MIDTERM 2-solutions - Version 018 MIDTERM 2A neitzke(55325...

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Version 018 – MIDTERM 2A – neitzke – (55325)1Thisprint-outshouldhave14questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.00110.0pointsFind an equation in parametric form for thetangent line to the graph ofr(s) =(s5, s6, s3)at the point (1,1,1).1.x(t) = 15t, y(t) = 1+6t, z(t) = 13t2.x(t) = 5t1, y(t) = 6t1, z(t) = 3t13.x(t) = 5t4, y(t) = 6t5, z(t) = 3t24.x(t) = 4t1, y(t) = 5t1,2t15.x(t) = 1 +5t, y(t) = 1 +6t, z(t) = 1 +3tcorrect6.x(t) = 1+4t, y(t) = 1+5t, z(t) = 1+2tExplanation:The graph ofr(s) has tangent vectorr(s) =(5s4,6s5,3s2).On the other hand, (1,1,1) =r(1).Thus,at (1,1,1) the equation of the tangent line invector form isL(t) =r(1) +tr(1)=(1 + 5t,1 + 6t,1 + 3t),which in parametric form becomesx(t) = 1 + 5t, y(t) = 1 + 6t, z(t) = 1 + 3t.keywords: tangent line, tangent vector, vectorfunction, derivative, vector form, parametricform,0021.fx>0,fy>02.fx<0,fy<0correct3.fx= 0,fy>04.fx= 0,fy<05.fx<0,fy>06.fx>0,fy<0Explanation:When we walk in thex-direction fromPweare walking downhill, sofx<0. On the otherhand, when we walk in they-direction fromPwe are again walking downhill, sofy<0 also.Consequently, atPfx<0,fy<0.keywords:contour map,contours,partialderivative, slope,00310.0pointsWhenCis parametrized byc(t) = (sin 3t)i+ 4tj+ (cos 3t)k,
10.0pointsFrom the contour map offshown below de-cide whetherfxandfyare positive, negative,or zero atP.00-2-2-4-4-6-6Pxy
Version 018 – MIDTERM 2A – neitzke – (55325)2find its arc length betweenc(0) andc(4).1.arc length = 162.arc length = 123.arc length = 84.arc length = 245.arc length = 20correctExplanation:The length of the curve betweenc(t0) andc(t1) is given by the integralL=integraldisplayt1t0bardblc(t)bardbldt .Now whenc(t) = (sin 3t)i+ 4tj+ (cos 3t)k,we see thatc(t) = (3 cos 3t)i+ 4j(3 sin 3t)bk .But then by the Pythagorean identity,bardblc(t)bardbl= (9 + 16)1/2= 5.ThusL=integraldisplay405dt=bracketleftBig5tbracketrightBig40.Consequently,arc length =L= 20.00410.0pointsDetermine∂z∂ywhenz= 4ex/3y.1.∂z∂y=4x3y2ex/3ycorrect2.∂z∂y=4x3y2ex/3y3.∂z∂y=43y2ex/3y4.∂z∂y=4x3yex/3y5.∂z∂y=43y2ex/3y6.∂z∂y=4x3yex/3yExplanation:Differentiatingzwith respect toykeepingxfixed we see that∂z∂y= 4ex/3y·(x/3y)∂y.Consequently,∂z∂y=4x3y2e

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