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Homework Help: Polar coordinates from rectangle

  1. Feb 7, 2006 #1
    Heres where Im struggling, I cant seem to change equations from rectangular to polar and vice versa

    an example

    x^2+y^2-2ax=0

    heres what I got when I tried
    r=2a cos theta
    and thats a graph of a rose curve, I think, Im about 10% sure on that answer

    heres an example of one I have no clue on

    (x^2+y^2)(arctan(y/x))^2=a^2

    heres what im thinkin on this one

    the x^2+y^2 can = r^2 and the arctan (y/x) can = theta
    so you would have (r^2)(theta^2)=a^2

    I dont know, I just cant get this
     
  2. jcsd
  3. Feb 8, 2006 #2

    Tide

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    What, exactly, are you having doubts about? It looks fine to me.
     
  4. Feb 8, 2006 #3

    HallsofIvy

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    x^2+y^2-2ax=0
    Okay, obviously [itex]x^2+ y^2= r^2[/itex] and [itex]2ax= 2ar cos(\theta)[/itex] so the is [itex]r^2- 2ar cos(\theta)= 0[/itex] which you can write as [itex]r^2= 2ar cos(\theta)[/itex] and, as long as r is not 0, divide by r to get [itex]r= 2a cos(\theta)[/itex] as you have.

    (x^2+y^2)(arctan(y/x))^2=a^2
    Again [itex]r^2= x^2+ y^2[/itex] and, essentially by definition, [itex]arctan(y/x)= \theta[/itex] so this is simply [itex]r2\theta^2= a^2[/itex] as you have.
    Assuming everything is positive, you can reduce that to [itex]r\theta= a[/itex].
     
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