Calculus: Early Transcendentals Single Variable: Student Solutions Manual

304 Pages·2012·3.173 MB·English

by  Tamas Wiandt, Howard Anton, Irl Bivens

(cid:2)
Student Solutions Manual

Tamas Wiandt
Rochester Institute of Technology

to accompany

CALCULUS
Early Transcendentals
Single Variable
Tenth Edition

Howard Anton
Drexel University

Irl C. Bivens
Davidson College

Stephen L. Davis
Davidson College

John Wiley& Sons, Inc.
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Printed in the United States of America

10 9 8 7 6 5 4 3 2 1
Contents

Chapter 0. Before Calculus ………..…………………………………………………………………………..……. 1
Chapter 1. Limits and Continuity ……………………………………………………………………………….. 21
Chapter 2. The Derivative ……………………………………………………………………………………..……. 39
Chapter 3. Topics in Differentiation ……………………………..………………………………………..……. 59
Chapter 4. The Derivative in Graphing and Applications ……………………………………..………. 81
Chapter 5. Integration …………………………………………………………………………………………..…… 127
Chapter 6. Applications of the Definite Integral in Geometry, Science, and Engineering… 159
Chapter 7. Principles of Integral Evaluation ……………………………………………………………….. 189
Chapter 8. Mathematical Modeling with Differential Equations …………………………………… 217
Chapter 9. Infinite Series ……………………………………………………………………………………..…….. 229
Chapter 10. Parametric and Polar Curves; Conic Sections ……………………………………….…….. 255
Appendix A. Graphing Functions Using Calculators and Computer Algebra Systems .………. 287
Appendix B. Trigonometry Review ……………………………………………………………………………….. 293
Appendix C. Solving Polynomial Equations …………………………………………………………………… 297
(cid:2)
Before Calculus
Exercise Set 0.1
1. (a) −2.9, −2.0, 2.35, 2.9 (b) None (c) y =0 (d) −1.75≤x≤2.15, x=−3, x=3
(e) ymax =2.8 at x=−2.6; ymin =−2.2 at x=1.2
3. (a) Yes (b) Yes (c) No (vertical line test fails) (d) No (vertical line test fails)
5. (a) 1999, $47,700 (b) 1993, $41,600
(c) The slope between 2000 and 2001 is steeper than the slope between 2001 and 2002, so the median income was
declining more rapidly during the first year of the 2-year period.
√
7. (a√) f(0) = 3(0)2−2 = −2; f(2) = 3(2)2−2 = 10; f(−2) = 3(−2)2−2 = 10; f(3) = 3(3)2−2 = 25; f( 2) =
3( 2)2−2=4; f(3t)=3(3t)2−2=27t2−2.
√ √
(b) f(0) = 2(0) = 0; f(2) = 2(2) = 4; f(−2) = 2(−2) = −4; f(3) = 2(3) = 6; f( 2) = 2 2; f(3t) = 1/(3t) for
t>1 and f(3t)=6t for t≤1.
9. (a) Natural domain: x(cid:4)=3. Range: y (cid:4)=0. (b) Natural domain: x(cid:4)=0. Range: {1,−1}.
√ √
(c) Natural domain: x≤− 3 or x≥ 3. Range: y ≥0.
√
(d) x2−2x+5=(x−1)2+4≥4. So G(x) is defined for all x, and is ≥ 4=2. Natural domain: all x. Range:
y ≥2.
(e)Naturaldomain: sinx(cid:4)=1,sox(cid:4)=(2n+1)π,n=0,±1,±2,.... Forsuchx,−1≤sinx<1,so0<1−sinx≤2,
2
and 1 ≥ 1. Range: y ≥ 1.
1−sinx 2 2
(f) Division by 0 occurs for x = 2. For all other x, x2−4 = x+2, which is nonnegative for x ≥ −2. Natural
√ x−2 √
domain: [−2,2)∪(2,+∞). The range of x+2 is [0,+∞). But we must exclude x = 2, for which x+2 = 2.
Range: [0,2)∪(2,+∞).
11. (a) The curve is broken whenever someone is born or someone dies.
(b) C decreases for eight hours, increases rapidly (but continuously), and then repeats.
h
t
13.
1
2 Chapter 0
√
15. Yes. y = 25−x2.
(cid:2) √
√25−x2, −5≤x≤0
17. Yes. y =
− 25−x2, 0<x≤5
19. False. E.g. the graph of x2−1 crosses the x-axis at x=1 and x=−1.
21. False. The range also includes 0.
23. (a) x=2,4 (b) None (c) x≤2; 4≤x (d) ymin =−1; no maximum value.
25. The cosine of θ is (L−h)/L (side adjacent over hypotenuse), so h=L(1−cosθ).
27. (a) If x(cid:2)<0, then |x|=−x so f(x)=−x+3x+1=2x+1. If x≥0, then |x|=x so f(x)=x+3x+1=4x+1;
2x+1, x<0
f(x)=
4x+1, x≥0
(b) If x<0, then |x|=−x and |x−1|=1−x so g(x)=−x+(1−x)=1−2x. If 0≤x<1, then |x|=x and
|x−1|=1−x so g(x)=x+(1−x)=1. If x≥1, then |x|=x and |x−1|=x−1 so g(x)=x+(x−1)=2x−1;
⎧
⎨ 1−2x, x<0
g(x)= 1, 0≤x<1
⎩
2x−1, x≥1
29. (a) V =(8−2x)(15−2x)x (b) 0<x<4
100
0 4
(c) 0 0<V ≤91, approximately
(d) As x increases, V increases and then decreases; the maximum value occurs when x is about 1.7.
31. (a) The side adjacent to the building has length x, so L=x+2y. (b) A=xy =1000, so L=x+2000/x.
120
20 80
(c) 0<x≤100 (d) 80 x≈44.72 ft, y ≈22.36 ft
500 500 10
2 2 2 2
33. (a) V = 500 = πr h, so h = . Then C = (0.02)(2)πr +(0.01)2πrh = 0.04πr +0.02πr = 0.04πr + ;
πr2 πr2 r
Cmin ≈4.39 cents at r ≈3.4 cm, h≈13.7 cm.
10
2 2
(b)C =(0.02)(2)(2r) +(0.01)2πrh=0.16r + . Since0.04π <0.16, thetopandbottomnowgetmoreweight.
r
Since they cost more, we diminish their sizes in the solution, and the cans become taller.
(c) r ≈3.1 cm, h≈16.0 cm, C ≈4.76 cents.
Exercise Set 0.2 3
35. (i) x=1,−2 causes division by zero. (ii) g(x)=x+1, all x.
◦ ◦ ◦
37. (a) 25 F (b) 13 F (c) 5 F
39. If v =48 then −60=WCT≈1.4157T −30.6763; thus T ≈15◦F when WCT =−10.
Exercise Set 0.2
y
y
1
2
x
1
–1 0 1 2
x
1. (a) –1 (b) 1 2 3
y y
1
2
x x
(c) –1 1 2 (d) –4 –2 2
y
1
y
1
x
x
1
–2 –1 1 2
–1 –1
3. (a) (b)
1 y y
1
x x
–1 1 2 3 –1 1 2 3
–1 –1
(c) (d)
5. Translate left 1 unit, stretch vertically by a factor of 2, reflect over x-axis, translate down 3 units.
y
x
–6 –2 2 6
–20
–60
7. y =(x+3)2−9; translate left 3 units and down 9 units.
4 Chapter 0
9. Translate left 1 unit, reflect over x-axis, translate up 3 units.
3
2
1
-1 0 1 2 3 4
1
11. Compress vertically by a factor of , translate up 1 unit.
2
y
2
x
1 2 3
13. Translate right 3 units.
y
10
x
4 6
–10
15. Translate left 1 unit, reflect over x-axis, translate up 2 units.
y
6
x
–3 –2 1 2
–6
17. Translate left 2 units and down 2 units.
y
x
–4 –2
–2