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Thread: Parametrize curve of intersection

  1. #1
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    Parametrize curve of intersection

    The problem is:
    "Parametrize the intersection of the hyperboloid $\displaystyle x^2+y^2=z^2+1 $ and the plane $\displaystyle z=ay $, where $\displaystyle a$ is a positive real number. Give the range of the values taken by the parameter. -Hint: Consider the cases 0<a<1 and a > 1 separately, and remember that x, y and z are all real."


    So here's what I did:
    I first plugged in the equation of the plane into the equation of the hyperboloid: $\displaystyle x^2+(1-a^2)y^2=1 $

    If a<1, I get an ellipse of the form: $\displaystyle \frac {x^2}{1^2} +\frac{y^2}{\frac{1}{(\sqrt{1-a^2})^2}}=1$, similar to $\displaystyle \frac{x^2}{c^2}+\frac{y^2}{d^2}=1$
    From an online source, I found that $\displaystyle x=\cos\phi$ and $\displaystyle y=\frac{1}{\sqrt{1-a^2}}\sin\phi $, and therefore $\displaystyle z=\frac{a}{\sqrt{1-a^2}}\sin\phi$ and that $\displaystyle 0\leq \phi \leq 2\pi $


    If a>1, the equation can be put in the form $\displaystyle \frac {x^2}{1^2} -\frac{y^2}{\frac{1}{(\sqrt{a^2-1})^2}}=1$ which is the equation of a hyperbola. Once again, from an online source (Wolfram MathWorld), I found that I could represent both branches of the hyperbola with $\displaystyle x=\sec\phi$, $\displaystyle y=\frac{1}{\sqrt{a^2-1}}\tan\phi $ and $\displaystyle z=\frac{a}{\sqrt{a^2-1}}\tan\phi$, where $\displaystyle -\pi < \phi < \pi $ with discontinuities at $\displaystyle \pm \frac{\pi}{2} $


    I've seriously spent a lot of time on this problem, and this is the best I can come up with. Can someone tell me if it's acceptable? If not, what should I do?
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  2. #2
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    It looks good to me. Of course, I can't speak for your teacher!
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  3. #3
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    Last edited by Nam007; Nov 5th 2009 at 01:36 PM. Reason: want to delete message
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  4. #4
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    I don't see why you have to do all that. If I substitute the plane into the hyperboloid, I get for $\displaystyle a\ne 1$:

    $\displaystyle x^2+y^2=a^2 y^2+1$ so that $\displaystyle y=\pm \sqrt{\frac{1-x^2}{1-a^2}}$ and $\displaystyle z=ay$. So a parametric representation of the intersection is:

    $\displaystyle \left\{x,\pm \sqrt{\frac{1-x^2}{1-a^2}},\pm a\sqrt{\frac{1-x^2}{1-a^2}}\right\}$

    with $\displaystyle -a\leq x\leq a$ for $\displaystyle a<1$

    and:

    $\displaystyle |x|>a$ for $\displaystyle a>1$

    And I think for the case $\displaystyle a=1$, then $\displaystyle z=y$ which then means $\displaystyle x^2=1$ so that the paramaterization is $\displaystyle \{\pm1,y,y\}$.

    The yellow contours below is this paramaterization for a=1/2 and a=2
    Attached Thumbnails Attached Thumbnails Parametrize curve of intersection-hypintersection.jpg  
    Last edited by shawsend; Nov 5th 2009 at 01:18 PM. Reason: corrected formula, need special case for a=1 too
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  5. #5
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    Quote Originally Posted by shawsend View Post
    I don't see why you have to do all that. If I substitute the plane into the hyperboloid, I get for $\displaystyle a\ne 1$:

    $\displaystyle x^2+y^2=a^2 y^2+1$ so that $\displaystyle y=\pm \sqrt{\frac{1-x^2}{1-a^2}}$ and $\displaystyle z=ay$. So a parametric representation of the intersection is:

    $\displaystyle \left\{x,\pm \sqrt{\frac{1-x^2}{1-a^2}},\pm a\sqrt{\frac{1-x^2}{1-a^2}}\right\}$

    with $\displaystyle -a\leq x\leq a$ for $\displaystyle a<1$

    and:

    $\displaystyle |x|>a$ for $\displaystyle a>1$

    And I think for the case $\displaystyle a=1$, then $\displaystyle z=y$ which then means $\displaystyle x^2=1$ so that the paramaterization is $\displaystyle \{\pm1,y,y\}$.

    The yellow contours below is this paramaterization for a=1/2 and a=2
    But how do you know that $\displaystyle -a\leq x\leq a$ for $\displaystyle a<1$? I can't see that...

    That thumbnail was really helpful! Did you use Maple to draw the surface and the intersection?
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  6. #6
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    Ok, I think that should be $\displaystyle |x|<1$ when $\displaystyle a<1$. Also then $\displaystyle |x|>1$ for $\displaystyle a>1$. Also, I drew that in Mathematica using this code for the a=1/2 one:


    Code:
    a = 1/2; 
    p1 = ParametricPlot3D[{x, Sqrt[1/(1 - a^2) - x^2/(1 - a^2)], 
        a*Sqrt[1/(1 - a^2) - x^2/(1 - a^2)]}, {x, -1, 1}, 
       PlotStyle -> {Thickness[0.008], Yellow}]
    p2 = ParametricPlot3D[{x, -Sqrt[1/(1 - a^2) - x^2/(1 - a^2)], 
        (-a)*Sqrt[1/(1 - a^2) - x^2/(1 - a^2)]}, {x, -1, 1}, 
       PlotStyle -> {Thickness[0.008], Yellow}]
    pic1 = Show[{ContourPlot3D[{x^2 + y^2 == z^2 + 1, z == (1/2)*y}, {x, -3, 3}, 
         {y, -3, 3}, {z, -3, 3}], p1, p2}]
    Last edited by shawsend; Nov 5th 2009 at 03:53 PM. Reason: corrected bounds on x
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  7. #7
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    Thank you so much!
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