1 if x 1692 f x 23 0 2 38 1 1 1 1 1 1 x 0

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Unformatted text preview: x2 2 -2 2 -2 2 -2 (d) f (1) = √ -6 √ 2 2 3 and f (1) = √ so the tangent line has the equation y − 3 = √ (x − 1). 3 3 3 0 2 0 54. (a) (c) 0.5 f (x) = 2x cos x2 cos x − sin x sin x2 1.2 ^ ^ 0 6 6 -1.2 87 Chapter 3 (d) f (1) = sin 1 cos 1 and f (1) = 2 cos2 1 − sin2 1, so the tangent line has the equation y − sin 1 cos 1 = (2 cos2 1 − sin2 1)(x − 1). 0.8 ^ 6 0 (a) dy/dt = −Aω sin ωt, d2 y/dt2 = −Aω 2 cos ωt = −ω 2 y (b) one complete oscillation occurs when ωt increases over an interval of length 2π , or if t increases over an interval of length 2π/ω (c) f = 1/T (d) 55. amplitude = 0.6 cm, T = 2π/15 s/oscillation, f = 15/(2π ) oscillations/s 56. dy/dt = 3A cos 3t, d2 y/dt2 = −9A sin 3t, so −9A sin 3t + 2A sin 3t = 4 sin 3t, −7A sin 3t = 4 sin 3t, −7A = 4, A = −4/7 57. (a) p ≈ 10 lb/in2 , dp/dh ≈ −2 lb/in2 /mi (b) dp dh dp = ≈ (−2)(0.3) = −0.6 lb/in2 /s dt dh dt 58. (a) (b) 59. 45 dF 45(− sin θ + 0.3 cos θ) , =− ; cos θ + 0.3 sin θ dθ (cos θ + 0.3 sin θ)2 if θ = 30◦ , then dF/dθ ≈ 10.5 lb/rad ≈ 0.18 lb/deg F= dF dθ dF = ≈ (0.18)(−0.5) = −0.09 lb/s dt dθ dt With u = sin x, d d du d d (| sin x|) = (|u|) = (|u|) = (|u|) cos x = dx dx du dx du cos x, sin x > 0 = − cos x, sin x < 0 = 60. (a) cos x, 0<x<π − cos x, −π < x < 0 d d (cos x) = [sin(π/2 − x)] = − cos(π/2 − x) = − sin x dx dx 61. cos x, − cos x, (b) (c) 62. (a) (b) for x = 0, f (x) = x cos lim x sin x→0 1 x − 1 x2 + sin 1 1 1 1 = − cos + sin x x x x 1 = 0 = f (0) x 1 h sin f (0 + h) − f (0) h = lim sin 1 , which does not exist = lim lim h→0 h→0 h→0 h h h f (x) = x2 cos lim x2 sin x→0 1 x − 1 = 0 = f (0) x 1 x2 + 2x sin 1 1 1 = − cos + 2x sin , x = 0 x x x u>0 u<0 Exercise Set 3.5 (c) (d) 88 1 h2 sin f (0 + h) − f (0) h = lim h sin 1 = 0 = lim f (0) = lim h→0 h→0 h→0 h h h 1 lim f (x) does not exist because cos oscillates between 1 and −1 x→0 x g (x) = 3[f (x)]2 f (x), g (2) = 3[f (2)]2 f (2) = 3(1)2 (7) = 21 h (x) = f (x3 )(3x2 ), h (2) = f (8)(12) = (−3)(12) = −36 (a) F (x) = f (g (x))g (x), F (−1) = f (g (−1))g (−1) = f (2)(−3) = (4)(−3) = −12 (b) 64. (a) (b) 63. G (x) = g (f (x))f (x), G (−1) = g (f (−1))f (−1) = g (2)(3) = (−5)(3) = −15 65. (f ◦ g ) (x) = f (g (x))g (x) so (f ◦ g ) (0) = f (g (0))g (0) = f (0)(3) = (2)(3) = 6 66. F (x) = f (g (x))g (x) = 67. √ √ 1 3x − 1 3 3 √ = = F (x) = f (g (x))g (x) = f ( 3x − 1) √ (3x − 1) + 1 2 3x − 1 2x 2 3x − 1 68. d [f (x2 )] = f (x2 )(2x), thus f (x2 )(2x) = x2 so f (x2 ) = x/2 if x = 0 dx 69. 70. √ 3(x2 − 1) + 4(2x) = 2x 3x2 + 1 d 2 d [f (3x)] = f (3x) (3x) = 3f (3x) = 6x, so f (3x) = 2x. Let u = 3x to get f (u) = u; dx dx 3 2 d [f (x)] = f (x) = x. dx 3 (a) If f (−x) = f (x), then d d [f (−x)] = [f (x)], f (−x)(−1) = f (x), f (−x) = −f (x) so dx dx f is odd. (b) d d [f (−x)] = − [f (x)], f (−x)(−1) = −f (x), f (−x) = f (x) so dx dx If f (−x) = −f (x), then f is even. 71. For an even function, the graph is symmetric about the y -axis; the slope of the tangent line at (a, f (a)) is the negative of the slope of the tangent line at (−a, f (−a). For an odd function, the graph is symmetric about the origin; the slope of the tangent line at (a, f (a)) is the same as the slope of the tangent line at (−a, f (−a). y y f(x) f(x) f '(x) x x f '(x) 72. dy du dv dw dy = dx du dv dw dx 73. d d [f (g (h(x)))] = [f (g (u))], dx dx = u = h(x) du du d [f (g (u))] = f (g (u))g (u) = f (g (h(x)))g (h(x))h (x) du dx dx 89 Chapter 3 EXERCISE SET 3.6 1. (a) dy = f (x)dx = 2xdx = 4(1) = 4 and ∆y = (x + ∆x)2 − x2 = (2 + 1)2 − 22 = 5 (b) y ∆y dy 4 2 2. (a) dy = 3x2 dx = 3(1)2 (1) = 3 and ∆y = (x + ∆x)3 − x3 = (1 + 1)3 − 13 = 7 (b) x 3 y 8 ∆y dy (1,1) 1 dx 1 3. (a) dy = (−1/x2 )dx = (−1)(−0.5) = 0.5 and ∆y = 1/(x + ∆x) − 1/x = 1/(1 − 0.5) − 1/1 = 2 − 1 = 1 x 2 y (b) 2 ∆y = 1 dy = 0.5 1 x 0.5 4. √ (a) dy = (1/2 x)dx = (1/(2 · 3))(−1) = −1/6 ≈√ 0.167 and − √ √ √ ∆y = x + ∆x − x = 9 + (−1) − 9 = 8 − 3 ≈ −0.172 1 y (b) 3 dy = – 0.167 ∆ y = – 0.171 x 8 5. dy = 3x2 dx; ∆y = (x + ∆x)3 − x3 = x3 + 3x2 ∆x + 3x(∆x)2 + (∆x)3 − x3 = 3x2 ∆x + 3x(∆x)2 + (∆x)3 6. dy = 8dx; ∆y = [8(x + ∆x) − 4] − [8x − 4] = 8∆x 7. dy = (2x − 2)dx; ∆y = [(x + ∆x)2 − 2(x + ∆x) + 1] − [x2 − 2x + 1] = x2 + 2x ∆x + (∆x)2 − 2x − 2∆x + 1 − x2 + 2x − 1 = 2x ∆x + (∆x)2 − 2∆x 8. dy = cos x dx; ∆y = sin(x + ∆x) − sin x 9 Exercise Set 3.6 9. (a) 90 dy = (12x2 − 14x)dx (b) dy = x d(cos x) + cos x dx = x(− sin x)dx + cos xdx = (−x sin x + cos x)dx 10. (a) dy = (−1/x2 )dx 11. (a) dy = (b) dy = −17(1 + x)−18 dx (a) dy = (x3 − 1)d(0) − (1)d(x3 − 1) (x3 − 1)(0) − (1)3x2 dx 3x2 = =− 3 dx (x3 − 1)2 (x3 − 1)2 (x − 1)2 (b) dy = (2 − x)(−3x2 )dx − (1 − x3 )(−1)dx 2x3 − 6x2 + 1 = dx (2 − x)2 (2 − x)2 (a) f (x) ≈ f (1) + f (1)(x − 1) = 1 + 3(x − 1) (c) From...
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