# UT M 408C - HW 14.5-solutions-1 (4 pages)

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## HW 14.5-solutions-1

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## HW 14.5-solutions-1

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Pages:
4
School:
University of Texas at Austin
Course:
M 408c - Differential and Integral Calculus
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lewis scl876 HW 14 5 radin 55315 This print out should have 8 questions Multiple choice questions may continue on the next column or page find all choices before answering 10 0 points dw Use the Chain Rule to find when dt 1 002 10 0 points Use the Chain Rule to find 001 z when t x y z and y z w xe x 5set y 3 se t and x t2 y 4 t z 4 t dw x 4xy y z 1 e t dt z z dw x 4xy y z 2 e t dt z z dw x xy 3 2t 2 ey z dt z z x xy y z dw e 2t 4 dt z z dw x 4xy 5 t 2 ey z dt z z dw x xy 6 2t 2 ey z correct dt z z Explanation By the Chain Rule for Partial Differentiation dw w dx w dy w dz dt x dt y dt z dt When z syet xs 1 t yet syet xs z 2 t y 2 et 3 5sye2t xs z correct t y 2 et 4 z 5sye2t xs t yet 5 5syet xs z t y 2 et 6 sye2t xs z t yet Explanation By the Chain Rule for Partial Differentiation z z x z y t x t y t But w xey z 1 z x y x 5set t while and x t2 y 4 t z 4 t therefore x xy dw 2tey z ey z 2 ey z dt z z Consequently x xy y z dw 2t 2 e dt z z z x 2 y y y se t t Consequently 5sye2t xs z t y 2 et lewis scl876 HW 14 5 radin 55315 keywords 003 10 0 points u Use the Chain Rule to find for p u 2 3t 2 z x 2 2 y z p keywords x y y z 004 10 0 points F when z z y F to find for y y z 0 Use the equation x p 5r 3t y p 5r 3t xe8y yz ze5x and z p 5r 3t 1 z e5x y y z 8xe8y 1 u 3t 2 correct p p 2 8e8y y z y z e5x 2 3 u 2 p p 3 z 8xe8y z y y 5e5x 3 3t2 u 3 p p 4 8xe8y z z correct y y e5x 4 u 3t 2 p p 5 3 u 2 p p 8e8y y z 5 y z 5e5x Explanation Explanation By the Chain Rule 005 u x u y u z u p x p y p z p But u 1 x y z x 1 p 10 0 points The temperature at a point x y in the plane is T x y C If a bug crawls on the plane so that its position in the plane after t minutes is given by x t y t where x 12 t while u z x y y z 2 y 1 p determine how fast the temperature is rising on the bug s path at t 4 when Tx 4 7 40 and x y u z y z 2 z 1 p Consequently 1 z x x y u 2 p y z y z y z 2 1 y 5 t 2 1 rate 8 C min 2 rate 11 C min 3 rate 9 C min Ty 4 7 4 lewis scl876 HW 14 5 radin 55315 Explanation The volume cylinder of a cylinder of height h and radius r is given by 4 rate 7 C min correct 5 rate 10 C min Explanation By the Chain Rule for partial differentiation the rate of change of temperature on the bug s path at x t y t is given by V r h r 2 h When h and r are changing with t then by the Chain Rule the rate of change of V with respect to t is given by dT dT x t y t dt dt dV V dr V dh dt r dt h dt T dx T dy x dt y dt 2 rh Now But dx 1 dt 2 12 t 1 dy dt 2 But when t 4 the bug is at the point 4 7 while dx dt t 4 1 2 12 t t 4 1 8 dy dt t 4 1 2 and we are told that Tx 4 7 40 Ty 4 7 4 Consequently after 4 minutes the temperature on the bug s path is changing at a rate 40 006 3 1 1 4 7 C min 8 2 dr 5 dt in which case dr dh r 2 dt dt dh 9 dt dV 10 rh 9 r 2 dt Consequently when r 5 and h 6 dV dlt r 5 h 6 007 75 cu in min 10 0 points The length and width w of the closed box shown in 10 0 points The radius of a right circular cylinder is increasing at a rate of 5 inches per minute while the height is decreasing at a rate of 9 inches per minute Determine the rate of change of the volume when r 5 and h 6 1 rate 67 cu in min h w 2 rate 75 cu in min correct 3 rate 79 cu in min 4 rate 63 cu in min 5 rate 71 cu in min are increasing at a rate of 4 ft min while its height h is decreasing at a rate of 8 ft min Find the rate at which the volume of the box is increasing when 6 w h 3 feet lewis scl876 HW 14 5 radin 55315 1 rate 38 cu ft min 008 3 rate 36 cu ft min F x y cos x 6y xe3y 0 4 rate 37 cu ft min 5 volume is decreasing not increasing correct 1 Explanation The box has volume 2 V w h 3 Thus by the Chain Rule for partial differentiation the rate at which the surface area is changing is given by 4 V V w V h V t t w t h t Now 5 6 V h w V w h and so we see that dy sin x 6y e3y dx 6xe3y 3 sin x 6y dy sin x 6y 3e3y dx 3 sin x 6y 6xe3y dy sin x 6y e3y dx 3xe3y 6 sin x 6y dy sin x 6y 3xe3y dx 6 sin x 6y e3y sin x 6y 3xe3y dy dx 3 sin x 6y 6e3y dy sin x 6y e3y correct dx 6 sin x 6y 3xe3y Explanation Applying the Chain Rule to both sides of the equation F x y 0 we see that V w h wh h w t t t t F dx F dy x dx y dx When 6 w h 3 Thus h 8 t therefore V 3 3 4 t 3 6 4 3 6 8 36 0 Consequently because this is …

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