# sol1 - Solutions to Test 1 Math 20C Oct 22,2008 Particle...

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Solutions to Test 1, Math 20C, Oct 22,2008 Particle path problem: (A) x = 3cos(2t), y=5sin(2t), z=4cos(2t) implies the derivatives are x' = -6sin(2t), y' = 10cos(2t), z' = -8sin(2t). Thus x'^2 + y'^2 + z'^2 = 100 sin(2t)^2 + 100 cos(2t)^2 = 100. The square root of this is 10, so we integrate 10 from t=0 to t=60 to get the arclength. The answer is 600 feet. (B) Choose t=Pi/4 in order to get the point x=0, y=5, z=0. The velocity vector for this value t=Pi/4 is (-6, 0, -8). The tangent line at (0,5,0) has parametric formula (0,5,0) + s*(-6, 0 , -8), where s is the parameter. Intersection problem: z=y-x^2 = 1-2x -3y. Solve for y to get y = (1/4)*(x-1)^2. Thus a parametric formula for the intersection curve C is (x, (1/4)*(x-1)^2, (1/4)*(x-1)^2 - x^2), where x is the parameter. There are other possible answers as well. Area of the triangle: The vector AB is (2,1,-1) and the vector AC is (0,1,1). The cross product of these two vectors is (2, -2, 2)

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sol1 - Solutions to Test 1 Math 20C Oct 22,2008 Particle...

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