M 0 0 y 0 1 m 1 m x2 z dv 881815 85 g xyz

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Unformatted text preview: 2 y0 b 2 z0 − z0 , i − c2 y0 / b2 z0 x0 x y0 y z0 z + 2+ 2= a2 b c x2 0 a2 j − k so the tangent plane is + 2 y0 z2 + 0 =1 b2 c2 50. ∂z/∂x = 2x/a2 , ∂z/∂y = 2y/b2 . At (x0 , y0 , z0 ) the vector 2x0 /a2 i + 2y0 /b2 j − k is normal 2 to the surface so the tangent plane is 2x0 /a2 x + 2y0 /b2 y − z = 2x2 /a2 + 2y0 /b2 − z0 , but 0 2 z0 = x2 /a2 + y0 /b2 so 2x0 /a2 x + 2y0 /b2 y − z = 2z0 − z0 = z0 , 2x0 x/a2 + 2y0 y/b2 = z + z0 0 51. n1 = fx (x0 , y0 ) i + fy (x0 , y0 ) j − k and n2 = gx (x0 , y0 ) i + gy (x0 , y0 ) j − k are normal, respectively, to z = f (x, y ) and z = g (x, y ) at P ; n1 and n2 are perpendicular if and only if n1 · n2 = 0, fx (x0 , y0 ) gx (x0 , y0 ) + fy (x0 , y0 ) gy (x0 , y0 ) + 1 = 0, fx (x0 , y0 ) gx (x0 , y0 ) + fy (x0 , y0 ) gy (x0 , y0 ) = −1. 52. n1 = fx i + fy j − k = x0 x2 0 + 2 y0 i+ y0 x2 0 + 2 y0 j − k; similarly n2 = − since a normal to the sphere is N = x0 i + y0 j + z0 k, and n1 · N = 2 n2 · N = − x2 + y0 − z0 = 0, the result follows. 0 x0 x2 0 + 2 y0 i− y0 x2 0 2 x2 + y0 − z0 = 0, 0 2 + y0 j − k; Exercise Set 15.6 548 EXERCISE SET 15.6 1. f increases the most in the direction of III. 2. The contour lines are closer at P , so the function is increasing more rapidly there, hence larger at P . 3. z = 4i − 8j 5. z= 6. z = e−5x sec x2 y 7. f (x, y ) = 3(2x + y ) x2 + xy 8. f (x, y ) = −x x2 + y 2 9. z = −4e−3y sin 4xi − 3e−3y cos 4xj f (x, y ) = [y/(x + y )]i + [y/(x + y ) + ln(x + y )]j, 10. 11. f is 4. y x i+ 2 j x2 + y 2 x + y2 2xy tan x2 y − 5 i + x2 tan x2 y j −3/2 2 i + 3x x2 + xy i − y x2 + y 2 f (x, y ) = 3y 2 tan2 x sec2 xi + 2y tan3 xj, −3/2 2 f (−1, −1) = −36i − 12j j, f (3, 4) = −(3/125)i − (4/125)j j, f (−3, 4) = 4i + 4j f (π/4, −3) = 54i − 6j f (x, y ) = (3y/2)(1 + xy )1/2 i + (3x/2)(1 + xy )1/2 j, √ √ Du f = f · u = 12/ 2 = 6 2 f (3, 1) = 3i + 9j, f · u = 32/5 12. f (x, y ) = 2ye2xy i + 2xe2xy j, 13. f (x, y ) = 2x/ 1 + x2 + y 14. f (x, y ) = − (c + d)y/(x − y )2 i + (c + d)x/(x − y )2 j, f (4, 0) = 8j, Du f = i + 1/ 1 + x2 + y j, √ f (0, 0) = j, Du f = −3/ 10 f (3, 4) = −4(c + d)i + 3(c + d)j, Du f = −(7/5)(c + d) f (2, 1) = 48i + 64j, u = (4/5)i − (3/5)j, Du f = 15. f (x, y ) = 12x2 y 2 i + 8x3 y j, 16. f (x, y ) = (2x − 3y )i + −3x + 12y 2 j, 17. f (x, y ) = y 2 /x i + 2y ln xj, 18. f (x, y ) = ex cos y i − ex sin y j, 19. f (x, y ) = − y / x2 + y 2 20. f (x, y ) = (ey − yex ) i + (xey − ex )j, f ·u=0 21. 22. √ √ f (−2, 0) = −4i + 6j, u = (i + 2j)/ 5, Du f = 8/ 5 √ √ f (1, 4) = 16i, u = (−i + j)/ 2, Du f = −8 2 √ √ √ f (0, π/4) = (i − j)/ 2, u = (5i − 2j)/ 29, Du f = 7/ 58 i + x/ x2 + y 2 j, √ √ f (−2, 2) = −(i + j)/4, u = −(i + j)/ 2, Du f = 2/4 √ √ f (0, 0) = i − j, u = (5i − 2j)/ 29, Du f = 7/ 29 f (x, y ) = (y/2)(xy )−1/2 i + (x/2)(xy )−1/2 j, f (1, 4) = i + (1/4)j, √ √ u = cos θi + sin θj = (1/2)i + 3/2 j, Du f = 1/2 + 3/8 f (x, y ) = [2y/(x + y )2 ]i − [2x/(x + y )2 ]j, f (−1, −2) = −(4/9)i + (2/9)j, u = j, Du f = 2/9 549 Chapter 15 23. f (x, y ) = 2 sec2 (2x + y )i + sec2 (2x + y )j, 24. f (x, y ) = cosh x cosh y i + sinh x sinh y j, √ √ f (π/6, π/3) = 8i + 4j, u = (i − j)/ 2, Du f = 2 2 f (0, 0) = i, u = −i, Du f = −1 25. f (1, 2) = 3, level curve 4x − 2y + 3 = 3, 2x − y = 0; f (x, y ) = 4i − 2j y f (1, 2) = 4i − 2j (1, 2) 4i – 2j 26. f (−2, 2) = 1/2, level curve y/x2 = 1/2, y = x2 /2 for x = 0. f (x, y ) = − 2y/x3 i + 1/x2 j f (−2, 2) = (1/2)i + (1/4)j x y 1 1 i+ j 2 4 (− 2, 2) x 27. f (−2, 0) = 4, level curve x2 + 4y 2 = 4, x2 /4 + y 2 = 1. f (x, y ) = 2xi + 8y j f (−2, 0) = −4i y 1 x 2 -4i 28. f (2, −1) = 3, level curve x2 − y 2 = 3. y f (x, y ) = 2xi − 2y j f (2, −1) = 4i + 2j 4i + 2j x (2, − 1) 29. f (x, y ) = 12x2 y 2 i + 8x3 y j, √ f (−1, 1) = 12i − 8j, u = (3i − 2j)/ 13, √ f (−1, 1) = 4 13 Exercise Set 15.6 550 30. f (x, y ) = 3i − (1/y )j, 31. f (x, y ) = x x2 + y 2 √ f (2, 4) = 3i − (1/4)j, u = (12i − j)/ 145, −1/2 i + y x2 + y 2 f (4, −3) = (4i − 3j)/5, u = (4i − 3j)/5, −1/2 33. f (x, y ) = −2xi − 2y j, 34. f (x, y ) = yexy i + xexy j; 35. f (x, y ) = −3 sin(3x − y )i + sin(3x − y )j, √ √ f (π/6, π/4) = (−3i + j)/ 2, u = (3i − j)/ 10, − 37. 38. 39. f (0, 2) = (1/2)i, u = i, f (0, 2) = 1/2 √ f (−1, −3) = 2i + 6j, u = −(i + 3j)/ 10, − √ f (−1, −3) = −2 10 √ f (2, 3) = e6 (3i + 2j), u = −(3i + 2j)/ 13, − x x+y x+y y i− j, 2 2 (x + y ) x−y (x + y ) x−y √ √ u = −(i − 3j)/ 10, − f (3, 1) = − 5/8 f (x, y ) = y (x + y )−2 i − x(x + y )−2 j, √ Du f = 1/ 5 145/4 f (4, −3) = 1 f (x, y ) = y (x + y )−2 i − x(x + y )−2 j, f (x, y ) = √ j, 32. 36. f (2, 4) = √ f (2, 3) = − 13e6 √ f (π/6, π/4) = − 5 √ f (3, 1) = ( 2/16)(i − 3j), −→ √ f (1, 0) = −j, P Q = −2i − j, u = (−2i − j)/ 5, f (x, y ) = −e−x sec y i + e−...
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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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