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Polar coordinates related (rose and limacon)

  1. May 22, 2010 #1
    1. The problem statement, all variables and given/known data

    I have some questions want to be answered.

    1. For rose, I believe there are two kinds, dealing with even peals and odd peals. My math professor confused himself in the lecture and could not tell us the right identification. The book is also helpless.

    For example, the form r = a +- b cos (k delta) and r = a +- b sin (k delta), in general how do you tell how many pedals such polar coordinate has based on the two general forms (sin and cos) Let r = sin 2 delta we have 4-leaf.

    2. When one graphs polar graphs like rose and limacon, one will often come across negative r. For example, r = sin 2 delta, the four-leaf rose will have many negative r values. For example, the interval [pi/2, 3pi/4] has r decreases from 0 to -1 and [3pi/4, pi] r increases from -1 to 0.
    What I don't understand is the statement made by the book author:
    Please help me to clarify this statement. Thank you

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. May 23, 2010 #2


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    Homework Helper

    in the usual definition of polar coordinates r is defined on [itex][0,\infty)[/itex]

    this to make the mapping [itex] (r,\theta) \to (x,y) [/itex] 1:1

    however i think you teacher wants to relax this for a bit to facilitate drawing the roses.

    Now to picture whats going on consider, a constant angle, say [itex]\theta = \frac{\pi}{2}[/itex]. As r changes from zero to infinity you trace out the line y = x, in the positive x, positive y quadrant.

    in normal polar coordinates teh part of the line y=x, in the negative x & y quadrant would be given by [itex]\theta = \frac{3 \pi}{2} [/itex].

    however if you allow negative r values, imagne as r goes to zero, the line goes through the origin then as r goes negative, the line traces intoe the negative x & y quadrant.

    so to be explicit [itex] (-r, \theta) = (r, \theta + \pi) [/itex], but in effect, a negative radius just points in the opposite direction to the poistive radius
  4. May 24, 2010 #3
    If theta=0 or pi corresponds to the line y=0, then pi/2 or 3/2 pi would correspond to the line x=0, wouldn't it? And y=x would be pi/4 or 5/4 pi.
  5. May 24, 2010 #4


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    yep good point ;) missed it
  6. May 24, 2010 #5
    thank you very much i understood now
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