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Polar equation to cartesian

  1. Jan 10, 2006 #1
    i have this equation: [tex]r=\sqrt{1+sin2\theta}[/tex]
    and am to convert to cartesian equation and from the equation see that it consists of two circles and directly note the radii of the cirlcles from the equation.

    so far i have manipulated it and gotten:

    [tex]x^6 + 3x^4 y^2 + 3 x^2 y^4 + y^6 - x^4 -6x^2 y^2 - 4x^3 y -y^4 - 4y^3 x=0[/tex]

    any suggestions on how to get it into the required form?
    thanks
     
  2. jcsd
  3. Jan 10, 2006 #2

    Tide

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    HINT: [tex]\sin \alpha = \cos \left (\frac {\pi}{2} - \alpha\right)[/tex]
     
  4. Jan 11, 2006 #3

    benorin

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    It should be a fourth order equation in x and y. I get:

    [tex]\left[ \left(x+\frac{1}{2}\right) ^2 + \left(y+\frac{1}{2}\right) ^2 -\frac{1}{2}\right] \left[ \left(x-\frac{1}{2}\right) ^2 + \left(y-\frac{1}{2}\right) ^2 -\frac{1}{2}\right] =0[/tex]
     
  5. Jan 11, 2006 #4

    dextercioby

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    Well, i get something like

    [tex] x^2 +y^2 =1+\frac{2xy}{x^2 +y^2} [/tex],

    from which the conclusion follows easily.

    Daniel.
     
  6. Jan 11, 2006 #5
    yes, i too found this but can you explain how the conclusion follows easily from this?
     
  7. Jan 11, 2006 #6

    arildno

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    For starters, rewrite it in the form:
    [tex](x^{2}+y^{2})^{2}=(x+y)^{2}[/tex]
     
  8. Jan 11, 2006 #7

    arildno

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    That was to be my next hint..:frown:

    Then, the grand finale would have been to refer to your earlier solution to this problem. I won't do that now.
     
  9. Jan 11, 2006 #8

    benorin

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    There :rolleyes: , or rather not there: all better?
     
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