13 - CHAPTER 13 Functions of Several Variables Section 13.1...

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Unformatted text preview: CHAPTER 13 Functions of Several Variables Section 13.1 Introduction to Functions of Several Variables . . . . . . . 160 Section 13.2 Limits and Continuity Section 13.3 Partial Derivatives . . . . . . . . . . . . . . . . . . . 168 . . . . . . . . . . . . . . . . . . . . . 176 Section 13.4 Differentials . . . . . . . . . . . . . . . . . . . . . . . . . 188 Section 13.5 Chain Rules for Functions of Several Variables . . . . . . 194 Section 13.6 Directional Derivatives and Gradients . . . . . . . . . . . 205 Section 13.7 Tangent Planes and Normal Lines . . . . . . . . . . . . . 214 Section 13.8 Extrema of Functions of Two Variables . . . . . . . . . . 227 Section 13.9 Applications of Extrema of Functions of Two Variables . . 236 Section 13.10 Lagrange Multipliers . . . . . . . . . . . . . . . . . . . . 247 Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 256 Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 267 CHAPTER 13 Functions of Several Variables Section 13.1 Introduction to Functions of Several Variables 1. x 2z yz xy 10 x2 y 10 xy 10 x2 xy y z z 2. x z 2 y2 2 xy 4 No, z is not a function of x and y. For example, x, y 1, 0 corresponds to both z ± 2. Yes, z is a function of x and y. 3. x2 4 y2 9 z2 4. z 1 z No, z is not a function of x and y. For example, x, y 0, 0 corresponds to both z ± 1. 5. f x, y 8 8 0 x ln y Yes, z is a function of x and y. x y 3 2 (a) f 3, 2 (b) f x ln y 1 4 1, 4 6. f x, y x2 4 (d) f 5, y 4y 2 (e) f x, 2 5 y (b) f 0, 1 4 0 4 (c) f 2, 3 4 4 36 4 1 4y 36 2 3 5e0 (b) f 3, 2 0 3e 2 (c) f 2, 4y 1 (e) f x, 0 4 x (f) f t, 1 4 t2 8. g x, y ln x 0 4 x 5e y (e) f x, 2 xe 2 te t xy z 9. h x, y, z y (a) g 2, 3 ln 2 3 ln 5 (b) g 5, 6 ln 5 6 ln 11 (c) g e, 0 ln e 0 1 (d) g 0, 1 ln 0 1 0 (e) g 2, (f) g e, e 3 ln 2 3 ln e e ln 2 ln e 0 23 9 2 3 (b) h 1, 0, 1 ln 1 (a) h 2, 3, 9 10 1 0 (c) h 2, 3, 4 ln 2e (d) h 5, 4, 160 2 e 1 2e (d) f 5, y 2 t2 4 5 2 (f) f t, t 2 5 t xe y (a) f 5, 0 4 x 2 (f) f 5, t 6 7. f x, y (a) f 0, 0 (d) f 1, y 30 5 (c) f 30, 5 ln 2 1 6 23 4 54 6 3 2 10 3 S ection 13.1 10. f x, y, z x z 0 (b) f 6, 8, 5 3 4 6 (c) f 4, 6, 2 (d) f 10, 11. f x, y y (a) f 0, 5, 4 4 4, Introduction to Functions of Several Variables 8 12 4 (b) f 3, 1 23 3 2 sin 4 11 2 10 x sin y (a) f 2, 3 6 3 3 (c) f 3 2 4 3 sin 1 3, (d) f 4, 3 sin 3 4 sin 2 2 13. g x, y 4 y 2t t2 3 dt 3 2 10 90 (b) V 5, 2 52 (c) V 4, 8 42 8 (d) V 6, 4 62 4 144 y 14. g x, y x 1 dt t 12 16 12 1 3 12 9 4 9 2 3 2, 4 16 (d) g 0, 3 2 9 4 (c) g y ln t 16 (b) g 1, 4 50 ln y ln x ln x ln 1 4 ln 4 (c) g 2, 5 (b) g 6, 3 ln 3 6 ln 2 (d) g (a) x2 fx (a) x, y x f x, y fx f x, y x y y f x, y x2 f x, y 9 4 5 2 7 12 ln x 2 2x x2 2y x ln 14 x x 2y 2 x2 2y 2y x 2x 2y x 2x x y y x2 x2 y2 x 3xy y2 y2 x 3xy y2 y y 2y 2y 2y y x2 y2 3xy x, y x f x, y 3x 3xy (b) ln 1 ,7 2 x 16. f x, y 3 2 6 25 4 2y x2 (b) 3 4 y x (a) g 4, 1 15. f x, y x2 3y x (a) g 0, 4 128 2 y2 3t x (a) V 3, 10 33 2 3 2 3 3 y r 2h 12. V r, h y y f x, y 3x y xy 3 xy y 2 3 xy x 3y, x y2 3xy y 3xy y 3x 3x y 2y y y2 y 2 2y y y y 3x y, y 2y 161 3xy 0 y2 0 x, x 2y 2y y 0 2, y 0 3x 162 Chapter 13 17. f x, y Functions of Several Variables x2 4 y2 18. f x, y 2 Domain: 4 19. f x, y 4y ≥ 0 4y ≤ 4 y2 ≤ 4 x, y : x 2 4y 2 x2 y2 ≤ 4 x x2 4 x2 y ≥0 x2 Domain: 4 2 2 Range: 0 ≤ z ≤ 2 x, y : Range: y2 ≤1 1 x2 4 y Domain: x, y : 1 ≤ x 2 x2 4 arcsin x 2 y≤1 ≤z≤ 2 y2 ≤1 1 Range: 0 ≤ z ≤ 2 20. f x, y arccos y x 21. f x, y ln 4 Domain: 4 Domain: y 1≤ ≤1 x x, y : x x x y>0 x y 24. z xy x, y : x Domain: 0 and y x 4 x, y : xy > 6 x2 y2 27. g x, y x, y : x is any real number, Domain: Domain: y is any real number 29. f x, y x2 4x y2 (c) View from the first octant: 20, 15, 25 (d) View from the line y x in the xy-plane: 20, 20, 0 z 5 5 28. g x, y 0 and y x, y : y ≥ 0 Range: all real numbers x, y : x is any real number, 2 2≤z≤2 (b) z 0 when x y-axis. 0 which represents points on the (c) No. When x is positive, z is negative. When x is negative, z is positive. The surface does not pass through the first octant, the octant where y is negative and x and z are positive, the octant where y is positive and x and z are negative, and the octant where x, y and z are all negative. 6 2x 3y z 6 Domain: entire xy-plane Range: <z< 2 4 y 4 x 0 xy Domain: 0 Range: 32. f x, y Plane 4 4 x, y : y Range: z > 0 1 xy x, y : x y y is any real number (b) View where x is negative, y and z are positive: 15, 10, 20 Plane: z Domain: y 30. (a) Domain: 1 (a) View from the positive x-axis: 20, 0, 0 31. f x, y x, y : x Range: all real numbers except zero Range: z ≥ 0 ex 25. f x, y Range: all real numbers Range: all real numbers 26. f x, y Range: all real numbers y Domain: 0 6>0 xy > 6 xy x 6 xy Domain: Range: all real numbers 23. z ln xy y<4 x, y : y < Range: 0 ≤ z ≤ 22. f x, y y x 3 2 3 4 y S ection 13.1 y2 33. f x, y 1 2x 1 2x 34. g x, y Since the variable x is missing, the surface is a cylinder with rulings parallel to the x-axis. The generating curve is z y 2. The domain is the entire xy-plane and the range is z ≥ 0. Introduction to Functions of Several Variables Plane: z 35. z x2 4 Paraboloid Domain: entire xy-plane z Range: z ≤ 4 4 3 2 z −4 4 x 4 −2 −3 −4 z 5 y2 4 y 4 −3 2 3 y 3 x 1 2 4 y 3 x 1 2 Cone x2 36. z y2 37. f x, y e x Since the variable y is missing, the surface is a cylinder with rulings parallel to the y-axis. The generating curve is z e x. The domain is the entire xy-plane and the range is z > 0. Domain of f : entire xy-plane Range: z ≥ 0 z 2 z 2 1 1 2 y 3 Range: z ≥ 0 z 25 20 15 8 3 xy, x ≥ 0, y ≥ 0 0, elsewhere Domain of f : entire xy-plane 38. f x, y 10 5 6 y x 4 5 x 2 4 y 4 x y2 39. z x2 1 40. f x, y 1 12 144 16x 2 Hyperbolic paraboloid 41. f x, y Semi-ellipsoid Domain: entire xy-plane 9y 2 x 2e xy 2 Domain: set of all points lying on or inside the ellipse x2 9 y 2 16 1 <z< Range: z z Range: 0 ≤ z ≤ 1 z y 4 x y x x 4 −2 −4 42. f x, y x sin y z 4 −4 −4 4 x y −4 4 y 163 164 Chapter 13 x2 43. f x, y (a) Functions of Several Variables y2 xy, x ≥ 0, y ≥ 0 44. f x, y (a) z z 5 25 4 20 15 10 5 −2 1 2 y y 2 5 x (b) g is a vertical translation of f three units downward. (b) g is a vertical translation of f two units upward. (c) g is a reflection of f in the xy-plane. (c) g is a horizontal translation of f two units to the right. The vertex moves from 0, 0, 0 to 0, 2, 0 . (d) The graph of g is lower than the graph of f. If z f x, y is on the graph of f, then 1 z is on the graph 2 of g. (d) g is a reflection of f in the xy-plane followed by a vertical translation 4 units upward. (e) (e) z z 5 z 25 5 4 x 4 20 15 10 5 y z = f (x, 1) z = f (1, y) 2 2 2 x y 5 2 x 45. z e1 x y x2 y2 2 2 e1 x y c 47. z ln c 1 y2 1 2 2 e1 x y c x2 ln c y2 ln c 1 x2 y2 1 c y2 x2 ln y ± ec ln c x2 x2 ln y Level curves: Level curves: Level curves: x2 x2 y2 e1 46. z y x2 x 2 ± ec y Circles centered at 0, 0 Hyperbolas centered at 0, 0 Parabolas Matches (c) Matches (d) Matches (b) 48. z cos 2y 2 x 49. z 4 cos x 2 1 c 2y Ellipses Matches (a) 2 y 50. f x, y Level curves are parallel lines of the form x y c. Level curves: c x cos x2 2y 2 y 4 x2 4 2x 3y The level curves are of the form 6 2x 3y c or 2x 3y 6 c. Thus, the level curves are straight lines with a slope of 2 . 3 y 2y 2 2 4 4 cos 6 3 1 c x −2 2 4 c=4 −2 c=2 c = −1 c=0 x −2 c = 10 c=8 c=0 c=2 c=4 c=6 S ection 13.1 51. f x, y x2 25 y2 x2 25 c x2 y2 y 2, c=5 c=4 c=3 c=2 −6 2y 2 x2 2y 2 c except x 2 2y 2 0 is the point 0, 0 . x −2 2 6 −2 Thus, the level curves are circles of radius 5 or less, centered at the origin. 165 The level curves are ellipses of the form 2 c 2. 25 x2 52. f x, y y 6 The level curves are of the form Introduction to Functions of Several Variables y c=1 c=8 c=6 c=4 c=2 c=0 3 c=0 −6 x −3 3 −3 53. f x, y xy e xy 54. f x, y The level curves are hyperbolas of the form xy c. 2 y y 1 −1 e xy c=6 c=5 c=4 c=3 c=2 c=1 x c = −1 c = −2 c = −3 c = −4 c = −5 c = −6 1 −1 x2 56. f x, y y2 x y2 c 2 y2 c= x c 2 −2 ln x y c = −2 1 2c c=1 ln x x 6 x −4 y y x 12 . 2c y2 1 c = −2 y ec c=2 0 2 2 c=1 4 c=1 3 −2 y 2 −3 2 1 c=1 2 −1 x e c. c=1 c= 3 c = −2 −6 c=2 c= 3 c = ±2 Thus, the level curves are parallel lines of slope 1 passing through the fourth quadrant. 2 c = −1 c=0 1 c= 2 c = −1 2 x x −1 y c = −1 x x2 1 xy . 2 The level curves are of the form The level curves are of the form c c, or ln c Thus, the level curves are hyperbolas. x 55. f x, y 2 c=4 c=3 c=2 2 The level curves are of the form 1 2 Thus, the level curves are circles passing through the origin and centered at ± 1 2c, 0 . 57. f x, y x2 y2 58. f x, y 2 6 xy 59. g x, y 8 x2 1 y2 4 4 −9 9 3 sin x y 1 −1 −6 6 −4 1 −1 6 −4 −6 60. h x, y −6 61. See Definition, page 884. 62. The graph of a function of two variables is the set of all points x, y, z for which z f x, y and x, y is in the domain of f. The graph can be interpreted as a surface in space. Level curves are the scalar fields f x, y c, for c, a constant. 166 Chapter 13 Functions of Several Variables 63. No, The following graphs are not hemispheres. x2 y2 z e z x2 x y 64. f x, y The level curves are the lines y2 c x or y y 1 x. c These lines all pass through the origin. 65. The surface is sloped like a saddle. The graph is not unique. Any vertical translation would have the same level curves. One possible function is f x, y 66. The surface could be an ellipsoid centered at 0, 1, 0 . One possible function is f x, y x2 xy . 67. V I, R 1 1000 0.10 1 1I R y 2 1 1. 4 10 68. A r, t 1000e rt Number of Years Inflation Rate 0 0.03 2593.74 1929.99 0.28 2004.23 1491.34 0.35 1877.14 1396.77 1152.40 69. f x, y, z x 2y 3z 70. f x, y, z 6 c 4 4x y 20 1105.17 1121.40 1349.86 1491.82 1221.40 1491.82 1822.12 2225.54 1349.86 1822.12 2459.60 3320.12 1491.82 2225.54 3320.12 4953.03 6 2y 3z c y x2 y2 z2 z2 Sphere z z z 4 3 3 y2 9 9 2z Plane Plane x2 71. f x, y, z 2z 4 x c 4x 15 0.06 1230.42 10 0.04 1592.33 5 0.02 0.05 0 Rate 0.08 Tax Rate 2 −3 y 2 3 −4 1 4 y 4 x y 4 x 6 −4 x c 12 4y x2 72. f x, y, z z x2 12 4y z2 z2 74. f x, y, z 4x 2 sin x z or z sin x z 2 z 1 z 0 0 sin x Elliptic cone Elliptic paraboloid Vertex: 0, 0, 4y 2 0 0 z 4y 2 c c 1 1 4x 2 73. f x, y, z 2 4 z x −2 5 −2 2 x 3 x 5 y 1 2 y 8 y S ection 13.1 d 75. N d, L 2 4 L 4 30 x y 12 243 board-feet 12 507 board-feet 1 (a) W 15, 10 (b) W 12, 9 15 2 4 4 1 600 0.75x 2 0.75y 2 c = 600 c = 500 c = 400 The level curves are of the form c x 2 y2 600 0.75x 2 f x, y f 2x, 2y 600 c . 0.75 C x a y1 100 2x 2y 0.6x 0.6 10 min 1 1 hr 2 2 5 x2 25 30 min y y2 c= 1 2 c= 1 3 c=1 25 x − 25 x 25 30 − 25 − 30 0.4 2 0.4y 0.4 100 2 0.6 2 0.4x 0.6y 0.4 a 2 100x 0.6y 0.4 2f x, y 81. C ln C a ln x ln z ln y ln C ln C a ln z y 1 hr 6 4 − 30 0.6 ln z ln 30 78. V x, y 6 100x 0.6 y 0.4 100 2 80. z c = 300 c = 200 c = 100 c=0 20 min 0.75y 2 The level curves are circles centered at the origin. 79. y 1 hr 3 1 4 12 min 9 12 (d) W 4, 2 77. T 1 hr 5 10 12 (c) W 12, 6 167 ,y<x 2 4 4 (b) N 30, 12 1 76. W x, y 22 (a) N 22, 12 Introduction to Functions of Several Variables 1 a ln x 0.75xy a ln y base a ln y 0.75xy 2 0.40 xz 2 0.40 yz front & back two ends 0.80 xz yz x y x z y 82. V r 2l 43 r 3 r2 3l 3 83. PV 4r kT, 20 2600 r (a) k 20 2600 300 (b) P kT V l k 300 520 3 520 T 3V The level curves are of the form: c V 520 3 T V 520 T 3c Thus, the level curves are lines through the origin 520 with slope . 3c 168 Chapter 13 84. (a) z Functions of Several Variables f x, y 0.156x 0.031y 1.66 Year 1998 1999 2000 2001 2002 2003 z 18.5 21.1 25.8 31.3 35.1 39.3 Model 18.16 21.36 26.26 30.60 34.91 39.42 (b) x has the greater influence because its coefficient 0.156 is larger than that of y 0.031 . (c) f x, 55 0.156x 0.031 55 1.66 0.156x 0.045 This function gives the shareholder’s equity in terms of net sales x, assuming a constant total assets of y 85. (a) Highest pressure at C 55 (billion). 86. Southwest (b) Lowest pressure at A (c) Highest wind velocity at B 88. (a) The different colors represent various amplitudes. 87. (a) The boundaries between colors represent level curves. (b) No, the level curves are uneven and sporadically spaced. (b) No, the colors represent intervals of different lengths, as indicated in the box (c) You could use more colors, which means using smaller intervals. 90. True 89. False. Let f x, y 2xy f 1, 2 f 2, 1 , but 1 2. 92. True 91. False. Let f x, y 5. Then, f 2x, 2y Section 13.2 1. lim x, y → 2, 3 x 22 f x, y . 5 Limits and Continuity 2. f x, y x, L > 0, there exists a -neighborhood about 2, 3 such that We need to show that for all f x, y L From 0 < 2, 3 lies in the neighborhood. x 2 So, choose 2. Let 2< x whenever x, y 2 2 y 3 2 < it follows that x whenever 0 < x Then if 0 < x x 2 2 ≤ x and the limit is verified. > 0 be given. We need to find x 2 x 4 2 < 4<. a 4 2 2 b y y > 0 such that f x, y 1 2 2 < x 4 2 , we have y L 1 2 x 4< < . Take . 2 2 y 3 2 <. S ection 13.2 3. lim x, y → 1, y 3 3. f x, y y, L f x, y L From 0 < 1, x So, choose x Then if 0 < a x 2 a < it follows that y b y 9. y 2 b 2 2 < y 2 > 0 such that f x, y y b < . Take lim f x, y lim 4 f x, y g x, y x, y → a, b x, y → a, b g x, y lim x, y → a, b f x, y lim g x, y x, y → a, b x, y → a, b lim f x, y g x, y lim f x, y g x, y f x, y x, y → a, b x, y → a, b lim x, y → 2, 1 lim x, y → 2, 4 lim x, y → 0, 1 lim x, y → y 2 y 3 0 0 2 <. b< lim x, y → a, b 3y 2 x 2 45 3 f x, y x, y → a, b lim x, y → a, b g x, y 5 3 lim x, y → a, b f x, y lim x, y → a, b 2 31 5 x x y y 2 2 g x, y lim x, y → a, b 53 g x, y 1, 2 e xy 4 4 lim 10. 0 14. 0, x y ≤ 1 1 e2 2 5 lim x, y → 0, 0 5x y 1 x 1 xy 11 Continuous for x y > 0 lim x, y → 1, 1 lim x, y → 4, 2 y cos xy 2 cos Continuous everywhere 16. lim x, y → 1, 1 xy x2 y2 1 2 Continuous except at 0, 0 x y z Continuous for x y z≥0 x, y, z → 1, 2, 5 3 5 12. 3 arcsin 0 2 5 1 Continuous everywhere 1, y e 15 f x, y y arcsin x y 1 xy 2 20 3 Continuous everywhere 17. 1 . f x, y lim 4 Continuous for xy 15. x b<. Continuous for x 13. ≤ 2 L Continuous everywhere 11. 3 , we have lim 8. 3 < y 7. 2 3 and the limit is verified. whenever 0 < 6. y > 0 be given. We need to find 4. Let 5. 3 lies in the neighborhood. 2 1 3 such that 3< y whenever x, y 3 > 0, there exists a -neighborhood about 1, We need to show that for all Limits and Continuity 8 22 18. lim x, y, z → 2, 0, 1 xe yz 2e0 Continuous everywhere 2 2 2 2 0 1 169 170 Chapter 13 Functions of Several Variables 19. The limit does not exist because along the line y you have lim x, y → 0, 0 x x2 y y lim x, 0 → 0, 0 x x2 0 20. The limit does not exist because along the line x you have 1 x lim x, 0 → 0, 0 which does not exist. 21. lim x, y → 1, 1 x2 lim y2 x, x → 0, 0 x x2 lim x2 x, x → 0, 0 Since the denominator is 0, the limit does not exist. 1 1 xy 1 1 xy x lim x, y → 0, 0 y 1 1 0 22. The limit does not exist because along the line x you have lim x, y → 0, 0 x x y y3 lim 0, y → 0, 0 y y3 lim 0, y → 0, 0 0 1 y2 which does not exist. 23. The limit does not exist because along the path x lim x, y, z → 0, 0, 0 xy x2 yz y2 whereas along the path x lim x, y, z → 0, 0, 0 xy x2 xz z2 y yz y2 lim 0, 0, z → 0, 0, 0 0, y 0 z2 0 x2 x2 x2 x2 z, you have xz z2 lim x, x, x → 0, 0, 0 24. The limit does not exist because along the path y yz2 y2 xz2 z2 However, along the path z 0, x yz2 xz2 y2 z2 lim x, y, z → 0, 0, 0 lim x, y, z → 0, 0, 0 25. lim x, y → 0, 0 e xy xy x2 xy x2 0, you have lim x, 0, 0 → 0, 0, 0 z 0 x2 x2 x2 1 0, you have 0 y, you have lim x, x, 0 → 0, 0, 0 x2 x2 x2 1 2 26. f x, y 1 x x2 1 y2 1 x2 x2 1 y2 1 2 Continuous everywhere lim x, y → 0, 0 Continuous everywhere 27. lim x, y → 0, 0 ln x2 y2 ln 0 28. lim x, y → 0, 0 1 cos x 2 y 2 x2 y2 The limit does not exist. The limit does not exist. Continuous except at 0, 0 Continuous except at 0, 0 0 0 10 1 0 x 0. S ection 13.2 29. f x, y Limits and Continuity xy x2 y2 Continuous except at 0, 0 Path: y 0 0.5, 0 0.1, 0 0.01, 0 0.001, 0 0 0 0 0 0 x, y 1, 1 0.5, 0.5 0.1, 0.1 0.01, 0.01 0.001, 0.001 f x, y x 1, 0 f x, y Path: y x, y 1 2 1 2 1 2 1 2 1 2 The limit does not exist because along the path y the function equals 1 . 2 30. f x, y 0 the function equals 0, whereas along the path y x y x2 y2 Continuous except at 0, 0 Path: y x 0.5, 0.5 0.1, 0.1 0.01, 0.01 0.001, 0.001 1 2 1 5 50 500 x, y 1, 0 0.5, 0 0.1, 0 0.01, 0 0.001, 0 f x, y 0 1, 1 f x, y Path: y x, y 0 0 0 0 0 The limit does not exist because along the path y the function tends to infinity. 0 the function equals 0, whereas along the path y x xy 2 31. f x, y x2 y4 Continuous except at 0, 0 Path: x y2 x, y 1, 1 y2 1 2 1, 1 x, y 0.25, 0.5 1 2 f x, y 2x 2x 2 0.0001, 0.01 1 2 0.000001, 0.001 1 2 0.01, 0.1 1 2 The limit does not exist because along the path x the function equals 1 . 2 32. f x, y 0.01, 0.1 1 2 f x, y Path: x 0.25, 0.5 1 2 0.0001, 0.01 1 2 0.000001, 0.001 1 2 y 2 the function equals 1 2, 1 2 whereas along the path x y2 y2 y Continuous except at 0, 0 Path: y 0 0.25, 0 0.01, 0 0.001, 0 0.000001, 0 1 4 100 1000 1,000,000 x, y 1, 1 0.25, 0.25 0.01, 0.01 0.001, 0.001 0.0001, 0.0001 f x, y x 1, 0 f x, y Path: y x, y 1 3 1.17 1.95 1.995 2.0 The limit does not exist because along the line y the function tends to 2. 0 the function tends to infinity, whereas along the line y x 171 172 33. Chapter 13 lim x, y → 0, 0 Functions of Several Variables f x, y x2 lim 2xy 2 y 2 x2 y 2 x, y → 0, 0 lim x, y → 0, 0 2xy2 x2 y2 1 lim 34. x, y → 0, 0 Hence, 1 4x2y2 x2 y 2 lim x, y → 0, 0 0 f x, y lim x, y → 0, 0 g x, y 0. f is continuous at 0, 0 , whereas g is not continuous at 0, 0 . (same limit for g) Thus, f is not continuous at 0, 0 , whereas g is continuous at 0, 0 . 35. lim x, y → 0, 0 sin x sin y 0 36. lim x, y → 0, 0 sin 1 x cos 1 x 37. x, y → 0, 0 x2y x 4y 2 4 Does not exist Does not exist z lim z z 2 y x y 4 6 x x 6 y 38. x2 lim y2 10xy 2x2 3y2 39. f x, y x2y x, y → 0, 0 Does not exist 40. f x, y The limit does not exist. Use the paths x 0 and x y. z x2 2xy y2 1 The limit equals 0. z 18 z 5 x 4 4 x y y x 41. 42. 43. 44. lim x, y → 0, 0 sin x 2 y 2 x2 y2 xy 2 lim lim r→0 lim x, y → 0, 0 x 2 y2 y3 y2 r→0 lim x, y → 0, 0 lim x 2y 2 y2 x, y → 0, 0 x 2 45. x r cos , y lim x, y → 0, 0 lim 2r cos r 2 2r sin3 r→0 r 4 cos2 sin2 r→0 r2 r→0 lim r cos3 r2 lim r 2 cos2 sin2 lim lim 1 0 r→0 r 3 cos3 x2 lim cos r 2 r→0 lim r cos sin2 r2 r sin , x2 y2 x2 y2 lim r→0 r 2 sin2 r cos r→0 x3 x2 sin r 2 r2 r→0 y2 r, x 2 r 2 cos2 sin2 r y2 r 2 cos2 lim r cos 2 r→0 sin3 0 0 sin2 sin2 0 5 5 y S ection 13.2 x2 46. y2 lim x, y → 0, 0 47. x 2 r sin x2 y 2 x2 y 2 y2 Limits and Continuity sin r r lim r→0 1 r2 lim x, y → 0, 0 x2 y 2 ln x 2 y2 lim r 2 ln r 2 lim 2r 2 ln r r→0 r→0 By L’Hôpital’s Rule, lim 2r 2 ln r lim r→0 48. x 2 y2 r→0 2r 2 r3 lim r→0 50. f x, y, z z y2 x2 lim 1 r→0 y2 cos r 2 r2 1 x2 y2 z2 Continuous except at 0, 0, 0 0, let z sin z z sin z ex ey 1 ft lim z→0 3x f g x, y f 3x 3x 9x 2 2y 2y ft g x, y f g x, y 1 ft 1 t x2 f g x, y 12xy y2 f x2 y2 1 4y 2 x2 58. 2y Continuous for y 3x 2 ft g x, y 2y f 3x y2 Continuous except at 0, 0 1 t 3x sin z z y 2. Then 2 Continuous everywhere 57. x2 implies that f is continuous for all x, y. 56. 2y y 2, let z g x, y t2 g x, y y 2, the function is clearly continuous. For x2 xy. Then xy sin z Continuous everywhere 54. For x2 implies that f is continuous for all x, y. 55. 52. f x, y, z Continuous everywhere 9 0, the function is clearly continuous. For xy 0 0 51. f x, y, z 9 Continuous for x 2 53. For xy r2 49. f x, y, z cos x 2 y 2 x2 y2 1 x, y → 0, 0 lim lim r→0 r2 lim z→0 2 ln r 1 r2 1 3x 2y f g x, y 1 4 x t 2 f x2 Continuous for x 2 y2 y2 y2 4 4 1 x2 y2 173 174 Chapter 13 x2 59. f x, y (a) lim Functions of Several Variables 4y fx x, y x x→0 f x, y x lim x x→0 f x, y y y y→0 f x, y x2 lim x2 (a) lim fx x, y x f x, y 4y f x, y y y y→0 y→0 x x fx x→0 x f x, y xy x2 lim x, y x f x, y y (b) lim y→0 f x, y y 2y y 2 x2 y y2 2 lim 2y y→0 y f x, y 2x lim x y y x xy x→0 f x, y 2x xy lim 2 x 2x lim 3y 2x xy 3y xy 3y x y x→0 xy y 2 3y y y y→0 2x y xy y→0 fx x→0 y lim x→0 lim 2x 3y y→0 (a) lim 2y y2 2 y→0 x→0 yy x2 x lim 62. f x, y y 4 y2 x 2x x y→0 2x 2 x→0 lim (b) lim 2x 4y 4 lim x→0 (a) lim 2x x x2 y y 4y y lim lim 61. f x, y x x→0 y2 x→0 (b) lim lim 2x y→0 y→0 4y 2 x lim 60. f x, y x2 4y x 2x x lim (b) lim 2 x x→0 3y lim x 3 y→0 y x 3 1 x, y x f x, y y y f x, y yy lim 1 x→0 yy y y 32 y y lim y 32 y3 y y y→0 lim y→0 31 y 2 1 0 x 2 y 1 y 2 12 12 y3 2 lim y 2 y1 y y→0 12 2 y1 2 y (L’Hôpital’s Rule) 3y 1 2y 63. True. Assuming f x, 0 exists for x 0. 64. False. Let f x, y See Exercise 29. xy x2 y2 . S ection 13.2 65. False. Let Limits and Continuity 175 66. True ln x2 0, f x, y y2 , x, y 0, 0 . x 0, y 0 See Exercise 27. 67. See the definition on page 897. Show that the value of 68. See the definition on page 854. lim f x, y is not the same x, y → x0 , y0 for two different paths to x0, y0 . 69. No. 70. No, f 2, 3 can equal any number, or not even be defined. The existence of f 2, 3 has no bearing on the existence of the limit as x, y → 2, 3 . 71. x2 lim y2 x, y → 0, 0 xy (a) Along y ax: x2 lim x, ax → 0, 0 If a ax x ax x2 : 2 x2 1 a2 lim x→0 ax2 a2 1 a ,a x If a 0, 0 and the limit does not exist. x2 lim 2 x2 2 lim x x2 x, x → 0, 0 x→0 x x yz y2 x2 sin lim cos sin sin lim →0 75. As x, y → 0, 1 , x 2 1 tan lim lim x, y → a, b 0< x lim x x 1 x2 f x, y a x, y → a, b 0< Let x4 2x 4 2 z2 2 2 0 1 2 → 0. y 76. lim x, y → 0, 0 lim r cos f x, y r→0 2 y Therefore, lim x, y → a, b 2 2 b 2 < y b 1 and L1 f x, y 2 < 2. . Hence, define f 0, 0 r 2 sin2 r2 cos2 sin2 0 0. 2 > 0, there corresponds 1 > 0 such that f x, y L1 < 2 whenever 2 > 0, there corresponds 2 > 0 such that g x, y L2 < 2 whenever 1. L2, then for 2 g x, y 1 L1, then for be the smaller of f x, y 1 y g x, y a r→0 r 2 cos2 r sin lim r 2 cos sin x, y → 0, 1 Since x2 x2 x2 1 →0 1 → 1 and x 2 2 a2 0. a2 1 y2 x2 Thus, 77. Since x4 1 lim tan cos cos sin cos x2 x2 1 2 tan lim x, y, z → 0, 0, 0 2 sin2 ax 2 (c) No, the limit does not exist. f approaches different numbers along different paths. 74. z2 →0 x2 ax ax ax lim x, ax → 0, 0 x4 2x 4 lim 2 (c) No, the limit does not exist. Different paths result in different limits. lim x4 x 2 : f x, x2 (b) y x2 1 x, x x, y, z → 0, 0, 0 y2 0 Limit does not exist. 73. 4 ax: f x, ax (a) y 0, then y (b) Along y x2 y 72. f x, y 2. By the triangle inequality, whenever L2 g x, y f x, y L1 L1 L 2. g x, y L2 x a 2 ≤ f x, y y L1 b 2 < , we have g x, y L2 < 2 2 . 176 Chapter 13 Functions of Several Variables 78. Given that f x, y is continuous, then a > 0 such that f x, y x 0< Let 2 a y b 2 f a, b f a, b ⇒ 2 < Section 13.3 > 0, there corresponds . ⇒ f a, b 2 < f x, y neighborhood since f a, b 2 f a, b < 3 1 f a, b < f x, y < f a, b < 0. 2 2 Partial Derivatives 2. fy 1. fx 4, 1 < 0 5. f x, y f a, b < 0, which means that for each whenever < < f x, y 2, then f x, y < 0 for every point in the corresponding f a, b f x, y f a, b lim x, y → a, b 1, 2 <0 3y 6. f x, y 5 x2 fx x, y 2x 4. fx 3y 2 2x 7 1, 8. z 3. fy 4, 1 > 0 2y2 x 1 2 fy x, y 3 x2 9. z z x 5x xe x xx e y y xe x z ln z x z y 17. y y y x y x2 x e y2 ln x 1 y y ln x x 1 2y x2 y 1 y x z x2 2y 2x 2y 4y 2 x2 2x x2 y y2 4y 2 x z x y2 z y x2 2y 2 8y x y z x y2 x x 2y z y 4y x 1 z x 2xe 2y z y 2x 2e 2y z z ln z x 1y 2 xy 1 2x y2 z y 1x 2 xy 1 2y ln x 2 x2 x2 8xy y2 14. 2x 2y z ln x2 z x x2 y2 z y 16. x2 4y 3 xy 1 ln xy 2 xy y2 2y y2 1 18. f x, y x3 x 2e 2y 11. z y2 3y 2 z y 1 1 y x ex x y2 y x x x y y 4y 2 z x 13. ex 4xy 2 z y y z y 15. z x 6y z y3 10. z xy z x 6y 5y z x 12. 3y2 5xy 2x z y fy x, y z z y fx x, y 7. 2 fx x, y x 3 16y 3 2xy 2 fy x, y 2x 2x x2 y2 xy x2 y2 x2 y2 y xy 2x x2 y2 2 y3 x2 x 2y y2 2 x2 y2 x xy 2y x2 y2 2 x3 x2 xy 2 y2 2 0 S ection 13.3 19. h x, y x2 y2 e 20. g x, y hx x, y 2xe 2ye x2 y2 1 ln x 2 2 y2 gx x, y 21. f x, y x2 y2 fx x, y 12 x 2 y2 12 fy x, y 12 x 2 y2 12 y2 y 2y x 2 y 1 2y 2 x2 y2 x2 2 24. y 2x 1 2x 2 y3 1 2x 2 y3 12 z 3 cos 3x cos 3y tan 2x 2 sec2 2x y z x z y sec 2 2x y z y e y sin xy z x ye y cos xy z y ey sin xy 26. z ey x cos xy cos x 2 y2 z y 2y sin x 2 y2 1 dt y y2 y3 3 t x x2 1 1 y x3 3 2t 2x lim 2t 1 dt x x y y 2 dt 2t fy x, y 2 2 3y fx f x, y x, y x f x, y y y f x, y lim 2x x x→0 lim y→0 2x 3y 3y x 2x y y 2x 3y lim 2x x 2 lim 3y y 3 x→0 3y y→0 2y x x fx x, y 1 dt y 2t x2 1 dt 2t y 1 x→0 y→0 1 dt y x [You could also use the Second Fundamental Theorem of Calculus.] f y x 28. f x, y x fx x, y lim y2 y t3 3 f x 3y 2 2 2x y 3 2x sin x 2 x 29. f x, y 3y 2 y3 sin xy t2 fy x, y 1 2x 3 sin 3x sin 3y y 27. f x, y 2 z x xey cos xy y2 sin 3x cos 3y z z y2 y 12 z x 25. x y3 f x, y f y x x2 x2 f x 22. 2x 1 2x 2 x2 y2 gy x, y 23. y2 x2 y2 hy x, y x2 ln Partial Derivatives 2x 177 178 Chapter 13 30. f x, y f x x2 Functions of Several Variables y2 2xy fx lim x, y x x→0 x lim x x 2 y f x, y xy x y2 y y 2x x→0 f y f x, y lim y y y→0 x2 lim 2 f x 2x y fx y 2 x2 x, y x f x, y x lim x y x x→0 x x x→0 f x, y y y y→0 f x, y x y x y y→0 y y 2y x x y y 1 2x y x y x y y y x x 1 y y x y y y x y x x y y y 1 2x y x y y y fx x, y x f x, y y y x→0 f x, y f x, y lim y→0 arctan fx x, y 2 : fx 2, 1 2 : fy 2, lim x 1 x x→0 lim y→0 y x x 1 y 1 y x x y y x y lim x→0 x 2 y x2 y x2 y lim y→0 x y 1 yx y 1 yx y arccos xy fx x, y y2 y 1 x2y2 At 1, 1 , fx is undefined. 1 4 1 1 y2 x2 x 1 2 4 x fy x, y x x2 y2 x 1 y 2 1 34. f x, y 1 y2 x2 1 fy x, y At 2, 2x 1 x lim At 2, x x x y lim y x y y y→0 y→0 33. f x, y 2y y x 1 y x x lim lim f y 2x x x lim f x 2y x y x→0 32. f x, y lim y2 2xy y→0 lim lim x x→0 y x→0 f y lim 2x y x lim y2 2xy f x, y y→0 31. f x, y x2 1 x x2y2 At 1, 1 , fy is undefined. x 1 y 2 S ection 13.3 xy 35. f x, y x yx fx x, y y xy x At 2, 2 : fx 2, y gx x, y fx x, y 2 x y fy x, y 2 1 4 y2 x2 38. h x, y hx x, y At 2 At 1, 1 : gy 1, 1 2 39. z 2, 1 e x z x 4 cos y x e 2, 1 2 z y cos 2x z x At 2 sin 2x At y z ,, 43 x z y 1 sin 2x y 45 At 2, 3, 6 : 1 2 6 x z y sin y 0 2, 45 y2 y z y 1 6 y sin y2, x Intersecting curve: z 2 sin sin 2x z ,, 43 y x2 41. z 49 2, 3, 6 y 1 e At 0, 0 : 40. z cos y z x At 0, 0 : 2y 2, 1 : hy 32 8 9 2x 2, 1 : hx At 10 9 24x 3 5y 2 4x 2 y2 hy x, y 2y 32 30 27 At 1, 1 , fy 1, 1 2x gy x, y 30y 3 5y 2 x2 xy At 1, 1 : gx 1, 1 4x 2 At 1, 1 , fx 1, 1 2 y 2 x2 4 y2 y 1 4 x 2 : fy 2, 37. g x, y x 2 xx fy x, y At 2, 2 y 179 6xy 4x 2 5y 2 36. f x, y y Partial Derivatives y2 z y 1 2 3 45 9 z x=2 10 8 8 x x2 42. z 4y 2, y 1, 2, 1, 8 Intersecting curve: z z x x2 4 43. z 9x2 z x z x 22 4 44. z 3, 1, 3, 0 9x 2 Intersecting curve: z 2x At 2, 1, 8 : y 2, y y 9 9x2 z y z x 18 1 z 18 At 1, 3, 0 : z y 9 23 z z 40 160 4 2 4 y 4 x 4 x 3 4 y y2 2y y=3 20 1, 1, 3, 0 Intersecting curve: z 18x At 1, 3, 0 : y 2, x x 4 y 6 180 Chapter 13 45. fx x, y fx 2x Functions of Several Variables 0: 2x fy 4, fy x, y 4y 4y 4x 4x 2y 16 fx 4 2y 6 and y fy 9x 2 12y, fy x, y 16 3y 2 12x 12y 0 ⇒ 3x 2 12x 0: 9x2 0⇒ 3y 2 4y y2 4x Solving for x in the second equation, x obtain 3 y 2 4 2 4y. Solving for x and y, x 46. fx x, y 4. 64y ⇒ y 0 or y 4 31 3 ⇒x 3y4 0 or x 1 16 4 32 3 Points: 0, 0 , 1 x2 47. fx x, y 1 y2 y, fy x, y 1 x2 fy 0: 1 y2 x y4 ⇒ y y y 1 0 and y fx 1 and x x2 1 y2 4 4 , 32 3 31 3 48. fx x, y x 0 y 2 4, you x2 2x y2 1 fy x, y x2 2y y2 1 0⇒x 0 0⇒y 0 Points: 0, 0 x Points: 1, 1 (b) The graph is that of fy. 52. w 3xz x w x x y 3z xy w y 3xz 3yz xy 3xz x y2 2 z2 x2 x y2 z2 x2 y y2 z2 x2 z y2 z2 51. 2 x2 y2 z2 1 ln x 2 2 53. F x, y, z y y2 w z (b) The graph is that of fx. x2 w x 50. (a) The graph is that of fx. w w y 49. (a) The graph is that of fy. y2 z2 ln x x y2 z2 Fy x, y, z 3x x2 x2 y y2 z2 Fz x, y, z w z Fx x, y, z x2 z y2 z2 y 1 54. G x, y, z x2 1 z2 x y2 z x2 y y2 z2 32 x2 z y2 z2 32 Gx x, y, z 1 x Gy x, y, z 1 Gz x, y, z 1 2 232 55. H x, y, z sin x 2y 3z Hx x, y, z y2 cos x 2y 3z Hy x, y, z 2 cos x 2y 3z Hz x, y, z 3 cos x 2y 3z S ection 13.3 3x 2 y 56. f x, y, z fx x, y, z 6xy fz x, y, z 3x 2 57. 5yz 3x 2 fy x, y, z 10yz 2 5xyz fx x, y, z 10z 2 5xz 5xy y2 fx 1, 20yz 2, 1 fy x, y, z fy 1, 2, 1 fz x, y, z fz 1, f x, y, z 58. fx x, y, z fx 3, 1, 1 fy x, y, z fy 3, 1, 1 fz x, y, z xy y x x zy x y 1 y y z y x z0 yz 3 9 x2y3 2xyz 2xy 3 y 4 2 3x2 2y y2 2 5 xz z 2 x 2z2 35 5 3 5 2 3x y2 2z2 3x2 y y2 2z2 2z2 3x2 2z y2 2z2 25 5 2 3x2 2 5 4z y2 25 5 y z cos x y fx 0, , 2 4 4 cos z cos x fy 0, , 2 2 xy yz xy 2yz 2, 1, 2 fy x, y, z 2, 1, 2 4 4 3x 2y 2 12 fz x, y, z fz 23 3x2 4 fz 0, , 2 4 4 cos 0 2 y 0 2 sin x y sin 1 2 2 1 3 fx x, y, z fy 6 z sin x fz x, y, z f x, y, z fx x 2z2 2 2 3 1 fz 3, 1, 60. x2 xy 2 3 32 x x 6x y2 fy x, y, z zx x 9 z yz yz 3x2 2 f x, y, z 0 32 x y2 xy 2 2xy 2, 1, 2 4 0 2xz 8 3z 6 2 3y 3 7 61. z x2 2xy z x 3yz 2x 2y 2z x2 2z y2 2z xy z x4 3x 2y 2 z x 4x 3 6xy 2 12x 2 6y 2 x2 2z yx 2 z y 62. 2z 2 2z yx 3y 2 2x 6y 12xy z y 6x 2 y 4y 3 6x 2 12y 2 2z 6 y2 2 181 2z2 fx x, y, z z y 1 59. 2, 1 Partial Derivatives 2z xy 12xy y4 182 Chapter 13 63. z x2 z x x2 y2 x2 z y y2 x2 x2 32 xy y2 2 z xy 1 z 2xe y z x 2e y 2z x2 2z yx z y 2z y2 2z xy x 3ye y 2 2z yx y 2 1 x y x 2z y2 1 x y 2 y 2 z yx 2 67. z z x 2xe y z y x 3e 2e y 3e x2 x2 y2 2 x 1 x x2 y2 2 x 2x y2 2 y2 x2 x2 y2 2 x sec y z x y2 y2 x2 y 2y y2 z sec y 69. 2 sin x 2y 2 sin x 2y z x2 0 2 z yx 2 cos x 4 sin x 2y z y 2y sec y tan y x sec y tan y 2 2 xy 2xy y2 x2 y x2 y2 2 1 y2 x2 1 xy 2 2z y2 x2 2y z y2 y x2 2 x2 2z x cos x z y x2 y2 z x yx 2xy y2 2z 2xe y 2y 2z x2 y x 1 y2 x2 yx sin x z x2 1 2z x z 68. arctan 2z 3ye e x sec2 y 2 z . xy x2 2e y 2e x sec2 y tan y 2z xy 1 x x 3ye e x sec2 y 1 y x 3ye e x sec2 y z y 1 x Therefore, 66. e x tan y y x 2 z y2 32 x2 y2 2z xy z y y 2z y2 2 z yx e x tan y 65. 1 2 z x2 32 xy y2 x2 x e x tan y 2z x2 y z z x ln x 32 y2 y2 x2 z z x y2 2z yx 64. x 2z x2 Functions of Several Variables z y2 x sec y sec2 y tan2 y 2 2 sin x 2y z xy sec y tan y Therefore, 2z 2z yx xy . There are no points for which zx z sec y 0. x 0 zy , because S ection 13.3 z 9 x2 y2 z x 70. 9 x x2 y2 2z 9 y2 x2 9 x2 x2 9 y2 2z z y 9 9 y2 32 yx x2 y2 y2 4x 2y 2 y 4 x2 y2 2 x2 xy y . 4xy y2 2 2y x2 y2 2 y2 x2 x2 y2 2 x2 4xy y2 2 There are no points for which zx 0 xy x 2 x fy x, y, z x y x y 2 y2 2 x y4 2y x xx y x 2x 2 xy x y 2 x y 0 0 0 fyxy Therefore, fxyy 2 x2 2 x y4 2x x y There are no points for which zx f x, y, z x2 3xy fx x, y, z 2x 3y fyy x, y, z 2 3 xy fy x, y, z 2xy x y3 z fxyy x, y, z 1 x2 xy y y z fyyx x, y, z 3 0 fyx x, y, z 2y 2 xz fxy x, y, z 2 yz fyxy x, y, z y2 y xy y xyz fyy x, y, z y x f x, y, z fx x, y, z 73. y yx z y 74. x2 y2 ln x 2 y2 x x2 2x 2z xy 2z yx 2z 2z y2 32 0 if x 2z y2 ln x y2 x2 x4 z y 2z z y 2z x2 2z y2 x2 Therefore, 2z z x2 xy xy x2 2 32 9 y2 2z z x 1 x yx x2 x2 9 y2 z x ln z x 32 y2 y x2 2z 72. z xy yx z x 71. Partial Derivatives 3x zy z3 4yz 0. 75. f x, y, z e fx x, y, z fy x, y, z 4z fyyx 2xy x y3 x sin yz x e x ze sin yz cos yz fyy x, y, z 0 z2e x sin yz fxy x, y, z 3 fxy x, y, z ze x cos yz fyx x, y, z 3 fyx x, y, z ze x cos yz fyyx x, y, z fyyx x, y, z 0 2 ze x sin yz sin yz sin yz fxyy x, y, z 0 fxyy x, y, z z2e x fyxy x, y, z 0 fyxy x, y, z z2e Therefore, fxyy fyxy fyyx 0. Therefore, fxyy x fyxy fyyx. 0. zy 0. 183 184 76. Chapter 13 Functions of Several Variables 2z f x, y, z x fx x, y, z 2z xy fy x, y, z 2z xy 5y 2 z x 2 2z x2 3 z y x y 2z 4z fxy x, y, z x y x y z y2 2z x2 y2 0 0 0. ey e y 2 e e sin x y e y y 2 Therefore, 2z x2 2z y2 sin x 12z x y4 fyxy x, y, z 2z y 2 2 2 0 y e ey ey sin x 12z x y4 fxyy x, y, z 12z x y4 z sin x 0. e x sin y z x e x sin y 2z x2 z y 2z y2 z z t 2z t2 z x 2z x2 80. z arctan y x From Exercise 67, we have 2z ex 2z x2 sin y y2 x2 2xy y2 2 x2 2xy y2 e x cos y e x sin y Therefore, 81. z y 5x e 2 sin x x2 3 fyyx x, y, z cos x 2z 0 Therefore, 4z fyx x, y, z 79. y2 3 sin x 78. 5xy 4z fyy x, y, z ey z z x z 77. y 2z x2 sin x 2z y2 ct 0. z ct t 2z t 2 16c 2 cos 4x 2 c2 4 sin 4x 2z x2 2z x 2 . 2z t 2 c2 16 cos 4x 2z 2x2 x 4ct 4ct 2z x2 ct c2 ct x 2 1 x ct x 1 ct x c2 ct 2z t2 ct c z t2 z x 4ct 16 cos 4x c2 4ct ln x z t 4ct 2 z x ct z 83. 4ct 4c sin 4x 2z ct sin x cos 4x z t ct c 2 sin x cos x e x sin y 82. c cos x Therefore, e x sin y c2 2z x2 2 2 2 0. ey e y 2 ey e 2 y S ection 13.3 84. 85. z sin wct sin wx z t z wc cos wct sin wx 2z z x e z x 1 e c 2z w sin wct cos wx 1 e c2 x 2z x2 w 2 sin wct sin wx 2z Therefore, 2z c2 t2 cos z t w 2c 2 sin wct sin wx t2 t e 2 t z t 86. z t x c sin t z cos c2 2z x c x2 2z x 1 e c sin t x c cos 1 e c2 Therefore, . 2 t e t sin z x x c e z t t x c x c sin x c c2 2z x2 . 88. If z f x, y , then to find fx you consider y constant and differentiate with respect to x. Similarly, to find fy , you consider x constant and differentiate with respect to y. z (x0, y0, z 0 ) 90. The plane z (x0, y0, z 0 ) x y f x, y satisfies f f < 0 and > 0. x y z 4 2 y x y x 4 Plane: x = x0 Plane: y = y0 2 4 y x f denotes the slope of the surface in the x-direction. x f denotes the slope of the surface in the y-direction. y 91. The plane z x y 92. In this case, the mixed partials are equal, fxy z f x, y satisfies 6 f f > 0 and > 0. x y fyx. See Theorem 13.3. y 8 x −6 93. (a) C C x C x C y C y 32 xy 16 175x y x 1 4 16 16 x y 1050 (b) The fireplace-insert stove results in the cost increasing at a faster rate because C > y 175 183 205 16 4 80, 20 205y 175 80, 20 185 . x2 87. See the definition on page 906. 89. cos z t Therefore, x c Partial Derivatives 205 237 C . x 186 Chapter 13 200x0.7y0.3 94. f x, y (a) f x 140x At x, y (b) Functions of Several Variables f x 0.3y0.3 140 f x 1000, 500 , 60x0.7y At x, y 0.7 0.3 y x x y 60 0.3 140 1 2 60 2 0.7 0.3 113.72 0.7 f x 1000, 500 , 500 1000 140 60 0.7 1000 500 97.47 95. An increase in either price will cause a decrease in demand. 1 V I, R 1000 VI I, R 96. 10,000 VI 0.03, 0.28 0.10 1 1I 1 10 R 0.10 1 1I R 0.10 1 1I R 9 1 0.10 1 R 1 I2 10,000 1 0.10 1 1I R 10 11 14,478.99 VR I, R 10,000 VR 0.03, 0.28 1 9 0.10 1I 1000 1 0.10 1 1I R 9 10 1391.17 The rate of inflation has the greater negative influence on the growth of the investment. (See Exercise 67 in Section 12.1.) 97. T 0.6x 2 500 1.5y 2 98. A T x 1.2x, T 2, 3 x 2.4 m T y 3y T 2, 3 y 9m (a) 0.885t A t 22.4h 0.885 0.544 1.20h A 30 , 0.80 t A h 1.20th 0.885 22.4 1.20 0.80 1.845 1.20 30 13.6 1.20t A 30 , 0.80 h 22.4 (b) The humidity has a greater effect on A since its coefficient 22.4 is larger than that of t. PV 99. P mRT ⇒ V P V V P V PV ⇒ mR mRT ⇒ P V T mR P mRT V2 mR P V T V mR T P mRT mRT x y 6y 17 and Uy 2, 3 16. The person (c) Ux 2, 3 should consume one more unit of y because the rate of decrease of satisfaction is less for y. (d) z 1 −2 x mRT VP 10x (b) Uy mRT V2 3y 2 xy (a) Ux V mR T T P 5x 2 100. U mRT 1 2 1 2 y S ection 13.3 101. z 0.04x z x (a) 0.64y (b) As the consumption of skim milk x increases, the consumption of whole milk z decreases. As the consumption of reduced-fat milk increases, the consumption of whole milk increases. 0.64 1.3520x 2 102. z z x (a) 0.0025y 2 2.704x 56.080x 1.537y 562.23 z x2 x2 <0 The rate of increase of Medicare expenses z is declining with respect to worker’s compensation expenses x . 2.704 z y 2z (b) Concave downward 56.08 2 0.005y 2 2z The rate of increase of Medicare expenses z is slightly declining with respect to public assistance expenses. 103. False 104. True x y x2 (a) fx x, y x, y x, y 0, 0 y 2 3x 2y x2 (b) fx 0, 0 lim fy 0, 0 lim y 2 x3 f y3 x3 y y 0, 0 x f y 0, 0 xy 3 2y x x4 x2 lim 0 2 x x→0 f 0, 0 lim 0 y y→0 2 0 fx 0, 0 fy x, 0 x→0 y x x x→0 y t 3 dt 1 x By the Second Fundamental Theorem of Calculus, f x d dx f y d dy y 1 t 3 dt x d dx x 1 y y 1 x t 3 dt 1 y 3. y lim fy 0, 0 x t 3 dt 1 y4 2 y y→0 (d) fyx or fxy or both are not continuous at 0, 0 . 108. f x, y 4x 2 y 2 y2 y4 2 0 lim y lim 4x 2y 2 y2 0 y fx 0, y lim 0 x y→0 y f x fyx 0, 0 f 0, 0 f 0, y y x4 x2 x y→0 xy 3 2x y2 2 3xy 2 x3y 2 22 x y x, 0 x→0 (c) fxy 0, 0 106. True 0, 0 x2 fy x, y 105. True 1. xy x 2 y 2 , x2 y2 0, 107. f x, y z <0 y2 (c) Concave downward 1.537 0.005 y2 Let z 187 3.4 0.04 z y Partial Derivatives x3 22 4 y x 22 lim y→0 1 1 4 x lim 1 x→0 1 188 Chapter 13 Functions of Several Variables x3 109. f x, y y3 (a) fy 0, 0 13 f 0, 0 lim y y y→0 y→0 (b) fy x, y y2 y3 x3 y2 For x, y f 0, 0 22 x 3 1 For x, y 23 fails to exist for y x, x 0. 23 0, 0 , fx x, y y y lim x2 110. f x, y y2 fx 0, 0 f0 lim z dz z 3x 2y 3 9x 2 y 2 dy 1 2 dx y2 2 x dx 2 x2 5. 2y y2 x2 x2 x4 x z x cos y lim x x→0 13 0 x z y2 2 1 dw dy y 2y z 2y z z dx 2x dy 2y 2 x z y dz 2y 2 y cos x cos y y sin x dx 1 x2 e 2 6. z dz z 2x ex 2 y2 e y2 e x sin y cos x dy x2 y2 dx 2 2y ex 2 y2 e 2 e x cos y dy dw 2z x sin y x2 y2 dy ex 2 e x2 y2 x dx w e y cos x z2 2z3 6z 2 y sin x dy sin x dz x 2 yz 2 dw 2xyz 2 4 f 1.05, 2.1 z x 2z 2 dx (b) dz 2x dx 2 0.05 f 1, 2 0.5125 2y dy 4 0.1 5 z (b) dz 0.5 z cos yz dy 2.2361 f 1.05, 2.1 3.4875 2z dz y cos yz dz 12. (a) f 1, 2 f 1.05, 2.1 e y cos x dy sin yz 2x 2yz 11. (a) f 1, 2 y dy e y sin x dx 10. w cos x dx cos x dy y2 dw 2z 3 y sin x 3y y sin x dx 8. e x sin y dx w cos y x2 y2 e x sin y dz 9. 3 y dy dz 7. f 0, 0 x x2 dy y2 2x dx y 4. w y2 2x dz 1 3. x2 y 2. z 6xy 3 dx x2 4x y2 Differentials dz 3. 3 x2 x x→0 x→0 1. 2x 0, 0 , use the definition of partial derivative. lim Section 13.4 13 5.5125 2.3479 0.11180 x x2 y2 dx x dx y dy x2 y2 y x2 0.05 y2 dy 2 0.1 5 0.11180 S ection 13.4 13. (a) f 1, 2 sin 2 f 1.05, 2.1 z 1.05 sin 2.1 f 1.05, 2.1 (b) dz sin y dx 0.00293 z x cos y dy 1 2 3 dx 1.05 2.1 f 1.05, 2.1 4 dy 3 0.05 z 4 0.1 0.25 (b) dz e 2 0.1 x2 y 2, x 2 3.1 5.05 x2 1 18. Let z 2 2.03 1 2 1 52 x dy y2 1 dx y 3 22 1 9 2 3, y 32 1 y2 , x 2 0.95 6, dx y 2 1, dx sin 2 3 9 32 dA A l dh 32 21 0.05 63 l h dl A dA dl h 2 1 cos 12 12 0.05 l dh lh ∆A dA h dh 9 2 0.1 y2 dy 0.094 y 3 dx 2x dx y2 3x 2 1 y 2 dy 0 2 1 x2 dy y3 0.012 2x cos x 2 0.05. Then: dz 0.05, dy f x, y at the point P ∆h h dl 1 x2 0 2 1 cos 12 12 y 2 dx 2y cos x 2 0.05 0 22. In general, the accuracy worsens as 23. The tangent plane to the surface z is a linear approximation of z. lh 2 2x 1 1 0.1 4 y dx 0.55 34 0.05. Then: dz 21. See the definition on page 916. 25. A y2 3 0.1 52 32 0.03 0.05, dy 23 0.05 62 62 sin x 2 22 1 x2 0.1. Then: dz 0.03, dy 3 x 0.1. Then: dz 5 0.05 52 32 9, dx x 2 y 2, x sin 1.05 0.05, dy 32 2, y 8.9 3.05 5.95 2 20. Let z 3, dx y 3, x 1 19. Let z 5, y 0.5 0 1 0.05 2 17. Let z 1.1084 0.5 5.25 0.25 (b) dz xe y dy e 2 0.05 16. (a) f 1, 2 f 1.05, 2.1 z e y dx 0.00385 5 8.5745 1.1854 (b) dz cos 2 0.1 dA dA 189 7.3891 1.05e 2.1 f 1.05, 2.1 f 1, 2 sin 2 0.05 15. (a) f 1, 2 e2 14. (a) f 1, 2 Differentials y 2 dy x and y increase. f x, y , then z dz is the propagated error, z dz and is the relative error. z z 24. If z 26. V dV r 2h 2 rh dr π r 2dh 2 r dh ∆V − dV ∆h 2πrhdr l ∆l ∆r dl dh dl dh r 190 Chapter 13 r 2h 3 V 27. Functions of Several Variables r r 2 h h r 2 6 h r r2 V V dV 0.1267 0.1 2.8274 2.8264 0.0010 0.002 0.0565 0.0566 0.0001 0.0002 0.0019 0.0019 0.0000 54 h2 3 3 28. S dV r2h 3 4.8391 0.001 r dh 4.7124 0.0001 r 2h dr 3 r V r r2 dh 3 2 rh dr 3 dV 0.1 0.1 6 h 0.1 3 h r 8, h 20 20 dS dr r2 h2 12 r 2 h2 r 2 h2 dS dh r r2 S 8, 20 29. z h2 12 2r 2 r2 r2 h2 h2 8 rh r 2 h2 r2 2r 2 h2 dr 10.0341 10.0768 0.0427 0.1 5.3671 5.3596 0.0075 0.001 rh dh 0.1 0.1 dh h2 h 0.1 rh r 0.002 0.12368 0.12368 0.683 10 5 0.0001 1 2h h2 dr h2 r2 h2 12 h2 2r 2 r2 dS r2 r2 dS 0.0002 0.00303 0.00303 0.286 10 7 541.3758 0.04x (a) dz 0.64y 0.04 dx z dx x (b) dz 3.4 0.64 dy z dy y 0.04 ± 0.25 0.64 ± 0.25 0.01 ± 0.16 Maximum error: ± 0.17 Relative error: dz z ± 0.17 0.04 6.2 ± 0.17 0.64 7.5 3.4 7.952 0.21 or 2.1% 8.5, 3.2 , dx ≤ 0.05, dy ≤ 0.05 30. x, y x2 r y 2 ⇒ dr x x2 y2 y dx x2 8.5 dx 8.52 3.22 dr ≤ 1.288 0.05 y arctan x y x2 y x ⇒d 1 y y2 Using the worst case scenario, dx ≤ 0.00194 dy 3.2 dy 8.52 3.22 0.9359 dx 0.3523 dy 0.064 x2 d y2 0.00515 2 1 x dx y x 1 dx x x2 y2 dy 0.05 and dy 0.0071. 2 dy 8.52 3.2 dx 3.22 0.05, you see that 8.5 8.52 3.22 dy S S dS S ection 13.4 31. r 2h ⇒ dV V dV V 2 dr r dh h 2 0.04 32. 1 2 ab A 1 2 dA 33. C 0.02 4 sin 45 3 sin 45 ab cos C dC 1 ± 16 35.75v0.16 0.6215T 12 cos 45 ± 0.02 ± 0.24 in.2 0.4275Tv0.16 0.4275v0.16 0.6215 C v 10% a sin C db 1 ± 16 35.74 C T 0.10 sin C b sin C da 1 2 r 2 dh 2 rh dr 0.84 5.72v 0.0684Tv 0.84 0.16 ±1 C dT T C dv v 0.6215 0.4275 23 ± 1.3275 1.1143 dC 5.72 23 0.84 0.84 0.0684 8 23 ±3 ± 2.4418 Maximum propagated error ± 2.4418 dC C 34. ± 0.19 or 19% 12.6807 v2 r a da da a 2 v2 dr r2 2v dv r dv v dr r 2 0.03 0.02 0.08 8% Note: The maximum error will occur when dv and dr differ in signs. 35. (a) V 1 bhl 2 18 sin b 2 18 cos 2 2 16 12 h 18 31,104 sin θ 2 in.3 18 18 sin ft3 V is maximum when sin (b) V s2 sin 2 dV 1 or 2. l s sin 18 sin l ds 2 1809 in3 s2 l cos 2 16 12 1 2 1.047 ft3 d s2 sin 2 dl 182 16 12 cos 2 2 90 182 sin 2 2 1 2 Differentials 191 192 Chapter 13 Functions of Several Variables 36. (a) Using the Law of Cosines: a2 b2 c2 2 a 420 ft 4202 12 b 2 da 12 4202 840b cos dP dP P 2E dE R 2E dE R E2 dR R2 12 2 P 1 R 1 R1 840b sin d 840 330 cos ± 1774.79 2 330 20 840 cos 20 6 840 330 sin 20 180 ± 8.27 ft dE E dR R 2 0.02 0.03 0.07 7% 1 R2 R1R 2 R1 R2 R1 0.5 dR2 R2 2 R L dL R dR R2 2 R dR R1 dR When R1 10 and R2 0.00021 ln dh h 0.00021 2 R22 R2 2 R1 R1 152 10 15 2 R12 R2 0.5 R2 2 102 10 15 2 2 0.14 ohm. 0.75 dr r 0.00021 ln 100 T R1 15, we have R 2h r L 40. db E2 dR R2 dR1 39. 840 cos E2 R R 38. 2b 12 4202 1 11512.79 2 P 9˚ 2b 420 cos 1 3302 2 37. 330 ft 2 330 420 cos 9 107.3 ft. b2 (b) a 2bc cos A 4202 330 0.00021 0.75 ± dL ± 1 100 ± 1 16 100 2 8.096 10 4 ± dL ± 6.6 8.096 10 10 4 6 ± 6.6 10 6 micro–henrys L g dg 32.23 dL 2.48 2.50 0.02 dT T dg g T dL L T When g 32.09 32.09 and L 0.14 g 2.50, T L dg g 32.09 Lg dL 2.5 0.14 32.09 2.5 32.09 0.02 0.0108 seconds. S ection 13.4 41. fx 2x y f x, y z x2 x, y z x2 2x 2x 2 42. z x x x z fx 1→0 y x x x2 z fx 5 x x2 y where 2 2 y y2 x and 1 y. 2 x 2y y 2x x yx x y 1 and x 3y 2 2x x 2 y 2 x x y y where 2 y x and 2 y 1 2 2x x x 2. y3 f x, y 10y y3 10 y 3y 2 y 10 y 0 fy x, y x y 2 → 0. 1→0 y 1 and 1→0 3x2y , y2 x, y 3y y x y 2 3 5x 10y y3 y where 2 y 0 and 1 3y y 2 y 2. 0, 0 x, y x 3y y 2 → 0. x, y → 0, 0 , 0, 0 x4 0, fy 0, 0 2 y 2 → 0. y x (a) fx 0, 0 and 10y 5x 45. f x, y 1 y x, y fx x, y As 0. 2 yy x2 y 5x 5x x fy x, y x f x, y 2 x x, y → 0, 0 , z x and 1 f x, y x fx x, y As 2y y y y x 2xy y where 2 2 → 0. y2 fy x, y 2x 2xy x and x x, y x2 y x2y f x, y z 2x 0y 1 2 2y y x x2 y f x, y x x, y → 0, 0 , As 44. y x fx x, y y y2 2x 2x x y 1→0 x y x fy x, y x, y 2 2 x y x2 fx 2x 2 x f x, y z 43. 2 x, y → 0, 0 , As 2 x x fx x, y f x, y x x 2x y lim f x, 0 f 0, 0 x→0 lim lim y→0 f 0, 0 lim 0 x 0 y y→0 y 0 4 x→0 x f 0, y 0 x 0 2 0 y Thus, the partial derivatives exist at 0, 0 . (b) Along the line y Along the curve y x: lim x, y → 0, 0 x2: lim f x, y x, y → 0, 0 f x, y lim x →0 3x3 x4 3x 4 2x 4 x2 lim x →0 3x x2 1 0 3 2 f is not continuous at 0, 0 . Therefore, f is not differentiable at 0, 0 . (See Theorem 12.5) Differentials 193 194 Chapter 13 Functions of Several Variables 5x2y , x3 y3 0, 46. f x, y (a) fx 0, 0 f lim 0, 0 x, y lim fy 0, 0 x, y 0, 0 x, 0 x→0 f 0, 0 x→0 f 0, y y0 0 lim x f 0, 0 0 lim y→0 y 0 0 x 0 (b) Along the line y x: Along the line x 0, 0 y F dm m dF F da a Section 13.5 w x2 x e y2 t w x2 x 2. cos t, y dw dt w dy y dt 2xet 2y e t 2 e2t w e 2t x sec y x et y2 et w dx w dy x dt y dt x sin t x2 y 2 w dy y dt et sec et 5. w xy, x dw dt sec y t1 sec t tan sin t (b) w 6. w 2 sin t, y (a) dw dt w dy y dt dw dt 1 x 2 sin t cos t tan t sin x cos t2 2y cos t t 2, y y 2t 1, sin t x 2 cos 2t sin 2t, y, x sin t t sin2 t 2t sin x (b) w 1 cos t w dx x dt cos x x sec y tan y sec t tan t 2 cos2 t (a) et dw dt 2 cos 2t 1 sin x y0 y 2t sin t 2 1 dw dt 2t sin t 2 1 2y cos t x sin t cot t y2 et cos t sin t e 2t cos2 t e 2t cos t y w dx x dt y x2 y x ln t dw dt 4. w x y 5x3 2x3 0. m da. x sin t yet x2 y2 3. f x, y Chain Rules for Functions of Several Variables w dx x dt dw dt lim x, y → 0, 0 lim x→0 ma: et y 1. a dm f x, y Thus, f is not continuous at 0, 0 . Therefore f is not differentiable at 0, 0 . Thus, the partial derivatives exist at 0, 0 . 47. Essay. For example, we can use the equation F lim x, y → 0, 0 1 cos t y 1 sin t cos t 5 . 2 S ection 13.5 x2 7. w y2 x et sin t z et (a) dw dt z2 et cos t y Chain Rules for Functions of Several Variables w dx x dt et cos t (b) w 8. w w dy y dt 2 w dz z dt et sin t 2 et sin t 2x et 2 et cos t dw dt 2e2t, 2y et cos t et sin t 2zet 4e2t 4e2t xy cos z x t y t2 z arccos t (a) dw dt y cos z 1 t2 t 9. w (a) xy dw dt w dx x dt t2 (b) (a) w 2t t 1 dw dt (b) w t dw dt 2 2t 3 11. Distance t t 1 t2 1t 2t 2t, z 2 x y 4t3 e 2t e t 3 2t e e t e x1 t 6t 2 2t e 3 y1 7 cos t 1 2 42 1 116 2 12 44 12 t y2 1 t t 3 2t 2 t2 e 2t 1 1 t t 2 2t 2 e t 3 2 2 7 cos t 2 3 2t 2 1 t 2t 2e 2 1 10 cos 2t 2 10 t2 t t x2 z 2t 1 3t 2 t 2t2e t x 1t 1 2 10 cos 2t f t3 t e xy 1 ft ft 2t 3 t z t2 2 1, z t 2t 1 xz 2 t t2 t3 t2 1 y 1 1 yz 2t 2t2e t t2 t2 w dz z dt t t 2, y xyz, x 1 1 t2 1 1, y w dy y dt 1 t dw dt 10. w t 1 xy sin z 4t 3 yz, x xz dw dt t t2 t t 2t t 4, (b) w x cos z 2t 10 cos 2t 6 sin 2t 20 sin 2t 2 10 7 22 2 29 4 sin t 7 sin t 2 11 29 29 4 7 cos t 2 2 6 sin 2t 4 sin t 2 12 2 6 sin 2t 12 2.04 4 sin t 12 cos 2t 4 cos t 195 196 Chapter 13 12. Distance Functions of Several Variables ft x2 48t ft 13. w 48 8 x1 8 22 2 y2 22 26 arctan 2xy , x 2 48 3 2 2 48t 1 2 2 26 f1 cos t, y dw dt y1 w dy y dt w dx x dt 1 2y 4x2y 2 1 sin t, t 2 sin t 4 cos2 t sin2 t sin t 1 sin t 0 2x cos t 4x2y 2 1 2 cos t cos t 4 cos2 t sin2 t 2 cos2 t 2 sin2 t 1 4 cos2 t sin2 t d 2w dt 2 4 cos2 t sin2 t 1 8 cos t sin t 1 2 cos2 t 2 sin2 t 8 cos3 t sin t 4 cos2 t sin2 t 2 8 sin3 t cos t sin4 8 cos t sin t 1 2 t 2 cos 4 t 2 22 1 4 cos t sin t At t 14. w t t 0. t2 y d 2w dt 2 x2 y x 0, 1 dw dt x 1 y2 2t 2 2t t1 t4 d 2w dt 2 t At t 1: t 1 1 4t 3 t 12 3t 4 t y2 s t 2x 2y 2x 2x 2y 1 y 2x 2 2x 2t y t t w t w dy y dt s w s w dx x dt x2 y 1 w x 15. When s w s 2 t4 4t 3 12 1 d 2w dt 2 2 12t 3 12t 2 t 4 24 3t 4 1 74 16 4t 3 2 t 4 68 16 4.25 1 2 and t 8 and 1, w t 4. 4s y 4t S ection 13.5 w y3 x es x et y s sin t w s 2x cos t 2y sin t 2s cos2 t 2s sin2 t 17. w w s 6xy e s 3y 2 3x2 0 w t 6xy 0 3y2 3x2 et 3et e2t When s 6e2s t w t e2s 0 and t 1, w s w t 6e and 3e e2 1. sin 2x s t y s t w s 2 cos 2x 3y 3 cos 2x 3y 5 cos 5s 3y 3 cos 2x (a) cos 2x When s 3y 0 and t cos 5s 2 , w s (a) w r 5x2 5y 2, x 5x 5x 2 25 5y 2 cos 5r cos2 5r sin2 2 25 5x 5y 2 w 5x 5x 2 25 25 5r 2 sin2 cos 5r 2 sin cos 2 25 5x 5y 2 (b) w w r 25 5r 2 5r 25 5r 2 ; w 0 sin 5r 25 r sin 5y 2 5y 2 5r 2 5y 5x 2 25 0 5y 2 r cos r 2y 1 2x 2y w 4y 0 8 2 4x 2r 2r 2 0 1 y r 2 r 4 5y 5x2 ,y 2x 0. r sin r 18. 2y 1 r2 r cos , y w t 2x w 25 0 and 2y 1 4r w t w s , 2x t 0 and 4 w r w r 20. w 3 and t 4x (b) 2s2 sin 2t 2y s cos t y2, x w w t s sin t 2s cos 2t 2xy 3y 3y y2 x2 t 2 cos 2x 2x 3y 5 cos 2x x2 When s 19. w w x 18. 197 s cos t y 16. 3x 2y Chain Rules for Functions of Several Variables 8 r 2 r2 2 r 2 r2 2r 2 198 Chapter 13 21. w (a) Functions of Several Variables y arctan , x x r cos , y y w r x2 y2 w (b) r sin cos r2 sin y2 x x2 y2 r sin r cos ,z r cos sin r2 r sin r cos 0 r sin r cos r2 r cos r2 1 arctan tan y w arctan 0 w 1 yz ,x x 2, w r r yz 0 x2 w (a) x2 r sin y2 w r 22. w x cos y x2 r sin z 1 x yz 2 x2 r yz x w r w s 2r 2 x w t r 2 2r 2 yz 1 r 2 2 2t t st2 s s2t2 yz 1 st2 t, z t4 xz 3s2t2 1 t st2 s t t2 ts t4 t2 3s2 t2 xy 2st t st2 s ts t 2st 2st3 2s3t s 2s3t 4st3 2st s2 s t, z st 2t2 1 s s xy t2 t st2 2st3 r2 t, y xz 1 s 3 r s 2s2t2 r 2 xyz, x s 2 r2 y 1 3 (b) w z y x 4 2 23. w y 1 x z 1 x r 2 r 2r 2 2r 2 w 3 24. w w s cos yz 2s cos st2 w t s2, y x cos yz, x t2, z xz sin yz 0 2t3 2s s2t2 sin st2 25. w w s xy sin yz 1 2t3 ze x y, x s z xy e1 y 2s2t s 6s2t2 xz sin yz 2t 2t sin st2 2t3 2s3t sin st2 xy sin yz 2s2t2 sin st2 s zx x y e1 y2 es t es cos yz 0 t s t es t s t zx e y y es t es t s t es t s t st t s 2 2t3 t, y s s t st s w t 4st st s s s2 s t t2 t 2 ts t 2 ex y s st s t s t2 t st s st2 t2 t2 st t t 2 zx x y e1 y2 1 s t st t2 s2t s t 2t3 t s2 ex y t st s t s t2 s ss t 2 S ection 13.5 x2 26. w w s y2 2x z2, x cos s t sin s, y 2y 2t2 sin s cos s w t t cos s, z st2 Chain Rules for Functions of Several Variables 27. x2 2t2 sin s cos s 2x sin s 2y cos s 2t cos2 s 4s2t3 y 5 0 2x 3y 2 3x 2y 1 3y 2y 2st 4 2x 3x 2 1 2z 2st 2t sin2 s 2st 4 2x Fx x, y Fy x, y dy dx 2z t2 t sin s y2 3xy 28. cos x tan xy 5 2t 4s2t3 x2 y2 xy 4 1 ln x2 2 y2 xy 4 29. ln 0 2 Fx x, y Fy x, y dy dx sin x y sec xy x sec2 xy 0 x dy dx Fx x, y Fy x, y x2 x x2 y2 y2 6 0 Fx y y2 x2 x2 2xy y2 2yx 4 2xy xz yz Fx z z x Fx Fz x z z y Fy Fz y z tan x y 2 x2 4x2y 3 2y 5 xy 33. F x, y, z z sec2 y z sec2 x Fy x y Fz sec2 y z sec2 x sec2 y sec2 x z x y x z y z x Fx Fz z y Fy Fz x x z y z y Fy Fz sec2 x sec2 y e x sin y z Fx ex Fy ex cos y z Fz ex cos y z z z 35. F x, y, z x2 y z y z y sec2 y 2 sec y z 1 z2 2yz 1 0 z x 1 z x Fx Fz e x sin y z 1 e x cos y x z y Fy Fz e x cos y z 1 e x cos y z 1 y sec2 Fx Fz sin y tan y Fx y 34. F x, y, z 25 2z z x x Fz 2y y Fy 2 z2 2y Fz y2 2 y2 y2 x2 2y x 2 y 2 2xy 32. F x, y, z x2 y2 x y2 x2y xy 2 x y 2x Fy Fx x, y Fy x, y dy dx x2 31. F x, y, z y y2 x2 30. 199 Fx x, y, z Fz x, y, z 2y 2x 2z y x z y Fy x, y, z Fz x, y, z 2y 2z 2z y z z z z y3 x3 200 Chapter 13 Functions of Several Variables sin y z 0 z cos y z (i) 1 x (ii) y2z 38. x ln y (i) (ii) z cos y y 1 z x z2 sec y z 8 Fx x, y, z Fz x, y, z Fy x, y, z Fz x, y, z x y y2 xzw 2z 1 e xz xz xw Fw xz yz 2w Fx 2x, Fy 2y w x 10w2 5yw 5w, Fz Fx Fw Fy Fw 5w 20w w z Fz Fw 5y 42. F x, y, z, w w x Fx Fw w y Fy Fw Fz Fw w z 44. y 1x 2 1 2y y 2y 2w Fz Fw xy xz xw yz yw 2w sin yz wz cos xy z w y Fy Fw x sin xy Fz Fw y f tx, ty z 12 z cos yz z y cos zy z 3xy 2 f tx, ty tx 3 3 tx ty 3 3 2 3xy x2 y2 tx ty tx 2 ty xy x2 y2 t 2 x fx x, y y fy x, y x z x2 y3 y2 xy 2 x y ty 3 3 3 f x, y ex f tx, ty 45. 2 e tx x fx x, y 3y 2 3x3 9xy 2 y2 32 y x2 x3 y2 1 f x, y y ty ex y f x, y Degree: 0 t f x, y Degree: 3 x 3x2 tf x, y Degree: 1 y3 y fy x, y w xy 43. f x, y 1 y 2 20 y sin xy z 0 1 2x 12 Fx Fw w z x3 x fx x, y yz zx 1 f x, y tx y xz w x 20w 2x 20w y 12 y 1 1 x 2 1 2x 5y 5y 2y 5y x Fy Fw zy w yz 2w w z 2z, Fw 2z 20w w xz 41. F x, y, z, w F x, y, z, w 2x 20w 5y w y 2 Fx Fw w z z2 5 yw w y y2 xz zw xy w2 yzw w x 40. x2 e zw Fz 2y2z 2yz x y3 yz Fy 2yz x xe xz xyz Fx ln y 2z y2 ze xz y xe xz 1. 39. F x, y, z, w Fy x, y, z Fz x, y, z z y z y 0 Fx x, y, z Fz x, y, z 0 z y z. 0 implies z xy z x 0 implies 1 cos y z x e xz 37. F x, y, z 36. x y 3y 3 6xy 3f x, y 3y 2 y fy x, y x 1x e y y y xx e y2 y 0 32 S ection 13.5 46. Chain Rules for Functions of Several Variables x2 f x, y x2 y2 tx f tx, ty 2 2 tx ty x2 t 2 x2 tf x, y y2 Degree: 1 x fx x, y y fy x, y x3 x2 x x4 x2 2xy 2 y2 3 2 x2y 2 y2 3 2 x2y y2 3 2 x 2 x2 x2 y 2 x2 y 2 3 2 x2 x2 y f x, y y2 w dx x dt dw dt w dy (Page 923) y dt w s 48. x s wy ys w x x t wy (Page 925) yt fx x, y fy x, y dy dx 50. 49. w f x, y is the explicit form of a function of two variables, as in z x2 y 2. The implicit form is of the form F x, y, z 0, as in z x2 y 2 0. w x w t 47. dV dt dS dt 52. V dV dt S fy x, y, z fz x, y, z r 2h V S fx x, y, z fz x, y, z z y 51. z x 2rh dr dt 2 rr 2 r2 dh dt r 2h dr dt r dh dt 12 2 36 6 12 4 4608 in.3 min h 2r dr dt h r dh dt 2 24 36 6 12 4 624 in.2 min 12 rh 3 1 3 2rh dr dt r r2 dS dt dh dt h2 r2 12 10 1 3 2 12 36 6 12 2 4 1536 in.3 min r 2 (Surface area includes base.) r2 h2 122 648 10 r2 r2 362 2r h2 144 122 362 12 6 10 144 in.2 min rh dh r 2 h2 dt 20 2 12 6 36 10 144 36 5 dr dt 36 12 122 362 4 9 10 in.2 min 4 201 202 53. Chapter 13 A 1 bh 2 Functions of Several Variables x sin A dx x dt dA dt x cos 2 Ad dt x2 sin 2 2 x2 cos 2 dx dt x sin b 2 d dt h x 6 sin 2 6 cos 2 4 1 2 4 32 2 90 2 10 m2 x θ 2 hr 2.566 m2 hr 54. (a) V dV dt r2 3 2r 3 dr dt Rh 2 15 3 R r dS dt R r R 2R h dR dt 15 r2 dh dt R2 rR 2 25 10 4 15 2 15 25 25 2 12 6,500 cm3 min R S r 25 10 4 19,500 3 (b) R2 h rR r r 2 2 h2 h2 h r R 25 15 2 25 15 2 R 2 r R R r r dr h2 dt 2 R r 2 h2 R r I dI dt 1 m r12 2 102 25 15 102 25 15 25 25 15 15 25 2 102 m62 2r2 15 15 2 4 25 mRT 102 dr2 dt 57. (a) T 2 x tan tan tan tan 1 tan tan x2 tan 6 4 x tan 2x 1 2 x tan 8 4 x 4 x x tan 2x —CONTINUED— 8 tan dT dt 28m cm2 sec 82 2 4 0 2 dR h2 dt 4 56. pV 25 r22 dr1 1 m 2r1 2 dt r r dh h2 dt 15 25 320 2 cm2 min 55. R R 4 θ φ x 8 tan x 1 pV mR 1 dp V mR dt p dV dt 10 15 2 102 12 S ection 13.5 Chain Rules for Functions of Several Variables 57. —CONTINUED— x2 (b) F x, Fx F d dx (c) d dx 1 x 8 1 x du dt f v 1f f x u x sin x⇒x ⇒ tan f y y w y sin y x w y u v v u w u w dx x du w dy y du w v y. Thus, w u x w y x, v f x, y y f y v nf x, y . x 1 x 2 2 ft. x, y . 1 and we have u w x 61. 8 x 0⇒ dv dt Now, let t w x ⇒ cos x ntn w 2x sin cos 8 59. and g t 60. x2 yt, then f u f x x 2 cos2 2 8 t n f x, y xt, v gt 0 2x sin cos 2x f xt, yt Let u 2x 2x tan sec2 x2 0 ⇒ 2 cos2 Thus, x2 58. g t 8 tan w dx x dv w dy y dv w v x y cos y x sin y x 0 w f x, y , x w r w cos x w w x y cos y r cos , y x sin y x r sin w sin y w r cos y r sin w r w r sin r r sin cos —CONTINUED— w r cos2 x w x w r cos r w x r cos w r sin2 x w sin w r cos2 x w r cos (a) w r sin cos y w cos r w r w r sin cos x w w x w r sin cos y r sin2 w sin w sin r r sin cos (First Formula) w r sin2 y w r cos2 y 0 w x w y w x w y 203 204 Chapter 13 Functions of Several Variables 61. —CONTINUED— w r w w y cos r w r (b) 2 1 r2 w w r sin2 y w r sin r w y r sin w sin r 2 y arctan , x x w arctan w x y2 w x w r 2 2 w y u x 2 2 w x u cos x v v x v cos x u u x v r < < 22 v ,x x 2 1 2 1 y2 x 2 x2 y2 1 r2 1 r2 r cos w 2 . and y v cos y v r cos y 2 r sin . v sin x v cos y r v sin x 1v . r v r Therefore, w 0, 1 1 r2 w r u sin y 2 v sin y r sin 1u . r u cos y u r cos y u sin x r u cos y u sin x 2 1 r2 w x sin2 w x cos2 x2 y 2 r sin u r w r , x w y for y2 0 2 w y cos w y cos r sin 2 v u and y y u r Therefore, y2 2 w 1 r2 Therefore, 63. Given x x2 (Second Formula) ww sin xy 2 arctan tan w y , cos w cos r ww sin xy r cos , y r sin r cos y x2 w w2 2 cos x 2 62. r cos2 2 w y 2 2 sin2 S ection 13.6 Directional Derivatives and Gradients 205 64. Note first that v y u x u y v x x u r x2 y y2 y x2 y2 r cos2 sin x x2 x cos x x2 x2 y2 y2 r sin2 1 r r2 r cos r 2 sin2 r 2 cos2 1 r2 x2 r sin cos sin y r sin y2 y2 r sin cos r2 r cos r 2 sin cos 0 r 2 sin cos 0 r2 1u . r v r Section 13.6 f x, y v f x, y Directional Derivatives and Gradients 3x u v 4y i 5i 4x 3 j 2 1 i 2 f 1, 2 48i u 1 2 5 f x, y yi xj f 2, 3 3i 2j u Du f 2, 3 v v f 4, 3 f 2, 3 2 j 2 u 52 2 2 j 2 u 24 2 u 1 x y j f x, y 2 i 2 27j 2 i 2 v f x, y 1 i y f 1, 1 i u Du f 1, 1 j 3y 2j 3 4. j 2 i 2 y 3, v v v Du f 4, 3 xy i 3x 2i u j v v x3 f 4, 3 5j f x, y f x, y 2. 5y 3j 3 Du f 1, 2 f x, y 4xy 1 i 2 f 1, 2 3. . r sin y2 x2 u 1. y2 1v . r u r v r Thus, y x2 y x2 y2 cos y2 v Thus, x x2 x j y2 j v v f 1, 1 j 27 2 2 21 2 2 206 5. Chapter 13 Functions of Several Variables x2 g x, y v y2 3i 4j g x2 y2 3 i 5 h x, y x2 h 1, j 4 j 5 Duh 1, 8. v v e x2 i h x, y v u 2 xy yz 2i j 10. xz f 1, 1, 1 2i 2j 6 i 3 v v Du f 1, 1, 1 f 1, 1, 1 h x, y, z zj x 5 26 26 y2 i x2 2ye y2 j 0 h 0, 0 u 1, 2, f 1, 2, 6 j 6 0 Du f 1, 2, i 4 v v 12. 2j xz yz 2j 1 xy k yz 2 2k 1 2 , , 6 6 h 4, 1, 1 u h h 2, 1, 1 u 1 6 8 46 Du h 2, 1, 1 8 24 6 z2 3k 2yj 4j v v 2z k 2k 1 i 14 f 1, 2, 1 2 j 14 u xyz v 1 2j 2i 1 h x, y, z y2 2x i u 26 3 u i 1 6 k 6 1 arctan yz i x2 f x, y, z yk x arctan yz Du h 4, 1, 1 x2 f 2k u 5 26 u v x h 4, 1, 1 5 j 26 y2 2xe k zi h x, y, z j e y v 2 i f x, y, z u xy j h e x cos yj h 0, 0 h 1, f x, y, z 1 1 i 26 g 1, 0 Du h 0, 0 2 x i 2 j v v v xy 5j 7 25 ei 2 1 Du g 1, 0 u e x sin yi i y g 1, 0 i u 11. y2 e x sin y h arccos xy, v u g 3, 4 v 9. i 3 i 5 Du g 3, 4 7. y 4 j 5 v v u g x, y g x, y x g 3, 4 6. 2, 1, 2 yz i i xz j 2j v v xy k 2k 2 i 3 h 2, 1, 1 1 j 3 u 2 k 3 8 3 3 k 14 6 7 14 S ection 13.6 x2 13. f x, y y2 f 1 j 2 2x i Du f y 14. f x, y 1 i 2 u Directional Derivatives and Gradients x y 3 i 2 u 1 j 2 2y j f 2 u x 2 y f 2 y 2 2x x y Du f f x 2i y u x 1 15. f x, y sin 2x u 1 i 2 f 2 cos 2x Du f 3 j 2 f u cos 2x 2 3 x2 v f u 18. f x, y 2j 2x i v v v 2 x 2 Du f At P 8 y 2 i j 2 sin x v v u 2x 4y yi 7 2. At 0, v h ln x 3i x Du h 21. h h j j z i u 7 19 10 3x 5y 2 f x, y 3i 10y j f 2, 1 3i 10 j ye z i u 3j k Du g 7 19 19 22. g x, y xe z j 4i v v xye z k 2j 8k. 1 i 5 g 2xe y 2 j 5 4 5 u 4 5 x 2y y e x g x, y g 2, 0 y 4j g At 2, 4, 0 , 1 3i 19 f x, y 2i k. 5 sin x 5 y y xye z g k 2 sin x 5 y 0. v k i v v 20. g x, y, z z 3j 1 y At 1, 0, 0 , u y , Du f yj 2 j 5 1 sin x 5 Du f sin x 1 i 5 1 sin x 5 19. h x, y, z ey 2 y f 1 j 2 3, 1 , Du f 1 x 3y xe 2 cos x 8yj 1 i 2 3x 2 xe y j 1y e 2 Dug y y y 4y 2 2i ey i g yj 3 cos 2x 2 y cos 2x 2 17. f x, y cos 2x 2x 3 j 2 1 i 2 u yi x 2 3y 2 y 2j xe y 16. g x, y y 3y y 2x 2x y 2i 2j x 2e y x i 2e y xj 8 5 207 208 Chapter 13 23. z cos x2 y2 4 z 24. 2x sin x2 z x, y z 3, Functions of Several Variables y2 i 6 sin 25i 2y sin x2 8 sin 25j y2 j 0.7941i 3x2 y w x, y, z w 1, 1, 5yz x2 4i 1.0588j 3x2 5z j 2 6i w 26. x tan y w x, y, z w 4, 3, 13j tan y 2 tan 2i 4j, u 1 i 5 2xi g 2 5 u 2i f x, y e f 0, 0 f cos y i 10 5 PQ tan y i h 2, h x, y h x, y h 0, h 0, 3 3 2 i j, u 25 5 i x sec2 y j 4j y cos x y y sin x yi 3 i 6 3 32 36 cos x 3 6 9 63 36 y y sin x yj j 3 2 32 2 23 6 3 f u 2 5 2j 14 53 1 i 5 2i Du f 7 j 53 18i 36 53 fu 2 cos 2x cos y i 17 4 2yj, f 3, 1 f x, y x tan y 4 7j, u \ 30. sin yj 2 5 u h x, y h 2, 2i 6x i Du f 25 1 j 5 x e PQ f x, y 4j 8 5 2 i 53 \ 28. f 0, 0 h x, y 32. zk i Du f 31. x x sec2 y zj 4 sec 2k 2i 2 i 5 j, u 5y k 2 j 5 2y j, g 1, 2 \ PQ j j y 2 4 sec 2j Du g 29. 2z x sec2 y zi 2i g x, y x2 z 1 PQ 1 i 9k \ 27. y z2 6xy i w y 2x z x, y z 2, 3 25. ln x2 2 j 5 sin 2x sin yj 25 5 50 53 50 53 53 S ection 13.6 3 x2 ln g x, y 1 2x i 3 x2 y 2 g 1, 2 12 i 35 4 j 5 y2 2 i 15 x2 f x, y, z y2 j x2i 2xye e x2 j j g 0, 5 2j x2 ye g 0, 5 z2 x2 f x, y, z 1 y2 z2 1 i 21 f 1, 4, 2 f 1, 4, 2 4j 36. 1 xi yj zk e yz i f 2, 0, i x 3 3 i w 2, 1, 1 For Exercises 39–46, f x, y 39. f x, y 2xyz2j w 2, 1, 1 65 3 x 3 y and D f x, y 2 y 2 3 yj zk 0 y 2z2i 8j 4 z2 xy 2z 2 w y2 xi 0 w xye yz k z2 1 x2 w 0, 0, 0 38. xze yz j y2 1 w 0, 0, 0 f x, y, z 4 x2 1 w 2k xe yz f 2, 0, 1 w 1 f x, y, z 37. g x, y 34. g x, y 2y x2 y2 209 25 15 g 1, 2 35. 1 ln x2 3 y2 g x, y 33. Directional Derivatives and Gradients 1 cos 3 2xy 2zk 4j 4k 33 1 sin . 2 40. (a) D 4 1 3 f 3, 2 2 2 1 2 2 2 52 12 z 3 (3, 2, 1) (b) D2 6 3 1 3 f 3, 2 1 2 1 2 3 2 y 9 x 41. (a) D4 3 1 3 f 3, 2 2 (b) D 6 3 1 2 3 2 42. (a) 3 2 23 12 1 2 1 i 2 u Du f 33 12 1 3 f 3, 2 1 2 f j u 1 3 1 2 3i 4j v 9 16 u 3 i 5 1 2 (b) v Du f f 1 2 1 2 2 5 3 5 5 4 j 5 u 1 5 52 12 2 33 12 210 Chapter 13 v Functions of Several Variables 3i 4j v 9 16 u 3 i 5 4 j 5 1 u 5 43. (a) Du f 44. f 1 i 3 f 2i i 3j v u 2 5 1 5 10 1 i 10 u f f 1 4 11 10 60 11 6 10 1 9 Du f 45. 3 j 10 1 6 13 1 j 2 1 13 f f v 5 1 j 2 1 i 3 f 46. (b) 3j 1 13 3i 2j and Du f 3, 2 Therefore, u f u 0. f is the direction of greatest rate of change of f. Hence, in a direction orthogonal to f, the rate of change of f is 0. For Exercises 47– 50, f x, y 47. f x, y 9 x2 x2 9 y2 and D f x, y z y2 2x cos 2y sin 48. (a) D 4 2 x cos f 1, 2 2 2 2 9 (b) D (1, 2, 4) 3 3 f 1, 2 2 y sin 1 2 . 2 2 3 1 23 y 3 x f 1, 2 49. 2i 4 f 1, 2 16 50. 20 f 1, 2 2i f 1, 2 f 1, 2 4j 1 5 25 4j i 2j Therefore, u 15 Du f 1, 2 51. (a) In the direction of the vector (b) 1 2x 10 f 3y i 1 10 4i 3x 1 1 4i 1j 10 10 (Same direction as in part (a)) f 1, 2 (c) 52. (a) In the direction of the vector i j (b) 2y j 2 i 5 1 j 10 2 1 i j, the direction opposite that of 5 10 the gradient f 2i j and f 1, 2 u 0. 1 1 y i 22x f f 1, 2 1 i 2 1 2 xj j y 4x i 1 2 xj 1 j 2 (Same direction as in part (a)) (c) 1 i 2 the gradient f 1 j, the direction opposite that of 2 S ection 13.6 x2 53. f x, y y 2, 4, 211 3, 7 z (a) Directional Derivatives and Gradients (b) Du f x, y f x, y Du f 4, x u 3 8 cos π 2π 2 x cos 2y sin 6 sin Du f y 12 8 4 −4 θ −8 − 12 Generated by Mathematica (c) Zeros: 2.21, 5.36 These are the angles (d) g Du f 4, g for which Du f 4, 3 equals zero. 3 8 cos 8 sin 6 sin 6 cos 0.64, 3.79 Critical numbers: These are the angles for which Du f 4, (e) f 4, 3 of Du f 4, 24i 2 3j 3 , at 0.64. 64 36 10, the maximum value y 6 4 x2 y2 8i f x, y (f ) 3 is a maximum 0.64 and minimum 3.79 . 6j is perpendicular to the level curve at 4, 7 2 f 4, 3 3. x −6 −4 −2 2 4 6 −4 −6 Generated by Mathematica 54. (a) f x, y 8y x2 1 2 y2 ⇒ 4y 4 y 1 x2 y2 4y x2 f (b) 8 1 16xy i x2 y2 2 1 8x2 x2 8y2 j y2 2 3 2 2 y2 4 Circle: center: 0, 2 , radius: x2 f 3 3, 2 2 1 y i 4 3 2 3 1 −2 (c) The directional derivative of f is 0 in the directions ± j. (d) z 6 −6 6 x y −6 x2 55. f x, y f x, y x2 2xi y2 6i c 6 2x 6, P 6 2x 2i 3j f 0, 0 3y 6 0 8j 3y 0, 0 f x, y 2yj 25 f 3, 4 56. f x, y 3, 4 25, P c y2 2x 2i 3y 3j −1 x 1 2 212 Chapter 13 x 57. f x, y x2 58. f x, y y2 xy 3, P c 1 ,P 2 c Functions of Several Variables 1, 1 y2 x2 f x, y x2 y2 y2 x2 i y2 2 2x x2 yi xy f 2xy j y2 2 1, 3 xj 3 3i j 1 2 x2 f x, y 1, 3 x 1 j 2 f 1, 1 59. 4x2 0 y 6 4x2 f x, y f x, y y 8xi f 2, 10 16i f x, y f 2, 1 x 4 4 6i 4 4y 2 1 9i 85 1 1 4 −2 2j xe y y ey i i 6 xe y 1j 4 4j f 5, 0 f 5, 0 2 1 i 17 2j −4 −1 y 4j 17 i 17 2j 1 2j 5 f 5, 0 x −1 4j 1 3i 13 f x, y x −4 8j y f x, y 2 8yj 36i 1 4y j f 1, 1 f 1, 1 62. xe y y 85 9i 85 f 2, f 2, 6xi j 18xi 2y 2 13 3i 13 j 40 9x2 y 3x2 f 1, 1 j 257 16i 257 f x, y 1 f x, y 8 1 16i 257 4y 2 2y 2 f x, y 12 j f 2, 10 f 2, 10 61. 9x2 60. 3x2 y 4j x 2 4 6 x 63. T x2 T 64. h x, y y2 y2 x2 T 3, 4 x2 i y2 2 7 i 625 x2 h 2xy j y2 2 24 j 625 1 7i 625 0.001x2 5000 0.002x i 0.008y j h 500, 300 24j 0.004y2 i 5i 12j 5h 2.4j or 65. See the definition, page 932. 66. The directional derivative gives the slope of a surface at a point in an arbitrary direction u cos i sin j. 67. Let f x, y be a function of two variables and u cos i sin j a unit vector. 68. See the definition, pages 934 and 935. (a) If 0 , then Du f (b) If 90 , then Du f f . x f . y S ection 13.6 z 69. Directional Derivatives and Gradients 213 70. The gradient vector is normal to the level curves. 3 See Theorem 13.12. 3 x y 5 P 72. The wind speed is greatest at B. 71. 18 00 1671 B 1994 A 00 18 73. T x, y 2x2 400 dx dt 10 x0 xt x C1 dx dt C2e xt C1e x2 2y 2, 2x yt y2 10 y2 2y 10 4t 10e 74. T x, y yt 4t 10, 10 dy dt C1e P 4x xt y 2, y2 t 2t y0 C2 4 2t 10e 100e 100 4y 2t yt C2e C1 3 e 2t 4e 3x 2 16 4t 4, 3 dy dt x0 xt P 4t 4t y0 yt 3e C2 4t 32 x 16 y⇒u 10x 75. (a) (b) The graph of D 250 30x2 would model the ocean floor. D 400 50 sin y2 300 1 2 1 x 2 y (c) D 1, 0.5 D y (e) 250 25 cos 30 1 50 sin D y and 1, 0.5 2 y 4 (d) 315.4 ft 25 cos 4 55.5 D x (f ) D 60x i D 1, 0.5 z 76. (a) (b) 500 D 1, 0.5 x 60x and T x, y T 3, 5 25 cos 60 i 400e 400e 60 y j 2 55.5j x2 7 y2 3i xi 1 2 1 2 j j There will be no change in directions perpendicular to the gradient: ± i 6j x 6 6 y (c) The greatest increase is in the direction of the gradient: 3i 1 j 2 214 Chapter 13 Functions of Several Variables 77. True 78. False 79. True Du f x, y u 80. True 2 > 1 when cos 4 i sin 4 j. f x, y, z Section 13.7 1. F x, y, z z2 2 ex cos y 81. Let f x, y, z ex x2 y2 C. Then ex cos y i sin y j z k. Tangent Planes and Normal Lines 3x 5y 3z 3x 15 5y 0 3z 2. F x, y, z z2 2 y x 15 Plane 25 2 0 2 z 25 Sphere, radius 5, centered at origin. 4x2 9y 2 4z2 0 2 3. F x, y, z 9y 2 4z2 4x 5. F x, y, z y z i F F n j F F n x2y4 F x, y, z 2xy4 i F F 32i 6i z2 2yj 2j 11 2zk 2k y2 j 2j j 3x2i 12i 2k 11 3i 11 k k F 2, 1, 8 k x3 k F F z 1 12i 145 k 145 12i 145 k 4j k 5k 4j 5k 4x2y3j k F x, y, z x2 F x, y, z 10. z 32j 1 6i 44 F x, y, z n 4 j 5 3i F x, y, z n y x2 k 2 3i 10 F 1, 2, 16 2xi F F y2 F x, y, z 8. i 53 i 5 25 52 9. z 4 j 5 1 0 Hyperbolic paraboloid x2 k y2 y2 3 i 5 144z 1 3i 11 x2 F 3, 4, 5 k j x F x, y, z 9y 2 F x, y, z n x2 F x, y, z 4 j 3 i 3 7. 0 F 3, 1, 1 6. k 1 i 3 144z 16x2 Elliptic cone 9y 2 F x, y, z x F 16x2 4. F x, y, z 2xi F 2, k 1 32i 2049 32j k 2049 32i 2049 32j k n 1, 2 F F 4i 3y z3 3z2k 3j 3j 1 4i 13 9 12k 3j 12k j k S ection 13.7 11. F x, y, z x ln F x, y, z 1 i y j 1 j z y z z 12. k j F F x sin y z F x, y, z sin y i 12j 1 12i 17 2 2yze x y2j 2 ex F F 4 k 12j k z 3 3j F x, y, z k F y cos x yi ,, 36 3 2 3 i 2 n k sin x F x, y, z 14. x cos y j 1 i 2 F 6, , 7 6 F F 2 10 3 i 2 3 j 2 k 3 j 2 2 cos x yj k 2 113 1 i 2 1 113 i 6 3j 2k 1 10 3i 3j 2k 113 113 n i 6 3j 2k 10 10 3i 3j 2k 3 3j f x, y 25 x2 25 x2 y2 k y 2, 3, 1, 15 F x, y, z 15. Fx x, y, z 6x z 2x Fx 3, 1, 15 Fy x, y, z 6 3 2y Fy 3, 1, 15 1 z 15 6x 1 Fz 3, 1, 15 2 1 0 6x z 2y Fz x, y, z 2y 0 16. y2 k k F x, y, z 12i 3 y 2i k 3 i 3 y2 2 2xze x n j 2 ze x F 2, 2, 3 k 1 i 3 F F n 13. ln y F x, y, z F x, y, z ln x z 1 i x F 1, 4, 3 y Tangent Planes and Normal Lines 2y z 35 35 y , 1, 2, 2 x f x, y y x F x, y, z z Fx x, y, z y x2 Fy x, y, z 1 x Fz x, y, z 1 Fx 1, 2, 2 2 Fy 1, 2, 2 1 Fz 1, 2, 2 1 2x 1 y 2 2x z 2 0 y z 2 0 2x y z 2 k 215 216 Chapter 13 Functions of Several Variables f x, y x2 y 2, 3, 4, 5 F x, y, z x2 y2 17. z x Fx x, y, z x2 3 5 Fx 3, 4, 5 3 4 y 5 3x 3 4y 4 z 5 5z 5 4 5 4y 5z 1 0 y arctan , 1, 0, 0 x g x, y G x, y, z y x arctan Gx x, y, z Gx 1, 0, 0 z y x2 y2 x2 y Gy x, y, z 0 z x2 y2 1 Gy 1, 0, 0 1 y 1 0 3x y2 0 4 Fz x, y, z Fz 3, 4, 5 x2 Fy 3, 4, 5 3 x 5 18. y Fy x, y, z y2 1x y 2 x2 x Gz x, y, z y2 0 g x, y x2 y 2, 5, 4, 9 G x, y, z x2 y2 19. z Gx x, y, z 2x Gy x, y, z 2y Gz x, y, z 1 Gx 5, 4, 9 10 Gy 5, 4, 9 8 Gz 5, 4, 9 1 10 x 8y 1, Fz x, y, z Fy x, y, z e x cos y 8y 2 3x y 2 3, 3 y z 2 0 3z 1 0 z y 1 6 3y 1, 0, , 2 2 e x sin y 1 z x e sin y Fx 0, , 2 2 z 1, 1 e x sin y Fx x, y, z 2x 9 z 1 2x F x, y, z 0 Fy x, y, z 2 3x z z y, 3, 2 3x Fx x, y, z 21. 9 2 F x, y, z x z 2 f x, y 2 3 4 10x 20. 5 2 2 1 Fy 0, , 2 2 0 Fz x, y, z 1 Fz 0, , 2 2 1 1 Gz 1, 0, 0 1 x2 1 S ection 13.7 x2 22. z y 2, 1, 2, 1 2xy x2 F x, y, z Fx x, y, z y2 2xy 2x Fx 1, 2, 1 2x Tangent Planes and Normal Lines z 2y Fy x, y, z 2 1 Fy 1, 2, 1 2y 2 2x z 2y 1 2y 2 1 1 0 z 1 2y 2x Fz x, y, z Fz 1, 2, 1 2x 0 z 1 h x, y ln x2 y 2, 3, 4, ln 5 H x, y, z ln x2 y2 23. 1 ln x2 2 z x Hx x, y, z x2 Hx 3, 4, ln 5 3 25 3 x 25 3x 3 z y Hy x, y, z x2 Hy 3, 4, ln 5 4 25 y2 4 y 25 3 y2 4 4y z ln 5 1 Hz 3, 4, ln 5 1 0 25 z 4 ln 5 Hz x, y, z y2 0 3x 4y h x, y cos y, cos y 25 1 ln 5 2 5, , 42 H x, y, z 25z z 24. Hx x, y, z 0 0 2 Hx 5, , 42 2 y 2 2 y 2 Hy x, y, z 2 Hy 5, , 42 2 2 z 4 2 8 z 4y 2 F x, y, z Fx x, y, z Fx 2, 4x 2, 4 2 x 2 2 2 Hz 5, , 42 0 8z z2 36, 2, 2, 4 x2 4y 2 z2 2x 2 4 36 Fy x, y, z 4 Fy 2, 8y 2, 4 16 16 y 2 2 2 0 4 2y 25. x2 Hz x, y, z sin y 2 8z 4 0 4y 2 2z 4 0 4y 2z 18 x Fz x, y, z Fz 2, 2, 4 2z 8 1 1 217 218 Chapter 13 26. x2 2z2 Functions of Several Variables y2, 1, 3, 2 x2 2z2 F x, y, z y2 Fx x, y, z Fx 1, 3, 2x 2 Fy x, y, z 2 Fz x, y, z 2y Fy 1, 3, 2 Fz 1, 3, 6 2x 1 6y 3 8z 2 1 3y 3 4z 2 4z 8 0 3y 2 0 x 4z 0 x 27. xy2 z2 3x 4, 2, 1, xy2 F x, y, z Fx x, y, z Fx 2, 1, 4x y 2 2 z2 3x 2 2 4 3 Fy x, y, z 4 1 4z x 28. x y 2z 2 y 4 Fz 2, 1, 3 Fz x, y, z 2y Fz 4, 4, 2 8 x z 1 2yz 3y 1 Fy x, y, z 2z Fx 4, 4, 2 1 Fy 4, 4, 2 1 1y z 2 y 8z x y2 8z x 29. x2 4 y 0 8z 16 16 9, 1, 2, 4 x2 F x, y, z y2 z 9 Fx x, y, z 2x Fy x, y, z 2y Fz x, y, z 1 Fx 1, 2, 4 2 Fy 1, 2, 4 4 Fz 1, 2, 4 1 Direction numbers: 2, 4, 1 Plane: 2 x Line: 30. x2 x 1 4y y 1 2 y2 2 2 z z 4 z2 4 0, 2x 4y z 14 4 1 9, 1, 2, 2 x2 F x, y, z y2 z2 9 Fx x, y, z 2x Fy x, y, z 2y Fz x, y, z 2z Fx 1, 2, 2 2 Fy 1, 2, 2 4 Fz 1, 2, 2 4 Direction numbers: 1, 2, 2 Plane: x Line: x 1 1 1 4 0 Fx x, y, z 4 2 3 , 4, 4, 2 F x, y, z x 2z 2 Fy 2, 1, 4y Fz x, y, z 2xy 2y 2 2 z y 2 2z 2 2 2 0, x 2y 2z 9 S ection 13.7 31. xy z 0, F x, y, z 2, xy 2, 3, 6 z Fx x, y, z Fx Tangent Planes and Normal Lines y Fy x, y, z 3, 6 3 Fy 2, Fz x, y, z 3, 6 1 2, x 1 Fz 2 3, 6 Direction numbers: 3, 2, 1 Plane: 3 x Line: 32. x2 x 2 2y 2 y 3 y2 Fx x, y, z 6 2y z 6 1 y2 12 z2 2x Fy x, y, z 12 0, 3x 6 0, 5, 13, x2 F x, y, z z z 2 z2 Fx 5, 13, 3 3 10 Fy 5, 13, Direction numbers: 5, 13, Fz x, y, z 2y 12 26 2z Fz x, y, z 24 12 Plane Line: 5x 5 13 y 13 12 z 12 12z 0 5x y arctan , x 33. z F x, y, z 1, 1, z y Fx x, y, z Fx 1, 1, 4 y x arctan x2 Fy x, y, z y2 1 2 4 Fy 1, 1, Line: 34. xyz x 1 y y 1 1 4 x x2 Fz x, y, z y2 1 2 Fz 1, 1, 1, 2 Direction numbers: 1, Plane: x 1 1 1 2z z 0, x 4 y 2z 2 4 2 10, 1, 2, 5 F x, y, z xyz 10 Fx x, y, z yz Fy x, y, z xz Fz x, y, z xy Fx 1, 2, 5 10 Fy 1, 2, 5 5 Fz 1, 2, 5 2 Direction numbers: 10, 5, 2 Plane: 10 x Line: x 1 10 1 5y y 2 5 5 5 0 13y x 2 z 2z 5 2 5 0, 10x 5y 2z 30 4 1 1 y 13 13 z 12 12 219 220 Chapter 13 35. z f x, y Functions of Several Variables 4xy 1 y2 x2 (a) Let F x, y, z 4xy 1 y2 x2 x2 4y F x, y, z 2 y 2 ≤ x ≤ 2, 0 ≤ y ≤ 3 , 1 2x2 1 2 1 z 1 x 1 4y 1 x2 i y 1 x2 1 2 1, y F 1 y 1 2 2 j k k 1 1, z Tangent plane: 0 x (b) x 2y2 1 2 k Direction numbers: 0, 0, Line: x 2 4x 1 y2 j x 1 y2 1 2 2 F 1, 1, 1 y2 4x i 2 4 5 1, 2, 1 t 1 0y 1 1z 43 j 2 52 0i 0⇒z 1 6 j 25 k k 1 (c) z z 1 Line: x 1, y 6 t, z 25 2 t 1 sin y , x 36. (a) f x, y sin y x F x, y, z 23 3 ,, 322 2 2 3 0 k 1 2 0y 4t 2 4z 1 2 0⇒x 4z 4 0 k 9 , 0, 4 9t, y Tangent plane: 9 x —CONTINUED— 32 1 or 1, 0, 4 ,z 2 9 i 4 Direction numbers: Line: x 25z 2 3 3 (d) At 1, 1, 1 , the tangent plane is parallel to the xy-plane, implying that the surface is level there. At 1, 2, 4 , 5 the function does not change in the x-direction. k t, y Tangent plane: 1 x F 0 cos y j x 1 , 0, 4 Direction numbers: (b) 20 2 −1 y z 1 i 4 2 25z x −2 0 12 sin y i x2 Line: x 4 5 y −1 3 ≤ x ≤ 3, 0 ≤ y ≤ 2 Let F x, y, z 1 F 2, , 22 1z 2 3 x 2 6y 1 2 6y Plane: 0 x 6 y 25 4 5 1 or 9, 0, 3 ,z 2 0y 3 2 3 2 4 4t 4z 3 2 0 ⇒ 9x 4z 12 0 S ection 13.7 Tangent Planes and Normal Lines 36. —CONTINUED— (c) (d) At both points the function does not change in the y-direction. z 3 x 3 −3 2π y 37. See the definition on page 944. 38. Fx x0, y0, z0 x x0 Fy x0, y0, z0 y y0 Fz x0, y0, z0 z z0 0 (Theorem 13.13) 40. Answers will vary. 39. For a sphere, the common object is the center of the sphere. For a right circular cylinder, the common object is the axis of the cylinder. F x, y, z x2 F x, y, z 41. 2x i F 2, 1, 2 (a) F y2 4i G x, y, z 2j i 4 1 G j 2 0 k 0 1 Direction numbers: 1, F F (b) cos F x, y, z x2 F x, y, z 42. 2xi F 2, (a) F 1, 5 4i k 4j 2 2j y 1 2 2 10 2i z G 2, k 1 1 i 4j x 4. F F 3 21 2 2 k 2 10 ; not orthogonal 5 4 G x, y, z k 2j 1 G x, y, z k j 2 1 2k 1 z 2yj G G i x Direction numbers: 1, 4, (b) cos k 4 20 2 y2 i 4 0 G i 2i 2, 1. G G z G 2, 1, 2 2y j x G x, y, z 5 y z j j 1, 5 k k 4k y 1 1 4 3 42 z 5 4 42 ; not orthogonal 14 221 222 Chapter 13 43. x2 F x, y, z F F z2 2x i F 3, 3, 4 (a) Functions of Several Variables 2z k 6i G 8k j 0 6 k 8 8 Direction numbers: 4, 4, F F (b) cos 44. G G x2 F x, y, z y2 x2 y2 Direction numbers: 1, 17, y F F cos 4 j 5 4 17 x2 F x, y, z 2x i F 2, 1, 1 F (a) z 46. x2 2xi F 1, 2, 5 (a) F F F 2j 3k 5i 2j 3k 26 k 5 8 Not orthogonal 5 76 y z G x, y, z i j k G 2, 1, 1 i j k 6j 1. x 6k 6j y 2, 1 k z 1 1 1 G x, y, z x y 6z G x, y, z i j 6k G 1, 2, 5 i j 6k k k j 4 1 G G 34 j 5 z k 1 6 Direction numbers: 25, (b) cos 5i 0; orthogonal 4j i 2 1 G G x, y, z k x k 2 1 2y j 2i j 2k y2 F x, y, z 3z 2z k G G F x, y, z 2y G x, y, z j 2 1 F F 5x 6 Direction numbers: 0, 1, (b) cos 3k 4 3 13 z2 2y j i 4 1 G 2 i 5 85 2 38 2j 4j 5 Tangent line 13 y2 4i y2 k G G F x, y, z z 12 4 i G 3, 4, 5 k 1 3 3 3 4 y i j 3545 5 2 1 36 k 16 ; not orthogonal 25 x2 F x y 3 4 8k G x, y, z i 3 i 5 G x 48 j 25 2z k 6j z F 3, 4, 5 45. 3. 64 10 10 x F x, y, z 48 i z2 2y j G 3, 3, 4 i 6 0 G y2 G x, y, z 25 13, 25i 2. 13j x 1 25 0; orthogonal 2k y 2 13 z 5 2 33 22 S ection 13.7 47. f x, y y2 , g x, y 4 x2 6 y2 4 x2 F x, y, z z F x, y, z (a) 2x i F 1, 2, 4 2x 2i j y G x, y, z 6 1 yj 2 Tangent Planes and Normal Lines k z 2x y G x, y, z j k G 1, 2, 4 k 2i 2i j k The cross product of these gradients is parallel to the curve of intersection. F 1, 2, 4 i 2 2 G 1, 2, 4 Using direction numbers 1, (b) k 1 1 2, 0, you get x G G F F cos j 1 1 4 1 6 1 2i 1 t, y 4 ⇒ 6 6 4j 2 2t, z 4. 48.2 z 8 (1, 2, 4) 6 8 y x 48. (a) f x, y x2 16 2x 3x2 1 y2 f x, y (b) x2 16 32 y2 2x2 4y 2y2 2x 4x 4y 6x 4y 1 1 2 3x2 3x2 1 4y 2x 1 42 3 y2 42 3y To find points of intersection, let x 3y y 2 2 2 5 5 y 2 6x 4y 4 2 1. Then 14 y f 1, 2 1 2j 4y 42 2 6x 12y 31 2 y2 y2 8y 2x 1 g x 3y2 x f g x, y x2 x2 z 5 2 2 g x, y y2 2± 14 14 2 j, g 1, k, which are orthogonal. 2 14 14 2 j and g 1, Similarly, f 1, 2 1 2 j k, which are also orthogonal. 1 2 2 j. The normals to f and g at this point are 14 1 2 j and the normals are 2j 2j k and (c) No, showing that the surfaces are orthogonal at 2 points does not imply that they are orthogonal at every point of intersection. k and 223 224 Chapter 13 Functions of Several Variables 3x2 49. F x, y, z 2y 2 z 15, 2, 2, 5 50. F x, y, z F x, y, z 6x i 4yj k F 2yi 2xj F 2, 2, 5 12i 8j k F 2, 2, 2 4i 1 209 F 2, 2, 5 k F 2, 2, 5 cos 1 209 arccos x2 51. F x, y, z y2 2x i 2yj F 1, 2, 3 2i 4j 1 21 F x, y, z F x, y, z 2x 0, x 6 z 4i 2j F 2, 1, 3 k F 2, 1, 3 arccos 0 6y 2y z 6j 0 90 z k 8 0, y 02 3 5, 2, 1, 3 y2 2yj cos 0 2y y2 2xi 25.24 2xi 77.40 x2 3 3 11 11 12 176 3 11 11 F 2, 1, 3 k 1 21 arccos 12k F x, y, z F 1, 2, 3 k F 1, 2, 3 cos 52. F x, y, z k 4j x2 arccos z, 1, 2, 3 3z2k F 2, 2, 2 k F 2, 2, 2 cos 86.03 F x, y, z 53. z3, 2, 2, 2 2xy 3 32 63 6 8 12 x 8 0, 3, 12 3x2 54. F x, y, z F x, y, z 2y2 6x 6x 3 0, x 4y 4 3 1 2, 12 2 2 55. T x, y, z 4j 5 z 30 k 25 1 2 1 1 2 −3 3 1 2 4 C1e x0 C1 4kt 2x2 y2 dy dt 4kx 4e 4y z 1 5 4kt 4 4z2, 4, 3, 10 yt C2e y0 C2 y dz dt 2ky 3e 2kt 2kt 3 −3 3 3 y 56. T x, y, z dx dt 8kz zt C3e z0 C3 z −2 31 4 x 400 xt x 4y 31 4 1, dx dt 3x 3i 0, y z y (vertex of paraboloid) 10e 8kt 8kt 10 100 x0 x 3t C1 C1 3t y z2, 2, 2, 5 dy dt 3 xt 3x 2 2 1 yt t y0 C2 y t dz dt C2 zt C3e 2 2 2z z0 C3 z 5e 2t 2t 5 S ection 13.7 57. F x, y, z x2 a2 Fx x, y, z y2 b2 z2 c2 Tangent Planes and Normal Lines 58. F x, y, z x2 a2 2x a2 Fx x, y, z 2x a2 Fy x, y, z 2y b2 Fy x, y, z 2y b2 Fz x, y, z 2z c2 Fz x, y, z Plane: 2x0 x a2 2y0 y b2 x0 x0 x a2 x02 a2 b2y 2 a2x2 Fx x, y, z z02 c2 1 y02 b2 0 2x0 x a2 Plane: 2y0 y b2 x0 z2 x0 x a2 y0 y b2 z0 z c2 2z Plane: 2a2x0 x 2b2y0 y x0 a2x0 x b2y0 y y0 2z0 z a2x02 z0 z z0 b2y02 0 z02 0 Hence, the plane passes through the origin. xf F x, y, z y x xf z 60. y x y x Fx x, y, z f Fy x, y, z xf Fx x, y, z z y x y x2 y x xf 1 x f f y x yy f xx y x 1 Tangent plane at x0, y0, z0 : f f y0 x0 y0 y0 f x0 x0 x y0 x0 y0 y0 f x0 x0 x0 f y0 x0 y0 f x x0 y0 x0 yf f Therefore, the plane passes through the origin x, y, z 61. f x, y ex ex y, fy x, y fxx x, y ex y, fyy x, y y0 x0 y0 y0 x0 z y0 f y0 y0 f x0 x0 0, 0, 0 . y fx x, y y0 x0 y0 y x0 f (a) P1 x, y f 0, 0 —CONTINUED— fx 0, 0 x ex ex y y, fy 0, 0 y ex fxy x, y 1 x y y x z z0 0 x0 f f 1 2z c2 2b2y Fz x, y, z z0 z2 c2 2a2x Fy x, y, z 2z0 z c2 y0 z0 z c2 y0 y b2 59. F x, y, z 1 y2 b2 225 y0 x0 0 y0 y x0 z 0 x02 a2 2z0 z c2 y0 y02 b2 z0 z02 c2 1 0 226 Chapter 13 Functions of Several Variables 61. —CONTINUED— (b) P2 x, y f 0, 0 fx 0, 0 x 1 y 12 2x x 1 2 fxx fy 0,0 y 0, 0 x2 1 2 fyy fxy 0, 0 xy 0, 0 y 2 12 2y xy (c) If x 0, P2 0, y 1 y 12 2y . This is the second–degree Taylor polynomial for e y. If y 0, P2 x, 0 1 x 12 2x . This is the second–degree Taylor polynomial for e x. (d) x y 0 f x, y 0 0 P1 x, y 1 1 0.1 0.9048 (e) P2 x, y 1 0.9000 z P1 0.1 1.1052 1.1000 0.5 0.7408 0.7000 0.5 1.6487 1.5000 −2 1 0.7450 1 2 1.1050 0.2 4 −2 0.9050 0.2 1.6250 62. f x, y cos x fxx x, y cos x y y fyy x, y y, sin x cos x y, y f 0, 0 (b) P2 x, y 2 −4 fy x, y sin x (a) P1 x, y −2 x y fx x, y fx 0, 0 x f 0, 0 0, P2 0, y 1 If y 0, P2 x, 0 1 y f x, y 0 0 0 1 2 xy (c) If x x fy 0, 0 y 1 2 fxx fxy x, y cos x y 1 fy 0,0 y fx 0, 0 x 12 2x 1 (d) f P2 0, 0 x2 fxy 0, 0 xy 1 2 fyy 0, 0 y2 y2 12 2 y . This is the second–degree Taylor polynomial for cos y. 12 2 x . This is the second–degree Taylor polynomial for cos x. P1 x, y P2 x, y 1 1 0.9950 1 0.9950 0.2 0.1 0.9553 1 0.9950 0.2 0.5 0.7648 1 0.7550 1 0.5 0.0707 1 z 1 0.1 (e) 0.1250 5 5 63. Given w x 5 y F x, y, z where F is differentiable at x0, y0, z0 and F x0, y0, z0 0, the level surface of F at x0, y0, z0 is of the form F x, y, z G x, y, z F x, y, z Then G x0, y0, z0 C C for some constant C. Let 0. F x0, y0, z0 where G x0, y0, z0 is normal to F x, y, z Therefore, F x0, y0z0 is normal to the level surface through x0, y0, z0 . C 0 at x0, y0, z0 . S ection 13.8 64. Given z Extrema of Functions of Two Variables f x, y , then: F x, y, z f x, y F x0, y0, z0 z 0 fx x0, y0 i fy x0, y0 j F x0, y0, z0 F x0, y0, z0 cos k k k 1 fx x0, y0 2 fx x0, y0 fy x0, y0 2 2 fy x0, y0 2 2 1 1 Section 13.8 1. g x, y x 1 Extrema of Functions of Two Variables 1 2 y 3 ≥0 2 z 5 Relative minimum: 1, 3, 0 gx 2x 1 0⇒x 1 gy 2y 3 0⇒y 3 1 1 2 3 x 2. g x, y 9 x 3 2 y Relative maximum: 3, 2 2 ≤9 z (3, − 2, 9) 2, 9 gx 2x 3 2y 2 8 0⇒x gy 0⇒y 6 3 4 2 2 y 1 6 x x2 3. f x, y y 4 (1, 3, 0) 1≥1 y2 z 5 Relative minimum: 0, 0, 1 x2 fy fxx x y2 x2 Check: fx y y2 x2 0⇒x 1 −3 3 0⇒y 1 y2 1 y2 1 0 32 , fyy 0 x2 x2 1 y2 1 At the critical point 0, 0 , fxx > 0 and fxx fyy 4. f x, y 25 x 2 fxy 32 2 , fxy xy y2 1 x2 2 y 3 32 > 0. Therefore, 0, 0, 1 is a relative minimum. y2 ≤ 5 2 (0, 0, 1) 2 x z (2, 0 , 5) 5 Relative maximum: 2, 0, 5 Check: fx 25 x x 2 2 2 y2 0⇒x 2 x fy fxx y 25 25 x 2 2 25 y 2 x 22 y2 y2 3 2 0⇒y At the critical point 2, 0 , fxx < 0 and fxx fyy 25 fxy y 0 25 , fyy 5 5 2 x x 2 2 2 2 y2 32 , fxy 25 yx 2 x 22 > 0. Therefore, 2, 0, 5 is a relative maximum. y2 32 227 228 Chapter 13 x2 5. f x, y Functions of Several Variables y2 2x Relative minimum: 6y 6 1, 3, x 0⇒x fy 2y 6 0⇒y x2 6. f x, y 2, fxy 4≥ z 4 x 1 3 4x −1 −2 −3 −4 1 y 7 (−1, 3, − 4) 0 1, 3 , fxx > 0 and fxx fyy y2 2 2 1 2 At the critical point 3 21 2x 2, fyy y 4 Check: fx fxx 2 1 8y 11 x 2 fxy 2 > 0. Therefore, 2 y 4 2 1, 3, 4 is a relative minimum. 9≤9 z Relative maximum: 2, 4 , 9 8 (2, 4 , 9) 6 Check: fx 2x 4 0⇒x 2 fy 2y 8 0⇒y 4 6 fxx 2, fyy 2, fxy 0 At the critical point 2, 4 , fxx < 0 and fxx fyy 7. f x, y 2x2 fx 4x 2y 2 fy 2x 2y 0 fxx y2 2xy 4, fyy fx fy 2x 10y fxx 2, fyy fx fy 10 30 10x 0 x 0 10x fxx 30y 2 > 0. Therefore, 1, 1, 4 is a relative minimum. 0 3 , fxx < 0 and fxx fyy 4xy y2 16 2 f xy > 0. 0 4y 2y x2 fx 2x 6y fy 6x 20y 2, fyy fxy 3 16x 10 Solving simultaneously yields x 0 2, fxy 6xy 10y2 8 and y 16. 4 At the critical point 8, 16 , fxx < 0 and fxx fyy fxx 1. 3, 8 is a relative maximum. 10, fyy 10. f x, y 1 and y 62 5, y 10, fxy 5x2 4x 3 1, 1 , fxx > 0 and fxx fyy At the critical point 5, 9. f x, y > 0. Therefore, 2, 4, 9 is a relative maximum. 2 5y 2 Therefore, 5, fxy 2 Solving simultaneously yields x 2, fxy x2 2 0 At the critical point 8. f x, y 2x 4 x 4y fxy 2 > 0. Therefore, 8, 16, 74 is a relative maximum. 4 0 4 20, fxy At the critical point 0 Solving simultaneously yields x 6 and y 2. 6 6, 2 , fxx > 0 and fxx fyy fxy 2 > 0. Therefore, 6, 2, 0 is a relative minimum. 8 y S ection 13.8 2x2 11. f x, y fx 4x 4 fy 6y 3y 2 12 fxx 4x 4x 6y 0 when y 2 2x fx y2 0 x2 y2 2y fy x2 3 4 x 4y 4 0 when y 0, y hx 0 hy x x2 2x y2 2y y2 3 x2 3 x2 16. f x, y y y2 13 2 23 0 23 0 x 1 Relative maximum: 18. f x, y x y 2 y3 3yx2 3y 2 1, 0, 2 3, ± Saddle points: 0, 2, −4 y 4 3, 1, y2 20 20. z z Relative minimum: 0, 0, 0 y 3 e xy z Saddle point: 0, 0, 1 6 100 5 Relative maxima: 0, ± 1, 4 Saddle points: ± 1, 0, 1 −4 −4 x 4 x x2 hx 2x y2 2 hy 2y hxx 2, hyy 2x 2x 4 1 4y 2 2, hxy 4 y 4 0 when x 2y 1. 0 when y 2. 0 At the critical point 1, 3 z 3 x2 1 40 x 4y 2 e1 3x2 5 −4 21. h x, y 0 Relative maximum: 0, 0, 1 2 x x2 0, y Since f x, y ≥ 2 for all x, y , the relative minima of f consist of all points x, y satisfying x y 0. 4 Relative minimum: 1, 0, 19. z 0 z 4x y2 2 1. Since h x, y ≥ 2 for all x, y , 0, 0, 2 is a relative minimum. 0, 0 is the only critical point. Since g x, y ≤ 4 for all x, y , 0, 0, 4 is a relative maximum. 17. z 4, fxy 6, fyy 5 1 2. 0 when x 14. h x, y Since f x, y ≥ 3 for all x, y , 0, 0, 3 is a relative minimum. 15. g x, y 3 4y 229 1 At the critical point 2 , 1 , fxx < 0 and 1 31 2 > 0. fxx fyy fxy Therefore, 2 , 1, 4 is a relative maximum. 0 y2 6x fxx x 3x fy 2. 0 2 x2 2y 2 fx 1. At the critical point 1, 2 , fxx > 0 and fxx fyy fxy 2 > 0. Therefore, 1, 2, 1 is a relative minimum. 13. f x, y 3x2 12. f x, y 13 0 when x 1 6, fxy 4, fyy 12y Extrema of Functions of Two Variables 2 , hxx hyy hxy 2 < 0. Therefore, 1, 2, 1 is a saddle point. 3 3 y 230 Chapter 13 22. g x, y 120x Functions of Several Variables 120y x2 xy gx 120 y 2x 0 gy 120 x 2y 0 y2 gxx Solving simultaneously yields x 2, gxy 2, gyy x2 hx 2x 3x hxx 2 > 0. Therefore, 40, 40, 4800 is a relative maximum. 0 2, hyy 2y Solving simultaneously yields x 0 2, hxy 0 and y 0. 3 At the critical point 0, 0 , hxx hyy 24. g x, y gxy y2 3xy 3y hy 40. 1 At the critical point 40, 40 , gxx < 0 and gxx gyy 23. h x, y 40 and y hxy 2 < 0. Therefore, 0, 0, 0 is a saddle point. gxy 2 < 0. Therefore, 0, 0, 0 is a saddle point. xy gx y gy x gxx x 0 and y 0, gyy 0 0, gxy 1 At the critical point 0, 0 , gxx gyy x3 25. f x, y fx fy 3 x2 3 fxx y3 3xy y 0 y2 x 6x, fyy Solving by substitution yields two critical points 0, 0 and 1, 1 . 0 6y, fxy 3 At the critical point 0, 0 , fxx fyy fxy 2 < 0. Therefore, 0, 0, 0 is a saddle point. At the critical point 1, 1 , fxx fxx fyy fxy 2 > 0. Therefore, 1, 1, 1 is a relative minimum. 26. f x, y fx fy 14 x 2 2xy 6x2, fyy 6y 2, fxy 27. f x, y e fy x e < 0 ⇒ 0, 0, 1 saddle point. 2 > 0 and fxx < 0 ⇒ 1, 1, 2 relative maximum. 1 , fxx fyy e fx 2 fxy At 1, 1 , fxx fyy 1, x x 2 fxy At 0, 0 , fxx fyy At 1 2x3 Solving by substitution yields 3 critical points: 2y 3 0, 0 , 1, 1 , 1, 1 2y 2x fxx y2 fxy 2 > 0 and fxx < 0 ⇒ 1, 1, 2 relative maximum. sin y sin y cos y 0 0 Since e x > 0 for all x and sin y and cos y are never both zero for a given value of y, there are no critical points. 6 > 0 and S ection 13.8 1 2 28. f x, y 2x3 fx x2 2xy 2 2x2y fy 4y 4 fyy 4x2y 2 x2 4xy 3 0 y2 3 e1 2y 2 8y 2 2xy e1 2 6 ,± ,0 . 2 2 Solving yields the critical points 0, 0 , 0, ± 0 12x2 2x2 231 y2 y2 x2 y e1 4x2y 2 4x3y fxy 3x e1 2y 3 4x 4 fxx x2 y 2 e1 Extrema of Functions of Two Variables x2 x2 y2 1 e1 x2 y2 y2 fxy 2 < 0. Therefore, 0, 0, e 2 is a saddle point. At the critical points 0, ± 2 2 , At the critical point 0, 0 , fxx fyy 2 > 0. fxx < 0 and fxx fyy fxy Therefore, 0, ± 2 2, e are relative maxima. At the critical points ± 6 2, 0 , fxx > 0 and fxx fyy fxy 2 > 0. Therefore, ± 6 2, 0, e e are relative minima. y4 ≥ 0. z y2 x x2 29. z 0 if x y Relative minimum at all points x, x , x x2 x2 30. z 0. y2 2 ≥ 0. z y2 0 if x2 y2 0. Relative minima at all points x, x and x, 0. x ,x z z 2 60 40 5 5 x y 3 3 x 31. fxx fyy fxy 2 y 94 62 0 32. fxx < 0 and fxx fyy Insufficient information. 33. fxx fyy fxy 2 96 fxy 2 3 8 22 > 0 f has a relative maximum at x0, y0 . 102 < 0 34. fxx > 0 and fxx fyy f has a saddle point at x0, y0 . fxy 2 25 8 102 > 0 f has a relative minimum at x0, y0 . 35. (a) The function f defined on a region R containing x0, y0 has a relative minimum at x0, y0 if f x, y ≥ f x0, y0 for all x, y in R. (b) The function f defined on a region R containing x0, y0 has a relative maximum at x0, y0 if f x, y ≤ f x0, y0 for all x, y in R. (c) A saddle point is a critical point which is not a relative extremum. (d) See definition page 953. 36. See Theorem 13.17. z 37. No extrema z 38. 75 4 60 Extrema at all x, y 3 45 30 2 2 x 2 4 y x 3 4 y 0. 232 Chapter 13 Functions of Several Variables z 39. Saddle point 40. x Relative maximum z 7 6 (2, 1, 4) 4 6 y 3 3 y 4 4 −3 x 42. A and B are relative extrema. C and D are saddle points. 41. In this case, the point A will be a saddle point. The function could be f x, y 43. d x y. fxx fyy fxy2 ⇒ fxy2 < 16 ⇒ x3 45. f x, y fx fy 3y 2 fxx 4 < fxy < 4 0 6y, fxy x 1 2x 2x 2 fxy 2 1y 1 2y y 4 y 2 fx fy fxx At 1, x x 2 1 2 1 2, fxy 2x 4 , fxx fyy 2 2 , fxx fyy 2 y 12x 12 18y 27 0 6x y 2 2 2 2 4x fxy 2 1y 12, fyy At 2, 3 , fxx fyy is a saddle point. 6y fxy 18, fxy 2 2 2 4, 0 ≥0 0 2 3 2, fyy x x 1 2 1 y 2 fxy 2 is undefined and the test fails. Absolute minimum: 1, 2, 0 2. 0 2 2 4. 0 and the test fails. 1 and y 2 2 3 2, fxy 27y Solving yields x 4 Solving yields x 12x 19 0 3y 2 Solving yields the critical points 1, a and b, 0 x1 12 y y2 12 y 1 3x2 0 4 y x 9y 2 ≥0 Absolute minima: 1, a, 0 and b, x 0 0 and the test fails. 0, 0, 0 4 4 2, fyy y 0 2 At both 1, a and b, 48. f x, y 6x2 fxx Solving yields x At 0, 0 , fxx fyy is a saddle point. fxx y3 fy 0 47. f x, y x3 46. f x, y fx 6x, fyy fy 44. d fxx fyy fxy2 < 0 if fxx and fyy have opposite signs. Hence, a, b, f a, b is a saddle point. For example, consider f x, y x2 y2 and a, b 0, 0 . fxy2 > 0 16 y3 3x2 fx fxy2 28 x x 1 2 1y y 2 2 232 2 and y 3. 0 0 and the test fails. 1, 2, 0 S ection 13.8 x2 49. f x, y fx fy 3 y2 Extrema of Functions of Two Variables ≥0 3 2 3 3 x fx and fy are undefined at x 2 3 3 0. The critical point is 0, 0 . y 2 fxx 0, y 3 9x x 2 , fyy 9y 3 , fxy y 0 fxy 2 is undefined and the test fails. At 0, 0 , fxx fyy Absolute minimum: 0 at 0, 0 x2 50. f x, y y2 4x y2 fx 3 x2 fy 4y 3 x2 y 2 4 x2 9 x2 fxx 23 ≥0 13 fx and fy are undefined at x 0, y 0. The critical point is 0, 0 . 13 3y 2 ,f y 2 4 3 yy 4 3x2 y 2 ,f 9 x2 y 2 4 3 xy 8xy y2 9 x2 43 fxy 2 is undefined and the test fails. At 0, 0 , fxx fyy Absolute minimum: 0, 0, 0 x2 51. f x, y, z y fx 2x fy 2y 3 0 fz 2z 1 2 3 z 1 2 ≥0 0 0 Solving yields the critical point 0, 3, Absolute minimum: 0 at 0, 3, 52. f x, y, z 4 xy fx 2x y 1 z 2 fy 2x2 y 1z 2 2 fz 2x y 2 2 53. f x, y f x, y 12 3x fx z 1 2 2 2 1 2 1z 1. 0 0 ≤4 Solving yields the critical points 0, a, b , c, 1, d , e, f, These points are all absolute maxima. 0 x 12 3x 2x 1 5x fx 12 3x 2 2x 4 x fx 12 3x 2 1 2 x and the maximum is 10, the minimum is 6. Absolute maximum: 10 at 0, 1 Absolute minimum: 5 at 1, 2 1 y=x+1 (1, 2) 2 4 2x y 3 1 2 and the maximum is 6, the minimum is 5. On the line y f x, y 4, 1 ≤ x ≤ 2, 10 and the maximum is 10, the minimum is 5. On the line y f x, y 1, 0 ≤ x ≤ 1, 2x 2y has no critical points. On the line y 2. x 1, 0 ≤ x ≤ 2, 1 y = −2x + 4 (0, 1) (2, 0) 10 1 y=− 2 1x 2 +1 x 3 233 234 Chapter 13 54. f x, y fx Functions of Several Variables 2x 4 2x fy 2 y 0 ⇒ 2x y 2 2x y 3 0 ⇒ 2x y On the line y f x, y y y (1, 2) 2 1, 0 ≤ x ≤ 1, x fx 2x x 2 1 x 1 2 1 2 and the maximum is 1, the minimum is 0. On the line y f x, y fx 1 2 2x x 2 1 5 2 x fx 2x 2x 2 4 1 4x (0, 1) 1, 0 ≤ x ≤ 2, x (2, 0) x 1 2 and the maximum is 16, the minimum is 0. On the line y f x, y y = 2x 1 2 3 4, 1 ≤ x ≤ 2, 2x 2 4 and the maximum is 16, the minimum is 0. Absolute maximum: 16 at 2, 0 Absolute minimum: 0 at 1, 2 and along the line y 55. f x, y 3x2 2y 2 fx 6x 0 ⇒x 0 fy 4y 4 0⇒y 1 2x. On the line y f x, y 4y y (−2, 4) f 0, 1 2 3 2 ≤ x ≤ 2, 4, 2 fx 3x 32 2 2 16 3x 16 1 f x, y fx 22 3x 2 2x 4 4x 2 2x 2 x 2 ≤ x ≤ 2, x2, and the maximum is 28, the minimum is 16. On the curve y 2 (2, 4) 2 x 2x −2 x −1 1 2 1 1 8. and the maximum is 28, the minimum is Absolute maximum: 28 at ± 2, 4 2 at 0, 1 Absolute minimum: y2 y 56. f x, y 2x 2xy fx 2 2y 0 ⇒y 1 fy 2y 2x 0⇒y x⇒x On the line y f x, y fx f x, y 2x 2x 1 2x x2 2x 11 16 Absolute minimum: 2x y fy x 0 f 0, 0 (1, 1) x −1 x2 2 x4 2x3 1 2x 11 16 . Absolute maximum: 1 at 1, 1 and on y fx (−1, 1) 1. and the maximum is 1, the minimum is x2 2 1 1≤x≤1 x2, fx 57. f x, y f 1, 1 1 ≤ x ≤ 1, 1, On the curve y 1 1 11 2, 4 0.6875 at x, y : x ≤ 2, y ≤ 1 0 xy, R 0 1, Thus, f 2, 1 Thus, f Along x Along x 2, f 2 0 1 2, x2 1 6, f 1 2, 2, 1 ≤ y ≤ 1, f Thus, the maxima are f 2, 1 2x f 2, 2x 1 2y ⇒ f 4 4 6 and f 2, 1 1 −2 1 0⇒x 1 2. 2. 2 2y ⇒ f x −1 1 2. 6. x, f 1 4, 0⇒x 1 and f 2, 1 x2 1 1 ≤ y ≤ 1, f 2, x, f 1 4 1 2 ≤ x ≤ 2, f 1, 2, y 2 ≤ x ≤ 2, f Along y Along y x y 0. 2 0. 6 and the minima are f 1 2, 1 1 4 and f 1 2, 1 1 4. S ection 13.8 x2 58. f x, y fx 2x 2y 2x 2y 0 f x, x2 x Along y f 2x x2 1, f 2x 2 1 ≤ y ≤ 1, f 2x 2y 0 fy 2x 2y 0 x x2 2, y ±8 x2 x2 2, f 0, 0 0 0 1 4 2y 9 and f 2, 1 9, f 1, 1 0, f 2, 1 4 x 1 ≤ x ≤ 1. 0, y x2 and 16 f 2, 16 and f −4 x2. Thus, x −2 2 4 −2 −4 16 4x2 . 8 x2 2 2, f 2 2, 2 0 16, and the minima are f x, x, y : 0 ≤ x ≤ 4, 0 ≤ y ≤ y 8 x2 ± 12 ± ± 2. 4x2 or x 5, R x 8 ± 2x 8 28 2x2 2 x2 and y 8 0, x ≤ 2. x x y 4 0 3 2 0, 0 ≤ x ≤ 4, f x2 5 and f 4, 0 21. Along x 4, 0 ≤ y ≤ 2, f 16 16y 16 x, 0 ≤ x ≤ 4, f Thus, the maximum is f 4, 0 x2 1. y2 ≤ 8 Along y 61. f x, y −2 0. 5 Along y 9. 0. 9, and the minima are f x, x2 8 12 4xy 4y 4x 2, 2y y 2, f 4y 8, we have y 2 0 implies 16 2x fy 0, f 2, 1 4 Thus, the maxima are f 2, 2 fx 1, 1 0 y2 f f 1, f y 2, f x, y : x2 x2 x2 ± 2x 8 x2 1, f 1 2 f 60. f x, y 2, 1 x 2x2 On the boundary x2 f 2, 2 4 1 y 2, R 2xy fx Then, f 1, f 0⇒x 4y Thus, the maxima are f f x, 0⇒x 2 4 1, f 2, x2 x −1 2x 2x 2, 59. f x, y 0 1 ≤ y ≤ 1, f x2 Along x x 2 ≤ x ≤ 2, 1, Along x y 2 ≤ x ≤ 2, x Along y f y 2x2 1, 2 235 2 0 fy x, y : x ≤ 2, y ≤ 1 y2, R 2xy Extrema of Functions of Two Variables 4xy 1 y2 4 1 x2 y 1 x2 1 fx y2 fy 4 1 y2 x 2 x 1 y2 1 4x3 2 5, f 2x 1 0 and f 4, 2 6x1 2 11. 11. x, y : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 0⇒x 1 or y x 1 0 on 0, 4 . 21 and the minimum is f 4, 2 ,R 1 2 x2 5, f y 1 0 R 2 ⇒x 0 or y 1 x 1 For x 0, y 0, also, and f 0, 0 For x 1, y 1, f 1, 1 0. 1. The absolute maximum is 1 f 1, 1 . The absolute minimum is 0 f 0, 0 . In fact, f 0, y f x, 0 0. 2 3 4 236 Chapter 13 62. f x, y Functions of Several Variables 4xy 1 y2 x2 fx y2 4 1 x2 y 1 x2 1 fy x2 4 1 y2 x 1 y2 1 2 0⇒x 1 or y 2 0⇒y 1 or x 0, y For x 1 and y y2 y2 ≤ 1 0 0 0, also, and f 0, 0 For x For x2 x, y : x ≥ 0, y ≥ 0, x2 ,R 1 0. 1, the point 1, 1 is outside R. 1, f x, y Absolute maximum is f x, 8 9 4x 1 2 x2 x2 1 x2 , and the maximum occurs at x x4 2 ,y 2 2 . 2 2 2 , . 2 2 f The absolute minimum is 0 f 0, 0 . In fact, f 0, y f x, 0 0 y 1 R x 1 63. False 64. False Let f x, y 1 x y. 0, 0, 1 is a relative maximum, but fx 0, 0 and fy 0, 0 do not exist. Section 13.9 x2 y2 12 2x 3y 2 2x 2 12 2x 3y Sy 2y 2 12 2x 3y 3. 0 and Sy From the equations Sx system 5x 6y 3x 5y 0, we obtain the 18. 12 7 54 7 12 18 7, 7, 2 6 7. 2 x 2x 1 Sy 2y 12 7, y 18 7 Therefore, the distance from 6 7 is 18 7 2 6 7 2 6 14 . 7 6x 9 2x 3y 2 29 2 2 29 1 5x Solving simultaneously, we have x the origin to 2. A point on the plane is given by x, y, 12 2x 3y . The square of the distance from 1, 2, 3 to a point on the plane is given by 2x 3y 3. y From the equations Sx system 24 24 7 1 Saddle point: 0, 0, 0 Sx 2 12 y 2. Relative minima: ± 1, 0, S Sx z 2x2 Applications of Extrema of Functions of Two Variables 1. A point on the plane is given by x, y, 12 2x 3y . The square of the distance from the origin to this point is S x4 Let f x, y 6y 0 and Sy 2x 3y 2 0, we obtain the 19 10y 2 29. Solving simultaneously, we have x and the distance is 16 14 2 1 31 14 2 2 16 14 , 43 14 31 14 , y 2 3 z 43 14 1 . 14 S ection 13.9 Applications of Extrema of Functions of Two Variables 4. A point on the paraboloid is given by x, y, x2 y 2 . The square of the distance from 5, 0, 0 to a point on the paraboloid is given by 3. A point on the paraboloid is given by x, y, x2 y 2 . The square of the distance from 5, 5, 0 to a point on the paraboloid is given by 2 2 x2 y2 x Sx 2x 5 4x x2 Sy 2y 5 4y x2 S x Sx 2x 0. y2 Sy 2y y2 5 y 5 From the equations Sx system 2x3 2xy 2 3 2x2y 2y 0 and Sy x 5 y 2 0 S 2x3 5 2 1 2 5 2 2 0 P xyz 30xy Px 30y 2xy 30x x2 Py y z y 30 2xy x 30 x 32xy 2 Px 32y 2 2xy 2 Py 64xy 2x2y x y3 3xy 2 Ignoring the solution y into Py 0, we have x2y 2 x2 y2 Sx 2x Sy 2y 1.235 30, z 30 x y y2 0. 4y x2 x 0 and Sy 5 2x y 0, we obtain the 0 y 2 0 0. 5 2 1.525 2 1.235, y 0, 4.06. y. 30 0x 2y 30 10, and z 10. 2y y. Therefore, y 2 32 2x y 2x2 y 64x 2x S 0 3xy 0. 32 2x x2 y2 Sx 2x Sy 2y y y2 z2. 2 30 x 2 30 x y 1 0 2x 2 30 x y 1 0x 10, y y 30 2y 30. 10, and z 10. 0 8 x2 7. Let x, y, and z be the numbers and let S Since x y z 30, we have xy 3 0. 16, and z 8. x2 8. Let x, y, and z be the numbers and let S Since x y z 1, we have S y2 Solving simultaneously yields x 3x 32 8, y 4x x2 2 Solving as in Exercise 3, we have x z 1.525 and the distance is 0 2x 0 and substituting y 4x x Therefore, x y 10, y 32 xy 2z 2x2 z 2x P 64x y y2 32, z 3 y2 xy 2 Solving simultaneously yields x 6. Since x x2 6. 5. Let x, y and z be the numbers. Since x x2y y2 5 2xy 2 2y 0. Multiply the first equation by y and the second equation by x, and subtract to obtain x y. Then, we have x 1, y 1, z 2 and the distance is 1 2 5 From the equations Sx system 0, we obtain the 0 5 237 y y2 z2. 2 1 x 21 x y 0 2x y 1 21 x y 0x 2y 1. Solving simultaneously yields x 1 3, y 1 3, and z 9. Let x, y, and z be the length, width, and height, respectively. Then the sum of the length and girth is given by x 2y 2z 108 or x 108 2y 2z. The volume is given by V 1 3. xyz 108zy 2zy 2 2yz2 Vy 108z 4yz 2z2 z 108 4y 2z 0 Vz 108y 2y 2 4yz y 108 2y 4z 0. Solving the system 4y 2z 108 and 2y 4z 108, we obtain the solution x 36 inches, y 18 inches, and z 18 inches. 238 Chapter 13 Functions of Several Variables 10. Let x, y, and z be the length, width, and height, respectively. Then C0 The volume is given by V y 2 2C0 3x2 4x y Vy x2 2C0 3y 2 4x y 2xz and z x2 2C0 9x2 16x2 6xy 6xy 2 . 0 and Vy 0, we note by the symmetry of the equations that y 1 3 9x2, x 0, 2C0 c 4 abc 3 4 3 ab k a kab a2b 2C0 , y 1 3 b2 x 0 2ab x. Substituting y 0. 2C0 , and z 1 4 2C0 . k. Then 4 3 b Va 4 kb 3 2ab Vb 4 ka 3 a2 b ab2 b2 0 kb 2ab 2ab 0 ka a2 Solving this system simultaneously yields a 12. Consider the sphere given by x2 Then the volume is given by V 2x 2y 2 r 2 Vx 8 xy Vy 8 xy x2 y2 z2 y2 r2 x x2 y2 r2 y x2 y2 y z b and substitution yields b k 3. Therefore, the solution is a r 2 and let a vertex of the rectangular box be x, y, 8xy r 2 x2 y r2 x2 y2 x r2 x2 y2 y2 8y r2 x2 y2 r2 8x x2 y2 r2 r2 2x2 x2 y2 2y 2 Solving the system 2x2 y2 r2 x2 2y 2 r2 yields the solution x r 3. 13. Let x, y, and z be the length, width, and height, respectively and let V0 be the given volume. Then V0 C0 1.5xy . 2x y 2 In solving the system Vx into Vx 0 yields V 2yz C0 xy 1.5x2y 2 2x y xyz Vx 11. Let a 1.5xy xyz and z V0 xy. The surface area is V0 x 2 xy S 2xy 2yz 2xz Sx 2y V0 x2 0 x2y V0 0 Sy 2x V0 y2 0 xy 2 V0 V0 y 0. Solving simultaneously yields x 3 V0, y 3 V0, and z 3 V0. 0 0. r2 x2 b y2 . c k 3. S ection 13.9 Applications of Extrema of Functions of Two Variables 239 14. Let x, y, and z be the length, width, and height, respectively. Then the sum of the two perimeters of the two cross sections is given by 2x 2z 2z 144 or x y 2z 2y 2yz2 72 y 2z. The volume is given by V xyz 72yz Vy 72z 2yz Vz 72y y2 2z2 4yz Solving the system 2y x 15. A x 2x2 sin 2x2 cos A x 30x 4z 72, we obtain the solution 18 inches. 2x cos x sin 2x sin cos 0 x2 2 cos2 0 we have 15 A From 0. x2 sin cos 4x sin 30 cos From 2x 0 4z 72 and y 30 30 sin A y 2z 24 inches, and z 2x 30x sin 2y y 72 2z 24 inches, y 1 30 2 A z 72 2x 1 0 2x 0 ⇒ cos x cos 15 x . 0 we obtain 2x 15 2x2 x 30 2x 2x 15 x2 2 x 15 2x 2x 15 2x 15 2 1 2 2x 15 2 3x2 0 x2 0 30x x 0 x 1 ⇒ 2 Then cos 10. 60 . 16. Let h be the height of the trough and r the length of the slanted sides. We observe that the area of a trapezoidal cross section is given by A h w 2r where x r cos A r, w w 2 and h 2r 2r r 2x w 2r xh w − 2r r sin . Substituting these expressions for x and h, we have r cos r sin 2r 2 sin wr sin r 2 sin cos . Now Ar r, w sin A r, wr cos 4r sin 2r sin cos 2r 2 cos sin r 2 cos 2 2 cos cos 2r cos 2 0⇒w 2r cos 0 into the equation A r, 2 1 r4 2 cos 1 0 or cos 0, we have 0 r 2 2 cos 4 2 4r 0. Substituting the expression for w from Ar r, r2 w r 2 cos 1 . 2 Therefore, the first partial derivatives are zero when 3 and r w 3. (Ignore the solution r trapezoid of maximum area occurs when each edge of width w 3 is turned up 60 from the horizontal. 0.) Thus, the θ x h 240 Chapter 13 Functions of Several Variables 5x12 17. R x1, x2 8x22 2x1x2 42x1 102x2 18. R 515p1 805p2 1.5p1 p2 Rx1 10x1 2x2 42 0, 5x1 x2 21 R p1 515 1.5p2 3p1 0 Rx2 16x2 2x1 102 0, x1 8x2 51 R p2 805 1.5p1 2p2 0 515 2p2 805 4p2 1610 2.5p2 2 Rx2x2 6. 10 Rx1x2 3 and x2 1.5p2 3p1 Rx1x1 2125 16 Rx1x1 < 0 and Rx1x1 Rx2 x2 Rx1x2 2 19. P x1, x2 15 x1 x2 C1 15x1 15x2 p2 3 and x2 6. C2 0.02x12 4x1 0.02x12 0.05x22 Px1 0.04x1 11 0, x1 275 Px2 0.10x2 11 0, x2 0.05x22 500 110 Px1x1 11x1 11x2 0.04 Px1x2 0 Px2x2 0.10 Px1x1 < 0 and Px1x1 Px2 x2 Px1x2 2 >0 Therefore, profit is maximized when x1 20. P p, q, r q 2pq r P p, q 2pr 2p 1 2pq 2pq 2q Solving P p p 2p 2 4p; P q q 2p 2q p 2q 1 2pq 2q 2p2 2q P q q. 2p p q 2pq 2q2 2q2 2 4q 0 gives 1 and hence p 110. 1 p 1 q 2p2 2p 275 and x2 2qr. 1 implies that r 2pq P p 775 q P 11 , 33 1 and 3 2 6 9 1 9 2 . 3 2 1 3 2 1 3 2 1 9 2 1 9 4x2 275 850 p1 >0 Thus, revenue is maximized when x1 p 3p1 1.5p1 Solving this system yields x1 5962 3 1.5p12 p22 S ection 13.9 x2 21. The distance from P to Q is C x2 3k Cx 2k x x2 3k Cy 4 2k y 2 x 2k 2 x x 3k x x2 d1 d2 x x x 0 y dS dy y 2 y 2 2 x x 2 1 3 0 2 d1 d2 x 2 1 2 1 1 1 1 2 2 23 32 6 23 32 1.284 kms. 6 y 2 0 y 2 0 when y 0 23 3 2 2 6 y 2 2 0 2 2 y 2 2 2 2 The sum of the distance is minimized when y 23. (a) S x, y y x 1 3 24 2y 2 4 y2 1 y 4 0.707 km and y d3 0 2 2 y 22. S 23 23 . 3 3 0.845. 3 d3 x 0 x2 2 y y2 0 x 2 2 x 2 y 2 2 2 2 y 2 x 2 4 x 2 y 4 2 2 y 2 2 2 From the graph we see that the surface has a minimum. 2 tan 2 y x y2 22 y y x2 2 Sx 1, 1 i 10 —CONTINUED— 1 1 2 2 Sy 1, 1 j 2 1 S x x y S 1, 1 2 2 y2 x Sy x, y (c) x x (b) Sx x, y y. 1 2 x 2 2 1 . The distance from R to S is 10 4 2 9x Therefore, x 2 y 1 0⇒ x2 2 y x 1 3 4 3x 4y 2 y 0 x 2y k 10 x x k 1 2 x2 1 y 1 2k 4 x 2 y x y 4. The distance from Q to R is y 4 Applications of Extrema of Functions of Two Variables 24 4 20 2 2 x 4 y 2 2 2 2 x y2 42 y 2 2 1 i 2 2 ⇒ 5 1 2 186.027 2 2 j 10 4 8 x 6 4 2 2 4 6 8 y 241 242 Chapter 13 Functions of Several Variables 23. —CONTINUED— (d) x2, y2 Sx x1, y1 t, y1 1 t, 1 2 S1 x1 2 10 Sy x1, y1 t 1 t 2 1 t, 1 2 1 2 10 5 2 2 10 2 2t 1 t 2 25 5 1 22 t 5 10 2 10 5 2 2t 1 25 5 22 t 5 10 2 10 5 4 2t 1 25 5 22 t 5 Using a computer algebra system, we find that the minimum occurs when t (e) x3, y3 x2 S 0.05 Sx x2, y2 t, y2 0.03t, 0.90 Sy x2, y2 t 0.26t 0.05 0.05 0.03t 0.03t, 0.90 2 2 0.26t 2.05 0.03t 3.95 Using a computer algebra system, we find that the minimum occurs when t x4, y4 x3 S 0.10 Sx x3, y3 t, y3 0.09t, 0.45 Sy x3, y3 t 0.01t 0.10 0.10 0.09t 1.10 2 2 0.01t 2.10 Using a computer algebra system, we find that the minimum occurs when t 0.26t 2 0.10, 0.44 . 0.09t 2 1.55 2 0.09t 0.01t 1.55 0.44. Thus, x4, y4 24. (a) S x 2 4 y2 x 1 0.06, 0.44 . 0.0555, 0.3992 2 y 6 2 x The surface appears to have a minimum near x, y (b) Sx Sy x x 4 4 2 x y2 x y x 4 (c) Let x1, y1 2 x y y6 12 y 6 6 2 2 2 z 30 x y 2 1, 5 . 1 2 2 x x 12 y 12 12 2 y 2 2 2 2 −2 x y 2 2 4 2 −4 2 4 6 8 1, 5 . Then S 1, 5 Direction y2 1 12 0.258i 0.03j. 6.6 1.24 y2 (d) t 0.94 x2 5.03 (e) t 3.56, x3 1.24, y3 5.06, t 1.04, x4 1.23, y4 5.06 Note: Minimum occurs at x, y 1.2335, 5.0694 S x, y points in the direction that S decreases most rapidly. 25. Write the equation to be maximized or minimized as a function of two variables. Set the partial derivatives equal to zero (or undefined) to obtain the critical points. Use the Second Partials Test to test for relative extrema using the critical points. Check the boundary points, too. 2 0.01t S x, y points in the direction that S decreases most rapidly. You would use S x, y for maximization problems. (f) (f ) 2 0.01t 0.45 0.26t 1.10 1.78. Thus x3, y3 0.09t, 0.44 2 2 0.03t 3.90 Note: The minimum occurs at x, y 0.05, 0.90 . 0.26t 0.90 1.344. Thus, x2, y2 y 2 S ection 13.9 Applications of Extrema of Functions of Two Variables 243 26. See pages 962 and 963. 27. (a) y xy 0 0 1 0 2 3 6 4 xi 0 yi 4 xi yi x2 0 0 9 1 x a y 04 02 3 x 4 3 ,b 4 3 0 4 1 4 3 1 1 3 8 4 , 3 2 6 9 0 yi a 04 02 y 3 x 10 2 4 3 2 4 3 0 3 2 1 1 10 0 2 4 3 3 1 6 29. (a) y xy 4 0 3 3 1 4 4 3 ,b 10 20 2 2 7 10 3 0 10 2 13 10 1 1, 19 10 1 1 1 0 0 1 x 2 2 5 4 2 2 3 2 7 4 2 2 2 3 4 4 2 1 0 1 2 4 1 1 4 xi 4 a 44 46 y 2x 8 xi yi 4 xi2 2, b 1 8 4 24 4, yi 48 42 6 4 x y xy x2 3 0 0 9 1 0 0 1 2 0 0 4 3 1 3 9 4 1 4 16 4 2 8 16 5 2 10 25 6 2 12 36 xi (b) S xi2 6 0 1 (b) S x2 0 a xi yi 2 2 1 5 x 30. (a) 4 1 (b) S (b) S 46 4 20 4 3 3 2 1 1 xi 36 38 1 1 xi2 6 xy 1 0 2 y 3 4 0 28. (a) x2 x 28 8 37 8 116 3 4 1 4 xi yi 37 xi2 72 144 28 8 28 2 2 0 8 yi 1 ,b 2 1 8 8 1 28 2 2 0 1 4 2 0 3 4 116 3 ,y 4 2 1 5 4 1 2 2 9 4 2 2 3 2 0 2 2 244 Chapter 13 Functions of Several Variables 31. 0, 0 , 1, 1 , 3, 4 , 4, 2 , 5, 5 xi 13, xi yi yi 12, xi2 46, a 5 46 5 51 13 12 13 2 b 1 12 5 37 13 43 y 37 x 43 32. 1, 0 , 3, 3 , 5, 6 xi 51 74 86 xi yi 37 43 3 39 3 35 b 1 9 3 y 7 y = 37 x + 43 43 7 yi 3 x 2 (5, 5) 9, xi2 39, a 7 43 7 43 9, 35 36 24 99 92 3 9 2 3 2 9 6 3 2 3 2 7 (3, 4) (4, 2) −1 (1, 1) (0, 0) −2 6 10 −1 −1 33. 0, 6 , 4, 3 , 5, 0 , 8, xi 27, 5 70 5 205 xi 42 xi yi 175 148 yi a 945 148 xi2 1 31 6 29 x 53 8 400 42 31 42 2 6 275 6 400 y 945 148 31 29 53 275 b 175 27 148 175 x 148 y 6 205 350 296 27 0 27 2 1 0 5 b 0, xi2 70, 34. 6, 4 , 1, 2 , 3, 3 , 8, 6 , 11, 8 , 13, 8 ; n 5 yi xi yi a 4 , 10, 29 42 53 425 318 0.5472 1.3365 (0, 6) (4, 3) −4 (5, 0) (8, − 4) −6 18 425 318 9 (10, − 5) y = − 175 x + 945 148 148 −1 14 −1 35. (a) y (b) 1.7236x 79.7334 36. (a) 1.00, 450 , 1.25, 375 , 1.50, 330 xi 240 3.75, xi yi yi a 100 (c) For each one-year increase in age, the pressure changes by 1.7236 (slope of line). 3 1,413.75 3 4.8125 b 1 1,155 3 240x (b) When x 37. 1.0, 32 , 1.5, 41 , 2.0, 48 , 2.5, 53 xi a 7, 14, b When x yi 19, y 1.6, y 174, 14x xiyi 322, 38. (a) y xi2 19 41.4 bushels per acre. 13.5 xi2 4.8125, 1,413.75 y 0 100 1,155, 1.85x 3.75 1,155 3.75 2 240 3.75 240 685 685 1.40, y 240 1.40 685 48.3 (b) For each 1 point increase in the percent x , y increases by about 1.85 (million). 349. S ection 13.9 Applications of Extrema of Functions of Two Variables 245 n 39. S a, b, c bxi 2 c 1 n S a 2xi2 yi i axi2 bxi c 0 1 n S b axi2 2xi yi i bxi c 0 1 n S c 2 axi2 yi i n xi4 i i i 1 n c 1 i n xi xi yi 1 i 1 n xi2 i i n xi2 b 1 xi2yi 1 n xi3 a 0 n xi2 c 1 n i c n xi3 b 1 a bxi 1 n a i axi2 yi i n b 1 xi i cn yi 1 i 1 n 40. S a, b axi i b n Sa a, b 2 yi 1 n xi2 2a i n 2b 1 xi i 2 1 n Sb a, b 1 n 2a xi i xi yi i 2nb 2 1 yi i 1 n Saa a, b xi2 2 i 1 Sbb a, b 2n Sab a, b 2 n xi i 1 0 for all i. (Note: If xi Saa a, b > 0 as long as xi n d Sab2 SaaSbb i As long as d 41. 2, 0 , 2 n xi2 4n 4 1 xi i 0 for all i, then x n 1 i 2 n xi2 4n 1 xi i 0 is the least squares regression line.) ≥ 0 since n 1 i n 1 xi . i 1 0, the given values for a and b yield a minimum. 1, 0 , 0, 1 , 1, 2 , 2, 5 xi 8 xi2 10 xi3 0 xi4 4, 5 , 42. 0 yi 2, 6 , 2, 6 , 4, 2 xi 0 yi 19 xi2 40 xi3 0 34 xi4 544 xi yi 12 xi yi xi2yi 22 xi2yi 8 (2, 5) (−1, 0) 8 (−2, 6) (− 4, 5) 34a b 6 (−2, 0) (0, 1) −2 10c 3 7, −9 22, 10b 6 5, c 26 35 , 12, 10a y 32 7x 5c 6 5 x 8 26 35 544a a (2, 6) (4, 2) (1, 2) a 2 n xi2 ≥ −9 9 −4 12 160 40c 5 24 , b 160, 40b 3 10 , c 12, 40a 41 6, y 52 24 x 4c 19 3 10 x 41 6 246 Chapter 13 Functions of Several Variables 44. 0, 10 , 1, 9 , 2, 6 , 3, 0 43. 0, 0 , 2, 2 , 3, 6 , 4, 12 xi 9 xi 14 yi 6 11 (0, 10) (4, 12) yi xi2 14 xi3 36 (1, 9) 25 20 (2, 6) (3, 6) xi2 29 (2, 2) −5 7 (0, 0) −2 (3, 0) −9 xi3 99 xi4 353 xi4 98 xi yi 70 xi yi 21 xi2yi 254 xi2yi 9 −1 33 353a 99b 29c 254 98a 36b 14c 33 99a 29b 9c 70 36a 14b 6c 21 29a 9b 4c 20 14a 4c 25 a 1, b 1, c x2 0, y 45. (a) 0, 0 , 2, 15 , 4, 30 , a x (b) 120 46. (a) y 6, 50 , 8, 65 , 10, 70 xi (c) −1 230 xi2 b 9 20 , 199 20 , c 0.075x 5.32 2 (b) y 30 yi 6b 5 4, 0.002x y x 0.10x 52 4x 9 20 x 199 20 4 is 1994 . 5.22 8 14 220 − 20 2 14 0 xi3 1800 xi4 15,664 xi yi 1670 xi2yi (d) For 2010 x 20 , the linear model gives 6.82 billion and the quadratic model gives 6.42 billion. The quadratic model is less because of the negative x 2-term. 13,500 15,664a 1800b 220c 13,500 1800a 220b 30c 1670 220a 30b 6c 230 25 2 112 x y 541 56 25 14 x 0.22x 2 47. (a) ln P 0.1499h 0.1499h 1.79 48. (a) 9.3018 (b) ln P 9.66x 1 y ax P (c) e 0.1499h 9.3018 10,957.7e (b) 14,000 −2 − 2,000 (d) Same answers y 0.1499h 24 0.0074x 1 0.0074x 9.3018 b 0.445 0.445 40 0 60 0 (c) No. For x 70, y 14, which is nonsense. y 1000 which seems inaccurate. Section 13.10 Section 13.10 f 2. Maximize f x, y xy. y f xj i y x y 4 g yi j x xy. Constraint: 2x 10 g yi 247 Lagrange Multipliers 1. Maximize f x, y Constraint: x Lagrange Multipliers xj 2i y y j 2 x x 10 ⇒ x y f 5, 5 y 5 2x 4⇒4 y 4 25 1, x y f 1, 2 1, y 2 2 12 10 Level curves 8 6 4 2 Constraint x 2 4 6 8 10 x2 3. Minimize f x, y Constraint: x f 12 y y 2. Constraint: 2x 4 f g 2x i 2x i x 2y f 2, 2 y 2. 4y 5 g 2y j y 8 2 2x 4j 2x y 4⇒x y 2x i j 2i 2 ⇒x 2y 2y j x x2 4. Minimize f x, y 4 ⇒y 2 5 ⇒ 10 4y y Constraint 4 5 1 2, f 1 2, x 1 2, y 1 5 4 1 x −4 4 Level curves −4 x2 5. Minimize f x, y Constraint: x f 2y y 2. 6 Constraint: 2y g 2x i f 2y j i 2j 2y x 2x i 2x ⇒x 2x 2 6 ⇒ 2 If x ⇒y 2y 6 4, x f 2, 4 12 2y j 4 f 0 2x 2x i ⇒x 2j 0 or 1 0 and f 0, 0 0. 2⇒y 0, then y 1 ⇒ x2 1, 2y 2, y x2 y 2. g If 3 2 x2 6. Maximize f x, y 2 2, 1 2 1 1, Maximum 2⇒x 2. 248 Chapter 13 Functions of Several Variables 7. Maximize f x, y 2x Constraint: 2x f y. 8. Minimize f x, y 100 f 2y i 2 2x 2x 2i ⇒y 2 1 x f 25, 50 3 2xy ⇒ 2 xy 1 x2 x2 ⇒ 1 100 25, y 10. x2 j 2 xy i 3 2x 2 100 ⇒ 4x y j 1 1y y 6 g 3i j ⇒x 2y 2x 1j 3x Constraint: x2y g 2 y 2 xy x 3x 2 6 ⇒ x2 x2y 50 2600 9. Note: f x, y is maximum. Maximize g x, y Constraint: x x 2y y x2 6 y y 2 is maximum when g x, y y 2. 2 4 Constraint: x2 ye xy 2x xe xy 2y y2 3 ,3 2 g Constraint: xy 2 y y 1 y x e 15 2x 3 ,3 2 12. Minimize f x, y 8 4 4y x 8 x y 2. 15 ⇒ 10 x 4y x xy y 16 3 ,y 2 35 2 y. 2x x f 4, 8 2x 15 32 32 ⇒ 2x2 2 4, y 33 4 2 y 2 is minimum when g x, y is x2 y 3 4 20 2 2 8 ⇒ 2x2 y2 4 2x e xy. 11. Maximize f x, y 2 2y 1 3 x2 Constraint: 2x y g 1, 1 9 Minimize g x, y f f 2, 2 3 2 10. Note: f x, y minimum. y 2⇒x f 1, 1 x2 3 2x 2x x x2 6 4, 4 x 3 3x 2 0 6 x3 f 2 xy ⇒ y 3x2 32 4, y 8 3 S ection 13.10 x2 13. Maximize or minimize f x, y Constraint: x2 y2 2x 3y 3x 2y Case 2: Inside the circle 1 2y x2 ± 2 ,y 2 14. Maximize or minimize f x, y Case 1: On the circle x 2 x2 15. Minimize f x, y, z Constraint: x y x y e1 8 e z 2. 0.8825 ≤0 2 1 2 2 , 2 at ± 2 . 2 16. Maximize f x, y, z yz xz xy z 2 y 1 2 14y x y y x z z 10 y 10 ⇒ 12 y z 2 8 g xz j y y 84 70 2 y y x 111 3, 3, 3 z 1 4 y z yz y z 32 x y z 0 y z 1 3 0 j k xy ⇒ x 2x 2z i j 1024 k z 32 ⇒ x y f 8, 16, 8 z 32 i xyk z 2. 1 1⇒x z y2 x yz. z x 36 x2 h yz xz xy 5. 40 x f 6⇒x f 2 1 xy 16 xy e 1 z x 14 0 < 0. 2x 2y 2z 10 14 fxy Constraint: x 6 yz i 1 2 2 fxy xy 4 , y 17. Minimize f x, y, z x yz. Constraints: x 12 12 x2 e 16 fyy 19. Maximize f x, y, z 70. 10 x y xy 4 , ⇒x 0 Combining the two cases, we have a maximum of e1 8 at 2 2 2 2 and a minimum of e 1 8 at ± , ,± ± . 2 2 2 2 Constraint: x y x 4e y2 e 16 0 xy 4 Saddle point: f 0, 0 18 6 y2 y 10 14 y 10 16 fxy 0 At 0, 0 , fxx fyy 1.1331 z 10x 2x 2y x f 4, 6 0 3, fxx fyy xy 4 y 4e fy f 2, 2, 2 Constraint: x 3, y fx y2 12 Then x 2, fxy y 2 2 and a minimum of ,± 2 2 fxx 2 2 y2 z 6⇒x 18. Minimize x2 2, fyy x Case 2: Inside the circle 2 2 ,± Minima: f ± 2 2 x ± 2 2 ± 2 , 2 Maxima: f ± f 2, 2, 2 0 By combining these two cases, we have a maximum of 5 at 2 ⇒ x2 2y 1⇒x y2 z 2y Saddle point: f 0, 0 1 2x xy 4 x 4e y 3x x y 4. e y2 xy 4 y 4e 2x 2y 2z 0 y2 ≤ 1 Constraint: x 2 x2 2 2 1 2 2 2 , 2 2 Minima: f ± ± 3y fxx 5 2 1⇒x 2x fy y2 2 2 ,± Maxima: f ± 2 2 x y 2. 3xy fx 2x y2 249 ≤1 y2 Case 1: On the circle x 2 x2 Lagrange Multipliers z 16 8 1 3 250 Chapter 13 Functions of Several Variables x2 20. Minimize f x, y, z Constraints: x y xy 2y 6 x 3z 0 12 h 2y j 2z k f i 2k 2x 2y 2z 6 ⇒z 6 y 12 ⇒ y 12 i j x 2x 2x 2 12 x 2y z x x x 2 3 g yi 2 2z x 21. Maximize f x, y, z 6 g 2x i z 2. Constraints: x 2z x f y2 h x y z y zj x x 2y 6⇒ y 3 3z 0⇒ z x 2 9 ⇒x 2 27 ⇒ x 3k 8 y 3 z x 2 6 0 x 2 x 3, y 3 ,z 2 x x yz. x2 z2 2y 0 g xz j 2y 2x i 2z k 2⇒ ⇒ x2 z2 5 ⇒z x 2y 0 ⇒y yz 2x 3y y 5 x2 x5 2 x2 x 3x 2 10 1 2 x 2 xz 2 x3 25 d x 2 x3 25 x2 x3 0 3x 3 x 0 or x 10 , 3 5 10x 5 3 10 ,z 3 5 15 9 0 does not yield a maximum. 5 3 x2 3x 2 1⇒x The point on the line is distance is 10 ,y 3 xy 2z x2 Note: f 0, 0, 3 2x xy 2z 2z 10 1 , 32 2j xy 2 xy x2 i 2 2x xz x2 6 2x h xyk 1 23. Minimize the square of the distance f x, y subject to the constraint 2x 3y 1. 5 yz f ⇒x 8 3 3 x Constraints: 2x 5 i x 3 x 3 22. Maximize f x, y, z x5 3 4 3 3 f 3, , 1 2 x5 2 y 2 2j 72 f 6, 6, 0 yz i i x 3 6, z f yk x 2 yz. 2 13 2 2 ,y 13 2 13 , 3 13 3 13 2 3 13 and the desired 13 . 13 y2 S ection 13.10 x2 24. Minimize the square of the distance f x, y 2x 2x 2y 10 x 4 4 x 2y 2 y x 4 8x 5 2 25 2 x 4 16 10 2 subject to the constraint x y ⇒y y 4 ⇒ x2 y2 10 x 58 ± x 58 ± 2 29 29 2 582 4 29 4 112 2 29 4 Using the smaller value, we have x The point on the circle is 4 1 16 1 4. 112 0 4.7428 or, by the Quadratic Formula, 4± 4 29 . 29 y 10 29 29 1.8570. 29 10 29 , 29 29 and the desired distance is d y2 4 and 29 29 41 100 58x 3.2572 and x 2 4 251 10 50x 29 2 x 4 Using a graphing utility, we obtain x Lagrange Multipliers 2 29 29 10 29 29 2 10 8.77. The larger x-value does not yield a minimum. 25. Minimize the square of the distance f x, y, z x 2 2 y y subject to the constraint x 2x 2y 2z x y 2 1 1 y 2 z z 1 2 f x, y, z 1. z and y x 2x 2x 1 1 1 x 1 2 2 0 1 1, y 2 z 0 x2 x2 2x y2 y z 0 1 2 y2 y2 z 3. d 0, x 2 36 x2 2 y 2 5y 2 5y 30y 2 20y 32 10y 16 2 y 4 2 , 4 2x 2y 02 2y 1 5± 265 15 265 3 0, z 2 22 2 2. 2x 2y ⇒y 2z 0 2 y2 z2 0 2z 4⇒x 4 Choosing the positive value for y we have the point 265 2x . 4 2z 02 z2 0 16z 16 0 3z 0 15 0. 3z2 0 15y 2 2 265 5 , 15 z x z x 36 y2 2 y2 2, y 0 2x z2 28. Maximize f x, y, z z subject to the constraints x2 y 2 z2 0 and x 2z 4. 2y 2⇒z x2 y2 y z y x2 1 z2 2y 10 x2 0 x y2 x 4 2 2x 2y 2z 2 The point on the plane is 2, 0, 2 and the desired distance is 27. Maximize f x, y, z z subject to the constraints x2 y 2 z2 36 and 2x y z 2. 0 0 1 4 2z The point on the plane is 1, 0, 0 and the desired distance is d x subject to the constraint 2y 1⇒x z 1 26. Minimize the square of the distance 4z 4 2z 2 z The maximum value of f occurs when z of 4, 0, 4 . 0 4 3 or z 4 4 at the point 252 Chapter 13 Functions of Several Variables 29. Optimization problems that have restrictions or constraints on the values that can be used to produce the optimal solution are called contrained optimization problems. 30. See explanation at the bottom of page 969. 31. Maximize V x, y, z x yz subject to the constraint x 2y 2z 108. 32. Maximize V x, y, z 1.5x y 2 xz 2yz yz xz xy x yz 2 2 2y y z and x 2y 108 ⇒ 6y 2z x 1.5y 2z 1.5x 2z 2x 2y xz xy 108, y 18 36, y z 1.5xy x yz subject to the constraint C. 2xz x 3 x 4 y and z 32 x 2 C ⇒ 1.5x 2 2yz 18 Volume is maximum when the dimensions are 36 18 18 inches. 32 x 2 x Volume is maximum when z x 2C 3 y and 2C . 4 z x y 8y xyz 6z 6z 6y xz xy x 480 ⇒ 4 y 3 3 y, 4y 2 rh r2 3 Dimensions: y 3 360 z 3z 360 3 43 3 360, z 43 3 360 4r 2r V0 ⇒ 2 r 3 r 2h 480 x 3 Dimensions: r 360 feet x y 35. Maximize V x, y, z 8yz 8xy 2z c2 y2 b2 z2 c2 1⇒ x x2 a2 2y b2 a ,y 3 8 xyz subject to the constraint 2x a2 8xz 2x 2y 2z x2 a2 y2 b2 3x2 a2 z2 c2 1, b ,z 3 3y 2 b2 1, 3z 2 c2 1 c 3 Therefore, the dimensions of the box are 2 3a 3 2 3b 3 2 r 2 subject to the 2 rh V0. yz 8x 6x xy subject 2yz 34. Minimize A , r constraint r 2h 2h 33. Minimize C x, y, z 5xy 3 2xz to the constraint x yz 480. 2 3c . 3 x2 a2 y2 b2 z2 c2 1. h 2r V0 V0 2 and h 2 3 V0 2 C 2C 3 S ection 13.10 36. (a) Maximize P x, y, z x y z n S. xi i yz xz xy x x y y y S 3 z n xyz ≤ S 3 i xyz ≤ xyz ≤ x3 x2 S n x1x2x3 . . . xn ≤ S n z n . d12 v1 Distance , minimize T x, y Rate 37. Using the formula Time x2 x1x2x3 . . . xn ≤ xn S n xn x1x2x3 . . . xn ≤ x1 S n S ... S ,x ≥ 0 n ni n S n n y 3 ... ... Therefore, x1x2x3 . . . xn ≤ x x3 1 S 3 3 x2 1 S ⇒ x1 xi S , x, y, z > 0 3 S3 xyz ≤ 27 3 x1 x1x2x3 . . . xn Therefore, S 3 S. 1 x2x3 . . . xn x1x3 . . . xn x1x2 . . . xn z S⇒x z x2 ... x3 n d22 y 2 subject to the constraint x v2 y xn . a. x x2 2 v1 d1 x v1 d12 y v2 d22 x y x y v2 d22 x2 Medium 1 P y2 y2 d1 a Since sin θ1 x 1 x2 d12 x2 2 d1 v1 l l or d22 sin v1 y θ2 a Medium 2 d2 Q sin 2 . v2 1 l2 8 l subject to the constraint lh 2 l y x , we have 2 A. l 4 2 ,1 l ⇒ 2 2h l 2h Case 2: Minimize A l, h h 2h h 2 2 y2 d2 v2 y y and sin 2 38. Case 1: Minimize P l, h 1 253 x1x2x3 . . . xn subject to the constraint (b) Maximize P x yz subject to the constraint Lagrange Multipliers 2 2 h l2 lh l 4 8 subject to the constraint 2h 2 l ,h 2 ⇒ l 2 h l or l 2 2h l 4 l 2 l 4 l l 2 P. l 254 Chapter 13 Functions of Several Variables 39. Maximize P p, q, r Constraint: p q 2pq r 2pr 2qr. 1 g 2x 2x 2q 2p 2p 2r 2r 2q 2y 0 2y 2z q r 1 q r 2 3 q r 1 P p p P 100 x2 y 2 subject to the z 2 50 and x z 0. 40. Maximize T x, y, z constraints x2 y 2 111 3, 3, 3 ⇒3 4p ⇒ 2 q r 41 4 3 If y 0, then Thus, x 1 3 ⇒p 1 3 2 1 3, 1 3 1 3 1 3, q 2 1 3 1 3 If y 2 3. 50, 0 0, then x2 50 , 0, 2 T 0, z 0 and y z T 0, 1 3 r 1 and 50. 100 z2 50 2 0. 50 150 2x 2 50 and x 50 4 100 z 50 2. 112.5 Therefore, the maximum temperature is 150. 41. Maximize P x, y 100x 0.25y 0.75 subject to the constraint 48x 36y 100,000. 0.75 48 75x0.25y ⇒ x y 0.25 36 0.25 ⇒ y x y x 25x 0.75y0.75 0.75 40x 36 75 60x0.4y 0.25 48 25 y x 3125 ,y 6 3125 6250 6,3 43. Minimize C x, y 48x 48x 36y 2x y 25x 36 75x0.25y 0.25 P 6250 3 2500 5000 , 3 3 147,314. 36y subject to the constraint 100x0.25y0.75 ⇒ y x 0.75 48 25 ⇒ x y 0.25 36 75 y x 0.75 0.75 48 0.75 y x 0.25 y x 100x0.25y0.75 0.4 ⇒ y x 0.6 48 48 40 ⇒ x y 0.4 36 36 60 0.6 100,000 ⇒ x 100,000 x Therefore, P 100,000. y x 0.6y0.6 75 36 4x 100,000 ⇒ 192x 36y 36y 4 y 48x Constraint: 48x 48 25 y x 100x0.4y0.6 42. Maximize P x, y 20,000 ⇒ x0.25 4x 0.75 48 25 4⇒y 4x 200 x 200 40.75 y Therefore, C 50 2, 200 2 75 36 4x $13,576.45. 200 22 200 2 50 2 20,000. $126,309.71. y x 0.4 48 40 60 36 y 2 ⇒ y 2x x 2500 5000 ,y 3 3 S ection 13.10 44. Minimize C x, y 36y subject to the constraint 100x0.6y0.4 48x ⇒ 36 40x0.6y 0.6 0.4 x y 0.6 36 40 0.4 y x 0.6 48 60 20,000 ⇒ x0.6 8 x 9 200 y 45. (a) Maximize g , , 8 x 9 0.4 x Therefore, C 209.65, 186.35 40 36 8 ⇒y 9 y x 100x0.6y0.4 20,000. 48 60 y x y 60x y x ⇒ 0.4 0.4 48 Lagrange Multipliers 200 8 9 0.4 209.65 8 200 9 8 9 0.4 186.35 $16,771.94. cos cos cos subject to the constraint . sin cos cos cos sin cos cos cos sin tan ⇒ g tan tan ⇒ 3 1 8 ,. 333 γ 3 ⇒ (b) g 2 cos cos cos cos cos cos cos cos 46. Let r radius of cylinder, and h S h2 2 rh V 2r r 2h 2 r 2h r2 5 3 3 C C2 rh r2 2Crh h f r, h h2 C2 r4 2Cr 5r 4 2C 0 5r 4 r2 C 2 C 5 —CONTINUED— height of cone. volume C2 r4 2Cr r2 α sin constant surface area r2 C2 Fr height of cylinder r4 Fr sin cos r 2h subject to g r, h We maximize f r, h 2 r 2h cos Cr 2 r5 2C rh r h2 r2 C. 3 3 β 255 256 Chapter 13 Functions of Several Variables 46. —CONTINUED— 10r 3 C Fr C2 r4 2Cr h C2 C2 5 2C C 2 5 1 4 4 5C 2 C2 5 1 4 2C 5r 2 5r 5r 2 25 r 5 Hence, h r 25 . 5 By the Second Derivative Test, this is a maximum. Review Exercises for Chapter 13 1. No, it is not the graph of a function. 2 y2 ex 3. f x, y 2. Yes, it is the graph of a function. 4. f x, y The level curves are of the form 2 ex c c x2 y2 x y. ec The level curves are circles centered at the origin. 1 x2 c y2 . c The level curves are hyperbolas. 1 2 y c= 2 The level curves are hyperbolas. 3 c= 2 y c=1 2 c = 10 c=0 −3 4 c = 12 1 x 2 −4 −2 −2 x x y c = −2 x 1 y c c c . y2 −3 3 3 −3 3 c= 2 c=2 −3 3 3 x. The level curves are passing through the origin with slope c x2 e z −2 y 7. f x, y 1 c= 2 c=1 x 1 1 4 Generated by Mathematica c=−1 3 c = −1 2 c=−2 2 The level curves are of the form c x −1 −4 Generated by Mathematica 6. f x, y c = −12 c = − 2 c=2 3 c=1 −2 y2 ln x y c y 2. x2 The level curves are of the form The level curves are of the form y2 x2 ln c 5. f x, y ln x y x −3 y R eview Exercises for Chapter 13 8. g x, y 1 y x x2 y z 60 y z2 x2 9. f x, y, z z2 9x2 10. f x, y, z 1 1 y2 Elliptic cone Elliptic paraboloid z 2 z 2 3 x 5 2 xy lim x, y → 1, 1 x2 y 3 x 11. 1 2 y2 12. Continuous except at 0, 0 . xy lim x2 x, y → 1, 1 y2 Does not exist. ± x. Continuous except when y 13. lim x, y → 0, 0 x 4 For y x 2, For y 0, 5 y 5 x 4x2y y2 x4 x4 14. 4x2 y y2 4x2 y y2 x4 4x 4 x4 0, for x 2, for x 0 y lim x, y → 0, 0 xe y 1 x2 2 0 1 0 0 0 Continuous everywhere. 0 Thus, the limit does not exist. Continuous except at 0, 0 . 15. f x, y e x cos y fx e x cos y fy ex 16. f x, y fx sin y fy yx x y2 xy y 2 x y x2 x y 2 z ln x 2 x2 2x y2 1 gx x2 2y y2 1 gy w x2 y2 z2 w x 12 x 2 y2 z2 w y x2 y y2 x2 z y2 w z 2x x y2 ey ye x xe y ex x2 y2 y x2 y2 xy 2x x2 y 2 2 x x2 x2 21. f x, y, z 12 y2 y2 2 z arctan 1 z y2 x2 z2 fy 1 z 1 y 2 x2 x z2 fz arctan z2 y x y y 2 x2 x2 y 2 2 y x fx x2 ye x xy 19. g x, y 1 2 xe y z y y z z x 17. y z y 20. y2 xy x z x 18. 9z2 y x2 x2 yz y2 xz x2 y2 y 257 0 258 Chapter 13 Functions of Several Variables 1 22. f x, y, z x2 1 1 1 2 fx y2 x2 23. u x, t z2 y2 z2 1 x2 x y2 1 x2 y y2 z2 1 x2 z y2 z2 cne u t sin nx n 2t cos nx n 2t cn2e sin nx 32 fz u x 2x z2 3 2 fy 32 n 2t ce 32 24. u x, t c sin akx cos kt z 25. 3 u x akc cos akx cos kt u t kc sin akx sin kt −1 y 3 3 x x2 ln y 26. z z x 1 2x ln y 27. f x, y z x 1 . At 2, 0, 0 , fx 0. z y 1 y fyy 4. hx x x y hy x x y hxx x 2y y hxy hyx x sin y sin y hy 2 y cos x x cos y y sin x cos x hxx y cos x hyy 2 x sin y hxy x y 3 x y 2 x y2 x 2y x y4 2y x y4 y x x y y y3 x x y y3 cos y sin x hyx 3 2x x 1 hx y x hyy 1 29. h x, y y 6y 2 12y fxy z y y 6 fyx . At 2, 0, 0 , 2y 3 xy x fxx Slope in y-direction 28. h x, y 6x fy Slope in x-direction x2 3x2 cos y sin x R eview Exercises for Chapter 13 30. g x, y cos x z gx sin x z x 2y gy 2 sin x 2y gxx cos x gyy x2 2x 31. 2y 4 cos x 2 z x2 2y y2 2 cos x z x gyx 2 cos x 2y z x 2z y2 2z x2 z y 2z x2 y2 x2 2 4x2 y2 1 3 2y y2 2 y2 2 2y y2 3x2 x2 y2 2 x2 x2 2z y2 y2 3 x2 z y y2 y2 2 2 x2 y 2 x2 2 x y2 4 2z x2 y2 Therefore, 2z x2 2z y2 0. e x sin y e x sin y e x cos y 2z y 2 2y e x sin y y2 y2 y2 3 2z Therefore, x2 3x2 x2 2y 6x e x sin y z x y2 x2 2y 0. 34. z x2 x2 2z Therefore, 2z x2 2z y2 0. 0. y x 35. z x sin dz z dx x z dy y sin y x cos y dy x x2 y y cos dx x x y 2x xy x2 y 2 36. z z dx x dz x2 37. 2z y2 2y 3y 2 6xy 2z y2 2 y 2xy x2 y 2 6x z y 2y Therefore, x2 3x2 2 z x2 2 z y 2y z 3xy 2 2y gxy 33. x3 32. z z2 2z dz dz x2 z dy y y2 y x2 y2 dx x2 y2 2x dx x dx z xy x x2 y 2 2y dy y dy z Percentage error: dz z 51 13 2 12 1 13 2 17 26 17 26 13 0.0503 5% 0.654 cm xy y y2 x2 y2 dy x2 y3 y2 32 dx x2 x3 y2 32 dy 259 260 Chapter 13 Functions of Several Variables 38. From the accompanying figure we observe h or h x tan x tan h h h dx x dh Letting x d x sec2 tan dx d. θ 1 ±, 2 100, dx 11 , and d 60 ± 180 x . (Note that we express the measurement of the angle in radians.) The maximum error is approximately dh 11 60 tan V 1 3 dV r h dr 1 2 11 60 100 sec2 ± ± 0.3247 ± 2.4814 180 ± 2.81 feet. r 2h 2 3 ± 39. 1 3 2 3 r 2 dh 2 5 ±1 8 5 ±6 A r r2 r2 h2 1 6 ± 2 2 1 ±8 in.3 h2 dA ± 1 3 40. 2r 2 r2 41. w ln x2 h2 dr h2 y2 , x Chain Rule: r2 r2 dw dt rh r2 2t h2 3, y w dx x dt ln x2 4 y2 dw dt 2 2t 2t 32 32 y2 x, x Chain Rule: du dt cos t, y ux xt 1 sin t sin t 1 Substitution: u du dt sin2 t 8 dh 32 4 Substitution: w 42. u r2 2x x2 y2 2 dh 25 1 ± 8 29 10 1 ± 29 8 24 t 4 t2 sin t 3 2y 2 2t t ln 2t x2 ± 43 8 29 24 32 2 4 t 4 t 10t 5t 2 1 y2 10t t 2 5t2 4 4t 25 2 4 4t 25 43. u uy yt sin t h2 t w dy y dt 2 2t 32 2t rh dr h2 x2 y2 Chain Rule: z2, x u r 2y cos t cos t ux xr 2x cos t u t ux xt sin t 1 2 cos t 2z 0 r sin2 t uy yt r sin t 2 sin t 2 t uz zr 2y sin t 2x 2 sin t cos t r sin t, z uy yr 2 r cos2 t 2 sin t cos t 2 cos t r cos t, y 2r uz zt 2y r cos t r sin t cos t 2z 2 r sin t cos t 2t 2t r 2 cos2 t Substitution: u r, t u r 2r u t 2t r 2 sin2 t t2 r2 t2 R eview Exercises for Chapter 13 xy ,x z 44. w 2r t, y w r Chain Rule: rt, z wx xr x t z 2rt 2r t 4r2t 2xy wz zr z x 2y 2yz z x x z2 xz z 2z 2 2r 2r tt t z x 0 2xy 2y t rt t2 x x2 4rt2 t 3 2r t 2 wy yt 0 z x xy 2 z2 2r 2r wx xt x2 y 45. t wy yr y 2 z w t 2r z y 2y 2z z y x 2z z y 2xy z 2y 2z x2 2y x r z 4r2t rt 2 2r xy z w Substitution: xy z2 t 2r t rt 2r t z y 4r2t 2r t y sin z z x z2 4rt2 t 3 2r t 2 rt2 y cos z y cos z z y sin z 12 y 4 f x, y f f 1, 4 u Du f 1, 4 4r 3 2 47. x2y f x, y f f 2, 1 2xz 4i 2i 1 v 5 1 v 2 2xz w y2 zi 2y j w 1, 2, 2 2i u 5 j 5 u 25 5 22 22 0 xz w 2j 45 5 u 2 j 2 sin z y cos z 1 yj 2 f 1, 4 2 i 2 f 2, 1 u 49. 25 i 5 4j 0 x2 2x i x 2j 2 xy i Du f 2, 1 z y 48. rt 2 t 0 z2 y cos z 2r2t 2r 0 z x 2xz 4r2t xz2 z x 2z 4r 3 w t 2xz 2y 2 w r 46. 2z 2z x x 1 2z x 2 y 1 z 2z 0 z y wz zt z x 25 5 Duw 1, 2, 2 4j 1 v 3 xk k 2 i 3 w 1, 2, 2 1 j 3 2 k 3 u 4 3 4 3 2 3 2 3 261 262 Chapter 13 6x2 w 50. Functions of Several Variables w 12x w 1, 0, 1 3y i 12 i Duw 1, 0, 1 3 i 3 w 1, 0, 1 x 8yz j z z 2, 1 3 j 3 3 k 3 53 y z 0, z 0, x2y z 2xy i z 2, 1 55. 9x2 4i z 2, 1 e j 2 4 z x e z x2 x 1 ,0 2 z 53. 2xy i y2 x2j x cos y i 4y2 y2 9x2 18x i f 3, 2 3 57. 54i f x, y 4y sin x f x, y 4y cos xi f 2 ,1 2y j 2 j 2 2 , 2 2 2 8yj 16 j 16 j 16 j F x, y, z x2y z F 4 sin x sin y j 4y2 54i 54i y2 x 65 Unit normal: 56. 4y sin x e 1 4 f x, y 42 cos y 2 i 2 4 f x, y 4j y2 j y2 2 1 2 z 1, 1 0 x2 x2 2xy i y2 2 1 i 2 z 1, 1 u 3 x2 4j z 2, 1 y2 4y k y x2 x x2 2 x2 z 54. 3x y z z 43 52. 51. 3j 1 v 3 u 4y 2z 3xy 4i 8j 0 x 2j 2xy i F 2, 1, 4 1 27i 793 4j k k Therefore, the equation of the tangent plane is 2j 4x 2 4x Normal vector: j 4y 4y z 1 z 4 0 or 8, and the equation of the normal line is x 58. y2 F x, y, z F z2 2 yj F 2, 3, 4 6j 25 59. 0 8k 3 4z F 2 3j 4k 4 0 or 3y and the equation of the normal line is x 2, y 3 3 z 4 4 2, y F x, y, z 2z k F 2, Therefore, the equation of the tangent plane is 3y 4t . 4z 3, 4 4t x2 1, z t 4x 6y y2 2x 4i 2y 4. z 6j 9 k k Therefore, the equation of the tangent plane is 25, z 4 0 or z 4, and the equation of the normal line is x 2, y 3, z 4 t. 0 R eview Exercises for Chapter 13 60. x2 F x, y, z F y2 2x i F 1, 2, 2 z2 2y j 2i 0 4k 2i x2 3 61. 2z k 4j F x, y, z G x, y, z 9 2j 2k 1 x 2y 2y 2 2z 2z 2 0 or y 1 F 2, 1, 3 F z 2 2 2 . y2 G x, y, z x i 4 0 y F y 2 G i F 4, 4, 9 F k j 8i G 2x i 4i n k 1 0 ,z 2i 2j 3. z2 2yj 2j k 14 2zk 6k 6 56 n k j 0 1 k 1 1 y2 x2 cos i 8 1 1 2 f x, y, z 63. 0 0 2y i k Therefore, the equation of the tangent line is f x, y, z 25 k j 2 0 f 2, 1, 3 z 2y j 2j 1 F x, y, z 0 0 4i G x 62. z k 9, 2 1 2x i G and the equation of the normal line is x z F Therefore, the equation of the tangent plane is x y2 Normal vector to plane. 3 14 14 36.7 i j 8k Therefore, the equation of the tangent line is x 4 y 4 1 z 1 64. (a) f x, y cos x 9 . 8 sin y, P1 x, y 1 1 (b) fxx cos x, fxx 0, 0 0 fyy sin y, fyy 0, 0 0 fy 0, 0 cos y, fy f 0, 0 fx 0, 0 sin x, fx 1 fxy fxy 0, 0 0 y 0, P2 x, y (c) If y 0, you obtain the 2nd degree Taylor polynomial for cos x. (d) 1 x 0 0 1 12 2x y y 0 P2 x, y 1.0 1.0 1.0 0.1 1.0998 1.1 1.1 0.1 1.0799 1.1 1.095 0.5 0.3 1.1731 1.3 1.175 1 z P1 x, y 0.2 (e) 0.5 1.0197 1.5 1.0 f x, y z z 3 2 2 2 −2 −1 1 1 y −1 −2 −1 1 x 263 y 1 x The accuracy lessens as the distance from 0, 0 increases. 2 x 1 −1 1 2 y 264 Chapter 13 x3 65. f x, y fy y2 3xy 3x2 fx Functions of Several Variables 3 x2 3y 3x z 2y y 0 0 x fxx − 30 3 From fx x2. Substituting this into fy 0, we have y 39 2, 4 At the critical point 2x2 66. f x, y fx 4x 6y fy 6x 18y 3y > 0 and fxx > 0. Therefore, 3, 9, 24 2 f xy 67. f x, y 0, x 14 3y 4 3, 4 18 x 1 x xy 2 6 4 4, 4, 3 36 > 0. Therefore, 2 is a relative minimum. 1 y z 20 y 1 x2 0, x2y 1 fy x 1 y2 0, x y 2 1 Thus, x2y 34 xy2 or x fxx 4 x − 20 − 24 y (1, 1, 3) 1 fyy y and substitution yields the critical point 1, 1 . 2 x3 fxy 2 y3 At the critical point 1, 1 , fxx 50 x 0.1x3 y 2 50 0.3x 50 0.15y 2 2 > 0 and fxx fyy 20x 20 20.6 zxx 0.6x, zyy At 10, 14 , zxx zyy 3 > 0. Thus, 1, 1, 3 is a relative minimum. 20.6y ± 14 0, y 14 , 2 2 ± 10 0.3y, zxy zxy fxy 0.05y3 150 0, x Critical Points: 10, 14 , 10, 6 10, 14 , 10, 14 0 4.2 02 > 0, zxx < 0. 10, 14, 199.4 is a relative maximum. At 10, 10, 14 , zxx zyy 14, zxy 2 6 4.2 02 < 0. 349.4 is a saddle point. 10, 14 , zxx zyy 10, 14, At is a relative minimum. 0 fx At 27 16 6 fxx fyy zy 0. Thus, x < 0. Therefore, 0, 0, 0 is a saddle point. 2 3 18 fxy x 2x 4 fyy 2x2 2 8x 8⇒y 6y fxx 8 3x fxy , fxx fyy 9y 2 6xy 0, we have fxy At the critical point 0, 0 , fxx fyy zx y 2 fxy 68. z 2 6x fyy 4 30 zxy 2 6 4.2 02 < 0. 200.6 is a saddle point. 10, 14 , zxx zyy 10, 14, xxy 2 6 4.2 02 > 0, zxx < 0. 749.4 is a relative minimum. 125 0 or 3 . 2 R eview Exercises for Chapter 13 265 69. The level curves are hyperbolas. There is a critical point at 0, 0 , but there are no relative extrema. The gradient is normal to the level curve at any given point x0, y0 . 70. The level curves indicate that there is a relative extremum at A, the center of the ellipse in the second quadrant, and that there is a saddle point at B, the origin. 71. P x1, x2 R C1 225 C2 0.4 x1 2 2 0.45x1 Px1 x2 x1 0.43x2 0.9x1 0.8x2 0.9x1 Px2 0.86x2 0.8x1x2 210 0.8x1 6100 11,500 157. 0.9 0.8 Px2x2 15x2 210 94 and x2 Px1x2 0.03x22 0 0.86x2 Px1x1 210x2 5400 210 210 Solving this system yields x1 210x1 15x1 0 0.8x2 0.8x1 0.05x12 x2 0.86 Px1x1 < 0 Px1x1 Px2x2 2 Px1x2 >0 Therefore, profit is maximum when x1 0.25x12 72. Minimize C x1, x2 0.50x1 10 0.30x2 12 5x1 10x1 3x2 94 and x2 0.15x22 1000 ⇒ 3x1 x2 3x2 3000 3x2 20 8x1 3020 x1 C 377.5, 622.5 377.5 x2 622.5 104,997.50 73. Maximize f x, y 4 y 20 x 2 4 4x 4y 5x xy 2y subject to the constraint 20x y 6 2000 ⇒ 5x y 500 5x 20x y 6 10x 494 x 13,201.8 49.4 y f 49.4, 253 12x2 subject to the constraint x1 20 5x1 x1 157. 253 4y 2000. x2 1000. 266 Chapter 13 Functions of Several Variables 74. Minimize the square of the distance: f x, y, z fx 2 2 2y 2 2 x2 2 2x fy x x2 2 y 2 y2 2x y2 75. (a) y 0 2. 2x3 2xy2 0 2 2y3 2x2y 0 2 0. Using a computer algebra system, x 0.6894 2 2 2 0.6894 2 2 4.3389. x 2 y2 2 0y 3 Distance x2 0x 2y Clearly x y and hence: 4x Thus, distance 2 0.6894 2 2 0.6894. 2.08 2.29t 2.34 (c) y 30 8.37 ln t (d) −2 1.54 y = 2.29t + 2.34 25 11 −1 −5 10 −5 (b) 20 y = 1.54 + 8.37 ln t The logarithmic model is a better fit. −1 3 −5 Yes, the data appears more linear. 76. (a) 25, 28 , 50, 38 , 75, 54 , 100, 75 , 125, 102 yi 375, xi xi4 382,421,875, xi2 297, xi yi xi2 yi 26,900, 382,421,875a 34,375c 34,375b 375c 26,900 34,375a 375b 5c 297 0.0045, b (b) When x y 80 km/hr, y z y x y x 111 3, 3, 3 xz subject to the constraint x x2 2y 2 z y z 1 3 1 3 x2y subject to the constraint x 78. Optimize f x, y 2 xy y x2 4 xy ⇒ x 0 or x 4y 2 0, y Maximum: f 1. If x 41 3, 3 Minimum: f 0, 1 4y, then y 16 27 0 0.0717x 23.2914 57.8 km. yz 1⇒x z Maximum: f If x xy 0.0045x2 z x x 23.2914, y 0.0717, c 77. Optimize f x, y, z x 2,760,000, 2,760,000 3,515,625a a 3,515,625b xi3 34,375, 1 3, x 4 3. 2y 2. y z 1. 3,515,625 P roblem Solving for Chapter 13 x2 79. PQ 3 x2 C 2 y2 4 Constraint: x C y2 4, QR y 1, RS 1 z z; x y z 10 z 10 g 2y 3x i x2 4 y2 3x x2 2y j k i j k 4 y2 1 1 1 9x2 x2 4 ⇒ x2 1 2 4y 2 y2 1 ⇒ y2 1 3 2 ,y 2 Hence, x 80. f x, y ax Constraint: 3 ,z 3 3 3 8.716 m. by, x, y > 0 x2 y2 36 64 1 (a) Level curves of f x, y 4 x 3 y 4x 3y are lines of form (b) Level curves of f x, y C. 4 x 3 Using y x 2 2 10 7, y y 12.3, you obtain 3, and f 7, 3 4 x 9 28 9 37. x 4, y 9y are lines of form C. 4 x 9 Using y 4x 7, you obtain 5.2, and f 4, 5.2 62.8. 8 − 10 10 −8 Constraint is an ellipse. Problem Solving for Chapter 13 1. (a) The three sides have lengths 5, 6, and 5. 16 2 Thus, s 8 and A (b) Let f a, b, c area 2 83 2 3 ss as bs 12. c , subject to the constraint a Using Lagrange multipliers, ss bs c ss as c ss as b . From the first 2 equations s Similarly, b c and hence a —CONTINUED— b s b a⇒a b. c which is an equilateral triangle. b c constant (perimeter). 267 268 Chapter 13 Functions of Several Variables 1. —CONTINUED— (c) Let f a, b, c a b c, subject to Area 2 ss as bs c constant. Using Lagrange multipliers, 1 ss bs c 1 ss as c 1 ss as b Hence, s 2. V a 43 r 3 b and a r 2h b c. 3. (a) F x, y, z 4 r2 M Material V b⇒a s M 4 2000 r 4 r2 dM dr 8r r3 43 r2 y0 z 0 x r3 x0 x0 z 0 y 16 r 3 x0 z 0 y 750 3 x0 y0 Tangent plane 5 3 x0 z0 13 6 . 3 3 1000 Then, h 43 r2 x 750 0. The tank is a sphere of radius r 4. (a) As x → ± lim x→ x3 ,f x fx gx 3 (b) Let x0, x0 5 1 lim 1 . 1 3→x and hence gx 4 0. −6 6 be a point on the graph of f. The line through this point perpendicular to g is y x Base 3 y0 z0 13 6 fx x→ 13 3 x0 x03 −4 1. This line intersects g at the point 1 x 20 3 x03 1, 1 x 20 3 x03 1. The square of the distance between these two points is h x0 1 x 20 3 h is a maximum for x0 9 2 3 x0 y0 z 3 ⇒r 2 x03 1. 1 . Hence, the point on f farthest from g is 2 3 1 , 2 3 x0 y0 z 3x0 y0 z 0 0 2000 r3 y0 x0 y0 z 11 3 3 3 2 y0 z0 x0 z0 2000 r2 8 3 xy 1 base height 3 (b) V 82 r 3 2000 r2 8r 16 r 3 1000 2r 0 xz, Fz y0 z 0 x Hence, r2 yz, Fy 1 Tangent plane: r3 43 r2 1000 1000 ⇒ h Fx 2 rh xyz 1 . 2 3 y z0 3 0 P roblem Solving for Chapter 13 5. We cannot use Theorem 13.9 since f is not a differentiable function of x and y. Hence, we use the definition of directional derivatives. Du f x, y fx lim t cos , y Du f 0, 0 f x, y t t f0 lim t ,0 2 t →0 f 0, 0 2 t t 4 1 lim t→0 t If f 0, 0 t sin t →0 t 2 2 t2 2 t2 2 lim t →0 1 2t2 t t2 lim t →0 2 which does not exist. t 2, then t f0 Du f 0, 0 2 lim t ,0 t →0 2 2 1 2t2 t →0 t t2 lim 2 3yz t 0 7. H which implies that the directional derivative exists. 6. Heat Loss H k 5xy k 6xy V xy 6xz Then H Setting Hx y2 2x2 x2 18 y2 y Ty 2y Inside: Tx y 1 4 z 1 2 2 1 4 12 14 4x y 2 1 0, Ty 2y T 0, 1 2 39 minimum 4 T± 3 , 2 1 2 49 maximum 4 150 and z 5 3 x − y2 y2 1 ,x 2 1 y3 1 2 21 0⇒y 3 1 2 − 1 2 1 2 ± 3 . 2 0⇒ 0, 1 2 y 10 12 y2 y 6000 . x 6000 y ⇒ x3 y 1 ellipse Ty 2 6000 y 0 ⇒ 5yx2 10. 1 4 1, T x, y k 5xy 6000 x2 5y 10 y 6yz By symmetry, x 10 y y 6xz 1000 ⇒H xy Hx 1000 . x y2 2x2 (b) On x2 k 5xy Thus, x 2x2 8. (a) T x, y z 0, you obtain x Hy 3yz 1000 . xy 1000 y 6k xy 3xz 6yz 1000 ⇒ z xyz 3xz 1200. 3 150. 269 270 Chapter 13 f x 9. (a) Cax a f x x Functions of Several Variables f y f y y a, a1 u u x u r cos y 2 x2 Similarly, a t f x, y a Similarly, 2 uy xy ux x2 2u r cos 2u a y1 u sin . y r sin 2 t Ct a x at1 uz z r sin 2 u a a z uy y u cos x u r 1 f r sin , z ux x ty C tx a axy a x ay1 a (b) f t x, ty a a1 C1 C1 Cx ay1 r cos , y a Cax y Ca 10. x a x ay C1 a Cx ay1 1y1 2u x 2u r 2 sin2 2u x2 2u cos2 z y2 2 2u xy u cos x r u sin y r yz r 2 cos2 y2 r2 2u y y2 yx 2u 2 uz xz u r cos x r 2 sin cos u r sin y 2 sin2 2 u cos sin . xy Now observe that 2u r2 1u rr 1 r2 2u 2u 2u 2 z2 x2 2u x2 cos2 sin2 2u y2 2u y2 2u 2u 2u x2 y2 z2 64 cos 45 t y 64 sin 45 t y (b) tan x 50 arctan (c) d dt (d) 1 2 cos2 2 2u xy 2u xy 1 r cos sin 1u cos rx sin cos 16t2 1 32 2t 32 2t 50 32 2t 2u 1u sin ry z2 2u r2 1u rr 1 r2 2u 2u 2 z2 0. 16t2 arctan y 32 2t 32 2t 16t2 2 16t2 50 64 8 2t2 25t 25 2 32 2t 50 2 64t4 16 8 2t2 25t 256 2t3 1024t2 50 30 4 −5 No. The rate of change of —CONTINUED— u sin y 32 2t x 0 u cos x . Thus, Laplace’s equation in cylindrical coordinates, is 11. (a) x sin2 is greatest when the projectile is closest to the camera. 25 2 800 2t 625 P roblem Solving for Chapter 13 271 11. —CONTINUED— (e) d dt 0 when 8 2t2 25t 25 2 0 252 25 t 48 2 No, the projectile is at its maximum height when dy dt x2 12. (a) d y2 32 2t 2 4096t2 t2 16t (c) When t 1024 2t3 4 2t 16t2 2 32 2t (b) 32t 0 or t 1.41 seconds. 2 32 t2 3 2t 8 t2 dd dt 256t4 4 2t 16 16 (d) 62 20 0.98 second. 32 2 2: 32 12 dd dt 25 2 28 2 82 32 t3 d 2d dt2 6 2t2 t2 4 2t 36t 16 32 12 0 32 when t 1.943 seconds. No. The projectile is at its 2. maximum height when t 38.16 ft sec 13. (a) There is a minimum at 0, 0, 0 , maxima at 0, ± 1, 2 e and saddle point at ± 1, 0, 1 e : z 1 x2 fx x2 y2 e x2 y2 2y2 e x2 2x 2y2 2x e 2x x2 y2 2x 2 y 2 x2 y2 e x2 fy 2x3 x2 y2 2y2 e e x2 y2 e x2 y2 4xy2 x2 2y 2y2 4y3 0 ⇒ x3 2x 4y e 2y 2x2y x 2xy2 x x2y 2y 0 x2 y2 4y 4y 0 ⇒ 2y3 0 Solving the two equations x 0 and 2y 0, you obtain the following critical points: 0, ± 1 , ± 1, 0 , 0, 0 . Using the second derivative test, you obtain the results above. x3 2xy2 fx fy y2 e e x2 y2 x2y (c) In general, for (b) As in part (a), you obtain x2 2y3 2x x2 1 2y2 0, 0, 0 minimum 2y 2 x2 2y2 0, ± 1, e maxima ± 1, 0, e saddle The critical numbers are 0, 0 , 0, ± 1 , ± 1, 0 . These yield For ± 1, 0, 1 e minima 0, 0, 0 saddle 1 2 −1 ± 1, 0, e minima e maxima 0, 0, 0 saddle z x < 0, you obtain 0, ± 1, 0, ± 1, 2 e maxima 1 > 0 you obtain y 272 Chapter 13 Functions of Several Variables 0, then fx x0, y0 0 and fy x0, y0 14. Given that f is a differentiable function such that f x0, y0 z z0 0 or z z0 f x0, y0 which is horizontal. Therefore, the tangent plane is 15. (a) (b) 6 cm 0. 6 cm 1 cm 1 cm hl ⇒ dA (d) A (c) The height has more effect since the shaded region in (b) is larger than the shaded region in (a). l dh h dl 5, 0.01 and dh 0, then dA 1 0.01 0.01. If dh 16. r, If dl 0.01 and dl 0, then dA 6 0.01 0.06. y 18 5 ± 0.05, d dr ± 0.05 4 3 x 5 cos r cos 4.924 18 2 1 y r sin 5 sin 0.868 18 π θ = 18 ( (r, θ ) = 5, cos dr r sin 0.985 dr 2 d 0.868 d dx is more effected by changes in r because 0.985 > 0.868. (b) dy should be more effected by changes in . dy sin dr r cos d 0.174 dr 4.924 d dy is more effected by 17. Let g x, y yf f gx x, y yf because 4.924 > 0.174. x . y x y gy x, y yf x y x y 1 y x y2 f x y f x f y x y x y Tangent plane at x0, y0, z 0 is f x0 x y0 ⇒f x0 x y0 x0 f f x0 y0 x0 y0 x0 f y0 x0 f y0 ) x 1 (a) dx should be more effected by changes in r. dx π 18 5 x0 y0 x0 y0 y y0 y z 0. 1z y0 f x0 y0 This plane passes through the origin, the common point of intersection. 0 3 4 5 P roblem Solving for Chapter 13 18. x 2 y2 2x 2 x 1 x2 y2 a2 y y2 b2 1 Circle 2 (x, y) 1 1 Ellipse x −1 The circle and ellipse intersect at x, y and x, 1 −1 y for a unique value of x. (x, − y) −2 2 y2 b2 a a2 x2 x2 b2 2 a a2 x2 2x 1 b2 2 x a2 2x b2 Ellipse Circle 0 Quadratic For these to be a unique x-value, the discriminant must be 0. 4 b2 2 b a2 41 a2 a 2b 2 0 b4 0 We use lagrange multipliers to minimize the area f a, b a 2 a 2b 2 b 4 0. constraint g a, b f g b, a b 2ab2, 2a 2a2b 4b3 a ⇒ 4b4 2a2b Using the constraint, a 2 a2 a2 a2 2 a 2 3 2 a x2 92 2u t2 u x 2u x2 Then, 1 2 1 2 b4 1 2 2a2b2 2a2 2a2b2 ⇒ 2b4 a 2 a2 ⇒ b2 0, 0 a 2 3 2 6 . 2 2, b y2 32 1 t cos x t sin x t sin x t 1 f x ct 2 x ct and s 20. u x, t Let r t 2u 2 x2 t t sin x u t t ur rt 2u t2 sin x 2u cos x 1 d 2f 2 dr 2 u x ur rx u x2 1 d 2f 1 2 dr 2 . 2 Thus, 2u t2 fx x 1 fr 2 Then u r, s 1 cos x 2 t a 2b 2 cos x Ellipse: u t 4b3 4b3 b 2ab2 2a 2a2b 2ab2 2a a 19. ab of the ellipse subject to the c2 2u x2 . 1 df 2 dr 1 d 2f c 2 ds2 2 us sx 2 ct. fs . us st c ct c 2 1 df 1 2 dr 1 d 2f 1 2 ds2 2 1 df c 2 ds c2 d 2f 2 dr 2 d 2f ds2 1 df 1 2 ds 1 d 2f 2 dr 2 d 2f ds2 273 ...
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This note was uploaded on 11/13/2010 for the course MATH MAT 231 taught by Professor Thurber during the Spring '08 term at Thomas Edison State.

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