Polar rose

  1. Nov 7, 2009 #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. Nov 7, 2009 #2

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

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

    Your parametrization is not.
     
  4. Nov 7, 2009 #3

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    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. Nov 7, 2009 #4
    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. Nov 8, 2009 #5

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hint: multiply both sides by r2. :wink:
     
  7. Nov 8, 2009 #6

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    And use the identity [itex]sin(\theta)= 2sin(\theta)cos(\theta)[/itex].
     
  8. Nov 8, 2009 #7

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    He knows that :rolleyes: :smile:
    (and have a theta: θ :wink:)
     
  9. Nov 8, 2009 #8
    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. Nov 8, 2009 #9

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    (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. Nov 8, 2009 #10
    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. Nov 8, 2009 #11

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hint: try it for cos(3θ) + isin(3θ) :wink:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Similar Discussions: Polar rose
  1. Derivative in polar (Replies: 1)

  2. Polar coordinates (Replies: 6)

  3. Polar coordinates (Replies: 7)

  4. Polar Integration (Replies: 7)

Loading...