ha (lvh262) – Homework 14.5 – karakurt – (56295) 1This print-out should have 15 questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.001 10.0 pointsDeterminedzdtwhenz = x ln(x + 9y)andx = sin t , y = cos t .1.dzdt=ln(x + 9y) sin t − 9x cos tx + 9y2.dzdt= ln(x + 9y) cos t −9x sin tx + 9y3.dzdt= ln(x + 9y) cos t +x(sin t − cos t)x + 9y4.dzdt= ln(x + 9y) sin t +x sin t − 9x cos tx + 9y5.dzdt= ln(x + 9y) cos t +x cos t − 9x sin tx + 9ycorrectExplanation:By the Chain Rule for Partial Differentia-tion,dzdt=∂z∂xdxdt+∂z∂ydydt.Here, we have that∂z∂x=xx + 9y+ ln(x + 9y) ,dxdt= cos tand∂z∂y=9xx + 9y,dydt= −sin t .It follows thatdzdt= ln(x + 9y) cos t +x cos t − 9x sin tx + 9y.keywords:002 10.0 pointsUse the Chain Rule to finddwdtwhenw = xey/zandx = t2, y = 1 − t , z = 1 + 2t .1.dwdt=2t −xz−2xyzey/z2.dwdt=t +xz+xyzey/z3.dwdt=t −xz−xyzey/z4.dwdt=t +xz+xyz2ey/z5.dwdt=2t +xz+2xyz2ey/z6.dwdt=2t −xz−2xyz2ey/zcorrectExplanation:By the Chain Rule for Partial Differentia-tion,dwdt=∂w∂xdxdt+∂w∂ydydt+∂w∂zdzdt.Whenw = xey/zandx = t2, y = 1 − t , z = 1 + 2t ,therefore,dwdt= 2tey/z−xzey/z−2xyz2ey/z.Consequently,dwdt=2t −xz−2xyz2ey/z.003 10.0 pointsha (lvh262) – Homework 14.5 – karakurt – (56295) 2Use the Chain Rule to find∂z∂swhenz = x2+ xy + y2,andx = 3s + t , y = st .1.∂z∂s= 6x + y + xt + 2yt2.∂z∂s= 2x + y + xt + 2yt3.∂z∂s= 2x + y + xs + 2ys4.∂z∂s= 6x + 3y + xs + 2ys5.∂z∂s= 6x + 3y + xt + 2yt correct6.∂z∂s= 2x + 3y + xs + 2ysExplanation:By the Chain Rule for Partial Differentia-tion,∂z∂s=∂z∂x∂x∂s+∂z∂y∂y∂s.Now∂z∂x= 2x + y,∂x∂s= 3while∂z∂y= +x + 2y,∂y∂s= t .Thus∂z∂s= 3(2x + y) + t(+x + 2 y) .Consequently,∂z∂s= 6x + 3y + xt + 2yt.004 10.0 pointsUse the Chain Rule to find∂z∂twhenz =xy,andx = 2set, y = 4 + se−t.1.∂z∂t=2sye2t− xsyet2.∂z∂t=syet− xsyet3.∂z∂t=2syet+ xsy2et4.∂z∂t=syet+ xsy2et5.∂z∂t=sye2t− xsyet6.∂z∂t=2sye2t+ xsy2etcorrectExplanation:By the Chain Rule for Partial Differentia-tion,∂z∂t=∂z∂x∂x∂t+∂z∂y∂y∂t.But∂z∂x=1y,∂x∂t= 2set,while∂z∂y= −xy2,∂y∂t= −se−t.Consequently,∂z∂t=2sye2t+ xsy2et.keywords:005 10.0 pointsUse the Chain Rule to find∂z∂uwhenz = ercos θha (lvh262) – Homework 14.5 – karakurt – (56295) 3andr = 4uv , θ =pu2+ v2.1.∂z∂u= uer4v cos θ +sin θ√u2+ v22.∂z∂u= er4v cos θ +sin θ√u2+ v23.∂z∂u= er4v cos θ −u sin θ√u2+ v2correct4.∂z∂u= er4v cos θ +u sin θ2√u2+ v25.∂z∂u= er4v cos θ −sin θ2√u2+ v2Explanation:By the Chain Rule for Partial Differentia-tion,∂z∂u=∂z∂r∂r∂u+∂z∂θ∂θ∂u.But∂z∂r= ercos θ,∂r∂u= 4vwhile∂z∂θ= −ersin θ,∂θ∂u=u√u2+ v2.Thus∂z∂u= er4v cos θ −u sin θ√u2+ v2.keywords: partial differentiation, Chain Rule,006 10.0 pointsIf z = f(x, y) andfx(4, 5) = 4, fy(4, 5) = −6 ,finddzdtat t = 2 when x = g(t), y = h(t) andg(2) = 4 , g′(2) = 2 .h(2) = 5 , h′(2) = 3 .1.dzdt= −82.dzdt= −123.dzdt= −44.dzdt= −10 correct5.dzdt= −6Explanation:By the Chain Rule for Partial Differentia-tion,dzdt= fx(g(t), h(t))g′(t)+ fy(g(t), h(t))h′(t) .When t = 2, therefore,dzdtt=2= fx(g(2), h(2))g′(2)+ fy(g(2), h(2))h′(2)= fx(4, 5)g′(2) + fy(4, 5)h′(2) .Consequently, given the values above,dzdt= (4)(2) − ( 6)(3) = −10.keywords:007 10.0 pointsUse the Chain Rule to find the parti alderivative∂w∂sforw = x2+ y2+ z2, x = st,y = s cos t, z = s sin twhen s = 5, t = 0.1.∂w∂s= 12ha (lvh262) – Homework 14.5 – karakurt – (56295) 42.∂w∂s= 113.∂w∂s= 134.∂w∂s= 145.∂w∂s= 10 correctExplanation:By the Chain Rule for Partial Differentia-tion∂w∂s=∂w∂x∂x∂s+∂w∂y∂y∂s+∂w∂z∂z∂s.Here, we have∂w∂x= 2x,∂x∂s= twhile∂w∂y= 2y,∂y∂s= cos tand∂w∂z= 2z,∂z∂s= sin t .Thus∂w∂s= 2xt + 2y cos t + 2z sin t .Note that when s = 5 and t = 0, it followsthat x = 0, y = 5, z = 0 . Consequently, forthese values,∂w∂s= 10.keywords:008 10.0 pointsUse the Chain Rule to find∂u∂pforu =x + yy + zwhenx = p + 7r + 8t, y = p − 7r + 8t ,andz = p + 7r − 8 t .1.∂u∂p=8tp22.∂u∂p= −8p23.∂u∂p= −8tp2correct4.∂u∂p=8p25.∂u∂p= −8t2p3Explanation:By the Chain Rule,∂u∂p=∂u∂x∂x∂p+∂u∂y∂y∂p+∂u∂z∂z∂p.But∂u∂x=1y + z,∂x∂p= 1while∂u∂y=z − x(y + z)2,∂y∂p= 1and∂u∂z=−x − y(y + z)2,∂z∂p= 1 .Consequently,∂u∂p=1y + z+z − x(y + z)2+−x − y(y + z)2.=2(z − x)(y + z)2=−8tp2.keywords:009 10.0 pointsha (lvh262) – Homework 14.5 – karakurt – (56295) 5Use partial differentiation and the ChainRule applied to F (x, y) = 0 to determinedy/dx whenF (x, y) = cos(x − 2y) − xe4y= 0 .1.dydx=sin(x − 2y) + e4y2xe4y− 4 sin(x − 2y)2.dydx=sin(x − 2y) + e4y4xe4y− 2 sin(x − 2y)3.dydx=sin(x − 2y) − 4e4y4 sin(x − 2y) − 2xe4y4.dydx=sin(x − 2y) + 4xe4y2 sin(x − 2y) − e4y5.dydx=sin(x − 2y) + e4y2 sin(x − 2y) − 4xe4ycorrect6.dydx=sin(x − 2y) − 4xe4y4 sin(x − 2y) − 2e4yExplanation:Applying the Chain Rule to both sides ofthe equation F (x, y) = 0, we see that∂F∂xdxdx+∂F∂ydydx=∂F∂x+∂F∂ydydx= 0 .Thusdydx= −∂F∂x∂F∂y= −FxFy.WhenF (x, y) = cos(x − 2y) − xe4y= 0 ,therefore,dydx= −−sin(x − 2y) − e4y2 sin(x − 2y) − 4xe4y.Consequently,dydx=sin(x − 2y) + e4y2 sin(x − 2y) − 4xe4y.010 10.0 pointsFind∂z∂xwhenxe4y− yz + ze2x= 0 .1.∂z∂x= −e4y+ 2ze2xe2x− ycorrect2.∂z∂x=e2x− ye4y+ 2ze2x3.∂z∂x= −e4y+ 2ze2xe2x+ y4.∂z∂x=e2x+ ye4y+ 2ze2x5.∂z∂x=e4y+ 2ze2xe2x− y6.∂z∂x= −e2x− ye4y+ 2ze2xExplanation:By Implici t Differentiation with respect tox,e4y− y∂z∂x+ 2ze2x+ e2x∂z∂x= 0 .Solving for∂z∂xwe thus get∂z∂xe2x− y+ e4y+ 2ze2x= 0 .Consequently,∂z∂x= −e4y+ 2ze2xe2x− y.keywords: Chain Rule, partial differentiation,implicit differentiation, exponential function,011 10.0 pointsha (lvh262) – Homework 14.5 – karakurt – (56295) 6Use the equation∂z∂y= −∂F∂y∂F∂zto find∂z∂yforxe8y+ yz + ze8x= 0 .1.∂z∂y= −e8x+ yz + 8xe8y2.∂z∂y= −8xe8y+ zy + 8e8x3.∂z∂y= −8e8y+ yz + 8e8x4.∂z∂y= −8e8y+ yz + e8x5.∂z∂y= −8xe8y+ zy + e8xcorrectExplanation:012 10.0