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Homework Help: Convert to Parametric Equation of surface?

  1. Sep 26, 2006 #1
    Convert to Parametric Equation of surface???

    How would I go about converting

    [tex] 9x^2+4y^2+z^2 = 640 [/tex]

    I tried just to solve for each variable but this doesn't seem right

    for example

    [tex] \sqrt{\frac{640-z^2-4y^2}{9}}=x [/tex]

    [tex] \sqrt{\frac{640-z^2-9x^2}{4}}=y [/tex]

    [tex] \sqrt{\frac{640-4y^2-9x^2}{1}}=z [/tex]
  2. jcsd
  3. Sep 26, 2006 #2


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    So, to range over a surface, you'll have two independent variables to run over. Pick those two, then solve for the third in terms of those two. If you are unhappy with this parametrization, you can convert to another set of coordinates.
  4. Sep 26, 2006 #3
    so I can just pick to random values for y and z and then solve for x

    i am not sure exactly what you mean... i mean if I define a certain point for two coordinates than the third coordinate is already defined..
    Last edited: Sep 26, 2006
  5. Sep 26, 2006 #4


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    As you say, given a y and z, x is determined (up to sign, in this case)... which can be visualized as a point, a height x above the yz-plane.
    So, by varying y and z, you get your surface. You'll probably note that, in this problem, you shouldn't range over all values of y and z.
  6. Sep 26, 2006 #5
    Ok so you are saying that

    [tex] z^2+4y^2 < 640 [/tex]

    [tex]z^2+9x^2 < 640 [/tex]

    [tex]4y^2+9x^2<640 [/tex]

    but how does this help me get the parametric equations

    I am still not getting how you go from

    [tex] 9x^2+4y^2+z^2 = 640 [/tex]

    to x=.... y=.... z=....
  7. Sep 26, 2006 #6


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    Based on what you wrote...
    [tex] x=\pm \sqrt{\frac{640-z^2-4y^2}{9}}[/tex]
    Now, vary y and z,... and you expect that, for points on your surface, you get real values of x.
    So, that constrains what y and z can be.
    If you want, you can set up a series of nested loops:
    for y=ymin to ymax
    * for z=zmin to zmax
    ** x=f(y,z) [this is a parameteric equation]
    ** plot(x,y,z)

    to determine ymin and ymax... consider the z=0 case.
    zmin and zmax are functions of y.
    * are indents to show python-like logical blocks.
  8. Sep 27, 2006 #7


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    [tex]9x^2+ 4y^2+ z^2= 640[/tex]
    [tex]\frac{9x^2}{640}+ \frac{4y^2}{640}+ \frac{z^2}{640}= 1[/tex]
    [tex]\frac{x^2}{\frac{640}{9}}+ \frac{y^2}{160}+ \frac{z^2}{640}= 1[/tex]
    [tex]\frac{x^2}{\left(\frac{\sqrt{640}}{3}\right)^2}+ \frac{y^2}{\left(\sqrt{160}\right)^2}+ \frac{z^2}{\left(\sqrt{640}\right)^2}= 1[/tex]
    That's an ellipsoid- a sphere "stretched" differently along each axis.

    Parametric equations for a sphere of radius R can be gotten by using spherical coordinates with [itex]\rho[/itex] set equal to R.

    [itex]x= Rsin(\phi)cos(\theta)[/itex]
    [itex]y= Rsin(\phi)cos(\theta)[/itex]
    [itex]z= R cos(\phi)[/itex]

    Your ellipse has "radius" along the x-axis of [itex]\frac{\sqrt{640}}{3}[/itex], along the y-axis of [itex]\sqrt{160}[/itex], and along the z-axis of [itex]\sqrt{640}[/itex]. Replace R above by each of those in the appropriate equation.
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