X y 1 x y u v 8xy so xy xy x y

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Unformatted text preview: 4k, n1 = −n2 so n1 and n2 are parallel. 33. n1 = 2i − 2j − k, n2 = 2i − 8j + 4k, n1 × n2 = 16i − 10j − 12k is tangent to the line, so x(t) = 1 + 8t, y (t) = −1 + 5t, z (t) = 2 + 6t 43 i+ j−k, n2 = i+2j+2k, n1 ×n2 = (16i−13j+5k)/5 55 is tangent to the line, x(t) = 4 + 16t, y (t) = 3 − 13t, z (t) = 5 + 5t 34. f (x, y, z ) = x2 + y 2 −z, n1 = f (4, 3, 5) = 35. f (x, y, z ) = x2 + z 2 − 25, g (x, y, z ) = y 2 + z 2 − 25, n1 = f (3, −3, 4) = 6i + 8k, n2 = g (3, −3, 4) = −6j + 8k, n1 × n2 = 48i − 48j − 36k is tangent to the line, x(t) = 3 + 4t, y (t) = −3 − 4t, z (t) = 4 − 3t 36. (a) f (x, y, z ) = z − 8 + x2 + y 2 , g (x, y, z ) = 4x + 2y − z, n1 = 4j + k, n2 = 4i + 2j − k, n1 × n2 = −6i + 4j − 16k is tangent to the line, x(t) = 3t, y (t) = 2 − 2t, z (t) = 4 + 8t 37. dw = 3x2 y 2 zdx + 2x3 yzdy + x3 y 2 dz, ∆w = (x + ∆x)3 (y + ∆y )2 (z + ∆z ) − x3 y 2 z 38. dw = yzexyz dx + xzexyz dy + xyexyz dz, ∆w = e(x+∆x)(y+∆y)(z+∆z) − exyz Exercise Set 15.7 556 39. dw = 8dx − 3dy + 4dz 40. dw = 8xy 3 z 7 − 3y dx + 12x2 y 2 z 7 − 3x dy + 28x2 y 3 z 6 + 1 dz 41. dw = yz xz xy dx + dy + dz 2 y2 z2 2 y2 z2 1+x 1+x 1 + x2 y 2 z 2 1 1 1 42. dw = √ dx + √ dy + √ dz 2y 2x 2z 43. df = 2y 2 z 3 dx + 4xyz 3 dy + 6xy 2 z 2 dz = 2(−1)2 (2)3 (−0.01) + 4(1)(−1)(2)3 (−0.02) + 6(1)(−1)2 (2)2 (0.02) = 0.96 44. df = xz (x + z ) xy (x + y ) yz (y + z ) dx + dy + dz 2 2 (x + y + z ) (x + y + z ) (x + y + z )2 = (−16)(−0.04) + (−12)(0.02) + (−6)(−0.03) = 0.58 45. V = wh, dV = wh d + h dw + w dh, |dV | ≤ wh|d | + h|dw| + w|dh| ≤ (4)(5)(0.05) + (3)(5)(0.05) + (3)(4)(0.05) = 2.35 cm3 1 2 = R2 /R1 , similarly 2 R1 (1/R1 + 1/R2 + 1/R3 )2 dR dR1 dR2 dR3 2 2 = (R/R1 ) + (R/R2 ) + (R/R3 ) , ∂R/∂R2 = R2 /R2 and ∂R/∂R3 = R2 /R3 so R R1 R2 R3 dR1 dR2 dR3 dR ≤ (R/R1 ) + (R/R2 ) + (R/R3 ) R R1 R2 R3 46. R = 1/ (1/R1 + 1/R2 + 1/R3 ), ∂R/∂R1 = ≤ (R/R1 ) (0.10) + (R/R2 ) (0.10) + (R/R3 ) (0.10) = R (1/R1 + 1/R2 + 1/R3 ) (0.10) = (1)(0.10) = 0.10 = 10% 1 1 1 b sin θda + a sin θdb + ab cos θdθ, 2 2 2 1 1 1 |dA| ≤ b sin θ|da| + a sin θ|db| + ab cos θ|dθ| 2 2 2 47. dA = √ 1 1 1 (50)(1/2)(1/2) + (40)(1/2)(1/4) + (40)(50) 3/2 (π/90) 2 2 2 √ = 35/4 + 50π 3/9 ≈ 39 ft2 ≤ 48. V = wh, dV = whd + hdw + wdh, |dV /V | ≤ |d / | + |dw/w| + |dh/h| ≤ 3(r/100) = 3r% 49. ∂f /∂v = 8vw3 x4 y 5 , ∂f /∂w = 12v 2 w2 x4 y 5 , ∂f /∂x = 16v 2 w3 x3 y 5 , ∂f /∂y = 20v 2 w3 x4 y 4 50. ∂w/∂r = cos st + ueu cos ur, ∂w/∂s = −rt sin st, ∂w/∂t = −rs sin st, ∂w/∂u = reu cos ur + eu sin ur 2 2 2 2 2 2 2 2 2 51. ∂f /∂v1 = 2v1 / v3 + v4 , ∂f /∂v2 = −2v2 / v3 + v4 , ∂f /∂v3 = −2v3 v1 − v2 / v3 + v4 , 2 2 2 2 ∂f /∂v4 = −2v4 v1 − v2 / v3 + v4 52. 2 ∂V ∂V ∂V ∂V = 2xe2x−y + e2x−y , = xe2x−y + w, = w2 ezw , = wzezw + ezw + y ∂x ∂y ∂z ∂w 557 Chapter 15 53. (a) 0 (b) 0 yw (e) 2(yw + 1)e (c) 0 (d) 0 yw sin z cos z (f ) 2xw(yw + 2)e sin z cos z 54. 128, −512, 32, 64/3 55. ∂z/∂r = (dz/dx)(∂x/∂r) = 2r cos2 θ/ r2 cos2 θ + 1 , ∂z/∂θ = (dz/dx)(∂x/∂θ) = −2r2 sin θ cos θ/ r2 cos2 θ + 1 56. ∂u/∂x = (∂u/∂r)(dr/dx) + (∂u/∂t)(∂t/∂x) = s2 ln t (2x) + rs2 /t y 3 = x(4y + 1)2 1 + 2 ln xy 3 ∂ u/∂y = (∂u/∂s)(ds/dy ) + (∂u/∂t)(∂t/∂y ) = (2rs ln t)(4) + rs2 /t 3xy 2 = 8x2 (4y + 1) ln xy 3 + 3x2 (4y + 1)2 /y 57. ∂w/∂ρ = 2ρ 4 sin2 φ + cos2 φ , ∂w/∂φ = 6ρ2 sin φ cos φ, ∂w/∂θ = 0 58. ∂w ∂w dy ∂w dz dw 1 = + + = 3y 2 z 3 + (6xyz 3 )(6x) + 9xy 2 z 2 √ dx ∂x ∂y dx ∂z dx 2 x−1 √ 9 2 2 3/2 2 2 3/2 = 3(3x + 2) (x − 1) + 36x (3x + 2)(x − 1) + x(3x2 + 2)2 x − 1 2 √ 3 = (3x2 + 2)(39x3 − 30x2 + 10x − 4) x − 1 2 59. (a) V = wh, ∂V d ∂V dw ∂V dh d dw dh dV = + + = wh + h +w dt ∂ dt ∂w dt ∂h dt dt dt dt = (3)(6)(1) + (2)(6)(2) + (2)(3)(3) = 60 in3 /s (b) D = 2 + w2 + h2 ; dD/dt = ( /D)d /dt + (w/D)dw/dt + (h/D)dh/dt = (2/7)(1) + (3/7)(2) + (6/7)(3) = 26/7 in/s 60. (a) ∂A/∂a = (1/2)b sin θ = (1/2)(10) √ √ 3/2 = 5 3/2 (b) ∂A/∂θ = (1/2)ab cos θ = (1/2)(5)(10)(1/2) = 25/2 (c) b = (2A csc θ)/a, ∂b/∂a = −(2A csc θ)/a2 = −b/a = −2 61. Let z = f (u) where u = x + 2y ; then ∂z/∂x = (dz/du)(∂u/∂x) = dz/du, ∂z/∂y = (dz/du)(∂u/∂y ) = 2dz/du so 2∂z/∂x − ∂z/∂y = 2dz/du − 2dz/du = 0 62. Let z = f (u) where u = x2 + y 2 ; then ∂z/∂x = (dz/du)(∂u/∂x) = 2x dz/du, ∂z/∂y = (dz/du)(∂u/∂y ) = 2ydz/du so y ∂z/∂x − x∂z/∂y = 2xydz/du − 2xydz/du = 0 63. ∂w/∂x = (dw/dρ)(∂ρ/∂x) = (x/ρ)dw/dρ, similarly ∂w/∂y = (y/ρ)dw/dρ and ∂w/∂z = (z/ρ)dw/dρ so (∂w/∂x)2 + (∂w/∂y )2 + (∂w/∂z )2 = (dw/dρ)2 64. Let w = f (r, s, t) where r = x − y , s = y − z , t = z − x; ∂w/∂x = (∂w/∂r)(∂r/∂x) + (∂w/∂t)(∂t/∂x) = ∂w/∂r − ∂w/∂t, similarly ∂w/∂y = −∂w/∂r + ∂w/∂s and ∂w/∂z = −∂w/∂s + ∂w/∂t so ∂w/∂x + ∂w/∂y + ∂w/∂z = 0 65. ∂w/∂ρ = sin φ cos θ∂w/∂x + sin φ sin θ∂w/∂y + c...
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