From: rusin@washington.math.niu.edu (Dave Rusin)
Newsgroups: sci.math
Subject: Re: A Mathematical Question
Date: 28 Apr 1995 19:47:19 GMT

In article <3nr7lr$smo@nova.sti.nasa.gov>,
matt smith <msmith@sti.nasa.gov> wrote:
>Given the equation y2 = R2 (x / L2) ^ n2, 0 <= x <= L2, can a second
>curve (of any form) be found inside the first curve (i.e., closer to
>the x-axis) such that the perpendicular distance between the two curves
>is a constant, thus forming a shell if the curves are revolved around
>the x-axis?  For instance, as a simpler case, for the curve y = x, the 
>second curve y = x - d (d>0) would be a satisfactory solution, forming a
>conical shell if revolved around the x-axis.

Of course. Just draw the first curve, proceed a distance  d  perpendicularly
"down" form each point, and "connect the dots". There's your new curve.

You can even get formulas for this new curve, but they're not pretty.
If your first curve is, for example, given by a polynomial curve
F(x,y)=0  (in your example  F(x,y) = -y + R2 (x/L2)^n2 ), then the
gradient vector points perpendicular to the curve, so all you have to do
is move your distance  d  in the direction of the gradient (or opposite it).
This will take you from the point (x,y) to the point
	(x,y) + d*(F_x, F_y)/sqrt(F_x^2+F_y^2)
If, as in your example, you can give the curve parametrically (or even
explicitly), you can also parameterize the new curve. I've been assuming
the original curve was given implicitly, so let's do the same with
the new curve: it consists of the set of points (x1,y1) for which there
are numbers  x  and  y  satisfying
	F(x,y)=0
	x1=x + d*F_x/sqrt(F_x^2+F_y^2)
	y1=y + d*F_y/sqrt(F_x^2+F_y^2)
You can think of this as three equations in 4 variables, which you can
reduce with elimination to one equation in two variables. Cross-multiplying
and squaring, for example, reduces these to 3 polynomial equations, which
you can process with maple or other packages:
>readlib(eliminate):
>eliminate({eq1,eq2,eq3},{x,y});
which will produce a number of pairs {blah,ugh} in which the second part
("ugh") shows the implicit relation between  x1 and y1.

Here's the result using F(x,y)=-y+x^2. I replaced the other two equations
above with  (x1-x)^2*(F_x^2+F_y^2) = d^2*F_x^2  and
(x1-x)^2+(y1-y)^2 = d^2 before elimination:

                2         2     4       2  2    4     3      2
        (- 1/2 d  + 1/2 x1  + x1  - 2 x1  d  + d  + x1  - 3 d  x1)

                      3       2         2      2
             + (- 2 x1  - 3 x1  - x1 + d  + 2 d  x1) y1

                         2       2           2                  3     4
             + (1/2 - 2 d  + 2 x1  + 3 x1) y1  + (- 2 x1 - 1) y1  + y1

(There's another factor to "ugh", which corresponds to another curve;
one is above the parabola, the other below it.)

As you can see, this expresses the curve implicitly as a polynomial
relation between  x1  and  y1. I would advise you not to try to solve
this expression to get  y1  explicitly in terms of  x1.

dave