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R^2 -> R^2 transformation of unit circle into square.

  1. Jul 16, 2008 #1
    I have a unit circle:
    x^2+y^2 <= 1

    And i'm asked to convert it to a square with verticies (0,0),(0,1),(1,0),(1,1).

    Now obviously I have to do this in polar coordinates, so I've rewritten the equation as:

    x = cos t
    y = sin t

    I'm sort of drawing a blank after setting up these equations. How can I convert this curve into straight lines. Any suggestions?
     
  2. jcsd
  3. Jul 16, 2008 #2
    Those aren't really polar coordinates, they're parametric. Converting your unit circle equation into a polar equation would yield, [tex]r=\pm\sqrt{\frac{1}{2cos^2(\theta)}[/tex], to make this into a square, you'll have to convert this polar equation to a parametric equation.
     
  4. Jul 16, 2008 #3
    Oops, I just noticed I did something wrong, I'll try to fix it.
     
  5. Jul 16, 2008 #4
    Okay, I think I found my problem...
    What you need to do is to convert the parametric equation you have into a polar parametric equation. The answer I got was:
    [tex]r=\pm\sqrt{cos^4(T)+sin^2(T)}[/tex]
    [tex]\theta=tan^{-1}(tan(T)[/tex]

    This simplifies to four x-y parametric equations with constraints:

    [tex]-1\leq T \leq1 for X_1,X_2,Y_3,Y_4[/tex]

    [tex]T=0 for Y_1,Y_2,X_3,X_4[/tex]

    [tex]X_1=tan^{-1}(tan(T))[/tex]
    [tex]Y_1=\sqrt{cos^4(T)+sin^2(T)}[/tex]

    [tex]X_2=tan^{-1}(tan(T))[/tex]
    [tex]Y_2=-\sqrt{cos^4(T)+sin^2(T)}[/tex]

    [tex]X_3=\sqrt{cos^4(T)+sin^2(T)}[/tex]
    [tex]Y_3=tan^{-1}(tan(T))[/tex]

    [tex]X_4=-\sqrt{cos^4(T)+sin^2(T)}[/tex]
    [tex]Y_4=tan^{-1}(tan(T))[/tex]
     
    Last edited: Jul 17, 2008
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