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Math Help - parametrization of curves

  1. #1
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    parametrization of curves

    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
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  2. #2
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    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}.
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  3. #3
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    [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
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  4. #4
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    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.
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  5. #5
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    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
    Last edited by Sogan; November 21st 2010 at 10:04 PM.
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  6. #6
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    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
    Last edited by Sogan; November 22nd 2010 at 07:15 AM.
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