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Parametrize this curve

  1. Apr 3, 2011 #1
    1. The problem statement, all variables and given/known data
    Parametrize the following equation using polar coordinates:


    2. Relevant equations
    $(x^2+y^2)^2 = r^2 (x^2 - y^2)$


    3. The attempt at a solution
    It seems that $x=r\cos{\theta}$ and $y = r\sin{\theta}$ don't work. any suggestions?
     
  2. jcsd
  3. Apr 3, 2011 #2

    tiny-tim

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    hi jakey! :smile:

    (have a theta: θ and try using the X2 icon just above the Reply box :wink:)
    yes it does …

    show us how far you've got :smile:
     
  4. Apr 3, 2011 #3
    hi tiny-tim, thanks for the reply :)

    well, the left hand side would equal to (r^2)^2= r^4.
    the right hand side would equal to r^2 (r^2 cos^2 θ - r^2 sin^2 θ)

    but these two don't equal...right?
     
  5. Apr 3, 2011 #4

    tiny-tim

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    hi jakey! :smile:
    d'oh! :rolleyes: if the question says they're equal, then they're equal! :smile:

    the equation has become r4 = r4(cos2θ - sin2θ) …

    what are the solutions to that, and what curve does it represent? :wink:
     
  6. Apr 3, 2011 #5
    Hi tiny-tim, I need the parametrization for "x" and "y" so that the equation holds. I need this to evaluate a line integral that's why I am not looking for the corresponding polar equation...
     
  7. Apr 3, 2011 #6

    tiny-tim

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    solve it anyway :smile:
     
  8. Apr 3, 2011 #7
    Well, r = 0 or cos (2\theta) = 1. how is this gonna help, tiny-tim?
     
  9. Apr 3, 2011 #8

    tiny-tim

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    and what curve is that? :wink:
     
  10. Apr 3, 2011 #9
    hi tiny-tim, i'm stuck. for one, r can't be 0 as the formula i'm dealing with has r=22. i just typed it as r as a generalization.

    well,for cos 2\theta = 1, a solution, for example would be theta = 0 or theta = \pi. that would just be a line in the polar axis.
     
  11. Apr 3, 2011 #10

    tiny-tim

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    well that's the parametrisation, isn't it? …

    0 ≤ r < ∞, along the line θ = 0 or π …

    in Cartesian coordinates: (x(t),y(t)) = (t, 0) for any t :wink:
     
  12. Apr 3, 2011 #11
    WOW, really?? But I couldn't find a period for this...? Or, is it t \in (-\infty, \infty)?
     
  13. Apr 3, 2011 #12

    tiny-tim

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    period? :confused:

    what's the question?
     
  14. Apr 3, 2011 #13
    Hmm, this is the hint given:

    "Parametrize the curve first using polar coordinates. Next, find the period which is to be done in Cartersian coordinates."

    you see, the equation i gave above is the curve for the line integral of \int |y| ds.
     
  15. Apr 3, 2011 #14

    tiny-tim

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    (btw, if you don't use polar coordinates, the original equation becomes 2y2(x2+y2) = 0, or y = 0 :wink:)

    ∫ |y| ds ? …

    well that's 0 :confused:
     
  16. Apr 3, 2011 #15
    it can't be. btw, it's ∫_C |y| ds where C is the curve I gave above. I need to parametrize it so I could use ds = ||r'(t)|| dt.
     
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