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Polar rose

  1. I'm trying to express the polar rose as an implicit function:
    r(t)=sin t

    x = sin t * cos t
    y = sin^2 t

    Since sin t * cos t = (1/2) * sin 2t and sin^2 t = (1/2) * (1-cos 2t)

    (2x)^2 + (1-2y)^2 = 1
    4x^2 -4y + 4y^2 = 0

    When I plot this, Maple plots a circle, where have I gone wrong?
     
  2. jcsd
  3. arildno

    arildno 12,015
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    "(2x)^2 + (1-2y)^2 = 1" This is an equation for a circle.

    Your parametrization is not.
     
  4. tiny-tim

    tiny-tim 26,054
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    Hi JanClaesen! :smile:

    (have a theta: θ :wink:)

    A rose is usually r = ksinθ or r = kcosθ … see http://en.wikipedia.org/wiki/Rose_(mathematics).

    For k = 1, it is a circle.

    (But you could have got the same equation if you'd just made it r2 = y :wink:)
     
  5. Do you have any hints on how to find the Cartesian equation for r(θ)=sin(2θ), I really can't seem to find it. :)
     
  6. tiny-tim

    tiny-tim 26,054
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    Hint: multiply both sides by r2. :wink:
     
  7. HallsofIvy

    HallsofIvy 40,413
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    And use the identity [itex]sin(\theta)= 2sin(\theta)cos(\theta)[/itex].
     
  8. tiny-tim

    tiny-tim 26,054
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    He knows that :rolleyes: :smile:
    (and have a theta: θ :wink:)
     
  9. Wow, that was clever, thank you :smile:
    For those interested:

    xy = 0.5(x^2+y^2)(x^2+y^2)^(1/2)

    (where x^2+y^2 = sin^2 (2θ) )
     
  10. tiny-tim

    tiny-tim 26,054
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    (try using the X2 tag just above the Reply box :wink:)

    That's it! :smile:

    And then expand it , and put it all on the left:

    (x2 + y2)3 - 2xy = 0. :wink:
     
  11. Yep, thanks again :smile:

    Is there a human way to do this also for sin(3θ)? Or would that be a computer job? :smile:
    I'm trying to do this now, but I have a feeling it's quite tough. :smile:
     
  12. tiny-tim

    tiny-tim 26,054
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    Hint: try it for cos(3θ) + isin(3θ) :wink:
     
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