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Thread: Polar equation conversion to rectangular equations

  1. #1
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    Polar equation conversion to rectangular equations

    Hi, I need help converting r=2-cos(theta) into rectangular form. My teacher gave us the correct answer: 0=(x^4)+(2x^3)-(3x^2)+(2x^2)(y^2)+(xy)-(4y^2)+(y^4), but I don't know how to get there. I understand how to solve basic polar conversions but this problem was above my understanding. Thanks for any help!

    This is as far as I got, but I don't even know if I started the problem correctly.
    Polar equation conversion to rectangular equations-math-help-small.jpg
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  2. #2
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    Re: Polar equation conversion to rectangular equations

    $r = \sqrt{x^2+y^2}$

    $\tan \theta = \dfrac{y}{x} = \dfrac{\text{opp}}{\text{adj}}$

    So, $\cos \theta = \dfrac{\text{adj}}{\text{hyp}} = \dfrac{x}{\sqrt{x^2+y^2}}$

    Or, you have $x = r\cos \theta$, so $\cos \theta = \dfrac{x}{r} = \dfrac{x}{\sqrt{x^2+y^2}}$ gives the same result.

    Plugging in, you have:

    $\sqrt{x^2+y^2} = 2-\dfrac{x}{\sqrt{x^2+y^2}}$

    Multiply out by $\sqrt{x^2+y^2}$

    $x^2+y^2 = 2\sqrt{x^2+y^2}-x$

    $(x^2+y^2+x)^2 = 2^2(\sqrt{x^2+y^2})^2$

    $x^4+y^4+x^2+2x^2y^2+2x^3+2xy^2=4x^2+4y^2$

    $x^4+2x^3-3x^2+2x^2y^2+2xy^2-4y^2+y^4=0$

    It appears your teacher made a mistake with the $xy^2$ term, writing $xy$ instead of $2xy^2$.
    Last edited by SlipEternal; May 25th 2018 at 11:20 AM.
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  3. #3
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    Re: Polar equation conversion to rectangular equations

    Thank you!
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