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Calculus Early Transcendentals 10th Edition Solution Manual by Anton Bivens Davis

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Calculus Early Transcendentals 10th Edition Solution Manual by Anton Bivens Davis (Comprehensive Guide| Ideal for Grade A+)

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SOLUTION MANUAL

, Contents
Chapter 0. Before Calculus..............................................................................................................................1

Chapter 1. Limits and Continuity............................................................................................. 39

Chapter 2. The Derivative ........................................................................................................ 71

Chapter 3. Topics in Differentiation ........................................................................................ 109

Chapter 4. The Derivative in Graphing and Applications ...................................................... 153

Chapter 5. Integration .............................................................................................................. 243

Chapter 6. Applications of the Definite Integral in Geometry, Science, and Engineering… 305

Chapter 7. Principals of Integral Evaluation ...........................................................................................363

Chapter 8. Mathematical Modeling with Differential Equations.....................................................413

Chapter 9. Infinite Series .................................................................................................................................437

Chapter 10. Parametric and Polar Curves; Conic Sections ..................................................................485

Chapter 11. Three-Dimensional Space; Vectors ....................................................................... 545

Chapter 12. Vector-Valued Functions ....................................................................................... 589

Chapter 13. Partial Derivatives .......................................................................................................................627

Chapter 14. Multiple Integrals .........................................................................................................................675

Chapter 15. Topics in Vector Calculus ..........................................................................................................713

Appendix A. Graphing Functions Using Calculators and Computer Algebra Systems ...............745

Appendix B. Trigonometry Review ................................................................................................................753

Appendix C. Solving Polynomial Equations ................................................................................................759

,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

2. (a) x = −1, 4 (b) None (c) y = −1 (d) x = 0, 3, 5

(e) ymax = 9 at x = 6; ymin = −2 at x = 0

3. (a) Yes (b) Yes (c) No (vertical line test fails) (d) No (vertical line test fails)

4. (a) The natural domain of f is x /= − 1, and for g it is the set of all x. f (x) = g(x) on the intersection of their
domains.

(b) The domain of f is the set of all x ≥ 0; the domain of g is the same, and f (x) = g(x).

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.
47.7 − 41.6 6.1
6. (a) In thousands, approximately = per yr, or $1017/yr.
6 6

(b) From 1993 to 1996 the median income increased from $41.6K to $44K (K for ‘kilodollars’; all figures approx-

imate); the average rate of increase during this time was (44 41.6)/3 K/yr = 2.4/3 K/yr = $800/year. From 1996
to 1999 the average rate of increase was (47.7 44)/3— K/yr = 3.7/3 K/yr $1233/year.
≈ The increase was larger
during the last 3 years of the period.

(c) 1994 and 2005.

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.

3 + 1 = 2; g(−1) = −1 + 1 = 0; g(π) = π + 1 ; g(−1.1) = −1.1 + 1 −0.1 = 1 ; g(t2 − 1) =
8. (a) g(3) = =

3−1 −1 − 1 π−1 −1.1 − 1 −2.1 21
t2 − 1 + 1 t2

= .
t2 − 1 − 1 t2 − 2
√ √
(b) g(3) = 3 + 1 = 2; g(−1) = 3; g(π) = π + 1; g(−1.1) = 3; g(t2 − 1) = 3 if t2 < 2 and g(t2 − 1) =

t2 − 1 + 1 = |t| if t2 ≥ 2.

, 1

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