# Calc 3 Final.pdf - PRACTICE PROBLEMS FOR THE FINAL EXAM(1(a...

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PRACTICE PROBLEMS FOR THE FINAL EXAM(1)(a) Find a vector perpendicular to the plane through (0,2,0), (2,3,-1) and (1,2,-4).(b) Give an equation for the plane in part (a).Solution:(a) First, we will find 2 vectors in the plane. Label the points asA(0,2,0),B(2,3,-1),andC(1,2,-4).Then--→AB=h2,1,-1iand-→AC=h1,0,-4i.To find a normalvector, i.e. a vector perpendicular, to the plane containing these 2 vectors, we mustcompute--→AB×-→AC. We find that--→AB×-→AC=h-4,7,-1i.(b) For the equation of the plane in part (a), we already found a normal vector. Thus,the plane is given by the equation-4x+ 7(y-2)-z= 0 or-4x+ 7y-z= 14.(2) Find a parametric equation of the line of intersection of the planesP1defined by 2x+3y-z= 0 andP2defined byx+ 3y+z= 1.Solution: There are two primary ways of finding the line of intersection. For the first method, labelthe equation ofP1ase1, 2x+ 3y-z= 0, and the equation ofP2ase2,x+ 3y+z= 1.Then takinge1+e2, we obtain the equation 3x+ 6y= 1. Lettingy=t, we get thatx=-2t+13. Finally, plugging intoe2, we get (-2t+13) + 3(t) +z= 1 orz=-t+23.Summarizing, the parametric equations for the line of intersection arex=-2t+13,y=t, andz=-t+23, fortR.Alternately, we note that ifn1=h2,3,-1iandn2=h1,3,1iare the normal vectors ofP1andP2, respectively, then their cross product,n1×n2will be parallel to both planes (asit is perpendicular to both normal vectors). Thus,n1×n2h6,-3,3iis in the directionof the line of intersection. Next, we need a point on both planes and hence, on the line.If we lety= 0, then thexandzcoordinate of the point should satisfy both 2x-z= 0andx+z= 1. Thus, the point is (13,0,23). Combining all of this information, we obtainthe parametric equations for the line arex= 6t+13,y=-3t, andz= 3t+23, fortR.(3) Find the arclength of the curve-→r(t) =etcost-→i+etsint-→jwitht[0,2π].Solution: To find the arc length of a curve parametrized by-→r(t) foratb, we com-puteRba|-→r0(t)|dt.Thus, we find that-→r0(t) =etcost-etsint, etsint+etcostand|-→r0(t)|=p(etcost-etsint)2+ (etsint+etcost)2=2et. Finally, the length of thecurve isRba|-→r0(t)|dt=R2π02etdt=2(e2π-1)1
(4) The position function for the motion of a particle is given by-→r(t) =ht,2t, t2i.(a) Find the velocity and acceleration vectors.(b) Find the tangential componentaTof acceleration.(c) Find the curvature whent= 0.Solution:(a) We see that-→v(t) =-→r0(t) =h1,2,2tiand-→a(t) =-→r00(t) =h0,0,2i.(b) There are two primary methods to computeaT. First, by definition,aT=ddt(|-→v(t)|),the derivative of speed. So,aT=ddt(5 + 4t2) =4t5+4t2. Alternately, we have thataT=-→v(t)·-→a(t)|v(t)|, soaT=h1,2,2ti·h0,0,2i5+4t2=4t5+4t2.
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