Transcription of Problem 1. - UCSD Mathematics
1 Problem the pointsP(1,1, 2),Q(2,0,1) andR(1, 1,0).(i) Find the area of the calculate~PQ= (1, 1,3),~RQ= (1,1,1).The area of the parallelogram spanned by~PQand~QRis the magnitude of the cross product(2, 1,3) (1,1,1) = ( 4,2,2).This vector has magnitude 2 6, so the trianglePQRhas area 6.(ii) Find the equation of the plane throughP, plane throughP, QandRhas as normal vector the cross product. The entries ofthe cross product are used as coefficients for the plane. We obtain the equation 4x+ 2y+ 2z= 6 2x+y+z= 3using the pointP(orQorR) to find the right hand 2.(i) Does there exist a constant such that the functionf(x,y,z) ={x4yx2+y2+z2if (x,y,z)6= (0,0,0)aif (x,y,z) = (0,0,0)is continuous?We must havea=lim(x,y,z) (0,0,0)x4yx2+y2+ that0 x4yx2+y2+z2 =x2x2+y2+z2 |x2y| |x2y| 0hencea= 0.(ii) Does the limitlim(x,y) (0,1)x2(y 1)2x4+ (y 1)4exist?We find the limit along the liney 1 =mx, keepingmfixed.}
2 Then, we obtainlimy 1=mxx 0x2(y 1)2x4+ (y 1)4= limx 0x2 (mx)2x4+ (mx)4=m2m4+ this depends onm, the original limit does not ~x,~y, and~zbe vectors whose magnitudes are 1, 2, and 1 respectively. Suppose that~xisparallel to (and in the same direction as)~y, and~xis perpendicular to~z. Find the angle betweenthe vectors~x+~zand~y+ 3~ have~x ~x= 1, ~y ~y= 4, ~z ~z= ~x ~y= 1 2 cos 0 = 2, ~x ~z= 0, ~y ~z= last two products are justified since the vectors involved are compute(~x+~z) (~y+ 3~z) =~x ~y+~z ~y+ 3~x ~z+ 3~z ~z= 2 + 0 + 0 + 3 = have||~x+~z||2= (~x+~z) (~x+~z) =~x ~x+ 2~x ~z+~z ~z= 1 + 0 + 1 = 2 = ||~x+~z||= 2||~y+ 3~z||2= (~y+ 3~z) (~y+ 3~z) =~y ~y+ 6~y ~z+ 9~z ~z= 4 + 0 + 9 = 13 = ||~y+ 3~z||= =6 2 13=5 26= = cos 15 line`is perpendicular to the planex+ 2y 3z= 2 and passes through the point (1,0, 1).Where does the line intersect the planex 2y+z= 4?
3 The line is parametrized by(x,y,z) = (1,0, 1) +t(1,2, 3).Thenx= 1 +t, y= 2t z= 1 3tand sincex 2y+z= 1we must have(1 +t) 4t+ ( 1 3t) = 4 = 6t= 4 = t= 2 gives (x,y,z) = (1,0, 1) +23(1,2, 3) = (53,43, 3). Problem the functionf(x,y) = 1 19(x 1)2 14y2.(i) Sketch the level diagram forfshowing at least three level level curves aref(x,y) =c 1 19(x 1)2 14y2=c 19(x 1)2+14y2= 1 level curves are ellipses centered at (1,0). We can pick the levelsc= 0,c= 3 andc= 8, for instance. We obtain the three ellipses19(x 1)2+14y2= 1,19(x 1)2+14y2= 4 and19(x 1)2+14y2= 9.(ii) Sketch the graph graph offwill be a paraboloid whosez-cross sections are ellipses. The highest pointon the paraboloid is (1,0,1). The paraboloid is concave down . Problem trajectory of a particle is given by the parametric curvex= cost+ 2 sint, y= 2 cost sint, z= 2t,0 t .(i) Find the speed and velocity of the velocity equals~r (t) = ( sint+ 2 cost, 2 sint cost,2),while the speed is||~r (t)||= ( sint+ 2 cost)2+ ( 2 sint cost)2+ 4= (sin2t+ 4 cos2t 4 costsint) + (4 sin2t+ cos2t+ 4 sintcost) + 4= 5(sin2t+ cos2t) + 4 = 9 = 3.
4 (ii) Calculate the tangent line to the trajectory att= calculate~r ( 2) = ( 1, 2,2).Alsor( /2) = (2, 1, ). Thus the tangent line passes through (2, 1, ) and is parallel to( 1, 2,2). The parametric equation is(x,y,z) = (2, 1, ) +s( 1, 2,2).(iii) Calculate the arclength parametrization of the solve for the arclength functions(t) = t03du= 3tands(t) =s= t=s/3. This gives the parametrization~R(s) =~r(s/3) = (coss3+ 2 sins3,2 coss3 sins3,2s3).(iv) Show that the trajectory stays on a cylinder whose central axis is thez-axis. Draw thetrajectory of the calculatex2+y2= (cost+ 2 sint)2+ (2 cost sint)2= 5(sin2t+ cos2t) = is the equation of a cylinder of radius 5 with central axis thez-axis. The trajectoryis a helix wrapping around the cylinder.
