# Since both of these quantities tend to it follows by

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Unformatted text preview: os φ∂w/∂z ∂w/∂φ = ρ cos φ cos θ∂w/∂x + ρ cos φ sin θ∂w/∂y − ρ sin φ∂w/∂z ∂w/∂θ = −ρ sin φ sin θ∂w/∂x + ρ sin φ cos θ∂w/∂y Exercise Set 15.7 558 66. ∂F ∂z ∂z ∂F/∂x ∂F ∂F ∂z ∂z ∂F/∂y ∂F + = 0 so =− , + = 0 so =− . ∂x ∂z ∂x ∂x ∂F/∂z ∂y ∂z ∂y ∂y ∂F/∂z 67. ∂z 2x + yz ∂z xz − 3z 2 = , = ∂x 6yz − xy ∂y 6yz − xy 68. ln(1 + z ) + xy 2 + z − 1 = 0; 69. yex − 5 sin 3z − 3z = 0; 70. 71. y 2 (1 + z ) ∂z 2xy (1 + z ) ∂z =− , =− ∂x 2+z ∂y 2+z yex yex ∂z ex ∂z =− = , = ∂x −15 cos 3z − 3 15 cos 3z + 3 ∂y 15 cos 3z + 3 ∂z zeyz cos xz − yexy cos yz ∂z zexy sin yz − xexy cos yz + zeyz sin xz = − xy , = − xy ∂x ye sin yz + xeyz cos xz + yeyz sin xz ∂y ye sin yz + xeyz cos xz + yeyz sin xz f (u, v, w) = ∂f ∂f ∂f i+ j+ k ∂x ∂y ∂z ∂f ∂w ∂ f ∂u ∂f ∂v + + ∂u ∂x ∂v ∂x ∂w ∂x = + i+ ∂f ∂w ∂ f ∂u ∂f ∂v + + ∂u ∂z ∂v ∂z ∂w ∂z ∂f ∂w ∂ f ∂u ∂f ∂v + + ∂u ∂y ∂v ∂y ∂w ∂y k= ∂f ∂f ∂z ∂w = + ∂x ∂x ∂z ∂x 72. (a) ∂f ∂f u+ ∂u ∂v (b) v+ j ∂f w ∂w ∂w ∂f ∂f ∂z = + ∂y ∂y ∂z ∂y 2 73. wr = er / (er + es + et + eu ), wrs = −er es / (er + es + et + eu ) , 3 wrst = 2er es et / (er + es + et + eu ) , wrstu = −6er es et eu / (er + es + et + eu ) = −6er+s+t+u /e4w = −6er+s+t+u−4w 4 74. ∂w/∂y1 = a1 ∂w/∂x1 + a2 ∂w/∂x2 + a3 ∂w/∂x3 , ∂w/∂y2 = b1 ∂w/∂x1 + b2 ∂w/∂x2 + b3 ∂w/∂x3 4 75. (a) dw/dt = (∂w/∂xi ) (dxi /dt) i=1 4 (b) ∂w/∂vj = (∂w/∂xi ) (∂xi /∂vj ) for j = 1, 2, 3 i=1 76. Let u = x2 + x2 + ... +x2 ; then w = uk , ∂w/∂xi = kuk1 (2xi ) = 2k xi uk−1 , n 1 2 ∂ 2 w/∂x2 = 2k (k − 1)xi uk−2 (2xi ) + 2kuk−1 = 4k (k − 1)x2 uk−2 + 2kuk−1 for i = 1, 2, . . . , n i i n n ∂ 2 w/∂x2 = 4k (k − 1) uk−2 i so i=1 x2 + 2kn uk−1 i i=1 = 4k (k − 1)uk−2 u + 2kn uk−1 = 2kuk−1 [2(k − 1) + n] which is 0 if k = 0 or if 2(k − 1) + n = 0, k = 1 − n/2. 77. dF/dx = (∂F/∂u)(du/dx) + (∂F/∂v )(dv/dx) = f (u)g (x) − f (v )h (x) = f (g (x))g (x) − f (h(x))h (x) 559 Chapter 15 78. f = fx i + fy j + fz k and g = gx i + gy j + gz k evaluated at (x0 , y0 , z0 ) are normal, respectively, to the surfaces f (x, y, z ) = 0 and g (x, y, z ) = 0 at (x0 , y0 , z0 ). The surfaces are orthogonal at (x0 , y0 , z0 ) if and only if f · g = 0 so fx gx + fy gy + fz gz = 0. 79. f (x, y, z ) = x2 + y 2 + z 2 − a2 = 0, g (x, y, z ) = z 2 − x2 − y 2 = 0, fx gx + fy gy + fz gz = −4x2 − 4y 2 + 4z 2 = 4g (x, y, z ) = 0 EXERCISE SET 15.8 1. (a) minimum at (2, −1), no maxima (b) maximum at (0, 0), no minima (c) no maxima or minima 2. (a) maximum at (−1, 5), no minima (b) no maxima or minima (c) no maxima or minima 3. f (x, y ) = (x − 3)2 + (y + 2)2 , minimum at (3, −2), no maxima 4. f (x, y ) = −(x + 1)2 − 2(y − 1)2 + 4, maximum at (−1, 1), no minima 5. fx = 6x + 2y = 0, fy = 2x + 2y = 0; critical point (0,0); D = 8 > 0 and fxx = 6 > 0 at (0,0), relative minimum. 6. fx = 3x2 − 3y = 0, fy = −3x − 3y 2 = 0; critical points (0,0) and (−1, 1); D = −9 < 0 at (0,0), saddle point; D = 27 > 0 and fxx = −6 < 0 at (−1, 1), relative maximum. 7. fx = 2x − 2xy = 0, fy = 4y − x2 = 0; critical points (0,0) and (±2, 1); D = 8 > 0 and fxx = 2 > 0 at (0,0), relative minimum; D = −16 < 0 at (±2, 1), saddle points. 8. fx = 3x2 − 3 = 0, fy = 3y 2 − 3 = 0; critical points (−1, ±1) and (1, ±1); D = −36 < 0 at (−1, 1) and (1, −1), saddle points; D = 36 > 0 and fxx = 6 > 0 at (1,1), relative minimum; D = 36 > 0 and fxx = −36 < 0 at (−1, −1), relative maximum. 9. fx = y + 2 = 0, fy = 2y + x + 3 = 0; critical point (1, −2); D = −1 < 0 at (1, −2), saddle point. 10. fx = 2x + y − 2 = 0, fy = x − 2 = 0; critical point (2, −2); D = −1 < 0 at (2, −2), saddle point. 11. fx = 2x + y − 3 = 0, fy = x + 2y = 0; critical point (2, −1); D = 3 > 0 and fxx = 2 > 0 at (2, −1), relative minimum. 12. fx = y − 3x2 = 0, fy = x − 2y = 0; critical points (0,0) and (1/6, 1/12); D = −1 < 0 at (0,0), saddle point; D = 1 > 0 and fxx = −1 < 0 at (1/6, 1/12), relative maximum. 13. fx = 2x − 2/ x2 y = 0, fy = 2y − 2/ xy 2 = 0; critical points (−1, −1) and (1, 1); D = 32 > 0 and fxx = 6 > 0 at (−1, −1) and (1, 1), relative minima. 14. fx = ey = 0 is impossible, no critical points. 15. fx = 2x = 0, fy = 1 − ey = 0; critical point (0, 0); D = −2 < 0 at (0, 0), saddle point. 16. fx = y − 2/x2 = 0, fy = x − 4/y 2 = 0; critical point (1,2); D = 3 > 0 and fxx = −4 > 0 at (1, 2), relative minimum. Exercise Set 15.8 560 17. fx = ex sin y = 0, fy = ex cos y = 0, sin y = cos y = 0 is impossible, no critical points. 18. fx = y cos x = 0, fy = sin x = 0; sin x = 0 if x = nπ for n = 0, ±1, ±2, . . . and cos x = 0 for these values of x so y = 0; critical points (nπ, 0) for n = 0, ±1, ±2, . . .; D = −1 < 0 at (nπ, 0), saddle points. 2 2 2 2 19. fx = −2(x + 1)e−(x +y...
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## This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.

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