X ln u y uv 7 x 17 x y u v 3u12 2v

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Unformatted text preview: he region then fx (x, y ) = fy (x, y ) = 0 throughout the region, the result follows from Exercise 51, Section 15.4. 66. Let u1 and u2 be nonparallel unit vectors for which the directional derivative is zero. Let u be any other unit vector, then u = c1 u1 + c2 u2 for some choice of scalars c1 and c2 , Du f (x, y ) = f (x, y ) · u = c1 f (x, y ) · u1 + c2 f (x, y ) · u2 = c1 Du1 f (x, y ) + c2 Du2 f (x, y ) = 0. EXERCISE SET 15.7 1. 165t32 4. 7. 1 − 512t5 − 2560t5 ln t √ 2t 1 + ln t − 512t5 ln t 2. 3 − (4/3)t−1/3 − 24t−7 3t − 2t2/3 + 4t−6 5. 3264 f (x, y, z ) = 20x4 y 2 z 3 i + 8x5 yz 3 j + 12x5 y 2 z 2 k, 3. −2t cos t2 6. 0 f (2, −1, 1) = 320i − 256j + 384k, Du f = −320 Exercise Set 15.7 8. 9. 10. 554 f (x, y, z ) = yzexz i + exz j + (xyexz + 2z ) k, f (0, 2, 3) = 6i + j + 6k, Du f = 45/7 4y 6z 2x i+ 2 j+ 2 k, 2 + 3z 2 2 + 3z 2 + 2y x + 2y x + 2y 2 + 3z 2 f (−1, 2, 4) = (−2/57)i + (8/57)j + (24/57)k, Du f = −314/741 f (x, y, z ) = x2 f (x, y, z ) = yz cos xyz i + xz cos xyz j + xy cos xyz k, √ √ √ f (1/2, 1/3, π ) = (π 3/6)i + (π 3/4)j + ( 3/12)k, Du f = (1 − π )/12 11. f (x, y, z ) = 3x2 z − 2xy i − x2 j + x3 + 2z k, √ √ u = (3i − j + 2k)/ 14, Du f = 72/ 14 12. i + j − z x2 + z 2 f (x, y, z ) = −x x2 + z 2 u = (2i − 2j − k)/3, Du f = 0 13. 14. −1/2 z−x y+x 1 i− j+ k, 2 z+y (z + y ) (z + y )2 u = (−6i + 3j − 2k)/7, Du f = −8/63 −1/2 k, f (−3, 1, 4) = (3/5)i + j−(4/5)k, f (1, 0, −3) = (1/3)i + (4/9)j + (1/9)k, f (x, y, z ) = − f (x, y, z ) = ex+y+3z (i + j+3k), Du f = (31/21)e−3 f (2, −1, 1) = 16i − 4j + 10k, f (−2, 2, −1) = e−3 (i + j+3k), u = (20i − 4j + 5k)/21, 15. √ f (1, 1, −1) = 3i − 3j, u = (i − j)/ 2, √ f (1, 1, −1) = 3 2 16. √ f (0, −3, 0) = (i−3j+4k)/6, u = (i−3j+4k)/ 26, 17. √ f (1, 2, −2) = (−i + j)/2, u = (−i + j)/ 2, 18. √ f (4, 2, 2) = (i − j − k)/8, u = (i − j − k)/ 3, 19. √ f (5, 7, 6) = −i + 11j − 12k, u = (i − 11j + 12k)/ 266, − 20. √ √ f (0, 1, π/4) = 2 2(i − k), u = −(i − k)/ 2, − 21. −→ √ √ f (2, 1, −1) = −i + j − k. P Q = −3i + j + k, u = (−3i + j + k)/ 11, Du f = 3/ 11 22. f (−1, −2, 1) = 13i + 5j − 20k, u = −k, Du f = 20 f (0, −3, 0) = √ 26/6 √ f (1, 2, −2) = 1/ 2 f (4, 2, 2) = √ 3/8 √ f (5, 7, 6) = − 266 f (0, 1, π/4) = −4 23. Let u be the unit vector in the direction of a, then f (3, −2, 1) cos θ = 5 cos θ = −5, cos θ = −1, θ = π so Du f (3, −2, 1) = f (3, −2, 1) · u = f (3, −2, 1) is oppositely directed to u; f (3, −2, 1) = −5u = −10/3i + 5/3j + 10/3k. √ √ 24. (a) T (1, 1, 1) = (i + j + k)/8, u = −(i + j + k)/ 3, Du T = − 3/8 √ √ (c) 3/8 (b) (i + j + k)/ 3 25. (a) f (x, y, z ) = x2 + y 2 + 4z 2 , f = 2xi + 2y j + 8z k, f (2, 2, 1) = 4i + 4j + 8k, n = i + j + 2k, x + y + 2z = 6 (b) r(t) = 2i + 2j + k + t(i + j + 2k), x(t) = 2 + t, y (t) = 2 + t, z (t) = 1 + 2t √ n·k 2 = √ , θ ≈ 35.26◦ (c) cos θ = n 3 555 Chapter 15 26. (a) f (x, y, z ) = xz − yz 3 + yz 2 , n = f (2, −1, 1) = i + 3k; tangent plane x + 3z = 5 (b) normal line x = 2 + t, y = −1, z = 1 + 3t (c) cos θ = 3 n·k = √ , θ ≈ 18.43◦ n 10 27. Set f (x, y ) = z + x − z 4 (y − 1), then f (x, y, z ) = 0, n = ± f (3, 5, 1) = ±(i − j − 19k), 1 (i − j − 19k) unit vectors ± 363 28. f (x, y, z ) = sin xz − 4 cos yz , f (π, π, 1) = −i − π k; unit vectors ± √ 1 (i + π k) 1 + π2 29. f (x, y, z ) = x2 + y 2 + z 2 , if (x0 , y0 , z0 ) is on the sphere then f (x0 , y0 , z0 ) = 2 (x0 i + y0 j + z0 k) is normal to the sphere at (x0 , y0 , z0 ), the normal line is x = x0 + x0 t, y = y0 + y0 t, z = z0 + z0 t which passes through the origin when t = −1. 30. f (x, y, z ) = 2x2 + 3y 2 + 4z 2 , if (x0 , y0 , z0 ) is on the ellipsoid then f (x0 , y0 , z0 ) = 2 (2x0 i + 3y0 j + 4z0 k) is normal there and hence so is n1 = 2x0 i + 3y0 j + 4z0 k; n1 must be parallel to n2 = i − 2j + 3k which is normal to the given plane so n1 = cn2 for some constant c. Equate corresponding components to get x0 = c/2, y0 = −2c/3, and z0 = 3c/4; substitute into the equation of the ellipsoid yields 2 c2 /4 + 3 4c2 /9 + 4 9c2 /16 = 9, √ √ √ √ c2 = 108/49, c = ±6 3/7. The points on the ellipsoid are 3 3/7, −4 3/7, 9 3/14 and √ √ √ −3 3/7, 4 3/7, −9 3/14 . 31. f (x, y, z ) = x2 + y 2 − z 2 , if (x0 , y0 , z0 ) is on the surface then f (x0 , y0 , z0 ) = 2 (x0 i + y0 j − z0 k) −→ is normal there and hence so is n1 = x0 i + y0 j − z0 k; n1 must be parallel to P Q = 3i + 2j − 2k so −→ n1 = c P Q for some constant c. Equate components to get x0 = 3c, y0 = 2c and z0 = 2c which when substituted into the equation of the surface yields 9c2 + 4c2 − 4c2 = 1, c2 = 1/9, c = ±1/3 so the points are (1, 2/3, 2/3) and (−1, −2/3, −2/3). 32. f1 (x, y, z ) = 2x2 + 3y 2 + z 2 , f2 (x, y, z ) = x2 + y 2 + z 2 − 6x − 8y − 8z + 24, n1 = f1 (1, 1, 2) = 4i + 6j + 4k, n2 = f2 (1, 1, 2) = −4i − 6j ...
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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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