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Thread: Curve parametrization

  1. #1
    Senior Member vincisonfire's Avatar
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    Curve parametrization

    Hi, I would like to parametrize the intersection between the unit sphere ( $\displaystyle x^2 + y^2 + z^2 =1 $ ) and the plane $\displaystyle x+y+z =0 $*
    I came with
    $\displaystyle *x(t) = \frac{cos(t)}{\sqrt{2}} $
    $\displaystyle y(t) = \frac{sin(t)}{\sqrt{2}} $
    $\displaystyle z(t) = -\frac{(cos(t)+sin(t)}{\sqrt{2}} $
    It almost works, but it seem there are scale problems and I can't solve it.
    Thanks for your help!
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  2. #2
    MHF Contributor
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    Hi

    u$\displaystyle (\frac{1}{\sqrt{2}} , 0 , -\frac{1}{\sqrt{2}})$ and v$\displaystyle (0 , \frac{1}{\sqrt{2}} , -\frac{1}{\sqrt{2}})$ are 2 direction vectors of the plane $\displaystyle x + y + z = 1$

    A parametrization of the plane is

    $\displaystyle x(t) = \frac{t}{\sqrt{2}}$

    $\displaystyle y(t) = \frac{t'}{\sqrt{2}}$

    $\displaystyle z(t) = -\frac{t}{\sqrt{2}} - \frac{t'}{\sqrt{2}}$

    Substituting in the equation of the sphere gives

    $\displaystyle t'^2 + t t' + t^2 - 1 = 0$

    Discriminant is $\displaystyle -3 t^2 +4$

    It is positive for $\displaystyle -\frac{2}{\sqrt{3}} \leq t \leq \frac{2}{\sqrt{3}}$

    Let $\displaystyle t = \frac{2}{\sqrt{3}} \cos(\theta)$

    Then $\displaystyle t' = -\frac{\cos(\theta)}{\sqrt{3}} \pm |\sin(\theta)|$

    You just have to substitute t and t' in the expression of x(t), y(t) and z(t)
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