ha lvh262 Homework 14 5 karakurt 56295 This print out should have 15 questions Multiple choice questions may continue on the next column or page find all choices before answering 001 10 0 points dz Determine when dt 002 10 0 points dw when Use the Chain Rule to find dt w xey z and x t2 z x ln x 9y and x sin t y cos t 1 ln x 9y sin t 9x cos t dz dt x 9y 2 dz 9x sin t ln x 9y cos t dt x 9y 3 dz x sin t cos t ln x 9y cos t dt x 9y 4 dz x sin t 9x cos t ln x 9y sin t dt x 9y dz x cos t 9x sin t ln x 9y cos t dt x 9y correct 5 Explanation By the Chain Rule for Partial Differentiation dz z dx z dy dt x dt y dt Here we have that z x ln x 9y x x 9y dx cos t dt y 1 t dy sin t dt It follows that x cos t 9x sin t dz ln x 9y cos t dt x 9y keywords z 1 2t dw x 2xy y z 1 e 2t dt z z x xy y z dw e t 2 dt z z dw x xy y z 3 e t dt z z x xy dw t 2 ey z 4 dt z z x 2xy y z dw 2t 2 e 5 dt z z dw x 2xy y z 6 2t 2 e correct dt z z Explanation By the Chain Rule for Partial Differentiation w dx w dy w dz dw dt x dt y dt z dt When w xey z and x t2 y 1 t and 9x z y x 9y 1 z 1 2t therefore x 2xy dw 2tey z ey z 2 ey z dt z z Consequently dw x 2xy y z 2t 2 e dt z z 003 10 0 points ha lvh262 Homework 14 5 karakurt 56295 Use the Chain Rule to find z when s and x 2set z x2 xy y 2 and x 3s t y st y 4 se t 1 2sye2t xs z t yet 1 z 6x y xt 2yt s 2 z syet xs t yet 2 z 2x y xt 2yt s 3 z 2syet xs t y 2 et 3 z 2x y xs 2ys s 4 z 6x 3y xs 2ys s 5 6 z 6x 3y xt 2yt correct s z 2x 3y xs 2ys s Explanation By the Chain Rule for Partial Differentiation z z x z y s x s y s Now x z 2x y 3 x s while z x 2y y y t s Thus 2 z syet xs 4 t y 2 et 5 sye2t xs z t yet 6 2sye2t xs z correct t y 2 et Explanation By the Chain Rule for Partial Differentiation z z x z y t x t y t But z 1 x y x 2set t while z x 2 y y y se t t Consequently z 3 2x y t x 2y s z 2sye2t xs t y 2 et Consequently z 6x 3y xt 2yt s 004 10 0 points z when Use the Chain Rule to find t x z y keywords 005 10 0 points Use the Chain Rule to find z er cos z when u ha lvh262 Homework 14 5 karakurt 56295 and r 4uv p u2 v 2 z sin 1 uer 4v cos u u2 v 2 z sin 2 er 4v cos u u2 v 2 z u sin r 3 correct e 4v cos u u2 v 2 z u sin r 4 e 4v cos u 2 u2 v 2 z sin 5 er 4v cos u 2 u2 v 2 Explanation By the Chain Rule for Partial Differentiation z z r z u r u u But z er cos r 1 dz 8 dt 2 dz 12 dt 3 dz 4 dt 4 dz 10 correct dt 5 dz 6 dt 3 Explanation By the Chain Rule for Partial Differentiation dz fx g t h t g t dt fy g t h t h t When t 2 therefore dz dt r 4v u t 2 fx g 2 h 2 g 2 fy g 2 h 2 h 2 while z er sin u u u2 v 2 fx 4 5 g 2 fy 4 5 h 2 Consequently given the values above Thus dz 4 2 6 3 10 dt z u sin er 4v cos u u2 v 2 keywords keywords partial differentiation Chain Rule 007 006 10 0 points If z f x y and fx 4 5 4 find fy 4 5 6 dz at t 2 when x g t y h t and dt g 2 4 g 2 2 h 2 5 h 2 3 10 0 points Use the Chain Rule to find the partial w for derivative s w x2 y 2 z 2 x st y s cos t z s sin t when s 5 t 0 1 w 12 s ha lvh262 Homework 14 5 karakurt 56295 4 when w 2 11 s x p 7r 8t y p 7r 8t 3 w 13 s 4 w 14 s 1 5 w 10 correct s u 8t 2 p p 2 u 8 2 p p Explanation By the Chain Rule for Partial Differentiation 3 u 8t 2 correct p p w w x w y w z s x s y s z s 4 u 8 2 p p 5 u 8t2 3 p p and z p 7r 8t Here we have w 2x x while and x t s w 2y y y cos t s w 2z z z sin t s Thus Explanation By the Chain Rule u x u y u z u p x p y p z p But u 1 x y z x 1 p while w 2xt 2y cos t 2z sin t s Note that when s 5 and t 0 it follows that x 0 y 5 z 0 Consequently for these values w 10 s u z x y y z 2 y 1 p x y u z y z 2 z 1 p and Consequently u 1 z x x y 2 p y z y z y z 2 keywords 008 10 0 points Use the Chain Rule to find u x y y z u for p 8t 2 z x 2 2 y z p keywords 009 10 0 points ha lvh262 Homework 14 5 karakurt 56295 Use partial differentiation and the Chain Rule applied to F x y 0 to determine dy dx when 4y F x y cos x 2y xe 1 0 010 Find 10 0 points z when x xe4y yz ze2x 0 2y e4y dy sin x dx 2xe4y 4 sin x 2y 2y …
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