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Math Help - Compute Curve Given Implicitly

  1. #1
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    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?
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  2. #2
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    Quote Originally Posted by bumcheekcity View Post
    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?

    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)}}.
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  3. #3
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    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
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