Question

Part 1:. Why the second derivative test works for extrema of functions of two variables We follow the Caleulus 1 example of making sure that the first derivative is 0. And you have seen us simply set the first two partials equal to 0: f-0 and y -0. Then we apply information about a version of the second derivative, namelyDy()in concert with the sign of Ju. Why does this work? Step 1 for this part of the explanation First, is it sufficient to say that -0 and , -0 when, in fact, we need that first derivative to be zero no matter what direction we use away from a point P on the surface zf(x,y)? When we choose a point P on the surface, the direction in which we move is determined by a vector that can be projected onto the x-y plane. A diagram is shown below. Recall that a directional derivative calculation requires that we use a unit vector in the given direction. 1 The vector vs(y, v,). The unit vector in this direction is given by-Sv , )-(cos( ), sin(a)) Imagine now the surface -f(x, y) hovering above (or below) the x-y plane and we want to calculate the first directional derivative off in the direction cos(α), sin (0), making certain that this is so that we have a first derivative that is 0 in all possible directions from the point P. This is calculated as follows: The first directional derivative off in the direction of v is Because cos(a) and sin(a) are never 0 at the same time, this expression for Dif is only 0 if both f, and J equal 0. The first derivative is 0 from all directions when f.-f, 0.

Answer 1

SOLUTION:

We know that nabla = (partial differential/partial differential x, partial differential/partial differential y) Second directional derivative = nabla (f_x cos alpha + f_y sin alpha). < (cos alpha, sin alpha) > = < partial differential/partial differential x (f_x cos alpha + f_y sin alpha), partial differential/partial differential y (f_x cos alpha + f_y sin alpha) > . < (cos alpha, sin alpha) > Now partial differential/partial differential x f_x = partial differential^2 f/partial differential x^2 = f_xx partial differential/partial differential y f_y = partial differential^2 f/partial differential y^2 = f_yy Therefore Second directional derivative = < (f_xx cos alpha, f_yy sin alpha) > . < (cos alpha, sin alpha) > = f_xx cos^2 alpha + f_yy sin^2 alpha

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