ODD12 - CHAPTER 12 Functions of Several Variables Section...

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Unformatted text preview: CHAPTER 12 Functions of Several Variables Section 12.1 Introduction to Functions of Several Variables . . . . . . . 76 Section 12.2 Limits and Continuity . . . . . . . . . . . . . . . . . . . . 80 Section 12.3 Partial Derivatives . . . . . . . . . . . . . . . . . . . . . . 83 Section 12.4 Differentials . . . . . . . . . . . . . . . . . . . . . . . . . 88 Section 12.5 Chain Rules for Functions of Several Variables . . . . . . . 92 Section 12.6 Directional Derivatives and Gradients . . . . . . . . . . . . 98 Section 12.7 Tangent Planes and Normal Lines . . . . . . . . . . . . . 103 Section 12.8 Extrema of Functions of Two Variables . . . . . . . . . . 109 Section 12.9 Applications of Extrema of Functions of Two Variables . 113 Section 12.10 Lagrange Multipliers . . . . . . . . . . . . . . . . . . . . 119 Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123 Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129 CHAPTER 12 Functions of Several Variables Section 12.1 Introduction to Functions of Several Variables Solutions to Odd-Numbered Exercises 1. x 2z z x2 x2 4 y2 9 z2 yz y z xy 10 10 x2 10 xy xy y 3. 1 No, z is not a function of x and y. For example, x, y 0, 0 corresponds to both z ± 1. Yes, z is a function of x and y. x y 3 2 5 y (b) f 1, 4 x 2 xy z 23 9 10 1 2 3 0 1 4 (c) f 30, 5 (f) f 5, t 5 t 30 5 6 5. f x, y (a) f 3, 2 (d) f 5, y (e) f x, 2 7. f x, y (a) f 5, 0 xe y 5e0 3e 2 1 2e 5e y xe 2 tet 1 9. h x, y, z 5 (b) f 3, 2 (c) f 2, (d) f 5, y (e) f x, 2 (f) f t, t (a) h 2, 3, 9 2 e (b) h 1, 0, 1 y 11. f x, y (a) f 2, x sin y 4 2 sin 3 sin 1 4 2 13. g x, y x 2t 4 3 dt 4 (a) g 0, 4 0 4 2t 2t 1 3 dt 3 dt t2 t2 3t 0 4 4 6 1 (b) f 3, 1 (b) g 1, 4 3t 15. f x, y (a) fx x2 2y x, y x f x, y x x2 2x 2y x 2 2y x x y y 2 x2 2y x2 2y x2 2y 2y x2 2y x 2x x 2y y x2 x x x (b) 76 f x, y y y f x, y x2 2x 2y x, x 2y y 0 2, y 0 S ection 12.1 17. f x, y Domain: 4 4 x2 x2 x 2 Introduction to Functions of Several Variables y y≤1 2 21. f x, y ln 4 x x y y>0 77 y2 y ≥0 2 2 19. f x, y arcsin x y ≤4 y2 ≤ 4 Domain: x, y : 1 ≤ x Range: 2 Domain: 4 x y<4 x, y : y < x 4 x, y : x 2 Range: 0 ≤ z ≤ 2 x xy x, y : x y ≤z≤ Range: all real numbers 1 xy x, y : x 0 and y 0 23. z 25. f x, y 0 and y 0 Domain: ex y 27. g x, y 0 Domain: Domain: x, y : y Range: all real numbers 4x y2 Range: z > 0 Range: all real numbers except zero 29. f x, y x2 1 (b) View where x is negative, y and z are positive: 15, 10, 20 (d) View from the line y 33. f x, y y2 x in the xy-plane: 20, 20, 0 (a) View from the positive x-axis: 20, 0, 0 (c) View from the first octant: 20, 15, 25 31. f x, y Plane: z 5 5 4 z 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. 2 4 2 4 z y 5 x 4 1 4 x 2 3 y 35. z 4 x2 y2 4 z 37. f x, y e x z 8 6 4 2 Paraboloid Domain: entire xy-plane Range: z ≤ 4 −3 3 x 2 3 y 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. 4 x 4 y 39. z y2 x2 1 z 41. f x, y x 2e z xy 2 Hyperbolic paraboloid Domain: entire xy-plane Range: <z< y x y x 78 Chapter 12 x2 z 5 4 Functions of Several Variables y2 (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) g is a reflection of f in the xy-plane followed by a vertical translation 4 units upward. (e) z 5 4 5 4 43. f x, y (a) z −2 2 x 1 2 y (b) g is a vertical translation of f two units upward z = f (1, y) 2 x 2 y z = f (x, 1) 2 x 2 y 45. z e1 x2 y2 47. z 2 2 e1 x y ln y x2 x2 x2 49. z x y Level curves: c ln c x2 y2 Level curves: c y2 ± ec ln y y Level curves are parallel lines of the form x y c. y 1 1 x2 ln c 4 y Parabolas x2 ± e c −2 2 Circles centered at 0, 0 Matches (c) Matches (b) 2 −2 x 4 c=4 c=2 c = −1 c=0 51. f x, y c x2 25 25 y2 x2 x2 25 y2 y 2, c 2. 53. f x, y xy c. The level curves are of the form The level curves are hyperbolas of the form xy y Thus, the level curves are circles of radius 5 or less, y centered at the origin. 6 1 c=5 c=4 c=3 c=2 x −1 −1 1 2 −6 −2 −2 2 6 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 c=1 c=0 −6 55. f x, y x x2 y2 y 57. f x, y x2 y2 2 −9 6 The level curves are of the form c x x 2 9 x x2 0 y2 c=− c=− 1 2 2 x c 1 2c 2 3 2 c=1 c=2 x −6 y 2 2 y2 1 2c 2 c = −2 c = −1 c= 1 c= 2 3 2 Thus, the level curves are circles passing through the origin and centered at 1 2c, 0 . S ection 12.1 8 x2 4 Introduction to Functions of Several Variables 79 59. g x, y 1 y2 61. See Definition, page 838. 63. No, The following graphs are not hemispheres. z e x2 x2 y2 −6 6 z y2 −4 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 x2 y 2. 67. V I, R 1000 1 0.10 1 1I R 10 Inflation Rate Tax Rate 0 0.28 0.35 0 2593.74 2004.23 1877.14 0.03 1929.99 1491.34 1396.77 0.05 1592.33 1230.42 1152.40 69. f x, y, z c 6 6 x 2y x 2y 3z 3z 71. f x, y, z c 9 9 x2 x2 y2 y2 z2 z2 73. f x, y, z c 0 0 4x 2 z 2 4x 2 4y 2 4y 2 z2 z2 Plane z 3 Sphere z 4 Elliptic cone −3 y −2 −4 4 x 4 y −2 2 x 1 2 y 6 x −4 75. N d, L d 4 4 2 L 22 4 4 2 (a) N 22, 12 12 243 board-feet (b) N 30, 12 30 4 4 2 12 507 board-feet 77. T 600 0.75x 2 0.75y 2 0.75y 2 79. C 0.75xy base 0.75xy 2 0.40 xz front & back 0.80 xz yz 2 0.40 yz two ends The level curves are of the form c x2 600 y2 0.75x 2 600 c . 0.75 The level curves are circles centered at the origin. c = 600 c = 500 c = 400 y 30 c = 300 c = 200 c = 100 c=0 x z y − 30 x 30 − 30 80 Chapter 12 kT, 20 2600 20 2600 300 kT V Functions of Several Variables k 300 520 3 81. PV (a) k (b) P 520 T 3V 520 3 520 T 3c 520 . 3c 85. (a) The boundaries between colors represent level curves (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 T V The level curves are of the form: c V Thus, the level curves are lines through the origin with slope 83. (a) Highest pressure at C (b) Lowest pressure at A (c) Highest wind velocity at B 87. False. Let f x, y f 1, 2 2xy f 2, 1 , but 1 2 89. False. Let f x, y 5. 5 22 f x, y . Then, f 2x, 2y Section 12.2 1. Let Limits and Continuity > 0 such that f x, y 2 > 0 be given. We need to find L y b< whenever 0 < Then if 0 < y y 3. lim b 2 x x < a a 2 2 y y b b 2 < . Take . < , we have b<. f x, y g x, y lim f x, y lim g x, y 5 3 2 x, y → a, b x, y → a, b x, y → a, b 5. x, y → a, b lim f x, y g x, y x, y → a, b lim f x, y x, y → a, b lim g x, y 53 15 7. x, y → 2, 1 lim x 3y 2 2 31 2 5 9. x, y → 2, 4 lim x x y y 2 2 y 4 4 3 Continuous everywhere arcsin x y 1 xy Continuous for x 11. x, y → 0, 1 lim arcsin 0 1, y y y z z≥0 0 13. x, y → lim 1, 2 e xy e 2 1 e2 Continuous for xy 15. lim x 0, x y ≤ 1 8 22 17. Continuous everywhere lim e xy 1 x, y, z → 1, 2, 5 x, y → 0, 0 Continuous for x Continuous everywhere S ection 12.2 19. lim ln x2 y2 ln 0 Limits and Continuity 81 x, y → 0, 0 The limit does not exist. Continuous except at 0, 0 xy x2 y2 21. f x, y Continuous except at 0, 0 Path: y 0 x, y f x, y Path: y x x, y f x, y 1, 0 0 1, 1 1 2 0.5, 0 0 0.5, 0.5 1 2 0.1, 0 0 0.1, 0.1 1 2 0.01, 0 0 0.001, 0 0 0.001, 0.001 1 2 0.01, 0.01 1 2 The limit does not exist because along the path y the function equals 1 . 2 xy 2 x2 y2 y4 0 the function equals 0, whereas along the path y x 23. f x, y Continuous except at 0, 0 Path: x x, y f x, y Path: x y2 x, y f x, y 1, 1 1 2 0.25, 0.5 1 2 0.01, 0.1 1 2 0.0001, 0.01 1 2 0.000001, 0.001 1 2 1, 1 1 2 0.25, 0.5 1 2 0.01, 0.1 1 2 0.0001, 0.01 1 2 1 2, 0.000001, 0.001 1 2 The limit does not exist because along the path x the function equals 1 . 2 x2 2xy2 y2 x2 y2 2xy2 x y2 2 y 2 the function equals whereas along the path x y2 25. x, y → 0, 0 lim f x, y x, y → 0, 0 lim 27. 1 x, y → 0, 0 lim sin x z sin y 0 x, y → 0, 0 lim 1 (same limit for g) Thus, f is not continuous at 0, 0 , whereas g is continuous at 0, 0 . x2y 4y 2 x y 29. x, y → 0, 0 lim x4 31. f x, y 10xy 2x2 3y2 0 and x y. Does not exist z The limit does not exist. Use the paths x z y x y x 82 Chapter 12 Functions of Several Variables sin r 2 r→0 r2 lim 2r cos r 2 r→0 2r lim 33. x, y → 0, 0 lim sin x 2 y 2 x2 y2 r→0 lim cos r 2 1 35. x, y → 0, 0 lim x3 x2 y3 y2 r→0 lim r 3 cos3 r2 sin3 r→0 lim r cos3 sin3 0 37. f x, y, z 1 x y2 z2 Continuous except at 0, 0, 0 2 39. f x, y, z sin z ex ey Continuous everywhere 41. ft g x, y f g x, y t2 3x f 3x 3x 9x 2 2y 2y 2y 2 43. ft g x, y f g x, y 1 t 3x f 3x 2y 2y 3x 2 1 3x 2y 12xy 4y 2 Continuous for y Continuous everywhere 45. f x, y (a) lim x→0 x2 fx 4y x, y x f x, y lim x 2x x x x2 4y y 4y y y lim y→0 x 2 x→0 4y x x 2 x2 4y lim (b) lim f x, y y y f x, y lim x→0 lim 2x x→0 x 4y 2x x2 y→0 y→0 lim y→0 4 4 47. f x, y (a) lim x→0 2x fx xy 3y x, y x f x, y lim 2x 2x x 2x xy y xy 3y x xy x xy x lim 2 x→0 3y 2x xy 3y x→0 lim (b) lim f x, y y y f x, y lim x→0 y y y 2 y 2x xy 3y y 3y y→0 y→0 lim y→0 lim x y→0 3 x 3 49. See the definition on page 851. Show that the value of x, y → x0 , y0 51. No. f x, y is not the same The existence of f 2, 3 has no bearing on the existence of the limit as x, y → 2, 3 . lim for two different paths to x0, y0 . Section 12.3 53. Since lim f x, y a 2 Partial Derivatives 2 whenever 83 x, y → a, b L1, then for y b 2 2 > 0, there corresponds 1 > 0 such that f x, y L1 < 0< Since x lim < 1. x, y → a, b g x, y a 2 L2, then for y 1 2 > 0, there corresponds 2 > 0 such that g x, y L2 < 2 whenever 0< Let x b and 2 < 2. be the smaller of f x, y g x, y x, y → a, b 2. By the triangle inequality, whenever f x, y L1 L1 L 2. g x, y L2 x a 2 y L1 b 2 < , we have L1 f x, y L2 g x, y ≤ f x, y g x, y L2 < 2 2 . Therefore, lim 55. True 57. False. Let f x, y ln x2 0, y2 , x, y x 0, y 0, 0 0 See Exercise 19. Section 12.3 1. fx 4, 1 < 0 Partial Derivatives 3. fy 4, 1 > 0 5. f x, y fx x, y fy x, y 2x 2 3 3y 5 7. z z x z y xy y x 2y 9. z z x z y x2 2x 5x 5xy 5y 6y 3y2 11. z z x z y x 2e 2y 2xe 2y 2x 2e 2y 13. z z x z y ln x 2 2x x2 2y x2 y2 y2 y2 15. z z x z y ln x x 1 y 1 y y y 1 x 1 x ln x y 2y ln x y x x y y x2 x2 2x y2 y2 17. z z x z y x2 2y 2x 2y x2 2y 2 4y 2 x 4y 2 x2 8y x x2 12 x 2 12 x 2 y2 y2 y2 12 19. h x, y x3 4y 3 2 xy x 3 16y 3 2xy 2 23. 2x 2y x x2 y x2 y2 y2 z z x z y hx x, y hy x, y e x2 y2 2xe 2ye x2 y2 x2 y2 21. f x, y fx x, y fy x, y tan 2x 2 sec2 2x sec 2 2x y y y 12 84 Chapter 12 Functions of Several Variables y 25. z z x z y e y sin xy ye y cos xy ey sin xy ey x cos xy xey cos xy sin xy 27. f x, y x t2 t3 3 y 1 dt t x y3 3 1 x2 y x3 3 x fx x, y fy x, y y2 x2 1 1 [You could also use the Second Fundamental Theorem of Calculus.] 29. f x, y f x f y 31. f x, y f x lim x→0 2x lim lim fx x→0 3y x, y x y y f x, y f x, y lim lim 2x 2x x 3y x→0 3y x y y 2x 2x 3y 3y lim lim x→0 2x x 3y y 2 3 f x, y y→0 y→0 y→0 x fx y x, y x f x, y lim x x x x x→0 y x y x x 1 y y y y y x 1 y x x y y x y 1 2x x x x y y y x y lim lim f y lim f x, y y y f x, y lim x→0 x→0 x x x y x x x x y y y y y→0 y→0 lim y x y 1 2x e z x x y x y y y x y y→0 lim y→0 x y x y 33. g x, y gx x, y 4 x2 2x y2 35. z cos y e x cos y 1 e x At 1, 1 : gx 1, 1 gy x, y At 1, 1 : gy 1, 1 2y 2 At 0, 0 : 2 z y At 0, 0 : z x sin y 0 z y xy x yx x y 37. f x, y fx x, y At 2, arctan y x y x2 1 4 x x 2 39. f x, y y x 2 1 1 y2 x2 2 y 2 fx x, y At 2, y y 2 y x 2 2 xy 1 4 xy y 2 y2 xy 2 2 : fx 2, 2 : fx 2, xx fy x, y At 2, 1 2 : fy 2, 1 1 y2 x2 x 1 2 4 y 2 fy x, y At 2, x2 x y 2 2 : fy 2, 1 4 S ection 12.3 49 41. z 2, 3, 6 x2 y2, x 2, 45 y 2 Partial Derivatives 85 43. z 9x2 y 2, y 3, 1, 3, 0 9x 2 9 Intersecting curve: z y 45 y2 3 45 9 1 2 At 1, 3, 0 : z x 18x z x z Intersecting curve: z z y z At 2, 3, 6 : y z 18 1 y=3 18 x=2 10 160 4 x 8 x 8 y 3 2 4 y 45. fx x, y fx fy 2x 0: 2x 4x 4y 4y 2y 4, fy x, y 4 16 4x 2y 16 47. fx x, y fx fy 0: 1 x2 1 x2 y, fy x, y y y 0 and 1 y2 1 y2 x x 1 y2 0 Solving for x and y, x 6 and y 4. y y4 ⇒ y 1 1 and x x2 x Points: 1, 1 49. (a) The graph is that of fy. (b) The graph is that of fx. 51. w w x w y w z x2 x2 x2 x2 y2 x y2 y y2 z y2 z2 z2 z2 z2 Fx x, y, z Fy x, y, z Fz x, y, z 53. F x, y, z ln x2 y2 y2 z2 z2 1 ln x 2 2 x2 x2 x2 x2 x x2 x2 x2 y x2 x2 x2 y2 x2 y2 xy y2 32 x y2 y y2 z y2 y2 y2 z2 z2 z2 55. H x, y, z Hx x, y, z Hy x, y, z Hz x, y, z sin x cos x 2 cos x 3 cos x 2y 2y 2y 2y 3z 3z 3z 3z 57. z z x 2z x2 2z x2 2x 2 2 2x 6 2 2xy 2y 3y 2 59. z z x 2z x2 2z y2 y2 xy y2 32 yx z y 2z yx 6y z y 2z 32 y2 2z xy y2 2z xy 32 86 Chapter 12 Functions of Several Variables y x y x2 y x2 y2 61. z z x 2z e x tan y e x tan y e x tan y e x sec2 y ex sec2 y 63. z z x 2z arctan 1 x2 1 y2 x2 2xy y2 x2 x2 2 x2 2z x2 2z yx z y 2z y2 1 y2 x2 y 2y y2 2 1 x x2 yx z y 2z y2 x2 x y2 x2 y2 2 1 x2 x2 y2 2z 2e x sec2 y tan y e x sec2 y y2 2z 2xy y2 y2 x2 x x2 x2 2 xy x 2x y2 2 xy y2 x2 x2 y2 2 65. z z x 2z x sec y sec y 0 sec y tan y x sec y tan y x sec y sec2 y tan2 y 67. z z x 2z ln 1 x x4 y2 2x y2 ln x y2 x x2 ln x 2 x2 y2 y2 x2 2z x2 2z x2 x2 4x 2y 2 y 4 x2 y2 2 2 yx z y 2z yx z y 2z y2 2z 2z 4xy y2 2y x2 y2 y2 2z 2 y2 x2 x2 y2 2 x2 4xy y2 2 xy sec y tan y 2z xy . 0 zy, because Therefore, yx xy There are no points for which zx zy 0. There are no points for which zx z x 69. f x, y, z fx x, y, z fy x, y, z fyy x, y, z fxy x, y, z fyx x, y, z fyyx x, y, z fxyy x, y, z fyxy x, y, z sec y 0. xyz yz xz 0 z z 0 0 0 fyxy fyyx 0. 71. f x, y, z fx x, y, z fy x, y, z fyy x, y, z fxy x, y, z fyx x, y, z fyyx x, y, z fxyy x, y, z fyxy x, y, z e x sin yz x x 73. z z x 2z 5xy 5y 0 5x 0 2z x2 2z y2 e ze sin yz cos yz x x x x z2e ze ze z2e sin yz x2 z y z y2 2 cos yz cos yz sin yz sin yz sin yz fyyx. z2e x z2e x Therefore, 0 0 0. Therefore, fxyy Therefore, fxyy fyxy S ection 12.3 75. z z x 2z Partial Derivatives 87 e x sin y e x sin y 77. z z t 2 sin x ct ct ct c cos x c 2 sin x cos x sin x 2 x2 z y 2z y2 e x sin y e x cos y e x sin y 2z 2z z t2 z x 2 z x2 ct ct c2 2 z . x2 Therefore, x2 x c x c x c y2 e x sin y e x sin y 0. Therefore, z t2 79. z z t z x 2z e t cos t 81. See the definition on page 859. e cos t 1 e c 1 e c2 z t sin t x2 cos c2 x c 2z Therefore, x2 . 83. z (x0, y0, z 0 ) z (x0, y0, z 0 ) 85. The plane z x y f x, y satisfies f f > 0 and > 0. x y z 6 x y x y Plane: y = y0 Plane: x = x0 x y 8 −6 f denotes the slope of the surface in the x-direction. x f denotes the slope of the surface in the y-direction. y 87. (a) C C x C x C y C y 32 xy 16 y x 16 80, 20 175x 175 1 4 205 205y 1050 (b) The fireplace-insert stove results in the cost increasing at a faster rate because C C > . y x 175 183 16 x y 16 4 80, 20 205 237 88 Chapter 12 Functions of Several Variables 91. T T x T y 500 1.2x, 3y z x z x mRT V2 mR P mR P 1 0.6x 2 T 2, 3 x T 2, 3 y 1.83 1.09 1.5y 2 2.4 m 9m 89. An increase in either price will cause a decrease in demand. 93. PV T P V T P P V V T mRT PV ⇒ mR mRT ⇒ V mRT ⇒ P V mR mRT VP T P P V V T mRT V2 mRT mRT V mR 95. (a) (b) As the consumption of skim milk x increases, the consumption of whole milk z decreases. Similarly, as the consumption of reduced-fat milk y increases, the consumption of whole milk z decreases. 97. f x, y xy x 2 y 2 , x2 y2 0, x2 x2 f x, y x, y y3 x2 0, 0 0, 0 x3 y y2 2 (a) fx x, y fy x, y (b) fx 0, 0 fy 0, 0 (c) fxy 0, 0 y 2 3x 2y y 2 x3 x, 0 x f 0, y y f x f y xy 3 2x xy 3 2y 0 0 x y fx 0, 0 y 2 y x4 x2 x x4 x2 0 0 4x 2y 2 y2 y4 2 3xy 2 x3y 2 22 x y f 0, 0 f 0, 0 lim lim 4x 2 y 2 y2 y4 2 lim x→0 x→0 x 2 0 0 y y x x 22 lim y→0 y→0 y lim y→0 y lim 0, 0 fx 0, y y 4 y→0 y 4 lim y→0 1 1 fyx 0, 0 x lim 0, 0 fy x, 0 x fy 0, 0 x→0 lim x 22 x→0 x lim 1 x→0 1 (d) fyx or fxy or both are not continuous at 0, 0 . 99. True 101. True Section 12.4 1. z dz 3x 2y 3 6xy dx 3 Differentials 3. z 9x 2 y 2 dy dz x2 2 x2 y2 2 1 x2 y2 2x y2 2 dx x dx 2y x2 y dy y2 2 dy Section 12.4 5. z dz 7. z dz x cos y cos y e x sin y e x sin y dx 4 3.4875 f 1, 2 2y dy 4 0.1 0.5 0.5125 e x cos y dy y cos x y sin x dx x sin y cos x dy cos y y sin x dx 9. w dw x sin y 2z 3 y sin x 2z 3 y cos x dx sin 2 1.05 sin 2.1 f 1, 2 x cos y dy cos 2 0.1 2z3 sin x dy cos x dy Differentials 89 6z 2 y sin x dz 11. (a) f 1, 2 13. (a) f 1, 2 f 1.05, 2.1 z (b) dz f 1.05, 2.1 z (b) dz f 1.05, 2.1 2x dx 2 0.05 f 1.05, 2.1 sin y dx 0.00293 sin 2 0.05 0.00385 15. (a) f 1, 2 f 1.05, 2.1 z (b) dz 5 5.25 0.25 3 dx 3 0.05 4 dy 4 0.1 0.25 x x2 y2 y x2 0.094 y2 17. Let z x2 5.05 2 y 2, x 3.1 2 5, y 52 3, dx 32 0.05, dy 0.1. Then: dz dx 0.55 34 dy 5 0.05 52 32 3 0.1 52 32 19. Let z 1 1 3.05 5.95 2 x 2 y 2, x 2 3, y 32 6, dx 0.05, dy 21 63 32 0.05. Then: dz 0.05 0.012 2x dx y2 2 1 x2 dy y3 1 62 23 0.05 62 21. See the definition on page 869. 23. The tangent plane to the surface z is a linear approximation of z. f x, y at the point P 25. A dA ∆h lh l dh dA dA l ∆l h dl ∆A dA h 27. V r h dV 3 6 r 2h 3 r 0.1 0.1 r2 dh 3 r 2h dr 3 0.001 r dh 0.0001 h 0.1 0.1 0.002 0.0002 dV 4.7124 2.8274 0.0565 0.0019 V 4.8391 2.8264 0.0566 0.0019 V dV 0.1267 0.0010 0.0001 0.0000 2 rh dr 3 90 Chapter 12 Functions of Several Variables 29. (a) dz (b) dz 1.83 dx z dx x 1.09 dy z dy y 1.09 ± 0.25 1.83 ± 0.25 ± 0.73 Maximum propagated error: ± 0.73 Relative error: dz z ± 0.73 ± 0.73 1.83 7.2 1.09 8.5 r 2 dh 28.7 6.259 ± 0.1166 11.67% 31. V dV V 2 r 2h dr r dV dh h 2 rh dr 2 0.04 1 2 ab 1 2 1 2 0.02 0.10 10% 33. A dA sin C a sin C db 3 sin 45 ab cos C dC ± 16 1 b sin C da 4 sin 45 1 ± 16 12 cos 45 ± 0.02 ± 0.24 in.2 35. (a) V 1 bhl 2 18 sin 2 18 cos 2 16 12 18 h b 2 31,104 sin in.3 18 sin ft3 V is maximum when sin (b) V s2 sin 2 dV s sin 18 sin l l ds s2 l cos 2 16 12 1 2 d s2 sin 2 dl 1 or 2. θ 2 18 2 182 16 12 cos 2 2 90 182 sin 2 2 1 2 1809 in3 1.047 ft3 37. P dP dP P E2 R 2E dE R 2 dE E E2 dR R2 dR R 2 0.02 0.03 0.07 7% S ection 12.4 2h r dr r Differentials 91 39. L dL L 41. z z 0.00021 ln 0.00021 dh h 0.75 0.00021 0.75 2x y x 2 ± 1 100 ± 1 16 100 10 4 2 ± dL ± 6.6 10 ± 6.6 6 0.00021 ln 100 f x, y fx x2 2x x 2x fx x, y 2 x x x2 x, y 2x x x 2 8.096 8.096 10 4 10 6 micro–henrys y f x, y 2x 2x y 0y x 2 → 0. 2 y y x2 2x y 2x y xx y and 1 fy x, y y where 1 x and 2 0. As 43. z z x, y → 0, 0 , f x, y fx x2 x2y x, y 2x x yx x x 2 1→0 y x 2 2 f x, y y y 2x x x 1 x2y y x x 2 2 2xy x 2xy x fx x, y As x2 y yx y and y y y fy x, y 2x x x 2 y where 1 y x and 2 2x x x 2. x, y → 0, 0 , 3x2y , y2 1→0 2 → 0. x, y x, y f x, 0 x f 0, y y 0, 0 0, 0 f 0, 0 0 x 0 y 4 45. f x, y x4 0, 0 x 0 0 (a) fx 0, 0 lim x→0 lim x→0 fy 0, 0 lim f 0, 0 2 y→0 lim y→0 y 0 Thus, the partial derivatives exist at 0, 0 . (b) Along the line y Along the curve y x: x, y → 0, 0 lim f x, y f x, y x →0 x 4 lim 3x3 x2 3 2 x →0 lim 3x x2 1 0 x2: x, y → 0, 0 lim 3x 4 2x 4 f is not continuous at 0, 0 . Therefore, f is not differentiable at 0, 0 . (See Theroem 12.5) 47. Essay. For example, we can use the equation F dF F dm m F da a a dm m da. ma: 92 Chapter 12 Functions of Several Variables Section 12.5 1. w x y dw dt x2 et e t Chain Rules for Functions of Several Variables 3. w x y x sec y et t sec y et et sec et sec t x sec y tan y t1 tan t 1 y2 2xet 2y e t 2 e2t e 2t dw dt sec t tan t 5. w (a) xy, x dw dt 2 sin t, y 2y cos t 2 cos2 t x cos t sin t 2y cos t 2 cos 2t dw dt 2 cos 2t x sin t sin2 t sin 2t, (b) w 2 sin t cos t 7. w x y z (a) x2 y2 z2 et cos t et sin t et dw dt 2x 2e2t, et sin t dw dt 4e2t et cos t 2y et cos t et sin t 2zet 4e2t (b) w 9. w (a) xy dw dt xz y t2 yz, x z 1 1 t2 1 x t t 1, y z 2t t 1 t 1 t2 x y 1, z t 1 2t 1t 2t t2 1 t 1t 3t 2 1 t2 1 3 2t 2 1 (b) w dw dt t 2t t 1 t2 1 3 2t 2 1 11. Distance ft ft x1 x2 2 y1 7 cos t 7 cos t 42 44 12 y2 2 2 10 cos 2t 6 sin 2t 4 sin t 7 sin t 2 11 29 20 4 7 cos t 2 12 2 6 sin 2t 4 sin t 2 1 10 cos 2t 2 2 10 cos 2t 20 sin 2t 2 10 7 2 6 sin 2t 12 2.04 4 sin t 12 cos 2t 4 cos t f 2 1 2 10 2 1 116 2 12 22 2 29 Section 12.5 13. w arctan 2xy , x dw dt w dx x dt 1 2y 4x2y2 cos t, y w dy y dt sin t 1 sin t 2x cos t 4x2y2 2 cos t cos t 4 cos2 t sin2 t sin t, t 0 Chain Rules for Functions of Several Variables 93 1 2 sin t 4 cos2 t sin2 t 1 2 cos2 t 2 sin2 t 1 4 cos2 t sin2 t d 2w dt 2 1 4 cos2 t sin2 t 8 cos t sin t 1 8 cos t sin t 1 2 sin4 t 2 cos 4 t 1 4 cos2 t sin2 t 2 At t 0, d 2w dt 2 y2 t t 2y 2y 2x 1 y 2x 1, w t y2, x 2y 1 2y 1 4y 4x r 2 2 cos2 t 2 sin2 t 8 cos3 t sin t 4 cos2 t sin2 t 2 8 sin3 t cos t 0. 15. w x y w s w t x2 s s 2x 2x 17. w x y 4s y 4t w t 4. w s x2 y2 s cos t s sin t 2x cos t 2s cos2 t 2x s sin t 3 and t 2y sin t 2s sin2 t 2s cos 2t 2s2 sin 2t w t 18. When s w s x2 w r w 2 and t 8 and 2y s cos t , w s 0 and When s 4 19. w (a) 2xy 2x 2x 4x 4r r ,y 2x 2x y 8 r 2 r2 r 2y 1 2y 1 0 (b) w r r2 4 2 2r 2 r 2 2 2r r2 2r 2 w r w 0 8 94 21. w Chapter 12 y arctan , x x w r w x2 (b) w w r w y x2 y2 y y2 Functions of Several Variables r cos , y cos r sin x x2 y2 x2 arctan tan r sin sin x y2 r cos r sin cos r2 r sin r2 r cos sin r2 r sin 0 r cos r cos r2 1 (a) arctan 0 1 r sin r cos 23. w w s xyz, x yz 1 s 2s2t2 s xz 1 t st2 s2t2 xz t st2 t, y s xy t2 t, z st2 25. w w s ze x y, x z xy e1 y es ts t s t, y s t, z st zx x y e1 y2 st s t t 4st st st s t s2 s s ex y t t st t2 s2t s t2 st2 t2 t ts t 2 s t4 1 s 2s3t t st2 3s2t2 xy 2st t st2 2st3 s t4 ts t t2 t2 t2 3s2 w t yz 1 s es s ts 4st3 t 2st 2st s2 2t2 w t es zx e y es es es ts t ts t 2 2st3 2s3t y 1 t zx x y e1 y2 st s t t t2 t 4 4 0 x 2 ex y s s ss t 2 ts st s t s t2 st s t s t2 ts t st s s s2 s xy xy ts t 27. x2 3xy dy dx y2 2x y 5 0 29. ln x2 y2 y2 Fx x, y Fy x, y 3y 2y 2x 3y 2 3x 2y 1 2x 3x 2 1 1 ln x2 2 dy dx Fx x, y Fy x, y x2 y x2 y2 y2 z y x x y x2y xy2 y3 x3 31. F x, y, z Fx Fy Fz z x z y x2 2x 2y 2z Fx Fz Fy Fz y2 z2 25 33. F x, y, z Fx Fy Fz z x z y tan x sec x sec x sec2 Fx Fz Fy Fz sec2 x sec2 y y 2 2 y y y z tan y 1 sec2 y z x z y z sec2 x sec2 y sec2 x y z y z y sec2 y 2y sec z 1 z S ection 12.5 35. x 2 2yz 2y z y z2 z x 2z 1 2z 2z 0 z x z y 0 implies 0 implies z x z y x y z y z z . . Chain Rules for Functions of Several Variables 37. exz xy z x z y 0 Fx x, y, z Fz x, y, z Fy x, y, z Fz x, y, z cos xy Fx Fw Fy Fw Fz Fw zexz y xexz x xexz 1 exz e xz 95 (i) 2x (ii) 2y 39. F x, y, z, w Fx Fy Fz Fw w x w y w z yz xz xy xz xyz zw zw xw yz Fx Fw Fy Fw Fz Fw xy x2 y2 xzw yzw w2 5 41. F x, y, z, w w x sin yz wz 20 y sin xy z x sin xy z y cos zy z w z cos yz yw 2w zy xz zx xz xy xz yz xw yz yz w 2w yw 2w w 2w w y w z 43. f x, y f tx, ty Degree: 1 x fx x, y tx ty tx 2 ty 2 t xy x2 y2 tf x, y y fy x, y x x2 y3 y2 32 y x2 x3 y2 32 xy x2 y2 1 f x, y 45. f x, y f tx, ty Degree: 0 ex etx y ty 47. ex y dw dt w dx x dt w dy (Page 876) y dt f x, y x fx x, y y fy x, y x 1x e y y y xx e y2 y 0 49. w f x, y is the explicit form of a function of two variables, as in z x2 0, as in z x2 y2 0. The implicit form is of the form F x, y, z 1 bh 2 x sin 6 sin x2 sin 2 y2. 51. A dA dt x sin dx dt 4 1 2 2 x cos 2 b 2 x2 cos 2 d dt x h 62 cos 2 4 90 32 2 2 10 m hr 2 2 x 96 Chapter 12 12 rh 3 1 3 2rh r r2 r2 Functions of Several Variables 53. (a) V dV dt dr dt h2 h2 r2 dh dt 1 3 2 12 36 6 12 2 4 1536 in.3 min (b) S dS dt r 2 (Surface area includes base.) r2 r2 h2 2r dr dt 2 12 36 10 20 rh dh r 2 h2 dt 6 4 36 12 122 362 4 122 12 10 648 10 1 m r12 2 362 144 122 362 144 36 5 12 6 10 144 in.2 min 9 10 in.2 min 55. I dI dt r22 2r2 dr2 dt m62 82 28m cm2 sec dr1 1 m 2r1 2 dt 57. (a) tan tan 2 x 4 x 4 x 4 x 2 4 0 2x 2 8 0 2 cos2 x2 8 tan x 8 6 4 θ φ tan tan 1 tan tan tan 2x 1 2 x tan x tan x2 tan (b) F x, d dx (c) d dx Fx F 2x x2 8 tan 8 tan 2x tan sec2 x2 x 2x sin cos 8 x sin ⇒ tan 1 x 0 ⇒ 2 cos2 1 x f x, y u v v u w dx x du w dx x dv 0 2x 8 2x sin cos 1 x 0⇒ 8 x ⇒ cos x⇒x Thus, x2 2 2 ft. 59. w x y w u w v w u w v w dy y du w dy y dv w x w x w y w y Section 12.5 61. w w r w f x, y , x w cos x w x r sin r cos sin r cos w r sin r r cos , y w sin y w r cos y w r w w w x w x r sin cos r sin w r cos r w r w w w y w y (b) w r 2 Chain Rules for Functions of Several Variables 97 r sin (a) w r cos2 x w r sin2 x w r cos2 x w r cos r w cos r w r sin cos y w r sin cos x r sin2 w sin w sin r w r sin2 y w r cos2 y w r sin cos x w x r sin cos w r sin2 y w r sin r w sin r w2 2 cos x 2 ww sin xy 2 r cos2 w cos w cos r ww sin xy cos cos w y 2 1 r2 w 2 w y 2 sin2 w x 2 w x w y 2 2 sin2 cos2 63. Given u r v u x v u and y y u cos x v x u r v cos x u x v r r sin 1u . r r sin 1v . r v sin y u sin y v ,x x r cos v cos y r and y v sin x v cos y r sin . v r cos y v sin x Therefore, v r u u cos y u r cos y r u sin x u cos y u sin x Therefore, 98 Chapter 12 Functions of Several Variables Section 12.6 1. f x, y v f x, y f 1, 2 u Du f 1, 2 v v 3x 1 i 2 3 5i Directional Derivatives and Gradients 4xy 3j 4y i j 1 i 2 u 3 j 2 1 2 5 3 u Du f 2, 3 4x 5j 5y 3. f x, y v f x, y f 2, 3 xy i yi 3i v v f 2, 3 j xj 2j 2 i 2 u 2 j 2 52 2 f 1, 2 5. g x, y v g g 3, 4 u Du g 3, 4 3 i 5 v v 3i x2 4j x x2 y2 7. h x, y v e x sin y i ex ei sin yi e x cos yj i y2 4 j 5 3 i 5 u 4 j 5 y x2 j y2 h 1, 2 h u Duh 1, v v h 1, 2 i u e g 3, 4 7 25 xz 11. 2 9. f x, y, z v f x, y, z f 1, 1, 1 u Du f 1, 1, 1 xy 2i y 2i v v yz j zi 2j k h x, y, z v x arctan yz 1, 2, 1 xz 1 2k 2 1 , , 6 6 u 1 6 8 46 8 24 6 yz 2 x 2k zj x yk h x, y, z h 4, 1, 1 u Du h 4, 1, 1 arctan yz i i 2j j 1 xy k yz 2 6 i 3 f 1, 1, 1 u 6 j 6 26 3 6 k 6 4 v v h 4, 1, 1 y 3 j 2 yi cos 2x 3 2 cos 2x 13. f x, y u f Du f x2 1 i 2 2x i f y2 1 j 2 2y j u 2 2 x 2 2 y 2x y 15. f x, y u f Du f sin 2x 1 i 2 2 cos 2x f 2 u cos 2x y y yj 3 cos 2x 2 y Section 12.6 17. f x, y v f u Du f At P x2 2i 2x i v v 2 x 2 3, 1 , Du f 4y2 2j 8yj 1 i 2 8 y 2 7 2. 1 j 2 2x 4y 19. h x, y, z v h Directional Derivatives and Gradients ln x 3i x h v v h u y 3j 1 y i z z k i j 1 3i 19 7 19 y2 y2 i 2y sin x2 y2 j j k. 3j k k 99 At 1, 0, 0 , u Du h 7 19 19 21. f x, y f x, y f 2, 1 3x 3i 3i 5y2 10y j 10 j 10 23. z z x, y z 3, 4 cos x2 2x sin x2 6 sin 25i 8 sin 25j 1 i 5 2 j 5 2i 8 5 0.7941i 1.0588j 25. w w x, y, z w 1, 1, 2 3x2y 6xy i 6i 5yz 3x2 13j z2 5z j 9k 2z 5y k \ 27. PQ g x, y Du g 2i 2xi g 4j, u 2y j, g 1, 2 u 2 5 4j 10 5 25 \ 29. PQ f x, y f 0, 0 Du f 2i e i f j, u x 2 i 5 e x 1 j 5 sin yj 31. h x, y h x, y h 2, 4 x tan y tan y i i 4j 17 x sec2 y j cos y i u 2 5 25 5 h 2, 4 33. g x, y g x, y g 1, 2 g 1, 2 ln 3 x2 y2 1 ln x2 3 2y x2 2 i 15 j y2 2j y2 35. f x, y, z f x, y, z f 1, 4, 2 f 1, 4, 2 1 x2 x2 1 i 21 y2 1 y2 4j z2 z2 xi 2k yj zk 1 2x i 3 x2 y2 12 i 35 25 15 xeyz eyz i i 8j 65 4 j 5 37. f x, y, z f x, y, z f 2, 0, f 2, 0, 4 4 xzeyz j xyeyz k 100 Chapter 12 Functions of Several Variables x 3 y and D f x, y 2 1 cos 3 1 sin . 2 1 3 2 6 y For Exercises 39–45, f x, y 3 39. f x, y 3 z 3 x 3 y 2 41. (a) D4 3 f 3, 2 1 2 1 2 3 2 (3, 2, 1) 33 12 1 3 3 2 1 2 1 2 9 x (b) D 6 f 3, 2 3 23 12 43. (a) v v u Du f 3i 9 3 i 5 f 4j 16 4 j 5 1 u 5 1 6 (b) 5 v v u i 3j 10 1 i 10 f u 3 j 10 11 6 10 11 10 60 2 5 1 5 Du f 45. f 1 9 1 4 13 x2 y2 and D f x, y For Exercises 47 and 49, f x, y 47. f x, y z 9 9 2x cos 2y sin 2 x cos y sin . 9 x2 y2 49. f 1, 2 f 1, 2 2i 4 4j 16 20 25 (1, 2, 4) 3 x 3 y 51. (a) In the direction of the vector (b) f 1 10 4i j. 2y j 2 5i 1 10 j 2x 1 10 3y i 4i 1 10 1 10 3x 1j f 1, 2 (Same direction as in part (a).) (c) 53. f x, y (a) f 2 5i 1 10 j, the direction opposite that of the gradient. 3, 7 x2 y2, 4, z x y —CONTINUED– Section 12.6 53. —CONTINUED— (b) Du f x, y Du f 4, y 12 8 4 Directional Derivatives and Gradients 101 f x, y 3 8 cos u 2x cos 6 sin 2y sin −4 −8 − 12 π 2π x Generated by Mathematica (c) Zeros: 2.21, 5.36 for which Du f 4, 3 equals zero. 8 cos 6 cos 0.64, 3.79 3 is a maximum 0.64 and minimum 3.79 . 36 10, the maximum value of Du f 4, 3 , at y These are the angles (d) g g Du f 4, 8 sin 3 6 sin Critical numbers: These are the angles for which Du f 4, (e) (f ) f 4, 3 x2 8i 24i y2 23j 7 64 0.64. f x, y f 4, 3 6j is perpendicular to the level curve at 4, 3. 6 4 2 −6 −4 x −2 −4 −6 2 4 6 Generated by Mathematica 55. f x, y c x2 y2 3, 4 57. f x, y c 1 ,P 2 f x, y x x2 y2 1, 1 y2 x2 1 2 2x 1 j 2 40 4y2 8yj 8j 1 9i 85 85 9i 85 2j 2j −4 −2 −4 4 2 x 25, P f x, y 2xi 25 6i 2yj x2 y2 x2 i y2 2 x2 2xy j y2 2 f 3, 4 8j x2 x2 x y2 y2 0 f 1, 1 59. 4x2 f x, y y 6 4x2 8xi 16i y j j 1 16i 257 257 16i 257 j 4 12 y 61. 9x2 f x, y 4y2 9x2 y f x, y f 2, 10 f 2, 10 f 2, 10 8 f x, y f 2, x 18xi 1 1 1 36i 4 4 f 2, f 2, j 102 Chapter 12 x x2 T y2 x2 y2 x2 i y2 2 7 i 625 Functions of Several Variables 63. T 65. See the definition, page 885. 2xy j y2 2 1 7i 625 24j x2 T 3, 4 24 j 625 67. Let f x, y be a function of two variables and u cos i sin j a unit vector. (a) If (b) If 0 , then Du f 90 , then Du f f . x f . y 69. 3 z 3 x P 5 y 71. 73. T x, y dx dt xt 1994 400 4x C1e x0 10e 4t 4t 2x2 y2, P dy dt yt 10, 10 2y C2e y0 10e 2t 2t 1671 18 00 B A 18 00 10 xt x y2 C1 10 yt y2 t C2 y2 10 10x 100e 4t 75. (a) 400 300 D (b) The graph of D 250 30x2 would model the ocean floor. 50 sin y2 1 2 x 1 2 y (c) D 1, 0.5 (e) D y 250 30 1 50 sin 4 315.4 ft 55.5 (d) (f ) D x D 60x and 60x i D 1, 0.5 x 25 cos y j 2 60 25 cos D y and 1, 0.5 2 y 25 cos 4 D 1, 0.5 77. True z2 2 79. True 60 i 55.5j 81. Let f x, y, z ex cos y C. Then f x, y, z ex cos y i ex sin y j z k. Section 12.7 Tangent Planes and Normal Lines 103 Section 12.7 1. F x, y, z 3x 5y Tangent Planes and Normal Lines 3z 5y 15 3z 0 15 Plane 3. F x, y, z 4x2 9y2 4x2 4z2 9y2 0 4z2 Elliptic cone 3x 5. F x, y, z F n F F x i y j 1 i 3 3 i 3 z k j j 4 7. F x, y, z F x, y, z x2 x x2 3 i 5 y2 y2 4 j 5 i k z y x2 y2 j k k F 3, 4, 5 k n F F 53 i 5 25 1 52 3i 4 j 5 4j 4j k 5k 5k 2 3i 10 x y y j 1 i 3 3 i 3 z 4 k k j j z 1 9. F x, y, z F x, y, z F 1, 2, 16 n F F x2y4 2xy4 i 32i z 4x2y3j 32j k 32j 32j k k k 11. F x, y, z F x, y, z F 1, 4, 3 n F F ln 1 i x i ln x j 1 y ln y k z z z 1 32i 2049 2049 32i 2049 k k 13. F x, y, z F x, y, z F 6, , 7 6 n F F x sin y sin y i 1 i 2 2 113 1 113 113 113 x cos y j 3 3j 1 i 2 i i k 3 3j 6 3j 6 3j k 2k 2k 15. f x, y F x, y, z Fx x, y, z Fx 3, 1, 15 6x 3 25 25 x2 x2 2x 6 y2, 3, 1, 15 y2 z Fy x, y, z Fy 3, 1, 15 1 z 15 0 6x 2y z 0 6x 35 2y z 35 2y 2 Fz x, y, z Fz 3, 1, 15 1 1 2y 104 17. Chapter 12 f x, y F x, y, z Fx x, y, z Fx 3, 4, 5 3 x 5 3x 3 3 3 5 4 y 5 4y x2 x2 x x2 Functions of Several Variables y2, 3, 4, 5 y2 y2 z Fy x, y, z Fy 3, 4, 5 4 4 3x z 5z 4y 5 5 5z 0 0 0 4 5 y x2 y2 Fz x, y, z Fz 3, 4, 5 1 1 19. g x, y G x, y, z Gx x, y, z Gx 5, 4, 9 10 x 5 x2 x2 2x 10 8y y2, 5, 4, 9 y2 z Gy x, y, z Gy 5, 4, 9 4 10x z 8y 9 z 0 9 2y 8 Gz x, y, z Gz 5, 4, 9 1 1 21. z F x, y, z Fx x, y, z Fx 0, , 2 2 2x z h x, y H x, y, z Hx x, y, z Hx 3, 4, ln 5 3 x 25 3x 3 3 ex sin y ex sin y ex sin y 2 2 ln ln x2 x2 x x2 3 25 4 y 25 4y 4 1, 1 1 0, , 2 2 z Fy x, y, z Fy 0, , 2 2 ex cos y 0 Fz x, y, z Fz 0, , 2 2 1 1 23. y2, 3, 4, ln 5 y2 z 1 ln x2 2 Hy x, y, z Hy 3, 4, ln 5 4 z 25 z 3x 4y 2, 4 36 Fy x, y, z Fy 2, 2 2 x 8z 2z 4y 2, 4 4 4 2z 0 0 18 8y 16 Fz x, y, z Fz 2, 2, 4 2z 8 ln 5 ln 5 25z 0 0 25 1 ln 5 y2 z y x2 4 25 y2 Hz x, y, z Hz 3, 4, ln 5 1 1 y2 25. x2 4y2 z2 x2 36, 2, 4y2 2x 4 z2 F x, y, z Fx x, y, z Fx 2, 4x x 2, 4 2 2 16 y 4y S ection 12.7 Tangent Planes and Normal Lines 105 27. xy2 3x z2 xy2 4, 2, 1, 3x y2 4 3 z2 2 4 Fy x, y, z Fy 2, 1, 2 0 1 2xy 4 Fz x, y, z FZ 2, 1, 2 4 2z F x, y, z Fx x, y, z Fx 2, 1, 4x 2 2 4y 1 x 4z y 2 z 29. x2 y2 z x 2 9, 1, 2, 4 y2 z 9 Fy x, y, z Fy 1, 2, 4 4y y 4 2 z 1 2 4 z 2y 4 Fz x, y, z Fz 1, 2, 4 0, 2x 4y z 1 1 F x, y, z Fx x, y, z Fx 1, 2, 4 Plane: 2 x Line: x 2 2x 2 Direction numbers: 2, 4, 1 1 1 4 14 31. xy z 0, xy 2, z y 3, 6 F x, y, z Fx x, y, z Fx 2, 3, 6 Fy x, y, z 3 Fy 3 z 1 6 z 2, 3, 6 x 2 Fz 2y z Fz x, y, z 2, 3, 6 1 1 Direction numbers: 3, 2, 1 Plane: 3 x Line: x 3 2 2 y 2 2y 3 6 0, 3x 6 33. z y arctan , x 1, 1, y x y x2 1 2 4 z Fy x, y, z Fy 1, 1, 1, 2 1 z 2 2z 4 4 0, x y 2z 2 x x2 1 2 y2 Fz x, y, z Fz 1, 1, 1 1 F x, y, z Fx x, y, z Fx 1, 1, arctan y2 4 4 4 Direction numbers: 1, Plane: x Line: x 1 1 1 y y 1 1 106 Chapter 12 Functions of Several Variables 4xy 1 y2 x2 4y y2 y2 1 35. z f x, y x2 1 , 2 ≤ x ≤ z, 0 ≤ y ≤ 3 z i 4x x2 1 y2 y2 1 1 2y2 2 (a) Let F x, y, z F x, y, z 4xy 1 y2 x2 x2 1 1 2x2 1 2 j k 4y 1 x2 i 1 x2 1 2 x2 4x 1 y2 j 1 y2 1 2 k F 1, 1, 1 k. 1. 1 t 0y 1 k 4 5 1z 25z 25z z 1 3 2 y x −1 Direction numbers: 0, 0, Line: x 1, y 4 5 1, y 1 1, z 1 0i 2 6 y 25 6y Tangent plane: 0 x (b) F 1, 2, 1z 1 6 j 25 k 0⇒z 1 43 j 2 52 6 t, z 25 2 12 6y Line: x Plane: 0 x t 4 5 20 32 0 0 0 (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. 3 y (c) 1 z −2 2 2 x −1 37. Fx x0, y0, z0 x (Theorem 12.13) 39. F x, y, z F x, y, z F 2, 1, 2 (a) F G x2 x0 Fy x0, y0, z0 y y0 F2 x0, y0, z0 z z0 0 y2 2y j 2j i 4 1 j 2 0 5 G x, y, z G x, y, z G 2, 1, 2 k 0 1 2, 1, x 1 2 10 G x, y, z G G 3, 3, 4 2i 2 4j y x i i 2k 1 2 z k k 2i z 1 2 2j k 2x i 4i Direction numbers: 1, (b) cos F F x2 2x i 6i G G z2 2z k 8k 25 4 20 2 10 ; not orthogonal 5 y2 2y j 6j z2 2z k 8k 25 41. F x, y, z F F 3, 3, 4 —CONTINUED— S ection 12.7 Tangent Planes and Normal Lines 107 41. —CONTINUED— (a) F G i 6 0 j 0 6 k 8 8 3, x 4 48 i 3 48 j y 4 3 36 k z 4 3 12 4 i 4j 3k Direction numbers: 4, 4, (b) cos F F G G 64 10 10 16 ; not orthogonal 25 43. F x, y, z F x, y, z F 2, 1, 1 (a) F G x2 2x i 4i i 4 1 y2 2y j 2j j 2 1 z2 6 2z k G x, y, z G x, y, z G 2, 1, 1 x i i k z 1 1 y j j z k k (b) cos F F G G 0; orthogonal 2k k 2 1 1, x 6j 6k y 1 6j 1 Direction numbers: 0, 1, 2, 45. f x, y 6 x2 z y2 , g x, y 4 x2 2x i 2i j y2 4 1 yj 2 k 6 2x y (a) F x, y, z F x, y, z F 1, 2, 4 G x, y, z k G x, y, z G 1, 2, 4 z 2x 2i 2i j j y k k The cross product of these gradients is parallel to the curve of intersection. F 1, 2, 4 G 1, 2, 4 i 2 2 j 1 1 k 1 1 1 2i t, y 4j 2 48.2 2t, z 4. Using direction numbers 1, cos (b) 8 2, 0, you get x 4 6 1 6 1 F F z G G 4 ⇒ 6 (1, 2, 4) 6 8 x y 47. F x, y, z F x, y, z F 2, 2, 5 cos 3x2 6x i 12i 2y2 4yj 8j z k k 15, 2, 2, 5 49. F x, y, z F x, y, z F 1, 2, 3 x2 2x i 2i y2 z, 1, 2, 3 2yj 4j k 1 21 k F 2, 2, 5 k F 2, 2, 5 arccos 1 209 1 209 86.03 cos F 1, 2, 3 k F 1, 2, 3 arccos 1 21 77.40 108 51. Chapter 12 F x, y, z F x, y, z 2x 2y z 3 0, x 6 02 3 x2 2xi 0 0, y 32 Functions of Several Variables y2 2y 6y 6j z k z 53. T x, y, z dx dt xt 4kx C1e C1 4e 4kt 400 2x2 dy dt y2 4z2, 4, 3, 10 2ky dz dt 2kt 8kz C3e C3 10e 8kt 8kt 4kt 3 63 12 8 x 8 y 6 8 yt y0 y C2e C2 3e 2kt zt z0 z x0 x 4 3 10 0, 3, 12 (vertex of paraboloid) 55. F x, y, z Fx x, y, z Fy x, y, z Fz x, y, z Plane: 2x0 x a2 x0x a2 59. f x, y fx x, y fxx x, y (a) P1 x, y (b) P2 x, y ex ex ex x2 a2 2x a2 2y b2 2z c2 x0 y0 y b2 y y, y, y2 b2 z2 c2 1 57. F x, y, z Fx x, y, z Fy x, y, z Fz x, y, z a2x2 2a2x 2b2y 2z x0 b2y2 z2 Plane: 2a2x0 x a2x0x 2y0 y b2 z0z c2 x02 a2 y0 y02 b2 2z0 z c2 z02 c2 z0 1 0 2b2y0 y z0z y0 2z0 z b2y02 z0 z0 2 0 0 b2y0y a2x02 Hence, the plane passes through the origin. fy x, y fyy x, y fx 0, 0 x fx 0, 0 x y 1 1 12 2x ex ex y, y fxy x, y 1 1 2 fxx ex y f 0, 0 f 0, 0 1 x fy 0, 0 y fy 0,0 y xy 1 2 x y fxy 0, 0 xy 1 2 fyy 0, 0 x2 0, 0 y2 y2 (c) If x If y (d) x 0 0 0.2 0.2 1 0, P2 0, y 0, P2 x, 0 y 0 0 0.1 0.5 0.5 f x, y 1 y x 12 y 2 y . This is the second–degree Taylor polynomial for e . 12 x. 2 x . This is the second–degree Taylor polynomial for e P1 x, y 1 0.9000 1.1000 0.7000 1.5000 P2 x, y 1 0.9050 1.1050 0.7450 1.6250 (e) P1 f P2 4 −2 1 −2 −4 −2 z 0.9048 1.1052 0.7408 1.6487 2 x 2 y 61. Given w F x, y, z where F is differentiable at 0, C for some constant C. Let x0, y0, z0 and F x0, y0, z0 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 C 0. Then G x0, y0, z0 F x0, y0, z0 where G x0, y0, z0 is normal to F x0, y0, z0 C. C 0. Therefore, F x0, y0z0 is normal to F x0, y0, z0 Section 12.8 Extrema of Functions of Two Variables 109 Section 12.8 1. g x, y x 1 2 Extrema of Functions of Two Variables y 3 2 ≥0 5 z Relative minimum: 1, 3, 0 gx gy 2x 2y 1 3 0⇒x 0⇒y 1 3 3 x 2 1 1 (1, 3, 0) 4 y 3. f x, y x2 y2 1≥1 5 z Relative minimum: 0, 0, 1 Check: fx fy fxx x x2 y x2 x2 y2 y2 1 y2 1 1 3 2, y2 1 0⇒x 0⇒y fyy 0 −3 3 2 2 (0, 0, 1) 3 y 0 x2 1 y2 1 fxy 2 3 2, 2 x x2 fxy x2 xy y2 1 32 At the critical point 0, 0 , fxx > 0 and fxx fyy 5. f x, y x2 y2 2x 6y 1, 3, 2 6 4 1 3 6 x 1 > 0. Therefore, 0, 0, 1 is a relative minimum. 3 2 y 4≥ 4 2 1 21 x −1 −2 −3 −4 z Relative minimum: Check: fx fy fxx 2x 2y 2, fyy 0⇒x 0⇒y 2, fxy 0 1 7 y (−1, 3, − 4) At the critical point 7. f x, y fx fy fxx 4x 2x 2x2 2y 2y 2xy 2 0 2, fxy 0 1, 3 , fxx > 0 and fxx fyy y2 2x 3 fxy 2 > 0. Therefore, 1, 3, 4 is a relative minimum. Solving simultaneously yields x 1 and y 1. 4, fyy 2 1, 1 , fxx > 0 and fxx fyy y2 0 Solving simultaneously yields x 8 and y 16. At the critical point fxy 2 > 0. Therefore, 1, 1, 4 is a relative minimum. 9. f x, y fx fy fxx 10x 4x 5x2 4y 2y 10, fyy 0 4xy 16 16x 10 2, fxy 4 fxy 2 At the critical point 8, 16 , fxx < 0 and fxx fyy 11. f x, y fx fy fxx 4x 6y 2x2 4 12 3y2 4x 6y 6, fxy 4x 1 2 0 12y 13 1. 2. > 0. Therefore, 8, 16, 74 is a relative maximum. 13. f x, y fx fy 2 x2 2x x2 2y x2 y2 y2 y2 0 0 x 3 0, y 0 0 when x 0 when y 4, fyy At the critical point 1, 2 , fxx > 0 and fxx fyy fxy 2 > 0. Therefore, 1, 2, 1 is a relative minimum. Since f x, y ≥ 3 for all x, y , 0, 0, 3 is relative minimum. 110 Chapter 12 4 x Functions of Several Variables y 15. g x, y 0, 0 is the only critical point. Since g x, y ≤ 4 for all x, y , 0, 0, 4 is relative maximum. 4x y2 17. z x2 1 2 1, 0, 2 19. z x2 4y2 e1 x2 y2 Relative minimum: 1, 0, Relative maximum: z 4 Relative minimum: 0, 0, 0 Relative maxima: 0, ± 1, 4 Saddle points: ± 1, 0, 1 z 6 5 −4 4 x 5 −4 x 4 4 y y −4 −4 21. h x, y hx hy hxx 2x x2 2 4 y2 2x 2x 1 2y 2, hxy 4y 4 1. 2. 0 when x 2 0 2 , hxx hyy 2y 2, hyy 0 when y At the critical point 1, 23. h x, y hx hy hxx 2x 3x 2, hyy x2 3y 2y 3xy 0 0 2, hxy y2 hxy 2 < 0. Therefore, 1, 2, 1 is a saddle point. Solving simultaneously yields x 0 and y 0. 3 hxy 2 At the critical point 0, 0 , hxx hyy 25. f x, y fx fy fxx 3 x2 3 x x3 y y2 3xy 0 0 y3 < 0. Therefore, 0, 0, 0 is a saddle point. Solving by substitution yields two critical points 0, 0 and 1, 1 . 6x, fyy 6y, fxy 3 6 > 0 and 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. 27. f x, y fx fy e x x2 e x e x x sin 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. sin y cos y 29. z y4 ≥ 0. z y2 0 if x y 0. 60 z Relative minimum at all points x, x , x 0. 40 3 x 3 y S ection 12.8 31. fxx fyy fxy 2 Extrema of Functions of Two Variables fxy 2 111 94 62 0 33. fxx fyy 96 102 < 0 Insufficient information. f has a saddle point 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 906. 37. 75 60 45 30 6 −3 3 y z No extrema 39. 7 6 z Saddle point 41. The point A will be a saddle point. The function could be f x, y x2 y2. 2 x 2 y x 43. d ⇒ fxx fyy fxy2 fxy2 28 fxy2 16 fxy2 > 0 45. f x, y fx fy fxx 3x2 3y2 x3 0 0 y3 Solving yields x y 0 < 16 ⇒ 4 < fxy < 4 6x, fyy 6y, fxy fxy 2 0 0 and the test fails. 0, 0, 0 At 0, 0 , fxx fyy is a saddle point. 47. f x, y fx fy fxx 2x 2x 2y x 2 1 2 y 4 4 2 4 2 ≥0 Solving yields the critical points 1, a and b, 4. 1y 1 y 4 2, fyy 0 0 2x 1 2, fxy 4x fxy 2 1y 4 At both 1, a and b, 4 , fxx fyy 0 and the test fails. Absolute minima: 1, a, 0 and b, 49. f x, y fx fy fxx 2 3 3 3 4, 0 x2 x 3 y2 3 ≥0 2 3 fx and fy are undefined at x 0, y 0. The critical point is 0, 0 . y 2 9x 3 3 , fyy 2 9y 3 y , fxy 0 At 0, 0 , fxx fyy fxy 2 is undefined and the test fails. Absolute minimum: 0 at 0, 0 51. f x, y, z fx fy fz 2x 2y 2z x2 0 3 1 0 0 1 Solving yields the critical point 0, 3, 1. y 3 2 z 1 2 ≥0 Absolute minimum: 0 at 0, 3, 112 Chapter 12 12 3x fx fx fx Functions of Several Variables 2y has no critical points. On the line y 12 12 12 3x 3x 3x 2x 2 2 2x 1 2 53. f x, y f x, y f x, y f x, y x 2x 1 2 1, 0 ≤ x ≤ 1, 3 y 1 4 1 5x x 10 4, 1 ≤ x ≤ 2, 4 x 1, 0 ≤ x ≤ 2, 2x 10 and the maximum is 10, the minimum is 5. On the line y and the maximum is 6, the minimum is 5. On the line y x and the maximum is 10, the minimum is 6. Absolute maximum: 10 at 0, 1 Absolute minimum: 5 at 1, 2 55. f x, y fx fy 6x 4y f x, y f x, y 3x2 0 4 2y2 4y 0 1 32 2 x2 2 f 0, 1 2 y=x+1 2 (1, 2) (0, 1) y = −2x + 4 (2, 0) 1 2 3 1 x y=− 1x 2 +1 y ⇒x 0⇒y 4, fx fx 3x 2 (−2, 4) (2, 4) 3 2 On the line y 2 ≤ x ≤ 2, 16 3x 4x2 2 16 x2, 1 2 ≤ x ≤ 2, −2 −1 1 x 1 2 and the maximum is 28, the minimum is 16. On the curve y 3x2 2x4 1 8. x2 x2 2x2 and the maximum is 28, the minimum is Absolute maximum: 28 at ± 2, 4 Absolute minimum: 57. f x, y fx fy 2x x 0 0 1, 2, 1 1, 2, 2, 2, 1 2 ≤ x ≤ 2, f 2, f 6, f 1 2, 1 2, 2 at 0, 1 x, y : x ≤ 2, y ≤ 1 y 0 −1 2 x2 y xy, R 0 x y f 0, 0 Along y Thus, f Along y Thus, f Along x Along x x x2 1 4 x, f x2 1 4, 2x 2x 1 1 6. 1 2. 2 0⇒x 0⇒x 0. 2 0. 1 2. 1 2. 1 1 1 4 and f 2, 1 x, f f 2, 2y ⇒ f −2 2 ≤ x ≤ 2, f 1 ≤ y ≤ 1, f 1 ≤ y ≤ 1, f 4 6 and f 2y ⇒ f 2, 1 Thus, the maxima are f 2, 1 x2 2y 2y x2 y2, R y x2 y2 x2 12 6 and the minima are f 1 2, 1 1 4 and f 1 2, 1 1 4. 59. f x, y fx fy f x, 2x 2x x 2xy 0 0 x, y : x2 y2 ≤ 8 4 2 y x 0 8, we have y2 8 2x x2 28 8 x2 and y x2 ± −4 −2 2x2 x On the boundary x2 f f Then, f f 2, 2 x2 ± 2x 8 ±8 8 x2. Thus, 2 −2 −4 4 8 ± 2x 8 x2 ± 2. 12 x2 ± 16 4x2 . 8 x2 0 implies 16 f 2, 2 4x2 or x 16 and f 2, 2, 2 2 f 2, 2 0 x 0, x ≤ 2. Thus, the maxima are f 2, 2 16 and f 16, and the minima are f x, Section 12.9 4xy 1 y2 Applications of Extrema of Functions of Two Variables 113 61. f x, y fx fy For x For x x2 41 y2 1 ,R x, y : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 1 or y 1 0 1 y x2 y 1 x2 1 2 0⇒x ⇒x 4 1 y2 x 2 x 1 y2 1 0, y 1, y R 0 or y 0. x 1 0, also, and f 0, 0 1, f 1, 1 1. The absolute maximum is 1 The absolute minimum is 0 63. False Let f x, y 1 x y. f 1, 1 . f 0, 0 . In fact, f 0, y f x, 0 0 0, 0, 1 is a relative maximum, but fx 0, 0 and fy 0, 0 do not exist. Section 12.9 Applications of Extrema of Functions of Two Variables 3. A point on the paraboloid is given by x, y, x2 y2 . The square of the distance from 5, 5, 0 to a point on the paraboloid is given by S Sx 0, we obtain the Sy x 2x 2y 5 2 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 Sx Sy x2 2x 2y y2 2 12 2 12 12 2x 2x 2x 3y 3y 3y 2 2 3 y 4x x2 4y x2 5 2 x2 y2 y2 0 0. y2 2 5 5 From the equations Sx system 5x 3x 6y 5y 24 7 0 and Sy 24 18. 12 7, From the equations Sx system 2x3 y 18 7 0 and Sy 5 5 0 0 0, we obtain the 2xy2 2x2y x y Solving simultaneously, we have x z 12 the origin to 12 7 54 7 12 18 7, 7, 2 2y3 6 7. 6 7 Therefore, the distance from is 2 2 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 5 2 18 7 6 7 6 14 . 7 y z 30, z 30 x y. 1 5 2 2 0 2 6. 5. Let x, y and z be the numbers. Since x P Px Py xyz 30y 30x 30xy 2xy x2 x2y y2 2xy xy2 y 30 x 30 2x x 10, y y 2y 0 2x 0x y 2y 10. 30 30 Solving simultaneously yields x 10, and z x2 y2 7. Let x, y, and z be the numbers and let S S Sx Sy x2 2x 2y y2 2 30 2 30 30 x x x y y y 2 z2. Since x y z 30, we have 1 1 0 2x 0x 10, y y 2y 30 30. 10. Solving simultaneously yields x 10, and z 114 Chapter 12 Functions of Several Variables 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 Vy Vz xyz 108z 108y 108zy 4yz 2y2 2zy2 2z2 4yz 2z 2yz2 z 108 y 108 4y 2y 2z 4z 4z 0 0. 108, we obtain the solution x 36 inches, y 18 inches, and z 18 inches. Solving the system 4y 11. Let a V b c 4 abc 3 108 and 2y k. Then 4 3 4 3 ab k kab b2 2ab a a2b b ab2 2ab a2 b2 2ab 0 0. 13. Let x, y, and z be the length, width, and height, respectively and let V0 be the given volume. Then V0 S Sx Sy xyz and z 2xy 2y 2x 2yz V0 x2 V0 y2 V0 xy. The surface area is 2xz 2 xy V0 V0 3 V0 x 0 0. V0, y V0 y Va Vb 4 kb 3 4 ka 3 2ab a2 0 kb 0 ka 0 x2y 0 xy2 Solving this system simultaneously yields a b and substitution yields b k 3. Therefore, the solution is a b c k 3. 15. The distance from P to Q is C Cx Cy 3k 3k x2 3k 2k x x2 x x2 y y x 4 x 2 Solving simultaneously yields x 3 V. and z 0 3 V0, x2 y 2k 4. The distance from Q to R is x y k 2 y x 2 1 . The distance from R to S is 10 y. 4 2k 1 x x 2 k 10 1 y y 0 y y 1 1 2 x 4 3x 9x2 x2 x 0⇒ x x 2 1 1 2 4 2k 0 1 3 x2 x2 1 2 2 2 y y 1 3 1 3 1 23 32 6 2 23 32 1.284 kms. 6 x x 2 2 x2 4 4 2y 4y y x x x 2 1 1 2 y Therefore, x 2 2 0.707 km and y S ection 12.9 Applications of Extrema of Functions of Two Variables 115 17. 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 where x A r, Now Ar r, A r, w sin wr cos 4r sin 2r 2 cos 2r sin cos r 2 cos 2 sin 0. 0 into the equation A r, 1 1 0 0 or cos 1 . 2 0.) Thus, the 0, we have w 4r 2r cos 0⇒w r4 2 cos h w r cos w 2r and h 2r w 2 2r 2x w 2r xh r sin . Substituting these expressions for x and h, we have r cos r sin wr sin 2r 2 sin r 2 sin cos Substituting the expression for w from Ar r, r2 4 2 cos cos 2r 2 cos r 2 2 cos2 r 2 2 cos 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. 19. R x1, x2 Rx1 Rx2 10x1 16x2 5x12 2x2 2x1 8x22 42 102 2x1x2 0, 5x1 0, x1 42x1 x2 8x2 6. 102x2 21 51 Solving this system yields x1 Rx1x1 Rx1x2 Rx2x2 10 2 16 Rx1x2 3 and x2 Rx1x1 < 0 and Rx1x1 Rx2x2 2 >0 3 and x2 6. Thus, revenue is maximized when x1 21. P x1, x2 15 x1 15x1 x2 15x2 C1 C2 4x1 500 11x2 0.05x22 775 4x2 275 0.02x12 0.05x22 0.02x12 Px1 Px2 Px1x1 Px1x2 Px2x2 0 0.10 0.04x1 0.10x2 0.04 11x1 275 110 11 11 0, x1 0, x2 Px1x1 < 0 and Px1x1 Px2x2 Px1x2 2 >0 275 and x2 110. Therefore, profit is maximized when x1 116 Chapter 12 d1 x x2 Functions of Several Variables d2 0 y2 2 23. (a) S x, y d3 y x 0 2 2 x y 2 2 2 2 y x 2 2 2 x y 4 2 2 2 y 2 2 2 4 From the graph we see that the surface has a minimum. (b) Sx x, y Sy x, y (c) tan (d) x2, y2 S 1, 1 2 x x2 y x2 y2 Sx 1, 1 i 10 1 x1 1 t, 1 2 1 2 x y2 x x 2 2 S 2 y 2 2 1 i 2 2 ⇒ 5 1 2 x x x 1 2 186.027 1 t, 1 2 2 10 5 2 10 5 2 10 5 4 2 4 y 2 2 2 24 20 y2 22 y 2 y2 42 y 2 j 10 2 8 x 6 4 2 4 2 4 6 8 y Sy 1, 1 j 2 1 Sx x1, y1 t, y1 2 10 1 2 Sy x1, y1 t 2 10 1 1 t 2 25 5 1 25 5 25 5 22 t 5 22 t 5 22 t 5 0.05, 0.90 . S1 t 2 2 2t 10 2 2t 10 4 2t 1 Using a computer algebra system, we find that the minimum occurs when t (e) x3, y3 S 0.05 x2 Sx x2, y2 t, y2 0.26t Sy x2, y2 t 0.05 0.05 0.03t 2 1.344. Thus, x2, y2 0.03t, 0.90 0.90 0.26t 0.26t 2 0.03t, 0.90 2.05 3.95 0.03t 2 2 1.10 1.10 0.26t 2 2 0.03t 0.26t Using a computer algebra system, we find that the minimum occurs when t x4, y4 S 0.10 x3 Sx x3, y3 t, y3 0.01t Sy x3, y3 t 0.10 0.10 0.09t 2 1.78. Thus x3, y3 0.10, 0.44 . 0.09t, 0.44 0.45 0.01t 0.01t 2 0.09t, 0.45 2.10 3.90 0.09t 2 2 1.55 1.55 0.01t 2 2 0.09t 0.01t Using a computer algebra system, we find that the minimum occurs when t Note: The minimum occurs at x, y (f) 0.0555, 0.3992 0.44. Thus, x4, y4 0.06, 0.44 . S x, y points in the direction that S decreases most rapidly. You would use S x, y for maximization problems. 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. S ection 12.9 Applications of Extrema of Functions of Two Variables 117 27. (a) x 2 0 2 xi 36 38 3 x 4 3 2 1 6 4 3 4 3 0 04 02 y 0 1 3 yi 4 3 ,b 4 xy 0 0 6 xiyi 1 4 3 6 3 0 4 x2 4 0 4 xi2 4 , 3 8 29. (a) x 0 1 1 2 xi 4 48 42 4 4 2 y 4 3 1 0 yi 8 xy 0 3 1 0 xiyi 1 8 4 4 x2 0 1 1 4 xi2 4, 6 a y (b) S a y 2 44 46 2x 4 2, b 24 0 4 3 2 1 3 2 4 3 2 3 (b) S 2 3 2 2 1 2 0 0 2 2 31. 0, 0 , 1, 1 , 3, 4 , 4, 2 , 5, 5 xi xiyi a b y 13, 46, 13 12 13 2 37 13 43 7 43 yi xi2 74 86 7 43 12, 51 37 43 33. 0, 6 , 4, 3 , 5, 0 , 8, xi xiyi a b y 8 4 , 10, yi 0, 5 27, 70, 70 5 205 27 0 27 2 175 27 148 945 148 xi2 205 175 148 5 46 5 51 1 12 5 37 x 43 7 5 350 296 945 148 1 0 5 175 x 148 (0, 6) (4, 3) 7 y = 37 x + 43 43 (5, 5) (3, 4) (4, 2) −2 −4 (1, 1) (0, 0) −1 (5, 0) (8, − 4) −6 18 10 (10, − 5) y = − 175 x + 945 148 148 35. (a) y (b) 240 1.7236x 79.7334 37. 1.0, 32 , 1.5, 41 , 2.0, 48 , 2.5, 53 xi a 7, 14, b yi 19, y 1.6, y 174, 14x xiyi 19 322, xi2 13.5 When x 0 100 100 41.4 bushels per acre. (c) For each one-year increase in age, the pressure changes by 1.7236 (slope of line). 118 Chapter 12 n Functions of Several Variables 39. S a, b, c i 1 yi 2xi2 yi axi2 axi2 axi2 bxi bxi bxi c c c n c 2 41. 0 0 2, 0 , xi yi 0 8 1, 0 , 0, 1 , 1, 2 , 2, 5 8 S a S b S c n n i n 1 (−1, 0) −9 (2, 5) (1, 2) 2xi yi i 1 n xi2 xi3 xi4 10 0 34 12 22 10c 22, 10b 6 5, (−2, 0) −2 6 (0, 1) 2 i 1 n yi b i 1 n axi2 n bxi xi2 1 n i 0 a i 1 n xi4 xi3 1 n xi3 xi2 1 n c i xi2yi 1 n xiyi xi2yi 34a a i b i c i 1 xi i n xiyi 1 12, 10a y 32 7x 6 5 5c x 26 35 8 a i 1 xi2 b i 1 xi cn i yi 1 a 3 7, b c 26 35 , 43. 0, 0 , 2, 2 , 3, 6 , 4, 12 xi yi xi2 xi3 xi4 xiyi xi2yi 353a 99a 29a a 1, b 14 45. 0, 0 , 2, 15 , 4, 30 , 6, 50 , 8, 65 , 10, 70 xi yi xi2 xi3 xi4 xiyi xi2yi 30, 230, 220, 1,800, 15,664, 1,670, 13,500 1,800b 220b 30b 541 56 9 20 29 99 353 70 254 99b 29b 9b 29c 9c 4c 1, c (4, 12) (3, 6) 254 70 20 0, y x2 x 15,664a 1,800a 220a y 120 220c 30c 6c x 25 14 13,500 1,670 230 0.22x 2 9.66x 1.79 25 2 112 x −5 (2, 2) (0, 0) −2 7 −1 − 20 14 47. (a) ln P (b) ln P P (c) 14,000 0.1499h 0.1499h e 0.1499h 9.3018 9.3018 10,957.7e 0.1499h 9.3018 −2 − 2,000 24 (d) Same answers. Section 12.10 Lagrange Multipliers 119 Section 12.10 1. Maximize f x, y Constraint: x f yi y x x f 5, 5 y 25 x2 2y xj x 10 g i y j y Lagrange Multipliers xy. 10 12 10 8 6 4 2 y 3. Minimize f x, y Constraint: x level curves x2 y 4 y 2. 4 y constraint f 2x i g 2y j 2x x y 8 2x y 2 xy 100 y. i y −4 j x −4 4 constraint x 2 4 6 8 10 12 2y x f 2, 2 level curves ⇒x y 5 4⇒x y 2 5. Minimize f x, y Constraint: x f 2x i g 2y j 2x 2y x 2y 2 6 i y 2. 6 7. Maximize f x, y Constraint: 2x f g 2y i 2 2 2x 1 y 2y 2x 2 2j ⇒x ⇒y ⇒ 3 2 6 4, x 2, y 4 2 1j ⇒y ⇒x 2i 1 j 1y 2 100 25, y 50 2x 2x 100 ⇒ 4x x f 25, 50 2600 f 2, 4 12 6 6 y x 2⇒x g 1, 1 y y 2 1 f 2, 2 e4 2 x2 x2 x2 y 2 is maximum when g x, y y 2. 11. Maximize f x, y Constraint: x2 ye xy xe xy y2 2x 2y y2 x y 8 y 2 e xy. 8 9. Note: f x, y is maximum. Maximize g x, y Constraint: x 2x 2y x y 8 ⇒ 2x2 x f 1, 1 13. Maximize or minimize f x, y Constraint: x2 y2 ≤1 x2 2x 2y y2 x2 ± x2 3xy y 2. Case 2: Inside the circle fx 2x 3x 2, fyy 3y 2y 2, fxy 0 0 x y 0 fxy 2 Case 1: On the circle 2x 3x x2 y2 3y 2y 1 y2 2 ,y 2 5 2 1 2 ± fy fxx 3, fxx fyy 0 ≤0 Saddle point: f 0, 0 2 2 1⇒x By combining these two cases, we have a maximum of 2 at ± 5 2 2 ,± Maxima: f ± 2 2 Minima: f ± 2 , 2 2 2 2 2 ,± 2 2 1 2 and a minimum of ± at 2 , 2 2 . 2 120 Chapter 12 Functions of Several Variables 15. Minimize f x, y, z Constraint: x 2x 2y 2z x y z y z x2 6 y2 z 2. 17. Minimize f x, y, z Constraint: x 2x 2y 2z y z x2 1 y2 z 2. x y z y z 2 x f x z 1 3 y z y z 1 3 6⇒x 12 y 111 3, 3, 3 1⇒x f 2, 2, 2 19. Maximize f x, y, z Constraints: x x f yz i yz xz xy x x y y z z g xz j y y h xyk x yz. z z 32 0 21. Maximize f x, y, z Constraints: x x f g x y z y 2y 3z zj 2y 3z h yk xy 6 0 yz. i j k i j k yi i 3 4 2j i 8 y 3 3k yz xy ⇒ x z x 2 3 y ⇒x z 32 0 2x 2z 32 ⇒ x y z 16 8 x x 6⇒ y 0⇒ z x x 3 x 3 x 3 8 3 3 3, y x 2 f 8, 16, 8 1024 x 2 3 ,z 2 1 3 f 3, , 1 2 6 23. Minimize the square of the distance f x, y 1. subject to the constraint 2x 3y 2x 2y 2x 3y 2 3 y 3x 2 2 ,y 13 2 13 , 3 13 x2 y2 25. Minimize the square of the distance f x, y, z x 2 2 y y 1 z 2 z 1. 1 2 subject to the constraint x 3 13 and the desired 13 . 13 x 2x 2y 2z y z 2 1 1 1⇒x y 1⇒x z and y 2x 1 x x 1 1 The point on the line is distance is d 2 13 2 3 13 2 1, y z 0 The point on the plane is 1, 0, 0 and the desired distance is d 1 2 2 0 1 2 0 1 2 3. S ection 12.10 z subject to the constraints 27. Maximize f x, y, z x2 y 2 z2 36 and 2x y z 2. 0 0 1 x2 2x y2 y 2x 2y 2z z2 z 2 x 36 2⇒z 2y 2 Lagrange Multipliers 121 29. Optimization problems that have restrictions or contstraints on the values that can be used to produce the optimal solution are called contrained optimization problems. 2y 2x y2 5y 20y 10y y 2 2 2 5y 36 0 0 5± 2 30y 2 15y 2 32 16 y 265 15 Choosing the positive value for y we have the point 10 2 265 5 , 15 265 15 , 1 3 265 . 31. Maximize V x, y, z x yz subject to the constraint x 2y 2z 108. yz xz xy x 2y 2 2 2z y z and x 108 ⇒ 6y x 2y 108, y 36, y z 18 18 33. Minimize C x, y, z 5xy 3 2xz to the constraint x yz 480. 8y 8x 6x xyz 6z 6z 6y yz xz xy x y, 4y 3z 2yz xy subject 480 ⇒ 4 y 3 3 x 480 y 3 3 360, z 43 3 43 3 360 Volume is maximum when the dimensions are 36 18 18 inches Dimensions: 3 360 360 360 feet z x y z x y 35. Maximize V x, y, z 8yz 8xz 8xy x2 a2 y2 b2 2x a2 2y b2 2z c2 z2 c2 x 1⇒ a ,y 3 x2 a2 2x 2y 2z 8 xyz subject to the constraint x2 a2 y2 b2 z2 c2 1. y2 b2 z2 c2 3x2 a2 1, b ,z 3 3y 2 b2 c 3 1, 3z 2 c2 1 Therefore, the dimensions of the box are 2 3a 3 2 3b 3 2 3c . 3 122 Chapter 12 Functions of Several Variables Distance , minimize T x, y Rate d12 v1 x2 d22 y 2 subject to the constraint x v2 37. Using the formula Time x v1 d22 y v2 d22 x y a x 1 y a. x2 y2 x v1 d12 x2 y v2 d22 y2 P Medium 1 d1 θ1 Since sin x d12 x2 y x2 and sin 2 y 2 y d22 sin v1 1 y , we have 2 x θ2 a Medium 2 d2 Q d1 v1 2 d2 v2 2pq y2 or sin 2 . v2 41. Maximize P x, y 100x 0.25y 0.75 36y y x x y y x 0.75 39. Maximize P p, q, r Constraint: p P 2q 2p 2p p q g 2r 2r 2q r q p P 111 3, 3, 3 2pr 2qr. q r 1 subject to the constraint 48x 25x 0.75y0.75 100,000. 48 25 36 75 y x 0.25 48 36 ⇒ ⇒ ⇒3 ⇒ 1 r r 2 1 3 1 3 2 3 4 3 4p q r 41 75x0.25y 0.25 0.25 0.75 48 25 4 4x 75 36 q 1 ⇒p 2 1 3 1 3 1 3, q 2 1 3 1 3, 1 3 r 1 3 2 3. y x y 48x 36y 100,000 ⇒ 192x x Therefore, P 3125 6250 6,3 100,000 3125 ,y 6 6250 3 147,314. 43. Minimize C x, y 48 36 25x y 48x 36y subject to the constraint 100x0.25y0.75 ⇒ ⇒ y x x y y x 0.75 20,000. 0.75 0.75 48 25 36 75 y x 0.25 0.25 75x0.25y 0.25 0.75 48 25 75 36 4x y x 100x0.25y0.75 20,000 ⇒ x0.25 4x 0.75 4⇒y 200 200 40.75 4x 200 22 200 2 50 2 x y Therefore, C 50 2, 200 2 $13,576.45. R eview Exercises for Chapter 12 45. (a) Maximize g , , sin cos cos cos sin cos cos cos sin ⇒ g (b) g ,. 333 1 8 3 123 cos cos tan cos tan subject to the constraint tan ⇒ . 3 γ ⇒ cos cos cos cos cos cos cos cos cos cos sin sin α 3 2 3 β Review Exercises for Chapter 12 1. No, it is not the graph of a function. 3. f x, y ex 2 y2 5. f x, y x2 y2 The level curves are of the form c ln c 2 ex The level curves are of the form c 1 x2 c x2 y2 y2 . c y2 x2 y 2. The level curves are circles centered at the origin. y 2 The level curves are hyperbolas. y c = 10 4 c = −12 c = − 2 c=2 c = 12 c=1 −2 x 2 −4 −2 1 x −1 1 4 Generated by Mathematica −4 Generated by Mathematica 7. f x, y e z 3 x2 y2 9. f x, y, z y x2 x2 y z2 z2 1 1 3 z −3 −3 Elliptic paraboloid 2 x 3 y 3 x −3 3 y 11. x, y → 1, 1 lim xy x 2 y 2 1 2 13. x, y → 0, 0 lim x4 x 2, 0, 4x2y y2 x4 4 Continuous except at 0, 0 . For y For y 4x 2y y2 x4 4x 4 x4 0 2, for x 0 4x 2y x y2 0, for x Thus, the limit does not exist. Continuous except at 0, 0 . 124 Chapter 12 ex cos y ex cos y e sin y x Functions of Several Variables 17. z z x z y xey ey xey yex yex ex 15. f x, y fx fy 19. g x, y gx gy xy x2 y x2 x x2 x2 y2 y2 x2 y2 y2 2 xy 2x y2 2 y y2 x2 x2 y 2 2 21. f x, y, z fx fy fz z arctan y x y x2 x2 x2 xz y2 yz y2 1 1 z y2 x2 z 1 y 2 x2 x y x arctan 23. u x, t u x u t ce cne n 2t sin nx cos nx n2t 25. z 3 n2t cn2e sin nx −1 3 3 x y 27. f x, y fx fy fxx fyy fxy fyx 3x2 6x x 6 12y 1 1 xy y 6y 2 2y 3 29. h x, y hx hy hxx hyy hxy hyx x sin y sin y x cos y y cos x x sin y cos y cos y y cos x y sin x cos x 31. z z x 2z x2 2x 2 2y 2 y2 x2 z y 2z sin x sin x y2 Therefore, z x2 2 z y2 2 0. 33. z z x 2 z x2 y x2 x2 y2 2xy y2 x2 y2 x2 x2 y2 3x2 x2 2z 2 35. z dz 4x2 y2 y2 2 2 x sin y x z dy y sin y x y y cos dx x x cos y dy x z dx x 2y x2 1 3 x2 x2 x2 y2 2 2y 3x2 x2 y2 y2 3 z y z y2 2 2y 2y y2 y2 3 2z y2 y2 2 y 2 2y 2 x2 y 2 x2 x2 y 2 4 2y Therefore, x2 y2 0. R eview Exercises for Chapter 12 37. z2 2z dx dz x2 2x dx x dx z y2 2y dy y dy z dz z 51 13 2 17 26 13 12 1 13 2 0.0503 17 26 5% 0.654 cm 39. V dV 1 3 2 3 125 r 2h r h dr 1 3 r 2 dh 2 3 2 5 ±1 8 5 1 3 2 in.3 2 ±8 1 ±6 ± 1 6 ± Percentage error: 41. w ln x2 y2 , x dw dt 2t w dx x dt 2x x2 2t 5t2 y2 3, y 4 t Chain Rule: w dy y dt 2 2y x2 32 4 4 4t y2 25 ln 2t 10t 5t2 4t 3 2 y2 t 2 1 24 32 t 4 2 2t 32 10t 2t t 2 Substitution: w dw dt 43. u 2 2t 2t x2 y2 32 32 ln x2 4 25 t t 2 24 t 4 t2 r cos t, y ux xr 2x cos t 2r cos2 t 4 z2, x u r r sin t, z uz zr 2z 0 t 2r uz zt 2y r cos t Chain Rule: uy yr 2y sin t r sin2 u t ux xt 2x 2 2t uy yt r sin t 2z 2t r 2 sin t cos t r 2 sin t cos t Substitution: u r, t u r u t 45. 2xy 2y z x x2y x 2r 2t r 2 cos2 t r 2 sin2 t t2 r2 t2 2yz z x z xz 2z z2 z x z x 0 0 2xy 2y 0 x2 2y x 2z 2z x x2 2y 2z 2z x z 2z x 2xy z 2y 2z x2 2y z y 2z x z y 2z z y z y 126 47. Chapter 12 f x, y f f 2, 1 u Du f 2, 1 x2y 2 xy i 4i 4j 1 v 2 Functions of Several Variables 49. x 2j w 1, 2, 2 2 i 2 u 2 j 2 22 22 0 u Duw 1, 2, 2 w w y2 zi 2i 1 v 3 xz 2y j 4j 2 i 3 k 1 j 3 u 2 k 3 4 3 4 3 2 3 2 3 xk f 2, 1 y x2 y2 w 1, 2, 2 51. z z z 1, 1 z 1, 1 53. x2 x2 y2 j y2 2 z z z 0, z 0, 4 e x cos y x 2xy i x2 y 2 2 1 i 2 1 2 65 9x2 18x i 54i 54i 54i 4y2 8yj 16 j 16 j 16 j 1 ,0 2 e cos y i 2 j 2 e x sin y j 2 , 2 2 2 2 i 2 1 4 55. 9x2 4y2 f x, y f x, y f 3, 2 57. F x, y, z F F 2, 1, 4 x2y 2xy i 4i z x 2j 4j 0 k k Therefore, the equation of the tangent plane is 1 27i 793 8j 4x 4x 2 4y 4y z 8, 1 z 4 0 or Unit normal: and the equation of the normal line is x 4 x2 2x k y2 4i x2 3 2x i k 4i i 4 0 2j j 2 0 k k 1 1 2i 2j 2 y 4 y2 z 0 2y j k 1 z 4 . 1 59. F x, y, z F F 2, 3, 4 4x 2y 6y 6j z 9 k 0 61. F x, y, z G x, y, z F G F 2, 1, 3 F G z 0 Therefore, the equation of the tangent plane is z 4 0 or z 4, and the equation of the normal line is x 2, y 3, z 4 t. Therefore, the equation of the tangent line is x 1 63. f x, y, z f x, y, z f 2, 1, 3 cos n n 36.7 x2 2x i 4i k y2 2yj 2j 6 56 z2 14 2zk 6k Normal vector to plane. 3 14 14 2 y 2 1 ,z 3. R eview Exercises for Chapter 12 65. f x, y fx fy fxx fyy fxy From fx 2 3 0, we have y 39 2, 4 127 x3 3x2 3x 6x 3xy 3y 2y y2 3 x2 0 x 2 − 30 z y 0 30 y x2. Substituting this into fy fxy fxy 2 2 0, we have 3x 39 2, 4, 2x2 27 16 x 2x 3 0. Thus, x 0 or 3 . 2 At the critical point 0, 0 , fxx fyy At the critical point 1 x 1 x2 1 y2 xy2 2 x3 1 2 y3 , fxx fyy < 0. Therefore, 0, 0, 0 is a saddle point. > 0 and fxx > 0. Therefore, is a relative minimum. 67. f x, y fx fy Thus, x2y xy y x 1 y 20 z 0, x2y 0, x y 2 or x 1 1 4 x − 20 − 24 34 y y and substitution yields the critical point 1, 1 . (1, 1, 3) fxx fxy fyy At the critical point 1, 1 , fxx 2 > 0 and fxx fyy fxy 2 3 > 0. Thus, 1, 1, 3 is a relative minimum. 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 at x0, y0 . 71. P x1, x2 R 225 C1 0.45x1 Px1 Px2 0.9x1 0.86x2 2 C2 0.4 x1 0.8x2 0.9x1 0.8x1 0.8x1 x2 x1 2 x2 0 210 0 210 0.05x12 210x1 15x1 210x2 5400 11,500 0.03x22 15x2 6100 0.43x2 0.8x1x2 210 0.8x2 210 0.86x2 0.9 0.8 0.86 Solving this system yields x1 Px1x1 Px1x2 Px2x2 Px1x1 < 0 Px1x1 Px2x2 Px1x2 2 94 and x2 157. >0 94 and x2 157. Therefore, profit is maximum when x1 128 Chapter 12 Functions of Several Variables 2y subject to the 75. (a) y 2.29t 30 4x xy 73. Maximize f x, y constraint 20x 4y 2000. 4 x 20x y 2 4y 20 4 5x y 2.34 6 5x 5x 10x x y y y 500 6 494 49.4 253 −1 −5 3 −2 −5 11 2000 ⇒ (b) 20 f 49.4, 253 13,201.8 Yes, the data appears more linear. (c) y (d) 8.37 ln t 25 1.54 −1 −5 10 The logarithmic model is a better fit. 77. Optimize f x, y, z y x x x z z y y z xy yz xz subject to the constraint x y z 1. x y 1⇒x 1 3 z y z 1 3 Maximum: f 79. PQ C 3 x2 x2 111 3, 3, 3 4, QR 4 y 2 z y2 y2 1 10 1, RS 2 z; x y z 10 Constraint: x C 3x i x2 4 3x 2y 1 9x2 4y2 x2 y2 x2 y 2 g 2y j y2 1 4 1 k i j k 4 ⇒ x2 1 ⇒ y2 2 ,y 2 1 2 1 3 3 ,z 3 10 2 2 3 3 8.716 m. Hence, x ...
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