= (cid:10)2xz + cos(y2 z), z 2xy sin(y2 z), x2 + y + x sin(y2 z)(cid:11) At the point p = (0, 1, 1) we have: F (0, 1, 1) = (cid:10)2(0)(1) + cos(12 1), 1 2(0)(1) sin(11 1), 02 + 1 + (0) sin(11 1)(cid:11) = h1, 1, 1i (b) the unit vector v that points from p = (0, 1, 1) towards q = (2, 3, 0) is: Thus, the directional derivative dvf (0, 1, 1) is: Dvf (0, 1, 1) = f (0, 1, 1) v. Solution: (a) in order for the vector eld h = hf (x, y), g(x, y)i to be conservative, it must be the case that: X: f : using f (x, y) = yexy + y2 and g(x, y) = xexy + 2xy we have: = exy + xyexy + 2y verifying that f is conservative: g: using f (x, y) = xexy and g(x, y) = yexy we have:
