**(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).
**