Math 241: Old Tests' Hints

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These are email responses to some questions students have emailed
to me. Students in my Math 241 classes may obtain further electronic
help by contacting me by email at filaseta@math.sc.edu .

Problem 3 (1992): The partial derivative of v with respect to t is 0 (since v does not depend on t). Also, since y only depends on v which does not depend on t, the partial derivative of y with respect to t is 0. Also, since the partial derivative of v with respect to t is 0, we get the partial derivative of x with respect to t is simply 3 times the partial derivative of u with respect to t. I'll let you handle it from there noting that even more is 0, the partial derivative of r with respect to t. (The point is that there's enough zeroes that you don't really have much work to do in this problem.)

Problem 4 (1992): Don't mess up the expression I have for z by expanding (x+y)^3. The partial derivative with respect to y of the function you want for the tangent plane is 3 x (x+y)^2 which equals 0 precisely when either x = 0 or x = -y. Use this information to decide also when the partial derivative with respect to x is 0.

Part II, Problem 3 (1998): The tip of the half-cone in the picture should be at (0,0,4) and the cone comes downward from there. You are interested in the volume of the part that is inside the cylinder x^2 + y^2 = 1. Even a bad picture would work here (say if you drew the cone going upward from (0,0,4)) for setting up the integral. Use cyclindrical coordinates (note the integrand and the cylinder in the problem). Because you are working in the cylinder x^2 + y^2 = 1, theta goes from 0 to 2*pi and r goes from 0 to 1. Then z goes from 0 to the cone z = 4 - r. The integrand is r^3 * r = r^4.