CALCULUS III SAMPLE FINALProblem 1:(20 points) Find parametric equations for the line of intersection of the planesz=x+y,2x-5y-z= 1.Solution:The normal vectors to the two planes aren1=1,1,-1andn2=2,-5,-1respectively. The direction vector of the intersecting line isv=n1×n2=1,1,-1×2,-5,-1=-6,-1,-7To find a point on the line of intersection, try to lety= 0, then we get a systemz=x,2x-z= 1, and by solving it we getx=z= 1, so we get a point (1,0,1) on the line ofintersection. The parametric equation of the line isx, y, z=1,0,1 +t-6,-1,-7=1-6t,-t,1-7tProblem 2:(20 points) Find the distance between skew linesx=y=zand-x-1 =y/2 =z/3.Solution:The pointP= (0,0,0) is on the first line, and the pointQ= (-1,0,0) is on thesecond line. The direction vectors of these two lines are1,1,1and-1,2,3respectively.The vectorn=1,1,1× -1,2,3=1,-4,3is orthogonal to both lines. The distanced=compnPQ=|PQ·n||n|=|-1,0,0·1,-4,3|√12+ 42+ 33=1√26Problem 3.(20 points)(a) Find (-2√3 + 2i)5using De Moivre’s Theorem:(b) Find the cube roots of-1-iand sketch the roots in the complex plane:Solution:(a)(-2√3 + 2i)5=4ei5π/65= 45ei25π6= 45eiπ/6(b) Cube roots of -1-i: write-1-i=√2ei5π/4, then the three cube roots are6√2ei5π/12,6√2ei(5π/12+2π/3)=6√2ei13π/12, and6√2ei(5π/12+4π/3)=6√2ei21π/121
2CALCULUS III SAMPLE FINALProblem 4.(20 points)Solve the initial-value problem2y+ 5y+ 3y= 0,y(0) = 3,y(0) =-4Solution:The auxiliary equation is 2r2+ 5r+ 3 = 0 and it has too roots-1,-3/2. The generalsolution to the differential equation isc1e-x+c2e-3x/2. Plug in the initial conditions,c1e-0+c2e-0= 3,-c1e-0-3c2e-0/2 =-4.Solve this system we getc1= 1,c2= 2. So the solution to the IVP ise-x+ 2e-3x/2.Problem 5.(20 points) Find the tangential and normal components of the accelerationvector:r(t) =t i+ cos2t j+ sin2t kat timet=π/8.