1. Write out the vector V=2i+3j at the point (x,y) = (1,2) in terms of the unit vectors in plane polar coordinates. Do this again for the same vector at the origin. Are your results different? Why? Answer: From my notes, I see that x,y = (r, theta) and x=r cos theta, y=r sin theta. So, for (1,2) the vector is 2 cos 2 i + 3 sin 2 j At the origin, (0,0), the vector is zero. Am I on the right track? 2.Find the gradient of the function phi(x,y)=2x^2y at the point (x,y)=(1,2), in plane polar coordinates. Answer: The gradient is 4xyi+2x^2j. At (1,2) it's 8i+2j. How would I put it in plane polar cooridnates? Would the answer be (8 cos 2) i + (2 sin 2) j? Thanks.