# X3 dd 2x 3 dt 2 x2 3x 4 3 3 unitss 4 24 b

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Unformatted text preview: x2 /9 (1/3) = 1/ 9 − x2 x7 7 1 (7x6 ) = √ 14 − 1 x x x14 − 1 1 (c) 25.8◦ (b) − 2x/(1 + x4 ) √ 33.6◦ 1 √ |x| x2 − 1 1 1+x 1 −1/2 x 2 =− 1 √ 2(1 + x) x √ (b) −1/ e2x − 1 1 1 + x2 sin x sin x = = 2x | sin x| 1 − cos 1, sin x &gt; 0 −1, sin x &lt; 0 (b) √ (b) 1 −√ −1 x(1 + x2 ) 2 cot ex √ + ex sec−1 x x x2 − 1 (b) 3x2 (sin−1 x)2 √ + 2x(sin−1 x)3 1 − x2 0 (b) 0 (−1/x2 ) = − 25. (a) 26. (a) − 27. (a) 28. (a) 29. x3 + x tan−1 y = ey , 3x2 + 1− 1/x2 (cos−1 1 √ x) 1 − x2 x (3x2 + tan−1 y )(1 + y 2 ) y + tan−1 y = ey y , y = 2 1+y (1 + y 2 )ey − x Exercise Set 4.5 30. 1 sin−1 (xy ) = cos−1 (x − y ), y= 31. 120 y 1 − (x − y )2 + 1− x2 y 2 (xy + y ) = − 1 1 − (x − y )2 (1 − y ), 1 − x2 y 2 1 − x2 y 2 − x 1 − (x − y )2 (a) 32. y (a) 3 y 1 2 1 x -1 33. 1 (a) 2 x y π −2π 4π x 35. (a) sin−1 0.9 &gt; 1, so it is not in the domain of sin−1 x (b) 34. −1 ≤ sin−1 x ≤ 1 is necessary, or −0.841471 ≤ x ≤ 0.841471 (a) (b) y y 6 6 x -0.5 x 0.5 -6 (a) x = 2π − cos−1 k (c) 36. 2x = sin−1 k or 2x = π − sin−1 k so x = -6 (b) x = π + tan−1 k 1 2 sin−1 k or x = π/2 − 1 sin−1 k 2 R 6378 = sin−1 ≈ 23◦ R+h 16, 378 37. (b) θ = sin−1 38. (a) If γ = 90◦ , then sin γ = 1, 1 − sin2 φ sin2 γ = 1 − sin2 φ = cos φ, D = tan φ tan λ = (tan 23.55◦ )(tan 65◦ ) ≈ 0.934684245 so h ≈ 21.2 hours. (b) If γ = 270◦ , then sin γ = −1, D = − tan φ tan λ ≈ −0.934684245 so h ≈ 2.8 hours. 39. sin 2θ = gR/v 2 = (9.8)(18)/(14)2 = 0.9, 2θ = sin−1 (0.9) or 2θ = 180◦ − sin−1 (0.9) so θ= 1 2 sin−1 (0.9) ≈ 32◦ or θ = 90◦ − 1 sin−1 (0.9) ≈ 58◦ . The ball will have a lower 2 parabolic trajectory for θ = 32◦ and hence will result in the shorter time of ﬂight. 40. 42 = 22 + 32 − 2(2)(3) cos θ, cos θ = −1/4, θ = cos−1 (−1/4) ≈ 104◦ 121 Chapter 4 41. √ y = 0 when x2 = 6000v 2 /g , x = 10v 60/g = 1000 30 for v = 400 and g = 32; √ √ tan θ = 3000/x = 3/ 30, θ = tan−1 (3/ 30) ≈ 29◦ . 42. θ = α − β , cot α = θ = cot−1 x x and cot β = so a+b b x −1 x − cot a+b b a θ b β α x 43. (a) Let θ = sin−1 (−x) then sin θ = −x, −π/2 ≤ θ ≤ π/2. But sin(−θ) = − sin θ and −π/2 ≤ −θ ≤ π/2 so sin(−θ) = −(−x) = x, −θ = sin−1 x, θ = − sin−1 x. (b) 45. (a) Let θ = cos−1 (−x) then cos θ = −x, 0 ≤ θ ≤ π . But cos(π − θ) = − cos θ and 0 ≤ π − θ ≤ π so cos(π − θ) = x, π − θ = cos−1 x, θ = π − cos−1 x (b) 44. proof is similar to that in part (a) Let θ = sec−1 (−x) for x ≥ 1; then sec θ = −x and π/2 &lt; θ ≤ π . So 0 ≤ π − θ &lt; π/2 and π − θ = sec−1 sec(π − θ) = sec−1 (− sec θ) = sec−1 x, or sec−1 (−x) = π − sec−1 x. (a) sin−1 x = tan−1 √ x 1 − x2 1 x sin−1x ÷1 − x2 (b) 46. sin−1 x + cos−1 x = π/2; cos−1 x = π/2 − sin−1 x = π/2 − tan−1 √ tan(α + β ) = tan α + tan β , 1 − tan α tan β tan(tan−1 x + tan−1 y ) = x+y tan(tan−1 x) + tan(tan−1 y ) = 1 − tan(tan−1 x) tan(tan−1 y ) 1 − xy so tan−1 x + tan−1 y = tan−1 (a) tan−1 (b) 47. x 1 − x2 2 tan−1 x+y 1 − xy 1 1 1/2 + 1/3 + tan−1 = tan−1 = tan−1 1 = π/4 2 3 1 − (1/2) (1/3) 2 tan−1 1 1 = tan−1 + tan−1 3 3 1 1 + tan−1 = tan−1 3 7 48. sin(sec−1 x) = sin(cos−1 (1/x)) = 1 1/3 + 1/3 3 = tan−1 = tan−1 , 3 1 − (1/3) (1/3) 4 3 1 3/4 + 1/7 + tan−1 = tan−1 = tan−1 1 = π/4 4 7 1 − (3/4) (1/7) 1− 1 x 2 = √ x2 − 1 |x| Exercise Set 4.6 122 EXERCISE SET 4.6 1. (b) (d) 2. (b) (d) A = x2 Find (c) dA dt x=3 Find dA dt r=5 Find (b) 2 dV dt h=6, r=10 = 2. From part (c), r=5 dh dt given that h=6, r=10 = 1 and h=6, r=10 5. (a) (b) d dt = x=3, y =4 dr dt h=6, r=10 x=3 dr dA = 2πr dt dt dA dt = 2π (5)(2) = 20π cm2 /s. r=5 = −1. From part (a), 1 d = dt given that x=3, y =4 x dx dy +y . dt dt 1 dy 1 dx = and =− . dt 2 dt 4 1 3 5 1 2 +4 − 1 4 = 5 when x = 3 and y = 4, = 1 ft/s; the diagonal is increasing. 10 dy dx x −y y cos2 θ dθ dθ tan θ = , so sec2 θ = dt 2 dt , = x dt x dt x2 Find = 2(3)(2) = 12 ft2 /min. = π [102 (1) + 2(10)(6)(−1)] = −20π in3 /s; the volume is decreasing. From part (a) and the fact that d dt dA dt dh dr dV = π r2 + 2rh . dt dt dt = x2 + y 2 , so Find dr dt given that V = πr2 h, so (a) x=3 (c) (a) dV dt 4. = 2. From part (c), A = πr2 (b) 3. dx dt given that dx dA = 2x dt dt dθ dt given that x=2, y =2 dx dt = 1 and x=2, y =2 dy dt x=2, y =2 x dx dy −y dt dt 1 =− . 4 π π 1 When x = 2 and y = 2, tan θ = 2/2 = 1 so θ = and cos θ = cos = √ . Thus 4 4 2 √ 5 1 (1/ 2)2 dθ − 2(1) = − rad/s; θ is decreasing. 2− = from part (a), dt x=2, 22 4 16 y =2 6. Find dz dt given that x=1, y =2 dx dt x=1, y =2 dy dx dz dz = 2x3 y + 3x2 y 2 , dt dt dt dt 7. = −2 and x=1, y =2 dy dt = 3. x=1, y =2 = (4)(3) + (12)(−2) = −12 units/s; z is decreasing Let A be the area swept out, and θ the angle through which the minute hand has rotated. dθ π 1 dθ 4π 2 dA dA given that = rad/min; A = r2 θ = 8θ, so =8 = in /min. Find dt dt 30 2 dt dt 15 123 8. Chapter 4 Let r be the radius and A the area enclosed by the ripple. We want dA dt given that t=10 dr = 3. We dt dr dA = 2πr . Becau...
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