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Converting from square to polar coordinants

  1. Sep 27, 2005 #1

    If I have the following equation:

    x^2 - 2y = 0

    I need to convert it into polar coordinants:

    r^2 * cos(phi)^2 - 2*r* sin(phi) = 0

    Is this correct ?

  2. jcsd
  3. Sep 27, 2005 #2


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    Yes, it is correct.
  4. Sep 27, 2005 #3


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    And, of course, as long as r is not 0, you can divide through by r to get
    [tex]r cos^2(\theta)- 2sin(\theta)= 0[/tex]
    [tex]r= \frac{2sin(\theta)}{cos^2(\theta)}[/tex]

    (It is customary to use "theta" in polar coordinates rather than "phi")
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