Since both of these quantities tend to it follows by

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.

Ask a homework question - tutors are online