(786, #97)

Let


and f(0,0) = 0.

a) Use a computer to graph f.

Solution: Here is the graph on the circular domain 0 <= x^2 + y^2 <= 1, using Maple. If your calculator or computer can't handle the singularity at (0,0), graph on the annular domain 1/10 <= x^2 + y^2 <= 1

> plot3d([r*cos(t), r*sin(t), f(r*cos(t), r*sin(t))], r = 0..1, t = 0..2*Pi, color = blue,

axes = frame, tickmarks = [2,2,2]);

b) Find when (x,y) is not equal to (0,0).

Solution: Straightforward calculation. Let Maple do the work:

> simplify(diff(f(x,y), x));


>simplify(diff(f(x,y), y));


c) Find .

Solution: Both partials are 0:



d) Show that


Solution: Using parts (a) and (b):



e) Does the result of part (d) contradict Clairaut's Theorem?

Solution. No, Clairaut's Theorem requires that be continuous on their domain. The fact that they are not can be seen by computing, say , and then plotting it on the annualar domain 1/10 <= x^2 + y^2 <= 1

> simplify(diff(f(x,y), x,y));


If you are using Maple you can finnesse the labor involved in typing in such a complicated function.

> plot3d([r*cos(t), r*sin(t), subs( {x = r*cos(t), y = r*sin(t)}, diff(f(x,y), x,y))],

r = 1/10..1, t = 0..2*Pi, color = red, axes = framed);


The graph shows a discontinuity of the second derivative at (0,0).