EVNREV12 - R eview Exercises for Chapter 12 44. Minimize C...

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Unformatted text preview: R eview Exercises for Chapter 12 44. Minimize C x, y 48 36 60x 0.4y0.4 355 48x ⇒ ⇒ 36y subject to the constraint 100x0.6y0.4 y x x y y x 0.4 20,000. 48 60 36 40 y x 0.6 0.6 40x0.6y 0.6 0.4 48 60 40 36 8 x 9 y x 100x0.6y0.4 20,000 ⇒ x0.6 8 x 9 0.4 8 ⇒y 9 200 x y Therefore, C 209.65, 186.35 46. f x, y ax by, x, y > 0 y2 36 1 4x 200 8 9 0.4 8 200 9 8 9 0.4 209.65 186.35 $16,771.94. x2 Constraint: 64 (a) Level curves of f x, y y Using y x 7, y 8 3y are lines of form (b) Level curves of f x, y y 4 x 9 4 x 9 4, y C. 4x 9y are lines of form 4 x 3 4 x 3 C. 12.3, you obtain 3, and f 7, 3 28 9 37. Using y x 7, you obtain 5.2, and f 4, 5.2 62.8. − 10 10 −8 Constraint is an ellipse. Review Exercises for Chapter 12 2. Yes, it is the graph of a function. x x y c=−1 3 c = −1 2 c=−2 2 c = −2 c= 2 1 4. f x, y ln x y 2 6. f x, y c= 1 2 The level curves are of the form c ec ln x y x y. −3 c= 3 2 c=1 c=0 3 The level curves are of the form c y 1 c c c x x c y x. c=1 −3 3 The level curves are hyperbolas. −2 −2 c= 2 c=2 3 The level curves are passing through the origin with slope 1 . 356 Chapter 12 Functions of Several Variables xy x2 y2 ± x. 8. g x, y z 60 y 1 x 10. f x, y, z Elliptic cone z 2 9x2 y2 9z2 0 12. x, y → 1, 1 lim Does not exist Continuous except when y 2 x 5 y 5 x 5 y 14. x, y → 0, 0 lim y xe y 1 x2 2 0 1 0 0 0 16. f x, y fx fy xy x yx x x2 x y 2 y y y 2 18. xy x y2 y 2 z z x z y ln x 2 x2 x2 2x y2 2y y2 y2 1 1 1 Continuous everywhere 20. w w x w y w z x2 12 x 2 x2 x2 y2 y2 y y2 z y2 z2 z2 12 22. f x, y, z 2x x2 x y2 z2 fx 1 1 1 1 2 1 x2 x2 x2 1 1 . At 2, 0, 0 , z x 0. x2 x2 x y2 y y2 z y2 y2 y2 z2 z2 32 2x z2 z2 fy fz z2 z2 z2 32 1 1 32 32 24. u x, t u x u t c sin akx cos kt akc cos akx cos kt kc sin akx sin kt 26. z z x x2 ln y 2x ln y Slope in x-direction. z y x2 1 y . At 2, 0, 0 , z y 4. Slope in y-direction. x x y x x x x x x y x y 2y y 2x y y 3 2 2 28. h x, y hx hy hxx hyy hxy hyx y 30. g x, y gx gy gxx 2 cos x sin x 2 sin x cos x 2y 2y 2y 2y 2y 2y 2y gyy gxy gyx 4 cos x 2 cos x 2 cos x 3 x y2 x 2y x y4 2y x y4 y y x x x x y y3 y y3 R eview Exercises for Chapter 12 32. z z x 2z x2 357 x3 3x2 6x 3xy 2 3y 2 34. z z x 2z e x sin y e x sin y e x sin y e x cos y e x sin y 2z 2z x2 z y 2z y2 2z x2 2z y2 z y 2z y2 6xy 6x 0. Therefore, Therefore, x2 y2 0. 36. z dz xy x2 y 2 z dx x x2 z dy y y2 y xy x x2 y 2 x2 y2 dx x2 y 2x x2 xy y y2 x2 y2 dy x2 y3 y2 32 dx x2 x3 y2 32 dy 38. From the accompanying figure we observe tan dh Letting x h or h x h dx x 100, dx x tan h h d tan dx x sec2 d. θ 1 ±, 2 11 , and d 60 ± 180 . x (Note that we express the measurement of the angle in radians.) The maximum error is approximately dh tan 11 60 h2 h2 h2 dr h2 cos t, y ux xt 1 sin t sin t 1 Substitution: u du dt sin2 t r2 r2 r2 h2 rh h2 dr dh rh r2 8 h2 dh 1 10 ± 29 8 ± ± 1 2 100 sec2 11 60 ± 180 ± 0.3247 ± 2.4814 ± 2.81 feet. 40. A dA r r2 r2 2r 2 r2 25 1 ± 8 29 43 8 29 42. u y2 x, x du dt sin t uy yt sin t 2y cos t Chain Rule: 2 sin t cos t 2 cos t cos t sin t sin t 1 2 cos t 2 sin t cos t 358 Chapter 12 xy ,x z Functions of Several Variables 44. w 2r w r t, y rt, z 2r wy yr t wz zr 46. 2xz z x z2 xz2 y sin z y cos z z x z x 0 0 z2 y cos z 0 sin z y cos z Chain Rule: wx xr y 2 z 2rt 2r t 4r2t x t z 2r 2r xy 2 z2 tt t 2 2r 2r t rt t2 2xz z y y cos z z y 2xz sin z z y 4rt2 t 3 2r t 2 wy yt x r z rt 2 2r t 2 2xz w t wx xt y 1 z 4r2t xy z 4r2t 4r2t 2r wz zt 1 xy z2 4r 3 Substitution: w w r w t 12 y 4 2r t rt 2r t 4rt2 t 3 2r t 2 rt2 t 2 2r2t 2r rt 2 t 4r 3 48. f x, y f f 1, 4 u Du f 1, 4 x2 1 yj 2 2j 25 i 5 u 5 j 5 45 5 25 5 25 5 50. w w w 1, 0, 1 u Duw 1, 0, 1 6x2 12x 12 i 1 v 3 3xy 3y i 3j 3 i 3 4y 2z 3x 8yz j 4y 2 k 2x i 2i 1 v 5 f 1, 4 3 j 3 u 3 k 3 w 1, 0, 1 43 3 0 53 52. z z z 2, 1 z 2, 1 x2 x x2 x 4j 4 y2 3 4y sin x 4y cos xi 2j y2 4 sin x 2y j y 2xy i y2 x2 x y j 2 54. z z z 2, 1 z 2, 1 x2y 2xy i 4i 42 x2j 4j 56. 4y sin x 58. F x, y, z F F 2, 3, 4 y2 2 yj 6j z2 2z k 8k 25 0 f x, y f x, y f 2 ,1 2 3j 4k Therefore, the equation of the tangent plane is 3y 3 y 3 4z 3 z 4 4 4 0 or 3y 4z 25, Normal vector: j and the equation of the normal line is x 2, . R eview Exercises for Chapter 12 60. F x, y, z F F 1, 2, 2 x2 2x i 2i y2 2y j 4j z2 9 2z k 4k 2i 2j 2k 0 62. F x, y, z G x, y, z F G F 4, 4, 9 F G y2 x 2y i i 8i i 8 1 j k j 0 1 k 1 0 i j 8k z y k 25 0 0 359 Therefore, the equation of the tangent plane is x x 1 2y 2y 2z 9, 2 2z 2 0 or and the equation of the normal line is x 1 1 y 2 2 z 2 2 . Therefore, the equation of the tangent line is x 1 4 y 1 cos x, sin y, 0, 1 y 0 0.1 0.1 0.3 0.5 z 3 4 z 9 . 8 1 0 0 64. (a) f x, y fx fy P1 x, y cos x sin x, cos y, 1 y sin y, f 0, 0 fx 0, 0 fy 0, 0 1 0 1 (b) fxx fyy fxy fxx 0, 0 fyy 0, 0 fxy 0, 0 y 12 2x P2 x, y (d) x 0 0 0.2 0.5 1 (c) If y 0, you obtain the 2nd degree Taylor polynomial for cos x. f x, y 1.0 1.0998 1.0799 1.1731 1.0197 P1 x, y 1.0 1.1 1.1 1.3 1.5 P2 x, y 1.0 1.1 1.095 1.175 1.0 (e) 2 z z 2 2 −2 −1 1 y 1 x −1 1 y −2 2 x 1 −1 1 −1 2 y 1 x The accuracy lessens as the distance from 0, 0 increases. 66. f x, y fx fy 4 3y fxx fyy fxy fxx fyy 2x2 4x 6x 6y 4 18 6 f xy 2 6xy 6y 18y 8 9y 2 0 0, x 8x 14 3y 4 3, 8⇒y x 4 4 18 6 2 36 > 0 Therefore, 4, 4 , 3 2 is a relative minimum. 360 68. z zx zy Chapter 12 50 x 50 50 zxx y 0.3x2 0.15y 2 Functions of Several Variables 0.1x3 20 20.6 20x 0, x 0, y 14 , 0 4.2 6 4.2 6 2 150 ± 10 ± 14 0.05y3 20.6y 125 Critical Points: 10, 14 , 10, 0.6x, zyy zxy 2 10, 14 , 10, 14 0.3y, zxy 6 2 At 10, 14 , zxx zyy At 10, 10, At At 14 , zxx zyy 14, 02 > 0, zxx < 0. 02 < 0. 02 < 0. 02 > 0, zxx < 0. 10, 14, 199.4 is a relative maximum. zxy zxy 349.4 is a saddle point. 2 10, 14 , zxx zyy 10, 14, 10, 10, 14, 14 , zxx zyy 4.2 6 4.2 200.6 is a saddle point. xxy 749.4 is a relative minimum. 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. 0.25x12 5x1 0.15x22 72. Minimize C x1, x2 0.50x1 0.30x2 x1 x2 10 12 10x1 3x2 3x2 3x2 x1 x2 20 12x2 subject to the constraint x1 x2 1000. 1000 ⇒ 3x1 5x1 8x1 3000 20 3020 377.5 622.5 C 377.5, 622.5 104,997.50 74. Minimize the square of the distance: f x, y, z fx fy 2x 2y 2 2 x 2 x2 2 x2 2 2 y y2 2x y2 2y 2 2 x2 2 2 y2 2x3 2y3 0 2. 2xy2 2x2y 0 0 0.6894. 0x 0y Clearly x y and hence: 4x3 x 0.6894 2 2 Thus, distance 2 distance 2.08 2 0. Using a computer algebra system, x 0.6894 2 2 2 0.6894 2 2 4.3389. 76. (a) 25, 28 , 50, 38 , 75, 54 , 100, 75 , 125, 102 xi xi4 375, 382,421,875, 3,515,625b 34,375b 375b 0.0717, c yi xi yi 297, 26,900, xi2 xi2 yi 34,375, 2,760,000, xi3 3,515,625 382,421,875a 3,515,625a 34,375a a 0.0045, b 34,375c 375c 5c 2,760,000 26,900 297 0.0045x2 0.0717x 23.2914 23.2914, y (b) When x 80 km/hr, y 57.8 km. P roblem Solving for Chapter 12 361 78. Optimize f x, y 2 xy x2 x If x 2y 2 2 x2 x2y subject to the constraint x 4 xy ⇒ x 0 or x 4y 2y 2. 0, y 1. If x 41 3, 3 16 27 4y, then y 1 3, x 4 3. Maximum: f Minimum: f 0, 1 0 Problem Solving for Chapter 12 2. V 43 r 3 M r 2h 4 r2 1000 2 rh 43 r2 r3 Material V 1000 ⇒ h Hence, M 4 r2 4 r2 dM dr 8r r3 16 r 3 8 3 r3 Then, h 1000 8r 2000 r2 2000 750 43 r2 ⇒r 750 5 6 13 2r 2000 r 2000 r2 1000 82 r 3 16 r 3 43 r2 r3 0 . 0. The tank is a sphere of radius r 5 6 13 . 4. (a) As x → ± x→ ,f x gx 1 13 x3 x→ 1 lim 1 3→x and hence gx 0. −6 4 lim fx fx 6 (b) Let x0, x03 be a point on the graph of f. −4 The line through this point perpendicular to g is y x x0 3 x0 3 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 x03 1. 1 . Hence, the point on f farthest from g is 2 1 , 2 1 . 2 2 h is a maximum for x0 3 3 3 362 Chapter 12 H k 6xy V xyz Functions of Several Variables k 5xy 6xz xy 6yz 1000 . xy 1000 . x y 10 1 4 1 4 1 ellipse Ty 0⇒y 2y 1 21 y2 1 ,x 2 0⇒ ± − 1 2 1 2 6. Heat Loss 3xz 3xz 3yz 3yz 1000 ⇒ z 6k xy Hy 2x2 2x2 2x2 x2 18 y 1000 y Then H Setting Hx 8. (a) T x, y 0, you obtain x y2 y2 y y y 1 2 10 1 4 2 z 10. y x − 1 2 1 2 12 14 2 (b) On x2 Ty y2 1, T x, y 2y 4x 1 0, Ty y2 3 . 2 1 2 y 10 12 y2 y Inside: Tx T 0, T± 1 2 3 , 2 0, 39 minimum 4 1 2 49 maximum 4 r sin , z uy y r sin z uz z u r cos y u sin . y 2u 10. x u r cos , y ux x u x u r 2u 2 Similarly, u cos x r sin r cos 2u x 2 2u y 2u z x2 2 ux yx 2u xy uy y2 xz 2 uz yz r r u cos x u sin y u r cos x u r sin y x2 Similarly, r2 sin2 2 y2 2 r 2 cos2 2 2 2u r 2 sin cos xy u r2 u cos2 x2 u2 sin y2 2 2 u cos sin . xy Now observe that 2u r2 1u rr 1 r2 2u 2 2u 2u z2 x2 2u cos2 sin2 2 2u y2 2u sin2 cos2 2 2 2u cos sin xy 2u sin cos xy 1 r u cos x u sin y 1u sin ry 2u x2 2 y2 2 1u cos rx z2 u x2 u y2 u . z2 2u Thus, Laplaces equation in cylindrical coordinates, is r2 1u rr 1 r2 2u 2 2u z2 0. P roblem Solving for Chapter 12 dd dt 32 t2 t2 3 2t 4 2t 8 16 363 12. (a) d x2 y2 32 2t 2 1024 2t3 4 2t 16 32 2t 256t4 16t2 2 (b) 4096t2 16t (c) When t dd dt t2 2: (d) 62 82 38.16 ft sec d 2d dt2 32 t3 6 2t2 t2 4 2t 36t 16 32 12 32 0 32 12 20 when t 1.943 seconds. No. The projectile is at its maximum height when t 2. 0. 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 16. r, dr x y 5, y 18 5 ± 0.05, d ± 0.05 4 3 r cos r sin 5 cos 5 sin 18 4.924 0.868 2 1 π θ = 18 5 (r, θ ) = 5, ( π 18 ) 18 x 1 2 3 4 5 (a) dx should be more effected by changes in r. dx cos dr r sin 0.868 d d (b) dy should be more effected by changes in . dy sin dr 0.174 dr r cos d 4.924 d because 4.924 > 0.174. 0.985 dr dx is more effected by changes in r because 0.985 > 0.868. dy is more effected by 18. u t 2u 1 2 1 2 cos x sin x t t t cos x sin x cos x t t t t2 u x 2u 1 cos x 2 1 2 2u x2 Then, sin x 2u t sin x t t2 x2 . ...
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