The last questions of the `Maple` "field trip" asked students
about the surfaces which are the graphs
*z=x ^{2}+y^{2}* and

Here is how I answered the questions. It is unfortunate that the
procedure is quite long and intricate, but I must admit that these
questions demand a good deal of analysis in order to be *certain*
that the answers are correct.

I used `Maple` to suggest the answers, and then used more
standard mathematical procedures to verify the suggestions. The
verfications are sometimes quite elaborate, but since `Maple`
only samples finitely many points, some logical analysis is
necessary to be sure.

`plot3d({x^2+y^2,x^3-x*y^2},x=-10..10,y=-10..10);`

and got the following picture

which I found difficult to understand. I rotated it, reshaded it, etc. I then used the following

`YELLOW:=plot3d(x^2+y^2,x=-10..10,y=-10..10,color=yellow,grid=[100,100]):
RED:=plot3d(x^3-x*y^2,x=-10..10,y=-10..10,color=red,grid=[100,100]):`

Before the next instruction I typed `with(plots);` which loads a
collection of programs for graphics. `Maple` has hundreds
(certainly) and maybe thousands (probably!) of commands. Only the
"basic" commands are loaded when the program is first invoked. In this
case, we will want to use several less common plotting commands, so we
must load the `plots` package. The result of

`display3d({YELLOW,RED});`

is shown:

I understood this picture better. It suggested that there were three space curves in the intersection.

`implicitplot(x^2+y^2=x^3-x*y^2,x=-5..5,y=-5..5,grid=[100,100]);`

In this picture, a version of the intersection curves is projected
onto the *(x,y)*-plane (the *z*-coordinate is "forgotten").

From this it appears that there are indeed three curves in the intersection.

`plot((x^2*((x-1)/(1+x))),x=-4..2,y=-10..15,thickness=2,color=black,labels=[`x`,`y^2`]);`

By the way, I use `thickness=2` to get a slightly thicker curve
which is easier to see. And, of course, I used `color=black`
because the contrast with the white background is stronger.

I tried various versions of this instruction, including the following.

`plot((x^2*((x-1)/(1+x))),x=-.5..2,thickness=2,color=black,labels=[`x`,`y^2`]);`

which was the first hint I had from
`Maple` that there was something happening at the
origin. None of the 3-dimensional plots had explicitly shown me an
intersection there. Plotting is done by sampling and connecting the
dots. I was not lucky enough to see something at (0,0,0).

Apparently there will be (*x*,*y*)'s satisfying
(1+*x*)*y*^{2}=*x*^{3}-*x*^{2}
for *x*'s in the range *x*<-1 and *x* not equal to
1 and *x*=0. The pairs are
(*x*,+/-*x**sqrt((*x*-1)/(*x*+1)) when *x* is
not equal to 0, and also the point (0,0). Note: sqrt(*A*) will
mean the non-negative square root of the non-negative number,
*A*. I'll refer to the parts away from the origin as "the
curves" and discuss them in the next step. I'll verify my suspicions
(?) about the origin in the last step.

`Q:=t*sqrt((t-1)/(t+1));`

`spacecurve([t,Q,t^2+Q^2],t=-4..-1.2,axes=normal,color=black,thickness=2);`

This is a static representation which really benefits by looking at a
`Maple` picture interactively, and rotating it. One can also
draw either (or both) of the surfaces with the space curve (I did
this!) and verify that the curve is actually in both surfaces.

Of course there is another curve which can be gotten by changing the sign of the second coordinate.

Here is a display of the third curve (corresponding to *x*>
1).

`PLUS:=spacecurve([t,Q,t^2+Q^2],t=1..5,axes=normal,color=black,thickness=2):
MINUS:=spacecurve([t,-Q,t^2+Q^2],t=1..5,axes=normal,color=black,thickness=2):
display3d(PLUS,MINUS);
`

Again the picture suffers because it is "static". Also, although this
is a "smooth" curve (it has no corners -- we will call such a curve
differentiable) the parameterization using *x* as "*t*"is
flawed because of the square root in the second coordinate. We could
in fact parameterize using *y* and get functions which appear
smooth, but the details are complicated. The `Maple`
instruction `solve(y=x*sqrt((x-1)/(x+1)),x);` shows you one
possible answer.

Here is one view of three curve fragments in space on the cup
*z*=*x*^{2}+*y*^{2}, and the other is
the curve fragments on the surface *z*=*x*^{3}-*xy*^{2}.

**
Maintained by
greenfie@math.rutgers.edu and last modified 8/21/2002.
**