lewis (scl876) – HW 14.5 – radin – (55315) 1This print-out should have 8 questions.Multiple-choice questions may continue o nthe next column or page – find all choicesbefore answering.001 10.0 pointsUse the Chain Rule to finddwdtwhenw = xey/zandx = t2, y = 4 − t , z = 4 + t .1.dwdt=t −xz−4xyzey/z2.dwdt=t +xz+4xyzey/z3.dwdt=2t +xz+xyz2ey/z4.dwdt=2t −xz−xyzey/z5.dwdt=t +xz+4xyz2ey/z6.dwdt=2t −xz−xyz2ey/zcorrectExplanation:By the Chain Rule for Partial Differentia-tion,dwdt=∂w∂xdxdt+∂w∂ydydt+∂w∂zdzdt.Whenw = xey/zandx = t2, y = 4 − t , z = 4 + t ,therefore,dwdt= 2tey/z−xzey/z−xyz2ey/z.Consequently,dwdt=2t −xz−xyz2ey/z.002 10.0 pointsUse the Chain Rule to find∂z∂twhenz =xy,andx = 5set, y = 3 + se−t.1.∂z∂t=syet− xsyet2.∂z∂t=syet+ xsy2et3.∂z∂t=5sye2t+ xsy2etcorrect4.∂z∂t=5sye2t− xsyet5.∂z∂t=5syet+ xsy2et6.∂z∂t=sye2t− xsyetExplanation:By the Chain Rule for Partial Differentia-tion,∂z∂t=∂z∂x∂x∂t+∂z∂y∂y∂t.But∂z∂x=1y,∂x∂t= 5set,while∂z∂y= −xy2,∂y∂t= −se−t.Consequently,∂z∂t=5sye2t+ xsy2et.lewis (scl876) – HW 14.5 – radin – (55315) 2keywords:003 10.0 pointsUse the Chain Rule to find∂u∂pforu =x + yy + zwhenx = p + 5r + 3t, y = p − 5r + 3t ,andz = p + 5r − 3t .1.∂u∂p= −3tp2correct2.∂u∂p=3p23.∂u∂p=3t2p34.∂u∂p=3tp25.∂u∂p= −3p2Explanation: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=−3tp2.keywords:004 10.0 pointsUse the equation∂z∂y= −∂F∂y∂F∂zto find∂z∂yforxe8y+ yz + ze5x= 0 .1.∂z∂y= −e5x+ yz + 8xe8y2.∂z∂y= −8e8y+ yz + e5x3.∂z∂y= −8xe8y+ zy + 5e5x4.∂z∂y= −8xe8y+ zy + e5xcorrect5.∂z∂y= −8e8y+ yz + 5e5xExplanation:005 10.0 pointsThe temperature at a point (x, y) in theplane is T (x, y)◦C. If a bug crawls on theplane so that its position in the plane after tminutes is given by (x(t), y(t)) wherex =√12 + t , y = 5 +12t ,determine how fast the temperature is risingon the bug’s path at t = 4 whenTx(4, 7) = 40 , Ty(4, 7) = 4 .1. rate = 8◦C/min2. rate = 11◦C/min3. rate = 9◦C/minlewis (scl876) – HW 14.5 – radin – (55315) 34. rate = 7◦C/min correct5. rate = 10◦C/minExplanation:By the Chain Rule for partial differentia-tion, the rate of change of temperature on thebug’s path at (x(t), y(t)) is gi ven bydTdt=dT (x(t), y(t))dt=∂T∂xdxdt+∂T∂ydydt.Nowdxdt=12√12 + t,dydt=12.But when t = 4, the bug is at the point (4, 7),whiledxdtt=4=12√12 + tt = 4=18,dydtt=4=12,and we are told thatTx(4, 7) = 40 , Ty(4, 7) = 4 .Consequently, after 4 minutes the tempera-ture on the bug’s path is changing at arate = 40 ×18+ 4 ×12= 7◦C/min.006 10.0 pointsThe radius of a right circular cylinder isincreasing at a rate of 5 inches per minutewhile the height is decreasing at a rat e of9 inches per minute. Determine the rate ofchange of the volume when r = 5 and h = 6.1. rate = 67 π cu.in./min.2. rate = 75 π cu.in./min. correct3. rate = 79 π cu.in./min.4. rate = 63 π cu.in./min.5. rate = 71 π cu.in./min.Explanation:The volume cylinder of a cylinder of heighth and radius r is given byV (r, h) = πr2h .When h and r are changing with t, then bythe Chain Rule the rate of change of V withrespect to t is given bydVdt=∂V∂rdrdt+∂V∂hdhdt= 2πrhdrdt+ πr2dhdt.Butdrdt= 5,dhdt= −9 ,in which casedVdt= 10 πrh − 9πr2.Consequently, when r = 5 and h = 6,dVdlt(r=5, h=6)= 75 π cu.in./min..007 10.0 pointsThe length ℓ and width w of the closed boxshown inhℓware increasing at a rate of 4 ft/mi n while itsheight h is decreasing at a rate of 8 ft/min.Find the rat e at which the volume of the boxis increasing whenℓ = 6 , w = h = 3 feet .lewis (scl876) – HW 14.5 – radin – (55315) 41. rate = 38 cu ft/min2. rate = 39 cu ft/min3. rate = 36 cu ft/min4. rate = 37 cu ft/min5. volume is decreasing, not increasing cor-rectExplanation:The box has volumeV = wℓh .Thus by t he Chain Rule for partial differen-tiation, the rate at which the surface a rea ischanging is given by∂V∂t=∂V∂ℓ∂ℓ∂t+∂V∂w∂w∂t+∂V∂h∂h∂t.Now∂V∂ℓ= wh ,∂V∂w= ℓh ,∂V∂h= wℓ ,and so we see that∂V∂t= wh∂ℓ∂t+ ℓ h∂w∂t+ wℓ∂h∂t.Whenℓ = 6 , w = h = 3 ,and∂ℓ∂t=∂w∂t= 4 ,∂h∂t= −8 ,therefore,∂V∂t= 3 · 3 · 4+ (3 · 6) · 4 − (3 · 6) · 8 = −36 < 0 .Consequently, because this is negative thevolume is decreasing, not increasing .008 10.0 pointsUse partial differentiation and the ChainRule applied to F (x, y) = 0 t o determinedy/dx whenF (x, y) = cos(x − 6y) − xe3y= 0 .1.dydx=sin(x − 6y) + e3y6xe3y− 3 sin(x − 6y)2.dydx=sin(x − 6y) − 3e3y3 sin(x − 6y) − 6xe3y3.dydx=sin(x − 6y) + e3y3xe3y− 6 sin(x − 6y)4.dydx=sin(x − 6y) + 3xe3y6 sin(x − 6y) − e3y5.dydx=sin(x − 6y) − 3xe3y3 sin(x − 6y) − 6e3y6.dydx=sin(x − 6y) + e3y6 sin(x − 6y) − 3xe3ycorrectExplanation:Applying the Chain Rule t o 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 − 6y) − xe3y= 0 ,therefore,dydx= −−sin(x − 6y) − e3y6 sin(x − 6y) − 3xe3y.Consequentl y,dydx=sin(x − 6y) + e3y6 sin(x − 6y) −