parametrization of curves

**Sogan** · November 19th 2010, 08:32 AM

Hello,

I have a question about parametrization of "submanifolds" or curves, respectively.
In particular in the following example:

"Let M $\subset \mathbb{R}^2$ be the set of all points which have the product of their distanes from (-1,0) and (1,0) equal 1."

Give a parametrization of M. Is M a submanifold of dimension 1?

so i thing M={(x,y): $(x^4-2x^2)/(2x^2+2) = -y^4-y^2$ }

so i think the curve in $\mathbb{R}$ is like two symmetric loops between $+-\sqrt{2}$ .

I don't think thant M is a submanifold because of the point (0,0). But i don't know how i can parametrize M. I never have done this before. How can this be done?

Regards

**HallsofIvy** · November 19th 2010, 11:51 AM

The distance from (x, y) to (-1, 0) is $\sqrt{(x+1)^2+ y^2}$ and the distance from (x, y) to (1, 0) is $\sqrt{(x- 1)^2+ y^2}$ / Saying that the product of those two distances is equal to 1 means that $\sqrt{((x+1)^2+ y^2)((x-1)^2+ y^2)}= \sqrt{(x+1)^2(x-1)^2+ 2y^2(x+1)^2+ 2y^2(x- 1)^2+ y^4}= 1$ .

That means that $(x+1)^2(x- 1)^2+ 4x^2y^2+ 4y^2+ y^4= 1$
I don't see how you can get your formula from that. It is, in any case, NOT "a curve in $\mathbb{R}$ ". It is, just as the problem says, a subset of $\mathbb{R}^2$ , not $\mathbb{R}$ .

**Sogan** · November 19th 2010, 09:29 PM

[QUOTE=HallsofIvy;585988]The distance from (x, y) to (-1, 0) is $\sqrt{(x+1)^2+ y^2}$ and the distance from (x, y) to (1, 0) is $\sqrt{(x- 1)^2+ y^2}$ / Saying that the product of those two distances is equal to 1 means that $\sqrt{((x+1)^2+ y^2)((x-1)^2+ y^2)}= \sqrt{(x+1)^2(x-1)^2+ 2y^2(x+1)^2+ 2y^2(x- 1)^2+ y^4}= 1$ .

Ok you are right. My Formula is not correct, but also yours is not correct:

$\sqrt{((x+1)^2+ y^2)((x-1)^2+ y^2)}= \sqrt{(x+1)^2(x-1)^2+ y^2(x+1)^2+ y^2(x- 1)^2+ y^4}= 1$

=> $(x^2-1)^2+y^2*2x^2+2y^2+y^4=1<=> x^4-2x^2=-y^4-y^2*(2x^2+2)$ ,

So its correct?
You are right. Our set is a subset of $\mathbb{R}^2$ . But it looks like a curve in the plane.

But what about the parametrization. I Think it must be locally or? How can i find such a parametrization?

Regards

**Opalg** · November 20th 2010, 12:40 AM

For the parametrisation, think in terms of polar coordinates. The equation of the curve reduces to $r^2 = 2\cos2\theta$ . But that is a bit problematical, because $\cos2\theta$ can be negative. A possible parametrisation would be
$\theta\mapsto (f(\theta)\cos\theta,f(\theta)\sin\theta)$ , where $f(\theta) = \begin{cases}\sqrt{2\cos2\theta} &\text{if }\cos2\theta\geqslant0, \\0 &\text{if }\cos2\theta<0,\end{cases}$
and $0\leqslant\theta\leqslant2\pi$ .

That doesn't look like a good solution, because of the unpleasant singularity at the origin. The curve looks like an $\infty$ symbol, and that parametrisation describes both loops in the same (anticlockwise) direction. It would be much better to have a parametrisation that went straight across the origin without stopping there, so as to go anticlockwise round the right loop and clockwise round the left loop. But as far as I can see, there is no simple formula for such a description of the curve.

**Sogan** · November 21st 2010, 08:56 AM

Thank you so much for your help!!!

I have one more question to our parametrisation. How can we compute the length of our path.

Regards

**Sogan** · November 21st 2010, 10:40 PM

So i have following formula, but i don't know how i can compute the integral:

i thing the length is defined as: L:= $\int_0^{\pi/4} \sqrt{f'(\theta)^2+f(\theta)^2 dt$ = $\int_0^{\pi/4} \sqrt{2/(cos(2\theta))} dt$

is this correct so far? and how can i solve this integral?

Regards

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