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When plotting graphs in polar coordinates, how does one know when to

  1. Aug 13, 2012 #1
    When plotting graphs in polar coordinates, how does one know when to make the graph sharp (at θ=0) (as in for the graph for r=1-cosθ) as opposed to a dimple (r=3/2 + cos θ) ?
     
  2. jcsd
  3. Aug 13, 2012 #2

    micromass

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    Re: Limacons

    A graph in polar coordinates is given by [itex]r=f(\theta)[/itex].
    Now, we can express such a graph in cartesian coordinates. So, if [itex]r=f(\theta)[/itex], then we can use the formulas

    [tex]x=r\cos(\theta),~y=r\sin(\theta)[/tex]

    to come up with the following form of the graph in cartesian coordinates:

    [tex](f(\theta)\cos(\theta),f(\theta)\sin(\theta)[/tex]

    For example, given [itex]r=a-\cos(\theta)[/itex] (with a constant), we can write this in cartesian coordinates as

    [tex]((a-\cos(\theta))\cos(\theta),(a-\cos(\theta))\sin(\theta))[/tex]

    Now, the use of this is simpy that we can now investigate our curve using analysis. So, we can find the "velocity vector" at a point by taking derivatives. The derivative of our above function now becomes

    [tex](\sin(\theta)(2\cos(\theta)-a),a\cos(\theta)-\cos(2\theta))[/tex]

    Now, if a=3/2, then our derivative in 0 is [itex](0,1/2)[/itex]

    So, we can deduce that in 0, our function is going up with a speed of 1/2.

    What if a=1? Then our derivative in 0 is (0,0). This is a weird result. It means that at 0, our velocity vector is zero and thus the curve just stands still. This is the explanation of why you get a sharp point (= a cusp) when a=0, but just a smooth line when a=3/2.
     
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