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Math Help - Converting to Polar co-ordinates

  1. #1
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    Converting to Polar co-ordinates

    Hi all,

    I've got to convert this equation to polar co-ordinates.

    <br />
x^2 + 2y^2 = y(y + 3x) <br />

    My working so far is as follows,

    <br />
x^2 + 2y^2 = y^2 + 3xy<br />

    <br />
x^2 - 3xy + y^2 = 0<br />

    Let y = (r \cos\theta)^2 and x = (r \sin\theta)^2

    <br />
 (r \cos \theta)^2 - 3(r \cos \theta)(r \sin \theta) + (r \sin \theta)^2 = 0<br />

    <br />
  r^2 \cos ^2\theta - 3(r \cos \theta)(r \sin \theta) + r^2 \sin ^2\theta = 0<br />

    <br />
   r^2 (\cos ^2\theta + \sin ^2\theta) = 3(r \cos\theta)(r \sin\theta)<br />

    <br />
r = \sqrt{3(r \cos\theta)(r \sin\theta)}<br />

    I can't see this being right with r on the RHS though, anyone able to help? Cheers.
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  2. #2
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    Quote Originally Posted by Peleus View Post
    Hi all,

    I've got to convert this equation to polar co-ordinates.

    <br />
x^2 + 2y^2 = y(y + 3x) <br />

    My working so far is as follows,

    <br />
x^2 + 2y^2 = y^2 + 3xy<br />

    <br />
x^2 - 3xy + y^2 = 0<br />

    Let y = (r \cos\theta)^2 and x = (r \sin\theta)^2

    <br />
(r \cos \theta)^2 - 3(r \cos \theta)(r \sin \theta) + (r \sin \theta)^2 = 0<br />

    <br />
r^2 \cos ^2\theta - 3(r \cos \theta)(r \sin \theta) + r^2 \sin ^2\theta = 0<br />

    <br />
r^2 (\cos ^2\theta + \sin ^2\theta) = 3(r \cos\theta)(r \sin\theta)<br />

    <br />
r = \sqrt{3(r \cos\theta)(r \sin\theta)}<br />

    I can't see this being right with r on the RHS though, anyone able to help? Cheers.
    Your equation simplifies to

    Case 1: r = 0

    Case 2: 1 = 3 \cos \theta \sin \theta = \frac{3}{2} \sin (2 \theta) \Rightarrow \sin (2 \theta) = \frac{2}{3}

    so you have four rays, all with terminus at the origin (alternatively, two lines passing through the origin). See http://www.wolframalpha.com/input/?i...%28y+%2B+3x%29
    Last edited by mr fantastic; October 19th 2009 at 02:12 AM.
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  3. #3
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    Any hints as to how it actually simplifies down to that? I'm not sure at what point I should be manipulating it to produce that result.
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  4. #4
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    Quote Originally Posted by Peleus View Post
    Any hints as to how it actually simplifies down to that? I'm not sure at what point I should be manipulating it to produce that result.
    Delete the last line of your calcultation (which should have had a \pm \sqrt{} in it by the way) and simplify the second last line.
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