ODDREV12 - R eview Exercises for(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

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Unformatted text preview: 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 Problem Solving for Chapter 12 129 Problem Solving for Chapter 12 1. (a) The three sides have lengths 5, 6, and 5. Thus, s 16 2 8 and A area 2 83 2 3 ss as bs 12 c , subject to the constraint a b c constant (perimeter). (b) Let f a, b, c Using Lagrange multipliers, ss ss ss bs as bs c c b b b s a⇒a b. From the first 2 equations s Similarly, b (c) Let f a, b, c c and hence a a b c which is an equilateral triangle. 2 c, subject to Area ss as bs c constant. Using Lagrange multipliers, 1 1 1 ss ss ss a bs as as s c c b b⇒a b and a b c. 1 base height 3 3 Hence, s 3. (a) F x, y, z Fx yz, Fy xyz 1 0 xy (b) V z xz, Fz 3 x0 y0 Tangent plane: y0 z 0 x y0 z 0 x x0 x0 z 0 y x0 z 0 y x0 y0 z y0 x0 y0 z 3 z0 0 11 3 3 3 2 y0 z0 x0 z0 3 x0 y0 9 2 3 x 0 y0 Tangent plane 3 x0 z0 3 y 3x0 y0 z 0 3 x 3 y0 z0 Base 5. We cannot use Theorem 12.9 since f is not a differentiable function of x and y. Hence, we use the definition of directional derivatives. Du f x, y t →0 lim fx t cos , y t t 2 t sin t 2 t f x, y f0 Du f 0, 0 t →0 ,0 f 0, 0 lim 1 lim t→0 t If f 0, 0 2, then 4 t2 2 t 2 t 2 t2 2 t →0 lim 1 2t2 t t2 t →0 lim 2 which does not exist. t f0 Du f 0, 0 t →0 t 2 ,0 t t 2 2 lim 1 2t2 t →0 t t2 lim 2 0 which implies that the directional derivative exists. 130 Chapter 12 Functions of Several Variables f x x f x f y a 7. H z Hx k 5xy 6xz 6yz k 5xy 6000 y 6000 . x 6000 1200. 9. (a) Caxa y f y 1y1 a, C1 C1 a x ay a 1000 ⇒H xy 5y 6000 x2 Cax ay1 Ca Cx ay1 a x ay1 a a 0 ⇒ 5yx2 y ⇒ x3 3 C1 a a a x ay1 f Ct a x at1 By symmetry, x Thus, x y 2 y3 5 3 3 (b) f t x, ty 150. C tx a ty a 1 a y1 a 150 and z Cx ay1 t t f x, y 11. (a) x y 64 cos 45 t 32 2t 16t2 32 2t 16t2 64 sin 45 t y (b) tan x 50 arctan (c) d dt x y 50 arctan 32 2t 32 2t 16t2 50 64t4 16 8 2t2 25t 256 2t3 1024t2 25 2 800 2t 625 1 1 32 2t 32 2t 16t2 2 50 64 8 2t2 25t 25 2 32 2t 50 2 (d) 30 0 −5 4 No. The rate of change of (e) d dt 0 when 8 2t2 25t 25 2 t is greatest when the projectile is closest to the camera. 0 25 252 48 2 25 2 28 2 32 2 0.98 second. 32t 0 or t 2 1.41 seconds. z 1 No, the projectile is at its maximum height when dy dt 13. (a) There is a minimum at 0, 0, 0 , maxima at 0, ± 1, 2 e and saddle point at ± 1, 0, 1 e : fx x2 e e fy x2 e e x2 x2 x2 2y2 e y2 y2 x2 y2 2x 2x 2x 2x e 2x x2 y2 x2 2x3 x2 2y2 4xy2 y2 2 2 y 0 ⇒ x3 4y e 4y x2 y2 2xy2 x 0 x 2y2 e y2 2y 2y 4y x2 4y3 2y2 2x2y x2 y2 0 ⇒ 2y3 x2y 2y 0 Solving the two equations x3 2xy2 x 0 and 2y3 x2y 2y 0, you obtain the following critical points: 0, ± 1 , ± 1, 0 , 0, 0 . Using the second derivative test, you obtain the results above. —CONTINUED— Problem Solving for Chapter 12 13. —CONTINUED— (b) As in part (a), you obtain fx fy e e x2 x2 y2 y2 131 (c) In general, for 1 x2 2y2 2y2 > 0 you obtain 2x x2 2y 2 0, 0, 0 minimum 0, ± 1, ± 1, 0, e maxima e saddle The critical numbers are 0, 0 , 0, ± 1 , ± 1, 0 . These yield ± 1, 0, 1 e minima 0, ± 1, 2 e maxima For < 0, you obtain ± 1, 0, e minima e maxima 0, 0, 0 saddle z 1 0, ± 1, 0, 0, 0 saddle 1 x 2 −1 y 15. (a) 6 cm 1 cm (b) 6 cm 1 cm (c) The height has more effect since the shaded region in (b) is larger than the shaded region in (a). (d) A If dl If dh hl ⇒ dA l dh h dl 0, then dA 0, then dA 1 0.01 6 0.01 0.01. 0.06. 0.01 and dh 0.01 and dl 17. Essay 1 fx 2 x u t 2 19. u x, t Let r ct fx x us st ct ct. Then u r, s 1 df 2 dr c 2 ct and s ur rt 1 d 2f 2 dr2 ur rx 1 d 2f 1 2 dr2 2 1 fr 2 fs . 1 df c 2 ds c2 d 2f 2 dr2 1 df 1 2 ds d 2f ds2 u t2 u x c 2 1 d 2f c 2 ds2 us sx 1 df 1 2 dr 2 2 u x2 2 1 d 2f 1 2 ds2 1 d 2f 2 dr2 d 2f ds2 Thus, u t2 2 c2 u . x2 ...
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This note was uploaded on 05/18/2011 for the course MAC 2311 taught by Professor All during the Fall '08 term at University of Florida.

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