Download Calculus Concepts and Applications Paul A Foerster...
Calculus C o n c e p t s an d Ap p l i c a t i o n s
Second Edition
Solutions Manual
P a u l A. F o e r s t e r
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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 and Applications. Solutions or key steps in the solutions are presented for all but the simplest problems. In most cases the solutions are presented in the form your students would be expected to use. For instance, decimal approximations are displayed as exact answers using ellipsis format for a mathematical-world answer, then rounded to an appropriate number of decimal places with units of measurement applied for the corresponding real-world answer. 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 for possible use in subsequent computations. The ellipses indicate that the student chooses not to write all the digits on his or her paper. Because the problems applying to the real world may be somewhat unfamiliar to both you and your students, fairly complete solutions are presented for these. Often commentary is included over and above what the student would be expected to write to further guide your evaluation of students’ solutions, and in some cases reference is provided to later sections in which more sophisticated solutions appear. Later in the text, the details of computing definite integrals by the fundamental theorem are omitted because students are usually expected to do these numerically. However, exact answers such as V = 8π/3 are presented where 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 each student. The “prompts” in most problems calling for journal entries should be sufficient to guide students in making their own responses. Where programs are called for, you may use as a model the programs in the Instructor’s Resource Book. Check the publisher’s Web page (see the address on the copyright page of this manual) for further information on programs for specific models of the graphing calculator. If you or your students find any mistakes, please report them to Key Curriculum Press by sending in the Correction/Comment Form in the back of this book.
Paul A. Foerster
v
Chapter 1—Limits, Derivatives, Integrals, and Integrals Problem Set 1-1 1. a. 95 cm b. From 5 to 5.1: average rate ≈ 26.34 cm/s From 5 to 5.01: average rate ≈ 27.12 cm/s From 5 to 5.001: average rate ≈ 27.20 cm/s So the instantaneous rate of change of d at t = 5 is about 27.20 cm/s. c. Instantaneous rate would involve division by zero. d. For t = 1.5 to 1.501, rate ≈ −31.42 cm/s. The pendulum is approaching the wall: The rate of change is negative, so the distance is decreasing. e. The instantaneous rate of change is the limit of the average rates as the time interval approaches zero. It is called the derivative. f. Before t = 0, the pendulum was not yet moving. For large values of t, the pendulum’s motion will die out because of friction. 2. a. x = 5: y = 305, price is $3.05 x = 10: y = 520, price is $5.20 x = 20: y = 1280, price is $12.80 b. x = 5.1, rate ≈ 46.822 ¢/ft x = 5.01, rate ≈ 46.9820… ¢/ft x = 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 the 10-ft board. The reason is that longer boards require taller trees, which are harder to find.
Problem Set 1-2 Q1. Power function, or polynomial function Q2. f (2) = 8 Q3. Exponential function Q4. g (2) = 9 Q5. h (x )
3. a. Decreasing fast 4. a. Decreasing slowly 5. a. Increasing fast c. Decreasing slowly 6. a. Decreasing fast c. Increasing fast 7. a. Increasing slowly c. Increasing slowly 8. a. Decreasing fast c. Decreasing fast 9. a. Increasing fast
b. b. b. d. b. d. b.
Decreasing slowly Increasing slowly Increasing slowly Increasing fast Increasing slowly Decreasing fast Increasing slowly
b. Decreasing fast b. Neither increasing nor decreasing d. Increasing slowly b. Decreasing fast d. Neither increasing nor decreasing
c. Increasing fast 10. a. Decreasing slowly c. Decreasing fast 11. a. T(x) (°C) 100
50
x (s) 100
200
x = 40: rate ≈ 1.1°/s x = 100: rate = 0°/s x = 140: rate ≈ −0.8°/s b. • Between 0 and 80 s the water is warming up, 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. v ( x ) (ft/s)
1
x 1
Q6. Q8. Q10. 1. 2.
h (5) = 25 Q7. y=x Q9. Derivative a. Increasing slowly a. Increasing fast
Calculus Solutions Manual © 2005 Key Curriculum Press
y = ax2 + bx + c, a ≠ 0 y = |x|
70 60 50 40 30 20
b. Increasing fast b. Decreasing slowly
10
x (s) 1
2
3
4
5
6
7
8
Problem Set 1-2
1
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. h (x)
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.
f (4.01) f (4) = 129.9697… 0.01 f (4) f (3.99) = 131.4833… 0.01 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. a(2.1) a(2) 15. a. Average rate = = 0.1 2 52.9902… mm /h d.
18
x
2 3 4
7
• Increasing at x = 3 • Decreasing at x = 7 b. h (3) = 17, h(3.1) = 17.19 0.19 = 1.9 ft/s Average rate = 0.1 c. From 3 to 3.01: 0. 0199 average rate = = 1. 99 ft/s 0.01 From 3 to 3.001: 0.001999 average rate = = 1.99 ft/s 0.001 The limit appears to be 2 ft/s. d. h (7) = 9, h(7.001) = 8.993999 0.006001 = 6.001 ft /s Average rate = 0.001 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. f (t )
Not much
500
300 Decrease
Increase
100
t 10
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
b. r (t) =
200(1.2 t ) 200(1.2 2 ) t2
r (t ) (mm2/hr) 60 40 20
t (mm) 2
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. 4 16. a. v(x ) = x 3 v(6) = 288 3 4 (6.13 63 ) b. 6 to 6.1: average rate = 3 = 0.1 146.4133… 4 (6 3 5. 93 ) 5.9 to 6: average rate = 3 = 0.1 141.6133… Estimate of instantaneous rate is (146.4133… + 141.6133…)/2 = 144.0133… = 452.4312… cm3/cm. 4 x 3 4 63 3 c. r (x ) = 3 x6 r(x ) (cm3/cm)
undefined
1.01
107.2171…
1.02
106.3703…
1.03
105.5200…
144π 48π
x 6
r(6) is undefined.
2
Problem Set 1-2
Calculus Solutions Manual © 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
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.
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)
Problem Set 1-3 Q1. 72 ft2
Q2. y = cos x
Q3. y = 2
x
Q4. y = 1/x
2
Q6. f(5) = 4
Q5. y = x Q7.
Q8. y
y
20. a. Quadratic (or polynomial) x
b. f(1) = 12
x
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.
Q10. x = 3
Q9. y
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.
x
1. f(x) = 0.1x2 + 7
24. a. Rational algebraic b. Increasing, because the rate of change from 2.01 to 1.99 is positive.
2. f(x) = 0.2x2 + 8
a. Approximately 30.8
a. Approximately 22.2
b. Approximately 41.8
b. Approximately 47.1
7
8 f(x)
f (x)
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)
x –2
28. a. Circular (or trigonometric) b. Decreasing, because the rate of change from 0.99 to 1.01 is negative.
• 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 Calculus Solutions Manual © 2005 Key Curriculum Press
3
5
4. g( x) = 2 x + 5
3. h(x) = sin x
b. Decreasing, because the rate of change from 1.99 to 2.01 is negative.
29. • Physical meaning of a derivative: instantaneous rate of change
x
5 6
–1
a. Approximately 2.0
a. Approximately 7.9
b. Approximately 1.0
b. Approximately 12.2 g ( x)
h(x)
1
6
x 3
x
–1
1
Problem Set 1-3
2
3
5. There are approximately 6.8 squares between the curve 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 the curve and the x-axis. Each square represents (0.5)(10) = 5 miles. So the distance is about (53.3)(5) = 266.5 miles. tan 1.01 – tan 0.99 = 3.42K 7. Derivative ≈ 1.01 – 0.99 8. Derivative = −7 (exactly, because that is the slope of the linear function) 9. a. v (t )
v(3.01) – v(2.99) = 1.8648K 3.01 − 2.99 About 1.86 (ft/s)/s The derivative represents the acceleration. From t = 0 to t = 5, the object travels about 11.4 cm. From t = 5 to t = 9, the object travels back about 4.3 cm. So the object is located about 11.4 − 4.3 = 7.1 cm from its starting point. See the text for the meaning of derivative. See the text for the meaning of definite integral. See the text for the meaning of limit. d. Rate ≈
11.
12. 13. 14.
Problem Set 1-4 Q1. y changes at 30 Q3.
100
Q2. Derivative ≈ −500 Q4. f (3) = 9
y 60
x
t 5
8.7 10
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. v(5.01) − v( 4.99) d. Rate ≈ = 3.1107K 5.01 − 4.99 About 3.1 (ft/s)/s The concept is the derivative. The rate of change of velocity is called acceleration. 10. a. 10
v (t)
5
t 1
2
3
4
5
b. v(4) = 9.3203… ≈ 9.3 ft/s Domain: 0 ≤ t ≤ 4 Range: 0 ≤ v(t) ≤ 9.3203… c. By counting squares, the integral from t = 0 to t = 4 is about 21.3 ft. The units of the integral are (ft/s) · s = ft. The integral tells the length of the slide.
4
Problem Set 1-4
Q5. Q7. Q9. 1.
100 366 days Definite integral a.
Q6. sin (π/2) = 1 Q8. Derivative Q10. f (x) = 0 at x = 4
v (t ) 20,000
t 30
b. Integral ≈ 5(0.5v(0) + v(5) + v(10) + v(15) + v(20) + v(25) + 0.5v(30)) = 5(56269.45…) = 281347.26… ≈ 281,000 ft The sum overestimates the integral because the trapezoids are circumscribed about the region and thus include more area. c. The units are (ft/s)(s), which equals feet, so the integral represents the distance the spaceship has traveled. d. Yes, it will be going fast enough, because v(30) = 27,919.04… , which is greater than 27,000. 2. a. v(t) = 4 + sin 1.4t v (t ) 5
t 3
Calculus Solutions Manual © 2005 Key Curriculum Press
3. 4.
5.
6.
7.
b. A definite integral has the units of the x-variable times the y-variable. Distance = rate × time. Because v(t) is distance/time and t is time, their product is expressed in units of distance. c. See graph in part a. Distance ≈ 0.5(0.5v(0) + v(0.5) + v(1) + v(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/h Maximum speed was 5 mi/h at about 1.12 h. Distance ≈ 0.6(150 + 230 + 150 + 90 + 40 + 0) = 396 ft Volume ≈ 3(2500 + 8000 + 12000 + 13000 + 11000 + 7000 + 4000 + 6000 + 4500) = 204,000 ft3 Programs will vary depending on calculator. See the program TRAPRULE in the Instructor’s Resource Book for an example. The program gives T20 = 23.819625. See the program TRAPDATA in the Instructor’s Resource Book for an example. The program gives T7 = 33, as in Example 2. a.
9.
f (x ) 7
x 1
4
b. T10 = 18.8955 T20 = 18.898875 T50 = 18.89982 These values underestimate the integral, because the trapezoids are inscribed in the region. c. T10: 0.0045 unit from the exact answer T20: 0.001125 unit from the exact answer T50: 0.00018 unit from the exact answer Tn is first within 0.01 unit of 18.9 when n = 7. T7 = 18.8908… , which is 0.0091… unit from 18.9. Because Tn is getting closer to 18.9 as n increases, Tn is within 0.01 unit of 18.9 for all n ≥ 7. 8. a.
10.
11.
12.
13.
b. T10 = 8.6700… T20 = 8.6596… Τ50 = 8.65672475… These values overestimate the integral, because the trapezoids are circumscribed about the region. c. T10: 0.01385… unit from answer T20: 0.003465… unit from answer T50: 0.0005545… unit from answer Tn is first within 0.01 of 8.65617024… when n = 12. T12 = 8.665795… , which is 0.009624… unit from 8.65617024… . Because Tn is getting closer to the exact answer as n increases, Tn is within 0.01 unit of the answer for all n ≥ 12. From the given equation, y = ±( 40/110) 110 2 – x 2 . Using the trapezoidal rule program on the positive branch with n = 100 increments gives 6904.190… for the top half of the ellipse. Doubling this gives an area of 13,808.38… cm 2 . The estimate is too low because the trapezoids are inscribed within the ellipse. The area of an ellipse is πab, where a and b are the x- and y-radii, respectively. So the exact area is π (110)(40) = 4400π = 13,823.007… cm 2 , which agrees both with the answer and with the conclusion that the trapezoidal rule underestimates the area. 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) = 190.6 The integral will have the units (in.2)(in.) = in.3, representing the volume of the football. n = 10: integral ≈ 21.045 n = 100: integral ≈ 21.00045 n = 1000: integral ≈ 21.0000045 Conjecture: integral = 21 The word is limit. The trapezoidal rule with n = 100 gives integral ≈ 156.0096. Conjecture: integral = 156 If the trapezoids are inscribed (graph concave down), the rule underestimates the integral. If the trapezoids are circumscribed (graph concave up), the rule overestimates the integral.
g (x )
x
1 1
Calculus Solutions Manual © 2005 Key Curriculum Press
3
Concave down Inscribed trapezoids Underestimates integral
Concave up Circumscribed trapezoids Overestimates integral
Problem Set 1-4
5
Problem Set 1-5 1. Answers will vary.
Problem Set 1-6
R3. By counting squares, the integral is approximately 23.2. Distance ≈ 23.2 ft (exact answer: 23.2422…) Concept: definite integral R4. a. f (x )
Review Problems
5
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/s From 4 to 4.1: average rate ≈ −29.3 ft/s Instantaneous rate ≈ −34.7 ft/s The distance from water is decreasing, so he is going down. d (5.01) − d ( 4.99) ≈ 70.8 0.02 d. Going up at about 70.8 ft/s e. Derivative
c. Instantaneous rate ≈
R2. a. Physical meaning: instantaneous rate of change of a function Graphical meaning: slope of a tangent line to a function at a given point b. x = − 4: decreasing fast x = 1: increasing slowly x = 3: increasing fast x = 5: neither increasing nor decreasing c. From 2 to 2.1: 52.1 − 52 average rate = = 43.6547K 0.1 From 2 to 2.01: 52.01 − 52 average rate = = 40.5614K 0.01 From 2 to 2.001: 52.001 − 52 average rate = = 40.2683K 0.001 Differences between average rates and instantaneous 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 the instantaneous rate as x approaches 2. The concept is the derivative. The concept used is the limit. d. t = 2: 3.25 m/s t = 18: 8.75 m/s t = 24: 11.5 m/s Her velocity stays constant, 7 m/s, from 6 s to 16 s. At t = 24, Mary is in her final sprint toward the finish line.
6
Problem Set 1-6
x 1
4
The graph agrees with Figure 1-6c. b. By counting squares, integral ≈ 15.0. (Exact answer is 15.) c. T 6 = 0.5(2.65 + 5.575 + 5.6 + 5.375 + 4.9 + 4.175 + 1.6) = 14.9375 The trapezoidal sum underestimates the integral because the trapezoids are inscribed in the region. d. T50 = 14.9991; Difference = 0.0009 T100 = 14.999775; Difference = 0.000225 The 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 = x 2 − 7x + 12 2 c. f ( x ) – f (3) = x – 7 x + 12 = ( x – 4)( x – 3) = x–3 x–3 x–3
x − 4, if x ≠ 3 d. The limit is found by substituting 3 for x in (x − 4). Limit = exact rate = 3 − 4 = −1 C2. The line through (3, f (3)) with slope −1 is y = −x + 2. f (x )
2
x 3
The line is tangent to the graph. Zooming in by a factor of 10 on the point (3, 2) shows that the graph becomes straighter and looks almost like the tangent line. (Soon students will learn that this property is called local linearity.)
Calculus Solutions Manual © 2005 Key Curriculum Press
T5. Concept: definite integral By counting squares, distance ≈ 466. (Exact answer is 466.3496… .) 2
T6. 3 Speed (ft/s)
4 x 2 − 19 x + 21 ( 4 x − 7)( x − 3) = = x −3 x −3 4x − 7, x ≠ 3 When x = 3, 4x − 7 = 4 ⋅ 3 − 7 = 5.
C3. a. f ( x ) =
25 20 15 10 5
b.
Time (s) 5
f (x ) (ft)
10
15
20
25
30
35
40
6 5.8 5
T 7 = 5(2.5 + 5 + 5 + 10 + 20 + 25 + 20 + 5) = 462.5 Trapezoidal rule probably underestimates the integral, but some trapezoids are inscribed and some circumscribed.
4.2 4 3 2
T7. Concept: derivative
1 2.8 1
2
x (s)
3.2 3
c. 5.8 = 4(3 + δ ) − 7 5.8 = 12 + 4δ − 7 4δ = 0.8 δ = 0.2 d. 4(3 + δ ) − 7 = 5 + ε 12 + 4δ − 7 = 5 + ε 4δ = ε δ = 14 ε
4
5
4.2 4.2 −4δ δ
6
= 4(3 − δ ) − 7 = 12 − 4δ − 7 = −0.8 = 0.2
There is a positive value of δ, namely 14 ε , for each positive value of ε, no matter how small ε is. e. L = 5, c = 3. “. . . but not equal to 3” is needed so that you can cancel the (x − 3) factors without dividing by zero. Chapter Test T1. Limit, derivative, definite integral, indefinite integral T2. See the text for the definition of limit. T3. Physical meaning: instantaneous rate T4. y 6
3
Speed (ft/s) 25 20 15 10 5 Time (s) 5
10
15
20
25
30
35
40
Slope ≈ −1.8 (ft/s)/s (Exact answer is −1.8137… .) Name: acceleration T8. The roller coaster is at the bottom of the hill at 25 s because that’s where it is going the fastest. The graph is horizontal between 0 and 10 seconds because the velocity stays constant, 5 ft/s, as the roller 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 and the 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 is getting closer to 416.349667… .
x 2
5
Calculus Solutions Manual © 2005 Key Curriculum Press
Problem Set 1-6
7
T12. From 30 to 31: y(31) − y(30) average rate = = −1.9098K 1 From 30 to 30.1: y(30.1) − y(30) = −1.8246K 0.1 From 30 to 30.01: y(30.01) − y(30) average rate = = −1.8148K 0.01 T13. The rates are negative because the roller coaster is slowing down. T14. The differences between the average rates and instantaneous 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… average rate =
8
Problem Set 1-6
The differences are getting smaller, so the average rates are getting closer to the instantaneous rate. y( x ) − y(30) = −1.81379936 + 1, getting x − 30 x = 30.092220… . So keep x within 0.092… unit of 30, on the positive side. T16. Concept: derivative f ( 4.3) − f (3.7) 35 − 29 T17. f ′( 4) ≈ = = 10 4.3 − 3.7 0.6 T18. Answers will vary. T15. Solve
Calculus Solutions Manual © 2005 Key Curriculum Press
Chapter 2—Properties of Limits Problem Set 2-1
h (x )
8 − 10 + 2 0 = 2−2 0 No value for f (2) because of division by zero.
1. a. f (2) =
2
x
b.
2.7
x
f (x)
1.997 1.998 1.999 2 2.001 2.002 2.003
2.994 2.996 2.998 undefined 3.002 3.004 3.006
Yes, f (x) stays close to 3 when x is kept close to 2, but not equal to 2. c. To keep f (x) within 0.0001 unit of 3, keep x within 0.00005 unit of 2. To keep f (x) within 0.00001 unit of 3, keep x within 0.000005 unit of 2. To keep f (x) arbitrarily close to 3, keep x within 12 that distance of 2. d. The discontinuity can be “removed” by defining f (2) to equal 3.
3
There appears to be no limit, because the graph cycles infinitely as it approaches x = 3.
Problem Set 2-2 Q1.
Q2. y
y
8
x π
x
–1
3
Q3.
Q4. y
y x –2
4
x 6
2
–4
2. Q5.
g (x )
Q6. Trapezoidal rule
3
y
2
4
x x
3 1
g (x )
2
x 3
The limit seems to be 2. 3. h (x )
2
x 3
Calculus Solutions Manual © 2005 Key Curriculum Press
Q7. Q8. Q9. Q10. 1. 2. 3. 5. 7. 9. 11. 13.
Counting squares Slope of the tangent line Instantaneous rate of change B See the text for the definition of limit. f (x) might be undefined at x = c, or might have a value at x = c that is different from the limit. Has a limit, 3 4. Has a limit, 2 Has a limit, 3 6. Has a limit, 5 Has no limit 8. Has no limit Has a limit, 7 10. Has a limit, 20 Has no limit 12. Has no limit lim f ( x ) = 5. For ε = 0.5, δ ≈ 0.2 or 0.3. x →3
Problem Set 2-2
9
14. lim f ( x ) = 3. For ε = 0.5, δ ≈ 0.8. x →2
15. lim f ( x ) = 4. For ε = 0.7, δ ≈ 0.5 or 0.6. x →6
(The right side is more restrictive.) 16. lim f ( x ) = 2. For ε = 0.8, δ ≈ 0.7 or 0.8. x→4
(The left side is more restrictive.) 17. lim f ( x ) = 2. For ε = 0.3, δ ≈ 0.5 or 0.6. x →5
(The right side is more restrictive.) 18. lim f ( x ) = 6. For ε = 0.4, δ ≈ 0.1. x →3
19. a. The graph should match Problem 13. b. lim f ( x ) = 5 x →3
c. Graph is symmetrical about x = 3. Let 5 − 2 sin (x − 3) = 5 + 0.5 = 5.5. ∴ sin (x − 3) = −0.25 x = 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/2 x = 3 + sin− 1 (−ε/2) Max. δ = 3 − [3 + sin− 1 (−ε/2)] = −sin− 1 (−ε/2) = sin− 1 (ε/2), which is positive for any positive value of ε. 20. a. The graph should match Problem 14. b. lim f ( x ) = 3 x →2
c. The graph is symmetrical about x = 2. Let (x − 2)3 + 3 = 3 + 0.5 = 3.5. ∴ x = 2 + 3 0.5 Max. δ = 2 + 3 0.5 − 2 = 3 0.5 = 0.7937K d. Let (x − 2) + 3 = 3 + ε . ∴ x = 2 + ε 1/3 Max. δ = 2 + ε 1/3 − 2 = ε 1/3, which is positive for any positive value of ε. 21. a. The graph should match Problem 15. b. lim f ( x ) = 4 3
x →6
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, set 1 + 3 3 7 − x = 4 − 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 f ( x ) = 2 x→4
c. The left side is more restrictive. Let 1 + 24− x = 2 + 0.8 = 2.8.
10
Problem Set 2-2
∴ 24− x = 1.8 log 1.8 x = 4− log 2 log 1.8 Max. δ = 4 − 4 − = 0.84799K log 2 d. Because the left side is more restrictive, set 1 + 24− x = 2 + ε . ∴ 24− x = 1 + ε x = 4−
log(1 + ε ) log 2
log(1 + ε ) log(1 + ε ) Max. δ = 4 − 4 − = , log 2 log 2 which is positive for all ε > 0. 23. a. The graph should match Problem 17. b. lim f ( x ) = 2 x →5
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+ ε Max. δ = (5 + ε ) − 5 = ε , which is positive for all ε > 0. 24. a. The graph should match Problem 18. b. lim f ( x ) = 6 x →3
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, which is positive for all ε > 0. 25. a. f (2) =
(52 − 6 ⋅ 5 + 13)(5 − 2) (5)(0) 0 = = 5−2 0 0
The graph has a removable discontinuity at x = 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 © 2005 Key Curriculum Press
c.
e. The limit of the average velocity is the instantaneous velocity.
f (x ) L=5
Problem Set 2-3 δ δ
Q1. 13 Q2.
Q3. y
y
3
x c=2
x
26. a.
x –4
2
y
Q4.
Q5. y
y
8
1
x 1
x
x
2
The graph is linear. There is a removable discontinuity at x = 2. The limit appears to be 9. 4(2 2 ) − 7(2) − 2 0 = 2–2 0 Indeterminate form ( 4 x + 1)( x − 2) c. f ( x ) = = 4 x + 1, x ≠ 2 x−2 Limit = 4(2) + 1 = 9 If x ≠ 2, then (x − 2) ≠ 0. Canceling is a division 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 = 0.00025 δ2 = 2 − 1.99975 = 0.00025 Largest number is 0.00025. e. L = 9, c = 2, ε = 0.001, δ = 0.00025
b. f (2) =
d (t ) − d ( 4) 3t 2 – 48 = t−4 t−4 b. Removable discontinuity at x = 4.
27. a. m(t ) =
m (t ) 30
t 4
c. Limit = 24 ft/s 3(t − 4)(t + 4) = 3t + 12, if t ≠ 4 d. m(t ) = t–4 3t + 12 = 24.12 ⇒ t = 4.04 3t + 12 = 23.88 ⇒ t = 3.96 Keep t within 0.04 s of 4 s.
Q6. (x − 10)(x + 10) Q7. 75% Q8. Product of x and y, where x varies and y may vary Q9. 3 1 –8 22 –21 3
–15
21
1 –5 x 2 − 5x + 7 Q10. D 1.
7
0
y g+h
10
h
g
x 2
lim f ( x ) = 10, lim g( x ) = 4, and lim h( x ) = 6 x →2
x →2
x →2
∴ lim f ( x ) = lim g( x ) + lim h( x ), Q .E.D . x →2
x 1.96 1.97 1.98 1.99 2.00 2.01 2.02 2.03 2.04
x →2
x →2
f (x) 9.9640… 9.9722… 9.9810… 9.9902… 10 10.0102… 10.0209… 10.0322… 10.0439…
All these f (x) values are close to 10.
Calculus Solutions Manual © 2005 Key Curriculum Press
Problem Set 2-3
11
2.
y 3 = f (x)
x y g
0.997
2.9739…
0.998
2.9825…
0.999
2.9912…
1
3
lim f ( x ) = 1.8 and lim g( x ) = 9
1.001
3.0087…
∴ lim f ( x ) = 0.2 lim g( x ), Q.E.D.
1.002
3.0174…
1.003
3.0262…
9
f
1.8
x
3
x →3
x →3
x →3
x →3
x
f (x)
2.96 2.97 2.98 2.99 3.00 3.01 3.02 3.03 3.04
All these f (x) values are close to 2(1.5) = 3.
1.75232 1.76418 1.77608 1.78802 1.8 1.81202 1.82408 1.83618 1.84832
3π 6. 2 3 = 8 and sin = 0.5 3.6 r(3) =
8 = 16 0.5
x
All these f (x) values are close to 1.8. 3. f (x ) Limit = 7 7
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…
x 3
The limit is 7 because f (x) is always close to 7, no matter what value x takes on. (It shouldn’t bother you that f (x) = 7 for x ≠ 3 if you think of the definition of limit for a while.) 4.
All these r(x) values are close to 16. 2 3.6 lim f ( x ) → , so the limit of a quotient x →3.6 0 cannot be applied because of division by zero. 7. lim f ( x ) = lim x 2 − 9 x + 5 x →3
x →3
= lim x − lim 9 x + lim 5 2
f (x) = x
x →3
Limit = 6
x →3
x →3
= lim x ⋅ lim x − 9 lim x + 5
x
x →3
6
lim f ( x ) = 6. The y-value equals the x-value. x →6
x →3
x →3
Limit of a product, limit of a constant Limit of x
= (3)(3) − 9(3) + 5 = 9 − 27 + 5 = −13
8. lim f ( x ) = lim x 2 + 3 x − 6
5.
x →−1
y 5
x →−1
= lim x 2 + lim 3 x − lim 6 x →−1
y1 y1 y2
x →−1
x →−1
Limit of a sum = lim x ⋅ lim x + 3 lim x − 6
y2
x →−1
x →−1
x →−1
x 1
lim y1 = 2, lim y2 = 1.5, and lim y1 ⋅ y2 = 3 x →1
x →1
x →1
2(1.5) = 3, ∴ lim y1 ⋅ lim y2 = lim y1 ⋅ y2 x →1
12
Limit of a sum (or difference)
Problem Set 2-3
x →1
= (−1)(−1) + 3(−1) − 6 = 1 − 3 − 6 = −8
Limit of a product, limit of a constant Limit of x
x →1
Calculus Solutions Manual © 2005 Key Curriculum Press
9.
r (x )
Proof:
–2
lim f ( x ) = lim ( x 2 + 2 x + 6) Because x ≠ 5
x
x →5
x →5
= lim x + lim (2 x ) + lim 6
Limit of a sum
= lim x ⋅ lim x + 2 lim x + 6
Limit of a product, limit of a constant times a function, limit of a constant
2
x →5
–8
x →5
x →5
( −2) − 4( −2) − 12 4 + 8 − 12 0 = = 0 0 ( −2) + 2 ( x – 6)( x + 2) r( x ) = = x − 6, x ≠ −2 x+2 lim r ( x ) = −2 − 6 = −8 2
x →5
x →5
x →5
r(–2) =
x →−2
= 5 ⋅ 5 + 2 · 5 + 6 = 41, Q .E .D . Limit of x 12. f (x )
Proof:
28
lim r ( x ) = lim ( x − 6)
Because x ≠ −2
= lim x + lim (–6)
Limit of a sum
= −2 − 6 = −8, Q .E .D .
Limit of x, limit of a constant
x →−2
x →−2
x →−2
x →−2
10.
x 3
f (x ) 13
x
f (3) =
33 + 32 − 5(3) − 21 27 + 9 − 15 − 21 0 = = 3−3 0 0
f ( x) =
( x 2 + 4 x + 7)( x – 3) = x 2 + 4 x + 7, x ≠ 3 x–3
5
lim f ( x ) = 32 + 4(3) + 7 = 28
52 + 3(5) − 40 25 + 15 − 40 0 f (5) = = = 5−5 0 0 ( x + 8)( x – 5) f ( x) = = x + 8, x ≠ 5 x–5 lim f ( x ) = 5 + 8 = 13
x →3
Proof: lim f ( x ) = lim ( x 2 + 4 x + 7) x →3
x →3
Because x ≠ 3 Limit of a sum
x →5
= lim x 2 + lim 4 x + lim 7
Proof:
= lim x ⋅ lim x + 4 lim x + 7
x →3
lim f ( x ) = lim ( x + 8)
Because x ≠ 5
= lim x + lim 8
Limit of a sum
= 5 + 8 = 13, Q .E .D .
Limit of x, limit of a constant
x →5
x →5
x →5
x →5
11.
x →3
x →3
Limit of a product, limit of a constant times a function, limit of a constant = 3 ⋅ 3 + 4 ⋅ 3 + 7 = 28, Q .E .D . Limit of x x →3
x →3
x →3
13.
f (x ) 41
f (x ) 9 10
x
x
5
53 − 3(52 ) − 4(5) − 30 5−5 125 − 75 − 20 − 30 0 = = 0 0
–1
f ( −5) =
( x 2 + 2 x + 6)( x – 5) = x 2 + 2 x + 6, x ≠ 5 x–5 lim f ( x ) = 52 + 2(5) + 6 = 41 f ( x) =
f ( −1) = =
( −1)3 − 4( −1) 2 − 2( −1) + 3 ( −1) + 1 −1 − 4 + 2 + 3 0 = 0 0
x →5
Calculus Solutions Manual © 2005 Key Curriculum Press
Problem Set 2-3
13
f ( x) =
( x 2 – 5 x + 3)( x + 1) = x 2 − 5 x + 3, x ≠ −1 x +1
15. x
lim f ( x ) = ( −1) − 5( −1) + 3 = 9 2
4.990 4.991 4.992 4.993 4.994 4.995 4.996 4.997 4.998 4.999 5 5.001 5.002 5.003 5.004 5.005 5.006 5.007 5.008 5.009
x →−1
Proof: lim f ( x ) = lim ( x 2 − 5 x + 3) x →−1
x →−1
Because x ≠ −1 = lim x 2 + lim (–5 x ) + lim 3 x → –1
x → –1
x → –1
Limit of a sum = lim x ⋅ lim x – 5 lim x + 3 x → –1
x → –1
x → –1
Limit of a product, limit of a constant times a function, limit of a constant = (−1)(−1) + (−5)(–1) + 3 = 9, Q.E.D. Limit of x 14. f (x) 2
x
–17
2 4 − 11(2 3 ) + 21(2 2 ) − 2 − 10 f (2) = 2−2 16 − 88 + 84 − 2 − 10 0 = = 0 0
f (x) 40.8801 40.8921… 40.9040… 40.9160… 40.9280… 40.9400… 40.9520… 40.9640… 40.9760… 40.9880… undefined 41.0120… 41.0240… 41.0360… 41.0480… 41.0600… 41.0720… 41.0840… 41.0960… 41.1080…
The table shows that f (x) will be within 0.1 unit of lim f ( x ) = 41 if we keep x within 0.008 unit x →5
of 5. 16. f (x )
( x – 9 x + 3 x + 5)( x – 2) x–2 3 2 = x − 9 x + 3 x + 5, x ≠ 2 3
f ( x) =
9
2
x –1
lim f ( x ) = 2 3 − 9(2 2 ) + 3(2) + 5 = −17 x →2
Proof: lim f ( x ) = lim ( x 3 – 9 x 2 + 3 x + 5) x →2
x →2
Because x ≠ 2 = lim x 3 + lim ( −9 x 2 ) + lim 3 x + lim 5 x →2
x →2
x →2
x →2
Limit of a sum lim x lim x lim x ( 9 ) lim x ⋅ lim x = ⋅ ⋅ + − x →2
x →2
+ 3 lim x + 5
x →2
x →2
x →2
x →2
Limit of a product, limit of a constant times a function, limit of a constant = 2 ⋅ 2 ⋅ 2 + (−9)(2 ⋅ 2) + 3 ⋅ 2 + 5 = −17, Q .E .D . Limit of x
14
Problem Set 2-3
When x is close to –1, f (x) is close to 9. x 2 − 5 x + 6 ( x − 2)( x − 3) x − 2 = = x 2 − 6 x + 9 ( x − 3)( x − 3) x − 3 You cannot find the limit by substituting into the simplified form because the denominator still becomes zero.
17. f ( x ) =
18. f ( x ) =
( x − 2)( x 2 + 2 x + 4) x3 − 8 = ( x − 2)( x − 2) x − 4x + 4 2
x2 + 2x + 4 x−2 You cannot find the limit by substituting into the simplified form because the denominator still goes to zero. =
Calculus Solutions Manual © 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)
21. By the symmetric difference quotient, 0.75.01 – 0.7 4.99 derivative ≈ = −0.05994 K . 2(0.01)
v(9.001) – v(9) = 0.8333101K 9.001 – 9 Conjecture: a(9) = 0.83 = 5/6 Units of a(t): (mi/h)/s
b. a(9) ≈
22. By the trapezoidal rule with n = 100, integral ≈ 11.8235K . 23. Prove that lim x n = c n for any positive integer n. x →c
Proof: Anchor:
v(t ) – v( 9) 5t 1/ 2 – 15 = lim t →9 t →9 t–9 t–9 1/ 2 5(t – 3) = lim 1/ 2 t →9 (t – 3)(t 1/ 2 + 3) 5 = lim 1/ 2 t →9 t +3 5 = , which agrees with the conjecture. 6 d. Distance = integral of v(t) from 1 to 9. By the trapezoidal rule with n = 100 increments, integral ≈ 86.6657… . The units are (mi/h) · s. To convert to ft, multiply by 5280 and divide by 3600, getting 127.1111… (exact: 127 19 ) . The truck went about 127 ft. c. a(9) = lim
3
Proof: x3 – 8 = lim ( x 2 + 2 x + 4) x →2 x – 2 x →2 Because x ≠ 2 = lim x 2 + lim 2 x + lim 4
lim
x →2
x →c
Induction Hypothesis: Assume that the property is true for n = k. ∴ lim x k = c k x →c
Verification for n = k + 1: lim x k +1 = lim ( x k ⋅ x ) x →c
x →2
x →c
= lim x ⋅ lim x = c k ⋅ c k
x →c
x →c
By the induction hypothesis
= c k +1 Conclusion: ∴ lim x n = c n for all integers n ≥ 1, Q.E.D.
3
2.1 – 2 = 12.61 2.1 – 2 x 3 – 8 ( x – 2)( x 2 + 2 x + 4) b. = = x–2 x–2 x 2 + 2 x + 4, provided x ≠ 2. This expression approaches 12 as x approaches 2.
20. a. Derivative ≈
If n = 1, lim x 1 = c = c1 by the limit of x.
x →c
24.
Answers will vary.
Problem Set 2-4 Q1. Instantaneous rate of change Q2. Product of x and y, where x varies and y can vary Q3. 0.0005 Q4.
x →2
Limit of a sum = lim x ⋅ lim x + 2 lim x + 4 x →2
x →2
x →2
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 is y = 12x − 16. The line appears to be tangent to the graph of f at point (2, 8). f (x )
12 8
1
x
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
2
Calculus Solutions Manual © 2005 Key Curriculum Press
Problem Set 2-4
15
2. a. b. c. 3. a. b. c. 4. a. b. c. 5. a. b. c. 6. a. b. c.
15. Answers may vary.
7.
19. Answers may vary.
8.
9.
10.
11.
Has left and right limits Has a limit Discontinuous. No f (3) Has left and right limits Has a limit Continuous Has left and right limits Has a limit Continuous Has no left or right limit Has no limit Discontinuous. No limit or f (2) Has left and right limits Has a limit Continuous (Note that the x-value 5 is not at the discontinuity.) a. Has left and right limits b. Has a limit c. Discontinuous. f (1) ≠ limit a. Has left and right limits b. Has no limit c. Discontinuous. No limit a. Has left and right limits b. Has a limit c. Discontinuous. No f (c) a. Has left and right limits b. Has no limit c. Discontinuous. No limit, no f (c) Answers may vary. 12. Answers may vary. f (x )
f (x )
16. Answers may vary. f (x )
f (x )
x
x
6
2
17. Answers may vary.
18. Answers may vary. f(x)
f (x )
10
x
5 –2
x –2
20. Answers may vary. f(x)
f (x ) 6 5
4
x
x
1
3
21. Discontinuous at x = −3 22. Discontinuous at x = 11 23. Discontinuous at x = π/2 + π n, where n is an integer 24. Nowhere discontinuous 25. f (x )
3 2 1
x 2
x
x 4
3
Discontinuous because lim f ( x ) = 2 and f (2) = 3 x →2
13. Answers may vary.
14. Answers may vary.
26. g (x )
f (x )
f (x )
2 1
f (–2) x
x 5
16
Problem Set 2-4
x 2
–2
Discontinuous because g(x) has no limit as x approaches 2
Calculus Solutions Manual © 2005 Key Curriculum Press
27.
33. a. d (x )
s (x )
3
3
x 2
x
Discontinuous because s(x) has no limit as x approaches 2 from the left (no real function values to the left of x = 2)
2
b. lim− d ( x ) = 3, lim+ d ( x ) = 3. Limit = 3. x →2
x →2
Continuous.
28. 34. a.
p (x )
h ( x)
3 2
1
x
x 1
2
Discontinuous because p(x) has no limit as x approaches 2 b. lim− h( x ) = 3, lim+ h( x ) = 2. No limit.
29.
x →1
x →1
Not continuous.
h(x )
35. a. 1
m (x ) 9
x
7
2
Discontinuous because there is no value of h(2)
x
30.
2
f(x)
b. lim− m( x ) = 9, lim+ m( x ) = 7. No limit. x →2
3
x →2
Not continuous.
x
36. a.
2
q (x)
Discontinuous because f (x) has no limit as x approaches 2 2
31. c
f (c)
1 2 4 5
4 1 5 none
lim− f ( x )
lim+ f ( x )
x →c
x →c
2 1 5 none
2 1 2 none
lim f ( x ) x →c
2 1 none none
x
Continuous? removable continuous step infinite
–1
b. lim− q( x ) = 2, lim+ q( x ) = 2. Limit = 2. x →−1
32. c 1 2 3 5
f (c) 3 1 5 5
x →−1
Continuous. 37. 9 – 22 = 2k ∴ k = 2.5 g (x )
lim− f ( x )
x →c
2 4 5 5
lim+ f ( x )
x →c
3 4 5 none
Calculus Solutions Manual © 2005 Key Curriculum Press
lim f ( x ) x →c
none 4 5 none
Continuous? step removable continuous infinite
5
x 2
Problem Set 2-4
17
38. 0.4(1) + 1 = k(1) + 2 ∴ k = −0.6 f (x )
1.4
43. Let T(θ ) = the number of seconds it takes to cross. if θ = 90° 24, T(θ ) = 40 , if 0° < θ < 90° or 90° < θ < 180° sin θ θ
x 1 40
39. (32)k = 3k − 3 ∴ k = −1/2.
θ 90
u (x ) 1
f (x )
44. a.
x 3
4
x
40. −k + 5 = (−1) k ∴ k = 5/2 2
1
v (x )
5
x –1
41. a. b − 1 = a(1 − 2)2 ⇒ b − 1 = a b. a = −1 ⇒ b = 0. Continuous at x = 1. f (x )
1
1
x
a = –1, b = 0
c. For example, a = 1 ⇒ b = 2. Continuous at x = 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. You must 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 addition and multiplication. Because the set of real numbers is closed under multiplication and addition, P(c) will be a unique, real number for any real value x = c. P(c) is the limit of P(x) as x approaches c by the properties of the limit of a product of functions (for powers of x), the limit of a constant times a function (for multiplication by the coefficients), and the limit of a sum (for the individual terms). Therefore, P is continuous for all values of x. 46. a. lim |sgn x| = 1 but f (0) = 0 x→0
lim f ( x ) ≠ f (0), so discontinuous
f (x )
x→0
b.
1
g(x )
x 1
3
e.g., a = 1, b = 2
x 2
42. lim− f ( x ) = lim− k 2 − x 2 = k 2 − 4 x →2
x →2
x →2
x →2
lim+ f ( x ) = lim+ 1.5kx = 1.5k (2) = 3k
For f (x) to be continuous at x = 2, these two limits must be equal, so find k such that k 2 − 4 = 3k k 2 − 3k − 4 = 0 (k − 4)(k + 1) = 0 so k = 4 and k = −1 are the two values of k that will make f (x) continuous at x = 2. 18
Problem Set 2-4
c. h(x )
1 –1
x
1
Calculus Solutions Manual © 2005 Key Curriculum Press
b. lim+ f ( x ) = ∞, lim− f ( x ) = – ∞,
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.
x →3
x →3
x →−∞
c. 2 +
2
x π
–2
Problem Set 2-5 Q1. Q3. Q5. Q7. Q9. 1.
No limit 4 2 No No • lim f ( x ) = ∞
Q2. Q4. Q6. Q8. Q10. •
x →−∞
•
lim f ( x ) = 3
3 3 No Yes Yes lim− f ( x ) = –4
• lim f ( x ) = 1
• lim− f ( x ) = ∞
• lim+ f ( x ) = 2
• lim f ( x ) does not
x →3
x →∞
x →3
2. • •
exist. lim− g( x ) = 4
lim g( x ) = 2
•
lim+ g( x ) = –3
• lim− g( x ) = ∞
x →−∞ x →−2
• lim g( x ) = 3
x →−2 x →1
• lim− g( x ) = 4
x →2
x →3
f (x) 2.00099… 2.00099… 2.00099…
All of these f (x) values are within 0.001 unit of 2. lim f ( x ) = 2 means that you can keep
x 2
x →∞
x
f (x) arbitrarily close to 2 by making the value of x arbitrarily large. y = 2 is a horizontal asymptote.
2
7
102 1002 10002
1004 1005 1006
4. Answers may vary. f (x )
5. Answers may vary.
3.01 3.001 3.0001
x
x →∞
f (x )
f (x)
arbitrarily far from zero just by keeping x close enough to 3 on the positive side. There is a vertical asymptote at x = 3. 1 = 2.001 d. 2 + x−3 1 = 0.001 x−3 x − 3 = 1000 x = 1003
• lim g( x ) = −2 3. Answers may vary.
x
x →3
x →1
x →2
1 = 100 x −3 1 = 98 x −3 1 x–3= 98 1 x = 3 = 3.0102 K 98
All of these f (x) values are greater than 100. lim+ f ( x ) = ∞ means that f (x) can be kept
x →−3
• lim f ( x ) = – ∞
x →−3+
x →∞
lim f ( x ) = 2
f (x )
–2
x →3
lim f ( x ), none, lim f ( x ) = 2,
6. Answers may vary. 8. a.
f (x )
f (x )
g (x ) x
x
1
x π/2
–5
7. a.
f (x )
b.
2
x 3
lim g( x ) = ∞, lim + g( x ) = −∞
x →π / 2 −
x →π / 2
The limit is infinite because |g(x)| can be kept arbitrarily far from zero. You can’t say lim g( x ) = ∞ because the left and right
x →π / 2
Calculus Solutions Manual © 2005 Key Curriculum Press
Problem Set 2-5
19
limits are not the same (one is positive and the other is negative). c. sec x = −1000 ∴ cos x = −0.001 x = arccos (−0.001) = 1.57179… x
g(x)
1.5717 1.5716 1.5715
–1106.5… –1244.2… –1421.1…
x →π / 2
g(x) can be kept arbitrarily far from zero in the negative direction by keeping x close enough to π2 on the positive side. The line x = π2 is a vertical asymptote. 9. a. r (x )
2
x 5
b. lim r ( x ) = 2 because (sin x)/x approaches zero. x →∞
sin (28) = 2.00967K , which is 28 within 0.01 unit of 2. sin (32) r(32) = 2 + = 2.01723K , which is 32 more than 0.01 unit away from 2.
c. r(28) = 2 +
y r 2.01 1.99
x 28
32
Keeping r(x) within 0.01 unit of 2 means you want to keep sinx x < 0.01, or |sin x| < 0.01 |x|. You are looking for a large value of x, so you know x will be positive, so you want |sin x| < 0.01x. You can’t get rid of the absolute value symbol on the sine because sine will keep alternating as x gets larger. You know |sin x| ≤ 1 for all values of x, so you need to make 0.01x > 1, or x > 100. So D = 100. d. The line y = 2 is an asymptote. Even though r (x) oscillates back and forth across this line, the limit of r(x) is 2 as x approaches infinity, satisfying the definition of asymptote.
20
Problem Set 2-5
x→0
(The exact value is e, 2.7182… .) 10. a. h( x ) = (1 + 1/ x ) x h(x ) 3 2
All of these f (x) values are less than −1000. lim + g( x ) = −∞ means that arbitrarily far
3
e. The graph suggests that lim r ( x ) = 3.
1
x 10
b. There is a compromise number (bigger than 1, but finite) that wins. (The exact limit is e.) 11. The limit is infinite. y is unbounded as x approaches infinity. If there were a number E such that log x < E for all x > 0, then you could let x = 102E so that log x = log 102E = 2E, which is greater than E, which was assumed to be an upper bound. 12. “Wanda, here’s what happens to a fraction when the denominator gets close to zero: 01.1 = 10, 1 1 0.0001 = 10, 000, 0.00001 = 100, 000. The answers just keep getting bigger and bigger. When the denominators get bigger and bigger, the fraction gets closer and closer to zero, like this: 1 1 1 10 = 0.1, 100 = 0.01, 1000 = 0.001.” 13. a. The definite integral is the product of the independent and dependent variables. Because distance = (rate)(time), the integral represents distance 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 because the sums can be made as close to it as you like, just by making the number of trapezoids large enough (and thus keeping their widths close to zero). The sums are smaller than the integral because each trapezoid is inscribed under the graph and thus leaves out a part of its 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 0.01 unit of 18 for any value of n ≥ 67. An alternative solution is to plot the graph of the difference between 18 and Tn as a function of the number of increments, n, or to do a regression analysis to find an equation. The best-fitting elementary function is an inverse power variation function, y = (5.01004…) Calculus Solutions Manual © 2005 Key Curriculum Press
(x− 1.48482…). The graph of this function and three of the four data points are shown here. Use TRACE or the solver feature of your grapher to find n ≈ 67. y
67
0.1
n
0.01 90
Q7. 1 Q8. + Q9. Indeterminate Q10. C 1. IVT applies on [1, 4] because f is a polynomial function, and polynomial functions are continuous for all x. f (1) = 18, f (4) = 3 ∴ There is a value x = c in (1, 4) for which f (c) = 8. Using the intersect or solver feature, c = 1.4349… , which is between 1 and 4. f (x )
14. a. Work = force × distance. Because a definite integral measures the y-variable times the x-variable, it represents work in this case. b. By the trapezoidal rule, T10 = 24.147775… and T100 = 24.004889… . The units are foot-pounds. c. The integer is 24. d. By experimentation, T289 = 24.001003… and T290 = 24.000998… . ∴ D = 290 15. Length = 100 sec x = 100/cos x Length > 1000 ⇒ 100/cos x > 1000 cos 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 the form 0 · ∞. g(2) = 5 · 2 · 0 · (1/0)2, which has the form 0 · ∞. h(2) = 5 · 2 · 02 · (1/0), which has the form 0 · ∞. 1 b. f ( x ) = 5 x ( x − 2) ⋅ = 5 x, x ≠ 2 x–2 ∴ lim f ( x ) = 10 x →2
1 5x ,x≠2 g( x ) = 5 x ( x − 2) ⋅ 2 = ( x – 2) x–2 ∴ lim g( x ) is infinite. x →2
h( x ) = 5 x ( x − 2 ) 2 ⋅ ∴ lim h( x ) = 0
1 = 5 x ( x – 2), x ≠ 2 x–2
x →2
c. The indeterminate form 0 · ∞ could approach zero, infinity, or some finite number.
Problem Set 2-6 Q1. 53 Q3. Undefined Q5. Undefined Calculus Solutions Manual © 2005 Key Curriculum Press
Q2. 53 Q4. 5 Q6. Does not exist
f (1)
8
f (4)
x 1 c
4
2. IVT applies on [0, 6] because f is a polynomial function, and polynomial functions are continuous for all x. f (0) = −8, f (6) = −0.224 ∴ There is a value x = c in (0, 6) for which f (c) = −1. Using the intersect or solver feature, c = 5.8751… , which is between 0 and 6. f (x ) 0 –0.224
c6 x
–8
3. a. For 1 ≤ y < 2 or for 5 < y ≤ 8, the conclusion would be true. But for 2 ≤ y ≤ 5, it would be false 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 because every number y in [4, 6] is a value of g(x) for some value of x in [1, 5]. 4. a. f (2) = 4, f (3) = 8, f (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 the sets of rational and irrational numbers are dense, there is a rational number between any two irrational numbers, and vice versa. So there is no limit of f (x) as x approaches any number other than 3. d. The conclusion is not true for all values of y between 1 and 4. For instance, if y = 3, then c would have to equal log2 3. But log2 3 is irrational, so f (c) = 8, which is not between 1 and 4. Problem Set 2-6
21
5. Let f (x) = x2. f is a polynomial function, so it is continuous and thus the intermediate value theorem applies. f (1) = 1 and f (2) = 4, so there is 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) is positive and f (b) is negative, then there is at least one zero of f (x) between x = a and x = b. Proof:
7.
8.
9.
10.
11.
12.
22
f is continuous, so the intermediate value theorem applies. f (a) is positive and f (b) is negative, so there is a number x = c between a and b for which f (c) = 0. Therefore, f has at least one zero between x = a and x = b, Q.E.D. The intermediate value theorem is called an existence theorem because it tells you that a number such as 3 exists. It does not tell you how to calculate that number. Telephone your sweetheart’s house. An answer to the call tells you the “existence” of the sweetheart at home. The call doesn’t tell such things as how to get there, and so on. Also, getting no answer does not necessarily mean that your sweetheart is out. 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 to be continuous (because of laws of physics), so f is also continuous and the intermediate value theorem applies. So there is a value of t between 1 and 3 for which f (t) = 0, meaning that Jesse and Kay are going at exactly the same speed at that time. The existence of the time tells you neither what that time is nor what the speed is. An existence theorem, such as the intermediate value theorem, does not tell these things. Let f (x) = number of dollars for x-ounce letter. f does not meet the hypothesis of the IVT on the interval [1, 9] because there is a step discontinuity at each integer value of x. There is no value of c for which f (c) = 2 because f (x) jumps from 1.98 to 2.21 at x = 8. You must assume that the cosine is function continuous. Techniques: • c = cos− 1 0.6 = 0.9272… • Using the solver feature, c = 0.9272... • Using the intersect feature, c = 0.9272... You must assume that 2x is continuous. f (0) = 20 = 1, because any positive number to the 0 power equals 1.
Problem Set 2-7
• c = log 2 3 =
log 3 = 1.5849... log 2
• Using the solver feature, c = 1.5849... • Using the intersect feature, c = 1.5849... 13. This means that a function graph has a high point and a low point on any interval in which the function is continuous. f (x )
x a
c1
c2
b
If the function is not continuous, there may be a point missing where the maximum or minimum would have been. f (x)
x a
b
Another possibility would be a graph with a vertical asymptote somewhere between a and b. 14. Prove that if f is continuous on [a, b], the image of [a, b] under f is all real numbers between the minimum and maximum values of f (x), inclusive. Proof: By the extreme value theorem, there are numbers x1 and x2 in [a, b] such that f (x1) and f (x2) are the minimum and maximum values of f (x) on [a, b]. Because x1 and x2 are in [a, b], f is continuous on the interval whose endpoints are x1 and x2. Thus, the intermediate value theorem applies on the latter interval. Thus, for any number y between f (x1) and f (x2), there is a number x = c between x1 and x2 for which f (c) = y, implying that the image of [a, b] under f is all real numbers between the minimum and maximum values of f (x), inclusive, Q.E.D.
Problem Set 2-7 Review Problems R0. Answers will vary. 36 − 51 + 15 0 R1. a. f (3) = = 3−3 0 Indeterminate form
Calculus Solutions Manual © 2005 Key Curriculum Press
b. f ( x ) = 4 x – 5, x ≠ 3
You can cancel the (x − 3) because the definition of limit says “but not equal to 3.”
y 9
• x –9
lim g( x ) x →3
= lim x 2 + lim (–10 x ) + lim 2
9
x →3
x →3
x →3
Limit of a sum = lim x ⋅ lim x – 10 lim x + 2
–9
x →3
At x = 3 there is a removable discontinuity. c. For 0.01, keep x within 0.0025 unit of 3. For 0.0001, keep x within 0.000025 unit of 3. To keep f (x) within ε unit of 7, keep x within 1 4 ε unit of 3. R2. a. L = lim f ( x ) if and only if for any number x →c
ε > 0, no matter how small, there is a number δ > 0 such that if x is within δ units of c, but x ≠ c, then f (x) is within ε units of L. b. lim f ( x ) = 2 x →1
x →3
x →3
Limit of a product, limit of a constant times a function, limit of a constant = 3 · 3 − 10(3) + 2 Limit of x = −19, which agrees with the graph. c. •
f (x) = 2x, x 2 − 8 x + 15 ( x − 3)( x − 5) g( x ) = = 3− x 3− x = −x + 5, x ≠ 3 lim f ( x ) = 8, lim g( x ) = 2 x →3
x →3
• p(x) = f (x) · g(x) lim p( x ) = 8 ⋅ 2 = 16
lim f ( x ) does not exist. x →2
lim f ( x ) = 4
x →3
x →3
lim f ( x ) does not exist.
x
x →4
lim f ( x ) = 3 x →5
2.997 2.998 2.999 3 3.001 3.002 3.003
c. lim f ( x ) = 3 x →2
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 + x −1 = 3 − ε x = (1 − ε )2 + 1 Let δ = 2 − [(1 − ε)2 + 1] = 1 − (1 − ε)2, which is positive for all positive ε < 1. If ε ≥ 1, simply take δ = 1. Then δ will be positive for all ε > 0. R3. a. See the limit property statements in the text. b. •
p(x) 15.9907… 15.9938… 15.9969… undefined 16.0030… 16.0061… 16.0092…
All these p(x) values are close to 16. •
r( x ) =
f ( x) g( x )
lim r ( x ) = x →3
8 =4 2
y 20 15
g (x ) 20
f 10
r
x 3
5
g –19
x
3
•
•
The limit of a quotient property does not apply because the limit of the denominator is zero. ( x − 3)( x 2 − 10 x + 2) x−3 g(x) = x 2 − 10x + 2, x ≠ 3 g( x ) =
Calculus Solutions Manual © 2005 Key Curriculum Press
d. For 5 to 5.1 s: average velocity = −15.5 m/s. f (t ) – f (5) Average velocity = = t–5 35t – 5t 2 – 50 –5(t – 2)(t – 5) = = t–5 t–5 Problem Set 2-7
23
−5(t − 2), for t ≠ 5. Instantaneous velocity = limit = −5(5 − 2) = −15 m/s. The rate is negative, so the distance above the starting point is getting smaller, which means 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 f ( x ) exists x →c
3. lim f ( x ) = f (c) x →c
•
f is continuous on [a, b] if and only if f is continuous at every point in (a, b), and lim+ f ( x ) = f ( a) and lim− f ( x ) = f (b).
x→a
4 2
x 2
The left limit is 4 and the right limit is 2, so f is discontinuous at x = 2, Q.E.D. Let 22 = 22 − 6(2) + k. ∴ k = 12 R5. a. lim f ( x ) = ∞ means that f (x) can be kept x→4
arbitrarily far from 0 on the positive side just by keeping x close enough to 4, but not equal to 4. lim f ( x ) = 5 means that f (x) can be made to
x→b
b. lim− f ( x )
lim+ f ( x )
lim f ( x )
f (x )
d.
Continuous?
x →∞
none
infinite
stay arbitrarily close to 5 just by keeping x large enough in the positive direction.
3
3
removable
2
5
none
step
3
3
3
3
continuous
• lim f ( x ) = 1
1
1
1
1
continuous
• lim− f ( x ) = ∞
c
f (c)
1
x →c
x →c
x →c
none
none
none
2
1
3
3
5
4 5
b. • lim f ( x ) does not exist. x →−∞ x →−2 x →2
c. i.
• lim+ f ( x ) = −∞
ii.
x →2
y
y
• lim f ( x ) = 2 x →∞
c. f (x) = 6 − 2− x lim f ( x ) = 6 x
x →∞
x
1
f (x) = 5.999 = 6 − 2− x 2− x = 0.001 log 0.001 x=− log 2 x = 9.965...
2
iii.
iv. y
y
x 3
10 20 30
4
v.
vi. y
y f (6)
d.
L
x 5
y
5.999023… 5.999999046… 5.99999999907…
All of these f (x) values are within 0.001 of 6. g(x) = x− 2 lim g( x ) = ∞ x→0
g(x) = 106 = x− 2 x2 = 10− 6 x = 10− 3 x
5
x 1 –2
24
x 6
vii.
f (x)
x
x
Problem Set 2-7
0.0009 0.0005 0.0001
g(x) 1.2345… ⋅ 106 4,000,000 1 ⋅ 108
Calculus Solutions Manual © 2005 Key Curriculum Press
All of these g(x) values are larger than 1,000,000. e. v(t) = 40 + 6 t n
Trapezoidal Rule
50 100 200 400
467.9074… 467.9669… 467.9882… 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 intermediate value theorem. The basis is the completeness axiom. See the text statement of the extreme value theorem. The word is corollary. b. f (x) = −x 3 + 5x 2 − 10x + 20 f (3) = 8, f (4) = −4 So f (x) = 0 for some x between 3 and 4 by the intermediate value theorem. The property is continuity. The value of x is approximately 3.7553. c.
f (x )
3
x –4
f (−6) = 1 and f (−2) = 5 by tracing on the graph or by simplifying the fraction to get f (x) = x + 7, then substituting. You will not always get a value of x if y is between 1 and 5. If you pick y = 3, there is no value of x. This fact does not contradict the intermediate value theorem. Function f does not meet the continuity hypothesis of the theorem.
x →1
∴ f is continuous at x = 4, Q.E.D. For the derivative, from the left side, f ( x ) – f (1) x 2 + 3 – 4 ( x – 1)( x + 1) = = = x –1 x –1 x –1 x + 1, x ≠ 1 ∴ lim− f ′( x ) = 1 + 1 = 2 x →1
For the derivative, from the right side, f ( x ) – f (1) x 2 – 6 x + 9 – 4 ( x – 1)( x – 5) = = = x –1 x –1 x –1 x − 5, x ≠ 1 ∴ lim+ f ′( x ) = 1 – 5 = −4 x →1
So f is continuous at x = 1, but does not have a value for the derivative there because the rate of change jumps abruptly from 2 to −4 at x = 1. In general, if a function has a cusp at a point, then the derivative does not exist, but the function is still continuous. C3. The graph is a y = x2 parabola with a step discontinuity 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 the discontinuity at first.) To create the discontinuity, use the signum function with argument (x − 1). Because there is no value for f (1), the absolute value form of the signum function can be used. | x –1| y = x2 + 2 − x –1 C4. The quantity | f ( x ) – L | is the distance between f (x) and L. If this distance is less than ε, then f (x) is within ε units of L. The quantity |x − c| is the distance between x and c. The right part of the inequality, |x – c| < δ, says that x is within δ units of c. The left part, 0 < |x − c|, says that x does not equal c. Thus, this definition of limit is equivalent to the other definition.
Chapter Test
Concept Problems C1.
C2. f (1) = 12 − 6 ⋅ 1 + 9 = 4 As x → 1 from the left, f (x) → 12 + 3 = 4. As x → 1 from the right, f (x) → 12 − 6 + 9 = 4. ∴ lim f ( x ) = 4 = f (1)
T1. f is continuous at x = c if and only if 1. f (c) exists
y
h
2. lim f ( x ) exists
7
x →c
f
3. lim f ( x ) = f (c) x →c
g x 4
Conjecture: lim f ( x ) = 7 x→4
Calculus Solutions Manual © 2005 Key Curriculum Press
f is continuous on [a, b] if and only if f is continuous at all points in (a, b), and lim+ f ( x ) = f ( a) and lim− f ( x ) = f (b) .
x→a
x→b
Problem Set 2-7
25
T2. a. • • •
lim f ( x ) = 3
•
x →2 −
lim f ( x ) does not exist. • x →2
lim f ( x ) = 2
lim f ( x ) = 4
= lim x 2 + lim ( −5 x ) + lim 8
lim− f ( x ) = 2
Limit of a sum = lim x ⋅ lim x + ( −5) ⋅ lim x + 8
x →2 + x →6
b. f is continuous on [2, 6] because it is continuous for all values in (2, 6) and lim+ f ( x ) = f (2) and lim− f ( x ) = f (6) . x →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.
x →2
5
x
T12. lim− f ( x ) = k ⋅ 2 2 x →2
lim f ( x ) = 2 + k
x →2 +
∴ 4k = 2 + k k = 2/3 T13. See graph in T11. T14. a. lim T ( x ) = 20
x
x →∞
b. g(x )
f (x )
1 x
x
d. h(x)
s (x ) 1
1
x
x 1
–1
(0 2 − 5 ⋅ 0 + 8)(0 − 3) 0 = , 0−3 0 an indeterminate form b. lim f ( x ) = lim ( x 2 − 5 x + 8), x ≠ 3
T10. a. f (3) =
x →3
Definition of limit “x ≠ c” 26
Problem Set 2-7
f(x)
2
4
x →3
x →2
∴ f is discontinuous at x = 3.
4
c.
x →3
x 2 , x≤2 T11. If k = 1, f ( x ) = x + 1, x > 2 lim− f ( x ) = 4, lim+ f ( x ) = 3
y
T9. a.
x →3
x →3
Limit of a product, limit of a constant times a function, limit of a constant = 3 · 3 + (−5) · 3 + 8 Limit of x = 2, Q .E .D .
x →6
x →2
x →3
x →3
• lim f ( x ) = 2
x →6 +
x →3
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 at T = 20. b. T = 20 + 8(0.97x) cos 0.5x. The amplitude of the cosine factor is 8(0.97 x ). Make this amplitude < 0.1. 8(0.97c) = 0.1 0.97c = 0.0125 log 0.0125 c= log 0.97 c = 143.8654… ∴ T is within 0.1 unit of 20 whenever x > 143.8654… . c. The time of day would be mid-afternoon, when the temperature of the surface is highest. 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. d (20.1) − d (20) b. Average rate = = 20.1 − 20 14.0901 – 14 = 0.901 cm/day 20.1 – 20 Calculus Solutions Manual © 2005 Key Curriculum Press
d (t ) − d (20) = t − 20 0.01t 2 + 0.5t – 14 0.01(t + 70)(t – 20) = = t – 20 t – 20 0.01t + 0.7, t ≠ 20. The limit as t approaches 20 is 0.01(20) + 0.7, which equals 0.9 cm/day. This instantaneous rate is called the derivative. d. The glacier seems to be speeding up because each 10-day period it moved farther than it had in the preceding 10-day period. T16. c(0) = p(0) = 10, so each has the same speed at t = 0. lim c(t ) = 16. lim p(t ) = ∞. Surprise for c. Average rate =
t →∞
t →∞
Phoebe! kx 2 , if x ≤ 2 T17. f ( x ) = 10 – kx, if x > 2
Make 4k = 10 − 2k ⇒ k = 5/3. There is a cusp at x = 2. 10
f(x)
x 2
T18. h(x) = x3. h(1) = 1 and h(2) = 8, so 7 is between h(1) and h(2). The intermediate value theorem allows you to conclude that there is a real number between 1 and 8 equal to the cube root of 7. T19. Answers will vary.
lim f ( x ) = k ⋅ 2 2 = 4 k
x →2 −
lim f ( x ) = 10 − 2 k
x →2 +
Calculus Solutions Manual © 2005 Key Curriculum Press
Problem Set 2-7
27
Chapter 3—Derivatives, Antiderivatives, and Indefinite Integrals Problem Set 3-1
Q4. y
1. The graph is correct. f (5.1) – f (5) 3.21 – 3 2. Average rate = = = 0.1 0.1 2.1 km/min 3. y
1
x
Q5. 9x 2 − 42x + 49 Q6. “−” sign Q7.
Q8.
y
2
y
x
2
5
x
x 5
3
4. The graph of r has a removable discontinuity at x = 5. f (5) − f (5) 0 = r (5) = 5−5 0 f ( x ) − f (5) x 2 − 8 x + 18 − 3 5. r ( x ) = = x−5 x−5 ( x − 5)( x − 3) = = x − 3, x ≠ 5 x−5 f ′(5) = lim r ( x ) x →5
( x − 5)( x − 3) = lim = lim ( x − 3) = 2 x →5 x →5 x−5 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
Q9. Q10. 1. 2.
Newton and Leibniz D See the text for the definition of derivative. Physical: Instantaneous rate of change of the dependent variable with respect to the independent variable Graphical: Slope of the tangent line to the graph of the function at that point 0.6 x 2 – 5.4 x →3 x–3 0.6( x – 3)( x + 3) = lim = 3.6 x →3 x–3 b. Graph of the difference quotient m(x)
3. a. f ′(3) = lim
m(x ) 3.6
y x
3 3
x
c., d.
Tangent line: y = 3.6x − 5.4
5
f (x )
This line is tangent to the graph of f (x) at (5, 3). 7. As you zoom in, the line and the graph appear to be the same. 8. Answers will vary.
Problem Set 3-2 Q1. Instantaneous rate of change Q2. x + 9 Q3. 18
28
Problem Set 3-2
1
x 3
– 0.2 x 2 + 7.2 x →6 x–6 –0.2( x – 6)( x + 6) = lim = −2.4 x →6 x–6
4. a. f ( x ) = – 0.2 x 2 , f ′(6) = lim
Calculus Solutions Manual © 2005 Key Curriculum Press
b. m (x )
x 6 –2.4
c., d.
Tangent line: y = −2.4x + 7.2 f (x) x 6
–7.2
x 2 + 5x + 1 + 5 x →−2 x+2 ( x + 2)( x + 3) = lim =1 x →−2 x+2
5. f ′(–2) = lim
x 2 + 6 x – 2 + 10 x →−4 x+4 ( x + 4)( x + 2) = lim = –2 x →−4 x+4
6. f ′(– 4) = lim
x3 – 4x2 + x + 8 – 6 7. f ′(1) = lim x →1 x –1 ( x – 1)( x 2 – 3 x – 2) = lim = –4 x →1 x –1 8.
9.
10.
11. 12.
x3 – x2 – 4x + 6 – 8 f ′(–1) = lim x →−1 x +1 ( x + 1)( x 2 – 2 x – 2) = lim =1 x →−1 x +1 –0.7 x + 2 + 0.1 f ′(3) = lim x →3 x–3 –0.7( x – 3) = lim = −0.7 x →3 x–3 1.3 x – 3 – 2.2 f ′( 4) = lim x→4 x–4 1.3( x – 4) = lim = 1.3 x→4 x–4 5–5 f ′(–1) = lim =0 x →−1 x + 1 –2 + 2 f ′(3) = lim =0 x →3 x – 3
13. The derivative of a linear function equals the slope. The tangent line coincides with the graph of a linear function.
Calculus Solutions Manual © 2005 Key Curriculum Press
14. The derivative of a constant function is zero. Constant functions are horizontal and don’t change! The tangent line coincides with the graph. 15. a. Find f ′ ( 1) = 2, then plot a line through point (1, f (1)) using f ′ ( 1) as the slope. The line is y = 2x − 1. b. Near the point (1, 1), the tangent line and the curve appear nearly the same. c. The curve appears to get closer and closer to the line. d. Near point (1, 1) the curve looks linear. e. If a graph has local linearity, the graph near that point looks like the tangent line. Therefore, the derivative at that point could be said to equal the slope of the graph at that point. 16. a. f ( x) = x 2 + 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 changes direction abruptly, not smoothly. d. If you draw a secant line through (1, 1) from a point just to the left of x = 1, it has a large negative slope. If you draw one from a point just to the right, it has a large positive slope. In both cases, the secant line becomes vertical as x approaches 1 and a vertical line has infinite slope. So there is no real number equal to the derivative. 17. a. 7
f (x )
5
x 3
b. First simplify the equation. x + 2, if x ≠ 3 f ( x) = if x = 3 7,
Problem Set 3-2
29
The difference quotient is ( x + 2) – 7 x – 5 m( x ) = = x–3 x–3 m (x )
= lim 0.25( x – 7) = –1, Q .E.D . x →3
The tangent line on the graph has slope −1. b. f (x)
5 Draw secant lines from here.
x 3
c. x
2
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 positive numbers on the left side of 3. On the right side, they are large negative numbers. For a derivative to exist, the difference quotient must approach the same number as x gets closer to 3.
x 3
As the x-distance between the point and 3 decreases, the secant lines (solid) approach the tangent line (dashed). c. The same thing happens with secant lines from the left of x = 3. See the graph in part b. d. g ( x) 4 Draw secant lines from here.
18. a. s (x ) x 3 2
x 1
b. m( x ) =
| sin ( x – 1) | x –1
m(x ) 1
x 1
e. A derivative is a limit. Because the left and right limits are unequal, there is no derivative at x = 3. π –6 cos x 6 f. m( x ) = . By table, x–3 x
c. As x approaches 1 from the left, m(x) approaches −1. As x approaches 1 from the right, m(x) approaches 1. Because the left and right limits are unequal, there is no derivative at x = 1. 0.25 x 2 – 2.5 x + 7.25 – 2 x →3 x–3 0.25( x – 3)( x – 7) = lim x →3 x–3
19. a. f ′( x ) = lim
30
Problem Set 3-2
2.9 2.99 3 3.01 3.1
m(x) 3.1401… 3.1415… undefined −3.1415… −3.1401…
Conjecture: The numbers are π and −π. 20. From Problem 19, parts b and c, the tangent line is the limit of the secant lines as x approaches c. Because the slope of the secant line is the average
Calculus Solutions Manual © 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 the limit of this average rate, the slope of the tangent line equals f ′(c).
2. 15
y g' x 3
g
Problem Set 3-3 Q1. 3 Q2.
The graph does not have the high and low points that are typical of a cubic function. As x increases, the graph starts to roll over and form a high point, but it starts going back up again before that happens. This behavior is revealed by the fact that the derivative is positive everywhere. Between x = 0 and x = 1, the derivative reaches a low point, indicating that the slope is a minimum, but the slope is still positive and the graph of g is still going up.
y 5
x –2
Q3. y x
3. a. y 200
Q4. Q5. Q6. Q7. Q8.
h
20% 3x 2 − 2x − 8 25x2 − 70x + 49 log 6
h´ x –2
Q10. “lim” is missing. x →c
1. a. y f x –2
2.5
b. The h′ graph looks like a cubic function graph. Conjecture: Seventh-degree function has 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 or a low point. This is reasonable because if h′(x) = 0, the rate of change of h(x) is zero, which would happen when the graph stops going up and starts going down, or vice versa. e. See the graph in part a.
Q9.
40
1
4.
2
y f´
q'
10
2
x
q
b. f ′(x) is positive for −2 < x < 2. The graph of f 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, the f graph has a high point or a low point. e. See the graph in part a. f. Conjecture: f ′ is quadratic.
Calculus Solutions Manual © 2005 Key Curriculum Press
The graph does not have the expected shape for a quartic function. The two high points and the low point all appear to occur as a high point at x = 2. The derivative graph crosses the x-axis just once, at x = 2, indicating that this is the only place where the function graph is horizontal.
Problem Set 3-3
31
5. a.
derivative equal the function value. So the base must be greater than 2. By experimenting, 3 is too large, but not by much. You can use trial and error with bases between 2 and 3, checking the results either by plotting the graph and the numerical derivative or by constructing tables. An ingenious method that some students come up with uses the numerical derivative and numerical solver features to solve nDeriv(bx, x, 1) = b1 at x = 1. The answer is about 2.718281… . (In Section 3-9, students will learn that this number is e, the base of natural logarithms.) The graph of f ( x) = 2.781…x and its numerical derivative are shown here.
y f
4
f´
x
5
b. Amplitude = 1, period = 2π = 6.283… c. The graph of f ′ also has amplitude 1 and period 2π. d. y f 4
g
y
x 5 f´ and g´
The graphs of f and g are the same shape, spaced 1 unit apart vertically. The graphs of f ′ and g′ are identical! This is to be expected because the shapes of the f and g graphs are the same. 6. f
f´
x
5
The function available on the grapher is y = cos x. The amplitude is 1, the period is 2π, and the shape is sinusoidal. cos 0 = 1, and the graph is at a high point, y = 1, when x = 0. 7.
8. y
y f 3
3
f'
x 4
x
4
f´
f
9.
10. y 5
1
x 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.
y 4
f and f'
y f
f 3
f´
f' x
x 1
2
b. Let x be the number of inches. Area = x2. The right side of 12 is more restrictive, so set x2 = 144.02. ∴ x = 144.021/2 = 12.000833… Keep the tile dimensions within 0.0008 in. of 12 in. c. The 0.02 in part b corresponds to ε, and the 0.0008 corresponds to δ. 14. The average of the forward and backwards difference quotients equals 1 f ( x + h) – f ( x ) f ( x ) – f ( x – h) + 2 h h 1 f ( x + h) – f ( x – h) = 2 h f ( x + h) – f ( x – h) = , Q .E .D . 2h 15. a. f ( x) = x 3 − x + 1 ⇒ f ( 1) = 1 ( x 3 – x + 1) – 1 x →1 x –1 x3 – x x ( x + 1)( x – 1) = lim = lim x →1 x – 1 x →1 x –1 = lim x ( x + 1) = 2
f ′(1) = lim 11. The derivative for f ( x) = 2x is consistently below that of the function itself. This fact implies that f ( x) does not increase rapidly enough to make the 32
Problem Set 3-3
x →1
Calculus Solutions Manual © 2005 Key Curriculum Press
f (1.1) – f (1) 1.231 – 1 = = 2.31 0.1 0.1 f (1) – f (0.9) 1 – 0.829 Backwards: = = 1.71 0.1 0.1 Symmetric:
b. Forward:
f (1.1) – f (0.9) 1.231 – 0.829 = = 2.01 2(0.1) 0.2 The symmetric difference quotient is closer to the actual derivative because it is the average of the other two, and the other two span the actual derivative. c. f ( 0) = 1 ( x 3 – x + 1) – 1 f ′(0) = lim x→0 x–0 x3 – x = lim = lim ( x 2 – 1) = −1 x→0 x→0 x f (0.1) – f (0) 0.901 – 1 d. Forward: = = –0.99 0.1 0.1 f (–0.1) – f (0) 1.099 − 1 Backwards: = = –0.99 0.1 0.1 Symmetric: f (0.1) – f (–0.1) 0.901 – 1.099 = = −0.99 2(0.1) 0.2 All three difference quotients are equal because f ( x) changes just as much from −0.1 to 0 as it does 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 becoming large and negative, while the forward difference quotients are becoming large and positive. Their average, the symmetric difference quotient, is always equal to 1. 17. Answers will vary.
Problem Set 3-4 Q1. 9x 2 − 24x + 16 Q2. a3 + 3a2b + 3ab2 + b3 Q3. See the text definition of derivative. f ( x + h) – f ( x – h) Q4. 2h Q5. No limit (infinite) Q6. log 73 Q7. 3 Q8. Pythagorean theorem Q9. 10 Q10. C 1. f ( x) = 5x 4 ⇒ f ′ ( x) = 20x3 Calculus Solutions Manual © 2005 Key Curriculum Press
2. y = 11x8 ⇒ dy/dx = 88x7 3. v = 0.007t− 83 ⇒ dv/dt = −0.581t− 84 x –9 1 4. v( x ) = ⇒ v ′( x ) = – x –10 18 2 5. M(x) = 1215 ⇒ M′ (x) = 0 (Derivative of a constant) 6. f (x) = 4.7723 ⇒ f ′ (x) = 0 (Derivative of a constant) 7. y = 0.3x 2 − 8x + 4 ⇒ dy/dx = 0.6x − 8 8. r = 0.2x2 + 6x − 1 ⇒ dr/dx = 0.4x + 6 d 9. (13 – x ) = −1 dx 10. f (x) = 4.5x 2 − x ⇒ f′ (x) = 9x − 1 11. y = x 2.3 + 5x − 2 − 100x + 4 ⇒ dy/dx = 2.3x1.3 − 10x− 3 − 100 2 d 2/5 12. ( x – 4 x 2 – 3 x –1 + 14) = x –3/ 5 − 8 x + 3 x –2 5 dx 13. v = (3x − 4)2 = 9x2 − 24x + 16 ⇒ dv/dx = 18x − 24 14. u = (5x − 7)2 = 25x 2 − 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 = 64x 3 − 48x 2 + 12x − 1 ⇒ f ′ (x) = 192x2 − 96x + 12 x2 17. P( x ) = – x + 4 ⇒ P ′( x ) = x – 1 2 x3 x2 – x + 1 ⇒ Q′( x ) = x 2 + x – 1 18. Q( x ) = + 3 2 19. f (x) = 7x4 7( x + h) 4 – 7 x 4 f ′( x ) = lim h→0 h 3 = lim (28 x + 42 x 2 h + 28 xh 2 + 7h 3 ) = 28 x 3 h→ 0
By formula, f ′(x) = 7 ⋅ 4x3 = 28x3, which checks. 20. g(x) = 5x3 5( x + h)3 – 5 x 3 g′( x ) = lim h→0 h 2 = lim (15 x + 15 xh + 5h 2 ) = 15 x 2 h→ 0
By formula, g ′ (x) = 5 ⋅ 3x2 = 15x2, which checks. 21. v(t) = 10t2 − 5t + 7 [10(t + h)2 − 5(t + h) + 7] − (10t 2 − 5t + 7) h→0 h 20th + 10 h 2 – 5h = lim h→0 h = lim (20t + 10 h – 5) = 20t – 5
v ′(t ) = lim
h→0
By formula, v ′ (t) = 10 ⋅ 2t − 5 = 20t − 5, which checks.
Problem Set 3-4
33
22. s(t ) = t 4 – 6t 2 + 3.7 [(t + h) 4 − 6(t + h)2 + 3.7] − [t 4 − 6t 2 + 3.7] s ′(t ) = lim h→0 h 4t 3 h + 6t 2 h 2 + 4th 3 + h 4 – 12th – 6h 2 = lim h→0 h = lim ( 4t 3 + 6t 2 h + 4th 2 + h 3 – 12t – 6h) h→0 3
= 4t – 12t By formula, s′(t ) = 4t 3 – 6 ⋅ 2t = 4t 3 – 12t , which checks. 23. Mae should realize that you differentiate functions, not values of functions. If you substitute a value for x into f (x) = x4, you get f (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 the velocity is zero. v(x) = 0 if and only if x = 2.
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. y 1
y1 x 1
y 2 and y3
b. h(x) = x1/ 2. Conjecture: g′(x) = 0.5x− 1/ 2. Conjecture is confirmed. y 2
y1
y2 and y3
25.
1
y
x
7
c. e(x) = 2x. Conjecture: e′ (x) = x ⋅ 2x− 1. Conjecture is refuted!
f´ f
x y
9
y1
26. 3
1
y
1
g x 6
g'
y3 x
29. f ( x ) = x 1/ 2 + 2 x – 13 f ′( x ) = 12 x –1/ 2 + 2, f ′( 4) = 94 Increasing by 9/4 y-units per x-unit at x = 4 30. f (x) = x− 2 − 3x + 11 f ′(x) = −2x− 3 − 3, f ′(1) = −5 Decreasing by 5 y-units per x-unit at x = 1
27. a. y 10
f´
1
x f
b. The graph of f ′ is shown dashed in part a. c. There appear to be only two graphs because the exact and the numerical derivative graphs almost coincide. 34
y2
Problem Set 3-4
31. f (x) = x1.5 − 6x + 30 f ′(x) = 1.5x0.5 − 6, f ′(9) = −1.5 Decreasing by 1.5 y-units per x-unit at x = 9 32. f ( x ) = –3 x + x + 1 f ′( x ) = – 23 x –1/ 2 + 1, f ′(2) = –0.0606K Decreasing by approximately 0.0607 y-unit per x-unit at x = 2 33. f ( x ) =
x3 – x 2 – 3 x + 5, f ′( x ) = x 2 – 2 x – 3 3 Calculus Solutions Manual © 2005 Key Curriculum Press
High and low points of the f graph are at the x-intercepts of the f ′ graph. y
x –1
3
f
x3 – 2 x 2 + 3 x + 9, f ′( x ) = x 2 – 4 x + 3 3 High and low points of the f graph are at the x-intercepts of the f ′ graph.
34. f ( x ) =
y 15
f
f' 1
x
3
35. If f (x) = k ⋅ g(x), then f ′(x) = k ⋅ g′(x). Proof: f ( x + h) – f ( x ) h k ⋅ g( x + h ) – k ⋅ g ( x ) = lim h→0 h g( x + h) – g( x ) = lim k ⋅ h→0 h g( x + h) – g( x ) = k ⋅ lim h→0 h = k ⋅ g ′( x ), Q .E .D .
f ′( x ) = lim h→0
Dilating a function f (x) vertically by a constant k results in the new function g(x) = k ⋅ f (x). What has been shown is that d d ( k ⋅ f ( x )) = k ⋅ f ( x) dx dx That is, dilating a function vertically by a constant k dilates the derivative function by a constant factor of k. 36. If f ( x ) = x 5 , then f ′(c) = 5c 4 . Proof: f ′(c) = lim x →c
= lim x →c
f ( x ) – f (c ) x–c x 5 – c5 x–c
( x − c)( x 4 + x 3c + x 2 c 2 + xc 3 + c 4 ) x →c x−c
= lim
= lim ( x 4 + x 3c + x 2 c 2 + xc 3 + c 4 ) x →c
= c4 + c4 + c4 + c4 + c4 = 5c4, Q .E.D . Calculus Solutions Manual © 2005 Key Curriculum Press
Proof: ( x + h) n – x n h→0 h x n + nx n−1h + 12 n(n − 1) x n−2 h 2 + L + h n − x n = lim h→ 0 h 1 = lim nx n−1 + n(n − 1) x n−2 h + L + h n−1 h→ 0 2 = nx n–1 + 0 + 0 + L + 0 = nx n –1 , which is from the second term in the binomial expansion of (x + h)n, Q.E.D. 38. If yn = u1 + u2 + u3 + L + un , where the ui are differentiable functions of x, prove that yn′ = u1′ + u2′ + u3′ + L + un′ for all integers n ≥ 2. f ′( x ) = lim
f´
5
37. If f (x) = xn, then f′(x) = nxn− 1.
Proof: Anchor: For n = 2, y2 = u1 + u2 . ∴ y2′ = u1′ + u2′ by the derivative of a sum of the two functions property, thus anchoring the induction. Induction hypothesis: Suppose that for n = k > 2, yk′ = u1′ + u2′ + u3′ + L + uk′ . Verification for n = k + 1: Let yk +1 = u1 + u2 + u3 + L + uk + uk +1 . Then yk +1 = (u1 + u2 + u3 + L + uk ) + uk +1 , which is a sum of two terms. ∴ yk′ +1 = (u1 + u2 + u3 + L + uk )′ + uk′ +1 , which, by the anchor, = (u1′ + u2′ + u3′ + L + uk′ ) + uk′ +1 = u1′ + u2′ + u3′ + L + uk′ + uk′ +1 , which completes the induction. Conclusion: ∴ yn′ = u1′ + u2′ + u3′ + L + un′ for all integers n ≥ 2, Q .E .D . 39. a. f ′( x ) = 3 x 2 – 10 x + 5 ⇒ f ( x ) = x 3 – 5 x 2 + 5 x b. g(x) = f (x) + 13 is also an answer to part a because it has the same derivative as f (x). The derivative of a constant is zero. c. The name antiderivative is chosen because it is an inverse operation of taking the derivative. d d d. [ g( x )] = [ f ( x ) + C] = dx dx d d d f ( x) + C = f ( x) dx dx dx The word indefinite is used because of the unspecified constant C. 40. a. f ′(x) = 5x 4 ⇒ f (x) = x 5 + C f(2) = 23 ⇒ (2)5 + C = 23 C = −9 ∴ f (x) = x 5 − 9 b. f ′(x) = 0.12x2 ⇒ f (x) = 0.04x3 + C f (1) = 500 ⇒ 0.04(1)3 + C = 500 Problem Set 3-4
35
C = 499.96 ∴ f (x) = 0.04x3 + 499.96 c. f ′( x ) = x 3 ⇒ f ( x ) = 14 x 4 + C f (5) = 2 ⇒ 14 (5) 4 + C = 2 C = −154.25 ∴ f ( x ) = 14 x 4 − 154.25
Problem Set 3-5 Q1. No values of t Q3. y ′ = −51x
Q2. dy/dx = 10 x
−4
Q4. f ′( x ) = 1.7 x 0.7
Q5. ( d/dx )(3 x + 5) = 3
Q6. f (3) = 45
Q7. f ′(3) = 30
Q8. 45
Q9. ε
Q10. C 4
2.4
1. y = 5t – 3t + 7t dy v= = 20t 3 – 7.2t 1.4 + 7, dt dv a= = 60t 2 – 10.08t 0.4 dt 2. y = 0.3t –4 – 5t dy dv v= = –1.2t –5 – 5, a = = 6t –6 dt dt 3. x = −t3 + 13t2 − 35t + 27. The object starts out at x = 27 ft when t = 0 s. It moves to the left to x ≈ 0.15 ft when t ≈ 1.7 s. It turns there and goes to the right to x = 70 ft when t = 7 s. It turns there and speeds up, going to the left for all higher values of t. y Turns at t = 7, x = 76 Turns at t = 1.7, x = 0.15 Starts at t = 0, x = 27
x 10
4. x = t 4 – 11t 3 + 38t 2 – 48t + 50. The object starts at x = 50 ft when t = 0 s. It moves to the left to x ≈ 30 ft when t ≈ 1.0 s. Then it moves to the right to x ≈ 34.8 ft when t ≈ 2.4 s. The object moves to the left again, turning at x ≈ 9.4 ft when t ≈ 4.8 s and then moving back to the right for higher values of t. y
5. a. x = – t 3 + 13t 2 – 35t + 27 (See Problem 3.) v = –3t 2 + 26t – 35, a = –6t + 26 b. v(1) = –3 + 26 – 35 = –12 So x is decreasing at 12 ft/s at t = 1. a(1) = –6 + 26 = 20 So the object is slowing down at 20 (ft/s)/s because the velocity and acceleration are in opposite directions when t = 1. v(6) = −3(6)2 + 26(6) − 35 = 13 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)/s because the velocity and acceleration are in opposite directions when t = 6. v(8) = –3(8)2 + 26(8) – 35 = –19 So x is decreasing at 19 ft/s at t = 8. a(1) = –6(8) + 26 = –22 So the object is speeding up at 22 (ft/s)/s because the velocity and acceleration are in the same directions when t = 8. c. At t = 7, x has a relative maximum because v(7) = 0 at that point and is positive just before t = 7 and negative just after. No, x is never negative for t in [0, 9]. It starts out at 27 ft, decreases to just above zero around t = 1.7, and does not become negative until some time between t = 9.6 and 9.7. 6. a. x = t 4 – 11t 3 + 38t 2 – 48t + 50 (See Problem 4.) v = 4t 3 – 33t 2 + 76t – 48, a = 12t 2 – 66t + 76 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) = 7 and a(5) = 46 . The object is slowing down at t = 1 because the velocity and acceleration are in opposite directions. The object is speeding up at t = 3 and t = 5 because velocity and acceleration 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 a minimum whenever v = 0. y
x 10
Turns at t ≈ 4.8, x ≈ 9.4
t 1
2
3
4
5
v
Turns at t ≈ 2.4, x ≈ 34.8 Turns at t ≈ 1.0, x ≈ 30.0 Starts at t ≈ 0, x ≈ 50
x 10
36
Problem Set 3-5
e. a = 0 when t = 1.6413… or 3.8586… . When a = 0, v is at a maximum or minimum
Calculus Solutions Manual © 2005 Key Curriculum Press
point and the graph of x is at its steepest for times around these values. y
9. a. d (t ) = 18t – 4.9t 2 ⇒ d ′(t ) = 18 – 9.8t d′(1) = 18 – 9.8 = 8.2 d′(3) = 18 – 9.8 ⋅ 3 = –11.4 d′ 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 at 11.4 m/s. The ball is going up when the derivative is positive and coming down when the derivative is negative. The ball is going up when the graph slopes up and coming down when the graph slopes down.
x 10
t 1
2
3
4
5
v
a
7. a. y 300
d
d´
t 10
b. v = d′ = 30 − 2t. Velocity is positive for 0 ≤ t < 15. Calvin is going up the hill for the first 15 s. c. At 15 seconds his car stopped. d(15) = 324, so distance is 324 feet. d. 99 + 30t – t 2 = 0 ⇒ (33 – t )(3 + t ) = 0 ⇒ t = 33 or t = −3. He’ll be back at the bottom when t = 33 s. e. d(0) = 99. The car runs out of gas 99 ft from the bottom. 8. a. y
c. d′( 4) = –21.2, which suggests that the ball is going down at 21.2 m/s. However, d( 4) = –6.4, which reveals that the ball has gone underground. The function gives meaningful answers in the real world only if the domain of t is restricted to values that make d(t) nonnegative. 10. a. y 2
t v
5
b. The acceleration at the bottom of the swing is 0. The acceleration is greatest at either end of its swing.
v = 251
200
v
a
11. y a = v'
t
2
v
30
b. Trace the v′ graph to find a(0) ≈ 32 . The acceleration decreases because the velocity is approaching a constant. In the real world, this occurs because the wind resistance increases as the velocity increases. c. The limit is 251 ft/s as t approaches infinity. The term 0.88t approaches zero as t gets very large, leaving only 1 inside the parentheses. d. 90% of terminal velocity is 0.9(251) = 225.9 ft/s. Algebraic solution: 251(1 – 0.88t ) = 225.9 ⇒ –0.88t = –0.1 log 0.1 t= = 18.012394... ≈ 18.0 s log 0.88 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 is approximately 10% of the initial acceleration. Calculus Solutions Manual © 2005 Key Curriculum Press
t 5
a
12. v(t) = 15t0.6 . Because v(t ) = x ′(t ) , x(t) must have had t 1.6 in it. The derivative of t1.6 can be assumed to be 1.6t0.6 . So the coefficient of t 1.6 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(101.6 ) + 50 = 423.225K . So the distance traveled is 423.225… − 50 = 373.225… , or about 373 ft. 13. The average rate is defined to be the change in the dependent variable divided by the change in the independent variable (such as total distance divided by total time). Thus, the difference quotient is an average rate. The instantaneous rate is the limit of this average rate as the change in the independent variable approaches zero. Problem Set 3-5
37
m ′′(5) = 14.6299K m ′′(10) = 23.5616K Both quantities are in the units ($/yr)yr. The quantities represent the instantaneous rate of change of the instantaneous rate of change of the amount of money in the account. For example, at t = 5, the rate of increase of the account (153.50 $/yr) is increasing at a rate of 14.63 ($/yr)/yr.
14. a. y 3
t 10
v
20. p ′(7) = − 4.9510K
15. 16. 17.
18.
19.
38
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 when t ≈ 3 or 7. The displacement graph at these times appears to be increasing the fastest. d. The equation used in the text is π y = 2 + 0.85t cos t 2 The student could observe that the period is 4, leading to the coefficient π/2. The amplitude decreases in a way that suggests an exponential function 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 graph seems to converge to 2 as t gets larger. The numerical derivative of the function shown in part a agrees with the graph of the velocity. Note that the actual maximum and minimum values occur slightly before the values of t read from the graph in part a. For instance, the maximum near t = 4 is actually at t = 3.9343… . dy d2y y = 5x 3 ⇒ = 15 x 2 ⇒ 2 = 30 x dx dx dy d2y y = 7x 4 ⇒ = 28 x 3 ⇒ 2 = 84 x 2 dx dx dy 2 5 y = 9x + x ⇒ = 18 x + 5 x 4 ⇒ dx d2y = 18 + 20 x 3 dx 2 y = 10 x 2 − 15 x + 42 ⇒ dy d2y = 20 x − 15 ⇒ 2 = 20 dx dx m′(5) = 153.4979K m′(10) = 247.2100K These numbers represent the instantaneous rate of change of the amount of money in the account. The second quantity is larger because the money grows at a rate proportional to the amount of money in the account. Because there is more money after 10 years, the rate of increase should also be larger. Problem Set 3-6
p ′(14) = −2.4755K ∴ p ′(14) = 12 p ′(7) The fact that these derivatives are negative tells us that the amount of nitrogen 17 is decreasing. p ′′(7) = 0.4902K p ′′(14) = 0.2451K Both quantities are in units (% of nitrogen 17/s)/s. The quantities represent the rate of change of the rate of change of the percentage of nitrogen 17 remaining. For example, at t = 7 s, the rate of decrease (−4.95%/s) is changing at a rate of 0.49 (%/s)/s. 21. y 2
f
f´ x 5
The graph of the derivative looks like the graph of y = cos (x). 22. y 2
f
f'
x
5
The graph of the derivative looks like the graph of y = −sin (x).
Problem Set 3-6 1. y 3
x 10
Calculus Solutions Manual © 2005 Key Curriculum Press
2. The graph confirms the conjecture. y
2
y2 and y 3
y1 x
5
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 3. g(x) = sin 3x Conjecture: g′(x) = 3 cos 3x The graph confirms the conjecture. 3
y g´
Q1. f (c) exists Q2. lim f ( x ) exists x →c Q3. lim f ( x ) = f (c) x →c
( Any order is acceptable.)
Q4. No (not continuous) Q5. dy/dx = 16 x –1/ 5 Q6. f (x) = −10x− 3 Q7. Antiderivative Q8.
g x 10
y = sin x x
4. h( x ) = sin x 2 Conjecture: h′( x ) = 2 x cos x 2 The graph confirms the conjecture.
Q9. y = cos x x
y h' h
1
Q10. C 1. a. Let y = f (u), u = g(x). dy dy du = ⋅ dx du dx b. y′ = f ′[ g( x )] ⋅ g′( x )
x 5
5. t ( x ) = sin x 0.7 Conjecture: t ′( x ) = 0.7 x –0.3 cos x 0.7 The graph confirms the conjecture. y
2. 1
t
x t´
10
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 to the inside function. Then multiply the answer by the derivative of the inside function with respect to 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.
Calculus Solutions Manual © 2005 Key Curriculum Press
3. 4. 5. 6. 7. 8. 9. 10.
c. To differentiate a composite function, differentiate the outside function with respect to the inside function, then multiply by the derivative of the inside function with respect to x. f ( x ) = ( x 2 − 1)3 a. f ′(x) = 3(x2 − 1)2(2x) = 6x(x2 − 1)2 b. (x 2 − 1)3 = x 6 − 3x 4 + 3x 2 − 1, so f ′(x) = 6x 5 − 12x 3 + 6x. c. From part a, f ′(x) = 6x(x2 − 1)2 = 6x(x 4 − 2x 2 + 1) = 6x 5 − 12x 3 + 6x, so the two answers are equivalent. f ( x ) = cos 3 x ⇒ f ′( x ) = − sin 3 x ⋅ 3 = −3 sin 3 x f ( x ) = sin 5 x ⇒ f ′( x ) = 5 cos 5 x g( x ) = cos ( x 3 ) ⇒ g′( x ) = −3 x 2 sin ( x 3 ) h( x ) = sin ( x 5 ) ⇒ h′( x ) = 5 x 4 cos ( x 5 ) y = (cos x)3 ⇒ y′ = 3(cos x)2 ⋅ (−sin x) = −3 cos2 x sin x f ( x ) = (sin x )5 ⇒ f ′( x ) = 5(sin x ) 4 ⋅ cos x = 5 sin 4 x cos x y = sin 6 x ⇒ y′ = 6 sin 5 x cos x f ( x ) = cos 7 x ⇒ f ′( x ) = 7 cos 6 x ⋅ ( − sin x ) = –7 cos 6 x sin x Problem Set 3-7
39
11. y = −6 sin 3x ⇒ y′ = −18 cos 3x 12. f (x) = 4 cos (−5x) ⇒ f′ (x) = 4[−sin (−5x)] ⋅ (−5) = 20 sin (−5x) d 13. (cos 4 7 x ) = 4 cos3 7 x ⋅ ( − sin 7 x ) ⋅ 7 dx = −28 cos3 7x sin 7x d 14. (sin 9 13 x ) = 9 sin 8 13 x cos 13 x ⋅ 13 dx = 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)
The line is tangent to the graph. y 5
x 3
28. y = 7 sin π t + 12t1.2 dy velocity = = 7π cos π t + 14.4t0.2 dt Yes, there are times when the beanstalk is shrinking. The velocity graph is negative for brief intervals, and the y-graph is decreasing in these intervals.
17. f (x) = (5x + 3)7 ⇒ f′ (x) = 7(5x + 3)6 ⋅ 5 = 35(5x + 3)6 18. f (x) = (x + 8) ⇒ f′ (x) = 9(x2 + 8)8 ⋅ 2x = 18x(x2 + 8)8 2
y
9
19. y = (4x3 − 7)− 6 ⇒ y′ = −6(4x3 − 7)− 7 ⋅ 12x2 = −72x2(4x3 − 7)− 7 20. y = (x 2 + 3x − 7)− 5 ⇒ y′ = −5(x 2 + 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 dy 23. y = 4 cos 5x ⇒ = 4(−sin 5x)5 = −20 sin 5x ⇒ dx d2y = −20(cos 5x)5 = −100 cos 5x dx 2 24. y = 7 sin (2x + 5) ⇒ dy = 7 cos (2x + 5)(2) = 14 cos (2x + 5) ⇒ dx 2 d y = 14[−sin (2x + 5)](2) = dx 2 −28 sin (2x + 5) 1 25. f′ (x) = cos 5x ⇒ f (x) = sin 5x + C 5 26. f′ (x) = 10 sin 2x ⇒ f (x) = −5 cos 2x + C 27. f (x) = 5 cos 0.2x f′ (x) = −5 sin 0.2x ⋅ 0.2 = −sin 0.2x f′ (3) = −sin 0.6 = −0.5646… and f (3) = 4.1266… The line has the equation y = −0.5646…x + 5.8205… . 40
y and v
Problem Set 3-7
50
v t 10
4π 3 dV r ⇒ = 4πr 2 3 dr dV/dr is in (cm3/cm), or cm2.
29. a. V =
b. r = 6t + 10 dr = 6 (not surprising!). Units: cm/min dt dV dV dr d. = ⋅ dt dr dt When t = 5, r = 40. So dV = 4π ( 40 2 ) = 6400π . dr dV ∴ = 6400π ⋅ 6 = 38,400π cm 3 /min. dt dV/dr has units cm 2, and dr/dt has units cm , cm/min, so dV/dt has units cm 2 ⋅ min which becomes cm3/min, Q.E.D. This matches the commonsense answer that rate of volume cm 3 change of volume is = . time min c.
4π (6t + 10)3 3 dV ∴ = 4π (6t + 10)2 (6) = 24π (6t + 10)2 dt dV When t = 5, = 24π [6(5) + 10]2 = 38,400π . dt
e. V =
Calculus Solutions Manual © 2005 Key Curriculum Press
y(t ) = 25 + 20 cos
30. a. u
π (t – 3) 10
π (t – 3) 10 π y ′(15) = –2π sin (15 – 3) = 3.69316 K 10 y(t) is increasing at about 3.7 ft/s. The fastest that y(t) changes is 2π, or 6.28… ft/s. The seat is at y(t) = 25 ft above the ground then. y = C + A cos B(x − D). B = 2π /6 = π /3 rad/s D = phase displacement = 1.3 s A = 0.5(110 − 50) = 30 cm C = 110 − 30 or 50 + 30, which equals 80 cm. π ∴ d = 80 + 30 cos (t – 1.3) 3 π d ′ = –10π sin (t – 1.3) 3 π d′(5) = −10π sin (5 – 1.3) = 21.02135K 3 π d′(11) = –10π sin (11 – 1.3) = 21.02135K 3 At both times, the pendulum is moving away from the wall at about 21.0 cm/s. The answers are the same because the times are exactly one period apart. π d′(20) = –10π sin (20 – 1.3) = −21.02135K 3 The pendulum is moving toward the wall. Because the derivative is negative, d is decreasing, which in this problem implies motion toward the wall. The fastest is 10π ≈ 31.4 cm/s, when d = 80. π π 0 = –10π sin (t – 1.3) ⇒ sin (t − 1.3) = 0 3 3 π π −1 ⇒ (t – 1.3) = sin 0 ⇒ (t − 1.3) = 0 + π n 3 3 ⇒ 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 velocity is zero, the pendulum is at its maximum height. The curb has slope (3.25 − 0.75)/44 = 2.5/44. ∴ equation is f (x) = 0.75 + (2.5/44)x. Sinusoid has period 8 ft, so B = 2π /8 = π /4. Amplitude = 0.5(0.75 − 0.25) = 0.25 ft. Low end of ramp is a low point on the sinusoid. ∴ sinusoidal axis is at y = 0.25 when x = 0 and goes up with slope 2.5/44. Sinusoid is at a low point when x = 0. So phase displacement is zero if the cosine is subtracted.
b. y ′(t ) = –2π sin ∆u
c. ∆x
x
d.
∆u does not approach zero as ∆x approaches zero from the left side. (∆u does approach zero as ∆x approaches zero from the left side.)
2. a.
b. u
∆u
b.
∆x
x
c. ∆u does approach zero as ∆x approaches zero from either side.
Problem Set 3-8 Q1. Q3. Q5. Q7. Q9.
f′(x) = 9x 8 y′ = 72x5 (5x6 + 11)1.4 12 Yes (continuous)
Q2. Q4. Q6. Q8. Q10.
dy/dx = −3 sin x s′ = 0 1 f (x) = −cos x 2 E
d.
y
4
f'
e. x
f.
4
1. a. 45
d ( x)
25 Increasing
x
5 3
15
23
y(t) = C + A cos B(t − D) Vertical displacement = 25 = C Amplitude = 0.5(40) = 20 = A Phase disp. (for cosine) = 3 = D Period = 60/3 = 20, so B = 2π/20 = π/10. (Note that B is the angular velocity in radians per second.)
Calculus Solutions Manual © 2005 Key Curriculum Press
3. a. b.
Problem Set 3-8
41
∴ equation is
2.5 π x − 0.25 cos x 44 4 (There are other correct forms.) π 2.5 π c. g ′( x ) = + sin x 44 16 4 π 2.5 π g ′( 9) = + sin (9) = 0.1956 K ft/ft 44 16 4 Going up at about 0.2 vertical ft per horizontal ft π 2.5 π g ′(15) = + sin (15) = −0.0820 K ft/ft 44 16 4 Going down at about 0.08 vertical ft per horizontal ft. A positive derivative implies g(x) is getting larger and thus the child is going up. A negative derivative implies g(x) is getting smaller and thus the child is going down. d. By tracing the g ′ graph, maximum value of g′ (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 of minutes. 14 hours 3 minutes is 843 minutes. 10 hours 15 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 2π ∴ L( d ) = 729 + 114 cos ( d – 172) 365 On August 7, d = 219. 2π L(219) = 729 + 114 cos (219 – 172) = 365 807.67… , or about 13 hours 28 minutes. 228π 2π b. L ′( d ) = − sin ( d − 172) 365 365 On August 7, d = 219. 228π 2π sin (219 – 172) = L ′(219) = − 365 365 –1.42009… Rate ≈ –1.42 min/day (Decreasing at about 1.42 min/day) c. The greatest rate occurs when the sine is 1 or −1. Rate is 228π/365 ≈ 1.96 min/day. 1/4 year is about 91 days. So greatest rate occurs at day 172 ± 91, which is day 263 or day 81 (September 20 or March 22). 5. In general, the period for a pendulum formed by a weight suspended by a string of negligible mass is 2π L/g , where L is the length from the pivot point to the center of mass (actually, the center of percussion) of the weight, and g is the g( x ) = 0.25 +
42
Problem Set 3-8
gravitational acceleration, about 9.8 m/s2. Consequently, if the pendulum is 1 meter long, its period will be 2π 1/9.8 = 2.007K , or about 2 s. This is the period for a complete back-andforth swing. You must quadruple the length of a pendulum to double its period. A pendulum hung from the ceiling will have a period slow enough to measure fairly precisely. A good way to get more accuracy is to count the total time for ten swings, then divide by 10. The period is roughly constant for any (moderate) amplitude, as long as the amplitude is not too big. This fact is not obvious to the uninitiated student and is worth spending time showing. It is quite dramatic to watch a pendulum take just as long to make ten swings with amplitude 2 cm as it does with amplitude 20 or 30 cm. 6. The following data were computed from actual sunrise and sunset times for San Antonio for each ten days. You can get similar information for your locality from the local weather bureau or newspaper office, from the Nautical Almanac Office, U.S. Naval Observatory, Washington, D.C., 20390, or from the Internet. Day
Min
Day
Min
Day
Min
0 10 20 30 40 50 60 70 80 90 100 110 120
617 623 632 645 660 676 693 711 729 747 764 780 797
130 140 150 160 170 180 190 200 210 220 230 240 250
811 823 833 840 842 842 836 828 816 803 789 772 755
260 270 280 290 300 310 320 330 340 350 360
738 720 703 686 669 653 639 628 620 615 615
L(d ) 800
700
600 0
d 100
200
300
The graph shows a good fit to the data. But there is 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 © 2005 Key Curriculum Press
is the fact that in the fall and winter, Earth is closer to the Sun and hence moves slightly more rapidly through its angle with the Sun than during the spring and summer. 7. a. y g
h
f
4
x 1
The limits are all equal to 4. b. f (x) < g(x) and lim f ( x ) = lim g( x ) = 4 x →1 x →1 f (x) ≤ h(x) ≤ g(x) c. x f(x) h(x) g(x) 0.95
3.795
3.8
3.805
0.96 0.97 0.98 0.99 1.00 1.01 1.02 1.03 1.04 1.05
3.8368 3.8782 3.9192 3.9598 4 4.0398 4.0792 4.1182 4.1568 4.195
3.84 3.88 3.92 3.96 4 4.04 4.08 4.12 4.16 4.2
3.8432 3.8818 3.9208 3.9602 4 4.0402 4.0808 4.1218 4.1632 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 of 4. From the table, δ = 0.01 or 0.02 will work, but 0.03 is too large. All the values of h(x) are between the corresponding values of f (x) and g(x), and the three functions all approach 4 as a limit. sin x 8. Prove that lim = 0. See the text proof. x→0 x 9. a. The numbers are correct. b. x (sin x)/x 0.05 0.04 0.03 0.02 0.01
0.99958338541… 0.99973335466… 0.99985000674… 0.99993333466… 0.99998333341…
c. Answers will vary according to calculator. For the TI-83 in TABLE mode, starting x at 0 and using ∆ x = 10−7 shows that all values round to 1 until x reaches 1.8 × 10−6, which registers as 0.999999999999. d. Answer will depend on calculator. For TI-83 in TABLE mode, (sin 0.001)/0.001 is 0.999999833333, which agrees exactly with the value published by NBS to 12 places. e. If students have studied Taylor series (Chapter 12) before taking this course, they will be able to see the reason. The Taylor series for sin 0.001 is 0.0013 0.0015 0.0017 + − +L 3! 5! 7! = 0.00100 00000 00000 00000 000… − 0.00000 00001 66666 66666 666… + 0.00000 00000 00000 00833 333… _______________________________ 0.001 −
= 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 that 0 < |x − c | < δ f ⇒ |f (x) − L | < ε, because lim f ( x ) = L. Similarly, there is a δ g > 0 x →c
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 | < δ g ⇒ |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 h( x ) = L . x →c
Q.E.D.
c. See Figure 3-8c or 3-8d. 13. a. The limit seems to equal 2. b. g
h
2
g
1
h
c. See the graph in part b. The lines have equations g(x) = x + 1 and h(x) = 3 − x. d. Prove that lim y = 2. x →1
Values are getting closer to 1.
Calculus Solutions Manual © 2005 Key Curriculum Press
Problem Set 3-8
43
Proof: lim( x + 1) = 1 + 1 = 2 x →1
lim(3 − x ) = 3 − 1 = 2 x →1
For x < 1, g(x) ≤ y ≤ h(x). ∴ the squeeze theorem applies, and lim− y = 2. x →1
For x > 1, h(x) ≤ y ≤ g(x). ∴ the squeeze theorem applies, and lim+ y = 2. x →1
Because both left- and right-hand limits equal 2, lim y = 2, Q .E.D . x →1
e. The word envelope (a noun) is used because the small window formed by the two lines “envelops” (a verb) the graph of the function. 1 f. As |x| becomes large, (x − 1) · sin = x –1 sin [1/( x – 1)] takes on the form 1/( x – 1) sin (argument) as the argument approaches (argument) zero. Thus the limit is 1 and y approaches 2 + 1, which equals 3. 14. Answers will vary.
Problem Set 3-9 Q1. 1
Q2. –sin x 2
d d (cos x ) = ( − sin x ) = − cos x dx 2 dx Q4. y = sin x + C Q5. x 8 Q6. x48 Q7. log 32 = 5 log 2 Q8. Q3.
y
c. 1061.84/1000 = 1.06184, so the APR for 0 to 1 year is approximately 6.184%. 1127.50/1061.84 = 1.061836… , so the APR for 1 to 2 years is approximately 6.184%. 1197.22/1127.50 = 1.06183… , so the APR for 2 to 3 years is approximately 6.184%. The APR is higher than the instantaneous rate. Savings institutions may prefer to advertise the APR instead of the instantaneous rate 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) = −3.4 f ′ (2) = −1.7224… f ′ (4) = −0.8726… f ′ (6) = −0.4420 Factor of change from 0 to 2: −1.7224…/−3.4 = 0.5066 Factor of change from 2 to 4: −0.8726…/−1.7224… = 0.5066 Factor of change from 4 to 6: −0.4420…/−0.8726… = 0.5066 b. f (0) = 10 f (2) = 5.0661… f (4) = 2.5666… f (6) = 1.3002… Factor of change from 0 to 2: 5.0661…/10 = 0.5066 Factor of change from 2 to 4: 2.5666…/5.0661… = 0.5066 Factor of change from 4 to 6: 1.3002…/2.5666… = 0.5066 The factors of change are the same in part a and part b. c. y
x
10
f t
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… $/yr M′(10) = 109.3271… $/yr M′(20) = 199.2070… $/yr b. M(0) = $1000 M(1) = $1061.84 M(2) = $1127.50 M(3) = $1197.22 Increase from year 0 to year 1: $61.84 Increase from year 1 to year 2: $65.66 Increase from year 2 to year 3: $69.72 No, the amount of money in the account does not change by the same amount each year. 44
Problem Set 3-9
5
f'
The values of f are negative because the amount of 18-F is decreasing as time goes on. −23.5 3. a. A( p) = 63 − 23.5 ln p ⇒ A′( p) = p y
50
A x A´
10
b. If the pressure is increasing, then the altitude is decreasing. A′(10) = −2.35, so the altitude is changing at −2.35 thousand feet/psi. That Calculus Solutions Manual © 2005 Key Curriculum Press
is, a change of 1 psi would indicate that the altitude had decreased by 2.35 thousand feet. The negative sign means that the altitude is decreasing. c. A′(5) = −4.7 = 4.7 A′(10) = −2.35 = 2.35 This shows that the altitude is changing faster at 5 psi than it is at 10 psi. d. A(p) = 0 63 − 23.5 ln p = 0 −23.5 ln p = −63 ln p = 2.6808… p = 14.5975… The fact that A(p) is negative for all values of p greater than 14.5975… means that if the air pressure is above 14.5975 psi, then the plane must 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 1 y= (ln x − ln 3000) 0.05 y = 20 ln x – 20 ln 3000 b. y(3000) = 20 ln 3000 − 20 ln 3000 = 0 y(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 $6000: 3.6464… The time intervals decrease as the amount of money increases because when there is more money in the account, it takes less time to earn the given amount of interest. 20 yr/$ c. y = 20 ln x − 20 ln 3000 ⇒ y ′ = x y′(3000) = 0.0066… y′(4000) = 0.005 y′(5000) = 0.004 y′(6000) = 0.0033… This shows that the number of years it takes to earn each dollar decreases as the amount of money increases. 5. f (x) = 5e3x ⇒ f ′ (x) = 15e3x 6. f (x) = 7e− 6x ⇒ f ′ (x) = −42e− 6x
Calculus Solutions Manual © 2005 Key Curriculum Press
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) 10 10 11. f ( x ) = 10 ln (7 x ) ⇒ f ′( x ) = ⋅7 = 7x x 9 9 12. g( x ) = 9 ln 4 x ⇒ g ′( x ) = ⋅4 = 4x x 18 54 13. T = 18 ln x 3 ⇒ T ′ = 3 (3 x 2 ) = x x 1000 700 14. P = 1000 ln x 0.7 ⇒ P ′ = 0.7 0.7 x −0.3 = x x 15. y = 3 ln (cos 5x) ⇒ 3 y′ = ( − sin 5 x )5 = −15 tan 5 x cos 5 x 16. y = 11 ln (sin 0.2x) ⇒ 11 y′ = (cos 0.2 x ) 0.2 = 2.2 cot 0.2 x sin 0.2 x 17. u = 6 ln (sin x0.5 ) ⇒ 6 u′ = (cos x 0.5 ) 0.5 x −0.5 = 3 x −0.5 cot x 0.5 sin x 0.5 18. v = 0.09 ln (cos x 8 ) ⇒ v′ =
0.09 ( − sin x 8 )8 x 7 = −0.72x7 tan x8 cos x 8
1 x ⋅e = 1 ex Not surprising because we could have first used the fact that natural log and exp are inverses: r(x) = ln ex = x ⇒ r′ (x) = 1 c(x) = eln x = x ⇒ c ′ (x) = 1 c ′ (2) = 1, c ′ (3) = 1, c ′ (4) = 1 f (x) = 3x ⇒ f ′ ( x ) = ( ln 3 ) 3 x g (x) = 0.007x ⇒ g′(x) = (ln 0.007) 0.007x y = 1.6cos x ⇒ y′ = (ln 1.6)1.6cos x (−sin x) = −ln 1.6 sin x (1.6cos x) y = sin 5x ⇒ y′ = cos 5x (ln 5)5x dy 1 5 y = ln x 5 ⇒ = ⋅ 5 x 4 = = 5 x −1 ⇒ dx x 5 x − 5 d2y = −5 x −2 = 2 dx 2 x
19. r ( x ) = ln e x ⇒ r ′( x ) =
20. 21. 22. 23. 24. 25.
dy d2y = 7e 7 x ⇒ 2 = 49e 7 x dx dx − 0.7 x − 0.7 x 27. y = e ⇒ y′ = −0.7e ⇒ y″ = 0.49e− 0.7 x
26. y = e 7 x ⇒
Problem Set 3-9
45
28. y = ln 8 x ⇒ y ′ =
1 1 ⋅ 8 = = x −1 ⇒ 8x x
−1 x2 2x 29. f ′ (x) = 12e ⇒ f ( x ) = 6 e 2 x + C
c. m( x ) =
y ′′ = −1x −2 =
30. y′ = 5x ln 5 ⇒ y = 5x + C
0.4 x 2 – x − 0.6 x–3
m (x ) 2
1
x 3
Problem Set 3-10
d. Line: y = 1.4x + 1.4
Review Problems
y
R0. Answers will vary. R1. a. x
Average rate of change from 2 to x
1.97
f (2) − f (1.97) = 11.82 0.03
1.98
f (2) − f (1.98) = 11.88 0.02
1.99
f (2) − f (1.99) = 11.94 0.01
2.01
f (2.01) − f (2) = 12.06 0.01
2.02
f (2.02) − f (2) = 12.12 0.02
2.03
f (2.03) − f (2) = 12.18 0.03
The derivative of f at x = 2 is approximately 12. f ( x ) − f (2) b. r ( x ) = x−2 r(2) is of the form 00 . lim r ( x ) appears to be 12.
4
x 3
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 that the graph looks more and more like the line. R3. a. y y2
50
y1 1
b. See the graph in part a. c. The y1 graph has a high point or a low point at each x-value where the y2 graph is zero. d. y
x →2
p
x 3 − 8 ( x − 2)( x 2 + 2 x + 4) c. r ( x ) = = x−2 ( x − 2) = x 2 + 2 x + 4, x ≠ 2 lim r ( x ) = lim x 2 + 2 x + 4 because x ≠ 2. x →2
x →2
∴ lim r ( x ) = 12 x →2
d. The answers in parts a, b, and c are the same. f ( x ) − f (c ) x–c 2 b. f (x) = 0.4x − x + 5
R2. a. f ′(c) = lim x →c
0.4 x 2 − x + 5 − 5.6 x →3 x–3 ( x – 3)(0.4 x + 0.2) = lim x →3 x–3 = lim (0.4 x + 0.2) = 1.4
f ′(3) = lim
x →3
46
x
Problem Set 3-10
20
1
t p´
Take the numerical derivative at t = 3, 6, and 0. p′(3) ≈ −2.688… . Decreasing at about 2.69 psi/h when t = 3. p′(6) ≈ −1.959… . Decreasing at about 1.96 psi/h when t = 6. p′(0) ≈ −3.687… . Decreasing at about 3.69 psi/h when t = 0. The units are psi/h. The sign of the pressure change is negative because the pressure is decreasing. Yes, the rate of pressure change is getting closer to zero. R4. a. See the text for the definition of derivative. b. Differentiate Calculus Solutions Manual © 2005 Key Curriculum Press
c. If y = x n, then y′ = nx n− 1. d. See solution to Problem 35 in Problem Set 3-4. e. See the proof in Section 3-4. dy f. is pronounced “d y, d x.” dx d ( y) is pronounced “d, d x, of y.” dx Both mean the derivative of y with respect to x. 63 4 / 5 g. i. f ( x ) = 7 x 9/ 5 ⇒ f ′( x ) = x 5 x2 ii. g( x ) = 7 x –4 − − x+7⇒ 6 x g′( x ) = −28 x –5 − − 1 3 iii. h(x) = 73 ⇒ h′ (x) = 0 h. f ′(32) = 635 (32) 4 / 5 = 201.6 exactly. The numerical derivative is equal to or very close to 201.6. i.
4
e.
y f x
i. y = −0.01t3 + 0.9t2 − 25t + 250 dy v= = −0.03t 2 + 1.8t − 25 dt dv a= = −0.06t + 1.8 dt 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 = 15 because the velocity and the acceleration have opposite signs. iii. v = −0.03t2 + 1.8t − 25 = 0 By using the quadratic formula or the solver feature of your grapher, t = 21.835… or t = 38.164… . The spaceship is stopped at about 21.8 and 38.2 seconds. iv. y = −0.01t3 + 0.9t2 − 25t + 250 = 0 By using TRACE or the solver feature of your grapher, t = 50. v(50) = −10 Because the spaceship is moving at 10 km/s when it reaches the surface, it is a crash landing!
R6. a.
4
y
f´
derivative
1
cosine
x 1
R5. a. v = a=
dx or x ′(t ). dt dv d2x or v ′(t ), a = 2 or x ′′(t ) dt dt
d2y means the second derivative of y with dx 2 respect to x. y = 10x 4 ⇒ y′ = 40x 3 ⇒ y″ = 120x 2 c. f ′( x ) = 12 x 3 ⇒ f ( x ) = 3 x 4 + C. f (x) is the antiderivative, or the indefinite integral, of f (x). d. The slope of y = f ( x ) is determined by the value of f ′( x ). So the slope of y = f ( x ) at x = 1 is f ′(1) = −1, at x = 5 is f ′(5) = 3, and at x = −1 is f ′( −1) = 0. b.
y
f
f´
5
x 5
Calculus Solutions Manual © 2005 Key Curriculum Press
b. The graph of the derivative is the same as the sine graph but inverted in the y-direction. Thus, (cos x )′ = − sin 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) dy dy du R7. a. i. = ⋅ dx du dx ii. f (x) = g(h(x)) ⇒ f ′(x) = g ′(h(x)) ⋅ h ′(x) iii. The derivative of a composite function is the derivative of the outside function with respect to the inside function times the derivative of the inside function with respect to x. b. See the derivation in the text. This derivation constitutes a proof. ∆u must be nonzero throughout the interval.
Problem Set 3-10
47
c.
d.
e.
f.
R8. a.
i. f (x) = (x2 − 4)3 f ′(x) = 3(x 2 − 4)2 ⋅ 2x = 6x(x2 − 4)2 ii. f (x) = x 6 − 12x 4 + 48x 2 − 64 f′(x) = 6x 5 − 48x 3 + 96x Expanding the answer to part i gives f′(x) = 6x5 − 48x3 + 96x, which checks. i. f′(x) = −3x 2 sin x 3 ii. g′(x) = 5 cos 5x iii. h′(x) = 6 cos5 x (−sin x) = −6 sin x cos5 x iv. k ′(x) = 0 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 indefinite integral, of f (x). W = 0.6x3 and dx/dt = 0.4 dW dW dx = ⋅ = 1.8 x 2 ⋅ 0.4 = 0.72 x 2 dt dx dt If x = 2, W = 0.6 ⋅ 23 = 4.8 lb dW/dt = 0.72(22) = 2.88 The shark is gaining about 2.88 lb/day. If x = 10, W = 0.6 ⋅ 103 = 600 lb. dW/dt = 0.72(102) = 72 The shark is gaining about 72 lb/day. The chain rule is used to get dW/dt from dW/dx by multiplying the latter by dx/dt. sin x lim =1 x→0 x x −0.05 −0.04 −0.03 −0.02 –0.01 0.00 0.01 0.02 0.03 0.04 0.05
(sin x)/x 0.99958338541… 0.99973335466… 0.99985000674… 0.99993333466… 0.99998333341… undefined 0.99998333341… 0.99993333466… 0.99985000674… 0.99973335466… 0.99958338541…
The values of (sin x)/x approach 1 as x approaches 0. b. See the text for the statement of the squeeze theorem. Squeeze (sin x)/x between cos x and sec x. c. See the proof in Section 3-8 of the text.
48
Problem Set 3-10
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 = 20 D = 0 for cosine because hand starts at a high point. B = 2π/60 = π/30 because period is 60 s. π d (t ) = 180 + 20 cos t 30 2π π d ′(t ) = − sin t 3 30 At 2, t = 10: d′(10) ≈ −1.81 cm/s At 3, t = 15: d′(15) ≈ −2.09 cm/s At 7, t = 35: d′(35) ≈ 1.05 cm/s At the 2 and 3, the tip is going down, so the distance from the floor is decreasing, which is implied by the negative derivatives. At the 7, the tip is going up, as implied by the positive derivative. R9. a. p( x ) = 100e −0.1x ⇒ p ′( x ) = 100( −0.1)e −0.1x = −10e− 0.1 x p′(0) = −10 p′(10) = −3.6787… p′(20) = −1.3533… The rates are negative because the amount of medication in your body is decreasing. To find the biological half-life, find x such that 1 p( x ) = p(0) = 50 2 100e −0.1x = 50 1 e −0.1x = 2 1 −0.1x = ln 2 1 x = −10 ln 2 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 the medicine remains in your body. b. i. f (x) = 5e2x ⇒ f ′(x) = 5(2)e2x = 10e2x dy ii. y = 7 x ⇒ = (ln 7)7 x dx d 1 iii. [ln (cos x )] = ( − sin x ) = − tan x dx cos x dy 1 = 8 ⋅ = 8 x −1 ⇒ iv. y = ln x 8 = 8 ln x ⇒ dx x − 8 d2y = −8 x −2 = 2 dx 2 x c. f ′( x ) = 12e 3x ⇒ f ( x ) = 4e 3x + C
Calculus Solutions Manual © 2005 Key Curriculum Press
Chapter Test
d. y
y1
y3
3
T1. See the definition of derivative in Section 3-2 or 3-4. T2. Prove that if f (x) = 3x4, then f ′ (x) = 12x3.
y2 x 3
Proof: y1 = ex is the inverse of y2 = ln x, so y1 is a reflection of y2 across the line y = x.
f ( x + h) – f ( x ) 3( x + h) 4 − 3 x 4 = lim h→ 0 h→ 0 h h 4 3 2 2 3 x + 12 x h + 18 x h + 12 xh 3 + 3h 4 − 3 x 4 = lim h→0 h = lim (12 x 3 + 18 x 2 h + 12 xh 2 + 3h 3 ) = 12 x 3 ,
f ′( x ) = lim
Concept Problems C1. a. f (x) = x7, g(x) = x9. So h(x) = f(x) ⋅ g(x) = x 16. b. h′(x) = 16x15 c. f ′(x) = 7x 6, g′(x) = 9x 8. So f ′(x) ⋅ g′(x) = 63x14 ≠ h ′(x). d. h′(x) = f ′ (x) ⋅ g(x) + f (x) ⋅ g′(x) = 7x 6 ⋅ x 9 + x 7 ⋅ 9x 8 = 16x 15 x – sin 2 x C2. a. f ( x ) = . f (0) has the form 0/0, sin x which is indeterminate. f is discontinuous at x = 0 because f (0) does not exist. b. By graph (below) or by TABLE , f (x) seems to approach −1 as x approaches zero. Define f (0) to be −1.
h→ 0
Q .E .D .
T3. If you zoom in on the point where x = 5, the graph appears to get closer and closer to the tangent line. The name of this property is local linearity. y 5
Slope = 2 x 5
f (x ) 5
x π
T4. Amos substituted before differentiating instead of 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) = −5x 4 sin x 5
c. Conjecture: The function is differentiable at x = 0. The derivative should equal zero because the graph is horizontal at x = 0. f ( x ) – f (0) d. f ′(0) = lim h→0 x–0 x – sin 2 x – (–1) = lim sin x x→0 x x – sin 2 x + sin x = lim x→0 x sin x Using TABLE for numerator, denominator, and quotient shows that the numerator goes to zero faster than the denominator. For instance, if x = 0.001, 1.1666 K × 10 –9 = 0.00116 K 9.999K × 10 –7 Thus, the limit appears to be zero. (The limit can be found algebraically to equal zero by l’Hospital’s rule after students have studied Section 6-5.) quotient =
Calculus Solutions Manual © 2005 Key Curriculum Press
T7.
d 1 [ln (sin x )] = ⋅ cos x = cot x dx sin x
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 = 60x 2/3 − x + 25 ⇒ y′ = 40x− 1/3 − 1 T11. y = e 9x ⇒
dy d2y = 9e 9x ⇒ 2 = 81e 9x dx dx
T12. y′ ≈ 0.6 (Function is y = −3 + 1.5x, for which the 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 the displacement function. T14. f ′ (x) = 72x 5/4 ⇒ f (x) = 32x 9/4
Problem Set 3-10
49
T15. f ′ (x) = 5 sin x and f (0) = 13 f (x) = −5 cos x + C 13 = −5 cos 0 + C ⇒ C = 18 f (x) = −5 cos x + 18
Proof:
T16. f (x) = cos 3x ⇒ f ′ (x) = −3 sin 3x f ′ (5) = −3 sin 15 = −1.95086… Decreasing at 1.95… y-units per x-unit. T17. f ( x ) =
sin x x f (x ) 1
x 1
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 g( x ) = lim h( x ) = L, and (3) f is a x →c
x →c
function for which g(x) ≤ f (x) ≤ h(x) for all x in that neighborhood of c, then lim f ( x ) = L. x →c
T18. h –0.0003 –0.0002 –0.0001 0 0.0001 0.0002 0.0003
5h − 1 h 1.6090… 1.6091… 1.6093… undefined 1.6095… 1.6097… 1.6098…
ln 5 = 1.6094… . The table shows that 5h − 1 lim = ln 5. h→0 h
50
Problem Set 3-10
5 x +h − 5 x d x Definition of derivative. (5 ) = lim h→ 0 dx h 5h − 1 = 5 x lim Factor out 5x. h→0 h = 5x · ln 5 Evaluate. t T19. v(t) = 251(1 − 0.88 ) a(t) = 251[− ln (0.88)] 0.88t = −251(ln 0.88)0.88t a(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 opposite signs for a particular value of t, then the object is slowing down at that time. T21. a. v(t) = t1.5 + 3 ⇒ a(t) = 1.5t0.5 t 2.5 b. d (t ) = + 3t + C 2.5 d(1) = 20 12.5 + 3(1) + C = 20 2.5 3.4 + C = 20 C = 16.6 ∴ d(t) = 0.4t2.5 + 3t + 16.6 c. d(9) − d(1) = 120.8 This represents the displacement between the first and ninth seconds. 2π T22. a. c(t ) = 300 + 2 cos t⇒ 365 4π 2π c ′(t ) = − sin t 365 365 4π 2π b. c′(273) = − sin ⋅ 273 365 365 = 0.03442… ppm/day 0.03442 K 1 c. Rate is (6 × 1015 ) ⋅ ⋅ = 1, 000, 000 24 ⋅ 60 ⋅ 60 2390.6627… , which is approximately 2390 tons per second! T23. Answers will vary.
Calculus Solutions Manual © 2005 Key Curriculum Press
Chapter 4—Products, Quotients, and Parametric Functions Problem Set 4-1 1. f (x) = 3 cos x ⇒ f ′ (x) = −3 sin x g(x) = 2 sin x ⇒ g′(x) = 2 cos x 2. 6
p (x )
x
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 f ′ ( x ) g( x ) – f ( x ) g ′ ( x ) q ′( x ) = [ g( x )]2
10
∴ q ′( 2 ) =
(–3 sin 2)(2 sin 2) – (3 cos 2)(2 cos 2) = (2 sin 2)2
−1.8141… , which agrees with Problem 3. p′(2) ≈ −3.9218… p(x) is decreasing at x = 2 because p ′(2) < 0. This fact corresponds with the graph, which slopes 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. 6
q (x )
Problem Set 4-2 Q1. y ′ = Q3.
3 –1/ 4 x 4
dy = −30(5 x − 7) –7 dx
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. y t = 2 here 2
x 3
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 ∆y –0.16618K ≈ = = 0.3051K dx ∆x –0.54466 K dy/dt 2 cos 2 At t = 2, = = 0.3051 K , dx/dt –3 sin 2 which agrees with the difference quotient.
Calculus Solutions Manual © 2005 Key Curriculum Press
Q4.
d (sin 2 x ) = 2 cos 2 x dx
Q5. v′ = −3 cos2 t sin t
Q6. L′ = 2m + 5
Q7. y = sin x + C
Q8. y ′ ≈ −3
3
10
Q2. y′ = 1/x
Q9. 1. 2. 3. 4.
Q10. B 4 ft/s 3 f ( x) = x cos x ⇒ f ′ (x) = 3x 2 cos x – x 3 sin x f ( x) = x 4 sin x ⇒ f ′ (x) = 4x 3 sin x + x 4 cos x g(x) = x1.5 e2x ⇒ g′ (x) = 1.5x0.5 e2x + 2x1.5 e2x 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) 8 6. y = x (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)
Problem Set 4-2
51
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 ⇒ d2y/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 2 x + 1 = − cot 2 x + 1 = − csc 2 x − sin 2 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 differentiable functions of x, then y′ = u′vw + uv′w + uvw′. 52
Problem Set 4-2
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 yn = u1u2 u3 …un where u1…un are differentiable functions of x, then yn′ = u1′u2 u3 Kun + u1u2′ u3 Kun + u1u2 u3′ Kun + L + u1u2 u3 Kun′ K . 23. z = x 5 cos6 x sin 7x ⇒ z′ = 5x4 cos6 x sin 7x + x5 6 cos5 x (−sin x) ⋅ sin 7x + x 5 cos6 x ⋅ 7 cos 7x = 5x 4 cos6 x sin 7x − 6x 5 cos5 x sin x sin 7x + 7x5 cos6 x cos 7x 24. y = 4x 6 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 = x 5 e2x cos 2x sin 3x ⇒ u′ = 5x 4 e2x cos 2x sin 3x + x5 · 2e2x · cos 2x sin 3x + x 5 e2x(−2 sin 2x) sin 3x + x 5 e2x cos 2x · 3 cos 3x = 5x 4 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 π t 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 because v(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 t Graph confirms Figure 4-2d. Calculus Solutions Manual © 2005 Key Curriculum Press
b. v exceeds 25.
25
v
25
c. In 1940, wind-induced vibrations in the Tacoma Narrows Bridge increased in amplitude until the bridge collapsed. 29. Prove that the derivative of an odd function is an even function and that the derivative of an even function is an odd function.
By the derivative of a product property, fk′+1 ( x ) = ( x k )′( x ) + ( x k )( x )′ = ( x k )′( x ) + x k . Substituting for (xk)′ from the induction hypothesis, fk′+1 ( x ) = ( kx k −1 )( x ) + x k = kx k + x k = ( k + 1) x k = (k + 1)x ( k+ 1 ) − 1, completing the induction. ∴ fn′ ( x ) = nx 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(x 2 − x − 12)7(2x − 1) = 8(x + 3)7(x − 4)7(2x − 1), which checks. 33. a.
Proof: For any function, the chain rule gives d f ( − x ) = f ′( − x ) ⋅ ( −1) = − f ′( − x ). dx For an odd function, d d f ( − x ) = [– f ( x )] = − f ′( x ). dx dx ∴ −f ′ (−x) = −f ′ (x) or f ′ (−x) = f ′ (x), and the derivative is an even function. For an even function, d d f (− x ) = f ( x ) = f ′( x ). dx dx ∴ −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 2x g(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 the uniqueness 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 2x g(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 the uniqueness theorem, Q.E.D. 31. Prove that if fn(x) = xn, then fn′ ( x ) = nx n−1 for all integers ≥ 1. Proof (by induction on n): If n = 1, then f1(x) = x1 = x, which implies that f1′( x ) = 1 = 1x 0, which anchors the induction. Assume that for some integer n = k > 1, fk′ ( x ) = kx k −1 . For n = k + 1, fk+ 1(x) = xk+ 1 = (xk)(x). Calculus Solutions Manual © 2005 Key Curriculum Press
f (x) 5
f x 1
f´
b. f ′ (x) = 3x 2 sin x + x 3 cos x The graph in part a is correct. c. The numerical derivative graph duplicates the algebraic derivative graph, as in part a, thus showing that the algebraic derivative is right. 34. a. f (x ) 599,128 500,000
x –1.5
1.4 –500,000
1/9
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 = 0 or 9x − 1 = 0 ∴ x = 7/5 = 1.4, or x = −3/2 = −1.5, or x = 1/9 See 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 it levels off and then continues changing in the same direction, as at x = −1.5 in part a. 35. a. A = L W dA dL dW = ⋅W + L ⋅ dt dt dt
Problem Set 4-2
53
dW = −2 sin t dt dL L = 3 + 2 sin 2t ⇒ = 4 cos 2t dt dA = ( 4 cos 2t )(2 + 2 cos t ) dt + (3 + 2 sin 2t )( −2 sin t ) At t = 4, dA/dt = 7.132… , so A is increasing. At t = 5, dA/dt = −4.949… , so A is decreasing.
b. W = 2 + 2 cos t ⇒
Problem Set 4-3 Q1. Q2. Q3. Q4. Q5. Q6. Q7. Q8. Q9. Q10.
1066x1065 f (x) = 12x5 + C y′ = 3x2 sin x + x3 cos x dy/dx = −sin (x7) ⋅ 7x6 = −7x6 sin (x7) f ′ (x) = 0 (derivative of a constant) 54e9t See the text for the definition of derivative. Instantaneous rate of change at a given x (x − 3)4(x − 3 + 2x) = 3(x − 3)4(x − 1) 4
y
x
cos 12 x ⇒ sin 18 x –12 sin 12 x sin 18 x – 18 cos 12 x cos 18 x y′ = sin 2 18 x
6. y =
7. y = y′ =
2. f ( x ) =
3 x 2 sin x – x 3 cos x x3 ⇒ f ′( x ) = sin x sin 2 x 4 x 3 cos x + x 4 sin x x4 ⇒ f ′( x ) = cos x cos 2 x
cos3 x ⇒ ln x 3 cos 2 x ( − sin x ) ⋅ ln x − cos3 x ⋅ (1/ x ) g ′( x ) = (ln x )2 −3 ln x sin x cos 2 x − (cos3 x )/ x = (ln x )2
3. g( x ) =
sin 5 x ⇒ e3 x 5 sin 4 x cos x e 3 x − sin 5 x ⋅ 3e 3 x h ′( x ) = (e 3 x ) 2 4 5 sin x cos x − 3 sin 5 x = e3 x sin 10 x 5. y = ⇒ cos 20 x 10 cos 10 x cos 20 x + 20 sin 10 x sin 20 x y′ = 2 cos 20 x
4. h( x ) =
54
Problem Set 4-3
⇒ 6x + 5 3(6 x + 5) – (3 x − 7)(6) (6 x + 5)
2
=
57 (6 x + 5)
2
10 x + 9 ⇒ 5x − 3 –75 10 ⋅ (5 x − 3) − (10 x + 9) ⋅ 5 y′ = = (5 x − 3) 2 (5 x – 3)2
8. y =
(8 x + 1)6 dz ⇒ (5 x – 2 ) 9 dx 6(8 x + 1)5 (8) ⋅ (5 x − 2) 9 − (8 x + 1)6 ⋅ (9)(5 x − 2)8 (5) = (5 x − 2)18
9. z =
=−
(8 x + 1) 5 (120 x + 141) (5 x – 2)10
( 4 x – 1) 7 dA 4 ⇒ (7 x + 2) dx 7( 4 x − 1)6 ( 4) ⋅ (7 x + 2) 4 − ( 4 x − 1) 7 ⋅ 4( 7 x + 2)3 ( 7) = ( 7 x + 2 )8 6 28( 4 x – 1) ( 7 x + 2)3 [(7 x + 2) − ( 4 x − 1)] = ( 7 x + 2 )8
10. A =
4
1. f ( x ) =
3x − 7
=
28( 4 x – 1)6 ( 7 x + 2)3 (3 x + 3) ( 7 x + 2 )8
=
84( 4 x – 1)6 ( x + 1) ( 7 x + 2)5 3
11. Q =
3
3
3 x 2 e x sin x − e x cos x ex ⇒ Q′ = sin x sin 2 x
ln x 4 ⇒ cos x 4 x 3 (1/ x 4 )cos x − (ln x 4 )( − sin x ) r′ = cos 2 x ( 4 cos x )/ x + (ln x 4 ) sin x 4 cos x + x ln x 4 sin x = = x cos 2 x cos 2 x d 13. (60 x –4 / 3 ) = −80 x −7/3 dx 12. r =
14.
d (24 x –7/ 3 ) = −56 x −10/3 dx
Problems 15–22 and 25–26 can be done using either the power rule or the quotient rule. 12 –36 = 12 x −3 ⇒ r ′( x ) = −36 x −4 = 4 x3 x 51 –867 16. t ( x ) = 17 = 51x −17 ⇒ t ′( x ) = −867 x −18 = 18 x x 15. r ( x ) =
Calculus Solutions Manual © 2005 Key Curriculum Press
14 = 14(cos 0.5 x ) −1 ⇒ cos 0.5 x v′(x) = −14(cos 0.5x)–2(−sin 0.5x)(0.5) 7 sin 0.5 x = cos 2 0.5 x 20 a( x ) = = 20(sin x ) −2 ⇒ sin 2 x a′ (x) = −40(sin x) −3 (cos x) –40 cos x = sin 3 x 1 −1 r ( x ) = = x −1 ⇒ r ′( x ) = − x −2 = 2 x x 1 –2 s( x ) = 2 = x −2 ⇒ s′( x ) = −2 x −3 = 3 x x 10 3 5 W ( x) = 3 = 10( x − 1) ⇒ ( x – 1) –5 W′(x) = 150x2(x3 − 1)4 1 T ( x) = = (cos x sin x ) −1 ⇒ cos x sin x T ′( x ) = −(cos x sin x ) −2 ( − sin x sin x sin 2 x – cos 2 x + cos x cos x ) = , cos 2 x sin 2 x which transforms to – cos 2 x T ′( x ) = 1 2 = −4 csc 2 x cot 2 x 4 sin 2 x sin x T ( x) = ⇒ cos x (cos x )(cos x ) – (sin x )(– sin x ) T ′( x ) = cos 2 x
17. v( x ) =
18.
19. 20. 21.
22.
23.
cos 2 x + sin 2 x 1 = = sec 2 x cos 2 x cos 2 x (T is for tangent function.) cos x ⇒ 24. C( x ) = sin x (– sin x )(sin x ) – (cos x )(cos x ) C ′( x ) = sin 2 x =
– sin 2 x – cos 2 x –1 = = − csc 2 x 2 sin x sin 2 x (C is for cotangent function.) 1 0 – cos x ⇒ C ′( x ) = 25. C( x ) = sin x sin 2 x
1000 3–t 1000 v(1) = = 500 mi/h 3 –1 1000 v(2) = = 1000 mi/h 3–2 1000 1000 = v(3) = . No value for v(3). 3–3 0 b. v(t) = 100(3 − t)− 1 ⇒ 1000 a(t) = −1000(3 − t)− 2 − 1 = (3 – t ) 2 1000 a(1) = = 250 (mi/h)/h (3 – 1)2 1000 a(2) = = 1000 (mi/h)/h (3 – 2 ) 2 1000 1000 = a(3) = . No value for a(3). 0 (3 – 3)2
27. a. v(t ) =
c. a or v
a 2000
t 3
d.
28. a.
b.
=
c.
1 cos x =− ⋅ = − csc x cot x sin x sin x (C is for cosecant function.) 1 = (cos x ) −1 ⇒ 26. S( x ) = cos x S′(x) = −(cos x)− 2(−sin x) sin x = = sec x tan x cos 2 x (S is for secant function.)
Calculus Solutions Manual © 2005 Key Curriculum Press
v
d.
1000 = 500 ⇒ 2 = (3 − t )2 ⇒ (3 – t ) 2 3−t = ± 2 ⇒ t = 3± 2 ⇒ t = 3 − 2 = 1.585K in the domain. Range is 0 ≤ t < 1.585… . Because they are walking in the same direction, their relative rate is the difference (x − 5). 300 t( x ) = , assuming Willie’s rate is x–5 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 domain is x > 5. t(x) = 300(x – 5)− 1 −300 t′(x) = −300(x − 5)− 2 = ( x − 5)2 t′(6) = −300 s/(ft/s) t′(5) does not exist because of division by zero. More fundamentally, t′(5) does not exist because t(5) does not exist.
29. f ( x ) =
3x + 7 ⇒ 2x + 5
Problem Set 4-3
55
3 ⋅ (2 x + 5) – (3 x + 7) ⋅ 2 1 = (2 x + 5)2 (2 x + 5) 2 1 f ′( 4 ) = = 0.005917159K 169 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. f ′( x ) =
x2 – 8 ⇒ x–3 2 x ( x – 3) – ( x 2 – 8)(1) x 2 − 6 x + 8 f ′( x ) = = ( x – 3)2 ( x − 3)2
b. f ( x ) =
31. If y = xn, where n is a negative integer, then y′ = nx n− 1. Proof: Let n = −p, where p is a positive integer. 1 ∴ y = x−p = p x 0 ⋅ x p – 1 ⋅ px p–1 ∴ y′ = because p is a x2p positive integer. px p–1 = − 2 p = − px p−1−2 p = − px − p−1 . x Replacing −p with n gives y′ = nxn− 1, Q .E .D . 32. y
y y' y'
1
f (x )
y
1
x f
x
1
1
y
5
f' 3
x
33. Answers will vary. y2 and y3 both agree with the graph of f ′. c. x
f (x)
2.95
−14.05
2.96 2.97 2.98 2.99 3.00 3.01 3.02 3.03 3.04 3.05
−19.04 −27.363… −44.02 −94.01 undefined 106.01 56.02 39.363… 31.04 26.05
f ′(x) −399 −624 −1110.11… −2499 −9999 undefined −9999 −2499 −1110.11… −624 −399
f ( x) changes faster and faster as x approaches 3, shooting off to negative infinity as x approaches 3 from the negative side and to positive infinity as x approaches 3 from the positive side. Note that the rates are symmetrical about x = 3. d. There is a relative minimum at x = 4 and a relative maximum at x = 2. 2 2 − 6(2) + 8 =0 (2 − 3)2 4 2 − 6( 4) + 8 f ′( 4 ) = =0 ( 4 − 3)2 f ′( 2 ) =
56
Problem Set 4-4
Problem Set 4-4 Q1. Q2. Q3. Q4. Q5. Q6. Q7. Q8. Q9.
(sin x)/(tan x) = cos x 1/(sec x) = cos x sin2 3 + cos2 3 = 1 f ′ (x) = ex sin x + ex cos x cos x + x sin x g ′( x ) = cos 2 x h ′ (x) = −(15/7)(3x)–12/7 dy/dx = 3(cos x)− 4 sin x Limit = −3 (Function is secant.) y y y' 1
x π
Q10. 1. 2. 3. 4.
C f (x) = tan 5x ⇒ f ′(x) = 5 sec2 5x f (x) = sec 3x ⇒ f ′(x) = 3 sec x tan x y = sec x7 ⇒ y′ = 7x6 sec x7 tan x7 z = tan x9 ⇒ z′ = 9x8 sec2 (x9)
Calculus Solutions Manual © 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) ⇒ 1 r ′( x ) = (−csc x cot x) = −cot x csc x 8. p(x) = ln (cot x) ⇒ 1 1 p ′( x ) = ( − csc 2 x ) = cot x cos x sin x 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 tan x 15. y = = sec x ⇒ y′ = sec x tan x sin x cot x 16. y = = csc x ⇒ y′ = −csc x cot x cos x 5 ln 7 x 17. y = ⇒ cot 14 x 5 ⋅ 7 ( 71x )cot 14 x − 5 ln 7 x ( −14 csc 2 14 x ) y′ = cot 2 14 x
22. m(x) = cot2 x − csc2 x = −1 ⇒ m′(x) = 0 (The differentiation formulas give the same.) 23. A(x) = sin x 2 ⇒ A′(x) = cos x 2 · 2x = 2x cos x 2 24. f (x) = cos x3 ⇒ f ′ (x) = −sin x 3 · 3x 2 = −3x 2 sin x 3 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 cos x ⇒ sin x – sin x ⋅ sin x – cos x ⋅ cos x y′ = sin 2 x –1 = = − csc 2 x or: sin 2 x 1 y= = ( tan x ) −1 ⇒ tan x y′ = −1 ⋅ (tan x)− 2 ⋅ sec2 x = −csc2 x
29. y = cot x =
1 = (sin x ) –1 ⇒ sin x − cos x y′ = −(sin x ) −2 cos x = = − csc x cot x sin 2 x 31. a. See graph in part b. b. f (x) = tan x ⇒ f ′(x) = sec2 x. Predicted graph should be close to actual one. 30. y = csc x =
y
(5 cot 14 x )/ x + 70 ln 7 x csc 2 14 x = cot 2 14 x 70 ln 7 x 5 = + x cot 14 x cos 2 14 x 18. y =
f
1
x
1
4 csc 10 x ⇒ e 40x 4( −10 csc10 x cot 10 x )e − 4 csc10 x ( 40e 40x (e ) 2 40x
y′ =
f´
−40 csc 10 x cot 10 x − 160 csc 10 x e 40x 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.) =
Calculus Solutions Manual © 2005 Key Curriculum Press
40x
)
tan 1.01 – tan 0.99 = 3.42646416 K 2(0.01) 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. c.
y
y1
5
y2
x
1
Problem Set 4-4
57
The formula is confirmed by the fact that the line is tangent to the graph. c. y
f
f'
1
x π/2
cannot be continuous. (Some texts restrict the range of the inverse cosecant to 0 ≤ y ≤ π/2 so that the function will be continuous, but doing so throws away the other half of the possible values.) 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 ⇒
33. a. b.
c.
34. a.
b. c.
35. a. b.
If f ′ (x) is negative, the graph of f is decreasing. y/10 = tan x ⇒ y = 10 tan x, Q .E .D . y′ = 10 sec2 x. At x = 1, y′ = 10 sec2 1 = 34.2551… . y is increasing at about 34.3 ft/radian. π (34.2551K) = 0.5978… ft/degree 180 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. opposite side y tan x = = adjacent side 500 ∴ y = 500 tan x, Q.E.D. dy/dt = 500 sec2 x · dx/dt dx/dt = 0.3 rad/s At y = 300, x = tan− 1 (300/500) = 0.5404… ∴ dy/dt = 500 (sec2 0.5404…)(0.3) = 500(1.36)(0.3) = 204 ft/s y = sin x + C y = − 12 cos 2 x + C
c. y = 13 tan 3 x + C d. y = − 14 cot 4 x + C e. y = 5 sec x + C 36. Answers will vary.
Problem Set 4-5 sin′ x = cos x Q2. cos′ x = −sin x 2 tan′ x = sec x Q4. cot′ x = −csc2 x sec′ x = sec x tan x Q6. csc′ x = −csc x cot x f ′ (1) is infinite. Q8. f ′ (3) is undefined. f ′ (4) = −1 Q10. f ′ (6) = 0 See Figure 4-5d. 2. See Figure 4-5d. See Figure 4-5d. 4. See Figure 4-5d. The principal branch of the inverse cotangent function goes from zero to π so that the function will be continuous. 6. There are no values of the inverse secant for x between −1 and 1, so the inverse secant function
Q1. Q3. Q5. Q7. Q9. 1. 3. 5.
58
Problem Set 4-5
1 1 = , Q .E .D . cos y 1 – x2
y′ =
1
x
y
√1 – x 2
[Because sin y = (opposite leg)/(hypotenuse), put x on the opposite leg and 1 on the hypotenuse. Adjacent leg = 1 – x 2 , and cos y = (adjacent)/(hypotenuse).] 10. y = cos− 1 x ⇒ cos y = x ⇒ −sin y · y′ = 1 ⇒ 1 1 y′ = − =− , Q .E .D . sin y 1 – x2
√1 – x 2
1
y x
[Because cos y = (adjacent leg)/(hypotenuse), put x on the adjacent leg and 1 on the hypotenuse. Opposite leg = 1 – x 2 , and sin y = (opposite)/(hypotenuse).] 11. y = csc− 1 x ⇒ csc y = x ⇒ −csc y cot y · y′ ⇒ 1 1 y′ = − =− if x > 0 csc y cot y x x2 – 1 If x < 0, then y is in Quadrant IV (see Figure 4-5d). So both csc y and cot y are negative, and thus their product must be positive. 1 ∴ y′ = − , Q .E .D . | x | x2 – 1
x
1
y
√x 2 – 1
[Because csc y = (hypotenuse)/(opposite leg), put x on the hypotenuse and 1 on the opposite leg.
Calculus Solutions Manual © 2005 Key Curriculum Press
Adjacent leg = x 2 – 1, and csc y = x and cot y = (adjacent)/(opposite).] 12. y = cot− 1 x ⇒ cot y = x ⇒ −csc2 y · y′ = 1 ⇒ 1 1 1 , Q .E .D . y′ = − =− 2 =− csc 2 y 1+ x2 2 1+ x
(
)
16. y = tan− 1 (ln x) ⇒ tan y = ln x ⇒ sec2 y · y′ = l/x ⇒ 1 1 1 y′ = = 2 = ( ln 2 x ) x sec 2 y x 1 + x 1 + ln 2 x
(
)
√ 1 + ln2x ln x
√1 + x 2
y 1
1
y x
[Because cot y = (adjacent leg)/(opposite leg), put x on the adjacent leg and 1 on the opposite leg. Hypotenuse = 1 + x 2 , and csc y = (hypotenuse)/(opposite).] Problems 13−18 are shown done “from scratch,” as in Example 1. If students practice doing them this way, they will not be dependent on memorized formulas. Problem 13 shows how an alternate solution could be found using the formulas and the chain rule. 13. y = sin− 1 4x ⇒ sin y = 4x ⇒ cos y · y′ = 4 ⇒ 4 4 y′ = = cos y 1 – 16 x 2
17. y = sec −1
1 x x ⇒ sec y = ⇒ sec y tan y ⋅ y′ = ⇒ 3 3 3 1
y′ =
3 sec y tan y
1
= 3⋅ ( x 3
, if x > 0
2
x – 9) / 3
If x < 0, then y is in Quadrant II, where both sec y and tan y are negative. So their product is positive. 3 ∴ y′ = 2 |x| x –9 x
√x 2 – 9
y 3
1
4x
√1 – 16x 2
Alternate solution by application of the formula: 1 4 y = sin −1 4 x ⇒ y ′ = ⋅4= 2 1 – (4 x ) 1 – 16 x 2 14. y = cos− 1 10x ⇒ cos y = 10x ⇒ 10 10 −sin y · y′ = 10 ⇒ y′ = − =− sin y 1 – 100 x 2 1
x x ⇒ csc y = ⇒ 10 10 1 − csc y cot y ⋅ y ′ = ⇒ 10 1 1 y′ = − =− 10 csc y cot y x x 2 – 100 10 ⋅ 10 100 If x < 0, then y is in Quadrant IV, where both csc y and cot y are negative. So their product is positive. 10 ∴ y′ = − | x | x 2 – 100
18. y = csc −1
y
√1 – 100x 2
y 10x
15. y = cot −1 e 0.5x ⇒ cot y = e 0.5x ⇒ − csc 2 y ⋅ y ′ = 0.5e 0.5x ⇒ 0.5e 0.5x 0.5e 0.5x 0.5e 0.5x y′ = − =− 2 =− 2 csc y 1+ ex 1+ ex
(
√ 1 + ex 1
y e 0.5x
Calculus Solutions Manual © 2005 Key Curriculum Press
)
x
10
y
√x 2 – 100
For Problems 19−24, a solution is shown using the appropriate formula. 19. y = cos− 1 5x2 1 10 x y′ = − ⋅ 10 x = − 2 2 1 – (5 x ) 1 – 25 x 4 20. f (x) = tan− 1 x3 1 3x 2 2 f ′( x ) = ⋅ 3 x = 1 + ( x 3 )2 1+ x6 Problem Set 4-5
59
21. g(x) = (sin− 1 x)2
d. Maximum is between x = 38 and 39. 1
−1
g ′( x ) = 2 sin x ⋅
θ
1 – x2
0.5
22. u = (sec− 1 x)2 u ′ = 2 sec −1 x ⋅
23.
24.
25.
26.
1
| x | x2 – 1 v = x sin− 1 x + (1 − x2)1/ 2 1 1 + (1 – x 2 ) −1/2 ⋅ ( −2 x ) v ′ = 1 ⋅ sin −1 x + x ⋅ 2 2 1– x x x = sin −1 x + − = sin −1 x 2 2 1– x 1– x The surprise is that you now have seen a formula for the antiderivative of the inverse sine function. f (x) = cot− 1 (cot x) = x ⇒ f ′ (x) = 1 (Surprise!!) Application of the formulas gives the same result. a. tan θ = x/100, so θ = tan− 1 (x/100), Q.E.D. 1 1 100 dθ b. = ⋅ = dx 1 + ( x/100)2 100 10000 + x 2 100 dθ dθ dx dx = ⋅ = ⋅ dt dx dt 10000 + x 2 dt c. If x = 500 ft and dθ/dt = −0.04 rad/s, then 100 dx −0.04 = 2 ⋅ 10000 + 500 dt dx (–0.04)(260000) = = −104 dt 100 The truck is going 104 ft/s. 104(3600/5280) = 70.909… ≈ 71 mi/h 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 the advantage that the function appears on the calculator. The inverse cotangent equation has the advantage that x is in the numerator of the argument, which makes the chain rule less complicated to use.) b.
–50 x –2 –30 x –2 dθ = 2 − dx 1 + (50/ x ) 1 + (30/ x )2 –50 30 = 2 + 2 x + 2500 x + 900 –20 x 2 + 30000 = 2 ( x + 2500)( x 2 + 900)
c. dθ/dx = 0 ⇒ −20x2 + 30000 = 0 ⇒ 20x2 = 30000 x = ± 1500 = ±38.729K About 38.7 ft
60
Problem Set 4-5
x 100
40
27. x
Num. Deriv.*
Alg. Deriv.
−0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8
−1.666671… −1.250000… −1.091089… −1.020620… −1.000000… −1.020620… −1.091089… −1.250000… −1.666671…
−1.666666… −1.25 −1.091089… −1.020620… −1 −1.020620… −1.091089… −1.25 −1.666666…
*The precise value for the numerical derivative will depend on the tolerance to which the grapher is set. The values given by numerical derivative and the formula are very close. dy 1 = dx | x | x 2 – 1 dy 1 At x = 2, = = 0.288675K . dx | 2 | 3 The answer is reasonable because the graph slopes up at x = 2, with slope significantly less than 1. b. At x = 2, y = sec− 1 2 = cos− 1 (1/2) = 1.04719… . d (sec y) = sec y tan y dy At y = 1.047… , d (sec y) = (sec 1.047K)(tan 1.047K) = dy 3.464101… . c. The answer to part b is the reciprocal of the 1 answer to part a. That is, 3.464101 K = 0.288675… . Thus, the derivative of the inverse secant at x = c is the reciprocal of the derivative of the secant at y = sec− 1 c. 29. a. y = sin− 1 x ⇒ sin y = x ⇒ cos y · y′ = 1 ⇒ 1 y′ = , Q .E .D . cos y
28. a. y = sec −1 x ⇒
Calculus Solutions Manual © 2005 Key Curriculum Press
1 1 = = 1.25 cos(sin –1 x ) cos(sin –1 0.6) 1 1 1 y′ = = = = 1.25, Q .E.D . 2 2 0.8 1– x 1 – 0.6 d c. y = f −1 ( x ) ⇒ f ( y) = x ⇒ f ′( y) ⋅ ( y) = 1 ⇒ dx d 1 d 1 ⇒ [ f –1 ( x )] = ( y) = , dx f ′( y ) dx f ′[ f −1 ( x )] 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) = 3x 2 + 1, 1 1 1 / . h ′(10) = = = = 113 f ′[h(10)] f ′(2) 3 ⋅ 2 2 + 1 b. y ′ =
30. The inverse trig cofunctions, cos− 1, cot− 1, and csc− 1, are the ones whose derivatives are preceded by a minus sign.
Problem Set 4-6
14. a. f (x) 4
x –2
b. f ( x) = x 2 15. a. f (x )
x 6
( x − 6)( x + 1) x−6
b. f ( x ) = 16. a.
f (x )
Q1. 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 Q7. Q8. Q9. Q10. 1. 3. 5. 7. 9. 11.
Q6. 1 | x | x 2 − 1
)
f ′ (x) = 4x 3; f ″ (x) = 12x2; f ″ (2) = 48 dy/dx = 15x2(x3 + 1)4 Integral ≈ 5.4 (Function is y = 2− x.) E Continuous 2. Neither Neither 4. Both Neither 6. Neither Both 8. Neither Neither 10. Neither Continuous 12. Both
For Problems 13−20, sample answers are given. Equations do not necessarily correspond to the graphs shown. 13. a. f (x )
2
x 1
x 2 ( x − 1) , if x ≠ 1 b. f ( x ) = x − 1 5, if x = 1 17. a. f (x )
x —5
x, if x ≤ −5 b. f ( x ) = 3 x, if x > −5 18. a.
5
f (x )
3
x 3
x
b. f ( x) = x + 2 Calculus Solutions Manual © 2005 Key Curriculum Press
–1
Problem Set 4-6
61
b = 1 − a = 1 − (−1.5) ⇒ b = 2.5
b. f ( x) = (x + 1) 2/3 + 3 19. a.
f (x )
f (x ) 7
1
x 1
x 4
f is differentiable at x = 1. –( x – 3)2 + 7, if x ≥ 2 26. f ( x ) = 3 if x < 2 ax + b, For f to be continuous at x = 2, lim− ( ax 3 + b) = lim+ [ −( x – 3) 2 + 7] ⇒
x 2 − 9, if x < 4 b. f ( x ) = 11 − x, if x ≥ 4 20. a. No such function
x →2
x →2
a ⋅ 23 + b = 6 ⇒ 8a + b = 6 ⇒ b = 6 − 8a For f to be differentiable at x = 2, lim− 3ax 2 = lim+ [ −2( x – 3)] ⇒ 3a ⋅ 2 2 = 2 ⇒
f(x )
Not possible. Differentiability implies continuity.
x →2
x →2
a = 1/6 b = 6 − 8(1/6) ⇒ b = 14/3
x
f (x ) 6
b. No such function 21. Continuous 22. Both f (x )
x
f (x )
2
f is differentiable at x = 2.
4
x 3
x 2
23. Both
x →2
24. Neither
x →2
4 – 12 + b ⇒ b = 4a + 18 For f to be differentiable at x = 2, lim− 2 ax = lim+ (2 x – 6) ⇒ 2 a ⋅ 2 = 2 ⋅ 2 – 6 ⇒
f (x )
y
ax 2 + 10, if x < 2 27. f ( x ) = 2 x − 6 x + b, if x ≥ 2 For f to be continuous at x = 2, lim− ( ax 2 + 10) = lim+ ( x 2 − 6 x + 6) ⇒ 4 a + 10 =
x →2 x
x π/2
1
x →2
a = −0.5 b = 4(–0.5) + 18 ⇒ b = 16 f (x )
10
x , if x < 1 25. f ( x ) = 2 a( x – 2) + b, if x ≥ 1 For f to be continuous at x = 1, a( x − 2 ) 2 + b ] ⇒ lim− x 3 = lim[ + 3
x →1
x →1
1 = a(1 − 2) + b ⇒ a + b = 1 ⇒ b = 1 − a For f to be differentiable at x = 1, lim− 3 x 2 = lim+ 2 a( x − 2) ⇒ 3 = 2 a(1 − 2) ⇒ 2
x →1
x →1
x 2
f is differentiable at x = 2. if x ≤ 1 a/ x, 28. f ( x ) = 2 12 − bx , if x > 1 For f to be continuous at x = 1,
a = −1.5 62
Problem Set 4-6
Calculus Solutions Manual © 2005 Key Curriculum Press
lim a/ x = lim+ (12 – bx 2 ) ⇒ a/1 = 12 – b ⋅ 12 ⇒
x →1−
f (x )
x →1
a + b = 12 For f to be differentiable at x = 1, lim− − ax −2 = lim+ −2 bx ⇒ − a ⋅ 1−2 = −2 b ⋅ 1 ⇒
x →1
0.5
x →1
a = 2b ∴ 2b + b = 12 ⇒ b = 4 a=2·4⇒a=8
3
f is differentiable at x = 2π/3. ax 3 + bx 2 + cx + d , if 0 ≤ x ≤ 0.5 31. a. y = if x > 0.5 x + k,
f (x )
10
x 1
f is differentiable at x = 1. e ax , if x ≤ 1 29. f ( x ) = b + ln x, if x > 1 For f to be continuous at x = 1, lim− e ax = lim+ (b + ln x ) ⇒ e a = b x →1
x →1
For f to be differentiable at x = 1, lim− ae ax = lim+ (1/ x ) ⇒ ae a = 1
x →1
x →1
Solve by grapher: a = 0.5671… and b = 1.7632…
lim ( − 43 x 3 + 2 x 2 ) = lim + ( x + k ) ⇒
x →0.5−
x →0.5
32. Equation of the linear part of the fork is y – 20 = 5(x – 10) ⇒ y = 5x – 30
2
x 1
f is differentiable at x = 1. a sin x, if x < 2π /3 30. f ( x ) = bx if x ≥ 2π /3 e , For f to be continuous at x = 2π/3, lim − a sin x = lim + e bx ⇒ x →( 2π / 3)
a 3 2e 2πb / 3 = e 2πb / 3 ⇒ a = 2 3 For f to be differentiable at x = 2π/3, lim − a cos x = lim + be bx ⇒ x→( 2π /3)
For y to contain the origin, a ⋅ 03 + b ⋅ 02 + c ⋅ 0 + d = 0 ⇒ d = 0 For y′ = 0 at x = 0, y′ = 3ax 2 + 2bx + c ⇒ 0 = 3a ⋅ 02 + 2b ⋅ 0 + c ⇒ c = 0 For 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 – 0.75a For y ′′ = 0 at x = 0.5, y ′′ = 6 ax + 2 b ⇒ 0 = 3a + 2b Solve 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, − 43 (0.5)3 + 2(0.5)2 = 0.5 + k ⇒ k = – 16 = −0.1666 K
f (x )
x →( 2π / 3)
x
x →( 2π /3)
−a = be 2πb / 3 ⇒ a = –2 be 2πb/3 2 2 −1 2e 2πb/3 So = –2 be 2πb/3 ⇒ = –2 b ⇒ b = 3 3 3 2 −2π /( 3 3 ) = –0.5773… and a = e = 0.3446 K 3
Calculus Solutions Manual © 2005 Key Curriculum Press
ax 3 + bx, if x < 10 ∴y = 5 x − 30, if x ≥ 10 For y to be continuous at x = 10, a ⋅ 10 3 + b ⋅ 10 = 5 ⋅ 10 – 30 1000a + 10b = 20 ⇒ b = 2 – 100a For y to be differentiable at x = 10, 3a ⋅ 10 2 + b = 5 300a + (2 – 100a) = 5 200a = 3 ⇒ a = 3/200 b = 2 – 100(3/200) ⇒ b = 0.5 2 x−2 , if x ≠ 2 x − 33. f ( x ) = x−2 4, if x = 2 Simplifying the equation for f (x) gives x 2 + 1, if x < 2 f ( x ) = x 2 − 1, if x > 2 4, if x = 2
Problem Set 4-6
63
Taking the derivative for each branch gives if x < 2 2 x, f ′( x ) = 2 x, if x > 2 undefined, if x = 2 Taking the left and right limits gives lim− f ′( x ) = 2 ⋅ 2 = 4; lim+ f ′( x ) = 2 ⋅ 2 = 4. x →2
x →2
Using the definition of derivative, taking the x2 +1− 4 1 → , limit from the left, f ′( x ) = lim− x →2 x−2 0 which is infinite. The same thing happens from the right. As the following graph shows, the secant lines become vertical as x approaches 2 from either side. f (x )
Secant slope becomes infinite.
4
x 2
Thus, f is not differentiable at x = 2, even though the right and left limits of f ′ (x) are equal to each other. The function must be continuous if it is to have a chance of being differentiable. 0.5 − t 60.5 0.5 + t , if t ≤ 0.5 34. a. d (t ) = 150 2 − 1 , if t ≥ 0.5 t −2 −60.5(0.5 + t ) , if t < 0.5 d ′(t ) = 150t −2, if t > 0.5 The inequality signs must be < and > because although the function is defined at x = 0.5, the derivative is not. b. d ′(1) = 150(1) –2 = 150 ⇒ d is continuous at x = 1 because it is differentiable there. c. lim − d ′(t ) = −60.5(0.5 + 0.5) −2 = −60.5 x →0.5
lim d ′(t ) = 150(0.5) −2 = 600
e. A regulation baseball diamond has the pitcher’s mound 60.5 feet from home plate. Substituting zero for t gives d(0) = 60.5, confirming that the pitcher was on the mound at that time. 35. a. y = mx + b ⇒ y′ = m, which is independent of x. ∴ linear functions are differentiable for all x. ∴ linear functions are continuous for all x. b. y = ax 2 + bx + c ⇒ y′ = 2ax + b, which exists for all x by the closure axioms. ∴ quadratic functions are differentiable for all x. ∴ quadratic functions are continuous for all x. c. y = 1/x = x–1 ⇒ y′ = –x–2, which exists for all x ≠ 0 by closure and multiplicative inverse axioms. ∴ the reciprocal function is differentiable for all x ≠ 0. ∴ the reciprocal function is continuous for all x ≠ 0. d. y = x ⇒ y′ = 1, which is independent of x. ∴ the identity function is differentiable for all x. ∴ the identity function is continuous for all x. e. y = k ⇒ y′ = 0, which is independent of x. ∴ constant functions are differentiable for all x. ∴ constant functions are continuous for all x. 36. See text proof.
Problem Set 4-7 Q1. 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 Q7. θ ′ = −1/ 1 − x 2
Q6. y′ = 0
Q8. v(t) is decreasing at t = 5. Q9. Q10. E
x →0.5+
As the ball was about to be hit, it was approaching the plate at 60.5 ft/s. Just after the ball was hit, it was going away from the plate at 600 ft/s. d. Function d is not differentiable at t = 0.5 because d′ (t) approaches different limits from both sides of x = 0.5. Function d is continuous at t = 0.5 because you get zero as the limit of d(t) as t approaches zero from either left or right.
64
Problem Set 4-7
y' 1
x π/2
y
1. x = t4 , y = sin 3t dy dy/dt 3 cos 3t d 2 y d 3 cos 3t = = ⇒ = dx dx/dt dx 2 dx 4t 3 4t 3
Calculus Solutions Manual © 2005 Key Curriculum Press
=
( −9 sin 3t )( dt/dx ) ⋅ 4t 3 − 3 cos 3t (12t 2 )( dt/dx ) ( 4t 3 ) 2
=
−36t 3 sin 3t − 36t 2 cos 3t dx ÷ 16t 6 dt
4. a. x = t 2 , y = t 3 t x –3 –2 –1 0 1 2 3
−36t 3 sin 3t − 36t 2 cos 3t 64t 9 −9t sin 3t − 9 cos 3t = 16t 7 =
2. x = 6 ln t, y = t 3 2
2
–27 –8 –1 0 1 8 27
y
b.
dy dy/dt 3t d y = = = 0.5t 3 ⇒ 2 dx dx/dt 6/t dx d dx (0.5t 3 ) = 1.5t 2 ( dt/dx ) = 1.5t 2 ÷ = dx dt
9 4 1 0 1 4 9
y
5
x
2
1.5t = = 0.25t 3 6/t
5
3. a. x = 2 + t, y = 3 – t 2 t –3 –2 –1 0 1 2 3
x
y
–1 0 1 2 3 4 5
–6 –1 2 3 2 –1 –6
c.
d.
b. y
e.
2
x 3
5. a. b. dy dy/dt −2t = = = −2t dx dx/dt 1 If t = 1, dy/dx = –2 and (x, y) = (3, 2). Line through (3, 2) with slope –2 is tangent to the graph. See part b. d. x = 2 + t ⇒ t = x – 2 ⇒ y = 3 – (x – 2)2 This is the Cartesian equation of a parabola because only one of the variables is squared. e. By direct differentiation, dy/dx = –2( x – 2). At (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. c.
Calculus Solutions Manual © 2005 Key Curriculum Press
c.
dy dy/dt 3t 2 = = = 1.5t dx dx/dt 2t If t = 1, dy/dx = 1.5 and (x, y) = (1, 1). Line through (1, 1) with slope 1.5 is tangent to the graph. See graph in part b. x = t 2 ⇒ t = x 1/ 2 ⇒ y = ( x 1/ 2 )3 ⇒ y = x 1.5 The name semicubical is picked because 1.5 is half of 3, the exponent for a cubic function. The name parabola is used because the equation looks similar to y = x2 for a parabola. 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, which agrees with part c. The graph confirms the figure in the text. dy 5 cos t −5 = = cot t dx −3 sin t 3 If t = π /4, x = 3 2 /2 and y = 5 2 /2. (x, y) = (2.121… , 3.535…) dy −5 π = cot = −5/3 dx 3 4 y 5
x –3
3
–5
The line is tangent to the graph. Problem Set 4-7
65
d. False. The line from (0, 0) to (2.1… , 3.5…) does not make an angle of 45° with the x-axis. (This shows that the t in parametric functions is not the same as the θ in polar coordinates.) 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 t y/5 = sin t ⇒ (y/5)2 = sin2 t Adding left and right sides of the equations gives (x/3)2 + (y/5)2 = cos2 t + sin2 t. ∴ (x/3)2 + (y/5)2 = 1, which is a standard form of the equation of an ellipse centered at the origin, with x-radius 3 and y-radius 5. 6. a. The graph confirms the figure in the text. dy 24 sin 2 t cos t sin t b. = =− = – tan t 2 dx 24 cos t ( − sin t ) cos t ∴ 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… . y 8
x –8
8
–8
The line is tangent to the graph. d. dx/dt = –24 cos2 t sin t dy/dt = 24 sin2 t cos t The 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 ( dy/dx ) = lim (– tan t ) = – tan 0 = 0 t→0
t→0
So the graph becomes horizontal at (8, 0). t = π/2 gives the cusp at (0, 8). lim ( dy/dx ) = lim ( − tan t ), which is infinite. t →π /2
t →π /2
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 ⇒ x 2/3 + y 2/3 = 4 66
Problem Set 4-7
7. a. x = 6 + 5 cos t, y = 3 + 5 sin t y dy /dx is infinite here.
3
x 6
dy 5 cos t = ⇒ dy/dx = − cot t dx −5 sin t
b.
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 must be zero. This happens where t = π /2 ± nπ , so dy/dx = 5 cos t = 0 at those points. See graph in part a. x−6 y−3 d. = cos t and = sin t 5 5 2 2 x − 6 + y − 3 = cos 2 t + sin 2 t 5 5 x − 6 + y − 3 = 1 5 5 This is an equation of a circle centered at (6, 3) with radius 5. e. The 6 and 3 added in the original equations are the x- and y-coordinates of the center. The coefficients, 5, for cosine and sine in the original equations are the x- and y-radii, respectively. Because the x- and y-radii are equal, the graph is a circle. 8. x = cos2 t, y = sin2 t dy 2 cos t ( − sin t ) = = –1 (cos t ≠ 0, sin t ≠ 0) dx 2 sin t cos t 2
2
y 1
x 1
The graph is a line segment with a slope of –1. x + y = cos2 t + sin2 t ⇒ x + y = 1 This is the equation of a line with slope –1, confirming what was observed on the graph. The parametric equations restrict the ranges of x and y to the first quadrant, no matter what is the domain of t. This is true because cos2 t and sin2 t are never negative. The Cartesian equation allows –∞ < x < ∞ and –∞ < y < ∞. Calculus Solutions Manual © 2005 Key Curriculum Press
9. a. The grapher confirms the figure in the text. b.
dy 2 cos t − 2 cos 2t cos t − cos 2t = = dx −2 sin t − 2 sin 2t − sin t − sin 2t
c. Cusps occur where both dx/dt and dy/dt = 0. A graphical solution shows that this occurs at t = 0, t = 2π /3, t = 4π /3, t = 2π , … . (A cusp could also happen if dx/dt = 0 and dy/dt ≠ 0, but for this figure there is no such place.) dx /dt or dy /dt dx /dt t __ 2 π 3
0
4π __ 3
2π
dy /dt
At t = 0, 2π, … , the tangent appears to be horizontal. At t = 2π /3, 4π /3, 8π /3, 10π /3, … , there appears to be a tangent line but not a horizontal one. A numerical solution shows the following values as t approaches 2π/3: t
dy/dx
2π /3 – 0.1 2π/3 – 0.01 2π/3 – 0.001 2π/3 2π/3 + 0.001 2π /3 + 0.01 2 π /3 + 0.1
–1.547849… –1.712222… –1.730052… indeterminate –1.734052… –1.752225… –1.951213…
dy/dx seems to be approaching about –1.732 as t approaches 2π/3. [The exact answer is − 3, which students will be able to prove easily with l’Hospital’s rule after they have studied Section 6-5. Joan Gell and Cavan Fang have shown clever trigonometric transformations that “remove” the removable discontinuity and lead to the same answer. These are 1. Use the sum and product properties on dy/dx: dy 2 sin 1.5t sin 0.5t = dx −2 sin 1.5t cos 0.5t = − tan 0.5t if dx/dt ≠ 0 As t → 2π /3, dy/dx → – tan(π /3) = – 3 .
Calculus Solutions Manual © 2005 Key Curriculum Press
2. Use the double argument properties on dy/dx: dy cos t – (2 cos 2 t – 1) = dx –(sin t + 2 sin t cos t ) (1 – cos t )(1 + 2 cos t ) 1 – cos t = = , –(sin t )(1 + 2 cos t ) – sin t which approaches − 3 as t → 2π/3.] 10. a. The grapher confirms the figure in the text. dy 4 a cos t (– sin t ) b. = = –2 cos3 t sin t dx 2 a sec 2 t (The answer is independent of a.) 4 a 2 sin 2 t 4 a 2 (1 – cos 2 t ) = cos 2 t cos 2 t 2 2 y = 2a cos t ⇒ cos t = y/(2a) 4 a 2 [1 – y/(2 a)] 4 a 2 (2 a − y) ∴ x2 = = y/(2 a) y x2y = 8a3 – 4a2y ⇒ (x2 + 4a2)y = 8a3 ⇒ 8a 3 y= 2 x + 4a 2 216 a = 3⇒ y = 2 x + 36 d. y = 8a 3 ( x 2 + 4 a 2 ) −1 ⇒ –16 a 3 x dy = −8a 3 ( x 2 + 4 a 2 ) −2 ⋅ 2 x = 2 ( x + 4a 2 )2 dx
c. x 2 = 4 a 2 tan 2 t =
e. At t = π/4, x = 2a tan (π/4) = 2a. From part d, dy –16 a 3 (2 a) –32 a 4 = = –1/2 2 2 2 = dx [(2 a) + 4 a ] 64 a 4 From part b, dy = –2 cos3 (π /4) sin (π /4) dx = –2( 2 /2)3 ( 2 /2) = –1/2, which agrees. At t = π/4, x = 2a tan (π/4) = 2a = 6 and y = 2a cos2 (π/4) = 2a(1/2) = a = 3. A line through (6, 3) with slope –1/2 is tangent to the graph at that point. 10
y
5
t = π/4 x
–15
–10
–5
0
5
10
15
–5
11. a. x = cos t + t sin t y = sin t – t cos t The grapher confirms the figure in the text. [Note: In the derivation of these equations from the geometric definition of involute,
Problem Set 4-7
67
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 the string. Because the circle is a unit circle, the length of the string is also t, the central angle in radians. The string makes an angle of (t – π/2) with the positive x-axis so that (t cos (t – π/2), t sin (t – π/2)) is a vector representing the unwound string. The cofunction properties and odd-even properties from trig are used to simplify the equations so that the calculus will be easier.] b.
dy cos t – [cos t + t (– sin t )] = dx – sin t + (sin t + t cos t ) t sin t = = tan t t cos t
c. At t = π, dy/dt = tan π = 0. The string will be pointing straight up from the x-axis. The diagram shows that the tangent to the graph is horizontal at this point. y
( x, y)
String 1
t=π
x 1
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 Bt The period is 60 seconds. So B = 2π /60 = π /30 π ∴ x = 25 + 15 sin t 30 π y = 20 + 15 cos t 30 π π t b. dx/dt = cos 2 30 π π dy/dt = – sin t 2 30 At t = 5, π π dx/dt = cos = π 3 /4 = 1.3603K 2 6 π π dy/dt = – sin = –π /4 = –0.7853K 2 6 c. The slope of the circular path is dy/dx. At t = 5, dy – π /4 = = –1 / 3 = –0.5773K dx π 3 /4 68
Problem Set 4-7
x – 25 π x – 25 π = sin = sin 2 t⇒ t 15 15 30 30 2 y – 20 π y – 20 π = cos = cos 2 t⇒ t 15 15 30 30 π π Because sin 2 t + cos 2 = 1, 30 30 2 2 x – 25 + y – 20 = 1. 15 15 This is an equation of a circle centered at (25, 20) with radius 15, confirming that the path really is a circle. 13. The actual solutions will vary depending on the period of the pendulum, as determined by the length of the string. The following solution supposes that the period turns out to be 3.1 seconds. 2π 2π x = 30 cos t y = 20 sin t 3.1 3.1 2π t ( 40π /3.1) cos dy 3.1 = – 2 cot 2π t = dx –(60π /3.1) sin 2π t 3 3.1 3.1 At t = 5, x ≈ –22.8, y ≈ –13.0, and dy/dx ≈ –0.78. If the measurements have been accurate, the pendulum will be above the coin when t = 5. 14. The graph looks like an ellipse that moves in the x-direction as t increases. Because y starts at a high point and varies between 5 and 1, the ellipse has center at y = 3 and y-radius 2. Thus, an equation for y would be y = 3 + 2 cos t. x starts at 0 and increases. If the ellipse had x-radius 0.5, an equation for x would be x = 0.5 sin t. The graph of this ellipse is 2
d.
y 5
x 10
The graph seems to move over 1 unit to the right each cycle. Thus, if t increases by 2π, x increases by 1. The equations are thus x = t/(2π ) + 0.5 sin t, y = 3 + 2 cos t The graph here duplicates the one in the text. y 5
x 10
Calculus Solutions Manual © 2005 Key Curriculum Press
d. n = 1. (x = cos t, y = sin t)
To locate “interesting” features, dy dy/dt –2 sin t . = = dx dx/dt 1/(2π ) + 0.5 cos t
1 y
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/π Solving numerically for t gives t = 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, then 1/(2π )t + 0.5 sin t = 0.5. Solving numerically for the value of t closest to 0, 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)
x 1
n = 2. (x = cos 2t, y = sin t) 1 y
x 1
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 in Massachusetts.
1 y
x 1
Problem Set 4-8 If n is an even number, the graph comes to endpoints and retraces its path, making two complete cycles as t goes from 0 to 2π. If n is an odd number, the graph does not come to endpoints. It makes one complete cycle as t goes from 0 to 2π. c. i. (x = cos 5t, y = sin t)
Q1. Q3. Q5. Q7. Q9.
y′ = 2001x2000 5 product x3 + C
y′ = ln (2001)2001x f ′(u) = –csc2 u 1/(1 + 9x2) Instantaneous rate –2.4033… ft/s
4 y
y'
1 y
Q2. Q4. Q6. Q8. Q10.
x 6
x 1
1. x 3 + 7y 4 = 13 ⇒ 3x 2 + 28y 3y′ = 0 ⇒ 3x 2 y′ = – 28 y 3
ii. (x = cos 6t, y = sin t)
2. 3x 5 − y 4 = 22 ⇒ 15x 4 − 4y 3y′ = 0 ⇒ y′ =
1 y
3. x ln y = 104 ⇒ 1 · ln y + x ⋅ x 1
Calculus Solutions Manual © 2005 Key Curriculum Press
15 x 4 4 y3
1 ⋅ y′ = 0 ⇒ y
− y ln y x ex 4. y = 213 ⇒ p xy′ + yex = 0 ⇒ y′ = –y y′ =
Problem Set 4-8
69
5. x + xy + y = sin 2x ⇒ 1 + y + xy′ + y′ = 2 cos 2x ⇒ y′(x + 1) = 2 cos 2x – 1 – y ⇒ 2 cos 2 x – 1 – y y′ = x +1 6. cos xy = x – 2y ⇒ (–sin xy) ( y + xy′) = 1 – 2y′ ⇒ y′(–x sin xy + 2) = 1 + y sin xy ⇒ 1 + y sin xy y′ = 2 – x sin xy 7. x0.5 – y0.5 = 13 ⇒ 0.5x–0.5 – 0.5y–0.5 y′ = 0 ⇒ y′ = y 0.5 /x0.5 8. x1.2 + y 1.2 = 64 ⇒ 1.2x0.2 + 1.2y0.2 y′ = 0 ⇒ y′ = –x 0.2 /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 − ye xy ⇒ y′(xexy – sec2 y) = –yexy ⇒ y′ = xe xy − sec 2 y 10. ln (xy) = tan x ⇒ tan (ln xy) = x ⇒ –1
1 2 sec (ln xy) · (1 · y + y′x) = 1 ⇒ y + y′x = xy 2 xy cos (ln xy) y′x = xy cos2 (ln xy) – y ⇒ xy cos 2 (ln xy) − y x 3 4 5 11. (x y ) = x – y ⇒ 5(x 3y 4)4(3x 2y 4 + x 3 · 4y 3y′) = 1 – y′ ⇒ y′(20x15y19 + 1) = 1 – 15x14y20 ⇒ 1 – 15 x 14 y 20 y′ = 1 + 20 x 15 y19 y′ =
12. (xy)6 = x + y ⇒ 6(xy)5(y + xy′) = 1 + y′ ⇒ y′(6x6y5 – 1) = 1 – 6x5y6 ⇒ 1 – 6 x 5 y6 y′ = 6 5 6x y – 1 13. cos 2 x + sin2 y = 1 ⇒ 2 cos x · (–sin x) + 2 sin y · cos y · y′ = 0 ⇒ cos x sin x y′ = cos y sin y 14. sec2 y – tan2 x = 1 ⇒ 2 sec y · sec y tan y · y′ – 2 tan x · sec2 x = 0 ⇒ sec 2 x tan x y′ = sec 2 y tan y 15. tan xy = xy ⇒ (sec 2 xy) · (y + xy′) = y + xy′ ⇒ y′(x sec2 xy – x) = y – y sec2 xy ⇒ y(1 – sec 2 xy) y y′ = ⇒ y′ = – 2 x (sec xy – 1) x
y′(–x – x sin xy) = y + y sin xy ⇒ y(1 + sin xy) y y′ = ⇒ y′ = – x (–1 – sin xy) x 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 ⇒ 1 1 =– y′ = – sin y 1 – x2 √1 – x 2
1
y x
1 1 = ey x 23. y = x 11/5 ⇒ y 5 = x 11 ⇒ 5y 4 · y′ = 11x 10 ⇒ 11x 10 11x 10 11x 10 11 6 / 5 y′ = = = = x , 5y 4 5( x 11/ 5 ) 4 5 x 44 / 5 5 22. y = ln x ⇒ e y = x ⇒ e y · y′ = 1 ⇒ y′ =
which is the answer obtained using the derivative of a power formula, Q.E.D. 24. Prove that if y = xn, where n = a/b and a and b are integers, then y′ = nan− 1. Proof: y = x n = x a/ b ⇒ y b = x a. Because a and b are integers, byb− 1 y′ = ax a− 1 ax a –1 ax a –1 ax a –1 a a−1−( a− a/b ) y′ = b –1 = = x a/b b –1 a – a/b = by b( x ) bx b a = x a/b –1 = nx n –1, Q .E.D . b 25. a. x 2 + y 2 = 100 At (–6, 8), (–6)2 + 82 = 100, which shows that (–6, 8) is on the graph, Q.E.D. b. x 2 + y 2 = 100 ⇒ 2x + 2y · dy/dx = 0 ⇒ dy/dx = –x/y At (–6, 8), dy/dx = –(–6)/8 = 0.75. A line at (–6, 8) with slope 0.75 is tangent to the graph, showing that the answer is reasonable. 10 y
x 10
16. cos xy = xy ⇒ (–sin xy) · (y + xy′) = y + xy′ ⇒
70
Problem Set 4-8
Calculus Solutions Manual © 2005 Key Curriculum Press
c. x = 10 cos t y = 10 sin t dy 10 cos t cos t = =– dx –10 sin t sin t –1 At x = –6, t = cos (–0.6). sin [cos–1 (–0.6)] = 0.8 dy –0.6 ∴ =– = 0.75, dx 0.8 which agrees with part b, Q.E.D. 26. a. x 2 – y 2 = 36 At (10, –8), 102 – (–8)2 = 36, which shows that (10, –8) is on the graph, Q.E.D. b. x2 – y2 = 36 ⇒ 2x – 2y · dy/dx = 0 ⇒ dy/dx = x/y At (10, –8), dy/dx = 10/(–8) = –1.25. A line at (10, –8) with slope –1.25 is tangent to the graph, showing that the answer is reasonable. 10 y
x 10
c. x = 6 sec t y = 6 tan t dy 6 sec t tan t tan t = = dx 6 sec 2 t sec t –1 At x = 10, t = ±sec (10/6). tan [±sec–1 (10/6)] = ±8/6. Choose the negative value because y < 0. dy –10/6 ∴ = = –1.25, dx 8/6 which agrees with part b, Q.E.D. 27. a. x 3 + y 3 = 64 ⇒ 3x 2 + 3y 2 · dy/dx = 0 ⇒ dy/dx = –x2/y2 x = 0: y 3 = 64 ⇒ y = 4 ∴ dy/dx = –0/16 = 0 The tangent is horizontal (see the next graph). x = 2: 8 + y 3 = 64 ⇒ y 3 = 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 + y 3 = 64 ⇒ y = 0 ∴ dy/dx = –42/0, which is infinite. The tangent line is vertical. 10 y
x 10
Calculus Solutions Manual © 2005 Key Curriculum Press
b. y = x: x 3 + x 3 = 64 ⇒ x 3 = 32 ⇒ x = 3.1748… dy/dx = –x2/y2 = –x2/x2 = –1 c. y = (64 – x 3)1/3 As x becomes infinite, (64 – x3)1/3 gets closer to (–x3)1/3, which equals –x. The graph has a diagonal asymptote at y = –x, and dy/dx → –1. d. By analogy with the equation of a circle, such as x 2 + y 2 = 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 = 1200 4 x − 72x 2 + 1296 + 2x 2y 2 + 72y 2 + y 4 = 1200 x 4 − 72x 2 + 2x 2y 2 + 72y 2 + y 4 = −96 Differentiate the simplified equation implicitly. 4x 3 − 144x + 4xy 2 + 4x 2y · dy/dx + 144y · dy/dx + 4y3 · dy/dx = 0 2 (4x y + 144y + 4y3) · dy/dx = −4x3 + 144x − 4xy 2 dy – x 3 + 36 x – xy 2 = 2 dx x y + 36 y + y 3 At x = 8: (4 + y2)(196 + y2) = 1200 784 + 200y2 + y4 = 1200 y 4 + 200y 2 − 416 = 0 –200 ± 41664 y2 = = 2.058806 K or 2 −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 steep negative and positive slopes, respectively. 5
y
x 10
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 each x-intercept. At x = 36 + 1200 = 8.4048K ,
Problem Set 4-8
71
dy –(8.4 K)3 + 36(8.4 K) – (8.4 K)(0) = dx (8.4 K)2 (0) + 36(0) + 0 3 896.29K = , which is infinite, as conjectured. 0 c. From part a, x 4 − 72x 2 + 2x 2y 2 + 72y 2 + y 4 = −96 ⇒ y 4 + (2x 2 + 72)y 2 + (x 4 − 72x 2 + 96) = 0 y2 = −(2 x 2 + 72) ± (2 x 2 + 72)2 − 4(1)( x 4 − 72 x 2 + 96) 2 2 2 y = − x − 36 ± 144 x 2 – 1200 Only the positive part of the ambiguous sign ± gives real solutions for y. y = ± – x 2 – 36 + 144 x 2 – 1200 Plot the graph letting y1 equal the positive branch and y2 equal the negative branch. The graph is as in the text. The two loops may not appear to close, depending on the window you use for x. d. Repeating the algebra of parts a and c with 1400 in place of 1200 gives y = ± – x 2 – 36 + 144 x 2 – 1400 Plot the graph as in part a. The two ovals in the original graph merge into a single closed figure resembling an (unshelled) peanut.
A = πr 2 ⇒ dr /dt
1
r 3
dr 2 = = 0.6366 K mm/h when r = 3 mm. dt π dr varies inversely with the radius. dt dr dV = 2 cm/s. Want: . dt dt 4 dV dr V = πr 3 ⇒ = 4πr 2 3 dt dt dV = 72π = 226.1946 K cm 3 /s at r = 3 cm dt dV = 288π = 904.7786 K cm 3 /s at r = 6 cm dt
2. Know:
dV/dt
500
r 3
y
dA dr dr 6 = 2πr ⇒ = dt dt dt πr
6
5
x 10
e. The two factors in the equation [(x − 6)2 + y2][(x + 6)2 + y2] = 1200 are the squares of the distances from (x, y) to the points (6, 0) and (−6, 0), respectively. The product of the distances is 1200, a constant.
Problem Set 4-9 Q1. y 2 + 2xyy′ Q3. Product rule Q5. Speeding up
Q2. Implicit differentiation Q4. Chain rule Q6. smaller 1 dx Q7. cos x − x sin x Q8. x dt −x −x Q10. E Q9. −2e + xe dA dr 1. Know: = 12 mm2/h. Want: . dt dt
72
Problem Set 4-9
The graph shows that the larger the balloon gets, the faster Phil must blow air to maintain the 2 cm/s rate of change of radius. dA da 3. Know: = −144 cm 2 /s. Want: . dt dt 1 A = πab and a = 2b ⇒ A = πa 2 2 dA da da 1 dA = πa ⇒ = dt dt dt πa dt da 6 b = 12 ⇒ a = 24 ⇒ = − = −1.9098K dt π ≈ −1.91 cm/s The length of the major axis is 2a, so the major axis is decreasing at 12/π cm/s. dK dm 4. Know: = 100, 000 MJ/s; = −20 kg/s. dt dt dV Want: . (Note: 1 megaJoule—MJ—is the dt energy required to accelerate a 1-kg mass by 1 km/s through a distance of 1 km; it can be expressed 1 MJ = 1 kg · km2/s2.) 1 dK 1 2 dm dV K = mV 2 ⇒ = V + mV ⇒ 2 dt 2 dt dt Calculus Solutions Manual © 2005 Key Curriculum Press
dV 1 dK V dm = − dt mV dt 2 m dt dV 100000 10(–20) = − = 2.02 (km/s)/s dt 5000 ⋅ 10 2 ⋅ 5000 5. Let y = Milt’s distance from home plate. Let x = Milt’s displacement from third base. dx dy Know: = −20 ft/s. Want: . dt dt dy dx 2 2 2 y = x + 90 ⇒ 2 y = 2x dt dt dy x dx –20 x ⇒ = ⋅ = dt y dt x 2 + 90 2 dy /dt 10
x 90
dy = −8.944 K ≈ −8.9 ft/s dt (exact: −4 5 ). dy At x = 0, = 0 ft/s, which is reasonable because dt Milt is moving perpendicular to his line from home plate. 6. Let y = displacement from stern to dock along pier. Let x = displacement from bow to pier along dock. dy dx Know: = −3 m/s. Want: . dt dt 2 2 2 x + y = 200 dx dy dx y dy 3y + 2y =0⇒ =− = 2x dt dt dt x dt 200 2 – y 2 At x = 45,
At y = 120,
dx 360 = = 2.25 m/s. dt 160
dx/dt
dH H dW H dL =− ⋅ − ⋅ dt W dt L dt 20 20 =− (0.1) − 2 (–0.3) LW 2 LW b.
dH 20 ⋅ 0.1 20 ⋅ 0.3 + 2 = 0.02 =− dt 5 ⋅ 22 5 ⋅2 Depth is increasing at 0.02 m/s.
8. Let L = distance between spaceships. dx dy Know: = 80 km/s; = −50 km/s. dt dt dL Want: . dt dL dx dy L2 = x 2 + y 2 ⇒ 2 L = 2x + 2y dt dt dt dL 1 ⇒ = (80 x – 50 y) dt x 2 + y2 dL 1 –200 (80 ⋅ 500 – 50 ⋅ 1200) = = = dt 1300 13 −15.3846… Distance is decreasing at about 15.4 km/s. 9. a. Let x = distance from bottom of ladder to wall. Let y = distance from top of ladder to floor. dx dy 20 2 = x 2 + y 2 ⇒ 0 = 2 x + 2y ⇒ dt dt dy x dx =− dt y dt Note that the velocity of the weight is −dy/dt, so x dx v= 2 dt 400 – x 4 6 ⋅ ( −3) = − = −0.6123K ft/s 4 384 dx 40 c. Here x = 20, = 2, so v = → ∞ (!!) dt 0 10. a. b. v =
10
D 1200 in.2
W
L y 120
200
dL dW . Want: . dt dt dL dW LW = 1200 ⇒ ⋅W + L⋅ =0 dt dt dW −W dL 1 dL ⇒ = ⋅ =− W2 ⋅ dt 1200/W dt 1200 dt Know:
7. a. Let L = length. Let W = width. Let H = depth (meters). dW dL Know: = 0.1 m/s; = −0.3 m/s. dt dt dH Want: . dt LWH = 20 ⇒ dL dW dH ⋅ WH + L ⋅ ⋅ H + LW ⋅ =0 dt dt dt
Calculus Solutions Manual © 2005 Key Curriculum Press
1 W 2 (6) ⇒ W = 20 in. ⇒ 1200 L = 60 in.
b. −2 = −
Problem Set 4-9
73
c. D2 = L2 + W 2 ⇒ 2 D ⇒
dD = dt
1
dD dL dW = 2L + 2W dt dt dt dL dW L +W dt dt
L + W2 At L = 60 and W = 20, dD 1 [60(6) + 20( −2)] = 2 dt 20 + 60 2 320 = = 1.6 10 = 5.0596 K 4000 Diagonal is increasing at about 5.06 in./min. 11. a. Let h = depth of water. Let r = radius of water at surface. Let V = volume of water. dh dV Know: = 5 m/h. Want: . dt dt 1 V = πr 2 h 3 r 3 3 By similar triangles, = ⇒r= h h 5 5 2 1 3 3 3 ∴ V = π h h = πh 3 5 25 dV 9 2 dh = πh dt 25 dt dV 81 At h = 3, = π = 16.2 π = 50.8938K ≈ dt 5 3 50.9 m /h. dV dh b. i. Know: = −2 m3/h. Want: . dt dt dV 9 dh dh 25 dV = πh 2 ⇒ = dt 25 dt dt 9πh 2 dt dh –50 = = −0.1105K ≈ dt 144π −0.11 m/h at h = 4 m dh ii. → −∞ as h → 0 m dt dV c. i. Know: = k h. dt dV = −0.5 at h = 4 ⇒ k = −0.25 dt dV = −0.25 h dt dV ii. = −0.25 0.64 = −0.2 m 3 /h dt at h = 0.64 m dV iii. = −0.2 at h = 0.64 m ⇒ dt dh 25 = (–0.2) = −0.4317K ≈ dt 9π (0.64)2 −0.43 m/h 12. Let h = altitude. Let r = radius. Let V = volume of cone. dh dr dV = −6 ft/min; = 7 ft/min. Want: Know: . dt dt dt
74
2
Problem Set 4-9
1 dV 2 dr 1 dh V = πr 2 h ⇒ = πr h + πr 2 3 dt 3 dt 3 dt dV 2 1 2 = π (8)(3)(7) + π (8) ( −6) = dt 3 3 −16π ft3/min = −50.2654… Volume is decreasing at about 50.3 ft3/min. 13. a. Let ω = angular velocity in radians per day. 2π 2π ωE = , ωM = 365 687 1 1 dθ = ω E − ω M = 2π − = 365 687 dt 644π = 0.008068K ≈ 0.00807 rad/day 250755 −1
1 1 b. T = − = 778.7422 K ≈ 365 687 778.7 days The next time after 27 Aug. 2003 when the two planets will be closest is 779 days later, on 14 Oct. 2005 (or 15 Oct., if the planets were aligned later than about 6:11 a.m. back on 27 Aug. 2003). Because the actual orbits of Earth and Mars are not as simple as previously assumed, the actual closest distances are not always the same. In fact, the approach on 27 Aug. 2003 was the closest one in nearly 60,000 years! Nor is the period between close approaches quite so simple. The next close approach will actually be on 30 Oct. 2005, not 15 Oct. c. By the law of cosines, D 2 = 932 + 1412 − 2 · 93 · 141 cos θ D = 28530 – 26226 cos θ million mi d.
dD 26226 sin θ dθ = dt 2 28530 – 26226 cos θ dt 1 1 26226 ⋅ – ⋅ 2π sin θ 365 687 = million mi/day 2 28530 – 26226 cos θ 1 1 1, 000, 000 ⋅ 26226 ⋅ – ⋅ 2π sin θ 365 687 = 24 ⋅ 2 28530 – 26226 cos θ 1 1 π ⋅ sin θ − 1, 092, 750, 000 ⋅ 365 687 = mi/h 28530 − 26226 cos θ 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 dθ 1 1 by = − 2π to find θ, then dt 365 687 substitute θ into the previous expressions.
Calculus Solutions Manual © 2005 Key Curriculum Press
dD , plot the variable part of dt sin θ . 28530 − 26226 cos θ
e. To maximize dD , y= dt y
0.01
θ π
2π
dl = −1.9963… units/s at x = −5 units. dt dl = 2.6610… units/s at x = 2 units. dt 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 stops decreasing and starts increasing.
Problem Set 4-10 From the graph, it is clear that the maximum occurs well before θ = π/2 (90°). Using the maximize feature, the maximum occurs at θ ≈ 0.8505… , or 48.7…°. (The exact value is cos− 1 (93/141). One can find this by finding (d/dt)(dD/dt) and setting it equal to zero. One can also see this by decomposing Earth’s motion vector into two components—one toward/away from Mars and the other perpendicular to the first. The rate of change in D is maximized when all of Earth’s motion is along the Earth-to-Mars component, which occurs when the Earth-Mars-Sun triangle has a right angle at Earth. 93 In this case, cos θ = 141 .) 1 1 2π t if t = days since f. θ = − 365 687 27 Aug. 2003. 1 1 D = 28530 – 26226 cos – 2π t 365 687 D 200
t 1000
The graph is not a sinusoid. The high and low points are not symmetric. 14. As B moves from negative values of x to positive values of x, the length of AB decreases to about 0.56 unit, then begins to increase when the xvalue of point B passes about −0.3. Let l = length of AB. dl dx . = 2 units/s. Want: Know: dt dt dl l = e 0.8 x + x 2 ⇒ dt 1 dx dx = (e 0.8 x + x 2 ) −1/ 2 ⋅ 2 x + 0.8e 0.8 x dt 2 dt =
0.8e 0.8 x + 2 x e 0.8 x + x 2
Calculus Solutions Manual © 2005 Key Curriculum Press
Review Problems R0. Answers will vary. R1. a. x = g(t) = t3 ⇒ g′(t) = 3t2 y = h(t) = cos t ⇒ h′(t) = −sin t If 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 t x = t3 ⇒ t = x1/3 ⇒ y = cos (x1/3 ) dy 1 = − sin ( x 1/ 3 ) ⋅ x −2 / 3 dx 3 dy 1 At x = 1, = −sin 1 ⋅ = −0.280490 K . dx 3 If x = 1, then t = 11/3 = 1. dy/dt – sin t – sin 1 ∴ = = = −0.280490 K , dx/dt 3t 2 3 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 ⇒ 3 = 7x6 ln 3x + x6 f ′(x) = 7x6 ln 3x + x 7 ⋅ 3x 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.
Problem Set 4-10
75
u′ v – uv ′ . v2 b. See proof of quotient formula in text. sin 10 x c. i. f ( x ) = ⇒ x5 10 cos 10 x ⋅ x 5 – sin 10 x ⋅ 5 x 4 f ′( x ) = x 10 10 x cos 10 x – 5 sin 10 x = x6
R3. a. If y = u/v, then y′ =
(2 x + 3)9 ⇒ g ′( x ) (9 x – 5) 4 9(2 x + 3)8 ⋅ 2(9 x − 5) 4 − (2 x + 3)9 ⋅ 4(9 x − 5)3 ⋅ 9 = (9 x − 5)8 18(2 x + 3)8 (5 x – 11) = (9 x – 5)5 ii. g( x ) =
iii. h(x) = (100x3 − 1)− 5 ⇒ h ′(x) = −5(100x3 − 1)− 6 · 300x2 = −1500x2(100x3 − 1)− 6 d. y = 1/x 10 As a quotient: 0 ⋅ x 10 – 1 ⋅ 10 x 9 –10 y′ = = 11 = −10 x −11 x 20 x As a power: y = x − 10 y′ = −10x − 11, which checks. sin x = tan x e. t ( x ) = cos x cos x cos x – sin x (– sin x ) t ′( x ) = cos 2 x cos 2 x + sin 2 x 1 = = = sec 2 x cos 2 x cos 2 x t ′ (1) = sec2 1 = 3.4255… t ( x ) – t (1) tan x – tan 1 f. m( x ) = = x –1 x –1
The values get closer to 3.4255… as x approaches 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, which is connected to the fact that tan′ x equals the square of a function and is thus always positive. y
1
x
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 at t = π /2 = 1.57079… . As t gets closer to this value, secant changes very rapidly! 3 R5. a. i. y = tan −1 3 x ⇒ y′ = 1 + 9x 2 d 1 ii. (sec –1 x ) = dx | x | x2 – 1 iii. c( x ) = (cos −1 x )2 ⇒ c′( x ) =
/2
1 – x2
1
b. y = sin −1 x ⇒ y ′ =
m (x)
–2 cos –1 x
1− x2
y
3.42...
x 1
1
x 1
76
x
m(x)
0.997 0.998 0.999 1 1.001 1.002 1.003
3.40959… 3.41488… 3.42019… undefined 3.43086… 3.43622… 3.44160…
Problem Set 4-10
y′(0) =
1 1 – 02
= 1 , which agrees with the
graph. y′(1) =
1 2
=
1 , which is infinite. 0
1–1 The graph becomes vertical as x approaches 1 from the negative side. y′(2) is undefined because y(2) is not a real number. R6. a. Differentiability implies continuity. Calculus Solutions Manual © 2005 Key Curriculum Press
b. i. Answers may vary. ii. Answers may vary. f (x )
f (x )
=
6t ( dt/dx ) ⋅ 2e 2 t − 3t 2 ⋅ 4e 2 t ( dt/dx ) (2e 2 t ) 2
3t − 3t 2 dx 3t − 3t 2 ÷ = 2e 4 t e2t dt b. x = (t/π) cos t y = (t/π ) sin t dy dy/dt (1/π ) sin t + (t/π )(cos t ) = = dx dx/dt (1/π ) cos t + (t/π )(– sin t ) sin t + t cos t = cos t – t sin t Where 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, then dy sin 6π + 6π cos 6π 0 + 6π = = = 6π . dx cos 6π – 6π sin 6π 1– 0 So the graph is not vertical where it crosses the 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 radius of the Ferris wheel. The phase displacement is 3 seconds. The period is 20 seconds, so the coefficient of the arguments of sine and cosine is 2 π /20 = π /10. π x = 20 sin (t – 3) 10 π y = 25 + 20 cos (t – 3) 10 π dx/dt = 2π cos (t – 3) 10 π dy/dt = −2π sin (t – 3) 10 When t = 0, dy/dt = 5.0832… . The Ferris wheel is going up at about 5.1 ft/s. When t = 0, dx/dt = 3.6931… . The Ferris wheel is going right at about 3.7 ft/s. dy dy/dt = dx dx/dt dy/dx will be infinite if dx/dt = 0 and dy/dt ≠ 0. π dx/dt = 0 if 2π cos (t – 3) = 0 . 10 π π (t – 3) = + π n (where n is an integer) 10 2 t = 8 + 5n The first positive time is t = 8 s. =
x
x c
c
iii. No such function. iv. Answers may vary. f (x )
x c
c. i. f (x )
2
x 1
ii. f is continuous at x = 1 because right and left limits both equal 2, which equals f (1). iii. f is differentiable. Left and right limits of f ′ (x) are both equal to 2, and f is continuous at x = 2. sin –1 x, if 0 ≤ x ≤ 1 d. g( x ) = 2 x + ax + b, if x ≤ 0 (1 − x ) −1/2, if 0 < x < 1 g ′( x ) = if x < 0 2 x + a, lim− g( x ) = 0 + 0 a + b = b
x→0
lim g( x ) = sin −1 0 = 0
x→0 +
∴b=0 lim– g′( x ) = 0 + a = a x→0
lim g′( x ) = 1−1/2 = 1
x→0 +
∴a=1 g (x ) 1
x 0
1
The graph appears to be differentiable and continuous at x = 0. dy dy/dt 3t 2 R7. a. x = e 2 t, y = t 3 ⇒ = = ⇒ dx dx/dt 2e 2 t d 2 y d 3t 2 = dx 2 dx 2e 2 t
Calculus Solutions Manual © 2005 Key Curriculum Press
Problem Set 4-10
77
R8. a. y = x 8/5 ⇒ y 5 = x 8 8x 7 8x 7 8 3/ 5 5 y 4 y′ = 8 x 7 ⇒ y′ = 4 = x 8/ 5 4 = 5y 5( x ) 5 Using the power rule directly: y = x 8/5 ⇒ y′ = 85 x 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] = 4.5x 3/5 − y 4 cos xy dy 4.5 x 3.5 – y 4 cos xy y′ = = 2 dx 3 y sin xy + xy 3 cos xy c. i. 4y 2 − xy 2 = x 3 ⇒ 8yy′ − y 2 − x · 2yy′ = 3x 2 y′(8y − 2xy) = 3x 2 + y 2 dy 3 x 2 + y 2 y′ = = dx 8 y – 2 xy At (2, 2), dy/dx = 2. At (2, −2), dy/dx = −2. Lines at these points with these slopes are tangent to the graph (see diagram). y 5
x 2
5
ii. At (0, 0), dy/dx has the indeterminate form 0/0, which is consistent with the cusp. iii. To find the asymptote, solve for y. (4 − x)y 2 = x 3 x3 y2 = 4–x 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.
At z = 200, x = 200 2 – 70 2 = 30 39 dz 20 ⋅ 30 39 = = 3 39 = 18.7349... dt 200 The glass moves at the same speed as the tablecloth, or about 18.7 cm/s, which is about 1.3 cm/s slower than Rover. Concept Problems C1. a. Let (x, y) be the coordinates of a point on the tangent line. y – y0 = m ⇒ y = m( x − x 0 ) + y0 x – x0 b. Substituting (x1, 0) for (x, y) gives y 0 = m( x1 − x 0 ) + y0 ⇒ x1 = x 0 − 0 , Q .E.D . m 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 y x 2 = x1 − 1 m Because y1 = f (x1) and m = f ′(x1), f ( x1 ) x 2 = x1 − , Q .E .D . f ′( x1 ) d. Programs will vary according to the kind of grapher 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 numerical derivative. • Display the new x. • Save the new x as the old x and repeat. For f (x) = x2 − 9x + 14, the program should give x = 2, x = 7. e. For g(x) = x3 − 9x2 + 5x + 10, first plot the graph to get approximations for the initial values of x. g (x )
R9. 20
z
70
x 1
x
Let x = Rover’s distance from the table. Let z = slant length of tablecloth. dx dz Know: at z = 200. = 20 cm/s. Want: dt dt 2 2 2 z = x + 70 dz dx 2z = 2x dt dt dz x dx 20 x = = dt z dt z
78
Problem Set 4-10
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-in solver feature. The same preliminary analysis is needed to find starting values of x. Calculus Solutions Manual © 2005 Key Curriculum Press
f. f (x) = sec x − 1.1 Starting with x0 = 1, it takes seven iterations to get x = 0.429699666… . C2. a. The connecting rod, the crankshaft, and the y-axis form a triangle with angle φ = θ − π /2 included between sides of 6 cm and (y − 8) cm.
a 500,000 θ 3
Solving graphically and numerically, a < −980 for θ ∈ (0.2712… , 2.8703…). The piston is going down (v < 0) for θ ∈ (π /2, 3π /2). So the piston is going down with acceleration greater than gravity for θ between π /2 and 2.8703… .
8
y–8
20 φ
θ
6
By the law of cosines, 20 2 = (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.
Chapter Test T1. y = uv ⇒ y′ = u′v + uv′ u + ∆u − u v + ∆v v (v + ∆v)v T2. y ′ = lim ⋅ ∆x →0 (v + ∆v)v ∆x (u + ∆u)v − u(v + ∆v) = lim ∆x →0 ∆x (v + ∆v)v
y − 8 = 6 sin θ + (36 ⋅ sin 2 θ + 364) . (The solution with the negative radical gives a triangle below the origin, which has no reallife meaning.)
uv + ∆uv − uv − u∆v = lim ∆x →0 ∆x (v + ∆v)v ∆u ⋅ v − u∆v 1 = lim ⋅ ∆x →0 ( v + ∆v )v ∆x ∆ ∆ 1 u v = lim ⋅ v − u ∆x →0 ( v + ∆v )v ∆x ∆x
y = 8 + 6 sin θ + 2 9 ⋅ sin 2 θ + 91 dy dθ 18 sin θ cos θ dθ b. v = = 6 cos θ + dt dt 9 ⋅ sin 2 θ + 91 dt v = 6 cos θ c. a =
dθ 9 sin 2θ dθ + 2 dt 9 ⋅ sin θ + 91 dt
d2y dt 2
dθ = −6 sin θ dt + 18
2
91 cos 2θ – 9 sin 4 θ dθ (9 sin 2 θ + 91)3/ 2 dt
2
91 cos 2θ – 9 sin 4 θ dθ 2 = 18 ⋅ 6 – sin θ (9 sin 2 θ + 91)3/ 2 dt (There are many other correct forms of the answer, depending on how you use the double-argument properties and Pythagorean properties from trigonometry.) Note that the angular velocity is constant at 6000π radians per minute, so dθ = 100π rad/s. dt d. See the graph. Note that a line at a = −980 is so close to the x-axis that it does not show up. Calculus Solutions Manual © 2005 Key Curriculum Press
1 du dv u ′v − uv ′ v−u = 2 v dx dx v2 [Because as ∆x → 0, u/x and ∆v/∆x become du/dx and dv/dx and v → 0, so (v + ∆v)v → v 2 ]. cos x T3. cot x = sin x − sin x sin x − cos x cos x = sin 2 x −(sin 2 x + cos 2 x ) 1 = = − 2 = − csc 2 x sin 2 x sin x T4. y = sin −1 x ⇒ sin y = x ⇒ y′ cos y = 1 ⇒ =
1 cos 2 y + sin 2 y = 1 ⇒ cos 2 y = 1 − sin 2 y ⇒ cos y 1 cos y = 1 − sin 2 y = 1 − x 2 ⇒ y = 1− x2
y′ =
T5.
dy dy/dt 4t 3 d2y d (2t 2 ) = = = = 2t 2 2 = dx dx/dt 2t dx dx dx 4t 4t ( dt/dx ) = 4t ÷ = =2 dt 2t Problem Set 4-10
79
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. 1/cos x – 1/cos 2 m( x ) = x–2 x
m(x)
1.997 1.998 1.999 2.000 2.001 2.002 2.003
5.28893631… 5.27611340… 5.26335022… undefined 5.23800134… 5.22541482… 5.21288638…
T8. Answers may vary. f (x ) 7
T15. y = 4 sin− 1 (5x3) 1 60 x 2 y′ = 4 ⋅ ⋅ 15 x 2 = 1 – (5 x 3 ) 2 1 – 25 x 6 T16. 9x 2 − 20xy + 25y 2 − 16x + 10y − 50 = 0 ⇒ 18x − 20y − 20xy ′ + 50yy ′ − 16 + 10y ′ = 0 y ′(−20x + 50y + 10) = −18x + 20y + 16 dy –18 x + 20 y + 16 y′ = = dx –20 x + 50 y + 10 –9 x + 10 y + 8 = –10 x + 25 y + 5 If x = −2, then 36 + 40y + 25y 2 + 32 + 10y − 50 = 0 25y 2 + 50y + 18 = 0 Solving numerically gives y = −0.4708… or y = −1.5291… , both of which agree with the graph. (Solving algebraically by the quadratic formula, y = −1 ± 7 /5 , which agrees with the numerical solutions.) At (−2, −0.4708…), dy/dx = 1.60948… . At (−2, −1.5291…), dy/dx = −0.80948… . The answers are reasonable, because lines of these slopes are tangent to the graph at the respective points, as shown here. 5 y
x –2
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′ = 37 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) e3 x T13. f ( x ) = ⇒ ln x 3e 3 x ln x − e 3 x (1/ x ) 3 xe 3 x ln x − e 3 x f ′( x ) = = (ln x )2 x (ln x )2 T14. x = sec 2t y = tan 2t3 dy dy/dt sec 2 2t 3 ⋅ 6t 2 3t 2 sec 2 2t 3 = = = dx dx/dt sec 2t tan 2t ⋅ 2 sec 2t tan 2t
80
Problem Set 4-10
–2
x 5
x 3 + 1, if x ≤ 1 T17. f ( x ) = 2 a( x – 2) + b, if x > 1 3 x 2 , if x < 1 f ′( x ) = 2 a( x – 2), if x > 1 For equal derivatives on both sides of x = 1, lim– f ′( x ) = 3 ⋅ 12 = 3 x →1
lim f ′( x ) = 2 a(1 − 2) = −2 a
x →1+
∴ −2 a = 3 ⇒ a = −1.5 For continuity at x = 1, lim– f ( x ) = 13 + 1 = 2 x →1
lim f ( x ) = a(1 − 2) 2 + b = a + b
x →1+
∴a + b = 2 Substituting a = −1.5 gives b = 3.5.
Calculus Solutions Manual © 2005 Key Curriculum Press
The graph shows differentiability at x = 1. f (x )
2
x 1
Values of b other than 3.5 will still cause the two branches to have slopes approaching 4 as x approaches 1 from either side as long as a = −1.5. However, f will not be continuous, and thus will not be differentiable, as shown here for b = 4.5.
T19. cot = adjacent/opposite = x/5 = cot− 1 (x/5) 1 1 1 dθ T20. =− ⋅ =− 1 + ( x/5)2 5 5 + 5( x 2 / 25) dx 1 5 =− =− 5 + ( x 2 / 5) 25 + x 2 dx T21. = −420 mi/h dt 5 2100 dθ dθ dx = ⋅ =− ⋅ ( −420) = T22. 25 + x 2 25 + x 2 dt dx dt T23. The plane is changing fastest when x approaches zero, when the plane is nearest the station. y
f (x ) 25
x 30
2
x
T24. Answers will vary.
1
T18. y = x 7/3 ⇔ y 3 = x 7 3y 2y′ = 7x 6 7x 6 7 x 6 7 7 y′ = 2 = = x 6−14 / 3 = x 4 / 3 3y 3 ( x 7/ 3 ) 2 3 3 This answer agrees with y′ = nxn− 1. 4/3 is 7/3 − 1.
Calculus Solutions Manual © 2005 Key Curriculum Press
Problem Set 4-10
81
Chapter 5—Definite and Indefinite Integrals Problem Set 5-1 1. f (1000) = 20 + (0.000004)(10002) = 24 $/ft f (4000) = 20 + (0.000004)(40002) = 84 $/ft The price increases because it is harder and slower to 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 the trapezoids are circumscribed above the curve. (Note that T6 can be found more easily by first factoring out the 500, then adding the function values.) 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 first factoring out the 500, then adding the function values.) 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)x 3 + C g (4,000) − g (1,000) = (165,333.3333… + C) − (21,333.3333… + C) = 144,000, which is the conjectured value of the definite integral! The other name for antiderivative is indefinite integral. 6. a. f (x) = x 7 + C b. y = − cos x + C 2x c. u = 0.5e + C d. v = 321 (4x + 5)8 + C
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
dx
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) 82
Problem Set 5-2
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) = x 2 ⇒ f ′ (x) = 2x ⇒ f ′ (1) = 2 Tangent line: y − 1 = 2(x − 1) ⇒ y = 2x − 1 The graph shows a zoom by factor of 10.
1 graph
tangent line 1
Local linearity describes the property of the function because if you keep x close to 1 (in the “locality” of 1), the curved graph of the function looks like the straight graph of the tangent line. x
x a
Error
b.
Area = product of x and y
y
y
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.
Problem Set 5-2 Q1. Answers may vary.
f (x)
0.97 0.98 0.99 1 1.01 1.02 1.03
f (x)
y
Error, f (x) − y
0.9409 0.9604 0.9801 1 1.0201 1.0404 1.0609
0.94 0.96 0.98 1 1.02 1.04 1.06
0.0009 0.0004 0.0001 0 0.0001 0.0004 0.0009
Calculus Solutions Manual © 2005 Key Curriculum Press
The table shows that for x-values close to 1 (the point of tangency), the tangent line is a close approximation to the function values. 4. f (x) = x2 − 0.1(x − 1)1/3 Zooming in on (1, 1) shows that the graph goes vertical at x = 1. This observation is confirmed algebraically. f ′ (x) = 2x − (1/3)(0.1)(x − 1) − 2/3 f ′ (1) = 2 − (1/3)(0.1)(0) − 2/3 , which is infinite.
V ≈ 43 π (63) + 4.32π = 288π + 4.32π = 292.32π ≈ 918.350 mm3 4 Actual volume is V = π (6.033) = 3 292.341636π ≈ 918.418 mm3. ∆V = 43 π (6.033 ) − 43 π (6 3 ) = 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 about 82 cents.
1 graph tangent line 1
f does not have local linearity at x = 1. Because the slope of the graph becomes infinite, no linear function can approximate the graph there. If f is differentiable at x = c, then f is locally linear there. The converse is also true. If f is locally linear at x = c, then f is differentiable there. 5. a. Let A be the number of radians in θ degrees. x ⇒ By trigonometry, tan A = 100 x A = tan− 1 . 100 Because 1 radian is 180/π degrees, 180 x , Q .E . D . θ= tan− 1 π 100 1 180 1 b. dθ = · dx 2 · 1 + ( x / 100) π 100 1.8 / π = dx 1 + ( x / 100)2 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 multiplying by it is approximately equivalent to dividing by 2. For a 20% grade, this estimate gives 10°, compared to the actual angle of 11.309…°, an error of about 11.6%. For a 100% grade, this estimate gives 50°, compared to the actual angle of 45°, an error of about 11.1%. 6. dV = 4π r 2 dr dr = 0.03 and r = 6, so dV = 4π (62)(0.03) = 4.32π ≈ 13.57 mm3 Calculus Solutions Manual © 2005 Key Curriculum Press
b. m = 6000e (0.05/365)t ⇒ dm = 6000(0.05/365)e(0.05/365)t dt Substituting 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 accurate approximation for ∆m. 8. a. dS = − 1.636 dt March 11: dS = − 1.636(10) = − 16.36 minutes Sunrise time ≈ 6:26 − 0:16 = 6:10 a.m., which agrees with the tabulated value. March 21: dS = − 1.636(20) = − 32.72 minutes Sunrise time ≈ 6:26 − 0:33 = 5:53 a.m., which agrees with the tabulated value. b. By September 1, t = 185, giving dS = − 1.636(185) = − 302.66, or 5:04 hours. So the predicted sunrise time would be 6:26 − 5:04 = 1:22 a.m. Because the sunrise reaches its earliest in mid-June, the time predicted 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 Problem Set 5-2
83
15 1 −2 / 3 5 ⋅ x = dx x 1/ 3 3 x y = sin 3x ⇒ dy = 3 cos 3x dx y = cos 4x ⇒ dy = − 4 sin 4x dx y = tan3 x ⇒ dy = 3 tan2 x sec2 x dx y = sec3 x ⇒ dy = 3 sec3 x tan x dx y = 4x cos x ⇒ dy = (4 cos x − 4x sin x) dx y = 3x sin x ⇒ dy = (3 sin x + 3x cos x) dx y = x 2 /2 − x/4 + 2 ⇒ dy = (x − 1/4) dx y = x 3 /3 − x/5 + 6 ⇒ dy = (x 2 − 1/5) dx − sin (ln x ) y = cos (ln x) ⇒ dy = dx x y = sin (e0.1 x) ⇒ dy = 0.1e0.1 x cos (e0.1 x) dx dy = 20x3 dx ⇒ y = 5x4 + C dy = 36x4 dx ⇒ y = 7.2x5 + C dy = sin 4x dx ⇒ y = − (1/4) cos 4x + C dy = cos 0.2x dx ⇒ y = 5 sin 0.2x + C dy = (0.5x − 1)6 dx ⇒ y = (2/7)(0.5x − 1)7 + C dy = (4x + 3)− 6 dx ⇒ y = (− 1/20)(4x + 3)− 5 + C dy = sec2 x dx ⇒ y = tan x + C dy = csc x cot x dx ⇒ y = − csc x + C dy = 5 dx ⇒ y = 5x + C dy = − 7 dx ⇒ y = − 7x + C dy = (6x2 + 10x − 4) dx ⇒ y = 2x3 + 5x 2 − 4x + C dy = (10x2 − 3x + 7) dx ⇒ y = (10/3)x3 − (3/2)x 2 + 7x + C dy = sin5 x cos x dx ⇒ y = (1/6) sin6 x + C dy = sec7 x tan x dx = sec6 x(sec x tan x dx) ⇒ y = (1/7) sec7 x + C 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… . 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... .
16. y = 15 ln x 1/3 ⇒ dy = 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39. 40. 41.
42.
84
Problem Set 5-3
Problem Set 5-3 Q1. Q2. Q3. Q5. Q7. Q9. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13.
Antiderivative = x3 + C Indefinite integral = (1/6)x6 + C y′ = 3x 2 Q4. y′ = ln 3(3x) dy = ln 3(3x) dx Q6. y ′′ = 5 x 4 Integral = sin x + C Q8. y′ = − sin x 1 Q10. E 1 x 10 dx = x 11 + C 11 1 21 20 x dx = x +C 21 4 4 x −6 dx = − x −5 + C 5 3 −6 −7 9 x dx = − x + C 2
∫ ∫ ∫ ∫ ∫ cos x dx = sin x + C ∫ sin x dx = − cos x + C 4 ∫ 4 cos 7x dx = 7 sin 7x + C 20 ∫ 20 sin 9 x dx = − 9 cos 9 x + C 5 ∫ 5e dx = 0.3 e + C ∫ 2e dx = −200e + C 0.3 x
0.3 x
−0.01 x
∫ ∫ ∫
−0.01 x
4m +C ln 4 8.4 r 8.4 r dr = +C ln 8.4 1 ( 4v + 9)2 dv = ( 4v + 9)2 ( 4 dv) 4 1 = ( 4 v + 9) 3 + C 12 1 (3 p + 17)5 dp = (3 p + 17)5 (3 dp) 3 1 = (3 p + 17) 6 + C 18 1 (8 − 5 x )3 dx = − (8 − 5 x )3 ( −5 dx ) 5 1 = − (8 – 5 x ) 4 + C 20 4 m dm =
∫
14.
∫
∫
15.
∫
16.
∫ (20 − x ) dx = (−1)∫ (20 − x ) (−dx )
∫
4
4
1 = − (20 – x ) 5 + C 5
Calculus Solutions Manual © 2005 Key Curriculum Press
17. 18. 19. 20. 21. 22. 23.
∫ (sin x ) cos x dx = 7 sin x + C 1 ∫ (cos x ) sin x dx = − 9 cos x + C 1 ∫ cos θ sin θ dθ = − 5 cos θ + C 1 ∫ sin θ cosθ dθ = 6 sin θ + C 1 3 ∫ ( x + 3x − 5) dx = 3 x + 2 x − 5x + C 1 ∫ ( x − 4 x + 1) dx = 3 x − 2 x + x + C ∫ ( x + 5) dx = ∫ ( x + 15x + 75x + 125) dx 7
8
9
4
5
5
6
2
3
2
3
2
= 24.
1
6
∫ (x
3
3
2
2
6
4
2
1 7 x + 3x 5 + 25x 3 + 125x + C 7
∫
− 6)2 dx = ( x 6 − 12 x 3 + 36) dx
1 = x 7 − 3x 4 + 36x + C 7
25. 26. 27. 28. 29. 30. 31.
∫ ∫e ∫ sec ∫ csc ∫ tan ∫ cot ∫ csc e
sec x
sec x tan x dx = e
tan x
sec 2 x dx = e tan x + C
2
2
∫ sec
Let h( x ) =
By the derivative of a sum property, d d h ′( x ) = f ( x ) dx + g( x ) dx. dx dx By the definition of indefinite integral applied twice to the right side of the equation, h ′ (x) = f (x) + g (x). By the definition of indefinite integral applied in the other direction,
∫
1 8 tan x + C 8 1 8 x csc 2 x dx = − cot 9 x + C 9 x sec 2 x dx =
9
x cot x dx = csc8 x (csc x cot x dx )
∫
∫
∫
1.5 2.5 3.5 Sum: Integral ≈ 50.75
1 sec7 x + C 7 33. v (t) = 40 + 5 t = 40 + 5t1/ 2 =
∫
D(t ) = ( 40 + 5t 1/ 2 ) dt = 40t +
10 3/ 2 t +C 3
10 3/ 2 0 +C⇒C=0 3
10 3/ 2 ∴ D (t) = 40t + t 3 D (10) = 505.4092... ≈ 505 ft 34. a. f (x) = 0.3x 2 + 1 T100 = 9.300135 b. g( x ) = (0.3 x 2 + 1) dx = 0.1x 3 + x + C
Calculus Solutions Manual © 2005 Key Curriculum Press
∫ x cos x dx = x sin x + cos x + C.
Phoebe checked this by differentiating: d ( x sin x + cos x + C ) dx = 1 ⋅ sin x + x ⋅ (cos x) − sin x + 0 = x cos x By the definition of indefinite integral, she knew that Calvin was right. 37. a. C v (t)
x tan x dx = sec 6 x (sec x tan x dx )
∫
∫
By the transitive property, then,
36. Calvin says
x dx = − cot x + C
D (0) = 0 ⇒ 0 = 40 ⋅ 0 +
∫ f ( x ) dx + ∫ g( x ) dx.
Q .E .D .
x dx = tan x + C
7
7
Proof:
∫ [ f ( x ) + g( x )] dx = ∫ f ( x ) dx + ∫ g( x ) dx,
1 = − csc9 x + C 9
32.
∫ [ f ( x ) + g( x )] dx = ∫ f ( x ) dx + ∫ g( x ) dx.
h( x ) = [ f ( x ) + g( x )] dx +C
sec x
c. g (4) − g (1) = 6.4 + 4 + C − 0.1 − 1 − C = 9.3, which is about equal to the definite integral! It is also interesting that the constant C drops out. 35. Prove that if f and g are functions that can be integrated, then
12.25 16.25 22.25 50.75
b. c
v (t)
1.25 1.75 2.25 2.75 3.25 3.75 Sum:
11.5625 13.0625 15.0625 17.5625 20.5625 24.0625 101.8750
Integral ≈ (101.8750)(0.5) = 50.9375 Problem Set 5-3
85
c. As shown in Figures 5-3c and 5-3d, the Riemann sum with six increments has smaller regions included above the graph and smaller regions excluded below the graph, so the Riemann sum should be closer to the integral. d. Conjecture: Exact value is 51. By the trapezoidal rule with n = 100, integral ≈ 51.00045, which agrees with the conjecture. e. The integral is the product of v (t) and t, and thus has the units (ft/min)(min), or ft. So the object went 51 ft. Average velocity = 51/3 = 17 ft/min. 38. Answers will vary.
3.
7.
4
2.
1
c
1.25 1.5625 1.75 3.0625 2.25 5.0625 2.75 7.5625 3.25 10.5625 3.75 14.0625 Sum = 41.8750 R6 = (0.5)(41.875) = 20.9375
86
Problem Set 5-4
0
f (c)
c
∫
f (c)
0.1 0.995004… 0.3 0.955336… 0.5 0.877582… 0.7 0.764842… 0.9 0.621609… Sum = 4.214375… R5 = (0.2)(4.21…) = 0.842875…
1.2
tan x dx
L4 = 0.73879… , U 4 = 1.16866… M 4 = 0.92270… , T4 = 0.95373… ∴ M4 and T4 are between L4 and U4, Q.E.D.
∫
6
3
x 3 dx
8.
c
L4 = 9.5, U 4 = 12.8333… M 4 = 10.89754… , T4 = 11.1666… ∴ M4 and T4 are between L4 and U4, Q.E.D.
f (c)
2.25 11.390625 2.75 20.796875 3.25 34.328125 3.75 52.734375 4.25 76.765625 4.75 107.171875 5.25 144.703125 5.75 190.109375 Sum = 638.000000 R8 = (0.5)(638) = 319
∫ 10/x dx: 1
2
f (c)
∫ cos x dx
0.4
Q9. If a + b = 5, then a = 2 and b = 3. Q10. No
∫
− 0.75 0.59460… −0.25 0.84089… 0.25 1.18920… 0.75 1.68179… 1.25 2.37841… 1.75 3.36358… Sum = 10.04849… R6 = (0.5)(10.04…) = 5.024249…
6.
1.1 0.891207… 1.3 0.963558… 1.5 0.997494… 1.7 0.991664… 1.9 0.946300… Sum = 4.790225… R5 = (0.2)(4.79…) = 0.958045…
x
f (c)
1
∫ sin x dx
c
2 x dx
c
1
y
x 2 dx
f (c)
2
−1
2
5.
y′ = sin x + x cos x tan x + C f ′(x) = sec2 x (1/4)x4 + C z′ = − 7 sin 7x − cos u + C Limit = 8
4
∫
4.
− 0.75 0.43869… − 0.25 0.75983… 0.25 1.31607… 0.75 2.27950… 1.25 3.94822… 1.75 6.83852… 2.25 11.84466… 2.75 20.51556… Sum = 47.94108… R8 = (0.5)(47.94…) = 23.97054…
7
1.
−1
3 x dx
c
Problem Set 5-4 Q1. Q2. Q3. Q4. Q5. Q6. Q7. Q8.
∫
3
5
9.
∫ ln x dx is underestimated by the trapezoidal 1
rule and overestimated by the midpoint rule. y
y 2
2
x
x 5
5
Calculus Solutions Manual © 2005 Key Curriculum Press
10.
∫
2
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. U n = (3/n)3(12 + 22 + 32 + ⋅ ⋅ ⋅ + n 2) = (3/n)3(n/6)(n + 1)(2n + 1) = (4.5/n2)(n + 1)(2n + 1) U100 = (4.5/1002)(101)(201) = 9.13545, which is correct. e. Using the formula, U1000 = 9.013504… , which does seem to be approaching 9 n + 1 2n + 1 f. Un = 4.5 ⋅ ⋅ n n = 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!
e x dx is overestimated by the trapezoidal rule
0
and underestimated by the midpoint rule. y
y
4
4
x
x 1
1
11. a. h (x) = 3 + 2 sin x For an upper sum, take sample points at x equals 1, π/2, 2, 3, 4, and 6. b. For a lower sum, take sample points at x equals 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… L 6 = 1[h (0) + h (1) + h (3) + h (4) + h (3π/2) + h (5)] = 14.53372… 12. Programs will vary depending on the type of grapher used. See the program in the Programs for Graphing Calculators section of the Instructor’s Resource Book. 13. a. For
∫
4
x 2 dx, the program should give the
1
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 the same limit as n approaches infinity. d. The trapezoids are circumscribed around the region under the graph and thus contain more area (see left diagram). For rectangles, the “triangular” part of the region that is left out has more area than the “triangular” part that is added, because the “triangles” have equal bases but unequal altitudes (see right diagram).
∫
2
x 3 dx
0
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 + ⋅ ⋅ ⋅ + n 3) = (2/n)4[(n/2)(n + 1)]2 = 4/n2 ⋅ (n + 1)2 = 4(1 + 1/n)2 lim Un = 4(1 + 0)2 = 4 n→∞
Problem Set 5-5 Q1. x 2/2 + 2x + C Q3. − cot x + C 5(ln x ) 4 Q5. +C x
10 t +C ln 10 Q4. − csc x cot x
Q2.
Q6. y
x 1 2
y Rectangle leaves out more area.
y Trapezoid includes more area.
15.
Q7. Answers may vary. Q8. Answers may vary. y
y 5
x
x x
3
14. a.
∫x
3
2
1
x 2
dx
0
U 100 = 9.13545, L100 = 8.86545 Conjecture: Integral equals 9 exactly. b. The sample points will be at the right of each interval, 1 ⋅ 3/n, 2 ⋅ 3/n, 3 ⋅ 3/n, . . . , n ⋅ 3/n. Calculus Solutions Manual © 2005 Key Curriculum Press
Q9. No limit (infinite) Q10. D 1. See the text for the statement of the mean value theorem. 2. See the text for the statement of Rolle’s theorem.
Problem Set 5-5
87
3. g (x) = 6/x; [1, 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]
g (x ) 6
2
f (x ) x c
1
1
4
x
6/4 – 6 m= = −3/2 4 –1 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 )
10
x c
–1
m=
2
f (x )
1
∴ 4c = 5 ⇒ c = 3 5 / 4 = 1.077K 3
Tangent at x = 1.077… parallels the secant line. π 5. c( x ) = 2 + cos x; 0, 2 c (x )
1
x c
_ π 2
c1
f ′ (x) = cos x − x sin x ∴ f is differentiable for all x. x cos x = 0 ⇔ x = 0 or cos x = 0 cos 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 at c = 0.86033… . Horizontal line at x = 0.86033… is tangent. 8. f (x) = x 2 sin x
16 – 1 =5 2 – (–1)
g ′ (x) = 4x3
m=
0
π /2
cos (π /2) – cos 0 = −2/π = 0.6366 K π /2 – 0
c′(x) = − sin x ∴ − sin c = − 2/π ⇒ c = 0.69010… Tangent at x = 0.690… parallels the secant line.
x c
0
π
f ′ (x) = 2x sin x + x 2 cos x ∴ f is differentiable for all x. x 2 sin x = 0 ⇔ x = 0 or sin x = 0 sin x = 0 at x = 0 + π n , where n is an integer Interval: [0, π] Using the solver feature, f ′ (c) = 0 at c = 2.28892… . Horizontal line at x = 2.288… is tangent. 9. f (x) = (6x − x 2) 1/2 f (x ) 3
6. h( x ) = 5 − x ; [1, 9]
x 0
c
6
h (x) 5
x 1
m=
88
c
2–4 = −1/4 9 –1
Problem Set 5-5
9
f ′( x ) = 12 (6 x – x 2 ) −1/ 2 ⋅ (6 − 2 x ) ∴ f is differentiable on (0, 6). f is continuous at x = 0 and x = 6. (6x − x 2)1/2 = 0 x(6 − x) = 0 ⇒ x = 0 or 6 Interval: [0, 6] (6c − c2)− 1/2(3 − c) = 0 ⇒ c = 3 Horizontal line at x = 3 is tangent. Calculus Solutions Manual © 2005 Key Curriculum Press
10. f (x) = x 4/3 − 4x 1/3 f (x )
1
c
0
x 4
f ′( x ) = 43 x 1/ 3 − 43 x −2 / 3 ∴ f is differentiable for all x ≠ 0. f is continuous at x = 0. f (x) = 0 ⇒ x 4/3 − 4x 1/3 = 0 ⇒ x 1/3 (x − 4) = 0 ⇒ x = 0 or 4 Interval: [0, 4] 4 3
c1/ 3 − 43 c −2 / 3 = 0
c −2 / 3 (c − 1) = 0 ⇒ c = 1 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!) 74357.5K – 1000 b. Average rate is = 50 1467.150… ≈ $1,467.15 per year. c. d′ (t) = ln (1.09)1000(1.09)t d′ (0) ≈ $86.18 per year d′ (50) ≈ $6,407.96 per year The 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 = 4 3
1000(1.09)50 − 1000 algebraically gives 50 50 1.09 − 1 (1.09)t = 50 ln 1.09 1.09 50 − 1 50 ln 1.09 = ln (1.0950 − 1) − ln 50 − ln (ln 1.09) ⇒ ⇒ t ln 1.09 = ln
intervals for which 1 is an endpoint, such as [1, 2]. The conclusion is true if the instantaneous velocity, d′ (t), ever equals the average velocity. The average velocity equals d (2) – d ( 0) m= = 28.5 ft/s for [0, 2], 2 d (2) – d (0.5) m= = 57.111K ft/s for [0.5, 2], 1.5 d (2) – d (1) m= = 100 ft/s for [1, 2]. 1 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 conclusion is 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 is true. See the right graph. d (t )
d (t )
200
200
c is outside (0, 2).
c is in (0.5, 2).
100
100
t 0.5
1
c?
2
t 0.5
c? 2
1
For d′ (c) = 100, 200c− 2 = 100 ⇒ c = 1.414… . Because 1.414… is in (1, 2), the conclusion is true, as is guaranteed by the mean value theorem. The fact that the conclusion is true when the hypotheses are met illustrates the fact that the hypotheses are sufficient. The fact that the conclusion can be true even if the hypotheses are not met proves that the hypotheses are not necessary. 13. See Figure 5-5d. 14. Answers may vary.
ln (1.09 50 − 1) − ln 50 – ln (ln 1.09) ln 1.09 = 32.893K years.
f (x )
t=
x a
This time is not halfway between 0 and 50. 1– t 43 1 + t , if t ≤ 1 12. d (t ) = 1 200 1 – , if t ≥ 1 t The hypotheses of the mean value theorem do not apply on any interval that contains t = 1 as an interior point, such as [0, 2] and [0.5, 2], because d is not differentiable there. The hypotheses do apply on any interval not containing 1 and on Calculus Solutions Manual © 2005 Key Curriculum Press
15. Answers may vary.
b
16. Answers may vary. f (x )
f (x )
a
b
x
d x a
Problem Set 5-5
b
89
f ′ (x) never equals 0.
17. Answers may vary.
f ′ (x) never equals 0. f (x )
f (x )
f (x ) 5
1
x 1
x
x a
c
5
b
18. Michel Rolle (1652− 1719) lived in France. Sources will vary. 19. f (x) = x 2 − 4x 20. f (x) = x 2 − 6x + 5 f (1) = − 3 ≠ 0 f (2) = − 3 ≠ 0 Conclusion is not true. Conclusion is not true. f ′ (2) = 0, but 2 is not f ′ (3) = 0, but 3 is not in the interval (0, 1). in the interval (1, 2). f (x )
f (x ) x 0
2
27. f (x) = 1 − (x − 3)2/3 28. f ( x ) =
f is not differentiable at x = 3. Conclusion is not true. f ′ (x) never equals 0.
x
1
1
x 3 – 6 x 2 + 11x – 6 x–2 f is not continuous or differentiable at 2. Conclusion is not true. There is no point at x = 2 to draw the tangent line. f (x )
f (x )
2 1
1
–3
x
–3 1
3
x
21. f (x) = x − 4x 22. f (2) = − 4 ≠ 0 Conclusion is not true. f ′ (2) = 0, but 2 is not in the open interval (0, 2). 2
f (x) = x − 6x + 5 f (4) = − 3 ≠ 0 Conclusion is true. f ′ (3) = 0, and 3 is in the interval (1, 4). f (x )
f (x ) 1
x
0
1
c 1
2
x 4
23. f (x) = x 2 − 4x f (3) = − 3 ≠ 0
24. f (x) = |x − 2| − 1 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). c
x 3 – 7 x 2 + 13 x – 6 x–2 ( x – 2)( x 2 – 5 x + 3) = = x 2 − 5 x + 3, x ≠ 2 x–2 Thus, g is discontinuous at x = 2, and the hypotheses of the mean value theorem are not met. 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. The conclusion is true for (1, 5), because the slope of the secant line is 1, and g ′ (x) = 1 at x = 3, which is in the interval (1, 5). See the right graph.
29. g( x ) =
g (x )
g (x )
3
3 1
2
3
x
1
2
x 3
5
x 3
1
x 1
3
–3
25. f (x) = 1/x f (0) does not exist. Conclusion is not true. 90
No point of tangency
4
f (x )
f (x )
0
2
2
Problem Set 5-5
26. f (x) = x − [x] f is discontinuous at 1 and 2. Conclusion is not true.
30. h (x) = x 2/3 ⇒ 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 is infinite. The hypotheses of the mean value theorem are met on the interval [0, 8], because the function need not be differentiable at an endpoint. The hypotheses are not met on [− 1, 8], because the point x = 0 where h is not differentiable is in the open interval (− 1, 8). To see if the conclusion of the mean value theorem
Calculus Solutions Manual © 2005 Key Curriculum Press
is true anywhere, find the slope of the secant line (see next graph). 4 –1 m= = 1/3 8 – (–1) 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 is not true because 8 is at the endpoint of the interval, not in the open interval (− 1, 8).
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) = 0 Because the derivative at x = 5 is 0, the tangent line at x = 5 is horizontal. This is consistent with x = 5 being a high point on the graph. c.
h(x)
m (x )
y1
4 20
y2 x 4
x –1
8
3 x – 3, if x ≥ 3 31. a. f ( x ) = x + 3, if x < 3 f (x )
6
x 3
b. f is continuous at x = 3 because the right and left limits both equal 6. f is not differentiable at x = 3 because the left limit of f ′ (x) is 1 and 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 line has slope either 1 or 3, and thus is never 11/5. d. f is integrable on [1, 6]. The integral equals 41.5, the sum of the areas of the two trapezoids shown in this diagram. 15
f (x )
6 4
x 1
3
6
32. a. f (t) = number of miles in t hours t = number of hours driven For the mean value theorem to apply on [a, b], f must be differentiable on (a, b) and continuous at t = a and t = b. b. The 60 mi/h equals the slope of the secant line. Therefore, there must be a tangent line at some value t = c in (a, b) with slope equal to 60. This tangent line’s slope is the instantaneous speed at t = c. Therefore, the speed was exactly 60 at some time between t = a and t = b, Q .E.D . Calculus Solutions Manual © 2005 Key Curriculum Press
x
m (x)
3.0 2 3.5 6.8333… 4.0 1 4.5 16.5 5.0 no value
x
m (x)
5.5 6.0 6.5 7.0
− 16.5 −1 − 6.833… −2
The difference quotient is positive when x is less than 5 and negative when x is greater than 5. d. In the proof of Rolle’s theorem, the left limit of the difference quotient was shown to be positive or zero and the right limit was shown to be negative or zero. The unmentioned hypothesis is differentiability on the interval (a, b). The function f is differentiable. Because there is a value of f ′(5), both the left and right limits of the difference quotient must be equal. This number can only be zero, which establishes the conclusion of the theorem. The conclusion of Rolle’s theorem can be true even if the hypotheses aren’t met. For instance, f (x) = 2 + cos x has zero derivatives every π units of x, although f (x) is never equal to zero. f ( 4.5) – f (2) 5.5 – 4 = = 0.6 2.5 2.5 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 x Using the solver feature, f ′(c) = 0.6 at c = 2.0406… , 2.9593… , and 4.0406… , all of which are in (2, 4.5).
34. a. m =
Problem Set 5-5
91
c. h (x) = f (x) − g (x)
∴ 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 theorem state that f should be differentiable on the open interval (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 because both f and g are given to be differentiable, and a linear combination of differentiable functions is also differentiable. b. By the mean value theorem, there is a number c in (a, b) for which
y y
2
5
y1 x 2
y3
For c1 = 2.0406… : x
h(x)
1.7906… 1.8406… 1.8906… 1.9406… 1.9906… 2.0406… 2.0906… 2.1406… 2.1906… 2.2406… 2.2906…
–0.2925… –0.1865… –0.1022… –0.0411… –0.0041… 0.0081… –0.0039… –0.0397… –0.0977… –0.1760… –0.2723…
For c = 2.9593… : For c = 4.0406… : x 2.7093… 2.7593… 2.8093… 2.8593… 2.9093… 2.9593… 3.0093… 3.0593… 3.1093… 3.1593… 3.2093…
h( 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. D – D1 ∴ h ′( c ) = 2 b–a c. If D1 ≠ D2, then h ′ (c) ≠ 0. But h ′ (x) = f ′ (x) − g ′ (x) by the derivative of a sum, and thus h ′(x) = 0 for all x in the domain. ∴ h ′(c) = 0, which contradicts h ′(c) ≠ 0. So the supposition that D1 ≠ D2 is false, meaning that D1 and D2 are equal, Q.E.D. h ′( c ) =
h(x)
x
h(x)
–1.3274… –1.4237… –1.5021… –1.5601… –1.5959… –1.6081… –1.5958… –1.5589… –1.4979… –1.4136… –1.3077…
3.7906… 3.8406… 3.8906… 3.9406… 3.9906… 4.0406… 4.0906… 4.1406… 4.1906… 4.2406… 4.2906…
0.5075… 0.6134… 0.6977… 0.7588… 0.7958… 0.8081… 0.7960… 0.7602… 0.7022… 0.6239… 0.5276…
37. By the definition of antiderivative (indefinite integral), g( x ) = 0 dx if and only if g ′ (x) = 0.
∫
Any other function f for which f ′ (x) = 0 differs from g (x) by a constant. Thus the antiderivative of zero is a constant function, Q.E.D. 38. f (x) = (cos x + sin x)2, and g (x) = sin 2x y f 1
x 1
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 value theorem, 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. 92
Problem Set 5-5
g
x
f (x)
g (x)
0 1 2 3 4
1 1.9092… 0.2431… 0.7205… 1.9893…
0 0.9092… − 0.7568… − 0.2794… 0.9893…
In each case, f (x) = g (x) + 1. Calculus Solutions Manual © 2005 Key Curriculum Press
Proof: (cos x + sin x) = cos x + 2 cos x sin x + sin x = 2 cos x sin x + 1 = sin 2x + 1, Q .E .D . 39. The hypotheses of Rolle’s theorem say that f is 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 of continuity at a and at b allows you to conclude that the function is continuous on the closed interval [a, b]. 40. The intermediate value theorem applies to continuous functions, whereas the mean value theorem applies to differentiable functions. Both are existence theorems, concluding that there is a value x = c in the open interval (a, b). For the intermediate value theorem, f (c) equals a pre-selected number v between f ( a) and f ( b). For the mean value theorem, f ′ (c) equals the slope of the secant line connecting (a, f ( a)) and (b, f ( b)). 41. Answers will vary. 2
2
Problem Set 5-6 Q1. Q2. Q3. Q4. Q5. Q6. Q7. Q8. Q9. Q10.
r′(x) = m(x) See the text for the definition of derivative. Increasing at 6 units/unit dy = sec x tan x dx y′ = 8x(x 2 + 3)3 d2z/dz2 = −25 sin u f ′(x) = 0 4.5 See Figure 5-5b. E
1. a. I =
9
∫ 10 x
−1.5
dx
= (−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. y
1
x 9
c. Pick sample points at left ends of subintervals for U5 and at right ends for L5.
Calculus Solutions Manual © 2005 Key Curriculum Press
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
1
5
0.89442719…
2
6
0.68041381…
3
7
0.53994924…
4
8
0.44194173…
5
9
0.37037037…
Sum =
f ( x)
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 the graph 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 =
∫
1.5
0
sin x dx = − cos 1.5 − ( − cos 0)
= 0.92926279… Using sample points at the midpoints, M10 = 0.93013455…
4
4
k 2
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 point is at the midpoint of the subinterval, the “triangles” have equal bases. Because the graph is concave down, the “triangle” below the horizontal line has a larger altitude, and thus a larger area, than the one above the line. So the rectangle includes more area on the left
Problem Set 5-6
93
than it leaves out on the right, and thus overestimates the integral. y Rectangle includes more area.
x
3. See the statement of the fundamental theorem in the text. 4. See the work in the text preceding the proof of the fundamental theorem for a derivation of a Riemann sum that is independent of the number of 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,
A(u + ∆u) − A(u) < h(u + ∆u) ∆u But the limits of h(u) and h(u + ∆u) both equal h(u) because h is continuous and h(u) is independent of ∆u. Therefore, by the squeeze theorem, A(u + ∆u) – A(u) lim = h(u). But the limit on ∆u→0 ∆u 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 M 10 = 12.667… . (Note also that A(9) < M10, which is expected because the graph of h is concave down.) 9. a. Answers may vary. b. Answers may vary. c. h(u) <
f (x )
f (x )
then the exact integral equals g(b) – g( a) ⋅ (b − a), which equals g(b) − g(a). (b – a) Distance =
8
∫ [100 – 20(t + 1)
1/ 2
] dt
c. Answers may vary.
0
d. Answers may vary. f (x )
f (x )
8
40 (t + 1)3/ 2 3 0 40 3/ 2 40 3/ 2 = 800 − ( 9) − 0 + (1) 3 3 1 = 453 ft 3 8. a. h(x) = x 1/2 = 100t −
x
x
7. v(t) = 100 − 20(t + 1)1/2
x
x
e. Answers may vary.
f. Answers may vary. f (x )
f (x )
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 lower sum and an upper sum, respectively, because h(x) is increasing. ∴ h(u)∆u < A(u + ∆u) – A(u) < h(u + ∆u)∆u
94
Problem Set 5-7
x
x
10. Answers will vary.
Problem Set 5-7 Q1.
x6 + C
1 6
Q2.
Q3. − 13 x −3 + C Q5.
Q4.
sin 5 x + C
1 5
Q6. x + C
Q7. tan x + C
Q8. y″ = −1/x 2
Q9. definite 1.
∫
4
∫
5
1
2.
2
1 x dx = x 3 3
4
2
x 3 dx =
6 1 18 (3 x + 7) + C 6 1 6 sin x + C
1 4 x 4
1 5 2
Q10. indefinite 1 1 = (64) − (1) = 21 3 3 =
1 1 609 1 (625) − (16) = = 152 4 4 4 4
Calculus Solutions Manual © 2005 Key Curriculum Press
3.
4.
3 1 (1 + 3 x )3 −2 9 −2 1 1 = (1000) − ( −125) = 125 9 9
∫
∫
3
(1 + 3 x )2 dx =
4
1 (5 x – 2) dx = (5 x – 2)3 −1 15
∫ 60 x
2/3
1
4
6.
∫ 24 x
8.
3/ 2
10.
11.
8
2
2
∫
50
∫
0
dx = x
50
−1
= 9.6(32 − 1) = 297.6
= 40 − 10 = 30
= 50 − 20 = 30
( x 2 + 3 x + 7) dx =
0 −2
1 3 x + 2 x 2 + 10 x −3 3 = 0 − (−9 + 18 − 30) = 21 1 1 1 4 x + 5 dx = ( 4 x + 5)1/ 2 ( 4 dx ) −1 4 −1 ( x 2 + 4 x + 10) dx =
∫ ∫
3
1 2 x + 10 dx = 2
−3
1 −1
∫
3
−3
∫ ∫
=
1 13 1 (27 – 1) = =4 6 3 3
π /2
0
15.
∫
π /3
π /6
−π / 2
6 cos x dx = 12 sin x
∫
22.
∫
23.
∫
5
−5
24.
∫
1
−1
π /2
= 12(1) − 0 = 12
∫
−1
5
x
−2
5
∫ 2 dx 0
= 20 − 0 = 20
0
(cos x + 10 x 3 – tan x ) dx = 2
1
1 0
1
∫ cos x dx 0
= 2 sin 1 − 2 sin 0 = 1.682941K
dx has no value because y = x− 2 has a
vertical asymptote at x = 0, which is within the interval. 26.
∫
2
−2
x dx has no value because the integrand is
not a real number for negative values of x. 27. Integral = −(area) 28. Integral = area f (x )
f (x )
(sec x + cos x ) dx π /6
1 4 2 sin x 4 1
( x 7 – 6 x 3 + 4 sin x + 2) dx = 2
= 2 sin x
3
2
= tan x + sin x
= − e − ln 3 + e 0
0.2
= 2( 2 x )
= 6(1) − 6( −1) = 12
0
π /3
ln 3 0
1 2 ln 4 1 0 e − e = 7.5 2 2
0.2 1 cos 3 x dx = sin 3 x 0.1 3 0.1 1 = (sin 0.6 – sin 0.3) = 0.0897074 K 3 0.4 0.4 1 sin 2 x dx = − cos 2 x 0 2 0 1 = − (cos 0.8 – cos 0) = 0.1516466K 2
21.
25.
π /2
or: Integral of an even function between symmetric limits.
∫
0
=
1 (sin 4 2 – sin 4 1) = 0.045566 K 4 3 3 1 20. (1 + cos x ) 4 sin x dx = − (1 + cos x )5 −3 5 −3 1 1 = − (1 + cos 3)5 + [1 + cos(–3)]5 = 0 5 5 or: Integral equals zero because an odd function is integrated between symmetric limits.
(2 x + 10) (2 dx )
6 cos x dx = 6 sin x
π /2
e − x dx = − e − x
ln 4
sin 3 x cos x dx =
1/ 2
0
−π / 2
2
1
4 sin x dx = −4 cos x = −4( −1) + 4(1) = 8
0
14.
∫
2
π
π
1 2x e 2
e 2 x dx =
1 2 = − +1 = 3 3
−3
3 1 2 1 56 2 = ⋅ (2 x + 10)3/ 2 = (64 – 8) = = 18 2 3 3 3 3 −3
13.
0
0
∫
1 2 ⋅ ( 4 x + 5)3/ 2 4 3
∫
ln 3
1 − 1 + 1 = 1.5 2
∫
20
0
18.
π /3
=
1 3 3 2 x + x + 7x −2 3 2 8 32 2 = 0 − − + 6 − 14 = = 10 3 3 3
∫
∫
ln 4
0
1
∫ 5 dx = 5x
= 12.
17.
19.
4
(sec x tan x + sin x ) dx = sec x − cos x
0
= 36(32 − 1) = 1116
5/ 3
dx = 9.6 x 5/ 2 8
20
9.
4
1
1
7.
dx = 36 x
8
π /3
=2−
1 6175 1235 2 [5832 – (–343)] = = = 411 15 15 3 3
8
∫
0
2
= 5.
16.
6
x
1
x
= 3 + 3 /2 − 1/ 3 − 1/2
-5
5
7
= (7/6) 3 − 1/2 = 1.52072K Calculus Solutions Manual © 2005 Key Curriculum Press
Problem Set 5-7
95
29. Integral ≠ area
30. Integral ≠ area
f (x )
39. Statement: “If f ( x) < g(x) for all x in [a, b],
f (x )
1
then
1
a
7
x 8
x
“If
12
∫
a
b
32.
f ( x ) dx = −
b
b
a
a
c
a
b
∫ g( x ) dx = ∫ g( x ) dx + ∫ g( x ) dx a
= 12 + 13 = 25 34.
∫
c
∫
c
∫
b
∫
b
b
∫ g( x ) dx. ” a
b
b
∫ g( x ) dx,
f ( x ) dx <
a
then f ( x) < g(x) for all x in [a, b].” The converse can be shown to be false by any counterexample in which the area of the region under the g graph is greater than the area under the f graph, but the g graph touches or crosses the f graph somewhere in [a, b]. One counterexample is f ( x) = 1.5 and g(x) = 2 + cos x on [0, 2π].
f ( x ) dx = 4(7) = 28
b
∫
a
f ( x ) dx = −7
a
∫ 4 f ( x ) dx = 4∫ c
33.
∫
b
f ( x ) dx <
Converse:
0
31.
∫
b
y 3
g
f ( x ) dx cannot be determined.
f
a
35.
a
36.
a
a
x
f ( x ) dx +
c
∫ g( x ) dx cannot be determined. a
[ f ( x ) + g( x )] dx =
40.
∫
4
1
f ( x ) dx +
2π
0
b
∫ g( x ) dx = 7 + 12 = 19 a
37. y
x 2 dx =
1 3 x +C 3
4 1
1 1 = ( 4 3 ) + C − (13 ) − C = 21 3 3 The two C’s will always cancel, so it is not necessary to write them.
f (3) = 7 7
y = f(x )
Problem Set 5-8 Q1. 30x2.4 + C Q2. 30(42.4 − 1) = 805.72…
y = f' (x) x 5
10
Q3. y′ = −1/ 1 – x 2 Q4. f ′ (x) = 3x2 sin x + x3 cos x Q5. f (x ) and f' (x )
38.
f y 8
f (1) = 8
1
f'
x
1
y = f (x )
Q6. Yes, continuous Q7. Increasing at x = 7 x 7
y = f' ( x )
Q8. f ( a) = f ( b) = 0 Q9. v(9) = 450 ft/s Q10. a(9) = 25 (ft/s)/s
96
Problem Set 5-8
Calculus Solutions Manual © 2005 Key Curriculum Press
1. a.
3. a. v
y
100
(x , y)
(t, v )
dx
5
50
x
dt 2
t t=a
1
t=b
b. dA = y dx = 10e0. 2x dx
dy = v dt = (55 + 12t0.6 ) dt b. Displacement = 1
∫ (55 + 12t
0.6
0
b
∫ (55 + 12t
∫
c. 0.6
0
) dt
2
a
) dt = 55t + 7.5t
1
∫
1
2
= 110 + 22.735… − 55 − 7.5 ≈ 70.2 mi 2
c.
∫ (55 + 12t
0.6
0
) dt = 55t + 7.5t 1.6
b
∫ (55 + 12t 0
55t
b + 7.5t 1.6 0
y
dx
4
∫
= 300
y 300
x 4.134
e. v(4.13372…) = 55 + 12(4.13372…)0.6 = 83.1181… . At the end of the trip, you were going about 83 mi/h. 2. v
t 10
dy = v dt = 15e −0.1t dt 0
15e −0.1t dt = −150e −0.1t
20 0
= 129.6997… ft Calculus Solutions Manual © 2005 Key Curriculum Press
x
6
6
The area of the circumscribed rectangle is 6 ⋅ 9 = 54. The area of the parabolic region is two-thirds this area. 5. a. dA = [x + 2 − (x2 − 2x − 2)] dx The top and bottom of the strip are not horizontal, so the area of the strip is slightly different from dA. As dx approaches zero, the differences in height at different values of x in the strip become smaller, so the difference between dA and the area of the strip gets smaller. b. y1 = y2 ⇒ x + 2 = x 2 − 2x − 2 ⇒ 0 = x 2 − 3x − 4 ⇒ 0 = (x − 4)(x + 1) ⇒ x = 4 or x = −1 4 4 1 3 ( − x 2 + 3 x + 4) dx = − x 3 + x 2 + 4 x −1 3 2 −1 1 3 3 2 = − ⋅4 + ⋅4 + 4⋅4 3 2 1 3 − − ⋅ ( −1)3 + ⋅ ( −1)2 + 4 ⋅ ( −1) 2 3 125 = = 20.83333K 6
∫
( t, v )
dv
∫
3
1 1 (6 x − x 2 ) dx = 3 x 2 − x 3 = 3 ⋅ 6 2 − ⋅ 6 3 = 36 0 3 0 3
55b + 7.5b = 300 b ≈ 4.13372… ≈ 4.134 h
20
( x, y )
dA = (6x − x2 ) dx
) dt = 300
1.6
10
dx = 24.59123K
2 0
0.6
0.2 x
4.
= 110 + 22.735… − 0 − 0 = 132.735… , which equals the sum of the two integrals above. d.
= 50e 0.4 − 50
The region is approximately a trapezoid with height 2 and bases 10 and y(2). y(2) = 14.9182… , so the area of the trapezoid is 2/2(10 + 14.9182…) = 24.9182… .
0
1
2 0
0
1.6
(55 + 12t 0.6 ) dt = 55t + 7.5t 1.6
10e 0.2 x dx = 50e 0.2 x
∫ 10e
d.
= 55 + 7.5 − 0 − 0 = 62.5 mi 2
2
= −150e −2 + 150
c. R100 = 20.834375 (Checks)
Problem Set 5-8
97
6.
9. a. y
T
( x, y )
1
1
20 (x, T )
dx ( x, y )
dx
x
2
0.5
π⁄4
The curves intersect at x = π /4. dA = (cos x − sin x) dx
∫
π /4
0
b. dD = T dx = [20 − 12 cos 2π(x − 0.1)] dx
(cos x − sin x ) dx = (sin x + cos x )
π /4
c. D =
0.5
0
[20 – 12 cos 2π ( x – 0.1)] dx 0.5 0
= 10 − (6 /π) sin 0.8π − 0 + (6 /π) sin (−0.2π) = 7.75482… ≈ 7.75 degree-days
= 2 −1 7. a.
d. From noon to midnight,
F
D=
(x , F )
9
b. dW = F dx = 0.6x dx
∫
9
0
0.5
[20 – 12 cos 2π ( x – 0.1)] dx
0.6 x dx = 0.3 x 2
9 0
= 24.3 − 0 = 24.3 inch-pounds c. The region under F from x = 0 to x = 9 is a triangle with base 9 and height F(9) = 5.4. So the area is 1/2 ⋅ 9 ⋅ 5.4 = 24.3. d. dW is found by multiplying F by dx. F is measured in pounds, and dx is measured in inches, so the units of dW are (pounds)(inches), or inch-pounds.
1 0.5
= 20 − (6 /π) sin 1.8π − 10 + (6 /π) sin 0.8π = 12.24517… ≈ 12.25 degree-days The total number of degree-days is D = 7.75482… + 12.24517… = 20 degree-days. Note that this answer can be found more easily by observing that in one full cycle of a sinusoid, there is just as much area above the sinusoidal axis, T = 20, as there is below it. So the average temperature difference for the day is 20 degrees, making the number of degree-days for one day equal to 20.
x 0
∫
1
= 20 x − (6/π ) sin 2π ( x − 0.1)
dx
W=
∫
= 20 x − (6/π ) sin 2π ( x − 0.1)
0
6
x 1
10. a. C 1000
8. F
dT
( x, F )
T
30 10
dx
30
x
dH = C dT = (−0.016T 3 + 0.678T 2 + 7.45T + 796) dT
5
dW = 50 cos
∫
10
50 cos
0
π x dx 20
1000 π π x dx = sin x 20 20 π
H= 10 0
1000 1000 π 1000 sin − sin 0 = 2 π π π = 318.3098… The midpoint Riemann sum R100 gives 318.313… , which is close to the answer found using integration. =
98
Problem Set 5-8
∫
30
10
(–0.016T 3 + 0.678T 2
+ 7.45T + 796) dT = −0.004T 4 + 0.226T 3 + 3.725T 2 30 + 796T 10 = −3240 + 6102 + 3352.5 + 23880 + 40 − 226 − 372.5 − 7960 = 21,576 Btu b. (2000)(21576) = 43,152,000 Btu The property is the integral of a constant
Calculus Solutions Manual © 2005 Key Curriculum Press
times a function. That is,
∫
30
10
2000
∫
The graph intersects the x-axis at x = 1 and x = 5. y = −x 2 + 6x − 5 = −(x − 1) ( x − 5) = 0 ⇒ x = 1, 5, which confirms the graphical solution. dA = (y − 0) dx = (−x2 + 6x − 5) dx
2000C dT =
30
C dT .
10
11. a.
P
A=
∫
5
1
(x , P ) 1000
=−
dx x
0
100
200
1 (– x 2 + 6 x – 5) dx = − x 3 + 3 x 2 − 5 x 3
y
2
C=
∫
b
0
∫
100
0
(100 + 0.06 x 2 ) dx = 100 x + 0.02 x 3
(x, y ) –6
The graph intersects the x-axis at x = −2 and x = 3. y = x 2 − x − 6 = (x + 2)(x − 3) = 0 ⇒ x = −2, 3, which confirms the graphical solution. dA = (0 − y) dx = (−x2 + x + 6) dx A=
= 100(200) + 0.02(2003) − 100(100) − 0.02(1003) = 150000
∫
200
0
P dx =
∫
100
0
P dx +
∫
1 1 (– x 2 + x + 6) dx = − x 3 + x 2 + 6 x −2 3 2 9 8 5 + 18 − − 2 + 12 = 20 2 3 6
(x , y )
(0, y )
100
x –2
The graph intersects the y-axis at y = 1 and y = 4. x = (y − 1) ( y − 4) = 0 ⇒ y = 1, 4, which confirms the graphical solution. dA = (0 − x) dy = −(y − 1) ( y − 4) = (−y2 + 5y − 4) dy A=
∫
4
1
=− 5
1 5 (– y 2 + 5 y – 4) dy = − y 3 + y 2 − 4 y 3 2
1
y (x, y )
x
x (x, 0)
4
64 1 5 1 + 40 − 16 + − + 4 = 4 3 3 2 2
(0, y )
1
−2
4
P dx,
16.
(x , y )
3
y
15.
200
y 4
3
= −9 +
which shows that the sum of integrals with the same integrand applies. 12. Using trapezoids, the area is approximately 10(0/2 + 38 + 50 + 62 + 60 + 55 + 51 + 30 + 3/2) = 3475 ft2. The fundamental theorem cannot be used because the function is specified only by data, not by an equation whose antiderivative can be found. Plan of attack for area problems: • Do geometry to get dA in terms of sample point (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.
∫
200 100
Thus,
x 3
b
= 100b + 0.02b 3 − 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 be 180,000 − 30,000 = $150,000. As a check, 200
(x, 0)
–2
(100 + 0.06 x 2 ) dx = 100 x + 0.02 x 3
1
125 1 2 + 75 − 25 + − 3 + 5 = 10 3 3 3
14.
b. dC = P dx = (100 + 0.06x ) dx
5
5
5 –1
Calculus Solutions Manual © 2005 Key Curriculum Press
Problem Set 5-8
99
The graph intersects the y-axis at y = −1 and y = 5. x = 5 + 4y − y 2 = (1 + y) ( 5 − y) = 0 ⇒ y = −1, 5, which confirms the graphical solution. dA = (x − 0) dy = (5 + 4y − y2) dy A=
∫
5
1 (5 + 4 y – y 2 ) dy = 5 y + 2 y 2 − y 3 −1 3
= 25 + 50 −
The graphs intersect at x = −2 and x = 2. 0.5x 2 + 2x = −x 2 + 2x + 6 ⇔ 1.5x 2 = 6 ⇔ x = −2, 2, which confirms the graphical solution. dA = (y2 − y1) dx = (−1.5x2 + 6) dx A=
5
∫
2
−2
(–1.5 x 2 + 6) dx = −0.5 x 3 + 6 x
2 −2
= −4 + 12 − 4 + 12 = 16
−1
20.
125 1 + 5 − 2 − = 36 3 3
y
17. 5
y
(x, y1 )
(x, y2)
x
2 0
(x, y2 )
x 4
–1
The graphs intersect at x = 0 and x = 5. 0.2x 2 + 3 = x 2 − 4x + 3 ⇔ 0.8x 2 − 4x = 0 ⇔ 0.8x(x − 4) = 0 ⇔ x = 0, 5, which confirms the graphical solution. dA = (y1 − y2) dx = (−0.8x2 + 4x) dx
(x, y1 )
The graphs intersect at x = −1 and x = 4. x 2 − 2x − 2 = x + 2 ⇔ x 2 − 3x − 4 = 0 ⇔ x = −1, 4, which confirms the graphical solution. dA = (y2 − y1) dx = (−x2 + 3x + 4) dx A=
∫
4
−1
5
1 3 ( − x 2 + 3 x + 4) dx = − x 3 + x 2 + 4 x 3 2
A=
4 −1
64 1 3 5 =− + 24 + 16 − − + 4 = 20 3 3 2 6
∫
5
0
=−
(–0.8 x 2 + 4 x ) dx = −
4 3 x + 2x2 15
5 0
500 2 + 50 + 0 − 0 = 16 15 3
21.
18.
y y
(x, y1 )
10 2 (x, y1)
4
x
x 0
–2 (x, y2 )
The graphs intersect at x = −2 and x = 4. −2x + 7 = x 2 − 4x − 1 ⇔ x 2 − 2x − 8 = 0 ⇔ (x + 2)(x − 4) = 0 ⇔ x = −2, 4, which confirms the graphical solution. dA = (y1 − y2) dx = (−x2 + 2x + 8) dx A=
∫
4
1 (– x 2 + 2 x + 8) dx = − x 3 + x 2 + 8 x −2 3
5
(x , y 2 )
The graphs intersect at x = 0 and x = 5. dA = (y1 − y2) dx = (2e0. 2x − cos x) dx A=
∫
5
0
5
(2e 0.2 x − cos x ) dx = 10e 0.2 x − sin x
0
= 10e − sin 5 − 10 + 0 = 18.1417… 22.
4
y
−2
(x, y2 )
64 8 =− + 16 + 32 − − 4 + 16 = 36 3 3
(x, y1 )
19.
x
1
y
(x , y 2 )
6
dA = (y2 − y1) dx = (e2x − sec2 x) dx ( x, y1 )
x –2
2
A=
∫
1
0
1
(e 2 x – sec 2 x ) dx = 0.5e 2 x − tan x 0
= 0.5e − tan 1 − 0.5 + 0 = 1.6371… 2
100
Problem Set 5-8
Calculus Solutions Manual © 2005 Key Curriculum Press
23.
A=
y
∫
2
−1
( x 3 – 3 x 2 + 4) dx =
1 4 x − x3 + 4x 4
2 −1
4
= 4−8+8−
(x 1 , y )
( x 2 , y)
1 3 −1+ 4 = 6 4 4
26.
1
x –2
2
y
The graphs intersect at y = 1 and y = 4. Write y = x + 3 as x = y − 3. y − 3 = −y 2 + 6y − 7 ⇒ y 2 − 5y + 4 = 0 ⇒ (y − 1)(y − 4) = 0 ⇒ y = 1, 4, which confirms the graphical solution. dA = (x2 − x1) dy = (−y2 + 5y − 4) dy A=
4
∫ (– y
2
1
=−
1 5 + 5 y – 4) dy = − y 3 + y 2 − 4 y 3 2
4 1
64 1 5 1 + 40 − 16 + − + 4 = 4 3 3 2 2
24. 5
y
(x1 , y)
(x2 , y)
x
–5
∫
−5
=−
x –1
(–0.25 y 2 + 6.25) dy = −
1 3 25 y + y 12 4
125 125 125 125 2 + − + = 41 12 4 12 4 3
5 −5
8
The graphs intersect at x = −1 and x = 8. x 2/3 = (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 step of squaring both sides. So x = −1, 8. dA = (y 2 − y 1) dx = [(x + 1)1/2 + 1 − x 2/3] dx
=
The graphs intersect at y = −5 and y = 5. Write y = −2x 1 + 11 as x 1 = 5.5 − 0.5y. 5.5 − 0.5y = 0.25y 2 − 0.5y − 0.75 ⇒ 0.25y2 = 6.25 ⇒ y = −5, 5, which confirms the graphical solution. dA = (x1 − x2) dy = (−0.25y2 + 6.25) dy 5
(x, y1 )
A=
8
A=
(x, y2 ) 2
∫
8
−1
[( x + 1)1/ 2 + 1 – x 2 / 3 ] dx
2 3 ( x + 1)3/ 2 + x − x 5/ 3 3 5
8 −1
96 3 1 − 0 +1− = 7 5 5 5 27. Wanda: You can always tell the right way because the altitude of the strip should be positive. 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). = 18 + 8 −
y line
25. y 4
curve x
(x , y1 )
x –1
2
(x , y 2 )
The graphs intersect at x = −1 and x = 2. x 3 − 4x = 3x 2 − 4x − 4 ⇒ x 3 − 3x 2 + 4 = (x + 1) ( x − 2)2 = 0 ⇒ x = −1, 2, which confirms the graphical solution. dA = (y1 − y2) dx = (x3 − 3x2 + 4) dx
Calculus Solutions Manual © 2005 Key Curriculum Press
For curve minus line, you’d get the opposite of the right answer. Note that if you slice horizontally, it would be curve minus line. 28. a. Peter: Horizontal slicing would be awkward because for some values of y the length of the strip would be given by line minus curve, but in others it would be boundary minus curve, and yet elsewhere it would be curve minus curve. If you use vertical slices, the length
Problem Set 5-8
101
For y = 7 cos 5x, width is 1/5 as much and altitude is 7 times as much. ∴ A = (2)(1/5)(7) = 2.8
of the strip will always be line minus curve. (See graphs.) y
y (x, y1)
31.
x
y
x 8
(x, y2 )
(x, y 2) (x, y1 )
b. Peter: In the graph on the right, y1 − y2 will be positive. Because y2 is negative, you will get (pos.) − (neg.), which is equivalent to (pos.) + (pos.). Thus, the altitude for the strip is positive.
–2
x 3
The graphs intersect at x = −2 and x = 3. dA = (y2 − y1) dx = (−x2 + x + 6) dx
29.
A=
y (x, ah2)
∫
3
1 1 (– x 2 + x + 6) dx = − x 3 + x 2 + 6 x −2 3 2
3 −2
9 8 5 + 18 − − 2 + 12 = 20 = 20.8333K 2 3 6 R 10 = 20.9375 R100 = 20.834375 R1000 = 20.83334375 The Riemann sums seem to be approaching the exact answer. = −9 +
x –h
h (x , y )
The graph shows the parabolic region from x = −h to x = h and a strip from the graph to a horizontal line at y = ah2. dA = (ah2 − y) dx = (ah2 − ax2) dx h h 1 A = ( ah 2 – ax 2 ) dx = 2 ah 2 x − ax 3 0 −h 3 1 4 = 2 a h 3 − h 3 = ah 3 3 3 Area of rectangle = 2h(ah2) = 2ah3 area of region ( 4/3)ah 3 2 ∴ = = , Q. E . D . area of rectangle 2 ah 3 3 2 The graph shows y = 67 − 0.6x and the line y = 7, with a circumscribed rectangle.
32. t (x )
∫
67
∫
0
sin x dx = − cos x
π
= −( −1) + 1 = 2, which
0
is a rational number. y 1
y = cos x
3π
4π
t′(x) = 1 + cos x t′(x) = 0 ⇒ cos x = −1 ⇒ x = π + 2π n = … , π , 3π , 5π , … t′(x) is never negative, so t′(x) does not change signs. These points are plateau points.
Q1.
1 3 1 2 x + x + x+C 3 2
Q3. y′ =
2 −1/ 3 x 3
Q2.
4 7/4 x +C 7
1 Q4. − e −3 x + C 3
Q5. −csc x + C
Q6. x− 1
Q7. See Section 5-4. Q9.
Q8. mean value Q10.
y
7 y
y
y = 7 cos 5x (x, y )
x π
102
2π
Problem Set 5-9
x
7 = 67 − 0.6x 2 ⇒ 0.6x 2 = 60 ⇒ x = ±10 Rectangle has width 10 − (−10) = 20 and length 67 − 7 = 60. Area of region = 23 (20)(60) = 800. 30. dA = sin x dx π
x π
y
y=7
A=
10
Problem Set 5-9
x x –π/10
π/10
4
1
x 1
Calculus Solutions Manual © 2005 Key Curriculum Press
Plan of attack for volume problems: • Do geometry to get dV in terms of sample point (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 dy y = 9 − x2 ⇒ x2 = 9 − y dV = π (9 − y) dy
∫
V=
4
1
π ( 4 x − x 2 )2 dx = π
4
∫ (16 x
2
1
16 1 =π x3 − 2x4 + x5 3 5
4
– 8 x 3 + x 4 ) dx
= 30.6π = 96.132 …
1
The midpoint Riemann sum R100 gives 96.1341… , which is close to the answer found using integration. 5. y = x 1.5 is rotated about the x-axis. y 27
b. V =
9
∫ π (9 − y) dy = π (9 y − 0.5y )
9
2
0
0
= π (81 − 40.5) − π (0 − 0) = 40.5π = 127.2345…
(x, y )
x
c. R100 = 127.2345… (Checks.)
9
1
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 dy y = 10 − 2x ⇒ x = 5 − 0.5y ∴ dV = π (5 − 0.5y)2 dy V=
∫
10
0
dV = π y 2 dx = π x3 dx
2 π (5 − 0.5 y)2 dy = − π (5 − 0.5 y)3 3
10 0
2 2 250π = − π (0) + π (125) = = 261.7993… 3 3 3 1 1 V cylinder = π r2h, so of that is V = πr 2 h. 3 3 1 Here, r = 5 and h = 10, so V = π (52 )(10) = 3 250π , as found by integrating. 3 3. a. dV = π y 2 dx = π (3e− 0.2 x)2 dx = 9π e− 0.4 x dx 5
b.
∫ 9πe
−0.4x
0
dx = −22.5πe −0.4x
0
(x, y )
x 1
4
dV = π y 2 dx = π (4x − x2)2 dx Calculus Solutions Manual © 2005 Key Curriculum Press
∫
9
1
π x 3 dx =
1 4 πx 4
9
= 1640π = 5152.2119…
1
6. y = ln x ⇒ x = ey is rotated about the y-axis. y
1 (x, y )
x
1
dV = π x 2 dy = π e2y dy
5
= −22.5π e− 2 + 22.5π e0 = 61.1195… The midpoint Riemann sum R100 gives 61.1185… , which is close to the answer found using integration. c. Slice perpendicular to the axis of rotation, so slice vertically if rotating about the x-axis and horizontally if rotating about the y-axis. 4. y = 4x − x2 is rotated about the x-axis. y
V=
V=
∫
1
0
πe 2 y dy =
π 2y e 2
1 0
=
π 2 (e − 1) = 10.0359K 2
7. y = x 3/4 ⇒ x = y 4/ 3 is rotated about the y-axis. y
8 (x, y ) 1
x 1
16
dV = π x 2 dy = π y 8/3 dy 8 8 3 6141 V = πy 8/3 dy = π ⋅ y11/3 = π 1 11 11 1 = 1753.8654… 8. y = x14 and y = 8x2, intersecting at (0, 0) and (2, 16), are rotated about the y-axis. Area of cross section is π x12 − π x 22 .
∫
Problem Set 5-9
103
y
x 1 = y 1/ 4, and x 2 =
1 y 8
4
∴ dV = π
1 2 dy = π y1/2 − y dy 64
3
∫
V=
16
0
−
( x12
x 22 )
(x 1 , y )
1 2 π y1/2 − y dy 64
2 1 3 = π y 3/2 − y 3 192
16
64 π = 67.0206 K 3 0 The midpoint Riemann sum R100 gives V ≈ 67.0341… , which is close to the answer found using integration. 9. y1 = e0.4 x and y2 = x + 1, from x = 0 to x = 3, are rotated about the x-axis. Area of cross section is πy22 − πy12 . dV = π ( y22 − y12 ) dx = π [( x + 1) 2 − e 0.8 x ] dx 3
∫ π [( x + 1)
V=
(x2 , y)
=
− e 0.8 x ] dx
2
0
3
1 = π ( x + 1)3 − 1.25e 0.8 x 3
0
64 1 = π − 1.25e 2.4 − + 1.25 3 3 = π (22.25 − 1.25e 2.4 ) = 26.6125… ft 3 The midpoint Riemann sum R100 gives V = 26.6127… , which is close to the answer found using integration. 10. y1 = x1/3 and y2 = 10e− 0.1 x are rotated about the x-axis. Only the back half of the solid is shown.
x 1
dV = π ( x 22 − x12 ) dy = π [( 4 − y) − ( 4 − y) 2 ] dy = π (−y2 + 7y − 12) dy V=
4
∫ π (− y
2
3
+ 7 y − 12) dy
1 7 = π − y 3 + y 2 − 12 y 3 2
4 3
64 63 = π− + 56 − 48 + 9 − + 36 3 2 1 = π = 0.523598… 6 12. y = ax2 ⇒ x = (y/a)1/2, from (0, 0) to (r, h), is rotated about the y-axis. y h
(x, y )
x r
dV = π x 2 dy = π (y/a) dy = (π /a)y dy h h 1 1 V = (π /a) y dy = (π /a) ⋅ y 2 = (π /a)(h) 2 − 0 0 2 2 0
y 10
∫
(x, y2) (x, y1)
x 8
0
dV = π ( y22 − y12 ) dx = π (100e −0.2 x − x 2/3 ) dx V=
8
∫ π (100e
−0.2 x
0
= π ( −500e = π ( −500e
− x ) dx
(x, y )
− 0.6 x ) 5/3
8
4
0
+ 480.8) = 1193.3394 …
11. y = 4 − x 1 ⇒ x 1 = 4 − y, and y = 4 − x 22 ⇒ x2 = 4 – y , intersecting at x = 0 and x = 1, are rotated about the y-axis. Only the back half of the solid is shown.
104
y 2
2/3
x
−0.2 x
−1.6
Because y = ax2, h = ar2. 1 1 ∴ V = (π /a)( ar 2 )2 = π ar 4 2 2 Volume of circumscribed cylinder is V c = π r2h = π r2(ar2) = π ar4. Thus, the volume of the paraboloid is half the volume of the circumscribed cylinder, Q.E.D. 13. a. y = 0.3x1.5 is rotated about the x-axis.
Problem Set 5-9
dV = π y 2 dx = π (0.3x1.5 )2 dx = π (0.09x3) dx V=
∫
4
0
π (0.09 x 3 ) dx = 0.0225πx 4
4 0
= 5.76π = 18.09557…
Calculus Solutions Manual © 2005 Key Curriculum Press
b. R10 = 5.7312π R100 = 5.75971…π R1000 = 5.7599971…π Values are getting closer to V = 5.76π. 14. y = 4 − x 2 ⇒ x = (4 − y)1/2 dy Inner 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=
4
∫ π [6(4 − y)
1/2
0
1
− x 2 )2 dx
2/5
− 2 x 11/ 5 + x 4 ) dx
0
5 5 1 = x 7/5 − x 16/5 + x 5 7 8 5
1
= 0
81 = 0.2892 K 280
17. Cross sections perpendicular to y-axis are squares of edge 2x, where (x, y) is a sample point on the line in the xy-plane. 15
4
y y = –(15/4)x + 15
(x, y)
0
= π (0 − 16 + 8 + 32 + 0 − 0) = 24π = 75.3982… 15. y = 4 − x2 is rotated about the line y = −5. Only the back half of the solid is shown. y (x, y )
x
2
1/ 5
0
− 4 + y] dy
= π [ −4( 4 − y)3/2 − 4 y + 0.5 y 2 ]
4
1
∫ (x = ∫ (x
V=
x 4
Equation of line is 15 4 y = − x + 15 ⇒ x = 4 − y. 4 15 2 4 2 dV = (2 x ) dy = 4 4 − y dy 15 2
1 = 64 1 − y dy 15
y = –5
∫
15
0
dV = π[(y + 5)2 − 52] dx = π[(9 − x2)2 − 52] dx = π (56 − 18x2 + x4) dx V=
2
∫ π (56 − 18x
2
0
+ x 4 ) dx
= π (56 x − 6 x 3 + 0.2 x 5 )
2 0
= π (112 − 48 + 6.4 − 0 + 0 − 0) = 70.4π = 221.168… 16. Cross sections perpendicular to the x-axis are squares with side length (y2 − y1). The curves intersect at (0, 0) and (1, 1).
5
∫ π (1 − 0.08x 0
2
1
x
dV = (y2 − y1)2 dx = (x1/5 − x2)2 dx
Calculus Solutions Manual © 2005 Key Curriculum Press
0
+ 0.0016 x 4 ) dx
0.08 3 0.0016 5 = π x − x + x 3 5
1
3 15
= 320 cm3 The circumscribed rectangular box has volume l · w · h = 8 · 8 · 15 = 960 = 3V, so the pyramid is 1/3 the volume of the circumscribed rectangular solid, Q.E.D. The volume of a pyramid is one-third the volume of the circumscribed rectangular box, just as the volume of a cone is one-third the volume of the circumscribed cylinder. 18. Center line: y = 0.2x2 Upper bound: y = 0.16x2 + 1 Radius of circular cross section is 1 − 0.04x2. The tip of the “horn” is where 0.2x2 = 0.16x2 + 1 with x ≥ 0, which is at x = 5. dV = π (1 − 0.04x2)2 dx = π (1 − 0.08x2 + 0.0016x4) dx V=
y
2
1 1 64 1 − y dy = −320 1 − y 15 15
5 0
10 = π 5 − + 1 − 0 + 0 − 0 3 8 = π = 8.3775… ≈ 8.4 cm 3 3 Problem Set 5-9
105
19. a. y = x 0.6 Pick sample point (x, y) on the curve within the slice. One leg of the isosceles triangle is y, so the other leg is also equal to y. 1 1 dV = y 2 dx = x 1.2 dx 2 2 b. V =
∫
4
0
1 1.2 1 2.2 4 1 x dx = x = ⋅ 4 2.2 − 0 2 4.4 4 . 4 0
= 4.7982… The midpoint Riemann sum R100 gives 4.7981… , which is close to the answer found using integration. c. If the cross sections were squares, they would have twice the area of the triangles, so dV 1 would be twice as much and V = ⋅ 4 2.2 = 2.2 9.5964… . 20. y = ex, y = 3, and x = 0. Cross sections perpendicular to the x-axis are rectangles with height equal to 4 times the base. Each base has length (3 − y).
√3
60° 1 2b
b
1 1 3 bh dx = ( y2 − y1 ) ( y2 − y1 ) dx 2 2 2 3 3 = ( y2 − y1 )2 dx = (2 − x 2 − x 2 )2 dx 4 4 3 = (2 − 2 x 2 )2 dx 4 1 3 V= (2 − 2 x 2 )2 dx 0 4 3 1 = ( 4 − 8 x 2 + 4 x 4 ) dx 4 0 3 8 4 1 = 4x − x3 + x5 4 3 5 0 dV =
∫
∫
3 8 4 4− + −0+0−0 4 3 5 8 3 = = 0.9237... 15
=
22. y = ln x and y = 1. Cross sections perpendicular to the y-axis are rectangles with height equal to 1/2 the base. Each base has length x.
y 2
1 2b
x
1
5
dV = (3 − y)[ 4(3 − y)] dx
y
= (3 − e x )[ 4(3 − e x )] dx = 4(3 − e x )2 dx V=
∫
ln 3
0
4(3 − e x )2 dx = 4
∫
ln 3
0
(9 − 6e x + e 2x ) dx
ln 3
1 = 4 9 x − 6e x + e 2x 2 0
1 1 = 4 9 ln 3 − 6 ⋅ 3 + ⋅ 9 − 0 + 6 − 2 2 = 7.5500 … 21. y = x2 and y = 2 − x2, intersecting at x = 1. Cross sections perpendicular to the x-axis are equilateral triangles. Each base has length (y2 − y1).
y 1
1
x
x
1
dV = x ⋅ 12 x dy = 12 x 2 dy Solve y = ln x for x, to get x = e y. dV = 12 (e y )2 dy = 12 e 2 y dy V=
11
∫2 0
e 2 y dy =
23. a. Line has equation y = 12 x, 0 ≤ x ≤ 6. b. The log has radius = 6, so the circle is x2 + z2 = 36, or z = 36 – x 2 = (36 − x 2 )1/2 . 1 c. dV = y ⋅ 2z ⋅ dx = x · 2(36 − x 2)1/2 ⋅ dx 2 = (36 − x2)1/2 (x dx) 6
∫ (36 – x ) ( x dx ) 1 = − ∫ (36 – x ) ( −2 x dx ) 2
V=
2 1/2
0
6
Using properties of special right triangles, you can find that an equilateral triangle with 3 base b has height b. 2 106
Problem Set 5-9
1 2y 1 1 2 1 e = e − = 1.5972 K 4 4 4 0
2 1/2
0
=−
6 1 2 ⋅ (36 – x 2 )3/2 = 72 in.3 2 3 0
Calculus Solutions Manual © 2005 Key Curriculum Press
24. The points (0, 0) and (r, h) in xy-coordinates are on the line running up the top surface, so the h line is y = x. The circle forming the boundary r for the bottom surface has radius = r and center (0, 0) in xz-coordinates, so the circle is x2 + z2 = r2, or z = r 2 – x 2 . The slab at x = x0 is hx rectangular of height y = 0 , width r 2 z = 2 r 2 – x 02 , and thickness dx, so 2 hx 2 dV = r − x 2 dx, and r 2h r V= x r 2 − x 2 dx r 0 h r 2 (r − x 2 )1/ 2 ( −2 x dx ) =− r 0 r 2 h 2 h = − ⋅ (r 2 − x 2 )3/2 = − ⋅ (0 3/ 2 − r 3 ) 3 r 3 r 0
Slicing perpendicular to the y-axis as in Problem 26 gives dV = πx2 dy = π(r2 − y2) dy. V=
1 π (r 2 − y 2 ) dy = π r 2 y − y 3 −r 3
∫
r
28. The graph shows slices perpendicular to x-axis with sample points (x, y) and (x, z). b
2
(x , y )
x
r2 2 x dx h2 h h πr 2 1 1 x 2 dx = 2 ⋅ x 3 = πr 2 h, Q .E.D . 0 h 3 0 3
2
dV = πy 2 dx = π
∫
−10
V=
∫
a
−a
π (c/b)(b/a)2 ( a 2 − x 2 ) dx
1 = π (c/b)(b/a)2 a 2 x − x 3 3
a −a
4 4 = π (c/b)(b/a) ⋅ a 3 = πabc 3 3 Note that the volume formula for a sphere is a special case of the volume formula for an ellipsoid in which a = b = c = r, the radius of the sphere. 2
10
Calculus Solutions Manual © 2005 Key Curriculum Press
2
∴ dV = π (c/b)(b/a)2(a2 − x2) dx
π (100 − y 2 ) dy
1 = π 100 y − y 3 3 −10 1 1 = π 1000 − (1000) + 1000 − (1000) 3 3 4 3 = π (1000) cm 3 b. Formula: V = 43 πr 3 = 43 π 10 3 = 43 π (1000) cm3, which agrees with the answer by calculus. 27. Sphere can be generated by rotating about the y-axis the circle x2 + y2 = r2.
2
x + y = 1, from which y2 = (b/a)2(a2 − x2). a b
πr 2 h2 26. a. Equation of circle in xy-plane is x 2 + y 2 = 100. dV = πx2 dy = π(100 − y2) dy
∫
2
For a fixed value of x, the x-term will be constant. Subtracting this term from both sides of the equation gives an equation of the form 2 2 y + z = k 2 , where k2 = 1 − (x/a)2. b c Dividing both sides by k2 gives 2 2 y + z = 1. Thus, the y- and z-radii are kb kc kb and kc, which have the original ratio b/c. Therefore, each elliptical cross section is similar to the ellipse at the yz-plane, Q.E.D. dV = π yz dx Because z = (c/b)y, dV = π (c/b)y2 dx. The ellipse in the xy-plane (z = 0) has equation
y
V=
a
(x, z )
x y z Equation of ellipsoid is + + = 1. a b c
2 2 r h 3 25. A cone of radius r and altitude h can be generated by rotating about the x-axis the line r y = x from x = 0 to h. h
10
(x, y )
z
=
V=
y
x c
∫
r
−r
1 1 4 = π r 3 − r 3 − π − r 3 + r 3 = πr 3, Q.E.D. 3 3 3
∫
h
r
29. 50 + 2 L
L
L y 50
Problem Set 5-9
107
Note that the top of each isosceles trapezoidal cross section has length 50 + 2L yards, where y 52π = tan (52°) ⇒ L = y cot (52°) = y cot . L 180 1 So each slab is dV = 2 (50 + 2 L + 50) y dx; 52π dV = 50 y + y 2 cot dx, and 180 19
V=
∑ 50 y
k
+ yk2 cot
k =0
7. y 2
x
52π ⋅ 30 180
∫
π
0
π sin 2 x dx = 4.9348K
We cannot compute this integral algebraically because we do not know an antiderivative for sin 2 x.
= 1, 649, 443.6 K ≈ 1, 649, 443 yd 3 . Cost = 12 ⋅ 1,649,443.6… ≈ $19,793,324 8.
y
Problem Set 5-10
2
Q1.
1 3 x + x+C 3
Q2. 24
Q3.
∫ sec
Q4. 2 sec2 x tan x
2
x dx = tan x + C
Q5. Answers may vary.
x 5
Q6. See graph in Q5.
5
∫ π (ln x ) 2
v
dx = 14.6673K
We cannot compute this integral algebraically because we do not know an antiderivative for (ln x)2. 9. a.
(t, v )
dt
2
t
2
Q8. d(disp) = v dt
Q7. See graph in Q5.
Si x
b
Q9.
Q10. A ∫ v dt ∫ cos x dx ≈ 0.6899295233K ∫ ( x – 3x + 5) dx = 13.5 ∫ 2 dx ≈ 10.0988652K ∫ tan x dx ≈ 1.76714178K
x
a 1.4
1.
0.3 4
2.
1 3
3.
–20
–2
2
x
0 1.4
4.
0.1
5.
∫
1.4
0.3
cos x dx = sin x
1.4
= sin 1.4 − sin 0.3 =
0.3
0.6899295233… For the ten digits of the answer shown by calculator, there is no difference between this solution and the solution to Problem 1. 6.
∫
4
1
1 3 ( x − 3 x + 5) dx = x 3 − x 2 + 5 x 3 2
b. (sin x)/(x) approaches 1 as x approaches zero. c. Answers will vary depending on the grapher used. The TI-83 gives Si 0.6 = 0.58812881 using TRACE or 0.588128809608 using TABLE, both of which are correct to as many decimal places as the NBS values. d. By TABLE, Si x seems to be oscillating between about 1.53 and 1.61 when x is between 20 and 30. The limit is somewhere between these two numbers, say about 1.57. The actual limit is π /2, which equals 1.570796… . e.
4
2
Si x
2
1
1 3 = ( 4)3 − ( 4)2 + 5( 4) 2 3 1 3 − (1)3 − (1)2 + 5(1) = 13.5 2 3 There is no difference between this answer and the solution to Problem 2. 108
20
Problem Set 5-10
x –20
20
–2
The f graph is positive when x is between −π and π (as well as elsewhere), and has its greatest values there, which agrees with the Calculus Solutions Manual © 2005 Key Curriculum Press
large positive slope of the Si x graph in this region. Each place where the Si x graph has a high or low point, the f ( x) graph has a zero, corresponding to the zero slope of the Si x graph. So f ( x) is the derivative of Si x. 2 x −t2 10. a. erf x = e dt π 0
c. The answer by Simpson’s rule should be closer, because the graph is represented by curved segments instead of straight ones. 12. (Data and CAT scans for this problem were provided by Dr. James Stewart of San Antonio.)
The integrand e − t is an even function integrated between symmetrical limits. Thus, rather than using the entire interval [−x, x], one may find the integral on [0, x] and double the result.
+ 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 within the normal range of 150 to 200 g. 13. a.
∫
2
b.
a.
∫
8
A dD ≈ (0.8/3)(6.8 + 4 ⋅ 6.8 + 2 ⋅ 20.1
0
y = erf x
Force (lb)
1 300
x
Distance (in.)
2
0.5
c. By TABLE, values of erf x are x erf x 1 2 3 4
0.842700792… 0.995322265… 0.999977909… 0.999999984…
The values approach 1, meaning that the fraction of the data between −x and x is virtually 100% when x is beyond 4. d. Answers will vary depending on the grapher used. The TI-83 gives erf 0.5 = 0.52049987781… using TABLE, which is correct to as many decimal places as the NBS value. e. y = erf x 1
x 2
The slope of y = erf x appears to equal about 1 when x = 0 and decreases toward zero as x increases, which agrees with the graph of f. 11. a.
∫
12
0
(speed) dt ≈ [(2/60)/3](33 + 4 ⋅ 25 + 2 ⋅ 27
+ 4 ⋅ 13 + 2 ⋅ 21 + 4 ⋅ 5 + 9) = (1/90)(310) = 3.444… ≈ 3.4 nautical miles b. T 6 = (1/30)(33 ⋅ 0.5 + 25 + 27 + 13 + 21 + 5 + 9 ⋅ 0.5) = (1/30)(112) = 3.7333… ≈ 3.7 nautical miles
Calculus Solutions Manual © 2005 Key Curriculum Press
b. Let F = force, x = displacement, W = work. W=
∫
0.5
F dx
0
≈ (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 are from the given figure. T 500 1000 1500 2000 2500 3000 3500 4000 4500
∫
C
Simpson’s factor
8.44 9.24 10.08 10.84 11.48 11.98 12.36 12.68 12.94
1 4 2 4 2 4 2 4 1
4500
500 [8.44 + (9.24)( 4) 3 + (10.08)(2) + (10.84)(4) + (11.48)(2) + (11.98)(4) + (12.36)(2) + (12.68)(4) + 12.94] 500 = (268.18) = 44696.6666 K ≈ 44,697 Btu 3 The answers students get will vary slightly. H=
500
C dT ≈
Problem Set 5-10
109
15. a. Simpson’s rule will give a more accurate answer because the function y = sin x is approximated better by quadratic functions than 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…
∫
π
0
π
sin x dx = − cos x
0
2 − x2 e . A Simpson’s rule program π
gives S50 = 0.5204998781… and S100 = 0.5204998778… . There is little difference between the two estimations, and both are close to the tabulated value. 19. a. y
2
t
x
1
As x varies, the area beneath the curve y = 1/t from t = 1 to t = x varies also.
∫
b. Using the power formula on t −1 dt gives
Problem Set 5-11 Review Problems
= cos π − cos 0 = 2
Simpson’s rule does give a better approximation of the integral because S4 is closer to 2 than is T4. 16. Programs will vary depending on the type of grapher used. See the program in the Programs for Graphing Calculators section of the Instructor’s Resource Book. 17. Using a Simpson’s rule program, the mass of the spleen is 171.6 cm3. 18. Enter Y1 =
d. f (2) = 0.6931… f ( 3) = 1.0986… f ( 6) = 1.7917… f ( 2) + f ( 3) = f ( 2 ⋅ 3). This is a property of logarithmic functions.
t0 . 0
Division by zero is undefined, so this approach does not work. c. Graph Y1 = fnint(x− 1, x, 1, x). (Entries may be different for different calculators.) The graph looks like y = ln x. The value of f (x) is negative for x < 1 because for these values the lower limit of integration is larger than the upper limit, resulting in negative values for dx.
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 integral because v(t) is concave down, so trapezoids are 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 trapezoidalrule sum. c. T50 = 343.9964… , and T100 = 343.9991… Conjecture: The exact value of the integral is 344. d. g ( t) = 10t + 4t1.5 g(16) − g(4) = 344 This is the value the trapezoidal-rule sums are approaching. R2. a. The slope of the linear function is the same as the slope of the curve at x = 1. So the slope is f ( x) = sin π x ⇒ f ′(x) = π cos π x ⇒ f ′ ( 1) = π cos π = − π At x = 1, y = sin π = 0 y − 0 = − π (x − 1) ⇒ l(x) = − π x + π y l (x)
1
f (x) x 1
f (x )
l (x )
0.2
x
y
0.8
1
1.2
1
x 5
As you zoom in, you see that f ( x) is very close to the line l(x) for values near x = 1. 110
Problem Set 5-11
Calculus Solutions Manual © 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 .
iii. M6 = 2.209073… iv. T6 = 2.359018… d. U6 L6 5
b. i. y = csc5 2x ⇒ dy = −10 csc5 2x cot 2x dx
y
y
5
ii. y = x 5 /5 − x − 3/3 ⇒ dy = (x4 + x− 4) dx iii. y = (7 − 3x)4 ⇒ dy = −12(7 − 3x)3 dx
x 0.2
iv. y = 5e− 0. 3x ⇒ dy = −1.5e− 0. 3x dx
M6
1 v. y = ln (2 x ) ⇒ dy = ⋅ 4(2 x )3 ⋅ 2 dx (2 x ) 4 4
5
= 4/x dx; or y = ln ( 2x)4 = 4 ln ( 2x) ⇒ 1 dy = 4 ⋅ ⋅ 2 dx = 4/x dx 2x c. i. dy = sec x tan x dx ⇒ y = sec x + C 1 ii. dy = (3 x + 7)5 dx ⇒ y = (3 x + 7)6 + C 18 iii. dy = 5 dx ⇒ y = 5x + C iv. dy = 0.2e− 0.2 x ⇒ y = −e− 0.2 x + C v. dy = 6 x dx ⇒ y =
6x +C ln 6
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 indefinite integral. b. i.
∫ 12 x
ii.
∫ sin
iii.
∫ (x
iv.
∫ 12e
v.
∫ 7 dx = ln 7 + C
6
2
2/3
dx = 7.2 x 5/3 + C
x cos x dx =
1 sin 7 x + C 7
− 8 x + 3) dx = 3x
1 3 x − 4 x 2 + 3x + C 3
dx = 4e 3x + C
x
7x
R4. a. See the text for the definition of integrability. b. See the text for the definition of definite integral. c.
∫
1.4
sec x dx
0.2
i. U6 = 2.845333… ii. L6 = 1.872703…
Calculus Solutions Manual © 2005 Key Curriculum Press
x
1.4
0.2
1.4
T6
y
5
y
x 0.2
1.4
x 0.2
1.4
e. y
f (x)
x
R5. a. The hypothesis is the “if ” part of a theorem, and the conclusion is the “then” part. (Hypomeans “under,” and -thesis means “theme.”) π b. d (t ) = 20 + 3 sin t 4 d (2) – d ( 0) Average velocity = = 1.5 m/s 2–0 π Instantaneous velocity = d′(t) = 0.75π cos t 4 π d ′(c) = 0.75π cos c = 1.5 4 π c = cos− 1 (2/π) = 0.880689… 4 c = 1.12132… ≈ 1.12 s c. g(x) = x 4/3 − 4x 1/3 = x 1/3(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 is infinite. Thus, g is not differentiable at x = 0. However, the function need not be differentiable at the endpoints of the interval, just continuous at the endpoints and differentiable at interior points. d. For a function to be continuous on a closed interval, the limit needs to equal the function value only as x approaches an endpoint from within the interval. This is true for function f
Problem Set 5-11
111
at both endpoints, but not true for function g at x = 2. The graphs show that the conclusion of the mean value theorem is true for f but not for g. f (x )
∫
b.
x c
7
g (x )
(10 – x 2 ) dx = 10 x − (1/3) x 3
No tangent parallels secant
∫
R7. a. i.
Secant
−1
5
5
x −2 dx = − x −1 = −5 −1 + 1−1 = 4/5
1
1 4
∫ (x
ii.
2
3
x 2
3
= 30 − 9 − (−10) + (−1/3) = 92/3 = 30.6666… c. T100 = 30.6656, which is close to 92/3. d. M 10 = 30.72 M 100 = 30.6672 M 1000 = 30.666672 These Riemann sums are approaching 92/3.
Secant
2
3
−1
Tangent parallels secant
4
4
R3 = (1.513…)1.5 + (2.508…)1.5 + (3.505…)1.5 = 12.4, which is the exact value of the integral.
7
(Middle branch has the equation y = 1.4 . 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, satisfying one hypothesis of Rolle’s theorem. The other two hypotheses are satisfied because f and g are differentiable and continuous at the appropriate places, and a difference of differentiable and continuous functions also has these properties. The c in (a, b) for which h′(c) = 0 turns out to be the c in (a, b) for which f ′ (c) equals the slope of the secant line, g′(c), which equals [f (b) − f (a)]/(b − a). f.
= (1/2)
( x−2 )
f (x )
+ 3) 5 ( x dx ) 4
∫ (x
2
3
+ 3)5 (2 x dx ) 4
= (112 / )( x 2 + 3)6 3
= (1/12)(19)6 − (1/12)(12)6 = 3,671,658.08…
∫
iii.
π
0
(sin x – 5) dx = −cos x − 5 x
π 0
= −cos π − 5π + cos 0 + 0 = 2 − 5π
∫
iv.
ln 5
0
∫
ln 5
4e 2 x dx = 2e 2 x 0
4
x
3 1 ln 3 = 70.9986…
v.
= 2e 2 ln 5 − 2e 0 = 48
3 x dx =
4
= 1
34 31 78 − = ln 3 ln 3 ln 3
b. y
8
x
x –5
1
Points are 18 , 14 , 83 , 12 , 85 , 43 , and 87 . g. If r′(x) = s′(x) for all x in an interval, then r(x) = s(x) + C for some constant C.
∫
R6. a. g( x ) = x 1.5 dx = 0.4 x 2.5 + C Without loss of generality, let C = 0. g(2) – g(1) g′(c1 ) = = 1.862741K 2 –1 ∴ c11.5 = 1.862741… ⇒ c1 = (1.862741…)1/1.5 = 1.513915927… Similarly, c2 = 2.50833898… . c3 = 3.505954424… For
∫
4
x 1.5 dx,
Integral is negative, because each y-value in the Riemann sum is negative. c.
∫
10
−10
( 4 sin x + 6 x 7 – 8 x 3 + 4) dx = 2
= 8x
10 0
∫
10
4 dx
0
= 80
d. Total area = sum of two areas f (x )
x a
c
b
1
112
Problem Set 5-11
Calculus Solutions Manual © 2005 Key Curriculum Press
c. y = x 1 + 2 ⇒ x 1 = y − 2 1 y = 3x 2 − 6 ⇒ x 2 = y + 2 3 Graphs intersect at y = 6. Diameter of circular cross section is (x2 − x1). dV = π [0.5(x2 − x1)]2 dy 2 2 π 1 π 2 = y + 2 − ( y − 2) dy = 4 − y dy 4 3 4 3
R8. a. v (t, v)
dt t
dy = v dt = 150t0.5 dt 9
∫ 150t
y=
9
dt = 100t 1.5 = 2700 ft
0.5
0
0 4
∫ 150t For [ 4, 9], y = ∫ 150t For [0, 4], y =
0 9
0.5
0.5
4
= 1900.
V=
dt = 800.
R10. a.
∫ v(t ) dt = 2700 = 800 + 1900 = ∫ v(t ) dt + ∫ v(t ) dt. 0
4
∫
0
e y dy = e y
ln 4
1
t 1
= e ln 4 − e 0 = 4 − 1 = 3
0
R9. a. y = e , from x = 0 to x = 4, is rotated about the x-axis.
10
b. dW = v · y · dx = (1000 + 50x)(4 − 0.2x2) dx 3
0.2 x
∫ (1000 + 50 x )(4 − 0.2 x ) dx = 10, 897.5 ∫ v(t ) dt = 1/3(0.2)[29 + 4(41) + 2(50) + 4(51) 2
0
5
y
c.
(x, y )
3
+ 2(44) + 4(33) + 2(28) + 4(20) + 2(11) + 4(25) + 39] = 67.6 Values of velocity are more likely to be connected by smooth curves than by straight lines, so the quadratic curves given by Simpson’s rule will be a better fit than the straight lines given by the trapezoidal rule.
1
x 0
4
dV = π y2 dx = πe0.4 x dx V=
log x dx = 6.0913K
v (t )
b. dA = x dy y = ln x ⇒ x = e y dA = ey dy ln 4
10
The integral is reasonable because counting squares gives approximately 6.
0
9
∫
1
So
4
0
The right circular cone of altitude 6 and radius 2 also has volume 13 π ⋅ 2 2 ⋅ 6 = 8π .
dt = 2700 − 800
9
6
∫ dV ≈ 25.1327… (exactly 8π)
∫
4
0
πe 0.4 x dx = 2.5πe 0.4 x
4 0
= 2.5π (e − 1) = 31.0470… b. y = x10.25 and y = x2, intersecting at (0, 0) and (1, 1) in Quadrant I, is rotated about the y-axis. Only the back half of the solid is shown. 1.6
y
Concept Problems C1. a. f (b) 5
(x 1 , y )
π/2
b 1
–5
5
(x 2 , y ) –5
x 1
y = x10.25 ⇒ x1 = y 4 y = x2 ⇒ x2 = y dV = π ( x 22 − x12 ) dy = π ( y 2 − y 8 ) dy 1 1 1 V = π ( y 2 − y 8 ) dy = π y 3 − y 9 3 0 9 2 = π = 0.6981K 9
∫
Calculus Solutions Manual © 2005 Key Curriculum Press
b. f (b)
1
b
1 3
0
Problem Set 5-11
113
The two lines are horizontal asymptotes. y4 is the inverse of y1. c. The graph of f in part a is a reflection of the graph of y = tan x in part b across the line y = x. Function f seems to be the inverse tangent function, f (b) = tan− 1 b. (In Chapter 9, students will learn that this is actually true.) C2. f (x) = ax2 + bx + c
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 + 1) = 0 2 + 12 + L + k 2 + ( k + 1)2 = (0 2 + 12 + L + k 2 ) + ( k + 1)2 = (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 of k, thus completing the induction. ∴ S(n) = (n/6)(n + 1)(2n + 1) for any positive integer n, Q.E.D.
y
x d
k
e
f (d) = ad2 + bd + c f (e) = ae2 + be + c ae 2 + be + c − ( ad 2 + bd + c) ∴m = e−d a( e 2 − d 2 ) + b ( e − d ) = = a( e + d ) + b e−d 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) = 0 2 + 12 + 2 2 + 32 + L + n 2 a. S(0) = 0, S(1) = 1, S(2) = 5, S(3) = 14 S(n) = an3 + bn2 + cn + d 0=0+0+0+d 1=a+b+c+d 5 = 8a + 4b + 2c + d 14 = 27a + 9b + 3c + d Solving 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) = 30 S(5) = (5/6)(6)(11) = 55 By addition, S(4) = 0 + 1 + 4 + 9 + 16 = 30, which checks. S(5) = 0 + 1 + 4 + 9 + 16 + 25 = 55, which checks. S(1000) = (1000/6)(1001)(2001) = 333,833,500 c. Prove that S(n) = (n/6)(n + 1)(2n + 1) for any positive integer n. Proof (by induction on n): Anchor: For n = 1, S(n) = (1/6)(2)(3) = 1, the correct answer, which anchors the induction.
114
Problem Set 5-11
d. S(n) = 0 3 + 13 + 2 3 + 33 + L + n 3 S(0) = 0 S(1) = 0 + 1 = 1 S(2) = 0 + 1 + 8 = 9 S(3) = 0 + 1 + 8 + 27 = 36 S(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+e 1=a+b+c+d+e 9 = 16a + 8b + 4c + 2d + e 36 = 81a + 27b + 9c + 3d + e 100 = 256a + 64b + 16c + 4d + e Solving this system gives a = 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) is a perfect square. By equation, S(5) = [(5/2)(6)]2 = 225 S(6) = [(6/2)(7)]2 = 441 By 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.
∫
π
4 sin x sin 10 x dx
0
=
π
∫ [–2 cos 11x + 2 cos (–9 x )] dx 0
π 2 2 sin 11x − sin ( −9 x ) 11 9 0 = −0 − 0 + 0 + 0 = 0, Q .E .D . There is just as much area below the x-axis as there is above it, so the integral is 0.
=−
Calculus Solutions Manual © 2005 Key Curriculum Press
b.
∫
π
4 sin x sin nx dx
0
=
π
∫ {−2 cos [(1 + n) x] + 2 cos [(1 − n) x]} dx 0
π –2 2 sin [(1 + n) x ] + sin [(1 − n) x ] 1+ n 1– n 0 –2 2 = sin (1 + n)π + sin (1 − n)π 1+ n 1– n –2 2 − sin 0 − sin 0 1+ n 1– n If n is an integer, the first two terms will involve sines of integer multiples of π, and are thus equal to 0. The last two terms are 0 unless n = ±1. Thus, the integral equals 0 for any integer n > 1, Q.E.D. C5. a. Algebraic solution: Pick sample points ck at the right end of each subinterval. Because f (x) is increasing on the interval [1, 9], the high points of f (x) are located at the right ends of the subintervals and the low points are at the left ends.
=
n
∴ Un =
∑
b. From part a, 0 ≤ U n − Ln ≤ ||P|| (1.29 − 1.21). As ||P|| approaches zero, the rightmost member of the inequality goes to zero. By the squeeze theorem, lim (Un − Ln ) = 0, which P →0
implies lim Un = lim Ln . So f is integrable P →0
Proof: Partition the interval [1, 4] into n subintervals whose widths are not necessarily equal. Let ||P|| be the norm of the partition. Pick sample points ck at the left end of each subinterval. Because g (x) is decreasing on [1, 4], the high points are located at the left ends of the subintervals and the low points are at the right ends (graph). y 1
n
f (ck )∆x k and Ln =
∑ f (c
k −1 ) ∆x k
k =1
k =1
Note that c0 = 1 and cn = 9. Subtracting gives
x 1
n
Un − Ln =
P →0
on [1, 9] by the definition of integrability, Q .E .D . c. Prove that g(x) = 1/x is integrable on [1, 4].
∑ [ f (c ) − f (c k
k −1 )]∆x k
k =1
= [ f (c1 ) − f (c0 )]∆x1 + [ f (c2 ) − f (c1 )]∆x 2 + L + [ f ( cn) − f ( cn− 1)]∆xn ≤ [ f ( c1) − f ( c0)] ||P|| + [ f ( c2) − f ( c1)] ||P|| + L + [ f (cn ) − f (cn−1 )] || P || = [ f (c1 ) − f (c0 ) + f (c2 ) − f (c1 ) + L + f (cn ) − 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 of the spaces between the lower and upper rectangles in Figure 5-11g. Imagine these spaces moved over to the left so that they align at x = 1 (graph). The spaces can be circumscribed with a rectangle of base ||P|| and altitude (1.29 − 1.21). Thus, Un − Ln ≤ ||P|| (1.29 − 1.21), Q.E.D.
4
By algebraic or graphical reasoning as in part 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 to h (x) = sin x on the interval [0, 3] because h (x) is both increasing and decreasing on different parts of the interval. Thus, the high points are not always at the same end of the subinterval and the high point at π/2 may not be at either end of a subinterval (graph). y = sin x 1
x 0
π/2
3
f (x ) 1.29
The reasoning could be applied indirectly by first splitting the interval [0, 3] into [0, π /2] and [π /2, 3] so that h(x) is increasing on one and decreasing on the other.
Slide them over.
Chapter Test
1.21
x 1 Norm
Calculus Solutions Manual © 2005 Key Curriculum Press
Norm = largest
∆x 9
T1. Indefinite integral: g( x ) =
∫ f ( x ) dx if and only if g′(x) = f (x). Problem Set 5-11
115
T2. Definite integral: Let Ln and Un be lower and upper sums of f (x) on the interval [a, b]. Then
∫
b
f ( x ) dx = lim Ln = lim Un , n→∞
a
n→∞
provided the two limits are equal. T3. Fundamental theorem: If f is an integrable function, and g( x ) =
∫
b
a
∫ f ( x ) dx, then
f ( x ) dx = g(b) − g( a).
T12.
−2
∫
(12 x 3 + 10 x 2 ) dx = 3 x 4 +
2
10 3 x 3
−2 2
80 80 1 = 48 − − 48 − = −53 3 3 3 d
∫ [( x )
T13. π
2
1
c
− ( x 2 )2 ] dy
T14. The slope of the linear function is the same as the slope of the curve at x = 1. So the slope is found by y = x3 ⇒ y′ = 3x2 ⇒ y′(1) = 3. At x = 1, y = 1. y − 1 = 3(x − 1) ⇒ y = 3x − 2 y
T4. The function f is continuous on the interval [3, 8] and differentiable on (3, 8). It does not matter that it is not differentiable at the endpoints. T5.
1
x 1
f (x ) 1.2 1
(1, 1)
x c 8
3
1
T6. Hypotheses: f (a) = f (b) = 0 Differentiable on (a, b). Continuous at x = a and x = b. Conclusion: There is a c in (a, b) such that f ′ (c) = 0. f(x )
c
x
a
b
T7. f (x )
x 8
3
As you zoom in on (1, 1), you see that the graph of y = x3 is locally linear. T15. a.
∫
3
0
3
12e 0.25 x dx = 48e 0.25 x 0
= 48e 0.75 − 48e 0 = 53.6160 K b. M 50 = 53.6154… T50 = 53.6170… S50 = 53.61600081… The midpoint Riemann sum error is 0.000502646712… . The trapezoidal-rule error is 0.0010052962… . The midpoint Riemann sum error is half of the trapezoidal-rule error, because 2(0.000502646712…) = 0.0010052962… . The Simpson’s rule error is 0.000000015077… , which is much smaller than the error for the other two methods. T16. a. y 1
y = cos x
T8. y = esin x ⇒ dy = cos x esin x dx T9. T10. T11.
∫ (4 x ∫
4
1
116
(x, y )
x
∫ 0.1 dx = ln 0.1 + C x
3
0.1
+ 13)5 ( x 2 dx ) =
x3 x dx = 3
1.2
4
=
2
1
Problem Set 5-11
π
1 ( 4 x 3 + 13)6 + C 72
4 3 13 − = 21 3 3
x
1 2 1 y dx ⇒ dV = cos 2 x dx 2 2 π /2 1 V= cos 2 x dx 0 2
b. dV =
∫
Calculus Solutions Manual © 2005 Key Curriculum Press
c. This integral cannot be evaluated algebraically because we do not know an antiderivative for cos2 x. π /2 1 V= cos 2 x dx = 0.3926 K 0 2
∫
∫
T17. a. g( x ) = 0.3 x 2 dx = 0.1x 3 + C = 0.1x 3 b. f (x )
g(x )
5
5
x 1
2.64...
4
Calculus Solutions Manual © 2005 Key Curriculum Press
x 1
2.64...
g( 4) – g(1) 6.4 – 0.1 = = 2.1 4 –1 3 g′(x) = f ( x) = 0.3x 2 ∴ 0.3c2 = 2.1 ⇒ c = 7 = 2.6457513… In the right graph in part b, the tangent line at x = 2.64… is parallel to the secant line from x = 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 of rectangle, as shown in the graph on the left in part b. T18. Answers will vary. c. m =
4
Problem Set 5-11
117
Chapter 6—The Calculus of Exponential and Logarithmic Functions Problem Set 6-1
Q5. differentiable
1. The integral would be division by zero. N 1000 1500 2000 2500 500 100
1 0
P 0 , which involves
Integral 0 0.4054… 0.6931… 0.9162… −0.6931… −2.3205…
1
N 2000
–1
7.
–2
The graph resembles a logarithmic function.
∫
10
0
0.05 dt = 0.05t
10
= 0.5 − 0 = 0.5, Q .E.D .
8. 9.
0
0.5 is between 0.4054… and 0.6931… , the values of the left integral for N = 1500 and N = 2000. N 1 By solver, dP = 0.5 when 1000 P N = 1648.7212… , or about 1649 people. 3. At 20 years, the integral on the right equals 1. At 0 years, the integral equals 0. Solving for N at these times gives N = 2718.2818… for 20 years, and N = 1000 (as expected!) for 0 years.
10.
∫
11.
Population 3000 2000
12.
1000 Time (yr) 5
10
15
20
The graph resembles an exponential function. 4. ln 1648.7212… − ln 1000 = 0.5, exactly. This is the value of the integral on the left!
Problem Set 6-2 1 0.7 x +C 0.7 Q3. f ′ ( x) = −2 cos x sin x Q1.
118
Problem Set 6-2
1 – x2
Q8. Riemann sum Q7. y′ = − csc x cot x Q9. indefinite integral, or antiderivative 0. 1. 2. 3. 4. 5. 6.
Integral
2.
1
Q10. log 12
2
1000
Q6. y′ =
Q2. 9 Q4. continuous
Answers will vary. y = ln 7x ⇒ y′ = 1/(7x) · 7 = 1/x y = ln 4x ⇒ y′ = 1/(4x) · 4 = 1/x f (x) = ln x5 ⇒ f ′ (x) = 1/(x5) · 5x4 = 5/x f (x) = ln x3 ⇒ f ′ (x) = 1/(x3) · 3x2 = 3/x h (x) = 6 ln x− 2 ⇒ h′ (x) = 6/(x− 2) · (−2x− 3) = −12/x g (x) = 13 ln x− 5 ⇒ g′ (x) = 13/(x− 5) · (−5x− 6) = −65/x r(t) = ln 3t + ln 4t + ln 5t ⇒ r′ (t) = 1/(3t) · 3 + 1/(4t) · 4 + 1/(5t) · 5 = 3/t v (z) = ln 6z + ln 7z + ln 8z ⇒ v′ (z) = 1/(6z) · 6 + 1/(7z) · 7 + 1/(8z) · 8 = 3/z y = (ln 6x)(ln 4x) ⇒ y′ = 1/(6x) · 6 · (ln 4x) + (ln 6x)[1/(4x) · 4] ln 24 x 2 = (1/x)(ln 4x + ln 6x) or x z = (ln 2x)(ln 9x) ⇒ z′ = 1/(2x) · 2 · (ln 9x) + (ln 2x)[1/(9x) · 9] ln 18 x 2 = (1/x)(ln 9x + ln 2x) or x ln 11x y = ⇒ ln 3 x 1/(11x ) ⋅ 11 ⋅ (ln 3 x ) – (ln 11x )1/(3 x ) ⋅ 3 y′ = (ln 3 x )2 ln 3 x – ln 11x ln(3/11) = or 2 x (ln 3 x )2 x (ln 3 x ) ln 9 x y= ⇒ ln 6 x 1/(9 x ) ⋅ 9 ⋅ (ln 6 x ) – (ln 9 x ) ⋅ 1/(6 x ) ⋅ 6 y′ = (ln 6 x )2 ln 6 x – ln 9 x ln(2/3) or = x (ln 6 x )2 x (ln 6 x )2
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 © 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 = x 4 ln 3x ⇒ y′ = 4x 3 ln 3x + x 4 · 1/(3x) · 3 = 4x 3 ln 3x + x 3 24. y = x 7 ln 5x ⇒ y′ = 7x 6 ln 5x + x 7 · 1/(5x) · 5 = 7x 6 ln 5x + x 6 25. y = ln (1/x) ⇒ y′ = 1/(1/x) · (−x −2) = −1/x 26. y = ln (1/x4) ⇒ y′ = 1/(1/x)4 · (−4x−5) = −4/x 27. 28. 29. 30. 31.
32.
33.
34.
35. 36. 37. 38.
∫ 7/x dx = 7 ln | x | + C ∫ 5/x dx = 5 ln | x | + C 1 ∫ 1/(3x ) dx = 3 ln | x | + C ∫ 1/(8x ) dx = 8 ln | x | + C 1
x2 1 1 (3 x 2 dx ) dx = x +5 3 x3 + 5 1 = ln | x 3 + 5 | + C 3
∫
3
∫
x5 1 1 dx = (6 x 5 dx ) 6 x –4 6 x6 – 4 1 = ln | x 6 – 4 | + C 6 x5 1 1 dx = − (–6 x 5 dx ) 9 – x6 6 9 – x6 1 = − ln | 9 – x 6 | + C 6 x3 1 1 (–4 x 3 dx ) 4 dx = − 10 – x 4 10 – x 4 1 = − ln | 10 – x 4 | + C 4 sec x tan x dx = ln | 1 + sec x | + C 1 + sec x sec 2 x dx = ln | 1 + tan x | + C 1 + tan x cos x dx = ln | sin x | + C sin x sin x dx – sin x dx =− = − ln | cos x | + C cos x cos x
∫
∫
∫
∫
∫
∫
39.
∫
Calculus Solutions Manual © 2005 Key Curriculum Press
4
0.5
4
(1/w ) dw = ln | w |
= ln 4 − ln 0.5
0.5
= ln 8 = 2.079441… 40.
∫
10
0.1
(1/v) dv = ln | v |
10
= ln 10 − ln 0.1
0.1
= ln 100 = 4.605170… 41.
∫
−3
−0.1
(1/ x ) dx = ln | x |
−3 − 0.1
= ln | −3 | − ln | −0.1 |
= ln 3 − ln 0.1 = ln 30 = 3.401197… 42.
∫
−4
−0.2
(1/ x ) dx = ln | x |
−4 −0.2
= ln | −4 | − ln | −0.2 |
= ln 4 − ln 0.2 = ln 20 = 2.995732… x 1/ 2 dx 2 9 1 3 1/ 2 x dx 3/ 2 = 3/ 2 ⋅ 4 1+ x 3 4 1+ x 2 9 2 2 = ln| 1 + x 3/ 2 | = (ln 28 – ln 9) = 0.756653K 3 4 3
43.
∫
44.
∫
45. 46.
9
∫
x −1/ 3 dx 3 8 1 2 ⋅ x −1/ 3 dx 2/3 = 1 2+x 2 1 2 + x 2/3 3 8 3 3 = ln | 2 + x 2 / 3 | = (ln 6 – ln 3) = 1.5 ln 2 2 2 1 = 1.039720… dx 1 (ln x )5 = (ln x )6 + C x 6 ln x dx 1 dx = (ln x )1 = (ln x )2 + C x x 2 8
∫
∫ ∫
∫
x
∫ cos 3t dt ⇒ f ′( x ) = cos 3x f ( x ) = ∫ (t + 10t – 17) dt ⇒
47. f ( x ) =
2
x
48.
2
5
f ′( x ) = x 2 + 10 x − 17 49. 50.
d dx
∫
x
d dx
∫
x
2
tan 3 t dt = tan 3 x
−1
2 t dt = 2 x
∫ g( x ) = ∫ h( x ) = ∫
51. f ( x ) =
x2
1
3t dt ⇒ f ′( x ) = 2 x ⋅ 3 x
cos x
52.
0 3 x −5
53.
0
∫
∫ ∫ ∫
∫
2
t dt ⇒ g′( x ) = cos x ⋅ ( − sin x ) 1 + t 2 dt ⇒
h′( x ) = 3 1 + (3 x – 5)2 54. p( x ) = 55.
∫
3
1
∫
x3
–1
(t 4 + 1) 7 dt ⇒ p′( x ) = ( x 12 + 1) 7 ⋅ 3 x 2
(5 / x ) dx = 5 ln | x |
3
= 5 ln 3 − 5 ln 1 = 5 ln 3
1
= 5.493061… Problem Set 6-2
119
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. y 7 6 5 4 3 2 1
g
b. 30
(h, F )
dh
1
2
3
5
6
7
b. h( x ) =
∫
1
dW = F dh = (600/h − 30) dh
8
f (t ) dt ⇒ h ′( x ) = f ( x − 1) ⋅ 2x ⇒
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. y
g
x –1 –2 –3 –4 –5 –6
1 2 3 4 5
b. h( x ) =
7 8 9 10
∫
N
∫
10
1000
0
∫
2
f (t ) dt ⇒ h′( x ) = f (2 x ) ⋅ 2 ⇒
(1/ P) dP = ln | P |
0.05 dt = 0.05t
N
= ln N − ln 1000
1000 10
= 0.5
0
ln N − ln 1000 = 0.5 N ln = 0.5 1000 N = e 0.5 1000 N = 1000e0.5 ≈ 1648.721… ≈ 1649 people 60. a. F + 30 = k/h 0 + 30 = k/20 ⇒ k = 600 ∴ F + 30 = 600/h ⇒ F = 600/h − 30 120
10
(600/h – 30) dh
20
Problem Set 6-2
10 20
= 600 ln 10 − 300 − 600 ln 20 + 600 = −115.8883… ≈ −116 inch-pounds This number is negative because each value of dh is negative and F is positive, making their product negative. e. Distance is measured in inches, force is measured in pounds, and we are finding their product. 61. a. d(f ) = a + b ln f 0 = a + b ln 53, 10 = a + b ln 160 10 = b ln 160 − b ln 53 ⇒ 10 b= = 9.050741... ln 160 – ln 53 a = −9.050741… ln 53 = −35.934084… d( f ) = –35.934084… + 9.050741… ln f b.
2x
h′(4) = f (8) ⋅ 2 = (0.5)(2) = 1 59.
∫
= 600 ln | h | − 30 h
2
h′(2) = f (3) ⋅ 4 = −1 ⋅ 4 = −4
4 3 2 1
20
c. Work equals force times displacement. Because the force varies, a definite integral must be used. d. The work done compressing the air a small amount, dh, is approximately equal to the force at the sample point (h, F ) times dh (see part b). ∴W=
x 2 −1
h 10
x –1 –2 –3
F
f 53 60 70 80 100 120 140 160
d cm 0 1.1227… 2.5197… 3.7265… 5.7461… 7.3962… 8.7914… 10.0
d′ (part c) 0.1707… 0.1508… 0.1292… 0.1131… 0.0905… 0.0754… 0.0646… 0.0565…
The measured distances will vary. They should 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 the numbers getting smaller as f increases.
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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 (ab ) = b ln a ln 2 d. = 2.30258… log 2 ln 3 = 2.30258… log 3 They seem to be the same. ln 10 = 2.30258… 1 = 2.30258… log e log 4 = 0.60205… and ln 4 1.3862 K = = 0.60205K ln 10 2.30258K 63. Answers will vary.
Problem Set 6-3 Q1. y′ = 1/(1 + x 2) Q3. 1 Q5. 35 Q7.
1 ln | 4 x + 1 | + C 4 Q4. 1/4 Q6. 8 Q2.
y 1
x 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…
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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 uniqueness theorem. 8. See the text for the proof. 9. Prove that ln (a/b) = ln a − ln b for all a > 0, b > 0. Proof: Let f (x) = ln (x/b), g(x) = ln x − ln b for x, b > 0 Then f ′ (x) = (b/x)(1/b) = 1/x, and g′ (x) = (1/x) − 0 = 1/x. ∴ f ′ (x) = g′ (x) for all x > 0. f (b) = ln (b/b) = ln 1 = 0 g(b) = ln b − ln b = 0 ∴ f (b) = g (b). ∴ f (x) = g (x) for all x > 0 by the uniqueness theorem. ∴ 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 (ab) = 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 and g′ (x) = b (1/x) = b/x. ∴ f ′ (x) = g′ (x) for all x > 0. f (1) = ln (1b) = ln 1 = 0 g(1) = b ln 1 = b · 0 = 0 ∴ f (1) = g (1). ∴ f (x) = g (x) for all x > 0 by the uniqueness theorem. ∴ 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 for all 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.
Problem Set 6-3
121
13. See the text definition of ln x.
15. f (x) = log3 x ⇒ f ′ (x) = 1/(x ln 3) f ′ (5) = 0.182047… The graph shows a tangent line with a small positive slope. f (x )
1
x 5
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. f (x ) 5
x
4 10
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 ′( x ) =
1 ⋅ cos x sec x ⋅ ln 5
T′ (x) = (cot x)/(ln 5) 20. R (x) = log4 (sec x) ⇒ 1 tan x R′( x ) = ⋅ sec x tan x = sec x ⋅ ln 4 ln 4 21. p (x) = (ln x)(log5 x) ⇒ p′ (x) = (1/x) · (log5 x) + (ln x) · [1/(x ln 5)] 1 ⋅ ln x ln x 2 ln x = + = x ln 5 x ln 5 x ln 5 ln x ln 3 ln 3 1 ⋅ = = ln 9 ln x 2 ln 3 2 ∴ q′ (x) = 0 because q(x) is constant.
22. q(x) = (log9 x)/(log3 x) =
x3 3 23. f ( x ) = ln = ln x − ln sin x sin x = 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) ⇒ 4 sec 2 x 4 1 f ′( x ) = + = + x tan x x sin x cos x
122
Problem Set 6-4
d d (ln x 3 x ) = (3 x ln x ) = dx dx 3x 3 ln x + = 3 ln x + 3 x d d (ln 5sec x ) = (sec x ln 5) 26. dx dx = 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/h x = 1: dy/dx = 0.663… mi/h x = 5: dy/dx = 0.435… mi/h x = 10: dy/dx = 0.257… mi/h The lava is slowing down. b. y/7 = 2 − 0.9x 0.9x = 2 − y/7 x ln 0.9 = ln (2 − y/7) x = (1/ln 0.9)[ln (2 − y/7)] dx 1 1 c. = (1/ln 0.9) ⋅ ⋅ − dy 2 – y/7 7 dx 9.491221K = dy 14 – y y = 10: dx/dy = 2.372… h/mi d. If x = 10, then y = 7(2 − 0.910), so dx 9.491221K = = 3.888651… . dy 14 – 7(2 – 0.910 ) 25.
ln x log a x 14. log b x = = ln b log a b
e. 3.88… is the reciprocal of 0.257… , the value 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) 1 1 b. dt/dM = ⋅ (ln 1.06)( M/1000) 1000 1 = M ln 1.06 1 = c. If M = 1000, dt/dM = 1000 ln 1.06 0.01716… yr/$. At this rate, with $1000 in the account, it would take 0.017… year, or about 6 days, to earn a dollar of interest. d. dt/dM gets smaller as M increases; more interest is earned when M is larger, so it takes less time to accumulate $1000. 29. The intersection point is at x = 2.7182818… , which is approximately e. 30. Answers will vary.
Problem Set 6-4 Q1. y′ = 3/x Q3. 15 ln 5 x + C
Q2. (−1/10)(5x)− 2 + C Q4. y′ = −1/ 1 – x 2 Calculus Solutions Manual © 2005 Key Curriculum Press
Q5. 5 tan (5x) Q7. Q9. 0. 1.
Q6.
ln 23 ln 17
36 Q8. 8 B Q10. E Answers will vary. a. R(t) = aekt 60,000 = aek·0 ⇒ a = 60,000 2,400,000 = 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 introduced about 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 depreciation in part b because the latter rate is an average for the year. The rate at the end will be lower than 736 to give the average of 700. d. 5,000 = 20,000e− 0.1 t 0.25 = e− 0.1 t ln (0.25) = −0.1t t = (ln 0.25)/(−0.1) = 13.8629… ≈ 14 yr. 3. a. m(t) = 1000(1.06)t ln m(t) = ln 1000 + t ln 1.06 1/m(t) · m′ (t) = 0 + ln 1.06 m′ (t) = m (t) · ln 1.06 m′ (t) = 1000(1.06)t (ln 1.06) m′ (0) = 58.27 $/yr m′ (5) = 77.98 $/yr m′ (10) = 104.35 $/yr b. m(0) = $1000.00 m(5) = $1338.23 m(10) = $1790.85 The rates are increasing. $338.23 is earned between 0 and 5 years; $452.62 is earned
Calculus Solutions Manual © 2005 Key Curriculum Press
between 5 and 10 years, which agrees with the increasing derivatives shown in part a. m′(t ) 1000(1.06)t (ln 1.06) c. = = ln 1.06 m( t ) 1000(1.06)t ∴ m′ (t)/m(t) = ln 1.06, a constant d. m(1) = 1060.00. So you earn $60.00. The rate starts out at only $58.27/year but has increased enough by year’s end to make the total for the year equal to $60.00. 4. d(t) = 200t · 2− t ⇒ ln d(t) = ln 200t − t ln 2 1 ⋅ d ′(t ) = 1/t − ln 2 ⇒ d (t ) d ′(t ) = (200t ⋅ 2 − t )(1/t − ln 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 at 1 second and closing at about 19.3°/s at 2 seconds, which agrees with the graph. d (t ) 100
t 1 2
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. n 1 5. e = lim 1 + and e = lim(1 + n)1/ n n→∞ n→0 n When you substitute ′ for n in the first equation, you get the indeterminate form 1∞ . When you substitute 0 for n in the second equation, you also get the indeterminate form 1∞ . n (1 + 1/n)n
6. 7. 8. 9.
100 1000 10000
2.70481… 2.71692… 2.71814…
n
(1 + n)1/ n
0.01 2.70481… 0.001 2.71692… 0.00001 2.71826… − 5x y = 17e ⇒ y′ = −85e− 5x y = 667e− 3x ⇒ y′ = −2001e− 3x h(x) = x3ex ⇒ h′ (x) = 3x2ex + x3ex = x2ex (3 + x) g (x) = x− 6ex ⇒ g′ (x) = −6x− 7ex + x− 6ex = x− 7ex (−6 + x)
Problem Set 6-4
123
f ′( x ) = 10 −0.2 x ( −0.2 ln 10) 30. g (x) = 4(7x) ⇒ ln g (x) = ln 4 + x ln 7 ⇒ 1 g ′( x ) = ln 7 ⇒ g ′( x ) = 4(7 x ) ln 7 g( x )
35. y = x ln x ⇒ ln y = ln x · ln x ⇒ ln y = (ln x)2 ⇒ 1 1 2 ln x ln x y ′ = 2 ln x ⋅ ⇒ y ′ = ⋅ x = 2 ln x · xln x−1 y x x 1 36. y = (csc 5x)2x ⇒ ln y = 2x ln (csc 5x) ⇒ y′ = y 1 2 ln (csc 5 x ) + 2 x (–5 csc 5 x cot 5 x ) ⇒ csc 5 x y′ = (csc 5x)2x [2 ln (csc 5x) − 10x cot 5x] 37. y = (cos 2x)3x ⇒ ln y = 3x ln (cos 2x) ⇒ 1 1 y′ = 3 ln (cos 2 x ) + 3 x (–2 sin 2 x ) ⇒ y cos 2 x y′ = (cos 2x)3x [3 ln (cos 2x) − 6x tan 2x] 38. Two solution methods are possible. Differentiate directly: 5x + 2 y = ln ⇒ 7x – 8 7 x – 8 5(7 x – 8) – (5 x + 2) ⋅ 7 y′ = (7 x – 8)2 5 x + 2 –54 = (5 x + 2)(7 x – 8) Or simplify using properties of logarithms first: y = ln (5x + 2) − ln (7x − 8) ⇒ 5 7 –54 y′ = − = 5 x + 2 7 x – 8 (5 x + 2)(7 x – 8) 39. Two solution methods are possible. Differentiate directly: y = ln [(4x − 7)(x + 10)] 1 y′ = ⋅ [ 4( x + 10) + ( 4 x − 7) ⋅ 1] ( 4 x – 7)( x + 10) 8 x + 33 = ( 4 x – 7)( x + 10) Or simplify using properties of logarithms first: y = ln (4x − 7) + ln (x + 10) 4 1 8 x + 33 y′ = + = 4 x − 7 x + 10 ( 4 x – 7)( x + 10)
31. h (x) = 1000(1.03x) ⇒ ln h (x) = ln 1000 + 1 x ln 1.03 ⇒ h ′( x ) = ln 1.03 ⇒ 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 c ′( x ) = 5/ x + ln 3 ⇒ c( x ) c ′( x ) = x5 · 3x(5/x + ln 3) 33. m(x) = 5x · x 7 ⇒ ln m(x) = x ln 5 + 7 ln x ⇒ 1 m ′( x ) = ln 5 + 7/ x ⇒ m( x ) m′ (x) = 5x · x7(ln 5 + 7/x) 34. y = (ln x)0.7 x ⇒ ln y = 0.7x ln (ln x) ⇒ 1 1 ⇒ y ′ = 0.7 ln (ln x ) + 0.7 x ⋅ y x ln x 0.7 y ′ = 0.7 ln (ln x ) + ⋅ (ln x ) 0.7 x ln x
40. y = (2x + 5)3 4 x − 1 ⇒ 1 ln y = 3 ln (2x + 5) + ln ( 4 x − 1) ⇒ 2 1 6 2 + ⇒ y′ = y 2x + 5 4x −1 6 2 y′ = + [(2 x + 5)3 4 x − 1 ] 2 x + 5 4 x − 1 (28 x + 4)(2 x + 5)2 = 4x −1 (10 + 3 x )10 ⇒ ln y = 10 ln (10 + 3 x ) − 41. y = ( 4 − 5 x )3 1 30 15 3 ln ( 4 − 5 x ) ⇒ y ′ = + ⇒ y 10 + 3 x 4 − 5 x 30 15 (10 + 3 x )10 y′ = + 10 + 3 x 4 − 5 x ( 4 − 5 x )3 (270 − 105 x )(10 + 3 x )9 = (4 − 5x )4
10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25.
r(t) = et sin t ⇒ r′ (t) = et sin t + et cos t s (t) = et tan t ⇒ s′ (t) = et tan t + et sec2 t u = 3exe− x = 3 ⇒ u′ = 0 v = e4xe− 4x = 1 ⇒ v′ = 0 ex e x ln x − e x (1/ x ) y= ⇒ y′ = ln x (ln x )2 ln x (1/ x )e x − ln x ⋅ e x y = x ⇒ y′ = e e2 x y = 4esec x ⇒ y′ = 4esec x sec x tan x y = 7ecos x ⇒ y′ = −7ecos x sin x y = 3 ln e 2x = 6x ln e = 6x ⇒ y′ = 6 y = 4 ln e 5x = 4 · 5x = 20x ⇒ y′ = 20 y = (ln e3x)(ln e4x) = 3x · 4x = 12x2 ⇒ y′ = 24x y = (ln e− 2x)(ln e5x) = −2x · 5x = −10x2 ⇒ y′ = −20x g(x) = 4eln 3x = 4 · 3x = 12x ⇒ g′ (x) = 12 h (x) = 6eln 7x = 6 · 7x = 42x ⇒ h′ (x) = 42 y = ex + e− x ⇒ y′ = ex − e− x y = ex − e− x ⇒ y′ = ex + e− x 3
3
26. y = e 5 x ⇒ y ′ = e 5 x ⋅ 15 x 2 = 15 x 2 e 5 x 5
5
27. y = 8e x ⇒ y ′ = 8e x ⋅ 5 x 4 = 40 x 4 e x
3
5
28. f (x) = 0.42x ⇒ ln f (x) = 2x ln 0.4 ⇒ 1 f ′( x ) = 2 ln 0.4 ⇒ f ′( x ) = 0.4 2 x ⋅ 2 ln 0.4 f ( x) 29. f (x) = 10− 0.2 x ⇒ ln f (x) = −0.2x ln 10 ⇒ 1 f ′( x ) = −0.2 ln 10 ⇒ f ( x)
124
Problem Set 6-4
Calculus Solutions Manual © 2005 Key Curriculum Press
42. 43. 44. 45. 46.
d dx
x
∫ 10 3
d dx
∫
x
d dx
∫
4x
∫
x2
d dx
t
3
dt = 10 x
ln t dt = ln x
5
6.3
log 2 t dt = 4 log 2 ( 4 x ) ln (cos t ) dt = 2 x (ln cos x 2 )
5 d2 d2 d 5 5 (ln ) = =− 2 x 2 2 (5 ln x ) = dx dx dx x x
e = lim(1 + n)1/ n n→0
Q3. Q5. Q7. Q9.
∫
51.
∫
52. 53. 54. 55. 56. 57. 58. 59.
72 x +C 2 ln 7 1.05 x 1.05 x dx = +C ln 1.05 72 x dx =
∫ 6e dx = 6e + C ∫ e dx = 5e + C ∫ e cos x dx = e + C ∫ e sec x dx = e + C 1 ∫ e dx = ∫ x dx = 4 x + C ∫ 60e dx = 60∫ 5x dx = 150 x + C 1 ∫ (1 + e ) e dx = 102 (1 + e ) + C 1 ∫ (1 − e ) e dx = − 404 (1 − e ) + C x
2
3 ln x
3
4
2
2 x 50 2 x
2 x 51
4 x 100 4 x
0
−x
61.
∫
2
−1
1
1
x
4 tan 3 x 0 → x→0 5x 0 2 12 sec 3 x 12 = lim = x→0 5 5
2. lim
y
4 x 101
− e ) dx = e + e x
−x
1
1
x
2 0
−2
= e + e − 1 − 1 = 5.524391... Numerically: integral ≈ 5.524391... (Checks.) 2
y
tan x
ln 5 x
∫ (e
x e ex E
sin x
tan x
x
1 Q4. x Q6. (ln x)/(ln b) Q8. −x −e + C Q10. 2 sin 5 x 0 1. lim → x→0 3x 0 10 cos 5 x 10 = lim = x→0 3 3
0.2 x
sin x
2
60.
x
0.2 x
1 Q2. e = lim 1 + or n→∞ n
Q1. e ≈ 2.71828
∫
∫
Problem Set 6-5 n
d 2 7x d 47. (7e 7 x ) = 49e 7 x 2 (e ) = dx dx 1 48. e 5 x dx = e 5 x + C 5 1 49. e 7 x dx = e 7 x + C 7 50.
Step 4: Write division as multiplication by the reciprocal; 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.
(e x + e – x ) dx = (e x − e – x )
2 −1
= 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 in reverse.
Calculus Solutions Manual © 2005 Key Curriculum Press
tan x 0 → x 0 sec 2 x =1 = lim x→0 1 sin x 0 → 4. lim x→0 x 0 cos x = 1, a “well-known” limit. = lim x→0 1
3. lim x→0
Problem Set 6-5
125
1 − cos x 0 → x→0 x2 0 sin x 0 → = lim x→0 2 x 0 cos x 1 = = lim x→0 2 2
5. lim
6. lim x→0
x2 0 → cos 3 x − 1 0
2x 0 → = lim x →0 −3 sin 3 x 0 = lim x→0
2 2 =− −9 cos 3 x 9
sin x 0 7. lim+ 2 → x→0 x 0 cos x = lim+ =∞ x→0 2x 1 − cos x 0 8. lim 2 → x→0 x + x 0 sin x =0 = lim x→0 1 + 2 x ln x −∞ → 9. lim+ x →0 1/ x ∞ = lim+ x→0
x −1 = lim ( − x ) = 0 − x −2 x→0 +
1 e3 x 10. lim 2 = ∞ Form is . x→0 x 0 ex − e 0 → x →1 5 ln x 0 ex e = lim −1 = x →1 5 x 5
11. lim
ln x − x + 1 0 12. lim 2 → x →1 x − 2 x + 1 0 x −1 − 1 0 = lim → x →1 2 x − 2 0 − x −2 1 = lim =− x →1 2 2 3x + 5 11 13. lim = = −26.43297K x →2 cos x cos 2 14. lim x →2
tan x tan 2 . = ∞ Form is x−2 0
∞ e 15. lim 2 → x →∞ x ∞ x
∞ ex → x →∞ 2 x ∞ ex = lim =∞ x →∞ 2 = lim
126
Problem Set 6-5
∞ x3 x → x →∞ e ∞ ∞ 3x 2 = lim x → x →∞ e ∞ ∞ 6x = lim x → x →∞ e ∞ 6 6 = lim x = 0 Form: . x →∞ e ∞
16. lim
3 x + 17 3 3 = lim = 4 x – 11 x→∞ 4 4 2 – 7x –7 7 18. lim = lim =− x →∞ 3 + 5 x x →∞ 5 5 17. lim
x →∞
x 3 – 5 x 2 + 13 x – 21 ∞ → 4 x 3 + 9 x 2 – 11x – 17 ∞
19. lim
x →∞
∞ 3 x 2 – 10 x + 13 → 2 x →∞ 12 x + 18 x – 11 ∞
= lim
∞ 6 x – 10 → ∞ 24 x + 18 6 1 = lim = x →∞ 24 4 = lim
x →∞
3x 5 + 2 ∞ → 5 x →∞ 7 x – 8 ∞ 15 x 4 15 3 = lim = 4 = lim x →∞ 35 x x →∞ 35 7
20. lim
21. L = lim+ x x → 0 0 x→0
ln L = lim+ ( x ln x ) = lim+ = lim+
x→0 −1
x→0
x→0
ln x −∞ –1 → x ∞
x = lim ( − x ) = 0 − x −2 x→0 +
∴ L = e0 = 1 22. L = lim+ (sin x )sin x → 0 0 x→0
ln sin x −∞ → csc x ∞ 1/(sin x ) ⋅ cos x 1 = lim+ = lim+ x→0 x →0 − csc x − csc x cot x
ln L = lim+ sin x (ln sin x ) = lim+ x→0
x→0
= lim+ ( − sin x ) = 0 x→0
∴ L = e0 = 1 23. L = lim − (sin x ) tan x → 1∞ x →π / 2
ln L = lim − tan x (ln sin x ) x →π / 2
ln sin x 0 → x →π /2 cot x 0 (1/sin x ) ⋅ cos x = lim − x →π /2 − csc 2 x = lim −
= lim − x →π / 2
− cos x sin 2 x sin x
Calculus Solutions Manual © 2005 Key Curriculum Press
cos x − 1 0 → x cos x + sin x 0 − sin x = lim =0 x →0 − x sin x + 2 cos x
= lim − ( − cos x sin x ) = 0
= lim
x →π / 2
x→0
∴ L = e0 = 1 24. L = lim+ x 1/( x −1) → 1∞ x →1
ln L = lim+ [1/( x − 1) ⋅ ln x ] = lim+ x →1
π π x − tan 2 x. Where secant and 2 2 tangent are defined, the Pythagorean properties tell us that f (x) = 1.
31. f ( x ) = sec 2
x →1
ln x 0 1/ x → = lim+ =1 0 x→1 1 x −1
∴ L = e1 = e
f (x )
25. L = lim(1 + ax )1/ x → ∞ 0 ( Note: a ≥ 0.) x →∞
ln (1 + ax ) ∞ → x →∞ x →∞ x ∞ 1/(1 + ax ) ⋅ a a = lim = lim =0 x →∞ x →∞ 1 + ax 1 ∴ L = e0 = 1
1
ln L = lim [1/ x ⋅ ln (1 + ax )] = lim
26. L = lim (1 + ax )1/ x → 1∞ x→0
ln (1 + ax ) 0 → x→0 x→0 x 0 1/(1 + ax ) ⋅ a a = lim = lim =a x→0 x →0 1 + ax 1 ∴ L = ea ln L = lim [1/ x ⋅ ln (1 + ax )] = lim
27. L = lim+ x 3/(ln x ) → 0 0 x→0
ln L = lim+ [3/(ln x ) ⋅ ln x ] = lim+ 3 = 3 x→0 3
x→0
∴ L = e = 20.08553... 28. L = lim+ (7 x )5/(ln x ) → 0 0 x→0
ln L = lim+ [5/(ln x ) ⋅ ln(7 x )] x→0
= lim+
−∞ 5 ln (7 x ) → ln x −∞
= lim+
5 ⋅ [1/(7 x )] ⋅ 7 =5 1/ x
x→0
x→0
∴ L = e 5 = 148.4131...
x –1
1
ex – 1 0 x → x →0 1(e – 1) + x ⋅ e 0
tan x sec 2 x = lim x →π /2 sec x x →π /2 sec x tan x sec x = lim , the original expression! x →π /2 tan x = lim
Using tan x = (sec x)/(csc x), the expression reduces to sec x lim = lim csc x = 1 x →π /2 (sec x )/(csc x ) x →π /2 33. L = lim+ x k /(ln x ) → 0 0 x→0
ln L = lim+ [k/(ln x ) ⋅ ln x ] = lim+ k = k x→0
ex – 1 0 x x → x →0 e – 1 + xe 0 ex 1 = lim x x x = x →0 e + e + xe 2 1 1 30. lim − →∞−∞ x→0 x sin x x→0
x→0
∴ L = ek The graph turns out to be a horizontal line, y = ek, defined for x > 0. y
y = ek
x
x
= lim
= lim
5
32. Using l’Hospital’s rule leads to sec x sec x tan x lim = lim x →π /2 tan x x →π /2 sec 2 x
1 1 29. lim − x → ∞ − ∞ x→0 x e − 1 x e –1− x 0 = lim → x x →0 x (e – 1) 0 = lim
3
sin x − x 0 → x sin x 0
Calculus Solutions Manual © 2005 Key Curriculum Press
By the definition of a power, x k /( ln x ) = ( x k )1/ln x = (e k ln x )1/ln x = e k g( x ) 0.3 x 2 – 2.7 = h( x ) 0.2 x 2 – 2 x + 4.2 g(3) = 0.3(9) − 2.7 = 0, h(3) = 0.2(9) − 2(3) + 4.2 = 0, Q .E.D .
34. a. f ( x ) =
b. g′ (x) = 0.6x ⇒ g′ (3) = 1.8 h′ (x) = 0.4x − 2 ⇒ h′ (3) = −0.8
Problem Set 6-5
127
Tangent lines at (3, 0) have these equations. For g: y1 = 1.8(x − 3) For h: y2 = −0.8(x − 3) y 1.8( x – 3) c. 1 = = −2.25, for x ≠ 3. y2 –0.8( x – 3) g′(3) 1.8 = = −2.25, which equals y1/y2, h′(3) –0.8 Q .E .D . d. Because the ratio g (x)/h(x) approaches the ratio y1/y2 as x approaches 3, and because y1/y2 equals g′ (3)/h′ (3) for all x ≠ 3, the ratio g (x)/h(x) also approaches g′ (3)/h′ (3) as x approaches 3. This is what l’Hospital’s rule concludes. If g (3) or h (3) were any number other than 0, the canceling of the (x − 3)’s in part c could not be done, and the ratio would almost certainly not equal 1.8/(−0.8). e. The graph shows a removable discontinuity at (3, −2.25): f (x )
x
1 3
35. a. For yearly compounding, m(t) = 1000(1 + 0.06)t. For semiannual compounding, m ( t) = 1000(1 + 0.06/2)2t because there are two compounding periods per year, each of which gets half the interest rate. b. m(t) = 1000(1 + 0.06/n)nt lim m(t ) = lim 1000(1 + 0.06/n) nt
n→∞
n→∞
= 1000 lim (1 + 0.06/n)
nt
n→∞
Let L = lim (1 + 0.06/n) nt .
c. t 5 20 50
m(t), m(t), Annual Continuous 1,338.23 1,349.86 3,207.14 3,320.12 18,420.15 20,085.54
Difference 11.63 112.98 1,665.38
d. For 7% interest, compounded continuously, m (t) = 1000e0.0 7t . 36. a. f (x) = x n, g (x) = ln x, h (x) = ex ∞ f ( x) xn lim = lim → x →∞ g( x ) x →∞ ln x ∞ nx n –1 = lim nx n = ∞, if n > 0 x →∞ 1/ x x →∞
= lim
∴ a power function is higher-order than the natural log function. ∞ f ( x) xn lim = lim x → x →∞ h( x ) x →∞ e ∞ ∞ nx n –1 → , if n − 1 > 0 x x →∞ e ∞ Eventually, the exponent of the power will become zero, in which case the limit takes the form constant/∞, which is 0. ∴ a power function is lower-order than an exponential function. Using “ 50 for 6.3245… − 0.2899… = 6.03… , or about 6 hours. iv. C 1(t) = 200t · 0.3t
c. lim x 3e − x → ∞ ⋅ 0 x →∞
∞ x3 x → x →∞ e ∞ 2 ∞ 3x = lim x → x →∞ e ∞ ∞ 6x = lim x → x →∞ e ∞ 6 = lim x = 0 (Form: 6/∞) x →∞ e = lim
d. L = lim x tan (πx/2 ) → 1∞ x →1
ln L = lim [tan (π x/2) ⋅ ln x ] x →1
= lim
ln x 0 → cot (π x/2) 0
= lim
−2 1/ x 1 = = 2 −(π /2)csc π x/2 −π /2 π
x →1
x →1
∴ L = e− 2/ π = 0.529077… e. lim 3 x 4 = 48 x →2
f.
lim (tan 2 x − sec 2 x ) = lim (1) = 1
x →π /2
x →π /2
g. Examples of indeterminate forms: 0/0, ∞/∞, 0 · ∞, 00, 1∞ , ∞ 0, ∞ − ∞
C (t )
100 50
t 1
From the graph, the maximum is about 60 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 iii gives C (t) > 50 for 0.409… < t < 1.473… , or for about 1.06 hours. In conclusion, the concentration peaks sooner at a lower concentration and stays above 50 ppm for a much shorter time. ln 0.5 = −0.025t
Calculus Solutions Manual © 2005 Key Curriculum Press
2x2 − 3 ∞ 2 → x →∞ 7 − 5 x −∞ 4x 2 = lim =− x →∞ −10 x 5 2 x − cos x + 1 0 b. lim → x x→0 e − x −1 0 2 x + sin x 0 = lim → x→0 ex − 1 0 2 + cos x 2 + 1 = lim = =3 x→0 ex 1
R5. a. lim
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 4 ln x 4 iv. y = log 3 x 4 = ⇒ y′ = ln 3 x ln 3 b. i. ii.
∫e ∫2 = ∫e
−1.7 x
sec x
dx = ( −1 / 1.7) e −1.7 x + C
sec x tan x dx
ln 2 sec x
sec x tan x dx
Problem Set 6-7
133
ii. E( x ) =
x
∫ 150e 0
−0.16 t
dt = 937.5( − e −0.16 x + 1)
E(5) = 937.5(−e− 0.8 + 1) = 516.25… ppm · days E(10) = 937.5(−e− 1.6 + 1) = 748.22… ppm · days As x grows very large, E(x) seems to approach 937.5. iii. E′ (x) = 150e− 0.16x = C (x) E′ (5) = 67.39… ppm (or ppm · days per day) E′ (10) = 30.28… ppm f. i. From Figure 6-7d, the maximum concentration is about 150 ppm at about 2 hours. (These values can be found more precisely by setting the numerical or algebraic derivative equal to zero, solving to get t = −1/ln 0.6 = 1.9576… . Then C (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/h when t = 1 and decreasing at about 24.2 ppm/h when t = 5. The concentration is increasing if C′ (t) is positive and decreasing if it is negative. iii. Solving 50 = 200t · 0.6t numerically for t gives 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 C (t )
c. lim x 3e − x → ∞ ⋅ 0 x →∞
x3 ∞ x → x →∞ e ∞ 2 3x ∞ = lim x → x →∞ e ∞ 6x ∞ = lim x → x →∞ e ∞ 6 = lim x = 0 (Form: 6/∞) x →∞ e = lim
d. L = lim x tan (πx/2 ) → 1∞ x →1
ln L = lim [tan (π x/2) ⋅ ln x ] x →1
= lim
0 ln x → 0 cot (π x/2)
= lim
1/ x 1 −2 = = 2 −(π /2)csc π x/2 −π /2 π
x →1
x →1
∴ L = e− 2/ π = 0.529077… e. lim 3 x 4 = 48 x →2
f.
lim (tan x − sec x ) = lim ( −1) = −1 2
2
x → π /2
x → π /2
g. Examples of indeterminate forms: 0/0, ∞/∞, 0 · ∞, 00, 1∞ , ∞ 0, ∞ − ∞
100 50
t 1
From the graph, the maximum is about 60 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 iii gives C (t) > 50 for 0.409… < t < 1.473… , or for about 1.06 hours. In conclusion, the concentration peaks sooner at a lower concentration and stays above 50 ppm for a much shorter time. ln 0.5 = −0.025t Calculus Solutions Manual © 2005 Key Curriculum Press
2x2 − 3 ∞ 2 → x →∞ 7 − 5 x −∞ 4x 2 = lim =− x →∞ −10 x 5 2 x − cos x + 1 0 b. lim → x x→0 e − x −1 0 2 x + sin x 0 = lim → x→0 ex − 1 0 2 + cos x 2 + 1 = lim = =3 x→0 ex 1
R5. a. lim
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 · e 2x + x− 3 · 2e2x = x− 4e2x (2x − 3) iii. y = cos (2x ) ⇒ y′ = −sin (2x ) · 2x ln 2 4 ln x 4 iv. y = log 3 x 4 = ⇒ y′ = ln 3 x ln 3 b. i. ii.
∫e ∫2 = ∫e
−1.7 x
sec x
dx = ( −1 / 1.7) e −1.7 x + C
sec x tan x dx
ln 2 sec x
sec x tan x dx
Problem Set 6-7
133
C4. a. Suppose there is a number M > 0 such that ln x ≤ M for all x > 0. Let x = e M +1. Then ln x = ln e M +1 = (M + 1) ln e = M + 1 > M. This contradicts ln x ≤ M for all x > 0. Thus the supposition is false, and there can be no such number M that is an upper bound for ln x, Q .E .D . b. If M were a lower bound for ln x, then −M would be an upper bound for ln (1/x), but part a shows no such number can exist. c. ln′ x = 1/x, which shows that ln is differentiable for all x > 0. Thus, ln is continuous for all x > 0 because differentiability implies continuity. d. Because ln is continuous for all x > 0, the intermediate value theorem applies. Thus, if k is between ln a and ln b, there is a number c between a and b such that ln c = k. y = ln x ln b
k ln a
x a
c
b
e. Part a shows k cannot be an upper bound for ln, so there must be some b > 0 such that ln b > k. Similarly, part b shows k is not a lower bound, so some a > 0 exists for which ln a < k. By part d there is some number c between a and b such that ln c = k, Q .E .D . f. The domain of ln is the positive reals, and the range is all reals; the domain of the inverse to ln (i.e., exp) is the range of ln (i.e., all reals), and the range of the inverse is the domain of ln (i.e., positive reals). C5. a. g( x ) =
∫
4
x2
sin t dt = −
g′ (x) = −2x sin b. g( x ) = =
∫
4
x
2
∫
∫
x2
4
sin t dt ⇒
x2
x2
sin t dt
∫
x1
1
t
dt n
1 T2. e = lim 1 + or e = lim (1 + n)1/n n→∞ n→0 n T3. If g (x) =
∫
x
f (t ) dt and f ( t ) is continuous in a
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, then f ( 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 and g′ (x) = (1/x) · loge e = (1/x) · 1 = 1/x ∴ f ′ ( x ) = g′ (x) for all x > 0 f ( 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 ) 1 a. f ′( x ) = 3 x ⋅ (3 x 2 e x + x 3e x ) = 3/ x + 1 x e 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′ = 6e 2 x ⋅ ln x + 3e 2 x ⋅ (1 / x ) = 3e 2 x (2 ln x + 1/x ) 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(log 2 4 x )6 = x ln 2 T10. t ( x ) = ln (cos2 x + sin2 x) = ln 1 = 0 ⇒ t′ (x) = 0
∫
tan x
4
T11. p( x ) = sin t dt ⇒
g′ (x) = −2x sin x2 + sin (tan x) sec2 x
∫
T1. ln x =
tan x
sin t dt +
c. g( x ) =
Chapter Test
v( x )
u( x )
Calculus Solutions Manual © 2005 Key Curriculum Press
1
e t sin t dt ⇒
p′ (x) = e ln x sin ln x · 1/x = sin ln x T12.
∫e
T13.
∫
T14.
∫
f (t ) dt ⇒
g′ (x) = f (v(x)) · v′ (x) − f ( u ( x ) ) · u′ (x) C6. log cabin (or log cabin + C, which equals “houseboat”)
∫
ln x
1 5x e +C 5 1 (ln x )6 ( dx/x ) = (ln x ) 7 + C 7 1 sec 5 x dx = ln | sec 5 x + tan 5 x | + C 5 5x
dx =
Problem Set 6-7
135
1 x2 1 5 = (25 – 1) 0 ln 5 0 ln 5 = 14.9120… −∞ 5 – 3x T16. lim → x →∞ ln 4 x ∞ –3 = lim = lim(–3 x ) = −∞ x →∞ [1/( 4 x )] ⋅ 4 x →∞ T15.
∫
2
5 x dx =
T17. L = lim − (tan x ) cot x → ∞ 0 x →π / 2
ln L = lim − [cot x ⋅ ln ( tan x )] → 0 ⋅ ∞ x →π / 2
ln (tan x ) ∞ = lim − → x →π / 2 tan x ∞ = lim − x →π / 2
∴L=e =1 T18. a. 0
(1 / tan x ) ⋅ sec 2 x = lim − cot x = 0 x →π / 2 sec 2 x
But h ′ ( x ) = f ′ ( x ) − g′ (x), which equals 0 for all values of x. ∴ h ′ (c) = 0 This result thus contradicts the mean value theorem, Q.E.D. T22. a. F ( x ) = 60e0.1 x ⇒ F ′ ( x ) = 6e0.1 x so F ′ ( 5 ) = 6e0.5 = 9.8923… lb/ft F ′ ( 1 0 ) = 6e = 16.3096… lb/ft b. Work equals force times displacement. But the force varies at different displacements. Thus, a definite integral has to be used. c. F
100
(x, F )
y 5 4 3 2 1
dx 5
dW = F dx = 60 e0.1 x dx
t or x 1 2 3 4 5 6 7 8 9 10
-1 -2 -3 -4 -5
x
g
W=
f
5
∫ 60e
0.1 x
dx
0
5
= 600e 0.1x = 600(e 0.5 − 1) 0
b. h( x ) =
∫
3 x −5
2
W ≈ 389.23 ft-lb f (t ) dt ⇒ h ′( x ) = f (3 x − 5) ⋅ 3,
T23. Answers will vary.
h′(3) = f (9 − 5) ⋅ 3 = f ( 4) ⋅ 3 = 1 ⋅ 3 = 3 T19. ln x =
∫
x1
dt, so ln 1.8 =
1.8 1
Problem Set 6-8
1
Cumulative Review, Chapters 1–6
∫
dt t 1 1 1 1 M4 = 0.2 + + + = 0.58664 K 1.1 1.3 1.5 1.7 t
1
From calculator, ln 1.8 = 0.58778… . T20. g( x ) =
∫
x2
2
sin t dt = − cos t
x2 2
= −cos x2 + cos 2 ⇒ g′(x) = 2x sin x2 2
d x g ′( x ) = sin t dt = 2 x sin x 2 dx 2 T21. Let h ( x ) = f ( x ) − g (x). Then h ( a) = f ( a) − g (a) = 0 and h ( b ) = f ( b ) − g (b) ≠ 0. h( b ) – h( a ) ∴ ≠0 b–a By the mean value theorem, there is a number c between a and b such that h( b ) – h( a ) h ′( c ) = . b–a ∴ h′ (c) ≠ 0
∫
136
Problem Set 6-8
1. f ( x ) = 2x 2 3.1 – 2 2.9 f ′(3) ≈ = 5.549618K 0.2 2. There are about 10.0 squares, each 20 units. ∴
∫
50
10
g( x ) dx ≈ 200
(Function is g( x ) = 2 + 0.1x + sin
2π x, so exact 15
answer is 200.) 3. L = lim f ( x ) if and only if for any ε > 0, there is x →c a δ > 0 such that if x is within δ units of c but not equal to c, f ( x ) is within ε units of L. 4. Answers may vary. f (x ) 4
x 3
Calculus Solutions Manual © 2005 Key Curriculum Press
f ( x + h) – f ( x ) h f ( x ) − f (c ) or f ′(c) = lim x →c x−c 6. f (x) = x 3 ( x + h)3 − x 3 f ′( x ) = lim h→0 h 3 x + 3 x 2 h + 3 xh 2 + h 3 – x 3 = lim h→0 h 2 = lim(3 x + 3 xh + h 2 ) = 3 x 2 , Q.E.D. 5. f ′ (x) = lim
12. Optional graph showing upper sum:
h→0
y=x2
10
x 1
∫
U 6 = 0.5(1.52 + 22 + 2.52 + 32 + 3.52 + 42) = 24.875
x3
8. f (x) =
x 2 dx
1
h→0
7. f (x) = ⇒ f ′ (x) = 3x 2 f ′ (5) = 3·52 = 75 5.013 – 4.99 3 = 75.0001 f ′ (5) ≈ 0.02 5.0013 – 4.999 3 f ′ (5) ≈ = 75.000001 0.002 The symmetric differences are getting closer to 75 as ∆ x gets closer to zero.
4
13. M 10 = 20.9775 M 100 = 20.999775 Sums seem to be approaching 21. 1 14. a. cos 5 x sin x dx = − cos 6 x + C 6
∫ ∫ (1/x ) dx = ln | x | + C ∫ tan x dx = ln | sec x | + C = − ln | cos x | + C ∫ sec x dx = ln | sec x + tan x | + C 1 ∫ (3x − 5) dx = 3 ∫ (3x − 5) (3 dx )
b.
3 x – 5 = (3 x − 5) 1 f ′ (x) = (3 x – 5) −1/ 2 ⋅ 3 = 1.5(3 x − 5) −1/ 2 2 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 at x = 7. 1/ 2
c. d.
1/ 2
e.
f (x )
=
5
8
3
15. x 5
10. a. y = e 2x cos 3x ⇒ y′ = 2 e2x cos 3x − 3e2x sin 3x ln x b. q( x ) = ⇒ q ′( x ) = tan x 2 (tan x )/x − ln x sec x 1 ln x = − 2 2 tan x x tan x sin x d2 x d [(ln 5)5 x ] = (ln 5)2 5 x c. 2 (5 ) = dx dx 11. For the function to be differentiable, lim− ( ax 2 + 1) = lim+ ( − x 2 + 6 x + b) and x →2
x →2
lim 2 ax = lim+ ( −2 x + 6) .
x →2 −
x →2
4a + 1 = 8 + b and 4a = 2 ⇒ a = y
1 and b = −5 2
4
1/ 2
2 1 2 ⋅ (3 x – 5)3/ 2 + C = (3 x – 5)3/ 2 + C 9 3 3
1 3 4 64 1 = − = 21, x 1 3 3 3 1 which agrees with the conjecture in Problem 13.
∫
4
x 2 dx =
16. The graph shows a tangent line at x = c parallel to the secant line. f (x )
x a
c b
Statement: If f is differentiable on (a, b) and continuous at x = a and x = b, then there is a number x = c in f (b) – f ( a) (a, b) such that f ′( x ) = . b–a 17. y = x9/7 y7 = x9 7y6 y′ = 9x 8 9x8 9x8 9 9 9 = x 8−54 / 7 = x 2 / 7 = x 9/ 7−1 6 = 7y 7( x 9/ 7 )6 7 7 7 as from the derivative of a power formula y′ =
2
x 2
Calculus Solutions Manual © 2005 Key Curriculum Press
Problem Set 6-8
137
18. If x− 1 were the derivative of a power, then the power would have to be x0. But x0 = 1, so its derivative equals 0, not x− 1. Thus, x− 1 is not the derivative of a power, Q.E.D. 19. f ( x ) =
∫
tan x
1
cos 3t dt ⇒
f ′ (x) = cos (3 tan x) · sec2 x 20. f ( x ) =
x
∫ (1/t ) dt ⇒ f ′( x ) = 1/x, Q.E.D. 1
21. Prove ln xa = a ln x for any constant a and all x > 0. Proof: Let f (x) = ln xa and g (x) = a ln x. 1 1 a Then f ′( x ) = a ⋅ ax a−1 = a ⋅ = and x x x 1 a g ′( x ) = a ⋅ = . x x ∴ f ′ (x) = g′ (x) for all x > 0 f (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 thus ln xa = a ln x for all x ≥ 0, Q .E .D . 22. x = 5 cos t, y = 3 sin t dy dy/dt 3 cos t 3 ∴ = = = − cos t dx dx/dt –5 sin t 5 23. At t = 2, (x, y) = (5 cos 2, 3 sin 2) = (−2.08… , 2.72…). dy 3 At t = 2, = − cot 2 = 0.2745… . dx 5 The graph shows that a line of slope 0.27… at point (−2.08… , 2.72…) is tangent to the curve. y
26. L = lim (1 + n)1/n → 1∞ n→0
1 ln L = lim ln (1 + n) → ∞ ⋅ 0 n→0 n ln (1 + n) 0 = lim → n→0 n 0 1/(1 + n) = lim =1 n→0 1 ∴ L = e1 = e, Q.E.D. dx dy 27. Know: = −30 ft/s, = 40 ft/s dt dt dz Want: when x = 200 and y = 100 dt x2 + y2 = z2 dx dy dz 2x + 2y = 2z dt dt dt When x = 200 and y = 100, z = 50, 000 = 100 5 . 2(200)( −30) + 2(100)( 40) = 2 ⋅ 100 5
dz dt
dz −20 = = −8.94427K ft/s dt 5 The distance z is decreasing. 28.
∫
5
2
f (x ) dx ≈ (1/3)(0.5)(100 + 4 · 150 + 2 · 170 +
4 · 185 + 2 · 190 + 4 · 220 + 300) = (1/3)(0.5)(3340) = 556 23 29. Area of cross section = πy2 Because the end of the radius is on a line through the origin with slope r/h, y = (r/h)x. πr 2 ∴ Area = π [(r/h) x ]2 = 2 x 2 h Area
3
x (x, Area)
5
dx
x h
24. y = tan− 1 t dy 1 v= = = (1 + t 2 ) −1 dt 1 + t 2 dv 2t a= = −1(1 + t 2 ) −2 ⋅ 2t = − dt (1 + t 2 )2 3x e –1 0 → 25. lim x →0 sin 5 x 0 3e 3 x 3 = lim = x →0 5 cos 5 x 5
138
Problem Set 6-8
dV = (Area) dx ∴V=
∫
h
0
( Area ) dx =
∫
h πr 2
0
h2
x 2 dx
πr 2 1 3 h 1 r 2 3 1 ⋅ x = π (h − 0 3 ) = πr 2 h, Q .E.D . h2 3 0 3 h2 3 30. Answers will vary. =
Calculus Solutions Manual © 2005 Key Curriculum Press
Chapter 7—The Calculus of Growth and Decay Problem Set 7-1
Q9.
1. D (0) = 500 D (10) = 895.4238482… D (20) = 1603.567736… 2. D ′ (t) = 500(ln 1.06)(1.06t ) $/yr D ′ (0) = 29.13445406… D ′ (10) = 52.17536994… D ′ (20) = 93.43814108… The rate of change, in $/yr, increases as the amount in the account increases. D′(t ) 500 ⋅ (ln 1.06) ⋅ (1.06 t ) = D(t ) 500 ⋅ (1.06 t ) = ln 1.06 = 0.0582689081… R(0) = ln 1.06 R(10) = ln 1.06 R(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. 3. R(t ) =
Problem Set 7-2 Q1.
Q2. y
y
1
1
x
Q3.
x
Q4. y
y
Q6. y
y x
3
x x
4 3
1. a. B = number of millions of bacteria; t = number of hours
∫
∫
dB/dt = kB ⇒ dB/ B = k dt ⇒ ln | B| = kt + C | B| = e kt +C = e kt ⋅ e C ⇒ B = C1e kt b. 5 = C1e k⋅0 ⇒ C1 = 5 7 = 5e3k ⇒ ln (7/5) = 3k 1 7 ⇒ k = ⋅ ln = 0.112157K 3 5 7 ∴ B = 5e (1/3) ln( 7/5)t = 5 5
t/3
= 5e 0.112157Kt
c. B
t
5 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) 3 ln 200 t= = 47.24001K ln (7/5) About 47 hours after start, so in a little less than 2 days 2. a. N = number of units of radiation from N17; t = number of seconds
∫
x
Q5.
y
∫
dN /dt = kN ⇒ dN / N = k dt
x 1
Q10. y
x
⇒ ln |N| = kt + C | N | = e kt +C ⇒ N = C1e kt
b. 3 × 1017 = C 1ek·0 ⇒ C 1 = 3 × 1017 5.6 × 1013 = 3 × 1017e60k ⇒ ln(1.866 K × 10 −4 ) = 60 k ⇒ k = −0.143103… ∴ N = 3 × 1017 e −0.143103Kt c. N
Q7.
Q8. y
3 × 1017
y 3
x x
Calculus Solutions Manual © 2005 Key Curriculum Press
t
Problem Set 7-2
139
d. t = 5(60) = 300 s N = 3 × 1017 e ( −0.143103K)(300 ) = 0.067991K It 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 ⇒ ln | F| = kt + C
∫
∫
| F| = e kt +C ⇒ F = C1e kt 50 = C 1ek·0 ⇒ C 1 = 50 30 = 50e20k ⇒ ln (30/50) = 20k ⇒ k = (1/20) ⋅ ln (0.6) = −0.025541… F = 50e ln ( 0.6 )t / 20 = 50(0.6)t / 20 = 50e −0.025541Kt b. F 50
t 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 ⇒ ln |V | = kt + C |V | = e kt +C ⇒ V = C1e kt
b. 4200 = C1e k⋅0 ⇒ C1 = 4200 4700 = 4200e ( k )( −3) ⇒ ln( 4700/4200) = −3k k = (−1/3) ln (4700/4200) = −0.037492… ∴ V = 4200e −0.037492Kt c. V
4200
t –30
30
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 = 4200e ( −0.037492K)( −34 ) = 15026.795K About $15,000 g. The difference between $16,000 and $15,000 is the dealer’s profit. 140
Problem Set 7-2
5. a. dC/dt = kC b.
∫ dC/C = ∫ k dt ⇒ ln |C| = kt + D
⇒ | C | = e kt + D ⇒ C = D1e kt 0.00372 = D1e k⋅0 ⇒ D1 = 0.00372 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…t t = −21.05… , which is before the poison was inhaled, 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. P 100
50
t 20,000
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 ⇒ ln | P | = kt + C
⇒ | P| = e kt +C ⇒ P = C1e kt 100 = C1e k⋅0 ⇒ C1 = 100 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 = 100e −0.0001205473Kt ln 0.4837 = −0.0001205473…t t = 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. 100
P
61.7 50
t 4000 5750
20,000
Calculus Solutions Manual © 2005 Key Curriculum Press
7. dM/dt = kM ⇒ M = Cekt by the techniques in Problems 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 = Ce it 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: $4055.20 Leaving the money in the account twice as long has the same effect as doubling the interest rate. Doubling the amount invested obviously doubles the money at any particular time, but that doesn’t tell us how that compares with doubling the time or the interest rate. Algebraically, Cei ·2t = Ce2i ·t shows that doubling the time is equivalent to doubling the interest rate. Solving Ce2it > 2Ceit gives Ce2it − 2Ceit > 0 ⇒ Ceit(eit − 2) > 0 ⇒ eit > 2 (because Ceit > 0, so C > 0, being an investment) ⇒ it > ln 2. So doubling either the time or the interest rate will always eventually yield more than doubling the investment, 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 more than 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 = $1402.55 Quarterly: M = 1000(1.0175)20 = $1414.78 Monthly: M = 1000(1.00583…)60 = $1417.63 Daily: M = 1000(1.0001917808…)1825 = $1419.02 Continuously (Problem 7): $1419.07 Note that compounding continuously is only 5 cents better than compounding daily for a $1000 investment in 5 years! M = M 0(1 + k/n)nt Let L = lim (1 + k/n) nt → 1∞. n→∞
ln L = lim [nt ⋅ ln (1 + k/n)] → ∞ ⋅ 0
9.
dy = 0.3 y dx
∫
20
n→∞
compounding equation. Calculus Solutions Manual © 2005 Key Curriculum Press
y
10
x 5 –10 –20
10.
dy = −0.2 y dx dy = −0.2 x dx y ln |y| = −0.2x + C | y | = e −0.2 x +C = e −0.2 x ⋅ e C y = ±eC ⋅ e− 0.2 x = C1e− 0.2 x 30 = C1e− 1.4
∫
∫
30 = 7.3979K e1.4 ∴ y = 7.3979…e− 0.2 x C1 =
∫
∫
11. dy/dx = ky ⇒ dy/y = k dx ⇒ ln | y | = kx + C1 | y | = e kx +C1 ⇒ y = Ce kx y(0) = Ce k⋅0 ⇒ C = y(0) ⇒ y = y0 e kx , Q.E.D.
Problem Set 7-3 Q1. Cekx Q3. kx + C
Q2. (kx2)/2 + C Q4. −cos x + C
Q5. 1/ 1 – x 2
Q6. 5 cos x
Q7. tan x Q8.
n→∞
∞ t ⋅ ln (1 + k/n) = lim → n→∞ ∞ n –1 1 −2 t⋅ ⋅ ( − kn ) 1 + k /n = lim n→∞ − n −2 kt = lim = kt n→∞ 1 + k /n ∴ L = ekt ∴ lim M = M0 e kt , which is the continuous
∫
dy = 0.3 dx y ln |y| = 0.3x + C | y | = e 0.3 x +C = e 0.3 x ⋅ e C y = ±eC ⋅ e0.3 x = C1e0.3 x −4 = C1e0 ⇒ C1 = −4, showing that C1 can be negative. ∴ y = −4e0.3 x
y' or y y
1
x 1
y'
Q9. lim Ln = lim Un ∆x →0
∆x →0
Q10. B 1. a. dM/dt = 100 − S b. S = kM ⇒ dM/dt = 100 − kM Problem Set 7-3
141
c.
− k dM
∫ 100 − kM = ∫ dt ⇒ − k ∫ 100 − kM = ∫ dt dM
1
1 ⇒ − ln |100 − kM | = t + C k ⇒ |100 − kM| = e− kte− kC ⇒ 100 − km = C 1e− kt ⇒ kM = 100 − C 1e− k t 1 ⇒ M = (100 – C1e – kt ) k Substitute M = 0 when t = 0. 1 0 = (100 – C1e 0 ) ⇒ C1 = 100 k 100 ∴M = (1 – e – kt ) k d. k = 0.02 ⇒ M = 5000(1 – e–0.02 t) e. M 5000
t 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 account dM/dt = 100 − 0.02(4996.622…) = 0.06755… Increasing at about $0.07 per day h. lim M = lim 5000(1 – e −0.02 t ) t →∞
t →∞
= 5000(1 − 0) = 5000
2. dM/dt = 100 + kM (k = daily interest rate) dM 1 k dM = dt ⇒ = dt ⇒ 100 + kM k 100 + kM 1 ln |100 + kM | = t + C ⇒ |100 + kM | = e kt e kC ⇒ k 100 + km = C 1e kt ⇒ kM = −100 + C 1e kt ⇒ 1 M = (C1e kt − 100) k Substitute M = 0 when t = 0. 1 0 = (C1e 0 – 100) ⇒ C1 = 100 k 100 kt ∴M = (e – 1) k Let k = 0.0002 (0.02% per day). ∴ M = (500000)(e0.0002 t – 1)
∫
142
∫
Problem Set 7-3
∫
∫
The graph is almost straight. The $100/day deposits far exceed the interest for the first few years. M
50,000
t 500
Make a table of M and dM/dt for various numbers of years. Neglect leap years. Years 0 1 10 20
M
dM/dt
0 37865 537540 1652980
100.00 107.57 207.51 430.60
After 1 year, the $100/day is putting more into the account. After 10 years, the interest has started putting in more than the $100/day. After 20 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 the account becomes infinite! 3. a. E = RI + L(dI/dt) b. L dI/dt = E – RI L dI L − R dI = dt ⇒ − = dt E – RI R E − RI L ⇒ – ln | E – RI | = t + C R ⇒ | E – RI | = e − ( R/ L )t e − ( R/ L )C 1 ⇒ E − RI = C1e − ( R/ L )t ⇒ I = [ E – C1e –( R/ L )t ] R Substitute I = 0 when t = 0. 1 0 = ( E – C1e 0 ) ⇒ C1 = E R E ∴ I = [1 – e –( R/ L )t ] R 110 c. I = [1 – e –(10 / 20 )t ] 10 I = 11(1 – e− 0.5 t)
∫
∫
∫
∫
I 11
t 5
10
Calculus Solutions Manual © 2005 Key Curriculum Press
d.
i. I = 11(1 – e− 0.5) = 4.3281… ≈ 4.33 amps
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
ii. I = 11(1 – e–5) = 10.9258… ≈ 10.93 amps iii. lim I = lim 11(1 – e −0.5t ) = 11(1 – 0) t →∞
t →∞
= 11 amps
e. I = 0.95(11) = 10.45 10.45 = 11(1 – e −0.5t )
5. a.
0.95 = 1 – e −0.5t b.
e −0.5t = 0.05 −0.5t = ln 0.05 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 − h dT
−
C h
−
C ln | R − hT | = t + D h
∫ R − hT = ∫ dt
−1/2
∫
dV = k dt ⇒
2V 1/2 = kt + C ⇒ V =
kt + C 2
2
V varies quadratically with t. c. Initial conditions t = 0; V = 196; dV/dt = −28: k⋅0+C 1961/2 = ⇒ C = 28 2 and −28 = k ⋅ 1961/ 2 ⇒ k = −2 2
| R − hT | = e e R − hT = D1e –( h /C )t T = (1/ h)[ R – D1e –( h /C )t ] Substitute T = 0 when t = 0. 1 0 = ( R – D1e 0 ) ⇒ D1 = R h R ∴ T = [1 – e –( h / C )t ] h − (0.04/ 2)t
c. T = (50/0.04)[1 – e T = 1250(1 – e− 0.02t ) d.
d. False. dV/dt = 2t − 28, so the water flows out at 28 ft3/min only when t = 0. For instance, at t = 5, dV/dt = −18, which means water flows out at only 18 ft3/min. So it takes longer than 7 min to empty the tub. e. 0 = (t − 14)2 ⇒ the tub is empty at t = 14 min. f. V
] 100
t 1250
14
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 T = lim 1250(1 – e −0.02 t ) = 1250(1 – 0) t →∞
∫V
−2 t + 28 ∴ V = ⇒ V = (14 − t ) 2 2
− ( h / C )t − ( h / C ) D
T
dV = kV 1/2 dt
t →∞
g. See the solution to Problem C4 in Problem Set 7-7. 6. The following data were gathered in the author’s class in December 1994. Times t are in seconds and volumes V are in mL. Note that a burette reads the amount of fluid delivered, so you must subtract the reading from 50 to find the volume remaining. Use food coloring in the water to make the liquid level easier to read. Read from the bottom of the meniscus (the curved surface of the liquid).
= 1250° g. T = 0.99(1250) = 1237.5 1237.5 = 1250(1 – e− 0.02t ) Calculus Solutions Manual © 2005 Key Curriculum Press
Problem Set 7-3
143
Seconds
Reading
Volume
0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180 190 200 210 220 230 240 250 260 M 320 360
0 2.4 4.4 6.4 8.5 10.5 12.4 14.3 16.1 17.8 19.9 21.2 22.8 24.5 25.6 27.4 28.6 30.0 31.3 32.6 33.8 35.1 36.4 37.4 38.5 39.5 40.6 M 46.1 49.3
50 47.6 45.6 43.6 41.5 39.5 37.6 35.7 33.9 32.2 30.1 28.8 27.2 25.6 24.4 22.6 21.4 20.0 18.7 17.4 16.2 14.9 13.6 12.6 11.5 10.5 9.4 M 3.9 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 the grapher, as shown.
position of the vertex can be used to predict the position of the stopcock and the time when the fluid would all be gone if the burette were of uniform diameter all the way down to the stopcock. For the preceding data, the vertex is at −0.20964 K t=− ≈ 500 s (2)(0.000209255K) V ≈ −3.0 mL So the stopcock should be found at a point corresponding to about 3 mL below the bottom mark. 7. a. n = 1, k = 1, C = −3: ∴ dy/dx = y ⇒ dy/ y = dx ⇒ ln | y | = x − 3
∫
∫
⇒ |y| = e x−3 = exe− 3 ⇒ y = ±0.04978…ex y 1
x 1
b. n = 0.5, k = 1, C = −3: ∴ dy/dx = y 0.5 ⇒ y −0.5 dy = dx
∫
∫
1 ⇒ 2 y 0.5 = x − 3 ⇒ y = ( x − 3)2 4 Note: x ≥ 3 because y0.5 is a positive number. y
1
x 3
∫
∫
c. n = −1 ⇒ dy/dx = ky −1 ⇒ y dy = k dx 1 ⇒ y 2 = kx + C ⇒ y = ± 2 kx + 2C 2 k = 1, C = −3 ⇒ y = ± 2 x − 6 5
y
V 50
x 3
t 100
The volume does seem to vary quadratically with time. Because there is still fluid in the burette when V = 0, the graph crosses the t-axis, unlike the graphs in Problem 5 and Example 1. The 144
Problem Set 7-3
∫
∫
n = −2 ⇒ dy/dx = ky −2 ⇒ y 2 dy = k dx 1 3 y = kx + C ⇒ y = 3 3kx + 3C 3 k = 1, C = −3 ⇒ y = 3 3 x − 9
⇒
Calculus Solutions Manual © 2005 Key Curriculum Press
5
⇒ | M − kB| = e − kct e − kcC ⇒ M − kB = C1e − kct
y
x
d. For n > 1,
∫
∫
dy = ky n ⇒ y − n dy = k dx dx
− ( n −1)
y = kx + C because n > 1, n −1 −1 so y = n −1 ( n − 1) ⋅ ( kx + C ) which has a vertical asymptote at x = −C/k because the denominator equals zero for this point. Note that the radical will involve a ± sign when the root index is even (for example, when n is odd). For n = 2, k = 1, C = −3: y = −( x − 3) −1 ⇒−
1 ( M − C1e − kct ) k Use the initial condition B = 0 when t = 0. 0 = (1/k) ( M − C 1e0) ⇒ C 1 = M M ∴ B = (1 – e – kct ) k Use the initial condition kB = 80 when B = 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=
3
y 2
B = ( M/0.08)[1 − e −0.08/( 500 / M )t ] B = 12.5 M[1 − e − ( 40 / M )t ] 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 )
x 3
B M = 10000 100,000
M = 5000
−1 For n = 3, k = 1, C = −3: y = ± 2x − 6
M = 1000 250
t 500
y 2
x 3
Note that the graph shows two branches. dy e. For n = 0, = ky 0 = k , so y = kx + C, dx a linear function. For k = 1, C = −3, y = x − 3.
As shown on the graph, the sales start out increasing at the same rate (500 bottles/day). As t increases, the number of bottles/day increases, approaching a steady state equal to 12.5M. To find the break-even time, first find the total number of bottles sold as a function of time. B is in bottles per day, so the total sales in x days, T(x), is T ( x) =
∫
x
B dt.
0
Use, for example, M = $10,000/day.
y
T ( x) =
2
x
x
∫ 125000(1 − e
−0.004 t
) dt
0
= 125000[t + (1/0.004)e −0.004 t ]
3
x 0
= 125000( x + 250e −0.004 x − 0 − 250) = 125000[ x − 250(1 − e −0.004 x )] 8. dB/dt = c ( M − kB), where k and c are constants. dB 1 − k dB = c dt ⇒ − = c dt M − kB k M − kB 1 ⇒ − ln | M − kB | = ct + C k
∫
∫
Calculus Solutions Manual © 2005 Key Curriculum Press
∫
∫
For selling prices of $0.25 and $0.50/bottle, the total numbers of dollars are D25(x) = 31250[x − 250(1 − e− 0.004 x)] D50(x) = 62500[x − 250(1 − e− 0.004 x)]
Problem Set 7-3
145
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) $0.25/bottle $0.50/bottle 2 1
x 90
207
9. The differential equation is dT/dt = k(1200 − L), and L = h (T − 70), where h is a proportionality constant. ∴ dT/dt = k(1200 + 70h − hT )
∫ dT/(1200 + 70h − hT ) = ∫ k dt (−1/h) ln |1200 + 70h − hT | = kt + C ln |1200 + 70h − hT | = −kht − h C |1200 + 70h − hT | = e− kht ⋅ e− hC 1200 + 70h − hT = C 1e− kht ⇒ hT = 1200 + 70h − C 1e− kht T = 1200/h + 70 − (C 1/h) ⋅ e− kht Use T = 70 when t = 0. 70 = 1200/h + 70 − C 1/h ⋅ e− kh⋅ 0 ⇒ C 1 = 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 the original 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 found by grapher or by substituting for T and solving for t. T Never reaches 180° 140°
160°
170°
100
t 39
146
72
Problem Set 7-3
130
T
t
140°
39 min 61 min 72 min 130 min Never!
155° 160° 170° 180°
The limit of T as t increases is 70 + 102.26…(1 + 0), which equals 172.26…°. Thus, the temperature never reaches 180°. When the heater turns off, the differential equation becomes dT = − kh (T − 70) ⇒ T = 70 + C2 e − kht . dt Using T = 160 at time t = 0 when the heater turns off, T = 70 + 90e− 0.02933…t. To find the time taken to drop to 155°, substitute 155 = 70 + 90e −0.02933Kt . Solving numerically or algebraically gives t = 1.9… . Thus, it takes only 2 minutes for the temperature to drop 5°! By contrast, from the preceding table, it takes 11 minutes (t = 61 to t = 72 in the table) to warm back up from 155° to 160°. The design of the heater is inadequate because it takes much longer to warm up by a certain amount than it does to cool back down again. Near 172°, a slight increase in the thermostat setting for the heater makes a great increase in the time taken to reach that setting. For instance, it takes an hour (72 minutes to 130 minutes) to warm the 10 degrees from 160° to 170°. These inadequacies could be corrected most easily by adding more insulation. The resulting decrease in h would make the heater cool more slowly, heat up faster, and reach the 180 degrees it currently cannot reach. Decreasing h would also reduce the power consumption. 10. a. dP/dT = kP/T 2 dP/P = k dT /T 2 ⇒ ln | P | = − kT −1 + C
∫
⇒ | P| = e
∫
− k /T + C
⇒ P = C1e − k / T
b. 0.054 = C1e − k /293 3.95 = C1e − k /343 (3.95 / 0.054) = e ( − k /343+ k /293) ln (3.95/0.054) = −k/343 + k/293 k = [ln (3.95 / 0.054)]/( −1/ 343 + 1/ 293) = 8627.812641K From ln |P| = −kT − 1 + C, C = ln 0.054 + 8627.812641…/293. C = 26.52768829… ⇒ C1 = e26.52768829… P = e 26.52768829…e −8627.812641…/T P = e ( 26.52768829K−8627.812641K/T ) Calculus Solutions Manual © 2005 Key Curriculum Press
c. Temperature
T
10 20 30 40 50 60 70 80
283 293 303 313 323 333 343 353
90 100 110 200
P
Actual*
0.0190… 0.054 0.142… 0.354… 0.832… 1.85… 3.95 8.05…
363 373 383 473
So the new equation models the data above the 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 5734.569702 K T = = 490.214 K 18.33140949K – ln 760 About 490 K, or 217°C (actual: 218°C) e. Answers will vary.
15.7… 29.8… 54.6… 3972.1…
0.021 0.054 0.133 0.320 0.815 1.83 3.95 7.4 (melting point) 12.6 18.5 27.3 496.5
Problem Set 7-4 Q1. Q3. Q5. Q6. Q7.
y ′ = 5x 4 Q2. y ′ = 5x ln 5 8 (1/8)x + C Q4. 7 x/ln 7 + C y ′ = −y/x or y ′ = −3x − 2 87.5 Q8. y
*Source: Lange’s Handbook of Chemistry, 1952, p. 1476.
The function models the data well up to the melting point, but not above it. The differences between the predicted and actual answers are most likely due to the fact that naphthalene changes from solid to liquid at 80°C; the constants for solid and liquid naphthalene differ. Use initial conditions for T = 90, 110 as in part 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
10 20 30 40 50 60 70 80
283 293 303 313 323 333 343 353
90 100 110 200
363 373 383 473
Calculus Solutions Manual © 2005 Key Curriculum Press
P 0.144… 0.289… 0.551… 1.01… 1.78… 3.03… 5.01… 8.05…
12.6 19.2… 28.7… 496.5
y
1
1
x
x
1
1
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 upward with a slope less than 1. At (−5, 1) the line slopes downward with a slope much steeper than −1. b. The figure looks like one branch of a hyperbola opening in the y-direction. (The lower branch shown on the graph is also part of the solution, but students would not be expected to find this graphically.)
Actual 0.021 0.054 0.133 0.320 0.815 1.83 3.95 7.4 (melting point) 12.6 18.5 27.3 496.5
y
(3, 5)
5
(–5, 1)
(1, 2)
(5, 1)
x
5
c. See graph in part b. The figure looks like the right branch of a hyperbola opening in the x-direction. (The left branch is also part of the
Problem Set 7-4
147
solution, but students would not be expected to find this graphically.) dy x 1 d. = ⇒ 2 y dy = x dx ⇒ y 2 = x 2 + C dx 2 y 2 x = 5, y = 1 ⇒ C = 1 − 12.5 = −11.5 By algebra, x2 − 2y2 = 23. This is the particular equation of a hyperbola opening in the x-direction, which confirms the observations in part c. 2. At (3, 3), dy/dx = 0.1(3) = 0.3, which is reasonable because the slope is positive and less than 1. At (0, −2), dy/dx = 0.1(−2) = −0.2, which is reasonable because the slope is negative and less than 1 in absolute value. The next graph shows the two particular solutions. For the first, y (6) ≈ 4.0. For the second, y (6) ≈ −3.6. dy = 0.1y ⇒ y = C1e 0.1x dx For (3, 3), the particular solution is y = 2.2224…e0.1 x, giving y(6) = 4.0495… . For (0, −2), the particular solution is y = −2e0.1 x, giving y(6) = −3.6442… . Both graphical answers are close to these actual answers.
∫
1 2 x +C 2 1 ( −1)2 = − (1)2 + C ⇒ C = 1.5 2 2 y = 1.5 − 0.5x 2 y2 = −
∫
y = − 1.5 − 0.5 x 2 (Use the negative square root because of the initial condition.) The graph agrees with part b. From the next-to-last line, add 0.5x2 to both sides, getting 0.5x2 + y2 = 1.5, which is the equation of an ellipse because x2 and y2 have the same sign but unequal coefficients. dy = 3(1 − 1) = 0. dx dy At (1, 2), = 1(1 − 2) = −1. dx dy At (0, − 1), = 0(1 + 1) = 0. dx
4. a. At (3, 1),
y 3
2
y 4
1
3 2
x
1 –2 –1 –1
–3
x 1
2
3
4
5
–2
–1
6
1
2
3
–1
–2 –3
b. See the graph from part a. Both graphs have a horizontal asymptote at y = 1.
–4
dy 3 =− = −0.75. dx (2)(2) dy 1 At (1, 0), =− , which is infinite. dx (2)(0)
3. a. At (3, 2),
y
dy
1
x –2
–1
1
2
3
–1 –2
b. See the graph from part a. The figures resemble half-ellipses. dy x c. =− dx 2y 2y dy = − x dx
∫
∫
2 y dy = − x dx
148
Problem Set 7-4
dy = x (1 − y) dx dy = x dx 1– y
∫ 1 – y = ∫ x dx
2
–3
c.
−ln |1 − y| = 0.5x 2 + C 2 1 − y = ± e −0.5 x ⋅ e − C 2 y = 1 + C1e −0.5 x (C1 can be positive or negative.) −1 = 1 + C 1e 0 ⇒ C 1 = −2 2 ∴ y = 1 − 2e −0.5 x The grapher confirms the graph in part b. 2 As | x | → ∞, e −0.5 x → 0. So y → 1, which agrees with the horizontal asymptote at y = 1.
Calculus Solutions Manual © 2005 Key Curriculum Press
9. a.
5. 2
y
y
1.5
(3, 2)
1
x
0.5
x –2 –1.5 –1 –0.5 –0.5
0.5
1
1.5
2
(1, –2)
–1 –1.5 –2
b. 6. 2
y
1.5 1 0.5
x –2 –1.5 –1 –0.5 –0.5
0.5
1
1.5
2
–1 –1.5
dy = −0.2 xy dx Evidence: At (1, 1) the slope was given to be −0.2, which is true for this differential equation. As x or y increases from this point, the slope gets steeper in the negative direction, which is also true for this differential equation. In Quadrants I and III the slopes are all negative, and in Quadrants II and IV they are all positive. (Note: The 2 algebraic solution is y = Ce −0.1x .)
–2
10. a. Initial condition (0, 2) y
7. 2
y 5
1.5 1
(0, 2)
0.5
x
x –2 –1.5 –1 –0.5 –0.5
0.5
1
1.5
2
–1
5 (0,–2.5) (0, –5)
–1.5 –2
8. 2
y
1.5 1 0.5
x –2 –1.5 –1 –0.5 –0.5
0.5
–1 –1.5 –2
Calculus Solutions Manual © 2005 Key Curriculum Press
1
1.5
2
b. See the graph in part a with initial condition (0, −5). The graph goes toward −∞ in the y-direction instead of toward +∞. c. If a ruler is aligned with the slope lines, the lines that form a straight line are the ones crossing the y-axis at −2.5 with slope −1/2. (In courses on differential equations, students will learn that the given equation is a firstorder linear equation that can be solved using an integrating factor. The general solution is y = Ce0.2 x − 0.5x − 2.5. For C = 0, the curve is the line y = −0.5x − 2.5, which intersects the y-axis at (0, −2.5).)
Problem Set 7-4
149
11. a. Initial condition (0, 2) P (0, 18)
d.
10.5
e.
f. (0, 2)
(4, 2)
t
b. See the graph in part a with initial condition (4, 2). The graph is the same as that in part a but shifted over 4 months. This behavior is to be expected because dP/dt depends only on P, not on t, and both initial conditions have the same value of P. c. See the graph in part a with initial condition (0, 18). The population is decreasing to the same asymptote, P = 10.5, as in parts a and b. d. The asymptote at P = 10.5 indicates that the island can sustain only 1050 rabbits. If the population is lower than that, it increases. If the population is higher than that, it decreases. The number 10.5 is a value of P that makes dP/dt equal zero. Note that there is another asymptote at P = 0, which also makes 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 are different because the graph is scaled by a factor of 10. b. Initial condition (0, 0) v (0,180)
146.4...
13. a.
b.
c.
The graph shows this velocity for times above about 15 seconds. See the graph in part b with initial condition (5, 0). The graph is identical to the one in part b except shifted 5 seconds to the right. This behavior is to be expected because the differential equation is independent of t. See the graph in part b. This graph decreases to the terminal velocity because the diver starts out going faster. Similarities include: Both models have a horizontal asymptote that the particular solutions approach from above or below. Both models decrease rapidly and gradually level off for values above the asymptotic limit. Differences include: For values below the asymptotic limit, one model starts with rapid increase and gradually slows its growth, whereas the other starts with a slow increase that becomes more rapid growth before slowing toward the asymptote. mg ma = 2 By hypothesis r dv dv g Divide by m; a = . = 2 dt dt r dv dr g Chain rule ⋅ = dr dt r 2 dr dv g v= (r = distance) ⋅v = 2 dt dr r dv g Divide by v. = 2 dr r v dv (5, 2) = −1.2488 dr dv (1, 10) = −6.244 dr dv (10, 4) = −0.1561 dr These slopes agree with those shown. Initial condition (r, v) = (1, 10) From the graph, the velocity is zero at r ≈ 5. So the spaceship is about 4 Earth radii, or about 25,000 km, above the surface. v
(5,120)
(1,18)
14.11... (1,12) 50 (1,10)
t (0, 0)
(5, 0)
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 150
Problem Set 7-4
(2,10)
6.12... 5 4.37... (10,4) (5,2)
r
Calculus Solutions Manual © 2005 Key Curriculum Press
The precise value of r can be found algebraically. dv –62.44 –62.44 = ⇒ v dv = dr dr r 2v r2 v 2 62.44 ⇒ = +C r 2 For the solution through (1, 10), C = 50 − 62.44 = −12.44, so the ship starts falling when 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 between 4 and 5 km/s. The precise value of v can be found algebraically. v 2 62.44 C = 72 − 62.44 = 9.56 ⇒ = + 9.56 2 r Because r > 0, v is never zero, so the spaceship never stops and falls back. As r approaches infinity, v2/2 approaches 9.56, and thus v approaches (2)(9.56) = 4.37… km/s.
∫
f. See the graph in part c with initial condition (r, v) = (2, 10). The graph levels off at about 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 found algebraically as in parts d and e. For the solution through (2, 10), C = 50 − 31.22 = 18.78. As r → ∞, v → (2)(18.78) = 6.12… km/s. 14. See the Programs for Graphing Calculators section of the Instructor’s Resource Book.
Problem Set 7-5
Calculus Solutions Manual © 2005 Key Curriculum Press
y
∫
e. See the graph in part c with initial condition (r, v) = (1, 18). The graph levels off at v ≈ 14 km/s. Here the spaceship loses about 4 km/s of velocity, whereas it loses 7 or 8 km/s when starting at 12 km/s. Both cases lose the same amount of kinetic energy, which is proportional to v2 (the change in v2 is the same in both cases). The precise value of 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.
Q1. k y Q3. 4.8 Q5. −ln |1 − v| + C
Q7.
Q2. y = Ce3x Q4. 100 Q6. sec x tan x
y'
1
y'
x
1
Q8. 3x 2y 5 + 5x 3y 4y′ = 1 + y′ Q9. continuous Q10. A 1. a. dy = −(x/y) dy For (1, 3), dy = −(1/3)(0.5) = −0.1666… , so new y ≈ 3 − 0.1666… = 2.8333… at x = 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 0 0.5 1 1.5 2 2.5 3
y 3.2456… 3.1666… 3 2.8333… 2.5686… 2.1793… 1.6057…
The Euler’s y-values overestimate the actual values because the tangent lines are on the convex side of the graph and the convex side is upward. b. dy = −(x/y) dy
∫ y dy = − ∫ x dx
0.5y 2 = −0.5x 2 + C 0.5(32) = −0.5(12) + C ⇒ C = 5 0.5y 2 = −0.5x 2 + 5 y = 10 − x 2 (Use the positive square root.)
At x = 3, y = 10 – 32 = 1. The particular solution stops at the x-axis because points on the circle below the x-axis would lead to two values of y for the same value of x, making the solution not a function. The Euler’s value of 1.6057… overestimates the actual value by 0.6057… . 2. a. dy = (x/y) dy For (1, 2), dy = (1/2)(0.5) = 0.25, so new y ≈ 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.
Problem Set 7-5
151
x
y
y
0
1.6071…
0.5
1.75
1
2
1.5
2.25
2
2.5833…
2.5
2.9704…
3
3.3912…
4 3 2 1
x 1
The Euler’s y-values underestimate the actual values because the tangent lines are on the convex side of the graph and the convex side is downward. The error is greater at x = 0 because the graph is more sharply curved between x = 0 and x = 1 than it is between x = 1 and x = 3. b. dy = x/y dy
∫
∫
y dy = x dx
0.5y = 0.5x 2 + C 0.5(22) = 0.5(12) + C ⇒ C = 1.5 0.5y 2 = 0.5x 2 + 1.5 y = x 2 + 3 (Use the positive square root.) At x = 0, y = 0 + 3 = 3 = 1.7320 … . The particular solution stops at the x-axis because points on the circle below the x-axis would lead to two values of y for the same value of x, making the solution not a function. The Euler’s value of 1.6071… underestimates the 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
152
2
dy/dx
dy
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
Problem Set 7-5
y
2
3
4
You cannot tell whether the last value of y is an overestimate or an underestimate because the convex side of the graph is downward in some places 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
y 5 4 3 2 1
x 1
2
3
4
The approximate values of y underestimate the actual values of y because the convex side of the graph is down. 5. See the Programs for Graphing Calculators section of the Instructor’s Resource Book. 6. See the Programs for Graphing Calculators section of the Instructor’s Resource Book. dy 7. a. and b. = −0.2 xy dx y (3, 2)
x
(1, –2)
Calculus Solutions Manual © 2005 Key Curriculum Press
8. a. and b.
dy = −0.1x + 0.2 y dx
putting the graph into a region of the slope field from which the spacecraft would not escape Earth’s gravity.
y
d. Let v1 be the initial velocity at r = 1. Solving for C gives 0.5v12 = 62.44 + C C = 0.5v12 − 62.44
5 (0, 4) (0, 2)
If v1 < 2(62.44), then C is negative,
x
5
making v = 128.88r + C an imaginary number when r is large enough. If v1 > 2(62.44), then C is positive, making v a positive real number for all positive values of r. (The asymptote is v = C .) –1
c. When the graph is observed, the slope lines seem to follow a straight path using (0, 2.5) as an initial condition. Euler’s method confirms this. (In differential equations, students will learn how to solve such first-order linears by multiplying both sides by the integrating factor e− 0.2x . The general solution is y = Ce 0.2x + 0.5x − 2.5. For C = 0, the particular solution is y = 0.5x + 2.5.) 9. a. Using dr = 0.6, v(13.6) ≈ 0.1414… and v(14.2) ≈ −1.2900… , so the spacecraft seems to reverse direction somewhere between these two values of r, as shown in the graph in part b. b. Using dr = 0.1, v(20) ≈ 4.5098… , and the values are leveling off, as shown in the graph.
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 because the graph is concave down (convex side up), so the Euler’s tangent lines will be above the actual graph, as in the next graph. 200 v
Euler
100
Actual
20 v
t
10
10
20
Actual
dr = 0.1 dr = 0.6
r 10
c.
20
dv –62.44 = dr r 2v
∫ v dv = −62.44 ∫ r
−2
dr
−1
0.5v = 62.44r + C 0.5(12)2 = 62.44(1) −1 + C ⇒ C = 9.56 0.5v 2 = 62.4r −1 + 9.56 2
v = 124.88r –1 + 19.12 When r = 20, v = 5.0362… . Because the graph is concave up (convex side down), the Euler’s solution underestimates the actual velocity. The first increment, where the graph is so steep, makes a large error that accumulates as the iterations continue, Calculus Solutions Manual © 2005 Key Curriculum Press
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 because the graph is concave up (convex side down), so the Euler’s tangent lines will be below the actual graph. c.
dv = 0 ⇔ 0.0015v 2 = 32.16 ⇔ dt 32.16 v= = 146.42404 K(store) 0.0015 The terminal velocity is about 146.4 ft/s.
d. The table shows the values of v and the errors, Euler minus actual. The errors increase only for a while, then approach zero because both the Euler’s solution and the actual solution approach the same asymptote. (It is not always true that values farther from the Problem Set 7-5
153
initial condition have a greater error in their Euler’s approximation.) t
Euler’s v
Actual v
Error
2 4 6 8 10 20
61.6831… 106.2850… 129.7139… 139.9323… 143.9730… 146.4066…
60.4791… 103.3298… 126.8383… 137.9573… 142.8466… 146.3792…
1.2040… 2.9552… 2.8756… 1.9749… 1.1264… 0.0274…
The graph in part a shows the Euler’s solutions from parts a and b, and the actual solution from part c, thus confirming graphically the numerical answers to this problem. 11. a. For x ≤ 5, the radicand 25 − x2 is nonnegative, giving a real-number answer for y. For x > 5, the radicand is negative, giving no real solution. b. The slope field shows that the graph will be concave up (convex side down), making the Euler’s tangent lines lie below the graph, leading to an underestimate. At x = 4.9, y = −0.6 25 – 4.9 2 = −0.5969… . The Euler’s solution at x = 4.9 is −0.8390… , which is an underestimate because −0.8390… < −0.5969… but is reasonably close to the actual value. c. The Euler’s solutions for the given points are
12. Answers will vary.
Problem Set 7-6 Q1. Q2. Q3. Q4. Q5. Q6. Q7. Q9. 1.
definition of definite integral fundamental theorem of calculus definition of indefinite integral the intermediate value theorem Rolle’s theorem the mean value theorem the chain rule Q8. general particular Q10. initial a. dB/dt is proportional to B, which means that the larger the population is, the faster it grows. But dB/dt is also proportional to (30 − B)/30, which means that the closer B is to 30, the slower it grows. dB/dt > 0 when 0 < B < 30 because when the population is less than 30 million the population will increase until it reaches the carrying capacity. dB/dt < 0 when B > 30 because when the population is greater than 30 million, the population will decrease until it reaches the carrying capacity. b. B
40
30 Actual 20 Euler
x
y
5.1 5.2 5.3 6.6
−0.3425… 0.1935… −0.7736… 26.9706…
(9)(5.1) (0.1) (25)(–0.3425K) = 0.5360… , indicating that the graph is still taking upward steps. (9)(5.2) (0.1) From 5.2 to 5.3, dy = − (25)(0.1935K) = −0.9672… , indicating that the graph takes a relatively large downward step. The sign change in dy happens whenever the prior Euler’s y-value changes sign. The graph starts over on another ellipse representing a different particular solution. d. Euler’s method can predict values that are outside the domain, which are inaccurate.
10
t 10
Problem Set 7-6
30
40
For the initial condition (0, 3), the population grows, leveling off at B = 30. For the initial condition (10, 40), the population drops because it is starting out above the maximum sustainable value (carrying capacity).
From 5.1 to 5.2, dy = −
154
20
c. t
B
0 10 20 30 40
3 13.8721… 26.4049… 29.5565… 29.9510…
See the graph in part b. The graph shows that the Euler’s points and graphical solution are close to each other. Calculus Solutions Manual © 2005 Key Curriculum Press
d.
dB 30 – B = 0.21B ⋅ dt 30 30 dB = 0.21 dt B(30 – B) Separate the variables. 1 1 + dB = 0.21 dt B 30 – B By partial fractions (see Example 1). ln |B| − ln |30 − B| = 0.21t + C Why the “−” sign? −ln |B| + ln |30 − B| = −0.21t − C To simplify later steps. 30 – B = C1e −0.21t B C1 = ± e − C 27 = C1e 0 ⇒ C1 = 9 3 Substitute the initial condition (0, 3) to find C1. 30 – B = 9e −0.21t B 30 − B = 9 Be −0.21t 30 Solve for B explicitly B = in terms of t. 1 + 9e −0.21t 30 At t = 20, B = = 26.4326 K . 1 + 9e –4.2 The Euler’s value, B ≈ 26.4049… , is very close to this precise value.
e.
d dB = −0.014 B + 30(0.007) dB dt 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 and B = 30. 30 15 = ⇒ t ≈ 10.4629… 1 + 9e –0.21t The point of inflection is (10.4629… , 15).
2. a. A logistic function is reasonable because the number of houses grows at an increasing rate for a while, then slows down as the number approaches 120, the “carrying capacity” of the subdivision. b.
dy 120 – y = 0.9 y ⋅ dx 120 120 dy = 0.9 dx y(120 – y) 1 1 + dy = 0.9 dx y 120 – y
Calculus Solutions Manual © 2005 Key Curriculum Press
ln |y| − ln |120 − y| = 0.9x + C −ln |y| + ln |120 − y| = −0.9x − C 120 – y = C1e −0.9 x C1 = ± e − C y 115 = C 1e0 ⇒ C 1 = 23 5 Substitute the initial condition (0, 5) to find C1. 120 – y = 23e −0.9 x y 120 − y = 23 ye −0.9 x 120 1 + 23e –0.9 x The graph confirms that the particular solution follows the slope lines.
y=
y (houses) 100
50
x (years) 5
10
c. 70% of 120 is 84. 120 84 = ⇒ x ≈ 4.4253K , or about 1 + 23e –0.9 x 4 years 5 months Solve numerically for x. 1 lot left means 119 lots built on. 120 119 = ⇒ x ≈ 8.7940 … , or about 1 + 23e –0.9 x 8 years 10 months. dy 120 – y 0.9 d. (120 y – y 2 ) = 0.9 y ⋅ = dx 120 120 d dy 0.9 = (120 – 2 y) dy dx 120 The derivative is zero if 120 − 2y = 0, which is true if and only if y = 60. This value is halfway between y = 0 and y = 120. If y < 60, the derivative is positive, so dy/dx is increasing. If y > 60, the derivative is negative, so dy/dx is decreasing. Therefore, dy/dx is a maximum when y = 60, and the number of houses is increasing the most rapidly at this point of inflection.
Problem Set 7-6
155
3. a.
dy M–y = ky ⋅ dx M k k 0.5 (10)( M – 10) ⇒ = M M 10( M – 10) k k 1.1 1.1 = (24)( M – 24) ⇒ = M M 24( M – 24)
0.5 =
0.5 1.1 = 10( M − 10) 24( M − 24) Eliminate k by equating the two values of k/M. 12(M − 24) = 11.0(M − 10) 12M − 11M = 288 − 110 ⇒ M = 178 Solve for M.
∴
k 0.5 89 = ⇒k= = 178 10(178 − 10) 1680 0.05297… (Store this.) Ajax expects to sell 178,000 CDs based on this mathematical model. dy 89 178 − y b. = ⋅y⋅ dx 1680 178 200
y (thousand CDs)
178
100
x (days) 50
100
The slope field has horizontal slope lines at about y = 178, thus confirming M = 178. M c. The general solution is y = . 1 + ae − kx Substitute M = 178 and k = 89/1680 = 0.05297… and the initial condition y = 10 at x = 0. 178 10 = ⇒ a = 16.8 1 + ae 0 178 The equation is y = . 1 + 16.8e −0.05297Kx See the graph in part b. The graph follows the slope 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 the 51st day. e. The point of inflection is halfway between y = 0 and y = 178, that is, at y = 89. 156
Problem Set 7-6
178 1 + 16.8e −0.05297Kx Solving numerically gives x ≈ 53.2574… , or on the 54th day. dy M−y 4. = ky ⋅ dx M M dy = k dx y( M − y ) 1 1 + dy = k dx y M − y See Section 9-7 for a quick way to resolve into partial fractions. ln |y| − ln |M − y| = kx + C The differential of the second denominator is −dy. −ln |y| + ln |M − y| = −kx − C M−y ln = − kx − C y 89 =
∫
∫
∫
∫
M−y = e − kx −C = e − C ⋅ e − kx y M−y = ± e − C ⋅ e − kx = C1e − kx C1 = ± e − C y M − y = C 1ye − kx y + C 1ye− kx = M M M y= − kx = 1 + C1e 1 + ae − kx a = C 1, Q .E .D . 5. a. At t = 5.5, F ≈ 1.7869… ≈ 2 fish left. At t = 5.6, F ≈ −11.0738… , meaning no fish are left. The fish are predicted to become extinct in just over 5.5 years. F (fish) Part b 1000
500
200
Parts c and d
Part a, dt = 0.1
t (years) 5
10
b. See the graph in part a with initial condition (3, 1200), showing that the fish population will decrease because the initial condition is above the 1000 maximum sustainable. c. See the graph in part a with initial condition (0, 300), showing that the population rises slowly at first, then faster, eventually
Calculus Solutions Manual © 2005 Key Curriculum Press
slowing down as the population approaches the 1000 maximum sustainable (carrying capacity). d. Let F = y + 200. dy y 800 − y = 130 ⋅ ⋅ dt 200 1000 200000 dy = 130 dt y(800 − y)
∫
The logistic function fits reasonably well (as shown in this graph), especially if you use several values of the maximum number of people as shown in the table. 25 20
∫
250 250 + dy = 130 dt 800 − y y See Section 9-7 for quick partial fractions. 250 ln |y| − 250 ln |800 − y| = 130t + C Why “−” ?
∫
15 10 5
∫
x
N
0 1 2 3 4 5 6 7 8
1 3 6 13 21 25 25 25 25
x 1
2
3
4
5
6
7
8
7. a. and b.
800 − y = e − (13/25)t ⋅ e − C y 800 − y = C1e − (13/25)t y Substitute y = 100 (F = 300) when t = 0. 800 − 100 = C1e 0 ⇒ C1 = 7 100 800 − y = 7e − (13/25)t y 800 y= 1 + 7e − (13/25)t 800 F= + 200 1 + 7e − (13/25)t See the graph in part a, showing that the sketch from part c reasonably approximates this precise algebraic solution. 6. Answers will vary. Here is a typical run with a class of 25 people.
N
Year
P
∆P/∆t
(∆P/∆t)/P
1940 1950 1960 1970 1980 1990
131.7 151.4 179.3 203.2 226.5 248.7
2.38 2.59 2.36 2.275
0.01571… 0.01444… 0.01161… 0.01004…
You can’t find ∆P/∆t for 1940 and 1990 because you don’t know values of P both before and after these values. c. Using linear regression on the values of (∆P/∆t)/P without round-off gives 1 ∆P ≈ 0.02802596 … − 0.0000792747… P. P ∆t The correlation coefficient is r = −0.98535… . For the other types of regression: r = −0.978… for logarithmic r = −0.981… for exponential r = −0.971… for power Thus, a linear function fits best because r is closest to −1. d.
1 dP ≈ 0.02802596 … − 0.0000792747… P P dt dP ⇒ = P(0.02802596 … − 0.0000792747… P) dt
e. 500
P
Stable population at 353.5 million
Logistic regression gives 25.5083K N= 1 + 43.1120 K e −1.3032Kx
Calculus Solutions Manual © 2005 Key Curriculum Press
t –50
0
50
100
Problem Set 7-6
157
f. Year
t
Euler
1890 1900 1910 1920 1930 1940 1950 1960 1970 1980 1990 2000 2010 2020 2030 2040
−50
44.6… 56.9… 71.7… 89.2… 109.3… 131.7 155.4… 180.1… 204.7… 228.2… 249.9… 269.3… 286.1… 300.2… 311.8… 321.1…
−40 −30 −20 −10
0 10 20 30 40 50 60 70 80 90 100
Actual*
Euler**
62.9 76.0 92.0 105.7 122.8 131.7 151.4 179.3 203.2 226.5 248.7 281.4
46.1… 58.3… 72.9… 90.1… 109.8… 131.7 155.0… 179.2… 203.5… 226.9… 248.8… 268.6
*Data from The World Book Encyclopedia . **Note that although linear regression gives the “best” fit for (∆ P/∆ t ) /P versus P, actually plotting the graph shows that the data point for 1960 is considerably out of line.
0.015 0.014
(∆ P/∆ t)/P 1950 1960
0.013 0.012 0.011 150
1980 200
Problem Set 7-6
Year
t
2010 2020 2030 2040 2050
70 80 90 100 110
Euler 486.1… 444.5… 417.7… 399.7… 387.1…
The logistic model predicts that the population will drop, approaching the ultimate value of 353.5 million from above. This behavior shows up in the slope field of part e because the slopes are negative for populations above 353.5. 1 A B A(10 – y) + By 8. a. = + = y(10 – y) y 10 – y y(10 – y)
1970
The numerator of the first fraction must equal that of the last fraction for all values of y. That is, 1 = 10A − Ay + By. The constant and linear coefficients on the left must equal the corresponding ones on the right. Thus, 1 = 10A and 0 = −Ay + By. So A = B = 0.1.
P
Using the two endpoints, 1950 and 1980, gives (∆P/∆t)/P = 0.002716… − 0.00007557…P. Using this equation gives populations much closer to the actual ones for the given years, as shown in the rightmost column of the table in part f. This is, of course, no guarantee that the later model fits any better in the future than the former one. g. The population growth rate is zero if dP/dt = 0. Let P(0.02802596… − 0.0000792747…P) = 0. P = 0 or P = (0.02802596…)/ (0.0000792747…) = 353.5… Predicted ultimate population ≈ 353.5 million Differential equation: P = 353.5… makes dP/dt = 0. Graph: P = 353.5… is a horizontal asymptote. 158
h. See the graph in part e. Data do follow the solution. i. Sample answer: The predicted populations agree fairly well with the data for the six given years. The fit is exact for 1940 because this point was used as an initial condition. For the other five years, the predicted populations are a bit higher than the actual population. j. Actual data are given in the table in part f. k. The predicted population for 2010 from part f is 286.1… million. Using 486.1 million as an initial condition in 2010 gives the following predictions:
b.
0.1 0.1 + dy y 10 – y
∫ y(10 – y) dy = ∫ 1
1 1 = 0.1 − dy, which equals 3 dx. y y – 10
∫
∫
∴ 0.1(ln |y| − ln |y − 10|) = 3x + C ln
y y – 10 = − ln = 30 x + 10C y – 10 y
y – 10 10 = 1− = e − ( 30 x +10 C ) y y 10 = 1 ± e −10 C e −30 x y 10 y= , where k = ± e −10 C , Q .E.D . 1 + ke –30 x Calculus Solutions Manual © 2005 Key Curriculum Press
c.
dP = P(0.02802 … − 0.00007927… P) dt = 0.00007927…P(353.5… − P) 1 dP = 0.00007927K dt P(353.5K – P) 1 1 1 − dP = 353.5K P P – 353.5K
∫
∫
∫
= 0.00007927K
∫ dt
d. Year
t
Algebraic
Euler
Actual
1940 1950 1960 1970 1980 1990
0 10 20 30 40 50
131.7 155.5… 180.2… 204.7… 228.2… 249.8…
131.7 155.4… 180.1… 204.7… 228.2… 249.9…
131.7 151.4 179.3 203.2 226.5 248.7
The two methods of evaluating the mathematical model agree almost perfectly. However, the fact that they agree with each other is no guarantee that they will fit the real world as closely as they match each other.
10.
11.
dR dR = k1 R ⇒ = k1dt ⇒ ln | R| = k1t + C dt R C k1t ⇒ | R| = e e ⇒ R = C1e k1t R is increasing because k1 > 0. dF dF = − k2 F ⇒ = − k2 dt dt F ⇒ ln | F| = − k2 t + C ⇒ | F| = e C e − k2t ⇒ F = C2 e − k2t F is decreasing because −k2 < 0. dR = k1 R − k3 RF dt dF = − k2 F + k4 RF dt
Calculus Solutions Manual © 2005 Key Curriculum Press
14. The slope at (70, 15) is about 0.4. F
ln |P| − ln |P − 353.5…| = 353.5…(0.00007927…t + C) P P – 353.5K ln = − ln P – 353.5K P = 0.02802…t + 353.5…C P – 353.5K = e − ( 0.02802…t +353.5…C ) P 353.5K = 1 + ke 0.02802…t P 353.5K P= 1 + ke –0.02802...t For the initial condition t = 0, P = 131.7, 353.5K k= − 1 = 1.684 … . 131.7
9.
dF dF/dt – k2 F + k4 RF = = dR dR/dt k1 R – k3 RF The dt cancels out. 13. R = 70, F = 15 dF –15 + 0.025 ⋅ 1050 11.25 ⇒ = = = 0.4017K dR 70 – 0.04 ⋅ 1050 28
12.
50
(70, 15)
R 100
dR dF = 28, and = 11.25, dt dt which are both positive. So both populations are increasing and the graph starts up and to the right. 15. The populations vary periodically and the graph is cyclical. The fox population reaches its maximum 1/4 cycle after the rabbit population reaches 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 dF/dt – F + 0.025 RF = = dR dR/dt R – 0.04 RF – 0.01R 2 dF 11.25 R = 70 and F = 15 ⇒ = = −0.5357K dR −21 18. At R = 70, F = 15,
F
50
(70, 15)
R 100
Note that the slope at (70, 15) is now negative. 19. The populations now spiral to a fixed point. The rabbit population stabilizes at the same value as in Problem 16, R = 40 (4000 rabbits), which is surprising. The stable fox population decreases from 25 to 15. Problem Set 7-6
159
20. Assume that dF/dt still equals = −F + 0.025RF. dF dF/dt – F + 0.025 RF = = dR dR/dt R – 0.04 RF – 0.01R 2 – 10 dF 11.25 R = 70 and F = 15 ⇒ = = −0.3629K dR −31
b.
∫V
dV
∫
= k dt
ln |V| = kt + C | V | = e kt +C = e C ⋅ e kt V = C 1e kt C1 can be positive or negative, so the absolute value sign is not needed for V. In the real world, V is positive, which also makes the absolute value sign unnecessary.
21. F
50
c. 400 = Ce k·0 ⇒ C = 400 500 = 400e k⋅ 40
(70, 30)
ln 1.25 = 0.005578… 40 V = 400e0.005578…t ⇒k=
(70, 15)
R 100
Note that the slope at (70, 15) is about −0.4. 22. The fox and rabbit populations spiral toward a fixed point. Again, and even more surprisingly, the rabbits stabilize at R = 40 (4000). But the stable fox population is reduced to 8 or 9. Along the way, the model shows that the foxes are reduced to about 1, thus becoming in danger of extinction! 23. See the graph in Problem 21 with initial condition (70, 30). With this many foxes and hunters chasing rabbits, the rabbits become extinct. At this point, the foxes have been reduced to just 5. After the rabbits become extinct, the foxes decrease exponentially with time, eventually becoming extinct themselves.
Review Problems R0. Answers will vary. R1. P(t) = 35(0.98 t ) P′(t) = 35(0.98t) ln 0.98 P(t)
0 10 20
35 28.5975… 23.3662…
Problem Set 7-7
∫y
−1/ 2
∫
dy = 6 dx ⇒ y = (3 x + C )2
b. y = (3x − 4) (y = (3x − 14)2 does not work because at (3, 5), dy/dx = −30 but 6y1/2 = 30.) 2
c. y
(3, 25)
10
x 1
2
3
e. i. dN/dt = 100 − kN dN = dt 100 – kN −(1/k) ln |100 − kN| = t + C Using (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 = −kt ln |1 − (k/100)N| = −kt 1 − ( k / 100) N = e − kt N = (100 / k )(1 − e − kt ) Using (7, 600) and solving numerically gives k ≈ 0.045236. ∴ N = 2210.6…(1 − e− 0.045236 t) ii. t = 30: About 1642 names
∫
P′(t) −0.7070… −0.5777… −0.4720…
P′(t)/P(t) −0.2020… −0.2020… −0.2020…
P′(t ) 35(0.98t ) ln 0.98 = = ln 0.98 P( t ) 35(0.98t ) = −0.2020… , which is a constant, Q.E.D. R2. a. V = speed in mi/h; t = time in s dV = kV dt 160
R3. a.
d. At x = 2, y′ = 12 and y = 4. See graph in part c. A line through (2, 4) with slope 12 is tangent to the graph.
Problem Set 7-7
t
d. 750 = 400e0.005578…t ln 1.875 ⇒t = = 112.68… ≈ 113 s 0.005578...
∫
Calculus Solutions Manual © 2005 Key Curriculum Press
iii. lim N = 2210.6 …(1 − 0) = 2210.6 … t →∞
Table with initial condition (1, 9), ∆x = 1:
The brain saturates at about 2211 names. iv. Let dN/dt = 30. 70 30 = 100 − kN ⇒ N = = 1547.4 … k names. Substituting this for N gives 1547.4 K = 2210.6 K(1 − e −0.045236 t ). 1547.4 K e −0.045236 t = 1 − = 0.3 (exactly) 2210.6 K ln 0.3 t= = 26.6 … ≈ 27 days –0.04523K or: 30 = N(t) − N(t − 1) = 2210.6 K[ − e −0.045236 t + e −0.045236( t −1) ] = 2210.6 Ke −0.045236 t ( t −1) ( − e −0.045236 + 1) ⇒ t ≈ 27 days 20 dy =− + 0.05 y dx xy At (2, 5), dy/dx = −1.75. At (10, 16), dy/dx = 0.675. The slopes at (2, 5) and (10, 16) agree with these numbers. b. Initial conditions (1, 8) and (1, 12)
R4. a.
y (10, 16)
y (∆x = 1)
x 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 M 28.9 29
9 7.227… 6.205… 5.441… 4.794… 4.200… 3.616… 3.007… 2.326… 1.488… 0.2185… −8.091…
y (∆x = 0.1) 9 7.707… 6.949… 6.413… 5.999… 5.662… 5.377… 5.130… 4.910… 4.712… 4.529… 4.359… 4.199… 4.045… 3.896… 3.750… 3.604… 3.457… 3.306… 3.150… M 0.1344… −0.3810…
y (1, 12) (1, 10) (1, 8) 5 (1, 9)
(2, 5)
x 5
∆x = 0.1
5
∆x = 1
The solution containing (1, 8) crosses the x-axis near x = 7, converges asymptotically to the y-axis as x approaches zero, and is symmetric across the x-axis. The solution containing (1, 12) goes to infinity as x goes to 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 in plotting may make it seem to go the other way. R5. a.
dy 20 =− + 0.05 y dx xy
Calculus Solutions Manual © 2005 Key Curriculum Press
x 5
For ∆x = 1, the graph crosses the x-axis at about 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 initial condition is very sensitive to the size of the increment. For instance, in part a the first step takes the graph so far down that it crosses the x-axis before running off the edge of the grid. The greater accuracy with ∆x = 0.1 shows that the graph actually does not cross the x-axis before x = 20.
Problem Set 7-7
161
d. Continuing the computations in part c, the graph crosses the x-axis close to x = 28.9. See the table in part a. R6. a. y (hundred beavers)
(Note that the general solution to the differential equation is (x − 6)2 + 2(y − 7)2 = C, and the specific solution for the given initial condition is (x − 6)2 + 2(y − 7)2 = 0, whose graph is a single point.) e. Initial condition (9, 7)
10
y
(6, 7) (9, 7)
5
(15, 7) (19, 7)
5
x
x (years) 5
The population is decreasing because it is above the maximum sustainable, 900 beavers (y = 9). By Euler’s method, y ≈ 9.3598… , or about 936 beavers, at x = 3 years. b. See the graph in part a with initial condition (3, 100), showing that the population is expected to increase slowly, then more rapidly, then more slowly again, leveling off asymptotically toward 900. This happens because the initial population of 100 is below the maximum sustainable. c.
dy 9– y = 0.6 y ⋅ dx 9 9 y= 1 + ae –0.6 x
Substitute into the general equation.
9 Substitute the initial condition (3, 1). 1 + ae –1.8 a = 8e1.8 = 48.3971… Solve for a. 9 9 y= = 1 + 8e1.8e −0.6 x 1 + 48.3971K e –0.6 x The point of inflection is halfway between the asymptotes at y = 0 and y = 9.
1=
4.5 =
9 1 + 8e1.8e −0.6 x
Substitute 4.5 for y.
x = ln (8e1.8 )/0.6 = 6.4657… ≈ 6.5 yr d.
162
dy –0.5( x – 6) = dx ( y – 7) dy = 0 when x = 6, and dx = 0 when y = 7. So the stable point is (6, 7), corresponding to the present population of 600 Xaltos natives and 7000 yaks.
Problem Set 7-7
5
10
Suddenly there are too many predators for the number of prey, so the yak population declines. 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 at x ≈ 14.4, indicating that the yaks are hunted to extinction. (The Xaltos would then starve or 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, indicating that the yak population becomes so sparse that the predators become extinct. (The yak population would then explode!)
Concept Problems C1. a.
dy = k ⋅ y1/2 ⇒ y −1/2 dy = k dx ⇒ dx 2 y1/2 = kx + C, so y = [0.5( kx + C )]2 .
b. The differential equation would have to become y1/3 after it is integrated. So the original equation would have to contain y −2 /3 after the variables have been separated. dy Conjecture: = ky 2/3 dx c. Confirmation: dy = ky 2 /3 ⇒ y −2 /3 dy = k dx ⇒ dx 3y1/ 3 = kx + C ⇒ y = [(1/3)(kx + C)]3, a cubic function, Q.E.D. dy d. For n ≠ 0, = k ⋅ y ( n−1)/n ⇒ dx y − ( n−1)/n dy = k dx ⇒ ny1/n = kx + C ⇒ y = [(1/n)(kx + C)]n
Calculus Solutions Manual © 2005 Key Curriculum Press
dy = y 7/8 ⇒ y −7/8 dy = dx dx = x + C ⇒ y = [(1/8)( x + C )]8
For example: ⇒ 8 y1/8 C2. a.
Ticket Price
People
2.00 2.50 3.00 4.00 4.50 5.50 6.00
460 360 320 260 140 120 80
b. N 500
P
Limit is 10, indicating maximum possible population. b. a = 421.3692… , c = 0.7303036… , and k = 0.01589546… , either by twice taking logarithms as suggested, or by this method: Taking ln once ⇒ ln a − ce− kt = ln P, so ln a − ce10k = ln 179 ln a − c = ln 203 ln a − ce −10 k = ln 226 Then substituting ln a = c + ln 203 into the first and third equations gives c(1 − e10k) = ln 179 − ln 203 c(1 − e −10 k ) = ln 226 − ln 203 Substituting c(1 − e −10 k ) = c(e10 k − 1)e −10 k = − e −10 k (ln 179 − ln 203) into the previous equation yields ln 226 − ln 203 ln 226 − ln 203 e −10 k = − = ln 179 − ln 203 ln 203 − ln 179 1 ln 226 − ln 203 so k = − ln = 0.01589K . 10 ln 203 − ln 179 Then find c using c(1 − e −10 k ) = ln 226 − ln 203 and find a using 203 = ae − c . c = 0.7303… and a = 421.3692…
6
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 ≈ −90.83P 2 + 605.4P d. Maximize M: M′ ≈ −181.66P + 605.4 605.4 = 3.332… 181.66 Maximum M at P ≈ 3.332… because M′ changes from positive to negative there (or because the graph of M is a parabola opening downward). Charge $3.30 or $3.35. e. M has a local maximum at this price because charging more than the optimum price reduces attendance enough to reduce the total amount made, whereas charging less than the optimum price increases attendance, but not enough to make up for the lower price per ticket. −0.5 t C3. a. g(t ) = 10e −0.8e The graph does look like Figure 7-7e. −0.5 t −0.5 t lim 10e −0.8e = 10e limt →∞ e M′ = 0 ⇔ P ≈
t →∞
= 10e −0.8⋅0 = 10
Calculus Solutions Manual © 2005 Key Curriculum Press
g (t )
431.3...
100
t 100
Note that this model predicts an ultimate population of lim P(t ) ≈ 421 million. t →∞
c. Now a = 551.1655… , c = 0.9988291… , k = 0.01186428… , and the ultimate population is lim P(t ) ≈ 551 million. Thus, t →∞
the Gompertz model is quite sensitive to a small change in initial conditions. The predicted ultimate population increased by 130 million with only a 1 million change in one data point! C4. dV/dt = −2V1/ 2 + F, where F is a constant. dV = dt F − 2V 1/2 The integral on the right is not the integral of the reciprocal function because the numerator cannot be made the differential of the denominator. A slope field gives information about the solutions. The following graph is for F = 20 ft3/min flowing in. (The dashed line shows the solution with F = 0, the original condition.) Starting with 196 ft3 in the tub, the volume levels off near 100 ft3. Starting below 100 ft3, the volume would increase toward 100.
∫
∫
Problem Set 7-7
163
V 196
F = 20
F=0 14
t
If the inflow rate is too high, the tub will overflow. The next graph is for F = 40 ft3/min. In this case, the stable volume is above the initial 196 ft3. V F = 40 196
F=0 14
t
It is possible to antidifferentiate the left side by the algebraic substitution method of Problem Set 9-11, Problems 101–106. The general solution is F t + C = − ln ( F − 2V 1/2 ) − V 1/2 2 and the particular solution for V = 196 at t = 0 is F F − 28 t − 14 = − ln − V 1/2 2 F − 2V 1/2 Unfortunately, it is difficult or impossible to solve for V. The volume will asymptotically approach F2/4, overflowing the tub if F2/4 > tub capacity. Chapter Test dy = ky dx T2. Solving a differential equation means finding the equation of the function whose derivative appears in the differential equation. T3. The general solution involves an arbitrary constant of integration, C. A particular solution has C evaluated at a given initial condition. T4.
T1.
y 5
x 5
(0, –4)
T5. The concave side of the graph is up, so the actual graph curves up from the Euler’s tangent lines, making the Euler’s method values an 164
Problem Set 7-7
underestimate. (Or: The convex side of the graph is down, so the Euler’s tangent lines are below the actual graph.) T6. General logistic differential equation: dy M−y = ky ⋅ dx M dy T7. = 0.4 y dx dy = 0.4 dx y ln |y| = 0.4x + C |y| = eCe0.4 x y = C1e0.4 x −5 = C 1 e 0.4(0) = C 1 y = −5e0.4 x dy T8. = 12 y1/2 ⇒ y −1/2 dy = 12 dx ⇒ dx 2 y1/ 2 = 12 x + C
∫
∫
∫
∫
dP = kP ⇒ P = Ce kt dt P = 3000 at t = 0 ⇒ P = 3000ekt b. P = 2300 at t = 5 ⇒ 1 2300 k = ln = −0.05314 K ⇒ 5 3000 P(25) = 794.6… Phoebe will not quite make it because the pressure has dropped just below 800 psi by time t = 25. or: 800 = 3000e −0.05314Kt 1 800 t= ln = 24.87K −0.05314 K 3000 Phoebe will not quite make it because the pressure has dropped to 800 just before t = 25. T10. a. y = number of grams of chlorine dissolved t = number of hours since chlorinator was started dy = 30 − ky dt dy = dt 30 – ky 1 − ln |30 − ky| = t + C k ln |30 − ky| = −kt + C 1 30 − ky = C 2e− k t y = 0 when t = 0 ⇒ C 2 = 30 ∴ ky = 30(1 − e− kt) 30 y= (1 – e – kt ) k The rate of escape is ky = 13 when y = 100. So k = 0.13. 30 ∴y = (1 – e –0.13t ) = 230.7K(1 − e −0.13t ) 0.13 T9. a.
∫
∫
Calculus Solutions Manual © 2005 Key Curriculum Press
b. 200 = 230.7…(1 − e −0.13t ) 200 e −0.13t = 1 − = 0.1333K 230.7K ln 0.1333K t= = 15.499K≈ 15.5 hr –0.13 dy 16 – y = 0.5 y ⋅ T11. a. dx 16 16 16 y= ⇒2= ⇒a=7 1 + ae –0.5 x 1 + ae 0 16 y= 1 + 7e –0.5 x b. At x = 0, y = 2: dy = 0.5(2)(16 − 2)/(16)(0.1) = 0.0875 At 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… . 16 = The precise solution is y = 1 + 7e –0.1 2.18166… , which is greater than 2.17825… , as expected because the graph is concave up (convex side downward). 16 c. 4 = ⇒ x = [ln (3/7)]/−0.5 = 1 + 7e –0.5 x 1.6945… About 1 month 21 days d. y (hundred lilies)
The graph starts going downward and to the right from (80, 700) because the coyote population is relatively high, thus decreasing the number of roadrunners. b. There can be two different values for the roadrunner population for a particular coyote population because the two events happen at two different times. For example, coyotes are increasing from 80 when there are 700 roadrunners, but later they are decreasing from 80 when there are about 200 roadrunners. T13. Answers will vary.
Problem Set 7-8 Cumulative Review, Chapters 1–7 1. v(t) 200 ( t , v ( t ))
t 8
v(t) dt represents the distance traveled in time dt. 2. Definite integral 8
3.
∫ (t
3
0
– 21t 2 + 100t + 80) dt
1 4 t − 7t 3 + 50t 2 + 80t 4 = 1280 mi 4. M 100 = 1280.0384 M1000 = 1280.000384 =
T12. a.
0
The Riemann sums seem to be approaching 1280 as n increases. Thus, the 1280 that was found by purely algebraic methods seems to give the correct value of the limit of the Riemann sum.
x (months)
The graph shows that the number of lilies is expected to decrease toward 1600 (y = 16) because of overcrowding.
8
5. v(t) 200 ( t , v ( t ))
R (roadrunners)
t (80, 700)
C (coyotes)
Calculus Solutions Manual © 2005 Key Curriculum Press
8
6. Any Riemann sum is bounded by the corresponding lower and upper sums. That is, L n ≤ R n ≤ U n. By the definition of integrability, the limits of Ln and Un are equal to each other and to the definite integral. By the squeeze theorem, then, the limit of Rn is also equal to the definite integral. Problem Set 7-8
165
7. Definition:
∫
b
18. z 2 = x 2 + y 2 ⇒ 2 z
f ( x ) dx = lim Ln = lim Un ∆x →0
a
dz 1 = [0.3 x + 6e –0.5 x ⋅ (–0.9e –0.5 x )] dt z At x = 2, z = 2 2 + 2.2072 K2 = 2.9786 … ,
∆x →0
provided that the two limits are equal. Fundamental theorem: If f is integrable on [a, b] and g( x ) = f ( x ) dx, then
∫
b
a
∫
f ( x ) dx = g(b) − g( a).
Or: If F( x ) =
∫
x
a
f (t ) dt, then F ′( x ) = f ( x ).
8. Numerically, the integral equals 1280. By counting, there are approximately 52 squares. Thus, the integral ≈ 52(25)(1) = 1300. v( 4.1) – v(3.9) (mi/min ) 9. v ′( 4) ≈ = −19.9 0.2 min v( 4.01) – v(3.99) (mi/min) v ′( 4 ) ≈ = −19.9999 0.02 min f ( x ) – f (c ) or 10. f ′(c) = lim x →c x–c f ( x + ∆x ) – f ( x ) f ′( x ) = lim ∆x →0 ∆x 2 v′(t) = − + 11. 3t 42t 100 ⇒ v′(4) = −20 12. Slowing down. v′(4) < 0 and v(4) = 208 > 0 ⇒ velocity is positive but decreasing ⇒ speed is slowing down. 13. The line has slope −20, and passes through (4, 208). The line is tangent to the graph.
∴ z is decreasing at about 0.044 unit per second. dm = km 19. dt dm = k dt ⇒ ln |m | = kt + C ⇒ 20. m | m | = e kt +C ⇒ m = C1e kt
∫
∫
21. Exponentially 22. General 23. 10000 = C1e0 ⇒ C1 = 10000 10900 = 10000e k⋅1 ⇒ k = ln 1.09 ⇒ m = 10000e ln(1.09) t = 10000(1.09)t 24. False. The rate of increase changes as the amount in 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,
∫
42
y dx
30
≈
5 Slope = –20
dz 1 = (0.6 – 0.7308K) = −0.04391K . dt 2.9786 K
so
v(t) 200
dz dx dy = 2x + 2y dt dt dt
2 (74 + 4 ⋅ 77 + 2 ⋅ 83 + 4 ⋅ 88 + 2 ⋅ 90 3 + 4 ⋅ 91 + 89) = 1022.
26. By symmetric difference quotient, at x = 36 90 – 83 y′ ≈ = 1.75. 2( 2 )
–100
100
27. If f is differentiable on (a, b) and continuous at x = a and x = b, and if f ( a) = f ( b) = 0, then there is a number x = c in (a, b) such that f ′ (c) = 0.
t 0
5
10
28. f(x)
14. Acceleration 15. At a maximum of v(t), v′(t) will equal zero. 3t2 − 42t + 100 = 0 ⇔ t =
42 ± 42 2 – 4 ⋅ 3 ⋅ 100 6
t = 3.041… or 10.958… So the maximum is not at exactly t = 3. 16. v″(t) = 6t − 42 dx dy dz 17. Know: = 0.3. Want: and . dt dt dt dy dx y = 6e −0.5 x ⇒ = −3e −0.5 x ⇒ = −0.9e −0.5 x dt dt dy At x = 2, = −0.9e −1 = −0.3310 K . dt y is decreasing at about 0.33 unit per second.
166
Problem Set 7-8
tangent
f (b )
secant
f (a )
x a c
b
29. f(x) and f´(x) 5
f
f´ x
0
5
10
f´
Calculus Solutions Manual © 2005 Key Curriculum Press
30.
36.
f(x)
x 1
Step discontinuity at x = 1. 31. g(x) = x1/ 3(x − 1) 1 1 g′( x ) = x 13/ + x −2/3 ( x − 1) = x −2/3 ( 4 x − 1) 3 3 g′(0) is undefined because 0 −2 /3 takes on the form 1 / 0 2 /3 or 1/0.
1
x 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=
∫ (0.6 x – e a
0.2x
) dx ≈ 0.8787K
(Integrate numerically.) 33. dV = π [(e0.2 x)2 − (0.6x)2] dx V =π
a
∫ (e 0
0.4x
– 0.36 x 2 ) dx ≈ 8.0554K (Integrate numerically.)
34.
dy x = 0.25 dx y Initial conditions: (0, 3) and (10, 4)
dy 10 = 0.25 ⋅ = 0.625. dx 4 Using ∆x = 0.5, y(10.5) ≈ 4 + (0.625)(0.5) = 4.3125, which is close to the exact value of 4.30842… .
38. At (10, 4),
y
d 3x 2 (sin –1 x 3 ) = dx 1 – x6
40. x = ln (cos t) and y = sec t 1 dx/dt = ⋅ ( − sin t ) = − tan t cos t dy/dt = sec t tan t dy dy/dt sec t tan t = = = − sec t = − y dx dx/dt – tan t dx 1 41. = − ln |4 − 3 x | + C 4 – 3x 3 x 42. h( x ) = 5 = e x⋅ln 5 ⇒ h′( x ) = ln 5 e 5⋅ln 5 = 5 x ln 5 sin 5 x + cos 3 x – 5 x – 1 0 43. lim → x→0 x2 0 5 cos 5 x – 3 sin 3 x – 5 0 = lim → x→0 2x 0 –25 sin 5 x – 9 cos 3 x = lim = −4.5 x→0 2 sin 5 x + cos 3 x – 5 x – 1 , showing a 44. y = x2 removable discontinuity at (0, −4.5).
∫
5
5
y
(10, 4)
(0, 3)
= 0.125 x 2 + C1
37. x = 10.5: y = 0.5 10.52 – 36 = 4.30842 …
39.
g (x )
b
2
⇒ x 2 − 4y 2 = C Initial condition: (10, 4) 100 − 64 = C ⇒ C = 36 ∴ x 2 − 4 y 2 = 36 ⇒ y = ±0.5 x 2 – 36
2
–1
∫ y dy = 0.25 ∫ x dx ⇒ 0.5y
x x
5
35. See the graph in Problem 34. Any initial condition for which y = 0.5x, such as (2, 1), gives the asymptote.
Calculus Solutions Manual © 2005 Key Curriculum Press
–1
1 (0, –4.5) –5
45. Answers will vary. 46. Answers will vary.
Problem Set 7-8
167
Chapter 8—The Calculus of Plane and Solid Figures Problem Set 8-1
Problem Set 8-2
1. f (x) = x 3 − 6x 2 + 9x + 3 f ′ (x) = 3x 2 − 12x + 9
Q1.
Q2. y
y
y
1
x
f
f´
1
1
x 1
x 1
2
3
Q3.
4
Q4. y
g (x) = x 3 − 6x 2 + 15x − 9 g ′ (x) = 3x 2 − 12x + 15 y
y
1
1
x
x
1
1
f f´
Q5.
Q6. y
y x 1
2
3
1
4
x 1
1
h (x) = x − 6x + 12x − 3 h ′ (x) = 3x 2 − 12x + 12 3
x 1
2
Q7.
Q8.
y
y f
y
f´ 1
1
x
x
1
1
x 1
2
3
4
Q9.
2.
3.
4.
5.
168
Positive derivative ⇒ increasing function Negative derivative ⇒ decreasing function Zero derivative ⇒ function could be at a high point or a low point, but not always. The functions have vertex points at values of x where their derivatives change sign. If the derivative is never zero, as for function g, the function graph has no vertex points. If the derivative is zero but does not change sign, as for function h, the function graph just levels off, then continues in the same direction, with no vertex. g ′′ (x) = (d/dx)(3x2 − 12x + 15) = 6x − 12 h ′′ (x) = (d/dx)(3x2 − 12x + 12) = 6x − 12 All the second derivatives are the same! The curves are concave up where the second derivative is positive and concave down where the second derivative is negative. Points of inflection occur where the first derivative graph reaches a minimum. Points of inflection occur where the second derivative graph crosses the x-axis.
Problem Set 8-2
Q10. y 1
y 1
x 1
x 2
1. max.
f (x ) f´(x)
+
x
0
–
2 no p.i.
f (x ) f´´(x)
–
x
–
–
2
2. min.
f ( x) f' ( x)
–
x
0
+
2 no p.i.
f(x ) f"(x) x
+
0
+
2
Calculus Solutions Manual © 2005 Key Curriculum Press
3.
9. plateau
no max./min.
f (x ) f´(x)
f (x ) +
x
0
f´(x)
+
–
undef.
x
2
no p.i.
p.i.
f (x )
f (x ) f´´(x)
–
2
–
x
0
f´´(x)
+
–
undef.
x
2
+
2
10.
4.
no max.min.
max.
f (x )
f (x) f' (x)
+
x
f'(x )
–
undef.
+
2 no p.i.
no p.i.
f (x )
f (x ) f"(x )
–
undef.
x
2
+
x
undef.
f"(x )
+
+
undef.
x
2
+
2
11.
5.
max.
min.
f (x ) f´(x)
min.
max.
f (x ) –
x
undef.
+
f´(x)
2
x
+
0
–
0 –2
+
1
–
0 3
no p.i.
f (x ) f´´(x)
p.i. 0
x
undef.
p.i.
0
f (x )
2
f´´(x)
6.
–
x
0
+
0
–1
–
2
no max./min.
f (x ) f'(x )
f (x ) –
x
undef.
–
2 p.i.
f (x ) f"(x )
–
x
undef.
x
+ –2
2
7.
–1
1
2
3
12. no max./min.
min.
f (x )
plateau
max.
f(x )
f´(x)
–
x
2
f'(x)
– 0
x
–3
+
0
+
0
–1
–
3
p.i.
f (x ) f´´(x)
+
x
p.i.
–
p.i.
p.i.
f (x )
2
f"(x)
8.
+
x
0 –2
–
0
+
–1
0
–
2
no max./.min.
f (x ) f'(x )
–
x
2
f (x )
p.i.
f (x ) f"(x)
–
x
Calculus Solutions Manual © 2005 Key Curriculum Press
0 2
+
x –3
–2
–1
2
3
Problem Set 8-2
169
13.
16. max.
plateau
min.
max.
min.
max.
f (x )
f (x ) f´(x)
∞
+
x
–
–2
f' (x )
–
0
no p.i.
–
e.p. + 0
x
1
0 + e.p.
2
1
6
p.i.
7
no p.i.
f(x)
f (x ) f´´(x)
∞
+
x
+
f"(x)
–
0
–2
e.p.
x
1
–
0
1
f (x )
0
zero
3
5
3
5
+
e.p. 7
f (x )
x –2
x
1
1
2
6
7
17. 14.
y 3 max.
none
f´
max.
f(x ) f'(x)
–
x
f´
0 +
∞ +
0
2
3
4
–
p.i.
18.
–
4
x
f´
–3
∞
+
6
f
f (x ) f"(x )
x 3
y
3
f (x )
f'
f
f' x
4
8
f' x 2
3
–4
4
19. y
15. min.
max.
4
min.
f
f (x ) f´(x)
e.p.
x
–1
+
0
zero
0
1
–
e.p.
f´
f´
5
3
x 4
8
no p.i.
f (x ) f´´(x)
e.p.
x
–1
–
0
zero
0
1
3
–
e.p. 5
20. y
f (x ) 4 2
f
f' x –1
170
1
Problem Set 8-2
3
x 4
5
Calculus Solutions Manual © 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 = 2 f ′′ (x) = 2ex − ex − xex = ex (1 − x) f ′′ (2) = e2 (1 − 2) = −7.3890... < 0 ∴ local maximum at x = 2 f (x )
f (x ) 1
x 2
The graph confirms a maximum at x = 2.
5
x 1
2
3
The graph confirms a maximum at x = 2. 22. f ( x ) = − sin
π x 4
25. f (x) = (x − 2)3 + 1 f ′ (x) = 3(x − 2)2 f ′ (2) = 3(2 − 2)2 = 0 ⇒ critical point at x = 2 f ″ (x) = 6(x − 2) f ″ (2) = 6(2 − 2) = 0, so the test fails. f ′ (x) goes from positive to positive as x increases through 2, so there is a plateau at x = 2.
π π cos x 4 4 π π f ′(2) = − cos (2) = 0 ⇒ critical point 4 4 at x = 2
f (x )
f ′( x ) = −
π2 π sin x 16 4 2 π π sin (2) = 0.6168K > 0 f ′′(2) = 16 4 ∴ local minimum at x = 2
1
x 2
f ′′( x ) =
f (x ) 1
x 2 –1
The graph confirms a plateau at x = 2. 26. f (x) = (2 − x)4 + 1 f ′ (x) = −4(2 − x)3 f ′ (2) = −4(2 − 2)3 = 0 ⇒ critical point at x = 2 f ″ (x) = 12(2 − x)2 f ″ (2) = 12(2 − 2)2 = 0, so the test fails. f ′ (x) changes from negative to positive as x increases through 2, so there is a local minimum at x = 2. f (x )
The graph confirms a minimum at x = 2. 23. f (x) = (2 − x)2 + 1 f ′ (x) = −2(2 − x) f ′ (2) = −2(2 − 2) = 0 ⇒ critical point at x = 2 f ″ (x) = 2 ⇒ f ″ (2) = 2 > 0 ∴ local minimum at x = 2 f (x )
1
x 2
The graph confirms a minimum at x = 2. 24. f (x) = −(x − 2)2 + 1 f ′ (x) = −2(x − 2) f ′ (2) = −2(2 − 2) = 0 ⇒ critical point at x = 2 f ″ (x) = −2 ⇒ f ″ (2) = −2 < 0 ∴ local maximum at x = 2
Calculus Solutions Manual © 2005 Key Curriculum Press
1
x 2
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 points for f (x)) f ′′( x ) = 120 x 3 − 60 x = 60 x ( 2 x + 1)( 2 x − 1) f ″ (x) = 0 ⇔ x = 0, ± 1/2 (critical points for f ′ (x)) b. The graph begins after the f-critical point at x = −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.
Problem Set 8-2
171
28. a. f (x) = 0.1x 4 − 3.2x + 7 f ′ (x) = 0.4x 3 − 3.2 = 0.4(x − 2)(x 2 + 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 − x + e − x = e − x (1 − x ) f ′ (x) = 0 ⇔ x = 1 (critical point for f (x)) f ′′( x ) = xe − x − 2e − x = e − x ( x − 2) f ″ (x) = 0 ⇔ x = 2 (critical point for f ′ (x)) b. Because f (x) approaches its horizontal asymptote (y = 0) from above, the graph must be concave up for large x; but the graph is concave down near x = 1, and the graph is smooth; somewhere the concavity must change from down to up. c. No. e − x ≠ 0 for all x, so xe − x = 0 ⇔ x = 0. 30. a. f (x) = x 2 ln x f ′ (x) = x + 2x ln x = x (1 + 2 ln x) f (x) and f ′ (x) are undefined at x = 0, so f ′( x ) = 0 ⇔ ln x = −0.5 ⇔ x = e −0.5 = 0.6065… (critical point for f (x)). f ″ (x) = 3 + 2 ln x f ′′( x ) = 0 ⇔ ln x = −1.5 ⇔ x = e −1.5 = 0.2231… (critical point for f ′ (x)).
c. There is no inflection point at x = 0 because concavity is down for both sides, but there is an inflection point at x = 1. 32. a. f ( x ) = x 1.2 − 3 x 0.2 f ′( x ) = 1.2 x 0.2 − 0.6 x −0.8 = 0.6 x −0.8 (2 x − 1) f ′ (x) = 0 ⇔ x = 0.5, and f ′ (x) is undefined at x = 0 (critical points for f (x)). f ′′( x ) = 0.24 x −0.8 + 0.48 x −1.8 = 0.24 x −1.8 ( x + 2) f ″ (x) = 0 ⇔ x = −2 (critical point for f ′ (x); f ′ (0) is undefined, so f ′ has no critical point at x = 0). b. f (0) = 01.2 − 3 ⋅ 00.2 = 0 has only one value. c. Curved concave up because f ′′(x) > 0 for x < −2 33. a. f (x) = −x 3 + 5x 2 − 6x + 7 f (x ) 7
x 1
b.
ln x 1/ x = lim x –2 x → 0 + – 2 x – 3 = lim+ − 0.5 x 2 = 0 by L’Hospital’s rule.
b. lim+ x 2 ln x = lim+ x→0
x→0
x→0
lim x 2 ln x does not exist because x2 ln x is
c.
x→0 −
undefined for x < 0. c. All critical points from part a appear, although the inflection point at x = e −1.5 is hard to see on the graph. 31. a. f (x) = x 5/3 + 5x 2/3 5 10 5 f ′( x ) = x 2/3 + x −1/3 = x −1/3 ( x + 2) 3 3 3 f ′ (x) = 0 ⇔ x = −2, and f ′ (x) is undefined at x = 0 (critical points for f (x)). 10 −1/3 10 −4/3 10 −4/3 = f ′′( x ) = x − x x ( x − 1) 9 9 9 f ″ (x) = 0 ⇔ x = 1 (critical point for f ′(x); f ′ (0) is undefined, so f ′ has no critical point at x = 0). b. The y-axis (x = 0) is a tangent line because the slope approaches −∞ from both sides.
172
Problem Set 8-2
d.
34. a.
Maximum (2.5, 7.6), minimum (0.8, 4.9), points of inflection (1.7, 6.3) No global maximum or minimum f ′(x) = −3x 2 + 10x − 6 1 f ′( x ) = 0 ⇔ x = (5 ± 7 ) = 2.5485K or 3 0.7847… 5 f ′′( x ) = −6 x + 10; f ′′( x ) = 0 ⇔ x = = 3 1.666… f ′′(0.7847…) = −6(0.7847…) + 10 = 5.2915… > 0, confirming local minimum. Critical and inflection points occur only where f, f ′, or f ′′ is undefined (no such points exist) or is zero (all such points are found above). f (x) = x 3 − 7x 2 + 9x + 10 f (x )
10
x 1
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 © 2005 Key Curriculum Press
b. f ′(x) = 3x 2 − 14x + 9 1 f ′( x ) = 0 at x = (7 ± 22 ) = 3.896 K or 3 0.769… 7 f ′′( x ) = 6 x − 14; f ′′( x ) = 0 at x = = 2.333K 3 c. f ′′(0.769…) = 6(0.769…) − 14 = −9.3808… < 0, confirming local maximum. d. Critical and inflection points occur only where f, f ′, or f ′′ is undefined (no such points exist) or is zero (all such points are found above). 35. a. f (x) = 3x 4 + 8x 3 − 6x 2 − 24x + 37, x ∈ [−3, 2] f (x ) 80
x –3
b.
c. d.
36. a.
2
Maximum (−3, 82), (−1, 50), (2, 77), minimum (−2, 45), (1, 18), points of inflection (−1.5, 45.7), (0.2, 32.0) Global maximum at (−3, 82) and global minimum at (1, 18) f ′(x) = 12x 3 + 24x 2 − 12x − 24 = 12(x + 2)(x − 1)(x + 1) f ′(x) = 0 ⇔ x = −2, −1, 1 f ′(x) is undefined ⇔ x = −3, 2. f ′′(x) = 36x 2 + 48x − 12 = 12(3x 2 + 4x − 1); 1 f ′′( x ) = 0 ⇔ x = − (2 ± 7 ) = 0.2152 … 3 or −1.5485… f ′′(x) is undefined ⇔ x = −3, 2. f ′′(−2) = 12[3(4) + 4(−2) − 1] = 36 > 0, confirming local minimum. Critical and inflection points occur only where f, f ′, or f ′′ is undefined (only at endpoints) or is zero (all such points are found above). f (x) = (x − 1)5 + 4, x ∈ [−1, 3]
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 only where f, f ′, or f ′′ is undefined (only at endpoints) or is zero (all such points are found above). 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 with nonzero 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 or minimum (if f ′(x) is never zero); if this is not the case, then the maximum and minimum occur where f ′(x) = 3ax2 + 2bx + c = 0, at –2 b ± 4b 2 – 4 ⋅ 3a ⋅ c – b b 2 – 3ac = ± , 6a 3a 3a and the maximum and minimum occur at x=
b 2 – 3ac /(3a) units on either side of the inflection point −b/(3a) (see Problem 33). 39. f (x) = ax3 + bx2 + cx + d f ′(x) = 3ax2 + 2bx + c; f ′′(x) = 6ax + 2b Points of inflection at (2, 3) ⇒ f ′′(3) = 0 ⇒ 18a + 2b = 0 Maximum 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 = 10 Solving this system of equations yields 1 9 15 5 f ( x) = − x 3 + x 2 − x − . 2 2 2 2 f (x )
(5, 10)
5 (3, 2) 3
x 5
f (x )
10 –1
x 1
3
Maximum (3, 36), minimum (−1, −28), plateau and points of inflection (1, 4) Global maximum at (3, 36) and global minimum at (−1, −28)
Calculus Solutions Manual © 2005 Key Curriculum Press
The graph confirms maximum (5, 10) and points of inflection (3, 2). 40. f (x) = ax3 + bx2 + cx + d f ′(x) = 3ax2 + 2bx + c; f ′′(x) = 6ax + 2b Points of inflection at (2, 7) ⇒ f ′′(2) = 0 ⇒ 12a + 2b = 0 Maximum at (−1, 61) ⇒ f ′(−1) = 0 ⇒ 3a − 2b + c = 0 Problem Set 8-2
173
(2, 7) and (−1, 61) are on the graph ⇒ 8a + 4b + 2c + d = 7. −a + b − c + d = 61 Solving this system of equations yields f (x) = x 3 − 6x 2 − 15x + 53. 80
d. f (x )
f (x ) x c x
–1
e.
2
f ( x)
The graph confirms maximum (−1, 61) and points of inflection (2, 7). 41. a. f (x) = x 3 ⇒ f ′(x) = 3x 2 f ′(−0.8) = 1.92 f ′(−0.5) = 0.75 f ′(0.5) = 0.75 f ′(0.8) = 1.92 b. The slope seems to be decreasing from −0.8 to −0.5; f ″(x) = 6x < 0 on −0.8 ≤ x ≤ −0.5, which confirms that the slope decreases. The slope seems to be increasing from 0.5 to 0.8; f ″(x) = 6x > 0 on 0.5 ≤ x ≤ 0.8, which confirms 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 not possibly be a straight line with slope = 1. 43. a.
(Locally constant)
x c
44. f (x) = 10(x − 1)4/3 + 2 f (1) = 2, so f (1) is defined. 40 f ′( x ) = ( x – 1)1/ 3 3 f ′(1) = 0, so f is differentiable at x = 1. 40 f ″( x ) = ( x – 1) −2 / 3 9 40 40 f ″(1) has the form (0 –2 / 3 ) or (1/0) , so 9 9 f ″(1) is infinite. There seems to be a cusp at (1, 2), but zooming in on this point reveals that the tangent is actually horizontal there. f (x )
f (x )
2
x x c
b. f (x )
x c
c. f (x )
x c
174
Problem Set 8-2
1
See Problem 20 in Problem Set 10-6 for calculation 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 + 0.000216x f (0) = 1 and g (0) = 1 f ′(0) = 0.06 and g ′(0) = 0.06 f ″(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) = 0 somewhere between x = −100 and x = 0, meaning that g does have an x-intercept. Calculus Solutions Manual © 2005 Key Curriculum Press
1. Let x = total width of pen, y = length of pen. Domains: 0 ≤ x ≤ 300, 0 ≤ y ≤ 200 Maximize A(x) = xy. 2 2 x + 3 y = 600 ⇒ y = 200 − x 3 2 2 ∴ A( x ) = 200 x − x 3
( x – 1)3 sin 1 + 2, if x ≠ 1 46. f ( x ) = x –1 2, if x = 1 2.01
f (x )
2
1.99
A (x)
1
1 + 2 = 2 = f (1) x –1 (The limit of the first term is zero because (x − 1)3 approaches zero and the sine factor is bounded.) ∴ f is continuous at x = 1. f ( x ) – f (1) f ′(1) = lim x →1 x –1 [( x – 1)3 sin(1/( x – 1))] + 2 – 2 = lim x →1 x –1 1 2 = lim ( x – 1) sin =0 x →1 x –1 (x − 1)2 → 0 and the sine factor is bounded. ∴ f ′(1) = 0
max.
lim f ( x ) = lim ( x – 1)3 ⋅ sin x →1
x →1
2.001
2
1.999
0.9
1
1.1
Problem Set 8-3
1
Q2. ln |x + 6| + C Q4. 3x 1/3 + C Q6. x + C
Q10. D Calculus Solutions Manual © 2005 Key Curriculum Press
x 14.28...
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 = 35/ 6 , y = 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. Minimum at x = 192.5 = 13.87… ,
y" x
The graph shows a maximum at x ≈ 150. 4 Algebraically, A′( x ) = 200 − x. 3 A′(x) = 0 ⇔ x = 150, confirming the graph. 2 x = 150 ⇒ y = 200 − ⋅ 150 = 100 3 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 ≥ 0 Minimize P (x) = 12x + 7y. xy = 350 ⇒ y = 350x − 1 ∴ P (x) = 12x + 2450x− 1
P ′ (x) = 20 − 3850x− 2 = 0 at x = 192.5
Q9.
y
1
300
P (x )
The graph is zoomed in by a factor of 10 both ways. The graph does appear to be locally linear at x = 2. Although the sine factor makes an infinite number of cycles in any neighborhood of x = 1, the (x − 1)3 factor approaches zero so rapidly that the graph is “flattened out.” The name pathological is used to describe the fact that the graph makes an infinite number of cycles in a bounded neighborhood of x = 1. 47. Answers will vary.
Q1. y′ = −3(3 x + 5) −2 2 Q3. − x −5/ 3 3 Q5. − x −1 + C Q7. ln |sin x| + C Q8.
x 150
1
x 2
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 Problem Set 8-3
175
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 of rectangle, y = width of square. A rect = 2x 2, A sq = y 2 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 A(x) 20,000
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 140 around the circular fence of radius 70 to enclose a total area of only 19,607 yd2 , but Big Bill might not like your solution!) b. The graph of A versus r shows that the maximum area occurs at the largest possible circle. Big Bill should use all 1000 yards for the circular fence and not build a corral. 5. a. Let x = length of square base, z = height of box. Domain of x: 0 ≤ x ≤ 120 = 10.954 … Maximize V(x) = x2z. Area = x2 + 4xz = 120 ⇒ z = 30/x − x/4 ∴ V(x) = 30x − x3/4 V
max.
x 20
93.3
c. The graph shows a maximum at endpoint x = 93.3333… . A′(x) = −450 + 8.5x A′(x) = 0 ⇔ x = 450/8.5 = 52.9411… Because A(52.9…) is a minimum, the maximum 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 square Diameter ≥ 50 ⇒ r ≥ 25 Circumference ≤ 1000 ⇒ 2πr ≤ 1000 ⇒ r ≤ 500/π Domain 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 A
max.
min.
r 25
70
160
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… 176
Problem Set 8-3
x 0
6.32...
10.95...
The graph shows a maximum at x ≈ 6.3. V′(x) = 30 − 3x 2/4 = 0 at x = ± 40 x = − 40 is out of the domain. Critical points at x = 0, x = 40 , x = 120 V (0) = 0, V ( 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 cm deep. b. Conjecture: An open box with square base of side length x and fixed surface area A will have maximal volume when the base length is 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 integer value of x) V(3) = 252 cm2 V(4) = 192 cm2 V(5) = 100 cm2 V(6) = 0 cm2
Calculus Solutions Manual © 2005 Key Curriculum Press
c. V(x) = (20 − 2x)(12 − 2x)x = 240x − 64x 2 + 4x 3 V (x )
A – 2 z(–2 z + 4 z 2 + A )
y=
–2 z + 4 z 2 + A + 2 z
= −2 z + 4 z 2 + A
200
x 6
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 maximum because it is positive and V(0) = V(6) = 0. Maximum volume ≈ 262.7 cm2 at x ≈ 2.43 cm 7. Let x = length, y = depth, C(x) = total cost. Domains: x > 0, y > 0 Area of bottom = 5x Total area of sides is (10 + 2x)y. Minimize C(x) = 10(5x) + 5(10 + 2x)y. Volume = 72 ⇒ 5xy = 72 ⇒ y = 72/(5 x ) = 14.4 x −1 ∴ C( x ) = 50 x + 5(10 + 2 x )(14.4 x −1 ) C( x ) = 50 x + 720 x −1 + 144
Therefore, x = y for maximum volume, Q .E .D . b. Let x = y. Maximize V = xyz = x 2z. Fixed area A = xy + 2xz + 2yz = x2 + 4xz ⇒ z = A/(4x) − x/4 ∴ V = (A/4)x − x3/4 dV = ( A/4) − 3 x 2 /4 = 0 at x = ± A/3 dx dV/dx goes from positive to negative at x = A/3 ⇒ maximum at x = A/3. 1 1 A/3 = x 2 2 c. For the maximal box in part b, the depth is half the length of the base. Thus, the box is short and fat. This makes sense because the problem is equivalent to maximizing the volume of two open boxes with the second box placed upside-down on the first. The resulting single closed box will have maximum volume when it is a cube, which will happen if each open box is half a cube. z = A/( 4 A/3 ) − A/3 /4 =
9. For y = ex, minimize D( x ) = x 2 + y 2 =
C
x 2 + e2 x . D (x) x 3.794...
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 ≈ 523.47 Minimum cost is $523.47. 8. a. Maximize V = xyz. Fixed area A = xy + 2xz + 2yz ⇒ y = (A − 2xz)/(x + 2z) ∴V =
Axz – 2 x 2 z 2 x + 2z
dV –2 z 2 x 2 – 8z 3 x + 2 Az 2 = dx ( x + 2 z )2 dV = 0 at x = −2 z + 4 z 2 + A dx
Calculus Solutions Manual © 2005 Key Curriculum Press
1
x –0.4263...
The graph shows a minimum at x ≈ −0.43. 1 2 ( x + e 2 x ) −1/2 (2 x + 2e 2 x ) 2 D′(x) = 0 ⇔ 2x + 2e2x = 0 ⇔ x = −e2x Because x appears both algebraically and exponentially, there is no analytic solution. Solving numerically gives x ≈ −0.4263. By graphing 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 r x ≥ 100 ⇒ r ≤ 100/π ∴ domain is 20 ≤ r ≤ 100/π. A(r) = π r2 + 2r(200 − π r) = 400r − π r2 D′( x ) =
Problem Set 8-3
177
A (r ) 10,000
r 20 31.83...
The graph shows a minimum at endpoint x = 20. A′ = 400 − 2π r A′ = 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 straight sections 137.17 m. 11.
L
y 8
x–1
1
x
L( x ) = x 2 + y 2 . Domains: x ≥ 1, y ≥ 8 Minimize L 2(x) = x2 + y2. 8x y 8 Using similar triangles, = . ⇒y= x –1 x x –1 64 x 2 ∴ L2(x) = x2 + ( x – 1)2
L( x ) = ( x + 7)2 + ( y + 5)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 L 2(x) = x 2 + 14x + 49 + 1225/x 2 + 350/x + 25 L 20
x 5.59...
The graph shows a minimum of L(x) at x ≈ 5.6. ( L2 ( x ))′ = 2 x + 14 − 350 x −2 − 2450 x −3 By numerical solution, (L2)′ = 0 at x ≈ 5.5934… . (Exact answer is x = 3 175 .) But a minimum distance L in the hall implies that the maximal ladder that will go through the hall is at x = 5.5934… . L2(5.5934…) = 285.3222… L(5.5934…) = 16.8914… No ladder longer than 16.8 ft (rounded down) can pass through the hall. 13. Let r = radius, h = height. V = π r2h 2r + 2h = 1200 ⇒ h = 600 − r ∴ V = πr2(600 − r) = π(600r2 − r3) V
L
10
r 400
x 1
5
The graph shows a minimum of L(x) at x ≈ 5. 128 x (L2)′(x) = 2x − ( x – 1)3 128 x ⇔ (L2)′(x) = 0 ⇔ 2 x = ( x – 1)3 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. L y
5 7
178
x
Problem Set 8-3
The graph shows a maximum at r ≈ 400. V′ = π(1200r − 3r2) V′ = 0 ⇔ r = 0 or r = 400 From graph, maximum is at r = 400. h = 600 − 400 = 200 Maximum volume occurs with rectangle 400 mm wide (radius), 200 mm high. 14. Rotating a square does not give the maximum volume. The solution to Problem 13 gives a counterexample. Repeating the calculations with perimeter P instead of 1200 gives r = (1/3)P and h = (1/6)P, showing that the proportions for maximum volume are with radius twice the height. 15. a. Let r = radius, h = height. V = πr2h = π(3.652)(10.6) = 141.2185π = 443.6510… cm3
Calculus Solutions Manual © 2005 Key Curriculum Press
b. A = 2πrh + 2πr2 V = πr2h = 141.2185π ⇒ h = 141.2185/r2 ∴ A = 2πr(141.2185/r2) + 2πr2 A = 2π (141.2185r −1 + r 2 ) c.
A
500
r 4.13...
The graph shows a minimum at x ≈ 4.1. A′ = 2π ( −141.2185r −2 + 2 r ) A′ = 2π/r2(−141.2185 + 2r3) A′ = 0 ⇔ r3 = 70.60925 ⇒ r = 3 70.60925 = 4.1332 … Minimum at r = 4.1… because A′ goes from negative to positive. h = 141.2815/(3 70.60925 )2 = 23 70.60925 = 8.2664… Radius ≈ 4.1 cm, height ≈ 8.3 cm Because height = 2 × radius, height = diameter. So minimal can is neither tall and narrow nor short and wide. d. Normally proportioned can is taller and narrower 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 ) = 2πr 2 k + 2πrh = 2πr 2 k + 282.437πr −1 C ′(r ) = 4πrk − 282.437πr −2 = 4πr −2 ( kr 3 − 70.60925) C′(r) = 0 at r = 3 70.60925/k C ″ (r ) = 4πk + 564.874πr −3 > 0 for all r > 0, so this is a local minimum. If the normal can is the cheapest to make, then 3.65 = 3 70.60925/k ⇒ k = 70.60925(3.65) −3 = 1.4520 … . This is reasonable because metal for the ends is cut into circles, so some must be wasted. b. Now it takes (2r)2 cm2 of metal to make each end of the can, so the function to minimize is C(r ) = 8r 2 k + 2πrh = 8r 2 k + 282.437πr −1 . C ′(r ) = 16rk − 282.437πr −2 C ′(r ) = 0 at r = 3 Calculus Solutions Manual © 2005 Key Curriculum Press
282.437π 16k
C ″ (r ) = 16k + 564.874πr −3 > 0 for all r > 0, so this is a local minimum. If the normal can is the cheapest to make, 282.437π 282.437π then 3.65 = 3 ⇒k= 16k 16(3.65)3 = 1.1404… . To minimize the area (not the cost) of the can, minimize 8r 2 + 2πrh = 8r 2 + 282.437πr −1. C ′(r ) = 16r – 282.437πr −2 = 0 ⇒ 282.437π r=3 = 3.8126 cm 16 141.2185 h= 2 = 9.7099 K cm. 3 282.437π /16
(
)
The proportions of this can are closer to those of the normal can. c. If the metal for the ends can be cut without waste, then it takes π(r + 0.6)2 to make each end and (2πr + 0.5)h to make the sides, so minimize C(r) = 2π(r + 0.6)2 + (2πr + 0.5)h = 2π (r + 0.6)2 + 141.2185(2πr + 0.5)r −2 C ′(r ) = 4π (r + 0.6) − 282.437πr −2 − 141.2185r −3 C′(r) = 0 at r ≈ 3.9966 by graphing calculator. C ″ (r ) = 4π + 564.874πr −3 + 423.6555r −4 > 0 for 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 from squares, then it takes 4(r + 0.6)2 to make each end and (2πr + 0.5)h to make the sides, so minimize: C(r) = 8(r + 0.6)2 + (2πr + 0.5)h = 8(r + 0.6)2 + 141.2185(2πr + 0.5)r −2 C ′(r ) = 16(r + 0.6) − 282.437πr −2 − 141.2185r −3 C′(r) = 0 at r ≈ 3.6776… by graphing calculator. C ′′(r ) = 16 + 564.874πr −3 + 423.6555r −4 > 0 for 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 = 43.75π Let r = radius of cup, h = height of cup. Minimize A(r) = πr2 + 2πrh. πr 2 h = 43.75π ⇒ h = 43.75r −2 ∴ A(r ) = πr 2 + 87.5πr −1 A
100
r 3.52...
Problem Set 8-3
179
The graph shows a minimum at r ≈ 3.5 cm. A′(r ) = 2πr − 87.5πr −2 = 2πr −2 (r 3 − 43.75) A′(r) = 0 at r = 3 43.75 = 3.5236… . There is a minimum at x = 3.5236… because A(r) goes from decreasing to increasing. (See graph.) h = 43.75( 43.75) −2/3 = 3 43.75 = r 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 per cup, 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 would save 754,299.93 ≈ $754,000 per year in paper costs (about 10% of the current annual paper bill), but would likely result in loss of sales because a cup of that shape is hard to drink from. d. Let r = radius of cup, h = height of cup. πr 2 h = V ⇒ h = (V /π )r −2 Minimize A(r ) = πr 2 + 2πrh = πr 2 + 2Vr −1 . A′(r ) = 2πr − 2Vr −2 = 0 at r = 3 V /π A′′(r ) = 2π + 4Vr −3 > 0 for all r > 0, so this is a minimum. Minimal cup has r = 3 V/π , h = (V /π )(V /π ) −2/3 = 3 V /π = r. 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 A (x ) c. 1000
x 25
100
The graph shows a maximum at x ≈ 25. A′(x) = 2 − 0.08x = 0 at x = 25. Critical points at x = 0, 25, 100 A(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 for x = 100 in. 19. Maximize A(x) = 2xy = 2x cos x. Use 0 ≤ x ≤ π/2 for the domain of x. 180
Problem Set 8-3
A (x ) 1
x 1
The graph shows a maximum at x ≈ 0.86. A′(x) = 2 cos x − 2x sin x A′(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. 200
x
50
y
Street
Let x = width of store, y = length of store. Minimize C(x) = 100x + 80(x + 2y). xy = 4000 ⇒ y = 4000 x −1 C( x ) = 180 x + 640000 x −1 y ≤ 200 ⇒ x ≥ 20, so domain of x is 20 ≤ x ≤ 50. Graph shows minimum at x endpoint x = 50. C (x ) 50,000
x 20
50
C ′( x ) = 180 − 640000 x −2 = 0 80 5 at x = = 59.628… , outside the domain. 3 C(20) = $35,600.00; C(50) = $21,800.00 Minimum 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 x 0 lim A = lim → x→0 x →0 2 tan x 0 1 1 = lim = 2 x →0 2 sec x 2 b. Domain of x is 0 < x ≤ π /2. A (x ) 0.5
x π/2
The graph shows that the area approaches a maximum as x approaches the endpoint x = 0 from the positive side. Calculus Solutions Manual © 2005 Key Curriculum Press
1 (cot x – x csc 2 x ) 2 A′(x) = 0 when x = cos x sin x or 2x = 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 get arbitrarily close to 1/2, but never achieve it. A′ ( x ) =
22.
y (x, y)
The graph shows a maximum at x ≈ 1. P′(x) = 4 − 4x = 0 at x = 1 P(0) = 18; P(1) = 20; P(3) = 12 Maximal 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) = π x2y = π x2(9 − x2) = 9 π x 2 − π x 4. Domain: 0 ≤ x ≤ 3 V(x ) 50
x x
3
Domain of x is 0 ≤ x ≤ 3. Maximize A = 0.5(3 − x) ( y) = 0.5(3 − x)ex = 1.5ex − 0.5xex. A (x ) 4
x 0
2
3
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 − 2x 3. Domain: 0 ≤ x ≤ 3
x 0
2.121...
The graph shows a maximum at x ≈ 2.1. V ′( x ) = 18πx − 4πx 3 = 0 at x = 0, ± 4.5 . − 4.5 is out of the domain. V (0) = V (3) = 0, V ( 4.5 ) = 20.25π = 63.6172… Maximum is at x = 4.5 , y = 9 − 4.5 = 4.5. Maximal cylinder has radius = 4.5 = 2.12132… and height = 4.5. b. Maximize L(x) = 2π xy = 2π x(9 − x2) = 18π x − 2π x 3 . L (x ) 50
x 0
A(x ) 20
x 0
1.732
3
The graph shows a maximum at x ≈ 1.7. A′( x ) = 18 − 6 x 2 = 0 at x = ± 3 = ±1.732… −1.732 is out of the domain. A(0) = A(3) = 0; A( 3) = 12 3 = 20.7846 K Maximal rectangle has width = 2 3, length = 9 − 3 = 6. b. Maximize P(x) = 4x + 2y = 4x + 18 − 2x2.
3
1.732...
3
The graph shows a maximum at x ≈ 1.7. L ′( x ) = 18π − 6πx 2 = 0 at x = ± 3 . − 3 is out of the domain. L(0) = L(3) = 0; L( 3 ) = 12π 3 = 65.2967K Maximum is at x = 3 , y = 9 − 3 = 6. Maximal cylinder has radius = 3 = 1.7320… and height = 6. c. Maximize A(x) = 2πx2 + 2π xy = 2π x 2 + 2π x(9 − x 2) = 2π x 2 + 18π x − 2π x 3 . A ( x)
50
P(x ) 20 x 0
x 0
1
Calculus Solutions Manual © 2005 Key Curriculum Press
3
2.097...
3
The graph shows a maximum at x ≈ 2.1. A′(x) = 18π + 4π x − 6π x 2 Problem Set 8-3
181
A′(x) = 0 at x =
−1.430… −1.430… is out of the domain. A(0) = 0; A(2.0971…) = 88.2727… ; A(3) = 18π = 56.5486… 1+ 2 7 Maximal cylinder has radius = = 3 52 – 4 7 2.0971… and height = = 9 4.6018… . d. No. The maximum-volume cylinder has dimensions different from both of the maximum-area cylinders in parts b and c. e. No. Rotating the maximum-area rectangle does not produce the maximum-volume cylinder. But it produces the cylinder with maximum lateral area. f. If y = a2 − x2, the paraboloid has radius = a. V = π x2(a2 − x2) = π (a2x2 − x4) V′ = π (2a 2x − 4x3) V ′ = 0 ⇔ x = 0 or x = ± a/ 2 . 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 not constant for the cylinder of maximum total area. 25. a. x 2 + y 2 = 100, 0 ≤ x ≤ 10 Maximize V ( x ) = πx 2 ⋅ y = 2πx 2 100 – x 2 . b.
V ( x) 2000
x 0
8.16... 10
The graph shows a maximum volume at x ≈ 8.2. V ′( x ) = =
–2πx 3
+ 4π x 100 – x 2
100 – x 2 –6π x 3 + 400π x
c. Height = radius ⋅ 2 4 4000π π ⋅ 1000 = 3 3 4000π 3 Volume of maximal cylinder Vc = 9 ∴ Vc = Vs / 3 Volume of sphere Vs =
26. Let r = radius of cone, h = height. Lateral area A(r) = π r · (slant height) = πr r 2 + h 2 1 V = πr 2 h = 5π ⇒ h = 15r −2 3 ∴ A(r ) = πr r 2 + 225r –4 h ≥ 2r ⇒ 2r ≤ 15r− 2 Domain of r is 0 < r ≤ 3 7.5 = 1.9574 K . A(r )
20
r 1
1.957...
The graph shows a minimum of A(r) at endpoint r = 1.957… . Minimize A 2 (r ) = π 2 (r 4 + 225r −2 ). ( A 2 (r ))′ = π 2 ( 4r 3 − 450 r −3 ) = 0 at r = 6 112.5 = 2.1971… , which is out of the domain. A(1.9574…) = 26.915… , lim+ A(r ) = ∞. r→0
Minimal cone has radius = 3 7.5 = 1.9574 … and height = 2 r = 23 7.5 = 3.9148K . Make r ≈ 1.96 ft and h ≈ 3.91 ft. 27. a. Lateral area L(x) = 2π x y Domains: 0 ≤ x ≤ 5 and 0 ≤ y ≤ 7 Equation of element of cone is 7 y = − x + 7 ⇒ y = −1.4 x + 7. 5 ∴ L(x) = 2πx(−1.4x + 7) = 2π(−1.4x2 + 7x) L (x ) 50
100 – x 2
V ′( x ) = 0 at x = 0,
200 10 6 = = 8.1649K 3 3
V(0) = V(10) = 0 10 6 4000π 3 V = 2418.399K = 9 3 182
Maximal cylinder has radius = 8.1649… , 20 3 height = = 11.5470 K , and volume = 3 2418.39… .
1± 2 7 = 2.0971K or 3
Problem Set 8-3
x 0
2.5
5
The graph shows a maximum of L(x) at x ≈ 2.5. L′(x) = 2π (−2.8x + 7) L′(x) = 0 at x = 2.5. Calculus Solutions Manual © 2005 Key Curriculum Press
L′(x) goes from positive to negative at x = 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) A (x ) 150
x 0
Maximum area at x =
rh if h ≥ 2r; 2(h – r )
x = r otherwise. c. From part b, the maximal cylinder degenerates to two circular bases if the radius of the cone is at least half the height. 29. Maximize V = π y2x. Ellipse equation is (x/9)2 + (y/4)2 = 1, from which y2 = (16/81)(81 − x2). ∴ V = (16π/81)(81x − x3) Domain: 0 ≤ x ≤ 9
5
V
The graph shows a maximum at endpoint x = 5. A′(x) = 2π(7 − 0.8x) = 0 at x = 8.75, out of domain. ∴ maximum is at an endpoint, x = 5. A(0) = 0; A(5) = 2π(52) = 50π = 157.07… Maximum area is with the degenerate cylinder consisting only of the top and bottom, radius 5 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 is y = (−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 at x = 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 –h x= 2(1 – h/r ) –h rh A′(x) = 0 at x = = 2 – 2 h/r 2(h – r ) If h ≤ 2r, then A′(x) ≠ 0 for all x ≤ r, so in this case the critical points are the endpoints, x = 0, r. A(0) = 0; A(r) = 2πr2 rh If h ≥ 2r, then 0 ≤ ≤ r, so this is a 2( h – r ) rh πrh 2 . critical point; A = 2( h – r ) 2( h – r ) A′(x) goes from positive to negative at rh x= . 2(h – r ) Calculus Solutions Manual © 2005 Key Curriculum Press
150
x 0
5.196...
10
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 (0) = V (9) = 0; V ( 27 ) = 32π 3 = 174.1… At 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 form x = ay 2 + 16. 0 = a ⋅ 16 + 16 ⇒ a = −1 ⇒ x = 16 − y2 V(y) = π y2(16 − y2) = π (16y2 − y4) Domain: 0 ≤ y ≤ 4 F (y ) 300
y 0
2.828...
4
The graph shows a maximum V(y) at y ≈ 2.8. C′(y) = π (32y − 4y 3) = 4π y(8 − y 2) = 0 at y = 0, ± 8 . y = − 8 is out of the domain. C(0) = C( 4) = 0, C( 8 ) = 64π = 201.0619… Maximum C(y) at y = 8. At y = 8 , x = 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 Problem Set 8-3
183
Vf = volume of frustum, h = height of frustum, R = larger radius, and r = smaller radius. 1 ∴ F( y) = πx (16 + y 2 + 4 y) 3 1 = π (16 − y 2 )( y 2 + 4 y + 16) 3 1 F( y) = π (256 + 64 y − 4 y 3 − y 4 ) 3 F (y )
false because f (c) could be a local minimum or a plateau point. 32. a. Let x = length of corral (parallel to wall), y = width of corral (perpendicular to wall). A = xy If x ≤ 600, then 1000 = x + 2y ⇔ y = 500 − 0.5x. If x ≥ 600, then 1000 = x + 2y + (x − 600) ⇔ y = 800 − x. 500 x – 0.5 x 2, x ≤ 600 ∴ A= 2 x > 600 800 x – x ,
300
A 150,000 y 0
1.821...
4
The graph shows a maximum F(y) at y ≈ 1.8. 1 F ′( y) = π (64 − 12 y 2 − 4 y 3 ) 3 F ′(y) = 0 ⇔ 64 − 12y 2 − 4y 3 = 0 Solving numerically for y close to 1.8 gives y ≈ 1.8216… . Substituting y = 1.8216… gives x = 16 − y 2 ≈ 12.6816… . 1 F(1.8216 K) = π x (16 + y 2 + 4 y) ≈ 3 353.318… . 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 more than the maximal cylinder, about 75.7% more. 31. a. If f (c) is a local maximum, then f (x) − f (c) ≤ 0 for x in a neighborhood of c. For x to the left of c, x − c < 0. f ( x ) – f (c ) Thus, ≥ 0 (neg./neg.) and x–c f ( x ) – f (c ) ≥ 0. f ′(c) = lim− x→0 x–c For x to the right of c, x − c > 0. f ( x ) – f (c ) Thus, ≤ 0 (neg./pos.) and x–c f ( x ) – f (c ) f ′(c) = lim+ ≤ 0. x→0 x–c Therefore, 0 ≤ f ′(c) ≤ 0. Because f ′(c) exists, f ′(c) = 0 by the squeeze theorem, 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 in part a shows only that f ′(x) changes sign at x = c. There could be a cusp, a removable discontinuity, or a step discontinuity at x = c. c. The converse would say that if f ′(c) = 0, then f (c) is a local maximum. This statement is 184
Problem Set 8-3
x 500
The graph shows a maximum A at x ≈ 500. 500 – x, x < 600 A′ = 800 – 2 x, x > 600 For x < 600, A′ = 0 ⇔ x = 500. For x > 600, A′ = 0 ⇔ x = 400 (out of the domain). A′ is undefined at the cusp, x = 600. Maximum at x = 500 because graph is parabola opening downward. Or: Check the critical points. A(500) = 500(500) − 0.5(500)2 = 125,000 A(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. 500 x – 0.5 x 2, x ≤ 400 ∴ A= 2 x > 400 700 x – x , A 150,000
x 400
The graph shows a maximum A at the cusp, x = 400. x < 400 500 – x, A′ = 700 – 2 x, x > 400 For x < 400, A′ = 0 ⇔ x = 500 (out of the domain). For x > 400, A′ = 0 ⇔ x = 350 (out of the domain). Calculus Solutions Manual © 2005 Key Curriculum Press
Maximum area is at the cusp, x = 400. A = 700(400) − 4002 = 120,000 Maximum 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. 500 x – 0.5 x 2, x ≤ 200 ∴ A= 2 x > 200 600 x – x ,
Q7.
Q8. Sample answer: y
y x 1
x 2
Q9. tan x + C
Q10. B
1. a. y = 4 − x dV = 2πxy ⋅ dx = 2π (4x − x3) dx b. 0 = 4 − x 2 = (2 − x)(2 + x) at x = ±2 2 1 V = 2π ( 4 x − x 3 ) dx = 2π 2 x 2 − x 4 0 4 = 8π = 25.1327… 2
A 150,000
2
∫
x 300
The graph shows a maximum A at x ≈ 300. 500 – x, x < 200 A′ = 600 – 2 x, x > 200 For x < 200, A′ = 0 ⇔ x = 500 (out of the domain). For x > 200, A′ = 0 ⇔ x = 300. A′ is undefined at the cusp, x = 200. Maximum area is at x = 300 because graph is a parabola opening downward. Or: Check critical points. A(300) = 600(300) − 3002 = 90,000 A(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 8-4 Q1.
y x
Q3.
c. y = 4 − x 2 ⇒ x 2 = 4 − y Upper bound of solid is at y = 4. dV = πx2 dy = π (4 − y) dy V=
4
∫ π (4 − y) dy = π (4 y − 0.5y )
x
0
=
0
8π = 25.1327… , which is the same answer as by cylindrical shells in part b. 2. a. Height of cylinder = 8 − x b. y = x 2/3 ⇒ x = y 3/2 dV = 2π (8 − x)y dy = 2π (8 − y3/2)y dy = 2π (8y − y 5/2) dy c. At x = 8, y = 82/3 = 4. 4 2 V = 2π (8 y − y 5/2 ) dy = 2π 4 y 2 − y 7/2 0 7 384 = π = 172.3387… 7 d. dV = π y 2 dx = π x 4/3 dx 8 8 3 384 V = π x 4/3 dx = π x 7/3 = π= 0 7 7 0
∫
4 0
172.3387… , which is the same as the volume by cylindrical shells in part c. 3. The graph shows y = −x2 + 4x + 3, from x = 1 to x = 4, sliced parallel to the y-axis, with sample point (x, y), rotated about the y-axis, showing back half of solid only.
Q4. y
4
2
∫
Q2. y
0
y
(x, y)
y x
x
1
x 1
Q5.
Q6. y
dV = 2πxy ⋅ dx = 2π (−x 3 + 4x 2 + 3x) dx
y x
4
x
V=
4
∫ 2π (− x 1
3
+ 4 x 2 + 3 x ) dx
≈ 268.6061… (exactly 85.5π)
Calculus Solutions Manual © 2005 Key Curriculum Press
Problem Set 8-4
185
Circumscribed hollow cylinder of radii 1 and 4 and height 7 has volume π(42 − 12) ⋅ 7 = 329.8… , which is a reasonable upper bound for the calculated volume.
6
y (0, y )
(x, y )
4. The graph shows y = x2 − 8x + 17, from x = 2 to x = 5, sliced parallel to the y-axis, with sample point (x, y), rotated about the y-axis, showing back half of solid only.
x
–1
y 5 (x, y)
dV = 2π y(0 − x) ⋅ dy = 2π (−y3 + 10y2 − 24y) dy x 2
V=
5
dV = 2πxy ⋅ dx = 2π (x3 − 8x2 + 17x) dx V=
5
∫ 2π ( x 2
3
− 8 x 2 + 17 x ) dx
≈ 117.8097… (exactly 37.5π) Circumscribed hollow cylinder of radii 2 and 5 and height 5 has volume π(52 − 22) ⋅ 5 = 329.8… , which is a reasonable upper bound for the calculated volume. Assuming that the part of the solid above y = 2 could be fit into the “trough,” the volume is approximately π(52 − 22) ⋅ 2 = 131.9… , which is close to the calculated volume. 5. The graph shows x = −y2 + 6y − 5, intersecting y-axis at y = 1 and y = 5, rotated about the x-axis, showing back half of solid only. y 5 (x, y)
(0, y )
1
x
6
∫ 2π (− y 4
3
+ 10 y 2 − 24 y) dy
40 ≈ 41.8879… exactly π 3 Circumscribed hollow cylinder of radii 4 and 6 and height 1 has volume π(62 − 42) ⋅ 1 = 62.83… , which is a reasonable upper bound for the calculated volume. 7. Figure 8-4h shows y = x3, intersecting the line y = 8 at x = 2 and the line x = 1. Rotate about the y-axis. Slice parallel to the y-axis. Pick sample points (x, y) on the graph and (x, 8) on the line y = 8. dV = 2π x(8 − y) ⋅ dx = 2π (8x − x4) dx V=
2
∫ 2π (8x − x
4
) dx
1
≈ 36.4424… (exactly 11.6π) Circumscribed hollow cylinder of radii 2 and 1 and height 7 has volume π(22 − 12) ⋅ 7 = 65.9… , which is a reasonable upper bound for the calculated volume. 8. The graph shows y = 1/x, intersecting line y = 4 at x = 0.25 and the line x = 3, rotated about the y-axis, showing back half of solid only.
4
y (x, 4) y=4
(x, y )
x
dV = 2π y(x − 0) ⋅ dy = 2π (−y + 6y − 5y) dy 3
V=
5
∫ 2π (− y 1
3
+ 6 y 2 − 5 y) dy
≈ 201.0619… (exactly 64π) Circumscribed hollow cylinder of radii 1 and 5 and height 4 has volume π(52 − 12) ⋅ 4 = 301.5… , which is a reasonable upper bound for the calculated volume. 6. The graph shows x = y2 − 10y + 24, intersecting y-axis at y = 4 and 6, rotated about the x-axis, showing back half of solid only. 186
Problem Set 8-4
0.25
2
3
dV = 2πx ⋅ (4 − y) · dx = 2π (4x − 1) dx V=
∫
3
0.25
2π ( 4 x − 1) dx
≈ 95.0331… (exactly 30.25π) Circumscribed hollow cylinder of radii 0.25 and 4 and height 3.7 has volume π(32 − 0.252)(3.7) = 103.8… , which is a reasonable upper bound for the calculated volume. 9. Figure 8-4i shows y = 1/x2, intersecting the line x = 5 at y = 0.04 and the line y = 4. Rotate Calculus Solutions Manual © 2005 Key Curriculum Press
about the x-axis. Slice parallel to the x-axis. Pick sample points (x, y) on the graph and (5, y) on the line x = 5. dV = 2π y(5 − x) ⋅ dy = 2π (5y − y1/2 ) dy
∫
V=
4
0.04
y 2
x 4
y 4 (x, y )
(8, y )
1
x 1
8
dV = 2π y(x2 − x1) dy = 2π (2y2 + 4y − y4) dy V=
dV = 2π y(8 − x) ⋅ dy = 2π (8y − y5/2 ) dy 4
∫ 2π (8y − y
5/2
) dy
1
3 ≈ 149.0012 … exactly 47 π 7 Circumscribed hollow cylinder of radii 1 and 4 and height 7 has volume π(42 − 12) ⋅ 7 = 329.8… , which is a reasonable upper bound for the calculated volume. 11. Figure 8-4j shows y1 = x2 − 6x + 7 and y2 = x + 1, intersecting at (1, 2) and (6, 7). Rotate about the y-axis. Slice parallel to y-axis. Pick sample points (x, y1) and (x, y2). dV = 2πx ⋅ (y2 − y1) ⋅ dx = 2π(−x3 + 7x2 − 6x) dx 6
∫ 2π (− x 1
3
+ 7 x 2 − 6 x ) dx
5 ≈ 458.1489… exactly 145 π 6 Circumscribed hollow cylinder of radii 1 and 6 and height 7 has volume π(62 − 12) ⋅ 7 = 769.6… , which is a reasonable upper bound for the calculated volume. 12. The graph shows y = x11/3 ⇒ x1 = y 3 and y = 0.5x2 − 2 ⇒ x2 = 2y + 4, intersecting at (8, 2) in Quadrant I and bounded by the x-axis, rotated about the x-axis, showing back half of solid only.
Calculus Solutions Manual © 2005 Key Curriculum Press
2
∫ 2π (2 y
2
0
+ 4 y − y 4 ) dy
13 ≈ 43.5634 K exactly 13 π 15 Circumscribed cylinder of radius 2 and height 8 has volume π ⋅ 22 ⋅ 8 = 100.5… , which is a reasonable upper bound for the calculated volume. 13. Figure 8-4k shows y = x3/2 , from x = 1 to x = 4. Rotate about the line x = 5. Slice parallel to the y-axis. Pick sample point (x, y). dV = 2π(5 − x)y ⋅ dx = 2π(5x3/2 − x5/2 ) dx V=
V=
8
2π (5 y − y1/2 ) dy
≈ 217.8254 … (exactly 69.336π ) Circumscribed cylinder of radius 4 and height 4.5 has volume π ⋅ 42 ⋅ 4.5 = 226.1… , which is a reasonable upper bound for the calculated volume. 10. The graph shows y = x2/3 , intersecting the line y = 1 and intersecting the line x = 8 at y = 4, rotated about the x-axis, showing back half of solid only.
V=
(x 2 , y ) (x1, y)
4
∫ 2π (5x
3/2
1
− x 5/2 ) dx
3 ≈ 161.5676 … exactly 51 π 7 Circumscribed cylinder of radius 4 and height 8 has volume π(42) ⋅ 8 = 402.1… , which is a reasonable upper bound for the calculated volume. 14. The graph shows y = x− 2, from x = 1 to x = 2, rotated about the line x = 3, showing back half of solid only. y
(x, y )
1
x 1
2
3
dV = 2π (3 − x ) y ⋅ dx = 2π (3 x −2 − x −1 ) dx V=
2
∫ 2π (3x 1
−2
− x −1 ) dx
≈ 5.0696 … (exactly π (3 − 2 ln 2)) Circumscribed hollow cylinder of radii 1 and 2 and height 1 has volume π(22 − 12) ⋅ 1 = 9.4… , which is a reasonable upper bound for the calculated volume. 15. The graph shows y1 = x4 and y2 = 5x + 6, intersecting at x = −1 and x = 2, rotated about the line x = 4, showing back half of solid only.
Problem Set 8-4
187
18. The graph shows y1 and y2 as described in Problem 17, but rotated about the line y = −1, showing back half of solid only. Slicing perpendicular to the x-axis is appropriate because slicing parallel to it would give strips of length (curve) minus (curve) at some values of y and (curve) minus (other curve) at other values of y.
y 16
(x, y2)
(x, y1 )
x –1
2
4
y (x, y1)
dV = 2π(4 − x)(y2 − y1) dx = 2π (4 − x)(5x + 6 − x4) dx V=
∫
2
2π ( 4 − x )(5 x + 6 − x ) dx 4
−1
≈ 390.1858… (exactly 124.2π ) Circumscribed hollow cylinder of radii 2 and 5 and height 16 has volume π(52 − 22) ⋅ 16 = 1055.5… , which is a reasonable upper bound for the calculated volume. 16. The graph shows y1 = x = x 1/2 and y 2 = 6 − x, intersecting at x = 4 in Quadrant I and bounded by the line x = 1, rotated about the line x = −1, showing back half of solid only. y 5
(x, y2 )
1 –1
x
4
dV = 2π(x + 1)(y2 − y1) dx = 2π (x + 1)(6 − x − x 1/2) dx V=
4
∫ 2π ( x + 1)(6 − x − x
1/2
) dx
1
13 ≈ 109.5368K exactly 34 π 15 Circumscribed hollow cylinder of radii 2 and 5 and height 4 has volume π(52 − 22) ⋅ 4 = 263.8… , which is a reasonable upper bound for the 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. Slice parallel 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=
∫
b
0
2π ( x + 2)( − x 2 + 4 x + 1 − 1.4 x ) dx
≈ 163.8592… Circumscribed hollow cylinder with radii 2 and 5.4 and height 4 has volume π(5.42 − 22)4 = 316.1… , a reasonable upper bound for calculated volume. 188
Problem Set 8-4
1
x
0
3
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, where b = 3.3740… , as in Problem 17. V=
(x, y1 ) 1
(x, y2 )
b
∫ π [(− x
2
0
+ 4 x + 2)2 − (1.4 x + 1)2 ] dx
≈ 181.0655… Circumscribed hollow cylinder of radii 2 and 6 and height 3.4 has volume π(62 − 22) · 3.4 = 341.8… , a reasonable upper bound for the calculated volume. 19. Slice perpendicular to the y-axis. Pick sample points (x, y) on the graph of y = x3 and (1, y) on the line x = 1. y = x 3 ⇒ x = y 1/3; y 1/3 = 1 at y = 1 dV = π ( x2 − 12) dy = π (y2/3 − 1) dy V=
8
∫ π (y 1
2/3
− 1) dy ≈ 36.4424 …
(exactly 11.6π ), which agrees with the answer to Problem 7. 20. See the graph for Problem 8. Slice perpendicular to the y-axis. Pick sample points (x, y) on the graph of y = 1/x and (3, y) on the line x = 3. dV = π (32 − x 2 ) dy = π (9 − y −2 ) dy V=
∫
4
1/ 3
π (9 − y −2 ) dy ≈ 95.0331… (exactly
30.25π ) , which agrees with the answer to Problem 8. 21. The graph shows y = x1/3, from x = 0 to x = 8, rotated about the x-axis, showing back half of solid only.
Calculus Solutions Manual © 2005 Key Curriculum Press
y 2
(x, y )
The integral can be found algebraically using the Pythagorean properties from trigonometry. sin3 t = (1 − cos2 t) sin t = sin t − cos2 t sin t
(8, y )
x 8
∫
V= y = x 1/3 ⇒ x = y 3 dV = 2π y ( 8 − x) ⋅ dy = 2π ( 8y − y4) dy 2 1 V = 2π (8 y − y 4 ) dy = 2π 4 y 2 − y 5 0 5
∫
2 0
2
∫ 2πx sin x dx ≈ 10.9427… 0
numerically (exactly 2π ( sin 2 − 2 cos 2), integrating by parts). b. The integrand, x sin x, is a product of two functions, for which the antiderivative cannot be found using techniques known so far. 23. a. x = 5 cos t, dx = −5 sin t dt y = 3 sin t, dy = 3 cos t dt Slice parallel to the x-axis, then rotate about the x-axis. Pick sample points (−x, y) at the left end of the strip and (x, y) at the right end. dV = 2π y [ x − (−x)] ⋅ dy = 4π xy 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=
∫
π /2
0
∫
0
π
π 3
45π cos 2 t ( − sin t dt ) 0
π
Problem Set 8-5 Q1.
Q2. y
y 16
16
x
x 0
Q3. A =
∫
4
1
4
Q4. A =
x 2 dx
1
4
1 34 x 3 1
Q5. A = 21 Q6. y 16
π /2 0
= −60π ( 0 − 1) = 60π = 188.4955… b. Slice the region in Quadrant I perpendicular to the x-axis, then rotate about the x-axis. Pick sample point (x, y) on the graph. dV = π y 2 dx = π ( 3 sin t)2(−5 sin t dt) = −45π sin3 t dt Limits of integration are from x = −5 to x = 5. At x = −5, t = π . At x = 5, t = 0. V=
∫
0
= 45π − 15π − (−45π ) + (−15π ) = 60π c. 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 lower branches. 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 in part b. V ≈ 2072.6169… (exactly 210π 2, using the half-argument properties for sin2 t, as in Problem 16 of Problem Set 5-9, or by using integration by parts as in Chapter 9). 24. Answers will vary.
180π cos 2 t sin t dt
= −60π cos3 t
π
−45π sin t dt −
= 45π cos t − 15π cos t
= 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 about the y-axis, as in Figure 8-4m. Slice parallel to the y-axis. Pick sample point (x, y) on the graph. dV = 2π xy ⋅ dx = 2π x sin x dx V=
0
x 4
Q7. V =
∫
4
1
2πx 3 dx
Q8. V =
Q9. V = 127.5π 1. a.
π 44 x 2 1
Q10. E
y
−45π sin 3 t dt ≈ 188.4955…
(exactly 60π ) , which agrees with the volume found in part a.
Calculus Solutions Manual © 2005 Key Curriculum Press
1 0
x 2
Problem Set 8-5
189
5
5
b. L ≈
∑
(0.4)2 + [e 0.4 n – e 0.4( n –1) ]2
dL = dx 2 + dy 2 = 1 + e 2 x dx
∫
L=
0
∑
(0.3)2 + [sec 0.3n – sec 0.3 (n – 1)]2
n =1
n =1
= 6.7848… c. dy = e x dx 2
b. L ≈
= 13.7141… c. dy = sec x tan x dx dL = dx 2 + dy 2 = 1 + tan 2 x sec 2 x dx L=
1 + e dx ≈ 6.7886 … numerically 2x
∫
1/5
1 + tan 2 x sec 2 x dx ≈ 13.7304 …
0
numerically
2. a.
5. a. y 9
1
1
x
0
∑
(0.6)2 + [2 0.6 n – 2 0.6( n –1) ]2
n =1
= 7.7853… c. dy = (2x ln 2) dx
6
∫
3
0
b. dy = (2x − 5) dx dL = dx 2 + dy 2 = 1 + (2 x – 5)2 dx L=
dL = dx 2 + dy 2 = 1 + (2 x ln 2)2 dx L=
x
–1
3
5
b. L ≈
y
1 + (2 x ln 2)2 dx ≈ 7.7920 K
numerically 3. a.
∫
6
1
1 + (2 x – 5)2 dx ≈ 15.8617K
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 reasonable lower bound for L. 6. a.
y 4
y
10
x 0
0
5
b. L ≈
∑
(0.3)2 + [tan 0.3n – tan 0.3(n – 1)]2
n =1
= 14.4394… c. dy = sec2 x dx
∫
1.5
0
1 + sec 4 x dx ≈ 14.4488K
numerically 4. a.
4
b. dy = (4 − 2x) dx dL = dx 2 + dy 2 = 1 + ( 4 – 2 x )2 dx L=
dL = dx 2 + dy 2 = 1 + sec 4 x dx L=
x
1.5
∫
4
0
1 + ( 4 – 2 x )2 dx ≈ 9.2935K
c. Chords from (0, 0) to (2, 4) and from (2, 4) to (4, 0) have combined length 2 20 = 8.9442 … , which is a reasonable lower bound for L. 7. a. 16 y
y
10
1 0
190
Problem Set 8-5
x
x 1.5
–1
2
Calculus Solutions Manual © 2005 Key Curriculum Press
b. dy = −4x3 dx
b. dy = (sin x + x cos x) dx
dL = dx 2 + dy 2 = 1 + 16 x 6 dx L=
∫
2
−1
dL = dx 2 + dy 2 = 1 + (sin x + x cos x )2 dx
1 + 16 x 6 dx ≈ 18.2470 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 bound for L. 8. a.
L=
∫
4π
0
1 + (sin x + x cos x )2 dx ≈ 54.1699K
c. Eight chords of ∆x = π/2 extend from middle to high to middle to low points on the graph. Lengths sum to 52.6109… , a reasonable lower bound for L. 11. a.
y
y 50 10
x –1
9
x 0
1.5
b. dy = (3x2 − 18x + 5) dx b. dy = sec2 x dx
dL = dx 2 + dy 2
dL = dx 2 + dy 2 = 1 + sec 4 x dx
= 1 + (3 x 2 – 18 x + 5) 2 dx L=
∫
9
−1
L=
1 + (3 x – 18 x + 5) dx ≈ 219.4873K 2
2
c. Using five chords with ∆x = 2, L ≈ 204.4605… , which is a reasonable lower bound for L. 9. a.
∫
1.5
0
1 + sec 4 x dx ≈ 14.4488K
c. Distance between the endpoints is 14.1809… , which is a reasonable lower bound for L. 12. a. y
y 10
1
0
e
b. dy = 2 ln x ⋅ x −1 dx = 2 x −1 ln x dx dL = dx 2 + dy 2 = 1 + (2 x –1 ln x )2 dx L=
∫
e
0.1
x
x 1
1 + (2 x –1 ln x )2 dx ≈ 7.6043…
c. Chords from x = 0.1 to x = 1 and from x = 1 to x = e have combined length 7.3658… , which is a reasonable lower bound for L. 10. a. y
1.5
b. dy = sec x tan x dx dL = 1 + (sec x tan x )2 dx L=
∫
1.5
0
1 + (sec x tan x )2 dx ≈ 13.7304K
c. The distance between the endpoints is 13.2221… , which is a reasonable lower bound for L. 13. a. 5
y
5
x 0
5
Calculus Solutions Manual © 2005 Key Curriculum Press
4π
x 5
Problem Set 8-5
191
b. dx = −15 cos2 t sin t dt, dy = 15 sin2 t cos t dt dL = dx 2 + dy 2 = (–15 cos 2 t sin t )2 + (15 sin 2 t cos t )2 dt L=
∫
2π
(–15 cos 2 t sin t )2 + (15 sin 2 t cos t ) 2 dt
0
≈ 30 To see why the answer is so simple, transform the radicand and use the fundamental theorem. L=
∫
2π
225(sin t cos t )2 (cos 2 t + sin 2 t ) dt
0
2π
= 7.5
∫
= 7.5
∫
2π
= 7.5
∫
2π
(2 sin t cos t )2 dt
0
= 10 2
∫
2π
= 10 2
∫
2π
= 20 2
∫
∫
π 0
= −40 2 (1 – 1) + 40 2 (1 + 1)1/2 = 80 c. Maximum/minimum values of y are ±7.5 3 . Circle of radius 7.5 3 has circumference 15π 3 = 81.6209K . 15. a. 1/2
π /2
x
sin 2t dt
4
0
= 30 (exactly!)
0
b. dx = (−5 sin t + 5 sin 5t) dt dy = (5 cos t − 5 cos 5t) dt dL = dx 2 + dy 2 = (–5 sin t + 5 sin 5t ) 2 + (5 cos t – 5 cos 5t ) 2 dt
y
L=
10
x 5
b. dx = 5(−2 sin t + 2 sin 2t) dt dy = 5(2 cos t − 2 cos 2t) dt dL = dx + dy = 2
2
[5(–2 sin t + 2 sin 2t )]2 + [5(2 cos t – 2 cos 2t )]2 dt
∫
2π
0
(–5 sin t + 5 sin 5t ) 2 + (5 cos t – 5 cos 5t ) 2 dt
≈ 40 To see why the answer is so simple, transform the radicand and use the fundamental theorem. L=5
∫
2π
[5(–2 sin t + 2 sin 2t )]2 + [5(2 cos t – 2 cos 2t )]2 dt ≈ 80 To see why the answer is so simple, transform the radicand algebraically and use the fundamental theorem. L = 10
∫
2π
2 – 2 sin t sin 2t – 2 cos t cos 2t dt
0
= 10 2
∫
2π
0
Problem Set 8-5
1 – cos t dt (using cos ( A − B))
2 – 2 sin t sin 5t – 2 cos t cos 5t dt
0
=5 2 =5 2
L=
192
(1 + cos t ) −1/2 (sin t dt )
= −40 2 (1 + cos t )1/2
14. a.
0
0
y
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.
∫
π
4
|sin 2t | dt = 7.5 ⋅ 4 π /2
|sin t | dt 1 + cos t
0
0
1 = 30 ( − cos 2t ) 2
2π
0
sin 2 2t dt
0
1 – cos 2 t dt 1 + cos t
=5 2
∫
2π
∫
2π
∫
2π
0
1 – cos 4t dt (using cos ( A − B))
0
0
= 40 2
∫
π /4
0
1 – cos 2 4t dt 1 + cos 4t |sin 4t | dt 1 + cos 4t (1 + cos 4t ) −1/2 (sin 4t dt )
= −20 2 (1 + cos 4t )1/2
π /4 0
= −0 + 20 2 ⋅ 2 = 40 c. Maximum/minimum values of x, y are ±3 3. Circle of radius 3 3 has circumference 32.6483… , which is close.
Calculus Solutions Manual © 2005 Key Curriculum Press
16. a.
c. Distance between endpoints is 1.006944 K = 1.00346 K , which is a reasonable lower bound for L. 19. a.
y 5
x 5
y t = 4π
b. dx = (−sin t + sin t + t cos t) dt = t cos t dt dy = (cos t − cos t + t sin t) dt = t sin t dt dL = dx 2 + dy 2 = (t cos t )2 + (t sin t )2 dt = |t | dt = t dt (because t ≥ 0) L=
∫
4π
0
t dt = 0.5t 2
4π 0
= 8π 2 = 78.9568K
c. Circle of radius 4π = 12.5663… would have circumference = 8π 2. 17. a.
5
x 1
b. dy = 2 x −1/ 3 dx dL = dx 2 + dy 2 = 1 + 4 x –2/3 dx
∫ 1 + 4x = ∫ ( x + 4)
x
b. dy = 6x1/2 dx dL = dx 2 + dy 2 = 1 + 36 x dx 0
1 + 36 x dx =
1 36
1/2
( x −1/3 dx )
∫
= 8 8 − 5 5 = 11.4470 K
4
4
2/3
dx
3 8 2/3 2 ( x + 4)1/2 x −1/3 dx 2 1 3 8 3 2 = ⋅ ( x 2/3 + 4)3/2 2 3 1 =
∫
–2/3
1 8 1
y
L=
8
L=
30
0
8
4
∫ (1 + 36 x )
1/2
(36 dx )
c. Distance between endpoints is 130 = 11.4017… , which is a reasonable lower bound for L. 20. a.
0
4 1 1 = (1 + 36 x )3/2 0 = (1453/ 2 – 1) 54 54 = 32.3153… c. The chord connecting the endpoints has length 32.2490… , which is a reasonable lower bound for L. 18. a.
y
5
x 0
3
b. dy =
y
1 2 ( x + 2)1/ 2 2 x dx = x ( x 2 + 2)1/ 2 dx 2
dL = dx 2 + dy 2 = 1 + x 2 ( x 2 + 2) dx
1
= 1 + x 4 + 2 x 2 dx = (1 + x 2 ) dx x 1
L=
2
−2
b. dy = ( x /4 − x ) dx 2
dL = dx + dy = 1 + ( x /4 – x ) dx 2
2
2
–2 2
= 1 + x 4 /16 – 1/2 + x –4 dx
1 3 (1 + x 2 ) dx = x + x 3 = 12 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-axis through the vertex of the parabola.
= ( x 2 /4 + x –2 )2 dx = | x 2 /4 + x −2 | dx L=
2
∫ ( x /4 + x 2
–2
1
2
750
) dx (because integrand > 0)
= x 3 /12 − x −1 1 = 1 Calculus Solutions Manual © 2005 Key Curriculum Press
y
1 = 1.0833K 12
–2100
220 0
x 2100
Problem Set 8-5
193
General equation is y − 220 = ax2. Substitute (2100, 750) for (x, y). 53 750 − 220 = a ⋅ 2100 2 ⇒ a = 441000 53 Equation of parabola is y = x 2 + 220 . 441000 106 dy = x dx 441000 dL = dx 2 + dy 2 = 1 + (106/441000)2 x 2 dx L=
∫
2100
≈ 4372.0861… numerically ≈ 4372 feet. The answer is reasonable because the 4200 feet between supports is a lower bound for L. 22. y = 0.2(e x + e − x ), dy = 0.2(e x − e − x ) dx dL = dx + dy 2
2
= 1 + 0.04(e x – e – x )2 dx L=
∫
≈ 24.1722 … ≈ 24.2 ft The parabola with vertex (0, 0.4) and endpoints (±4, 0.2(e4 + e− 4)) = (±4, 10.9232…) has equation y = ax2 + 0.4. Substituting (4, 10.9232…) gives 10.9232… = 16a + 0.4 ⇒ a = 0.6577… . y = 0.6577… x2 + 0.4 ⇒ dy = 1.3154… x dx dL = dx 2 + dy 2 = 1 + 1.7303... x 2 dx L=
∫
4
(–100 sin t )2 + (50 cos t )2 dt
24. 5
y
x 8
–8
–5
dx = −16 sin 2t, dy = 5 cos t dL = dx 2 + dy 2 = (–16 sin 2t )2 + (5 cos t )2 dt Curve appears to have length
∫
2π
0
(–16 sin 2t )2 + (5 cos t )2 dt
= 68.7694… Length should be less than the lengths of three circumscribing segments, 16 + 16 + 10 = 42. The discrepancy is explained by the fact that the parabola 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. 2 25. 9 x 2 = 4 y 3 ⇔ x = ± y 3/2 . 3
1 + 1.7303K x 2 dx ≈ 23.2193… ≈ 23.2 ft,
−4
0
≈ 484.4224 … ≈ 484.4 m
L=
1 + 0.04(e x – e – x )2 dx
−4
∫
2π
1 + (106/441000)2 x 2 dx
–2100
4
L=
y
3
which is about a foot shorter than the catenary, as shown by graph: y 10
x 0
3
4
dx = y 1/2 dy dL = dx 2 + dy 2 = ( y + 1)1/2 dy x
L=
4
–4
23. Outer ellipse: x = 120 cos t, dx = −120 sin t dt y = 100 sin t, dy = 100 cos t dt dL = dx 2 + dy 2 = (–120 sin t )2 + (100 cos t )2 dt L=
∫
2π
0
(–120 sin t )2 + (100 cos t )2 dt
0
dy =
2 ( y + 1)3/2 3
= (–100 sin t ) + (50 cos t ) dt 2
0
y 4
1
2
3
2 = 4.6666 … 3 26. x 2 = y 3 ⇔ x = ±y 1.5 2x dx = 3y2 dy ⇒ 4x2 dx2 = 9y4 dy2 9 ⇒ 4 y 3 dx 2 = 9 y 4 dy 2 ⇒ dx 2 = y dy 2 4 Note that dy < 0 between (−1, 1) and (0, 0):
dL = dx 2 + dy 2
Problem Set 8-5
1/2
=4
≈ 692.5791… ≈ 692.6 m Inner ellipse: x = 100 cos t, dx = −100 sin t dt y = 50 sin t, dy = 50 cos t dt
194
3
∫ ( y + 1)
–1
x 8
Calculus Solutions Manual © 2005 Key Curriculum Press
For x in [−1, 0], x = −y 1.5 , dx = −1.5y0.5 dy, dL = − dx + dy = − 2.25 y + 1 dy. For x in [0, 8], x = y1.5 , dx = 1.5y0.5 dy, 2
2
dL = dx 2 + dy 2 = 2.25 y + 1 dy. L=
∫
0
1
− 2.25 y + 1 dy +
∫
4
2.25 y + 1 dy
0
0 4 8 8 (2.25 y + 1)3/2 + (2.25 y + 1)3/2 27 1 27 0 8 3/2 3/2 = (–1 + 3.25 + 10 – 1) = 10.5131… 27 t 1 27. x = cos t, dx = (cos t – t sin t ) dt π π t 1 y = sin t, dy = (sin t + t cos t ) dt π π
=−
dL = dx 2 + dy 2 1 = (cos t – t sin t )2 + (sin t + t cos t )2 dt π 1 = 1 + t 2 dt π The curve crosses the x-axis exactly when sin t = 0, when t is a multiple of π. There are seven crossings after the beginning, so t should run between 0 and 7π. To check this, note that the curve ends at (−7, 0), so solve (t/π) cos t = −7 with t = nπ ⇒ (nπ /π ) cos nπ = −7 ⇒ n cos n π = −7 ⇒ n = 7 ⇒ 0 ≤ t ≤ 7π . 1 7π L= 1 + t 2 dt ≈ 77.6508… π 0 The integral can be evaluated algebraically by trigonometric substitution as in Section 9-6, giving 1 1 + t 2 dt = t t 2 + 1 + ln t + 1 + t 2 + C. 2
∫
(
∫
)
28. x = r cos t, dx = −r sin t dt y = r sin t, dy = r cos t dt
Doubling A doubles the amplitude of the sinusoid. However, it less than doubles the length of the sinusoid for much the same reason that doubling one leg of a right triangle does not double the hypotenuse. In the limit as A approaches infinity, doubling A approaches doubling the length. 30. x = cos t, dx = −sin t dt y = A sin t, dy = A cos t dt dL = dx 2 + dy 2 = sin 2 t + A 2 cos 2 t dt The entire ellipse is generated as t increases from 0 to 2π . L=
∫
2
Circumference =
2
∫
2π
A
L
0 1 2 3
4 (a double line segment) 6.283185… (= 2π) 9.688448… 13.364893…
Doubling A doubles one axis of the ellipse without changing the other axis. That is why the length does not double when A doubles. The reasoning is similar to that in the solution to Problem 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 partition of the interval skips over the discontinuity, as shown in the graph. y 25 Mae's error
0
2
r dt = rt
2
2π
2
= 2πr, Q .E.D .
0
29. y = A sin x, dy = A cos x dx dL = dx 2 + dy 2 = 1 + A 2 cos 2 x dx Pick a convenient interval for x such as [0, 2π]. L=
∫
2π
0
x 1
0 1 2 3
6.283185… (= 2π) 7.640395… 10.540734… 13.974417…
Calculus Solutions Manual © 2005 Key Curriculum Press
3
y Amos's sample points 1
x 0
L
2
32. The sample points are all of the form (n/2, sin nπ), which all lie on the x-axis and therefore fail to measure the wiggly bits.
1 + A 2 cos 2 x dx
A
sin 2 t + A 2 cos 2 t dt
0
dL = dx + dy = r sin t + r cos t dt = r dt (for r ≥ 0) The range 0 ≤ t ≤ 2π generates the entire circle. 2
2π
10
The length of the curve is 40 times the length of the 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 Problem Set 8-5
195
multiply his answer by 40 to find the total length. 33. See the Programs for Graphing Calculators section of the Instructor’s Resource Book.
Q3. Q5. Q6. Q7.
Q8. Q9. Q10. 1.
dL = dx 2 + dy 2 = 1 + cos 2 x dx dS = 2πy ⋅ dL = 2π sin x 1 + cos 2 x dx S=
Problem Set 8-6 Q1.
b. dy = cos x dx
1 + 9 x 4 dx
Q2.
1 + sec 4 x dx
1 6 sin x + C Q4. 156 6 xex + ex Maximum y = 7 (at x = 1) f ( x + ∆x ) – f ( x ) f ′( x ) = lim ∆x →0 ∆x f ( x ) – f (c ) or f ′(c) = lim x →c x–c Instantaneous rate of change 1 ln | sec 2 x + tan 2 x | + C 2 D a. The graph shows y = 0.5x2, from x = 0 to x = 3, rotated about the y-axis.
(x, y ) 1
x
0
0
2π sin x 1 + cos 2 x dx ≈ 14.4235…
c. The circumscribed cylinder of length π and radius 1 has lateral area = 2π 2 = 19.7392… , which is a reasonable upper bound for S. 3. The graph shows y = ln x, from x = 1 to x = 3, rotated about the x-axis. y (x, y )
1
x 1
3
dy = x −1 dx dL = dx 2 + dy 2 = 1 + x –2 dx dS = 2π y ⋅ dL = 2π ln x 1 + x –2 dx S=
y
∫
π
3
∫ 2π ln x 1
1 + x –2 dx ≈ 9.0242 …
4. The graph shows y = ln x, from x = 1 to x = 3, rotated about the y-axis, showing back half of surface only.
3
y
dy = x dx dL = dx 2 + dy 2 = 1 + x 2 dx
1 (x, y )
dS = 2πx ⋅ dL = 2πx 1 + x 2 dx S=
3
∫ 2πx 0
x 3
1 + x 2 dx ≈ 64.1361…
b. The inscribed cone of height 4.5 and radius 3 has lateral surface area = πrL = π · 3 · 32 + 4.52 = 50.9722 … , which is a reasonable lower bound for S. c. S =
1
3
∫ π (1 + x
2 1/2
) (2 x dx )
0
3 2 2 π (1 + x 2 )3/2 = π (10 10 − 1) 3 3 0 = 64.1361… , agreeing with the answer found numerically. 2. a. The graph shows y = sin x, from x = 0 to x = π, rotated about the x-axis.
=
dL = 1 + x −2 dx, from Problem 3. dS = 2π x ⋅ dL = 2π x 1 + x –2 dx S=
3
∫ 2π x 1
1 + x −2 dx ≈ 28.3047K
5. The graph shows y = 1/ x = x −1,from x = 0.5 to x = 2, rotated about the y-axis. 2
y (x, y )
x 2
y 1
(x, y )
x 0
196
Problem Set 8-6
π
dy = − x −2 dx dL = dx 2 + dy 2 = 1 + x –4 dx Calculus Solutions Manual © 2005 Key Curriculum Press
dS = 2πx ⋅ dL = 2πx 1 + x –4 dx S=
∫
2
0.5
2πx 1 + x –4 dx ≈ 15.5181K
6. The graph shows y = 1/x = x , from x = 0.5 to x = 2, rotated about the x-axis. −1
y
3
∫ 2πx
S=
1 + (–3 x 2 + 10 x – 8)2 dx
0
≈ 58.7946… 9. The graph shows y = x = x 1/2, from x = 0 to x = 1, rotated about the x-axis. y 1 (x, y )
2 (x, y )
x
x
2
0
dL = 1 + x –4 dx, from Problem 5. dS = 2πy ⋅ dL = 2πx S=
∫
2
0.5
2πx
−1
1+ x
–1 –4
1+ x
–4
dx
dx ≈ 15.5181K
(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. y
1
dy = 0.5 x −1/ 2 dx dL = dx 2 + dy 2 = 1 + 0.25 x –1 dx dS = 2πy ⋅ dL = 2πx 1/ 2 1 + 0.25 x –1 dx = 2π x + 0.25 dx = 2π ( x + 0.25)1/ 2 dx S=
∫
1
0
2π ( x + 0.25)1/2 dx =
1 0
4π = (1.253/ 2 − 0.125) = 5.3304 K 3 10. The graph shows y = x3, from x = 1 to x = 2, rotated about the x-axis, showing back half of surface only.
8
(x, y )
x 0
8
y
2
dy = 3x2 dx dL = dx 2 + dy 2 = 1 + 9 x 4 dx dS = 2πx ⋅ dL = 2πx 1 + 9 x 4 dx S=
4π ( x + 0.25)3/2 3
2
∫ 2π x 0
(x, y )
1
x 1
1 + 9 x 4 dx ≈ 77.3245K
2
8. The graph shows y = −x3 + 5x2 − 8x + 6, from x = 0 to x = 3, rotated about the y-axis. y 6
dy = 3x2 dx (x, y )
x 3
dy = (−3x2 + 10x − 8) dx dL = dx 2 + dy 2 = 1 + (–3 x 2 + 10 x – 8)2 dx dS = 2π x ⋅ dL = 2π x 1 + (–3 x 2 + 10 x – 8)2 dx Graph intersects x-axis where y = 0. −x 3 + 5x 2 − 8x + 6 = −(x − 3)(x 2 − 2x + 2) = 0 at x = 3. Calculus Solutions Manual © 2005 Key Curriculum Press
dL = dx 2 + dy 2 = 1 + 9 x 4 dx dS = 2π y ⋅ dL = 2π x3 (1 + 9x4)1/2 dx S=
∫
2
1
2πx 3 (1 + 9 x 4 )1/2 dx
π 2 (1 + 9 x 4 )1/ 2 (36 x dx ) 18 1 2 π 2 = ⋅ (1 + 9 x 4 )3/ 2 18 3 1 π = (1453/ 2 – 10 3/ 2 ) = 199.4804 K 27 11. The graph shows y = x 4 / 8 + x −2 / 4, from x = 1 to x = 2, rotated about the x-axis, showing back side of surface only. =
∫
Problem Set 8-6
197
2
dL = (1 + x2) dx, from Problem 20 in Section 8-5 dS = 2π x ⋅ dL = 2π ( x + x3) dx
y
x 1
S=
2
1 2π ( x + x 3 ) dx = π x 2 + x 4 0 2
∫
3
= 1 + 0.25 x 6 – 0.5 + 0.25 x –6 dx
y
−3
= 0.25( x + x ) dx = 0.5( x + x ) dx dS = 2π y ⋅ dL = 2π ( x 4/8 + x − 2/4)[0.5(x3 + x− 3)] dx π = ( x 7 + 3 x + 2 x –5 ) dx 8 π 2 7 S= ( x + 3 x + 2 x –5 ) dx 8 1 2 π 1 3 1 = x 8 + x 2 – x –4 1 8 8 2 2 –3 2
3
4
x 1
y
dL = dx 2 + dy 2 = 1 +
S = 2π = 2π
(x, y )
=
2
dy = 2 x dx dS = 2π x ⋅ dL = 2π x 1 + 4 x dx 2
∫ x 1 + 4x π = ∫ (1 + 4 x ) 4
S = 2π
0 2
2
2 1/ 2
0
π (1 + 4 x 2 )3/2 6
∫x 1
∫
3 π 2
8
1
∫
4 –4 / 3 x dx 9
4 x 1/ 3 x 4 / 3 + 9
8
1
1+
4 –4 / 3 x dx 9
x 4/3 + 4 9
1/2
⋅
1/ 2
dx 4 1/3 x dx 3
3/2 8
1
π (1483/2 − 133/2 ) = 204.0435K 27 1 1 15. The graph shows y = x 3 + x −1 , from x = 1 to 3 4 x = 3, rotated about the line y = −1, showing back half of surface only. =
dx (8 x dx ) =
8
4 = π x 4/3 + 9
dL = dx 2 + dy 2 = 1 + 4 x 2 dx
4 –4 / 3 x dx 9
dS = 2π x ⋅ dL = 2π x 1 +
4
x
8
2 −2/3 x dx 3
dy =
π 1 1 3 1 155 = 32 + 6 – =4 – – + π 8 32 8 2 2 256 = 14.4685… 12. The graph shows y = x2, from x = 0 to x = 2, rotated about the y-axis.
2
(x, y )
2
∫
0
0
= 49.5π = 155.5088… 14. The graph shows y = 2x1/3, from x = 1 to x = 8, rotated about the y-axis, showing back half of surface only.
dy = (x3/2 − x− 3/2) dx = 0.5(x3 − x− 3) dx dL = dx 2 + dy 2 = 1 + 0.25( x 3 – x –3 )2 dx 3
3
2 0
π = (173/ 2 − 1) = 36.1769K 6 1 13. The graph shows y = ( x 2 + 2)3/ 2 , from x = 0 to 3 x = 3, rotated about the y-axis.
9
y
(x, y)
y 12
x 1
3
(x, –1) (x, y )
x
1 0
198
3
Problem Set 8-6
Calculus Solutions Manual © 2005 Key Curriculum Press
1 dx = x 2 − x −2 dx 4
b. i. S0,1 = 2
1 dL = dx 2 + dy 2 = 1 + x 2 − x −2 dx 4 1 1 = 1 + x 4 − + x −4 dx 2 16
ii. S1,2 = iii. S2,3 =
2
1 1 = x 2 + x −2 dx = x 2 + x −2 dx 4 4 dS = 2π ( y + 1) ⋅ dL 1 1 1 = 2π x 3 + x −1 + 1 x 2 + x −2 dx 3 4 4 1 1 1 1 = 2π x 5 + x 2 + x + x −2 + x −3 dx 3 3 4 16 S = 2π
3
1
∫ 3x
5
1
+ x2 +
iv. S3,4 = v. S4,5 =
1 1 1 x + x −2 + x −3 dx 3 4 16
1 1 1 1 1 −2 x = 2π x 6 + x 3 + x 2 − x −1 − 18 3 6 4 32
3 1
5 = 101 π = 318.1735K 18 1 1 16. The graph shows y = x 3 + x −1 , from x = 1 to 3 4 x = 3, rotated about line x = 4. 9
y
1
0
0
∫
2
1
10π dx = 10πx
= 10π
∫
∫
4
2
3
10π dx = 10πx
= 10π
2
10π dx = 10πx
5
4
3
4
= 10π
3
10π dx = 10πx
5
= 10π
4
c. The two features exactly balance each other. The area of a zone of a sphere is a function of the height of the zone only, and is independent of where the zone is located on the sphere. 18. Suppose that the sphere is centered at the origin, as in Problem 17. The equation of a great circle in the xy-plane is x2 + y2 = r2, from which y = r 2 − x 2 = (r 2 − x 2 )1/2 . dy = −x(r2 − x2)− 1/2dx dL = dx 2 + dy 2 = 1 + x 2 (r 2 − x 2 ) −1 dx dS = 2πy ⋅ dL
= 2π r 2 − x 2 + x 2 dx = 2πr dx (if r > 0)
(4, y )
∫
S=
x
r
−r
2πr dx = 2πrx
3 4
1 dL = x 2 + x −2 dx, from Problem 15 4 1 dS = 2π ( 4 − x ) ⋅ dL = 2π ( 4 − x ) x 2 + x −2 dx 4 1 −1 2 3 −2 = 2π 4 x − x + x − x dx 4 3 1 S = 2π 4 x 2 − x 3 + x −2 − x −1 dx 1 4
∫
4 1 1 = 2π x 3 − x 4 − x −1 − ln | x| 3 4 4
3 1
1 1 = 2π 15 − ln 3 = 94.6164 K 3 4 17. a. x 2 + y 2 = 25 ⇒ y = 25 − x 2 2 −1/ 2
dy = − x (25 − x )
dx
2
2
= 2π 25 − x + x dx = 10π dx 2
Calculus Solutions Manual © 2005 Key Curriculum Press
2
r −r
= 4πr 2 , Q .E.D .
19. Pick a sample point in the spherical shell at radius r from the center. Surface area at the sample point is 4π r2. Volume of shell is approximately (surface area)(thickness). dV = 4πr 2 ⋅ dr V=
∫
R
0
4 4π r 2 dr = π r 3 3
R
=
0
4 π R 3 , Q .E .D . 3
4 dV 20. V = πr 3 ⇒ = 4πr 2 = S, Q.E.D. 3 dr dV or: V = S dr ⇒ = S by the definition of dr indefinite integral. 21. y = ax2, dy = 2ax dx dL = dx 2 + dy 2 = (1 + 4 a 2 x 2 )1/ 2 dx
∫
dS = 2πx ⋅ dL = 2πx (1 + 4 a 2 x 2 )1/ 2 dx 2 −1
dL = dx + dy = 1 + x (25 − x ) dx dS = 2π y ⋅ dL = 2π 25 − x 2 1 + x 2 (25 − x 2 ) −1 dx 2
2 1
3
∫
= 10π
= 2π (r 2 − x 2 )1/ 2 1 + x 2 (r 2 − x 2 ) −1 dx
(x, y)
1
1
∫ 10π dx = 10πx
r
∫ x(1 + 4a x ) π = (1 + 4 a x ) 4a ∫
S = 2π
2 2 1/ 2
r
2
=
dx
0
2 2 1/ 2
(8a 2 x dx )
0
r π π 2 2 3/2 ( + ) = 2 [(1 + 4 a 2 r 2 )3/ 2 − 1] 1 4 a x 2 6a 6a 0
Problem Set 8-6
199
22. Let h be the height of the paraboloid from the vertex to the center of the base. Because h is the value of y when x = r, h = ar2. Substituting into the formula for S from Problem 21 gives π S = 2 [(1 + 4 ah)3/ 2 − 1] 6a Let a = 1 and evaluate S for various h. Find the zone areas by subtracting. Use the TABLE feature. h
S
Zone
π (0) 6 π (10.1803K) 6 π (26) 6 π ( 45.8721K) 6 π (69.0927K) 6 π (95.2340 K) 6 π (124) 6
0 1 2 3 4 5 6
N.A.
π (10.1803K) 6 π (15.8196 K) 6 π (19.8721K) 6 π (23.2206 K) 6 π (26.1412 K) 6 π (28.7659K) 6
The property is not true for paraboloids. The areas of zones of equal height are greater if the zone is farther away from the vertex. 23. x = 5 cos t, dx = −5 sin t dt y = 3 sin t, dy = 3 cos t dt 3 y
(x, y )
x
dL = dx 2 + dy 2 = ( −5 sin t )2 + (3 cos t )2 dt dS = 2πy ⋅ dL = 2π (3 sin t ) ( −5 sin t )2 + (3 cos t )2 dt
∫
π
0
6π sin t ( −5 sin t ) 2 + (3 cos t )2 dt
≈ 165.7930 … From ( x/5)2 + ( y/3)2 = 1, y = ±0.6 25 − x 2 . Using the upper branch of the graph, dy = −0.6 x (25 − x 2 ) −1/ 2 dx. dL = dx 2 + dy 2 = 1 + 0.36 x 2 (25 − x 2 ) −1 dx At x = ±5, dL involves division by zero, which is awkward, and makes the Cartesian equation
200
Problem Set 8-6
dS = 0.24π 252 – 16 x 2 dx, which is defined at x = ±5. 24. a. x = 35 sec t, dx = 35 sec t tan t dt y = 100 + 80 tan t, dy = 80 sec2 t dt y = 0 ⇔ 100 + 80 tan t = 0 ⇒ tan t = −5/4 t = tan −1 ( −5 / 4) Radius at base is x = 35 sec [tan −1 ( −5/4)] = 56.0273… ≈ 56.0 ft. b. At top, t = 0.5. Radius: x = 35 sec 0.5 = 39.8822… ≈ 39.9 ft Height: 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. dx Minimize x: = 35 sec t tan t = 0 at t = 0 dt (or, because cos t has a max at t = 0, sec t = 1/cos t has a minimum there). Minimum radius = 35 ft Height = y = 100 + tan 0 = 100 ft d. dL = dx 2 + dy 2 = 352 sec 2 t ⋅ tan 2 t + 80 2 sec 4 t dt dS = 2π x ⋅ dL = 2π (35 sec t ) 352 sec 2 t ⋅ tan 2 t + 80 2 sec 4 t dt S=
∫
0.5
tan −1 ( −5/4 )
dS ≈ 37, 756.5934 … ≈ 37, 757 ft 2
4 = 12, 585.5311… ft 3 ≈ 12 466.1307… yd3 25. From Figure 8-6m, a circle of radius L has area πL2 and circumference 2πL. The circumference of the cone’s base is 2πR. The arc length of the sector of the circle of radius L must be equal to this, so the sector is (2πR)/(2πL) = R/L of the circle and has surface area S = πL2(R/L) = πRL, Q .E .D . 26. S = π RL − π rl The objective is to get the lateral area in terms of the slant height of the frustum, L − l. r S = πR L − ⋅ l R l r l = πR L − ⋅ l , because = . L R L πR 2 2 = (L – l ) L e. Volume ≈ S ⋅
5
S=
inappropriate for finding the arc length of an ellipse. For the surface area, however, the offending denominator cancels out, giving
Calculus Solutions Manual © 2005 Key Curriculum Press
πR ( L + l )( L – l ) L l = π R + R ⋅ ( L − l) L
c. dr = 3 cos θ dθ
=
dL = dr 2 + (r dθ )2 = (3 cos θ )2 + ( 4 + 3 sin θ )2 dθ
R + r = 2π ( L − l ), Q .E.D . 2
Problem Set 8-7
Q5. − e
Q2. 12(4x − 9)2 Q4. 3 sec 3x tan 3x
Q6. −1/x 1 2 Q7. ln |x| + C Q8. x +C 2 Q9. 3x + C Q10. x + C 1. a. r = 10 sin θ ⇒ dA = 50 sin2 θ dθ
∫
2π
0
50 sin 2 θ dθ ≈ 157.0796 …
(exactly 50π ) b. The area of the circle is π ⋅ 5 = 25π. The calculated area is twice this because the circle is traced out twice as θ increases from 0 to 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 2
dL = dr + (r dθ ) 2
2
= 100 cos 2 θ + 100 sin 2 θ dθ = 10 dθ L=
∫
2π
0
10 dθ = 10θ
2π
= 20π
0
The circumference is 2π ⋅ 5 = 10π. The calculated length is twice this value because the circle is traced out twice as θ increases from 0 to 2π. The calculus of this section always gives the dynamic answer as the distance traveled by a point on the curve as θ increases from one value to another. This path length does not necessarily equal the length of the curve. 3. a. r = 4 + 3 sin θ. The calculator graph confirms that the text figure is traced out once as θ increases from 0 to 2π. 1 b. dA = ( 4 + 3 sin θ )2 dθ 2 A=
∫
2π
0
dA ≈ 64.4026 … (exactly 20.5π)
Calculus Solutions Manual © 2005 Key Curriculum Press
dL ≈ 28.8141K
∫
2π
0
dA ≈ 92.6769… (exactly 29.5π )
c. dr = 3 sin θ dθ dL = dr 2 + (r dθ )2 = (3 sin θ )2 + (5 – 3 cos θ )2 dθ
2
A=
0
A=
−x
2π
4. a. r = 5 − 3 cos θ. The calculator graph confirms that the text figure is traced out once as θ increases from 0 to 2π. 1 b. dA = (5 – 3 cos θ )2 dθ 2
= π ( R + r)(L − l)
Q1. 15x 2 − 14x + 4 Q3. 3 sin2 x cos x
∫
L=
r = π R + R ⋅ ( L − l) R
∫
L=
2π
0
dL ≈ 34.3136 …
5. a. r = 7 + 3 cos 2θ. The calculator graph confirms that the text figure is traced out once as θ increases from 0 to 2π. 1 b. dA = (7 + 3 cos 2θ )2 dθ 2 A=
∫
2π
0
dA ≈ 168.0752 … (exactly 53.5π)
c. dr = −6 sin 2θ dθ dL = dr 2 + (r dθ )2 = (–6 sin 2θ )2 + (7 + 3 cos 2θ )2 dθ
∫
L=
2π
0
dL ≈ 51.4511…
6. a. r = 8 cos 2θ. The calculator graph confirms that the text figure is traced out once as θ increases from 0 to 2π. 1 b. dA = (8 cos 2θ )2 dθ 2 A=
∫
2π
0
dA ≈ 100.5309K (exactly 32π )
c. dr = −16 sin θ dθ dL = dr 2 + (r dθ )2 = (–16 sin 2θ )2 + (8 cos 2θ )2 dθ L=
∫
2π
0
dL ≈ 77.5075K
7. a. 5 = 5 + 5 cos θ. The calculator graph confirms that the text figure is traced out once as θ increases from 0 to 2π. 1 b. dA = (5 + 5 cos θ )2 dθ 2 A=
∫
2π
0
dA ≈ 117.8097K (exactly 37.5π )
Problem Set 8-7
201
c. dr = −5 sin θ dθ dL = dr 2 + (r dθ )2
A=
= (–5 sin θ ) + (5 + 5 cos θ ) dθ 2
2π
∫
L=
0
2
10 . The calculator graph 3 – 2 cos θ confirms that the text figure is traced out once as θ increases from 0 to 2π.
8. a. r =
A=
∫
2π
0
dA ≈ 84.2977K (exactly 12 5 π )
dL = dr 2 + (r dθ )2 2
∫
L=
2
–20 sin θ 10 (3 – 2 cos θ )2 + 3 – 2 cos θ dθ
= 2π
0
1
−1
dA ≈ 4.5557K
dL = dr 2 + (r dθ )2 = 4 (sec θ tan θ + sin θ )2 + (sec θ – cos θ )2 dθ L=
2
–20 sin θ dθ (3 – 2 cos θ )2
c. dr =
∫
(exactly 16 tan 1 − 24 + 4 sin 2) c. dr = 4(sec θ tan θ + sin θ) dθ
dL = 40 (exactly)
1 10 b. dA = 2 3 – 2 cos θ
1 ( 4 sec θ – 4 cos θ )2 dθ 2
b. dA =
dL ≈ 33.0744 K
∫
1
−1
dL ≈ 10.9534 K
11. r = 49 cos 2θ r = 0 ⇔ 2θ = cos −1 0 = ± π /2 + 2π n (n an integer) θ = ± π /4 + π n The right-hand loop corresponds to nonnegative values of the integrand, − π /4 ≤ θ ≤ π /4. 1 dA = ( 49 cos 2θ ) dθ 2 A=
∫
π /4
1 ( 49 cos 2θ ) dθ = 12.25 sin 2θ − π /4 2
π /4 −π /4
= 24.5 Area of both loops is 49. 12. The graph of r = csc θ + 4 shows a closed loop from θ ≈ 3.4 to θ ≈ 6.
9. a. 1
1 5
r = sin 3θ makes one complete cycle as θ increases from 0 to π. 1 b. dA = (sin 3θ )2 dθ 2 A=
∫
π
0
dA ≈ 0.7853K (exactly 0.25π )
c. dr = 3 cos 3θ dθ dL = dr 2 + (r dθ )2 = (3 cos 3θ )2 + (sin 3θ )2 dθ L=
∫
π
0
dL ≈ 6.6824 K
10. a. 4
4
202
Problem Set 8-7
5
The graph passes through the pole where r = 0. csc θ + 4 = 0 ⇔ θ = csc −1 ( −4) = sin −1 ( −0.25) = −0.2526 … + 2π n or [π − (−0.2526…)] + 2π n Desired range is 3.3942… ≤ θ ≤ 6.0305… . 1 dA = (csc θ + 4)2 dθ 2 A=
∫
6.0305K
3.3942K
dA ≈ 8.4553…
13. r1 = 4 + 4 cos θ and r2 = 10 cos θ intersect where 4 + 4 cos θ = 10 cos θ θ = cos −1 (2/3) = ±0.8410 … + 2π n. (The graphs also touch at the pole, but not for the 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 © 2005 Key Curriculum Press
the circle is generated as θ goes from −0.841… to 0.841… . 1 dA = (r22 – r12) dθ 2 1 = [(10 cos θ )2 − ( 4 + 4 cos θ )2 ] dθ 2 A=
∫
0.841K
−0.841K
dA ≈ 18.8863K −1
(exactly 26 cos (2/3) − (4/3) 5) 14. r1 = 5 and r2 = 5 − 5 cos θ intersect at θ = −π /2 and π /2. 5
4
10
r = 4 + 6 cos θ = 0 ⇔ cos θ = −2/3 θ = cos −1 ( −2/3) = ±2.3005… + 2πn 1 dA = ( 4 + 6 cos θ )2 dθ 2 The outer loop is swept out as θ increases from −2.3005… to 2.3005. A1 =
A2 = 1 dA = (r12 – r22 ) dθ 2 1 = [52 − (5 − 5 cos θ )2 ] dθ 2 Integrate from −π /2 to π /2, because in Quadrants II and III the cardioid lies outside the circle. A=
∫
–2.3005K
dA ≈ 105.0506 K
∫
3.926K
2.3005K
dA ≈ 1.7635K
Area of the region between the loops is A 1 − A 2 ≈ 103.2871… . 17. a. 4
2
dA ≈ 30.3650 K (exactly 50 − 6.25π )
− π /2
15. a. r = 0.5θ. The graph starts at θ = 0 and makes three revolutions, so θ increases from 0 to 6π. dr = 0.5 dθ dL = dr 2 + (r dθ )2 = 0.52 + (0.5θ )2 dθ L=
2.3005K
The inner loop is generated as θ increases from 2.3005… to 3.9826… .
5
π /2
∫
∫
6π
0
dL ≈ 89.8589…
1 1 (0.5θ )2 dθ = θ 2 dθ 2 8 Area swept out for third revolution in Quadrant I is
dr = −2.5θ −1.5 dθ dL = dr 2 + (r dθ )2 = 6.25θ –3 + 25θ –1 dθ
∫
6π
π /2
dL ≈ 31.0872 …
b. The graph shows sectors of central angles 1, 2, and 3 radians.
b. dA =
4.5π 1 2 1 217 3 θ dθ = θ 3 π = 4π 8 24 192 4π Area swept out for second revolution in Quadrant I is
A3 =
A2 =
∫
4.5π
∫
2.5π
2π
1 2 1 θ dθ = θ 3 8 24
2.5π 2π
61 3 π = 192
Area of region between second and third revolution in Quadrant I is A3 = A2 = 13 3 π = 25.1925… . 16 16. The graph of r = 4 + 6 cos θ shows a closed loop from θ ≈ 2.3 to θ ≈ 4.0.
Calculus Solutions Manual © 2005 Key Curriculum Press
Area of sector is A(θ ) =
1 2 r θ. 2
1 2 (5) (1) = 12.5 2 1 A(2) = (3.5355...)2 (2) = 12.5 2 1 A(3) = (2.8867K)2 (3) = 12.5 2 1 In general, A(θ ) = (5θ –1/ 2 )2 (θ ) = 12.5, which 2 is independent of the value of θ. A(1) =
Problem Set 8-7
203
18. The graph shows r = sec θ and a segment from θ = 0 to 1.5.
10
1
The point with polar coordinates (r, θ) has xycoordinates x = r cos θ, y = r sin θ. The graph given by r = sec θ can be written x = r cos θ = sec θ cos θ = 1 y = r sin θ = sec θ sin θ = tan θ (i.e., −∞ < y < ∞). Thus, this graph is the line x = 1. By calculus, the segment from θ = 0 to θ = 1.5 has length as follows: dr = sec θ tan θ dθ dL = dr 2 + (r dθ )2 = (sec θ tan θ )2 + sec 2 θ dθ = sec θ tan 2 θ + 1 dθ = sec 2 θ dθ L=
∫
1.5
0
sec 2 θ dθ = tan θ
1.5 0
= 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 radius 6.6 cm and outer radius 14.6 cm, and takes about 24 minutes to play. There are thus (33.333…)(24) or about 800 grooves in a space of (14.6 − 6.6) or 8.0 cm. Thus, the grooves decrease in radius by about 8.0/800 = 0.01 cm per revolution. A simple equation of the spiral is 0.01 0.005 r= θ= θ 2π π which assumes that the grooves start at the center and have a pitch of 0.01 cm. The innermost actual 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) dθ dL = dr 2 + (r dθ )2 = (0.005/π ) 2 + [(0.005/π )θ ]2 dθ 0.005 = 1 + θ 2 dθ π L=
∫
2920 π
1320 π
dL ≈ 53, 281.4120 … cm
= 16,960.0002…π cm Rough check: Average radius = 10.6 cm L should equal approximately the sum of 800
204
Problem Set 8-7
circles of radius 10.6 cm. L ≈ 800(2π · 10.6) = 16,960π cm, which is very close to the calculated 16,960.0002…π cm. (The integral can be evaluated algebraically by the tangent trigonometry substitution from Chapter 9. The result, 16,960.00021…π, is remarkably close both to the numerical answer and to the sum of the lengths of the 800 circles of average radius 10.6 cm.) 100 = 100(3 − 2 cos θ ) −1 3 – 2 cos θ 1 dA = [100(3 – 2 cos θ ) –1 ]2 dθ 2 = 5000(3 − 2 cos θ ) −2 dθ
20. a. r =
A=
∫
0.2
0
b. Solving
dA ≈ 974.3071… ≈ 974 (kilo-mi)2
∫
θ
0.8
5000(3 − 2 cos t ) −2 dt =
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 is 120 thousand miles, so a = 60. P = k · 601.5 = 81.9 hours (precise answer) e. The total area of the ellipse is A=
∫
2π
0
5000(3 − 2 cos θ ) −2 dθ
= 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 the period. 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 to 1.88976… . f. dr = −100(3 − 2 cos θ ) −2 ⋅ (2 sin θ ) dθ = −200 sin θ (3 − 2 cos θ ) −2 dθ dL = dr 2 + (r dθ )2 = [( −200 sin θ (3 − 2 cos θ ) −2 ) 2 + (100(3 − 2 cos θ ) −1 )2 ]1/2 dθ From θ = 0 to θ = 0.2, L=
∫
0.2
0
dL ≈ 20.2228K ≈ 20.2 kilo-mi.
From θ = 0.8 to θ = 1.88976… , L=
∫
1.88K
0.8
dL ≈ 56.7896 K ≈ 56.8 kilo-mi.
g. Average speed from θ = 0 to θ = 0.2 is 20.2228K = 2.1363K , or about 2136 mi/h. 9.4659K Calculus Solutions Manual © 2005 Key Curriculum Press
Average speed from θ = 0.8 to 56.7896 K θ = 1.88976… is = 5.9993K , 9.4659K or about 5999 mi/h. h. When the spaceship is farthest from Earth, its radial velocity (toward the Earth) is zero. As it proceeds in its orbit, it can be thought of as falling toward the Earth, thus picking up speed. The reverse is true on the other side of the Earth, where it is moving away and is thus being slowed by gravity. 21. a. Count 5 spaces to the right and about 7.5 spaces down from the given point. Slope ≈ −1.5. b. r = θ x = θ cos θ ⇒ dx = dθ · cos θ − θ sin θ dθ y = θ sin θ ⇒ dy = dθ · sin θ + θ cos θ dθ dy dy/dθ sin θ + θ cos θ = = dx dx/dθ cos θ – θ sin θ At θ = 7, dy/dx = −1.54338… , thus confirming the answer found graphically. 22. a. x = r cos θ, y = r sin θ ⇒ y/x = sin θ /cos θ = tan θ slope =
r sin r cos
= tan
r
dr dx dy = 2x + 2y dθ dθ dθ dr dx dy ⇒r =x +y dθ dθ dθ Substitute these expressions in parts d and e into the top and bottom of the expression in part c to show the property.
e. r 2 = x 2 + y 2 ⇒ 2 r
r a – a cos θ 1 – cos θ = = = dr/dθ a sin θ sin θ tan θ/2, using the half-angle formula. Then ψ = θ/2 + nπ. But 0 ≤ θ ≤ 2π, and 0 ≤ ψ ≤ π , which implies n = 0, so ψ = θ/2. r dr g. tan ψ = ⇒ = const ⋅ r ⇒ r = Cekθ dr/dθ dθ dr dr/dθ Note that = kCe kθ = kr ⇒ k = = dθ r 1 = cot ψ . tan ψ f. tan ψ =
Equations for the spiral will vary.
Problem Set 8-8 Review Problems R0. Answers will vary. R1. a.
r sin 50 y
f
r cos
b. The slope of any line is tan φ, where φ is the angle between the x-axis and the line. And, because the tangent line has slope dy dy/dθ = ( by the chain rule), dx dx/dθ dy/dθ tan φ = . dx/dθ tan φ – tan θ c. tan ψ = tan (φ − θ) = 1 + tan φ tan θ dy/dθ y – dx/dθ x x dx/dθ = ⋅ dy/dθ y x dx/dθ 1+ ⋅ dx/dθ x dy dx x –y d d θ θ = dx dy x +y dθ dθ d. dx/dθ = −r sin θ; dy dx dy/dθ = r cos θ x −y dθ dθ = r cos θ ⋅ (r cos θ ) − r sin θ ⋅ ( − r sin θ ) = r2 cos2 θ + r2 sin2 θ = r2
Calculus Solutions Manual © 2005 Key Curriculum Press
g h x –10
3
b. f ′ (x) = 3x 2 − 18x + 30; f ″(x) = 6x − 18 g′ (x) = 3x2 − 18x + 27; g″(x) = 6x − 18 h′ (x) = 3x2 − 18x + 24; h″(x) = 6x − 18 c. h′ (x) = 3(x − 2)(x − 4) = 0 at x = 2 and 4 h″(2) = −6 < 0, so h has a local maximum at x = 2. h″(4) = 6 > 0, so h has a local minimum at 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 is neither a maximum nor a minimum point. e. From the graphs, each point of inflection appears at x = 3. Because each second derivative equals 6x − 18, each one equals zero when x = 3.
Problem Set 8-8
205
R2. a. no max. or min.
f (x ) +
f´(x)
undef.
x
+
2 p.i.
f (x ) +
f´´(x)
undef.
x
–
2
b. f (x )
Minimum at (0, 0). Maximum at (2, 0.5413…). f ″(x) = 0 at x = 2 + 2 = 3.4142 … and at x = 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 = 10 ⇒ y = 10 x −1 ; 0 < x Minimize L( x ) = 12 x + 7 y = 12 x + 70 x −1 . The graph shows minimum L (x) at x ≈ 2.4. L (x )
x –2
1
3
5 100
2 –2 −4/3 c. i. f ′( x ) = x −1/3 − 1, f ′′( x ) = x 3 9 ii. Zooming in shows that there is a local minimum cusp at (0, 0) and a local maximum with zero derivative at x ≈ 0.3. 0.5
f (x )
x 0
1
Algebraically, f ′ (x) = 0 at x = (2/3)3 = 8/27, and f ′ (x) is undefined at x = 0, thus locating precisely the minimum and maximum found by graphing. Because there are no other critical values of x, there are no other maximum or minimum points. iii. f ″(x) is undefined at x = 0, and f ″(x) < 0 everywhere else; f ″ never changes sign, so there 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.
x 2
L ′( x ) = 12 − 70 x −2 L ′( x ) = 0 at x = 70/12 = 2.4152 … At x = 70/12 , y = 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 typical battery, 9″ by 6.7″. Thus, minimal wall length does not seem to be a major consideration in battery design. b. The graph shows y = 8 − x3, from x = 0 to x = 2, with rectangle touching sample point (x, y) on the graph, rotated about the y-axis, generating cubic paraboloid and inscribed cylinder. 8
y
(x, y )
f (x ) 1
x 2
x 2
The graph shows that f ( x ) = x 2 e − x has local minimum at x = 0, local maximum at x ≈ 2, and points of inflection at x ≈ 3.4 and at x ≈ 0.6. f ′( x ) = 2 xe − x − x 2 e − x = x (2 − x )e − x f ″ ( x ) = 2e − x − 4 xe − x + x 2 e − x = (2 − 4 x + x 2 )e − x f ′ (x) = 0 at x = 0, 2
206
Problem Set 8-8
Domain of x is 0 ≤ x ≤ 2. Maximize V (x) = π r2h = π x 2y = 8π x2 − π x 5 . The graph shows that V (x) has a maximum at x ≈ 1.5. V(x ) 30
x 0
2
Calculus Solutions Manual © 2005 Key Curriculum Press
V ′ (x) = 16π x − 5π x 4 = π x(16 − 5x3) V ′( x ) = 0 at x = 0 and x = 3 16/5 = 1.4736 … Maximal rectangle has x = 3 16/5 = 1.4736 … , y = 8 − 16/5 = 4.8. R4. a. The graph shows y = x11/3 and y = x 22 , intersecting at (0, 0) and (1, 1), rotated about the x-axis, sliced parallel to the x-axis, showing back half of solid only. y
ii. The graph shows the region described in part i, rotated about the x-axis, showing back half of solid only. y (x, 4) 4
(x, y )
x –2
2
(x 1 , y )
1 (x 2 , y )
dV = π (42 − y2) dx = π (16 − x4) dx
x 0
V=
1
x 1 = y 3 , x 2 = y 1/2 dV = 2π y (x 2 − x 1) · dy = 2π y (y 1/2 − y 3) dy V=
∫
2
–2
dV ≈ 160.8495… (exactly 51.2π)
(Cylindrical shells can also be used.) iii. The graph shows the region described in part i, rotated about the line y = 5, showing back half of solid only. y
1
∫ dV ≈ 1.2566… (exactly 0.4π ) 0
b. The graph shows y1 and y2 as in part a, but sliced perpendicular to the x-axis, generating plane washer slices.
(x, 4)
5 4
(x, y )
y (x, y1 )
1
x 2
–2 (x, y2 ) 0
x
1
dV = π [(5 − y)2 − 12] dx = π [(5 − x2)2 − 1] dx 2 7 V = dV ≈ 174.2536 … exactly 55 π –2 15 (Cylindrical shells can also be used.) iv. The graph shows the region described in part i, rotated about the line x = 3, showing back half of solid only.
∫
dV = π ( y12 − y22 ) dx = π ( x 2/3 − x 4 ) dx V=
1
∫ dV ≈ 1.2566… (exactly 0.4π), which is
y
0
(x, 4)
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), rotated about the y-axis, showing back half of solid only. y (x, 4)
(x, y )
x 2
dV = 2π x (4 − y) · dx = 2π x (4 − x2) dx V ≈ 25.1327… (exactly 8π) (Disks can also be used.) Calculus Solutions Manual © 2005 Key Curriculum Press
(x, y ) (3, y)
x 2
–2
3
dV = 2π (3 − x) · (4 − y) · dx = 2π (3 − x)(4 − x2) dx V=
4
–2
4
∫
2
–2
dV ≈ 201.0619… (exactly 64π)
(Washers can also be used.) R5. a. y = x 2 from x = −1 to x = 2. dy = 2x dx dL = dx 2 + dy 2 = 1 + (2 x )2 dx L=
∫
2
–1
dL ≈ 6.1257… Problem Set 8-8
207
b. y = x3/2 from x = 0 to x = 9. dy = 1.5x1/2 dx dL = dx 2 + dy 2 = 1 + (1.5 x 1/2 )2 dx 9
∫ (1 + 2.25x ) dx 1 (1 + 2.25 x ) = 2.25 ∫
L=
1/2
0
9
1/2
(2.25 dx )
0
9 2 (1 + 2.25 x )3/2 6.75 0 2 3/ 2 (21.25 – 1) = 28.7281… = 6.75 Distance between the endpoints is 10 2 + 26 2 = 27.8567… , so the answer is reasonable. c. x = t cos π t ⇒ dx = (cos π t − π t sin π t ) dt y = t sin π t ⇒ dy = (sin π t + π t cos π t ) dt The graph shows t increases from 0 to 4.
=
3 π 2
=
1/2
x 4 / 3 + 1 4 x 1/3 dx 0 9 3
∫
8
1 = π x 4/3 + 9
3/2 8
0
π = (1453/ 2 – 1) = 203.0436 K 27 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 to x = 1, rotated about the line y = −1, showing the back half of the solid only. y 1
(x, y)
x 0
1 (x,–1)
4 y
x 4
dy = sec2 x dx
dL = dx 2 + dy 2 = (cos π t – π t sin π t )2 + (sin π t + π t cos π t )2 dt = 1 + (π t )2 dt L=
∫
4
0
1 + π 2 t 2 dt ≈ 25.7255K
S=
(x, y )
x
0
8
1 −2/3 x dx 3 2
1 dL = dx 2 + dy 2 = 1 + x –2/3 dx 3 2
1 dS = 2πx ⋅ dL = 2πx 1 + x –2/3 dx 3
= 2π 208
8
0
∫
8
0
0
R7. a. r = θ ⇒ dr = d θ
L=
(0, y) 2
∫x
1
∫ dS ≈ 20.4199K
dL = dr 2 + (rdθ )2 = 1 + θ 2 dθ
y
S = 2π
dS = 2π (y + 1) · dL = 2π ( tan x + 1) 1 + sec 4 x dx
R6. a. The graph shows y = x1/3 , from x = 0 to x = 8, rotated about the y-axis, showing the back half of the solid only.
dy =
dL = dx 2 + dy 2 = 1 + sec 4 x dx
2
1 1 + x –2/3 dx 3
∫
5π / 2
0
dL ≈ 32.4706 K
1 2 1 r dθ = θ 2 dθ . Area of the region 2 2 between the curves equals the area traced out from t = 2π to t = 5π /2 minus the area traced out from t = 0 to t = π /2. π /2 1 5π / 2 1 A= θ 2 dθ − θ 2 dθ 2π 0 2 2
b. dA =
∫
1 = θ3 6
∫
5π /2 2π
1 − θ3 6
π /2 0
1 3 7.5 3 π (2.53 − 2 3 − 0.53 + 0 3 ) = π 6 6 = 38.7578… =
1/2
1 x 1/3 x 4/3 + dx 9
Problem Set 8-8
Calculus Solutions Manual © 2005 Key Curriculum Press
y
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 = 0 and T ≈ 2.0 and minima at T ≈ 0.5 and T = 3. µ (T )
f g
1 0
x
1
2
130
T 0
3
To maximize µ(T): µ′ (T ) = −12 + 30T − 12T 2 = −6(2T − 1)(T − 2) 1 µ′ (T ) = 0 at T = , 2 2 1 µ(0) = 130; µ = 127.25; 2 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 the graph straightens out at x = 1 but does not change concavity. y
1
x 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 not change sign at x = 1. Thus, the graph is straight at x = 1, but not horizontal. Zooming in on (1, 1) shows that the graph resembles y = x when x is close to 1, although it is actually concave up slightly. 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 vertical asymptote at x = 0 for function g.
Calculus Solutions Manual © 2005 Key Curriculum Press
π C4. a. y = 3 + 1.25 1 + cos ( x – 5) 3 y′ = π π π 2.5 1 + cos ( x – 5) − sin ( x – 5) 3 3 3 –2.5π 3
=
π π 1 + cos 3 ( x – 5) sin 3 ( x – 5) 2
–2.5π (1 + cos χ )sin χ dx dL = 1 + 3 where χ temporarily stands for π ( x – 5) 3 L=
∫
7.5
5
dL ≈ 5.7726…
b. y ′′ = –2.5π dχ [(1 + cos χ ) cos χ + ( − sin χ ) sin χ ] 3 dx π –2.5π = (2 cos 2 χ − 1 + cos χ ) ⋅ 3 3 2 –2.5π = (cos χ + 1)(2 cos χ – 1) 9 y ″ = 0 ⇔ cos χ = −1 or cos χ = 0.5 χ = π + 2π n or χ = ± π /3 + 2π n x = 8 + 6n, 4 + 6n, or 6 + 6n The only zero of y ′′ within the domain is x = 6, so the point of inflection must be at x = 6. c. dS = 2π (x − 4) dL, where dL is as in part a. S=
∫
7.5
5
dS ≈ 78.2373…
5π d. x = 7.5 ⇒ y = 3 + 1.25 1 + cos 6
2
= 3 + 1.25(1 − 3 /2)2 = 3 + 1.25(1.75 − 3 ) = 3.0224 … dV = 2π (x − 4) · (y − 3.0224…) · dx V=
∫
7.5
5
2π ( x − 4)1.25[(1 + cos χ )2 − (1.75 − 3 )] dx
= 58.8652 … C5. The 2000 World Almanac and Book of Facts lists the area of Brazil as 3,286,478 square miles. Individual answers will vary. Problem Set 8-8
209
Because the spike goes from x = 0 to x =
C6. Let the cylinder lie on the x-axis and the hole lie on the y-axis so that the z-axis is perpendicular to both the cylinder and the hole. The cylinder is thus described by y2 + z2 ≤ 25, and the hole by x 2 + z2 ≤ 9. Slice the hole with planes perpendicular to the z-axis. Then for −3 ≤ z ≤ 3, the cross section at z of the hole is a rectangle with height 2 y = 2 25 – z 2 and width 2 x = 2 9 – z 2 . Area of cross-section rectangle is
Vs =
∫
−3
= 8 + 8 2 − 6π = 0.46415… cm 3
4 225 – 34 z 2 + z 4 dz
= 269.3703… cm3 According to the CRC Handbook, the density of uranium is 19.1 g/cm3. So the mass of the uranium drilled out is m = (269.3703…)(19.1) = 5144.97… g. Value is 200(5144.97…) ≈ $1,029,000.
dVs ≈ 0.0109642 … .
11 2 π 1 – – = 8(1 − 2 /2)3 + 24 4 2 12
so dV = 4 225 – 34 z 2 + z 4 dz. Thus, the volume of the hole (and thus of the uranium that once filled the hole) is 3
0
2 /2,
(The integral can be evaluated algebraically using trigonometry substitution, as in Chapter 9. The 11 2 π 1 exact value is Vs = − − .) 12 4 2 The 24 spikes (3 for each of the eight corners) are identical. Thus, the total volume remaining is V = 8Vc + 24Vs
4 225 – 34 z 2 + z 4 ,
V=
∫
2 /2
Chapter Test T1. y 3
C7. Draw x- and y-axes with origin at the center of the circle on one face of the cube.
x 0
1
2
3
4
y
T2.
1
y 6
(x, y )
Function 5 4
x 1
3 2
The solid remaining consists of eight identical corner pieces. Each corner piece consists of a cube and three identical spikes. The spikes have square cross sections when sliced perpendicular to the appropriate axes. The hole perpendicular to the xy-plane cuts a circle in that plane with equation x2 + y2 = 1. The cube shown in the diagram begins 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 = (1 − 2 /2)3 = 0.0251262 … cm 3 . Consider the leftmost spike in the preceding diagram. Pick a sample point (x, y) on the part of the circle in that spike. The cross section perpendicular to the x-axis for this spike is a square of side (1 − y) = 1 − 1 – x 2 . Thus,
1
x 1
2
3
4
5
6
7
Derivative
T3. A = xy = x(500 − 0.5x) = 500x − 0.5x2 A′ = 500 − x A′ = 0 ⇔ x = 500 A′ goes from positive to negative at x = 500 ⇒ local maximum at x = 500. A(0) = A(1000) = 0 ⇒ global maximum at x = 500. Maximum at x = 500, y = 250.
dVs = (1 − 1 – x 2 )2 dx. 210
Problem Set 8-8
Calculus Solutions Manual © 2005 Key Curriculum Press
T4. V = 2π
∫
b
a
T11. Vcyl = π · 22 · 8 = 32π
x ( y1 − y2 ) dx
Vsolid 19.2π = = 0.6 32π Vcyl
1 T5. a. dA = r 2 dr 2 b. dL = ( dr ) + (r dθ ) 2
So, V solid = 0.6V cyl . T12. a.
2
y
c. dL = dx 2 + dy 2
2
d. dS = 2π ( x − 1) dL T6. f (x) = x 3 − 7.8x 2 + 20.25x − 13 f ′ (x) = 3x2 − 15.6x + 20.25 = 3(x − 2.5)(x − 2.7) f ′ (x) changes from positive to negative at x = 2.5 and from negative back to positive at x = 2.7. So there is a local maximum at x = 2.5 and a local minimum at x = 2.7. f ″(x) = 6x − 15.6 = 6(x − 2.6) f ″(x) = 0 at x = 2.6 f ′ (2.6) = −0.03, so the graph is not horizontal at the inflection point. T7. y = x3 ⇒ dy = 3x2 dx dL = dx 2 + dy 2 = 1 + (3 x 2 )2 dx L=
∫
2
0
1 + 9 x 4 dx = 8.6303…
T8. dS = 2π x · dL = 2π x 1 + 9 x 4 dx S=
2
∫ 2π x 0
1 + 9 x 4 dx = 77.3245…
b. dx = −5 sin t dt, dy = 2 cos t dt dL = dx 2 + dy 2 = (–5 cos t )2 + (2 sin t )2 dt L=
y
0
1
=
2
V′(x) = 16π x − 5π x 4 = 0 at x = 0 or 3.21/3 V(0) = V(2) = 0 V(3.2 1/3) = 4.8 · 3.22/3π > 0, so this is a maximum. Maximal cylinder has V = 4.8 · 3.22/3π cm3 = 32.7459… cm3. T10. Slicing parallel to the y-axis generates cylindrical shells of radius x extending from the sample point (x, y) to the line y = 8. dV = 2π x · (8 − y) · dx = 2π x (8 − x3) dx V=
∫
2
0
2π (8 x − x 4 ) dx = 2π ( 4 x 2 − 0.2 x 5 )
= 19.2π = 60.3185…
Calculus Solutions Manual © 2005 Key Curriculum Press
2 0
0
dL ≈ 23.0131…
∫
0
∫
0
∫
0
π
−20π sin 3 t dt = 83.7758…
= 26.6666 …π (numerically) V can be evaluated algebraically by transforming two of the three sin t factors into cosines. V=
x
∫
2π
c. Slicing perpendicular to the x-axis generates circular slices of radius y, where sample point (x, y) is on the upper branch of the ellipse. dV = π y2 dx = 4π sin2 t (−5 sin t dt) = −20π sin3 t dt Leftmost slice is at t = π, and rightmost slice is at t = 0. V=
T9. V(x) = π x 2(8 − y) = 8π x2 − π x 5 The graph shows a maximum V (x) at x ≈ 1.5. 40
x 5
π
π
−20π (1 − cos 2 t ) sin t dt −20π sin t dt +
0
∫ 20π cos π
2
t sin t dt
0 20 2 = 20π cos t − π cos3 t = 26 π π 3 3 The x-radius is 5, and the y-radius is 2. 4 4 π (x-radius)(y-radius)2 = π (5)(2)2 = 3 3 2 26 π , Q .E.D . 3 (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 4 solution gives V = πab 2 .) 3
T13. r = 5e0.1 θ dr = 0.5e0.1 θ dθ
Problem Set 8-8
211
dL = dr 2 + (rdθ )2 = (0.5e 0.1θ )2 + (5e 0.1θ )2 dθ = e 0.1θ 25.25 dθ The spiral starts at r = 5 = 5e0.1·0 and makes three complete revolutions, so 0 ≤ θ ≤ 6π. L=
∫
6π
0
e 0.1θ 25.25 dθ = 10e 0.1θ 25.25
6π 0
= 10 25.25 (e 0.6π − 1) = 280.6961… 1 T14. dA = r 2 dθ = 12.5e 0.2θ dθ 2 The area between the second and third revolutions equals the area swept out for the third revolution
212
Problem Set 8-8
minus the area swept out for the second revolution. In Quadrant I, the third revolution extends from θ = 4π to θ = 4.5π and the second revolution extends from θ = 2π to θ = 2.5π . A=
∫
4.5π
4π
12.5e 0.2θ dθ − 4.5π
∫
2.5π
2π
12.5e 0.2θ dθ 2.5π
= 62.5e 0.2θ 4π − 62.5e 0.2θ 2π = 62.5(e0.9 π − e0.8 π − e0.5 π + e0.4 π ) = 203.7405… T15. Answers will vary.
Calculus Solutions Manual © 2005 Key Curriculum Press
Chapter 9—Algebraic Calculus Techniques for the Elementary Functions Problem Set 9-1 1. V = 2π
∫
π /2
0
2.
x ⋅ cos x dx ≈ 3.5864 …
2. f (x) = x sin x ⇒ f ′(x) = x cos x + sin x 3. 4.
∫ ∫ ∫ ∫ x cos x dx = ∫ f ′( x ) dx − ∫ sin x dx
= f (x) + cos x + C (by definition of indefinite integral) = x sin x + cos x + C
∫
3.
0
= 2π x sin x + 2π cos x π0 /2 = π 2 − 2π 6. V = π2 − 2π = 3.5864… , which is the same as the approximation, to the accuracy shown. 7. The method involves working separately with the different “parts” of the integrand. The function f (x) = x sin x was chosen because one of the terms in its derivative is x cos x, which is the original integrand. See Section 9-2.
4.
∫ xe
4x
dx
∫ 6 xe
−3 x
u = 6x
dx
∫ ( x + 4)e
−5 x
dx
∫ ( x + 7)e
2x
dx
y
7.
∫
u=x
dv = sin x dx
du = dx
v = −cos x
= − x cos x − ( − cos x ) dx = −x cos x + sin x + C Calculus Solutions Manual © 2005 Key Curriculum Press
u=x+4
dv = e− 5x dx
du = dx
1 v = − e −5 x 5
u=x+7
dv = e2x dx
du = dx
v=
1 2x e 2
∫
f ( x + h) – f ( x ) h
Q10. C
∫ x sin x dx
dv = e− 3x dx
1 1 2x = ( x + 7) ⋅ e 2 x − e dx 2 2 7 1 1 = e 2 x + xe 2 x − e 2 x + C 2 2 4 13 2 x 1 2 x = e + xe + C 4 2
x
1.
1 4x e 4
∫
Q6.
h→0
v=
∫
6.
Q9. ≈ 110/6
du = dx
1 1 = −( x + 4) ⋅ e −5 x + e −5 x dx 5 5 4 1 1 = − e −5 x − xe −5 x − e −5 x + C 5 5 25 21 −5 x 1 −5 x = − e − xe + C 25 5
x
Q8. lim
dv = e4x dx
∫
y
Q7. r′(x) = t (x)
u=x
1 v = − e −3 x 3 1 1 = (6 x ) − e −3 x − (6) − e −3 x dx 3 3 2 = −2 xe −3 x − e −3 x + C 3
1 11 Q2. x +C 11 1 Q4. sin 3 x + C 3
Q5. 5 cos2 5x − 5 sin2 5x
1 v = sin 3 x 3
du = 6 dx
5.
Problem Set 9-2
Q3.
du = dx
∫
=
x cos x dx
Q1. y ′ = x sec x + tan x
dv = cos 3x dx
1 4x 1 4x xe − e dx 4 4 1 1 = xe 4 x − e 4 x + C 4 16
π /2
2
u=x
1 1 x sin 3 x − sin 3 x dx 3 3 1 1 = x sin 3 x + cos 3 x + C 3 9 =
f ′( x ) dx = x cos x dx + sin x dx
5. V = 2π
∫ x cos 3x dx
∫x
3
ln x dx
u = ln x
dv = x 3 dx
du = x− 1 dx
v=
1 4 x 4
∫
1 4 1 3 x ln x − x dx 4 4 1 1 = x 4 ln x − x 4 + C 4 16 =
Problem Set 9-2
213
8.
∫x
5
u = ln 3x
dv = x 5 dx
du = x− 1 dx
v=
ln 3 x dx
1.
1 6 x 6
∫x e
3 2x
dx
∫
9.
∫x e
2 x
u = 2x du = 2 dx
dv = ex dx v = ex
∫
∫x
2
u = x2
sin x dx
u = 2x du = 2 dx
dv = −cos x dx v = −sin x
∫
∫ ln x dx
u = ln x du = x
∫x
5 −x
e
dx
v = −cos x
= − x 2 cos x − −2 x sin x − ( −2 sin x ) dx 2 = −x cos x + 2x sin x + 2 cos x + C 11.
2.
1 3 2x 3 2 2x 3 2x 3 2x x e − x e + xe − e + C 2 4 4 8
dv = sin x dx
du = 2x dx 2 = − x cos x − ( −2 x cos x ) dx
∫
=
v = ex
= x 2 e x − 2 xe x − 2e x dx = x2ex − 2xex + 2ex + C 10.
−1
3.
∫x
4
sin x dx
dv = dx dx
v=x
∫
= x ln x − x + C
4.
Q4.
1 4 Q5. x + 11x + C 4 Q7. 1/2 Q8. V = π
b
∫ [ f ( x) a
2
– + – + – +
+ – + – + –
dv sin x –cos x –sin x cos x sin x –cos x
∫x
2
u x2 2x 2 0
cos x dx
1 3 ( x + 11)6 + C 18
+ – + –
dv cos x sin x –cos x –sin x
= x2 sin x + 2x cos x − 2 sin x + C
Q6. 1 5. − g( x )2 ] dx
Q9.
Q10. B y 4
x 2
214
u x4 4x 3 12x 2 24x 24 0
Q2. 2m cos 2m + sin 2m
Q3. tan x + C
+
dv e –x –e –x e –x –e –x e –x –e –x e –x
= −x4 cos x + 4x3 sin x + 12x2 cos x − 24x sin x − 24 cos x + C
Problem Set 9-3 1 6 r +C 6
u x5 5x 4 20x 3 60x 2 120x 120 0
= −x5e− x − 5x4e− x − 20x3e− x − 60x2e− x − 120xe− x − 120e− x + C
= x ln x − x ⋅ x −1 dx
Q1.
– +
0
du = 2x dx 2 x x = x e − 2 xe dx
∫
+
6 dv = ex dx
1 2x 2e 1 2x 4e 1 2x 8e 1 2x 16 e
–
6x
u = x2
dx
dv e 2x
+
3x2
1 6 1 x ln 3 x − x 5 dx 6 6 1 1 6 = x 6 ln 3 x − x +C 6 36 =
u x3
Problem Set 9-3
∫x
5
cos 2 x dx
u x5 5x 4 20x 3 60x 2 120x 120 0
+ – + – + – +
dv cos 2x 1 2 sin 2x 1 – 4 cos 2x 1 – 8 sin 2x 1 16 cos 2x 1 32 sin 2x 1 – 64 cos 2x
Calculus Solutions Manual © 2005 Key Curriculum Press
6.
=
1 5 5 5 x sin 2 x + x 4 cos 2 x − x 3 sin 2 x 2 4 2 15 15 15 − x 2 cos 2 x + x sin 2 x + cos 2 x + C 4 4 8
∫
x 3 sin 5 x dx
u x3 3x 2 6x 6 0
+ – + – +
dv sin 5x 1 – 5 cos 5x 1 – 25 sin 5x 1 125 cos 5x 1 625 sin 5x
∫
10.
1 3 2 6 = − x 3 cos 5 x + x sin 5 x + x cos 5 x 5 25 125 6 − sin 5 x + C 625 7.
∫ e sin x dx x
u ex ex ex
+ – +
∫
34 3x e cos 5 x dx 25 1 3 = e 3x sin 5 x + e 3x cos 5 x + C1 5 25 ⇒ e 3x cos 5 x dx ⇒
∫e
=
5 3x 3 e sin 5 x + e 3x cos 5 x + C 34 34
4x
sin 2 x dx
1 = − e 4x cos 2 x + e 4x sin 2 x + C1 2 ⇒ e 4x sin 2 x dx
∫
∫
= − e x cos x + e x sin x − e x sin x dx
∫
11.
= − e x cos x + e x sin x + C1
∫x
7
1 4x 1 e cos 2 x + e 4x sin 2 x + C 10 5 u dv ln 3x + x 7 1 8 1/x 8x ----------------1 1 – 8 x7 1 0 + 64 x 8
ln 3 x dx
∫
⇒ e x sin x dx 1 1 = − e x cos x + e x sin x + C 2 2
8.
∫e
x
u ex ex ex
cos x dx
+ – +
dv cos x sin x –cos x
= 12.
1 8 1 8 x ln 3 x − x +C 8 64
∫x
5
u dv ln 6x + x 5 1 6 1/x 6x ------------------------1 1 – 6x5 1 0 + 36 x 6
ln 6 x dx
∫
= e x sin x + e x cos x − e x cos x dx
∫
⇒ 2 e cos x dx x
= ex sin x + ex cos x + C 1 ⇒ e x cos x dx
∫
9.
∫e
=
1 x 1 e sin x + e x cos x + C 2 2
3x
cos 5 x dx
u e 3x 3e 3x 9e 3x
= 13.
+ – +
dv cos 5x 1 5 sin 5x 1 – 25 cos 5x
14. 15. 16.
1 3 = e 3x sin 5 x + e 3x cos 5 x 5 25 9 − e 3x cos 5 x dx 25
∫
Calculus Solutions Manual © 2005 Key Curriculum Press
– +
dv sin 2x 1 – 2 cos 2x 1 – 4 sin 2x
∫
∫
=−
⇒ 2 e sin x dx
+
1 = − e 4x cos 2 x + e 4x sin 2 x − 4 e 4x sin 2 x dx 2 ⇒ 5 e 4x sin 2 x dx
dv sin x –cos x –sin x
x
u e 4x 4e 4x 16e 4x
∫ ∫ ∫ ∫
1 6 1 6 x ln 6 x − x +C 6 36 ln 7 5 x 4 ln 7 dx = x + C (ln 7 is a constant!) 5 cos 5 7x e 7x cos 5 dx = e +C 7 1 sin 5 x cos x dx = sin 6 x + C 6 1 2 2/3 x (3 − x ) dx = − (3 − x 2 )2/3 ( −2 x dx ) 2 3 = − (3 – x 2 )5/3 + C 10
∫
Problem Set 9-3
215
17.
∫ x ( x + 5) 3
=
18.
1/2
u x3 3x 2 6x 6 0
dx
dv (x + 5) 1/2 2 3/2 3(x + 5) 4 5/2 15(x + 5) 8 7/2 105(x + 5) 16 9/2 945(x + 5)
+ – + – +
23.
∫ x (ln x )
3
u dv (ln x) 3 + x 1 2 3 (ln x)2/x 2x -------------------------1 3 (ln x) 2 – 2 x 1 2 6 (ln x)/x 4x -------------------------1 6 ln x + 4 x 1 2 6/x 8x
dx
2 3 4 x ( x + 5)3/2 − x 2 ( x + 5)5/2 3 5 16 32 7/2 + x ( x + 5) − ( x + 5)9/2 + C 35 315
∫x
2
--------------------------
=
∫
2 − x dx = x 2 (2 − x )1/2 dx u x2 2x 2 0
dv (2 – x)1/2 2 – 3(2 – x)3/2 4 5/2 15(2 – x) 8 – 105(2 – x)7/2
+ – + –
24.
6
–
0
+
1 8x 1 2 16 x
1 2 3 x (ln x )3 − x 2 (ln x )2 2 4 3 3 + x 2 ln x − x 2 + C 4 8
∫ x (ln x ) dx 3
2
u (ln x) 2 2(ln x)/x
+
dv x3 1 4 4x
--------------------------
1 3 4x 1 4 16 x -------------------------1 2 + 16 x 3 1 0 – 64 x 4
2 ln x 2/x
2 8 = − x 2 (2 − x )3/2 − x (2 − x )5/2 3 15 16 7/2 − (2 – x ) + C 105 19.
20. 21.
∫ ln x ∫e
5
ln 7 x
∫x e
5 x2
∫
dx = 5 ln x dx = 5 x ln x − 5 x + C
∫
dx = 7 x dx =
=
7 2 x +C 2
dx
25. u x4 4x 3
1 4 1 1 4 x (ln x )2 − x 4 ln x + x +C 4 8 32
∫ x (x 3
2
+ 1) 4 dx
u x2
dv x(x 2 + 1)4 1 2 5 2x 10 (x + 1) ---------------------------------1 2 5 5 – x(x + 1) 1 2 6 + 0 12 (x + 1)
dv xe x 2 1 x2 2e
+
--------------------------
2x 2 – xe x 2 1 x2 4x 2e -------------------------2 + xe x 2 1 0 – 2 e x2
= 26.
= 22.
2 2 1 4 x2 x e − x 2e x + e x + C 2
∫x e
5 x
3
dx
u x3
+
3x 2
3 1 1 3 = x 3e x − e x + C 3 3
216
Problem Set 9-3
– +
1 2 2 1 x ( x + 1)5 − ( x 2 + 1)6 + C 10 60
∫x
∫
x 2 − 3 dx = x 3 ( x 2 − 3)1/2
3
+
dv x(x 2 – 3) 1/2 1 2 3/2 3(x – 3)
----------------------------------2
dv x 2 ex 3 1 x3 3e
3
0
x 2 ex 3 1 3
+
u x2 2x
---------------------------
1 0
–
=
ex3 27.
– +
x(x 2 – 3) 3/2 1 2 5/2 5(x – 3)
1 2 2 2 x ( x − 3)3/2 − ( x 2 – 3)5/2 + C 3 15
∫ cos
2
x dx
u cos x –sin x
+ –
dv cos x sin x
Calculus Solutions Manual © 2005 Key Curriculum Press
∫ = cos x sin x + ∫ (1 − cos x ) dx = cos x sin x + x − ∫ cos x dx ⇒ 2 ∫ cos x dx = cos x sin x + x + C 1 1 ⇒ ∫ cos x dx = cos x sin x + x + C 2 2 = cos x sin x − ( − sin 2 x ) dx
32.
2
1
2
∫
33. 34.
u sin 0.4x 0.4 cos 0.4x
+ –
dv sin 0.4x –2.5 cos 0.4x
∫ = −2.5 sin 0.4 x cos 0.4 x + ∫ (1 − sin 0.4 x ) dx = −2.5 sin 0.4 x cos 0.4 x + ∫ dx − ∫ sin 0.4 x dx ⇒ 2 ∫ sin 0.4 x dx = −2.5 sin 0.4 x cos 0.4 x + cos 2 0.4 x dx 2
2
2
35. 36. 37. 38. 39.
∫
+ –
dv sec2 x tan x
∫ = sec x tan x − ∫ sec x (sec x − 1) dx = sec x tan x − ∫ sec x dx + ln | sec x + tan x | ⇒ 2 ∫ sec x dx = sec x tan x − sec x tan 2 x dx 2
3
3
40.
∫x
2
cos x dx
∫
∫
∫
∫
= − e x cos x + e x sin x − e x sin x dx
3
but after two more integrations,
∫ e sin x dx = −e x
cos x + e x sin x + e x cos x
∫
Two integrations produced the original integral with the opposite sign (which is useful), and two more integrations reversed the sign again to give the original integral with the same sign (which is not useful).
∫
∫
Calculus Solutions Manual © 2005 Key Curriculum Press
x
− e x sin x + e x sin x dx
sec 2 x tan x dx
1 = (sec x )1 ⋅ (sec x tan x dx ) = sec 2 x + C 2 1 31. log 3 x dx = ln x dx ln 3 1 = ( x ln x – x ) + C ln 3 1 = x log 3 x − x+C ln 3
∫
∫
cos x dx = x 2 sin x − x 2 sin x + x 2 cos x dx,
Amos’s choice of u and dv transforms 1 1 x 2 cos x dx into x 3 cos x + x 3 sin x dx, 3 3 which is more complicated than the original expression. 41. After two integrations by parts, e x sin x dx
1 1 = sec x tan x + ln | sec x + tan x | + C 2 2
∫
2
which is true but not very useful!
⇒ sec x dx
30.
2
∫x
= sec x tan x + ln | sec x + tan x | + C 1
∫
∫ sin x dx = − cos x + C ∫ cos x dx = sin x + C ∫ csc x dx = − ln | csc x + cot x | + C ∫ sec x dx = ln | sec x + tan x | + C ∫ tan x dx = − ln | cos x | + C ∫ cot x dx = ln | sin x | + C ∫ x cos x dx effectively canceling out what she did in the first part. She will get
= −1.25 sin 0.4x cos 0.4x + 0.5x + C u sec x sec x tan x
∫ ln x dx
∫
∫
⇒ sin 2 0.4 x dx
sec 3 x dx
1 ln 10
For the first integral, Wanda integrated cos x and differentiated x2, but in the second integral she plans to differentiate cos x dx and integrate 2x,
= −2.5 sin 0.4x cos 0.4x + x + C 1
29.
x dx =
1 ( x ln x – x ) + C ln 10 1 = x log10 x − x+C ln 10
2
sin 2 0.4x dx
10
=
2
28.
∫ log
42.
∫ cos
2
∫
1 (1 + cos 2 x ) dx 2 1 1 = x + sin 2 x + C 2 2
x dx =
Problem Set 9-3
217
By the double-argument properties from trigonometry, 1 1 1 x + sin 2 x + C = ( x + sin x cos x ) + C 2 2 2 which is equivalent to the answer in Problem 27 found using integrating by parts.
47. For integration by parts,
Applying limits of integration gives
∫
d
c
u dv = uv
(1,1/e)
x 3
y′ = −xe − x + e− x = e− x(1 − x) Critical points at x = 0, 1, 3; maximum at x = 1. −x
∫ 12 x e
2 −x
0
dx
0
= −12 x 2 e − x − 24 xe − x − 24e − x
b 0
= −12b2e− b − 24be− b − 24e− b + 24 The first two terms approach zero as b → ∞ by L’Hospital’s rule. The third term also approaches 0. ∴ lim Ab = 24 b→∞
45. y = ln x dV = πy2 dx = π (ln x)2 dx 5
∫ π (ln x ) dx
u dv (ln x)2 + 1 2 (ln x)/x x ----------------------2 ln x – 1 2/x x ----------------------2 + 1 0 – x
2
1
= π x (ln x )2 − 2π x ln x + 2π x
5 1
= 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.
218
b
∫ v du a
∫ ln ax dx = ∫ (ln a + ln x ) dx
= x ln a + x ln x − x + C = x ln ax − x + C 49.
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) =
a
dx
= ( − xe − x − e − x )
V=
48.
0
b
b
∫ v du
The quantity (bd − ac) is the area of the “L-shaped” region, which is the area of the larger rectangle minus the area of the smaller one. Thus, the integral of u dv equals the area of the L-shaped region minus the area represented by the integral of v du.
1
∫ xe
−
= (bd − ac) −
y
A=
u=b u =a
43.
3
∫ u dv = uv − ∫ v du.
Problem Set 9-4
∫ sin
7
u sin 6 x 6 sin 5 x cos x
x dx
+ –
dv sin x –cos x
∫ = − sin x cos x + 6 ∫ sin x (1 − sin x ) dx = − sin x cos x + 6 ∫ sin x dx − 6 ∫ sin x dx 7 ∫ sin x dx = − sin x cos x + 6 ∫ sin x dx 1 6 ∫ sin x dx = − 7 sin x cos x + 7 ∫ sin x dx = − sin 6 x cos x + 6 sin 5 x cos 2 x dx 6
5
6
5
7
7
2
7
6
5
6
5
The fractions are 1/(old exponent) and (old exponent − 1)/(old exponent). The new exponent is 2 less than the old exponent. So 1 sin 7 x dx = − sin 6 x cos x 7 6 1 4 4 + – sin x cos x + sin 3 x dx 7 5 5 1 6 6 4 = − sin x cos x − sin x cos x 7 35 24 1 2 + – sin 2 x cos x + sin x dx 35 3 3 1 6 6 4 = − sin x cos x − sin x cos x 7 35 8 16 − sin 2 x cos x − cos x + C 35 35 50. Answers will vary.
∫
∫
∫
Problem Set 9-4
∫
Q1. uv − v du
Calculus Solutions Manual © 2005 Key Curriculum Press
Q2.
Q3. y
4.
y
3
∫ tan x dx = ∫ tan x tan = ∫ tan x (sec x − 1) dx = ∫ tan x sec x dx − ∫ tan 1 = tan x − ∫ tan x dx 19 20
18
18
1
x
x 2π
2π
2
19
Q4. y′ = 1 + ln 5x Q6. ln |x| + C
5.
∫ sec
13
x dx
2
18
1 sin 6 x + C Q5. 6
2
18
x dx
18
x dx
Q7.
u sec 11 x 11 sec 10 x sec x tan x
dv sec 2 x tan x
+ –
y 1 3
= sec
11
1 Q8. 1+ x2 Q10. D 1.
∫
∫ x tan x − 11 ∫ sec x tan x − 11 ∫ sec
= sec11 x tan x − 11 sec11 x tan 2 x dx
x
= sec11
Q9. ln | sec x + tan x | + C
sin 9 x dx
u 8
∫ x cos x + 8 ∫ sin x cos x + 8 ∫ sin
sin7
sin 8 x x cos x
+ –
dv sin x –cos x
∫
= − sin
7
∫ 6.
9
∫ csc
100
∫
∫
dv cos x
9
8
9
8
10
10
8
9
8
∫ cot x dx = ∫ cot x cot x dx = ∫ cot x (csc x − 1) dx = ∫ cot x csc x dx − ∫ cot x dx 1 = − cot x − ∫ cot x dx 11 12
10
10
2
2
10
2
11
Calculus Solutions Manual © 2005 Key Curriculum Press
10
∫ x cot x − 98 ∫ csc x cot x − 98 ∫ csc
100
dv csc 2 x –cot x
+ –
7.
98
100
x (csc 2 x − 1) dx
∫
x dx + 98 csc 98 x dx
∫ ∫
x dx = − csc x cot x + 98 csc 98 x dx 98
1 98 csc100 x dx = − csc 98 x cot x + csc 98 x dx 99 99
∫ cos
n
x dx
2
9
10
∫
sin x
∫ = cos x sin x + 9 ∫ cos x (1 − cos x ) dx = cos x sin x + 9 ∫ cos x dx − 9 ∫ cos x dx 10 ∫ cos x dx = cos x sin x + 9 ∫ cos x dx 1 9 ∫ cos x dx = 10 cos x sin x + 10 ∫ cos x dx = cos 9 x sin x + 9 cos8 x sin 2 x dx
3.
98
99 csc + –
∫
x dx
= − csc 98
7
u cos 9 x –9 cos 8 x sin x
∫
1 11 sec x dx = sec11 x tan x + sec11 x dx 12 12 13
= − csc
7
cos10 x dx
∫
= − csc 98 x cot x − 98 csc 98 x cot 2 x dx
x dx − 8 sin 9 x dx
8
x dx + 11 sec11 x dx
11
x (1 − sin 2 x ) dx
8
9
2.
7
13
u csc 98 x –98 csc 97 x csc x cot x
∫ 9 ∫ sin x dx = − sin x cos x + 8 ∫ sin x dx 1 8 ∫ sin x dx = − 9 sin x cos x + 9 ∫ sin x dx = − sin 8
x (sec 2 x − 1) dx
12 sec x dx = sec x tan x + 11 sec11 x dx 13
= − sin 8 x cos x + 8 sin 7 x cos 2 x dx 8
11
u cos n – 1 x –(n –1) cosn – 2 x sin x
dv cos x sin x
+ –
∫ = cos x sin x + (n − 1) ∫ cos x (1 − cos x ) dx = cos x sin x + (n − 1) ∫ cos x dx −(n − 1) ∫ cos x dx n ∫ cos x dx = cos x sin x + (n − 1) ∫ cos x dx n –1 1 ∫ cos x dx = n cos x sin x + n ∫ cos x dx = cos n−1 x sin x + (n − 1) cos n−2 x sin 2 x dx n −1
n−2
n −1
n−2
2
n
n
n
n −1
n−2
n −1
n−2
10
Problem Set 9-4
219
8.
∫ sin
n
u sin n –1 x (n –1) sinn –2 x cos x
x dx
∫ = − sin x cos x + (n − 1) ∫ sin x (1 − sin = − sin x cos x + (n − 1) ∫ sin x dx − (n − 1) ∫ sin x dx n ∫ sin x dx = − sin x cos x + (n − 1)∫ sin n –1 1 ∫ sin x dx = − n sin x cos x + n ∫ sin ∫ tan x dx = ∫ tan x tan x dx = ∫ tan x (sec x − 1) dx = ∫ tan x sec x dx − ∫ tan x dx 1 = tan x − ∫ tan x dx n –1 ∫ cot x dx = ∫ cot x cot x dx = ∫ cot x (csc x − 1) dx = ∫ cot x csc x dx − ∫ cot x dx 1 =− cot x − ∫ cot x dx n –1 ∫ csc x dx
= − sin n−1 x cos x + (n − 1) sin n−2 x cos 2 x dx n −1
n−2
n −1
n−2
2
x ) dx
n −1
9.
n−2
n−2
2
14.
n −1
n
u –(n – 2) csc n – 3x csc x cot x
+ –
15.
∫ = − csc x cot x − (n − 2)∫ csc x (csc x − 1) dx = − csc x cot x − (n − 2)∫ csc x dx + (n − 2)∫ csc x dx (n − 1)∫ csc x dx = − csc x cot x + (n − 2)∫ csc x dx ∫ csc x dx 1 n–2 =− csc x cot x + csc x dx n –1 n –1 ∫ ∫ sec x dx n−2
n−2
n
2
n−2
n
n−2
n−2
n
n−2
12.
n−2
n
u sec n – 2x (n – 2) sec n – 3 x sec x tan x
220
Problem Set 9-4
+ –
5
x dx
dv sec 2 x tan x
∫ cot
6
x dx
1 1 = − cot 5 x − − cot 3 x − ( − cot x − x ) + C 5 3 1 1 3 5 = − cot x + cot x − cot x − x + C 5 3
dv csc 2 x –cot x
= − csc n−2 x cot x − (n − 2) csc n−2 x cot 2 x dx n−2
∫ cos
1 4 1 2 cos 4 x sin x + cos 2 x sin x + sin x + C 5 53 3 1 4 8 = cos 4 x sin x + cos 2 x sin x + sin x + C 5 15 15
n−2
csc n – 2x
5
=
n−2
2
n−2
1 4 1 2 = − sin 4 x cos x + − sin 2 x cos x − cos x + C 5 5 3 3 1 4 8 = − sin 4 x cos x − sin 2 x cos x − cos x + C 5 15 15
2
n−2
11.
13.
n−2
n−2
n−2
n−2
n−2
n
n
2
n
x dx
2
2
n −1
10.
x dx
2
n−2
n−2
n−2
n−2
n−2
n
n−2
n
n−2
n −1
n
n−2
n−2
n
n
∫ = sec x tan x − (n − 2) ∫ sec x (sec x − 1) dx = sec x tan x − (n − 2) ∫ sec x dx + (n − 2)∫ sec x dx (n − 1)∫ sec x dx = sec x tan x + (n − 2)∫ sec x dx ∫ sec x dx 1 n–2 = sec x dx sec x tan x + n –1 n –1 ∫ ∫ sin x dx = sec n−2 x tan x − (n − 2) sec n−2 x tan 2 x dx
dv sin x –cos x
+ –
16.
∫ tan
7
x dx
1 1 1 = tan 6 x − tan 4 x − tan 2 x + ln | cos x | + C 6 4 2 1 1 1 6 4 2 = tan x − tan x + tan x + ln | cos x | + C 6 4 2 1 2 17. sec 4 x dx = sec 2 x tan x + tan x + C 3 3 1 2 4 2 18. csc x dx = − csc x cot x − cot x + C 3 3 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. A1 =
∫
π /2
– π /2
cos x dx = sin x
π /2 − π /2
sin (π/2) − sin (−π/2) = 2
Calculus Solutions Manual © 2005 Key Curriculum Press
A3 =
∫
π /2
– π /2
ε π ε < x < , then cos x < cos 4 2 4 ε ⇒ cos N x < . 2π Now, for any n > N,
cos3 x dx
Note that if
π /2 1 2 = cos 2 x sin x + sin x 3 3 − π /2 1 2 = cos 2 (π /2) sin (π /2) + sin (π /2) 3 3 1 2 2 − cos ( −π /2) sin ( −π /2) − sin ( −π /2) 3 3 2 2 4 = 0 + − 0 + = = 1.3333… 3 3 3 2 Observe that A3 = A1 . 3
A5 =
∫
π /2
– π /2
∫
π /2 − π /2
=2
ε /4
∫ And 2 ∫
+
∫
4 5
π /2
− π /2
cos3 x dx
N )
dx =
ε N ε cos x < . 2 2π
∫
π /2
ε /4
cos n x dx < ε .
cos n x dx = 0, Q .E.D .
∫ cos x dx = ∫ cos x cos x dx = ∫ (1 − sin x ) cos x dx = ∫ (1 − 2 sin x + sin x ) cos x dx = ∫ cos x dx − 2 ∫ sin x cos x dx + ∫ sin x cos x dx 5
4
2
2
2
y = cos100 x
∫
π /2
cos n x dx + 2
0
∴ lim
∫
ε /4
cos n x dx
− π /2
=2
x
4
4
0.5π
n→∞ – π /2
∫
cos n x dx.
ε /4
cos n x dx < 2
ε ε dx < 2π π
π /2
π /2
2
–0.5π
π /2
∫
0
1
∫
0
cos n x dx + 2
ε /4
4 4 4 4 2 16 A3 = ⋅ = ⋅ ⋅ 2 = 5 5 3 5 3 15 = 1.066666 … 4 Observe that A5 = A3 . 5 d. Based on the graphs, the area under cos x should be greater than that under cos3 x, which in turn is greater than the area under cos5 x. This is exactly what happens with the calculated answers: A1 > A3 > A5. e.
f. Yes, lim
∫
ε /4
But 2
=0+
y
cos n x dx
− π /2
cos 5 x dx
1 = cos 4 x sin x 5
π /2
2 1 = sin x − sin 3 x + sin 5 x + C 3 5
cos n x dx = 0.
∴ A5 =
Following the pattern in part c, for odd n, (n – 1)(n – 3)(n – 5)K( 4)(2)(1) An = ⋅ 2, (n)(n – 2)(n – 4)K(5)(3) the denominator gets large faster than the numerator. However, because both go to infinity, this observation is not decisive. The following epsilon proof by Cavan Fang establishes the fact rigorously, using the definition of limit in the form “For any ε > 0, there is an N > 0 such that whenever n > N, A n < ε .” Proof: Pick 0 < ε < 2π . ε Then 0 < cos < 1, so there exists N > 0 4 ε N ε such that cos < . 4 2π Calculus Solutions Manual © 2005 Key Curriculum Press
∫
π /2
− π /2
sin 5 x dx
2 1 = 2 sin x − sin 3 x + sin 5 x 3 5
π /2 0
4 2 16 + = = 1.0666 … , which agrees with 3 5 15 the answer from Problem 19. =2−
21.
∫ sec
3
u sec x sec x tan x
x dx
+ –
dv sec 2 x tan x
∫ = sec x tan x − ∫ sec x (sec x − 1) dx = sec x tan x − ∫ sec x dx + ∫ sec x dx = sec x tan x − sec x tan 2 x dx 2
3
Problem Set 9-4
221
1 = (sin 3 ax + 2 sin ax – 2 sin ax cos 2 ax ) 3 1 = [sin 3 ax + 2 sin ax (1 – cos 2 ax )] 3 1 = [sin 3 ax + 2 sin ax (sin 2 ax )] 3 = sin3 ax
∫
2 sec 3 x dx = sec x tan x + ln |sec x + tan x| + C 1
∫ sec
3
x dx
1 1 sec x tan x + ln |sec x + tan x| + C 2 2 Note that the answer is half the derivative of secant plus half the integral of secant. =
22.
∫ sin
n
ax dx
∴
∫ sin
3
ax dx
1 (cos ax )(sin 2 ax + 2) + C, Q .E.D . 3a 24. Use integration by parts, or use the technique of Problem 20, as shown here. =−
u sin n –1 ax a(n – 1) sin n – 2ax cos ax
+ –
dv sin ax 1 – a cos ax
∫ cos ax dx = ∫ cos ax cos ax dx = ∫ (1 − sin ax ) cos ax dx = ∫ cos ax dx − ∫ sin ax cos ax dx 3
1 = − sin n−1 ax cos ax a
∫
+ (n − 1) sin
n−2
2
2
ax cos ax dx
2
1 = − sin n−1 ax cos ax a
1 1 sin ax − sin 3 ax + C 3a a 1 = (sin ax )(3 – sin 2 ax ) + C 3a 1 = (sin ax )(2 + cos 2 ax ) + C, Q .E .D . 3a Or: Differentiate, as in the alternate solution for Problem 23. =
∫
+ (n − 1) sin n−2 ax (1 − sin 2 ax ) dx =−
∫
1 sin n−1 ax cos ax + (n − 1) sin n−2 ax dx a
∫
− (n − 1) sin n ax dx
∫
n sin n ax dx =−
∫
1 sin n−1 ax cos ax + (n − 1) sin n−2 ax dx a
∫ sin
n
ax dx
∫
1 n –1 sin n−1 ax cos ax + sin n−2 ax dx an n 1 sin 5 3 x dx = − sin 4 3 x cos 3 x 15 4 8 − sin 2 3 x cos 3 x − cos 3 x + C 45 45 1 sin 2 ax cos ax 23. sin 3 ax dx = − 3a 2 + sin ax dx (From Problem 22) 3 1 2 = − sin 2 ax cos ax − cos ax + C 3a 3a 1 2 = − cos ax (sin ax + 2) + C, Q .E.D . 3a d 1 Or: – (cos ax )(sin 2 ax + 2) dx 3a 1 2 = − (– a sin ax )(sin ax + 2) 3a 1 − (cos ax )(2 a sin ax cos ax ) 3a =−
∫ ∫
Problem Set 9-5 Q1. f ′(1) = −4 Q3. h′(3) = −12 Q5. p′(5) = 6e5 Q7. 5
Problem Set 9-5
g′(2) = 1/2 t′(4) = π/24 x = 83 = 512 integration by parts
Q2. Q4. Q6. Q8.
y
x 3
Q9.
Q10. E f'(x) and f (x )
∫
222
2
f f'
1 –1
1.
x 1
3
5
6
8
∫ sin x dx = ∫ (1 − cos x ) sin x dx = ∫ (1 − 2 cos x + cos x ) sin x dx 5
2
2
= − cos x +
2
4
2 1 cos3 x − cos 5 x + C 3 5 Calculus Solutions Manual © 2005 Key Curriculum Press
2.
∫ cos x dx = ∫ (1 − sin x ) cos x dx = ∫ (1 − 3 sin x + 3 sin x − sin x ) cos x dx 7
2
2
3
4
12.
6
∫ cos 9 x dx = ∫ (1 − sin 9 x ) = ∫ (1 − 3 sin 9 x + 3 sin 9 x 7
2
2
3
cos 9 x dx
4
− sin6 9x) cos 9x dx 1 1 = sin 9 x − sin 3 9 x 9 9 1 1 5 + sin 9 x − sin 7 9 x + C 15 63 4.
∫
15.
∫
sin 3 10 x dx = (1 − cos 2 10 x ) sin 10 x dx
6. 7.
1 1 cos 10 x + cos3 10 x + C 10 30 1 4 sin 3 x cos 3 x dx = sin 5 3 x + C 15 1 cos8 7 x sin 7 x dx = − cos 9 7 x + C 63
=−
17.
∫ sec
4
18.
2
∫ csc x dx = ∫ (cot x + 1) csc x dx = ∫ (cot x + 2 cot x + 1) csc x dx 6
19.
1 1 = sin 5 2 x − sin 7 2 x + C 10 14
∫ sin x cos x dx = ∫ sin x (1 − cos x ) cos = ∫ (cos x − 2 cos x + cos x ) sin x dx 2
2
4
2
20. 2
x dx
∫ cos x sin x dx = ∫ cos x (1 − sin = ∫ (sin x − sin x ) cos x dx 2
2
21.
6
1 2 1 = − cos3 x + cos 5 x − cos 7 x + C 3 5 7 3
∫
Calculus Solutions Manual © 2005 Key Curriculum Press
2
2
2
2
8
2
3
4
2
2
2
6 x dx
∫ sec 100 x dx = ∫ (tan 100 x + 1) sec 100 x dx 4
2
22. 2
x ) sin x dx
2
2
1 1 tan 3 100 x + tan 100 x + C 300 100 1 tan10 x sec 2 x dx = tan11 x + C 11 1 cot 8 x csc 2 x dx = − cot 9 x + C 9
∫ ∫ ∫ sec
10
=
∫
x tan x dx = sec 9 x (sec x tan x dx )
1 sec10 x + C 10
∫ csc
8
∫
x cot x dx = csc 7 x (csc x cot x dx )
1 = − csc8 x + C 8
4
1 1 = sin 3 x − sin 5 x + C 3 5 1 2 11. cos x dx = (1 + cos 2 x ) dx 2 1 1 = x + sin 2 x + C 2 4
2
∫ csc 6 x dx = ∫ (cot 6 x + 1) csc 6 x dx = ∫ (cot 6 x + 3 cot 6 x + 3 cot 6 x + 1) csc
=
6
2
∫
x dx = ( tan 2 x + 1) sec 2 x dx
1 1 1 cot 7 6 x − cot 5 6 x − cot 3 6 x 42 10 6 1 − cot 6 x + C 6
3
5
4
=−
∫ sin 2 x cos 2 x dx = ∫ sin 2 x (1 − sin 2 x ) cos 2 x dx = ∫ (sin 2 x − sin 2 x ) cos 2 x dx
∫
∫
1 1 cos 7 8 x + cos 9 8 x + C 56 72
4
10.
∫
6
8
4
9.
∫
1 2 = − cot 5 x − cot 3 x − cot x + C 5 3
2
6
∫
4
3
6
8.
16.
∫ ∫ ∫ cos 8x sin 8x dx = ∫ cos 8 x (1 − cos 8 x ) sin 8 x dx = ∫ (cos 8 x − cos 8 x ) sin 8 x dx 6
∫
1 (1 − cos 2 x ) dx 2
1 = tan 3 x + tan x + C 3
=− 5.
x dx =
2
1 1 = x − sin 2 x + C 2 4 1 13. sin 2 5 x dx = (1 − cos 10 x ) dx 2 1 1 = x − sin 10 x + C 2 20 1 14. cos 2 6 x dx = (1 + cos 12 x ) dx 2 1 1 = x + sin 12 x + C 2 24
3 1 = sin x − sin 3 x + sin 5 x − sin 7 x + C 5 7 3.
∫ sin
23. 24. 25.
∫ sec 20 dx = sec 20∫ dx = (sec 20) x + C ∫ csc 12 dx = csc 12∫ dx = (csc 12) x + C 1 ∫ (cos x − sin x ) dx = ∫ cos 2 x dx = 2 sin 2 x + C 10
10
8
2
8
10
8
2
Problem Set 9-5
223
26. 27. 28. 29.
∫ (cos x + sin x ) dx = ∫ dx = x + C ∫ (sin x ) dx = ∫ csc x dx = − cot x + C 1 ∫ (cos 3x ) dx = ∫ sec 3x dx = 3 tan 3x + C ∫ sec x dx 2
2
−2
2
−2
2
3
1 1 = sec x tan x + ln | sec x + tan x | + C 2 2 30.
∫ csc
3
x dx
∫
1
∫ (sec x + 6 sec x ) dx = π ∫ [(1 + tan x ) sec x + 6 sec x ] dx = π ∫ (tan x sec x + 7 sec x ) dx 4
2
0
1
2
2
2
0
+ – +
dv cos 5x 1 5 sin 5x 1 – 25 cos 5x
1 3 = sin 5 x sin 3 x + cos 5 x cos 3 x 5 25 9 + cos 5 x sin 3 x dx 25 16 cos 5 x sin 3 x dx 25 1 3 = sin 5 x sin 3 x + cos 5 x cos 3 x + C1 5 25
∫
∫
1
2
2
2
0
1 π tan 3 x + 7π tan x 3 0 π = tan 3 1 + 7π tan 1 ( = 38.2049K) 3 b. dV = 2π (x + 3) y dx = 2π (x + 3)(sec2 x) dx
=
V = 2π
1
∫ x sec 0
= 2π x tan x
2
x dx + 6π
1 0
− 2π
∫
1
0
1
∫ sec
2
x dx
0
tan x dx + 6π tan x
= 8π tan 1 + 2π ln |cos x |
5 3 = sin 5 x sin 3 x + cos 5 x cos 3 x + C 16 16
= 8π tan 1 + 2π ln (cos 1) (= 35.2738…)
∫
35. dA =
2π
cos 5 x sin 3 x dx
2π 5 3 = sin 5 x sin 3 x + cos 3 x cos 5 x = 0 16 16 0 Because the integral finds the area above minus the area below, this calculation shows the two areas are equal.
32. a. y 1
x π
b. A =
∫
π
0
1 sin 3 x dx = cos3 x − cos x 3
1 0
1 2 1 r dθ = (5 + 4 cos θ )2 dθ 2 2
1 2
∫
π /4
0
(16 cos 2 θ + 40 cos θ + 25) dθ
= 4θ + 2 sin 2θ + 20 sin θ +
25 θ 2
π /4 0
25 = π + 2 + 10 2 + π = 29.1012 … , 8 which agrees with the numerical answer. 1 1 36. dA = r 2 dθ = a 2 (1 + cos θ )2 dθ 2 2 π 2 1 A = a2 (1 + 2 cos θ + cos 2 θ ) dθ 0 2
∫
π 0
=
4 3
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 Problem Set 9-5
A=
1 0
∫ cos 5x sin 3x dx 0
224
∫
34. a. y = sec2 x dV = π[(y + 3)2 − 32] dx = π (sec4 x + 6 sec2 x) dx
cos 5 x sin 3 x dx u sin 3x 3 cos 3x –9 sin 3x
b.
∫
V =π
1 1 = − csc x cot x − ln | csc x + cot x | + C 2 2 31. a.
odd function between symmetrical limits is equal to zero. 33. dV = π y2 dx = π sin2 x dx π π π V = π sin 2 x dx = (1 – cos 2 x ) dx 0 2 0 π π π = x − sin 2 x = π 2 /2 2 4 0
=
1 2 1 1 a θ + 2 sin θ + θ + sin 2θ 2 2 4
2π 0
3 2 πa , which is 1.5 times A circle. 2 37. Answers will vary. =
Calculus Solutions Manual © 2005 Key Curriculum Press
Problem Set 9-6 Q1. Q3. Q5. Q6. Q7. Q8. Q9. Q10.
1 1 sin 3 x + C Q2. − cos 4 x + C 3 4 1 1 − ln | cos 5 x | + C Q4. ln | sin 6 x | + C 6 5 1 ln | sec 7 x + tan 7 x | + C 7 5 sec2 5x y′ = 4 cos 4x d See the text for the statement of fundamental theorem of calculus. See the text for the definition of indefinite integral, Section 5-3.
substitution to use and the ensuing calculus, and because algebraic techniques are of less importance now that technology is used for evaluating integrals, the student is not expected to carry along the absolute value just to eliminate it later. 1.
7
x θ
√ 49 –
x = sin θ . x = 7 sin θ , dx = 7 cos θ dθ , 7 x 49 – x 2 = 7 cos θ , θ = sin −1 7 ∴ 49 – x 2 dx = 7 cos θ (7 cos θ dθ )
∫
= = =
a 2 + x 2 = a |sec θ |, and θ ∈ Quadrant I or IV
=
x 2 – a 2 = a |tan θ |, and θ ∈ Quadrant I or II 2.
∫
v
10
x θ
ln x + x 2 + a 2 for x > 0
The second form can be transformed into the first by taking advantage of the property −ln n = ln (1/n). Thus, 1 x – a2 – x 2
= ln x + a 2 – x 2 which can be shown by rationalizing the denominator of the fraction and incorporating the constant ln a2 (or 2 ln a) into the constant of integration. Because the major focus of this section is on the correct
Calculus Solutions Manual © 2005 Key Curriculum Press
x2
x = sin θ . x = 10 sin θ , dx = 10 cos θ dθ , 10 x 100 – x 2 = 10 cos θ , θ = sin −1 10 2 ∴ 100 – x dx = 10 cos θ (10 cos θ dθ ) Let
− ln x − x 2 + a 2 for x < 0
= ln
u
√ 100 –
Where the integral of sec θ occurs, one gets
− ln x − a – x
∫
49 (1 + cos 2θ ) dθ 2
100 – x 2 dx
x – a = − a tan θ
2
∫
49 49 θ + sin 2θ + C 2 4 49 49 θ + sin θ cos θ + C 2 2 49 −1 x 49 1 1 sin + ⋅ x⋅ 49 – x 2 + C 2 7 2 7 7 49 −1 x 1 sin + x 49 − x 2 + C 2 7 2
2
2
∫
= 49 cos 2 θ dθ =
a – x = a |cos θ |, and θ ∈ Quadrant I or IV
2
u x2
Let
2
For the first two, the absolute value is unnecessary because cos θ ≥ 0 and sec θ ≥ 0 in the respective quadrants. For the secant substitution, if x is negative, then θ is in Quadrant II, where tan θ < 0. Thus, the radical equals the opposite of a tan θ, and one should write
49 – x 2 dx v
Note: A radical without a sign in front of it means the positive root. Because trigonometric functions can be positive or negative, the radical should technically be replaced by the absolute value of the appropriate trigonometric function. Fortunately, this turns out to be unnecessary. If x has been replaced by a sin θ, a tan θ, or a sec θ, it is assumed that θ is the corresponding inverse trigonometric function. So θ is restricted to the range of that inverse trigonometric function. Thus, respectively, 2
∫
∫
∫
∫
∫
100 = 100 cos θ dθ = (1 + cos 2θ ) dθ 2 = 50θ + 25 sin 2θ + C = 50θ + 50 sin θ cos θ + C x 1 1 = 50 sin −1 + 50 ⋅ x ⋅ 100 – x 2 + C 10 10 10 x 1 = 50 sin −1 + x 100 – x 2 + C 10 2 2
Problem Set 9-6
225
3.
∫
x 2 + 16 dx
5.
∫
9 x 2 – 1 dx
v
v
√ x 2 + 16 3x
x θ
√ 9x 2 – 1
θ
u 4
x = tan θ . x = 4 tan θ , dx = 4 sec 2 θ dθ , 4 x x 2 + 16 = 4 sec θ , θ = tan −1 4 ∴ x 2 + 16 dx = 4 sec θ ( 4 sec 2 θ dθ )
3x 1 = sec θ . x = sec θ , 1 3 1 dx = sec θ tan θ dθ , 3
Let
∫
Let
∫
∫
9 x 2 – 1 = tan θ , θ = sec −1 3 x
= 16 sec θ dθ (Compare Problem 21 in 3
∴
Problem Set 9-4.) 16 16 = sec θ tan θ + ln | sec θ + tan θ | + C1 2 2 1 = x x 2 + 16 + 8 ln 2
x 2 + 16 x + + C1 4 4
1 x x 2 + 16 + 8 ln 2 1 = x x 2 + 16 + 8 ln 2
x 2 + 16 + x − 8 ln 4 + C1
=
4.
∫
x 2 + 16 + x + C
v
√ 81 + x 2
6.
x θ
u
1 9 x 2 – 1 dx = tan θ sec θ tan θ dθ 3
∫
∫
=
∫ ∫
∫
16 x 2 – 1 dx
1 sec θ tan 2 θ dθ 3 1 = (sec 3 θ − sec θ ) dθ 3 1 1 = sec θ tan θ + ln | sec θ + tan θ | 6 6 1 − ln | sec θ + tan θ | + C 3 1 1 = sec θ tan θ − ln | sec θ + tan θ | + C 6 6 1 1 2 = x 9 x – 1 − ln 3 x + 9 x 2 – 1 + C 2 6
81 + x 2 dx
v
9
x = tan θ . x = 9 tan θ , dx = 9 sec 2 θ dθ , 9 x 81 + x 2 = 9 sec θ , θ = tan −1 9 ∴ 81 + x 2 dx = 9 sec θ (9 sec 2 θ dθ )
4x
Let
∫
∫
∫
= 81 sec 3 θ dθ (Compare Problem 21 in Problem Set 9-4.) =
81 81 sec θ tan θ + ln | sec θ + tan θ | + C1 2 2
1 81 = x 81 + x 2 + ln 2 2
81 + x 2 x + + C1 9 9
1 81 81 x 81 + x 2 + ln 81 + x 2 + x − ln 9 + C1 2 2 2 1 81 = x 81 + x 2 + ln 81 + x 2 + x + C 2 2 =
226
u 1
Problem Set 9-6
√ 16x 2 – 1
θ
u 1
4x 1 = sec θ . x = sec θ , 1 4 1 dx = sec θ tan θ dθ , 4 16 x 2 – 1 = tan θ , θ = sec −1 4 x 1 ∴ 16 x 2 – 1 dx = tan θ sec θ tan θ dθ 4 1 2 = sec θ tan θ dθ 4 1 = (sec 3 θ − sec θ ) dθ 4 1 1 = sec θ tan θ + ln | sec θ + tan θ | 8 8 1 − ln | sec θ + tan θ | + C 4 Let
∫
∫
∫ ∫
Calculus Solutions Manual © 2005 Key Curriculum Press
7.
1 1 = sec θ tan θ − ln |sec θ + tan θ | + C 8 8 1 1 2 = x 16 x – 1 − ln 4 x + 16 x 2 – 1 + C 2 8 dx
∫
∴
x2 +1
∫
x – 121
x u
17 – x 2 = 17 cos θ , θ = sin −1
x = sec θ . x = 11 sec θ , 11 dx = 11 sec θ tan θ dθ , Let
x 17
2
∫
=
∫ ∫ = ∫ sec θ dθ = ln | sec θ + tan θ | + C
∴
∫
1
= ln
2
v
= ln x + x 2 – 121 + C
x
θ
u
11.
√13 – x 2
13 – x = 13 cos θ , θ = sin 2
−1
13 cos θ dθ 13 cos θ 13 – x x = dθ = θ + C = sin −1 +C 13 dx
∫
dx
2
=
∫
∫
x2 +1
∫x
2
x 2 – 9 dx
v
Let x/ 13 = sin θ . x = 13 sin θ , dx = 13 cos θ dθ ,
∫
x 2 – 121 + C1 11
x + 11
= ln x + x 2 – 121 − ln 11 + C1 √ 13
9.
x 11 dx 11 sec θ tan θ dθ = 11 tan θ x 2 – 121
x 2 – 121 = 11 tan θ , θ = sec −1
17 cos θ dθ 17 cos θ 17 – x x = dθ = θ + C = sin −1 +C 17 dx dx
13 – x
u 11
Let x/ 17 = sin θ . x = 17 sin θ , dx = 17 cos θ dθ ,
∫
√x 2 – 121
θ
√ 17 – x 2
∴
x2 +1 + x + C
dx
x
θ
∫
∫
v
√17
8.
∫
sec 2 θ dθ = sec θ dθ sec θ
2
v
∴
=
= ln | sec θ + tan θ | + C = ln 10.
17 – x 2
∫
dx
x
x 13
√x 2 – 9
θ
u 3
x = sec θ . x = 3 sec θ , 3 dx = 3 sec θ tan θ dθ, Let
x 2 – 9 = 3 tan θ , θ = sec −1
x 3
∫ = ∫ (9 sec θ ) (3 tan θ ) (3 sec θ tan θ dθ ) = 81 ∫ sec θ tan θ dθ = 81 ∫ sec θ dθ − ∫ sec θ dθ 1 3 = 81 sec θ tan θ + ∫ sec θ dθ − ∫ sec θ dθ 4 4 81 81 = sec θ tan θ − ∫ sec θ dθ 4 4
∴ x 2 x 2 – 9 dx
v
2
√ x2 + 1
3
x θ
u 1
x = tan θ . dx = sec 2 θ dθ , 1 x 2 + 1 = sec θ , θ = tan −1 x
Let
Calculus Solutions Manual © 2005 Key Curriculum Press
2
5
3
3
3
3
3
3
Problem Set 9-6
227
81 3 81 sec θ tan θ − sec θ tan θ 4 8 81 − ln | sec θ + tan θ | + C1 8 81 x 3 x 2 – 9 81 x x2 – 9 = ⋅ ⋅ − ⋅ ⋅ 4 27 3 8 3 3 2 81 x x –9 − ln + + C1 8 3 3 =
1 3 2 9 x x –9 − x 4 8 81 2 − ln x + x – 9 8 1 3 2 9 = x x –9 − x 4 8 81 2 − ln x + x – 9 8 =
12.
∫x
2
13.
Let x = sin θ. dx = cos θ dθ, 1 – x 2 = cos θ , θ = sin −1 x
81 ln 3 + C1 8
∫ = ∫ cos θ dθ
x2 – 9
4
+C
=
= x
=
u x2
=
x Let = sin θ . x = 3 sin θ , dx = 3 cos θ dθ , 3 x 9 – x 2 = 3 cos θ , θ = sin −1 3 ∴ x 2 9 – x 2 dx
∫ = ∫ (9 sin θ ) (3 cos θ ) (3 cos θ dθ ) = 81 ∫ sin θ cos θ dθ = 81 ∫ (cos θ − cos θ ) dθ 81 = 81∫ cos θ dθ − cos θ sin θ 4 3 ⋅ 81 − cos θ dθ 4 ∫ 81 81 = ∫ (1 + cos 2θ ) dθ − cos θ sin θ 8 4
14.
∫ ∫
1 3 cos3 θ sin θ + cos 2 θ dθ 4 4 1 3 cos3 θ sin θ + (1 + cos 2θ ) dθ 4 8 1 3 3 cos3 θ sin θ + θ + sin 2θ + C 4 8 16 1 3 3 cos3 θ sin θ + θ + sin θ cos θ + C 4 8 8 1 3 3 x (1 − x 2 )3/2 + sin −1 x + x 1 – x 2 + C 4 8 8
∫ (x
2
− 81) −3/2 dx
v
x
2
u 9
4
2
3
x = sec θ . x = 9 sec θ , dx = 9 sec θ tan θ dθ , 9 1 x 2 – 81 = 9 tan θ , θ = sec −1 x 9
Let
2
3
81 81 81 θ + sin 2θ − cos3 θ sin θ + C 8 16 4 81 81 81 = θ + sin θ cos θ − cos3 θ sin θ + C 8 8 4 81 81 = θ + sin θ cos θ (1 − 2 cos 2 θ ) + C 8 8 81 −1 x = sin 8 3 81 x 9 – x 2 2( 9 – x 2 ) − ⋅ ⋅ ⋅ 1 – +C 8 3 3 9 81 −1 x 1 = sin − x ( 2 x 2 − 9) 9 – x 2 + C 8 3 8 =
228
√ x 2 – 81
θ
2
2
∫
∴ (1 − x 2 )3/2 dx = cos3 θ (cos θ dθ )
=
2
u
√1 – x2
v
√9 –
x
θ
9 – x dx
θ
) dx
1
2
3
2 3/2
v
x2 – 9 +
∫ (1 − x
Problem Set 9-6
∫ = ∫ (9 tan θ ) (9 sec θ tan θ dθ ) 1 sec θ dθ = ∫ 81 tan θ
∴ ( x 2 − 81) −3/ 2 dx −3
2
∫
1 cot θ csc θ dθ 81 1 = − csc θ + C 81 –x = +C 81 x 2 – 81 =
Calculus Solutions Manual © 2005 Key Curriculum Press
15.
∫ 81 + x dx
∫
b.
2
v
√ 81 +
x u 9
∫
18. a. x Let = tan θ . x = 9 tan θ , dx = 9 sec 2 θ dθ , 9 x 81 + x 2 = 9 sec θ , θ = tan −1 9
16.
∴
∫ 81 + x = ∫
=
x 1 tan −1 + C 9 9 dx 25 x 2 + 1
∫
dx
2
x = sec θ . x = 7 sec θ , 7 dx = 7 sec θ tan θ dθ, x 7 7 sec θ (7 sec θ tan θ dθ ) 7 tan θ
x 2 – 49 = 7 tan θ , θ = sec −1 x dx x 2 – 49
=
∫
∫
1
∫
b. 5x 1 1 = tan θ . x = tan θ , dx = sec 2 θ dθ , 1 5 5 19.
25 x 2 + 1 = sec θ , θ = tan −1 5 x sec θ dθ 2
∫ 5 ⋅ sec θ
+1 1 −1 = tan 5 x + C 5 x dx 17. a. x 2 + 25
∫
= 7 sec 2 θ dθ = 7 tan θ + C = x 2 – 49 + C
u
=
u 7
Let
θ
2
√ x 2 – 49
θ
∫
∴
dx
x 2 – 49
9 sec 2 θ dθ 1 1 = dθ = θ + C 81 sec 2 θ 9 9
5x
∫ 25x
∫
1 ( x 2 + 25) −1/2 (2 x dx ) 2
x
√ 25x 2 + 1
∴
=
v
v
Let
x + 25 2
= x 2 + 25 + C, which agrees with part a. Moral: Always check for an easy way to integrate before trying a more sophisticated technique! x dx
x2
θ
x dx
2
=
∫
2
=
x – 49
∫
1 ( x 2 − 49) −1/2 (2 x dx ) 2
= x 2 – 49 + C,which agrees with part a. dx
∫
1 1 dθ = θ + C 5 5
x dx
9 – ( x – 5)2 v
3
x–5
θ
∫
√ 9 – (x –
u 5)2
x–5 = sin θ . x = 5 + 3 sin θ , dx = 3 cos θ dθ , 3 x–5 9 – ( x – 5)2 = 3 cos θ , θ = sin −1 3 dx 3 cos θ dθ ∴ = 3 cos θ 9 – ( x – 5)2 x–5 = dθ = θ + C = sin −1 +C 3 dx Let
v
√ x 2 + 25 x θ
∫
u 5
x = tan θ . x = 5 tan θ , dx = 5 sec 2 θ dθ , 5 1 x 2 + 25 = 5 sec θ , θ = tan −1 x 5 x dx 5 tan θ (5 sec 2 θ dθ ) ∴ = 5 ⋅ sec θ x 2 + 25
∫
Let
∫ ∫ = 5∫ tan θ sec θ dθ = 5 sec θ + C = x + 25 + C 2
Calculus Solutions Manual © 2005 Key Curriculum Press
∫
20.
∫
36 – ( x + 2)2 v
6
x+2
θ
√ 36 – (x +
u 2)2
Problem Set 9-6
229
x+2 = sin θ . x = 6 sin θ − 2, dx = 6 cos θ dθ , 6 x+2 36 – ( x + 2)2 = 6 cos θ , θ = sin −1 6 dx 6 cos θ dθ ∴ = 6 cos θ 36 – ( x + 2)2 Let
∫
dx x + 8 x – 20 2
v
=
10
∫
( x + 4)2 – 36
x = sin θ . x = 10 sin θ , dx = 10 cos θ dθ , 10 x 100 – x 2 = 10 cos θ , θ = sin −1 10
Let
∴
u 6
=
x+4 = sec θ . x = 6 sec θ − 4, 6 dx = 6 sec θ tan θ dθ,
Let
8
∫
sin –1 0.8
100 – x 2 dx
−3
sin –1 (–0.3)
= 50
∫
x 2 + 8 x – 20 + C1 6
∫
∫
10 cos θ ⋅ 10 cos θ dθ
sin –1 0.8
sin –1 (–0.3)
sin –1 0.8
sin –1 (–0.3)
cos 2 θ dθ
(1 + cos 2θ dθ )
= 50θ + 25 sin 2θ
sin –1 0.8 sin –1 (–0.3)
= 50 sin −1 0.8 + 25 sin (2 sin −1 0.8) − 50 sin− 1 (−0.3) − 25 sin [2 sin− 1 (−0.3)] = 99.9084… Numerical integration: 99.9084… (Checks.)
= ln | sec θ + tan θ | + C 1 x+4 + 6
∫
= 100
x 2 + 8 x – 20 = ( x + 4)2 – 36 = 6 tan θ , x+4 θ = sec −1 6 dx 6 sec θ tan θ dθ ∴ = 2 6 tan θ x + 8 x – 20
∫ = ∫ sec θ dθ
u
√ 100 – x 2
√ (x + 4) 2 –36
x+4
x
θ
v
= ln
100 – x 2 dx
−3
x+2 +C 6 dx
∫
∫
∫
8
∫
= dθ = θ + C = sin −1 21.
23.
24.
∫
4
x 2 + 25 dx
–1
= ln x + 4 + x 2 + 8 x – 20 − ln 6 + C1
v
= ln x + 4 + x + 8 x – 20 + C 2
√ x 2 + 25 x
22.
∫
dx x 2 – 14 x + 50
=
∫
dx
θ
( x – 7)2 + 1
u 5
v
√(x – 7) 2 + 1
x = tan θ . x = 5 tan θ , dx = 5 sec 2 θ dθ , 5
Let x–7
θ
u
x 2 + 25 = 5 sec θ , θ = tan −1 0.2 x
1
x–7 Let = tan θ . x = 7 + tan θ , dx = sec 2 θ dθ , 1 x 2 – 14 x + 50 = ( x – 7) 2 + 1 = sec θ , θ = tan− 1 (x − 7) dx sec 2 θ dθ ∴ = = sec θ dθ sec θ x 2 – 14 x + 50 = ln | sec θ + tan θ | + C
∫
= ln
230
∫
∫
∴ =
∫
4
∫
tan –1 0.8
tan –1 (–0.2 )
= 25 =
x 2 + 25 dx
–1
∫
5 sec θ ⋅ 5 sec 2 θ dθ
tan –1 0.8
tan –1 (–0.2 )
sec 3 θ dθ
25 25 sec θ tan θ + ln |sec θ + tan θ | 2 2
tan −1 0.8 tan −1 ( −0.2 )
x – 14 x + 50 + x − 7 + C 2
Problem Set 9-6
Calculus Solutions Manual © 2005 Key Curriculum Press
25 sec ( tan −1 0.8) ⋅ 0.8 2 25 + ln |sec ( tan −1 0.8) + 0.8 | 2 25 − sec [tan −1 ( −0.2)] ⋅ ( −0.2) 2 25 − ln | sec [tan −1 ( −0.2)] − 0.2 | 2 = 26.9977… Numerical integration: 26.9977… (Checks.) 25. y = 3x 2 dL = 1 + ( y ′)2 dx = 1 + 36 x 2 dx L=
∫
5
1 + 36 x 2 dx
0
v
√ 1 + 36x2 6x θ
u 1
6x 1 1 = tan θ . x = tan θ , dx = sec 2 θ dθ , 1 6 6 2 −1 1 + 36 x = sec θ , θ = tan 6 x x =5 1 ∴L = sec θ ⋅ sec 2 θ dθ x =0 6 1 x =5 3 = sec θ dθ 6 x =0 x =5 1 1 = sec θ tan θ + ln |sec θ + tan θ | 12 12 x =0 5 1 1 = x 1 + 36 x 2 + ln 1 + 36 x 2 + 6 x 2 12 0 5 1 = 901 + ln 901 + 30 = 75.3828K 2 12 Numerical integration: L = 75.3828… (Checks.) 3 26. a. 9 x 2 + 25 y 2 = 225 ⇒ y = ± 25 – x 2 5 Slice the region vertically. Pick a sample point (x, y) on the positive branch of the graph, within the strip. 6 dA = 2 y dx = 25 – x 2 dx 5 6 4 A= 25 – x 2 dx 5 –3 Let
∴A=
= 15θ +
cos 2 θ dθ
(1 + cos 2θ ) dθ
15 sin 2θ 2
x =4 x =−3
∫
∴ A = 30
∫
π /2
–π /2
cos 2 θ dθ
π /2 15 sin 2θ 2 − π /2 15π 15 15π 15 = + sin π + − sin ( −π ) 2 2 2 2 = 15π = 47.1238… The area is π (x-radius)(y-radius).
= 15 θ +
27. x 2 + y 2 = r 2 ⇒ y = ± r 2 – x 2 , x = 0 at y = ± r Slice the region inside the circle perpendicular to the x-axis. Pick sample point (x, y) on the positive branch of the circle, within the strip. dA = 2 y dx = 2 r 2 − x 2 dx A=2
∫
r
r 2 – x 2 dx
–r
v
r
x
θ
u
√r 2 – x 2
v
x = sin θ . x = r sin θ, dx = r cos θ dθ, r x r 2 – x 2 = r cos θ , θ = sin −1 r x = r ⇒ θ = π /2, x = −r ⇒ θ = − π /2
Let
x u
√ 25 – x2
x = sin θ . x = 5 sin θ , dx = 5 cos θ dθ , 5 x 25 – x 2 = 5 cos θ , θ = sin −1 5
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5 cos θ ⋅ 5 cos θ dθ
= 15θ + 15 sin θ cos θ xx == 4–3 x =4 x 3 = 15 sin −1 + x 25 – x 2 x =−3 5 5 3 −1 = 15 sin 0.8 + ( 4) 9 5 3 −1 − 15 sin ( −0.6) − (–3) 16 5 = 15[sin −1 0.8 − sin −1 ( −0.6)] + 14.4 15π = + 14.4 = 37.9619… 2 Numerical integration: A = 37.9619… (Checks.) 6 5 b. A = 25 – x 2 dx 5 –5 x = 5 ⇒ θ = π /2, x = −5 ⇒ θ = − π /2
∫
Let
x = –3 x =4
x = –3
∫
θ
6 5
x = –3 x =4
∫
5
x =4
∫ = 30 ∫ = 15∫
=
∴A = 2
∫
= 2r
π /2
r cos θ ⋅ r cos θ dθ
– π /2 π /2 2
∫
− π /2
cos 2 θ dθ
Problem Set 9-6
231
= r2
∫
π /2
− π /2
(1 + cos 2θ ) dθ
π /2 1 2 r sin 2θ 2 − π /2 π 1 π = r 2 ⋅ + r 2 sin π + r 2 ⋅ 2 2 2 1 2 2 − r sin ( −π ) = πr 2 ∴ A = π r 2, Q .E .D .
= r 2θ +
2
30. x 2 − y 2 = 9 ⇒ y = ± x 2 – 9 Slice the region perpendicular to the x-axis. Pick a sample point (x, y) on the positive branch of the hyperbola, within the strip. dA = 2 y dx = 2 x 2 – 9 dx A=2
2
x y b 2 28. + = 1 ⇒ y = ± a – x2 a b a Slice the region inside the ellipse perpendicular to the x-axis. Pick sample point (x, y) on the positive branch of the ellipse, within the strip. 2b 2 dA = 2 y dx = a – x 2 dx a 2b a A= a 2 – x 2 dx a –a
∫
v
a
u
√a 2 – x 2
∫
= ab
∫
∫
π /2
– π /2
π /2
– π /2
cos 2 θ dθ
(1 + cos 2θ ) dθ π /2
ab sin 2θ 2 − π /2 abπ ab abπ ab = + − sin ( −π ) = πab sin π + 2 2 2 2 ∴ A = πab = abθ +
Note that if a = b = r, then πab = πr2, the area of a circle. 29. dV = πx 2 dy = π V =π⋅
a2 b2
∫
b
5
x 2 – 9 dx
3
v
x
√x 2 – 9
θ
u 3
x = sec θ . x = 3 sec θ, dx = 3 sec θ tan θ dθ, 3 x x 2 – 9 = 3 tan θ , θ = sec −1 3
Let
x =5
∫ = 18 ∫ = 18 ∫
3 tan θ ⋅ 3 sec θ tan θ dθ
x =3 x =5 x =3
x Let = sin θ . x = a sin θ, dx = a cos θ dθ, a x 2 a – x 2 = a cos θ , θ = sin −1 a x = a ⇒ θ = π /2, x = −a ⇒ θ = − π /2 2 b π /2 ∴A = a cos θ ⋅ a cos θ dθ a – π /2 = 2 ab
∫
∴ A=2
x
θ
a2 2 (b – y 2 ) dy, − b ≤ y ≤ b b2
(b 2 – y 2 ) dy
–b
a2 y3 = π ⋅ 2 b2 y – 3 b 4 2 ∴ V = πa b 3 232
Rotating instead about the x-axis is equivalent to interchanging the a and b, giving V = 43 πab 2 .
Problem Set 9-6
b
= −b
4 2 πa b 3
x =5
x =3
tan 2 θ sec θ dθ (sec 3 θ – sec θ ) dθ
= 9 sec θ tan θ + 9 ln | sec θ + tan θ | x =5 − 18 ln | sec θ + tan θ | x =3 = 9 sec θ tan θ − 9 ln | sec θ + tan θ | =x
x 2 – 9 − 9 ln
1 1 2 x+ x –9 3 3
x =5 x =3 5 3
= 20 − 9 ln 3 = 10.1124… Numerical integration: A = 10.1124… (Checks.) 31. dV = 2πx (2 y) dx = 4πx x 2 – 9 dx
∫ = 2π ∫
V = 4π
5
x 2 – 9 x dx
3
5
x 2 – 9 (2 x dx )
3
5 2 = 2π ⋅ ( x 2 – 9)3/2 3 3 4 256 = π ⋅ 64 = π = 268.0825… 3 3 32. From Problems 30 and 31, A = 20 − 9 ln 3, 256 V= π. 3 128 V = 2π x ⋅ A ⇒ x = = 4.2192 K 3(20 – 9 ln 3) x is a little more than halfway through the region.
Calculus Solutions Manual © 2005 Key Curriculum Press
33. x = a cos t ⇒ dx = −a sin t dt y = b sin t dA = 2y dx = 2(b sin t)(−a sin t dt) = −2ab sin2 t dt x = −a ⇒ t = π , x = a ⇒ t = 0 ∴ A = −2 ab
∫
0
π
sin 2 t dt = − ab
Q7. Q8. Q9. Q10.
0
∫ (1 – cos 2t ) dt π
1.
0
ab = − abt + sin 2t = 0 + 0 + ab(π ) − 0 = πab 2 π ∴ A = πab, as in Problem 28. With this method, you get sin 2 t dt, directly.
2.
∫
With trigonometric substitution in Problem 28, you get cos 2 t dt, indirectly.
∫
34. r = 0.5θ ⇒ dr/dθ = 0.5 dL = r 2 + ( dr/dθ )2 dθ = 0.25θ 2 + 0.25 dθ
3.
4.
= 0.5 θ 2 + 1 dθ L = 0.5
∫
6π
θ 2 + 1 dθ
0
5.
v
√ θ2 + 1
θ
φ
6.
u 1
Let θ = tan φ ⇒ dθ = sec2 φ dφ. θ 2 + 1 = sec φ , φ = tan −1 θ
∫ = 0.5 ∫
∴ L = 0.5
θ =6π
θ =0 θ =6π
θ =0
7.
sec 3 φ dφ 8.
θ =6π θ =0
= 0.25θ θ 2 + 1 + 0.25 ln
θ 2 +1 +θ
= 1.5π 36π + 1 + 0.25 ln
36π + 1 + 6π 2
= 89.8589… , same as numerical integration. 35. See the note preceding the solutions for this section. For the sine and tangent substitution, the range of the inverse sine and inverse tangent make the corresponding radical positive. For the secant substitution, the situation is more complicated but still gives an answer of the same algebraic form as if x had been only positive.
Calculus Solutions Manual © 2005 Key Curriculum Press
9.
∫
∫
∫
∫
∫
∫
∫
∫
∫
∫
9 x 2 – 25 x – 50 dx ( x + 1)( x – 7)( x + 2) 2 3 4 = + + dx x +1 x – 7 x + 2 = 2 ln |x + 1| + 3 ln |x − 7| + 4 ln |x + 2| + C
∫
Q2. x 2 + 2x − 15 Q4. x 2 + 14x + 49 Q6. x 2 − 64
7 x 2 + 22 x – 54 dx ( x – 2)( x + 4)( x – 1) 3 –1 5 = + + dx x – 2 x + 4 x – 1 = 3 ln |x − 2| − ln |x + 4| + 5 ln |x − 1| + C
∫
4 x 2 + 15 x – 1 dx x 3 + 2 x 2 – 5x – 6 –1 2 3 = + + dx x + 3 x +1 x – 2 = −ln |x + 3| + 2 ln |x + 1| + 3 ln |x − 2| + C
∫
∫
10.
–3 x 2 + 22 x – 31
∫ x – 8x + 19 x – 12 dx –2 –4 3 = ∫ + + dx x –1 x – 3 x – 4 3
2
= −2 ln |x − 1| − 4 ln |x − 3| + 3 ln |x − 4| + C 11.
Problem Set 9-7 Q1. (x + 5)(x − 5) Q3. (x + 2)(x − 6) Q5. (x + 4)2
∫
∫
6π 0
2
∫
∫
sec φ ⋅ sec 2 φ dφ
= 0.25 sec φ tan φ + 0.25 ln | sec φ + tan φ |
ex b2 − 4ac = −1500, so x2 + 50x + 1000 is prime. b2 − 4ac = −144, so x2 + 36 is prime. B 11x – 15 4 7 + dx = dx x –1 x – 2 x 2 – 3x + 2 = 4 ln |x − 1| + 7 ln |x − 2| + C 7 x + 25 –2 9 dx = + dx x + 1 x – 8 x 2 – 7x – 8 = −2 ln |x + 1| + 9 ln |x − 8| + C (5 x – 11) dx 7/2 3/2 = + dx x + 2 x – 4 x2 – 2x – 8 7 3 = ln | x + 2 | + ln | x − 4 | + C 2 2 (3 x – 12) dx 9/5 6/5 = + dx 2 x – 5 x – 50 x+5 x – 10 9 6 = ln | x + 5 | + ln | x − 10 | + C 5 5 21 dx –7 7 = + dx x + 5 x + 2 x 2 + 7 x + 10 = −7 ln |x + 5| + 7 ln |x + 2| + C 10 x dx 2 8 = + dx x + 3 x – 12 x 2 – 9 x – 36 = 2 ln |x + 3| + 8 ln |x − 12| + C
3 x 3 + 2 x 2 – 12 x + 9 dx x –1 2 = 3x 2 + 5x − 7 + dx x – 1 5 = x 3 + x 2 − 7 x + 2 ln | x − 1 | + C 2
∫
∫
Problem Set 9-7
233
12.
13.
x 3 – 7 x 2 + 5 x + 40 dx x2 – 2x – 8 3x dx = x −5+ 2 x – 2x – 8 1 2 = x −5+ + dx x + 2 x – 4 1 = x 2 − 5 x + ln | x + 2 | + 2 ln | x − 4 | + C 2
∫
∫
∫
∫ y + 1000 – y dy = ∫ 2 dt
4 x + 6 x + 11 dx 2 + 1)( x + 4) x+2 3 = 2 + dx x +1 x + 4 1 2 x dx dx 3 dx = +2 2 + 2 x2 +1 x +1 x+4 1 = ln | x 2 + 1| + 2 tan −1 x + 3 ln | x + 4 | + C 2 2
∫ (x
∫
∫
∫
4 x 2 – 15 x – 1 dx x 3 – 5x 2 + 3x + 1 3 x–2 = + dx x – 1 x 2 – 4 x – 1 1 = 3 ln | x − 1| + ln | x 2 − 4 x − 1| + C 2 Note that x–2 1/2 1/2 = + , x2 – 4x – 1 x – 2 + 5 x – 2 – 5 1 1 but ln x − 2 + 5 + ln x − 2 − 5 2 2 1 2 = ln | x − 4 x − 1|, so the answer comes out 2 the same.
∫
∫
15.
4 x 2 + 18 x + 6
∫ ( x + 5)( x + 1)
2
dx
1 3 –2 = + + 2 dx x + 5 x + 1 ( x + 1) = ln |x + 5| + 3 ln |x + 1| + 2(x + 1)− 1 + C
∫
16.
3 x 2 – 53 x + 245 dx x 3 – 14 x 2 + 49 x 5 –2 3 = + + 2 dx x x – 7 ( x – 7) = 5 ln |x| − 2 ln |x − 7| − 3(x − 7)− 1 + C
∫
∫
∫x
∫
dx dx = – 6 x + 12 x – 8 ( x – 2)3 1 = − ( x – 2) −2 + C 2 1 dx 18. dx = ( x + 1) 4 x 4 + 4x3 + 6x2 + 4x + 1 1 = − ( x + 1) −3 + C 3 17.
3
2
∫
234
dy 1000 – y 1000 dy = 2y ⇒ = 2 dt dt 1000 y(1000 – y) 1000 dy = 2 dt y(1000 – y)
∫
∫
14.
19. a.
∫
Problem Set 9-7
∫
1
1
ln |y| − ln |1000 − y| = 2t + C ln
y = 2t + C 1000 – y
y = e 2 t + C (Note that 0 ≤ y < 1000.) 1000 – y 1000 − y 1000 = e −2 t −C ⇒ − 1 = e −2 t −C y y 1000 = 1 + e −2 t −C = 1 + ke −2 t ( k = e − C ) y 1000 y= 1 + ke –2 t Initial condition y = 10 when t = 0 ⇒ k = 99. 1000 y= 1 + 99e –2 t 1000 b. y(1) = = 69.4531… ≈ 69 students 1 + 99e –2 have heard the rumor after one hour. 1000 y( 4 ) = = 967.8567… ≈ 968 1 + 99e –8 students have heard by lunchtime. 1000 y(8) = = 999.9888… ≈ 1000 1 + 99e –16 students—everyone knows by the end of the day! c. It is quicker to analyze the original differential equation, which already refers to the derivative, than to analyze the equation found in part a. 1000 – y Maximize y′ = 2 y 1000 1 2 = (1000 y – y ). 500 1 y ′′ = (1000 y′ – 2 yy′) = 0 when y = 500. 500 This is the maximum point because y″ > 0 for y < 500 and y″ < 0 for y > 500 (and y′ > 0 for all t). So the rate of spreading (y′) is greatest when 500 students have heard the news. This occurs when 1000 500 = 1 + 99e –2 t 99e −2 t + 1 = 2 1 e −2 t = 99 1 t = ln 99 = 2.2975… hr 2 Calculus Solutions Manual © 2005 Key Curriculum Press
d. The graph follows the slope-field pattern. y 1000
2
4
6
t
20. a. Assume that an infected person and an uninfected person have about the same chance of meeting any other infected person (i.e., infected people are not quarantined). An infected person can meet N − P uninfected people out of the total population, so the chance of meeting an uninfected person will be (N – 1)/N, so of an average infected person’s three contacts per day, 3(N − 1)/N of them will be with uninfected persons. (Actually (N − P)/(N − 1) because the total population that someone can meet is N − 1—people don’t meet themselves outside the Twilight Zone— but (N − P)/N is reasonably close enough for now.) So there are P infected people, each meeting an average of 3(N − P)/N uninfected people per day, for a grand total of 3P(N − P)/N contacts between infected and uninfected people per day. b. If 10% of the contacts with infected people per day result in infection, then the number of new infections per day should be 0.1 times the number of contacts between infected and uninfected people, that is, dP 3 P( N – P ) N–P = 0.1 ⋅ = 0.3P . dt N N dP N–P N dP c. = 0.3P ⇒ = 0.3 dt dt N P( N – P ) N dP = 0.3 dt P( N – P ) 1 + 1 dP = 0.3 dt P N – P ln |P| − ln |N − P| = 0.3t + C P ln = 0.3t + C N–P P (Note that 0 ≤ P < N.) = e 0.3t +C N–P N = 1 + e −0.3t −C = 1 + ke −0.3t ( k = e − C ) P N P( t ) = 1 + ke –0.3t Initial condition P(0) = P0 N N ⇒ P0 = ⇒k= − 1. P0 1+ k N ∴ P( t ) = 1 + ( N / P0 – 1)e –0.3t
∫
∫
∫
Calculus Solutions Manual © 2005 Key Curriculum Press
∫
d. N = 1000 and P0 = 10 1000 ⇒ P( t ) = 1 + 99e –0.3t 1000 P(7) = = 76.2010… ≈ 76 people 1 + 99e –2.1 infected after 1 week e. Solve P(t) = 990. 1000 990 = 1 + 99e –0.3t 100 1 + 99e −0.3t = 99 e0.3 t = 992 2 t= ln 99 = 30.6341K ≈ 31 days 0.3 b b 25 5 –5 21. A = dx = + dx 2 2 x + 3x – 4 2 x –1 x + 4
∫
∫
= 5 ln | x − 1| −5 ln | x + 4|
b 2
= 5 ln
x –1 x+4
b 2
b –1 1 b –1 = 5 ln − 5 ln = 5 ln + 5 ln 6 b+4 6 b+4 6 A(7) = 5 ln + 5 ln 6 = 5.9281… 11 1 lim A(b) = lim 5 ln + 5 ln 6 b→∞ b→∞ 1 (l’Hospital’s rule) = 5 ln 6 = 8.9587… So the area does approach a finite limit. 50π x 22. dV = 2π x y dx = 2 dx x + 3x – 4 b b 10π 50π x 40π V= dx = + dx 2 2 x + 3x – 4 2 x –1 x + 4
∫
∫
= 10π ln | x − 1| + 40π ln | x + 4| 2b = 10π ln |b − 1| + 40π ln |b + 4| − 40π ln 6 V(7) = 40π ln 11 − 30π ln 6 = 132.4590… lim V (b) = ∞ because both ln terms become
b→∞
infinite and are added. (Note that if the region were rotated about the x-axis, the limit of the volume would be 35 finite. The answer would be 5π – 2 ln 6 6 = 35.3400… . x–3 1/2 1/2 23. a. + dx = dx 2 x – 6x + 8 x – 2 x – 4 1 1 = ln | x – 2| + ln | x – 4| + C 2 2 b. x 2 – 6x + 8 = (x – 3)2 – 1
∫
∫
v
x–3
√ (x – 3) 2 – 1
θ
u 1
Problem Set 9-7
235
Let x – 3 = sec θ. dx = sec θ tan θ dθ, ( x – 3) – 1 = tan θ , θ = sec ( x – 3) 2
∴ =
∫x ∫
2
–1
∫
x–3 x–3 dx dx = – 6x + 8 ( x – 3)2 – 1
sec θ (sec θ tan θ dθ ) = tan 2 θ
∫
sec 2 θ dθ tan θ
= ln | tan θ | + C = ln ( x – 3)2 – 1 + C = ln x 2 – 6 x + 8 + C c.
∫x
2
x–3 1 dx = – 6x + 8 2
∫x
2
2x – 6 dx – 6x + 8
1 ln | x 2 – 6 x + 8| + C 2 d. From part a, 1 1 ln | x – 2| + ln | x – 4| + C 2 2 1 = ln |( x – 2)( x – 4)| + C 2 1 = ln | x 2 – 6 x + 8| + C 2 which is the answer in part c. This equals =
ln | x 2 – 6 x + 8|1/2 + C = ln x 2 – 6 x + 8 + C, which is the answer from part b. So all three answers are equivalent, Q.E.D. 24. a. When the population is very much smaller than the maximum, (m – p) behaves like a constant, and dp/dt = k(m – p) · p is approximately proportional to p. But when p is 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 a quadratic function of p. Thus, the turning point is at m p=– = m/2. 2(–1) If k > 0, the graph of dp/dt versus p opens downward and the turning point is a maximum. So the population grows fastest when p = m/2. dp dp c. = kp( m – p) ⇒ = k dt dt p( m – p)
∫ p(m – p) = ∫ k dt dp
1/m 1/m + dp = k dt p m – p
∫
∫
1 1 ln | p| – ln |m – p| = kt + C1 m m 236
Problem Set 9-7
p = kmt + C2 (C 2 = mC 1) m– p p (C3 = e C2 ) = e kmt +C2 = C3e kmt m– p p > 0. Note that m > p > 0 ⇒ m– p m– p m – kmt – kmt (b = 1/C3 ) = be ⇒ – 1 = be p p m m = 1 + be – kmt ⇒ p = 1 + be – kmt p At time t = 0, p = p 0. m ∴ p0 = ⇒ m = p0 (1 + b) 1+ b p (1 + b) ∴ p = 0 – kmt 1 + be 1+ b Letting K = km, p = p0 , Q .E .D . 1 + be − Kt e. Let p denote millions of people. Then p0 = 179.3. Substitute p(10) = 203.2. 1+ b 203.2 = 179.3 1 + be –10 K ⇒ 203.2 + 203.2be–10K = 179.3 + 179.3b ⇒ b(203.2e–10K – 179.3) = –23.9 –23.9 ⇒b= 203.2e –10 K – 179.3 By substituting p(20) = 226.5 and transforming, –47.2 b= . 226.5e –20 K – 179.3 Equating the two values of b and solving numerically for K gives K = 0.0259109… . –23.9 ∴b= = 1.0630436 … 203.2e –0.259109... – 179.3 2.063036... ∴ p = 179.3 1 + 1.063036...e –0.0259109... Check that this equation gives a good approximation for 1990. 2.0630... p(30) = 179.3 ⋅ 1 + 1.0630... ⋅ e –30⋅0.0259... = 248.4892… ≈ 248.5 million people, which is close to the actual population, 248.7 million. 2.0630... f. p( 40) = 179.3 ⋅ 1 + 1.0630... ⋅ e –40⋅0.0259... = 268.6144… ≈ 268.6 million people, which is lower than the actual population by about 13 million people. 1+ b g. k > 0 ⇒ lim p = lim p0 = p0 (1 + b) t →∞ t →∞ 1 + be – kt = 179.3 · (1 + 1.0630…) = 369.9024… ≈ 369.9 million people d. ln
Calculus Solutions Manual © 2005 Key Curriculum Press
h. If p(10) had been 204.2, then K would have been given by –24.9 –47.2 = 204.2e –10 K – 179.3 226.5e –20 K – 179.3 ⇒ K = 0.0343965… –24.9 ⇒b= –0.0343965... – 179.3 204.2e = 0.721075… So the ultimate population would have been lim p = p0 (1 + b) = 179.3(1 + 0.7210 …)
3.
4.
= x tan −1
∫ cot
−1
+ –
x
x dx 1− x2
∫
−1
u sin –1x dx 1 √ – x2
x dx
∫
dv 1
+ –
x
x dx 1− x2
∫
1 (1 − x 2 ) −1/2 ( −2 x dx ) 2 1 = x sin −1 x + ⋅ 2(1 − x 2 )1/2 + C 2 −1 = x sin x + 1 − x 2 + C = x sin −1 x +
5.
+ –
dv 1 x
∫ sec
−1
u sec –1 x dx |x| √ x 2 – 1
x dx
= x sec −1 x −
∫ | x|
+ –
dv 1 x
x dx
x2 −1 dx = x sec −1 x − sgn x ( x/| x | = sgn x ) x2 −1
∫
∫ x +1 1 2 x dx x− ∫ 2 x +1 x dx 2
v
2
= x tan −1 x − 2.
u tan –1 x dx 1 + x2
x dx
= x tan −1 x −
∫ sin
= x sin −1 x −
integration by parts Q2. partial fractions x = tan θ or θ = tan–1 x x = sec θ or θ = sec–1 x x = sin θ or θ = sin–1 x 1 2 Q6. ( x + 1)8 + C 16 Q7. 7 (at f (1)). Q8. 3 (at f (5)). Q9. undefined Q10. B −1
∫
dv 1
1 (1 − x 2 ) −1/2 ( −2 x dx ) 2 1 = x cos −1 x − ⋅ 2(1 − x 2 )1/2 + C 2 = x cos −1 x − 1 − x 2 + C (Checks.)
Q1. Q3. Q4. Q5.
∫ tan
u cos –1 x dx √1 – x 2
x dx
= x cos −1 x −
Problem Set 9-8
1.
−1
= x cos −1 x +
t →∞
= 308.5888… ≈ 308.6 million people. An increase of 1 million in one of the initial conditions causes a decrease of over 61 million in the predicted maximum population! So this model does have a fairly sensitive dependence on the initial conditions.
∫ cos
1 ln | x 2 + 1| + C (Checks.) 2
√x 2 – 1
x θ
u 1
u cot –1 x dx 1 + x2
x dx
∫ x +1 1 2 x dx x+ ∫ 2 x +1
= x cot −1 x + = x cot −1
= x cot −1 x +
x dx 2
2
1 ln | x 2 + 1| + C (Checks.) 2
+ –
dv 1 x
x = sec θ . 1 dx = sec θ tan θ dθ x 2 – 1 = tan θ θ = sec–1 x ∴ sec –1 x dx Let
∫
= x sec –1
Calculus Solutions Manual © 2005 Key Curriculum Press
sec θ tan θ dθ
∫ tan θ x – ∫ sgn x sec θ dθ
= x sec –1 x – sgn x
Problem Set 9-8
237
= x sec–1 x – sgn x ln |sec θ + tan θ| + C
Simpson’s rule for y = sec− 1 x: n = 10 ⇒ ∆x = 0.2 0.2 A≈ ( y1 + 4 y1.2 + 2 y1.4 + 4 y1.6 + L + 4 y2.8 + y3 ) 3 = 1.919692K
= x sec–1 x – sgn x ln x + x 2 – 1 + C (Checks.) Note: This answer can be transformed to x sec –1 x – ln (| x | + x 2 – 1 ) + C. 6.
∫ csc
−1
x dx
u x
csc –1 dx |x| √ x 2 – 1
= x csc –1 x +
∫ | x|
x dx
x2 – 1 dx = x csc –1 x + sgn x x2 – 1
∫
1
θ
u
√ x2 – 1
csc θ cot θ dθ cot θ
= x csc x + sgn x ln | csc θ + cot θ | + C
x
= 3 sec −1 3 − 1 ⋅ ln (3 + 8 ) − sec −11 + 1 ⋅ ln 1
π /2 0
∫
4
1 ln | x 2 + 1| 1 2 1 1 = 4 tan −1 4 − ln 17 − tan −11 + ln 2 2 2 π 1 17 −1 = 4 tan 4 − − ln = 3.4478K 4 2 2 tan −1 x dx = x tan −1 x −
Numerically,
∫
4
1
π /2
y
x 1
238
4 1
tan −1 x dx = 3.4478K .
8.
Problem Set 9-8
3
∫
π /2
sin y dy
0
= − cos
1
V = 2π
∫ x tan
∫ x tan
x dx
−1
x dx
0
−1
u tan –1 x dx 1 + x2
2
(Checks.) Note: This answer can be transformed to x sec −1 x + ln (| x | + x 2 – 1 ) + C. 7.
1
= 1.930131… . The Simpson’s rule answer differs from this by 0.0104… , or about 0.5%. 9. By vertical slices, 1 π A = – sin –1 x dx 0 2 1 π = x − x sin −1 x − 1 – x 2 2 0 π −1 = − sin 1 − 0 − 0 + 0 + 1 = 1 2
= x csc x + sgn x ln x + x – 1 + C –1
3
= x sec −1 x − sgn x ln ( x + x 2 – 1 )
π + cos 0 = 1, which is the 2 same answer as by vertical slices. 10. By cylindrical shells, dV = 2π x tan− 1 x dx.
∫ x – ∫ sgn x csc θ dθ
= x csc –1 x – sgn x
–1
x dx
dv 1
= − cos y
∴ csc –1 x dx
= x csc –1
−1
1
By horizontal slices, A =
Let x = csc θ. dx = –csc θ cot θ dθ, x 2 – 1 = cot θ , θ = csc –1 x
∫
+ –
3
∫ sec
∫
v
x
A=
+ –
dv x 1 2 2x
x2 1 2 1 x tan −1 x − dx 2 2 1+ x2 1 1 1 = x 2 tan −1 x − dx 1− 2 2 1+ x2 1 1 1 = x 2 tan −1 x − x + tan −1 x + C 2 2 2 1 2 −1 ∴ V = π x tan x − π x + π tan −1 x 0 =
∫ ∫
= π tan −1 1 − π + π tan −1 1 − 0 + 0 − 0 π = 2π tan −1 1 − π = 2π ⋅ − π 4 1 2 = π − π = 1.7932 K 2 Compare this with a cylinder (π r2h) minus a cone (π r2h/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 is expected because the cylinder minus the cone is generated by rotating a line that lies below the graph.
Calculus Solutions Manual © 2005 Key Curriculum Press
Problem Set 9-9
7. x 5 1 Q3. tan 3 x + C 3 Q5. ln |x| + C Q7. False
Q1. x = 5 tan θ or θ = tan −1 Q2. xex − ex + C Q4. 2x1/ 2 + C Q6. reduction formula Q8.
8. 9. 10.
dx 2 + dy 2 or dr 2 + (rd θ )2
Q9. (1 − x2)−1/ 2 1.
Q10. C
11.
y
y
12.
cosh 1
1
x
d (csch x sin x ) dx = −csch x coth x sin x + csch x cos x d (tan x tanh x ) = sec 2 x tanh x + tan x sech2 x dx 1 sech 2 4 x dx = tanh 4 x + C 4 1 sech 7 x tanh 7 x dx = − sech 7 x + C 7 d 3 2 ( x coth x ) = 3 x coth x − x 3 csch 2 x dx d 2.5 ( x csch 4 x ) dx = 2.5x1.5 csch 4x − 4x2.5 csch 4x coth 4x
∫ ∫
x
1
1
tanh
13.
sinh
∫
3
∫
4
1
y
14.
sech 1
x 1
csch
2. y
y
1
cosh–1
–1
1
–1
x
3 1
= ln (cosh 3) − ln (cosh 1) = 1.875547…
coth
1
tanh x dx = ln (cosh x )
coth–1 x 1 tanh
−4
sinh x dx = cosh x
4 −4
=0
(Note that sinh is an odd function.) d sinh 5 x 15. dx ln 3 x 5 cosh 5 x ln 3 x – x –1 sinh 5 x = (ln 3 x )2 d cosh 6 x 16. dx cos 3 x 6 sinh 6 x cos 3 x + 3 cosh 6 x sin 3 x = cos 2 3 x
sinh
17.
y –1 sech –1
∫ x sinh x dx
u x 1 0
csch–1
1 1
+ – +
dv sinh x cosh x sinh x
x
= x cosh x − sinh x + C ∴ d tanh 3 x = 3 tanh 2 x sec h 2 x dx d 4. 5 sec h 3 x = −15 sec h 3 x tanh 3 x dx 1 5. cosh 5 x sinh x dx = cosh 6 x + C 6 1 −3 6. (sinh x ) cosh x dx = − (sinh x ) −2 + C 2 1 2 = − csch x + C 2 Or: (sinh x ) −3 cosh x dx = csch 2 x coth x dx
∫
∫
1 1 = − coth 2 x + C1 = − (csch 2 x + 1) + C1 2 2 1 2 = − csch x + C 2 Calculus Solutions Manual © 2005 Key Curriculum Press
0
= x cosh x − sinh x
3.
∫ ∫
1
∫ x sinh x dx
=e 18.
−1
∫x
2
1 0
= cosh 1 − sinh 1
= 0.36787… u x2 2x 2 0
cosh x dx
+ – + –
dv cosh x sinh x cosh x sinh x
= x 2 sinh x – 2x cosh x + 2 sinh x + C ∴
∫
b
x 2 cosh x dx
a
= x 2 sinh x − 2 x cosh x + 2 sinh x
b a
Problem Set 9-9
239
= b2 sinh b – 2b cosh b + 2 sinh b – a2 sinh a + 2a cosh a – 2 sinh a 12 d 19. (3 sinh –1 4 x ) = dx 16 x 2 + 1
∫ tanh =
22.
∫
−1
∫ ∴ 25 ∫ sinh
4 4 x cosh −1 6 x − [(6 x )2 – 1]1/2 + C 6 6 2 2 −1 = x cosh 6 x − 36 x 2 – 1 + C 3 3 23. Let x = 3 sinh t, dx = 3 cosh t dt, x 2 + 9 = 9 ⋅ sinh 2 t + 9 = 3 cosh t, x t = sinh −1 . 3 ∴ x 2 + 9 dx = 3 cosh t ⋅ 3 cosh t dt =
∫
∫
= 9 cosh 2 t dt
u cosh t sinh t
dv cosh t sinh t
+ –
∫ = 9 cosh t sinh t − 9 ∫ (cosh t − 1) dt = 9 cosh t sinh t − 9 ∫ cosh t + 9 ∫ dt ∴ 18 ∫ cosh t dt = 9 cosh t sinh t + 9t + C ∴ 9 ∫ cosh t dt = 4.5 cosh t sinh t + 4.5t + C
= 1 + ( y′)2 dx w w ⇒ d ( y′) = ds = 1 + ( y′)2 dx h h d.
2
∫ = 25 sinh t cosh t − 25 ∫ (sinh
u sinh t cosh t
= 25 sinh t cosh t − 25 cosh 2 t dt 2
t + 1) dt
+ –
−1/2
1
2
w x+C h At x = 0, y ′ = 0, so w sinh −1 0 = 0 + C ⇒ C = 0. h w w sinh −1 y′ = x ⇒ y′ = sinh x h h dy w w = sinh x ⇒ dy = sinh x dx dx h h h w ⇒ y = cosh x + C w h 1 y = 2 when x = 0 ⇒ 2 = k cosh 0 + C k ⇒2=k+C⇒C=2−k 4 y = 5 when x = 4 ⇒ 5 = k cosh + 2 − k k Using the solver feature of your grapher, k ≈ 3.0668… . ⇒ sinh −1 y′ =
e.
f.
∫
∫
−1/2
−1
1
2
= 25 sinh 2 t dt
2 −1/2
2
2
x2 + 9 x x = 4.5 ⋅ ⋅ + 4.5 sinh −1 + C 3 3 3 x = 0.5 x x 2 + 9 + 4.5 sinh −1 + C 3 24. Let x = 5 cosh t, dx = 5 sinh t dt, x 2 – 25 = 25 ⋅ cosh 2 t – 25 = 5 sinh t, x t = cosh −1 . 5 ∴ x 2 – 25 dx = 5 sinh t ⋅ 5 sinh t dt
∫ [1 + ( y′) ] d( y′) = ∫ (1 + sinh t ) d (sinh t ) = ∫ (cosh t ) (cosh t dt ) = ∫ dt = t + C = sinh y′ + C w w ∫ h dx = h x + C 2
2
Problem Set 9-9
t dt = 12.5 sinh t cosh t − 12.5t + C
c. ds = dx 2 + dy 2 = dx 2 [1 + ( dy/dx )2 ]
= 9 cosh t sinh t − 9 sinh 2 t dt
240
2
x 2 – 25 x x ⋅ − 12.5 cosh −1 5 5 5 x = 0.5 x x 2 – 25 − 12.5 cosh −1 + C 5 25. a. Figure 9-9g shows that the horizontal force is given by the vector (h, 0) and the vertical force is the vector (0, v), so their sum, the tension vector, is the vector (h, v), which has v slope . Because the tension vector points h along the graph, the graph’s slope, y ′, also v equals . h b. v = weight of chain below (x, y) = s ⋅ w v s⋅w w ⇒ y′ = = = ⋅s h h h
1 1 x tanh −1 5 x + ln |1 − (5 x )2 | + C 5 10 4 cosh −1 6 x dx
∫
2
= 12.5 ⋅
5x dx
∫
∫
∴ 50 sinh t dt = 25 sinh t cosh t − 25t + C1
15 x 2 d 20. (5 tanh –1 x 3 ) = 1 – x6 dx 21.
∫
= 25 sinh t cosh t − 25 sinh 2 t dt − 25 dt
g. dv sinh t cosh t
26. a.
∫
∫
Calculus Solutions Manual © 2005 Key Curriculum Press
1 x + 2 − 3.0668… 3.0668… 1 y = 3.0668… cosh x − 1.0668… 3.0668… y (20) = 1040.9739… y = 4: 1 4 = 3.0668… cosh x − 1.0668… 3.0668… 1 5.0668K cosh x= 3.0668K 3.0668K −1 5.0668… x = 3.0668… cosh = 3.3355… 3.0668… By symmetry, x = ±3.3355… . The answer can be found numerically using the solver feature of your grapher. 1 y′ = sinh x; y′(3) = 1.1418… k 3 1 A = k cosh x + 2 – k dx –1 k y = 3.0668… cosh
b. c.
d. e.
∫
= k 2 sinh
1 x + (2 − k ) x k
3 −1
3 –1 = (3.0668…)2 sinh − sinh 3.0668… 3.0668… + 4(2 − 3.0668…) = 9.5937… f.
3
∫ =∫
L=
−1 3
−1
=
∫
1 + ( y′)2 dx 1 + sinh 2 ( x/k ) dx
3
−1
cosh
1 1 x dx = k sinh x k k
3 −1
3 1 = k sinh + sinh = 4.5196 … k k 27. a. The vertex is midway between the poles, so y = 110 ft when x = 150 ft. h w y = cosh x + C w h 400 lb 0.8 = cosh x+C 0.8 lb / ft 400 1 110 = 500 cosh ⋅ 150 + C 500 ⇒ C = 110 − 500 cosh 0.3 1 y = 500 cosh x + 110 − 500 cosh 0.3 500 The cable comes closest to the ground at x = 0. y (0) = 500 cosh 0 + 110 − 500 cosh 0.3 = 610 − 500 cosh 0.3 = 87.3307… ≈ 87.3 ft
Calculus Solutions Manual © 2005 Key Curriculum Press
b. y′ = sinh
∫ =∫
L=
150
–150 150
1 x 500 1 + sinh 2 ( x/500) dx
1 x dx 500 150 1 = 500 sinh x 500 −150 = 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) (Compare Problem 25.) 1 Weight = 2 ⋅ 400 sinh 150 = 800 sinh 0.3 500 = 243.6162… ≈ 243.6 lb (Note: Because w ⋅ L = weight, either of these methods could give both the weight and the length.) −150
cosh
c. T = h 2 + v 2 ; h is constant and v is greatest at the ends, so the maximum tension is at x = 150 ft. T (150) = h 2 + [hy′(150)]2 = 400 1 + sinh 2 0.3 = 400 cosh 0.3 = 418.1354… ≈ 418.1 lb h w d. The general equation is y = cosh x + C. w h If y (0) = 100 and y (150) = 110, find h such that y (150) − y (0) = 10. Solve: h w h cosh 150 − = 10, or w h w 120 8 cosh −1 = h h By grapher, h = 901.3301... ≈ 901.3 lb. 28. The answers will depend on the dimensions of the chain used. Note that the answer is independent of the kind of chain. You might show students how a heavy chain and a light chain of equal length will hang in the same catenary if they are suspended from the same points. Assume that the dimensions are the same as in Example 5. a. Vertex: (0, 20). Supports: (±90, 120). 1 x – 31.78K b. y = 51.78… cosh 51.78K c. Note: To conserve class time, you might have students plot only each 20 cm for x, as shown here for Example 5. Use the TABLE feature.
Problem Set 9-9
241
x
y
0 ±20 ±40 ±60 ±80
20.0 23.9 36.2 58.8 95.1
c. Slice perpendicular to the y-axis. dV = π x 2 dy = π (sinh–1 y)2 dy Top of bowl is at 1 y = sinh 1 − = 1.133534K . 24 ∴ V =π
d. A clever way to make sure the measurements are vertical is to hold a book against the board with its bottom edge along the chalk tray. Then hold the meterstick against the vertical edge of the book. It is crucial that the points be plotted accurately to get the dramatic impact of “perfect fit.” e. For a quadratic function with vertex on the y-axis, y = ax2 + c. Using the data for Example 5, 20 = a(0) + c ⇒ c = 20 1 120 = a(90 2 ) + 20 ⇒ a = 81 1 2 y= x + 20 81 y
∫
1.133K
= 1.25317… by numerical integration ≈ 1.253 ft3 1 x+C k Inner catenary: yinner(0) = 612, yinner(260) = 0 0 612 = − ki cosh + Ci ⇒ Ci = 612 + ki ki 260 0 = − ki cosh + 612 + ki ki ⇒ ki = 97.1522… (numerically) 1 yinner = 97.1522… − cosh x + 1 + 612 97.1522 K Outer catenary: youter (0) = 630, youter (315) = 0 0 630 = −ko cosh + C o ⇒ C o = 630 + k o ko
30. a. y = − k cosh
315 + 630 + k o ko ⇒ ko = 127.7114… (numerically) 1 youter = 127.7114 K − cosh x + 1 + 630 127.7114 K 0 = −ko cosh
parabola
catenary 10
x 20
b. The graphs are the same as in Figure 9-9k:
The parabola is more curved at the vertex. f. For Example 5, dL = 1 + ( y′)2 dx = 1 + sinh 2
y
1 x dx 51.78K
1 x dx 51.78K 90 1 L= cosh x dx −90 51.78K 90 1 = 51.78Ksinh x = 285.349K 51.78K −90 ≈ 285.3 cm The actual length should be close to this. 29. a. y = sinh x
100
= cosh
∫
dS = 2π x dL = 2π x 1 + cosh 2 x dx S = 2π
1
∫x 0
1 + cosh 2 x dx
= 5.07327… by numerical integration ≈ 5.07 ft2 b. Cost = 2(57)(5.07327…) = 578.3532… ≈ $578.35
242
Problem Set 9-9
(sinh –1 y)2 dy
0
x 100
c. A = =
∫
315
−315
− ko2 +
youter dx −
∫
260
−260
yinner dx
x sinh + ko x + 630 x ko
ki2
x sinh − ki x − 612 x ki
315 −315 260 −260
= 54323.2729… ≈ 54,323 ft2 dyouter 1 d. = − sinh x, so dx ko L= =
∫
∫
315
−315
1 + sinh 2
315
−315
cosh
1 x dx ko
1 1 x dx = ko sinh x ko ko
315 −315
Calculus Solutions Manual © 2005 Key Curriculum Press
315 127.7114 K = 1493.7422… ≈ 1494 ft. 315 −315 e. youter = sinh ′ ( −315) = − sinh 127.7144K ko
exponential form or by transforming the second to hyperbolic form, as shown here.) 1 3x 1 − x 1 1 e + e + C = e x e 2 x + e −2 x + C 6 6 2 2
= 2 ⋅ 127.7114 K ⋅ sinh
= 5.8481… is the spider’s starting slope. f. José can fly through at altitude yinner(x) if x ≥ 50 + 120/2 = 110. yinner(110) = 542.7829… , so the plane can fly through at heights between 0 and 542 feet. (Because of the curvature of the arch and the vertical thickness of the plane, the closest distance is slightly less than 50 feet when the horizontal distance is 50 feet. The plane can fly through at slightly higher altitudes by banking slightly.) 31. a. H(x) = csch x ⇒ H′(x) = −csch x coth x H′(1) = −csch 1 coth 1 = −1.1172855… csch (1.01) – csch (0.99) b. H′(1) ≈ 0.02 = −1.11738505… . The answers differ by 0.0000995… , which is about 0.0089% of the actual answer. 32.
∫
2
1
sech x dx = sin −1 ( tanh x )
2
∫ sech x dx = 0.435990K 1
(Checks.) 33. By parts: x
u sinh 2x 2 cosh 2x 4 sinh 2x
+ – +
x
u ex ex ex
sinh x dx
∫
⇒ −3 e sinh 2 x dx x
∫
= e sinh 2 x − 2e cosh 2 x + C1 ⇒ e sinh 2 x dx x
x
2 1 = e x cosh 2 x − e x sinh 2 x + C 3 3 By transforming to exponential form: 1 x 2 x −2 x e x sinh 2 x dx = e (e − e ) dx 2 1 1 1 = (e 3 x − e − x ) dx = e 3 x + e − x + C 2 6 2 Transforming to exponential form is easier! (Note that the two answers can be shown to be equivalent either by transforming the first to
∫
∫
∫
Calculus Solutions Manual © 2005 Key Curriculum Press
+ – +
dv sinh x cosh x sinh x
∫
= e x cosh x − e x sinh x + e x sinh x dx
∫
⇒ 0 e sinh x dx = e (cosh x − sinh x ) + C x
x
The original integral reappeared with the same coefficient, so when it was added again to the left side, it exactly canceled out the desired integral. Use the exponential form of sinh x. 1 x x e x sinh x dx = e (e − e − x ) dx 2 1 = (e 2 x − 1) dx 2 1 1 = e2 x − x + C 4 2 2 35. a. cosh x − sinh2 x
∫
ex + e– x ex – e– x = − 2 2
dv ex ex ex
= e x sinh 2 x − 2e x cosh 2 x + 4 e x sinh 2 x dx
∫
∫e
2
sinh 2 x dx
x
34.
∫
= sin–1 (tanh 2) − sin–1 (tanh 1) = 0.435990…
∫e
2 x e 2 x + e –2 x 1 x e 2 x – e –2 x e − e +C 3 2 3 2 2 1 = e x cosh 2 x − e x sinh 2 x + C 3 3
=
∫
2 1
Numerically,
1 1 1 1 = e x e 2 x + e −2 x − e 2 x + e −2 x + C 3 3 6 6
2
e 2 x + 2 + e –2 x e 2 x – 2 + e –2 x − =1 4 4 ∴ cosh2 x − sinh2 x = 1, Q .E.D . =
b.
1 1 (cosh 2 x – sinh 2 x ) = cosh 2 x cosh 2 x
⇒ 1 − tanh2 x = sech2 x 1 1 c. (cosh 2 x – sinh 2 x ) = sinh 2 x sinh 2 x ⇒ coth2 x − 1 = csch2 x 36. a. Substitute 2x for x in the definition of sinh x. 1 b. sinh 2 x = (e 2 x – e –2 x ) 2 1 1 = 2 ⋅ (e x – e – x ) ⋅ ⋅ (e x + e − x ) 2 2 = 2 sinh x cosh x
Problem Set 9-9
243
1 2x (e + e –2 x ) 2 1 + 1 + e –2 x ) + ( e 2 x – 1 + e –2 x ) 4
c. cosh 2 x = =
1 2x (e 4
d. Slice as in part b.
∫
Let l = sinh 2 t dt
u sinh t cosh t
2
2 1 1 = (e x + e – x ) + (e x – e – x ) 2 2 2 = cosh 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 1 ⇒ sinh 2 x = (cosh 2 x – 1) 2 37. a. On the circle, u2 + v2 = 1 ⇒ 2u du + 2v dv = 0 ⇒ dv = (−u/v) du.
dL = du 2 + dv 2 = du 2 + (u/v)2 du 2 = L=
∫
1 1 u2 + v2 du = du = du 2 v v 1 – u2 1 du
cos 2
1 – u2
= − cos −1 u
1 cos 2
= − cos −1 1 + 2 = 2
The curve along the hyperbola from u = 1 to u = cosh 2 has length greater than the line segment along the horizontal axis from (1, 0) to (cosh 2, 0). This segment has length L = cosh 2 − 1 = 2.762… . So the length of the curve is greater than 2, Q.E.D. b. The area of the triangle that circumscribes the sector is 0.5(2 sinh 2 cosh 2) = sinh 2 cosh 2. The area of the sector is the area of this triangle minus the area of the region between the upper and lower branches of the hyperbola from 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 the sample point. 0 ≤ t ≤ 2. dA = 2v du = 2 sinh t d(cosh t) = 2 sinh2 t dt A=2
2
∫ sinh
2
0
t dt ≈ 11.644958K
Thus, the area of the sector is cosh 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, and its area is π. Thus, 2x Asector = π = x , Q .E .D . 2π
244
Problem Set 9-9
+ –
dv sinh t cosh t
∫ = sinh t cosh t − ∫ (1 + sinh t ) dt = sinh t cosh t − t − ∫ sinh t dt = sinh t cosh t − cosh 2 t dt 2
2
∫
∴ 2 sinh 2 t dt = sinh t cosh t − t + C
∫ sinh
2
t dt = 0.5 sinh t cosh t − 0.5t + C1
Slicing as in part b, the area A between the upper and lower branches of the hyperbola is A=2
x
∫ sinh
2
t dt
0
= 2(0.5 sinh x cosh x − 0.5 x ) − 0 = sinh x cosh x − x Thus, the area of the sector is cosh 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 1 1 1 , Q. E . D . y′ = = = 2 2 cosh y x +1 sinh y + 1 b. y = tanh− 1 x ⇒ tanh y = x ⇒ sech2 y y′ = 1 1 1 1 y′ = , Q.E.D. 2 = 2 = sec h y 1 – tan h y 1 – x 2 c. y = coth− 1 x ⇒ coth y = x ⇒ −csch2 y y′ = 1 1 1 1 y′ = = , 2 = 2 – csch y –(coth y – 1) 1 – x 2 Q .E .D . d. y = sech− 1 x ⇒ sech y = x ⇒ −sech y tanh y y′ = 1 1 y′ = – sech y tanh y 1 = – sech y 1 – sech 2 y =−
1 x 1 – x2
, Q. E . D .
e. y = csch− 1 x ⇒ csch y = x ⇒ −csch y coth y y′ = 1 1 1 = y′ = – csch y coth y – csch y 1 + csch 2 y =−
1 | x| 1 + x 2
, Q .E .D .
Calculus Solutions Manual © 2005 Key Curriculum Press
Problem Set 9-10
4. a.
y
Q2. sinh x + C
Q1. y
1
x 1
1
x 1
Q3. Q5. Q7. Q9. 1.
sinh x sin x + C y = tan x y = ex y a.
Q4. Q6. Q8. Q10.
−sin x y = x3 y = sinh x or x 3 + x A
It might converge because the integrand becomes infinite only as x approaches zero. 1
b.
1
∫ (1/x ) dx = lim ∫ (1/x ) dx 0
= lim+ ln | x | a→ 0
1
a→ 0 1 a
+
a
= lim+ (0 – ln a) = ∞ a→ 0
The integral diverges. 5. a.
y
x 2
It might converge because the integrand approaches zero as x approaches infinity. b.
∫
∞
b
∫ (x
(1/ x ) dx = lim 2
−2
b→∞ 2
2
b
2
b→∞
1 2
1 Integral converges to . 2 2. a.
x
1
) dx
= lim −x −1 = lim( −1/b + 1/2) = b→∞
1
It might converge because the integrand approaches zero as x approaches infinity. b.
∫
∞
1
1/ x 0.2 dx = lim
= lim 1.25 x
y
b→∞ 1 b 0.8
b→∞
1
∫
b
x −0.2 dx
= lim (1.25b 0.8 – 1.25) = ∞ b→∞
1
The integral diverges. 6. a.
y
x 3 1
It might converge because the integrand approaches zero as x approaches infinity. b.
∫
∞
3
1/ x 4 dx = lim
∫
b
b→∞ 3 b −3
x −4 dx
1 1 1 = lim – 3 + = b →∞ 3 b 81 81 3 1 The integral converges to . 81 1 = lim − x b→∞ 3
3. a.
x 1
It might converge because the integrand approaches zero as x approaches infinity. b.
∫
∞
1
1/ x 1.2 dx = lim
b
b→∞ 1
= lim − 5 x −0.2
y
∫
b→∞
b 1
x −1.2 dx
= lim (–5b –0.2 + 5) = 5 b→∞
The integral converges to 5. 7. a.
y
1
x 1
It might converge because the integrand approaches zero as x approaches infinity. b.
∫
∞
1
(1/ x ) dx = lim
b→∞
b 1
x 1
b
∫ (1/x ) dx
b→∞ 1
= lim ln | x |
1
= lim (ln b – 0) = ∞ b→∞
It might converge because the integrand becomes infinite only as x approaches zero.
The integral diverges. Calculus Solutions Manual © 2005 Key Curriculum Press
Problem Set 9-10
245
b.
∫
1
0
1/ x 0.2 dx = lim+ a→0 1
= lim+ 1.25 x 0.8 a→ 0
a
1
∫x
−0.2
11. a.
dx
a
y 1
x
0
1
= lim+ (1.25 – 1.25a 0.8 ) a→ 0
= 1.25 The integral converges to 1.25. 8. a.
It might converge because the integrand becomes infinite only as x approaches 0 or 1. b. To determine whether this converges, split the integral into two pieces. Each piece must converge in order for the integral to converge. The integral can be written
y
1
x 1
1
∫ 1/( x ln x ) dx = ∫ 1/( x ln x ) dx + ∫ 1/( x ln x ) dx = lim ∫ 1/( x ln x ) dx + lim ∫ 1/( x ln x ) dx 0
It might converge because the integrand becomes infinite only as x approaches zero. 1
b.
∫ 1/x
1.2
dx = lim+ a→0
0
a→ 0
∫x
−1.2
c
dx
a→0
a
+
b
−
b→ 1
a
a
c
= lim+ ln | ln x | + lim− ln | ln x |
= lim+ (–5 + 5a –0.2 ) = ∞
a→0
a→0
b→1
a
= lim+(ln |ln c | – ln |ln a |)
b c
a→ 0
+ lim−(ln |ln b | – ln |ln c |)
y
b→1
=∞+∞ For the integral to converge, both limits must exist. Because neither exists, the integral diverges.
1
x 0
c
0
The integral diverges. 9. a.
1
c
1
= lim+ − 5 x −0.2
1
c
12. a.
1
y
It might converge because the integrand approaches zero as x approaches infinity. b.
∫
∞
1/(1 + x 2 ) dx = lim
∫
b
b→∞ 0
0
= lim tan −1 x b→∞
b 0
1
1/(1 + x 2 ) dx
= lim (tan –1 b – 0) = b→∞
x
π 2
The integral converges to π /2. 10. a.
0
1
3
It might converge because the integrand approaches zero as x approaches infinity. b. The indefinite integral can be written (ln x ) −2 ( dx/ x ) = −(ln x ) −1 + C.
∫ ∫
1
∞
3
1/[ x (ln x )2 ] dx = lim
b→∞
b 3
= lim [–(ln b) –1 + (ln 3) −1 ] = (ln 3) −1
x
1
2
b→∞ 3
= lim − (ln x ) −1 0
b
∫ 1/[ x(ln x ) ] dx
b→∞
It might converge because the integrand approaches zero as x approaches infinity. b.
∫
∞
0
1/(1 + x ) dx = lim
∫
b
b→∞ 0 b
1/(1 + x ) dx
The integral converges to (ln 3)− 1 = 0.910239… . 13. a. y 1
= lim ln | x + 1| = lim [ln (b + 1) – 0] = ∞ b→∞
0
b→∞
The integral diverges.
x 2
It might converge because the integrand approaches zero as x approaches infinity. 246
Problem Set 9-10
Calculus Solutions Manual © 2005 Key Curriculum Press
b.
∫
∞
2
b
∫e
e −0.4 x dx = lim
= lim − 2.5e
−0.4 x
b→∞ 2 b −0.4 x
b→∞
17. a.
dx
y 1
2
= lim(–2.5e −0.4 b + 2.5e –0.8 ) = 2.5e −0.8 b→∞
The integral converges to 2.5e− 0.8 = 1.1233… .
x 0
14. a.
It might converge because the integrand seems to approach zero as x approaches infinity. b. Integrate by parts: xe − x dx = − e − x ( x + 1) + C
y 1
0
1
∫ ∫
xe − x dx = lim
∫
= lim − e − x ( x + 1) b→∞
It diverges because the integrand does not approach zero as x approaches infinity. b. (Not applicable) 15. a.
b
b→∞ 0
0
x
1
∞
xe − x dx
b
= lim [– e – b (b + 1) + 1] = 1 b→∞
0
(The first term is zero by l’Hospital’s rule.) Integral converges to 1. 18. a. y
y
1
1
x
x –1
0
1
0
2
It does not converge because the integrand is undefined for x < 0. b. (Not applicable) 16. a.
3
It might converge because the integrand becomes infinite only as x approaches 1. 3
b.
∫ ( x – 1) dx = lim ∫ ( x – 1) −2
0
b
y
b→1
–
−2
0
dx + lim+ a→1
b
3
∫ ( x – 1)
= lim– − ( x − 1) −1 + lim+ − ( x − 1) −1 b→1
0
a→1 –1
−2
dx
a
3 a
= lim– [–(b – 1) + (–1) ] –1
1
b→1
+ lim+ [–2 –1 + ( a – 1) –1 ]
x 1
0
3
a→1
7
=∞+∞ For the integral to converge, both limits must exist. Because neither exists, the integral diverges.
It might converge because the integrand becomes infinite only as x approaches 3. 7
b.
∫ ( x – 3) dx = lim ∫ ( x – 3) −2/3
19. a.
1
b
−2/3
b→3− 1
= lim− 3( x − 3)
1/3
b→3
dx + lim+ a→3
b 1
7
∫ ( x – 3)
dx
a
+ lim+ 3( x − 3)1/3 a→3
−2/3
y 1
7 a
x 20
= lim− [3(b – 3)1/3 − 3( −2)1/3 ] b→ 3
+ lim+ [3( 4)1/3 − 3( a − 3)1/3 ] a→3
= 3 ⋅ 21/3 + 3 ⋅ 41/3 The integral converges to 3 ⋅ 21/3 + 3 ⋅ 41/3 = 8.5419… .
Calculus Solutions Manual © 2005 Key Curriculum Press
It diverges because the integrand does not approach zero as x approaches infinity. b. (Not applicable)
Problem Set 9-10
247
24. y = −1/x ⇒ x = −y − 1 Slice the vertical cross section horizontally. dA = x dy = −y − 1 dy
20. a. y 1
x
∫
A = lim
20
–1
a→−∞ a
− y −1 dy = lim − ln | y| a→ – ∞
−1 a
= lim (– ln | – 1| + ln |a|) = ∞ a→−∞
It diverges because the integrand does not approach zero as x approaches infinity. b. (Not applicable) b
∫ cos x dx oscillates between −1 and
21. As b → ∞,
0
1 and never approaches a limit. Similarly, b
∫ sin x dx oscillates between 0 and 2. a. I = ∫ 1/ x dx = lim ∫ x dx 0
∞
22.
1.001
b
1.001
b→∞ 1 b
1
= lim −1000 x
−1.001
−0.001
b→∞
= lim (–1000 b
–0.001
∫
1
b→∞
∫
b
b→∞ 1 b
1
b→∞
x −0.999 dx
f (2) = lim
1
b→∞
c. “Ip converges if p > 1 and diverges if p ≤ 1.” 23. a. y = 1/x = x − 1 dA = y dx = x− 1 dx
∫
1
−1
x dx = ∞ (See Problem 3)
∫
b
b→∞ 1
π x −2 dx = lim −π x −1 b→∞
b 1
= lim (–π b –1 + π ) = π b→∞
V = lim
∫
b
b→∞ 1
2π dx = lim 2π x b→∞
= lim (2πb – 2π ) = ∞
b 1
b→∞
Volume diverges. d. False. The volume could approach a constant as in part b or become infinite as in part c.
248
Problem Set 9-10
0
+2
∫
∞
te − t dt
0
b→∞
(Using l’Hospital’s rule on b2e− b gives 2b 2 lim b 2 e − b = lim b = lim b = 0.) b→∞ b→∞ e b→∞ e f (3) = lim
∫
∞
b→∞ 0
t 3e − t dt
= lim − t 3e − t b→∞
b 0
+3
∫
∞
t 2 e − t dt
0
= lim (– b 3e – b + 0) + 3(2) = 6 b→∞
(Using l’Hospital’s rule on b3e− b gives lim b 3e − b = lim
The volume converges to π. c. By cylindrical shells, dV = 2π xy dx = 2π x(x− 1) dx = 2π dx.
b
= lim (– b 2 e – b + 0) + 2(1) = 2
The area does not approach a finite limit. b. By plane slices, dV = πy2 dx = πx− 2 dx. V = lim
0
+ 0 + 1) = 1
t 2 e − t dt
= lim − t 2 e − t
diverges.
A=
∫
∞
b→∞ 0
1/ x dx = ∞ (see Problem 3), so I1
∞
b→∞
–b
b
(Using l’Hospital’s rule on be− b gives b 1 lim be − b = lim b = lim b = 0. ) b→∞ b→∞ e b→∞ e
= ∞ (diverges), Q.E.D. ∞
te − t dt = lim (– te – t – e – t )
= lim (– be – e
b→∞
∫
∫
b
–b
= lim (1000 b 0.001 – 1000) b. I1 =
t x e − t dt
0
f (1) = lim
+ 1000)
1/ x 0.999 dx = lim
= lim 1000 x 0.001
∫
∞
b→∞ 0
= 1000 (converges), Q.E.D. ∞
25. a. f ( x ) =
1
b→∞
I 0.999 =
The area of the bucket’s surface is greater than the area of the cross section, and the crosssectional area diverges. Thus, the bucket has infinite surface area. The bucket is congruent to the solid in Problem 23b, which has volume approaching π. Thus, π cubic units of paint would fill the bucket but could not coat the whole surface!
b→∞
b→∞
3b 2 6b 6 b = lim b = lim b = 0.) b→∞ e b→∞ e e
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 ) =
∫
∞
0
t x e − t dt
u tx
xt x–1
+ –
dv e –t –e –t
Calculus Solutions Manual © 2005 Key Curriculum Press
= lim − t x e − t b→∞
b 0
+x
∫
∞
t x −1e − t dt
0
= lim (– b x e – b + 0) + x f ( x − 1) b→∞
= 0 + 0 + x f (x − 1) = x f (x − 1), Q .E .D . ( lim (– b x e – b ) = 0 can be proved by b→∞
mathematical induction using l’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.
∫
1000
0
3 −t
t e dt ≈ 6
b
3 −t
0
dt
3 −b
2 −b
−b
−b
= − b e − 3b e − 6be − 6e + 6 |b3e− b + 3b2e− b + 6be− b + 6e− b| < 0.000001 for b > 23.4050… , say, b ≈ 24.
∫
f. 0.5! =
∞
0
t 0.5e − t dt ≈
∫
b→∞
The value of b that makes the integral come within 0.000001 of 6 can be found numerically (though it will be slow), or algebraically:
∫te
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. π = 0.886226925K , which agrees j. 0.5! = 2 with the tabulated value. 26. dW = F dr = 1000r− 2 dr At the earth’s surface, r = 1. ∞ b W = 1000 r −2 dr = lim –1000 r −1 1 b→∞ 1 = lim (–1000 b –1 + 1000)
∫
24
0
t 0.5e − t dt ≈
= 1000 radius-pounds Thus, the amount of work does not increase without bound as r goes to infinity. 3 x−2 x 27. a. 2 − dx 1 x−2
∫
b.
Error =
∫
t 0.5e − t dt <
24
∫
∞
t 3e − t dt < 0.000001
24
The difference between the tabulated value of 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 negative because the calculated value is larger than the tabulated value. This observation means that either the tabulated value is incorrect or there is more inaccuracy in the numerical integration algorithm than there is in the error caused by dropping the tail of the integral. (Using a smaller tolerance in the numerical integrator 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
b t 0 e − t dt = lim – e – t 0 b→∞
= lim(– e – b + 1) = 1, Q .E.D . b→∞
Calculus Solutions Manual © 2005 Key Curriculum Press
∫
2
∫
∫
∫
x−2 dx b→2 1 x−2 3 x−2 + lim+ 2 x − dx a→2 a x−2
0.886227311… From the graphs, t0.5 e− t < t3e − t for x ≥ 24. The error in 0.5! from stopping at b = 24 is the area under the “tail” of the graph from b = 24. ∞
3 x−2 x−2 x x dx 2 − dx + 2 − 1 2 x−2 x−2 2 3 x−2 x−2 x x dx dx + 2 − 2 − 1 2 x−2 x−2
=
= lim−
b
∫ 2
x
−
∫
c.
x−2 dx x−2 3 x−2 + lim+ 2 x − dx a→ 2 a x−2
lim−
b→2
b
∫ 2
x
1
−
∫
= lim− b→2
b
∫ (2 1
x
+ 1) dx + lim+ a→2
3
∫ (2 a
x
− 1) dx
b
= lim− (2 x /ln 2 + x ) + lim+ (2 x /ln 2 – x ) b→2
1
a→2
3 a
= lim− (2 /ln 2 + b – 2/ln 2 − 1) b
b→2
+ lim+ (2 3 /ln 2 – 3 – 2 a /ln 2 + a) a→ 2
= 4/ln 2 + 2 − 2/ln 2 − 1 + 8/ln 2 − 3 − 4/ln 2 + 2 = 6/ln 2 The integral converges to 6/ln 2 = 8.6561… . d. The integral is defined by dividing the interval into Riemann partitions and summing the subintervals. But the Riemann partitions may be chosen so that the discontinuities are at endpoints of subintervals. Then the subintervals corresponding to each continuous piece may be summed separately. e. False. Some discontinuous functions (notably, piecewise continuous functions) are integrable. 28. Answers will vary. Problem Set 9-10
249
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 3x 2. y = sinh 5x tanh 5x ⇒ y′ = (5 cosh 5x) tanh 5x + sinh 5x (5 sech2 5x) = 5 sinh 5x + 5 sinh 5x sech2 5x or 5 sinh 5x + 5 tanh 5x sech 5x 3.
4.
∫ x cosh 4 x dx
u x 1 0
=
1 1 x sinh 4 x − cosh 4 x + C 4 16
∫
x cos x dx
+ – +
dv cosh 4x 1 4 sinh 4x 1 16 cosh 4x
+ – +
∫
∫
250
∫
∫
∫
∫
18. g(t ) = t 2 – 1 = (t 2 − 1)1/2 ⇒ 1 t g′(t ) = (t 2 – 1) −1/2 (2t ) = 2 2 t –1
∫
1 + t 2 dt =
∫
∫
= sec 3 θ dθ =
dv cos x sin x –cos x
= x sin x + cos x + C 5. f (x) = (3x + 5)− 1 ⇒ f ′(x) = −3(3x + 5)− 2 6. f (x) = (5 − 2x− 1) ⇒ f ′(x) = 2(5 − 2x)− 2 1 7. (3 x + 5) −1 dx = ln |3 x + 5| + C 3 1 8. (5 − 2 x ) −1 dx = − ln |5 – 2 x | + C 2 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 1 11. sin 2 x dx = (1 − cos 2 x ) dx 2 1 1 = x − sin 2 x + C 2 4 1 1 = x − sin x cos x + C (or integrate by parts) 2 2 1 12. cos 2 x dx = (1 + cos 2 x ) dx 2 1 1 = x + sin 2 x + C 2 4 1 1 = x + sin x cos x + C (or integrate by parts) 2 2 6 x – 11 13. y = ⇒ x+2 6( x + 2) – (6 x – 11)(1) 23 y′ = = 2 ( x + 2) ( x + 2)2 5x + 9 ⇒ 14. y = x–4 5( x – 4) – (5 x + 9)(1) –29 y′ = = ( x – 4) 2 ( x – 4) 2
∫
17. f (t ) = 1 + t 2 = (1 + t 2 )1/2 ⇒ 1 t f ′(t ) = (1 + t 2 ) −1/2 (2t ) = 2 1+ t2
19. u x 1 0
∫ ∫
6 x – 11 23 dx = 6 − dx x+2 x + 2 = 6x – 23 ln | x + 2 | + C 5x + 9 29 16. dx = 5 + dx x–4 x – 4 = 5x + 29 ln | x − 4 | + C
15.
1 + tan 2 θ d ( tan θ ) 1 sec θ tan θ 2
1 + ln | sec θ + tan θ | + C 2 1 1 = t 1 + t 2 + ln 1 + t 2 + t + C 2 2 20.
∫
t 2 – 1 dt =
∫
∫
sec 2 θ – 1 d (sec θ )
∫
= sec θ tan 2 θ dθ = (sec 3 θ − sec θ ) dθ 1 1 sec θ tan θ + ln | sec θ + tan θ | 2 2 − ln | sec θ + tan θ | + C 1 1 = sec θ tan θ − ln | sec θ + tan θ | + C 2 2 1 1 2 = t t – 1 − ln t + t 2 – 1 + C 2 2 3 x 21. y = x e ⇒ y′ = 3x2ex + x3ex = x2ex(3 + x) 22. y = x4e− x ⇒ y′ = 4x3e− x − x4e− x = x3e− x(4 − x) =
23.
∫x e
3 x
u x3 3x 2 6x 6 0
dx
∫
Problem Set 9-11
+ – + – +
dv ex ex ex ex ex
= x3ex − 3x2 ex + 6xex − 6ex + C 24.
∫x e
4 −x
dx
u x4 4x 3 12x 2 24x 24 0
+ – + – + –
dv e –x –e –x e –x –e –x e –x –e –x
= −x4 e− x − 4x3 e− x −12x2 e− x − 24xe− x − 24e− x + C 1 25. f ( x ) = sin −1 x ⇒ f ′( x ) = = (1 − x 2 ) −1/2 2 1– x Calculus Solutions Manual © 2005 Key Curriculum Press
26. g( x ) = tan −1 x ⇒ g′( x ) = 27.
∫ sin
−1
1 x +1 2
u sin –1 x (1 – x 2)–1/2
x dx
+ –
dv 1 x
= x sin− 1 x − (−0.5)(2)(1 − x2)1/2 + C = x sin −1 x + 1 – x 2 + C
∫ tan
−1
u tan –1 x 1 1 + x2
x dx
= x tan −1 x −
∫ 1+ x 1
2
∫
(Absolute value is optional.) cosh x dx 36. coth x dx = = ln | sinh x | + C sinh x
( x dx )
+ –
dv 1 x
39.
1 ln |1 + x 2 | + C 2 1 –1/6 1/6 29. dx = + dx x + 5 x – 1 x2 + 4x – 5 1 1 = − ln | x + 5| + ln | x − 1| + C 6 6 1 1/8 –1/8 30. dx = + dx 2 x – 6x – 7 x + 1 x – 7 1 1 = − ln | x + 1| + ln | x − 7| + C 8 8 1 1 dx = dx 31. 2 x + 4x – 5 ( x + 2)2 – 9 1 = (3 sec θ tan θ dθ ) (3 sec θ )2 – 9 1 = (3 sec θ tan θ dθ ) = sec θ dθ 3 tan θ = ln | sec θ + tan θ | + C 1 1 = ln ( x + 2) + ( x + 2)2 − 9 + C1 3 3
∫
∫
∫
∫
2x
cos 3 x dx
∫ =
1
∫
x2 − 6x − 7 1
dx =
∫
1 ( x − 3)2 − 16
dx
( 4 sec θ tan θ dθ )
( 4 sec θ ) – 16 1 = ( 4 sec θ tan θ dθ ) = sec θ dθ 4 tan θ = ln |sec θ + tan θ | + C 2
∫
= ln
∫
1 1 ( x − 3) + ( x – 3)2 – 16 + C1 4 4
= ln x − 3 + x 2 – 6 x – 7 + C
Calculus Solutions Manual © 2005 Key Curriculum Press
+ – +
dv cos 3x 1 3 sin 3x 1 – 9 cos 3x
∫
∫e
= 40.
2x
cos 3 x dx
3 2x 2 e sin 3 x + e 2 x cos 3 x + C 13 13
∫e
−3 x
cos 4 x dx
= ln x + 2 + x 2 + 4 x − 5 + C 32.
u e 2x 2e 2x 4e 2x
∫
∫
∫
∫e
1 2 = e 2 x sin 3 x + e 2 x cos 3 x 3 9 4 2x − e cos 3 x dx 9 13 2 x e cos 3 x dx 9 1 2 = e 2 x sin 3 x + e 2 x cos 3 x + C1 3 9
∫
∫
∫
(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)
= x tan −1 x −
∫
∫
∫
∫
= x sin −1 x − (1 − x 2 ) −1/2 ( x dx )
28.
33. f (x) = tanh x ⇒ f ′(x) = sech2 x 34. f (x) = coth x ⇒ f ′(x) = −csch2 x sinh x dx 35. tanh x dx = = ln |cosh x | + C cosh x
u e –3x –3e –3x 9e –3x
+ – +
dv cos 4x 1 4 sin 4x 1 – 16 cos 4x
1 −3 x 3 e sin 4 x − e −3 x cos 4 x 4 16 9 − e −3 x cos 4 x dx 16 25 −3 x e cos 4 x dx 16 1 3 = e −3 x sin 4 x − e −3 x cos 4 x + C1 4 16 =
∫
∫
∫e =
−3 x
cos 4 x dx
4 −3 x 3 e sin 4 x − e −3 x cos 4 x + C 25 25 Problem Set 9-11
251
Note: As a check for integrals such as Problems 39 and 40, the numerators of the coefficients equal the 3 and 4 in the arguments of e−3x and cos 4x. The denominators equal 32 + 42, or 25. 41. g (x) = x 3 ln 5x ⇒ g ′ (x) = (3x2) ln 5x + x3 (5/5x) = x 2 (3 ln 5x + 1) 42. h (x) = x 2 ln 8x ⇒ h ′(x) = (2x) ln 8x + x2 (8/8x) = x(2 ln 8x + 1) 43.
∫ x ln 5x dx 3
u ln 5x 1 x
dv x3 1 4 4x
+
48.
1 0
– +
49.
50.
51. 52. 53.
= 44.
1 4 1 x ln 5 x − x 4 + C 4 16
∫x
2
ln 8 x dx
u ln 8x
∫ ∫ ∫ cos x dx = ∫ (1 − sin x ) cos x dx = ∫ cos x dx − ∫ sin x (cos x dx ) 3
2
+
1 = sin x − sin 3 x + C 3 1 2 Or: cos3 x dx = cos 2 x sin x + cos x dx 3 3 1 2 = cos 2 x sin x + sin x + C 3 3
dv x2 1 3 3x
∫
------------------------
1 0
– +
1 2 3x 1 3 9x
1 1 = x 3 ln 8 x − x 3 + C 3 9 x 45. y = ⇒ ( x + 2)( x + 3)( x + 4) ln 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 −1 − ( x + 2) −1 ( x + 2)( x + 3)( x + 4) − (x + 3)− 1 − (x + 4)− 1] x 46. y = ⇒ ( x – 1)( x – 2)( x – 3) ln 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 −1 − ( x − 1) −1 ( x – 1)( x – 2)( x – 3)
∫
1/2 2 3/2 = − + dx x – 1 x – 2 x – 3 1 3 = ln | x − 1| −2 ln | x − 2| + ln | x − 3| + C 2 2 y = cos3 x sin x ⇒ y′ = (−3 cos2 x sin x) sin x + cos3 x (cos x) = −3 cos2 x sin2 x + cos4 x y = sin5 x cos x ⇒ y′ = (5 sin4 x cos x) cos x + sin5 x (−sin x) = 5 sin4 x cos2 x − sin6 x 1 cos3 x (sin x dx ) = − cos 4 x + C 4 1 6 5 sin x (cos x dx ) = sin x + C 6
2
1 x
47.
x
∫
------------------------
1 3 4x 1 4 16 x
∫ ( x – 1)( x – 2)( x – 3) dx
− (x − 2)− 1 − (x − 3)− 1] x dx ( x + 2)( x + 3)( x + 4)
–1 3 2 = + − dx x + 2 x + 3 x + 4
∫
54.
∫
∫ sin x dx = ∫ (1 − cos x ) (sin x dx ) = ∫ (1 − 2 cos x + cos x )(sin x dx ) = ∫ sin x dx − 2 ∫ cos x sin x dx + ∫ cos x sin x dx 5
2
2
2
4
2
4
2 1 cos3 x − cos 5 x + C 3 5 1 4 Or: sin 5 x dx = − sin 4 x cos x + sin 3 x dx 5 5 1 4 8 = − sin 4 x cos x − sin 2 x cos x + sin x dx 5 15 15 1 4 = − sin 4 x cos x − sin 2 x cos x 5 15 8 − cos x + C 15 1 3 55. cos 4 x dx = cos3 x sin x + cos 2 x dx 4 4 1 3 3 = cos3 x sin x + cos x sin x + dx 4 8 8 1 3 3 = cos3 x sin x + cos x sin x + x + C 4 8 8 = − cos x +
∫
∫
∫
∫
∫
∫
= −ln |x + 2| + 3 ln |x + 3| − 2 ln |x + 4| + C 252
Problem Set 9-11
Calculus Solutions Manual © 2005 Key Curriculum Press
56.
∫
∫
1 5 sin 6 x dx = − sin 5 x cos x + sin 4 x dx 6 6 1 5 = − sin 5 x cos x − sin 3 x cos x 6 24 15 + sin 2 x dx 24 1 5 = − sin 5 x cos x − sin 3 x cos x 6 24 15 15 − sin x cos x + dx 48 48 1 5 = − sin 5 x cos x − sin 3 x cos x 6 24 5 5 − sin x cos x + x+C 16 16
68.
= − xe − x − e − x
72. r ( x ) =
58. f (x) = (x − 1) ⇒ f ′(x) = 4(x3 − 1)3(3x2) = 12x2(x3 − 1)3 59.
∫ (x
∫ ( x − 1) dx = ∫ (x − 4x 3
=
∫
62.
∫
63.
∫
64.
∫
x
∫ (t
66. h( x ) =
∫ (t
∫
9
+ 6 x 6 − 4 x 3 + 1) dx
74.
1 13 2 10 6 7 4 x − x + x − x + x+C 13 5 7 1 ( x 4 + 3)3 x 3 dx = ( x 4 + 3) 4 + C 16 1 ( x 3 − 1) 4 x 2 dx = ( x 3 − 1)5 + C 15 1 ( x 4 + 3) dx = x 5 + 3 x + C 5 1 ( x 3 − 1) dx = x 4 − x + C 4
65. f ( x ) =
67.
∫
4
12
61.
3(ln x )2 − (ln x )3 − 4 x2 ln x + 2 dx 73. dx = (ln x + 2) x x 1 = (ln x + 2)2 + C 2
4
1 13 9 27 5 x +x + x + 27 x + C 13 5
= 60.
∫
2
(ln x )3 + 4 dx dx = (ln x )3 + x x 1 = (ln x ) 4 + 4 ln | x | + C 4
∫
∫
2
75. f ( x ) = e x ⇒ f ′( x ) = 2 xe x 3
3
− 1) 4 dt ⇒ h ′( x ) = ( x 3 − 1) 4
79.
2 1
∫
+ – +
= 2e 2 − e 2 − e + e = e 2 = 7.3890 K
Calculus Solutions Manual © 2005 Key Curriculum Press
3
∫
u x 1 0
xe x dx
4
2
+ 3)3 dt ⇒ f ′( x ) = ( x 4 + 3)3
1
∫ x dx
(The absolute values are optional because ln x appears in the original integrand, so only positive values of x can be used.)
4
5
= xe x − e x
∫
76. f ( x ) = e x ⇒ f ′( x ) = 3 x 2 e x 2 1 2 77. xe x dx = e x + C 2 3 1 3 78. x 2 e x dx = e x + C 3
1
x
3(ln x )2 (1/ x ) ⋅ x − [(ln x )3 + 4] ⋅ 1 x2
=
+ 3) dx = ( x + 9 x + 27 x + 27) dx 8
= −2e −2 − e −2 + 0 + 1
(ln x )3 + 4 ⇒ x
r ′( x ) =
4
12
0
70. s(x) = xe− x ⇒ s′(x) = −xe− x + e− x ln x + 2 71. q( x ) = ⇒ x (1/ x ) ⋅ x − (ln x + 2) ⋅ 1 −1 − ln x q ′( x ) = = x2 x2
3
3
2
+ – +
dv e –x –e –x e –x
69. r(x) = xex ⇒ r′(x) = xex + ex
57. g (x) = (x + 3) ⇒ g ′(x) = 3(x4 + 3)2(4x3) = 12x3(x4 + 3)2
4
u x 1 0
= −3e− 2 + 1 = 0.59399…
∫
3
xe − x dx
0
∫
4
∫
2
∫x e
3 x2
dx
dv ex ex ex
u x2 2x
+
dv xe x 2 1 x2 2e
---------------------
2 0 =
– +
1 x2 2 xe 1 x2 4e
1 2 x2 1 x2 x e − e +C 2 2
Problem Set 9-11
253
80.
∫x e
5 x3
u x3 3x 2
dx
dv x 2e x 3 1 x3 3e
+
84.
=
---------------------
1 0
x 2e x 3
– +
1 x3 3e
∫e
ax
u e ax ae ax
cos bx dx
∫
1 ax a a2 e sin bx + 2 e ax cos bx − 2 b b b a2 + b2 e ax cos bx dx b2 1 a = e ax sin bx + 2 e ax cos bx + C1 b b ax cos e bx dx
=
dv cos bx 1 b sin bx 1 –b 2 cos bx
+ – +
a 2e ax
∫e
ax
cos bx dx
∫
87.
∫
ax
u e ax ae ax a 2e ax
sin bx dx
+ – +
∫ ∫
=
83.
=
∫ 254
a b e ax sin bx − 2 e ax cos bx + C a2 + b2 a + b2 (for a, b not both 0) ax e sin bx dx = C (for a = b = 0)
∫ ∫ sin
2
cx dx =
∫
1 (1 − cos 2cx ) dx 2
1 1 x− sin 2cx + C 2 4c sin 2 cx dx = C
Problem Set 9-11
∫ (1 + cos 2cx ) dx
1 1 x+ sin 2cx + C 2 4c cos 2 cx dx = x + C
ax + b
∫ 88.
(for c ≠ 0) (for c = 0)
∫ ∫
dv sinbx 1 – b cosbx 1 – 2 sinbx b
1 a = − e ax cos bx + 2 e ax sin bx b b a2 ax − 2 e sin bx dx b a2 + b2 e ax sin bx dx b2 1 a = − e ax cos bx + 2 e ax sin bx + C1 b b e ax sin bx dx
∫
1 2
89.
a
∫ cx + d dx = ∫ c + =
b a = 2 e ax sin bx + 2 e ax cos bx + C a + b2 a + b2 (for a, b not both 0) ax e cos bx dx = x + C (for a = b = 0)
∫e
cx dx =
(for c ≠ 0) (for c = 0)
ax + b cx + d a(cx + d ) – c( ax + b) ad – bc f ′( x ) = = 2 (cx + d ) (cx + d )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)
∫
82.
2
85. f ( x ) =
3 1 1 3 = x 3e x − e x + C 3 3
81.
∫ cos
∫ ∫
b − ( a/c)d dx cx + d
ax bc – ad + ln |cx + d | + C (for c ≠ 0) c c2 ax + b a 2 b dx = x + x+C cx + d 2d d (for c = 0, d ≠ 0) (undefined for c = d = 0) ( ax + b) n+1 +C a(n + 1) (for n ≠ −1, a ≠ 0) 1 n ( ax + b) dx = ln |ax + b| + C a (for n = −1, a ≠ 0) n n (for a = 0) ( ax + b) dx = b x + C ( ax + b) n dx =
x dx x +a 2
2
=
∫
1 ( x 2 + a 2 ) −1/2 (2 x dx ) 2
1 ⋅ 2( x 2 + a 2 )1/2 + C = x 2 + a 2 + C 2 x dx 1 =− 90. ( a 2 − x 2 ) −1/2 ( −2 x dx ) 2 2 2 a −x 1 = − ⋅ 2( a 2 − x 2 )1/ 2 + C = − a 2 – x 2 + C 2 (for a ≠ 0) (undefined for a = 0) dx d ( a tan θ ) = 91. 2 2 x +a a 2 tan 2 θ + a 2 =
∫
∫
∫
∫
a sec 2 θ dθ = sec θ dθ a sec θ = ln |sec θ + tan θ | + C1 1 2 1 = ln x + a 2 + x + C1 a a =
∫
= ln
∫
x 2 + a2 + x + C
Calculus Solutions Manual © 2005 Key Curriculum Press
92.
∫
dx
=
d ( a sin θ )
∫
a −x a − a sin θ a cos θ dθ x = = dθ = θ + C = sin −1 + C a cos θ a (for a ≠ 0) (undefined for a = 0) 2 93. f (x) = x sin ax ⇒ f ′(x) = 2x sin ax + ax2 cos ax 94. f (x) = x2 cos ax ⇒ f ′(x) = 2x cos ax − ax2 sin ax 2
2
∫
95.
∫x
2
2
2
∫
u x2 2x 2
sin ax dx
0
+ – + –
∫ 100.
dv sinax 1 – a cosax 1 – 2 sinax a
96.
∫x
2
u x2 2x 2
cos ax dx
0 =
97.
98.
99.
+ – + –
–
1 sinax a3
1 2 2 2 x sin ax + 2 x cos ax − 3 sin ax + C a a a (for a ≠ 0) 1 3 2 x cos ax dx = x + C (for a = 0) 3 1 sinh ax dx = cosh ax + C (for a ≠ 0) a (for a = 0) sinh ax dx = C
∫ ∫ ∫ 1 ∫ cosh ax dx = a sinh ax + C ∫ cosh ax dx = x + C ∫ cos
−1
–
= x cos −1 ax +
∫
= x cos −1 ax −
1 [1 − ( ax )2 ]−1/ 2 ( −2 a 2 x dx ) 2a
ax dx 1 − ( ax )2
∫
x
ax dx 1 – ( ax )2
∫
∫ 1 ∫ 1 + x dx
Let u = 1 + x . x = (u − 1)2 dx = 2(u − 1) du
∫
∫
∫
∫
∫
∫
∫
∫
2(u – 1) du = 2 du − (2/u) du u = 2u − 2 ln |u| + C = 2(1 + x ) − 2 ln |1 + x | + C Or: 2 x − 2 ln |1 + x | + C1 Absolute values are optional because 1 + x > 0. 1 102. Let u = 1 − x . dx 1– x x = (1 − u)2 dx = 2(u − 1) du 2(u – 1) du = = 2 du − (2/u) du u = 2u − 2 ln |u| + C = 2(1 − x ) − 2 ln |1 − x | + C Or: −2 x − 2 ln |1 − x | + C1 1 dx 103. Let u = 1 + 4 x . 1+ 4 x x = (u − 1)4 dx = 4(u − 1)3 du 3 4(u – 1) du = = ( 4u 2 − 12u + 12 − 4/u) du u 4 = u 3 − 6u 2 + 12 u − 4 ln |u| + C 3 4 = (1 + 4 x )3 − 6(1 + 4 x )2 + 12(1 + 4 x ) 3 − 4 ln 1 + 4 x + C =
dv 1 x
Or:
Calculus Solutions Manual © 2005 Key Curriculum Press
+ –
∫
(for a = 0)
+
∫
dv 1
∫
(for a ≠ 0)
u cos –1ax –a √1 – (ax) 2
ax dx
u sin –1ax a √ 1 – (ax) 2
ax dx
1 [1 − ( ax )2 ]−1/2 ( −2 a 2 x dx ) 2a 1 = x sin −1 ax + 1 – ( ax )2 + C (for a ≠ 0) a (for a = 0) sin −1 ax dx = C 101.
dv cosax 1 a sinax 1 – 2 cosax a
−1
= x sin −1 ax +
=−
∫
∫ sin
= x sin −1 ax −
1 cosax a3
1 2 2 2 x cos ax + 2 x sin ax + 3 cos ax + C a a a (for a ≠ 0) (for a = 0) x 2 sin ax dx = C
1 1 – ( ax )2 + C a (for a ≠ 0) π −1 cos ax dx = x + C (for a = 0) 2
= x cos −1 ax −
2
4 4 3 ( x ) − 2( 4 x )2 + 4 4 x − 4 ln |1 + 4 x | + C1 , 3
Problem Set 9-11
255
by expanding the powers or by starting with u = 4 x. Absolute values are optional because 1 + 4 x > 0. 1 dx 104. Let u = x 1/ 6 . x+3x x = u6 dx = 6u5 du 5 3 6u du 6u du = 3 2 = u +u u +1 6 2 = 6u − 6u + 6 − du u + 1 (by long division) = 2u 3 − 3u 2 + 6u − 6 ln |u + 1| + C = 2 x − 33 x + 66 x − 6 ln (6 x + 1) + C 1 105. Let u = e x + 1. dx x e +1 ex = u2 − 1 x = ln (u 2 − 1) 2u du dx = 2 u −1 2 du 1 1 = = − du u – 1 u + 1 u2 – 1 (by partial fractions) = ln |u − 1| − ln |u + 1| + C
∫
∫ ∫
∫
∫
∫
∫
= ln ( e + 1 − 1) − ln ( e + 1 + 1) + C 1 dx 106. Let u = e x – 1. x e –1 ex = u2 + 1 x = ln (u 2 + 1) 2u du dx = 2 u +1 2 du −1 = = 2 tan u + C = 2 tan −1 e x – 1 + C u2 + 1 107. a. Let t = x/2 and substitute, getting cos x = 2 cos2 (x/2) − 1 and sin x = 2 sin (x/2) cos (x/2). 2 b. cos x = −1 sec 2 ( x/2) 2 – sec 2 ( x/2) = sec 2 ( x/2) 2 – [1 + tan 2 ( x/2)] = 1 + tan 2 ( x/2) 1 – tan 2 ( x/2) = , Q .E .D . 1 + tan 2 ( x/2) sin ( x/2) sin x = 2 cos 2 ( x/2) cos ( x/2) 1 = 2 tan ( x/2) sec 2 ( x/2) 2 tan ( x/2) = , Q .E .D . 1 + tan 2 ( x/2) x
x
c. u = tan ( x/2) ⇒ 2 tan −1 u = x ⇒ dx = cos x = 1
1+
2
2
1+ u 2 du = = du (1 + u 2 ) + (1 – u 2 ) 1 e. dx = du = u + C = tan ( x/2) + C 1 + cos x 1 dx 108. a. sec x dx = cos x 1 + u 2 2 du 2 ⋅ = = du 1 – u2 1 + u2 1 – u2 1 1 + du b. sec x dx = 1– u 1+ u = −ln | 1 − u | + ln |1 + u| + C 2
∫
∫
∫ ∫
∫
∫
∫
∫
∫
∫
1+ u 1 + tan ( x/2) + C = ln +C 1– u 1 – tan ( x/2)
= ln
∫ sec x dx = ln
c.
1 + tan ( x/2) +C 1 – 1 ⋅ tan ( x/2)
tan (π /4) + tan ( x/2) +C 1 – tan (π /4) tan ( x/2) = ln |tan (π /4 + x/2)| + C
= ln
d. i.
∫
1
0
sec x dx = ln | tan (π /4 + x/2) |
1 0
= ln |tan (π/4 + 1/2)| − ln |tan π/4| = ln |tan (π/4 + 1/2)| = 1.226191…
∫
Problem Set 9-11
1 – u2 2u from part b. 2 and sin x = 1+ u 1 + u2
∫ 1 + cos x dx 1 2 du =∫ ⋅ 1– u 1+ u
d.
∫
256
2 du 1 + u2
ii.
1
1
0
0
∫ sec x dx = ln | sec x + tan x|
= ln |sec 1 + tan 1| − ln |1 + 0| = ln |sec 1 + tan 1| = 1.226191… , which agrees with the answer in part i. 1 1 2 du ⋅ 109. dx = 2 1 – u 1 + u2 1 – cos x 1– 1 + u2 2 du du –1 = = +C 2 2 = (1 + u ) – (1 – u ) u2 u
∫
∫
∫
∫
= −cot (x/2) + C 1 1 2 du ⋅ 110. dx = 2 u 1 + sin x 1 + u2 1+ 2 1+ u 2 du du –1 = = +C 2 2 = (1 + u ) + 2u (u + 1) u +1 –1 = +C tan ( x/2) + 1
∫
∫
∫
∫
Calculus Solutions Manual © 2005 Key Curriculum Press
111.
1 – u2 cos x 1 + u 2 ⋅ 2 du dx = 1 – u2 1 + u2 1 – cos x 1– 1 + u2
∫
R3. a.
∫
1 – u2 1 2 = du = 2 − du 2 u u (1 + u 2 ) 1 + u2 1 = − − 2 tan −1 u + C u 1 =− − 2 tan −1[tan ( x/2)] + C tan ( x/2)
∫
∫ x cos 2 x dx 3
∫
=
= −cot (x/2) − x + C Or:
∫ 1 – cos x dx = ∫ −1 + 1 – cos x dx cos x
∫
= − dx +
1
b.
∫e
4x
sin 3 x dx
1
Review Problems R0. Answers will vary. R1. f ( x ) = x cos x ⇒ f ′( x ) = x(−sin x) + (1) cos x = cos x − x sin x ⇒ x cos x dx + C = f ′( x ) dx
∫
=− c.
∫ x (ln x ) dx 2
∫
1 0
4
x sin x dx = sin x − x cos x
u dv (ln x)2 + x 1 1 2 2 ln x · x 2x -----------------------ln x x 1 – 1 2 x 2x ------------------------
⇒ x sin x dx = sin x − x cos x + C
+ –
1 2x 1 2 4x
1
= sin 4 − 4 cos 4 − sin 1 + cos 1 = 1.5566…
∫
4
1
R2.
+
3 4x 4 e cos 3 x + e 4x sin 3 x + C 25 25
= sin x − x sin x dx
Numerically,
–
4x
dv sin 3x 1 – 3 cos 3x 1 – 9 sin 3x
+
∫
= (cos x − x sin x ) dx
1
– +
∫ ∫
Problem Set 9-13
∫
+
1 4 = – e 4x cos 3 x + e 4x sin 3 x 3 9 16 4x – e sin 3 x dx 9 25 4x ⇒ e sin 3 x dx 9 1 4 = − e 4x cos 3 x + e 4x sin 3 x + C1 3 9 ⇒ e 4x sin 3 x dx
1. Answers will vary.
4
–
u e 4x 4e 4x 16e
Problem Set 9-12
∫
dv cos 2x 1 2 sin 2x 1 – 4 cos 2x 1 – 8 sin 2x 1 16 cos 2x
+
1 3 3 x sin 2 x + x 2 cos 2 x 2 4 3 3 − x sin 2 x − cos 2 x + C 4 8
∫ 1 – cos x dx
= −x − cot (x/2) + C (using Problem 109)
∫
u x3 3x 2 6x 6 0
x sin x dx ≈ 1.5566 K .
∫ 5x sin 2 x dx
u = 5x
dv = sin 2x dx
du = 5 dx
1 v = − cos 2 x 2
1 1 = 5 x − cos 2 x + cos 2 x (5 dx ) 2 2 5 5 = − x cos 2 x + sin 2 x + C 2 4
∫
Calculus Solutions Manual © 2005 Key Curriculum Press
1 2 1 1 x (ln x )2 − x 2 ln x + x 2 + C 2 2 4 d. Slice parallel to the y-axis. Pick a sample point (x, y) on the graph, within the slice. dV = 2π x · y · dx = 2π x(x ln x) dx = 2π x 2 ln x dx =
V = 2π
∫
2
1
x 2 ln x dx
u ln x
+
1 x
dv x2 1 3 3x
----------------------
1 0
– +
Problem Set 9-13
1 2 3x 1 3 9x
257
2 1 1 = 2π x 3 ln x − x 3 3 9 1 16 16 2 16 14 = π ln 2 − π + π = π ln 2 − π 3 9 9 3 9 = 6.7268K
R4. a.
∫ cos
d.
∫ sec
3
u secx secx tanx
x dx
+ –
dv sec 2 x tanx
∫ = sec x tan x − ∫ (sec − 1) sec x dx = sec x tan x + ∫ sec x dx − ∫ sec x dx 2 ∫ sec x dx = sec x tan x − tan 2 x sec x dx 2
30
u cos 29x – 29 cos 28x sinx
dx
+ –
dv cosx sinx
3
3
∫ x sin x + 29 ∫ cos
= cos x sin x + 29 cos x sin x dx 29
= cos
29
∫
28 28
2
= sec x tan x + ln |sec x + tan x | + C
∫ sec
x (1 − cos x ) dx 2
⇒ 30 cos dx 30
=
∫
= cos 29 x sin x + 29 cos 28 x dx
∫
e.
⇒ cos dx 30
∫
1 29 cos 29 x sin x + cos 28 x dx 30 30 1 4 b. sec 6 x dx = sec 4 x tan x + sec 4 x dx 5 5 1 4 = sec 4 x tan x + sec 2 x tan x 5 15 8 2 + sec x dx 15 1 4 = sec 4 x tan x + sec 2 x tan x 5 15 8 + tan x + C 15 =
∫
∫
∫
c.
∫ tan x dx = ∫ tan x (tan x dx ) = ∫ tan x (sec x − 1) dx = ∫ tan x sec x dx − ∫ tan x dx 1 = tan x − ∫ tan x dx n −1 ∫ cos x dx = ∫ (1 − sin x ) (cos x dx ) = ∫ (1 − 2 sin x + sin x )(cos x dx ) n−2
n
n−2
2
n−2
n−2
2
n −1
R5. a.
2
2
2
6
2
4
∫
32 dx = ( tan 9 32) dx = ( tan 9 32) x + C
f. r = 9 + 8 sin θ 1 1 dA = r 2 dθ = (9 + 8 sin θ )2 dθ 2 2 1 π /4 A= (64 sin 2 θ + 144 sin θ + 81) dθ 2 0 1 π /4 = [32(1 − cos 2θ ) + 144 sin θ + 81] dθ 2 0 1 81 π /4 = 16 θ − sin 2θ − 72 cos θ + θ 2 2 0 81 = 4π − 8 − 36 2 + π + 72 8 113 = π + 64 − 36 2 = 57.4633K 8
∫ ∫
R6. a.
∫
x 2 − 49 dx v
√x 2 – 49
x θ
u 7
2
2
∫
Problem Set 9-13
x = sec θ . x = 7 sec θ , 7 dx = 7 sec θ tan θ dθ,
Let 2
2
1 2 = tan 5 x + tan 3 x + tan x + C 5 3 1 c. sin 2 7 x dx = (1 − cos 14 x ) dx 2 1 1 = x − sin 14 x + C 2 28
258
9
4
∫ sec x dx = ∫ (tan x + 1) (sec x dx ) = ∫ (tan x + 2 tan x + 1)(sec x dx ) ∫
∫ tan
2
2 1 = sin x − sin 3 x + sin 5 x + C 3 5 b.
x dx
1 1 sec x tan x + ln |sec x + tan x | + C 2 2
n−2
5
3
x 2 − 49 = 7 tan θ , θ = sec −1
x 7
∫ x − 49 dx = ∫ (7 tan θ )(7 sec θ tan θ dθ ) = 49 ∫ sec θ tan θ dθ = 49 ∫ sec θ dθ − ∫ sec θ dθ
∴
2
2
3
Calculus Solutions Manual © 2005 Key Curriculum Press
1 1 = 49 sec θ tan θ + ln |sec θ + tan θ | 2 2 − ln |sec θ + tan θ | + C1
c.
∫
1 – 0.25 x 2 dx v
1
49 49 = sec θ tan θ − ln |sec θ + tan θ | + C1 2 2 49 x ⋅ ⋅ 2 7
=
−
x 2 − 49 7
1 x x 2 − 49 − 2 49 + ln 7 + C1 2 1 = x x 2 − 49 − 2 =
b.
∫
u
√ 1 – 0.25x 2
0.5 x = sin θ . 1 x = 2 sin θ, dx = 2 cos θ dθ,
Let
x 2 − 49 + C1 7
49 x ln + 2 7
0.5x
θ
x 2 1 – 0.25 x 2 dx = (cos θ )(2 cos θ dθ )
1 – 0.25 x 2 = cos θ , θ = sin −1
49 ln x + x 2 − 49 2
∫
∫
∫
∫
= 2 cos θ dθ = (1 + cos 2θ ) dθ 49 ln x + x 2 − 49 + C 2
x 2 − 10 x + 34 dx =
∫
( x − 5)2 + 9 dx
v
√ (x – 5)2 + 9 x–5 θ
2
1 = θ + sin 2θ + C = θ + sin θ cos θ + C 2 1 −1 x = sin + x 1 − 0.25 x 2 + C 2 2 d. Slice region vertically. Pick sample point (x, y) on the upper branch of the circle, within the strip. dA = 2 y dx = 2 25 – x 2 dx v
u 3
5
x−5 = tan θ . 3 x = 5 + 3 tan θ, dx = 3 sec2 θ dθ,
θ
Let
( x − 5)2 + 9 = 3 sec θ , θ = tan −1
x−5 3
x = sin θ . x = 5 sin θ , dx = 5 cos θ dθ , 5 x 25 – x 2 = 5 cos θ , θ = sin −1 5
Let
2
2
3
=
9 9 sec θ tan θ + ln |sec θ + tan θ | + C1 2 2
=
9 ( x − 5)2 + 9 x − 5 2 3 3 +
9 ln 2
( x − 5) + 9 x − 5 + + C1 3 3
4
∫2 = 2∫ = 25∫
A=
3
x =4
25 − x 2 dx 5 cos θ (5 cos θ dθ )
x =3 x =4 x =3
(1 + cos 2θ ) dθ
= 25θ + 12.5 sin 2θ
2
1 = ( x − 5)2 + 9 ⋅ ( x − 5) 2 9 9 + ln ( x − 5)2 + 9 + x − 5 − ln 3 + C1 2 2 1 2 = ( x – 5) x – 10 x + 34 2 9 + ln x 2 – 10 x + 34 + x − 5 + C 2
Calculus Solutions Manual © 2005 Key Curriculum Press
u
√ 25 – x 2
∫ ( x – 5) + 9 dx = ∫ (3 sec θ )(3 sec θ dθ ) = 9 ∫ sec θ dθ
∴
x
x =4 x =3
= 25θ + 25 sin θ cos θ
x =4 x =3
4 x x 1 + 25 ⋅ ⋅ 25 − x 2 5 5 5 3 = 25 sin −1 0.8 + 4 9 − 25 sin −1 0.6 − 3 16 = 25(sin− 1 0.8 − sin− 1 0.6) = 7.0948… (6 x + 1) dx (6 x + 1) dx R7. a. = x 2 – 3x – 4 ( x + 1)( x – 4) 1 5 = + dx x +1 x − 4
= 25 sin −1
∫
∫
∫
= ln |x + 1| + 5 ln |x − 4| + C Problem Set 9-13
259
b.
∫
5 x 2 – 21x – 2 dx ( x – 1)( x + 2)( x – 3)
R8. a. y
3 4 2 = + − dx x – 1 x + 2 x – 3
∫
= 3 ln |x − 1| + 4 ln |x + 2| − 2 ln |x − 3| + C c.
5 x 2 + 3 x + 45 dx = x3 + 9x
∫
∫
5 x 2 + 3 x + 45 dx x ( x 2 + 9)
5 3 x = + 2 dx = 5 ln | x | + tan −1 + C x x + 9 3
∫
(The second integral may be found by inspection or by trigonometric substitution.) d.
5 x + 27 x + 32 dx x ( x + 4) 2 2
∫
2
x 1
b. f ( x ) = sec− 1 3x 3 1 f ′( x ) = = 2 |3 x | (3 x ) – 1 | x | 9 x 2 – 1 c.
2 3 1 = + − dx x x + 4 ( x + 4) 2
∫
= 2 ln | x | + 3 ln | x + 4 | + (x + 4)− 1 + C 1 = ln | x 2 ( x + 4)3 | + +C x+4 dy e. = 0.1( y – 3)( y – 8) dx dy = 0.1 dx ( y – 3)( y – 8)
∫
∫
−1/5
1/5
∫ y − 3 + y − 8 dy = ∫ 0.1 dx
+ –
dv 1 x
∫ 1 + 25x dx 1 1 5x − ∫ (50 x dx ) 10 1 + 25 x 5x
2
2
1 ln |1 + 25 x 2 | + C 10 (Absolute values are optional because 1 + 25x 2 > 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 = x tan −1 5 x −
A=
∫
1
0
cos −1 x dx = x cos −1 x − 1 − x 2
1 0
−1
= cos 1 − 0 − 0 + 1 = 1 Easier way: Slice horizontally. Pick a sample point (x, y) on the graph within the strip. dA = x dy = cos y dy A=
y–8 − = 0.25e 0.5 x y–3 ((y – 8)/(y – 3) < 0 because (0, 7) is on the graph) 5 y = 3+ 1 + 0.25e 0.5 x The graph shows that solution fits slope field.
u tan –1 5x 5 1 + 25x 2
5x dx
= x tan −1
y–8 = 0.5 x − ln 4 y−3
y–8 = e 0.5 x −ln 4 = 0.25e 0.5 x y−3
−1
= x tan −1 5 x −
1 1 − ln | y − 3 | + ln | y − 8 | = 0.1x + C1 5 5 −ln | y − 3 | + ln | y − 8 | = 0.5x + C Substituting (0, 7) gives C = –ln 4 + ln 1 = –ln 4. ln
∫ tan
∫
π /2
0
cos y dy = sin y
π /2
= 1− 0 = 1
0
R9. a. y
1
x 1
b. y
y 1
x
7 1
x
260
Problem Set 9-13
Calculus Solutions Manual © 2005 Key Curriculum Press
c. h(x) = x2 sech x h′(x) = –x2 sech x tanh x + 2x sech x d. f ( x ) = sinh− 1 5x 5 f ′( x ) = 25 x 2 + 1 e.
∫
tanh 3 x dx =
∫
b.
f.
∫ cosh
u cosh –1 7x 7 √ 49x 2 – 1
π /2
= x cosh 7 x −
∫
7x 49 x 2 − 1
a→π /2
R10. a.
∫
∞
3
= lim − 5( x − 2)
b→∞ b −0.2
b→∞
∫
b
c.
∫
1
−1
3
3
= lim [–5(b – 2) –0.2 + 5] = 5 b→∞
The integral converges to 5.
Calculus Solutions Manual © 2005 Key Curriculum Press
a
x −2/3 dx
= lim− b→ 0
+ –
∫
b
−1
x −2/3 dx + lim+ a→ 0
= lim− 3 x 1/3
dv 1
b→ 0
b −1
1
∫x
−2/3
dx
a
+ lim+ 3 x 1/3 a→ 0
1 a
= lim− [3b − ( −3)] + lim+ (3 − 3a1/3 ) = 6 1/3
b→0
x
dx
( x − 2) −1.2 dx
a
a→π /2
a→0
The integral converges to 6. d.
x − | x − 1| dx x −1
∫ 0
= lim−
∫
( x − 2) −1.2 dx = lim
∫ tan x dx
The integral diverges.
1 ( 49 x 2 − 1) −1/2 (98 x dx ) 14 1 = x cosh −1 7 x − ⋅ 2( 49 x 2 − 1)1/2 + C 14 1 = x cosh −1 7 x − 49 x 2 − 1 + C 7 g. cosh2 x – sinh2 x 1 1 = (e x + e – x ) 2 − (e x − e – x ) 2 2 2 1 2x 1 –2 x = (e + 2 + e ) − (e 2 x − 2 + e –2 x ) 4 4 = 1, Q .E .D . 1 h. The general equation is y = k cosh x + C. k y = 5 at x = 0 ⇒ 5 = k cosh 0 + C ⇒C=5–k 3 y = 7 at x = 3 ⇒ 7 = k cosh + 5 − k k 3 ⇒ 2 = k cosh − k k ⇒ k = 2.5269… (solving numerically) t y = 2.5269K cosh + 5 − 2.5269K 2.5269K y(10) = 68.5961… 20 x = 2.5269K cosh + 5 − 2.5269K 2.5269K ⇒ x = ±6.6324… (solving numerically) = x cosh −1 7 x −
0
= lim − (ln |sec 0 | − ln |sec a |) = −∞
4
−1
a→π /2
= lim − ln |sec x |
1 sinh 3 x dx cosh 3 x
7x dx
∫
tan x dx = lim − 0
1 = ln |cosh 3 x | + C 3 (Absolute values are optional because cosh 3x > 0.) −1
0
b→1
∫
b
0
( x + 1) dx + lim+ a→1
4
∫( a
x − 1) dx
b
2 = lim− x 3/2 + x b→1 3
2 + lim+ x 3/2 − x → 1 a 3 0
4 a
2 = lim− b 3/ 2 + b − 0 b→1 3 2 2 + lim+ ⋅ 4 3/2 − 4 − a 3/2 + a a→1 3 3 2 3/2 2 3/2 2 3/2 = ⋅1 +1 + ⋅ 4 − 4 − ⋅1 +1 3 3 3 16 10 = 1+ − 4 +1 = 3 3 10 The integral converges to = 3.333K . 3 e.
∫
∞
1
x − p dx converges if p > 1 and diverges
otherwise. R11. a. f ( x ) = x sin −1 x ⇒ f ′( x ) = sin −1 x +
∫
b. I = x sin −1 x dx
=
1 2 −1 1 x sin x − 2 2
Let I1 =
∫
u sin –1 x 1 √1 – x2
∫
x 2 dx 1 – x2
∴ dx = cos θ dθ , θ = sin− 1 x
x 1− x2
+ –
dv x 1 2 2x
x 2 dx 1 – x2 and x = sin θ.
1 − x 2 = cos θ ,
Problem Set 9-13
261
sin 2 θ cos θ dθ = sin 2 θ dθ cos θ 1 1 1 = (1 − cos 2θ ) dθ = θ − sin 2θ + C 2 2 4 1 1 = θ − sin θ cos θ + C 2 2 1 1 −1 = sin x − x 1 − x 2 + C 2 2 1 2 −1 1 1 ∴ I = x sin x − sin −1 x + x 1 − x 2 + C 2 4 4 d x x 2 x c. tanh e = e sech e dx 1 d. ( x 3 − x ) −1 dx = 3 dx x −x 1 = dx x ( x – 1)( x + 1) 1 1/2 1/2 = − + + dx x x − 1 x + 1 1 1 = − ln | x | + ln | x − 1| + ln | x + 1| + C 2 2 e. f ( x ) = (1 − x 2)1/2 1 f ′( x ) = (1 − x 2 ) −1/2 ( −2 x ) = − x (1 − x 2 ) −1/2 2 I1 =
∫
∫
For (9 − x 2 ) −1/2 dx, transforming the dx to the
∫
differential of the inside function, −2x dx, requires multiplying by a variable. Because the integral of a product does not equal the product of the two integrals, you can’t divide on the outside of the integral by −2x. So a more sophisticated technique must be used, in this case, trigonometric substitution. As a result, an inverse sine appears in the answer: x (9 − x 2 ) −1/2 dx = sin −1 + C 3
∫
∫
∫
∫ ∫
∫
f. I = (1 − x 2 )1/2 dx Let x = sin θ . ∴ dx = cos θ dθ , (1 − x 2 )1/2 = cos θ , θ = sin− 1 x ∴ I = cos θ ⋅ cos θ dθ = cos 2 θ dθ
∫
∫
∫ x ln x dx
1 x
+
dv x 1 2 2x
-------------------1 1 – 2x 0
=
+
1 2 4x
1 2 1 x ln x − x 2 + C 2 4
∫
R12. For (9 − x 2 ) −1/2 x dx, the x dx can be transformed to the differential of the inside function by multiplying by a constant, 1 − (9 − x 2 ) −1/2 ( −2 x dx ) = −(9 − x 2 )1/2 + C, 2 and thus has no inverse sine.
∫
262
Problem Set 9-13
C1.
∫ sech x dx = ∫
1 − tanh 2 x dx
v
1
tanh x
θ
√1–
u tanh2x
Let tanh x = sin θ. ∴ x = tanh− 1 (sin θ) and θ = sin− 1 (tanh x) 1 1 dx = dθ 2 ⋅ cos θ dθ = 1 − sin θ cos θ
∫
∫
u ln x
Concept Problems
1 – tanh 2 x = 1 – sin 2 θ = cos θ 1 ∴ sech x dx = cos θ ⋅ dθ = dθ cos θ = θ + C = sin −1 (tanh x ) + C, Q .E.D .
1 1 1 = (1 + cos 2θ ) dθ = θ + sin 2θ + C 2 2 4 1 1 = θ + sin θ cos θ + C 2 2 1 1 −1 = sin x + x 1 − x 2 + C 2 2 1 g. g( x ) = (ln x )2 ⇒ g′( x ) = 2 ln x ⋅ x h.
∫
∫
1
0
∫
∫
sech x dx = sin −1 (tanh x )
1 0
= 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, so sinh x = 2 sinh (x/2) cosh (x/2) ⇒ csch x 1 1 = = sinh x 2 ⋅ sinh ( x/2) ⋅ cosh ( x/2) 1 sech 2 ( x/2) csch x = ⋅ 2 ⋅ sinh ( x/2) ⋅ cosh ( x/2) sech 2 ( x/2) 1 sech 2 ( x/2) sech 2 ( x/2) 2 = = tanh ( x/2) 2 ⋅ tanh ( x/2) 1 1 ∴ csch x dx = ⋅ sech 2 ( x/2) dx tanh ( x/2) 2 = ln | tanh (x/2) | + C, Q .E .D .
∫
∫
2
1
∫
csch x dx = ln |tanh ( x/2)|
2 1
= ln |tanh 1| – ln |tanh(1/2) | = 0.49959536 K Numerical integration gives 0.49959536… . Calculus Solutions Manual © 2005 Key Curriculum Press
C3. From sin 2A = 2 sin A cos A, let A = x/2, so sin x = 2 sin (x/2) cos (x/2) 1 1 csc x = = sin x 2 sin ( x/2) cos ( x/2) 1 sec 2 ( x/2) ⋅ 2 sin ( x/2) cos ( x/2) sec 2 ( x/2) 1 sec 2 ( x/2) sec 2 ( x/2) 2 = = tan ( x/2) 2 tan ( x/2) 1 1 ∴ csc x dx = ⋅ sec 2 ( x/2) dx tan ( x/2) 2 = ln | tan (x/2) | + C, Q .E .D . Or: Let u = tan (x/2), as in Problem 107 of Problem Set 9-11. 2 du 1 + u2 Then dx = 2 and csc x = 1+ u 2u 1 + u 2 2 du ∴ csc x dx = ⋅ 2u 1 + u 2 = (1/u) du = ln | u | + C = ln |tan ( x/2)| + C, =
∫
Chapter Test T1.
∫ sin
T2.
∫x
∫
=
T3.
Q .E .D . Confirmation: 1
0.5
1 6 sin x + C 6 u x3 3x 2 6x 6 0
sinh 6 x dx
∞
1 2 dx −∞ 1 + x 0 1 = lim 2 dx + lim a→−∞ a 1 + x b→∞
∫
= lim tan −1 x a→−∞
−1
0 a
∫ 1+ x 0
+ lim tan −1 x b→∞
1
2
+
x 1 – x2
+ –
dv 1 x
dx
∫
T4.
∫ sec x dx 3
=
1 1 sec x tan x + ln |sec x + tan x | + C 2 2
∫e
2x
cos 5 x dx
u e 2x 2e 2x 4e 2x
+ – +
dv cos 5x 1 5 sin 5x 1 – 25 cos 5x
b 0
b→∞
= −(− π /2) + (π /2) = π C5. 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 that ln 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 unbounded above, Q.E.D.
Calculus Solutions Manual © 2005 Key Curriculum Press
∫
dx
= lim (0 − tan −1 a) + lim (tan −1 b − 0) a→−∞
+ –
1 (1 − x 2 ) −1/2 ( −2 x dx ) 2 1 = x cos −1 x − (2)(1 − x 2 )1/2 + C 2 −1 = x cos x − 1 − x 2 + C
T5. b
–
u cos –1x 1 – 1 – √ x2
x dx
= x cos −1 x −
0.5
∫
∫ cos
1
1 1 = ln tan − ln tan = 0.7605K 2 4 Numerical integration gives 0.7605… . 2 sin ( x/2) cos ( x/2) Note that tan ( x/2) = 2 cos 2 ( x/2) sin x 1 = = , so 1 + cos x csc x + cot x ln |tan ( x/2)| = − ln |csc x + cot x | .
dv sinh 6x 1 6 cosh 6x 1 36 sinh 6x 1 216 cosh 6x 1 1296 sinh 6x
+
1 3 1 x cosh 6 x − x 2 sinh 6 x 6 12 1 1 + x cosh 6 x − sinh 6 x + C 36 216
= x cos −1 x +
csc x dx = ln |tan ( x/2)|
C4. A =
x cos x dx =
∫
∫ ∫
∫
3
5
1 2x 2 e sin 5 x + e 2 x cos 5 x 5 25 4 − e 2 x cos 5 x dx 25 29 2 x e cos 5 x dx 25 1 2 = e 2 x sin 5 x + e 2 x cos 5 x + C 5 25 =
∫
∫
∫e =
2x
cos 5 x dx
5 2x 2 e sin 5 x + e 2 x cos 5 x + C 29 29
Problem Set 9-13
263
T6.
∫ ln 3x dx
u ln 3x 1/x
+ –
= ln
dv 1 x
=
∫
= x ln 3 x − dx = x ln 3 x − x + C ii.
T7. f (x) = sech (e ) ⇒ f ′(x) = 3 sech2 (e5x) ⋅ [−sech (e5x) tanh (e5x)] ⋅ 5e5x = −15e5x sech3 (e5x) tanh (e5x) 1 T8. g( x ) = sin −1 x ⇒ g ′( x ) = 1 – x2 3
5x
T9. f (x) = tanh− 1 x ⇒ tanh f (x) = x, |x| ≤ 1 sech2 f(x) ⋅ f ′(x) = 1 [1 − tanh2 f (x)] ⋅ f ′(x) = 1 (1 – x2) ⋅ f ′(x) = 1 1 f ′( x ) = , |x | < 1 1 – x2 1 1 f ′(0.6) = = = 1.5625 1 – 0.36 0.64 Numerically, f ′(0.6) ≈ 1.5625… (depending on the tolerance of the calculator). 1 T10. General equation is y = k cosh x + C. k y = 1 at x = 0 ⇒ 1 = k cosh 0 + C ⇒ C=1−k 5 y = 3 at x = 5 ⇒ 3 = k cosh + 1 – k k Solving numerically, k ≈ 6.5586… . 1 y = 6.5586 K cosh x + 1 − 6.5586 K 6.5586 K x–3 x–3 T11. a. i. I = dx = dx 2 x – 6x + 5 ( x – 3)2 – 4
∫
∫
v
x–3
√ (x – 3)2 – 4
θ
x–3 = sec θ . x – 3 = 2 sec θ , 2 dx = 2 sec θ tan θ dθ, Let
x–3 2 (2 sec θ )(2 sec θ tan θ dθ ) 4 tan 2 θ
( x – 3)2 – 4 = 2 tan θ , θ = sec –1
=
∫ tan θ ⋅ sec θ dθ = ln |tan θ | + C
= ln
264
∫ 1
2
1
1 ( x − 3)2 − 4 + C1 2
Problem Set 9-13
1 ln | x 2 − 6 x + 5 | + C 2
x −3 1/2 1/2 + dx = dx − 6x + 5 x − 1 x − 5 1 1 = ln | x – 1| + ln | x – 5 | + C 2 2 1 = ln | ( x – 1)( x – 5) | + C 2 1 = ln | x 2 – 6 x + 5 | + C, 2 which agrees with part a.
∫x
∫x
∫
2
x–3 dx – 6x + 5 1 1 = ⋅ (2 x − 6) dx 2 x2 – 6x + 5 1 = ln | x 2 – 6 x + 5| + C, as in parts a and b. 2 b. See parts i, ii, and iii. iii.
2
∫
T12.
∫ cos
2
x dx =
∫ 2 (1 + cos 2 x ) dx
= T13. a. i.
1
1 1 x + sin 2 x + C 2 4
∫ cos x dx = ∫ (1 – sin x ) cos x dx = ∫ (1 – 2 sin x + sin x ) cos x dx 5
2
2
2
4
2 3 1 sin x + sin 5 x + C 3 5 1 4 ii. cos 5 x dx = cos 4 x sin x + cos3 x dx 5 5 1 4 = cos 4 x sin x + cos 2 x sin x 5 15 8 + cos x dx 15 1 4 = cos 4 x sin x + cos 2 x sin x 5 15 8 + sin x + C 15 1 4 8 4 b. cos x sin x + cos 2 x sin x + sin x 5 15 15 1 = (1 – sin 2 x )2 sin x 5 4 8 + (1 – sin 2 x ) sin x + sin x 15 15 1 2 3 1 5 = sin x – sin x + sin x 5 5 5 4 4 8 + sin x – sin 3 x + sin x 15 15 15 = sin x –
∫
∫
∫
u 2
∴ I=
( x − 3)2 − 4 + C
Calculus Solutions Manual © 2005 Key Curriculum Press
T14.
∫
∞
1 4 8 2 4 = + + sin x − + sin 3 x 5 15 15 5 15
= lim (–10 xe –0.1x – 100e –0.1x )
1 + sin 5 x 5 2 1 = sin x – sin 3 x + sin 5 x 3 5
= lim (–10 be
xe −0.1x dx
0
Calculus Solutions Manual © 2005 Key Curriculum Press
b→∞ b→∞
–0.1b
– 100e
–0.1b
b 0
+ 0 + 100)
10 b + 100 + 100 e 0.1b 10 = lim – 0.1b + 100 (by l’Hospital’s rule) b→∞ 0.1e = 100 T15. Answers will vary. = lim – b→∞
u x 1 0
+ – –
dv e –0.1x –10e –0.1x 100e –0.1x
Problem Set 9-13
265
Chapter 10—The Calculus of Motion—Averages, Extremes, and Vectors Problem Set 10-1 1. v(t) = 100(0.8) − 30 = 100e − 30 = 0 ln 0.3 t ln 0.8 ⇒ e = 0.3 ⇒ t = = 5.3955K ln 0.8 v becomes negative after t0 ≈ 5.40 min. t
2. sup =
∫
t0
0
v dt =
t ln 0.8
∫
t0
(100e t ln 0.8 – 30) dt
0
= 151.8341… (numerically) ≈ 151.8 ft sdown = −
∫
10
t0
v dt = −
∫
10
(100e t ln 0.8 – 30) dt
t0
= 51.8110… (numerically) ≈ 51.8 ft Distance = sup + sdown = 203.6452… ≈ 203.6 ft 3. Displacement = sup − sdown = 100.0231… ≈ 100.0 ft The displacement is positive, so Calvin is upstream of his starting point. 4. Displacement =
∫
10
(100e t ln 0.8 – 30) dt
5. Distance =
∫
0
Distance =
6
∫ |t
2
0
6
∫ (t 0
2
– 10t + 16) dt = −12 ft
1 − 10t + 16 | dt = 41 ft 3
2 2 + −26 = −12 ft 3 3 2 2 1 Distance = 14 + 26 = 41 ft 3 3 3
d. Displacement = 14
e. a(t) = v′(t) = 2t − 10 a(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].
0
= 100.0231… (numerically) ≈ 100.0 ft 10
c. Displacement =
|v| dt =
∫
10
0
|100e
t ln 0.8
− 30 | dt
= 203.6452… (numerically) ≈ 203.6 ft
b. For [10, 5π), displacement =
Q1. 120 mi Q3. 1.25 h Q5. x ln x − x + C Q7. g′(t) = sech2 t Q9.
1 x 2 +C ln 2
Q2. 25 mi/h Q4. f ′(x) = 1/x Q6. f ′(t) = sec2 t 1 3 Q8. x +C 3 Q10. ln 2 e x ln 2 = 2 x ln 2
1. a. v(t) = t2 − 10t + 16 on [0, 6] v(t) = (t − 2)(t − 8) = 0 ⇔ t = 2 or 8 s v(t) > 0 for t in [0, 2). v(t) < 0 for t in (2, 6]. b. For [0, 2), displacement 2 2 = (t 2 – 10t + 16) dt = 14 0 3 2 Distance = 14 ft 3 For (2, 6], displacement 6 2 = (t 2 – 10t + 16) dt = −26 2 3 2 Distance = 26 ft 3
∫ ∫
266
Problem Set 10-2
5π
tan 0.2t dt
10
= 5 ln | sec π | − 5 ln | sec 2 | = −4.3835… Distance = 4.3835… ≈ 4.38 cm For (5π, 20], displacement =
Problem Set 10-2
∫
∫
20
5π
tan 0.2t dt
= 5 ln |sec 4| − 5 ln | sec π | = 2.1259… Distance = 2.1259… ≈ 2.13 cm c. Displacement = −2.26 cm Distance =
∫
20
10
∫
20
10
tan 0.2t dt = −2.2576 K ≈
|tan 0.2t | dt = 6.5095K ≈
6.51 cm d. Displacement = −4.3835… + 2.1259… = −2.2576… ≈ −2.26 cm Distance = −(−4.3835…) + 2.1259… = 6.5095… ≈ 6.51 cm e. a( t) = v′(t) = 0.2 sec2 t a( 15) = 0.2 sec2 3 = 0.2040… ≈ 0.20 (cm/s)/s
π t − 2 on [1, 11] 24 v(t) = 0 when π cos t = 0.5 ⇔ t = 8 in [1, 11] 24 v(t) < 0 for t in [1, 8). v(t) > 0 for t in (8, 11].
3. a. v(t ) = sec
Calculus Solutions Manual © 2005 Key Curriculum Press
b. For [1, 8), displacement =
8
π
∫ sec 24 t – 2 dt 1
π π 24 = ln sec + tan − 16 π 3 3 π π 24 − ln sec + tan +2 π 24 24 = −4.9420… Distance ≈ 4.94 km For (8, 11], displacement 11 π = sec t – 2 dt 8 24
∫
=
24 11 11 ln sec π + tan π − 22 π 24 24
π π 24 ln sec + tan + 16 π 3 3 = 4.7569… Distance ≈ 4.76 km 11 π c. Displacement = sec t – 2 dt = 1 24 −0.1850 K ≈ −0.19 km 11 π Distance = sec t − 2 dt = 9.6990 K ≈ 1 24 9.70 km −
∫
∫
d. Displacement = −4.9420… + 4.7569… = −0.1850… ≈ −0.19 km Distance = −(−4.9420…) + 4.7569… = 9.6990… ≈ 9.70 km π π π t tan t sec e. a(t ) = v ′(t ) = 24 24 24 a( 6) = 0.1851… ≈ 0.19 (km/h)/h exactly π 2 24 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 = 3 3 (t 3 – 5t 2 + 8t – 6) dt = −6 0 4 3 Distance = 6 mi 4 For (3, 5], displacement = 5 2 (t 3 – 5t 2 + 8t – 6) dt = 24 3 3 2 Distance = 24 mi 3 c. Displacement = 5 11 (t 3 – 5t 2 + 8t – 6) dt = 17 mi 0 12 5 5 3 2 Distance = | t − 5t + 8t − 6 | dt = 31 mi 0 12
∫ ∫ ∫
∫
Calculus Solutions Manual © 2005 Key Curriculum Press
3 2 11 + 24 = 17 mi 4 3 12 3 2 5 Distance = − −6 + 24 = 31 mi 4 3 12 e. a( t) = v′(t) = 3t2 − 10t + 8 a(2.5) = 1.75 (mi/min)/min d. Displacement = −6
5. a(t ) = t 1/2, v(0) = −18, on [0, 16] 2 v(t ) = t 1/2 dt = t 3/2 + C; v(0) = −18 ⇒ C = −18 3 2 3/2 v(t ) = t − 18 3 16 2 t 3/2 − 18 dt = −14 14 ft Displacement = 0 3 15 16 2 7 Distance = t 3/2 − 18 dt = 179 ft 0 3 15
∫
∫
∫
6. a(t ) = t −1 , v(1) = 0, on [0.4, 1.6]
∫
v(t ) = t −1 dt = ln t + C (t > 0); v(1) = 0 ⇒ C = 0 v(t ) = ln t Displacement =
∫
1.6
0.4
ln t dt = −0.0814 …
≈ −0.081 cm Distance =
∫
1.6
0.4
|ln t | dt = 0.3854 …
≈ 0.385 cm 7. a(t ) = 6 sin t, v(0) = −9, on [0, π ] v(t ) = 6 sin t dt = −6 cos t + C; v(0)
∫
= −9 ⇒ C = −3 v(t) = − 6 cos t − 3 Displacement =
∫
π
0
(–6 cos t – 3) dt =
−9.4247… ≈ −9.42 km (exact: −3π km) Distance =
∫
π
0
|− 6 cos t − 3| dt = 13.5338…
≈ 13.53 km (exact: 6 3 + π ) 8. a(t) = sinh t, v(0) = −2, on [0, 5] v(t ) = sinh t dt = cosh t + C
∫
v(0) = −2 ⇒ C = −3 v(t) = cosh t − 3 Displacement =
5
∫ (cosh t – 3) dt = 0
59.2032… ≈ 59.20 mi (exact: sinh 5 − 15) Distance =
5
∫ |cosh t − 3| dt 0
= 64.1230 K ≈ 64.12 mi 9. a. v = t − 2 = 0 ⇔ t = 4 s; v < 0 if t < 4, v > 0 if t > 4 1/2
b. Displacement = c. Distance =
9
∫ |t 1
9
∫ (t 1
1/2
1/2
1 – 2) dt = 1 ft 3
− 2| dt = 4 ft
Problem Set 10-2
267
1 1 10. a. v = sin 2t = 0 at t = … , − π , − π , 0, π , 2 2 3 π, π, … 2 1 sin 2t ≥ 0 on 0, π , so 2 Distance =
∫
π /2
0
∫
4.5π
0
∫
4.5π
0
Displacement = b. Distance =
∫
40
10
∫
40
10
(60 – 2t ) dt = 300 ft
|60 − 2t | dt = 500 ft
∫
v(t ) = ( 40 cos 0.015t − 9.8) dt 40 = sin 0.015t − 9.8t + C 0.015 v(0) = 0 ⇒ C = 0 For t > 100, v(t ) = −9.8 dt = −9.8t + C
∫
40 sin 1.5 − 980 = 1679.986 … 0.015 ⇒ C = 1679.986… + 980 = 2659.986… 40 sin 0.015t − 9.8t, t in [0, 100] v(t ) = 0.015 −9.8t + 2659.986 K , t > 100 v(100) =
a (t ) 30
t 100
v (t )
1000
t
268
Problem Set 10-2
∫
100
0
+
sin 2t dt = 9 cm
40 cos 0.015t – 9.8, t in [0, 100] 12. a. a(t ) = t > 100 –9.8, For t in [0, 100],
100
=
sin 2t dt = 1 cm
Or: The regions where the graph is below the x-axis cancel out the regions where the graph is above the axis, leaving only one uncancelled region above the graph, so Displacement = area from part a = 1 cm. The absolute values of the regions above and below the graph are the same, so Distance = 9 times the area from part a = 9 cm. 11. a. v = 60 − 2t
∫
c. Displacement =
sin 2t dt = 1 cm
b. Displacement = Distance =
1 9.8 cos −1 = 0.015 40 88.2184 K ≈ 88.2 s v = 0 at t = 2,659.986…/9.8 = 271.4272… ≈ 271.4 s
b. a = 0 at t =
∫
300
v(t ) dt
0
40 sin 0.015t – 9.8t dt 0.015
300
100
(–9.8t + 2659.986 K) dt
≈ 116202.27… + 139997.32… ≈ 256,200 m Distance = 116202.27… +
∫
300
100
|− 9.8t + 2659.986 K | dt
= 116202.27… + 147998.09… ≈ 264,200 m The distance is greater than the displacement, which agrees with the fact that the velocity becomes negative at t = 271.4… s. d. v(300) = −9.8(300) + 2,659.986… = −280.0133… , so the rocket is moving downward (falling) at about 280 m/s. 13. a. tend s
aav (mi/h)/s
vend mi/h
vav mi/h
0
—
0
—
5
7.375
send mi 0
2.95
14.75
10
3.8
33.75
24.25
0.0102… 0.0439…
15
1.75
42.5
38.125
0.0968…
20
0.3
44
43.25
0.1569…
25
0
44
44
0.2180…
30
0
44
44
0.2791…
35
0
44
44
0.3402…
40
−0.2
43
43.5
0.4006…
45
−0.9
38.5
40.75
0.4572…
50
−2.6
25.5
32
0.5017…
55
−3.5
8
16.75
0.525
60
−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; its acceleration must be greater than zero to get it moving, even though it is stopped at t = 0. Acceleration and velocity are different quantities; the velocity can be zero but changing, which means the acceleration is nonzero. d. Zero acceleration means the velocity is constant, but not necessarily zero. e. The last entry in the last column is the displacement at time t = 60. Thus, it is 0.5305… ≈ 0.53 mi between stations. Calculus Solutions Manual © 2005 Key Curriculum Press
14. a.
v (t )
tend s
aav (mi/h)/s
v end mi/h
vav mi/h
send mi
0 10 20 30 40 50 60 70 80 90 100
— 8.5 22 33 39.5 42.5 53 71 83.5 47.5 3
6000 6085 6305 6635 7030 7455 7985 8695 9530 10005 10035
— 6042.5 6195 6470 6832.5 7242.5 7720 8340 9112.5 9767.5 10020
400 416.7847… 433.9930… 451.9652… 470.9444… 491.0625… 512.5069… 535.6736… 560.9861… 588.1180… 615.9513…
b. According to these calculations, the spaceship is only about 620 mi from the launchpad and moving at only about 10,000 mi/h. So the specifications are definitely not met, and the project should be sent back to the drawing board. dv 15. a. a = ⇒ v = a dt = at + C; dt v = v 0 when t = 0 ⇒ C = v 0 ⇒ v = v 0 + at ds b. v = ⇒ s = v dt = (v0 + at ) dt = dt 1 v0 t + at 2 + C 2 s = s0 when t = 0 ⇒ C = s0 ⇒ 1 s = v0 t + at 2 + s0 2 16. Use s(t) for displacement. Assume v(0) = s(0) = 0. 2, if 0 ≤ t < 6 a. a(t ) = 0, if t ≥ 6
10
t 0
10
t 0
b. The acceleration suddenly jumps from 0 to 2 at t = 0 and drops back to 0 at t = 6. (The velocity graph has cusps in both places.) π c. a(t ) = 2 − 2 cos t 3
π lim+ a(t ) = lim+ 2 – 2 cos t t→0 t→0 3 = 2 − 2 cos 0 = 0
π lim a(t ) = lim– 2 – 2 cos t t →6 3
t →6 –
= 2 − 2 cos 2π = 0 Because a(t) is continuous at t = 0 and 6, there are no sudden changes in acceleration.
∫
2 – 2 cos π t, if 0 ≤ t ≤ 6 d. a(t ) = 3 0, if t ≥ 6 2t – 6 sin π t, if 0 ≤ t ≤ 6 v(t ) = 3 π if t ≥ 6 12, e. v (t ) 10
t 2 , if 0 ≤ t < 6 s( t ) = – , t 12 36 if t > 6
t 0
−36 comes from the initial condition, s(6) = 36. a (t )
6
10
There are no step discontinuities in a(t), and thus the graph of v(t) is smooth. f.
2
6
2t − 6 sin π t dt = t 2 + 18 cos π t 0 3 3 0 π π2
∫
6
18 18 2 − 0 − 2 = 36 π π Elevator goes 36 ft.
= 36 + t
Calculus Solutions Manual © 2005 Key Curriculum Press
6
50
2t, if 0 ≤ t < 6 v(t ) = 12, if t > 6
0
10
s(t )
∫
∫
6
6
10
Problem Set 10-2
269
g. The elevator will take another 36 ft to slow down and stop. So the deceleration should start where the elevator is 564 ft up, about the 47th floor (from part h, one floor = 12 ft). h. The elevator takes a total of 12 s to accelerate and decelerate. During these intervals it travels a total of 72 ft, leaving 528 ft for the constant velocity portion. At 12 ft/s, this part of the trip will take 44 s. Thus, the total trip takes 56 s. i. The elevator must start to decelerate halfway through the trip, where s(t) = 6 ft. Solving b
π
∫ 2t – π sin 3 t dt = 6 6
0
numerically for b gives b ≈ 3.1043… ≈ 3.1 s. a(3.1043…) = 3.9880… ≈ 4.0 ft/s2 By symmetry, the deceleration process must start at this time, meaning the acceleration jumps to −3.9880… ft/s2. The graph looks like this:
b. The rectangle has the same area as the shaded region. f (x )
y = 41
50
x 1
5
c. 41 = c3 − c + 5 c = 3.4028… , which is in [1, 5]. 1 9 1/2 1 2. a. yav = ( x – x + 7) dx = 4 8 1 6 b. The rectangle has the same area as the shaded region.
∫
f (x ) 5
y = 4.1666...
x 1
9
5 a (t )
1 = c1/2 − c + 7 6 c = 5.0892… , which is in [1, 9]. 1 7 3. a. yav = 3 sin 0.2 x dx = 2.0252 … 6 1 b. The rectangle has the same area as the shaded region. c. 4
t 0
6.2
∫
Thus, the passengers get a large jerk at the midpoint of the trip. One way to remedy the problem is to reduce the acceleration so that the elevator goes only 6 ft instead of 36 ft in the first 6 seconds. That is, 1 1 π a(t ) = − cos t 3 3 3 You may think of other ways.
50 mi/h 20 min No local maximum f (x) = 16 (at x = 1) Mean value theorem
1. a. yav =
270
1 4
∫
5
1
Q2. Q4. Q6. Q8. Q10.
( x 3 – x + 5) dx =
Problem Set 10-3
y = 2.0252...
x 1
7
c. 2.0252… = 3 sin 0.2c c = 3.7053… , which is in [1, 7]. 1 1.5 4. a. yav = tan x dx = 2.5181… 1 0.5 b. The rectangle has the same area as the shaded region.
∫
Problem Set 10-3 Q1. Q3. Q5. Q7. Q9.
g (x ) 3
30 mi 2π 1.5 infinite D
1 (164) = 41 4
h(x ) 10
y = 2.5181... x 0.5
1.5
Calculus Solutions Manual © 2005 Key Curriculum Press
c. 2.5181… = tan c c = 1.1927… , which is in [0.5, 1.5]. 1 9 1 5. a. yav = t dt = 2 8 1 6 b. The rectangle has the same area as the shaded region.
∫
v (t) 3
y = 2.1666...
t 1
9
1 = c 6 25 c = 4 , which is in [1, 9]. 36 1 3 100 6. a. yav = (2 + e –3 ) 100(1 − e − t ) dt = 3 0 3 = 68.3262… b. The rectangle has the same area as the shaded region. c. 2
∫
v (t ) 100 y = 68.32...
t 0
7. 8. 9. 10. 11.
12.
3
c. 68.3262… = 100(1 − e− c) c = 1.1496… , which is in [0, 3]. 1 k 2 1 yav = ax dx = ak 2 k 0 3 1 k 3 1 3 yav = ax dx = ak k 0 4 1 k x 1 yav = ae dx = a(e k − 1) k 0 k 1 k 1 yav = tan x dx = ln | sec k | k 0 k − 1/2 a(t) = 6t 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 The general equation of a parabola with vertex (h, k) is v − k = a(t − h)2. Vertex is at (t, v) = (2, 50), so v − 50 = a(t − 2)2 . v = 30 when t = 0, so −20 = a(−2)2 ⇒ a = −5.
∫
∫ ∫ ∫
Calculus Solutions Manual © 2005 Key Curriculum Press
v = 50 − 5(t − 2)2 1 4 1 vav = [50 – 5(t – 2)2 ] dt = 43 mi/h 4 0 3 1 This is just 13 mi/h above the speed limit. 3 If Ida wins her appeal, her fine will be 1 1 7 ⋅ 13 = $93 ≈ $93.33, which is $46.67 less 3 3 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) = v 0 ⇒ v 0 = at0 + C ⇒ C = v 0 − at0. ∴ v(t) = at + v0 − at0 = v0 + a(t − t0) t1
vav
[v ∫ = t0
0
+ a(t – t0 )] dt
t1 – t0 1 1 1 = v0 t1 + a(t1 – t0 )2 – v0 t0 – a(t0 – t0 )2 2 2 t1 – t0
1 a(t1 − t0 ) 2 The average of v0 and v1 is 1 1 (v0 + v1 ) = [v0 + v0 + a(t1 – t0 )] 2 2 1 = v0 + a(t1 − t0 ) 2 ∴ vav = the average of v0 and v1, Q.E.D. 14. Counterexample: In Problem 11, the car’s acceleration is a = 6/ t .The initial velocity is v(0) = 60 ft/s; the final velocity after 25 seconds is v(25) = 120 ft/s; and the average velocity is vav = 100 ft/s. But the average of the initial and 1 final velocities is [v(0) + v(25)] = 90 ft/s ≠ vav . 2 15. a. Integral = area = 12(100 + 70)/2 + 6(40) + 12(40 + 10)/2 = 1560 yav = 1560/30 = 52, or $52,000 Cost of inventory = 0.50(52000)/100 = $260.00 b. At x = 12, they may have had a single, large sale, dropping the inventory from $70,000 to $40,000. There is no day on which the inventory is worth $52,000. = v0 +
y (thousand dollars) 100 No x where y = 52 50
10
20
x (days) 30
Problem Set 10-3
271
16. y (ft) Water surface 5
10
15
20
25
x (ft) 30
19. v = A sin 120π t and y = | A sin 120π t | 1 1/60 yav = | A sin 120π t | dt 1/60 0
∫
= 60 y av = – 6.5142
–5
=− –10
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 about 6.51 feet deep. The volume would equal 6.5142… times the area of the horizontal cross section times the number of 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 = 309 yav = 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 the actual average. Averaging high and low temperatures is easier than finding the average by calculus, but the latter is more realistic for such applications as determining heating and air conditioning needs. 18. a. At x = 3, y = 81.3139… ≈ 81.3 mg. 1 3 1 yav = 200e −0.3 x dx = (395.6202 K) 3 0 3 = 131.8734 … ≈ 131.9 mg
∫
b. k = 81.3139… , so the equation is y = 281.3139…e− 0.3(x− 3 ) . 6 1 yav = 395.6202 K + 281.3139K e –0.3( x –3) dx 3 6 1 = (395.6202 K + 556.4674 K) = 158.6812 K 6 ≈ 158.7 mg
∫
c. As the graph shows, there are two times in [0, 6] at which there are 158.7 mg. So the conclusion of the mean value theorem is true, in spite of the discontinuity.
∫
/ 1120
0
A sin 120π t dt − 60 1/120
A cos 120πt 2π
0
+
∫
1/60
/ 1120
A sin 120π t dt
A cos 120πt 2π
1/60 1/120
A 2A (– cos π + cos 0 + cos 2π – cos π ) = 2π π 2A If yav = 110, then = 110 ⇒ A = 55π π = 172.78… V. The average value of one arc of π 1 2 y = sin x is sin x dx = , and π −0 0 π y = sin x has a maximum value of 1. A horizontal stretch does not affect the average value. Write a proportion to find the maximum of a sinusoidal curve with an average value 2/π 110 of 110. = , so m = 55π . m 1 =
∫
20. a. d = k sin x 1 2π 2 2 2 dav k sin x dx = 2π 0
∫
= =
k2 1 1 x – sin 2 x 2π 2 4
2π
0
2
k k2 (π – 0 – 0 + 0) = 2π 2
∴ rms = k/ 2 = 0.7071K k b. cos 2 x = 1 − 2 sin 2 x ⇒ sin 2 x =
1 1 − cos 2 x 2 2
Thus, sin2 x is a sinusoid. y 1
x π
2π
c. By symmetry across the line y =
1 , the 2
1 1 − cos 2 x (and hence 2 2 1 y = sin2 x) over [0, 2π] is . Thus, the 2 1 2 2 average of y = k sin x is k 2 . 2 ∴ rms = k/ 2 , as in part a. average of y =
y (mg) 300
y av = 158.68... 200
100 Two times
x (h) 1
272
2
Problem Set 10-3
3
4
5
6
7
Calculus Solutions Manual © 2005 Key Curriculum Press
d. By symmetry, it suffices to find the average and rms for one arch of the graph, that is, over [0, π]. 1 π yav = |sin x | dx π 0 1 π = sin x dx (because sin x ≥ 0 in [0, π ]) π 0 π 1 2 = − cos x = π π 0 1 π 2 dav = (|sin x | – 2/π ) 2 dx ≈ 0.094715K π 0 ∴ rms ≈ 0.094715…1/2 = 0.3077… The maximum distance between high and low points for this curve is 1; a sinusoidal curve with maximum distance 1 between high and 1 low points has equation y = sin x, with 2 rms = 2 /4 = 0.3535K(using part a). This number is greater than the rms for |sin x|, so |sin x| is smoother.
Ann should swim toward a point about 21.8 m downstream. 2. 100
∫
x
100 – x
30
∫
1 1 (100 – x ) + 30 2 + x 2 13 12 The graph shows a minimum T at x ≈ 72. T =
∫
T 10
x 100
Algebraic solution: 1 1 T ′ = − + (30 2 + x 2 ) −1/2 ⋅ 2 x 13 24 1 1 = T′ = 0 ⇔ x (30 2 + x 2 ) −1/ 2 13 12 13x = 12(302 + x2)1/2 169x2 = 144 · 302 + 144x2 x = ±72 The diver should swim for 100 − 72 = 28 m, then dive.
Problem Set 10-4 Q1. x = 81 1 Q3. − (100 – x 2 )3/2 + C 3 1 2x 1 2x Q5. xe − e + C 2 4 Q7. 1.5 Q9. t = 4
Q2. y′ = −x(100 − x 2)− 1/ 2 Q4. y′ = 3 ⋅ (1 − 9x2)− 1/ 2 Q6. y′ = sech2 x Q8. t = 1 and t = 4 Q10. A
3. 1000
x
1.
1000 – x
300
50
x
C = 40(1000 − x ) + 50 300 2 + x 2 The graph shows a minimum C at x ≈ 400 (exactly x = 400).
100 – x 100
1 1 50 2 + x 2 + (100 – x ) 2 5 The graph shows a minimum T at x ≈ 22 m. T=
C 100,000
T 100
x 1000
x 100
Algebraic solution: 1 1 T ′ = (50 2 + x 2 ) −1/2 ⋅ 2 x − 4 5 1 1 2 2 −1/2 T ′ = 0 ⇔ (50 + x ) x = 2 5 5x = 2(502 + x 2)1/2 25x 2 = 4 · 502 + 4x 2 x = ±100/ 21 = ±21.8217K
Calculus Solutions Manual © 2005 Key Curriculum Press
The pipeline should be laid 600 m along the road from the storage tanks, then straight across the field to meet the well. 4. 120 400 – x
x
400
Problem Set 10-4
273
W = 3000( 400 − x ) + 4000 120 2 + x 2 The graph shows a minimum W at x ≈ 136 (exactly x = 360/ 7 = 136.067K).
9.
W 2,000,000
x 400
The walkway should go 400 −136.067… ≈ 263.9 m parallel to the street, then cross the street. 5. a. For minimal path, x = 100/ 21. ∴ sin θ =
x 50 + x 2 2
= 0.4 = 2/5, Q .E.D .
b. For minimal path, x = 400. x sin θ = = 0.8 = 40/50, Q .E.D . 300 2 + x 2
7. sin θ =
49,213 49,002 49,000 49,002 49,155
T (x )
x 500
900
The minimum occurs at an endpoint of the domain. Because Calvin can walk entirely along pavement when x = 0, there is a removable discontinuity in the above function and T(0) = 100 + 240 = 340 s. Because T(500) = 433.333… , which is greater than 340, the minimum time is at x = 0. Calvin’s time is minimized by staying on the sidewalks. If road construction (for instance) prevented Calvin from walking on Heights Street, his time would be minimized by walking directly to Phoebe’s house.
12 13
16x2 = 9(1202 + x2) 7 x 2 = 9 ⋅ 120 2 ⇔ x = ±360/ 7 = 136.067… The walkway should go 400 − 136.067… ≈ 263.9 m parallel to the street, then cross the street. The algebraic solution is easier than before because no algebraic calculus needs to be done. Mathematicians find general solutions to gain insight, and to find patterns and methods to allow easier solution of similar problems.
300 390 400 410 500
500
s , Q .E .D . w
12 x = 30 tan sin −1 = 72 13 The diver should swim 100 − 72 = 28 m, then dive. The algebraic solution is easier than before because no algebraic calculus needs to be done. Mathematicians find general solutions to gain insight, and to find patterns and methods to allow easier solution of similar problems. x 3000 3 8. sin θ = = = 2 2 4000 4 120 + x
C(x), approximate
The table shows that a near miss will have virtually no effect on the minimal cost. For instance, missing the optimal value of x by 10 m will make about a $2 difference in cost, and missing by 100 m makes only a $150 to $200 difference. 1 1 10. T ( x ) = (500 – x ) + 1200 2 + x 2 5 3 The graph shows a local minimum at x ≈ 900 ft (exact: 900 ft), which is out of the domain.
6. Distance swimming = p 2 + x 2 . Distance walking = k − x. 1 2 1 T= p + x 2 + (k – x ) s w x 1 1 1 T′ = − = sin θ − 2 2 w s w s p +x T ′ = 0 ⇔ sin θ =
x
11. 120 300 – x
x 70
1 1 120 2 + x 2 + 70 2 + (300 – x )2 50 130 The graph shows a minimum T at x ≈ 48 yd. T=
T
5
x 300
274
Problem Set 10-4
Calculus Solutions Manual © 2005 Key Curriculum Press
Algebraically: x 300 – x T′ = − 2 2 50 120 + x 130 70 2 + (300 – x )2 Setting T ′ = 0 and simplifying leads to a fourth-degree equation, which must be solved numerically. Minimum is at x = 47.8809… ≈ 47.9 yd.
b. Light always takes the path requiring the least time between two points. c. When you look at the object, your mind tells you that the light rays go straight. Actually, they are bent, as shown in the diagram. So the object is deeper than it appears to be. Because θ water < θ air, v water < v air.
12.
Apparent path of light rays
θAir
θ 120 1
θ1 300 – x
x
θ2
θ2
Apparent depth
Actual path of light rays θ Water 70
Actual depth
15. Answers will vary. From Problem 11, x 300 – x T′ = − 2 2 50 120 + x 130 70 2 + (300 – x )2 By trigonometry, x sin θ1 = , 120 2 + x 2 300 – x sin θ 2 = 2 70 + (300 – x )2 1 1 ∴ T′ = sin θ1 − sin θ 2 50 130 For minimal path, T ′ = 0. Thus, 1 1 sin θ1 = sin θ 2 50 130 sin θ1 50 = , Q. E . D . sin θ 2 130 1 1 a2 + x 2 + b 2 + (k – x )2 v1 v2 x k–x T′ = − 2 2 2 v1 a + x v2 b + ( k – x ) 2
13. T =
sin θ1 = sin θ 2 = ∴ T′ =
x
Problem Set 10-5 Q1. f ′(x) = sin x + x cos x Q3. xex − ex + C Q5. x Q6.
Q2. g″(x) = x− 1 Q4. Snell’s law Q7. y 1
x 1
Q8. Total distance Q9. Newton and Leibniz
Q10. C
1 1. D = t + D′ = 1 − t −2 t The graphs show zero derivative and local minimum of D at t = 1, and maximum of D at t = 3. 3 D or D'
D D' t
,
3
a2 + x 2 k–x
b 2 + (k – x )2 1 1 sin θ1 − sin θ 2 v1 v2
For minimal path, T ′ = 0. Thus, 1 1 sin θ1 = sin θ 2 v1 v2 sin θ1 v1 = , Q .E .D . sin θ 2 v2 14. a. The light rays take the minimal time to get from one point to another, just as Robinson Crusoe wanted to take the minimal time to get from hut to wreck. Calculus Solutions Manual © 2005 Key Curriculum Press
D′ = 0 ⇔ t 2 = 1 ⇔ t = ±1, confirming the graph. Minimum is D(1) = 2, or 2000 mi. 1 Maximum is D(3) = 3 , or about 3333 mi. 3 2. Fuel cost per mile = k · v2. At v = 30, cost = 0.18. 1 0.18 = k ⋅ 30 2 ⇒ k = 5000 100 2000 Driver cost is 20t = 20 ⋅ = . v v 2000 v2 2000 v 2 ∴ C= + ⋅ 100 = + v 5000 v 50 2000 v C′ = − 2 + v 25 Problem Set 10-5
275
The graphs show minimum C at v ≈ 37 mi/h; C′ is multiplied by 10 so that it is easier to see its behavior around C′ = 0. C ′ = 0 at v = 103 50 = 36.8403… C or C' times 100 C 100
v 85 100
37 C' times 10
3. Maximize f(x) = x − x2. f ′(x) = 1 − 2x; f ′(x) = 0 at x = 0.5; f ″(x) = −2, so graph is concave down everywhere. Maximum of f (x) is at x = 0.5. 4. Maximize f (x) = x − x 2 for x ≥ 2. The graph shows maximum at endpoint x = 2. f (x )
1
x
ii. G(2) = 54.5454… , ≈ 55, about 1 point less. 6. a. µ = 130 − 12T + 15T 2 − 4T 3 , 0 ≤ T ≤ 3 dµ = −12 + 30T − 12T 2 = −6(2T − 1)(T − 2) dT dµ = 0 at T = 0.5 or T = 2 dT µ(0) = 130 µ(0.5) = 127.25 µ(2) = 134 µ(3) = 121 Maximum viscosity occurs at T = 2, or 200°. b. Minimum viscosity = 121 centipoise at T = 3, or 300°. c.
dµ dµ dT = ⋅ dt dT dt Because T = t ,
2
When T = 1,
dT 1 1 = = . dt 2 t 2T
dT dµ = 0.5 and = 6. dt dT
dµ = 0.5(6) = 3 dt Viscosity is increasing at 3 centipoise/min. 7. a. Put a coordinate system with origin at the center of the cone’s base. Pick a sample point (x, y) where the cylinder touches the element of the cone. Thus, x is the radius of the cylinder and y is its altitude. The volume and surface area are V = π x 2y A = 2π x 2 + 2π x y The cone element has equation y = −0.6x + 6. V = π x 2(−0.6x + 6) = π (−0.6x3 + 6x2) A = 2π x 2 + 2π x(−0.6x + 6) = π (0.8x2 + 12x) ∴
Because f (2) is the maximum and it is negative, there is no number greater than 2 that exceeds its square. 100t 9 900t 5. a. S = ;F= ;G = t +1 t+9 (t + 1)(t + 9) The graph shows a maximum of G at t = 3 hours. y S G 50 F times 20
t 10
900t 900t = 2 (t + 1)(t + 9) t + 10t + 9 900(t 2 + 10t + 9) – 900t (2t + 10) G′ = (t 2 + 10t + 9)2 900(9 – t 2 ) = 2 (t + 10t + 9)2 G ′ = 0 ⇔ t = ±3 Because G′ changes from positive to negative at t = 3, there is a local maximum there, as in the graph. Fran should study for 3 hours. c. Optimum grade = G(3) = 56.25 ≈ 56 (Not good!) i. G(4) = 55.3846… ≈ 55, about 1 point less.
b. G =
276
Problem Set 10-5
y 600 A
V x 10
b. From the graphs, the maximum volume occurs where the radius x ≈ 6.7 in. The maximum area occurs at x = 10, where all of the area is in the two bases of the cone. Algebraically, V′ = π ( −1.8x2 + 12x) 2 V ′ = 0 ⇔ x = 0 or x = 6 3
Calculus Solutions Manual © 2005 Key Curriculum Press
Maximum V is at x = 6
2 in., as shown on 3
the graph, and y = 2 in. A′ = π (1.6x + 12) A′ = 0 ⇔ x = −7.5, which is out of the domain. A(0) = 0 and A(10) = 200π, so maximum A is at x = 10 in., as shown on the graph, and y = 0 in. The maximum volume and maximum area do not occur at the same radius. Note that the radius of the cone is large compared to its altitude. Thus, the increase in areas of the two bases of the cylinder offsets the decrease in its lateral area as x increases, making the maximum area that of the degenerate cylinder of altitude zero. 8. a. Put a coordinate system with origin at the center of the cone’s base. Pick a sample point (x, y) where the cylinder touches the element of the cone. Thus, x is the radius of the cylinder and y is its altitude. dy dV Know: = 2 in./min. Want: . dt dt V = π x 2y The cone element has equation y = −3x + 18, 1 from which x = 6 − y. 3 2 1 1 V = π 6 − y ⋅ y = π 36 y − 4 y 2 + y 3 3 9 1 1 dV = π 36 − 8 y + y 2 = π ( y − 6)( y − 18) 3 3 dy dV dV dy 2 = ⋅ = π ( y − 6)( y − 18) dt dy dt 3 dV When y = 12, = −24π . dt V is decreasing at 24π = 75.3982… ≈ 75.4 in.3/min. b. If t ∈ [0, 9], then y ∈ [0, 18]. dV = 0 ⇔ y = 6 or y = 18 dt V(0) = 0; V(6) = 96π ; V(18) = 0 Maximum 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) = 48π Maximum V is 88.8888…π ≈ 279.3 in.3 at t = 4 min. dV dy = −0.7 m 3 /min. Want: . 9. Know: dt dt dV = πx2 dy dV dy dy – 0.7 ∴ = πx 2 ⇒ = dt dt dt π x 2 Calculus Solutions Manual © 2005 Key Curriculum Press
When the water is 3 m deep, y = 8. Because y = x 4 + 5, x = 4 3 . dy – 0.7 = = − 0.1286 K ≈ 0.129 m/min dt π 3 10. a. Pick a sample point (x, y) where the cylinder touches the parabola. Thus, the radius of the cylinder is x and its altitude is y. dx dV Know: = 0.3. Want: . dt dt V = π x 2y = π x 2(4 − x 2) = π (4x 2 − x 4) dV dx = π (8 x − 4 x 3 ) ⋅ = 1.2π (2 x − x 3 ) dt dt dV When x = 1.5, = −0.45π . dt dV b. = 4π (2 x − x 3 ) dx dV = 0 ⇔ x = 0, ± 2 (– 2 is out of domain) dx V (0) = V (2) = 0; V ( 2 ) = 4π Maximum volume = 4π ≈ 12.6 units3 at radius = 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 ($) dA 35 2 b. = 3.5 − 0.3t = 0 at t = = 11 days dt 3 3 2 Maximum A at t = 11 , not a minimum, 3 dA because goes from positive to negative dt there. 2 5 c. A 11 = 920 ≈ $920.42 3 12 12. a. 0 ≤ D ≤ 130 ⇒ 0 ≤ 20x + 10 ≤ 130 ⇒ 1 − ≤x≤6 2 0 ≤ W ≤ 310 ⇒ 0 ≤ 10(x 2 − 8x + 22) ≤ 310 ⇒ −1 ≤ x ≤ 9 Given x ≥ 1, the domain of x is [1, 6]. b. Minimize/maximize W on x ∈ [1, 6]. dW = 10(2 x − 8) = 0 at x = 4. dx W(1) = 150; W(4) = 60; W(6) = 100 Minimum: W = 60 ft (at x = 4 mi) Maximum: W = 150 ft (at x = 1 mi) c. C = k · D · W = k · (20x + 10) · 10(x 2 − 8x + 22) = 100k(2x3 − 15x2 + 36x + 22) (k > 0) dC 2 = 100 k (6 x − 30 x + 36) dx = 600 k ( x – 2)( x – 3) dC = 0 ⇔ x = 2 or x = 3 dx Problem Set 10-5
277
C(1) = 4500k; C(2) = 5000k; C(3) = 4900k; C(6) = 13,000k Cheapest bridge at x = 1 mi. d. No. The shortest bridge at x = 4 mi would cost C(4) = 5400k, which is 900k more than the cheapest bridge at x = 1.
Problem Set 10-6 Q1. −x cos x + sin x + C Q2. 2xe3x + 3x2e3x 2x Q3. +C Q4. 53 = 125 ln 2 1 4 6 sec 2 6t 2 x +7 Q6. = sec 2 ln x 2 4 3e 3t x Q7. parametric Q8. x ln x − x + C Q9. x2 Q10. E 1. The velocity is tangent to the path, the acceleration is toward the concave side of the path, and there is an obtuse angle between acceleration and velocity. Q5.
y
→ v
→ r
→ a
x
2. The velocity is tangent to the path, the acceleration is toward the concave side of the path, and there is an acute angle between the acceleration vector and the velocity vector. y
→ v → a → r
x
r r r 3. a. r = (e t cos t ) i + (e t sinr t ) j r r v = (e t cos t − e t sin t ) i + (e t sin t + e t cos t ) j r r r v (1) = −0.8186 Ki + 3.7560 … j x is decreasing at t = 1 because dx/dt is negative. Speed = 0.8186 K2 + 3.7560...2 = 3.8442… ≈ 3.84 cm/s b. L = 1
∫ =∫
(e cos t − e sin t ) + (e sin t + e cos t ) dt t
0
1
0
t
2
t
2
s1 (1) = 12 + 2 2 ≈ 2.24 cm/s and s2 (1) = 1.52 + 1.52 ≈ 2.12 cm/s b. Distance =
∫
4
1
12 + [2(t – 2)]2 dt ≈
6.1257… ≈ 6.13 m c. The paths cross at (x, y) = (−1, r 1) and (2, 4). By tracing on the grapher, r1 is at (−1, 1) r when t = 1, but r2 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. r r r 5. a. r (t ) = (10 sin 0.6t )i + ( 4 cos 1.2t ) j r r r v (t ) = (6 cos 0.6t )i + ( −4.8 sin 1.2t ) j r r r a (t ) = ( −3.6 sin 0.6t )i + ( −5.76 cos 1.2t ) j r r r b. r (0.5) = (10 sin 0.3)i + ( 4 cos 0.6) j r r = 2.9552 …i + 3.3013… j r r r v (0.5) = (6 cos 0.3)i + ( −4.8 sin 0.6) j r r = 5.7320 …i − 2.7102 … j r r r a (0.5) = ( −3.6 sin 0.3)i + ( −5.76 cos 0.6) j r r = −1.0638…i − 4.7539… j r r r The graph shows r , v , and a at t = 0.5. y
t = 0.5 at v
a
x an a
t=7 v
2e 2 t dt = 2 (e – 1) = 2.4300 K ≈ 2.43
Distance from origin = (e1 cos1)2 + (e1 sin 1)2 = e = 2.7812… ≈ 2.78 cm 278
t
(Note that the distance traveled is less than the distance from the origin because the particle started at (1, 0), not at (0, 0).) r r r c. a = ( −2e t sin t )i + (2e t cos t ) j . r r r a (1) = ( −4.5747...) i + (2.9373...) j r r r 4. a. r1 = (t − 2)i + (t − 2) 2 j and r r r r2 = (1.5t − 4)i + (1.5t − 2) j r r r r r r v1 = 1i + 2(t − 2) j and v2 = 1.5i + 1.5 j r r r r r r a1 = 0i + 2 j and a2 = 0i + 0 j r r r r r r v1 (1) = i − 2 j and v2 (1) = 1.5i + 1.5 j r r r r r r a1 (1) = 0i + 2 j and a2 (1) = 0i + 0 j
Problem Set 10-6
These vectors make sense because the head of r r r is on the graph, v is tangent to the graph, r and a points to the concave side of the graph. c. The object is speeding up because the angle r r between a and v is acute. Calculus Solutions Manual © 2005 Key Curriculum Press
r d. | v(0.5) | = (6 cos 0.3)2 + (–4.8 sin 0.6)2 = 6.3404… r r a (0.5) ⋅ v (0.5) = ( −3.6 sin 0.3)(6 cos 0.3) + (−5.76 cos 0.6)(−4.8 sin 0.6) = 6.7863… , so the angle is acute. r r a (0.5) ⋅ v (0.5) P= = 1.0703… r |v (0.5)| r r v (0.5) at (0.5) = P r | v (0.5)| r r (6 cos 0.3)i + (–4.8 sin 0.6) j =P r | v (0.5)| r r = 0.9676 …i − 0.4575… j r r r an (0.5) = a (0.5) − at (0.5) r r = −2.0314 …i − 4.2964 … j See the graph in part b. r e. The object is speeding up at | at (0.5)| = P = 1.0703… ≈ 1.07 (ft/s)/s. r r r f. r (7) = (10 sin 4.2)ri + ( 4 cos 8.4r) j = −8.7157…i − 2.0771… j r r r v (7) = (6 cos 4.2)i + ( −4.8 sin 8.4) j r r = −2.9415…i − 4.1020 … j r r r a (7) = ( −3.6 sin 4.2)i + ( −5.76 cos 8.4) j r r = 3.1376 …i + 2.9911… j See the graph in part b. The object is slowing down because the r r angle between ra andr v is obtuse. a (7) ⋅ v (7) (Note that P = = −4.2592 … , so r | v (7)| the object is slowing down at 4.2592… ≈ 4.26 (ft/s)/s.) r r r r r g. r (0) = (10 sin 0)i + ( 4 cos 0) j = 0i + 4 j r r r r r v (0) = (6 cos 0)i + ( −4.8 sin 0) j = 6i + 0 j r r r a (0) = ( −3.6 sin 0)i + ( −5.76 cos 0) j r r = 0i − 5.76 j r r a (0) ⋅ v (0) = (0)(6) + ( −5.76)(0) = 0 r r ∴ a(0) and v(0) are perpendicular, Q.E.D. This means the object is neither slowing down nor speeding up at t = 0. r r r 6. a. r (t ) = (8 cos 0.8t )i + (6 sin 0.4t ) j r r r v (t ) = ( −6.4 sin 0.8t )i + (2.4 cos 0.4t ) j r r r a (t ) = ( −5.12 cos 0.8t )i + ( −0.96 sin 0.4t ) j r r r b. r (1) = (8 cos 0.8)i + (6 sin 0.4) j r r = 5.5736 …i + 2.3365… j r r r v (1) = ( −6.4 sin 0.8)i + (2.4 cos 0.4) j r r = −4.5910 …i + 2.2105… j r r r a (1) = ( −5.12 cos 0.8)i + ( −0.96 sin 0.4) j r r = −3.5671…i − 0.3738… j
Calculus Solutions Manual © 2005 Key Curriculum Press
r r r The graph shows r , v , and a at t = 1. y t = 1.25π a
v
at a
r
t=1 an
x
r a v t = 10.5
c. d.
e. f.
These vectors make sense because the head of r r r is on the graph, v is tangent to the graph, r and a points to the concave side of the graph. The object is speeding up because the angle r r between a and v is acute. r | v(1) | = (–6.4 sin 0.8)2 + (2.4 cos 0.4)2 = 5.0955… r r a (1) ⋅ v (1) = ( −5.12 cos 0.8)( −6.4 sin 0.8) + (−0.96 sin 0.4)(2.4 cos 0.4) = 15.5506… , so the angle is acute. r r a (1) ⋅ v (1) P= r = 3.0518… | v (1)| r r v (1) at (1) = P r | v (1)| r r (–6.4 sin 0.8)i + (2.4 cos 0.4) j =P r | v (1)| r r = −2.7496 …i + 1.3239… j r r r an (1) = a (1) − at (1) r r = −0.8174 …i − 1.6977… j r r See the graph in part b, showing at and an at t = 1. The r object is speeding up at | at (1)| = P(1) = 3.0518… ≈ 3.05 (ft/s)s. r r r r (10.5) = (8 cos 8.4)i + (6 sin 4.2) j r r = −4.1543…i − 5.2294 … j r r r v (10.5) = ( −6.4 sin 8.4) i + (2.4 cos 4.2) j r r = −5.4694 …i − 1.1766 … j r r r a (10.5) = ( −5.12 cos 8.4)i + ( −0.96 sin 4.2) j r r = 2.6587…i + 0.8367… j r r r See the graph in part b, showing r , v , and a at t = 10.5. The object is slowing downr at t =r10.5 because the angle between a and v is obtuse at that time. (Note that r r a (10.5) ⋅ v (10.5) P= = −2.7552 … , r |v (10.5)| so the object is slowing down at about 2.8 (ft/s)/s.) Problem Set 10-6
279
g. The object is stoppedrwhen r r r v (t ) = ( −6.4 sin 0.8t )i + (2.4 cos 0.4t ) j = 0 ⇔ −6.4 sin 0.8t = 0 and 2.4 cos 0.4t = 0. Using −6.4 sin 0.8t = −1.28 rsin 0.4t r cos 0.4t, you see that v (t ) = 0 exactly when cos 0.4t = 0; the first time this happens is at t = 1.25π s. r r r r (1.25π ) = (8 cos π )i + (6 sin 0.5π ) j r r = −8i + 6 j r r r a (1.25π ) = ( −5.12 cos π )i + ( −0.96 sin 0.5π ) j r r = 5.12i − 0.96 j r See the graph in part b, showing a at t = 1.25π . The acceleration vector points along the path at t = 1.25π. So the object is stopped, but it has a nonzero acceleration. At first glance, this fact may be surprising to you! r π r π r 7. a. r (t ) = 10 cos t i + 6 sin t j 6 6 r r π π r 10π 6π v (t ) = − sin t i + cos t j 6 6 6 6 r r π π r 10π r (t ) + v (t ) = 10 cos t − sin t i 6 6 6 π π r 6π + 6 sin t + cos t j 6 6 6 r r The graph shows the path of r + v . y
3 2
4 5
1 0 x 12
6 7
11 10
8
9
r b. See the graph in part a, showing vectors v . r r c. For r + v , x π π π = cos t − sin t 10 6 6 6 y π π π = sin t + cos t 6 6 6 6 2 x = cos 2 π t − π cos π t sin π t 10 6 3 6 6
π π + sin 2 t 6 6 2
y = sin 2 π t + π sin π t cos π t 6 6 3 6 6 2
π π + cos 2 t 6 6 2
280
Problem Set 10-6
π π x y ∴ + = cos 2 t + sin 2 t 10 6 6 6 2
2
π π π π + sin 2 t + cos 2 t 6 6 6 6 2
2
x + y = 1+ π 10 6 6 2
2
2
x2 y2 + =1 100(1 + π 2 /36) 36(1 + π 2 /36) This is the equation of an ellipse centered at the origin with x-radius 11.2878… and y-radius 6.7727… . 10π 2 r π r 6π 2 π r d. a (t ) = − cos t i + − sin t j 6 6 36 36 r r r (t ) + a (t ) 10π 2 6π 2 π r π r = 10 − sin t j cos t i + 6 − 36 6 36 6 See the graph in part a, showing an elliptical path followed by the heads of the acceleration vectors. e. The direction of each acceleration vector is the opposite of the corresponding position vector and is thus directed toward the origin. r π2 r Note that a (t ) = − r (t ). 36 r r r 8. a. r (t ) = (0.5t cos t )i + (0.5t sin t ) j r r v (t ) = (0.5 cos t − 0.5t sin t )i r + (0.5 sin t + 0.5t cos t ) j r r a (t ) = ( − sin t − 0.5t cos t )i r + (cos t − 0.5t sin t ) j r r r b. r (8.5) = 4.25 cos 8.5 i + 4.25 sin 8.5 j r r = −2.5585…i + 3.3935… j r r v (8.5) = (0.5 cos 8.5 − 4.25 sin 8.5)i r + (0.5 sin 8.5 + 4.25 cos 8.5) j r r = −3.6945…i − 2.1593… j r r a (8.5) = ( − sin 8.5 − 4.25 cos 8.5)i r + (cos 8.5 − 4.25 sin 8.5) j r r = 1.7600 …i − 3.9955… j r r r r (12) = 6 cos 12i + 6 sin 12 j r r = 5.0631…i − 3.2194 … j r r v (12) = (0.5 cos 12 − 6 sin 12)i r + (0.5 sin 12 + 6 cos 12) j r r = 3.6413…i + 4.7948… j r r a (12) = ( − sin 12 − 6 cos 12)i r + (cos 12 − 6 sin 12) j r r = −4.5265…i + 4.0632 … j Calculus Solutions Manual © 2005 Key Curriculum Press
r r The graph shows that r (8.5) and r (12) really do terminate on the path. y
t = 8.5 v
r
r At x = 2, speed = | v (2) | = 153 = 12.3693… ≈ 12.4 cm/s. r r c. The graph shows r (2) and v(2). r This is reasonable because v(2) points along the curve to the left, indicating that x is decreasing.
v
a
y
x
a
a r
c.
d.
e.
f.
g.
9. a.
b.
at t = 12
r See the graph in part b, showing v(8.5), r r r v (12), a (8.5), and a(12). The velocity vectors point along the path as it spirals outward, and the acceleration vectors point inward to the concave side of the graph. r r In both cases, the angle between a and v appears to be acute. Check using dot products. r r a (8.5) ⋅ v (8.5) = 2.125, which is positive. r r a (12) ⋅ v (12) = 3, which is also positive. Thus, the angles are acute, and the object is speeding up at both times. r r At t = 12 h, a (12) ⋅ v (12) = 3. r | v(12) | = 36.25 = 6.0207… r r r r a (12) ⋅ v (12) v (12) at (12) = ⋅ r r | v (12)| | v (12)| r r 3 = (3.6413...i + 4.7948... j ) 36.25 r r = 0.3013…i + 0.3968… j r r r an (12) = a (12) − at (12) r r = −4.8279…i + 3.6664 … j r r See the graph in part b, showing an and at at t = 12. r Speed = | v(12) | = 36.25 = 6.0207… ≈ 6.02 mi/h 3 = Speed is increasing by P(12) = 36.25 0.4982… ≈ 0.498 (mi/h)/h. r r See the graph in part b, showing r (t ) + a (t ). The heads seem to lie on a unit circle. Algebraic verification: r r r r r (t ) + a (t ) = − sin t i + cos t j , which is a circle. r r r r r r ( x ) = xi + yj = xi + x 2 j r dx r dy r dx r dx r v( x) = i+ j= i + 2x j dt dt dt dt r r r dx = −3 ⇒ v ( x ) = −3i − 6 xj dt r r r v (2) = −3i − 12 j
Calculus Solutions Manual © 2005 Key Curriculum Press
at
20
an
x=2 x 5
v
r r r d. From part b, v ( x ) = −3i − 6 xj , r r r dx r ∴ a ( x ) = 0i − 6 j = 18 j dt r r a (2) = 18 j . See the graph in part c. r r r r a (2) ⋅ v (2) v (2) e. at (2) = ⋅ r r | v (2)| | v (2)| r r –216(–3i – 12 j ) = 153 r r 72 r 288 r = i+ j = 4.2352 …i + 16.9411… j r 17 r 17 r an (2) = a (2) − at (2) r r 72 r 18 r =− i + j = −4.2352 …i + 1.0588… j 17 r 17 r at (2) is parallel to the curve. an (2) is normal to the curve and points inward to the concave side. f. The object is slowing downr when xr= 2 because the angle between a(2) and v(2) is obtuse, as shown by the graph and by the r fact that the dot product is negative. Also, at (2) r points in the opposite direction of v(2). g. dL = 1 + ( dy/dx )2 dx = 1 + 4 x 2 dx dL dx = 1 + 4x2 =5 dt dt dx dx At x = 2, 5 = 1 + 4(2)2 = 17 dt dt dx 5 ⇒ = = 1.2126 … ≈ 1.21 cm/s. dt 17 r r r 10. r (1) = 8.8615…i + 4.8410 … j r r r q (2) = −2.9659…i + 4.3406 … j r r r q (1.5) = −0.2065…i + 6.6362 … j r r r q (1.1) = 2.5579…i + 7.4880 … j r r v (t ) = (12 cos t cos 0.5t − 6 sin t sin 0.5t )i r + (12 cos t sin 0.5t + 6 sin t cos 0.5t ) j r r r v (1) = 3.2693…i + 7.5391… j Problem Set 10-6
281
The graph shows average velocity vectors approaching the instantaneous velocity vector r v(1) as t approaches 1. The instantaneous velocity vector is tangent to the graph and points in the direction of motion.
e. When x = 400, t = 2 sec 15° = 2.0705… s y(2.0705…) = 41.5846… . Phyllis makes the home run because 41.5… > 10. y
y
t = 1.1
100
v
t = 1.5 10
x 400
t=2
x 10
r r r 11. r (t ) = (10 sin 0.6t )i + ( 4 cos 1.2t ) j dL = ( dx/dt )2 + ( dy/dt )2 dt = (6 cos 0.6t )2 + (–4.8 sin 1.2t )2 dt L=
2
∫ dL ≈ 12.0858… ft (numerically) 0
r π r π r 12. r (t ) = 10 cos t i + 6 sin t j 6 6 dL = ( dx/dt )2 + ( dy/dt )2 dt
π 6π π –10π = ⋅ sin t + ⋅ cos t dt 6 6 6 6 One complete cycle of the curve is 0 ≤ t ≤ 12, so 2
L=
∫
2
12
dL = 51.0539… (numerically) ≈ 51.1 ft. r r r 13. a. a (t ) = 0i − 32 j r r r r r v (t ) = (0i − 32 j ) dt = C1i + ( −32t + C2 ) j r r r v (0) = −130i + 0 j ⇔ C1 = −130 and C2 = 0 r r r ∴ v (t ) = −130i − 32tj r r r b. r (t ) = ( −130i − 32tj ) dt r r = ( −130t + C3 )i + ( −16t 2 + C4 ) j r r r r (0) = 60.5i + 8 j ⇔ C3 = 60.5 and C4 = 8 r r r ∴ r (t ) = ( −130t + 60.5)i + ( −16t 2 + 8) j 0
∫
∫
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 the strike zone. d. At t = 0, dx/dt = 200 cos 15°, dy/dt = 200 sin 15°. r r r As in part a, v (t ) = C1i + ( −32t + C2 ) j = r r (200 cos 15°)i + ( −32t + 200 sin 15°) j r r As in part b, r (t ) = (200t cos 15°)i + r ( −16t 2 + 200t sin 15° + 3) j . 282
Problem Set 10-6
r r r 14. a. a (t ) = 3i + y ′′(t ) j r r r v (t ) = (3t + C1 )i + ( y′(t )) j r r r v (0) = 0i + 0 j ⇒ C1 = 0 r r r r (t ) = (1.5t 2 + C2 )i + ( y(t )) j r r r r (0) = 0i + 0 j ⇒ C2 = 0 r r r r r ∴ r (t ) = (1.5t 2 )i + ( y(t )) j = (1.5t 2 )i + (sin x (t )) j r r r r (t ) = (1.5t 2 )i + (sin 1.5t 2 ) j r r r b. v (t ) = (3t )i + (3t cos 1.5t 2 ) j If x = 6, t = 2. r r r v (2) = 6i + 5.7610 K j r Speed = | v(2) | = 6 2 + 5.76 K2 = 8.3180 … ≈ 8.32 m/ min 15. a. d = a + b cos t t = 0: 240 = a + b cos 0 = a + b t = π : −60 = a + b cos π = a − b 2a = 180 ⇒ a = 90 2b = 300 ⇒ b = 150 ∴ d = 90 + 150 cos t r r b. r (t ) = (90 cos t + 150 cos 2 t ) i r + (90 sin t + 150 sin t cos t ) j r r v (t ) = ( −90 sin t − 300 cos t sin t ) i r + (90 cos t + 150 cos 2 t − 150 sin 2 t ) j r r v (t ) = ( −90 sin t − 150 sin 2t ) i r + (90 cos t + 150 cos 2t ) j r r r v (1) = −212.1270 Ki − 13.7948K j Speed = (–212.1K)2 + (–13.7K)2 = 212.5750 K ≈ 212.6 cm/s r r c. a (t ) = ( −90 cos t − 300 cos 2t ) i r + ( −90 sin t − 300 sin 2t ) j r r r a (1) = 76.2168Ki − 348.5216 K j r r a (1) ⋅ v (1) P(1) = r = −53.4392 K | v (1)| r r v (1) at (1) = P(1) r | v (1)| r r = 53.3266 Ki + 3.4678K j Calculus Solutions Manual © 2005 Key Curriculum Press
r r r an (1) = a (1) − at (1) r r = 22.8902 Ki − 351.9894 K j Annie is slowing down. The angle between the acceleration and velocity vector is obtuse, as revealed by the negative dot product. She is slowing down at 53.4392… ≈ 53.4 cm/s. r r r 16. a. r = (0.5t + sin t ) i + ( 4 cos 0.5t ) j r r r v (t ) = (0.5 + cos t ) i + ( −2 sin 0.5t ) j r r r a (t ) = ( − sin t )i + ( − cos 0.5t ) j r r r v (14) = 0.6367…i − 1.3139… j r r r a (14) = −0.9906 …i − 0.7539… j r Speed = | v(14) | = 1.4601… mi/h r r a (14) ⋅ v (14) P(14) = r | v (14)| 0.3598K = 0.2464 K 1.4601K r r v (14) at (14) = P(14) r | v (14)| r r = 0.1074 …i − 0.2217… j r r r an (14) = a (14) − at (14) r r = −1.0980 …i − 0.5312 … j =
The log is speeding up at t = 14. You can tell by the fact that rP(14) is positive, and thus the r angle between a(14) and v(14) is acute, which r means that at (14) points in the same direction r as v(14) . It is speeding up at 0.2464… ≈ 0.246 (mi/h)/h. b. dL = dx 2 + dy 2 = (0.5 + cos t )2 + (–2 sin 0.5t )2 dt L=
∫
14
0
dL ≈ 22.7185… (numerically)
≈ 22.7 mi 1 (22.7185K) 14 = 1.6227… ≈ 1.62 mi/h r r r 17. a. r (t ) = (5t − 12 sin t ) i + (15 + 12 cos t ) j r r r v (t ) = (5 − 12 cos t ) i + ( −12 sin t ) j r r r a (t ) = (12 sin t ) i + ( −12 cos t ) j r r r b. v (2.5) = 14.6137…i − 7.1816 … j r r r a (2.5) = 7.1816 Ki + 9.6137K j Average speed ≈
Calculus Solutions Manual © 2005 Key Curriculum Press
r r The graph shows v (2.5) and a (2.5). y
a
5
x 5
v
r r a (2.5) ⋅ v (2.5) r | v (2.5)| 60 sin 2.5 = = 2.2052 … 169 – 120 cos 2.5 r r v (2.5) at (2.5) = P(2.5) r | v (2.5)| r r = 1.9791…i − 0.9726 … j r r r an (2.5) = a (2.5) − at (2.5) r r = 5.2024 …i + 10.5863… j r d. v(2.5) is reasonable because its graph points along r the path in the direction of motion. a(2.5) is reasonable because it points toward the concave side of therpath. The roller coaster is traveling at | v(2.5) | = 16.2830… ft/s. Its speed is increasing at P(2.5) = 2.2052… ft/s2, as shown by the fact that P(2.5) r is positive, r meaning that the angle between a(2.5) and v(2.5) is acute. e. The path is at a high r point when the y-component of r is a maximum. This happens when cos t = 1, or t = 0 + 2π n. r r r a (0 + 2π n) = 0i − 12 j , pointing straight down. Similarly, the path is at a low point when cos t = −1, or t = π + 2π n. r r r a (π + 2π n) = 0i + 12 j , pointing straight up, Q .E .D . c. P(2.5) =
f. dL = dx 2 + dy 2 = (5 – 12 cos t )2 + (–12 sin t )2 dt L=
∫
2π
0
dL ≈ 78.7078… (numerically)
≈ 78.7 ft
r r 18. Recall that |i | = | j | = 1. r r The angle between i and i is 0, so r r r r i ⋅ i = |i ||i | cos 0 = 1. r r Similarly, j ⋅ j = 1.
Problem Set 10-6
283
r r π The angle between i and j is , so 2 r r r r π i ⋅ j = | i | | j | cos = 0. r 2 r r r r r ∴ v1 ⋅ v2 = ( x1i + y1 j ) ⋅ ( x2 i + y2 j ) = r r r r r r r r x1 x 2 i ⋅ i + x1 y2 i ⋅ j + y1 x 2 j ⋅ i + y1 y2 j ⋅ j = x1 x 2 (1) + x1 y2 (0) + y1 x 2 (0) + y1 y2 (1) = x1 x 2 + y1 y2 , Q .E.D . r r r 19. r (t ) = (10 sin 0.8t )i + (10 cos 0.6t ) j r + (6t 0.5 )k r r r v (t ) = (8 cos 0.8t )i + ( −6 sin 0.6t ) j r + (3t −0.5 )k r r a (t ) = ( −6.4 sin 0.8t )i r r + ( −3.6 cos 0.6t ) j + ( −1.5t −1.5 )k r r r r v (1) = (8 cos 0.8)i + ( −6 sin 0.6) j + 3k r r r r a (1) = ( −6.4 sin 0.8)i + ( −3.6 cos 0.6) j − 1.5k To determine whether the object is speeding up or slowing down, find the dot product. r r a (1) ⋅ v (1) = ( −6.4 sin 0.8)(8 cos 0.8) + ( −3.6 cos 0.6)( −6 sin 0.6) + ( −1.5)(3) = −20.0230… ∴ the object is slowing down because the angle r r between a(1) and v(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 . dy/dt y′(t ) c. tan φ = = dx/dt x ′(t ) ⇒
d d y ′(t ) (tan φ ) = ds ds x ′(t )
dφ x ′ y′′ – x ′′ y′ dt = ds x′2 ds dφ x ′ y′′ – x ′′ y′ ∴ = ds x ′ 2 (sec 2 φ )( ds/dt )
⇒ sec 2 φ
∴
dφ 15 sin 2 t + 15 cos 2 t = dt (25 sin 2 t + 9 cos 2 t )3/ 2 15 = (16 sin 2 t + 9)3/ 2
which is maximized for sin2 t = 0 at (± 5 , 0), the ends of the major axis, and is minimized for sin2 t = 1 at (0, ±3), the ends of the minor axis, Q.E.D. e. x = r cos t ⇒ x′ = −r sin t ⇒ x′′ = −r cos t y = r sin t ⇒ y′= r cos t ⇒ y′′ = −r sin t r |v | = r 2 cos 2 t + r 2 sin 2 t = r dφ r 2 sin 2 t + r 2 cos 2 t 1 = = ds | r |3 |r | 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 2t y = 3 sin2 t ⇒ y′ = 6 sin t cos t = 3 sin 2t ⇒ y′′ = 6 cos 2t r |v | = 25 sin 2 2t + 9 sin 2 2t = 34 | sin 2t | ∴
dφ –30 sin 2t cos 2t + 30 sin 2t cos 2t = ds 34 |sin 2t | = 0, Q.E.D. 15 5 dφ g. At (5, 0), sin t = 0, so = 3/ 2 = . 9 ds 9 9 Radius of curvature = = 1.8 5 h. The osculating circle has radius 1.8 and center on the x-axis at x = 4 − 1.8 = 3.2. Equations are x = 3.2 + 1.8 cos t y = 1.8 sin t The graph shows the osculating circle. The name is appropriate because the circle “kisses” the ellipse at the point (5, 0). ∴
3 y
x 3.2
But sec φ = 1 + tan φ = 1 + y′ /x′ , so r ds/dt = | v |. 2
∴
2
2
2
dφ x ′ y′′ – x ′′ y′ = 2 r ds x ′ (1 + y′ 2 / x ′ 2 ) |v | =
x ′ y′′ – x ′′ y′ x ′ y′′ – x ′′ y′ , Q .E .D . r 2 2 r = ( x ′ + y ′ ) |v | | v |3
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 r |v | = 25 sin 2 t + 9 cos 2 t
284
5
Problem Set 10-7
Problem Set 10-7 Review Problems R0. Answers will vary. R1. v = t − 3 = 0 at t = 9 s v > 0 for t > 9 s
Calculus Solutions Manual © 2005 Key Curriculum Press
Displacement from t = 0 to t = 9 is 9
∫( 0
ii. The rectangle has the same area as the shaded region.
t – 3) dt = −9.
f (x )
They have moved 9 ft closer to the sawmill. From t = 0 to t = 25: 25 1 Displacement = ( t – 3) dt = 8 ft 0 3 25 1 Distance = | t − 3| dt = 26 ft 0 3 R2. a. i.
18
∫
∫
5
iii. The average of the two values of f(x) at the endpoints is 0, not 18. R4. a.
v (t )
5
ii. Displacement =
4
700
∫ (2
t
1
– 8) dt ≈ −3.8022 K
≈ −3.8 cm (exactly 14/ln 2 − 24) iii. Distance =
4
∫ |2 1
t
− 8| dt ≈ 10.8853…
≈ 10.9 cm (exactly 2/ln 2 + 8) b. tend
2 — 30 8 5 55 1 4.5 77.5 0 0.5 80 −10 −5 55 −20 −15 −20
speeding up speeding up speeding up neither slowing down slowing down
b. i.
3
∫ sin (π t/6) dt = 2/π = 0.6366K 0
9
∫ sin (π t/6) dt = 0
1 12
3
∫
12
0
sin (π t/6) dt = 0
f ( x ) = 6 x 2 − x 3 = x 2 (6 − x ) = 0 at x = 0, 6 Average =
1 6
T
100
x
(Note that the object is speeding up, slowing down, or neither, exactly when aend > 0, aend < 0, or aend = 0, respectively, in the original table.)
iii. vav =
Let x = distance from intersection to cutoff. 0 ≤ x ≤ 700 Let T = total time taken. 1 1 T= 200 2 + x 2 + (700 – x ) 5.7 6.2
aav
a
1 3 1 ii. vav = 6
x
700 – x
3
R3. a. i. vav =
θ 200
θ
t
0 5 10 15 20 25
x 6
6
∫ (6 x
Calculus Solutions Manual © 2005 Key Curriculum Press
0
2
– x 3 ) dx = 18
700
1 1 1 ⋅ (200 2 + x 2 ) −1/2 (2 x ) − 5.7 2 6.2 1 1 2 2 −1/2 = ⋅ x (200 + x ) − 5.7 6.2 T′ = 0 ⇔ 6.2x = 5.7(2002 + x 2)1/2 38.44x2 = 32.49(2002) + 32.49x2 32.49(200 2 ) x2 = ⇔ x = ±467.3544 K 5.95
T′ =
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 along the sidewalk. 1 T(0) = ⋅ 900 = 145.1612 K 6.2 T(467.3544…) = 126.7077… T(700) = 127.7212… Turning at a point about 467 ft from the intersection of the two sidewalks takes the minimum time, although it takes only a second longer to head straight for the English building.
Problem Set 10-7
285
b. 6 θ
x θ 10 10 – x
Let x = distance from closest point on the beach to the cutoff point, 0 ≤ x ≤ 10. Let C = total cost of the road. C = 5(10 − x ) + 13 36 + x 2 The graph shows a minimum C at x ≈ 2.5. C 150
x 10
Let θ be the angle of incidence. By the minimal path property, the cost is minimized when x 5 sin θ = = bridge length 13 x = 6 tan [sin− 1 (5/13)] = 2.5 C(2.5) = 122 C(10) = 151.6047… The minimum cost is $122,000, obtained by going 7.5 km along the beach, then cutting across 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 − 2t a′(t) = 0 ⇔ t = 3 a(0) = 0; a(3) = 9; a(10) = −40 Maximum acceleration = 9 at t = 3. Minimum acceleration = −40 at t = 10. 1 ii. v(t ) = (6t − t 2 ) dt = 3t 2 − t 3 + C 3 v(0) = 0 ⇒ C = 0 1 ∴ v(t ) = 3t 2 − t 3 3 v′(t) = a(t) = t(6 − t) v′(t) = 0 ⇔ t = 0 or t = 6 1 v(0) = 0; v(6) = 36; v(10) = −33 3 Maximum velocity = 36 at t = 6. 1 Minimum velocity = −33 at t = 10. 3 1 iii. s(t ) = v(t ) dt = t 3 − t 4 + C 12
Because s(t) measures distance from the starting point, s(0) = 0, which implies that C = 0. 1 ∴ s( t ) = t 3 − t 4 12 1 s ′(t ) = v(t ) = t 2 ( 9 − t ) 3 s′(t) = 0 ⇔ t = 0 or t = 9 1 2 s(0) = 0; s(9) = 182 ; s(10) = 166 4 3 1 Maximum displacement = 182 at t = 9. 4 Minimum displacement = 0 at t = 0. b. i. Let t = number of days Dagmar has been saving, P(t) = number of pillars in Dagmar’s account, and V(t) = real value (in constant day-zero pillars) of money in account after t days. P(t) = 50t (assuming continuous depositing) 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. V (t ) 6000
t 500
V′(t) = 50(0.50.005 t ) + 50t[0.005(0.50.005 t )] ln 0.5 V′(t) = 0 ⇔ 1 = −0.005t ln 0.5 200 t=− = 288.5390 K ln 0.5 Dagmar’s greatest purchasing power will be after about 289 days. R6. a. i. and ii. v
∫
∫
286
Problem Set 10-7
a Speeding up Slowing down
a
r r r b. i. r (t ) = (5 cosh t )i + (3 sinh t ) j r r r v (t ) = (5 sinh t )i + (3 cosh t ) j r r r a (t ) = (5 cosh t )i + (3 sinh t ) j r r r r (1) = (5 cosh 1)i + (3 sinh 1) j r r = 7.7154 Ki + 3.5256 K j r r r v (1) = 5.8760 Ki + 4.6292 K j r r r a (1) = 7.7154 Ki + 3.5256 K j
Calculus Solutions Manual © 2005 Key Curriculum Press
r r r ii. The graph shows r (1), v (1), and a(1). r r Note that a (1) = r (1) so that the acceleration vector points directly away from ther origin when it is drawn from the head of r (1). (For an elliptical path, the acceleration vector points directly toward the origin.) y
Concept Problems C1. a. 5
v (t ) t
0
4
–5
v a Asymptote
5
r x 5
10
r iii. Speed = | v (1) | = 25 sinh 2 1 + 9 cosh 2 1 = 7.4804… ≈ 7.48 units/min r r a (1) ⋅ v (1) = 34 sinh 1 cosh 1 = 61.6566 K The object is speeding up at t = 1, as shown by the positive dot product and by r r the acute angle between a(1) and v(1). r r r r a (1) ⋅ v (1) v (1) at (1) = r r | v (1)| | v (1)| 34 sinh 1 cosh 1 r = v (1) 25 sinh 2 1 + 9 cosh 2 1 r = 1.1018… v (1) r r = 6.4744 …i + 5.1007… j r r r ( an (1) = 1.2409…i − 1.5751… j ) r r r | a (1) ⋅ v (1)| 61.6566 … | at (1)| = = r | v (1)| 7.4804 … = 8.2423… r r at (1) points in the same direction as v(1), as indicated by the positive dot product and r r by the acute angle between a(1) and v(1), so the object is speeding up at about 8.24 units/min2. iv. Distance = =
∫
1
0
∫
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 5 32 5 v(0) = −9; v = =1 ; 3 27 27 v(3) = 0; v( 4) = 3 Maximum velocity at t = 4, minimum velocity at t = 0. d. v″(t) = 6t − 14 v″(t) = 0 ⇔ t = 7/3 v″(t) changes from negative to positive at t = 7/3, so there is a point of inflection at that point. e. At t = 7/3, the particle’s acceleration stops decreasing and starts increasing. Thus, the minimum acceleration is at that time. 1 7 15 f. y = v(t ) dt = t 4 − t 3 + t 2 − 9t + C 4 3 2 y(0) = 4 ⇒ C = 4 1 7 15 ∴ y = t 4 − t 3 + t 2 − 9t + 4 4 3 2 g.
∫
y (t ) 4
t
1
ds
0
4
0
25 sinh 2 t + 9 cosh 2 t dt
= 4.5841… (numerically) ≈ 4.58 units r r r v. r (t ) + v (t ) = (5 cosh t + 5 sinh t )i r + (3 sinh t + 3 cosh t ) j Note that the y-coordinate is 0.6 times the x-coordinate, so the head lies on y = 0.6x, one asymptote of the hyperbola.
Calculus Solutions Manual © 2005 Key Curriculum Press
h. y′(t) = v(t) = 0 when t = 1, 3. 5 7 8 y(0) = 4; y(1) = ; y(3) = ; y( 4) = 12 4 3 Maximum y at t = 0, minimum y at t = 1. 5 i. y ′′ = v ′ = (3t − 5)(t − 3) = 0 at t = , 3 3 5 y″ changes sign at t = and at t = 3, so there 3 are points of inflection at these values of t.
Problem Set 10-7
287
j. At t = 5/3, the particle stops accelerating and starts decelerating, so the velocity at that time is a local maximum. At t = 3, the particle stops decelerating and starts accelerating, so the velocity is a local minimum. k. y is never negative because its minimum value is 5/12 at t = 1. l. Displacement =
4
∫ v ( t ) = y( 4 ) − y( 0 ) 0
8 4 − 4 = − ft 3 3 4 5 m. Distance = | v(t )| dt = 5 ft 0 6 1 4 1 1 n. vav = v dt = ⋅ (displacement) = − ft/s 4 0 4 3 4 1 1 35 o. |v|av = |v| dt = ⋅ (distance) = ft/s 4 0 4 24 C2. Assume that the maximum g a human can withstand is A and that the distance from New York to Los Angeles is D km. Recall that 1 g = 9.81 (m/s)/s. For the fastest trip, the passenger accelerates at A g for the first D/2 km, then decelerates at −A g for the last D/2 km. Starting at rest, the velocity t seconds after leaving New York, when accelerating at the maximum rate, is v(t) = A · 9.81 t and the distance from New York is 1 s(t ) = A ⋅ 9.81 t 2 . 2 The passenger reaches the halfway point of 1 the trip when s(t ) = 1000 ⋅ D (because D is km 2 and s is m), so the first half of the trip takes =
∫
∫ ∫
1000 D seconds. By symmetry, the second 9.81 A half takes exactly as long, so the minimum time
t=
1000 D seconds. 9.81 A For example, suppose that it is 4000 km from New York to Los Angeles and that the human body can withstand A = 5 g. Then the minimal
for the trip is t = 2
1000( 4000) = 571.1372 … , or 9.81(5) about 9.5 min. C3. Let x = distance from center along clock hand, L = length of web, and θ = central angle. dx dθ π Know: = −0.7 cm/s; = rad/s. dt dt 30 dL Want: at t = 10 s. dt time is t = 2
288
Problem Set 10-7
By the law of cosines, L 2 = x 2 + 252 − 2 · x · 25 · cos θ dL dx dx dθ = 2x − 50 cos θ + 50 x sin θ dt dt dt dt π 1 At t = 10, x = 18 cm, θ = , cos θ = , 3 2
2L
sin θ =
3 . 2
So L = 182 + 252 – 25 ⋅ 18 = 499 1 1 −18 ⋅ 0.7 + 25 ⋅ ⋅ 0.7 2 499
dL = dt
+ 25 ⋅ 18 ⋅
3 π ⋅ 2 30
= 1.6545… cm/s C4. a. Let t = time since vertex of cone touched water, y = distance from vertex of cone to bottom of cylinder (0 ≤ y ≤ 15), h = altitude of submerged part of cone, r = radius of submerged part of cone, and D = depth of water in cylinder. dy dD Know: = −2 cm/min . Want: . dt dt Volume of water is 15 · 72π = 735π cm3. 1 Volume of submerged part of cone is πr 2 h. 3 Volume of submerged part of cone plus water is π · 72 · D. ∴ 49πD = 735π + 49 D = 735 +
1 2 πr h 3
1 2 r h 3
But D = h + y, and r = 49 D = 735 + 49
5 h, so 12
1 25 ⋅ ( D – y)3 3 144
dD 25 dD dy = ( D – y)2 − dt dt dt 144
Find D when y = 10. 25 ( D – 10)3 432 Solving numerically gives D ≈ 15.1624… . Substitute this for D, 10 for y, and −2 for dy/dt. dD 25 dD 49 = +2 (5.1624 K)2 dt dt 144 49 D = 735 +
Calculus Solutions Manual © 2005 Key Curriculum Press
Solving algebraically or numerically gives dD = 0.2085… ≈ 0.21 cm/min . dt b. When the cone is completely submerged, the total volume is 1 735π + π ⋅ 52 ⋅ 12 = 835π . 3 835π In this case, D = = 17.0408… . 49π When the cone first becomes completely submerged, y = 17.0408… − 12 = 5.0408… . Thus when y = 1 the cone is already completely submerged and the depth D is not changing. C5. Let h(t) = f(t) − g(t), so h′(t) = f ′(t) − g′(t). Then h(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( b ) – h( a ) h ′( c ) = = 0. b–a But because h′(c) = f ′(c) − g′(c) at time c, f ′(c) = g′(c), so the knights have the same velocity at this time. C6. Let L = length of track, z = vertical coordinate of a point on the track, and T( θ ) = number of minutes to reach the top. v = 30 − 60 sin θ Domain of θ is 0 ≤ θ ≤ π / 6 because v would be negative for acute angles θ > π /6. Because θ is constant, v is constant. Thus, L L T (θ ) = = . v 30 – 60 sin θ To find L for any value of θ, consider z to be an independent variable. By trigonometry, dL = csc θ , and thus dL = csc θ dz. dz L=
∫
1000
z =0
csc θ dz = 1000 csc θ (θ is constant)
(Another way to find L is to “unroll” the track into a vertical plane. Because the track always makes an angle of θ with the horizontal, this will result in a right triangle with hypotenuse = L, altitude = 1000, and base angle θ. Then sin θ = 1000/L so that L = 1000 csc θ.) 1000 csc θ ∴ T (θ ) = 30 – 60 sin θ 100 = (sin θ – 2 sin 2 θ ) −1 3 100 T′(θ ) = − (sin θ − 2 sin 2 θ ) −2 ⋅ 3 (cos θ − 4 sin θ cos θ ) Calculus Solutions Manual © 2005 Key Curriculum Press
T′(θ ) = 0 ⇔ cos θ ( 1 − 4 sin θ ) = 0 cos θ = 0 only for values of θ outside the domain. ∴ 1 − 4 sin θ = 0 ⇔ θ = sin− 1 0.25 T(θ ) approaches positive infinity as θ approaches either end of the domain. So T(θ ) is a minimum for θ = sin− 1 0.25. Optimal trip takes T(sin− 1 0.25) 100 2 = [0.25 – 2(0.25)2 ]−1 = 266 s, 3 3 2 or 4 minutes 26 seconds. 3 Chapter Test T1. If the acceleration and velocity have the same sign, the object is speeding up. If the acceleration and velocity have opposite signs, the object is slowing down. T2. Displacement =
∫
60
y dx
0
20 + 10 25 + 15 + 15 ⋅ 0 − 10 ⋅ 2 2 = 120 + 0 − 200 = −80 ft = 8⋅
Distance =
∫
60
0
| y| dx = 120 + 0 + 200 = 320 ft
T3. Average value of f ( x) ≈ 2.8. The graph shows equal areas above and below y = 2.8, and the point x = c where f ( c) ≈ 2.8 by the mean value theorem for integrals. f (x ) Equal areas 5
Average ¯ 2.8
x c 5
10
r r r T4. r = 7i + 3 j The velocity vector points in the negative x-direction and the acceleration vector makes an obtuse angle with the velocity vector, indicating that the object is slowing down. y → v 5
→ a
x 5
10
T5. v = t + 60 Displacement =
∫
25
0
( t + 60) dt = 1583
1 ft 3
Problem Set 10-7
289
T6.
Because v(0) is negative and a(0) is positive, the object is slowing down.
3 θ
θ 7–x
x
T9.
7
By the minimal path property, 360 sin θ = = 0.45. 800 ∴ 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… mi along the road, then turning toward Ima’s house. This path costs 360 ⋅ 5.4882 K + 800 ⋅ 9 + 1.5117K2 = 4663.2685… ≈ $4663, which is about $257 cheaper than the proposal. T7. f(x) = x 3 − 4x + 5, x ∈ [1, 3] f ′(x) = 3x 2 − 4 f ′( x ) = 0 ⇔ x = ± 4/3 = ±1.1547K (The negative value is out of the domain.) f(1) = 2 f(1.1547…) = 1.9207… , the minimum. f(3) = 20, the maximum. 1 3 3 ( x – 4 x + 5) dx = 7 2 1 The graph shows a minimum at x = 1.1547… , a maximum at x = 3, and an average of 7. The area of the rectangle of altitude 7 equals the area of the region under the graph. Average =
∫
f (x) 20
7
x 1
3
T8. a. v(t) = 10(0.5 − 2− t) Distance =
∫
2
0
| v(t ) | dt = 3.6067K ≈ 3.61 ft
(exactly 2.5/ln 2) Displacement =
2
∫ v(t ) dt 0
= −0.8202 K ≈ −0.82 ft (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
290
Problem Set 10-7
t
a
aav
0
4
—
vend
vav
50
displ end
—
0
7
6
5
85
14
10
8
141
113
67.5
1263.5
472.5
21
13
11.5
221.5
181.25
2532.25
The object traveled about 2532.25 cm. Average velocity was about 1 ⋅ (2532.25) = 120.583K ≈ 120.6 cm/s. 21 r r r T10. r (t ) = (10 cos 0.4t )i + (10 sin 0.6t ) j r r r v (t ) = ( −4 sin 0.4t )i + (6 cos 0.6t ) j r r r T11. a (t ) = ( −1.6 cos 0.4t )i + ( −3.6 sin 0.6t ) j r r r T12. r (2) = (10 cos 0.8)i + (10 sin 1.2) j = 6.9670 Ki + 9.3203K j r The graph shows r (2). y v t=2 r
a
5
x 0
5
r r r T13. v (2) = ( −4 sin 0.8)i + (6 cos 1.2) j r r = −2.8694 Ki + 2.1741K j See the graph in Problem T12. The velocity vector is tangent to the path, pointing in the direction of motion. r r r T14. a (2) = ( −1.6 cos 0.8)i + ( −3.6 sin 1.2) j r r = −1.1147Ki − 3.3553K j See the graph in Problem T12. r r T15. a (2) ⋅ v (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 r | v (2) | = 16 sin 2 0.8 + 36 cos 2 1.2 = 3.6000… mi/h r r a (2) ⋅ v (2) P= = −1.1378K ≈ −1.14 (mi/h)/h r | v (2) | r r r P( 2 ) r at (2) = r v (2) = 0.9069Ki − 0.6871K j | v (2) |
Calculus Solutions Manual © 2005 Key Curriculum Press
r r r an (2) = a (2) − at (2) r r = −2.0216 Ki − 2.6681K j v
t=2
r T17. Object is slowing down at | at (2)| = | P(2)| = 1.1378K (mi/h)/h. r T18. an (2) points inward to the concave side because r an is the component of acceleration that pulls the object out of the straight path into a curve. r T19. dL = dx 2 + dy 2 = | v (t ) | dt
at
= 16 ⋅ sin 2 0.4t + 36 ⋅ cos 2 0.6t dt L=
an a
r T16. The tangential component at (2) has direction the r r opposite of v(2), so v is decreasing and the object is slowing down at t = 2.
Calculus Solutions Manual © 2005 Key Curriculum Press
2
∫ dL = 10.0932K (numerically) ≈ 10.09 mi 0
T20. Answers will vary.
Problem Set 10-7
291
Chapter 11—The Calculus of Variable-Factor Products Problem Set 11-1
2 dW = 30 − y dy 7 70 30 – 2 y dy = 1400 mi-tons W= 0 7
1.
∫
F (x , F ) 10 ∆x = 0.2
x 4
F ≈ F(4) = 80e− 2 = 10.8268… ≈ 10.83 lb in the strip. W ≈ F(4) · ∆x = 16e− 2 = 2.1653… ≈ 2.17 ft-lb 2. dW = 20xe− 0.5 x dx 3. W =
∫
7
0
20 xe −0.5 x dx = 69.1289K
(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=
∫
b
b. W total = 90 tons · 70 mi = 6300 mi-tons Excess energy becomes kinetic energy of rocket and spent fuel. 3. Hooke’s law: F = k · s dW = ks ds W=
∫
10
0
4. a. F = −x 3 + 6x 2 − 12x + 16 The graph starts at the high force of 16 lb, levels off, then drops to F = 0 at x = 4. F 15
20 x −0.5 x dx
0
x
= −40 be −0.5b − 80e −0.5b + 80 lim W = 0 + 0 + 80 = 80 ft-lb
b→∞
− 0.5 b
(Use l’Hospital’s rule for be
4
b. dW = F dx = (−x3 + 6x2 − 12x + 16) dx .)
W=
2 3 Q3. v(t) = ln | sec t | + C Q4. a( t) = t− 1 Fundamental theorem of calculus Riemann Integration by parts Implicit differentiation Heaviside method Q10. A 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 − 0.16y dW = (weight) dy = (20 − 0.16y) dy W=
∫
50
0
(20 – 0.16 y) dy = 800 ft-lb
2. a. Let y = number of miles up. 2 Slope of linear function is −20/70 = − . 7 2 Weight = 30 − y (tons) 7
292
Problem Set 11-2
0
(– x 3 + 6 x 2 – 12 x + 16) dx = 32 ft-lb
17
Q2. 10
Q1. 2
∫
4
5.
Problem Set 11-2
Q5. Q6. Q7. Q8. Q9. 1.
ks ds = 50 k
10 (x, y ) 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=
∫
10
0
dW = 1,396, 752.0937...
≈ 1.4 million ft-lb (exactly 444,600π) 6. dV = π x2 dy At 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=
∫
16
0
dW = 337,891.2751… ≈ 337,891 ft-lb
Calculus Solutions Manual © 2005 Key Curriculum Press
Initial condition V = 1 at p = 1000 ⇒ k1 = 1000. ∴ dW = 1000V − 1.4 dV
7. a. Draw x- and y-axes with the origin at the center of the sphere. To fill the sphere half full, the water must be pumped from −120 to y, where y is negative. Integration is from y = −20 to y = 0. dV = π x2 dy x 2 + y 2 = 202 ⇒ x 2 = 400 − y 2 ∴ dW = [y − (−120)](62.4)[π (400 − y2) dy] = 62.4π (y + 120)(400 − y2) dy W=
∫
0
−20
W=
∫
−20
W=
dW = 250, 925, 288.4 K
= 62.4π ⋅ 1, 280, 000
W=
∫
0
dW ≈ 9,134, 602 ft-lb
4 1/4 exactly (20100)(0.5) 540 9 9. a. If x is the distance between the piston and the cylinder head and F is the force exerted by the hot gases, then dW = F dx. F = pA, where p is the pressure and A is the area of the piston. ∴ dW = pA dx A dx = dV ∴ dW = p dV p = k 1V − 1.4 Calculus Solutions Manual © 2005 Key Curriculum Press
∫
1
10
15 ⋅ 101.4 V −1.4 dV ≈ − 566.9574 K
So about 567 in.-lb of work is done in compressing the gases (exactly 37.5 ⋅ 101.4 (10− 0.4 − 1)). (Mathematically, the work is negative because the force is positive and dx is negative. Physically, the work is negative because energy is taken out of the surroundings to put into the gases. Positive work indicates that energy is put into the surroundings by the expanding 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 a pioneer in the field of thermodynamics.
≈ 250.9 million ft-lb (exactly 62.4π · 1,280,000) This answer can be found quickly by lifting the entire weight of the water through the distance the center of the sphere moves, namely 120 ft. 4 W = (62.4) (π ⋅ 20 3 )(120) 3
16
1000V −1.4 dV ≈ 1504.7320 K
b. Initial condition p = 15 at V = 10 ⇒ k2 = 15 ⋅ 101.4 dW = 15 ⋅ 101.4 V − 1.4 dV
dW = 117, 621, 229
(Note that the amount of work to fill the entire tank is more than twice the amount needed to half-fill it. The work to fill the top hemisphere is greater than that to fill the bottom hemisphere because the same amount of water has to be lifted through a greater displacement.) 8. Slice the water horizontally. Pick sample point (x, y) on the curve y = 0.0002x4 within the slice. 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
1
≈ 1504.7 in.-lb (exactly 2500(1 − 10− 0.4 ))
≈ 117.6 million ft-lb (exactly 62.4π · 600000) b. For filling the tank, the limits of integration are from −20 to 20. 20
∫
10
Problem Set 11-3 Q1. 2 cm3 Q3.
Q2. 3 cm3 Q4. y y
x
1
x
1
π
Q5.
Q6. y
y
1
x
x
Q7. (mass)/(volume) 1 Q9. 1 – x2
Q8. (force)(displacement) Q10. B
1. a. The graph shows y = ln x, rotated about x = 0, showing back half of solid only. y 1
(x, y)
x 1
3
Problem Set 11-3
293
Slice the region parallel to the axis of rotation, generating cylindrical shells. Pick sample point (x, y) on the curve, within the slice. ρ = kx− 1 dm = ρ dV = (kx− 1)(2π x ln x dx) = 2π k ln x dx m=
∫
3
1
2π k ln x dx ≈ 8.1419K k
(exactly 2π k(3 ln 3 − 2)) b. Slice perpendicular to the axis of rotation, generating plane washers. ρ = 5 + 2y dm = ρ dV = (5 + 2y) ⋅ π ( 32 − x2) dy = π ( 5 + 2y)(9 − e2y) dy m=
∫
ln 3
0
b. Slice perpendicular to the axis of rotation. ρ = ky 2 dm = ρ dV = ky 2 ⋅ π ( 9 − y) dy m=
∫
9
0
dm = 546.75π k
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=
∫
3
0
dm = 105.3π k
d. The solid in part b has the largest mass. 4. a. The graph shows y1 = x and y2 = 0.5x, intersecting at (0, 0) and (4, 2), rotated about the x-axis, showing back half of solid only.
dm ≈ 108.1103K
y
(exactly π [36 ln 3 + 9 (ln 3) − 16]) 2. The graph shows y = sin x, rotated about the y-axis, showing back half of solid only. 2
2
y1 y2
x 4
y 1
x π
Slice the region parallel to the axis of rotation, generating cylindrical shells. ρ = kx dm = ρ dV = kx ⋅ 2π xy dx = 2π kx 2 sin x dx m=
∫
π
0
dm ≈ 36.8798K k (exactly 2π (π 2 − 4)k )
3. a. The graph shows y = 9 − x2, rotated about the y-axis. 9
y
(x, y )
x 3
Slice the region perpendicular to the axis of rotation, generating plane disks. ρ = k, dm = k dV = k ⋅ π x2 dy = k ⋅ π ( 9 − y) dy m=
9
∫ dm = 40.5πk 0
Or: Slice parallel to the axis of rotation. dm = k ⋅ 2π xy dx = 2π kx(9 − x2) dx m=
∫
3
0
dm = 40.5π k
Or: Volume of paraboloid is half the volume of the circumscribed cylinder, or 0.5(π ⋅ 32)(9) = 40.5π, so m = 40.5π k. 294
Problem Set 11-3
Slice perpendicular to the axis of rotation, generating plane washers. Pick sample points (x, y1) and (x, y2). ρ = kx dm = ρ dV = kx ⋅ π ( y12 − y22 ) dx = kxπ ( x − 0.25 x 2 ) dx 16 π k = 16.7551K k 3 b. Slice parallel to the axis of rotation, generating cylindrical shells. Pick sample points (x1, y) and (x2, y). ρ = ky 2 dm = ρ dV = ky 2 ⋅ 2π y(x2 − x1) dy = 2 π ky3(2y − y2) dy 2 64 m = dm = π k = 13.4041K k 0 15 5. a. Prediction: The cone on the left, with higher density at its base, has greater mass because higher density is in the larger part of the cone. b. Set up a coordinate system with the origin at the center of the base. Slice each cone perpendicular to its axis, generating plane disks. Pick sample point (x, y) on the element of the cone, y = 6 − 2x. dV = π x 2 dy = π ( 3 − 0.5y)2 dy For the cone on the left, ρ = 80 − 5y. dm = (80 − 5y) · π(3 − 0.5y)2 dy m=
∫
m=
6
∫ dm = 1305π oz 0
Calculus Solutions Manual © 2005 Key Curriculum Press
For the cone on the right, ρ = 50 + 5y. dm = (50 + 5y) · π ( 3 − 0.5y)2 dy m=
∫
6
0
dm = 1035π oz
∴ the cone on the left has the greater mass, as predicted in part a. 6. a. Prediction: The cylinder on the left, with higher density at walls, has greater mass because higher density is in the larger part of the cylinder. b. Set up a coordinate system with the origin at the center of the bottom base. Slice each cylinder parallel to its axis, generating cylindrical shells. Pick sample point (x, 6). dV = 2π x · 6 · dx = 12π x dx For the cylinder on the left, ρ = 50 + 10x. dm = (50 + 10x) · 12π x dx = 12π (50x + 10x2) dx m=
∫
3
0
ρ is given in the table in the text. π dV = π 0.52 dy = dy 4 π dm = ρ dV = ρ dy 4 π 2 m= ρ dy 4 0 Simpson’s rule cannot be used because there is an odd number of increments. Use the trapezoidal rule. π 1 1 m ≈ (0.4) (10) + 9.9 + 9.8 + 9.6 + 9.4 + (9.0) 4 2 2 = 4.82π ≈ 15.14 g 10. a. The graph shows y1 = 4 − x2 and y2 = 4x − x2, intersecting at (1, 3), rotated about the y-axis, showing back half of solid only.
∫
4 y
dm = 3780π oz
(x, y2)
For the cylinder on the right, ρ = 80 − 10x. dm = (80 − 10x) · 12π x dx = 12π ( 80x − 10x2) dx m=
∫
3
0
dm = 3240π oz
∴ the cylinder on the left has the greater mass, as predicted in part a. 7. y1 = 4 − 2x2 and y2 = 3 − x2, rotated about the 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 = kx 2, dV = π ( y12 − y22 ) dx dm = ρ dV = π kx2(7 − 10x2 + 3x4) dx 1 16 m = dm = π k = 2.3935K k 0 21 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 = ρ dV = 2πxe − x (1 − x 2 ) dx
x 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) dx dm = ρ dV = 2π k x2(4 − 4x) dx 1 2 m = dm = π k = 2.0943K k 0 3 b. The graph shows the curves in part a, rotated about the x-axis, showing back half of solid only.
∫
4
y (x, y ) 1
∫
m=
1
∫ dm = 0.9444K (exactly 2π (14e 0
−1
− 5))
9. Set up axes with the origin at the center of the lower base and the y-axis coaxial with the cylinder’s axis. Slice perpendicular to the axis of the cylinder, generating plane disks of constant radius 0.5.
Calculus Solutions Manual © 2005 Key Curriculum Press
(x, y 1)
(x, y 2)
x 1
Slice perpendicular to the axis of rotation, generating plane washers. Pick sample points (x, y1) and (x, y2). ρ = kx, dV = π ( y12 − y22 ) dx dm = ρ dV = π kx(16 − 24x2 + 8x3) dx m=
1
∫ dm = 3.6πk = 11.3097Kk 0
Problem Set 11-3
295
Let x = r sin θ.
c. The region is rotated about the y-axis as in 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
Then dx = r cos θ dθ , r 2 – x 2 = r cos θ . x = r ⇒ θ = sin −1 1 = π /2 ∴ m = 4π k
∫
= πr 4 k
= π (2 − 4 – y ) dy. 2
∫
0
+ =1
∫
∫
π /2
0
π /2
sin 2 θ cos 2 θ dθ
sin 2 2θ dθ
0
(half-argument property)
For x in [3, 4], dV = π x12 dx = π ( 4 − y) dy. m=
r 2 sin 2 θ ⋅ r cos θ ⋅ r cos θ dθ
0
= 4π r 4 k
For x in [0, 3], dV = π x 22 dx
3
π /2
=
ky ⋅ π (2 − 4 – y ) dy 2
1 4 πr k 2
∫
π /2
0
(1 – cos 4θ ) dθ (half-argument property)
∫
4
3
ky ⋅ π ( 4 − y) dy
=
14 2 π k + 1 π k = 3.6π k = 11.3097K k 15 3
(Coincidentally, this answer equals the answer to part b.) 11. a. The graph shows a sphere with origin at its center. y
r
(x, y )
1 4 1 π r k θ − sin 4θ 2 4
m = 4π k
r
m = πk
∫
r
| x | (r 2 − x 2 ) dx
–r
= 2π k
∫
r
0
1 2 4 π r k 4
∫
r
r
x 3 dx = π kx 4 0 = π kr 4
0
12. Assume Earth is spherical, with radius 3960 mi = 3960 ⋅ 5280 ⋅ 12 ⋅ 2.54 cm = 637,300,224 cm, and slice into spherical shells with radius x and dV = 4π x2 dx. 8x ρ = 12 − g/cm 3 637300224 32π x 3 dm = ρ dV = 48π x 2 − dx 637300224 m=
∫
637300224
dV
0 637300224
(r 2 x – x 3 ) dx
0
1 1 = 2π k r 2 x 2 − x 4 2 4
=
c. Slice into spherical shells. Pick a sample point on the x-axis within the shell. Then x is the radius of the shell and 4π x 2 is the area of the shell at the sample point. ∴ dV = 4π x2 dx ρ = kx dm = ρ dV = 4π kx3 dx
x
Slice the upper semicircular region perpendicular to the x-axis and rotate it to get plane disks. Pick a sample point (x, y). Equation of the circle in the xy-plane is x 2 + y 2 = r2. ρ = k|x|, dV = π y 2 dx = π (r2 − x2) dx dm = ρ dV = k|x| π (r2 – x2) dx
π /2
r
1 4 πr k 2 0 b. Slice the right semicircular region parallel to the y-axis, and rotate it to get cylindrical shells coaxial to the y-axis. =
8π x 4 = 16π x 3 − 637300224 0 = 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 of solid only. y
ρ = kx, dV = 2π x ⋅ 2 y ⋅ dx = 4π x r 2 – x 2 dx .
dm = ρ dV = 4π kx 2 r 2 – x 2 dx
(x, y ) 1
m = 4π k
296
∫
r
x 2 r 2 – x 2 dx
x 1
0
Problem Set 11-3
Calculus Solutions Manual © 2005 Key Curriculum Press
Slice parallel to the axis of rotation, generating cylindrical shells. Pick a sample point (x, y). ρ = cos x, dV = 2π x ⋅ y ⋅ dx = 2π xex dx dm = ρ dV = cos x 2π xex dx π /2 π m= dm ≈ 8.6261K exactly πeπ /2 − 1 2 0
∫
14. a = 4 mi, b = 1 mi, c = 0.5 mi a.
2
2
2
x y z + + = 1, a b c
where a = 4, b = 1, c = 0.5. The cross section at z = z0 < c has equation 2
2
–1/3 1/3 + ( x – 2) ( x – 5) 1 1 Q6. ln | x − 5| − ln | x − 2| + C 3 3 1/3 –1/3 Q7. + ( x – 2) 2 ( x – 5)2 Q5.
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 about the y-axis. y
9
2
x + y = 1 − z0 . a b c
(x, y )
This is the equation of an ellipse with x
x -radius a 1 – ( z0 /c)2 and
3
y -radius b 1 – ( z0 /c)2 . b. Slice horizontally into plane elliptical disks. The area of the cross section is π (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=
∫
0.5
0
dm ≈ 0.32 ⋅ 5280 3 ⋅ 1.008953K
≈ 4.7525… × 1010 lb, or about 23,762,540 tons (exactly 0.32 · 52803π (1100e− 0.1 − 995)) 2 4 c. Volume of semi-ellipsoid = π abc = π mi 3 3 3 3 4 Weight = 0.08 ⋅ 5280 ⋅ π 3 ≈ 49,326,507,160 lb, 1,801,427,783 lb more (≈ 3.8% more than actual) d. V = 2
∫
c
0
πab[1 − ( z/c)2 ] dz c
1 = 2πab z − ⋅ c( z/c)3 3 0 2 4 = 2πab c = πabc, Q .E.D . 3 3
Slice the region perpendicular to the axis of rotation, generating plane disks. dV = π x 2 dy = π (9 − y) dy V=
∫
9
0
π (9 − y) dy = 40.5π
b. Each point in a disk is about y units from the xz-plane, where y is at the sample point (x, y). dMxz = y dV = π (9y − y2) dy M xz =
∫
9
0
π (9 y − y 2 ) dy = 121.5π
121.5π =3 40.5π x = z = 0 by symmetry. The centroid is at (0, 3, 0).
c. y ⋅ V = M xz ⇒ y =
2
2
x y 1 2 2. a. + = 1 ⇒ y 2 = 25 1 − x 12 5 144 Slice the ellipsoidal region above the x-axis perpendicular to the x-axis, generating plane disks as the region rotates. Pick sample point (x, y). 1 2 dV = πy 2 dx = 25π 1 − x dx 144 V = 25π
∫
12
0
1 – 1 x 2 dx 144 12
Problem Set 11-4 Q1. −5 2 = −25 Q2. (−11)2 = 121 Q3. sin 2x = 2 sin x cos x 1 Q4. cos 2 x = (1 + cos 2 x ) 2 Calculus Solutions Manual © 2005 Key Curriculum Press
1 3 = 25π x − x = 200π 432 0 2 This answer equals ⋅ π ⋅ 12 ⋅ 52 , which is 3 expected because the volume of a (whole) 4 ellipsoid is V = πabc. 3 Problem Set 11-4
297
b. Each point in a disk is about x units from the yz-plane, where x is at the sample point (x, y). 1 2 dM yz = x dV = 25π x 1 − x dx 144
∫
M yz =
12
0
dM yz = 900π
900π = 4.5 200π 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 the xz-plane, where y is at the sample point (x, y), so each point has about the same density. ρ = ky 1/3, dm = ρ dV = k π (9y 1/3 − y 4/3) dy c. x ⋅ V = M yz ⇒ x =
m=
∫
9
0
dm = 170.1375K k
93 3 π 9k exactly = 28 b. Each point in a disk is about y units from the xz-plane, where y is at the sample point (x, y). dMxz = y dm = k π (9y 4/3 − y 7/3) dy
∫
M xz =
9
0
dM xz
94 3 π 9k = 612.4952 K k exactly 28 612.4952 K k 170.1375K 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 the center 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 from the yz-plane, where x is at the sample point (x, y), so each point has about the same density as at the sample point. 1 2 ρ = kx, dV = π y 2 dx = 25π 1 − x dx 144 c. y ⋅ m = M xz ⇒ y =
1 2 dm = ρ dV = 25πkx 1 − x dx 144 m=
∫
12
0
dm = 2827.4333K k
(exactly 900π k)
298
Problem Set 11-4
b. Each point in a disk is about x units from the yz-plane, where x is at the sample point (x, y). 1 2 dM yz = x dm = 25πkx 2 1 − x dx 144 M yz =
∫
12
0
dM yz = 5760πk
5760πk = 6.4 900πk 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 the center 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 c. x ⋅ m = M yz ⇒ x =
A=
∫
2
0
e x dx = e 2 − 1 = 6.3890 K
Each point in a strip is about x units from the y-axis, where x is at the sample point (x, y). ∴ dMy = x dA = xex dx My =
∫
2
0
xe x dx = e 2 + 1 = 8.3890 K
x ⋅ A = My ⇒ e2 + 1 x= 2 = 1.3130… (= coth 1) e –1 b. Strips in part a generate plane disks. Each point in a disk is about x units from the yz-plane, where x is at the sample point (x, y). dV = πy2 dx = πe2x dx 2 π V = πe 2 x dx = (e 4 – 1) = 84.1917K 0 2 dMyz = x dV = πxe2x dx 2 3 1 M yz = dM yz = πe 4 + π = 129.4292 K 0 4 4 3e 4 + 1 x ⋅ V = M yz ⇒ x = = 1.5373K 2(e 4 – 1)
∫
∫
c. False. For the solid, x is farther from the yz-plane. 6. a. Slice the region parallel to the y-axis. dA = sec x dx A=
∫
π /3
0
sec x dx = ln (2 + 3 )
= 1.3169… dM y = x dA = x sec x dx My =
∫
π /3
x sec x dx
0
= 0.7684… (numerically) 0.7684 K x ⋅ A = My ⇒ x = = 0.5835K 1.3169K
Calculus Solutions Manual © 2005 Key Curriculum Press
b. Strips in part a generate plane disks. Each point in a disk is about x units from the yz-plane, where x is at the sample point (x, y). dV = π y2 dx = π sec2 x dx V=
∫
π /3
π sec x dx = π 3 = 5.4413K 2
0
dM yz = x dV = π x sec2 x dx M yz =
∫
π /3
0
π x sec 2 x dx = 3.5206 K
π2 3 1 exactly 3 + π ln 2 x ⋅ V = M yz ⇒ x = 0.6470 K
π ln 2 exactly − 3 3
c. Each point in a strip of part a is approximately x units from the y-axis, where x is the value in the sample point (x, y). dM y = x dA = x 5/3 dx
∫
My =
8
x 5/3 dx = 96
0
96 =5 19.2 27.4285K y ⋅ A = Mx ⇒ y = = 1.4285… 19.2 exactly 1 3 7 Centroid is at (5, 1.4285…). e. The balance point is shown on the graph.
d. x ⋅ A = M y ⇒ x =
y 4
c. False. For the solid, x is farther from the yz-plane. 7. Construct axes with the origin at a vertex and the x-axis along the base, b. Slice the triangle parallel to the x-axis. b The width of a strip is b − y. h b dA = b − y dy h b dM x = y dA = by − y 2 dy h b M x = by – y 2 dy 0 h h 1 b 1 = by 2 − y 3 = bh 2 2 3h 6 0 1 2 bh 1 y ⋅ A = Mx ⇒ y = 6 = h, Q.E.D. 1 bh 3 2 8. a. y = x 2/3 from x = 0 to x = 8. Slice the region vertically. Pick a sample point (x, y) on the graph within the strip. (See the graph in part e.) dA = y dx = x 2/3 dx
∫
A=
h
∫
8
x 2/3 dx = 19.2
0
b. Slice the region parallel to the x-axis so that each point in a strip is about y units from the x-axis, where y is at the sample point (x, y). dM x = y(8 − x) dy = (8y − y5/2) dy Mx =
∫
4
0
(8 y – y 5/2 ) dy = 27.4285K
exactly 27 3 7
Calculus Solutions Manual © 2005 Key Curriculum Press
x 8
9. a. Slice the region parallel to the y-axis so that each point in a strip will be about x units from the y-axis, where x is at the sample point (x, y). dA = y dx = sin x dx A=
∫
π
sin x dx = 2 (exactly)
0
(This may be “well-known” by now.) dM y = x dA = x sin x dx My =
∫
π
0
x sin x dx = 3.1415… = π (exactly)
π , Q .E .D . 2 (Or just note the symmetry.) b. dM 2y = x2 dA = x2 sin x dx x ⋅ A = My ⇒ x =
M2y =
∫
π
0
x 2 sin x dx = 5.8696 …
(exactly π 2 − 4)
π2 – 4 = 1.7131… 2 10. a. Set up axes with the x-axis along the base, B. dM 2B = y2 dA = y2B dy c. x 2 ⋅ A = M2y ⇒ x =
∫
H
1 BH 3 3 b. Set up axes with the x-axis through the centroid. dM 2c = y2 dA = y2B dy M2B =
M2c =
0
∫
y 2 B dy =
0.5 H
–0.5 H
y 2 B dy =
1 BH 3 12 Problem Set 11-4
299
c. Set up axes with the x-axis along the base, B. B dM2B = y 2 dA = y 2 B − y dy H H B 1 M2B = By 2 – y 3 dy = BH 3 0 H 12 d. Use the axes in part c. The distance from the centroidal axis to a sample point (x, y) is 1 y − H. 3 2 2 B 1 1 dM2 c = y − H dA = y − H B − y dy H 3 3 2 H/3 – 1 y 3 + 5 y 2 – 7 Hy + 1 H 2 dy M2 c = B – H/3 H 3 9 9 H B 4 5 3 7 1 =− y + By − BHy 2 + BH 2 y 4H 9 18 9 0 1 3 = BH 36 11. a. Slice into cylindrical shells so that each point in 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 R 1 M = 2πHr 3 dr = πHR 4 0 2 1 πHR 4 1 2 2 2 ⋅ = ⇒ = ⇒r = r V M r R πR 2 H 2 b. Slice the cone into cylindrical shells so that each point in a shell will be about r units from the axis. H The altitude of a shell is H − r. R H dM = r 2 dV = r 2 2πr H − r dr R R 1 1 Mh = 2πH r 3 – r 4 dr = πHR 4 0 R 10 1 πHR 4 2 2 10 r ⋅V = M ⇒ r = ⇒ r = 0.3 R 1 2 πR H 3 c. Slice the sphere into cylindrical shells so that each point in a shell is about r units from the axis. The equation of the sphere is r2 + y2 = R2. The altitude of a shell is 2y.
∫
∫
∫
∫
dM = r 2 dV = r 2 2πr ⋅ 2 R 2 – r 2 dr M = 4π
∫
R
r 3 R 2 – r 2 dr
0
Let r = R sin θ. dr = R cos θ dθ ,
R 2 – r 2 = R cos θ π r = 0 ⇒ θ = 0, r = R ⇒ θ = 2
300
Problem Set 11-4
∫
M = 4π
π /2
R3 sin 3 θ ⋅ R cos θ ⋅ R cos θ dθ
0
= 4π R 5
∫
π /2
0
(cos 2 θ – cos 4 θ ) sin θ dθ
1 1 = 4πR 5 − cos3 θ + cos 5 θ 3 5
π /2 0
1 1 8 = 4πR 5 −0 + 0 + − = πR 5 3 5 15 8 πR 5 2 2 15 ⇒ r = 0.4 R r ⋅V = M ⇒ r = 4 3 πR 3 12. Assume the clay has uniform density ρ. Cylinder: H = 2 RC , V = π RC2 H = 2π RC3 = 1000 500 RC = π
1/3
cm
Second moment of volume =
1 π (2 RC ) RC4 = π RC5 2
(from Problem 11a) Second moment of mass 500 = ρπRC5 = 500 ρ π
2/3
= 14, 684.1932 … ρ 1/3
4 3 750 πRS = 1000 ⇒ RS = cm π 3 8 Second moment of volume = πRS5 15 (from Problem 11c) Second moment of mass 2/3 8 750 = ρ πRS5 = 400 ρ π 15 = 15,393.3892…ρ The sphere has higher moment of mass. 13. a. Set up axes with the x-axis through the centroid. dM2 = y dA = y2 · B dy Sphere: V =
M2 = B
∫
0.5 H
−0.5 H
y 2 dy =
1 3 y 3
0.5 H −0.5 H
1 = BH 3 , Q.E.D. 12 (Same answer as in Problem 10b) b. i. B = 2, H = 12; M 2 = 288; stiffness = 288k ii. B = 12, H = 2; M 2 = 8; stiffness = 8k A board on its edge is 36 times stiffer. c. i. Set up axes with the x-axis through the centroid. From y = 0 to y = 2, dM 2 = y2 · 2 dy. From y = 2 to y = 4, dM 2 = y2 · 4 dy.
Calculus Solutions Manual © 2005 Key Curriculum Press
By symmetry, M2 = 2 =2
∫
2
∫
2 y 2 dy + 2
0
4
2
∫
4
0
y
dM2
r
4 y 2 dy = 160.
f (r )
Stiffness = 160k ii. From y = 0 to y = 4, dM 2 = y2 · 1 dy. From y = 4 to y = 6, dM 2 = y2 · 4 dy. By symmetry, M2 = 2 =2
∫
4
0
y dy + 2 2
∫
6
4
∫
r a
Let f(r) be the length of the strip or the sum of the lengths if the region has S-shaped parts. Let A be the area of the region. dMy = r dA = r f (r) dr
6
0
dM2
4 y dy = 448. 2
Stiffness = 448k (2.8 times stiffer!) d. Increasing the depth does seem to increase stiffness greatly, but making the beam very tall would also make the web very thin, perhaps too thin to withstand much force. 14. a. dA = y dx = x 3 dx A=
∫
2
My =
V=
∫
2
0
2π x 4 dx = 12.8π
c. dM y = x dA = x4 dx My =
∫
2
0
x 4 dx = 6.4
The volume integral is 2π times the moment integral. 6.4 d. x ⋅ A = M y ⇒ x = = 1.6 4 e. The centroid travels 2π x = 3.2π . (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 to the centroid is R. Thus, the theorem of Pappus implies V = 2π RA = 2π R(π r2) = 2π 2r2R 1 b. Area of a semicircle = πr 2 2 4 3 Volume of a sphere = πr 3 4 3 πr 4 2πr ⋅ A = V ⇒ r = 3 = r 1 2 3π 2π ⋅ πr 2 16. Pick a closed region that does not lie on both sides of the y-axis. Slice the region parallel to the y-axis so that each point in the strip will be about r units from the y-axis (see graph).
Calculus Solutions Manual © 2005 Key Curriculum Press
b
r f (r ) dr
a
∫
b
a
2πr f (r ) dr = 2π
∫
b
r f (r ) dr
a
= 2π M y But My also equals r ⋅ A. ∴ V = 2πrA = (2πr )( A) = (distance traveled by centroid)(area of region), Q .E .D .
x dx = 4
b. dV = 2π x · y · dx = 2π x4 dx V=
∫
Rotate the region about the y-axis. The strips generate cylindrical shells. dV = 2π r f (r) dr
3
0
b
Problem Set 11-5 Q1. Q3. Q5. Q6. Q7. Q9. 1.
centroid Q2. center of mass radius of gyration Q4. definite integration indefinite integration (or antidifferentiation) ρ = (mass) ÷ (volume) x1/ 2 Q8. ln | sec x + tan x | + C 2 −1 y′ = (x + 1) Q10. A a. Slice the trough face horizontally so that each point in a strip is about the same distance below the surface as at the sample point (x, y). y = 2x 4 ⇒ x = (0.5y)1/4 p = k(2 − y), dA = 2x dy = 2(0.5y)1/4 dy dF = p dA = 2k(2 − y)(0.5y)1/4 dy 2 128 F = dF = 2.8444 K k exactly k 0 45
∫
b. dM x = y dF = y · 2k(2 − y)(0.5y)1/4 dy 2 256 M x = dM x = 2.1880 K k exactly k 0 117 2.1880 K k c. y ⋅ F = M x ⇒ y = 2.8444 K k 10 = 0.7692 K exactly 13 x = 0 by symmetry. 10 The center of pressure is at 0, . 13
∫
2. a. The graph shows y = x2, between y = 0 and y = 100. Problem Set 11-5
301
100 y
x
Width at y = 100 ft is 2 y = 20 ft, Q .E .D . b. Slice the dam face horizontally so that each point in a strip is the same distance below the surface as the sample point (x, y). dA = 2x dy = 2y1/2 dx 100 1 A= 2 y1/2 dx = 1333 ft 2 0 3 (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 100 1 F= dF = 53, 333 k = 3, 328, 000 0 3 Force is 3,328,000 lb, or 1664 tons. d. dMx = y dF = 2ky(100 − y)(y1/2) dy 100 16000000 Mx = dM x = k 0 7 = 142, 628, 571.4285K ≈ 142.6… million lb-ft e. y ⋅ F = M x ⇒ 300 7 y= = 42 = 42.8571… ≈ 42.86 ft 7 8 3. a. Slice the bulkhead horizontally so that each point in a strip is the same distance below the surface as the sample point (x, y).
∫
∫
4 1 dM = y dA = y ⋅ 40 1 − y − 1 32
M=
4
4
x + y – 32 = 1 20 32 4 1/4 1 x = 20 1 − y −1 32
0
∫
0
16
dA ≈ 548.6345K ≈ 548.6 ft 2
0
∫
16
0
dM ≈ 4749.3398K ≈ 4749.3 ft 3
y ⋅ Aw = Mw ⇒ y ≈ 8.6566 K ft x = 0 by symmetry. The center of buoyancy is at about (0, 8.66) ft. 2
2
x y 4. a. Equation of ellipse is + = 1. 6 3 b. Slice the ellipse horizontally so that each point in a strip is y units from the surface where y is at the sample point and y is negative. Surface of oil is at y = 0 ⇒ p = −50y.
dF = p dA = −50 y ⋅ 4 9 – y 2 dy = −200 y 9 – y 2 dy F=
4 1 = y ⋅ 67(32 − y) ⋅ 40 1 − y − 1 32
∫
0
−3
dF = 1800 lb (exactly)
5. a.
1/ 4
dy
dF ≈ 1,199, 294.1645K
Problem Set 11-5
∫
Mw =
dA ≈ 1186.6077K ≈ 1186.6 ft 2
≈ 1.199 million lb c. dMx = y dF
302
dM ≈ 20, 071.5364 K
First moment of area below waterline:
dy
4 1 = 67(32 − y) ⋅ 40 1 − y − 1 32 32
dy
x = 2 9 – y 2 , dA = 2 x dy = 4 9 – y 2 dy
b. p = 67(32 − y) dF = p dA
F=
0
Aw =
4 1/ 4
1 dA = 2 x dy = 40 1 − y − 1 32
∫
32
1/ 4
≈ 20.07 thousand ft3 y ⋅ A = M ⇒ y ≈ 16.9150 K ft x = 0 by symmetry. The centroid is at about (0, 16.92) ft. The centroid is different from center of pressure. f. Area below waterline:
∫
∫
dM x ≈ 13, 992, 028.2564 K
0
x = 0 by symmetry. Center of pressure is at about (0, 11.67) ft. e. Moment of area:
10
32
32
≈ 13.992 million lb-ft d. y ⋅ F = M x ⇒ y ≈ 11.6668K ft
(x, y )
A=
∫
Mx =
60
y
(x, y )
1/ 4
x
dy
10
Calculus Solutions Manual © 2005 Key Curriculum Press
Slice the wing parallel to the y-axis. Pick sample point (x, y) within the strip. π dA = y dx = 60 cos x dx 20 A=
∫
10
−10
dA =
7. a. Slice the region as shown in Figure 11-5f. At a sample point (x, t), d(dM2x) = t2 dx dt.
F=
∫
10
−10
dF = 4, 863.4168K k
exactly 48000 k π2 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. Pick sample point (x, y) within the strip. p = 90 − 7x dA = y dx = (100 − x2) dx dF = p dA = (90 − 7x)(100 − x ) dx 2
∫
F=
10
−10
dF = 120, 000 lb (exactly)
b. dMy = x dF = x(90 − 7x)(100 − x2) dx 10 560000 My = dM y = − lb-ft −10 3 5 c. x ⋅ F = M y ⇒ x = −1 ft 9
∫
6 d. p = ky = y (because p = 60 at y = 50) 5 dA = 2 x dy = 2 100 – y dy dF = p dA =
12 y 100 – y dy 5
∫
t =0
t 2 dt dx
1 1 = t3 dx = y 3 dx 3 t =0 3 1 = [0.25( x – 4) – ( x – 4)1/ 3 ]3 dx 3
≈ 763.9 ft2
π x dx 20
∫
t=y
t=y
2400 = 763.9437K π
b. dF = p dA = k (10 − | x |) ⋅ 60 cos
t=y
t 2 dx dt = t =0
dM2 x =
∫
b. M2 x =
x =4
x = –4
8 dM2 x = 0.5333K exactly 15
8. a. Slice the region parallel to the y-axis so that each point in a strip will have about the same pressure as at the sample point (x, y). y = e− x p = kx 2, dA = (1 − e − x) dx dF = p dA = kx2(1 − e− x) dx F=
∫
ln 5
0
dF = 0.9514 K k
2 8 1 3 1 2 exactly (ln 5) + (ln 5) + ln 5 − k 3 5 5 5 b. Slice the region parallel to the x-axis so that each point in a strip will have about the same pressure 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) dy x = ln 5 ⇒ y = e − l n 5 = 0.2 1 1 F= dF = 1.2951K k exactly (ln 5)2 k 0.2 2
∫
c. Slice the region parallel to the y-axis. Then slice a strip parallel to the x-axis as shown in Figure 11-5g. At sample point (x, t), p = kx2t− 1. d(dF) = p dA = kx2t− 1 dx dt t =1 t =1 −1 2 dF = kx 2 t −1 dx dt = kt dt x dx t=y t=y
∫
∫
dF = 64,000 lb (exactly)
= ( k ln t tt ==1y ) x 2 dx = k (0 − ln y) x 2 dx = kx 3 dx
12 2 e. dM x = y dF = y 100 – y dy 5
1 k (ln 5) 4 = 1.6774 K k 4 9. The integrals in Problems 7 and 8 can be written in the form
F=
∫
Mx =
100
0
∫
100
0
dM x = 3, 657,142.8K
25600000 ≈ 3.657 million lb-ft exactly 7 f. y ⋅ F = M x ⇒ 400 y = 57.1428K ≈ 57.14 ft exactly 7
Calculus Solutions Manual © 2005 Key Curriculum Press
F=
∫
x =b
t =d
x =a
t =c
∫ ∫
ln 5
0
kx 3 dx =
f ( x, t ) dt dx
The result is called a double integral because two integrals appear. (Hiding inside each integral is a second integral!) π 10. a. y = 5 tan 2 x 8 π y = 5 ⇒ tan 2 x = 1 ⇒ x = ±2 8 Problem Set 11-5
303
Slice the floodgate parallel to the y-axis. π dA = 5 − 5 tan 2 x dx 8 A=
∫
2
−2
dA = 14.5352 K ≈ 14.54 ft 2
exactly 40 − 80 π b. Slice the floodgate parallel to the x-axis so that each point in a strip has about the same pressure as at the sample point (x, y). p = k(20 − y) with k = 62.4 lb/ft2 8 π y = 5 tan 2 x ⇒ x = tan −1 0.2 y 8 π 16 dA = 2 x dy = tan −1 0.2 y dy π 16 dF = p dA = k (20 − y) ⋅ tan −1 0.2 y dy π F=
∫
5
0
dF = 248.2628K k
= 15491.6027… ≈ 15.49 thousand lb 5200 k exactly 800 − 3π (The force can also be found by slicing parallel to the y-axis as in part a, then slicing the strip horizontally and using a double integral. In this case, the pressure at a sample point (x, t) is p = k(20 − t) d(dF) = p dA = k(20 − t) dt dx The first integration is from t = y to t = 5. The second integration is from x = −2 to x = 2.) c. Let µ (Greek letter mu) = coefficient of friction. 10000 µ ⋅ F = 10000 ⇒ µ = 15491.6027K = 0.6455K
2. a. v(t) = 55 + 6 t − t2 v(0) = 55 + 6 · 0 − 02 = 55 mi/h v(3) = 55 + 6 · 3 − 32 = 64 mi/h v(6) = 55 + 6 · 6 − 62 = 55 mi/h, Q .E .D . b. Cost of a short time, dt, at speed v is dC = 3(v − 55) dt = 18t − 3t2 dt. Total ticket cost is C=
Q3. Q5. Q7. Q9.
1 101 x +C Q2. 101 x ln x − x + C Q4. (force)(displacement) Q6. x = −2 Q8. y ′′ = −9 cos 3 x Q10.
1. Partition the interval into small subintervals of width dT so that C is about the same at any point in a subinterval. The amount of heat, dH, to raise the temperature by dT is dH = C dT = (10 + 0.3T1/2) dT. H=
304
∫
900
100
dH = 13, 200 calories (exactly)
Problem Set 11-6
0
dC = 108 (exactly).
a ⋅ 100 2 + b ⋅ 100 + 500 = 820 a = 0.002 ⇒ a ⋅ 200 2 + b ⋅ 200 + 500 = 1180 b = 3 b. c.
P(x) = 0.002x2 + 3x + 500 P(700) = 0.002 · 7002 + 3 · 700 + 500 = $3580/ft Cost to dig a short distance, dx, is dC = P dx = (0.002x2 + 3x + 500) dx. Cost to dig 1000 feet is 1000 8 (0.002 x 2 + 3 x + 500) dx = ⋅ 1000 2. 0 3 Cost is about $2,666,667. Cost to dig 500 feet twice (once from each end) is 500 17 C=2 (0.002 x 2 + 3 x + 500) dx = ⋅ 500 2. 0 3 Cost is about $1,416,667. Savings is about $1,250,000! in. velocity · area has the units ⋅ in.2 , s which is in.3/s, correct for flow rate. v = 4 − x 2 ⇒ v′ = −2x v′ changes from positive to negative at x = 0. ∴ there is a maximum flow rate at the center of the pipe where x = 0. (Or simply observe that the graph of v is a parabola opening downward with vertex at x = −b/(2a) = 0.) v(2) = 4 − 22 = 0, Q .E .D . Slice the water in the pipe into cylindrical shells. Each point in a shell has about the same water velocity as at the sample point x units from the axis. Let F = flow rate in in.3/s. dF = v dA = (4 − x2) · 2π x · dx = 2π (4x − x3) dx
∫
d.
∫
4. a. b.
0 2 sin x cos x = sin 2x y′ = 3(1 + 9x2)− 1 2 sec2 x tan x D
6
Fine should be $108.00. 3. a. Cost per foot, P, = ax2 + bx + c a · 02 + b · 0 + c = 500 ⇒ c = 500
Problem Set 11-6 Q1.
∫
c.
F=
∫
2
0
2π ( 4 x − x 3 ) dx = 8π = 25.1327…
≈ 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 © 2005 Key Curriculum Press
e. 4 in./s · π · 22 in.2 = 16π = 13.0559… ≈ 13.06 gal/min (exactly twice the actual rate) f. The problem is equivalent to finding the volume of a solid of rotation by cylindrical shells. The velocity takes the place of the altitude of a shell. 5. a. F 600
x 5
b. F has a step discontinuity at x = 2. c. dW = F dx Because the graph is linear on [0, 2], the work equals the area of the triangle. 1 W = ⋅ 2 ⋅ 600 = 600 in.-lb 2 d. W =
∫
5
F dx
2
By Simpson’s rule, 1 W ≈ (0.5)( 450 + 4 ⋅ 470 + 2 ⋅ 440 + 4 ⋅ 420 3 + 2 ⋅ 410 + 4 ⋅ 390 + 330) 2 = 1266 in.-lb 3 2 2 e. Total work ≈ 600 + 1266 = 1866 in.-lb 3 3 f. Yes, a piecewise continuous function such as this one can be integrable. See Problem 27 in Problem Set 9-10. 6. a. Slice the solid into disks parallel to the xz-plane so that each point in a disk has about the 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=
∫
4
0
kπ ( 4 − y) dy = 8π k g
b. Each point in a disk of part a is also about the same distance from the xz-plane as the sample point (x, y). Let K stand for the constant. dF = K · dm · y− 1/2 = K · k π (4 − y) dy · y− 1 / 2 = K kπ (4y− 1/2 − y 1/2) dy 4 32 F = dF = Kkπ = 33.5103K Kk 0 3
∫
Calculus Solutions Manual © 2005 Key Curriculum Press
7. Slice the solid into cylindrical shells so that each point in a shell is about the same distance from the y-axis as the sample point (x, y). dM 2y = x 2 dm = x 2 · ρ dV = x 2 · k · 2π xy dx = 2π kx3(4 − x2) dx 2 32 M2 y = dM2 y = π k = 33.5103K kg-cm 2 0 3 8. a. T(D) = 20 sin 2π D T(0) = 20 sin 0 = 0 T(1/4) = 20 sin π /2 = 20, which checks. b. Partition the time interval into short increments of width dD so that T is about the same at any time in the increment as it is at the sample point (D, T ) . Let H = number of degree-days. dH = T dD = 20 sin 2π D dD 1/4 10 H= dH = = 3.1830 K 0 π ≈ 3.18 degree-days 9. a. m = 2000 − 5t (mass in kilograms, time in seconds) b. a = F/m = 7000(2000 − 5t)− 1 = 1400(400 − t)− 1 dv 1400 c. a = = dt 400 – t 1400 dv = dt 400 – t v = −1400 ln | 400 − t | + C Assume the car starts at rest at t = 0. 0 = −1400 ln 400 + C ⇒ C = 1400 ln 400 400 v(t ) = 1400 ln |400 – t | 20 d. v(20) = 1400 ln = 71.8106 K ≈ 71.81 m/s 19 ds 400 v= = 1400 ln dt 400 – t 20 400 s= 1400 ln dt = 711.9673K 0 400 – t 20 ≈ 712.0 m exactly 28000 1 − 19 ln 19
∫
∫
∫
∫
∫
10. Slice the tract parallel to the tracks so that each point in the strip will have about the same value per square kilometer as at the sample point (x, y). Let v = thousands of dollars per square kilometer and 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) dx The curves intersect at x = 1. 1 1 W = 800( x − x 2 ) dx = 133 0 3
∫
Problem Set 11-6
305
The land is worth about $133,333. If all the land were worth $200,000 per km2, W = 200 A = 200
1
∫ (4 – 4 x ) dx = 400. 0
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 that each point in a strip will be about the same value per square unit as at the sample point (x, y). y = cos x Let v = value of land per square unit and W = worth of the land. v = kx, dA = y dx = cos x dx dW = v dA = kx cos x dx π /2 π W= dA = − 1 k = 0.5707K k 2 0
∫
b. Slice the tract parallel to the x-axis so that each point in a strip will be about the same value per square unit as at the sample point (x, y). v = ky dW = v dA = v · x dy = ky · cos− 1 y dy 1 π W = dW = k = 0.3926 K k 0 8
∫
y
9
(x, y )
x 3
Slice the wall parallel to the ground so that each point in the slice will cost about the same to paint per square meter. Let r = rate in dollars per square meter and C = cost in dollars to paint the wall. r = ky2 = 3y2 (r = 12 when y = 2) dA = 2 x dy = 2 9 – y dy dC = r dA = 3 y ⋅ 2 9 – y dy 2
∫
9
0
19 dC ≈ $1999.54 exactly 1999 35
13. a. Let v = value of land per square kilometer, W = worth of the land in dollars, and r = distance from center of town. Slice the city into circular rings of width dr so that each point in a ring will be about r units from the center. v = 10 − 3r, dA = 2π r dr dW = v dA = (10 − 3r ) ⋅ 2π r dr W=
∫
3
0
dW = 36π = 113.0973K
≈ 113.1 million dollars 306
Problem Set 11-6
∫
W=
3
dW = 71.4328K ≈ 71.4 million
0
18π dollars exactly (9 – ln 10) (ln 10)2 c. By Simpson’s rule,
∫
3
1 v ⋅ 2πr dr ≈ (0.3)(2π ) (10 + 4 ⋅ 12 0 3 + 2 ⋅ 15 + 4 ⋅ 14 + 2 ⋅ 13 + 4 ⋅ 10 + 2 ⋅ 8 + 4 ⋅ 5 + 2 ⋅ 3 + 4 ⋅ 2 + 1) = 52.2π = 163.9911… ≈ 164.0 million dollars d. This problem is equivalent to volume by cylindrical shells, where the value of the land per square unit takes the place of the altitude of the cylinder. It is also equivalent to the water 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 W=
F=
12.
C=
1 b. v = 10e kr , v(3) = 1 ⇒ k = − ln 10 ⇒ 3 v = 10e − (ln 10 ) r/3 dW = v dA = 10e − ( ln 10 ) r/3 ⋅ 2πr dr
∫
10
8
dF = 177.1236 K ≈ 177 lb
800 2 exactly 3 − 200 b. Average pressure = total force/total area 177.1236 K = = 44.2809K ≈ 44.3 lb/ft 2 4 200 2 exactly 3 − 50 c. dMyz = x dF = 2px dx = 200x[(x − 8)1/2 − 0.5(x − 8)] dx M yz =
∫
10
8
dM yz = 1602.8706 K ≈ 1603 lb-ft
7360 2 5600 − exactly 3 3 1602.8706 K 177.1236 K = 9.0494… ft Calvin should stand about 10 − 9.0494 ft 1 ≈ 11 in. from the end. 2 15. a. f (x) = 9 − x2 = (3 − x)(3 + x) = 0 only at x = ±3. 1 g( x ) = − x 3 − x 2 + 3 x + 9 3 1 = − ( x – 3)( x + 3)2 = 0 only at x = ±3. 3 d. x ⋅ F = M yz ⇒ x =
Calculus Solutions Manual © 2005 Key Curriculum Press
∫ =∫
b. A f =
3
(9 – x 2 ) dx = 36
–3 3
– 1 x 3 – x 2 + 3 x + 9 dx = 36 –3 3 To simplify algebraic integration, you could use Ag
3
∫ (9 – x ) dx = 2 ∫ (9 – x ) dx, where the odd terms
Af = 2 Ag
2
0 3
y
New graph
g
2
0
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) = −x 2 − 2x + 3 = −(x + 3)(x − 1) g′(x) = 0 ⇔ x = −3 or x = 1 The high point is at x = 1. d. Slice the region under the g graph parallel to the y-axis so that each point in a strip will be about the same distance from the y-axis as the sample point (x, y). dM y = x dA = x g(x) dx 1 = − x 4 − x 3 + 3 x 2 + 9 x dx 3 My =
∫
3
–3
dM y = 21.6
21.6 = 0.6 36 e. False. For the symmetrical region under the graph of f, the centroid is on the line through the high point. But for the asymmetrical region under the graph of g, the high point is at x = 1 and the centroid is at x = 0.6. f. False x ⋅ A = My ⇒ x =
Area to left =
∫
0.6
–3
∫
Area to right =
g( x ) dx = 17.1072 (exactly)
3
0.6
g( x ) dx = 18.8928 (exactly)
(or 36 − 17.1072 = 18.8928) g. Let S stand for skewness. 3
3
3 3 dS = x − dA = x − g( x ) dx 5 5 3
x – 3 − 1 x 3 − x 2 + 3 x + 9 dx S= –3 5 3
∫
3
–64 ⋅ 35 = −17.7737K exactly 7 ⋅ 125 h. By symmetry, the centroid of the area under f is on the y-axis, so x = 0. Then dS = x3 dA = x3(9 − x2) dx S=
The “skewness” being zero reflects the symmetry of this region. It is not skewed at all. i. For example, graph 1 g( − x ) = x 3 − x 2 − 3 x + 9. 3
∫
3
–3
x (9 − x ) dx = 0 (odd function 3
2
integrated between symmetrical limits) Calculus Solutions Manual © 2005 Key Curriculum Press
x –3
3
16. a. y = x 2 dL = dx 2 + dy 2 = 1 + 4 x 2 dx dM y = x dL = x 1 + 4 x 2 dx
∫
b. M y =
2
0
dM y = 5.7577K
exactly 1 (17 17 – 1) 12 c. L =
∫
2
0
1 + 4 x 2 dx = 4.6467K
exactly 1 ln ( 17 + 4) + 17 4 5.7577K = 1.2390 K d. x ⋅ L = M y ⇒ x = 4.6467K e. dS = 2π x dL = 2π x 1 + 4 x 2 dx
π dS = 36.1769K exactly (17 17 – 1) 6 f. Integral for S is 2π times the integral for My! 17. In Problem 16, R = x = 1.2390 K and L = 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 S=
∫
2
0
Problem Set 11-7 Review Problems R0. Answers will vary. R1. Slice the region parallel to the force axis so that each point in a strip has about the same force as at the sample point (x, F ) . dW = F dx = 30e− 0.2 x dx W=
∫
10
0
30e −0.2 x dx = 150(1 − e −2 )
= 129.6997… ≈ 129.7 ft-lb
Problem Set 11-7
307
R2. a. dW = F dx = kx− 2 dx 1 2 W = kx −2 dx = − k ft-lb 3 3 (Mathematically, the answer is negative because dx is negative. Physically, the answer is negative because the magnets absorb energy from their surroundings rather than releasing energy to their surroundings.) b. Construct axes with the origin at the vertex of the cone. An element of the cone in the xy7 3 plane has the equation y = x or x = y. 3 7 Slice the water horizontally into disks so that each point in a disk is lifted about the same distance as the sample point (x, y) on the element of the cone.
∫
F = 0.036 dV = 0.036 ⋅ π x2 dy 9 2 = 0.036 ⋅ π y dy 49 Each disk is lifted (10 − y) cm. dW = (10 − y) dF 9 2 = (10 − y)(0.036)π y dy 49 W=
∫
7
0
y
(x, y )
x 2
8
∫ kπy(8 − y) 0
2/3
dy = 57.6π k
b. Slice the region parallel to the y-axis, generating cylindrical shells, so that each point in a shell is about the same distance from the y-axis as the sample point (x, y).
ρ = ex, dV = 2π xy dx = 2π (8x − x4) dx m=
308
∫
2
0
b dM x = y dA = by − y 2 dy h Mx =
h
2
∫ by – h y dy = 2 by b
1
2
0
−
b 3 y 3h
h 0
1 2 b 3 1 bh − h − 0 + 0 = bh 2 2 3h 6 1 y ⋅ A = y ⋅ bh = M x 2 1 2 bh 1 ⇒y= 6 = h, Q .E.D . 1 bh 3 2 =
b. The graph shows the region under y = ex rotated about the y-axis, showing back half of solid only. y (x, y )
2π e x (8 x − x 4 ) dx = 64π
Problem Set 11-7
1
x 1
Slice the region parallel to the y-axis, generating cylindrical shells, so that each point in a shell will be about the same distance from the y-axis as the sample point (x, y). dV = 2π x ⋅ y ⋅ dx = 2π xex dx dM2y = x2 dV = 2πx3ex dx M2 y =
Slice the region parallel to the x-axis, generating disks, so that each point in a disk is about the same distance from the xz-plane as the sample point (x, y). ρ = ky, dV = π x2 dy = π (8 − y)2/3 dy m=
b b y ⇒ dA = b − y dy h h
w =b−
dW = 3.591π = 11.2814 …
≈ 11.28 in.-lb R3. a. The graph shows the region in Quadrant I under the graph of y = 8 − x3 rotated about the y-axis. 8
R4. a. The width of a strip at the sample point (x, y) is
1
∫ 2πx e 0
3 x
dx = 3.5401K
(exactly 12π − 4πe) R5. Draw axes with the x-axis at ground level and the y-axis through the upper vertex of the triangle. Slice the face of the building horizontally so that the wind pressure at any point in a strip is about equal to the pressure at the sample point (x, y). 150 dA = 150 − y dy 400 1 dF = p dA = 200 ⋅ 150(1 − e −0.01y )1 − y dy 400 F=
∫
400
0
dF = 3736263.2708…
≈ 3.736 million lb (exactly 30000(125 − 25e− 4))
Calculus Solutions Manual © 2005 Key Curriculum Press
R6. a. Let x = number of feet at which drill is operating, and r(x) = number of dollars per foot to drill at x feet. r(x) = a · bx r(0) = 30 ⇒ a = 30 5 50 = 30 ⋅ b10000 ⇒ b = 3 5 ∴ r ( x ) = 30 3 (or r ( x ) = 30e
5 b. dC = r ( x ) dx = 30 3 C=
∫
5 30 3
50000
0
= 30e
x/10000
dx
x/10000
dx = 6965243.17K
≈ 6.965 million dollars 5 10000 5 exactly − 30 ⋅ ln 0.6 3 – 1
Concept Problems C1. a. Either slice the region parallel to the y-axis, dA = (8 − y) dx = (8 − x3) dx A=
∫
2
0
(8 – x 3 ) dx = 12
or slice parallel to the x-axis, A=
∫
8
0
y1/3 dy = 12
b. i. Use slices parallel to the x-axis so that each point in a strip will be about the same distance from the x-axis as the sample point (x, y). dM x = y dA = y(y1/3 dy) 8 384 M x = y 4/3 dy = = 54.8571K 0 7 ii. Use slices parallel to the y-axis so that each point in a strip will be about the same distance from the y-axis as the sample point (x, y). dM y = x dA = x(8 − x3) dx
∫
My =
∫
2
0
(8 x – x 4 ) dx = 9.6
9.6 = 0.8 12 384/7 32 y ⋅ A = Mx ⇒ y = = = 4.5714 K 12 7 Centroid is at (0.8, 4.5714…).
c. x ⋅ A = M y ⇒ x =
d. i. With slices parallel to the x-axis, dV = 2π y · x · dy = 2π y 4/3 dy
0
2π y 4/3 dy =
∫
x/10000
0.00005108256Kx
8
768 π = 344.6775K 7 With slices perpendicular to the x-axis, dV = π (82 − y2) dx = π (64 − x6) dx 2 768 V = π (64 − x 6 ) dx = π = 344.6775K 0 7 ii. With slices parallel to the y-axis, dV = 2π x · (8 − y) · dx = 2π x(8 − x3) dx
/ 110000
− ln 0.6⋅x /10000
∫
V=
)
V=
∫
2
0
2π (8 x − x 4 ) dx = 19.2π
= 60.3185… With slices perpendicular to the y-axis, dV = π x 2 dy = π y 2/3 dy V=
∫
8
0
πy 2/3 dy = 19.2π = 60.3185K
iii. With slices parallel to the line x = 3, dV = 2π (3 − x) · (8 − y) · dx = 2π (3 − x)(8 − x3) dx V=
∫
2
0
2π (3 − x )(8 − x 3 ) dx = 52.8π
= 165.8760… With slices perpendicular to the line x = 3, dV = π [32 − (3 − x)2] dy = π [9 − (3 − y1/3)2] dy V=
∫
8
0
π [9 − (3 − y1/3 )2 ] dy = 52.8π
= 165.8760… e. i. The centroid is 32/7 units from the x-axis. 32 768 ∴ V = 2π ⋅ ⋅ 12 = π 7 7 = 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 from the line x = 3. V = 2π · 2.2 · 12 = 52.8π = 165.8760… (Checks.) f. Use horizontal slices so that each point in a disk will be about the same distance from the xz-plane as the sample point (x, y). dMxz = y dV = y(π x 2 dy) = y · π y 2/3 dy M xz =
∫
8
0
πy 5/3 dy = 96π = 301.5928K
96π =5 19.2π x = z = 0 by symmetry.
g. y ⋅ V = M xz ⇒ y =
Centroid is at (0, 5, 0). Calculus Solutions Manual © 2005 Key Curriculum Press
Problem Set 11-7
309
h. No. For the solid, y = 5, but for the region, y = 4.5714 … . i. Use slices of the region parallel to the y-axis so that each point in a resulting cylindrical shell will be about the same distance from the y-axis as the sample point (x, y). ρ = kx 2, dV = 2π x(8 − y) dx = 2π (8x − x4) dx dm = ρ dV = kx2 · 2π (8x − x4) dx = 2π k(8x3 − x6) dx 2 192 m = 2πk (8 x 3 − x 6 ) dx = πk 0 7 = 86.1693…k j. Use cylindrical shells as in part i so that each point in a shell will be about the same distance from the y-axis as the sample point (x, y). dM 2 = x2 · dm = 2π k(8x5 − x8) dx 2 512 M2 = 2πk (8 x 5 − x 8 ) dx = πk 0 9 = 178.7217… k. Use vertical slices of the region so that each point in a strip will have about the same pressure acting on it as at the sample point (x, y). p = 3 − x, dA = (8 − y) dx = (8 − x3) dx dF = p dA = (3 − x)(8 − x3) dx
∫
∫
F=
∫
2
0
(3 – x )(8 – x ) dx = 26.4 3
(Note the similarity to the integral in part 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=
∫
3
1
26.4 z −2 dz = 17.6
m. Use horizontal slices so that each point in a resulting disk will be at about the same temperature as the sample point (x, y). dH = CT dm = 0.3(10 − y)(5.8π y2/3 dy) H=
8
∫ 1.74(10 − y)(π y 0
2/3
) dy = 167.04π
= 524.7716… ≈ 524.8 cal C2. Let f (x) be the height of a vertical strip at x (or combined heights if the region being rotated is not convex). Let x = a and x = b be the left and right boundaries of the region. dV = 2π x dA = 2π x f ( x ) dx ⇒ V = 2π dM y = x dA = x f ( x ) dx ⇒ M y = 310
Problem Set 11-7
∫
∫
b
x f ( x ) dx
a
b
Note that V = 2π M y, thus showing that the two problems are mathematically equivalent, Q.E.D. C3. dMxz = y dVxz = y π x 2 dy = π y(9 − y) dy M xz = π
∫
9
0
π y(9 − y) dy = 121.5π
dM 2y = x 2 dV y = x 2 2π xy dx = 2π x3(9 − x2) dx M2 y =
∫
3
0
2π x 3 (9 − x 2 ) dx = 121.5π , Q .E.D .
This is not true in general. Counterexample: Rotate the region under y = 2 − 2x2. 1 dM xz = y dVxz = y π x 2 dy = π y1 − y dy 2 2 1 2 M xz = π y 1 − y dy = π 2 0 3 dM 2y = x 2 dV y = x 2 2π xy dx = 2π x3(2 − 2x2) dx 1 1 2 M2 y = 2π x 3 (2 − 2 x 2 ) dx = π , not π . 0 3 3 General proof: For any paraboloid of height H and base radius R, let h = distance (along the axis) from the base and r = radius. Then a H generating parabola is given by h = H − 2 ⋅ r 2. R R2 dMbase = h dV = h π r 2 dh = h π ( H − h) dh H h= H R2 Mbase = π ( Hh – h 2 ) dh h=0 H
∫
∫
∫
=
2
R H 2 1 3 π h − h H 2 3
h= H
= h=0
1 2 2 πR H 6
dM2axis = r dV = r 2π rh dr H = r 2 2π r H − 2 r 2 dr R r= R 1 r3 – M2 axis = 2πH r 5 dr r =0 R2 2
2
∫
1 1 = 2πH r 4 − 2 r 6 4 6R
r= R r =0
=
1 4 πR H 6
In the original example, H = R , so the two moments turned out to be equal. C4. a. Assume m ≠ 0. The area of the trapezoid is b +b ma + mb A= 1 2 ⋅h = (b – a) 2 2 1 = m( b 2 − a 2 ) 2 b b b 1 Integrating, y dx ≈ mx dx = mx 2 a a 2 a 1 2 2 = m(b − a ) = A, Q .E.D . 2 The length is L = (b − a) 1 + m 2 . 2
∫
Integrating,
∫
∫
b
a
dL ≈
∫
b
a
dx = (b − a) ≠ L,
Q .E .D .
x f ( x ) dx
a
Calculus Solutions Manual © 2005 Key Curriculum Press
b. Note that r = mh. The volume of the cone is 1 1 V = πr 2 h = πm2 h3 . 3 3 Integrating dV ≈ π y 2 dx = π m 2x2 dx, h 1 π m 2 x 2 dx = π m 2 h 3 = V , Q.E.D. 0 3 The surface area is S = π r r 2 + h 2 =
∫
π mh 2 1 + m 2 . Integrating dS ≈ 2π y dx = 2π mx dx,
∫
h
0
2π mx dx = π mh 2 ≠ S, Q.E.D.
c. Exact area of a strip: 1 1 ∆ A = ( mx + m( x + ∆x ))∆x = y∆x + ∆y∆x 2 2 Exact volume of frustum: π ∆V = ( m 2 ( x + ∆x )2 + m 2 x ( x + ∆x ) + m 2 x 2 )∆x 3 π = m 2 ∆x (3 x 2 + 3 x ( ∆x ) + ( ∆x )2 ) 3 1 = π y 2 ∆x + y∆y∆x + ( ∆y)2 ∆x 3 1 d. dA − y dx = y dx + ∆y dx − y dx 2 1 = ∆y dx 2 2 dV − π y dx 1 = π y 2 dx + y ∆y dx + ∆y 2 dx − π y 2 dx 3 1 2 = π y ∆y dx + ∆y dx 3 Both differences contain only higher-order infinitesimals. e. If dQ = ∆Q leaves out only infinitesimals of higher order, then
∫
b
dQ is exactly equal to Q.
a
f. Reasons: i. 0.5 and ∆y are constant with respect to the summation, 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 Test T1. a. force · displacement b. mass c. force d. area · displacement e. second moment of volume f. x
Calculus Solutions Manual © 2005 Key Curriculum Press
T2. dW = (40x − 10x2) dx
∫
W=
4
( 40 x − 10 x 2 ) dx = 90
1
T3. y ⋅ m = M xz ⇒ y ⋅ 200 = 3000 ∴ y = 3000/200 = 15 cm T4. The center of the circle is (8, 9) and the radius is 7, so the circle is on just one side of the axis of rotation (the y-axis). So the solid satisfies the hypothesis of the theorem of Pappus. The centroid of the circle is (8, 9), the displacement from the y-axis is R = 8, and the area of the circle is 49π. ∴ V = 2π RA = (2π )(8)( 49π ) = 784π = 2463.0086… T5. Using exponential regression, F ≈ 29.9829… (1.0626…)x dW = F dx W=
∫
10
F dx ≈ 412.4652 … ≈ 412.5 ft-lb
0
(By the trapezoidal rule, W ≈ 413 ft-lb. Simpson’s rule cannot be used because there is an odd number of increments.) T6. a. y
(x, y )
1
x 2
Slice the region parallel to the y-axis so that each point in a strip will be about the same distance from the y-axis as the sample point (x, y). dMy = x dA = xex dx My =
∫
2
xe x dx = e 2 + 1 = 8.3890 … ≈ 8.39 in.3
0
b. dM2y = x2 dA = x2ex dx M2 y =
∫
2
0
x 2 e x dx = 2e 2 − 2 = 12.7781…
≈ 12.78 in.4 c. A =
∫
2
0
e x dx = e 2 − 1 = 6.3890 … ≈ 6.39 in.2
x ⋅ A = My ⇒ x =
e2 + 1 = 1.3130 … ≈ 1.31 in. e2 – 1
T7. The graph shows y = x1/2 from x = 0 to x = 16, rotated about the x-axis. y 4
(x, y )
x 16
Problem Set 11-7
311
Slice the region parallel to the x-axis so that each point in a resulting cylindrical shell will be about the same distance from the x-axis as the sample point (x, y). ρ = 3 y, dm = ρ dV = 3 y ⋅ 2π (16 − x ) y dy = 6π y2(16 − y2) dy m=
∫
4
0
6π (16 y 2 − y 4 ) dy = 819.2π
= 2573.5927… ≈ 2573.6 g T8. a. Slice the end of the trough parallel to the x-axis so that each point in a strip has about the same pressure acting on it as the sample point (x, y), where x ≥ 0. p = 62.4(8 − y), dA = 2x dy = 2y1/3 dy dF = p dA = 62.4(8 − y) · 2y1/3 dy F=
∫
8
0
62.4(8 − y) ⋅ 2 y1/3 dy
64 ⋅ 9 = 62.4 = 5134.6285… ≈ 5134.6 lb 7
312
Problem Set 11-7
b. dM x = y dF = 62.4(8 − y) · 2y4/3 dy 8 9 ⋅ 210 F = 62.4(8 − y) ⋅ 2 y 4/3 dy = ⋅ 62.4 0 35 = 16430.8114… ≈ 16.43 thousand lb-ft 16430.8114 K y ⋅ F = Mx ⇒ y = = 3.2 ft 5134.6285K x = 0 by symmetry. Center of pressure is at (0, 3.2). T9. a. Slice the seating area into concentric rings of width dr. Each point in a ring will be about the same distance from the center as the sample point. Let W = worth of the seating and v = value per square foot. dW = v dA = 150r− 1 · 2π r dr = 300π dr
∫
W=
∫
b
30
300π dr = 300π (b − 30) dollars
b. 300π (b − 30) = 60000 ⇒ 200 b = 30 + = 93.6619… ≈ 93.7 ft π
Calculus Solutions Manual © 2005 Key Curriculum Press
Chapter 12—The Calculus of Functions Defined by Power Series Problem Set 12-1 6 , 1– x P5 ( x ) = 6 + 6 x + 6 x 2 + 6 x 3 + 6 x 4 + 6 x 5
1. f ( x ) =
100 y
f
P5
f
x –1
1
P5 –100
The graph of P5 fits the graph of f reasonably well for about −0.8 < x < 0.6. The graph of P5 bears no resemblance to the graph of f at x = 2 and at x = −2, for example. 2. P6(x) = P5(x) + 6x6 100 y
P6
f
P5
f
x –1
1
P5 –100
3.
4. 5.
6.
The graph of P5 fits the graph of f slightly better, perhaps for −0.9 < x < 0.7. 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. Possible conjecture: P(x) converges to f (x) for −1 < x < 1, or perhaps for −1 ≤ x ≤ 1. P0(1) = 6 P0(−1) = 6 P1(1) = 12 P1(−1) = 0 P2(1) = 18 P2(−1) = 6 P3(1) = 24 P3(−1) = 0 P4(1) = 30 P4(−1) = 6 For x = 1, the sums just keep getting larger and larger as more terms are added. For x = −1, the sums oscillate between 0 and 6. In neither case does the series converge. If the answer to Problem 4 includes x = 1 or x = −1, the conjecture will have to be modified. P5(0.5) = 11.8125 and f (0.5) = 12 The values differ by 0.1875, and 6(0.5)6 = 0.09375. P5(−0.5) = 3.9375 and f (−0.5) = 4
Calculus Solutions Manual © 2005 Key Curriculum Press
The values differ by 0.0625, and 6(−0.5)6 = −0.09375. For P5(0.5), the difference is greater than the value of the next term of the series. This result is to be expected because the rest of the series is formed by adding more positive terms. For P5(−0.5), the difference is less in absolute value than the absolute value of the next term. This result is reasonable because the terms alternate in sign so that you are adding and subtracting ever smaller quantities. 7. A geometric series; the common ratio
Problem Set 12-2 Q1. . . . 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 theorem Q5. derivative Q6. cos x − x sin x 1 Q7. x sin x + cos x + C Q8. dA = r 2 dθ 2 Q9. f (x) = e− x Q10. D 1. Series: 200 − 120 + 72 − 43.2 + 25.92 − 15.552 + L Sums: 200, 80, 152, 108.8, 134.72, 119, 168, … Sn
125 100
n 10
1 = 125 1 + 0.6 The series converges to 125. 1 – (–0.6) n |125 − Sn | = 125 − 200 ⋅ 1 + 0.6 S = 200 ⋅
= 125 1 − 1.6 ⋅
1 – (–0.6) n 1.6
= 125|1 − (1 − (−0.6)n)| = 125|−0.6n| = 125(0.6n) ln 0.0001125 / 125(0.6 n ) = 0.0001 ⇔ n = = 27.48K ln 0.6 Make n ≥ 28. Sn will be within 0.0001 unit of 125 for all values of n ≥ 28. Problem Set 12-2
313
2. Series: 30 + 33 + 36.3 + 39.93 + 43.923 + 48.3153 + K Sums: 30, 63, 99.3, 139.23, 183.153, 231.4683, … Sn
100
n
c. The total perimeter converges to 1 16 ⋅ = 54.6274 K cm. 1 – 0.51/2 d. The sum of the areas is 16 + 8 + 4 + 2 + L , which is a convergent geometric series with r = 0.5. 1 S = 16 ⋅ = 32 cm2 1 – 0.5 5. a. The interest rate for one month is 0.09/12 = 0.0075.
10
Months The graph shows divergence. 1 – 1.1100 S100 = 30 ⋅ = 4,133,883.70 K (Wow!) 1 – 1.1 The formula S = t1/(1 − r) gives −300 for S, but it has no meaning because the series does not converge. 3. a. Series: ∞
∑ 7(0.8
n −1
) = 7 + 5.6 + 4.48 + 3.584 + L
n =1
Sums: 7, 12.6, 17.08, 20.664, 23.5312, … S 4 = 20.664, so the amount first exceeds 20 µg at the fourth dose. 1 S = 7⋅ = 35, so the total amount 1 – 0.8 never exceeds 40 µg. The graph confirms that the partial sums of the series approach 35 asymptotically and first exceed 20 µg at the fourth dose. 40 Asymptote
30 Goes > 20 20
Stays > 20
10
n 1
2
3
4
5
6
7
8
9
10
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 show the partial sums just before a dose. t7 = 20.6599… . The amount remains above 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 = 16 and r = 0.5 . b. S10 = 16 ⋅
314
1 – 0. 5 5 = 52.9203K cm 1 – 0.51/2
Problem Set 12-2
0 1 2 3
Dollars 1,000,000.00 1,007,500.00 1,015,056.25 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, the second at time t = 1, and so forth, so at time t = 12, the term index is 13. d. Meg earned $93,806.90 the first year. 93806.90 APR = ⋅ 100 = 9.3806 K% 1000000 e. (1,000,000)(1.0075n) = 2,000,000 ln 2 n= = 92.7657K ln 1.0075 After 93 months 6. a. The interest rate for one month is 0.108/12 = 0.009. S5 = 100 + 100(1.009) + 100(1.009)2 + 100(1.009)3 + 100(1.009)4 + 100(1.009)5 1 – 1.009 6 = 100 ⋅ = $613.66 1 – 1.009 b. There are six terms because the term index of the first term is zero. c. 10 years equals 120 months. There will have been 121 deposits after 10 years because the initial deposit was made at time 0. So there are 121 terms. 1 – 1.009121 S120 = 100 ⋅ = $21,742.92 1 – 1.009 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. S 4 = 20 + 18 + 16.2 + 14.58 = 68.78 ft 1 c. S = 20 ⋅ = 200 1 – 0.9 So the ball travels 200 ft before it comes to rest. Calculus Solutions Manual © 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 is 2(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 first term 1.5762… and common ratio equal to 0.91/2 = 0.9486… . So the series of times converges to 1 S = 1.5762 K ⋅ = 30.7155K 1 – 0.91/2 The model predicts that the ball comes to rest after about 30.7 s. 8. a. Iteration Total Length 0 1 2 3
27 36 48 64
Because each segment is divided into four pieces, each of which is 1/3 of the original length, the length at the next iteration can be calculated by multiplying the previous length by 4/3. b. The sequence of lengths diverges because the common ratio, 4/3, is greater than 1. Thus, the total length of the snowflake curve is infinite! c. From geometry, the area of an equilateral 3 2 triangle of side s is A = s . 4 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 [3(3) 2 + 12(1) 2 + 48(1/3)2 + 192(1/9)2 + L ] 4 3 = ⋅ 3 ⋅ 9 2 [ 4 0 (1/3)2 + 41 (1/3) 4 + 4 2 (1/3)6 4 + 4 3 (1/3)8 + L ] 3 ⋅ 3 ⋅ 9 2 [ 4/9 + ( 4/9)2 + ( 4/9)3 + ( 4/9) 4 + L ] 16 3 1 = ⋅ 3 ⋅ 9 2 ( 4/9) ⋅ = 12.15 3 16 1 – 4/9 The area of the pre-image is 3 2 ⋅ 9 = 20.25 3 . 4 The total area is 32.4 3 = 56.1184 K cm 2 .
=
Calculus Solutions Manual © 2005 Key Curriculum Press
6 1– x P( x ) = 6 + 6 x + 6 x 2 + 6 x 3 + 6 x 4 + 6 x 5 + L P ′( x ) = 6 + 12 x + 18 x 2 + 24 x 3 + 30 x 4 + L P ′′( x ) = 12 + 36 x + 72 x 2 + 120 x 3 + L P ′′′( x ) = 36 + 144 x + 360 x 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) = 6 P″(0) = 12 and f ″(0) = 12 P″′(0) = 36 and f ″′(0) = 36 Conjecture: P (n)(0) = f (n)(0) for all values of n.
9. f ( x ) =
Problem Set 12-3 Q1. 0.3333… Q2. 0.4444… 2 4 Q3. Q4. 3 9 Q5. 13 Q6. 125 Q8. centroid Q7. mass ⋅ displacement Q9. ln x + C Q10. D 1. f (x) = 5e2x f ′(x) = 10e2x f ″(x) = 20e2x f ′′′(x) = 40e2x f (4)(x) = 80e 2x 2. P1(x) = c0 + c1x and P1′ ( x ) = c1 P1(0) = c0 and f (0) = 5 ⇒ c0 = 5 P1′ (0) = c1 and f ′(0) = 10 ⇒ c1 = 10 ∴ P1(x) = 5 + 10x 3. P2(x) = c0 + c1x + c2x2 P2′ ( x ) = c1 + 2c2 x and P2′′( x ) = 2c2 P2(0) = c0 and f (0) = 5 ⇒ c0 = 5 P2′ (0) = c1 and f ′(0) = 10 ⇒ c1 = 10 P2′′(0) = 2c2 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. P 3(x) = c0 + c1x + c2x2 + c3x3 P 4(x) = c0 + c1x + c2x2 + c3x3 + c4x4 P4′′′ ( x ) = 6c3 + 24c4 x and P4( 4 ) ( x ) = 24c4 P4′′′ (0) = 6c3 and f ′′′(0) = 40 20 ⇒ 6c3 = 40 ⇒ c3 = 3 P (44 ) (0) = 24c4 and f ( 4 ) (0) = 80 10 ⇒ 24c4 = 80 ⇒ c4 = 3 c0, c1, and c2 are the same as before. 20 3 5. P3 ( x ) = 5 + 10 x + 10 x 2 + x 3 20 3 10 4 P4 ( x ) = 5 + 10 x + 10 x 2 + x + x 3 3 Problem Set 12-3
315
y
f
100
1. a.
P4 P3
P4 x
f (n) ( x )
0 1
ex ex
1 1
c0 c1
2
ex
1
2!c2
3
ex
1
3!c3
1
P3
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.
∑ n=0
d. y
ex
2
x 3
Problem Set 12-4 Q2. y
y x
Q3.
x
Q4. y
y
x
e. The two graphs are indistinguishable for approximately −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 cn f (n) ( x ) f (n) (0) P (n) (0) 0 1
cos x −sin x
1 0
2
−cos x
−1
2!c2
3
sin x
0
3!c3
4
cos x
1
4!c4
5
−sin x
0
5!c5
6
−cos x
−1
6!c6
7
sin x
0
7!c7
8
cos x
1
8!c8
x
Q5.
Q6. y
y
x
x
Q7. exponent Q9. power
316
Problem Set 12-4
Q8. coefficient Q10. D
S3
5
5 ⋅ 2n n x n!
Q1.
1 1 1 2! 1 3!
∑
20 5 ⋅ 2 20 5 ⋅ 2 9. c3 = = , c2 = = , 6 3! 2 2! 10 5 ⋅ 21 5 ⋅ 20 c1 = = , c0 = 5 = (0! = 1) 1 1! 0! 10. Conjecture: 5 ⋅ 2 5 160 4 5 ⋅ 2 6 320 4 c5 = = = , c6 = = = 5! 120 3 6! 720 9 ∞
cn
1 2 1 3 x + x + L , Q .E .D . 2! 3! 1 4 1 5 b. Next two terms: L + x + x + L 4! 5! ∞ 1 n c. x n! n=0
80 5 ⋅ 2 4 = 24 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. 3
f (n) (0) P (n) (0)
∴ P( x ) = 1 + x +
8. c4 =
11. P( x ) =
n
c0 c1
1 0 1 − 2! 0 1 4! 0 1 − 6! 0 1 8!
Calculus Solutions Manual © 2005 Key Curriculum Press
∴ P( x ) = 1 −
1 2 1 4 1 6 1 8 x + x − x + x −L , 2! 4! 6! 8!
Q .E .D .
b. L −
1 10 1 12 1 14 x + x − x +L 10! 12! 14!
∞
c.
1 3 1 5 1 7 x + x + x +L 3! 5! 7! b. By equating derivatives:
4. a. P( x ) = x +
∑ (–1)
n
n=0
1 x 2n (2 n)!
d. y = cos x y S4 cos
x
n
f (n)( x )
f (n)(0)
P (n) (0)
cn
0 1 2
sinh x cosh x sinh x
0 1 0
c0 c1 2!c2
3
cosh x
1
3!c3
4
sinh x
0
4!c4
5
cosh x
1
5!c5
6
sinh x
0
6!c6
7
cosh x
1
7!c7
0 1 0 1 3! 0 1 5! 0 1 7!
S7
e. See the graph in part d, showing S7(x) (eighth partial sum). The graphs are indistinguishable for approximately −5.5 < x < 5.5. f. Solve S7(x) − cos x = 0.0001 for x close to 5.5. x ≈ 4.5414… (Note that some solvers may give an error message. 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 so forth.) By symmetry, the interval is −4.5414… < x < 4.5414… . g. Both functions are even. P(x) is even because it has only even powers of x. 1 1 1 3. a. S3 (0.6) = 0.6 − (0.6 3 ) + (0.6 5 ) − (0.6 7 ) 3! 5! 7! = 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 = 0.000648 sin 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 than the absolute value of the first term of the tail, Q .E .D . c. Make | tn+1 | < 0.5 × 10 −20. 1 (0.6 2 n+3 ) < 5 × 10 −21 (2 n + 3)! Inequality is first true for n = 8. Use at least nine terms (n = 8). Calculus Solutions Manual © 2005 Key Curriculum Press
1 3 1 5 1 7 x + x + x + L , Q .E .D . 3! 5! 7! S3(0.6) = 0.636653554… sinh 0.6 = 0.636653582… ∴ S3(0.6) ≈ sinh 0.6, Q.E.D. Solve S3(x) − sinh x = 0.0001 for x close to 1. x ≈ 1.4870… By symmetry, the interval is −1.4870… < x < 1.4870… . 1 1 1 P ′( x ) = 1 + ⋅ 3 x 2 + ⋅ 5 x 4 + ⋅ 7 x 6 + L 3! 5! 7! 1 2 1 4 1 6 = 1+ x + x + x +L 2! 4! 6! Find S3(0.6) for the P′ series. S3(0.6) = 1.1854648 cosh 0.6 = 1.18546521… ∴ S3(0.6) ≈ cosh 0.6, and thus the P′(x) series seems to represent cosh x, Q.E.D.
∴ P( x ) = x + c.
d.
e.
f.
g.
∫ P( x ) dx
1 2 1 1 4 1 1 6 x + ⋅ x + ⋅ x +L+ C 2 3! 4 5! 6 Simplifying and letting C = 1 gives 1 1 1 1+ x2 + x4 + x6 + L , 2! 4! 6! which is the series for cosh x, Q.E.D. 5. a. f (x) = ln x f (1) = 0 f ′(x) = x − 1 f ′(1) = 1 f ′′( x ) = − x −2 f ′′(1) = −1 f ′′′(1) = 2 f ′′′( x ) = 2 x −3 1 1 P( x ) = ( x − 1) − ( x – 1)2 + ( x – 1)3 2 3 1 4 − ( x – 1) + L 4 =
Problem Set 12-4
317
P′( x ) = 1 − ( x − 1) + ( x − 1)2 − ( x − 1)3 + K P ′′( x ) = −1 + 2( x − 1) − 3( x − 1) 2 + K P ′′′( x ) = 2 − 6( x − 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 . 1 1 b. L + ( x – 1)5 − ( x – 1)6 + L 5 6 c. P( x ) =
∞
∑
1 (–1) n+1 ⋅ ( x – 1) n n n =1
d. y ln x 1
x 1
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… S 10(x) fits ln x in about 0 < x < 2. This is a wider interval of agreement than that for the fourth partial sum, which looks like about 0.3 < x < 1.7. S10(1.2) and ln 1.2 agree through the eighth decimal place. The values of S10(1.95) and ln 1.95 agree only to one decimal place. The values of S10(3) and ln 3 bear no resemblance to each other. 1 1 6. a. P( x ) = ( x − 1) − ( x – 1)2 + ( x – 1)3 2 3 1 4 K − ( x – 1) + 4
2n ∞ → x →∞ x →∞ n ∞ ∞ 2 n ln 2 = lim → x →∞ 1 1 =∞ The series cannot possibly converge because the terms do not approach zero as n approaches infinity.
b. lim | tn | = lim
c. n
tn(1.2)
1 2 3 4 5 6
0.2 −0.02 0.0026666… −0.0004 0.000064 −0.00001066…
The absolute values of the terms are approaching 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 = −0.02 ln 1.2 − S2(1.2) = 0.002321… t3 = 0.002666… ln 1.2 − S3(1.2) = −0.0003451… t4 = −0.0004 In each case, the tail is less in magnitude than the absolute value of the first term of the tail. 7. a. f (x) = tan− 1 x ∞ 1 P( x ) = (–1) n x 2 n+1 2 n + 1 n=0 1 1 1 = x − x3 + x5 − x7 + L 3 5 7 b. y = tan− 1 x, y = S 5(x), and y = S 6(x) (sixth and seventh partial sums)
∑
y
S6
S5
f (x ) 1
x
n 1 2 3 4 5 6
tn(3) 2 −2 2.6666… −4 6.4 −10.6666…
The absolute values of the terms are getting larger as n increases.
318
1
Problem Set 12-5
S5 S6
Both partial sums fit the graph of f very well for about −0.9 < x < 0.9. For x > 1 and x < −1, the partial sums bear no resemblance to the graph of f.
Problem Set 12-5 Q1. 4! = 24 Q3. 4!/4 = 6
Q2. 3! = 6 Q4. n = 3
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Q5. n = m − 1
Q6. m = 1
Q7. 0! = 1!/1 = 1
Q8. (−1)! = 0!/0 = 1/0 = ∞
Q9. x/ x 2 – 7 1. f (u) = e
Q10. A
u
1 2 1 3 1 4 1 5 K u + u + u + u + 2! 3! 4! 5! 2. f (u) = ln u = 1+ u +
1 1 1 = (u − 1) − (u – 1)2 + (u – 1)3 − (u – 1) 4 + K 2 3 4 3. f (u) = sin u =u−
1 3 1 5 1 7 1 9 1 11 u + u − u + u − u +L 3! 5! 7! 9! 11!
1 1 13. ln x 2 = ( x 2 − 1) − ( x 2 – 1)2 + ( x 2 – 1)3 − L 2 3 2 (Or: ln x = 2 ln x = 2(x − 1) − (x − 1)2 2 + ( x – 1)3 − L ) 3 2 1 1 14. e − x = 1 + (– x 2 ) + (– x 2 )2 + (– x 2 )3 2! 3! 1 + (– x 2 ) 4 + L 4! 1 1 1 = 1 − x2 + x 4 − x6 + x8 − L 2! 3! 4!
4. f (u) = cos u = 1−
1 2 1 4 1 6 1 8 1 10 u + u − u + u − u +L 2! 4! 6! 8! 10!
16.
∫
x
1 3 1 5 1 7 1 9 1 11 u + u + u + u + u +L 3! 5! 7! 9! 11!
1 4 1 6 1 8 1 10 x + x − x + x −L 3! 5! 7! 9! 1 1 1 1 10. x sinh x = x x + x 3 + x 5 + x 7 + x 9 + L 3! 5! 7! 9! = x2 +
+
(3t − 1) − 1 (3t – 1)2 1/3 2
∫
x
x
1/3
1 = 1 − x 4 + x 8 − x 12 + x 16 − L x +1
18.
9 3 = x 2 + 3 1 + ( x 2 /3)
4
1 1 1 = 3 1 − x 2 + 2 x 4 − 3 x 6 + L 3 3 3 1 4 1 6 x − 2 x +L 3 3
= 3 − x2 + 19.
∫
x
1 dt = t +1 4
0
x
∫ (1 – t
20.
∫
x
9 dt = t +3 2
4
0
= x−
0
Calculus Solutions Manual © 2005 Key Curriculum Press
2
17.
11. cosh x 3
1 1 1 1 = 1 − ( x 2 ) 2 + ( x 2 ) 4 − ( x 2 ) 6 + ( x 2 )8 − L 2! 4! 6! 8! 1 4 1 8 1 12 1 16 = 1− x + x − x + x −L 2! 4! 6! 8!
ln (3t ) dt =
1 4 1 6 1 8 1 10 x + x + x + x +L 3! 5! 7! 9!
1 1 1 1 = 1 + (x 3)2 + (x 3) 4 + (x 3)6 + (x 3)8 + L 2! 4! 6! 8! 1 1 1 1 = 1 + x 6 + x 12 + x 18 + x 24 + L 2! 4! 6! 8! 12. cos x 2
x
∫ 1 – t
1 1 1 (3 x – 1)2 − (3 x – 1)3 + (3 x – 1) 4 6 18 36 1 − (3 x – 1)5 + L 60
=
8. f (u) = tan −1 u
= x2 −
dt =
1 1 + (3t – 1) 4 − (3t – 1)5 + L 3⋅ 4 ⋅3 3⋅5⋅ 4
7. f (u) = (1 − u) −1 = 1 + u + u 2 + u 3 + u 4 + u 5 + L
1 1 1 1 1 = u − u 3 + u 5 − u 7 + u 9 − u11 + L 3 5 7 9 11 1 1 1 1 9. x sin x = x x − x 3 + x 5 − x 7 + x 9 − L 3! 5! 7! 9!
−t2
1 1 + (3t – 1)3 − (3t – 1) 4 + … dt 3 4 1 1 = (3t – 1)2 − (3t – 1)3 3 ⋅ 2 ⋅1 3⋅3⋅ 2
1 2 1 4 1 6 1 8 1 10 u + u + u + u + u +L 2! 4! 6! 8! 10!
6. f (u) = sinh u =u+
∫e
1/3
5. f (u) = cosh u = 1+
1 4 1 6 1 8 t – t + t – L dt 0 0 2! 3! 4! 1 3 1 1 5 1 1 7 1 1 9 = x − x + ⋅ x − ⋅ x + ⋅ x −L 3 5 2! 7 3! 9 4! x
15.
x
1 5 1 9 1 13 1 17 K x + x − x + x − 5 9 13 17
∫ 3 – t 0
+ t 8 – t 12 + t 16 – K) dt
2
1 1 + t 4 – 2 t 6 + K dt 3 3
1 1 5 1 1 = 3x − x 3 + x − 2 x7 + 3 x9 −K 3 3⋅5 3 ⋅7 3 ⋅9
Problem Set 12-5
319
21.
d (sinh x 2 ) dx d 2 1 6 1 10 1 14 K x + x + x + x + = dx 3! 5! 7! 6 10 14 = 2 x + x 5 + x 9 + x 13 + K 3! 5! 7! 2 5 2 9 2 13 K = 2x + x + x + x + 2! 4! 6! 1 9 1 13 … 5 = 2x + x + x + x + 12 360 Alternate solution:
d (sinh x 2 ) = 2 x cosh x 2 dx 1 1 1 = 2 x 1 + x 4 + x 8 + x 12 + K 2! 4! 6! 2 2 2 = 2 x + x 5 + x 9 + x 13 + … 2! 4! 6! d (cos x 0.5 ) 22. dx d 1 1 1 1 1 – x + x 2 – x 3 + x 4 – K = dx 2! 4! 6! 8! 1 2 3 4 = − + x − x2 + x3 −K 2! 4! 6! 8! 1 1 1 2 1 =− + x − x + x3 − … 2 12 240 10080 Alternate solution: d 1 (cos x 0.5 ) = − x −0.5 sin x 0.5 dx 2 1 −0.5 0.5 1 1.5 1 2.5 1 3.5 =− x x − x + x − x + L 2 3! 5! 7! 1 1 1 2 1 3 x− x + x −L =− + 2! 2 ⋅ 3! 2 ⋅ 5! 2 ⋅ 7! 1 2 3 4 = − + x − x2 + x3 −L 2! 4! 6! 8! Multiply by 1/1, 2/2, 3/3, 4/4, … and simplify. 0.7 23. P4 ( x ) = −8 + 3( x − 2) + ( x − 2)2 2! 0.51 0.048 ( x − 2)3 − ( x − 2) 4 + 3! 4! = −8 + 3(x − 2) + 0.35(x − 2)2 + 0.085(x − 2)3 − 0.002(x − 2)4 0.48 0 ( x + 1) 2 + ( x + 1)3 2! 3! 0.36 0.084 4 ( x + 1) − ( x + 1)5 + 4! 5! = 7 + 2(x + 1) − 0.24(x + 1)2 + 0.015(x + 1)4 − 0.0007(x + 1)5
24. P5 ( x ) = 7 + 2( x + 1) −
320
Problem Set 12-5
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 = 1.694912 We must assume that the series converges for x = 0.4. b. f (−1) = c 0 = 2 f ′(−1) = c 1 = 0.5 f ″(−1) = 2!c2 = 2(−0.3) = −0.6 f ″′(−1) = 3!c3 = 6(−0.18) = −1.08 f (4)(−1) = 4!c4 = 24(0.02) = 0.48 c. g(x) = f (x2 − 1) ≈ P 4(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 powers of x. ∴ g′(0) = 0 g″(x) = 1 − 3.2x2 + terms in higher powers of x. ∴ g″(0) = 1 > 0. ∴ (0, g(0)) is a local minimum. e.
∫
x
0
g (t ) dt ≈
∫
x
0
(2 + 0.5t − 0.3t ) dt
= 2t + = 2x +
2
0.5 3 1 6
t − 3
4
0.3 5
x t
x − 0.06 x 3
5
0 5
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 = −6.47 We must assume that the series converges for x = 1. b. f (2) = c 0 = −4 f ′(2) = c 1 = 3 f ″(2) = 2!c2 = 2(0.5) = 1 f ′″(2) = 3!c3 = 6(−0.09) = −0.54 f (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 © 2005 Key Curriculum Press
d. g′(x) = 6x + 2x3 + terms in higher powers of x. ∴ g′(0) = 0 g″(x) = 6 + 6x2 + terms in higher powers of x. ∴ g″(0) = 6 > 0 ∴ (0, g(0)) = (0, 4) is a local minimum. e. h( x ) =
∫
x
0
g(t ) dt ≈
x
∫ (−4 + 3t 0
2
+ 0.5t 4 ) dt
x
= −4t + t + 0.1t 5 0 = −4 x + x 3 + 0.1x 5 3
27. f (x) = sin x, about x = π/4: f ( x) =
π π 2 2 2 + x– − x– 2 2 4 2 ⋅ 2! 4 −
π π 2 2 x– + x– 2 ⋅ 3! 4 2 ⋅ 4! 4
+
π 2 x – −L 2 ⋅ 5! 4
3
2
n
f (n)( x )
f (n)(0)
P (n)(0)
cn
0
cos 3x
1
c0
1
1
−3 sin 3x
0
c1
2
−9 cos 3x
−9
2!c2
3
27 sin 3x
0
3!c3
4
81 cos 3x
81
4!c4
5
−243 sin 3x
0
5!c5
6
−729 cos 3x
−729
6!c6
0 9 − 2! 0 81 4! 0 729 − 6!
4
5
28. f (x) = cos x, about x = π/4: 2 π π 2 2 2 f ( x) = − x– − x– 2 2 4 2 ⋅ 2! 4 3 4 π π 2 2 x– + x– + 2 ⋅ 3! 4 2 ⋅ 4! 4 5 π 2 x – −L − 2 ⋅ 5! 4 29. f (x) = ln x, about x = 1: 1 1 f ( x ) = ( x − 1) − ( x – 1)2 + ( x – 1)3 3 2 1 4 − ( x – 1) + L 4 30. f (x) = log x, about x = 10: ( x – 10) 2 1 ( x – 10) 1 f ( x) = 1 + ⋅ − ⋅ ln 10 10 2 ln 10 10 2 ( x – 10)3 ( x – 10) 4 1 1 + ⋅ − ⋅ +L 3 3 ln 10 10 4 ln 10 10 4 31. f (x) = (x − 5)7/3 , about x = 4: 7 7⋅4 f ( x ) = −1 + ( x – 4) − 2 ( x – 4)2 3 3 2! 7 ⋅ 4 ⋅1 7 ⋅ 4 ⋅ 1 ⋅ (–2) ( x – 4) 4 + 3 ( x – 4)3 − 3 3! 34 4! 7 ⋅ 4 ⋅ 1 ⋅ (–2)(–5) ( x – 4) 5 − L + 355! 32. f (x) = (x + 6)4.2, about x = −5: 4.2 ⋅ 3.2 ( x + 5)2 f ( x ) = 1 + 4.2( x + 5) + 2! 4.2 ⋅ 3.2 ⋅ 2.2 ( x + 5)3 + 3! 4.2 ⋅ 3.2 ⋅ 2.2 ⋅ 1.2 ( x + 5) 4 + L + 4!
Calculus Solutions Manual © 2005 Key Curriculum Press
33. By equating derivatives:
9 2 81 4 729 6 x + x − x +L 2! 4! 6! By substitution: 1 1 1 cos 3 x = 1 − (3 x )2 + (3 x ) 4 − (3 x )6 + L 4! 6! 2! 9 2 81 4 729 6 x +L = 1− x + x − 2! 4! 6! The two answers are equivalent. Substitution gives the answer much more easily in this case.
∴ cos 3 x = 1 −
34. By equating derivatives: n
f (n)( x )
f (n)(0)
P (n)(0)
cn
0 1 2
ln (1 + x) (1 + x)− 1 −(1 + x)− 2
0 1 −1
c0 c1 2!c2
3
2(1 + x)− 3
2
3!c3
4
−6(1 + x)− 4
−6
4!c4
0 1 1 − 2! 2 3! 6 − 4!
1 2 1 3 1 4 x + x − x +L 2 3 4 By substitution, substitute (1 + x) for u in 1 1 1 ln u = (u − 1) − (u – 1)2 + (u – 1)3 − (u – 1) 4 + L 2 3 4 1 2 1 3 1 4 ln (1 + x ) = x − x + x − x + L . 2 3 4 The two answers are equivalent. Substitution gives the answer much more easily in this case.
∴ ln (1 + x ) = x −
35. S4(1.5) = 0.40104166… ; ln 1.5 = 0.40546510… Error = 0.00442344… 1 Fifth term = (1.5 – 1)5 = 0.00625 5 The error is smaller in absolute value than the first term of the tail. Problem Set 12-5
321
36. Solve numerically for x close to 2: S 4(x) − ln x = 0.0001 x ≈ 1.2263… Solve numerically for x close to 0.1: ln x − S 4(x) = 0.0001 x ≈ 0.7896… Interval is about 0.7896… < x < 1.2263… . 1 1 1 37. a. tan −1 x = x − x 3 + x 5 − x 7 3 5 7 1 9 1 11 + x − x +L 9 11 1 1 1 1 1 ∴ tan −1 1 = 1 − + − + − + L 3 5 7 9 11 The tenth partial sum is S9(1). 9
S9 (1) =
∑ (–1)
n
n=0
1 1 tan tan –1 + tan tan –1 2 3 = 1 1 1 – tan tan –1 ⋅ tan tan –1 2 3 1 1 + = 2 3 =1 1 1 1– ⋅ 2 3 1 1 ∴ tan −1 1 = tan −1 + tan −1 , Q .E.D . 2 3 9 2 n +1 1 1 p = 4 S9 = 4 (–1) n 2n + 1 2 n=0
∑ 9
∑ (–1)
n
n=0
9
=4
(−1) 1 n
∑ 2n + 1 2 n=0
= 3.14159257… π = 3.14159265…
322
Problem Set 12-5
38. sin x = x −
1 3 1 5 1 7 x + x − x +L 3! 5! 7!
cos x = 1 −
1 2 1 4 1 6 x + x − x +L 2! 4! 6!
1 = 0.760459904 K (2 n + 1)
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 that although 4S49(1) differs from π in the second decimal place, several other decimal places later on do match up!) c. By the composite argument properties from trig, 1 1 tan tan −1 + tan −1 2 3
+
The answer differs from π by only 1 in the seventh decimal place. The improvement in accuracy is accounted for by the fact that the inverse tangent series converges much more rapidly for x = 1/2 and x = 1/3 than it does for x = 1. In Problem 17 of Problem Set 12-6, you will see that the interval of convergence for the inverse tangent series is −1 ≤ x ≤ 1. In general, power series converge slowly at the endpoints of the convergence interval.
1 1 2n + 1 3 2 n +1
1 + 3
2 n +1
2 n +1
x+
1−
1 3
x3 +
2 15
17
x 5+
315
x 7+ L
1 2 1 4 1 6 1 1 1 x + x − x + L x – x 3 + x 5 – x 7 +L 2! 4! 6! 3! 5! 7! 1 3 1 5 1 7 x – x + x – x +L 2! 4! 6! 1 3 4 5 6 7 x − x + x −L 3 5! 7! 1 3 1 5 1 7 x – x + x –L 3 6 72 2 5 64 7 x − x −L 15 7! 2 5 1 7 x – x +L 15 15 17 7 x −L 315 1 3 2 5 17 7 x + x + x +L 3 15 315 S4(0.2) = 0.202710024… tan 0.2 = 0.202710035… ∴ tan x = x +
39. Define ai ( x ) =
f
(i )
( a) ( x − a)i, the ith term of the i!
general Taylor series. So, f ( x ) =
∞
∑ a ( x ). i
i=0
We must assume
dn dx n
∞
∑ i=0
ai ( x ) =
∞
dn
∑ dx
n
ai ( x );
i=0
that is, the nth derivative of an infinite series is the infinite sum of the nth derivatives of the individual terms. For i < n,
dn f ( i ) ( a) d n ⋅ n ( x − a)i n ai ( x ) = dx i! dx (i ) f ( a) = ⋅0 = 0 i!
Calculus Solutions Manual © 2005 Key Curriculum Press
dn dn a ( x ) = an ( x ) i dx n dx n f ( n ) ( a) d n = ⋅ n ( x − a) n n! dx f ( n ) ( a) = ⋅ n!( x − a) 0 n! dn f ( i ) ( a) d n For i > n, n ai ( x ) = ⋅ n ( x − a)i dx i! dx f (i ) ( x ) = i ⋅ (i − 1)(i − 2)K(i − n + 1)( x − a)i −n = 0 i! for x = a. dn So n ai ( a) = 0 for i < n and i > n, and dx dn an ( a) = f ( n ) ( a). dx n ∞ dn Thus, n ai ( x ) evaluated at x = a dx i =0 For i = n,
∑
dn an ( a) = f ( n ) ( a). dx n 40. Brook Taylor: 1685−1731 Colin Maclaurin: 1698−1746 Sir Isaac Newton: 1642−1727 Gottfried Wilhelm von Leibniz: 1646−1716 1 (–1) n+2 ( x – 1) n+1 tn+1 n 1 n + | x − 1| 41. a. = = 1 tn n +1 (–1) n+1 ( x – 1) n n 2 b. r10 = for x = 1.2 11 9.5 r10 = for x = 1.95 11 20 r10 = for x = 3 11 n n | x − 1| = | x − 1| lim = | x − 1| c. r = lim n→∞ n + 1 n→∞ n + 1 d. r = 1.1 for x = −0.1 r = 1 for x = 0 r = 0.9 for x = 0.1 r = 0.9 for x = 1.9 r = 1 for x = 2 e. Possible conjecture: The series converges to ln x whenever the value of x makes r < 1, and diverges 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 k, then n tn is within ε units of L.
∞
25.
c. For all integers n > k, 0 ≤ n tn < L + ε ⇒ 0 ≤ tn < ( L + ε ) n and (L + ε )n < (L + ε )n− k for all n > k because L + ε < 1, so 0 ≤ tn < (L + ε ) n− k, Q .E .D . d. Because 0 ≤ tn < (L + ε)n− k for all n > k, it follows that the tail after tn satisfies 0 ≤ tn+1 + tn+2 + tn+3 + 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 is bounded above by
Calculus Solutions Manual © 2005 Key Curriculum Press
n n
n=1
L = lim n |n n x n | = lim |nx | = | x | ⋅ ∞
Thus, n tn < L + ε , Q .E .D . b. L < 1 ⇒ 1 − L > 0 So take any ε < 1 − L ⇒L+ε 1000 if
∫
n +1
1
(1/ x ) dx > 1000 ⇔ ln x
n +1
=
1
ln (n + 1) > 1000 ⇔ n > e1000 − 1 ≈ 1.97 · 10434 s. It would take approximately 6.24 · 10420 yr. 1 1 1 5. a. 1 − + − + L 2 3 4 The series converges because (1) strictly alternating signs, (2) strictly decreasing |tn|, (3) tn → 0. Calculus Solutions Manual © 2005 Key Curriculum Press
b.
∫
∞
1
(1/ x ) dx = lim (ln b – ln 1) = ∞ b→∞
Τhe series
∞
∑ |t | diverges, so the given series n
n=1
does not converge absolutely. c. S1000 = 0.692647… , S1001 = 0.693646… , ln 2 = 0.693147… | S 1000 − ln 2| = 0.0004997… , |S 1001 − 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 more than once. 1 1 1 1 1 1 Series is − + − + − + K 2 4 6 8 10 12 1 1 1 1 = 1 – + – + K . 2 2 3 4 1 ∴ the series converges to ln 2. 2 Conditional convergence means that whether the series converges, and, if so, what value it converges to, depends on the condition that you do not rearrange the terms. 1 1 1 6. a. 1 − + − + K 4 9 16 The series converges because (1) strictly alternating signs, (2) strictly decreasing |tn|, (3) tn → 0. ∞
b.
∑ |t | converges by the p-series test n
n=1
because p > 1. 1 n2 c. L = lim ⋅ = 1, so the ratio test 2 n→∞ ( n + 1) 1 fails because L is not less than 1. ∞
∑ (–1)
1 x 2 n+1 ( 2 n + 1 )! n=0 1 t3 = ( −1)3 ⋅ ⋅ 0.6 2⋅3+1 (2 ⋅ 3 + 1)! 1 = − 0.6 7 = −0.00000555428571K 7! 1 b. S1 (0.6) = 0.6 − 0.6 3 = 0.564 3! 1 1 S2 (0.6) = 0.6 − 0.6 3 + 0.6 5 = 0.564648 3! 5! c. R 1 = sin 0.6 − S1(0.6) = 0.0006424… R 2 = sin 0.6 − S2(0.6) = −0.0000055266… |R1| = 0.0006424… |t2| = 0.000648 ∴ |R 1| < |t2| |R2| = 0.0000055266… |t3| = 0.0000055542… ∴ |R 2| < |t3|
7. a. sin x =
n
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d. The terms are strictly alternating in sign, the terms are strictly decreasing in absolute value, and the terms approach zero for a limit as n → ∞. Thus, the series converges by the alternating series test. Or: | Rn | < |tn+1 | for all n ≥ 1, as shown by example in part c. 0 lim |tn+1 | = 0 because it takes the form . n→∞ ∞ ∴ lim | Rn | = 0, and thus the series converges. n→∞
Or: Use the ratio technique. L = lim
n→∞
0.6 2 n+3 (2 n + 1)! ⋅ (2 n + 3)! 0.6 2 n+1
= 0.36 lim
n→∞
1 =0 (2 n + 3)(2 n + 2)
Because L < 1, the series converges. 8. The sequence converges because lim tn = 2, a n→∞
(finite) real number. The series does not converge because lim tn ≠ 0. n→∞
1 1 1 1 K + + + + 3 8 15 24 Compare with the p-series with p = 2: 1/(n 2 − 1) n2 lim lim = = 1, a positive 2 n→∞ n→∞ n − 1 1/n 2 real number. ∴ the series converges by the limit comparison test. b. The p-series with p = 2 begins 1 1 1 1 + + + . These terms form a lower 4 9 16 25 bound, not an upper bound, so the direct comparison test fails. c. If n started at 1, the first term would be 1/0, which is infinite.
9. a.
∞
10. a. The seventh term of
∑ n! 0.6 1
n
is
n=0
t6 =
1 0.6 6 = 0.0000648. 6! 4
b. S4 =
∑ n! 0.6 1
n
= 1.8214
n=0
e0.6 = 1.8221188… S4 differs from e0.6 by 0.00071880… , which is greater than t5 = 0.000648, but not much greater. The difference is greater than t5 because all subsequent terms are added, not subtracted. It is not much greater than t5 because the subsequent terms are very small.
Problem Set 12-7
329
c. n
5
6
7
Tail:
0.000648
0.0000648
0.000005554…
Geometric series:
0.000648
0.0000648
0.00000648
d.
e.
11. a.
b.
c.
Terms of the e0.6 series are formed by multiplying the previous term: 0.6 tn = tn−1 n Terms of the geometric series are formed by multiplying the previous term by 0.1: tn = 0.1tn−1 For n ≥ 7, 0.6/n is smaller than 0.1, so the terms of the e0.6 series are smaller than the corresponding terms of the geometric series. Thus, the geometric series forms an upper bound for the tail of the e0.6 series after term t6. Geometric series converges to 1 0.0000648 ⋅ = 0.000072. 1 – 0.1 The tail of the series after t6 is bounded by 0.000072. The entire series is bounded by S6(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 . 1 2 3 4 5 K + + + + + 1 1 2 6 24 n + 1 (n – 1)! n +1 1 L = lim ⋅ = lim ⋅ =0 n→∞ n→∞ n n! n n ∴ the series converges because L < 1. 1 2 3 4 K 1 1 + + + + = 1+1+ + + K 1 2 6 24 2! 3! This is the Maclaurin series that converges to e1. n/(n – 1)! n! L = lim = lim n→∞ n →∞ n/n! (n – 1)! = lim n = ∞
The terms of V are bounded below by the corresponding terms of G, and so the direct comparison test fails in this case.
13. 14. 15. 16.
∞
17.
330
Problem Set 12-7
∑4
3 n
= 3+
n=0
3 3 3 K + + + 4 16 64
Converges because it is a geometric series with common ratio 1/4, which is less than 1 in absolute value ∞
18.
∑ n=0
3n 3 9 27 + + +L n = 1+ 4 4 16 64
Converges because it is a geometric series with common ratio 3/4, which is less than 1 in absolute value ∞
19.
∑ (2n + 1)! = 1! + 3! + 5! + 7! + K 1
1
1
1
1
n=0
1 1 1 + + +K 6 120 5040 Converges by comparison with geometric series with t0 = 1 and r = 1/6 = 1+
∞
20.
∑ (–3) 1
n
= 1−
n=0
1 1 1 K + − + 3 9 27
Converges by the alternating series test. (Terms are strictly alternating. Terms are strictly decreasing in absolute value. tn approaches zero as n approaches infinity.)
n→∞
∴ the test fails because the limit of the ratio is infinite. 2 4 8 16 K + + 12. a. U: + + 4 10 28 82 2 4 8 16 K G: + + + + 3 9 27 81 2 4 8 16 K V: + + + + 2 8 26 80 The terms of U are bounded above by the corresponding terms of G, and so U converges by the direct comparison test.
2 n /(3n – 1) 3n = lim n n n n→∞ 3 – 1 n→∞ 2 /3 n 3 ln 3 = lim n =1 n→∞ 3 ln 3 (using l’Hospital’s rule) ∴ the V series converges because the G series converges and L is a (finite) positive number. Divergent harmonic series Convergent p-series, p > 1 Convergent alternating series meeting the three hypotheses Divergent p-series, p ≤ 1
b. L = lim
∞
21.
n3 8 27 64 125 K = + + + + – 1 15 80 255 624 n=2 Diverges. Use the integral test.
∑n ∫
∞
2
4
x3 1 dx = lim ln | x 4 − 1| 4 b→∞ 4 x –1
b
2
1 = lim ln (b 4 − 1) − 0 = ∞ b→∞ 4 Or: Compare with a harmonic series. ∞ ∞ n3 1 > →∞ 4 n – 1 n=2 n n=2
∑
∑
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1 2 3 4 K + + + + , divergent because tn does 2 3 4 5 not approach zero –1 1 –1 1 K 23. + + + + , a convergent alternating 1 2 3 4 series meeting the three hypotheses
22.
∞
24.
∑ sin n = sin 0 + sin 1 + sin 2 + sin 3 + K n=0
= 0 + 0.8414 … + 0.9092 … + 0.1411… + K Diverges. tn does not approach 0 as n → ∞. tn+1 ( 4/3) n+1 /(n + 1) = lim = 4/3 > 1. n→∞ t n→∞ ( 4/3) n /n n Diverges by the ratio test 26. Convergent geometric series with | r| = 7/11 < 1 27. Diverges because tn does not approach zero
25. L = lim
1 1 1 1 K + + + + 1 2 3 4 Diverges. p series with p = 0.5, which is less than 1. Complete interval is [−4, −2). 1 36. x = −2 : − 1 + 1 − 1 + 1 − 1 + K 3 Diverges by the nth term test 2 x = −1 : 1 + 1 + 1 + 1 + 1 + K 3 Diverges by the nth term test 1 2 Complete interval is −2 , – 1 . 3 3 x = −2:
∞
37.
L = lim
n→∞
n +1 = | x − 3| ⋅ 1 n L < 1 ⇔ |x − 3| < 1 ⇔ 2 < x < 4 At x = 2 the series is −1 + 2 − 3 + 4 − L , which diverges because the terms do not approach zero. At x = 4 the series is 1 + 2 + 3 + 4 + L , which diverges because the terms do not approach zero. Interval of convergence is (2, 4). n→∞
=
∫
∫
b→∞
∞
38.
Diverges by the integral test 30. Converges by direct comparison with the ∞ 2 convergent p-series 2 n n=3
32. 33.
34.
35.
Calculus Solutions Manual © 2005 Key Curriculum Press
∑ n =1
5n ⋅ x n n2
L = lim
∑
1 2 6 24 K 1 + 2 + 3 + 4 + e e e e Diverges because tn does not approach zero Converges to e by the definition of e x = 1: − 1 + 1 − 1 + 1 − 1 + K Diverges by the nth term test x = 9: 1 + 1 + 1 + 1 + 1 + K Diverges by the nth term test Complete interval is (1, 9). 1 1 1 1 K x = −1: − + − + − 3 18 81 324 Converges by the alternating series test 1 1 1 1 K x = 5: + + + + 3 18 81 324 Converges by comparison with the geometric 1 1 1 1 series + + + +K 3 9 27 81 Complete interval is [−1, 5]. 1 1 1 1 K x = −4: − + − + − 1 2 3 4 Converges by the alternating series test
(n + 1)( x – 3) n+1 n( x – 3) n
= | x − 3 | lim
[(n + 1)2 – 1]/2 n+1 = lim n→∞ [n 2 – 1]/2 n (n + 1)2 + 1 1 1 lim = 1 | x| 1 At x = −1 the series is −1 + 1 − 1 + 1 − L , which diverges because the terms do not approach zero. At x = 1 the series is 1 + 1 + 1 + 1 + L , which diverges because the terms do not approach zero. 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 the block, and equal length = L. Write Hn = the distance the nth block overhangs the (n + 1)th block. (n = 1 for the top block.) Note that according to the rule, Hn = the distance between the rightmost edge of the nth block and the center of mass of the pile of the top n blocks. Now, the center of mass of the nth block is 1 2 L units from its rightmost edge, and the center of mass of the pile of the top n − 1 blocks is 0 units from (i.e., right on top of ) the edge of the nth block according to the rule. Therefore, the center of mass of the pile of the 1 1 top n blocks is L ⋅ m + 0 ⋅ (n − 1)m nm 2 units from the edge of the nth block; that is, 1 Hn = L , Q .E .D . 2n b. The total distance the top (first) block overhangs the nth block is H1 + H2 + L + Hn−1 . So for a pile of n blocks, the top block will project entirely beyond the bottom block if L < H1 + L + Hn−1 = 1 1 1 1 L + L + L +L+ L 2 4 6 2(n – 1) L (this is 2 3 n –1 L possible because the harmonic series diverges to infinity). Then a stack of n blocks will have total overhang H1 + L + Hn−1 =
1 1 1 1 L + L + L +K+ L 2 4 6 2(n – 1)
=
1 1 1 K 1 1 2H L 1+ + + + < L⋅ =H 2 2 3 n + 1 2 L
(The achieved overhang is greater than H, so one may pull blocks slightly back—moving blocks back can only make the pile more stable—until the overhang equals H exactly.) d. The theoretical overhang for a stack of 52 objects is H = H1 + L + H51 1 1 1 1 L + L + L +L+ L 2 4 6 102 1 1 1 1 = L + + +K+ 2 4 6 102 = 2.2594 K L =
slightly more than two-and-a-quarter card lengths. 48. The least upper bound postulate says that any non-empty set of real numbers that has an upper bound has a least upper bound. In particular, any number less than this least upper bound cannot be an upper bound for the set. The set of real numbers {t1, t2, t3, …} is nonempty and is bounded above (given). Therefore, this set has a least upper bound L. Any number less than L is also less than some tD in the set. Claim: L = lim tn b→ ∞
Proof: Pick a number ε > 0. Because L is an upper bound for tn, L + ε is also an 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 that tD > L − ε . But the values of tn are increasing. ∴ tn > tD > L − ε for all n > D.
Problem Set 12-7
333
Keep n > D. Then L − ε < t n < L + ε . Thus, tn is within ε units of L for all n > D. ∴ L = lim tn by the definition of limit as n→∞
n → ∞, Q .E.D .
3. a. e x =
Problem Set 12-8 Q1. ratio Q2. |common ratio| < 1 Q3. for all values of x Q4. radius = 1 1 1 1 Q5. x − x 3 + x 5 − x 7 + K 3 5 7 Q6. S4
sin
Q7. ln |sec x + tan x| + C Q8. y′ = sec2 x Q9. tan x + C Q10. Newton and Leibniz 1. a. cosh x =
∞
∑ (2n)! x 1
2n
n=0
f ( 2.5+2 ) (c) 2⋅5+2 b. R5 ( 4) = ⋅4 (2 ⋅ 5 + 2)! f
( x ) = cosh x
1 4 –4 (3 + 2 ) 2 = 40.5312… < 41 41 12 | R5 ( 4) | ≤ ⋅ 4 = 1.4360 K 12! 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 well within the 1.4360… upper bound found by Lagrange form. ∴ M = cosh 4 <
2. a. sinh x =
∞
∑ (2n + 1)! x 1
2 n +1
n=0
S9(5) = 74.2032007… f ( 2⋅9+3) (c) 2⋅9+3 ⋅5 (2 ⋅ 9 + 3)! f (21)(x) = cosh x 1 ∴ M = cosh 5 < (35 + 2 –5 ) 2 = 121.5156… < 122 122 21 | R9 (5)| ≤ ⋅ 5 = 0.001138K 21! S9(5) is within 2 units of sinh 5 in the third decimal place.
b. R9 (5) =
334
Problem Set 12-8
∞
∑ n=0
xn n!
The fifteenth partial sum is S14(3) = 20.0855234… . f (15) (c) 15 ⋅3 15! f (15)(x) = e x ∴ M = e3 < 33 = 27 27 15 | R14 (3)| ≤ ⋅ 3 = 0.0002962 K 15! S14(3) is within 3 units of e3 in the fourth decimal place. c. e3 = 20.085536923… S14(3) = 20.085523458… e3 − S14(3) = 0.00001346… , which is within the 0.0002962… found by Lagrange form.
b. R14 (3) =
4. a. ln x =
S5 ( 4) = 27.2699118K
(12 )
c. sinh 5 = 74.2032105… S9(5) = 74.2032007… sinh 5 − S9(5) = 0.00000981… , which is well within the 0.001138… upper bound found by Lagrange form.
∞
∑ (–1) n =1
n +1
1 ⋅ ( x – 1) n n
S8(0.7) = −0.356671944… f ( 9 ) (c ) ⋅ (0.7 − 1)9 9! f (9)(x) = 8!x − 9 ∴ M = 8!(0.7)− 9
b. R8 (0.7) =
8!(0.7) –9 1 ⋅ (0.3) 9 = (3/7)9 9! 9 = 5.4195… × 10− 5 S8(0.7) is within 6 units of ln 0.7 in the fifth decimal place. (Note that for ln x, the Lagrange form of the remainder simplifies to
| R8 (0.7)| ≤
n +1
1 | x – 1| ⋅ n +1 x For x < 0.5, the fraction |x − 1|/x is greater than 1. The Lagrange form of the remainder becomes infinite 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 is within the 5.4195… × 10− 5 found by Lagrange form. 5. For sinh 2, all derivatives are bounded by cosh 2. 1 cosh 2 < (32 + 2 –2 ) = 4.625 2 1 The general term is tn = ⋅ 2 2 n+1. (2 n + 1)! | Rn ( x )| ≤
Calculus Solutions Manual © 2005 Key Curriculum Press
For six-place accuracy, 4.625 | Rn (2)| ≤ ⋅ 2 2 n+3 < 0.5 × 10 −6. (2 n + 3)! 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. 1 cosh 3 < (33 + 2 –3 ) = 13.5625 2 1 The general term is tn = ⋅ 32 n . (2 n)! For eight-place accuracy, 13.5625 2 n+2 | Rn (3)| ≤ ⋅3 < 0.5 × 10 −8. (2 n + 2)! 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 (0.6)| ≤ =
n!(0.6) –( n+1) ⋅ | 0.6 − 1|n+1 (n + 1)!
1 2 ⋅ n +1 3
n +1
For seven-place accuracy, n +1
1 2 ⋅ < 0.5 × 10 −7. n +1 3 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, 59049 | Rn (10)| ≤ ⋅ 10 n+1 < 0.5 × 10 −5. (n + 1)! 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… 1 The general term is tn = ⋅ 2 2 n. (2 n)! R4 (2) =
f ( 2⋅4+2 ) (c) 2⋅4+2 210 ⋅2 = cosh c ⋅ 10! (2 ⋅ 4 + 2)!
210 = 0.000290929K 10! 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 ∴ cosh c ⋅
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The general term is R19 (5) =
1 n 5. n!
f ( 20 ) (c) 20 520 ⋅ 5 = ec ⋅ 20! 20!
520 = 5.1234 K × 10 −5 20! ec = 1.30702776… c = ln 1.3070… = 0.2677… , which is between 0 and 5. 1 1 1 11. cos 2.4 = 1 − (2.4)2 + (2.4) 4 − (2.4)6 2! 4! 6! 1 1 8 10 + (2.4) − (2.4) + L 8! 10! = 1 − 2.88 + 1.3824 − 0.2654208 + 0.0273004 L − L The terms are strictly alternating. They are decreasing in absolute value after t1, and they approach zero for a limit as n → ∞. Therefore, the hypotheses of the alternating series test apply, and | Rn(2.4) | < | tn+ 1| = 1 (2.4)2 n+2 . (2 n + 2)! 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). 1 1 1 12. e −2 = 1 + ( −2) + (–2)2 + (–2)3 + (–2) 4 2! 3! 4! 1 1 5 6 + (–2) + (–2) + L 5! 6! = 1 − 2 + 2 − 1.3333… + 0.6666 … − 0.2666 … + L The terms are strictly alternating. They are decreasing in absolute value after t2, and they approach zero for a limit as n → ∞. Therefore, the hypotheses of the alternating series 1 test apply, and | Rn ( −2)| < |tn+1 | = ⋅ 2 n+1. (n + 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). ∴ ec ⋅
10
13. a. S10 =
∑n
1 3
= 1.19753198…
n =1 ∞
b x −3 dx = lim –0.5 x –2 10 = →∞ b 10 0.5(10− 2) = 0.005 ∞ b R10 > x −3 dx = lim –0.5 x –2 11 = b→∞ 11 0.5(11− 2) = 0.00413223… R10 ≈ 0.5(0.005 + 0.00413223…) = 0.00456611…
R10 <
∫ ∫
Problem Set 12-8
335
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 x 3 dx − x 3 dx n n +1 −2 −2 = 0.25n − 0.25(n + 1) . Solve 0.25n− 2 − 0.25(n + 1)− 2 = 0.000005 to get n = 45.9194… . Use 46 terms. Using only the upper bound, Rn < 0.000005
∫
if
∫
∞
∫
x 3 dx < 0.000005.
n
Set 0.5n− 2 = 0.000005n = (0.000005/0.5)− 1 / 2 = 316.2277… , which rounds up to 317 terms, considerably more that the 46 terms to give this accuracy by comparing upper and lower bounds of Rn. 100
14. a. S100 =
∑n
1
1.05
= 4.698244 …
n =1 ∞
b x −1.05 dx = lim –20 x –0.05 100 = b→∞ 20(100− 0.05 ) = 15.886564… ∞ b R100 > x −1.05 dx = lim –20 x –0.05 101 = b→∞ 101 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)
R100 <
∫
100
∫
∞ ∞ b. Error < 0.5 x 1.05 dx − x 1.05 dx n n +1 = 20n− 0.05 − 20(n + 1)− 0.05 Solve 20n− 0.05 − 20(n + 1)− 0. 05 = 0.000005 to get n = 111840.2309… . Use 111,841 terms. With a value of p such as 1.05, which is closer to 1 than 3 is, it takes more terms because the terms approach zero more slowly.
∫
10
15. a. S10 =
∑n n=0
∫
1 = 1.9817928… +1
2
The series converges because the terms of the tail starting at t1 are bounded above by the convergent p-series with p = 2. b 1 b dx = lim tan –1 x 10 = b. R10 < 2 b →∞ 10 x + 1 π /2 − tan− 110 = 0.0996686… b 1 b R10 > dx = lim tan –1 x 11 = 2 b→∞ 11 x + 1 π /2 − tan− 111 = 0.0906598…
∫
∫
336
Problem Set 12-8
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 two decimal places. Make R n < 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) = 0.00005 to get n = 99.4962… . Use 100 terms. 16. In this p-series, p = 0.5, which is not greater than or equal to 1. Thus, the series diverges and the remainder is infinite. 17. e 2 =
∞
∑ n! ⋅ 2 1
n
n=0
From Example 1, S10 = 7.38899470… . By Lagrange form, | R 10 | < 0.0004617… . Use a geometric series as an upper bound. 1 11 1 t11 = ⋅ 2 and t12 = ⋅ 212 11! 12! t 1 Common ratio r = 12 = t11 6 1 11 1 ⋅2 ⋅ = 0.00006156 … 1 11! 1– 6 The geometric series gives a better estimate of the remainder than does the Lagrange form. ∴ | R10 | <
∞
∑ (–1)
1 n ⋅2 n! n=0 S10 = 0.135379188…
18. e −2 =
n
211 = 0.000051306 … 11! This number appears to be a better estimate of the error. However, it represents an error of ≈ | R 10 |/S(10) = 0.03789…%. e2 ≈ 1/S10 = 7.38665971… A 0.037…% error for this value would be 0.002799… , which is a worse estimate of the error than that by Lagrange or by geometric series. (In general, an error of ε% in1/f (x) gives a ε maximum error of in the value of f (x). 1 – ε /100 So an error of 0.03789…% in 1/e2 means an 0.03789K error of = 0.03788K% in e 2 .) 1 – 0.0003789K | R10 | < |t11 | =
Calculus Solutions Manual © 2005 Key Curriculum Press
250 = 39.7887357K 2π Thus, 250 radians is 39 complete cycles plus 0.7887… additional cycle, or b = (2π)(0.7887…) = 4.9557730… radians. sin b = −0.970528019… sin 250 = −0.970528019… (Checks.) The value of b can be calculated efficiently using the fraction part command. For a typical grapher, b = f Part(250/(2π))2π. b. From Figure 12-8c, you can tell that the value of 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 the value of d is a quarter-cycle ahead of the value of c. The value of the sine is the opposite of the corresponding value of cos d. π d = + c = 0.243384039K 2 Check: −cos d = −0.970528019… sin 250 = −0.970528019… (Checks.) In general: π π π c ∈ − , − : d = + c and sin x = − cos d 4 2 2 π c ∈ − , 0 : d = − c and sin x = − sin d 4 π c ∈ 0, : d = c and sin x = sin d 4 π π π c ∈ , : d = − c and sin x = cos d 2 4 2
The second inequality is first true when n is 348. Because both numerator and denominator may overflow most computers, you can calculate values of ln | Rn(250) | as follows:
19. a.
d. For x in [0, π/4], both the sine and cosine series meet the hypotheses of the alternating series test. Thus, the error in S5(x) is bounded by | t6 |, the first term of the tail. | t6 | is greater for the cosine series than for the sine series. The maximum of | t6 | in the interval is at x = π /4. 1 ∴ | R5 ( x )| < | t6 (π /4)| = (π /4)2⋅6+2 (2 ⋅ 6 + 2)! = 3.8980… × 10− 13, which is small enough to guarantee that sin x will be correct to ten decimal places. For direct calculation, 1 | Rn (250)| < ⋅ 250 2 n+3 < 0.5 ⋅ 10 −10 (2 n + 3)!
Calculus Solutions Manual © 2005 Key Curriculum Press
ln | Rn (250)| = (2 n + 3) ln 250 −
2 n +3
∑ ln i i =2
Then compare the values with ln (0.5 ·10− 10). So you would need to use 349 terms. Unfortunately, even this procedure would not be practical because the terms themselves would have to be calculated to ten or more decimal places, and they are so large that each term overflows most computers’ capacities. e. The program will have the following steps. The particular commands will depend on the grapher 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 shown in part c. • Calculate and display the answer. 1 20. For sin 1, | Rn (1)| < | tn+1 | = ⋅ 12 n+3 = (2 n + 3)! 1 . (2 n + 3)! 1 Set < 0.5 × 10 −23. (2 n + 3)! This inequality is first true for n = 11. Use at least 12 terms (n = 11). Using the technique in Problem 19, 1 | Rn (1)| < (π /4)2 n+2 < 0.5 ⋅ 10 −23. (2 n + 2)! The second inequality is first true for n = 10. You would save only one term by the method of Problem 19. 21. a. Apply the mean value theorem to f ′(x) on [a, x]. There is a number x = c in (x, a) such that f ′( x ) – f ′( a ) f ′′(c) = 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
1 f ( x ) = f ′( a) x + f ′′(c) ⋅ ( x – a)2 + C 2 Substituting the initial condition (a, f (a)) gives f (a) = f ′(a)a + f ″(c)(0) + C ⇒ C = f (a) − f ′(a)a
Problem Set 12-8
337
f ( x) = 1 f ′( a) x + f ′′(c) ⋅ ( x – a)2 + f ( a) − f ′( a)a 2 1 f ( x ) = f ( a) + f ′( a)( x − a) + f ′′(c)( x − a) 2 , 2 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 ′′( x ) – f ′′( a) f ′′′(c) = 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 1 f ′( x ) = f ′′( a) x + f ′′′(c) ⋅ ( x – a)2 + C 2 Use (a, f ′(a)) as an initial condition. 1 f ′( a) = f ′′( a)a + f ′′′(c)(0) + C ⇒ 2 C = f ′( a) − f ′′( a)a 1 f ′( x ) = f ′′( a) x + f ′′′(c)( x − a)2 2 + f ′( a) − f ′′( a)a 1 f ′( x ) = f ′( a) + f ′′( a)( x − a) + f ′′′(c)( x − a) 2 2 Integrate again to get f(x).
∫ f ′( x ) dx = ∫ f ′(a) dx + ∫ f ′′(a)( x − a) dx 1 + ∫ f ′′′(c)( x − a) dx 2 2
f ( x ) = f ′( a ) x +
1 f ′′( a)( x − a) 2 2
1 f ′′′(c)( x − a)3 + C 6 Use (a, f ″(a)) as an initial condition. 1 1 f ( a) = f ′( a)a + f ′′( a)(0) + f ′′′(c)(0) + C 2 6 ⇒ C = f ( a ) − f ′( a ) a 1 f ( x ) = f ′( a) x + f ′′( a)( x − a) 2 2! 1 + f ′′′(c)( x − a)3 + f ( a) − f ′( a)a 3! 1 f ( x ) = f ( a) + f ′( a)( x − a) + f ′′( a)( x − a) 2 2! 1 3 + f ′′′(c)( x − a) , Q .E.D . 3! d. The technique is mathematical induction. +
e – x –2 , if x ≠ 0 22. a. f ( x ) = if x = 0 0, It is given that f (n)(0) = 0 for all n > 0. c0 = f (0) = 0 c1 = f ′(0) = 0 338
Problem Set 12-9
2!c2 = f ″(0) = 0 ⇒ c2 = 0 3!c3 = f ″′(0) = 0 ⇒ c3 = 0… ∴ series is 0 + 0x + 0x2 + 0x3 + … , Q .E .D . b. Each partial sum of the Maclaurin series equals zero for any value of x. Thus, the sequence of partial sums converges to zero for all x. But f (x) does not equal zero except at x = 0. Thus, the series converges to f (x) only at x = 0. −2 1 1 c. e − x = 1 + ( − x −2 ) + (– x –2 )2 + (– x –2 )3 + L 2! 3! 1 −4 1 −6 −2 = 1− x + x − x +L 2! 3! d. The fourth partial sum, S3(2) = 0.7786458333… . −2 f (2) = e −2 = e −0.25 = 0.7788007830 K The partial sum is close to f (2), so it is reasonable to make the conjecture that the Laurent series converges to f (2). 23. Using the Lagrange form of the remainder, the value of ex is given exactly by k 1 n ex = x + Rk ( x ), where n! n=0
∑
f ( k +1) (c) k +1 x and c is between 0 and x. ( k + 1)! M | Rk ( x )| ≤ | x |k +1 ( k + 1)! Because all derivatives of e x equal e x, the value of M for any particular value of x is also e x, which is less than 3x, if x ≥ 0; or 1, if x < 0. 3x lim | Rk ( x )| < lim | x |k +1 k →∞ k →∞ ( k + 1)! which approaches 0 as k → ∞ by the ratio technique. Because the remainder approaches zero as n approaches infinity, ex is given exactly by ∞ 1 n ex = x , Q.E.D. n ! n=0
Rk ( x ) =
∑
Problem Set 12-9 Review Problems R0. Answers will vary. 9 R1. f ( x ) = and P( x ) = 9 + 9 x + 9 x 2 + 9 x 3 + L 1– x y P6 20
f
x 1
P5
Calculus Solutions Manual © 2005 Key Curriculum Press
The graph shows that P5(x) and P6(x) are close to f (x) for x between about −0.7 and 0.6, and bear little resemblance to f (x) beyond ±1. P5(0.4) = 14.93856 P6(0.4) = 14.975424 f (0.4) = 15 ∴ P6(0.4) is closer to f (0.4) than P5(0.4) is, Q .E .D . P 5(x) = 9 + 9x + 9x 2 + 9x 3 + 9x 4 + 9x 5; P5(0) = 9 P5′ ( x ) = 9 + 18 x + 27 x 2 + 36 x 3 + 45 x 4 ; P5′ (0) = 9 P5′′( x ) = 18 + 54 x + 108 x 2 + 180 x 3 ; P5′′(0) = 18 P5′′′( x ) = 54 + 216 x + 540 x 2 ; P5′′′(0) = 54 f (x) = 9(1 − x)− 1; f (0) = 9 f ′(x) = 9(1 − x)− 2; f ′(0) = 9 f ″(x) = 18(1 − x)− 3; f ″(0) = 18 f ′′′( x ) = 54(1 − x ) −4 ; f ′′′(0) = 54 ∴ P5′ (0) = f ′(0), P5′′(0) = f ′(0), and P5′′′(0) = f ′′′(0) Pn(x) is a subseries of a geometric series. R2. a. Series is 3 + 2.7 + 2.43 + 2.187 + L . 1 – 0.910 After 10 days, S10 = 3 ⋅ = 1 – 0.9 19.5396… . About 19.5 mm increase in 10 days 1 S = 3⋅ = 30 1 – 0.9 About 30 mm increase eventually b. Let x be the amount invested to have 0.5 million dollars at the end of 19 years. Interest rate is 10% per year, so the amount at the end of a year is 1.1 times the amount at the beginning of the year. x(1.119) = 0.5 x = 0.5(1.1− 19) = 0.081753995… They must invest $81,754.00 now in order to make the last payment. The total to invest is the sum 0.5(1.1−1 ) + 0.5(1.1−2 ) + L + 0.5(1.1−19 ). This is the nineteenth partial sum of the geometric series with first term 0.5(1.1)− 1 and common ratio 1.1− 1. 1 – 1.1–19 S19 = 0.5(1.1−1 ) ⋅ = 4.182460045… 1 – 1.1–1 They must invest $4,182,460.05 now to make all 19 payments. R3. P(x) = c0 + c1x + c2x2 + c3x3 + c4x4 + L f (x) = 7e3x ⇒ f (0) = 7 ⇒ c0 = 7 f ′(x) = 21e3x ⇒ f ′(0) = 21 ⇒ c1 = 21 f ″(x) = 63e3x ⇒ f ″(0) = 63 ⇒ 2!c2 = 63 ⇒ c2 = 31.5 f ′′′( x ) = 189e 3 x ⇒ f ′′′(0) = 189 ⇒ c3 = 189/3! = 31.5 Calculus Solutions Manual © 2005 Key Curriculum Press
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 is close. 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 not close. R5. a. A Maclaurin series is a Taylor series expanded about x = 0. b. Substitute t = x + 1 into 1 1 ln t = (t − 1) − (t – 1)2 + (t – 1)3 2 3 1 − (t – 1) 4 + L . 4 1 1 1 ln ( x + 1) = x − x 2 + x 3 − x 4 + L 2 3 4 =
∞
∑ n =1
(–1) n+1 n x n
c. Assume one may integrate this function term by term. 1 1 3 1 4 ln ( x + 1) dx = x 2 − x + x −L+ C 2 3⋅2 4⋅3
∫
=
∞
(–1) n+1
∑ (n + 1)n x
n +1
+C
n =1
d.
∫ ln ( x + 1) dx = ( x + 1) ln ( x + 1) − ( x + 1) + C
1
= x ln ( x + 1) + ln ( x + 1) − x + C (C = C1 − 1) 1 1 1 = x2 − x3 + x4 − x5 + L 2 3 4 1 2 1 3 1 4 + x − x + x − x +L− x + C 2 3 4 1 2 1 3 1 4 = x − x + x − L + C, 2 3⋅2 4⋅3 which is the same as the series in part c. x x 1 1 e. t cos t 2 dt = t 1 − (t 2 )2 + (t 2 ) 4 − L dt 0 0 2! 4! x 1 1 1 = t – t 5 + t 9 – t 13 + L dt 0 2! 4! 6!
∫
∫
∫
1 2 1 6 1 1 x − x + x 10 − x 14 + L 2 6 ⋅ 2! 10 ⋅ 4! 14 ⋅ 6! (Note that the series can be transformed to 1 1 1 1 = x 2 − ( x 2 )3 + ( x 2 )5 − ( x 2 ) 7 + L 2 3! 5! 7! x 1 1 = sin x 2 , and t cos t 2 dt = sin x 2. ) 0 2 2 =
∫
Problem Set 12-9
339
f. tan −1 x = =
x
∫
x
∫ [1 − t
2
+ (t 2 )2 − (t 2 )3 + (t 2 ) 4 − L ] dt
0
(| t | ≤ 1) 1 1 1 1 = x − x3 + x5 − x7 + x9 − L 3 5 7 9 g. f(3) = 5 ⇒ c0 = 5 f ′(3) = 7 ⇒ c1 = 7 f ″(3) = −6 ⇒ c2 = −6/2! = −3 f ′′′(3) = 0.9 ⇒ c3 = 0.9 / 3! = 0.15 ∴ f (x) = 5 + 7(x − 3) − 3(x − 3)2 + 0.15(x − 3)3 + L ∞
R6. a.
∑ = (−3)
interval. Above x = 2 the partial sums diverge rapidly to ±∞. Below x = 0 the partial sums give answers, but there are no real values for ln x.
1 dt 1+ t2
0
−n
( x − 5) n
R7. a. S10 = 1000 ⋅ b. S = 1000
c. d. e.
1 1 1 = − ( x – 5) + ( x – 5)2 − ( x – 5)3 + L 3 9 27 n→∞
∞
∑ (2n)! x 1
2n
n=0
x 2 n+2 (2 n)! ⋅ 2n n→∞ ( 2 n + 2 )! x 1 = x 2 lim = x2 ⋅ 0 n→∞ ( 2 n + 2 )(2 n + 1) L < 1 for all x. Series converges for all x, Q.E.D. 1 1 d. e1.2 = 1 + 1.2 + (1.2)2 + (1.2)3 2! 3! 1 4 + (1.2) + L 4! S4(1.2) = 3.2944 (the fifth partial sum) e1.2 = 3.32011692… Error = e1.2 − S4(1.2) = 0.02571692… 1 The first term of the tail is t5 = (1.2)5 = 5! 0.020736. The error is greater than t5, but not much greater. e. L = lim
n
=
1000 = 5000 1 – 0. 8
S − S10 = 536.870912, which differs from the limit by about 10.7%. “Tail” “Remainder” ∞ 1 1 R10 < x −3 dx = lim − b −2 + (10 –2 ) b→∞ 2 10 2 = 0.005 ∞ 1 1 R10 > x −3 dx = lim − b −2 + (11–2 ) b→∞ 2 11 2 = 0.004132… Series converges because the tail is bounded above by 0.005. S = S 10 + R 10 ≈ 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 about three decimal places. 1 1 1 1 − + + + +L 4 3 22 59 1/(n 3 – 5) n3 L = lim = lim 3 =1 3 n→∞ n→∞ n – 5 1/n (Apply l’Hospital’s rule three times.) ∴ the series converges because L is a positive real number. The terms of the F series begin 1 1 1 1 + + + +L . 1 8 27 64 Although the F series converges, its terms (after t1) are less, not greater, than the corresponding terms of the S series, so the comparison test is inconclusive. 2/1! + 4/2! + 8/3! + 16/4! + 32/5! + L = 2 + 2 + 1.3333… + 0.6666…
∫
f.
g.
+ 0.2666 … + L =
∞
∑ 2 /n! n
n =1
y
S11 ln
1
x 1
2
S10
The open interval of convergence is (0, 2). Both partial sums fit ln well within this 340
∑ 0.8
∫
(–3) –( n+1) ( x – 5) n+1 1 = | x − 5| (–3) – n ( x – 5) n 3
L < 1 ⇔ |x − 5| < 3 ⇔ 2 < x < 8 Open interval of convergence is (2, 8). Radius of convergence = 3 c. cosh x =
∞
n=0
n =1
b. L = lim
1 – 0.810 = 4463.129088 (exactly) 1 – 0.8
Problem Set 12-9
The terms are decreasing starting at t2, which can be seen numerically, above, or algebraically by the fact that the next term is formed by multiplying the numerator by 2 and the denominator by more than 2. R1 is bounded by the geometric series with first term 2 and common ratio 1.3333…/2 = 2/3. Calculus Solutions Manual © 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 is bounded above by a convergent geometric series, Q.E.D. h. The given series converges because, as written, it meets the three hypotheses of the alternating series test. It does not converge absolutely because replacing all minus signs with plus signs gives the divergent harmonic series. The given series is the Taylor series for ln x expanded about x = 1 and evaluated at x = 2. The remainders approach zero, so the series converges to ln 2. Rearrange the series this way: 1 – 1 − 1 + 1 – 1 − 1 + 1 – 1 2 4 3 6 8 5 10 1 +L 12 Each term of the series appears exactly once. Simplifying inside the parentheses and factoring out 1/2 gives 1 1 1 1 1 1 − + − + − +L 2 4 6 8 10 12 1 1 1 1 1 1 1 = – + – + – + L 2 1 2 3 4 5 6
which converges because it is a p-series with p = 2. Interval of convergence is [2.9, 3.1]. ∞
ii.
∞
j. i.
10 n ( x – 3) n2 n =1
∑
L = lim
n→∞
n
10 n+1 ( x – 3) n+1 n2 ⋅ (n + 1)2 10 n ( x − 3) n 2
n = 10 | x − 3 | lim = 10 | x − 3 | ⋅ 1 n→∞ n + 1 L < 1 ⇔ 10 |x − 3| < 1 ⇔ 2.9 < x < 3.1 At x = 2.9 the series is 1 1 1 −1 + − + − L , 4 9 16 which converges by the alternating series test. At x = 3.1 the series is 1 1 1 1+ + + + L , 4 9 16 Calculus Solutions Manual © 2005 Key Curriculum Press
∑ n =1
n→∞
=
n
( x + 1) n+1 n ⋅ 2n n +1 ⋅ (n + 1) ⋅ 2 ( x + 1) n
L = lim
1 n 1 | x + 1| lim = | x + 1| ⋅ 1 n→∞ n + 1 2 2
1 | x + 1| < 1 ⇔ | x + 1| < 2 2 ⇔ −3< x
∫
∞
x
−1.5
dx
101
= lim [–2 b –0.5 + 2(101–0.5 )] b→∞
= 0.19900743…
Calculus Solutions Manual © 2005 Key Curriculum Press
R100 ≈ 0.5(0.2 + 0.19900743…) = 0.19950371… S = S 100 + R 100 ≈ 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 two decimal places.) T20. Answers will vary.
Problem Set 12-10 Cumulative 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 sequence converges if and only if lim tn exists. n→∞
d. Convergence of a series: A series converges if and only if the sequence of partial sums converges. 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: See Section 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: See Section 5-6. h. Lagrange form of the remainder: See Section 12-8. i. Parametric chain rule: See Section 4-7. j. Polar differential of arc length: See Section 8-7. 4. a. f ( x ) =
∫
x
3
1 + sech t dt ⇒
f ′( x ) = 1 + sech x b. f (x) = ax ⇒ f ′(x) = ax ln a c. f (x) = x a ⇒ f ′(x) = axa− 1 d. f (x) = xx ⇒ ln f (x) = x ln x 1/f (x) · f ′(x) = ln x + x · (1/x) f (x) = (ln x + 1) – f (x) f ′(x) = x x ln x + x x Problem Set 12-10
345
e. e6x cos 3x dx
u e 6x 6e 6x 36e 6x
+ – +
dv cos 3x 1 3 sin 3x 1 – 9 cos 3x
1 2 = e 6 x sin 3 x + e 6 x cos 3 x 3 3
∫
F=
− 4 e cos 3 x dx 6x
∫
1 2 = e 6 x sin 3 x + e 6 x cos 3 x + C1 3 3 6x e cos 3 x dx = f. g.
1 6x 2 e sin 3 x + e 6x cos 3 x + C 15 15 1 cosh 5 x sinh x dx = cosh 6 x + C 6
∫ ∫ sec
3
∫
40
0
2(10 y)1/2 dy =
dF = 17066.6 … k lb (exactly
256,000k/15) d. dM = y dF = y ⋅ k(40 − y) ⋅ 2(10y)1/2 dy
5 e 6 x cos 3 x dx
∫
40 2 2 ⋅ (10 y)3/2 0 10 3 0 2 = 3200/3 = 1066.6666… yd (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 40
∫
A=
x dx
1 1 sec x tan x + ln | sec x + tan x | + C 2 2 1 −1 h. (sin 5 x ) cos 5 x dx = ln | sin 5 x | + C 5 cos 7 x – 1 –7 sin 7 x i. lim = lim x→0 x→0 13 x 2 26 x –49 cos 7 x 49 = lim =− x→0 26 26 j. L = lim (1 – x )3/ x =
∫
x→0
–3 3 ln (1 – x ) = lim = −3 x→0 x→0 1 – x x L = e− 3 = 0.0497… dy 5. a. = 0.2 x − 0.3 y + 0.3, (1, 8) dx ln L = lim
M=
∫
40
0
dM = 292, 571.4 … k lb-yd
(exactly 10,240,000k/35) M 10240000 k/35 1 = = 17 yd F 256000 k/15 7 By symmetry, x = 0. 1 Center of pressure is at 0, 17 . 7 7. a. z = 30 − 0.5y b. For a cross section, A = 2xz = 2(10y)1/2(30 − 0.5y). A = 101/2(60y 1/2 − y 3/2) A′ = 101/ 2(30y− 1/ 2 − 1.5y1/ 2) = (101/ 2)(y− 1/ 2)(30 − 1.5y) A′ = 0 ⇔ 30 − 1.5y = 0 ⇔ y = 20 A′ is infinite ⇔ y = 0. A(0) = 0 A(20) = 565.6854… (exactly 400 2 ) A(40) = 400 Maximum at y = 20; minimum at y = 0 c. dV = 2xz dy = 101/2(60y 1/2 − y 3/2) dy e. F ⋅ y = M , y =
∫
V=
40
0
dV = 19200 (exactly)
Use 19200/5 = 3840 truckloads. d. dL = dx 2 + dy 2 = 1 + 0.04 x 2 dx
10 y
∫
L=
20
dL = 92.9356 … ≈ 92.9 yd r r r 8. r = (100 cos 0.03t )i + (50 sin 0.03t ) j r r r v = ( −3 sin 0.03t )i + (1.5 cos 0.03t ) j r Speed = | v | = (–3 sin 1.5)2 + (1.5 cos 1.5) 2 x 10
b. If x = 9, y ≈ 5.413… , which agrees with the graph. 6. a. p = k(40 − y) b. y = 0.1x 2 ⇒ x = (10y)1/2 dA = 2x dy = 2(10y)1/2 dy
346
Problem Set 12-10
−20
= 2.9943… ≈ 2.99 ft/s t sin u 9. Si t = du 0 u
∫
= =
t
1
∫ u u – 3! u 1
3
∫ 1 – 3! u
+
0
t
0
1
2
+
1 5 1 7 u – u + L du 5! 7!
1 4 1 6 u – u + L du 5! 7!
Calculus Solutions Manual © 2005 Key Curriculum Press
=t−
1 3 1 5 1 7 t + t − t +L 3 ⋅ 3! 5 ⋅ 5! 7 ⋅ 7!
6.
t 2 n +3 (2 n + 1)(2 n + 1)! ⋅ (2 n + 3)(2 n + 3)! t 2 n+1
L = lim
n→∞
(2 n + 1) = t2 ⋅ 0 n→∞ ( 2 n + 3)(2 n + 3)(2 n + 2 ) ∴ L < 1 for all values of t, and the series converges for all values of t. Third partial sum is 1 1 S2 (0.6) = 0.6 − (0.6 3 ) + (0.6 5 ) 3 ⋅ 3! 5 ⋅ 5! = 0.5881296 1 | R2 | < | t3 | = (0.6 7 ) = 0.0000007934 … 7 ⋅ 7! The answer is correct to within ±1 in the sixth decimal place. 0.6 sin u Si 0.6 = du ≈ 0.588128809… 0 u Note that this answer agrees with the third partial sum to within 1 in the sixth decimal place. 10. r = 5 + 4 cos θ 1 dA = (5 + 4 cos θ )2 dθ 2 = t 2 lim
11.
2π
0
dA ≈ 103.6725… ≈ 103.7 ft 2 (exactly 33π )
Cumulative Review Number 2 1. 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 ) = 5.
∫ tanh
∫
5
x
3
g(t ) dt ⇒ f ′( x ) = g( x )
x sech 2 x dx =
Calculus Solutions Manual © 2005 Key Curriculum Press
1 tanh 6 x + C 6
– +
dv sinh 2x 1 2 cosh 2x 1 4 sinh 2x
1 1 x cosh 2 x − sinh 2 x + C 2 4 3 x + 14 –1 4 7. + dx = dx x + 3 x – 2 ( x + 3)( x – 2) = −ln | x + 3 | + 4 ln | x − 2 | + C
∫
8.
∫
sinh x 1 1 1 1 dx = x + x 3 + x 5 + x 7 + L dx x x 3! 5! 7! 1 1 1 = 1 + x 2 + x 5 + x 6 + K dx 3! 5! 7! 1 3 1 5 1 7 … = x+ x + x + x + +C 3 ⋅ 3! 5 ⋅ 5! 7 ⋅ 7!
∫
∫
∫
∞
9.
n( x − 5) n 3n
∑ n =1
L = lim
n→∞
(n + 1)( x – 5) n+1 3n ⋅ 3n+1 n( x – 5) n
1 n +1 1 1 = | x − 5| ⋅ 1 = | x − 5| | x − 5 | lim n→∞ n 3 3 3 1 L < 1 ⇔ | x − 5| < 1 ⇔ − 3 < x − 5 < 3 3 Open interval of convergence is 2 < x < 8. =
1
10.
∫x
−0.998
dx = lim+ a→ 0
0
dV dV = kV ⇒ = k dt dt V ln | V | = kt + C V = C 1e kt At t = 0, V = 300. 300 = C 1 dV = −5 when V = 300 dt 1 −5 = 300 k ⇒ k = − 60 ∴ V = 300e− ( 1 / 6 0 ) t At t = 10, V = 300e− 1/6 = 253.9445… ≈ 253.9 million gal.
+
=
∫
∫
u x 1 0
(–1) n t 2 n+1 + L (2 n + 1)(2 n + 1)!
+
A=
∫ x sinh 2 x dx
= lim+ a→ 0
1
∫x
−0.998
dx
a
1 x 0.002 0.002
= lim+ (500 − 500 a a→0
1 a 0.002
)
= 500 11. y = x
2
∫ y=
9
x 2 dx
13. V =
∫
=
1 1 3 ⋅ x 6 3
9
= 39 9–3 3 12. f ( x) = x 2 f ( 4) = 16 f ( 3.99) = 15.9201, which is within 0.08 unit of 16. f ( 4.01) = 16.0801, which is not within 0.08 unit of 16. Thus, δ = 0.01 is not small enough to keep f (x) within 0.08 unit of 4. 3
10
A dx
2
2 ≈ [153 + 4(217) + 2(285) + 4(319) + 343] 3 = 2140 ft3
Problem Set 12-10
347
14. r = 4 sin 2θ 1 2 r dθ = 8 sin 2 2θ dθ 2
dA =
π /2
∫
A=
1 1 23. ln x = ( x − 1) − ( x – 1)2 + ( x – 1)3 − L 2 3
0
+
8 sin 2 2θ dθ ≈ 6.2831 ≈ 6.28 ft 2
(exactly 2π) 15. (x/5)2 + (y/3)2 = 1
n→∞
y = ± 0.6 25 – x ( Use + .) dA = 2 y dx = 1.2 25 – x 2 dx 5
∫ dA ≈ 17.6021K ≈ 17.6 square units 1
(exactly 15(sin −1 1 − sin −1 0.2) − 1.2 6 ) 16. y = 0.0016x4 dA = (16 − 0.0016x4) dx
∫
A=
10
−10
∫
L=
−10
(16 – 0.0016 x 4 ) dx = 256 ft 2
dL ≈ 42.5483… ≈ 42.55 ft
18. p = 62.4(16 − y) dA = 2x dy = 10y1/4 dy dF = p dA = 62.4(16 − y) ⋅ 10y1/4 dy F=
∫
16
0
dF ≈ 113595.73… ≈ 113, 600 lb
exactly 113595 11 15 ln x ∞ → x →∞ x ∞ 1/ x = lim =0 x →∞ 1
19. lim y = lim x →∞
n = | x − 1| ⋅ 1 = | x − 1| n→∞ n + 1 L < 1 ⇔ | x − 1 | < 1 ⇔ −1 < x − 1 < 1 ⇔ 0 45.817… . Use 46 terms. 25. If the velocity is 0 ft/s at time t = 0, the ship speeds up, approaching approximately 34 ft/s asymptotically as t increases. If the velocity is 50 ft/s at time t = 0, the ship slows down, again approaching 34 ft/s asymptotically as t increases. (The graphs are shown here. The differential equation is dv/dt = 0.7(34 − v).) v
21. y′′ =
t
r r r 26. r = (ln t )i + (sin 2t ) j r r r v = (1/t )i + (2 cos 2t ) j r r r a = ( −1/t 2 )i + ( −4 sin 2t ) j
22. y 1
x 40
348
Problem Set 12-10
Calculus Solutions Manual © 2005 Key Curriculum Press
Cumulative Review Number 3 1. δ is clearly smaller than necessary.
11.
f (x ) L+ε
12.
L
3 x – 11 5 –2 + dx = dx + 2x – 3 x + 3 x – 1 = 5 ln | x + 3 | − 2 ln | x − 1| + C
∫x
∫
2
∫ sin
−1
u sin –1 x 1 √1 – x 2
x dx
L–ε x c – δ c c+δ
2. See Sections 3-2 and 3-4 for definitions of derivative. Graphical meaning: slope of tangent line Physical meaning: instantaneous rate of change 3. g( x ) = 4.
∫
s
r
∫ f ( x ) dx if and only if g′( x ) = f ( x ).
f (t ) dt = lim Ln = lim Un , where Ln and Un are ∆t →0
∆t →0
lower and upper Riemann sums, respectively, provided the two limits are equal. 5. l’Hospital’s rule x cos x 0 lim 5x → x→0 1 – e 0 cos x – x sin x 1 = lim → 5x x→0 –5e –5 = −0.2 6. y = tan (sin 5x) y′ = sec2 (sin 5x) ⋅ 5 cos 5x Chain rule 7. y = (5x − 3)(2x + 7)4(x − 9) ln y = ln (5x − 3) + 4 ln (2x + 7) + ln (x − 9) 5 8 1 y′ = y + + 5x – 3 2 x + 7 x – 9 8. y = tan x tan y = x, sec2 y y′ = 1 1 1 y′ = = 2 sec y 1 + tan –1 y 1 y′ = 1+ x2 1 9. sin 7 x cos x dx = sin 8 x + C 8 10.
x 2 + 9 dx
x = 3 tan θ dx = 3 sec2 θ dθ x 2 + 9 = 3 sec θ
∫
= 9 sec 3 θ dθ 9 9 sec θ tan θ + ln | sec θ + tan θ | + C1 2 2 9 x2 + 9 x 9 x2 + 9 x = ⋅ + ln + + C1 2 3 3 2 3 3 =
=
1 9 x x 2 + 9 + ln 2 2
Calculus Solutions Manual © 2005 Key Curriculum Press
x2 + 9 + x + C
–
dv 1 x
∫
= x sin −1 x − (1 − x 2 ) −1/2 ( x dx ) = x sin −1 x +
∫
1 (1 − x 2 ) −1/2 ( −2 x dx ) 2
= x sin −1 x + 1 − x 2 + C 13. Fundamental theorem of calculus See Section 5-6 for statement. 14. See Figure 5-5b. 15. f ( x ) =
∫
x
3
h(t ) dt ⇒ f ′( x ) = h( x )
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 = 2 f ″(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 2 + dy 2 = 1 + cos 2 x dx 2
∫ dL ≈ 2.3516K
L= 18. a.
0
∫
16
x −3/4 dx = lim+ 4 x 1/4 a→0
0
16 a
= lim+ (8 – 4 a1/4 ) = 8 a→ 0
−1
∫ ∫
+
b. Average value =
8 1 = 16 – 0 2
19. r = 10 cos θ dA = 50 cos2 θ dθ A=
∫
1
0.5
50 cos 2 θ dθ ≈ 13.3478…
(exactly 12.5(1 + sin 2 − sin 1)) r r r 20. r = t 2 i + 3t −1 j r r r v = 2ti − 3t −2 j r r r v (1) = 2i − 3 j Speed = 13 = 3.6055… r Distance from origin is |r | = t 4 + 9t –2 . r d |r | 1 4 = (t + 9t –2 ) −1/2 ( 4t 3 − 18t −3 ) dt 2 = −7/ 10 at t = 1 Distance is decreasing at 2.2135… .
Problem Set 12-10
349
21. y = cos x dV = 2π x ⋅ y ⋅ dx V=
∫
π /2
0
h→0
2π x cos x dx ≈ 3.5864 …
(exactly 2π (π/2 − 1)) 22. a. y = −1.5x + 6 A = xy = −1.5x 2 + 6x A′ = −3x + 6 A′ = 0 ⇔ −3x + 6 = 0 ⇔ x = 2 A(0) = 0, A(4) = 0, A(2) > 0 Thus, maximum area is at x = 2, Q.E.D. b. V = π x 2y = π (−1.5x 3 + 6x 2) V′ = π (−4.5x 2 + 12x) 2 V ′ = 0 ⇔ x = 0 or x = 2 3 2 V (0) = V ( 4) = 0 and V 2 > 0 3 2 Thus, maximum volume is at x = 2 . 3 1 23. V ≈ (2)[51 + 4(37) + 2( 41) + 4(63) + 59] 3 2 = 394 ft 3 3 24. a. erf x = (2π −1/2 ) f ( x) =
∫
x
x
∫e
−t
2
e − t dt 2
∞
∑ (–1)
n
n=0
1 x 2 n+1 (2 n + 1)n!
x 2 n +3 (2 n + 1)n! ⋅ n→∞ ( 2 n + 3)( n + 1)! x 2 n+1 (2 n + 1) = x 2 lim = x2 ⋅ 0 n→∞ ( 2 n + 3)( n + 1) L < 1 for all values of x, and thus the series converges for all values of x, Q.E.D. L = lim
Final Examination sin 1.1 – sin 1 = 0.497363752… 0.1 sin 1.01 – sin 1 = 0.536085981… 0.01 sin 1.001 – sin 1 = 0.539881480… 0.001 2. f ′(1) = cos 1 = 0.540302305… The quotients in Problem 1 are converging to cos 1. 1.
350
x →2
If f (x) = e 2 + 0.1, x = ln (e 2 + 0.1) = 2.01344… . If f (x) = e 2 − 0.1, x = ln (e 2 − 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 = lim f ( x ) if and only if for any ε > 0 there is x →c
a δ > 0 such that if x is within δ units of c but not 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 r
10
0
x
b. f ( x ) =
4. f (x) = ex lim f ( x ) = e 2
dt
0
1 1 1 = 1 – t 2 + t 4 – t 6 + t 8 – L dt 0 2! 3! 4! 1 3 1 5 1 7 1 9 = x− x + x − x + x −L 3 5 ⋅ 2! 7 ⋅ 3! 9 ⋅ 4!
∫
f ( x + h) – f ( x ) h f ( x ) – f (c ) f ′(c) = lim x →c x–c
3. f ′( x ) = lim
Problem Set 12-10
t 1
1.3
1.6
1.9
2.2
2.5
2.8
1 9. Distance ≈ (0.3)[7 + 4(9) + 2(13) + 4(12) + 3 2(10) + 4(8) + 5] = 17.4, which agrees with Problem 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 =
2
∫ te 0
−t
dt = −te − t − e − t
2 0
= −2e − e + 0 + 1 = 1 − 3e − 2 = 0.593994… The difference is 0.00748… , which is about 1.26%. −2
−2
12. If f is integrable on [a, b] and g( x ) = then
∫
b
a
∫ f ( x ) dx,
f ( x ) dx = g(b) − g( a).
13. f (x) = x 2/3 2 f ′( x ) = x −1/3 3 f is differentiable everywhere except at x = 0. But f 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 © 2005 Key Curriculum Press
f meets the hypotheses of the mean value theorem because it is differentiable on (0, 1) and continuous at 0 and 1. Slope of the secant line from (0, 0) to (1, 1) is 1.
17. Cross section of solid at any point in the slice is essentially the same as at the sample point. y 4 (x, y )
2 −1/3 8 c =1⇔ c = 3 27 Tangent at x = 8/27 is parallel to secant. f ′( c ) =
x 2
1
f (x )
18. Height at any point in the slice is essentially the same as at the sample point. y 4
x 8/27
(x, y)
1
14. a. Example:
∫ ( x – 2)( x + 3) dx 5x – 3
1
x 2
4
∫ x – 2 + x + 3 dx
=
19. dM y = x dA = x(4 − x2) dx My =
= ln | x − 2 | + 4 ln | x + 3 | + C b. Example:
∫
A=
9 – x 2 dx x = 3 sin θ dx = 3 cos θ dθ
∫
9 2
∫ (1 + cos 2θ ) dθ
∫ sec
3
u sec x sec x tan x
x dx
+ –
dv sec 2 x tan x
3
3
16.
x dx =
1 1 sec x tan x + ln | sec x + tan x | + C 2 2
∫
∫
dy dy = ky ⇒ = k dx ⇒ ln | y | = kx + C ⇒ dx y | y | = ekx+ C = ekxeC ⇒ y = C1ekx
Calculus Solutions Manual © 2005 Key Curriculum Press
0
2
) dx =
16 3
4 3 = 16/3 4
20. Let H = number of calories added. dH = C dT = (10 + 0.3T1/2) dT
∫
900
100
(10 + 0.3T 1/2 ) dT = 13, 200 cal
∫
x
sin u sin x du ⇒ Si ′ x = u x t 1 1 1 b. Si x = 1 – u 2 + u 4 – u 6 + K du 0 3! 5! 7! 1 3 1 5 1 7 K =t− t + t − t + 3 ⋅ 3! 5 ⋅ 5! 7 ⋅ 7! 1 (0.73 ) = c. Si 0.7 ≈ S1 (0.7) = 0.7 − 3 ⋅ 3! 0.68094444… 1 d. | R1 (0.7)| < | t2 (0.7)| = (0.75 ) = 5 ⋅ 5! 0.0002801… S1(0.7) equals Si 0.7 correct to three decimal places and is within ±0.3 in the fourth decimal place. e. See Cumulative Review Number 1, Problem 9. r r r 22. r = (t 3 )i + (t 2 ) j r r r r r r v = (3t 2 )i + (2t ) j ⇒ v (0.5) = 0.75i + 1 j r r r r r r a = (6t )i + (2) j ⇒ a (0.5) = 3i + 2 j 0
∫
∫ = sec x tan x − ∫ sec x dx + ∫ sec x dx 2 ∫ sec x dx = sec x tan x + ∫ sec x dx 3
2
21. a. Si x =
= sec x tan x − sec x tan 2 x dx
∫ sec
x ( 4 − x 2 ) dx = 4
∫ (4 – x
H=
9 9 = θ + sin 2θ + C 2 4 9 9 = θ + sin θ cos θ + C 2 2 x 1 9 = sin −1 + x 9 – x 2 + C 2 3 2 15.
0
xA = M y ⇒ x =
9 – x 2 = 3 cos θ = 9 cos 2 θ dθ =
∫
2
Problem Set 12-10
351
→ a
y → v 1
x 1
The object is speeding up at t = 0.5 because the angle between the velocity and acceleration vectors is acute, indicating that the tangential component of acceleration acts in the same direction as the rvelocity. r (Algebraically, v ⋅ a = 4.25, which is positive, again indicating an acute angle.)
352
Problem Set 12-10
23. r = cos θ 1 1 dA = r 2 dθ = cos 2 θ dθ 2 2 π /6 1 2 A= cos θ dθ ≈ 0.2391K 0 2 π 3 exactly 24 + 16
∫
Calculus Solutions Manual © 2005 Key Curriculum Press
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