# Compute Curve Given Implicitly

• Oct 14th 2009, 12:35 AM
bumcheekcity
Compute Curve Given Implicitly
So I have a curve given implicitly by the two equasions:

$\displaystyle \frac{x}{a}^2 + \frac{y}{b}^2 + \frac{z}{c}^2 = 1$

$\displaystyle a \sqrt{b^2 - c^2}z = c\sqrt{a^2 - b^2}$

So I've defined functions such that:

$\displaystyle F_{1}(x,y,z) = \frac{x}{a}^2 + \frac{y}{b}^2 + \frac{z}{c}^2 - 1$

$\displaystyle F_{2}(x,y,z) = a\sqrt{b^2 - c^2}z - c\sqrt{a^2 - b^2}$

For the NEW curve (which I'll call C for argument's sake, and I appreciate here a sketch might be useful, but I don't know how to do one online), I'll let $\displaystyle p_{0}$ be the "centre" of the curve (which a sketch will show you is an ellipse/circle) and will define the vectors $\displaystyle e_{1}, e_{2}$ to be an orthonormal basis in this new curve.

I'm doing this to make it easier to paramatrise, as we will effectively take it down to 2-Dimensions, in this new basis.

$\displaystyle p_{0} = (0,0,0)$

$\displaystyle e_{1} = (a\sqrt{b^2 - c^2}, 0, c\sqrt{a^2 - b^2})$

$\displaystyle e_{2} = (0,1,0)$

Now for the bit I'm stuck on. I'd LIKE to substitute $\displaystyle p_{0} + \lambda e_{1} + \mu e_{2}$ into $\displaystyle F_{1}$ or $\displaystyle F_{2}$, but it just doesn't work. I don't get the equasion of a circle (our lecturer told us it would be a circle, because we have to arc length parametrise it later.

I've tried "swapping round" the values in $\displaystyle e_{2}$, but I'm sure they're right, and that gives me something even worse than what I get the other way. Any help?
• Oct 16th 2009, 04:29 AM
HallsofIvy
Quote:

Originally Posted by bumcheekcity
So I have a curve given implicitly by the two equasions:

$\displaystyle \frac{x}{a}^2 + \frac{y}{b}^2 + \frac{z}{c}^2 = 1$

$\displaystyle a \sqrt{b^2 - c^2}z = c\sqrt{a^2 - b^2}$

So z is a constant?

Quote:

So I've defined functions such that:

$\displaystyle F_{1}(x,y,z) = \frac{x}{a}^2 + \frac{y}{b}^2 + \frac{z}{c}^2 - 1$

$\displaystyle F_{2}(x,y,z) = a\sqrt{b^2 - c^2}z - c\sqrt{a^2 - b^2}$

For the NEW curve (which I'll call C for argument's sake, and I appreciate here a sketch might be useful, but I don't know how to do one online), I'll let $\displaystyle p_{0}$ be the "centre" of the curve (which a sketch will show you is an ellipse/circle) and will define the vectors $\displaystyle e_{1}, e_{2}$ to be an orthonormal basis in this new curve.

I'm doing this to make it easier to paramatrise, as we will effectively take it down to 2-Dimensions, in this new basis.

$\displaystyle p_{0} = (0,0,0)$

$\displaystyle e_{1} = (a\sqrt{b^2 - c^2}, 0, c\sqrt{a^2 - b^2})$

$\displaystyle e_{2} = (0,1,0)$

Now for the bit I'm stuck on. I'd LIKE to substitute $\displaystyle p_{0} + \lambda e_{1} + \mu e_{2}$ into $\displaystyle F_{1}$ or $\displaystyle F_{2}$, but it just doesn't work. I don't get the equasion of a circle (our lecturer told us it would be a circle, because we have to arc length parametrise it later.

I've tried "swapping round" the values in $\displaystyle e_{2}$, but I'm sure they're right, and that gives me something even worse than what I get the other way. Any help?
The original surface is a sphere so taking z= constant gives a circle in the z= constant plane.

From $\displaystyle a \sqrt{b^2 - c^2}z = c\sqrt{a^2 - b^2}$[/quote]
$\displaystyle a^2(b^2- c^2)z^2= c^2(a^2- b^2)$ so $\displaystyle \frac{z^2}{c^2}= \frac{a^2- b^2}{a^2(b^2- c^2)}$

$\displaystyle \frac{x}{a}^2 + \frac{y}{b}^2 + \frac{z}{c}^2 = 1$
$\displaystyle \frac{x}{a}^2 + \frac{y}{b}^2 = 1- \frac{z}{c}^2$
$\displaystyle \frac{x}{a}^2 + \frac{y}{b}^2 = 1- \frac{a^2- b^2}{a^2(b^2- c^2)}$

That's a circle with center on the z-axis at $\displaystyle (0, 0, z_0)= (0, 0, \frac{c\sqrt{a^2- b^2}}{a\sqrt{b^2- c^2}})$ and radius $\displaystyle r= \sqrt{1- \frac{a^2- b^2}{a^2(b^2- c^2)}}$ so a good parameterization would be

$\displaystyle x= r cos(\theta)$, $\displaystyle y= r sin(\theta)$, $\displaystyle z= z_0$, with $\displaystyle \theta$ going from 0 to $\displaystyle 2\pi$.

I don't see any good simplification of $\displaystyle z_0= \frac{c\sqrt{a^2- b^2}}{a\sqrt{b^2- c^2}}$ or $\displaystyle r= \sqrt{1- \frac{a^2- b^2}{a^2(b^2- c^2)}}$.
• Oct 17th 2009, 11:24 AM
bumcheekcity
Bah, of course. All I had to do was sub in the new value of z in terms of x. Thanks, you're a lifesaver :D:D:D