# Find the parametric equations for the line l passing

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Find theparametric equationsfor the lineLpassing through thepointsAandC.Solution:A vector parallel to the lineLis:v=-→AC=h1-1,2-0,3-0,i=h0,2,3i.A point on the line isA= (1,0,0).Thereforeparametric equationsfor the lineLare:x= 1y= 2tz= 3t.
Problem 1(b) - Fall 2008Consider the pointsA= (1,0,0),B= (2,1,0)andC= (1,2,3).Find an equation of the plane inR3which contains the pointsA,B,C.Solution:Since a plane is determined by its normal vectornand a point onit, say the pointA, it suffices to findn. Note that:n=-→AB×-→AC=ijk110023=h3,-3,2i.So theequation of the planeis:h3,-3,2i · hx-1,y-0,z-0i= 3(x-1)-3y+ 2z= 0.
Problem 1(c) - Fall 2008Consider the pointsA= (1,0,0),B= (2,1,0)andC= (1,2,3).Find the area of the triangleΔwith verticesA,BandC.Solution:Consider the pointsA= (1,0,0),B= (2,1,0) andC= (1,2,3).Then the area of the triangleΔwith these vertices can be foundby taking the area of the parallelogram spanned by-→ABand-→ACand dividing by 2. Thus:Area(Δ) =|-→AB×-→AC|2=12ijk110023=12|h3,-3,2i|=129 + 9 + 4 =1222.
Problem 2 - Fall 2008Find the volume under the graph off(x,y) =x+ 2xyand overthe bounded region in the first quadrant{(x,y)|x0,y0}bounded by the curvey= 1-x2and thexandy-axes.Solution:Z10Z1-x20(x+ 2xy)dy dx=Z10(xy+xy2)1-x20dx=Z10x(1-x2) +x(1-x2)2dx=Z10x5-3x3+ 2x dx= (x66-3x44+x2)10=16-34+ 1 =512.
Problem 3 - Fall 2008LetI=Z10Z22xsin(y2)dy dx.1Sketch the region of integration.2Write the integralIwith the order of integration reversed.3Evaluate the integralI.Show your work.Solution:1See the blackboard for a sketch.2I=Z10Z12y0sin(y2)dx dy.3By Fubini’s Theorem,Z10Z12y0sin(y2)dx dy=Z10sin(y2)x12y0dy=Z10(sin(y2)·12y)dy=14Z10sin(y2)·2y dy=cos(y2)410=14(cos 1-1).
Problem 4(a, b, c) - Fall 2008Consider the functionF(x,y,z) =x2+xy2+z.1What is the gradientF(x,y,z) ofFat the point (1,2,-1)?2Calculate the directional derivative ofFat the point (1,2,-1)in the directionh1,1,1i?3What is the maximal rate of change ofFat the point(1,2,-1)?Solution:f=h2x+y2,2xy,1i.So,f(1,2,-1) =h6,4,1i.The unit vectoruin the directionh1,1,1iisu=h1,1,1i3.Duf(1,2,-1) =f(1,2,-1)·u=h6,4,1i ·13h1,1,1i=113.Themaximum rate of changeis the length of thegradient:MRC(f) =|f(1,2,-1)|=|h6,4,1i|=53..
Problem 4(d) - Fall 2008Consider the functionF(x,y,z) =x2+xy2+z. Find the equationof the tangent plane to the level surfaceF(x,y,z) = 4 at the point(1,2,-1).Solution:Recall that the gradient ofF(x,y,z) =x2+xy2+zis normalnto the surface.Calculating, we obtain:F(x,y,z) =h2x,2xy,1in=F(1,2,-1) =h6,4,1i.The equation of thetangent planeis:h6,4,1i·hx-1,y-2,z+1i= 6(x-1)+4(y-2)+(z+1) = 0.
Problem 5 - Fall 2008Find the volumeVof the solid under the surfacez= 1-x2-y2and above thexy-plane.
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