n C 1/ " 1
!
"
1
D ln
" n C 12 ln n
nC1
"
!
C n C 21 ln.n C 1/ C ln.n C 1/ " 1
!
" nC1
D n C 12 ln
"1
n
1
" 1 C 2nC1
!
" 1:
D n C 21 ln
1
1 " 2nC1
22n .nŠ/2
p
D
.2n/Š 2n C 1
45.
We have:
(a) By part (a) of Exercise 44,
n ln n " n < ln.nŠ/ < .n C 1/ ln.n C 1/ " n:
Therefore,
1<
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.n C 1/ ln.n C 1/ " "n
ln.nŠ/
<
:
n ln n " n
n ln n " n
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 9.7 (PAGE 554)
METHOD II. (using Taylor’s Theorem) Since
P5 .x/ D P6 .x/ (Maclaurin polynomials for sin have only
odd degree terms) we are better off using the remainder
E6 .
We divide the numerator and denominator of the
fraction on the right by n and take the limit as
n ! 1 by l’H^opital’s rule:
$
#
1
ln.n C 1/ " 1
1C
n
lim
n!1
ln n " 1
1
.n C 1/ 1
" 2 ln.n C 1/ C
n
n
nC1
D lim
1
n!1
n
ln.n C 1/
D lim "
C 1 D 1:
n!1
n
ln.nŠ/
D 1, and
n ln n " n
ˇ
ˇ
ˇ ln.nŠ/ " .n ln n " n/ ˇ
ˇD0
lim ˇˇ
ˇ
n!1
ln.nŠ/
Hence lim
n!1
jf .0:2/ " P5 .0:2/j D jE6 .0:2/j D
for some s between 0 and 0:2. Now f .7/ .x/ D " cos x, so
jf .0:2/ " P5 .0:2/j <
2.
so that the relative error in the approximation
ln.nŠ/ ' n ln "n approaches 0as n ! 1.
ln.10Š/
% p
&
ln.20Š/ " ln 40# C 20 ln.20/ " 20
ln.20Š/
' 0:000552
3.
' 0:137613
ln.20Š/
' 0:057185
Section 9.7 Applications of Taylor and
Maclaurin Series (page 554)
1.
x5
x3
C
.
6
120
METHOD I. (using an alternating series bound)
If f .x/ D sin x, then P5 .x/ D x "
.0:2/n
.0:2/2
C !!! C
D sn
2Š
nŠ
.0:2/nC2
.0:2/nC1
0 < e 0:2 " sn D
C
C !!!
.n C 1/Š
.n C 2/Š
)
(
.0:2/nC1
0:2
.0:2/2
$
1C
C
C
!
!
!
.n C 1/Š
nC2
.n C 2/2
.0:2/nC1 10n C 20
D
!
< 5 & 10!5 if n D 4:
.n C 1/Š 10n C 18
.0:2/2
.0:2/3
.0:2/4
e 0:2 ' 1 C 0:2 C
C
C
2Š
3Š
4Š
' 1:221400
4.
Evidently, Stirling’s formula gives a significantly
better approximation than the modified version.
e 0:2 ' 1 C 0:2 C
Error estimate:
For the modified Stirling formula,
ln.nŠ/ ' n ln n " n, these errors are
ln.10Š/
%
&
ln.20Š/ " 20 ln.20/ " 20
If f .x/ D ln x, then f 0 .x/ D 1=x, f 00 .x/ D "1=x 2 ,
f 000 .x/ D 2=x 3 , f .4/ .x/ D "6=x 4 , and f .5/ .x/ D 24=x 5 .
If P4 .x/ is the Taylor polynomial for f about x D 2, then
for some s between 1.95 and 2 we have (using Taylor’s
Theorem)
24 .0:05/5
!
s5
5Š
24.0:05/5
< 2:22 & 10!9 :
$
.1:95/5 120
' 0:000098
%
&
ln.10Š/ " 10 ln.10/ " 10
1
& .0:2/7 < 2:6 & 10!9 :
7Š
jf .1:95/ " P4 .1:95/j D
(b) From Stirling’s
p
Formula, ln.nŠ/ ' ln 2# n C n ln n " n. For
this approximation we have the following relative
errors (using a calculator)
&
% p
ln.10Š/ " ln 20# C 10 ln.10/ " 10
jf .7/ .s/j
.0:2/7 ;
7Š
We have
1
1
1
1
1
D e !1 D 1 "
C
"
C
" !!!
e
1Š
2Š 3Š
4Š
which satisfies the conditions for the alternating series test,
and the error incurred in using a partial sum to approximate e !1 is less than the first omitted term in absolute
1
< 5 & 10!5 if n D 7, so
value. Now
.n C 1/Š
1 1
1
1
1
1
1
' " C
"
C
"
' 0:36786
e
2 6
24 120
720 5040
jf .0:2/ " P5 .0:2/j $
.0:2/7
< 2:6 & 10!9 :
7Š
with error less than 5 & 10!5 in absolute value.
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SECTION 9.7 (PAGE 554)
ADAMS and ESSEX: CALCULUS 9
5. e 1:2 D ee 0:2 . From Exercise 1: e 0:2 ' 1:221400,
.0:2/5 60
with error less than
!
' 0:000003. Since
5Š
58
e D 2:718281828 ! ! !, it follows that e 1:2 ' 3:3201094 ! ! !,
1
with error less than 3 & 0:000003 D 0:000009 <
.
20; 000
1:2
Thus e ' 3:32011 with error less than 1/20,000.
10. We have
%# &
sin 80ı D cos 10ı D cos
18
1 % # &4
1 % # &2
C
" !!!:
D1"
2Š 18
4Š 18
6. We have
Since
sin.0:1/ D 0:1 "
.0:1/3
.0:1/5 .0:1/7
C
"
C !!!:
3Š
5Š
7Š
1 % # &4
< 5 & 10!5 , therefore
4Š 18
sin 80ı ' 1 "
5
Since
.0:1/
D 8:33 & 10!8 < 5 & 10!5 , therefore
5Š
with error less than 5 & 10!5 in absolute value.
.0:1/3
sin.0:1/ D 0:1 "
' 0:09983
3Š
7.
with error less than 5 & 10!5 in absolute value.
5#
#
cos 5ı D cos
D cos
180
36
1 % # &2
1 % # &4
."1/n % # &2n
'1"
C
" !!! C
2Š 36
4Š 36
.2n/Š 36
% # &2nC2
1
jErrorj <
.2n C 2/Š 36
1
< 0:00005 if n D 1:
<
.2n C 2/Š92nC2
%
&
1 # 2
cos 5ı ' 1 "
' 0:996192
2Š 36
with error less than 0:00005.
8. We have
#
$
# $
1
6
D ln 1 C
ln
5
5
# $
# $
# $
1 1 1 2 1 1 3 1 1 4
C
"
C !!!:
D "
5 2 5
3 5
4 5
# $
1 1 n
Since
< 5 & 10!5 if n D 6, therefore
n 5
# $
# $
# $
# $
# $
1 1 1 2 1 1 3 1 1 4 1 1 5
6
C
"
C
' "
ln
5
5 2 5
3 5
4 5
5 5
' 0:18233
11.
5#
5#
1
D
"
180
180 3Š
cos 65ı '
#
5#
180
$3
' 0:0871557
55 # 5
< 0:00000005. Thus
5Š1805
0:996192
"
2
p
3.0:0871557/
' 0:42262
2
with error less than 0.00005.
12. We have
tan!1 .0:2/ D 0:2 "
Since
.0:2/3
.0:2/5 .0:2/7
C
"
C !!!:
3
5
7
.0:2/7
< 5 & 10!5 , therefore
7
tan!1 .0:2/ ' 0:2 "
' "0:1 "
376
#
with error less than
ln.0:9/ D ln.1 " 0:1/
.0:1/2 .0:1/3
.0:1/n
"
" !!! "
2
3
n
.0:1/nC1
.0:1/nC2
jErrorj <
C
C !!!
nC1
nC2
.0:1/nC1 ,
<
1 C 0:1 C .0:1/2 C ! ! !
nC1
.0:1/nC1 10
!
< 0:00005 if n D 3:
D
nC1
9
.0:1/2 .0:1/3
ln.0:9/ ' "0:1 "
"
' "0:10533
2
3
with error less than 0:00005.
$
#
5#
C
3
180
p
1
5#
3
5#
D cos
"
sin
2
180
2
180
From Exercise 5, cos.5#=180/ ' 0:996192 with error less
than 0:000003. Also
cos 65ı D cos
sin
with error less than 5 & 10!5 in absolute value.
9.
1 % # &2
' 0:98477
2Š 18
.0:2/5
.0:2/3
C
' 0:19740
3
5
with error less than 5 & 10!5 in absolute value.
13.
cosh 1 ' 1 C
1
1
1
C
C !!! C
with error less than
2Š
4Š
.2n/Š
(
)
1
1
1
1C
C
C
!
!
!
.2n C 2/Š
.2n C 3/2
.2n C 3/4
1
1
D
!
< 0:00005 if n D 3:
1
.2n C 2/Š
1"
.2n C 3/2
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lOMoARcPSD|6566483
INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 9.7 (PAGE 554)
1
1
1
Thus cosh 1 ' 1 C C
C
' 1:54306 with error
2
24
720
less than 0:00005.
18.
L.x/ D
D
14.
We have
ln
15.
Z
x
0
19.
Z
D
Z
Z
20.
We have
x
L.0:5/ D 0:5 "
Since
x
0
x
0
.0:5/4nC1
< 5 & 10!4 if n D 2, therefore
.2n/Š ! .4n C 1/
!
t2
t3
t
C
C
C ! ! ! dt
1C
2Š
3Š
4Š
21.
x5
x3
C
" !!!
3Š3
5Š5
1
1
1
C
" ! ! ! C ."1/n
I.1/ ' 1 "
3Š3
5Š5
.2n C 1/Š.2n C 1/
1
jErrorj $
< 0:0005 if n D 2:
.2n C 3/Š.2n C 3/
1
1
C
' 0:946 correct to three
Thus I.1/ ' 1 "
3Š3
5Š5
decimal places.
1Cx
ln t
dt
let u D t " 1
K.x/ D
t
"1
Z1 x
ln.1 C u/
D
du
u
0
)
Z x(
u
u2 u3
D
1" C
"
C ! ! ! du
2
3
4
0
2
3
4
x
x
x
D x " 2 C 2 " 2 C !!!
2
3
4
1
nC1
X
n x
."1/
D
."1 $ x $ 1/
.n C 1/2
From Exercise 15:
I.x/ D x "
nD1
nD0
.0:5/5
' 0:497
2Š ! 5
rounded to 3 decimal places.
et " 1
dt
t
x2
x3
x4
C
C
C !!!
2Š ! 2
3Š ! 3
4Š ! 4
1
X xn
D
:
nŠ ! n
Z
.0:5/9 .0:5/13
.0:5/5
C
"
C !!!:
2Š ! 5
4Š ! 9
6Š ! 13
L.0:5/ ' 0:5 "
DxC
17.
x
tan!1 .t 2 /
dt
t2
0
)
Z x(
t4
t 8 t 12
D
1"
C
"
C ! ! ! dt
3
5
7
0
x9
x 13
x5
C
"
C !!!
Dx"
3&5
5 & 9 7 & 13
1
X
x 4nC1
D
."1/n
."1 $ x $ 1/
.2n C 1/.4n C 1/
M.x/ D
nD0
sin t
I.x/ D
dt
t
0
)
Z x(
t4 t6
t2
C
"
C ! ! ! dt
D
1"
3Š
5Š
7Š
0
3
5
x
x
Dx"
C
" !!!
3 & 3Š
5 & 5Š
1
X
x 2nC1
D
."1/n
for all x:
.2n C 1/.2n C 1/Š
J.x/ D
!
t4
t 8 t 12
1"
C
"
C ! ! ! dt
2Š
4Š
6Š
nD0
nD0
16.
cos.t 2 / dt
0
x9
x 13
x5
C
"
C !!!
2Š ! 5
4Š ! 9 6Š ! 13
1
X
x 4nC1
D
."1/n
:
.2n/Š ! .4n C 1/
# $
# $
# $
# $
3
1 1 10
1 1 1 2 1 1 3
C
" !!! "
' "
2
2 2 2
3 2
10 2
' 0:40543
with error less than 5 & 10!5 in absolute value.
Z
x
Dx"
#
$
# $
3
1
ln
D ln 1 C
2
2
# $
# $
# $
1 1 1 2 1 1 3 1 1 4
D "
C
"
C !!!:
2 2 2
3 2
4 2
# $
1
1 1 n
<
if n D 11, therefore
Since
n 2
20000
Z
22.
x 10
x6
C
" !!!
x2 "
sin.x 2 /
3Š
5Š
lim
D lim
x!0 sinh x
x!0
x3
x5
xC
C
C !!!
3Š
5Š
x9
x5
C
" !!!
x"
3Š
5Š
D 0:
D lim
4
2
x!0
x
x
C
C !!!
1C
3Š
5Š
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SECTION 9.7 (PAGE 554)
ADAMS and ESSEX: CALCULUS 9
x4 x8
1"1C
"
C !!!
1 " cos.x 2 /
2Š
4Š
D
lim
23. lim
#
$2
x!0
x!0 .1 " cos x/2
x2 x4
1"1C
"
C !!!
2Š
4Š
1
C O.x 2 /
2Š
D 2:
D lim
x!0 1
C O.x 2 /
4
24. We have
% x2
&2
x3
x4
C
C
C !!!
.e x " 1 " x/2
4Š
lim
D lim 2Š4 3Š6
x!0 x 2 " ln.1 C x 2 /
x!0
x
x
x8
"
C
" !!!
2
3
4
&
&
%
2
4%
2
x
x
1
x
1C C
C !!!
1
4
3
12
4
D % & D :
D lim
1
x!0
2
x8
x4 x6
"
C
" !!!
2
2
3
4
2 sin 3x " 3 sin 2x
5x " tan!1 5x
#
$
#
$
23 x 3
33 x 3
2 3x "
C ! ! ! " 3 2x "
C !!!
3Š
3Š
D lim
#
$
x!0
53 x 3
5x " 5x "
C !!!
3
"9 C 4 C O.x 2 /
3
5&3
D lim
D" :
D"
x!0 125
125
25
2
C O.x /
3
26. We have
25.
Section 9.8 The Binomial Theorem and
Binomial Series (page 559)
1.
p
1 C x D .1 C x/1=2
# $ 2
# $# $ 3
1
1 x
1
1
3 x
x
"
C
"
"
C !!!
D1C C
2
2
2 2Š
2
2
2 3Š
1
X
x
1 ! 3 ! 5 ! ! ! .2n " 3/ n
D1C C
."1/n!1
x
2
2n nŠ
nD2
1
X
.2n " 2/Š
x
."1/n!1 2n!1
xn
D1C C
2
2
.n " 1/ŠnŠ
."1 < x < 1/:
nD2
lim
x!0
2.
sin.sin x/ " x
lim
x!0 xŒcos.sin x/ " 1!
%
&
1
1
sin x " sin3 x C sin5 x " ! ! ! " x
3Š
5Š
D lim
i
h
1
1
x!0
2
x 1 " sin x C sin4 x " ! ! ! " 1
2Š
4Š
%
& 1%
&3
&5
x3
x3
1%
x"
C !!! "
x"
C !!! C
x " !!! " !!! " x
3Š
3Š
3Š
5Š
D lim
h 1%
&2
&4
i
3.
x!0
x3
1%
x"
C !!! C
x " !!! " !!!
x "
2Š
3Š
4Š
2 3
2
" x C higher degree terms
2
D lim 3Š
D 3Š D :
1 3
1
x!0
3
" x C higher degree terms
2Š
2Š
27.
lim
sinh x " sin x
" cos x
$ #
$
#
x3
x3
C !!! " x "
C !!!
xC
3Š
3Š
D lim #
$ #
$
x!0
x2
x2
1C
C !!! " 1 "
C !!!
2Š
2Š
x3
C O.x 5 /
D lim 32
D 0:
x!0 x C O.x 4 /
p
x 1 " x D x.1 " x/1=2
# $
1
1 ."1/2 x 3
x2
C
"
Dx"
2
2
2
2Š
# $# $
1
1
3 ."1/3 x 4
C
"
"
C !!!
2
2
2
3Š
1
x 2 X 1 ! 3 ! 5 ! ! ! .2n " 3/ nC1
Dx"
"
x
2
2n nŠ
nD2
1
x2 X
.2n " 2/Š
Dx"
"
."1/n!1 2n!1
x nC1
2
2
.n " 1/ŠnŠ
p
r
x
4Cx D2 1C
4
2
# $
1
1
"
6
1
x
2
2 % x &2
D 26
41 C 2 ! 4 C
2Š
4
# $# $
3
1
3
1
"
"
%
&
7
x 3
2
2
2
C ! ! !7
C
5
3Š
4
x!0 cosh x
378
."1 < x < 1/:
nD2
D2C
1
X
x
1 ! 3 ! 5 ! ! ! .2n " 3/ n
C2
."1/n!1
x
4
23n nŠ
nD2
1
X
x
.2n " 1/Š
D2C C2
."1/n!1 4n!1
xn
4
2
nŠ.n " 1/Š
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."4 < x < 4/:
lOMoARcPSD|6566483
INSTRUCTOR’S SOLUTIONS MANUAL
4.
SECTION 9.8 (PAGE 559)
"
#
% x &2 !1=2
1
p
D r
1
C
D
% x &2
2
2
4 C x2
2 1C
2
"
# $% &
# $# $% &
1
1
x 2
1
1
3
x 4
D
1C "
C
"
"
C
2
2
2
2Š
2
2
2
#
# $# $# $% &
1
3
5
x 6
1
"
"
"
C !!!
3Š
2
2
2
2
1
1
9.
i)
%n&
0
%n&
nŠ
D 1;
0ŠnŠ
D
n
ii) If 0 $ k $ n, then
nŠ
D 1.
nŠ0Š
D
% n & %n&
nŠ
nŠ
C
D
C
k"1
k
.k " 1/Š.n " k C 1/Š
kŠ.n " k/Š
%
&
nŠ
k C .n " k C 1/
D
kŠ.n " k C 1/Š
$
#
.n C 1/Š
nC1
D
:
D
k
kŠ.n C 1 " k/Š
1
1
3
3&5
" x 2 C 7 x 4 " 10 x 6 C ! ! !
2 24
2 2Š
2 3Š
1
1 X
1 & 2 & 3 & ! ! ! & .2n " 1/ 2n
D C
."1/n
x
2
23nC1 nŠ
D
nD1
."2 $ x $ 2/:
5.
.1 " x/!2
."2/."3/
."2/."3/."4/
D 1 " 2."x/ C
."x/2 C
."x/3 C ! ! !
2Š
3Š
1
X
D 1 C 2x C 3x 2 C 4x 3 C ! ! ! D
nx n!1
."1 < x < 1/:
10. The formula .a C b/n D
.1 C x/!3
."3/."4/ 2 ."3/."4/."5/ 3
x C
x C !!!
D 1 " 3x C
2Š
3Š
.3/.4/ 2 .4/.5/ 3
D 1 " 3x C
x "
x C !!!
2
2
1
X
.n C 2/.n C 1/ n
D
."1/n
x
."1 < x < 1/:
2
.a C b/mC1 D .a C b/
nD0
7.
!1
Using the Maclaurin series for sin
section, we have
D
x obtained in this
kD0
k
k
an!k b k
k
kD0
amC1!k b k C
am!k b k
m % &
X
m
kD0
k
am!k b kC1
kD1
D amC1 C
#
x
3 5
"x"
"
x " !!!
2
6
40
D amC1 C
for "1 < x < 1
d
1
, we have, using the series
8. Since
sinh!1 x D p
1 C x2
p dx
for 1 C x 2 ,
Z
m % &
X
m
m % &
X
m
kD0
nD1
3
D
x
dt
p
1
C t2
#
Z x"
1
X
n 1 & 3 & 5 & ! ! ! & .2n " 1/ 2n
dt
t
D
1C
."1/
2n nŠ
0
sinh!1 x D
%n&
(replace k by k " 1 in the latter sum)
m % &
mC1
X
X% m &
m mC1!k k
D
a
b C
amC1!k b k
k
k"1
#
cos!1 x D " sin!1 x
2
1
X
#
1 & 3 & 5 & ! ! ! & .2n " 1/ 2nC1
D "x"
x
2
2n nŠ.2n C 1/
D
kD0
holds for n D 1; it says a C b D a C b in this case.
Suppose the formula holds for n D m, where m is some
positive integer. Then
nD1
6.
Pn
0
mC1
X#
kD0
m h% &
X
m
kD1
k
% m &i
C
amC1!k b k C b mC1
k"1
(by #9i))
$
m #
X
mC1
kD1
k
amC1!k b k C b mC1
$
mC1
amC1!k b k
k
(by #9(ii))
(by #9(i) again).
Thus the formula holds for n D m C 1. By induction it
holds for all positive integers n.
nD1
DxC
1
X
nD1
3
."1/n
1 & 3 & 5 & ! ! ! & .2n " 1/ 2nC1
x
2n nŠ.2n C 1/
11.
Consider the Leibniz Rule:
x
3 5
Dx"
C
x " !!!
6
40
.fg/.n/ D
for "1 < x < 1
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n % &
X
n
kD0
k
f .n!k/ g .k/ :
379
lOMoARcPSD|6566483
SECTION 9.8 (PAGE 559)
ADAMS and ESSEX: CALCULUS 9
This holds for n D 1; it says .fg/0 D f 0 g C fg 0 in this
case. Suppose the formula holds for n D m, where m is
some positive integer. Then
d
.fg/.mC1/ D
.fg/.m/
dx
m
d X % m & .m!k/ .k/
D
f
g
k
dx
completing the induction and thus the proof.
13.
.f1 f2 ! ! ! fn /.k/ D
kD0
D
m % &
X
m
kD0
k
f .mC1!k/ g .k/ C
m % &
X
m
kD0
k
f .m!k/ g .kC1/
D f .mC1/ g .0/ C
&f
Df
D
g
mC1
X#
% m &i
k"1
g C f .0/ g .mC1/
(by Exercise 9(i))
.mC1/ .0/
kD0
k
kD1
.mC1!k/ .k/
C
C
m #
X
kD1
v .k/ D .ufnC1 /.k/ D
$
mC1
f .mC1!k/ g .k/ C f .0/ g .mC1/
k
(by Exercise 9(ii))
$
mC1
f .mC1!k/ g .k/
k
D
(by 9(i) again).
.x1 Cx2 C! ! !Cxn / D
jmjDk
D
kŠ
x m1 x m2 ! ! ! xnmn ;
m1 Š m2 Š ! ! ! mn Š 1 2
holds for some n # 2 and all k, and we apply the Binomial Theorem to
%
&k
.x1 C ! ! ! C xm C xmC1 /k D .x1 C ! ! ! C xn / C xnC1
D
D
k
X
j D0
kŠ
k!j
.x1 C ! ! ! jxn /j xnC1
j Š .k " j /Š
j D0
X
kŠ
jŠ
k!j
x
x m1 ! ! ! xnmn :
j Š .k " j /Š nC1
m1 Š ! ! ! mn Š 1
k
X
Now let k " j D mnC1 and let m" D .m1 ; ! ! ! mn ; mnC1 /,
so that jm" j D j C .k " j / D k. The result above then
simplifies to
jm! jDk
380
kŠ
mnC1
; x m1 ! ! ! xnmn xnC1
m1 Š ! ! ! mn Š mnC1 Š 1
j D0
kŠ
.k!j /
u.j / fnC1
j Š.k " j /Š
X
kŠ
jŠ
.k!j /
.m / .m /
fnC1
f 1 f2 2 ! ! ! fn.mn / :
j Š.k " j /Š
m1 Š ! ! ! mn Š 1
jmjDj
k
X
j D0
X
kŠ
jŠ
.m
/
.m /
f 1 ! ! ! fn.mn / fnC1nC1
j ŠmnC1 Š
m1 Š ! ! ! mn Š 1
jmjDj
X
jm! jDk
kŠ
.m
/
.m /
f 1 ! ! ! fn.mn / fnC1nC1 ;
m1 Š ! ! ! mn ŠmnC1 Š 1
which completes the induction and the proof.
Section 9.9 Fourier Series
1.
2.
(page 565)
f .t / D sin.3t / has fundamental period 2#=3 since sin t
has fundamental period 2#:
!
f tC
jmjDj
.x1 C ! ! ! C xn C xnC1 /k
X
D
j D0
v .k/ D
12. As suggested in the hint, we assume that
X
k
X
k
X
Now let m" D .m1 ; m2 ; : : : ; mn ; mnC1 /, where
mnC1 D k " j to ensure that jm" j D k. Then we have
Thus the Rule holds for n D m C 1. By induction, it holds
for all positive integers n.
k
jmjDk
kŠ
m1 Šm2 Š ! ! ! mn Š
Observe that the case n D 2 has been proved in Exercise 11. To complete the induction on n, we assume
that the formula above holds for some n and all k. If
u D f1 f2 ! ! ! fn and v D f1 f2 ! ! ! fn fnC1 D ufnC1
then the Product rule and the induction hypothesis show
that
kD1
m h% &
X
m
X
holds for all n # 2 and all positive integers k. The sum
is taken over all multiindices m or order n satisfying
jmj D k.
i
(replace k by k " 1 in the latter sum)
m % &
mC1
X
X% m &
m
D
f .mC1!k/ g .k/ C
f .mC1!k/ g .k/
k
k"1
kD0
We want to prove that
2!
3
"
! !
D sin 3 t C
2!
3
""
D sin.3t / D f .t /:
D sin.3t C 2#/
g.t / D cos.3 C # t / has fundamental period 2 since cos t
has fundamental period 2#:
%
&
g.t C 2/ D cos 3 C #.t C 2/ D cos.3 C # t C 2#/
D cos.3 C # t / D g.t /:
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 9.9 (PAGE 565)
3. h.t / D cos2 t D 21 .1 C cos 2t / has fundamental period #:
h.t C #/ D
7.
1 C cos 2t /
1 C cos.2t C 2#/
D
D h.t /:
2
2
4. Since sin 2t has periods #, 2#, 3#, : : : , and cos 3t
4! 6!
has periods 2!
D 2#, 8!
3 , 3 , 3
3 , : : : , the sum
k.t / D sin.2t / C cos.3t / has periods 2#, 4#, : : : . Its
fundamental period is 2#.
0
."1/n
D"
:
n#
The Fourier series of f is
1
1
%
1
1 X ."1/n
1 2 X
" 2
cos .2n"1/# t /"
sin.n# t /:
2
4 #
.2n " 1/
#
n
5. Since f .t / D t is odd on ."#; #/ and has period 2#, its
cosine coefficients are 0 and its sine coefficients are given
by
2
bn D
2#
Z
!
2
t sin.nt / dt D
#
!!
Z
t sin.nt / dt:
0
8.
2
2
bn D " cos.n#/ D ."1/nC1 :
n
n
1
X
."1/nC1
nD1
2
sin.nt /.
n
'
0 if 0 $ t < 1
, f has period 2.
1 if 1 $ t < 2
The Fourier coefficients of f are as follows:
6. f .t / D
Z
Z
a0
1 2
1 2
1
D
f .t / dt D
dt D
2
2 0
2 1
2
Z 2
Z 2
an D
f .t / cos.n# t / dt D
cos.n# t / dt
0
1
ˇ2
ˇ
1
ˇ
sin.n# t /ˇ D 0;
.n # 1/
D
ˇ
n#
1
ˇ2
Z 2
ˇ
1
ˇ
cos.n# t /ˇ
bn D
sin.n# t / dt D "
ˇ
n#
1
1
(
2
1 " ."1/n
D"
D " n# if n is odd
n#
0
if n is even
The Fourier series of f is
1
%
&
1 X
2
"
sin .2n " 1/# t :
2
.2n " 1/#
nD1
nD1
nD1
!
This integral can be evaluated by a single integration by
parts. Instead we used Maple to do the integral:
The Fourier series of f is
'
0 if "1 $ t < 0
, f has period 2.
t if 0 $ t < 1
The Fourier coefficients of f are as follows:
Z
Z
"1 1
1 1
1
a0
D
f .t / dt D
t dt D
2
1 !1
2 0
4
Z 1
Z 1
an D
f .t / cos.n# t / dt D
t cos.n# t / dt
!1
0
'
2
."1/n " 1
D "2=.n#/ if n is odd
D
n2 # 2
0
if n is even
Z 1
t sin.n# t / dt
bn D
f .t / D
(
t
if 0 $ t < 1
1
if 1 $ t < 2 , f has period 3.
3 " t if 2 $ t < 3
f is even, so its Fourier sine coefficients are all zero. Its
cosine coefficients are
Z
2
a0
1 2 3
2
f .t / dt D .2/ D
D !
2
2 3 0
3
3
Z
2n# t
2 3
f .t / cos
dt
an D
3 0
3
"Z
Z 2
1
2n# t
2n# t
2
t cos
dt C
cos
dt
D
3 0
3
3
1
#
Z 3
2n# t
C
.3 " t / cos
dt
3
2
(
)
4n#
2n#
3
cos
" 1 " cos.2n#/ C cos
:
D
2n2 # 2
3
3
The latter expression was obtained using Maple to evaluate the integrals. If n D 3k, where k is an integer, then
an D 0. For other integers n we have an D "9=.2# 2 n2 /.
Thus the Fourier series of f is
1
1
2
9 X 1
2n# t
1 X 1
"
cos
C
cos.2n# t /:
3 2# 2
n2
3
2# 2
n2
f .t / D
nD1
9.
nD1
The even extension of h.t / D 1 on Œ0; 1! to Œ"1; 1! has the
value 1 everywhere. Therefore all the coefficients an and
bn are zero except a0 , which is 2. The Fourier series is
a0 =2 D 1.
10. The Fourier sine series of g.t / D # " t on Œ0; #! has
coefficients
Z
2
2 !
.# " t / sin nt dt D :
bn D
# 0
n
The required Fourier sine series is
1
X
2
sin nt:
n
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381
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SECTION 9.9 (PAGE 565)
11.
ADAMS and ESSEX: CALCULUS 9
The Fourier sine series of f .t / D t on Œ0; 1! has coefficients
Z 1
."1/n
:
bn D 2
t sin.n# t / dt D "2
n#
0
The required Fourier sine series is
1
X
Similarly,
an D
2
D
T
2."1/n
sin.n# t /:
n#
nD1
2
T
D
12. The Fourier cosine series of f .t / D t on Œ0; 1! has coefficients
Z 1
a0
1
D
t dt D
2
2
0
Z 1
an D 2
t cos.n# t / dt
0
(
0
if n is even
2."1/n " 2
"4
D
D
if
n is odd.
n2 # 2
n2 # 2
4
T
"Z
"Z
Z
0
f .t / cos
!T =2
2n# t
dt C
T
0
Z
T =2
f .t / cos
0
2n# t
f .t / cos
."dt / C
T
T =2
T =2
f .t / cos
0
Z
T =2
0
2n# t
dt
T
2n# t
f .t / cos
dt
T
2n# t
dt:
T
The corresponding result for an odd function f states
that an D 0 and
bn D
4
T
Z
T =2
f .t / sin
0
2n# t
dt;
T
and is proved similarly.
The required Fourier cosine series is
%
&
1
1
4 X cos .2n " 1/# t
:
"
2 #2
.2n " 1/2
Review Exercises 9 (page 566)
nD1
13. From Example 3,
#
C
2
1
X
nD1
4
#.2n " 1/2
%
&
cos .2n " 1/# t D # " jt j
for "# $ t $ #. Putting t D #, we obtain
1.
lim
n!1
2.
1
X
#
4
."1/ D 0:
C
2
#.2n " 1/2
."1/n e n
D 0. The sequence converges.
nŠ
#
$
n100 C 2n #
n100
D
lim
#
C
D #.
n!1
n!1
2n
2n
The sequence converges.
lim
nD1
Thus
1
X
nD1
14.
1
# #
#
D ! D .
.2n " 1/2
2 4
8
3.
In the first integral in the line above replace t with "t .
Since f ."t / D f .t / and sine is odd, we get
"Z
#
$
0
2n# t
2
f .t / " sin
."dt /
bn D
T T =2
T
#
Z T =2
2n# t
C
f .t / sin
dt
T
0
" Z
#
Z T =2
T =2
2n# t
2n# t
2
"
dt C
f .t / sin
dt
f .t / sin
D
T
T
T
0
0
D 0:
382
ln n
ln n
# lim
D 1:
tan!1 n n!1 #=2
The sequence diverges to infinity.
lim
n!1
If f is even and has period T , then
Z
2n# t
2 T =2
f .t / sin
dt
bn D
T !T =2
T
"Z
#
Z T =2
0
2
2n# t
2n# t
D
f .t / sin
dt C
dt :
f .t / sin
T
T
T
!T =2
0
4.
."1/n n2
."1/n
D lim
does not exist.
n!1 # n.n " #/
n!1 1 " .#=n/
The sequence diverges (oscillates).
5.
Let a1 >
lim
p
2 and anC1 D
an
1
C
.
2
an
p
1
1
1
x
If f .x/ D C , then f 0 .x/ D " 2 > 0 if x > 2.
2
x
2
x
p
p
p
p
Since f . 2/ D 2,
> 2.
p we have f .x/ > 2 if x p
Therefore, ifpan > 2, then anC1 D f .an / > 2.
Thus an > 2 for all n # 1, by induction.
p
an > 2 ) 2 < an2 ) an2 C 2 < 2an2
an2 C 2
< an ) anC1 < an :
2an
p
Thus fan g is decreasing and an > 2 for all n.
)
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#
lOMoARcPSD|6566483
INSTRUCTOR’S SOLUTIONS MANUAL
REVIEW EXERCISES 9 (PAGE 566)
p
Being decreasing and bounded below by 2, fan g must
converge by
p the completeness axiom. Let limn!1 an D a.
Then a # 2, and
lim anC1 D lim
n!1
aD
n!1
1
a
C :
2
a
#
an
1
C
2
an
12.
$
Thus a=2 D 1=a, so a2 D 2, and limn!1 an D a D
nD1
p
n C 2n
.n=2n / C 1
1 C 3n
D lim
D 1:
lim
n!1 .2=3/n
n!1 .1=3n / C 1
2.
6. By l’H^opital’s Rule,
ln.x C 1/
1=.x C 1/
x
lim
D lim
D lim
D 1:
x!1
x!1
x!1 x C 1
ln x
1=x
13.
Thus
%
&
ln.n C 1/
lim ln ln.nC1/"ln ln n D lim ln
D ln 1 D 0:
n!1
n!1
ln n
7.
8.
$
#
1
1
2!.n!5/=2 D 22 1 C p C C ! ! !
2
2
nD1
p
4
4 2
D
p D p
:
1 " .1= 2/
2"1
$n
1 #
1
X
4
1X
4n!1
D
.# " 1/2n
4
.# " 1/2
9.
1
X
nD1
1
4
1"
.# " 1/2
since .# " 1/2 > 4.
1
n2 "
1
4
D
1
X
nD1
1
n"
D 2 " lim
N !1
10.
11.
14.
1
X
1
2
n "
nD1
Since 0 $
9
4
1
2
"
1
NC
D
1
2
.# " 1/2
;
4.# " 1/2 " 16
1
2
!
n2
p converges by comparison with the
C n n/
1 p
X
n
(which converges by the ratio
convergent series
2n
nD1
test) because
.1 C
2n /.1
n2
.1 C
n!1
(telescoping)
D 2:
15.
!
1
X
1
1
1
D
"
(telescoping)
3 n " 32
n C 32
nD1
"
1
1
1
1
1
D
"
C
"
3 "1=2 5=2
1=2 7=2
#
1
1
1
1
"
C
"
C !!!
C
3=2 9=2
5=2 11=2
(
)
1
2
2
D
"2 C 2 C
D :
3
3
9
1
X
n"1
1
1
$
for
n
#
1
and
converges,
n3
n2
n2
1
X
n"1
must also converge.
n3
1
X
lim
1
nC
n
p converges by comparison with the
.1
C
n/.1
C n n/
nD1
1
X
1
because
convergent p-series
3=2
n
nD1
nD1
nD0
1
D !
4
1
X
n
p
1
.1 C n/.1 C n n/
$ D 1:
$#
lim
D lim #
1
n!1
n!1
1
1
C
1
n3=2
n
n3=2 C 1
1
X
nD0
1
X
n C 2n
converges by comparison with the convergent
1 C 3n
nD1
1 # $n
X
2
geometric series
because
3
nD1
p
1
C n n/
$ D 1:
$#
D lim #
n!1
1
1
n
C
1
2n
n3=2 C 1
2n
2n /.1
p
1
X
32nC1
converges by the ratio test, because
nŠ
nD1
lim
n!1
16.
32.nC1/C1
9
nŠ
D lim
!
D 0 < 1:
n!1 n C 1
.n C 1/Š 32nC1
1
X
nŠ
converges by comparison with the con.n C 2/Š C 1
nD1
1
X
1
, because
vergent p-series
n2
nD1
0$
nŠ
1
1
nŠ
<
D
< 2:
.n C 2/Š C 1
.n C 2/Š
.n C 2/.n C 1/
n
nD1
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REVIEW EXERCISES 9 (PAGE 566)
17.
ADAMS and ESSEX: CALCULUS 9
1
X
."1/n!1
converges absolutely by comparison with the
1 C n3
nD1
1
X
1
convergent p-series
, because
n3
22.
ˇ
ˇ
ˇ ."1/n!1 ˇ
ˇ$ 1 :
0 $ ˇˇ
1 C n3 ˇ n3
1
X
."1/n
converges absolutely by comparison with the
2n " n
nD1
1
X
1
, because
convergent geometric series
2n
nD1
23.
Let s D
ˇ
ˇ
ˇ ."1/n ˇ
ˇ
ˇ
ˇ 2n " n ˇ
1
lim
D lim
n D 1:
1
n!1
n!1
1" n
2
2n
19.
1
X
jx " 2j
.x " 2/n
< 1, that is,
p
converges absolutely if
n n
3
3
nD1
if "1 < x < 5, and diverges if x < "1 or x > 5.
X ."1/n
If x D "1 the series is
p , which converges condin
tionally.
X 1
If x D 5 the series is
p , which diverges (to 1).
n
384
kD1
1
Let
sn" D
(
)
1
1
1
n2 C .n C 1/2
C
s
C
:
sn C
D sn C
n
2
2
2
2.n C 1/
2n
4n2 .n C 1/2
Then s ' sn" with error satisfying
js " sn" j <
nD1
21.
kD1
Z 1
dt
dt
<
s
"
s
<
n
3
t3
nC1 t
n
1
1
sn C
< s < sn C 2 :
2.n C 1/2
2n
1
X
n2 cos.n#/
20.
converges by the alternating series test
1 C n3
nD1
(note that cos.n#/ D ."1/n ), but the convergence is only
conditional because
1
X
1
for n # 1, and
is a divergent harmonic series.
2n
nD1
ˇ
ˇ
ˇ .x " 2/nC1 ˇ
ˇ
ˇ
r
ˇ
ˇ nC1 p
n
jx " 2j
jx " 2j
nC1ˇ
ˇ3
D
:
lim ˇ
ˇ D lim
n
n!1 ˇ
.x " 2/
ˇ n!1 3
nC1
3
ˇ
ˇ
p
ˇ
ˇ
3n n
1
n
X
X
1
1
and
s
D
. Then
n
k3
k3
Z
1
X
."1/n!1
converges by the alternating series test, but
ln ln n
nD1
1
X
1
dithe convergence is only conditional since
ln ln n
nD1
verges to infinity by comparison with the divergent har1
X
1
monic series
. (Note that ln ln n < n for all n # 1.)
n
ˇ 2
ˇ
ˇ n cos.n#/ ˇ
n2
1
ˇ
ˇ
ˇ 1 C n3 ˇ D 1 C n3 # 2n
ˇ
ˇ
ˇ
ˇ
ˇ D lim j5 " 2xj n D j5 " 2xj:
ˇ n!1
nC1
ˇ
ˇ
1
X
.5 " 2x/n
converges absolutely if j5 " 2xj < 1, that is,
n
nD1
if 2 < x < 3, and diverges if x < 2 or x > 3.
X1
If x D 2 the series is
, which diverges.
n
X ."1/n
If x D 3 the series is
, which converges condin
tionally.
nD1
18.
ˇ
ˇ .5 " 2x/nC1
ˇ
ˇ
nC1
lim ˇˇ
n!1 ˇ .5 " 2x/n
ˇ
n
(
)
1
1
1
2n C 1
"
:
D
2 2n2 2.n C 1/2
4n2 .n C 1/2
This error is less than 0.001 if n # 8. Hence
s'
24.
1
1
1
1
1
1
1
1
C 3 C 3 C 3 C 3 C 3 C 3 C 3
13
2
3
4
5
6
7
8
64 C 81
' 1:202
C
4.64/.81/
with error less than 0:001.
1
n
X
X
1
1
Let s D
and
s
D
. Then
n
2
4Ck
4 C k2
kD1
Z
kD1
Z
1
dt
dt
< s " sn <
2
4 C t2
n
nC1 4 C t
#
1
nC1
#
1
n
sn C " tan!1
< s < sn C " tan!1 :
4
2
2
4
2
2
1
Let
sn" D sn C
(
)
#
1
nC1
n
"
tan!1
C tan!1
:
4
4
2
2
Then s ' sn" with error satisfying
js " sn" j <
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)
nC1
n
1
tan!1
" tan!1
:
4
2
2
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INSTRUCTOR’S SOLUTIONS MANUAL
REVIEW EXERCISES 9 (PAGE 566)
This error is less than 0.001 if n # 22. Hence
s'
22
X
kD1
(Remark: Examining the ln of the absolute value of the
nth term at x D 8 shows that this term ! 0 as n ! 1.
Therefore the series also converges at x D 8.)
(
)
#
1
1
!1 23
!1
C
"
tan
C
tan
.11/
' 0:6605
4 C k2
4
4
2
with error less than 0:001.
1
1
25.
D %
x&
3"x
3 1"
3
1
1
X
xn
1 X % x &n
D
D
3
3
3nC1
nD0
32.
."3 < x < 3/:
nD0
26. Replace x with x 2 in Exercise 25 and multiply by x to get
1
X
x
x 2nC1
D
2
3"x
3nC1
# $# $
2
1
"
1
3
3
x2
.1 C x/1=3 D 1 C x C
# $ # 3$ # $ 2Š
1
2
5
"
"
3
3
3
x3 C ! ! !
C
3Š
1
X
x
2 ! 5 ! 8 ! ! ! .3n " 4/ n
D1C C
."1/n!1
x
3
3n nŠ
."1 < x < 1/:
nD2
p
p
." 3 < x < 3/:
(Remark: the series also converges at x D 1.)
nD0
27.
$
#
x2
ln.e C x / D ln e C ln 1 C
e
1
X
x 2n
D ln e C
."1/n!1 n
ne
nD1
28.
1
X
."2x/n
1"1"
nŠ
1
1 " e !2x
D
x
x
D
29.
33.
2
nD1
1
X
."1/n!1
nD1
2n x n!1
nŠ
p
p
." e < x $ e/:
1
1
1
1
D
D !
x
# C .x " #/
# 1C x"#
#
1
% x " # &n
1 X
D
."1/n
#
#
nD0
!
D
(for all x ¤ 0).
34.
1
X
nD0
nD0
30.
."1/n
nD1
22n!1 x 2nC1
.2n/Š
(for all x).
%
#&
#
#
sin x C
D sin x cos C cos x sin
3
3
3
p 1
1
2nC1
3X
x 2n
1X
n x
."1/
."1/n
D
C
2
.2n C 1/Š
2
.2n/Š
nD0
nD0
!
p 2n
1
2nC1
n
X ."1/
3x
x
D
C
(for all x).
2
.2n/Š
.2n C 1/Š
nD0
1
p X
."1/n %
# &2n
x"
D 2
.2n/Š
4
x &!1=3
1%
1C
2
# 8 $# $
1
4
"
"
"
1
1 %x &
3
3 % x &2
D
1"
C
2
3 8
2Š
8
# $# $# $
4
7
1
#
"
"
"
3
3
3 % x &3
C !!!
C
3Š
8
35.
1
1 X
1 ! 4 ! 7 ! ! ! .3n " 2/ n
."1/n
C
x
2
2 ! 3n ! 8n ! nŠ
nD1
36.
."8 < x < 8/:
ex
2 C2x
2
D e x e 2x
#
$
4x 2
8x 3
D .1 C x 2 C ! ! !/ 1 C 2x C
C
C !!!
2Š
3Š
4
D 1 C 2x C 2x 2 C x 3 C x 2 C 2x 3 C ! ! !
3
10
2
P3 .x/ D 1 C 2x C 3x C x 3 :
3
.8 C x/!1=3 D
D
(for all x).
nD0
nD0
31.
.0 < x < 2#/:
%
%
#&
#&
sin x C cos x D sin u C
C cos u C
4
4
&
1 %
D p .sin u C cos u/ C .cos u " sin u/
2
1
p
p X
u2n
D 2 cos u D 2
."1/n
.2n/Š
x
.1 C cos.2x//
2
!
1
X
x
.2x/2n
D
1C
."1/n
2
.2n/Š
DxC
.x " #/n
# nC1
Let u D x " .#=4/, so x D u C .#=4/. Then
x cos2 x D
1
X
."1/n
sin.1 C x/ D sin.1/ cos x C cos.1/ sin x
#
$
#
$
x2
x3
D sin.1/ 1 "
C ! ! ! C cos.1/ x "
C !!!
2Š
3Š
sin.1/ 2 cos.1/ 3
P3 .x/ D sin.1/ C cos.1/x "
x "
x :
2
6
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REVIEW EXERCISES 9 (PAGE 566)
37.
38.
ADAMS and ESSEX: CALCULUS 9
#
$2
x3
x"
C !!!
.x " ! ! !/4
3Š
cos.sin x/ D 1 "
C
" !!!
2Š
4Š
#
$
4
4
x
1
x
x2 "
D1"
C !!! C
C !!!
2
3
24
1
5
P4 .x/ D 1 " x 2 C x 4 :
2
24
# $# $
1
1
"
p
1
2
2
1 C sin x D 1 C sin x C
.sin x/2
2
2Š
# $# $# $
1
1
3
"
"
2
2
2
C
.sin x/3
3Š
# $# $# $# $
1
1
3
5
"
"
"
2
2
2
2
C
.sin x/4 C ! ! !
4Š
#
$
#
$2
x3
1
x3
1
x"
C !!! "
x"
C !!!
D1C
2
6
8
6
1
5
C .x " ! ! !/3 "
.x " ! ! !/4 C ! ! !
16
128
x2
x4
x 3 5x 4
x x3
"
C
C
"
C !!!
D1C "
2
12
8
24
16
128
x x2
x3
x4
P4 .x/ D 1 C "
"
C
:
2
8
48
384
39. The series
1
X
."1/n x n
is the Maclaurin series for cos x
.2n/Š
nD0
1
X
'
p
cos p
x
f .x/ D
cosh jxj
xn D
nx n!1 D
nD0
1
X
nD0
nx n D
1
1"x
1
X
nD1
nx n!1 D
1
.1 " x/2
x
.1 " x/2
1
1
X
1
nC1
#
D #
$2 C
1
#n
1
nD0
1"
1"
#
#
% # &2
#
#
C
D
:
D
2
.# " 1/
# "1
# "1
42.
1
X
nD0
1
X
nx n D
n2 x n!1 D
nD0
1
X
x
.1 " x/2
as in Exercise 23
x
1Cx
d
D
dx .1 " x/2
.1 " x/3
x.1 C x/
.1 " x/3
nD0
#
$
1
1
1
C
1
2
X n
#.# C 1/
#
#
D #
:
$3 D
#n
.# " 1/3
1
nD0
1"
#
43.
nD0
with x 2 replaced by x. For x > 0 the series therefore
1
X
p
jxjn
represents cos x. For x < 0, the series is
,
.2n/Š
nD0
p
which is the Maclaurin series for cosh jxj. Thus the
given series is the Maclaurin series for
1
X
41.
n2 x n D
1
X
xn
D " ln.1 " x/
n
nD1
#
$
1
X
1
1
D
"
ln
1
"
D 1 " ln.e " 1/:
ne n
e
nD1
44.
1
X
."1/n!1 x 2n!1
D sin x
.2n " 1/Š
nD1
1
X
if x # 0
if x < 0.
."1/n # 2n!1
D " sin # D 0
.2n " 1/Š
nD1
#
$
1
X
."1/n # 2n!4
1
."1/#
1
D 3 0"
D 2:
.2n " 1/Š
#
1Š
#
nD2
40. Since
1
1
X
X
x 2n
f .k/ .0/ k
1C
D
x
2
n
kŠ
nD1
kD0
for x near 0, we have, for n D 1, 2, 3, : : :
f .2n/ .0/ D
386
.2n/Š
;
n2
f .2n!1/ .0/ D 0:
45.
Z
x
sin.t 2 / dt
$
Z x#
t6
C ! ! ! dt
D
t2 "
3Š
0
3
7
x
x
D
"
C !!!
3
7 ! 3Š
S.x/ D
0
x7
" !!!
x3 " x3 C
x 3 " 3S.x/
1
14
lim
D
lim
D
:
x!0
x!0
x7
x7
14
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INSTRUCTOR’S SOLUTIONS MANUAL
46.
47.
CHALLENGING PROBLEMS 9 (PAGE 567)
.x " tan!1 x/.e 2x " 1/
x!0 2x 2 " 1 C cos.2x/
#
$#
$
x5
4x 2
x3
x"xC
"
C !!!
2x C
C !!!
3
5
2Š
D lim
x!0
4x 2
16x 4
2x 2 " 1 C 1 "
C
" !!!
2Š
4Š
#
$
2
x4
C !!!
3
#
$ D 1:
D lim
2
x!0
4
x
C !!!
3
Z 1=2
Z 1=2 X
1
."x 4 /n
4
dx
e !x dx D
nŠ
0
0
nD0
ˇ1=2
1
X
."1/n x 4nC1 ˇˇ
D
ˇ
.4n C 1/nŠ ˇ
lim
D
nD0
1
X
nD0
49.
bn D
The series is
50.
The series satisfies the conditions of the alternating series
test, so if we truncate after the term for n D k " 1, then
the error will satisfy
1
C 1/kŠ
:
This is less than 0:000005 if 24kC1 .4k C 1/kŠ > 200; 000,
which happens if k # 3. Thus, rounded to five decimal
places,
Z
1=2
0
4
e !x dx '
1
1
1
"
C
' 0:49386:
2 ! 1 ! 1 32 ! 5 ! 1 512 ! 9 ! 2
jf
2
:
n
'
1 if "# < t $ 0
has period 2#. Its Fourier
t if 0 < t $ #
coefficients are
f .t / D
"
nD1
#.2n " 1/2
C
%
&
.2 " #/ sin .2n " 1/t
#.2n " 1/
#
sin.2nt /
C
:
2n
Challenging Problems 9 (page 567)
4
.x/j $ 24 csc .1:5/ cot.1:5/ $ 2
for 1:5 $ x $ #=2. Therefore, the error in the approximation
ln.sin 1:5/ ' P4 .x/;
1.
If an > 0 and
2 ˇˇ
# ˇˇ5
ˇ1:5 " ˇ $ 3 & 10!8 :
5Š
2
anC1
n
>
for all n, then
an
nC1
1
a2
>
a1
2
a3
2
>
a2
3
where P4 is the 4th degree Taylor polynomial for f .x/
about x D #=2, satisfies
jerrorj $
0
.# " t / sin.nt / dt D
1
X
2
sin.nt /.
n
2C#
4 %
&
"
1
2
X cos .2n " 1/t
Observe that 1:5 < #=2 ' 1:5708, that csc x # 1 and
cot x # 0, and that both functions are decreasing on that
interval. Thus
.5/
!
The required Fourier series is, therefore,
48. If f .x/ D ln.sin x/, then calculation of successive derivatives leads to
f .5/ .x/ D 24 csc4 x cot x " 8 csc2 cot x:
Z
Z !
a0
1
D
f .t / dt
2
2# !!
(Z 0
)
Z !
1
#
1
dt C
t dt D C
D
2# !!
2
4
0
)
(Z 0
Z !
1
t cos.nt / dt
cos.nt / dt C
an D
# !!
0
Z
1 !
D
.1 C t / cos.nt / dt
# 0
'
."1/n " 1
"2=.# n2 / if n is odd
D
D
2
#n
0
if n is even
(Z 0
)
Z !
1
bn D
sin.nt / dt C
t sin.nt / dt
# !!
0
Z
1 !
.t " 1/ sin.nt / dt
D
# 0
'
1 C ."1/n .# " 1/
.# " 2=.# n/ if n is odd
D"
D
".1=n/
if n is even.
#n
0
24kC1 .4k
2
#
nD1
."1/n
:
24nC1 .4n C 1/nŠ
jerrorj $
The Fourier sine series of f .t / D # " t on Œ0; #! has
coefficients
)
)
a1
2
2a2
a1
a3 >
>
3
3
a2 >
::
:
n"1
an
>
an!1
n
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)
an >
a1
:
n
387
lOMoARcPSD|6566483
CHALLENGING PROBLEMS 9 (PAGE 567)
ADAMS and ESSEX: CALCULUS 9
If x ¤ m# for any integer m, then sin.x=2/ ¤ 0. Using
the addition formulas we obtain
&
%
&i
1h %
sin.nx/ sin.x=2/ D
cos .n " 12 /x " cos .n C 21 /x :
2
Therefore, using the telescoping property of these terms,
h %
&
%
&i
N
N
cos .n " 12 /x " cos .n C 21 /x
X
X
sin.nx/ D
2 sin.x=2/
nD1
nD1
%
&
cos.x=2/ " cos .N C 21 /x
D
:
2 sin.x=2/
P
Therefore, the partial sums of 1
nD1 sin.nx/ are bounded.
Since the sequence f1=ng is positive, decreasing,
and has
P
limit 0, part (b) of Problem 2 shows that 1
sin.nx/=n
nD1
converges in this case too. Therefore the series converges
for all x.
(This can be
P verified by induction.)
Therefore 1
nD1 an diverges by comparison with the harP
1
monic series 1
nD1 .
n
Pn
2.
a) If sn D kD1 vk for n # 1, and s0 D 0, then
vk D sk " sk!1 for k # 1, and
n
X
kD1
uk vk D
n
X
kD1
uk sk "
n
X
uk sk!1 :
kD1
In the second sum on the right replace k with k C 1:
n
X
kD1
uk vk D
D
n
X
kD1
n
X
uk sk "
n!1
X
ukC1 sk
kD0
.uk " ukC1 /sk " u1 s0 C unC1 sn
4.
kD1
D unC1 sn C
n
X
.uk " ukC1 /sk :
kD1
b) If fun g is positive and decreasing, and
limn!1 un D 0, then
n
X
nD1
.uk " ukC1 / D u1 " u2 C u2 " u3 C ! ! ! C un " unC1
kD1
Thus
D u1 " unC1 ! u1 as n ! 1.
n
X
.uk " ukC1 / is a convergent, positive, tele-
kD1
scoping series.
If the partial sums sn of fvn g are bounded, say jsn j $ K
for all n, then
j.un " unC1 /sn j $ K.un " unC1 /;
P
so 1
nD1 .un " unC1 /sn is absolutely convergent (and
therefore convergent) by the comparison test. Therefore, by
part (a),
1
X
kD1
uk vk D lim
n!1
D
1
X
unC1 sn C
n
X
.uk " ukC1 /sk
kD1
!
.uk " ukC1 /sk
Let an be the nth integer that has no zeros in its decimal
representation. The number of such integers that have m
digits is 9m . (There are nine possible choices for each of
the m digits.) Also, each such m-digit number is greater
than 10m!1 (the smallest m-digit number). Therefore the
sum of all the terms 1=an for which an has m digits is
less than 9m =.10m!1 /. Therefore,
$
1
1 #
X
X
1
9 m!1
<9
D 90:
an
10
5.
Z
kC1=2
f .x/ dx " f .k/ D
k!1=2
a) By the Mean-Value Theorem,
!
"
!
" !
"
f 0 k C 23 " f 0 k C 21 D 23 " 12 f 00 .u/ D f 00 .u/
for some u in Œk C 12 ; k C 23 !. Similarly,
"
!
" !
"
!
f 0 k " 12 " f 0 k " 32 D " 21 C 23 f 00 .v/ D f 00 .v/
for some v in Œk " 32 ; k " 12 !. Since f 00 is decreasing
and v $ c $ u, we have f 00 .u/ $ f 00 .c/ $ f 00 .v/,
and so
!
!
!
"
"
!
"
"
0
f k C 23 " f 0 k C 12 $ f 00 .c/ $ f 0 k " 12 " f 0 k " 32 :
b) If f 00 is decreasing,
0
R1
1
NC2
f .x/ dx converges, and
f .x/ ! 0 as x ! 1, then
Z 1
1
X
f .x/ dx
f .n/ "
1
nDN C1
kD1
f 00 .c/
, for some c in the
24
interval Œk " 21 ; k C 12 !.
D
converges.
3. If x DPm# for some integer m, then all the terms of the
series 1
nD1 .1=n/ sin.nx/ are 0, so the series converges to
0.
388
mD1
1
X
nDN C1
D"
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1
24
0
NC2
@f .n/ "
1
X
nDN C1
Z
1
nC 2
1
n! 2
f 00 .cn /;
1
f .x/ dx A
lOMoARcPSD|6566483
INSTRUCTOR’S SOLUTIONS MANUAL
CHALLENGING PROBLEMS 9 (PAGE 567)
for some numbers cn in Œn " 12 ; n C 12 !. Using the
result of part (a), we see that
1
X
, 0
f .n C 23 / " f 0 .n C 12 / $
nDN C1
1
X
$
nDN C1
,
f 0 .n " 21 / " f 0 .n " 23 /
" f 0 .N C 32 / $
0
f .N "
24
1
/
2
$
1
X
nDN C1
1
X
nDN C1
1
X
b) Suppose e is rational, say e D M=N where M and
N are
P positive integers. Then N Še is an integer and
N Š jND0 .1=j Š/ is an integer (since each j Š is a factor of N Š). Therefore the number
1
0
N
X
1
A
Q D N Š @e "
jŠ
f 00 .cn /
nDN C1
j D0
is a difference of two integers and so is an integer.
f 00 .cn / $ "f 0 .N " 21 /
f .n/ "
Z
1
1
NC2
f .x/ dx $
f 0 .N C 32 /
:
24
7.
1
$ 1. By part (b), Q is
c) By part (a), 0 < Q <
N
an integer. This is not possible; there are no integers
between 0 and 1. Therefore e cannot be rational.
1
X
22k kŠ
Let f .x/ D
ak x 2kC1 , where ak D
.
.2k C 1/Š
kD0
2
0
3
a) Since
c) Let f .x/ D 1=x . Then f .x/ D "2=x ! 0 as
x ! 1, f 00 .x/ D 6=x 4 is decreasing, and
Z
1
1
NC2
f .x/ dx D
Z
1
1
NC2
dx
1
D
x2
NC
ˇ
ˇ
ˇa
2kC3 ˇ
ˇ
ˇ kC1 x
lim ˇ
ˇ
k!1 ˇ ak x 2kC1 ˇ
22kC2 .k C 1/Š .2k C 1/Š
!
!
kŠ
.2k C 3/Š
k!1 22k
4k
C
4
D jxj2 lim
D0
k!1 .2k C 3/.2k C 2/
1
2
D jxj2 lim
converges. From part (b) we obtain
ˇ
ˇ 1
ˇ X 1
1
ˇ
"
ˇ
2
NC
ˇnDN C1 n
ˇ
ˇ
ˇ
1
ˇ$
"3 :
!
1ˇ
ˇ
12 N " 12
2
for all x, the series for f .x/ converges for all x. Its
radius of convergence is infinite.
b) f 0 .x/ D
The right side is less than 0:001 if N D 5. Therefore
kD0
kD0
(replace k with k " 1)
1
X
22k!1 .k " 1/Š 2k
D1C
x
.2k " 1/Š
nD1
correct to within 0.001.
6.
a) Since e D
1
X
j D0
1
, we have
jŠ
D1C
1
n
X
X
1
1
D
jŠ
jŠ
j DnC1
j D0
#
$
1
1
1
D
1C
C
C !!!
.n C 1/Š
nC2
.n C 2/.n C 3/
#
$
1
1
1
1C
C
C
!
!
!
$
.n C 1/Š
nC2
.n C 2/2
1
1
nC2
1
D
!
D
<
:
1
.n C 1/Š
.n C 1/Š.n C 1/
nŠn
1"
nC2
is, n2 C 2n < n2 C 2n C 1.
1
nC2
< , that
.n C 1/2
n
kD1
1
X
kD1
22k kŠ 2k
x D f 0 .x/:
.2k/Š
&
%
&
d % !x 2
2
2
c)
e
f .x/ D e !x f 0 .x/ " 2xf .x/ D e !x .
dx
d) Since f .0/ D 0, we have
Z x
Z x
&
d % !t 2
2
2
e !x f .x/ " f .0/ D
e f .t / dt D
e !t dt
dt
0
Z x 0
x2
!t 2
f .x/ D e
e
dt:
0<e"
The last inequality follows from
kD1
1
X
22kC1 kŠ 2kC2
1 C 2xf .x/ D 1 C
x
.2k C 1/Š
1
5
X
X
1
1
1
D
C
' 1:6454
n2
n2
5:5
nD1
1
1
X
X
22k kŠ
22k kŠ 2k
.2k C 1/x 2k D 1 C
x
.2k C 1/Š
.2k/Š
0
8.
Let f be a polynomial and let
g.x/ D
1
X
."1/j f .2j / .x/:
j D0
This “series” is really just a polynomial since sufficiently
high derivatives of f are all identically zero.
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CHALLENGING PROBLEMS 9 (PAGE 567)
ADAMS and ESSEX: CALCULUS 9
a) By replacing j with j " 1, observe that
g 00 .x/ D
D
1
X
j D0
1
X
j D1
This constant
is
$ an integer because the binomial coef#
k
is an integer and i Š=kŠ is an integer.
ficient
i "k
(The other factors are also integers.) Hence f .i/ .0/ is
an integer, and so g.0/ is an integer.
."1/j f .2j C2/ .x/
%
&
."1/j !1 f .2j / .x/ D " g.x/ " f .x/ :
e) Observe that f .# " x/ D f ..m=n/ " x/ D f .x/ for all
x. Therefore f .i/ .#/ is an integer (for each i ), and
so g.#/ is an integer. Thus g.#/ C g.0/ is an integer,
which contradicts the conclusion of part (b). (There
is no integer between 0 and 1.) Therefore, # cannot
be rational.
Also
&
d % 0
g .x/ sin x " g.x/ cos x
dx
D g 00 .x/ sin x C g 0 .x/ cos x " g 0 .x/ cos x C g.x/ sin x
%
&
D g 00 .x/ C g.x/ sin x D f .x/ sin x:
9.
Let x > 0, and let
Thus
ˇ
Z !
%
&ˇ!
ˇ
0
f .x/ sin x dx D g .x/ sin x " g.x/ cos x ˇ D g.#/ C g.0/:
ˇ
0
f .x/ D
j D0
The sum is just the binomial expansion.
For 0 < x < # D m=n we have
Thus 0 <
R!
0
1
# k mk
< :
kŠ
2
f .x/ sin x dx <
so 0 < g.#/ C g.0/ < 1.
1
2
Z
!
0
sin x dx D 1, and
j D0
.i/
d) Evidently f .0/ D 0 if i < k or if i > 2k.
If k $ i $ 2k, the only term in the sum for f .i/ .0/
that is not zero is the term for which j D i " k. This
term is the constant
1
kŠ
390
k
i "k
mk!j ."n/j
Z
x
e !1=t dt D I0 D x 2 e !1=x " 2I1
%
&
D x 2 e !1=x " 2 x 3 e !1=x " 3I2
%
&
D e !1=x Œx 2 " 2Šx 3 ! C 3Š x 4 e !1=x " 4I3
%
&
D e !1=x Œx 2 " 2Šx 3 C 3Šx 4 ! " 4Š x 5 e !1=x " 5I4
0
::
:
N
X
."1/n .n " 1/Šx n
nD2
C ."1/
j Ck!i
& .j C k/.j C k " 1/ ! ! ! .j C k " i C 1/x
k # $
.j C k/Š
1 X k
mk!j ."n/j
x j Ck!i :
D
j
kŠ
.j C k " i /Š
$
Therefore,
N C1
j D0
#
Ik D x kC2 e !1=x " .k C 2/IkC1 :
D e !1=x
k # $
1 X k
mk!j ."n/j
c) f .i/ .x/ D
kŠ
j
t k e !1=t dt
0
0
k # $
x k .m " nx/k
1 X k
D
mk!j ."n/j x j Ck :
kŠ
kŠ
j
0 < f .x/ <
x
1
U D t kC2
d V D 2 e !1=t dt
t
d U D .k C 2/t kC1 dt
V D e !1=t
ˇx
Z x
ˇ
ˇ
D t kC2 e !1=t ˇ " .k C 2/
t kC1 e !1=t dt
ˇ
0
0
b) Suppose that # D m=n, where m and n are positive
integers. Since limk!1 x k =kŠ D 0 for any x, there
exists an integer k such that .# m/k =kŠ < 1=2. Let
Z
Ik D
iŠ
:
0Š
NŠ
Z
x
t N !1 e !1=t dt:
0
The Maclaurin series for e !1=t does not exist. The function is not defined at t D 0.
For x D 0:1 and N D 5, the approximation
I D
Z
0
0:1
e !1=t dt ' e !10
5
X
."1/n .n " 1/Š.0:1/n
nD2
%
&
D e !10 .0:1/2 " 2.0:1/3 C 6.0:1/4 " 24.0:1/5
' 0:00836e !10
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INSTRUCTOR’S SOLUTIONS MANUAL
CHALLENGING PROBLEMS 9 (PAGE 567)
has error E given by
E D ."1/6 5Š
Z
0:1
t 4 e !1=t dt:
which is about 0:4% of the size of I .
For N D 20, the error estimate is
0
Since e !1=t $ e !10 for 0 $ t $ 0:1, we have
Z 0:1
t 4 dt ' 2:4 & 10!4 e !10 ;
jEj $ 120e !10
0
jEj $ 20Še !10
Z
0:1
0
t 19 dt ' 1:2 & 10!3 e !10 ;
which is about 3% of the size of I .
For N D 10, the error estimate is
Z 0:1
jEj $ 10Še !10
t 9 dt ' 3:6 & 10!5 e !10 ;
0
which is about 15% of the size of I .
Observe, therefore, that the sum for N D 10 does a better
job of approximating I than those for N D 5 or N D 20.
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SECTION 10.1 (PAGE 574)
ADAMS and ESSEX: CALCULUS 9
CHAPTER 10. VECTORS AND
COORDINATE GEOMETRY IN 3-SPACE
8.
If A D .1; 2; 3/, B D .1; 3; 4/, and C D .0; 3; 3/, then
p
p
.1 " 1/2 C .3 " 2/2 C .4 " 3/2 D 2
p
p
jAC j D .0 " 1/2 C .3 " 2/2 C .3 " 3/2 D 2
p
p
jBC j D .0 " 1/2 C .3 " 3/2 C .3 " 4/2 D 2:
jABj D
Section 10.1 Analytic Geometry in Three
Dimensions (page 574)
1.
The distance between .0; 0; 0/ and .2; "1; "2/ is
p
22
C
."1/2
C
."2/2
All three sides being equal, the triangle is equilateral.
9.
D 3 units.
2. The distance between ."1; "1; "1/ and .1; 1; 1/ is
p
p
.1 C 1/2 C .1 C 1/2 C .1 C 1/2 D 2 3 units:
3. The distance between .1; 1; 0/ and .0; 2; "2/ is
p
.0 "
1/2
C .2 "
1/2
C ."2 "
0/2
D
p
10. The distance from the origin to .1; 1; 1; : : : ; 1/ in Rn is
6 units:
4. The distance between .3; 8; "1/ and ."2; 3; "6/ is
p
p
."2 " 3/2 C .3 " 8/2 C ."6 C 1/2 D 5 3 units.
5.
IfA D .1; 1; 0/, B D .1; 0; 1/, and C D .0; 1; 1/, then
p
jABj D jAC j D jBC j D 2:
p
Thus the triangle ABC is equilateral with sides 2. Its
area is, therefore,
r
p
1 p
1
3
& 2& 2" D
sq. units.
2
2
2
a) The shortest distance from .x; y; z/ to the xy-plane is
jzj units.
b) The
p shortest distance from .x; y; z/ to the x-axis is
y 2 C z 2 units.
p
p
12 C 12 C 12 C ! ! ! C 1 D n units.
11.
The point on the x1 -axis closest to .1; 1; 1; : : : ; 1/ is
.1; 0; 0; : : : ; 0/. The distance between these points is
p
p
02 C 12 C 12 C ! ! ! C 12 D n " 1 units.
12. z D 2 is a plane, perpendicular to the z-axis at .0; 0; 2/.
z
2
6. If A D .1; 2; 3/, B D .4; 0; 5/, and C D .3; 6; 4/, then
zD2
p
p
jABj D 32 C ."2/2 C 22 D 17
p
p
jAC j D 22 C 42 C 12 D 21
p
p
jBC j D ."1/2 C 62 C ."1/2 D 38:
Since jABj2 C jAC j2 D 17 C 21 D 38 D jBC j2 , the
triangle ABC has a right angle at A.
7.
y
x
Fig. 10.1-12
13.
If A D .2; "1; "1/, B D .0; 1; "2/, and C D .1; "3; 1/,
then
y # "1 is the half-space consisting of all points on the
plane y D "1 (which is perpendicular to the y-axis at
.0; "1; 0/) and all points on the same side of that plane as
the origin.
p
.0 " 2/2 C .1 C 1/2 C ."2 C 1/2 D 3
p
b D jAC j D .1 " 2/2 C ."3 C 1/2 C .1 C 1/2 D 3
p
p
a D jBC j D .1 " 0/2 C ."3 " 1/2 C .1 C 2/2 D 26:
z
c D jABj D
yD!1
!1
By the Cosine Law,
x
a2 D b 2 C c 2 " 2bc cos †A
26 D 9 C 9 " 18 cos †A
26 " 18
' 116:4ı :
†A D cos!1
"18
392
y
Fig. 10.1-13
14.
z D x is a plane containing the y-axis and making 45ı
angles with the positive directions of the x- and z-axes.
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 10.1
z
(PAGE 574)
z
.1;0;1/
x 2 Cz 2 D4
2
zDx
y
x
y
x
Fig. 10.1-14
Fig. 10.1-20
21.
15.
x C y D 1 is a vertical plane (parallel to the z-axis)
passing through the points .1; 0; 0/ and .0; 1; 0/.
z D y 2 is a “parabolic cylinder” — a surface all of whose
cross-sections in planes perpendicular to the x-axis are
parabolas.
z
z
xCyD1
zDy 2
1
y
x
1
y
x
Fig. 10.1-15
22.
16. x 2 C y 2 C z 2 D 4 is a sphere centred at the origin and having radius 2 (i.e., all points at distance 2 from the origin).
17.
.x " 1/2 C .y C 2/2 C .z " 3/2 D 4 is a sphere of radius 2
with centre at the point .1; "2; 3/.
Fig. 10.1-21
p
z # x 2 C y 2 represents every point whose distance
above the xy-plane is not less than its horizontal distance
from the z-axis. It therefore consists of all points inside
and on a circular cone with axis along the positive z-axis,
vertex at the origin, and semi-vertical angle 45ı .
z
18. x 2 C y 2 C z 2 D 2z can be rewritten
x 2 C y 2 C .z " 1/2 D 1;
p
45ı
zD
and so it represents a sphere with radius 1 and centre at
.0; 0; 1/. It is tangent to the xy-plane at the origin.
x 2 Cy 2
y
x
Fig. 10.1-22
z
x 2 Cy 2 Cz 2 D2z
23.
.0;0;1/
x C 2y C 3z D 6 represents the plane that intersects the
coordinate axes at the three points .6; 0; 0/, .0; 3; 0/, and
.0; 0; 2/. Only the part of the plane in the first octant is
shown in the figure.
z
.0;0;2/
y
x
Fig. 10.1-18
.0;3;0/
y
19. y 2 C z 2 $ 4 represents all points inside and on the circular
cylinder of radius 2 with central axis along the x-axis (a
solid cylinder).
20. x 2 C z 2 D 4 is a circular cylindrical surface of radius 2
with axis along the y-axis.
x
.6;0;0/
Fig. 10.1-23
24.
'
xD1
represents the vertical straight line in which the
yD2
plane x D 1 intersects the plane y D 2.
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SECTION 10.1 (PAGE 574)
ADAMS and ESSEX: CALCULUS 9
z
27.
yD2
xD1
y
x
.1;2;0/
Fig. 10.1-24
28.
25.
'
xD1
is the straight line in which the plane z D 1
yDz
intersects the plane y D z. It passes through the points
.1; 0; 0/ and .1; 1; 1/.
'
x2 C y2 C z2 D 4
is the circle in which the sphere
x 2 C y 2 C z 2 D 4z
of radius 2 centred at the origin intersects the sphere of
radius 2 centred at .0; 0; 2/. (The second equation can be
rewritten x 2 C y 2 C .z " 2/2 D 4 for easier recognition.)
Subtracting the equations of the two spheres we get z D 1,
so the circle must lie in the plane z D 1 as well. Thus it
is the same circle as in the previous exercise.
'
x 2 C y 2 C z 2 D 4 represents the two circles in
x2 C z2 D 1
which the cylinder x 2 C z 2 " 1 intersects the sphere
x 2 C y 2 C z 2 D 4. Subtracting the two equations, we
p
get y 2 D 3. Thus,pone circle lies in the plane y D 3
and haspcentre .0; 3; 0/ and the
p other lies in the plane
y D " 3 and has centre .0; " 3; 0/. Both circles have
radius 1.
z
z
xD1
.1;1;1/
2
1
zDy
p
y
3
x
.1;0;0/
x
y
Fig. 10.1-25
26.
Fig. 10.1-28
29.
'
x 2 C y 2 C z 2 D 4 is the circle in which the horizontal
zD1
plane z D 1 intersects the sphere of radius 2 centred at the
origin.
Thepcircle has centre .0; 0; 1/ and radius
p
4 " 1 D 3.
'
x 2 C y 2 D 1 is the ellipse in which the slanted plane
zDx
z D x intersects the vertical cylinder x 2 C y 2 D 1.
z
x 2 Cy 2 D1
z
zDx
p
3
1
y
zD1
.0;0;1/
x
2
y
Fig. 10.1-29
x
x 2 Cy 2 Cz 2 D4
Fig. 10.1-26
394
30.
'
y#x
is the quarter-space consisting of all points lying
z$y
on or on the same side of the planes y D x and z D y as
does the point .0; 1; 0/.
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INSTRUCTOR’S SOLUTIONS MANUAL
31.
SECTION 10.2
'
x2 C y2 $ 1
represents all points which are inside or on
z#y
the vertical cylinder x 2 C y 2 D 1, and are also above or on
the plane z D y.
z
35.
S D f.x; y/ W x C y D 1g
The boundary of S is S. The interior of S is the empty
set. S is closed, but not bounded. There are points on the
line x C y D 1 arbitrarily far away from the origin.
36.
S D f.x; y/ W jxj C jyj $ 1g
The boundary of S consists of all points on the edges of
the square with vertices .˙1; 0/ and .0; ˙1/.
The interior of S consists of all points inside that square.
S is closed since it contains all its boundary points. It is
bounded since all points in it are at distance not greater
than 1 from the origin.
zDy
x
y
x 2 Cy 2 D1
Fig. 10.1-31
32.
37. S D f.x; y; z/ W 1 $ x 2 C y 2 C z 2 $ 4g
Boundary: the spheres of radii 1 and 2 centred at the origin.
Interior: the region between these spheres. S is closed.
38.
S D f.x; y; z/ W x # 0; y > 1; z < 2g
Boundary: the quarter planes x D 0; .y # 1; z $ 2/,
y D 1; .x # 0; z $ 2/, and z D 2; .x # 0; y # 1/.
Interior: the set of points .x; y; z/ such that x > 0, y > 1,
z < 2.
S is neither open nor closed.
39.
S D f.x; y; z/ W .x " z/2 C .y " z/2 D 0g
The boundary of S is S, that is, the line x D y D z. The
interior of S is empty. S is closed.
40.
S D f.x; y; z/ W x 2 C y 2 < 1; y C z > 2g
Boundary: the part of the cylinder x 2 C y 2 D 1 that lies
on or above the plane y C z D 2 together with the part of
that plane that lies inside the cylinder.
Interior: all points that are inside the cylinder x 2 C y 2 D 1
and above the plane y C z D 2. S is open.
'
2
2
2
x
p C y C z $ 1 represents all points which are in2
2
x Cy $z
side or on the sphere of radius 1 centred at the origin and
which are also inside or on the upper half of the circular
cone with axis along the z-axis, vertex at the origin, and
semi-vertical angle 45ı .
z
p
zD
(PAGE 583)
x 2 Cy 2
y
Section 10.2 Vectors
x
x 2 Cy 2 Cz 2 D1
1.
Fig. 10.1-32
34. S D f.x; y/ W x # 0; y < 0g
The boundary of S consists of points .x; 0/ where x # 0,
and points .0; y/ where y $ 0.
The interior of S consists of all points of S that are not
on the y-axis, that is, all points .x; y/ satisfying x > 0
and y < 0.
S is neither open nor closed; it contains some, but not all,
of its boundary points.
S is not bounded; .x; "1/ belongs to S for 0 < x < 1.
A D ."1; 2/; B D .2; 0/; C D .1; "3/; D D .0; 4/.
""!
"!
(a) AB D 3i " 2j
(b) BA D "3i C 2j
""!
(c) AC D 2i " 5j
33. S D f.x; y/ W 0 < x 2 C y 2 < 1g
The boundary of S consists of the origin and all points on
the circle x 2 C y 2 D 1. The interior of S is S, which
is therefore open. S is bounded; all points in it are at
distance less than 1 from the origin.
(page 583)
""!
(d) BD D "2i C 4j
""!
""! ""!
(e) DA D "i " 2j
(f) AB " BC D 4i C j
""!
""!
""!
(g) AC " 2AB C 3CD D "7i C 20j
5
1 %""! ""! ""!&
AB C AC C AD D 2i " j
(h)
3
3
2.
uDi"j
v D j C 2k
a)
u C v D i C 2k
u " v D i " 2j " 2k
2u " 3v D 2i " 5j " 6k
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SECTION 10.2 (PAGE 583)
b) juj D
jvj D
p
p
1C1D
1C4D
p
p
ADAMS and ESSEX: CALCULUS 9
Thus MN is parallel to and half as long as BC .
2
C
5
1
c) uO D p .i " j/
2
1
vO D p .j C 2k/
5
d) u + v D 0 " 1 C 0 D "1
N
A
e) The angle between u and v is
"1
cos!1 p ' 108:4ı .
10
f) The scalar projection of u in the direction of v is
u+v
"1
D p .
jvj
5
g) The vector projection of v along u is
.v + u/u
1
D " .i " j/.
juj2
2
3.
B
Fig. 10.2-5
6.
We have
""! "
"
! ""!
PQ D PB C BQ D
"! "! ""!
SR D SD C DR D
u D 3i C 4j " 5k
v D 3i " 4j " 5k
"!
1"
AB C
2
"
"
1 !
2 AD C
"!
1"
BC D
2
"
!
1 "
2 DC D
"!
1"
AC I
2
"
!
1 "
2 AC :
""! "!
Therefore, PQ D SR. Similarly,
u C v D 6i " 10k
u " v D 8j
2u " 3v D "3i C 20j C 5k
p
p
b) juj D 9 C 16 C 25 D 5 2
p
p
jvj D 9 C 16 C 25 D 5 2
a)
""! ""! "!
""!
QR D QC C CR D 12 BDI
"
"
!
"! "! "!
PS D PA C AS D 21 BD:
""! "!
Therefore, QR D PS . Hence, PQRS is a parallelogram.
1
c) uO D p .3i C 4j " 5k/
5 2
1
vO D p .3i " 4j " 5k/
5 2
d) u + v D 9 " 16 C 25 D 18
B
Q
P
e) The angle between u and v is
18
' 68:9ı .
cos!1
50
f) The scalar projection of u in the direction of v is
u+v
18
D p .
jvj
5 2
g) The vector projection of v along u is
9
.v + u/u
D
.3i C 4j " 5k/.
juj2
25
4. If a D ."1; 1/, B D .2; 5/ and C D .10; "1/, then
""!
""!
""! ""!
AB D 3i C 4j and BC D 8i " 6j. Since AB + BC D 0,
""! ""!
therefore, AB ? BC . Hence, 4ABC has a right angle at
B.
A
""!
""! ""!
BC
AC " AB
""! ""! ""!
D
:
MN D AN " AM D
2
2
C
R
S
D
Fig. 10.2-6
7.
Let the parallelogram be ABCO. Take the origin at O.
The position vector of the midpoint of OB is
""! ""!
""! "!
""!
OB C CB
OC C OA
OB
D
D
:
2
2
2
The position vector of the midpoint of CA is
"!
"! ""!
""! CA
""! OA " OC
OC C
D OC C
2
2
""! "!
OC C OA
:
D
2
5. Let the triangle be ABC . If M and N are the midpoints
""!
""!
of AB and AC respectively, then AM D 12 AB, and
""!
"
"
!
AN D 12 AC . Thus
396
M
Thus the midpoints of the two diagonals coincide, and the
diagonals bisect each other.
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 10.2
9.
B
C
(PAGE 583)
Let i point east and j point north. Let the wind velocity
be
vwind D ai C bj:
Now vwind D vwind rel car C vcar .
When vcar D 50j, the wind appears to come from the west,
so vwind rel car D ,i. Thus
ai C bj D ,i C 50j;
A
O
Fig. 10.2-7
8. Let X be the point of intersection of the medians AQ and
BP as shown. We must show that CX meets AB in the
""!
"
"
!
""!
""!
midpoint of AB. Note that PX D ˛ PB and QX D ˇ QA
for certain real numbers ˛ and ˇ. Then
#
$
1 ""!
1 ""!
1 ""! "!
""!
""!
CX D CB C ˇ QA D CB C ˇ CB C BA
2
2
2
1 C ˇ ""!
"!
D
CB C ˇ BAI
2
#
$
1 "!
1 "!
1 "! ""!
""!
"
"
!
CX D CA C ˛ PB D CA C ˛ CA C AB
2
2
2
1 C ˛ "!
""!
CA C ˛ AB:
D
2
so a D , and b D 50.
When vcar D 100j, the wind appears to come from the
northwest, so vwind rel car D -.i-j). Thus
ai C bj D -.i " j/ C 100j;
so a D - and b D 100 " -.
Hence 50 D 100 " -, so - D 50.pThus a D b D 50. The
wind is from the southwest at 50 2 km/h.
10. Let the x-axis point east and the y-axis north. The velocity of the water is
vwater D 3i:
If you row through the water with speed 5 in the direction
making angle & west of north, then your velocity relative
to the water will be
vboat
Thus,
1 C ˇ ""!
1 C ˛ "!
"!
""!
CB C ˇ BA D
CA C ˛ AB
2
2
1 C ˛ "! 1 C ˇ ""!
"!
.ˇ C ˛/BA D
CA "
CB
2
2
1 C ˛ "! 1 C ˇ ""!
"! ""!
.ˇ C ˛/.CA " CB/ D
CA "
CB
2$
$
# 2
#
1 C ˇ ""!
1 C ˛ "!
CA D ˇ C ˛ "
CB:
ˇC˛"
2
2
"!
""!
Since CA is not parallel to CB,
1C˛
1Cˇ
DˇC˛"
D0
2
2
1
)˛DˇD :
3
ˇC˛"
Since ˛ D ˇ, x divides AQ and BP in the same ratio.
By symmetry, the third median CM must also divide the
other two in this ratio, and so must pass through X and
MX D 31 M C .
rel water
D "5 sin & i C 5 cos & j:
Therefore, your velocity relative to the land will be
vboat
rel land
D vboat rel water C vwater
D .3 " 5 sin & /i C 5 cos & j:
To make progress in the direction j, choose & so that
3 D 5 sin & . Thus & D sin!1 .3=5/ ' 36:87ı . In this case,
your actual speed relative to the land will be
4
5 cos & D & 5 D 4 km/h.
5
To row from A to B, head in the direction 36:87ı west of
north. The 1=2 km crossing will take .1=2/=4 D 1=8 of
an hour, or about 7 12 minutes.
B
vboat
rel water
A
"
M
vwater
P
X
B
j
i
A
Q
C
Fig. 10.2-10
Fig. 10.2-8
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SECTION 10.2 (PAGE 583)
11.
ADAMS and ESSEX: CALCULUS 9
If this velocity is true easterly, then
We use the notations of the solution to Exercise 4. You
1
now want to make progress in the direction ki C j, that
2
is, in the direction making angle
/ D tan!1
100
750 sin & D p ;
2
1
2k
so & ' 5:41ı . The speed relative to the ground is
100
750 cos & " p ' 675:9 km/h:
2
with vector i. Head at angle & upstream of this direction.
Since your rowing speed is 2, the triangle with angles &
and / has sides 2 and 3 as shown in the figure. By the
3
2
Sine Law,
D
, so
sin &
sin /
The time for the 1500 km trip is
1500
' 2:22 hours.
675:9
y
3
1
3
sin & D sin / D
q
2
2 2 k2 C
3
1
4
:
D p
2 4k 2 C 1
3
This is only possible if p
$ 1, that is, if
2C1
2
4k
p
5
k#
.
4
3
Head in the direction & D sin!1 p
upstream of
2 4k 2 C 1
the direction of AC p
, as shown in the figure. The trip is
not possible if k < 5=4.
B
750
"
x
100
C
k
Fig. 10.2-12
)
13.
3
2
1
2
The two vectors are perpendicular if their dot product is
zero:
.2t i C 4j " .10 C t /k/ + .i C t j C k/ D 0
2t C 4t " 10 " t D 0 ) t D 2:
"
3i
The vectors are perpendicular if t D 2.
14.
)
A
Fig. 10.2-11
The cube with edges i, j, and k has diagonal iCjCk. The
angle between i and the diagonal is
cos!1
12. Let i point east and j point north. If the aircraft heads in
a direction & north of east, then its velocity relative to the
air is
750 cos & i C 750 sin & j:
i + .i C j C k/
1
p
D cos!1 p ' 54:7ı :
3
3
z
The velocity of the air relative to the ground is
100
100
" p i C " p j:
2
2
"
Thus the velocity of the aircraft relative to the ground is
#
$
#
$
100
100
750 cos & " p
i C 750 sin & " p
j:
2
2
398
x
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iCjCk
y
i
Fig. 10.2-14
lOMoARcPSD|6566483
INSTRUCTOR’S SOLUTIONS MANUAL
15.
SECTION 10.2
The cube of Exercise 10 has six faces, each with 2 diagonals. The angle between i C j C k and the face diagonal
i C j is
cos!1
2
.i C j/ + .i C j C k/
D cos!1 p ' 35:26ı :
p p
2 3
6
20.
If a ¤ 0, then a + r D 0 implies that the position vector
r is perpenducular to a. Thus the equation is satisfied by
all points on the plane through the origin that is normal
(perpendicular) to a.
21.
If r + a D b, then the vector projection of r along a is the
constant vector
r+a a
b
D 2 a D r0 ; say.
jaj jaj
jaj
Six of the face diagonals make this angle with i C j C k.
The face diagonal i " j (and five others) make angle
cos!1
Thus r + a D b is satisfied by all points on the plane
through r0 that is normal to a.
.i " j/ + .i C j C k/
D cos!1 0 D 90ı
p p
2 3
with the cube diagonal i C j C k.
In Exercises 22–24, u D 2i C j " 2k, v D i C 2j " 2k, and
w D 2i " 2j C k.
22.
u+i
u1
16. If u D u1 i C u2 j C u3 k, then cos ˛
D
.
juj
juj
u3
u2
and cos % D
.
Similarly, cos ˇ D
juj
juj
Thus, the unit vector in the direction of u is
uO D
u
D cos ˛i C cos ˇj C cos % k;
juj
If uO makes equal angles ˛ D ˇ D % with
p the coordinate
axes, then 3 cos2 ˛ D 1, and cos ˛ D 1= 3. Thus
uO D
23.
†BCA D cos!1
†CAB D cos!1
Let x D xi C yj C zk. Then
x+uD9
x+vD4
x+w D 6
iCjCk
:
p
3
"! ""!
BA + BC
4
D cos!1 p p ' 60:26ı
jBAjjBC j
5 13
""! "!
CB + CA
9
D cos!1 p p ' 37:87ı
jCBjjCAj
10 13
""! ""!
1
AC + AB
D cos!1 p p ' 81:87ı :
jAC jjABj
10 5
19. Since r " r1 D ,r1 C .1 " ,/r2 " r1 D .1 " ,/.r1 " r2 /,
therefore r " r1 is parallel to r1 " r2 , that is, parallel to the
line P1 P2 . Since P1 is on that line, so must P be on it.
1
1
If , D , then r D .r1 C r2 /, so P is midway between
2
2
P1 and P2 .
2
1
2
If , D , then r D r1 C r2 , so P is two-thirds of the
3
3
3
way from P2 towards P1 along the line.
If , D "1, the r D "r1 C 2r2 D r2 C .r2 " r1 /, so P is
such that P2 bisects the segment P1 P .
If , D 2, then r D 2r1 " r2 D r1 C .r1 " r2 /, so P is such
that P1 bisects the segment P2 P .
,
,
,
2x C y " 2z D 9
x C 2y " 2z D 4
2x " 2y C z D 6:
This system of linear equations has solution x D 2,
y D "3, z D "4. Thus x D 2i " 3j " 4k.
18. If A D .1; 0; 0/, B D .0; 2; 0/, and C D .0; 0; 3/, then
†ABC D cos!1
Vector x D xi C yj C zk is perpendicular to both u and v
if
u + x D 0 , 2x C y " 2z D 0
v + x D 0 , x C 2y " 2z D 0:
Subtracting these equations, we get x " y D 0, so x D y.
The first equation now gives 3x D 2z. Now x is a unit
vector if x 2 C y 2 C z 2 D 1, that is, if x 2 C x 2 C 49 x 2 D 1,
p
or x D ˙2= 17. The two unit vectors are
$
#
2
3
2
xD˙ p iC p jC p k :
17
17
17
O 2 D 1.
and so cos2 ˛ C cos2 ˇ C cos2 % D juj
17.
(PAGE 583)
24.
Since u, v, and w all have the same length (3), a vector
x D xi C yj C zk will make equal angles with all three if
it has equal dot products with all three, that is, if
2x C y " 2z D x C 2y " 2z
2x C y " 2z D 2x " 2y C z
,
,
xDyD0
3y " 3z D 0:
Thus x D y D z. Two unit vectors satisfying this condition are
$
#
1
1
1
xD˙ p iC p jC p k :
3
3
3
25.
Let uO D u=juj and vO D v=jvj.
Then uO C vO bisects the angle between u and v. A unit
vector which bisects this angle is
u
v
C
uO C vO
juj
jvj
ˇ
D ˇ
ˇ u
v ˇˇ
juO C vO j
ˇ
C
ˇ juj
jvj ˇ
jvju C jujv
ˇ:
D ˇ
ˇ
ˇ
ˇjvju C jujvˇ
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SECTION 10.2 (PAGE 583)
ADAMS and ESSEX: CALCULUS 9
Similarly, r + v D b and r + w D c.
u
uO
31.
uO C vO
vO
w+a
a, (the vector projection of w along a). Let
jaj2
v D w " u. Then w D u C v. Clearly u is parallel to a,
and
Let u D
v
v+aDw+a"
Fig. 10.2-25
w+a
a + a D w + a " w + a D 0;
jaj2
so v is perpendicular to a.
26. If u and v are not parallel, then neither is the zero vector,
and the origin and the two points with position vectors u
and v lie on a unique plane. The equation r D ,u C -v
(,; - real) gives the position vector of an arbitrary point
on that plane.
27.
w
v
a
a) ju C vj2 D .u C v/ + .u C v/
Du+uCu+vCv+uCv+v
u
D juj2 C 2u + v C jvj2 :
Fig. 10.2-31
b) If & is the angle between u and v, then cos & $ 1, so
u + v D jujjvj cos & $ jujjvj:
32.
c) ju C vj2 D juj2 C 2u + v C jvj2
$ juj2 C 2jujjvj C jvj2
Let nO be a unit vector that is perpendicular to u and lies
in the plane containing the origin and the points U , V ,
and P . Then uO D u=juj and nO constitute a standard basis in that plane, so each of the vectors v and r can be
expressed in terms of them:
D .juj C jvj/2 :
Thus ju C vj $ juj C jvj.
28.
v D s uO C t nO
O
r D x uO C y n:
a) u, v, and u C v are the sides of a triangle. The triangle inequality says that the length of one side cannot
exceed the sum of the lengths of the other two sides.
Since v is not parallel to u, we have t ¤ 0. Thus
O and
nO D .1=t /.v " s u/
b) If u and v are parallel and point in the same direction, (or if at least one of them is the zero vector),
then ju C vj D juj C jvj.
29. u D 53 i C 45 j, v D 54 i " 35 j, w D k.
q
q
9
16
a) juj D 25
C 16
D
1,
jvj
D
25
25 C
u+v D
12
25
"
12
25
9
25
D 1, jwj D 1,
D 0, u + w D 0, v + w D 0.
r D x uO C
where , D .tx " ys/=.t juj/ and - D y=t .
33.
Let jaj2 " 4rst D K 2 , where K > 0. Now
jaj2 D a + a D .rx C sy/ + .rx C sy/
b) If r D xi C yj C zk, then
.r + u/u C .r + v/v C .r + w/w
$#
$
#
4
3
4
3
xC y
iC j
D
5
5
5
5
#
$#
$
4
3
4
3
C
x" y
i " j C zk
5
5
5
5
16y C 9y
9x C 16x
iC
j C zk
D
25
25
D xi C yj C zk D r:
30. Suppose juj D jvj D jwj D 1, and u+v D u+w D v+w D 0,
and let r D au C bv C ww. Then
D r 2 jxj2 C s 2 jyj2 C 2rsx + y
K 2 D jaj2 " 4rsx + y
D jrx " syj2
(since x + y D t ).
O for some unit vector u.
O
Therefore rx " sy D K u,
Since rx C sy D a, we have
2rx D a C K uO
O
2sy D a " K u:
Thus
r + u D au + u C bv + u C cw + u D ajuj2 C 0 C 0 D a:
400
y
O D ,u C -v;
.v " s u/
t
xD
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a C K uO
;
2r
yD
a " K uO
;
2s
lOMoARcPSD|6566483
INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 10.3
p
where K D jaj2 " 4rst, and uO is any unit vector. (The
solution is not unique.)
(PAGE 592)
The length of the cable between x D 0 and x D 10 m is
Z 10 q
LD
1 C sinh2 .ax/ dx
0
ˇ10
Z 10
ˇ
1
ˇ
D
cosh.ax/ dx D sinh.ax/ˇ
ˇ
a
0
34. The derivation of the equation of the hanging cable given
in the text needs to be modified by replacing W D "ıgsj
with W D "ıgxj. Thus Tv D ıgx, and the slope of the
cable satisfies
dy
ıgx
D
D ax
dx
H
0
1
D sinh.10a/ ' 12:371 m:
a
where a D ıg=H . Thus
yD
1 2
ax C C I
2
Section 10.3 The Cross Product in 3-Space
(page 592)
the cable hangs in a parabola.
1
35. If y D cosh.ax/, then y 0 D sinh.ax/, so
a
Z
Z x
q
sD
1 C sinh2 .au/ du D
cosh.au/ du
0
0
ˇx
sinh.au/ ˇˇ
1
D
ˇ D sinh.ax/:
ˇ
a
a
1.
.i " 2j C 3k/ & .3i C j " 4k/ D 5i C 13j C 7k
2.
.j C 2k/ & ."i " j C k/ D 3i " 2j C k
3.
If A D .1; 2; 0/, B D .1; 0; 2/, and C D .0; 3; 1/, then
""!
""!
AB D "2j C 2k, AC D "i C j C k, and the area of triangle
ABC is
x
""! ""!
p
j " 4i " 2j " 2kj
jAB & AC j
D
D 6 sq. units.
2
2
0
As shown in the text, the tension T at P has horizontal
ıg
and vertical components that satisfy Th D H D
and
a
ıg
Tv D ıgs D
sinh.ax/. Hence
a
q
jTj D
50 D
45
0
."ai C bj/ & ."ai C ck/ D bci C acj C abk:
A unit vector in this direction is
p
&
1%
cosh.45a/ " 1 ' 19:07 m:
a
1
cosh.ax/.
a
At the point P where x D 10 m, the slope of the cable is
sinh.10a/ D tan.55ı /. Thus
37. The equation of the cable is of the form y D
bci C acj C abk
b 2 c 2 C a2 c 2 C a2 b 2
:
1p 2 2
b c C a2 c 2 C a2 b 2 .
2
A vector perpendicular to i C j and j C 2k is
The triangle has area
5.
˙.i C j/ & .j C 2k/ D ˙.2i " 2j C k/;
q
1
1 C sinh2 .ax/ dx D sinh.45a/:
a
The equation sinh.45a/ D 50a has approximate solution
a ' 0:0178541. The vertical distance between the lowest
point on the cable and the support point is
aD
A vector perpendicular to the plane containing the three
given points is
ıg
cosh.ax/ D ıgy:
Th2 C Tv2 D
a
1
36. The cable hangs along the curve y D cosh.ax/, and its
a
length from the lowest point at x D 0 to the support tower
at x D 45 m is 50 m. Thus
Z
4.
which has length 3. A unit vector in that direction is
#
$
2
2
1
˙
i" jC k :
3
3
3
6.
A vector perpendicular to u D 2i " j " 2k and to
v D 2i " 3j C k is the cross product
ˇ
ˇ
ˇi
j
k ˇˇ
ˇ
u & v D ˇˇ 2 "1 "2 ˇˇ D "7i " 6j " 4k;
ˇ 2 "3 1 ˇ
p
which has length 101. A unit vector with positive k
component that is perpenducular to u and v is
1
sinh!1 .tan.55ı / ' 0:115423:
10
1
"1
p
u&v D p
.7i C 6j C 4k/:
101
101
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SECTION 10.3 (PAGE 592)
ADAMS and ESSEX: CALCULUS 9
Since u makes zero angle with itself, ju & uj D 0 and
u & u D 0.
ˇ
ˇ
ˇ i
j
k ˇˇ
ˇ
8. u & v D ˇˇ u1 u2 u3 ˇˇ
ˇ v1 v2 v3 ˇ
ˇ
ˇ
ˇ i
j
k ˇˇ
ˇ
D " ˇˇ v1 v2 v3 ˇˇ D "v & u:
ˇ u1 u2 u3 ˇ
7.
9.
10.
13.
Suppose that u C v C w D 0. Then
u & v C v & v C w & v D 0 & v D 0:
Thus u & v C w & v D 0.
Thus u & v D "w & v D v & w.
By symmetry, we also have v & w D w & u.
14.
ˇ
ˇ
ˇ
ˇ
i
j
k
ˇ
ˇ
.u C v/ & w D ˇˇ u1 C v1 u2 C v2 u3 C v3 ˇˇ
ˇ w1
w2
w3 ˇ
ˇ
ˇ
ˇ ˇ
ˇ i
j
k ˇˇ ˇˇ i
j
k ˇˇ
ˇ
D ˇˇ u1 u2 u3 ˇˇ C ˇˇ v1 v2 v3 ˇˇ
ˇ w1 w2 w3 ˇ ˇ w1 w2 w3 ˇ
The base of the tetrahedron is a triangle spanned by v and
w, which has area
AD
The altitude h of the tetrahedron (measured perpendicular
to the plane of the base) is equal to the length of the projection of u onto the vector v & w (which is perpendicular
to the base). Thus
D u & w C v & w:
ˇ
ˇ
ˇ i
j
k ˇˇ
ˇ
.t u/ & v D ˇˇ t u1 t u2 t u3 ˇˇ
ˇ v1
v2
v3 ˇ
ˇ
ˇ
ˇ i
j
k ˇˇ
ˇ
D t ˇˇ u1 u2 u3 ˇˇ D t .u & v/;
ˇ v1 v2 v3 ˇ
hD
u + .u & v/
ˇ
ˇ
ˇ
ˇ u2 u3 ˇ
ˇ
ˇ " u2 ˇ u1
D u1 ˇˇ
ˇ v1
v2 v3 ˇ
ˇ
ˇ
ˇ u1
u3 ˇˇ
ˇ
C
u
3ˇ
v3 ˇ
v1
D u1 u2 v3 " u1 v2 u3 " u2 u1 v3
C u2 v1 u3 C u3 u1 v2 " u3 v1 u2 D 0;
v + .u & v/ D "v + .v & u/ D 0:
V D
u
h
h
ˇ
k ˇˇ
0 ˇˇ D .sin ˛ cos ˇ " cos ˛ sin ˇ/k:
0ˇ
Because v is displaced counterclockwise from u, the cross
product above must be in the positive k direction. Therefore its length is the k component. Therefore
sin.˛ " ˇ/ D sin ˛ cos ˇ " cos ˛ sin ˇ:
402
w
v
Fig. 10.3-14
15.
But
j
sin ˇ
sin ˛
1
1
Ah D ju + .v & w/j
3
6 ˇ
ˇ
ˇu
u2 u3 ˇˇ
1 ˇˇ 1
D j ˇ v1 v2 v3 ˇˇ j:
6 ˇw w w ˇ
1
2
3
u&v
ˇ
u2 ˇˇ
v2 ˇ
12. Both u D cos ˇ i C sin ˇ j and v D cos ˛ i C sin ˛ j
are unit vectors. They make angles ˇ and ˛, respectively,
with the positive x-axis, so the angle between them is
j˛ " ˇj D ˛ " ˇ, since we are told that 0 $ ˛ " ˇ $ #.
They span a parallelogram (actually a rhombus) having
area
ju & vj D jujjvj sin.˛ " ˇ/ D sin.˛ " ˇ/:
ˇ
ˇ i
ˇ
u & v D ˇˇ cos ˇ
ˇ cos ˛
ju + .v & w/j
:
jv & wj
The volume of the tetrahedron is
u & .t v/ D ".t v/ & u
D "t .v & u/ D t .u & v/:
11.
1
jv & wj:
2
16.
The tetrahedron with vertices .1; 0; 0/, .1; 2; 0/, .2; 2; 2/,
and .0; 3; 2/ is spanned by u D 2j, v D i C 2j C 2k, and
w D "i C 3j C 2k. By Exercise 14, its volume is
ˇ
ˇ
ˇ 0 2 0ˇ
ˇ
4
1 ˇˇ
V D j ˇ 1 2 2 ˇˇ j D cu. units.
6 ˇ "1 3 2 ˇ
3
Let the cube be as shown in the figure. The required parallelepiped is spanned by ai C aj, aj C ak, and ai C ak. Its
volume is
ˇ
ˇ
ˇa a 0ˇ
ˇ
ˇ
V D j ˇˇ 0 a a ˇˇ j D 2a3 cu. units.
ˇa 0 aˇ
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 10.3
(PAGE 592)
z
20.
.0;a;a/
If v & w ¤ 0, then .v & w/ + .v & w/ ¤ 0. By the previous
exercise, there exist constants ,, - and 0 such that
.a;0;a/
u D ,v C -w C 0.v & w/:
But v & w is perpendicular to both v and w, so
u + .v & w/ D 0 C 0 C 0.v & w/ + .v & w/:
If u + .v & w/ D 0, then 0 D 0, and
y
x
u D ,v C -w:
.a;a;0/
Fig. 10.3-16
17.
The points A D .1; 1; "1/, B D .0; 3; "2/, C D ."2; 1; 0/,
""! ""! ""!
and D D .k; 0; 2/ are coplanar if .AB & AC / + AD D 0.
Now
ˇ
ˇ i
j
""! ""! ˇˇ "1 2
AB & AC D ˇ
ˇ "3 0
ˇ
k ˇˇ
"1 ˇˇ D 2i C 4j C 6k:
1 ˇ
Thus the four points are coplanar if
21.
u D i C 2j C 3k
v D 2i " 3j
wDj"k
u & .v & w/ D u & .3i C 2j C 2k/ D "2i C 7j " 4k
.u & v/ & w D .9i C 6j " 7k/ & w D i C 9j C 9k:
u & .v & w/ lies in the plane of v and w;
.u & v/ & w lies in the plane of u and v.
22.
u + v & w makes sense in that it must mean u + .v & w/.
(.u + v/ & w makes no sense since it is the cross product of
a scalar and a vector.)
2.k " 1/ C 4.0 " 1/ C 6.2 C 1/ D 0;
that is, if k D "6.
18.
23.
ˇ
ˇ
ˇ u1 u2 u3 ˇ
ˇ
ˇ
u + .v & w/ D ˇˇ v1 v2 v3 ˇˇ
ˇ w1 w2 w3 ˇ
ˇ
ˇ
ˇ v1 v2 v3 ˇ
ˇ
ˇ
D " ˇˇ u1 u2 u3 ˇˇ
ˇ w1 w2 w3 ˇ
ˇ
ˇ
ˇ v1 v2 v3 ˇ
ˇ
ˇ
D ˇˇ w1 w2 w3 ˇˇ
ˇ u1 u2 u3 ˇ
D v + .w & u/
D w + .u & v/
u&v&w makes no sense. It is ambiguous, since .u&v/&w
and u & .v & w/ are not in general equal.
v D v1 i;
u & .v & w/ D .u1 i C u2 j C u3 k/ & .v1 w2 k/
D u1 v1 w2 i & k C u2 v1 w2 j & k
D "u1 v1 w2 i " u1 v1 w2 j:
But
.by symmetry/:
x + .v & w/
D ,u + .v & w/ C -v + .v & w/ C 0w + .v & w/
D ,u + .v & w/:
,D
x + .v & w/
:
u + .v & w/
Since u + .v & w/ D v + .w & u/ D w + .u & v/, we have, by
symmetry,
-D
x + .w & u/
;
u + .v & w/
0D
w D w1 i C w2 j:
Thus v & w D v1 w2 i & j D v1 w2 k, and
.u + w/v " .u + v/w
D .u1 w1 C u2 w2 /v1 i " u1 v1 .w1 i C w2 j/
D u2 v1 w2 i " u1 v1 w2 j:
19. If u + .v & w/ ¤ 0, and x D ,u C -v C 0w, then
Thus
As suggested in the hint, let the x-axis lie in the direction
of v, and let the y-axis be such that w lies in the xyplane. Thus
x + .u & v/
:
u + .v & w/
Thus u & .v & w/ D .u + w/v " .u + v/w.
24.
If u, v, and w are mutually perpendicular, then v & w is
parallel to u, so u & .v & w/ D 0. In this case,
u + .v & w/ D ˙jujjvjjwj; the sign depends on whether u
and v & w are in the same or opposite directions.
25.
Applying the result of Exercise 23 three times, we obtain
u & .v & w/ C v & .w & u/ C w & .u & v/
D .u + w/v " .u + v/w C .v + u/w " .v + w/u
C .w + v/u " .w + u/v
D 0:
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SECTION 10.3 (PAGE 592)
ADAMS and ESSEX: CALCULUS 9
26. If a D "i C 2j C 3k and x D xi C yj C zk, then
ˇ
ˇ i
ˇ
a & x D ˇˇ "1
ˇ x
j
2
y
ˇ
k ˇˇ
3 ˇˇ
zˇ
D .2z " 3y/i C .3x C z/y " .y C 2x/k
D i C 5j " 3k;
3.
The plane through the origin having normal i " j C 2k has
equation x " y C 2z D 0.
4.
The plane passing through .1; 2; 3/, parallel to the plane
3x C y " 2z D 15, has equation 3z C y " 2z D 3 C 2 " 6,
or 3x C y " 2z D "1.
5.
The plane through .1; 1; 0/, .2; 0; 2/, and .0; 3; 3/ has normal
.i " j C 2k/ & .i " 2j " 3k/ D 7i C 5j " k:
It therefore has equation
provided 2z " 3y D 1, 3x C z D 5, and "y " 2x D "3.
This system is satisfied by x D t , y D 3 " 2t , z D 5 " 3t ,
for any real number t . Thus
x D t i C .3 " 2t /j C .5 " 3t /k
7.x " 1/ C 5.y " 1/ " .z " 0/ D 0;
or 7x C 5y " z D 12.
6.
The plane passing through ."2; 0; 0/, .0; 3; 0/, and .0; 0; 4/
has equation
x
y
z
C C D 1;
"2
3
4
or 6x " 4y " 3z D "12.
7.
The normal n to a plane through .1; 1; 1/ and .2; 0; 3/
must be perpendicular to the vector i " j C 2k joining
these points. If the plane is perpendicular to the plane
x C 2y " 3z D 0, then n must also be perpendicular to
i C 2j " 3k, the normal to this latter plane. Hence we can
use
gives a solution of a & x D i C 5j " 3k for any t . These
solutions constitute a line parallel to a.
27.
Let a D "i C 2j C 3k and b D i C 5j. If x is a solution of
a & x D b, then
a + b D a + .a & x/ D 0:
However, a + b ¤ 0, so there can be no such solution x.
28. The equation a & x D b can be solved for x if and only
if a + b D 0. The “only if” part is demonstrated in the
previous solution. For the “if” part, observe that if
a + b D 0 and x0 D .b & a/=jaj2 , then by Exercise 23,
a & x0 D
n D .i " j C 2k/ & .i C 2j " 3k/ D "i C 5j C 3k:
The plane has equation
.a + a/b " .a + b/a
1
a & .b & a/ D
D b:
jaj2
jaj2
The solution x0 is not unique; as suggested by the example
in Exercise 26, any multiple of a can be added to it and
the result will still be a solution. If x D x0 C t a, then
".x " 1/ C 5.y " 1/ C 3.z " 1/ D 0;
or x " 5y " 3z D "7.
8.
a & x D a & x0 C t a & a D b C 0 D b:
Since ."2; 0; "1/ does not lie on x " 4y C 2z D "5, the
required plane will have an equation of the form
2x C 3y " z C ,.x " 4y C 2z C 5/ D 0
for some ,. Thus
Section 10.4 Planes and Lines
1.
2
2
2
(page 599)
"4 C 1 C ,."2 " 2 C 5/ D 0;
2
a) x Cy Cz D z represents a line in 3-space, namely
the z-axis.
b) x C y C z D x C y C z is satisfied by every point in
3-space.
c) x 2 C y 2 C z 2 D "1 is satisfied by no points in (real)
3-space.
2. The plane through .0; 2; "3/ normal to 4i " j " 2k has
equation
4.x " 0/ " .y " 2/ " 2.z C 3/ D 0;
so , D 3. The required plane is 5x " 9y C 5z D "15.
9.
A plane through the line x Cy D 2, y "z D 3 has equation
of the form
x C y " 2 C ,.y " z " 3/ D 0:
This plane will be perpendicular to 2x C 3y C 4z D 5 if
.2/.1/ C .1 C ,/.3/ " .,/.4/ D 0;
that is, if , D 5. The equation of the required plane is
x C 6y " 5z D 17:
or 4x " y " 2z D 4.
404
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 10.4
10. Three distinct points will not determine a unique plane
through them if they all lie on a straight line. If the points
have position vectors r1 , r2 , and r3 , then they will all lie
on a straight line if
.r2 " r1 / & .r3 " r1 / D 0:
11.
or in standard form
y
z"1
xC1
D
D
:
2
"1
7
17.
i
.r2 " r1 / + .r3 " r1 / & .r4 " r1 / D 0
.i C 2j " k/ & .2i " j C 4k/ D 7i " 6j " 5k:
(or if they satisfy any similar such condition that asserts
that the tetrahedron whose vertices they are has zero volume).
Since the line passes through the origin, it has equations
r D 7t i " 6t j " 5t k
x D 7t; y D "6t;
y
z
x
D
D
7
"6
"5
12. x C y C z D , is the family of all (parallel) planes normal
to the vector i C j C k.
The distance from the planes
,x C
p
p
1 " ,2 y D 1
(standard form).
Since the line passes through .2; "1; "1/, its equations
are, in vector parametric form
r D .2 C t /i " .1 C t /j " .1 C t /k;
or in scalar parametric form
r D .1 C 2t /i C .2 " 3t /j C .3 " 4t /k;
x D 2 C t;
or in scalar parametric form by
y D 2 " 3t;
(vector parametric)
(scalar parametric)
.i C j/ & .i " j C 2k/ D 2.i " j " k/:
The line through .1; 2; 3/ parallel to 2i " 3j " 4k has
equations given in vector parametric form by
x D 1 C 2t;
z D "5t
18. A line parallel to x C y D 0 and to x " y C 2z D 0 is
parallel to the cross product of the normal vectors to these
two planes, that is, to the vector
to the origin is 1= ,2 C 1 " ,2 D 1. Hence the equation
represents the family of all vertical planes at distance 1
from the origin. All such planes are tangent to the cylinder x 2 C y 2 D 1.
15.
2x " y C 4z D 5
is parallel to the vector
h
14.
A line parallel to the line with equations
x C 2y " z D 2;
If the four points have position vectors ri , .1 $ i $ 4/,
then they are coplanar if, for example,
13. x C ,y C ,z D , is the family of all planes containing the
line of intersection of the planes x D 0 and y C z D 1,
except the plane y C z D 1 itself. All these planes pass
through the points .0; 1; 0/ and .0; 0; 1/.
(PAGE 599)
y D ".1 C t /;
z D ".1 C t /;
or in standard form
z D 3 " 4t;
x " 2 D ".y C 1/ D ".z C 1/:
or in standard form by
19.
y"2 z"3
x"1
D
"
:
2
"3
"4
16. The line through ."1; 0; 1/ perpendicular to the plane
2x " y C 7z D 12 is parallel to the normal vector
2i " j C 7k to that plane. The equations of the line are, in
vector parametric form,
r D .1 C t /i C .2 C t /j C ."1 C t /k
(vector parametric)
x D 1 C t; y D 2 C t; z D "1 C t (scalar parametric)
x"1Dy "2DzC1
(standard form).
r D ."1 C 2t /i " t j C .1 C 7t /k;
20.
or in scalar parametric form,
x D "1 C 2t;
y D "t;
z D 1 C 7t;
A line making equal angles with the positive directions
of the coordinate axes is parallel to the vector i C j C k.
If the line passes through the point .1; 2; "1/, then it has
equations
The line r D .1 " 2t /i C .4 C 3t /j C .9 " 4t /k has standard
form
y "4
z"9
x"1
D
D
:
"2
3
"4
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SECTION 10.4 (PAGE 599)
21.
The line
(
x D 4 " 5t
has standard form
y D 3t
zD7
x"4
y
D ;
"5
3
22. The line
'
ADAMS and ESSEX: CALCULUS 9
26.
p
z D 7:
27.
x " 2y C 3z D 0
is parallel to the vector
2x C 3y " 4z D 4
4
12
28.
A vector parallel to the line x Cy Cz D 0, 2x "y "5z D 1
is
We need a point on this line: if z D 0 then x C y D 0
and 2x " y D 1, so x D 1=3 and y D "1=3. The position
vector of this point is
2x C 3y D 4:
r1 D
29.
23. The equations
8̂
ˆ
ˆ
ˆ
ˆ
<
x D x1 C t .x2 " x1 /
ˆ
ˆ
ˆ
ˆ y D y1 C t .y2 " y1 /
:̂
z D z1 C t .z2 " z1 /
'
x C 2y D 3
contains the points .1; 1; 1/ and
y C 2z D 3
1
.3; 0; 3=2/, so is parallel to the vector 2i " j C k, or to
2
4i " 2j Cnk.
xCy Cz D6
The line
contains the points ."5; 11; 0/
x " 2z D "5
and ."1; 5; 2/, and so is parallel to the vector 4i " 6j C 2k,
or to 2i " 3j C k.
Using the values
The line
r1 D i C j C k
r2 D "i C 5j C 2k
certainly represent a straight line. Since
.x; y; z/ D .x1 ; y1 ; z1 / if t D 0, and
.x; y; z/ D .x2 ; y2 ; z2 / if t D 1, the line must pass through
P1 and P2 .
j.r1 " r2 / + .a1 & a2 /j
ja1 & a2 j
j.2i " 4j " k/ + .i " 2j " 8k/j
D
ji " 2j " 8kj
18
D p
units.
69
sD
30.
y C3
z"1
D
passes through the point
2
4
.2; "3; 1/, and is parallel to a D i C 2j C 4k.
The plane 2y " z D 1 has normal n D 2j " k.
Since a + n D 0, the line is parallel to the plane.
The distance from the line to the plane is equal to the
distance from .2; "3; 1/ to the plane 2y " z D 1, so is
The line x " 2 D
(Other similar answers are possible.)
406
a1 D 4i " 2j C k
a2 D 2i " 3j C k;
we calculate the distance between the two lines by the
formula in Section 10.4 as
The point on the line corresponding to t D "1 is the point
P3 such that P1 is midway between P3 and P2 .
The point on the line corresponding to t D 1=2 is the
midpoint between P1 and P2 .
The point on the line corresponding to t D 2 is the point
P4 such that P2 is the midpoint between P1 and P4 .
.r2 " r1 / & .r4 " r3 / ¤ 0; and
h
i
.r3 " r1 / + .r2 " r1 / & .r4 " r3 / D 0:
1
1
i " j:
3
3
The distance from the origin to the line is
r
jr1 & aj
ji C j C kj
3
D
sD
D
units.
p
jaj
74
74
4
8
y"
7 D
7 D z;
"1
10
7
x"
25. Let ri be the position vector of Pi (1 $ i $ 4). The
line P1 P2 intersects the line P3 P4 in a unique point if
the four points are coplanar, and P1 P2 is not parallel to
P3 P4 . It is therefore sufficient that
4
units:
D p
14
The distance from .1; 2; 0/ to 3x " 4y " 5z D 2 is
The solution of this system is y D 4=7, x D 8=7. A
possible standard form for the given line is
24.
C
32
a D .i C j C k/ & .2i " j " 5k/ D "4i C 7j " 3k:
We need a point on this line. Putting z D 0, we get
though, of course, this answer is not unique as the coordinates of any point on the line could have been used.
C
22
j3 " 8 " 0 " 2j
7
p
D p units:
5 2
32 C 42 C 52
.i " 2j C 3k/ & .2i C 3j " 4k/ D "i C 10j C 7k:
x " 2y D 0;
The distance from .0; 0; 0/ to x C 2y C 3z D 4 is
DD
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j " 6 " 1 " 1j
p
D p units.
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 10.5
31. .1 " ,/.x " x0 / D ,.y " y0 / represents any line in the
xy-plane passing through .x0 ; y0 /. Therefore, in 3-space
the pair of equations
.1 " ,/.x " x0 / D ,.y " y0 /;
4.
.y " 1/2
z2
.x C 2/2
C
C
D1
42
22
.4=3/2
This is an ellipsoid with centre ."2; 1; 0/ and semi-axes 4,
2, and 4/3.
z D z0
x " x0
y " y0
p
D
D z " z0 represents all lines through
2
,
1",
.x0 ; y0 ; z0 / parallel to the vectors
aD
p
x 2 C 4y 2 C 9z 2 C 4x " 8y D 8
.x C 2/2 C 4.y " 1/2 C 9z 2 D 8 C 8 D 16
represents all straight lines in the plane z D z0 which pass
through the point .x0 ; y0 ; z0 /.
32.
(PAGE 603)
5.
1 " ,2 i C ,j C k:
All such lines are generators of the circular cone
z D x 2 C 2y 2 represents an elliptic paraboloid with vertex
at the origin and axis along the positive z-axis. Crosssections
in
p
p planes z D k > 0 are ellipses with semi-axes
k and k=2.
z
zDx 2 C2y 2
.z " z0 /2 D .x " x0 /2 C .y " y0 /2 ;
so the given equations specify all straight lines lying on
that cone.
33. The equation
.A1 x C B1 y C C1 z C D1 /.A2 x C B2 y C C2 z C D2 / D 0
is satisfied if either A1 x C B1 y C C1 z C D1 D 0 or
A2 x C B2 y C C2 z C D2 D 0, that is, if .a; y; z/ lies on
either of these planes. It is not necessary that the point
lie on both planes, so the given equation represents all the
points on each of the planes, not just those on the line of
intersection of the planes.
Fig. 10.5-5
6.
Section 10.5 Quadric Surfaces
1.
y
x
(page 603)
z D x 2 " 2y 2 represents a hyperbolic paraboloid.
z
zDx 2 !2y 2
x 2 C 4y 2 C 9z 2 D 36
x2
y2
z2
C
C
D1
62
32
22
This is an ellipsoid with centre at the origin and semi-axes
6, 3, and 2.
2. x 2 Cy 2 C4z 2 D 4 represents an oblate spheroid, that is, an
ellipsoid with its two longer semi-axes equal. In this case
the longer semi-axes have length 2, and the shorter one
(in the z direction) has length 1. Cross-sections in planes
perpendicular to the z-axis between z D "1 and z D 1 are
circles.
3.
y
x
Fig. 10.5-6
2x 2 C 2y 2 C 2z 2 " 4x C 8y " 12z C 27 D 0
2.x 2 " 2x C 1/ C 2.y 2 C 4y C 4/ C 2.z 2 " 6z C 9/
D "27 C 2 C 8 C 18
1
.x " 1/2 C .y C 2/2 C .z " 3/2 D
2
p
This is a sphere with radius 1= 2 and centre .1; "2; 3/.
7.
x 2 " y 2 " z 2 D 4 represents a hyperboloid of two sheets
with vertices at .˙2; 0; 0/ and circular cross-sections in
planes x D k, where jkj > 2.
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SECTION 10.5 (PAGE 603)
ADAMS and ESSEX: CALCULUS 9
z
z
x 2 !y 2 !z 2 D4
x 2 C4z 2 D4
x
y
y
x
Fig. 10.5-10
Fig. 10.5-7
8. "x 2 C y 2 C z 2 D 4 represents a hyperboloid of one sheet,
with circular cross-sections in all planes perpendicular to
the x-axis.
11.
x 2 " 4z 2 D 4 represents a hyperbolic cylinder with axis
along the y-axis.
z
z
!x 2 Cy 2 Cz 2 D4
y
y
x
x
x
y
x 2 !4z 2 D4
Fig. 10.5-11
Fig. 10.5-8
9. z D xy represents a hyperbolic paraboloid containing the
x- and y-axes.
12. y D z 2 represents a parabolic cylinder with vertex line
along the x-axis.
z
z
yDz 2
y
y
x
zDxy
x
Fig. 10.5-12
Fig. 10.5-9
2
2
10. x C4z D 4 represents an elliptic cylinder with axis along
the y-axis.
408
13.
#
$
1 2 1
" represents a parabolic cylinder
x D z 2 Cz D z C
2
4
with vertex line along the line z D "1=2, x D "1=4.
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 10.5
(PAGE 603)
z
z
.z!1/2 D.x!2/2 C.y!3/2 C4
xDz 2 Cz
.2;3;1/
y
y
x
x
Fig. 10.5-16
Fig. 10.5-13
14.
x 2 D y 2 C 2z 2 represents an elliptic cone with vertex at
the origin and axis along the x-axis.
17.
z
x 2 Dy 2 C2z 2
'
x2 C y2 C z2 D 4
represents the circle of intersection of
xCy Cz D1
a sphere and a plane. The circle lies in the plane
x
pC y C z D 1,pand has centre .1=3; 1=3; 1=3/ and radius
4 " .3=9/ D 11=3.
z
%
xCyCzD1
x
1 1 1
3;3;3
&
y
x 2 Cy 2 Cz 2 D4
x
y
Fig. 10.5-14
15.
.z " 1/2 D .x " 2/2 C .y " 3/2 represents a circular
cone with axis along the line x D 2, y D 3, and vertex at
.2; 3; 1/
Fig. 10.5-17
z
.z!1/2 D.x!2/2 C.y!3/2
18.
'
x2 C y2 D 1
is the ellipse of intersection of the plane
z DxCy
z D x C y and the circular cylinder x 2 C y 2 D 1. The
centre of the ellipse p
is at the
p origin,
p and the ends of the
major axis are ˙.1= 2; 1= 2; 2/.
z
.2;3;1/
y
zDxCy
x
y
x 2 Cy 2 D1
Fig. 10.5-15
x
16. .z " 1/2 D .x " 2/2 C .y " 3/2 C 4 represents a hyperboloid
of two sheets with centre at .2; 3; 1/, axis along the line
x D 2, y D 3, and vertices at .2; 3; "1/ and .2; 3; 3/.
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Fig. 10.5-18
409
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SECTION 10.5 (PAGE 603)
19.
ADAMS and ESSEX: CALCULUS 9
'
z 2 D x 2 C y 2 is the parabola in which the plane
z D1Cx
z D 1 C x intersects the circular cone z 2 D x 2 C y 2 . (It
is a parabola because the plane is parallel to a generator
of the cone, namely the line z D x, y D 0.) The vertex
of the parabola is ."1=2; 0; 1=2/, and its axis is along the
line y D 0, z D 1 C x.
z
Family 2:
The cylinder 2x 2 C y 2 D 1 intersects horizontal planes
p
in ellipses with semi-axes 1 in the y direction and 1= 2
in the x direction. Tilting the plane in the x direction
will cause the shorter semi-axis to increase in length.
The plane z D cx intersects the cylinder in an ellipse
with principal
axes
p
p through the points .0; ˙1; 0/ and
.˙1= 2; 0; ˙c= 2/. The semi-axes will be equal (and
the ellipse will be a circle) if .1=2/ C .c 2 =2/ D 1, that is,
if c D ˙1. Thus cross-sections of the cylinder perpendicular to the vectors a D i ˙ k are circular.
24.
The plane z D cx C k intersects the elliptic cone
z 2 D 2x 2 C y 2 on the cylinder
zD1Cx
c 2 x 2 C 2ckx C k 2 D 2x 2 C y 2
.2 " c 2 /x 2 " 2ckx C y 2 D k 2
#
$2
ck
c2k2
2k 2
D
.2 " c 2 / x "
C y2 D k2 C
2
2
2"c
2"c
2 " c2
2
2
.x " x0 /
y
C 2 D 1;
a2
b
y
Fig. 10.5-19
20.
'
x 2 C 2y 2 C 3z 2 D 6
is an ellipse in the plane
yD1
y D 1. Its projection onto the xz-plane is the ellipse
x 2 C 3z 2 D 4. One quarter of the ellipse is shown in the
figure.
where x0 D
z
p
2
yD1
.x0 " a; 0; c.x0 " a/ C k/
and .x0 ; "b; cx0 C k/
3
x 2 C2y 2 C3z 2 D6
a2 C c 2 a2 D b 2 ;
Fig. 10.5-20
21.
x2
y2 z2
C 2 " 2 D1
2
a
b
c
y2
x2 z2
" 2 D1" 2
2
a
% x cz & % x bz & %
y&%
y&
C
"
D 1C
1"
a
c
a c
b
b
%
8 x
z
y&
<
C D, 1C
a% c &
b
Family 1:
: , x " z D 1" y.
a c
b
%
8 x
z
y&
<
C D- 1"
a% c &
b :
Family 2:
: - x " z D 1C y.
a c
b
22. z D xy
Family 1:
410
'
z D ,x
, D y:
to .x0 C a; 0; c.x0 C a/ C k/;
to .x0 ; b; cx0 C k/:
The centre of this ellipse is .x0 ; 0; cx0 C k/. The ellipse is
a circle if its two semi-axes have equal lengths, that is, if
y
p
6
2k 2
2k 2
ck
2
2
,
a
D
,
and
b
D
.
2 " c2
.2 " c 2 /2
2 " c2
As in the previous exercise, z D cx C k intersects the
cylinder (and hence the cone) in an ellipse with principal
axes joining the points
p
x
z D -y
- D x:
23.
z 2 Dx 2 Cy 2
x
n
that is,
2k 2
2k 2
D
;
2
2
.2 " c /
2 " c2
p
or 1 C c 2 D 2 " c 2 . Thus
p c D ˙1= 2. Apvector normal
to the plane z D ˙.x= 2/ C k is a D i ˙ 2k.
.1 C c 2 /
Section 10.6 Cylindrical and Spherical
Coordinates (page 607)
1.
Cartesian: .2; "2;
p 1/;
Cylindrical: Œ2 2; "#=4; 1!;
Spherical: Œ3; cos!1 .1=3/; "#=4!.
2.
Cylindrical: p
Œ2; #=6; "2!;
p
Cartesian: . 3; 1; "2!; Spherical: Œ2 2; 3#=4; #=6!.
3.
Spherical: Œ4;p
#=3; 2#=3!;
p
Cartesian: ." 3; 3; "2/; Cylindrical: Œ2 3; 2#=3; 2!.
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lOMoARcPSD|6566483
INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 10.7
4. Spherical: Œ1; /; & !; Cylindrical: Œr; #=4; r!.
4.
p
x D sin / cos & D r cos #=4 D r= 2
p
y D sin / sin & D r sin #=4 D r= 2
z D cos / D r:
Thus x D y, & D #=4,pand r D sin / D cos /. Hence
/ D #=4,
p so r D 1= 2. Finally: x D y D 1=2,
z D 1= 2.
p
Cartesian: .1=2; 1=2; 1= 2/.
5.
5. & D #=2 represents the half-plane x D 0, y > 0.
6. / D 2#=3 represents the lower half of the right-circular
cone with vertex at the origin, axis along the z-axis,
and semi-vertical
angle #=3. Its Cartesian equation is
p
z D " .x 2 C y 2 /=3.
7.
/ D #=2 represents the xy-plane.
8. R D 4 represents the sphere of radius 4 centred at the
origin.
9. r D 4 represents the circular cylinder of radius 4 with axis
along the z-axis.
10. R D z represents the positive half of the z-axis.
11.
w
y
x
z
$#
a
c
b
d
1
B0
T
B
AA D @
0
0
0
4
B3
DB
@2
1
0
1
B0
2
B
A D@
0
0
0
1
B0
DB
@0
0
1
1
0
0
3
3
2
1
1
1
0
0
2
1
0
0
0
$
D
1
1
1
0
2
2
2
1
1
1
1
0
3
2
1
0
#
aw C cx
ay C cz
10
1
1
B
1C
CB1
1A@1
1
1
1
1
1C
C
1A
1
10
1
1
B
1C
CB0
1A@0
0
1
1
4
3C
C
2A
1
bw C dx
by C dz
0
1
1
1
0
0
1
1
1
0
0C
C
0A
1
1
1
0
0
1
1
1
0
1
1
1C
C
1A
1
$
R D r represents the xy-plane.
12. R D 2x represents the half-cone with vertex at the origin, axis along the positive x-axis, and
p semi-vertical angle
#=3. Its Cartesian equation is x D .y 2 C z 2 /=3.
6.
2
13. If R D 2 cos /, then R D 2R cos /, so
x 2 C y 2 C z 2 D 2z
x 2 C y 2 C z 2 " 2z C 1 D 1
x 2 C y 2 C .z " 1/2 D 1:
Thus R D 2 cos / represents the sphere of radius 1 centred
at .0; 0; 1/.
14.
#
(PAGE 617)
r D 2 cos & ) x 2 C y 2 D r 2 D 2r cos & D 2x, or
.x " 1/2 C y 2 D 1. Thus the given equation represents the
circular cylinder of radius 1 with axis along the vertical
line x D 1, y D 0.
D ax 2 C by 2 C cz 2 C 2pxy C 2qxz C 2ryz
Section 10.7 A Little Linear Algebra
(page 617)
0
10
1 0
3 0 "2
2 1
6
1. @ 1 1 2 A @ 3 0 A D @ 5
"1 1 "1
0 "2
1
0
10
1 0
1 1 1
1 1 1
1
2. @ 0 1 1 A @ 0 1 1 A D @ 0
0 0 1
0 0 1
0
#
$#
$ #
a b
w x
aw C by
3.
D
c d
y z
cw C dy
1
7
"3 A
1
1
2 3
1 2A
0 1
7.
ax C bz
cx C dz
$
0 1
0
1
x
a p q
x D @y A;
A D @p b r A
z
q r c
0 1
0 2
1
x
x
xy xz
xxT D @ y A .x; y; z/ D @ xy y 2 yz A
z
xz yz z 2
0 1
x
xT x D .x; y; z/ @ y A D .x 2 C y 2 C z 2 /
z
0
10 1
a p q
x
T
x Ax D .x; y; z/ @ p b r A @ y A
q r c
z
0
1
ax C py C qz
D .x; y; z/ @ px C by C rz A
qx C ry C cz
ˇ
ˇ 2
ˇ
ˇ 4
ˇ
ˇ 1
ˇ
ˇ "2
ˇ
ˇ
ˇ
3 "1 0 ˇˇ
ˇ 4
2 1 ˇˇ
ˇ
0 2 1 ˇˇ
D "3 ˇˇ 1 "1 1 ˇˇ
0 "1 1 ˇˇ
ˇ "2 0 1 ˇ
0 0 1ˇ
ˇ
ˇ$
ˇ
ˇ
#
ˇ 4 1ˇ
ˇ 1 1ˇ
ˇ
ˇ
ˇ
ˇ
" 1ˇ
D "3 "2 ˇ
"2 1 ˇ
"2 1 ˇ
D 6.3/ C 3.6/ D 36
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411
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SECTION 10.7 (PAGE 617)
8.
ˇ
ˇ 1
ˇ
ˇ 1
ˇ
ˇ "2
ˇ
ˇ 3
ˇ
1 1 ˇˇ
3 4 ˇˇ
2 4 ˇˇ
2 "2 ˇ
ˇ
ˇ
ˇ
ˇ1 1
1 1 ˇˇ
1 ˇˇ
ˇ
3 4 ˇˇ C 2 ˇˇ 1 2
4 ˇˇ
ˇ 3 "3 "2 ˇ
2 "2 ˇ
ˇ
ˇ
ˇ1 1 1ˇ
ˇ
ˇ
" 4 ˇˇ 1 2 3 ˇˇ
ˇ 3 "3 2 ˇ
ˇ
ˇ
ˇ
ˇ1 1
1 1 ˇˇ
1 ˇˇ
ˇ
3 ˇˇ
1 2 ˇˇ C 2 ˇˇ 0 1
ˇ 0 "6 "5 ˇ
5 1ˇ
ˇ
ˇ
ˇ1 1
1 ˇˇ
ˇ
2 ˇˇ
" 4 ˇˇ 0 1
ˇ 0 "6 "1 ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ 1
ˇ 1
2 ˇˇ
3 ˇˇ
2 ˇˇ
ˇ
ˇ
C
2
"
4
ˇ "6 "5 ˇ
ˇ "6 "1 ˇ
1ˇ
1
2
0
"3
ˇ
ˇ 1
ˇ
D "2 ˇˇ 2
ˇ "3
ˇ
ˇ1
ˇ
D "2 ˇˇ 0
ˇ0
ˇ
ˇ1
D "2 ˇˇ
5
9.
ADAMS and ESSEX: CALCULUS 9
D "2."9/ C 2.13/ " 4.11/ D 0
ˇ
ˇ
ˇ a11 a12 a13 ! ! ! a1n ˇ
ˇ
ˇ
ˇ 0 a22 a23 ! ! ! a2n ˇ
ˇ
ˇ
ˇ 0
0 a33 ! ! ! a3n ˇˇ
ˇ
::
::
:: ˇ
ˇ ::
::
ˇ :
:
:
:
: ˇˇ
ˇ
ˇ 0
0
0
! ! ! ann ˇ
ˇ
ˇ
ˇ a22 a23 ! ! ! a2n ˇ
ˇ
ˇ
ˇ 0 a33 ! ! ! a3n ˇ
ˇ
ˇ
::
:: ˇ
D a11 ˇ ::
::
ˇ :
ˇ
:
:
:
ˇ
ˇ
ˇ 0
0
! ! ! ann ˇ
ˇ
ˇ
ˇ a33 ! ! ! a3n ˇ
ˇ
ˇ
:: ˇ
ˇ :
::
D a11 a22 ˇ ::
:
: ˇˇ
ˇ
ˇ 0
! ! ! ann ˇ
Generalization:
ˇ
ˇ 1
ˇ
ˇ x1
ˇ 2
ˇ x1
ˇ
ˇ ::
ˇ :
ˇ
ˇ x n!1
1
11.
ˇ
ˇ 1
ˇ
f .x; y; z/ D ˇˇ x
ˇ x2
Let A D
#
ˇ
1 ˇ
ˇ
xn ˇ
Y
2 ˇˇ
xn
.xj " xi /:
ˇD
:: ˇ
1$i<j
$n
ˇ
: ˇ
xnn!1 ˇ
!!!
!!!
!!!
::
:
!!!
$
#
$
#
b
` m
w
, CD
, BD
d
n p
y
a` C bn am C bp
.AB/C D
c` C d n cm C dp
#
a`w C bnw C amy C bpy
D
c`w C d nw C cmy C dpy
#
$#
a b
`w C my
A.BC/ D
c d
nw C py
#
a`w C amy C bnw C bpy
D
c`w C cmy C d nw C dpy
12. If A D
#
$
w x
y z
$
a`x C bnx C amz C bpz
c`x C d nx C cmz C dpz
$
`x C mz
nx C pz
$
a`x C amz C bnx C bpz
c`x C cmz C d nx C dpz
$
#
b
a
, then AT D
d
b
a
c
$
x
. Then
z
$#
Thus .AB/C D A.BC/:
$
c
, and
d
det.A/ D ad " bc D det.AT /:
We generalize this by induction.
Suppose det(BT )=det(B) for any .n " 1/ & .n " 1/ matrix,
where n # 3. Let
0
1
y
y2
a11
B a21
B
A D B ::
@ :
an1
ˇ
1 ˇˇ
z ˇˇ ;
z2 ˇ
then f is a polynomial of degree 2 in z.
Since f .x; y; x/ D 0 and f .x; y; y/ D 0, we must have
f .x; y; z/ D A.z " x/.z " y/ for some A independent of z.
But
ˇ
ˇ
ˇ 1
1 1 ˇˇ
ˇ
y 0 ˇˇ D xy.y " x/;
Axy D f .x; y; 0/ D ˇˇ x
ˇ x2 y2 0 ˇ
a12
a22
::
:
!!!
!!!
::
:
an2
!!!
1
a1n
a2n C
C
:: C
: A
ann
be an n & n matrix. If det(A) is expanded in minors about
the first row, and det(AT ) is expanded in minors about the
first column, the corresponding terms in these expansions
are equal by the induction hypothesis. (The .n"1/&.n"1/
matrices whose determinants appear in one expansion are
the transposes of those in the other expansion.) Therefore
det(AT )=det(A) for any square matrix A.
13.
Let A D
#
so A D y " x and
f .x; y; z/ D .y " x/.z " x/.z " y/:
412
a
c
1
x3
x32
::
:
x3n!1
#
D a11 a22 a33 ! ! ! ann
(or use induction on n)
ˇ
ˇ
ˇ1 1ˇ
ˇ D y " x. If
10. ˇˇ
x yˇ
1
x2
x22
::
:
x2n!1
a
c
b
d
$
and B D
AB D
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#
#
w
y
aw C by
cw C dy
$
x
. Then
z
ax C bz
cx C dz
$
:
lOMoARcPSD|6566483
INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 10.7
Therefore,
det.A/det.B/ D .ad " bc/.wz " xy/
D adwz " adxy " bcwz C bcxy
det.AB/ D .aw C by/.cx C dz/ " .ax C bz/.cw C dy/
D awcx C awdz C bycx C bydz
" axwc " axdy " bzcw " bzdy
D adwz " adxy " bcwz C bcxy
D det.A/det.B/:
14.
If A" D
#
A!" D
cos &
" sin &
#
$
sin &
, then
cos &
cos."& /
" sin."& /
and
A" A!" D
#
1
18. Let A D @ "1
2
AA!1 D I we
0
1
a
0 "1
!1
1 0 A, A D @ d
g
1 3
must have
a"g D1
"a C d D 0
2a C d C 3g D 0
b"hD0
"b C e D 1
2b C e C 3h D 0
A
D
1 0
0 1
$
#
cos &
sin &
" sin &
cos &
$
If D D ad " bc,
0
d
#
$
a b B
D
c d @ "c
19.
D
1
"ab C ab
#
$
1 0
C
D
D
A
"cb C ad
0 1
D
20.
16. Since D D detB D xy 2 " x 2 y D xy.y " x/, B is
nonsingular, and therefore invertible, provided x ¤ 0,
y ¤ 0, and x ¤ y. Using the formula for the inverse of a
general 2 & 2 matrix, we have
1
0 2
1 0
y
"1
"y
y
B x.y " x/ x.y " x/ C
B
D C
C
B!1 D @ D2
ADB
@
A
"x
1
"x
x
y.y " x/ y.y " x/
D
D
17.
1
Let A D @ 0
0
!1
AA D I we
1
0
1 1
a
1 1 A, A!1 D @ d
0 1
g
must have
aCd Cg D1
d Cg D0
gD0
bCeChD0
eChD1
hD0
Thus a D 1, d D g D 0, h D 0, e
f D "1, c D 0, and so
0
1 "1
A!1 D @ 0 1
0 0
b
e
h
D 1, b D "1, i D 1,
Thus
0
1
2
1
@
2
" 21
" 16
5
6
" 16
1
6
1
6
1
6
1
A:
0 1
1 0 1
0
1
x
"2
@ y A D A!1 @ 1 A D @ 2 A ;
3
z
13
so x D 1, y D 2, and z D 3.
If A is the matrix of Exercises 16 and 17 then
det.A/ D 6. By Cramer’s Rule,
1
xD
6
1
c
f A. Since
i
cCf Ci D0
f Ci D0
i D 1:
1
0
"1 A :
1
c"i D0
"c C f D 0
2c C f C 3i D 1:
0 1 0
1
x
"2
A@y A D @ 1 A:
z
13
with a similar calculation for the product with a reversed
order of factors.
0
1
c
f A. Since
i
The given system of equations is
D I:
we have
1 0
ad " bc
"b
B
D
D C
a A D @ cd " cd
D
D
D
;
Thus A!" D .A" /!1 :
15.
b
e
h
Solving these three systems of equations, we get
!1
$
sin."& /
cos."& /
0
(PAGE 617)
21.
yD
1
6
zD
1
6
ˇ
ˇ "2
ˇ
ˇ 1
ˇ
ˇ 13
ˇ
ˇ 1
ˇ
ˇ "1
ˇ
ˇ 2
ˇ
ˇ 1
ˇ
ˇ "1
ˇ
ˇ 2
ˇ
0 "1 ˇˇ
6
1 0 ˇˇ D D 1
6
1 3 ˇ
ˇ
"2 "1 ˇˇ
12
1
0 ˇˇ D
D2
6
13 3 ˇ
ˇ
0 "2 ˇˇ
18
1 1 ˇˇ D
D 3:
6
1 13 ˇ
1
1 1
1
1
B1 1
1 "1 C
C
ADB
@ 1 1 "1 "1 A
1 "1 "1 "1
ˇ
ˇ
ˇ0 0
0
2 ˇˇ
ˇ
ˇ0 0
2
0 ˇˇ
det.A/ D ˇˇ
0
2
0
0 ˇˇ
ˇ
ˇ 1 "1 "1 "1 ˇ
ˇ
ˇ
ˇ0 0
ˇ
ˇ
2 ˇˇ
ˇ
ˇ0 2 ˇ
ˇD8
0 ˇˇ D "4 ˇˇ
D "2 ˇˇ 0 2
1 "1 ˇ
ˇ 1 "1 "1 ˇ
0
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SECTION 10.7 (PAGE 617)
ADAMS and ESSEX: CALCULUS 9
ˇ
ˇ
ˇ0 1
1
1 ˇˇ
ˇ
1 ˇ4 1
1 "1 ˇˇ
x1 D ˇˇ
8 ˇ 6 1 "1 "1 ˇˇ
ˇ 2 "1 "1 "1 ˇ
ˇ
ˇ
ˇ0 1 1 1 ˇ
ˇ
ˇ
1 ˇ 4 0 0 "2 ˇˇ
D ˇˇ
8 ˇ 6 2 0 0 ˇˇ
ˇ2 0 0 0 ˇ
ˇ
ˇ
ˇ1 1 1 ˇ
ˇ
4
2 ˇˇ
D " ˇ 0 0 "2 ˇˇ D "
8 ˇ2 0 0 ˇ
8
ˇ
ˇ1
ˇ
1 ˇˇ 1
x2 D ˇ
8 ˇ1
ˇ1
ˇ
ˇ2
ˇ
1 ˇˇ 0
D ˇ
8 ˇ0
ˇ0
ˇ
ˇ4
2ˇ
D ˇˇ 6
8 ˇ2
0
4
6
2
0
4
6
2
2
0
0
ˇ
1
1 ˇˇ
1 "1 ˇˇ
"1 "1 ˇˇ
"1 "1 ˇ
ˇ
0 1 ˇˇ
2 "1 ˇˇ
0 "1 ˇˇ
0 "1 ˇ
ˇ
"1 ˇˇ
"4
"1 ˇˇ D
8
ˇ
"1
$# $
pa C qc pb C qd
x1
ra C sc rb C sd
x2
$#
$# $
#
x1
p q
a b
D
x2
r s
c d
# $
x1
D GF
:
x2
D
23.
ˇ
ˇ
ˇ1 1 ˇ
ˇ
ˇ
ˇ 0 "2 ˇ D 1
ˇ
ˇ
ˇ 6 "1 ˇ
ˇ
ˇ
ˇ 2 "1 ˇ D 2
$
$
#
a b
x1
.
, where F D
c d
x
# 2$
#
$
y1
p q
Let G.y1 ; y2 / D G
, where G D
.
y2
r s
If y1 D ax1 C bx2 and y2 D cx1 C dx2 , then
#
G ı F .x1 ; x2 / D G.y1 ; y2 /
$
#
$#
ax1 C bx2
p q
D
cx1 C dx2
r s
#
$
pax1 C pbx2 C qcx1 C qdx2
D
rax1 C rbx2 C scx1 C sdx2
414
Thus, G ı F is represented by the matrix GF.
#
$
"1 1
A D
. Use Theorem 8. D1 D "1 < 0,
1 "2
ˇ
ˇ
ˇ "1 1 ˇ
ˇ D 1 > 0. Thus A is negative definite.
D2 D ˇˇ
1 "2 ˇ
0
1
1 2 0
A D @ 2 1 0 A. Use Theorem 8.
0 0 1
D1 D 1 > 0;
ˇ
ˇ
ˇ1 1 0 1 ˇ
ˇ
ˇ
1 ˇ 1 1 4 "1 ˇˇ
x3 D ˇˇ
8 ˇ 1 1 6 "1 ˇˇ
ˇ 1 "1 2 "1 ˇ
ˇ
ˇ
ˇ0 2 0 1 ˇ
ˇ
ˇ
1 ˇ 0 0 4 "1 ˇˇ
D ˇˇ
8 ˇ 0 0 6 "1 ˇˇ
ˇ 2 "2 2 "1 ˇ
ˇ
ˇ
ˇ2 0 1 ˇ
ˇ
ˇ
ˇ
2 ˇˇ
4 ˇ 4 "1 ˇˇ
D " ˇ 0 4 "1 ˇˇ D " ˇˇ
D "1
8 ˇ 0 6 "1 ˇ
8 6 "1 ˇ
x4 D ".x1 C x2 C x3 / D "2:
22. Let F .x1 ; x2 / D F
24.
#
25.
Thus A is
0
2
A D @1
1
ˇ
ˇ1
ˇ
D3 D ˇˇ 2
ˇ0
ˇ
ˇ1
D2 D ˇˇ
2
ˇ
2 0 ˇˇ
1 0 ˇˇ D "3 < 0:
0 1ˇ
indefinite.
1
1 1
2 1 A. Use Theorem 8.
1 2
D1 D 2 > 0;
ˇ
ˇ2
ˇ
D3 D ˇˇ 1
ˇ1
26.
ˇ
2 ˇˇ
D "3 < 0;
1ˇ
ˇ
ˇ2
D2 D ˇˇ
1
ˇ
1 ˇˇ
D 3 > 0;
2ˇ
ˇ
1 1 ˇˇ
2 1 ˇˇ D 4 > 0:
1 2ˇ
Thus A is positive definite.
1
0
ˇ
ˇ
1 1 0
ˇ1 1ˇ
ˇ D 0, we cannot
A D @ 1 1 0 A. Since D2 D ˇˇ
1 1ˇ
0 0 1
use Theorem 8. The corresponding quadratic form is
Q.x; y; z/ D x 2 C y 2 C 2xy C z 2 D .x C y/2 C z 2 ;
27.
which is positive semidefinite. (Q.1; "1; 0/ D 0.). Thus
A is positive semidefinite.
0
1
1 0
1
A D @ 0 1 "1 A. Use Theorem 8.
1 "1 1
D1 D 1 > 0;
ˇ
ˇ1
ˇ
D3 D ˇˇ 0
ˇ1
Thus A is indefinite.
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ˇ
ˇ1
D2 D ˇˇ
0
ˇ
0 ˇˇ
D 1 > 0;
1ˇ
ˇ
0
1 ˇˇ
1 "1 ˇˇ D "1 < 0:
"1 1 ˇ
lOMoARcPSD|6566483
INSTRUCTOR’S SOLUTIONS MANUAL
0
2 0
28. A D @ 0 4
1 "1
SECTION 10.8
1
1
11 A. Use Theorem 8.
1
D1 D 2 > 0;
ˇ
ˇ
ˇ2 0ˇ
ˇ
ˇ D 8 > 0;
D2 D ˇ
0 4ˇ
ˇ
ˇ2 0
ˇ
D3 D ˇˇ 0 4
ˇ 1 "1
Thus A is positive definite.
ˇ
1 ˇˇ
11 ˇˇ D 2 > 0:
1 ˇ
Section 10.8 Using Maple for Vector and
Matrix Calculations (page 626)
> a := DotProduct(U,(V
W),conjugate=false):
&x
> b := DotProduct(V,(W
U),conjugate=false):
&x
> c := DotProduct(W,(U
V),conjugate=false):
&x
(PAGE 626)
> simplify(a-b); simplify(a-c);
0
0
4.
These calculations verify the identity:
> U := Vector[row](3,symbol=u): V := Vector[row](3,symbol=v):
It is assumed that the Maple package LinearAlgebra has
been loaded for all the calculations in this section.
> W := Vector[row](3,symbol=w):
1.
We use the result of Example 9 of Section 10.4.
> LHS := (U &x V) &x (U &x W):
> RHS := (DotProduct(U,(V &x
W),conjugate=false))*U:
> r1 := <3|0|2>: v1 := <2,1,-2>:
> r2 := <1|2|4>: v2 := <1,3,4>:
> v1xv2 := v1 &x v2:
> dist
:=
abs((r2-r1).v1xv2)/Norm(v1xv2,2);
d i st WD 2
> simplify(LHS-RHS);
Œ0; 0; 0!
The distance between the two lines is 2 units.
2. The plane P through the origin containing the vectors
v1 D i " 2j " 3k and v2 D 2i C 3j C 4k has normal
n D v1 & v2 .
>
n := <1|-2|-3> &x <2|3|4>;
n WD Œ1; "10; 7!
5.
sp := (U,V) -> DotProduct(U,Normalize(V,2),conjugate=false)
6.
vp := (U,V) -> DotProduct(U,Normalize(V,2),
conjugate=false)*Normalize(V,2)
7.
ang
:=
(u,v)
->
evalf((180/Pi)*VectorAngle(U,V))
8.
unitn := (U,V)->Normalize((U &x V),2)
9.
VolT
:=
(U,V,W)->(1/6)*abs(DotProduct(U,(V &x W),
conjugate=false))
The angle between v D i " j C 2k and n (in degrees) is
> angle
:=
evalf((180/Pi)*VectorAngle(n,<1,-1,2>));
angvn WD 33:55730975
Since this angle is acute, the angle between v and the
plane P is its complement.
>
angle := 90 - angvn;
angle WD 56:44269025
10. dist:=(A,B)->Norm(A-B,2)
3. These calculations verify the identity:
> dist(<1,1,1,1>,<3,-1,2,5>);
5
> U := Vector[row](3,symbol=u): V := Vector[row](3,symbol=v):
>
W := Vector[row](3,symbol=w):
11.
We use LinearSolve.
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SECTION 10.8 (PAGE 626)
ADAMS and ESSEX: CALCULUS 9
> A := Matrix([[1,2,3,4,5],
> [6,-1,6,2,-3],[2,8,-8,-2,1],
> [1,1,1,1,1],[10,-3,3,-2,2]]):
> X
:=
LinearSolve(A,<20,0,6,5,5>,free=t);
2
3
1
6 0 7
6
7
7
X WD 6
6 "1 7
4 3 5
2
16.
> A := Matrix([[1,1/2,1/3],
> [1/2,1/3,1/4],[1/3,1/4,1/5]]):
> Ainv := MatrixInverse(A):
> Digits := 10: evalf(Eigenvalues(A));
2
3
1:408318927 " 4 10!11 I
!10
4 0:00268734034 " 5:673502694 10 I 5
0:1223270659 C 5:873502694 10!10 I
> evalf(Eigenvalues(Ainv));
2
3
372:1151279 " 2 10!9 I
!8
4 0:710066409 " 5:096152424 10 I 5
8:174805711 C 5:296152424 10!8 I
The solution is u D 1, v D 0, x D "1, y D 3, z D 2.
12. We use LinearSolve.
> B := Matrix([[1,1,1,1,0],
> [1,0,0,1,1],[1,0,1,1,0],
> [1,1,1,0,1],[0,1,0,1,-1]]):
> X
:=
LinearSolve(B,<10,10,8,11,1>,free=t);
2
3
11 " 2t5
6
7
2
6
7
"2
C
t
X WD 6
5 7
6
7
4 "1 C t5 5
t5
The small imaginary parts are due to round-off errors in
the solution process. The eigenvalues are real since the
matrix and its inverse are real and symmetric.
Although they appear in different orders, each eigenvalue
of A!1 is the reciprocal of an eigenvalue of A. This is to
be expected since
A!1 x D ,x
)
.1=,/x D Ax:
There is a one-parameter family of solutions: u D 11 " 2t ,
v D 2, x D "2 C t , y D "1 C t , z D t , for arbitrary t .
13.
14.
15.
> A := Matrix([[1,2,3,4,5],
> [6,-1,6,2,-3],[2,8,-8,-2,1],
> [1,1,1,1,1],[10,-3,3,-2,2]]):
> Determinant(A);
"935
> B := Matrix([[1,1,1,1,0],
> [1,0,0,1,1],[1,0,1,1,0],
> [1,1,1,0,1],[0,1,0,1,-1]]):
> Digits := 5: evalf(Eigenvalues(B));
2
3
0
6 3:3133 " 0:0000053418I 7
6
7
6 0:8693 C 0:0000073520I 7
6
7
4 "1:2728 " 0:0000025143I 5
"1:9098 C 5:041 10!7 I
Review Exercises 10 (page 627)
1.
x C 3z D 3 represents a plane parallel to the y-axis and
passing through the points .3; 0; 0/ and .0; 0; 1/.
2.
y " z # 1 represents all points on or below the plane parallel to the x-axis that passes through the points .0; 1; 0/
and .0; 0; "1/.
3.
x C y C z # 0 represents all points on or above the plane
through the origin having normal vector i C j C k.
4.
x " 2y " 4z D 8 represents all points on the plane passing
through the three points .8; 0; 0/, .0; "4; 0/, and .0; 0; "2/.
5.
y D 1 C x 2 C z 2 represents the circular paraboloid obtained
by rotating about the y-axis the parabola in the xy-plane
having equation y D 1 C x 2 .
y D z 2 represents the parabolic cylinder parallel to the
x-axis containing the curve y D z 2 in the yz-plane.
The tiny imaginary parts are due to roundoff error in the
calculations. They should all be 0. Since B is a real,
symmetric matrix, its eigenvalues are all real. The eigenvalues, rounded to 5 decimal places are 0, 3.3133, 0.8693,
"1:2728, and "1:9098.
6.
7.
x D y 2 " z 2 represents the hyperbolic paraboloid whose
intersections with the xy- and xz-planes are the parabolas
x D y 2 and x D "z 2 , respectively.
> A := Matrix([[1,1/2,1/3],
> [1/2,1/3,1/4],[1/3,1/4,1/5]]):
> Ainv := MatrixInverse(A);
2
3
9
"36
30
Ai nv WD 4 "36 192 "180 5
30 "180 180
8.
z D xy is the hyperbolic paraboloid containing the x- and
y-axes that results from rotating the hyperbolic paraboloid
z D .x 2 " y 2 /=2 through 45ı about the z-axis.
416
9.
x 2 C y 2 C 4z 2 < 4 represents the interior of the circular
ellipsoid (oblate spheroid) centred at the origin with semiaxes 2, 2, and 1 in the x, y, and z directions, respectively.
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INSTRUCTOR’S SOLUTIONS MANUAL
REVIEW EXERCISES 10
10. x 2 C y 2 " 4z 2 D 4 represents a hyperboloid of one sheet
with circular cross-sections in planes perpendicular to the
z-axis, and asymptotic to the cone obtained by rotating the
line x D 2z about the z-axis.
11.
21.
x 2 " y 2 " 4z 2 D 0 represents an elliptic cone with axis
along the x-axis whose cross-sections in planes x D k
are ellipses with semi-axes jkj and jkj=2 in the y and z
directions, respectively.
14.
15.
.x " z/2 C y 2 D z 2 represents an elliptic cone with oblique
axis along the line z D x in the xz-plane, having circular
cross-sections of radius jkj in horizontal planes z D k.
The z-axis lies on the cone.
17.
or 2x C 5y C 3z D 2.
22.
23.
19. The given line is parallel to the vector a D 2i"jC3k. The
plane through the origin perpendicular to a has equation
2x " y C 3z D 0.
20. A plane through .2; "1; 1/ and .1; 0; "1/ is parallel to
b D .2 " 1/i C ."1 " 0/j C .1 " ."1//k D i " j C 2k. If
it is also parallel to the vector a in the previous solution,
then it is normal to
ˇ
ˇi
j
ˇ
a & b D ˇˇ 2 "1
ˇ 1 "1
ˇ
k ˇˇ
3 ˇˇ D i " j " k:
2ˇ
The plane has equation .x " 1/ " .y " 0/ " .z C 1/ D 0, or
x " y " z D 2.
A plane containing the line of intersection of the planes
x C y C z D 0 and 2x C y " 3z D 2 has equation
2x C y " 3z " 2 C ,.x C y C z " 0/ D 0:
This plane passes through .2; 0; 1/ if "1 C 3, D 0. In this
case, the equation is 7x C 4y " 8z D 6.
24.
x 2 C y 2 C z 2 D 4, x C y C z D 3 together represent the
circle in which the sphere of radius 2 centred at the origin
intersects the plane through .1; 1;p1/ with normal i C j C k.
Since this plane lies
p at distance 3 from the origin, the
circle has radius 4 " 3 D 1.
18. x 2 C z 2 $ 1, x " y # 0 together represent all points
that lie inside or on the circular cylinder of radius 1 and
axis along the y-axis and also either on the vertical plane
x " y D 0 or on the side of that plane containing the
positive x-axis.
The plane through A D ."1; 1; 0/, B D .0; 4; "1/ and
C D .2; 0; 0/ has normal
ˇ
ˇ
ˇi
j
k ˇ
""! ""! ˇˇ 3 "1 0 ˇˇ
AC & AB D ˇ
ˇ D i C 3j C 10k:
ˇ 1 3 "1 ˇ
Its equation is .x "2/C3y C10z D 0, or x C3y C10z D 2.
x C2y D 0, z D 3 together represent the horizontal straight
line through the point .0; 0; 3/ parallel to the vector 2i " j.
16. x C y C 2z D 1, x C y C z D 0 together represent the
straight line through the points ."1; 0; 1/ and .0; "1; 1/.
A plane perpendicular to x "y Cz D 0 and 2x Cy "3z D 2
has normal given by the cross product of the normals of
these two planes, that is, by
ˇ
ˇ
ˇi
j
k ˇˇ
ˇ
ˇ 1 "1 1 ˇ D 2i C 5j C 3k:
ˇ
ˇ
ˇ 2 1 "3 ˇ
If the plane also passes through .2; "1; 1/, then its equation is
2.x " 2/ C 5.y C 1/ C 3.z " 1/ D 0;
12. x 2 " y 2 " 4z 2 D 4 represents a hyperboloid of two sheets
asymptotic to the cone of the previous exercise.
13. .x " z/2 C y 2 D 1 represents an elliptic cylinder with
oblique axis along the line z D x in the xz-plane, having circular cross-sections of radius 1 in horizontal planes
z D k.
(PAGE 627)
A plane containing the line of intersection of the planes
x C y C z D 0 and 2x C y " 3z D 2 has equation
2x C y " 3z " 2 C ,.x C y C z " 0/ D 0:
This plane is perpendicular to x " 2y " 5z D 17 if their
normals are perpendicular, that is, if
1.2 C ,/ " 2.1 C ,/ " 5."3 C ,/ D 0;
or 9x C 7y " z D 4.
25.
The line through .2; 1; "1/ and ."1; 0; 1/ is parallel to the
vector 3i C j " 2k, and has vector parametric equation
r D .2 C 3t /i C .1 C t /j " .1 C 2t /k:
26.
A vector parallel to the planes x " y D 3 and
x C 2y C z D 1 is .i " j/ & .i C 2j C k/ D "i " j C 3k. A
line through .1; 0; "1/ parallel to this vector is
x"1
y
zC1
D
D
:
"1
"1
3
27.
The line through the origin perpendicular to the plane
3x " 2y C 4z D 5 has equations x D 3t , y D "2t ,
z D 4t .
28.
The vector
%
&
a D .1 C t /i " t j " .2 C 2t /k " 2si C .s " 2/j " .1 C 3s/k
D .1 C t " 2s/i " .t C s " 2/j " .1 C 2t " 3s/k
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REVIEW EXERCISES 10 (PAGE 627)
ADAMS and ESSEX: CALCULUS 9
joins points on the two lines and is perpendicular to both
lines if a + .i " j " 2k/ D 0 and a + .2i C j " 3k/ D 0, that
is, if
32.
The tetrahedron with vertices A D .1; 2; 1/,
B D .4; "1; 1/, C D .3; 4; "2/, and D D .2; 2; 2/ has
volume
1
1 ""! ""! ""!
j.AB & AC / + ADj D j.9i C 9j C 12k/ + .i C k/j
6
6
7
9 C 12
D cu. units:
D
6
2
1 C t " 2s C t C s " 2 C 2 C 4t " 6s D 0
2 C 2t " 4s " t " s C 2 C 3 C 6t " 9s D 0;
or, on simplification,
33.
6t " 7s D "1
7t " 14s D "7:
This system has solution t D 1, s D 1. We would expect
to use a as a vector perpendicular to both lines, but, as it
happens, a D 0 if t D s D 1, because the two given lines
intersect at .2; "1; "4/. A nonzero vector perpendicular to
both lines is
ˇ
ˇ
ˇi
j
k ˇˇ
ˇ
ˇ 1 "1 "2 ˇ D 5i " j C 3k:
ˇ
ˇ
ˇ 2 1 "3 ˇ
Thus the required line is parallel to this vector and passes
through .2; "1; "4/, so its equation is
The inverse of A satisfies
10
0
1 0 0 0
a b
B2 1 0 0CB e f
B
CB
@3 2 1 0A@ i j
4 3 2 1
m n
a D 1;
b D 0;
c D 0;
d D 0;
2a C e D 0; 3a C 2e C i
2b C f D 1; 3b C 2f C j
2c C g D 0; 3c C 2g C k
2d C h D 0; 3d C 2h C l
a D 1; e D "2;
b D 0; f D 1;
c D 0; g D 0;
d D 0; h D 0;
Thus
A!1
.r2 " r1 / & .r3 " r1 / D 0:
h
i
.r2 " r1 / & .r3 " r1 / + .r4 " r1 / D 0:
(Any permutation of the subscripts 1, 2, 3, and 4 in the
above equation will do as well.)
31. The triangle with vertices A D .1; 2; 1/, B D .4; "1; 1/,
and C D .3; 4; "2/ has area
ˇ
ˇi
1 ""! ""!
1 ˇˇ
jAB & AC j D j ˇ 3
2
2 ˇ2
D
ˇ
j
k ˇˇ
"3 0 ˇˇ j
2 "3 ˇ
p
1
3 34
j9i C 9j C 12kj D
sq. units:
2
2
34.
0
0
i
j
k
l
D 0;
D 0;
D 1;
D 0;
D 1;
D "2;
D 1;
D 0;
1
0
B "2 1
B
D@
1 "2
0
1
1
0
0C
C:
0A
1
4a C 3e C 2i C m D 0:
4b C 3f C 2j C n D 0:
4c C 3g C 2k C o D 0:
4d C 3h C 2l C p D 1:
0
0
1
"2
m D 0;
n D 1;
o D "2;
p D 1:
1
0
0C
C:
0A
1
1
0 1
0 1
1 1 1
x1
b1
Let A D @ 2 1 0 A, x D @ x2 A, and b D @ b2 A.
1 0 "1
x3
b3
Then
Ax D b ,
x1 C x2 C x3 D b1
2x1 C x2
D b2
x1
" x3 D b3 :
The sum of the first and third equations is
2x1 C x2 D b1 C b3 , which is incompatible with the second
equation unless b2 D b1 C b3 , that is, unless
b + .i " j C k/ D 0:
If b satisfies this condition then there will be a line of
solutions; if x1 D t , then x2 D b2 " 2t , and x3 D t " b3 ,
so
0
1
t
x D @ b2 " 2t A
t " b3
is a solution for any t .
418
0
0
1
0
These systems have solutions
29. The points with position vectors r1 , r2 , and r3 are
collinear if the triangle having these points as vertices has
zero area, that is, if
30. The points with position vectors r1 , r2 , r3 , and r4 are
coplanar if the tetrahedron having these points as vertices
has zero volume, that is, if
0
1
0
0
Expanding the product on the left we get four systems of
equations:
r D .2 C 5t /i " .1 C t /j C ."4 C 3t /k:
(Any permutation of the subscripts 1, 2, and 3 in the
above equation will do as well.)
1 0
d
1
B0
hC
CDB
l A @0
p
0
c
g
k
o
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INSTRUCTOR’S SOLUTIONS MANUAL
0
3 "1
35. A D @ "1 1
1 "1
CHALLENGING PROBLEMS 10
1
1
"1 A. We use Theorem 8.
2
D1 D 3 > 0;
ˇ
ˇ 3
D2 D ˇˇ
"1
ˇ
ˇ 3 "1
ˇ
D3 D ˇˇ "1 1
ˇ 1 "1
Thus A is positive definite.
1
jx2 y3 " x2 y1 " x1 y3 " x3 y2 C x3 y1 C x1 y2 j
2 ˇ
ˇ
ˇ x y1 1 ˇ
ˇ
1 ˇˇ 1
D j ˇ x2 y2 1 ˇˇ j:
2 ˇx y 1ˇ
3
3
D
ˇ
"1 ˇˇ
D 2 > 0;
3 ˇ
ˇ
1 ˇˇ
"1 ˇˇ D 2 > 0:
2 ˇ
4.
Challenging Problems 10 (page 628)
1.
If d is the distance from P to the line AB, then d is the
altitude of the triangle APB measured perpendicular to the
base AB. Thus the area of the triangle is
"!
.1=2/d jBAj D .1=2/d jrA " rB j:
On the other hand, the area is also given by
"! ""
!
.1=2/jPA & PBj D .1=2/j.rA " rP / & .rB " rP /j:
a) Let Q1 and Q2 be the points on lines L1 and L2 ,
respectively, that are closest together. As observed in
""""!
Example 9 of Section 1.4, Q1 Q2 is perpendicular to
both lines.
Therefore, the plane P1 through Q1 having normal
""""!
Q1 Q2 contains the line L1 . Similarly, the plane P2
""""!
through Q2 having normal Q1 Q2 contains the line
L2 . These planes are parallel since they have the
same normal. They are different planes because
Q1 ¤ Q2 (because the lines are skew).
b) Line L1 through .1; 1; 0/ and .2; 0; 1/ is parallel to
i " j C k, and has parametric equation
Equating these two expressions for the area of the triangle
and solving for d we get
dD
r1 D .1 C t /i C .1 " t /j C t k:
j.rA " rP / & .rB " rP /j
:
jrA " rB j
Line L2 through .0; 1; 1/ and .1; 2; 2/ is parallel to
i C j C k, and has parametric equation
r2 D si C .1 C s/j C .1 C s/k:
2. By the formula for the vector triple product given in Exercise 23 of Section 1.3,
Now r2 " r1 D .s " t " 1/i C .s C t /j C .1 C s " t /k.
.u & v/ & .w & x/ D Œ.u & v/ + x!w " Œ.u & v/ + w!x
.u & v/ & .w & x/ D ".w & x/ & .u & v/
D "Œ.w & x/ + v!u C Œ.w & x/ + u!v:
To find the points Q1 on L1 and Q2 on L2 for
""""!
which Q1 Q2 is perpendicular to both lines, we solve
In particular, if w D u, then, since .u & v/ + u D 0, we have
.s " t " 1/ " .s C t / C .1 C s " t / D 0
.s " t " 1/ C .s C t / C .1 C s " t / D 0:
.u & v/ & .u & x/ D Œ.u & v/ + x!u;
Subtracting these equations gives s C t D 0, so
t D "s. Then substituting into either equation
gives 2s " 1 C 1 C 2s D 0, so s D "t D 0.
Thus Q1 D .1; 1; 0/ and Q2 D .0; 1; 1/, and
""""!
Q1 Q2 D "i C k. The required planes are x " z D 1
(containing L1 ) and x " z D "1 (containing L2 ).
or, replacing x with w,
.u & v/ & .u & w/ D Œ.u & v/ + w!u:
3. The triangle with vertices .x1 ; y1 ; 0/, .x2 ; y2 ; 0/, and
.x3 ; y3 ; 0/, has two sides corresponding to the vectors
.x2 " x1 /i C .y2 " y1 /j and .x3 " x1 /i C .y3 " y1 /j.
Thus the triangle has area given by
ˇ
ˇ
ˇ
i
j
k ˇˇ
1 ˇˇ
A D j ˇ x2 " x1 y2 " y1 0 ˇˇ j
2 ˇx " x y " y
0ˇ
3
1
3
1
(PAGE 628)
5.
This problem is similar to Exercise 28 of Section 1.3. The
equation a & x D b has no solution x unless a + b D 0. If
this condition is satisfied, then x D x0 C t a is a solution
for any scalar t , where x0 D .b & a/=jaj2 .
1
D jŒ.x2 " x1 /.y3 " y1 / " .x3 " x1 /.y2 " y1 /!kj
2
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SECTION 11.1 (PAGE 635)
ADAMS and ESSEX: CALCULUS 9
CHAPTER 11. VECTOR FUNCTIONS
AND CURVES
9.
Section 11.1 Vector Functions of
One Variable (page 635)
1.
10. Position: r D 3 cos t i C 4 sin t j C t k
Velocity: v D
p"3 sin t i C 4 cos t j C k
p
Speed: v D 9 sin2 t C 16 cos2 t C 1 D 10 C 7 cos2 t
Acceleration : a D "3 cos t i " 4 sin t j D t k " r
Path: a helix (spiral) wound around the elliptic cylinder
.x 2 =9/ C .y 2 =16/ D 1.
Position: r D i C t j
Velocity: v D j
Speed: v D 1
Acceleration : a D 0
Path: the line x D 1 in the xy-plane.
Position: r D ae t i C be t j C ce t k
Velocity and acceleration:
vDaDr
p
Speed: v D e t a2 C b 2 C c 2
x
y
z
Path: the half-line
D D > 0.
a
b
c
12. Position: r D at cos !t i C at sin !t j C b ln t k
Velocity: v D a.cos !t " !t sin !t /i
C a.sin !t C !t cos !t /j C .b=t /k
p
Speed: v D a2 .1 C ! 2 t 2 / C .b 2 =t 2 /
Acceleration: a D "a!.2 sin !t C ! cos !t /i
11.
2
2. Position: r D t i C k
Velocity: v D 2t i
Speed: v D 2jt j
Acceleration : a D 2i
Path: the line z D 1, y D 0.
3. Position: r D t 2 j C t k
Velocity: v D
p2t j C k
Speed: v D 4t 2 C 1
Acceleration : a D 2j
Path: the parabola y D z 2 in the plane x D 0.
C a!.2 cos !t " ! sin !t /j " .b=t 2 /k
Path: a spiral on the surface x 2 C y 2 D a2 e z=b .
13.
4. Position: r D i C t j C t k
Velocity: v D
pj C k
Speed: v D 2
Acceleration : a D 0
Path: the straight line x D 1, y D z.
5. Position: r D t 2 i " t 2 j C k
Velocity: v Dp
2t i " 2t j
Speed: v D 2 2t
Acceleration: a D 2i " 2j
Path: the half-line x D "y # 0, z D 1.
Position: r D a cos t i C a sin t j C ct k
Velocity: v D
p"a sin t i C a cos t j C ck
Speed: v D a2 C c 2
Acceleration: a D "a cos t i " a sin t j
Path: a circular helix.
Position: r D a cos t sin t i C a sin2 t j C a cos t k
&
a%
a
1 " cos 2t j C a cos t k
D sin 2t i C
2
2
Velocity: v Dp
a cos 2t i C a sin 2t j " a sin t k
Speed: v D a 1 C sin2 t
Acceleration: a D "2a sin 2t i C 2a cos 2t j " a cos t k
Path: the path lies on the sphere x 2 C y 2 C z 2 D a2 , on
the surface defined in terms of spherical polar coordinates
by / D & , on the circular cylinder x 2 C y 2 D ay, and on
the parabolic cylinder ay C z 2 D a2 . Any two of these
surfaces serve to pin down the shape of the path.
15.
The position of the particle is given by
r D 5 cos.!t /i C 5 sin.!t /j;
8. Position: r D a cos !t i C bj C a sin !t k
Velocity: v D "a! sin !t i C a! cos !t k
Speed: v D ja!j
Acceleration: a D "a! 2 cos !t i " a! 2 sin !t k
Path: the circle x 2 C z 2 D a2 , y D b.
420
t
Position: r D e !t% cos.e t /i C e !t sin.e t /j
& "e k
!t
t
t
Velocity: v D " e cos.e / C sin.e / i
%
&
" e !t sin.e t / " cos.e t / j " e t k
p
Speed: v D 1 C e%!2t C e 2t
&
Acceleration: a D .e !t " e t / cos.e t / C sin.e t / i
%
&
C .e !t " e t / sin.e t / " cos.e t / j " e t k
p
Path: a spiral on the surface z x 2 C y 2 D "1.
14.
6. Position: r D t i C t 2 j C t 2 k
Velocity: v D
pi C 2t j C 2t k
Speed: v D 1 C 8t 2
Acceleration: a D 2j C 2k
Path: the parabola y D z D x 2 .
7.
Position: r D 3 cos t i C 4 cos t j C 5 sin t k
Velocity: v D
p"3 sin t i " 4 sin t j C 5 cos t k
Speed: v D 9 sin2 t C 16 sin2 t C 25 cos2 t D 5
Acceleration : a D "3 cos t i " 4 cos t j " 5 sin t k D "r
Path: the circle of intersection of the sphere
x 2 C y 2 C z 2 D 25 and the plane 4x D 3y.
where ! D # to ensure that r has period 2#=! D 2 s.
Thus
d 2r
a D 2 D "! 2 r D "# 2 r:
dt
At .3; 4/, the acceleration is "3# 2 i " 4# 2 j.
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 11.1
16. When its x-coordinate is x, the particle is at position
r D xi C .3=x/j, and its velocity and speed are
Thus
du
2
D
, and
dt
1 C 2u2
"2
du
"16u
d 2u
D
4u
:
D
dt 2
.1 C 2u2 /2
dt
.1 C 2u2 /3
dr
dx
3 dx
D
i" 2
j
dt
dt
x dt
ˇ ˇr
ˇ dx ˇ
9
v D ˇˇ ˇˇ 1 C 4 :
dt
x
vD
The particle is at .3; 3; 2/ when u D 1. At this point
du=dt D 2=3 and d 2 u=dt 2 D "16=27, and so
We know that dx=dt > 0 since the particle is moving to
the right. When x D
p2, we have
10 D v D .dx=dt / 1 C .9=16/ D .5=4/.dx=dt /. Thus
dx=dt D 8. The velocity at that time is v D 8i " 6j.
17.
(PAGE 635)
2
.3i C 6uj C 6u2 k/ D 2i C 4j C 4k
3
# $2
2
"16
.6j C 12k/
.3i C 6j C 6k/ C
aD
27
3
8
D ."2i " j C 2k/:
9
vD
The particle moves along the curve z D x 2 , x C y D 2, in
the direction of increasing y. Thus its position at time t is
r D .2 " y/i C yj C .2 " y/2 k;
where y is an increasing function of time t . Thus
20.
i
dy h
"i C j " 2.2 " y/k
dt
dy p
vD
1 C 1 C 4.2 " y/2 D 3
dt
vD
since the speed
p When y D 1, we have
p is 3.
dy=dt D 3= 6 D 3=2. Thus
vD
r
3
."i C j " 2k/:
2
p
16x 3
d 2x
d
jvj D 1 C 8x 4 2 C p
dt
dt
1 C 8x 4
18. The position of the object when its x-coordinate is x is
i
dx h
i C 2xj C 3x 2 k . Since
dt
dz=dt D 3x 2 dx=dt D 3, when x D 2 we have
12 dx=dt D 3, so dx=dt D 1=4. Thus
so its velocity is v D
19.
Since u is increasing and the speed of the particle is 6,
6 D jvj D 3
du p
du
1 C 4u2 C 4u4 D 3.1 C 2u2 / :
dt
dt
dx
dt
$2
:
a D "15.i " 2j C 2k/ C 9."2j C 2k/ D "15i C 12j " 12k:
21.
r D 3ui C 3u2 j C 2u3 k
du
vD
.3i C 6uj C 6u2 k/
dt
# $2
d 2u
du
2
aD
.3i C 6uj C 6u k/ C
.6j C 12uk/:
dt 2
dt
#
The left side is 3 when x D 1, so 3.d 2 x=dt 2 / C 48 D 3,
and d 2 x=dt 2 D "15 at that point, and the acceleration
there is
r D xi C x 2 j C x 3 k;
1
v D i C j C 3k:
4
r D xi " x 2 j C Cx 2 k
dx
vD
.i " 2xj C 2xk/
dt
# $2
d 2x
dx
aD
.i
"
2xj
C
2xk/
C
."2j C 2k/:
dt 2
dt
ˇ ˇ
p
ˇ dx ˇ p
dx
Thus jvj D ˇˇ ˇˇ 1 C 4x 4 C 4x 4 D 1 C 8x 4
,
dt
dt
since x is increasing. At .1; "1; 1/, x D 1 and jvj D 9,
so dx=dt D 3, and the velocity at that point is
v D 3i " 6j C 6k. Now
22.
d 2
d
jvj D
v + v D 2v + a.
dt
dt
If v + a > 0 then the speed v D jvj is increasing.
If v + a < 0 then the speed is decreasing.
If u.t / D u1 .t /i C u2 .t /j C u3 .t /k
v.t / D v1 .t /i C v2 .t /j C v3 .t /k
then u + v D u1 v2 C u2 v2 C u3 v3 , so
du1
dv1
du2
dv2
d
u+vD
v1 C u1
C
v2 C u2
dt
dt
dt
dt
dt
du3
dv3
C
v3 C u3
dt
dt
du
dv
D
+vCu+
:
dt
dt
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SECTION 11.1 (PAGE 635)
23.
24.
25.
ADAMS and ESSEX: CALCULUS 9
ˇ
ˇ
ˇ a11 a12 a13 ˇ
ˇ
ˇ
ˇ a21 a22 a23 ˇ
ˇ
ˇ
ˇ a31 a32 a33 ˇ
d h
D
a11 a22 a33 C a12 a23 a31 C a13 a21 a32
dt
i
" a11 a23 a32 " a12 a21 a33 " a13 a22 a31
d
dt
0
0
0
D a11
a22 a33 C a11 a22
a33 C a11 a22 a33
0
0
0
C a12 a23 a31 C a12 a23 a31 C a12 a23 a31
0
0
0
C a13
a21 a32 C a13 a21
a32 C a13 a21 a32
0
0
0
" a11
a23 a32 " a11 a23
a32 " a11 a23 a32
0
0
0
" a12 a21 a33 " a12 a21 a33 " a12 a21 a33
0
0
0
" a13 a22 a31 " a13 a22 a31 " a13 a22 a31
ˇ 0
ˇ ˇ
ˇ
0
0 ˇ
ˇ a11 a12
ˇ
a13
a12 a13 ˇˇ
ˇ
ˇ ˇ a11
0
0
0
D ˇˇ a21 a22 a23 ˇˇ C ˇˇ a21 a22 a23 ˇˇ
ˇ a31 a32 a33 ˇ ˇ a31 a32 a33 ˇ
ˇ
ˇ
ˇ a11 a12 a13 ˇ
ˇ
ˇ
C ˇˇ a21 a22 a23 ˇˇ
0
0
0 ˇ
ˇ a31
a32
a33
31.
32.
33.
30.
d
d 2
jrj D
r + r D 2r + v D 0 implies that jrj is constant.
dt
dt
Thus r.t / lies on a sphere centred at the origin.
34.
d
d
jr " r0 j2 D
.r " r0 / + .r " r0 /
dt
dt
dr
D 2.r " r0 / +
D0
dt
implies that jr " r0 j is constant. Thus r.t / lies on a sphere
centred at the point P0 with position vector r0 .
&
d %
u & .v & w/
dt
D u0 & .v & w/ C u & .v0 & w/ C u & .v & w0 /:
#
#
$$
d
d u d 2u
u&
& 2
dt
dt
dt
#
$
# 2
$
du
d u d 2u
d u d 2u
D
&
& 2 Cu&
& 2
dt
dt
dt
dt 2
dt
#
3 $
du d u
Cu&
& 3
dt
dt
#
#
$
2 $
du d u
d u d 3u
du
&
& 2 Cu&
& 3 :
D
dt
dt
dt
dt
dt
422
Since
dr
D v.t / D 2r.t / and r.0/ D r0 , we have
dt
r.t / D r.0/e 2t D r0 e 2t ;
dv
dr
a.t / D
D2
D 4r0 e 2t :
dt
dt
The path is the half-line from the origin in the direction of
r0 .
%v &
0
r D r0 cos !t C
sin !t
!
dr
D "!r0 sin !t C v0 cos !t
dt
2
d r
D "! 2 r0 cos !t " !v0 sin !t D "! 2 r
dt 2
ˇ
d r ˇˇ
D v0 :
r.0/ D r0 ;
ˇ
dt ˇ
tD0
Observe that r + .r0 & v0 / D 0 for all t . Therefore the
path lies in a plane through the origin having normal
N D r0 & v0 .
Let us choose our coordinate system so that r0 D ai
(a > 0) and v0 D !bi C !cj (c > 0). Therefore, N is in
the direction of k. The path has parametric equations
26. If r + v > 0 then jrj is increasing. (See Exercise 16 above.)
Thus r is moving farther away from the origin. If r+ v < 0
then r is moving closer to the origin.
#
$
d 2u d 2u
d u d 3u
d d u d 2u
27.
& 2 D
& 2 C
& 3
2
dt dt
dt
dt
dt
dt
dt
d u d 3u
& 3:
D
dt
dt
%
&
d
28.
u + .v & w/
dt
D u0 + .v & w/ C u + .v0 & w/ C u + .v & w0 /:
29.
i
d h
.u C u00 / + .u & u0 /
dt
D .u0 C u000 / + .u & u0 / C .u C u00 / + .u0 & u0 /
C .u C u00 / + .u & u00 /
000
D u + .u & u0 /:
i
d h
.u & u0 / + .u0 & u00 /
dt
D .u0 & u0 / + .u0 & u00 / C .u & u00 / + .u0 & u00 /
C .u & u0 / + .u00 & u00 / C .u & u0 / + .u0 & u000 /
D .u & u00 / + .u0 & u00 / C .u & u0 / + .u0 & u000 /:
x D a cos !t C b sin !t
y D c sin !t:
The curve is a conic section since it has a quadratic equation:
#
$
by 2 y 2
1
C 2 D 1:
x
"
a2
c
c
Since the path is bounded (jr.t /j $ jr0 j C .jv0 j=!/), it
must be an ellipse.
If r0 is perpendicular to v0 , then b D 0 and the path is
the ellipse .x=a/2 C .y=c/2 D 1 having semi-axes a D jr0 j
and c D jv0 j=!.
35.
d 2r
dr
D "gk " c
dt 2
dtˇ
d r ˇˇ
r.0/ D r0 ;
ˇ
dt ˇ
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tD0
D v0 :
lOMoARcPSD|6566483
INSTRUCTOR’S SOLUTIONS MANUAL
Let w D e ct
SECTION 11.2
dr
. Then
dt
2.
dr
d 2r
dw
D ce ct
C e ct 2
dt
dt
dt
dr
ct d r
ct
D ce
" e gk " ce ct
dt
dt
D "e ct gk
Z
e ct
w.t / D " e ct gk dt D "
gk C C:
c
Let v.t / be the speed of the tank car at time t seconds.
The mass of the car at time t is m.t / D M " k t kg.
At full power, the force applied to the car is F D Ma
(since the motor can accelerate the full car at a m/s2 ).
By Newton’s Law, this force is the rate of change of the
momentum of the car. Thus
i
d h
.M " k t /v D Ma
dt
dv
.M " k t /
" kv D Ma
dt
dv
dt
D
Ma C kv
M " kt
1
1
1
ln.Ma C kv/ D " ln.M " k t / C ln C
k
k
k
C
Ma C kv D
:
M " kt
g
Put t D 0 and get v0 D " k C C, so
c
e ct
g
dr
D w D v0 C .1 " e ct /k
dt
c
dr
g
D e !ct v0 " .1 " e !ct /k
dt
c #
$
e !ct
g
e !ct
rD"
v0 "
tC
kCD
c
c
c
g
1
r0 D r.0/ D " v0 " 2 k C D:
c
c
At t D 0 we have v D 0, so Ma D C =M . Thus
C D M 2 a and
kv D
Thus we have
r D r0 C
v.t / D
The limit of this solution, as c ! 0, is calculated via
l’H^opital’s Rule:
t e !ct
t " t e !ct
" gk lim
c!0
c!0
1
2c
t 2 e !ct
D r0 C v0 t " gk lim
c!0
2
1 2
D r0 C v0 t " gt k;
2
c!0
which is the solution obtained in Example 4.
3.
dr
D k & r, r.0/ D i C k.
dt
Let r.t / D x.t /i C y.t /j C z.t /k. Then x.0/ D z.0/ D 1
and y.0/ D 0.
Since k + .d r=dt / D k + .k & r/ D 0, the velocity is always
perpendicular to k, so z.t/ is constant: z.t / D z.0/ D 1
for all t . Thus
dx
dy
dr
iC
jD
D k & r D xj " yi:
dt
dt
dt
Separating this equation into components,
It was shown in the text that
v.T / " v.0/ D " ln
Mat
m/s:
M " kt
Given:
Section 11.2 Some Applications of Vector
Differentiation (page 642)
1.
Mak t
M 2a
" Ma D
:
M " kt
M " kt
The speed of the tank car at time t (before it is empty) is
g
1 " e !ct
v0 " 2 .ct C e !ct " 1/k:
c
c
lim r.t / D r0 C v0 lim
(PAGE 642)
#
$
m.0/
ve :
m.T /
dx
D "y;
dt
If v.0/ D 0 and v.T / D "ve then ln.m.0/=m.T // D 1
and m.T / D .1=e/m.0/. The rocket must therefore burn
e"1
fraction
of its initial mass to accelerate to the speed
e
of its exhaust gases.
2
Similarly, if v.T / D "2ve , then m.T / D .1=e /m.0/, so
e2 " 1
the rocket must burn fraction
of its initial mass to
e2
accelerate to twice the speed of its exhaust gases.
dy
D x:
dt
Therefore,
dy
d 2x
D"
D "x;
dt 2
dt
and x D A cos t C B sin t . Since x.0/ D 1 and y.0/ D 0,
we have A D 1 and B D 0. Thus x.t / D cos t and
y.t / D sin t . The path has equation
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r D cos t i C sin t j C k:
423
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SECTION 11.2 (PAGE 642)
ADAMS and ESSEX: CALCULUS 9
Remark: This result also follows from comparing the
given differential equation with that obtained for circular
motion in the text. This shows that the motion is a rotation with angular velocity k, that is, rotation about the
z-axis with angular speed 1. The initial value given for r
then forces
r D cos t i C sin t j C k:
6.
4. First observe that
We use the fixed and rotating frames as described in
the text. Assume the satellite is in an orbit in the plane
spanned by the fixed basis vectors I and K. When the
satellite passes overhead an observer at latitude 45ı , its
position is
ICK
RDR p ;
2
where R is the radius of the earth, and since it circles the
earth in 2 hours, its velocity at that point is
%
&
d
dr
jr " bj2 D 2.r " b/ +
D 2.r " b/ + a & .r " b/ D 0;
dt
dt
I"K
V D #R p :
2
The angular velocity of the earth is ! D .#=12/K.
so jr " bj is constant; for all t the object lies on the sphere
centred at the point with position vector b having radius
r0 " b.
Next, observe that
The rotating frame with origin at the observer’s position
has, at the instant in question, its basis vectors satisfying
1
1
I D "p j C p k
2
2
JDi
1
1
K D p j C p k:
2
2
%
&
d
.r " r0 / + a D a & .r " b/ + a D 0;
dt
so r " r0 ? a; for all t the object lies on the plane through
r0 having normal a. Hence the path of the object lies
on the circle in which this plane intersects the sphere described above. The angle between r " b and a must therefore also be constant, and so the object’s speed jd r=dtj is
constant. Hence the path must be the whole circle.
As shown in the text, the velocity v of the satellite as it
appears to the observer is given by V D v C ! & R. Thus
vD V"!&R
R
pi
#R
K & p .I C K/
D p .I " K/ "
12
2
2
#R
#R
D p .I " K/ " p J
2
12 2
#R
D "#Rj " p i:
12 2
5. Use a coordinate system with origin at the observer, i
pointing east, and j pointing north. The angular velocity
of the earth is 2#=24 radians per hour northward:
!D
#
j:
12
Because the earth is rotating west to east, the true north to
south velocity of the satellite will appear to the observer
to be shifted to the west by #R=12 km/h, where R is the
radius of the earth in kilometres. Since the satellite circles
the earth at a rate of # radians/h, its velocity, as observed
at the moving origin, is
vR D "#Rj "
v makes angle
p !
p
!1 #R=12 2
tan
D tan!1 .1=.12 2/ ' 3:37ı with
#R
the southward direction. Thus the satellite appears to the
observer to be moving in a direction 3:37ı west of south.
The apparent Coriolis force is
#
$
#R
#R
#
"2! & v D "2 K & p .I " K " p J
12
2
12 2
$
#
# 2R
1
D" p
JC I
12
6 2
$
#
1
# 2R
i C p ."j C k/ :
D" p
6 2
12 2
#R
i:
12
$
#R=12
D tan!1 .1=12/ ' 4:76ı
#R
with the southward direction. Thus the satellite appears
to the observer to be moving in a direction 4:76ı west of
south.
vR makes angle tan!1
#
The apparent Coriolis force is
7.
#
$
#R
#2R
2#
i D"
k;
"2! & vR D " j & "#Rj "
12
12
72
which is pointing towards the ground.
424
The angular velocity of the earth is !, pointing due north.
For a particle moving with horizontal velocity v, the tangential and normal components of the Coriolis force C,
and of !, are related by
CT D "2!N & v;
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CN D "2!T & v:
lOMoARcPSD|6566483
INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 11.3
At the north or south pole, !T D 0 and !N D !. Thus
CN D 0 and CT D "2! & v. The Coriolis force is
horizontal. It is 90ı east of v at the north pole and 90ı
west of v at the south pole.
Section 11.3 Curves and Parametrizations
(page 649)
1.
At the equator, !N D 0 and !T D !. Thus CT D 0 and
CN D "2! & v. The Coriolis force is vertical.
8. We continue with the same notation as in Example
4. Since j points northward at the observer’s position,
the angle - between the direction vector of the sun,
S D cos )I C sin )J and north satisfies
On the first quadrant
part of the circle x 2 C y 2 D a2
p
2
we have x D a " y 2 , 0 $ y $ a. The required
parametrization is
p
r D r.y/ D a2 " y 2 i C yj; .0 $ y $ a/:
2.
On the first quadrant
part of the circle x 2 C y 2 D a2
p
we have y D a2 " x 2 , 0 $ x $ a. The required
parametrization is
p
r D r.x/ D xi C a2 " x 2 j; .0 $ x $ a/:
3.
From the figure we see that
cos - D S + j D " cos ) cos / cos & C sin ) sin /:
For the sun, & D 0 and at sunrise and sunset we have, by
Example 4, cos & D " tan )= tan /, so that
#
#
;
0$& $
2
2
%
#&
D a sin /
x D a cos & D a cos / "
2
%
#&
y D a sin & D a sin / "
D "a cos /:
2
/D&C
tan )
C sin ) sin /
tan /
cos2 /
D sin )
C sin ) sin /
sin /
sin )
:
D
sin /
cos - D cos ) cos /
The required parametrization is
%#
r D a sin /i " a cos /j;
a
#
$
24 !1
tan 23:3ı
cos
"
' 16
#
tan 40:8ı
.x;y/
)
hours between sunrise and sunset. By Exercise 8, the sun
will rise and set at an angle
cos!1
ı$
sin 23:3
sin 40:8ı
"
a
' 52:7ı
4.
10. At Umea, / D 90ı " 63:5ı D 26:5ı . On June 21st,
) D 23:3ı . By Example 4 there will be
s
s
s
#
x D a sin ; y D a cos ;
0$ $
a
a %
a &2
s
s
a#
r D a sin i C a cos j;
0$s$
:
a
a
2
#
$
tan 23:3ı
24 !1
cos
"
' 20
#
tan 26:5ı
y
sin 23:3ı
sin 26:5ı
$
s
a
hours between sunrise and sunset. By Exercise 8, the sun
will rise and set at an angle
#
x
Fig. 11.3-3
to the east and west of north.
cos!1
2
&
$/$# :
y
9. At Vancouver, / D 90ı " 49:2ı D 40:8ı . On June 21st,
) D 23:3ı . Ignoring the mountains and the rain, by Example 4 there will be
#
(PAGE 649)
.x;y/
s
a
' 27:6ı
a
x
Fig. 11.3-4
to the east and west of north.
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SECTION 11.3 (PAGE 649)
ADAMS and ESSEX: CALCULUS 9
5. z D x 2 , z D 4y 2 . If t D y, then z D 4t 2 , so x D ˙2t .
The curve passes through .2; "1; 4/ when t D "1, so
x D "2t . The parametrization is r D "2t i C t j C 4t 2 k.
6. z D x 2 , x C y C z D 1. If t D x, then
z D t 2 and y D 1 " t " t 2 . The parametrization is
r D t i C .1 " t " t 2 /j C t 2 k.
12. By symmetry, the centre of the circle C of intersection of
the plane x C y C z D 1 and the sphere x 2 C y 2 C z 2 D 1
must lie on the plane and must have its three coordinates
equal. Thus the centre has position vector
r0 D
z D x C y, x 2 C y 2 D 9. One possible parametrization is
r D 3 cos t i C 3 sin t j C 3.cos t C sin t /k.
p
8. x C y D 1, z D 1 " x 2 " y 2 . If x D t , then y D 1 " t
and p
p
z D 1 " t 2 " .1 " t /2 D 2.t " t 2 /. One possible
parametrization is
7.
Since C passes through the point .0; 0; 1/, its radius is
s
#
x C y D 2x " 4y " 1;
2
1
3
$2
r
#
#
$
$
2
1 2
1 2
C 0"
C 1"
D
:
3
3
3
v2 D .i C j C k/ & .i " j/ D i C j " 2k:
9. z D x 2 C y 2 , 2x " 4y " z " 1 D 0. These surfaces intersect
on the vertical cylinder
2
0"
Any vector v that satisfies v + .i C j C k/ D 0 is parallel to
the plane x C y C z D 1 containing C. One such vector is
v1 D i " j. A second one, perpendicular to v1 , is
p
r D t i C .1 " t /j C 2.t " t 2 /k:
2
1
.i C j C k/:
3
Two perpendicular unit vectors that are parallel to the
plane of C are
that is
2
i"j
vO 1 D p ;
2
.x " 1/ C .y C 2/ D 4:
vO 2 D
i C j " 2k
:
p
6
One possible parametrization is
Thus one possible parametrization of C is
x D 1 C 2 cos t
y D "2 C 2 sin t
z D "1 C 2.1 C 2 cos t / " 4."2 C 2 sin t / D 9 C 4 cos t " 8 sin t
r D .1 C 2 cos t /i " 2.1 " sin t /j C .9 C 4 cos t " 8 sin t /k:
10. yz C x D 1, xz " x D 1. One possible parametrization is
x D t , z D .1 C t /=t , and y D .1 " t /=z D .1 " t /t =.1 C t /,
that is,
1Ct
t " t2
jC
k:
r D ti C
1Ct
t
11.
2
.cos t vO 1 C sin t vO 2 /
3
cos t
iCjCk
sin t
C p .i " j/ C
.i C j " 2k/:
D
3
3
3
r D r0 C
13.
z 2 D x 2 C y 2 , z D 1 C x.
2
2
2
a) If t D x, then
p z D 1 C t , so 1 C 2t C t D t C y ,
and y D ˙ 1 C 2t . Two parametrizations are needed
to get the whole parabola, one for y $ 0 and one for
y # 0.
r D t i C ,t 2 j C t 3 k; .0 $ t $ T /
p
p
v D 1 C .2,t /2 C 9t 4 D .1 C 3t 2 /2
p
if 4,2 D 6, that is, if , D ˙ 3=2. In this case, the
length of the curve is
s.T / D
t2 C 1
t2 " 1
i C tj C
k:
2
2
15.
2
2
2
c) If t D z, then
p x D t " 1 and t D t " 2t C 1 C y ,
so y D ˙ 2t " 1. Again two parametrizations are
needed to get the whole parabola.
426
r D t 2 i C t 2 j C t 3 k; .0 $ t $ 1/
p
p
v D .2t /2 C .2t /2 C .3t 2 /2 D t 8 C 9t 2
Z 1 p
Length D
t 8 C 9t 2 dt Let u D 8 C 9t 2
0
du D 18t dt
ˇ17
p
p
17 17 " 16 2
1 2 3=2 ˇˇ
u ˇ D
units.
D
ˇ
18 3
27
8
14.
b) If t D y, then x 2 C t 2 D z 2 D 1 C 2x C x 2 , so
2x C 1 D t 2 , and x D .t 2 " 1/=2. Thus
z D 1 C x D .t 2 C 1/=2. The whole parabola is
parametrized by
rD
r
Length D
D
Z
Z
T
1
T
1
Z
T
0
ˇ ˇ
ˇdrˇ
ˇ ˇ dt
ˇ dt ˇ
s
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.1 C 3t 2 / dt D T C T 3 :
4a2 t 2 C b 2 C
c2
dt units.
t2
lOMoARcPSD|6566483
INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 11.3
If b 2 D
r%then
Z T4ac
c &2
Length D
dt
2at C
t
1
Z T%
c&
D
dt
2at C
t
1
D a.T 2 " 1/ C c ln T units.
Since the first octant part of C lies in the plane y D z, it
must be a quarter of a circle of radius 1. Thus the length
of all of C is 8 & .#=2/ D 4# units.
If you wish to use an integral, the length is
Z !=2 r
1
1
sin2 t C cos2 t C cos2 t dt
8
2
2
0
Z !=2
dt D 4# units.
D8
a
16. x D a cos t sin t D sin 2t ,
2
a
y D a sin2 t D .1 " cos 2t /,
2
z D bt .
The curve is a circular helix lying on the cylinder
0
z
x2 C y2 C z2 D 1
x 2 C 2z 2 D 1
%
a &2
a2
x C y"
D
:
2
4
2
(PAGE 649)
C
Its length, from t D 0 to t D T , is
Z
LD
p
a2 cos2 2t C a2 sin2 2t C b 2 dt
T
0
DT
y
x
p
a2 C b 2 units.
Fig. 11.3-18
19.
17.
r D t cos t i C t sin t j C t k;
0 $ t $ 2#
v D .cos t " t sin t /i C .sin t C t cos t /j C k
p
p
v D jvj D .1 C t 2 / C 1 D 2 C t 2 :
The length of the curve is
LD
Z
D2
2!
0
Z
p
2 C t 2 dt
tD2!
x D e t cos t;
p
Let t D 2 tan &
p
dt D 2 sec2 & d&
sec3 & d&
tD0
%
&ˇˇtD2!
D sec & tan & C ln j sec & C tan & j ˇˇ
tD0
!ˇ2!
t 2C
2C
t ˇˇ
D
p
C ln
Cp ˇ
2
2
2 ˇ0
%p
p
p &
D # 2 C 4# 2 C ln 1 C 2# 2 C 2# units.
p
t2
p
t2
The curve is called a conical helix because it is a spiral
lying on the cone x 2 C y 2 D z 2 .
18. One-eighth of the curve C lies in the first octant. That
part can be parametrized
1
z D p sin t; .0 $ t $ #=2/
2
1
1
1 " cos2 t " sin2 t D p sin t:
2
2
x D cos t;
r
yD
If C is the curve
y D e t sin t;
z D t;
.0 $ t $ 2#/;
then the length of C is
s
Z 2! # $2 # $2 # $2
dy
dz
dx
C
C
dt
LD
dt
dt
dt
0
Z 2! p
D
e 2t .cos t " sin t /2 C e 2t .sin t C cos t /2 C 1 dt
0
Z 2! p
D
2e 2t C 1 dt Let 2e 2t C 1 D v 2
0
2e 2t dt D v dv
$
Z tD2! 2
Z tD2! #
v dv
1
D
D
dv
1
C
v2 " 1
v2 " 1
tD0
tD0
ˇ
ˇ$ˇtD2!
#
1 ˇˇ v " 1 ˇˇ ˇˇ
D v C ln ˇ
ˇ
2
v C 1ˇ ˇ
tD0
ˇ2!
p
p
p
1
2e 2t C 1 " 1 ˇˇ
4!
D 2e C 1 " 3 C ln p
ˇ
2
2e 2t C 1 C 1 ˇ0
ˇ2!
p
p
p
2e 2t C 1 " 1 ˇˇ
4!
p
D 2e C 1 " 3 C ln
ˇ
ˇ
2e t
0
%p
&
p
p
D 2e 4! C 1 " 3 C ln 2e 4! C 1 " 1
p
" 2# " ln. 3 " 1/ units.
Remark: This answer appears somewhat different
from that given in the answers section of the text. The
two are, however, equal. Somewhat different simplifications were used in the two.
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427
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SECTION 11.3 (PAGE 649)
20.
ADAMS and ESSEX: CALCULUS 9
r D t 3i C t 2j
v D 3t 2 i C 2t j
p
p
v D jvj D 9t 4 C 4t 2 D jt j 9t 2 C 4
2a cos &
L
The length L between t D "1 and t D 2 is
LD
Z
0
!1
p
."t / 9t 2 C 4 dt C
Z
2
t
0
p
a
a
9t 2 C 4 dt:
&
2#.a C b/
one revolution
Making the substitution u D 9t 2 C 4 in each integral, we
obtain
(Z
Z
13
40
1
u1=2 du C
18 4
4
&
1 % 3=2
D
13
C 403=2 " 16
27
LD
21.
Fig. 11.3-22
Observe that tan & D
)
u1=2 du
sin & D
units.
cos & D
2a
1
&
. Therefore
cos & 2#.a C b/
a
#.a C b/
s
a2
1" 2
D
# .a C b/2
p
# 2 .a C b/2 " a2
:
#.a C b/
The total length of spool required is
r1 D t i C t j, .0 $ t $ 1/ represents the straight line
segment from the origin to .1; 1/ in the xy-plane.
H D L sin & C 2a cos &
%
&
p
a
L C 2 # 2 .a C b/2 " a2 units.
D
#.a C b/
r2 D .1 " t /i C .1 C t /j, .0 $ t $ 1/ represents the straight
line segment from .1; 1/ to .0; 2/.
Thus C D C1 C C2 is the 2-segment polygonal line from
the origin to .1; 1/ and then to .0; 2/.
L sin &
2a
cos "
23.
22. (Solution due to Roland Urbanek, a student at Okanagan College.) Suppose the spool is vertical and the cable
windings make angle & with the horizontal at each point.
r D At i C Bt j C C t k.
The arc length from the point where t D 0 to the point
corresponding to arbitrary t is
s D s.t / D
Z tp
p
A2 C B 2 C C 2 du D A2 C B 2 C C 2 t:
0
p
Thus t D s= A2 C B 2 C C 2 : The required parametrization is
Asi C Bsj C C sk
:
rD p
A2 C B 2 C C 2
24.
H
&
2a
b
s D s.t / D
a
Fig. 11.3-22
The centreline of the cable is wound around a cylinder of
2a
in
radius a C b and must rise a vertical distance
cos &
one revolution. The figure below shows the cable unwound
from the spool and inclined at angle & . The total length of
spool required is the total height H of the cable as shown
in that figure.
428
p
r D e t i C p2t j " e !t k
v D e t i C p2j C e !t k
v D jvj D e 2t C 2 C e !2t D e t C e !t :
The arc length from the point where t D 0 to the point
corresponding to arbitrary t is
Z
t
0
.e u C e !u / du D e t " e !t D 2 sinh t:
!
p
s C s2 C 4
,
Thus t D sinh!1 .s=2/ D ln
2
p
s C s2 C 4
. The required parametrization is
and e t D
2
!
p
p
s C s2 C 4 p
2k
s C s2 C 4
rD
iC 2 ln
j"
p
:
2
2
s C s2 C 4
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INSTRUCTOR’S SOLUTIONS MANUAL
25.
SECTION 11.4
#
2
v D "3a cos2 t sin t i C 3a sin2 t cos t j " 4b sin t cos t k
p
v D 9a2 C 16b 2 sin t cos t
Z tp
sD
9a2 C 16b 2 sin u cos u du
r D a cos3 t i C a sin3 t j C b cos 2t k;
which gives the (signed) arc length s measured from r.t0 /
along the curve, is an increasing function:
0$t $
ds
D g 0 .t / D jv.t /j > 0
dt
on Œa; b!, by the Fundamental Theorem of Calculus.
Hence g is invertible, and defines t as a function of arc
length s:
t D g !1 .s/ , s D g.t /:
0
1p 2
9a C 16b 2 sin2 t D K sin2 t
2
1p 2
where K D
9a C 16b 2
2
r
r
s
s
, cos t D 1 " ,
Therefore sin t D
K
K
2s
cos 2t D 1 " 2 sin2 t D 1 " .
K
The required parametrization is
D
Then
Section 11.4 Curvature, Torsion, and the
Frenet Frame (page 658)
1.
p p
v D 3.cos t " t sin t /i C 3.sin t C t cos t /j C 3 2 t k
p
v D jvj D 3 1 C t 2 C 2t D 3.1 C t /
#
$
Z t
t2
sD
3.1 C u/ du D 3 t C
2
0
r
2s
2s
2
Thus t C 2t D
, so t D "1 C 1 C
since t # 0.
3
3
The required parametrization
is
the
given
one
with t rep
placed by "1 C 1 C .2s/=3.
As claimed %in the& statement of the problem,
r1 .t / D r2 u.t / , where u is a function from Œa; b! to
Œc; d !, having u.a/ D c and u.b/ D d . We assume u
is differentiable. Since u is one-to-one and orientationpreserving, du=dt # 0 on Œa; b!. By the Chain Rule:
v
i " 4t j C 9t 2 k
TO D D p
:
v
1 C 16t 2 C 81t 4
2.
3.
4.
d
du
d
r1 .t / D
r2 .u/
;
dt
du
dt
and so
a
ˇ
ˇ
Z
bˇ
ˇ
ˇ d r1 .t /ˇ dt D
ˇ
ˇ dt
ˇ
Z d
%
&ˇˇ
bˇ
ˇ d r2 u.t / ˇ du dt D
ˇ du
ˇ dt
a
c
ˇ
ˇ
ˇ
ˇ d
ˇ r2 .u/ˇ du:
ˇ
ˇ du
28. If r D r.t / has nonvanishing velocity v D d r=dt on Œa; b!,
then for any t0 in Œa; b!, the function
s D g.t / D
Z
t
t0
jv.u/j du;
r D t i " 2t 2 j C 3t 3 k
v D i " 4t j C 9t 2 k
p
v D 1 C 16t 2 C 81t 4
1p 2
9a C 16b 2 .
for 0 $ s $ K, where K D
2
p
26. r D 3t cos t i C 3t sin t j C 2 2t 3=2 k; .t # 0/
Z
%
&
r D r2 .s/ D r g !1 .s/
is a parametrization of the curve r D r.t / in terms of arc
length.
#
$
%
% s &3=2
s &3=2
2s
k
rDa 1"
iCa
Cb 1"
K
K
K
27.
(PAGE 658)
5.
6.
r D a sin !t i C a cos !t k
v D a! cos !t i " a! sin !t k; v D ja!j
h
i
TO D sgn.a!/ cos !t i " sin !t k :
r D cos t sin t i C sin2 t C cos t k
1
1
D sin 2t i C .1 " cos 2t /j C cos t k
2
2
v D cos 2t i C sin 2t j " sin t k
p
v D jvj D 1 C sin2 t
%
&
1
TO D p
cos 2t i C sin 2t j " sin t k :
1 C sin2 t
r D a cos t i C b sin t j C t k
v D "a sin t i C b cos t j C k
p
v D a2 sin2 t C b 2 cos2 t C 1
"a sin t i C b cos t j C k
v
:
TO D D p
v
a2 sin2 t C b 2 cos2 t C 1
d TO
O D 0, so
D 1N
ds
dr
O
O
O
T.s/
D T.0/
is constant. This says that
D T.0/,
so
ds
O
r D T.0/s
C r.0/, which is the vector parametric equation
of a straight line.
If 1.s/ D 0 for all s, then
If +.s/ D 0 for all s, then
O
dB
O D 0, so B.s/
O
O
D "+ N
D B.0/
is constant. Therefore,
ds
&
dr O
d %
O
O
O
r.s/ " r.0/ + B.s/
D
+ B.s/ D T.s/
+ B.s/
D 0:
ds
ds
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SECTION 11.4 (PAGE 658)
ADAMS and ESSEX: CALCULUS 9
Hence 1.0/ D 1 and 1.#=2/ D 0. The radius of curvature
at x D 0 is 1. The radius of curvature at x D #=2 is
infinite.
It follows that
%
&
%
&
O
O
r.s/ " r.0/ + B.0/
D r.s/ " r.0/ + B.s/
D0
for all s. This says that r.s/ lies in the plane through r.0/
O
having normal B.0/.
7.
3.
The circle C1 given by
r D 2t i C .1=t /j " 2t k
v D 2i " .1=t 2 /j " 2k
a D .2=t 3 /j
1
1
rD
cos C si C sin C sj
C
C
v & a D .4=t 3 /i C .4=t 3 /k
At .2; 1; "2/, that is, at t D 1, we have
is parametrized in terms of arc length, and has curvature
C and torsion 0. (See Examples 2 and 3.)
If curve C has constant curvature 1.s/ D C and constant
torsion +.s/ D 0, then C is congruent to C1 by Theorem 3.
Thus C must itself be a circle (with radius 1=C ).
1 D 1.1/ D
p
jv & aj
4 2
D
:
v3
27
p
Thus the radius of curvature is 27=.4 2/.
8. The circular helix C1 given by
r D a cos t i C a sin t j C bt k
4.
has curvature and torsion given by
1.s/ D
a
;
a2 C b 2
+.s/ D
v D 3t 2 i C 2t j C k
a D 6t i C 2j
v.1/ D 3i C 2j C k; a.1/ D 6i C 2j
v.1/ & a.1/ D "2i C 6j " 6k
p
p
2 19
4 C 36 C 36
D 3=2
1.1/ D
.9 C 4 C 1/3=2
14
p
At t D 1 the radius of curvature is 143=2 =.2 19/.
b
;
a2 C b 2
by Example 3.
if a curve C has constant curvature 1.s/ D C > 0, and
constant torsion +.s/ D T ¤ 0, then we can choose a and
b so that
a
D C;
a2 C b 2
b
D T:
a2 C b 2
5.
r D t i C t 2 j C 2k
v D i C 2t j
a D 2j
v & a D 2k
At .1; 1; 2/, where t D 1, we have
p
TO D v=jvj D .i C 2j/= 5
BO D .v & a/=jv & aj D k
p
O D BO & TO D ."2i C j/= 5:
N
6.
r D t i C t 2j C t k
v D i C 2t j C k
a D 2j
v & a D "2i C 2k
At .1; 1; 1/, where t D 1, we have
p
TO D v=jvj D .i C 2j C k/= 6
p
BO D .v & a/=jv & aj D ".i " k/= 2
p
O D BO & TO D ".i " j C k/= 3:
N
T
C
, and b D 2
.) By
2
CT
C CT2
Theorem 3, C is itself a circular helix, congruent to C1 .
(Specifically, a D
C2
Section 11.5 Curvature and Torsion
for General Parametrizations (page 664)
1.
For y D x 2 we have
1.x/ D
jd 2 y=dx 2 j
2
D
:
2
3=2
.1 C .dy=dx/ /
.1 C 4x 2 /3=2
p
Hence 1.0/ D 2 and 1. 2/
p D 2=27. The radii of curvature at x D 0 and x D 2 are 1=2 and 27=2, respectively.
2. For y D cos we have
1.x/ D
430
jd 2 y=dx 2 j
j cos xj
D
:
2
3=2
.1 C .dy=dx/ /
.1 C sin2 x/3=2
r D t 3i C t 2j C t k
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INSTRUCTOR’S SOLUTIONS MANUAL
7.
t3
t2
jC k
2
3
v D i C t j C t 2k
da
a D j C 2t k;
D 2k
dt
v & a D t 2 i " 2t j C k
p
v D jvj D 1 C t 2 C t 4 ;
da
.v & a/ +
D2
dt
SECTION 11.5
9.
r D ti C
jv & aj D
p
1 C 4t 2 C t 4
i C t j C t 2k
O D v D p
T
v
1 C t2 C t4
2
O D v & a D pt i " 2t j C k
B
jv & aj
1 C 4t 2 C t 4
3
C t /i C .1 " t 4 /j C .t 3 C 2t /k
O DB
O & TO D ".2t
N
p
.1 C t 2 C t 4 /.1 C 4t 2 C t 4 /
p
1 C 4t 2 C t 4
jv & aj
D
1D
3
v
.1 C t 2 C t 4 /3=2
da
.v & a/ +
2
dt D
+D
:
jv & aj2
1 C 4t 2 C t 4
8.
r D e t cos t i C e t sin t j C e t k
v D e t .cos t " sin t /i C e t .sin t C cos t /j C e t k
a D "2e t sin t i C 2e t cos t j C e t k
da
D "2e t .cos t C sin t /i C 2e t .cos t " sin t /j C e t k
dt
v & a D e 2t .sin t " cos t /i " e 2t .cos t C sin t /j C 2e 2t k
p
p
jv & aj D 6e 2t
v D jvj D 3e t ;
da
.v & a/ +
D 2e 3t
dt
v
.cos t " sin t /i C .cos t C sin t /j C k
p
TO D D
v
3
v
&
a
.sin
t
"
cos
t
/i
" .cos t C sin t /j C 2k
p
BO D
D
jv & aj
6
.cos
t
C
sin
t
/i
"
.cos
t " sin t /j
O D BO & T
O D"
p
N
2
p
2
jv & aj
D t
1D
v3
3e
da
.v & a/ +
dt D 1 :
+D
jv & aj2
3e t
10.
(PAGE 664)
p
r D .2 C 2 cos t /i C .1 " sin t /j C .3 C sin t /k
p
v D " 2 sin t i " cos t j C cos t k
p
p
v D 2 sin2 t C cos2 t C cos2 t D 2
p
a D " 2 cos t i C sin t j " sin t k
p
da
D 2 sin t i C cos t j " cos t k
dt
p
p
v & a D " 2j " 2k
1
jv & aj
2
1D
D p D p
v3
2 2
2
p
p
da
.v & a/ +
D " 2 cos t C 2 cos t D 0
dt
+ D 0:
p
Since 1 D 1= 2 is constant, and + D 0, thepcurve is a
circle. Its centre is .2; 1; 3/ andpits radius is 2. It lies in
O
a plane with normal j C k.D " 2B/.
r D xi C sin xj
dx
dx
vD
i C cos x
j D k.i C cos xj/
dt
dt
p
v D k 1 C cos2 x
dx
a D "k sin x
j D "k 2 sin xj
dt
v & a D "k 3 sin xk
j sin xj
jv & aj
D
:
1D
v3
.1 C cos2 x/3=2
The tangential and normal components of acceleration are
k
dx
k 2 cos x sin x
dv
D p
2 cos x/." sin x/
D"p
2
dt
dt
2 1 C cos x
1 C cos2 x
2
k j sin xj
:
v21 D p
1 C cos2 x
11.
r D sin t cos t i C sin2 t j C cos t k
v D cos 2t i C sin 2t j " sin t k
a D "2 sin 2t i C 2 cos 2t j " cos t k
da
D "4 cos 2t i " 4 sin 2t j C sin t k:
dt
At t D 0 we have v D i, a D 2j " k,
da
D "4i,
dt
da
D 0.
dtp
p
O D i, BO D .j C 2k/= 5, N
O D .2j " k/= 5,
ThuspT
1 D 5, and + D 0.
v & a D j C 2k, .v & a/ +
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SECTION 11.5 (PAGE 664)
ADAMS and ESSEX: CALCULUS 9
1
1
At t D #=4 we have v D j " p k, a D "2i " p k,
2
2
p
1
1
da
D "4j C p k, v & a D " p i C 2j C 2k,
dt
2
2
p
da
.v & a/ +
D "3 2.
dt
Thus
p
O D p1 . 2j " k/
T
3
p
OB D p1 ."i C 2j C 2 2k/
13
p
1
O
N D " p .6i C j C 2k/
39
p
p
2 39
6 2
1D
;
+ D"
:
9
13
12.
The rate of change of the speed, dv=dt , is the tangential component of the acceleration, and is due entirely
to the tangential component of the gravitational force since
there is no friction:
dv
O
D g cos & D g."j/ + T;
dt
O and "j. (See the figwhere & is the angle between T
ure.) Since the p
slope of y D x 2 at .1; 1/ is 2, p
we have
O D ".i C 2j/= 5, and therefore dv=dt D 2g= 5.
T
y
r D a cos t i C b sin t j
v D "a sin t i C b cos t j
a D "a cos t i " b sin t j
v & a D abk
p
v D a2 sin2 t C b 2 cos2 t :
.1; 1/
&
"gj
dv O
T
dt
The tangential component of acceleration is
2
2
.a " b / sin t cos t
dv
D p
;
dt
a2 sin2 t C b 2 cos2 t
15.
Curve: r D xi C e x j.
p
Velocity: v D i C e x j. Speed: v D 1 C e 2x .
Acceleration: a D e x j. We have
v & a D e x k;
jv & aj
ab
v21 D v2
D p
:
2
v3
2
a sin t C b 2 cos2 t
ex
. Therefore, the radius
.1 C e 2x /3=2
.1 C e 2x /3=2
.
of curvature is * D
ex
The unit normal is
ab
.a2 sin2 t C b 2 cos2 t /3=2
ab
D %
&3=2 :
.a2 " b 2 / sin2 t C b 2
1D
"e x i C j
O DB
O & TO D .v & a/ & v D p
N
:
j.v & a/ & vj
1 C e 2x
The centre of curvature is
If a > b > 0, then the maximum curvature occurs when
sin t D 0, and is a=b 2 . The minimum curvature occurs
when sin t D ˙1, and is b=a2 .
O
rc D r C *N
#
$
1
D xi C e x j C .1 C e 2x / "i C x j
e
D .x " 1 " e 2x /i C .2e x C e !x /j:
By Example 2, the curvature of y D x 2 at .1; 1/ is
xD1
2
D p :
5 5
432
This is the equation of the evolute.
16.
Thus the magnitude of the normal
p acceleration of the bead
at that point is v 2 1 D 2v 2 =.5 5/.
jv & aj D e x :
The curvature is 1 D
13. The ellipse is the same one considered in Exercise 16, so
its curvature is
ˇ
ˇ
2
ˇ
1D
ˇ
2
3=2
.1 C 4x / ˇ
x
Fig. 11.5-14
which is zero if t is an integer multiple of #=2, that is, at
the ends of the major and minor axes of the ellipse.
The normal component of acceleration is
14.
y D x2
O
v21 N
The curve with polar equation r D f .& / is given parametrically by
r D f .& / cos & i C f .& / sin & j:
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 11.5
Thus we have
%
&
v D f 0 .& / cos & " f .& / sin & i
%
&
C f 0 .& / sin & C f .& / cos & j
%
&
a D f 00 .& / cos & " 2f 0 .& / sin & " f .& / cos & i
%
&
C f 00 .& / sin & C 2f 0 .& / cos & " f .& / sin & j
r
%
&2 %
&2
f 0 .& / C f .& /
v D jvj D
h %
&2 %
&2
i
v & a D 2 f 0 .& / C f .& / " f .& /f 00 .& / k:
The curvature is, therefore,
%
&2 %
&2
j2 f 0 .& / C f .& / " f .& /f 00 .& /j
:
h%
&2 %
&2 i3=2
f 0 .& / C f .& /
17.
Therefore
r.t / D
3a2 .1 " cos & /
3
:
&3=2 D p
2 2ar
2a2 .1 " cos & /
r.t / D
19.
Given that
$
t " sin t
1 " cos t
t C sin t
p
C1 iC
p k:
jC
2
2 2
2 2
dr
D c & r.t /, we have
dt
%
&
Thus jr.t /j D jr.0/j is constant, and r.t /"r.0/ +c D 0 is
constant. Thus r.t / lies on the sphere centred at the origin
with radius jr.0/j, and also on the plane through r.0/ with
normal c. The curve is the circle of intersection of this
sphere and this plane.
20.
For r D a cos t i C a sin t j C bt k, we have, by Example 3 of
Section 2.4,
O D " cos t i " sin t j;
N
1
1
1
cos t i1 C sin t j1 C t k1 C r0
2
2
2
1D
a
:
a2 C b 2
The centre of curvature rc is given by
for some right-handed basis fi1 ; j1 ; k1 g, and some constant
vector r0 . Example 3 of Section 2.4 provides values for
O
O
O
T.0/,
N.0/,
and B.0/,
which we can equate to the given
values of these vectors:
O DrC
rc D r C *N
1O
N:
1
Thus the evolute has equation
1
1
O
i D T.0/
D p j1 C p k1
2
2
O
j D N.0/
D "i1
1
1
O
k D B.0/
D " p j1 C p k1 :
2
2
r D a cos t i C a sin t j C bt k
a2 C b 2
.cos t i C sin t j/
a
2
b
b2
D " cos t i "
sin t j C bt k:
a
a
"
Solving these equations for i1 , j1 , and k1 in terms of the
given basis vectors, we obtain
i1 D "j
1
1
j1 D p i " p k
2
2
1
1
k1 D p i C p k:
2
2
#
d 2
d
jrj D
r + r D 2r + .c & r/ D 0
dt
dt
%
&
d
dr
r.t / " r.0/ + c D
+ c D .c & r/ + c D 0:
dt
dt
18. By Exercise 8 of Section 2.4, the required curve must be
a circular helix with parameters a D 1=2 (radius), and
b D 1=2. Its equation will be
rD
cos t
t " sin t
t C sin t
p i"
p k C r0 :
jC
2
2 2
2 2
1
We also require that r.0/ D i, so r0 D iC j. The required
2
equation is, therefore,
If r D a.1 " cos & /, then r 0 D a sin & , and r 00 D a cos & .
By the result of Exercise 20, the curvature of this cardioid
is
ˇ
1
ˇ 2 2
1D %
&3=2 & ˇ2a sin &
a2 sin2 & C a2 .1 " cos & /2
ˇ
ˇ
C a2 .1 " cos & /2 " a2 .cos & " cos2 & /ˇ
D %
(PAGE 664)
The evolute is also a circular helix.
21.
The parabola y D x 2 has curvature
1D
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;
.1 C 4x 2 /3=2
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SECTION 11.5 (PAGE 664)
ADAMS and ESSEX: CALCULUS 9
by Exercise 18. The normal at .x; x 2 / is perpendicular to
the tangent, so has slope "1=.2x/. Since the unit normal
points upward (the concave side of the parabola), we have
The conditions at x D 1 become
A C
A C
"2xi C j
O D p
N
:
1 C 4x 2
.1 C 4x 2 /3=2
2
#
"2xi C j
p
1 C 4x 2
1 C 4x 2
D xi C x 2 j " .1 C 4x 2 /xi C
j
2
#
$
1
D "4x 3 i C 3x 2 C
j:
2
C
C
C
$
is one possible solution.
y
.1;1/
2
, so the radius of
.3 sin t C 1/3=2
.3 sin2 t C 1/3=2
curvature is * D
. We have
2
x
yD!1
.!1;!1/
Fig. 11.5-23
2
O Dk
B
24.
We require
f .0/ D 1;
f ."1/ D 1;
D
3 sin2 t C 1
.cos t i C 2 sin t j/
2
3
cos3 i " 3 sin3 t j:
2
f 00 .0/ D "1;
f 00 ."1/ D 0:
ˇ
ˇ
d2 p
2 ˇˇ
1
"
x
2
ˇ
dx
xD0
D "1:
As in Example 5, we try
f .x/ D A C Bx C C x 2 C Dx 3 C Ex 4 C F x 5
f 0 .x/ D B C 2C x C 3Dx 2 C 4Ex 3 C 5F x 4
23. We require that
f .1/ D 1;
f ."1/ D "1;
f 0 .1/ D 0;
f 0 ."1/ D 0;
f 00 .1/ D 0;
f 00 ."1/ D 0:
f 00 D 2C C 6Dx C 12Ex 2 C 20F x 3 :
The required conditions force the coefficients to satisfy the
system of equations
As in Example 5, we try a polynomial of degree 5. However, here it is clear that an odd function will do, and we
need only impose the conditions at x D 1. Thus we try
f .x/ D Ax C Bx 3 C C x 5
f 0 .x/ D A C 3Bx 2 C 5C x 4
f 00 .x/ D 6Bx C 20C x 3 :
434
f 0 .0/ D 0;
f 0 ."1/ D 0;
The condition f 00 .0/ D "1 follows from the fact that
Therefore the evolute has equation
r D 2 cos t i C sin t j "
yD1
yDf .x/
v D "2 sin t i C cos t j
a D "2 cos t i " sin t j
v & a D 2k
p
p
v D 4 sin2 t C cos2 t D 3 sin2 t C 1:
"2 sin t i C cos t j
TO D p
;
3 sin2 t C 1
t i C 2 sin t j
O D " cos
p
:
N
3 sin2 t C 1
1
0
0:
15
5
3
x " x3 C x5
8
4
8
f .x/ D
22. For the ellipse r D 2 cos t i C sin t j, we have
The curvature is 1 D
D
D
D
C
5C
20C
This system has solution A D 15=8, B D "5=4, and
C D 3=8. Thus
Thus the evolute of the parabola has equation
r D xi C x 2 j C
B
3B
6B
A"B CC "DCE "F D1
B " 2C C 3D " 4E C 5F D 0
2C " 6D C 12E " 20F D 0
AD1
BD0
2C D "1
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 11.5
which has solution A D 1, B D 0, C D "1=2, D D "3=2,
E D "3=2, F D "1=2. Thus we can use a track section
in the shape of the graph of
This leads to the values
p
2
p
.cos.t /2 C 1/ 2 cos.t /2 C 2
1
3
3
1
1
f .x/ D 1 " x 2 " x 3 " x 4 " x 5 D 1 " x 2 .1 C x/3 :
2
2
2
2
2
and
0
for the curvature and torsion, respectively. Maple
doesn’t seem to recognize that the curvature simplifies
to 1=.cos2 t C 1/3=2 . The torsion is zero because the curve
is lies in the plane z D x. It is the ellipse in which this
plane intersects the ellipsoid 2x 2 C y 2 C 2z 2 D 4. The
maximum and minimum values of the curvature are 1 and
1=23=2 , respectively, at the ends of the major and minor
axes of the ellipse.
y
.!1;1/
yD1
(PAGE 664)
yDf .x/
x
27.
>
>
>
>
>
>
>
>
>
>
>
>
>
>
x 2 Cy 2 D1
Fig. 11.5-24
25. Given: a.t / D ,.t /r.t / C -.t /v.t /, v & a ¤ 0. We have
v & a D ,v & r C -v & v D ,v & r
da
D ,0 r C ,v C -0 v C -a
dt
D ,0 r C ., C -0 /v C -.,r C -v/
D .,0 C -,/r C ., C -0 C -2 /v:
Since v & r is perpendicular to both v and r, we have
.v & a/ +
da
D 0:
dt
R := t -> <t-sin(t), 1-cos(t), t>;
assume(t::real):
interface(showassumed=0):
V := t -> diff(R(t),t):
A := t -> diff(V(t),t):
v := t -> Norm(V(t),2):
VxA := t -> V(t) &x A(t):
vxa := t -> Norm(VxA(t),2):
Ap := t -> diff(A(t),t):
Curv := t ->
simplify(vxa(t)/(v(t))^3):
Tors := t -> simplify(
(VxA(t).Ap(t))/(vxa(t))^2):
Curv(t); Tors(t);
This leads to the values
p
cos.t /2 C 2 " 2 cos.t /
.3 " 2 cos.t //3=2
Thus the torsion +.t/ of the curve is identically zero.
It remains zero when expressed in terms of arc length:
+.s/ D 0. By Exercise 6 of Section 2.4, r.t / must be a
plane curve.
and
"
26. After loading the LinearAlgebra and VectorCalculus packages, issue the following commands:
>
>
>
>
>
>
>
>
>
>
>
>
>
>
After loading the LinearAlgebra and VectorCalculus packages, issue the following commands:
R := t -> <cos(t), 2*sin(t), cos(t)>;
assume(t::real):
interface(showassumed=0):
V := t -> diff(R(t),t):
A := t -> diff(V(t),t):
v := t -> Norm(V(t),2):
VxA := t -> V(t) &x A(t):
vxa := t -> Norm(VxA(t),2):
Ap := t -> diff(A(t),t):
Curv := t ->
simplify(vxa(t)/(v(t))^3):
Tors := t -> simplify(
(VxA(t).Ap(t))/(vxa(t))^2):
Curv(t); Tors(t);
2 cos.t /2
1
C sin.t /2 " 2 cos.t / C 1
for the curvature and torsion, respectively. Each of these
formulas can be simplified somewhat:
p
2 " 2 cos t C cos2 t
.3 " 2 cos t /3=2
"1
Tors.t / D
:
2 " 2 cos t C cos2 t
Curv.t / D
Since 3 " 2 cos t
>
0
and 2 " 2 cos t C cos2 t D 1 C .1 " cos t /2 > 0 for all
t , the curvature and torsion are both continuous for all t .
The curve appears to be some sort of helix (but not a circular one) with central axis along the line x D z, y D 1.
28.
After loading the LinearAlgebra and VectorCalculus packages, issue the following commands:
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SECTION 11.5 (PAGE 664)
ADAMS and ESSEX: CALCULUS 9
> R := t -> <cos(t)*cos(2*t),
cos(t)*sin(2*t), sin(t)>;
> assume(t::real):
> interface(showassumed=0):
> V := t -> diff(R(t),t):
> A := t -> diff(V(t),t):
> v := t -> Norm(V(t),2):
> VxA := t -> V(t) &x A(t):
> vxa := t -> Norm(VxA(t),2):
> Ap := t -> diff(A(t),t):
> Curv := t ->
> simplify(vxa(t)/(v(t))^3):
> Tors := t -> simplify(
> (VxA(t).Ap(t))/(vxa(t))^2):
> Curv(t); Tors(t);
> simplify(%,trig);
This leads to the values
p
2 cos.t /2 C cos t C 1
Curv.t / D !
"3=2
5 " cos.t /2 C 2 cos t
1
Tors.t / D
2.cos.t /2 / C cos t C 1
This appears to be an elliptical helix with central axis
along the line x D y D z " 1.
30.
evolute := R -> (t ->
R(t)+TNBFrame(R)[2](t)
*(1/Curvature(R)(t)));
31.
tanline := R ->
((t,u) -> R(t)+TNBFrame(R)[1](t)*u);
The last line simplifies the rather complicated expression
that Tors(t) returns by applying some trigonometric
identities. The values for the curvature and torsion are
p
17 C 60 cos.t /2 C 48 cos.t /4
Curv.t / D
!
"3=2
4 cos.t /2 C 1
Tors.t / D
12 cos t .2 cos.t /2 C 3/
:
17 C 60 cos.t /2 C 48 cos.t /4
Section 11.6 Kepler’s Laws of
Planetary Motion (page 673)
1.
436
r C $x D `
.1 " $ 2 /x 2 C 2`$x C y 2 D `2
$2
#
`2
`2 $ 2
`$
D
C y 2 D `2 C
.1 " $ 2 / x C
2
2
1"$
1"$
1 " $2
#
$2
`$
xC
y2
1 " $2
$2 D 1:
#
$2 C #
`
`
p
1 " $2
1 " $2
The command simplify(Norm(R(t),2));
gives output 1, indicating that the curve lies on the sphere
x 2 C y 2 C z 2 D 1.
> R := t -> <t+cos(t), t+sin(t), 1+tcos(t)>;
> assume(t::real):
> interface(showassumed=0):
> V := t -> diff(R(t),t):
> A := t -> diff(V(t),t):
> v := t -> Norm(V(t),2):
> VxA := t -> V(t) &x A(t):
> vxa := t -> Norm(VxA(t),2):
> Ap := t -> diff(A(t),t):
> Curv := t ->
> simplify(vxa(t)/(v(t))^3):
> Tors := t -> simplify(
> (VxA(t).Ap(t))/(vxa(t))^2):
> Curv(t); Tors(t);
%
x 2 C y 2 D r 2 D `2 " 2`$x C $ 2 x 2
Plotting the curvature as a function of t ,
(plot(Curv(t),t=-2*Pi..2*Pi)), shows that the
minimum curvature occurs at t D 0 (and
p any integer multiple of #). The minimum curvature is 125=53=2 D 1.
29. After loading the LinearAlgebra and VectorCalculus packages, issue the following commands:
`
1 C $ cos &
r D ` " $x
rD
2.
Position: r D r rO D k rO .
Velocity: v D k rPO D k &P "O ; speed: v D k &P .
P
Acceleration: k &R "O C k &P "O D "k &P 2 rO C k &R "O .
Radial component of acceleration: "k &P 2 .
Transverse component of acceleration: k &R D vP (the rate of
change of the speed).
3.
Position: on the curve r D e " .
Radial velocity: rP D e " &P .
Transverse velocity:
r &P D e " &P .
p
p "
&P D 1 % &P D .1= 2/e !" .
Speed v D 2ep
Thus &R D ".1= 2/e !" &P D "e !2" =2.
p
Radial velocity = transverse velocity D 1= 2.
Radial acceleration:
rR " r &P 2 D e " &P 2 C e " &R " e " &P 2 D e " &R D "e !" =2.
Transverse acceleration:
r &R C 2rP &P D ".e !" /=2 C e !" D e !" =2.
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 11.6
4. Path: r D & .qThus rP D &P , rR D &R .
p
Speed: v D .Pr/2 C .r &P /2 D &P 1 C r 2 .
Transverse acceleration = 0 (central force). Thus
r &R C 2rP &P D 0, or &R D "2&P 2 =r.
Radial acceleration:
The satellite has a circular orbit of radius aS and period
TS D 1 day. (If the orbit is in the plane of the equator, the satellite will remain above the same point on the
earth.) By Kepler’s third law,
TS2
aS3
rR " r &P 2 D &R " r &P 2
$
#
.2 C r 2 /v 2
2
C r &P 2 D "
:
D"
r
r.1 C r 2 /
The magnitude of the acceleration is, therefore,
.2 C r 2 /v 2
.
r.1 C r 2 /
For r D &
!2
8.
rP D "2& !3 &P D "2& !3
h
D "2h&:
r2
Thus rR D "2h&P D "2h2 =r 2 . The speed v is given by
v 2 D rP 2 C r 2 &P 2 D 4h2 & 2 C .h2 =r 2 /:
2
TM
3
aM
:
The period T (in years) and radius R (in km) of the asteroid’s orbit satisfies
2
Tearth
T2
12
D
D
:
3
R3
.150 & 106 /3
Rearth
Thus the radius of the asteroid’s orbit is
R ' 150 & 106 T 2=3 km.
9.
, we have
D
Thus aS D 385; 000 & .1=27/2=3 ' 42; 788. The satellite’s
orbit should have radius about 42,788 km, and should lie
in the equatorial plane.
5. For a central force, r 2 &P D h (constant), and the acceleration is wholly radial, so
jaj D jRr " r &P 2 j:
(PAGE 673)
If R is the radius and T is the period of the asteroid’s
circular orbit, then almost stopping the asteroid causes it to
drop into a very eccentric elliptical orbit with major axis
approximately R. (Thus, a D R=2.) The period Te of the
new elliptical orbit satisfies
Te2
.R=2/3
1
D
D :
2
3
T
R
8
p
Thus Te D T =.2 2/. The time the asteroid will
p take to
fall into the sun is half of Te . Thus it is T =.4 2/.
Since the speed is v0 when &pD 1 (and so r D 1), we
have v02 D 5h2 , and h D v0 = 5. Hence the magnitude of
the acceleration at any point on the path is
R
ˇ
ˇ
#
$
ˇ h2
v2 2
h2 ˇ
1
C
:
jaj D ˇˇ"2 2 " r 4 ˇˇ D 0
r
r
5 r2
r3
6. Let the period and the semi-major axis of the orbit of Halley’s comet be TH D 76 years and aH km respectively.
Similar parameters for the earth’s orbit are TE D 1 year
and aE D 150 & 106 km. By Kepler’s third law
TH2
3
aH
D
TE2
3
aE
:
Thus
aH D 150 & 106 & 762=3 ' 2:69 & 109 :
The major axis of Halley’s comet’s orbit is
2aH ' 5:38 & 109 km.
7.
Fig. 11.6-9
10. At perihelion, r D a " c D .1 " $/a.
At aphelion r D a C c D .1 C $/a.
Since rP D 0 at perihelion and aphelion, the speed is
v D r &P at each point. Since r 2 &P D h is constant over the
orbit, v D h=r. Therefore
vperihelion D
vaphelion D
h
:
a.1 C $/
If vperihelion D 2vaphelion then
h
2h
D
:
a.1 " $/
a.1 C $/
The period and semi-major axis of the moon’s orbit
around the earth are
TM ' 27 days,
h
;
a.1 " $/
aM ' 385; 000 km.
Hence 1 C $ D 2.1 " $/, and $ D 1=3. The eccentricity of
the orbit is 1/3.
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SECTION 11.6 (PAGE 673)
11.
ADAMS and ESSEX: CALCULUS 9
The orbital speed v of a planet satisfies (by conservation
of energy)
v2 k
" DK
2
r
14.
(total energy).
If v is constant so must be r, and the orbit will therefore
be circular.
As in Exercise 12, rP vP D rA vA , where rA D `=.1 " $/
and rP D `=.1 C $/, $ being the eccentricity of the orbit.
Thus
vP
rA
1C$
D
D
:
vA
rP
1"$
Solving this equation for $ in terms of vP and vA , we get
12. Since r 2 &P D h D constant for the planet’s orbit, and since
the speed is v D r &P at perihelion and at aphelion (the
radial velocity is zero at these points), we have
rp vp D ra va ;
By conservation of energy the speed v at the ends of the
minor axis of the orbit (where r D a) satisfies
where the subscripts p and a refer to perihelion and aphelion, respectively. Since rp =ra D 8=10, we must have
vp =va D 10=8 D 1:25. Also,
rp D
`
`
D
;
1 C $ cos 0
1C$
ra D
v2
v2
v2 k
k
k
" D P "
D A " :
2
a
2
rP
2
rA
`
`
D
:
1 C $ cos #
1"$
The latter equality shows that
Thus `=.1C$/ D .8=10/`=.1"$/, and so 10"10$ D 8C8$.
Hence 2 D 18$. The eccentricity of the orbit is $ D 1=9.
vP2
13. Let the radius of the circular orbit be R, and let the parameters of the new elliptical orbit be a and c, as shown
in the figure. Then R D a C c. At the moment of the collision, r does not change (r D R), but the speed r &P is cut
in half. Therefore &P is cut in half, and so h D r 2 &P is cut
in half. Let H be the value of r 2 &P for the circular orbit,
and let h be the value for the new elliptical orbit. Thus
h D H=2. We have
H2
RD
;
k
Thus v D
15.
c
a
p
#
D
4k$
:
`
1
1
"
a rP
$
Since the radial line from the sun to the planet sweeps
out equal areas in equal times, the fraction of the planet’s
period spend on the same side of the minor axis as the
sun is equal to the shaded area in the figure to the total
area of the ellipse, that is,
" 12 .2bc/
D
#ab
438
$
vP vA .
1
2 #ab
Fig. 11.6-13
1
1
"
rP
rA
&
2k %
1 " $ 2 " .1 C $/
`
2k$
2
.1 C $/
D vP "
`
#
$
2
v " vA2
vP " vA
1C
D vP2 " P
2
vP C vA
vP " vA
2
D vP "
.2vP / D vP vA :
2
It follows that 1 D 4 " 4$, so $ D 3=4. The new elliptical
orbit has eccentricity $ D 3=4.
S
#
D vP2 C
.1 C $/R
R
D
:
4.1 " $ 2 /
4.1 " $/
R
D 2k
v 2 D vP2 C 2k
$R
, so
4.1 " $ 2 /
R DcCa D
"
vA2
Using this result and the parameters of the orbit given in
the text, we obtain
h2
H2
R
aD
D
D
:
2
2
k.1 " $ /
4k.1 " $ /
4.1 " $ 2 /
Similarly, c D $a D
vP " vA
:
vP C vA
$D
1
2 #ab
" $ab
1
$
D " ;
#ab
2 #
where $ D c=a is the eccentricity of the orbit.
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 11.6
2#
Thus T D p
k
#
v2
2
" 0
r0
k
$!3=2
:
y
a
b
(PAGE 673)
A
c
a
b
a
c
S
P
x
Fig. 11.6-15
Fig. 11.6-16
17.
16. By conservation of energy, we have
#
$
k 1 2 h2
"
rP C 2 D "K
r
2
r
where K is a constant for the orbit (the total energy). The
term in the parentheses is v 2 , the square of the speed.
Thus
k
1
k 1 2
" v D "K D
" v2;
r
2
r0 2 0
where r0 and v0 are the given distance and speed. We
evaluate "K at perihelion.
The parameters of the orbit are
`D
h2
;
k
aD
h2
;
k.1 " $ 2 /
bD
p
h2
k 1 " $2
;
Let r1 .s/ and r2 .s/ be the distances from the point
P D r.s/ on the ellipse E to the two foci. (Here s denotes
arc length on E, measured from any convenient point.) By
symmetry
Z
Z
E
r1 .s/ ds D
E
r2 .s/ ds:
But r1 .s/ C r2 .s/ D 2a for any s. Therefore,
Z
E
r1 .s/ ds C
Z
E
r2 .s/ ds D
Z
E
2a ds D 2ac.E/:
R
Hence E r1 .s/ ds D ac.E/, and
Z
1
r1 .s/ ds D a:
c.E/ E
c D $a:
y
P
At perihelion P we have
r2
h2
r D a " c D .1 " $/a D
:
k.1 C $/
Since rP D 0 at perihelion, the speed there is v D r &P . By
Kepler’s second law, r 2 &P D h, so v D h=r D k.1 C $/= h.
Thus
k v2
"K D "
r
2
k2
1 k2
D 2 .1 C $/ "
.1 C $/2
h
2 h2
h
i
k2
D 2 .1 C $/ 2 " .1 C $/
2h
k2
k
D 2 .1 " $ 2 / D
:
2h
2a
Thus a D
k
. By Kepler’s third law,
"2K
T2 D
4# 2
4# 2 3
a D
k
k
#
k
"2K
$3
r1
F2
F1
x
E
Fig. 11.6-17
18. Start with
rR "
h2
k
D " 2:
r3
r
1
, where & D &.t /. Since r 2 &P D h
u.& /
(constant), we have
Let r.t / D
1 du P
du h
du
& D "r 2
D "h
u2 d&
d& r 2
d&
2
d 2u P
h2 d 2 u
2 2d u
rR D "h 2 & D " 2
D
"h
u
:
d&
r d& 2
d& 2
rP D "
:
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SECTION 11.6 (PAGE 673)
Thus "h2 u2
ADAMS and ESSEX: CALCULUS 9
d 2u
" h2 u3 D "ku2 , or
d& 2
Such orbits are bounded away from zero and infinity only
if A D 0, in which case they are circular.
k
d 2u
Cu D 2:
2
d&
h
Thus, the only possible orbits which are bounded
away from zero and infinity (i.e., which do not escape
to infinity or plunge into the sun) in a universe with an
inverse cube gravitational attraction are some circular orbits for which h2 D k. Such orbits cannot be considered
“stable” since even slight loss of energy would result in
decreased h and the condition h2 D k would no longer
be satisfied. Now aren’t you glad you live in an inverse
square universe?
This is the DE for simple harmonic motion with a constant forcing term (nonhomogeneous term) on the righthand side. It is easily verified that
uD
&
k %
1
C
$
cos.&
"
&
/
0
h2
is a solution for any choice of the constants $ and &0 .
Expressing the solution in terms of r, we have
rD
20.
h2 =k
;
1 C $ cos.& " &0 /
which is an ellipse if j$j < 1.
19. For inverse cube attraction, the equation of motion is
rR "
k
h2
D " 3;
r3
r
where r 2 &P D h is constant, since the force is central.
Making the same change of variables used in Exercise 18,
we obtain
d 2u
"h2 u2 2 " h2 u3 D "ku3 ;
d&
or
d 2 u k " h2
"
u D 0:
d& 2
h2
There are three cases to consider.
d 2u
C ! 2 u D 0, where
d& 2
! 2 D .h2 " k/= h2 . This has solution u D A cos !.& " &0 /.
Thus
1
:
rD
A cos !.& " &0 /
#
Note that r ! 1 as & ! &0 C
. There are no bounded
2!
orbits in this case.
CASE I. If k < h2 the DE is
21.
1
k
D v 2 " K by conservation of energy, if K < 0,
Since
r
2
then
k
# "K > 0;
r
k
so r $ " . The orbit is, therefore, bounded.
K
`
rD
, .$ > 1/.
1 C $ cos &
See the following figure.
Vertices: At V1 , & D 0 and r D `=.1 C $/.
At V2 , & D # and r D `=.1 " $/ D "`=.$ " 1/.
Semi-focal separation:
#
$
`
`
`$
1
C
D 2
:
cD
2 1C$
1"$
$ "1
The centre is .c; 0/.
Semi-transverse axis:
`$
`
`
aD 2
"
D 2
:
$ "1 $C1
$ "1
Semi-conjugate axis:
p
`
:
b D c 2 " a2 D p
$2 " 1
Direction of asymptotes (see figure):
b
a
1
& D tan!1 D cos!1 D cos!1 :
a
c
$
d 2u
" ! 2 u D 0, where
d& 2
! 2 D .k " h2 /= h2 . This has solution u D Ae !" C Be !!" .
Since u ! 0 or 1 as & ! 1, the corresponding solution
r D 1=u cannot be both bounded and bounded away from
zero. (Note that &P D h=r 2 # K > 0 for any orbit which
is bounded away from zero, so we can be sure & ! 1 on
such an orbit.)
y
CASE II. If k > h2 the DE is
c
"
F1
V1
b
F2
"
C
a
d 2u
D 0, which has
d& 2
solutions u D A& C B, corresponding to
CASE III. If k D h2 the DE is
rD
440
1
:
A& C B
Fig. 11.6-21
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INSTRUCTOR’S SOLUTIONS MANUAL
REVIEW EXERCISES 11
22. By Exercise 17, the asymptotes make angle
& D cos!1 .1=$/ with the transverse axis, as shown in the
figure. The angle of deviation ı satisfies 2& C ı D #, so
#
ı
& D " , and
2
2
ı
cos & D sin ;
2
(PAGE 675)
Therefore
2
v2 a
ı
ı
Dv1
D 1 cot D cot :
k
k
2
2
ı
sin & D cos :
2
Review Exercises 11 (page 675)
y
1.
Given that a + r D 0 and a + v D 0, we have
d
jr.t / " t v.t /j2
dt %
& %
&
D 2 r.t / " t v.t / + v.t / " v.t / " t a.t /
%
&
D 2 r.t / " t v.t / + a.t / D 0 " 0 D 0:
.c;0/
D
2"
"
rp
S
a
x
ı
2.
Fig. 11.6-22
r D t cos t i C t sin t j C .2# " t /k, .0 $ t p
$ 2#/ is a conical
helix wound around the cone z D 2# " x 2 C y 2 starting
at the vertex .0; 0; 2#/, and completing one revolution to
end up at .2#; 0; 0/. Since
By conservation of energy,
v D .cos t " t sin t /i C .sin t C t cos t /j " k;
v2
v2 k
" D constant D 1
2
r
2
the length of the curve is
for all points on the orbit. At perihelion,
LD
`
;
$C1
h.$ C 1/
h
D
:
v D vp D rp &P D
rp
`
r D rp D c " a D .$ " 1/a D
Since h2 D k`, we have
2
v1
D
vp2
Z
0
2!
!
p
p
p
2# C 2 C 4# 2
2
2
2 C t dt D # 2 C 4# Cln
p
2
units.
3.
The position of the particle at time t is
2k
"
rp
r D xi C x 2 j C 23 x 3 k;
2
h
2k
.$ C 1/2 "
.$ C 1/
`2
`
h
i
k
.$ C 1/2 " 2.$ C 1/
D
`
k 2
k
D .$ " 1/ D :
`
a
D
where x is an increasing function of t . Thie velocity is
vD
&
dx %
i C 2xj C 2x 2 k :
dt
Since the speed is 6, we have
2
Thus av1
D k.
If D is the perpendicular distance from the sun S to
an asymptote of the orbit (see the figure) then
D D c sin & D $a sin & D a
Da
cos.ı=2/
ı
D a cot :
sin.ı=2/
2
sin &
cos &
6D
dx
dx p
1 C 4x 2 C 4x 4 D .2x 2 C 1/ ;
dt
dt
so that dx=dt D 6=.2x 2 C 1/. The particle is at .1; 1; 32 /
when x D 1. At this time its velocity is
v.1/ D 2.i C 2j C 2k/:
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REVIEW EXERCISES 11 (PAGE 675)
ADAMS and ESSEX: CALCULUS 9
Also
5.
At x D 1, we have
D
16
.i C 2j C 2k/ C 2.4j C 8k/
3
8
."2i " j C 2k/:
3
6.
4. The position, velocity, speed, and acceleration of the particle are given by
7.
ˇ ˇp
ˇ dx ˇ
v D ˇˇ ˇˇ 1 C 4x 2
dt
# $2
d 2x
dx
aD
.i C 2xj/ C 2
j:
dt 2
dt
ks 2
ks 2
i C sin
j
2
2
2
ks
ks 2
a D "ks sin
i C ks cos
j
2
2
v & a D ksk:
Therefore the curvature at position s is
1 D jv & aj=v 3 D ks.
8.
p
p
3C4 2
aD"
.i C 2 2j/ C 2j:
9
442
If r D e !" , and &P D k, then rP D "e !" &P D "kr, and
rR D k 2 r. Since r D r rO , we have
v D rP rO C r &P "O D "kr rO C kr "O
a D .Rr " r &P 2 /Or C .r &R C 2rP &P /"O
Hence the acceleration is
If the particle is moving to the left, so that dx=dt < 0, a
similar calculation shows that at t D 3 its acceleration is
t
!t
Tangential acceleration: dv=dt
pD e "e .
2
Normal acceleration: v 1 D 2.
Since v D 2 cosh t , the minimum speed is 2 at time t D 0.
Z s
Z s
kt2
kt2
For x.s/ D
cos
dt , y.s/ D
sin
dt , we have
2
2
0
0
v D cos
dx
t
D p
dt
1 C 4x 2
p
4tx
dx
1 C 4x 2 " p
2
d x
1 C 4x 2 dt
D
:
dt 2
1 C 4x 2
p
p
3"4 2
.i C 2 2j/ C 2j:
9
2t j C e !t k
Since x.0/ D y.0/ D 0, the arc length along the curve,
measured from the origin, is s. Also,
Let us assume that the particle is moving to the right, so
that dx=dt > 0. Since the speed is t , we have
aD
p
dx
ks 2
dy
ks 2
D cos
;
D sin
;
ds
2
ds
2
so that the speed is unity:
s
# $2 # $2
dx
dy
C
D 1:
vD
ds
ds
r D xi C x 2 j
dx
vD
.i C 2xj/;
dt
p
If the particle is at . 2; 2/ at t D 3, then dx=dt D 1 at
that time, and
p
d 2x
3"4 2
D
:
dt 2
9
p
v D e t i C 2j " e !t k
a D e t i C e !t k
da
D e t i " e !t k
dt
p
p
v & a D 2e !t i " 2j " 2e t k
p
v D e 2t C 2 C e !2t D e t C e !t
p
jv & aj D 2.e t C e !t /
p
jv & aj
2
1D
D
v3
.e t C e !t /2
da
p
.v & a/ +
2
dt
+D
D t
D 1:
jv & aj2
.e C e !t /2
d 2x
6
dx
144x
D"
.4x/
D"
dt 2
.2x 2 C 1/2
dt
.2x 2 C 1/3
d 2x
.i C 2xj C 2x 2 k/
aD
dt 2
#
$
dx
dx
dx
C
2 j C 4x
k :
dt
dt
dt
a.1/ D "
r D et i C
D .k 2 r " k 2 r/Or C .0 " 2k 2 r/"O D "2k 2 r "O :
9.
r D a.t " sin t /i C a.1 " cos t /j
v D a.1 " cos t /i C a sin t j
p
v D a 1 " 2 cos t C cos2 t C sin2 t
p p
t
D a 2 1 " cos t D 2a sin if 0 $ t $ 2#.
2
The length of the cycloid from t D 0 to t D T $ 2# is
#
$
Z T
T
t
units.
s.T / D
2a sin dt D 4a 1 " cos
2
2
0
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INSTRUCTOR’S SOLUTIONS MANUAL
REVIEW EXERCISES 11
#
$
%
t
s &
10. s D 4a 1 " cos
) t D 2cos!1 1 "
D t .s/.
2
4a
Thus rQ .t / represents the same cycloid as r.t /, but translated #a units to the left and 2a units upward. From Exercise 11, the given cycloid is the evolute of its involute.
y
The required arc length parametrization of the cycloid is
%
&
%
&
r D a t .s/ " sin t .s/ i C a 1 " cos t .s/ j:
11.
(PAGE 675)
Q
A
From Exercise 9 we have
P
O / D v D .1 " cos t /i C sin t j
T.t
v
2 sin.t =2/
t
t
D sin i C cos j
2
2
1
t
1
t
cos i " sin j
O
O
dT
1 dT
2
2
2
2
D
D
t
ds
v dt
2a sin
2
#
$
t
1
cot i " j
D
4a
2
ˇ
ˇ
ˇ d TO ˇ
1
ˇ
ˇ
1.t / D ˇ
ˇD
ˇ ds ˇ
4a sin.t =2/
O
x
Fig. R-11-12
13.
The position vector of P is given by
r D R sin / cos & i C R sin / sin & j C R cos /k:
Mutually perpendicular unit vectors in the directions of
increasing R, / and & can be found by differentiating r
with respect to each of these coordinates and dividing the
resulting vectors by their lengths. They are
O
1 dT
O / D r.t / C
rC .t / D r.t / C *.t /N.t
2
.1.t // ds
$
#
t
16a2 sin2 .t =2/
cot i " j
D r.t / C
4a
2
t
t
t
D r.t / C 4a cos sin i " 4a sin2 j
2
2
2
D a.t " sin t /i C a.1 " cos t /j
C 2a sin t i " 2a.1 " cos t /j
D a.t C sin t /i " a.1 " cos t /j (let t D u " #)
D a.u " sin u " #/i C a.1 " cos u " 2/j:
O D d r D sin / cos & i C sin / sin & j C cos /k
R
dR
1 dr
O
#D
D cos / cos & i C cos / sin & j " sin /k
R d/
dr
1
D " sin & i C cos & j:
"O D
R sin / d&
The triad fOr; #O ; "O g is right-handed. This is the reason for
ordering the spherical polar coordinates .R; /; & / rather
than .R; &; //.
14.
This is the same cycloid as given by r.t / but translated
#a units to the right and 2a units downward.
12. Let P be the point with position vector r.t /
on the cycloid. By Exercise 9, the arc OP has
length 4a " 4a cos.t =2/, and so PQ has length
4a - arc OP D 4a cos.t =2/ units. Thus
t O
""!
PQ D 4a cos T.t
/
2#
$
t
t
t
sin i C cos j
D 4a cos
2
2
2
D 2a sin t i C 2a.1 C cos t /j:
By Kepler’s Second Law the position vector r from the
origin (the sun) to the planet sweeps out area at a constant
rate, say h=2:
dA
h
D :
dt
2
As observed in the text, dA=dt D r 2 &P =2, so r 2 &P D h, and
r & v D .r rO / & .Pr rO C r &P "O / D r 2 &P rO & "O D hk D h
is a constant vector.
15.
By Exercise 14, r & rP D r & v D h is constant, so, by
Newton’s second law of motion,
r & F.r/ D mr & rR D m
It follows that Q has position vector
""!
rQ D r C PQ
D a.t " sin t /i C a.1 " cos t /j C 2a sin t i C 2a.1 C cos t /j
D a.t C sin t /i C a.1 C cos t C 2/j (let t D u C #)
D a.u " sin u C #/i C a.1 " cos u C 2/j:
d
.r & rP / D 0:
dt
Thus F.r/ is parallel to r, and therefore has zero transverse component:
F.r/ D "f .r/Or
for some scalar function f .r/.
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REVIEW EXERCISES 11 (PAGE 675)
ADAMS and ESSEX: CALCULUS 9
16. By Exercise 15, F.r/ D m.Rr " r &P 2 /Or D "f .r/Or. We are
given that r D `=.1 C $ cos & /. Thus
which is appropriate since 2 is much smaller than g,
then
p dz
dr
2! &
' 22
i:
dt
dt
Breaking the DE into its components, we get
`
."$ sin & /&P
.1 C $ cos & /2
$` sin &
&P
D
.1 C $ cos & /2
h$
$ sin & 2 P
r &D
sin &
D
`
`
2
h$
h $ cos &
rR D
.cos & /&P D
:
`
`r 2
rP D "
p dz
d 2x
D 22 ;
2
dt
dt
z.t/ D 100 "
h2
h2 $ cos &
" 3
2
`r#
r $
h2
`
h2
D 2 $ cos & "
D " 2;
`r
r
`r
tD
(because .`=r/ D 1 C $ cos & ). Hence
mh2
:
`r 2
x'"
Challenging Problems 11 (page 676)
a) The angular velocity ! of the earth points northward
in the direction of the earth’s axis; in terms of the basis vectors defined at a point P at 45ı north latitude,
it points in the direction of j C k:
jCk
2
2D
2#
rad/s:
24 & 3;600
b) If v D "vk, then
c) If r.t / D x.t /i C y.t /j C z.t/k is the position of the
falling object at time t , then r.t / satisfies the DE
d 2r
dr
D "gk C 2! &
dt 2
dt
and the initial conditions r.0/ D 100k, r0 .0/ D 0. If
we use the approximation
444
2gt 3
x.t / D " p :
3 2
s
200
' 4:52;
g
2.
2#
9:8
p .4:52/3 ' "0:0155 m:
24 & 3;600 3 2
The object strikes the ground about 15.5 cm west of
P.
8
d
< v D k & v " 32k
dt
:
v.0/ D 70i
a) If v D v1 i C v2 j C v3 k, then k & v D v1 j " v2 i. Thus
the initial-value problem breaks down into component
equations as
8
8
8
< dv1 D "v
< dv2 D v
< dv3 D "32
2
1
dt
dt
dt
:
:
:
v1 .0/ D 70
v2 .0/ D 0
v3 .0/ D 0:
b) If r D xiCyjCzk denotes the position of the baseball
t s after it is thrown, then x.0/ D y.0/ D z.0/ D 0
and we have
p
22v
aC D 2! & v D " p .j C k/ & k D " 22vi:
2
dz
dr
'
k;
dt
dt
y.t / D 0;
at which time we have
This says that the magnitude of the force on the planet is
inversely proportional to the square of its distance from
the sun. Thus Newton’s law of gravitation follows from
Kepler’s laws and the second law of motion.
!D2 p ;
gt 2
;
2
Since g ' 9:8 m/s2 , the time of fall is
rR " r &P 2 D
1.
d 2z
D "g:
dt 2
Solving these equations (beginning with the last one),
using the initial conditions, we get
It follows that
f .r/ D
d 2y
D 0;
dt 2
dz
D v3 D "32t ) z D "16t 2 :
dt
dv2
d 2 v1
D"
Also,
D "v1 (the equation of simple
dt 2
dt
harmonic motion), so
v1 .t / D A cos t C B sin t;
v2 .t / D A sin t " B cos t:
Since v1 .0/ D 70, v2 .0/ D 0, x.0/ D 0, and
y.0/ D 0, we have
dx
D v1 D 70 cos t
dt
x.t / D 70 sin t
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dy
D v2 D 70 sin t
dt
y.t / D 70.1 " cos t /:
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INSTRUCTOR’S SOLUTIONS MANUAL
CHALLENGING PROBLEMS 11
d) If d r=dt D v and r.0/ D 0, then
v0 " .v0 + k/k
sin.!t /
r.t / D
!
%
&
v0 & k
C
1 " cos.!t / C .v0 + k/t k:
!
Since the three constant vectors
v0 " .v0 + k/k
v0 & k
;
; and .v0 + k/k
!
!
are mutually perpendicular, and the first two have the
same length because
At time t seconds after it is thrown, the ball is at
position
r D 70 sin t i C 70.1 " cos t /j " 16t 2 k:
c) At t D 1=5 s, the ball is at about .13:9; 1:40; "0:64/.
If it had been thrown without the vertical spin, its
position at time t would have been
r D 70t i " 16t 2 k;
jv0 " .v0 + k/kj D jv0 j sin & D jv0 & kj;
where & is the angle between v0 and k, the curve
r.t / is generally a circular helix with axis in the z
direction. However, it will be a circle if v0 + k D 0,
that is, if v0 is horizontal, and it will be a straight
line if v0 & k D 0, that is, if v0 is vertical.
so its position at t D 1=5 s would have been
.14; 0; "0:64/. Thus the spin has deflected the ball
approximately 1.4 ft to the left (as seen from above)
of what would have been its parabolic path had it not
been given the spin.
8
< d v D !v & k;
3.
dt
:
v.0/ D v0
a)
!D
qB
m
4.
d
dv
.v + k/ D
+ k D !.v & k/ + k D 0:
dt
dt
Thus v + k D constant D v0 + k.
dv
d
jvj2 D 2
+ v D 2!.v & k/ + v D 0,
dt
dt
so jvj D constant D jv0 j for all t .
Also,
b) If w.t / D v.t / " .v0 + k/k, then w + k D 0 by part (a).
Also, using the result of Exercise 23 of Section 1.3,
we have
d 2w
d 2v
dv
D
D!
& k D ! 2 .v & k/ & k
2
2
dt
dt h
dt
i
D "! 2 .k + k/v " .k + v/k
h
i
D "! 2 v " .v0 + k/k D "! 2 w;
the equation of simple harmonic motion. Also,
w.0/ D v0 " .v0 + k/k
w0 .0/ D !v0 & k:
c) Solving the above initial-value problem for w, we get
w D A cos.!t / C B sin.!t /;
A D w.0/ D v0 " .v0 + k/k;
!B D w0 .0/ D ! & k:
where
and
Therefore,
v.t / D w.t / C .v0 + k/k
h
i
D v0 " .v0 + k/k cos.!t / C .v0 & k/ sin.!t /
C .v0 + k/k:
(PAGE 676)
The arc length element on x D a.& " sin & /,
y D a.cos & " 1/ is (for & $ #)
q
ds D a .1 " cos & /2 C sin2 & d&
p
D a 2.1 " cos & / d& D 2a sin.&=2/ d&:
If the bead slides downward from rest at height y.&0 /
to height y.& /, its gravitational potential energy has decreased by
h
i
mg y.&0 / " y.& / D mga.cos &0 " cos & /:
Since there is no friction, all this potential energy is converted to kinetic energy, so its speed v at height y.& / is
given by
1 2
mv D mga.cos &0 " cos & /;
2
p
and so v D 2ga.cos &0 " cos & /. The time required for the bead to travel distance ds at speed v is
dt D ds=v, so the time T required for the bead to
slide from its starting position at & D &0 to the lowest
point on the wire, & D #, is
Z !
Z "D!
ds
1 ds
T D
D
d&
"0 v d&
"D"0 v
s
Z
sin.&=2/
2a !
d&
D
p
g "0
cos &0 " cos &
s
Z
2a !
sin.&=2/
D
p
d&
g "0
2 cos2 .&0 =2/ " 2 cos2 .&=2/
Let u D cos.&=2/
du D " 12 sin.&=2/ d&
r Z cos."0 =2/
a
du
p
D2
2
g 0
cos .&0 =2/ " u2
#
$ˇˇcos."0 =2/
r
a !1
u
ˇ
D2
sin
ˇ
g
cos.&0 =2/ ˇ
0
p
D # ag
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CHALLENGING PROBLEMS 11 (PAGE 676)
ADAMS and ESSEX: CALCULUS 9
A
which is independent of &0 .
vertical section
y
x
& D &0 starting point
E
y
horizontal section
p
g D .g= 2/.i " j/
& D#
B
D
."2; 2/
.2; 2/
C
Fig. C-11-4
5.
x
a) The curve BCD is the graph of an even function; a
fourth degree polynomial with terms of even degree
only will enable us to match the height, slope, and
curvature at D, and therefore also at C . We have
f .x/ D ax 4 C bx 2 C c
f 0 .x/ D 4ax 3 C 2bx
f 00 .x/ D 12ax 2 C 2b:
At D we have x D 2, so we need
Fig. C-11-5
6.
a) At time t%, the hare is
& at P D .0; vt / and the fox is
at Q D x.t /; y.t / , where x and y are such that
the slope dy=dx of the fox’s path is the slope of tline
y " vt
dy
D
.
PQ:
dx
x
b) Since
2 D f .2/ D 16a C 4b C c
1 D f 0 .2/ D 32a C 4b
0 D f 00 .2/ D 48a C 2b:
These equations yield a D "1=64, b D 3=8, c D 3=4,
so the curved track BCD is the graph of
1
y D f .x/ D
."x 4 C 24x 2 C 48/.
64
b) Since we are ignoring friction, the
p speed v of the car
during its drop is given by v D 2gs, where s is the
vertical distance dropped. (See the previous solution.)
At B the car p
has dropped about 7.2 m, so its speed
there is v ' 2.9:8/.7:2/ ' 11:9 m/s. At C the car
p
has dropped 10 " .c= 2/ ' 9:47 m, so its speed there
is v D 13:6 m/s. At D the car has dropped 10 m, so
its speed is v D 14:0 m/s.
c) At C we have x D 0, f 0 .0/ D 0, and
f 00 .0/ D 2b D 3=4. Thus the curvature of the track
at C is
jf 00 .0/j
3
1D
D :
4
.1 C .f 0 .0//2 /3=2
The normal accelerationpis v 2 1 ' 138:7 m/s2 (or
about 14g). Since v D 2gs, we have
p
p
p
2g ds
2g
19:6
dv
.13:6/ ' 9:78 m/s2 ;
D p
D p v' p
dt
2 s dt
2 s
2 9:47
so the total acceleration has magnitude approximately
p
.138:7/2 C .9:78/2 ' 139 m/s2 ;
#
$
dx d 2 y
d y " vt
D
dt dx 2
dt
x
$
#
dx
dy
" v " .y " vt /
x
dt
dt
D
2
x
#
$
1 dy dx
1
dx
D
" v " 2 .y " vt /
x dx dt
x
dt
1
dx
v
1
dx
v
D 2 .y " vt /
" " 2 .y " vt /
D" :
x
dt
x x
dt
x
d 2y
v
D"
.
dx 2
dx=dt
Since the fox’s speed is also v, we have
Thus x
#
dx
dt
$2
C
#
dy
dt
$2
D v2:
Also, the fox is always running to the left (towards the
y-axis from points where x > 0), so dx=dt < 0. Hence
v
# $ D
dx
"
dt
s
.dy=dt /2
1C
D
.dx=dt /2
s
1C
#
dy
dx
and so the fox’s path y D y.x/ satisfies the DE
which is again about 14g.
446
d 2 y dx
d dy
D
, we have
dt dx
dx 2 dt
d 2y
D
x
dx 2
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s
1C
#
dy
dx
$2
:
$2
;
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INSTRUCTOR’S SOLUTIONS MANUAL
CHALLENGING PROBLEMS 11
c) If u D dy=dx, then u D 0 and y D 0 when x D a,
and
p
du
x
D 1 C u2
Z dx
Z
du
dx
p
D
Let u D tan &
2
x
1Cu
du D sec2 & d&
Z
sec & d& D ln x C ln C
ln.tan & C sec & / D ln.C x/
p
u C 1 C u2 D C x:
b) Since yourpvelocity at any point has a northward component v= 2, and progress northward is measured
along a circle of radius a (a meridian), your colatitude /.t / satisfies
a
p
x
1 C u2 D " u
a
2
x
2xu
1 C u2 D 2 "
C u2
a
a
x2
2xu
D 2 "1
a
a
dy
x
a
DuD
"
dx
2a 2x
a
x2
" ln x C C1 :
yD
4a 2
a a
Since y D 0 when x D a, we have C1 D " C ln a,
4 2
so
x 2 " a2 a x
" ln
yD
4
2 a
Since /.0/ D #=2, it follows that
a) Since you are always travelling northeastpat speed v,
you are always moving north at rate v= 2. Therefore
you will reach the north pole in finite time
T D
#a
#a=2
p D p :
v= 2
2v
vt
#
" p :
2
a 2
Since
p your velocity also has an eastward component
v= 2 measured along a parallel of latitude that is a
circle of radius a sin /, your longitude coordinate &
satisfies
d&
v
.a sin //
D p
dt
2
$
#
d&
v
vt
D p
cos p
a 2 dt# a $2
Z
v
vt
&D p
sec
p
dt
a #2
a 2
$
vt
vt
C C:
D ln sec p C tan p
a 2
a 2
As & D 0 at t D 0, we have C D 0, and so
$
#
vt
vt
&.t / D ln sec p C tan p :
a 2
a 2
is the path of the fox.
7.
d/
v
D "p :
dt
2
/.t / D
Since u D 0 when x D a, we have C D 1=a.
(PAGE 676)
p
c) As t ! T D #a=. 2v/, the expression for
&.t / ! 1, so your path spirals around the north
pole, crossing any meridian infinitely often.
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SECTION 12.1 (PAGE 684)
ADAMS and ESSEX: CALCULUS 9
z
CHAPTER 12. PARTIAL
DIFFERENTIATION
.2;0;2/
.2;3;2/
Section 12.1 Functions of Several Variables
(page 684)
1.
zDx
xCy
.
f .x; y/ D
x"y
The domain consists of all points in the xy-plane not on
the line x D y.
p
2. f .x; y/ D xy.
Domain is the set of points .x; y/ for which xy # 0, that
is, points on the coordinate axes and in the first and third
quadrants.
x
.
C y2
The domain is the set of all points in the xy-plane except
the origin.
3. f .x; y/ D
x2
Fig. 12.1-11
12. f .x; y/ D sin x; 0 $ x $ 2#; 0 $ y $ 1
z
z D sin x
xy
4. f .x; y/ D 2
.
x " y2
The domain consists of all points not on the lines
x D ˙y.
p
5. f .x; y/ D 4x 2 C 9y 2 " 36.
The domain consists of all points .x; y/ lying on or outside the ellipse 4x 2 C 9y 2 D 36.
p
6. f .x; y/ D 1= x 2 " y 2 .
The domain consists of all points in the part of the plane
where jxj > jyj.
7.
f .x; y/ D ln.1 C xy/.
The domain consists of all points satisfying xy > "1, that
is, points lying between the two branches of the hyperbola
xy D "1.
2!
x
1
13.
z D f .x; y/ D y 2
z
z D y2
xyz
.
C y2 C z2
The domain consists of all points in 3-dimensional space
except the origin.
x2
y
e xyz
10. f .x; y; z/ D p
.
xyz
The domain consists of all points .x; y; z/ where xyz > 0,
that is, all points in the four octants x > 0, y > 0, z > 0;
x > 0, y < 0, z < 0; x < 0, y > 0, z < 0; and x < 0,
y < 0, z > 0.
11.
z D f .x; y/ D x
448
y
Fig. 12.1-12
8. f .x; y/ D sin!1 .x C y/.
The domain consists of all points in the strip
"1 $ x C y $ 1.
9. f .x; y; z/ D
y
3
x
x
Fig. 12.1-13
14.
f .x; y/ D 4 " x 2 " y 2 ; .x 2 C y 2 $ 4; x # 0; y # 0/
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 12.1
(PAGE 684)
z
z
4
2
z D4"x "y
2
x
2
y
y
2
x
Fig. 12.1-14
15.
z D f .x; y/ D
p
Fig. 12.1-17
18. f .x; y/ D 6 " x " 2y
x2 C y2
z
zD
p
x2 C y2
z
6
z D 6 " x " 2y
y
3
y
6
x
x
Fig. 12.1-18
Fig. 12.1-15
19.
16. f .x; y/ D 4 " x
f .x; y/ D x " y D C , a family of straight lines of slope 1.
y
2
cD!3
cD!2
z
z D 4 " x2
cD!1
cD0
x
cD1
cD2
y
cD3
x
x"y Dc
Fig. 12.1-19
Fig. 12.1-16
20.
17.
z D f .x; y/ D jxj C jyj
f .x; y/ D x 2 C 2y 2 D C , a family of similar ellipses
centred at the origin.
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SECTION 12.1 (PAGE 684)
ADAMS and ESSEX: CALCULUS 9
y
23.
x 2 C 2y 2 D c
x"y
D C , a family of straight lines through
xCy
the origin, but not including the origin.
f .x; y/ D
y
cD16
cD9
cD!1
cD4
cD!:5
cD1
cD0
x
cD:5
cD1
x
cD2
cD!2
Fig. 12.1-20
21.
x!y
xCy
f .x; y/ D xy D C , a family of rectangular hyperbolas
with the coordinate axes as asymptotes.
Dc
Fig. 12.1-23
y
cD9
cD4
cD1
24.
cD0
x
cD!1
cD!4
y
D C.
x2 C y2
!
"
1 2
This is the family x 2 C y " 2C
D 4C1 2 of circles
passing through the origin and having centres on the yaxis. The origin itself is, however, not on any of the level
curves.
f .x; y/ D
y
cD!9
cD1
xy D c
cD2
Fig. 12.1-21
cD3
x2
D C , a family of parabolas, y D x 2 =C ,
22. f .x; y/ D
y
with vertices at the origin and vertical axes.
x
y
cD1
cD!3
cD0:5
cD!2
cD2
cD!1
Fig. 12.1-24
x
x2
y
Dc
cD!2
cD!0:5
Fig. 12.1-22
450
cD!1
25.
f .x; y/ D xe !y D C .
x
This is the family of curves y D ln .
C
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y
x 2 Cy 2
Dc
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INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 12.1
y
29.
cD!1
(PAGE 684)
The graph of the function whose level curves are as shown
in part (a) of Figure 12.1.29 is a plane containing the yaxis and sloping uphill to the right. It is consistent with,
say, a function of the form f .x; y/ D y.
(a)
(b)
y
y
cD1
C D10
cD2
cD!2
x
x
cD4
cD!4
xe !y D c
C D!5
(c)
Fig. 12.1-25
26. f .x; y/ D
s
f .x; y/ D
s
y
C D5
C D5
1
1
" x2 D C ) y D 2
.
y
x C C2
y
1
" x2 D C
y
x
C D0
(d)
y
C D3
x
C D 0:8
C D0
x
C D!5
Fig. 12.1-29
C D1
30.
The graph of the function whose level curves are as shown
in part (b) of Figure 12.1.29 is a cylinder parallel to the
x-axis, rising from height zero first steeply and then more
and more slowly as y increases. Itpis consistent with, say,
a function of the form f .x; y/ D y C 5.
31.
The graph of the function whose level curves are as shown
in part (c) of Figure 12.1.29 is an upside down circular
cone with vertex at height 5 on the z-axis and base circle
in the xy-plane. It is consistent
with, say, a function of
p
the form f .x; y/ D 5 " x 2 C y 2 .
C D2
x
Fig. 12.1-26
27.
The landscape is steepest at B where the level curves are
closest together.
500
A
500
600
C
32.
The graph of the function whose level curves are as
shown in part (d) of Figure 12.1.29 is a cylinder (possibly parabolic) with axis in the yz-plane, sloping upwards
in the direction of increasing y. It is consistent with, say,
a function of the form f .x; y/ D y " x 2 .
33.
The curves y D .x " C /2 are all horizontally shifted
versions of the parabola y D x 2 , and they all lie in the
half-plane y # 0. Since each of these curves intersects
all of the others, they cannot be level curves of a function
f .x; y/ defined in y # 0. To be a family of level curves
of a function f .x; y/ in a region, the various curves in the
family cannot intersect one another in that region.
34.
4z 2 D .x " z/2 C .y " z/2 .
If z D c > 0, we have .x " c/2 C .y " c/2 D 4c 2 , which is
a circle in the plane z D c, with centre .c; c; c/ and radius
2c.
N
B
W
E
S
400
300
200 100
Fig. 12.1-27
28. C is a “pass” between two peaks to the east and west.
The land is level at C and rises as you move to the east
or west, but falls as you move to the north or south.
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451
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SECTION 12.1 (PAGE 684)
ADAMS and ESSEX: CALCULUS 9
y
40.
cD3
x2 C y2
.
z2
The equation f .x; y; z/ D c can be rewritten
x 2 C y 2 D C 2 z 2 . The level surfaces are circular cones
with vertices at the origin and axes along the z-axis.
f .x; y; z/ D
cD2
41. f .x; y; z/ D jxj C jyj C jzj.
The level surface f .x; y; z/ D c > 0 is the surface of
the octahedron with vertices .˙c; 0; 0/, .0; ˙c; 0/, and
.0; 0; ˙c/. (An octahedron is a solid with eight planar
faces.)
cD1
x
42.
.x " c/2 C .y " c/2 D 4c 2
Fig. 12.1-34
The graph of the function z D z.x; y/ # 0 defined by
the given equation is (the upper half of) an elliptic cone
with axis along the line x D y D z, and circular crosssections in horizontal planes.
35.
f .x; y; z; t / D x 2 C y 2 C z 2 C t 2 .
The “level hypersurface”
p f .x; y; z; t / D c > 0 is the
“4-sphere” of radius c centred at the originpin R4 . That
is, it consists of all points in R4 at distance c from the
origin.
43.
z
2
2
2
a) f .x; y/ D C
p is x C y D C implies that
f .x; y/ D x 2 C y 2 .
zD
b) f .x; y/ D C is x 2 C y 2 D C 4 implies that
f .x; y/ D .x 2 C y 2 /1=4 .
1
1 C x2 C y2
c) f .x; y/ D C is x 2 C y 2 D C implies that
f .x; y/ D x 2 C y 2 .
d) f .x; y/ D Cpis x 2 C y 2 D .ln C /2 implies that
f .x; y/ D e
x 2 Cy 2
y
.
36. If the level surface f .x; y; z/ D C is the plane
x
Fig. 12.1-43
x
y
z
C
C
D 1;
C3
2C 3
3C 3
that is, x C
z
y
C D C 3 , then
2
3
44.
z
zD
%
z &1=3
y
f .x; y; z/ D x C C
:
2
3
cos x
1 C y2
37. f .x; y; z/ D x 2 C y 2 C z 2 .
The
p level surface f .x; y; z/ D c > 0 is a sphere of radius
c centred at the origin.
38. f .x; y; z/ D x C 2y C 3z.
The level surfaces are parallel planes having common normal vector i C 2j C 3k.
y
x
39. f .x; y; z/ D x 2 C y 2 .
The level p
surface f .x; y; z/ D c > 0 is a circular cylinder
of radius c with axis along the z-axis.
452
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"5 $ x $ 5;
"5 $ y $ 5
Fig. 12.1-44
lOMoARcPSD|6566483
INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 12.2
(PAGE 689)
z
45.
z
zD
zD
y
1 C x2 C y2
1
xy
y
y
"4 $ x $ 4
"4 $ y $ 4
x
Fig. 12.1-48
x
Fig. 12.1-45
Section 12.2 Limits and Continuity
(page 689)
46.
z
1.
x
zD 2
.x " 1/2 C y 2
2.
3.
y
lim
.x;y/!.0;0/
xy C x 2 D 2."1/ C 22 D 2
p
x2 C y2 D 0
x2 C y2
does not exist.
y
.x;y/!.0;0/
lim
If .x; y/ ! .0; 0/ along x D 0, then
If .x; y/ ! .0; 0/ along y
x2 C y2
D 1 C x 2 ! 1.
y
x
Fig. 12.1-46
47.
lim
.x;y/!.2;!1/
4.
x
.
x2 C y2
Then jf .x; 0/j D j1=xj ! 1 as x ! 0.
But jf .0; y/j D 0 ! 0 as y ! 0.
Thus lim.x;y/!.0;0/ f .x; y/ does not exist.
Let f .x; y/ D
z
5.
z D xy
6.
lim
.x;y/!.1;!/
lim
cos.xy/
cos #
D
D "1
1 " x " cos y
1 " 1 " cos #
x 2 .y " 1/2
D 0, because
C .y " 1/2
.x;y/!.0;1/ x 2
y
ˇ 2
ˇ
ˇ x .y " 1/2 ˇ
ˇ $ x2
0 $ ˇˇ 2
x C .y " 1/2 ˇ
7.
x
Fig. 12.1-47
and x 2 ! 0 as .x; y/ ! .0; 1/.
ˇ
ˇ
ˇ y3 ˇ
y2
ˇ
ˇ
ˇ x 2 C y 2 ˇ $ x 2 C y 2 jyj $ jyj ! 0
as .x; y/ ! .0; 0/. Thus
48. The graph is asymptotic to the coordinate planes.
D
x2 C y2
D y ! 0.
y
x 2 , then
8.
lim
.x;y/!.0;0/
lim
.x;y/!.0;0/ x 2
y3
D 0.
C y2
sin 0
sin.x " y/
D
D 0:
cos.x C y/
cos 0
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