4 x 17 2x 1 x2 dx 0 0 2 0 1 x3 16xdx 576 0 16 1 y

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Unformatted text preview: = 2 dx −6xy + 3y 2xy − y 2 yexy dy = − xy dx xe + yey + ey √ 2 xy − y 1 − (1/2)(xy )−1/2 y dy = =− + 3y − 4, √ dx x − 6 xy −(1/2)(xy )−1/2 x + 3 1/2 where x and y are the distances of cars A and B , respectively, from the 25. D = x2 + y 2 intersection and D is the distance between them. dD/dt = x/ x2 + y 2 1/2 (dx/dt) + y / x2 + y 2 1/2 (dy/dt), dx/dt = −25 and dy/dt = −30 when x = 0.3 and y = 0.4 so dD/dt = (0.3/0.5)(−25) + (0.4/0.5)(−30) = −39 mph. 26. T = (1/10)P V , dT /dt = (V /10)(dP/dt) + (P/10)(dV /dt), dV /dt = 4 and dP/dt = −1 when V = 200 and P = 5 so dT /dt = (20)(−1) + (1/2)(4) = −18 K/s. 1 1 ab sin θ but θ = π/6 when a = 4 and b = 3 so A = (4)(3) sin(π/6) = 3. 2 2 1 6 , 0 ≤ θ ≤ π/2. Solve ab sin θ = 3 for θ to get θ = sin−1 2 ab 27. A = ∂θ da ∂θ db dθ + = = ∂a dt ∂b dt dt = −√ 6 a2 b2 − 36 1 − 6 a2 b da + dt 1 36 36 1− 2 2 a2 b2 ab da db 1 da 1 db + , = 1 and =1 a dt b dt dt dt when a = 4 and b = 3 so 1− 6 dθ = −√ dt 144 − 36 11 + 43 =− − 6 ab2 db dt 7 7√ √ =− 3 radians/s 36 12 3 Exercise Set 15.4 540 √ 28. From the law of cosines, c = a2 + b2 − 2ab cos θ where c is the length of the third side. √ θ = π/3 so c = a2 + b2 − ab, ∂c da ∂c db 1 da 1 2 dc = + = (a2 + b2 − ab)−1/2 (2a − b) + a + b2 − ab dt ∂a dt ∂b dt 2 dt 2 −1/2 (2b − a) db dt db da db 1 da + (2b − a) , = 2 and = 1 when a = 5 and b = 10 =√ (2a − b) dt dt dt dt 2 a2 + b2 − ab √ 1 dc = √ [(0)(2) + (15)(1)] = 3/2 cm/s. The third side is increasing. so dt 2 75 29. V = (π/4)D2 h where D is the diameter and h is the height, both measured in inches, dV /dt = (π/2)Dh(dD/dt) + (π/4)D2 (dh/dt), dD/dt = 3 and dh/dt = 24 when D = 30 and h = 240, so dV /dt = (π/2)(30)(240)(3) + (π/4)(30)2 (24) = 16,200π in3 /year. 30. ∂T dx ∂T dy y 2 dx dy dT = + = + 2y ln x , dx/dt = 1 and dy/dt = −4 at (3,2) so dt ∂x dt ∂y dt x dt dt dT /dt = (4/3)(1) + (4 ln 3)(−4) = 4/3 − 16 ln 3◦ C/s. 31. (a) xy -plane, fx = 12x2 y + 6xy, fy = 4x3 + 3x2 , fxy = fyx = 12x2 + 6x (b) y = 0, fx = 3x2 /y, fy = −x3 /y 2 , fxy = fyx = −3x2 /y 2 32. (a) x2 + y 2 > 1, (the exterior of the circle of radius 1 about the origin); fx = x/ x2 + y 2 − 1, fy = y/ x2 + y 2 − 1, fxy = fyx = −xy x2 + y 2 − 1 −3/2 (b) xy -plane, fx = 2x cos(x2 + y 3 ), fy = 3y 2 cos(x2 + y 3 ), fxy = fyx = −6xy 2 sin x2 + y 3 33. (a) 4: fxxx , fxxy = fxyx = fyxx , fxyy = fyxy = fyyx , fyyy (b) 5: fxxxx , fxxxy = fxxyx = fxyxx = fyxxx , fxxyy = fxyxy = fxyyx = fyxyx = fyyxx = fyxxy , fxyyy = fyxyy = fyyxy = fyyyx , fyyyy 34. (a) Since ew has infinitely many continuous derivatives, as does xy 2 , by the Chain Rule the 2 function exy has infinitely many continuous derivatives, hence by Theorem 15.4.6, fxyx = fxxy ; since fxy = fyx if follows that fxyx = fyxx . 2 (b) fxyx = fxxy = fyxx = 2xy 5 exy + 4y 3 exy 2 35. (a) f (tx, ty ) = 3t2 x2 + t2 y 2 = t2 f (x, y ); n = 2 (b) f (tx, ty ) = t2 x2 + t2 y 2 = tf (x, y ); n = 1 (c) f (tx, ty ) = t x y − 2t3 y 3 = t3 f (x, y ); n = 3 32 (d) f (tx, ty ) = 5/ t2 x2 + 2t2 y 2 36. (a) If f (u, v ) = tn f (x, y ), then let t = 1 to get x 2 = t−4 f (x, y ); n = −4 ∂f ∂f du ∂f dv ∂f + = ntn−1 f (x, y ), x +y = ntn−1 f (x, y ); ∂u dt ∂v dt ∂u ∂v ∂f ∂f +y = nf (x, y ). ∂x ∂y 541 Chapter 15 (b) If f (x, y ) = 3x2 + y 2 then xfx + yfy = 6x2 + 2y 2 = 2f (x, y ); If f (x, y ) = x2 + y 2 then xfx + yfy = x2 / x2 + y 2 + y 2 / x2 + y 2 = x2 + y 2 = f (x, y ); If f (x, y ) = x2 y − 2y 3 then xfx + yfy = 3x2 y − 6y 3 = 3f (x, y ); 5(−2)2x 5(−2)4y 5 then xfx + yfy = x 2 +y 2 = −4f (x, y ) If f (x, y ) = 2 2 )2 2 )3 (x + 2y (x + 2y (x + 2y 2 )3 37. (a) (b) dz ∂u ∂z dz ∂u ∂z = , = ∂x du ∂x ∂y du ∂y ∂2z dz ∂ 2 u ∂ = + 2 2 ∂x du ∂x ∂x dz du dz ∂ 2 u ∂ ∂2z = + ∂y∂x du ∂y∂x ∂y ∂z ∂ 2 u ∂ ∂2z = + ∂y 2 ∂u ∂y 2 ∂y ∂z ∂u dz ∂ 2 u d2 z ∂u = +2 ∂x du ∂x2 du dz du ∂u ∂x 2 ; dz ∂ 2 u d2 z ∂u ∂u ∂u = +2 ∂x du ∂y∂x du ∂x ∂y dz ∂ 2 u d2 z ∂u = +2 ∂y du ∂y 2 du ∂u ∂z 2 38. (a) z = f (u), u = x2 − y 2 ; ∂z/∂x = (dz/du)(∂u/∂x) = 2xdz/du ∂z/∂y = (dz/du)(∂u/∂y ) = −2ydz/du, y∂z/∂x + x∂z/∂y = 2xydz/du − 2xydz/du = 0 dz ∂u dz ∂z dz ∂u dz ∂z = =y , = =x , ∂x du ∂x du ∂y du ∂y du ∂z ∂z dz dz x −y = xy − xy = 0. ∂x ∂y du du (b) z = f (u), u = xy ; (c) yzx + xzy = y (2x cos(x2 − y 2 )) − x(2y cos(x2 − y 2 )) = 0 (d) xzx − yzy = xyexy − yxexy = 0 ∂r ∂θ ∂r ∂θ + cos θ and 0 = r cos θ + sin θ ; solve for ∂r/∂x and ∂θ/∂x. ∂x ∂x ∂x ∂x ∂r ∂θ ∂r ∂θ + cos θ and 1 = r cos θ + sin θ ; solve for ∂r/∂y and ∂θ/∂y . (b) 0 = −r sin θ ∂y ∂y ∂y ∂y ∂z ∂z ∂r ∂z ∂θ ∂z 1 ∂z (c) = + = cos θ − sin θ. ∂x ∂r ∂x ∂θ ∂x ∂r r ∂θ ∂z ∂r ∂z ∂θ ∂z 1 ∂z ∂z = + = sin...
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This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.

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