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C o n c e p t s a n d A p p l i c a t i o n s Second Edition
P a u l A . F o e r s t e r
Solutions Manual
Calculus
Project Editor: Josephine Noah
Project Administrator: Shannon Miller
Consulting Editor: Christopher David
Accuracy Checkers: Jenn Berg, Dudley Brooks
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Production Editor: Angela Chen
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Text and Cover Designer: Jenny Somerville
Art Editors: Jason Luz, Laura Murray Productions
Art and Design Coordinator: Kavitha Becker
Cover Photo Credit: Alec Pytlowany/Masterfile
Compositor: Interactive Composition Corporation
Printer: Alonzo Printing
Executive Editor: Casey FitzSimons
Publisher: Steven Rasmussen
© 2005 by Key Curriculum Press. All rights reserved.
Limited Reproduction Permission
The publisher grants the teacher who purchases Calculus: Concepts and Applications Solutions Manual the right to reproduce material for use in his or her own classroom. Unauthorized copying of Calculus: Concepts and Applications Solutions Manual constitutes copyright infringement and is a violation of federal law.
The Geometer’s Sketchpad, Dynamic Geometry, and Key Curriculum Press are registered trademarks of Key Curriculum Press. Sketchpad is a trademark of Key Curriculum Press. Fathom Dynamic Statistics is a trademark of KCP Technologies, Inc. All other registered trademarks and trademarks in this book are the property of their respective holders.
Key Curriculum Press 1150 65th Street Emeryville, CA 94608 editorial@keypress.com www.keypress.com
Printed in the United States of America
10 9 8 7 6 5 4 12 11 10 09 08
ISBN: 978-1-55953-657-8
Contents
Chapter 1 Limits, Derivatives, Integrals, and Integrals .................................................... 1
Chapter 2 Properties of Limits ................................................................................................... 9
Chapter 3 Derivatives, Antiderivatives, and Indefinite Integrals .............................. 28
Chapter 4 Products, Quotients, and Parametric Functions ......................................... 51
Chapter 5 Definite and Indefinite Integrals ....................................................................... 82
Chapter 6 The Calculus of Exponential and Logarithmic Functions ..................... 118
Chapter 7 The Calculus of Growth and Decay ............................................................ 139
Chapter 8 The Calculus of Plane and Solid Figures ................................................... 168
Chapter 9 Algebraic Calculus Techniques for the Elementary Functions .......... 213
Chapter 10 The Calculus of Motion—Averages, Extremes, and Vectors ............. 266
Chapter 11 The Calculus of Variable-Factor Products ................................................. 292
Chapter 12 The Calculus of Functions Defined by Power Series ............................. 313
iii
Overview
This Solutions Manual contains the answers to all problems in Calculus: Concepts andApplications. Solutions or key steps in the solutions are presented for all but the simplestproblems.
In most cases the solutions are presented in the form your students would be expected touse. For instance, decimal approximations are displayed as exact answers using ellipsisformat for a mathematical-world answer, then rounded to an appropriate number ofdecimal places with units of measurement applied for the corresponding real-worldanswer. An answer such as f(3) = 13.7569... ≈ 13.8 cm indicates that the precise answer,13.7569... , has been retained in memory in the student’s calculator without round-off forpossible use in subsequent computations. The ellipses indicate that the student chooses notto write all the digits on his or her paper.
Because the problems applying to the real world may be somewhat unfamiliar to both youand your students, fairly complete solutions are presented for these. Often commentary isincluded over and above what the student would be expected to write to further guide yourevaluation of students’ solutions, and in some cases reference is provided to later sectionsin which more sophisticated solutions appear. Later in the text, the details of computingdefinite integrals by the fundamental theorem are omitted because students are usuallyexpected to do these numerically. However, exact answers such as V = 8π/3 are presentedwhere possible in case you choose to have your students do the algebraic integration.
Solutions are not presented for journal entries because these are highly individual for eachstudent. The “prompts” in most problems calling for journal entries should be sufficient toguide students in making their own responses.
Where programs are called for, you may use as a model the programs in the Instructor’sResource Book. Check the publisher’s Web page (see the address on the copyright page ofthis manual) for further information on programs for specific models of the graphingcalculator.
If you or your students find any mistakes, please report them to Key Curriculum Press bysending in the Correction/Comment Form in the back of this book.
Paul A. Foerster
v
Calculus Solutions Manual Problem Set 1-2 1© 2005 Key Curriculum Press
Chapter 1—Limits, Derivatives, Integrals, and Integrals
Problem Set 1-11. a. 95 cm
b. From 5 to 5.1: average rate ≈ 26 34. cm/sFrom 5 to 5.01: average rate ≈ 27 12. cm/sFrom 5 to 5.001: average rate ≈ 27 20. cm/s So the instantaneous rate of change of d att = 5 is about 27.20 cm/s.
c. Instantaneous rate would involve divisionby zero.
d. For t = 1.5 to 1.501, rate ≈ −31.42 cm/s.The pendulum is approaching the wall: Therate of change is negative, so the distance isdecreasing.
e. The instantaneous rate of change is the limitof the average rates as the time intervalapproaches zero. It is called the derivative.
f. Before t = 0, the pendulum was not yetmoving. For large values of t, the pendulum’smotion will die out because of friction.
2. a. x = 5: y = 305, price is $3.05x = 10: y = 520, price is $5.20x = 20: y = 1280, price is $12.80
b. x = 5.1, rate ≈ 46 822. ¢/ftx = 5.01, rate ≈ 46.9820… ¢/ftx = 5.001, rate ≈ 46.9982… ¢/ft
c. 47 ¢/ft. It is called the derivative.
d. x = 10: 44 ¢/ft. x = 20: 128 ¢/ft
e. The 20-ft board costs more per foot than the10-ft board. The reason is that longer boardsrequire taller trees, which are harder to find.
Problem Set 1-2Q1. Power function, or polynomial function
Q2. f (2) = 8 Q3. Exponential function
Q4. g (2) = 9 Q5.
1
1x
h(x)
Q6. h (5) = 25 Q7. y = ax2 + bx + c, a ≠ 0
Q8. y = x Q9. y = |x|
Q10. Derivative
1. a. Increasing slowly b. Increasing fast
2. a. Increasing fast b. Decreasing slowly
3. a. Decreasing fast b. Decreasing slowly
4. a. Decreasing slowly b. Increasing slowly
5. a. Increasing fast b. Increasing slowly
c. Decreasing slowly d. Increasing fast
6. a. Decreasing fast b. Increasing slowly
c. Increasing fast d. Decreasing fast
7. a. Increasing slowly b. Increasing slowly
c. Increasing slowly
8. a. Decreasing fast b. Decreasing fast
c. Decreasing fast
9. a. Increasing fast b. Neither increasingnor decreasing
c. Increasing fast d. Increasing slowly
10. a. Decreasing slowly b. Decreasing fast
c. Decreasing fast d. Neither increasingnor decreasing
11. a.
100 200
50
100
T(x) (°C)
x (s)
x = 40: rate ≈ °1 1. /sx = 100: rate = 0°/sx = 140: rate ≈ − °0 8. /s
b. • Between 0 and 80 s the water is warmingup, but at a decreasing rate.
• Between 80 and 120 s the water is boiling,thus staying at a constant temperature.
• Beyond 120 s the water is cooling down,rapidly at first, then more slowly.
12. a.
1 2 3 4 5 6 7 8
10
20
30
40
50
60
70
v(x) (ft/s)
x (s)
2 Problem Set 1-2 Calculus Solutions Manual © 2005 Key Curriculum Press
x = 2: rate 18 (ft/s)/s x = 5: rate = 0 (ft/s)/s x = 6: rate 11 (ft/s)/s
b. Units are (ft/s)/s, sometimes written as ft/s2. The physical quantity is acceleration.
13. a.
3 4 7
2
18
h(x)
x
• Increasing at x = 3
• Decreasing at x = 7
b. h (3) = 17, h(3.1) = 17.19
Average rate =0.19
0.1= 1.9 ft/s
c. From 3 to 3.01:
average rate =0.0199
0.01= 1.99 ft/s
From 3 to 3.001:
average rate =0.001999
0.001= 1.99 ft/s
The limit appears to be 2 ft/s.
d. h (7) = 9, h(7.001) = 8.993999
Average rate =0.006001
0.001= 6.001 ft /s
The derivative at x = 7 appears to be 6 ft/s. The derivative is negative because h(x) is decreasing at x = 7.
14. a.
10
100
300
500f(t)
t
DecreaseIncrease
Not much
b. Enter y2 =y1 (x ) y1(1)
x 1
t r(t) = y2 (foxes/year)
0.97 110.5684…
0.98 109.7361…
0.99 108.9001…
1 undefined
1.01 107.2171…
1.02 106.3703…
1.03 105.5200…
c. Substituting 1 for t causes division by zero, so r(1) is undefined. Estimate: r approaches the average of r(0.99) and r(1.01), 108.0586… foxes/year. (Actual is 108.0604… .) The instantaneous rate is called the derivative.
d.
f (4.01) f (4)
0.01= 129.9697…
f (4) f (3.99)
0.01= 131.4833…
Instantaneous rate = ( 129.9697… 131.4833…)/2 = 130.7265… foxes/year (actual: 130.7287…) The answer is negative because the number of foxes is decreasing.
15. a. Average rate =a(2.1) a(2)
0.1=
52.9902… mm2/h
b. r (t) =200(1.2 t ) 200(1.22 )
t 2
2
20
40
60
r(t ) (mm2/hr)
t (mm)
r(2) is undefined.
c. r(2.01) = 52.556504… 52.556504… 52.508608… = 0.04789… Use the solver to find t when r(t) = 52.508608… + 0.01 = 52.518608… . t = 2.002088… , so keep t within 0.002 unit of 2.
16. a. v(x ) =4
3x 3 v(6) = 288
b. 6 to 6.1: average rate =
43 (6.13 63 )
0.1=
146.4133…
5.9 to 6: average rate =43 (63 5.93)
0.1=
141.6133…
Estimate of instantaneous rate is (146.4133… + 141.6133… )/2 = 144.0133… = 452.4312… cm3/cm.
c. r (x ) =43 x3 4
3 63
x 6
) (cm /cm)
6
r(x
144π�
48π�
3
x
r(6) is undefined.
Calculus Solutions Manual Problem Set 1-3 3 © 2005 Key Curriculum Press
d. r(6.1) = 146.4133… = 459.9710… r(6.1) is 7.5817… units from the derivative. Use the solver feature to find x if r(x) = 144 + 0.1. x = 6.001326… , so keep x within 0.00132… unit of 6.
17. a. i. 1.0 in./s ii. 0.0 in./s iii. 1.15 in./s
b. 1.7 s, because y = 0 at that time
18. a. i. 0.395 in./min ii. 0.14 in./min
iii. 0.105 in./min
b. The rate is negative, because y is decreasing as the tire goes down.
19. a. Quadratic (or polynomial)
b. f(3) = 30
c. Increasing at about 11.0 (2.99 to 3.01)
20. a. Quadratic (or polynomial)
b. f(1) = 12
c. Increasing at about 6.0 (0.99 to 1.01)
21. a. Exponential
b. Increasing, because the rate of change from 1.99 to 2.01 is positive.
22. a. Exponential
b. Increasing, because the rate of change from 3.01 to 2.99 is positive.
23. a. Rational algebraic
b. Decreasing, because the rate of change from 3.99 to 4.01 is negative.
24. a. Rational algebraic
b. Increasing, because the rate of change from 2.01 to 1.99 is positive.
25. a. Linear (or polynomial)
b. Decreasing, because the rate of change from 4.99 to 5.01 is negative.
26. a. Linear (or polynomial)
b. Increasing, because the rate of change from 7.99 to 8.01 is positive.
27. a. Circular (or trigonometric)
b. Decreasing, because the rate of change from 1.99 to 2.01 is negative.
28. a. Circular (or trigonometric)
b. Decreasing, because the rate of change from 0.99 to 1.01 is negative.
29. • Physical meaning of a derivative: instantaneous rate of change
• To estimate a derivative graphically: Draw a tangent line at the point on the graph and measure its slope.
• To estimate a derivative numerically: Take a small change in x, find the corresponding
change in f(x), then divide. Repeat, using a smaller change in x. See what number these average rates approach as the change in x approaches zero.
• The numerical method illustrates the fact that the derivative is a limit.
30. Problems 13 and 14 involve estimating the value of a limit.
Problem Set 1-3 Q1. 72 ft2 Q2. y = cos x
Q3. y = 2x Q4. y = 1/x
Q5. y = x2 Q6. f(5) = 4
Q7. Q8.
x
y
x
y
Q9. Q10. x = 3
x
y
1. f(x) = 0.1x2 + 7 2. f(x) = 0.2x2 + 8
a. Approximately 30.8 a. Approximately 22.2
b. Approximately 41.8 b. Approximately 47.1
x
f (x)7
–1 5 6
f(x)
x
–2 3 5
8
3. h(x) = sin x 4. g( x) = 2 x+ 5
a. Approximately 2.0 a. Approximately 7.9
b. Approximately 1.0 b. Approximately 12.2
1
3
x
h(x)
–1 1 2
6
x
g(x)
4 Problem Set 1-4 Calculus Solutions Manual© 2005 Key Curriculum Press
5. There are approximately 6.8 squares between thecurve and the x-axis. Each square represents(5)(20) = 100 feet. So the distance is about(6.8)(100) = 680 feet.
6. There are approximately 53.3 squares between thecurve and the x-axis. Each square represents(0.5)(10) = 5 miles. So the distance is about(53.3)(5) = 266.5 miles.
7. Derivative
≈ =tan . – tan .
. – ..
1 01 0 99
1 01 0 993 42K
8. Derivative = −7 (exactly, because that is theslope of the linear function)
9. a.
100
5 108.7
t
v(t)
60
The range is 0 ≤ y ≤ 32.5660… .
b. Using the solver, x = 8.6967… ≈ 8.7 s.
c. By counting squares, distance ≈ 150 ft.The concept used is the definite integral.
d. Rate ≈ −−
=v v( . ) ( . )
. ..
5 01 4 99
5 01 4 993 1107K
About 3.1 (ft/s)/sThe concept is the derivative.The rate of change of velocity is calledacceleration.
10. a.
2
10
5
1 3 4 5
t
v(t)
b. v(4) = 9.3203… ≈ 9.3 ft/s
Domain: 0 ≤ t ≤ 4Range: 0 ≤ v(t) ≤ 9.3203…
c. By counting squares, the integral from t = 0to t = 4 is about 21.3 ft. The units of theintegral are (ft/s) · s = ft. The integral tellsthe length of the slide.
d. Rate ≈
−=v v( . ) – ( . )
. ..
3 01 2 99
3 01 2 991 8648K
About 1.86 (ft/s)/sThe derivative represents the acceleration.
11. From t = 0 to t = 5, the object travels about11.4 cm. From t = 5 to t = 9, the object travelsback about 4.3 cm. So the object is located about11.4 − 4.3 = 7.1 cm from its starting point.
12. See the text for the meaning of derivative.
13. See the text for the meaning of definite integral.
14. See the text for the meaning of limit.
Problem Set 1-4Q1. y changes at 30 Q2. Derivative ≈ −500
Q3. Q4. f (3) = 9
x
y
Q5. 100 Q6. sin (π/2) = 1
Q7. 366 days Q8. Derivative
Q9. Definite integral Q10. f (x) = 0 at x = 4
1. a.
30
20,000
t
v(t)
b. Integral ≈ + + + +5 0 5 0 5 10 15( . ( ) ( ) ( ) ( )v v v vv(20) + v(25) + 0.5v(30)) = 5(56269.45…) =281347.26… ≈ 281 000, ftThe sum overestimates the integral becausethe trapezoids are circumscribed about theregion and thus include more area.
c. The units are (ft/s)(s), which equals feet, sothe integral represents the distance thespaceship has traveled.
d. Yes, it will be going fast enough, becausev(30) = 27,919.04… , which is greater than27,000.
2. a. v(t) = 4 + sin 1.4t
5
t
3
v(t)
Calculus Solutions Manual Problem Set 1-4 5© 2005 Key Curriculum Press
b. A definite integral has the units of thex-variable times the y-variable. Distance =rate × time. Because v(t) is distance/time and t is time, their product is expressed inunits of distance.
c. See graph in part a.Distance ≈ + + +0 5 0 5 0 0 5 1. ( . ( ) ( . ) ( )v v vv(1.5) + v(2) + v(2.5) + 0.5v(3)) =0.5(26.041…) = 13.02064… ≈ 13.0 ft
d. v(3) = 3.128… ≈ 3.1 mi/hMaximum speed was 5 mi/h at about 1.12 h.
3. Distance ≈ + + + + + =0 6 150 230 150 90 40 0. ( )396 ft
4. Volume ≈ 3(2500 + 8000 + 12000 + 13000 +11000 + 7000 + 4000 + 6000 + 4500) =204,000 ft3
5. Programs will vary depending on calculator. Seethe program TRAPRULE in the Instructor’sResource Book for an example. The programgives T20 = 23.819625.
6. See the program TRAPDATA in the Instructor’sResource Book for an example. The programgives T7 = 33, as in Example 2.
7. a.
1 4
7
f(x)
x
b. T10 = 18.8955T20 = 18.898875T50 = 18.89982These values underestimate the integral,because the trapezoids are inscribed in theregion.
c. T10: 0.0045 unit from the exact answerT20: 0.001125 unit from the exact answerT50: 0.00018 unit from the exact answerTn is first within 0.01 unit of 18.9 whenn = 7.T7 = 18.8908… , which is 0.0091… unitfrom 18.9.Because Tn is getting closer to 18.9 as nincreases, Tn is within 0.01 unit of 18.9 forall n ≥ 7.
8. a.
1 3
1
g(x)
x
b. T10 = 8.6700…T20 = 8.6596…Τ50 = 8.65672475…These values overestimate the integral,because the trapezoids are circumscribedabout the region.
c. T10: 0.01385… unit from answerT20: 0.003465… unit from answerT50: 0.0005545… unit from answerTn is first within 0.01 of 8.65617024… whenn = 12.T12 = 8.665795… , which is 0.009624… unitfrom 8.65617024… .Because Tn is getting closer to the exactanswer as n increases, Tn is within 0.01 unitof the answer for all n ≥ 12.
9. From the given equation,
y x= ±( / ) – .40 110 1102 2 Using the trapezoidal
rule program on the positive branch with n = 100increments gives 6904.190… for the top half ofthe ellipse. Doubling this gives an area of13,808.38… cm2. The estimate is too lowbecause the trapezoids are inscribed within theellipse. The area of an ellipse is πab, where aand b are the x- and y-radii, respectively. Sothe exact area is π (110)(40) = 4400π =13,823.007… cm2, which agrees both with theanswer and with the conclusion that thetrapezoidal rule underestimates the area.
10. Integral = 1(0.0 + 2.1 + 7.9 + 15.9 + 23.8 +29.7 + 31.8 + 29.7 + 23.8 + 15.9 + 7.9 +2.1 + 0) = 190.6The integral will have the units (in.2)(in.) = in.3,representing the volume of the football.
11. n = 10: integral ≈ 21.045n = 100: integral ≈ 21.00045n = 1000: integral ≈ 21.0000045Conjecture: integral = 21The word is limit.
12. The trapezoidal rule with n = 100 givesintegral ≈ 156.0096.Conjecture: integral = 156
13. If the trapezoids are inscribed (graph concavedown), the rule underestimates the integral.If the trapezoids are circumscribed (graph concaveup), the rule overestimates the integral.
Concave downInscribed trapezoids
Underestimates integral
Concave upCircumscribed trapezoids
Overestimates integral
6 Problem Set 1-6 Calculus Solutions Manual© 2005 Key Curriculum Press
Problem Set 1-51. Answers will vary.
Problem Set 1-6
Review Problems
R1. a. When t = 4, d = 90 − 80 sin [1, 2(4 − 3)] ≈15.4 ft.
b. From 3.9 to 4: average rate ≈ –40.1 ft/sFrom 4 to 4.1: average rate ≈ −29 3. ft/sInstantaneous rate ≈ −34.7 ft/sThe distance from water is decreasing, so he isgoing down.
c. Instantaneous rate ≈ − ≈d d( . ) ( . )
..
5 01 4 99
0 0270 8
d. Going up at about 70.8 ft/s
e. Derivative
R2. a. Physical meaning: instantaneous rate ofchange of a functionGraphical meaning: slope of a tangent line toa function at a given point
b. x = − 4: decreasing fastx = 1: increasing slowlyx = 3: increasing fastx = 5: neither increasing nor decreasing
c. From 2 to 2.1:
average rate = − =5 5
0 143 6547
2 1 2.
.. K
From 2 to 2.01:
average rate = − =5 5
0 0140
2 01 2.
..5614K
From 2 to 2.001:
average rate = − =5 5
0 00140 2683
2 001 2.
.. K
Differences between average rates andinstantaneous rates, respectively:43.6547… − 40.235947… = 3.4187…40.5617… − 40.235947… = 0.3255…40.2683… − 40.235947… = 0.03239…The average rates are approaching theinstantaneous rate as x approaches 2.The concept is the derivative.The concept used is the limit.
d. t = 2: 3.25 m/st = 18: 8.75 m/st = 24: 11.5 m/sHer velocity stays constant, 7 m/s, from 6 sto 16 s. At t = 24, Mary is in her final sprinttoward the finish line.
R3. By counting squares, the integral isapproximately 23.2.Distance ≈ 23 2. ft (exact answer: 23.2422…)Concept: definite integral
R4. a.
1 4
x
5
f (x)
The graph agrees with Figure 1-6c.
b. By counting squares, integral ≈ 15.0.(Exact answer is 15.)
c. T6 = 0.5(2.65 + 5.575 + 5.6 + 5.375 + 4.9 +4.175 + 1.6) = 14.9375The trapezoidal sum underestimates theintegral because the trapezoids are inscribed inthe region.
d. T50 = 14.9991; Difference = 0.0009T100 = 14.999775; Difference = 0.000225The trapezoidal sums are getting closer to 15.Concept: limit
R5. Answers will vary.
Concept Problems
C1. a. f (3) = 32 − 7.3 + 11 − 1
b. f (x) − f (3) = x2 − 7x + 11 + 1 = x2 − 7x + 12
c. f x f
x
x x
x
x x
x
( ) – ( )
–
–
–
( – )( – )
–
3
3
7 12
3
4 3
3
2
= + = =
x − 4, if x ≠ 3
d. The limit is found by substituting 3 for xin (x − 4).Limit = exact rate = 3 − 4 = −1
C2. The line through (3, f (3)) with slope −1 isy = −x + 2.
3
2 x
f(x)
The line is tangent to the graph. Zooming in bya factor of 10 on the point (3, 2) shows that thegraph becomes straighter and looks almost likethe tangent line. (Soon students will learn thatthis property is called local linearity.)
Calculus Solutions Manual Problem Set 1-6 7© 2005 Key Curriculum Press
2
3
C3. a. f xx x
x
x x
x( )
( )( )= − +−
= − −−
=4 19 21
3
4 7 3
3
2
4x − 7, x ≠ 3When x = 3, 4x − 7 = 4 ⋅ 3 − 7 = 5.
b.
1 2 3 4 5 6
1
2
3
4
5
6f(x) (ft)
x (s)
5.8
4.2
2.8 3.2
c. 5.8 = 4(3 + δ ) − 7 4.2 = 4(3 − δ ) − 75.8 = 12 + 4δ − 7 4.2 = 12 − 4δ − 74δ = 0.8 −4δ = −0.8δ = 0.2 δ = 0.2
d. 4(3 + δ ) − 7 = 5 + ε12 + 4δ − 7 = 5 + ε
4δ = εδ ε= 1
4
There is a positive value of δ, namely 14 ε, for
each positive value of ε, no matter how smallε is.
e. L = 5, c = 3. “. . . but not equal to 3” isneeded so that you can cancel the (x − 3)factors without dividing by zero.
Chapter Test
T1. Limit, derivative, definite integral, indefiniteintegral
T2. See the text for the definition of limit.
T3. Physical meaning: instantaneous rate
T4.
2 5
3
6
x
y
T5. Concept: definite integralBy counting squares, distance ≈ 466.(Exact answer is 466.3496… .)
T6.
5 10 15 20 25 30 35 40
5
10
15
20
25
Speed (ft/s)
Time (s)
T7 = 5(2.5 + 5 + 5 + 10 + 20 + 25 + 20 + 5) =462.5Trapezoidal rule probably underestimates theintegral, but some trapezoids are inscribed andsome circumscribed.
T7. Concept: derivative
5 10 15 20 25 30 35 40
5
10
15
20
25
Speed (ft/s)
Time (s)
Slope ≈ −1.8 (ft/s)/s(Exact answer is −1.8137… .)Name: acceleration
T8. The roller coaster is at the bottom of the hill at25 s because that’s where it is going the fastest.The graph is horizontal between 0 and 10 secondsbecause the velocity stays constant, 5 ft/s, as theroller coaster climbs the ramp.
T9. Distance = (rate)(time) = 5(10) = 50 ft
T10. T5 = 412.5; T50 = 416.3118… ;T100 = 416.340219…
T11. The differences between the trapezoidal sum andthe exact sum are:For T5: difference = 3.8496…For T50: difference = 0.03779…For T100: difference = 0.009447…The differences are getting smaller, so Tn isgetting closer to 416.349667… .
8 Problem Set 1-6 Calculus Solutions Manual© 2005 Key Curriculum Press
T12. From 30 to 31:
average rate = − = −y y( ) ( )
.31 30
11 9098K
From 30 to 30.1:
average rate = − = −y y( . ) ( )
..
30 1 30
0 11 8246K
From 30 to 30.01:
average rate
= − = −y y( . ) ( )
..
30 01 30
0 011 8148K
T13. The rates are negative because the roller coaster isslowing down.
T14. The differences between the average rates andinstantaneous rate are:For 30 to 31: difference = 0.096030…For 30 to 31.1: difference = 0.010833…For 30 to 30.01: difference = 0.001095…
The differences are getting smaller, so the averagerates are getting closer to the instantaneous rate.
T15. Solve . gettingy x y
x
( ) ( ),
−−
= − +30
301 81379936 1
x = 30.092220… . So keep x within 0.092…unit of 30, on the positive side.
T16. Concept: derivative
T17. ′ ≈ −−
= − =ff f
( )( . ) ( . )
. . .4
4 3 3 7
4 3 3 7
35 29
0 610
T18. Answers will vary.
Calculus Solutions Manual Problem Set 2-2 9© 2005 Key Curriculum Press
Chapter 2—Properties of Limits
Problem Set 2-1
1. a. f ( )28 10 2
2 2
0
0= − +
−=
No value for f (2) because of division by zero.
b.
x f (x)
1.997 2.994
1.998 2.996
1.999 2.998
2 undefined
2.001 3.002
2.002 3.004
2.003 3.006
Yes, f (x) stays close to 3 when x is keptclose to 2, but not equal to 2.
c. To keep f (x) within 0.0001 unit of 3, keepx within 0.00005 unit of 2. To keep f (x)within 0.00001 unit of 3, keep x within0.000005 unit of 2. To keep f (x) arbitrarilyclose to 3, keep x within 1
2 that distanceof 2.
d. The discontinuity can be “removed” bydefining f (2) to equal 3.
2.
3
2
3
x
g(x)
3
2
g(x)
x
The limit seems to be 2.
3.
3
2
x
h(x)
2
32.7
h(x)
x
There appears to be no limit, because the graphcycles infinitely as it approaches x = 3.
Problem Set 2-2Q1. Q2.
3
8
x
y
π
–1
x
y
Q3. Q4.
6
4
x
y
2–2
–4
x
y
Q5. Q6. Trapezoidal rule
1
4
x
y
Q7. Counting squares
Q8. Slope of the tangent line
Q9. Instantaneous rate of change
Q10. B
1. See the text for the definition of limit.
2. f (x) might be undefined at x = c, or might have avalue at x = c that is different from the limit.
3. Has a limit, 3 4. Has a limit, 2
5. Has a limit, 3 6. Has a limit, 5
7. Has no limit 8. Has no limit
9. Has a limit, 7 10. Has a limit, 20
11. Has no limit 12. Has no limit
13. lim ( ) .x
f x→
=3
5 For ε = 0.5, δ ≈ 0.2 or 0.3.
10 Problem Set 2-2 Calculus Solutions Manual© 2005 Key Curriculum Press
14. lim ( ) .x
f x→
=2
3 For ε = 0.5, δ ≈ 0.8.
15. lim ( ) .x
f x→
=6
4 For ε = 0.7, δ ≈ 0.5 or 0.6.
(The right side is more restrictive.)
16. lim ( ) .x
f x→
=4
2 For ε = 0.8, δ ≈ 0.7 or 0.8.
(The left side is more restrictive.)
17. lim ( ) .x
f x→
=5
2 For ε = 0.3, δ ≈ 0.5 or 0.6.
(The right side is more restrictive.)
18. lim ( ) .x
f x→
=3
6 For ε = 0.4, δ ≈ 0.1.
19. a. The graph should match Problem 13.
b. lim ( )x
f x→
=3
5
c. Graph is symmetrical about x = 3.Let 5 − 2 sin (x − 3) = 5 + 0.5 = 5.5.∴ sin (x − 3) = −0.25x = 3 + sin− 1 (−0.25)Max. δ = 3 − [3 + sin− 1 (−0.25)] = 0.25268…
d. Let 5 − 2 sin (x − 3) = 5 + ε.∴ sin (x − 3) = −ε/2x = 3 + sin− 1 (−ε/2)Max. δ = 3 − [3 + sin− 1 (−ε/2)] =−sin− 1 (−ε/2) = sin− 1 (ε/2), which is positivefor any positive value of ε.
20. a. The graph should match Problem 14.
b. lim ( )x
f x→
=2
3
c. The graph is symmetrical about x = 2.Let (x − 2)3 + 3 = 3 + 0.5 = 3.5.
∴ = + .x 2 0 53
Max. . . .δ = + − = =2 0 5 2 0 5 0 79373 3 K
d. Let (x − 2)3 + 3 = 3 + ε.∴ x = 2 + ε1/3
Max. δ = 2 + ε1/3 − 2 = ε1/3, which ispositive for any positive value of ε.
21. a. The graph should match Problem 15.
b. lim ( )x
f x→
=6
4
c. The right side is more restrictive.Let 1 + 3(7 − x)1/3 = 4 − 0.7 = 3.3.∴ x = 7 − (2.3/3)3
Max. δ = [7 − (2.3/3)3] − 6 = 0.5493…
d. Because the right side is more restrictive, set1 3 7 43+ − = − x ε.∴ x = 7 − [(3 − ε)/3]3
Max. δ = 7 − [(3 − ε)/3)3] − 6 = 1 − [(3 − ε)/3]3,which is positive for all positive values of ε.
22. a. The graph should match Problem 16.
b. lim ( )x
f x→
=4
2
c. The left side is more restrictive.Let 1 + 24− x = 2 + 0.8 = 2.8.
∴ 24− x = 1.8
x = −41 8
2
log .
log
Max .
log .
log.δ = − −
=4 41 8
20 84799K
d. Because the left side is more restrictive, set1 + 24− x = 2 + ε.∴ 24− x = 1 + ε
x = − +4
1
2
log( )
log
ε
Max. log(1 ) log(1 )δ ε ε= − − +
= +
4 42 2log log
,
which is positive for all ε > 0.
23. a. The graph should match Problem 17.
b. lim ( )x
f x→
=5
2
c. The right side is more restrictive.Let (x − 5)2 + 2 = 2 + 0.3 = 2.3.
∴ = + .x 5 0 3
Max = ( . ) .. δ 5 0 3 5 0 54772+ − = K
d. Because the right side is more restrictive, set(x − 5)2 + 2 = 2 + ε.∴ = + x 5 εMax. ( ) ,δ ε ε= + − =5 5 which ispositive for all ε > 0.
24. a. The graph should match Problem 18.
b. lim ( )x
f x→
=3
6
c. The graph is symmetrical about x = 3.Let 6 − 2(x − 3)2/3 = 6 − 0.4 = 5.6.∴ x = 3 + 0.23/2
Max. δ = (3 + 0.23/2) − 3 = 0.08944…
d. Let 6 − 2(x − 3)2/3 = 6 − ε.∴ x = 3 + (ε/2)3/2
Max. δ = [3 + (ε/2)3/2] − 3 = (ε/2)3/2, whichis positive for all ε > 0.
25. a. f ( )( )( ) ( )( )
25 6 5 13 5 2
5 2
5 0
0
0
0
2
= − ⋅ + −−
= =
The graph has a removable discontinuity atx = 2.Limit = 22 − 6(2) + 13 = 5
b. When f (x) = 5.1, x = 1.951191… .δ1 = 2 − 1.951191… = 0.048808…When f (x) = 4.9, x = 2.051316… .δ2 = 2.051316… − 2 = 0.051316…∴ max. δ = 0.048808…
Calculus Solutions Manual Problem Set 2-3 11© 2005 Key Curriculum Press
c.f(x)
x
c = 2
L = 5
δδ
26. a.
2
8
x
y
The graph is linear.There is a removable discontinuity at x = 2.The limit appears to be 9.
b. f ( )( ) ( )
–2
4 2 7 2 2
2 2
0
0
2
= − − =
Indeterminate form
c. f xx x
xx x( )
( )( ),= + −
−= + ≠4 1 2
24 1 2
Limit = 4(2) + 1 = 9If x ≠ 2, then (x − 2) ≠ 0. Canceling is adivision process, but because (x − 2) ≠ 0,you do not risk dividing by zero.
d. If f (x) = 9.001, x = 2.00025.If f (x) = 8.999, x = 1.99975.δ1 = 2.00025 − 2 = 0.00025δ2 = 2 − 1.99975 = 0.00025Largest number is 0.00025.
e. L = 9, c = 2, ε = 0.001, δ = 0.00025
27. a. m td t d
t
t
t( )
( ) ( ) –= −−
=−
4
4
3 48
4
2
b. Removable discontinuity at x = 4.
30
4
m(t)
t
c. Limit = 24 ft/s
d. m tt t
tt t( )
( )( )
–,= − + = + ≠3 4 4
43 12 4 if
3t + 12 = 24.12 ⇒ t = 4.043t + 12 = 23.88 ⇒ t = 3.96Keep t within 0.04 s of 4 s.
e. The limit of the average velocity is theinstantaneous velocity.
Problem Set 2-3Q1. 13
Q2. Q3.
2
3
y
x
–4
y
x
Q4. Q5.
x
y
x
y
1
1
Q6. (x − 10)(x + 10) Q7. 75%Q8. Product of x and y, where x varies and y may
varyQ9.
3 1 –8 22 –21
3 –15 21
1 –5 7 0
x2 − 5x + 7
Q10. D
1.
10
2
x
y
g h
g + h
lim ( ) , lim ( ) , lim ( )x x x
f x g x h x→ → →
= = =2 2 2
10 4 6 and
∴ = +→ → →
lim ( ) lim ( ) lim ( ),x x x
f x g x h x2 2 2
Q.E.D.
x f (x)
1.96 9.9640…
1.97 9.9722…
1.98 9.9810…
1.99 9.9902…
2.00 10
2.01 10.0102…
2.02 10.0209…
2.03 10.0322…
2.04 10.0439…
All these f (x) values are close to 10.
12 Problem Set 2-3 Calculus Solutions Manual© 2005 Key Curriculum Press
2.
9
1.8 f
g
3
x
y
lim ( ) . lim ( )x x
f x g x→ →
= =3 3
1 8 9 and
∴ =→ →
lim ( ) . lim ( ),x x
f x g x3 3
0 2 Q.E.D.
x f (x)
2.96 1.75232
2.97 1.76418
2.98 1.77608
2.99 1.78802
3.00 1.8
3.01 1.81202
3.02 1.82408
3.03 1.83618
3.04 1.84832
All these f (x) values are close to 1.8.
3.
3
7Limit = 7
x
f(x)
The limit is 7 because f (x) is always close to 7,no matter what value x takes on. (It shouldn’tbother you that f (x) = 7 for x ≠ 3 if you think ofthe definition of limit for a while.)
4.
x
f(x) = x
6
Limit = 6
lim ( ) .x
f x→
=6
6 The y-value equals the x-value.
5.
1
5y
1
y1
y2 y
2
y
x
lim , lim . , limx x x
y y y y→ → →
= = ⋅ =1
11
21
1 22 1 5 3 and
2 1 5 31
11
21
1 2( . ) , lim lim lim= ∴ ⋅ = ⋅→ → →
x x x
y y y y
x y3 = f (x)
0.997 2.9739…
0.998 2.9825…
0.999 2.9912…
1 3
1.001 3.0087…
1.002 3.0174…
1.003 3.0262…
All these f (x) values are close to 2(1.5) = 3.
6. 2 83
3 60 53 =
= and sin
..
π
r( ).
38
0 516= =
x r (x)
2.9997 15.9894…
2.9998 15.9929…
2.9999 15.9964…
3 16
3.0001 16.0035…
3.0002 16.0070…
3.0003 16.0105…
All these r (x) values are close to 16.
lim ( ).
.
xf x
→→
3 6
3 62
0 , so the limit of a quotient
cannot be applied because of division by zero.
7. lim ( ) limx x
f x x x→ →
= − +3 3
2 9 5
= − +→ → →
lim lim limx x x
x x3
2
3 39 5 Limit of a sum
(or difference)
= ⋅ − +→ → →
lim lim limx x x
x x x3 3 3
9 5
Limit of a product,limit of a constant
= (3)(3) − 9(3) + 5 Limit of x= 9 − 27 + 5 = −13
8. lim ( ) limx x
f x x x→− →−
= + −1 1
2 3 6
= lim lim limx x x
x x→− →− →−
+ −1
2
1 13 6
Limit of a sum= lim lim lim
x x xx x x
→− →− →−⋅ + −
1 1 13 6
Limit of a product,limit of a constant
= (−1)(−1) + 3(−1) − 6 Limit of x= 1 − 3 − 6 = −8
Calculus Solutions Manual Problem Set 2-3 13© 2005 Key Curriculum Press
9.
–8
–2 x
r(x)
r(– ) =22 4 2 12
2 2
4 8 12
0
0
0
2( ) ( )
( )
− − − −− +
= + − =
r xx x
xx x( )
( – )( ),= +
+= − ≠ −6 2
26 2
lim ( )x
r x→−
= − − = −2
2 6 8
Proof:
lim ( ) lim ( )x x
r x x→− →−
= −2 2
6 Because x ≠ −2
= +→− →−lim lim
2 2x xx (– )6 Limit of a sum
= −2 − 6 = −8, Q.E.D. Limit of x, limit of aconstant
10.13
5
x
f(x)
f ( )( )
55 3 5 40
5 5
25 15 40
0
0
0
2
= + −−
= + − =
f xx x
xx x( )
( )( – )
–,= + = + ≠8 5
58 5
lim ( )x
f x→
= + =5
5 8 13
Proof:
lim ( ) lim ( )x x
f x x→ →
= +5 5
8 Because x ≠ 5
= lim limx x
x→ →
+5 5
8 Limit of a sum
= 5 + 8 = 13, Q.E.D. Limit of x, limit of aconstant
11.41
10
5
x
f(x)
f ( )( ) ( )− = − − −
−
= − − − =
55 3 5 4 5 30
5 5125 75 20 30
0
0
0
3 2
f xx x x
xx x x( )
( )( – )
–,= + + = + + ≠
222 6 5
52 6 5
lim ( ) ( )x
f x→
= + + =5
25 2 5 6 41
Proof:
lim ( ) lim ( )x x
f x x x→ →
= + +5 5
2 2 6 Because x ≠ 5
= lim lim ( ) limx x x
x x→ → →
+ +5
2
5 52 6 Limit of a sum
= lim lim limx x x
x x x→ → →
⋅ + +5 5 5
2 6 Limit of a product,limit of a constanttimes a function,limit of a constant
= 5 ⋅ 5 + 2 · 5 + 6 = 41, Q.E.D.Limit of x
12.
28
3
x
f(x)
f ( )( )
33 3 5 3 21
3 3
27 9 15 21
0
0
0
3 2
= + − −−
= + − − =
f xx x x
xx x x( )
( )( – )
–,= + + = + + ≠
224 7 3
34 7 3
lim ( ) ( )x
f x→
= + + =3
23 4 3 7 28
Proof:
lim ( ) lim )x x
f x x x→ →
= + +3 3
2 4 7 (
Because x ≠ 3= + +
→ → →lim lim limx x x
x x3
2
3 34 7 Limit of a sum
= ⋅ + +→ → →
lim lim limx x x
x x x3 3 3
4 7 Limit of a product,limit of a constanttimes a function,limit of a constant
= 3 ⋅ 3 + 4 ⋅ 3 + 7 = 28, Q.E.D.Limit of x
13.
9
f(x)
x
–1
f ( )( ) ( ) ( )
( )− = − − − − − +
− +1
1 4 1 2 1 3
1 1
3 2
= − − + + =1 4 2 3
0
0
0
14 Problem Set 2-3 Calculus Solutions Manual© 2005 Key Curriculum Press
f xx x x
xx x x( )
( – )( ),= + +
+= − + ≠ −
225 3 1
15 3 1
lim ( ) ( ) ( )x
f x→−
= − − − + =1
21 5 1 3 9
Proof:
lim ( ) lim ( )x x
f x x x→− →−
= − +1 1
2 5 3
Because x ≠ −1
= + +→ → →lim lim (– ) lim
– – –x x xx x
1
2
1 15 3
Limit of a sum= ⋅ +
→ → →lim lim – lim
– – –x x xx x x
1 1 15 3
Limit of a product,limit of a constanttimes a function,limit of a constant
= (−1)(−1) + (−5)(–1) + 3 = 9, Q.E.D.Limit of x
14.
–17
2 xf(x)
f ( )( ) ( )
22 11 2 21 2 2 10
2 216 88 84 2 10
0
0
0
4 3 2
= − + − −−
= − + − − =
f xx x x x
x( )
( – )( – )
–= + +3 29 3 5 2
2
= − + + ≠x x x x3 29 3 5 2,
limx
f x→
= − + + = −2
3 22 9 2 3 2 5 17( ) ( ) ( )
Proof:
lim ( ) lim ( – )x x
f x x x x→ →
= + +2 2
3 29 3 5
Because x ≠ 2= + − + +
→ → → →lim lim ( ) lim limx x x x
x x x2
3
2
2
2 29 3 5
Limit of a sum= ⋅ ⋅ + − ⋅
→ → → → →lim lim lim ( ) lim limx x x x x
x x x x x2 2 2 2 2
9
+ +→
3 52
limx
x
Limit of a product,limit of a constanttimes a function,limit of a constant
= 2 ⋅ 2 ⋅ 2 + (−9)(2 ⋅ 2) + 3 ⋅ 2 + 5 = −17,Q.E.D. Limit of x
15.
x f (x)
4.990 40.8801
4.991 40.8921…
4.992 40.9040…
4.993 40.9160…
4.994 40.9280…
4.995 40.9400…
4.996 40.9520…
4.997 40.9640…
4.998 40.9760…
4.999 40.9880…
5 undefined
5.001 41.0120…
5.002 41.0240…
5.003 41.0360…
5.004 41.0480…
5.005 41.0600…
5.006 41.0720…
5.007 41.0840…
5.008 41.0960…
5.009 41.1080…
The table shows that f (x) will be within 0.1 unitof lim ( )
xf x
→=
541 if we keep x within 0.008 unit
of 5.
16.f (x)
x
9
–1
When x is close to –1, f (x) is close to 9.
17. f xx x
x x
x x
x x
x
x( )
( )( )
( )( )= − +
− += − −
− −= −
−
2
2
5 6
6 9
2 3
3 3
2
3You cannot find the limit by substituting intothe simplified form because the denominator stillbecomes zero.
18. f xx
x x
x x x
x x
x x
x
( )( )( )
( )( )= −
− += − + +
− −
= + +−
3
2
2
2
8
4 4
2 2 4
2 2
2 4
2
You cannot find the limit by substituting intothe simplified form because the denominator stillgoes to zero.
Calculus Solutions Manual Problem Set 2-4 15© 2005 Key Curriculum Press
19. a. 5(0)1/2 = 0 = v(0)5(1)1/2 = 5 = v(1)5(4)1/2 = 10 = v(4)5(9)1/2 = 15 = v(9)5(16)1/2 = 20 = v(16)
b. a
v v( ) .9
9 001 9
9 001 90 8333101≈ =( . ) – ( )
. –K
Conjecture: a( ) . /9 0 83 5 6= =Units of a(t): (mi/h)/s
c. av t v
t
t
tt t( ) lim
( ) – ( )
–lim
–
–
/
99
9
5 15
99 9
1 2
= =→ →
=+
=+
→
→
lim( – )
( – )( )
lim
/
/ /
/
t
t
t
t t
t
9
1 2
1 2 1 2
9 1 2
5 3
3 3
5
3
= 5
6, which agrees with the conjecture.
d. Distance = integral of v(t) from 1 to 9. By thetrapezoidal rule with n = 100 increments,integral ≈ 86.6657… . The units are(mi/h) · s. To convert to ft, multiply by 5280and divide by 3600, getting 127.1111…(exact: 127 1
9 ) . The truck went about 127 ft.
20. a. Derivative ≈ =2 1 2
2 1 212 61
3 3. –
. –.
b.x
x
x x x
x
3 28
2
2 2 4
2
–
–
( – )( )
–= + + =
x x2 2 4+ + , provided x ≠ 2. This expressionapproaches 12 as x approaches 2.
Proof:
lim–
–lim ( )
x x
x
xx x
→ →= + +
2
3
2
28
22 4
Because x ≠ 2= + +
→ → →lim lim limx x x
x x2
2
2 22 4
Limit of a sum= ⋅ + +
→ → →lim lim limx x x
x x x2 2 2
2 4
Limit of a product,limit of a constant
= 2 · 2 + 2 · 2 + 4 = 12, Q.E.D.Limit of x
c. The line through point (2, 8) with slope 12 isy = 12x − 16. The line appears to be tangentto the graph of f at point (2, 8).
12
1
2
x
f(x)
8
21. By the symmetric difference quotient,
derivative ≈ = −0 7 0 7
2 0 010 05994
5 01 4 99. – .
( . ). .
. .
K
22. By the trapezoidal rule with n = 100,integral ≈ 11 8235. K .
23. Prove that limx c
n nx c→
= for any positive integer n.
Proof:
Anchor:
If n = 1, limx c
x c c→
=1 1= by the limit of x.
Induction Hypothesis:
Assume that the property is true for n = k.∴ =
→limx c
k kx c
Verification for n = k + 1:lim lim ( )x c
k
x c
kx x x→
+
→= ⋅1
= ⋅ = ⋅→ →
lim limx c
k
x c
kx x c c By the inductionhypothesis
= +ck 1
Conclusion:
∴ =→
limx c
n nx c for all integers n ≥ 1, Q.E.D.
24. Answers will vary.
Problem Set 2-4Q1. Instantaneous rate of change
Q2. Product of x and y, where x varies and ycan vary
Q3. 0.0005
Q4.
Q5. Exponential function
Q6.y = cos x
x
Q7. (x + 6)(x − 1) Q8. 53
Q9. 120 Q10. 103
1. a. Has left and right limits
b. Has no limit
c. Discontinuous. Has no limit
16 Problem Set 2-4 Calculus Solutions Manual© 2005 Key Curriculum Press
2. a. Has left and right limits
b. Has a limit
c. Discontinuous. No f (3)
3. a. Has left and right limits
b. Has a limit
c. Continuous
4. a. Has left and right limits
b. Has a limit
c. Continuous
5. a. Has no left or right limit
b. Has no limit
c. Discontinuous. No limit or f (2)
6. a. Has left and right limits
b. Has a limit
c. Continuous (Note that the x-value 5 is not atthe discontinuity.)
7. a. Has left and right limits
b. Has a limit
c. Discontinuous. f (1) ≠ limit
8. a. Has left and right limits
b. Has no limit
c. Discontinuous. No limit
9. a. Has left and right limits
b. Has a limit
c. Discontinuous. No f (c)
10. a. Has left and right limits
b. Has no limit
c. Discontinuous. No limit, no f (c)
11. Answers may vary. 12. Answers may vary.
3
x
f(x)
4
f(x)
x
13. Answers may vary. 14. Answers may vary.
5
x
f(x)
–2
x
f(x)
f(–2)
15. Answers may vary. 16. Answers may vary.
x
f(x)
6
2
x
f(x)
17. Answers may vary. 18. Answers may vary.
10
f(x)
x
–2
–2
5 x
f(x)
19. Answers may vary. 20. Answers may vary.
x
f(x)
6
4
1
x
f(x)
5
3
21. Discontinuous at x = −3
22. Discontinuous at x = 11
23. Discontinuous at x = π/2 + π n, where n is aninteger
24. Nowhere discontinuous
25.
x
f(x)
2
1
2
3
Discontinuous because limx
f x→
=2
2( ) and f (2) = 3
26.
2
1
2
x
g(x)
Discontinuous because g(x) has no limit as xapproaches 2
Calculus Solutions Manual Problem Set 2-4 17© 2005 Key Curriculum Press
27.
2
3
x
s(x)
Discontinuous because s(x) has no limit as xapproaches 2 from the left (no real functionvalues to the left of x = 2)
28.
2
x
p(x)
1
Discontinuous because p(x) has no limit as xapproaches 2
29.
2
1
x
h(x)
Discontinuous because there is no value of h(2)
30.
2
3
x
f(x)
Discontinuous because f (x) has no limit as xapproaches 2
31.
c f c( )lim ( )
x cf x
→ −lim ( )
x cf x
→ +lim ( )x c
f x→ Continuous?
1 4 2 2 2 removable
2 1 1 1 1 continuous
4 5 5 2 none step
5 none none none none infinite
32.
c f c( )lim ( )
x cf x
→ −lim ( )
x cf x
→ +lim ( )x c
f x→ Continuous?
1 3 2 3 none step
2 1 4 4 4 removable
3 5 5 5 5 continuous
5 5 5 none none infinite
33. a.
x
d(x)
2
3
b. lim ( ) , lim ( ) .x x
d x d x→ →− +
= =2 2
3 3 Limit = 3.
Continuous.
34. a.
3
2
1
x
h(x)
b. lim ( ) , lim ( ) .x x
h x h x→ →− +
= =1 1
3 2 No limit.
Not continuous.
35. a.
2
x
m(x)
9
7
b. lim ( ) , lim ( ) .x x
m x m x→ →− +
= =2 2
9 7 No limit.
Not continuous.
36. a.
–1
2
q(x)
x
b. lim ( ) , lim ( ) .x x
q x q x→− →−− +
= =1 1
2 2 Limit = 2.
Continuous.
37. 9 – 22 = 2k∴ k = 2.5
g(x)
x
2
5
18 Problem Set 2-4 Calculus Solutions Manual© 2005 Key Curriculum Press
38. 0.4(1) + 1 = k(1) + 2∴ k = −0.6
x
f(x)
1
1.4
39. (32)k = 3k − 3∴ k = −1/2.
3
x
u(x)
1
40. −k + 5 = (−1)2k∴ k = 5/2
x
v(x)
–1
5
41. a. b − 1 = a(1 − 2)2 ⇒ b − 1 = a
b. a = −1 ⇒ b = 0. Continuous at x = 1.
x
f(x)
1 1
a = –1, b = 0
c. For example, a = 1 ⇒ b = 2. Continuousat x = 1.
1
1 x
f(x)
e.g., a = 1, b = 2
42. lim ( ) limx x
f x k x k→ →− −
= − = −2 2
2 2 2 4
lim ( ) lim . . ( )x x
f x kx k k→ →+ +
= = =2 2
1 5 1 5 2 3
For f (x) to be continuous at x = 2, these twolimits must be equal, so find k such thatk2 − 4 = 3kk2 − 3k − 4 = 0(k − 4)(k + 1) = 0so k = 4 and k = −1 are the two values of k thatwill make f (x) continuous at x = 2.
43. Let T(θ ) = the number of seconds it takes tocross.
T( )if
40
sin, if 0 or 90θ
θ
θθ θ=
= °° < < ° ° < < °
24 90
90 180
,
90
40
θ
θ
44. a.
x
f(x)
4
1
b. f (x) seems to approach 4 as x approaches 1.
c. f (1.0000001) = 1.0000001 + 3 + 10− 13 ≈4.0000001, which is close to 4.
d. There is a vertical asymptote at x = 0. Youmust get x much closer to 1 than x =1.0000001 for the discontinuity to show up.
45. For any value of c, P(c) is determined by additionand multiplication. Because the set of realnumbers is closed under multiplication andaddition, P(c) will be a unique, real number forany real value x = c. P(c) is the limit of P(x) asx approaches c by the properties of the limit of aproduct of functions (for powers of x), the limitof a constant times a function (for multiplicationby the coefficients), and the limit of a sum (forthe individual terms). Therefore, P is continuousfor all values of x.
46. a. limx→0
|sgn x| = 1 but f (0) = 0
lim ( ) ( ),x
f x f→
≠0
0 so discontinuous
b.
x
g(x)
2
3
c.
x
h(x)
1
–1 1
Calculus Solutions Manual Problem Set 2-5 19© 2005 Key Curriculum Press
d. For x > 0, a(x) = x/x = 1 = sgn x.For x < 0, a(x) = (−x)/x = −1 = sgn x.For x = 0, a(0) is not defined.∴ a(x) = sgn x for all x ≠ 0, Q.E.D.
e.
2
–2
–2x
f(x)
π
Problem Set 2-5Q1. No limit Q2. 3
Q3. 4 Q4. 3
Q5. 2 Q6. No
Q7. No Q8. Yes
Q9. No Q10. Yes
1. • limx
f x→−∞
∞ ( ) = • limx
f x→− −3
4( ) = –
• limx
f x→− +
=3
3( ) • limx
f x→
= ∞1
( ) –
• limx
f x→
=2
1( ) • limx
f x→ −
= ∞3
( )
• limx
f x→ +
=3
2( ) • limx
f x→∞
( ) does not
exist.
2. • limx
g x→−∞
=( ) 2 • limx
g x→− −
=2
4( )
• limx
g x→− +
=2
3( ) – • limx
g x→ −
= ∞1
( )
• limx
g x→
=2
3( ) • limx
g x→ −
=3
4( )
• lim ( )x
g x→∞
= −2
3. Answers may vary. 4. Answers may vary.
x
f(x)
2
2
x
f(x)
5. Answers may vary. 6. Answers may vary.
7
–5
x
f(x)
x
f(x)
7. a.
3
2
x
f(x)
b. limx
f x→ +
∞3
( ) = , limx
f x→ −
∞3
( ) = – ,
lim ( ),x
f x→3
none, limx
f x→∞
( ) = ,2
limx
f x→−∞
( ) = 2
c. 21
3100+
−=
x1
398
x −=
x – 31
98=
x = =3
1
983 0102. K
x f (x)
3.01 102
3.001 1002
3.0001 10002
All of these f (x) values are greater than 100.lim ( )x
f x→ +
= ∞3
means that f (x) can be kept
arbitrarily far from zero just by keeping xclose enough to 3 on the positive side.There is a vertical asymptote at x = 3.
d. 21
32 001+
−=
x.
1
30 001
3 1000
1003
xx
x
−=
− ==
.
x f (x)
1004 2.00099…
1005 2.00099…
1006 2.00099…
All of these f (x) values are within 0.001 unitof 2. lim
xf x
→∞=( ) 2 means that you can keep
f (x) arbitrarily close to 2 by making the valueof x arbitrarily large. y = 2 is a horizontalasymptote.
8. a.
π/2
1x
g(x)
b. lim ( ) , lim ( )/ /x x
g x g x→ →− +
= ∞ = −∞π π2 2
The limit is infinite because |g(x)| can be keptarbitrarily far from zero. You can’t saylim ( )
/xg x
→= ∞
π 2 because the left and right
20 Problem Set 2-5 Calculus Solutions Manual© 2005 Key Curriculum Press
limits are not the same (one is positive and theother is negative).
c. sec x = −1000∴ cos x = −0.001x = arccos (−0.001) = 1.57179…
x g(x)
1.5717 –1106.5…
1.5716 –1244.2…
1.5715 –1421.1…
All of these f (x) values are less than −1000.lim ( )
/xg x
→ += −∞
π 2 means that arbitrarily far
g(x) can be kept arbitrarily far from zero in thenegative direction by keeping x close enoughto π
2 on the positive side.The line x = π
2 is a vertical asymptote.
9. a.
5
3r(x)
x
2
b. lim ( )x
r x→∞
= 2 because (sin x)/x approaches zero.
c. r( )
sin( ). ,28 2
28
282 00967= + = K which is
within 0.01 unit of 2.
r( )
sin( )32 2
32
32= + = 2.01723 ,K which is
more than 0.01 unit away from 2.
y
x
1.99
2.01
28 32
r
Keeping r(x) within 0.01 unit of 2 means you
want to keep sin . ,xx < 0 01 or |sin x| <
0.01 |x|. You are looking for a large value ofx, so you know x will be positive, so youwant |sin x| < 0.01x. You can’t get rid of theabsolute value symbol on the sine becausesine will keep alternating as x gets larger.You know |sin x| ≤ 1 for all valuesof x, so you need to make 0.01x > 1, orx > 100. So D = 100.
d. The line y = 2 is an asymptote. Even thoughr (x) oscillates back and forth across this line,the limit of r(x) is 2 as x approaches infinity,satisfying the definition of asymptote.
e. The graph suggests that lim ( ) .x
r x→
=0
3
(The exact value is e, 2.7182… .)
10. a. h x x x( ) ( / )= +1 1
10
1
2
3
x
h(x)
b. There is a compromise number (bigger than1, but finite) that wins. (The exact limit is e.)
11. The limit is infinite. y is unbounded as xapproaches infinity. If there were a number Esuch that log x < E for all x > 0, then you couldlet x = 102E so that log x = log 102E = 2E, whichis greater than E, which was assumed to be anupper bound.
12. “Wanda, here’s what happens to a fraction whenthe denominator gets close to zero: 1
0 1 10. = ,1
0.00011
0.00001, , ,= =10 000 100 000. The answers
just keep getting bigger and bigger. When thedenominators get bigger and bigger, the fractiongets closer and closer to zero, like this:
110
1100
110000 1 01 001= = =. , 0. , 0. .”
13. a. The definite integral is the product of theindependent and dependent variables. Becausedistance = (rate)(time), the integral representsdistance in this case.
b. T9 = 17.8060052… T45 = 17.9819616… T90 = 17.9935649…T450 = 17.9994175…
c. The exact answer is 18. It is a limit becausethe sums can be made as close to it as youlike, just by making the number of trapezoidslarge enough (and thus keeping their widthsclose to zero). The sums are smaller than theintegral because each trapezoid is inscribedunder the graph and thus leaves out a part ofits respective strip of the region.
d. Tn is 0.01 unit from 18 when it equals 17.99.From part b, this occurs between n = 45 and n= 90. By experimentation,T66 = 17.9897900… and T67 = 17.9900158… .Therefore, the approximation is within 0.01unit of 18 for any value of n ≥ 67.An alternative solution is to plot the graph ofthe difference between 18 and Tn as a functionof the number of increments, n, or to do aregression analysis to find an equation. Thebest-fitting elementary function is an inversepower variation function, y = (5.01004…)
Calculus Solutions Manual Problem Set 2-6 21© 2005 Key Curriculum Press
(x− 1.48482…). The graph of this function and threeof the four data points are shown here. UseTRACE or the solver feature of your grapher tofind n ≈ 67.
0.01
0.1
y
n
67
90
14. a. Work = force × distance. Because a definiteintegral measures the y-variable times thex-variable, it represents work in this case.
b. By the trapezoidal rule, T10 = 24.147775…and T100 = 24.004889… . The units arefoot-pounds.
c. The integer is 24.
d. By experimentation, T289 = 24.001003… andT290 = 24.000998… .∴ D = 290
15. Length = 100 sec x = 100/cos xLength > 1000 ⇒ 100/cos x > 1000cos x < 0.1 (because cos x is positive)x > cos–1 0.1 (because cos is decreasing)x > 1.4706289…π/2 − 1.4706289… = 0.100167…x must be within 0.100167… radian of π/2.The limit is (positive) infinity.
16. a. f (2) = 5 · 2 · 0 · (1/0), which has theform 0 · ∞.g(2) = 5 · 2 · 0 · (1/0)2, which has theform 0 · ∞.h(2) = 5 · 2 · 02 · (1/0), which has theform 0 · ∞.
b. f x x xx
x x( ) ( )= − ⋅ = ≠5 21
25 2
–,
∴ =→
( )limx
f x2
10
g x x xx
x
xx( ) ( )= − ⋅ = ≠5 2
1
2
5
222( – ) –
,
∴→
( )limx
g x2
is infinite.
h x x xx
x x x( ) ( ) = ( – ), = − ⋅ ≠5 21
25 2 22
–∴
→ ( ) =lim
xh x
20
c. The indeterminate form 0 · ∞ could approachzero, infinity, or some finite number.
Problem Set 2-6
Q1. 53 Q2. 53
Q3. Undefined Q4. 5
Q5. Undefined Q6. Does not exist
Q7. 1 Q8. +
Q9. Indeterminate Q10. C
1. IVT applies on [1, 4] because f is a polynomialfunction, and polynomial functions arecontinuous for all x.f (1) = 18, f (4) = 3∴ There is a value x = c in (1, 4) for whichf (c) = 8.Using the intersect or solver feature,c = 1.4349… , which is between 1 and 4.
f (x)
x
f(1)
f(4)
8
1 4c
2. IVT applies on [0, 6] because f is a polynomialfunction, and polynomial functions arecontinuous for all x.f (0) = −8, f (6) = −0.224∴ There is a value x = c in (0, 6) for whichf (c) = −1.Using the intersect or solver feature,c = 5.8751… , which is between 0 and 6.
–0.224
–8
0 6c xf(x)
3. a. For 1 ≤ y < 2 or for 5 < y ≤ 8, the conclusionwould be true. But for 2 ≤ y ≤ 5, it would befalse because there are no values of x in [1, 5]that give these values for f (x).
b. The conclusion of the theorem is true becauseevery number y in [4, 6] is a value of g(x) forsome value of x in [1, 5].
4. a. f f f( ) , ( ) , ( . ) . ,2 4 3 8 0 5 2 1 414= = = = K
f ( )5 8=b. f is continuous at x = 3 because it has a limit
and a function value and they both equal 8.
c. f is continuous nowhere else. Because thesets of rational and irrational numbers aredense, there is a rational number between anytwo irrational numbers, and vice versa. Sothere is no limit of f (x) as x approaches anynumber other than 3.
d. The conclusion is not true for all values of ybetween 1 and 4. For instance, if y = 3, thenc would have to equal log2 3. But log2 3 isirrational, so f (c) = 8, which is not between1 and 4.
22 Problem Set 2-7 Calculus Solutions Manual© 2005 Key Curriculum Press
5. Let f (x) = x2. f is a polynomial function, so it iscontinuous and thus the intermediate valuetheorem applies. f (1) = 1 and f (2) = 4, so thereis a number c between 1 and 2 such that f (c) = 3.By the definition of square root, c = 3 , Q.E.D.
6. Prove that if f is continuous, and if f (a) ispositive and f (b) is negative, then there is at leastone zero of f (x) between x = a and x = b.
Proof:
f is continuous, so the intermediate valuetheorem applies. f (a) is positive and f (b) isnegative, so there is a number x = c between aand b for which f (c) = 0. Therefore, f has at leastone zero between x = a and x = b, Q.E.D.
7. The intermediate value theorem is called anexistence theorem because it tells you that anumber such as 3 exists. It does not tell youhow to calculate that number.
8. Telephone your sweetheart’s house. An answer tothe call tells you the “existence” of thesweetheart at home. The call doesn’t tell suchthings as how to get there, and so on. Also,getting no answer does not necessarily mean thatyour sweetheart is out.
9. Let f (t) = Jesse’s speed − Kay’s speed. f (1) =20 − 15 = 5, which is positive. f (3) = 17 − 19 =−2, which is negative. The speeds are assumed tobe continuous (because of laws of physics), so fis also continuous and the intermediate valuetheorem applies.So there is a value of t between 1 and 3 forwhich f (t) = 0, meaning that Jesse and Kay aregoing at exactly the same speed at that time.The existence of the time tells you neither whatthat time is nor what the speed is. An existencetheorem, such as the intermediate value theorem,does not tell these things.
10. Let f (x) = number of dollars for x-ounce letter.f does not meet the hypothesis of the IVT on theinterval [1, 9] because there is a stepdiscontinuity at each integer value of x. There isno value of c for which f (c) = 2 because f (x)jumps from 1.98 to 2.21 at x = 8.
11. You must assume that the cosine is functioncontinuous. Techniques:
• c = cos− 1 0.6 = 0.9272…
• Using the solver feature, c = 0.9272...
• Using the intersect feature, c = 0.9272...
12. You must assume that 2x is continuous.f (0) = 20 = 1, because any positive number to the0 power equals 1.
• c = = =loglog
log2 33
21 5849. ...
• Using the solver feature, c = 1.5849...
• Using the intersect feature, c = 1.5849...
13. This means that a function graph has a highpoint and a low point on any interval in whichthe function is continuous.
x
f(x)
a bc1 c2
If the function is not continuous, there may be apoint missing where the maximum or minimumwould have been.
x
f(x)
a b
Another possibility would be a graph with avertical asymptote somewhere between a and b.
14. Prove that if f is continuous on [a, b], the imageof [a, b] under f is all real numbers between theminimum and maximum values of f (x),inclusive.
Proof:
By the extreme value theorem, there are numbersx1 and x2 in [a, b] such that f (x1) and f (x2) are theminimum and maximum values of f (x) on [a, b].Because x1 and x2 are in [a, b], f is continuous onthe interval whose endpoints are x1 and x2. Thus,the intermediate value theorem applies on thelatter interval. Thus, for any number y betweenf (x1) and f (x2), there is a number x = c betweenx1 and x2 for which f (c) = y, implying that theimage of [a, b] under f is all real numbersbetween the minimum and maximum values off (x), inclusive, Q.E.D.
Problem Set 2-7Review Problems
R0. Answers will vary.
R1. a. f ( )336 51 15
3 3
0
0= − +
−=
Indeterminate form
Calculus Solutions Manual Problem Set 2-7 23© 2005 Key Curriculum Press
b. f x x x( ) = – ,4 5 3≠
9–9
9
–9
y
x
At x = 3 there is a removable discontinuity.
c. For 0.01, keep x within 0.0025 unit of 3. For0.0001, keep x within 0.000025 unit of 3. Tokeep f (x) within ε unit of 7, keep x within14 ε unit of 3.
R2. a. L f xx c
=→
lim ( ) if and only if for any number
ε > 0, no matter how small,there is a number δ > 0 such that if x iswithin δ units of c, but x ≠ c, then f (x) iswithin ε units of L.
b. limx
f x→
=1
2( )
limx
f x→2
( ) does not exist.
limx
f x→
=3
4( )
limx
f x→4
( ) does not exist.
limx
f x→5
( ) = 3
c. limx
f x→
=2
3( )
Maximum δ: 0.6 or 0.7
d. The left side of x = 2 is the more restrictive.
Let 2 + x – 1 = 3 − 0.4 = 2.6.∴ x = 1 + 0.62 = 1.36∴ maximum value of δ is 2 − 1.36 = 0.64.
e. Let f x( ) = 3 − ε.
2 1 3+ − = −x εx = (1 − ε)2 + 1Let δ = 2 − [(1 − ε)2 + 1] = 1 − (1 − ε)2,which is positive for all positive ε < 1. Ifε ≥ 1, simply take δ = 1. Then δ will bepositive for all ε > 0.
R3. a. See the limit property statements in the text.
b. •
20
3
x
g(x)
–19
• The limit of a quotient property does notapply because the limit of the denominatoris zero.
• g xx x x
x( ) = − − +
−( )( )3 10 2
3
2
g(x) = x2 − 10x + 2, x ≠ 3
You can cancel the (x − 3) because thedefinition of limit says “but not equal to 3.”
• lim ( )x
g x→3
= + +→ → →
lim lim (– ) limx x x
x x3
2
3 310 2
Limit of a sum= ⋅ +
→ → →lim lim – limx x x
x x x3 3 3
10 2
Limit of a product,limit of a constanttimes a function,limit of a constant
= 3 · 3 − 10(3) + 2 Limit of x= −19, which agrees with the graph.
c. • f (x) = 2x,
g xx x
x
x x
x( ) = − +
−= − −
−
2 8 15
3
3 5
3
( )( )
= −x + 5, x ≠ 3lim ( ) , limx x
f x g x→ →
= =3 3
8 2 ( )
• p(x) = f (x) · g(x)limx
p x→
= ⋅ =3
8 2 16( )
x p(x)
2.997 15.9907…
2.998 15.9938…
2.999 15.9969…
3 undefined
3.001 16.0030…
3.002 16.0061…
3.003 16.0092…
All these p(x) values are close to 16.
• r xf x
g x( ) = ( )
( )
lim ( )x
r x→
= =3
8
24
3
5
10
15
20y
x
f
g
r
d. For 5 to 5.1 s: average velocity = −15.5 m/s.
Average velocity = =f t f
t
( ) – ( )
–
5
535 5 50
5
5 2 5
5
2t t
t
t t
t
– –
–
– ( – )( – )
–= =
24 Problem Set 2-7 Calculus Solutions Manual© 2005 Key Curriculum Press
−5(t − 2), for t ≠ 5. Instantaneous velocity =limit = −5(5 − 2) = −15 m/s.The rate is negative, so the distance above thestarting point is getting smaller, whichmeans the rock is going down.Instantaneous velocity is a derivative.
R4. a. • f is continuous at x = c if and only if
1. f (c) exists
2. lim ( )x c
f x→
exists
3. limx c
f x f c→
=( ) ( )
• f is continuous on [a, b] if and only if f iscontinuous at every point in (a, b), andlim ( ) ( )x a
f x f a→ +
= and limx b
f x f b→ −
=( ) ( ).
b.
c f c( )lim
x cf x
→ −( ) lim ( )
x cf x
→ +lim ( )x c
f x→ Continuous?
1 none none none none infinite
2 1 3 3 3 removable
3 5 2 5 none step
4 3 3 3 3 continuous
5 1 1 1 1 continuous
c. i. ii.
x
y
1
x
y
2
iii. iv.
x
y
3
x
y
4
v. vi.
5
x
y
L
f(6)
x
y
6
vii.
1
5
–2
x
y
d.
2
4
x
f(x)
2
The left limit is 4 and the right limit is 2, sof is discontinuous at x = 2, Q.E.D.Let 22 = 22 − 6(2) + k.∴ k = 12
R5. a. limx
f x→
= ∞4
( ) means that f (x) can be kept
arbitrarily far from 0 on the positive side justby keeping x close enough to 4, but not equalto 4.limx
f x→∞
=( ) 5 means that f (x) can be made to
stay arbitrarily close to 5 just by keeping xlarge enough in the positive direction.
b. • lim ( )x
f x→−∞
does not exist.
• limx
f x→−
=2
1( )
• limx
f x→ −
= ∞2
( )
• limx
f x→ +
= −∞2
( )
• limx
f x→∞
=( ) 2
c. f (x) = 6 − 2− x
limx
f x→∞
=( ) 6
f (x) = 5.999 = 6 − 2− x
2− x = 0.001
x = − log
log
0 001
2
.
x = 9.965...
x f (x)
10 5.999023…
20 5.999999046…
30 5.99999999907…
All of these f (x) values are within 0.001 of 6.
d. g(x) = x− 2
lim ( )x
g x→
= ∞0
g(x) = 106 = x− 2
x2 = 10− 6
x = 10− 3
x g(x)
0.0009 1.2345… ⋅ 106
0.0005 4,000,000
0.0001 1 ⋅ 108
Calculus Solutions Manual Problem Set 2-7 25© 2005 Key Curriculum Press
All of these g(x) values are larger than1,000,000.
e. v(t) = 40 + 6 t
n Trapezoidal Rule
50 467.9074…
100 467.9669…
200 467.9882…
400 467.9958…
The limit of these sums seems to be 468.
By exploration,T222 = 467.98995…T223 = 467.99002…∴ D = 223
R6. a. See the text statement of the intermediatevalue theorem.The basis is the completeness axiom.See the text statement of the extreme valuetheorem.The word is corollary.
b. f (x) = −x3 + 5x2 − 10x + 20f (3) = 8, f (4) = −4So f (x) = 0 for some x between 3 and 4 bythe intermediate value theorem.The property is continuity.The value of x is approximately 3.7553.
c.
–4
3
x
f(x)
f (−6) = 1 and f (−2) = 5 by tracing on thegraph or by simplifying the fraction to getf (x) = x + 7, then substituting. You will notalways get a value of x if y is between 1 and5. If you pick y = 3, there is no value of x.This fact does not contradict the intermediatevalue theorem. Function f does not meet thecontinuity hypothesis of the theorem.
Concept Problems
C1.
hh
f
g
y
x
7
4
Conjecture: lim ( )x
f x→
=4
7
C2. f (1) = 12 − 6 ⋅ 1 + 9 = 4As x → 1 from the left, f (x) → 12 + 3 = 4.As x → 1 from the right, f (x) → 12 − 6 + 9 = 4.∴ lim
xf x f
→14 1( ) = = ( )
∴ f is continuous at x = 4, Q.E.D.For the derivative, from the left side,f x f
x
x
x
x x
x
( ) – ( )
–
–
–
( – )( )
–
1
1
3 4
1
1 1
1
2
= + = + =
x + 1, x ≠ 1∴ ′ = + =
→ −limx
f x1
1 1 2( )
For the derivative, from the right side,f x f
x
x x
x
x x
x
( ) – ( )
–
– –
–
( – )( – )
–
1
1
6 9 4
1
1 5
1
2
= + = =
x − 5, x ≠ 1∴ lim
xf x
→ +′ −
11 5 4( ) = – =
So f is continuous at x = 1, but does not have avalue for the derivative there because the rate ofchange jumps abruptly from 2 to −4 at x = 1. Ingeneral, if a function has a cusp at a point, thenthe derivative does not exist, but the function isstill continuous.
C3. The graph is a y = x2 parabola with a stepdiscontinuity at x = 1. (Use the “rise-run”property. Start at the vertex. Then run 1, rise 1;run 1, rise 3; run 1, rise 5; . . . . Ignore thediscontinuity at first.) To create thediscontinuity, use the signum function withargument (x − 1). Because there is no value forf (1), the absolute value form of the signumfunction can be used.
y xx
x= 2 2
1
1+ − | – |
–
C4. The quantity | ( ) – |f x L is the distance betweenf (x) and L. If this distance is less than ε, thenf (x) is within ε units of L. The quantity |x − c|is the distance between x and c. The right part ofthe inequality, |x – c| < δ, says that x is within δunits of c. The left part, 0 < |x − c|, says that xdoes not equal c. Thus, this definition of limit isequivalent to the other definition.
Chapter Test
T1. f is continuous at x = c if and only if
1. f (c) exists
2. lim ( )x c
f x→
exists
3. limx c
f x f c→
=( ) ( )
f is continuous on [a, b] if and only if f iscontinuous at all points in (a, b), andlim ( ) ( )x a
f x f a→ +
= and lim ( ) ( )x b
f x f b→ −
= .
26 Problem Set 2-7 Calculus Solutions Manual© 2005 Key Curriculum Press
T2. a. • limx
f x→ −
=2
3( ) • limx
f x→ +
=2
4( )
• lim ( )x
f x→2
does not exist. • lim ( )x
f x→ −
=6
2
• lim ( )x
f x→ +
=6
2 • limx
f x→
=6
2( )
b. f is continuous on [2, 6] because it iscontinuous for all values in (2, 6) andlim ( ) ( )x
f x f→ +
=2
2 and lim ( ) ( )x
f x f→ −
=6
6 .
T3. See the text statement of the quotient property.
T4. a. Left: −4; right: −4
b. Limit: −4
c. Discontinuous
T5. a. Left: none; right: none
b. Limit: none
c. Discontinuous
T6. a. Left: 6; right: 6
b. Limit: 6
c. Continuous
T7. a. Left: −2; right: 3
b. Limit: none
c. Discontinuous
T8.
4
4
x
y
T9. a. b.
x
f(x)
x
g(x)
1
c. d.
1
–1
h(x)
x
1
1
x
s(x)
T10. a. f ( )( )( )
,30 5 0 8 0 3
0 3
0
0
2
= − ⋅ + −−
=
an indeterminate form
b. lim ( ) lim ( ),x x
f x x x x→ →
= − + ≠3 3
2 5 8 3
Definition of limit“x ≠ c”
= + − +→ → →
lim lim ( ) limx x x
x x3
2
3 35 8
Limit of a sum= ⋅ + − ⋅ +
→ → →lim lim ( ) limx x x
x x x3 3 3
5 8
Limit of a product,limit of a constanttimes a function,limit of a constant
= 3 · 3 + (−5) · 3 + 8 Limit of x= 2, Q.E.D.
T11. If k f xx x
x x= =
≤+ >
12
1 2
2
, ( ),
,
lim limx x
f x f x→ →− +
= =2 2
4 3( ) , ( )
∴ f is discontinuous at x = 3.
2
5f(x)
x
T12. limx
f x k→ −
= ⋅2
22( )
limx
f x k→ +
+2
2 ( ) =
∴ 4k = 2 + kk = 2/3
T13. See graph in T11.
T14. a. limx
T x→∞
=( ) 20
From the graph, it appears that if x > 63 ft,then T(x) is within 1° of the limit.The graph of T has a horizontal asymptote atT = 20.
b. T = 20 + 8(0.97x) cos 0.5x. The amplitude ofthe cosine factor is 8 0 97( . ).x Make thisamplitude < 0.1.8(0.97c) = 0.1
0.97c = 0.0125
c = log .
log .
0 0125
0 97
c = 143.8654…∴ T is within 0.1 unit of 20 wheneverx > 143.8654… .
c. The time of day would be mid-afternoon,when the temperature of the surface ishighest.
T15. a. Use either TRACE or TABLE to show:d(0) = 0, d(10) = 6, d(20) = 14, d(30) = 24,d(40) = 36, and d(50) = 50.
b. Average rate = −−
=d d( . ) ( )
.
20 1 20
20 1 2014 0901 14
20 1 200 901
. –
. –= . cm/day
Calculus Solutions Manual Problem Set 2-7 27© 2005 Key Curriculum Press
c. Average rate = −−
=d t d
t
( ) ( )20
200 01 0 5 14
20
0 01 70 20
20
2. . –
–
. ( )( – )
–
t t
t
t t
t
+ = + =
0.01t + 0.7, t ≠ 20. The limit as t approaches20 is 0.01(20) + 0.7, which equals0.9 cm/day. This instantaneous rate is calledthe derivative.
d. The glacier seems to be speeding up becauseeach 10-day period it moved farther than it hadin the preceding 10-day period.
T16. c(0) = p(0) = 10, so each has the same speed att = 0. lim ( ) . lim ( ) .
t tc t p t
→∞ →∞= = ∞16 Surprise for
Phoebe!
T17. f xkx x
kx x( )
,
– ,=
≤>
2 2
10 2
if
if
limx
f x k k→ −
= ⋅ =2
22 4( )
limx
f x k→ +
= −2
10 2( )
Make 4k = 10 − 2k ⇒ k = 5/3. There is a cuspat x = 2.
2
10f(x)
x
T18. h(x) = x3. h(1) = 1 and h(2) = 8, so 7 is betweenh(1) and h(2). The intermediate value theoremallows you to conclude that there is a realnumber between 1 and 8 equal to the cube rootof 7.
T19. Answers will vary.
28 Problem Set 3-2 Calculus Solutions Manual© 2005 Key Curriculum Press
Chapter 3—Derivatives, Antiderivatives,and Indefinite Integrals
Problem Set 3-11. The graph is correct.
2. Average rate = = =f f( . ) – ( )
.
. –
.
5 1 5
0 1
3 21 3
0 12.1 km/min
3.
5
2
x
y
4. The graph of r has a removable discontinuityat x = 5.
rf f
( )( ) ( )
55 5
5 5
0
0= −
−=
5. r xf x f
x
x x
x( )
( ) ( )= −−
= − + −−
5
5
8 18 3
5
2
= − −−
= − ≠( )( ),
x x
xx x
5 3
53 5
′ =→
f r xx
( ) lim ( )55
= − −−
= − =→ →
lim( )( )
lim( )x x
x x
xx
5 5
5 3
53 2
The derivative is the velocity of the spaceship,in km/min.
6. Find the equation of the line through (5, f (5)),or (5, 3), with slope 2.y − 3 = 2(x − 5) ⇒ y = 2x − 7
5
3
x
y
This line is tangent to the graph of f (x) at (5, 3).
7. As you zoom in, the line and the graph appear tobe the same.
8. Answers will vary.
Problem Set 3-2Q1. Instantaneous rate of change
Q2. x + 9 Q3. 18
Q4.
x
y
1
Q5. 9x2 − 42x + 49
Q6. “−” sign
Q7. Q8.
x
y
3
x
y
5
2
Q9. Newton and Leibniz
Q10. D
1. See the text for the definition of derivative.
2. Physical: Instantaneous rate of change of thedependent variable with respect to theindependent variableGraphical: Slope of the tangent line to the graphof the function at that point
3. a. fx
xx′
→( ) =3
0 6 5 4
33
2
lim. – .
–
= lim. ( – )( )
–.
x
x x
x→
+ =3
0 6 3 3
33 6
b. Graph of the difference quotient m(x)
3
3.6
x
m(x)
c., d. Tangent line: y = 3.6x − 5.4
3
1 x
f(x)
4. a. f x x fx
xx( ) = – . , ( ) =0 2 6
0 2 7 2
62
6
2
′ +→
lim– . .
–
= + = −→
lim– . ( – )( )
–.
x
x x
x6
0 2 6 6
62 4
Calculus Solutions Manual Problem Set 3-2 29© 2005 Key Curriculum Press
b.
6
–2.4
x
m(x)
c., d. Tangent line: y = −2.4x + 7.2
6
–7.2
xf(x)
5. fx x
xx′ + + +
+→−(– ) =2
5 1 5
22
2
lim
= lim( )( )
x
x x
x→−
+ ++
=2
2 3
21
6. (– ) =fx x
xx′ + +
+→−4
6 2 10
44
2
lim–
= + ++→−
lim( )( )
x
x x
x4
4 2
42= –
7. ( ) =fx x x
xx′ + +
→1
4 8 6
11
3 2
lim– –
–
= = –lim( – )( – – )
–x
x x x
x→1
21 3 2
14
8. fx x x
xx′ +
+→−(– ) =1
4 6 8
11
3 2
lim– – –
= ++
=→−lim
( )( – – )x
x x x
x1
21 2 2
11
9. fx
xx′ + +
→( ) =3
0 7 2 0 1
33lim
– . .
–
= = −→
lim– . ( – )
–.
x
x
x3
0 7 3
30 7
10. fx
xx′
→( ) =4
1 3 3 2 2
44lim
. – – .
–
= =→
lim. ( – )
–.
x
x
x4
1 3 4
41 3
11. fxx
′+
=→−
(– ) =15 5
10
1lim
–
12. fxx
′ + =→
( ) =32 2
30
3lim
–
–
13. The derivative of a linear function equals theslope. The tangent line coincides with the graphof a linear function.
14. The derivative of a constant function is zero.Constant functions are horizontal and don’tchange! The tangent line coincides with thegraph.
15. a. Find f ′ ( 1) = 2, then plot a line through point(1, f (1)) using f ′ ( 1) as the slope. The line isy = 2x − 1.
b. Near the point (1, 1), the tangent line and thecurve appear nearly the same.
c. The curve appears to get closer and closer tothe line.
d. Near point (1, 1) the curve looks linear.
e. If a graph has local linearity, the graph nearthat point looks like the tangent line.Therefore, the derivative at that point could besaid to equal the slope of the graph at thatpoint.
16. a. f ( x) = x2 + 0.1 (x − 1)2/3
f ( 1) = 12 + 0.1(1 − 1)2/3 = 1 + 0 = 1, Q.E.D.The graph appears to be locally linear at(1, 1), because it looks smooth there.
b. Zoom in by a factor of 10,000.
1
1
c. The graph has a cusp at x = 1. It changesdirection abruptly, not smoothly.
d. If you draw a secant line through (1, 1) from apoint just to the left of x = 1, it has a largenegative slope. If you draw one from a pointjust to the right, it has a large positive slope.In both cases, the secant line becomes verticalas x approaches 1 and a vertical line hasinfinite slope. So there is no real numberequal to the derivative.
17. a.
3
5
x
f(x)7
b. First simplify the equation.
f xx x
x( )
,
,=
+ ≠=
2 3
7 3
if
if
30 Problem Set 3-2 Calculus Solutions Manual© 2005 Key Curriculum Press
The difference quotient is
m xx
x
x
x( ) =
( ) –
–
–
–
+ =2 7
3
5
3
3
5
x
m(x)
c.
x f (x)
2.997 667.66…
2.998 1001
2.999 2001
3.000 undefined
3.001 −1999
3.002 −999
3.003 −665.66…
The difference quotients are all large positivenumbers on the left side of 3. On the rightside, they are large negative numbers. For aderivative to exist, the difference quotientmust approach the same number as x getscloser to 3.
18. a.
1
2
x
s(x)
b. m xx
x( ) =
| sin ( – ) |
–
1
1
1
1
x
m(x)
c. As x approaches 1 from the left, m(x)approaches −1. As x approaches 1 from theright, m(x) approaches 1. Because the left andright limits are unequal, there is no derivativeat x = 1.
19. a. ′ +→
f xx x
xx( ) = lim
. – . . –
–3
20 25 2 5 7 25 2
3
=→
lim. ( – )( – )
–x
x x
x3
0 25 3 7
3
=→
lim ,x
x3
0 25 7 1 . ( – ) = – Q.E.D.
The tangent line on the graph has slope −1.
b.
Draw secant linesfrom here.
f (x)
3
2
x
As the x-distance between the point and 3decreases, the secant lines (solid) approach thetangent line (dashed).
c. The same thing happens with secant linesfrom the left of x = 3. See the graph in part b.
d.
3
4
x
Draw secant linesfrom here.
g(x)
e. A derivative is a limit. Because the left andright limits are unequal, there is no derivativeat x = 3.
f. m xx
x( ) =
– cos
–
66
3
π
. By table,
x m(x)
2.9 3.1401…
2.99 3.1415…
3 undefined
3.01 −3.1415…
3.1 −3.1401…
Conjecture: The numbers are π and −π.
20. From Problem 19, parts b and c, the tangent lineis the limit of the secant lines as x approaches c.Because the slope of the secant line is the average
Calculus Solutions Manual Problem Set 3-3 31© 2005 Key Curriculum Press
rate of change of f (x) for the interval from x to c(or from c to x) and the derivative, f ′(c), is thelimit of this average rate, the slope of the tangentline equals f ′(c).
Problem Set 3-3Q1. 3
Q2.
–2
y
x
5
Q3.y
x
Q4. 20%
Q5. 3x2 − 2x − 8
Q6. 25x2 − 70x + 49
Q7. log 6
Q8.
Q9.
Q10. “lim”x c→
is missing.
1. a.40
f´
–2 2
f
x
y
b. f ′(x) is positive for −2 < x < 2. The graph off is increasing for these x-values.
c. f ( x) is decreasing for x satisfying |x| > 2.f ′(x) < 0 for these values of x.
d. Where the f ′ graph crosses the x-axis, thef graph has a high point or a low point.
e. See the graph in part a.
f. Conjecture: f ′ is quadratic.
2.15
3
x
y
g
g'
The graph does not have the high and lowpoints that are typical of a cubic function. As xincreases, the graph starts to roll over and forma high point, but it starts going back up againbefore that happens. This behavior is revealed bythe fact that the derivative is positive everywhere.Between x = 0 and x = 1, the derivative reachesa low point, indicating that the slope is aminimum, but the slope is still positive andthe graph of g is still going up.
3. a.
200h
h´
x
y
–2 1 2.5
b. The h′ graph looks like a cubic functiongraph. Conjecture: Seventh-degree functionhas a sixth-degree function for its derivative.
c. By plotting the graph using a friendly window,then tracing, the zeros of h′ are −2, 1, 2.5.
d. If h′(x) = 0, the h graph has a high point ora low point. This is reasonable because ifh′(x) = 0, the rate of change of h(x) is zero,which would happen when the graph stopsgoing up and starts going down, or viceversa.
e. See the graph in part a.
4.
102 x
y
q
q'
The graph does not have the expected shape for aquartic function. The two high points and the lowpoint all appear to occur as a high point atx = 2. The derivative graph crosses the x-axis justonce, at x = 2, indicating that this is the onlyplace where the function graph is horizontal.
32 Problem Set 3-3 Calculus Solutions Manual© 2005 Key Curriculum Press
5. a.
4
5
x
y
f
f´
b. Amplitude = 1, period = 2π = 6.283… c. The graph of f ′ also has amplitude 1 and
period 2π.d.
f
g
f´ and g´5
4
x
y
The graphs of f and g are the same shape,spaced 1 unit apart vertically. The graphs off ′ and g′ are identical! This is to be expectedbecause the shapes of the f and g graphs arethe same.
6.
4
5
x
y
f
f´
The function available on the grapher is y = cos x.The amplitude is 1, the period is 2π, and theshape is sinusoidal. cos 0 = 1, and the graph is ata high point, y = 1, when x = 0.
7. 8.
x
y
4
3f
f´
x
y
4
3f'
f
9. 10.
x
f
f´
y
1
5
x
y
2
3f'
f
11. The derivative for f ( x) = 2x is consistently belowthat of the function itself. This fact implies thatf ( x) does not increase rapidly enough to make the
derivative equal the function value. So the basemust be greater than 2. By experimenting, 3 istoo large, but not by much. You can use trialand error with bases between 2 and 3, checkingthe results either by plotting the graph and thenumerical derivative or by constructing tables.An ingenious method that some students comeup with uses the numerical derivative andnumerical solver features to solvenDeriv(bx, x, 1) = b1 at x = 1. The answer isabout 2.718281… . (In Section 3-9, students willlearn that this number is e, the base of naturallogarithms.) The graph of f ( x) = 2.781…x
and its numerical derivative are shown here.
f and f '
x
y
1
1
12. Answer will vary depending on calculator.
13. a. Maximum area = (12.01)2 = 144.2401 in.2
Minimum area = (11.99)2 = 143.7601 in.2
Range is 143.7601 ≤ area ≤ 144.2401.Area is within 0.2401 in.2 of the ideal.
b. Let x be the number of inches.Area = x2.The right side of 12 is more restrictive, so setx2 = 144.02.∴ x = 144.021/2 = 12.000833…Keep the tile dimensions within 0.0008 in. of12 in.
c. The 0.02 in part b corresponds to ε, and the0.0008 corresponds to δ.
14. The average of the forward and backwardsdifference quotients equals
1
2
f x h f x
h
f x f x h
h
( ) – ( ) ( ) – ( – )+ +
= +
1
2
f x h f x h
h
( ) – ( – )
= +f x h f x h
h
( ) – ( – )
2, Q.E.D.
15. a. f ( x) = x3 − x + 1 ⇒ f ( 1) = 1
fx x
xx′ +
→( ) =1
1 1
11
3
lim( – ) –
–
= = +→ →
lim–
–lim
( )( – )
–x x
x x
x
x x x
x1
3
11
1 1
1=
→limx
x x1
1 2( + ) =
Calculus Solutions Manual Problem Set 3-4 33© 2005 Key Curriculum Press
b. Forward: f f( . ) – ( )
.
. –
..
1 1 1
0 1
1 231 1
0 12 31= =
Backwards: f f( ) – ( . )
.
– .
..
1 0 9
0 1
1 0 829
0 11 71= =
Symmetric:
f f( . ) – ( . )
( . )
. – .
..
1 1 0 9
2 0 1
1 231 0 829
0 22 01= =
The symmetric difference quotient is closer tothe actual derivative because it is the averageof the other two, and the other two span theactual derivative.
c. f ( 0) = 1
fx x
xx′ +
→( ) =0
1 1
00
3
lim( – ) –
–
= lim–
lim( – )x x
x x
xx
→ →= = −
0
3
0
2 1 1
d. Forward: f f( . ) – ( )
.
. –
.
0 1 0
0 1
0 901 1
0 1= = –0.99
Backwards:
f f(– . ) – ( )
.
.
.
0 1 0
0 1
1 099 1
0 1= − = –0.99
Symmetric:
f f( . ) – (– . )
( . )
. – .
..
0 1 0 1
2 0 1
0 901 1 099
0 20 99= = −
All three difference quotients are equal becausef ( x) changes just as much from −0.1 to 0 as itdoes from 0 to 0.1.
16.
h Backwards Forward Symmetric
0.1 −1.1544… 3.1544… 1
0.01 −3.6415… 5.6415… 1
0.001 −9 11 1
The backwards difference quotients are becominglarge and negative, while the forward differencequotients are becoming large and positive. Theiraverage, the symmetric difference quotient, isalways equal to 1.
17. Answers will vary.
Problem Set 3-4Q1. 9x2 − 24x + 16
Q2. a3 + 3a2b + 3ab2 + b3
Q3. See the text definition of derivative.
Q4.f x h f x h
h
( ) – ( – )+2
Q5. No limit (infinite) Q6. log 73
Q7. 3 Q8. Pythagorean theorem
Q9. 10 Q10. C
1. f ( x) = 5x4 ⇒ f ′ ( x) = 20x3
2. y = 11x8 ⇒ dy/dx = 88x7
3. v = 0.007t− 83 ⇒ dv/dt = −0.581t− 84
4. v xx
v x x( ) = ( ) = – ––9
10
18
1
2⇒ ′
5. M(x) = 1215 ⇒ M′ (x) = 0 (Derivative of aconstant)
6. f (x) = 4.7723 ⇒ f ′ (x) = 0 (Derivative of aconstant)
7. y = 0.3x2 − 8x + 4 ⇒ dy/dx = 0.6x − 8
8. r = 0.2x2 + 6x − 1 ⇒ dr /dx = 0.4x + 6
9.d
dxx( – )13 1= −
10. f (x) = 4.5x2 − x ⇒ f′ (x) = 9x − 1
11. y = x2.3 + 5x− 2 − 100x + 4 ⇒dy/dx = 2.3x1.3 − 10x− 3 − 100
12.d
dxx x x x x x( – – )/ –2 5 2 1 3 5 24 3 14
2
58 3+ = − +– / –
13. v = (3x − 4)2 = 9x2 − 24x + 16 ⇒ dv/dx = 18x − 24
14. u = (5x − 7)2 = 25x2 − 70x + 49 ⇒du/dx = 50x − 70
15. f (x) = (2x + 5)3 = 8x3 + 60x2 + 150x + 125 ⇒f ′ (x) = 24x2 + 120x + 150
16. f (x) = (4x − 1)3 = 64x3 − 48x2 + 12x − 1 ⇒f ′ (x) = 192x2 − 96x + 12
17. P xx
x P x x( ) – ( ) –= + ⇒ ′ =2
24 1
18. Q xx x
x Q x x x( ) = – + ( ) = + –3 2
2
3 21 1+ ⇒ ′
19. f (x) = 7x4
′ = +→
f xx h x
hh( ) lim
( ) –0
4 47 7
= + + +→
lim ( )h
x x h xh h x0
3 2 2 3 328 42 28 7 28=
By formula, f ′(x) = 7 ⋅ 4x3 = 28x3, which checks.
20. g(x) = 5x3
g xx h x
hh′ +
→( ) = lim
0
3 35 5( ) –
= + + =→
limh
x xh h x0
2 2 215 15 5 15( )
By formula, g ′ (x) = 5 ⋅ 3x2 = 15x2, which checks.
21. v(t) = 10t2 − 5t + 7
v tt h t h t t
hh′ + − + + − − +
→( ) = lim
[ ( ) ( ) ] ( )0
2 210 5 7 10 5 7
= +→
lim–
h
th h h
h0
220 10 5
= +→
lim ( – )h
t h t0
20 10 5 20 5= –
By formula, v ′ (t) = 10 ⋅ 2t − 5 = 20t − 5, whichchecks.
34 Problem Set 3-4 Calculus Solutions Manual© 2005 Key Curriculum Press
22. s t t t( ) = – + .4 26 3 7
′ = + − + + − − +→
s tt h t h t t
hh( ) lim
[( ) ( ) . ] [ . ]0
4 2 4 26 3 7 6 3 7
= + + +→
lim– –
h
t h t h th h th h
h0
3 2 2 3 4 24 6 4 12 6
= + + +→
lim ( – – )h
t t h th h t h0
3 2 2 34 6 4 12 6
= 4 123t t– By formula, s t t t t t′ ⋅( ) = – = –4 6 2 4 123 3 , which checks.
23. Mae should realize that you differentiatefunctions, not values of functions. If yousubstitute a value for x into f (x) = x4, you getf (3) = 34 = 81, which is a new function, g(x) =81. It is the derivative of g that equals zero.Moral: Differentiate before you substitute for x.
24. a. v(x) = h′(x) = −10x + 20
b. The book was going down at 10 m/s.The velocity is −10, so h(x) is decreasing.
c. The book was 15 m above where he threw it.
d. 2 s. The book is at its highest point when thevelocity is zero. v(x) = 0 if and only if x = 2.
25.
x
f´
f
y
9
7
26.
6
x
g
g'
y3
27. a.
f´
f
y
x
10
1
b. The graph of f ′ is shown dashed in part a.
c. There appear to be only two graphs becausethe exact and the numerical derivative graphsalmost coincide.
d. f (3) = −6.2
f ′(3) = 3.8 (by formula)
f ′(3) ≈ 3.8000004 (depending on grapher)
The two values of f ′( )3 are almost identical!
28. a. g(x) = x− 1. Conjecture: g′(x) = −1 ⋅ x− 2.Conjecture is confirmed.
y1
y 2 and y3
1
1
x
y
b. h(x) = x1/ 2. Conjecture: g′(x) = 0.5x− 1/ 2.Conjecture is confirmed.
y1
y2 and y3
1
2
x
y
c. e(x) = 2x. Conjecture: e′ (x) = x ⋅ 2x− 1.Conjecture is refuted!
y1
y2
y3
111
x
y
29. f x x x( ) = –/1 2 2 13+f x x f′ ′ =( ) = + , ( ) 1
2– / 9
41 2 2 4
Increasing by 9/4 y-units per x-unit at x = 4
30. f (x) = x− 2 − 3x + 11f ′(x) = −2x− 3 − 3, f ′(1) = −5Decreasing by 5 y-units per x-unit at x = 1
31. f (x) = x1.5 − 6x + 30f ′(x) = 1.5x0.5 − 6, f ′(9) = −1.5Decreasing by 1.5 y-units per x-unit at x = 9
32. f x x x( ) = – + +3 1
f x x f′ = ′ =( ) – + , ( ) – .32
– /1 2 1 2 0 0606K
Decreasing by approximately 0.0607 y-unit perx-unit at x = 2
33. f xx
x x f x x x( ) = ′3
2 2
33 5 2 3– – + , ( ) = – –
Calculus Solutions Manual Problem Set 3-4 35© 2005 Key Curriculum Press
High and low points of the f graph are at thex-intercepts of the f ′ graph.
–1 3
x
y
5 f´
f
34. f xx
x x f x x x( ) = – + + , ( ) = – +3
2 2
32 3 9 4 3′
High and low points of the f graph are at thex-intercepts of the f ′ graph.
1 3
15f
f ' x
y
35. If f (x) = k ⋅ g(x), then f ′(x) = k ⋅ g′(x).
Proof:
f xf x h f x
hk g x h k g x
h
kg x h g x
h
kg x h g x
h
h
h
h
h
′ +
= ⋅ + ⋅
= ⋅ +
= ⋅ +
→
→
→
→
( ) = lim( ) – ( )
lim( ) – ( )
lim( ) – ( )
lim( ) – ( )
0
0
0
0
= ⋅ ′k g x( ) , Q.E.D.
Dilating a function f (x) vertically by a constant kresults in the new function g(x) = k ⋅ f (x). Whathas been shown is that
d
dxk f x k
d
dxf x( ( )) ( )⋅ = ⋅
That is, dilating a function vertically by aconstant k dilates the derivative function by aconstant factor of k.
36. If f x x( ) = 5 , then f c c′ =( ) 5 4 .
Proof:
f cf x f c
x c
x c
x c
x c x x c x c xc c
x c
x x c x c xc c
c c c c c
x c
x c
x c
x c
′ =
=
= − + + + +−
= + + + +
=
→
→
→
→
( )
+ + + +
lim( ) – ( )
–
lim–
–
lim( )( )
lim ( )
5 5
4 3 2 2 3 4
4 3 2 2 3 4
4 4 4 4 4
= 5c4, Q.E.D.
37. If f (x) = xn, then f′(x) = nxn− 1.
Proof:
f xx h x
hh
n n
′ +→
( ) = lim( ) –
0
= + + − + + −
→
− −
lim( )
h
n n n n nx nx h n n x h h x
h0
1 12
2 21 L
= + − + +
→
− − −lim ( )h
n n nnx n n x h h0
1 2 11
21 L
= + + + +nxn–1 0 0 0L = ,–nxn 1 which is from the second term in the
binomial expansion of (x + h)n, Q.E.D.
38. If y u u u un n= + + + + ,1 2 3 L where the ui aredifferentiable functions of x, prove that
′ ′ ′ ′ ′y u u u un n= + + + +1 2 3 L for all integers n ≥ 2.
Proof:
Anchor: For n = 2, y u u2 1 2= + .∴ ′ ′ ′y u u2 1 2= + by the derivative of a sum of thetwo functions property, thus anchoring theinduction.Induction hypothesis:Suppose that for n = k > 2,
′ ′ ′ ′ ′y u u u uk k= + + + +1 2 3 L .Verification for n = k + 1:Let y u u u u uk k k+ + .1 1 2 3 1= + + + + +LThen y u u u u uk k k+ += ( + + + + ) +1 1 2 3 1L , whichis a sum of two terms.∴ ′ ′ ′y u u u u uk k k+ += ( + + + + ) +1 1 2 3 1L ,which, by the anchor,= ( + + + +′ ′ ′ ′ + ′+u u u u uk k1 2 3 1L )= ′ + ′ + ′ + + ′ + ′+u u u u uk k1 2 3 1L ,which completes the induction.
Conclusion:∴ ′ ′ ′ ′ ′y u u u un n= + + + +1 2 3 L for all integersn ≥ 2, Q.E.D.
39. a. f x x x f x x x x′ ⇒( ) = – + ( ) = – +3 10 5 5 52 3 2
b. g(x) = f (x) + 13 is also an answer to part abecause it has the same derivative as f (x). Thederivative of a constant is zero.
c. The name antiderivative is chosen because itis an inverse operation of taking the derivative.
d.d
dxg x
d
dxf x C[ ( )] [ ( ) ]= + =
d
dxf x
d
dxC
d
dxf x( ) ( )+ =
The word indefinite is used because of theunspecified constant C.
40. a. f ′(x) = 5x4 ⇒ f (x) = x5 + Cf(2) = 23 ⇒ (2)5 + C = 23C = −9∴ f (x) = x5 − 9
b. f ′(x) = 0.12x2 ⇒ f (x) = 0.04x3 + Cf (1) = 500 ⇒ 0.04(1)3 + C = 500
36 Problem Set 3-5 Calculus Solutions Manual© 2005 Key Curriculum Press
C = 499.96∴ f (x) = 0.04x3 + 499.96
c. ′ = ⇒ = +f x x f x x C( ) ( )3 14
4
f C( ) ( )5 2 5 214
4= ⇒ + =C = −154.25
∴ = − f x x( ) .14
4 154 25
Problem Set 3-5Q1. No values of t Q2. dy dx x/ = 10
Q3. ′ = − −y x51 4 Q4. ′ =f x x( ) . .1 7 0 7
Q5. ( / )( )d dx x3 5 3+ = Q6. f ( )3 45=
Q7. f ′( ) =3 30 Q8. 45
Q9. ε Q10. C
1. y t t t= – +.5 3 74 2 4
vdy
dtt t= = – . + ,.20 7 2 73 1 4
adv
dtt t= = – . .60 10 082 0 4
2. y t t= . ––0 3 54
vdy
dtt a
dv
dtt= = – . – , = =– –1 2 5 65 6
3. x = −t3 + 13t2 − 35t + 27. The object starts outat x = 27 ft when t = 0 s. It moves to the left tox ≈ 0.15 ft when t ≈ 1.7 s. It turns there andgoes to the right to x = 70 ft when t = 7 s. Itturns there and speeds up, going to the left for allhigher values of t.
10
x
y
Starts at t = 0, x = 27
Turns at t = 1.7, x = 0.15
Turns at t = 7, x = 76
4. x t t t t= – + – +4 3 211 38 48 50. The object startsat x = 50 ft when t = 0 s. It moves to the leftto x ≈ 30 ft when t ≈ 1.0 s. Then it moves to theright to x ≈ 34.8 ft when t ≈ 2.4 s. The objectmoves to the left again, turning at x ≈ 9.4 ftwhen t ≈ 4.8 s and then moving back to the rightfor higher values of t.
Starts at t ≈ 0, x ≈ 50
x
y
10
Turns at t ≈ 4.8, x ≈ 9.4
Turns at t ≈ 2.4, x ≈ 34.8
Turns at t ≈ 1.0, x ≈ 30.0
5. a. x t t t= – + – +3 213 35 27 (See Problem 3.)v t t a t= – + – , = – +3 26 35 6 262
b. v( ) = – + – = –1 3 26 35 12So x is decreasing at 12 ft/s at t = 1.a( ) = – + =1 6 26 20So the object is slowing down at 20 (ft/s)/sbecause the velocity and acceleration are inopposite directions when t = 1.v( ) ( ) ( )6 3 6 26 6 35 132= − + − =So x is increasing at 13 ft/s at t = 6.a( ) ( )6 6 6 26 10= − + = −So the object is slowing down at 10 (ft/s)/sbecause the velocity and acceleration are inopposite directions when t = 6.v( ) = – ( ) + ( ) – = –8 3 8 26 8 35 192
So x is decreasing at 19 ft/s at t = 8.a( ) = – ( ) + = –1 6 8 26 22So the object is speeding up at 22 (ft/s)/sbecause the velocity and acceleration are in thesame directions when t = 8.
c. At t = 7, x has a relative maximum becausev( ) =7 0 at that point and is positive justbefore t = 7 and negative just after. No, x isnever negative for t in [0, 9]. It starts out at27 ft, decreases to just above zero aroundt = 1.7, and does not become negative untilsome time between t = 9.6 and 9.7.
6. a. x t t t t= – + – +4 3 211 38 48 50 (See Problem 4.)v t t t a t t= – + – , = – +4 33 76 48 12 66 763 2 2
b. At t = 1, v( ) = –1 1 and a( ) =1 22. At t = 3,v( ) = –3 9 and a( ) = –3 14 . At t = 5, v( ) =5 7and a( ) =5 46 . The object is slowing down att = 1 because the velocity and acceleration arein opposite directions. The object is speedingup at t = 3 and t = 5 because velocity andacceleration are in the same direction.
c. v = 0 when t = 1.0475… , 2.3708… ,or 4.8315… .
d. The displacement is at a maximum or aminimum whenever v = 0.
1 2 3 4 5
10 t
y
x
v
e. a = 0 when t = 1.6413… or 3.8586… .When a = 0, v is at a maximum or minimum
Calculus Solutions Manual Problem Set 3-5 37© 2005 Key Curriculum Press
point and the graph of x is at its steepest fortimes around these values.
1 2 3 4 5
10 t
y
x
va
7. a.
300
10
t
y
d
d´
b. v = d′ = 30 − 2t. Velocity is positive for0 ≤ t < 15. Calvin is going up the hill forthe first 15 s.
c. At 15 seconds his car stopped. d(15) = 324,so distance is 324 feet.
d. 99 30 0 33 3 0 332+ – = ( – )( + ) = =t t t t t⇒ ⇒or t = −3. He’ll be back at the bottom whent = 33 s.
e. d( ) =0 99. The car runs out of gas 99 ft fromthe bottom.
8. a.
30
200
v = 251y
ta = v'
v
b. Trace the v′ graph to find a( )0 32≈ . Theacceleration decreases because the velocity isapproaching a constant. In the real world, thisoccurs because the wind resistance increases asthe velocity increases.
c. The limit is 251 ft/s as t approaches infinity.The term 0.88t approaches zero as t gets verylarge, leaving only 1 inside the parentheses.
d. 90% of terminal velocity is 0.9(251) =225.9 ft/s.Algebraic solution:251 1 0 88 225 9 0 88 0 1( – . ) = . – . = – .t t⇒
t = = ≈log .
log .
0 1
0 8818 012394 18 0. ... . s
Numerical solution gives the same answer.Graphical solution: Trace to v(t) = 225.9.T is between 18 and 18.5.
e. Find the numerical derivative.v′(18.0123…) ≈ 3.2086… , which isapproximately 10% of the initial acceleration.
9. a. d t t t d t t( ) = – . ( ) = – .18 4 9 18 9 82 ⇒ ′d′( ) = – . = .1 18 9 8 8 2d′ ⋅( ) = – . = – .3 18 9 8 3 11 4d′ is called velocity in physics.
b. At t = 1 the football is going up at 8.2 m/s.At t = 3 the football is going down at11.4 m/s. The ball is going up when thederivative is positive and coming down whenthe derivative is negative. The ball is goingup when the graph slopes up and comingdown when the graph slopes down.
c. d′( ) = – .4 21 2, which suggests that theball is going down at 21.2 m/s. However,d( ) = – .4 6 4, which reveals that the ballhas gone underground. The function givesmeaningful answers in the real world only ifthe domain of t is restricted to values thatmake d(t) nonnegative.
10. a.
5
2
t
y
v
a
b. The acceleration at the bottom of the swingis 0. The acceleration is greatest at either endof its swing.
11.
5
2
t
y
a
v
12. v(t) = 15t0.6 . Because v t x t( ) = ( )′ , x(t) musthave had t1 6. in it. The derivative of t1.6 can beassumed to be 1.6t0.6 . So the coefficient of t1 6.
must be 15 1 6/ . , or 9.375. But x(0) was 50.Thus, x(t) = 9.375t1.6 + 50. The derivative x′(t)really does equal v(t). Using this equation,x( ) = . ( ) + = . ..10 9 375 10 50 423 2251 6 KSo the distance traveled is 423.225… − 50 =373.225… , or about 373 ft.
13. The average rate is defined to be the change inthe dependent variable divided by the change inthe independent variable (such as total distancedivided by total time). Thus, the difference quotient is an average rate. The instantaneousrate is the limit of this average rate as the changein the independent variable approaches zero.
38 Problem Set 3-6 Calculus Solutions Manual© 2005 Key Curriculum Press
14. a.
10
y
t
3
v
b. y is a relative maximum when t ≈ 0, 4,8, … .y is a relative minimum when t ≈ 2, 6, 10, … .
c. The velocity is a relative maximum whent ≈ 3 or 7. The displacement graph at thesetimes appears to be increasing the fastest.
d. The equation used in the text is
y tt= +2 0 852
. cosπ
The student could observe that the period is 4,leading to the coefficient π/2. The amplitudedecreases in a way that suggests an exponentialfunction with base close to, but less than, 1.The additive 2 raises the graph up two units,as can be ascertained by the fact that the graphseems to converge to 2 as t gets larger. Thenumerical derivative of the function shown inpart a agrees with the graph of the velocity.Note that the actual maximum and minimumvalues occur slightly before the values of tread from the graph in part a. For instance,the maximum near t = 4 is actually att = 3.9343… .
15. y xdy
dxx
d y
dxx= ⇒ = ⇒ =5 15 303 2
2
2
16. y xdy
dxx
d y
dxx= ⇒ = ⇒ =7 28 844 3
2
22
17. y x xdy
dxx x= + ⇒ = + ⇒9 18 52 5 4
d y
dxx
2
2318 20= +
18. y x x= − + ⇒10 15 422
dy
dxx
d y
dx= − ⇒ =20 15 20
2
2
19. m′( ) = .5 153 4979K
m′( ) = .10 247 2100K
These numbers represent the instantaneous rate ofchange of the amount of money in the account.The second quantity is larger because the moneygrows at a rate proportional to the amount ofmoney in the account. Because there is moremoney after 10 years, the rate of increase shouldalso be larger.
′′ =m ( ) .5 14 6299K
′′ =m ( ) .10 23 5616KBoth quantities are in the units ($/yr)yr.The quantities represent the instantaneous rate ofchange of the instantaneous rate of change of theamount of money in the account. For example,at t = 5, the rate of increase of the account(153.50 $/yr) is increasing at a rate of14.63 ($/yr)/yr.
20. ′ = −p ( ) .7 4 9510K
′ = −p ( ) .14 2 4755K∴ ′ = ′p p( ) ( )14 71
2
The fact that these derivatives are negative tellsus that the amount of nitrogen 17 is decreasing.
′′ =p ( ) .7 0 4902K
′′ =p ( ) .14 0 2451KBoth quantities are in units (% of nitrogen17/s)/s. The quantities represent the rate ofchange of the rate of change of the percentage ofnitrogen 17 remaining. For example, at t = 7 s,the rate of decrease (−4.95%/s) is changing at arate of 0.49 (%/s)/s.
21.
5
x
y
2f´f
The graph of the derivative looks like the graphof y = cos (x).
22.
5
x
y
2
f' f
The graph of the derivative looks like the graphof y = −sin (x).
Problem Set 3-61.
y
x
3
10
Calculus Solutions Manual Problem Set 3-7 39© 2005 Key Curriculum Press
2. The graph confirms the conjecture.
5
x
y
2y1
y2 y 3and
3. g(x) = sin 3xConjecture: g′(x) = 3 cos 3xThe graph confirms the conjecture.
3
10
x
y
g
g´
4. h x x( ) = sin 2
Conjecture: h x x x′( ) = 2 2cosThe graph confirms the conjecture.
5
x
y
1 hh'
5. t x x( ) = .sin 0 7
Conjecture: t x x x′( ) = . – . .0 7 0 3 0 7cosThe graph confirms the conjecture.
10
1x
y
t
t´
6. f (x) = sin [g(x)]f is a composite function.g is the inside function.sine is the outside function.Differentiate the outside function with respect tothe inside function. Then multiply the answer bythe derivative of the inside function with respectto x.
7. a. f (x) = sin 3x. Inside: 3x. Outside: sine.
b. h(x) = sin3 x. Inside: sine. Outside: cube.
c. g(x) = sin x3. Inside: cube. Outside: sine.
d. r(x) = 2cos x. Inside: cosine. Outside:exponential.
e. q(x) = 1/(tan x). Inside: tangent. Outside:reciprocal.
f. L(x) = log (sec x). Inside: secant. Outside:logarithm.
8. Answers will vary.
Problem Set 3-7
Q1 exists
Q2 exists
Q3
Any order is acceptable.
. ( )
. lim ( )
. lim ( ) ( )
( )
f c
f x
f x f cx c
x c
→
→=
Q4. No (not continuous) Q5. dy dx x/ = – /16 1 5
Q6. f (x) = −10x− 3 Q7. Antiderivative
Q8.y = sin x
x
Q9.y = cos x
x
Q10. C
1. a. Let y = f (u), u = g(x).dy
dx
dy
du
du
dx= ⋅
b. y f g x g x′ ′ ⋅ ′= ( )] ( )[
c. To differentiate a composite function,differentiate the outside function with respect tothe inside function, then multiply by the deriva-tive of the inside function with respect to x.
2. f x x( ) ( )= −2 31
a. f ′(x) = 3(x2 − 1)2(2x) = 6x(x2 − 1)2
b. (x2 − 1)3 = x6 − 3x4 + 3x2 − 1,
so f ′(x) = 6x5 − 12x3 + 6x.
c. From part a, f ′(x) = 6x(x2 − 1)2 =6x(x4 − 2x2 + 1) = 6x5 − 12x3 + 6x, so thetwo answers are equivalent.
3. f x x f x x x( ) ( )= ⇒ ′ = − ⋅ = −cos sin sin3 3 3 3 3
4. f x x f x x( ) ( )= ⇒ ′ =sin cos5 5 5
5. g x x g x x x( ) ( ) ( ) ( )= ⇒ ′ = −cos sin3 2 33
6. h x x h x x x( ) ( ) ( ) ( )= ⇒ ′ =sin cos5 4 55
7. y = (cos x)3 ⇒y′ = 3(cos x)2 ⋅ (−sin x) = −3 cos2
x sin x
8. f x x( ) ( )= ⇒sin 5
f x x x x x′ = ⋅ =( ) ( )5 54 4sin cos sin cos
9. y x y x x= ⇒ ′ =sin sin cos6 56
10. f x x( ) = ⇒cos7
f x x x x x′ = ⋅ − =( ) ( ) –7 76 6cos sin cos sin
40 Problem Set 3-7 Calculus Solutions Manual© 2005 Key Curriculum Press
11. y = −6 sin 3x ⇒ y′ = −18 cos 3x
12. f (x) = 4 cos (−5x) ⇒f′ (x) = 4[−sin (−5x)] ⋅ (−5) = 20 sin (−5x)
13.d
dxx x x(cos ) cos ( sin )4 37 4 7 7 7= ⋅ − ⋅
= −28 cos3 7x sin 7x
14.d
dxx x x(sin ) sin cos9 813 9 13 13 13= ⋅
= 117 sin8 13x cos 13x
15. f (x) = 24 sin5/3 4x ⇒f′ (x) = 40 sin2/3 4x ⋅ cos 4x ⋅ 4
= 160 sin2/3 4x cos 4x
16. f (x) = −100 sin6/5 (−9x) ⇒f′ (x) = −120 sin1/5 (−9x) ⋅ cos (−9x) ⋅ (−9)
= 1080 sin1/5 (−9x) cos (−9x)
17. f (x) = (5x + 3)7 ⇒f′ (x) = 7(5x + 3)6
⋅ 5 = 35(5x + 3)6
18. f (x) = (x2 + 8)9 ⇒f′ (x) = 9(x2 + 8)8
⋅ 2x = 18x(x2 + 8)8
19. y = (4x3 − 7)− 6 ⇒y′ = −6(4x3 − 7)− 7
⋅ 12x2 = −72x2(4x3 − 7)− 7
20. y = (x2 + 3x − 7)− 5 ⇒y′ = −5(x2 + 3x − 7)− 6
⋅ (2x + 3)= −5(2x + 3)(x2 + 3x − 7)− 6
21. y = [cos (x2 + 3)]100 ⇒y′ = 100 [cos (x2 + 3)]99 ⋅ [−sin (x2 + 3)] ⋅ 2x
= −200x cos99(x2 + 3) sin (x2 + 3)
22. y = [cos (5x + 3)4]5 ⇒ y′ = 5[cos (5x + 3)4]4 ⋅[−sin (5x + 3)4] ⋅ 4(5x + 3)3
⋅ 5 =−100(5x + 3)3 cos4 (5x + 3)4 sin (5x + 3)4
23. y = 4 cos 5x ⇒ dy
dx = 4(−sin 5x)5 = −20 sin 5x ⇒
d y
dx
2
2 = −20(cos 5x)5 = −100 cos 5x
24. y = 7 sin (2x + 5) ⇒dy
dx = 7 cos (2x + 5)(2) = 14 cos (2x + 5) ⇒
d y
dx
2
2 = 14[−sin (2x + 5)](2) =
−28 sin (2x + 5)
25. f′ (x) = cos 5x ⇒ f (x) = 1
5sin 5x + C
26. f′ (x) = 10 sin 2x ⇒ f (x) = −5 cos 2x + C
27. f (x) = 5 cos 0.2xf′ (x) = −5 sin 0.2x ⋅ 0.2 = −sin 0.2xf′ (3) = −sin 0.6 = −0.5646… andf (3) = 4.1266…
The line has the equationy = −0.5646…x + 5.8205… .
The line is tangent to the graph.
3
5y
x
28. y = 7 sin π t + 12t1.2
velocity = dy
dt= 7π cos π t + 14.4t0.2
Yes, there are times when the beanstalk isshrinking. The velocity graph is negative forbrief intervals, and the y-graph is decreasing inthese intervals.
10
50
t
y and v
y
v
29. a. V rdV
drr= ⇒ =4
343 2π π
dV/dr is in (cm3/cm), or cm2.
b. r = 6t + 10
c.dr
dt= 6 (not surprising!). Units: cm/min
d.dV
dt
dV
dr
dr
dt= ⋅
When t = 5, r = 40. So
dV
dr= =4 40 64002π π( ) .
∴ = ⋅ =dV
dt6400 6 38 400 3π π, min. cm /
dV/dr has units cm2, and dr/dt has units
cm/min, so dV/dt has units cmcm2 ⋅min
,
which becomes cm3/min, Q.E.D. Thismatches the commonsense answer that rate of
change of volume is volume
time
cm
min
3
= .
e. V t= +4
36 10 3π
( )
∴ = + = +dV
dtt t4 6 10 6 24 6 102 2π π( ) ( ) ( )
When t = 5, dV
dt= + =24 6 5 10 38 4002π π[ ] ,( ) .
Calculus Solutions Manual Problem Set 3-8 41© 2005 Key Curriculum Press
30. a.
∆u
∆xx
u
∆u does not approach zero as ∆x approacheszero from the left side. (∆u does approach zeroas ∆x approaches zero from the left side.)
b.
∆u
∆ xx
u
∆u does approach zero as ∆x approaches zerofrom either side.
Problem Set 3-8Q1. f′(x) = 9x8 Q2. dy/dx = −3 sin x
Q3. y′ = 72x5 (5x6 + 11)1.4 Q4. s′ = 0
Q5. 12 Q6. 1
Q7. Yes (continuous) Q8. f (x) = −cos x2
Q9. Q10. E
4
4 y
x
f'
1. a.
15
Increasingx
d(x)
25
5
3
45
23
y(t) = C + A cos B(t − D)Vertical displacement = 25 = CAmplitude = 0.5(40) = 20 = APhase disp. (for cosine) = 3 = DPeriod = 60/3 = 20, so B = 2π/20 = π/10.(Note that B is the angular velocity in radiansper second.)
y t t( ) = +25 2010
3cos ( – )π
b. ′ =y t t( ) – 210
3π πsin ( – )
c.
′ = =y ( ) – 3.6931615 210
15 3π πsin ( – ) K
y(t) is increasing at about 3.7 ft/s.
d. The fastest that y(t) changes is 2π, or6.28… ft/s. The seat is at y(t) = 25 ft abovethe ground then.
2. a. y = C + A cos B(x − D).B = 2π /6 = π /3 rad/sD = phase displacement = 1.3 sA = 0.5(110 − 50) = 30 cmC = 110 − 30 or 50 + 30, which equals 80 cm.
∴ = +d t80 303
1 3cos ( – . )π
b. d t′ = –103
1 3π πsin ( – . )
c. d′ = − =( ) 5 10
35 1 3 21 02135π π
sin ( – . ) . K
d′ = =( ) –11 10
311 1 3 21 02135π π
sin ( – . ) . K
At both times, the pendulum is moving awayfrom the wall at about 21.0 cm/s. Theanswers are the same because the times areexactly one period apart.
d. d′ = = −( ) –20 10
320 1 3 21 02135π π
sin ( – . ) . K
The pendulum is moving toward the wall.Because the derivative is negative, d isdecreasing, which in this problem impliesmotion toward the wall.
e. The fastest is 10π ≈ 31.4 cm/s, when d = 80.
f. 0 103
1 33
1 3 0= ⇒ − =– π π πsin ( – . ) sin ( . )t t
⇒ = ⇒ − = +−π π π3
1 3 03
1 3 01( – . ) sin ( . )t t n
⇒ t − 1.3 = 3n ⇒ t = 1.3 + 3n.
The first positive time occurs when n = 0,that is, when t = 1.3 s. When the velocityis zero, the pendulum is at its maximumheight.
3. a. The curb has slope (3.25 − 0.75)/44 = 2.5/44.∴ equation is f (x) = 0.75 + (2.5/44)x.
b. Sinusoid has period 8 ft, so B = 2π /8 = π /4.Amplitude = 0.5(0.75 − 0.25) = 0.25 ft. Lowend of ramp is a low point on the sinusoid.∴ sinusoidal axis is at y = 0.25 when x = 0and goes up with slope 2.5/44.Sinusoid is at a low point when x = 0. Sophase displacement is zero if the cosine issubtracted.
42 Problem Set 3-8 Calculus Solutions Manual© 2005 Key Curriculum Press
∴ equation is
g x x x( ) .= + −0 252 5
440 25
4
.. cos
π
(There are other correct forms.)
c. ′ = +g x x( )2 5
44 16 4
.sin
π π
′ = + =g ( ) ft/ft9
2 5
44 16 49 0 1956
.sin ( ) .
π πK
Going up at about 0.2 vertical ft perhorizontal ft
′ = + = −g (15) ft/ft
2 5
44 16 415 0 0820
.sin ( ) .
π πK
Going down at about 0.08 vertical ft perhorizontal ft. A positive derivative impliesg(x) is getting larger and thus the child isgoing up. A negative derivative implies g(x)is getting smaller and thus the child is goingdown.
d. By tracing the ′g graph, maximum value ofg′ (x) is 0.2531… ft/ft (about 14.2° up).Minimum is −0.1395… ft/ft (about 7.9°down).
4. a. Let d = day number and L(d ) = number ofminutes.14 hours 3 minutes is 843 minutes. 10 hours15 minutes is 615 minutes.∴ amplitude = (1/2)(843 − 615) = 114 min.Sinusoidal axis is at L(d) = 615 + 114 =729 min.Assuming a 365-day year, B = 2π/365.Phase displacement = 172
∴ + ( ) =L d d729 1142
365172cos ( – )
π
On August 7, d = 219.
L( ) = 219 729 1142
365219 172+ =cos ( – )
π
807.67… , or about 13 hours 28 minutes.
b. L d d′ − −( ) =228
365
2
365172
π πsin ( )
On August 7, d = 219.
′ − =L ( ) =219228
365
2
365219 172
π πsin ( – )
–1.42009…Rate ≈ –1.42 min/day(Decreasing at about 1.42 min/day)
c. The greatest rate occurs when the sine is1 or −1.Rate is 228π/365 ≈ 1.96 min/day.1/4 year is about 91 days. So greatest rateoccurs at day 172 ± 91, which is day 263 orday 81 (September 20 or March 22).
5. In general, the period for a pendulum formed bya weight suspended by a string of negligiblemass is 2π L g/ , where L is the length from the
pivot point to the center of mass (actually, thecenter of percussion) of the weight, and g is the
gravitational acceleration, about 9.8 m/s2.Consequently, if the pendulum is 1 meter long,its period will be 2 1 9 8 2 007π / . . ,= K or about
2 s. This is the period for a complete back-and-forth swing. You must quadruple the length of apendulum to double its period. A pendulum hungfrom the ceiling will have a period slow enoughto measure fairly precisely. A good way to getmore accuracy is to count the total time for tenswings, then divide by 10. The period is roughlyconstant for any (moderate) amplitude, as long asthe amplitude is not too big. This fact is notobvious to the uninitiated student and is worthspending time showing. It is quite dramatic towatch a pendulum take just as long to make tenswings with amplitude 2 cm as it does withamplitude 20 or 30 cm.
6. The following data were computed from actualsunrise and sunset times for San Antonio foreach ten days. You can get similar informationfor your locality from the local weather bureau ornewspaper office, from the Nautical AlmanacOffice, U.S. Naval Observatory, Washington,D.C., 20390, or from the Internet.
Day Min Day Min Day Min
0 617 130 811 260 738
10 623 140 823 270 720
20 632 150 833 280 703
30 645 160 840 290 686
40 660 170 842 300 669
50 676 180 842 310 653
60 693 190 836 320 639
70 711 200 828 330 628
80 729 210 816 340 620
90 747 220 803 350 615
100 764 230 789 360 615
110 780 240 772
120 797 250 755
100 200 300
d
L(d )
800
700
0
600
The graph shows a good fit to the data. But thereis a noticeable deviation in the fall and winter,here the day is slightly longer than predicted.The main reason for the discrepancy, apparently,
Calculus Solutions Manual Problem Set 3-8 43© 2005 Key Curriculum Press
is the fact that in the fall and winter, Earth iscloser to the Sun and hence moves slightly morerapidly through its angle with the Sun thanduring the spring and summer.
7. a.
f
hg
1
4
y
x
The limits are all equal to 4.
b. f (x) < g(x) and lim ( ) lim ( )x x
f x g x→ →
= =1 1
4
f (x) ≤ h (x) ≤ g(x)
c.
x f(x) h(x) g(x)
0.95 3.795 3.8 3.805
0.96 3.8368 3.84 3.8432
0.97 3.8782 3.88 3.8818
0.98 3.9192 3.92 3.9208
0.99 3.9598 3.96 3.9602
1.00 4 4 4
1.01 4.0398 4.04 4.0402
1.02 4.0792 4.08 4.0808
1.03 4.1182 4.12 4.1218
1.04 4.1568 4.16 4.1632
1.05 4.195 4.2 4.205
d. From the table, if x is within 0.02 unit of 2,then both f (x) and g(x) are within 0.1 unit of4. From the table, δ = 0.01 or 0.02 willwork, but 0.03 is too large. All the values ofh(x) are between the corresponding values off (x) and g(x), and the three functions allapproach 4 as a limit.
8. Prove that limsin
.x
x
x→=
00 See the text proof.
9. a. The numbers are correct.
b.
x (sin x)/x
0.05 0.99958338541…
0.04 0.99973335466…
0.03 0.99985000674…
0.02 0.99993333466…
0.01 0.99998333341…
Values are getting closer to 1.
c. Answers will vary according to calculator. Forthe TI-83 in TABLE mode, starting x at 0 andusing ∆ x = 10−7 shows that all values roundto 1 until x reaches 1.8 × 10−6, whichregisters as 0.999999999999.
d. Answer will depend on calculator. For TI-83in TABLE mode, (sin 0.001)/0.001 is0.999999833333, which agrees exactly withthe value published by NBS to 12 places.
e. If students have studied Taylor series(Chapter 12) before taking this course, theywill be able to see the reason. The Taylorseries for sin 0.001 is
0 001
0 001
3
0 001
5
0 001
7
3 5 7
..
!
.
!
.
!− + − + L
= 0.00100 00000 00000 00000 000…
− 0.00000 00001 66666 66666 666…
+ 0.00000 00000 00000 00833 333…_______________________________
= 0.00099 99998 33333 34166 666…
10. See the text proof.
11. See the text proof.
12. a. See the text statement of the theorem.
b. Proof:
Given any ε > 0, there is a δ f > 0 such that0 < |x − c | < δ f ⇒ |f (x) − L | < ε, becauselim ( ) .x c
f x L→
= Similarly, there is a δ g > 0
such that 0 < |x − c | < δ g ⇒ |g(x) − L| < ε.Let δ be the smaller of δ f and δ g. Then 0 <|x − c | < δ ⇒ 0 < |x − c | < δ f = |f (x) − L| <ε ⇒ f (x) < L + ε , and also 0 < |x − c | <δ ⇒ 0 < |x − c | < δg ⇒ |g(x) − L| < ε ⇒L − ε < g(x). Then L − ε < g(x) < h(x) < f(x)< L + ε, so |h(x) − L | < ε, so lim ( )
x ch x L
→= .
Q.E.D.
c. See Figure 3-8c or 3-8d.
13. a. The limit seems to equal 2.
b.
2
1g
g
h
h
c. See the graph in part b. The lines haveequations g(x) = x + 1 and h(x) = 3 − x.
d. Prove that limx
y→1
2= .
44 Problem Set 3-9 Calculus Solutions Manual© 2005 Key Curriculum Press
Proof:
lim( )x
x→
+ = + =1
1 1 1 2
lim( )x
x→
− = − =1
3 3 1 2
For x < 1, g(x) ≤ y ≤ h(x).∴ the squeeze theorem applies, and lim
xy
→ −12= .
For x > 1, h(x) ≤ y ≤ g(x).∴ the squeeze theorem applies, and lim
xy
→ +12= .
Because both left- and right-hand limits equal 2, lim
xy
→12= , Q.E.D.
e. The word envelope (a noun) is used becausethe small window formed by the two lines“envelops” (a verb) the graph of the function.
f. As |x| becomes large, (x − 1) · sin–
1
1x=
sin [ /( – )]
/( – )
1 1
1 1
x
x takes on the form
sin (argument)
(argument) as the argument approaches
zero. Thus the limit is 1 and y approaches2 + 1, which equals 3.
14. Answers will vary.
Problem Set 3-9Q1. 1 Q2. –sin x
Q3.d
dxx
d
dxx x
2
2 (cos ) ( sin ) cos= − = −
Q4. y = sin x + C
Q5. x8 Q6. x48
Q7. log 32 = 5 log 2 Q8.
x
y
Q9. Cube function Q10. C
1. M(x) = 1000e0.06 x
a. M′(x) = 1000(0.06)e0.06 x = 60e0.06 x
M′(1) = 63.7101… $/yrM′(10) = 109.3271… $/yrM′(20) = 199.2070… $/yr
b. M(0) = $1000M(1) = $1061.84M(2) = $1127.50M(3) = $1197.22Increase from year 0 to year 1: $61.84Increase from year 1 to year 2: $65.66Increase from year 2 to year 3: $69.72No, the amount of money in the account doesnot change by the same amount each year.
c. 1061.84/1000 = 1.06184, so the APR for0 to 1 year is approximately 6.184%.1127.50/1061.84 = 1.061836… , so the APRfor 1 to 2 years is approximately 6.184%.1197.22/1127.50 = 1.06183… , so the APRfor 2 to 3 years is approximately 6.184%.The APR is higher than the instantaneousrate. Savings institutions may prefer toadvertise the APR instead of the instantaneousrate because the APR is higher.
2. a. f (t) = 10e− 0.34 t ⇒f ′ (t) = 10(−0.34)e− 0.34 t = −3.4e− 0.34 t
f ′ (0) = −3.4f ′ (2) = −1.7224…f ′ (4) = −0.8726…f ′ (6) = −0.4420Factor of change from 0 to 2:−1.7224…/−3.4 = 0.5066Factor of change from 2 to 4:−0.8726…/−1.7224… = 0.5066Factor of change from 4 to 6:−0.4420…/−0.8726… = 0.5066
b. f (0) = 10f (2) = 5.0661…f (4) = 2.5666…f (6) = 1.3002…Factor of change from 0 to 2:5.0661…/10 = 0.5066Factor of change from 2 to 4:2.5666…/5.0661… = 0.5066Factor of change from 4 to 6:1.3002…/2.5666… = 0.5066The factors of change are the same in part aand part b.
c.
5
10
t
y
f
f '
The values of f are negative because theamount of 18-F is decreasing as time goes on.
3. a. A p p A pp
( ) . = − ⇒ ′ = −63 23 5
23 5ln ( )
.
10
50
x
y
A
A´
b. If the pressure is increasing, then the altitudeis decreasing. A′(10) = −2.35, so the altitudeis changing at −2.35 thousand feet/psi. That
Calculus Solutions Manual Problem Set 3-9 45© 2005 Key Curriculum Press
is, a change of 1 psi would indicate that thealtitude had decreased by 2.35 thousand feet.The negative sign means that the altitude isdecreasing.
c. ′ = − =A ( ) . .5 4 7 4 7
′ = − =A ( ) .10 2 35 2.35
This shows that the altitude is changing fasterat 5 psi than it is at 10 psi.
d. A(p) = 063 − 23.5 ln p = 0−23.5 ln p = −63ln p = 2.6808…p = 14.5975…The fact that A(p) is negative for all values ofp greater than 14.5975… means that if the airpressure is above 14.5975 psi, then the planemust be beneath sea level.
4. x = 3000e0.05 y
a. ln x = ln (3000e0.05 y)ln x = ln 3000 + ln e0.05 y
ln x − ln 3000 = 0.05y
y x= −( )1
0 053000
.ln ln
y = 20 ln x – 20 ln 3000
b. y(3000) = 20 ln 3000 − 20 ln 3000 = 0y(4000) = 20 ln 4000 − 20 ln 3000 =5.7536…y(5000) = 20 ln 5000 − 20 ln 3000 =10.2165…y(6000) = 20 ln 6000 − 20 ln 3000 =13.8629…Number of years to get from $3000 to $4000:5.7536…Number of years to get from $4000 to $5000:4.4628…Number of years to get from $5000 to $6000:3.6464…The time intervals decrease as the amount ofmoney increases because when there is moremoney in the account, it takes less time toearn the given amount of interest.
c. y x yx
= − ⇒ ′ =20 20 300020
ln ln yr/$
y′(3000) = 0.0066…y′(4000) = 0.005y′(5000) = 0.004y′(6000) = 0.0033…This shows that the number of years it takesto earn each dollar decreases as the amount ofmoney increases.
5. f (x) = 5e3x ⇒ f ′ (x) = 15e3x
6. f (x) = 7e− 6x ⇒ f ′ (x) = −42e− 6x
7. g(x) = −4ecos x ⇒g′(x) = −4(ecos x)(−sin x) = 4(sin x) ecos x
8. h(x) = 8e− sin x ⇒h ′ (x) = 8e− sin x (−cos x) = −8(cos x)e− sin x
9. y = 2 sin (e4x) ⇒ y′ = 2 cos (e4x) 4e4x = 8e4x cos (e4x)
10. y = 6 cos (e− 0.5 x) ⇒y′ = 6[−sin (e− 0.5 x)](−0.5) e− 0.5 x = 3e− 0.5 x sin (e− 0.5 x)
11. f x x f xx x
( ) ln ( ) ( )= ⇒ ′ = ⋅ =10 710
77
10
12. g x x g xx x
( ) ln ( )= ⇒ ′ = ⋅ =9 49
44
9
13. T x Tx
xx
= ⇒ ′ = =1818
3543
32ln ( )
14. P x Px
xx
= ⇒ ′ = =−10001000
0 77000 7
0 70 3ln ..
..
15. y = 3 ln (cos 5x) ⇒
′ = − = −yx
x x3
55 5 15 5
cos( sin ) tan
16. y = 11 ln (sin 0.2x) ⇒
′ = =yx
x x11
0 20 2 0 2 2 2 0 2
sin .(cos . ) . . cot .
17. u = 6 ln (sin x0.5 ) ⇒
′ = =− −ux
x x x x6
0 5 30 50 5 0 5 0 5 0 5
sin(cos ) . cot.
. . . .
18. v x= ⇒0 09 8. ln (cos )
′ = −vx
x x0 09
888 7.
cos( sin ) = −0.72x7 tan x8
19. r x e r xe
exx
x( ) ln ( )= ⇒ ′ = ⋅ =11
Not surprising because we could have first usedthe fact that natural log and exp are inverses:r (x) = ln ex = x ⇒ r ′ (x) = 1
20. c(x) = eln x = x ⇒ c ′ (x) = 1c ′ (2) = 1, c ′ (3) = 1, c ′ (4) = 1
21. f (x) = 3x ⇒ f ′ ( x ) = ( ln 3 ) 3 x
22. g (x) = 0.007x ⇒ g′(x) = (ln 0.007) 0.007x
23. y = 1.6cos x ⇒ y′ = (ln 1.6)1.6cos x (−sin x) =−ln 1.6 sin x (1.6cos x)
24. y = sin 5x ⇒ y′ = cos 5x (ln 5)5x
25. y xdy
dx xx
xx= ⇒ = ⋅ = = ⇒−ln 5
54 11
55
5
d y
dxx
x
2
22
255= − = −−
26. y edy
dxe
d y
dxex x x= ⇒ = ⇒ =7 7
2
277 49
27. y = e− 0.7 x ⇒ y′ = −0.7e− 0.7 x ⇒ y″ = 0.49e− 0.7 x
46 Problem Set 3-10 Calculus Solutions Manual© 2005 Key Curriculum Press
28. y x yx x
x= ⇒ ′ = ⋅ = = ⇒−ln 81
88
1 1
′′ = − = −−y xx
1122
29. f ′ (x) = 12e2x ⇒ f ( x ) = 6 e 2 x + C
30. y′ = 5x ln 5 ⇒ y = 5x + C
Problem Set 3-10Review Problems
R0. Answers will vary.
R1. a.
x Average rate of change from 2 to x
1.97f f( ) ( . )
.
2 1 97
0 0311 82
− = .
1.98f f( ) ( . )
.
2 1 98
0 0211 88
− = .
1.99f f( ) ( . )
.
2 1 99
0 0111 94
− = .
2.01f f( . ) ( )
.
2 01 2
0 0112 06
− = .
2.02f f( . ) ( )
.
2 02 2
0 0212 12
− = .
2.03f f( . ) ( )
.
2 03 2
0 0312 18
− = .
The derivative of f at x = 2 is approximately 12.
b. r xf x f
x( )
( ) ( )= −−
2
2r(2) is of the form 0
0 .lim ( )x
r x→2
appears to be 12.
c. r xx
x
x x x
x( )
( )( )
( )= −
−= − + +
−
3 28
2
2 2 4
2= + + ≠x x x2 2 4 2,
lim ( ) limx x
r x x x→ →
= + +2 2
2 2 4 because x ≠ 2.
∴ =→
lim ( )x
r x2
12
d. The answers in parts a, b, and c are the same.
R2. a. f cf x f c
x cx c′ −
→( ) = lim
( ) ( )
–b. f (x) = 0.4x2 − x + 5
′ = − + −→
fx x
xx( ) lim
. .
–3
0 4 5 5 6
33
2
= +→
lim( – )( . . )
–x
x x
x3
3 0 4 0 2
3= +
→lim( . . )x
x3
0 4 0 2 1 4= .
c. m xx x
x( ) =
0 4 0 6
3
2. – .
–
−
3
1
2
x
m(x)
d. Line: y = 1.4x + 1.4
3
4
x
y
e. The line is tangent to the graph.
f. Yes, f does have local linearity at x = 3.Zooming in on the point (3, 5.6) shows thatthe graph looks more and more like the line.
R3. a.
50y1
y2
1 x
y
b. See the graph in part a.
c. The y1 graph has a high point or a low pointat each x-value where the y2 graph is zero.
d.
20
p
p´
t
y
1
Take the numerical derivative at t = 3, 6,and 0.p′(3) ≈ −2.688… . Decreasing at about2.69 psi/h when t = 3.p′(6) ≈ −1.959… . Decreasing at about1.96 psi/h when t = 6.p′(0) ≈ −3.687… . Decreasing at about3.69 psi/h when t = 0.The units are psi/h. The sign of the pressurechange is negative because the pressure isdecreasing. Yes, the rate of pressure change isgetting closer to zero.
R4. a. See the text for the definition of derivative.
b. Differentiate
Calculus Solutions Manual Problem Set 3-10 47© 2005 Key Curriculum Press
c. If y = xn, then y′ = nxn− 1.
d. See solution to Problem 35 in Problem Set3-4.
e. See the proof in Section 3-4.
f.dy
dx is pronounced “d y, d x.”
d
dxy( ) is pronounced “d, d x, of y.”
Both mean the derivative of y with respect to x.
g. i. f x x f x x( ) = ( ) =/ /763
59 5 4 5⇒ ′
ii. g x xx
x( ) = –76
742
− − + ⇒
g x xx′ − − −( ) = –283
15
iii. h(x) = 73 ⇒ h′ (x) = 0
h. f ′ = =( ) ./32 32 201 6635
4 5( ) exactly. The
numerical derivative is equal to or very closeto 201.6.
i.
4
4y
f´
f
x
R5. a. vdx
dtx t= ′or ( ).
adv
dtv t a
d x
dtx t= or ( ), = or ( )′ ′′
2
2
b. d y
dx
2
2 means the second derivative of y with
respect to x.
y = 10x4 ⇒ y′ = 40x3 ⇒ y″ = 120x2
c. ′ = ⇒ = +f x x f x x C( ) ( ) .12 33 4 f (x) isthe antiderivative, or the indefinite integral,of f (x).
d. The slope of y f x= ( ) is determined by thevalue of ′f x( ). So the slope of y f x= ( ) atx = 1 is ′ = −f ( ) ,1 1 at x = 5 is ′ =f ( ) ,5 3 andat x = −1 is ′ − =f ( ) .1 0
5
5
x
y f´f
e. i. y = −0.01t3 + 0.9t2 − 25t + 250
vdy
dtt t= = − + −0 03 1 8 252. .
adv
dtt= = − +0 06 1 8. .
ii. a ( 15) = −0.06(15) + 1.8 = 0.9 (km/s)/s
v(15) = −0.03(152) + 1.8(15) − 25= −4.75 km/s
The spaceship is slowing down at t = 15because the velocity and the accelerationhave opposite signs.
iii. v = −0.03t2 + 1.8t − 25 = 0By using the quadratic formula or thesolver feature of your grapher,t = 21.835… or t = 38.164… .The spaceship is stopped at about 21.8and 38.2 seconds.
iv. y = −0.01t3 + 0.9t2 − 25t + 250 = 0By using TRACE or the solver feature of your grapher, t = 50.v(50) = −10Because the spaceship is moving at10 km/s when it reaches the surface, it isa crash landing!
R6. a.
1
1
x
cosinederivative
y
b. The graph of the derivative is the same as thesine graph but inverted in the y-direction.Thus, ( )cos sinx x′ = − is confirmed.
c. −sin 1 = −0.841470984…Numerical derivative ≈ −0.841470984…The two are very close!
d. Composite function
f′(x) = −2x sin (x2)
R7. a. i.dy
dx
dy
du
du
dx= ⋅
ii. f (x) = g(h(x)) ⇒ f ′(x) = g ′(h(x)) ⋅ h ′(x)
iii. The derivative of a composite function isthe derivative of the outside function withrespect to the inside function times thederivative of the inside function withrespect to x.
b. See the derivation in the text. This derivationconstitutes a proof. ∆u must be nonzerothroughout the interval.
48 Problem Set 3-10 Calculus Solutions Manual© 2005 Key Curriculum Press
c. i. f (x) = (x2 − 4)3
f ′(x) = 3(x2 − 4)2 ⋅ 2x = 6x(x2 − 4)2
ii. f (x) = x6 − 12x4 + 48x2 − 64f′(x) = 6x5 − 48x3 + 96xExpanding the answer to part i givesf′(x) = 6x5 − 48x3 + 96x, which checks.
d. i. f′(x) = −3x2 sin x3
ii. g′(x) = 5 cos 5x
iii. h′(x) = 6 cos5 x (−sin x)= −6 sin x cos5 x
iv. k′(x) = 0
e. f ′(x) = 12 cos 3x ⇒ f ″ (x) =12(−sin 3x)3 = −36 sin 3x f ′(x) = 12 cos 3x ⇒ f (x) = 4 sin 3x + C f (x) is the second derivative of f (x). f (x) is the antiderivative, or indefiniteintegral, of f (x).
f. W = 0.6x3 and dx/dt = 0.4dW
dt
dW
dx
dx
dtx x= = . . .⋅ ⋅ =1 8 0 4 0 722 2
If x = 2, W = 0.6 ⋅ 23 = 4.8 lbdW/dt = 0.72(22) = 2.88The shark is gaining about 2.88 lb/day.If x = 10, W = 0.6 ⋅ 103 = 600 lb.dW/dt = 0.72(102) = 72The shark is gaining about 72 lb/day.The chain rule is used to get dW/dt fromdW/dx by multiplying the latter by dx/dt.
R8. a. limsin
x
x
x→=
01
x (sin x)/x
−0.05 0.99958338541…
−0.04 0.99973335466…
−0.03 0.99985000674…
−0.02 0.99993333466…
–0.01 0.99998333341…
0.00 undefined
0.01 0.99998333341…
0.02 0.99993333466…
0.03 0.99985000674…
0.04 0.99973335466…
0.05 0.99958338541…
The values of (sin x)/x approach 1 as xapproaches 0.
b. See the text for the statement of the squeezetheorem. Squeeze (sin x)/x between cos x andsec x.
c. See the proof in Section 3-8 of the text.
d. cos x = sin (π/2 − x)cos′ x = cos (π/2 − x) (−1)
= −sin x, Q.E.D.
e. d(t) = C + A cos B(t − D)C = 180, A = 20D = 0 for cosine because hand starts at a highpoint.B = 2π/60 = π/30 because period is 60 s.
d t t( ) = +180 2030
cosπ
′ = −d t t( )2
3 30
π πsin
At 2, t = 10: d ′(10) ≈ −1.81 cm/sAt 3, t = 15: d ′(15) ≈ −2.09 cm/sAt 7, t = 35: d ′(35) ≈ 1.05 cm/sAt the 2 and 3, the tip is going down, so thedistance from the floor is decreasing, which isimplied by the negative derivatives. At the 7,the tip is going up, as implied by the positivederivative.
R9. a. p x e p x ex x( ) ( ) ( . ). .= ⇒ ′ = −− −100 100 0 10 1 0 1
= −10e− 0.1 x
p′(0) = −10p′(10) = −3.6787…p′(20) = −1.3533…
The rates are negative because the amount ofmedication in your body is decreasing.To find the biological half-life, find x such that
p x p( ) ( )= =1
20 50
100 500 1e x− =.
e x− =0 1 1
2.
− =
0 1
1
2. lnx
x = −101
2ln
x = 6.9314…The half-life is 6.9314… h.p(2(6.9314…)) = p(13.8629…) =100e− 0.1(13.8629…) = 25
After two half-lives have elapsed, 25% of themedicine remains in your body.
b. i. f (x) = 5e2x ⇒ f ′(x) = 5(2)e2x = 10e2x
ii. ydy
dxx x= ⇒ =7 7 7(ln )
iii. d
dxx
xx x[ln (cos )]
cos( sin ) tan= − = −1
iv. y x xdy
dx xx= = ⇒ = ⋅ = ⇒−ln ln8 18 8
18
d y
dxx
x
2
22
288= − = −−
c. ′ = ⇒ = +f x e f x e Cx x( ) ( )12 43 3
Calculus Solutions Manual Problem Set 3-10 49© 2005 Key Curriculum Press
d.
x
y
3
3 y2
y3y1
y1 = ex is the inverse of y2 = ln x, so y1 is areflection of y2 across the line y = x.
Concept Problems
C1. a. f (x) = x7, g(x) = x9. So h(x) = f(x) ⋅ g(x) = x16.
b. h′(x) = 16x15
c. f ′(x) = 7x6, g′(x) = 9x8. So f ′(x) ⋅ g′(x) =63x14 ≠ h ′(x).
d. h′(x) = f ′ (x) ⋅ g(x) + f (x) ⋅ g′(x) =7x6 ⋅ x9 + x7 ⋅ 9x8 = 16x15
C2. a. f xx x
x( ) =
– sin
sin.
2 f (0) has the form 0/0,
which is indeterminate. f is discontinuous atx = 0 because f (0) does not exist.
b. By graph (below) or by TABLE , f (x) seems toapproach −1 as x approaches zero. Define f (0)to be −1.
π
5
x
f(x)
c. Conjecture: The function is differentiable atx = 0. The derivative should equal zerobecause the graph is horizontal at x = 0.
d. ′ =→
ff x f
xh( ) lim
( ) – ( )
–0
0
00
=→
lim
– sin
sin– (– )
x
x x
xx0
21
= +→
lim– sin sin
sinx
x x x
x x0
2
Using TABLE for numerator, denominator, andquotient shows that the numerator goes tozero faster than the denominator. For instance,if x = 0.001,
quotient = ×
×=1 1666 10
9 999 10
9
7
.
.
–
–
K
KK0.00116
Thus, the limit appears to be zero. (The limitcan be found algebraically to equal zero byl’Hospital’s rule after students have studiedSection 6-5.)
Chapter Test
T1. See the definition of derivative in Section 3-2or 3-4.
T2. Prove that if f (x) = 3x4, then f ′ (x) = 12x3.
Proof:
′ = + = + −→ →
f xf x h f x
h
x h x
hh h( ) lim
( ) – ( )lim
( )0 0
4 43 3
= + + + + −→
limh
x x h x h xh h x
h0
4 3 2 2 3 4 43 12 18 12 3 3
= + + + =→
lim ( ) ,h
x x h xh h x0
3 2 2 3 312 18 12 3 12
Q.E.D.
T3. If you zoom in on the point where x = 5, thegraph appears to get closer and closer to thetangent line. The name of this property is locallinearity.
Slope = 2Slope = 2
5
x
y
5
T4. Amos substituted before differentiating insteadof after. Correct solution is f (x) = 7x ⇒f ′ (x) = 7 ⇒ f ′ (5) = 7.
T5. f (x) = (7x + 3)15 ⇒ f ′ (x) = 105(7x + 3)14
T6. g(x) = cos (x5) ⇒ g ′ (x) = −5x4 sin x5
T7. d
dxx
xx x[ln (sin )]
sincos cot= ⋅ =1
T8. y = 36x ⇒ y′ = (ln 3)36x(6) = 6(ln 3)36x
T9. f (x) = cos (sin5 7x) ⇒f ′(x) = −sin (sin5 7x) ⋅ 5 sin4 7x ⋅ cos 7x ⋅ 7
= −35 sin (sin5 7x) sin4 7x cos 7x
T10. y = 60x2/3 − x + 25 ⇒ y′ = 40x− 1/3 − 1
T11. y edy
dxe
d y
dxex x x= ⇒ = ⇒ =9 9
2
299 81
T12. y′ ≈ 0.6 (Function is y = −3 + 1.5x, for whichthe numerical derivative is 0.6081… .)
T13. y = 3 + 5x− 1.6
v(x) = 5(−1.6)x− 2.6 = −8x− 2.6
a(x) = −8(−2.6)x− 3.6 = 20.8x− 3.6
Acceleration is the second derivative of thedisplacement function.
T14. f ′ (x) = 72x5/4 ⇒ f (x) = 32x9/4
50 Problem Set 3-10 Calculus Solutions Manual© 2005 Key Curriculum Press
T15. f ′ (x) = 5 sin x and f (0) = 13f (x) = −5 cos x + C
13 = −5 cos 0 + C ⇒ C = 18f (x) = −5 cos x + 18
T16. f (x) = cos 3x ⇒ f ′ (x) = −3 sin 3xf ′ (5) = −3 sin 15 = −1.95086…Decreasing at 1.95… y-units per x-unit.
T17. f xx
x( )
sin=
1
1
x
f(x)
As x approaches zero, f (x) approaches 1.The squeeze theorem states:
If (1) g(x) ≤ h(x) for all x in a neighborhood of c,
(2) lim ( ) lim ( ) ,x c x c
g x h x L→ →
= = and (3) f is a
function for which g(x) ≤ f (x) ≤ h(x) for all x inthat neighborhood of c, then lim ( ) .
x cf x L
→=
T18.
h5 1h
h
−
–0.0003 1.6090…
–0.0002 1.6091…
–0.0001 1.6093…
0 undefined
0.0001 1.6095…
0.0002 1.6097…
0.0003 1.6098…
ln 5 = 1.6094… . The table shows that
lim ln .h
h
h→
− =0
5 15
Proof:
d
dx hx
h
x h x
( ) lim55 5
0= −
→
+ Definition of derivative.
= −→
55 1
0
x
h
h
hlim Factor out 5x.
= 5x · ln 5 Evaluate.
T19. v(t) = 251(1 − 0.88t)a(t) = 251[− ln (0.88)] 0.88t = −251(ln 0.88)0.88ta(10) = −251(ln 0.88)(0.88)(10) = 8.9360…
Numerical derivative gives 8.9360… as well.
T20. If the velocity and the acceleration have oppositesigns for a particular value of t, then the object isslowing down at that time.
T21. a. v(t) = t1.5 + 3 ⇒ a(t) = 1.5t0.5
b. d tt
t C( ).
.
=
+ +2 5
2 53
d(1) = 20
1
2 53 1 20
2 5.
.( )+ + =C
3.4 + C = 20C = 16.6∴ d (t) = 0.4t2.5 + 3t + 16.6
c. d (9) − d (1) = 120.8
This represents the displacement between thefirst and ninth seconds.
T22. a. c t t( ) = +300 22
365cos
π ⇒
c t t′ −( ) =4
365
2
365
π πsin
b. c′ = − ⋅
( )273
4
365
2
365273
π πsin
= 0.03442… ppm/day
c. Rate is ( ).
, ,6 10
0 03442
1 000 000
1
24 60 6015× ⋅ ⋅
⋅ ⋅=K
2390.6627… , which is approximately 2390tons per second!
T23. Answers will vary.
Calculus Solutions Manual Problem Set 4-2 51© 2005 Key Curriculum Press
Chapter 4—Products, Quotients, and Parametric Functions
Problem Set 4-11. f (x) = 3 cos x ⇒ f ′ (x) = −3 sin x
g(x) = 2 sin x ⇒ g′(x) = 2 cos x
2.
10
x
p(x)6
p′(2) ≈ −3.9218…p(x) is decreasing at x = 2 because ′p ( ) < .2 0This fact corresponds with the graph, whichslopes steeply in the negative direction at x = 2.f ′ (2) ⋅ g′(2) = (−3 sin 2)(2 cos 2) = 2.2704…So p′ (2) ≠ f ′ (2) ⋅ g′ (2).
3.
x
10
6q(x)
q is the cotangent function.
q ′ (2) = −1.8141…
q x( ) is decreasing at x = 2.
f ′ (2)/g′ (2) = (−3 sin 2)/(2 cos 2) = 3.2775…
So q ′ (2) ≠ f ′ (2)/g′ (2).
4.
3
2x
y
t = 2 here
The geometric figure seems to be an ellipse.
5. See graph in Problem 4.∆x = 3 cos 2.1 − 3 cos 1.9 = −0.54466…∆y = 2 sin 2.1 − 2 sin 1.9 = −0.16618…
dy
dx
y
x≈ ∆
∆= = 0.3051
– .
– .
0 16618
0 54466
K
KK
At , . ,tdy dt
dx dt= = =2
2 2
3 20 3051
/
/
cos
– sinK
which agrees with the difference quotient.
6. You’ll see in Section 4-2 that
p′(x) = f ′ (x)g(x) + f (x)g′(x).∴ p′(2) = (–3 sin 2)(2 sin 2) + (3 cos 2)(2 cos 2) =–3.9218… , which agrees with Problem 2.
You’ll see in Section 4-3 that
′ ′ ′q x
f x g x f x g x
g x( ) =
( ) ( ) – ( ) ( )
[ ( )]2
∴ ′ = =q ( )23 2 2 2 3 2 2 2
2 2 2
(– sin )( sin ) – ( cos )( cos )
( sin )
−1.8141… , which agrees with Problem 3.
Problem Set 4-2
Q1. ′ =y x3
41 4– / Q2. y′ = 1/x
Q3.dy
dxx= − −30 5 7 7( )– Q4.
d
dxx x(sin ) cos2 2 2=
Q5. v′ = −3 cos2 t sin t Q6. L′ = 2m + 5
Q7. y = sin x3 + C Q8. ′ ≈ −y 3
Q9. 4 ft/s Q10. B
1. f ( x) = x3 cos x ⇒ f ′ (x) = 3x2 cos x – x3 sin x
2. f ( x) = x4 sin x ⇒ f ′ (x) = 4x3 sin x + x4 cos x
3. g(x) = x1.5e2x ⇒ g′ (x) = 1.5x0.5e2x + 2x1.5 e2x
4. h(x) = x− 6.3 ln 4x ⇒h′ (x) = −6.3x− 7.3 ln 4x + x− 6.3 (1/4x)4
= −6.3x− 7.3 ln 4x + x− 7.3
5. y = x7(2x + 5)10 ⇒dy/dx = 7x6(2x + 5)10 + x7(10)(2x + 5)9 ⋅ 2
= x6(2x + 5)9(34x + 35)
6. y = x8(3x + 7)9 ⇒dy/dx = 8x7(3x + 7)9 + x8(9)(3x + 7)8(3)
= x7(3x + 7)8 (51x + 56)
7. z = ln x sin 3x ⇒z′ = (1/x) sin 3x + 3 ln x cos 3x
8. v = e5x cos 2x ⇒ v′ = 5e5x cos 2x − 2e5x sin 2x
9. y = (6x + 11)4(5x − 9)7 ⇒y ′ = 4(6x + 11)3(6)(5x − 9)7
+ (6x + 11)4(7)(5x − 9)6(5)
= (6x + 11)3(5x − 9)6(330x + 169)
10. y = (7x – 3)9(6x − 1)5 ⇒y′ = 9(7x – 3)8(7)(6x − 1)5
+ (7x − 3)9(5)(6x − 1)4(6)
= (7x − 3)8(6x − 1)4(588x − 153)
52 Problem Set 4-2 Calculus Solutions Manual© 2005 Key Curriculum Press
11. P = (x2 − 1)10(x2 + 1)15 ⇒P′ = 10(x2 − 1)9(2x)(x2 + 1)15
+ (x2 − 1)10(15)(x2 + 1)14(2x)
= 10x(x2 − 1)9(x2 + 1)14[2(x2 + 1) + 3(x2 – 1)]
= 10x(x2 − 1)9(x2 + 1)14(5x2 − 1)
12. P(x) = (x3 + 6)4(x3 + 4)6 ⇒P′ (x) = 4(x3 + 6)3(3x2)(x3 + 4)6
+ (x3 + 6)4 ⋅ 6(x3 + 4)5
⋅ 3x2
= 6x2(x3 + 6)3(x3 + 4)5[2(x3 + 4) + 3(x3 + 6)]
= 6x2(x3 + 6)3(x3 + 4)5(5x3 + 26)
13. a ( t) = 4 sin 3t cos 5t ⇒a′(t) = 12 cos 3t cos 5t + 4 sin 3t(−5 sin 5t)
= 12 cos 3t cos 5t − 20 sin 3t sin 5t
14. v = 7 cos 2t sin 6t ⇒v′ = −14 sin 2t sin 6t + 7 cos 2t(6 cos 6t)
= −14 sin 2t sin 6t + 42 cos 2t cos 6t
15. y = cos (3 sin x) ⇒y′ = −sin (3 sin x) · 3 cos x
= −3 sin (3 sin x) cos x
16. y = sin (5 cos x) ⇒ y′ = cos (5 cos x) · (−5 sin x)
= −5 cos (5 cos x) sin x
17. y = cos e6x ⇒dy/dx = 6e6x(−sin e6x) = −6e6x sin e6x ⇒d 2y/dx2 = −36e6x sin e6x − 6e6x(6e6x cos e6x)
= −36e6x sin e6x − 36e12x cos e6x
18. y = ln (sin x) ⇒ dy/dx = (1/sin x) cos x
= (cos x)(sin x)− 1 ⇒ d 2y/dx2
= (cos x) [−(sin x)]− 2 (cos x) + (−sin x) ⋅ (sin x)− 1
= −−
+ = − + = −cos
sincot csc
2
22 21 1
x
xx x
19. z = x3(5x − 2)4 sin 6x ⇒z′ = 3x2(5x − 2)4 sin 6x + x3[(4)(5x − 2)3(5) sin 6x
+ (5x − 2)4(6 cos 6x)]
= 3x2(5x − 2)4 sin 6x + 20x3(5x − 2)3 sin 6x
+ 6x3(5x − 2)4 cos 6x
20. u = 3x5(x2 − 4) cos 10x
u′ = 15x4(x2 − 4) cos 10x + 3x5[2x cos 10x
+ (x2 − 4) ⋅ (−10 sin 10x)]
= 15x4 ⋅ (x2 − 4) cos 10x + 6x6 cos 10x
− 30x5(x2 − 4) sin 10x
21. If y = uvw, where u, v, and w are differentiablefunctions of x, then y′ = u′vw + uv′w + uvw′.
Proof:
y = uvw = (uv)w
∴ y′ = (uv)′w + (uv)w′ = (u′v + uv′)w + (uv)w′= u′vw + uv′w + uvw′, Q.E.D.
22. If y u u u un n= …1 2 3 where u un1… are differentiablefunctions of x, then
′ = ′ + ′y u u u u u u u un n n1 2 3 1 2 3K K
+ ′ + + ′u u u u u u u un n1 2 3 1 2 3K L K K .
23. z = x5 cos6 x sin 7x ⇒z′ = 5x4 cos6 x sin 7x + x5 6 cos5 x (−sin x) ⋅
sin 7x + x5 cos6 x ⋅ 7 cos 7x
= 5x4 cos6 x sin 7x − 6x5 cos5 x sin x sin 7x
+ 7x5 cos6 x cos 7x
24. y = 4x6 sin3 x cos 5x ⇒y′ = 24x5 sin3 x cos 5x + 4x6 3 sin2 x cos x ⋅
cos 5x + 4x6 sin3 x(−5 sin 5x)
= 24x5 sin3 x cos 5x + 12x6 sin2 x cos x ⋅cos 5x − 20x6 sin3 x sin 5x
25. y = x4 (ln x)5 sin x cos 2x ⇒y′ = 4x3(ln x)5 sin x cos 2x + x4 ⋅ 5(ln x)4(1/x) ⋅
sin x cos 2x + x4(ln x)5 cos x cos 2x
+ x4(ln x)5 sin x ⋅ (−2 sin 2x)
= 4x3(ln x)5 sin x cos 2x + 5x3(ln x)4 sin x ⋅cos 2x + x4(ln x)5 cos x cos 2x
− 2x4(ln x)5 sin x sin 2x
26. u = x5 e2x cos 2x sin 3x ⇒u′ = 5x4 e2x cos 2x sin 3x + x5 · 2e2x ·
cos 2x sin 3x + x5 e2x(−2 sin 2x) sin 3x
+ x5 e2x cos 2x · 3 cos 3x
= 5x4 e2x cos 2x sin 3x + 2x52e2x cos 2x sin 3x
− 2x5 e2x sin 2x sin 3x + 3x5 e2x cos 2x cos 3x
27. a. y(t) = 4 + 3e− 0.1 t cos πt
v(t) = y′(t) = 3(−0.1)e− 0.1 t cos π t+ 3e− 0.1 t(−π sin π t)
= e− 0.1 t(−0.3 cos π t − 3π sin π t )
b. v(2) = e− 0.2 (−0.3 cos 2π − 3π sin 2π)
= e−0.2(−0.3 − 0) = −0.2456…
There is not a high point at t = 2 becausev(2) ≠ 0.
v(t) = 0 ⇒ e− 0.1 t(−0.3 cos π t − 3π sin π t) = 0
⇒ −0.3 cos π t = 3π sin π t ⇒ t = 1.9898…
28. a. y(t) = t sin t ⇒ v(t) = y′(t) = sin t + t cos tGraph confirms Figure 4-2d.
Calculus Solutions Manual Problem Set 4-2 53© 2005 Key Curriculum Press
b. v exceeds 25.
v
25
25
c. In 1940, wind-induced vibrations in theTacoma Narrows Bridge increased inamplitude until the bridge collapsed.
29. Prove that the derivative of an odd function is aneven function and that the derivative of an evenfunction is an odd function.
Proof:
For any function, the chain rule givesd
dxf x f x f x( ) ( ) ( ) ( ).− = ′ − ⋅ − = − ′ −1
For an odd function,d
dxf x
d
dxf x f x( ) = ( ).− = − ′[– ( )]
∴ −f ′ (−x) = −f ′ (x) or f ′ (−x) = f ′ (x),
and the derivative is an even function.For an even function,d
dxf x
d
dxf x f x( ) ( ) ( ).− = = ′
∴ −f ′ (−x) = f ′ (x) or f ′ (−x) = −f ′ (x),and the derivative is an odd function, Q.E.D.
30. f (x) = 2 sin x cos x ⇒f ′ (x) = 2 cos x · cos x + 2 sin x(−sin x)
= 2 cos2 x − 2 sin2 x = 2 cos 2xg(x) = sin 2x ⇒ g′(x) = 2 cos 2x = f ′ (x), Q.E.D.f(0) = 0 and g(0) = 0∴ f (x) = 2 sin x cos x = sin 2x = g(x), by theuniqueness theorem for derivatives, Q.E.D.f (x) = cos2 x − sin2 x ⇒f ′ (x) = 2 cos x(−sin x) − 2 sin x cos x
= −4 sin x cos x = −3 sin 2xg(x) = cos 2x ⇒g′(x) = (−2 sin 2x) = −sin 2x = f ′ (x), Q.E.D.f(0) = 1 and g(0) = 1
∴ f (x) = cos2 x − sin2 x = cos 2x = g(x) by theuniqueness theorem, Q.E.D.
31. Prove that if fn(x) = xn, then ′ = −f x nxnn( ) 1
for allintegers ≥ 1.
Proof (by induction on n):
If n = 1, then f1(x) = x1 = x, which implies that
′ = =f x x101 1( ) , which anchors the induction.
Assume that for some integer n = k > 1,′ = −f x kxk
k( ) .1
For n = k + 1, fk+ 1(x) = xk+ 1 = (xk)(x).
By the derivative of a product property,′ = ′ + ′ = ′ ++f x x x x x x x xk
k k k k1( ) ( ) ( ) ( )( ) ( ) ( ) .
Substituting for (xk)′ from the inductionhypothesis,
′ = + = + = + =+−f x kx x x kx x k xk
k k k k k1
1 1( ) ( )( ) ( )(k + 1)x( k+ 1 ) −1, completing the induction.∴ ′ = −f x nxn
n( ) 1 for all integers ≥ 1, Q.E.D.
32. Way 1: y = (x + 3)8(x − 4)8
y′ = 8(x + 3)7 · (x − 4)8 + (x + 3)8 · 8(x − 4)7
= 8(x + 3)7(x − 4)7(x + 3 + x − 4)= 8(x + 3)7(x − 4)7(2x − 1)
Way 2: y = (x2 − x − 12)8
y′ = 8(x2 − x − 12)7(2x − 1) = 8(x + 3)7(x − 4)7(2x − 1), which checks.
33. a.
f
f´
x
f(x)
5
1
b. f ′ (x) = 3x2 sin x + x3 cos xThe graph in part a is correct.
c. The numerical derivative graph duplicates thealgebraic derivative graph, as in part a, thusshowing that the algebraic derivative is right.
34. a.
1/9
1.4
500,000599,128
–500,000
–1.5
x
f(x)
b. f ′ (x) = 4(5x − 7)3(5) · (2x + 3)5
+ (5x − 7)4 · (5)(2x + 3)4(2)= 10(5x − 7)3(2x + 3)4[2(2x + 3)
+ 5x − 7]= 10(5x − 7)3(2x + 3)4(9x − 1)
c. f ′ (x) = 0 ⇔ 5x − 7 = 0 or 2x + 3 = 0or 9x − 1 = 0∴ x = 7/5 = 1.4, or x = −3/2 = −1.5,or x = 1/9See graph in part a.
d. f (1.4) = 0, f (−1.5) = 0, f (1/9) = 599,127.6… .See graph in part a.
e. False. The graph may have a point where itlevels off and then continues changing in thesame direction, as at x = −1.5 in part a.
35. a. A = L WdA
dt
dL
dtW L
dW
dt= ⋅ + ⋅
54 Problem Set 4-3 Calculus Solutions Manual© 2005 Key Curriculum Press
b. W tdW
dtt= + ⇒ = −2 2 2cos sin
L tdL
dtt= + ⇒ =3 2 2 4 2sin cos
dA
dtt t= +( )( )4 2 2 2cos cos
+ + −( )( )3 2 2 2sin sint tAt t = 4, dA/dt = 7.132… , so A is increasing.At t = 5, dA/dt = −4.949… , so A isdecreasing.
Problem Set 4-3Q1. 1066x1065
Q2. f (x) = 12x5 + C
Q3. y′ = 3x2 sin x + x3 cos x
Q4. dy/dx = −sin (x7) ⋅ 7x6 = −7x6 sin (x7)
Q5. f ′ (x) = 0 (derivative of a constant)
Q6. 54e9t
Q7. See the text for the definition of derivative.
Q8. Instantaneous rate of change at a given x
Q9. (x − 3)4(x − 3 + 2x) = 3(x − 3)4(x − 1)Q10.
4
4 y
x
1. f xx
xf x
x x x x
x( ) ( )= ⇒ ′ =
3 2 3
2
3
sin
sin – cos
sin
2. f xx
xf x
x x x x
x( ) ( )= ⇒ ′ = +4 3 4
2
4
cos
cos sin
cos
3. g xx
x( )
cos
ln= ⇒
3
′ = − ⋅ − ⋅g x
x x x x x
x( )
cos ( sin ) ln cos ( / )
(ln )
3 12 3
2
= − −3 2 3
2
ln sin cos (cos )/
(ln )
x x x x x
x
4. h xx
e x( )sin= ⇒
5
3
′ = − ⋅h x
x x e x e
e
x x
x( )sin cos sin
( )
5 34 3 5 3
3 2
= −5 34 5
3
sin cos sinx x x
e x
5. yx
x= ⇒sin
cos
10
20
′ =+
yx x x x
x
10 10 20 20 10 20
202
cos cos sin sin
cos
6. yx
x= ⇒cos
sin
12
18
′ =yx x x x
x
– sin sin – cos cos
sin
12 12 18 18 12 18
182
7. y3x 7
6x 5=
−
+⇒
yx x
x x′ =
+ −
+=
+
3 6 5 3 7 6
6 5
57
6 52 2
( ) – ( )( )
( ) ( )
8. yx
x= +
−⇒10 9
5 3
′ = ⋅ − − + ⋅−
=yx x
x x
10 5 3 10 9 5
5 3
75
5 32 2
( ) ( )
( )
–
( – )
9. zx
x
dz
dx= + ⇒( )
( – )
8 1
5 2
6
9
= + ⋅ − − + ⋅ −−
6 8 1 8 5 2 8 1 9 5 2 5
5 2
5 9 6 8
18
( ) ( ) ( ) ( ) ( )( ) ( )
( )
x x x x
x
= − + +( ) ( )
( – )
8 1 120 141
5 2
5
10
x x
x
10. Ax
x
dA
dx=
+⇒( – )
( )
4 1
7 2
7
4
= − ⋅ + − − ⋅ ++
7 4 1 4 7 2 4 1 4 7 2 7
7 2
6 4 7 3
8
( ) ( ) ( ) ( ) ( ) ( )
( )
x x x x
x
= + + − −+
= + ++
= ++
28 4 1 7 2 7 2 4 1
7 2
28 4 1 7 2 3 3
7 2
84 4 1 1
7 2
6 3
8
6 3
8
6
5
( – ) ( ) [( ) ( )]
( )
( – ) ( ) ( )
( )
( – ) ( )
( )
x x x x
x
x x x
x
x x
x
11. Qe
xQ
x e x e x
x
x x x
= ⇒ ′ = −3 3 3
3 2
2sin
sin cos
sin
12. rx
x= ⇒ln
cos
4
′ = − −r
x x x x x
x
4 13 4 4
2
( / )cos (ln )( sin )
cos
= + = +( cos )/ (ln ) sin
cos
cos ln sin
cos
4 44
2
4
2
x x x x
x
x x x x
x x
13.d
dxx x( )– /60 804 3 7 3= − − /
14.d
dxx x( )– /24 7 3 = − − /56 10 3
Problems 15–22 and 25–26 can be done using eitherthe power rule or the quotient rule.
15. r xx
x r x xx
( ) ( )= = ⇒ ′ = − =− −1212 36
363
3 44
–
16. t xx
x t x xx
( ) ( )= = ⇒ ′ = − =− −5151 867
86717
17 1818
–
Calculus Solutions Manual Problem Set 4-3 55© 2005 Key Curriculum Press
17. v xx
x( ) = = ⇒−14
0 514 0 5 1
cos .(cos . )
v′(x) = −14(cos 0.5x)–2(−sin 0.5x)(0.5)
= 7 0 5
0 52
sin .
cos .
x
x
18. a xx
x( ) = = ⇒−20202
2
sin(sin )
a ′ (x) = −40(sin x) −3 (cos x)
= – cos
sin
403
x
x
19. r xx
x r x xx
( ) ( )= = ⇒ ′ = − = −− −1 11 22
20. s xx
x s x xx
( ) ( )= = ⇒ ′ = − =− −12
22
2 33
–
21. W xx
x( ) ( )5= = − ⇒10
110 13 5
3
( – )–
W′(x) = 150x2(x3 − 1)4
22. T xx x
x x( ) = = ⇒−1 1
cos sin(cos sin )
′ = − −−T x x x x x( ) ( ) (cos sin sin sin2
+ =cos cos )sin – cos
cos sinx x
x x
x x
2 2
2 2 ,
which transforms to
T xx
xx x′ = = −( )
– cos
sincsc cot
2
24 2 2
14
2
23. T xx
x( ) = ⇒sin
cos
T xx x x x
x
x x
x xx
′ =
= + = =
( )(cos )(cos ) – (sin )(– sin )
cos
cos sin
cos cossec
2
2 2
2 221
(T is for tangent function.)
24. C xx
x( ) = ⇒cos
sin
C xx x x x
x
x x
x xx
′ =
= = = −
( )(– sin )(sin ) – (cos )(cos )
sin
– sin – cos
sin
–
sincsc
2
2 2
2 221
(C is for cotangent function.)
25. C xx
C xx
x( ) ( )= ⇒ ′ =1 0
2sin
– cos
sin
= − ⋅ = −1
sin
cos
sincsc cot
x
x
xx x
(C is for cosecant function.)
26. S xx
x( ) ( )= = ⇒−1 1
coscos
S′(x) = −(cos x)− 2(−sin x)
= =sin
cossec tan
x
xx x2
(S is for secant function.)
27. a. v tt
( ) = 1000
3 –
v( ) mi h11000
3 1500= = /
–
v( ) mi h21000
3 21000= = /
–
v( ) .31000
3 3
1000
0= =
– No value for v(3).
b. v(t) = 100(3 − t)− 1 ⇒
a(t) = −1000(3 − t)− 2 − 1 = 1000
3 2( – )t
a( ) (mi h) h11000
3 12502= = / /
( – )
a( ) (mi h) h21000
3 210002= = / /
( – )
a( ) = .31000
3 3
1000
02( – )= No value for a (3).
c.
2000v
a
t
a or v
3
d.1000
3500 2 32( – )t
t= ⇒ = − ⇒( )2
3 2 3 2− = ± ⇒ = ± ⇒t t
t = − =3 2 1 585. K in the domain.Range is 0 ≤ t < 1.585… .
28. a. Because they are walking in the samedirection, their relative rate is the difference(x − 5).
b. t xx
( ) = 300
5–, assuming Willie’s rate is
constant.t(6) = 300 s, t(8) = 100 s, t(10) = 60 s,t(5) = 300/0, which is infinite, t(4) = −300,which is not reasonable in the real world,and t(5.1) = 3000 s. A reasonable domainis x > 5.
c. t(x) = 300(x – 5)− 1
t′(x) = −300(x − 5)− 2 = −−300
5 2( )xt′(6) = −300 s/(ft/s)
d. t′(5) does not exist because of division byzero. More fundamentally, t′(5) does not existbecause t(5) does not exist.
29. f xx
x( ) = +
+⇒3 7
2 5
56 Problem Set 4-4 Calculus Solutions Manual© 2005 Key Curriculum Press
f xx x
x x′ = ⋅ + + ⋅
+=
+( )
3 2 5 3 7 2
2 5
1
2 52 2
( ) – ( )
( ) ( )
f ′ = =( ) .4
1
1690 005917159K
Using 4.1, f ′(4) ≈ 0.005827505… .
Using 4.01, f ′(4) ≈ 0.005908070… .
Using 4.001, f ′(4) ≈ 0.005916249… .
f ′(4) (exact) = 0.005917159…
Difference quotients are approaching f ′(4).
30. a. Sketch. See accurate plot in part b.
b. f xx
x( ) = ⇒
2 8
3
–
–
′ = = − +−( )
f xx x x
x
x x
x( )
2 3 8 1
3
6 8
3
2
2
2
2
( – ) – ( – )( )
( – )
f
f'
x3
5
f(x)
y2 and y3 both agree with the graph of f ′.
c.
x f (x) f ′(x)
2.95 −14.05 −399
2.96 −19.04 −624
2.97 −27.363… −1110.11…
2.98 −44.02 −2499
2.99 −94.01 −9999
3.00 undefined undefined
3.01 106.01 −9999
3.02 56.02 −2499
3.03 39.363… −1110.11…
3.04 31.04 −624
3.05 26.05 −399
f ( x) changes faster and faster as x approaches3, shooting off to negative infinity as xapproaches 3 from the negative side and topositive infinity as x approaches 3 from thepositive side. Note that the rates aresymmetrical about x = 3.
d. There is a relative minimum at x = 4 and arelative maximum at x = 2.
′ = − +−
=f ( )( )
( )2
2 6 2 8
2 30
2
2
′ = − +−
=f ( )( )
( )4
4 6 4 8
4 30
2
2
31. If y = xn, where n is a negative integer, theny′ = nxn− 1.
Proof:
Let n = −p, where p is a positive integer.
∴ = =− y xx
pp
1
∴ ′ = ⋅ ⋅ y
x px
x
p p
p
0 1 1
2
– –
because p is a
positive integer.
= − = − = −− − − −px
xpx px
p
pp p p
–
.1
21 2 1
Replacing −p with n gives y′ = nxn− 1, Q.E.D.32.
1
1
x
y
y
y '
1
1
x
y
y'
y
33. Answers will vary.
Problem Set 4-4Q1. (sin x)/(tan x) = cos x
Q2. 1/(sec x) = cos x
Q3. sin2 3 + cos2 3 = 1
Q4. f ′ (x) = ex sin x + ex cos x
Q5. g xx x x
x′ = +( )
cos sin
cos2
Q6. h ′ (x) = −(15/7)(3x)–12/7
Q7. dy/dx = 3(cos x)− 4 sin x
Q8. Limit = −3
Q9. (Function is secant.)
1
π
x
y
yy'
Q10. C
1. f (x) = tan 5x ⇒ f ′(x) = 5 sec2 5x
2. f (x) = sec 3x ⇒ f ′(x) = 3 sec x tan x
3. y = sec x7 ⇒ y′ = 7x6 sec x7 tan x7
4. z = tan x9 ⇒ z′ = 9x8 sec2 (x9)
Calculus Solutions Manual Problem Set 4-4 57© 2005 Key Curriculum Press
5. g(x) = cot e11x ⇒ g ′ (x) = −11e11x csc2 (e11x)
6. h(x) = csc e10x ⇒ h ′ (x) = −10e10x csc (e10x) cot (e10x)
7. r (x) = ln (csc x) ⇒
′ =r xx
( )1
csc(−csc x cot x) = −cot x
8. p(x) = ln (cot x) ⇒
p xx
xx x
′ = − =( ) ( )1 12
cotcsc
cos sin
9. y = tan5 4x ⇒(d/dx)(y) = 5 tan4 4x · sec2 4x · 4
= 20 tan4 4x sec2 4x
10. y = tan7 9x ⇒(d/dx)(y) = 7 tan6 9x · sec2 9x · 9
= 63 tan6 9x sec2 9x
11. (d/dx)(sec x tan x) = sec x tan x · tan x +sec x · sec2 x = sec x tan2 x + sec3 x
12. (d/dx)(csc x cot x) = −csc x cot x · cot x +csc x · (−csc2 x) = −csc x cot2 x − csc3 x
13. y = sec x csc x ⇒y′ = sec x tan x · csc x + sec x · (−csc x cot x) = sec2 x − csc2 x
14. y = tan x cot x = 1 for all x ⇒ y′ = 0
15. yx
x= tan
sin = sec x ⇒ y′ = sec x tan x
16. yx
x= cot
cos = csc x ⇒ y′ = −csc x cot x
17. yx
x= ⇒5 7
14
ln
cot
yx x x
x
x x x x
x
x x
x
x
x′ = ⋅ − −
= +
= +
5 7 14 5 7 14 14
14
5 14 70 7 14
145
14
70 7
14
17
2
2
2
2
2
( )cot ln ( csc )
cot
( cot )/ ln csc
cot
cot
ln
cos
18. yx
e x= ⇒4 1040
csc
yx x e x e
e
x x
x′ = − −4 10 10 10 4 10 4040 40
40 2
( csc cot ) csc ( )
( )
= − −40 10 10 160 1040
csc cot cscx x x
e x
19. w = tan (sin 3x) ⇒w′ = sec2 (sin 3x) · 3 cos 3x
= 3 sec2 (sin 3x) · cos 3x
20. t = sec (cos 4x) ⇒t′ = sec (cos 4x) tan (cos 4x) · (−4 sin 4x)
= −4 sec (cos 4x) tan (cos 4x) sin 4x
21. S(x) = sec2 x − tan2 x = 1 ⇒ S′(x) = 0(The differentiation formulas give the same.)
22. m(x) = cot2 x − csc2 x = −1 ⇒ m′(x) = 0(The differentiation formulas give the same.)
23. A(x) = sin x2 ⇒ A′(x) = cos x2 · 2x = 2x cos x2
24. f (x) = cos x3 ⇒ f ′ (x) = −sin x3 · 3x2
= −3x2 sin x3
25. F(x) = sin2 x ⇒ F ′ (x) = 2 sin x cos x
26. g(x) = cos3 x ⇒g′(x) = 3 cos2 x · (−sin x) = −3 cos2 x sin x
27. y = tan x ⇒ dy/dx = sec2 x ⇒d2y/dx2 = 2 sec x(sec x tan x) = 2 sec2 x tan x
28. y = sec x ⇒ y′ = sec x tan x ⇒y″ = (sec x tan x) · tan x + sec x · sec2 x
= sec x tan2 x + sec3 x
29. y xx
x= = ⇒cot
cos
sin
yx x x x
x′ = ⋅ ⋅– sin sin – cos cos
sin2
= = −–
sincsc
12
2
xx or:
yx
x= = ⇒−1
tantan( ) 1
y′ = −1 ⋅ (tan x)− 2 ⋅ sec2 x = −csc2 x
30. y xx
x= = = ⇒cscsin
sin1 1( )–
y x xx
xx x′ = − = − = −−( )sin cos
cos
sincsc cot2
2
31. a. See graph in part b.
b. f (x) = tan x ⇒ f ′(x) = sec2 x. Predicted graphshould be close to actual one.
f
f´
1
1 x
y
c.
tan . – tan .
( . )
1 01 0 99
2 0 013 42646416= . K
tan′ 1 = sec2 1 = (1/cos 1)2 = 3.42551882…Difference quotient is within 0.001 of actual.
32. a. f (x) = sec x ⇒ f ′ (x) = sec x tan x ⇒f ′ (1) = sec 1 tan 1 = 2.8824…
b.
1
5
x
yy1
y 2
58 Problem Set 4-5 Calculus Solutions Manual© 2005 Key Curriculum Press
The formula is confirmed by the fact that theline is tangent to the graph.
c.
f f '
x
y
π/2
1
If f ′ (x) is negative, the graph of f isdecreasing.
33. a. y/10 = tan x ⇒ y = 10 tan x, Q.E.D.
b. y′ = 10 sec2 x. At x = 1, y′ = 10 sec2 1 =34.2551… . y is increasing at about34.3 ft/radian.
( . )34 2551
180K
π = 0.5978… ft/degree
c. y = 535 ⇒ x = tan− 1 53.5 = 1.55210…∴ y′ = 10 sec2 1.55210… = 28632.5…y is increasing at about 28,632.5 ft/radian.
34. a. tan xy= =opposite side
adjacent side 500∴ y = 500 tan x, Q.E.D.
b. dy/dt = 500 sec2 x · dx/dt
c. dx/dt = 0.3 rad/sAt y = 300, x = tan− 1 (300/500) = 0.5404…∴ dy/dt = 500 (sec2 0.5404…)(0.3)
= 500(1.36)(0.3) = 204 ft/s35. a. y = sin x + C
b. y x C= − +12 2cos
c. y x C= +13 3tan
d. y x C= − +14 4cot
e. y = 5 sec x + C
36. Answers will vary.
Problem Set 4-5Q1. sin′ x = cos x Q2. cos′ x = −sin x
Q3. tan′ x = sec2 x Q4. cot′ x = −csc2 x
Q5. sec′ x = sec x tan x Q6. csc′ x = −csc x cot x
Q7. f ′ (1) is infinite. Q8. f ′ (3) is undefined.
Q9. f ′ (4) = −1 Q10. f ′ (6) = 0
1. See Figure 4-5d. 2. See Figure 4-5d.
3. See Figure 4-5d. 4. See Figure 4-5d.
5. The principal branch of the inverse cotangentfunction goes from zero to π so that the functionwill be continuous.
6. There are no values of the inverse secant for xbetween −1 and 1, so the inverse secant function
cannot be continuous. (Some texts restrict therange of the inverse cosecant to 0 ≤ y ≤ π/2 sothat the function will be continuous, but doingso throws away the other half of the possiblevalues.)
7. sin (sin− 1 0.3) = 0.3
8. cos− 1 (cos 0.8) = 0.8
9. y = sin− 1 x ⇒ sin y = x ⇒ cos y · y′ = 1 ⇒
yy x
′ = =1 1
1 2cos –, Q.E.D.
y
x1
[Because sin y = (opposite leg)/(hypotenuse), putx on the opposite leg and 1 on the hypotenuse.
Adjacent leg = 1 2– ,x and cos y =(adjacent)/(hypotenuse).]
10. y = cos− 1 x ⇒ cos y = x ⇒ −sin y · y′ = 1 ⇒
yy x
′ = − = −1 1
1 2sin –, Q.E.D.
y
1
x
1 – x 2√
[Because cos y = (adjacent leg)/(hypotenuse), putx on the adjacent leg and 1 on the hypotenuse.
Opposite leg = 1 2– ,x and sin y =(opposite)/(hypotenuse).]
11. y = csc− 1 x ⇒ csc y = x ⇒ −csc y cot y · y′ ⇒
yy y x x
′ = − = −1 1
12csc cot – if x > 0
If x < 0, then y is in Quadrant IV (see Fig-ure 4-5d). So both csc y and cot y are negative,and thus their product must be positive.
∴ ′ = − ,yx x
1
12| | – Q.E.D.
y
1x
x 2 – 1√
[Because csc y = (hypotenuse)/(opposite leg), putx on the hypotenuse and 1 on the opposite leg.
1 – x 2√
Calculus Solutions Manual Problem Set 4-5 59© 2005 Key Curriculum Press
Adjacent leg ,= x2 1– and csc y = x and cot y =(adjacent)/(opposite).]
12. y = cot− 1 x ⇒ cot y = x ⇒ −csc2 y · y′ = 1 ⇒
yy x x
′ = − = −+( )
= −+
1 1
1
1
122
2 2csc, Q.E.D.
y
x
1
1 + x 2√
[Because cot y = (adjacent leg)/(opposite leg),put x on the adjacent leg and 1 on the opposite
leg. Hypotenuse = +1 2x , and csc y =(hypotenuse)/(opposite).]
Problems 13−18 are shown done “from scratch,” asin Example 1. If students practice doing them thisway, they will not be dependent on memorizedformulas. Problem 13 shows how an alternatesolution could be found using the formulas and thechain rule.
13. y = sin− 1 4x ⇒ sin y = 4x ⇒ cos y · y′ = 4 ⇒
′ = =yy x
4 4
1 16 2cos –
y
4x1
1 – 16x2√
Alternate solution by application of the formula:
y x yx x
= ⇒ ′ = ⋅ =−sin– ( ) –
1
2 24
1
1 44
4
1 16
14. y = cos− 1 10x ⇒ cos y = 10x ⇒
−sin y · y′ = 10 ⇒ yy x
′ = − = −10 10
1 100 2sin –
y
1
10x
1 – 100x2√
15. y e y ex x= ⇒ = ⇒−cot cot1 0 5 0 5. .
− ⋅ ′ = ⇒csc2 0 50 5y y e x. .
′ = − = −+( )
= −+
ye
y
e
e
e
e
x x
x
x
x
0 5 0 5
1
0 50 5
2
0 5
2
0 5. . .
1
. . .
csc
y
1 + ex
e
1
0.5x
√
16. y = tan− 1 (ln x) ⇒ tan y = ln x ⇒sec2 y · y′ = l/x ⇒
′ = =+( )
=+
yx y x x x x
1 1
1
1
122
2 2sec ln ( ln )
y
1
1 + ln2x√ln x
17. yx
yx
y y y= ⇒ = ⇒ ⋅ ′ = ⇒−sec sec sec tan1
3 3
1
3
′ = =⋅
y xy y
xx
1
3
1
32
9 33sec tan
( – ) /
, if > 0
If x < 0, then y is in Quadrant II, where bothsec y and tan y are negative. So their product ispositive.
∴ ′ =y
x x
3
2 9| | –
y
x
3
x 2 – 9√
18. yx
yx= ⇒ = ⇒−csc csc1
10 10
− ⋅ ′ = ⇒csc coty y y1
10
yy y x x
′ = − = −⋅
1
10
1
1010
100100
2csc cot –
If x < 0, then y is in Quadrant IV, where bothcsc y and cot y are negative. So their product ispositive.
∴ ′ = −yx x
10
1002| | –
y
10
x 2
– 100
x
√
For Problems 19−24, a solution is shown using theappropriate formula.
19. y = cos− 1 5x2
′ = − ⋅ = −yx
xx
x
1
1 510
10
1 252 2 4– ( ) –
20. f (x) = tan− 1 x3
′ =+
⋅ =+
f xx
xx
x( )
1
13
3
13 22
2
6( )
60 Problem Set 4-5 Calculus Solutions Manual© 2005 Key Curriculum Press
21. g(x) = (sin− 1 x)2
′ = ⋅−g x xx
( ) 21
1
1
2sin
–
22. u = (sec− 1 x)2
′ = ⋅−u xx x
21
1
1
2 sec
| | –
23. v = x sin− 1 x + (1 − x2)1/ 2
′ = ⋅ + ⋅ + ⋅ −
= + − =
− − /
− −
v x xx
x x
xx
x
x
xx
11
1
1
21 2
1 1
1
2
2 1 2
1
2 2
1
sin–
( – )
sin– –
sin
( )
The surprise is that you now have seen a formulafor the antiderivative of the inverse sine function.
24. f (x) = cot− 1 (cot x) = x ⇒ f ′ (x) = 1 (Surprise!!)Application of the formulas gives the sameresult.
25. a. tan θ = x/100, so θ = tan− 1 (x/100), Q.E.D.
b.d
dx x x
θ =+
⋅ =+
1
1 100
1
100
100
100002 2( / )
d
dt
d
dx
dx
dt x
dx
dt
θ θ= ⋅ =+
⋅100
10000 2
c. If x = 500 ft and dθ/dt = −0.04 rad/s, then
− =+
⋅0 04100
10000 5002.dx
dtdx
dt= = −(– . )( )0 04 260000
100104
The truck is going 104 ft/s.
104(3600/5280) = 70.909… ≈ 71 mi/h26. a. θ = tan− 1 (50/x) − tan− 1 (30/x) or
θ = cot− 1 (x/50) − cot− 1 (x/30)The inverse tangent equation has theadvantage that the function appears on thecalculator. The inverse cotangent equation hasthe advantage that x is in the numerator of theargument, which makes the chain rule lesscomplicated to use.)
b.d
dx
x
x
x
x
θ =+
−+
–
( / )
–
( / )
– –50
1 50
30
1 30
2
2
2
2
=+
++
–50
2500
30
9002 2x x
= ++ +–
( )( )
20 30000
2500 900
2
2 2
x
x x
c. dθ/dx = 0 ⇒ −20x2 + 30000 = 0 ⇒20x2 = 30000
x = ± = ±1500 38 729. KAbout 38.7 ft
d. Maximum is between x = 38 and 39.
0.5
100
x
40
θ
27.
x Num. Deriv.* Alg. Deriv.
−0.8 −1.666671… −1.666666…
−0.6 −1.250000… −1.25
−0.4 −1.091089… −1.091089…
−0.2 −1.020620… −1.020620…
0 −1.000000… −1
0.2 −1.020620… −1.020620…
0.4 −1.091089… −1.091089…
0.6 −1.250000… −1.25
0.8 −1.666671… −1.666666…
*The precise value for the numericalderivative will depend on the tolerance towhich the grapher is set. The values givenby numerical derivative and the formula arevery close.
28. a. y xdy
dx x x= ⇒ =−sec
| | –
1
2
1
1
At x = 2,
dy
dx= =1
2 30 288675
| |. .K
The answer is reasonable because the graphslopes up at x = 2, with slope significantlyless than 1.
b. At x = 2, y = sec− 1 2 = cos− 1 (1/2) =1.04719… .d
dyy y y(sec sec tan) =
At y = 1.047… ,
d
dyy(sec ) (sec . )(tan . )= =1 047 1 047K K
3.464101… .
c. The answer to part b is the reciprocal of theanswer to part a. That is,
13 464101. K =
0.288675… . Thus, the derivative of theinverse secant at x = c is the reciprocal ofthe derivative of the secant at y = sec− 1 c.
29. a. y = sin− 1 x ⇒ sin y = x ⇒ cos y · y′ = 1 ⇒
′ =yy
1
cos, Q.E.D.
Calculus Solutions Manual Problem Set 4-6 61© 2005 Key Curriculum Press
b. ′ = = =yx
1 1
0 61 251 1cos(sin ) cos(sin . )– – .
yx
′ = = = =1
1
1
1 0 6
1
0 81 25
2 2– – . .. , Q.E.D.
c. y f x f y x f yd
dxy= ⇒ = ⇒ ′ ⋅ = ⇒−1 1( ) ( ) ( ) ( )
d
dxy
f y
d
dxf x
f f x( )
( )[ ( )]
[ ( )],–=
′⇒ =
′ −1 11
1
Q.E.D.
d. f ( x) = x3 + x = 10 ⇒ (x − 2)(x2 + 2x + 5) = 0
⇒ x = 2 (only)
∴ h(10) = 2
Because h(x) = f −1(x) and f ′(x) = 3x2 + 1,
′ =′
=′
=⋅ +
=hf h f
( ) .101
10
1
2
1
3 2 11132[ ( )] ( )/
30. The inverse trig cofunctions, cos− 1, cot− 1, andcsc− 1, are the ones whose derivatives are precededby a minus sign.
Problem Set 4-6Q1. See the text for the definition of continuity.
Q2. See the text for the definition of derivative.
Q3. y′ = −6x− 2 + C Q4. cos′ x = −sin x
Q5. dy/dx = sec2 x Q6. 1 12| |x x −( )Q7. f ′ (x) = 4x3; f ″ (x) = 12x2; f ″ (2) = 48
Q8. dy/dx = 15x2(x3 + 1)4
Q9. Integral ≈ 5.4 (Function is y = 2− x.)
Q10. E
1. Continuous 2. Neither
3. Neither 4. Both
5. Neither 6. Neither
7. Both 8. Neither
9. Neither 10. Neither
11. Continuous 12. Both
For Problems 13−20, sample answers are given.Equations do not necessarily correspond to the graphsshown.
13. a.
5
f(x)
x
3
b. f ( x) = x + 2
14. a.
x
f(x)
4
–2
b. f ( x) = x2
15. a.
6
x
f(x)
b. f xx x
x( )
( )( )= − +−
6 1
616. a.
1
2
f(x)
x
b. f xx x
xx
x( )
( ) ,
,=
−−
≠
=
2 1
15
if 1
if 1
17. a.
—5
x
f(x)
b. f xx x
x x( )
,
,=
≤ −> −
if
if
5
3 5
18. a.
–1
3
f(x)
x
62 Problem Set 4-6 Calculus Solutions Manual© 2005 Key Curriculum Press
b. f ( x) = (x + 1)2/3 + 3
19. a.
4
x
f(x)
7
b. f xx x
x x( )
,
,=
− <− ≥
2 9 4
11 4
if
if
20. a. No such function
x
f(x)
Not possible.Differentiabilityimpliescontinuity.
b. No such function
21. Continuous 22. Both
3
x
f(x)
4
x
f(x)
2
23. Both 24. Neither
1
x
y
x
f(x)
π/2
25. f xx x
a x b x( )
,
( – ) ,=
<+ ≥
3
2
1
2 1
if
if For f to be continuous at x = 1,
lim lim[ ( ) ]x x
x a x b→ →− +
= − + ⇒1
3
1
22
1 = a(1 − 2)2 + b ⇒ a + b = 1 ⇒ b = 1 − a
For f to be differentiable at x = 1,
lim lim ( )x x
x a x a→ →− +
= − ⇒ = − ⇒1
2
12 23 3 2 (1 2)
a = −1.5
b = 1 − a = 1 − (−1.5) ⇒ b = 2.5
1
1
x
f(x)
f is differentiable at x = 1.
26. f xx x
ax b x( )
if
if =
+ ≥+ <
–( – ) ,
,
3 7 2
2
2
3
For f to be continuous at x = 2,
lim ( ) limx x
ax b x→ →− +
+ = − + ⇒2
3
2
23 7[ ( – ) ]
a ⋅ 23 + b = 6 ⇒ 8a + b = 6 ⇒ b = 6 − 8a
For f to be differentiable at x = 2,
lim lim )x x
ax x a→ →− +
= − ⇒ ⋅ = ⇒2
2
2
23 2 3 3 2 2[ ( – ]
a = 1 6/
b = 6 − 8(1/6) ⇒ b = 14/3
2
x
f(x)
6
f is differentiable at x = 2.
27. f xax x
x x b x( ) =
+ <− + ≥
2
2
10 2
6 2
, if
, if For f to be continuous at x = 2,
lim ( ) lim ( )x x
ax x x a→ →− +
+ = − + ⇒ + =2
2
2
210 6 6 4 10
4 – 12 + b ⇒ b = 4a + 18For f to be differentiable at x = 2,lim lim ( )x x
ax x a→ →− +
= ⇒ ⋅ = ⋅ ⇒2 2
2 2 6 2 2 2 2 6– –
a = −0.5b = 4(–0.5) + 18 ⇒ b = 16
2
10
f(x)
x
f is differentiable at x = 2.
28. f xa x x
bx x( )
if
, if =
≤
− >
/ , 1
12 2 1
For f to be continuous at x = 1,
Calculus Solutions Manual Problem Set 4-6 63© 2005 Key Curriculum Press
lim lim ( )x x
a x bx a b→ →− +
= ⇒ = ⋅ ⇒1 1
2 212 1 12 1/ – / –
a + b = 12For f to be differentiable at x = 1,lim limx x
ax bx a b→
−
→
−− +− = − ⇒ − ⋅ = − ⋅ ⇒
1
2
1
22 1 2 1
a = 2b∴ 2b + b = 12 ⇒ b = 4a = 2 · 4 ⇒ a = 8
1
10
x
f(x)
f is differentiable at x = 1.
29. f xe x
b x x
ax
( )ln
=≤
+
, if
, if >
1
1For f to be continuous at x = 1,lim lim ( ln )x
ax
x
ae b x e b→ →− +
= + ⇒ =1 1
For f to be differentiable at x = 1,lim limx
ax
x
aae x ae→ →− +
= ⇒ =1 1
1 1( / )
Solve by grapher: a = 0.5671… andb = 1.7632…
1
2
x
f(x)
f is differentiable at x = 1.
30. f xa x x
e xbx( )sin
,=
≥
, if < /
if /
2 3
2 3
ππ
For f to be continuous at x = 2π/3,
lim sin lim( / ) ( / )x x
bxa x e→ →− +
= ⇒2 3 2 3π π
ae a
ebb3
2
2
32 3
2 3
= ⇒ =ππ
//
For f to be differentiable at x = 2π/3,lim( / )x→ −2 3π
a x bex
bxcos lim( / )
= ⇒→ +2 3π
− = ⇒ =abe a beb b
222 3 2 3π π/ /–
So 2
32
2
32
1
3
2 32 3e
be b bb
bπ
π/
= ⇒ = ⇒ = −– –/
= … = =−– . and 0 5773
2
30 34462 3 3a e π /( ) . K
3
0.5
x
f(x)
f is differentiable at x = 2π/3.
31. a. yax bx cx d x
x k x=
+ + + ≤ ≤+
3 2 0 0 5
0 5
, .
,
if
if > .
For y to contain the origin,a ⋅ 03 + b ⋅ 02 + c ⋅ 0 + d d= ⇒ =0 0For y′ = 0 at x = 0, y′ = 3ax2 + 2bx + c ⇒ 0 = 3a ⋅ 02 + 2b ⋅ 0 + c ⇒ c = 0For y′ = 1 at x = 0.5, y′ = 3ax2 + 2bx + c ⇒ 1 = 3a(0.5)2 + 2b(0.5) + c ⇒ 1 = 0.75a + b ⇒b = 1 – 0.75aFor ′′ = =y x0 0 5 at . , ′′ = + ⇒y ax b6 20 = 3a + 2bSolve for a and b:3a + 2(1 – 0.75a) = 0 ⇒ 1.5a = –2 ⇒a = – 4/3 b = 2
b. For the function to be continuous,
lim ( ) lim ( ). .x x
x x x k→ →− +
− + = + ⇒0 5
43
3 2
0 52
− + = + ⇒43
3 20 5 2 0 5 0 5( . ) ( . ) . kk = = −– . 6661
6 0 1 K
32. Equation of the linear part of the fork isy – 20 = 5(x – 10) ⇒ y = 5x – 30
∴ =+ <
− ≥
yax bx x
x x
3 if
if
,
,
10
5 30 10
For y to be continuous at x = 10,a b⋅ + ⋅ = ⋅10 10 5 10 303 –1000a + 10b = 20 ⇒ b = 2 – 100aFor y to be differentiable at x = 10,3 10 52a b⋅ + =300a + (2 – 100a) = 5200a = 3 ⇒ a = 3/200b = 2 – 100(3/200) ⇒ b = 0.5
33. f xx
x
xx
x
( ),
,=
− −−
≠
=
2 2
22
4 2
if
if
Simplifying the equation for f (x) gives
f x
x x
x x
x
( )
,
,=
+ <
−=
2
2
1 2
1
2
if
if > 2
4, if
64 Problem Set 4-7 Calculus Solutions Manual© 2005 Key Curriculum Press
Taking the derivative for each branch gives
′ =<>=
f x
x x
x x
x
( )
,
,
2 2
2 2
2
if
if
undefined, if
Taking the left and right limits giveslim ( ) ; lim ( ) .x x
f x f x→ →− +
′ = ⋅ = ′ = ⋅ =2 2
2 2 4 2 2 4
Using the definition of derivative, taking the
limit from the left, ′ = + −−
→→ −
f xx
xx( ) lim ,
2
2 1 4
2
1
0which is infinite. The same thing happens fromthe right. As the following graph shows, thesecant lines become vertical as x approaches 2from either side.
4
2
x
f(x) Secantslopebecomesinfinite.
Thus, f is not differentiable at x = 2, eventhough the right and left limits of f ′ (x) are equalto each other. The function must be continuousif it is to have a chance of being differentiable.
34. a. d t
t
tt
tt
( )
..
., .
, .
=
−+
≤
−
≥
60 50 5
0 50 5
150 21
0 5
if
if
′ =− + <
>
−
−d t
t t
t t( )
. ( . ) , .
, .
60 5 0 5 0 5
150 0 5
2
2
if
if
The inequality signs must be < and > becausealthough the function is defined at x = 0.5,the derivative is not.
b. d d′ = = ⇒( ) ( –1 150 1 1502) is continuous atx = 1 because it is differentiable there.
c. lim ( ) . ( . . ) ..x
d t→
−−
′ = − + = −0 5
260 5 0 5 0 5 60 5
lim ( ) ( . ).x
d t→
−+
′ = =0 5
2150 0 5 600
As the ball was about to be hit, it wasapproaching the plate at 60.5 ft/s.Just after the ball was hit, it was going awayfrom the plate at 600 ft/s.
d. Function d is not differentiable at t = 0.5because d ′ (t) approaches different limits fromboth sides of x = 0.5.Function d is continuous at t = 0.5 becauseyou get zero as the limit of d (t) as tapproaches zero from either left or right.
e. A regulation baseball diamond has thepitcher’s mound 60.5 feet from home plate.Substituting zero for t gives d (0) = 60.5,confirming that the pitcher was on the moundat that time.
35. a. y = mx + b ⇒ y′ = m, which is independentof x.∴ linear functions are differentiable for all x.∴ linear functions are continuous for all x.
b. y = ax2 + bx + c ⇒ y′ = 2ax + b, whichexists for all x by the closure axioms.∴ quadratic functions are differentiable forall x.∴ quadratic functions are continuous for all x.
c. y = 1/x = x–1 ⇒ y′ = –x–2, which exists forall x ≠ 0 by closure and multiplicative inverseaxioms.∴ the reciprocal function is differentiable forall x ≠ 0.∴ the reciprocal function is continuous for allx ≠ 0.
d. y = x ⇒ y′ = 1, which is independent of x.∴ the identity function is differentiable forall x.∴ the identity function is continuous forall x.
e. y = k ⇒ y′ = 0, which is independent of x.∴ constant functions are differentiable forall x.∴ constant functions are continuous for all x.
36. See text proof.
Problem Set 4-7Q1. y′ = 243x1214 Q2. dy/dx = 2/(x–1)2
Q3. f ′ (x) = 1 + ln x Q4. y′(x) = 5e5x cos e5x
Q5. (d /dx)(y) = 3x2, x ≠ 0; d2y/dx2 = 6x, x ≠ 0
Q6. y′ = 0 Q7. ′ = − −θ 1 1 2/ x
Q8. v(t) is decreasing at t = 5.
Q9. Q10. E
1
π/2
x
y
y'
1. x = t4, y = sin 3tdy
dx
dy dt
dx dt
t
t
d y
dx
d
dx
t
t= = ⇒ =
/
/
cos cos3 3
4
3 3
43
2
2 3
Calculus Solutions Manual Problem Set 4-7 65© 2005 Key Curriculum Press
= − ⋅ −( sin )( / ) cos ( )( / )
( )
9 3 4 3 3 12
4
3 2
3 2
t dt dx t t t dt dx
t
= − − ÷36 3 36 3
16
3 2
6
t t t t
t
dx
dt
sin cos
= − −36 3 36 3
64
3 2
9
t t t t
t
sin cos
= − −9 3 9 3
16 7
t t t
t
sin cos
2. x = 6 ln t, y = t3
dy
dx
dy dt
dx dt
t
tt
d y
dx
d
dxt t dt dx t
dx
dt
t
tt
= = = ⇒
= = = ÷
= =
/
/ /
) .
.
/
3
60 5
0 5 1 5 1 5
1 5
60 25
23
2
2
3 2 2
23
.
( . ) . ( /
.
3. a. x = 2 + t, y = 3 – t2
t x y
–3 –1 –6
–2 0 –1
–1 1 2
0 2 3
1 3 2
2 4 –1
3 5 –6
b.
3
2
x
y
c.dy
dx
dy dt
dx dt
tt= = − = −/
/
2
12
If t = 1, dy dx/ –= 2 and (x, y) = (3, 2).Line through (3, 2) with slope –2 is tangentto the graph. See part b.
d. x = 2 + t ⇒ t = x – 2 ⇒ y = 3 – (x – 2)2
This is the Cartesian equation of a parabolabecause only one of the variables is squared.
e. By direct differentiation, dy dx x/ – ( – ).= 2 2At (x, y) = (3, 2), dy/dx = –2(3 – 2) = –2,which agrees with part c.dy/dx = –2(x – 2) = –2(2 + t – 2) = –2t,which agrees with part c.
4. a. x = t2, y = t3
t x y
–3 9 –27–2 4 –8–1 1 –1 0 0 0 1 1 1 2 4 8 3 9 27
b.
x
y
5
5
c.dy
dx
dy dt
dx dt
t
tt= = =/
/.
3
21 5
2
If t = 1, dy/dx = 1.5 and (x, y) = (1, 1).Line through (1, 1) with slope 1.5 is tangentto the graph. See graph in part b.
d. x t t x y x y x= ⇒ = ⇒ = ⇒ =2 1 2 1 2 3 1 5/ / .( )The name semicubical is picked because 1.5is half of 3, the exponent for a cubic function.The name parabola is used because the equationlooks similar to y = x2 for a parabola.
e. By direct differentiation, dy/dx = 1.5x0.5 .At (x, y) = (1, 1), dy/dx = 1.5 ⋅ 10.5 = 1.5,which agrees with part c.dy/dx = 1.5x0.5 = 1.5(t2)0.5 = 1.5t, whichagrees with part c.
5. a. The graph confirms the figure in the text.
b.dy
dx
t
tt=
−= −5
3
5
3
cos
sincot
c. If t = π/4, x y= =3 2 2 5 2 2/ and / .(x, y) = (2.121… , 3.535…)dy
dx= − = −5
3 45 3cot /
π
5
–5
3–3
y
x
The line is tangent to the graph.
66 Problem Set 4-7 Calculus Solutions Manual© 2005 Key Curriculum Press
d. False. The line from (0, 0) to (2.1… , 3.5…)does not make an angle of 45° with thex-axis. (This shows that the t in parametricfunctions is not the same as the θ in polarcoordinates.)
e. The tangent line is horizontal if dy/dx = 0.
∴ cos t = 0 and sin t ≠ 0.This happens at t = π/2, 3π/2, … .Points are (0, 5), (0, –5).Tangent line is vertical if dy/dx is infinite.
∴ sin t = 0 and cos t ≠ 0.This happens at t = 0, π, 2π, … .Points are (3, 0), (–3, 0). See graph in part c.
f. x/3 = cos t ⇒ (x/3)2 = cos2 ty/5 = sin t ⇒ (y/5)2 = sin2 tAdding left and right sides of the equationsgives (x/3)2 + (y/5)2 = cos2 t + sin2 t.∴ (x/3)2 + (y/5)2 = 1, which is a standard formof the equation of an ellipse centered at theorigin, with x-radius 3 and y-radius 5.
6. a. The graph confirms the figure in the text.
b.dy
dx
t t
t t
t
tt=
−= − =24
24
2
2
sin cos
cos ( sin )
sin
costan–
∴ dy/dx = –tan t
c. If t = 1, x = 8 cos3 1 = 1.2618… , and
y = 8 sin3 1 = 4.7665… ,
(x, y) = (1.2618… , 4.7665…).
At t = 1, dy/dx = –tan 1 = –1.5574… .
x
y
8–8
8
–8
The line is tangent to the graph.
d. dx/dt = –24 cos2 t sin tdy/dt = 24 sin2 t cos tThe cusps occur where t is a multiple of π/2.At each such value, dx/dt and dy/dt equal zero.t = 0 gives the cusp at (8, 0).lim( / ) lim(– tan ) tant t
dy dx t→ →
= = =0 0
0 0–
So the graph becomes horizontal at (8, 0).t = π/2 gives the cusp at (0, 8).
lim ( / ) lim ( tan ),/ /t t
dy dx t→ →
= −π π2 2
which is infinite.
So the graph becomes vertical at (0, 8).
e. x/8 = cos3 t ⇒ (x/8)2/3 = cos2 t
y/8 = sin3 t ⇒ (y/8)3/2 = sin2 t
∴ (x/8)2/3 + (y/8)2/3 = cos2 t + sin2 t
⇒ x2/3 + y2/3 = 4
7. a. x = 6 + 5 cos t, y = 3 + 5 sin t
6
3
dy/dx isinfinitehere.
x
y
b.dy
dx
t
tdy dx t=
−⇒ = −5
5
cos
sin/ cot
c. dy/dx = 0 if cos t = 0 and sin t ≠ 0.∴ t = 0.5π, 1.5π, 2.5π, …dy/dx is infinite if sin t = 0 and cos t ≠ 0.∴ t = 0, π, 2π, …At a point where dy/dx is infinite, dx/dt mustbe zero. This happens where t n= ±π π/ ,2so dy/dx = 5 cos t = 0 at those points. Seegraph in part a.
d.x
ty
t− = − =6
5
3
5cos sin and
x yt t
−
+ −
= +6
5
3
5
2 22 2cos sin
x y−
+ −
=6
5
3
51
2 2
This is an equation of a circle centered at(6, 3) with radius 5.
e. The 6 and 3 added in the original equationsare the x- and y-coordinates of the center.The coefficients, 5, for cosine and sine in theoriginal equations are the x- and y-radii,respectively. Because the x- and y-radii areequal, the graph is a circle.
8. x = cos2 t, y = sin2 t
dy
dx
t t
t tt t= − = ≠ ≠2
21 0 0
cos ( sin )
sin coscos sin– ( , )
1
1
x
y
The graph is a line segment with a slope of –1.x + y = cos2 t + sin2 t ⇒ x + y = 1This is the equation of a line with slope –1,confirming what was observed on the graph.The parametric equations restrict the ranges of xand y to the first quadrant, no matter what is thedomain of t. This is true because cos2 t and sin2 tare never negative.The Cartesian equation allows–∞ < x < ∞ and –∞ < y < ∞.
Calculus Solutions Manual Problem Set 4-7 67© 2005 Key Curriculum Press
9. a. The grapher confirms the figure in the text.
b.dy
dx
t t
t t
t t
t t= −
− −= −
− −2 2 2
2 2 2
2
2
cos cos
sin sin
cos cos
sin sin
c. Cusps occur where both dx/dt and dy/dt = 0.A graphical solution shows that this occurs att = 0, t = 2π/3, t = 4π/3, t = 2π, … .(A cusp could also happen if dx/dt = 0 anddy/dt ≠ 0, but for this figure there is no suchplace.)
0 2 2π4π3 3__ __π
dx/dt
dy/dt
t
dx/dt or dy/dt
At t = 0, 2π, … , the tangent appears tobe horizontal. At t = 2π/3, 4π/3, 8π/3,10π/3, … , there appears to be a tangentline but not a horizontal one.A numerical solution shows the followingvalues as t approaches 2π/3:
t dy/dx
2π/3 – 0.1 –1.547849…
2π/3 – 0.01 –1.712222…
2π/3 – 0.001 –1.730052…
2π/3 indeterminate
2π/3 + 0.001 –1.734052…
2π/3 + 0.01 –1.752225…
2π/3 + 0.1 –1.951213…
dy/dx seems to be approaching about –1.732as t approaches 2π/3.
[The exact answer is − 3, which studentswill be able to prove easily with l’Hospital’srule after they have studied Section 6-5. JoanGell and Cavan Fang have shown clevertrigonometric transformations that “remove”the removable discontinuity and lead to thesame answer. These are
1. Use the sum and product properties ondy/dx:
dy
dx
t t
t t=
−2 1 5 0 5
2 1 5 0 5
sin . sin .
sin . cos .
= − ≠tan . /0 5 0t dx dt if
As t dy dx→ → =2 3 3π π/ , / – ( / ) – 3tan .
2. Use the double argument properties ondy/dx:dy
dx
t t
t t t=
+cos – ( cos – )
–(sin sin cos )
2 1
2
2
= ++
=( – cos )( cos )
–(sin )( cos )
– cos
– sin
1 1 2
1 2
1t t
t t
t
t,
which approaches − 3 as t → 2π/3.]
10. a. The grapher confirms the figure in the text.
b.dy
dx
a t t
a tt t= =4
222
3cos (– sin )
seccos sin
–
(The answer is independent of a.)
c. x a ta t
t
a t
t2 2 2
2 2
2
2 2
244 4 1= = =tan
sin
cos
( – cos )
cosy = 2a cos2 t ⇒ cos2 t = y/(2a)
∴ = = −x
a y a
y a
a a y
y2
2 24 1 2
2
4 2[ – /( )]
/( )
( )
x2y = 8a3 – 4a2y ⇒ (x2 + 4a2)y = 8a3 ⇒
ya
x a=
+8
4
3
2 2
a yx
= ⇒ =+
3216
362
d. y a x a= + ⇒−8 43 2 2 1( )
dy
dxa x a x
a x
x a= − + ⋅ =
+−8 4 2
16
43 2 2 2
3
2 2 2( )–
( )
e. At t = π/4, x = 2a tan (π/4) = 2a.From part d,
dy
dx
a a
a a
a
a=
+= =– ( )
[( ) ]
–16 2
2 4
32
64
3
2 2 2
4
4 –1/2
From part b,dy
dx= – ( / ) ( / )2 4 43cos sinπ π
= =– ( / ) ( / ) – / ,2 2 2 2 2 1 23 which agrees.
At t = π/4, x = 2a tan (π/4) = 2a = 6 andy = 2a cos2
(π/4) = 2a(1/2) = a = 3.A line through (6, 3) with slope –1/2 istangent to the graph at that point.
5
y
x
10
–5
–5–10–15 0 5 10 15
t = π/4
11. a. x = cos t + t sin ty = sin t – t cos tThe grapher confirms the figure in the text.[Note: In the derivation of these equationsfrom the geometric definition of involute,
68 Problem Set 4-7 Calculus Solutions Manual© 2005 Key Curriculum Press
x = cos t + t cos (t – π/2)y = sin t + t sin (t – π/2)(cos t, sin t) is the point of tangency of thestring.Because the circle is a unit circle, the lengthof the string is also t, the central angle inradians.The string makes an angle of (t – π/2) withthe positive x-axis so that(t cos (t – π/2), t sin (t – π/2)) is a vectorrepresenting the unwound string.The cofunction properties and odd-evenproperties from trig are used to simplify theequations so that the calculus will be easier.]
b.dy
dx
t t t t
t t t t= +
+ +cos – [cos (– sin )]
– sin (sin cos )
= =t t
t tt
sin
costan
c. At t = π, dy/dt = tan π = 0. The string willbe pointing straight up from the x-axis. Thediagram shows that the tangent to the graph ishorizontal at this point.
x
1
1
y
String
(x, y)
t = π
12. a. x starts at a middle point and increases.y starts at a high point and decreases.∴ x = 25 + 15 sin Bt
y = 20 + 15 cos BtThe period is 60 seconds.So B = 2π/60 = π/30
∴ = +x t25 1530
sinπ
y t= +20 1530
cosπ
b. dx dt t/ = π π2 30
cos
dy dt t/ –= π π2 30
sin
At t = 5,
dx dt/ / .= = =π π π
2 63 4 1 3603cos K
dy dt/ – – / – .= = =π π π
2 64 0 7853sin K
c. The slope of the circular path is dy/dx.At t = 5,dy
dx= = =– /
/
ππ
4
3 41 3 0 5773– / – . K
d.x
tx
t–
sin–
sin25
15 30
25
15 30
22= ⇒
=π π
yt
yt
–cos
–cos
20
15 30
20
15 30
22= ⇒
=π π
Because sin cos2 2
30 301
π πt + = ,
x y– –.
25
15
20
151
2 2
+
=
This is an equation of a circle centered at(25, 20) with radius 15, confirming that thepath really is a circle.
13. The actual solutions will vary depending on theperiod of the pendulum, as determined by thelength of the string. The following solutionsupposes that the period turns out to be 3.1seconds.
x t y t= =302
3 120
2
3 1cos
.sin
.
π π
dy
dx
t
tt= =
( / . ) cos.
–( / . ) sin .
40 3 123 1
60 3 12
2
3
2
3 1
π π
π ππ
3.1
– cot
At t = 5, x ≈ –22.8, y ≈ –13.0,and dy/dx ≈ –0.78.
If the measurements have been accurate, thependulum will be above the coin when t = 5.
14. The graph looks like an ellipse that moves in thex-direction as t increases. Because y starts at ahigh point and varies between 5 and 1, the ellipsehas center at y = 3 and y-radius 2. Thus, anequation for y would be y = 3 + 2 cos t.x starts at 0 and increases. If the ellipse hadx-radius 0.5, an equation for x would bex = 0.5 sin t. The graph of this ellipse is
x
y5
10
The graph seems to move over 1 unit to the righteach cycle. Thus, if t increases by 2π, x increasesby 1. The equations are thusx = t/(2π) + 0.5 sin t, y = 3 + 2 cos tThe graph here duplicates the one in the text.
x
y5
10
Calculus Solutions Manual Problem Set 4-8 69© 2005 Key Curriculum Press
To locate “interesting” features,dy
dx
dy dt
dx dt
t
t= =
+/
/
– sin
/( ) . cos
2
1 2 0 5π.
For horizontal tangents, dy/dt = 0 and dx/dt ≠ 0.∴ 2 sin t = 0 ⇔ t = 0 + πn (n an integer)Thus, x = 0, 0.5, 1, 1.5, … .For vertical tangents, dx/dt = 0 and dy/dt ≠ 0.∴ 1/(2π) + 0.5 cos t = 0 ⇔ cos t = –1/πSolving numerically for t givest = 1.8947… + 2π n or 4.3884… + 2πn.For crossing points, x = 0.5, 1.5, 2.5, …from symmetry on the graph. If x = 0.5, then1/(2π)t + 0.5 sin t = 0.5.Solving numerically for the value of t closest to0, t = 0.8278… .y(0.8278…) = 3 + 2 cos 0.8278… = 4.3529…A crossing point is (0.5, 4.3529…) at t =0.8278… .
15. a. The grapher confirms the figure in the text.
b. (x = cos 4t, y = sin t)
1
1
x
y
If n is an even number, the graph comes toendpoints and retraces its path, making twocomplete cycles as t goes from 0 to 2π.If n is an odd number, the graph does notcome to endpoints. It makes one completecycle as t goes from 0 to 2π.
c. i. (x = cos 5t, y = sin t)
1
1 y
x
ii. (x = cos 6t, y = sin t)
1
1 y
x
d. n = 1. (x = cos t, y = sin t)
1
1 y
x
n = 2. (x = cos 2t, y = sin t)
1
1 y
x
If n = 1, the graph is a circle.If n = 2, the graph is a parabola.
e. Jules Lissajous (1822–1880) lived in France.Nathaniel Bowditch (1773–1838) lived inMassachusetts.
Problem Set 4-8Q1. y′ = 2001x2000 Q2. y′ = ln (2001)2001x
Q3. 5 Q4. f ′(u) = –csc2 u
Q5. product Q6. 1/(1 + 9x2)
Q7. x3 + C Q8. Instantaneous rateQ9. Q10. –2.4033… ft/s
6
4
x
y
y'
1. x3 + 7y4 = 13 ⇒ 3x2 + 28y3y′ = 0 ⇒
yx
y′ = –
3
28
2
3
2. 3x5 − y4 = 22 ⇒ 15x4 − 4y3y′ = 0 ⇒ yx
y′ = 15
4
4
3
3. x ln y = 104 ⇒ 1 · ln y + xy
y⋅ ⋅ ′1 = 0 ⇒
′ = −y
y y
x
ln
4. yex = 213 ⇒ pxy′ + yex = 0 ⇒ y′ = –y
70 Problem Set 4-8 Calculus Solutions Manual© 2005 Key Curriculum Press
5. x + xy + y = sin 2x ⇒1 + y + xy′ + y′ = 2 cos 2x ⇒y′(x + 1) = 2 cos 2x – 1 – y ⇒
yx y
x′ =
+2 2 1
1
cos – –
6. cos xy = x – 2y ⇒(–sin xy) ( y + xy′) = 1 – 2y′ ⇒y′(–x sin xy + 2) = 1 + y sin xy ⇒
yy xy
x xy′ = +1
2
sin
– sin
7. x0.5 – y0.5 = 13 ⇒0.5x–0.5 – 0.5y–0.5y′ = 0 ⇒ y′ = y0.5 /x0.5
8. x1.2 + y1.2 = 64 ⇒ 1.2x0.2 + 1.2y0.2y′ = 0 ⇒y′ = –x0.2 /y0.2
9. e xy = tan y ⇒ exy(1 · y + x · y′) = y′sec2 y ⇒ ye xy
+ xy′e xy = y′sec2 y ⇒ xy′e xy – y′sec2 y = –ye xy
⇒ y′(xexy – sec2 y) = –yexy ⇒ y′ = −−ye
xe y
xy
xy sec2
10. ln (xy) = tan–1 x ⇒ tan (ln xy) = x ⇒
sec2 (ln xy) · 1
xy
(1 · y + y′x) = 1 ⇒ y + y′x =
xy cos2 (ln xy) y′x = xy cos2 (ln xy) – y ⇒
′ = −y
xy xy y
x
cos (ln )2
11. (x3y4)5 = x – y ⇒5(x3y4)4(3x2y4 + x3 · 4y3y′) = 1 – y′ ⇒y′(20x15y19 + 1) = 1 – 15x14y20 ⇒
yx y
x y′ =
+1 15
1 20
14 20
15 19
–
12. (xy)6 = x + y ⇒6(xy)5(y + xy′) = 1 + y′ ⇒y′(6x6y5 – 1) = 1 – 6x5y6 ⇒
yx y
x y′ = 1 6
6 1
5 6
6 5
–
–
13. cos2 x + sin2 y = 1 ⇒2 cos x · (–sin x) + 2 sin y · cos y · y′ = 0 ⇒
yx x
y y′ = cos sin
cos sin
14. sec2 y – tan2 x = 1 ⇒2 sec y · sec y tan y · y′ – 2 tan x · sec2 x = 0 ⇒
yx x
y y′ = sec tan
sec tan
2
2
15. tan xy = xy ⇒(sec2 xy) · (y + xy′) = y + xy′ ⇒y′(x sec2 xy – x) = y – y sec2 xy ⇒
yy xy
x xyy
y
x′ = ⇒ ′ =( – sec )
(sec – )
1
1
2
2 –
16. cos xy = xy ⇒(–sin xy) · (y + xy′) = y + xy′ ⇒
y′(–x – x sin xy) = y + y sin xy ⇒
yy xy
x xyy
y
x′ = + ⇒ ′ =( sin )
(– – sin )
1
1–
17. sin y = x ⇒ cos y · y′ = 1 ⇒ y′ = sec y
18. cos y = x ⇒ –sin y · y′ = 1 ⇒ y′ = –csc y
19. csc y = x ⇒ –csc y cot y · y′ = 1 ⇒y′ = –sin y tan y
20. cot y = x ⇒ –csc2 y · y′ = 1 ⇒ y′ = –sin2 y
21. y = cos− 1 x ⇒ cos y = x ⇒ –sin y · y′ = 1 ⇒
yy x
′ = =– –1 1
1 2sin –
y
1
x
√1 – x 2
22. y = ln x ⇒ ey = x ⇒ ey · y′ = 1 ⇒ y′ = =1 1
e xy
23. y = x11/5 ⇒ y5 = x11 ⇒ 5y4 · y′ = 11x10 ⇒
yx
y
x
x
x
xx′ = = = =11
5
11
5
11
5
11
5
10
4
10
11 5 4
10
44 56 5
( )/ // ,
which is the answer obtained using the derivativeof a power formula, Q.E.D.
24. Prove that if y = xn, where n = a/b and a and bare integers, then y′ = nan− 1.
Proof:
y = xn = xa/ b ⇒ yb = xa.Because a and b are integers,byb− 1 y′ = ax a− 1
yax
by
ax
b x
ax
bx
a
bx
a
b
a
a b b
a
a a ba a a b′ = = = = − − −
–
–
–
/ –
–
– /( / )
( )
1
1
1
1
11
= =a
bx nxa b n/ – – ,1 1 Q.E.D.
25. a. x2 + y2 = 100At (–6, 8), (–6)2 + 82 = 100, which shows that (–6, 8) is on the graph, Q.E.D.
b. x2 + y2 = 100 ⇒ 2x + 2y · dy/dx = 0 ⇒ dy/dx = –x/yAt (–6, 8), dy/dx = –(–6)/8 = 0.75.A line at (–6, 8) with slope 0.75 is tangent tothe graph, showing that the answer isreasonable.
x
y10
10
Calculus Solutions Manual Problem Set 4-8 71© 2005 Key Curriculum Press
c. x = 10 cos t y = 10 sin tdy
dx
t
t
t
t= =10
10
cos
– sin
cos
sin–
At x = –6, t = cos–1 (–0.6).sin [cos–1 (–0.6)] = 0.8
∴ = =dy
dx– . ,
– .
.
0 6
0 80 75
which agrees with part b, Q.E.D.
26. a. x2 – y2 = 36At (10, –8), 102 – (–8)2 = 36, which showsthat (10, –8) is on the graph, Q.E.D.
b. x2 – y2 = 36 ⇒ 2x – 2y · dy/dx = 0 ⇒dy/dx = x/yAt (10, –8), dy/dx = 10/(–8) = –1.25.A line at (10, –8) with slope –1.25 is tangentto the graph, showing that the answer isreasonable.
x
y
10
10
c. x = 6 sec t y = 6 tan tdy
dx
t t
t
t
t= =6
6 2
sec tan
sec
tan
sec
At x = 10, t = ±sec–1 (10/6).tan [±sec–1 (10/6)] = ±8/6.Choose the negative value because y < 0.
∴ = =dy
dx
– /
/
10 6
8 61 25– . ,
which agrees with part b, Q.E.D.
27. a. x3 + y3 = 64 ⇒ 3x2 + 3y2 · dy/dx = 0 ⇒dy/dx = –x2/y2
x = 0: y3 = 64 ⇒ y = 4∴ dy/dx = –0/16 = 0The tangent is horizontal (see the next graph).x = 2: 8 + y3 = 64 ⇒ y3 = 56 ⇒y = 3.8258…∴ dy/dx = –22/(3.8258…)2 = –0.2732…The tangent line has a small negative slope,which agrees with the graph.x = 4: 64 + y3 = 64 ⇒ y = 0∴ dy/dx = –42/0, which is infinite.The tangent line is vertical.
x
y
10
10
b. y = x: x3 + x3 = 64 ⇒ x3 = 32 ⇒x = 3.1748…dy/dx = –x2/y2 = –x2/x2 = –1
c. y = (64 – x3)1/3
As x becomes infinite, (64 – x3)1/3 gets closerto (–x3)1/3, which equals –x. The graph hasa diagonal asymptote at y = –x, anddy/dx → –1.
d. By analogy with the equation of a circle, suchas x2 + y2 = 64
28. a. First simplify the equation.[(x – 6)2 + y2][(x + 6)2 + y2] = 1200(x – 6)2(x + 6)2 + (x – 6)2y2 + (x + 6)2y2 + y4
= 1200(x2 – 36)2 + (x2 – 12x + 36 + x2 + 12x + 36)y2
+ y4 = 1200x4 − 72x2 + 1296 + 2x2y2 + 72y2 + y4 = 1200x4 − 72x2 + 2x2y2 + 72y2 + y4 = −96Differentiate the simplified equationimplicitly.4x3 − 144x + 4xy2 + 4x2y · dy/dx
+ 144y · dy/dx + 4y3 · dy/dx = 0(4x2 y + 144y + 4y3) · dy/dx = −4x3
+ 144x − 4xy2
dy
dx
x x xy
x y y y= +
+ +– –3 2
2 3
36
36At x = 8: (4 + y2)(196 + y2) = 1200784 + 200y2 + y4 = 1200y4 + 200y2 − 416 = 0
y2 200 41664
22 058806= ± =–. K or
−202.0…y = ±1.4348542… (No other real solutions)At (8, 1.434…), dy/dx = −1.64211… .At (8, –1.434…), dy/dx = 1.64211… .Both answers agree with the moderately steepnegative and positive slopes, respectively.
x
y
10
5
b. At the x-intercepts, y = 0.∴ (x − 6)2 (x + 6)2 = 1200(x2 − 36)2 = 1200
x = ± ± = ±36 1200 8.4048K or±1.1657…Derivative appears to be infinite at eachx-intercept.
At x = + =36 1200 8.4048K ,
72 Problem Set 4-9 Calculus Solutions Manual© 2005 Key Curriculum Press
dy
dx= +
+ +–( . ) ( . ) – ( . )( )
( . ) ( ) ( )
8 4 36 8 4 8 4 0
8 4 0 36 0 0
3
2 3
K K K
K
= 896 29
0
.,
K which is infinite, as conjectured.
c. From part a,x4 − 72x2 + 2x2y2 + 72y2 + y4 = −96 ⇒y4 + (2x2 + 72)y2 + (x4 − 72x2 + 96) = 0
y2 =
− + ± + − − +( ) ( ) ( )( )2 72 2 72 4 1 72 96
2
2 2 2 4 2x x x x
y x x2 2 236 144 1200= − − ± –
Only the positive part of the ambiguoussign ± gives real solutions for y.
y x x= ± +– – –2 236 144 1200
Plot the graph letting y1 equal the positivebranch and y2 equal the negative branch. Thegraph is as in the text. The two loops maynot appear to close, depending on the windowyou use for x.
d. Repeating the algebra of parts a and c with1400 in place of 1200 gives
y x x= ± +– – –2 236 144 1400
Plot the graph as in part a. The two ovals inthe original graph merge into a single closedfigure resembling an (unshelled) peanut.
x
y
10
5
e. The two factors in the equation[(x − 6)2 + y2][(x + 6)2 + y2] = 1200are the squares of the distances from (x, y) tothe points (6, 0) and (−6, 0), respectively.The product of the distances is 1200, aconstant.
Problem Set 4-9Q1. y2 + 2xyy′ Q2. Implicit differentiation
Q3. Product rule Q4. Chain rule
Q5. Speeding up Q6. smaller
Q7. cos x − x sin x Q8.1
x
dx
dt
Q9. −2e− x + xe− x Q10. E
1. Know: dA
dt= 12 mm2/h. Want:
dr
dt.
A rdA
dtr
dr
dt
dr
dt r= ⇒ = ⇒ =π π
π2 2
6
3
r
dr/dt
1
dr
dt= =2
0 6366π
. K mm/h when r = 3 mm.
dr
dt varies inversely with the radius.
2. Know: dr
dt= 2 cm/s. Want:
dV
dt.
V rdV
dtr
dr
dt= ⇒ =4
343 2π π
dV
dt= =72 226 1946 3π . /K cm s at r = 3 cm
dV
dt= =288 904 7786π . /K cm s3 at r = 6 cm
3 6
dV/dt
r
500
The graph shows that the larger the balloon gets,the faster Phil must blow air to maintain the2 cm/s rate of change of radius.
3. Know: dA
dt= −144 cm /s.2 Want:
da
dt.
A = πab and a = 2b ⇒ A a= 1
22π
dA
dta
da
dt
da
dt a
dA
dt= ⇒ =π
π1
b a
da
dt= ⇒ = ⇒ = − = −12 24
61 9098
π. K
≈ −1 91. cm/sThe length of the major axis is 2a, so the majoraxis is decreasing at 12/π cm/s.
4. Know: dK
dt= 100 000, MJ/s;
dm
dt= −20 kg/s.
Want: dV
dt. (Note: 1 megaJoule—MJ—is the
energy required to accelerate a 1-kg mass by1 km/s through a distance of 1 km; it can beexpressed 1 MJ = 1 kg · km2/s2.)
K mVdK
dtV
dm
dtmV
dV
dt= ⇒ = + ⇒1
2
1
22 2
Calculus Solutions Manual Problem Set 4-9 73© 2005 Key Curriculum Press
dV
dt mV
dK
dt
V
m
dm
dt= −1
2dV
dt=
⋅−
⋅=100000
5000 10
10 20
2 50002 02
(– ). (km/s)/s
5. Let y = Milt’s distance from home plate.Let x = Milt’s displacement from third base.
Know:dx
dt= −20 ft/s. Want:
dy
dt.
y x ydy
dtx
dx
dt2 2 290 2 2= + ⇒ =
⇒ = ⋅ =+
dy
dt
x
y
dx
dt
x
x
–20
902 2
90
10
x
dy/dt
At xdy
dt= = − ≈ −45 8 944 8 9, . K . ft/s
(exact: −4 5 ).
At xdy
dt= =0 0, ft/s, which is reasonable because
Milt is moving perpendicular to his line fromhome plate.
6. Let y = displacement from stern to dock alongpier. Let x = displacement from bow to pieralong dock.
Know: dy
dt= −3 m/s. Want:
dx
dt.
x2 + y2 = 2002
2 2 03
2002 2x
dx
dty
dy
dt
dx
dt
y
x
dy
dt
y
y+ = ⇒ = − =
–
At ydx
dt= = =120
360
1602 25, . m/s.
200
10
y
dx/dt
120
7. a. Let L = length. Let W = width. Let H = depth(meters).
Know: dW
dt
dL
dt= = −0 1 0 3. m/s; . m/s.
Want: dH
dt.
LWH = ⇒20dL
dtWH L
dW
dtH LW
dH
dt⋅ + ⋅ ⋅ + ⋅ = 0
dH
dt
H
W
dW
dt
H
L
dL
dt
LW L W
= − ⋅ − ⋅
= − −200 1
200 32 2( . ) (– . )
b.dH
dt= − ⋅
⋅+ ⋅
⋅=20 0 1
5 2
20 0 3
5 20 022 2
. ..
Depth is increasing at 0.02 m/s.
8. Let L = distance between spaceships.
Know:dx
dt
dy
dt= = −80 50 km/s; km/s.
Want: dL
dt.
L x y LdL
dtx
dx
dty
dy
dt2 2 2 2 2 2= + ⇒ = +
⇒ =+
dL
dt x yx y
180 50
2 2( – )
dL
dt= ⋅ ⋅ = =1
130080 500 50 1200
200
13( – )
–
−15.3846…
Distance is decreasing at about 15.4 km/s.
9. a. Let x = distance from bottom of ladder towall. Let y = distance from top of ladder tofloor.
20 0 2 22 2 2= + ⇒ = + ⇒x y xdx
dty
dy
dtdy
dt
x
y
dx
dt= −
Note that the velocity of the weight is −dy/dt,so
vx
x
dx
dt=
400 2–
b. v = ⋅ − = − = −4
3843
6
40 6123( ) . ft/sK
c. Here xdx
dtv= = = → ∞20 2
40
0, , so (!!)
10. a.
1200 in.2W
L
D
Know: dL
dt. Want:
dW
dt.
LWdL
dtW L
dW
dt= ⇒ ⋅ + ⋅ =1200 0
⇒ = − ⋅ = − ⋅dW
dt
W
W
dL
dtW
dL
dt1200
1
12002
/
b. − = − ⇒ = ⇒21
12006 202W W( ) in.
L = 60 in.
74 Problem Set 4-9 Calculus Solutions Manual© 2005 Key Curriculum Press
c. D L W DdD
dtL
dL
dtW
dW
dt2 2 2 2 2 2= + ⇒ = +
⇒ =+
+
dD
dt L WL
dL
dtW
dW
dt
12 2
At L = 60 and W = 20,dD
dt=
++ −1
20 6060 6 20 2
2 2[ ( ) ( )]
= = =320
40001 6 10 5 0596. . K
Diagonal is increasing at about 5.06 in./min.
11. a. Let h = depth of water. Let r = radius of waterat surface. Let V = volume of water.
Know: dh
dt = 5 m/h. Want:
dV
dt.
V r h= 1
32π
By similar triangles, r
h =
3
5 ⇒ r =
3
5h
∴ =
=V h h h
1
3
3
5
3
25
23π π
dV
dth
dh
dt= 9
252π
At h = 3,
dV
dt= = =81
516 2 50 8938π π. . K ≈
50.9 m3/h.
b. i. Know: dV
dt = −2 m3/h. Want:
dh
dt.
dV
dth
dh
dt
dh
dt h
dV
dt= ⇒ =9
25
25
92
2ππ
dh
dt= = −–50
1440 1105
π. K ≈
−0.11 m/h at h = 4 m
ii. dh
dt → −∞ as h → 0 m
c. i. Know: dV
dtk h= .
dV
dth k= − = ⇒ = −0 5 4 0 25. at .
dV
dth= −0 25.
ii. dV
dt= − = −0 25 0 64 0 2. . m /h3.
at mh = 0 64.
iii. dV
dth= − = ⇒0 2 0 64. at . m
dh
dt= = − ≈25
9 0 640 2 0 43172π ( . )
(– . ) . K
−0.43 m/h
12. Let h = altitude. Let r = radius. Let V = volumeof cone.
Know: dh
dt = −6 ft/min;
dr
dt = 7 ft/min. Want:
dV
dt.
V r hdV
dtr
dr
dth r
dh
dt= ⇒ = +1
3
2
3
1
32 2π π π
dV
dt= + − =2
38 3 7
1
38 62π π( )( )( ) ( ) ( )
−16π ft3/min = −50.2654…Volume is decreasing at about 50.3 ft3/min.
13. a. Let ω = angular velocity in radians per day.
ω π ω πE M, = =2
365
2
687
d
dt
θ ω ω π= − = −
=E M 2
1
365
1
687
644
2507550 008068
π = . K ≈ 0.00807 rad/day
b. T = −
= ≈
−1
365
1
687778 7422
1
. K
778.7 days
The next time after 27 Aug. 2003 when thetwo planets will be closest is 779 days later,on 14 Oct. 2005 (or 15 Oct., if the planetswere aligned later than about 6:11 a.m. backon 27 Aug. 2003). Because the actual orbitsof Earth and Mars are not as simple aspreviously assumed, the actual closestdistances are not always the same. In fact, theapproach on 27 Aug. 2003 was the closestone in nearly 60,000 years! Nor is the periodbetween close approaches quite so simple.The next close approach will actually be on30 Oct. 2005, not 15 Oct.
c. By the law of cosines,
D2 = 932 + 1412 − 2 · 93 · 141 cos θD = 28530 26226– cos θ million mi
d.dD
dt
d
dt= 26226
2 28530 26226
sin
cos
θθ
θ–
=⋅
⋅26226
1365
1687
2
2 28530 26226
–
–
π θ
θ
sin
cosmillion mi/day
=⋅ ⋅
⋅
⋅
1 000 000 262261
3651
6872
24 2 28530 26226
, , sin
cos
–
–
π θ
θ
=⋅ −
⋅
−
1 092 750 0001
3651
68728530 26226
, , , sin
cos
π θ
θmi/h
To find out how fast D is changing today,first determine how many days after 27 Aug.2003 it is today, then multiply that number
by d
dt
θ π= −
1
365
1
6872 to find θ, then
substitute θ into the previous expressions.
Calculus Solutions Manual Problem Set 4-10 75© 2005 Key Curriculum Press
e. To maximize dD
dt, plot the variable part of
dD
dt, y =
−sin
cos.
θθ28530 26226
y0.01
π 2π
θ
From the graph, it is clear that the maximumoccurs well before θ = π/2 (90°). Using themaximize feature, the maximum occurs atθ ≈ 0.8505… , or 48.7…°.(The exact value is cos− 1 (93/141). One canfind this by finding (d/dt)(dD/dt) and setting itequal to zero. One can also see this bydecomposing Earth’s motion vector into twocomponents—one toward/away from Mars andthe other perpendicular to the first. The rate ofchange in D is maximized when all of Earth’smotion is along the Earth-to-Mars component,which occurs when the Earth-Mars-Suntriangle has a right angle at Earth.In this case, cosθ = 93
141 .)
f. θ π= −
1
365
1
6872 t if t = days since
27 Aug. 2003.
D t=
28530 262261
365
1
6872– cos – π
1000
200
t
D
The graph is not a sinusoid. The high and lowpoints are not symmetric.
14. As B moves from negative values of x to positivevalues of x, the length of AB decreases to about0.56 unit, then begins to increase when the x-value of point B passes about −0.3.Let l = length of AB.
Know: dx
dt= 2 units/s. Want:
dl
dt.
l e xdl
dt
e x xdx
dte
dx
dt
e x
e x
x
x x
x
x
= + ⇒
= + ⋅ +
= +
+
−
0 8 2
0 8 2 1 2 0 8
0 8
0 8 2
1
22 0 8
0 8 2
.
. / .
.
.
( ) .
.
dl
dt = −1.9963… units/s at x = −5 units.
dl
dt = 2.6610… units/s at x = 2 units.
The length of AB is at a minimum when dl/dt = 0.Use your grapher to solve 0.8e0.8x + 2x = 0.At x = −0.3117… , the length of AB stopsdecreasing and starts increasing.
Problem Set 4-10
Review Problems
R0. Answers will vary.
R1. a. x = g(t) = t3 ⇒ g′(t) = 3t2
y = h(t) = cos t ⇒ h′(t) = −sin tIf f (t) = g(t) · h(t) = t3 cos t, then, for example,f ′(1) = 0.7794… by numerical differentiation.g′(1) · h′(1) = 3(12) · (−sin 1) = −2.5244…∴ f ′(t) ≠ g′(t) · h′(t), Q.E .D.
b. If f (t) = g(t)/h(t) = t3/cos t, then, for example,f ′(1) = 8.4349… by numerical differentiation.g′(1)/h′(1) = 3(12)/(−sin 1) = 3.5651…∴ f ′(t) ≠ g′(t)/h′(t), Q.E .D.
c. y = cos tx = t3 ⇒ t = x1/3 ⇒ y = cos (x1/3 )dy
dxx x= − ⋅ −sin ( )/ /1 3 2 31
3
At x = 1,
dy
dx= − ⋅ = −sin 1 0.280490
1
3K .
If x = 1, then t = 11/3 = 1.
∴ = = = −dy dt
dx dt
t
t
/
/
– sin – sin
3
1
30 2804902 . ,K
which equals dy/dx, Q.E.D.
R2. a. If y = uv, then y′ = u′v + uv′.b. See the proof of the product formula in the
text.
c. i. f (x) = x7 ln 3x ⇒f ′(x) = 7x6 ln 3x + x
x7 3
3⋅ = 7x6 ln 3x + x6
ii. g(x) = sin x cos 2x ⇒g′(x) = cos x cos 2x − 2 sin x sin 2x
iii. h(x) = (3x − 7)5(5x + 2)3
h′(x) = 5(3x − 7)4(3) · (5x + 2)3
+ (3x − 7)5(3)(5x + 2)2(5)= 15(3x − 7)4(5x + 2)2(5x + 2
+ 3x − 7)= 15(3x − 7)4(5x + 2)2(8x − 5)
iv. s(x) = x8e− x ⇒ s′(x) = −x8e− x + 8x7e− x
d. f (x) = (3x + 8)(4x + 7)i. f ′(x) = 3(4x + 7) + (3x + 8)(4) = 24x + 53
ii. f (x) = 12x2 + 53x + 56
f ′(x) = 24x + 53, which checks.
76 Problem Set 4-10 Calculus Solutions Manual© 2005 Key Curriculum Press
R3. a. If y = u/v, then yu v uv
v′ = ′ ′–
2 .
b. See proof of quotient formula in text.
c. i. f xx
x( ) = ⇒sin 10
5
f xx x x x
xx x x
x
′ = ⋅ ⋅
=
( )10 10 10 5
10 10 5 10
5 4
10
6
cos – sin
cos – sin
ii. g xx
xg x( ) ( )= + ⇒ ′( )
( – )
2 3
9 5
9
4
= + ⋅ − − + ⋅ − ⋅−
9 2 3 2 9 5 2 3 4 9 5 9
9 5
8 4 9 3
8( ) ( ) ( ) ( )
( )
x x x x
x
= +18 2 3 5 11
9 5
8
5
( ) ( – )
( – )
x x
x
iii. h(x) = (100x3 − 1)− 5 ⇒h ′(x) = −5(100x3 − 1)− 6 · 300x2
= −1500x2(100x3 − 1)− 6
d. y = 1/x10
As a quotient:
yx x
x xx′ = ⋅ ⋅ = = − −0 1 10 10
1010 9
20 1111– –
As a power:
y = x− 10
y′ = −10x− 11, which checks.
e. t xx
xx( ) = =sin
costan
′ =t xx x x x
x( )
cos cos – sin (– sin )
cos2
= + = =cos sin
cos cossec
2 2
2 221x x
x xx
t ′ (1) = sec2 1 = 3.4255…
f. m xt x t
x
x
x( ) = =( ) – ( )
–
tan – tan
–
1
1
1
1
1
1
x
m(x)
3.42...
x m(x)
0.997 3.40959…
0.998 3.41488…
0.999 3.42019…
1 undefined
1.001 3.43086…
1.002 3.43622…
1.003 3.44160…
The values get closer to 3.4255… as xapproaches 1 from either side, Q.E.D.
R4. a. i. y = tan 7x ⇒ y′ = 7 sec2 7x
ii. y = cot (x4) ⇒ y′ = −4x3 csc2
(x4)
iii. y = sec e x ⇒ y′ = e x sec e x tan e x
iv. y = csc x ⇒ y′ = −csc x cot x
b. See derivation in text for tan′x = sec2 x.
c. The graph is always sloping upward, whichis connected to the fact that tan′ x equals thesquare of a function and is thus alwayspositive.
�
x
y
1
d. f (t) = 7 sec t ⇒ f ′(t) = 7 sec t tan t
f ′ (1) = 20.17…
f ′ (1.5) = 1395.44…
f ′ (1.57) = 11038634.0…
There is an asymptote in the secant graph att = π /2 = 1.57079… . As t gets closer to thisvalue, secant changes very rapidly!
R5. a. i. y x yx
= ⇒ ′ =+
−tan 123
3
1 9
ii. d
dxx
x x(sec )
| | –
–1
2
1
1=
iii. c x x c xx
x( ) ( ) ( )2= ⇒ ′ =−cos
– cos
–
–1
1
2
2
1
b. y x yx
= ⇒ ′ =−
−sin 1
2
1
1
1
�/2
x
y
y′ = =( )01
1 01
2–, which agrees with the
graph.
y′ = =( )11
1 1
1
02–, which is infinite.
The graph becomes vertical as x approaches 1from the negative side. y′(2) is undefinedbecause y(2) is not a real number.
R6. a. Differentiability implies continuity.
Calculus Solutions Manual Problem Set 4-10 77© 2005 Key Curriculum Press
b. i. Answers may vary. ii. Answers may vary.
c
x
f(x)
c
x
f(x)
iii. No such function. iv. Answers may vary.
c
x
f(x)
c. i.
1
2
x
f(x)
ii. f is continuous at x = 1 because right andleft limits both equal 2, which equals f (1).
iii. f is differentiable. Left and right limitsof f ′ (x) are both equal to 2, and f iscontinuous at x = 2.
d. g xx x
x ax b x( )
if
if =
≤ ≤+ + ≤
sin ,
,
–1
2
0 1
0
g xx x
x a x′ =
− < <+ <
−
( )if
if
( ) ,
,
/1 0 1
2 0
1 2
limx
g x a b b→ −
= + + =0
0 0( )
lim ( ) sinx
g x→
−+
= =0
1 0 0
∴ b = 0lim
–xg x a a
→′ = + =
00( )
lim /
xg x
→
−+
′ = =0
1 21 1( )
∴ a = 1
10
1
x
g(x)
The graph appears to be differentiable andcontinuous at x = 0.
R7. a. x e y tdy
dx
dy dt
dx dt
t
et
t= = ⇒ = = ⇒2 32
2
3
2,
/
/
d y
dx
d
dx
t
e t
2
2
2
2
3
2=
= ⋅ − ⋅
= − ÷ = −
6 2 3 4
2
3 3 3 3
2
2 2 2
2 2
2
2
2
4
t dt dx e t e dt dx
e
t t
e
dx
dt
t t
e
t t
t
t t
( / ) ( / )
( )
b. x = (t/π) cos t
y = (t/π) sin tdy
dx
dy dt
dx dt
t t t
t t t= = +
+/
/
( / ) sin ( / )(cos )
( / ) cos ( / )(– sin )
1
1
π ππ π
= +sin cos
cos – sin
t t t
t t tWhere the graph crosses the positive x-axis,t = 0, 2π, 4π, 6π, … .If t = 6π, x = 6 and y = 0.∴ (6, 0) is on the graph.If t = 6π, thendy
dx= + = + =sin cos
cos – sin –
6 6 6
6 6 6
0 6
1 06
π π ππ π π
π π .
So the graph is not vertical where it crossesthe x-axis. It has a slope of 6π = 18.84… .
c. At a high point, y is a maximum and x is zero.Use cosine for y and sine for x.For y, the sinusoidal axis is at 25 ft.For x, the sinusoidal axis is at 0 ft.Both x and y have amplitude 20 ft, the radiusof the Ferris wheel.The phase displacement is 3 seconds.The period is 20 seconds, so the coefficientof the arguments of sine and cosine is2π/20 = π /10.
x t= 2010
3sin ( – )π
y t= +25 2010
3cos ( – )π
dx dt t/ = 210
3π πcos ( – )
dy dt t/ = −210
3π πsin ( – )
When t = 0, dy/dt = 5.0832… .The Ferris wheel is going up at about5.1 ft/s.When t = 0, dx/dt = 3.6931… .The Ferris wheel is going right at about3.7 ft/s.dy
dx
dy dt
dx dt= /
/dy/dx will be infinite if dx/dt = 0 anddy/dt ≠ 0.
dx/dt = 0 if 210
3 0π πcos ( – )t = .
π π π10
32
( – )t n= + (where n is an integer)
t = 8 + 5n
The first positive time is t = 8 s.
78 Problem Set 4-10 Calculus Solutions Manual© 2005 Key Curriculum Press
R8. a. y = x8/5 ⇒ y5 = x8
5 88
5
8
5
8
54 7
7
4
7
8 5 43 5y y x y
x
y
x
xx′ = ⇒ ′ = = =
( )//
Using the power rule directly:
y x y x= ⇒ ′ =8 5 85
3 5/ /
b. y3 sin xy = x4.5 ⇒
3y2y′ · sin xy + y3(cos xy)(y + xy′) = 4.5x3.5
y′[3y2 sin xy + xy3
cos xy]
= 4.5x3/5 − y4 cos xy
ydy
dx
x y xy
y xy xy xy′ = =
+4 5
3
3 5 4
2 3
. – cos
sin cos
.
c. i. 4y2 − xy2 = x3 ⇒8yy′ − y2 − x · 2yy′ = 3x2
y′(8y − 2xy) = 3x2 + y2
ydy
dx
x y
y xy′ = = +3
8 2
2 2
–At (2, 2), dy/dx = 2. At (2, −2), dy/dx = −2.Lines at these points with these slopes aretangent to the graph (see diagram).
x
y
5
5
2
ii. At (0, 0), dy/dx has the indeterminate form0/0, which is consistent with the cusp.
iii. To find the asymptote, solve for y.(4 − x)y2 = x3
yx
x2
3
4=
–As x approaches 4 from the negative side,y becomes infinite. If x > 4, y2 is negative,and thus there are no real values of y.Asymptote is at x = 4.
R9.
70
x
z
Let x = Rover’s distance from the table.Let z = slant length of tablecloth.
Know:dx
dt= 20 cm/s. Want:
dz
dt at z = 200.
z2 = x2 + 702
2 2zdz
dtx
dx
dt =
dz
dt
x
z
dx
dt
x
z= = 20
At z = 200, x = =200 70 30 392 2–
dz
dt= = = . .
20 30 39
2003 39 18 7349
⋅..
The glass moves at the same speed as thetablecloth, or about 18.7 cm/s, which is about1.3 cm/s slower than Rover.
Concept Problems
C1. a. Let (x, y) be the coordinates of a point on thetangent line.y y
x xm y m x x y
–
–0
00 0= ⇒ = − +( )
b. Substituting (x1, 0) for (x, y) gives
0 1 0 0 1 00= − + ⇒ = −m x x y x x
y
m( ) , Q.E.D.
c. The tangent line intersects the x-axis at (x2, 0).Repeating the above reasoning with x2 and x1
in place of x1 and x0 gives
x xy
m2 11= −
Because y1 = f (x1) and m = f ′(x1),
x xf x
f x2 11
1
= −′( )
( ), Q.E.D.
d. Programs will vary according to the kind ofgrapher used. The following steps are needed:
• Store f (x) in the Y= menu.
• Input a starting value of x.
• Find the new x using the numericalderivative.
• Display the new x.
• Save the new x as the old x and repeat.
For f (x) = x2 − 9x + 14, the program shouldgive x = 2, x = 7.
e. For g(x) = x3 − 9x2 + 5x + 10, first plot thegraph to get approximations for the initialvalues of x.
20
1
x
g(x)
Run the program three times with x0 = −1, 1,and 8. The values of x are
x = −0.78715388…
x = 1.54050386…
x = 8.24665002…
The answers are the same using the built-insolver feature. The same preliminary analysisis needed to find starting values of x.
Calculus Solutions Manual Problem Set 4-10 79© 2005 Key Curriculum Press
f. f (x) = sec x − 1.1
Starting with x0 = 1, it takes seven iterationsto get x = 0.429699666… .
C2. a. The connecting rod, the crankshaft, and they-axis form a triangle with angle φ = θ − π /2included between sides of 6 cm and(y − 8) cm.
8
y – 820
6
θφ
By the law of cosines,
202 = (y − 8)2 + 62 − 2 · 6 ·(y − 8) cos (θ − π /2)
202 = (y − 8)2 + 62 − 12(y − 8) sin θ(y − 8)2 − 12 sin θ (y − 8) − 364 = 0
Solve for y − 8 using the quadratic formula.
y − = + ⋅8 6 36 3642sin sin .θ θ( + ) (The
solution with the negative radical gives atriangle below the origin, which has no real-life meaning.)
y = + + ⋅8 6 2 9 912sin sinθ θ +
b. vdy
dt
d
dt
d
dt= = +
⋅6
18
9 912cos
sin cos
sinθ θ θ θ
θθ
+
vd
dt
d
dt= +
⋅6
9 2
9 912cos
sin
sinθ θ θ
θθ
+
c. ad y
dt=
2
2
= −
6 sin θ θd
dt
2
+
18
91 2 9
9 91
4
2 3 2
2cos sin
sin
θ θθ
θ–
( + ) /
d
dt
= ⋅
18
91 2 9
9 916
4
2 3 2
2cos sin
sinsin
θ θθ
θ θ–
( + )–/
d
dt
(There are many other correct forms of theanswer, depending on how you use thedouble-argument properties and Pythagoreanproperties from trigonometry.)Note that the angular velocity is constant at6000π radians per minute, sod
dt
θ π= 100 rad/s.
d. See the graph. Note that a line at a = −980is so close to the x-axis that it does notshow up.
500,000
3
θ
a
Solving graphically and numerically,a < −980 for θ ∈ (0.2712… , 2.8703…). Thepiston is going down (v < 0) forθ ∈ (π /2, 3π /2).So the piston is going down with accelerationgreater than gravity for θ between π /2 and2.8703… .
Chapter Test
T1. y = uv ⇒ y′ = u′v + uv′
T2. ′ =
+ ∆+ ∆
−
∆⋅ + ∆
+ ∆
∆ →
y
u uv v
uv
x
v v v
v v vxlim
( )
( )0
= + ∆ − + ∆∆ + ∆
∆ →
lim( ) ( )
( )x
u u v u v v
x v v v0
= + ∆ − − ∆∆ + ∆
∆ →
lim( )x
uv uv uv u v
x v v v0
=+ ∆
⋅ ∆ ⋅ − ∆∆
∆ →
lim( )x v v v
u v u v
x0
1
=+ ∆
⋅ ∆∆
− ∆∆
∆ →
lim( )x v v v
u
xv u
v
x0
1
= −
= ′ − ′1
2 2v
du
dxv u
dv
dx
u v uv
v[Because as ∆x → 0, u/x and ∆ ∆v x/ become
du/dx and dv/dx and v → 0, so ( ) ]v v v v+ →∆ 2 .
T3. cotcos
sinx
x
x=
= − −sin sin cos cos
sin
x x x x
x2
= − + = − = −(sin cos )
sin sincsc
2 2
2 221x x
x xx
T4. y x y x y y= ⇒ = ⇒ ′ = ⇒−sin sin cos1 1
yy
y y y y′ = + = ⇒ = − ⇒11 12 2 2 2
coscos sin cos sin
cos siny y x yx
= − = − ⇒ =−
1 11
1
2 2
2
T5.dy
dx
dy dt
dx dt
t
tt
d y
dx
d
dxt= = = = =/
/
4
22 2
32
2
22( )
4 44
22t dt dx t
dx
dt
t
t( / ) = ÷ = =
80 Problem Set 4-10 Calculus Solutions Manual© 2005 Key Curriculum Press
T6. c(x) = cot 3x
c′(x) = −3 csc2 3x, which is negative for all
permissible values of x.c′(5) = −3 csc2 15 = −3/sin2 15 = −7.0943…
c(t) is decreasing at about 7.1 y-units/x-unit.
T7. f (x) = sec x ⇒ f ′ (x) = sec x tan x
f ′ (2) = sec 2 tan 2 = 5.25064633…
Use m(x) for the difference quotient.
m xx
x( ) = 1 1 2
2
/cos – /cos
–
x m(x)
1.997 5.28893631…
1.998 5.27611340…
1.999 5.26335022…
2.000 undefined
2.001 5.23800134…
2.002 5.22541482…
2.003 5.21288638…
T8. Answers may vary.
x
f(x)
2
7
1–2
T9. f (x) = mx + b
f ′ (x) = m for all x
∴ f is differentiable for all x.
∴ f is continuous for all x, Q.E.D.
T10. f (x) = sec 5x ⇒ f ′(x) = 5 sec 5x tan 5x
T11. y = tan7/3 x ⇒ y′ = 73 tan4/3 x
T12. f (x) = (2x – 5)6(5x – 1)2
f ′ (x) = 6(2x – 5)5(2) ⋅ (5x – 1)2
+ (2x – 5)6 ⋅ 2(5x – 1) ⋅ 5= 2(2x – 5)5(5x – 1)[6(5x – 1) + 5(2x – 5)]
= 2(2x – 5)5(5x – 1)(40x – 31)
T13. ( )f xe
x
x
= ⇒3
ln
′ = − = −f x
e x e x
x
xe x e
x x
x x x x
( )ln ( / )
(ln )
ln
(ln )
3 1 33 3
2
3 3
2
T14. x = sec 2t
y = tan 2t3
dy
dx
dy dt
dx dt
t t
t t
t t
t t= = ⋅
⋅=/
/
sec
sec tan
sec
sec tan
2 3 2 2 2 32 6
2 2 2
3 2
2 2
T15. y = 4 sin− 1 (5x3)
yx
xx
x′ = ⋅ ⋅ =4
1
1 515
60
1 253 2
22
6– ( ) –T16. 9x2 − 20xy + 25y2 − 16x + 10y − 50 = 0 ⇒
18x − 20y − 20xy ′ + 50yy ′ − 16 + 10y ′ = 0y ′(−20x + 50y + 10) = −18x + 20y + 16
ydy
dx
x y
x y′ = = + +
+ +–
–
18 20 16
20 50 10
= + ++ +
–
–
9 10 8
10 25 5
x y
x y
If x = −2, then
36 + 40y + 25y2 + 32 + 10y − 50 = 0
25y2 + 50y + 18 = 0
Solving numerically givesy = −0.4708… or y = −1.5291… , both ofwhich agree with the graph.
(Solving algebraically by the quadratic formula,
y = − ±1 7 5 / , which agrees with the numericalsolutions.)
At (−2, −0.4708…), dy/dx = 1.60948… .
At (−2, −1.5291…), dy/dx = −0.80948… .
The answers are reasonable, because lines ofthese slopes are tangent to the graph at therespective points, as shown here.
x
y
5
5
–2
T17. f xx x
a x b x( )
if
if=
+ ≤+ >
3
2
1 1
2 1
,
( – ) ,
f xx x
a x x′ =
<>
( )if
if
3 1
2 2 1
2,
( – ),
For equal derivatives on both sides of x = 1,
lim–x
f x→
′ = ⋅ =1
23 1 3( )
limx
f x a a→ +
′ = − = −1
2 1 2 2( ) ( )
∴− = ⇒ = −2 3 1 5a a .
For continuity at x = 1,
lim–x
f x→
= + =1
31 1 2( )
limx
f x a b a b→ +
= − + = +1
1 2( ) ( )2
∴ + =a b 2
Substituting a = −1.5 gives b = 3.5.
Calculus Solutions Manual Problem Set 4-10 81© 2005 Key Curriculum Press
The graph shows differentiability at x = 1.
1
2
x
f(x)
Values of b other than 3.5 will still cause thetwo branches to have slopes approaching 4as x approaches 1 from either side as long asa = −1.5. However, f will not be continuous,and thus will not be differentiable, as shownhere for b = 4.5.
1
2
x
f(x)
T18. y = x7/3 ⇔ y3 = x7
3y2y′ = 7x6
yx
y
x
xx x′ = = = =−7
3
7
3
7
3
7
3
6
2
6
7 3 26 14 3 4 3
( )// /
This answer agrees with y′ = nxn− 1. 4/3 is7/3 − 1.
T19. cot = adjacent/opposite = x/5 = cot− 1 (x/5)
T20.d
dx x x
θ = −+
⋅ = −+
1
1 5
1
5
1
5 5 252 2( / ) ( / )
= −+
= −+
1
5 5
5
252 2( / )x x
T21.dx
dt= −420 mi/h
T22.d
dt
d
dx
dx
dt x x
θ θ= ⋅ = −+
⋅ − =+
5
25420
2100
252 2( )
T23. The plane is changing fastest when x approacheszero, when the plane is nearest the station.
30
25
x
y
T24. Answers will vary.
82 Problem Set 5-2 Calculus Solutions Manual© 2005 Key Curriculum Press
Chapter 5—Definite and Indefinite Integrals
Problem Set 5-11. f (1000) = 20 + (0.000004)(10002) = 24 $/ft
f (4000) = 20 + (0.000004)(40002) = 84 $/ft
The price increases because it is harder and slowerto drill at increasing depths.
2. T6 = 500(24 + 29)/2 + 500(29 + 36)/2
+ 500(36 + 45)/2 + 500(45 + 56)/2
+ 500(56 + 69)/2 + 500(69 + 84)/2= 13,250 + 16,250 + 20,250 + 25,250
+ 31,250 + 38,250 = 144,500
About $144,500, an overestimate because thetrapezoids are circumscribed above the curve.(Note that T6 can be found more easily by firstfactoring out the 500, then adding the functionvalues.)
3. R6 = 500(26.25) + 500(32.25) + 500(40.25) +500(50.25) + 500(62.25) + 500(76.25) = 143,750
About $143,750(Note that R6 can be found more easily by firstfactoring out the 500, then adding the functionvalues.)R6 is close to T6. (They differ by less than 1%.)
4. T100 = 144,001.8, T500 = 144,000.072
Conjecture: Exact value is $144,000.
5. g (x) = 20x + 13 0 000004( . )x3 + C
g (4,000) − g (1,000) = (165,333.3333… + C) − (21,333.3333… + C) = 144,000,
which is the conjectured value of the definiteintegral!The other name for antiderivative is indefiniteintegral.
6. a. f (x) = x7 + C b. y = − cos x + C
c. u = 0.5e2x + C d. v = 132 (4x + 5)8 + C
Problem Set 5-2Q1. Answers may vary.
y
x
Area = productof x and y
a b
Q2. Instantaneous rate of change
Q3. f ′ (x) = −(ln 2)2− x Q4. y = sin x + C
Q5. y′ = 4 m/s Q6. sec x tan x (derivative)
Q7. 1 Q8. That constant.
Q9. 0 Q10. B
1. f (x) = 0.2x4 ⇒ f ′ (x) = 0.8x3 ⇒ f ′ (3) = 21.6;f (3) = 0.2(34) = 16.2
∴ y − 16.2 = 21.6(x − 3) ⇒ y = 21.6x − 48.6
x f (x) y Error
3.1 18.47042 18.36 0.11042
3.001 16.22161… 16.2216 0.0000108…
2.999 16.17841… 16.1784 0.0000107…
2. g (x) = sec x ⇒ g ′ (x) = sec x tan x ⇒g ′ (π /3) = 2 3 = 3.464…
g (π /3) = sec (π /3) = 2
Linear function is y − 2 = 2 3 (x − π /3) ⇒y = 2 3 (x − π /3) + 2.
dx f (x) y Error
0.04 2.15068… 2.13856… 0.01212…
− 0.04 1.87184… 1.86143… 0.01041…
0.001 2.003471… 2.003464… 7.01 × 10− 6
3. a. f (x) = x2 ⇒ f ′ (x) = 2x ⇒ f ′ (1) = 2
Tangent line: y − 1 = 2(x − 1) ⇒ y = 2x − 1The graph shows a zoom by factor of 10.
1
1
graph tangent line
Local linearity describes the property of thefunction because if you keep x close to 1 (inthe “locality” of 1), the curved graph of thefunction looks like the straight graph of thetangent line.
b.
x f (x) y Error, f (x) − y
0.97 0.9409 0.94 0.0009
0.98 0.9604 0.96 0.0004
0.99 0.9801 0.98 0.0001
1 1 1 0
1.01 1.0201 1.02 0.0001
1.02 1.0404 1.04 0.0004
1.03 1.0609 1.06 0.0009
Calculus Solutions Manual Problem Set 5-2 83© 2005 Key Curriculum Press
The table shows that for x-values close to 1 (thepoint of tangency), the tangent line is a closeapproximation to the function values.
4. f (x) = x2 − 0.1(x − 1)1/3
Zooming in on (1, 1) shows that the graph goesvertical at x = 1. This observation is confirmedalgebraically.f ′ (x) = 2x − (1/3)(0.1)(x − 1)−2/3
f ′ (1) = 2 − (1/3)(0.1)(0)−2/3 , which is infinite.
graphtangent line
1
1
f does not have local linearity at x = 1. Becausethe slope of the graph becomes infinite, no linearfunction can approximate the graph there. If f isdifferentiable at x = c, then f is locally linearthere. The converse is also true. If f is locallylinear at x = c, then f is differentiable there.
5. a. Let A be the number of radians in θ degrees.
By trigonometry, tan A = x
100 ⇒
A = tan− 1 x
100.
Because 1 radian is 180/π degrees,
θ = 180
π tan− 1
x
100, Q.E .D.
b. dθ = 180
π ·
1
1 100 2+ ( / )x ·
1
100 dx
= 1 8
1 100 2
. /
( / )
π+ x
dx
x = 0: dθ = 0.5729… dx
x = 10: dθ = 0.5672… dx
x = 20: dθ = 0.5509… dx
c. At x = 0, θ = 0. For x = 20, dx = 20.∴ θ ≈ (0 + 0.5729…)(20) = 11.459…°The actual value is (180/π)(tan− 1 0.2) =11.309… .The error is 0.1492…°, which is about 1.3%.
d. 0.5729… is approximately 0.5, so multiplyingby it is approximately equivalent to dividingby 2. For a 20% grade, this estimate gives 10°,compared to the actual angle of 11.309…°, anerror of about 11.6%. For a 100% grade, thisestimate gives 50°, compared to the actualangle of 45°, an error of about 11.1%.
6. dV = 4π r 2 drdr = 0.03 and r = 6, so dV = 4π (62)(0.03) =4.32π ≈ 13.57 mm3
V ≈ 43 π (63) + 4.32π = 288π + 4.32π =
292.32π ≈ 918.350 mm3
Actual volume is V = 4
3π (6.033) =
292.341636π ≈ 918.418 mm3.∆V = 4
33 4
336 03 6π π( . ) ( )− = 4.341636π ≈
13.640 mm3
Error is 292.32π − 292.341636π = − 0.021636π,or about 0.068 mm3 too low.
7. a. (6000 · 0.05)/365 = 0.8219… , or about82 cents.
b. m = 6000e (0.05/365)t ⇒dm = 6000(0.05/365)e(0.05/365)t dtSubstituting t = 0 and dt = 1 gives dm =0.8219… , the same as part a.Substituting t = 0 and dt = 30 gives dm =24.6575… ≈ $24.66.Substituting t = 0 and dt = 60 gives dm =49.3150… ≈ $49.32.
c. t = 1: ∆m = 6000e (0.05/365)(1) − 6000 =0.8219… , almost exactly equal to dm.t = 30: ∆m = 6000e (0.05/365)(30) − 6000 =24.7082… , about 5 cents higher than dm.t = 60: ∆m = 6000e (0.05/365)(60) − 6000 =49.5182… , about 20 cents higher than dm.As t increases, dm is a less accurateapproximation for ∆m.
8. a. dS = − 1.636 dtMarch 11: dS = − 1.636(10) = − 16.36minutesSunrise time ≈ 6:26 − 0:16 = 6:10 a.m.,which agrees with the tabulated value.March 21: dS = − 1.636(20) = − 32.72 minutesSunrise time ≈ 6:26 − 0:33 = 5:53 a.m.,which agrees with the tabulated value.
b. By September 1, t = 185, givingdS = − 1.636(185) = − 302.66, or 5:04 hours.
So the predicted sunrise time would be6:26 − 5:04 = 1:22 a.m. Because the sunrisereaches its earliest in mid-June, the timepredicted by dS is not reasonable.
9. y = 7x3 ⇒ dy = 21x2 dx
10. y = − 4x11 ⇒ dy = − 44x10 dx
11. y = (x4 + 1)7 ⇒ dy = 28x3(x4 + 1)6 dx
12. y = (5 − 8x)4 ⇒ dy = − 32(5 − 8x)3 dx
13. y = 3x2 + 5x − 9 ⇒ dy = (6x + 5) dx
14. y = x2 + x + 9 ⇒ dy = (2x + 1) dx
15. y = e− 1.7x ⇒ dy = − 1.7e− 1.7x dx
84 Problem Set 5-3 Calculus Solutions Manual© 2005 Key Curriculum Press
16. y = 15 ln x1/3 ⇒ dy = 15 1
3
51 3
2 3
xx
xdx/
/⋅ =−
17. y = sin 3x ⇒ dy = 3 cos 3x dx
18. y = cos 4x ⇒ dy = − 4 sin 4x dx
19. y = tan3 x ⇒ dy = 3 tan2 x sec2 x dx
20. y = sec3 x ⇒ dy = 3 sec3 x tan x dx
21. y = 4x cos x ⇒ dy = (4 cos x − 4x sin x) dx
22. y = 3x sin x ⇒ dy = (3 sin x + 3x cos x) dx
23. y = x2/2 − x/4 + 2 ⇒ dy = (x − 1/4) dx
24. y = x3/3 − x/5 + 6 ⇒ dy = (x2 − 1/5) dx
25. y = cos (ln x) ⇒ dy = −sin(ln )x
xdx
26. y = sin (e0.1 x) ⇒ dy = 0.1e0.1 x cos (e0.1 x) dx
27. dy = 20x3 dx ⇒ y = 5x4 + C
28. dy = 36x4 dx ⇒ y = 7.2x5 + C
29. dy = sin 4x dx ⇒ y = − (1/4) cos 4x + C
30. dy = cos 0.2x dx ⇒ y = 5 sin 0.2x + C
31. dy = (0.5x − 1)6 dx ⇒ y = (2/7)(0.5x − 1)7 + C
32. dy = (4x + 3)− 6 dx ⇒ y = (− 1/20)(4x + 3)− 5 + C
33. dy = sec2 x dx ⇒ y = tan x + C
34. dy = csc x cot x dx ⇒ y = − csc x + C
35. dy = 5 dx ⇒ y = 5x + C
36. dy = − 7 dx ⇒ y = − 7x + C
37. dy = (6x2 + 10x − 4) dx ⇒ y = 2x3 + 5x2 − 4x + C
38. dy = (10x2 − 3x + 7) dx ⇒y = (10/3)x3 − (3/2)x2 + 7x + C
39. dy = sin5 x cos x dx ⇒ y = (1/6) sin6 x + C
40. dy = sec7 x tan x dx = sec6 x(sec x tan x dx) ⇒ y = (1/7) sec7 x + C
41. a. y = (3x + 4)2(2x − 5)3 ⇒y′ = 2(3x + 4)(3)(2x − 5)3
+ (3x + 4)2 ⋅ 3(2x − 5)2 ⋅ 2= 6(3x + 4)(2x − 5)2[2x − 5 + 3x + 4]
∴ dy = 6(3x + 4)(2x − 5)2(5x − 1) dx
b. dy = 6(7)(− 3)2(4)(− 0.04) = − 60.48
c. x = 1 ⇒ y = (7)2(− 3)3 = − 1323
x = 0.96 ⇒ y = − 1383.0218…
∴ ∆y = − 1383.0218… − (− 1323)
= − 60.0218…
d. − 60.48 is close to − 60.0218… .
42. a. y = sin 5x ⇒ dy = 5 cos 5x dx
b. dy = 5 cos (5π /3) ⋅ 0.06 = 0.15
c. x = π /3 ⇒ y = sin (5π /3) = − 3 /2
= − 0.86602…
x = π /3 + 0.06 ⇒ y = − 0.679585565…
∴ ∆y = − 0.679… − (− 0.866…)
= 0.186439…
d. 0.15 is (fairly) close to 0.186439... .
Problem Set 5-3Q1. Antiderivative = x3 + C
Q2. Indefinite integral = (1/6)x6 + C
Q3. y′ = 3x2 Q4. y′ = ln 3(3x)
Q5. dy = ln 3(3x) dx Q6. ′′ =y x5 4
Q7. Integral = sin x + C Q8. y′ = − sin x
Q9. 1 Q10. E
1. x dx x C10 111
11= +∫
2. x dx x C20 211
21= +∫
3. 44
56 5x dx x C− −= − +∫
4. 9 7 6x dx x C− −= − +∫ 3
2
5. cos sinx dx x C = +∫6. sin cosx dx x C = − +∫7. 4 7
4
77cos sinx dx x C = +∫
8. 20 920
99sin cosx dx x C = − +∫
9. 55
0 30 3 0 3e dx e Cx x. .
.= +∫
10. 2 2000 01 0 01e dx e Cx x− −= − +∫ . .
11. 44
4m
m
dm C= +∫ ln
12. 8 48 4
8 4.
.rr
dr C= +∫ ln .
13. ( ) ( ) ( )4 91
44 9 42 2v dv v dv+ = +∫∫
= +1
124 9( )v 3 + C
14. ( ) ( ) ( )5 53 171
33 17 3p dp p dp+ = +∫∫
= +1
183 17( )p 6 + C
15. ( ) ( ) ( )38 51
58 5 53− = − − −∫ ∫x dx x dx
= − +1
208 5 4( – )x C
16. ( ) ( ) ( ) ( )4 420 1 20− = − − −∫∫ x dx x dx
= − 1
520( – )x 5 + C
Calculus Solutions Manual Problem Set 5-3 85© 2005 Key Curriculum Press
17. ( ) 6sin cos sinx x dx x C= +∫ 1
77
18. ( ) 8cos sin cosx x dx x C= − +∫ 1
99
19. cos sin cos4 51
5θ θ θ θ d C= − +∫
20. sin cos sin5 61
6θ θ θ θ d C= +∫
21. ( )x x dx x x x C2 3 23 51
3
3
25+ − = + − +∫
22. ( ) x x dx x x x C2 3 24 11
32− + = − + +∫
23. ( ) ( ) x dx x x x dx2 3 6 4 25 15 75 125+ = + + +∫∫= 1
7x7 + 3x5 + 25x3 + 125x + C
24. ( ) ( )x dx x x dx3 2 6 36 12 36− = − +∫∫= 1
7x7 − 3x4 + 36x + C
25. e x x dx e Cx xsec secsec tan = +∫26. e x dx e Cx xtan tansec2 = +∫27. sec tan2 x dx x C = +∫28. csc cot2 x dx x C = − +∫29. tan sec tan7 2 81
8x x dx x C = +∫
30. cot csc cot8 2 91
9x x dx x C = − +∫
31. csc cot csc csc cot9 8x x dx x x x dx ( )= ∫∫= − 1
9csc9 x + C
32. sec tan sec sec tan7 6x x dx x x x dx ( )= ∫∫= 1
7sec7 x + C
33. v (t) = 40 + 5 t = 40 + 5t1/ 2
D t t dt t t C( ) ( )= + = + +∫ 40 5 4010
31 2 3 2/ /
D (0) = 0 ⇒ 0 = 40 ⋅ 0 + 10
303/ 2 + C ⇒ C = 0
∴ D (t) = 40t + 10
3t3/ 2
D (10) = 505.4092... ≈ 505 ft
34. a. f (x) = 0.3x2 + 1T100 = 9.300135
b. g x x dx x x C( ) ( . ) .= + = + +∫ 0 3 1 0 12 3
c. g (4) − g (1) = 6.4 + 4 + C − 0.1 − 1 − C =9.3, which is about equal to the definiteintegral! It is also interesting that the constantC drops out.
35. Prove that if f and g are functions that can beintegrated, then
[ ( ) ( )] ( ) ( ) .f x g x dx f x dx g x dx+ = + ∫∫∫Proof:
Let h x f x dx g x dx( ) ( ) ( ) .= + ∫∫By the derivative of a sum property,
h xd
dxf x dx
d
dxg x dx′ = + ∫∫( ) ( ) ( ) .
By the definition of indefinite integral appliedtwice to the right side of the equation,h ′ (x) = f (x) + g (x).By the definition of indefinite integral applied inthe other direction,
h x f x g x dx( ) [ ( ) ( )] = +∫By the transitive property, then,
[ ( ) ( )] ( ) ( ) , f x g x dx f x dx g x dx+ = +∫ ∫ ∫Q.E.D.
36. Calvin says x x dx x x x Ccos sin cos= + +∫ .
Phoebe checked this by differentiating:d
dxx x x C( sin cos )+ +
= 1 ⋅ sin x + x ⋅ (cos x) − sin x + 0 = x cos x
By the definition of indefinite integral, she knewthat Calvin was right.
37. a.
C v (t)
1.5 12.25
2.5 16.25
3.5 22.25
Sum: 50.75
Integral ≈ 50.75
b.
c v (t)
1.25 11.5625
1.75 13.0625
2.25 15.0625
2.75 17.5625
3.25 20.5625
3.75 24.0625
Sum: 101.8750
Integral ≈ (101.8750)(0.5) = 50.9375
86 Problem Set 5-4 Calculus Solutions Manual© 2005 Key Curriculum Press
c. As shown in Figures 5-3c and 5-3d, theRiemann sum with six increments hassmaller regions included above the graph andsmaller regions excluded below the graph, sothe Riemann sum should be closer to theintegral.
d. Conjecture: Exact value is 51.By the trapezoidal rule with n = 100,integral ≈ 51.00045, which agrees with theconjecture.
e. The integral is the product of v (t) and t, andthus has the units (ft/min)(min), or ft. So theobject went 51 ft. Average velocity = 51/3 =17 ft/min.
38. Answers will vary.
Problem Set 5-4Q1. y′ = sin x + x cos x
Q2. tan x + C
Q3. f ′(x) = sec2 x
Q4. (1/4)x4 + C
Q5. z′ = − 7 sin 7x
Q6. − cos u + C
Q7. Limit = 8
Q8.
x
y
4
7
Q9. If a + b = 5, then a = 2 and b = 3.
Q10. No
1. x dx2
1
4
∫ 2. x dx3
2
6
∫c f (c) c f (c)
1.25 1.5625 2.25 11.390625
1.75 3.0625 2.75 20.796875
2.25 5.0625 3.25 34.328125
2.75 7.5625 3.75 52.734375
3.25 10.5625 4.25 76.765625
3.75 14.0625 4.75 107.171875
Sum = 41.8750 5.25 144.703125
R6 = (0.5)(41.875) 5.75 190.109375
= 20.9375 Sum = 638.000000
R8 = (0.5)(638) = 319
3. 31
3x dx
−∫ 4. 21
2x dx
−∫c f (c) c f (c)
− 0.75 0.43869… − 0.75 0.59460…
− 0.25 0.75983… −0.25 0.84089…
0.25 1.31607… 0.25 1.18920…
0.75 2.27950… 0.75 1.68179…
1.25 3.94822… 1.25 2.37841…
1.75 6.83852… 1.75 3.36358…
2.25 11.84466… Sum = 10.04849…
2.75 20.51556… R6 = (0.5)(10.04…)
Sum = 47.94108… = 5.024249…
R8 = (0.5)(47.94…)
= 23.97054…
5. sin x dx1
2
∫ 6. cos x dx 0
1
∫ c f (c) c f (c)
1.1 0.891207… 0.1 0.995004…
1.3 0.963558… 0.3 0.955336…
1.5 0.997494… 0.5 0.877582…
1.7 0.991664… 0.7 0.764842…
1.9 0.946300… 0.9 0.621609…
Sum = 4.790225… Sum = 4.214375…
R5 = (0.2)(4.79…) R5 = (0.2)(4.21…)
= 0.958045… = 0.842875…
7. tan.
.
x dx0 4
1 2
∫L4 = 0.73879… , U4 = 1.16866…M4 = 0.92270… , T4 = 0.95373…∴ M4 and T4 are between L4 and U4, Q.E.D.
8. 101
3
∫ / :x dx
L4 = 9.5, U4 = 12.8333…M4 = 10.89754… , T4 = 11.1666…∴ M4 and T4 are between L4 and U4, Q.E.D.
9. ln x dx1
5
∫ is underestimated by the trapezoidal
rule and overestimated by the midpoint rule.
5
2
y
x
5
2
y
x
Calculus Solutions Manual Problem Set 5-5 87© 2005 Key Curriculum Press
10. e dxx 0
2
∫ is overestimated by the trapezoidal rule
and underestimated by the midpoint rule.
1
4
y
x
1
4
y
x
11. a. h (x) = 3 + 2 sin xFor an upper sum, take sample points at xequals 1, π/2, 2, 3, 4, and 6.
b. For a lower sum, take sample points at xequals 0, 1, 3, 4, 3π/2, and 5.
c. U6 = 1[h (1) + h (π/2) + h (2) + h (3) + h (4)+ h (6)] = 21.71134…
L6 = 1[h (0) + h (1) + h (3) + h (4) + h (3π/2)+ h (5)] = 14.53372…
12. Programs will vary depending on the type ofgrapher used. See the program in the Programsfor Graphing Calculators section of theInstructor’s Resource Book.
13. a. For x dx2
1
4
,∫ the program should give the
values listed in the text.
b. L100 = 20.77545, L500 = 20.955018.Ln seems to be approaching 21.
c. U100 = 21.22545, U500 = 21.045018.Un also seems to be approaching 21.f is integrable on [1, 4] if Ln and Un have thesame limit as n approaches infinity.
d. The trapezoids are circumscribed around theregion under the graph and thus contain morearea (see left diagram). For rectangles, the“triangular” part of the region that is left outhas more area than the “triangular” part that isadded, because the “triangles” have equal basesbut unequal altitudes (see right diagram).
Trapezoidincludesmore area.
x
y
Rectangleleaves outmore area.
x
y
14. a. x dx2
0
3
∫U100 = 9.13545, L100 = 8.86545
Conjecture: Integral equals 9 exactly.
b. The sample points will be at the right of eachinterval, 1 ⋅ 3/n, 2 ⋅ 3/n, 3 ⋅ 3/n, . . . ,n ⋅ 3/n.
c. Un = (3/n)(1 ⋅ 3/n)2 + (3/n)(2 ⋅ 3/n)2
+ (3/n)(3 ⋅ 3/n)2 + ⋅ ⋅ ⋅ + (3/n)(n ⋅ 3/n)2
d. Un = (3/n)3(12 + 22 + 32 + ⋅ ⋅ ⋅ + n2)
= (3/n)3(n/6)(n + 1)(2n + 1)
= (4.5/n2)(n + 1)(2n + 1)
U100 = (4.5/1002)(101)(201) = 9.13545, whichis correct.
e. Using the formula, U1000 = 9.013504… ,which does seem to be approaching 9
f. Un
n
n
nn = ⋅ + ⋅ +4 5
1 2 1.
= 4.5(1 + 1/n)(2 + 1/n)
As n approaches infinity, 1/n approaches zero.∴ Un approaches 4.5(1 + 0)(2 + 0),which equals 9, exactly!
15. x dx3
0
2
∫Find an upper sum using the sample points
1 ⋅ 2/n, 2 ⋅ 2/n, 3 ⋅ 2/n, . . . , n ⋅ 2/n.
Un = (2/n)(1 ⋅ 2/n)3 + (2/n)(2 ⋅ 2/n)3
+ (2/n)(3 ⋅ 2/n)3 + ⋅ ⋅ ⋅ + (2/n)(n ⋅ 2/n)3
= (2/n)4(13 + 23 + 33 + ⋅ ⋅ ⋅ + n3)
= (2/n)4[(n/2)(n + 1)]2
= 4/n2 ⋅ (n + 1)2 = 4(1 + 1/n)2
lim ( )n
nU→∞
= + =4 1 0 42
Problem Set 5-5
Q1. x2/2 + 2x + C Q2.10
10
t
Cln
+
Q3. − cot x + C Q4. − csc x cot x
Q5.5 4(ln )x
xC+ Q6.
1 2
x
y
Q7. Answers may vary. Q8. Answers may vary.
x
y5
3
x
y
1
2
Q9. No limit (infinite) Q10. D
1. See the text for the statement of the mean valuetheorem.
2. See the text for the statement of Rolle’stheorem.
88 Problem Set 5-5 Calculus Solutions Manual© 2005 Key Curriculum Press
3. g (x) = 6/x; [1, 4]
1 4
6
2
g(x)
x
c
m = = −6 4 6
4 13 2
/ –
–/
g ′ (x) = − 6x− 2
∴ − 6c− 2 = − 3/2 ⇒ c = 2
Tangent at x = 2 parallels the secant line.
4. f (x) = x4; [− 1, 2]
f(x)
–1 2
10
x
c
m = =16 1
2 15
–
– (– )
g ′ (x) = 4x3
∴ = ⇒ = =4 5 5 4 1 0773 3c c / . K
Tangent at x = 1.077… parallels the secant line.
5. c x x( ) ; , = +
2 02
cosπ
1
x
c(x)
/2c
m = = − =cos ( / ) – cos
/ –
ππ
π2 0
2 02 0 6366/ . K
c′(x) = − sin x
∴ − sin c = − 2/π ⇒ c = 0.69010…
Tangent at x = 0.690… parallels the secant line.
6. h x x( ) ; [ , ]= −5 1 9
c1 9
5h(x)
x
m = = −2 4
9 11 4
–
–/
h ′ (x) = − (1/2)x− 1/2
∴ (− 1/2)c− 1/2 = − 1/4 ⇒ c = 4
Tangent at x = 4 parallels the secant line.
7. f (x) = x cos x on [0, π/2]
1
1
2_0
f(x)
x
c
f ′ (x) = cos x − x sin x∴ f is differentiable for all x.x cos x = 0 ⇔ x = 0 or cos x = 0cos x = 0 at ±π/2 + 2π n, where n is an integer∴ hypotheses are met on [0, π/2].Using the solver feature, f ′ (c) = 0 atc = 0.86033… .Horizontal line at x = 0.86033… is tangent.
8. f (x) = x2 sin x
π0
1
f(x)
x
c
f ′ (x) = 2x sin x + x2 cos x∴ f is differentiable for all x.x2 sin x = 0 ⇔ x = 0 or sin x = 0sin x = 0 at x = 0 + π n , where n is an integerInterval: [0, π]Using the solver feature, f ′ (c) = 0 atc = 2.28892… .Horizontal line at x = 2.288… is tangent.
9. f (x) = (6x − x2)1/2
60
3
f(x)
x
c
f x x x x′ = ⋅ −−( ) ( )/12
2 1 26 6 2( – )
∴ f is differentiable on (0, 6).f is continuous at x = 0 and x = 6.(6x − x2)1/2 = 0x(6 − x) = 0 ⇒ x = 0 or 6Interval: [0, 6](6c − c2)− 1/2(3 − c) = 0 ⇒ c = 3Horizontal line at x = 3 is tangent.
π
π
Calculus Solutions Manual Problem Set 5-5 89© 2005 Key Curriculum Press
10. f (x) = x4/3 − 4x1/3
c
4
1
f(x)
x
0
f x x x′ = − −( ) / /43
1 3 43
2 3
∴ f is differentiable for all x ≠ 0.f is continuous at x = 0.f (x) = 0 ⇒ x4/3 − 4x1/3 = 0 ⇒x1/3(x − 4) = 0 ⇒ x = 0 or 4Interval: [0, 4]43
1 3 43
2 3 0c c/ /− =−
43
2 3 1 0 1c c c− − = ⇒ =/ ( )
Horizontal line at x = 1 is tangent.
11. a. d (t) = 1000(1.09t )
d (50) = 1000(1.0950) = 74,357.520…
= $74,357.52 (Surprising!)
b. Average rate is
74357 5 1000
50
. –K =
1467.150… ≈ $1,467.15 per year.
c. d ′ (t) = ln (1.09)1000(1.09)t
d ′ (0) ≈ $86.18 per yeard ′ (50) ≈ $6,407.96 per yearThe average of these is $3,247.07 per year,which does not equal the average in part b.
d. Solving 1000(1.90)t ln 1.09 =1000 1 09 1000
50
50( . ) − algebraically gives
( . )1 091 09 1
50 1 09
50t = −.
ln .
⇒ = −t ln ln
.
ln .1 09
1 09 1
50 1 09
50
.
= ln (1.0950 − 1) − ln 50 − ln (ln 1.09) ⇒
t = − −ln ( . ) ln – ln (ln . )
ln .
1 09 1 50 1 09
1 09
50
= 32.893 years.K
This time is not halfway between 0 and 50.
12. d t
t
tt
tt
( )if
if = +
≤
≥
431
11
200 11
1
–,
– ,
The hypotheses of the mean value theorem do notapply on any interval that contains t = 1 as aninterior point, such as [0, 2] and [0.5, 2], becaused is not differentiable there. The hypotheses doapply on any interval not containing 1 and on
intervals for which 1 is an endpoint, suchas [1, 2].The conclusion is true if the instantaneousvelocity, d ′ (t), ever equals the average velocity.The average velocity equals
md d= =( ) – ( )2 0
228 5 0 2. ft/s for [ , ],
m
d d= =( ) – ( . )
..
2 0 5
1 557 111 0 5 2K ft/s for [ . , ],
md d= =( ) – ( )2 1
1100 1 2 ft/s for [ , ].
Between t = 0 and t = 1, d ′ (t) is negative.Above t = 1, d ′ (t) = 200t− 2.For d ′(c) = 28.5 ft/s,200c− 2 = 28.5 ⇒ c = 2.649… .But 2.649… is outside (0, 2), so the conclusionis not true. See the left graph.For d ′ (c) = 57.111… ft/s,200c− 2 = 57.111… ⇒ c = 1.871… .Because 1.871… is in (0.5, 2), the conclusion istrue. See the right graph.
100
200
0.5 1 2
t
d(t)
c?
c is outside(0, 2).
100
200
0.5 1 2
t
d(t)
c?
c is in(0.5, 2).
For d ′ (c) = 100, 200c− 2 = 100 ⇒ c = 1.414… .Because 1.414… is in (1, 2), the conclusion istrue, as is guaranteed by the mean value theorem.The fact that the conclusion is true when thehypotheses are met illustrates the fact that thehypotheses are sufficient. The fact that theconclusion can be true even if the hypotheses arenot met proves that the hypotheses are notnecessary.
13. See Figure 5-5d. 14. Answers may vary.
a b
f(x)
x
15. Answers may vary. 16. Answers may vary.
f(x)
xa
d
b
a b
x
f(x)
90 Problem Set 5-5 Calculus Solutions Manual© 2005 Key Curriculum Press
17. Answers may vary.
x
a c b
f(x)
18. Michel Rolle (1652− 1719) lived in France.Sources will vary.
19. f (x) = x2 − 4x 20. f (x) = x2 − 6x + 5
f (1) = − 3 ≠ 0 f (2) = − 3 ≠ 0Conclusion is not true. Conclusion is not true.f ′ (2) = 0, but 2 is not f ′ (3) = 0, but 3 is notin the interval (0, 1). in the interval (1, 2).
10
x
f(x)
–3
x
f(x)
1 2
–3
21. f (x) = x2 − 4x 22. f (x) = x2 − 6x + 5
f (2) = − 4 ≠ 0 f (4) = − 3 ≠ 0Conclusion is not true. Conclusion is true.f ′ (2) = 0, but 2 is not in f ′ (3) = 0, and 3 is inthe open interval (0, 2). the interval (1, 4).
0 2
1x
f(x)
1 4
xc
f(x)1
23. f (x) = x2 − 4x 24. f (x) = |x − 2| − 1f (3) = − 3 ≠ 0 f is not differentiable
at x = 2.Conclusion is true. Conclusion is not
true.f ′ (2) = 0, and 2 is in f ′ (x) never equals 0.the interval (0, 3).
30
–3
x
f(x)
c
1 3
1x
f(x)
25. f (x) = 1/x 26. f (x) = x − [x]
f (0) does not exist. f is discontinuous at1 and 2.
Conclusion is Conclusion isnot true. not true.
f ′ (x) never equals 0. f ′ (x) never equals 0.
5
5
x
f(x)
1 2
1x
f(x)
27. f (x) = 1 − (x − 3)2/3 28. f x( ) =
x x x
x
3 26 11 6
2
– –
–
+
f is not differentiable f is not continuous orat x = 3. differentiable at 2.Conclusion is Conclusion isnot true. not true.f ′ (x) never equals 0. There is no point at x = 2
to draw the tangent line.
f(x)
x
1
2 4
1 3
x
f(x)
1
No point oftangency
29. g xx x x
x( ) = +3 27 13 6
2
– –
–
= + = − + ≠( – )( – )
–
x x x
xx x x
2 5 3
25 3 2
22 ,
Thus, g is discontinuous at x = 2, and thehypotheses of the mean value theorem are notmet. The conclusion is not true for [1, 3],because the tangent line would have to contain(2, g (2)), as shown in the left graph. Theconclusion is true for (1, 5), because the slope ofthe secant line is 1, and g ′ (x) = 1 at x = 3, whichis in the interval (1, 5). See the right graph.
2 x
g(x)
3
1 3 2 x
g(x)
1
3
3
5
30. h (x) = x2/3 ⇒ h ′ (x) = (2/3)x− 1/3
∴ h is differentiable for all x ≠ 0.h ′ (0) would be 0− 1/3 = 1/(01/3) = 1/0, which isinfinite. The hypotheses of the mean valuetheorem are met on the interval [0, 8], becausethe function need not be differentiable at anendpoint. The hypotheses are not met on [− 1, 8],because the point x = 0 where h is notdifferentiable is in the open interval (− 1, 8). Tosee if the conclusion of the mean value theorem
Calculus Solutions Manual Problem Set 5-5 91© 2005 Key Curriculum Press
is true anywhere, find the slope of the secant line(see next graph).
m = =4 1
8 11 3
–
– (– )/
The tangent line has slope h ′ (c) = 1/3. Therefore,(2/3)c− 1/3 = 1/3 ⇒ c− 1/3 = 1/2 ⇒ c1/3 = 2 ⇒ c = 8.So the conclusion of the mean value theorem isnot true because 8 is at the endpoint of theinterval, not in the open interval (− 1, 8).
–1 8
4
x
h(x)
31. a. f xx x
x x( )
if
if=
≥+ <
3 3 3
3 3
– ,
,
3
6
x
f(x)
b. f is continuous at x = 3 because the right andleft limits both equal 6. f is not differentiableat x = 3 because the left limit of f ′ (x) is 1and the right limit is 3.
c. f is not differentiable at x = 3, which is in (1, 6).The secant line has slope 11/5. The tangent linehas slope either 1 or 3, and thus is never 11/5.
d. f is integrable on [1, 6]. The integral equals41.5, the sum of the areas of the two trapezoidsshown in this diagram.
3
6
x
f(x)
1 6
4
15
32. a. f (t) = number of miles in t hourst = number of hours drivenFor the mean value theorem to apply on[a, b], f must be differentiable on (a, b) andcontinuous at t = a and t = b.
b. The 60 mi/h equals the slope of the secantline. Therefore, there must be a tangent lineat some value t = c in (a, b) with slope equalto 60. This tangent line’s slope is theinstantaneous speed at t = c. Therefore, thespeed was exactly 60 at some time betweent = a and t = b, Q.E.D.
33. a. f (x) = 25 − (x − 5)2 + 4 cos 2π (x − 5)The graph agrees with Figure 5-5l.
b. f ′ (x) = − 2(x − 5) − 4 sin 2π (x − 5) ⋅ 2π =− 2x + 10 − 8π sin 2π (x − 5) f ′ (5) = 0Because the derivative at x = 5 is 0, thetangent line at x = 5 is horizontal. This isconsistent with x = 5 being a high point onthe graph.
c.
20
4
x
y1
y2
m(x)
x m (x) x m (x)
3.0 2 5.5 − 16.5
3.5 6.8333… 6.0 − 14.0 1 6.5 − 6.833…
4.5 16.5 7.0 − 25.0 no value
The difference quotient is positive when x isless than 5 and negative when x is greaterthan 5.
d. In the proof of Rolle’s theorem, the left limitof the difference quotient was shown to bepositive or zero and the right limit wasshown to be negative or zero.The unmentioned hypothesis isdifferentiability on the interval (a, b). Thefunction f is differentiable. Because there is avalue of f ′(5), both the left and right limitsof the difference quotient must be equal. Thisnumber can only be zero, which establishesthe conclusion of the theorem. Theconclusion of Rolle’s theorem can be trueeven if the hypotheses aren’t met. Forinstance, f (x) = 2 + cos x has zero derivativesevery π units of x, although f (x) is neverequal to zero.
34. a. mf f= = =( . ) – ( )
.
. –
.
4 5 2
2 5
5 5 4
2 50 6.
g (x) − 4 = 0.6(x − 2) ⇒g (x) = 0.6x + 2.8
Your graph should agree with Figure 5-5m.
b. f ′ (x) = 1 − π sin π xUsing the solver feature, f ′(c) = 0.6 atc = 2.0406… , 2.9593… , and 4.0406… ,all of which are in (2, 4.5).
92 Problem Set 5-5 Calculus Solutions Manual© 2005 Key Curriculum Press
c. h (x) = f (x) − g (x)
2
5
x
y
y1
y2
y3
For c1 = 2.0406… :
x h(x)
1.7906… –0.2925…
1.8406… –0.1865…
1.8906… –0.1022…
1.9406… –0.0411…
1.9906… –0.0041…
2.0406… 0.0081…
2.0906… –0.0039…
2.1406… –0.0397…
2.1906… –0.0977…
2.2406… –0.1760…
2.2906… –0.2723…
For c = 2.9593… : For c = 4.0406… :
x h(x) x h(x)
2.7093… –1.3274… 3.7906… 0.5075…
2.7593… –1.4237… 3.8406… 0.6134…
2.8093… –1.5021… 3.8906… 0.6977…
2.8593… –1.5601… 3.9406… 0.7588…
2.9093… –1.5959… 3.9906… 0.7958…
2.9593… –1.6081… 4.0406… 0.8081…
3.0093… –1.5958… 4.0906… 0.7960…
3.0593… –1.5589… 4.1406… 0.7602…
3.1093… –1.4979… 4.1906… 0.7022…
3.1593… –1.4136… 4.2406… 0.6239…
3.2093… –1.3077… 4.2906… 0.5276…
h(c1) = h(2.0406…) = 0.0081…So h(c1) is an upper bound for h(x).h (c2) = h (2.9593…) = − 1.60811669…So h (c2) is a lower bound for h (x).h (c3) = h (4.0406…) = 0.808116698…So h (c3) is an upper bound for h (x).
d. f meets the hypotheses of the mean valuetheorem, because f is differentiable for all x.h (x) = f (x) − g (x)∴ h ′(x) = f ′(x) − g ′(x)∴ h ′(c) = f ′(c) − g ′(c)For each of the values of c in part b, h ′(c) = 0.
∴ f ′(c) − g ′(c) = 0∴ f ′(c) = g ′(c)∴ f ′(c) = the slope of the secant line, Q.E.D.
35. The hypotheses of the mean value theoremstate that f should be differentiable on the openinterval (a, b) and continuous at x = a and x = b.If f is differentiable on the closed interval [a, b],it is automatically continuous at x = a and x = b,because differentiability implies continuity.
36. a. h (x) = f (x) − g (x)The mean value theorem applies to h becauseboth f and g are given to be differentiable, anda linear combination of differentiablefunctions is also differentiable.
b. By the mean value theorem, there is a numberc in (a, b) for which
′ = −−
h ch b h a
b a( )
( ) ( ).
If f (a) = g (a) + D1 and f (b) = g (b) + D2,then h (a) = D1 and h (b) = D2.
∴ ′ =h cD D
b a( )
–
–2 1
c. If D1 ≠ D2, then h ′ (c) ≠ 0.But h ′ (x) = f ′ (x) − g ′ (x) by the derivative ofa sum, and thus h ′(x) = 0 for all x in thedomain.∴ ′ =h c( ) ,0 which contradicts ′ ≠h c( ) .0So the supposition that D1 ≠ D2 is false,meaning that D1 and D2 are equal, Q.E.D.
37. By the definition of antiderivative (indefinite
integral), g x dx( ) = ∫ 0 if and only if g ′ (x) = 0.
Any other function f for which f ′ (x) = 0 differsfrom g (x) by a constant. Thus the antiderivativeof zero is a constant function, Q.E.D.
38. f (x) = (cos x + sin x)2, and g (x) = sin 2x
y
x
1
1
g
f
x f (x) g (x)
0 1 0
1 1.9092… 0.9092…
2 0.2431… − 0.7568…
3 0.7205… − 0.2794…
4 1.9893… 0.9893…
In each case, f (x) = g (x) + 1.
Calculus Solutions Manual Problem Set 5-6 93© 2005 Key Curriculum Press
Proof:
(cos x + sin x)2 = cos2 x + 2 cos x sin x + sin2 x
= 2 cos x sin x + 1 = sin 2x + 1, Q.E.D.
39. The hypotheses of Rolle’s theorem say that fis differentiable on the open interval (a, b).Because differentiability implies continuity,f is also continuous on the interval (a, b).Combining this fact with the hypothesis ofcontinuity at a and at b allows you to concludethat the function is continuous on the closedinterval [a, b].
40. The intermediate value theorem applies tocontinuous functions, whereas the mean valuetheorem applies to differentiable functions. Bothare existence theorems, concluding that thereis a value x = c in the open interval (a, b). Forthe intermediate value theorem, f (c) equals apre-selected number v between f ( a) and f ( b).For the mean value theorem, f ′ (c) equals theslope of the secant line connecting (a, f ( a))and (b, f ( b)).
41. Answers will vary.
Problem Set 5-6Q1. r′(x) = m(x)
Q2. See the text for the definition of derivative.
Q3. Increasing at 6 units/unit
Q4. dy = sec x tan x dx
Q5. y′ = 8x(x2 + 3)3
Q6. d 2z/dz2 = −25 sin u
Q7. f ′(x) = 0
Q8. 4.5
Q9. See Figure 5-5b.
Q10. E
1. a. I x dx= −∫ 10 1 5
4
9.
= (−10/0.5)(9− 0.5 ) − (−10/0.5)(4− 0.5 )
= −20/3 + 20/2 = 10/3 = 3.33333…
The +C and −C add up to zero.
b.
4 9
x
y
1
c. Pick sample points at left ends of subintervalsfor U5 and at right ends for L5.
k x f ( x)
1 4 1.25
2 5 0.89442719…
3 6 0.68041381…
4 7 0.53994924…
5 8 0.44194173…
Sum = 3.80673199…
U5 = (1)(3.80673199…) = 3.80673199…
k x f ( x)
1 5 0.89442719…
2 6 0.68041381…
3 7 0.53994924…
4 8 0.44194173…
5 9 0.37037037…
Sum = 2.92710236…
L5 = (1)(2.92710236…) = 2.92710236…
Average = (U5 + L5)/2 = 3.36691717… .Average overestimates the integral,3.33333… .This fact is consistent with the fact that thegraph is concave up.
d. Use sample points at the midpoints.
M10 = 3.32911229…
M100 = 3.33329093…
M1000 = 3.33333290…
Sums are converging toward 10/3.
2. I x dx= = − − −∫ sin cos . ( cos ).
1 5 00
1 5
= 0.92926279…
Using sample points at the midpoints,
M10 = 0.93013455…
M100 = 0.92927151…
M1000 = 0.92926288…
Integral = 0.92926279…
The sums are converging toward the integral.The rectangle and the region differ by the two“triangular” regions. Because the sample pointis at the midpoint of the subinterval, the“triangles” have equal bases. Because the graphis concave down, the “triangle” below thehorizontal line has a larger altitude, andthus a larger area, than the one above the line.So the rectangle includes more area on the left
94 Problem Set 5-7 Calculus Solutions Manual© 2005 Key Curriculum Press
than it leaves out on the right, and thusoverestimates the integral.
Rectangleincludesmore area.
y
x
3. See the statement of the fundamental theorem inthe text.
4. See the work in the text preceding the proof ofthe fundamental theorem for a derivation of aRiemann sum that is independent of the numberof subintervals.
5. See the text proof of the fundamental theorem.
6. If c is picked as the point in (a, b) where the
mean value theorem is true for g x f x dx( ) ( ) ,= ∫then the exact integral equalsg b g a
b ab a
( ) – ( )
( – )( ),⋅ − which equals g(b) − g(a).
7. v(t) = 100 − 20(t + 1)1/2
Distance = +∫ [ – ( ) ] /
0
81 2100 20 1t dt
= − +10040
31 3 2
0
8
t t( ) /
= − − +80040
39 0
40
313 2 3 2( ) ( )/ /
= 4531
3ft
8. a. h(x) = x1/2
k x f ( x)
1 4.25 2.0615528…
2 4.75 2.1794494…
3 5.25 2.2912878…
4 5.75 2.3979157…
5 6.25 2.5
6 6.75 2.5980762…
7 7.25 2.6925824…
8 7.75 2.7838821…
9 8.25 2.8722813…
10 8.75 2.9580398…
Sum = 25.3350679…
M10 = (0.5)(25.3350679…) = 12.66753…
b. h(u)∆u and h(u + ∆u)∆u are terms in a lowersum and an upper sum, respectively, becauseh(x) is increasing.
∴ h(u)∆u < A(u + ∆u) – A(u) < h(u + ∆u)∆u
c. h uA u u A u
uh u u( )
( ) ( )( )< + − < +∆
∆∆
But the limits of h(u) and h(u + ∆u) both equalh(u) because h is continuous and h(u) isindependent of ∆u. Therefore, by the squeezetheorem,
lim( ) – ( )
( ).∆u
A u u A u
uh u
→
+ ∆∆
=0
But the limit on
the left is defined to be dA/du.∴ dA/du = h(u), Q.E.D.
d. dA = h(u) du
∴ A(u) = ∫ h(u) du = ∫ u1/2 du = (2/3)u3/2 + C∴ A(u) = (2/3)u3/2 − 16/3
e. A( )9 12 23= , which agrees with M10 = 12.667… .
(Note also that A(9) < M10, which is expectedbecause the graph of h is concave down.)
9. a. Answers may vary. b. Answers may vary.
x
f(x)
x
f(x)
c. Answers may vary. d. Answers may vary.
x
f(x)
x
f(x)
e. Answers may vary. f. Answers may vary.
x
f(x)
x
f(x)
10. Answers will vary.
Problem Set 5-7
Q1. 16
6x C+ Q2. 118
63 7( )x C+ +Q3. − +−1
33x C Q4. 1
66sin x C+
Q5. 15 5sin x C+ Q6. x + C
Q7. tan x + C Q8. y″ = −1/x2
Q9. definite Q10. indefinite
1. x dx x2 3
1
4
1
41
3
1
364
1
31 21= = − =∫ ( ) ( )
2. x dx x3 4
2
5
2
51
4
1
4625
1
416
609
4152
1
4= = − = =∫ ( ) ( )
Calculus Solutions Manual Problem Set 5-7 95© 2005 Key Curriculum Press
3. ( ) ( )1 31
91 32 3
2
3
2
3
+ = +− −∫ x dx x
= − − =1
91000
1
9125 125( ) ( )
4. ( – ) ( – )5 21
155 22
1
43
1
4
x dx x− −∫ =
= = = =1
155832 343
6175
15
1235
3411
2
3[ – (– )]
5. 60 36 36 32 1 11162 3 5 3
1
8
1
8
x dx x/ /= = − =∫ ( )
6. 24 9 6 9 6 32 1 297 63 2 5 2
1
4
1
4
x dx x/ /. . .= = − =∫ ( )
7. 5 5 40 10 302
8
2
8
dx x= = − =∫8. dx x= = − =∫20
50
20
50
50 20 30
9. ( )x x dx x x x2
2
03 2
2
0
3 71
3
3
27+ + = + +
− −∫= − − + −
= =0
8
36 14
32
310
2
3
10. ( )x x dx x x x2
3
03 2
3
0
4 101
32 10+ + = + +
− −∫= 0 − (−9 + 18 − 30) = 21
11. 4 51
44 5 4
1
11 2
1
1
x dx x dx+ = +− −∫ ∫ ( ) / ( )
= ⋅ + = = =−
1
4
2
34 5
1
627 1
13
34
1
33 2
1
1
( ) ( – )/x
12. 2 101
22 10 2
3
31 2
3
3
x dx x dx+ = +− −∫ ∫ ( ) / ( )
= ⋅ + = = =−
1
2
2
32 10
1
364 8
56
318
2
33 2
3
3
( ) ( – )/x
13. 4 4 4 1 4 1 80 0
sin cosx dx x= − = − − + =∫π π
( ) ( )
14. 6 6 6 1 6 1 122
2
2
2
cos sin/
/
/
/
x dx x= = − − =− −∫ π
π
π
π
( ) ( )
or: Integral of an even function betweensymmetric limits.
2 6 12 12 1 0 120
2
0
2
cos sin/ /
x dx x= = − =∫π π
( )
15. (sec cos )/
/2
6
3
x x dx+∫π
π
= + = + − −tan sin / //
/x x
π
π
6
33 3 2 1 3 2/ 1
= − =( / ) / .7 6 3 1 2 1 52072K
16. (sec tan sin ) sec cos/ /
x x x dx x x+ = −∫0
3
0
3π π
= − − + =21
21 1 1 5.
17. e dx e e ex x2
0
42
0
42 4 01
2
1
2
1
27 5
ln lnln .∫ = = − =
18. e dx e e ex x− − −= − = − +∫0
3
0
3 3 0ln ln ln
= − + =1
31
2
3
19. sin cos sin3 4
1
2
1
21
4x x dx x =∫
= =1
42 1 0 0455664 4(sin – sin ) . K
20. ( cos ) sin ( cos )11
514 5
3
3
3
3
+ = − +− −∫ x x dx x
= − + + + =1
51 3
1
51 3 05 5( cos ) [ cos(– )]
or: Integral equals zero because an odd function isintegrated between symmetric limits.
21. cos sin.
.
.
.
31
33
0 1
0 2
0 1
0 2
x dx x=∫= =1
30 6 0 3 0 0897074(sin . – sin . ) . K
22. sin cos. .
21
22
0
0 4
0
0 4
x dx x = −∫= − =1
20 8 0(cos . – cos ) 0.1516466K
23. ( – sin )x x x dx dx7 3
5
5
6 4 2 2 2+ + =−∫ ∫
0
5
= = − =2 2 20 0 20( )0
5x
24. (cos – tan ) cosx x x dx x dx+ =−∫ ∫10 23
1
1
0
1
= = − =2 2 1 2 0 1 682941
0
1sin sin sinx . K
25. x dx−
−∫2
1
1
has no value because y = x− 2 has a
vertical asymptote at x = 0, which is within theinterval.
26. x dx−∫ 2
2
has no value because the integrand is
not a real number for negative values of x.
27. Integral = −(area) 28. Integral = area
x3 6
-5
f(x)
5 7
1
f(x)
x
96 Problem Set 5-8 Calculus Solutions Manual© 2005 Key Curriculum Press
29. Integral ≠ area 30. Integral ≠ area
7 x
0
f(x)
1
x8
2
1 f(x)
1
31. f x dx f x dxb
a
a
b
( ) ( ) = − = −∫ ∫ 7
32. 4 4 4 7 28 ( ) ( ) ( )f x dx f x dxa
b
a
b
= = =∫ ∫33. g x dx g x dx g x dx
a
c
a
b
b
c
( ) ( ) ( )= +∫ ∫ ∫= 12 + 13 = 25
34. f x dxa
c
( ) ∫ cannot be determined.
35. f x dx g x dxa
c
a
c
( ) ( )∫ ∫+ cannot be determined.
36. [ ( ) ( )]f x g x dxa
b
+ =∫f x dx g x dx
a
b
a
b
( ) ( ) 7 12 19∫ ∫+ = + =
37.
5 10
7
y
x
y = f' (x)
f(3) = 7
y = f(x)
38.
7
8
y
x
y = f ' (x)
f(1) = 8
y = f(x )
39. Statement:“If f ( x) < g(x) for all x in [a, b],
then f x dx g x dxa
b
a
b
( ) ( ) .∫ ∫< ”
Converse:
“If f x dx g x dxa
b
a
b
( ) ( ) ,∫ ∫<
then f ( x) < g(x) for all x in [a, b].”The converse can be shown to be false by anycounterexample in which the area of the regionunder the g graph is greater than the area underthe f graph, but the g graph touches or crossesthe f graph somewhere in [a, b]. Onecounterexample isf ( x) = 1.5 and g(x) = 2 + cos x on [0, 2π].
g
f
y3
20
x
40. x dx x C2
1
43
1
41
3∫ = +
= + − − =1
34
1
31 213 3( ) ( )C C
The two C’s will always cancel, so it is notnecessary to write them.
Problem Set 5-8Q1. 30x2.4 + C
Q2. 30(42.4 − 1) = 805.72…
Q3. y x′ = −1/ –1 2
Q4. f ′ (x) = 3x2 sin x + x3
cos x
Q5.
x1
1
f(x) and f'(x)
f
f'
Q6. Yes, continuous
Q7. Increasing at x = 7
Q8. f ( a) = f ( b) = 0
Q9. v(9) = 450 ft/s
Q10. a(9) = 25 (ft/s)/s
π
Calculus Solutions Manual Problem Set 5-8 97© 2005 Key Curriculum Press
1. a.
t
50
100
v
t = a t = b1
(t, v)
dt
dy = v dt = (55 + 12t0.6 ) dt
b. Displacement = +∫ ( ).55 12 0 6t dta
b
( ). .55 12 55 7 50 6 1 6
0
1
0
1
+ = +∫ t dt t t.
= 55 + 7.5 − 0 − 0 = 62.5 mi
( ). .55 12 55 7 50 6 1 6
1
2
1
2
+ = +∫ t dt t t.
= 110 + 22.735… − 55 − 7.5 ≈ 70.2 mi
c. ( ). .55 12 55 7 50 6 1 6
0
2
0
2
+ = +∫ t dt t t.
= 110 + 22.735… − 0 − 0 = 132.735… ,which equals the sum of the two integralsabove.
d. ( ).55 12 0 6
0+ =∫ t dt
b
300
55 7 5 3001 6
0t t
b+ =. .
55b + 7.5b1.6 = 300b ≈ 4.13372… ≈ 4.134 h
300
x
y
4.134
e. v(4.13372…) = 55 + 12(4.13372…)0.6 =83.1181… . At the end of the trip, you weregoing about 83 mi/h.
2.
10
10
v
t
dv
(t, v)
dy v dt e dtt= = −15 0 1.
15 150 150 1500 1 0 1
0
20
0
202e dt e et t− − −= − = − +∫ . .
= 129.6997… ft
3. a.
2
5
y
x
dx
(x, y)
b. dA = y dx = 10e0. 2x dx
c. 10 50 50 500 2 0 2
0
2
0
20 4e dx e ex x. . .= = −∫
d.
10 0 2
0
2
e dxx.∫ = 24.59123K
The region is approximately a trapezoid withheight 2 and bases 10 and y(2). y(2) =14.9182… , so the area of the trapezoid is2/2(10 + 14.9182…) = 24.9182… .
4.
3
4
y
x
dx
(x, y )
dA = (6x − x2 ) dx
( )6 31
33 6
1
36 362
0
62 3
0
62 3x x dx x x− = − = ⋅ − ⋅ =∫
The area of the circumscribed rectangle is 6 ⋅ 9 =54. The area of the parabolic region is two-thirdsthis area.
5. a. dA = [x + 2 − (x2 − 2x − 2)] dx
The top and bottom of the strip are nothorizontal, so the area of the strip is slightlydifferent from dA. As dx approaches zero, thedifferences in height at different values of x inthe strip become smaller, so the differencebetween dA and the area of the strip getssmaller.
b. y1 = y2 ⇒ x + 2 = x2 − 2x − 2 ⇒0 = x2 − 3x − 4 ⇒ 0 = (x − 4)(x + 1) ⇒x = 4 or x = −1
( )− + + = − + +− −∫ x x dx x x x2
1
43 2
1
4
3 41
3
3
24
= − ⋅ + ⋅ + ⋅
1
34
3
24 4 43 2
− − ⋅ − + ⋅ − + ⋅ −
1
31
3
21 4 13 2( ) ( ) ( )
= =125
620 83333. K
c. R100 = 20.834375 (Checks)
98 Problem Set 5-8 Calculus Solutions Manual© 2005 Key Curriculum Press
6.
π⁄4
1y
x
dx
(x, y )1
(x, y )2
The curves intersect at x = π /4.
dA = (cos x − sin x) dx
(cos sin ) (sin cos )/ /
x x dx x x− = +∫0
4
0
4π π
= −2 1
7. a.
90
6
x
F
dx
(x, F )
b. dW = F dx = 0.6x dx
W x dx x= =∫ 0 6 0 3 2
0
9
0
9
. .
= 24.3 − 0 = 24.3 inch-pounds
c. The region under F from x = 0 to x = 9 is atriangle with base 9 and height F(9) = 5.4. Sothe area is 1/2 ⋅ 9 ⋅ 5.4 = 24.3.
d. dW is found by multiplying F by dx. F ismeasured in pounds, and dx is measured ininches, so the units of dW are(pounds)(inches), or inch-pounds.
8.
5
30
F
x
dx
(x, F )
dW x dx= 5020
cosπ
5020
1000
200
10
0
10
cos sinπ
ππ
x dx x=∫= − =
= …
1000
2
10000
1000
318 3098π
ππ π
sin sin
.
The midpoint Riemann sum R100 gives318.313… , which is close to the answer foundusing integration.
9. a.
dx
(x, T )
0.5 1
x
T
20
b. dD = T dx = [20 − 12 cos 2π(x − 0.1)] dx
c. D x dx= ∫ [ – cos ( – . )].
20 12 2 0 10
0 5
π
= − −20 6 2 0 1x x( ) ( . )0
0.5/ sinπ π
= 10 − (6 /π) sin 0.8π − 0 + (6 /π) sin (−0.2π)= 7.75482… ≈ 7.75 degree-days
d. From noon to midnight,
D x dx= ∫ [ – cos ( – . )].
20 12 2 0 10 5
1
π
= − −20 6 2 0 1x x( ) ( . )0.5
1/ sinπ π
= 20 − (6 /π) sin 1.8π − 10 + (6 /π) sin 0.8π= 12.24517… ≈ 12.25 degree-days
The total number of degree-days isD = 7.75482… + 12.24517… =20 degree-days.Note that this answer can be found moreeasily by observing that in one full cycle of asinusoid, there is just as much area above thesinusoidal axis, T = 20, as there is below it.So the average temperature difference for theday is 20 degrees, making the number ofdegree-days for one day equal to 20.
10. a.
10 30
T
C
1000
dT
dH = C dT= (−0.016T 3 + 0.678T 2 + 7.45T + 796) dT
H T T= +∫ (– . .0 016 0 6783 2
10
30
+ +7 45 796. )T dT
= − + +0 004 0 226 3 7254 3 2. . .T T T
+ 796T 10
30= −3240 + 6102 + 3352.5
+ 23880 + 40 − 226 − 372.5 − 7960= 21,576 Btu
b. (2000)(21576) = 43,152,000 BtuThe property is the integral of a constant
Calculus Solutions Manual Problem Set 5-8 99© 2005 Key Curriculum Press
times a function. That is, 200010
30
C dT ∫ =
200010
30
C dT .∫11. a.
xdx
P
1000
1000 200
(x, P)
b. dC = P dx = (100 + 0.06x2) dx
C x dx x xb b
= + = +∫ ( . )100 0 06 100 0 022 3
0.
0
= 100b + 0.02b3 − 0 − 0
∴ C = 100b + 0.02b3
c. b = 100: C = 100(100) + 0.02(1003) =$30,000
b = 200: C = 100(200) + 0.02(2003) =$180,000
For 100 m to 200 m, the cost should be180,000 − 30,000 = $150,000.
As a check,
( . )100 0 06 100 0 022 3
100
200
100
200
+ = +∫ x dx x x.
= 100(200) + 0.02(2003) − 100(100)− 0.02(1003) = 150000
Thus, P dx P dx P dx0
200
0
100
100
200
∫ ∫ ∫= + ,
which shows that the sum of integrals withthe same integrand applies.
12. Using trapezoids, the area is approximately10(0/2 + 38 + 50 + 62 + 60 + 55 + 51 + 30 +3/2) = 3475 ft2. The fundamental theorem cannotbe used because the function is specified only bydata, not by an equation whose antiderivative canbe found.
Plan of attack for area problems:
• Do geometry to get dA in terms of samplepoint (x, y).
• Do algebra to get dA in terms of one variable.
• Do calculus to sum the dA’s and take the limit(i.e., integrate).
13. y
x
(x, y)
(x, 0)
4
1 5
The graph intersects the x-axis at x = 1 and x = 5.y = −x2 + 6x − 5 = −(x − 1) ( x − 5) = 0 ⇒ x = 1,5, which confirms the graphical solution.
dA = (y − 0) dx = (−x2 + 6x − 5) dx
A x x dx x x x= + = − + −∫ (– – )2 3 2
1
5
1
5
6 51
33 5
= − + − + − + =125
375 25
1
33 5 10
2
3
14.–2 3
–6
xy
(x, 0)
(x, y)
The graph intersects the x-axis at x = −2and x = 3.y = x2 − x − 6 = (x + 2)(x − 3) = 0 ⇒ x = −2,3, which confirms the graphical solution.dA = (0 − y) dx = (−x2 + x + 6) dx
A x x dx x x x= + + = − + +− −∫ (– )2 3 2
2
3
2
3
61
3
1
26
= − + + − − + =99
218
8
32 12 20
5
6
15.(x, y)
y
x
–2
(0, y)
4
The graph intersects the y-axis at y = 1and y = 4.x = (y − 1) ( y − 4) = 0 ⇒ y = 1, 4, whichconfirms the graphical solution.
dA = (0 − x) dy = −(y − 1) ( y − 4) =(−y2 + 5y − 4) dy
A y y dy y y y= + = − + −∫ (– – )2 3 2
1
4
1
4
5 41
3
5
24
= − + − + − + =64
340 16
1
3
5
24 4
1
2
16. y
x
–1
5
5
(x, y)(0, y)
100 Problem Set 5-8 Calculus Solutions Manual© 2005 Key Curriculum Press
The graph intersects the y-axis at y = −1and y = 5.
x = 5 + 4y − y2 = (1 + y) ( 5 − y) = 0 ⇒ y =−1, 5, which confirms the graphical solution.
dA = (x − 0) dy = (5 + 4y − y2) dy
A y y dy y y y= + = + −− −∫ ( – )5 4 5 2
1
32 2 3
1
5
1
5
= + − + − − =25 50125
35 2
1
336
17.y
x
–1 4
2
(x, y1)
(x, y2)
The graphs intersect at x = −1 and x = 4.x2 − 2x − 2 = x + 2 ⇔ x2 − 3x − 4 = 0 ⇔ x =−1, 4, which confirms the graphical solution.dA = (y2 − y1) dx = (−x2 + 3x + 4) dx
A x x dx x x x= − + + = − + +− −∫ ( )2 3 2
1
4
1
4
3 41
3
3
24
= − + + − − + =64
3
1
3
3
2
5
624 16 4 20
18.
y
x
10
–2
4(x, y1)
(x, y2)
The graphs intersect at x = −2 and x = 4.−2x + 7 = x2 − 4x − 1 ⇔ x2 − 2x − 8 = 0 ⇔(x + 2)(x − 4) = 0 ⇔ x = −2, 4, which confirmsthe graphical solution.
dA = (y1 − y2) dx = (−x2 + 2x + 8) dx
A x x dx x x x= + + = − + +− −∫ (– )2 3 2
2
4
2
4
2 81
38
= − + + − − + =64
316 32
8
34 16 36
19.y
x
(x, y2)
(x, y1)
2–2
6
The graphs intersect at x = −2 and x = 2.0.5x2 + 2x = −x2 + 2x + 6 ⇔ 1.5x2 = 6 ⇔ x =−2, 2, which confirms the graphical solution.dA = (y2 − y1) dx = (−1.5x2 + 6) dx
A x dx x x= + = − +− −∫ (– . )1 5 6 0 5 62 3
2
2
2
2
.
= −4 + 12 − 4 + 12 = 16
20.
0 5
5
x
y
(x, y2)
(x, y1)
The graphs intersect at x = 0 and x = 5.0.2x2 + 3 = x2 − 4x + 3 ⇔ 0.8x2 − 4x = 0 ⇔0.8x(x − 4) = 0 ⇔ x = 0, 5, which confirms thegraphical solution.
dA = (y1 − y2) dx = (−0.8x2 + 4x) dx
A x x dx x x= + = − +∫ (– . )0 8 44
1522 3 2
0
5
0
5
= − + + − =500
15
2
350 0 0 16
21.
(x, y2)
x
2
0
(x, y1)
5
y
The graphs intersect at x = 0 and x = 5.dA = (y1 − y2) dx = (2e0. 2x − cos x) dx
A e x dx e xx x= − = −∫ ( cos ) sin. .2 100 2 0 2
0
5
0
5
= 10e − sin 5 − 10 + 0 = 18.1417…
22.
x
y
1
(x, y2)
(x, y1)
dA = (y2 − y1) dx = (e2x − sec2 x) dx
A e x dx e xx x= = −∫ ( – sec ) tan2 2 2
0
1
0
1
0 5.
= 0.5e2 − tan 1 − 0.5 + 0 = 1.6371…
Calculus Solutions Manual Problem Set 5-8 101© 2005 Key Curriculum Press
23.
4
(x1, y)(x2, y)
–2 2
x
y
1
The graphs intersect at y = 1 and y = 4.Write y = x + 3 as x = y − 3.y − 3 = −y2 + 6y − 7 ⇒ y2 − 5y + 4 = 0 ⇒(y − 1)(y − 4) = 0 ⇒ y = 1, 4, which confirmsthe graphical solution.
dA = (x2 − x1) dy = (−y2 + 5y − 4) dy
A y y dy y y y= + = − + −∫ (– – )2 3 2
1
4
1
4
5 41
3
5
24
= − + − + − + =64
340 16
1
3
5
24 4
1
224.
(x2, y) (x1, y)
y
x
5
–5
8
The graphs intersect at y = −5 and y = 5.Write y = −2x1 + 11 as x1 = 5.5 − 0.5y.
5.5 − 0.5y = 0.25y2 − 0.5y − 0.75 ⇒0.25y2 = 6.25 ⇒ y = −5, 5, which confirms thegraphical solution.
dA = (x1 − x2) dy = (−0.25y2 + 6.25) dy
A y dy y y= + = − +− −∫ (– . . )0 25 6 25
1
12
25
42 3
5
5
5
5
= − + − + =125
12
125
4
125
12
125
4
2
341
25.
(x, y1)y
x
(x, y2)
–1 2
4
The graphs intersect at x = −1 and x = 2.x3 − 4x = 3x2 − 4x − 4 ⇒ x3 − 3x2 + 4 =(x + 1) ( x − 2)2 = 0 ⇒ x = −1, 2, whichconfirms the graphical solution.
dA = (y1 − y2) dx = (x3 − 3x2 + 4) dx
A x x dx x x x= + = − +− −∫ ( – )3 2 4 3
1
2
1
2
3 41
44
= − + − − + =4 8 8 1 4 61
4
3
426.
(x, y2)
(x, y1)
y
x
–1 8
2
The graphs intersect at x = −1 and x = 8.x2/3 = (x + 1)1/2 + 1 ⇒ x = −1, 8 numerically,which confirms the graphical solution.Or x2/ 3 − 1 = (x + 1)1/ 2 ⇒ (x2/ 3 − 1)2 = x + 1.Write t = x1/ 3, so (t2 − 1)2 = t3 + 1 ⇒t4 − t3 − 2t2 = t2(t + 1)(t − 2) = 0 ⇒t = 0, −1, 2 ⇒ x = t3 = 0, −1, 8.
But x = 0 is extraneous from the irreversible stepof squaring both sides. So x = −1, 8.
dA = (y2 − y1) dx = [(x + 1)1/2 + 1 − x2/3] dx
A x x dx= + +−∫ [( ) – ]/ /1 11 2 2 3
1
8
= + + −−
2
31
3
53 2 5 3
1
8
( ) /x x x /
= + − − + − =18 8 0 1 796
5
3
5
1
5
27. Wanda: You can always tell the right waybecause the altitude of the strip should bepositive. This will happen if you take(larger value) minus (smaller value). In this case,if you slice vertically, it’s line minus curve(see graph).
curve
line
x
y
For curve minus line, you’d get the opposite ofthe right answer. Note that if you slicehorizontally, it would be curve minus line.
28. a. Peter: Horizontal slicing would be awkwardbecause for some values of y the length of thestrip would be given by line minus curve, butin others it would be boundary minus curve,and yet elsewhere it would be curve minuscurve. If you use vertical slices, the length
102 Problem Set 5-9 Calculus Solutions Manual© 2005 Key Curriculum Press
of the strip will always be line minus curve.(See graphs.)
x
y
x
y
(x, y1)
(x, y2)
b. Peter: In the graph on the right, y1 − y2 willbe positive. Because y2 is negative, you willget (pos.) − (neg.), which is equivalent to(pos.) + (pos.). Thus, the altitude for the stripis positive.
29.
x
y
h(x, y)
(x, ah2)
–h
The graph shows the parabolic region fromx = −h to x = h and a strip from the graph to ahorizontal line at y = ah2.
dA = (ah2 − y) dx = (ah2 − ax2) dx
A ah ax dx ah x axh
h h
= = −
−∫ ( – )2 2 2 3
0
21
3
= −
=2
1
3
4
33 3 3a h h ah
Area of rectangle = 2h(ah2) = 2ah3
∴ = =area of region
area of rectangle,
( / )4 3
2
2
3
3
3
ah
ah Q.E.D.
The graph shows y = 67 − 0.6x2 and the liney = 7, with a circumscribed rectangle.
67
y = 7
y
x
7 = 67 − 0.6x2 ⇒ 0.6x2 = 60 ⇒ x = ±10Rectangle has width 10 − (−10) = 20 and length67 − 7 = 60. Area of region = 2
3 20 60 800( )( ) = .
30. dA = sin x dx
A x dx x= = − = − − + =∫ sin cos0 0
1 1 2π π
( ) , which
is a rational number.
1
π
x
y
(x, y)y = cos x
–π/10 π/10
x
y = 7 cos 5x
y7
For y = 7 cos 5x, width is 1/5 as much andaltitude is 7 times as much.∴ A = (2)(1/5)(7) = 2.8
31.
–2 3
8
x
y(x, y
2)
(x, y1)
The graphs intersect at x = −2 and x = 3.dA = (y2 − y1) dx = (−x2 + x + 6) dx
A x x dx x x x= + + = − + +− −∫ (– )2 3 2
2
3
2
3
61
3
1
26
= − + + − − + = =9
9
218
8
32 12 20
5
620 8333. K
R10 = 20.9375R100 = 20.834375R1000 = 20.83334375The Riemann sums seem to be approaching theexact answer.
32.
10
π 2π 3π 4π
x
t(x)
t′(x) = 1 + cos xt′(x) = 0 ⇒ cos x = −1 ⇒x = π + 2π n = … , π, 3π, 5π, …t′(x) is never negative, so t′(x) does not changesigns. These points are plateau points.
Problem Set 5-9
Q1.1
3
1
23 2x x x C+ + + Q2.
4
77 4x C/ +
Q3. y x′ = −2
31 3/ Q4. − +−1
33e Cx
Q5. −csc x + C Q6. x− 1
Q7. See Section 5-4. Q8. mean value
Q9. Q10.
4
x
y
x
y
1
1
Calculus Solutions Manual Problem Set 5-9 103© 2005 Key Curriculum Press
Plan of attack for volume problems:
• Do geometry to get dV in terms of samplepoint (x, y).
• Do algebra to get dV in terms of one variable.
• Do calculus to add up the dV’s and take the limit(i.e., integrate).
1. a. dV = π x 2 dyy = 9 − x2 ⇒ x2 = 9 − ydV = π (9 − y) dy
b. V y dy y y= − = −∫ π π( ) ( . ) 9 9 0 5 2
0
9
0
9
= π (81 − 40.5) − π (0 − 0) = 40.5π= 127.2345…
c. R100 = 127.2345… (Checks.)
d. Volume of circumscribed cylinder is 9(π ⋅ 32)= 254.4690… . Half of this is 127.2345… ,equal to the volume of the paraboloid.
2. dV = π x2 dyy = 10 − 2x ⇒ x = 5 − 0.5y∴ dV = π (5 − 0.5y)2 dy
V y dy y= − = − −∫ π π( . ) ( . )2 3
0
10
5 0 52
35 0 5
0
10
= − + = = …2
30
2
3125
250
3261 7993π π π
( ) ( ) .
Vcylinder = πr2h, so 1
3 of that is V r h= 1
32π .
Here, r = 5 and h = 10, so V = =1
35 102π ( )( )
250
3
π, as found by integrating.
3. a. dV = π y2 dx = π (3e− 0.2 x)2 dx = 9π e−0.4 x dx
b. 9 22 50 4 0 4
0
5
0
5
π πe dx ex x− −= −∫ . ..
= −22.5π e−2 + 22.5π e0 = 61.1195…The midpoint Riemann sum R100 gives61.1185… , which is close to the answerfound using integration.
c. Slice perpendicular to the axis of rotation, soslice vertically if rotating about the x-axis andhorizontally if rotating about the y-axis.
4. y = 4x − x2 is rotated about the x-axis.
41
x
y(x, y)
dV = π y2 dx = π (4x − x2)2 dx
V x x dx x x x dx= − = +∫ ∫π π( )4 16 82 2
1
42 3 4
1
4
( – )
= − +
= = …π π16
32
1
530 6 96 1323 4 5
1
4
x x x . .
The midpoint Riemann sum R100 gives96.1341… , which is close to the answer foundusing integration.
5. y = x1.5 is rotated about the x-axis.
(x, y)
19
27
x
y
dV = π y2 dx = π x3 dx
V x dx x= = = = …∫ π π π3 4
1
9
1
9 1
41640 5152 2119.
6. y = ln x ⇒ x = ey is rotated about the y-axis.
(x, y) x
y
1
1
dV = π x2 dy = π e2y dy
V e dy e ey y= = = − =∫ π π π2 2
0
12
0
1
2 21 10 0359( ) . K
7. y = x3/4 ⇒ x = y4/ 3 is rotated about the y-axis.
(x, y)
x
y
1
8
1 16
dV = π x2 dy = π y8/3 dy
V y dy y= = ⋅ =∫ π π π8 3 11 3
1
83
11
6141
11/ /
1
8
= 1753.8654…
8. y x= 14
and y = 8x2, intersecting at (0, 0) and(2, 16), are rotated about the y-axis. Area of crosssection is π πx x1
222− .
104 Problem Set 5-9 Calculus Solutions Manual© 2005 Key Curriculum Press
x1 = y1/ 4, and x y21
8=
∴ = − = −
dV x x dy y y dyπ π( ) /
12
22 1 2 21
64
V y y dy= −
∫ π 1 2 2
0
16 1
64/
= −
= =π π2
3
1
192
64
367 02063 2 3
0
16
y y/ . K
The midpoint Riemann sum R100 gives V ≈67.0341… , which is close to the answer foundusing integration.
9. y1 = e0.4 x and y2 = x + 1, from x = 0 to x = 3, arerotated about the x-axis.Area of cross section is π πy y2
212− .
dV y y dx x e dxx= − = + −π π( ) [( ) ]222
12 0 81 .
V x e dx
x e
e
e
x
x
= + −
= + −
= − − +
= − = …
∫ π
π
π
π
[( ) ]
.
1.25 1.25
( . . ) . ft
2
.
1
1
31 1 25
64
3
1
3
22 25 1 25 26 6125
0 8
0
3
3 0 8
0
3
2 4
2 4 3
.
.
.
( )
The midpoint Riemann sum R100 givesV = 26.6127… , which is close to the answerfound using integration.
10. y1 = x1/3 and y2 = 10e− 0.1 x are rotated about thex-axis. Only the back half of the solid is shown.
80
10y
x
(x, y2)
(x, y1)
dV y y dx e x dxx= − = −−π π( ) ( )22
12 0 2 2 3100 . /
V e x dx
e x
e
x
x
= −
= − −
= − + = …
−
−
−
∫ π
π
π
( )
( . )
( . ) .
0
8
.
100
500 0 6
500 480 8 1193 3394
0 2 2 3
0
8
0 2 5 3
1 6
. /
. /
11. y = 4 − x1 ⇒ x1 = 4 − y, and y = 4 − x22
⇒ x2 =4 – ,y intersecting at x = 0 and x = 1, are
rotated about the y-axis. Only the back half ofthe solid is shown.
x
y4
3
1
(x1, y)
(x2, y)
dV x x dy y y dy= − = − − −π π( ) [( ) ( ) ]222
12 4 4
= π (−y2 + 7y − 12) dy
V y y dy= − + −∫ π ( ) 4
2
37 12
= − + −
π 1
3
7
2123 2
3
4
y y y
= − + − + − +
π 64
356 48 9
63
236
= = …1
60 523598π .
12. y = ax2 ⇒ x = (y/a)1/2, from (0, 0) to (r, h), isrotated about the y-axis.
r
h
(x, y)x
y
dV = π x2 dy = π (y/a) dy = (π /a)y dy
V a y dy a y a hh h
= = ⋅ = −∫ ( / ) ( / )( )π π π( / )1
2
1
202
0 0
2
Because y = ax2, h = ar2.
∴ = =V a ar ar1
2
1
22 2 4( / )( )π π
Volume of circumscribed cylinder isV c = π r2h = π r2(ar2) = π ar4.Thus, the volume of the paraboloid is half thevolume of the circumscribed cylinder, Q.E.D.
13. a. y = 0.3x1.5 is rotated about the x-axis.
2
4
(x, y)
x
y
dV = π y2 dx = π (0.3x1.5 )2 dx = π (0.09x3) dx
V x dx x= =
= = …
∫ π π
π
( . ) .
5.76 18.09557
0 09 0 02253 4
0
4
0
4
Calculus Solutions Manual Problem Set 5-9 105© 2005 Key Curriculum Press
b. R10 = 5.7312πR100 = 5.75971…πR1000 = 5.7599971…πValues are getting closer to V = 5.76π.
14. y = 4 − x2 ⇒ x = (4 − y)1/2 dyInner radius is 3 − x; outer radius is 3.
dV = π[32 − (3 − x)2] dy
= π{9 − [3 − (4 − y)1/2]2} dy
= π [6(4 − y)1/2 − 4 + y] dy
V y y dy= − − +∫ π[ ( ) ]1/26 4 40
4
= − − − +π[ ( ) . ]3/2
0
44 4 4 0 5 2y y y
= π (0 − 16 + 8 + 32 + 0 − 0) = 24π= 75.3982…
15. y = 4 − x2 is rotated about the line y = −5. Onlythe back half of the solid is shown.
(x, y)4
y = –5
x
y
2
dV = π[(y + 5)2 − 52] dx
= π[(9 − x2)2 − 52] dx = π (56 − 18x2 + x4) dx
V x x dx
x x x
= − +
= − +
= − + − + −= = …
∫ π
π
ππ
( )
( . )
( . )
. .
0
2
56 18
56 6 0 2
112 48 6 4 0 0 0
70 4 221 168
2 4
0
2
3 5
16. Cross sections perpendicular to the x-axis aresquares with side length (y2 − y1). The curvesintersect at (0, 0) and (1, 1).
x1
y
1
dV = (y2 − y1)2 dx = (x1/5 − x2)2 dx
V x x dx
x x x dx
= −
= − +
∫∫
( )
( )
/
/ /
1 5 2
0
12
2 5 11 5 4
0
1
2
= − +
= =5
7
5
8
1
5
81
2800 28927 5 16 5 5
0
1
x x x/ / . K
17. Cross sections perpendicular to y-axis are squaresof edge 2x, where (x, y) is a sample point on theline in the xy-plane.
y = –(15/4)x + 15
4
15
(x, y)
x
y
Equation of line is
y x x y= − + ⇒ = −15
415 4
4
15.
dV x dy y dy
y dy
= = −
= −
( )22
2
2 4 44
15
64 11
15
64 11
15320 1
1
150
15
0
15
−
= − −
∫ y dy y
2 3
= 320 3 cm
The circumscribed rectangular box has volumel · w · h = 8 · 8 · 15 = 960 = 3V, so the pyramidis 1/3 the volume of the circumscribed rectangularsolid, Q.E.D.
The volume of a pyramid is one-third the volumeof the circumscribed rectangular box, just as thevolume of a cone is one-third the volume of thecircumscribed cylinder.
18. Center line: y = 0.2x2
Upper bound: y = 0.16x2 + 1Radius of circular cross section is 1 − 0.04x2.The tip of the “horn” is where 0.2x2 = 0.16x2 + 1with x ≥ 0, which is at x = 5.
dV = π (1 − 0.04x2)2 dx
= π (1 − 0.08x2 + 0.0016x4) dx
V x x dx
x x x
= − +
= − +
= − + − + −
= = … ≈
∫ π
π
π
π
( . . )
. cm
0
5
3
1 0 08 0 0016
0 08
3
0 0016
5
510
31 0 0 0
8
38 3775 8 4
2 4
0
5
3 5. .
.
106 Problem Set 5-9 Calculus Solutions Manual© 2005 Key Curriculum Press
19. a. y = x0.6
Pick sample point (x, y) on the curve withinthe slice. One leg of the isosceles triangle isy, so the other leg is also equal to y.
dV y dx x dx= =1
2
1
22 1 2.
b. V x dx x= = = ⋅ −∫ 1
2
1
4 4
1
4 44 01 2 2 2
0
4
0
42 2. .
. ..
= 4.7982…
The midpoint Riemann sum R100 gives4.7981… , which is close to the answer foundusing integration.
c. If the cross sections were squares, they wouldhave twice the area of the triangles, so dV
would be twice as much and V = ⋅ =1
2 242 2
..
9.5964… .
20. y = ex, y = 3, and x = 0. Cross sectionsperpendicular to the x-axis are rectangles withheight equal to 4 times the base. Each base haslength (3 − y).
x
y
12
dV y y dx
e e dx e dxx x x
= − −
= − − = −
( )[ ( )]
( )[ ( )] ( )
3 4 3
3 4 3 4 3 2
V e dx e e dx
x e e
x x x
x x
= − = − +
= − +
= − ⋅ + ⋅ − + −
= …
∫ ∫4 3 4 9 6
4 9 61
2
4 9 3 6 31
29 0 6
1
2
7 5500
2
0
32
0
3
2
0
3
( ) ( )
ln
ln ln
ln
.
21. y = x2 and y = 2 − x2, intersecting at x = 1.Cross sections perpendicular to the x-axis areequilateral triangles. Each base has length (y2 − y1).
x
y
1
1
Using properties of special right triangles, youcan find that an equilateral triangle with
base b has height3
2b.
bb
b √ 312
60°12
dV bh dx y y y y dx
y y dx x x dx
x dx
= = − −
= − = − −
= −
1
2
1
2( )
3
2( )
( ) ( )
( )
2 1 2 1
2 12 2 2 2
2 2
3
4
3
42
3
42 2
V x dx
x x dx
x x x
= −
= − +
= − +
= − + − + −
= =
∫∫
3
42 2
3
44 8 4
3
44
8
3
4
5
3
44
8
3
4
50 0 0
8 3
15
0
12 2
2 4
0
1
3 5
0
1
( )
( )
0.9237...
22. y = ln x and y = 1. Cross sections perpendicularto the y-axis are rectangles with height equal to1/2 the base. Each base has length x.
x
y
1
5
dV x x dy x dy= ⋅ =12
12
2
Solve y = ln x for x, to get x = ey.
dV e dy e dyy y= =12
2 12
2( )
V e dy e ey y= = = − =∫ 1
2
1
4
1
4
1
41 59722 2
0
12
0
1
. K
23. a. Line has equation y x= 12 , 0 ≤ x ≤ 6.
b. The log has radius = 6, so the circle is
x2 + z2 = 36, or z x x= = −36 362 2– ( ) .1/2
c. dV = y ⋅ 2z ⋅ dx = 1
2x · 2(36 − x2)1/2 ⋅ dx
= (36 − x2)1/2 (x dx)
V x x dx
x x dx
=
= − −
∫∫
( – )
( – )
36
1
236 2
2 1 2
0
6
2 1 2
0
6
/
/
( )
( )
= − ⋅ =1
2
2
336 722 3 2
0
6
( – )x / 3 in.
Calculus Solutions Manual Problem Set 5-9 107© 2005 Key Curriculum Press
24. The points (0, 0) and (r, h) in xy-coordinates areon the line running up the top surface, so the
line is yh
rx= . The circle forming the boundary
for the bottom surface has radius = r and center(0, 0) in xz-coordinates, so the circle is x2 + z2 =r2, or z r x= 2 2– . The slab at x = x0 is
rectangular of height yhx
r= 0 , width
2 2 202z r x= – , and thickness dx, so
dVhx
rr x dx= −2 2 2 , and
Vh
rx r x dx
h
rr x x dx
r
r
= −
= − − −
∫∫
2
2
2 2
0
2 2 1 2
0( ) ( )/
= − ⋅ − = − ⋅ −
=
2
3
2
30
2
3
2 2 3 2
0
3 2 3
2
h
rr x
h
rr
r h
r
( ) ( )/ /
25. A cone of radius r and altitude h can be generatedby rotating about the x-axis the line
yr
hx= from x = 0 to h.
(x, y)
rh x
y
dV y dxr
hx dx= =π π2
2
22
Vr
hx dx
r
hx r h
h h
= = ⋅ =∫π π π2
22
2
23
0 0
21
3
1
3, Q.E.D.
26. a. Equation of circle in xy-plane is x2 + y2 = 100.
dV = πx2 dy = π(100 − y2) dy
V y dy= −−∫ π ( ) 100 2
10
10
= −
−
π 1001
33
10
10
y y
= − + −
π 10001
31000 1000
1
31000( ) ( )
= 4
3π (1000) cm3
b. Formula: V r= = =43
3 43
3 4310 1000π π π ( ) cm3,
which agrees with the answer by calculus.
27. Sphere can be generated by rotating about they-axis the circle x2 + y2 = r2.
Slicing perpendicular to the y-axis as in Problem26 gives dV = πx2 dy = π(r2 − y2) dy.
V r y dy r y yr
r
r
r
= − = −
− −
∫ π π( )2 2 2 31
3
= −
− − +
=π π πr r r r r3 ,
1
3
1
3
4
33 3 3 3
Q.E.D.
28. The graph shows slices perpendicular to x-axiswith sample points (x, y) and (x, z).
a
by
x
z
(x, y)
(x, z)c
Equation of ellipsoid is x
a
y
b
z
c
+
+
=
2
.2 2
1
For a fixed value of x, the x-term will beconstant. Subtracting this term from both sidesof the equation gives an equation of the form
y
b
z
ck
+
=
2 22, where k2 = 1 − (x/a)2.
Dividing both sides by k2 givesy
kb
z
kc
+
=
2 2
1. Thus, the y- and z-radii are
kb and kc, which have the original ratio b/c.Therefore, each elliptical cross section is similarto the ellipse at the yz-plane, Q.E.D.dV = πyz dxBecause z = (c/b)y, dV = π (c/b)y2 dx.The ellipse in the xy-plane (z = 0) has equation
x
a
y
b
+
=
2 2
1, from which y2 = (b/a)2(a2 − x2).
∴ dV = π (c/b)(b/a)2(a2 − x2) dx
V c b b a a x dxa
a
= −−∫ π ( / )( / ) ( )2 2 2
= −
−
π ( / )( / )2c b b a a x xa
a2 31
3
= ⋅ =π π( / )( / )2c b b a a abc4
3
4
33
Note that the volume formula for a sphere is aspecial case of the volume formula for anellipsoid in which a = b = c = r, the radius of thesphere.
29.50 + 2L
L
y
L
50
108 Problem Set 5-10 Calculus Solutions Manual© 2005 Key Curriculum Press
Note that the top of each isosceles trapezoidalcross section has length 50 + 2L yards, wherey
LL y y= ° ⇒ = ° =tan cot cot( ) ( ) .52 52
52
180
π
So each slab is dV L y dx= + +12 50 2 50( ) ;
dV y y dx= +5052
1802 cot
π, and
V y yk k
k
= +
⋅
= ≈=
∑ 5052
18030
1 649 443 6 1 649 443
2
0
19
cot
, , . , , .
π
K yd3
Cost = 12 ⋅ 1,649,443.6… ≈ $19,793,324
Problem Set 5-10
Q1.1
33x x C+ + Q2. 24
Q3. sec2 x dx x C= +∫ tan Q4. 2 sec2 x tan x
Q5. Answers may vary. Q6. See graph in Q5.
dt t
v
(t, v)
Q7. See graph in Q5. Q8. d (disp) = v dt
Q9. v dt a
b
∫ Q10. A
1. cos x dx .0.3
1.4
≈∫ 0 6899295233K
2. ( – )x x dx2
1
4
3 5 13 5+ =∫ .
3. 2 10 0988652x dx ≈∫ .0
3
K
4. tan 0.1
1.4
x dx∫ ≈ 1 76714178. K
5. cos sin sin sin.
.
.
.
x dx x0 3
1 4
0 3
1 4
1 4 0 3∫ = = − =. .
0.6899295233…
For the ten digits of the answer shown bycalculator, there is no difference between thissolution and the solution to Problem 1.
6. ( )x x dx x x x2
1
43 2
1
4
3 51
3
3
25− + = − +
∫
= − +
1
34
3
24 5 43 2( ) ( ) ( )
− − +
=1
31
3
21 5 1 13 53 2( ) ( ) ( ) .
There is no difference between this answer andthe solution to Problem 2.
7.
x
y
2
π
πsin2
04 9348∫ =x dx . K
We cannot compute this integral algebraicallybecause we do not know an antiderivative forsin2 x.
8.
x
y
5
2
π (ln ) .x dx2
2
5
14 6673∫ = K
We cannot compute this integral algebraicallybecause we do not know an antiderivative for(ln x)2.
9. a.
–20 20
–2
2 Si x
x
b. (sin x)/(x) approaches 1 as x approaches zero.
c. Answers will vary depending on the grapherused. The TI-83 gives Si 0.6 = 0.58812881using TRACE or 0.588128809608 using TABLE,both of which are correct to as many decimalplaces as the NBS values.
d. By TABLE, Si x seems to be oscillating betweenabout 1.53 and 1.61 when x is between 20and 30. The limit is somewhere betweenthese two numbers, say about 1.57. Theactual limit is π /2, which equals 1.570796… .
e.
–20 20
–2
2 Si x
x
The f graph is positive when x is between −πand π (as well as elsewhere), and has itsgreatest values there, which agrees with the
Calculus Solutions Manual Problem Set 5-10 109© 2005 Key Curriculum Press
large positive slope of the Si x graph in thisregion. Each place where the Si x graph has ahigh or low point, the f ( x) graph has a zero,corresponding to the zero slope of the Si xgraph. So f ( x) is the derivative of Si x.
10. a. erf x e dttx
= −∫2 2
0πThe integrand e t− 2
is an even functionintegrated between symmetrical limits. Thus,rather than using the entire interval [−x, x],one may find the integral on [0, x] and doublethe result.
b.y = erf x
x
1
2
c. By TABLE, values of erf x are
x erf x
1 0.842700792…
2 0.995322265…
3 0.999977909…
4 0.999999984…
The values approach 1, meaning that thefraction of the data between −x and x isvirtually 100% when x is beyond 4.
d. Answers will vary depending on thegrapher used. The TI-83 gives erf 0.5 =0.52049987781… using TABLE, which iscorrect to as many decimal places as the NBSvalue.
e.y = erf x
x
1
2
The slope of y = erf x appears to equal about1 when x = 0 and decreases toward zero as xincreases, which agrees with the graph of f.
11. a. (speed) [( / )/ ](dt ≈ + ⋅ + ⋅∫ 2 60 3 33 4 25 2 270
12
+ 4 ⋅ 13 + 2 ⋅ 21 + 4 ⋅ 5 + 9)= (1/90)(310) = 3.444… ≈ 3.4 nautical miles
b. T6 = (1/30)(33 ⋅ 0.5 + 25 + 27 + 13 + 21 + 5 + 9 ⋅ 0.5)= (1/30)(112) = 3.7333… ≈ 3.7 nautical miles
c. The answer by Simpson’s rule should becloser, because the graph is represented bycurved segments instead of straight ones.
12. (Data and CAT scans for this problem wereprovided by Dr. James Stewart of San Antonio.)
a. A dD ( . / )( .≈ + ⋅ + ⋅∫ 0 8 3 6 8 4 6 8 2 20 10
8
. .
+ 4 ⋅ 25.3 + 2 ⋅ 29.5 + 4 ⋅ 34.6 + 2 ⋅ 38.4+ 4 ⋅ 33.9 + 2 ⋅ 15.8 + 4 ⋅ 6.1 + 2.3)
= (0.8/3)(643.5) = 171.6 cm3
b. The mass will be 171.6 g, which is withinthe normal range of 150 to 200 g.
13. a.
0.5
Distance (in.)
Force (lb)
300
b. Let F = force, x = displacement, W = work.
W F dx= ∫ 0
0 5.
≈ (0.05/3)(0 + 4 ⋅ 120 + 2 ⋅ 240 + 4 ⋅ 360+ 2 ⋅ 370 + 4 ⋅ 330 + 2 ⋅ 290 + 4 ⋅ 280+ 2 ⋅ 270 + 4 ⋅ 270 + 190)
= (0.05/3)(7970) = 132.8333…≈ 132.8 inch-pounds
14. Let C = heat capacity (Btu/lb mole)/ºF,T = temperature (ºF), H = heat (Btu/lb-mole).Approximate values of C to the nearest 0.02 arefrom the given figure.
T C Simpson’s factor
500 8.44 1
1000 9.24 4
1500 10.08 2
2000 10.84 4
2500 11.48 2
3000 11.98 4
3500 12.36 2
4000 12.68 4
4500 12.94 1
H C dT= ≈ +∫ 500
38 44 9 24 4
500
4500
[ . ( . )( )
+ (10.08)(2) + (10.84)(4) + (11.48)(2)+ (11.98)(4) + (12.36)(2)+ (12.68)(4) + 12.94]
= = ≈500
3268 18 44696 6666( . ) . K 44,697 Btu
The answers students get will vary slightly.
110 Problem Set 5-11 Calculus Solutions Manual© 2005 Key Curriculum Press
15. a. Simpson’s rule will give a more accurateanswer because the function y = sin x isapproximated better by quadratic functionsthan by straight lines.
b. S4 = (1/3)(π /4)[sin 0 + 4 sin (π /4) +2 sin (π/2) + 4 sin (3π /4) + sin π] = 2.0045…
T4 = (1/2)(π /4)[sin 0 + 2 sin (π/4) +2 sin (π /2) + 2 sin (3π /4) + sin π] = 1.8961…
sin cos0
x dx xπ π
π∫ = − = − =0
0 2cos cos
Simpson’s rule does give a betterapproximation of the integral because S4 iscloser to 2 than is T4.
16. Programs will vary depending on the type ofgrapher used. See the program in the Programsfor Graphing Calculators section of theInstructor’s Resource Book.
17. Using a Simpson’s rule program, the mass of thespleen is 171.6 cm3.
18. Enter Y12 2
= −
πe x . A Simpson’s rule program
gives S50 = 0.5204998781… and S100 =0.5204998778… . There is little differencebetween the two estimations, and both are closeto the tabulated value.
19. a.
1
2
y
tx
As x varies, the area beneath the curve y = 1/tfrom t = 1 to t = x varies also.
b. Using the power formula on t dt−∫ 1 gives
t0
0.
Division by zero is undefined, so thisapproach does not work.
c. Graph Y1 = fnint(x− 1, x, 1, x). (Entries maybe different for different calculators.) Thegraph looks like y = ln x. The value of f (x)is negative for x < 1 because for these valuesthe lower limit of integration is larger than theupper limit, resulting in negative values for dx.
5
1
y
x
d. f (2) = 0.6931…
f ( 3) = 1.0986…
f ( 6) = 1.7917…
f ( 2) + f ( 3) = f ( 2 ⋅ 3). This is a property oflogarithmic functions.
Problem Set 5-11Review Problems
R0. Answers will vary.
R1. a. The width of each region is 4. So
T3 = (4/2)[v(4) + 2v(8) + 2v(12) + v(16)] =2[22 + 2(26.9705…) + 2(30.7846…) + 34] =343.0206… . T3 underestimates the integralbecause v(t) is concave down, so trapezoidsare inscribed under the curve.
b. R3 = 4[v(6) + v(10) + v(14)] = 4(24.6969… +28.9736… + 32.4499…) = 344.4821…
This Riemann sum is close to the trapezoidal-rule sum.
c. T50 = 343.9964… , and T100 = 343.9991…Conjecture: The exact value of the integralis 344.
d . g ( t) = 10t + 4t1.5
g(16) − g(4) = 344
This is the value the trapezoidal-rule sums areapproaching.
R2. a. The slope of the linear function is the sameas the slope of the curve at x = 1. So theslope is
f ( x) = sin π x ⇒ f ′(x) = π cos π x ⇒f′( 1) = π cos π = −πAt x = 1, y = sin π = 0
y − 0 = −π (x − 1) ⇒ l(x) = −π x + π
1
1y
xf(x)
l (x)
0.8 1 1.2
x
f(x)l (x)
0.2
As you zoom in, you see that f ( x) is veryclose to the line l(x) for values near x = 1.
Calculus Solutions Manual Problem Set 5-11 111© 2005 Key Curriculum Press
For x = 1.1, the error is sin [π ( 1.1)] −[−π ( 1.1) + π] = 0.0051…For x = 1.001, the error is sin [π ( 1.001)] −[−π(1.001) + π] = 5.1677… × 10− 9.
b. i. y = csc5 2x ⇒ dy = −10 csc5
2x cot 2x dx
ii. y = x5/5 − x− 3/3 ⇒ dy = (x4 + x− 4) dx
iii. y = (7 − 3x)4 ⇒ dy = −12(7 − 3x)3 dx
iv. y = 5e− 0. 3x ⇒ dy = −1.5e− 0. 3x dx
v. y x dyx
x dx= ⇒ = ⋅ ⋅ln( )
( )( )421
24 2 24
3
= 4/x dx; or y = ln ( 2x)4 = 4 ln ( 2x) ⇒
dy = 41
22⋅ ⋅
xdx = 4/x dx
c. i. dy = sec x tan x dx ⇒ y = sec x + C
ii. dy x dx y x C= + ⇒ = + +( )53 71
183 7 6( )
iii. dy = 5 dx ⇒ y = 5x + C
iv. dy = 0.2e− 0.2 x ⇒ y = −e− 0.2 x + C
v. dy dx y Cxx
= ⇒ = +66
6
ln
d. i. y = (2x + 5)1/2 ⇒ dy = (2x + 5)− 1/2 dx
ii. x = 10 and dx = 0.3 ⇒dy = 25− 1/2 ⋅ 0.3 = 0.06
iii. ∆y = (2 ⋅ 10.3 + 5)1/2 − (2 · 10 + 5)1/2
= 0.059644…
iv. 0.06 is close to 0.059644… .
R3. a. See the text for the definition of indefiniteintegral.
b. i. 12 7 22 3 5 3x dx x C/ /= +∫ .
ii. sin cos sin6 71
7x x dx x C= +∫
iii. ( )x x dx x x x C2 3 28 31
34 3− + = − + +∫
iv. 12 43 3e dx e Cx x∫ = +
v. 77
7x
x
dx C∫ = +ln
R4. a. See the text for the definition of integrability.
b. See the text for the definition of definiteintegral.
c. sec.
.
x dx0 2
1 4
∫i. U6 = 2.845333…
ii. L6 = 1.872703…
iii. M6 = 2.209073…
iv. T6 = 2.359018…
d. U6 L6
0.2 1.4
x
5y
0.2 1.4
x
5y
M6 T6
0.2 1.4
x
5y
0.2 1.4
x
5y
e.
y
x
f(x)
R5. a. The hypothesis is the “if ” part of a theorem,and the conclusion is the “then” part. (Hypo-means “under,” and -thesis means “theme.”)
b. d t t( ) = +20 34
sinπ
Average velocity 1.5 m/s= =d d( ) – ( )
–
2 0
2 0
Instantaneous velocity = d ′(t) = 0.75π cos π4
t
d c c′ = =( ) . .0 754
1 5π πcos
π4
c = cos− 1 (2/π) = 0.880689…
c = 1.12132… ≈ 1.12 s
c. g(x) = x4/3 − 4x1/3 = x1/3(x − 4)
g(x) = 0 ⇒ x = 0 or x = 4. Interval is [0, 4].
g′(x) = (4/3)x1/3 − (4/3)x− 2/3 = (4/3)x− 2/3(x − 1)
g′(c) = 0 ⇒ c = 1
At x = 0, g′(0) takes the form 1/0, which isinfinite.Thus, g is not differentiable at x = 0.However, the function need not bedifferentiable at the endpoints of the interval,just continuous at the endpoints anddifferentiable at interior points.
d. For a function to be continuous on a closedinterval, the limit needs to equal the functionvalue only as x approaches an endpoint fromwithin the interval. This is true for function f
112 Problem Set 5-11 Calculus Solutions Manual© 2005 Key Curriculum Press
at both endpoints, but not true for function gat x = 2. The graphs show that the conclusionof the mean value theorem is true for f butnot for g.
2 7
4
f(x)
x
Tangent parallels secant
c
Secant
2 7
4
g(x)
x
Secant
No tangentparallels secant
(Middle branch has the equation y = 1.4( x−2 ) .Point c = 4.4825… .)
e. g is the linear function containing the points(a, f (a)) and (b, f (b)). h is the function h (x) =f ( x) − g(x). Thus, h(a) = h ( b) = 0, satisfyingone hypothesis of Rolle’s theorem. The othertwo hypotheses are satisfied because f and gare differentiable and continuous at theappropriate places, and a difference ofdifferentiable and continuous functions alsohas these properties. The c in (a, b) for whichh′(c) = 0 turns out to be the c in (a, b) forwhich f ′ (c) equals the slope of the secant line,g′(c), which equals [f (b) − f (a)]/(b − a).
f.
x
f(x)8
1
Points are 18
14
38
12
58
34
78, , , , , , and .
g. If r′(x) = s′(x) for all x in an interval, thenr(x) = s(x) + C for some constant C.
R6. a. g x x dx x C( ) .= = +∫ 1 5 2 50 4. .
Without loss of generality, let C = 0.
g c
g g′ = =( ) .12 1
2 11 862741
( ) – ( )
–K
∴ = …⇒c11 5 1 862741. .
c1 = (1.862741…)1/1.5 = 1.513915927…
Similarly, c2 = 2.50833898… .
c3 = 3.505954424…
For x dx1 5
1
.4
,∫
R3 = (1.513…)1.5 + (2.508…)1.5 + (3.505…)1.5
= 12.4, which is the exact value of theintegral.
b. ( – )10 10 1 32 3
1
3
1
3
x dx x x= −− −∫ ( / )
= 30 − 9 − (−10) + (−1/3) = 92/3 = 30.6666…
c. T100 = 30.6656, which is close to 92/3.
d. M10 = 30.72
M100 = 30.6672
M1000 = 30.666672
These Riemann sums are approaching 92/3.
R7. a. i. x dx x− − − −= − = − + =∫ 2 1
1
5
1
51 15 1 4 5 /
ii. ( )x x dx2 5
3
4
3+∫ ( )
= +∫( / ) ( )1 2 3 22 5
3
4
( )x x dx
= +( / )( )112 32 63
4x
= (1/12)(19)6 − (1/12)(12)6
= 3,671,658.08…
iii. (sin – ) cosx dx x x5 50 0
= − −∫π π
= −cos π − 5π + cos 0 + 0 = 2 − 5π
iv. 4 2 2 2 482 2
0
5
0
52 5 0e dx e e ex x= = − =∫
ln lnln
v. 33
3
3
3
3
3
78
31
4
1
4 4 1x
x
dx∫ = = − =ln ln ln ln
= 70.9986…
b.
x
y
–5
Integral is negative, because each y-value inthe Riemann sum is negative.
c. ( sin – )4 6 8 4 2 47 3
10
10
x x x dx dx+ + =−∫ ∫0
10
= =8 80x 0
10
d. Total area = sum of two areas
x
f(x)
a c b
Calculus Solutions Manual Problem Set 5-11 113© 2005 Key Curriculum Press
R8. a.
dt t
v
(t, v)
dy = v dt = 150t0.5 dt
y t dt t= = =∫ 150 100 27000 5 1 5
0
9
0
9. . ft
For [ , ], ..0 4 150 8000 5
0
4
y t dt= =∫For [ , ], .4 9 150 2700 8000 5
4
9
y t dt= = −∫= 1900.
So v t dt( ) = = +∫ 2700 800 19000
9
= + ∫∫ v t dt v t dt( ) ( ) .4
9
0
4
b. dA = x dyy = ln x ⇒ x = ey
dA = ey dy
e dy e e ey y
0
4
0
44 0 4 1 3
ln lnln∫ = = − = − =
R9. a. y = e0.2 x, from x = 0 to x = 4, is rotated aboutthe x-axis.
0 4
1
(x, y)
x
y
dV = π y2 dx = πe0.4 x dx
V e dx ex x= =∫ π π0 4 0 4
0
4
0
4
2 5. . .
= 2.5π (e1.6 − 1) = 31.0470…
b. y x= 10 25. and y = x2, intersecting at (0, 0) and
(1, 1) in Quadrant I, is rotated about they-axis. Only the back half of the solid isshown.
1
1
x
y
(x2, y)
(x1, y)
y x x y= ⇒ =10 25
14.
y = x2 ⇒ x2 = y
dV x x dy y y dy= − = −π π( ) ( ) 22
12 2 8
V y y dy y y= − = −
= =
∫ π π
π
( )
.
2 8 3 9
0
1
0
11
3
1
9
2
90 6981K
c. y = x1 + 2 ⇒ x1 = y − 2
y = 3x2 − 6 ⇒ x2 = 1
3y + 2
Graphs intersect at y = 6. Diameter of circularcross section is (x2 − x1).
dV = π [0.5(x2 − x1)]2 dy
= +
− −
= −
π π4
1
32 2
44
2
3
2 2
y y dy y dy( )
V = dV0
6
∫ ≈ 25.1327… (exactly 8π)
The right circular cone of altitude 6 andradius 2 also has volume 1
3 π π⋅ ⋅ =2 6 82 .
R10. a.
log x dx =∫ 6 09131
10
. K
The integral is reasonable because countingsquares gives approximately 6.
1 10
1t
v(t)
b. dW = v · y · dx = (1000 + 50x)(4 − 0.2x2) dx
( )( . ) , .1000 50 4 0 2 10 897 52
0
3
+ − =∫ x x dx
c. v t dt( )3
5
∫ = 1/3(0.2)[29 + 4(41) + 2(50) + 4(51)
+ 2(44) + 4(33) + 2(28) + 4(20)+ 2(11) + 4(25) + 39] = 67.6
Values of velocity are more likely to beconnected by smooth curves than by straightlines, so the quadratic curves given bySimpson’s rule will be a better fit than thestraight lines given by the trapezoidal rule.
Concept Problems
C1. a.
5b
f(b)
π/2
5–
5
–5
b.
3
1 b
f (b)
114 Problem Set 5-11 Calculus Solutions Manual© 2005 Key Curriculum Press
The two lines are horizontal asymptotes. y4 isthe inverse of y1.
c. The graph of f in part a is a reflection ofthe graph of y = tan x in part b across the liney = x. Function f seems to be the inversetangent function, f (b) = tan− 1 b. (In Chapter 9,students will learn that this is actually true.)
C2. f (x) = ax2 + bx + cy
x
d k e
f (d ) = ad 2 + bd + cf (e) = ae2 + be + c
∴ = + + − + +−
mae be c ad bd c
e d
2 2( )
= − + −−
= + +a e d b e d
e da e d b
( ) ( )( )
2 2
f ′(x) = 2ax + b ⇒ f ′(k) = 2ak + b∴ 2ak + b = a(e + d) + b
2ak = a(e + d)
k = (1/2)(e + d)∴ k is at the midpoint of [d, e], Q.E.D.
C3. S n n( ) = + + + + +0 1 2 32 2 2 2 2L
a. S(0) = 0, S(1) = 1, S(2) = 5, S(3) = 14S(n) = an3 + bn2 + cn + d0 = 0 + 0 + 0 + d1 = a + b + c + d5 = 8a + 4b + 2c + d14 = 27a + 9b + 3c + dSolving this system gives a = 1/3, b = 1/2,c = 1/6, d = 0.∴ S(n) = (1/3)n3 + (1/2)n2 + (1/6)n
= (n/6)(n + 1)(2n + 1)
b. By equation,S(4) = (4/6)(5)(9) = 30S(5) = (5/6)(6)(11) = 55By addition,S(4) = 0 + 1 + 4 + 9 + 16 = 30, which checks.S(5) = 0 + 1 + 4 + 9 + 16 + 25 = 55, whichchecks.S(1000) = (1000/6)(1001)(2001) =333,833,500
c. Prove that S(n) = (n/6)(n + 1)(2n + 1) for anypositive integer n.
Proof (by induction on n):
Anchor:
For n = 1, S(n) = (1/6)(2)(3) = 1, the correctanswer, which anchors the induction.
Induction hypothesis:
Assume that for some integer n = k > 1,S(k) = (k/6)(k + 1)(2k + 1).Verification for n = k + 1:
S k k k( ) ( )2+ = + + + + +1 0 1 12 2 2L
= + + + + +( ) ( )20 1 12 2 2L k k
= (k/6)(k + 1)(2k + 1) + (k + 1)2
= [(k + 1)/6][k(2k + 1) + 6(k + 1)]= [(k + 1)/6][2k2 + 7k + 6]= [(k + 1)/6][(k + 2)(2k + 3)]= [(k + 1)/6][(k + 1) + 1] ⋅
[2(k + 1) + 1],which is the formula with (k + 1) in place ofk, thus completing the induction.∴ S(n) = (n/6)(n + 1)(2n + 1) for any positiveinteger n, Q.E.D.
d. S n n( ) = + + + + +0 1 2 33 3 3 3 3LS(0) = 0S(1) = 0 + 1 = 1S(2) = 0 + 1 + 8 = 9S(3) = 0 + 1 + 8 + 27 = 36S(4) = 0 + 1 + 8 + 27 + 64 = 100(The answers are perfect squares!)Assume that S(n) = an4 + bn3 + cn2 + dn + e.0 = 0 + 0 + 0 + 0 + e1 = a + b + c + d + e9 = 16a + 8b + 4c + 2d + e36 = 81a + 27b + 9c + 3d + e100 = 256a + 64b + 16c + 4d + eSolving this system givesa = 1/4, b = 1/2, c = 1/4, d = 0, e = 0.∴ S(n) = (1/4)n4 + (1/2)n3 + (1/4)n2
= (1/4)n2(n2 + 2n + 1)= [(n/2)(n + 1)]2,
which agrees with the observation that S(n) isa perfect square.By equation,S(5) = [(5/2)(6)]2 = 225S(6) = [(6/2)(7)]2 = 441By addition,S(5) = 03 + 13 + 23 + 33 + 43 + 53 = 225,which checks.S(6) = 03 + 13 + 23 + 33 + 43 + 53 + 63 =441, which checks.
C4. a. 4 100
sin sinx x dxπ
∫= +∫ [– cos cos (– )]2 11 2 9
0x x dx
π
= − − −2
1111
2
99
0sin sinx x( )
π
= −0 − 0 + 0 + 0 = 0, Q.E.D.There is just as much area below the x-axis asthere is above it, so the integral is 0.
Calculus Solutions Manual Problem Set 5-11 115© 2005 Key Curriculum Press
b. 40
sin sinx nx dxπ
∫= − + + −∫ { cos [( ) ] cos [( ) ]}2 1 2 1
0n x n x dx
π
=+
+ + −–sin ]
–sin
2
11
2
11
0nn x
nn x[( ) [( ) ]
π
=+
+ + −–sin
–sin
2
11
2
11
nn
nn( ) ( )π π
−+
−–sin
–sin
2
10
2
10
n n
If n is an integer, the first two terms willinvolve sines of integer multiples of π, andare thus equal to 0. The last two terms are 0unless n = ±1. Thus, the integral equals 0 forany integer n > 1, Q.E.D.
C5. a. Algebraic solution:
Pick sample points ck at the right end of eachsubinterval. Because f (x) is increasing on theinterval [1, 9], the high points of f (x) arelocated at the right ends of the subintervalsand the low points are at the left ends.
∴ ==
∑U f c xn k k
k
n
( )∆1
and L f c xn k k
k
n
= −=
∑ ( )1
1
∆
Note that c0 = 1 and cn = 9. Subtracting gives
U L f c f c xn n k k k
k
n
− = − −=
∑[ ( ) ( )]1
1
∆
= − + − +[ ( ) ( )] [ ( ) ( )]f c f c x f c f c x1 0 1 2 1 2∆ ∆ L+ [ f ( cn) − f ( cn− 1)]∆xn
≤ [ f ( c1) − f ( c0)] ||P|| + [ f ( c2) − f ( c1)] ||P||
+ + − −L [ ( ) ( )] || ||f c f c Pn n 1
= − + − + +[ ( ) ( ) ( ) ( )f c f c f c f c f cn1 0 2 1 L ( ) − f ( cn− 1)] ||P||= [ f ( cn) − f ( c0)] ||P|| = ||P||(1.29 − 1.21)∴ Un − Ln ≤ ||P||(1.29 − 1.21), Q.E.D.
Graphical solution:
The difference Un − Ln is equal to the area ofthe spaces between the lower and upperrectangles in Figure 5-11g. Imagine thesespaces moved over to the left so that theyalign at x = 1 (graph). The spaces can becircumscribed with a rectangle of base ||P||and altitude (1.29 − 1.21). Thus, Un − Ln ≤||P|| (1.29 − 1.21), Q.E.D.
1 9
x
f(x)
Norm = largest ∆ x
Slide them over.
Norm
1.21
1.29
b. From part a, 0 ≤ Un − Ln ≤ ||P|| (1.29 − 1.21).As ||P|| approaches zero, the rightmostmember of the inequality goes to zero. By the
squeeze theorem, lim ( ) ,P
n nU L→
− =0
0 which
implies lim lim .P
nP
nU L→ →
=0 0
So f is integrable
on [1, 9] by the definition of integrability,Q.E.D.
c. Prove that g(x) = 1/x is integrable on [1, 4].
Proof:
Partition the interval [1, 4] into nsubintervals whose widths are not necessarilyequal. Let ||P|| be the norm of the partition.Pick sample points ck at the left end of eachsubinterval. Because g (x) is decreasing on[1, 4], the high points are located at the leftends of the subintervals and the low pointsare at the right ends (graph).
1 4
x
y1
By algebraic or graphical reasoning as inpart a, Un − Ln ≤ ||P||(1 − 1/4). As ||P||approaches zero, Un − Ln is squeezed to zero.Thus, Un and Ln approach the same limit,which implies that g is integrable on [1, 4],Q.E.D.
d. This reasoning cannot be applied directly toh (x) = sin x on the interval [0, 3] becauseh (x) is both increasing and decreasing ondifferent parts of the interval. Thus, the highpoints are not always at the same end of thesubinterval and the high point at π/2 may notbe at either end of a subinterval (graph).
x
y = sin x
0 3
1
π/2
The reasoning could be applied indirectly byfirst splitting the interval [0, 3] into [0, π /2]and [π /2, 3] so that h(x) is increasing on oneand decreasing on the other.
Chapter Test
T1. Indefinite integral:
g x f x dx( ) ( ) = ∫ if and only if g′(x) = f (x).
116 Problem Set 5-11 Calculus Solutions Manual© 2005 Key Curriculum Press
T2. Definite integral:Let Ln and Un be lower and upper sums of f (x)on the interval [a, b]. Then
f x dx L Un
na
b
nn( ) lim lim ,= =
→∞ →∞∫provided the two limits are equal.
T3. Fundamental theorem:If f is an integrable function, and
g x f x dx( ) ( ) , = ∫ then
f x dx g b g aa
b
( ) ( ) ( ). = −∫T4. The function f is continuous on the interval
[3, 8] and differentiable on (3, 8). It does notmatter that it is not differentiable at theendpoints.
T5.
3
x
f (x)
8c
T6. Hypotheses: f (a) = f (b) = 0Differentiable on (a, b).Continuous at x = a and x = b.Conclusion: There is a c in (a, b) such thatf ′ (c) = 0.
a
c
b
x
f(x)
T7.
3
x
f (x)
8
T8. y = esin x ⇒ dy = cos x esin x dx
T9. 0 10 1
0 1.
.
ln .x
x
dx C= +∫T10. ( ) ( ) ( )4 13
1
724 133 5 2 3 6x x dx x C+ = + +∫
T11. x dxx2
3
1
4
1
4 3 3
3
4
3
1
321= = − =∫
T12. ( )12 10 310
33 2 4
2
23
2
2
x x dx x x+ = +− −
∫= − − − = −48
80
348
80
353
1
3
T13. π [( ) ( ) ]x x dyc
d
12
22−∫
T14. The slope of the linear function is the same asthe slope of the curve at x = 1. So the slope isfound by y = x3 ⇒ y′ = 3x2 ⇒ y′(1) = 3.At x = 1, y = 1.y − 1 = 3(x − 1) ⇒ y = 3x − 2
1
1
x
y
(1, 1)
1
1
1.2
1.2
As you zoom in on (1, 1), you see that the graphof y = x3 is locally linear.
T15. a. 12 480 25 0 25
0
3
0
3
e dx ex x. .=∫= − =48 48 53 61600 75 0e e. . K
b. M50 = 53.6154…T50 = 53.6170…S50 = 53.61600081…The midpoint Riemann sum error is0.000502646712… .The trapezoidal-rule error is 0.0010052962… .The midpoint Riemann sum error is half ofthe trapezoidal-rule error, because2(0.000502646712…) = 0.0010052962… .The Simpson’s rule error is0.000000015077… , which is much smallerthan the error for the other two methods.
T16. a.
y = cos x
x
(x, y)
y
π
1
b. dV y dx dV x dx= ⇒ =1
2
1
22 2cos
V x dx= ∫ 1
22
0
2
cos/π
Calculus Solutions Manual Problem Set 5-11 117© 2005 Key Curriculum Press
c. This integral cannot be evaluated algebraicallybecause we do not know an antiderivative forcos2 x.
V x dx= =∫ 1
20 39262
0
2
cos ./
Kπ
T17. a. g x x dx x C x( ) . . .= = + =∫ 0 3 0 1 0 12 3 3
b.
1 2.64... 4
x
f(x)
5
1 2.64... 4
x
g(x)
5
c. mg g= = =( ) – ( )
–
. – .4 1
4 1
6 4 0 1
32 1.
g′(x) = f ( x) = 0.3x2
∴ 0.3c2 = 2.1 ⇒ c = 7 = 2.6457513…
In the right graph in part b, the tangent line atx = 2.64… is parallel to the secant line fromx = 1 to x = 4.
d. f ( 2.645…) = 0.3(2.645…)2 = 2.1 (exactly)The point is (2.645… , 2.1).Area of region under graph equals area ofrectangle, as shown in the graph on the left inpart b.
T18. Answers will vary.
118 Problem Set 6-2 Calculus Solutions Manual© 2005 Key Curriculum Press
Chapter 6—The Calculus of Exponentialand Logarithmic Functions
Problem Set 6-11. The integral would be 1
00P , which involves
division by zero.
N Integral
1000 01500 0.4054…2000 0.6931…2500 0.9162…500 −0.6931…100 −2.3205…
1000 2000
1
2
–1
–2
Integral
N
The graph resembles a logarithmic function.
2. 0 05 0 05 0 5 0 0 50
10
0
10
.∫ = = − =dt t. . . , Q.E.D.
0.5 is between 0.4054… and 0.6931… , thevalues of the left integral for N = 1500 andN = 2000.
By solver, 1
1000 P
N
∫ dP = 0.5 when
N = 1648.7212… , or about 1649 people.
3. At 20 years, the integral on the right equals 1. At0 years, the integral equals 0. Solving for N atthese times gives N = 2718.2818… for 20 years,and N = 1000 (as expected!) for 0 years.
5 10 15 20
1000
2000
3000Population
Time (yr)
The graph resembles an exponential function.
4. ln 1648.7212… − ln 1000 = 0.5, exactly. Thisis the value of the integral on the left!
Problem Set 6-2
Q1.1
0 70 7
.x C. + Q2. 9
Q3. f ′ ( x) = −2 cos x sin x Q4. continuous
Q5. differentiable Q6. yx
′ = 1
1 2–
Q7. y′ = − csc x cot x Q8. Riemann sum
Q9. indefinite integral, or antiderivative
Q10. log 12
0. Answers will vary.
1. y = ln 7x ⇒ y′ = 1/(7x) · 7 = 1/x2. y = ln 4x ⇒ y′ = 1/(4x) · 4 = 1/x3. f (x) = ln x5 ⇒ f ′ (x) = 1/(x5) · 5x4 = 5/x4. f (x) = ln x3 ⇒ f ′ (x) = 1/(x3) · 3x2 = 3/x5. h (x) = 6 ln x− 2 ⇒ h′ (x) = 6/(x− 2) · (−2x− 3) = −12/x6. g (x) = 13 ln x− 5 ⇒
g′ (x) = 13/(x− 5) · (−5x− 6) = −65/x7. r (t) = ln 3t + ln 4t + ln 5t ⇒
r ′ (t) = 1/(3t) · 3 + 1/(4t) · 4 + 1/(5t) · 5 = 3/t8. v (z) = ln 6z + ln 7z + ln 8z ⇒
v′ (z) = 1/(6z) · 6 + 1/(7z) · 7 + 1/(8z) · 8 = 3/z9. y = (ln 6x)(ln 4x) ⇒
y′ = 1/(6x) · 6 · (ln 4x) + (ln 6x)[1/(4x) · 4]
= (1/x)(ln 4x + ln 6x) or ln 24 2x
x
10. z = (ln 2x)(ln 9x) ⇒z′ = 1/(2x) · 2 · (ln 9x) + (ln 2x)[1/(9x) · 9]
= (1/x)(ln 9x + ln 2x) or ln 18 2x
x
11. yx
x = ln
ln
11
3 ⇒
yx x x x
x′ = ⋅ ⋅ ⋅1 11 11 3 11 1 3 3
3 2
/( ) (ln ) – (ln ) /( )
(ln )
= ln – ln
(ln )
3 11
3 2
x x
x x or
ln( / )
(ln )
3 11
3 2x x
12. yx
x= ⇒ln
ln
9
6
yx x x x
x′ = ⋅ ⋅ ⋅ ⋅1 9 9 6 9 1 6 6
6 2
/( ) (ln ) – (ln ) /( )
(ln )
= ln – ln
(ln )
6 9
6 2
x x
x x or
ln( / )
(ln )
2 3
6 2x x
13. p = (sin x)(ln x) ⇒ p′ = (cos x)(ln x) + (sin x)(1/x)
14. m = (cos x)(ln x) ⇒m′ = (−sin x)(ln x) + (cos x)(1/x)
15. y = cos (ln x) ⇒ y′ = −sin (ln x) · (1/x)
16. y = sin (ln x) ⇒ y′ = cos (ln x) · (1/x)
Calculus Solutions Manual Problem Set 6-2 119© 2005 Key Curriculum Press
17. y = ln (cos x), where cos x > 0 ⇒ y′ = (1/cos x) · (−sin x) = −tan x (Surprise!)
18. y = ln (sin x), where sin x > 0 ⇒ y′ = (1/sin x) · (cos x) = cot x (Surprise!)
19. T (x) = tan (ln x) ⇒ T ′ (x) = sec2 (ln x) · (1/x)
20. S (x) = sec (ln x) ⇒ S ′ (x) = sec (ln x) · tan (ln x) · (1/x)
21. y = (3x + 5)−1 ⇒ y′ = −(3x + 5)−2 · 3 = −3(3x + 5)−2
22. y = (x3 − 2)−1 ⇒y′ = −(x3 − 2)−2 · 3x2 = −3x2(x3 − 2)
23. y = x4 ln 3x ⇒ y′ = 4x3 ln 3x + x4 · 1/(3x) · 3 = 4x3 ln 3x + x3
24. y = x7 ln 5x ⇒ y′ = 7x6 ln 5x + x7 · 1/(5x) · 5 = 7x6 ln 5x + x6
25. y = ln (1/x) ⇒ y′ = 1/(1/x) · (−x−2) = −1/x26. y = ln (1/x4) ⇒ y′ = 1/(1/x)4 · (−4x−5) = −4/x
27. 7 7/ | | x dx x C= +∫ ln
28. 5 5/ x dx x C= +∫ ln | |
29. 1 31
3/( ) x dx x C= +∫ ln | |
30. 1 81
8/( ) | | x dx x C= +∫ ln
31.x
xdx
xx dx
2
3 32
5
1
3
1
53
+=
+∫∫ ( )
= + +1
353ln || x C
32.x
xdx
xx dx
5
6 65
4
1
6
1
46
– –( )= ∫∫
= +1
646ln –| |x C
33.x
xdx
xx dx
5
6 65
9
1
6
1
96
– –(– )= −∫ ∫
= − +1
69 6ln –| |x C
34.x
xdx
xx dx
3
4 43
10
1
4
1
104
– –(– )= −∫ ∫
= − +1
410 4ln –| |x C
35.sec tan
secln sec
x x dx
xx C
11
+= + +∫ | |
36.sec
tanln | tan |
2
11
x dx
xx C
+= + +∫
37.cos
sinln | sin |
x dx
xx C= +∫
38.sin
cos
– sin
cosln | cos |
x dx
x
x dx
xx C= − = − +∫∫
39. ( / ) ln ln ln. .
1 4 0 50 5
4
0 5
4
w dw w= = −∫ | | .
= ln 8 = 2.079441…
40. ( / ) ln ln ln. .
1 10 0 10 1
10
0 1
10
v dv v= = −∫ | | .
= ln 100 = 4.605170…
41. ( / ) ln ln ln. .
1 3 0 10 1
3
0 1
3
x dx x | | | | | . |= = − − −−
−
−
−
∫= ln 3 − ln 0.1 = ln 30 = 3.401197…
42. ( / ) ln ln ln. .
1 4 0 20 2
4
0 2
4
x dx x= = − − −−
−
−
−
∫ | | | | | . |
= ln 4 − ln 0.2 = ln 20 = 2.995732…
43.x dx
x xx dx
1 2
3 24
9
3 24
91 2
1
2
3
1
1
3
2
/
/ /+=
+⋅∫ ∫ /
= + = =2
31
2
328 9 0 7566533 2
4
9
ln | | (ln – ln )/x . K
44.x dx
x xx dx
−−
+=
+⋅∫ ∫
1 3
2 31
8
2 31
81 3
2
3
2
1
2
2
3
/
/ //
= + = =3
22
3
26 3 1 5 22 3
1
8
ln | | (ln – ln ) ln/x .
= 1.039720…
45. ( )5ln (ln )xdx
xx C= +∫ 1
66
46.ln
ln (ln )x
xdx x
dx
xx C= = +∫∫ ( ) 1 21
2
47. f x t dt f x xx
( ) ( )= ⇒ ′ =∫ cos cos3 32
48. f x t t dtx
( ) = + ⇒∫ ( – )2
510 17
′ = + −f x x x( ) 2 10 17
49.d
dxt dt x
x
tan tan3
2
3∫
=
50.d
dxdtt
xx2
1−∫
= 2
51. f x dt f x xt xx
( ) ( )= ⇒ ′ = ⋅∫ 3 2 32
2
1
52. g x t dtx
( ) = ⇒∫0
cos
g x x x′ = ⋅ −( ) ( )cos sin
53. h x t dtx
( ) = + ⇒−
∫ 1 2
0
3 5
h x x′ = +( ) 3 1 3 5 2( – )
54. p x t dt p x x xx
( ) ( ) ( )= + ⇒ ′ = + ⋅∫ ( )–
4 7 12 7 2
11 1 3
3
55. ( / ) ln ln ln5 5 5 3 5 11
3
1
3
x dx x= = −∫ | | = 5 ln 3
= 5.493061…
120 Problem Set 6-2 Calculus Solutions Manual© 2005 Key Curriculum Press
Midpoint Riemann sum: M100 = 5.492987…Trapezoidal rule: T100 = 5.493209…Numerical integration: 5.493061…
56. Answers will vary.
57. a. By finding areas, g (0) ≈ −2.7, g (1) = 0,g(2) = 1, g(3) ≈ 0.3, g(4) ≈ −0.3, g(5) ≈ 0.7,g (6) ≈ 3.3, g (7) = 6, and g (8) = 7.
1 2 3 5 6 7 8
1
2
3
4
5
6
7
–1
–2
–3
g
x
y
b. h x f t dt h x f xx
( ) ( )= ⇒ ′ = − ⋅−
∫ ( ) ( )1
12
2
1 2x ⇒
h′(2) = f (3) ⋅ 4 = −1 ⋅ 4 = −4
58. a. By finding areas, g (0) = −6, g (1) = −2.5,
g (2) = 0, g (3) = 1.5, g (4) = 2, g (5) = 1.5,
g (6) = 0.75, g (7) = 0.5, g (8) = 0.75,
g (9) = 1.5, and g (10) = 2.75.
1 2 3 4 5 7 8 9 10
1
2
3
4
–1
2–
3–
4–
5–
6–
g
x
y
b. h x f t dt h x f xx
( ) ( ) ( ) ( )= ⇒ ′ = ⋅ ⇒∫ 2 22
2
h′(4) = f (8) ⋅ 2 = (0.5)(2) = 1
59. ( / ) ln ln ln1 10001000 1000
P dP P NN N
= = −∫ | |
0 05 0 05 0 50
10
0
10
. . dt t= =∫ .
ln N − ln 1000 = 0.5
lnN
10000 5= .
Ne
10000 5= .
N = 1000e0.5 ≈ 1648.721…≈ 1649 people
60. a. F + 30 = k/h0 + 30 = k/20 ⇒ k = 600∴ F + 30 = 600/h ⇒ F = 600/h − 30
b.
h
F
30
10 20
dh
(h, F)
c. Work equals force times displacement.Because the force varies, a definite integralmust be used.
d. The work done compressing the air a smallamount, dh, is approximately equal to theforce at the sample point (h, F ) times dh(see part b).
dW = F dh = (600/h − 30) dh
∴ = ∫W h dh( / – )600 3020
10
= −600 30ln | | 20
10h h
= 600 ln 10 − 300 − 600 ln 20 + 600= −115.8883… ≈ −116 inch-pounds
This number is negative because each value ofdh is negative and F is positive, making theirproduct negative.
e. Distance is measured in inches, force ismeasured in pounds, and we are finding theirproduct.
61. a. d (f ) = a + b ln f0 = a + b ln 53, 10 = a + b ln 16010 = b ln 160 − b ln 53 ⇒
b = =10
160 539 050741
ln – ln. ...
a = −9.050741… ln 53 = −35.934084…d ( f ) = –35.934084… + 9.050741… ln f
b. f d cm d ′ (part c)
53 0 0.1707…
60 1.1227… 0.1508…
70 2.5197… 0.1292…
80 3.7265… 0.1131…
100 5.7461… 0.0905…
120 7.3962… 0.0754…
140 8.7914… 0.0646…
160 10.0 0.0565…
The measured distances will vary. Theyshould be close to the calculated distances.
c. d ′ ( f ) = b/f = 9.050…/f. See table in part b.d. d ′ ( f ) is in cm/10 kHz.e. d ′ ( f ) decreases as f gets larger; this is
consistent with the spaces between thenumbers getting smaller as f increases.
Calculus Solutions Manual Problem Set 6-3 121© 2005 Key Curriculum Press
62. a. ln 2 = 0.693147…
ln 3 = 1.098612…
ln 6 = 1.791759…
ln 2 + ln 3 = ln 6
Conjecture: ln (ab) = ln a + ln b
b. ln (10/2) = ln 5 = 1.609437…
ln 10 = 2.302585…
ln 2 = 0.693147…
ln (10/2) = ln 10 − ln 2
Conjecture: ln (a/b) = ln a − ln b
c. ln (210) = ln 1024 = 6.931471…
ln 2 = 0.6931471…
ln (210) = 10 ln 2
Conjecture: ln (a b ) = b ln a
d.ln
log
2
2= …2.30258
ln
log
3
32 30258= ….
They seem to be the same.ln 10 = 2.30258…
12 30258
loge= ….
log 4 = 0.60205… and
ln
ln
.
.
4
10
1 3862
2 302580 60205= =K
KK.
63. Answers will vary.
Problem Set 6-3
Q1. y′ = 1/(1 + x2) Q2.1
44 1ln | |x C+ +
Q3. 1 Q4. 1/4
Q5. 35 Q6. 8
Q7.
1x
y
2
Q8. f is differentiable on (a, b).
Q9. f is continuous at x = a and x = b.
Q10. B
1. ln 6 + ln 4 = 1.79175… + 1.38629… =3.17805…ln 24 = 3.17805…
2. ln 5 + ln 7 = 1.60943… + 1.94591… =3.55534…ln 35 = 3.55534…
3. ln 2001 − ln 667 = 7.60140… − 6.50279… =1.09861…ln (2001/667) = ln 3 = 1.09861…
4. ln 1001 − ln 77 = 6.90875… − 4.34380… =2.56494…ln (1001/77) = ln 13 = 2.56494…
5. 3 ln 1776 = 3(7.48211…) = 22.44635…ln (17763) = ln 5,601,816,576 = 22.44635…
6. 4 ln 1066 = 4(6.97166…) = 27.88667…ln (10664) = ln 1,291,304,958,736 = 27.88667…
7. See the text for the proof of the uniquenesstheorem.
8. See the text for the proof.
9. Prove that ln (a/b) = ln a − ln b forall a > 0, b > 0.
Proof:
Let f (x) = ln (x/b), g(x) = ln x − ln b for x, b > 0Then f ′ (x) = (b/x)(1/b) = 1/x, andg′ (x) = (1/x) − 0 = 1/x.∴ f ′ (x) = g′ (x) for all x > 0.f (b) = ln (b/b) = ln 1 = 0g(b) = ln b − ln b = 0∴ f (b) = g (b).∴ f (x) = g (x) for all x > 0 by the uniquenesstheorem.∴ ln (x/b) = ln x − ln b for all x > 0.∴ ln (a/b) = ln a − ln b for all a > 0 and b > 0,Q.E.D.
10. Prove that ln (a b) = b ln a for all a > 0 and all b.
Proof:
Let f (x) = ln (xb); g (x) = b ln x for x > 0.Then f ′ (x) = 1/(xb) · bxb− 1 = b/x andg′ (x) = b (1/x) = b/x.∴ f ′ (x) = g′ (x) for all x > 0.f (1) = ln (1b) = ln 1 = 0g(1) = b ln 1 = b · 0 = 0∴ f (1) = g (1).∴ f (x) = g (x) for all x > 0 by the uniquenesstheorem.∴ ln (xb) = b ln x for all x > 0.∴ ln (ab) = b ln a for all a > 0 and all b, Q.E.D.
11. Prove that ln (a/b) = ln a − ln b forall a > 0, b > 0.
Proof:
ln (a/b) = ln (a · b− 1) = ln a + ln b− 1 =ln a + (−1) ln b = ln a − ln b∴ ln (a/b) = ln a − ln b, Q.E.D.
12. Example: ln (2 + 3) = ln 5 = 1.60943…ln 2 + ln 3 = 0.69314… + 1.09861… =1.79175…∴ ln (2 + 3) ≠ ln 2 + ln 3.∴ ln (a + b) = ln a + ln b is false, Q.E.D.
122 Problem Set 6-4 Calculus Solutions Manual© 2005 Key Curriculum Press
13. See the text definition of ln x.
14. logln
lnlog
logb
a
a
xx
bx
b= =
15. f (x) = log3 x ⇒ f ′ (x) = 1/(x ln 3)f ′ (5) = 0.182047…The graph shows a tangent line with a smallpositive slope.
5
1x
f(x)
16. f (x) = log0.8 x ⇒ f ′ (x) = 1/(x ln 0.8)f ′ (4) = −1.120355…The graph shows a tangent line with slope ≈ −1.
x
f(x)
4
10
5
17. g (x) = 8 ln (x5) = 40 ln x ⇒ g′ (x) = 40/x
18. h (x) = 10 ln (x0.4 ) = 4 ln x ⇒ h′ (x) = 4/x
19. T(x) = log5 (sin x) ⇒ ′ =⋅
⋅ T xx
x( )sec ln
cos1
5T′ (x) = (cot x)/(ln 5)
20. R (x) = log4 (sec x) ⇒
R xx
x xx′ =
⋅⋅ =( )
1
4 4sec lnsec tan
tan
ln
21. p (x) = (ln x)(log5 x) ⇒p′ (x) = (1/x) · (log5 x) + (ln x) · [1/(x ln 5)]
= ⋅ + =1
5 5
2
5
ln
ln
ln
ln
ln
ln
x
x
x
x
x
x
22. q (x) = (log9 x)/(log3 x) = ⋅ = =ln
ln
ln
ln
ln
ln
x
x9
3 3
2 3
1
2∴ q ′ (x) = 0 because q (x) is constant.
23. f xx
xx x( ) =
= −lnsin
ln ln sin3
3
= 3 ln x − ln sin x ⇒ f ′ (x) = 3/x − cot x
24. f (x) = ln (x4 tan x) = ln x4 + ln (tan x)= 4 ln x + ln (tan x) ⇒
f xx
x
x′ = +( )
4 2sec
tan= +4 1
x x xsin cos
25.d
dxx
d
dxx xx( ) ( )ln ln3 3= =
33
3 3ln lnxx
xx+ = +
26.d
dx
d
dxxx(ln ) (sec ln )sec5 5=
= ln 5 sec x tan x
27. a. y = 7 · (2 − 0.9x)dy/dx = 7(−0.9x)(ln 0.9)dy/dx = 0.737523…(0.9x)x = 0: dy/dx = 0.737… mi/hx = 1: dy/dx = 0.663… mi/hx = 5: dy/dx = 0.435… mi/hx = 10: dy/dx = 0.257… mi/hThe lava is slowing down.
b. y/7 = 2 − 0.9x
0.9x = 2 − y/7x ln 0.9 = ln (2 − y/7)x = (1/ln 0.9)[ln (2 − y/7)]
c.dx
dy y= ⋅ ⋅ −
( / . )1 0 9
1
2 7
1
7ln
– /dx
dy y= 9 491221
14
.
–
K
y = 10: dx/dy = 2.372… h/mi
d. If x = 10, then y = 7(2 − 0.910), so
dx
dy= = …9 491221
14 7 2 0 93 88865110
.
– ( – . )
K. .
e. 3.88… is the reciprocal of 0.257… , thevalue of dy/dx when x = 10, not when y = 10.
28. a. 1000(1.06t) = M ⇒ 1.06t = M/1000 ⇒log1.06 1.06t = log1.06 (M/1000) ⇒t = log1.06 (M/1000)
b. dt dMM
/ = ⋅1
1 06 1000
1
1000(ln . )( / )
= 1
1 06M ln .
c. If , /M dt dM= = =10001
1000 1 06ln .0.01716… yr/$. At this rate, with $1000 inthe account, it would take 0.017… year, orabout 6 days, to earn a dollar of interest.
d. dt/dM gets smaller as M increases; moreinterest is earned when M is larger, so it takesless time to accumulate $1000.
29. The intersection point is at x = 2.7182818… ,which is approximately e.
30. Answers will vary.
Problem Set 6-4Q1. y′ = 3/x Q2. (−1/10)(5x)− 2 + C
Q3. 15 5ln x C+ Q4. y x′ = −1 1 2/ –
Calculus Solutions Manual Problem Set 6-4 123© 2005 Key Curriculum Press
Q5. 5 tan (5x) Q6.ln
ln
23
17
Q7. 36 Q8. 8
Q9. B Q10. E
0. Answers will vary.
1. a. R(t) = aekt
60,000 = aek·0 ⇒ a = 60,0002,400,000 = aek·2 = 60,000e2k
40 = e2k ⇒ 2k = ln 40 ⇒k = (ln 40)/2 = 1.844…(Store 1.844… without round-off as k.)∴ R(t) = 60,000e1.844…t
b. R ( 5 ) = 60,000e1.844…(5) = 607,157,310.7…About 607 million rabbits.
c. 2 = 60,000e1.844…t ⇒ 1/30,000 = e1.844…t ⇒−ln 30,000 = 1.844…t ⇒ t = −5.589…So the first pair of rabbits was introducedabout 5.6 years earlier, or in about 1859.
2. a. v (t) = 20,000e− 0.1 t
v(0) = 20,000e0 = 20,000$20,000 when built.
b. v (10) = 20,000e− 1 = 7357.588…v(11) = 20,000e− 1.1 = 6657.421…At 10 years, value is $7357.59.At 11 years, value is $6657.42.So depreciation is 7357.59 − 6657.42 =$700.17.
c. v′ (t) = −2000e− 0.1 t
v′ (10) = −2000e− 1 = −735.758… ,or about $736 per year.This rate is higher than the actual depreciationin part b because the latter rate is an averagefor the year. The rate at the end will be lowerthan 736 to give the average of 700.
d. 5,000 = 20,000e− 0.1 t
0.25 = e− 0.1 t
ln (0.25) = −0.1tt = (ln 0.25)/(−0.1) = 13.8629… ≈14 yr.
3. a. m(t) = 1000(1.06)t
ln m(t) = ln 1000 + t ln 1.061/m(t) · m′ (t) = 0 + ln 1.06m ′ (t) = m (t) · ln 1.06m ′ (t) = 1000(1.06)t (ln 1.06)m ′ (0) = 58.27 $/yrm ′ (5) = 77.98 $/yrm ′ (10) = 104.35 $/yr
b. m(0) = $1000.00m(5) = $1338.23m(10) = $1790.85The rates are increasing. $338.23 is earnedbetween 0 and 5 years; $452.62 is earned
between 5 and 10 years, which agrees withthe increasing derivatives shown in part a.
c.m t
m t
t
t
′ = =( )
( )
( . ) (ln . )
( . )ln
1000 1 06 1 06
1000 1 061 06.
∴ m′ (t)/m(t) = ln 1.06, a constantd. m(1) = 1060.00. So you earn $60.00.
The rate starts out at only $58.27/year buthas increased enough by year’s end to makethe total for the year equal to $60.00.
4. d (t) = 200t · 2− t ⇒ ln d (t) = ln 200t − t ln 21
1 2d t
d t t( )
ln⋅ ′ = − ⇒( ) /
′ = ⋅ / −−d t t tt( ) ( )(1 ln 2)200 2
d ′ (1) = (200 · 2− 1)(1 − ln 2) = 30.685…d ′ (2) = (400 · 2− 2)(1/2 − ln 2) = −19.314…So the door is opening at about 30.7°/s at1 second and closing at about 19.3°/s at2 seconds, which agrees with the graph.
1 2
100
t
d(t)
The widest opening occurs when d ′ (t) = 0.Solving numerically for t in(200t · 2− t)(1/t − ln 2) = 0,t = 1.44269… .d (1.44269…) = 106.147…So the widest is about 106° at t ≈ 1.4 s.
5. enn
n
= +
→∞
lim 11
and e nn
n= +→
lim( ) /
0
11
When you substitute ′ for n in the first equation,you get the indeterminate form 1∞ . When yousubstitute 0 for n in the second equation, youalso get the indeterminate form 1∞ .
n (1 + 1/n)n
100 2.70481…
1000 2.71692…
10000 2.71814…
n (1 + n)1/ n
0.01 2.70481…
0.001 2.71692…
0.00001 2.71826…
6. y = 17e− 5x ⇒ y′ = −85e− 5x
7. y = 667e− 3x ⇒ y′ = −2001e− 3x
8. h(x) = x3ex ⇒ h′ (x) = 3x2ex + x3ex = x2ex (3 + x)
9. g (x) = x− 6ex ⇒ g′ (x) = −6x− 7ex + x− 6ex = x− 7ex (−6 + x)
124 Problem Set 6-4 Calculus Solutions Manual© 2005 Key Curriculum Press
10. r (t) = et sin t ⇒ r ′ (t) = et sin t + et cos t
11. s (t) = et tan t ⇒ s′ (t) = et tan t + et sec2 t
12. u = 3exe−x = 3 ⇒ u′ = 0
13. v = e4xe− 4x = 1 ⇒ v′ = 0
14. ye
xy
e x e x
x
x x x
= ⇒ ′ = −ln
ln ( / )
(ln )
12
15. yx
ey
x e x e
ex
x x
x= ⇒ ′ = − ⋅ln ( / ) ln12
16. y = 4esec
x ⇒ y′ = 4esec
x sec x tan x
17. y = 7ecos
x ⇒ y′ = −7ecos
x sin x
18. y = 3 ln e2x = 6x ln e = 6x ⇒ y′ = 6
19. y = 4 ln e5x = 4 · 5x = 20x ⇒ y′ = 20
20. y = (ln e3x)(ln e4x) = 3x · 4x = 12x2 ⇒ y′ = 24x
21. y = (ln e− 2x)(ln e5x) = −2x · 5x = −10x2 ⇒y′ = −20x
22. g(x) = 4eln 3x = 4 · 3x = 12x ⇒ g′ (x) = 12
23. h (x) = 6eln 7x = 6 · 7x = 42x ⇒ h′ (x) = 42
24. y = ex + e− x ⇒ y′ = ex − e− x
25. y = ex − e− x ⇒ y′ = ex + e− x
26. y e y e x x ex x x= ⇒ ′ = ⋅ =5 5 2 2 53 3 3
15 15
27. y e y e x x ex x x= ⇒ ′ = ⋅ =8 8 5 405 5 54 4
28. f (x) = 0.42x ⇒ ln f (x) = 2x ln 0.4 ⇒1
2 0 4 0 4 2 0 42
f xf x f x x
( )ln ln′ = ⇒ ′ = ⋅( ) . ( ) . .
29. f (x) = 10− 0.2 x ⇒ ln f (x) = −0.2x ln 10 ⇒1
0 2 10f x
f x( )
ln′ = − ⇒( ) .
′ = −−f x x( ) ( ..10 0 2 100 2 ln )
30. g (x) = 4(7x) ⇒ ln g (x) = ln 4 + x ln 7 ⇒1
7 4 7 7g x
g x g x x
( )ln ln′ = ⇒ ′ =( ) ( ) ( )
31. h (x) = 1000(1.03x) ⇒ ln h (x) = ln 1000 +
x ln( )
ln1 031
1 03. ( ) .⇒ ′ = ⇒h x
h x
′h x( ) = 1000(1.03x) ln 1.03
32. c (x) = x5 · 3x ⇒ ln c (x) = 5 ln x + x ln 3 ⇒1
5 3c x
c x x( )
ln′ = + ⇒( ) /
′c x( ) = x5 · 3x(5/x + ln 3)
33. m(x) = 5x · x7 ⇒ ln m(x) = x ln 5 + 7 ln x ⇒1
5 7m x
m x x( )
ln′ = + ⇒( ) /
m ′ (x) = 5x · x7(ln 5 + 7/x)
34. y = (ln x)0.7 x ⇒ ln y = 0.7x ln (ln x) ⇒1
0 7 0 71
yy x x
x x′ = + ⋅ ⇒. ln (ln ) .
ln
′ = +
⋅y x
xx x0 7
0 7 0 7. ( )ln ln.
ln(ln ) .
35. y = xln x ⇒ ln y = ln x · ln x ⇒ ln y = (ln x)2 ⇒1
21
yy x
x′ = ⋅ ⇒ln ′ =
⋅y
x
xx x2 ln ln = 2 ln x · xln x−1
36. y = (csc 5x)2x ⇒ ln y = 2x ln (csc 5x) ⇒ 1
yy′ =
2 5 21
5ln csc
csc( )
x x
x+ (– csc cot )5 5 5x x ⇒
y′ = (csc 5x)2x [2 ln (csc 5x) − 10x cot 5x]37. y = (cos 2x)3x ⇒ ln y = 3x ln (cos 2x) ⇒
13 2 3
1
22 2
yy x x
xx′ = +ln cos
cos(– sin )( ) ⇒
y′ = (cos 2x)3x [3 ln (cos 2x) − 6x tan 2x]38. Two solution methods are possible.
Differentiate directly:
yx
x= + ⇒ln
–
5 2
7 8
′ =+
+ ⋅
yx
x
x x
x
7 8
5 2
5 7 8 5 2 7
7 8 2
– ( – ) – ( )
( – )
=+
–
( )( – )
54
5 2 7 8x xOr simplify using properties of logarithms first:
y = ln (5x + 2) − ln (7x − 8) ⇒
′ =+
− =+
yx x x x
5
5 2
7
7 8
54
5 2 7 8–
–
( )( – )39. Two solution methods are possible.
Differentiate directly:y = ln [(4x − 7)(x + 10)]
′ =+
⋅ + + − ⋅yx x
x x1
4 7 104 10 4 7 1
( – )( )[ ( ) ( ) ]
= ++
8 33
4 7 10
x
x x( – )( )Or simplify using properties of logarithms first:y = ln (4x − 7) + ln (x + 10)
′ =−
++
= ++
yx x
x
x x
4
4 7
1
10
8 33
4 7 10( – )( )
40. y = (2x + 5)3 4 1x − ⇒
ln y = 3 ln (2x + 5) +
1
24 1ln ( )x − ⇒
1 6
2 5
2
4 1yy
x x′ =
++
−⇒
′ =+
+−
+ −y
x xx x
6
2 5
2
4 12 5 4 13[( ) ]
= + +−
( )( )28 4 2 5
4 1
2x x
x
41. yx
xy x= +
−⇒ = + −( )
( )ln ln ( )
10 3
4 510 10 3
10
3
3 4 51 30
10 3
15
4 5ln ( )− ⇒ ′ =
++
−⇒x
yy
x x
′ =+
+−
+−
yx x
x
x
30
10 3
15
4 5
10 3
4 5
10
3
( )
( )
= − +−
( )( )
( )
270 105 10 3
4 5
9
4
x x
x
Calculus Solutions Manual Problem Set 6-5 125© 2005 Key Curriculum Press
42.d
dxdtt
xx10 10
3∫
=
43.d
dxt dt x
x
ln ln3∫
=
44.d
dxt dt x
x
log log ( )25
4
24 4∫
=
45.d
dxt dt x x
x
ln (cos ) (ln cos ).6 3
22
2∫
=
46.d
dxx
d
dxx
d
dx x x
2
25
2
2 255 5
(ln ) ( ln )= =
= −
47.d
dxe
d
dxe ex x x
2
27 7 77 49( ) ( )= =
48. e dx e Cx x5 51
5= +∫
49. e dx e Cx x7 71
7= +∫
50. 77
2 72
2x
x
dx C= +∫ ln
51. 1 051 05
1 05.
.
ln .x
x
dx C= +∫52. 6 6e dx e Cx x= +∫53. e dx e Cx x0 2 0 25. .= +∫54. e x dx e Cx xsin sincos = +∫55. e x dx e Cx xtan tansec2 = +∫56. e dx x dx x Cx3 3 41
4ln∫ ∫= = +
57. 60 60 5 1505 2e dx x dx x Cxln∫ ∫= = +
58. ( ) ( )11
10212 50 2 2 51+ = + +∫ e e dx e Cx x x
59. ( ) ( )11
40414 100 4 4 101− = − − +∫ e e dx e Cx x x
60. ( )e e dx e ex x x x− = +− −∫0
2
0
2
= + − − =−e e2 2 1 1 5 524391. ...
Numerically: integral ≈ 5.524391... (Checks.)
61. ( ) ( )– –e e dx e ex x x x+ = −− −∫ 1
2
1
2
= e2 − e−2 − e−1 + e1 = 9.604123…Numerically: integral ≈ 9.604123... (Checks.)
62. Step 2: Definition of derivative.
Step 3: Logarithm of a quotient, applied inreverse.
Step 4: Write division as multiplication by thereciprocal; distribute division over addition.
Step 6: 1/x does not depend on h, so it is a“constant” with respect to h.
Step 7: Logarithm of a power, applied in reverse.
Step 9: The expression in parentheses has the form(1 + n)1/n, whose limit is e as n approaches zero.
63. Answers will vary.
64. Answers will vary.
Problem Set 6-5
Q1. e ≈ 2.71828 Q2. enn
n
= +
→∞
lim 11
or
e nn
n= +→
lim( ) /
0
11
Q3. 1 Q4. x
Q5. x Q6. e
Q7. (ln x)/(ln b) Q8. ex
Q9. −e−x + C Q10. E
1. limsin
x
x
x→→
0
2 5
3
0
0
= =→
limcos
x
x0
10 5
3
10
3
11
x
y
2. limtan
x
x
x→→
0
4 3
5
0
0
= =→
limsec
x
x0
212 3
5
12
5
11 x
y
3. limtan
x
x
x→→
0
0
0
= =→
limsec
x
x0
2
11
4. limsin
x
x
x→→
0
0
0
= =→
limcos
,x
x0 1
1 a “well-known” limit.
126 Problem Set 6-5 Calculus Solutions Manual© 2005 Key Curriculum Press
5. limcos
x
x
x→
− →0 2
1 0
0
= →→
limsin
x
x
x0 2
0
0
= =→
limcos
x
x0 2
1
2
6. limcosx
x
x→ −→
0
2
3 1
0
0
=−
→→
limsinx
x
x0
2
3 3
0
0
=−
= −→
limcosx x0
2
9 3
2
9
7. limsin
x
x
x→ +→
02
0
0
= = ∞→ +lim
cosx
x
x0 2
8. limcos
x
x
x x→
−+
→0 2
1 0
0
=+
=→
limsin
x
x
x0 1 20
9. limln
/x
x
x→ +→ −∞
∞0 1
= −= − =
→
−
− →+ +lim lim ( )x x
x
xx
0
1
20
0
10. lim .x
xe
x→= ∞
0
3
2 Form is 1
0
11. limlnx
xe e
x→
− →1 5
0
0
= =→ −lim
x
xe
x
e1 15 5
12. limln
x
x x
x x→
− +− +
→1 2
1
2 1
0
0
= −−
→→
−
limx
x
x1
1 1
2 2
0
0
= − = −→
−
limx
x1
2
2
1
2
13. lim
cos cos.
x
x
x→
+ = = −2
3 5 11
226 43297K
14. limtan
.x
x
x→ −= ∞
2 2
Form is tan 2
0
15. limx
xe
x→∞→ ∞
∞2
= → ∞∞→∞
limx
xe
x2
= = ∞→∞
limx
xe
2
16. limx x
x
e→∞→ ∞
∞
3
= → ∞∞→∞
limx x
x
e
3 2
= → ∞∞→∞
limx x
x
e
6
= =∞
→∞
lim .x xe
60
6 Form:
17. lim–
limx x
x
x→∞ →∞
+ = =3 17
4 11
3
4
3
4
18. lim limx x
x
x→∞ →∞+= = −2 – 7
3 5
–7
5
7
5
19. lim– –
– –x
x x x
x x x→∞
++
→ ∞∞
3 2
3 2
5 13 21
4 9 11 17
= ++
→ ∞∞→∞
lim–
–x
x x
x x
3 10 13
12 18 11
2
2
=+
→ ∞∞→∞
lim–
x
x
x
6 10
24 18
= =→∞
limx
6
24
1
4
20. limx
x
x→∞
+ → ∞∞
3 2
7 8
5
5 –
= = =→∞ →∞
lim limx x
x
x
15
35
15
35
3
7
4
4
21. L xx
x= →→ +lim
0
00
ln lim ( ln ) limln
–L x xx
xx x= = → −∞
∞→ →+ +0 01
=−
= − =→
−
− →+ +lim lim ( )x x
x
xx
0
1
20
0
∴ = =L e0 1
22. L xx
x= →→ +lim (sin )sin
0
00
ln lim sin (ln sin ) limlnsin
cscL x x
x
xx x= = → −∞
∞→ →+ +0 0
= ⋅−
=−→ →+ +
lim/(sin ) cos
csc cotlim
cscx x
x x
x x x0 0
1 1
= − =→ +lim ( sin )x
x0
0
∴ = =L e0 1
23. L xx
x= →→
∞−
lim (sin )/
tan
π 21
ln lim tan ln sin/
L x xx
=→ −π 2
( )
= →
= ⋅−
= −
→
→
→
−
−
−
limln sin
cot
lim( /sin ) cos
csc
limcos sin
sin
/
/
/
x
x
x
x
x
x x
x
x x
x
π
π
π
2
22
2
2
0
0
1
Calculus Solutions Manual Problem Set 6-5 127© 2005 Key Curriculum Press
= − =→ −lim ( cos sin )
/xx x
π 20
∴ = =L e0 1
24. L xx
x= →→
−+
∞lim /( )
1
1 1 1
ln lim [ /( ) ln ]L x xx
= − ⋅→ +1
1 1
=−
→ = =→ →+ +lim
lnlim
/x x
x
x
x1 11
0
0
1
11
∴ = =L e e1
25. L ax ax
x= + → ∞ ≥→∞
lim( ) ( .)/1 01 0 Note:
ln lim [ / ln ( )] limln ( )
L x axax
xx x= ⋅ + = + → ∞
∞→∞ →∞1 1
1
= + ⋅ =+
=→∞ →∞
lim/( )
limx x
ax a a
ax
1 1
1 10
∴ = =L e0 1
26. L axx
x= + →→
∞lim ( ) /
0
11 1
ln lim [ / ln ( )] limln ( )
L x axax
xx x= ⋅ + = + →
→ →0 01 1
1 0
0
= + ⋅ =+
=→ →
lim/( )
limx x
ax a a
axa
0 0
1 1
1 1
∴ =L ea
27. L xx
x= →→ +lim /(ln )
0
3 00
ln lim ] limL x xx x
= ⋅ = =→ →+ +0 0
3 3[3/(ln ) ln
∴ = =L e3 20 08553. ...
28. L xx
x= →→ +lim ( ) /(ln )
0
5 07 0
ln lim [ /(ln ) ln( )]L x xx
= ⋅→ +0
5 7
= → −∞−∞
= ⋅ ⋅ =
→
→
+
+
limln ( )
ln
lim[ /( )]
/
x
x
x
x
x
x
0
0
5 7
5 1 7 7
15
∴ = =L e5 148 4131. ...
29. limx xx e→
−−
→ ∞ − ∞
0
1 1
1
= − →→
lim–
( – )x
x
x
e x
x e0
1
1
0
0
=+ ⋅
→→
lim–
( – )x
x
x x
e
e x e0
1
1 1
0
0
=+
→→
lim–
–x
x
x x
e
e xe0
1
1
0
0
=+ +
=→
limx
x
x x x
e
e e xe0
1
2
30. limsinx x x→
−
→ ∞ − ∞0
1 1
= − →→
limsin
sinx
x x
x x0
0
0
= −+
→→
limcos
cos sinx
x
x x x0
1 0
0
= −− +
=→
limsin
sin cosx
x
x x x0 20
31. f x x x( ) .= −sec tan2 2
2 2
π π Where secant and
tangent are defined, the Pythagorean propertiestell us that f (x) = 1.
–1 1 3 5
1
x
f(x)
32. Using l’Hospital’s rule leads to
limsec
tanlim
sec tan
sec/ /x x
x
x
x x
x→ →=
π π2 2 2
= =→ →lim
tan
seclim
sec
sec tan/ /x x
x
x
x
x xπ π2 2
2
=→lim
sec
tan,
/x
x
xπ 2 the original expression!
Using tan x = (sec x)/(csc x), the expressionreduces to
limsec
(sec )/(csc )lim csc
/ /x x
x
x xx
→ →= =
π π2 21
33. L xx
k x= →→ +lim /(ln )
0
00
ln L k x x k kx x
= ⋅ = =→ →+ +lim ln ln lim
0 0[ /( ) ]
∴ L = ek
The graph turns out to be a horizontal line,y = ek, defined for x > 0.
x
yy = e k
By the definition of a power,
x x e ek x k x k x x k/ / /= = =( ) ( ) ( )ln ln ln ln1 1
34. a. f xg x
h x
x
x x( ) = =
+( )
( )
. – .
. – .
0 3 2 7
0 2 2 4 2
2
2
g(3) = 0.3(9) − 2.7 = 0,h(3) = 0.2(9) − 2(3) + 4.2 = 0, Q.E.D.
b. g′ (x) = 0.6x ⇒ g′ (3) = 1.8h′ (x) = 0.4x − 2 ⇒ h′ (3) = −0.8
128 Problem Set 6-6 Calculus Solutions Manual© 2005 Key Curriculum Press
Tangent lines at (3, 0) have these equations.
For g: y1 = 1.8(x − 3)For h: y2 = −0.8(x − 3)
c.y
y
x
xx1
2
1 8 3
0 8 32 25= = − ≠. ( – )
– . ( – ). , for 3.
g
h
′′
= = −( )
( )
.
– .
3
3
1 8
0 82 25. , which equals y1/y2,
Q.E.D.
d. Because the ratio g (x)/h(x) approaches theratio y1/y2 as x approaches 3, and becausey1/y2 equals g′ (3)/h′ (3) for all x ≠ 3, the ratiog (x)/h(x) also approaches g′ (3)/h′ (3) as xapproaches 3. This is what l’Hospital’s ruleconcludes.
If g (3) or h (3) were any number other than 0,the canceling of the (x − 3)’s in part c couldnot be done, and the ratio would almostcertainly not equal 1.8/(−0.8).
e. The graph shows a removable discontinuity at(3, −2.25):
3
1 x
f(x)
35. a. For yearly compounding, m(t) =1000(1 + 0.06)t. For semiannual compound-ing, m ( t) = 1000(1 + 0.06/2)2t because thereare two compounding periods per year, eachof which gets half the interest rate.
b. m(t) = 1000(1 + 0.06/n)nt
lim limn n
ntm t n→∞ →∞
= +( ) ( . / )1000 1 0 06
= +→∞
1000 1 0 06lim ( . / )n
ntn
Let L nn
nt= +→∞
lim ( . / )1 0 06 .
ln L nt nn
= +→∞
lim [ ln ( . / )]1 0 06
= + →→∞
limln ( . / )
/( )n
n
nt
1 0 06
1
0
0
= + ⋅ −−→∞
−
−lim/( . / ) ( . )
/n
n n
n t
1 1 0 06 0 06 2
2
=+
=→∞
lim.
( . / )n
t
nt
0 06
1 0 060 06.
∴ = ⇒ =→∞
L e m t et
n
t0 06 0 061000. .lim ( )
When interest is compounded continuously,m(t) = 1000e0. 06t.
c.
t m(t), Annual m(t),Continuous Difference
5 1,338.23 1,349.86 11.63
20 3,207.14 3,320.12 112.98
50 18,420.15 20,085.54 1,665.38
d. For 7% interest, compounded continuously,m (t) = 1000e0.0 7t.
36. a. f (x) = xn, g (x) = ln x, h (x) = ex
lim( )
( )lim
lnx x
nf x
g x
x
x→∞ →∞= → ∞
∞
= = = ∞→∞ →∞
lim/
lim ,–
x
n
x
nnx
xnx n
1
10if >
∴ a power function is higher-order than thenatural log function.
lim( )
( )lim
x x
n
x
f x
h x
x
e→∞ →∞= → ∞
∞
= → ∞∞→∞
lim ,–
x
n
x
nx
e
1
if n − 1 > 0
Eventually, the exponent of the power willbecome zero, in which case the limit takes theform constant/∞, which is 0.∴ a power function is lower-order than anexponential function.Using “<” to represent “is lower-order than,”natural log < power < exponential.
b. i. limln
x
x
x→∞=3
05 ii. lim .x x
x
e→∞=
100
0 01 0
iii. limln
.
x
xe
x→∞= ∞
0 3
100
iv. lim limx x
x
x x→∞ →∞= =1
0
v. lim lim.x
x
x x
xe
ee
→∞ →∞= = ∞0 2
0 8.
37. Answers will vary.
Problem Set 6-61. y = ln (3x + 4) ⇒ y′ = 3/(3x + 4)
2. y = ln (3x5) = ln 3 + 5 ln x ⇒ y′ = 5/x3. y = ln (e3x) = 3x ⇒ y′ = 3
4. y = ln (sin 4x) ⇒ y′ = =4 4
4
cos
sin
x
x 4 cot 4x
5. y = ln (cos5 x) = 5 ln (cos x) ⇒
′ = − = −yx
xx
55
sin
costan
6. y = ln (e5) = 5 ⇒y′ = 0 (Derivative of a constant!)
Calculus Solutions Manual Problem Set 6-6 129© 2005 Key Curriculum Press
7. y = ln [cos (tan x)] ⇒
y ′ = ⋅– sin (tan )
cos (tan )
x
x sec2 x = −tan (tan x) sec2 x
8. y x x x x= + = − +ln – ln2 22 31
22 3( ) ⇒
′ =
−− +
= −− +
yx
x x
x
x x
1
2
2 2
2 3
1
2 32 2
9. y = cos (ln x) ⇒ y′ = −(1/x) sin (ln x)
10. y = sin x · ln x ⇒ y′ = cos x · ln x + (1/x) sin x
11. y = e7x ⇒ y′ = 7e7x
12. y e y x ex x= ⇒ ′ =3 3
3 2
13. y e e x y xx x= = = ⇒ ′ =5 5 45
5ln ln
14. y = ecos x ⇒y′ = ecos x · (−sin x) = −ecos x sin x
15. y = cos (ex) ⇒ y′ −sin (ex) · ex = −ex sin ex
16. y = (cos3 x)(e3x) ⇒y′ = 3 cos2 x (−sin x) · e3x + cos3 x · e3x · 3
= −3e3x cos2 x sin x + 3e3x cos3 x= 3e3x cos2 x (−sin x + cos x) (Factoringoptional)
17. y e y x ex x= ⇒ ′ =5 5
5 4
18. y e y e ee e xx x
= ⇒ ′ = ⋅19. sin y = ex ⇒ cos y · y′ = ex ⇒
y′ = e
y
e
e
x x
xcos –=
1 2(See sketch.)
1sin y = e x
y
(Showing cos – y e x= 1 2 )
20. y = ex · ln x ⇒y′ = ex · ln x + ex · (1/x) = ex (ln x + 1/x)
21. y = 11
/tx
∫ dt ⇒ y′ = 1/x
22. tan y = ex ⇒ sec2 y · y′ = ex ⇒
y′ = e
y
e
e
x x
xsec2 21=
+(See sketch.)
1
ex
y
√1 + e 2x
23. y = ln (eln x) = ln x ⇒ y′ = 1/x24. y = 2x ⇒ ln y = x ln 2 ⇒ (1/y)y′ = ln 2 ⇒
y′ = y ln 2 = 2x ln 2
25. y e ex xx
= = =ln ln2 2 2 ⇒y′ = 2x ln 2 (See Problem 24.)
26. y = e2 ln x = e xln 2
= x2 ⇒ y′ = 2x
27. y = x2 ⇒ y′ = 2x
28. y = ex ln x (which equals xx) ⇒y′ = ex ln x [ln x + x (1/x)] = xx (ln x + 1)
29. y = xx = (eln x)x = ex ln x ⇒y′ = xx (ln x + 1) (See Problem 28.)
30. y = x ln x − x ⇒ y′ = ln x + x · (1/x) − 1 = ln x(Note: This answer reveals that the integral ofln x is x ln x − x.)
31. y = ex(x − 1) ⇒ y′ = ex(x − 1) + ex · 1 = xex
32. y = 1
2( )–e ex x+ ⇒ y′ =
1
2( – )–e ex x
33. y = 1
2( – )–e ex x
⇒ y′ = 1
2( )–e ex x+
(Problems 32 and 33 are the hyperbolic cosineand sine functions, respectively. See Chapter 8.)
34. y = e
e
x
x1+ ⇒
y′ = e e e e
e
x x x x
x
⋅ + ⋅+
( ) – ( )
( )
1
1 2 =+e
e
x
x( )1 2
35. y = 5x ⇒ ln y = x ln 5 ⇒ (1/y)y′ = ln 5 ⇒ y′ = y ln 5 = 5x ln 5
36. y = log5 x = ln
ln
x
5 ⇒ y′ =
1
5
/
ln
x = 1
5x ln
37. y = x− 7 log2 x = x− 7 · ln
ln
x
2 ⇒
y′ = 1
27
18 7
ln– ln– –x x x
x⋅ + ⋅
= xx
–
ln(– ln )
8
27 1+
38. y = 2− x cos x ⇒y′ = 2− x (−ln 2) · cos x + 2− x (−sin x)
= −2− x(ln 2 · cos x + sin x) (Factoring optional)
39. y = e− 2x ln 5x ⇒y′ = −2e− 2x · ln 5x + e− 2x · (1/x)
= e− 2x (−2 ln 5x + 1/x)
40. y yx
x x= ⇒ ′ = ⋅ =7
7
1
77 7 7
ln lnln
41. yx
ex x y
xe= = = ⇒ ′ =log
loglog ln3
3
1
42. yx
ex x y
xe= = = ⇒ ′ =log
loglog ln10
10
1
43. y xx= = ⋅( )( )log ln
ln
lnln8 8
88 = ln x ⇒
y′ = 1
x
44. y = (log4 x)10 ⇒
y′ = 10(log4 x)9 · 1
4x ln=
10
44
9(log )
ln
x
x
45. y = log5 x7 = 7 log5 x =
7
5
ln
ln
x ⇒ y′ =
7
5x ln
130 Problem Set 6-6 Calculus Solutions Manual© 2005 Key Curriculum Press
46. y = tan ex ⇒ y′ = sec2 ex · ex = ex sec2 ex
47. y = esin x ⇒ y′ = esin x cos x
48. y = ln csc x ⇒y′ = (1/csc x) · (−csc x cot x) = −cot x
49. y = 35 ⇒ y′ = 0 (Derivative of a constant!)
50. y = ln (cos2 x + sin2 x) = ln 1 = 0 ⇒ y′ = 0
51. y = sin x ⇒ y′ = cos x
52. y = sin− 1 x ⇒ y′ = 1
1 2– x
53. y = csc x ⇒ y′ = −csc x cot x
54. y = tan− 1 x ⇒ y′ = 1
1 2+ x
55. y = tan x ⇒ y′ = sec2 x
56. y = cot x ⇒ y′ = −csc2 x
57. e dx e Cx x4 41
4 = +∫
58. e dx e x C4 4= +∫59. x e dx e x dx e Cx x x3 34 4 41
44
1
4∫ ∫= = +( )
60. cos cossin sinx e dx e x dxx x⋅ =∫ ∫ ( )
= +e Cxsin
61.(ln )
(ln ) (ln )x
xdx x
xdx x C
55 61 1
6= = +∫∫
62. 5 5x xdx e dx=∫ ∫ ln
= ∫( / ) 1 5 55ln lnlne dxx
= + = +1
5
5
55
ln lnlne C Cx
x
63. e dx dx Cx xx
ln
ln5 5
5
5 = = +∫ ∫
(See Problem 62.)
64.1
2
1
2( ) ( – )– –e e dx e e Cx x x x+ = +∫
65.1
1 t
x
∫ dt = ln x (By definition!)
66. e dx e Cx x− −= − +∫
67. 22
2x
x
dx C= +∫ ln
68. ( ) .x dx x Cxx
− + = + +∫ 0 2 0 83 1 253
3. .
ln
69. ( / ) | | 3 3x dx x C= +∫ ln
70.
44
4
12
48 656170
1
2
1
2x
x
dx = = =∫ ln ln. K
71. (ln ) (ln )xx
dx x C9 101 1
10= +∫
72. cos sin x dx x C= +∫73. e dx x dx x Cxln = = +∫ ∫ 1
22
74. ln ln( ) e dx x e dx x dx x Cx3 23 33
2= = = +∫ ∫∫
75. 0 dx C=∫ (Integral of zero is a constant.)
76. cos secx x dx dx x C = = +∫ ∫1
77. sec sec21
22 2x dx x dx=∫ ∫ ( )
= + +1
22 2ln | sec tan |x x C
78. tan tan31
33 3x dx x dx∫ ∫= ( )
= +1
33ln | sec |x C
79. cot cot41
44 4x dx x dx=∫ ∫ ( )
= +1
44ln | sin |x C
80. csc csc51
55 5x dx x dx=∫ ∫ ( )
= − + +1
55 5ln | csc cot |x x C
81. limcos
x
x
x→
− →0
1 0
0
= limsin
x
x→
=0 1
0
82. lim– cosx
x
x→→
0 1
0
0
= = ∞→
lim0x x
1
sin (Reciprocal of Problem 81.)
83. lim– cos
/
– cos ( / )/x
x
x→= =
π
ππ
π2 1
2
1 2 2
84. limcosx
x
x→ += ∞
π
π1 0
Form:
85. lim– sin
x
x x
x→ →
0 3
5 5 0
0
= →→
lim– cos
x
x
x0 2
5 5 5
3
0
0
Calculus Solutions Manual Problem Set 6-7 131© 2005 Key Curriculum Press
= →→
limsin
x
x
x0
25 5
6
0
0
= = =
→lim
cosx
x0
125 5
6
125
620 8333. K
86. lim( . / )x
xx→∞
∞+ →1 0 03 1
Let L xx
x= +→∞
lim( . / )1 0 03 .
ln L = + = +→∞ →∞
ln lim ( . / ) lim lnx
x
x
xx x1 0 03 1 0 03( . / )
= +→∞
lim [ lnx
x x( . / )] 1 0 03
= + →→∞
limln ( . )–
–x
x
x
1 0 03 0
0
1
1
= + ⋅ =→∞
lim/( . / ) (– . )
–
–
–x
x x
x
1 1 0 03 0 030 03
2
2 .
∴ L = e0.03 = 1.03045…
87. lim ( . ) /
x
xx→∞
+ → ∞1 0 03 1 0
Let L xx
x= +→∞
lim( . )1 0 03 1/ .
ln lim lnL x xx
= + → ⋅ ∞→∞
[( / ) ( . )]1 1 0 03 0
= + → ∞∞→∞
limln ( . )
x
x
x
1 0 03
= + ⋅ =→∞
lim/( . ) .
x
x1 1 0 03 0 03
10
∴ L = e0 = 1
88. limx
x
x→∞→ ∞
∞2
2
= → ∞∞→∞
limln
x
x
x
2 2
2
= = ∞→∞
lim(ln )
x
x2 2
2
2
or: limx
x
x→∞= ∞2
2 by (exponential)/(power)
89. lim ( . ) /( )
x
xx→
− ∞→2
3 20 5 1
Let L = ln (0.5x)3/(2 −x) .
ln lim–
ln .Lx
xx
= ⋅
→ ∞ ⋅
→2
3
20 5 0
= →→
limx
x
x2
3 0 5
2
0
0
ln .
–
= ⋅ = −→
lim/( . ) .
–x
x2
3 0 5 0 5
1
3
2
∴ L = e− 3/2 = 0.22313…
90. lim–
–x xe x→
⇒ ∞ − ∞
0 3
1
1
1
3
= →→
lim– ( – )
( – )x
x
x
x e
x e0
3
3
3 1
3 1
0
0
=+ ⋅
→→
lim–
( – )x
x
x x
e
e x e0
3
3 3
3 3
3 1 3 3
0
0
=+ +
= −→
lim–
x
x
x x x
e
e e xe0
3
3 3 3
9
9 9 27
1
2
Problem Set 6-7Review Problems
R0. Answers will vary.
R1. a. dM/dt = 0.06M ⇒ M− 1 dM = 0.06 dt
∴ =− ∫∫ M dM dtx
1
0
5
1000 06. , Q.E.D.
0.06 0.06 0.30
5dt t= =∫0
5
b. Solving numerically for x in
M dMx
− =∫ 1
1000 3.
gives x ≈ 134.9858… .
c. There will be $134.99 in the account, so theinterest will be $34.99.
R2. a. Integrating x− 1 by the power rule results in
division by zero:x
C–1 1
1 1
+
− ++ .
b. If g x f t dta
x
( ) = ∫ ( ) and f (x) is continuous in
a neighborhood of a, then g′ (x) = f (x).
ln xt
dtx
= ∫ 11
d
dxx
d
dx tdt
x
x
(ln ) =
=∫ 1 1
1
c. i. y = (ln 5x)3 ⇒ y′ = (3/x)(ln 5x)2
ii. f (x) = ln x9 = 9 ln x ⇒ f ′ (x) = 9/x
iii. y = csc (ln x) ⇒y′ = −csc (ln x) cot (ln x) · (1/x)
iv. g x t dt g x x xx
( ) ( )= ⇒ ′ =∫ csc csc2 2
1
2
d. i.sec tan
sec secsec tan
x x
xdx
xx x dx= ∫∫ 1
= +ln sec || x C
ii.10
102
3
2
3
xdx x=∫ ln
–
–
–
–
| |
= − −10 3 10 2ln | – | ln | |= 10(ln 3 − ln 2) = 4.054651…
iii. x x dx x x dx2 3 1 3 1 241
34 3( ) ( ) ( )− = −− −∫∫
= +1
343ln | – |x C
132 Problem Set 6-7 Calculus Solutions Manual© 2005 Key Curriculum Press
e. By finding areas, h (1) = −2.5, h (2) = 0,h (6) = 2.7, h (10) = 0, and h (11) = −2.5.
1 2 6 11
4
–4
y
x or t
y = f (t )
y = h(x)
f. i. y (100) ≈ 70 names; 70% rememberedy(1) = 1 name; 100% remembered
ii. yx
′ =+
101
100y′ (100) = 101/(200) = 0.505 names/persony′ (1) = 101/101 = 1 name/person
iii. Paula has probably not forgotten anynames as long as x − y < 0.5. Aftermeeting 11 people, she remembers about10.53… ≈ 11 names, but after meeting12 people, she remembers about 11.44… ≈11 names.
R3. a. i. See the text for the definition of logarithm.
ii. See the text for the definition of ln x.
iii. See the text for the statement of theuniqueness theorem.
iv. See the text for the proof.
v. See the solution to Problem 10 inLesson 6-3.
b. i. e n e nn
n
n
n= + = +→ →∞
lim ( ) lim ( / )/
0
11 1 1or
ii. logln
lnb xx
b=
c. i. y xx
yx
= = ⇒ ′ =logln
ln ln4 4
1
4
ii. f x xx
( ) ( )= = ⇒log cosln(cos )
ln2 2
′ = ⋅ −f xx
x( ) ( )1
2(cos )(ln )sin
= − tan
ln
x
2
iii. y x yx= = ⇒ ′ =log log5 59 9 9log5
R4. a. i.
2
10
y
x
ii.
–2
10
y
x
iii.
5
1
y
x
b. i. f (x) = x1.4e5x ⇒f ′ (x) = 1.4x0.4e5x + 5x1.4 e5x
ii. g (x) = sin (e− 2x) ⇒ g′ (x) = −2e− 2x cos(e− 2x)
iii.d
dxe
d
dxxx( ) ( )ln = = 1
iv. y = 100x ⇒ y′ = (ln 100)100 x
v. f (x) = 3.7 · 100.2 x ⇒ f ′ (x) = 0.74 ln 10 · 100.2 x
v i . r (t) = t tan t ⇒ ln r = tan t ln t ⇒1 2
rr t t
t
t′ = + ⇒sec ln
tan
′ = +
r t t t
t
tttan sec ln
tan2
c. y = (5x − 7)3 (3x + 1)5 ⇒ln y = 3 ln (5x − 7) + 5 ln (3x + 1) ⇒1 15
5 7
15
3 1yy
x x′ =
−+
+⇒
′ =−
++
− +y
x xx x
15
5 7
15
3 15 7 3 13 5( ) ( )
= (120x − 90)(5x − 7)2 (3x + 1)4
d. i. 10 52 2e dx e Cx x− −= − +∫ii. e x dx e Cx xcos cossin = − +∫
iii. e dx ex x− −
− −= −∫ 0 1 0 1
2
2
2
2
10. .
= − + =−10 10 4 026720 2 0 2e e. . . K
iv. 10
10
0 2 100 2
0 2.
.
. lnx
x
dx C= +∫e. i. The exposure is the product of C (t) and t,
where C (t) varies. Thus, a definite integralmust be used.
Calculus Solutions Manual Problem Set 6-7 133© 2005 Key Curriculum Press
ii. E x e dt et xx
( ) . ( ). .= = − +− −∫ 150 937 5 10 16 0 16
0
E(5) = 937.5(−e− 0.8 + 1) =516.25… ppm · days E (10) =937.5(−e− 1.6 + 1) = 748.22… ppm · daysAs x grows very large, E (x) seems toapproach 937.5.
iii. E ′ (x) = 150e− 0.16 x = C (x)E ′ (5) = 67.39… ppm (or ppm · daysper day)E′ (10) = 30.28… ppm
f. i. From Figure 6-7d, the maximumconcentration is about 150 ppm at about2 hours. (These values can be found moreprecisely by setting the numerical oralgebraic derivative equal to zero, solvingto get t = −1/ln 0.6 = 1.9576… . ThenC (1.9576…) = −200/(e ln 0.6) =144.0332… .)
ii. C (t) = 200t · 0.6t
C′ (t) = 200t · 0.6t ln 0.6 + 200 · 0.6t
C′ (1) = 200 · 0.61(ln 0.6 + 1) = 58.70…
C′(5) = 200 · 0.65(5 ln 0.6 + 1)
= −24.16… < 0
C(t) is increasing at about 58.7 ppm/hwhen t = 1 and decreasing at about24.2 ppm/h when t = 5. The concentrationis increasing if C′ (t) is positive anddecreasing if it is negative.
iii. Solving 50 = 200t · 0.6t numerically for tgives t ≈ 0.2899… and t ≈ 6.3245… .So C(t) > 50 for 6.3245… − 0.2899… =6.03… , or about 6 hours.
iv. C1(t) = 200t · 0.3t
100
C(t)
1
50t
From the graph, the maximum is about60 ppm around t = 1. (Exactly, t = −1/ln 0.3 =0.8305… , for which C (0.8305…) =−200/(e ln 0.3) = 61.11092… ≈ 61.1 ppm.)Repeating the computations of part iiigives C (t) > 50 for 0.409… < t < 1.473… ,or for about 1.06 hours.In conclusion, the concentration peakssooner at a lower concentration and staysabove 50 ppm for a much shorter time.ln 0.5 = −0.025t
R5. a. limx
x
x→∞
−−
→ ∞−∞
2 3
7 5
2
2
=−
= −→∞
limx
x
x
4
10
2
5
b. limcos
x x
x x
e x→
− +− −
→0
2 1
1
0
0
= +−
→→
limsin
x x
x x
e0
2
1
0
0
= + = + =→
limcos
x x
x
e0
2 2 1
13
c. limx
xx e→∞
− → ∞ ⋅3 0
= → ∞∞→∞
limx x
x
e
3
= → ∞∞→∞
limx x
x
e
3 2
= → ∞∞→∞
limx x
x
e
6
= = ∞→∞
limx xe
60 (Form: 6/ )
d. L xx
x= →→
∞lim tan( / )
1
2 1π
ln lim [tan( / ) ln ]L x xx
= ⋅→1
2π
= →→
limln
cot ( / )x
x
x1 2
0
0π
=−
=−
= −→
lim/
( / )csc / /x
x
x1 2
1
2 2
1
2
2
π π π π
∴ L = e− 2/ π = 0.529077…
e. limx
x→
=2
43 48
f. lim (tan sec ) lim ( )/ /x x
x x→ →
− = =π π2
2 2
21 1
g. Examples of indeterminate forms:
0/0, ∞/∞, 0 · ∞, 00, 1∞ , ∞0, ∞ − ∞
R6. a. i. y = ln (sin4 7x) = 4 ln sin 7x ⇒y′ = 4(1/sin 7x) · cos 7x · 7 = 28 cot 7x
ii. y = x− 3e2x ⇒y′ = −3x− 4 · e2x + x− 3 · 2e2x
= x− 4e2x (2x − 3)
iii. y = cos (2x) ⇒ y′ = −sin (2x) · 2x ln 2
iv. y xx
yx
= = ⇒ ′ =logln
ln ln34 4
3
4
3
b. i. e dx e Cx x− −= − / +∫ 1 7 1 71 1 7. .( . )
ii. 2sec sec tanx x x dx∫= ∫ e x x dxxln sec sec tan2
Calculus Solutions Manual Problem Set 6-7 133© 2005 Key Curriculum Press
ii. E x e dt et xx
( ) . ( ). .= = − +− −∫ 150 937 5 10 16 0 16
0
E(5) = 937.5(−e− 0.8 + 1) =516.25… ppm · days E(10) =937.5(−e− 1.6 + 1) = 748.22… ppm · daysAs x grows very large, E(x) seems toapproach 937.5.
iii. E ′ (x) = 150e− 0.16x = C (x)E ′ (5) = 67.39… ppm (or ppm · daysper day)E′ (10) = 30.28… ppm
f. i. From Figure 6-7d, the maximumconcentration is about 150 ppm at about2 hours. (These values can be found moreprecisely by setting the numerical oralgebraic derivative equal to zero, solvingto get t = −1/ln 0.6 = 1.9576… . ThenC (1.9576…) = −200/(e ln 0.6) =144.0332… .)
ii. C (t) = 200t · 0.6t
C′ (t) = 200t · 0.6t ln 0.6 + 200 · 0.6t
C′ (1) = 200 · 0.61(ln 0.6 + 1) = 58.70…
C′(5) = 200 · 0.65(5 ln 0.6 + 1)
= −24.16… < 0
C(t) is increasing at about 58.7 ppm/hwhen t = 1 and decreasing at about24.2 ppm/h when t = 5. The concentrationis increasing if C′ (t) is positive anddecreasing if it is negative.
iii. Solving 50 = 200t · 0.6t numerically for tgives t ≈ 0.2899… and t ≈ 6.3245… .So C(t) > 50 for 6.3245… − 0.2899… =6.03… , or about 6 hours.
iv. C1(t) = 200t · 0.3t
100
C(t)
1
50t
From the graph, the maximum is about60 ppm around t = 1. (Exactly, t = −1/ln 0.3= 0.8305… , for which C (0.8305…) =−200/(e ln 0.3) = 61.11092… ≈ 61.1ppm.)Repeating the computations of part iiigives C (t) > 50 for 0.409… < t < 1.473…, or for about 1.06 hours.In conclusion, the concentration peakssooner at a lower concentration and staysabove 50 ppm for a much shorter time.ln 0.5 = −0.025t
R5. a. limx
x
x→∞
−−
→ ∞−∞
2 3
7 5
2
2
=−
= −→∞
limx
x
x
4
10
2
5
b. limcos
x x
x x
e x→
− +− −
→0
2 1
1
0
0
= +−
→→
limsin
x x
x x
e0
2
1
0
0
= + = + =→
limcos
x x
x
e0
2 2 1
13
c. limx
xx e→∞
− → ∞ ⋅3 0
= → ∞∞→∞
limx x
x
e
3
= → ∞∞→∞
limx x
x
e
3 2
= → ∞∞→∞
limx x
x
e
6
= = ∞→∞
limx xe
60 (Form: 6/ )
d. L xx
x= →→
∞lim tan( / )
1
2 1π
ln lim [tan( / ) ln ]L x xx
= ⋅→1
2π
= →→
limln
cot ( / )x
x
x1 2
0
0π
=−
=−
= −→
lim/
( / )csc / /x
x
x1 2
1
2 2
1
2
2
π π π π
∴ L = e− 2/ π = 0.529077…
e. limx
x→
=2
43 48
f. lim (tan sec ) lim ( )/ /x x
x x→ →
− = − = −π π2
2 2
2
1 1
g. Examples of indeterminate forms:
0/0, ∞/∞, 0 · ∞, 00, 1∞ , ∞0, ∞ − ∞
R6. a. i. y = ln (sin4 7x) = 4 ln sin 7x ⇒y′ = 4(1/sin 7x) · cos 7x · 7 = 28 cot 7x
ii. y = x− 3e2x ⇒y′ = −3x− 4 · e2x + x− 3 · 2e2x
= x− 4e2x (2x − 3)
iii. y = cos (2x ) ⇒ y′ = −sin (2x ) · 2x ln 2
iv. y xx
yx
= = ⇒ ′ =logln
ln ln34 4
3
4
3
b. i. e dx e Cx x− −= − / +∫ 1 7 1 71 1 7. .( . )
ii. 2sec sec tanx x x dx∫= ∫ e x x dxxln sec sec tan2
Calculus Solutions Manual Problem Set 6-7 135© 2005 Key Curriculum Press
C4. a. Suppose there is a number M > 0 such thatln x ≤ M for all x > 0. Let x = eM +1. Thenln x = ln eM +1 = (M + 1) ln e = M + 1 > M.This contradicts ln x ≤ M for all x > 0. Thusthe supposition is false, and there can be nosuch number M that is an upper bound forln x, Q.E.D.
b. If M were a lower bound for ln x, then −Mwould be an upper bound for ln (1/x), butpart a shows no such number can exist.
c. ln′ x = 1/x, which shows that ln isdifferentiable for all x > 0. Thus, ln iscontinuous for all x > 0 becausedifferentiability implies continuity.
d. Because ln is continuous for all x > 0, theintermediate value theorem applies. Thus, if kis between ln a and ln b, there is a number cbetween a and b such that ln c = k.
ln a
ln b
a bc
k
x
y = ln x
e. Part a shows k cannot be an upper bound forln, so there must be some b > 0 such thatln b > k. Similarly, part b shows k is nota lower bound, so some a > 0 exists forwhich ln a < k. By part d there is somenumber c between a and b such that ln c = k,Q.E.D.
f. The domain of ln is the positive reals, and therange is all reals; the domain of the inverse toln (i.e., exp) is the range of ln (i.e., all reals),and the range of the inverse is the domain ofln (i.e., positive reals).
C5. a. g x t dt t dtx
x( ) = = − ⇒∫∫ sin sin
4
4 2
2
g′ (x) = −2x sin x2
b. g x t dtx
x
( ) = ∫ sintan
2
= + ⇒∫∫ sin sintan
t dt t dtx
x 4
4
2
g′ (x) = −2x sin x2 + sin (tan x) sec2 x
c. g x f t dtu x
v x
( ) = ⇒∫ ( )( )
( )
g′ (x) = f (v(x)) · v′ (x) − f ( u ( x ) ) · u′ (x)
C6. log cabin(or log cabin + C, which equals “houseboat”)
Chapter Test
T1. ln xt
dtx
= ∫ 11
T2. enn
n
= +
→∞
lim 11
or e nn
n= +→
lim ( ) /
0
11
T3. If g (x) = f t dta
x
( )∫ and f ( t ) is continuous in a
neighborhood of a, then g′ (x) = f ( x ) .
T4. If (1) f ′ ( x ) = g′ (x) for all x in the domain and(2) f ( a ) = g ( a ) for some a in the domain, thenf ( x ) = g ( x ) for all x in the domain.
T5. Prove that ln x = loge x for all x > 0.
Proof:
Let f ( x ) = ln x, and g (x) = loge x.f ′ ( x ) = 1/x andg′ (x) = (1/x) · loge e = (1/x) · 1 = 1/x∴ f ′ ( x ) = g′ (x) for all x > 0f ( 1 ) = ln 1 = 0 and g (1) = loge 1 = 0∴ f ( 1 ) = g (1)∴ by the uniqueness theorem,f ( x ) = g (x) for all x > 0.∴ ln x = loge x for all x > 0, Q.E.D.
T6. f ( x ) = ln (x3ex )
a. ′ = ⋅ + = +f xx e
x e x e xxx x( ) ( ) /
13 3 13
2 3
b. f ( x ) = 3 ln x + x ln e = 3 ln x + x ⇒ f ′ ( x ) = 3/x + 1 (Checks.)
T7. y = e2x ln x3 = 3e2x ln x ⇒y e x e xx x′ = ⋅ + ⋅ /6 3 12 2ln ( )
= + /3 2 12e x xx ( )ln
T8. v = ln (cos 10x) ⇒v′ = 1/(cos 10x) · (−10 sin 10x) = −10 tan 10x
T9. f ( x ) = (log2 4x)7 = [(ln 4x)/(ln 2)]7 ⇒f ′ ( x ) = 7[(ln 4x)/(ln 2)]6 · [(1/4x) · 4 · (1/ln 2)]
= 7 4
22
6(log )
ln
x
x
T10. t ( x ) = ln (cos2 x + sin2 x) = ln 1 = 0 ⇒t′ (x) = 0
T11. p x etx
( )ln
= ∫1sin t dt ⇒
p′ (x) = eln x sin ln x · 1/x = sin ln x
T12. e dx e Cx x5 51
5= +∫
T13. ( ) ( )6ln (ln )x dx x x C/ = +∫ 1
77
T14. sec ln sec tan51
55 5x dx x x C= + +∫ | |
136 Problem Set 6-8 Calculus Solutions Manual© 2005 Key Curriculum Press
T15. 51
55
1
525 1
0
2
0
2x xdx = =∫ ln ln
( – )
= 14.9120…
T16. lim–
lnx
x
x→∞→ −∞
∞5 3
4
=⋅
= = −∞→∞ →∞
lim–
[ /( )]lim(– )
x xxx
3
1 4 43
T17. L xx
x= → ∞→ −lim (tan )
/
cot
π 2
0
ln lim cot ln tan/
L x xx
= ⋅ → ⋅∞→ −π 2
0[ ( )]
= → ∞∞→ −
limln (tan )
tan/x
x
xπ 2
= ⋅ = =→ →− −lim
( / tan ) sec
seclim cot
/ /x x
x x
xx
π π2
2
22
10
∴ L = e0 = 1
T18. a.
1 2 3 4 5 6 7 8 9 10
1
2
3
4
5
-1
-2
-3
-4
-5
y
t or x
f
g
b. h x f t dt h x f xx
( ) ( ) ( ) ( ) ,= ⇒ ′ = − ⋅−
∫ 3 5 32
3 5
h f f′ = − ⋅ = ⋅ = ⋅ =( ) ( ) ( )3 9 5 3 4 3 1 3 3
T19. ln ,xt
dtx
= ∫ 11
so ln ..
1 81
1
1 8
= ∫ tdt
M4 0 2
1
1 3
1
1 5
1
1 70 58664= + + +
=.
1
1.1.
. . .K
From calculator, ln 1.8 = 0.58778… .
T20. g x t dt tx x
( ) sin cos= = −∫2 2
2 2
= −cos x2 + cos 2 ⇒ g′(x) = 2x sin x2
′ = =∫g xd
dxt dt x x
x
( ) sin sin2 2
2
2
T21. Let h ( x ) = f ( x ) − g (x).
Then h ( a ) = f ( a ) − g (a) = 0 andh ( b ) = f ( b ) − g (b) ≠ 0.
∴ ≠h b h a
b a
( ) – ( )
–0
By the mean value theorem, there is a number cbetween a and b such that
′ =h ch b h a
b a( )
( ) – ( )
–.
∴ h′ (c) ≠ 0
But h ′ ( x ) = f ′ ( x ) − g′ (x), which equals 0 forall values of x.
∴ h ′ ( c ) = 0
This result thus contradicts the mean valuetheorem, Q.E.D.
T22. a. F ( x ) = 60e0.1 x ⇒ F ′ ( x ) = 6e0.1 x so
F ′ ( 5 ) = 6e0.5 = 9.8923… lb/ftF ′ ( 1 0 ) = 6e = 16.3096… lb/ft
b. Work equals force times displacement. But theforce varies at different displacements. Thus, adefinite integral has to be used.
c.
dx
F
x
5
100 (x, F)
dW = F dx = 60 e0.1 x dx
W e dxx= ∫ 60 0 1
0
5.
= = −600 600 10 1
0
5 0 5e ex. .( )
W ≈ 389.23 ft-lb
T23. Answers will vary.
Problem Set 6-8Cumulative Review, Chapters 1–6
1. f ( x ) = 2x
f ′ ≈ =( ) .3
2 2
0 25 549618
3 1 2 9. .–
.K
2. There are about 10.0 squares, each 20 units.
∴ ≈∫ g x dx( ) 20010
50
(Function is g x x x( ) .= + +2 0 12
15sin ,
π so exact
answer is 200.)
3. L f xx c
= →lim ( ) if and only if for any ε > 0, there is
a δ > 0 such that if x is within δ units of c butnot equal to c, f ( x ) is within ε units of L.
4. Answers may vary.
3
4
x
f(x)
Calculus Solutions Manual Problem Set 6-8 137© 2005 Key Curriculum Press
5. f ′ (x) = +→
lim( ) – ( )
h
f x h f x
h0
or ′ = −−→
f cf x f c
x cx c( ) lim
( ) ( )
6. f (x) = x3
f xx h x
hh′ = + −
→( ) lim
( )0
3 3
= + + +→
lim–
h
x x h xh h x
h0
3 2 2 3 33 3
= + + =→
lim( )h
x xh h x0
2 2 23 3 3 , Q.E.D.
7. f (x) = x3 ⇒ f ′ (x) = 3x2
f ′ (5) = 3·52 = 75
f ′ (5) ≈ 5 01 4 99
0 0275 0001
3 3. – .
..=
f ′ (5) ≈ 5 001 4 999
0 00275 000001
3 3. – .
..=
The symmetric differences are getting closer to75 as ∆ x gets closer to zero.
8. f (x) = 3 5 3 5 1 2x x– ( ) /= −
f ′ (x) = 1
23 5 3 1 5 3 51 2 1 2( – )x x− −⋅ = −/ /. ( )
f ′ (7) = 1.5(21 − 5)−1/2 = 1.5/4 = 0.375 = 3/8
9. Line with slope of 3/8 is tangent to the graph atx = 7.
5
5
x
f(x)
8
3
10. a. y = e2x cos 3x ⇒y′ = 2 e2x cos 3x − 3e2x sin 3x
b. q xx
xq x( )
ln
tan( )= ⇒ ′ =
(tan )/ ln sec
tan tan sin
x x x x
x x x
x
x
−= −
2
2 2
1 ln
c. d
dx
d
dxx x x
2
225 5 5 5 5( ) [(ln ) ] (ln )= =
11. For the function to be differentiable,lim ( ) lim ( )x x
ax x x b→ →− +
+ = − + +2
2
2
21 6 and
lim lim ( )x x
ax x→ →− +
= − +2 2
2 2 6 .
4a + 1 = 8 + b and 4a = 2 ⇒ a = 1
2 and b = −5
2
2
y
x
12. Optional graph showing upper sum:
1 4
10
x
y = x 2
x dx2
1
4
∫U6 = 0.5(1.52 + 22 + 2.52 + 32 + 3.52 + 42)
= 24.875
13. M10 = 20.9775M100 = 20.999775Sums seem to be approaching 21.
14. a. cos sin cos5 61
6x x dx x C= − +∫
b. ( / ) | | 1 x dx x C= +∫ ln
c. tan ln sec ln cosx dx x C x C= + = − +∫ | | | |
d. sec ln sec tanx dx x x C= + +∫ | |
e. ( ) ( ) ( )/ /3 51
33 5 31 2 1 2x dx x dx− = −∫∫
= ⋅ + = +1
3
2
33 5
2
93 53 2 3 2( – ) ( – )x C x C/ /
15. x dx x2 3
1
4
1
4 1
3
64
3
1
321= = − =∫ ,
which agrees with the conjecture in Problem 13.
16. The graph shows a tangent line at x = c parallelto the secant line.
a c b
x
f(x)
Statement:
If f is differentiable on (a, b) and continuous atx = a and x = b, then there is a number x = c in
(a, b) such that f xf b f a
b a′ =( )
( ) – ( )
–.
17. y = x9/7
y7 = x9
7y6 y′ = 9x8
yx
y
x
xx x x′ = = = = =− −9
7
9
7
9
7
9
7
9
7
8
6
8
9 7 68 54 7 2 7 9 7 1
( )// / /
as from the derivative of a power formula
138 Problem Set 6-8 Calculus Solutions Manual© 2005 Key Curriculum Press
18. If x− 1 were the derivative of a power, then thepower would have to be x0. But x0 = 1, so itsderivative equals 0, not x− 1. Thus, x− 1 is not thederivative of a power, Q.E.D.
19. f x t dtx
( ) = ⇒∫ costan
31
f ′ (x) = cos (3 tan x) · sec2 x
20. f x t dt f x xx
( ) ,1
= ⇒ ′ =∫ ( / ) ( ) /1 1 Q.E.D.
21. Prove ln xa = a ln x for any constant a and allx > 0.
Proof:
Let f (x) = ln xa and g (x) = a ln x.
Then f xx
ax ax
a
xaa′ = ⋅ = ⋅ =−( )
1 11 and
g x ax
a
x′ = ⋅ =( ) .
1
∴ f ′ (x) = g′ (x) for all x > 0f (1) = ln (1a) = ln 1 = 0 and g(1) = a ln 1 = 0∴ f (1) = g (1)∴ f (x) = g (x) for all x > 0, and thusln xa = a ln x for all x ≥ 0, Q.E.D.
22. x = 5 cos t, y = 3 sin t
∴ = = = −dy
dx
dy dt
dx dt
t
tt
/
/
cos
– sincos
3
5
3
5
23. At t = 2, (x, y) = (5 cos 2, 3 sin 2) = (−2.08… , 2.72…).
At t = 2, dy
dx= − = …3
52 0 2745cot . .
The graph shows that a line of slope 0.27…at point (−2.08… , 2.72…) is tangent to thecurve.
5
3
x
y
24. y = tan− 1 t
vdy
dt tt= =
+= + −1
112
2( ) 1
adv
dtt t
t
t= = − + ⋅ = −
+−1 1 2
2
12 2
2 2( )( )
25. lim–
sinx
xe
x→→
0
3 1
5
0
0
= =→
limx
xe
x0
33
5 5
3
5cos
26. L nn
n= + →→
∞lim ( ) /
0
11 1
ln lim ln ( )Ln
nn
= +
→ ∞ ⋅→0
11 0
= + →→
limln ( )
n
n
n0
1 0
0
= + =→
lim/( )
n
n0
1 1
11
∴ L = e1 = e, Q.E.D.
27. Know: dx
dt
dy
dt= − =30 40 ft/s, ft/s
Want: dz
dt when x = 200 and y = 100
x2 + y2 = z2
2 2 2xdx
dty
dy
dtz
dz
dt+ =
When x = 200 and y = 100, z = =50 000 100 5, .
2 200 30 2 100 40 2 100 5( )( ) ( )( )− + = ⋅ dz
dt
dz
dt= − = −20
58 94427. K ft/s
The distance z is decreasing.
28. f x( )2
5
∫ dx ≈ (1/3)(0.5)(100 + 4 · 150 + 2 · 170 +
4 · 185 + 2 · 190 + 4 · 220 + 300)= (1/3)(0.5)(3340) = 556 2
3
29. Area of cross section = πy2
Because the end of the radius is on a line throughthe origin with slope r/h, y = (r/h)x.
∴ = =Area [( / ) ]2π πr h x
r
hx
2
22
dx
(x, Area)
Area
x
h
dV = (Area) dx
∴ = = ∫∫V dxr
hx dx
hh
( )Areaπ 2
22
00
= ⋅ = − =π π πr
hx
r
hh r h
h2
23
0
2
23 3 21
3
1
30
1
3( ) , Q.E.D.
30. Answers will vary.
Calculus Solutions Manual Problem Set 7-2 139© 2005 Key Curriculum Press
Chapter 7—The Calculus of Growth and Decay
Problem Set 7-1
1. D (0) = 500D (10) = 895.4238482…D (20) = 1603.567736…
2. D ′ (t) = 500(ln 1.06)(1.06t ) $/yrD ′ (0) = 29.13445406…D ′ (10) = 52.17536994…D ′ (20) = 93.43814108…The rate of change, in $/yr, increases as theamount in the account increases.
3. R tD t
D t
t
t( )( )
( )
(ln . ) ( . )
( . )= ′ = ⋅ ⋅
⋅500 1 06 1 06
500 1 06= ln 1.06 = 0.0582689081…
R(0) = ln 1.06R(10) = ln 1.06R(20) = ln 1.06
4. The percent interest rate stays the same:approximately 5.83%.
5. f (x) = a ⋅ bx ⇒f ′(x) = a ⋅ (ln b) ⋅ bx = (ln b)(a ⋅ bx )
= (ln b) ⋅ f (x)
So f ′(x) is directly proportional to f (x).
6. See Problem 11 in Section 7-2.
Problem Set 7-2Q1. Q2.
1
y
x1
y
x
Q3. Q4.
x
y
1
y
x
Q5. Q6.
x
y y
x
Q7. Q8.y
x
3
x
y
Q9. Q10.
4
x
y
3
3
x
y
1. a. B = number of millions of bacteria;t = number of hours
dB dt kB dB B k dt B kt C/ = ⇒ = ⇒ = +∫ ∫/ ln | |
| |B e e e B C ekt C kt C kt= = ⋅ ⇒ =+1
b. 5 510
1= ⇒ =⋅C e Ck
7 = 5e3k ⇒ ln (7/5) = 3k
⇒ = ⋅ =k
1
3
7
50 112157ln . K
∴ = =
=B e et
tt5 5
7
551 3 7 5
30 112157( / ) ln( / )
/. K
c.B
t5
10
d. B = 5(7/5)24/3 = 73.78945…About 74 million
e. 1000 = 5(7/5)t/3 ⇒ ln (1000/5) = t/3 ⋅ ln (7/5)
t = =3 200
7 547 24001
ln
ln ( / ). K
About 47 hours after start, so in a little lessthan 2 days
2. a. N = number of units of radiation from N17;t = number of seconds
dN dt kN dN N k dt/ / = ⇒ = ∫∫⇒ ln |N| = kt + C| | N e N C ekt C kt= ⇒ =+
1
b. 3 × 1017 = C1ek·0 ⇒ C1 = 3 × 1017
5.6 × 1013 = 3 × 1017e60k
⇒ × =−ln )( .1 866 10 604K k⇒ k = −0.143103…
∴ = × −N e t3 1017 0 143103. K
c.N
t
3 × 1017
140 Problem Set 7-2 Calculus Solutions Manual© 2005 Key Curriculum Press
d. t = 5(60) = 300 s
N e= × =−3 10 0 06799117 0 143103 300( . )( ) .K KIt will not be safe because 0.067… > 0.007.
3. a. F = number of mg; t = number of minutes
dF dt kF dF F k dt F kt C/ /= ⇒ = ⇒ = +∫∫ ln | |
| |F e F C ekt C kt= ⇒ =+1
50 = C1ek·0 ⇒ C1 = 50
30 = 50e20k ⇒ ln (30/50) = 20k⇒ k = (1/20) ⋅ ln (0.6) = −0.025541…
( . )( . ) / / .F e et t t= = = −50 50 0 6 500 6 20 20 0 025541ln K
b.F
t
50
50
c. F = 50(0.6)(60/20) = 10.8 mg (exactly)
d. 0.007 = 50(0.6)t/20
⇒ ln (0.007/50) = ln (0.6)t/20⇒ t = 347.4323…
About 5 h 47 min
4. a. V = number of dollars trade-in value;t = number of months from the present
dV dt kV dV V k dt V kt C/ /= ⇒ = ⇒ = +∫ ∫ ln | |
| |V e V C ekt C kt= ⇒ =+1
b. 4200 420010
1= ⇒ =⋅C e Ck
4700 4200 4700 4200 33= ⇒ = −−e kk( )( ) ( / )ln
k = (−1/3) ln (4700/4200) = −0.037492…
∴ = −V e t4200 0 037492. K
c.
–30 30
t
V
4200
d. At 1 year after V = 4700, t = 9 months.V = 4200e( −0.037492…)(9) = 2997.116…About $3000
e. 1200 = 4200e− 0.037492…t
⇒ ln (1200/4200) = −0.037492…t⇒ t = (−1/0.037492…) ⋅ ln (1200/4200)= 33.4135…About 33 months from the present
f. 31 months before V = 4700, t = −34.
∴ = =− −V e4200 15026 7950 037492 34( . )( ) .K KAbout $15,000
g. The difference between $16,000 and $15,000is the dealer’s profit.
5. a. dC/dt = kC
b. dC C k dt C kt D/ = ⇒ = +∫ ∫ ln | |
⇒ = ⇒ =+| |C e C D ekt D kt1
0 00372 0 0037210
1. .= ⇒ =⋅D e Dk
0.00219 = 0.00372e8k
⇒ ln (0.00219/0.00372) = 8k⇒ k = (1/8) ⋅ ln (219/372) = −0.0662277…∴ C = 0.00372e− 0.0662277…t
c. Either: C = 0.015⇒ 0.015 = 0.00372e− 0.0662277…t
ln 4.0322… = −0.0662277…tt = −21.05… , which is before the poison wasinhaled,or: t = −20 ⇒ C = 0.00372e− 0.0662277…( −20)
C = 0.0139… , which is less than 0.015∴ the concentration never was that high.
d.
20,000
t
P
50
100
e. (1/2)(0.00372) = 0.00372e− 0.0662277…t
ln (1/2) = −0.0662277…t ⇒ t = 10.4661…About 10.5 hours
6. a. dP/dt = kP
b. dP P k dt P kt C/ = ⇒ | | = +∫ ∫ ln
⇒ = ⇒ =+| |P e P C ekt C kt1
100 10010
1= ⇒ =⋅C e Ck
50 = 100e5750k ⇒ ln 0.5 = 5750k⇒ k = −0.0001205473…∴ P = 100e− 0.0001205473…t
c. P = 100e( −0.0001205473…)(4000) = 61.74301…About 61.7%
d. 48 37 100 0 0001205473. .= −e tK
ln 0.4837 = −0.0001205473…tt = 6024.939…The wood is about 6025 years old. For 1996,the flood would have been 1996 − (−4004) =6000 years ago, so the wood is old enough.
e.
20,000
t
P
5061.7
4000 5750
100
Calculus Solutions Manual Problem Set 7-3 141© 2005 Key Curriculum Press
7. dM/dt = kM ⇒ M = Cekt by the techniques inProblems 1–6, where C is the initial investment.∴ M varies exponentially with t.Let i = the interest rate as a decimal.dM/dt = Ck ⋅ ekt
At t = 0, dM/dt = Ci.∴ Ci = Ck ⋅ e0 ⇒ k = i ⇒ M = Ceit
Examples:
$1000 at 7% for 5 yr: $1419.07$1000 at 7% for 10 yr: $2013.75$1000 at 14% for 5 yr: $2013.75$1000 at 14% for 10 yr: $4055.20Leaving the money in the account twice as longhas the same effect as doubling the interest rate.Doubling the amount invested obviously doublesthe money at any particular time, but that doesn’ttell us how that compares with doubling the timeor the interest rate.Algebraically, Cei ·2t = Ce2i ·t shows that doublingthe time is equivalent to doubling the interestrate. Solving Ce2it > 2Ceit gives Ce2it − 2Ceit > 0⇒ Ceit(eit − 2) > 0 ⇒ eit > 2 (because Ceit > 0, soC > 0, being an investment) ⇒ it > ln 2. Sodoubling either the time or the interest rate willalways eventually yield more than doubling theinvestment, once t is high enough. For example,at 7%, 0.07t > ln 2 ⇔ t > (ln 2)/0.07 =9.9021… ⇒ t, so by 9 years 11 months,doubling the time or interest rate will yield morethan doubling the investment.
8. Assume an investment of $1000 at 7% per year.For 5 years, as in Problem 7:Annually: M = 1000(1.07)5 = $1402.55Quarterly: M = 1000(1.0175)20 = $1414.78Monthly: M = 1000(1.00583…)60 = $1417.63Daily: M = 1000(1.0001917808…)1825 =$1419.02Continuously (Problem 7): $1419.07Note that compounding continuously is only5 cents better than compounding daily for a$1000 investment in 5 years!
M = M0(1 + k/n)nt
Let L k nn
nt= + →→∞
∞lim ( / ) .1 1
ln lim [ lnL nt k nn
= ⋅ + / → ∞ ⋅→∞
( )] 01
= ⋅ + → ∞∞→∞
limln( / )
–n
t k n
n
11
=⋅
+⋅ −
−→∞
−
−lim/
( )
n
tk n
kn
n
11
2
2
=+
=→∞
lim/n
kt
k nkt
1∴ L = ekt
∴ =→∞
limn
ktM M e0 , which is the continuous
compounding equation.
9.dy
dxy= 0 3.
dy
ydx=∫ ∫0 3.
ln |y| = 0.3x + C
| | = = ⋅+y e e ex C x C0 3 0 3. .
y = ±eC ⋅ e0.3 x = C1e0.3 x
−4 = C1e0 ⇒ C1 = −4, showing that C1 can be
negative.∴ y = −4e0.3 x
5
10
20
–10
–20
y
x
10.dy
dxy= −0 2.
dy
yx dx= − ∫∫ 0 2.
ln |y| = −0.2x + C
| | = = ⋅− + −y e e ex C x C0 2 0 2. .
y = ±eC ⋅ e− 0.2 x = C1e− 0.2 x
30 = C1e− 1.4
C
e1 1 4
307 3979= =. . K
∴ y = 7.3979…e− 0.2 x
11. dy dx ky dy y k dx y kx C/ = ⇒ / = ⇒ | | = +∫ ∫ ln 1
| |y e y Cekx C kx= ⇒ =+ 1
y Ce C y y y ek kx( ) ( ) ,0 000= ⇒ = ⇒ =⋅
Q.E.D.
Problem Set 7-3Q1. Cekx Q2. (kx2)/2 + C
Q3. kx + C Q4. −cos x + C
Q5. 1 1 2/ – x Q6. 5 cos x
Q7. tan x
Q8.
x1
1
y' or y
y
y'
Q9. lim lim∆ ∆x
nx
nL U→ →
=0 0
Q10. B
1. a. dM/dt = 100 − S
b. S = kM ⇒ dM/dt = 100 − kM
142 Problem Set 7-3 Calculus Solutions Manual© 2005 Key Curriculum Press
c.dM
kMdt
k
k dM
kMdt
100
1
100−= ⇒ − −
−= ∫∫∫∫
⇒ − | − | = +1100
kkM t Cln
⇒ |100 − kM| = e− kte− kC
⇒ 100 − km = C1e− kt ⇒ kM = 100 − C1e
− k t
⇒ =Mk
C e kt1100 1( – )–
Substitute M = 0 when t = 0.
01
100 10010
1= ⇒ =k
C e C( – )
∴ =Mk
e kt1001( – )–
d. k = 0.02 ⇒ M = 5000(1 – e–0.02 t)
e.
M
t
5000
30 60 90
f. t = 30: $2255.94 ($3000 in, $744.06 spent)t = 60: $3494.03 ($6000 in, $2505.97 spent)t = 90: $4173.51 ($9000 in, $4826.49 spent)
g. t = 365: (365.23 or 366 could be used.)M = 5000(1 – e− 7.30 ) = 4996.622…≈ $4996.62 in the accountdM/dt = 100 − 0.02(4996.622…) = 0.06755…Increasing at about $0.07 per day
h. lim lim .
t t
tM e→∞ →∞
−= 5000 1 0 02( – )
= 5000(1 − 0) = 5000
2. dM/dt = 100 + kM (k = daily interest rate)dM
kMdt
k
k dM
kMdt
100
1
100+= ⇒
+= ⇒∫∫∫∫
1100 100
kkM t C kM e ekt kCln | || |+ = + ⇒ + = ⇒
100 + km = C1ekt ⇒ kM = −100 + C1e
kt ⇒
Mk
C ekt= −11001( )
Substitute M = 0 when t = 0.
01
100 10010
1= ⇒ =k
C e C( – )
∴ =Mk
ekt1001( – )
Let k = 0.0002 (0.02% per day).∴ M = (500000)(e0.0002 t – 1)
The graph is almost straight. The $100/daydeposits far exceed the interest for the first fewyears.
t
500
50,000
M
Make a table of M and dM/dt for variousnumbers of years. Neglect leap years.
Years M dM/dt
0 0 100.00
1 37865 107.57
10 537540 207.51
20 1652980 430.60
After 1 year, the $100/day is putting more intothe account. After 10 years, the interest hasstarted putting in more than the $100/day. After20 years, the interest puts in about $331 a day,while the winnings still put in only $100 a day.As t approaches infinity, the amount in theaccount becomes infinite!
3. a. E = RI + L(dI/dt)
b. L dI/dt = E – RIL dI
E RIdt
L
R
RdI
E RIdt
–= ⇒ − −
−=∫ ∫∫∫
⇒ = +– | – |L
RE RI t Cln
⇒ = − −| – |E RI e eR L t R L C( / ) ( / )
⇒ − = ⇒ =−E RI C e IR
E C eR L t R L t1 1
1( / ) –( / )[ – ]
Substitute I = 0 when t = 0.
01
10
1= ⇒ =R
E C e C E( – )
∴ =IE
Re R L t[ – ]–( / )1
c. I e t= 110
101 10 20[ – ]–( / )
I = 11(1 – e− 0.5 t)
I
t
5 10
11
Calculus Solutions Manual Problem Set 7-3 143© 2005 Key Curriculum Press
d. i. I = 11(1 – e− 0.5) = 4.3281… ≈ 4.33 amps
ii. I = 11(1 – e–5) = 10.9258… ≈ 10.93 amps
iii. lim lim .
t t
tI e→∞ →∞
−= =11 1 11 1 00 5( – ) ( – )
= 11 amps
e. I = 0.95(11) = 10.45
10 45 11 1
0 95 1
0 05
0 5 0 05
0 5
0 5
0 5
. ( – )
. –
.
. .
.
.
.
=
=
=− =
−
−
−
e
e
e
t
t
t
t
ln
t = −2 ln 0.05 = 5.9914…About 6 seconds
4. a. R = C(dT/dt) + hT
b. C dT/dt = R – hT
C dT R hT dt/( )− = ∫∫− −
−= ∫∫C
h
h dT
R hTdt
− − = +C
hR hT t Dln | |
| | ( / ) ( / )R hT e eh C t h C D− = − −
R hT D e h C t− = /1
–( )
T h R D e h C t= / /( )[ – ]–( )1 1
Substitute T = 0 when t = 0.
01
10
1= ⇒ =h
R D e D R( – )
∴ = TR
he h C t[ – ]–( / )1
c. T = (50/0.04)[1 – e− (0.04/ 2)t]T = 1250(1 – e− 0.02t)
d.
1250T
t
100 200
e. Use TRACE or TABLE.
t = 10: T = 226.586… ≈ 227°t = 20: T = 412.099… ≈ 412°t = 50: T = 790.150… ≈ 790°t = 100: T = 1080.830… ≈ 1081°t = 200: T = 1227.105… ≈ 1227°
f. lim lim .
t t
tT e→∞ →∞
−= =1250 1 1250 1 00 02( – ) ( – )
= 1250°g. T = 0.99(1250) = 1237.5
1237.5 = 1250(1 – e− 0.02t)
0.99 = 1 – e− 0.02t
e− 0.02t = 0.01
−0.02t = ln 0.01
t = −50 ln 0.01 = 230.258…
About 230 seconds
5. a.dV
dtkV= 1 2/
b. V dV k dt− = ⇒∫∫ 1 2/
22
1 22
V kt C Vkt C/ = + ⇒ = +
V varies quadratically with t.
c. Initial conditions t = 0; V = 196;dV/dt = −28:
1960
2281 2/ = ⋅ + ⇒ =k C
C
and −28 = k ⋅ 1961/ 2 ⇒ k = −2
∴ =− +
⇒ = −
V
tV t
2 28
214
2
2( )
d. False. dV/dt = 2t − 28, so the water flowsout at 28 ft3/min only when t = 0. Forinstance, at t = 5, dV/dt = −18, whichmeans water flows out at only 18 ft3/min.So it takes longer than 7 min to emptythe tub.
e. 0 = (t − 14)2 ⇒ the tub is empty att = 14 min.
f.
14
100
V
t
g. See the solution to Problem C4 in ProblemSet 7-7.
6. The following data were gathered in the author’sclass in December 1994. Times t are in secondsand volumes V are in mL. Note that a burettereads the amount of fluid delivered, so youmust subtract the reading from 50 to find thevolume remaining. Use food coloring in thewater to make the liquid level easier to read.Read from the bottom of the meniscus (thecurved surface of the liquid).
144 Problem Set 7-3 Calculus Solutions Manual© 2005 Key Curriculum Press
Seconds Reading Volume
0 0 50
10 2.4 47.6
20 4.4 45.6
30 6.4 43.6
40 8.5 41.5
50 10.5 39.5
60 12.4 37.6
70 14.3 35.7
80 16.1 33.9
90 17.8 32.2
100 19.9 30.1
110 21.2 28.8
120 22.8 27.2
130 24.5 25.6
140 25.6 24.4
150 27.4 22.6
160 28.6 21.4
170 30.0 20.0
180 31.3 18.7
190 32.6 17.4
200 33.8 16.2
210 35.1 14.9
220 36.4 13.6
230 37.4 12.6
240 38.5 11.5
250 39.5 10.5
260 40.6 9.4
M M M
320 46.1 3.9
360 49.3 0.7
Using quadratic regression with these data,V = 0.000209255…t2 − 0.20964…t + 49.54… .The data and the equation can be plotted on thegrapher, as shown.
t
V
100
50
The volume does seem to vary quadratically withtime. Because there is still fluid in the burettewhen V = 0, the graph crosses the t-axis, unlikethe graphs in Problem 5 and Example 1. The
position of the vertex can be used to predict theposition of the stopcock and the time when thefluid would all be gone if the burette were ofuniform diameter all the way down to thestopcock. For the preceding data, the vertex is at
t = − − ≈0 20964
2 0 000209255500
.
( )( . )
K
K s
V ≈ −3.0 mL
So the stopcock should be found at a pointcorresponding to about 3 mL below the bottommark.
7. a. n = 1, k = 1, C = −3:
∴ = ⇒ = ⇒ = −∫ ∫ dy dx y dy y dx y x/ / ln | | 3
⇒ |y| = e x−3 = exe− 3 ⇒ y = ±0.04978…ex
1
1
x
y
b. n = 0.5, k = 1, C = −3:
∴ = ⇒ =−∫ ∫ dy dx y y dy dx/ . .0 5 0 5
⇒ = − ⇒ = −2 31
430 5y x y x. ( )2
Note: x ≥ 3 because y0.5 is a positive number.
3
1
x
y
c. n dy dx ky y dy k dx= − ⇒ = ⇒ =− ∫∫1 1/
⇒ = + ⇒ = ± +1
22 22y kx C y kx C
k C y x= = − ⇒ = ± −1 3 2 6,
5
3
x
y
n dy dx ky y dy k dx= − ⇒ = ⇒ =− ∫∫2 2 2/
⇒ = + ⇒ = +1
33 33 3y kx C y kx C
k C y x= = − ⇒ = −1 3 3 93,
Calculus Solutions Manual Problem Set 7-3 145© 2005 Key Curriculum Press
5
3
x
y
d. For ndy
dxky y dy k dxn n> = ⇒ =− ∫∫1,
⇒ −−
= + >− −y
nkx C n
n( )
,1
11 because
so yn kx Cn
= −− ⋅ +−
1
11 ( ) ( )
which has a vertical asymptote at x = −C/kbecause the denominator equals zero for thispoint.
Note that the radical will involve a ± signwhen the root index is even (for example,when n is odd).
For , , : ( )n k C y x= = = − = − − −2 1 3 3 1
2
3
x
y
For n k C yx
= = = − = −± −
3 1 31
2 6, , :
2
3
x
y
Note that the graph shows two branches.
e. For so ndy
dxky k y kx C= = = = +0 0, , ,
a linear function. For k = 1, C = −3, y = x − 3.
2
3
x
y
8. dB/dt = c ( M − kB), where k and c are constants.dB
M kBc dt
k
k dB
M kBc dt
−= ⇒ − −
−=∫ ∫ ∫∫1
⇒ − − = +1
kM kB ct Cln | |
⇒ − = − −| |M kB e ekct kcC
⇒ − =
⇒ = −
−
−
M kB C e
Bk
M C e
kct
kct
1
11
( )
Use the initial condition B = 0 when t = 0.
0 = (1/k) ( M − C1e0) ⇒ C1 = M
∴ =BM
ke kct( – )–1
Use the initial condition kB = 80 whenB = 1000.
80 = k(1000) ⇒ k = 0.08
Use the initial condition dB/dt = 500 when t = 0.From dB/dt = c (M − kB), 500 = c (M − 0) ⇒c = 500/M.∴ particular equation is
B M e
B M e
M t
M t
= −= −
−
−( / . )[ ]
. [ ]
. /( / )
( / )
0 08 1
12 5 1
0 08 500
40
Assume various values of M:M = 1000: B = 12500(1 − e− 0.04 t)M = 5000: B = 62500(1 − e− 0.008 t )M = 10000: B = 125000(1 − e− 0.004 t )
250
100,000
M = 1000
M = 5000
M = 10000
500
B
t
As shown on the graph, the sales start outincreasing at the same rate (500 bottles/day). Ast increases, the number of bottles/day increases,approaching a steady state equal to 12.5M.
To find the break-even time, first find the totalnumber of bottles sold as a function of time. B isin bottles per day, so the total sales in x days,T(x), is
T x B dtx
( ) .= ∫0
Use, for example, M = $10,000/day.
T x e dt
t e
x e
x e
tx
t x
x
x
( ) ( )
( / . )
( )
[ ( )]
.
.
.
.
= −
= +
= + − −
= − −
−
−
−
−
∫ 125000 1
125000 1 0 004
125000 250 0 250
125000 250 1
0 004
0
0 004
0
0 004
0 004
[ ]
For selling prices of $0.25 and $0.50/bottle, thetotal numbers of dollars are
D25(x) = 31250[x − 250(1 − e− 0.004 x)]D50(x) = 62500[x − 250(1 − e− 0.004 x)]
146 Problem Set 7-3 Calculus Solutions Manual© 2005 Key Curriculum Press
The total amount spent on advertising is M ⋅ x,or A(x) = 10000x.The three graphs can be plotted by grapher. For$0.25/bottle, the break-even time is 207 days.For $0.50/bottle, the break-even time is 90 days(less than half!).
Dollars (millions)
x
2
1
20790
$0.50/bottle
$0.25/bottle
9. The differential equation is dT/dt = k(1200 − L),and L = h (T − 70), where h is a proportionalityconstant.
∴ dT/dt = k(1200 + 70h − hT )
dT h hT k dt/( ) 1200 70+ − = ∫ ∫(−1/h) ln |1200 + 70h − hT | = kt + C
ln |1200 + 70h − hT | = −kht − hC
|1200 + 70h − hT | = e− kht ⋅ e− hC
1200 + 70h − hT = C1e− kht ⇒
hT = 1200 + 70h − C1e− kht
T = 1200/h + 70 − (C1/h) ⋅ e− kht
Use T = 70 when t = 0.
70 = 1200/h + 70 − C1/h ⋅ e− kh⋅ 0
⇒ C1 = 1200
∴ T = 1200/h + 70 − 1200/h ⋅ e− kht
T = 70 + (1200/h)(1 − e− kht)
Substitute t = 0, L = 0, and dT/dt = 3 into theoriginal differential equation.3 = k(1200 − 0) ⇒ k = 0.0025
∴ T = 70 + (1200/h)(1 − e− 0.0025 ht)Using T = 96 when t = 10,26 = (1200/h)(1 − e− 0.025 h).Solving numerically gives h ≈ 11.7347… .
∴ equation is T ≈ 70 + 102.26…(1 − e− 0.02933…t ).
Time data for various temperatures can be foundby grapher or by substituting for T andsolving for t.
100
180°
170°160°140°
Never reaches
39 72 130
t
T
T t
140° 39 min
155° 61 min
160° 72 min
170° 130 min
180° Never!
The limit of T as t increases is 70 +102.26…(1 + 0), which equals 172.26…°. Thus,the temperature never reaches 180°. When theheater turns off, the differential equation becomesdT
dtkh T T C e kht= − − ⇒ = + −( ) .70 70 2
Using T = 160 at time t = 0 when the heaterturns off, T = 70 + 90e− 0.02933…t.To find the time taken to drop to 155°, substitute155 70 90 0 02933= + −e t. .K
Solving numerically or algebraically givest = 1.9… . Thus, it takes only 2 minutes for thetemperature to drop 5°! By contrast, from thepreceding table, it takes 11 minutes(t = 61 to t = 72 in the table) to warm back upfrom 155° to 160°.
The design of the heater is inadequate because ittakes much longer to warm up by a certainamount than it does to cool back down again.Near 172°, a slight increase in the thermostatsetting for the heater makes a great increase inthe time taken to reach that setting. For instance,it takes an hour (72 minutes to 130 minutes) towarm the 10 degrees from 160° to 170°. Theseinadequacies could be corrected most easily byadding more insulation. The resulting decrease inh would make the heater cool more slowly, heatup faster, and reach the 180 degrees it currentlycannot reach. Decreasing h would also reduce thepower consumption.
10. a. dP/dT = kP/T 2
dP P k dT T P kT C/ = / ⇒ = − +∫ ∫ −2 1ln | |
⇒ = ⇒ =− / + −| |P e P C ek T C k T1
/
b. 0 054 1293. = − /C e k
3 95 1343. = − /C e k
( . . ) ( )3 95 0 054 343 293/ = − / + /e k k
ln (3.95/0.054) = −k/343 + k/293k = / / − / + /
=[ ( . . )] ( )
.ln 3 95 0 054 1 343 1 293
8627 812641KFrom ln |P| = −kT −1 + C,C = ln 0.054 + 8627.812641…/293.C = 26.52768829… ⇒ C1 = e26.52768829…
P e e T= … − …/26 52768829 8627 812641. .
P e T= − /( . . )26 52768829 8627 812641K K
Calculus Solutions Manual Problem Set 7-4 147© 2005 Key Curriculum Press
c.
Temperature T P Actual*
10 283 0.0190… 0.021
20 293 0.054 0.054
30 303 0.142… 0.133
40 313 0.354… 0.320
50 323 0.832… 0.815
60 333 1.85… 1.83
70 343 3.95 3.95
80 353 8.05… 7.4(meltingpoint)
90 363 15.7… 12.6
100 373 29.8… 18.5
110 383 54.6… 27.3
200 473 3972.1… 496.5*Source: Lange’s Handbook of Chemistry, 1952, p. 1476.
The function models the data well up to themelting point, but not above it. Thedifferences between the predicted and actualanswers are most likely due to the fact thatnaphthalene changes from solid to liquid at80°C; the constants for solid and liquidnaphthalene differ.Use initial conditions for T = 90, 110 as inpart b to get a better equation for the liquid:12.6 = C1e
− k/ 363
496.5 = C1e− k/ 473
k = [ln (496.5/12.6)]/(−1/473 + 1/363)= 5734.569702…
C = ln 12.6 + 5734.569702…/363= 18.33140949…
⇒ C1 = e18.33140949…
∴ P = e(18.33140949… −5734.569702…/ T )
With the new equation,
Temperature T P Actual
10 283 0.144… 0.021
20 293 0.289… 0.054
30 303 0.551… 0.133
40 313 1.01… 0.320
50 323 1.78… 0.815
60 333 3.03… 1.83
70 343 5.01… 3.95
80 353 8.05… 7.4(meltingpoint)
90 363 12.6 12.6
100 373 19.2… 18.5
110 383 28.7… 27.3
200 473 496.5 496.5
So the new equation models the data abovethe melting point, but not below it.
d. Using the equation for liquid naphthalene,760 = e(18.33140949… −5734.569702…/ T ) ,
ln 760 = 18.33140949… − 5734.569702…/T
T .= =5734 569702
18 33140949 760490 214
.
. – ln
K
KK
About 490 K, or 217°C (actual: 218°C)
e. Answers will vary.
Problem Set 7-4Q1. y ′ = 5x4 Q2. y ′ = 5x ln 5
Q3. (1/8)x 8 + C Q4. 7 x/ln 7 + C
Q5. y ′ = −y/x or y ′ = −3x− 2
Q6. 87.5
Q7. Q8.
1
1
y
x1
1
x
y
Q9. g(5) − g(1) Q10. E
1. a. dy/dx = x/(2y)At (3, 5), dy/dx = 3/10 = 0.3.At (−5, 1), dy/dx = −5/2 = −2.5.On the graph, the line at (3, 5) slopes upwardwith a slope less than 1. At (−5, 1) the lineslopes downward with a slope much steeperthan −1.
b. The figure looks like one branch of ahyperbola opening in the y-direction. (Thelower branch shown on the graph is also partof the solution, but students would not beexpected to find this graphically.)
y
5
5
x(–5, 1) (1, 2) (5, 1)
(3, 5)
c. See graph in part b. The figure looks like theright branch of a hyperbola opening in thex-direction. (The left branch is also part of the
148 Problem Set 7-4 Calculus Solutions Manual© 2005 Key Curriculum Press
solution, but students would not be expectedto find this graphically.)
d.dy
dx
x
yy dy x dx y x C= ⇒ = ⇒ = +∫ ∫2
21
22 2
x = 5, y = 1 ⇒ C = 1 − 12.5 = −11.5
By algebra, x2 − 2y2 = 23. This is theparticular equation of a hyperbola opening inthe x-direction, which confirms theobservations in part c.
2. At (3, 3), dy/dx = 0.1(3) = 0.3, which isreasonable because the slope is positive and lessthan 1. At (0, −2), dy/dx = 0.1(−2) = −0.2, whichis reasonable because the slope is negative andless than 1 in absolute value. The next graphshows the two particular solutions. For the first,y (6) ≈ 4.0. For the second, y (6) ≈ −3.6.dy
dxy y C e x= ⇒ =0 1 1
0 1. .
For (3, 3), the particular solution isy = 2.2224…e0.1 x, giving y(6) = 4.0495… .
For (0, −2), the particular solution isy = −2e0.1 x, giving y(6) = −3.6442… .
Both graphical answers are close to these actualanswers.
1 2 3 4 5 6–1–2
1
2
3
4
–1
–2
–3
–4
y
x
3. a. At ( , ), . .3 23
2 20 75
dy
dx= − = −
( )( )
At ( , ), ,1 01
2 0
dy
dx= −
( )( ) which is infinite.
1 2 3–1–2–3
1
2
–1
–2
y
x
b. See the graph from part a. The figuresresemble half-ellipses.
c.dy
dx
x
y= −
2
2y dy = −x dx
2 y dy x dx = −∫ ∫
y x C2 21
2= − +
( ) .− = − + ⇒ =11
21 1 52 2( ) C C
y2 = 1.5 − 0.5x2
y x . . = − −1 5 0 5 2
(Use the negative square root because of theinitial condition.)
The graph agrees with part b.
From the next-to-last line, add 0.5x2 to bothsides, getting 0.5x2 + y2 = 1.5, which is theequation of an ellipse because x2 and y2 havethe same sign but unequal coefficients.
4. a. At ( , ), ( ) .3 1 3 1 1 0dy
dx= − =
At ( , ), ( ) .1 2 1 1 2 1dy
dx= − = −
At ( , ), ( ) .0 1 0 1 1 0− = + =dy
dx
1 2 3–1–2–3
1
2
3
–1
y
x
b. See the graph from part a. Both graphs have ahorizontal asymptote at y = 1.
c.dy
dxx y= −( )1
dy
yx dx
1–=
dy
yx dx
1–=∫ ∫
−ln |1 − y| = 0.5x2 + C
1 0 5 2
− = ± ⋅− −y e ex C.
y C e x= + −1 10 5 2. (C1 can be positive or
negative.)
−1 = 1 + C1e0 ⇒ C1 = −2
∴ = − − .y e x1 2 0 5 2
The grapher confirms the graph in part b.
As | | , ..x e x→ ∞ →−0 5 2
0 So y → 1,which agrees with the horizontal asymptoteat y = 1.
Calculus Solutions Manual Problem Set 7-4 149© 2005 Key Curriculum Press
5.
0.5 1 1.5 2–0.5–1–1.5–2
0.5
1
1.5
2
–0.5
–1
–1.5
–2
y
x
6.
0.5 1 1.5 2–0.5–1–1.5–2
0.5
1
1.5
2
–0.5
–1
–1.5
–2
y
x
7.
0.5 1 1.5 2–0.5–1–1.5–2
0.5
1
1.5
2
–0.5
–1
–1.5
–2
y
x
8.
0.5 1 1.5 2–0.5–1–1.5–2
0.5
1
1.5
2
–0.5
–1
–1.5
–2
y
x
9. a.
x
y
(3, 2)
(1, –2)
b.dy
dxxy= −0 2.
Evidence: At (1, 1) the slope was given tobe −0.2, which is true for this differentialequation. As x or y increases from thispoint, the slope gets steeper in the negativedirection, which is also true for thisdifferential equation. In Quadrants I and III theslopes are all negative, and in Quadrants IIand IV they are all positive. (Note: The
algebraic solution is y Ce x= −0 1 2. .)
10. a. Initial condition (0, 2)
(0, –5)
(0,–2.5)
y
(0, 2)
x5
5
b. See the graph in part a with initial condition(0, −5). The graph goes toward −∞ in they-direction instead of toward +∞.
c. If a ruler is aligned with the slope lines, thelines that form a straight line are the onescrossing the y-axis at −2.5 with slope −1/2.(In courses on differential equations, studentswill learn that the given equation is a first-order linear equation that can be solved usingan integrating factor. The general solution isy = Ce0.2 x − 0.5x − 2.5. For C = 0, the curveis the line y = −0.5x − 2.5, which intersectsthe y-axis at (0, −2.5).)
150 Problem Set 7-4 Calculus Solutions Manual© 2005 Key Curriculum Press
11. a. Initial condition (0, 2)
P
t
(0, 18)
(4, 2)(0, 2)
10.5
b. See the graph in part a with initial condition(4, 2). The graph is the same as that in part abut shifted over 4 months. This behavior isto be expected because dP/dt depends only onP, not on t, and both initial conditions havethe same value of P.
c. See the graph in part a with initial condition(0, 18). The population is decreasing to thesame asymptote, P = 10.5, as in parts aand b.
d. The asymptote at P = 10.5 indicates that theisland can sustain only 1050 rabbits. If thepopulation is lower than that, it increases. Ifthe population is higher than that, itdecreases. The number 10.5 is a value of Pthat makes dP/dt equal zero. Note that there isanother asymptote at P = 0, which alsomakes dP/dt equal zero.
12. a. dv/dt = 32.16 − 0.0015v2
The slope at (5, 120) appears to be about 1,but dv/dt actually equals 32.16 −0.0015(120)2 = 10.56. The answers aredifferent because the graph is scaled by afactor of 10.
b. Initial condition (0, 0)
t
v
(5,120)
(0,180)
146.4...
(0, 0) (5, 0)
50
c. Terminal velocity occurs when dv/dt = 0.0 = 32.16 − 0.0015v2
v = (32.16/0.0015)1/2 = 146.424… ≈ 146 ft/s
The graph shows this velocity for timesabove about 15 seconds.
d. See the graph in part b with initial condition(5, 0). The graph is identical to the one inpart b except shifted 5 seconds to the right.This behavior is to be expected because thedifferential equation is independent of t.
e. See the graph in part b. This graph decreasesto the terminal velocity because the diverstarts out going faster.
f. Similarities include: Both models have ahorizontal asymptote that the particularsolutions approach from above or below.Both models decrease rapidly and graduallylevel off for values above the asymptoticlimit.Differences include: For values below theasymptotic limit, one model starts with rapidincrease and gradually slows its growth,whereas the other starts with a slow increasethat becomes more rapid growth beforeslowing toward the asymptote.
13. a. mamg
r= 2 By hypothesis
dv
dt
g
r= 2 Divide by m; a
dv
dt= .
dv
dr
dr
dt
g
r⋅ = 2 Chain rule
dv
drv
g
r⋅ = 2 v
dr
dtr= =( )distance
dv
dr
g
r v= 2 Divide by v.
b.dv
dr( , )5 2 1 2488 .= −
dv
dr( , )1 10 6 244 .= −
dv
dr( , )10 4 0 1561 .= −
These slopes agree with those shown.c. Initial condition (r, v) = (1, 10)
From the graph, the velocity is zero at r ≈ 5.So the spaceship is about 4 Earth radii, orabout 25,000 km, above the surface.
r
v(1,18)
5
14.11...
(1,12)
(2,10)(1,10)
6.12...
4.37...(10,4)
(5,2)
Calculus Solutions Manual Problem Set 7-5 151© 2005 Key Curriculum Press
The precise value of r can be foundalgebraically.dv
dr r vv dv
rdr= ⇒ = ∫∫– . – .62 44 62 44
2 2
⇒ = +v
rC
2
2
62 44.
For the solution through (1, 10), C = 50 −62.44 = −12.44, so the ship starts fallingwhen v = 0 at r = 62.44/12.44 ≈ 5.
d. See the graph in part c with initial condition(r, v) = (1, 12). The graph levels off between4 and 5 km/s. The precise value of v can befound algebraically.
Cv
r= − = ⇒ = +72 62 44 9 56
2
62 449 56
2
. . ..
Because r > 0, v is never zero, so thespaceship never stops and falls back.As r approaches infinity, v2/2 approaches9.56, and thus v approaches
( )( . )2 9 56 4 37= …. km/s.
e. See the graph in part c with initial condition(r, v) = (1, 18). The graph levels off atv ≈ 14 km/s. Here the spaceship loses about4 km/s of velocity, whereas it loses 7 or8 km/s when starting at 12 km/s. Both caseslose the same amount of kinetic energy,which is proportional to v2 (the change in v2
is the same in both cases). The precise valueof v can be found algebraically as in part d.For the solution through (1, 18),C = 162 − 62.44 = 99.56. As r → ∞,v → = …( )( . )2 99 56 14 11. km/s.
f. See the graph in part c with initial condition(r, v) = (2, 10). The graph levels off atabout 6 km/s, so the spaceship does escape.Alternatively, note that the solution through(2, 10) lies above the solution through(1, 12). The precise value of v can be foundalgebraically as in parts d and e. For thesolution through (2, 10), C = 50 − 31.22 =18.78. As r → ∞, v → =( )( . )2 18 786.12… km/s.
14. See the Programs for Graphing Calculatorssection of the Instructor’s Resource Book.
Problem Set 7-5Q1. ky Q2. y = Ce3x
Q3. 4.8 Q4. 100
Q5. −ln |1 − v| + C Q6. sec x tan x
Q7.
1
1x
y
y' y'
Q8. 3x2y5 + 5x3y4y′ = 1 + y′Q9. continuous Q10. A
1. a. dy = −(x/y) dyFor (1, 3), dy = −(1/3)(0.5) = −0.1666… ,so new y ≈ 3 − 0.1666… = 2.8333… atx = 1.5.For (1.5, 2.8333…), dy =−(1.5/2.8333…)(0.5) = −0.2647… ,so new y ≈ 2.8333… − 0.2647… = 2.5686…at x = 2.
x y
0 3.2456…
0.5 3.1666…
1 3
1.5 2.8333…
2 2.5686…
2.5 2.1793…
3 1.6057…
The Euler’s y-values overestimate the actualvalues because the tangent lines are on theconvex side of the graph and the convex sideis upward.
b. dy = −(x/y) dy
y dy x dx= −∫∫0.5y2 = −0.5x2 + C0.5(32) = −0.5(12) + C ⇒ C = 50.5y2 = −0.5x2 + 5
y x= −10 2 (Use the positive square root.)
At x y= = =3 10 3 12, – .
The particular solution stops at the x-axisbecause points on the circle below the x-axiswould lead to two values of y for the samevalue of x, making the solution not afunction.
The Euler’s value of 1.6057… overestimatesthe actual value by 0.6057… .
2. a. dy = (x/y) dyFor (1, 2), dy = (1/2)(0.5) = 0.25, so newy ≈ 2 + 0.25 = 2.25 at x = 1.5.For (1.5, 2.25), dy = (1.5/2.25)(0.5) =0.3333… , so new y ≈ 2.25 + 0.3333… =2.5833… at x = 2.
152 Problem Set 7-5 Calculus Solutions Manual© 2005 Key Curriculum Press
x y
0 1.6071…
0.5 1.75
1 2
1.5 2.25
2 2.5833…
2.5 2.9704…
3 3.3912…
The Euler’s y-values underestimate the actualvalues because the tangent lines are on theconvex side of the graph and the convex sideis downward. The error is greater at x = 0because the graph is more sharply curvedbetween x = 0 and x = 1 than it is betweenx = 1 and x = 3.
b. dy = x/y dy
y dy x dx= ∫∫0.5y2 = 0.5x2 + C0.5(22) = 0.5(12) + C ⇒ C = 1.50.5y2 = 0.5x2 + 1.5
y x= +2 3 (Use the positive square root.)
At x y= = + = = …0 0 3 3 1 7320, . .
The particular solution stops at the x-axisbecause points on the circle below the x-axiswould lead to two values of y for the samevalue of x, making the solution not afunction.
The Euler’s value of 1.6071… underestimatesthe actual value by 0.1249… unit.
3. dx = 0.2. Make a table showing values of dy =0.2(dy/dx) and new y = old y + dy.
x dy/dx dy y
2 3 0.6 1
2.2 5 1.0 1.6
2.4 4 0.8 2.6
2.6 1 0.2 3.4
2.8 −3 −0.6 3.6
3 −6 −1.2 3.0
3.2 −5 −1.0 1.8
3.4 −3 −0.6 0.8
3.6 −1 −0.2 0.2
3.8 1 0.2 0.0
4 2 0.4 0.2
1 2 3 4
1
2
3
4
y
x
You cannot tell whether the last value of y is anoverestimate or an underestimate because theconvex side of the graph is downward in someplaces and upward in other places.
4. dx = 0.3. Make a table showing values of dy =0.3(dy/dx) and new y = old y + dy.
x dy/dx dy y
1 −3 −0.9 2
1.3 −2 −0.6 1.1
1.6 −1 −0.3 0.5
1.9 0 0 0.2
2.2 1 0.3 0.2
2.5 2 0.6 0.5
2.8 3 0.9 1.1
3.1 4 1.2 2
3.4 5 1.5 3.2
3.7 6 1.8 4.7
3.9 7 2.1 6.5
1 2 3 4
1
2
3
4
5y
x
The approximate values of y underestimate theactual values of y because the convex side of thegraph is down.
5. See the Programs for Graphing Calculatorssection of the Instructor’s Resource Book.
6. See the Programs for Graphing Calculatorssection of the Instructor’s Resource Book.
7. a. and b. dy
dxxy= −0 2.
x
y
(3, 2)
(1, –2)
Calculus Solutions Manual Problem Set 7-5 153© 2005 Key Curriculum Press
8. a. and b. dy
dxx y= − +0 1 0 2. .
x
5
5
y
(0, 2)
(0, 4)
c. When the graph is observed, the slope linesseem to follow a straight path using (0, 2.5)as an initial condition. Euler’s methodconfirms this.(In differential equations, students will learnhow to solve such first-order linears bymultiplying both sides by the integratingfactor e− 0.2x . The general solution isy = Ce0.2x + 0.5x − 2.5. For C = 0, theparticular solution is y = 0.5x + 2.5.)
9. a. Using dr = 0.6, v(13.6) ≈ 0.1414… andv(14.2) ≈ −1.2900… , so the spacecraft seemsto reverse direction somewhere between thesetwo values of r, as shown in the graph inpart b.
b. Using dr = 0.1, v(20) ≈ 4.5098… , and thevalues are leveling off, as shown in thegraph.
dr = 0.1
dr = 0.6
10 20
10
20 v
r
Actual
c.dv
dr r v= – .62 44
2
v dv r dr= − −∫∫ 62 44 2.
0 5 62 442 1. .v r C= +−
0 5 12 62 44 1 9 562 1. ( ) . ( )= + ⇒ =− C C .0 5 62 4 9 562 1. . .v r= +−
v r= +124 88 19 121. .–
When r = 20, v = 5.0362… .Because the graph is concave up (convex sidedown), the Euler’s solution underestimatesthe actual velocity. The first increment, wherethe graph is so steep, makes a large error thataccumulates as the iterations continue,
putting the graph into a region of the slopefield from which the spacecraft would notescape Earth’s gravity.
d. Let v1 be the initial velocity at r = 1.Solving for C gives0 5 62 441
2. .v C= +C v= −0 5 62 441
2. .
If v1 2 62 44< ( . ), then C is negative,
making v r C= +128 88 1. – an imaginarynumber when r is large enough. If
v1 > 2 62 44( . ), then C is positive, making va positive real number for all positive
values of r. (The asymptote is v C= .)
10. a. v(2) = 61.6831… , v(4) = 106.2850… ,v(6) = 129.7139… , v(8) = 139.9323… ,v(10) = 143.9730… , v(20) = 146.4066…These values will be overestimates becausethe graph is concave down (convex side up),so the Euler’s tangent lines will be above theactual graph, as in the next graph.
v
t
Actual
Euler
100
10 20
200
b. v(2) = 157.7979… , v(4) = 150.5128… ,v(6) = 147.9234… , v(8) = 146.9777… ,v(10) = 146.6290… , v(20) = 146.4254…These values will be underestimates becausethe graph is concave up (convex side down),so the Euler’s tangent lines will be below theactual graph.
c.dv
dtv= ⇔ = ⇔0 0 0015 32 162. .
v = =32 16
0 0015146 42404
.
.. (store)K
The terminal velocity is about 146.4 ft/s.
d. The table shows the values of v and theerrors, Euler minus actual. The errors increaseonly for a while, then approach zero becauseboth the Euler’s solution and the actualsolution approach the same asymptote. (It isnot always true that values farther from the
154 Problem Set 7-6 Calculus Solutions Manual© 2005 Key Curriculum Press
initial condition have a greater error in theirEuler’s approximation.)
t Euler’s v Actual v Error
2 61.6831… 60.4791… 1.2040…
4 106.2850… 103.3298… 2.9552…
6 129.7139… 126.8383… 2.8756…
8 139.9323… 137.9573… 1.9749…
10 143.9730… 142.8466… 1.1264…
20 146.4066… 146.3792… 0.0274…
The graph in part a shows the Euler’ssolutions from parts a and b, and the actualsolution from part c, thus confirminggraphically the numerical answers to thisproblem.
11. a. For x ≤ 5, the radicand 25 − x2 is non-negative, giving a real-number answer for y.For x > 5, the radicand is negative, giving noreal solution.
b. The slope field shows that the graph will beconcave up (convex side down), making theEuler’s tangent lines lie below the graph,leading to an underestimate.
At x y= = − = − …4 9 0 6 25 4 9 0 59692. , . . .– .
The Euler’s solution at x = 4.9 is−0.8390… , which is an underestimatebecause −0.8390… < −0.5969… but isreasonably close to the actual value.
c. The Euler’s solutions for the given points are
x y
5.1 −0.3425…
5.2 0.1935…
5.3 −0.7736…
6.6 26.9706…
From 5.1 to 5.2,
dy = − ( )( . )
( )(– . )( . )
9 5 1
25 0 34250 1
K
= 0.5360… , indicating that the graph is stilltaking upward steps.
From 5.2 to 5.3,
dy = − ( )( . )
( )( . )( . )
9 5 2
25 0 19350 1
K
= −0.9672… , indicating that the graph takesa relatively large downward step. The signchange in dy happens whenever the priorEuler’s y-value changes sign. The graph startsover on another ellipse representing a differentparticular solution.
d. Euler’s method can predict values that areoutside the domain, which are inaccurate.
12. Answers will vary.
Problem Set 7-6Q1. definition of definite integral
Q2. fundamental theorem of calculus
Q3. definition of indefinite integral
Q4. the intermediate value theorem
Q5. Rolle’s theorem
Q6. the mean value theorem
Q7. the chain rule Q8. general
Q9. particular Q10. initial
1. a. dB/dt is proportional to B, which means thatthe larger the population is, the faster itgrows. But dB/dt is also proportional to(30 − B)/30, which means that the closer B isto 30, the slower it grows. dB/dt > 0 when0 < B < 30 because when the population isless than 30 million the population willincrease until it reaches the carrying capacity.dB/dt < 0 when B > 30 because when thepopulation is greater than 30 million, thepopulation will decrease until it reaches thecarrying capacity.
b.B
t
Euler
Actual
10 20 30 40
10
20
30
40
For the initial condition (0, 3), the populationgrows, leveling off at B = 30. For the initialcondition (10, 40), the population dropsbecause it is starting out above the maximumsustainable value (carrying capacity).
c.
t B
0 3
10 13.8721…
20 26.4049…
30 29.5565…
40 29.9510…
See the graph in part b. The graph shows thatthe Euler’s points and graphical solution areclose to each other.
Calculus Solutions Manual Problem Set 7-6 155© 2005 Key Curriculum Press
d.dB
dtB
B= ⋅0 2130
30.
–
30
300 21
B BdB dt
( – )= .
Separate the variables.
1 1
300 21
B BdB dt+
=
–.
By partial fractions(see Example 1).
ln |B| − ln |30 − B| = 0.21t + CWhy the “−” sign?
−ln |B| + ln |30 − B| = −0.21t − CTo simplify latersteps.
30
27
39
10 21
1
10
– B
BC e
C e
C e C
t
C
=
= ±
= ⇒ =
−
−
.
1
Substitute the initialcondition (0, 3) tofind C1.
309
30 9
0 21
0 21
– B
Be
B Be
t
t
=
− =
−
−
.
.
Be
Btt Solve for explicitly
in terms of .=
+ −30
1 9 0 21.
At , 26.4326 .t B
e= =
+=20
30
1 9 4 2– . K
The Euler’s value, B ≈ 26.4049… , is veryclose to this precise value.
e.d
dB
dB
dtB
= − +0 014 30 0 007. ( . )
Derivative = 0 if −0.014B + 30(0.007) = 0,which is true if and only if B = 15. This value is halfway between B = 0 andB = 30.
1530
1 9 0 21=+ e t– . ⇒ t ≈ 10.4629…
The point of inflection is (10.4629… , 15).
2. a. A logistic function is reasonable because thenumber of houses grows at an increasing ratefor a while, then slows down as the numberapproaches 120, the “carrying capacity” of thesubdivision.
b.dy
dxy
y= ⋅0 9120
120.
–
120
1200 9
y ydy dx
( – )= .
1 1
120y y+
–
dy = 0.9 dx
ln |y| − ln |120 − y| = 0.9x + C−ln |y| + ln |120 − y| = −0.9x − C120
10 9
1– y
yC e C ex C= = ±− −.
115
5= C1e
0 ⇒ C1 = 23
Substitute the initialcondition (0, 5) tofind C1.
12023
120 23
0 9
0 9
– y
ye
y ye
x
x
=
− =
−
−
.
.
ye x=
+120
1 23 0 9– .
The graph confirms that the particularsolution follows the slope lines.
y (houses)
50
5 10
100
x (years)
c. 70% of 120 is 84.
84
120
1 234 42530 9=
+⇒ ≈
exx– . . ,K or about
4 years 5 months Solve numericallyfor x.
1 lot left means 119 lots built on.
119120
1 23 0 9=+
⇒ ≈ …e
xx– . 8.7940 , or about
8 years 10 months.
d.dy
dxy
yy y= ⋅ =0 9
120
120
0 9
120120 2.
– .( – )
d
dy
dy
dxy
= 0 9
120120 2
.( – )
The derivative is zero if 120 − 2y = 0, whichis true if and only if y = 60. This value ishalfway between y = 0 and y = 120.If y < 60, the derivative is positive, so dy/dxis increasing. If y > 60, the derivative isnegative, so dy/dx is decreasing. Therefore,dy/dx is a maximum when y = 60, and thenumber of houses is increasing the mostrapidly at this point of inflection.
156 Problem Set 7-6 Calculus Solutions Manual© 2005 Key Curriculum Press
3. a.dy
dxky
M y
M= ⋅ –
0 5 10 100 5
10 10. = ⇒ =k
MM
k
M M( )( – )
.
( – )
1 1 24 241 1
24 24. = ⇒ =k
MM
k
M M( )( – )
.
( – )
∴−
=−
0 5
10 10
1 1
24 24
.
( )
.
( )M MEliminate k byequating the twovalues of k/M.
12(M − 24) = 11.0(M − 10)12M − 11M = 288 − 110 ⇒ M = 178
Solve for M.
kk
178
0 5
10 178 10
89
1680=
−⇒ = =.
( )0.05297… (Store this.)Ajax expects to sell 178,000 CDs based onthis mathematical model.
b.dy
dxy
y= ⋅ ⋅ −89
1680
178
178
y (thousand CDs)
50 100
100
178
200
x (days)
The slope field has horizontal slope lines atabout y = 178, thus confirming M = 178.
c. The general solution is yM
ae kx=+ −1
.
Substitute M = 178 and k = 89/1680 =0.05297… and the initial condition y = 10 atx = 0.
10178
116 80=
+⇒ =
aea .
The equation is y
e x=+ −
178
1 16 8 0 05297. . K .
See the graph in part b. The graph follows theslope lines.
d. At x = 50, y = 81.3396… .At x = 51, y = 83.6844… .83.6844… − 81.3396… = 2.35447…They expect to sell about 2354 CDs on the51st day.
e. The point of inflection is halfway betweeny = 0 and y = 178, that is, at y = 89.
89
178
1 16 8 0 05297=+ −. .e xK
Solving numerically gives x ≈ 53.2574… ,or on the 54th day.
4.dy
dxky
M y
M= ⋅ −
M dy
y M yk dx
( )−= ∫∫
1 1
y M ydy k dx+
−
= ∫∫See Section 9-7 for aquick way to resolveinto partial fractions.
ln |y| − ln |M − y| = kx + CThe differential of thesecond denominatoris −dy.
−ln |y| + ln |M − y| = −kx − C
lnM y
ykx C
− = − −
M y
ye e ekx C C kx− = = ⋅− − − −
M y
ye e C e C eC kx kx C− = ± ⋅ = = ±− − − −
1 1
M − y = C1ye− kx
y + C1ye− kx = M
yM
C e
M
aekx kx=+
=+− −1 11
a = C1, Q.E.D.
5. a. At t = 5.5, F ≈ 1.7869… ≈ 2 fish left.At t = 5.6, F ≈ −11.0738… , meaning nofish are left.
The fish are predicted to become extinct injust over 5.5 years.
Part a, dt = 0.1
F (fish)
t (years)
Part b
Parts c and d
5 10
200
500
1000
b. See the graph in part a with initial condition(3, 1200), showing that the fish populationwill decrease because the initial condition isabove the 1000 maximum sustainable.
c. See the graph in part a with initial condition(0, 300), showing that the population risesslowly at first, then faster, eventually
Calculus Solutions Manual Problem Set 7-6 157© 2005 Key Curriculum Press
slowing down as the population approachesthe 1000 maximum sustainable (carryingcapacity).
d. Let F = y + 200.dy
dt
y y= ⋅ ⋅ −130
200
800
1000200000
800130
y ydy dt
( )−= ∫∫
250 250
800130
y ydy dt+
−
= ∫∫See Section 9-7 forquick partial fractions.
250 ln |y| − 250 ln |800 − y| = 130t + CWhy “−” ?
800 13 25− = ⋅− −y
ye et C( / )
8001
13 25− = −y
yC e t( / )
Substitute y = 100 (F = 300) when t = 0.800 100
10071
01
− = ⇒ =C e C
8007 13 25− = −y
ye t( / )
ye t=
+ −800
1 7 13 25( / )
Fe t=
++−
800
1 720013 25( / )
See the graph in part a, showing that thesketch from part c reasonably approximatesthis precise algebraic solution.
6. Answers will vary. Here is a typical run with aclass of 25 people.
x N
0 1
1 3
2 6
3 13
4 21
5 25
6 25
7 25
8 25
Logistic regression gives
N
e x=+ −
25 5083
1 43 1120 1 3032
.
. .
K
K K
The logistic function fits reasonably well (asshown in this graph), especially if you useseveral values of the maximum number ofpeople as shown in the table.
1 2 3 4 5 6 7 8
5
10
15
20
25N
x
7. a. and b.
Year P ∆P/∆t (∆P/∆t)/P
1940 131.7
1950 151.4 2.38 0.01571…
1960 179.3 2.59 0.01444…
1970 203.2 2.36 0.01161…
1980 226.5 2.275 0.01004…
1990 248.7
You can’t find ∆P/∆t for 1940 and 1990because you don’t know values of P bothbefore and after these values.
c. Using linear regression on the values of(∆P/∆t)/P without round-off gives
10 02802596 0 0000792747
P
P
tP
∆∆
≈ …− …. . .
The correlation coefficient is r = −0.98535… .For the other types of regression:r = −0.978… for logarithmicr = −0.981… for exponentialr = −0.971… for powerThus, a linear function fits best because r isclosest to −1.
d.1
0 02802596 0 0000792747P
dP
dtP≈ …− …. .
⇒ = …− …dP
dtP P( . .0 02802596 0 0000792747 )
e.
100
500P
0
t
–50 50
Stable population at 353.5 million
158 Problem Set 7-6 Calculus Solutions Manual© 2005 Key Curriculum Press
f.
Year t Euler Actual* Euler**
1890 −50 44.6… 62.9 46.1…
1900 −40 56.9… 76.0 58.3…
1910 −30 71.7… 92.0 72.9…
1920 −20 89.2… 105.7 90.1…
1930 −10 109.3… 122.8 109.8…
1940 0 131.7 131.7 131.7
1950 10 155.4… 151.4 155.0…
1960 20 180.1… 179.3 179.2…
1970 30 204.7… 203.2 203.5…
1980 40 228.2… 226.5 226.9…
1990 50 249.9… 248.7 248.8…
2000 60 269.3… 281.4 268.6
2010 70 286.1…
2020 80 300.2…
2030 90 311.8…
2040 100 321.1…*Data from The World Book Encyclopedia .
**Note that although linear regression gives the“best” fit for (∆ P/∆ t ) /P versus P , actually plottingthe graph shows that the data point for 1960 isconsiderably out of line.
0.015
0.014
0.013
0.012
0.011
150 200
(∆P/∆t)/P
P
1950
1960
1980
1970
Using the two endpoints, 1950 and 1980,gives (∆P/∆t)/P = 0.002716… −0.00007557…P. Using this equation givespopulations much closer to the actual onesfor the given years, as shown in therightmost column of the table in part f. Thisis, of course, no guarantee that the latermodel fits any better in the future than theformer one.
g. The population growth rate is zeroif dP/dt = 0.LetP(0.02802596… − 0.0000792747…P) = 0.P = 0 or P = (0.02802596…)/(0.0000792747…) = 353.5…Predicted ultimate population ≈ 353.5 millionDifferential equation: P = 353.5… makesdP/dt = 0.Graph: P = 353.5… is a horizontalasymptote.
h. See the graph in part e. Data do follow thesolution.
i. Sample answer: The predicted populationsagree fairly well with the data for the sixgiven years. The fit is exact for 1940 becausethis point was used as an initial condition.For the other five years, the predictedpopulations are a bit higher than the actualpopulation.
j. Actual data are given in the table in part f.
k. The predicted population for 2010 from part fis 286.1… million. Using 486.1 million asan initial condition in 2010 gives thefollowing predictions:
Year t Euler
2010 70 486.1…
2020 80 444.5…
2030 90 417.7…
2040 100 399.7…
2050 110 387.1…
The logistic model predicts that thepopulation will drop, approaching theultimate value of 353.5 million from above.This behavior shows up in the slope field ofpart e because the slopes are negative forpopulations above 353.5.
8. a.1
10 10
10
10y y
A
y
B
y
A y By
y y( – ) –
( – )
( – )= + = +
The numerator of the first fraction mustequal that of the last fraction for all valuesof y. That is, 1 = 10A − Ay + By. Theconstant and linear coefficients on the leftmust equal the corresponding ones on theright. Thus, 1 = 10A and 0 = −Ay + By.So A = B = 0.1.
b.1
10
0 1 0 1
10y ydy
y ydy
( – )
. .
–= +
∫∫
= −
∫0 1
1 1
10. ,
y ydy
– which equals 3 dx.∫
∴ 0.1(ln |y| − ln |y − 10|) = 3x + C
ln–
ln–y
y
y
yx C
10
1030 10= − = +
y
y ye x C–10
110 30 10= − = − +( )
101 10 30
ye eC x= ± − −
yke
k exC=
+= ± −10
1 3010
– , where , Q.E.D.
Calculus Solutions Manual Problem Set 7-6 159© 2005 Key Curriculum Press
c.dP
dtP P= …− …( . . )0 02802 0 00007927
= 0.00007927…P(353.5… − P)
1
353 50 00007927
P PdP dt
( . – )KK= ∫∫ .
= −
∫1
353 5
1 1
353 5. – .K KP PdP
= ∫0 00007927. K dt
ln |P| − ln |P − 353.5…|= 353.5…(0.00007927…t + C)
ln
– .ln
– .P
P
P
P353 5
353 5
K
K= −
= 0.02802…t + 353.5…C
P
Pe t C– .353 5 0 02802 353 5K = − … + …( . . )
353 51 0 02802. K
Pke t= + ….
P
ke t=+
353 5
1 0 02802
.– . ...
K
For the initial condition t = 0, P = 131.7,
k = − = …353 5
131 71 1 684
.
.
K. .
d.
Year t Algebraic Euler Actual
1940 0 131.7 131.7 131.7
1950 10 155.5… 155.4… 151.4
1960 20 180.2… 180.1… 179.3
1970 30 204.7… 204.7… 203.2
1980 40 228.2… 228.2… 226.5
1990 50 249.8… 249.9… 248.7
The two methods of evaluating themathematical model agree almost perfectly.However, the fact that they agree with eachother is no guarantee that they will fit the realworld as closely as they match each other.
9.dR
dtk R
dR
Rk dt R k t C= ⇒ = ⇒ = +1 1 1ln | |
⇒ = ⇒ =| |R e e R C eC k t k t1 11
R is increasing because k1 > 0.
10.dF
dtk F
dF
Fk dt= − ⇒ = −2 2
⇒ = − +ln | |F k t C2
⇒ = ⇒ =− −| |F e e F C eC k t k t2 22
F is decreasing because −k2 < 0.
11.dR
dtk R k RF= −1 3
dF
dtk F k RF= − +2 4
12.dF
dR
dF dt
dR dt
k F k RF
k R k RF= = +/
/
–
–2 4
1 3
The dt cancels out.
13. R = 70, F = 15
⇒ = + ⋅
⋅= =dF
dR
– .
– .
..
15 0 025 1050
70 0 04 1050
11 25
280 4017K
14. The slope at (70, 15) is about 0.4.
100
50
F
R(70, 15)
At R FdR
dt
dF
dt= = = =70 15 28 11 25, , , and . ,
which are both positive. So both populations areincreasing and the graph starts up and to theright.
15. The populations vary periodically and the graphis cyclical. The fox population reaches itsmaximum 1/4 cycle after the rabbit populationreaches its maximum.
16. Neither population changes when dR/dt =dF/dt = 0.dF/dt = 0 ⇔ F = 0 or R = 1/0.025 =40 (4000 rabbits)dR/dt = 0 ⇔ R = 0 or F = 1/0.04 = 25 foxes
17. Assume that dF/dt still equals −F + 0.025RF.dF
dR
dF dt
dR dt
F RF
R RF R= = +/
/
– .
– . – .
0 025
0 04 0 01 2
R FdF
dR= = ⇒ =
−= −70 15
11 25
210 5357 and
.. K
18.
100
50
F
R(70, 15)
Note that the slope at (70, 15) is now negative.
19. The populations now spiral to a fixed point. Therabbit population stabilizes at the same value asin Problem 16, R = 40 (4000 rabbits), which issurprising. The stable fox population decreasesfrom 25 to 15.
160 Problem Set 7-7 Calculus Solutions Manual© 2005 Key Curriculum Press
20. Assume that dF/dt still equals = −F + 0.025RF.
dF
dR
dF dt
dR dt
F RF
R RF R= = +/
/
– .
– . – . –
0 025
0 04 0 01 102
R F
dF
dR= = ⇒ =
−= −70 15
11 25
310 3629 and
.. K
21.
100
50
F
R(70, 15)
(70, 30)
Note that the slope at (70, 15) is about −0.4.
22. The fox and rabbit populations spiral toward afixed point. Again, and even more surprisingly,the rabbits stabilize at R = 40 (4000). But thestable fox population is reduced to 8 or 9. Alongthe way, the model shows that the foxes arereduced to about 1, thus becoming in danger ofextinction!
23. See the graph in Problem 21 with initialcondition (70, 30). With this many foxes andhunters chasing rabbits, the rabbits becomeextinct. At this point, the foxes have beenreduced to just 5. After the rabbits becomeextinct, the foxes decrease exponentiallywith time, eventually becoming extinctthemselves.
Problem Set 7-7Review Problems
R0. Answers will vary.
R1. P(t) = 35(0.98 t )P′(t) = 35(0.98t) ln 0.98
t P(t) P′(t) P′(t)/P(t)
0 35 −0.7070… −0.2020…
10 28.5975… −0.5777… −0.2020…
20 23.3662… −0.4720… −0.2020…
P t
P t
t
t
′ = =( )
( )
( . ) ln .
( . )ln
35 0 98 0 98
35 0 980 98.
= −0.2020… , which is a constant, Q.E.D.
R2. a. V = speed in mi/h; t = time in sdV
dtkV=
b.dV
Vk dt= ∫∫
ln |V | = kt + C
| | V e e ekt C C kt= = ⋅+
V = C1ekt
C1 can be positive or negative, so theabsolute value sign is not needed for V. In thereal world, V is positive, which also makesthe absolute value sign unnecessary.
c. 400 = Cek·0 ⇒ C = 400
500 400
1 25
400 005578
40=
⇒ = = …
⋅e
k
k
ln ..
V = 400e0.005578…t
d. 750 = 400e0.005578…t
⇒ = = … ≈tln .
. ...
1 875
0 005578112 68 113. s
R3. a. y dy dx y x C− = ⇒ = +∫∫ 1 2 6 3/ 2( )
b. y = (3x − 4)2 (y = (3x − 14)2 does not workbecause at (3, 5), dy/dx = −30 but 6y1/2 = 30.)
c.
(3, 25)y
x
10
1 2 3
d. At x = 2, y′ = 12 and y = 4.See graph in part c.A line through (2, 4) with slope 12 is tangentto the graph.
e. i. dN/dt = 100 − kNdN
kNdt
100 –∫ ∫=
−(1/k) ln |100 − kN| = t + CUsing (0, 0) gives −(1/k) ln 100 = C.Substituting this value for C gives−(1/k) ln |100 − kN| = t − (1/k) ln 100.ln |100 − kN| − ln 100 = −ktln |1 − (k/100)N| = −kt1 100− = −( / )k N e kt
N k e kt= − −( / )( )100 1Using (7, 600) and solving numericallygives k ≈ 0.045236.∴ N = 2210.6…(1 − e− 0.045236 t)
ii. t = 30: About 1642 names
Calculus Solutions Manual Problem Set 7-7 161© 2005 Key Curriculum Press
iii. limt
N→∞
= … − = …2210.6 (1 0) 2210.6
The brain saturates at about 2211 names.
iv. Let dN/dt = 30.
30 10070
1547 4= − ⇒ = = …kN Nk
.
names. Substituting this for N gives
1547 4 2210 6 1 0 045236. . ( )..K K= − −e t
e t− = − =0 045236 1
1547 4
2210 60 3. .
.
.
K
K (exactly)
t = = … ≈ln .
– .
0 3
0 0452326 6 27
K. days
or:
30 = N(t) − N(t − 1)
= − +− − −2210 6 0 045236 0 045236 1. [ ]. . ( )K e et t
= − +− − −2210 6 10 045236 1 0 045236. ( ). ( ) .Ke et t
⇒ t ≈ 27 days
R4. a.dy
dx xyy= − +20
0 05.
At (2, 5), dy/dx = −1.75.At (10, 16), dy/dx = 0.675.The slopes at (2, 5) and (10, 16) agree withthese numbers.
b. Initial conditions (1, 8) and (1, 12)
5
5
y
x
(10, 16)
(2, 5)
(1, 8)
(1, 10)
(1, 12)
The solution containing (1, 8) crosses thex-axis near x = 7, converges asymptoticallyto the y-axis as x approaches zero, and issymmetric across the x-axis. The solutioncontaining (1, 12) goes to infinity as x goesto infinity.
c. See the graph in part b with initial condition(1, 10). The solution containing (1, 10)behaves more like the one containing(1, 12), although a slight discrepancy inplotting may make it seem to go theother way.
R5. a.dy
dx xyy= − +20
0 05.
Table with initial condition (1, 9), ∆x = 1:
x y (∆x = 1) y (∆x = 0.1)
1 9 9
2 7.227… 7.707…
3 6.205… 6.949…
4 5.441… 6.413…
5 4.794… 5.999…
6 4.200… 5.662…
7 3.616… 5.377…
8 3.007… 5.130…
9 2.326… 4.910…
10 1.488… 4.712…
11 0.2185… 4.529…
12 −8.091… 4.359…
13 4.199…
14 4.045…
15 3.896…
16 3.750…
17 3.604…
18 3.457…
19 3.306…
20 3.150…
M M
28.9 0.1344…
29 −0.3810…
5
5
y
x
(1, 9)
∆x = 1
∆x = 0.1
For ∆x = 1, the graph crosses the x-axis atabout x = 11.
b. See the table in part a for ∆x = 0.1.See the graph in part a.
c. The accuracy far away from the initialcondition is very sensitive to the size of theincrement. For instance, in part a the firststep takes the graph so far down that itcrosses the x-axis before running off theedge of the grid. The greater accuracy with∆x = 0.1 shows that the graph actually doesnot cross the x-axis before x = 20.
162 Problem Set 7-7 Calculus Solutions Manual© 2005 Key Curriculum Press
d. Continuing the computations in part c, thegraph crosses the x-axis close to x = 28.9.See the table in part a.
R6. a.
y (hundred beavers)
x (years)
5 10
5
10
The population is decreasing because it isabove the maximum sustainable, 900 beavers(y = 9). By Euler’s method, y ≈ 9.3598… , orabout 936 beavers, at x = 3 years.
b. See the graph in part a with initial condition(3, 100), showing that the population isexpected to increase slowly, then morerapidly, then more slowly again, leveling offasymptotically toward 900. This happensbecause the initial population of 100 is belowthe maximum sustainable.
c.dy
dxy
y= ⋅0 69
9.
–
yae x=
+9
1 0 6– .
Substitute into thegeneral equation.
19
1 1 8=+ ae– .
Substitute the initialcondition (3, 1).
a = 8e1.8 = 48.3971… Solve for a.
y
e e ex x=+
=+−
9
1 8
9
1 48 39711 8 0 6 0 6. . – .. K
The point ofinflection is halfwaybetween theasymptotes at y = 0and y = 9.
4 59
1 8 1 8 0 6. =+ −e e x. .
Substitute 4.5 for y.
x = ln (8e1.8 )/0.6 = 6.4657… ≈ 6.5 yr
d.dy
dx
x
y= – . ( – )
( – )
0 5 6
7dy = 0 when x = 6, and dx = 0 when y = 7. Sothe stable point is (6, 7), corresponding to thepresent population of 600 Xaltos natives and7000 yaks.
(Note that the general solution to the differ-ential equation is (x − 6)2 + 2(y − 7)2 = C,and the specific solution for the given initialcondition is (x − 6)2 + 2(y − 7)2 = 0, whosegraph is a single point.)
e. Initial condition (9, 7)
5
5
x
y
(6, 7) (9, 7) (15, 7) (19, 7)
Suddenly there are too many predators for thenumber of prey, so the yak populationdeclines. Because y is decreasing from (9, 7),the graph follows a clockwise path.
f. See the graph in part e with initial condition(19, 7). The graph crosses the x-axis atx ≈ 14.4, indicating that the yaks are huntedto extinction. (The Xaltos would then starveor become vegetarian!)
g. See the graph in part e with initial condition(15, 7). The graph never crosses the x-axis,but crosses the y-axis at y ≈ 2.3, indicatingthat the yak population becomes so sparsethat the predators become extinct. (The yakpopulation would then explode!)
Concept Problems
C1. a.dy
dxk y y dy k dx= ⋅ ⇒ = ⇒/ − /1 2 1 2
2 0 51 2 2y kx C y kx C/ = + = +, so [ . ( )] .
b. The differential equation would have tobecome y1 3/
after it is integrated. So theoriginal equation would have to containy− /2 3
after the variables have been separated.
Conjecture: dy
dxky= 2 3/
c. Confirmation:dy
dxky y dy k dx= ⇒ = ⇒/ − /2 3 2 3
3y1/ 3 = kx + C ⇒ y = [(1/3)(kx + C)]3, acubic function, Q.E.D.
d. For n ≠ 0, dy
dxk y n n= ⋅ ⇒− /( )1
y dy k dx ny kx Cn n n− − / /= ⇒ = + ⇒( ) 1 1
y = [(1/n)(kx + C)]n
Calculus Solutions Manual Problem Set 7-7 163© 2005 Key Curriculum Press
For example: dy
dxy y dy dx= ⇒ =/ − /7 8 7 8
⇒ = + ⇒ = / +/8 1 81 8 8y x C y x C[( )( )]
C2. a.
Ticket Price People
2.00 460
2.50 360
3.00 320
4.00 260
4.50 140
5.50 120
6.00 80
b.
500
6
P
N
Function behaves (more or less) linearly.Let N = number of tickets and P = number of$/ticket.By linear regression, N ≈ −90.83P + 605.4,with correlation coefficient r = −0.9747… .
c. Let M = total number of dollars.M ≈ P ⋅ N = P(−90.83P + 605.4)M ≈ −90.83P2 + 605.4P
d. Maximize M: M′ ≈ −181.66P + 605.4
M ′ = 0 ⇔ P ≈ 605 4
181 66
.
. = 3.332…
Maximum M at P ≈ 3.332… because M′changes from positive to negative there (orbecause the graph of M is a parabola openingdownward).Charge $3.30 or $3.35.
e. M has a local maximum at this price becausecharging more than the optimum pricereduces attendance enough to reduce the totalamount made, whereas charging less than theoptimum price increases attendance, but notenough to make up for the lower price perticket.
C3. a. g t e e t
( ) . .
= − −10 0 8 0 5
The graph does look like Figure 7-7e.
lim . lim. .
t
e ee et
tt
→∞
− −→∞
−=10 100 8 0 5 0 5
= =− ⋅10 100 8 0e .
Limit is 10, indicating maximum possiblepopulation.
b. a = 421.3692… , c = 0.7303036… , andk = 0.01589546… , either by twice takinglogarithms as suggested, or by this method:Taking ln once ⇒ ln a − ce− kt = ln P, soln a − ce10k = ln 179ln a − c = ln 203ln lna ce k− =−10 226Then substituting ln a = c + ln 203 into thefirst and third equations gives
c(1 − e10k) = ln 179 − ln 203c e k( )1 226 20310− = −− ln ln
Substituting c e c e ek k k( ) ( )1 110 10 10− = − =− −
− −−e k10 179 203( )ln ln into the previousequation yields
e k− = − −−
= −−
10 226 203
179 203
226 203
203 179
ln ln
ln ln
ln ln
ln ln
so k = − −
−
=1
10
226 203
203 1790 01589ln
ln ln
ln ln. .K
Then find c using c e k( )1 22610− = −− lnln 203 and find a using 203 = −ae c .c = 0.7303… and a = 421.3692…
431.3...
t
g(t)
100
100
Note that this model predicts an ultimatepopulation of lim ( )
tP t
→∞≈ 421 million.
c. Now a = 551.1655… , c = 0.9988291… ,k = 0.01186428… , and the ultimatepopulation is lim ( )
tP t
→∞≈ 551 million. Thus,
the Gompertz model is quite sensitive to asmall change in initial conditions. Thepredicted ultimate population increased by130 million with only a 1 million change inone data point!
C4. dV/dt = −2V1/ 2 + F, where F is a constant.dV
F Vdt
−=∫ ∫2 1 2/
The integral on the right is not the integral ofthe reciprocal function because the numeratorcannot be made the differential of thedenominator. A slope field gives informationabout the solutions. The following graph is forF = 20 ft3/min flowing in. (The dashed lineshows the solution with F = 0, the originalcondition.) Starting with 196 ft3 in the tub, thevolume levels off near 100 ft3. Starting below100 ft3, the volume would increase toward 100.
164 Problem Set 7-7 Calculus Solutions Manual© 2005 Key Curriculum Press
14
196 F = 20
F = 0
t
V
If the inflow rate is too high, the tub willoverflow. The next graph is for F = 40 ft3/min.In this case, the stable volume is above theinitial 196 ft3.
14
196
F = 0
t
VF = 40
It is possible to antidifferentiate the left side bythe algebraic substitution method of ProblemSet 9-11, Problems 101–106. The generalsolution is
t CF
F V V+ = − − −2
2 1 2 1 2ln ( )/ /
and the particular solution for V = 196 at t = 0 is
tF F
F VV− = − −
−−14
2
28
2 1 21 2ln //
Unfortunately, it is difficult or impossible tosolve for V. The volume will asymptoticallyapproach F2/4, overflowing the tub if F2/4 > tubcapacity.
Chapter Test
T1.dy
dxky=
T2. Solving a differential equation means finding theequation of the function whose derivative appearsin the differential equation.
T3. The general solution involves an arbitraryconstant of integration, C. A particular solutionhas C evaluated at a given initial condition.
T4.
5
5
y
x
(0, –4)
T5. The concave side of the graph is up, so theactual graph curves up from the Euler’s tangentlines, making the Euler’s method values an
underestimate. (Or: The convex side of the graphis down, so the Euler’s tangent lines are belowthe actual graph.)
T6. General logistic differential equation:dy
dxky
M y
M= ⋅ −
T7.dy
dxy= 0 4.
dy
ydx∫ ∫= 0 4.
ln |y| = 0.4x + C|y| = eCe0.4 x
y = C1e0.4 x
−5 = C1e0.4(0) = C1
y = −5e0.4 x
T8.dy
dxy y dy dx= ⇒ = ⇒− ∫∫12 121 2 1 2/ /
2 121 2y x C/ = +
T9. a.dP
dtkP P Cekt= ⇒ =
P = 3000 at t = 0 ⇒ P = 3000ekt
b. P = 2300 at t = 5 ⇒
k = = − ⇒1
5
2300
30000 05314ln . K
P(25) = 794.6…Phoebe will not quite make it because thepressure has dropped just below 800 psi bytime t = 25.or:800 3000 0 05314= −e t. K
t =−
=1
0 05314
800
300024 87
.ln .
KK
Phoebe will not quite make it because thepressure has dropped to 800 just before t = 25.
T10. a. y = number of grams of chlorine dissolvedt = number of hours since chlorinator wasstarteddy
dtky= −30
dy
kydt
30 –∫ ∫=
− − = +130
kky t Cln | |
ln |30 − ky| = −kt + C1
30 − ky = C2e− k t
y = 0 when t = 0 ⇒ C2 = 30∴ ky = 30(1 − e− kt)
yk
e kt= 301( – )–
The rate of escape is ky = 13 when y = 100.So k = 0.13.
∴ = = − −y e et t30
0 131 230 7 10 13 0 13
.( – ) .– . K( ).
Calculus Solutions Manual Problem Set 7-8 165© 2005 Key Curriculum Press
b. 200 230 7 1 0 13= … − −. ( e t. )
e t− = − =0 13 1
200
230 70 1333. .
. KK
t = =ln .
– .
0 1333
0 1315 499
KK. ≈ 15.5 hr
T11. a.dy
dxy
y= ⋅0 516
16.
–
yae ae
ax=+
⇒ =+
⇒ =16
12
16
170 5 0– .
ye x=
+16
1 7 0 5– .
b. At x = 0, y = 2:dy = 0.5(2)(16 − 2)/(16)(0.1) = 0.0875At x = 0.1, y ≈ 2 + 0.0875 = 2.0875,so dy = 0.5(2.0875)(16 − 2.0875)/(16)(0.1) =0.09075… .At x = 0.2, y ≈ 2.0875 + 0.09075… =2.17825… .
The precise solution is ye
=+
16
1 7 0 1– . =
2.18166… , which is greater than2.17825… , as expected because thegraph is concave up (convex side downward).
c. 416
1 7 0 5=+ e x– . ⇒ x = [ln (3/7)]/−0.5 =
1.6945…About 1 month 21 days
d.y (hundred lilies)
x (months)
The graph shows that the number of lilies isexpected to decrease toward 1600 (y = 16)because of overcrowding.
T12. a.
R (roadrunners)
C (coyotes)
(80, 700)
The graph starts going downward and to theright from (80, 700) because the coyotepopulation is relatively high, thus decreasingthe number of roadrunners.
b. There can be two different values for theroadrunner population for a particular coyotepopulation because the two events happen attwo different times. For example, coyotes areincreasing from 80 when there are 700 road-runners, but later they are decreasing from 80when there are about 200 roadrunners.
T13. Answers will vary.
Problem Set 7-8Cumulative Review, Chapters 1–7
1.
200
8
t
v(t)
(t, v(t))
v(t) dt represents the distance traveled in time dt.
2. Definite integral
3. ( – )t t t dt3 2
0
8
21 100 80+ +∫= − + +1
47 50 804 3 2
0
8t t t t
= 1280 mi
4. M100 = 1280.0384
M1000 = 1280.000384
The Riemann sums seem to be approaching 1280as n increases. Thus, the 1280 that was found bypurely algebraic methods seems to give thecorrect value of the limit of the Riemann sum.
5.
200
8
t
v(t)
(t, v(t))
6. Any Riemann sum is bounded by thecorresponding lower and upper sums. That is,Ln ≤ Rn ≤ Un.
By the definition of integrability, the limits of Ln
and Un are equal to each other and to the definiteintegral. By the squeeze theorem, then, the limitof Rn is also equal to the definite integral.
166 Problem Set 7-8 Calculus Solutions Manual© 2005 Key Curriculum Press
7. Definition:
f x dx L Ua
b
xn
xn( ) lim lim
0 0= =∫ → →∆ ∆
provided that the two limits are equal.Fundamental theorem: If f is integrable on
[ , ] and ( ) ( ) , a b g x f x dx= ∫then f x dx g b g a
a
b
( ) ( ) ( ). = −∫Or: If F x f t dt
a
x
( ) ( ) ,= ∫ then ′ =F x f x( ) ( ).
8. Numerically, the integral equals 1280. Bycounting, there are approximately 52 squares.Thus, the integral ≈ 52(25)(1) = 1300.
9. vv v′ ≈ = − /
( ) . (mi )
44 1 3 9
0 219 9
( . ) – ( . )
.
min
min
vv v′ ≈ = −( ) .
(mi/min)
min4
4 01 3 99
0 0219 9999
( . ) – ( . )
.
10. f cf x f c
x cx c′ =
→( ) lim
( ) – ( )
– or
f xf x x f x
xx′ = + ∆
∆→( ) lim
( ) – ( )∆ 0
11. v′(t) = 3t2 − 42t + 100 ⇒ v′(4) = −20
12. Slowing down. v′(4) < 0 and v(4) = 208 > 0 ⇒velocity is positive but decreasing ⇒ speed isslowing down.
13. The line has slope −20, and passes through(4, 208). The line is tangent to the graph.
100
200
100 5
v(t)
t
5
–100
Slope= –20
14. Acceleration
15. At a maximum of v(t), v′(t) will equal zero.
3t2 − 42t + 100 = 0 ⇔ t =± ⋅ ⋅42 42 4 3 100
6
2 –
t = 3.041… or 10.958…
So the maximum is not at exactly t = 3.16. v″(t) = 6t − 42
17. Know: dx
dt= 0 3. . Want:
dy
dt and
dz
dt.
y edy
dte
dx
dtex x x= ⇒ = − ⇒ = −− − −6 3 0 90 5 0 5 0 5. . ..
At x = 2, dy
dte= − = −−0 9 0 33101. . .K
y is decreasing at about 0.33 unit per second.
18. z x y zdz
dtx
dx
dty
dy
dt2 2 2 2 2 2= + ⇒ = +
dz
dt zx e ex x= + ⋅1
0 3 6 0 90 5 0 5[ . (– . )]– . – .
At x z= = + = …2 2 2 2072 2 97862 2, . . ,K
so
dz
dt= = −
1
2 97860 6 0 7308 0 04391
.( . – . )
KK K. .
∴ z is decreasing at about 0.044 unit per second.
19.dm
dtkm=
20.dm
mk dt m kt C= ⇒ = + ⇒∫∫ ln | |
| |m e m C ekt C kt= ⇒ =+1
21. Exponentially
22. General
23. 10000 = C1e0 ⇒ C1 = 10000
10900 10000 1= ⋅ek
⇒ k = ln 1.09 ⇒ m = 10000eln(1.09) t
= 10000(1.09)t
24. False. The rate of increase changes as the amountin the account increases. At t = 10,m = 10000(1.09)10 ≈ 23673.64.The amount of money would grow by$13,673.64, not just $9,000.
25. By Simpson’s rule, y dx30
42
∫≈ + ⋅ + ⋅ + ⋅ + ⋅2
374 4 77 2 83 4 88 2 90(
+ ⋅ + =4 91 89 1022) .
26. By symmetric difference quotient, at x = 36
y′ ≈ =90 83
2 21 75
–
( ). .
27. If f is differentiable on (a, b) and continuous atx = a and x = b, and if f ( a) = f ( b) = 0, then thereis a number x = c in (a, b) such that f ′ (c) = 0.
28.
a c b
f(b)
f(a)
f(x)
x
tangent
secant
29.f(x) and f´(x)
x
5
0 5 10
f f´
f´
Calculus Solutions Manual Problem Set 7-8 167© 2005 Key Curriculum Press
30.
x
f(x)
1
2
Step discontinuity at x = 1.
31. g(x) = x1/ 3(x − 1)
g x x x x x x′ = + − = −/ − / − /( ) ( ) ( )13 2 3 2 31
31
1
34 1
g′(0) is undefined because 0 2 3− / takes on the form
1 02 3/ / or 1/0.
g(x)
x
1
–1 1
32. e0.2 x = 0.6x ⇒ x ≈ 3.0953… (Store as a.)Or x ≈ 7.5606… (Store as b.)dA = (0.6x − e0.2 x) dx
A x e dxx
a
b
= ≈∫ ( . – ).0 6 0 87870 2 . K
(Integrate numerically.)
33. dV = π [(e0.2 x)2 − (0.6x)2] dx
V e x dxx
a
= ≈∫π .( – . ).0 4 2
00 36 8 0554K
(Integrate numerically.)
34.dy
dx
x
y= 0 25.
Initial conditions: (0, 3) and (10, 4)
5
5
x
y
(0, 3)(10, 4)
35. See the graph in Problem 34. Any initialcondition for which y = 0.5x, such as (2, 1),gives the asymptote.
36. y dy x dx y x C= ⇒ = +∫∫ 0 25 0 5 0 1252 21. . .
⇒ x2 − 4y2 = CInitial condition: (10, 4)100 − 64 = C ⇒ C = 36
∴ − = ⇒ = ±x y y x2 2 24 36 0 5 36. –
37. x y= = = …10 5 0 5 10 5 36 4 308422. : . .. –
38. At ( , ), . . .10 4 0 2510
40 625
dy
dx= ⋅ =
Using ∆x = 0.5, y(10.5) ≈ 4 + (0.625)(0.5) =4.3125, which is close to the exact value of4.30842… .
39.d
dxx
x
x(sin )
–
–1 32
6
3
1=
40. x = ln (cos t) and y = sec t
dx dtt
t t/ = ⋅ − = −1
cossin tan( )
dy/dt = sec t tan tdy
dx
dy dt
dx dt
t t
tt y= = = − = −/
/
sec tan
– tansec
41.dx
xx C
4 3
1
34 3
–ln | |= − − +∫
42. h x e h x ex x x( ) ( ) 5= = ⇒ ′ = =⋅ ⋅5 5 55 5 5ln lnln ln
43. limsin cos – –
x
x x x
x→
+ →0 2
5 3 5 1 0
0
= →→
limcos – sin –
x
x x
x0
5 5 3 3 5
2
0
0
= = −→
lim– sin – cos
x
x x0
25 5 9 3
24 5.
44. yx x x
x= +sin cos – –5 3 5 1
2 , showing a
removable discontinuity at (0, −4.5).
5
–5
–1 1
x
y
(0, –4.5)
45. Answers will vary.
46. Answers will vary.
168 Problem Set 8-2 Calculus Solutions Manual© 2005 Key Curriculum Press
Chapter 8—The Calculus of Plane and Solid Figures
Problem Set 8-1
1. f (x) = x3 − 6x2 + 9x + 3f ′ (x) = 3x2 − 12x + 9
ff´
x
y
1 32 4
g (x) = x3 − 6x2 + 15x − 9g ′ (x) = 3x2 − 12x + 15
x
y
1 32 4
ff´
h (x) = x3 − 6x2 + 12x − 3h ′ (x) = 3x2 − 12x + 12
f´
x
y
1 32 4
f
Positive derivative ⇒ increasing functionNegative derivative ⇒ decreasing functionZero derivative ⇒ function could be at a highpoint or a low point, but not always.
2. The functions have vertex points at values of xwhere their derivatives change sign. If thederivative is never zero, as for function g, thefunction graph has no vertex points. If thederivative is zero but does not change sign, asfor function h, the function graph just levelsoff, then continues in the same direction, withno vertex.
3. g ′′ (x) = (d/dx)(3x2 − 12x + 15) = 6x − 12h ′′ (x) = (d/dx)(3x2 − 12x + 12) = 6x − 12All the second derivatives are the same!
4. The curves are concave up where the secondderivative is positive and concave down where thesecond derivative is negative.
5. Points of inflection occur where the firstderivative graph reaches a minimum.Points of inflection occur where the secondderivative graph crosses the x-axis.
Problem Set 8-2Q1. Q2.
1
1x
y
1
1x
y
Q3. Q4.
1
1x
y
1
1x
y
Q5. Q6.
1
1 x
y
1
1x
y
Q7. Q8.
1
1x
y
1
1x
y
Q9. Q10.
1
1 x
y
2
1x
y
1.
x
f´(x)
f (x)
2
+ –0
max.
x 2
– ––
no p.i.
f´´(x)
f (x)
2.
x
f'(x)
f(x)
2
– +0
min.
x
f"(x)
f(x)
2
+ +0
no p.i.
Calculus Solutions Manual Problem Set 8-2 169© 2005 Key Curriculum Press
3.
x 2
+ 0 +
plateau
f´(x)
f (x)
x 2
– 0 +
p.i.
f´´(x)
f (x)
4.
x
f'(x)
f(x)
2
undef.+ –
max.
x
f"(x)
f(x)
2
undef.+ +
no p.i.
5.
x 2
– +undef.
min.
f´(x)
f (x)
x 2
0 0undef.
no p.i.
f´´(x)
f (x)
6.
x
f'(x)
f(x)
2
– –undef.
no max./min.
x
f"(x)
f(x)
2
– +undef.
p.i.
7.
x 2
no max./min.
–f´(x)
f (x)
x 2
p.i.
+ –f´´(x)
f (x)
8.
x
f'(x)
f(x)
2
–
no max./.min.
x
f"(x)
f(x)
2
0– +
p.i.
9.
x 2
– –undef.
no max./min.
f´(x)
f (x)
x 2
– +undef.
no p.i.
f´´(x)
f (x)
10.
x
f'(x)
f(x)
2
undef.+ –
no max.min.
x
f"(x)
f(x)
2
undef.+ +
no p.i.
11.
x –2 1 3
0 0 0+ – –+
max. min. max.
f´(x)
f (x)
x –1 2
00– + –
p.i. p.i.
f´´(x)
f (x)
x
f(x)
–2 –1 1 2 3
12.
–3 –1 3
0 0 0– + + –
min. plateau max.
f'(x)
f(x)
x
x –2 –1 2
0 0 0+ – + –
p.i. p.i. p.i.
f(x)
f"(x)
x
f(x)
–3 –2 –1 2 3
170 Problem Set 8-2 Calculus Solutions Manual© 2005 Key Curriculum Press
13.
x 1
0+ – –
–2
∞
max. plateau
f´(x)
f (x)
x 1
0+ + –
–2
∞
no p.i. p.i.
f´´(x)
f (x)
x
f(x)
–2 1
14.
2 3 4
0 ∞ 0– + + –
f(x)
max. max.none
f'(x)
x
3
∞+ –
f(x)
p.i.
f"(x)
x
x
f(x)
2 3 4
15.
3
zero+ –
–1 1 5
00 e.p.e.p.
min. max. min.
x
f´(x)
f (x)
3
zero– –
–1 1 5
00 e.p.e.p.
no p.i.
x
f´´(x)
f (x)
x
f(x)
–1 1 3 5
16.
6
–+ +
2 7
00 e.p.e.p.
1
f(x)
min. max.max. min.
x
f'(x)
x 5
– +
3 7
00 e.p.e.p.
1
zero
f(x)
no p.i.
f"(x)
f(x)
1 2 3 5 6 7
x
17.
3 6
3
–3
y
xf´
f´
f´f
18.
4 8
4
–4
y
x
f'
f' f' f
19.
4 8
4
y
x
f´ f´
f
20.
4
4
y
x
2
f'
f
Calculus Solutions Manual Problem Set 8-2 171© 2005 Key Curriculum Press
21. f (x) = 3ex − xex
f ′ (x) = 3ex − ex − xex = ex (2 − x)f ′ (2) = e2 (2 − 2) = 0 ⇒ critical point at x = 2f ′′ (x) = 2ex − ex − xex = ex (1 − x)f ′′ (2) = e2 (1 − 2) = −7.3890... < 0∴ local maximum at x = 2
1 2 3
5
f(x)
x
The graph confirms a maximum at x = 2.
22. f x x( ) = −sinπ4
f x x′ = −( )π π4 4
cos
f ′ = − = ⇒( )24 4
2 0π π
cos ( ) critical point
at x = 2
′′ =f x x( )π π2
16 4sin
′′ = = >f ( ) .216 4
2 0 6168 02π π
sin ( ) K
∴ local minimum at x = 2
2
1
–1
f (x )
x
The graph confirms a minimum at x = 2.
23. f (x) = (2 − x)2 + 1f ′ (x) = −2(2 − x)f ′ (2) = −2(2 − 2) = 0 ⇒ critical point at x = 2f ″ (x) = 2 ⇒ f ″ (2) = 2 > 0∴ local minimum at x = 2
2
1
f(x)
x
The graph confirms a minimum at x = 2.
24. f (x) = −(x − 2)2 + 1f ′ (x) = −2(x − 2)f ′ (2) = −2(2 − 2) = 0 ⇒ critical point at x = 2f ″ (x) = −2 ⇒ f ″ (2) = −2 < 0∴ local maximum at x = 2
2
1
f(x)
x
The graph confirms a maximum at x = 2.
25. f (x) = (x − 2)3 + 1f ′ (x) = 3(x − 2)2
f ′ (2) = 3(2 − 2)2 = 0 ⇒ critical point at x = 2f ″ (x) = 6(x − 2)f ″ (2) = 6(2 − 2) = 0, so the test fails.f ′ (x) goes from positive to positive asx increases through 2, so there is a plateauat x = 2.
2
1
f(x)
x
The graph confirms a plateau at x = 2.
26. f (x) = (2 − x)4 + 1f ′ (x) = −4(2 − x)3
f ′ (2) = −4(2 − 2)3 = 0 ⇒ critical point at x = 2f ″ (x) = 12(2 − x)2
f ″ (2) = 12(2 − 2)2 = 0, so the test fails.f ′ (x) changes from negative to positive asx increases through 2, so there is a localminimum at x = 2.
2
1
f(x)
x
The graph confirms a minimum at x = 2.
27. a. f (x) = 6x5 − 10x3
f ′ (x) = 30x4 − 30x2 = 30x2(x + 1)(x − 1)f ′ (x) = 0 ⇔ x = −1, 0, or 1 (critical pointsfor f (x))
′′ = − = + −f x x x x x x( ) 120 603 60 2 1 2 1( )( )
f ″ (x) = 0 ⇔ x = 0, ± 1 2/ (critical points for
f ′ (x))
b. The graph begins after the f-critical point atx = −1; the f ′ -critical point at x = − 1 2/ is
shown, but is hard to see.
c. f ′ (x) is negative for both x < 0 and x > 0.
172 Problem Set 8-2 Calculus Solutions Manual© 2005 Key Curriculum Press
28. a. f (x) = 0.1x4 − 3.2x + 7f ′ (x) = 0.4x3 − 3.2 = 0.4(x − 2)(x2 + 2x + 4)x2 + 2x + 4 has discriminant = 22 − 4 · 4 < 0,so f ′ (x) = 0 ⇔ x = 2 (critical point for f (x)).f ″ (x) = 1.2x2
f ″ (x) = 0 ⇔ x = 0 (critical point for f ′ (x))
b. f ″ (x) does not change sign at x = 0.( f ″ (x) ≥ 0 for all x)
c. f ″ (c) = 0, but f ′ (c) ≠ 0
29. a. f x xe x( ) = −
f x xe e e xx x x′ = − + = −− − −( ) ( )1
f ′ (x) = 0 ⇔ x = 1 (critical point for f (x))′′ = − = −− − −f x xe e e xx x x( ) 2 2( )
f ″ (x) = 0 ⇔ x = 2 (critical point for f ′ (x))
b. Because f (x) approaches its horizontalasymptote (y = 0) from above, the graph mustbe concave up for large x; but the graph isconcave down near x = 1, and the graph issmooth; somewhere the concavity mustchange from down to up.
c. No. e x− ≠ 0 for all x, so xe xx− = ⇔ =0 0.
30. a. f (x) = x2 ln xf ′ (x) = x + 2x ln x = x (1 + 2 ln x)f (x) and f ′ (x) are undefined at x = 0, sof x x x e′ = ⇔ = − ⇔ = =−( ) .0 0 5 0 5ln .
0.6065… (critical point for f (x)).f ″ (x) = 3 + 2 ln x
′′ = ⇔ = − ⇔ = =−f x x x e( ) ln .0 1 5 1 5.0.2231… (critical point for f ′ (x)).
b. lim ln limln
lim/
–– –x x x
x xx
x
x
x→ → →+ + += =
0
2
02
03
1
2
= − =→ +lim .x
x0
20 5 0 by L’Hospital’s rule.
limx
x→ −0
2 ln x does not exist because x2 ln x is
undefined for x < 0.
c. All critical points from part a appear,although the inflection point at x e= −1 5.
ishard to see on the graph.
31. a. f (x) = x5/3 + 5x2/3
′ = + = +− −f x x x x x( ) ( )/ / /5
3
10
3
5
322 3 1 3 1 3
f ′ (x) = 0 ⇔ x = −2, and f ′ (x) is undefinedat x = 0 (critical points for f (x)).
′′ = − = −− − −f x x x x x( ) ( )/ / /10
9
10
9
10
911 3 4 3 4 3
f ″ (x) = 0 ⇔ x = 1 (critical point for f ′(x);f ′ (0) is undefined, so f ′ has no critical pointat x = 0).
b. The y-axis (x = 0) is a tangent line becausethe slope approaches −∞ from both sides.
c. There is no inflection point at x = 0 becauseconcavity is down for both sides, but there isan inflection point at x = 1.
32. a. f x x x( ) = −1 2 0 23. .
f x x x x x′ = − = −− −( ) . . . ( )1 2 0 6 0 6 2 10 2 0 8 0 8. . .
f ′ (x) = 0 ⇔ x = 0.5, and f ′ (x) is undefinedat x = 0 (critical points for f (x)).′′ = + = +− − −f x x x x x( ) . . .0 24 0 48 0 24 20 8 1 8 1 8. . . ( )
f ″ (x) = 0 ⇔ x = −2 (critical point for f ′ (x);f ′ (0) is undefined, so f ′ has no critical pointat x = 0).
b. f (0) = 01.2 − 3 ⋅ 00.2 = 0 has only one value.
c. Curved concave up because f ′′(x) > 0 forx < −2
33. a. f (x) = −x3 + 5x2 − 6x + 7
7
1
x
f(x)
Maximum (2.5, 7.6), minimum (0.8, 4.9),points of inflection (1.7, 6.3)No global maximum or minimum
b. f ′(x) = −3x2 + 10x − 6
′ = ⇔ = ± =f x x( ) .01
35 7 2 5485( ) K or
0.7847…
′′ = − + ′′ = ⇔ = =f x x f x x( ) ; ( )6 10 05
31.666…
c. f ′′(0.7847…) = −6(0.7847…) + 10 =5.2915… > 0, confirming local minimum.
d. Critical and inflection points occur onlywhere f, f ′, or f ′′ is undefined (no such pointsexist) or is zero (all such points are foundabove).
34. a. f (x) = x3 − 7x2 + 9x + 10
f(x)
10
1
x
Maximum (0.8, 13.2), minimum (3.9, −2.1),points of inflection (2.3, 5.6)No global maximum or minimum
Calculus Solutions Manual Problem Set 8-2 173© 2005 Key Curriculum Press
b. f ′(x) = 3x2 − 14x + 9
′ = = ± =f x x( ) at .0
1
37 22 3 896( ) K or
0.769…
′′ = − ′′ = = =f x x f x x( ) ; ( ) at .6 14 0
7
32 333K
c. f ′′(0.769…) = 6(0.769…) − 14 =−9.3808… < 0, confirming local maximum.
d. Critical and inflection points occur onlywhere f, f ′, or f ′′ is undefined (no such pointsexist) or is zero (all such points are foundabove).
35. a. f (x) = 3x4 + 8x3 − 6x2 − 24x + 37,x ∈ [−3, 2]
–3 2
80
x
f(x)
Maximum (−3, 82), (−1, 50), (2, 77),minimum (−2, 45), (1, 18), points ofinflection (−1.5, 45.7), (0.2, 32.0)Global maximum at (−3, 82) and globalminimum at (1, 18)
b. f ′(x) = 12x3 + 24x2 − 12x − 24= 12(x + 2)(x − 1)(x + 1)
f ′(x) = 0 ⇔ x = −2, −1, 1f ′(x) is undefined ⇔ x = −3, 2.f ′′(x) = 36x2 + 48x − 12 = 12(3x2 + 4x − 1);
′′ = ⇔ = − ± = …f x x( ) 0.215201
32 7( )
or −1.5485…f ′′(x) is undefined ⇔ x = −3, 2.
c. f ′′(−2) = 12[3(4) + 4(−2) − 1] = 36 > 0,confirming local minimum.
d. Critical and inflection points occur onlywhere f, f ′, or f ′′ is undefined (only atendpoints) or is zero (all such points are foundabove).
36. a. f (x) = (x − 1)5 + 4, x ∈ [−1, 3]
1
10
3–1
x
f(x)
Maximum (3, 36), minimum (−1, −28), plateau and points of inflection (1, 4)Global maximum at (3, 36) and globalminimum at (−1, −28)
b. f ′(x) = 5(x − 1)4
f ′(x) = 0 ⇔ x = 1; f ′(x) is undefined ⇔x = −1, 3.f ′′(x) = 20(x − 1)3;f ′′(x) = 0 ⇔ x = 1; f ′′(x) is undefined ⇔x = −1, 3.
c. f ′′(1) = 20(1 − 1)3 = 0, so the test fails.
d. Critical and inflection points occur onlywhere f, f ′, or f ′′ is undefined (only atendpoints) or is zero (all such points are foundabove).
37. f (x) = ax3 + bx2 + cx + d; f ′(x) = 3ax2 + 2bx + c;f ′′(x) = 6ax + 2b ⇒ f ′′(x) = 0 at x = −b/(3a)Because the equation for f ′′(x) is a line withnonzero slope, f ′′(x) changes sign at x = −b/(3a),so there is a point of inflection at x = −b/(3a).
38. f (x) may not have a local maximum orminimum (if f ′(x) is never zero); if this is notthe case, then the maximum and minimum occurwhere f ′(x) = 3ax2 + 2bx + c = 0, at
xb b a c
a
b
a
b ac
a= ± ⋅ ⋅ = ±– – – –
,2 4 4 3
6 3
3
3
2 2
and the maximum and minimum occur at
b ac a2 3 3– /( )units on either side of the
inflection point −b/(3a) (see Problem 33).
39. f (x) = ax3 + bx2 + cx + df ′(x) = 3ax2 + 2bx + c; f ′′(x) = 6ax + 2bPoints of inflection at (2, 3) ⇒ f ′′(3) = 0 ⇒18a + 2b = 0Maximum at (5, 10) ⇒ f ′(5) = 0 ⇒ 75a + 10b +c = 0(3, 2) and (5, 10) are on the graph ⇒27a + 9b + 3c + d = 2.125a + 25b + 5c + d = 10Solving this system of equations yields
f x x x x( ) = − + − −1
2
9
2
15
2
5
23 2 .
3 5
5
(5, 10)
x
f(x)
(3, 2)
The graph confirms maximum (5, 10) and pointsof inflection (3, 2).
40. f (x) = ax3 + bx2 + cx + df ′(x) = 3ax2 + 2bx + c; f ′′(x) = 6ax + 2bPoints of inflection at (2, 7) ⇒ f ′′(2) = 0 ⇒12a + 2b = 0Maximum at (−1, 61) ⇒ f ′(−1) = 0 ⇒3a − 2b + c = 0
174 Problem Set 8-2 Calculus Solutions Manual© 2005 Key Curriculum Press
(2, 7) and (−1, 61) are on the graph ⇒8a + 4b + 2c + d = 7.−a + b − c + d = 61Solving this system of equations yieldsf (x) = x3 − 6x2 − 15x + 53.
2–1
80f(x)
x
The graph confirms maximum (−1, 61) andpoints of inflection (2, 7).
41. a. f (x) = x3 ⇒ f ′(x) = 3x2
f ′(−0.8) = 1.92f ′(−0.5) = 0.75f ′(0.5) = 0.75f ′(0.8) = 1.92
b. The slope seems to be decreasing from −0.8to −0.5; f ″(x) = 6x < 0 on −0.8 ≤ x ≤ −0.5,which confirms that the slope decreases. Theslope seems to be increasing from 0.5 to 0.8;f ″(x) = 6x > 0 on 0.5 ≤ x ≤ 0.8, whichconfirms that the slope increases.
c. The curve lies above the tangent line.
42. Ima could notice that y ′ = 0 at x = 0(or y ′ = 3 at x = ±1), so the graph could notpossibly be a straight line with slope = 1.
43. a.
cx
f(x)
b.
cx
f(x)
c.
cx
f(x)
d.
cx
f(x)
e.
(Locallyconstant)
cx
f(x)
44. f (x) = 10(x − 1)4/3 + 2f (1) = 2, so f (1) is defined.
f x x′ =( ) /40
31 1 3( – )
f ′(1) = 0, so f is differentiable at x = 1.
f x x″ = −( ) /40
91 2 3( – )
f ″(1) has the form40
90
40
91 02 3( ) ( / )– / or , so
f ″(1) is infinite.There seems to be a cusp at (1, 2), but zoomingin on this point reveals that the tangent isactually horizontal there.
f(x)
2
1
x
See Problem 20 in Problem Set 10-6 forcalculation of curvature.
45. f (x) = e0.06 x, f ′(x) = 0.06e0.06 x,f ″(x) = 0.0036e0.06 x
g (x) = 1 + 0.06x + 0.0018x2 + 0.000036x3
g ′(x) = 0.06 + 0.0036x + 0.000108x2
g ″(x) = 0.0036 + 0.000216xf (0) = 1 and g (0) = 1f ′(0) = 0.06 and g ′(0) = 0.06f ″(0) = 0.0036 and g ″(0) = 0.0036(In fact, f ′″(0) = g ′″(0).)But f (10) = e0.6 = 1.822… ≠ g (10) = 1.816;f ′(10) = 0.109… ≠ g ′(10) = 0.1068.Because f (x) > 0 for all x, f has no x-intercept.But g (0) = 1 and g (−100) = −23.By the intermediate value theorem, g (x) = 0somewhere between x = −100 and x = 0, meaningthat g does have an x-intercept.
Calculus Solutions Manual Problem Set 8-3 175© 2005 Key Curriculum Press
46. f x xx
x
x( ) if
if= + ≠
=
( – ) sin
–,
,
11
12 1
2 1
3
2.01
2
1.991
f(x)
lim lim ( – )–
( )x x
f x xx
f→ →
= ⋅ + = =1 1
311
12 2 1( ) sin
(The limit of the first term is zero because(x − 1)3 approaches zero and the sine factor isbounded.)∴ f is continuous at x = 1.
ff x f
x
x x
x
xx
x
x
x
′ =
= +
= =
→
→
→
( )
sin 0
11
1
1 1 1 2 2
1
11
1
1
1
3
1
2
lim( ) – ( )
–
lim[( – ) sin( /( – ))] –
–
lim ( – )–
(x − 1)2 → 0 and the sine factor is bounded.∴ f ′(1) = 0
1.999
2
2.001
1 1.10.9
The graph is zoomed in by a factor of 10 bothways. The graph does appear to be locally linearat x = 2. Although the sine factor makes aninfinite number of cycles in any neighborhood ofx = 1, the (x − 1)3 factor approaches zero sorapidly that the graph is “flattened out.” Thename pathological is used to describe the fact thatthe graph makes an infinite number of cycles in abounded neighborhood of x = 1.
47. Answers will vary.
Problem Set 8-3Q1. y x′ = − + −3 3 5( ) 2 Q2. ln |x + 6| + C
Q3. − −2
35 3x / Q4. 3x1/3 + C
Q5. − +−x C1 Q6. x + CQ7. ln |sin x| + CQ8. Q9.
x
y
1
1
2
1x
y"
Q10. D
1. Let x = total width of pen, y = length of pen.Domains: 0 ≤ x ≤ 300, 0 ≤ y ≤ 200Maximize A(x) = xy.
2 3 600 2002
3x y y x+ = ⇒ = −
∴ = −A x x x( ) 2002
32
300150
max.A(x)
x
The graph shows a maximum at x ≈ 150.
Algebraically, A x x′ = −( ) .2004
3A′(x) = 0 ⇔ x = 150, confirming the graph.
x y= ⇒ = − ⋅ =150 2002
3150 100
Make the total width 150 ft and length 100 ft.(Note: The maximum area was not asked for.)
2. a. Let x = width of a room across the front,y = depth of a room from front to back.Domains: x ≥ 0, y ≥ 0Minimize P (x) = 12x + 7y.xy = 350 ⇒ y = 350x− 1
∴ P (x) = 12x + 2450x− 1
14.28...
x
P(x)
The graph shows a minimum at x ≈ 14.Algebraically, P ′ (x) = 12 − 2450x− 2.P ′ (x) = 0 ⇔ 2450x− 2 = 12 ⇔x = ± = ± = ± …2450 12 35 6 14 288/ / .
Minimum is at x y= = =35 6 10 6/ , 24.49… .Make rooms 14.3 ft across and 24.5 ft deep.
b. For 10 rooms, P (x) = 20x + 11y =20x + 3850x− 1.
P ′ (x) = 20 − 3850x− 2 = 0 at x = 192 5.
Minimum at x = = …192 5 13 87. . ,
y = = …350 192 5 25 22/ ..
Make rooms 13.9 ft across and 25.2 ft deep.For 3 rooms, P (x) = 6x + 4y = 6x + 1400x− 1.
P ′(x) = 6 − 1400x− 2 = 0 at x = =1400 6/
10 7 3/
176 Problem Set 8-3 Calculus Solutions Manual© 2005 Key Curriculum Press
Minimum at x = = …10 7 3 15 27/ . ,
y = = …5 21 22 91.
Make rooms 15.3 ft across and 22.9 ft deep.
3. a. Let x = width of rectangle, 2x = length ofrectangle, y = width of square.A rect = 2x2, Asq = y2
For minimal rectangle, 2x2 ≥ 800 ⇒ x ≥ 20.For minimal square, y2 ≥ 100 ⇒ y ≥ 10.Perimeter P = 6x + 4y = 600 ⇒y = 150 − 1.5x∴ 150 − 1.5x ≥ 10 ⇒ x ≤ 140/1.5 =93.3333…Domain: 20 ≤ x ≤ 93.3333…
b. Total area A(x) = 2x2 + y2
= 2x2 + (150 − 1.5x)2
= 22500 − 450x + 4.25x2
20,000
93.3
x
A(x)
20
c. The graph shows a maximum at endpointx = 93.3333… .A′(x) = −450 + 8.5xA′(x) = 0 ⇔ x = 450/8.5 = 52.9411…Because A(52.9…) is a minimum, themaximum occurs at an endpoint.A(20) = 15200, A(93.3333…) =17522.2222…Greatest area ≈ 17,522 ft2
4. a. Let r = radius of circle, s = width of squareDiameter ≥ 50 ⇒ r ≥ 25Circumference ≤ 1000 ⇒ 2πr ≤ 1000 ⇒r ≤ 500/πDomain of r : 25 ≤ r ≤ 500/π = 159.154…Minimize A(r) = πr2 + s2.Perimeter 2πr + 4s = 1000 ⇒ s = 250 − πr/2∴ A(r) = πr2 + (250 − πr/2)2
7025 160
r
Amax.
min.
The graph shows minimum area at x ≈ 70.A′(r) = 2πr + 2(250 − πr/2)(−π/2)A′(r) = 0 ⇔ 2πr − π(250 − πr/2) = 0 ⇒r = 500/(4 + π) = 70.012…A(25) = 46370.667…A(70.012…) = 35006.197…
A(159.154…) = 79577.471…Minimum area at r = 70.012… ,s = 1000/(4 + π) = 140.024…For square, 4(140.024…) ≈ 560.For circle, 2π(70.012…) ≈ 440.Use 440 yd for square and 560 yd for circle.(You could build a square corral with side 140around the circular fence of radius 70 toenclose a total area of only 19,607 yd 2 , butBig Bill might not like your solution!)
b. The graph of A versus r shows that themaximum area occurs at the largest possiblecircle. Big Bill should use all 1000 yards forthe circular fence and not build a corral.
5. a. Let x = length of square base, z = heightof box.Domain of x x: .0 120 10 954≤ ≤ = …Maximize V(x) = x2z.Area = x2 + 4xz = 120 ⇒ z = 30/x − x/4∴ V(x) = 30x − x3/4
6.32... 10.95...0
max.V
x
The graph shows a maximum at x ≈ 6.3.
V′(x) = 30 − 3x2/4 = 0 at x = ± 40
x = − 40 is out of the domain.
Critical points at x = 0, x x= =40 120,
V V( ) , ( )0 0 120 0= =V( ) .40 20 40 126 49= = …Maximum at x = = …40 6 324. ,
z = = …40 2 3 162/ .
Make the box 6.32 cm square by 3.16 cmdeep.
b. Conjecture: An open box with square base ofside length x and fixed surface area A willhave maximal volume when the base lengthis twice the height, which occurs when
x A= /3 (see the solution to Problem 8b).
6. a. Domain of x is 0 ≤ x ≤ 6.
b. V(0) = 0 cm2
V(1) = 180 cm2
V(2) = 256 cm2 (largest volume for an integervalue of x)V(3) = 252 cm2
V(4) = 192 cm2
V(5) = 100 cm2
V(6) = 0 cm2
Calculus Solutions Manual Problem Set 8-3 177© 2005 Key Curriculum Press
c. V(x) = (20 − 2x)(12 − 2x)x= 240x − 64x2 + 4x3
200
6
x
V(x)
The graph shows a maximum at x ≈ 2.4.V′(x) = 240 − 128x + 12x2 = 0 at
x = ±( )128 4864 24/ = 2.427… or 8.239…
x = 8.239… is out of the domain.V(2.427…) = 262.68… is a maximumbecause it is positive and V(0) = V(6) = 0.Maximum volume ≈ 262.7 cm2 atx ≈ 2.43 cm
7. Let x = length, y = depth, C(x) = total cost.Domains: x > 0, y > 0Area of bottom = 5xTotal area of sides is (10 + 2x)y.Minimize C(x) = 10(5x) + 5(10 + 2x)y.Volume = 72 ⇒ 5xy = 72 ⇒y x x= = −72 5 14 4 1/( ) .
∴ = + + −C x x x x( ) ( )( . )50 5 10 2 14 4 1
C x x x( ) = + +−50 720 1441
3.794...
x
C
The graph shows a minimum at x ≈ 3.8.C′(x) = 50 − 720x− 2 = 0 ⇔ x = ± 72 5/ =±3.7947…x = −3.7947… is out of the domain.Minimum is at x = 3.7947 because C′(x)changes from negative to positive there.C(3.7947…) = 120 10 + 144 ≈ 523.47Minimum cost is $523.47.
8. a. Maximize V = xyz.Fixed area A = xy + 2xz + 2yz
⇒ y = (A − 2xz)/(x + 2z)
∴ =+
VAxz x z
x z
– 2
2
2 2
dV
dx
z x z x Az
x z= +
+– –
( )
2 8 2
2
2 2 3 2
2
dV
dxx z z A= = − + +0 2 4 2 at
yA z z z A
z z A z
z z A
= + +
+ + +
= − + +
– (– )
–
2 2 4
2 4 2
2 4
2
2
2
Therefore, x = y for maximum volume,Q.E.D.
b. Let x = y. Maximize V = xyz = x2z.Fixed area A = xy + 2xz + 2yz = x2 + 4xz⇒ z = A/(4x) − x/4∴ V = (A/4)x − x3/4
dV
dxA x x A= − = = ±( / ) at 4 3 4 0 32/ /
dV/dx goes from positive to negative atx A= /3 ⇒ maximum at x A= /3.
z A A A A x= − = =/( ) /4 3 3 41
23
1
2/ / /
c. For the maximal box in part b, the depth ishalf the length of the base. Thus, the box isshort and fat. This makes sense because theproblem is equivalent to maximizing thevolume of two open boxes with the secondbox placed upside-down on the first. Theresulting single closed box will havemaximum volume when it is a cube, whichwill happen if each open box is half a cube.
9. For y = ex, minimize D x x y( ) = + =2 2
x e x2 2+ .
–0.4263...
1
D(x)
x
The graph shows a minimum at x ≈ −0.43.
D x x e x ex x′ = + +−( ) /1
22 22 2 1 2 2( ) ( )
D′(x) = 0 ⇔ 2x + 2e2x = 0 ⇔ x = −e2x
Because x appears both algebraically andexponentially, there is no analytic solution.Solving numerically gives x ≈ −0.4263. Bygraphing D(x), D(−0.4263) is a minimum.Closest point to the origin is(x, y) = (−0.4263… , 0.6529…).
10. Minimize A(r) = πr2 + 2rx, r ≥ 20.2π r + 2x = 400 ⇒ x = 200 − π rx ≥ 100 ⇒ r ≤ 100/π∴ domain is 20 ≤ r ≤ 100/π.A(r) = π r2 + 2r(200 − π r) = 400r − π r2
178 Problem Set 8-3 Calculus Solutions Manual© 2005 Key Curriculum Press
10,000
20 31.83...
r
A(r)
The graph shows a minimum at endpoint x = 20.A′ = 400 − 2πrA′ = 0 ⇔ r = 200/π = 63.6… (out of domain)A′ > 0 for all r in the domain.∴ minimum occurs at left end of domain, r = 20.x = 200 − 20π = 137.168…Make radius of semicircles 20 m and straightsections 137.17 m.
11.
y
x
x – 1
L
1
8
L x x y( ) = +2 2 .
Domains: x ≥ 1, y ≥ 8Minimize L2(x) = x2 + y2.
Using similar triangles, y
x x= 8
1–⇒ y
x
x= 8
1–.
∴ L2(x) = x2 + 64
1
2
2
x
x( – )
10
1 5
x
L
The graph shows a minimum of L(x) at x ≈ 5.
(L2)′(x) = 2x − 128
1 3
x
x( – )
(L2)′(x) = 0 ⇔ 2128
1 3xx
x=
( – )⇔
x = 0 (out of domain) or (x − 1)3 = 64 ⇔ x = 5
By graph, L(x) is a minimum at x = 5.Shortest ladder has length L( )5 5 5= ≈ 11.18 ft.
12. Let x and y be the segments shown.
x
yL
7
5
L x x y( ) = + + +( ) ( )7 52 2
Maximize L2(x) = (x + 7)2 + (y + 5)2.Using similar triangles, y/7 = 5/x ⇒ y =35/x.∴ L2(x) = (x + 7)2 + (35/x + 5)2
L2(x) = x2 + 14x + 49 + 1225/x2 + 350/x + 25
5.59...
x
L
20
The graph shows a minimum of L(x) at x ≈ 5.6.( ( ))L x x x x2 2 32 14 350 2450′ = + − −− −
By numerical solution, (L2)′ = 0 at x ≈ 5.5934… .(Exact answer is x = 1753 .)But a minimum distance L in the hall impliesthat the maximal ladder that will go through thehall is at x = 5.5934… .L2(5.5934…) = 285.3222…L(5.5934…) = 16.8914…No ladder longer than 16.8 ft (rounded down) canpass through the hall.
13. Let r = radius, h = height.V = πr2h2r + 2h = 1200 ⇒ h = 600 − r∴ V = πr2(600 − r) = π(600r2 − r3)
V
400
r
The graph shows a maximum at r ≈ 400.V′ = π(1200r − 3r2)V′ = 0 ⇔ r = 0 or r = 400From graph, maximum is at r = 400.h = 600 − 400 = 200Maximum volume occurs with rectangle400 mm wide (radius), 200 mm high.
14. Rotating a square does not give the maximumvolume. The solution to Problem 13 gives acounterexample. Repeating the calculations withperimeter P instead of 1200 gives r = (1/3)P andh = (1/6)P, showing that the proportions formaximum volume are with radius twice theheight.
15. a. Let r = radius, h = height.V = πr2h = π(3.652)(10.6) = 141.2185π
= 443.6510… cm3
Calculus Solutions Manual Problem Set 8-3 179© 2005 Key Curriculum Press
b. A = 2πrh + 2πr2
V = πr2h = 141.2185π ⇒ h = 141.2185/r2
∴ A = 2πr(141.2185/r2) + 2πr2
A r r= +−2 141 2185 1 2π ( . )
c. A
r
4.13...
500
The graph shows a minimum at x ≈ 4.1.A r r′ = − +−2 141 2185 22π ( . )A′ = 2π/r2(−141.2185 + 2r3)A′ = 0 ⇔ r3 = 70.60925 ⇒r = = …70 60925 4 13323 . .Minimum at r = 4.1… because A′ goes fromnegative to positive.
h = =141 2815 70 60925 2 70 609253 3. /( )2. .
= 8.2664…Radius ≈ 4.1 cm, height ≈ 8.3 cmBecause height = 2 × radius, height = diameter.So minimal can is neither tall and narrow norshort and wide.
d. Normally proportioned can is taller andnarrower than minimal can. For normal can,A = 2π(3.65)(10.6) + 2π(3.65)2 =326.8041… .For minimal can, A = 2π(4.13…)(8.26…) +2π(4.13…)2 = 322.014… .Difference is 4.78… cm2.Percent: (4.789…)(100)/326.80… = 1.465…≈ 1.5% of metal in normal can
e. Savings = (0.06)(20 × 106)(0.01465…)(365) =6.419… × 106, or about $6.4 million!
16. a. C r r k rh r k r( ) .= + = + −2 2 2 282 4372 2 1π π π πC r rk r′ = − −( ) .4 282 437 2π π
= −−4 70 609252 3πr kr( . )
C′(r) = 0 at r k= 70 609253 . /
C r k r″ = + >−( ) .4 564 874 03π π for all r > 0,so this is a local minimum.If the normal can is the cheapest to make,then 3 65 70 609253. = ⇒. /k
k = = …−70 60925 3 65 1 45203. ( . ) . .
This is reasonable because metal for the endsis cut into circles, so some must be wasted.
b. Now it takes (2r)2 cm2 of metal to make eachend of the can, so the function to minimize isC r r k rh r k r( ) . .= + = + −8 2 8 282 4372 2 1π πC r rk r′ = − −( ) .16 282 437 2π
C r rk
′ = =( ) at0282 437
163
. π
C r k r″ = + >−( ) .16 564 874 03π for all r > 0,so this is a local minimum.If the normal can is the cheapest to make,
then 3 65282 437
16
282 437
16 3 653
3. = ⇒ =. .
( . )
π πk
k
= 1.1404… .To minimize the area (not the cost) of thecan, minimize 8 2 8 282 4372 2 1r rh r r+ = + −π π. .
C r r r′ = = ⇒−( ) – .16 282 437 02π
r = =282 437
163
. π3.8126 cm
h =( )
=141 2185
282 437 169 7099
32
.
. /.
πK cm.
The proportions of this can are closer to thoseof the normal can.
c. If the metal for the ends can be cut withoutwaste, then it takes π(r + 0.6)2 to make eachend and (2πr + 0.5)h to make the sides, sominimizeC(r) = 2π(r + 0.6)2 + (2πr + 0.5)h
= + + + −2 0 6 141 2185 2 0 5 2π π( . ) . ( . )2r r rC r r r′ = + − −( ) ( . ) .4 0 6 282 437 2π π
− − .141 2185 3rC′(r) = 0 at r ≈ 3.9966 by graphing calculator.C r r r″ = + + >− −( ) . .4 564 874 423 6555 03 4π πfor all r > 0, so this is a minimum point.Minimal can has r ≈ 3.9966… ,h ≈ 8.8411… cm.But if the metal for the ends is cut fromsquares, then it takes 4(r + 0.6)2 to make eachend and (2πr + 0.5)h to make the sides, sominimize:C(r) = 8(r + 0.6)2 + (2πr + 0.5)h
= + + + −8 0 6 141 2185 2 0 5 2( . ) . ( . )2r r rπC r r r′ = + − −( ) ( . ) .16 0 6 282 437 2π
− − .141 2185 3rC′(r) = 0 at r ≈ 3.6776… by graphingcalculator.
′′ = + + >− −C r r r( ) 16 564 874 423 6555 03 4. .πfor all r > 0, so this is a minimum point.Minimal can has r ≈ 3.6776… ,h ≈ 10.4411… .This is close to the normal can!
17. a. Volume of cup = π(2.5)2 · 7 = 43.75πLet r = radius of cup, h = height of cup.Minimize A(r) = πr2 + 2πrh.π πr h h r2 243 75 43 75= ⇒ = −. .∴ = + −A r r r( ) .π π2 187 5
100
3.52...
r
A
180 Problem Set 8-3 Calculus Solutions Manual© 2005 Key Curriculum Press
The graph shows a minimum at r ≈ 3.5 cm.A r r r r r′ = − = −− −( ) . ( . )2 87 5 2 43 752 2 3π π πA′(r) = 0 at r = 43 753 . = 3.5236… .There is a minimum at x = 3.5236… becauseA(r) goes from decreasing to increasing.(See graph.)h r= = =−43 75 43 75 43 752 3 3. ( . ) / .Minimal cup has r ≈ 3.52 cm, h ≈ 3.52 cm.
b. Ratio is d : h = 2r : h = 2 : 1.
c. Current cup design uses π(2.5)2 + π · 5 · 7 =41.25π = 129.59… cm2 = 0.012959… m2 percup, which costs(300,000,000)(0.012959…)(2.00)≈ $7,775,441.82 per year.Minimal cup design uses 3π(43.75)2/3 =117.01… cm2 = 0.011701… m2 per cup,which costs (300,000,000)(0.011701…)(2.00)≈ $7,021,141.88 per year.Switching to minimal cup design wouldsave 754,299.93 ≈ $754,000 per year inpaper costs (about 10% of the current annualpaper bill), but would likely result in loss ofsales because a cup of that shape is hard todrink from.
d. Let r = radius of cup, h = height of cup.π πr h V h V r2 2= ⇒ = −( / )
Minimize ( ) .A r r rh r Vr= + = + −π π π2 2 12 2
A r r Vr r V′ = − = =−( ) at2 2 02 3π π/′′ = + >−A r Vr( ) 2 4 03π for all r > 0, so this is
a minimum.Minimal cup has r = V/π3 ,
h V V V r= = =−( / )( / )π π π2 3 3/ / .
18. a. A = yz = (30 + 0.2x)(40 − 0.2x)A(x) = 1200 + 2x − 0.04x2
Left rectangle: A(0) = 1200 in.2
Right rectangle: A(100) = 1000 in.2
b. A(80) = 1104 in.2
c.
25 100
1000
x
A(x)
The graph shows a maximum at x ≈ 25.A′(x) = 2 − 0.08x = 0 at x = 25.Critical points at x = 0, 25, 100A(25) = 1225 in.2; A(0) = 1200 in.2;A(100) = 1000 in.2 (from part a)Maximum area at x = 25 in., minimum area forx = 100 in.
19. Maximize A(x) = 2xy = 2x cos x.Use 0 ≤ x ≤ π/2 for the domain of x.
1
1
x
A(x)
The graph shows a maximum at x ≈ 0.86.A′(x) = 2 cos x − 2x sin xA′(x) = 0 when x = cot x.Solving numerically gives x ≈ 0.8603… .A(0) = A(π/2) = 0; A(0.8603…) = 1.1221…Maximum area = 1.1221…
20.
50
200
Street
x y
Let x = width of store, y = length of store.Minimize C(x) = 100x + 80(x + 2y).xy y x= ⇒ = −4000 4000 1
C x x x( ) = + −180 640000 1
y ≤ 200 ⇒ x ≥ 20, so domain of x is20 ≤ x ≤ 50.Graph shows minimum at x endpoint x = 50.
20 50
50,000C(x)
x
C x x′ = − =−( ) 180 640000 02
at x = 80 5
3= 59.628… , outside the domain.
C(20) = $35,600.00; C(50) = $21,800.00Minimum cost is at x = 50, y = 4000/50 = 80.Bill should build the store 50 ft × 80 ft.
21. a. A = 0.5xy = 0.5x cot x
lim limtanx x
Ax
x→ →= →
0 0 2
0
0
= =→
limsecx x0 2
1
2
1
2b. Domain of x is 0 < x ≤ π/2.
π/2
0.5
A(x)
x
The graph shows that the area approaches amaximum as x approaches the endpoint x = 0from the positive side.
Calculus Solutions Manual Problem Set 8-3 181© 2005 Key Curriculum Press
A x x x x′ =( )1
22(cot – csc )
A′(x) = 0 when x = cos x sin x or2x = 2 sin x cos x = sin 2x,which happens at x = 0.A(π/2) = 0, so the “maximum” occurs at x = 0.But x = 0 is not in the domain; A(x) can getarbitrarily close to 1/2, but never achieve it.
22.y
x
(x, y)
3x
Domain of x is 0 ≤ x ≤ 3.Maximize A = 0.5(3 − x) ( y) = 0.5(3 − x)ex =1.5ex − 0.5xex.
4
2 30
x
A(x)
The graph shows a maximum at x ≈ 2.A′(x) = 1.5ex − 0.5ex − 0.5xex = 0.5ex(2 − x)A′(x) = 0 at x = 2, confirming the graph.A′(x) > 0 for x < 2, and A′(x) < 0 for x > 2,confirming maximum point at x = 2.Maximum area A(2) = e2/2 = 3.69452… .
23. a. Maximize A(x) = 2xy = 2x(9 − x2) =18x − 2x3.Domain: 0 ≤ x ≤ 3
1.7320 3
20
x
A(x)
The graph shows a maximum at x ≈ 1.7.′ = − = = ± =A x x x( ) at 18 6 0 32 ±1.732…
−1.732 is out of the domain.A A A( ) ( ) ; ( 3) .0 3 0 12 3 20 7846= = = = K
Maximal rectangle has width = 2 3,
length = 9 − 3 = 6.
b. Maximize P(x) = 4x + 2y = 4x + 18 − 2x2.
10 3
20
x
P(x)
The graph shows a maximum at x ≈ 1.P′(x) = 4 − 4x = 0 at x = 1P(0) = 18; P(1) = 20; P(3) = 12Maximal rectangle has width = 2,length = 9 − 1 = 8.
c. No. The maximum-area rectangle is 2 3 by 6.The maximum-perimeter rectangle is 2 by 8.
24. a. Maximize V(x) = πx2y = πx2(9 − x2) =9πx2 − πx4.Domain: 0 ≤ x ≤ 3
2.121...0 3
50
x
V(x)
The graph shows a maximum at x ≈ 2.1.V x x x x′ = − = = ±( ) at ,18 4 0 0 4 53π π . .
− 4 5. is out of the domain.
V V V( ) ( ) , ( ) .0 3 0 4 5 20 25= = = =. π63.6172…Maximum is at x y= = − =4 5 9 4 5 4 5. , . . .
Maximal cylinder has radius = =4 5.2.12132… and height = 4.5.
b. Maximize L(x) = 2πxy = 2πx(9 − x2) =18π x − 2π x3.
1.732...0 3
50
x
L(x)
The graph shows a maximum at x ≈ 1.7.L x x x′ = − = = ±( ) at 18 6 0 32π π .
− 3 is out of the domain.
L L L( ) ( ) ; ( ) .0 3 0 3 12 3 65 2967= = = =π K
Maximum is at x y= = − =3 9 3 6, .
Maximal cylinder has radius = =31.7320… and height = 6.
c. Maximize A(x) = 2πx2 + 2π xy = 2π x2 +2π x(9 − x2) = 2π x2 + 18π x − 2π x3.
2.097...0 3
50
x
A(x)
The graph shows a maximum at x ≈ 2.1.A′(x) = 18π + 4π x − 6π x2
182 Problem Set 8-3 Calculus Solutions Manual© 2005 Key Curriculum Press
A′(x) = 0 at x = ± =1 2 7
32 0971. K or
−1.430…−1.430… is out of the domain.A(0) = 0; A(2.0971…) = 88.2727… ;A(3) = 18π = 56.5486…
Maximal cylinder has radius = + =1 2 7
3
2.0971… and height = =52 4 7
9
–
4.6018… .
d. No. The maximum-volume cylinder hasdimensions different from both of themaximum-area cylinders in parts b and c.
e. No. Rotating the maximum-area rectangledoes not produce the maximum-volumecylinder. But it produces the cylinder withmaximum lateral area.
f. If y = a2 − x2, the paraboloid has radius = a.V = πx2(a2 − x2) = π(a2x2 − x4)V′ = π(2a2x − 4x3)V x x a′ = ⇔ = = ± /0 0 2 or .
V is maximum at x a= / 2.
For the cylinder of maximum volume,(cylinder radius):(paraboloid radius) = /1 2,a constant.Note: This ratio is also constant (1 3/ ) for the
cylinder of maximum lateral area, but is notconstant for the cylinder of maximum totalarea.
25. a. x2 + y2 = 100, 0 ≤ x ≤ 10
Maximize V x x y x x( ) .= ⋅ =π π2 2 22 100 –
b.
100
2000
x
V(x)
8.16...
The graph shows a maximum volume atx ≈ 8.2.
V xx
xx x′ = +( )
–
––
2
1004 100
3
2
2π π
= +–
–
6 400
100
3
2
π πx x
x
V x x′ = = = =( ) at , .0 0
200
3
10 6
38 1649K
V(0) = V(10) = 0
V10 6
3
4000 3
92418 399
= =π. K
Maximal cylinder has radius = 8.1649… ,
height
= =20 3
311 5470. ,K and volume =
2418.39… .
c. Height radius= ⋅ 2
Volume of sphere Vs = ⋅ =4
31000
4000
3π π
Volume of maximal cylinder Vc = 4000 3
9
π
∴ = / V Vc s 3
26. Let r = radius of cone, h = height.
Lateral area A(r) = π r · (slant height) = +πr r h2 2
V r h h r= = ⇒ = −1
35 152 2π π
∴ = +A r r r r( ) π 2 4225 –
h ≥ 2r ⇒ 2r ≤ 15r− 2
Domain of r is 0 7 5 1 95743< ≤ =r . .. K
1
20
r
A(r)
1.957...
The graph shows a minimum of A(r) at endpointr = 1.957… .Minimize ( ) ( ).A r r r2 2 4 2225= + −π( ( )) ( ) atA r r r r2 2 3 3 64 450 0 112 5′ = − = = =−π .
2.1971… , which is out of the domain.A(1.9574…) = 26.915… , lim
rA r
→ += ∞
0( ) .
Minimal cone has radius = = …7 5 1 95743 . .
and height = = =2 2 7 5 3 91483r . .. KMake r ≈ 1.96 ft and h ≈ 3.91 ft.
27. a. Lateral area L(x) = 2π xyDomains: 0 ≤ x ≤ 5 and 0 ≤ y ≤ 7Equation of element of cone is
y x y x= − + ⇒ = − +7
57 1 4 7. .
∴ L(x) = 2πx(−1.4x + 7) = 2π(−1.4x2 + 7x)
5
50
x
L(x)
2.50
The graph shows a maximum of L(x) atx ≈ 2.5.L′(x) = 2π(−2.8x + 7)L′(x) = 0 at x = 2.5.
Calculus Solutions Manual Problem Set 8-3 183© 2005 Key Curriculum Press
L′(x) goes from positive to negative atx = 2.5.∴ maximum lateral area at radius x = 2.5 cm.
b. Total area A(x) = 2πxy + 2πx2
= 2πx(−1.4x + 7) + 2πx2
A(x) = 2π(7x − 0.4x2)
150
A(x)
x
50
The graph shows a maximum at endpointx = 5.A′(x) = 2π(7 − 0.8x) = 0 at x = 8.75, out ofdomain.∴ maximum is at an endpoint, x = 5.A(0) = 0; A(5) = 2π(52) = 50π = 157.07…Maximum area is with the degenerate cylinderconsisting only of the top and bottom, radius5 and height 0.
28. a. Let r = radius of cone, h = height of cone(constants).Let (x, y) be a sample point on cone element.Domain of x is 0 ≤ x ≤ r.L(x) = 2πxy.Equation of element of cone isy = (−h/r)x + h.∴ L(x) = 2πx[(−h/r)x + h] = 2πh(−x2/r + x)L′(x) = 2πh(−2x/r + 1)L′(x) = 0 at x = r/2.L′(x) goes from positive to negative atx = r/2.∴ maximum lateral area at radius x = r/2.
b. A(x) = 2πxy + 2πx2
= 2πx[h − (h/r)x] + 2πx2
A(x) = 2π[(1 − h/r)x2 + hx]A′(x) = 2π[2(1 − h/r)x + h] = 0 at
xh
h r= –
( – / )2 1
A′(x) = 0 at xh
h r
rh
h r= =–
– / ( – )2 2 2If h ≤ 2r, then A′(x) ≠ 0 for all x ≤ r, so inthis case the critical points are the endpoints,x = 0, r.A(0) = 0; A(r) = 2πr2
If h ≥ 2r, then 02
≤ ≤rh
h rr
( – ), so this is a
critical point; Arh
h r
rh
h r2 2
2
( – ) ( – )
= π.
A′(x) goes from positive to negative at
xrh
h r=
2( – ).
Maximum area at xrh
h r=
2( – ) if h ≥ 2r;
x = r otherwise.
c. From part b, the maximal cylinder degeneratesto two circular bases if the radius of the coneis at least half the height.
29. Maximize V = π y2x.Ellipse equation is (x/9)2 + (y/4)2 = 1, fromwhich y2 = (16/81)(81 − x2).∴ V = (16π/81)(81x − x3)Domain: 0 ≤ x ≤ 9
150
0 5.196... 10
x
V
The graph shows a maximum V at x ≈ 5.2.V′ = (16π/81)(81 − 3x2) = (16π/27)(27 − x2)V′ = 0 at x = ± = ± …27 5 196.−5.196… is out of the domain.V V V( ) ( ) ; ( ) .0 9 0 27 32 3 174 1= = = = …πAt x = 5.196… , y2 = (16/81)(81 − 27) =32/3 ⇒ y = 32 3/ = 3.2659…
∴ maximum volume ≈ 174.1 cm3 at radius ≈3.27 m and height ≈ 5.20 m.
30. Maximize C(y) = πy2x, the volume of the cylinder.The parabola has an equation of the formx = ay2 + 16.0 = a ⋅ 16 + 16 ⇒ a = −1 ⇒ x = 16 − y2
V(y) = πy2(16 − y2) = π(16y2 − y4)Domain: 0 ≤ y ≤ 4
0 4
300
2.828...
y
F(y)
The graph shows a maximum V(y) at y ≈ 2.8.C′(y) = π(32y − 4y3) = 4πy(8 − y2) = 0 aty = ±0 8, .
y = − 8 is out of the domain.
C C C( ) ( ) , ( ) .0 4 0 8 64 201 0619= = = = …πMaximum C(y) at y = 8.
At y x= =8 8, .
Maximal cylinder has radius = ≈8 2 83. m,height = 8 m, and volume = 64π ≈ 201.1 m3.Maximize F(y), the volume of the frustum.Note that Vf = (1/3)πh(R2 + r2 + Rr), where
184 Problem Set 8-3 Calculus Solutions Manual© 2005 Key Curriculum Press
Vf = volume of frustum, h = height of frustum,R = larger radius, and r = smaller radius.
∴ = + +F y x y y( ) ( )1
316 42π
= − + +1
316 4 162 2π ( )( )y y y
F y y y y( ) ( )= + − −1
3256 64 4 3 4π
0 4
300
1.821...
y
F(y)
The graph shows a maximum F(y) at y ≈ 1.8.
′ = − −F y y y( ) ( )1
364 12 42 3π
F ′(y) = 0 ⇔ 64 − 12y 2 − 4y 3 = 0Solving numerically for y close to 1.8 givesy ≈ 1.8216… .Substituting y = 1.8216… givesx = 16 − y 2 ≈ 12.6816… .
F x y y( . ) ( )1 8216
1
316 42K = + + ≈π
353.318… .Maximal frustum has radii = 4 m and ≈1.82 m,height ≈ 12.68 m, and volume ≈ 353.3 m3.The maximal frustum contains ≈ 152.3 m3 morethan the maximal cylinder, about 75.7% more.
31. a. If f (c) is a local maximum, thenf (x) − f (c) ≤ 0 for x in a neighborhood of c.For x to the left of c, x − c < 0.
Thus, f x f c
x c
( ) – ( )
–≥ 0 (neg./neg.) and
′ = ≥→ −
f cf x f c
x cx( ) lim
( ) – ( )
–.
00
For x to the right of c, x − c > 0.
Thus, f x f c
x c
( ) – ( )
–≤ 0 (neg./pos.) and
′ = ≤→ +
f cf x f c
x cx( ) lim
( ) – ( )
–.
00
Therefore, 0 ≤ f ′(c) ≤ 0.Because f ′(c) exists, f ′(c) = 0 by the squeezetheorem, Q.E.D.
b. If f is not differentiable at x = c, then f ′(c)does not exist and thus cannot equal zero.Without this hypothesis, the reasoning inpart a shows only that f ′(x) changes sign atx = c. There could be a cusp, a removablediscontinuity, or a step discontinuity at x = c.
c. The converse would say that if f ′(c) = 0, thenf (c) is a local maximum. This statement is
false because f (c) could be a local minimumor a plateau point.
32. a. Let x = length of corral (parallel to wall), y =width of corral (perpendicular to wall).A = xyIf x ≤ 600, then 1000 = x + 2y ⇔ y = 500 − 0.5x.If x ≥ 600, then 1000 = x + 2y + (x − 600) ⇔y = 800 − x.
∴ =≤>
Ax x x
x x x
500 0 5 600
800 600
2
2
– . ,
– ,
500
x
150,000A
The graph shows a maximum A at x ≈ 500.
Ax x
x x′ =
<>
500 600
800 2 600
– ,
– ,
For x < 600, A′ = 0 ⇔ x = 500.For x > 600, A′ = 0 ⇔ x = 400 (out of thedomain).A′ is undefined at the cusp, x = 600.Maximum at x = 500 because graph isparabola opening downward.Or: Check the critical points.A(500) = 500(500) − 0.5(500)2 = 125,000A(600) = 500(600) – 0.5(600)2 = 120,000 ft2
Maximum area is 125,000 f t 2 at x = 500 ft.
b. If x ≤ 400, then 1000 = x + 2y ⇔y = 500 − 0.5x.If x ≥ 400, then 1000 = x + 2y + (x − 400)⇔ y = 700 − x.
∴ =≤>
Ax x x
x x x
500 0 5 400
700 400
2
2
– . ,
– ,
400
x
150,000A
The graph shows a maximum A at the cusp,x = 400.
′ =<>
Ax x
x x
500 400
700 2 400
– ,
– ,
For x < 400, A′ = 0 ⇔ x = 500 (out of thedomain).For x > 400, A′ = 0 ⇔ x = 350 (out of thedomain).
Calculus Solutions Manual Problem Set 8-4 185© 2005 Key Curriculum Press
Maximum area is at the cusp, x = 400.A = 700(400) − 4002 = 120,000Maximum area is 120,000 f t 2.
c. If x ≤ 200, then 1000 = x + 2y ⇔y = 500 − 0.5x.If x ≥ 200, then 1000 = x + 2y + (x − 200) ⇔y = 600 − x.
∴ =≤>
Ax x x
x x x
500 0 5 200
600 200
2
2
– . ,
– ,
300
x
150,000A
The graph shows a maximum A at x ≈ 300.
′ =<>
Ax x
x x
500 200
600 2 200
– ,
– ,
For x < 200, A′ = 0 ⇔ x = 500 (out of thedomain).For x > 200, A′ = 0 ⇔ x = 300.A′ is undefined at the cusp, x = 200.Maximum area is at x = 300 because graph isa parabola opening downward.Or: Check critical points.A(300) = 600(300) − 3002 = 90,000A(200) = 500(200) − 0.5(200)2 = 80,000 ft2
Maximum area is 90,000 ft2 at x = 300 ft.
33. Answers will vary.
Problem Set 8-4Q1. Q2.
x
y
x
y
Q3. Q4.
x
y
x
y
Q5. Q6.
x
y
x
y
Q7. Q8. Sample answer:
x
y
2
1
y
x
Q9. tan x C+ Q10. B
1. a. y = 4 − x2
dV = 2πxy ⋅ dx = 2π(4x − x3) dx
b. 0 = 4 − x2 = (2 − x)(2 + x) at x = ±2
V x x dx x x= − = −
∫ 2 4 2 2
1
43
0
22 4
0
2
π π( )
= 8π = 25.1327…
c. y = 4 − x2 ⇒ x2 = 4 − yUpper bound of solid is at y = 4.dV = πx2 dy = π(4 − y) dy
V y dy y y= − = − =∫ π π( ) ( . )4 4 0 50
42
0
4
8π = 25.1327… , which is the same answeras by cylindrical shells in part b.
2. a. Height of cylinder = 8 − x
b. y = x2/3 ⇒ x = y3/2
dV = 2π(8 − x)y dy = 2π(8 − y3/2)y dy= 2π(8y − y5/2) dy
c. At x = 8, y = 82/3 = 4.
V y y dy y y= − = −
∫ 2 8 2 4
2
75 2 2 7 2
0
4
0
4
π π( )/ /
= = …384
7172 3387π .
d. dV = π y2 dx = π x4/3 dx
V x dx x= = = =∫ π π π4 3 7 3
0
8
0
83
7
384
7/ /
172.3387… , which is the same as thevolume by cylindrical shells in part c.
3. The graph shows y = −x2 + 4x + 3, from x = 1 tox = 4, sliced parallel to the y-axis, with samplepoint (x, y), rotated about the y-axis, showingback half of solid only.
(x, y)
1 4
1 x
y
dV = 2πxy ⋅ dx = 2π(−x3 + 4x2 + 3x) dx
V x x x dx= − + +∫ 2 4 33 2
1
4
π ( )
≈ 268.6061… (exactly 85.5π)
186 Problem Set 8-4 Calculus Solutions Manual© 2005 Key Curriculum Press
Circumscribed hollow cylinder of radii 1 and4 and height 7 has volume π(42 − 12) ⋅ 7 =329.8… , which is a reasonable upper bound forthe calculated volume.
4. The graph shows y = x2 − 8x + 17, from x = 2 tox = 5, sliced parallel to the y-axis, with samplepoint (x, y), rotated about the y-axis, showingback half of solid only.
(x, y)
x
y
2 5
5
dV = 2πxy ⋅ dx = 2π(x3 − 8x2 + 17x) dx
V x x x dx= − +∫ 2 8 173 2
2
5
π ( )
≈ 117.8097… (exactly 37.5π)Circumscribed hollow cylinder of radii 2 and 5and height 5 has volume π(52 − 22) ⋅ 5 =329.8… , which is a reasonable upper boundfor the calculated volume. Assuming that thepart of the solid above y = 2 could be fit intothe “trough,” the volume is approximatelyπ(52 − 22) ⋅ 2 = 131.9… , which is close to thecalculated volume.
5. The graph shows x = −y2 + 6y − 5, intersectingy-axis at y = 1 and y = 5, rotated about thex-axis, showing back half of solid only.
(x, y)
x
y
1
5
4
(0, y)
dV = 2πy(x − 0) ⋅ dy = 2π(−y3 + 6y2 − 5y) dy
V y y y dy= − + −∫ 2 6 53 2
1
5
π ( )
≈ 201.0619… (exactly 64π)Circumscribed hollow cylinder of radii 1 and 5and height 4 has volume π(52 − 12) ⋅ 4 =301.5… , which is a reasonable upper bound forthe calculated volume.
6. The graph shows x = y2 − 10y + 24, intersectingy-axis at y = 4 and 6, rotated about the x-axis,showing back half of solid only.
x
y6
–1
(x, y)(0, y)
dV = 2πy(0 − x) ⋅ dy = 2π(−y3 + 10y2 − 24y) dy
V y y y dy= − + −∫ 2 10 243 2
4
6
π ( )
≈ 41.8879… exactly 40
3π
Circumscribed hollow cylinder of radii 4 and 6and height 1 has volume π(62 − 42) ⋅ 1 =62.83… , which is a reasonable upper bound forthe calculated volume.
7. Figure 8-4h shows y = x3, intersecting the liney = 8 at x = 2 and the line x = 1. Rotate aboutthe y-axis. Slice parallel to the y-axis. Picksample points (x, y) on the graph and (x, 8) onthe line y = 8.dV = 2πx(8 − y) ⋅ dx = 2π(8x − x4) dx
V x x dx= −∫ 2 8 4
1
2
π ( )
≈ 36.4424… (exactly 11.6π)Circumscribed hollow cylinder of radii 2 and 1and height 7 has volume π(22 − 12) ⋅ 7 = 65.9… ,which is a reasonable upper bound for thecalculated volume.
8. The graph shows y = 1/x, intersecting line y = 4at x = 0.25 and the line x = 3, rotated about they-axis, showing back half of solid only.
3
x(x, y)
y = 4
0.25
y(x, 4)
dV = 2πx ⋅ (4 − y) · dx = 2π(4x − 1) dx
V x dx= −∫ 2 4 10 25
3
π ( ).
≈ 95.0331… (exactly 30.25π)Circumscribed hollow cylinder of radii 0.25 and 4and height 3.7 has volume π(32 − 0.252)(3.7) =103.8… , which is a reasonable upper bound forthe calculated volume.
9. Figure 8-4i shows y = 1/x2, intersecting the linex = 5 at y = 0.04 and the line y = 4. Rotate
Calculus Solutions Manual Problem Set 8-4 187© 2005 Key Curriculum Press
about the x-axis. Slice parallel to the x-axis. Picksample points (x, y) on the graph and (5, y) onthe line x = 5.dV = 2πy(5 − x) ⋅ dy = 2π(5y − y1/2 ) dy
V y y dy= −
≈ …∫ 2 5 1 2
0 04
4
π
π
( )
217.8254 (exactly 69.336 )
/
.
Circumscribed cylinder of radius 4 and height 4.5has volume π ⋅ 42 ⋅ 4.5 = 226.1… , which is areasonable upper bound for the calculatedvolume.
10. The graph shows y = x2/3 , intersecting the liney = 1 and intersecting the line x = 8 at y = 4,rotated about the x-axis, showing back half ofsolid only.
(8, y)(x, y)
x
y
81
1
4
dV = 2πy(8 − x) ⋅ dy = 2π(8y − y5/2 ) dy
V y y dy= −
≈ …
∫ 2 8
473
7
5 2
1
4
π
π
( )
149.0012 exactly
/
Circumscribed hollow cylinder of radii 1 and 4and height 7 has volume π(42 − 12) ⋅ 7 =329.8… , which is a reasonable upper bound forthe calculated volume.
11. Figure 8-4j shows y1 = x2 − 6x + 7 and y2 =x + 1, intersecting at (1, 2) and (6, 7). Rotateabout the y-axis. Slice parallel to y-axis. Picksample points (x, y1) and (x, y2).dV = 2πx ⋅ (y2 − y1) ⋅ dx = 2π(−x3 + 7x2 − 6x) dx
V x x x dx= − + −
≈ …
∫ 2 7 6
1455
6
3 2
1
6
π
π
( )
458.1489 exactly
Circumscribed hollow cylinder of radii 1 and 6and height 7 has volume π(62 − 12) ⋅ 7 =769.6… , which is a reasonable upper bound forthe calculated volume.
12. The graph shows y x x y= ⇒ =11 3
13/ and y =
0.5x2 − 2 ⇒ x2 = 2y + 4, intersecting at (8, 2)in Quadrant I and bounded by the x-axis, rotatedabout the x-axis, showing back half of solidonly.
2
84
(x1, y)
(x2 , y)
x
y
dV = 2πy(x2 − x1) dy = 2π(2y2 + 4y − y4) dy
V y y y dy= + −
≈
∫ 2 2 4
1313
15
2 4
0
2
π
π
( )
43.5634 exactly K
Circumscribed cylinder of radius 2 and height 8has volume π ⋅ 22 ⋅ 8 = 100.5… , which is areasonable upper bound for the calculatedvolume.
13. Figure 8-4k shows y = x3/2 , from x = 1 to x = 4.Rotate about the line x = 5. Slice parallel to they-axis. Pick sample point (x, y).dV = 2π(5 − x)y ⋅ dx = 2π(5x3/2 − x5/2 ) dx
V x x dx= −
≈ …
∫ 2 5
161 5676 513
7
3 2 5 2
1
4
π
π
( )
. exactly
/ /
Circumscribed cylinder of radius 4 and height 8has volume π(42) ⋅ 8 = 402.1… , which is areasonable upper bound for the calculatedvolume.
14. The graph shows y = x− 2, from x = 1 to x = 2,rotated about the line x = 3, showing back half ofsolid only.
(x, y)
1
1 2 3
x
y
dV x y dx x x dx= − ⋅ = −− −2 3 2 3 2 1π π( ) ( )
V x x dx= −
≈ … −
− −∫ 2 3 2 1
1
2
π
π
( )
5.0696 (exactly (3 2 ln 2))Circumscribed hollow cylinder of radii 1 and 2and height 1 has volume π(22 − 12) ⋅ 1 = 9.4… ,which is a reasonable upper bound for thecalculated volume.
15. The graph shows y1 = x4 and y2 = 5x + 6,intersecting at x = −1 and x = 2, rotated aboutthe line x = 4, showing back half of solidonly.
188 Problem Set 8-4 Calculus Solutions Manual© 2005 Key Curriculum Press
(x, y2)
(x, y1)
–1 2 4
16
x
y
dV = 2π(4 − x)(y2 − y1) dx= 2π(4 − x)(5x + 6 − x4) dx
V x x x dx= − + −
≈ …−∫ 2 4 5 6 4
1
2
π
π
( )( )
390.1858 (exactly 124.2 )Circumscribed hollow cylinder of radii 2 and 5and height 16 has volume π(52 − 22) ⋅ 16 =1055.5… , which is a reasonable upper bound forthe calculated volume.
16. The graph shows y1 = x = x1/2 and y2 = 6 − x,intersecting at x = 4 in Quadrant I and boundedby the line x = 1, rotated about the line x = −1,showing back half of solid only.
(x, y2)
(x, y1) x
41
1
5
y
–1
dV = 2π(x + 1)(y2 − y1) dx= 2π(x + 1)(6 − x − x1/2) dx
V x x x dx= + − −
≈
∫ 2 1 6
109 5368 3413
15
1 2
1
4
π
π
( )( )
. exactly
/
K
Circumscribed hollow cylinder of radii 2 and 5and height 4 has volume π(52 − 22) ⋅ 4 =263.8… , which is a reasonable upper bound forthe calculated volume.
17. Figure 8-4l shows y1 = −x2 + 4x + 1 and y2 =1.4x, intersecting at x = 0 and x = 3.3740…(store as b). Rotate about the line x = −2. Sliceparallel to the y-axis. Pick sample points (x, y1)and (x, y2).dV = 2π(x + 2) ⋅ (y1 − y2) dx
= 2π(x + 2)(−x2 + 4x + 1 − 1.4x) dx
V x x x dxxb
= + − + + −∫ 2 2 4 1 1 42
0π ( )( . )
≈ 163.8592…Circumscribed hollow cylinder with radii 2 and5.4 and height 4 has volume π(5.42 − 22)4 =316.1… , a reasonable upper bound for calculatedvolume.
18. The graph shows y1 and y2 as described inProblem 17, but rotated about the line y = −1,showing back half of solid only. Slicingperpendicular to the x-axis is appropriatebecause slicing parallel to it would give stripsof length (curve) minus (curve) at some valuesof y and (curve) minus (other curve) at othervalues of y.
0 3
1
(x, y1)
(x, y2)
x
y
y = –1
dV = π [(y1 + 1)2 − (y2 + 1)2] dx= π [(−x2 + 4x + 2)2 − (1.4x + 1)2] dx
Limits of integration are 0 to b, whereb = 3.3740… , as in Problem 17.
V x x dxxb
= − + + − +∫ π[( ) ( . ) ] 22 2
04 2 1 4 1
≈ 181.0655…Circumscribed hollow cylinder of radii 2 and 6and height 3.4 has volume π(62 − 22) · 3.4 =341.8… , a reasonable upper bound for thecalculated volume.
19. Slice perpendicular to the y-axis. Pick samplepoints (x, y) on the graph of y = x3 and (1, y)on the line x = 1.y = x3 ⇒ x = y1/3; y1/3 = 1 at y = 1dV = π ( x2 − 12) dy = π (y2/3 − 1) dy
V y dy= − ≈ …∫ π ( ) ./2 3
1
8
1 36 4424
(exactly 11.6π ), which agrees with the answer toProblem 7.
20. See the graph for Problem 8. Sliceperpendicular to the y-axis. Pick sample points(x, y) on the graph of y = 1/x and (3, y) on theline x = 3.dV x dy y dy= − = − −π π( ) ( ) 3 92 2 2
V y dy= − ≈ …−∫ π ( ) .9 95 03312
1 3
4
/(exactly
30.25π ) , which agrees with the answer toProblem 8.
21. The graph shows y = x1/3, from x = 0 to x = 8,rotated about the x-axis, showing back half ofsolid only.
Calculus Solutions Manual Problem Set 8-5 189© 2005 Key Curriculum Press
(8, y)(x, y)
8
2
x
y
y = x1/3 ⇒ x = y3
dV = 2πy ( 8 − x) ⋅ dy = 2π ( 8y − y4) dy
V y y dy y y= − = −
∫ 2 8 2 4
1
54 2 5
0
2
0
2
π π( )
= 2π ( 16 − 6.4) = 19.2π = 60.3185789…R8 = 19.3662109… π = 60.8407460…R100 = 19.2010666… π = 60.3219299…R1000 = 19.2000106… π = 60.3186124…Rn is approaching 19.2π as n increases.
22. a. y = sin x from x = 0 to x = 2, rotated aboutthe y-axis, as in Figure 8-4m. Slice parallelto the y-axis. Pick sample point (x, y) onthe graph.dV = 2πxy ⋅ dx = 2πx sin x dx
V x x dx= ≈ …∫ 2 10 94270
π sin .2
numerically (exactly 2π ( sin 2 − 2 cos 2),integrating by parts).
b. The integrand, x sin x, is a product of twofunctions, for which the antiderivative cannotbe found using techniques known so far.
23. a. x = 5 cos t, dx = −5 sin t dty = 3 sin t, dy = 3 cos t dtSlice parallel to the x-axis, then rotate aboutthe x-axis. Pick sample points (−x, y) at theleft end of the strip and (x, y) at the right end.dV = 2πy [ x − (−x)] ⋅ dy = 4πxy dy
= 4π ( 5 cos t)(3 sin t)(3 cos t dt)= 180π cos2 t sin t dt
Limits of integration are y = 0 to y = 3.At y = 0, t = 0. At y = 3, t = π/2.
V t t dt= ∫ 180 2
0
2
ππ
cos sin/
= −60 30
2π πcos
/t
= −60π ( 0 − 1) = 60π = 188.4955…b. Slice the region in Quadrant I perpendicular to
the x-axis, then rotate about the x-axis. Picksample point (x, y) on the graph.dV = πy2 dx = π ( 3 sin t)2(−5 sin t dt)
= −45π sin3 t dtLimits of integration are from x = −5 tox = 5.At x = −5, t = π. At x = 5, t = 0.
V t dt= − ≈ …∫ 45 188 495530
ππ
sin .
(exactly 60π ) , which agrees with the volumefound in part a.
The integral can be found algebraicallyusing the Pythagorean properties fromtrigonometry.sin3 t = (1 − cos2 t) sin t = sin t − cos2 t sin t
V t dt t t dt= − − −∫∫ 45 45 20
π πππ
sin cos sin0
( )
= −45 15 3 0π π πcos cost t
= 45π − 15π − (−45π ) + (−15π ) = 60πc. Slice the region parallel to the line x = 7 and
rotate about that line. Pick sample points(x, y) and (x, −y) on the upper and lowerbranches.dV = 2π ( 7 − x)[y − (−y)] dx
= 4π ( 7 − 5 cos t)(3 sin t)(−5 sin t dt)= −60π ( 7 − 5 cos t)(sin2 t) dt
Limits of integration are t = π to t = 0, as inpart b.V ≈ 2072.6169… (exactly 210π 2, using thehalf-argument properties for sin2 t, as inProblem 16 of Problem Set 5-9, or by usingintegration by parts as in Chapter 9).
24. Answers will vary.
Problem Set 8-5Q1. Q2.
4
16
x
y
0 1 4
16
x
y
Q3. A x dx= ∫ 2
1
4
Q4. A x= 1
33
1
4
Q5. A = 21
Q6.
4
16
x
y
Q7. V x dx= ∫ 2 3
1
4
π Q8. V x= π2
4
1
4
Q9. V = 127.5π Q10. E
1. a.y
x
2
1
0
190 Problem Set 8-5 Calculus Solutions Manual© 2005 Key Curriculum Press
b. L e en n
n
≈ +
= …=
∑ ( . ) [ – ]. . ( – )0 4 2 0 4 0 4 1 2
1
5
6.7848c. dy = e x dx
dL dx dy e dxx= + = +2 2 21
L e dxx= + ≈ …∫ 1 6 78862
0.
2
numerically
2. a.y
x
3
1
0
b. L n n
n
≈ +
= …=
∑ ( . ) [ – ]. . ( – )0 6 2 22 0 6 0 6 1 2
1
5
7.7853c. dy = (2x ln 2) dx
dL dx dy dxx= + = +2 2 21 2 2( ln )
L dxx= + ≈∫ 1 2 2 7 79202
0
3
( ln ) . K
numerically
3. a.
1.50
10
x
y
b. L n nn
≈ +=
∑ ( . ) [tan . – tan . ( – )]0 3 0 3 0 3 12 2
1
5
= 14.4394…
c. dy = sec2 x dx
dL dx dy x dx= + = +2 2 41 sec
L x dx= + ≈∫ 1 14 44884
0
1 5
sec.
. K
numerically
4. a.
1.50
10
x
y
1
b. L n nn
≈ +=
∑ ( . ) [sec . – sec . ( – )]0 3 0 3 0 3 12 2
1
5
= 13.7141…
c. dy = sec x tan x dx
dL dx dy x x dx= + = +2 2 2 21 tan sec
L x x dx= + ≈ …∫ 1 2 2
0
1 5
tan sec/
13.7304
numerically
5. a.
6–1
9
1 x
y
b. dy = (2x − 5) dx
dL dx dy x dx= + = +2 2 21 2 5( – )
L x dx= + ≈∫ 1 2 5 15 86172
1
6
( – ) . K
c. Low point is (2.5, −3.25). Chords from(1, −1) to (2.5, −3.25) and from (2.5, −3.25)to (6, 9) have combined length 7 3125. +
162 3125 15 4. = …. , which is a reasonablelower bound for L.
6. a.
0 4
4
x
y
b. dy = (4 − 2x) dx
dL dx dy x dx= + = +2 2 21 4 2( – )
L x dx= + ≈∫ 1 4 2 9 29352
0
4
( – ) . K
c. Chords from (0, 0) to (2, 4) and from(2, 4) to (4, 0) have combined length2 20 8 9442= …. , which is a reasonablelower bound for L.
7. a.
2
16
–1
x
y
Calculus Solutions Manual Problem Set 8-5 191© 2005 Key Curriculum Press
b. dy = −4x3 dx
dL dx dy x dx= + = +2 2 61 16
L x dx= + ≈
−∫ 1 16 18 24706
1
2
. K
c. Chords from (−1, 15) to (0, 16) and (0, 16) to
(2, 0) have combined length 2 260+ =17.5… , which is a reasonable lower boundfor L.
8. a.
–1 9
50
x
y
b. dy = (3x2 − 18x + 5) dx
dL dx dy= +2 2
= + +1 3 18 52 2( – )x x dx
L x x dx= + + ≈−∫ 1 3 18 5 219 48732 2
1
9
( – ) . K
c. Using five chords with ∆x = 2, L ≈204.4605… , which is a reasonable lowerbound for L.
9. a.
1 e
1x
y
b. dy x x dx x x dx= ⋅ =− −2 21 1ln ln
dL dx dy x x dx= + = +2 2 1 21 2( ln )–
L x x dxe
= + ≈ …∫ 1 2 7 60431 2
0 1( ln )–
..
c. Chords from x = 0.1 to x = 1 and from x = 1to x = e have combined length 7.3658… ,which is a reasonable lower bound for L.
10. a.
5
5 4π0
x
y
b. dy = (sin x + x cos x) dx
dL dx dy= +2 2
= + +1 2(sin cos )x x x dx
L x x x dx= + + ≈∫ 1 2
0
4
(sin cos )π
54.1699K
c. Eight chords of ∆x = π/2 extend from middleto high to middle to low points on the graph.Lengths sum to 52.6109… , a reasonablelower bound for L.
11. a.
10
1.50
x
y
b. dy = sec2 x dx
dL dx dy x dx= + = +2 2 41 sec
L x dx= + ≈∫ 1 14 44884
0
1 5
sec.
. K
c. Distance between the endpoints is14.1809… , which is a reasonable lowerbound for L.
12. a.
10
1.50
x
y
b. dy = sec x tan x dx
dL x x dx= +1 2(sec tan )
L x x dx= + ≈∫ 1 2
0
1 5
(sec tan ).
13.7304K
c. The distance between the endpoints is13.2221… , which is a reasonable lowerbound for L.
13. a.
5
5y
x
192 Problem Set 8-5 Calculus Solutions Manual© 2005 Key Curriculum Press
b. dx = −15 cos2 t sin t dt, dy = 15 sin2 t cos t dt
dL dx dy= +2 2
= +(– cos sin ) ( sin cos )15 152 2 2 2t t t t dt
L t t t t dt= +∫ (– cos sin ) ( sin cos )15 152 2 2 2
0
2π
≈ 30
To see why the answer is so simple,transform the radicand and use thefundamental theorem.
L t t t t dt= +∫ 225 2 2 2
0
2
(sin cos ) (cos sin )π
= ∫7 5 2 2. 0
2
( sin cos )t t dtπ
= ∫7 5 22
0
2
. sin t dtπ
= = ⋅∫ ∫7 5 2 7 5 4 20
2
0
2
. | | ./
sin sint dt t dtπ π
=
− =30
1
22 30
0
2
( ) (exactly!)cos/
tπ
c. Circle of radius 5 (i.e., x = 5 sin t, y =5 cos t) has circumference 10π = 31.4152… ,which is close to the calculated value of L.
14. a.
5
10
x
y
b. dx = 5(−2 sin t + 2 sin 2t) dtdy = 5(2 cos t − 2 cos 2t) dt
dL dx dy= + =2 2
[ (– sin sin )] [ ( cos – cos )]5 2 2 2 5 2 2 22 2t t t t+ + dt
L =
[ (– sin sin )] [ ( cos – cos )]5 2 2 2 5 2 2 22 2
0
2
t t t t dt+ +∫π
≈ 80To see why the answer is so simple,transform the radicand algebraically anduse the fundamental theorem.
L t t t t dt= ∫10 2 2 2 2 2– sin sin – cos cos0
2π
= −∫10 2 10
2
– cos cost dt A Bπ
(using ( ))
=+∫10 2
1
1
2
0
2 – cos
cos
t
tdt
π
=+∫10 2
10
2 |sin |
cos
t
tdt
π
= + −∫20 2 1 1 2
0( cos ) sint t dt/ ( )
π
= − +40 2 1 1 2
0( cos )t / π
= − + + =40 2 1 1 40 2 1 1 801 2 1 2( – ) ( )/ /
c. Maximum/minimum values of y are±7 5 3. . Circle of radius 7.5 3 has
circumference 15 3 81 6209π = . .K
15. a.
4
4
x
y
b. dx = (−5 sin t + 5 sin 5t) dtdy = (5 cos t − 5 cos 5t) dt
dL dx dy= +2 2 =
(– sin sin ) ( cos – cos )5 5 5 5 5 52 2t t t t dt+ +
L t t t t dt= + +∫ (– sin sin ) ( cos – cos )5 5 5 5 5 52 2
0
2π
≈ 40To see why the answer is so simple,transform the radicand and use thefundamental theorem.
L t t t t dt= ∫5 2 2 5 2 5 0
2
– sin sin – cos cosπ
= −∫5 2 1 40
2
– cos cos ( )t dt A B(using )π
=+∫5 2
1 4
1 4
2
0
2 – cos
cos
t
tdt
π
=+∫5 2
4
1 40
2 |sin |
cos
t
tdt
π
= + −∫40 2 1 4 41 2
0
4
( cos ) sin/
t t dt/ ( )π
= − +20 2 1 4 1 20
4( cos )
/t / π
= − + ⋅ =0 20 2 2 40
c. Maximum/minimum values of x, y are±3 3. Circle of radius 3 3 has circumference32.6483… , which is close.
Calculus Solutions Manual Problem Set 8-5 193© 2005 Key Curriculum Press
16. a.
t = 4π
5
5
x
y
b. dx = (−sin t + sin t + t cos t) dt = t cos t dtdy = (cos t − cos t + t sin t) dt = t sin t dt
dL dx dy t t t t dt
t dt t dt t
= + = += = ≥
2 2 2 2( cos ) ( sin )
| | (because 0)
L t dt t= = = =∫ 0 5 8 78 95682
0
42. .
0
4π ππ K
c. Circle of radius 4π = 12.5663… would havecircumference = 8π 2.
17. a.
40
30
x
y
b. dy = 6x1/2 dx
dL dx dy x dx= + = +2 2 1 36
L x dx x dx= + = +∫ ∫1 361
361 36 36
0
41 2
4
( ) /
0 ( )
= + =1
541 36
1
54145 13 2
0
4 3 2( ) ( – )/x /
= 32.3153…
c. The chord connecting the endpoints has length32.2490… , which is a reasonable lowerbound for L.
18. a.
1
1 2
x
y
b. dy x x dx= − −( / ) 2 24
dL dx dy x x dx= + = +2 2 2 2 21 4( / – )–
= + +1 16 1 24 4x x dx/ – / –
= + = + −( / )–x x dx x x dx2 2 2 2 24 4| / |
L x x= +∫ ( / )–2 2
1
2
4 dx (because integrand > 0)
= − = =−x x3 1
1
212 1
1
121 0833/ . K
c. Distance between endpoints is
1 006944 6. K K= 1.0034 , which is areasonable lower bound for L.
19. a.
5
1 8
x
y
b. dy x dx= −2 1 3/
dL dx dy x dx= + = +2 2 2 31 4 – /
L x dx= +∫ 1 4 2 3
1
8– /
= + −∫ ( )/ /x x dx2 3 1 2 1 3
1
8
4 / ( )
= +
∫ −3
24
2
32 3 1 2
1
81 3( )/ /x x dx/
= ⋅ +3
2
2
342 3 3 2
1
8
( )/x /
= − =8 8 5 5 11 4470. K
c. Distance between endpoints is 130 =11.4017… , which is a reasonable lowerbound for L.
20. a.
0 3
5
x
y
b. dy x x dx x x dx= + = +1
22 2 22 1 2 2 1 2( ) / /( )
dL dx dy x x dx= + = + +2 2 2 21 2( )
= + + = +1 2 14 2 2x x dx x dx( )
L x dx x x= + = + =∫ ( )11
3122 3
0
3
0
3
c. Distance between endpoints is 11.6123… ,which is a reasonable lower bound for L.
21. Construct an x-axis at water level and a y-axisthrough the vertex of the parabola.
–2100 2100
x
y
750
220
0
194 Problem Set 8-5 Calculus Solutions Manual© 2005 Key Curriculum Press
General equation is y − 220 = ax2.Substitute (2100, 750) for (x, y).
750 220 210053
4410002− = ⋅ ⇒ =a a
Equation of parabola is y x= +53
4410002202 .
dy x dx= 106
441000
dL dx dy x dx= + = +2 2 2 21 106 441000( / )
L x dx= +∫ 1 106 441000 2 2
2100
2100
( / )–
≈ 4372.0861… numerically ≈ 4372 feet.The answer is reasonable because the 4200 feetbetween supports is a lower bound for L.
22. y e e dy e e dxx x x x= + = −− −0 2 0 2. ( ), . ( )
dL dx dy
e e dxx x
= +
= +
2 2
21 0 04. ( – )–
L e e dxx x= +
≈ … ≈−∫ 1 0 04 2
4
4
. ( – )–
24.1722 24.2 ftThe parabola with vertex (0, 0.4) and endpoints(±4, 0.2(e4 + e−4)) = (±4, 10.9232…) has equationy = ax2 + 0.4. Substituting (4, 10.9232…) gives10.9232… = 16a + 0.4 ⇒ a = 0.6577… .y = 0.6577… x2 + 0.4 ⇒ dy = 1.3154… x dx
dL dx dy x dx= + = +2 2 21 1 7303. ...
L x dx= + ≈ … ≈−∫ 1 1 7303 23 21932
4. K . 23.2 ft,
4
which is about a foot shorter than the catenary,as shown by graph:
4–4
10
x
y
23. Outer ellipse:x = 120 cos t, dx = −120 sin t dty = 100 sin t, dy = 100 cos t dt
dL dx dy
t t dt
= +
= +
2 2
2 2120 100(– sin ) ( cos )
L t t dt= +
≈ … ≈∫ (– sin ) ( cos )120 1002 2
0
2π
692.5791 692.6 mInner ellipse:x = 100 cos t, dx = −100 sin t dty = 50 sin t, dy = 50 cos t dt
dL dx dy
t t dt
= +
= +
2 2
2 2100 50(– sin ) ( cos )
L t t dt= +
≈ … ≈∫ (– sin ) ( cos )100 502 2
0
2π
484.4224 484.4 m
24.
8
x
y5
–8
–5
dx = −16 sin 2t, dy = 5 cos t
dL dx dy
t t dt
= +
= +
2 2
2 216 2 5(– sin ) ( cos )
Curve appears to have length
L t t dt= +∫ (– sin ) ( cos )16 2 52 2
0
2π
= 68.7694…Length should be less than the lengths of threecircumscribing segments, 16 + 16 + 10 = 42.The discrepancy is explained by the fact that theparabola is traced twice as t goes from 0 to 2π.Actual length ≈ (0.5)(68.7694…) = 34.384… ,for which 42 is a reasonable upper bound.
25. 9 42
32 3 3 2x y x y= ⇔ = ± / .
3
0 3 4
x
y
dx = y1/2 dy
dL dx dy y dy= + = +2 2 1 21( ) /
L y dy y= + = +
= = …
∫ ( ) ( )12
31
42
34 6666
1 2 3 2
0
3
0
3/ /
.
26. x2 = y3 ⇔ x = ±y1.5
2x dx = 3y2 dy ⇒ 4x2 dx2 = 9y4 dy2
⇒ = ⇒ =4 99
43 2 4 2 2 2y dx y dy dx y dy
Note that dy < 0 between (−1, 1) and (0, 0):
–1 8
x
y
1
4
Calculus Solutions Manual Problem Set 8-5 195© 2005 Key Curriculum Press
For x in [−1, 0], x = −y1.5 , dx = −1.5y0.5 dy,
dL dx dy y dy= − + = − +2 2 2 25 1. .
For x in [0, 8], x = y1.5 , dx = 1.5y0.5 dy,
dL dx dy y dy= + = +2 2 2 25 1. .
L y dy y dy= − + + +∫∫ 2 25 1 2 25 10
4
. .1
0
= − + + +8
272 25 1
8
272 25 13 2 3 2( . ) ( . )y y/
1
0/
0
4
= + + = …8
271 3 25 10 1 10 51313 2 3 2(– . – )/ / .
27. xt
t dx t t t dt= =π π
cos (cos – sin ), 1
yt
t dy t t t dt= = +π π
sin (sin cos ), 1
dL dx dy= +2 2
= + +1 2 2
π(cos – sin ) (sin cos )t t t t t t dt
= +11 2
πt dt
The curve crosses the x-axis exactly when sin t= 0, when t is a multiple of π. There are sevencrossings after the beginning, so t should runbetween 0 and 7π. To check this, note that thecurve ends at (−7, 0), so solve (t/π) cos t = −7with t = nπ ⇒ (nπ/π ) cos nπ = −7 ⇒n cos nπ = −7 ⇒ n = 7 ⇒ 0 ≤ t ≤ 7π.
L t dt= + ≈ …∫11 77 65082
0
7
π
π.
The integral can be evaluated algebraically bytrigonometric substitution as in Section 9-6,giving
11
21 12 2 2+ = + + + +( )
+∫ t dt t t t t Cln .
28. x = r cos t, dx = −r sin t dty = r sin t, dy = r cos t dt
dL dx dy r t r t dt= + = +2 2 2 2 2 2sin cos
= r dt (for r ≥ 0)The range 0 ≤ t ≤ 2π generates the entire circle.
Circumference = = =∫ r dt rt r0
2
0
2
2π π
π , Q.E.D.
29. y = A sin x, dy = A cos x dx
dL dx dy A x dx= + = +2 2 2 21 cos
Pick a convenient interval for x such as [0, 2π].
L A x dx= +∫ 1 2 2cos0
2π
A L
0 6.283185… (= 2π)
1 7.640395…
2 10.540734…
3 13.974417…
Doubling A doubles the amplitude of thesinusoid. However, it less than doubles thelength of the sinusoid for much the same reasonthat doubling one leg of a right triangle does notdouble the hypotenuse. In the limit as Aapproaches infinity, doubling A approachesdoubling the length.
30. x = cos t, dx = −sin t dty = A sin t, dy = A cos t dt
dL dx dy t A t dt= + = +2 2 2 2 2sin cos
The entire ellipse is generated as t increases from0 to 2π.
L t A t dt= +∫ sin cos2 2 2
0
2π
A L
0 4 (a double line segment)
1 6.283185… (= 2π)
2 9.688448…
3 13.364893…
Doubling A doubles one axis of the ellipsewithout changing the other axis. That is why thelength does not double when A doubles. Thereasoning is similar to that in the solution toProblem 29.
31. The function y x= − −( )2 1 has a vertical
asymptote at x = 2, which is in the interval[1, 3]. So the length is infinite. Mae’s partitionof the interval skips over the discontinuity, asshown in the graph.
1 2 3
x
y25
Mae'serror
32. The sample points are all of the form (n/2,sin nπ), which all lie on the x-axis and thereforefail to measure the wiggly bits.
100
x
y
1
Amos's sample points
The length of the curve is 40 times the length ofthe part from x = 0 to x = 0.25 (by symmetry),so Amos could use five subintervals of [0, 0.25]to estimate the length of half of one arch, then
196 Problem Set 8-6 Calculus Solutions Manual© 2005 Key Curriculum Press
multiply his answer by 40 to find the totallength.
33. See the Programs for Graphing Calculatorssection of the Instructor’s Resource Book.
Problem Set 8-6
Q1. 1 9 4+ x dx Q2. 1 4+ sec x dx
Q3.1
66sin x C+ Q4. 156
Q5. xex + ex
Q6. Maximum y = 7 (at x = 1)
Q7. f xf x x f x
xx′ = + ∆
∆→( ) lim
( ) – ( )∆ 0
or ( )f cf x f c
x cx c′ =
→lim
( ) – ( )
–
Q8. Instantaneous rate of change
Q9.1
22 2ln sec tan| |x x C+ +
Q10. D
1. a. The graph shows y = 0.5x2, from x = 0 tox = 3, rotated about the y-axis.
(x, y)
30
x1
y
dy = x dx
dL dx dy x dx= + = +2 2 21
dS x dL x x dx= ⋅ = +2 2 1 2π π
S x x dx= + ≈ …∫ 2 1 64 13612
0
3
π .
b. The inscribed cone of height 4.5 and radius3 has lateral surface area = πrL =π · 3 · 3 4 5 50 97222 2+ = …. . , which is
a reasonable lower bound for S.
c. S x x dx= +∫ π ( ) ( )1/21 22
0
3
= + = −2
31
2
310 10 12
0
3
π π( ) ( )3/2x
= 64.1361… , agreeing with the answerfound numerically.
2. a. The graph shows y = sin x, from x = 0 tox = π, rotated about the x-axis.
0 π
(x, y)
x
y
1
b. dy = cos x dx
dL dx dy x dx= + = +2 2 21 cos
dS y dL x x dx= ⋅ = +2 2 1 2π π sin cos
S x x dx= + ≈ …∫ 2 1 14 42352
0π
πsin cos .
c. The circumscribed cylinder of length πand radius 1 has lateral area = 2π 2 =19.7392… , which is a reasonable upperbound for S.
3. The graph shows y = ln x, from x = 1 to x = 3,rotated about the x-axis.
31
1
x
y
(x, y)
dy x dx= −1
dL dx dy x dx= + = +2 2 21 –
dS y dL x x dx= ⋅ = +2 2 1 2π π ln –
S x x dx= + ≈ …∫ 2 1 9 02422
1
3
π ln – .
4. The graph shows y = ln x, from x = 1 to x = 3,rotated about the y-axis, showing back half ofsurface only.
1 3
1
x
y
(x, y)
dL x dx= + −1 2 , from Problem 3.
dS x dL x x dx= ⋅ = +2 2 1 2π π –
S x x dx= + ≈−∫ 2 1 28 30472
1
3
π . K
5. The graph shows y x x= = −1 1/ ,from x = 0.5 to
x = 2, rotated about the y-axis.
2
2
x
y
(x, y)
dy x dx= − −2
dL dx dy x dx= + = +2 2 41 –
Calculus Solutions Manual Problem Set 8-6 197© 2005 Key Curriculum Press
dS x dL x x dx= ⋅ = +2 2 1 4π π –
S x x dx= + ≈∫ 2 1 15 51814
0 5
2
π –
.. K
6. The graph shows y = 1/x = x− 1, from x = 0.5 tox = 2, rotated about the x-axis.
2
2
x
y
(x, y)
dL x dx= +1 4– , from Problem 5.
dS y dL x x dx= ⋅ = +2 2 11 4π π – –
S x x dx= + ≈−∫ 2 1 15 51811 4
0 5
2
π –
.. K
(Note that surfaces 5 and 6 are congruent.)
7. The graph shows y = x3, from x = 0 to x = 2,rotated about the y-axis.
(x, y)
2
8
0
x
y
dy = 3x2 dx
dL dx dy x dx= + = +2 2 41 9
dS x dL x x dx= ⋅ = +2 2 1 9 4π π
S x x dx= + ≈∫ 2 1 9 77 32454
0
2
π . K
8. The graph shows y = −x3 + 5x2 − 8x + 6, fromx = 0 to x = 3, rotated about the y-axis.
(x, y)
x
y
3
6
dy = (−3x2 + 10x − 8) dx
dL dx dy= +2 2
= + +1 3 10 82 2(– – )x x dx
dS = 2π x ⋅ dL
= + +2 1 3 10 82 2π x x x dx(– – )
Graph intersects x-axis where y = 0.−x3 + 5x2 − 8x + 6 = −(x − 3)(x2 − 2x + 2) = 0at x = 3.
S x x x dx= + +∫ 2 1 3 10 82 2
0
3
π (– – )
≈ 58.7946…9. The graph shows y x= = x1/2, from x = 0 to
x = 1, rotated about the x-axis.
(x, y)
0 1
1
x
y
dy x dx
dL dx dy x dx
=
= + = +
−0 5
1 0 25
1 2
2 2 1
. /
. –
dS y dL x x dx= ⋅ = +2 2 1 0 251 2 1π π / –.
= + = +2 0 25 2 0 25 1 2π πx dx x dx. ( . ) /
S x dx x= + = +∫ 2 ( 0.25)4
3( 0.25) 1/2 3/2
0
1
0
1
π π
= − =4
31 25 0 125 5 33043 2π
( . . ) ./ K
10. The graph shows y = x3, from x = 1 to x = 2,rotated about the x-axis, showing back half ofsurface only.
(x, y)
1 2
1
8 y
x
dy = 3x2 dx
dL dx dy x dx= + = +2 2 41 9
dS = 2πy ⋅ dL = 2πx3 (1 + 9x4)1/2 dx
S x x dx= +∫ 2 (1 9 )1/2π 3 4
1
2
= +∫π18
( ) ( )/1 9 364 1 2
1
2
x x dx
= ⋅ +π18
2
31 9 4 3 2
1
2
( ) /x
= =π
27145 10 199 48043 2 3 2( – )/ / . K
11. The graph shows y x x= + −4 28 4/ / , from x = 1
to x = 2, rotated about the x-axis, showing backside of surface only.
198 Problem Set 8-6 Calculus Solutions Manual© 2005 Key Curriculum Press
1 2
2
x
y
dy = (x3/2 − x− 3/2) dx = 0.5(x3 − x− 3) dx
dL dx dy x x dx= + = +2 2 3 3 21 0 25. ( – )–
= + +1 0 25 0 5 0 256 6. – . . –x x dx
= + = + −0 25 0 53 3 2 3 3. ( )–x x dx x x dx. ( )
dS = 2π y ⋅ dL= 2π ( x4/8 + x− 2/4)[0.5(x3 + x− 3)] dx
= + +π8
3 27 5( )–x x x dx
S x x x dx= + +∫π8
3 27 5
1
2
( )–
= +
π8
1
8
3
2
1
28 2 4
1
2
x x x– –
= + +
=
= …
π π8
32 61
32
1
8
3
2
1
24
155
256
14 4685
– – –
.12. The graph shows y = x2, from x = 0 to x = 2,
rotated about the y-axis.
(x, y)
x
y
4
0 2
dy x dx
dL dx dy x dx
=
= + = +
2
1 42 2 2
dS x dL x x dx= ⋅ = +2 2 1 4 2π π
S x x dx= +∫2 1 4 2
0
2
π
= + = +∫π π4
1 4 86
1 42 1 2
0
22 3 2
0
2
( ) ( ( )/ /x x dx x)
= − =π
617 1 36 17693 2( ) ./ K
13. The graph shows y x= +1
322 3 2( ) ,/ from x = 0 to
x = 3, rotated about the y-axis.
30
1
12
(x, y)
x
y
dL = (1 + x2) dx, from Problem 20 in Section 8-5
dS = 2πx ⋅ dL = 2π ( x + x3) dx
S x x dx x x= + = +
∫ 2
1
23 2 4
0
3
0
3
π π( )
= 49.5π = 155.5088…
14. The graph shows y = 2x1/3, from x = 1 tox = 8, rotated about the y-axis, showing backhalf of surface only.
4
2
1 8
x
y
(x, y)
dy x dx= −2
32 3/
dL dx dy x dx= + = +2 2 4 314
9– /
dS x dL x x dx= ⋅ = +2 2 14
94 3π π – /
S x x dx= +∫2 14
94 3
1
8
π – /
= +
∫2
4
91 3 4 3
1
8 1 2
π x x dx/ //
= +
⋅∫3
2
4
9
4
34 3
1 21 3
1
8
π x x dx//
/
= +
π x 4 3
3 2
1
84
9/
/
= − =π27
148 13 204 04353 2 3 2( ) ./ / K
15. The graph shows y x x= + −1 1
43 1
3, from x = 1 to
x = 3, rotated about the line y = −1, showingback half of surface only.
(x, y)
1
9
(x, –1)
3
x
y
Calculus Solutions Manual Problem Set 8-6 199© 2005 Key Curriculum Press
dx x x dx= −
−2 21
4
dL dx dy x x dx= + = + −
−2 2 2 22
11
4
= + − + −11
2
1
164 4x x dx
= +
= +
− −x x dx x x dx2 22
2 21
4
1
4dS = 2π ( y + 1) ⋅ dL
= + +
+
− −21
3
1
41
1
43 1 2 2π x x x x dx
= + + + +
− −21
3
1
3
1
4
1
165 2 2 3π x x x x x dx
S x x x x x dx
x x x x x
= + + + +
= + + − −
= =
− −
− −
∫21
3
1
3
1
4
1
16
21
18
1
3
1
6
1
4
1
32
1015
18318 1735
5 2 2 3
1
3
6 3 2 1 2
1
3
π
π
π . K
16. The graph shows y x x= + −1
3
1
43 1, from x = 1 to
x = 3, rotated about line x = 4.
(x, y)
(4, y)
1 3 4
9
x
y
dL x x dx= +
−2 21
4, from Problem 15
dS x dL x x x dx= − ⋅ = − +
−2 4 2 41
42 2π π( ) ( )
= − + −
− −2 41
42 3 2 1π x x x x dx
S x x x x dx= − + −
− −∫2 41
42 3 2 1
1
3
π
= − − −
−24
3
1
4
1
43 4 1
1
3
π x x x xln | |
= −
=2 15
1
3
1
43 94 6164π ln . K
17. a. x y y x2 2 225 25+ = ⇒ = −dy x x dx= − − −( )25 2 1 2/
dL dx dy x x dx= + = + − −2 2 2 2 11 25( )
dS = 2πy ⋅ dL
= − + − −2 25 1 252 2 2 1π x x x dx( )
= − + =2 25 102 2π πx x dx dx
b. i. S dx x0 10
1
0
1
10 10 10, = = =∫ π π π
ii. S dx x1 21
2
1
2
10 10 10, = = =∫ π π π
iii. S dx x2 32
3
2
3
10 10 10, = = =∫ π π π
iv. S dx x3 43
4
3
4
10 10 10, = = =∫ π π π
v. S dx x4 54
5
4
5
10 10 10, = = =∫ π π π
c. The two features exactly balance each other.The area of a zone of a sphere is a functionof the height of the zone only, and isindependent of where the zone is locatedon the sphere.
18. Suppose that the sphere is centered at the origin,as in Problem 17. The equation of a greatcircle in the xy-plane is x2 + y2 = r2, from
which y r x r x= − = −2 2 2 2 1 2( ) ./
dy = −x(r2 − x2)− 1/2dx
dL dx dy x r x dx= + = + − −2 2 2 2 2 11 ( )dS y dL
r x x r x dx
r x x dx r dx r
= ⋅
= − + −
= − + = >
−
2
2 1
2 2 0
2 2 1 2 2 2 2 1
2 2 2
π
π
π π
( ) ( )
( )
/
if
S r dx rx rr
r
r
r
= = =− −∫ 2 2 4 2π π π , Q.E.D.
19. Pick a sample point in the spherical shell atradius r from the center. Surface area at thesample point is 4π r2. Volume of shell isapproximately (surface area)(thickness).dV r dr= ⋅4 2π
V r dr r RR R
= = =∫ 44
3
4
32
0
3
0
3π π π , Q.E.D.
20. V rdV
drr S= ⇒ = =4
343 2π π , Q.E.D.
or: V S drdV
drS= ⇒ =∫ by the definition of
indefinite integral.
21. y = ax2, dy = 2ax dx
dL dx dy a x dx= + = +2 2 2 2 1 21 4( ) /
dS x dL x a x dx= ⋅ = +2 2 1 4 2 2 1 2π π ( ) /
S x a x dxr
= +∫2 1 4 2 2 1 2
0π ( ) /
= +∫π4
1 4 822 2 1 2 2
0aa x a x dx
r
( ) ( )/
= + = + −π π6
1 46
1 4 122 2 3 2
02
2 2 3 2
aa x
aa r
r
( ) [( ) ]/ /
200 Problem Set 8-6 Calculus Solutions Manual© 2005 Key Curriculum Press
22. Let h be the height of the paraboloid from thevertex to the center of the base. Because h is thevalue of y when x = r, h = a r 2. Substituting intothe formula for S from Problem 21 gives
Sa
ah= + −π6
1 4 123 2[( ) ]/
Let a = 1 and evaluate S for various h. Find thezone areas by subtracting. Use the TABLE feature.
h S Zone
0π6
0( ) N.A.
1
π6
10 1803( . )K
π6
10 1803( . )K
2π6
26( )π6
15 8196( . )K
3
π6
45 8721( . )K
π6
19 8721( . )K
4
π6
69 0927( . )K
π6
23 2206( . )K
5π6
95 2340( . )Kπ6
26 1412( . )K
6 π6
124( )
π6
28 7659( . )K
The property is not true for paraboloids. Theareas of zones of equal height are greater if thezone is farther away from the vertex.
23. x = 5 cos t, dx = −5 sin t dty = 3 sin t, dy = 3 cos t dt
5
3(x, y)
x
y
dL dx dy t t dt= + = − +2 2 2 25 3( sin ) ( cos )dS y dL= ⋅2π
= − +2 3 5 32 2π ( sin ) ( sin ) ( cos )t t t dt
S t t t dt= − +
≈ …∫ 6 5 32 2
0π
πsin ( sin ) ( cos )
165.7930
From ( / ) ( / ) , . .x y y x5 3 1 0 6 252 2 2+ = = ± − Using the upper branch of the graph,dy x x dx= − − −0 6 25 2 1 2. ( ) ./
dL dx dy x x dx= + = + − −2 2 2 2 11 0 36 25. ( )
At x = ±5, dL involves division by zero, whichis awkward, and makes the Cartesian equation
inappropriate for finding the arc length of anellipse.For the surface area, however, the offendingdenominator cancels out, giving
dS x dx= 0 24 25 162 2. ,π – which is definedat x = ±5.
24. a. x = 35 sec t, dx = 35 sec t tan t dty = 100 + 80 tan t, dy = 80 sec2 t dty = 0 ⇔ 100 + 80 tan t = 0 ⇒ tan t = −5/4t = −−tan 1 5 4 ( / )Radius at base is x = −5/4 =−35 1sec tan[ ( )]56.0273… ≈ 56.0 ft.
b. At top, t = 0.5.Radius: x = 35 sec 0.5 = 39.8822… ≈ 39.9 ftHeight: y = 100 + 80 tan 0.5 = 143.7041…≈ 143.7 ft
c. From the information given in parts a and b,it can be assumed that −π /2 < t < π /2.
Minimize xdx
dtt t t: at= = =35 0 0sec tan
(or, because cos t has a max at t = 0,sec t = 1/cos t has a minimum there).Minimum radius = 35 ftHeight = y = 100 + tan 0 = 100 ft
d. dL dx dy= +2 2
= ⋅ +35 802 2 2 2 4sec tan sect t t dt
dS = 2πx ⋅ dL
= ⋅ +2 35 35 802 2 2 2 4π ( )sec sec tan sect t t t dt
S dS= ≈ … ≈− −∫ 37 756 5934 37 757 2
5 4
0 5
1, . , ft
tan ( / )
.
e. Volume , . ft3≈ ⋅ = … ≈S4
1212 585 5311
466.1307… yd3
25. From Figure 8-6m, a circle of radius L has areaπL2 and circumference 2πL. The circumference ofthe cone’s base is 2πR. The arc length of thesector of the circle of radius L must be equal tothis, so the sector is (2πR)/(2πL) = R/L of thecircle and has surface area S = πL2(R/L) = πRL,Q.E.D.
26. S = πRL − πrlThe objective is to get the lateral area in terms ofthe slant height of the frustum, L − l.
S R Lr
Rl= − ⋅
π
= − ⋅
πR L
l
Ll , because
r
R
l
L= .
= πR
LL l( – )2 2
Calculus Solutions Manual Problem Set 8-7 201© 2005 Key Curriculum Press
= +πR
LL l L l( )( – )
= + ⋅
−π R R
l
LL l( )
= + ⋅
−π R R
r
RL l( )
= π ( R + r)(L − l)
= +
−2
2π R r
L l( ), Q.E.D.
Problem Set 8-7Q1. 15x2 − 14x + 4 Q2. 12(4x − 9)2
Q3. 3 sin2 x cos x Q4. 3 sec 3x tan 3x
Q5. − −e x Q6. −1/x2
Q7. ln |x| + C Q8.1
22x C+
Q9. 3x + C Q10. x + C
1. a. r = 10 sin θ ⇒ dA = 50 sin2 θ dθ
A d= ≈ …∫ 50 157 07962
0
2
sin θ θπ
.
(exactly 50π )b. The area of the circle is π ⋅ 52 = 25π.
The calculated area is twice this because thecircle is traced out twice as θ increases from 0to 2π. Although r is negative for π < θ < 2π,dA is positive because r is squared.
2. a. r = 10 sin θ ⇒ dr = 10 cos θ dθ
dL dr r d
d d
= +
= + =
2 2
2 2100 100 10
( )
cos sin
θ
θ θ θ θ
L d= = =∫ 10 10 200
2
0
2
θ θ ππ π
The circumference is 2π ⋅ 5 = 10π. Thecalculated length is twice this value becausethe circle is traced out twice as θ increasesfrom 0 to 2π. The calculus of this sectionalways gives the dynamic answer as thedistance traveled by a point on the curve asθ increases from one value to another. Thispath length does not necessarily equal thelength of the curve.
3. a. r = 4 + 3 sin θ. The calculator graph confirmsthat the text figure is traced out once as θincreases from 0 to 2π.
b. dA d= +1
24 3 2( sin )θ θ
A dA= ≈ …∫ 64 40260
2
.π
(exactly 20.5π)
c. dr = 3 cos θ dθ
dL dr r d= +2 2( )θ
= + +( cos ) ( sin )3 4 32 2θ θ θd
L dL= ≈∫ 28 8141
0
2
.π
K
4. a. r = 5 − 3 cos θ. The calculator graph confirmsthat the text figure is traced out once as θincreases from 0 to 2π.
b. dA d= 1
25 3 2( – cos )θ θ
A dA= ≈ …∫ 92 6769 29 50
2
. (exactly . )ππ
c. dr = 3 sin θ dθdL dr r d= +2 2( )θ
= +( sin ) ( – cos )3 5 32 2θ θ θd
L dL= ≈ …∫ 34 31360
2
.π
5. a. r = 7 + 3 cos 2θ. The calculator graphconfirms that the text figure is traced out onceas θ increases from 0 to 2π.
b. dA d= +1
27 3 2 2( cos )θ θ
A dA= ≈ …∫ 168 07520
2
.π
(exactly 53.5π)
c. dr = −6 sin 2θ dθ
dL dr r d= +2 2( )θ
= + +(– sin ) ( cos )6 2 7 3 22 2θ θ θd
L dL= ≈ …∫ 51 45110
2
.π
6. a. r = 8 cos 2θ. The calculator graph confirmsthat the text figure is traced out once asθ increases from 0 to 2π.
b. dA d= 1
28 2 2( cos )θ θ
A dA= ≈∫ 100 5309 32
0
2
. (exactly )K ππ
c. dr = −16 sin θ dθ
dL dr r d= +2 2( )θ
= +(– sin ) ( cos )16 2 8 22 2θ θ θd
L dL= ≈∫ 77 5075
0
2
. Kπ
7. a. 5 = 5 + 5 cos θ. The calculator graphconfirms that the text figure is traced out onceas θ increases from 0 to 2π.
b. dA d= +1
25 5 2( cos )θ θ
A dA= ≈∫ 117 8097 37 5
0
2
. (exactly . )K ππ
202 Problem Set 8-7 Calculus Solutions Manual© 2005 Key Curriculum Press
c. dr = −5 sin θ dθ
dL dr r d= +2 2( )θ
= + +(– sin ) ( cos )5 5 52 2θ θ θd
L dL= =∫ 400
2
(exactly)π
8. a. r = 10
3 2– cos θ. The calculator graph
confirms that the text figure is traced out onceas θ increases from 0 to 2π.
b. dA =
1
2
10
3 2
2
– cosθ
A dA= ≈∫ 84 2977 12 5
0
2
. (exactly )K ππ
c. dr d= – sin
( – cos )
20
3 2 2
θθ
θ
dL dr r d= +2 2( )θ
=
+
– sin
( – cos ) – cos
20
3 2
10
3 22
2 2θθ θ
θd
L dL= ≈∫ 33 0744
0
2
. Kπ
9. a.
1
1
r = sin 3θ makes one complete cycle as θincreases from 0 to π.
b. dA d= 1
23 2(sin )θ θ
A dA= ≈∫ 0 7853 0 25
0. (exactly .K π
π)
c. dr = 3 cos 3θ dθdL dr r d= +2 2( )θ
= +( cos ) (sin )3 3 32 2θ θ θd
L dL= ≈∫ 6 68240
. Kπ
10. a.
4
4
b. dA d= 1
24 4 2( sec – cos )θ θ θ
A dA= ≈
−∫ 4 55571
1
. K
(exactly 16 tan 1 − 24 + 4 sin 2)
c. dr = 4(sec θ tan θ + sin θ) dθ
dL dr r d= +2 2( )θ
= + +4 2 2(sec tan sin ) (sec – cos )θ θ θ θ θ θd
L dL= ≈
−∫ 10 95341
1
. K
11. r = 49 2cos θr = 0 ⇔ 2θ = cos−1 0 = ±π /2 + 2π n (n an integer)θ = ±π /4 + π nThe right-hand loop corresponds tononnegative values of the integrand,−π /4 ≤ θ ≤ π /4.
dA d= 1
249 2( cos )θ θ
A d= =− −∫ 1
249 2 12 25 2
4
4
4
4
( cos ) . sin/
/
/
/
θ θ θπ
π
π
π
= 24 5.
Area of both loops is 49.
12. The graph of r = csc θ + 4 shows a closed loopfrom θ ≈ 3.4 to θ ≈ 6.
5
5
The graph passes through the pole where r = 0.csc 4 0 csc ( 4)θ θ+ = ⇔ = − =−1
sin− − = − … +1 0 25 0 2526 2( . ) . orπ n[π − (−0.2526…)] + 2π nDesired range is 3.3942… ≤ θ ≤ 6.0305… .
dA d= +1
24 2(csc )θ θ
A dA= ≈ …∫ 8 4553
3 3942
6 0305
. .
.
K
K
13. r1 = 4 + 4 cos θ and r2 = 10 cos θ intersect where4 + 4 cos θ = 10 cos θθ π= = ± …+−cos 1 2 3 0 8410 2( / ) . .n
(The graphs also touch at the pole, but not forthe same value of θ. For the cardioid,θ = π + 2πn. For the circle, θ = π /2 + 2π n.)Region outside the cardioid and inside
Calculus Solutions Manual Problem Set 8-7 203© 2005 Key Curriculum Press
the circle is generated as θ goes from−0.841… to 0.841… .
dA r r d= 1
2 22
12( – ) θ
= − +1
210 4 4[( ) ( ) ] 2 2cos cosθ θ θd
A dA= ≈
−∫ 18 88630 841
0 841
. KK
K
.
.
(exactly 26 cos (2/3) (4/3)− −1 5)
14. r1 = 5 and r2 = 5 − 5 cos θ intersect at θ = −π /2and π /2.
5
5
dA r r d= 1
2 12
22( – ) θ
= − −1
25 5 52[ ( ) ] 2cos θ θd
Integrate from −π /2 to π /2, because in QuadrantsII and III the cardioid lies outside the circle.
A dA= ≈ −
−∫ 30 3650 50 6 252
2
. (exactly .K ππ
π)
/
/
15. a. r = 0.5θ. The graph starts at θ = 0 and makesthree revolutions, so θ increases from 0 to 6π.
dr = 0.5 dθ
dL dr r d d= + = +2 2 2 20 5 0 5( ) . ( . )θ θ θ
L dL= ≈ …∫ 89 85890
6
.π
b. dA d d= =1
20 5
1
82 2( . )θ θ θ θ
Area swept out for third revolution inQuadrant I is
A d32 3
4
4 53
4
4 5 1
8
1
24
217
192= = =∫ θ θ θ π
π
π
π
π ..
Area swept out for second revolution inQuadrant I is
A d22 3
2
2 53
2
2 5 1
8
1
24
61
192= = =∫ θ θ θ π
π
π
π
π ..
Area of region between second and thirdrevolution in Quadrant I is A3 = A2 =13
1625 19253π = …. .
16. The graph of r = 4 + 6 cos θ shows a closed loopfrom θ ≈ 2.3 to θ ≈ 4.0.
10
4
r = 4 + 6 cos θ = 0 ⇔ cos θ = −2/3θ π= − = ± … +−cos 1 2 3 2 3005 2( / ) . n
dA d= +1
24 6 2( cos )θ θ
The outer loop is swept out as θ increases from−2.3005… to 2.3005.
A dA1
2 3005
2 3005
105 0506= ≈∫ . KK
K
– .
.
The inner loop is generated as θ increases from2.3005… to 3.9826… .
A dA2
2 3005
3 926
1 7635= ≈∫ . KK
K
.
.
Area of the region between the loops isA1 − A2 ≈ 103.2871… .
17. a.
4
2
dr = −2.5θ −1.5 dθdL dr r d d= + = +2 2 3 16 25 25( ) . – –θ θ θ θ
dL ≈ …∫ 31 08722
6
.π
π
/
b. The graph shows sectors of central angles 1,2, and 3 radians.
Area of sector is A r( ) .θ θ= 1
22
A( ) ( ) .11
25 1 12 52= =( )
A( ) ( ) .21
23 5355 2 12 52= =( . ...)
A( ) ( ) .31
22 8867 3 12 52= =( . )K
In general, A( ) ( ) . ,θ θ θ= =1
25 12 51 2 2( )– / which
is independent of the value of θ.
204 Problem Set 8-7 Calculus Solutions Manual© 2005 Key Curriculum Press
18. The graph shows r = sec θ and a segment fromθ = 0 to 1.5.
1
10
The point with polar coordinates (r, θ) has xy-coordinates x = r cos θ, y = r sin θ. The graphgiven by r = sec θ can be writtenx = r cos θ = sec θ cos θ = 1y = r sin θ = sec θ sin θ = tan θ(i.e., −∞ < y < ∞). Thus, this graph is theline x = 1.By calculus, the segment from θ = 0 to θ = 1.5has length as follows:dr = sec θ tan θ dθdL dr r d= +2 2( )θ
= +(sec tan ) secθ θ θ θ2 2 d
= + =sec tan secθ θ θ θ θ2 21 d d
L d= =∫ sec tan.
2
0
1 5
0
1 5
θ θ θ.
= tan 1.5 − 0 = 14.1014…As shown above, y = tan θ.At θ = 1.5, y = tan 1.5, confirming the calculus.
19. A typical record has grooves of inner radius6.6 cm and outer radius 14.6 cm, and takesabout 24 minutes to play. There are thus(33.333…)(24) or about 800 grooves in a spaceof (14.6 − 6.6) or 8.0 cm. Thus, the groovesdecrease in radius by about 8.0/800 = 0.01 cmper revolution. A simple equation of the spiral is
r = =0 01
2
0 005. .
πθ
πθ
which assumes that the grooves start at the centerand have a pitch of 0.01 cm. The innermostactual groove is at θ = 6.6π/0.005 = 1320π,and the outermost groove is at θ = 14.6π/0.005= 2920π.dr = −(0.005/π) dθdL dr r d= +2 2( )θ
= +( . / ) [( . / ) ]0 005 0 0052 2π π θ θd
= +0 0051 2.
πθ θd
L dL= ≈ …∫ 53 281 41201320
2920
, . cmπ
π
= 16,960.0002…π cmRough check: Average radius = 10.6 cmL should equal approximately the sum of 800
circles of radius 10.6 cm. L ≈ 800(2π · 10.6) =16,960π cm, which is very close to the calculated16,960.0002…π cm.(The integral can be evaluated algebraically bythe tangent trigonometry substitution fromChapter 9. The result, 16,960.00021…π, isremarkably close both to the numerical answerand to the sum of the lengths of the 800 circlesof average radius 10.6 cm.)
20. a. r = = − −100
3 2100 3 2 1
– coscos
θθ( )
dA d=
= − −
1
2100 3 2 1 2
2
[ ( – cos ) ]–θ θ
θ θ5000(3 2 cos ) d
A dA= ≈ … ≈∫ 974 3071 9740
0 2
. (kilo-mi)2.
b. Solving ( )5000 3 2 2
0 8− =−∫ cos
.t dt
θ
974.3071… gives θ ≈ 1.88976… .
c. P = ka1.5
(27.3)(24) = k(240)1.5
k = 0.17622…
d. The major axis of the spaceship’s orbit is120 thousand miles, so a = 60.P = k · 601.5 = 81.9 hours (precise answer)
e. The total area of the ellipse is
A d= − −∫ 5000 3 2 2
0
2
( )cos θ θπ
= 8429.7776… (kilo-mi)2
Fraction of area from θ = 0 to θ = 0.2 is(974.3071…)/(8429.7776…) = 0.1155… .This fraction is the same as the fraction of theperiod. Thus, the time is 0.1155…(81.9) =9.4659… hours to go from θ = 0 to θ = 0.2,and the same for θ to go from 0.8 to1.88976… .
f. dr d
d
= − − ⋅
= − −
−
−
100 3 2 2
200 3 2
2
2
( )
( )
cos ( sin )
sin cos
θ θ θ
θ θ θ
dL dr r d= +
= − − −
2 2
2
( )θ
θ θ[( 200 sin (3 2 cos ) )2
+ − −( ( ) ) ]2 1/2100 3 2 1cos θ θd
From θ = 0 to θ = 0.2,
L dL= ≈∫ 20 22280
0 2
. K.
≈ 20.2 kilo-mi.
From θ = 0.8 to θ = 1.88976… ,
L dL= ≈∫ 56 78960 8
1 88
. KK
.
.
≈ 56.8 kilo-mi.
g. Average speed from θ = 0 to θ = 0.2 is
20 2228
9 46592 1363
.
.
K
KK= . , or about 2136 mi/h.
Calculus Solutions Manual Problem Set 8-8 205© 2005 Key Curriculum Press
Average speed from θ = 0.8 to
θ = 1.88976… is
56 7896
9 46595 9993
.
.
K
KK= . ,
or about 5999 mi/h.
h. When the spaceship is farthest from Earth, itsradial velocity (toward the Earth) is zero. Asit proceeds in its orbit, it can be thought of asfalling toward the Earth, thus picking upspeed. The reverse is true on the other side ofthe Earth, where it is moving away and isthus being slowed by gravity.
21. a. Count 5 spaces to the right and about 7.5spaces down from the given point.Slope ≈ −1.5.
b. r = θx = θ cos θ ⇒ dx = dθ · cos θ − θ sin θ dθy = θ sin θ ⇒ dy = dθ · sin θ + θ cos θ dθdy
dx
dy d
dx d= = +/
/
sin cos
cos – sin
θθ
θ θ θθ θ θ
At θ = 7, dy/dx = −1.54338… , thusconfirming the answer found graphically.
22. a. x = r cos θ, y = r sin θ⇒ y/x = sin θ/cos θ = tan θ
r
r cos
r sin
slope = r sin
= tan
r cos
b. The slope of any line is tan φ, where φ is theangle between the x-axis and the line.And, because the tangent line has slopedy
dx
dy d
dx d= /
/( )
θθ
by the chain rule ,
tan/
/φ θ
θ .= dy d
dx d
c. tan ψ = tan (φ − θ) =+
tan – tan
tan tan
φ θφ θ1
=+ ⋅
⋅
dy ddx d
yx
dy ddx d
yx
x dx d
x dx d
//
–
//
/
/
θθ
θθ
θθ1
=+
xdyd
ydxd
xdxd
ydyd
θ θ
θ θ
–
d. dx/dθ = −r sin θ;
dy/dθ = r cos θ xdy
dy
dx
d
θ θ−
= ⋅ − ⋅ −r r r r ( ) ( )cos cos sin sinθ θ θ θ= r2 cos2 θ + r2 sin2 θ = r2
e. r x y rdr
dx
dx
dy
dy
d2 2 2 2 2 2= + ⇒ = +
θ θ θ
⇒ = +rdr
dx
dx
dy
dy
d
θ θ θSubstitute these expressions in parts d and einto the top and bottom of the expression inpart c to show the property.
f. tan/
– cos
sin
– cos
sinψ
θθ
θθ
θ = = = =r
dr d
a a
a
1
tan θ/2, using the half-angle formula. Thenψ = θ/2 + nπ. But 0 ≤ θ ≤ 2π, and 0 ≤ ψ ≤ π,which implies n = 0, so ψ = θ/2.
g. tan/
ψθ θ
= ⇒ = ⋅r
dr d
dr
drconst ⇒ r = Cekθ
Note that dr
dkCe kr k
dr d
rk
θθθ= = ⇒ = =/
1
tancot
ψψ= .
Equations for the spiral will vary.
Problem Set 8-8Review Problems
R0. Answers will vary.
R1. a.
3–10
50
x
yf
g
h
b. f ′ (x) = 3x2 − 18x + 30; f ″(x) = 6x − 18g′ (x) = 3x2 − 18x + 27; g″(x) = 6x − 18h′ (x) = 3x2 − 18x + 24; h″(x) = 6x − 18
c. h′ (x) = 3(x − 2)(x − 4) = 0 at x = 2 and 4h″(2) = −6 < 0, so h has a local maximumat x = 2.h″(4) = 6 > 0, so h has a local minimumat x = 4.
d. g′ (x) = 3(x − 3)2 = 0 only at x = 3.g′ (x) > 0 on both sides of x = 3, so this isneither a maximum nor a minimum point.
e. From the graphs, each point of inflectionappears at x = 3. Because each secondderivative equals 6x − 18, each one equalszero when x = 3.
206 Problem Set 8-8 Calculus Solutions Manual© 2005 Key Curriculum Press
R2. a.
x 2
+ undef. +
no max.or min.
f´(x)
f (x)
x 2
+ undef. –
p.i.
f´´(x)
f (x)
b.f (x)
x
–2 1 3 5
c. i. f x x f x x′ = − ′′ =− −( ) , ( )/ /2
31
2
91 3 4 3–
ii. Zooming in shows that there is a localminimum cusp at (0, 0) and a localmaximum with zero derivative at x ≈ 0.3.
1
0.5
0
x
f(x)
Algebraically, f ′ (x) = 0 at x = (2/3)3 =8/27, and f ′ (x) is undefined at x = 0, thuslocating precisely the minimum andmaximum found by graphing.Because there are no other critical valuesof x, there are no other maximum orminimum points.
iii. f ″(x) is undefined at x = 0, and f ″(x) < 0everywhere else; f ″ never changes sign, sothere are no inflection points.
iv. f (0) = 0, f (8/27) = 4/27, f (5) = −2.0759…Global maximum at (8/27, 4/27).Global minimum at (5, −2.0759…).
d. f (x)
x
2
1
The graph shows that f x x e x( ) = −2 has local
minimum at x = 0, local maximum at x ≈ 2,and points of inflection at x ≈ 3.4 and atx ≈ 0.6.f x xe x e x x ex x x′ = − = −− − −( ) ( )2 22
f x e xe x e x x ex x x x″ = − + = − +− − − −( ) ( )2 4 2 42 2
f ′ (x) = 0 at x = 0, 2
Minimum at (0, 0). Maximum at (2, 0.5413…).f ″(x) = 0 at x = + = …2 2 3 4142. and atx = − = …2 2 0 5857.
f ″(x) changes sign at each of these x-values,which implies points of inflection at(0.5857… , 0.1910…), (3.4142… ,0.3835…).
R3. a. Let x = width of a cell, y = length of the cell.xy y x x= ⇒ = <−10 10 01;
Minimize ( ) .L x x y x x= + = + −12 7 12 70 1
The graph shows minimum L (x) at x ≈ 2.4.
100
2
x
L(x)
L x x′ = − −( ) 12 70 2
L x x′ = = = …( ) at .0 70 12 2 4152/
At , . .x y= = = …70 12 10 12 70 4 1403/ /
Overall length of battery is 6(2.4152…) =14.4913… .Optimal battery is about 14.5″ by 4.1″,which is longer and narrower than the typicalbattery, 9″ by 6.7″. Thus, minimal walllength does not seem to be a majorconsideration in battery design.
b. The graph shows y = 8 − x3, from x = 0 tox = 2, with rectangle touching sample point(x, y) on the graph, rotated about the y-axis,generating cubic paraboloid and inscribedcylinder.
2
8
x
y
(x, y)
Domain of x is 0 ≤ x ≤ 2.Maximize V (x) = π r 2h = π x 2y = 8π x2 − π x5.The graph shows that V (x) has a maximum atx ≈ 1.5.
2
30
0
V(x)
x
Calculus Solutions Manual Problem Set 8-8 207© 2005 Key Curriculum Press
V ′ (x) = 16π x − 5π x4 = π x(16 − 5x3)V x x x′ = = = = …( ) at and .0 0 16 5 1 47363 /
Maximal rectangle has x = = …16 5 1 47363 / . ,
y = 8 − 16/5 = 4.8.R4. a. The graph shows y x y x= =1
1 322/ ,and
intersecting at (0, 0) and (1, 1), rotated aboutthe x-axis, sliced parallel to the x-axis,showing back half of solid only.
1
0 1
x
y
(x2, y)
(x1, y)
x1 = y3, x2 = y1/2
dV = 2π y (x2 − x1) · dy = 2π y (y1/2 − y3) dy
V dV= ≈ …∫ 1 25660
1
. (exactly 0.4π )
b. The graph shows y1 and y2 as in part a, butsliced perpendicular to the x-axis, generatingplane washer slices.
10
1
x
y
(x, y1 )
(x, y2 )
dV y y dx x x dx= − = −π π( ) ( )12
22 2 3 4/
V dV= ≈ …∫ 1 2566.0
1
(exactly 0.4π), which is
the same answer as in part a, Q.E.D.
c. i. The graph shows y = x2 and y = 4,intersecting at (2, 4) and (−2, 4), rotatedabout the y-axis, showing back half ofsolid only.
(x, y)
x
y
4
2–2
(x, 4)
dV = 2π x (4 − y) · dx = 2π x (4 − x2) dxV ≈ 25.1327… (exactly 8π)(Disks can also be used.)
ii. The graph shows the region described inpart i, rotated about the x-axis, showingback half of solid only.
x
y
4
2–2
(x, 4)
(x, y)
dV = π (42 − y2) dx = π (16 − x4) dx
V dV= ≈ …∫ 160 84952
.2
– (exactly 51.2π)
(Cylindrical shells can also be used.)
iii. The graph shows the region described inpart i, rotated about the line y = 5,showing back half of solid only.
x
y
4
2–2
(x, y)
(x, 4)5
dV = π [(5 − y)2 − 12] dx= π [(5 − x2)2 − 1] dx
V dV= ≈ …
∫ 174 2536
7
152. exactly 55
2
–π
(Cylindrical shells can also be used.)
iv. The graph shows the region described inpart i, rotated about the line x = 3,showing back half of solid only.
(x, y)x
y
4
2–2
(x, 4)
3
(3, y)
dV = 2π (3 − x) · (4 − y) · dx = 2π (3 − x)(4 − x2) dx
V dV= ≈ …∫ 201 06192
2
.–
(exactly 64π)
(Washers can also be used.)R5. a. y = x2 from x = −1 to x = 2.
dy = 2x dx
dL dx dy x dx= + = +2 2 21 2( )
L dL= ≈ …∫ 6 12571
.2
–
208 Problem Set 8-8 Calculus Solutions Manual© 2005 Key Curriculum Press
b. y = x3/2 from x = 0 to x = 9.dy = 1.5x1/2 dx
dL dx dy x dx= + = +2 2 1 2 21 1 5( . )/
L x dx= +∫ ( . )1 2 25 1 2
0
9/
= +∫1
2 251 2 25 2 251 2
0
9
.( . )x dx/ ( . )
= +2
6 751 2 25 3 2
0
9
.( . )x /
= = …2
6 7521 25 1 28 72813 2
.( . – )/ .
Distance between the endpoints is
10 26 27 85672 2+ = …. , so the answer
is reasonable.
c. x = t cos π t ⇒ dx = (cos π t − π t sin π t ) dty = t sin π t ⇒ dy = (sin π t + π t cos π t ) dtThe graph shows t increases from 0 to 4.
4
4
x
y
dL dx dy= +2 2
= + +(cos – sin ) (sin cos )π π π π π πt t t t t t dt2 2
= +1 2( )π t dt
L t dt= + ≈∫ 1 25 72552 2
0
4
π . K
R6. a. The graph shows y = x1/3 , from x = 0 tox = 8, rotated about the y-axis, showing theback half of the solid only.
(0, y)(x, y)
2
80
x
y
dy x dx= −1
32 3/
dL dx dy x dx= + = +
2 2 2 32
11
3– /
dS x dL x x dx= ⋅ = +
2 2 1
1
32 3
2
π π – /
S x x dx= +
∫2 1
1
32 3
2
π 0
8– /
= +
∫2
1
91 3 4 3
1 2
π / //
0
8
x x dx
= +
∫3
2
1
9
4
34 3
1 21 3π
//
0
8
x x dx/
= +
π x 4 3
3 2
0
81
9/
/
= =π
27145 1 203 04363 2( – )/ . K
The disk of radius 8 has area 64π =201.0619… , so the answer is reasonable.
b. The graph shows y = tan x, from x = 0 tox = 1, rotated about the line y = −1, showingthe back half of the solid only.
(x, y)
(x,–1)
1
1
0
x
y
dy = sec2 x dx
dL dx dy x dx= + = +2 2 41 sec
dS = 2π (y + 1) · dL
= + +2 1 1 4π ( )tan secx x dx
S dS= ≈∫ 20 4199
0
1
. K
R7. a. r = θ ⇒ dr = dθdL dr rd d= + = +2 2 21( )θ θ θ
L dL= ≈∫ 32 4706
0
5 2
. Kπ /
b. dA r d d= = .1
2
1
22 2θ θ θ Area of the region
between the curves equals the area traced outfrom t = 2π to t = 5π /2 minus the area tracedout from t = 0 to t = π /2.
A d d= −∫ ∫1
2
1
22
2
5 22
0
2
θ θ θ θπ
π π/ /
= −1
6
1
63
2
5 23
0
2
θ θπ
π π/ /
= − − + =
= …
1
62 5 2 0 5 0
7 5
63 3 3 3 3 3π π( . . )
38.7578
.
Calculus Solutions Manual Problem Set 8-8 209© 2005 Key Curriculum Press
Concept Problems
C1. a. The graph of µ(t) = 130 − 12T + 15T 2 − 4T 3
from T = 0 to T = 3 shows maxima at T = 0and T ≈ 2.0 and minima at T ≈ 0.5 andT = 3.
130
30
T
µ(T)
To maximize µ(T):
µ′ (T ) = −12 + 30T − 12T 2
= −6(2T − 1)(T − 2)
µ′ (T ) = 0 at T = 1
22,
µ(0) = 130; µ 1
2127 25
= . ;
µ(2) = 134; µ(3) = 121
Maximum viscosity occurs at T = 2, or 200°.
b. Minimum viscosity occurs at endpoint,T = 3, or 300°.
c.
C2. The graph of f (x) = (x − 1)4 + x shows that thegraph straightens out at x = 1 but does notchange concavity.
x
y
1
1
f ′ (x) = 4(x − 1)3 + 1; f″(x) = 12(x − 1)2,so f ′ (1) = 1 and f″(1) = 0.f″(x) > 0 for all x ≠ 1. In particular, f″(x) does notchange sign at x = 1. Thus, the graph is straightat x = 1, but not horizontal. Zooming in on (1,1) shows that the graph resembles y = x when xis close to 1, although it is actually concave upslightly.
C3. The graphs of f (x) = x2/3 and g x x( ) = −1 3/ show a
cusp at x = 0 for function f and a verticalasymptote at x = 0 for function g.
y
x
f
g
0 1
1
C4. a. y x= + +
3 1 25 13
52
. cos ( – )π
y′ =
2 5 13
53
53
. +
−
cos ( – ) sin ( – )π π π
x x
= +
– .cos ( – ) sin ( – )
2 5
31
35
35
π π πx x
dL dx= + +
12 5
31
2– .( cos )sin
π χ χ
where temporarily stands forχ π3
5( – )x
L dL= ≈ …∫ 5 77265
7 5
..
b. ′′ =y
– .cos cos sin sin
2 5
31
π χ χ χ χ χ[( ) ( ) ]+ + − d
dx
= − + ⋅– .cos cos
2 5
32 1
32π χ χ π
( )
= +– .(cos )( cos – )
2 5
91 2 1
2π χ χ
y ″ = 0 ⇔ cos χ = −1 or cos χ = 0.5χ = π + 2π n or χ = ±π /3 + 2π nx = 8 + 6n, 4 + 6n, or 6 + 6nThe only zero of y ′′ within the domain isx = 6, so the point of inflection must beat x = 6.
c. dS = 2π (x − 4) dL, where dL is as in part a.
S dS= ≈ …∫ 78 23735
7 5
..
d. x y= ⇒ = + +
7 5 3 1 25 1
5
6
2
. . cosπ
= + −3 1 25 1 3 2 2. ( / )
= + − = …3 1 25 1 75 3 3 0224. . .( )dV = 2π (x − 4) · (y − 3.0224…) · dx
V x dx= − + − −
= …∫ 2 4 1 25 1 1 75 3
58 8652
2
5
7 5
π χ( ) . [( ) ( . )]
.
cos.
C5. The 2000 World Almanac and Book of Factslists the area of Brazil as 3,286,478 square miles.Individual answers will vary.
210 Problem Set 8-8 Calculus Solutions Manual© 2005 Key Curriculum Press
C6. Let the cylinder lie on the x-axis and the hole lieon the y-axis so that the z-axis is perpendicularto both the cylinder and the hole. The cylinder isthus described by y2 + z2 ≤ 25, and the hole byx2 + z2 ≤ 9.Slice the hole with planes perpendicular to thez-axis. Then for −3 ≤ z ≤ 3, the cross section atz of the hole is a rectangle with height
2 and width .y z x z= =2 25 2 2 92 2– –
Area of cross-section rectangle is
4 225 34 2 4– z z+ ,
so . dV z z dz= +4 225 34 2 4–Thus, the volume of the hole (and thus of theuranium that once filled the hole) is
V z z dz= +−∫ 4 225 34 2 4
3
3
–
= 269.3703… cm3
According to the CRC Handbook, the densityof uranium is 19.1 g/cm3. So the mass of theuranium drilled out ism = (269.3703…)(19.1) = 5144.97… g.Value is 200(5144.97…) ≈ $1,029,000.
C7. Draw x- and y-axes with origin at the center ofthe circle on one face of the cube.
x
y
1
1
(x, y)
The solid remaining consists of eight identicalcorner pieces. Each corner piece consists of a cubeand three identical spikes. The spikes have squarecross sections when sliced perpendicular to theappropriate axes. The hole perpendicular to thexy-plane cuts a circle in that plane with equationx2 + y2 = 1. The cube shown in the diagrambegins at x = y so that 2x2 = 1,from which x = 2 2/ . Each cube is thus
( / )1 2 2− cm on a side, and thus has volume
Vc ( / ) . cm= − = …1 2 2 0 02512623 3.
Consider the leftmost spike in the precedingdiagram. Pick a sample point (x, y) on the partof the circle in that spike. The cross sectionperpendicular to the x-axis for this spike is a
square of side ( ) . 1 1 1 2− = −y x– Thus,
dV x dxs2( ) .= −1 1 2–
Because the spike goes from x = 0 to x = 2 2/ ,
V dVs s . .= ≈ …∫ 0 01096420
2 2/
(The integral can be evaluated algebraically usingtrigonometry substitution, as in Chapter 9. The
exact value is Vs .= − −11 2
12 4
1
2
π)
The 24 spikes (3 for each of the eight corners) areidentical.Thus, the total volume remaining isV V V= +
= − +
= + − = …
8 24
8 1 2 2 2411 2
12 4
1
2
8 8 2 6
3
c s
3
( / )
0.46415 cm
– –π
π
Chapter Test
T1.
x
y
3
0 1 2 3 4
T2.
1 2 3 4 5 6 7
1
2
3
4
5
6y
x
Derivative
Function
T3. A = xy = x(500 − 0.5x) = 500x − 0.5x2
A′ = 500 − xA′ = 0 ⇔ x = 500A′ goes from positive to negative at x = 500⇒ local maximum at x = 500.A(0) = A(1000) = 0 ⇒ global maximumat x = 500.Maximum at x = 500, y = 250.
Calculus Solutions Manual Problem Set 8-8 211© 2005 Key Curriculum Press
T4. V x y y dxa
b
= −∫2 1 2π ( )
T5. a. dA r dr= 1
22
b. dL dr r d= +( ) ( )2 2θ
c. dL dx dy= +2 2
d. dS x dL= −2 1π ( )
T6. f (x) = x3 − 7.8x2 + 20.25x − 13
f ′ (x) = 3x2 − 15.6x + 20.25
= 3(x − 2.5)(x − 2.7)
f ′ (x) changes from positive to negative atx = 2.5 and from negative back to positive atx = 2.7. So there is a local maximum at x = 2.5and a local minimum at x = 2.7.f ″(x) = 6x − 15.6 = 6(x − 2.6)f ″(x) = 0 at x = 2.6f ′ (2.6) = −0.03, so the graph is not horizontal atthe inflection point.
T7. y = x3 ⇒ dy = 3x2 dx
dL dx dy x dx= + = +2 2 2 21 3( )
L x dx= + = …∫ 1 9 8 63034
0
2
.
T8. dS = 2πx · dL = 2πx 1 9 4+ x dx
S x x dx= + = …∫ 2 1 9 77 32454
0
2
π .
T9. V(x) = π x2(8 − y) = 8π x2 − π x5
The graph shows a maximum V (x) at x ≈ 1.5.
0 1 2
40
x
y
V′(x) = 16π x − 5π x4 = 0 at x = 0 or 3.21/3
V(0) = V(2) = 0V(3.21/3) = 4.8 · 3.22/3π > 0, so this is amaximum.Maximal cylinder has V = 4.8 · 3.22/3π cm3 =32.7459… cm3.
T10. Slicing parallel to the y-axis generates cylindricalshells of radius x extending from the samplepoint (x, y) to the line y = 8.dV = 2πx · (8 − y) · dx = 2πx (8 − x3) dx
V x x dx x x= − = −∫ 2 8 2 4 0 24 2 5
0
2
π π( ) ( . )0
2
= 19.2π = 60.3185…
T11. Vcyl = π · 22 · 8 = 32πV
Vsolid
cyl
.= =19 2
320 6
. ππ
So, Vsolid = 0.6Vcyl .
T12. a.
5
2
x
y
b. dx = −5 sin t dt, dy = 2 cos t dt
dL dx dy= +2 2
= +(– cos ) ( sin )5 22 2t t dt
L dL= ≈ …∫ 23 01310
2
.π
c. Slicing perpendicular to the x-axis generatescircular slices of radius y, where sample point(x, y) is on the upper branch of the ellipse.dV = πy2 dx = 4π sin2 t (−5 sin t dt)
= −20π sin3 t dtLeftmost slice is at t = π, and rightmost sliceis at t = 0.
V t dt= − = …
= …∫ 20 83 7758
26 6666
3π
ππ
sin .0
. (numerically)V can be evaluated algebraically bytransforming two of the three sin t factorsinto cosines.
V t t dt= − −∫ 20 1 2ππ
( )0
cos sin
= − + ∫∫ 20 20 200
π πππ
sin cos sint dt t t dt
= −
=20
20
326
2
33
0
π π ππ
cos cost t
The x-radius is 5, and the y-radius is 2.4
3π (x-radius)(y-radius)2 = 4
3π (5)(2)2 =
262
3π , Q.E.D.
(In general, if a = x-radius and b = y-radius,the parametric functions are x = a cos t, y =b sin t. Repeating the preceding algebraic
solution gives V ab= 4
32π .)
T13. r = 5e0.1 θ
dr = 0.5e0.1 θ dθ
212 Problem Set 8-8 Calculus Solutions Manual© 2005 Key Curriculum Press
dL dr rd= +2 2( )θ
= +( . ) ( ). .0 5 50 1 2 0 1 2e e dθ θ θ= e d0 1 25 25. θ θ.
The spiral starts at r = 5 = 5e0.1·0 and makes threecomplete revolutions, so 0 ≤ θ ≤ 6π.
L e d e= = ∫ 0 1 0 1
0
6
0
6
25 25 10 25 25. .θ θππ
θ. .
= − = …10 25 25 1 280 69610 6. ( ) ..e π
T14. dA r d e d= =1
212 52 0 2θ θθ. .
The area between the second and third revolutionsequals the area swept out for the third revolution
minus the area swept out for the secondrevolution. In Quadrant I, the third revolutionextends from θ = 4π to θ = 4.5π and the secondrevolution extends from θ = 2π to θ = 2.5π.
A e d e d= − ∫∫ 12 5 12 50 2 0 2
2
2 5
4
4 5
. .. .θ θ
π
π
π
πθ θ
..
= −62 5 62 50 24
4 5 0 22
2 5. .. .e eθ
ππ θ
ππ. .
= 62.5(e0.9 π − e0.8 π − e0.5 π + e0.4 π )= 203.7405…
T15. Answers will vary.
Calculus Solutions Manual Problem Set 9-2 213© 2005 Key Curriculum Press
Chapter 9—Algebraic Calculus Techniquesfor the Elementary Functions
Problem Set 9-1
1. V x x dx= ⋅ ≈ …∫2 3 58640
2
ππ
cos/
.
2. f (x) = x sin x ⇒ f ′(x) = x cos x + sin x
3. f x dx x x dx x dx′ = + ∫∫∫ ( ) cos sin
4. x x dx f x dx x dxcos sin= ′ − ∫∫∫ ( )
= f (x) + cos x + C (by definition of indefiniteintegral)
= x sin x + cos x + C
5. V x x dx= ∫20
2
ππ
cos/
= +2 2 02π π πx x xsin cos /
= π2 − 2π6. V = π2 − 2π = 3.5864… , which is the same as
the approximation, to the accuracy shown.
7. The method involves working separately with thedifferent “parts” of the integrand. The functionf (x) = x sin x was chosen because one of theterms in its derivative is x cos x, which is theoriginal integrand. See Section 9-2.
Problem Set 9-2
Q1. y′ = x sec2 x + tan x Q2.1
1111x C+
Q3. Q4.1
33sin x C+
x
y
Q5. 5 cos2 5x − 5 sin2 5x Q6.
x
y
Q7. r′(x) = t (x) Q8. lim( ) – ( )
h
f x h f x
h→
+0
Q9. ≈ 110/6 Q10. C
1. x x dxsin∫ u = x dv = sin x dx
du = dx v = −cos x
= − − −∫x x x dxcos cos( )
= −x cos x + sin x + C
2. x x dxcos 3∫ u = x dv = cos 3x dx
du = dx v x= 1
33sin
= − ∫1
33 3x x x dxsin sin
1
3
= + +1
33
1
93x x x Csin cos
3. xe dxx4∫ u = x dv = e4x dx
du = dx v e x= 1
44
= − ∫1
4
1
44 4xe e dxx x
= − +1
4
1
164 4xe e Cx x
4. 6 3xe dxx−∫ u = 6x dv = e− 3x dx
du = 6 dx v e x= − −1
33
= −
− −
− −∫( ) ( )61
36
1
33 3x e e dxx x
= − − +− −22
33 3xe e Cx x
5. ( )x e dxx+ −∫ 4 5 u = x + 4 dv = e− 5x dx
du = dx v e x= − −1
55
= − + ⋅ +− −∫( )x e e dxx x41
5
1
55 5
= − − − +− − −4
5
1
5
1
255 5 5e xe e Cx x x
= − − +− −21
25
1
55 5e xe Cx x
6. ( )x e dxx+∫ 7 2 u = x + 7 dv = e2x dx
du = dx v e x= 1
22
= + ⋅ − ∫( )x e e dxx x71
2
1
22 2
= + − +7
2
1
2
1
42 2 2e xe e Cx x x
= + +13
4
1
22 2e xe Cx x
7. x x dx3 ln ∫ u = ln x dv = x3 dx
du = x− 1 dx v x= 1
44
= − ∫1
4
1
44 3x x x dxln
= − +1
4
1
164 4x x x Cln
214 Problem Set 9-3 Calculus Solutions Manual© 2005 Key Curriculum Press
8. x x dx5 3ln∫ u = ln 3x dv = x5 dx
du = x− 1 dx v x= 1
66
= − ∫1
63
1
66 5x x x dxln
= − +1
63
1
366 6x x x Cln
9. x e dxx2∫ u = x2 dv = ex dx
du = 2x dx v = ex
= − ∫x e xe dxx x2 2
u = 2x dv = ex dx
du = 2 dx v = ex
= − −
∫x e xe e dxx x x2 2 2
= x2ex − 2xex + 2ex + C
10. x x dx2 sin∫ u = x2 dv = sin x dx
du = 2x dx v = −cos x
= − − −∫x x x x dx2 2cos cos( )
u = 2x dv = −cos x dx
du = 2 dx v = −sin x
= − − − − −
∫x x x x x dx2 2 2cos sin sin( )
= −x2 cos x + 2x sin x + 2 cos x + C
11. ln x dx∫ u = ln x dv = dx
du = x− 1 dx v = x
= − ⋅ −∫x x x x dxln 1
= x ln x − x + C
Problem Set 9-3
Q1.1
66r C+ Q2. 2m cos 2m + sin 2m
Q3. tan x + C Q4.1
18113 6( )x C+ +
Q5.1
4114x x C+ + Q6. 1
Q7. 1/2
Q8. V f x g x dxa
b
= −∫π [ ( ) ( ) ]2 2
Q9. Q10. B
2
4
x
y
1. x e dxx3 2∫
e
e
u dv
x3 e2x
3x2 12 e
2x
6x14
2x
618
2x
01
16 e2x
+
+
–
–
+
= − + − +1
2
3
4
3
4
3
83 2 2 2 2 2x e x e xe e Cx x x x
2. x e dxx5 −∫ u dv
x 5 e –x
5x 4 –e –x
20x 3 e –x
60x 2 –e –x
120x e –x
120 –e –x
0 e –x
+
+
+
–
–
–
+
= −x5e− x − 5x4e− x − 20x3e− x − 60x2e− x − 120xe− x
− 120e− x + C
3. x x dx4 sin∫ u dvx 4 sin x
4x 3 –cos x12x 2 –sin x24x cos x24 sin x0 –cos x
+
+
+
–
–
–
= −x4 cos x + 4x3
sin x + 12x2 cos x
− 24x sin x − 24 cos x + C
4. x x dx2 cos∫ u dvx 2 cos x2x sin x2 –cos x0 –sin x
+
+
–
–
= x2 sin x + 2x cos x − 2 sin x + C
5. x x dx5 2cos∫ u dvx 5 cos 2x
5x 4 12 sin 2x
20x 3 –14 cos 2x
60x 2 –18 sin 2x
120x1
16 cos 2x
1201
32 sin 2x
0 –164 cos 2x
+
+
+
–
–
–
+
Calculus Solutions Manual Problem Set 9-3 215© 2005 Key Curriculum Press
= + −1
22
5
42
5
225 4 3x x x x x xsin cos sin
− + + +15
42
15
42
15
822x x x x x Ccos sin cos
6. x x dx3 5sin∫ u dvx3 sin 5x
3x2 –15 cos 5x
6x –125 sin 5x
61
125 cos 5x
01
625 sin 5x
+
+
–
–
+
= − + +1
55
3
255
6
12553 2x x x x x xcos sin cos
− +6
6255sin x C
7. e x dxx sin∫ u dvex sin xex –cos xex –sin x
+
–+
= − + −∫e x e x e x dxx x xcos sin sin
⇒ ∫2 e x dxx sin
= − + +e x e x Cx xcos sin 1
⇒ ∫ e x dxx sin
= − + +1
2
1
2e x e x Cx xcos sin
8. e x dxx cos∫ u dve x cos xe x sin xe x –cos x
+
–+
= + − ∫e x e x e x dxx x xsin cos cos
⇒ ∫2 e x dxx cos
= ex sin x + ex cos x + C1
⇒ ∫ e x dxx cos
= + +1
2
1
2e x e x Cx xsin cos
9. e x dxx3 5cos∫ u dve3x cos 5x
3e3x 15 sin 5x
9e 3x 125 cos 5x
+
–
+ –
= +1
55
3
2553 3e x e xx xsin cos
− ∫9
2553e x dxx cos
⇒ ∫34
2553e x dxx cos
= + +1
55
3
2553 3
1e x e x Cx xsin cos
⇒ ∫ e x dxx3 5cos
= + +5
345
3
3453 3e x e x Cx xsin cos
10. e x dxx4 2sin∫ u dve 4x sin 2x
4e 4x –12 cos 2x
16e 4x –14 sin 2x
+
–
+
= − + − ∫1
22 2 4 24 4 4e x e x e x dxx x xcos sin sin
⇒ ∫5 24e x dxx sin
= − + +1
22 24 4
1e x e x Cx xcos sin
⇒ ∫ e x dxx4 2sin
= − + +1
102
1
524 4e x e x Cx xcos sin
11. x x dx7 3ln∫-----------------
u dvln 3x x7
1/x18 x8
118 x7
0164 x8
+
–
+
= − +1
83
1
648 8x x x Cln
12. x x dx5 6ln∫-------------------------
u dvln 6x x 5
1/x16 x 6
116 x 5
0136 x 6
+
–
+
= − +1
66
1
366 6x x x Cln
13. x dx x C4 577
5ln
ln= +∫ (ln 7 is a constant!)
14. e dx e Cx x7 755
7cos
cos= +∫15. sin cos sin5 61
6x x dx x C= +∫
16. x x dx x x dx( ) ( ) ( ) /31
23 22 2 3 2 2 3− = − − −∫∫ /
= − +3
103 2 5 3( – )x C/
216 Problem Set 9-3 Calculus Solutions Manual© 2005 Key Curriculum Press
17. x x dx3 5( )1/2+∫ u dvx 3 (x + 5)1/2
3x 2 23(x + 5)3/2
6x415(x + 5)5/2
68
105(x + 5)7/2
016945(x + 5)9/2
+
+
–
–
+
= + − +2
35
4
553 3 2 2 5 2x x x x( ) ( )/ /
+ + − + +16
355
32
31557 2 9 2x x x C( ) / /( )
18. x x dx x x dx2 2 1 22 2∫ ∫− = −( ) /
u dvx2 (2 – x)1/2
2x –23(2 – x)3/2
2415(2 – x)5/2
08
105(2 – x)7/2+
+
–
– –
= − − − −2
32
8
1522 3 2 5 2x x x x( ) ( )/ /
− +16
1052 7 2( – )x C/
19. ln ln lnx dx x dx x x x C5 5 5 5= = − +∫ ∫
20. e dx x dx x Cxln 7 277
2= = +∫ ∫
21. x e dxx5 2
∫--------------------------
--------------------------
2
2
u dvx 4 xex2
4x 3 1 ex2
2x2 xex2
4x1 ex2
2 xex2
012 e x2
+
+
–
–
= − + +1
24 22 2 2
x e x e e Cx x x
22. x e dxx5 3
∫---------------------------
u dv
x 3 x 2 ex3
3x2 13 ex3
1 x 2 ex3
0 ex3
+
–
+ 13
= − +1
3
1
33 3 3
x e e Cx x
23. x x dx( )3ln∫--------------------------
--------------------------
--------------------------
2
8
8
16
u dv(ln x)3 x
3 (ln x)2/x12 x2
3 (ln x)2 1 x
6 (ln x)/x14 x2
6 ln x14 x
6/x1 x2
61 x
01 x2
+
+
–
–
+
= −1
2
3
42 3 2x x x x( ) ( )2ln ln
+ − +3
4
3
82 2x x x Cln
24. x x dx3( )2ln∫--------------------------
--------------------------
u dv(ln x)2 x 3
2(ln x)/x14 x4
2 ln x14 x3
2/x116 x 4
2116 x3
0164 x 4
+
+
–
–
= − + +1
4
1
8
1
324 4 4x x x x x C( )2ln ln
25. x x dx3 2 1( )4+∫----------------------------------
u dvx2 x(x2 + 1)4
2x110 (x2 + 1)5
15 x(x2 + 1)5
0112 (x2 + 1)6
+
–
+
= + − + +1
101
1
6012 2 5 2 6x x x C( ) ( )
26. x x dx x x3 2 3 2 1 23 3− = −∫ ∫ ( ) /
-----------------------------------
u dvx 2 x(x2 – 3)1/2
2x13(x2 – 3)3/2
23 x(x2 – 3)3/2
015(x2 – 3)5/2
+
–
+
= − − +1
33
2
1532 2 3 2 2 5 2x x x C( ) / /( – )
27. cos2 x dx∫ u dvcos x cos x–sin x sin x
+–
Calculus Solutions Manual Problem Set 9-3 217© 2005 Key Curriculum Press
= − −∫cos sin sinx x x dx( ) 2
= + −∫cos sin cosx x x dx( )1 2
= + − ∫cos sin cosx x x x dx2
⇒ = + +∫2 21cos cos sinx dx x x x C
⇒ = + +∫ cos cos sin2 1
2
1
2x dx x x x C
28. sin .2 0 4x dx∫ u dvsin 0.4x sin 0.4x
0.4 cos 0.4x –2.5 cos 0.4x+
–
= − + ∫2 5 0 4 0 4 0 42. . . .sin cos cosx x x dx
= − + −∫2 5 0 4 0 4 1 0 42. . . ( . ) sin cos sinx x x dx
= − + −∫ ∫2 5 0 4 0 4 0 42. . . . sin cos sinx x dx x dx
⇒ ∫2 0 42sin . x dx
= −2.5 sin 0.4x cos 0.4x + x + C1
⇒ ∫ sin2 0 4. x dx
= −1.25 sin 0.4x cos 0.4x + 0.5x + C
29. sec3 x dx∫ u dvsec x sec2 x
sec x tan x tan x+
–
= − ∫sec tan sec tanx x x x dx2
= − −∫sec tan sec secx x x x dx( ) 2 1
= − + +∫sec tan sec ln sec tanx x x dx x x3 | |
⇒ ∫2 3sec x dx
= sec x tan x + ln | sec x + tan x | + C1
⇒ ∫ sec3 x dx
= + + +1
2
1
2sec tan ln sec tanx x x x C| |
30. sec tan2 x x dx∫= ⋅ = +∫ ( ) ( )1sec sec tan secx x x dx x C
1
22
31. logln
ln31
3x dx x dx= ∫∫
= +1
3ln( ln – )x x x C
= − +x x x Clogln31
3
32. logln
ln101
10x dx x dx =∫ ∫
= +1
10ln( ln – )x x x C
= − +x x x Clogln10
1
10
33. sin cosx dx x C= − +∫34. cos sinx dx x C= +∫35. csc ln csc cotx dx x x C= − + +∫ | |
36. sec ln sec tanx dx x x C= + +∫ | |
37. tan ln cosx dx x C= − +∫ | |
38. cot ln sinx dx x C= +∫ | |
39. x2∫ cos x dx
For the first integral, Wanda integrated cos x anddifferentiated x2, but in the second integral she
plans to differentiate cos∫ x dx and integrate 2x,
effectively canceling out what she did in the firstpart. She will get
x x dx x x x x x x dx2 2 2 2cos sin sin cos ,= − + ∫∫which is true but not very useful!
40. x x dx2 cos∫Amos’s choice of u and dv transforms
x x dx x x x x dx2 3 31
3
1
3∫ ∫+cos cos sininto ,
which is more complicated than the originalexpression.
41. After two integrations by parts,
e x dxx sin ∫= − + − ∫e x e x e x dxx x xcos sin sin
but after two more integrations,
e x dx e x e x e xx x x xsin cos sin cos= − + +∫− + ∫e x e x dxx xsin sin
Two integrations produced the original integralwith the opposite sign (which is useful), and twomore integrations reversed the sign again to givethe original integral with the same sign (whichis not useful).
42. cos cos2 1
21 2x dx x dx= +∫ ∫ ( )
= +
+1
2
1
22x x Csin
218 Problem Set 9-4 Calculus Solutions Manual© 2005 Key Curriculum Press
By the double-argument properties fromtrigonometry,1
2
1
22
1
2x x C x x x C+
+ = + +sin ( sin cos )
which is equivalent to the answer in Problem 27found using integrating by parts.
43.
x
y
(1,1/e)
3
1
y′ = −xe− x + e− x = e− x(1 − x)Critical points at x = 0, 1, 3; maximum atx = 1.
A xe dxx= −∫0
3
= − −− −( )xe ex x
0
3
= −3e− 3 − e− 3 + 1 = −4e− 3 + 1 = 0.8008…
44. y = 12x2e− x
Area from x = 0 to x = b is
A b x e dxxb
( ) = −∫ 12 2
0
= − − −− − −12 24 242
0x e xe ex x x b
= −12b2e− b − 24be− b − 24e− b + 24
The first two terms approach zero as b → ∞by L’Hospital’s rule. The third term alsoapproaches 0.∴ =
→∞limb
bA 24
45. y = ln xdV = πy2
dx = π (ln x)2 dx
V x dx= ∫ π (ln )2
1
5
-----------------------
-----------------------
u dv(ln x)2 1
2 (ln x)/x x2 ln x 1
2/x x
2 10 x
+
–
–
+
= − +π π πx x x x x( )2ln ln2 21
5
= 5π ( ln 5)2 − 10π ln 5 + 10π − 0 + 0 − 2π= 15.2589…
46. Consider u dv ,∫ and write dv v C= +∫ . Then
u dv u v C v C du= + − +∫ ∫( ) ( )
= + − − ∫∫uv Cu v du C du
= + − − = −∫ ∫uv Cu v du Cu uv v du
Thus, the constant cancels out later, Q.E.D.
47. For integration by parts, u dv uv v du .= −∫ ∫Applying limits of integration gives
u dv uv v duc
d
u a
u b
a
b
= −∫ ∫=
=
= − − ∫( ) bd ac v dua
b
The quantity (bd − ac) is the area of the“L-shaped” region, which is the area of the largerrectangle minus the area of the smaller one.Thus, the integral of u dv equals the area of theL-shaped region minus the area represented by theintegral of v du.
48. ln ln lnax dx a x dx= +∫ ∫ ( )
= x ln a + x ln x − x + C= x ln ax − x + C
49. sin7 x dx∫ u dvsin6 x sin x
6 sin5 x cos x –cos x+
–
= − + ∫sin cos sin cos6 5 26x x x x dx
= − + −∫sin cos sin sin6 5 26 1x x x x dx( )
= − + − ∫∫sin cos sin sin6 5 76 6x x x dx x dx
7 67 6 5sin sin cos sinx dx x x x dx= − + ∫∫sin sin cos sin7 6 51
7
6
7x dx x x x dx = − + ∫∫
The fractions are 1/(old exponent) and(old exponent − 1)/(old exponent). The newexponent is 2 less than the old exponent. So
sin sin cos7 61
7x dx x x= −∫
+ +
∫6
7
1
5
4
54 3– sin cos sinx x x dx
= − −1
7
6
356 4sin cos sin cosx x x x
+ +
∫24
35
1
3
2
32– sin cos sinx x x dx
= − −1
7
6
356 4sin cos sin cosx x x x
− − +8
35
16
352sin cos cosx x x C
50. Answers will vary.
Problem Set 9-4
Q1. uv v du− ∫
Calculus Solutions Manual Problem Set 9-4 219© 2005 Key Curriculum Press
Q2. Q3.
x
y
2π
3
x
y
2π
1
Q4. y′ = 1 + ln 5x Q5.1
66sin x C+
Q6. ln |x| + C
Q7.
x
y
3
1
Q8.1
1 2+ xQ9. ln | sec x + tan x | + C
Q10. D
1. sin9 x dx∫ u dvsin8 x sin x
8 sin7 x cos x –cos x+
–
= − + ∫sin cos sin cos8 7 28x x x x dx
= − + −∫sin cos sin sin8 7 28 1x x x x dx ( )
= − + − ∫∫sin cos sin sin8 7 98 8x x x dx x dx
9 89 8 7 = − + ∫ ∫sin sin cos sinx dx x x x dx
sin sin cos sin9 8 71
9
8
9x dx x x x dx= − +∫ ∫
2. cos10 x dx∫ u dvcos 9 x cos x
–9 cos 8 x sin x sin x–+
= + ∫cos sin cos sin9 8 29x x x x dx
= + −∫cos sin cos cos9 8 29 1x x x x dx ( )
= + − ∫ ∫cos sin cos cos9 8 109 9x x x dx x dx
10 910 9 8cos cos sin cosx dx x x x dx= +∫ ∫cos cos sin cos10 9 81
10
9
10x dx x x x dx= + ∫∫
3. cot cot cot12 10 2x dx x x dx= ∫ ∫= −∫ cot csc10 2 1x x dx ( )
= − ∫ ∫cot csc cot10 2 10x x dx x dx
= − − ∫1
1111 10cot cotx x dx
4. tan tan tan20 18 2x dx x x dx = ∫ ∫= −∫ tan sec18 2 1x x dx( )
= − ∫ ∫tan sec tan18 2 18x x dx x dx
= − ∫1
1919 18tan tanx x dx
5. sec13 x dx∫ u dvsec 11 x sec 2 x
11 sec 10 x sec x tan x tan x+
–
= − ∫sec tan sec tan11 11 211x x x x dx
= − − ∫sec tan sec sec11 11 211 1x x x x dx( )
= − + ∫∫sec tan sec sec11 13 1111 11x x x dx x dx
12 1113 11 11sec sec tan secx dx x x x dx= + ∫∫sec sec tan sec13 11 111
12
11
12x dx x x x dx= + ∫∫
6. csc100 x dx∫u dv
csc 98 x csc 2 x–98 csc 97 x csc x cot x –cot x–
+
= − − ∫csc cot csc cot98 98 298x x x x dx
= − − −∫csc cot csc csc98 98 298 1x x x x dx( )
= − − + ∫∫csc cot csc csc98 100 9898 98x x x dx x dx
99 98100 98 98csc csc cot cscx dx x x x dx= − + ∫∫csc csc cot csc100 98 981
99
98
99x dx x x x dx= − + ∫∫
7. cosn x dx∫ u dvcos n – 1 x cos x
–(n –1) cosn – 2 x sin x sin x–+
= + −− −∫cos sin cos sinn nx x n x x dx1 2 21( )
= + − −− −∫cos sin cos cosn nx x n x x dx1 2 21 1( ) ( )
= + −− −∫cos sin cosn nx x n x dx1 21( )
− − ∫( )n x dxn1 cos
n x dx x x n x dxn n ncos cos sin cos= + −− −∫∫ 1 21( )
cos cos sin–
cosn n nx dxn
x xn
nx dx= +− −∫∫ 1 11 2
220 Problem Set 9-4 Calculus Solutions Manual© 2005 Key Curriculum Press
8. sinn x dx∫ u dvsinn –1 x sin x
(n –1) sinn –2 x cos x –cos x+
–
= − + −− −∫sin cos sin cosn nx x n x x dx1 2 21( )
= − + − −− −∫sin cos sin sinn nx x n x x dx1 2 21 1( ) ( )
= − + −− −∫sin cos sinn nx x n x dx1 21( )
− − ∫( )n x dxn1 sin
n x dx x x n x dxn n nsin sin cos sin= − + −− −∫∫ 1 21( )
sin sin cos–
sinn n nx dxn
x xn
nx dx= − +− −∫∫ 1 11 2
9. tan tan tann nx dx x x dx= −∫∫ 2 2
= −−∫ tan secn x x dx2 2 1( )
= −− −∫∫ tan sec tann nx x dx x dx2 2 2
= −− −∫1
11 2
nx x dxn n
–tan tan
10. cot cot cotn nx dx x x dx= −∫∫ 2 2
= −−∫ cot cscn x x dx2 2 1( )
= −− −∫∫ cot csc cotn nx x dx x dx2 2 2
= − −− −∫1
11 2
nx x dxn n
–cot cot
11. cscn x dx∫u dv
csc n – 2 x csc 2x–(n – 2) cscn – 3 x csc x cot x –cot x–
+
= − − −− −∫csc cot csc cotn nx x n x x dx2 2 22( )
= − − − −− −∫csc cot csc cscn nx x n x x dx2 2 22 1( ) ( )
= − − −− ∫csc cot cscn nx x n x dx2 2( )
+ − −∫( )n x dxn2 2csc
( )n x dxn− ∫1 csc
= − + −− −∫csc cot cscn nx x n x dx2 22( )
cscn x dx∫= − +− −∫1
1
2
12 2
nx x
n
nx dxn n
–csc cot
–
–csc
12. secn x dx∫u dv
sec n – 2 x sec 2 x(n – 2) secn – 3 x sec x tan x tan x–
+
= − −− −∫sec tan sec tann nx x n x x dx2 2 22( )
= − − −− −∫sec tan sec secn nx x n x x dx2 2 22 1( ) ( )
= − −− ∫sec tan secn nx x n x dx2 2( )
+ − −∫( )n x dxn2 2sec
( )n x dxn− ∫1 sec
= + −− −∫sec tan secn nx x n x dx2 22( )
secn x dx∫= +− −∫1
1
2
12 2
nx x
n
nx dxn n
–sec tan
–
–sec
13. sin5 x dx∫= − + − −
+1
5
4
5
1
3
2
34 2sin cos sin cos cosx x x x x C
= − − − +1
5
4
15
8
154 2sin cos sin cos cosx x x x x C
14. cos5 x dx∫= + +
+1
5
4
5
1
3
2
34 2cos sin cos sin sinx x x x x C
= + + +1
5
4
15
8
154 2cos sin cos sin sinx x x x x C
15. cot6 x dx∫= − − − − − −
+1
5
1
35 3cot cot cotx x x x C( )
= − + − − +1
5
1
35 3cot cot cotx x x x C
16. tan7 x dx∫= − − +
+1
6
1
4
1
26 4 2tan tan tan ln cosx x x x C| |
= − + + +1
6
1
4
1
26 4 2tan tan tan ln cosx x x x C| |
17. sec sec tan tan4 21
3
2
3x dx x x x C= + +∫
18. csc csc cot cot4 21
3
2
3x dx x x x C= − − +∫
19. a. y = cos x is on top; y = cos3 x is in the
middle; y = cos5 x is on the bottom.
b. For y = cos x, area ≈ 2.0000… .For y = cos3 x, area ≈ 1.3333… .For y = cos5 x, area ≈ 1.06666… .
c. A x dx x12
2
2
2
= =∫ −cos sin
– /
/
/
/
π
π
π
π
sin (π/2) − sin (−π/2) = 2
Calculus Solutions Manual Problem Set 9-4 221© 2005 Key Curriculum Press
A x dx33
2
2
= ∫ cos– /
/
π
π
= +−
1
3
2
32
2
2
cos sin sin/
/
x x xπ
π
= +1
32 2
2
322cos / ) sin / ) sin / )( ( (π π π
− − − − −1
32 2
2
32cos ( sin ( sin (2 π π π/ ) / ) / )
= + − + = = …02
30
2
3
4
31 3333.
Observe that A A3 12
3= .
A x dx
x x x dx
A
55
2
2
4
2
23
2
21
5
4
5
4
5
4
5
4
3
4
5
2
3
16
151 066666
=
= +
= + = ⋅ = ⋅ ⋅ =
= …
∫
∫− −
cos
cos sin cos
– /
/
/
/
/
/
π
π
π
π
π
π
0 2
.
3
Observe that A A5 34
5= .
d. Based on the graphs, the area under cos xshould be greater than that under cos3 x,which in turn is greater than the area undercos5 x. This is exactly what happens with thecalculated answers: A1 > A3 > A5.
e.
x
y
1
y = cos100 x
–0.5π 0.5π
f. Yes, lim cos– /
/
n
n x dx→∞
=∫ 02
2
.π
π
Following the pattern in part c, for odd n,
An n n
n n nn = ⋅( – )( – )( – ) ( )( )( )
( )( – )( – ) ( )( )
1 3 5 4 2 1
2 4 5 32
K
K,
the denominator gets large faster than thenumerator. However, because both go toinfinity, this observation is not decisive.The following epsilon proof by Cavan Fangestablishes the fact rigorously, using thedefinition of limit in the form “For anyε > 0, there is an N > 0 such that whenevern > N, An < ε.”Proof:
Pick 0 < ε < 2π.
Then 04
1< <cos ,ε
so there exists N > 0
such that cosN ε επ4 2
< .
Note that if ε π4 2
< < ,x then cos cosx <ε4
⇒ cosN x < .επ2
Now, for any n > N,
cos/
/n x dx
−∫ π
π
2
2
= +∫ ∫2 2cos cosn nx dx x dx0
/4
/4
/2
.ε
ε
π
But 2 < ( < ) . /
cos cos/
n nx dx dx x22
10
4
0
4 εε ε∫∫ =
And 2 < ( > )/4
/2
cos cos/
/n Nx dx x dx n N2
4
2
ε
π
ε
π
∫ ∫< < < .
/
22 2 24
2
0
2επ
επ
ε επε
π πdx dx xN
/
/
cos∫ ∫ =
So/2
/2
cosn x dx−∫ π
π
= + <∫ ∫2 2cos cos .n nx dx x dx0
/4
/4
/2ε
ε
πε
∴ =→∞ −∫lim cos ,
/
/
n
n x dxπ
π
2
2
0 Q.E.D.
20. cos cos cos5 4x dx x x dx=∫ ∫= −∫ ( )21 2sin cosx x dx
= − +∫ ( )1 2 2 4sin sin cosx x x dx
= −∫ ∫cos sin cosx dx x x dx2 2
+ ∫ sin cos4 x x dx
= − + +sin sin sinx x x C2
3
1
53 5
∴ =−∫A x dx5
5
2
2
sin/
/
π
π
= − +
2
2
3
1
53 5
0
2
sin sin sin/
x x xπ
= − + = = …24
3
2
5
16
151 0666. , which agrees with
the answer from Problem 19.
21. sec3 x dx∫ u dvsec x sec 2 x
sec x tan x tan x+
–
= − ∫sec tan sec tanx x x x dx2
= − −∫sec tan sec secx x x x dx ( )2 1
= − + ∫∫sec tan sec secx x x dx x dx3
222 Problem Set 9-5 Calculus Solutions Manual© 2005 Key Curriculum Press
2 3sec x dx ∫= sec x tan x + ln |sec x + tan x| + C1
sec3 x dx∫= +1
2
1
2sec tanx x ln |sec x + tan x| + C
Note that the answer is half the derivative ofsecant plus half the integral of secant.
22. sinn ax dx∫u dv
sinn –1 ax sin axa(n – 1) sinn – 2 ax cos ax –
1a cos ax–
+
= − −1 1
aax axnsin cos
+ − −∫( )n ax ax dxn1 2 2sin cos
= − −1 1
aax axnsin cos
+ − −−∫( ) ( )n ax ax dxn1 12 2sin sin
= − + −− −∫111 2
aax ax n ax dxn nsin cos sin( )
− − ∫( )n ax dxn1 sin
n ax dxnsin∫= − + −− −∫1
11 2
aax ax n ax dxn nsin cos sin( )
sinn ax dx∫= − +− −∫1 11 2
anax ax
n
nax dxn nsin cos
–sin
sin sin cos5 431
153 3x dx x x= −∫
− − +4
453 3
8
4532sin cos cosx x x C
23. sin sin cos3 21
3ax dx
aax ax= −∫
+ ∫2
322sin ax dx (From Problem )
= − − +1
3
2
32
aax ax
aax Csin cos cos
= − + +1
322
aax ax Ccos sin( ) , Q.E.D.
Or: d
dx aax ax– (cos )(sin )
1
322 +
= − +1
322
aa ax ax(– sin )(sin )
− 1
32
aax a ax ax(cos )( sin cos )
= +1
32 23 2(sin sin – sin cos )ax ax ax ax
= +1
32 13 2[sin sin ( – cos )]ax ax ax
= +1
323 2[sin sin (sin )]ax ax ax
= sin3 ax
∴ ∫ sin3 ax dx
= − + +1
322
aax ax C(cos )(sin ) , Q.E.D.
24. Use integration by parts, or use the technique ofProblem 20, as shown here.
cos cos cos3 2ax dx ax ax dx= ∫∫= −∫ ( )1 2sin cosax ax dx
= − ∫∫ cos sin cosax dx ax ax dx2
= − +1 1
33
aax
aax Csin sin
= +1
33 2
aax ax C(sin )( – sin )
= +1
32 2
aax ax(sin )( cos ) + C, Q.E.D.
Or: Differentiate, as in the alternate solution forProblem 23.
Problem Set 9-5Q1. f ′(1) = −4 Q2. g′(2) = 1/2
Q3. h′(3) = −12 Q4. t′(4) = π/24
Q5. p′(5) = 6e5 Q6. x = 83 = 512
Q7. Q8. integration by parts
x
y
3
5
Q9. Q10. E
1 3 5 6 8
x
–1
1
f'(x) and f(x)
f
f'
1. sin cos sin5 21x dx x x dx= −∫ ∫ ( )2
= − +∫ ( )1 2 2 4cos cos sinx x x dx
= − + − +cos cos cosx x x C2
3
1
53 5
Calculus Solutions Manual Problem Set 9-5 223© 2005 Key Curriculum Press
2. cos sin cos7 21x dx x x dx= −∫ ∫ ( )3
= − + −∫ ( )1 3 32 4 6sin sin sin cosx x x x dx
= − + − +sin sin sin sinx x x x C3 5 73
5
1
7
3. cos sin cos7 29 1 9 9x dx x x dx ( )3= −∫ ∫= − +∫ (1 3 9 3 92 4sin sinx x
− sin6 9x) cos 9x dx
= −1
99
1
993sin sinx x
+ − +1
159
1
6395 7sin sinx x C
4. sin cos sin3 210 1 10 10x dx x x dx= −∫∫ ( )
= − + +1
1010
1
30103cos cosx x C
5. sin cos sin4 53 31
153x x dx x C= +∫
6. cos sin cos8 97 71
637x x dx x C= − +∫
7. cos sin6 38 8x x dx∫= −∫ cos cos sin6 28 1 8 8x x x dx( )
= −∫ ( )cos cos sin6 88 8 8x x x dx
= − + +1
568
1
7287 9cos cosx x C
8. sin cos4 32 2x x dx∫= −∫ sin sin cos4 22 1 2 2x x x dx( )
= −∫ ( )sin sin cos4 62 2 2x x x dx
= − +1
102
1
1425 7sin sinx x C
9. sin cos sin cos cos5 2 2 21x x dx x x x dx= −∫ ∫ ( )2
= − +∫ ( )cos cos cos sin2 4 62x x x x dx
= − + − +1
3
2
5
1
73 5 7cos cos cosx x x C
10. cos sin cos sin sin3 2 2 21x x dx x x x dx∫ ∫= −( )
= −∫ ( )sin sin cos2 4x x x dx
= − +1
3
1
53 5sin sinx x C
11. cos cos2 1
21 2x dx x dx= +∫∫ ( )
= + +1
2
1
42x x Csin
12. sin cos2 1
21 2x dx x dx= −∫∫ ( )
= − +1
2
1
42x x Csin
13. sin cos2 51
21 10x dx x dx= −∫∫ ( )
= − +1
2
1
2010x x Csin
14. cos cos2 61
21 12x dx x dx= +∫∫ ( )
= + +1
2
1
2412x x Csin
15. sec tan sec4 2 21x dx x x dx= +∫∫ ( )
= + +1
33tan tanx x C
16. csc cot csc6 2 21x dx x x dx= +∫ ∫ ( )2
= + +∫ ( )cot cot csc4 2 22 1x x x dx
= − − − +1
5
2
35 3cot cot cotx x x C
17. csc cot csc8 2 26 6 1 6x dx x x dx= +∫ ∫ ( )3
= + + +∫ ( )cot cot cot csc6 4 2 26 3 6 3 6 1 6x x x x dx
= − − −1
426
1
106
1
667 5 3cot cot cotx x x
− +1
66cot x C
18. sec tan sec4 2 2100 100 1 100x dx x x dx∫ ∫= +( )
= + +1
300100
1
1001003tan tanx x C
19. tan sec tan10 2 111
11x x dx x C= +∫
20. cot csc cot8 2 91
9x x dx x C= − +∫
21. sec tan sec sec tan10 9x x dx x x x dx∫ ∫= ( )
= +1
1010sec x C
22. csc cot csc csc cot8 7x x dx x x x dx∫ ∫= ( )
= − +1
88csc x C
23. sec sec (sec )10 10 1020 20 20dx dx x C= = +∫ ∫24. csc csc csc8 8 812 12 12dx dx x C= = +∫ ∫ ( )
25. ( )cos sin cos sin2 2 21
22x x dx x dx x C− = = +∫ ∫
224 Problem Set 9-5 Calculus Solutions Manual© 2005 Key Curriculum Press
26. ( )cos sin2 2x x dx dx x C+ = = + ∫ ∫27. ( ) 2sin csc cotx dx x dx x C− = = − +∫ ∫ 2
28. ( ) 2cos sec tan3 31
332x dx x dx x C−∫ ∫= = +
29. sec3 x dx∫= + + +1
2
1
2sec tan ln sec tanx x x x C| |
30. csc3 x dx∫= − − + +1
2
1
2csc cot ln csc cotx x x x C| |
31. a. cos sin5 3x x dx∫
u dvsin 3x cos 5x
3 cos 3x15 sin 5x
–9 sin 3x – 125 cos 5x
+
–
+
= +1
55 3
3
255 3sin sin cos cosx x x x
+ ∫9
255 3cos sinx x dx
16
255 3cos sinx x dx∫
= + +1
55 3
3
255 3 1sin sin cos cosx x x x C
cos sin5 3x x dx∫= + +5
165 3
3
165 3sin sin cos cosx x x x C
b. cos sin5 30
2
x x dxπ
∫= + =5
165 3
3
163 5 0
0
2
sin sin cos cosx x x xπ
Because the integral finds the area aboveminus the area below, this calculation showsthe two areas are equal.
32. a.y
πx
1
b. A x dx x x= = − =∫ sin cos cos3 3
0 0
1
3
4
3
π π
c. Numerically: A ≈ 1.3333… (Checks.)
d. A = 0 because sin3 x is an odd function[sin3 (−x) = −sin3 x] and the integral of an
odd function between symmetrical limits isequal to zero.
33. dV = π y2 dx = π sin2 x dx
V x dx x dx= =∫ ∫π ππ πsin ( – cos )2
0 021 2
= − =π π ππ
2 42 2
0
2x xsin /
34. a. y = sec2 x
dV = π[(y + 3)2 − 32] dx
= π (sec4 x + 6 sec2 x) dx
V x x dx
x x x dx
x x x dx
x x
= +
= + +
= +
= +
= + =
∫∫∫
π
π
π
π π
π π
(sec sec )
tan sec sec
(tan sec sec )
tan tan
tan tan ( . )
4 2
2 2 2
2 2 2
0
1
3
0
1
3
6
1 6
7
37
31 7 1 38 2049
0
1
0
1
[( ) ]
K
b. dV = 2π (x + 3) y dx = 2π (x + 3)(sec2 x) dx
V x x dx x dx= +∫ ∫2 60
12 2
0
1
π πsec sec
= − +∫2 2 60
1
0
1
0
1
π π πx x x dx xtan tan tan
= +8 1 20
1π πtan ln cos| |x
= 8π tan 1 + 2π ln (cos 1) (= 35.2738…)
35. dA r d d= = +1
2
1
25 42 2θ θ θ( cos )
A d= + +∫1
216 40 252
0
4
( cos cos )/
θ θ θπ
= + + +4 2 2 2025
2 0
4
θ θ θ θπ
sin sin/
= + + + = …π π2 10 225
829 1012. ,
which agrees with the numerical answer.
36. dA r d a d= = +1
2
1
212 2θ θ θ( )2cos
A a d
a
= (1+ 2 +
= + + +
∫1
2
1
22
1
2
1
42
2 2
0
2
2
0
2
cos cos
sin sin
θ θ θ
θ θ θ θ
π
π
)
= 3
22πa , which is 1.5 times Acircle.
37. Answers will vary.
Calculus Solutions Manual Problem Set 9-6 225© 2005 Key Curriculum Press
Problem Set 9-6
Q1.1
33sin x C+ Q2. − +1
44cos x C
Q3. − +1
55ln cos| |x C Q4.
1
66ln sin| |x C+
Q5.1
77 7ln sec tan| |x x C+ +
Q6. 5 sec2 5x
Q7. y′ = 4 cos 4x
Q8. d
Q9. See the text for the statement of fundamentaltheorem of calculus.
Q10. See the text for the definition of indefiniteintegral, Section 5-3.
Note: A radical without a sign in front of it means thepositive root. Because trigonometric functions can bepositive or negative, the radical should technically bereplaced by the absolute value of the appropriatetrigonometric function. Fortunately, this turns out to beunnecessary. If x has been replaced by a sin θ, a tan θ,or a sec θ, it is assumed that θ is the correspondinginverse trigonometric function. So θ is restricted to therange of that inverse trigonometric function. Thus,respectively,
a x a2 2– cos= ∈| |, and θ θ Quadrant I or IV
a x a2 2+ = ∈| |, and sec θ θ Quadrant I or IV
x a a2 2– tan= ∈| |, and θ θ Quadrant I or II
For the first two, the absolute value is unnecessarybecause cos θ ≥ 0 and sec θ ≥ 0 in the respectivequadrants. For the secant substitution, if x is negative,then θ is in Quadrant II, where tan θ < 0. Thus, theradical equals the opposite of a tan θ, and one shouldwrite
x a a2 2– tan= − θWhere the integral of sec θ occurs, one gets
ln x x a x+ +2 2 0for >
− − + <ln x x a x2 2 0for
The second form can be transformed into the first bytaking advantage of the property −ln n = ln (1/n). Thus,
− − =ln – ln– –
x a xx a x
2 2
2 2
1
= +ln –x a x2 2
which can be shown by rationalizing the denominatorof the fraction and incorporating the constant ln a2
(or 2 ln a) into the constant of integration. Becausethe major focus of this section is on the correct
substitution to use and the ensuing calculus, andbecause algebraic techniques are of less importance nowthat technology is used for evaluating integrals, thestudent is not expected to carry along the absolute valuejust to eliminate it later.
1. 49 2– x dx∫
u
v
θ
√ 49 – x2
7x
Let . , ,x
x dx d7
7 7= = =sin sin cosθ θ θ θ
49 77
2 1– cos sinxx= = −θ θ,
∴ = ∫ ∫49 7 72– cos cosx dx dθ θ θ( )
= = +∫ ∫4949
21 22cos cosθ θ θ θd d( )
= + +49
2
49
42θ θsin C
= + +49
2
49
2θ θ θsin cos C
= + ⋅ ⋅ +−49
2 7
49
2
1
7
1
7491 2sin –
xx x C
= + − +−49
2 7
1
2491 2sin
xx x C
2. 100 2– x dx∫
u
v
θ
√ 100 – x2
10x
Let . , ,x
x dx d10
10 10= = =sin sin cosθ θ θ θ
100 1010
2 1– cos sinxx= = −θ θ,
∴ = ∫ ∫100 10 102– cos cosx dx dθ θ θ( )
= = + )∫ ∫100100
21 22cos cosθ θ θ θd d(
= 50θ + 25 sin 2θ + C= 50θ + 50 sin θ cos θ + C
= + ⋅ ⋅ +−5010
501
10
1
101001 2sin –
xx x C
= + +−5010
1
21001 2sin –
xx x C
226 Problem Set 9-6 Calculus Solutions Manual© 2005 Key Curriculum Press
3. x dx2 16+∫
u
v
θ
4
x
√x2 + 16
Let . , ,x
x dx d4
4 4 2= = =tan tan secθ θ θ θ
xx2 116 44
+ = = −sec tanθ θ,
∴ + =∫ ∫x dx d2 216 4 4sec secθ θ θ( )
= ∫16 3sec θ θd (Compare Problem 21 in
Problem Set 9-4.)
= + + +16
2
16
2 1sec tan ln sec tanθ θ θ θ| | C
= + + + + +1
216 8
16
4 42
2
1x xx x
Cln
= + + + + − +1
216 8 16 8 42 2
1x x x x Cln ln
= + + + + +1
216 8 162 2x x x x Cln
4. 81 2+∫ x dx
u
v
θ
9
x
√81 + x2
Let . , ,x
x dx d9
9 9 2= = =tan tan secθ θ θ θ
81 99
2 1+ = = −xx
sec tanθ θ,
∴ + =∫ ∫81 9 92 2x dx dsec secθ θ θ( )
= ∫81 3sec θ θd (Compare Problem 21 in
Problem Set 9-4.)
= + + +81
2
81
2 1sec tan ln sec tanθ θ θ θ| | C
= + + + + +1
281
81
2
81
9 92
2
1x xx x
Cln
= + + + + − +1
281
81
281
81
292 2
1x x x x Cln ln
= + + + + +1
281
81
2812 2x x x x Cln
5. 9 12x dx–∫
u
v
θ
1
3x √9x2 – 1
Let . ,3
1
1
3
xx= =sec secθ θ
dx d= 1
3sec tanθ θ θ,
9 1 32 1x x– tan sec= = −θ θ,
∴ =
∫∫ 9 1
1
32x dx d– tan sec tanθ θ θ θ
= ∫1
32sec tanθ θ θd
= − )∫1
33(sec secθ θ θd
= + +1
6
1
6sec tan ln sec tanθ θ θ θ| |
− + +1
3ln sec tan| |θ θ C
= − + +1
6
1
6sec tan ln sec tanθ θ θ θ| | C
= − + +1
29 1
1
63 9 12 2x x x x C– ln –
6. 16 12x dx–∫
u
v
θ
1
4x √16x2 – 1
Let . ,4
1
1
4
xx= =sec secθ θ
dx d= 1
4sec tanθ θ θ,
16 1 42 1x x– tan sec= = −θ θ,
∴ =
∫ ∫16 1
1
42x dx d– tan sec tanθ θ θ θ
= ∫1
42sec tanθ θ θd
= −∫1
43( )sec secθ θ θd
= + +1
8
1
8sec tan ln sec tanθ θ θ θ| |
− + +1
4ln sec tan| | θ θ C
Calculus Solutions Manual Problem Set 9-6 227© 2005 Key Curriculum Press
= − + +1
8
1
8sec tan ln sec tanθ θ θ θ| | C
= − + +1
216 1
1
84 16 12 2x x x x C– ln –
7.dx
x17 2–∫
u
v
θ
x√17
√17 – x 2
Let / .x 17 = sin θx dx d= =17 17sin cosθ θ θ, ,
17 1717
2 1– cos sinxx= = −θ θ,
∴ =∫ ∫dx
x
d
17
17
172–
cos
cos
θ θθ
= = + = +−∫ d Cx
Cθ θ sin 1
17
8.dx
x13 2–∫
u
v
θ
x√ 13
√13 – x 2
Let / . ,x x13 13= =sin sinθ θdx d= ,13 cos θ θ
13 1313
2 1– cos sinxx= = −θ θ,
∴ =∫ ∫dx
x
d
13
13
132–
cos
cos
θ θθ
= = + = +−∫ d Cx
Cθ θ sin 1
13
9.dx
x2 1+∫
u
v
θ
x
1
√ x2 + 1
Let . ,x
dx d1
2= =tan secθ θ θ
x x2 11+ = = −sec tanθ θ,
∴+
= =∫ ∫ ∫dx
x
dd
2
2
1
sec
secsec
θ θθ
θ θ
= + + = + + +ln sec tan ln| | θ θ C x x C2 1
10.dx
x2 121–∫
u
v
θ
x
11
√x 2 – 121
Let . ,x
x11
11= =sec secθ θ
dx d= 11sec tanθ θ θ,
xx2 1121 11
11– tan sec= , = −θ θ
∴ = ∫∫ dx
x
d2 121
11
11–
sec tan
tan
θ θ θθ
= = + +∫ sec ln sec tanθ θ θ θ | |d C1
= + +ln–x x
C11
121
11
2
1
= + − +ln – lnx x C21121 11
= + +ln –x x C2 121
11. x x dx2 2 9–∫
u
v
θ
x
3
√x 2 – 9
Let . ,x
x3
3= =sec secθ θ
dx = 3 sec θ tan θ dθ,
xx2 19 33
– tan sec= = −θ θ,
∴∫ x x dx2 2 9–
= ∫ ( ) ( ) ( ) 9 3 32sec tan sec tanθ θ θ θ θd
= ∫81 3 2sec tanθ θ θd
= −
∫ ∫81 5 3sec secθ θ θ θd d
= + −
∫ ∫81
1
4
3
43 3 3sec tan sec secθ θ θ θ θ θd d
= − ∫81
4
81
4sec tan sec3 3θ θ θ θd
228 Problem Set 9-6 Calculus Solutions Manual© 2005 Key Curriculum Press
= −81
4
81
83sec tan sec tanθ θ θ θ
− + +81
8 1ln sec tan| |θ θ C
= ⋅ ⋅ − ⋅ ⋅81
4 27
9
3
81
8 3
9
3
3 2 2x x x x– –
− + +81
8 3
9
3
2
1ln–x x
C
= −1
49
9
893 2 2x x x x– –
− + + +81
89 32
1ln – lnx x C81
8
= −1
49
9
893 2 2x x x x– –
− + +81
892ln –x x C
12. x x dx2 29 –∫
u
v
θ
x3
√9 – x2
Let . , ,x
x dx d3
3 3= = =sin sin cosθ θ θ θ
9 33
2 1– cos sinxx= = −θ θ,
∴∫ x x dx2 29 –
= ∫ ( sin ) cos cos9 3 32 θ θ θ θ( ) ( )d
= ∫81 2 2sin cosθ θ θd
= −∫81 2 4( )cos cosθ θ θd
= −∫8181
42 3cos cos sinθ θ θ θd
− ⋅ ∫3 81
42cos θ θd
= + −∫81
81 2
81
43( )cos cos sinθ θ θ θd
= + − +81
8
81
162
81
43θ θ θ θ sin cos sin C
= + − +81
8
81
8
81
43θ θ θ θ θsin cos cos sin C
= + − +81
8
81
81 2 2θ θ θ θ ( )sin cos cos C
= −81
8 31sin
x
− ⋅ ⋅ ⋅
+81
8 3
9
31
2 9
9
2 2x x xC
––
( – )
= − − +−81
8 3
1
82 9 91 2 2sin –
xx x x C( )
13. ( ) /1 2 3 2−∫ x dx
u
v
θ
x1
√1 – x2
Let x = sin θ. dx = cos θ dθ,
1 2 1– cos sinx x= = −θ θ,
∴ − = ∫∫ ( ) ( )/1 2 3 2 3x dx dcos cosθ θ θ
= ∫ cos4 θ θd
= + ∫1
4
3
43 2cos sin cosθ θ θ θd
= + +∫1
4
3
81 23cos sin cosθ θ θ θ( ) d
= + + +1
4
3
8
3
1623cos sin sinθ θ θ θ C
= + + +1
4
3
8
3
83cos sin sin cosθ θ θ θ θ C
= − + + +−1
41
3
8
3
812 3 2 1 2x x x x x C( ) / sin –
14. ( ) /x dx2 3 281− −∫
u
v
θ
9
x√x 2 – 81
Let . , ,x
x dx d9
9 9= = =sec sec sec tanθ θ θ θ θ
x x2 181 91
9– tan sec= = −θ θ,
∴ − −∫ ( ) /x dx2 3 281
= ∫ −( tan ) sec tan9 93θ θ θ θ( )d
= ∫1
81 2
sec
tan
θ θθd
= ∫1
81cot cscθ θ θd
= − +1
81csc θ C
= +–
–
x
xC
81 812
Calculus Solutions Manual Problem Set 9-6 229© 2005 Key Curriculum Press
15.dx
x81 2+∫
u
v
θ
9
x
√81 + x 2
Let . , ,x
x dx d9
9 9 2= = =tan tan secθ θ θ θ
81 99
2 1+ = = −xx
sec tanθ θ,
∴+
= = = +∫ ∫∫dx
x
dd C
81
9
81
1
9
1
92
2
2
sec
sec
θ θθ
θ θ
= +−1
9 91tan
xC
16.dx
x25 12 +∫
u
v
θ
1
5x
√25x2 + 1
Let . , ,5
1
1
5
1
52x
x dx d= = =tan tan secθ θ θ θ
25 1 52 1x x+ = = −sec tanθ θ,
∴+
=⋅
= = +∫∫∫ dx
x
dd C
25 1 5
1
5
1
52
2
2
sec
sec
θ θθ
θ θ
= +−1
551tan x C
17. a.x dx
x2 25+∫
u
v
θ
5
x√x 2 + 25
Let . , ,x
x dx d5
5 5 2= = =tan tan secθ θ θ θ
x x2 125 51
5+ = = −sec tanθ θ,
∴+
=⋅∫∫ x dx
x
d2
2
25
5 5
5
tan ( sec )
sec
θ θ θθ
= = +∫5 5tan sec secθ θ θ θd C
= + +x C2 25
b.x dx
xx x dx
2
2 1 2
25
1
225 2
+= + − /∫∫ ( ) ( )
= + +x C2 25 , which agrees with part a.Moral: Always check for an easy way tointegrate before trying a more sophisticatedtechnique!
18. a.x dx
x2 49–∫
u
v
θ
7
x√x 2 – 49
Let .x
7= sec θ x = 7 sec θ,
dx = 7 sec θ tan θ dθ,
xx2 149 77
– tan sec= = −θ θ,
∴ = ∫∫ x dx
x
d2 49
7 7
7–
sec ( sec tan )
tan
θ θ θ θθ
= = + = +∫7 7 492 2sec tan –θ θ θd C x C
b.x dx
xx x dx
2
2 12
49
1
249 2
–= − − /∫∫ ( ) ( )
= +x C2 49– ,which agrees with part a.
19.dx
x9 5 2– ( – )∫
u
v
θ
3 x – 5
√ 9 – (x – 5)2
Let . , ,x
x dx d–
sin sin cos5
35 3 3= = + =θ θ θ θ
9 5 35
32 1– ( – ) cos sin
–x
x= = −θ θ,
∴ = ∫∫ dx
x
d
9 5
3
32– ( – )
cos
cos
θ θθ
= = + = +−∫ d Cx
Cθ θ sin–1 5
3
20.dx
x36 2 2– ( )+∫
u
v
θ
6 x + 2
√ 36 – (x + 2)2
230 Problem Set 9-6 Calculus Solutions Manual© 2005 Key Curriculum Press
Let . , ,x
x dx d+ = = − =2
66 2 6sin sin cosθ θ θ θ
36 2 62
62 1– ( ) cos sinx
x+ = = +−θ θ,
∴+
= ∫∫ dx
x
d
36 2
6
62– ( )
cos
cos
θ θθ
= = + = + +−∫ d Cx
Cθ θ sin 1 2
6
21.dx
x x
dx
x2 28 20 4 36+=
+∫∫ – ( ) –
u
v
6
x + 4 √ (x + 4)2 –36
Let . ,x
x+ = = −4
66 4sec secθ θ
dx = 6 sec θ tan θ dθ,
x x x2 28 20 4 36 6+ = + =– ( ) – tan θ,
θ = +−sec 1 4
6
x
∴+
= ∫∫ dx
x x
d2 8 20
6
6–
sec tan
tan
θ θ θθ
= ∫ sec θ θd
= ln | sec θ + tan θ | + C1
= + + + +ln–x x x
C4
6
8 20
6
2
1
= + + + − +ln – lnx x x C4 8 20 621
= + + + +ln –x x x C4 8 202
22.dx
x x
dx
x2 214 50 7 1– ( – )+=
+∫∫
u
v
θ
1
x – 7
√(x – 7)2 + 1
Let . , ,x
x dx d–
tan tan sec7
17 2= = + =θ θ θ θ
x x x2 214 50 7 1– ( – ) sec+ = + = θ,
θ = tan− 1 (x − 7)
∴+
= = ∫∫∫ dx
x x
dd
2
2
14 50–
sec
secsec
θ θθ
θ θ
= ln | sec θ + tan θ | + C
= + + − +ln –x x x C2 14 50 7
23. 100 2
3
8
– x dx−∫
u
v
θ
10 x
√100 – x 2
Let . , ,x
x dx d10
10 10= = =sin sin cosθ θ θ θ
100 1010
2 1– cos sinxx= = −θ θ,
∴−∫ 100 2
3
8
– x dx
= ⋅∫ 10 101
1
0 3
0 8
cos cossin (– . )
sin .
–
–
θ θ θd
= ∫100 2
0 3
0 8
1
1
cossin (– . )
sin .
–
–
θ θd
= +∫50 ( cos )sin (– . )
sin .
–
–
1 21
1
0 3
0 8
θ θd
= +50 25 sin 2θ θsin (– . )
sin .–
–
1
1
0 3
0 8
= +− −50 sin 0.8 25 sin (2 sin 0.8)11
− 50 sin− 1 (−0.3) − 25 sin [2 sin− 1 (−0.3)]= 99.9084…Numerical integration: 99.9084… (Checks.)
24. x dx2
1
4
25+∫–
u
v
θ
5
x√x 2 + 25
Let . , ,x
x dx d5
5 5 2= = =tan tan secθ θ θ θ
x x2 125 5 0 2+ = = −sec tanθ θ, .
∴ +∫ x dx2
1
4
25–
= ⋅∫ 5 5 2
0 2
0 8
1
1
sec sectan (– . )
tan .
–
–
θ θ θd
= ∫25 3
0 2
0 8
1
1
sectan (– . )
tan .
–
–
θ θd
= + +−
−
−
25
2
25
2 1
1
0 2
0 8sec tan
tan ( . )
tan .θ θ θ θln |sec tan |
Calculus Solutions Manual Problem Set 9-6 231© 2005 Key Curriculum Press
= ⋅−25
20 8 0 81sec tan( . ) .
+ +−25
20 8 0 81ln sec tan| ( . ) . |
− − ⋅ −−25
20 2 0 21sec [tan ]( . ) ( . )
− − −−25
20 2 0 21ln sec [tan ]| ( . ) . |
= 26.9977…Numerical integration: 26.9977… (Checks.)
25. y = 3x2
dL y dx x dx= + ′ = +1 1 362 2( )
L x dx= +∫ 1 36 2
0
5
u
v
θ
1
6x√ 1 + 36x2
Let . , ,6
1
1
6
1
62x
x dx d= = =tan tan secθ θ θ θ
1 36 62 1+ = = −x xsec tanθ θ,
∴ L dx
x
= ⋅=
=
∫ sec secθ θ θ1
62
0
5
==
=
∫1
63
0
5
sec θ θdx
x
= + +=
=1
12
1
12 0
5
sec tan ln |sec tan |θ θ θ θx
x
= + + + +1
21 36
1
121 36 62 2
0
5
x x x xln
= + + =5
2901
1
12901 30 75 3828ln . K
Numerical integration: L = 75.3828… (Checks.)
26. a. 9 25 2253
5252 2 2x y y x+ = ⇒ = ± –
Slice the region vertically. Pick a samplepoint (x, y) on the positive branch of thegraph, within the strip.
dA y dx x dx= =26
525 2–
A x dx= ∫6
525 2
3
4
––
u
v
θ
5x
√ 25 – x2
Let , ,x
x dx d5
5 5= . = =sin sin cosθ θ θ θ
25 55
2 1– cos sinxx= = −θ θ,
∴ A dx
x
= ⋅=
=
∫6
55 5
3
4
cos cos–
θ θ θ
==
=
∫30 2
3
4
cos–
θ θdx
x
= +=
=
∫15 1 23
4
( cos )–
θ θdx
x
= +=−
=15
15
22
3
4
θ θsinx
x
= + ==15 15 3
4θ θ θsin cos –xx
= +−
=−
=15
5
3
5251 2
3
4sin –
xx x
x
x
= +−15 0 83
54 91sin ( ).
− − −−15 0 63
53 161sin (– )( . )
= − − +− −15 0 8 0 6 14 41 1[sin sin ]. ( . ) .
= + = …15
214 4 37 9619
π. .
Numerical integration: A = 37.9619…(Checks.)
b. A x dx= ∫6
525 2
5
5
––
x = 5 ⇒ θ = π/2, x = −5 ⇒ θ = −π/2
∴ A d= ∫30 2
2
2
cos– /
/
θ θπ
π
= +−
1515
22
2
2
θ θπ
πsin
/
/
= + + − −15
2
15
2
15
2
15
2
π π π πsin sin ( )
= 15π = 47.1238…The area is π (x-radius)(y-radius).
27. x y r y r x x y r2 2 2 2 2 0+ = ⇒ = ± = = ±– , at
Slice the region inside the circle perpendicular tothe x-axis. Pick sample point (x, y) on thepositive branch of the circle, within the strip.
dA y dx r x dx= = −2 2 2 2
A r x dxr
r
= ∫2 2 2––
u
v
θ
rx
√r 2 – x 2
Let x
r= sin θ. x = r sin θ, dx = r cos θ dθ,
r x rx
r2 2 1– cos sin= = −θ θ,
x = r ⇒ θ = π/2, x = −r ⇒ θ = −π/2
∴ = ⋅∫A r r d22
2
cos cos– /
/
θ θ θπ
π
=− /
/
∫2 2 2
2
2
r dcos θ θπ
π
232 Problem Set 9-6 Calculus Solutions Manual© 2005 Key Curriculum Press
= +− /
/
∫r d2
2
2
1 2( cos )θ θπ
π
= +−
r r2 2
2
21
22θ θ
π
πsin
/
/
= ⋅ + + ⋅r r r2 2 2
2
1
2 2
π π πsin
− − =1
22 2r rsin ( )π π
∴ A = πr2, Q.E.D.
28.x
a
y
by
b
aa x
+
= ⇒ = ±
2 22 21 –
Slice the region inside the ellipse perpendicularto the x-axis. Pick sample point (x, y) on thepositive branch of the ellipse, within the strip.
dA y dxb
aa x dx= =2
2 2 2–
Ab
aa x dx
a
a
= ∫2 2 2––
u
v
θ
ax
√a 2 – x 2
Let x
a= sin θ. x = a sin θ, dx = a cos θ dθ,
a x ax
a2 2 1– cos sin= = −θ θ,
x = a ⇒ θ = π/2, x = −a ⇒ θ = −π/2
∴ = ⋅∫Ab
aa a d
22
2
cos cos– /
/
θ θ θπ
π
= ∫2 2
2
2
ab dcos– /
/
θ θπ
π
= +∫ab d( cos )– /
/
1 22
2
θ θπ
π
= +−
ababθ θ
π
π
22
2
2
sin/
/
= + + − − =ab ab ab abab
π π π π π2 2 2 2
sin sin ( )
∴ A = πab
Note that if a = b = r, then πab = πr2, the area ofa circle.
29. dV x dya
bb y dy b y b= = − ≤ ≤π π2
2
22 2( – ) ,
Va
bb y dy
b
b
= ⋅ ∫π2
22 2( – )
–
= ⋅
=−
π πa
bb y
ya b
b
b2
22
32
3
4
3–
∴ =V a b4
32π
Rotating instead about the x-axis is equivalent tointerchanging the a and b, giving V ab= 4
32π .
30. x y y x2 2 29 9− = ⇒ = ± –
Slice the region perpendicular to the x-axis. Picka sample point (x, y) on the positive branch ofthe hyperbola, within the strip.
dA y dx x dx= =2 2 92 –
A x dx= ∫2 92
3
5
–
u
v
θ
3
x√x 2 – 9
Let x
3= secθ. x = 3 sec θ, dx = 3 sec θ tan θ dθ,
xx2 19 33
– tan sec= = −θ θ,
∴ = ⋅=
=
∫A dx
x
2 3 33
5
tan sec tanθ θ θ θ
==
=
∫18 2
3
5
tan secθ θ θdx
x
==
=
∫18 3
3
5
(sec – sec )θ θ θdx
x
= 9 sec θ tan θ + 9 ln | sec θ + tan θ |− + =
= |18
3
5ln sec tan |θ θ
x
x
= − + ==9 9 3
5sec tan ln sec tanθ θ θ θ| | xx
= − +x x x x 2 2
3
5
9 91
3
1
39– ln –
= 20 − 9 ln 3 = 10.1124…Numerical integration: A = 10.1124… (Checks.)
31. dV x y dx x x dx= =2 2 4 92π π( ) –
V x x dx= ∫4 92
3
5
π –
= ∫2 9 22
3
5
π x x dx– ( )
= ⋅ /2 3 2π 2
392
3
5
( – )x
= ⋅ = = …4
364
256
3268 0825π π .
32. From Problems 30 and 31, A = 20 − 9 ln 3,
V = 256
3π .
V x A x= ⋅ ⇒ = =2
128
3 20 9 34 2192π
( – ln ). K
x is a little more than halfway through theregion.
Calculus Solutions Manual Problem Set 9-7 233© 2005 Key Curriculum Press
33. x = a cos t ⇒ dx = −a sin t dty = b sin tdA = 2y dx = 2(b sin t)(−a sin t dt)
= −2ab sin2 t dtx = −a ⇒ t = π, x = a ⇒ t = 0
∴ = − = − ∫∫A ab t dt ab t dt2 1 2200
sin ( – cos )ππ
= − + = + + − =abtab
t ab ab2
2 0 0 00
sinπ
π π( )
∴ A = πab, as in Problem 28.
With this method, you get sin2 t dt,∫ directly.
With trigonometric substitution in Problem 28,
you get cos2 t dt,∫ indirectly.
34. r = 0.5θ ⇒ dr/dθ = 0.5
dL r dr d d d= + = +2 2 20 25 0 25( / ) . .θ θ θ θ
= +0 5 12. θ θd
L d= +∫0 5 12
0
6
. θ θπ
u
v
θ
1
φ
√ θ2 + 1
Let θ = tan φ ⇒ dθ = sec2 φ dφ.
θ φ φ θ2 11+ = = −sec tan,
∴ = ⋅=
=
∫L d0 5 2
0
6
. sec secφ φ φθ
θ π
==
=
∫0 5 3
0
6
. sec φ φθ
θ πd
= +0 25 0 25. . |sec tan ln secφ φ φ+ =
=tan φ θ
θ π|
0
6
= + + + +0 25 1 0 25 12 2
0
6
. .θ θ θ θπ
ln
= + + + +1 5 36 1 0 25 36 1 62 2. .π π π πln
= 89.8589… , same as numericalintegration.
35. See the note preceding the solutions for thissection. For the sine and tangent substitution,the range of the inverse sine and inverse tangentmake the corresponding radical positive. For thesecant substitution, the situation is morecomplicated but still gives an answer of the samealgebraic form as if x had been only positive.
Problem Set 9-7Q1. (x + 5)(x − 5) Q2. x2 + 2x − 15
Q3. (x + 2)(x − 6) Q4. x2 + 14x + 49
Q5. (x + 4)2 Q6. x2 − 64
Q7. e x
Q8. b2 − 4ac = −1500, so x2 + 50x + 1000 is prime.
Q9. b2 − 4ac = −144, so x2 + 36 is prime.
Q10. B
1.11 15
3 2
4
1
7
22
x
x xdx
x xdx
–
– – –+= +
∫∫
= 4 ln |x − 1| + 7 ln |x − 2| + C
2.7 25
7 8
2
1
9
82
x
x xdx
x xdx
+ =+
+
∫∫ – –
–
–= −2 ln |x + 1| + 9 ln |x − 8| + C
3.( – )
– –
/ /
–
5 11
2 8
7 2
2
3 2
42
x dx
x x x xdx=
++
∫∫
= + + − +7
22
3
24ln ln| | | |x x C
4.( – )
– –
/ /
–
3 12
5 50
9 5
5
6 5
102
x dx
x x x xdx=
++
∫∫
= + + − +9
55
6
510ln ln| | | |x x C
5.21
7 10
7
5
7
22
dx
x x x xdx
+ +=
++
+
∫∫ –
= −7 ln |x + 5| + 7 ln |x + 2| + C
6.10
9 36
2
3
8
122
x dx
x x x xdx
– – –=
++
∫∫
= 2 ln |x + 3| + 8 ln |x − 12| + C
7.9 25 50
1 7 2
2x x
x x xdx
– –
( )( – )( )+ +∫=
++ +
+
∫ 2
1
3
7
4
2x x xdx
–= 2 ln |x + 1| + 3 ln |x − 7| + 4 ln |x + 2| + C
8.7 22 54
2 4 1
2x x
x x xdx
++∫ –
( – )( )( – )
= ++
+
∫ 3
2
1
4
5
1x x xdx
–
–
–= 3 ln |x − 2| − ln |x + 4| + 5 ln |x − 1| + C
9.4 15 1
2 5 6
2
3 2
x x
x x xdx
++∫ –
– –
=+
++
+
∫ –
–
1
3
2
1
3
2x x xdx
= −ln |x + 3| + 2 ln |x + 1| + 3 ln |x − 2| + C
10.– –
– –
3 22 31
8 19 12
2
3 2
x x
x x xdx
++∫
= + +
∫ –
–
–
– –
2
1
4
3
3
4x x xdx
= −2 ln |x − 1| − 4 ln |x − 3| + 3 ln |x − 4| + C
11.3 2 12 9
1
3 2x x x
xdx
+ +∫ –
–
= + − +
∫ 3 5 7
2
12x x
xdx
–
= + − + − +x x x x C3 25
27 2 1ln | |
234 Problem Set 9-7 Calculus Solutions Manual© 2005 Key Curriculum Press
12.x x x
x xdx
3 2
2
7 5 40
2 8
–
– –
+ +∫= − +
∫ x
x
x xdx5
3
2 82 – –
= − ++
+
∫ x
x xdx5
1
2
2
4–
= − + + + − +1
25 2 2 42x x x x Cln ln| | | |
13.4 6 11
1 4
2
2
x x
x xdx
+ ++ +∫ ( )( )
= ++
++
∫ x
x xdx
2
1
3
42
=+
++
++∫∫∫1
2
2
12
1
3
42 2
x dx
x
dx
x
dx
x
= + + + + +−1
21 2 3 42 1ln tan ln| | | |x x x C
14.4 15 1
5 3 1
2
3 2
x x
x x xdx
– –
– + +∫= +
∫ 3
1
2
4 12x
x
x xdx
–
–
– –
= − + − − +3 11
24 12ln ln| | | |x x x C
Note thatx
x x x x
–
– –
/
–
/
– –,
2
4 1
1 2
2 5
1 2
2 52 =+
+
but1
22 5
1
22 5ln lnx x− + + − −
= − −1
24 12ln | |,x x so the answer comes out
the same.
15.4 18 6
5 1
2
2
x x
x xdx
+ ++ +∫ ( )( )
=+
++
++
∫ 1
5
3
1
2
1 2x x xdx
–
( )
= ln |x + 5| + 3 ln |x + 1| + 2(x + 1)− 1 + C
16.
= + +
∫ 5 2
7
3
7 2x x xdx
–
– ( – )
= 5 ln |x| − 2 ln |x − 7| − 3(x − 7)− 1 + C
17.dx
x x x
dx
x3 2 36 12 8 2– – ( – )+=∫ ∫
= − +−1
22 2( – )x C
18.1
4 6 4 1 14 3 2 4x x x xdx
dx
x+ + + +=
+∫ ∫ ( )
= − + +−1
31 3( )x C
19. a.dy
dty
y dy
y ydt= ⇒ =2
1000
1000
1000
10002
–
( – )
1000
10002
dy
y ydt
( – )=∫ ∫
1 1
10002
y ydy dt+
=∫ ∫–
ln |y| − ln |1000 − y| = 2t + C
ln–
y
yt C
10002= +
y
ye t C
10002
–= + (Note that 0 ≤ y < 1000.)
1000 100012 2− = ⇒ − =− − − −y
ye
yet C t C
10001 12 2
ye ke k et C t C= + = + =− − − − ( )
yke t=
+1000
1 2–
Initial condition y = 10 when t = 0 ⇒ k = 99.
ye t=
+1000
1 99 2–
b. ye
( ) . students11000
1 9969 4531 692=
+= … ≈–
have heard the rumor after one hour.
ye
( ) .41000
1 99967 8567 9688=
+= … ≈–
students have heard by lunchtime.
ye
( ) .–81000
1 99999 9888 100016=
+= … ≈
students—everyone knows by the end ofthe day!
c. It is quicker to analyze the original differentialequation, which already refers to the derivative,than to analyze the equation found in part a.
Maximize y yy′ = 2
1000
1000
–
= 1
5001000 2( – )y y .
′′ = ′ ′ = =y y yy y1
5001000 2 0 500( – ) when .
This is the maximum point because y″ > 0for y < 500 and y″ < 0 for y > 500 (andy′ > 0 for all t).So the rate of spreading (y′) is greatest when500 students have heard the news. This occurswhen
5001000
1 99 2=+ e t–
99 1 22e t− + =
e t− =2 1
99
t = = …1
299 2 2975ln . hr
3 53 245
14 49
2
3 2
x – x
x – x xdx
++∫
Calculus Solutions Manual Problem Set 9-7 235© 2005 Key Curriculum Press
d. The graph follows the slope-field pattern.
t
y
2 4 6
1000
20. a. Assume that an infected person and anuninfected person have about the same chanceof meeting any other infected person (i.e.,infected people are not quarantined). Aninfected person can meet N − P uninfectedpeople out of the total population, so thechance of meeting an uninfected person will be(N – 1)/N, so of an average infected person’sthree contacts per day, 3(N − 1)/N of themwill be with uninfected persons. (Actually(N − P)/(N − 1) because the total populationthat someone can meet is N − 1—people don’tmeet themselves outside the Twilight Zone—but (N − P)/N is reasonably close enough fornow.) So there are P infected people, eachmeeting an average of 3(N − P)/N uninfectedpeople per day, for a grand total of 3P(N − P)/Ncontacts between infected and uninfectedpeople per day.
b. If 10% of the contacts with infected peopleper day result in infection, then the numberof new infections per day should be 0.1 timesthe number of contacts between infected anduninfected people, that is,dP
dt
P N P
NP
N P
N= ⋅ =0 1
30 3. . .
( – ) –
c.dP
dtP
N P
N
N dP
P N Pdt= ⇒ =0 3 0 3. .
–
( – )
N dP
P N Pdt
( – )∫ ∫= 0.3
1 10 3
P N PdP dt+
=∫ ∫–
.
ln |P| − ln |N − P| = 0.3t + C
ln–
P
N Pt C= +0 3.
P
N Pe t C
–= +0.3 (Note that 0 ≤ P < N.)
N
Pe ke k et C t C= + = + =− − − −1 10 3 0 3. . ( )
P tN
ke t( ) =+1 0 3– .
Initial condition P(0) = P0
⇒ =+
⇒ = −PN
kk
N
P001
1.
∴ =+
( )P tN
N P e t1 100 3( / – ) – .
d. N = 1000 and P0 = 10
⇒ =+
P te t( )
1000
1 99 0 3– .
Pe
( )71000
1 99 2 1=+ – . = 76.2010… ≈ 76 people
infected after 1 week
e. Solve P(t) = 990.
9901000
1 99 0 3=+ e t– .
1 99100
990 3+ =−e t.
e0.3 t = 992
t = = ≈2
0 399 30 6341 31
.ln . daysK
21. Ax x
dxx x
dxb b
=+
= ++
∫ ∫25
3 4
5
1
5
422 2– –
–
= − − + =+
5 1 5 4 51
422
ln ln ln–
| | | |x xx
xb
b
=+
− =+
+51
45
1
65
1
45 6ln
–ln ln
–ln
b
b
b
b
A( ) .7 56
115 6 5 9281= + = …ln ln
lim lim ln lnb b
A b→∞ →∞
= +( ) 51
15 6
(l’Hospital’s rule)
= 5 ln 6 = 8.9587…So the area does approach a finite limit.
22. dV x y dxx
x xdx= =
+2
50
3 42π π–
Vx
x xdx
x xdx
b b
=+
= ++
∫ ∫50
3 4
10
1
40
422
π π π– –2
= − + +10 1 40 4 2π πln ln| | | |x x b
= 10π ln |b − 1| + 40π ln |b + 4| − 40π ln 6V(7) = 40π ln 11 − 30π ln 6 = 132.4590…
limb
V b→∞
= ∞( ) because both ln terms become
infinite and are added.(Note that if the region were rotated about thex-axis, the limit of the volume would be
finite. The answer would be 535
62 6π – ln
= 35.3400… .
23. a.x
x xdx
x xdx
–
–
/
–
/
–
3
6 8
1 2
2
1 2
42 +∫ ∫= +
= + +1
22
1
24ln ln| – | | – |x x C
b. x2 – 6x + 8 = (x – 3)2 – 1
u
v
θ
1
x – 3 √ (x – 3)2 – 1
236 Problem Set 9-7 Calculus Solutions Manual© 2005 Key Curriculum Press
Let x – 3 = sec θ. dx = sec θ tan θ dθ,
( – ) – tan secx x3 1 32 1= =θ θ, ( – )–
∴+
=∫ ∫x
x xdx
x
xdx
–
–
–
( – ) –
3
6 8
3
3 12 2
= = ∫∫ sec
tan(sec tan )
sec
tan
θθ
θ θ θ θθ
θ2
2
d d
= ln | tan θ | + C
= +ln ( – ) –x C3 12
= + +ln –x x C2 6 8
c.x
x xdx
x
x xdx
–
–
–
–
3
6 8
1
2
2 6
6 82 2+=
+∫ ∫= + +1
26 82ln | – |x x C
d. From part a,1
22
1
24ln ln| – | | – |x x C+ +
= +1
22 4ln |( – )( – )|x x C
= + +1
26 82ln | – |x x C
which is the answer in part c. This equals
ln ln –| – | ,/x x C x x C2 1 2 26 8 6 8+ + = + +which is the answer from part b. So all threeanswers are equivalent, Q.E.D.
24. a. When the population is very much smallerthan the maximum, (m – p) behaves like aconstant, and dp/dt = k(m – p) · p isapproximately proportional to p. But when pis approaching m, then (m – p) goes to zero,so dp/dt = kp(m – p) goes to zero.
b. dp/dt = kp(m – p) = k(mp – p2). So dp/dt is aquadratic function of p. Thus, the turningpoint is at
pm
m= =– / .2 1
2(– )
If k > 0, the graph of dp/dt versus p opensdownward and the turning point is amaximum.So the population grows fastest whenp = m/2.
c.dp
dtkp m p
dp
p m pk dt= ⇒ =( – )
( – )
dp
p m pk dt
( – )= ∫∫
1 1/ /
–
m
p
m
m pdp k dt+
=∫ ∫
1 11m
pm
m p kt Cln ln| | – | – | = +
d. ln–
p
m pkmt C= + 2 (C2 = mC1)
p
m pe C ekmt C kmt
–= =+ 2
3 ( )C eC3
2=
Note that > > > .m pp
m p0 0⇒
–
m p
pbe
m
pbe b Ckmt kmt– = ⇒ = =– –– ( / )1 1 3
m
pbe p
m
bekmt
kmt= + ⇒ =+
11
––
At time t = 0, p = p0.
∴ =+
⇒ = +pm
bm p b0 01
1( )
∴ = ++
pp b
be kmt0 1
1
( )–
Letting K = km, p pb
be Kt= ++ −01
1, Q.E.D.
e. Let p denote millions of people. Thenp0 = 179.3.Substitute p(10) = 203.2.
203 2 179 31
1 10. .= ++
b
be K–
⇒ 203.2 + 203.2be–10K = 179.3 + 179.3b⇒ b(203.2e–10K – 179.3) = –23.9
⇒ =be K
– .
. – .–
23 9
203 2 179 310
By substituting p(20) = 226.5 andtransforming,
be K= – .
. – ..–
47 2
226 5 179 320
Equating the two values of b and solvingnumerically for K gives K = 0.0259109… .
∴ = = …be
– .
. – .– . ...
23 9
203 2 179 31 06304360 259109 .
∴ =+
pe
179 32 063036
1 1 063036 0 0259109.. ...
. ... – . ...
Check that this equation gives a goodapproximation for 1990.
pe
( ) .30 179 32 0630
1 1 0630 30 0 0259= ⋅+ ⋅ ⋅
. ...
. ... – . ...
= 248.4892… ≈ 248.5 million people,which is close to the actual population,248.7 million.
f. pe
( ) .40 179 32 0630
1 1 0630 40 0 0259= ⋅+ ⋅ ⋅
. ...
. ... – . ...
= 268.6144… ≈ 268.6 million people, whichis lower than the actual population by about13 million people.
g. k p pb
bep b
t t kt> ( )01
110 0⇒ = +
+= +
→∞ →∞lim lim –
= 179.3 · (1 + 1.0630…)= 369.9024… ≈ 369.9 million people
Calculus Solutions Manual Problem Set 9-8 237© 2005 Key Curriculum Press
h. If p(10) had been 204.2, then K would havebeen given by
– .
. – .
– .
. – .– –
24 9
204 2 179 3
47 2
226 5 179 310 20e eK K=
⇒ K = 0.0343965…
⇒ =be
– .
. – .– . ...
24 9
204 2 179 30 0343965
= 0.721075…
So the ultimate population would have beenlimt
p p b→∞
= + = + …0 1 179 3 1 0 7210( ) . ( . )
= 308.5888… ≈ 308.6 million people.An increase of 1 million in one of the initialconditions causes a decrease of over61 million in the predicted maximumpopulation! So this model does have a fairlysensitive dependence on the initial conditions.
Problem Set 9-8Q1. integration by parts Q2. partial fractions
Q3. x = tan θ or θ = tan–1 x
Q4. x = sec θ or θ = sec–1 x
Q5. x = sin θ or θ = sin–1 x
Q6.1
1612 8( )x C+ +
Q7. 7 (at f (1)). Q8. 3 (at f (5)).
Q9. undefined Q10. B
1. tan−∫ 1 x dx u dvtan–1 x 1
dx1 + x2 x–
+
= −+
− ∫x xx dx
xtan 1
2 1
= −+
− ∫x xx dx
xtan 1
2
1
2
2
1
= − + +−x x x Ctan ln | |1 21
21 (Checks.)
2. cot−∫ 1 x dx u dvcot–1 x 1
dx1 + x2 x
+
–
= ++
− ∫x xx dx
xcot 1
2 1
= ++
− ∫x xx dx
xcot 1
2
1
2
2
1
= + + +−x x x Ccot ln | |1 21
21 (Checks.)
3. cos−∫ 1 x dx
√
u dvcos –1 x 1dx
1 – x2x–
+
= +−
− ∫x xx dx
xcos 1
21
= − − −− −∫x x x x dxcos ( ) ( )/1 2 1 21
21 2
= − ⋅ − +−x x x Ccos ( ) /1 2 1 21
22 1
= − − +−x x x Ccos 1 21 (Checks.)
4. sin−∫ 1 x dx
√
u dvsin–1 x 1dx
1 – x2x
+
–
= −−
− ∫x xx dx
xsin 1
21
= + − −− −∫x x x x dxsin ( ) ( )/1 2 1 21
21 2
= + ⋅ − +−x x x Csin ( ) /1 2 1 21
22 1
= + − +−x x x Csin 1 21
5. sec−∫ 1 x dx
√
sec –1 x 1dx
|x| x2 – 1x–
+
u dv
= −−
− ∫x xx dx
x xsec
| |
1
2 1
= −−
=− ∫x x xdx
xx x xsec sgn ( /| | sgn )1
2 1
u
v
θ
1
x √x 2 – 1
Let .x
1= sec θ
dx = sec θ tan θ dθx2 1– = tan θ
θ = sec–1 x
∴ ∫ sec–1 x dx
= ∫x x xd
sec sgnsec tan
tan– –1 θ θ θ
θ= ∫x x x dsec sgn sec– –1 θ θ
238 Problem Set 9-8 Calculus Solutions Manual© 2005 Key Curriculum Press
= x sec–1 x – sgn x ln |sec θ + tan θ| + C
= x sec–1 x – sgn x ln –x x C+ +2 1
(Checks.)Note: This answer can be transformed to
x x x x Csec ln ( – ) .– – | |1 2 1+ +
6. csc−∫ 1 x dx
√
u dvcsc –1 x 1dx
|x| x2 – 1x
+
–
= + ∫x xx dx
x xcsc
| | –
– 1
2 1
= + ∫x x xdx
xcsc sgn
–
–1
2 1
u
v
θ
1x
√ x2 – 1
Let x = csc θ. dx = –csc θ cot θ dθ,
x x2 11– cot csc= =θ θ, –
∴∫ csc– 1 x dx
= ∫x x xd
csc sgncsc cot
cot– –1 θ θ θ
θ= ∫x x x dcsc sgn csc– –1 θ θ
= x csc–1 x + sgn x ln | csc θ + cot θ | + C
= + + +x x x x x Ccsc sgn ln ––1 2 1
(Checks.)Note: This answer can be transformed to
x x x x Csec ln –− + + +1 2 1(| | ) .
7. tan tan ln− −= − +∫ 1 1 2
1
4
1
41
21x dx x x x| |
= − − +− −4 41
217 1
1
221 1tan ln tan ln
= − − =−4 44
1
2
17
23 44781tan ln
π. K
Numerically, tan− =∫ 1
1
4
3 4478x dx . .K
8.
π /2
1 3
x
y
Simpson’s rule for y = sec− 1 x: n = 10 ⇒∆x = 0.2
A y y y y y y≈ + + + + + +0 2
34 2 4 41 1 2 1 4 1 6 2 8 3
.( ). . . .L
= 1.919692K
A x dx= −∫ sec 1
1
3
= − +−x x x x xsec sgn ln –1 2
1
31( )
= − ⋅ + − + ⋅− −3 3 1 3 8 1 1 11 1sec ln sec ln( )
= 1.930131… .
The Simpson’s rule answer differs from thisby 0.0104… , or about 0.5%.
9. By vertical slices,
A x dx=
∫ π
21
0
1
– sin–
= − −−π2
11 2
0
1
x x x xsin –
= − − − + + =−π2
1 0 0 0 1 11sin
By horizontal slices, A y dy= ∫ sin/
0
2π
= − = − + =cos cos cos/y 02
20 1π π
, which is the
same answer as by vertical slices.
10. By cylindrical shells, dV = 2π x tan− 1 x dx.
V x x dx= −∫2 1
0
1
π tan
x x dxtan−∫ 1
12 x
u dvtan–1 x x
dx1 + x 2 2 –
+
= −+
− ∫1
2
1
2 12 1
2
2x xx
xdxtan
= − −+
− ∫1
2
1
21
1
12 1
2x xx
dxtan
= − + +− −1
2
1
2
1
22 1 1x x x x Ctan tan
∴ = − +− −V x x x xπ π π2 1 10
1tan tan
= − + − + −− −π π πtan tan1 11 1 0 0 0
= − = ⋅ −−2 1 24
1π π π π πtan
= − =1
21 79322π π . K
Compare this with a cylinder (π r2h) minus acone (π r 2h/3), both of radius 1 and altitude π / 4,which has volume 2π ( π / 4)/3 = π2/6 = 1.6449… ;the volume is slightly less than V, which isexpected because the cylinder minus the cone isgenerated by rotating a line that lies below thegraph.
Calculus Solutions Manual Problem Set 9-9 239© 2005 Key Curriculum Press
Problem Set 9-9
Q1. xx= = −55
1tan tanθ θ or
Q2. xex − ex + C Q3.1
33tan x C+
Q4. 2x1/ 2 + C Q5. ln |x| + C
Q6. reduction formula Q7. False
Q8. dx dy dr rd2 2 2 2+ +or ( )θ
Q9. (1 − x2)−1/ 2 Q10. C
1.
x
y
cosh
sinh
1
1x
y
tanh
coth
1
1
x
y
sech
csch
1
1
2.
x
y
1
1cosh–1
sinh
–1
x
y
–1coth
tanh–1 1
1
x
y
1
1 csch–1–1sech
–1
3.d
dxx x xtanh tanh sec3 23= h2
4.d
dxx x x5 3 15 3sec sec tanhh h 3= −
5. cosh sinh cosh5 61
6x x dx x C= +∫
6. ( )sinh cosh (sinh )x x dx x C− −= − +∫ 3 21
2
= − +1
2csch2 x C
Or: ( ) h2sinh cosh csc cothx x dx x x dx− =∫ ∫3
= − + = − + +1
2
1
212
1coth (csc )x C x Ch21
= − +1
2csch2 x C
7.d
dxx x(csc sin )h
= −csch x coth x sin x + csch x cos x
8.d
dxx x x x x x(tan tanh ) sec tanh tan sec= +2 2h
9. sec tanhh2 41
44x dx x C= +∫
10. sec tanh sech h7 71
77x x dx x C= − +∫
11.d
dxx x x x x x( coth ) coth csc3 2 3 23= − h
12.d
dxx x( csc ).2 5 4h
= 2.5x1.5csch 4x − 4x2.5 csch 4x coth 4x
13. tanh ln coshx dx x=∫ ( )1
3
1
3
= ln (cosh 3) − ln (cosh 1) = 1.875547…
14. sinh coshx dx x= =− −∫ 4
4
4
4
0
(Note that sinh is an odd function.)
15.d
dx
x
x
sinh
ln
5
3
= 5 5 3 5
3
1
2
cosh ln – sinh
(ln )
–x x x x
x
16.d
dx
x
x
cosh
cos
6
3
= +6 6 3 3 6 3
32
sinh cos cosh sin
cos
x x x x
x
17. x x dxsinh∫ u dvx sinh x1 cosh x0 sinh x
+
–+
= x cosh x − sinh x + C
∴∫ x x dxsinh0
1
= − = −x x xcosh sinh cosh sinh0
11 1
= e− 1 = 0.36787…
18. x x dx2 cosh∫ u dvx 2 cosh x2x sinh x
2 cosh x0 sinh x–
–
+
+
= x2 sinh x – 2x cosh x + 2 sinh x + C
∴∫ x x dxa
b2 cosh
= − +x x x x xa
b2 2 2sinh cosh sinh
240 Problem Set 9-9 Calculus Solutions Manual© 2005 Key Curriculum Press
= b2 sinh b – 2b cosh b + 2 sinh b– a2 sinh a + 2a cosh a – 2 sinh a
19.d
dxx
x( sinh )–3 4
12
16 1
1
2=
+
20.d
dxx
x
x( tanh )
––5
15
11 3
2
6=
21. tanh−∫ 1 5x dx
= + − +−1
55
1
101 51x x x Ctanh ln | ( ) |2
22. 4 61cosh−∫ x dx
= − +−4
66
4
66 11 2 1 2x x x Ccosh [( ) – ] /
= − +−2
36
2
336 11 2x x x Ccosh –
23. Let x = 3 sinh t, dx = 3 cosh t dt,
x t t2 29 9 9 3+ = ⋅ + =sinh cosh ,
tx= −sinh .1
3
∴ + = ⋅∫ ∫x dx t t dt2 9 3 3cosh cosh
= ∫9 2cosh t dt u dvcosh t cosh tsinh t sinh t
+–
= − ∫9 9 2cosh sinh sinht t t dt
= − −∫9 9 12cosh sinh (cosh )t t t dt
= − + ∫∫9 9 92cosh sinh cosht t t dt
∴ = + +∫18 9 921cosh cosh sinht dt t t t C
∴ = + +∫9 4 5 4 52cosh . cosh sinh .t dt t t t C
= ⋅ + ⋅ + +−4 59
3 34 5
3
21. . sinh
x x xC
= + + +−0 5 9 4 53
2 1. . sinhx xx
C
24. Let x = 5 cosh t, dx = 5 sinh t dt,
x t t2 225 25 25 5– cosh – sinh= ⋅ = ,
tx= −cosh 1
5.
∴ = ⋅∫∫ x dx t t dt2 25 5 5– sinh sinh
= ∫25 2sinh t dt u dvsinh t sinh t
cosh t cosh t–+
= − ∫25 25 2sinh cosh cosht t t dt
= − +∫25 25 12sinh cosh sinht t t dt( )
= − − ∫∫25 25 252sinh cosh sinht t t dt dt
∴ = − +∫50 25 2521sinh sinh cosht dt t t t C
∴ = − +∫25 12 5 12 52sinh sinh cosht dt t t t C. .
= ⋅ ⋅ − −12 525
5 512 5
5
21. .
x x x–cosh
= − +−0 5 25 12 55
2 1. .x xx
C– cosh
25. a. Figure 9-9g shows that the horizontal forceis given by the vector (h, 0) and the verticalforce is the vector (0, v), so their sum, thetension vector, is the vector (h, v), which has
slope v
h. Because the tension vector points
along the graph, the graph’s slope, y ′, also
equals v
h.
b. v = weight of chain below (x, y) = s ⋅ w
⇒ ′ = = ⋅ = ⋅yv
h
s w
h
w
hs
c. ds dx dy dx dy dx= + = +2 2 2 21[ ( / ) ]
= + ′1 2( )y dx
⇒ ′ = = + ′d yw
hds
w
hy dx( ) 1 2( )
d. [ ( ) ] ( )1 2 12+ ′ ′− /∫ y d y
= + − /∫ ( ) ( )1 2 12sinh sinht d t
= − /∫ ( ) ( )cosh cosh2 12t t dt
= = + = ′ +−∫ dt t C y Csinh 11
w
hdx
w
hx C= +∫ 2
⇒ ′ = +−sinh 1 yw
hx C
e. At x = 0, y ′ = 0, so
sinh− = + ⇒ =1 0 0 0w
hC C .
f. sinh sinh− ′ = ⇒ ′ =1 yw
hx y
w
hx
g.dy
dx
w
hx dy
w
hx dx= ⇒ = ∫∫sinh sinh
⇒ = +yh
w
w
hx Ccosh
26. a. y x kk
C= = ⇒ = +2 0 21
0when cosh
⇒ 2 = k + C ⇒ C = 2 − k
y x kk
k= = ⇒ = + −5 4 54
2when cosh
Using the solver feature of your grapher,k ≈ 3.0668… .
Calculus Solutions Manual Problem Set 9-9 241© 2005 Key Curriculum Press
y x= ……
+ − …3 06681
3 06682 3 0668. .cosh
.
y x= ……
− …3 06681
3 06681 0668. .cosh
.b. y (20) = 1040.9739…
c. y = 4:
4 3 06681
3 06681 0668= …
…− …. .cosh
.x
cosh
.
.
.
1
3 0668
5 0668
3 0668K
K
Kx =
x = … ……
= …−3 06685 0668
3 06683 33551. .cosh
.
.
By symmetry, x = ±3.3355… .The answer can be found numerically usingthe solver feature of your grapher.
d. yk
x y′ = ′ = …sinh1
3 1 1418; ( ) .
e. A kk
x k dx= +
∫ cosh –
–
12
1
3
= + −−
kk
x k x2
1
312sinh ( )
= ……
−…
( . )3 0668
3
3 0668
1
3 06682 sinh
.sinh
–
.
+ 4(2 − 3.0668…)= 9.5937…
f. L y dx= + ′−∫ 1 2
1
3
( )
= +−∫ 1 2
1
3
sinh ( / )x k dx
= =− −∫ cosh sinh
1 11
3
1
3
kx dx k
kx
= +
= …k
k ksinh sinh
3 14 5196.
27. a. The vertex is midway between the poles,so y = 110 ft when x = 150 ft.
yh
w
w
hx C= +cosh
= +400
0 8
0 8
400
lb
lb ft. /cosh
.x C
110 5001
500150= ⋅
+cosh C
⇒ C = 110 − 500 cosh 0.3
y x= + −5001
500110 500 0 3cosh cosh .
The cable comes closest to the ground atx = 0.y (0) = 500 cosh 0 + 110 − 500 cosh 0.3= 610 − 500 cosh 0.3 = 87.3307… ≈ 87.3 ft
b. y x′ = sinh1
500
L x dx
x dx
= +
=
∫∫−
1 500
1
500
2
150
150
150
150
sinh ( / )
cosh
–
=−
5001
500 150
150
sinh x
= 500 sinh 0.3 − 500 sinh (−0.3)= 1000 sinh 0.3 = 304.5202… ≈ 304.5 ft
A faster method is:Half weight of cable = vertical tension at(150, 110) = h ⋅ y ′ (150) (CompareProblem 25.)
Weight .= ⋅ =2 4001
500150 800 0 3sinh sinh
= 243.6162… ≈ 243.6 lb(Note: Because w ⋅ L = weight, either of thesemethods could give both the weight and thelength.)
c. T h v= +2 2 ; h is constant and v is greatestat the ends, so the maximum tension is atx = 150 ft.
T h hy( )150 1502 2= + ′[ ( )]
= + =400 1 0 3 400 0 32sinh . cosh .
= 418.1354… ≈ 418.1 lb
d. The general equation is yh
w
w
hx C= +cosh .
If y (0) = 100 and y (150) = 110, find h suchthat y (150) − y (0) = 10. Solve:h
w
w
h
h
wcosh 150 10− = , or
cosh120
18
h h− =
By grapher, h = 901.3301... ≈ 901.3 lb.
28. The answers will depend on the dimensionsof the chain used. Note that the answer isindependent of the kind of chain. You mightshow students how a heavy chain and a lightchain of equal length will hang in the samecatenary if they are suspended from the samepoints.Assume that the dimensions are the same as inExample 5.
a. Vertex: (0, 20). Supports: (±90, 120).
b. y = 51.78… cosh
.
1
51 7831 78
KKx – .
c. Note: To conserve class time, you might havestudents plot only each 20 cm for x, as shownhere for Example 5. Use the TABLE feature.
242 Problem Set 9-9 Calculus Solutions Manual© 2005 Key Curriculum Press
x y
0 20.0
±20 23.9
±40 36.2
±60 58.8
±80 95.1
d. A clever way to make sure the measurementsare vertical is to hold a book against the boardwith its bottom edge along the chalk tray.Then hold the meterstick against the verticaledge of the book. It is crucial that the pointsbe plotted accurately to get the dramaticimpact of “perfect fit.”
e. For a quadratic function with vertex on they-axis, y = ax2 + c. Using the data forExample 5,20 = a (0) + c ⇒ c = 20
120 90 201
812= + ⇒ =a a( )
y x= +1
81202
20
10
x
yparabola
catenary
The parabola is more curved at the vertex.
f. For Example 5,
dL y dx x dx= + ′ = +1 11
51 782 2( ) sinh
. K
= cosh
.
1
51 78Kx dx
L x dx=
−∫ cosh.
1
51 7890
90
K
= =
−51 78
1
51 78285 349
90
90
. .KK
Ksinh.
x
≈ 285.3 cmThe actual length should be close to this.
29. a. y = sinh x
dS x dL x x dx= = +2 2 1 2π π cosh
S x x dx= +∫2 1 2
0
1
π cosh
= 5.07327… by numerical integration≈ 5.07 ft2
b. Cost = 2(57)(5.07327…) = 578.3532… ≈$578.35
c. Slice perpendicular to the y-axis.dV = π x2 dy = π (sinh–1 y)2 dyTop of bowl is at
y = − =sinh . .1
1
241 133534K
∴ = ∫ V y dyπ (sinh )–
.1 2
0
1 133K
= 1.25317… by numerical integration≈ 1.253 ft3
30. a. y kk
x C= − +cosh1
Inner catenary: yinner(0) = 612, yinner(260) = 0
6120
612= − + ⇒ = +kk
C C kii
i i icosh
0260
612= − + +kk
kii
icosh
⇒ ki = 97.1522… (numerically)
yinner = 97.1522… − +
+cosh
.
1
97 15221 612
Kx
Outer catenary: youter (0) = 630, youter (315) = 0
630 = −ko cosh0
ko
+ Co ⇒ Co = 630 + ko
0 = −ko cosh315
ko
+ 630 + ko
⇒ ko = 127.7114… (numerically)
y xouter .= − +
+127 7114
1
127 71141 630K
Kcosh
.
b. The graphs are the same as in Figure 9-9k:
100
100 x
y
c. A y dx y dx= −− −∫ ∫outer inner
315
315
260
260
= − + +−
kx
kk x xo
oo
2
315
315
630sinh
+ − −−
kx
kk x xi
ii
2
260
260
612sinh
= 54323.2729… ≈ 54,323 ft2
d.dy
dx kxouter
o
so= −sinh ,1
Lk
x dx= +−∫ 1
12
315sinh
o
315
= =− −∫ cosh sinh
1 1
315
315
kx dx k
kx
o315
315
oo
Calculus Solutions Manual Problem Set 9-9 243© 2005 Key Curriculum Press
= ⋅ ⋅2 127 7114
315
127 7114. K
Ksinh
.= 1493.7422… ≈ 1494 ft.
e.
′ − = − − =ykouter
o
( ) sinh sinh.
315315 315
127 7144K
= 5.8481… is the spider’s starting slope.
f. José can fly through at altitude yinner(x) ifx ≥ 50 + 120/2 = 110.yinner(110) = 542.7829… , so the plane canfly through at heights between 0 and 542 feet.(Because of the curvature of the arch and thevertical thickness of the plane, the closestdistance is slightly less than 50 feet when thehorizontal distance is 50 feet. The plane canfly through at slightly higher altitudes bybanking slightly.)
31. a. H(x) = csch x ⇒ H′(x) = −csch x coth xH′(1) = −csch 1 coth 1 = −1.1172855…
b. H′ ≈( )h h
11 01 0 99
0 02
csc ( . ) – csc ( . )
.
= −1.11738505… . The answers differ by0.0000995… , which is about 0.0089% ofthe actual answer.
32. sec sin tanhh ( )x dx x= −∫ 1
1
2
1
2
= sin–1 (tanh 2) − sin–1 (tanh 1)= 0.435990…
Numerically,
sech .x dx =∫ 0 4359901
2
K
(Checks.)
33. By parts:
e x dxx sinh 2∫ u dvsinh 2x ex
2 cosh 2x ex 4 sinh 2x ex +
–
+
= − + ∫e x e x e x dxx x xsinh cosh sinh2 2 2 4 2
⇒ − ∫3 2e x dxx sinh
= − + ⇒ ∫e x e x C e x dxx x xsin cosh sinhh 2 2 2 21
= − +2
32
1
32e x e x Cx xcosh sinh
By transforming to exponential form:
e x dx e e e dxx x x xsinh 21
22 2= − −∫∫ ( )
= − = + +− −∫1
2
1
6
1
23 3( )e e dx e e Cx x x x
Transforming to exponential form is easier!(Note that the two answers can be shown to beequivalent either by transforming the first to
exponential form or by transforming the secondto hyperbolic form, as shown here.)
1
6
1
2
1
6
1
23 2 2e e C e e e Cx x x x x+ + = +
+− −
= + − +
+− −e e e e e Cx x x x x1
3
1
3
1
6
1
62 2 2 2
= + − +2
3 2
1
3 2
2 2 2 2
ee e
ee e
Cxx x
xx x– ––
= − +2
32
1
32e x e x Cx xcosh sinh
34. e x dxx sinh∫ u dve x sinh xe x cosh xe x sinh x
+
–+
= − + ∫e x e x e x dxx x xcosh sinh sinh
⇒ = − +∫0 e x dx e x x Cx xsinh cosh sinh )(
The original integral reappeared with the samecoefficient, so when it was added again to the leftside, it exactly canceled out the desired integral.Use the exponential form of sinh x.
e x dx e e e dxx x x xsinh )= − −∫∫ 1
2(
= −∫1
212( )e dxx
= − +1
4
1
22e x Cx
35. a. cosh2 x − sinh2 x
= +
−
e e e ex x x x– ––
2 2
2 2
= + + − + =e e e ex x x x2 2 2 22
4
2
41
– ––
∴ cosh2 x − sinh2 x = 1, Q.E.D.
b.1 1
22 2
2cosh(cosh – sinh )
coshxx x
x=
⇒ 1 − tanh2 x = sech2 x
c.1 1
22 2
2sinh(cosh – sinh )
sinhxx x
x=
⇒ coth2 x − 1 = csch2 x
36. a. Substitute 2x for x in the definition of sinh x.
b. sinh ( – )–21
22 2x e ex x=
= ⋅ ⋅ ⋅ +
=
−21
2
1
22
( – )
sinh cosh
–e e e e
x x
x x x x( )
244 Problem Set 9-9 Calculus Solutions Manual© 2005 Key Curriculum Press
c. cosh ( )–21
22 2x e ex x= +
= + + + +1
41
1
412 2 2 2( ) ( – )– –e e e ex x x x
= +
+
1
2
1
2
2 2
( ) ( – )– –e e e ex x x x
= cosh2 x + sinh2 x
d. cosh2 x − sinh2 x = 1 ⇒ cosh2 x = 1 + sinh2 x⇒ cosh 2x = cosh2 x + sinh2 x= (1 + sinh2 x) + sinh2 x = 1 + 2 sinh2 x
e. 1 + 2 sinh2 x = cosh 2x⇒ 2 sinh2 x = cosh 2x − 1
⇒ =sinh (cosh – )2 1
22 1x x
37. a. On the circle, u2 + v2 = 1 ⇒2u du + 2v dv = 0 ⇒ dv = (−u/v) du.
dL du dv du u v du
u v
vdu
vdu
udu
Ldu
u
= + = +
= + = =
= ∫
2 2 2 2 2
2 2
2 2
22
1
1 1
1
1
( / )
–
–cos
= − = − + =− −cos coscos
12
1 11 2 2u
The curve along the hyperbola from u = 1 tou = cosh 2 has length greater than the linesegment along the horizontal axis from (1, 0)to (cosh 2, 0). This segment has lengthL = cosh 2 − 1 = 2.762… . So the length ofthe curve is greater than 2, Q.E.D.
b. The area of the triangle that circumscribes thesector is 0.5(2 sinh 2 cosh 2) = sinh 2 cosh 2.The area of the sector is the area of thistriangle minus the area of the region betweenthe upper and lower branches of the hyperbolafrom u = 1 to u = cosh 2.Slice this region vertically. Pick sample point(u, v) on the upper branch, within the strip.Let t be the argument of sinh and cosh at thesample point. 0 ≤ t ≤ 2.dA = 2v du = 2 sinh t d(cosh t) = 2 sinh2 t dt
A t dt= ≈∫2 11 6449582
0
2
sinh . K
Thus, the area of the sector iscosh 2 sinh 2 − 11.644958… = 2, Q.E.D.
c. By definition of the circular functions,x is the length of the arc from (1, 0) to(cos x sin x). So the total arc has length 2x.The circumference of a unit circle is 2π, andits area is π. Thus,
Ax
xsector ,= =2
2ππ Q.E.D.
d. Slice as in part b.
Let l t dt= ∫ sinh2 u dvsinh t sinh t
cosh t cosh t–+
= − ∫sinh cosh cosht t t dt2
= − +∫sinh cosh ( sinh )t t t dt1 2
= − − ∫sinh cosh sinht t t t dt2
∴ = − +∫2 2sinh sinh cosht dt t t t C
sinh sinh cosh210 5 0 5t dt t t t C= − +∫ . .
Slicing as in part b, the area A between theupper and lower branches of the hyperbola is
A t dtx
= ∫2 2
0sinh
= − −2 0 5 0 5 0( . . )sinh coshx x x = sinh x cosh x − x
Thus, the area of the sector iscosh x sinh x − (sinh x cosh x − x) = x,Q.E.D.
38. a. y = sinh− 1 x ⇒ sinh y = x ⇒ cosh y y′ = 1
′ = =+
=+
yy y x
1 1
1
1
12 2cosh sinh, Q.E.D.
b. y = tanh− 1 x ⇒ tanh y = x ⇒ sech2 y y′ = 1
′ = = =yy y x
1 1
1
1
12 2 2sec – tan –h h, Q.E.D.
c. y = coth− 1 x ⇒ coth y = x⇒ −csch2 y y′ = 1
yy y x
′ = = =1 1
1
1
12 2 2– –(coth – ) –,
cschQ.E.D.
d. y = sech− 1 x ⇒ sech y = x⇒ −sech y tanh y y′ = 1
yy y
′ = 1
– tanhsech
= 1
1 2– –sech sechy y
= − 1
1 2x x–, Q.E.D.
e. y = csch− 1 x ⇒ csch y = x⇒ −csch y coth y y′ = 1
yy y y y
′ = =+
1 1
1 2– coth –csch csch csch
= −+
1
1 2| |x x, Q.E.D.
Calculus Solutions Manual Problem Set 9-10 245© 2005 Key Curriculum Press
Problem Set 9-10Q1. Q2. sinh x + C
1
1 x
y
Q3. sinh x Q4. −sin x
Q5. sin x + C Q6. y = x3
Q7. y = tan x Q8. y = sinh x or x3 + x
Q9. y = ex Q10. A
1. a.1
2
x
y
It might converge because the integrandapproaches zero as x approaches infinity.
b. ( / ) lim ( )1 2 2
22x dx x dx
b
b
=→∞
−∞
∫∫= − = − + =
→∞
−
→∞lim lim( / / )b
b
bx b1
21 1 2
1
2
Integral converges to 1
2.
2. a.
3
1
x
y
It might converge because the integrandapproaches zero as x approaches infinity.
b. 1 4 4
33/ lim
bx dx x dx
b
=→∞
−∞
∫∫= − = +
=
→∞
−
→∞lim lim –b
b
bx
b
1
3
1
3
1
81
1
813
33
The integral converges to 1
81.
3. a.
x
y
1
1
It might converge because the integrandapproaches zero as x approaches infinity.
b. ( / ) ( / )1 11 1
x dx x dxb
=→∞
∞
∫ ∫limb
= = = ∞→∞ →∞
lim ln lim (ln – )b
b
bx b| |
10
The integral diverges.
4. a.
1
1x
y
It might converge because the integrandbecomes infinite only as x approaches zero.
b. ( / ) lim ( / )1 10
1
0
1
x dx x dxa a
=→ + ∫∫
= = = ∞→ →+ +lim ln lim ( – ln )
a a ax a
0
1
00| |
The integral diverges.
5. a. y
x1
1
It might converge because the integrandapproaches zero as x approaches infinity.
b. 1 0 2 0 2
11/ . .x dx x dx
b
b
=→∞
−∞
∫∫ lim
= = = ∞→∞ →∞
lim lim ( . – . ).
b
b
bx b1 25 1 25 1 250 8
1
0 8. .
The integral diverges.
6. a.
x
y
1
1
It might converge because the integrandapproaches zero as x approaches infinity.
b. 1 1 2
1
1 2
1/ . .x dx x dx
b
b
=∞
→∞
−∫ ∫lim
= − = + =→∞
−
→∞lim lim (– )– .
b
b
bx b5 5 5 50 2
1
0 2.
The integral converges to 5.
7. a.
1
1
x
y
It might converge because the integrandbecomes infinite only as x approaches zero.
246 Problem Set 9-10 Calculus Solutions Manual© 2005 Key Curriculum Press
b. 1 0 2
0
0 21
0
1
/ .x dx x dxa a
=→
−+ ∫∫ lim .
= =→ →+ +lim lim ( . – . ).
a a ax a
0
0 81
0
0 81 25 1 25 1 25. .
= 1.25The integral converges to 1.25.
8. a. y
x
1
1
It might converge because the integrandbecomes infinite only as x approaches zero.
b. 1 1 2
0
1
0
1 21
/ . .x dx x dxa a
=∫ ∫→
−+
lim
= − = + = ∞→
−
→+ +lim lim (– )– .
a a ax a
0
0 21
0
0 25 5 5.
The integral diverges.
9. a.
0 1
1
x
y
It might converge because the integrandapproaches zero as x approaches infinity.
b. 1 1 1 12 2
00/( ) /( )+ = +
→∞
∞
∫∫ x dx x dxb
b
lim
= = =→∞
−
→∞lim tan lim (tan – )–
b
b
bx b1
0
1 02
π
The integral converges to π /2.
10. a.
0 x1
1
It might converge because the integrandapproaches zero as x approaches infinity.
b. 1 1 1 10 0
/( ) /( )+ = +∞
→∞∫ ∫x dx x dxb
b
lim
= + = + = ∞→∞ →∞
lim ln lim [ln ( ) – ]b
b
bx b| |1 1 0
0
The integral diverges.
11. a.
0
y
x
1
1
It might converge because the integrandbecomes infinite only as x approaches 0 or 1.
b. To determine whether this converges, split theintegral into two pieces. Each piece mustconverge in order for the integral to converge.The integral can be written
10
1
/( ) x x dxln∫= +∫ ∫1 1
0/( ) /( )
1
x x dx x x dxc
cln ln
= +→ →+ − ∫∫lim ln lim ln
a b c
b
a
c
x x dx x x dx0 1
1 1/( ) /( )
= +→ →+ −lim ln ln lim ln ln
a a
c
b c
b
x x0 1
| | | |
=
+→
→
+
−
lim (ln |ln | – ln |ln |)
lim (ln |ln | – ln |ln |)a
b
c a
b c0
1
= ∞ + ∞For the integral to converge, both limits mustexist. Because neither exists, the integraldiverges.
12. a.
0 3
y
x
1
1
It might converge because the integrandapproaches zero as x approaches infinity.
b. The indefinite integral can be written
( ) ( / ) .ln (ln )x dx x x C− −= − +∫ 2 1
1 12 2
33/[ ( ) ] /[ ( ) ]x x dx x x dx
b
b
ln lim ln=→∞
∞
∫∫= −
→∞
−lim lnb
b
x( ) 1
3
= + =→∞
− −lim [–(ln ) (ln ) ]–
bb 1 13 (ln 3) 1
The integral converges to(ln 3)− 1 = 0.910239… .
13. a.
2
x
1
y
It might converge because the integrandapproaches zero as x approaches infinity.
Calculus Solutions Manual Problem Set 9-10 247© 2005 Key Curriculum Press
b. e dx e dxx
b
xb
−
→∞
−∞
= ∫∫ 0 4 0 4
22
. .lim
= −→∞
−limb
xb
e2 5 0 4
2. .
= + =→∞
− −lim(– . . ). – .
b
be e e2 5 2 5 2 50 4 0 8 0 8. .
The integral converges to 2.5e− 0.8
= 1.1233… .
14. a.
0 1
1
x
y
It diverges because the integrand does notapproach zero as x approaches infinity.
b. (Not applicable)
15. a.
0–1 2
1
x
y
It does not converge because the integrand isundefined for x < 0.
b. (Not applicable)
16. a.
0 3 7
1
1
x
y
It might converge because the integrandbecomes infinite only as x approaches 3.
b. ( – )x dx3 2 3
1
7−∫ /
= +→
−
→
−− + ∫∫lim ( – ) lim ( – )
b a a
b
x dx x dx3
2 3
3
2 37
13 3/ /
= − + −→ →− +lim limb
b
a ax x
3
1 3
1 3
1 37
3 3 3 3( ) ( )/ /
= − −→ −
lim [ ( – ) ( ) ]/ /
bb
3
1 3 1 33 3 3 2
+ − −→ +lim [ ( ) ( ) ]/ /
aa
3
1 3 1 33 4 3 3
= 3 ⋅ 21/3 + 3 ⋅ 41/3
The integral converges to 3 ⋅ 21/3 + 3 ⋅ 41/3
= 8.5419… .
17. a.
0 1
x
y1
It might converge because the integrand seemsto approach zero as x approaches infinity.
b. Integrate by parts:
xe dx e x Cx x− −= − + +∫ ( 1)
xe dx xe dxx
b
xb
−
→∞
−∞
= ∫∫ lim00
= − + = + + =→∞
−
→∞lim lim [– ( ) ]–
b
xb
b
be x e b( )1 1 1 10
(The first term is zero by l’Hospital’s rule.)Integral converges to 1.
18. a.
0 3
x
y
1
1
It might converge because the integrandbecomes infinite only as x approaches 1.
b. ( – )x dx1 2
0
3−∫
= +→
−
→
−+ ∫∫lim ( – ) lim ( – )
–b a a
b
x dx x dx1
2
1
23
01 1
= − − + − −→
−
→
−+
lim lim–b
b
a ax x
1
1
0 1
13
1 1( ) ( )
= +→lim [–( – ) (– ) ]
–
– –
bb
1
1 11 1
+ +→ +lim [– ( – ) ]– –
aa
1
1 12 1
= ∞ + ∞For the integral to converge, both limits mustexist. Because neither exists, the integraldiverges.
19. a.
20
1
x
y
It diverges because the integrand does notapproach zero as x approaches infinity.
b. (Not applicable)
248 Problem Set 9-10 Calculus Solutions Manual© 2005 Key Curriculum Press
20. a.
20
1
x
y
It diverges because the integrand does notapproach zero as x approaches infinity.
b. (Not applicable)
21. As b x dxb
→ ∞ ∫, cos0
oscillates between −1 and
1 and never approaches a limit. Similarly,
sin x dxb
0∫ oscillates between 0 and 2.
22. a. I /.. .
1 0011 001 1 001
111= =
→∞
−∞
∫∫ x dx x dxb
b
lim
= −→∞
−limb
bx1000 0 001
1
.
= +→∞
lim (– )– .
bb1000 10000 001
= 1000 (converges), Q.E.D.
I /.. .
0 9990 999 0 999
111= =
→∞
−∞
∫∫ x dx x dxb
b
lim
=→∞
limb
bx1000 0 001
1
.
=→∞
lim ( – ).
bb1000 10000 001
= ∞ (diverges), Q.E.D.
b. I /11
1= = ∞∞
∫ x dx (see Problem 3), so I1
diverges.
c. “Ip converges if p > 1 and diverges if p ≤ 1.”
23. a. y = 1/x = x− 1
dA = y dx = x− 1 dx
A x dx= = ∞−∞
∫ 1
1 (See Problem 3)
The area does not approach a finite limit.
b. By plane slices, dV = πy2 dx = πx− 2 dx.
V x dx xb b
bb
= = −→∞
−
→∞
−∫lim limπ π2 1
11
= + =→∞
lim (– )–
bbπ π π1
The volume converges to π.c. By cylindrical shells, dV = 2π xy dx
= 2π x(x− 1) dx = 2π dx.
V dx xb
b
b
b
= =→∞ →∞∫lim lim2 2
1 1π π
= = ∞→∞
lim ( – )b
b2 2π π
Volume diverges.
d. False. The volume could approach a constantas in part b or become infinite as in part c.
24. y = −1/x ⇒ x = −y− 1
Slice the vertical cross section horizontally.dA = x dy = −y− 1 dy
A y dy ya a a a
= − = −→−∞
−
→ ∞
−
∫lim lim ln–
–
11 1
| |
= + = ∞→−∞lim (– ln |– | ln | |)
aa1
The area of the bucket’s surface is greater thanthe area of the cross section, and the cross-sectional area diverges. Thus, the bucket hasinfinite surface area. The bucket is congruent tothe solid in Problem 23b, which has volumeapproaching π. Thus, π cubic units of paintwould fill the bucket but could not coat thewhole surface!
25. a. f x t e dtx t( ) = −∞
∫0
f te dt te eb
tb
b
t tb
( )10 0
= =→∞
−
→∞∫lim lim (– – )– –
= + + =→∞
lim (– – )– –
b
b bbe e 0 1 1
(Using l’Hospital’s rule on be− b gives
lim lim limb
b
b b b bbeb
e e→∞
−
→∞ →∞= = =1
0.)
f t e dtb
t( )2 2
0=
→∞
−∞
∫lim
= − +→∞
− −∞
∫limb
tb
tt e te dt2
0 02
= + + =→∞
lim (– )–
b
bb e2 0 2 1 2( )
(Using l’Hospital’s rule on b2e− b gives
lim lim limb
b
b b b bb eb
e e→∞
−
→∞ →∞= = =2 2 2
0.)
f t e dtb
t( )3 3
0=
→∞
−∞
∫lim
= − +→∞
− −∞
∫limb
tb
tt e t e dt3
0
2
03
= + + =→∞
lim (– )–
b
bb e3 0 3 2 6( )
(Using l’Hospital’s rule on b3e− b gives
lim lim lim limb
b
b b b b b bb eb
e
b
e e→∞
−
→∞ →∞ →∞= = = =3
23 6 60.)
b. Conjecture:
f ( 4) = 4 f (3) = 24 = 4!
f ( 5) = 5 f (4) = 120 = 5!
f ( 6) = 6 f (5) = 720 = 6!
c. f x t e dtx t( ) = −∞
∫0u dvtx e–t
xtx–1 –e–t +
–
Calculus Solutions Manual Problem Set 9-10 249© 2005 Key Curriculum Press
= − +→∞
− − −∞
∫limb
x tb
x tt e x t e dt0
1
0
= + + −→∞
lim (– )–
b
x bb e x f x0 1 ( )
= 0 + 0 + x f (x − 1)= x f (x − 1), Q.E.D.
( lim (– )–
b
x bb e→∞
= 0 can be proved by
mathematical induction usingl’Hospital’s rule.)
d. Part a shows that f (1) = 1 = 1!.Part c shows that f (n) = nf (n − 1) =n(n − 1) f (n − 2) = n(n − 1)(n − 2)…(2)(1)= n!, Q.E.D.
e. t e dtt3
0
1000
6− ≈∫The value of b that makes the integral comewithin 0.000001 of 6 can be found numerically(though it will be slow), or algebraically:
t e dttb
3
0
−∫= − − − − +− − − −b e b e be eb b b b3 23 6 6 6
|b3e− b + 3b2e− b + 6be− b + 6e− b| < 0.000001for b > 23.4050… , say, b ≈ 24.
f. 0 5 0 5
0
0 5
0
24
. ! .= ≈ ≈−∞
−∫ ∫t e dt t e dtt t.
0.886227311…
From the graphs, t0.5e− t < t3e− t for x ≥ 24. Theerror in 0.5! from stopping at b = 24 is thearea under the “tail” of the graph from b = 24.
Error < < .= −∞
−∞
∫ ∫t e dt t e dtt t0 5
24
3
240 000001.
The difference between the tabulated valueof 0.5! and the value calculated here is
0.8862269255 − 0.866227311…
= −0.000000386
which is less in absolute value than 0.000001.Note, however, that the difference is negativebecause the calculated value is larger thanthe tabulated value. This observation meansthat either the tabulated value is incorrect orthere is more inaccuracy in the numericalintegration algorithm than there is in the errorcaused by dropping the tail of the integral.(Using a smaller tolerance in the numericalintegrator gives a value of 0.8862269252… .)
g. Using the tabulated value of 0.5!,1.5! = 1.5(0.5!) = 1.3293…2.5! = 2.5(1.5!) = 3.3233…3.5! = 3.5(2.5!) = 11.6317…
h. 0 0
0 0! = =
−∞
→∞∫ t e dt et
b
t blim – –
= + =→∞
lim(– )–
b
be 1 1, Q.E.D.
i. (−1)! = 0!/0, which is infinite. So (−2)! and(−3)!, which equal (−1)!/(−1) and (−2)!/(−2),are also infinite. However,(−0.5)! = 0.5!/(0.5) = 1.77245…(−1.5)! = (−0.5)!/(−0.5) = −3.54490…(−2.5)! = (−1.5)!/(−1.5) = 2.36327…all of which are finite.
j. 0 52
0 886226925. ! . ,= =πK which agrees
with the tabulated value.26. dW = F dr = 1000r− 2 dr
At the earth’s surface, r = 1.
W r dr rb
b= =
−
→∞
∞−∫ 1000 10002
1
11
lim –
= +→∞
lim (– )–
bb1000 10001
= 1000 radius-poundsThus, the amount of work does not increasewithout bound as r goes to infinity.
27. a. 22
21
3x x
xdx− −
−
∫
= − −−
+ − −−
∫ ∫2
2
22
2
21
2
2
3x xx
xdx
x
xdx
b. 22
22
2
21
2
2
3x xx
xdx
x
xdx− −
−
+ − −−
∫ ∫
= − −−
+ − −−
→
→
−
+
∫∫
lim
lim
b
xb
a
x
a
x
xdx
x
xdx
2 1
2
3
22
2
22
2
c. limb
xb x
xdx
→ −− −
−
∫2 1
22
2
+ − −−
→ + ∫lim
a
x
a
x
xdx
2
3
22
2
= + + −→ →− + ∫∫lim ( ) lim ( )
b
x
a
x
a
b
dx dx2 2
3
12 1 2 1
= + +→ →− +lim ( /ln ) lim ( /ln – )b
xb
a
x
ax x
2 1 2
3
2 2 2 2
= + −→ −lim ( /ln – /ln )b
b b2
2 2 2 2 1
+ +→ +lim ( /ln – – /ln )a
a a2
32 2 3 2 2
= + − − + −4 2 2 2 2 1 8 2 3/ / /ln ln ln− + = / /4 2 2 6 2ln ln
The integral converges to 6/ln 2 = 8.6561… .d. The integral is defined by dividing the
interval into Riemann partitions and summingthe subintervals. But the Riemann partitionsmay be chosen so that the discontinuitiesare at endpoints of subintervals. Then thesubintervals corresponding to each continuouspiece may be summed separately.
e. False. Some discontinuous functions (notably,piecewise continuous functions) are integrable.
28. Answers will vary.
250 Problem Set 9-11 Calculus Solutions Manual© 2005 Key Curriculum Press
Problem Set 9-11
1. y = sec 3x tan 3x ⇒y′ = (3 sec 3x tan 3x) tan 3x + sec 3x (3 sec2 3x)
= 3 sec 3x tan2 3x + 3 sec3 3x2. y = sinh 5x tanh 5x ⇒
y′ = (5 cosh 5x) tanh 5x + sinh 5x (5 sech2 5x)= 5 sinh 5x + 5 sinh 5x sech2 5x or5 sinh 5x + 5 tanh 5x sech 5x
3. x x dxcosh 4∫
16
u dvx cosh 4x1
14 sinh 4x
01
cosh 4x
+
–
+
= − +1
44
1
164x x x Csinh cosh
4. x x dxcos∫ u dvx cos x1 sin x0 –cos x+
–
+
= + +x x x Csin cos5. f (x) = (3x + 5)− 1 ⇒ f ′(x) = −3(3x + 5)− 2
6. f (x) = (5 − 2x− 1) ⇒ f ′(x) = 2(5 − 2x)− 2
7. ( )3 51
33 51x dx x C+ = + +−∫ ln | |
8. ( ) | |15 21
25 2− = − +−∫ x dx x Cln –
9. t(x) = tan5 4x ⇒t′(x) = 5 tan4 4x (4 sec2 4x) = 20 tan4 4x sec2 4x
10. h(x) = sech3 7x ⇒h′(x) = 3 sech2 7x (−7 sech 7x tanh 7x)
= −21 sech3 7x tanh 7x
11. sin cos2 1
21 2x dx x dx ( )= −∫ ∫
= − +1
2
1
42x x Csin
= − +1
2
1
2x x x Csin cos (or integrate by parts)
12. cos cos2 1
21 2x dx x dx= +∫∫ ( )
= + +1
2
1
42x x Csin
= + +1
2
1
2x x x Csin cos (or integrate by parts)
13. yx
x=
+⇒6 11
2
–
yx x
x x′ = +
+=
+6 2 6 11 1
2
23
22 2
( ) – ( – )( )
( ) ( )
14. yx
x= + ⇒5 9
4–
yx x
x x′ = + =5 4 5 9 1
4
29
42 2
( – ) – ( )( )
( – )
–
( – )
15.6 11
26
23
2
x
xdx
xdx
–
+= −
+
∫ ∫
= 6x – 23 ln | x + 2 | + C
16.5 9
45
29
4
x
xdx
xdx
+ = +
∫ ∫– –
= 5x + 29 ln | x − 4 | + C
17. f t t t( ) ( )1/2= + = + ⇒1 12 2
f t t tt
t′ = + =
+−( ) ( )
1
21 2
1
2 1 2
2( ) /
18. g t t t( ) ( )1/2= = − ⇒2 21 1–
g t t tt
t′ = =−( ) ( )
1
21 2
1
2 1 2
2( – )
–
/
19. 1 12 2+ = +∫ ∫t dt dtan tanθ θ( )
= =∫ sec sec tan3 1
2θ θ θ θd
+ + +1
2ln sec tan| |θ θ C
= + + + + +1
21
1
212 2t t t t Cln
20. t dt d2 21 1– sec – sec=∫ ∫ θ θ( )
= = −∫∫ sec tan sec secθ θ θ θ θ θ2 3d d( )
= + +1
2
1
2sec tan ln sec tanθ θ θ θ| |
− ln | sec θ + tan θ | + C
= − + +1
2
1
2sec tan ln sec tanθ θ θ θ| | C
= − + +1
21
1
212 2t t t t C– ln –
21. y = x 3 e x ⇒y′ = 3x2ex + x3ex = x2ex(3 + x)
22. y = x4e− x ⇒y′ = 4x3e− x − x4e− x = x3e− x(4 − x)
23. x e dxx3∫ u dvx 3 e x
3x 2 e x
6x e x
6 e x
0 e x
+
+
–
–+
= x3ex − 3x2 ex + 6xex − 6ex + C
24. x e dxx4 −∫ u dvx 4 e –x
4x 3 –e –x
12x 2 e –x
24x –e –x
24 e –x
0 –e –x–
–
–
+
+
+
= −x4 e− x − 4x3
e− x −12x2 e− x − 24xe− x − 24e− x + C
25. f x x f xx
x( ) ( ) ( ) /= ⇒ ′ = = −− −sin–
1
2
2 1 21
11
Calculus Solutions Manual Problem Set 9-11 251© 2005 Key Curriculum Press
26. g x x g xx
( ) ( )= ⇒ ′ =+
−tan 12
1
1
27. sin−∫ 1 x dx u dvsin–1 x 1
(1 – x2)–1/2 x+
–
= − −− −∫x x x x dxsin 1 2 1 21( ) ( )/
= x sin− 1 x − (−0.5)(2)(1 − x2)1/2 + C
= + +−x x x Csin –1 21
28. tan−∫ 1 x dx u dvtan–1 x 1
11 + x2 x–
+
= −+
− ∫x xx
x dxtan ( )12
1
1
= − + +−x x x Ctan ln1 21
21| |
29.1
4 5
1 6
5
1 6
12x xdx
x xdx
+=
++
∫∫ –
– / /
–
= − + + − +1
65
1
61ln ln| | | |x x C
30.1
6 7
1 8
1
1 8
72x xdx
x xdx
– –
– / /
–=
++
∫∫
= − + + − +1
81
1
87ln ln| | | |x x C
31.1
4 5
1
2 92 2x xdx
xdx
+=
+∫ ∫– ( ) –
= ∫ 1
3 93
2( sec ) –( sec tan )
θθ θ θd
= =∫ ∫1
33
tansec tan sec
θθ θ θ θ θ( )d d
= ln | sec θ + tan θ | + C
= + + + − +ln ( ) ( )1
32
1
32 92
1x x C
= + + + − +ln x x x C2 4 52
32.1
6 7
1
3 162 2x xdx
xdx
− −=
− −∫ ∫ ( )
= ∫ 1
4 164
2( sec ) –( sec tan )
θθ θ θd
= = ∫∫ 1
44
tan( sec tan ) sec
θθ θ θ θ θd d
= + +ln |sec tan |θ θ C
= − + +ln ( ) ( – ) –1
43
1
43 162
1x x C
= − + +ln – –x x x C3 6 72
33. f (x) = tanh x ⇒ f ′(x) = sech2 x
34. f (x) = coth x ⇒ f ′(x) = −csch2 x
35. tanhsinh
coshln coshx dx
x dx
xx C∫ ∫= = +| |
(Absolute value is optional.)
36. cothcosh
sinhln sinhx dx
x dx
xx C= = +∫ ∫ | |
(Absolute value is necessary.)
37. y = e2x cos 3x
y′ = (2e2x) cos 3x + e2x(−3 sin 3x)
= e2x(2 cos 3x − 3 sin 3x)
38. y = e−3x cos 4x
y′ = (−3e−3x) cos 4x + e−3x(−4 sin 4x)
= −e−3x(3 cos 4x + 4 sin 4x)
39. e x dxx2 3cos∫
9
u dve 2x cos 3x
2e 2x 13 sin 3x
4e 2x –1 cos 3x
+
–
+
= +1
33
2
932 2e x e xx xsin cos
− ∫4
932e x dxx cos
13
932e x dxx cos∫
= + +1
33
2
932 2
1e x e x Cx xsin cos
e x dxx2 3cos∫= + +3
133
2
1332 2e x e x Cx xsin cos
40. e x dxx−∫ 3 4cos u dve –3x cos 4x
–3e –3x 14 sin 4x
9e –3x –116 cos 4x+
–
+
= −− −1
44
3
1643 3e x e xx xsin cos
− −∫9
1643e x dxx cos
25
1643e x dxx−∫ cos
= − +− −1
44
3
1643 3
1e x e x Cx xsin cos
e x dxx−∫ 3 4cos
= − +− −4
254
3
2543 3e x e x Cx xsin cos
252 Problem Set 9-11 Calculus Solutions Manual© 2005 Key Curriculum Press
Note: As a check for integrals such asProblems 39 and 40, the numerators of thecoefficients equal the 3 and 4 in the argumentsof e−3x and cos 4x. The denominators equal32 + 42, or 25.
41. g (x) = x3 ln 5x ⇒g ′ (x) = (3x2) ln 5x + x3 (5/5x)
= x 2 (3 ln 5x + 1)
42. h (x) = x2 ln 8x ⇒h ′(x) = (2x) ln 8x + x2 (8/8x)
= x(2 ln 8x + 1)
43. x x dx3 5ln∫------------------------
u dvln 5x x3
1x
14 x4
114 x 3
0116x 4
+
–
+
= − +1
45
1
164 4x x x Cln
44. x x dx2 8ln∫------------------------
9
3
3
u dvln 8x x2
1x
1x3
11x2
01x3+
–
+
= − +1
38
1
93 3x x x Cln
45. yx
x x x=
+ + +⇒
( )( )( )2 3 4ln y = ln x − ln (x + 2)
− ln (x + 3) − ln (x + 4) ⇒y′ = y[x− 1 − (x + 2)− 1 − (x + 3)− 1 − (x + 4)− 1]
=+ + +
⋅ − +− −x
x x xx x
( )( )( )2 3 421 1[ ( )
− (x + 3)− 1 − (x + 4)− 1]
46. yx
x x x= ⇒
( – )( – )( – )1 2 3ln y = ln x − ln (x − 1)
− ln (x − 2) − ln (x − 3) ⇒y′ = y[x− 1 − (x − 1)− 1 − (x − 2)− 1 − (x − 3)− 1]
= ⋅ − −− −x
x x xx x
( – )( – )( – )1 2 311 1[ ( )
− (x − 2)− 1 − (x − 3)− 1]
47.x
x x xdx
( )( )( )+ + +∫ 2 3 4
=+
++
−+
∫ –1
2
3
3
2
4x x xdx
= −ln |x + 2| + 3 ln |x + 3| − 2 ln |x + 4| + C
48.x
x x xdx
( – )( – )( – )1 2 3∫= − +
∫ 1 2
1
2
2
3 2
3
/
– –
/
–x x xdx
= − − − + − +1
21 2 2
3
23ln ln ln| | | | | |x x x C
49. y = cos3 x sin x ⇒y′ = (−3 cos2 x sin x) sin x + cos3 x (cos x)
= −3 cos2 x sin2 x + cos4 x
50. y = sin5 x cos x ⇒y′ = (5 sin4
x cos x) cos x + sin5 x (−sin x)= 5 sin4 x cos2 x − sin6 x
51. cos sin cos3 41
4x x dx x C( ) = − +∫
52. sin cos sin5 61
6x x dx x C( ) = +∫
53. cos sin cos3 21x dx x x dx= −∫ ∫ ( )
= −∫ ∫cos sin cosx dx x x dx2 ( )
= − +sin sinx x C1
33
Or: cos cos sin cos3 21
3
2
3x dx x x x dx= +∫ ∫
= + +1
3
2
32cos sin sinx x x C
54. sin cos sin5 21x dx x x dx= −∫ ∫ ( ) ( )2
= − +∫ ( )( )1 2 2 4cos cos sinx x x dx
= −∫ ∫sin cos sinx dx x x dx2 2
+ ∫ cos sin4 x x dx
= − + − +cos cos cosx x x C2
3
1
53 5
Or: sin sin cos sin5 4 31
5
4
5x dx x x x dx= − + ∫∫
= − − + ∫1
5
4
15
8
154 2sin cos sin cos sinx x x x x dx
= − −1
5
4
154 2sin cos sin cosx x x x
− +8
15cos x C
55. cos cos sin cos4 3 21
4
3
4x dx x x x dx= + ∫∫
= + + ∫1
4
3
8
3
83cos sin cos sinx x x x dx
= + + +1
4
3
8
3
83cos sin cos sinx x x x x C
Calculus Solutions Manual Problem Set 9-11 253© 2005 Key Curriculum Press
56. sin sin cos sin6 5 41
6
5
6x dx x x x dx= − +∫ ∫
= − −1
6
5
245 3sin cos sin cosx x x x
+ ∫15
242sin x dx
= − −1
6
5
245 3sin cos sin cosx x x x
− + ∫15
48
15
48sin cosx x dx
= − −1
6
5
245 3sin cos sin cosx x x x
− + +5
16
5
16sin cosx x x C
57. g (x) = (x4 + 3)3 ⇒g ′(x) = 3(x4 + 3)2(4x3) = 12x3(x4 + 3)2
58. f (x) = (x3 − 1)4 ⇒f ′(x) = 4(x3 − 1)3(3x2) = 12x2(x3 − 1)3
59. ( ) ( )3x dx x x x dx4 12 8 43 9 27 27+ = + + +∫∫= + + + +1
13
27
52713 9 5x x x x C
60. ( )4x dx3 1−∫= − + − +∫ ( )x x x x dx12 9 6 34 6 4 1
= − + − + +1
13
2
5
6
713 10 7 4x x x x x C
61. ( )3x x dx x C4 3 4 431
163+ = + +∫ ( )
62. ( ) ( )x x dx x C3 4 2 3 511
151− = − +∫
63. ( )x dx x x C4 531
53+ = + +∫
64. ( )x dx x x C3 411
4− = − +∫
65. f x t dt f x xx
( ) ( ) ( ) ( )= + ⇒ ′ = +∫ 4 3
1
4 33 3
66. h x t dt h x xx
( ) ( ) ( ) ( )= − ⇒ ′ = −∫ 3
5
4 3 41 1
67. xe dxx
1
2
∫ u dvx ex
1 ex
0 ex
+
–+
= − = − − + = =xe e e e e e ex x
1
2 2 2 22 7 3890. K
68. xe dxx−∫0
2 u dvx e–x 1 –e–x 0 e–x
+–+
= − − = − − + +− − − −xe e e ex x0
2 2 22 0 1
= −3e− 2 + 1 = 0.59399…
69. r(x) = xex ⇒ r ′(x) = xex + ex
70. s(x) = xe− x ⇒ s′(x) = −xe− x + e− x
71. q xx
x( )
ln= + ⇒2
′ = ⋅ − + ⋅ = − −q x
x x x
x
x
x( )
( / ) (ln ) ln1 2 1 12 2
72. r xx
x( )
(ln )= + ⇒3 4
′ = ⋅ − + ⋅r x
x x x x
x( )
(ln ) ( / ) [(ln ) ]3 1 4 12 3
2
= − −3 42 3
2
(ln ) (ln )x x
x
73.ln
(ln )x
xdx x
dx
x
+ = +∫ ∫22
= + +1
22 2(ln )x C
74.(ln )
(ln )x
xdx x
dx
x xdx
334 4+ = +∫ ∫ ∫
= + +1
444(ln ) ln | |x x C
(The absolute values are optional because ln xappears in the original integrand, so onlypositive values of x can be used.)
75. f x e f x xex x( ) ( )= ⇒ ′ =2 2
2
76. f x e f x x ex x( ) ( )= ⇒ ′ =3 3
3 2
77. xe dx e Cx x2 21
2∫ = +
78. x e dx e Cx x2 3 31
3∫ = +
79. x e dxx3 2
∫---------------------
2
u dvx 2 xe x2
2x12e x2
21xe x2
014ex2 +
–
+
= − +1
2
1
22 2 2
x e e Cx x
254 Problem Set 9-11 Calculus Solutions Manual© 2005 Key Curriculum Press
80. x e dxx5 3
∫---------------------
u dvx3 x2 ex 3
3x2 13ex3
1 x2 ex3
013e x3
+
–
+
= − +1
3
1
33 3 3
x e e Cx x
81. e bx dxax∫ cos u dveax cos bx
aeax 1b sin bx
a2 e ax –1b2 cos bx+
–
+
= + − ∫12
2
2be bx
a
be bx
a
be bx dxax ax axsin cos cos
a b
be bx dxax
2 2
2
+ ∫ cos
= + +12 1b
e bxa
be bx Cax axsin cos
e bx dxax∫ cos
=+
++
+b
a be bx
a
a be bx Cax ax
2 2 2 2sin cos
(for a, b not both 0)
e bx dx x Cax cos = +∫ (for a = b = 0)
82. e bx dxax sin∫
b2
u dve ax sinbx
ae ax –1b cosbx
a2 eax –1
sinbx
+
–
+
= − +12b
e bxa
be bxax axcos sin
− ∫a
be bx dxax
2
2 sin
a b
be bx dxax
2 2
2
+ ∫ sin
= − + +12 1b
e bxa
be bx Cax axcos sin
e bx dxax sin∫=
+−
++a
a be bx
b
a be bx Cax ax
2 2 2 2sin cos
(for a, b not both 0)
e bx dx Cax sin =∫ (for a = b = 0)
83. sin ( cos )2 1
21 2cx dx cx dx∫ ∫= −
= − +1
2
1
42x
ccx Csin (for c ≠ 0)
sin2 cx dx C=∫ (for c = 0)
84. cos cos2 1
21 2cx dx cx dx= +∫ ∫ ( )
= + +1
2
1
42x
ccx Csin (for c ≠ 0)
cos2 cx dx x C= +∫ (for c = 0)
85. f xax b
cx d( ) = +
+
f xa cx d c ax b
cx d
ad bc
cx d′ = + +
+=
+( )
( ) – ( )
( )
–
( )2 2
(for c, d not both 0)(undefined for c = d = 0)
86. f (x) = (ax + b)n
f x na ax b n′ = + −( ) ( ) 1
(for a, b not both 0, or n ≥ 1)f ′(x) = 0 (for a = b = 0 and 0 ≤ n ≤ 1)
(undefined for a = b = 0 and n < 0)
87.ax b
cx ddx
a
c
b a c d
cx ddx
++
= + −+
∫ ∫ ( / )
= + + +ax
c
bc ad
ccx d C
–ln | |2 (for c ≠ 0)
ax b
cx ddx
a
dx
b
dx C
++
= + +∫ 22
(for c = 0, d ≠ 0)(undefined for c = d = 0)
88. ( )ax b dxax b
a nCn
n
+ = ++
++
∫ ( )
( )
1
1(for n ≠ −1, a ≠ 0)
( )ax b dxa
ax b Cn+ = + +∫ 1ln | |
(for n = −1, a ≠ 0)
( )ax b dx b x Cn n+ = +∫ (for a = 0)
89.x dx
x ax a x dx
2 2
2 2 1 21
22
+= +∫ ∫ −( ) ( )/
= ⋅ + + = + +1
22 2 2 1 2 2 2( ) /x a C x a C
90.x dx
a xa x x dx
2 2
2 2 1 21
22
−= − − −−∫∫ ( ) ( )/
= − ⋅ − + = − +1
22 2 2 1 2 2 2( ) /a x C a x C–
(for a ≠ 0)(undefined for a = 0)
91.dx
x a
d a
a a2 2 2 2 2+=
+∫∫ ( tan )
tan
θθ
= =∫ ∫a d
ad
sec
secsec
2 θ θθ
θ θ
= ln |sec θ + tan θ | + C1
= + + +ln1 12 2
1ax a
ax C
= + + +ln x a x C2 2
Calculus Solutions Manual Problem Set 9-11 255© 2005 Key Curriculum Press
92.dx
a x
d a
a a2 2 2 2 2−=
−∫ ∫ ( sin )
sin
θθ
= = = + = +∫ ∫ −a d
ad C
x
aC
cos
cossin
θ θθ
θ θ 1
(for a ≠ 0)(undefined for a = 0)
93. f (x) = x2 sin ax ⇒ f ′(x) = 2x sin ax + ax2 cos ax
94. f (x) = x2 cos ax ⇒ f ′(x) = 2x cos ax − ax2 sin ax
95. x ax dx2 sin∫ u dvx 2 sinax2x –
1a cosax
2 –1a2 sinax
01a3 cosax–
–
+
+
= − + + +1 2 222 3a
x axa
x axa
ax Ccos sin cos
(for a ≠ 0)
x ax dx C2 sin =∫ (for a = 0)
96. x ax dx2 cos∫ u dvx 2 cosax2x 1
a sinax
2 –1a2 cosax
0 –1a3 sinax
+
+
–
–
= + − +1 2 222 3a
x axa
x axa
ax Csin cos sin
(for a ≠ 0)
x ax dx x C2 31
3cos = +∫ (for a = 0)
97. sinh coshax dxa
ax C= +∫ 1 (for a ≠ 0)
sinh ax dx C=∫ (for a = 0)
98. cosh sinhax dxa
ax C= +∫ 1(for a ≠ 0)
cosh ax dx x C= +∫ (for a = 0)
99. cos−∫ 1 ax dx u dvcos –1 ax 1–a
√1 – (ax)2x–
+
= +−
− ∫x axax dx
axcos
( )
1
21
= − − −− −∫x axa
ax a x dxcos 1 2 1 2 21
21 2[ ( ) ] ( )/
= − +−x axa
ax Ccos – ( )1 211
(for a ≠ 0)
cos− = +∫ 1
2ax dx x C
π(for a = 0)
100. sin−∫ 1 ax dx u dvsin–1 ax 1a
1 – (ax)2x
+
–
√
= −− ∫x axax dx
axsin
– ( )
1
21
= + − −− −∫x axa
ax a x dxsin 1 2 1 2 21
21 2[ ( ) ] ( )/
= + +−x axa
ax Csin – ( )1 211 (for a ≠ 0)
sin− =∫ 1 ax dx C (for a = 0)
101.1
1+∫ xdx Let u x= +1 .
x = (u − 1)2
dx = 2(u − 1) du
= = − ∫∫∫ 2 12 2
( – )u du
udu u du( / )
= 2u − 2 ln |u| + C= + − + +2 1 2 1( ) |x x Cln |
Or: 2 2 1 1x x C− + +ln | |Absolute values are optional because1 0+ x > .
102.1
1– xdx∫ Let u x= −1 .
x = (1 − u)2
dx = 2(u − 1) du
= = − ∫∫∫ 2 12 2
( – )u du
udu u du ( / )
= − +2 2u u Cln | |
= − − − +2 1 2 1( ) | |x x Cln
Or: − − − +2 2 1 1x x Cln | |
103.1
1 4+∫ xdx Let u x= +1 4 .
x = (u − 1)4
dx = 4(u − 1)3 du
= = − + −∫∫ 4 14 12 12 4
32( – )u du
uu u u du( / )
= − + − +4
36 12 43 2u u u u Cln | |
= + − + + +4
31 6 1 12 14 3 4 2 4( )x x x( ) ( )
− + +4 1 4ln x C
Or: 4
32 4 4 14 3 4 2 4 4
1( ) ln ,x x x x C− + − + +( ) | |
256 Problem Set 9-11 Calculus Solutions Manual© 2005 Key Curriculum Press
by expanding the powers or by starting withu x= 4 .
Absolute values are optional because 1 04+ x > .
104.1
3x xdx
+∫ Let u x= 1 6/ .
x = u6
dx = 6u5 du
=+
=+∫∫ 6 6
1
5
3 2
3u du
u u
u du
u
= − + −+
∫ 6 6 6
6
12u u
udu
(by long division)= − + − + +2 3 6 6 13 2u u u u Cln | |
= − + − + +2 3 6 6 13 6 6x x x x Cln ( )
105.1
1edx
x +∫ Let u ex= + 1.
ex = u2 − 1 x = ln (u2 − 1)
dxu du
u=
−2
12
= = −+
∫∫ 2
1
1
1
1
12
du
u u udu
– –(by partial fractions)
= − − + +ln ln| | | |u u C1 1
= + − − + + +ln ln( ) ( )e e Cx x1 1 1 1
106.1
1edx
x –∫ Let u ex= – .1
ex = u2 + 1 x = ln (u2 + 1)
dxu du
u=
+2
12
=+
= + = +− −∫ 2
12 2 12
1 1du
uu C e Cxtan tan –
107. a. Let t = x/2 and substitute, gettingcos x = 2 cos2 (x/2) − 1 andsin x = 2 sin (x/2) cos (x/2).
b. cossec ( / )
xx
= −2
212
= 2 2
2
2
2
– sec ( / )
sec ( / )
x
x
= ++
2 1 2
1 2
2
2
– [ tan ( / )]
tan ( / )
x
x
=+
1 2
1 2
2
2
– tan ( / )
tan ( / )
x
x, Q.E.D.
sinsin ( / )
cos ( / )cosx
x
xx= 2
2
222 ( / )
= 2 21
22tansec ( / )
( / )xx
=+2 2
1 22
tan ( / )
tan ( / )
x
x, Q.E.D.
c. u x u x dxdu
u= ⇒ = ⇒ =
+−tan tan( / )2 2
2
11
2
cos–
xu
u=
+1
1
2
2 and sin xu
u=
+2
1 2 from part b.
d.1
1+∫ cos xdx
=+
+
⋅+∫ 1
111
2
12
2
2– uu
du
u
=+ +
= ∫∫ 2
1 12 2
du
u udu
( ) ( – )
e.1
12
+= = + = +∫∫ cos
tanx
dx du u C x C( / )
108. a. seccos
x dxx
dx= ∫∫ 1
= + ⋅+
=∫ ∫1
1
2
1
2
1
2
2 2 2
u
u
du
u udu
– –
b. sec–
x dxu u
du= ++
∫∫ 1
1
1
1 = −ln | 1 − u | + ln |1 + u| + C
= + + = + +ln–
lntan ( / )
– tan ( / )
1
1
1 2
1 2
u
uC
x
xC
c. sec lntan ( / )
– tan ( / )x dx
x
xC= +
⋅+∫ 1 2
1 1 2
= + +lntan ( / ) tan ( / )
– tan ( / ) tan ( / )
ππ4 2
1 4 2
x
xC
= ln |tan (π/4 + x/2)| + C
d. i. sec ln tan |x dx x= +∫ | ( / / )π 4 20
1
0
1
= ln |tan (π/4 + 1/2)| − ln |tan π/4|= ln |tan (π/4 + 1/2)| = 1.226191…
ii. sec ln sec tan |x dx x x= +∫ |0
1
0
1
= ln |sec 1 + tan 1| − ln |1 + 0| =ln |sec 1 + tan 1| = 1.226191… , whichagrees with the answer in part i.
109.1
1
1
111
2
12
2
2– cos–
–xdx
uu
du
u=
+
⋅+∫∫
=+
= = +∫∫ 2
1 1
12 2 2
du
u u
du
u uC
( ) – ( – )
–
= −cot (x/2) + C
110.1
1
1
12
1
2
12
2+=
++
⋅+∫ ∫sin x
dx uu
du
u
=+ +
=+
=+
+∫∫ 2
1 2 1
1
12 2
du
u u
du
u uC
( ) ( )
–
=+
+–
tan ( / )
1
2 1xC
Calculus Solutions Manual Problem Set 9-13 257© 2005 Key Curriculum Press
111.cos
– cos
–
––
x
xdx
uu
uu
du
u1
11
111
2
1
2
2
2
2
2= +
+
⋅+∫ ∫
=+
= −+
∫ ∫1
1
1 2
1
2
2 2 2 2
–
( )
u
u udu
u udu
= − − +−12 1
uu Ctan
= − − +−1
22 21
tan ( / )tan [tan ( / )]
xx C
= −cot (x/2) − x + C
Or: cos
– cos – cos
x
xdx
xdx
11
1
1= − +
∫ ∫
= − +∫ ∫dxx
dx1
1– cos= −x − cot (x/2) + C (using Problem 109)
Problem Set 9-121. Answers will vary.
Problem Set 9-13
Review ProblemsR0. Answers will vary.
R1. f ( x ) = x cos x ⇒f ′( x ) = x(−sin x) + (1) cos x = cos x − x sin x
⇒ + = ′∫x x dx C f x dxcos ( )
= −∫ (cos sin )x x x dx
= − ∫sin sinx x x dx
⇒ = − +∫ x x dx x x x Csin sin cos
x x dx x x xsin sin cos= −∫1
4
1
4
= sin 4 − 4 cos 4 − sin 1 + cos 1 = 1.5566…
Numerically,
x x dxsin . .≈∫ 1 55661
4
K
R2. 5 2x x dxsin∫ u = 5x dv = sin 2x dx
du = 5 dx v x= − 1
22cos
= −
+ ∫5
1
22
1
22 5x x x dxcos cos ( )
= − + +5
22
5
42x x x Ccos sin
R3. a. x x dx3 2cos∫ u dvx3 cos 2x
3x2 12 sin 2x
6x –14 cos 2x
6 –18 sin 2x
0116 cos 2x
+
+
–
–
+
= +1
22
3
423 2x x x xsin cos
− − +3
42
3
82x x x Csin cos
b. e x dxx4 3sin∫13
9
u dve4x sin 3x
4e 4x – cos 3x
16e –1 sin 3x+
–
+
4x
= +–1
33
4
934 4e x e xx xcos sin
–16
934e x dxx∫ sin
⇒ ∫25
934e x dxx sin
= − + +1
33
4
934 4
1e x e x Cx xcos sin
⇒ ∫ e x dxx4 3sin
= − + +3
253
4
2534 4e x e x Cx xcos sin
c. x x dx(ln )2∫------------------------2
------------------------
u dv(ln x)2 x
2 ln x · 1x
1x2
ln x x1x
12x2
112x
014x2
+
+
–
–
= − + +1
2
1
2
1
42 2 2 2x x x x x C( )ln ln
d. Slice parallel to the y-axis. Pick a samplepoint (x, y) on the graph, within the slice.dV = 2πx · y · dx = 2πx(x ln x) dx
= 2πx2 ln x dx
V x x dx= ∫2 2
1
2
π ln
----------------------
u dvln x x2
1x
13 x3
113 x2
019 x3 +
–
+
258 Problem Set 9-13 Calculus Solutions Manual© 2005 Key Curriculum Press
= −
2
1
3
1
93 3
1
2
π x x xln
= − + = −16
32
16
9
2
9
16
32
14
9π π π π πln ln
= 6 7268. K
R4. a. cos30 dx∫ u dvcos 29x cosx
29 cos 28x sinx sinx+
––
= + ∫cos sin cos sin29 28 229x x x x dx
= + −∫cos sin cos )29 2829 1x x x x dx( cos2
⇒ ∫30 30cos dx
= + ∫cos sin cos29 2829x x x dx
⇒ ∫ cos30 dx
= + ∫1
30
29
3029 28cos sin cosx x x dx
b. sec sec tan sec6 4 41
5
4
5x dx x x x dx= +∫ ∫
= +1
5
4
154 2sec tan sec tanx x x x
+ ∫8
152sec x dx
= +1
5
4
154 2sec tan sec tanx x x x
+ +8
15tan x C
c. tan tan tann nx dx x x dx=∫ ∫ − ( )2 2
= −−∫ tan secn x x dx2 2 1( )
= −− −∫ ∫tan sec tann nx x dx x dx2 2 2
=−
−− −∫1
11 2
nx x dxn ntan tan
R5. a. cos ( sin ) (cos )5 2 21x dx x x dx= −∫ ∫= − +∫ ( )( )1 2 2 4sin sin cosx x x dx
= − + +sin sin sinx x x C2
3
1
53 5
b. sec (tan ) (sec )6 2 2 21x dx x x dx= +∫∫= + +∫ (tan tan )(sec )4 2 22 1x x x dx
= + + +1
5
2
35 3tan tan tanx x x C
c. sin ( cos )2 71
21 14x dx x dx= −∫ ∫
= − +1
2
1
2814x x Csin
d. sec3 x dx∫ u dvsecx sec 2x
secx tanx tanx–+
= − ∫sec tan tan secx x x x dx2
= − −∫sec tan sec secx x x dx( )2 1
= + −∫ ∫sec tan sec secx x x dx x dx3
2 3sec x dx∫= + + +sec tan |sec tan |x x x x Cln
sec3 x dx∫= + + +1
2
1
2sec tan | tan |x x x x Cln sec
e. tan tan tan9 9 932 32 32dx dx x C= = +∫ ∫( ) ( )
f. r = 9 + 8 sin θ
dA r d d= = +1
2
1
29 82 2θ θ θ( sin )
A d= + +∫1
264 144 812
0
4
( sin sin )/
θ θ θπ
= − + +∫1
232 1 2 144 81
0
4
[ ( cos ) sin ]/
θ θ θπ
d
= −
− +16
1
22 72
81
2 0
4
θ θ θ θπ
sin cos/
= − − + +4 8 36 281
872π π
= + − =113
864 36 2 57 4633π . K
R6. a. x dx2 49−∫
u
v
θ
7
x √x 2 – 49
Let x
x7
7= =sec . secθ θ,
dx = 7 sec θ tan θ dθ,
xx2 149 77
− = = −tan , secθ θ
∴ −∫ x dx2 49
= ∫ ( tan )( sec tan )7 7θ θ θ θd
= ∫49 2sec tanθ θ θd
= −
∫ ∫49 3sec secθ θ θ θd d
Calculus Solutions Manual Problem Set 9-13 259© 2005 Key Curriculum Press
= + +49
1
2
1
2sec tan |sec tan |θ θ θ θln
− +
+ln |sec tan |θ θ C1
= − + +49
2
49
2 1sec tan | tan |θ θ θ θln sec C
= ⋅ ⋅ −49
2 7
49
7
2x x
− + − +49
2 7
49
7
2
1lnx x
C
= − − + −1
249
49
2492 2x x x xln
+ +49
27 1ln C
= − − + − +1
249
49
2492 2x x x x Cln
b. x x dx x dx2 210 34 5 9− + = − +∫∫ ( )
u
v
θ
3
x – 5
√ (x – 5)2 + 9
Let x − =5
3tan .θ
x = 5 + 3 tan θ, dx = 3 sec2 θ dθ,
( ) sec , tanxx− + = = −−5 9 3
5
32 1θ θ
∴ +∫ ( – )x dx5 92
= =∫ ∫( sec )( sec ) sec3 3 92 3θ θ θ θ θd d
= + + +9
2
9
2 1sec tan ln |sec tan |θ θ θ θ C
=− + −9
2
5 9
3
5
3
2( )x x
+− +
+ − +9
2
5 9
3
5
3
2
1ln( )x x
C
= − + ⋅ −1
25 9 52( ) ( )x x
+ − + + − − +9
25 9 5
9
232
1ln ( ) lnx x C
= +1
25 10 342( – ) –x x x
+ + + − +9
210 34 52ln –x x x C
c. 1 0 25 2– . x dx∫
u
v
θ
10.5x
√ 1 – 0.25x2
Let 0 5
1
. x = sin θ.
x = 2 sin θ, dx = 2 cos θ dθ,
1 0 252
2 1– . cos , sinxx= = −θ θ
1 0 25 22– . (cos )( cos )x dx d∫ ∫= θ θ θ
= = +∫ ∫2 1 22cos ( cos )θ θ θ θd d
= + + = + +θ θ θ θ θ1
22sin sin cosC C
= + − +−sin .1 2
2
1
21 0 25
xx x C
d. Slice region vertically. Pick sample point(x, y) on the upper branch of the circle,within the strip.
dA y dx x dx= =2 2 25 2–
u
v
θ
5x
√ 25 – x 2
Let x
x dx d5
5 5= = =sin sin cosθ θ θ θ. , ,
25 55
2 1– cos , sinxx= = −θ θ
A x dx= −∫ 2 25 2
3
4
==
=
∫2 5 53
4
cos ( cos )θ θ θdx
x
= +=
=
∫25 1 23
4
( cos )θ θdx
x
= + ==25 12 5 2 3
4θ θ. sin xx
= + ==25 25 3
4θ θ θsin cos xx
= + ⋅ ⋅ −−255
255
1
5251 2
3
4
sinx x
x
= + − −− −25 0 8 4 9 25 0 6 3 161 1sin . sin . = 25(sin− 1 0.8 − sin− 1 0.6) = 7.0948…
R7. a.( )
– –
( )
( )( – )
6 1
3 4
6 1
1 42
x dx
x x
x dx
x x
+ = ++∫ ∫
=+
+−
∫ 1
1
5
4x xdx
= ln |x + 1| + 5 ln |x − 4| + C
260 Problem Set 9-13 Calculus Solutions Manual© 2005 Key Curriculum Press
b.5 21 2
1 2 3
2x x
x x xdx
– –
( – )( )( – )+∫= +
+−
∫ 3
1
4
2
2
3x x xdx
– –
= 3 ln |x − 1| + 4 ln |x + 2| − 2 ln |x − 3| + C
c.5 3 45
9
5 3 45
9
2
3
2
2
x x
x xdx
x x
x xdx
+ ++
= + ++∫ ∫ ( )
= ++
= + +∫ −5 3
95
321
x xdx x
xCln | | tan
(The second integral may be found byinspection or by trigonometric substitution.)
d.5 27 32
4
2
2
x x
x xdx
+ ++∫ ( )
= ++
−+
∫ 2 3
4
1
4 2x x xdx
( )
= 2 ln | x | + 3 ln | x + 4 | + (x + 4)− 1 + C
= + ++
+ln | ( ) |x xx
C2 341
4
e.dy
dxy y= 0 1 3 8. ( – )( – )
dy
y ydx
( – )( – ).
3 80 1=∫ ∫
−−
+−
=∫ ∫1 5
3
1 5
80 1
/ /.
y ydy dx
− − + − = +1
53
1
58 0 1 1ln | | ln | | .y y x C
−ln | y − 3 | + ln | y − 8 | = 0.5x + C
Substituting (0, 7) gives
C = –ln 4 + ln 1 = –ln 4.
ln–
. lny
yx
8
30 5 4
−= −
y
ye ex x–
.. ln .8
30 250 5 4 0 5
−= =−
− =y
ye x–
–. .8
30 25 0 5
((y – 8)/(y – 3) < 0 because (0, 7) is on thegraph)
ye x= +
+3
5
1 0 25 0 5. .
The graph shows that solution fits slope field.
x
y
7
R8. a.
2
1
x
y
b. f ( x ) = sec− 1 3x
′ = =f xx x x x
( )| | ( ) – | | –
3
3 3 1
1
9 12 2
c. tan−∫ 1 5x dx u dvtan–1 5x 1
51 + 25x2 x
+
–
= −+
− ∫x xx
xdxtan 1
255
1 25
= −+
− ∫x xx
x dxtan ( )125
1
10
1
1 2550
= − + +−x x x Ctan ln | |1 251
101 25
(Absolute values are optional because1 + 25x2 > 0.)
d. “Obvious” way: Slice the region vertically.Pick a sample point (x, y) on the graph,within the strip.dA = y dx = cos− 1 x dx
A x dx x x x= = − −− −∫ cos cos1 1 2
0
1
0
1
1
= − − + =−cos 11 0 0 1 1Easier way: Slice horizontally. Pick a samplepoint (x, y) on the graph within the strip.dA = x dy = cos y dy
A y dy y= = = − =∫ cos sin/ /
0
2
0
2
1 0 1π π
R9. a.
x
y
1
1
b.
x
y
1
1
Calculus Solutions Manual Problem Set 9-13 261© 2005 Key Curriculum Press
c. h(x) = x2 sech xh′(x) = –x2 sech x tanh x + 2x sech x
d. f ( x ) = sinh− 1 5x
′ =+
f xx
( )5
25 12
e. tanhcosh
sinh31
33x dx
xx dx=∫ ∫
= +1
33ln |cosh |x C
(Absolute values are optional becausecosh 3x > 0.)
f. cosh−∫ 1 7x dx u dvcosh –1 7x 1
749x2 – 1
x+
–
√
= −−
− ∫x xx
xdxcosh 1
27
7
49 1
= − −− −∫x x x x dxcosh ( ) ( )/1 2 1 271
1449 1 98
= − ⋅ − +−x x x Ccosh ( ) /1 2 1 271
142 49 1
= − − +−x x x Ccosh 1 271
749 1
g. cosh2 x – sinh2 x
= + − −1
2
1
22 2( ) ( )– –e e e ex x x x
= + + − − +1
42
1
422 2 2 2( ) ( )– –e e e ex x x x
= 1, Q.E.D.
h. The general equation is y kk
x C= +cosh .1
y = 5 at x = 0 ⇒ 5 = k cosh 0 + C⇒ C = 5 – k
y x kk
k= = ⇒ = + −7 3 73
5at cosh
⇒ = −23
kk
kcosh
⇒ k = 2.5269… (solving numerically)
yt= + −2 5269
2 52695 2 5269. cosh
..K
KK
y( ) .10 68 5961 20= …
= + −2 52692 5269
5 2 5269. cosh.
.KK
Kx
⇒ x = ±6.6324… (solving numerically)
R10. a. ( ) lim ( ). .x dx x dxb
b
− = −−
→∞
∞−∫ ∫2 21 2
3
1 2
3
= − −→∞
−lim ( ) .
b
b
x5 2 0 2
3
= + =→∞
lim [– ( – ) ]– .
bb5 2 5 50 2
The integral converges to 5.
b. tan lim tan//
x dx x dxa a
=→ −∫ ∫ππ 22
0 0
=→ −lim ln |sec |
/a a
xπ 2
0
= − = −∞→ −lim (ln |sec | ln |sec |)
/aa
π 20
The integral diverges.
c. x dx−
−∫2 3
1
1/
= +→
−
− →
−− +∫ ∫lim lim/ /
b
b
a ax dx x dx
0
2 3
1 0
2 31
= +→ − →− +lim lim/ /
b
b
a ax x
0
1 3
1 0
1 31
3 3
= − − + − =→ →− +lim [ ( )] lim ( )/ /
b ab a
0
1 3
0
1 33 3 3 3 6
The integral converges to 6.
d. xx
xdx− −
−
∫ | |1
10
4
= + + −→ →− +∫ ∫lim ( ) lim ( )
b
b
a ax dx x dx
1 0 1
4
1 1
= +
+ −
→ →− +
lim lim/ /
b
b
a a
x x x x1
3 2
0 1
3 242
3
2
3
= + −
→ −
lim /
bb b
1
3 22
30
+ ⋅ − − +
→ +
lim / /
aa a
1
3 2 3 22
34 4
2
3
= ⋅ + + ⋅ − − ⋅ +2
31 1
2
34 4
2
31 13 2 3 2 3 2/ / /
= + − + =116
34 1
10
3
The integral converges to 10
33 333= . .K
e. x dxp−∞
∫1 converges if p > 1 and diverges
otherwise.
R11. a. f x x x f x xx
x( ) sin ( ) sin= ⇒ ′ = +
−− −1 1
21
b. I = −∫ x x dxsin 1 u dvsin–1 x x
11 – x2
12x2 –
+
√
= −− ∫1
2
1
2 1
2 12
2x x
x dx
xsin
–
Let I1
2
21= ∫ x dx
x– and x = sin θ.
∴ = − =dx d xcos , cos ,θ θ θ 1 2
θ = sin− 1 x
262 Problem Set 9-13 Calculus Solutions Manual© 2005 Key Curriculum Press
I1
22= =∫ ∫sin cos
cossin
θ θ θθ
θ θdd
= − = − +∫1
21 2
1
2
1
42( cos ) sinθ θ θ θd C
= − +1
2
1
2θ θ θsin cos C
= − − +−1
2
1
211 2sin x x x C
∴ = − + − +− − I1
2
1
4
1
412 1 1 2x x x x x Csin sin
c.d
dxe e ex x xtanh sec= h2
d. ( )x x dxx x
dx3 13
1− =−
−∫ ∫=
+∫ 1
1 1x x xdx
( – )( )
= − +−
++
∫ 1 1 2
1
1 2
1x x xdx
/ /
= − + − + + +ln | | ln | | ln | |x x x C1
21
1
21
e. f ( x ) = (1 − x2)1/2
′ = − − = − −− −f x x x x x( ) ( ) ( ) ( )/ /1
21 2 12 1 2 2 1 2
f. I = −∫ ( ) /1 2 1 2x dx
Let x = sin θ.∴ = − = dx d xcos , ( ) cos ,/θ θ θ1 2 1 2
θ = sin− 1 x
∴ = ⋅ =∫ ∫ I cos cos cosθ θ θ θ θd d2
= + = + +∫1
21 2
1
2
1
42( cos ) sinθ θ θ θd C
= + +1
2
1
2θ θ θsin cos C
= + − +−1
2
1
211 2sin x x x C
g. g x x g x xx
( ) ( ) ( )2= ⇒ ′ = ⋅ln ln21
h. x x dxln∫--------------------
u dvln x x
1x
12x2
112x
014x2
+
–
+
= − +1
2
1
42 2x x x Cln
R12. For ( ) ,/9 2 1 2− −∫ x x dx the x dx can be
transformed to the differential of the insidefunction by multiplying by a constant,
− − − = − − +−∫1
29 2 92 1 2 2 1 2( ) ( ) ( ) ,/ /x x dx x C
and thus has no inverse sine.
For ( ) ,/9 2 1 2− −∫ x dx transforming the dx to the
differential of the inside function, −2x dx,requires multiplying by a variable. Because theintegral of a product does not equal the product ofthe two integrals, you can’t divide on the outsideof the integral by −2x. So a more sophisticatedtechnique must be used, in this case,trigonometric substitution. As a result, aninverse sine appears in the answer:
( ) sin/93
2 1 2 1− = +− −∫ x dxx
C
Concept Problems
C1. sech x dx x dx∫ ∫= −1 2tanh
u
v
θ
1tanh x
√ 1 – tanh2x
Let tanh x = sin θ.∴ x = tanh− 1 (sin θ) and θ = sin− 1 (tanh x)
dx d d=−
⋅ =1
1
12sin
coscosθ
θ θθ
θ
1 12 2– tanh – sin cosx = =θ θ
∴ = ⋅ =∫ ∫ ∫sec coscos
h x dx d dθθ
θ θ1
= + = +−θ C x Csin (tanh ) ,1 Q.E.D.
sec sin (tanh )h0
11
0
1
∫ = −x dx x
= sin− 1 (tanh 1) − sin− 1 (tanh 0)= sin− 1 (tanh 1) = 0.86576948…Numerical integration gives 0.86576948… ,which agrees with the exact answer.
C2. From sinh 2A = 2 sinh A cosh A,let A = x/2, sosinh x = 2 sinh (x/2) cosh (x/2) ⇒ csch x
= =⋅ ⋅
1 1
2 2 2sinh sinh ( / ) cosh ( / )x x x
cscsinh ( / ) cosh ( / )
sec ( / )
sec ( / )h
h
hx
x x
x
x=
⋅ ⋅⋅1
2 2 2
2
2
2
2
=⋅
=sec ( / )
tanh ( / )
sec ( / )
tanh ( / )
h h22
2
2 2
12
2
2
x
x
x
x
∴ = ⋅
∫ ∫csc
tanh ( / )sec ( / )h h2x dx
xx dx
1
2
1
22
= ln | tanh (x/2) | + C, Q.E.D.
csc ln | tanh ( / )|h1
2
1
2
2∫ =x dx x
= =ln | tanh | ln | tanh |1 1 2 0 49959536– ( / ) . K
Numerical integration gives 0.49959536… .
Calculus Solutions Manual Problem Set 9-13 263© 2005 Key Curriculum Press
C3. From sin 2A = 2 sin A cos A, let A = x/2, sosin x = 2 sin (x/2) cos (x/2)
cscsin sin ( / ) cos ( / )
xx x x
= =1 1
2 2 2
= ⋅1
2 2 2
2
2
2
2sin ( / ) cos ( / )
sec ( / )
sec ( / )x x
x
x
= =sec ( / )
tan ( / )
sec ( / )
tan ( / )
22
2
2 2
12
2
2
x
x
x
x
∴ = ⋅
∫ ∫csc
tan ( / )sec ( / )x dx
xx dx
1
2
1
222
= ln | tan (x/2) | + C, Q.E.D.Or:Let u = tan (x/2), as in Problem 107 of ProblemSet 9-11.
Then anddxdu
ux
u
u=
+= +2
1
1
22
2
csc
∴ = + ⋅+∫ ∫csc x dx
u
u
du
u
1
2
2
1
2
2
= = + = +∫ ( / ) ln | | ln | tan ( / )| ,1 2u du u C x C
Q.E.D.Confirmation:
csc ln | tan ( / )|. .0 5
1
0 5
1
2∫ =x dx x
= − =ln tan ln tan .1
2
1
40 7605K
Numerical integration gives 0.7605… .
Note that tan ( / )sin ( / ) cos ( / )
cos ( / )x
x x
x2
2 2 2
2 22=
=+
=+
sin
cos csc cot,
x
x x x1
1 so
ln | tan ( / )| ln |csc cot | .x x x2 = − +
C4. Ax
dx=+−∞
∞
∫ 1
1 2
=+
++→−∞ →∞∫ ∫lim lim
a a b
b
xdx
xdx
1
1
1
12
0
20
= +→−∞
−
→∞
−lim tan lim tana a b
bx x1 0 1
0
= − + −→−∞
−
→∞
−lim ( tan ) lim (tan )a b
a b0 01 1
= −(−π/2) + (π/2) = πC5. Prove that f (x) = ln x is unbounded above.
Proof:Assume f (x) = ln x is not unbounded above.Then there is a number M > 0 such thatln x < M for all x > 0.Let x = eM+ 1.Then ln x = ln eM+ 1 = M + 1.∴ ln x > M, which is a contradiction.∴ the assumption is false, and ln x is unboundedabove, Q.E.D.
Chapter Test
T1. sin cos sin5 61
6x x dx x C= +∫
T2. x x dx3 6sinh∫ u dvx3 sinh 6x
3x2 16 cosh 6x
6x 136 sinh 6x
61
216 cosh 6x
01
1296 sinh 6x+
–
–
+
+
= −1
66
1
1263 2x x x xcosh sinh
+ − +1
366
1
2166x x x Ccosh sinh
T3. cos−∫ 1 x dx
√
u dvcos –1 x 1
–1
1 – x2x
+
–
= +− ∫x xx
xdxcos
–
1
21
= − − −− −∫x x x x dxcos ( ) ( )/1 2 1 21
21 2
= − − +−x x x Ccos ( )( ) /1 2 1 21
22 1
= − − +−x x x Ccos 1 21
T4. sec3∫ x dx
= + + +1
2
1
2sec tan ln |sec tan |x x x x C
T5. e x dxx2 5cos∫ u dve 2x cos 5x
2e 2x 15 sin 5x
4e 2x –125 cos 5x
+
–
+
= +1
55
2
2552 2e x e xx xsin cos
− ∫4
2552e x dxx cos
29
2552e x dxx∫ cos
= + +1
55
2
2552 2e x e x Cx xsin cos
e x dxx2 5cos∫= + +5
295
2
2952 2e x e x Cx xsin cos
264 Problem Set 9-13 Calculus Solutions Manual© 2005 Key Curriculum Press
T6. ln 3x dx∫ u dvln 3x 1
1/x x–+
= − = − +∫x x dx x x x Cln ln3 3
T7. f (x) = sech3 (e5x) ⇒f ′(x) = 3 sech2 (e5x) ⋅ [−sech (e5x)
tanh (e5x)] ⋅ 5e5x
= −15e5x sech3 (e5x) tanh (e5x)
T8. g x x g xx
( ) sin ( )–
= ⇒ ′ =−1
2
1
1
T9. f (x) = tanh− 1 x ⇒ tanh f (x) = x, |x| ≤ 1sech2 f(x) ⋅ f ′(x) = 1[1 − tanh2 f (x)] ⋅ f ′(x) = 1(1 – x2) ⋅ f ′(x) = 1
′ = <f xx
x( )–
, | |1
112
′ = = =f ( . )– . .
.0 61
1 0 36
1
0 641 5625
Numerically, f ′(0.6) ≈ 1.5625… (depending onthe tolerance of the calculator).
T10. General equation is y kk
x C= +cosh .1
y = 1 at x = 0 ⇒ 1 = k cosh 0 + C ⇒C = 1 − k
y = 3 at x kk
k= ⇒ = +5 35
1cosh –
Solving numerically, k ≈ 6.5586… .
y x= + −6 5586
1
6 55861 6 5586. K
KKcosh
..
T11. a. i. I =+
=∫ ∫x
x xdx
x
xdx
–
–
–
( – ) –
3
6 5
3
3 42 2
u
v
θ
x – 3
2
√ (x – 3)2 – 4
Let . – ,x
x–
sec sec3
23 2= =θ θ
dx = 2 sec θ tan θ dθ,
( – ) – tan sec–
xx
3 4 23
22 1= =θ θ, –
∴ = ∫ I( sec )( sec tan )
tan
2 2
4 2
θ θ θ θθ
d
= ⋅ = +∫ 1 21tan
sec ln | tan |θ
θ θ θd C
= − − +ln ( )1
23 42
1x C
= − − +ln ( )x C3 42
= − + +1
26 52ln | |x x C
ii. x
x xdx
x xdx
−− +
=−
+−
∫ ∫3
6 5
1 2
1
1 2
52
/ /
= + +1
21
1
25ln | | ln | |x x C– –
= +1
21 5ln | |( – )( – )x x C
= + +1
26 52ln | | ,x x C–
which agrees with part a.
iii.x
x xdx
–
–
3
6 52 +∫
=+
⋅ −∫1
2
1
6 52 62x x
x dx–
( )
= + +1
26 52ln | |x x C– , as in parts a and b.
b. See parts i, ii, and iii.
T12. cos ( cos )2 1
21 2x dx x dx= +∫ ∫
= + +1
2
1
42x x Csin
T13. a. i. cos sin cos5 2 21x dx x x dx=∫ ∫ ( – )
= +∫ ( – sin sin ) cos1 2 2 4x x x dx
= + +sin sin sinx x x C–2
3
1
53 5
ii. cos cos sin cos5 4 31
5
4
5x dx x x x dx= +∫ ∫
= +1
5
4
154 2cos sin cos sinx x x x
+ ∫8
15cos x dx
= +1
5
4
154 2cos sin cos sinx x x x
+ +8
15sin x C
b.1
5
4
15
8
154 2cos sin cos sin sinx x x x x+ +
= 1
51 2 2( – sin ) sinx x
+ +4
151
8
152( – sin ) sin sinx x x
= +1
5
2
5
1
53 5sin sin sinx x x–
+ +4
15
4
15
8
153sin sin sinx x x–
Calculus Solutions Manual Problem Set 9-13 265© 2005 Key Curriculum Press
= + +
− +
1
5
4
15
8
15
2
5
4
153sin sinx x
+ 1
55sin x
= +sin sin sinx x x– 2
3
1
53 5
T14. xe dxx−∞
∫ 0 1
0
. u dvx e –0.1x 1 –10e –0.1x 0 100e –0.1x
+
––
=→∞
lim (– – )– . – .
b
x xb
xe e10 1000 1 0 1
0
= + +→∞
lim (– – )– . – .
b
b bbe e10 100 0 1000 1 0 1
= + +
→∞
lim .b b
b
e–
10 1001000 1
= +
→∞
lim. .b be
–10
0 11000 1 (by l’Hospital’s rule)
= 100
T15. Answers will vary.
266 Problem Set 10-2 Calculus Solutions Manual© 2005 Key Curriculum Press
Chapter 10—The Calculus of Motion—Averages,Extremes, and Vectors
Problem Set 10-11. v(t) = 100(0.8)t − 30 = 100et ln 0.8 − 30 = 0
⇒ e tt ln . ln .
ln ..0 8 0 3
0 3
0 85 3955= ⇒ = =. K
v becomes negative after t0 ≈ 5.40 min.
2. s v dt e dtttt
up = = ∫∫ ( – )ln .100 300 8
00
00
= 151.8341… (numerically) ≈ 151.8 ft
s v dt e dtt
ttdown = − = −∫∫ ( – )ln .100 300 8
1010
00
= 51.8110… (numerically) ≈ 51.8 ftDistance = sup + sdown = 203.6452… ≈ 203.6 ft
3. Displacement = sup − sdown = 100.0231… ≈100.0 ftThe displacement is positive, so Calvin isupstream of his starting point.
4. Displacement = ∫ ( – )ln .100 300 8
0
10
e dtt
= 100.0231… (numerically) ≈ 100.0 ft
5. Distance = = −∫∫ | | | | .v dt e dtt100 300 8
0
10
0
10ln
= 203.6452… (numerically) ≈ 203.6 ft
Problem Set 10-2
Q1. 120 mi Q2. 25 mi/hQ3. 1.25 h Q4. f ′(x) = 1/xQ5. x ln x − x + C Q6. f ′(t) = sec2 t
Q7. g′(t) = sech2 t Q8.1
33x C+
Q9.1
22
lnx C+ Q10. ln lnln2 2 22ex x=
1. a. v(t) = t2 − 10t + 16 on [0, 6]v(t) = (t − 2)(t − 8) = 0 ⇔ t = 2 or 8 sv(t) > 0 for t in [0, 2). v(t) < 0 for t in (2, 6].
b. For [0, 2), displacement
= + =∫ ( – )t t dt2
0
2
10 16 142
3
Distance = 142
3ft
For (2, 6], displacement
= + = −∫ ( – )t t dt2
2
6
10 16 262
3
Distance = 2 ft62
3
c. Displacement = + = −∫ ( – )t t dt2
0
6
10 16 12 ft
Distance = − + =∫ | | ftt t dt2
0
6
10 16 411
3
d. Displacement = + −
= −14
2
326
2
312 ft
Distance = + =142
326
2
341
1
3ft
e. a(t) = v′(t) = 2t − 10a(3) = 2(3) − 10 = −4 (ft/s)/s
2. a. v(t) = tan 0.2t on [10, 20]v(t) = 0 ⇔ t = … 0, 5π, 10π, … = 5π in[10, 20]v(t) is infinite ⇔ t = … 2.5π, 7.5π, … ,none of which is in [10, 20].v(t) < 0 for t in [10, 5π). v(t) > 0 for t in(5π, 20].
b. For [10, 5π), displacement = ∫ tan 0 210
5
. t dtπ
= 5 ln | sec π | − 5 ln | sec 2 | = −4.3835…Distance = 4.3835… ≈ 4.38 cm
For (5π, 20], displacement = ∫ tan 0 25
20
. t dtπ
= 5 ln |sec 4| − 5 ln | sec π | = 2.1259…Distance = 2.1259… ≈ 2.13 cm
c. Displacement = = −∫ tan 0 2 2 257610
20
. .t dt K ≈
−2.26 cm
Distance = = ≈∫ | . | .tan 0 2 6 509510
20
t dt K
6.51 cm
d. Displacement = −4.3835… + 2.1259… =−2.2576… ≈ −2.26 cmDistance = −(−4.3835…) + 2.1259… =6.5095… ≈ 6.51 cm
e. a ( t) = v′(t) = 0.2 sec2 ta ( 15) = 0.2 sec2 3 = 0.2040… ≈ 0.20 (cm/s)/s
3. a. v t t( ) = −secπ24
2 on [1, 11]
v(t) = 0 when
cos .π24
0 5 8t t= ⇔ = in [1, 11]
v(t) < 0 for t in [1, 8). v(t) > 0 for t in(8, 11].
Calculus Solutions Manual Problem Set 10-2 267© 2005 Key Curriculum Press
b. For [1, 8), displacement =
∫ sec –
π24
21
8
t dt
= + −24
3 316
ππ π
ln sec tan
− + +24
24 242
ππ π
ln sec tan
= −4.9420…Distance ≈ 4.94 kmFor (8, 11], displacement
=
∫ sec –
π24
28
11
t dt
= + −24 11
24
11
2422
ππ πln sec tan
− + +24
3 316
ππ π
ln sec tan
= 4.7569…Distance ≈ 4.76 km
c. Displacement =
=∫ sec –
π24
21
11
t dt
− ≈ −0 1850 0 19. kmK .
Distance = − = ≈∫ secπ24
2 9 69901
11
t dt . K
9.70 km
d. Displacement = −4.9420… + 4.7569… =−0.1850… ≈ −0.19 kmDistance = −(−4.9420…) + 4.7569… =9.6990… ≈ 9.70 km
e. a t v t t t( ) ( ) = ′ = π π π24 24 24
sec tan
a ( 6) = 0.1851… ≈ 0.19 (km/h)/h
exactly π24
2
4. a. v(t) = t3 − 5t2 + 8t − 6 on [0, 5]v(t) = (t − 3)(t2 − 2t + 2) = 0 ⇔ t = 3 in [0, 5]v < 0 for t in [0, 3). v > 0 for t in (3, 5].
b. For [0, 3), displacement =
( – – )t t t dt3 2
0
3
5 8 6 63
4+ = −∫
Distance = 63
4mi
For (3, 5], displacement =
( – – )t t t dt3 2
3
5
5 8 6 242
3+ =∫
Distance = 242
3mi
c. Displacement =
( – – )t t t dt3 2
0
5
5 8 6 1711
12+ =∫ mi
Distance = − + − =∫ | | mit t t dt3 2
0
5
5 8 6 315
12
d. Displacement = − + =63
424
2
317
11
12mi
Distance = − −
+ =6
3
424
2
331
5
12mi
e. a ( t) = v′(t) = 3t2 − 10t + 8a(2.5) = 1.75 (mi/min)/min
5. a t t v( ) , ( ) , on [ , ]/= = −1 2 0 18 0 16
v t t dt t C v C( ) ; ( ) 8/ /= = + = − ⇒ = −∫ 1 2 3 22
30 18 1
v t t( ) /= −2
3183 2
Displacement ft= −
= −∫ 2
318 14
14
153 2
0
16
t dt/
Distance ft/= − =∫ 2
318 179
7
153 2
0
16
t dt
6. a t t v( ) , ( ) , on [ . , . ]= =−1 1 0 0 4 1 6
v t t dt t C t v C( ) ( ); ( )= = + > = ⇒ =−∫ 1 0 1 0 0ln
v t t( ) = ln
Displacement .= = − …∫ ln.
.
t dt 0 08140 4
1 6
≈ −0 081. cm
Distance | | .= = …∫ ln.
.
t dt 0 38540 4
1 6
≈ 0 385. cm
7. a t t v( ) , ( ) , on [ , ]= = −6 0 9 0sin πv t t dt t C v( ) ; ( )= = − +∫ 6 6 0sin cos
= − ⇒ = −9 3Cv(t) = −6 cos t − 3
Displacement = =∫ (– cos – )6 30
t dtπ
−9.4247… ≈ −9.42 km (exact: −3π km)
Distance | | .= − − = …∫ 6 3 13 53380
cos t dtπ
≈ 13.53 km (exact: 6 3 + π )
8. a(t) = sinh t, v(0) = −2, on [0, 5]
v t t dt t C( ) = = +∫ sinh cosh
v(0) = −2 ⇒ C = −3v(t) = cosh t − 3
Displacement = =∫ (cosh – )t dt30
5
59.2032… ≈ 59.20 mi (exact: sinh 5 − 15)
Distance | |
. . mi
= −
= ≈∫ cosh t dt3
64 1230 64 120
5
K
9. a. v = t1/2 − 2 = 0 ⇔ t = 4 s;v < 0 if t < 4, v > 0 if t > 4
b. Displacement ft= =∫ ( – )/t dt1 2
1
9
2 11
3
c. Distance | ft= − =∫ | /t dt1 2
1
9
2 4
268 Problem Set 10-2 Calculus Solutions Manual© 2005 Key Curriculum Press
10. a. v t t= = = … − −sin 2 01
20
1
2at , , , , ,π π π
π π, , 3
2…
sin ,2 0 01
2t ≥
on , soπ
Distance cm= =∫ sin/
2 10
2
t dtπ
b. Displacement cm= =∫ sin.
2 10
4 5
t dtπ
Distance cm= =∫ sin.
2 90
4 5
t dtπ
Or: The regions where the graph is belowthe x-axis cancel out the regions where thegraph is above the axis, leaving only oneuncancelled region above the graph, soDisplacement = area from part a = 1 cm. Theabsolute values of the regions above andbelow the graph are the same, so Distance =9 times the area from part a = 9 cm.
11. a. v = 60 − 2t
Displacement ft= =∫ ( – )60 2 30010
40
t dt
b. Distance | | ft= − =∫ 60 2 50010
40
t dt
12. a. a tt t
t( )
in =
>
40 0 015 9 8 0 100
9 8 100
cos . – . , [ , ]
– . ,
For t in [0, 100],
v t t dt
t t C
( ) ( . . )
. .
= −
= − +
∫ 40 0 015 9 8
40
0 0150 015 9 8
cos
.sin
v(0) = 0 ⇒ C = 0
For , ( ) . .t v t dt t C> = − = − +∫100 9 8 9 8
v( ) . .10040
0 0151 5 980 1679 986= − = …
.sin
⇒ C = 1679.986… + 980 = 2659.986…
v t t t t
t t( ) .
sin . . , [ ,
. . ,= −
− + >
40
0 0150 015 9 8 0
9 8 2659 986 100
in 100]
K
3030
100
t
a(t)
100
1000
t
v(t)
b. a t= = =−01
0 015
9 8
401at
.cos
.
88 2184 88 2. . sK ≈v = 0 at t = 2,659.986…/9.8 = 271.4272… ≈271.4 s
c. Displacement ( )= ∫ v t dt0
300
=
∫ 40
0 0150 015 9 8
0
100
.sin . – .t t dt
+ +∫ (– . . )9 8 2659 986
300
t dtK100
≈ 116202.27… + 139997.32… ≈ 256,200 m
Distance = 116202.27… +
| . . |− +∫ 9 8 2659 986
100
300
t dtK
= 116202.27… + 147998.09… ≈ 264,200 mThe distance is greater than the displacement,which agrees with the fact that the velocitybecomes negative at t = 271.4… s.
d. v(300) = −9.8(300) + 2,659.986… =−280.0133… , so the rocket is movingdownward (falling) at about 280 m/s.
13. a.tend
saav
(mi/h)/svend
mi/hvav
mi/hsend
mi
0 — 0 — 0
5 2.95 14.75 7.375 0.0102…
1 0 3.8 33.75 24.25 0.0439…
1 5 1.75 42.5 38.125 0.0968…
2 0 0.3 4 4 43.25 0.1569…
2 5 0 4 4 4 4 0.2180…
3 0 0 4 4 4 4 0.2791…
3 5 0 4 4 4 4 0.3402…
4 0 −0.2 4 3 43.5 0.4006…
4 5 −0.9 38.5 40.75 0.4572…
5 0 −2.6 25.5 3 2 0.5017…
5 5 −3.5 8 16.75 0.525
6 0 −1.6 0 4 0.5305…
b. At t = 60, vend = 0, ∴ the train is at rest.
c. The train is just starting at t = 0; itsacceleration must be greater than zero to get itmoving, even though it is stopped at t = 0.Acceleration and velocity are differentquantities; the velocity can be zero butchanging, which means the acceleration isnonzero.
d. Zero acceleration means the velocity isconstant, but not necessarily zero.
e. The last entry in the last column is thedisplacement at time t = 60. Thus, it is0.5305… ≈ 0.53 mi between stations.
Calculus Solutions Manual Problem Set 10-2 269© 2005 Key Curriculum Press
14. a.
tend
saav
(mi/h)/svend
mi/hvav
mi/hsend
mi
0 — 6000 — 400
10 8.5 6085 6042.5 416.7847…
20 22 6305 6195 433.9930…
30 33 6635 6470 451.9652…
40 39.5 7030 6832.5 470.9444…
50 42.5 7455 7242.5 491.0625…
60 53 7985 7720 512.5069…
70 71 8695 8340 535.6736…
80 83.5 9530 9112.5 560.9861…
90 47.5 10005 9767.5 588.1180…
100 3 10035 10020 615.9513…
b. According to these calculations, the spaceshipis only about 620 mi from the launchpad andmoving at only about 10,000 mi/h. So thespecifications are definitely not met, and theproject should be sent back to the drawingboard.
15. a. adv
dtv a dt at C= ⇒ = = +∫ ;
v = v0 when t = 0 ⇒ C = v0 ⇒ v = v0 + at
b. vds
dts v dt v at dt= ⇒ = = + =∫ ∫ ( )0
v t at C021
2+ +
s = s0 when t = 0 ⇒ C = s0 ⇒
s v t at s= + +02
01
2
16. Use s(t) for displacement. Assume v(0) =s(0) = 0.
a. a tt
t( )
,
,=
≤ <≥
2 0 6
0 6
if
if
v tt t
t( )
,
,=
≤ <>
2 0 6
12 6
if
if
s tt t
t t( )
,
– ,=
≤ <>
2 0 6
12 36 6
if
if
−36 comes from the initial condition,s(6) = 36.
t
a(t)
2
0 6 10
10
0 6 10
t
v(t)
0 6 10
50
t
s(t)
b. The acceleration suddenly jumps from 0 to 2at t = 0 and drops back to 0 at t = 6. (Thevelocity graph has cusps in both places.)
c. a t t( ) = −2 23
cosπ
lim lim – cos
cos
t ta t t
→ →+ +=
= − =0 0
2 23
2 2 0 0
( )π
lim lim – cos
cos
– –t ta t t
→ →=
= − =6 6
2 23
2 2 2 0
( )π
π
Because a(t) is continuous at t = 0 and 6,there are no sudden changes in acceleration.
d. a t t t
t( ) – cos ,
,= ≤ ≤
≥
2 2
30 6
0 6
πif
if
v t t t t
t( ) – sin ,
,= ≤ ≤
≥
2
6
30 6
12 6π
πif
if
e.
10
0 6 10
t
v(t)
There are no step discontinuities in a(t), andthus the graph of v(t) is smooth.
f. 26
3
18
30
62
20
6
t t dt t t−
= +
∫ π
ππ
πsin cos
= + − − =3618
018
362 2π πElevator goes 36 ft.
270 Problem Set 10-3 Calculus Solutions Manual© 2005 Key Curriculum Press
g. The elevator will take another 36 ft to slowdown and stop. So the deceleration shouldstart where the elevator is 564 ft up, aboutthe 47th floor (from part h, one floor = 12 ft).
h. The elevator takes a total of 12 s to accelerateand decelerate. During these intervals ittravels a total of 72 ft, leaving 528 ft for theconstant velocity portion. At 12 ft/s, this partof the trip will take 44 s. Thus, the total triptakes 56 s.
i. The elevator must start to decelerate halfwaythrough the trip, where s(t) = 6 ft. Solving
26
36
0t t dt
b
– sinπ
π
=∫
numerically for b gives b ≈ 3.1043… ≈3.1 s.a(3.1043…) = 3.9880… ≈ 4.0 ft/s2
By symmetry, the deceleration process muststart at this time, meaning the accelerationjumps to −3.9880… ft/s2. The graph lookslike this:
6.2
5
0
t
a(t)
Thus, the passengers get a large jerk at themidpoint of the trip.One way to remedy the problem is to reducethe acceleration so that the elevator goes only6 ft instead of 36 ft in the first 6 seconds.That is,
a t t( ) = −1
3
1
3 3cos
π
You may think of other ways.
Problem Set 10-3
Q1. 50 mi/h Q2. 30 mi
Q3. 20 min Q4. 2πQ5. No local maximum Q6. 1.5
Q7. f (x) = 16 (at x = 1) Q8. infinite
Q9. Mean value theorem Q10. D
1. a. y x x dxav = + = =∫1
45
1
4164 413
1
5
( – ) ( )
b. The rectangle has the same area as the shadedregion.
50
1 5
y = 41
x
f(x)
c. 41 = c3 − c + 5c = 3.4028… , which is in [1, 5].
2. a. y x x dxav = + =∫1
87 4
1
61 2
1
9
( – )/
b. The rectangle has the same area as the shadedregion.
y = 4.1666...5
1 9
x
f(x)
c. 41
671 2= − +c c/
c = 5.0892… , which is in [1, 9].
3. a. y x dxav . .= = …∫1
63 0 2 2 0252
1
7
sin
b. The rectangle has the same area as the shadedregion.
1 7
3
x
g(x)
y = 2.0252...
c. 2.0252… = 3 sin 0.2cc = 3.7053… , which is in [1, 7].
4. a. y x dxav .= = …∫1
12 5181
0 5
1 5
tan.
.
b. The rectangle has the same area as the shadedregion.
0.5 1.5
10
x
h(x)
y = 2.5181...
Calculus Solutions Manual Problem Set 10-3 271© 2005 Key Curriculum Press
c. 2.5181… = tan cc = 1.1927… , which is in [0.5, 1.5].
5. a. y t dtav = =∫1
82
1
61
9
b. The rectangle has the same area as the shadedregion.
1 9
t
v(t)
3y = 2.1666...
c. 21
6= c
c = 425
36, which is in [1, 9].
6. a. y e dt etav ( )= − = +−∫1
3100 1
100
32 3
0
3
( )–
= 68.3262…
b. The rectangle has the same area as the shadedregion.
t
v(t)
y = 68.32...
100
0 3
c. 68.3262… = 100(1 − e− c)c = 1.1496… , which is in [0, 3].
7. yk
ax dx akk
av = =∫1 1
32 2
0
8. yk
ax dx akk
av = =∫1 1
43 3
0
9. yk
ae dxk
a ex kk
av ( )= = −∫1 11
0
10. yk
x dxk
kk
av | |= =∫1 10
tan ln sec
11. a(t) = 6t− 1/2
v(t) = 12t1/2 + C; v(0) = 60 ⇒ C = 60
v(t) = 12t1/2 + 60
s(t) = 8t3/2 + 60t + s0
v(25) = 120 ft/s
Displacement = s(25) − s(0) = 2500 ft
vav = 2500/25 = 100 ft/s
12. The general equation of a parabola with vertex(h, k) is v − k = a(t − h)2. Vertex is at(t, v) = (2, 50), sov − 50 = a(t − 2)2
. v = 30 when t = 0, so−20 = a(−2)2 ⇒ a = −5.
v = 50 − 5(t − 2)2
v t dtav mi/h= =∫1
450 5 2 43
1
32
0
4
[ – ( – ) ]
This is just 131
3mi/h above the speed limit.
If Ida wins her appeal, her fine will be
7 131
393
1
393 33⋅ = ≈$ $ . , which is $46.67 less
than what she now faces.
13. Consider an object with constant acceleration a,for a time interval [t0, t1].
v t a dt at C( ) = = +∫At t = t0, v(t) = v0 ⇒ v0 = at0 + C ⇒C = v0 − at0.∴ v(t) = at + v0 − at0 = v0 + a(t − t0)
vv a t t dt
t tt
t
av =+∫ [ ( – )]
–
0 0
1 0
0
1
= +
1 1
2
1
21 00 1 1 0
20 0 0 0
2
t tv t a t t v t a t t
–( – ) – – ( – )
= + −v a t t0 ( )1
2 1 0
The average of v0 and v1 is1
2
1
21
2
0 1 0 0 1 0
0 1 0
( ) [ ( – )]v v v v a t t
v a t t
+ = + +
= + −( )
∴ vav = the average of v0 and v1, Q.E.D.
14. Counterexample: In Problem 11, the car’sacceleration is a t= 6/ .The initial velocity is
v(0) = 60 ft/s; the final velocity after 25 secondsis v(25) = 120 ft/s; and the average velocity isvav = 100 ft/s. But the average of the initial and
final velocities is 1
20 25 90[ ( ) ( )]v v v+ = ≠ft/s .av
15. a. Integral = area = 12(100 + 70)/2 + 6(40) +12(40 + 10)/2 = 1560yav = 1560/30 = 52, or $52,000Cost of inventory = 0.50(52000)/100 =$260.00
b. At x = 12, they may have had a single, largesale, dropping the inventory from $70,000 to$40,000. There is no day on which the inventoryis worth $52,000.
10 20 30
50
100
y (thousand dollars)
x (days)
No x where y = 52
272 Problem Set 10-3 Calculus Solutions Manual© 2005 Key Curriculum Press
16.
–5
–10
5 10 15 20 25 30x (ft)
yWater surface
(ft)
y = – 6.5142av
Integral = −(area of 4 rectangles, 2 trapezoids,and 2 quarter-circles)2(−8) + 8(−10) + 7(−3) + 1(−2) +7[−10 + (−5)]/2 + 5[−5 + (−3)]/2 − π(22) −π(1)2/4 = −195.4269…yav = −195.4269…/30 = −6.5142… , or about6.51 feet deep.The volume would equal 6.5142… times the areaof the horizontal cross section times the numberof gallons in a cubic foot.
17. Integral ≈ 3(16/2 + 15 + 15 + 17/2) +2(17 + 20)/2 + 1(20 + 14)/2 + 3(14/2 + 10 +9 + 8 + 9/2) = 139.5 + 37 + 17 + 115.5 = 309yav = 309/24 = 12.875 ≈ 12.9°C
The average of the high and low temperatures is(20 + 8)/2 = 14°C, which is higher than theactual average. Averaging high and lowtemperatures is easier than finding the average bycalculus, but the latter is more realistic for suchapplications as determining heating and airconditioning needs.
18. a. At x = 3, y = 81.3139… ≈ 81.3 mg.
y e dxxav
. mg
= =
= … ≈
−∫1
3200
1
3395 6202
131 8734 131 9
0 3
0
3. ( . )
.
K
b. k = 81.3139… , so the equation isy = 281.3139…e− 0.3(x− 3 ) .
y e dxx
av = +
∫1
6395 6202 281 3139 0 3 3
3
6
. . – . ( – )K K
= + =1
6395 6202 556 4674 158 6812( . . )K K K.
≈ 158 7. mg
c. As the graph shows, there are two times in[0, 6] at which there are 158.7 mg. So theconclusion of the mean value theorem is true,in spite of the discontinuity.
1 2 3 4 5 6 7
100
200
300
y (mg)
x (h)Two times
y = 158.68...av
19. v = A sin 120π t and y = | A sin 120π t |
y A t dtav | |= ∫1
1 60120
0
1 60
/sin
/
π
= −∫ ∫60 120 60 1200
1120
1120
1 60
A t dt A t dtsin sin/
/
/
π π
= − +At
At
2120
2120
1120
1 60
ππ
ππcos cos
/
/
0
1/120
= + + =A A
20 2
2
ππ π π
π(– cos cos cos – cos )
If , thenavyA
A= = ⇒ =1102
110 55π
π
= 172.78… V.The average value of one arc of
y x x dx=−
=∫sin sin , is and1
0
20π π
π
y = sin x has a maximum value of 1. Ahorizontal stretch does not affect the averagevalue. Write a proportion to find the maximumof a sinusoidal curve with an average value
of 110. 2
1
110/,
π =m
so m = 55π.
20. a. d = k sin x
d k x dx
kx x
k k
av2 2 2
0
2
2
0
2
2 2
1
2
2
1
2
1
42
20 0 0
2
=
=
= + =
∫π
π
ππ
π
π
sin
– sin
( – – )
∴ = = rms / .k k2 0 7071K
b. cos sin sin cos2 1 21
2
1
222 2x x x x= − ⇒ = −
Thus, sin2 x is a sinusoid.
π 2π
1x
y
c. By symmetry across the line y = 1
2, the
average of y x= −1
2
1
22cos (and hence
y = sin2 x) over [0, 2π] is1
2. Thus, the
average of is .y k x k= 2 2 21
2sin
∴ = rms k/ ,2 as in part a.
Calculus Solutions Manual Problem Set 10-4 273© 2005 Key Curriculum Press
d. By symmetry, it suffices to find the averageand rms for one arch of the graph, that is,over [0, π].
y x dxav | |= ∫10π
πsin
= ≥∫10 0
0ππ
πsin sinx dx x (because in [ , ])
= − =1 2
0π π
π
cos x
d x dxav .2 2
0
12 0 094715= ≈∫π
ππ
(|sin | – / ) K
∴ rms ≈ 0.094715…1/2 = 0.3077…
The maximum distance between high and lowpoints for this curve is 1; a sinusoidal curvewith maximum distance 1 between high and
low points has equation y x= 1
2sin , with
rms /= =2 4 0 3535. K(using part a). This
number is greater than the rms for |sin x|,so |sin x| is smoother.
Problem Set 10-4Q1. x = 81 Q2. y′ = −x(100 − x2)− 1/ 2
Q3. − +1
3100 2 3 2( – ) /x C Q4. y′ = 3 ⋅ (1 − 9x2)− 1/ 2
Q5.1
2
1
42 2xe e Cx x− + Q6. y′ = sech2 x
Q7. 1.5 Q8. t = 1 and t = 4
Q9. t = 4 Q10. A
1.
x
50
100
100 – x
T x x= + +1
250
1
51002 2 ( – )
The graph shows a minimum T at x ≈ 22 m.
100
100
x
T
Algebraic solution:
′ = + ⋅ −−T x x1
450 2
1
52 2 1 2( ) /
′ = ⇔ + =−T x x01
250
1
52 2 1 2( ) /
5x = 2(502 + x2)1/2
25x2 = 4 · 502 + 4x2
x = ± = ±100 21 21 8217/ . K
Ann should swim toward a point about 21.8 mdownstream.
2.100
100 – x x
30
T x x = + +1
13100
1
12302 2( – )
The graph shows a minimum T at x ≈ 72.
100
10
x
T
Algebraic solution:
′ = − + + ⋅−T x x1
13
1
2430 22 2 1 2( ) /
′ = ⇔ = + −T x x01
13
1
12302 2 1 2( ) /
13x = 12(302 + x2)1/2
169x2 = 144 · 302 + 144x2
x = ±72The diver should swim for 100 − 72 = 28 m,then dive.
3.
x 1000 – x1000
300
C x x= − + +40 1000 50 3002 2( )The graph shows a minimum C at x ≈ 400(exactly x = 400).
100,000C
x
1000
The pipeline should be laid 600 m along the roadfrom the storage tanks, then straight across thefield to meet the well.
4.
120
400
400 – x x
274 Problem Set 10-4 Calculus Solutions Manual© 2005 Key Curriculum Press
W x x= − + +3000 400 4000 1202 2( )The graph shows a minimum W at x ≈ 136(exactly x = =360 7 136 067/ . K).
2,000,000
400
x
W
The walkway should go 400 −136.067…≈ 263.9 m parallel to the street, then crossthe street.
5. a. For minimal path, x = 100 21/ .
∴ =+
= = . / ,sin θ x
x500 4 2 5
2 2 Q.E.D.
b. For minimal path, x = 400.
sin θ =+
= =x
x3000 8 40 50
2 2. / , Q.E.D.
6. Distance swimming = +p x2 2 . Distance
walking = k − x.
Ts
p xw
k x= + +1 12 2 ( – )
Tx
s p x w s w′ =
+− = −
2 2
1 1 1sin θ
Ts
w′ = ⇔ =0 sin θ , Q.E.D.
7. sin θ = 12
13
x =
=−30
12
13721tan sin
The diver should swim 100 − 72 = 28 m, thendive. The algebraic solution is easier than beforebecause no algebraic calculus needs to be done.Mathematicians find general solutions to gaininsight, and to find patterns and methods to alloweasier solution of similar problems.
8. sin θ =+
= =x
x120
3000
4000
3
42 2
16x2 = 9(1202 + x2)7 9 120 360 7 136 0672 2x x= ⋅ ⇔ = ± = …/ .
The walkway should go 400 − 136.067… ≈263.9 m parallel to the street, then cross thestreet.The algebraic solution is easier than beforebecause no algebraic calculus needs to be done.Mathematicians find general solutions to gaininsight, and to find patterns and methods to alloweasier solution of similar problems.
9.
x C(x), approximate
300 49,213
390 49,002
400 49,000
410 49,002
500 49,155
The table shows that a near miss will havevirtually no effect on the minimal cost. Forinstance, missing the optimal value of x by10 m will make about a $2 difference in cost,and missing by 100 m makes only a $150 to$200 difference.
10. T x x x( ) = + +1
5500
1
312002 2( – )
The graph shows a local minimum at x ≈ 900 ft(exact: 900 ft), which is out of the domain.
500
500 900
x
T(x)
The minimum occurs at an endpoint of thedomain. Because Calvin can walk entirely alongpavement when x = 0, there is a removablediscontinuity in the above function andT(0) = 100 + 240 = 340 s. Because T(500) =433.333… , which is greater than 340, theminimum time is at x = 0. Calvin’s time isminimized by staying on the sidewalks. If roadconstruction (for instance) prevented Calvin fromwalking on Heights Street, his time would beminimized by walking directly to Phoebe’shouse.
11.
x300 – x
70
120
T x x= + + +1
50120
1
13070 3002 2 2 2( – )
The graph shows a minimum T at x ≈ 48 yd.
300
5
x
T
Calculus Solutions Manual Problem Set 10-5 275© 2005 Key Curriculum Press
Algebraically:
Tx
x
x
x′ =
+−
+50 120
300
130 70 3002 2 2 2
–
( – )
Setting T ′ = 0 and simplifying leads to afourth-degree equation, which must be solvednumerically. Minimum is at x = 47.8809…≈ 47.9 yd.
12.
x300 – x
70
120θ
θ
1
2
θ
θ1
2
From Problem 11,
′ =+
−+
Tx
x
x
x50 120
300
130 70 3002 2 2 2
–
( – )By trigonometry,
sin θ1 2 2120=
+
x
x,
sin–
( – )θ2 2 2
300
70 300=
+
x
x
∴ ′ = − T1
50
1
1301 2sin sinθ θ
For minimal path, T ′ = 0. Thus,
1
50
1
1301 2sin sinθ θ=
sin
sin
θθ
1
2
50
130= , Q.E.D.
13. Tv
a xv
b k x= + + +1 1
1
2 2
2
2 2( – )
′ =+
−+
Tx
v a x
k x
v b k x12 2
22 2
–
( – )
sin θ1 2 2=
+
x
a x,
sin–
( – )θ2 2 2
=+
k x
b k x
∴ ′ = − Tv v
1 1
11
22sin sinθ θ
For minimal path, T ′ = 0. Thus,
1 1
11
22v v
sin sinθ θ=
sin
sin
θθ
1
2
1
2
= v
v, Q.E.D.
14. a. The light rays take the minimal time to getfrom one point to another, just as RobinsonCrusoe wanted to take the minimal time toget from hut to wreck.
b. Light always takes the path requiring the leasttime between two points.
c. When you look at the object, your mind tellsyou that the light rays go straight. Actually,they are bent, as shown in the diagram. Sothe object is deeper than it appears to be.Because θwater < θair, vwater < vair.
Apparentdepth
Actualdepth
θAir
θ Water
Actual path of light rays
Apparent path of light rays
15. Answers will vary.
Problem Set 10-5Q1. f ′(x) = sin x + x cos x Q2. g″(x) = x− 1
Q3. xex − ex + C Q4. Snell’s law
Q5. x
Q6. Q7.
x
y
1
1
Q8. Total distance
Q9. Newton and Leibniz Q10. C
1. D tt
D t= + ′ = − −11 2
The graphs show zero derivative and localminimum of D at t = 1, and maximum of D att = 3.
D
D'
3
3
t
D or D'
D t t′ = ⇔ = ⇔ = ±0 1 12 , confirming the graph.Minimum is D(1) = 2, or 2000 mi.
Maximum is D( ) ,3 31
3= or about 3333 mi.
2. Fuel cost per mile = k · v2.At v = 30, cost = 0.18.
0 18 301
50002. = ⋅ ⇒ =k k
Driver cost is 20 20100 2000
tv v
= ⋅ = .
∴ = + ⋅ = + Cv
v
v
v2000
5000100
2000
50
2 2
Cv
v′ = − +2000
252
276 Problem Set 10-5 Calculus Solutions Manual© 2005 Key Curriculum Press
The graphs show minimum C at v ≈ 37 mi/h;C′ is multiplied by 10 so that it is easier to seeits behavior around C′ = 0.C v′ = = = …0 10 50 36 83 at . 403
C or C' times 100
C
C' times 1037
100
85 100
v
3. Maximize f(x) = x − x2.f ′(x) = 1 − 2x; f ′(x) = 0 at x = 0.5;f ″(x) = −2, so graph is concave downeverywhere.Maximum of f (x) is at x = 0.5.
4. Maximize f (x) = x − x2 for x ≥ 2.The graph shows maximum at endpoint x = 2.
2
1 xf(x)
Because f (2) is the maximum and it is negative,there is no number greater than 2 that exceeds itssquare.
5. a. St
tF
tG
t
t t=
+=
+=
+ +100
1
9
9
900
1 9; ;
( )( )
The graph shows a maximum of G at t = 3hours.
S
G
F times 20
10
50
t
y
b. Gt
t t
t
t t=
+ +=
+ +900
1 9
900
10 92( )( )
Gt t t t
t t′ = + + +
+ +900 10 9 900 2 10
10 9
2
2 2
( ) – ( )
( )
=+ +
900 9
10 9
2
2 2
( – )
( )
t
t tG t′ = ⇔ = ±0 3
Because G′ changes from positive to negativeat t = 3, there is a local maximum there, asin the graph.Fran should study for 3 hours.
c. Optimum grade = G(3) = 56.25 ≈ 56 (Notgood!)
i. G(4) = 55.3846… ≈ 55, about 1 pointless.
ii. G(2) = 54.5454… , ≈ 55, about 1 pointless.
6. a. µ = 130 − 12T + 15T 2 − 4T 3, 0 ≤ T ≤ 3d
dTT T T T
µ = − + − = − − −12 30 12 6 2 1 22 ( )( )
d
dTT T
µ = = =0 0 5 2 at . or
µ(0) = 130µ(0.5) = 127.25µ(2) = 134µ(3) = 121
Maximum viscosity occurs at T = 2, or 200°.
b. Minimum viscosity = 121 centipoise atT = 3, or 300°.
c.d
dt
d
dT
dT
dt
µ µ= ⋅
Because , .T tdT
dt t T= = =1
2
1
2
When , . and .TdT
dt
d
dT= = =1 0 5 6
µ
∴ = = . ( )d
dt
µ0 5 6 3
Viscosity is increasing at 3 centipoise/min.
7. a. Put a coordinate system with origin at thecenter of the cone’s base. Pick a sample point(x, y) where the cylinder touches the elementof the cone. Thus, x is the radius of thecylinder and y is its altitude. The volume andsurface area areV = π x 2yA = 2π x 2 + 2π xyThe cone element has equation y = −0.6x + 6.V = π x 2(−0.6x + 6) = π(−0.6x3 + 6x2)A = 2π x2 + 2π x(−0.6x + 6)
= π(0.8x2 + 12x)
10
600y
A
V
x
b. From the graphs, the maximum volumeoccurs where the radius x ≈ 6.7 in. Themaximum area occurs at x = 10, where all ofthe area is in the two bases of the cone.Algebraically,
V ′ = π ( −1.8x2 + 12x)
V x x′ = ⇔ = =0 0 62
3 or
Calculus Solutions Manual Problem Set 10-5 277© 2005 Key Curriculum Press
Maximum V is at x = 62
3 in., as shown on
the graph, and y = 2 in.
A ′ = π (1.6x + 12)
A′ = 0 ⇔ x = −7.5, which is out of thedomain.
A(0) = 0 and A(10) = 200π, so maximum Ais at x = 10 in., as shown on the graph, andy = 0 in.
The maximum volume and maximum area donot occur at the same radius.Note that the radius of the cone is largecompared to its altitude. Thus, the increase inareas of the two bases of the cylinder offsetsthe decrease in its lateral area as x increases,making the maximum area that of thedegenerate cylinder of altitude zero.
8. a. Put a coordinate system with origin at thecenter of the cone’s base. Pick a sample point(x, y) where the cylinder touches the elementof the cone. Thus, x is the radius of thecylinder and y is its altitude.
Know: dy
dt= 2 in./min. Want:
dV
dt.
V = πx2yThe cone element has equation y = −3x + 18,
from which x y= −61
3.
V y y y y y= −
⋅ = − +
π π6
1
336 4
1
9
22 3
dV
dyy y y y= − +
= − −π π36 8
1
3
1
36 182 ( )( )
dV
dt
dV
dy
dy
dty y= ⋅ = − −2
36 18π ( )( )
When ydV
dt= = −12 24, . π
V is decreasing at 24π = 75.3982… ≈75.4 in.3/min.
b. If t ∈ [0, 9], then y ∈ [0, 18].dV
dty y= ⇔ = =0 6 18 or
V(0) = 0; V(6) = 96π; V(18) = 0Maximum V is 96π in.3 at t = 3 min.
c. If t ∈ [4, 6], then y ∈ [8, 12].No critical points for V are in [8, 12].V(8) = 88.8888…π; V(12) = 48πMaximum V is 88.8888…π ≈ 279.3 in.3 att = 4 min.
9. Know: . m /min. Want: .dV
dt
dy
dt= −0 7 3
dV = πx2 dy
∴ = ⇒ =
dV
dtx
dy
dt
dy
dt xπ
π2
2
0 7– .
When the water is 3 m deep, y = 8.
Because y x x= + =4 45 3, .
dy
dt= = − ≈– .0 7
30 1286 0 129
π. . m/minK
10. a. Pick a sample point (x, y) where the cylindertouches the parabola. Thus, the radius of thecylinder is x and its altitude is y.
Know: . . Want: .dx
dt
dV
dt= 0 3
V = π x2y = πx2(4 − x2) = π(4x2 − x4)dV
dtx x
dx
dtx x= − ⋅ = −π π( ) . ( )8 4 1 2 23 3
When . , . .xdV
dt= = −1 5 0 45π
b.dV
dxx x= −4 2 3π ( )
dV
dxx= ⇔ = ±0 0 2 2, is out of domain)(–
V V V( ) ( ) ; ( )0 2 0 2 4= = = πMaximum volume = 4π ≈ 12.6 units3 atradius = 2 units.
11. a. w = 1000 + 15t (lb); p = 0.90 − 0.01t ($/lb)A = (1000 + 15t)(0.90 − 0.01t)
= 900 + 3.5t − 0.15t2 ($)
b.dA
dtt t= − = = =3 5 0 3 0
35
311
2
3. . at days
Maximum A at t = 112
3, not a minimum,
because dA
dtgoes from positive to negative
there.
c. A 112
3920
5
12920 42
= ≈ $ .
12. a. 0 ≤ D ≤ 130 ⇒ 0 ≤ 20x + 10 ≤ 130 ⇒
− ≤ ≤1
26x
0 ≤ W ≤ 310 ⇒ 0 ≤ 10(x2 − 8x + 22) ≤310 ⇒ −1 ≤ x ≤ 9Given x ≥ 1, the domain of x is [1, 6].
b. Minimize/maximize W on x ∈ [1, 6].dW
dxx x= − = =10 2 8 0 4( ) at .
W(1) = 150; W(4) = 60; W(6) = 100Minimum: W = 60 ft (at x = 4 mi)Maximum: W = 150 ft (at x = 1 mi)
c. C = k · D · W= k · (20x + 10) · 10(x2 − 8x + 22)= 100k(2x3 − 15x2 + 36x + 22) (k > 0)dC
dxk x x
k x xdC
dxx x
= − +
=
= ⇔ = =
100 6 30 36
600 2 3
0 2 3
2( )
( – )( – )
or
278 Problem Set 10-6 Calculus Solutions Manual© 2005 Key Curriculum Press
C(1) = 4500k; C(2) = 5000k; C(3) = 4900k;C(6) = 13,000kCheapest bridge at x = 1 mi.
d. No. The shortest bridge at x = 4 mi wouldcost C(4) = 5400k, which is 900k more thanthe cheapest bridge at x = 1.
Problem Set 10-6Q1. −x cos x + sin x + C Q2. 2xe3x + 3x2e3x
Q3.2
2
x
Cln
+ Q4. 53 = 125
Q5.1
474x + Q6.
6 6
3
22
32 2sec
sec lnt
e xxt =
Q7. parametric Q8. x ln x − x + C
Q9. x2 Q10. E
1. The velocity is tangent to the path, theacceleration is toward the concave side of thepath, and there is an obtuse angle betweenacceleration and velocity.
y
x
r
v
a→
→
→
2. The velocity is tangent to the path, theacceleration is toward the concave side of thepath, and there is an acute angle between theacceleration vector and the velocity vector.
y
x
r
v
a
→
→
→
3. a.r r rr e t i e t jt t= +( ) ( )cos sin
r r rv e t e t i e t e t jt t t t= − + +( ) ( )cos sin sin cos
r
Kr r
v i j( )1 0 8186 3 7560= − + …. .
x is decreasing at t = 1 because dx/dt isnegative.
Speed = + =0 8186 3 75602 2. . ...K
3.8442… ≈ 3.84 cm/s
b. L =
( cos sin ) ( sin cos )e t e t e t e t dtt t t t− + +∫ 2 2
0
1
= = = ≈∫ 2 2 1 2 4300 2 432
0
1
e dt et ( – ) . .K
Distance from origin
= +( cos ) ( sin )e e1 2 1 21 1
= e = 2.7812… ≈ 2.78 cm
(Note that the distance traveled is less than thedistance from the origin because the particlestarted at (1, 0), not at (0, 0).)
c. r r ra e t i e t jt t= − +( ) ( )2 2sin cos .
r r ra i j( ) . ... . ...1 4 5747 2 9373= − +( ) ( )
4. a. r r rr t i t j1 2 2= − + −( ) ( ) and2
r r rr t i t j2 1 5 4 1 5 2= − + −( . ) ( . )
r r r r r rv i t j v i j1 21 2 2 1 5 1 5= + − = +( ) and . .
r r r r r ra i j a i j1 20 2 0 0= + = +and
r r r r r rv i j v i j1 21 2 1 1 5 1 5( ) and ( )= − = +. .
r r r r r ra i j a i j1 21 0 2 1 0 0( ) and ( )= + = +
s12 21 1 2 2 2( ) . 4 cm/s and= + ≈
s22 21 1 5 1 5 2 12( ) . cm/s= + ≈. .
b. Distance1
4
= + ≈∫ 1 2 22 2[ ( – )]t dt
6.1257… ≈ 6.13 m
c. The paths cross at (x, y) = (−1, 1) and (2, 4).By tracing on the grapher,
rr1 is at (−1, 1)
when t = 1, but rr2 is not at (−1, 1) until
t = 2.By further tracing, both paths are at (2, 4)when t = 4.So the particles collide only at (x, y) = (2, 4)when t = 4.
5. a.r r rr t t i t j( ) sin cos= +( . ) ( . )10 0 6 4 1 2r r rv t t i t j( ) cos sin= + −( . ) ( . . )6 0 6 4 8 1 2r r ra t t i t j( ) sin cos= − + −( . . ) ( . . )3 6 0 6 5 76 1 2
b.r r rr i j( . ) sin cos0 5 10 0 3 4 0 6= +( . ) ( . )
= … + …2 9552 3 3013. .r ri j
r r rv i j( . ) cos sin0 5 6 0 3 4 8 0 6= + −( . ) ( . . )
= … − …5 7320 2 7102. .r ri j
r r ra i j( . ) sin cos0 5 3 6 0 3 5 76 0 6= − + −( . . ) ( . . )
= − … − …1 0638 4 7539. .r ri j
The graph shows r rr v, , and
ra at t = 0.5.
x
y
t = 7
t = 0.5at
an a
va
v
These vectors make sense because the head ofrr is on the graph,
rv is tangent to the graph,
and ra points to the concave side of the graph.
c. The object is speeding up because the anglebetween
ra and
rv is acute.
Calculus Solutions Manual Problem Set 10-6 279© 2005 Key Curriculum Press
d. | |rv( . ) ( cos . ) (– . sin . )0 5 6 0 3 4 8 0 62 2= +
= 6.3404…
r ra v( . ) ( . ) sin cos0 5 0 5 3 6 0 3 6 0 3⋅ = −( . . )( . )
+ (−5.76 cos 0.6)(−4.8 sin 0.6)= 6.7863… , so the angle is acute.
P
a v
v= ⋅ = …
r r
r( . ) ( . )
| ( . )|
0 5 0 5
0 51 0703.
rr
ra Pv
vt ( . )0 50 5
0 5= ( . )
| ( . )|
= +
Pi j
v
( cos . ) (– . sin . )
| ( . )|
6 0 3 4 8 0 6
0 5
r r
r
= … − …0 9676 0 4575. .r ri j
r r ra a an t( . ) ( . )0 5 0 5 0 5= −( . )
= − … − …2 0314 4 2964. .r ri j
See the graph in part b.
e. The object is speeding up at | ( . )|ra Pt 0 5 =
= 1.0703… ≈ 1.07 (ft/s)/s.
f.r r rr i j( ) sin cos7 10 4 2 4 8 4= +( . ) ( . )
= − … − …8 7157 2 0771. .r ri j
r r r
r rv i j
i j
( ) cos sin7 6 4 2 4 8 8 4
2 9415 4 1020
= + −
= − … − …
( . ) ( . . )
. .r r ra i j( ) sin cos7 3 6 4 2 5 76 8 4= − + −( . . ) ( . . )
= … + …3 1376 2 9911. .r ri j
See the graph in part b.The object is slowing down because theangle between
ra and
rv is obtuse.
(Note that Pa v
v= ⋅ = − …
r r
r( ) ( )
| ( )|
7 7
74 2592. , so
the object is slowing down at 4.2592…≈ 4.26 (ft/s)/s.)
g. r r r r rr i j i j( ) sin cos0 10 0 4 0 0 4= + = +( ) ( )
r r r r rv i j i j( ) cos sin0 6 0 4 8 0 6 0= + − = +( ) ( . )
r r ra i j( ) sin cos0 3 6 0 5 76 0= − + −( . ) ( . )
= −0 5 76r ri j.
r ra v( ) ( )0 0 0 6 5 76 0 0⋅ = + − =( )( ) ( . )( )∴
ra( )0 and
rv( )0 are perpendicular, Q.E.D.
This means the object is neither slowingdown nor speeding up at t = 0.
6. a. r r rr t t i t j( ) cos sin= +( . ) ( . )8 0 8 6 0 4
r r rv t t i t j( ) sin cos= − +( . . ) ( . . )6 4 0 8 2 4 0 4
r r ra t t i t j( ) cos sin= − + −( . . ) ( . . )5 12 0 8 0 96 0 4
b. r r rr i j( ) cos sin1 8 0 8 6 0 4= +( . ) ( . )
= … + …5 5736 2 3365. .r ri j
r r rv i j( ) sin cos1 6 4 0 8 2 4 0 4= − +( . . ) ( . . )
= − … + …4 5910 2 2105. .r ri j
r r ra i j( ) cos sin1 5 12 0 8 0 96 0 4= − + −( . . ) ( . . )
= − … − …3 5671 0 3738. .r ri j
The graph shows r rr v, , and
ra at t = 1.
x
y
t = 10.5
t = 1
r
v
a
v
a
r
at
an
t = 1.25πa
These vectors make sense because the head of rr is on the graph,
rv is tangent to the graph,
and ra points to the concave side of the graph.
c. The object is speeding up because the anglebetween
ra and
rv is acute.
d. | |rv( ) (– . sin . ) ( . cos . )1 6 4 0 8 2 4 0 42 2= +
= 5.0955…r ra v( ) ( ) cos sin1 1 5 12 0 8 6 4 0 8⋅ = − −( . . )( . . )
+ (−0.96 sin 0.4)(2.4 cos 0.4)= 15.5506… , so the angle is acute.
Pa v
v= ⋅ = …
r r
r( ) ( )
| ( )|
1 1
13 0518.
rr
ra Pv
vt ( )11
1= ( )
| ( )|
= +P
i j
v
(– . sin . ) ( . cos . )
| ( )|
6 4 0 8 2 4 0 4
1
r r
r
= − … + …2 7496 1 3239. .r ri j
r r r
r ra a a
i j
n t( ) ( )
. .
1 1 1
0 8174 1 6977
= −
= − … − …
( )
See the graph in part b, showing rat and
ran
at t = 1.
e. The object is speeding up at
| ( )| ( ) . . (ft/s)s.ra Pt 1 1 3 0518 3 05= = … ≈
f. r r rr i j( . ) cos sin10 5 8 8 4 6 4 2= +( . ) ( . )
= − … − …4 1543 5 2294. .r ri j
r r r
r rv i j
i j
( . )10 5
5 4694 1 1766
= − +
= − … − …
( 6.4 sin 8.4) (2.4 cos 4.2)
. .
r r r
r ra i j
i j
( . ) cos sin10 5 5 12 8 4 0 96 4 2
2 6587 0 8367
= − + −
= … + …
( . . ) ( . . )
. .See the graph in part b, showing
r rr v, , and
ra
at t = 10.5.
The object is slowing down at t = 10.5because the angle between
ra and
rv is obtuse
at that time. (Note that
Pa v
v= ⋅ = − …
r r
r( . ) ( . )
| ( . )|
10 5 10 5
10 52 7552. ,
so the object is slowing down at about2.8 (ft/s)/s.)
280 Problem Set 10-6 Calculus Solutions Manual© 2005 Key Curriculum Press
g. The object is stopped when
r r r rv t t i t j( ) sin cos= − + =( . . ) ( . . )6 4 0 8 2 4 0 4 0⇔ −6.4 sin 0.8t = 0 and 2.4 cos 0.4t = 0.Using −6.4 sin 0.8t = −1.28 sin 0.4tcos 0.4t, you see that
r rv t( ) = 0 exactly when
cos 0.4t = 0; the first time this happens is att = 1.25π s.
r r rr i j( . ) cos sin1 25 8 6 0 5π π π= +( ) ( . )
= − +8 6r ri j
r r r
r ra i j
i j
( . ) cos sin1 25 5 12 0 96 0 5
5 12 0 96
π π π= − + −
= −
( . ) ( . . )
. .See the graph in part b, showing
ra at
t = 1.25π.The acceleration vector points along the pathat t = 1.25π. So the object is stopped, but ithas a nonzero acceleration. At first glance,this fact may be surprising to you!
7. a.r r rr t t i t j( ) cos sin=
+
10
66
6
π π
r r rv t t i t j( ) sin cos= −
+
10
6 6
6
6 6
π π π π
r r rr t v t t t i( ) ( ) cos sin+ = −
10
6
10
6 6
π π π
+ +
6
6
6
6 6sin cos
π π πt t j
r
The graph shows the path of r rr v+ .
x
y
0
1
23
4
5
6
7
89
10
11
12
b. See the graph in part a, showing vectorsrv .
c. For r rr v+ ,
xt t
10 6 6 6= −cos sin
π π π
yt t
6 6 6 6= +sin cos
π π π
xt t t
10 6 3 6 6
22
= −cos cos sin
π π π π
+
π π6 6
22sin t
yt t t
6 6 3 6 6
22
= +sin sin cos
π π π π
+
π π6 6
2
cos2 t
∴
+
= +
x yt t
10 6 6 6
2 22 2cos sin
π π
+
+
π π π π6 6 6 6
22 2sin cost t
2
x y
10 6 6
2
+
= +
2 2
1π
x y2
2
2
2100 1 36 36 1 36( / ) ( / )++
+=
π π1
This is the equation of an ellipse centered atthe origin with x-radius 11.2878… andy-radius 6.7727… .
d.
r r ra t t i t j( ) cos sin= −
+ −
10
36 6
6
36 6
2 2π π π π
r rr t a t( ) ( )+
= −
+ −
1010
36 66
6
36 6
2 2π π π πcos sint i t j
r r
See the graph in part a, showing an ellipticalpath followed by the heads of the accelerationvectors.
e. The direction of each acceleration vector is theopposite of the corresponding position vectorand is thus directed toward the origin.
Note that r ra t r t( ) ( )= − π 2
36.
8. a.r r rr t t t i t t j( ) cos sin= +( . ) ( . )0 5 0 5r rv t t t t i( ) cos sin= −( . . )0 5 0 5
+ +( . . )0 5 0 5sin cost t t jr
r ra t t t t i( ) sin cos= − −( . )0 5
+ −( . )cos sint t t j0 5r
b.r r rr i j( . ) cos sin8 5 4 25 8 5 4 25 8 5= +. . . .
= − … + …2 5585 3 3935. .r ri j
r rv i( . ) cos sin8 5 0 5 8 5 4 25 8 5= −( . . . . )
+ +( . . . . )0 5 8 5 4 25 8 5sin cosrj
= − … − …3 6945 2 1593. .r ri j
r ra i( . ) sin cos8 5 8 5 4 25 8 5= − −( . . . )
+ −( . . . )cos sin8 5 4 25 8 5rj
= … − …1 7600 3 9955. .r ri j
r r rr i j( )12 = +6 cos12 6 sin 12
= … − …5 0631 3 2194. .r ri j
r rv i( ) cos sin12 0 5 12 6 12= −( . )
+ +( . )0 5 12 6 12sin cosrj
= … + …3 6413 4 7948. .r ri j
r ra i( ) sin cos12 12 6 12= − −( )
+ −( )cos sin12 6 12rj
= − … + …4 5265 4 0632. .r ri j
Calculus Solutions Manual Problem Set 10-6 281© 2005 Key Curriculum Press
The graph shows that rr ( . )8 5 and
rr ( )12 really
do terminate on the path.
x
y
t = 8.5
t = 12at
va
r
r
a
v
c. See the graph in part b, showing rv( . ),8 5
r rv a( ), ( . ),12 8 5 and
ra( )12 . The velocity
vectors point along the path as it spiralsoutward, and the acceleration vectors pointinward to the concave side of the graph.
d. In both cases, the angle between ra and
rv
appears to be acute. Check using dotproducts.
r ra v( . ) ( . )8 5 8 5 2 125⋅ = . , which is positive.
r ra v( ) ( )12 12 3⋅ = , which is also positive.Thus, the angles are acute, and the object isspeeding up at both times.
e. At h, .t a v= ⋅ =12 12 12 3r r( ) ( )
| | .rv( ) .12 36 25 6 0207= = …r
r r
r
r
r
r r
r r
aa v
v
v
v
i j
i j
t ( )
. .
1212 12
12
12
12
3
36 253 6413 4 7948
0 3013 0 3968
= ⋅ ⋅
= +
= … + …
( ) ( )
| ( )|
( )
| ( )|
.( . ... . ... )
r r r
r ra a a
i j
n t( ) ( )
. .
12 12 12
4 8279 3 6664
= −
= − … + …
( )
See the graph in part b, showing ran and r
at att = 12.
f. Speed | | .= = = …rv( ) .12 36 25 6 0207
≈ 6.02 mi/h
Speed is increasing by P( )123
36 25= =
.0.4982… ≈ 0.498 (mi/h)/h.
g. See the graph in part b, showing r rr t a t( ) ( ).+
The heads seem to lie on a unit circle.Algebraic verification:r r r rr t a t t i t j( ) ( ) sin cos+ = − + , which is acircle.
9. a. r r r r rr x xi yj xi x j( ) = + = + 2
r r r r rv x
dx
dti
dy
dtj
dx
dti x
dx
dtj( ) = + = + 2
b.
dx
dtv x i xj= − ⇒ = − −3 3 6r r r( )
r r rv i j( )2 3 12= − −
At , speed | |x v= = = =2 2 153r( )
12.3693… ≈ 12.4 cm/s.
c. The graph shows rr ( )2 and
rv( )2 .
This is reasonable because rv( )2 points along
the curve to the left, indicating that x isdecreasing.
x = 2
at
a
an
v
5
20
x
y
d. From part b, r r rv x i xj( ) ,= − −3 6
∴ = − =
r r r ra x i
dx
dtj j( ) 0 6 18
r ra j( )2 18= . See the graph in part c.
e.r
r r
r
r
raa v
v
v
vt ( )22 2
2
2
2= ⋅ ⋅( ) ( )
| ( )|
( )
| ( )|
= – (– – )216 3 12
153
r ri j
= + = … + …72
17
288
174 2352 16 9411
r r r ri j i j. .
r r ra a an t( ) ( )2 2 2= −( )
= − + = − … + …72
17
18
174 2352 1 0588
r r r ri j i j. .
rat ( )2 is parallel to the curve.
ran ( )2 is
normal to the curve and points inward to theconcave side.
f. The object is slowing down when x = 2because the angle between
ra( )2 and
rv( )2 is
obtuse, as shown by the graph and by the factthat the dot product is negative. Also,
rat ( )2
points in the opposite direction of rv( )2 .
g. dL dy dx dx x dx= + = +1 1 42 2( / )dL
dtx
dx
dt= + =1 4 52
At ,xdx
dt
dx
dt= = + =2 5 1 4 2 172( )
⇒ = = … ≈dx
dt
5
171 2126 1 21. . cm/s.
10. r r rr i j( )1 8 8615 4 8410= … + …. .
r r rq i j( )2 2 9659 4 3406= − … + …. .
r r rq i j( . )1 5 0 2065 6 6362= − … + …. .
r r rq i j( . )1 1 2 5579 7 4880= … + …. .
r rv t t t t t i( ) cos cos sin sin= −( . . )12 0 5 6 0 5
+ +( . . )12 0 5 6 0 5cos sin sin cost t t t jr
r r rv i j( )1 3 2693 7 5391= … + …. .
282 Problem Set 10-6 Calculus Solutions Manual© 2005 Key Curriculum Press
The graph shows average velocity vectorsapproaching the instantaneous velocity vector
rv( )1 as t approaches 1. The instantaneousvelocity vector is tangent to the graph and pointsin the direction of motion.
10
10
x
y
vt = 1.1t = 1.5
t = 2
11. r r rr t t i t j( ) sin cos= +( . ) ( . )10 0 6 4 1 2
dL dx dt dy dt dt
t t dt
= +
= +
( / ) ( / )
( cos . ) (– . sin . )
2 2
2 26 0 6 4 8 1 2
L dL= ≈ …∫ 12 0858. ft (numerically)0
2
12.r r rr t t i t j( ) cos sin=
+
10
66
6
π π
dL dx dt dy dt dt
t t dt
= +
= ⋅
+ ⋅
( / ) ( / )
–sin cos
2 2
2 210
6 6
6
6 6
π π π π
One complete cycle of the curve is 0 ≤ t ≤ 12, so
L dL= = … ≈∫ 51 0539 51 1. (numerically) . ft.0
12
13. a.r r ra t i j( ) = −0 32r r r r rv t i j dt C i t C j( ) = − = + − +∫ ( ) ( )0 32 321 2
r r rv i j C C( )0 130 0 130 01 2= − + ⇔ = − =and
∴ = − − r r rv t i tj( ) 130 32
b.
r r rr t i tj dt( ) = − −∫ ( )130 32
= − + + − +( ) ( )130 1632
4t C i t C jr r
r r rr i j C C( )0 60 5 8 60 5 83 4= + ⇔ = =. . and
∴ = − + + − + ( . ) ( )r r rr t t i t j( ) 130 60 5 16 82
c. When the ball passes over the plate, x(t) = 0,so t = 60.5/130 = 0.4653… . At that time,y(t) = 4.5346… , which is slightly above thestrike zone.
d. At t = 0, dx/dt = 200 cos 15°,dy/dt = 200 sin 15°.As in part a,
r r rv t C i t C j( ) = + − + =1 232( )
( ) ( )200 15 32 200 15cos sin° + − + °r ri t j
As in part b, r rr t t i( ) cos= ° +( )200 15
( ) .− + ° +16 200 15 32t t jsinr
e. When x = 400, t = 2 sec 15° = 2.0705… sy(2.0705…) = 41.5846… .Phyllis makes the home run because41.5… > 10.
100
400
x
y
14. a. r r ra t i y t j( ) = + ′′3 ( )
r r rv t t C i y t j( ) = + + ′( ) ( ( ))3 1
r r rv i j C( )0 0 0 01= + ⇒ =
r r rr t t C i y t j( ) = + +( . ) ( ( ))1 5 2
2
r r rr i j C( )0 0 0 02= + ⇒ =
∴ = + = + ( . ) ( ( )) ( . ) ( ( ))r r r r rr t t i y t j t i x t j( ) sin1 5 1 52 2
r r rr t t i t j( ) sin= +( . ) ( . )1 5 1 52 2
b.r r rv t t i t t j( ) cos= +( ) ( . )3 3 1 5 2
If x = 6, t = 2.r r
Kr
v i j( )2 = +6 5.7610
Speed | | .
. m/
= = + = …≈
rKv( ) .
min
2 6 5 76 8 3180
8 32
2 2
15. a. d = a + b cos tt = 0: 240 = a + b cos 0 = a + bt = π: −60 = a + b cos π = a − b2a = 180 ⇒ a = 902b = 300 ⇒ b = 150∴ d = 90 + 150 cos t
b. r rr t t t i( ) cos cos= +( )90 150 2
+ +( )90 150sin sin cost t t jr
r rv t t t t i( ) sin cos sin= − −( )90 300
+ + −(90 cos t t t j150 1502 2cos sin )r
r rv t t t i( ) sin sin= − −( )90 150 2
+ +( )90 150 2cos cost t jr
r
Kr
Kr
v i j( )1 212 1270 13 7948= − −. .
Speed
. . cm/s
= += ≈
(– . ) (– . )212 1 13 7
212 5750 212 6
2 2K K
K
c.r ra t t t i( ) cos cos= − −( )90 300 2
+ − −( )90 300 2sin sint t jr
r
Kr
Kr
a i j( )1 76 2168 348 5216= −. .
P
a v
v( ) .1
1 1
153 4392= ⋅ = −
r r
r K( ) ( )
| ( )|
rr
r
Kr
Kr
a Pv
v
i j
t ( ) ( )
. .
1 11
1
53 3266 3 4678
=
= +
( )
| ( )|
Calculus Solutions Manual Problem Set 10-6 283© 2005 Key Curriculum Press
r r r
Kr
Kr
a a a
i j
n t( ) ( )
. .
1 1 1
22 8902 351 9894
= −
= −
( )
Annie is slowing down. The angle betweenthe acceleration and velocity vector is obtuse,as revealed by the negative dot product. She isslowing down at 53.4392… ≈ 53.4 cm/s.
16. a. r r rr t t i t j= + +( . ) ( . )0 5 4 0 5sin cosr r rv t t i t j( ) cos sin= + + −( . ) ( . )0 5 2 0 5
r r ra t t i t j( ) sin cos= − + −( ) ( . )0 5
r r rv i j( )14 0 6367 1 3139= … − …. .
r r ra i j( )14 0 9906 0 7539= − … − …. .
Speed | | . mi/h= = …rv( )14 1 4601
P
a v
v( )14
14 14
14= ⋅
r r
r( ) ( )
| ( )|
= =0 3598
1 46010 2464
.
.
K
KK.
rr
r
r r
a Pv
v
i j
t ( ) ( )
. .
14 1414
14
0 1074 0 2217
=
= … − …
( )
| ( )|
r r r
r ra a a
i j
n t( ) ( )
. .
14 14 14
1 0980 0 5312
= −
= − … − …
( )
The log is speeding up at t = 14. You can tellby the fact that P(14) is positive, and thus theangle between
ra( )14 and
rv( )14 is acute, which
means that rat ( )14 points in the same direction
as rv( )14 . It is speeding up at 0.2464… ≈
0.246 (mi/h)/h.
b. dL dx dy= +2 2
= + +( . cos ) (– sin . )0 5 2 0 52 2t t dt
L dL= ≈ …
≈∫ 22 7185
0
14
. (numerically)
22.7 mi
Average speed ≈ 1
1422 7185( . )K
= … ≈1 6227 1 62. . mi/h
17. a. r r rr t t t i t j( ) sin cos= − + +( ) ( )5 12 15 12
r r rv t t i t j( ) cos sin= − + −( ) ( )5 12 12r r ra t t i t j( ) sin cos= + −( ) ( )12 12
b. r r rv i j( . )2 5 14 6137 7 1816= … − …. .
r
Kr
Kr
a i j( . )2 5 7 1816 9 6137= +. .
The graph shows and .r rv a( . ) ( . )2 5 2 5
5
5
x
y
v
a
c. P
a v
v( . )2 5
2 5 2 5
2 5= ⋅
r r
r( . ) ( . )
| ( . )|
= = …60 2 5
169 120 2 52 2052
sin .
– cos ..
rr
r
r r
a Pv
v
i j
t ( . ) ( . )
. .
2 5 2 52 5
2 5
1 9791 0 9726
=
= … − …
( . )
| ( . )|
r r r
r ra a a
i j
n t( . ) ( . )
. .
2 5 2 5 2 5
5 2024 10 5863
= −
= … + …
( . )
d. rv( . )2 5 is reasonable because its graphpoints along the path in the direction ofmotion.
ra( . )2 5 is reasonable because it points
toward the concave side of the path. Theroller coaster is traveling at | |
rv( . )2 5 =
16.2830… ft/s. Its speed is increasing atP(2.5) = 2.2052… ft/s2, as shown by the factthat P(2.5) is positive, meaning that the anglebetween
ra( . )2 5 and
rv( . )2 5 is acute.
e. The path is at a high point when they-component of
rr is a maximum. This
happens when cos t = 1, or t = 0 + 2π n.r r ra n i j( )0 2 0 12+ = −π , pointing straightdown.Similarly, the path is at a low point whencos t = −1, or t = π + 2π n.r r ra n i j( )π π+ = +2 0 12 , pointing straight up,Q.E.D.
f. dL dx dy= +2 2
= +( – cos ) (– sin )5 12 122 2t t dt
L dL= ≈ …
≈∫ 78 7078
78 70
2
. (numerically)
. ft
π
18. Recall that | | | | .r ri j= = 1
The angle between r ri iand is 0, so
r r r ri i i i⋅ = =| || | .cos 0 1
Similarly, 1.r rj j⋅ =
284 Problem Set 10-7 Calculus Solutions Manual© 2005 Key Curriculum Press
The angle between
r ri jand is ,
π2
so
r r r ri j i j⋅ = = | | | | .cos
π2
0
∴ ⋅ = + ⋅ + = ( ) ( )r r r r r rv v x i y j x i y j1 2 1 1 2 2
x x i i x y i j y x j i y y j j1 2 1 2 1 2 1 2
r r r r r r r r⋅ + ⋅ + ⋅ + ⋅
= ( ) + ( ) + ( ) + ( )x x x y y x y y1 2 1 2 1 2 1 21 0 0 1= ,x x y y1 2 1 2+ Q.E.D.
19. r r rr t t i t j( ) sin cos= +( . ) ( . )10 0 8 10 0 6
+ ( ).6 0 5t kr
r r rv t t i t j( ) cos sin= + −( . ) ( . )8 0 8 6 0 6
+−( ).3 0 5t k
r
r ra t t i( ) sin= −( . . )6 4 0 8
+ − + − −( . cos . )3 6 0 6 1 5 1 5t j t kr r
( . ).
r r r rv i j k( ) ( . ) ( . )1 8 0 8 6 0 6 3= + − +cos sinr r r ra i j k( ) sin cos .1 6 4 0 8 3 6 0 6 1 5= − + − −( . . ) ( . . )
To determine whether the object is speeding upor slowing down, find the dot product.r ra v( ) ( ) sin cos1 1 6 4 0 8 8 0 8⋅ = − +( . . )( . )( . . )( . ) ( . )( )− − + −3 6 0 6 6 0 6 1 5 3cos sin
= −20.0230…∴ the object is slowing down because the anglebetween
ra( )1 and
rv( )1 is obtuse.
20. a. This is an example of the chain rule.
b. dy/dx equals the slope of the velocity vector,and tan φ also equals the slope of this vector.Thus, tan φ = dy/dx.
By the chain rule, dy/dx = (dy/dt)/(dx/dx),Q.E.D.
c. tan/
/
( )
( )φ = = ′
′dy dt
dx dt
y t
x t
⇒ = ′′
d
ds
d
ds
y t
x t(tan )
( )
( )φ
⇒ = ′ ′′ ′′ ′′
sec–2
2φ φd
ds
x y x y
x
dt
ds
∴ = ′ ′′ ′′ ′′
d
ds
x y x y
x ds dt
φφ–
(sec )( / )2 2
But sec2 φ = 1 + tan2 φ = 1 + y′2/x′2, so
ds dt v/ | |.=r
∴ = ′ ′′ ′′ ′
′ + ′ ′
d
ds
x y x y
x y x v
φ –
( / ) | |2 2 21r
= ′ ′′ ′′ ′
′ + ′= ′ ′′ ′′ ′x y x y
x y v
x y x y
v
–
( ) | |
–
| |2 2 3r r , Q.E.D.
d. x = 5 cos t ⇒ x′ = −5 sin t ⇒ x″ = −5 cos t
y = 3 sin t ⇒ y′ = 3 cos t ⇒ y′′ = −3 sin t
| |rv t t= +25 92 2sin cos
∴ = ++
d
dt
t t
t t
φ 15 15
25 9
2 2
2 2 3 2
sin cos
( sin cos ) /
=+
15
16 92 3 2( sin ) /t
which is maximized for sin2 t = 0 at (± 5 , 0),the ends of the major axis, and is minimizedfor sin2 t = 1 at (0, ±3), the ends of the minoraxis, Q.E.D.
e. x = r cos t ⇒ x′ = −r sin t ⇒ x′′ = −r cos ty = r sin t ⇒ y′= r cos t ⇒ y′′ = −r sin t
| |rv r t r t r= + =2 2 2 2cos sin
∴ = + = d
ds
r t r t
r r
φ 2 2 2 2
3
1sin cos
| | | |a constant equal to the reciprocal of the radius.
f. x = 5 cos2 t ⇒ x′ = −10 cos t sin t = −5 sin 2t⇒ x′′ = −10 cos 2ty = 3 sin2 t ⇒ y′ = 6 sin t cos t = 3 sin 2t⇒ y′′ = 6 cos 2t
| | | |rv t t t= + =25 2 9 2 34 22 2sin sin sin
∴ = +d
ds
t t t t
t
φ – sin cos sin cos
|sin |
30 2 2 30 2 2
34 2 = 0, Q.E.D.
g. At (5, 0), sin t = 0, so d
ds
φ = =15
9
5
93 2/ .
Radius of curvature = =9
51 8.
h. The osculating circle has radius 1.8 and centeron the x-axis at x = 4 − 1.8 = 3.2. Equationsare
x = 3.2 + 1.8 cos t
y = 1.8 sin t
The graph shows the osculating circle. Thename is appropriate because the circle “kisses”the ellipse at the point (5, 0).
3.2 5
3
x
y
Problem Set 10-7Review Problems
R0. Answers will vary.
R1. v t= − =3 0 at t = 9 s
v > 0 for t > 9 s
Calculus Solutions Manual Problem Set 10-7 285© 2005 Key Curriculum Press
Displacement from t = 0 to t = 9 is
( – )t dt3 90
9
= −∫ .
They have moved 9 ft closer to the sawmill.
From t = 0 to t = 25:
Displacement = =∫ ( – )t dt3 81
30
25
ft
Distance = − =∫ | | ftt dt3 261
30
25
R2. a. i.
3
5
t
v(t)
ii. Displacement = ≈ −∫ ( – )2 8 3 80221
4t dt . K
≈ −3.8 cm (exactly 14/ln 2 − 24)
iii. Distance = −∫ | |2 81
4t dt ≈ 10.8853…
≈ 10.9 cm (exactly 2/ln 2 + 8)
b.
tend a aav
0 2 — 30 speeding up
5 8 5 55 speeding up
10 1 4.5 77.5 speeding up
15 0 0.5 80 neither
20 −10 −5 55 slowing down
25 −20 −15 −20 slowing down
(Note that the object is speeding up, slowingdown, or neither, exactly when aend > 0,aend < 0, or aend = 0, respectively, in theoriginal table.)
R3. a. i. v t dtav ( / ) / .= = =∫1
36 2 0 6366
0
3
sin π π K
ii. v t dtav ( / )= =∫1
66 0
3
9
sin π
iii. v t dtav ( / )= =∫1
126 0
0
12
sin π
b. i. f x x x x x x( ) ( ) at= − = − = =6 6 0 0 62 3 2 ,
Average = =∫1
66 182 3
0
6
( – )x x dx
ii. The rectangle has the same area as theshaded region.
5
18
6
x
f(x)
iii. The average of the two values of f(x) at theendpoints is 0, not 18.
R4. a.
θ θ
x
200
700
700 – x
Let x = distance from intersection to cutoff.0 ≤ x ≤ 700Let T = total time taken.
T x x= + +1
5 7200
1
6 27002 2
. .( – )
700
100
x
T
T x x
x x
′ = ⋅ + −
= ⋅ + −
−
−
1
5 7
1
2200 2
1
6 21
5 7200
1
6 2
2 2 1 2
2 2 1 2
.( )
.
. .
/
/
( )
( )
T′ = 0 ⇔ 6.2x = 5.7(2002 + x2)1/2
38.44x2 = 32.49(2002) + 32.49x2
x x2
232 49 200
5 95467 3544= ⇔ = ±. ( )
.. K
Or: Let θ = angle of incidence.Minimal path occurs for θ = sin− 1 (5.7/6.2).x = 200 tan θ = 467.3544…Note that at x = 0, Juana goes entirely alongthe sidewalk.
T( ) .0
1
6 2900 145 1612= ⋅ =
.K
T(467.3544…) = 126.7077…T(700) = 127.7212…
Turning at a point about 467 ft from theintersection of the two sidewalks takes theminimum time, although it takes only asecond longer to head straight for the Englishbuilding.
286 Problem Set 10-7 Calculus Solutions Manual© 2005 Key Curriculum Press
b.
θ
θx
6
10
10 – x
Let x = distance from closest point on thebeach to the cutoff point, 0 ≤ x ≤ 10.Let C = total cost of the road.
C x x= − + +5 10 13 36 2( )The graph shows a minimum C at x ≈ 2.5.
10
150
x
C
Let θ be the angle of incidence.By the minimal path property, the cost isminimized when
sin θ = =x
bridge length
5
13x = 6 tan [sin− 1 (5/13)] = 2.5C(2.5) = 122C(10) = 151.6047…The minimum cost is $122,000, obtained bygoing 7.5 km along the beach, then cuttingacross to the island. This path saves about$29,600 over the path straight to the island.
R5. a. i. a(t) = 6t − t2, t in [0, 10]a′(t) = 6 − 2ta′(t) = 0 ⇔ t = 3a(0) = 0; a(3) = 9; a(10) = −40Maximum acceleration = 9 at t = 3.Minimum acceleration = −40 at t = 10.
ii. v t t t dt t t C( ) ( )= − = − +∫ 6 31
32 2 3
v(0) = 0 ⇒ C = 0
∴ = − ( )v t t t31
32 3
v′(t) = a(t) = t(6 − t)
v′(t) = 0 ⇔ t = 0 or t = 6
v v v( ) ; ( ) ; ( )0 0 6 36 10 331
3= = = −
Maximum velocity = 36 at t = 6.
Minimum velocity at .= − =331
310t
iii. s t v t dt t t C( ) ( )= = − +∫ 3 41
12
Because s(t) measures distance from thestarting point, s(0) = 0, which implies thatC = 0.
∴ = − ( )s t t t3 41
12
s t v t t t′ = = −( ) ( ) ( )1
392
s′(t) = 0 ⇔ t = 0 or t = 9
s s s( ) ; ( ) ; ( )0 0 9 1821
410 166
2
3= = =
Maximum displacement at .= =1821
49t
Minimum displacement = 0 at t = 0.
b. i. Let t = number of days Dagmar has beensaving, P(t) = number of pillars inDagmar’s account, and V(t) = real value(in constant day-zero pillars) of money inaccount after t days.P(t) = 50t (assuming continuousdepositing)V(t) = P(t)(0.50.005 t ) = 50t(0.50.005 t )
ii. The graph shows a maximum V(t) at t ≈289 days.
6000
500
t
V(t)
V′(t) =50(0.50.005 t ) + 50t[0.005(0.50.005 t )] ln 0.5V′(t) = 0 ⇔ 1 = −0.005t ln 0.5
t = − =200
0 5288 5390
ln .. K
Dagmar’s greatest purchasing power willbe after about 289 days.
R6. a. i. and ii.
v
a
a
Speeding up
Slowing down
b. i. r r rr t t i t j( ) cosh sinh= +( ) ( )5 3r r rv t t i t j( ) sinh cosh= +( ) ( )5 3r r ra t t i t j( ) cosh sinh= +( ) ( )5 3
r r r
Kr
Kr
r i j
i j
( ) cosh sinh1 5 1 3 1
7 7154 3 5256
= +
= +
( ) ( )
. .
r
Kr
Kr
v i j( )1 5 8760 4 6292= +. .
r
Kr
Kr
a i j( )1 7 7154 3 5256= +. .
Calculus Solutions Manual Problem Set 10-7 287© 2005 Key Curriculum Press
ii. The graph shows r rr v( ) ( )1 1, , and
ra( )1 .
Note that r ra r( ) ( )1 1= so that the
acceleration vector points directly awayfrom the origin when it is drawn from thehead of
rr ( ).1 (For an elliptical path, the
acceleration vector points directly towardthe origin.)
5 10
5
x
y
r
v
aAsymptote
iii. Speed | |= = + =rv ( ) sinh cosh1 25 1 9 12 2
7.4804… ≈ 7.48 units/minr r
Ka v( ) ( ) sinh cosh1 1 34 1 1 61 6566⋅ = = .The object is speeding up at t = 1, asshown by the positive dot product and bythe acute angle between
ra( )1 and
rv( )1 .
rr r
r
r
r
r
r
r r
aa v
v
v
v
v
v
i j
t ( )
.
. .
11 1
1
1
1
34 1 1
25 1 9 11
1 1018 1
6 4744 5 1007
2 2
= ⋅
=+
= …
= … + …
( ) ( )
| ( )|
( )
| ( )|
sinh cosh
sinh cosh( )
( )
( ( ) . . )r r ra i jn 1 1 2409 1 5751= … − …
| ( )|r
r r
raa v
vt 11 1
1
61 6566
7 4804= ⋅ = …
…| ( ) ( )|
| ( )|
.
.= …8 2423.
rat ( )1 points in the same direction as
rv( ),1
as indicated by the positive dot product andby the acute angle between
ra( )1 and
rv( ),1
so the object is speeding up at about8.24 units/min2.
iv. Distance = ∫ ds0
1
= +∫ 25 92 2
0
1
sinh cosht t dt
= 4.5841… (numerically) ≈ 4.58 units
v. r r rr t v t t t i( ) ( ) cosh sinh+ = +( )5 5
+ +( )3 3sinh cosht t jr
Note that the y-coordinate is 0.6 times thex-coordinate, so the head lies on y = 0.6x,one asymptote of the hyperbola.
Concept Problems
C1. a.
–5
0 4
t
v(t)5
b. v = t3 − 7t2 + 15t − 9 = (t − 1)(t − 3)2
v = 0 ⇔ t = 1, 3
Particle is stopped at 1 s and 3 s.
c. v′ = 3t2 − 14t + 15 = (3t − 5)(t − 3)v′ = 0 at t = 5/3, 3
v v( ) ; ;0 95
3
32
271
5
27= −
= =
v v( ) ; ( )3 0 4 3= = Maximum velocity at t = 4, minimumvelocity at t = 0.
d. v″(t) = 6t − 14v″(t) = 0 ⇔ t = 7/3v″(t) changes from negative to positive att = 7/3, so there is a point of inflection atthat point.
e. At t = 7/3, the particle’s acceleration stopsdecreasing and starts increasing. Thus, theminimum acceleration is at that time.
f. y v t dt t t t t C= = − + − +∫ ( )1
4
7
3
15
294 3 2
y(0) = 4 ⇒ C = 4
∴ = − + − + y t t t t1
4
7
3
15
29 44 3 2
g.
4
0 4
t
y(t)
h. y′(t) = v(t) = 0 when t = 1, 3.
y y y y( ) ; ( ) ; ( ) ; ( )0 4 15
123
7
44
8
3= = = =
Maximum y at t = 0, minimum y at t = 1.
i. ′′ = ′ = − − = =y v t t t( )( ) at ,3 5 3 05
33
y″ changes sign at t = 5
3and at t = 3, so there
are points of inflection at these values of t.
288 Problem Set 10-7 Calculus Solutions Manual© 2005 Key Curriculum Press
j. At t = 5/3, the particle stops accelerating andstarts decelerating, so the velocity at that timeis a local maximum. At t = 3, the particlestops decelerating and starts accelerating, sothe velocity is a local minimum.
k. y is never negative because its minimumvalue is 5/12 at t = 1.
l. Displacement ( ) ( ) ( )= = −∫ v t y y4 00
4
= − = −8
34
4
3ft
m. Distance | ( )| ft= =∫ v t dt 55
60
4
n. v v dtav (displacement) ft/s= = ⋅ = −∫1
4
1
4
1
30
4
o. | | | | (distance) ft/sav0
4
v v dt= = ⋅ =∫1
4
1
4
35
24
C2. Assume that the maximum g a human canwithstand is A and that the distance fromNew York to Los Angeles is D km.Recall that 1 g = 9.81 (m/s)/s.For the fastest trip, the passenger accelerates atA g for the first D/2 km, then decelerates at −A gfor the last D/2 km.Starting at rest, the velocity t seconds afterleaving New York, when accelerating atthe maximum rate, is v(t) = A · 9.81 tand the distance from New York is
s t A t( ) . .= ⋅1
29 81 2
The passenger reaches the halfway point of
the trip when s t D( ) = ⋅10001
2 (because D is km
and s is m), so the first half of the trip takes
tD
A= 1000
9 81.seconds. By symmetry, the second
half takes exactly as long, so the minimum time
for the trip is tD
A= 2
1000
9 81.seconds.
For example, suppose that it is 4000 km fromNew York to Los Angeles and that the humanbody can withstand A = 5 g. Then the minimal
time is t = = …21000 4000
9 81 5571 1372
( )
. ( ). , or
about 9.5 min.
C3. Let x = distance from center along clock hand,L = length of web, and θ = central angle.
Know: dx
dt
d
dt= − =0 7
30. cm/s; rad/s.
θ π
Want:dL
dttat s.= 10
By the law of cosines,
L2 = x2 + 252 − 2 · x · 25 · cos θ
2 2 50 50LdL
dtx
dx
dt
dx
dtx
d
dt = − +cos sinθ θ θ
At , cm, , ,t x= = = =10 183
1
2θ π θcos
sin θ = 3
2.
So L = + ⋅ =18 25 25 18 4992 2 –
dL
dt= − ⋅ + ⋅ ⋅
1
49918 0 7 25
1
20 7. .
+ ⋅ ⋅ ⋅
25 183
2 30
π
= 1.6545… cm/s
C4. a. Let t = time since vertex of cone touchedwater, y = distance from vertex of cone tobottom of cylinder (0 ≤ y ≤ 15), h = altitudeof submerged part of cone, r = radius ofsubmerged part of cone, and D = depth ofwater in cylinder.
Know: dy
dt
dD
dt= −2 cm/ . Want: .min
Volume of water is 15 · 72π = 735π cm3.
Volume of submerged part of cone is 1
32πr h.
Volume of submerged part of cone plus wateris π · 72 · D.
∴ = + 49 7351
32π π πD r h
49 7351
32D r h= +
But , and , soD h y r h= + = 5
12
49 7351
3
25
1443D D y= + ⋅ ( – )
4925
1442dD
dtD y
dD
dt
dy
dt= −
( – )
Find D when y = 10.
49 73525
43210 3D D= + ( – )
Solving numerically gives D ≈ 15.1624… .
Substitute this for D, 10 for y, and −2 fordy/dt.
49
25
1445 1624 22dD
dt
dD
dt= +
( . )K
Calculus Solutions Manual Problem Set 10-7 289© 2005 Key Curriculum Press
Solving algebraically or numerically givesdD
dt= … ≈0 2085 0 21. . cm/ .min
b. When the cone is completely submerged, thetotal volume is
7351
35 12 8352π π π+ ⋅ ⋅ = .
In this case, D = = …835
4917 0408
ππ
. .
When the cone first becomes completelysubmerged,y = 17.0408… − 12 = 5.0408… .Thus when y = 1 the cone is alreadycompletely submerged and the depth D is notchanging.
C5. Let h(t) = f(t) − g(t), so h′(t) = f ′(t) − g′(t). Thenh(a) = h(b) = 0 because f(a) = g(a), f(b) = g(b).
By the mean value theorem (or Rolle’s theorem),there exists an x = c in (a, b) such that
h ch b h a
b a′ = =( ) .
( ) – ( )
–0
But because h′(c) = f ′(c) − g′(c) at time c,f ′(c) = g′(c), so the knights have the samevelocity at this time.
C6. Let L = length of track, z = vertical coordinate ofa point on the track, and T( θ ) = number ofminutes to reach the top.v = 30 − 60 sin θDomain of θ is 0 6≤ ≤θ π / because v would be
negative for acute angles θ > π/6.Because θ is constant, v is constant. Thus,
TL
v
L( ) .θ
θ= =
30 60– sin
To find L for any value of θ, consider z to be anindependent variable. By trigonometry,dL
dzdL dz= =csc cscθ θ, and thus .
L dzz
= ==∫ csc cscθ θ θ1000
0
1000
( is constant)
(Another way to find L is to “unroll” the trackinto a vertical plane. Because the track alwaysmakes an angle of θ with the horizontal, thiswill result in a right triangle with hypotenuse =L, altitude = 1000, and base angle θ. Thensin θ = 1000/L so that L = 1000 csc θ.)
∴ = ( )T θ θθ
1000
30 60
csc
– sin
= −100
32 2 1(sin – sin )θ θ
T ′ = − − ⋅−( )θ θ θ100
32 2 2(sin sin )
(cos )θ θ θ− 4 sin cos
T′(θ ) = 0 ⇔ cos θ ( 1 − 4 sin θ ) = 0cos θ = 0 only for values of θ outside thedomain.∴ 1 − 4 sin θ = 0 ⇔ θ = sin− 1 0.25T(θ ) approaches positive infinity as θ approacheseither end of the domain. So T(θ ) is a minimumfor θ = sin− 1 0.25.Optimal trip takes T(sin− 1 0.25)
= =−100
30 25 2 0 25 266
2
32 1[ . – ( . ) ] s,
or 4 minutes 262
3seconds.
Chapter Test
T1. If the acceleration and velocity have the samesign, the object is speeding up. If the accelerationand velocity have opposite signs, the object isslowing down.
T2. Displacement = ∫ y dx0
60
= ⋅ + + ⋅ − ⋅ +
= + − = −
820 10
215 0 10
25 15
2120 0 200 80 ft
Distance | | ft= = + + =∫ y dx 120 0 200 3200
60
T3. Average value of f ( x) ≈ 2.8. The graph showsequal areas above and below y = 2.8, and thepoint x = c where f ( c) ≈ 2.8 by the mean valuetheorem for integrals.
5 10
5
f(x)
x
c
Average ¯ 2.8
Equal areas
T4. r r rr i j= +7 3The velocity vector points in the negativex-direction and the acceleration vector makes anobtuse angle with the velocity vector, indicatingthat the object is slowing down.
5 10
5
y
x
v
a
→
→
T5. v t= + 60
Displacement ft= + =∫ ( )t dt60 15831
30
25
290 Problem Set 10-7 Calculus Solutions Manual© 2005 Key Curriculum Press
T6.
θθ
x 7 – x
3
7
By the minimal path property,
sin θ = =360
8000 45. .
∴ θ = sin− 1 0.45
x = 3 tan θ = 3 tan (sin− 1 0.45) = 1.5117…
So the cheapest path is 7 − x = 5.4882… mialong the road, then turning toward Ima’s house.
This path costs
360 5 4882 800 9 1 5117 2⋅ + ⋅ +. K K.= 4663.2685… ≈ $4663, which is about $257cheaper than the proposal.
T7. f(x) = x3 − 4x + 5, x ∈ [1, 3]
f ′(x) = 3x2 − 4
′ = ⇔ = ± = ±f x x( ) .0 4 3 1 1547/ K
(The negative value is out of the domain.)
f(1) = 2
f(1.1547…) = 1.9207… , the minimum.
f(3) = 20, the maximum.
Average = + =∫1
24 5 73
1
3
( – )x x dx
The graph shows a minimum at x = 1.1547… ,a maximum at x = 3, and an average of 7. Thearea of the rectangle of altitude 7 equals the areaof the region under the graph.
7
20
1 3
x
f(x)
T8. a. v(t) = 10(0.5 − 2− t)
Distance = = ≈∫ | ( ) | . . ftv t dt 3 6067 3 610
2
K
(exactly 2.5/ln 2)
Displacement ( )
. . ft
=
= − ≈ −∫ v t dt
0
2
0 8202 0 82K
(exactly 10 − 7.5/ln 2)
b. a(0) = v′(0) = 10 ⋅ 2− 0 ln 2 = 10 ln 2
= 6.9314… ≈ 6.93 (ft/s)/s
c. v(0) = −5
Because v(0) is negative and a(0) is positive,the object is slowing down.
T9.
t a aav vend vav displend
0 4 — 50 — 0
7 6 5 85 67.5 472.5
14 10 8 141 113 1263.5
21 13 11.5 221.5 181.25 2532.25
The object traveled about 2532.25 cm.
Average velocity was about
1
212532 25 120 583 120 6⋅ = ≈( . ) . . cm/s.K
T10. r r rr t t i t j( ) cos sin= +( . ) ( . )10 0 4 10 0 6
r r rv t t i t j( ) sin cos= − +( . ) ( . )4 0 4 6 0 6
T11. r r ra t t i t j( ) cos sin= − + −( . . ) ( . . )1 6 0 4 3 6 0 6
T12. r r rr i j( ) cos sin2 10 0 8 10 1 2= +( . ) ( . )
= +6 9670 9 3203. .K Ki j
The graph shows rr ( )2 .
5
5
0x
y
r
a
v
t = 2
T13.r r rv i j( ) sin cos2 4 0 8 6 1 2= − +( . ) ( . )
= − +2 8694 2 1741. .Kr
Kr
i j
See the graph in Problem T12.
The velocity vector is tangent to the path,pointing in the direction of motion.
T14. r r ra i j( ) cos sin2 1 6 0 8 3 6 1 2= − + −( . . ) ( . . )
= − −1 1147 3 3553. .Kr
Kr
i j
See the graph in Problem T12.
T15. r ra v( ) ( )2 2⋅
= 6.4 cos 0.8 sin 0.8 − 21.6 cos 1.2 sin 1.2= −4.0963… ≈ −4.10 (mi/h)2/h
| ( ) sin . cos .rv 2 16 0 8 36 1 22 2| = +
= 3.6000… mi/h
Pa v
v= ⋅ = − ≈ −
r r
r K( ) ( )
| ( ) |
2 2
21 1378 1 14. . (mi/h)/h
rr
rK
rK
ra
P
vv i jt ( ) . .2
2
22 0 9069 0 6871= = −( )
| ( ) |( )
Calculus Solutions Manual Problem Set 10-7 291© 2005 Key Curriculum Press
r r ra a an t( ) ( )2 2 2= −( )
= − −2 0216 2 6681. .Kr
Kr
i j
t = 2
an
at
a
v
T16. The tangential component rat ( )2 has direction the
opposite of rv( ),2 so
rv is decreasing and the object
is slowing down at t = 2.
T17. Object is slowing down at |ra Pt ( )| | ( )|2 2= =
1 1378. (mi/h)/h.K
T18. ran ( )2 points inward to the concave side because
ran is the component of acceleration that pulls theobject out of the straight path into a curve.
T19. dL dx dy v t dt= + =2 2 | |r( )
= ⋅ + ⋅16 0 4 36 0 62 2sin . cos .t t dt
L dL= =∫ 10 0932
0
2
. K (numerically) ≈ 10.09 mi
T20. Answers will vary.
292 Problem Set 11-2 Calculus Solutions Manual© 2005 Key Curriculum Press
Chapter 11—The Calculus of Variable-Factor Products
Problem Set 11-11.
4
10
x
F
(x, F)
∆x = 0.2
F ≈ F(4) = 80e− 2 = 10.8268… ≈ 10.83 lbin the strip.W ≈ F(4) · ∆x = 16e− 2 = 2.1653… ≈ 2.17 ft-lb
2. dW = 20xe− 0.5 x dx
3. W xe dxx= =−∫ 20 69 12890 5
0
7. . K
(exactly 80 − 360e− 3.5 )
4. W ≈ 69.13 ft-lb
5. The amount of work done from x = 0 to x = b is
W x dx
be e
xb
b b
=
= − − +
−
− −
∫ 20
40 80 80
0 5
0
0 5 0 5
.
. .
limb
W→∞
= + + =0 0 80 80 ft-lb
(Use l’Hospital’s rule for be− 0.5 b.)
Problem Set 11-2
Q1. 2 Q2. 102
3
Q3. v(t) = ln | sec t | + C Q4. a ( t) = t− 1
Q5. Fundamental theorem of calculus
Q6. Riemann
Q7. Integration by parts
Q8. Implicit differentiation
Q9. Heaviside method Q10. A
1. Ignore the weight of the rope.Let y = the distance from the bottom of the well.Slope of linear function is −8/50 = −0.16.Weight = 20 − 0.16ydW = (weight) dy = (20 − 0.16y) dy
W y dy= =∫ ( – . )20 0 16 8000
50
ft-lb
2. a. Let y = number of miles up.
Slope of linear function is − = −20 702
7/ .
Weight (tons)= −302
7y
dW y dy= −
30
2
7
W y dy=
=∫ 30
2
71400
0
70
– mi-tons
b. W total = 90 tons · 70 mi = 6300 mi-tonsExcess energy becomes kinetic energy ofrocket and spent fuel.
3. Hooke’s law: F = k · sdW = ks ds
W ks ds k= =∫ 500
10
4. a. F = −x3 + 6x2 − 12x + 16
The graph starts at the high force of 16 lb,levels off, then drops to F = 0 at x = 4.
4
15
x
F
b. dW = F dx = (−x3 + 6x2 − 12x + 16) dx
W x x x dx= + + =∫ (– – )3 2
0
4
6 12 16 32 ft-lb
5.
10
(x, y)
17
15
dV = π x2 dy∴ dW = (17 − y)(62.4)(π x2 dy)
By similar triangles, x = 1.5y.∴ dW = (17 − y)(62.4)(π · 2.25y2 dy)
= 140.4π (17y2 − y3) dy
W dW= =∫ 1 396 752 09370
10
, , . ...
≈ 1.4 million ft-lb (exactly 444,600π)
6. dV = π x2 dyAt x = 4, y = 16.∴ dW = (26 − y)(54.8)(π x2 dy)
Because y = x2,dW = (26 − y)(54.8)(π y dy)
= 54.8π (26y − y2) dy
W dW= ∫0
16
= 337,891.2751… ≈ 337,891 ft-lb
Calculus Solutions Manual Problem Set 11-3 293© 2005 Key Curriculum Press
7. a. Draw x- and y-axes with the origin at thecenter of the sphere. To fill the sphere halffull, the water must be pumped from −120 toy, where y is negative. Integration is fromy = −20 to y = 0.
dV = π x2 dyx2 + y2 = 202 ⇒ x2 = 400 − y2
∴ dW = [y − (−120)](62.4)[π (400 − y2) dy]= 62.4π (y + 120)(400 − y2) dy
W dW= =−∫ 117 621 229
20
0
, ,
≈ 117.6 million ft-lb(exactly 62.4π · 600000)
b. For filling the tank, the limits ofintegration are from −20 to 20.
dW =
−∫ 250 925 288 420
20
, , . K
≈ 250.9 million ft-lb
(exactly 62.4π · 1,280,000)This answer can be found quickly by liftingthe entire weight of the water through thedistance the center of the sphere moves,namely 120 ft.
W =
⋅
= ⋅
( . )
. , ,
62 44
320 120
62 4 1 280 000
3( )( )π
π
(Note that the amount of work to fill theentire tank is more than twice the amountneeded to half-fill it. The work to fill the tophemisphere is greater than that to fill thebottom hemisphere because the same amountof water has to be lifted through a greaterdisplacement.)
8. Slice the water horizontally. Pick sample point(x, y) on the curve y = 0.0002x4 within theslice.dV = 15 · 2x · dy = 30(5000y)1/4 dy
= 300(0.5y)1/4 dy∴ dW = (30 − y)(67)[300(0.5y)1/4 dy]
= 20100(30 − y)(0.5y)1/4 dy
W dW= ≈∫ 9 134 6020
16
, , ft-lb
exactly ( )( . ) /20100 0 5 5404
91 4
9. a. If x is the distance between the piston and thecylinder head and F is the force exerted by thehot gases, then dW = F dx.F = pA, where p is the pressure and A is thearea of the piston.∴ dW = pA dxA dx = dV∴ dW = p dVp = k1V
− 1.4
Initial condition V = 1 at p = 1000 ⇒k1 = 1000.∴ dW = 1000V −1.4 dV
W V dV= ≈−∫ 1000 1504 73201 4
1
10. . K
≈ 1504.7 in.-lb (exactly 2500(1 − 10− 0.4 ))
b. Initial condition p = 15 at V = 10 ⇒k2 = 15 ⋅ 101.4
dW = 15 ⋅ 101.4V −1.4 dV
W V dV= ⋅ ≈ −−∫ 15 10 566 95741 4 1 4
10
1. . . K
So about 567 in.-lb of work is done incompressing the gases (exactly37.5 ⋅ 101.4 (10− 0.4 − 1)).(Mathematically, the work is negative becausethe force is positive and dx is negative.Physically, the work is negative becauseenergy is taken out of the surroundings to putinto the gases. Positive work indicates thatenergy is put into the surroundings by theexpanding gases.)
c. Net amount of work ≈ 1504.7320… −566.9574 = 937.7746… ≈ 937.8 in.-lb
d. Carnot (kar-NO), Nicolas Léonard Sadi,1796−1832, was a French physicist and apioneer in the field of thermodynamics.
Problem Set 11-3Q1. 2 cm3 Q2. 3 cm3
Q3. Q4.
π
1 x
y
1
x
y
Q5. Q6.
1
x
y
x
y
Q7. (mass)/(volume) Q8. (force)(displacement)
Q9.1
1 2– xQ10. B
1. a. The graph shows y = ln x, rotated aboutx = 0, showing back half of solid only.
(x, y)
1 3
1x
y
294 Problem Set 11-3 Calculus Solutions Manual© 2005 Key Curriculum Press
Slice the region parallel to the axis ofrotation, generating cylindrical shells. Picksample point (x, y) on the curve, within theslice.ρ = kx− 1
dm = ρ dV = (kx− 1)(2π x ln x dx)= 2π k ln x dx
m k x dx k= ≈∫ 2 8 14191
3
π .ln K
(exactly 2π k(3 ln 3 − 2))
b. Slice perpendicular to the axis of rotation,generating plane washers.ρ = 5 + 2ydm = ρ dV = (5 + 2y) ⋅ π ( 32 − x2) dy
= π ( 5 + 2y)(9 − e2y) dy
m dm= ≈∫ 108 1103
0
3
. Kln
(exactly π [36 ln 3 + 9 (ln 3)2 − 16])
2. The graph shows y = sin x, rotated about they-axis, showing back half of solid only.
π
1
y
x
Slice the region parallel to the axis of rotation,generating cylindrical shells.ρ = kxdm = ρ dV = kx ⋅ 2π xy dx = 2π kx2 sin x dx
m dm k k= ≈ −∫ 36 8798 2 42
0. (exactly ( ) )K π π
π
3. a. The graph shows y = 9 − x2, rotated about they-axis.
3
9
x
y
(x, y)
Slice the region perpendicular to the axis ofrotation, generating plane disks.ρ = k,dm = k dV = k ⋅ π x2 dy = k ⋅ π ( 9 − y) dy
m dm k= =∫ 40 50
9
. π
Or: Slice parallel to the axis of rotation.dm = k ⋅ 2π xy dx = 2π kx(9 − x2) dx
m dm k= =∫ 40 50
3
. π
Or: Volume of paraboloid is half the volumeof the circumscribed cylinder, or0.5(π ⋅ 32)(9) = 40.5π, so m = 40.5π k.
b. Slice perpendicular to the axis of rotation.ρ = ky2
dm = ρ dV = ky2 ⋅ π ( 9 − y) dy
m dm k= =∫ 546 750
9
. π
c. Slice parallel to the axis of rotation.ρ = k(1 + x)dm = ρ ⋅ 2 π xy dx = 2π kx(1 + x)(9 − x2) dx
m dm k= =∫ 105 30
3
. π
d. The solid in part b has the largest mass.
4. a. The graph shows y x1 = and y2 = 0.5x,intersecting at (0, 0) and (4, 2), rotated aboutthe x-axis, showing back half of solid only.
4
2
x
y
y1
y2
Slice perpendicular to the axis of rotation,generating plane washers.Pick sample points (x, y1) and (x, y2).ρ = kxdm dV kx y y dx
kx x x dx
= = ⋅ −
= −
ρ π
π
( )
( . )
12
22
20 25
m k k= =16
316 7551π . K
b. Slice parallel to the axis of rotation,generating cylindrical shells.
Pick sample points (x1, y) and (x2, y).ρ = ky2
dm = ρ dV = ky2 ⋅ 2π y(x2 − x1) dy= 2 π ky3(2y − y2) dy
m dm k k= = =∫ 64
1513 4041
0
2
π . K
5. a. Prediction: The cone on the left, with higherdensity at its base, has greater mass becausehigher density is in the larger part of the cone.
b. Set up a coordinate system with the originat the center of the base. Slice each coneperpendicular to its axis, generating planedisks.Pick sample point (x, y) on the element ofthe cone, y = 6 − 2x.dV = π x 2 dy = π ( 3 − 0.5y)2 dyFor the cone on the left, ρ = 80 − 5y.dm = (80 − 5y) · π(3 − 0.5y)2 dy
m dm= =∫ 13050
6
π oz
Calculus Solutions Manual Problem Set 11-3 295© 2005 Key Curriculum Press
For the cone on the right, ρ = 50 + 5y.dm = (50 + 5y) · π ( 3 − 0.5y)2 dy
m dm= =∫ 10350
6
π oz
∴ the cone on the left has the greater mass, aspredicted in part a.
6. a. Prediction: The cylinder on the left, withhigher density at walls, has greater massbecause higher density is in the larger partof the cylinder.
b. Set up a coordinate system with the origin atthe center of the bottom base. Slice eachcylinder parallel to its axis, generatingcylindrical shells. Pick sample point (x, 6).dV = 2π x · 6 · dx = 12π x dxFor the cylinder on the left, ρ = 50 + 10x.dm = (50 + 10x) · 12π x dx
= 12π (50x + 10x2) dx
m dm= =∫ 37800
3
π oz
For the cylinder on the right, ρ = 80 − 10x.dm = (80 − 10x) · 12π x dx
= 12π ( 80x − 10x2) dx
m dm= =∫ 32400
3
π oz
∴ the cylinder on the left has the greatermass, as predicted in part a.
7. y1 = 4 − 2x2 and y2 = 3 − x2, rotated aboutthe x-axis.The graphs intersect at (1, 2) in Quadrant I.Slice perpendicular to the axis of rotation,generating plane washers.
Pick sample points (x, y1) and (x, y2).
ρ π= = −kx dV y y dx212
22, ( )
dm = ρ dV = π kx2(7 − 10x2 + 3x4) dx
m dm k k= = =∫ 16
212 3935
0
1
π . K
8. Rotate the region in Problem 7 about the y-axis.Slice the region parallel to the axis of rotation,generating cylindrical shells.Pick sample points (x, y1) and (x, y2).
ρ = −e x , dV = 2π x(y1 − y2) dx = 2π x (1 − x2) dx
dm dV xe x dxx= = −−ρ π ( ) 2 1 2
m dm e= = −−∫ 0 9444 2 14 51
0
1
. (exactly ( ))K π
9. Set up axes with the origin at the center of thelower base and the y-axis coaxial with thecylinder’s axis.Slice perpendicular to the axis of the cylinder,generating plane disks of constant radius 0.5.
ρ is given in the table in the text.
dV dy dy= =π π0 5
42.
dm dV dy= =ρ π ρ4
m dy= ∫π ρ4 0
2
Simpson’s rule cannot be used because there isan odd number of increments. Use the trapezoidalrule.
m ≈ + + + + +
π4
0 41
210 9 9 9 8 9 6 9 4
1
29 0( . ) ( ) . . . . ( . )
= 4.82π ≈ 15.14 g
10. a. The graph shows y1 = 4 − x2 and y2 =4x − x2, intersecting at (1, 3), rotated aboutthe y-axis, showing back half of solid only.
(x, y )1
(x, y )2
x
y4
1
Slice parallel to the axis of rotation,generating cylindrical shells.Pick sample points (x, y1) and (x, y2).ρ = kx, dV = 2π x (y1 − y2) dx
= 2π x (4 − 4x) dxdm = ρ dV = 2π k x2(4 − 4x) dx
m dm k k= = =∫ 2
32 0943
0
1
π . K
b. The graph shows the curves in part a, rotatedabout the x-axis, showing back half of solidonly.
(x, y )1
(x, y )2
4
1
x
y
Slice perpendicular to the axis of rotation,generating plane washers.
Pick sample points (x, y1) and (x, y2).
ρ π= = −kx dV y y dx , ( )12
22
dm = ρ dV = π kx(16 − 24x2 + 8x3) dx
m dm k k= = =∫ 3 6 11 3097
0
1
. . π K
296 Problem Set 11-3 Calculus Solutions Manual© 2005 Key Curriculum Press
c. The region is rotated about the y-axis asin part a.Slice perpendicular to the axis of rotation,generating plane disks.Pick sample points (x1, y) and (x2, y).Below y = 3, disks have radius x2.Above y = 3, disks have radius x1.
ρ = ky
For in [ , ], x dV x dx0 3 22= π
= −π ( ) .2 4 2– y dy
For x in [3, 4], dV x dx y dy= = −π π ( ) .12 4
m ky y dy= ⋅ −∫ π ( )2 4 2
0
3
–
+ ⋅ −∫ ky y dyπ ( )43
4
= + = =114
151
2
33 6 11 3097π π π . .k k k kK
(Coincidentally, this answer equals the answerto part b.)
11. a. The graph shows a sphere with origin at itscenter.
(x, y)
r
r y
x
Slice the upper semicircular regionperpendicular to the x-axis and rotate it to getplane disks.Pick a sample point (x, y).Equation of the circle in the xy-plane isx2 + y2 = r2.ρ = k|x|, dV = π y2 dx = π (r2 − x2) dxdm = ρ dV = k|x| π (r2 – x2) dx
m k x r x dx
k r x x dx
k r x x r k
r
r
r
r
= −
=
= −
=
∫∫
π
π
π π
| | ( )2 2
2 3
0
2 2 4
0
4
2
21
2
1
4
1
2
–
( – )
b. Slice the right semicircular region parallel tothe y-axis, and rotate it to get cylindricalshells coaxial to the y-axis.
ρ π π = = ⋅ ⋅ =kx dV x y dx x r x dx , 2 2 4 2 2–
. dm dV kx r x dx= =ρ π4 2 2 2 –
m k x r x dxr
= ∫4 2 2 2
0π –
Let x = r sin θ.
Then , .dx r d r x r= =cos – cosθ θ θ2 2
x r= ⇒ = =−θ πsin 1 1 2/
∴ = ⋅ ⋅∫ m k r r r d4 2 2
0
2
π θ θ θ θπ
sin cos cos/
= ∫4 4 2 2
0
2
π θ θ θπ
r k dsin cos/
= ∫π θ θπ
r k d4 2
0
2
2sin/
(half-argument property)
= ∫1
21 44
0
2
π θ θπ
r k d( – cos )/
(half-argument property)
= −
=1
2
1
44
1
44
0
22 4π θ θ π
π
r k r ksin/
c. Slice into spherical shells. Pick a samplepoint on the x-axis within the shell.Then x is the radius of the shell and 4π x2
is the area of the shell at the sample point.
∴ dV = 4π x2 dx
ρ = kx
dm = ρ dV = 4π kx3 dx
m k x dx kx krr r
= = =∫4 3
0
40
4π π π
12. Assume Earth is spherical, with radius3960 mi = 3960 ⋅ 5280 ⋅ 12 ⋅ 2.54 cm =637,300,224 cm, and slice into sphericalshells with radius x and dV = 4π x2 dx.
ρ = −128
6373002243x
g cm/
dm dV xx
dx= = −
ρ π π48
32
6373002242
3
m dV= ∫0
637300224
= −
168
6373002243
4
0
637300224
π π
xx
= 8π ⋅ 6373002243 ≈ 6.505 × 1027 g
Mass is about 6.505 × 1021 metric tons!
13. The graph shows y = ex, from x = 0 to π/2,rotated about the y-axis, showing back half ofsolid only.
(x, y)
1
1x
y
Calculus Solutions Manual Problem Set 11-4 297© 2005 Key Curriculum Press
Slice parallel to the axis of rotation, generatingcylindrical shells.Pick a sample point (x, y).ρ = cos x, dV = 2π x ⋅ y ⋅ dx = 2π xex dxdm = ρ dV = cos x 2π xex dx
m dm e= ≈ −
∫ 8 6261
212
0
2
. exactly /K π πππ /
14. a = 4 mi, b = 1 mi, c = 0.5 mi
a.x
a
y
b
z
c
+
+
=
2 2 2
1,
where a = 4, b = 1, c = 0.5.
The cross section at z = z0 < c has equation
x
a
y
b
z
c
+
= −
2 20
2
1 .
This is the equation of an ellipse with
x a z c-radius and1 02– ( / )
y b z c-radius .1 02– ( / )
b. Slice horizontally into plane elliptical disks.The area of the cross section isπ (x-radius)(y-radius) = π ab(1 − (z/c)2)= 4π (1 − (z/0.5)2) = 4π (1 − 4z2).
ρ = 0.08(5280)3 e− 0.2 z lb/mi3
dm = ρ dV = 0.08 ⋅ 52803e− 0.2 z ⋅ 4π (1 – 4z2) dz
= 0.32 ⋅ 52803π e−0.2 z(1 – 4z2) dz
m dm= ≈ ⋅ ⋅∫ 0 32 5280 1 0089533
0
0 5
. . K.
≈ 4.7525… × 1010 lb, or about23,762,540 tons
(exactly 0.32 · 52803π (1100e− 0.1 − 995))
c. Volume of semi-ellipsoid = =2
3
4
33π π miabc
Weight
49,326,507,160 lb,
= ⋅ ⋅
≈
0 08 52804
33. π
1,801,427,783 lb more (≈ 3.8% more thanactual)
d. V ab z c dzc
= −∫2 1 2
0 [ ( / ) ]π
= − ⋅
21
33
0
πab z c z cc
( / )
=
=2
2
3
4
3π πab c abc, Q.E.D.
Problem Set 11-4
Q1. −52 = −25 Q2. (−11)2 = 121
Q3. sin 2x = 2 sin x cos x
Q4. cos ( cos )2 1
21 2x x= +
Q5.– /
( – )
/
( – )
1 3
2
1 3
5x x+
Q6.1
35
1
32ln ln| | | |x x C− − − +
Q7.1 3
2
1 3
52 2
/
( – )
– /
( – )x x+
Q8. g x f x dx g x f x( ) ( ) ( )= ⇔ ′ =∫ ( )
Q9. f ′(2) > 0: increasing Q10. E
1. a. The graph shows y = 9 − x2, rotated aboutthe y-axis.
(x, y)
x
y
3
9
Slice the region perpendicular to the axis ofrotation, generating plane disks.
dV = π x2 dy = π (9 − y) dy
V y dy= − =∫ π π( ) .9 40 50
9
b. Each point in a disk is about y units from thexz-plane, where y is at the sample point (x, y).
dMxz = y dV = π (9y − y2) dy
M y y dyxz = − =∫ π π( ) .9 121 52
0
9
c. y V M yxz⋅ = ⇒ = =121 5
40 53
.
.
ππ
x z= = 0 by symmetry.The centroid is at (0, 3, 0).
2. a.x y
y x12 5
1 25 11
144
2 22 2
+
= ⇒ = −
Slice the ellipsoidal region above the x-axisperpendicular to the x-axis, generating planedisks as the region rotates.Pick sample point (x, y).
dV y dx x dx= = −
π π2 225 1
1
144
V x dx=
∫25 1
1
1442
0
12
π –
= −
=25
1
4322003
0
12
π πx x
This answer equals 2
312 52⋅ ⋅ ⋅π , which is
expected because the volume of a (whole)
ellipsoid is V abc= 4
3π .
298 Problem Set 11-4 Calculus Solutions Manual© 2005 Key Curriculum Press
b. Each point in a disk is about x units from theyz-plane, where x is at the sample point(x, y).
dM x dV x x dxyz = = −
25 1
1
1442π
M dMyz yz= =∫ 9000
12
π
c. x V M xyz⋅ = ⇒ = =900
2004 5
ππ
.
y z= = 0 by symmetry.
The centroid is at (4.5, 0, 0).
3. a. See the graph in Problem 1.Each point in a disk is about y units from thexz-plane, where y is at the sample point(x, y), so each point has about the samedensity.
ρ = ky1/3, dm = ρ dV = kπ(9y1/3 − y4/3) dy
m dm k= =∫ 170 1375
0
9
. K
exactly =
9
289
33π k
b. Each point in a disk is about y units from thexz-plane, where y is at the sample point(x, y).dMxz = y dm = kπ (9y4/3 − y7/3) dy
M dM
k k
xz xz=
=
∫0
9
43612 4952
9
289. exactlyK π
c. y m M yk
kxz⋅ = ⇒ = 612 4952
170 1375
.
.
K
K
= 3 6. (exactly)x z= = 0 by symmetry.The center of mass is at (0, 3.6, 0).
d. False. The centroid is at (0, 3, 0), but thecenter of mass is at (0, 3.6, 0).
4. a. Slice the ellipsoid as in Problem 2.Each point in a disk is about x units fromthe yz-plane, where x is at the sample point(x, y), so each point has about the samedensity as at the sample point.
ρ π π= = = −
kx dV y dx x dx, 2 225 1
1
144
dm dV kx x dx= = −
ρ π 25 1
1
1442
m dm k= =∫ 2827 4333
0
12
. K
(exactly 900π k)
b. Each point in a disk is about x units from theyz-plane, where x is at the sample point(x, y).
dM x dm kx x dxyz = = −
25 1
1
1442 2π
M dM kyz yz= =∫ 57600
12
π
c. x m M xk
kyz⋅ = ⇒ = =5760
9006 4
ππ
.
y z= = 0 by symmetry.Center of mass is at (6.4, 0, 0).
d. False. The centroid is at (4.5, 0, 0), but thecenter of mass is at (6.4, 0, 0).
5. a. y = ex
Slice the region parallel to the y-axis.dA = y dx = ex dx
A e dx ex= = − =∫ 2
0
2
1 6 3890. K
Each point in a strip is about x units from they-axis, where x is at the sample point (x, y).∴ dMy = x dA = xex dx
M xe dx ey
x= = + =∫ 2
0
2
1 8 3890. K
x A My⋅ = ⇒
xe
e= +2
2
1
1– = 1.3130… (= coth 1)
b. Strips in part a generate plane disks. Eachpoint in a disk is about x units from theyz-plane, where x is at the samplepoint (x, y).dV = πy2 dx = πe2x dx
V e dx ex= = =∫ π π2 4
0
2
21 84 1917 . ( – ) K
dMyz = x dV = πxe2x dx
M dM eyz yz= = + =∫ 3
4
1
4129 42924
0
2
π π . K
x V M xe
eyz⋅ = ⇒ = + =3 1
2 11 5373
4
4( – ). K
c. False. For the solid, x is farther from theyz-plane.
6. a. Slice the region parallel to the y-axis.dA = sec x dx
A x dx= = +∫ sec ln ( )/
2 30
3
π
= 1.3169…dMy = x dA = x sec x dx
M x x dxy = ∫ sec/
0
3π
= 0.7684… (numerically)
x A M xy⋅ = ⇒ = =0 7684
1 3169
.
.
K
KK0.5835
Calculus Solutions Manual Problem Set 11-4 299© 2005 Key Curriculum Press
b. Strips in part a generate plane disks. Eachpoint in a disk is about x units from theyz-plane, where x is at the samplepoint (x, y).
dV = π y2 dx = π sec2 x dx
V x dx= = =∫ π π
πsec
/2
0
3
3 5 4413. K
dMyz = x dV = π x sec2 x dx
M x x dxyz = =∫ π
πsec
/2
0
3
3 5206. K
exactlyπ π
2 3
3
1
2+
ln
x V M xyz⋅ = ⇒ = 0 6470. K
exactlyπ3
2
3−
ln
c. False. For the solid, x is farther from theyz-plane.
7. Construct axes with the origin at a vertex and thex-axis along the base, b.Slice the triangle parallel to the x-axis.
The width of a strip is bb
hy− .
dA bb
hy dy= −
dM y dA byb
hy dyx = = −
2
M byb
hy dyx
h
=
∫ – 2
0
= − =1
2 3
1
62 3
0
2byb
hy bh
h
y A M ybh
bhhx⋅ = ⇒ = =
1612
1
3
2
, Q.E.D.
8. a. y = x2/3 from x = 0 to x = 8.Slice the region vertically. Pick a samplepoint (x, y) on the graph within the strip.(See the graph in part e.)
dA = y dx = x2/3 dx
A x dx= =∫ 2 3
0
8
19 2/ .
b. Slice the region parallel to the x-axis so thateach point in a strip is about y units from thex-axis, where y is at the sample point (x, y).
dMx = y(8 − x) dy = (8y − y5/2) dy
M y y dyx = =∫ ( – )/8 27 42855 2
0
4
. K
exactly 273
7
c. Each point in a strip of part a isapproximately x units from the y-axis, wherex is the value in the sample point (x, y).
dMy = x dA = x5/3 dx
M x dxy = =∫ 5 3
0
8
96/
d. x A M xy⋅ = ⇒ = =96
19 25
.
y A M yx⋅ = ⇒ = = …27 4285
19 21 4285
.
.
K.
exactly 13
7
Centroid is at (5, 1.4285…).
e. The balance point is shown on the graph.
8
4
x
y
9. a. Slice the region parallel to the y-axis so thateach point in a strip will be about x unitsfrom the y-axis, where x is at the samplepoint (x, y).
dA = y dx = sin x dx
A x dx= =∫ sin 20
(exactly)π
(This may be “well-known” by now.)dMy = x dA = x sin x dx
M x x dxy = = … =∫ sin0
3 1415π
π. (exactly)
x A M xy⋅ = ⇒ = π2
, Q.E.D.
(Or just note the symmetry.)
b. dM2y = x2 dA = x2 sin x dx
M x x dxy22
05 8696= = …∫ sin
π.
(exactly )π 2 4−
c. x A M xy2
2
2 4
2⋅ = ⇒ = = …π –
1.7131
10. a. Set up axes with the x-axis along the base, B.
dM2B = y2 dA = y2B dy
M y B dy BHB
H
22 3
0
1
3= =∫
b. Set up axes with the x-axis through thecentroid.
dM2c = y2 dA = y2B dy
M y B dy BHcH
H
22
0 5
0 531
12= =∫– .
.
300 Problem Set 11-4 Calculus Solutions Manual© 2005 Key Curriculum Press
c. Set up axes with the x-axis along the base, B.
dM y dA y BB
Hy dyB2
2 2= = −
M ByB
Hy dy BHB
H
22 3 3
0
1
12=
=∫ –
d. Use the axes in part c. The distance from thecentroidal axis to a sample point (x, y) is
y H− 1
3.
dM y H dA y H BB
Hy dyc2
2 21
3
1
3= −
= −
−
M BH
y y Hy H dycH
H
2 = + +
∫ – –
– /
/ 1 5
3
7
9
1
93 2 2
3
2 3
= − + − +B
Hy By BHy BH y
H
4
5
9
7
18
1
94 3 2 2
0
= 1
363BH
11. a. Slice into cylindrical shells so that each pointin a shell will be about r units from the axis.
The altitude of a shell is a constant, H.dM = r2 dV = r2 2πrH dr
M Hr dr HRR
= =∫ 21
23 4
0π π
r V M rHR
R Hr R2 2
4
2
12 1
2⋅ = ⇒ = ⇒ =
π
πb. Slice the cone into cylindrical shells so that
each point in a shell will be about r unitsfrom the axis.
The altitude of a shell is HH
Rr− .
dM r dV r r HH
Rr dr= = −
2 2 2π
M H rR
r dr HRh
R
=
=∫2
1 1
103 4
0
4π π–
r V M rHR
R Hr R2 2
4
2
11013
0 3⋅ = ⇒ = ⇒ =π
π.
c. Slice the sphere into cylindrical shells so thateach point in a shell is about r units from theaxis.The equation of the sphere is r2 + y2 = R2.The altitude of a shell is 2y.
dM r dV r r R r dr= = ⋅2 2 2 22 2π –
M r R r drR
= ∫4 3 2 2
0π –
Let r = R sin θ.
dr R d R r R= =cos – cosθ θ θ, 2 2
r = 0 ⇒ θ = 0, r = R ⇒ θ π=2
M R R R d= ⋅ ⋅∫4 3 3
0
2
π θ θ θ θπ
sin cos cos/
= ∫4 5 2 4
0
2
π θ θ θ θπ
R d(cos – cos ) sin/
= − +
4
1
3
1
55 3 5
0
2
π θ θπ
R cos cos/
= − + + −
=4 0 0
1
3
1
5
8
155 5π πR R
r V M rR
Rr R2 2
5
3
81543
0 4⋅ = ⇒ = ⇒ =π
π.
12. Assume the clay has uniform density ρ.
Cylinder: H R V R H R= = = =2 2 10002 3C C C, π π
RC
/
cm=
500 1 3
π
Second moment of volume = =1
22 4 5π π( )C C CR R R
(from Problem 11a)
Second moment of mass
= =
= …ρπ ρ
πρRC
/
, .52 3
500500
14 684 1932
Sphere: V R R= = ⇒ =
4
31000
75031 3
ππS S
/
cm
Second moment of volume = 8
155πRS
(from Problem 11c)Second moment of mass
= =
ρ π ρ
π8
15400
75052 3
RS
/
= 15,393.3892…ρThe sphere has higher moment of mass.
13. a. Set up axes with the x-axis through thecentroid.
dM2 = y dA = y2 · B dy
M B y dy yH
H
H
H
22 3
0 5
0 5
0 5
0 51
3= =
− −∫ .
.
.
.
= 1
123BH , Q.E.D.
(Same answer as in Problem 10b)
b. i. B = 2, H = 12; M2 = 288;stiffness = 288k
ii. B = 12, H = 2; M2 = 8; stiffness = 8kA board on its edge is 36 times stiffer.
c. i. Set up axes with the x-axis through thecentroid.From y = 0 to y = 2, dM2 = y2
· 2 dy.From y = 2 to y = 4, dM2 = y2
· 4 dy.
Calculus Solutions Manual Problem Set 11-5 301© 2005 Key Curriculum Press
By symmetry, M dM2 20
4
2= ∫= + =∫ ∫2 2 2 4 1602
0
22
2
4
y dy y dy .
Stiffness = 160k
ii. From y = 0 to y = 4, dM2 = y2 · 1 dy.
From y = 4 to y = 6, dM2 = y2 · 4 dy.
By symmetry, M dM2 20
6
2= ∫= + =∫∫2 2 4 4482 2
4
6
0
4
y dy y dy .
Stiffness = 448k (2.8 times stiffer!)
d. Increasing the depth does seem to increasestiffness greatly, but making the beam verytall would also make the web very thin,perhaps too thin to withstand much force.
14. a. dA = y dx = x3 dx
A x dx= =∫ 3
0
2
4
b. dV = 2π x · y · dx = 2π x4 dx
V x dx= =∫ 2 12 84
0
2
π π.
c. dMy = x dA = x4 dx
M x dxy = =∫ 4
0
2
6 4.
The volume integral is 2π times the momentintegral.
d. x A M xy⋅ = ⇒ = =6 4
41 6
..
e. The centroid travels 2 3 2π π . .x =(Area)(Distance traveled by centroid) =(4)(3.2π) = 12.8π, which equals the volume.Thus, the theorem of Pappus is confirmed.
15. a. Area of a small circle = π r2
The centroid of the small circle is its center,so the distance from the axis of rotation tothe centroid is R. Thus, the theorem ofPappus implies
V = 2π RA = 2π R(π r 2) = 2π 2r 2R
b. Area of a semicircle = 1
22πr
Volume of a sphere = 4
33πr
2
43
212
4
3
3
2π
π
π π πr A V r
r
rr⋅ = ⇒ =
⋅=
16. Pick a closed region that does not lie on bothsides of the y-axis.Slice the region parallel to the y-axis so that eachpoint in the strip will be about r units from they-axis (see graph).
f(r)
r
r
y
a b
Let f(r) be the length of the strip or the sum ofthe lengths if the region has S-shaped parts.Let A be the area of the region.dMy = r dA = r f (r) d r
M r f r drya
b
= ∫ ( )
Rotate the region about the y-axis. The stripsgenerate cylindrical shells.
dV = 2π r f (r) d r
V r f r dr r f r dra
b
a
b
= = ∫∫ 2 2π π( ) ( )
= 2π My
But My also equals r A⋅ .∴ = = ( )( )V rA r A2 2π π= (distance traveled by centroid)(area of region),Q.E.D.
Problem Set 11-5Q1. centroid Q2. center of mass
Q3. radius of gyration Q4. definite integration
Q5. indefinite integration (or antidifferentiation)
Q6. ρ = (mass) ÷ (volume)
Q7. x1/ 2 Q8. ln | sec x + tan x | + C
Q9. y′ = (x2 + 1)− 1 Q10. A
1. a. Slice the trough face horizontally so that eachpoint in a strip is about the same distancebelow the surface as at the sample point(x, y).y = 2x4 ⇒ x = (0.5y)1/4
p = k(2 − y), dA = 2x dy = 2(0.5y)1/4 dydF = p dA = 2k(2 − y)(0.5y)1/4 dy
F dF k k= =
∫ 2 8444
128
450
2
. exactlyK
b. dMx = y dF = y · 2k(2 − y)(0.5y)1/4 dy
M dM k kx x= =
∫ 2 1880
256
1170
2
. exactlyK
c. y F M y
k
kx⋅ = ⇒ = 2 1880
2 8444
.
.
K
K
=
0 7692
10
13. exactlyK
x = 0 by symmetry.
The center of pressure is at 010
13,
.
2. a. The graph shows y = x2, between y = 0 andy = 100.
302 Problem Set 11-5 Calculus Solutions Manual© 2005 Key Curriculum Press
10
100
x
y
(x, y)
Width at y = 100 ft is 2 y = 20 ft, Q.E.D.
b. Slice the dam face horizontally so that eachpoint in a strip is the same distance below thesurface as the sample point (x, y).dA = 2x dy = 2y1/2 dx
A y dx= =∫ 2 13331
31 2 2
0
100/ ft
(2/3 the area of the circumscribed rectangle)
c. p = k(100 − y) with k = 62.4 lb/ft3
dF = p dA = 2k(100 − y)(y1/2) dy
F dF k= = =∫ 53 3331
33 328 000
0
100
, , ,
Force is 3,328,000 lb, or 1664 tons.
d. dMx = y dF = 2ky(100 − y)(y1/2) dy
M dM kx x= =
=∫ 16000000
7142 628 571 4285
0
100
, , . K
≈ 142.6… million lb-fte. y F Mx⋅ = ⇒
y = =300
742
7
8 = 42.8571… ≈ 42.86 ft
3. a. Slice the bulkhead horizontally so that eachpoint in a strip is the same distance belowthe surface as the sample point (x, y).
x y
20
32
321
4 4
+
=–
x y= − −
20 1
1
321
4 1/4
dA x dy y dy= = − −
2 40 1
1
321
4 1 4
/
A dA= ≈ ≈∫ 1186 6077 1186 6 2
0
32
. . ftK
b. p = 67(32 − y)
dF = p dA
= − ⋅ − −
67 32 40 1
1
321
4 1 4
( )
/
y y dy
F dF= ≈∫ 1 199 294 1645
0
32
, , . K
≈ 1 199. million lbc. dMx = y dF
= ⋅ − ⋅ − −
y y y dy67 32 40 1
1
321
4 1 4
( )
/
M dMx x= ≈∫ 13 992 028 2564
0
32
, , . K
≈ 13.992 million lb-ft
d. y F M yx⋅ = ⇒ ≈ 11 6668. ftK
x = 0 by symmetry.Center of pressure is at about (0, 11.67) ft.
e. Moment of area:
dM y dA y y dy= = ⋅ − −
40 1
1
321
4 1 4/
M dM= ≈∫ 20 071 5364
0
32
, . K
≈ 20.07 thousand ft3
y A M y⋅ = ⇒ ≈ 16 9150. ftK
x = 0 by symmetry.The centroid is at about (0, 16.92) ft.The centroid is different from center ofpressure.
f. Area below waterline:
A dAw = ≈ ≈∫ 548 6345 548 6 2
0
16
. . ftK
First moment of area below waterline:
M dMw = ≈ ≈∫ 4749 3398 4749 3 3
0
16
. . ftK
y A M yw w⋅ = ⇒ ≈ 8 6566. ftK
x = 0 by symmetry.The center of buoyancy is at about(0, 8.66) ft.
4. a. Equation of ellipse is x y
6 31
2 2
+
= .
b. Slice the ellipse horizontally so that eachpoint in a strip is y units from the surfacewhere y is at the sample point and y isnegative.Surface of oil is at y = 0 ⇒ p = −50y.
x y dA x dy y dy= = =2 9 2 4 92 2– –,
dF p dA y y dy= = − ⋅50 4 9 2–
= −200 9 2y y dy–
F dF= =−∫ 1800
3
0
lb (exactly)
5. a.
10
60y
x
(x, y)
Calculus Solutions Manual Problem Set 11-5 303© 2005 Key Curriculum Press
Slice the wing parallel to the y-axis. Picksample point (x, y) within the strip.
dA y dx x dx= = 6020
cosπ
A dA= = =
−∫2400
763 943710
10
π. K
≈ 763.9 ft2
b. dF p dA k x x dx= = − ⋅( | |)10 6020
cosπ
F dF k= =
−∫ 10
10
4 863 4168, . K
exactly48000
2πk
c. Make 4863.4168…k ≥ 96.k ≥ 0.0197… tons/ft2 (exactly 0.002π 2 )
6. a. y = 100 − x2 intersects the x-axis at x = ±10.Slice the wing parallel to the y-axis. Picksample point (x, y) within the strip.
p = 90 − 7x
dA = y dx = (100 − x2) dx
dF = p dA = (90 − 7x)(100 − x2) dx
F dF= =−∫ 120 000
10
10
, lb (exactly)
b. dMy = x dF = x(90 − 7x)(100 − x2) dx
M dMy y= = −−∫
560000
310
10
lb-ft
c. x F M xy⋅ = ⇒ = −15
9ft
d. p ky y p y= = = =6
560 50 (because at )
dA x dy y dy= =2 2 100 –
dF p dA y y dy= = 12
5100 –
F dF= =∫ 64 0000
100
, lb (exactly)
e. dM y dF y y dyx = = 12
51002 –
M dMx x= =∫ 3 657 142 80
100
, , . K
≈ 3.657 million lb-ft exactly25600000
7
f. y F Mx⋅ = ⇒
y = ≈
57 1428 57 14
400
7. . ft exactlyK
7. a. Slice the region as shown in Figure 11-5f.At a sample point (x, t), d(dM2x) = t2 dx dt.
dM t dx dt t dt dxxt
t y
t
t y
22 2
00= =
=
=
=
=
∫∫= =
=
=1
3
1
33
0
3t dx y dxt
t y
= 1
30 25 4 4 1 3 3[ . ( – ) – ( – ) ]/x x dx
b. M dMx xx
x
2 24
4
0 53338
15= =
=
=
∫ . exactlyK–
8. a. Slice the region parallel to the y-axis so thateach point in a strip will have about the samepressure as at the sample point (x, y).y = e− x
p = kx2, dA = (1 − e− x) dxdF = p dA = kx2(1 − e− x) dx
F dF k= =∫ 0 95140
5
. Kln
exactly1
35
1
55
2
55
8
53 2(ln ) (ln ) ln+ + −
k
b. Slice the region parallel to the x-axis so thateach point in a strip will have about the samepressure as at the sample point (x, y).x = −ln y, p = ky− 1
dA = (ln 5 − x) dy = (ln 5 + ln y) dy= ln (5y) dy
dF = p dA = ky− 1ln (5y) dyx = ln 5 ⇒ y = e− l n 5 = 0.2
F dF k k= =
∫ 1 2951
1
25 2. exactly
0.2
1
K (ln )
c. Slice the region parallel to the y-axis. Thenslice a strip parallel to the x-axis as shown inFigure 11-5g.At sample point (x, t), p = kx2t− 1.d(dF) = p dA = kx2t− 1 dx dt
dF kx t dx dt kt dt x dxt y
t
t y
t
= =
− −
=
=
=
=
∫∫ 2 1 11
21
= = − ===( ) ( )k t x dx k y x dx kx dxt y
tln ln1 2 2 30
F kx dx k k= = =∫ 3 4
0
5 1
45 1 6774( ) .ln
ln
K
9. The integrals in Problems 7 and 8 can be writtenin the form
f x t dt dxt c
t d
x a
x b
( , )=
=
=
=
∫∫The result is called a double integral because twointegrals appear. (Hiding inside each integral is asecond integral!)
10. a. y x= 58
2tanπ
y x x= ⇒ = ⇒ = ±58
1 22tanπ
304 Problem Set 11-6 Calculus Solutions Manual© 2005 Key Curriculum Press
Slice the floodgate parallel to the y-axis.
dA x dx= −
5 5
82tan
π
A dA= = ≈
−∫ 14 5352 14 54 2
2
2
. .K ft
exactly 4080−
π
b. Slice the floodgate parallel to the x-axis sothat each point in a strip has about the samepressure as at the sample point (x, y).p = k(20 − y) with k = 62.4 lb/ft2
y x x y= ⇒ = −58
80 22 1tan tan .
ππ
dA x dy y dy= = −216
0 21
πtan .
dF p dA k y y dy= = − ⋅ −( )2016
0 21
πtan .
F dF k= =∫ 248 2628
0
5
. K
= 15491.6027… ≈ 15.49 thousand lb
exactly 8005200
3−
π
k
(The force can also be found by slicingparallel to the y-axis as in part a, then slicingthe strip horizontally and using a doubleintegral. In this case, the pressure at asample point (x, t) isp = k(20 − t)d(dF) = p dA = k(20 − t) dt dxThe first integration is from t = y to t = 5.The second integration is from x = −2 tox = 2.)
c. Let µ (Greek letter mu) = coefficient offriction.
µ µ⋅ = ⇒ =F 10000
10000
15491 6027. K= 0 6455. K
Problem Set 11-6
Q1.1
101101x C+ Q2. 0
Q3. x ln x − x + C Q4. 2 sin x cos x = sin 2x
Q5. (force)(displacement) Q6. y′ = 3(1 + 9x2)− 1
Q7. x = −2 Q8. 2 sec2 x tan x
Q9. ′′ = −y x9 3cos Q10. D
1. Partition the interval into small subintervalsof width dT so that C is about the same at anypoint in a subinterval. The amount of heat,dH, to raise the temperature by dT isdH = C dT = (10 + 0.3T1/2) dT.
H dH= =∫ 13 200100
,900
calories (exactly)
2. a. v(t) = 55 + 6 t − t2
v(0) = 55 + 6 · 0 − 02 = 55 mi/hv(3) = 55 + 6 · 3 − 32 = 64 mi/hv(6) = 55 + 6 · 6 − 62 = 55 mi/h, Q.E.D.
b. Cost of a short time, dt, at speed v isdC = 3(v − 55) dt = 18t − 3t2 dt.Total ticket cost is
C dC= =∫ 108 (exactly).0
6
Fine should be $108.00.
3. a. Cost per foot, P, = ax2 + bx + ca · 02 + b · 0 + c = 500 ⇒ c = 500
a b
a b
a
b
⋅ + ⋅ + =⋅ + ⋅ + =
⇒==
100 100 500 820
200 200 500 1180
0 002
3
2
2
.
P(x) = 0.002x2 + 3x + 500
b. P(700) = 0.002 · 7002 + 3 · 700 + 500= $3580/ft
c. Cost to dig a short distance, dx, isdC = P dx = (0.002x2 + 3x + 500) dx.Cost to dig 1000 feet is
( .0 002 3 5008
310002
0
10002x x dx+ + = ⋅∫ ) .
Cost is about $2,666,667.
d. Cost to dig 500 feet twice (once fromeach end) is
C x x dx= + + = ⋅∫2 0 002 3 50017
35002 2
500
( . ) .0
Cost is about $1,416,667.Savings is about $1,250,000!
4. a. velocity · area has the units in.
sin. ,2⋅
which is in.3/s, correct for flow rate.
b. v = 4 − x2 ⇒ v′ = −2xv′ changes from positive to negative at x = 0.∴ there is a maximum flow rate at the centerof the pipe where x = 0.(Or simply observe that the graph of v isa parabola opening downward with vertexat x = −b/(2a) = 0.)v(2) = 4 − 22 = 0, Q.E.D.
c. Slice the water in the pipe into cylindricalshells.Each point in a shell has about the samewater velocity as at the sample point x unitsfrom the axis.Let F = flow rate in in.3/s.dF = v dA = (4 − x2) · 2πx · dx
= 2π(4x − x3) dx
F x x dx= − = = …∫ 2 4 8 25 13273π π( ) .0
2
≈ 25.13 in.3/s
d. 25.1327… in.3/s · 60 s/min · 1 gal/231 in.3 =6.5279… ≈ 6.53 gal/min
Calculus Solutions Manual Problem Set 11-6 305© 2005 Key Curriculum Press
e. 4 in./s · π · 22 in.2 = 16π = 13.0559…≈ 13.06 gal/min (exactly twice the actualrate)
f. The problem is equivalent to finding thevolume of a solid of rotation by cylindricalshells. The velocity takes the place of thealtitude of a shell.
5. a.
600
5
x
F
b. F has a step discontinuity at x = 2.
c. dW = F dxBecause the graph is linear on [0, 2], the workequals the area of the triangle.
W = ⋅ ⋅ =1
22 600 600 in.-lb
d. W F dx= ∫2
5
By Simpson’s rule,
W ≈ + ⋅ + ⋅ + ⋅1
30 5 450 4 470 2 440 4 420( . )(
+ ⋅ + ⋅ +2 410 4 390 330)
= 12662
3in.-lb
e. Total work ≈ + =600 12662
31866
2
3 in.-lb
f. Yes, a piecewise continuous function such asthis one can be integrable. See Problem 27 inProblem Set 9-10.
6. a. Slice the solid into disks parallel to thexz-plane so that each point in a disk has aboutthe same density as at the sample point (x, y).
y = 4 − x2 ⇒ x2 = 4 − y
dm = ρ dV = k · π x 2 dy = kπ(4 − y) dy
m k y dy k= − =∫ π π( ) g4 80
4
b. Each point in a disk of part a is also about thesame distance from the xz-plane as the samplepoint (x, y).Let K stand for the constant.
dF = K · dm · y− 1/2 = K · kπ (4 − y) dy · y− 1/2
= Kkπ (4y− 1/2 − y1/2) dy
F dF Kk Kk= = =∫ 32
333 5103
0
4
π . K
7. Slice the solid into cylindrical shells so that eachpoint in a shell is about the same distance fromthe y-axis as the sample point (x, y).
dM2y = x2 dm = x2 · ρ dV = x2 · k · 2π xy dx= 2π kx3(4 − x2) dx
M dM ky y2 2
0
2 32
333 5103= = =∫ π . kg-cm2K
8. a. T(D) = 20 sin 2π DT(0) = 20 sin 0 = 0T(1/4) = 20 sin π /2 = 20, which checks.
b. Partition the time interval into shortincrements of width dD so that T is about thesame at any time in the increment as it is atthe sample point (D, T ) .Let H = number of degree-days.dH = T dD = 20 sin 2π D dD
H dH= = =∫ 10
3 18300
1 4
π. K
/
≈ 3.18 degree-days
9. a. m = 2000 − 5t (mass in kilograms, time inseconds)
b. a = F/m = 7000(2000 − 5t)− 1
= 1400(400 − t)− 1
c. adv
dt t= = 1400
400 –
dvt
dt=∫ ∫ 1400
400 –v = −1400 ln | 400 − t | + CAssume the car starts at rest at t = 0.0 = −1400 ln 400 + C ⇒ C = 1400 ln 400
v tt
( ) = 1400400
400ln
| – |
d. v( ) . . m/s20 1400
20
1971 8106 71 81= = ≈ln K
vds
dt t= = 1400
400
400ln
–
st
dt= =∫ 1400400
400711 9673
0
20
ln–
. K
≈ −
712 0 28000 1 1920
19. m exactly ln
10. Slice the tract parallel to the tracks so that eachpoint in the strip will have about the samevalue per square kilometer as at the samplepoint (x, y).Let v = thousands of dollars per square kilometerand W = thousands of dollars the land is worth.v = kx = 200x (v = 200 at x = 1)dW = v dA = 200x[(4 − x2) − (4x − x2)] dx
= 800(x − x2) dxThe curves intersect at x = 1.
W x x dx= − =∫ 800 1331
32
0
1
( )
306 Problem Set 11-6 Calculus Solutions Manual© 2005 Key Curriculum Press
The land is worth about $133,333.If all the land were worth $200,000 per km2,
W A x dx= = =∫200 200 4 4 4000
1
( – ) .
The land would be worth $400,000.Actual value is about $267,000 less.
11. a. Slice the tract parallel to the y-axis so thateach point in a strip will be about the samevalue per square unit as at the samplepoint (x, y).y = cos xLet v = value of land per square unit andW = worth of the land.v = kx, dA = y dx = cos x dxdW = v dA = kx cos x dx
W dA k k= = −
=∫ ππ
21 0 5707
0
2
. K/
b. Slice the tract parallel to the x-axis so thateach point in a strip will be about the samevalue per square unit as at the samplepoint (x, y).v = kydW = v dA = v · x dy = ky · cos− 1 y dy
W dW k k= = =∫ π8
0 39260
1
. K
12.
(x, y)
x
3
9y
Slice the wall parallel to the ground so thateach point in the slice will cost about thesame to paint per square meter.Let r = rate in dollars per square meter andC = cost in dollars to paint the wall.r = ky2 = 3y2 (r = 12 when y = 2)dA x dy y dy= =2 2 9 –
dC r dA y y dy= = ⋅3 2 92 –
C dC= ≈
∫ $ . exactly 1999 54 1999
19
350
9
13. a. Let v = value of land per square kilometer,W = worth of the land in dollars, andr = distance from center of town.Slice the city into circular rings of width dr sothat each point in a ring will be about r unitsfrom the center.v r dA r dr= − =10 3 2, πdW v dA r r dr= = − ⋅ ( ) 10 3 2π
W dW= = =∫ 36 113 0973
0
3
π . K
≈ 113.1 million dollars
b. v e v kkr= = ⇒ = − ⇒10 3 11
310, ( ) ln
v e r= −10 10 3(ln ) /
dW v dA e r drr= = ⋅−10 210 3( ) /ln π
W dW= = ≈∫ 71.4328 71.4 millionK
0
3
dollars exactly ( )18
109 102
π(ln )
– ln
c. By Simpson’s rule,
W v r dr= ⋅ ≈∫ 21
30 3 2
0
3
π π( . )( ) (10 + 4 ⋅ 12
+ 2 ⋅ 15 + 4 ⋅ 14 + 2 ⋅ 13 + 4 ⋅ 10+ 2 ⋅ 8 + 4 ⋅ 5 + 2 ⋅ 3 + 4 ⋅ 2 + 1)
= 52.2π = 163.9911… ≈ 164.0 milliondollars
d. This problem is equivalent to volume bycylindrical shells, where the value of the landper square unit takes the place of the altitudeof the cylinder. It is also equivalent to thewater flow in Problem 4 of this problem set.
e. Answers will vary.
14. a. p = 100[(x − 8)1/2 − 0.5(x − 8)]dF = p dA = 2p dx
= 200[(x − 8)1/2 − 0.5(x − 8)] dx
F dF= = ≈∫ 177 1236 177
8
10
. lbK
exactly800 2
3200−
b. Average pressure = total force/total area
= = ≈177 1236
444 2809 44 3 2. K
K. . lb/ft
exactly200 2
350−
c. dMyz = x dF = 2px dx= 200x[(x − 8)1/2 − 0.5(x − 8)] dx
M dMyz yz= = ≈∫ 1602 8706 1603
8
10
. lb-ftK
exactly7360 2
3
5600
3−
d. x F M xyz⋅ = ⇒ = 1602 8706
177 1236
.
.
K
K= 9.0494… ft
Calvin should stand about 10 − 9.0494 ft
≈ 111
2in. from the end.
15. a. f (x) = 9 − x2 = (3 − x)(3 + x) = 0 only atx = ±3.
g x x x x
x x x
( )
only at .
= − − + +
= − + = = ±
1
33 9
1
33 3 0 3
3 2
2( – )( )
Calculus Solutions Manual Problem Set 11-7 307© 2005 Key Curriculum Press
b. A x dxf = =∫ ( – )–
9 362
3
3
A x x x dxg = + +
=∫ – –
–
1
33 9 363 2
3
3
To simplify algebraic integration, you coulduse
A x dxf = ∫2 9 2
0
3
( – )
A x dxg = ∫2 9 2
0
3
( – ) , where the odd terms
integrate to zero between symmetrical limits.Thus, the two integrals are identical.
c. The high point of f comes at x = 0.The high point of g comes where g′(x) = 0.g′(x) = −x2 − 2x + 3 = −(x + 3)(x − 1)g′(x) = 0 ⇔ x = −3 or x = 1The high point is at x = 1.
d. Slice the region under the g graph parallel tothe y-axis so that each point in a strip will beabout the same distance from the y-axis as thesample point (x, y).
dMy = x dA = x g(x) dx
= − − + +
1
33 94 3 2x x x x dx
M dMy y= =∫ 21 63
3
.–
x A M xy⋅ = ⇒ = =21 6
360 6
..
e. False. For the symmetrical region under thegraph of f, the centroid is on the line throughthe high point. But for the asymmetricalregion under the graph of g, the high point isat x = 1 and the centroid is at x = 0.6.
f. False
Area to left = =∫ g x dx( ) .17 10723
0 6
–
.
(exactly)
Area to right = =∫ g x dx( ) .18 89280 6
3
. (exactly)
(or 36 − 17.1072 = 18.8928)
g. Let S stand for skewness.
dS x dA x g x dx= −
= −
3
5
3
5
3 3
( )
S x x x x dx=
− − + +
= − ⋅⋅
∫ –
–
–
3
5
1
33 9
17 773764 3
7 125
33 2
3
3
5
. exactlyK
h. By symmetry, the centroid of the area under fis on the y-axis, so x = 0. ThendS = x3 dA = x3(9 − x2) dx
S x x dx= − =∫ 3 2
3
3
9 0( )–
(odd function
integrated between symmetrical limits)
The “skewness” being zero reflects thesymmetry of this region. It is not skewedat all.
i. For example, graph
g x x x x( ) .− = − − +1
33 93 2
gNewgraph
y
x
3–3
16. a. y = x2
dL dx dy x dx= + = +2 2 21 4
dM x dL x x dxy = = + 1 4 2
b. M dMy y= =∫ 5 7577
0
2
. K
exactly1
1217 17 1( – )
c. L x dx= + =∫ 1 4 4 64672
0
2
. K
exactly1
417 4 17ln ( )+ +
d. x L M xy⋅ = ⇒ = =5 7577
4 64671 2390
.
.
K
KK.
e. dS x dL x x dx= = +2 2 1 4 2π π
S dS= =
∫ 36 1769
617 17 1
0
2
. exactlyKπ
( – )
f. Integral for S is 2π times the integral for My!
17. In Problem 16 1 2390, .R x= = K andL = 4.6467… .2π RL = 2π (1.2390…)(4.6467…) = 36.1769… ,which equals S, Q.E.D.
18. The centroid of the small circle is at its center,R units away from the axis.The arc length L of the small circle is 2πr.Surface area S = 2π R(2πr) = 4π 2rR
Problem Set 11-7Review Problems
R0. Answers will vary.
R1. Slice the region parallel to the force axis so thateach point in a strip has about the same force asat the sample point (x, F ) .
dW = F dx = 30e− 0.2 x dx
W e dx ex= = −− −∫ 30 150 10 2 2
0
10. ( )
= 129.6997… ≈ 129.7 ft-lb
308 Problem Set 11-7 Calculus Solutions Manual© 2005 Key Curriculum Press
R2. a. dW = F dx = kx− 2 dx
W kx dx k= = −−∫ 2
3
1 2
3ft-lb
(Mathematically, the answer is negativebecause dx is negative. Physically, the answeris negative because the magnets absorbenergy from their surroundings rather thanreleasing energy to their surroundings.)
b. Construct axes with the origin at the vertex ofthe cone. An element of the cone in the xy-
plane has the equation y x x y= =7
3
3
7 or .
Slice the water horizontally into disks so thateach point in a disk is lifted about the samedistance as the sample point (x, y) on theelement of the cone.
F = 0.036 dV = 0.036 ⋅ πx2 dy
= ⋅0 0369
492. π y dy
Each disk is lifted (10 − y) cm.dW = (10 − y) dF
= −( )( . )10 0 0369
492y y dyπ
W dW= = = …∫0
7
3.591 11.2814π
≈ 11.28 in.-lb
R3. a. The graph shows the region in Quadrant Iunder the graph of y = 8 − x3 rotated aboutthe y-axis.
2
8
(x, y)
x
y
Slice the region parallel to the x-axis,generating disks, so that each point in a diskis about the same distance from the xz-planeas the sample point (x, y).
ρ = ky, dV = π x2 dy = π(8 − y)2/3 dy
m k y y dy k= − =∫ π π ( ) ./8 57 62 3
0
8
b. Slice the region parallel to the y-axis,generating cylindrical shells, so that eachpoint in a shell is about the same distancefrom the y-axis as the sample point (x, y).
ρ = ex, dV = 2π xy dx = 2π(8x − x4) dx
m e x x dxx= − =∫ 2 8 644
0
2
π π ( )
R4. a. The width of a strip at the samplepoint (x, y) is
w bb
hy dA b
b
hy dy= − ⇒ = −
dM y dA byb
hy dyx = = −
2
M byb
hy dy by
b
hyx
h h
=
= −∫ – 2
0
2 3
0
1
2 3
= − − + =1
2 30 0
1
62 3 2bh
b
hh bh
y A y bh Mx⋅ = ⋅ =1
2
⇒ = =ybh
bhh
1612
1
3
2
, Q.E.D.
b. The graph shows the region under y = ex
rotated about the y-axis, showing back half ofsolid only.
(x, y)
x
y
1
1
Slice the region parallel to the y-axis,generating cylindrical shells, so that eachpoint in a shell will be about the samedistance from the y-axis as the samplepoint (x, y).
dV = 2πx ⋅ y ⋅ dx = 2πxex dx
dM2y = x2 dV = 2πx3ex dx
M x e dxyx
23
0
1
2 3 5401= =∫ π . K
(exactly 12π − 4πe)
R5. Draw axes with the x-axis at ground level and they-axis through the upper vertex of the triangle.Slice the face of the building horizontally so thatthe wind pressure at any point in a strip is aboutequal to the pressure at the sample point (x, y).
dA y dy= −
150
150
400
dF p dA e y dyy= = ⋅ − −
− ( ).200 150 1 11
4000 01
F dF= = …∫0
400
3736263.2708
≈ 3.736 million lb(exactly 30000(125 − 25e− 4))
Calculus Solutions Manual Problem Set 11-7 309© 2005 Key Curriculum Press
R6. a. Let x = number of feet at which drill isoperating, and r(x) = number of dollars perfoot to drill at x feet.r(x) = a · bx
r (0) = 30 ⇒ a = 30
50 305
310000
110000
= ⋅ ⇒ =
b b
/
∴ =
( )
/
r xx
305
3
10000
(or ( ) . / .r x e ex x) ln= =− ⋅30 300 6 10000 0 00005108256K
b. dC r x dx dxx
= =
( )
/
305
3
10000
C dx
x
=
=∫ 30
5
36965243 17
10000
0
50000 /
. K
≈ 6.965 million dollars
exactly − ⋅
30
10000
0 6
5
31
5
ln .–
Concept ProblemsC1. a. Either slice the region parallel to the y-axis,
dA = (8 − y) dx = (8 − x3) dx
A x dx= =∫ ( – )8 123
0
2
or slice parallel to the x-axis,
A y dy= =∫ 1 3
0
8
12/
b. i. Use slices parallel to the x-axis so thateach point in a strip will be about thesame distance from the x-axis as thesample point (x, y).
dMx = y dA = y(y1/3 dy)
M y dyx = = =∫ 4 3
0
8 384
754 8571/ . K
ii. Use slices parallel to the y-axis so thateach point in a strip will be about thesame distance from the y-axis as thesample point (x, y).
dMy = x dA = x(8 − x3) dx
M x x dxy = =∫ ( – )8 9 64 .0
2
c. x A M xy⋅ = ⇒ = =9 6
120 8
..
y A M yx⋅ = ⇒ = = =384 7
12
32
74 5714
/. K
Centroid is at (0.8, 4.5714…).
d. i. With slices parallel to the x-axis,
dV = 2π y · x · dy = 2π y4/3 dy
V y dy= = =∫ 2
768
7344 67754 3
0
8
π π/ . K
With slices perpendicular to the x-axis,
dV = π (82 − y2) dx = π (64 − x6) dx
V x dx= − = =∫ π π( ) .64
768
7344 67756
0
2
K
ii. With slices parallel to the y-axis,
dV = 2π x · (8 − y) · dx = 2π x(8 − x3) dx
V x x dx= − =∫ 2 8 19 24
0
2
π π( ) .
= …60 3185.
With slices perpendicular to the y-axis,
dV = π x2 dy = π y2/3 dy
V y dy= = =∫ π π2 3
0
8
19 2 60 3185/ . . K
iii. With slices parallel to the line x = 3,
dV = 2π (3 − x) · (8 − y) · dx= 2π (3 − x)(8 − x3) dx
V x x dx= − − =∫ 2 3 8 52 83
0
2
π π( )( ) .
= 165.8760…
With slices perpendicular to the linex = 3,
dV = π [32 − (3 − x)2] dy
= π [9 − (3 − y1/3)2] dy
V y dy= − − =∫ π π[9 3 52 81 3
0
8
( ) ] . / 2
= 165.8760…
e. i. The centroid is 32/7 units from the x-axis.
∴ = ⋅ ⋅ = V 232
712
768
7π π
= 344.6775… (Checks.)
ii. The centroid is 0.8 unit from the y-axis.
V = 2π · 0.8 · 12 = 19.2π= 60.3185… (Checks.)
iii. The centroid is 3 − 0.8 = 2.2 units fromthe line x = 3.V = 2π · 2.2 · 12 = 52.8π = 165.8760…(Checks.)
f. Use horizontal slices so that each point in adisk will be about the same distance from thexz-plane as the sample point (x, y).
dMxz = y dV = y(π x2 dy) = y · π y2/3 dy
M y dyxz = = =∫ π π5 3
0
8
96 301 5928/ . K
g. y V M yxz⋅ = ⇒ = =96
19 25
ππ.
x z= = 0 by symmetry.
Centroid is at (0, 5, 0).
310 Problem Set 11-7 Calculus Solutions Manual© 2005 Key Curriculum Press
h. No. For the solid, y = 5, but for the region,y = …4.5714 .
i. Use slices of the region parallel to the y-axisso that each point in a resulting cylindricalshell will be about the same distance from they-axis as the sample point (x, y).
ρ = kx2,dV = 2πx(8 − y) dx = 2π(8x − x4) dx
dm = ρ dV = kx2 · 2π (8x − x4) dx
= 2π k(8x3 − x6) dx
m k x x dx k= − =∫ 2 8192
73 6
0
2
π π( )
= 86.1693…k
j. Use cylindrical shells as in part i so that eachpoint in a shell will be about the samedistance from the y-axis as the samplepoint (x, y).
dM2 = x2 · dm = 2π k(8x5 − x8) dx
M k x x dx k25 8
0
2
2 8512
9= − =∫ π π( )
= 178.7217…
k. Use vertical slices of the region so that eachpoint in a strip will have about the samepressure acting on it as at the samplepoint (x, y).
p = 3 − x, dA = (8 − y) dx = (8 − x3) dx
dF = p dA = (3 − x)(8 − x3) dx
F x x dx= =∫ .0
2
( – )( – )3 8 26 43
(Note the similarity to the integral inpart d.iii.)
l. F = kz− 2
F = 26.4 at z = 1 ⇒ k = 26.4 ⇒ F = 26.4z− 2
dW = F dz = 26.4z− 2 dz
W z dz= =−∫ 26 4 17 62
1
3
. .
m. Use horizontal slices so that each point in aresulting disk will be at about the sametemperature as the sample point (x, y).dH = CT dm = 0.3(10 − y)(5.8π y2/3 dy)
H y y dy= − =∫ 1 74 10 167 042 3. ( )( ) .
/
0
8
π π
= 524.7716… ≈ 524.8 cal
C2. Let f (x) be the height of a vertical strip at x (orcombined heights if the region being rotated isnot convex). Let x = a and x = b be the left andright boundaries of the region.
dV x dA x f x dx V x f x dxa
b
= = ⇒ = ∫2 2 2π π π ( ) ( )
dM x dA x f x dx M x f x dxy ya
b
= = ⇒ = ∫ ( ) ( )
Note that V = 2π My, thus showing that the twoproblems are mathematically equivalent, Q.E.D.
C3. dMxz = y dVxz = yπ x2 dy = π y(9 − y) dy
M y y dyxz = − =∫π π π ( ) .0
9
9 121 5
dM2y = x2 dVy = x2 2π xy dx = 2π x3(9 − x2) dx
M x x dxy23 22 9 121 5= − =∫ π π ( ) . ,
0
3
Q.E.D.
This is not true in general. Counterexample:Rotate the region under y = 2 − 2x2.
dM y dV y x dy y y dyxz xz= = = −
π π
2 11
2
M y y dyxz = −
=∫ π π
0
2
11
2
2
3dM2y = x2 dVy = x2 2π xy dx = 2π x3(2 − 2x2) dx
M x x dxy23 22 2 2
1
3
2
3= − =∫ π π π ( ) , not .
0
1
General proof: For any paraboloid of height Hand base radius R, let h = distance (along theaxis) from the base and r = radius. Then a
generating parabola is given by h HH
Rr= − ⋅2
2.
dM h dV h r dh h H hR
Hdhbase ( )= = = −π π2
2
MR
HHh h dh
h
h H
base ==
=
∫2
2
0π ( – )
= −
=
=
=R
H
Hh h R H
h
h H22 3
0
2 2
2
1
3
1
6π π
dM r dV r rh dr2axis = =2 2 2π
= −
r r H
H
Rr dr2
222π
M H rR
r drr
r R
23
25
02
1axis =
=
=
∫π –
= −
=
=
=
21
4
1
6
1
64
26
0
4π πH rR
r R Hr
r R
In the original example, H = R2, so the twomoments turned out to be equal.
C4. a. Assume m ≠ 0.The area of the trapezoid is
Ab b
hma mb
b a= + ⋅ = +1 2
2 2( – )
= −1
22 2m b a( )
Integrating, y dx mx dx mxa
b
a
b
a
b
≈ =∫∫ 1
22
= − =1
22 2m b a A( ) , Q.E.D.
The length is L b a m= − +( ) .1 2
Integrating, dL dx b a La
b
a
b
≈ = − ≠∫∫ ( ) ,
Q.E.D.
Calculus Solutions Manual Problem Set 11-7 311© 2005 Key Curriculum Press
b. Note that r = mh.The volume of the cone is
V r h m h= =1
3
1
32 2 3π π .
Integrating dV ≈ π y2 dx = π m2x2 dx,
π π ,m x dx m h Vh
2 2 2 3
0
1
3= =∫ Q.E.D.
The surface area is S r r h= + =π
2 2
π .mh m2 21+Integrating dS ≈ 2π y dx = 2π mx dx,
2 2
0π π ,mx dx mh S
h
= ≠∫ Q.E.D.
c. Exact area of a strip:
∆ ∆ ∆ ∆ ∆A mx m x x x y x y x= + + ∆ = +1
2
1
2( ( ))
Exact volume of frustum:
∆ ∆
∆ ∆
∆ ∆ ∆ ∆
V m x x m x x x m x x
m x x x x x
y x y y x y x
= + ∆ + + ∆ +
= + + ∆
= + + ∆
π
π
π
3
33 3
1
3
2 2 2 2 2
2 2 2
2 2
( ( ) ( ) )
( )
( )
( ( ) )
d. dA y dx y dx y dx y dx− = +
−1
2∆
= 1
2∆y dx
dV − π y2 dx
= + +
−π πy dx y y dx y dx y dx2 2 21
3∆ ∆
= +π y y dx y dx∆ ∆1
32
Both differences contain only higher-orderinfinitesimals.
e. If dQ = ∆Q leaves out only infinitesimals of
higher order, then dQa
b
∫ is exactly equal to Q.
f. Reasons:
i. 0.5 and ∆y are constant with respect to thesummation, so they can be pulled out.
ii. The sum of all the subsegments ∆x of[a, b] must be b − a, the whole interval.
iii. ∆y has limit zero as ∆x goes to zero.
Chapter TestT1. a. force · displacement
b. mass
c. force
d. area · displacement
e. second moment of volume
f. x
T2. dW = (40x − 10x2) dx
W x x dx= − =∫ ( )40 10 902
1
4
T3. y m M yxz⋅ = ⇒ ⋅ =200 3000∴ = = 3000/200 15 cmy
T4. The center of the circle is (8, 9) and the radiusis 7, so the circle is on just one side of the axisof rotation (the y-axis). So the solid satisfies thehypothesis of the theorem of Pappus.The centroid of the circle is (8, 9), thedisplacement from the y-axis is R = 8, and thearea of the circle is 49π.∴ = = = = ( )( )( ) V RA2 2 8 49 784π π π π2463.0086…
T5. Using exponential regression,F ≈ 29.9829… (1.0626…)x
dW = F dx
W F dx= ≈ … ≈∫ 412 4652 412 5. . ft-lb0
10
(By the trapezoidal rule, W ≈ 413 ft-lb.Simpson’s rule cannot be used because there isan odd number of increments.)
T6. a.
(x, y)
x
y
2
1
Slice the region parallel to the y-axis so that eachpoint in a strip will be about the same distancefrom the y-axis as the sample point (x, y).dMy = x dA = xex dx
M xe dx eyx= = + = … ≈∫ 2 3
0
2
1 8 3890 8 39. . in.
b. dM2y = x2 dA = x2ex dx
M x e dx eyx
22 2
0
2
2 2 12 7781
12 78
= = − = …
≈
∫ .
. in.4
c. A e dx ex= = − = … ≈∫ 2 1 6 3890 6 39. . in.20
2
x A M xe
ey⋅ = ⇒ = + = … ≈2
2
1
11 3130 1 31
–. . in.
T7. The graph shows y = x1/2 from x = 0 to x = 16,rotated about the x-axis.
4
16x
y(x, y)
312 Problem Set 11-7 Calculus Solutions Manual© 2005 Key Curriculum Press
Slice the region parallel to the x-axis so that eachpoint in a resulting cylindrical shell will beabout the same distance from the x-axis as thesample point (x, y).ρ ρ π= = = ⋅ −3 3 2 16y dm dV y x y dy, ( )= 6π y2(16 − y2) dy
m y y dy= − =∫ 6 16 819 22 4π π( ) . 0
4
= 2573.5927… ≈ 2573.6 g
T8. a. Slice the end of the trough parallel to thex-axis so that each point in a strip has aboutthe same pressure acting on it as the samplepoint (x, y), where x ≥ 0.p = 62.4(8 − y), dA = 2x dy = 2y1/3 dydF = p dA = 62.4(8 − y) · 2y1/3 dy
F y y dy= − ⋅
= ⋅
= … ≈
∫ 62 4 8 2
62 464 9
75134 6285 5134 6
1 3. ( )
. . . lb
/
0
8
b. dMx = y dF = 62.4(8 − y) · 2y4/3 dy
F y y dy= − ⋅ = ⋅ ⋅∫ 62 4 8 29 2
3562 44 3
10
. ( ) ./
0
8
= 16430.8114… ≈ 16.43 thousand lb-ft
y F M yx⋅ = ⇒ = =16430 8114
5134 62853 2
.
.
K
K. ft
x = 0 by symmetry.Center of pressure is at (0, 3.2).
T9. a. Slice the seating area into concentric rings ofwidth dr. Each point in a ring will be aboutthe same distance from the center as thesample point.Let W = worth of the seating and v = valueper square foot.dW = v dA = 150r− 1 · 2π r dr = 300π dr
W dr bb
= = −∫ 300 300 3030
π π ( ) dollars
b. 300π (b − 30) = 60000 ⇒
b = + = … ≈30200
93 6619 93 7π
. . ft
Calculus Solutions Manual Problem Set 12-2 313© 2005 Key Curriculum Press
Chapter 12—The Calculus of Functions Definedby Power Series
Problem Set 12-1
1. f xx
( ) ,= 6
1–P x x x x x x5
2 3 4 56 6 6 6 6 6( ) = + + + + +
1–1
100
–100
x
yf P5
f
P5
The graph of P5 fits the graph of f reasonablywell for about −0.8 < x < 0.6.The graph of P5 bears no resemblance to thegraph of f at x = 2 and at x = −2, for example.
2. P6(x) = P5(x) + 6x6
1–1
100
–100
x
yf P5
f
P5
P6
The graph of P5 fits the graph of f slightly better,perhaps for −0.9 < x < 0.7.
3. P5(0.5) = 11.8125, P6(0.5) = 11.90625,f (0.5) = 12∴ P6(0.5) is closer to f (0.5) than P5(0.5) is.P5(2) = 378, P6(2) = 762, f (2) = −6∴ P6(2) is not closer to f (2) than P5(2) is.
4. Possible conjecture: P(x) converges to f (x) for−1 < x < 1, or perhaps for −1 ≤ x ≤ 1.
5. P0(1) = 6 P0(−1) = 6P1(1) = 12 P1(−1) = 0P2(1) = 18 P2(−1) = 6P3(1) = 24 P3(−1) = 0P4(1) = 30 P4(−1) = 6
For x = 1, the sums just keep getting larger andlarger as more terms are added. For x = −1, thesums oscillate between 0 and 6. In neither casedoes the series converge. If the answer toProblem 4 includes x = 1 or x = −1, theconjecture will have to be modified.
6. P5(0.5) = 11.8125 and f (0.5) = 12The values differ by 0.1875, and6(0.5)6 = 0.09375.P5(−0.5) = 3.9375 and f (−0.5) = 4
The values differ by 0.0625, and6(−0.5)6 = −0.09375.For P5(0.5), the difference is greater than thevalue of the next term of the series. This result isto be expected because the rest of the seriesis formed by adding more positive terms.For P5(−0.5), the difference is less in absolutevalue than the absolute value of the next term.This result is reasonable because the termsalternate in sign so that you are adding andsubtracting ever smaller quantities.
7. A geometric series; the common ratio
Problem Set 12-2Q1. . . . for any ε > 0 there is a D > 0 such that
if x > D, then f (x) is within ε units of L.
Q2. the fundamental theorem of calculus
Q3. the fundamental theorem of calculus
Q4. the mean value theoremQ5. derivative Q6. cos x − x sin x
Q7. x sin x + cos x + C Q8. dA r d= 1
22 θ
Q9. f (x) = e− x Q10. D
1. Series: 200 − 120 + 72 − 43.2 + 25.92 −15 552. + LSums: 200, 80, 152, 108.8, 134.72, 119,168, …
100
10
125
Sn
n
S = ⋅+
=2001
1 0 6125
.The series converges to 125.
| |125 125 2001 0 6
1 0 6− = − ⋅
+Sn
n– (– . )
.
= − ⋅125 1 1 61 0 6
1 6.
– (– . )
.
n
= 125|1 − (1 − (−0.6)n)|
= 125|−0.6n| = 125(0.6n)
125 0 6 0 0001
0 0001125
0 627 48( . ) . .n n= ⇔ = =ln . /
ln .K
Make n ≥ 28.Sn will be within 0.0001 unit of 125 for allvalues of n ≥ 28.
314 Problem Set 12-2 Calculus Solutions Manual© 2005 Key Curriculum Press
2. Series: 30 + 33 + 36.3 + 39.93 + 43.923 +48.3153 + K
Sums: 30, 63, 99.3, 139.23, 183.153,231.4683, …
10
100
Sn
n
The graph shows divergence.
S 30 4,133,883.70 (Wow!)100 = ⋅ =1 1 1
1 1 1
100– .
– .K
The formula S = t1/(1 − r) gives −300 for S, butit has no meaning because the series does notconverge.
3. a. Series:
7(0.8 ) 7 5.6 4.48 3.584n
n
−
=
∞
= + + + +∑ 1
1
L
Sums: 7, 12.6, 17.08, 20.664, 23.5312, …S4 = 20.664, so the amount first exceeds20 µg at the fourth dose.
S = ⋅ =71
1 0 835
– ., so the total amount
never exceeds 40 µg.The graph confirms that the partial sums ofthe series approach 35 asymptotically andfirst exceed 20 µg at the fourth dose.
1 2 3 4 5 6 7 8 9 10
10
20
30
40
Asymptote
n
Goes > 20
Stays > 20
b. tn = Sn − 7, so the sequence is 0, 5.6, 10.08,13.664, 16.5312, … .See the graph in part a. The open circles showthe partial sums just before a dose.t7 = 20.6599… . The amount remainsabove 20 µg for n ≥ 7.
c. See the graph in part a.
4. a. Perimeters are 16, 16 0 5. , 16(0.5), … ,which is a geometric sequence with t1 = 16and .r = 0 5.
b. S10
5
1 2161 0 5
1 0 552 9203= ⋅ =– .
– . / . cmK
c. The total perimeter converges to
16
1
1 0 554 62741 2⋅ =
– . / . cm.K
d. The sum of the areas is 16 8 4 2+ + + +L ,which is a convergent geometric series withr = 0.5.
S = ⋅ =161
1 0 532
– . cm2
5. a. The interest rate for one month is0.09/12 = 0.0075.
Months Dollars
0 1,000,000.00
1 1,007,500.00
2 1,015,056.25
3 1,022,669.17
b. Worth is (1,000,000)(1.007512) =$1,093,806.90; interest is $93,806.90.
c. The first deposit is made at time t = 0, thesecond at time t = 1, and so forth, so at timet = 12, the term index is 13.
d. Meg earned $93,806.90 the first year.
APR . %= ⋅ =93806 90
1000000100 9 3806
.K
e. (1,000,000)(1.0075n) = 2,000,000
n = =ln
ln .
2
1 007592 7657. K
After 93 months
6. a. The interest rate for one month is0.108/12 = 0.009.S5 = 100 + 100(1.009) + 100(1.009)2
+ 100(1.009)3 + 100(1.009)4
+ 100(1.009)5
= ⋅ =1001 1 009
1 1 009613 66
6– .
– .$ .
b. There are six terms because the term index ofthe first term is zero.
c. 10 years equals 120 months. There will havebeen 121 deposits after 10 years because theinitial deposit was made at time 0. So thereare 121 terms.
S120
121
1001 1 009
1 1 00921 742 92= ⋅ =– .
– .$ , .
The principal is 121(100) = $12,100.The interest is 21,742.92 − 12,100 =$9,642.92.
7. a. Sequence: 20, 18, 16.2, 14.58, 13.122, …
b. S4 = 20 + 18 + 16.2 + 14.58 = 68.78 ft
c. S = ⋅ =201
1 0 9200
– .So the ball travels 200 ft before it comes torest.
Calculus Solutions Manual Problem Set 12-3 315© 2005 Key Curriculum Press
d. For the 10-ft first drop, 10 = 0.5(32.2)t2.t = (10/16.1)1/2 = 0.7881…The total time for the 20-ft first cycle is2(0.7881…) = 1.5762… s.For the 18-ft second cycle,t = 2(9/16.1)1/2 = 1.4953… s.
e. The times form a geometric series with firstterm 1.5762… and common ratio equal to0.91/2 = 0.9486… . So the series of timesconverges to
S = ⋅ =1 5762
1
1 0 930 71551 2. .K K
– . /
The model predicts that the ball comes to restafter about 30.7 s.
8. a.
Iteration Total Length
0 27
1 36
2 48
3 64
Because each segment is divided into fourpieces, each of which is 1/3 of the originallength, the length at the next iteration can becalculated by multiplying the previous lengthby 4/3.
b. The sequence of lengths diverges because thecommon ratio, 4/3, is greater than 1. Thus,the total length of the snowflake curve isinfinite!
c. From geometry, the area of an equilateral
triangle of side s is A s= 3
42.
The number of triangles added is 3, 12, 48,192, … .The side of each added triangle is 3, 1, 1/3,1/9, … .The added areas form the series
3
43 3 12 1 48 1 3 192 1 9[ ( ) ( ) ( / ) ( / ) ]2 2 2 2+ + + + L
= ⋅ ⋅ + +3
43 9 4 1 3 4 1 3 4 1 32 0 1 2[ ( / ) ( / ) ( / )2 4 6
+ +4 1 33 8( / ) ]L
= ⋅ ⋅ + + + +3
163 9 4 9 4 9 4 9 4 92[ / ( / ) ( / ) ( / ) ]2 3 4 L
= ⋅ ⋅ ⋅ =3
163 9 4 9
1
1 4 912 15 32 ( / ) .
– /The area of the pre-image is
3
49 20 25 32⋅ = . .
The total area is 32 4 3 56 1184. . cm .2= K
9. f xx
( ) = 6
1–
P x x x x x x( ) = + + + + + +6 6 6 6 6 62 3 4 5 L
′ = + + + + +P x x x x x( ) 6 12 18 24 302 3 4 L
′′ = + + + +P x x x x( ) 12 36 72 1202 3 L
′′′ = + + +P x x x( ) 36 144 360 2 L
f ′(x) = 6(1 − x)− 2
f ″(x) = 12(1 − x)− 3
f ″′(x) = 36(1 − x)− 4
P′(0) = 6 and f ′(0) = 6P″(0) = 12 and f ″(0) = 12P″′(0) = 36 and f ″′(0) = 36Conjecture: P fn n( ) ( )( ) ( )0 0= for all values of n.
Problem Set 12-3Q1. 0.3333… Q2. 0.4444…
Q3.2
3Q4.
4
9Q5. 13 Q6. 125
Q7. mass ⋅ displacement Q8. centroid
Q9. ln x + C Q10. D
1. f (x) = 5e2x
f ′(x) = 10e2x
f ″(x) = 20e2x
f ′′′(x) = 40e2x
f (4)(x) = 80e2x
2. P1(x) = c0 + c1x and ′ =P x c1 1( )P1(0) = c0 and f (0) = 5 ⇒ c0 = 5
′ =P c1 10( ) and f ′(0) = 10 ⇒ c1 = 10
∴ P1(x) = 5 + 10x
3. P2(x) = c0 + c1x + c2x2
′ = + ′′ =P x c c x P x c2 1 2 2 22 2( ) and ( )
P2(0) = c0 and f (0) = 5 ⇒ c0 = 5′ =P c2 10( ) and f ′(0) = 10 ⇒ c1 = 10′′ =P c2 20 2( ) and f ′′(0) = 20 ⇒ 2c2 = 20
⇒ c2 = 10∴ P2(x) = P1(x) = 5 + 10x + 10x2
c0 and c1 are the same as for P1(x).
4. P3(x) = c0 + c1x + c2x2 + c3x
3
P4(x) = c0 + c1x + c2x2 + c3x
3 + c4x4
′′′ = + =P x c c x P x c4 3 4 44
46 24 24( ) and ( )( )
′′′ = ′′′ =P c f4 30 6 0 40( ) and ( )
⇒ = ⇒ =6 4020
33 3c c
P c f44
440 24 0 80( ) ( )( ) and ( )= =
⇒ = ⇒ =24 8010
34 4c c
c0, c1, and c2 are the same as before.
5. P x x x x32 35 10 10
20
3( ) = + + +
P x x x x x42 3 45 10 10
20
3
10
3( ) = + + + +
316 Problem Set 12-4 Calculus Solutions Manual© 2005 Key Curriculum Press
f P4
P3
P4
P31
100y
x
6. P4 is indistinguishable from f for about−1 < x < 0.9.
7. P3(1) = 31.6666666…P4(1) = 35.0000000…f (1) = 5e2 = 36.9452804…∴ P4(1) is closer to f (1) than P3(1), Q.E.D.
8. c4
480
24
5 2
4= = ⋅
!The 5 is the coefficient in 5e2x.The 2 is the exponential constant.The 4 is the exponent of x in the last term.
9. c c3
3
2
220
6
5 2
3
20
2
5 2
2= = ⋅ = = ⋅
! !, ,
c c1
1
0
010
1
5 2
15
5 2
00 1= = ⋅ = = ⋅ =
! !( ! ),
10. Conjecture:
c c5
5
6
65 2
5
160
120
4
3
5 2
6
320
720
4
9= ⋅ = = = ⋅ = =
! !,
11. P xn
xn
n
n( ) = ⋅
=
∞
∑ 5 2
0!
Problem Set 12-4Q1. Q2.
x
y
x
y
Q3. Q4.
x
y
x
y
Q5. Q6.
x
y
y
x
Q7. exponent Q8. coefficient
Q9. power Q10. D
1. a.
n f xn( ) ( ) f n( ) ( )0 P n( ) ( )0 cn
0 e x 1 c0 1
1 e x 1 c1 1
2 e x 1 2!c21
2!
3 e x 1 3!c31
3!
∴ = + + + +P x x x x( ) ,1
1
2
1
32 3
! !L Q.E.D.
b. Next two terms: L L+ + +1
4
1
54 5
! !x x
c.1
0n
xn
n!=
∞
∑d.
S3
ex
5
y
x
3
e. The two graphs are indistinguishable forapproximately −1 < x < 1.
f. Solve ex − S3(x) = 0.0001 for x close to 1.x ≈ 0.2188…Solve ex − S3(x) = 0.0001 for x close to −1.x = −0.2237…The interval is −0.2237… < x < 0.2188… .
g. The ninth partial sum is S8(x).Solve ex − S8(x) = 0.0001 for x close to 1.x ≈ 1.4648…Solve S8(x) − ex = 0.0001 for x close to −1.x = −1.5142…The interval is −1.5142… < x < 1.4648… .
2. a. By equating derivatives:
n f xn( ) ( ) f n( ) ( )0 P n( ) ( )0 cn
0 cos x 1 c0 1
1 −sin x 0 c1 0
2 −cos x −1 2!c2 − 1
2!
3 sin x 0 3!c3 0
4 cos x 1 4!c41
4!
5 −sin x 0 5!c5 0
6 −cos x −1 6!c6 − 1
6!
7 sin x 0 7!c7 0
8 cos x 1 8!c81
8!
Calculus Solutions Manual Problem Set 12-4 317© 2005 Key Curriculum Press
∴ = − + − + −P x x x x x( )
! ! ! !1
1
2
1
4
1
6
1
82 4 6 8 L ,
Q.E.D.
b. L L− + − +1
10
1
12
1
1410 12 14
! ! !x x x
c. (– )( )!
11
22
0
n n
nn
x=
∞
∑d. y = cos x
cos
S4
S7
x
y
e. See the graph in part d, showing S7(x) (eighthpartial sum).The graphs are indistinguishable forapproximately −5.5 < x < 5.5.
f. Solve S7(x) − cos x = 0.0001 for x closeto 5.5.x ≈ 4.5414…(Note that some solvers may give an errormessage. In this case, zoom in by table, starting at x = 5 and using increments of 0.1;then x = 4.5, and increments of 0.01, and soforth.)By symmetry, the interval is−4.5414… < x < 4.5414… .
g. Both functions are even. P(x) is even becauseit has only even powers of x.
3. a. S33 5 70 6 0 6
1
30 6
1
50 6
1
70 6( . ) .= − + −
!( . )
!( . )
!( . )
= 0.564642445…sin 0.6 = 0.564642473…∴ S3(0.6) ≈ sin 0.6, Q.E.D.
b. sin 0.6 = 0.564642473…Tail = sin 0.6 − Sn(0.6)First term of the tail is tn+1.
sin 0.6 − S1(0.6) = 0.0006424733…t2 = 0.000648sin 0.6 − S2(0.6) = −0.00000552660…t3 = −0.00000555428…sin 0.6 − S3(0.6) = 0.0000000276807…t4 = 0.0000000277714…In each case, the tail is less in magnitude thanthe absolute value of the first term of the tail,Q.E.D.
c. Make | | .tn+−< ×1
200 5 10 .1
2 30 6 5 102 3 21
( )!( . )
nn
+< ×+ −
Inequality is first true for n = 8.Use at least nine terms (n = 8).
4. a. P x x x x x( ) = + + + +1
3
1
5
1
73 5 7
! ! !L
b. By equating derivatives:
n f xn( )( ) f n( )( )0 P n( ) ( )0 cn
0 sinh x 0 c0 0
1 cosh x 1 c1 1
2 sinh x 0 2!c2 0
3 cosh x 1 3!c3
1
3!
4 sinh x 0 4!c4 0
5 cosh x 1 5!c5
1
5!
6 sinh x 0 6!c6 0
7 cosh x 1 7!c7
1
7!
∴ = + + + +P x x x x x( ) ,1
3
1
5
1
73 5 7
! ! !L Q.E.D.
c. S3(0.6) = 0.636653554…sinh 0.6 = 0.636653582…
∴ S3(0.6) ≈ sinh 0.6, Q.E.D.
d. Solve S3(x) − sinh x = 0.0001 for xclose to 1.x ≈ 1.4870…By symmetry, the interval is−1.4870… < x < 1.4870… .
e. P x x x x′ = + ⋅ + ⋅ + ⋅ +( ) 1
1
33
1
55
1
772 4 6
! ! !L
= + + + +11
2
1
4
1
62 4 6
! ! !x x x L
f. Find S3(0.6) for the P′ series.S3(0.6) = 1.1854648cosh 0.6 = 1.18546521…∴ S3(0.6) ≈ cosh 0.6, and thus the P′(x)series seems to represent cosh x, Q.E.D.
g. P x dx( )∫ = + ⋅ + ⋅ + +1
2
1
3
1
4
1
5
1
62 4 6x x x C
! !L
Simplifying and letting C = 1 gives
1
1
2
1
4
1
62 4 6+ + + +
! ! !x x x L ,
which is the series for cosh x, Q.E.D.
5. a. f (x) = ln x f (1) = 0f ′(x) = x− 1 f ′(1) = 1
′′ = − −f x x( ) 2 ′′ = −f ( )1 1
f x x′′′ = −( ) 2 3 f ′′′ =( )1 2
P x x x x( ) ( )= − − +11
21
1
312 3( – ) ( – )
− +1
41 4( – )x L
318 Problem Set 12-5 Calculus Solutions Manual© 2005 Key Curriculum Press
P x x x x′ = − − + − − − +( ) ( ) ( ) ( )1 1 1 12 3 K
′′ = − + − − − +P x x x( ) ( ) ( )1 2 1 3 1 2 K
′′′ = − − +P x x( ) ( )2 6 1 K
P(1) = 0 = f (1)P′(1) = 1 = f ′(1)P″(1) = −1 = f ″(1)P′′′(1) = 2 = f ′′′(2), Q.E.D.
b. L L+ − +1
51
1
615 6( – ) ( – )x x
c. P xn
xn n
n
( ) = ⋅+
=
∞
∑(– ) ( – )11
11
1
d.
1
1x
y
ln x
S10
e. S10(1.2) = 0.182321555…ln 1.2 = 0.182321556…S10(1.95) = 0.640144911…ln 1.95 = 0.667829372…S10(3) = −64.8253968…ln 3 = 1.0986122…S10(x) fits ln x in about 0 < x < 2.
This is a wider interval of agreement than thatfor the fourth partial sum, which looks likeabout 0.3 < x < 1.7.S10(1.2) and ln 1.2 agree through the eighthdecimal place. The values of S10(1.95) andln 1.95 agree only to one decimal place. Thevalues of S10(3) and ln 3 bear no resemblanceto each other.
6. a. P x x x x( ) ( )= − − +11
21
1
312 3( – ) ( – )
− +1
41 4( – )x K
n tn(3)
1 2
2 −2
3 2.6666…
4 −4
5 6.4
6 −10.6666…
The absolute values of the terms are gettinglarger as n increases.
b. lim limx
nx
n
tn→∞ →∞
= → ∞∞
| |2
= → ∞→∞
limln
x
n2 2
1 1= ∞
The series cannot possibly converge becausethe terms do not approach zero as napproaches infinity.
c.
n tn(1.2)
1 0.22 −0.023 0.0026666…4 −0.00045 0.0000646 −0.00001066…
The absolute values of the terms areapproaching zero as n increases.
d. Tail = ln 1.2 − Sn(1.2)First term of the tail is tn+1.
ln 1.2 − S1(1.2) = −0.01767…t2 = −0.02ln 1.2 − S2(1.2) = 0.002321…t3 = 0.002666…ln 1.2 − S3(1.2) = −0.0003451…t4 = −0.0004In each case, the tail is less in magnitude thanthe absolute value of the first term of the tail.
7. a. f (x) = tan− 1 x
P xn
xn n
n
( ) =+
+
=
∞
∑(– )11
2 12 1
0
= − + − +x x x x1
3
1
5
1
73 5 7 L
b. y = tan− 1 x, y = S5(x), and y = S6(x) (sixthand seventh partial sums)
1
1
y
x
f(x)
S6
S5
S5
S6
Both partial sums fit the graph of f very wellfor about −0.9 < x < 0.9. For x > 1 andx < −1, the partial sums bear no resemblanceto the graph of f.
Problem Set 12-5Q1. 4! = 24 Q2. 3! = 6
Q3. 4!/4 = 6 Q4. n = 3
Calculus Solutions Manual Problem Set 12-5 319© 2005 Key Curriculum Press
Q5. n = m − 1 Q6. m = 1
Q7. 0! = 1!/1 = 1 Q8. (−1)! = 0!/0 = 1/0 = ∞
Q9. x x/ 2 7– Q10. A
1. f u eu( ) =
= + + + + + +1
1
2
1
3
1
4
1
52 3 4 5u u u u u
! ! ! !K
2. f u u( ) = ln
= − − + − +( )u u u u1
1
21
1
31
1
412 3 4( – ) ( – ) ( – ) K
3. f u u( ) = sin
= − + − + − +u u u u u u
1
3
1
5
1
7
1
9
1
113 5 7 9 11
! ! ! ! !L
4. f u u( ) = cos
= − + − + − +1
1
2
1
4
1
6
1
8
1
102 4 6 8 10
! ! ! ! !u u u u u L
5. f u u( ) = cosh
= + + + + + +11
2
1
4
1
6
1
8
1
102 6 8 10
! ! ! ! !u u u u u4 L
6. f u u( ) = sinh
= + + + + + +u u u u u u
1
3
1
5
1
7
1
9
1
113 5 7 9 11
! ! ! ! !L
7. f u u u u u u u( ) ( )= − = + + + + + +−1 11 2 3 4 5 L
8. f u u( ) = −tan 1
= − + − + − +u u u u u u1
3
1
5
1
7
1
9
1
113 5 7 9 11 L
9. x x x x x x x xsin
! ! ! != − + − + −
1
3
1
5
1
7
1
93 5 7 9 L
= − + − + −x x x x x2 4 6 8 101
3
1
5
1
7
1
9! ! ! !L
10. x x x x x x x xsinh! ! ! !
= + + + + +
1
3
1
5
1
7
1
93 5 7 9 L
= + + + + +x x x x x2 4 6 8 101
3
1
5
1
7
1
9! ! ! !L
11. cosh x3
= + + + + +1
1
2
1
4
1
6
1
83 2 3 4 3 6 3 8
!( )
!( )
!( )
!( )x x x x L
= + + + + +1
1
2
1
4
1
6
1
86 12 18 24
! ! ! !x x x x L
12. cos x2
= − + − + −1
1
2
1
4
1
6
1
82 2 2 2
!( )
!( )
!( )
!( )x x x x2 4 6 8 L
= − + − + −11
2
1
4
1
6
1
84 8 12 16
! ! ! !x x x x L
13. ln ( ) ( – ) ( – )x x x x2 2 2 2 2 31
1
21
1
31= − − + −L
(Or: ln x2 = 2 ln x = 2(x − 1) − (x − 1)2
+ −2
31 3( – ) )x L
14. e x x xx− = + + +2
11
2
1
32 2 2 2 3(– )
!(– )
!(– )
+ +1
42 4
!(– )x L
= − + − + −1
1
2
1
3
1
42 4 6 8x x x x
! ! !L
15.
e dt t t t t dttx x
− = + +
∫ ∫
2
0
2 4 6 8
01
1
2
1
3
1
4–
!–
! !–L
= − + ⋅ − ⋅ + ⋅ −x x x x x
1
3
1
5
1
2
1
7
1
3
1
9
1
43 5 9
! ! !7 L
16. ln ( – )//
( ) ( ) 3 3 11
23 1 2
1 31 3t dt t t
xx
= − −∫∫
+ − +…
1
33 1
1
43 13 4( – ) ( – )t t dt
=⋅ ⋅
−⋅ ⋅
1
3 2 13 1
1
3 3 23 12 3( – ) ( – )t t
+⋅ ⋅
−⋅ ⋅
+1
3 4 33 1
1
3 5 43 14 5
1 3
( – ) ( – )/
t tx
L
= − +1
63 1
1
183 1
1
363 12 3 4( – ) ( – ) ( – )x x x
− +1
603 1 5( – )x L
17.
1
114
4 8 12 16
xx x x x
+= − + − + −L
18.9
3
3
1 32 2x x+=
+ ( / )
= − + − +
3 1
1
3
1
3
1
32
24
36x x x L
= − + − +3
1
3
1
32 4
26x x x L
19.
1
114
0
4 8 12 16
0tdt t t t t dt
x x
+= + +∫ ∫ ( – – – )K
= − + − + −x x x x x
1
5
1
9
1
13
1
175 9 13 17 K
20.
9
33
1
3
1
322 4
26
00 tdt t t t dt
xx
+= + +
∫∫ – – K
= − +
⋅−
⋅+
⋅−3
1
3
1
3 5
1
3 7
1
3 93 5
27
39x x x x x K
320 Problem Set 12-5 Calculus Solutions Manual© 2005 Key Curriculum Press
21.d
dxx(sinh )2
= + + + +
d
dxx x x x2 6 10 141
3
1
5
1
7! ! !K
= + + + +2
6
3
10
5
14
75 9 13x x x x
! ! !K
= + + + +2
2
2
2
4
2
65 9 13x x x x
! ! !K
= + + + +…21
12
1
3605 9 13x x x x
Alternate solution:
d
dxx x x(sinh ) cosh2 22=
= + + + +
2 1
1
2
1
4
1
64 8 12x x x x
! ! !K
= + + + +…22
2
2
4
2
65 9 13x x x x
! ! !
22. d
dxx(cos ).0 5
= + +
d
dxx x x x1
1
2
1
4
1
6
1
82 3 4–
! !–
! !–K
= − + − + −1
2
2
4
3
6
4
82 3
! ! ! !x x x K
= − + − + −…1
2
1
12
1
240
1
100802 3x x x
Alternate solution:
d
dxx x x(cos ) sin.0 5 0 5 0 51
2= − − . .
= − − + − +
−1
2
1
3
1
5
1
70 5 0 5 1 5 2 5 3 5x x x x x. . . . .
! ! !L
= − +⋅
−⋅
+⋅
−1
2
1
2 3
1
2 5
1
2 72 3
! ! ! !x x x L
= − + − + −1
2
2
4
3
6
4
82 3
! ! ! !x x x L
Multiply by 1/1, 2/2, 3/3, 4/4, … and simplify.
23. P x x x428 3 2
0 7
22( ) ( )
.
!( )= − + − + −
+ − − −0 51
32
0 048
423 4.
!( )
.
!( )x x
= −8 + 3(x − 2) + 0.35(x − 2)2
+ 0.085(x − 2)3 − 0.002(x − 2)4
24. P x x x x52 37 2 1
0 48
21
0
31( ) ( )
.
!( )
!( )= + + − + + +
+ + − +0 36
41
0 084
514 5.
!( )
.
!( )x x
= 7 + 2(x + 1) − 0.24(x + 1)2
+ 0.015(x + 1)4 − 0.0007(x + 1)5
25.a. f (0.4) ≈ 2 + 0.5(1.4) − 0.3(1.4)2 − 0.18(1.4)3
+ 0.02(1.4)4 = 1.694912We must assume that the series converges forx = 0.4.
b. f (−1) = c0 = 2f ′(−1) = c1 = 0.5f″(−1) = 2!c2 = 2(−0.3) = −0.6f″′(−1) = 3!c3 = 6(−0.18) = −1.08f (4)(−1) = 4!c4 = 24(0.02) = 0.48
c. g(x) = f (x2 − 1) ≈ P4(x2 − 1)= 2 + 0.5(x2 − 1 + 1) − 0.3(x2 − 1 + 1)2
− 0.18(x2 − 1 + 1)3 + 0.02(x2 − 1 + 1)4
= 2 + 0.5x2 − 0.3x4 − 0.18x6 + 0.02x9
Sixth-degree polynomial:
g(x) ≈ 2 + 0.5x2 − 0.3x6
g(1) ≈ 2.2
d. g ′(x) = x − 1.2x3 + terms in higher powersof x.∴ g ′(0) = 0g″(x) = 1 − 3.2x2 + terms in higher powersof x.∴ g″(0) = 1 > 0. ∴ (0, g(0)) is a localminimum.
e. g t dt t t dtx x
( ) ( . . )0
2 4
02 0 5 0 3∫ ∫≈ + −
= + −
= + −
20 5
3
0 3
5 0
21
60 06
3 5
3 5
t t tx
x x x
. .
.
26. a. f (1) ≈ P4(1)= −4 + 3(1 − 2) + 0.5(1 − 2)2
− 0.09(1 − 2)3 − 0.06(1 − 2)4
= −6.47We must assume that the series converges forx = 1.
b. f (2) = c0 = −4f ′(2) = c1 = 3f″(2) = 2!c2 = 2(0.5) = 1f ′″(2) = 3!c3 = 6(−0.09) = −0.54f (4)(2) = 4!c4 = 24(−0.06) = −1.44
c. g(x) = f (x2 + 2) ≈ P(x2 + 2)= −4 + 3(x2 + 2 − 2) + 0.5(x2 + 2 − 2)2
−0.09(x2 + 2 − 2)3 − 0.06(x2 + 2 − 2)4
= −4 + 3x2 + 0.5x4 − 0.09x6 − 0.06x8
Fourth-degree polynomial:
g(x) ≈ −4 + 3x2 + 0.5x4
Calculus Solutions Manual Problem Set 12-5 321© 2005 Key Curriculum Press
d. g′(x) = 6x + 2x3 + terms in higher powersof x.∴ g′(0) = 0g″(x) = 6 + 6x2 + terms in higher powersof x.∴ g″(0) = 6 > 0∴ (0, g(0)) = (0, 4) is a local minimum.
e. h x g t dt t t dtx x
( ) ( ) ( . )= ≈ − + +∫ ∫0
2 4
04 3 0 5
= − + + = − + +4 0 1 4 0 13 50
3 5t t t x x xx
. .
27. f (x) = sin x, about x = π/4:
f x x x( ) = +
−
⋅
2
2
2
2 4
2
2 2 4
2
–!
–π π
−⋅
+
⋅
2
2 3 4
2
2 4 4
3
!–
!–x x
π π 4
+
⋅
−2
2 5 4!–x
π 5
L
28. f (x) = cos x, about x = π/4:
f x x x( ) = −
−
⋅
2
2
2
2 4
2
2 2 4
2
–!
–π π
+⋅
+
⋅
2
2 3 4
2
2 4 4
3 4
!–
!–x x
π π
−
⋅
−2
2 5 4
5
!–x
πL
29. f (x) = ln x, about x = 1:
f x x x x( ) ( )= − − +11
21
1
312 3( – ) ( – )
− +1
41 4( – )x L
30. f (x) = log x, about x = 10:
f xx x
( ) = + ⋅ − ⋅11
10
10
10
1
2 10
10
10
2
2ln
( – )
ln
( – )
+ ⋅ − ⋅ +1
3 10
10
10
1
4 10
10
10
3
3
4
4ln
( – )
ln
( – )x xL
31. f (x) = (x − 5)7/3 , about x = 4:
f x x x( ) = − + − ⋅1
7
34
7 4
3 242
2( – )!( – )
+ ⋅ ⋅ − ⋅ ⋅ ⋅7 4 1
3 34
7 4 1 2
3 443
34
4
!( – )
(– )
!( – )x x
+ ⋅ ⋅ ⋅ −7 4 1 2 5
3 545
5(– )(– )
!( – )x L
32. f (x) = (x + 6)4.2, about x = −5:
f x x x( ) . ( )= + + + ⋅ +1 4 2 54 2 3 2
25 2. .
!( )
+ ⋅ ⋅ + 4 2 3 2 2 2
35 3. . .
!( )x
+ ⋅ ⋅ ⋅ + +4 2 3 2 2 2 1 2
45 4. . . .
!( )x L
33. By equating derivatives:
n f xn( )( ) f n( )( )0 P n( )( )0 cn
0 cos 3x 1 c0 1
1 −3 sin 3x 0 c1 0
2 −9 cos 3x −9 2!c2 − 9
2!
3 27 sin 3x 0 3!c3 0
4 81 cos 3x 81 4!c4
81
4!
5 −243 sin 3x 0 5!c5 0
6 −729 cos 3x −729 6!c6 − 729
6!
∴ = − + − + cos
! ! !3 1
9
2
81
4
729
62 4 6x x x x L
By substitution:
cos
!( )
!( )
!( )3 1
1
23
1
43
1
632 4 6x x x x= − + − +L
= − + − +1
9
2
81
4
729
62 4 6
! ! !x x x L
The two answers are equivalent. Substitutiongives the answer much more easily in this case.
34. By equating derivatives:
n f xn( )( ) f n( )( )0 P n( )( )0 cn
0 ln (1 + x) 0 c0 01 (1 + x)− 1 1 c1 12 −(1 + x)− 2 −1 2!c2 − 1
2!3 2(1 + x)− 3 2 3!c3 2
3!4 −6(1 + x)− 4 −6 4!c4 − 6
4!
∴ + = − + − + ( )ln 1
1
2
1
3
1
42 3 4x x x x x L
By substitution, substitute (1 + x) for u in
ln ( – ) ( – ) ( – )u u u u u= − − + − +( )1
1
21
1
31
1
412 3 4 L
ln ( ) .1
1
2
1
3
1
42 3 4+ = − + − +x x x x x L
The two answers are equivalent. Substitutiongives the answer much more easily in this case.
35. S4(1.5) = 0.40104166… ;ln 1.5 = 0.40546510…Error = 0.00442344…
Fifth term = =1
51 5 1 0 006255( . – ) .
The error is smaller in absolute value than thefirst term of the tail.
322 Problem Set 12-5 Calculus Solutions Manual© 2005 Key Curriculum Press
36. Solve numerically for x close to 2:S4(x) − ln x = 0.0001x ≈ 1.2263…Solve numerically for x close to 0.1:ln x − S4(x) = 0.0001x ≈ 0.7896…Interval is about 0.7896… < x < 1.2263… .
37. a. tan− = − + −1 3 5 71
3
1
5
1
7x x x x x
+ − +1
9
1
119 11x x L
∴ = − + − + − +− tan 11 1
1
3
1
5
1
7
1
9
1
11L
The tenth partial sum is S9(1).
Sn
n
n
9
0
9
1 11
2 10 760459904( ) .=
+=
=∑(– )
( )K
4S9(1) = 3.04183961…π = 3.14159265…The error is about 3%.
b. The fiftieth partial sum is S49(1).4S49(1) = 3.12159465…π = 3.14159265…The error is about 0.6%.(It is merely an interesting coincidence thatalthough 4S49(1) differs from π in the seconddecimal place, several other decimal placeslater on do match up!)
c. By the composite argument propertiesfrom trig,
tan tan tan− −+
1 11
2
1
3
=
+
⋅
tan tan tan tan
– tan tan tan tan
– –
– –
1 1
1 1
12
13
112
13
=+
⋅=
12
13
112
13
1–
∴ = +− − − ,tan tan tan1 1 111
2
1
3 Q.E.D.
p Sn
n
n
n
= =+
=
+
∑4 4 11
2 1
1
29
0
9 2 1
(– )
++
=
+
∑(– )11
2 1
1
30
9 2 1n
n
n
n
= −( )+
+
= …
+ +
=∑4
1
2 1
1
2
1
3
3 14159257
2 1 2 1
0
9 n n n
nn
.π = 3.14159265…
The answer differs from π by only 1 in theseventh decimal place. The improvement inaccuracy is accounted for by the fact that theinverse tangent series converges much morerapidly for x = 1/2 and x = 1/3 than it does forx = 1. In Problem 17 of Problem Set 12-6,you will see that the interval of convergencefor the inverse tangent series is −1 ≤ x ≤ 1.In general, power series converge slowly atthe endpoints of the convergence interval.
38. sin
! ! !x x x x x= − + − +1
3
1
5
1
73 5 7 L
cos
! ! !x x x x= − + − +1
1
2
1
4
1
62 4 6 L
x x x x+ + + +
1
3
2
15
17
3153 5 7 L
1
1
2
1
4
1
6
1
3
1
5
1
72 4 6 3 5 7− + − + + +
! ! !–
! !–
!x x x x x x xL L
x x x x–! !
–!
1
2
1
4
1
63 5 7+ +L
1
3
4
5
6
73 5 7x x x− + −
! !L
1
3
1
6
1
723 5 7x x x– –+ L
2
15
64
75 7x x− −
!L
2
15
1
155 7x x– +L
17
3157x −L
∴ = + + + + tan x x x x x
1
3
2
15
17
3153 5 7 L
S4(0.2) = 0.202710024…tan 0.2 = 0.202710035…
39. Define a xf a
ix ai
ii( ) = −
( )( )
!( ) , the ith term of the
general Taylor series. So, f x a xi
i
( ) ==
∞
∑ ( ).0
We must assume d
dxa x
d
dxa x
n
n i
i
n
n i
i
( )=
∞
=
∞
∑ ∑=0 0
( );
that is, the nth derivative of an infinite series isthe infinite sum of the nth derivatives of theindividual terms.
For , ( )i nd
dxa x
f a
i
d
dxx a
n
n i
i n
ni< = ⋅ −
( ) ( )
!( )
= ⋅ =f a
i
i( ) ( )
!0 0
Calculus Solutions Manual Problem Set 12-6 323© 2005 Key Curriculum Press
For ,i nd
dxa x
d
dxa x
n
n i
n
n n= =( ) ( )
= ⋅ −f a
n
d
dxx a
n n
nn
( ) ( )
!( )
= ⋅ −f a
nn x a
n( ) ( )
!!( )0
For , ( )i nd
dxa x
f a
i
d
dxx a
n
n i
i n
ni> = ⋅ −
( ) ( )
!( )
= ⋅ − − − + − =−f x
ii i i i n x a
ii n
( ) ( )
!( )( ) ( )( )1 2 1 0K
for x = a.
So ( ) for and , andd
dxa a i n i n
n
n i = < >0
d
dxa a f a
n
n nn( ) ( ).( )=
Thus, d
dxa x
n
n i
i
( )=
∞
∑0
evaluated at x = a
is d
dxa a f a
n
n nn( ) ( ).( )=
40. Brook Taylor: 1685−1731Colin Maclaurin: 1698−1746Sir Isaac Newton: 1642−1727Gottfried Wilhelm von Leibniz: 1646−1716
41. a.t
tn
x
nx
n
nxn
n
n n
n n
+
+ +
+= + =
+−1
2 1
1
11
11
11
1 11
(– ) ( – )
(– ) ( – )| |
b. r x102
111 2= =for .
r x109 5
111 95= =.
for .
r x1020
113= =for
c. rn
nx x
n
nx
n n=
+− = −
+= −
→∞ →∞lim lim
11 1
11| | | | | |
d. r = 1.1 for x = −0.1r = 1 for x = 0r = 0.9 for x = 0.1r = 0.9 for x = 1.9r = 1 for x = 2
e. Possible conjecture: The series converges toln x whenever the value of x makes r < 1, anddiverges whenever the value of x makes r > 1.
f. The series should converge for r < 1.r = |x − 1| < 1 ⇒ −1 < (x − 1) < 1 ⇒0 < x < 2
42. Answers will vary.
Problem Set 12-6Q1. sin x Q2. sinh x
Q3. e−x Q4. ex
Q5. (1 − x)− 1 (−1 < x ≤ 1)
Q6.1
22sin x C+
Q7. 3 sec2 3x Q8. 1
Q9. e Q10. B
1. a.n
x x x x xnn
n4
1
4
2
16
3
64
4
2562 3 4
1
= + + +=
∞
∑
+ +5
10245x L
b. Lt
t
n x
nxn
n
nn
n
n
n
n= = + ⋅→∞
+→∞
+
+lim lim11
1
1
4
4( )
= + =→∞
x n
n
xn4
1
4lim
Lx x
x< ⇔ < ⇔ − < < ⇔ − < <14
1 14
1 4 4
Open interval of convergence is (−4, 4).
c. Radius of convergence = 4.
2. a.x
nx x x x
n
nn
⋅= +
⋅+
⋅+
⋅=
∞
∑ 2
1
2
1
2 4
1
3 8
1
4 162 3 4
1
+⋅
+1
5 325x L
b. Lt
t
x
n
n
xn
n
nn
n
n
n
n= =+ ⋅
⋅ ⋅→∞
+→∞
+
+lim lim( )
11
11 2
2
=+
=→∞
x n
n
xn2 1 2lim
Lx x
x< ⇔ < ⇔ − < < ⇔ − < <12
1 12
1 2 2
Open interval of convergence is (−2, 2).
c. Radius of convergence = 2.
3. a.( ) ( )2 3
2 32 3
21
2x
nx
xn
n
+ = + + +
=
∞
∑ ( )
+ + + + +( ) ( )2 3
3
2 3
4
3 4x xL
b. Lx
n
n
xn
n
n= ++
⋅+→∞
+
lim( )
( )
2 3
1 2 3
1
= ++
= +→∞
| | | |2 31
2 3xn
nx
xlim
L < 1 ⇔ |2x + 3| < 1 ⇔ −1 < 2x + 3 < 1⇔ −2 < x < −1Open interval of convergence is (−2, −1).
c. Radius of convergence = 1
2.
324 Problem Set 12-6 Calculus Solutions Manual© 2005 Key Curriculum Press
4. a.( – )5 7
21
x
n
n
n=
∞
∑= + +( – ) ( – ) ( – )5 7
2
5 7
4
5 7
6
2 3x x x
+ +( – )5 7
8
4xL
b. Lx
n
n
xn
n
n=+
⋅→∞
+
lim( – )
( ) ( – )
5 7
2 1
2
5 7
1
= −+
= −→∞
| | | |5 71
5 7xn
nx
nlim
L < 1 ⇔ |5x − 7| < 1 ⇔ −1 < 5x − 7 < 1⇔ 1.2 < x < 1.6
Open interval of convergence is (1.2, 1.6).
c. Radius of convergence = 0.2.
5. a.n
nx n
n
3
1
8!
( – )=
∞
∑= − + +( )x x x8
8
28
27
682 3( – ) ( – )
+ +64
248 4( – )x L
b. Ln x
n
n
n xn
n
n= ++
⋅−→∞
+
lim( ) ( – )
( )!
!
( )
1 8
1 8
3 1
3
= − +
⋅
+
→∞| | lim
nx
n
n n8
1 1
1
3
= − ⋅ ⋅ =| |x 8 1 0 0
L < 1 for all values of x.
Series converges for all values of x.
c. Radius of convergence is infinite.
6. a.n
nx n
n
!( )4
1
2+=
∞
∑= + + + + +( )x x x2
2
162
6
8122 3( ) ( )
+ + +24
2562 4( )x L
b. Ln x
n
n
n xn
n
n= + ⋅ ++
⋅⋅ +→∞
+
lim( )! ( )
( ) ! ( )
1 2
1 2
1
4
4
= + + ⋅+
→∞| | ( ) x n
n
nn2 1
1
4
lim
= + + ⋅ = ∞→∞
| | [( ) ]x nn
2 1 1lim
The series converges only for |x + 2| = 0⇔ x = −2.
c. Radius of convergence = 0.
7. sin(– )
( )!x
nx
nn
n
=+
+
=
∞
∑ 1
2 12 1
0
Note that |(−1)n| can be left out of the ratio.
Lx
n
n
xn
n
n=+
⋅ +→∞
+
+lim( )!
( )!2 3
2 12 3
2 1
=+ +
= ⋅→∞
xn n
xn
2 21
2 3 2 20lim
( )( )∴ L < 1 for all x and the series converges forall x.
8. cos(– )
( )!x
nx
nn
n
==
∞
∑ 1
22
0
Note that |(−1)n| can be left out of the ratio.
Lx
n
n
xn
n
n=+
⋅→∞
+
lim( )!
( )!2 2
22 2
2
=+ +
= ⋅→∞
xn n
xn
2 21
2 2 2 10lim
( )( )∴ L < 1 for all x and the series converges forall x.
9. sinh( )!
xn
x n
n
=+
+
=
∞
∑ 1
2 12 1
0
Lx
n
n
xn
n
n=+
⋅ +→∞
+
+lim( )!
( )!2 3
2 12 3
2 1
=+ +
= ⋅→∞
xn n
xn
2 21
2 3 2 20lim
( )( )∴ L < 1 for all x and the series converges forall x.
10. cosh( )!
xn
x n
n
==
∞
∑ 1
22
0
Lx
n
n
xn
n
n=+
⋅→∞
+
lim( )!
( )!2 2
22 2
2
=+ +
= ⋅→∞
xn n
xn
2 21
2 2 2 10lim
( )( )∴ L < 1 for all x and the series converges forall x.
11. en
xx n
n
==
∞
∑ 1
0!
Lx
n
n
xx
nx
n
n
n n=
+⋅ =
+= ⋅
→∞
+
→∞lim
( )!
!lim
1
1
1
10| | | |
∴ L < 1 for all x and the series converges forall x.
12. en
xxn
n
n
−
=
∞
= ∑ (– )
!
1
0
Lx
n
n
xx
nx
n
n
n n=
+⋅ =
+= ⋅
→∞
+
→∞lim
( )!
!lim
1
1
1
10| | | |
∴ L < 1 for all x and the series converges for all x.
13. tn = xnn!
Lx n
x nx n x
n
n
n n= + = + = ⋅ ∞
→∞
+
→∞lim
( )!
!| | lim ( ) | |
1 11
L = ∞ for all x ≠ 0; L = 0 at x = 0.∴ the series converges only for x = 0.
Calculus Solutions Manual Problem Set 12-6 325© 2005 Key Curriculum Press
14. tn
xn nn= !
100
Ln x
n xn
n
n
n
n= + ⋅→∞
+
+lim( )!
!
1
100
1001
1
= + → ⋅ ∞→∞
| | | |xn
xnlim
1
100L = ∞ for all x ≠ 0; L = 0 at x = 0.∴ the series converges only for x = 0.
15. cosh( )!
101
2102
0
==
∞
∑ nn
n
Ln
nn
n
n=+
⋅→∞
+
lim( )!
( )!10
2 2
2
10
2 2
2
=+ +
=→∞
101
2 2 2 102 lim
( )( )n n nL = 0 < 1 ⇒ series converges.
16. ln (– ) (– . )0 1 11
0 91
1
. = +
=
∞
∑ n n
nn
tn = −(0.9)n/n
n tn t tn n+1/
1 −0.9 0.45
2 −0.405 0.6
3 −0.243 0.675
4 −0.164025 0.72
5 −0.118098 0.75
9 −0.043046721 0.81
15 −0.0137260754… 0.84375
35 −0.0007151872… 0.875
Ratio seems to approach 0.9.
Proof:
Ln
n
n
n
n
n
n
n
=+
⋅
=+
→∞
+
→∞
lim(– . )
(– . )
lim
0 9
1 0 9
0 91
1
.
= 0.9(1) = 0.9, Q.E.D.
17. a. P xn
xn n
n
( ) =+
+
=
∞
∑(– )11
2 12 1
0
Note that |(−1)n| = 1 for all n.
Lx
n
n
xn
n
n=+
⋅ +→∞
+
+lim2 3
2 12 3
2 1
= ++
= ⋅→∞
xn
nx
n
2 22 1
2 31lim by l’Hospital’s rule
L < 1 ⇔ x2 < 1 ⇔ −1 < x < 1Open interval of convergence is (−1, 1).
b.
tan–1
S4
S3S4
S3
x
y
1
1
The graphs of the partial sums of P(x) andtan− 1 x fit very well for −1 < x < 1. Thepartial sums diverge from tan− 1 x for x outsidethis interval.
c. S3(0.1) = 0.09966865238095…
d. tan− 1 0.1 = 0.09966865249116… ;Tail = 0.00000000011021…
e. First term of tail = =1
90 1 9( . )
0.00000000011111… ,which is larger than the tail.
18. a. y = x2 sin 2x, from x = 0 to x = 1.5, rotated
about the y-axis.A slice of the region parallel to the axis ofrotation generates a cylindrical shell.dV = 2π x · y · dx = 2πx3
sin 2x dx
V x x dx= ∫ 2 23
0
1 5
π sin.
Integrate by parts.
u dvx3 sin 2x
3x 2 –12 cos 2x
6x –14 sin 2x
618 cos 2x
0116 sin 2x
+
+
–
–
+
V x x x x= − +2
1
22
3
423 2π cos sin
+ −
3
42
3
82
0
1 5
x x xcos sin.
= − +
2
9
163
21
163π cos sin
= …4 662693947.
V x x dx= ≈ …∫ 2 2 4 6626939473π sin .0
1.5
The answers are the same to at least ninedecimal places.
b. Omitting the 2 23π , x x dxsin∫= − +
∫ x x x x3 3 521
32
1
52( )
!( )
!( )
− +
1
72 7
!( )x dxL
326 Problem Set 12-6 Calculus Solutions Manual© 2005 Key Curriculum Press
= − + − +
∫ 2
2
3
2
5
2
74
36
58
710x x x x dx
! ! !L
= −⋅
+⋅
2
5
2
7 3
2
9 55
37
59x x x
! !
−
⋅+ +2
11 7
711
!x CL
=+ ⋅ +
++
+
=
∞
∑(– )( ) ( )!
12
2 5 2 1
2 12 5
0
nn
n
nn n
x C
For this series,
Lx
n n
n n
xn
n n
n n= ⋅+ ⋅ +
⋅ + ⋅ +⋅→∞
+ +
+ +lim( ) ( )!
( ) ( )!2
2 7 2 3
2 5 2 1
2
2 3 2 7
2 1 2 5
= ++
⋅+ +→∞ →∞
42 5
2 7
1
2 3 2 22x
n
n n nn nlim lim
( )( )= 4x2
⋅ 1 ⋅ 0∴ the series converges for all x and thusconverges for x = 1.5.So,
V ≈ −⋅
+
⋅2
2
51 5
2
7 31 5
2
9 51 55
37
59π ( . )
!( . )
!( . )
−⋅
+⋅
2
11 71 5
2
13 91 5
711
913
!( . )
!( . )
= 4.67164363…The answer is within 0.01 of the answerfound in part a.
c. tn nn
nn
n+
++= − ⋅
+ ⋅ +1
2 32 71
2 2
2 7 2 31 5( )
π( ) ( )!
( . )
By table search, | tn+ 1 | < 0.5 × 10− 10 forn ≥ 10.Because n starts at zero for the first term, youwould need 11 terms to estimate the volumeto 10 decimal places.
19. a. Assume this series can be integrated term byterm.
f x e dttx
( ) = −∫2
0
= − + − + − +
∫ 1
1
2
1
3
1
4
1
52 4 6 8 10
0t t t t t dt
x
! ! ! !K
= − +⋅
−⋅
+⋅
x x x x x1
3
1
5 2
1
7 3
1
9 43 5 7 9
! ! !
−⋅
+1
11 511
!x K
b.
1
1
x
y
f(x)
S5
The partial sum is reasonably close forapproximately −1.5 < x < 1.5.
c. tn n
xn
nn=
++(– )
( ) !
1
2 12 1
Note that | (−1)n | = 1.
Lx
n n
n n
xn
n
n=+ +
⋅ +→∞
+
+lim( )( )!
( ) !2 3
2 12 3 1
2 1
= ++
⋅+→∞ →∞
xn
n nn n
2 2 1
2 3
1
1lim lim
= x2 ⋅ 1 ⋅ 0 < 1 for all x.
d. Erf x does seem to be approaching 1 as xincreases, as shown by the following tablegenerated by numerical integration.
x erf x
1 0.8427007929…
2 0.9953222650…
3 0.9999779095…
4 0.9999999845…
5 0.9999999999…
20. a. Assume this series can be integrated term byterm.
sin
! ! !
t
t tt t t t= ⋅ − + −
1 1
3
1
5
1
73 5 7
+ − +
1
9
1
119 11
! !t t K
= − + − + − +11
3
1
5
1
7
1
9
1
112 4 6 8 10
! ! ! ! !t t t t t K
=+=
∞
∑(– )( )!
12 1
2
0
nn
n
t
n
Si x =
11
3
1
5
1
7
1
9
1
112 4 6 8 10
0− + − + − +
∫ ! ! ! ! !
t t t t t dtx
K
= −⋅
+⋅
−⋅
x x x x1
3 3
1
5 5
1
7 75
! ! !3 7
+⋅
−⋅
+1
9 9
1
11 119
! !x x11 K
=+ +
+
=
∞
∑(– )( )( )!
11
2 1 2 12 1
0
n n
nn n
x
b. Ratio for is sin t
t
Lt
n
n
tn
n
n=+
⋅ +→∞
+
lim( )!
( )!2 2
22 3
2 1
=+ +
= ⋅→∞
tn n
tn
2 lim( )( )
1
2 3 2 202
∴ L < 1 for all values of t.Series for (sin t)/t converges for all valuesof t.
Calculus Solutions Manual Problem Set 12-7 327© 2005 Key Curriculum Press
Ratio for Si x is
Lx
n n
n n
x
xn
n n n
x
n
n
n
n n
=+ +
⋅ + +
= ++
⋅+ +
= ⋅ ⋅
→∞
+
+
→∞ →∞
lim( )( )!
( )( )!
lim lim( )( )
2 3
2 1
2
2
2 3 2 3
2 1 2 1
2 1
2 3
1
2 3 2 2
1 0∴ L < 1 for all values of x.Series for Si x converges for all values of x.The radii of convergence for both series areinfinite.
c. The third partial sum is S2(x).
S23 50 6 0 6
1
3 30 6
1
5 50 6( . ) .= −
⋅+
⋅!( . )
!( . )
= 0.5881296Si 0.6 = 0.5881288…The answers are quite close!
d.
x
y
Si (x)
TenthPartial Sum
π 2π 3π−π−2π−3π
1
−1
S9(x) is reasonably close to Si x for−3π < x < 3π .
21. a. Given L tn
nn=
→∞lim where L < 1.
By the definition of limit as n → ∞, thereis a number k > 0 for any ε > 0 such thatif n > k, then tn
n is within ε units of L.
Thus, tnn < L + ε, Q.E.D.
b. L < 1 ⇒ 1 − L > 0So take any ε < 1 − L⇒ L + ε < L + 1 − L⇒ L + ε < 1.
c. For all integers n > k,
0 ≤ < + ⇒ ≤ < +t L t Lnn
nnε ε0 ( ) and
(L + ε)n < (L + ε)n− k for all n > k becauseL + ε < 1, so 0 ≤ tn < (L + ε)n− k, Q.E.D.
d. Because 0 ≤ tn < (L + ε)n− k for all n > k, itfollows that the tail after tn satisfies0 1 2 3≤ + + ++ + +t t tn n n L
< (L + ε)n+ 1− k + (L + ε)n+ 2− k + (L + ε)n+ 3− k + K
= (L + ε)n+ 1− k[1 + (L + ε) + (L + ε)2 + K ],
which converges because L + ε < 1.
e. The tail of the series is increasing and isbounded above by
( ) [ ( ) ( ) ]L L Ln k+ + + + + + =+ −ε ε ε1 21 K
( )
– ( )
–L
L
n k++
+εε
1
1So the series converges, Q.E.D.
22. L nn
n=→∞
lim
Because ln is a continuous function,ln ln lim lim lnL n n
n
n
n
n= =→∞ →∞
( )
= = = → ∞∞→∞ →∞ →∞
lim ln limln
limln
n n xnn
n
n
x
x
1
=→∞
lim/
x
x1
1 by l’Hospital’s rule
= 0∴ L = e0 = 1, Q.E.D.
23. ln(– )
( – )xn
xn
n
n
=+
=
∞
∑ 11
1
1
L tx
nnn
nn
nn= =
→∞ →∞lim | | lim
( – )1
= = = −→∞
| – |
lim
| – |x
n
xx
n
n
1 1
11| |
L < 1 ⇔ |x − 1| < 1 ⇔ 0 < x < 2Open interval of convergence is (0, 2).
24.1
1n
xnn
n=
∞
∑L
x
n
x
nx
n
n
nn
n= = = ⋅
→∞ →∞lim
| |lim
| || | 0
∴ L < 1 for all values of x, and thus the seriesconverges for all values of x, Q.E.D.
25. n xn n
n=
∞
∑1
L n x nx xn
n nn
n= = = ⋅ ∞
→∞ →∞lim | | lim | | | |
L = 0 if x = 0 and is infinite if x ≠ 0.∴ the series converges only if x = 0.
26.n
nxn
n
n
!
=
∞
∑1
Use the ratio technique.
Ln x
n
n
n xn
n
n
n
n= + ⋅+
⋅⋅→∞
+
+lim( )!
( ) !
1
1
1
1
=+
=+
= ⋅→∞ →∞
| | | | | |xn
nx
nx
en
n
n n nlim( )
lim( / )1
1
1 1
1
L < 1 ⇔ | |xe
⋅ 1 < 1 ⇔ −e < x < e
Open interval of convergence is (−e, e).
Problem Set 12-7Q1. geometric Q2. multiplying by 2
Q3. common ratio Q4. ln x
328 Problem Set 12-7 Calculus Solutions Manual© 2005 Key Curriculum Press
Q5. 14
2
16
42 4− +
! !x x Q6. −1/3!
Q7. 2.5 Q8. 1 < x < 7
Q9. ( ) ( )2 3 32e i t jtr r
+ cos
Q10. D
1. a. Sn
n
n
51
1
5
16
66
2
6
3
6
4
6
5= = − + − ++
=∑(– )
! ! ! ! !
= − + − + =6 3 11
4
1
203 8.
b.
Tail = = = − + − ++
=
∞
∑Rn
n
n
51
6
16 6
6
6
7
6
8(– )
! ! ! !K
c. Hypotheses: (1) signs are strictly alternating,(2) |tn| are strictly decreasing, and(3) lim .
nnt→∞
= 0
| | | | / ! /R t5 6 6 6 1120< = =
d. Absolute convergence means that | |tn
n=
∞
∑1
converges.If the convergent series were not absolutelyconvergent, it would be called “conditionallyconvergent.”
e. When you show absolute convergence, youfind the partial sums of |tn|. The partial sumsmust be increasing because |tn| is positive.|tn| is decreasing because the series isconvergent.
2. a. Comparison test (direct)
b. Integral test
c. nth term test
d. Geometric series test
e. Ratio test (or ratio technique)
f. Limit comparison test
g. p-series test
3. a. Sn
n
5 21
51
11
4
1
9
1
16
1
25= = + + + +
=∑
= = …11669
36001 463611.
b.
Tail = = = + + +=
∞
∑Rn
n
5 26
1 1
36
1
49
1
64L
The graph shows the tail bounded above by
( / ) .1 0 22
5x dx =
∞
∫ .
5 6 7 8 9
0.03
tn
n
The series converges because the sequence ofpartial sums is increasing and bounded aboveby 0.2.
c. ( / ) ( / )1 121000
2
1001x dx R x< <
∞∞
∫∫ 1000
1/1001 < R1000 < 1/1000R1000 ≈ 0.5(1/1001 + 1/1000)
= 0.00099950049…S = S1000 + R1000
≈ 1.643934… + 0.00099950049…= 1.644934…
The answer is correct to at least three decimalplaces.(The exact answer is π 2/6 = 1.64493406… .The estimate is actually correct to ninedecimal places.)
d. 0.5[1/(n + 1) + 1/n] = 0.0000005 (sixdecimal places)n ≈ 1,999,999.5, so use about 2 millionterms.
4. a. Sn
n
5
1
51
11
2
1
3
1
4
1
5= = + + + +
=∑
= = …217
602 2833.
p = 1, a harmonic series.
b. Tail = = = + + +=
∞
∑Rn
n
5
6
1 1
6
1
7
1
8L . The
graph shows the tail for R5 bounded below by
( / )16
x dx∞
∫ .
5
0.1
0.2
6 7 8 9
t
n
n
( / ) lim (ln – ln )1 66
x dx bb
= = ∞→∞
∞
∫∴ the series diverges because a lower bound isinfinite.
c. Graphing the rectangles to the left of then-values leads to R5 < ∞, which does notimply that the tail is finite.
d. S x dx xn
n n
> > ⇔ =+ +
∫1000 1 10001
1
1
1
if ( / ) ln
ln (n + 1) > 1000 ⇔ n > e1000 − 1≈ 1.97 · 10434 s. It would take approximately6.24 · 10420 yr.
5. a. 1
1
2
1
3
1
4− + − + L
The series converges because(1) strictly alternating signs, (2) strictlydecreasing |tn|, (3) tn → 0.
Calculus Solutions Manual Problem Set 12-7 329© 2005 Key Curriculum Press
b. ( / ) lim (ln – ln )1 11
x dx bb
= = ∞→∞
∞
∫Τhe series | |tn
n=
∞
∑1
diverges, so the given series
does not converge absolutely.
c. S1000 = 0.692647… , S1001 = 0.693646… ,ln 2 = 0.693147…
| S1000 − ln 2| = 0.0004997… , |S1001 − ln 2| =0.0004992… , |t1001| = 1/1001 =0.00009900…∴ both partial sums are within |t1001| of ln 2.
d. No term is left out. No term appears morethan once.
Series is
1
2
1
4
1
6
1
8
1
10
1
12− + − + − +K
= + +
1
21
1
2
1
3
1
4– – K .
∴ the series converges to 1
22ln .
Conditional convergence means that whetherthe series converges, and, if so, what value itconverges to, depends on the condition thatyou do not rearrange the terms.
6. a. 11
4
1
9
1
16− + − +K
The series converges because(1) strictly alternating signs, (2) strictlydecreasing |tn|, (3) tn → 0.
b. | |tn
n=
∞
∑1
converges by the p-series test
because p > 1.
c. Ln
nn
=+
⋅ =→∞
lim( )
1
1 112
2
, so the ratio test
fails because L is not less than 1.
7. a. sin (– )( )!
xn
xn n
n
=+
+
=
∞
∑ 11
2 12 1
0
t32 3 11
1
2 3 10 6= − ⋅
⋅ +⋅ ⋅ +( ) .3
( )!
= − = −1
70 6 0 000005554285717
!. . K
b. S130 6 0 6
1
30 6 0 564( . ) . . .= − =
!
S23 50 6 0 6
1
30 6
1
50 6 0 564648( . ) . . . .= − + =
! !
c. R1 = sin 0.6 − S1(0.6) = 0.0006424…R2 = sin 0.6 − S2(0.6) = −0.0000055266…|R1| = 0.0006424…|t2| = 0.000648∴ |R1| < |t2||R2| = 0.0000055266…|t3| = 0.0000055542…∴ |R2| < |t3|
d. The terms are strictly alternating in sign, theterms are strictly decreasing in absolute value,and the terms approach zero for a limit asn → ∞. Thus, the series converges by thealternating series test.
Or:| | | |R tn n< +1 for all n ≥ 1, as shown by examplein part c.
limn
nt→∞ + =| |1 0 because it takes the form 0
∞.
∴ =→∞
| | ,limn
nR 0 and thus the series converges.
Or: Use the ratio technique.
Ln
n
n n
n
n
n
n
=+
⋅ +
=+ +
=
→∞
+
+
→∞
lim.
( )!
( )!
.
lim( )( )
0 6
2 3
2 1
0 6
0 361
2 3 2 20
2 3
2 1
.
Because L < 1, the series converges.
8. The sequence converges because limn
nt→∞= 2, a
(finite) real number.The series does not converge because lim
nnt→∞
≠ 0.
9. a.
1
3
1
8
1
15
1
24+ + + +K
Compare with the p-series with p = 2:
lim/( )
/lim ,
n n
n
n
n
n→∞ →∞
− =−
=1 1
1 11
2
2
2
2 a positive
real number.∴ the series converges by the limitcomparison test.
b. The p-series with p = 2 begins1
4
1
9
1
16
1
25+ + + . These terms form a lower
bound, not an upper bound, so the directcomparison test fails.
c. If n started at 1, the first term would be 1/0,which is infinite.
10. a. The seventh term of 1
0 60
nn
n!
.=
∞
∑ is
t661
60 6 0 0000648= =
!. . .
b. Sn
n
n
4
0
41
0 6 1 8214= ==
∑ !. .
e0.6 = 1.8221188…S4 differs from e0.6 by 0.00071880… , whichis greater than t5 = 0.000648, but not muchgreater. The difference is greater than t5
because all subsequent terms are added, notsubtracted. It is not much greater than t5
because the subsequent terms are very small.
330 Problem Set 12-7 Calculus Solutions Manual© 2005 Key Curriculum Press
c.
n 5 6 7
Tail: 0.000648 0.0000648 0.000005554…
Geometricseries: 0.000648 0.0000648 0.00000648
Terms of the e0.6 series are formed bymultiplying the previous term:
tn
tn n= −0 6
1.
Terms of the geometric series are formed bymultiplying the previous term by 0.1:t tn n= −0 1 1.For n ≥ 7, 0.6/n is smaller than 0.1, so theterms of the e0.6 series are smaller than thecorresponding terms of the geometric series.Thus, the geometric series forms an upperbound for the tail of the e0.6 series afterterm t6.
d. Geometric series converges to
0 00006481
1 0 10 000072. . .⋅ =
– .
e. The tail of the series after t6 is bounded by0.000072.The entire series is bounded byS6(0.6) + 0.000072 = 1.8221128 + 0.000072= 1.8221848.e0.6 = 1.8221188…So the upper bound is just above e0.6 , Q.E.D.
11. a.
1
1
2
1
3
2
4
6
5
24+ + + + + K
Ln
n
n
n
n
n nn n= + ⋅ = + ⋅
=
→∞ →∞lim
!
( – )!lim
1 1 1 10
∴ the series converges because L < 1.
b.
1
1
2
2
3
6
4
241 1
1
2
1
3+ + + + = + + + +K K
! !This is the Maclaurin series that convergesto e1.
c. Ln n
n n
n
nn n= =
→∞ →∞lim
/( – )!
/ !lim
!
( – )!
1
1= = ∞
→∞limn
n
∴ the test fails because the limit of the ratiois infinite.
12. a. U:
2
4
4
10
8
28
16
82+ + + + K
G:
2
3
4
9
8
27
16
81+ + + + K
V:
2
2
4
8
8
26
16
80+ + + + K
The terms of U are bounded above by thecorresponding terms of G, and so U convergesby the direct comparison test.
The terms of V are bounded below by thecorresponding terms of G, and so the directcomparison test fails in this case.
b. Ln
n n
n n n
n
n= =→∞ →∞
lim/( – )
/lim
–
2 3 1
2 3
3
3 1
= =→∞
limln
lnn
n
n
3 3
3 31
(using l’Hospital’s rule)∴ the V series converges because the G seriesconverges and L is a (finite) positive number.
13. Divergent harmonic series
14. Convergent p-series, p > 115. Convergent alternating series meeting the three
hypotheses
16. Divergent p-series, p ≤ 1
17.3
43
3
4
3
16
3
640
nn
= + + + +=
∞
∑ K
Converges because it is a geometric series withcommon ratio 1/4, which is less than 1 inabsolute value
18.
3
41
3
4
9
16
27
640
n
nn
= + + + +=
∞
∑ L
Converges because it is a geometric series withcommon ratio 3/4, which is less than 1 inabsolute value
19.
1
2 1
1
1
1
3
1
5
1
70
( )! ! ! ! !nn
+= + + + +
=
∞
∑ K
= + + + +1
1
6
1
120
1
5040K
Converges by comparison with geometricseries with t0 = 1 and r = 1/6
20.
1
31
1
3
1
9
1
270
(– )nn
= − + − +=
∞
∑ K
Converges by the alternating series test. (Termsare strictly alternating. Terms are strictlydecreasing in absolute value. tn approaches zeroas n approaches infinity.)
21.
n
nn
3
42
1
8
15
27
80
64
255
125
624–= + + + +
=
∞
∑ K
Diverges. Use the integral test.
x
xdx x
b
b3
44
2 21
1
41
–lim ln= −→∞
∞
∫ | |
= − −
= ∞→∞
lim lnb
b1
41 04( )
Or: Compare with a harmonic series.n
n nn n
3
42 2
1
1
–=
∞
=
∞
∑ ∑> → ∞
Calculus Solutions Manual Problem Set 12-7 331© 2005 Key Curriculum Press
22.
1
2
2
3
3
4
4
5+ + + + K , divergent because tn does
not approach zero
23.
– –1
1
1
2
1
3
1
4+ + + + K , a convergent alternating
series meeting the three hypotheses
24.
sin sin sin sin sinnn
= + + + +=
∞
∑ 0 1 2 30
K
= + …+ …+ …+0 0 8414 0 9092 0 1411. . . K
Diverges. tn does not approach 0 as n → ∞.
25. Lt
t
n
nn
n
nn
n
n= = + = >→∞
+→∞
+
lim lim( / ) /( )
( / ) /1
14 3 1
4 34 3 1/ .
Diverges by the ratio test
26. Convergent geometric series with | r | = 7/11 < 1
27. Diverges because tn does not approach zero
28. Lt
tn
n
n
=→∞
+lim 1
= +→∞
+
lim[( ) – ]/
[ – ]/n
n
n
n
n
1 1 2
1 2
2 1
2
= + ++
= <→∞
1
2
1 1
1
1
21
2
2lim( )
n
n
nConverges by the ratio test
29.1 1
22 x xdx x dx x
b
b
lnlim (ln )=→∞
−∞
∫∫ ( / )
= − = ∞→∞
lim ln ln ln lnb
b[ ( ) ( )]2
Diverges by the integral test
30. Converges by direct comparison with the
convergent p-series 22
3n
n=
∞
∑31.
1 2 6 241 2 3 4e e e e
+ + + + K
Diverges because tn does not approach zero
32. Converges to e by the definition of e
33. x = − + − + − +1 1 1 1 1 1: K
Diverges by the nth term testx = + + + + +9 1 1 1 1 1: K
Diverges by the nth term testComplete interval is (1, 9).
34. x = − − + − + −1
1
3
1
18
1
81
1
324: K
Converges by the alternating series test
x = + + + +51
3
1
18
1
81
1
324: K
Converges by comparison with the geometric
series
1
3
1
9
1
27
1
81+ + + + K
Complete interval is [−1, 5].
35. x = − − + − + −41
1
1
2
1
3
1
4: K
Converges by the alternating series test
x = − + + + +2
1
1
1
2
1
3
1
4: K
Diverges. p series with p = 0.5, which is lessthan 1.Complete interval is [−4, −2).
36. x = − − + − + − +2
1
31 1 1 1 1: K
Diverges by the nth term test
x = − + + + + +12
31 1 1 1 1: K
Diverges by the nth term test
Complete interval is −
2
1
31
2
3, – .
37. n x n
n
( )−=
∞
∑ 31
Ln x
n xn
n
n= +→∞
+
lim( )( – )
( – )
1 3
3
1
= − + = − ⋅→∞
| | lim | |xn
nx
n3
13 1
L < 1 ⇔ | x − 3 | < 1 ⇔ 2 < x < 4At x = 2 the series is − + − + −1 2 3 4 L , whichdiverges because the terms do not approach zero.At x = 4 the series is 1 2 3 4+ + + +L , whichdiverges because the terms do not approach zero.Interval of convergence is (2, 4).
38.5
21
n n
n
x
n
⋅
=
∞
∑L
x
n
n
xn
n n
n n= ⋅+
⋅⋅→∞
+ +
lim( )
5
1 5
1 1
2
2
=+
= ⋅
→∞5
15 1
2
| | lim | |xn
nx
n
L < 1 ⇔ 5| x | < 1 ⇔ −0.2 < x < 0.2
At x = −0.2 the series is − + − + −1
1
4
1
9
1
16L ,
which is a convergent alternating series.
At x = 0.2 the series is 1
1
4
1
9
1
16+ + + +K ,
which is a convergent p-series with p = 2.Interval of convergence is [−0.2, 0.2].
39.x
n
n
n=
∞
∑1
Lx
n
n
xn
n
n=+
⋅→∞
+
lim1
1
=+
= ⋅→∞
| | | |xn
nx
nlim
11
L < 1 ⇔ | x | < 1 ⇔ −1 < x < 1
At x = −1 the series is − + − + −1
1
2
1
3
1
4K ,
which is a convergent alternating series.
332 Problem Set 12-7 Calculus Solutions Manual© 2005 Key Curriculum Press
At x = 1 the series is 1
1
2
1
3
1
4+ + + +K ,
which is a divergent harmonic series (p-serieswith p = 1).Interval of convergence is [−1, 1).
40.(– ) ( – )1 6
24
n n
nn
x
n ⋅=
∞
∑
Lx
n
n
xn
n
n
n
n=+ ⋅
⋅ ⋅→∞
+
+lim( – )
( ) ( – )
6
1 2
2
6
1
1
= −+
= − ⋅→∞
1
26
1
1
26 1| | limx
n
nx
n| |
L x x< ⇔ − < ⇔ < <11
26 1 4 8| |
At x = 4 the series is
1
4
1
5
1
6
1
7+ + + +K ,
which is a divergent harmonic series (p-serieswith p = 1).
At x = 8 the series is
1
4
1
5
1
6
1
7− + − +K ,
which is a convergent alternating series.Interval of convergence is (4, 8].
41. (– ) ( )1 5
2
1 2
1
n n
n
x
n
+
=
∞ ⋅ +∑L
x
n
n
xn
n
n= ++
⋅+→∞
+
lim( )
( ) ( )
5
2 1
2
5
2 2
2
= ++
= + ⋅→∞
( ) ( )xn
nx
n5
15 12 2lim
L < 1 ⇔ (x + 5)2 < 1 ⇔ −6 < x < −4
At x = −6 the series is 1
2
1
4
1
6
1
8− + − +K ,
which is a convergent alternating series.
At x = −4 the series is
1
2
1
4
1
6
1
8− + − +K ,
which is a convergent alternating series.Interval of convergence is [−6, −4].
42. ( )x
n
n
n
+
=
∞
∑ 12
1
Lx
n
n
xn
n
n= ++
⋅+→∞
+
lim( )
( ) ( )
1
1 1
1
2
2
= ++
= + ⋅
→∞| | | |x
n
nx
n1
11 1
2
lim
L < 1 ⇔ | x + 1 | < 1 ⇔ −2 < x < 0
At x = −2 the series is − + − + −1
1
4
1
9
1
16K ,
which is a convergent alternating series.
At x = 0 the series is 11
4
1
9
1
16+ + + +K , which
is a convergent p-series with p = 2.Interval of convergence is [−2, 0].
43.ln ( )n
nxn
n
++=
∞
∑ 1
10
Ln x
n
n
n xn
n
n= + ⋅+
⋅ ++ ⋅→∞
+
limln ( )
ln ( )
2
2
1
1
1
= ++
⋅ ++→∞ →∞
| |xn
n
n
nn nlim
ln ( )
ln ( )lim
2
1
1
2
= ++
⋅ ++→∞ →∞
| |xn
n
n
nn nlim
/( )
/( )lim
1 2
1 1
1
2
(by l’Hospital’s rule)
= ++
⋅ ++
= ⋅ ⋅→∞ →∞
| | | |xn
n
n
nx
n nlim lim
1
2
1
21 1
L < 1 ⇔ | x | < 1 ⇔ −1 < x < 1
At x = −1 the series isln ln ln ln
.1
1
2
2
3
3
4
4− + − +K
By l’Hospital’s rule,
limln ( )
lim/
n n
n
n
n→∞ →∞
= =1
10.
Because the terms decrease in absolute value andapproach zero for a limit, the series converges bythe alternating series test.At x = 1 the series is
ln ln ln ln1
1
2
2
3
3
4
4+ + + +L .
ln ( )lim (ln )
x
xdx x
b
b
=
→∞
∞
∫ 1
22
11
= −
= ∞→∞
lim (ln )b
b1
202
Thus, the series diverges by the integral test.Interval of convergence is [−1, 1].
44. 5 31
( )x n
n
−=
∞
∑L
x
xx
n
n
n= = −→∞
+
lim( – )
( – )
5 3
5 33
1
| |
L < 1 ⇔ | x − 3 | < 1 ⇔ 2 < x < 4At x = 2 the series is − + − + −5 5 5 5 L , whichdiverges because the terms do not approach zero.At x = 4 the series is 5 5 5 5+ + + +L , whichdiverges because the terms do not approach zero.Interval of convergence is (2, 4).
45.4
0
n
nn
x=
∞
∑
Lx
x
xn
n
n
n
n= ⋅ =→∞
+
+lim| |
4
4
41
1
Lx
xx x< ⇔ < ⇔ > ⇔ < − >1
41
41 4 4
| |
| | or
Calculus Solutions Manual Problem Set 12-7 333© 2005 Key Curriculum Press
At x = −4 the series is 1 1 1 1− + − +L , whichdiverges because the terms do not approach zero.At x = 4 the series is 1 1 1 1+ + + +L , whichdiverges because the terms do not approach zero.Intervals of convergence are (−∞, −4) and (4, ∞).(Note that the series in Problems 45 and 46 havenegative powers of x and are called Laurent seriesrather than Taylor series.)
46.1
1xn
n=
∞
∑
Lx
x xn
n
n= =
→∞ +lim
| |1
1
Lx
x< ⇔ < ⇔ >11
11| |
| |1 ⇔ x < −1 or x > 1
At x = −1 the series is − + − + −1 1 1 1 L ,which diverges because the terms do not approachzero.At x = 1 the series is 1 1 1 1+ + + +L ,which diverges because the terms do not approachzero.Intervals of convergence are (−∞, −1) and (1, ∞).
47. a. Assume all the blocks have equal mass = m,with the center of mass at the center of theblock, and equal length = L.Write Hn = the distance the nth blockoverhangs the (n + 1)th block. (n = 1 forthe top block.)Note that according to the rule, Hn = thedistance between the rightmost edge of thenth block and the center of mass of the pileof the top n blocks.
Now, the center of mass of the nth block is12 L units from its rightmost edge, and the
center of mass of the pile of the top n − 1blocks is 0 units from (i.e., right on top of )the edge of the nth block according to therule.
Therefore, the center of mass of the pile of the
top n blocks is 1 1
20 1
nmL m n m⋅ + ⋅ −
( )
units from the edge of the nth block; that is,
Hn
Ln = 1
2, Q.E.D.
b. The total distance the top (first) blockoverhangs the nth block is H H1 2+ + +LHn−1. So for a pile of n blocks, the top block
will project entirely beyond the bottom blockif
L H Hn< + + =−1 1L
1
2
1
4
1
6
1
2 1L L L
nL+ + + +L
( – )
The first n for which
1
1
2
1
4
1
6
1
2 1< + + + + =L
( – )nnis 5.
c. To make a pile with overhang H, find an n
such that 1
1
2
1
3
1
1
2+ + + + >Ln
H
L– (this is
possible because the harmonic series divergesto infinity). Then a stack of n blocks willhave total overhang
H Hn1 1+ + −L
= + + + +1
2
1
4
1
6
1
2 1L L L
nLK
( – )
= + + + +
+
< ⋅ =1
21
1
2
1
3
1
1
1
2
2L
nL
H
LHK
(The achieved overhang is greater than H, soone may pull blocks slightly back—movingblocks back can only make the pile morestable—until the overhang equals H exactly.)
d. The theoretical overhang for a stack of 52objects is
H H H
L L L L
L
L
= + +
= + + + +
= + + + +
=
1 51
1
2
1
4
1
6
1
102
1
2
1
4
1
6
1
102
2 2594
L
L
K
K.
slightly more than two-and-a-quarter cardlengths.
48. The least upper bound postulate says that anynon-empty set of real numbers that has an upperbound has a least upper bound. In particular, anynumber less than this least upper bound cannotbe an upper bound for the set.The set of real numbers {t1, t2, t3, …} is non-empty and is bounded above (given). Therefore,this set has a least upper bound L. Any numberless than L is also less than some tD in the set.
Claim: L tb
n=→ ∞lim
Proof:
Pick a number ε > 0.Because L is an upper bound for tn, L + ε is alsoan upper bound.Because L is the least upper bound for tn,L − ε is not an upper bound.∴ there exists an integer D > 0 such thattD > L − ε.But the values of tn are increasing.∴ tn > tD > L − ε for all n > D.
334 Problem Set 12-8 Calculus Solutions Manual© 2005 Key Curriculum Press
Keep n > D.Then L − ε < tn < L + ε.Thus, tn is within ε units of L for all n > D.∴ L t
nn=
→∞lim by the definition of limit as
n → ∞, Q.E.D.
Problem Set 12-8Q1. ratio Q2. |common ratio| < 1Q3. for all values of x Q4. radius = 1
Q5. x x x x− + − +1
3
1
5
1
73 5 7 K
Q6.S4
sin
Q7. ln |sec x + tan x| + CQ8. y′ = sec2 xQ9. tan x + C Q10. Newton and Leibniz
1. a. cosh xn
x n
n
==
∞
∑ 1
22
0( )!
S5 4( ) = 27.2699118K
b. Rf c
5
2 5 22 5 24
2 5 24( )
=⋅ +
⋅+
⋅ +( . ) ( )
( )!
f x x( ) ( ) =12 cosh
∴ = < + M cosh ( )–41
23 24 4
= 40.5312… < 41
| ( ) | .R5
12441
124 1 4360≤ ⋅ =
!K
S5(4) is within 2 of cosh 4 in the units digit.
c. cosh 4 = 27.3082328…S5(4) = 27.2699118…cosh 4 − S5(4) = 0.0383… , which is wellwithin the 1.4360… upper bound found byLagrange form.
2. a. sinh( )!
xn
x n
n
=+
+
=
∞
∑ 1
2 12 1
0
S9(5) = 74.2032007…
b. Rf c
9
2 9 32 9 35
2 9 35( ) =
⋅ +⋅
⋅ +⋅ +
( ) ( )
( )!f (21)(x) = cosh x
∴ = < + M cosh ( )–51
23 25 5
= 121.5156… < 122
| ( )| .R9
215122
215 0 001138≤ ⋅ =
!K
S9(5) is within 2 units of sinh 5 in the thirddecimal place.
c. sinh 5 = 74.2032105…S9(5) = 74.2032007…sinh 5 − S9(5) = 0.00000981… , which iswell within the 0.001138… upper boundfound by Lagrange form.
3. a. ex
nx
n
n
==
∞
∑ !0
The fifteenth partial sum isS14(3) = 20.0855234… .
b. Rf c
14
15153
153( ) = ⋅
( ) ( )
!f (15)(x) = ex
∴ M = e3 < 33 = 27
| ( )| .R14
15327
153 0 0002962≤ ⋅ =
!K
S14(3) is within 3 units of e3 in the fourthdecimal place.
c. e3 = 20.085536923…S14(3) = 20.085523458…e3 − S14(3) = 0.00001346… , which is withinthe 0.0002962… found by Lagrange form.
4. a. ln (– ) ( – )xn
xn n
n
= ⋅+
=
∞
∑ 11
11
1
S8(0.7) = −0.356671944…
b. Rf c
8
990 7
90 7 1( . ) = ⋅ −
( ) ( )
!( . )
f (9)(x) = 8!x− 9
∴ M = 8!(0.7)− 9
| ( . )| ( . ) 9R8
990 7
8 0 7
90 3
1
93 7≤ ⋅ =!( . )
!( / )
–
= 5.4195… × 10− 5
S8(0.7) is within 6 units of ln 0.7 in the fifthdecimal place.(Note that for ln x, the Lagrange form of theremainder simplifies to
| ( )|R xn
x
xn
n
≤+
⋅
+1
1
1 1| – |
For x < 0.5, the fraction |x − 1|/x is greaterthan 1.The Lagrange form of the remainder becomesinfinite as n → ∞ and is thus not useful.)
c. ln 0.7 = −0.356674943…S8(0.7) = −0.356671944…|ln 0.7 − S8(0.3)| = 2.9998 × 10− 6, which iswithin the 5.4195… × 10− 5 found byLagrange form.
5. For sinh 2, all derivatives are bounded by cosh 2.
cosh ( )–21
23 2 4 6252 2< + = .
The general term is tnn
n=+
⋅ +1
2 122 1
( )!.
Calculus Solutions Manual Problem Set 12-8 335© 2005 Key Curriculum Press
For six-place accuracy,
| ( )| . .Rnn
n24 625
2 32 0 5 102 3 6≤
+⋅ < ×+ −.
( )!The second inequality is first true for n = 6.Use at least seven terms (n = 6).
6. For cosh 3, all derivatives are bounded by cosh 3.
cosh ( )–31
23 2 13 56253 3< + = .
The general term is tnn
n= ⋅1
232
( )!.
For eight-place accuracy,
| ( )| . .Rnn
n313 5625
2 23 0 5 102 2 8≤
+⋅ < ×+ −.
( )!The second inequality is first true for n = 10.Use at least 11 terms (n = 10).
7. For ln x, the nth derivative (n ≥ 1) on [x, 1]is bounded by M = (−1)n+ 1(n − 1)!(0.6)− n.
| ( . )| | . |Rn
nn
nn0 6
0 6
10 6 1
11≤
+⋅ −
++!( . )
( )!
–( )
=+
⋅
+1
1
2
3
1
n
n
For seven-place accuracy,
1
1
2
30 5 10
17
n
n
+⋅
< ×
+−. .
This inequality is first true for n = 32.Use at least 32 terms.
8. For e10, all derivatives are bounded by e10.e10 < 310 = 59049
For five-place accuracy,
| ( )| . .Rnn
n1059049
110 0 5 101 5≤
+⋅ < ×+ −
( )!The second inequality is first true for n = 43.Use 44 terms (n = 43).
9. cosh 2 = 3.76219569…S4(2) = 3.76190476…cosh 2 − S4(2) = 0.000290929…
The general term is tnn
n= ⋅1
222
( )!.
Rf c
c4
2 4 22 4 2
10
22 4 2
22
10( ) =
⋅ +⋅ = ⋅
⋅ +⋅ +
( ) ( )
( )!cosh
!
∴ ⋅ = .cosh!
c2
100 000290929
10
K
cosh c = 1.03098027…c = cosh− 11.0309… = 0.2482… ,which is between 0 and 2.
10. e5 = 148.413159…S19(5) = 148.413107…e5 − S19(5) = 5.1234… × 10− 5
The general term is 1
5n
n
!.
Rf c
ec19
2020
20
520
55
20( ) = ⋅ = ⋅
( ) ( )
! !
∴ ⋅ = × − .ec 5
205 1234 10
205
!K
ec = 1.30702776…c = ln 1.3070… = 0.2677… ,which is between 0 and 5.
11. cos!( . )
!( . )
!( . )2 4 1
1
22 4
1
42 4
1
62 42 4. 6= − + −
+ − +1
82 4
1
102 48 10
!( . )
!( . ) L
= 1 − 2.88 + 1.3824 − 0.2654208
+ −0 0273004. L LThe terms are strictly alternating. They aredecreasing in absolute value after t1, and theyapproach zero for a limit as n → ∞.Therefore, the hypotheses of the alternating seriestest apply, and | Rn(2.4) | < | tn+ 1| =
1
2 22 4 2 2
( )!( . ) .
nn
++
For six-place accuracy, make| Rn(2.4) | < 0.5 × 10− 6.The inequality is first true for n = 7.Use 8 terms (n = 7).
12. e− = + − + + +2 2 3 41 21
22
1
32
1
42( )
!(– )
!(– )
!(– )
+ + +1
52
1
62 6
!(– )
!(– )5 L
= − + − …+ …− …+1 2 2 1 3333 0 6666 0 2666. . . LThe terms are strictly alternating. They aredecreasing in absolute value after t2, and theyapproach zero for a limit as n → ∞.Therefore, the hypotheses of the alternating series
test apply, and | ( )| | | .R tnn n
n− < =+
⋅++2
1
121
1
( )!For seven-place accuracy, make| Rn(−2) | < 0.5 × 10− 7.The inequality is first true for n = 14.Use 15 terms (n = 14).
13. a. Sn
n
10 31
101
1 19753198= = …=
∑ .
R x dx xb
b10
3 21010
0 5< =
=−
→∞
∞
∫ lim – . –
0.5(10− 2) = 0.005
R x dx xb
b10
3 21111
0 5> =
=−
→∞
∞
∫ lim – . –
0.5(11− 2) = 0.00413223…R10 ≈ 0.5(0.005 + 0.00413223…) =0.00456611…
336 Problem Set 12-8 Calculus Solutions Manual© 2005 Key Curriculum Press
S ≈ 1.19753198… + 0.00456611… =1.20209810…Error < 0.5(0.005 − 0.00413223…) =0.00043388… (about three decimal places)
b. Using both the upper and lower bounds,
error < −
+
∞∞
∫∫0 5 3 3
1. x dx x dx
nn
= − +− −0 25 0 25 12 2. . ( ) .n nSolve 0.25n− 2 − 0.25(n + 1)− 2 = 0.000005 toget n = 45.9194… .Use 46 terms.Using only the upper bound, Rn < 0.000005
if . .x dxn
3 0 000005<∞
∫Set 0.5n− 2 = 0.000005n = (0.000005/0.5)− 1/2
= 316.2277… , which rounds up to 317terms, considerably more that the 46 terms togive this accuracy by comparing upper andlower bounds of Rn.
14. a. Sn
n
100 1 051
1001
4 698244= = …=
∑ . .
R x dx xb
b100
1 05 0 05100100
20< =
=−
→∞
∞
∫ . lim – – .
20(100− 0.05 ) = 15.886564…
R x dx xb
b100
1 05 0 05101101
20> =
=−
→∞
∞
∫ . lim – – .
20(101− 0.05 ) = 15.878662…R100 ≈ 0.5(15.886564… + 15.878662…) =15.882613…S ≈ 4.698244… + 15.882613… =20.580858…Error < 0.5(15.886564… − 15.878662…) =0.003950… (about two decimal places)
b. Error . . .< −
+
∞∞
∫∫0 5 1 05 1 05
1x dx x dx
nn
= 20n− 0.05 − 20(n + 1)− 0.05
Solve 20n− 0.05 − 20(n + 1)− 0. 05 = 0.000005 toget n = 111840.2309… .Use 111,841 terms.With a value of p such as 1.05, which iscloser to 1 than 3 is, it takes more termsbecause the terms approach zero more slowly.
15. a. Sn
n
10 20
101
11 9817928=
+= …
=∑ .
The series converges because the terms of thetail starting at t1 are bounded above by theconvergent p-series with p = 2.
b. Rx
dx xb
bb
10 21
1010
1
1<
+=
=
→∞∫ lim tan–
π /2 − tan− 110 = 0.0996686…
Rx
dx xb
bb
10 21
1111
1
1>
+=
=
→∞∫ lim tan–
π /2 − tan− 111 = 0.0906598…
R10 ≈ 0.5(0.0996686… + 0.0906598…) =0.0951642…S ≈ 1.9817928… + 0.0951642… =2.0769570…Error < 0.5(0.0996686… − 0.0906598…) =0.004504…S ≈ 2.0769570… is correct to at least twodecimal places.Make Rn < 0.00005.Rn ≈ 0.5[(π/2 − tan− 1 n) − (π /2 − tan− 1 (n + 1))]= 0.5(tan− 1 (n + 1) − tan− 1 n)Solve 0.5(tan− 1 (n + 1) − tan− 1 n) = 0.00005to get n = 99.4962… .Use 100 terms.
16. In this p-series, p = 0.5, which is not greaterthan or equal to 1. Thus, the series diverges andthe remainder is infinite.
17. en
n
n
2 = ⋅=
∞
∑ 12
0!
From Example 1, S10 = 7.38899470… .By Lagrange form, | R10 | < 0.0004617… .Use a geometric series as an upper bound.
t t1111
12121
112
1
122= ⋅ = ⋅
! !and
Common ratio rt
t= =12
11
1
6
∴ < ⋅ ⋅ = … | | .R10111
112
1
116
0 00006156! –
The geometric series gives a better estimate ofthe remainder than does the Lagrange form.
18. en
n n
n
−
=
∞
= ⋅∑2
0
11
2(– )!
S10 = 0.135379188…
| | | | .R t10 11
112
110 000051306< = = …
!This number appears to be a better estimate ofthe error. However, it represents an error of ≈| R10 |/S(10) = 0.03789…%.e2 ≈ 1/S10 = 7.38665971…A 0.037…% error for this value would be0.002799… , which is a worse estimate of theerror than that by Lagrange or by geometricseries.(In general, an error of ε% in1/f (x) gives a
maximum error of εε1 100– /
in the value of f (x).
So an error of 0.03789…% in 1/e2 means an
error of 0 03789
1 0 00037890 03788 2.
– .
K
KK= . % in .)e
Calculus Solutions Manual Problem Set 12-8 337© 2005 Key Curriculum Press
19. a.
250
239 7887357
π= . K
Thus, 250 radians is 39 complete cycles plus0.7887… additional cycle, orb = (2π)(0.7887…) = 4.9557730… radians.sin b = −0.970528019…sin 250 = −0.970528019… (Checks.)The value of b can be calculated efficientlyusing the fraction part command. For atypical grapher, b = f Part(250/(2π))2π.
b. From Figure 12-8c, you can tell that the valueof c is one cycle back from the value of b.c = b − 2π = −1.32741228…Check:sin c = −0.970528019…sin 250 = −0.970528019… (Checks.)In general:If b is in [0, π/2], then c = b.If b is in (π/2, 3π/2], then c = π − b.If b is in (3π/2, 2π], then c = 2π − b.
c. From Figure 12-8c, you can tell that thevalue of d is a quarter-cycle ahead of the valueof c. The value of the sine is the opposite ofthe corresponding value of cos d.
d c= + =π2
0 243384039. K
Check:−cos d = −0.970528019…sin 250 = −0.970528019… (Checks.)In general:
c d c x d∈ − −
= + = −π π π
2 4 2, : and sin cos
c d c x d∈ −
= − = −π
40, and : sin sin
c d c x d∈
= =0
4, : sin sin and
π
c d c x d∈
= − =π π π4 2 2
, : and sin cos
d. For x in [0, π/4], both the sine and cosineseries meet the hypotheses of the alternatingseries test. Thus, the error in S5(x) is boundedby | t6 |, the first term of the tail. | t6 | is greaterfor the cosine series than for the sine series.The maximum of | t6 | in the interval is atx = π/4.
∴ < =⋅ +
⋅ + | ( )| | ( / )|R x t5 62 6 24
1
2 6 24π π
( )!( / )
= 3.8980… × 10− 13,which is small enough to guarantee that sin xwill be correct to ten decimal places.For direct calculation,
| ( )| .Rnn
n2501
2 3250 0 5 102 3 10<
+⋅ < ⋅+ −
( )!
The second inequality is first true when n is348. Because both numerator and denominatormay overflow most computers, you cancalculate values of ln | Rn(250) | as follows:
ln ln ln| ( )| ( )R n in
i
n
250 2 3 2502
2 3
= + −=
+
∑Then compare the values with ln (0.5 ·10− 10).So you would need to use 349 terms.Unfortunately, even this procedure would notbe practical because the terms themselveswould have to be calculated to ten or moredecimal places, and they are so large that eachterm overflows most computers’ capacities.
e. The program will have the following steps.The particular commands will depend on thegrapher or computer used.
• Put in a value of x.
• Find b, as shown in part a.
• Find c, as shown in part b.
• Find d, as shown in part c.
• Choose the function and sign, as shownin part c.
• Calculate and display the answer.
20. For sin 1, | ( )| | |R tnn n
n11
2 311
2 3< =+
⋅ =++
( )!1
2 3( )!n +.
Set . .1
2 30 5 10 23
( )!n +< × −
This inequality is first true for n = 11.Use at least 12 terms (n = 11).Using the technique in Problem 19,
| ( )| . .Rnn
n11
2 24 0 5 102 2 23<
+< ⋅+ −
( )!( / )π
The second inequality is first true for n = 10.You would save only one term by the method ofProblem 19.
21. a. Apply the mean value theorem to f ′(x) on[a, x]. There is a number x = c in (x, a) suchthat
′′ = ′ ′f c
f x f a
x a( )
( ) – ( )
–⇒ f ′(x) = f ′(a) + f ″(c)(x − a), Q.E.D.
b. f x dx f a dx f c x a dx′ = ′ + ′′ −∫ ∫ ∫( ) ( ) ( )( )
f x f a x f c x a C( ) ( )= ′ + ′′ ⋅ +( ) ( – )1
22
Substituting the initial condition (a, f (a))gives
f (a) = f ′(a)a + f ″(c)(0) + C ⇒C = f (a) − f ′(a)a
338 Problem Set 12-9 Calculus Solutions Manual© 2005 Key Curriculum Press
f x( ) =
′ + ′′ ⋅ + − ′f a x f c x a f a f a a( ) ( ) ( )( ) ( – )1
22
f x f a f a x a f c x a( ) ( ) ( )( ) ( ) ,= + ′ − + ′′ −1
22( )
Q.E.D.
c. Apply the mean value theorem to f ″(x)on [a, x].There is a number x = c in (a, x) such that
′′′ = ′′ ′′f c
f x f a
x a( )
( ) – ( )
–⇒ ′′ = ′′ + ′′′ −f x f a f c x a( ) ( ) ( )( )
Integrate once to get f ′(a).
′′ = ′′ + ′′′ −∫∫∫ f x dx f a dx f c x a dx( ) ( ) ( )( )
′ = ′′ + ′′′ ⋅ +f x f a x f c x a C( ) ( ) ( )1
22( – )
Use (a, f ′(a)) as an initial condition.
f a f a a f c C′ = ′′ + ′′′ + ⇒( ) ( ) ( )( )1
20
C f a f a a= ′ − ′′( ) ( )
f x f a x f c x a′ = ′′ + ′′′ −( ) ( ) ( )( )21
2+ ′ − ′′f a f a a( ) ( )
f x f a f a x a f c x a′ = ′ + ′′ − + ′′′ −( ) ( ) ( )( ) ( )( )21
2
Integrate again to get f(x).
f x dx f a dx f a x a dx′ = ′ + ′′ −∫ ∫ ∫( ) ( ) ( )( )
+ ′′′ −∫ 1
2f c x a dx( )( )2
f x f a x f a x a( ) ( ) ( )( )2= ′ + ′′ −1
2
+ ′′′ − +1
6f c x a C( )( )3
Use (a, f ″(a)) as an initial condition.
f a f a a f a f c C( ) ( ) ( )( ) ( )( )= ′ + ′′ + ′′′ +1
20
1
60
⇒ = − ′C f a f a a( ) ( )
f x f a x f a x a( ) ( ) ( )( )2= ′ + ′′ −1
2!
+ ′′′ − + − ′1
33
!f c x a f a f a a( )( ) ( ) ( )
f x f a f a x a f a x a( ) ( ) ( )( ) ( )( )2= + ′ − + ′′ −1
2!
+ ′′′ −1
3!,f c x a( )( )3 Q.E.D.
d. The technique is mathematical induction.
22. a. f xe x
x
x
( )if
if = ≠
=
– –,
,
20
0 0
It is given that f n( )( )0 0= for all n > 0.c0 = f (0) = 0c1 = f ′(0) = 0
2!c2 = f ″(0) = 0 ⇒ c2 = 03!c3 = f ″′(0) = 0 ⇒ c3 = 0…∴ series is 0 + 0x + 0x2 + 0x3 + … , Q.E.D.
b. Each partial sum of the Maclaurin seriesequals zero for any value of x. Thus, thesequence of partial sums converges to zerofor all x. But f (x) does not equal zero exceptat x = 0. Thus, the series converges to f (x)only at x = 0.
c. e x x xx− −−
= + − + + +2
11
2
1
32 2 2 2 3( )
!(– )
!(– )– – L
= − + − +− − −1
1
2
1
32 4 6x x x
! !L
d. The fourth partial sum, S3(2) =0.7786458333… .
f e e( ) .2 0 77880078302 0 252
= = =− −− . K
The partial sum is close to f (2), so it isreasonable to make the conjecture that theLaurent series converges to f (2).
23. Using the Lagrange form of the remainder, thevalue of ex is given exactly by
en
x R xx nk
n
k
= +=
∑ 1
0!
( ), where
R xf c
kxk
kk( ) =
+
++
( ) ( )
( )!
11
1 and c is between 0 and x.
| ( )| | |R xM
kxk
k≤+
+
( )!11
Because all derivatives of e x equal e x, the value ofM for any particular value of x is also e x, whichis less than 3x, if x ≥ 0; or 1, if x < 0.
lim lim( )!k
kk
xkR x
kx
→∞ →∞
+<+
| ( )| | |3
11
which approaches 0 as k → ∞ by the ratiotechnique.Because the remainder approaches zero as napproaches infinity, ex is given exactly by
en
xx n
n
==
∞
∑ 1
0!
, Q.E.D.
Problem Set 12-9Review Problems
R0. Answers will vary.
R1. f x
xP x x x x( ) and ( )= = + + + +9
19 9 9 92 3
–L
1
20
x
yP6
P5
f
Calculus Solutions Manual Problem Set 12-9 339© 2005 Key Curriculum Press
The graph shows that P5(x) and P6(x) are close tof (x) for x between about −0.7 and 0.6, and bearlittle resemblance to f (x) beyond ±1.P5(0.4) = 14.93856P6(0.4) = 14.975424f (0.4) = 15∴ P6(0.4) is closer to f (0.4) than P5(0.4) is,Q.E.D.P5(x) = 9 + 9x + 9x2 + 9x3 + 9x4 + 9x5;P5(0) = 9
′ = + + + + ′ =P x x x x x P52 3 4
59 18 27 36 45 0 9( ) ; ( )
′′ = + + + ′′ =P x x x x P52 3
518 54 108 180 0 18( ) ; ( )
′′′ = + + ′′′ =P x x x P52
554 216 540 0 54( ) ; ( )
f (x) = 9(1 − x)− 1; f (0) = 9f ′(x) = 9(1 − x)− 2; f ′(0) = 9f ″(x) = 18(1 − x)− 3; f ″(0) = 18
′′′ = − ′′′ =−f x x f( ) ( ) ; ( )54 1 0 544
∴ ′ = ′ ′′ = ′ ( ) ( ), ( ) ( ),P f P f5 50 0 0 0and ( ) ( )′′′ = ′′′P f5 0 0
Pn(x) is a subseries of a geometric series.
R2. a. Series is 3 + 2.7 + 2.43 + 2.187 + L .
After 10 days, S10
10
31 0 9
1 0 9= ⋅ =– .
– .19.5396… .About 19.5 mm increase in 10 days
S = ⋅ =31
1 0 930
– .About 30 mm increase eventually
b. Let x be the amount invested to have0.5 million dollars at the end of 19 years.Interest rate is 10% per year, so the amountat the end of a year is 1.1 times the amount atthe beginning of the year.x(1.119) = 0.5x = 0.5(1.1− 19) = 0.081753995…They must invest $81,754.00 now in orderto make the last payment.The total to invest is the sum0 5 1 1 0 5 1 1 0 5 1 11 2 19. ( . ) . ( . ) . ( . ).− − −+ + +L
This is the nineteenth partial sum of thegeometric series with first term 0.5(1.1)− 1
and common ratio 1.1− 1.
S191
19
10 5 1 11 1 1
1 1 14 182460045= ⋅ = …−. ( . ) .
– .
– .
–
–
They must invest $4,182,460.05 now tomake all 19 payments.
R3. P(x) = c0 + c1x + c2x2 + c3x
3 + c4x4 + L
f (x) = 7e3x ⇒ f (0) = 7 ⇒ c0 = 7f ′(x) = 21e3x ⇒ f ′(0) = 21 ⇒ c1 = 21f ″(x) = 63e3x ⇒ f ″(0) = 63 ⇒ 2!c2 = 63 ⇒ c2 = 31.5
′′′ = ⇒ ′′′ = ⇒f x e fx( ) ( )189 0 1893
c3 = 189/3! = 31.5
R4. a. e0.12 = 1.127496851…S3(0.12) = 1.127488, which is close to e0.12 .
b. cos 0.12 = 0.9928086358538…S3(0.12) = 0.9928086358528, which is close.
c. sinh (0.12) = 0.1202882074311…S3(0.12) = 0.1202882074310… , which isclose.
d. ln 1.7 = 0.530628251…S20(1.7) = 0.530612301… , which is close.ln 2.3 = 0.83290912…S20(2.3) = −4.42067878… , which is notclose.
R5. a. A Maclaurin series is a Taylor series expandedabout x = 0.
b. Substitute t = x + 1 into
ln ( – ) ( – )t t t t= − − +( )11
21
1
312 3
− +1
41 4( – )t L .
ln ( )x x x x x+ = − + − +1
1
2
1
3
1
42 3 4 L
=+
=
∞
∑ (– )1 1
1
nn
nn
x
c. Assume one may integrate this function termby term.
ln ( )x dx x x x C+ = −
⋅+
⋅− +∫ 1
1
2
1
3 2
1
4 32 3 4 L
=+
++
+
=
∞
∑ (– )
( )
1
1
11
1
nn
nn n
x C
d. ln ln( ) ( ) ( ) ( )x dx x x x C+ = + + − + +∫ 1 1 1 1 1
= + + + − + = −x x x x C C Cln ln( ) ( ) ( )1 1 11
= − + − +x x x x2 3 4 51
2
1
3
1
4L
+ − + − + − +x x x x x C
1
2
1
3
1
42 3 4 L
= −⋅
+⋅
− +1
2
1
3 2
1
4 32 3 4x x x CL ,
which is the same as the series in part c.
e.
t t dt t t t dtxx
cos!( )
!( )2 2 2 2 4
001
1
2
1
4= − + −
∫∫ L
= + +
∫ t t t t dt
x
–! !
–!
1
2
1
4
1
65 9 13
0L
= −⋅
+⋅
−⋅
+1
2
1
6 2
1
10 4
1
14 62 6 10 14x x x x
! ! !L
(Note that the series can be transformed to
= − + − +
1
2
1
3
1
5
1
72 2 3 2 5 2 7x x x x
!( )
!( )
!( ) L
= =∫1
2
1
22 2 2
0sin cos sin )x t t dt x
x
, and .
340 Problem Set 12-9 Calculus Solutions Manual© 2005 Key Curriculum Press
f. tan− =+∫1
20
1
1x
tdt
x
= − + − + −∫ [ ( ) ( ) ( ) ]1 2 2 2 2 3 2 4
0t t t t dt
x
L
(| | )t ≤ 1
= − + − + −x x x x x
1
3
1
5
1
7
1
93 5 7 9 L
g. f(3) = 5 ⇒ c0 = 5f ′(3) = 7 ⇒ c1 = 7f ″(3) = −6 ⇒ c2 = −6/2! = −3
′′′ = ⇒ = =f c( ) . . / ! .3 0 9 0 9 3 0 153
∴ f (x) = 5 + 7(x − 3) − 3(x − 3)2 +0.15(x − 3)3 + L
R6. a. = − −−
=
∞
∑ ( ) ( )3 51
n n
n
x
= − + − +1
35
1
95
1
2752 3( – ) ( – ) ( – )x x x L
b. Lx
xx
n
n n
n n= = −→∞
+ +
lim(– ) ( – )
(– ) ( – )| |
–( )
–
3 5
3 5
1
35
1 1
L < 1 ⇔ |x − 5| < 3 ⇔ 2 < x < 8Open interval of convergence is (2, 8).Radius of convergence = 3
c. cosh( )!
xn
x n
n
==
∞
∑ 1
22
0
Lx
n
n
xn
n
n=+
⋅→∞
+
lim( )!
( )!2 2
22 2
2
=+ +
= ⋅→∞
xn n
xn
2 21
2 2 2 10lim
( )( )L < 1 for all x.Series converges for all x, Q.E.D.
d. e1 2 2 31 1 21
21 2
1
31 2. .= + + +
!( . )
!( . )
+ +1
41 2 4
!( . ) L
S4(1.2) = 3.2944 (the fifth partial sum)e1.2 = 3.32011692…Error = e1.2 − S4(1.2) = 0.02571692…
The first term of the tail is t551
51 2= =
!( . )
0.020736.The error is greater than t5, but not muchgreater.
e.
x
ln
S11
S10
y
1
1 2
The open interval of convergence is (0, 2).Both partial sums fit ln well within this
interval. Above x = 2 the partial sumsdiverge rapidly to ±∞. Below x = 0 the partialsums give answers, but there are no realvalues for ln x.
R7. a. S10
10
10001 0 8
1 0 84463 129088= ⋅ =– .
– .. (exactly)
b. S n
n
= = ==
∞
∑1000 0 81000
1 0 85000
0
.– .
S − S10 = 536.870912, which differs from thelimit by about 10.7%.
c. “Tail”
d. “Remainder”
e. R x dx bb
1010
< = − +
−
→∞
−∞
∫ 3 2 21
2
1
210lim ( )–
= 0.005
R x dx bb
103 2 21
2
1
211> = − +
−
→∞
−∞
∫ lim ( )–
11
= 0.004132…Series converges because the tail is boundedabove by 0.005.S = S10 + R10 ≈ 1.197531… + 0.5(0.005 +0.004132…) = 1.202098…R10 is approximately 0.5(0.005 − 0.004132…)= 0.0004338… , so S10 is correct to aboutthree decimal places.
f. − + + + +1
4
1
3
1
22
1
59L
Ln
n
n
nn n= = =
→∞ →∞lim
/( – )
/lim
–
1 5
1 51
3
3
3
3
(Apply l’Hospital’s rule three times.)∴ the series converges because L is a positivereal number.The terms of the F series begin1
1
1
8
1
27
1
64+ + + + L .
Although the F series converges, its terms(after t1) are less, not greater, than thecorresponding terms of the S series, sothe comparison test is inconclusive.
g. 2/1! + 4/2! + 8/3! + 16/4! + 32/5! + L= 2 + 2 + 1.3333… + 0.6666…
+ …+ ==
∞
∑0 2666 21
. / !L n
n
n
The terms are decreasing starting at t2, whichcan be seen numerically, above, oralgebraically by the fact that the next term isformed by multiplying the numerator by 2and the denominator by more than 2.R1 is bounded by the geometric series withfirst term 2 and common ratio1.3333…/2 = 2/3.
Calculus Solutions Manual Problem Set 12-9 341© 2005 Key Curriculum Press
Because | common ratio | is less than 1,the geometric series converges(to 2/(1 − 2/3) = 6).
Thus, the tail after the first partial sum isbounded above by a convergent geometricseries, Q.E.D.
h. The given series converges because, aswritten, it meets the three hypotheses of thealternating series test. It does not convergeabsolutely because replacing all minus signswith plus signs gives the divergent harmonicseries.
The given series is the Taylor series for ln xexpanded about x = 1 and evaluated at x = 2.The remainders approach zero, so the seriesconverges to ln 2.
Rearrange the series this way:
11
2
1
4
1
3
1
6
1
8
1
5
1
10– – –
− +
− +
− +1
12L
Each term of the series appears exactly once.Simplifying inside the parentheses andfactoring out 1/2 gives1
2
1
4
1
6
1
8
1
10
1
12− + − + − + L
= + + +
1
2
1
1
1
2
1
3
1
4
1
5
1
6– – – L
The series in parentheses is the original seriesthat converges to ln 2. So the series asrearranged converges to 0.5 ln 2, Q.E.D.
i. |R10000| < |t10001| = 1/10001
= 0.0000999900…
Upper bound is 1/10001.
j. i. 10 3
21
n n
n
x
n
( – )
=
∞
∑
Lx
n
n
x
xn
nx
n
n n
n n
n
=+
⋅−
= −+
= − ⋅
→∞
+ +
→∞
lim( – )
( ) ( )
10 3
1 10 3
10 31
10 3 1
1 1
2
2
2
| | lim | |
L < 1 ⇔ 10 |x − 3| < 1 ⇔ 2.9 < x < 3.1
At x = 2.9 the series is
− + − + −1
1
4
1
9
1
16L ,
which converges by the alternating seriestest.
At x = 3.1 the series is
1
1
4
1
9
1
16+ + + + L ,
which converges because it is a p-serieswith p = 2.
Interval of convergence is [2.9, 3.1].
ii. (– ) ( )1 1
21
n
n
n
n
x
n
+⋅=
∞
∑
Lx
n
n
x
xn
nx
n
n
n
n
n
n
= ++ ⋅
⋅ ⋅+
= ++
= + ⋅
→∞
+
+
→∞
lim( )
( ) ( )
lim
1
1 2
2
1
1
21
1
1
21 1
1
1
| | | |
L x x< ⇔ + < ⇔ + <11
21 1 1 2| | | |
⇔ − < <3 1x
At x = −3 the series is
1
1
2
1
3
1
4+ + + + L , which is a divergent
harmonic series.
At x = 1 the series is
− + − + −1
1
2
1
3
1
4L , which converges
by the alternating series test.
Interval of convergence is (−3, 1].
k. i. 10
10 10 5 1 66660
nn
!= + + +
=
∞
∑ . K
+ +0 4166. K L
The tail after S0 is bounded above by theconvergent geometric series with firstterm 10 and common ratio 0.5. Thus, theseries converges.
(Other justifications are possible.)
ii. ( )– –nn
3 1
1
5+=
∞
∑ = 1.2 + 0.325 + 0.2370… +
0.215625 + 0.208 + 0.2046… +
0 2029. ... + L
Diverges because tn → 0.2, not 0,as n → ∞
iii. Converges. The general term can berewritten 3(2/5)n, so the series is aconvergent geometric series with commonratio r = 2/5.
iv. Diverges. p-series with p = 1/3, which isnot greater than 1.
v. Converges by the ratio technique
lim( )!
! ( )!
! !
( )!n n
nn
n
n
n→∞ ++
⋅ + ⋅⋅ ⋅ ⋅
+4
3 1 3
3 3
31
= ++
= ⋅ <→∞
1
3
4
1
1
31 1lim
n
n
n
342 Problem Set 12-9 Calculus Solutions Manual© 2005 Key Curriculum Press
R8. a. cosh( )!
21
222
0
= ⋅=
∞
∑ nn
n
Fourth partial sum is S3(2).The (2n)th derivative of cosh x is cosh x,so all derivatives are bounded by
cosh ( )–21
23 2 4 6252 2< + = . .
| ( )| .R3
824 625
82 0 02936≤ ⋅ =.
!K
Error is less than 0.03.
b. en
n
n
3
0
13= ⋅
=
∞
∑ !
All derivatives of ex are equal to ex, so allderivatives are bounded by e3 < 33 = 27.For 20-place accuracy,
| ( )| . .Rnn
n327
13 0 5 101 20≤
+⋅ < ×+ −
( )!
The second inequality is first true for n = 33.Use at least 34 terms (n = 33).
c. Using the Lagrange form of the remainder, thevalue of cosh 4 is given exactly by
cosh( )!
41
24 42
0
= ⋅ +=
∑ nRn
k
n
k
( ), where
Rf c
kk
kk( )4
2 24
2 22 2=
+⋅
++
( ) ( )
( )! and c is between
0 and 4.
| ( )|RM
kkk4
2 242 2≤
+⋅ +
( )!Because all even derivatives of cosh x equalcosh x, for any value of x between 0 and 4 wecan use cosh 4 for M, and cosh 4 is less than1
23 2 40 531254 4( )–+ = . .
Use M = 41.
lim lim( )!k
kk
kRk→∞ →∞
+<+
⋅| ( )|441
2 242 2
=+
=→∞
+
414
2 20
2 2
lim( )!k
k
k
By the ratio technique, this fractionapproaches zero as k approaches infinity.Therefore, because the remainder approacheszero as k approaches infinity, cosh 4 is given
exactly by cosh( )!
,41
242
0
= ⋅=
∞
∑ nn
n
Q.E.D.
d. sinh( )!
0 61
2 10 62 1
0
. .=+
⋅ +
=
∞
∑ nn
n
S3(0.6) = 0.636653554…sinh 0.6 = 0.636653582…
sinh 0.6 − S3(0.6) = 2.7862… × 10− 8
Rf c c
3
2 3 32 3 3 90 6
2 3 30 6
90 6( . ) =
⋅ +⋅ = ⋅
⋅ +⋅ +
( ) ( ).
cosh
!.
= 2.7862… × 10− 8
cosh c = 1.00328…c = cosh− 1
1.00328… = 0.0809… , which isin the interval (0, 0.6).
e. ln (– ) ( – )xn
xn n
n
= +
=
∞
∑ 11
11
1
| Rn(1.3) | < | tn+ 1 |For 20-place accuracy, make
1
11 3 1 0 5 101 20
nn
+< ×+ −( . – ) . .
This inequality is first true for n = 35.Use at least 35 terms.
f. Sn
n
50 41
501
1 08232064= ==
∑ . K
R x dx xb
b50
4 3
501 3< = −
=−
→∞
−∞
∫ lim ( / )50
(1/3)(50− 3) = 0.000002666…
R x dx xb
b50
4 3
511 3> = −
=−
→∞
−∞
∫ lim ( / )51
(1/3)(51− 3) = 0.000002512…
The series converges because the sequence ofpartial sums is increasing, and the tail afterS50 is bounded above by 0.000002512… .R50 ≈ 0.5(0.000002666… + 0.000002512…)
= 0.000002589…S ≈ 1.082232064… + 0.000002589…
= 1.082323235…Error < 0.5(0.000002666… − 0.000002512…)= 0.0000000769… (about seven decimalplaces)
Concept Problems
C1. Recall that i i i i i= = − = − =–1 1 12 3 4, , , ,
so i4n = 1 and i4n+ 2 = −1 for all n.
a. cos!( )
!( )
!( )ix ix ix ix= − + −1
1
2
1
4
1
62 4 6
+ −1
88
!( )ix L
= − + − + −1
2 4 6 8
22
44
66
88i
xi
xi
xi
x! ! ! !
L
= − + − + −1
1
2
1
4
1
6
1
82 4 6 8–
! !
–
! !x x x x L
= + + + + +11
2
1
4
1
6
1
82 4 6 8
! ! ! !x x x x L
= cosh x, Q.E.D.
Calculus Solutions Manual Problem Set 12-9 343© 2005 Key Curriculum Press
b. sin
!( )
!( )
!( )ix ix ix ix ix= − + − +1
3
1
5
1
73 5 7 L
= − + − +ix i
ix i
ix i
ix
23
45
67
3 5 7! ! ! L
= − + − +ix i x i x i x
–
! !
–
!
1
3
1
5
1
73 5 7 L
= + + + +
i x x x x
1
3
1
5
1
73 5 7
! ! !L
= i sinh x, Q.E.D.
c. e ix ix ix ixix = + + + +11
2
1
3
1
42 3 4
!( )
!( )
!( )
+ +1
55
!( )ix L
= + + + + + +1
2 3 4 5
22
33
44
55ix
ix
ix
ix
ix
! ! ! !L
= + + + + + +11
2 3
1
4 52 3 4 5ix x
ix x
ix
–
!
–
! ! !L
= − + −1
1
2
1
42 4
! !x x L
+ − + −
i x x x
1
3
1
53 5
! !L
= cos x + i sin x, Q.E.D.
d. Using the formula in part c (Euler’s formula):
e iπ = cos π + i sin π = −1 + i ⋅ 0 = −1,Q.E.D.
C2. tan tan tan41
5
1
2391 1− −−
=
+
⋅
tan tan – tan tan
tan tan tan tan
– –
– –
415
1239
1 415
1239
1 1
1 1
=
+
⋅
tan tan –
tan tan
–
–
415
1239
1 415
1239
1
1
To evaluate tan 41
51tan−
, recall that
tantan
– tan2
2
1 2AA
A= .
Therefore,
tan tan
–
21
5
25
115
5
121
2−
=
=
and tan tan
–
2 21
5
1012
15
12
120
1191
2⋅
=
=−
Substituting this value gives
tan tan tan–
41
5
1
239
120119
1239
1120119
1239
11 1− −−
=
+ ⋅=
Thus, 41
5
1
2391
41 1 1tan tan tan− − −− = = π
, Q.E.D.
The two series are
41
54
1
5
4
3
1
5
4
5
1
5
4
7
1
51
3 5 7
tan− = ⋅ −
+
−
+K
tan− = ⋅ −
+
−1
3 51
2394
1
239
4
3
1
239
4
5
1
239L
Rnn
n1
5
4
2 3
1
5
2 3
<
+
+
Rnn
n1
239
1
2 3
1
239
2 3
<
+
+
| Total remainder |
<+
⋅
+
+ +1
2 34
1
5
1
239
2 3 2 3
n
n n
To get π accurate to 50 places, as shown, theremainder for π must be less than 0.5 × 10− 50.For π/4, the remainder must be less than0.125 × 10− 50.The inequality
1
2 34
1
5
1
2390 125 10
2 3 2 350
n
n n
+⋅
+
< ×
+ +−.
is first true for n = 34.Use at least 35 terms.
C3. a. y″ + 9xy = 0; y = 5 and y′ = 7 when x = 0.y = c0 + c1x + c2x
2 + c3x3 + c4x
4 + c5x5
+ +c x66 L
y′ = c1 + 2c2x + 3c3x2 + 4c4x
3 + 5c5x4
+ +6 65c x L
y″ = 2c2 + 6c3x + 12c4x2 + 20c5x
3
+ +30 64c x L
b. Substitute y = 5 into the y-equation: c0 = 5Substitute y′ = 7 into the y′-equation: c1 = 7
c. Constant term: 2c2 = 0 ⇒ c2 = 0
x c c c-term: .6 9 01
69 5 7 53 0 3+ = ⇒ = ⋅ = −(– )
x c c24 112 9 0-term: + = ⇒
c41
129 7 5 25= ⋅ = −(– ) .
x c c c35 2 520 9 0
1
209 0 0-term: + = ⇒ = ⋅ =(– )
x c c46 330 9 0-term: + = ⇒
c61
309 7 5 2 25= =(– )(– . ) .
d. y = 5 + 7x + 0x2 − 7.5x3 − 5.25x4 + 0x5
+ +2 25 6. x L
S6(0.3) = 5 + 7(0.3) − 7.5(0.3)3 − 5.25(0.3)4
+ 2.25(0.3)6 = 6.85661525
344 Problem Set 12-9 Calculus Solutions Manual© 2005 Key Curriculum Press
e. To ascertain convergence or divergence, noticethat y can be written as three separate series.
y c c x c x= +
⋅+
⋅ ⋅ ⋅+0 0
32
069
2 3
9
2 3 5 6
(– ) (– )L
+ +
⋅+
⋅ ⋅ ⋅+c x c x c x1 1
42
179
3 4
9
3 4 6 7
(– ) (– )L
+ +
⋅+
⋅ ⋅ ⋅+c x c x c x2
22
52
289
4 5
9
4 5 7 8
(– ) (– )L
If 0.3 is substituted for x, the first and secondseries have terms that are strictly alternating,decreasing in absolute value, and approachingzero for a limit as n approaches infinity.Thus, these series converge by the alternatingseries test. The third series is zero becausec2 = 0. Thus, the entire series for y convergeswhen x = 0.3.
Chapter Test
T1. e x x x
nxx n n− = − + − + + − ⋅ +1
1
2
1
31
12 3
! ! !L L( )
Note: The last ellipsis mark is necessary or thiswould stand for a Taylor polynomial (finitenumber of terms), not a Maclaurin series(infinite number of terms).
T2.
5
15 5 5 52 3
+= − + − +
xx x x L
Geometric series, common ratio r = −x
T3. cos!( – )
!( – )x x x= − + −1
1
2
1
42 4π π
+ −1
66
!( – )x π L
T4. R xf c
x5
66
6( ) =
( ) ( )
!, where c is between 0 and x.
f x n x f c cn n n( ) ( )( ) ( ) ( !)( ) ! = − + ⇒ =− + −1 1 61 6 7( ) ( )
∴ = =R xc
xx
c5
76
6
7
6
6( ) ,
!
!
–
where 0 < c ≤ x.
T5. sin
!( )
!( )
!( )( )x x x x x2 2 2 3 2 5 2 71
3
1
5
1
7= − + − +L
= − + − +x x x x2 6 10 141
3
1
5
1
7! ! !L
=+
+
=
∞
∑(– )( )!
11
2 14 2
0
n n
nn
x
T6. The alternating harmonic series
1
1
2
1
3
1
4− + − + L converges conditionally, but
not absolutely. The condition is that the termsremain in the order presented and not berearranged.
T7. f(x) = ln x, f(1) = 0, c0 = 0f ′(x) = x− 1, f ′(1) = 1, c1 = 1
′′ = − ′′ = − = − = −−f x x f c c ( ) , ( ) , ! ,22 21 1 2 1
1
2
′′′ = ′′′ = = =−f x x f c c( ) 2 1 2 3 21
33
3 3, ( ) , ! ,
f x x f c( ) ( )( ) ! , ( ) !, ! !,4 4 443 1 3 4 3= − = − = −−
c4
1
4= − L
∴ = − − + ( )ln ( – ) ( – )x x x x11
21
1
312 3
− +1
41 4( – )x L , Q.E.D.
T8. 1000 999+ +L converges becauser = 0.999 < 1.(It converges to 1000/(1 − 0.999) = 1,000,000.) 0 0001 0 0002. .+ +L diverges because r = 2 ≥ 1.
T9.( – )2 5
31
x
n
n
n=
∞
∑
Lx
n
n
xn
n
n=+
⋅→∞
+
lim( – )
( – )
2 5
3 3
3
2 5
1
= −+
= − ⋅→∞
| | | |2 51
2 5 1xn
nx
nlim
L < 1 ⇔ |2x − 5| < 1 ⇔ −1 < 2x − 5 < 1⇔ 4 < 2x < 6 ⇔ 2 < x < 3Open interval of convergence is (2, 3).Radius of convergence is 0.5.
T10. At x = 2 the series is
− + − + −1
3
1
6
1
9
1
12L , which converges by
the alternating series test.At x = 3 the series is
1
3
1
6
1
9
1
12+ + + +L , which is a divergent
harmonic series (or 1/3 of p-series withp = 1).Converges at x = 2, diverges at x = 3
T11. f xt
dtx
( ) =+∫ 1
1 20
= + +∫ ( – – )1 2 4 6
0t t t dt
x
L
= − + − +x x x x
1
3
1
5
1
73 5 7 L
=+
+
=
∞
∑(– )11
2 12 1
0
n n
nn
x
(The same as tan− 1 x)
T12. Lx
n
n
xn
n
n=+
⋅ +→∞
+
+lim2 3
2 12 3
2 1
= ++
= ⋅→∞
xn
nx
n
2 22 1
2 31lim
L < 1 ⇔ x2 < 1 ⇔ −1 < x < 1
Calculus Solutions Manual Problem Set 12-10 345© 2005 Key Curriculum Press
At x = −1 the series is
− + − + −1
1
3
1
5
1
7L , which converges by the
alternating series test.
At x = 1 the series is 1
1
3
1
5
1
7− + − +L , which
converges by the alternating series test.Interval of convergence is [−1, 1].
T13. f (0.6) ≈ S19(0.6) (the 20th partial sum)
= 0.540419500…
T14. f (0.6) ≈ 0.540419500… numerically
T15. f (0.6) = tan− 1 0.6 = 0.540419500… exactlyThe answers to Problems T13 and T14 are correctto at least ten decimal places.
T16. The first term of the tail for S19(0.6) is
t2041 111
410 6 1 9562 10= = … × −( . ) . , which agrees
with the observation that S19(0.6) is correct to atleast ten decimal places.
T17. cosh! ! !
x x x x= + + + +11
2
1
4
1
62 4 6 L
==
∞
∑ 1
22
0( )!n
x n
n
T18. All even derivatives of cosh x equal cosh x.
Derivatives are bounded by
cosh . .31
2
1
23 2 13 56253 3 3 3= + < + = =( ) ( )– –e e M
For ten-place accuracy,
| ( )| . .Rnn
n313 5625
2 23 0 5 102 2 10≤
+⋅ < ×+ −.
( )!
The second inequality is first true for n = 11.Use at least 12 terms (n = 11).
T19. a. p-series, with p = 1.5
b. x dx bb
−
→∞
∞− −= − + =∫ 1 5
1
0 5 0 52 2 1 2. lim [ ( )]. .
∴ the series converges because the integralconverges, Q.E.D.(As additional information, this calculationalso proves that R1 is bounded above by 2and thus that S is bounded above by 3.)
c. S100 = 2.41287409…
d. R x dx1001 5
100< −
∞
∫ .
= − + =→∞
− −lim [ ( )] .. .
bb2 2 100 0 20 5 0 5
R x dx1001 5
101> −
∞
∫ .
= +→∞
lim [– ( )]– . – .
bb2 2 1010 5 0 5
= 0.19900743…
R100 ≈ 0.5(0.2 + 0.19900743…) =0.19950371…
S = S100 + R100 ≈ 2.41287409… +0.19950371… = 2.61237781…
(As additional information, the error in R100
is less than 0.5(0.2 − 0.1900743…) =0.0004962… , making S correct to about twodecimal places.)
T20. Answers will vary.
Problem Set 12-10Cumulative Review Number 1
1. Limit: See Sections 2-2 and 2-5.Derivative: See Sections 3-2 and 3-4.Indefinite integral: See Section 5-3.Definite integral: See Section 5-4.
2. a. Continuity at a point: See Section 2-4.
b. Continuity on an interval: See Section 2-4.
c. Convergence of a sequence: A sequenceconverges if and only if lim
nnt→∞ exists.
d. Convergence of a series: A series convergesif and only if the sequence of partial sumsconverges.
e. Natural logarithm: See Section 3-9.
f. Exponential: ax = ex ln a
3. a. Mean value theorem: See Section 5-5.
b. Intermediate value theorem: See Section 2-6.
c. Squeeze theorem: See Section 3-8.
d. Uniqueness theorem for derivatives: SeeSection 6-3.
e. Limit of a product property: See Section 2-3.
f. Integration by parts formula: See Section 9-2.
g. Fundamental theorem of calculus: SeeSection 5-6.
h. Lagrange form of the remainder: SeeSection 12-8.
i. Parametric chain rule: See Section 4-7.
j. Polar differential of arc length: SeeSection 8-7.
4. a. f x t dtx
( ) sech= + ⇒∫ 13
f x x′ = +( ) sech1
b. f (x) = ax ⇒ f ′(x) = ax ln a
c. f (x) = xa ⇒ f ′(x) = axa− 1
d. f (x) = xx ⇒ ln f (x) = x ln x1/f (x) · f ′(x) = ln x + x · (1/x)f (x) = (ln x + 1) – f (x)f ′(x) = xx ln x + xx
346 Problem Set 12-10 Calculus Solutions Manual© 2005 Key Curriculum Press
e. e6x cos 3x dx dv ue6x cos 3x
6e6x 13sin 3x
36e6x – 19cos 3x
+
+
–
= +1
33
2
336 6e x e xx xsin cos
− ∫4 36e x dxx cos
5 36e x dxx cos ∫= + +1
33
2
336 6
1e x e x Cx xsin cos
e x dxx6 3cos∫= + +1
153
2
1536 6e x e x Cx xsin cos
f. cosh sinh cosh5 61
6x x dx x C= +∫
g. sec3 x dx∫= + + +1
2
1
2sec tan ln sec tanx x x x C| |
h. ( ) | |sin cos ln sin5 51
551x x dx x C− = +∫
i. limcos –
lim– sin
x x
x
x
x
x→ →=
0 2 0
7 1
13
7 7
26
= = −→
lim– cos
x
x0
49 7
26
49
26
j. L xx
x=→
lim ( – ) /
0
31
ln limln ( – )
lim–
–L
x
x xx x= = = −
→ →0 0
3 1 3
13
L = e− 3 = 0.0497…
5. a.dy
dxx y= − +0 2 0 3 0 3 1 8. . . , ( , )
x
y10
10
b. If x = 9, y ≈ 5.413… , which agrees with thegraph.
6. a. p = k(40 − y)
b. y = 0.1x2 ⇒ x = (10y)1/2
dA = 2x dy = 2(10y)1/2 dy
A y dy y= = ⋅∫ 2 102
10
2
310 3 2
0
40
( )1/2
0
40
( ) /
= 3200/3 = 1066.6666… yd2
(Or: Area = 2/3 of circumscribed rectangle =(2/3)(1600) = 3200/3, etc.)
c. dF = p dA = k(40 − y) ⋅ 2(10y)1/2 dy
F dF k= = …∫ 17066 60
40
. lb (exactly
256,000k/15)
d. dM = y dF = y ⋅ k(40 − y) ⋅ 2(10y)1/2 dy
M dM k= = …∫ 292 571 4, . lb-yd0
40
(exactly 10,240,000k/35)
e. F y M yM
F
k
k⋅ = = = =, yd
10240000 35
256000 1517
1
7
/
/By symmetry, x = 0.
Center of pressure is at 0 171
7,
.
7. a. z = 30 − 0.5y
b. For a cross section,A = 2xz = 2(10y)1/2(30 − 0.5y).A = 101/2(60y1/2 − y3/2)A′ = 101/ 2(30y− 1/ 2 − 1.5y1/ 2)
= (101/ 2)(y− 1/ 2)(30 − 1.5y)A′ = 0 ⇔ 30 − 1.5y = 0 ⇔ y = 20A′ is infinite ⇔ y = 0.A(0) = 0A(20) = 565.6854… (exactly 400 2 )A(40) = 400Maximum at y = 20; minimum at y = 0
c. dV = 2xz dy = 101/2(60y1/2 − y3/2) dy
V dV= =∫ 192000
40
(exactly)
Use 19200/5 = 3840 truckloads.
d. dL dx dy x dx= + = +2 2 21 0 04.
L dL= = … ≈−∫ 92 9356 92 9
20. . yd
20
8. r r rr t i t j= +( . ) ( . )100 0 03 50 0 03cos sin
r r rv t i t j= − +( . ) ( . . )3 0 03 1 5 0 03sin cos
Speed | |= = +rv (– sin . ) ( . cos . )3 1 5 1 5 1 52 2
= 2.9943… ≈ 2.99 ft/s
9. Si 0
tu
udu
t
= ∫ sin
= + +
∫ 1 1
3
1
5
1
73 5 7
uu u u u du
t
–! !
–!
L0
= + +
∫ 1
1
3
1
5
1
72 4 6
0–
! !–
!u u u du
t
L
Calculus Solutions Manual Problem Set 12-10 347© 2005 Key Curriculum Press
= −
⋅+
⋅−
⋅+t t t t
1
3 3
1
5 5
1
7 73 5 7
! ! !L
+
+ +++(– )
( )( )!
1
2 1 2 12 1
nn
n nt L
Lt
n n
n n
tn
n
n=+ +
⋅ + +→∞
+
+lim( )( )!
( )( )!2 3
2 12 3 2 3
2 1 2 1
= ++ + +
= ⋅→∞
tn
n n nt
n
2 22 1
2 3 2 3 2 20lim
( )
( )( )( )∴ L < 1 for all values of t, and the seriesconverges for all values of t.Third partial sum is
S23 50 6 0 6
1
3 30 6
1
5 50 6( . ) .= −
⋅+
⋅!( . )
!( . )
= 0.5881296
| | | | .R t2 371
7 70 6 0 0000007934< =
⋅= …
!( . )
The answer is correct to within ±1 in the sixthdecimal place.
Si . .0 6 0 5881288090
0 6
= ≈ …∫ sin. u
udu
Note that this answer agrees with the third partialsum to within 1 in the sixth decimal place.
10. r = 5 + 4 cos θ
dA d= +1
25 4 2( cos )θ θ
A dA= ≈ … ≈∫ 103 6725 103 7 332
0
2
. . ft (exactly )ππ
11.dV
dtkV
dV
Vk dt= ⇒ =
ln | V | = kt + CV = C1e
kt
At t = 0, V = 300.300 = C1
dV
dt= −5 when V = 300
− = ⇒ = −5 3001
60k k
∴ V = 300e− (1 /60) t
At t = 10, V = 300e− 1/6 = 253.9445… ≈ 253.9 million gal.
Cumulative Review Number 21. Derivative: See Sections 3-2 and 3-4.
2. Definite integral: See Section 5-4.
3. Mean value theorem: See Section 5-5.
4. f x g t dt f x g xx
( ) ( ) ( )= ⇒ ′ =∫3( )
5. tanh sec tanh5 61
6x x dx x Ch2 = +∫
6. x x dxsinh 2∫ dv u x sinh 2x1
12 cosh 2x
014 sinh 2x
+
–
+
= − +1
22
1
42x x x Ccosh sinh
7.3 14
3 2
1
3
4
2
x
x xdx
x xdx
++
=+
+
∫ ∫( )( – )
–
–= −ln | x + 3 | + 4 ln | x − 2 | + C
8.
sinh
! ! !
x
xdx
xx x x x dx∫ ∫= + + + +
1 1
3
1
5
1
73 5 7 L
= + + + +
∫ 1
1
3
1
5
1
72 5 6
! ! !x x x dxK
= +⋅
+⋅
+⋅
+…+x x x x C1
3 3
1
5 5
1
7 73 5 7
! ! !
9.n x n
nn
( )−
=
∞
∑ 5
31
Ln x
n xn
n
n
n
n= + ⋅→∞
+
+lim( )( – )
( – )
1 5
3
3
5
1
1
= − + = − ⋅ = −→∞
1
35
1 1
35 1
1
35| | | | | |x
n
nx x
nlim
L x x< ⇔ − < ⇔ − < − <11
35 1 3 5 3| |
Open interval of convergence is 2 < x < 8.
10. x dx x dxa a
−
→
−∫ ∫=+
0 998
0
1
0
0 9981
. .lim
=→ +lim
..
a ax
0
0 00211
0 002
= −→ +lim ( ).
aa
0
0 002500 500
= 500
11. y x= 2
yx dx
x= = ⋅ =∫ 2
3
9
3
3
9
9 3
1
6
1
339
–12. f ( x) = x2
f ( 4) = 16f ( 3.99) = 15.9201, which is within 0.08 unitof 16.f ( 4.01) = 16.0801, which is not within 0.08unit of 16.Thus, δ = 0.01 is not small enough to keepf (x) within 0.08 unit of 4.
13. V A dx= ∫2
10
≈ + + + +2
3153 4 217 2 285 4 319 343[ ( ) ( ) ( ) ]
= 2140 ft3
348 Problem Set 12-10 Calculus Solutions Manual© 2005 Key Curriculum Press
14. r = 4 sin 2θ
dA r d d= =1
28 22 2θ θ θsin
A d= ≈ ≈∫ 8 2 6 2831 6 282 2
0
2
sin/
θ θπ
. . ft
(exactly 2π)
15. (x/5)2 + (y/3)2 = 1
y x= ± +0 6 25 2. Use– ( .)
dA y dx x dx= =2 1 2 25 2. –
A dA= ≈ ≈∫ 17 6021 17 6
1
5
. .K square units
( sin sin )exactly ( . ) .15 1 0 2 1 2 61 1− −− −
16. y = 0.0016x4
dA = (16 − 0.0016x4) dx
A x dx= =−∫ ( – . )16 0 0016 2564 2
10
10
ft
17. dL dx dy x dx= + = +2 2 3 21 0 0064( . )
L dL= ≈ … ≈−∫ 42 5483 42 55
10
10
. . ft
18. p = 62.4(16 − y)dA = 2x dy = 10y1/4 dy
dF = p dA = 62.4(16 − y) ⋅ 10y1/4 dy
F dF= ≈ … ≈∫ 113595 73 113 6000
16
. , lb
exactly 11359511
15
19. lim limln
x xy
x
x→∞ →∞= → ∞
∞
= =→∞
lim/
x
x1
10
20. yx x x
x
x
x′ = =( / )( ) – (ln )( ) – ln1 1 1
2 2
y′ = 0 ⇔ ln x = 1 ⇔ x = e = 2.718… ftThere is a maximum at x = e because y′ goesfrom positive to negative there.
21. yx x x x
x
x
x′′ = = +(– / )( ) – ( – ln )( ) – ln1 1 2 3 22
4 3
y″ = 0 ⇔ ln x = 1.5 ⇔ x = e1.5 = 4.4816…≈ 4.48 ftThere is a point of inflection at x ≈ 4.48 ftbecause y″ changes sign there.
22.
1
40
x
y
23. ln ( – ) ( – )x x x x= − − + −( )1
1
21
1
312 3 L
+ +
+(– )( – )
11
1nn
nx L
Lx
n
n
xn
n
n=+
⋅→∞
+
lim( – )
( ) ( – )
1
1 1
1
= −+
= − ⋅ = −→∞
| | | | | |xn
nx x
n1
11 1 1lim
L < 1 ⇔ | x − 1 | < 1 ⇔ −1 < x − 1 < 1
⇔ 0 < x < 2
At x = 0 the series is − − − − −1
1
2
1
3
1
4L ,
which is a divergent harmonic series.
At x = 2 the series is 11
2
1
3
1
4− + − +L , which
converges because it meets the three hypothesesof the alternating series test.∴ interval of convergence is 0 < x ≤ 2, Q.E.D.
24. | | | |R tn
xn nn< =
+++
111
11( – )
For ln 1.4 to 20 places, make
0 4
10 5 10
120. n
n
+−
+< ×. .
Solving numerically gives n > 45.817… .Use 46 terms.
25. If the velocity is 0 ft/s at time t = 0, the shipspeeds up, approaching approximately 34 ft/sasymptotically as t increases.
If the velocity is 50 ft/s at time t = 0, the shipslows down, again approaching 34 ft/sasymptotically as t increases.(The graphs are shown here. The differentialequation is dv/dt = 0.7(34 − v).)
t
v
26.r r rr t i t j= +( ) ( )ln sin 2r r rv t i t j= +( / ) ( )1 2 2cosr r ra t i t j= − + −( / ) ( )1 4 22 sin
Calculus Solutions Manual Problem Set 12-10 349© 2005 Key Curriculum Press
Cumulative Review Number 31. δ is clearly smaller than necessary.
L+ε
L–ε
L
c – δ c c+δ
x
f(x)
2. See Sections 3-2 and 3-4 for definitions ofderivative.Graphical meaning: slope of tangent linePhysical meaning: instantaneous rate of change
3. g x f x dx g x f x( ) ( ) if and only if ( ) ( ). = ′ =∫4. f t dt L U
tn
tn
r
s
( ) ,= =→ →∫ lim lim
∆ ∆0 0 where Ln and Un are
lower and upper Riemann sums, respectively,provided the two limits are equal.
5. l’Hospital’s rule
limcos
–x x
x x
e→→
0 51
0
0
= →→
limcos – sin
– –x x
x x x
e0 55
1
5= −0.2
6. y = tan (sin 5x)y′ = sec2 (sin 5x) ⋅ 5 cos 5xChain rule
7. y = (5x − 3)(2x + 7)4(x − 9)ln y = ln (5x − 3) + 4 ln (2x + 7) + ln (x − 9)
y yx x x
′ = ++
+
5
5 3
8
2 7
1
9– –
8. y = tan− 1 xtan y = x, sec2 y y′ = 1
yy y
′ = =+
1 1
12 1sec tan–
yx
′ =+1
1 2
9. sin cos sin7 81
8x x dx x C= +∫
10. x dx2 9+∫ x = 3 tan θ
dx = 3 sec2 θ dθx2 9 3+ = sec θ
= ∫ 9 3sec θ θd
= + + +9
2
9
2 1sec tan ln sec tanθ θ θ θ| | C
= + ⋅ + + + +9
2
9
3 3
9
2
9
3 3
2 2
1x x x x
Cln
= + + + + +1
29
9
292 2x x x x Cln
11.3 11
2 3
5
3
2
12
x
x xdx
x xdx
–
–
–
–+=
++
∫∫
= 5 ln | x + 3 | − 2 ln | x − 1| + C
12. sin−∫ 1 x dx u dvsin–1 x 1
11 – x2
x–
+
√
= − −− −∫x x x x dxsin ( ) ( )/1 2 1 21
= + − −− −∫x x x x dxsin ( ) ( )/1 2 1 21
21 2
= + − +−x x x Csin 1 21
13. Fundamental theorem of calculusSee Section 5-6 for statement.
14. See Figure 5-5b.
15. f x h t dt f x h xx
( ) ( ) ( ) ( )= ⇒ ′ =∫3
16. f (x) = xe− x
f ′(x) = e− x − xe− x
f ″(x) = −e− x − e− x + xe− x = e− x(x − 2)f ″(x) = 0 ⇔ x = 2f ″(x) changes sign at x = 2.∴ the only point of inflection is at x = 2.
17. y = sin x from x = 0 to x = 2.
dL dx dy x dx= + = +2 2 21 cos
L dL= ≈∫ 2 3516
0
2
. K
18. a. x dx xa a
−
→=
+∫ 3 4
0
1 4
0
16 16
4/ /lim
= =→ +lim ( – )/
aa
0
1 48 4 8
b. Average value = =8
16 0
1
2–
19. r = 10 cos θdA = 50 cos2 θ dθ
A d= ≈ …∫ 50 13 34782cos θ θ .0.5
1
(exactly 12.5(1 + sin 2 − sin 1))
20.r r rr t i t j= + −2 13r r rv ti t j= − −2 3 2
r r rv i j( )1 2 3= −Speed .= = …13 3 6055
Distance from origin is | | .rr t t= +4 29 –
d r
dtt t t t
| |( )–
r
= + −− −1
29 4 184 2 1 2 3 3/ ( )
= −7 10/ at t = 1
Distance is decreasing at 2.2135… .
350 Problem Set 12-10 Calculus Solutions Manual© 2005 Key Curriculum Press
21. y = cos xdV = 2πx ⋅ y ⋅ dx
V x x dx= ≈ …∫ 2 3 58640
2
ππ
cos/
.
(exactly 2π (π/2 − 1))
22. a. y = −1.5x + 6A = xy = −1.5x2 + 6xA′ = −3x + 6A′ = 0 ⇔ −3x + 6 = 0 ⇔ x = 2A(0) = 0, A(4) = 0, A(2) > 0Thus, maximum area is at x = 2, Q.E.D.
b. V = πx2y = π(−1.5x3 + 6x2)V′ = π(−4.5x2 + 12x)
V x x′ = ⇔ = =0 0 22
3or
V V V( ) ( ) and0 4 0 22
30= =
>
Thus, maximum volume is at x = 22
3.
23. V ≈ + + + +1
32 51 4 37 2 41 4 63 59( )[ ( ) ( ) ( ) ]
= 3942
33ft
24. a. erf ( )/x e dttx
= − −∫2 1 2
0
2
π
f x e dttx
( ) = −∫2
0
= + +
∫ 1
1
2
1
3
1
42 4 6 8
0–
!–
! !–t t t t dt
x
L
= − +
⋅−
⋅+
⋅−x x x x x
1
3
1
5 2
1
7 3
1
9 43 5 7 9
! ! !L
b. f xn n
xn n
n
( ) =+
+
=
∞
∑(– )( ) !
11
2 12 1
0
Lx
n n
n n
xn
n
n=+ +
⋅ +→∞
+
+lim( )( )!
( ) !2 3
2 12 3 1
2 1
= ++ +
= ⋅→∞
xn
n nx
n
2 22 1
2 3 10lim
( )
( )( )L < 1 for all values of x, and thus the seriesconverges for all values of x, Q.E.D.
Final Examination
1.sin . – sin
.
1 1 1
0 1= 0.497363752…
sin . – sin
.
1 01 1
0 01= 0.536085981…
sin . – sin
.
1 001 1
0 001= 0.539881480…
2. f ′(1) = cos 1 = 0.540302305…The quotients in Problem 1 are converging tocos 1.
3. f xf x h f x
hh′ = +
→( ) lim
( ) – ( )0
f cf x f c
x cx c′ =
→( ) lim
( ) – ( )
–
4. f (x) = ex
limx
f x e→
=2
2( )
If f (x) = e2 + 0.1, x = ln (e2 + 0.1) =2.01344… .If f (x) = e2 − 0.1, x = ln (e2 − 0.1) =1.98637… .On the left, keep x within 0.01362… unit of 2.On the right, keep x within 0.01344… unit of 2.So you must keep x within 0.01344… unit of 2.
5. L f xx c
=→
lim ( ) if and only if for any ε > 0 there is
a δ > 0 such that if x is within δ units of c butnot equal to c, then f (x) is within ε units of L.
6. ε = 0.1, δ = 0.01344…
7. See Figure 1-3a.
8. Distance ≈ 17.4 m
10
1 1.3 1.6 1.9 2.2 2.5 2.8
t
r
9. Distance ≈ + + + +1
30 3 7 4 9 2 13 4 12( . )[ ( ) ( ) ( )
2 10 4 8 5 17 4( ) ( ) ] . ,+ + = which agrees withProblem 8.
10. v(t) = te− t
Distance ≈ 0.4[v(0.2) + v(0.6) + v(1) + v(1.4) +v(1.8)] = 0.601474…
11. Distance = = − −− − −∫ te dt te et t t
0
2
0
2
= −2e− 2 − e− 2 + 0 + 1 = 1 − 3e− 2 = 0.593994…
The difference is 0.00748… , which is about1.26%.
12. If f is integrable on [a, b] and g x f x dx( ) ( ) , = ∫then f x dx g b g a
a
b
( ) ( ) ( ).= −∫13. f (x) = x2/3
f x x′ = −( ) /2
31 3
f is differentiable everywhere except at x = 0. Butf is continuous at x = 0 because the limit of f (x)as x → 0 is zero, the same as f (0). Thus,
Calculus Solutions Manual Problem Set 12-10 351© 2005 Key Curriculum Press
f meets the hypotheses of the mean valuetheorem because it is differentiable on (0, 1) andcontinuous at 0 and 1.Slope of the secant line from (0, 0) to (1, 1) is 1.
f c c c′ = = ⇔ =−( )2
31
8
271 3/
Tangent at x = 8/27 is parallel to secant.
1
18/27
x
f(x)
14. a. Example: 5 3
2 3
x
x xdx
–
( – )( )+∫= +
+
∫ 1
2
4
3x xdx
–
= ln | x − 2 | + 4 ln | x + 3 | + C
b. Example: 9 2– x dx∫x = 3 sin θdx = 3 cos θ dθ
9 32– cosx = θ
= = +∫ ∫99
21 22cos cosθ θ θ θd d( )
= + +
= + +
9
2
9
42
9
2
9
2
θ θ
θ θ θ
sin
sin cos
C
C
= + +−9
2 3
1
291 2sin –
xx x C
15. sec3 x dx∫ dvusec x sec 2 x
sec x tan x tan x+
–
= − ∫sec tan sec tanx x x x dx2
= − +∫ ∫sec tan sec secx x x dx x dx3
2 3sec sec tan secx dx x x x dx= +∫ ∫sec sec tan ln sec tan3 1
2
1
2x dx x x x x C∫ = + + +| |
16.dy
dxky
dy
yk dx y kx C= ⇒ = ⇒ = + ⇒∫ ∫ ln | |
| y | = ekx+ C = ekxeC ⇒ y = C1ekx
17. Cross section of solid at any point in the slice isessentially the same as at the sample point.
2
4
x
y
(x, y)
18. Height at any point in the slice is essentially thesame as at the sample point.
2
4
x
y
(x, y)
19. dMy = x dA = x(4 − x2) dx
M x x dxy = − =∫ ( )4 42
0
2
A x dx= =∫ ( – )416
32
0
2
xA M xy= ⇒ = =4
16 3
3
4/
20. Let H = number of calories added.dH = C dT = (10 + 0.3T1/2) dT
H T dT= + =∫ ( . )/10 0 3 13 2001 2
100
900
, cal
21. a. Si Sixu
udu x
x
x
x
= ⇒ ′ =∫ sin sin0
b. Si x u u u dut
= + +
∫ 1
1
3
1
5
1
72 4 6
0–
! !–
!K
= −
⋅+
⋅−
⋅+t t t t
1
3 3
1
5 5
1
7 73 5 7
! ! !K
c. Si . ( . ) .0 7 0 7 0 71
3 30 71
3≈ = −⋅
=S!( . )
0.68094444…
d. | ( . )| | ( . )|R t1 250 7 0 7
1
5 50 7< =
⋅=
!( . )
0.0002801…S1(0.7) equals Si 0.7 correct to three decimalplaces and is within ±0.3 in the fourthdecimal place.
e. See Cumulative Review Number 1,Problem 9.
22. r r rr t i t j= +( ) ( )3 2
r r r r r rv t i t j v i j= + ⇒ = +( ) ( ) .3 2 0 5 0 75 12 ( . )r r r r r ra t i j a i j= + ⇒ = +( ) ( )6 2 0 5 3 2( . )
352 Problem Set 12-10 Calculus Solutions Manual© 2005 Key Curriculum Press
1
1
y
x
a
v→
→
The object is speeding up at t = 0.5 because theangle between the velocity and accelerationvectors is acute, indicating that the tangentialcomponent of acceleration acts in the samedirection as the velocity.(Algebraically,
r rv a⋅ = 4 25. , which is positive,
again indicating an acute angle.)
23. r = cos θ
dA r d d= =1
2
1
22 2θ θ θcos
A d= ≈∫ 1
20 23912
0
6
cos/
θ θπ
. K
exactlyπ24
3
16+
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