Results 1 to 6 of 6

Thread: parametrization of curves

  1. #1
    Member
    Joined
    Oct 2010
    Posts
    131

    parametrization of curves

    Hello,

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

    "Let M $\displaystyle \subset \mathbb{R}^2$ 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): $\displaystyle (x^4-2x^2)/(2x^2+2) = -y^4-y^2$}

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

    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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,714
    Thanks
    3002
    The distance from (x, y) to (-1, 0) is $\displaystyle \sqrt{(x+1)^2+ y^2}$ and the distance from (x, y) to (1, 0) is $\displaystyle \sqrt{(x- 1)^2+ y^2}$/ Saying that the product of those two distances is equal to 1 means that $\displaystyle \sqrt{((x+1)^2+ y^2)((x-1)^2+ y^2)}= \sqrt{(x+1)^2(x-1)^2+ 2y^2(x+1)^2+ 2y^2(x- 1)^2+ y^4}= 1$.

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

  3. #3
    Member
    Joined
    Oct 2010
    Posts
    131
    [QUOTE=HallsofIvy;585988]The distance from (x, y) to (-1, 0) is $\displaystyle \sqrt{(x+1)^2+ y^2}$ and the distance from (x, y) to (1, 0) is $\displaystyle \sqrt{(x- 1)^2+ y^2}$/ Saying that the product of those two distances is equal to 1 means that $\displaystyle \sqrt{((x+1)^2+ y^2)((x-1)^2+ y^2)}= \sqrt{(x+1)^2(x-1)^2+ 2y^2(x+1)^2+ 2y^2(x- 1)^2+ y^4}= 1$.



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

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

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

    So its correct?
    You are right. Our set is a subset of $\displaystyle \mathbb{R}^2$. 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
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    10
    For the parametrisation, think in terms of polar coordinates. The equation of the curve reduces to $\displaystyle r^2 = 2\cos2\theta$. But that is a bit problematical, because $\displaystyle \cos2\theta$ can be negative. A possible parametrisation would be
    $\displaystyle \theta\mapsto (f(\theta)\cos\theta,f(\theta)\sin\theta)$, where $\displaystyle f(\theta) = \begin{cases}\sqrt{2\cos2\theta} &\text{if }\cos2\theta\geqslant0, \\0 &\text{if }\cos2\theta<0,\end{cases}$
    and $\displaystyle 0\leqslant\theta\leqslant2\pi$.

    That doesn't look like a good solution, because of the unpleasant singularity at the origin. The curve looks like an $\displaystyle \infty$ 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.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Oct 2010
    Posts
    131
    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; Nov 21st 2010 at 10:04 PM.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Oct 2010
    Posts
    131
    So i have following formula, but i don't know how i can compute the integral:

    i thing the length is defined as: L:=$\displaystyle \int_0^{\pi/4} \sqrt{f'(\theta)^2+f(\theta)^2 dt $=$\displaystyle \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; Nov 22nd 2010 at 07:15 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. parametrization
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Dec 13th 2009, 05:31 PM
  2. parametrization
    Posted in the Calculus Forum
    Replies: 5
    Last Post: Oct 26th 2009, 12:26 PM
  3. Parametrization
    Posted in the Differential Equations Forum
    Replies: 0
    Last Post: May 4th 2009, 09:48 PM
  4. Parametrization of Intersecting Curves
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Apr 24th 2008, 12:48 PM
  5. sketching curves-transformation curves f(x)!!!!
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: Dec 2nd 2006, 06:55 AM

Search Tags


/mathhelpforum @mathhelpforum