The proof is similar for the function f x0 y exercise

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Unformatted text preview: ) all points not on the plane x + y + z = 0 z 21. z 22. 3 (0, 3, 0) y y x x z 23. z 24. x x y z 25. y z 26. (0, 0, 4) (0, 2, 0) x x y (2, 0, 0) y Exercise Set 15.1 526 z 27. z 28. (0, 0, 1) x (0, 1, 0) y (1, 0, 0) y x z 29. z 30. (0, 0, 1) (0, –1, 0) y y x x 31. y k = -2 32. k=01234 y k=2 k = -1 k=1 k=0 x x 33. y 34. y k=0 1234 x k k k k k = = = = = 2 1 0 –1 –2 x 527 Chapter 15 35. k = –1 y k = –2 2 y 36. k=0 k=1 k=2 k = -2 2 c o k=1 k=2 x -2 k = -1 k=0 x 2 -2 k=2 k=1 k = –2 k=0 k = –1 z 37. z 38. (0, 0, 2) y x (0, 4, 0) x (2, 0, 0) y z 39. z 40. (0, 0, 1) (0, − 1 , 0 ) 2 (0, 0, 3) x y y ( 1 , 0, 0 ) 4 x 41. concentric spheres, common center at (2,0,0) 42. parallel planes, common normal 3i − j + 2k 43. concentric cylinders, common axis the y -axis 44. circular paraboloids, common axis the z -axis, all the same shape but with different vertices along z -axis. 45. (a) f (−1, 1) = 0; x2 − 2x3 + 3xy = 0 (c) f (2, −1) = −18; x2 − 2x3 + 3xy = −18 (b) f (0, 0) = 0; x2 − 2x3 + 3xy = 0 Exercise Set 15.1 528 46. (a) f (ln 2, 1) = 2; yex = 2 (c) f (1, −2) = −2e; yex = −2e (b) f (0, 3) = 3; yex = 3 47. (a) f (1, −2, 0) = 5; x2 + y 2 − z = 5 (c) f (0, 0, 0) = 0; x2 + y 2 − z = 0 (b) f (1, 0, 3) = −2; x2 + y 2 − z = −2 48. (a) f (1, 0, 2) = 3; xyz + 3 = 3, xyz = 0 (c) f (0, 0, 0) = 3; xyz = 0 (b) f (−2, 4, 1) = −5; xyz + 3 = −5, xyz = −8 y 49. (a) (b) At (1, 4) the temperature is T (1, 4) = 4 so the temperature will remain constant along the path xy = 4. 4 T=1 T=2 T=3 x 4 50. V = 8/ 16 + x2 + y 2 16 + x2 + y 2 y = 8/V x + y = 64/V − 16 2 2 2 20 the equipotential curves are circles. 51. (a) f (x, y ) = 1 − x2 − y 2 , because f = c is a circle of radius radii in (a) decrease as c increases. √ V = 2.0 V = 1.0 V = 0.5 x 10 20 1 − c (provided c ≤ 1), and the (b) f (x, y ) = x2 + y 2 because f = c is a circle of radius c, and the radii increase uniformly. √ (a) is the contour plot of f (x, y ) = 1 − x2 − y 2 , because f = c is a circle of radius 1 − c (provided c ≤ 1), and the radii in (a) decrease as c increases. √ (c) f (x, y ) = x2 + y 2 because f = c is a circle of radius c and the radii in the plot grow like the square root function. 52. (a) III 53. (a) A (d) decrease (b) IV (c) I (b) B (e) increase (d) II (c) increase (f ) decrease 54. (a) Calgary, since the contour lines are closer together near Calgary than they are near Chicago. (b) The change in atmospheric pressure is about ∆p ≈ 1012 − 999 = 13, so the average rate of change is ∆p/1600 ≈ 0.0081. 529 Chapter 15 55. (a) (b) 2 1 -4 -5 4 5 -3 56. (a) -3 (b) 10 -10 10 10 -10 10 -10 -10 z 57. (a) (b) 2 1 0 x y -1 -2 -2 y 06 58. (a) ci 5 (b) o -1 0 1 2 o l i f z0 c -5 4 3 2 1 x 0 9 6 3 0 0 1 2 3 4 59. (a) The graph of g is the graph of f shifted one unit in the positive x-direction. (b) The graph of g is the graph of f shifted one unit up the z -axis. (c) The graph of g is the graph of f shifted one unit down the y -axis and then inverted with respect to the plane z = 0. Exercise Set 15.2 530 z 60. (a) x y (b) If a is positive and increasing then the graph of g is more pointed, and in the limit as a → +∞ the graph approaches a ’spike’ on the z -axis of height 1. As a decreases to zero the graph of g gets flatter until it finally approaches the plane z = 1. EXERCISE SET 15.2 y 1. y 2. y 3. x x x 5 -1 y=x y 4. 5. 1 y x y 6. x x y = 2x + 1 y 7. xy = –1 8. xy = 1 x xy = 1 9. all of 3-space y xy = –1 x 531 Chapter 15 10. all points inside the sphere with radius 2 and center at the origin 11. all points not on the cylinder x2 + z 2 = 1 12. all of 3-space 13. 35 14. π 2 /2 15. −8 16. e−7 17. 0 18. 0 19. (a) Along x = 0 lim (x,y )→(0,0) (b) Along x = 0, 3 3 = lim 2 does not exist. y →0 2y x2 + 2y 2 lim (x,y )→(0,0) x+y 1 = lim does not exist. y →0 y x + y2 1 x 1 → +∞ as x → 0 so the original = lim does not exist because x→0 x x2 x limit does not exist. 20. (a) Along y = 0 : lim x→0 (b) Along y = 0 : lim x→0 1 does not exist, so the original limit does not exist. x 21. Let z = x2 + y 2 , then sin x2 + y 2 sin z =1 = lim+ x2 + y 2 z (x,y )→(0,0) z →0 22. Let z = x2 + y 2 , then 1 − cos x2 + y 2 1 − cos z sin z = lim =0 = lim+ x2 + y 2 z (x,y )→(0,0) z →0 z →0 + 1 23. Let z = x2 + y 2 , then 24. With z = 25. 26. lim lim lim e−1/(x 2 (x,y )→(0,0) +y 2 ) = lim+ e−1/z = 0 z →0 √ 1 1 1 w , lim √ e−1/ z ; let w = √ , lim w = 0 x2 + y 2 z→+∞ z z w→+∞ e lim x2 + y 2 x2 − y 2 = lim x2 − y 2 = 0 x2 + y 2 (x,y )→(0,0) lim x2 + 4y 2 x2 − 4y 2 = lim x2 − 4y 2 = 0 x2 + 4y 2 (x,y )→(0,0) (x,y )→(0,0) (x,y )→(0,0) 0 x2 = lim 0 = 0; along y = x : lim 2 = lim 1/5 = 1/5 x→0 3x2 x→0 x→0 5x x→0 so the limit does not exist. 27. along y = 0 : lim 28. Let z = x2 + y 2 , then 1 − x2 − y 2 1 − z2 = lim = +∞ so the limit does not exist. x2 + y 2 z2 (x,y )→(0,0) z →0 + lim 29. 8/3 31. Let t = 30. ln 5 x2 + y 2 + z 2 , then lim (x,y,z )→(0,0,0) sin x2 + y 2 + z 2 x2 + y 2 + z 2 = lim t→0+ sin t2 =0 t Exercise Set 15.2 532 x2 + y 2 + z 2 , lim 32. With t = t→0+ sin t cos t = +∞ so the limit...
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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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