Compute Curve Given Implicitly

Printable View

October 14th 2009, 12:35 AM
bumcheekcity

Compute Curve Given Implicitly

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

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

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

So I've defined functions such that:

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

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

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 $p_{0}$ be the "centre" of the curve (which a sketch will show you is an ellipse/circle) and will define the vectors $e_{1}, e_{2}$ 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.

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

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

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

Now for the bit I'm stuck on. I'd LIKE to substitute $p_{0} + \lambda e_{1} + \mu e_{2}$ into $F_{1}$ or $F_{2}$ , 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 $e_{2}$ , but I'm sure they're right, and that gives me something even worse than what I get the other way. Any help?
October 16th 2009, 04:29 AM
HallsofIvy

Quote:

Originally Posted by bumcheekcity

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

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

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

So z is a constant?

Quote:

So I've defined functions such that:

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

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

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 $p_{0}$ be the "centre" of the curve (which a sketch will show you is an ellipse/circle) and will define the vectors $e_{1}, e_{2}$ 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.

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

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

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

Now for the bit I'm stuck on. I'd LIKE to substitute $p_{0} + \lambda e_{1} + \mu e_{2}$ into $F_{1}$ or $F_{2}$ , 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 $e_{2}$ , 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 $a \sqrt{b^2 - c^2}z = c\sqrt{a^2 - b^2}$ [/quote]
$a^2(b^2- c^2)z^2= c^2(a^2- b^2)$ so $\frac{z^2}{c^2}= \frac{a^2- b^2}{a^2(b^2- c^2)}$

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

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

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

I don't see any good simplification of $z_0= \frac{c\sqrt{a^2- b^2}}{a\sqrt{b^2- c^2}}$ or $r= \sqrt{1- \frac{a^2- b^2}{a^2(b^2- c^2)}}$ .
October 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