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Polar-parametric transformation

  1. Dec 17, 2004 #1
    Is it possible to transform a parametric "equation" into a polar equation? If so how would I go about it?

    Thanks for reading.
     
  2. jcsd
  3. Dec 17, 2004 #2

    dextercioby

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    If the fomulae relating various coordinate systems are diffemorphisms,then why not??Bring an example.A (plane) curve in parametric coordinates.And tell us what u gave to do to express it in (plane) polar coordinates.

    Daniel.
     
  4. Dec 17, 2004 #3
    If I'm given the parametric equations for a hypocycloid:
    x(t)=(a/n)[(n-1)cos(t)-cos[(n-1)t]
    y(t)-(a/n)[(n-1)sin(t)+sin[(n-1)t]

    how would I go about putting it into a function form [tex]r(\theta)[\tex].
    There has to be some way to do it. What would it be?

    Thanks
     
  5. Dec 17, 2004 #4
    If I'm given the parametric equations for a hypocycloid:
    x(t)=(a/n)[(n-1)cos(t)-cos[(n-1)t]
    y(t)-(a/n)[(n-1)sin(t)+sin[(n-1)t]

    how would I go about putting it into a function form [tex]r(\theta)[/tex].
    There has to be some way to do it. What would it be?

    Thanks
     
  6. Dec 17, 2004 #5

    dextercioby

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    That's a horrible curve... :yuck: Anyway'ill let u do the calculations of eliminating the parameter. :biggrin:
    [tex] \rho (t)=\sqrt{x^{2}(t)+y^{2}(t)} [/tex]
    [tex] \theta(t)=\arctan({\frac{y(t)}{x(t)}}) [/tex]

    Express [itex] t(\theta) [/itex] and plug it into [itex] \rho(t) [/itex].

    Daniel.

    PS.My advice:GIVE UP!!It's enough to know that it's possible. :wink:
     
    Last edited: Dec 17, 2004
  7. Dec 17, 2004 #6
    Thanks Kurt? I assume that is your name. That is all I needed. Oh, and I'm not going to give it up. :wink:
     
  8. Dec 17, 2004 #7

    dextercioby

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    :rofl: :rofl: :rofl: :rofl: :rofl: My name is Daniel.I write it all the time.
    That is a "signature".It's edited from the "USER CP" box.Kurt Lewin was a theorist and i loved his idea and decided to quote him.

    Daniel.Really,no bull****.
     
  9. Dec 18, 2004 #8

    arildno

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    It's much easier to find [tex]t(\rho)[/tex] than [tex]t(\theta)[/tex]
    Then, you might invert [tex]\theta(\rho)[/tex] into [tex]\rho(\theta)[/tex]
    the inversion is practically impossible to perform, so I concur with Daniel's advice.
     
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