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SalasSV_07_01_ex_ans

# SalasSV_07_01_ex_ans - ANSWERS TO ODD-NUMBERED EXERCISES 19...

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Unformatted text preview: ANSWERS TO ODD-NUMBERED EXERCISES 19. ( 10 , 3 40 ) 21 A-53 21. (2, 4) 23. (− 3 , 0) 5 25. (a) (0, 0) (b) 14 14 , 5π 5π 1 π R3 3 16 ) 35 (c) 0, 14 5π 27. V = π ab(2c + (b) √ a2 + b 2 ) 29. (a) ( 2 a, 1 h) 3 3 (b) ( 2 a + 1 b, 1 h) 3 3 3 (c) ( 1 a + 1 b, 1 h) 3 3 3 35. (a) A = 1 2 31. (a) (b) ( 16 , 35 sin2 θ (2 sin θ + cos θ ) (c) V = 16 π 35 2R sin θ (2 sin θ + cos θ ) 3(π sin θ + 2 cos θ ) 33. An annular region; see Exercise 25(a). 37. (a) A = 250 3 (d) V = 16 π 35 (b) (− 9 , 8 290 ) 21 ∼ (−1. 125, 13. 8095) = SECTION 6.5 1. 817. 5 ft-lb 3. 1 (64 3 − 73/2 ) ft-lb 5. 35π 2 1 − newton-meters 72 4 15. (a) 7. 625 ft-lb ft-lb 9. (a) 25-ft-lb (b) 225 4 ft-lb ft-lb 11. 1.95 ft 13. (a) (6480π + 8640) ft-lb 17. (a) 384π σ newton-meters 25. (a) 1 σ l2 2 (b) (15, 120π + 8640) ft-lb (b) 480π σ newton-meters ft-lb 27. 20,800 ft-lb 11 π r 2 h2 σ 192 11 (b) ( 192 π r 2 h2 σ + 7 π r 2 hk σ ) 24 19. 48,000 ft-lb 21. (a) 20,000 ft-lb (b) 30,000 ft-lb 23. 788 ft-lb ft-lb (b) 32 lσ 2 29. Let λ(x) be the mass density of the chain at the point x units above the ground. Let g be the gravitational constant. The work done to pull the chain to the top of the building is given by W= 0 H (H − x)g λ(x) dx = Hg 0 H λ(x) dx − g 0 H xλ(x) dx = HgM − g xM = (H − x)gM = (weight of chain) × (distance from center of mass to top of building). 31. Total weight 400 lbs.; center of mass (chain plus bucket): x = 25 ft from ground; W = 400(100 − 25) = 400(75) = 30, 000 ft-lbs . 33. W = a b ma dx = a b mv dv = 1212 mv − mv 2b2a 35. 94.8 ft-lb 37. 9. 714 × 109 ft-lb 39. (a) 670 sec or 11 min, 10 sec (b) 1116 sec or 18 min, 36 sec SECTION 6.6 1. 9000 lb 3. 1. 437 × 108 newtons (b) 41,250 lb 5. 1. 7052 × 106 newtons 7. 2160 lb 9. 8000 3 √ 2 lb 11. 333.33 lb 13. 2560 lb 15. (a) 41,250 lb 17. (a) 297,267 newtons (b) 39,200 newtons at the shallow end; 352,800 newtons at the deep end 19. 2. 21749 × 106 newtons 21. F = σ xA where A is the area of the submerged surface and x is the depth of its centroid. CHAPTER 7 SECTION 7.1 1. f −1 (x) = 1 (x − 3) 5 3. not one-to-one 5. f −1 (x) = (x − 1)1/5 7. f −1 (x) = [ 1 (x − 1)]1/3 3 9. f −1 (x) = 1 − x1/3 11. f −1 (x) = (x − 2)1/3 − 1 19. f −1 (x) = 1/x 29. 13. f −1 (x) = x5/3 15. f −1 (x) = 1 (2 − x1/3 ) 3 1−x x 1/3 17. f −1 (x) = sin−1 (x) (to be studied in Section 7.7) 27. they are equal 21. not one-to-one 23. f −1 (x) = 25. f −1 (x) = (2 − x)/(x − 1) 33. (a) k ≥ 1 √ √ (b) − 3 ≤ k ≤ 3 y 31. y x x 35. f (x) = 3x2 ≥ 0 on ( − ∞, ∞), f (x) = 0 only at x = 0; ( f −1 ) (9) = 1 12 1 37. f (x) = 1 + √ > 0 on (0, ∞); ( f −1 ) (8) = x 2 3 A-54 ANSWERS TO ODD-NUMBERED EXERCISES ( f −1 ) ( π ) = 1 1 6 39. f (x) = 2 − sin x > 0 on (−∞, ∞); 43. f (x) = 3x2 + 49. (a) f (x) = √ 41. f (x) = sec2 x > 0 on (−π/2, π/2); ( f −1 ) ( 3) = 45. ( f −1 ) (x) = 1 x 1 47. ( f −1 ) (x) = √ 1 − x2 51. (a) f (x) = (b) g (x) = 0 √ 1 + x2 > 0 1 4 3 > 0 on (0, ∞); ( f −1 ) (2) = x4 iff ad − bc =0 (cx + d )2 ad − bc = 0 (b) f −1 (x) = dx − b a − cx 1 (b) ( f −1 ) (0) = √ 5 1 g (x ) 53. (a) g (t ) ≥ 0 or g (t ) ≤ 0 on [a, b], with g (t ) = 0 only at “isolated" points. 55. (a) g (x) = 1 f ( g (x))g (x) f ( g (x)) =− ; g (x ) = − f ( g (x)) [ f ( g (x))]2 [ f ( g (x))]3 (c) ( f −1 ) (x) = (b) If f is increasing, then the graphs of f and g have opposite concavity; if f is decreasing, then the graphs of f and g have the same concavity. 1 57. ( f −1 ) (x) = √ . 1 − x2 65. y f 59. f −1 (x) = −1 + (2 − x)1/3 67. 61. f −1 (x) = x2 − 8x + 25 , x≥4 9 63. f −1 (x) = 1−x 1+x y 4 2 f –1 0 2 x –4 –4 π 4 π 4 x –4 SECTION 7.2 1. ln 2 + ln 10 ∼ 2. 99 = 3. 2 ln 4 − ln 10 ∼ 0. 48 = 5. −ln 10 ∼ −2. 30 = 7. ln 8 + ln 9 − ln 10 ∼ 1. 98 = 9. 1 2 area = In n m mn x 11. y y= 1 x ln 2 ∼ 0. 35 = 13. 0.406 15. (a) 1.65 (b) 1.57 (c) 1.71 17. x = e2 19. x = 1, e2 21. x = 1 23. lim x→1 d (ln x) ln x = x−1 dx x =1 =1 25. (a) P = {1, 2, . . . , n} is a regular partition of [1, n]; Lf (P ) = n 1 1 1 1 1 1 + + ··· + < dt = ln n < 1 + + · · · + = Uf (P ). 2 3 n 2 n−1 1t n 1 1 1 (b) Sum of shaded areas = Vf (P ) − dt = 1 + + · · · + − ln n. 2 n−1 1t 1 (c) Connect the points (1,1), (2, 1 ),. . . , n, by straight-line segments. The sum of the areas of the triangles that are formed is 2 n 1 ·1 2 so 1− 1 2 + 1 1 − 2 3 + ··· + 1 1 − n−1 n = 1 1 1− , 2 n 1 1 1− 2 n The sum of the areas of the indicated rectangles is 1 1− 1 2 + 1 1 − 2 3 + ··· + < γ. 1 1 − n−1 n =1− 1 , n ...
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