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Math Help - Need some help with polar!

  1. #1
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    May 2010
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    Need some help with polar!

    I've been having alot of trouble with this equation. I need to convert it into standard y intercept form...


    Thanks in advance for your help!
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  2. #2
    Senior Member
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    May 2010
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    Wow, are you sure you have this right? Looks like something your professor made up while drunk. Assuming it is correct, rewrite it as:

    r=8\cos^4 \theta [1 - \sqrt{(1- \cos \theta )/2}]^3+3\sin^2 \theta <br />

    Multiplying both sides by r^{11/2} gives:

    r^{13/2}=8r^4\cos^4 \theta [\sqrt r - \sqrt{(r- r\cos \theta )/2}]^3+3r^{7/2}r^2\sin^2 \theta <br />

    Now substitute x=r\cos \theta, y=r\sin \theta, and r=\sqrt{x^2+y^2}.

    Note: I took the posive square root in the half-angle identity \sin \theta/2 = \pm \sqrt{(1-\cos \theta)/2}. Convince yourself that this is justified.
    Last edited by ojones; May 13th 2010 at 02:14 PM.
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