APPM 2350 Final: appm2350fall2014final_sol_0
Document Summary
Instructions: electronic devices, books, and crib sheets are not permitted. Write your name and your instructor"s name on the front of your bluebook. The derivatives can be found as follows: fx = 2xy fy = x2 + z sin(y) fz = cos(y) Solution: we know from directional derivatives that df ds f df. Then, the above formula can be rewritten: ds = f u. In our case, u is the vector pointing straight towards the origin, whcih is given by h 4, 0, 3i. Normalizing this, we get that u = 1. The gradient of f is computed as usual. Afterwards, the point (4, 0, 3) is plugged in to obtain a vector: F = h2xy, x2 + z sin(y), cos(y)i f|(4,0,3) = h0, 16, 1i. S is given to be 0. 1 in the problem statement; thus, all that is left to do is to plug in all of our information to obtain f :
