\fequation ex=l+y2+z2 (a) There are two ways we can visualize this surface. First, since we can isolate the variable :r, we can think of S as the graph of the function y, z) 2 111(1 + y2 + 22). The vector Vf(P) in this case represents the direction in the yzplane for which the corresponding path on the surface S at the point P sees the largest increase in the m-direction. Sketch the surface 3, mark the point P = (ln(2)1 0, l) and draw the path on S in the direction of foP). (b) Now. let's think of 5 as the zero set of the function 9(31913) :63 _ 1 _.9,2 _z2' As we saw in Class, a normal vector to S at the point P is given by Vg{P). Use this to nd the cartesian equation for the tangent plane to S at the point P : (111(2), 0., 1)