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Traversetime of r=cos(n*θ)

  1. Feb 4, 2013 #1
    I have the function in polar coordinates r=cos(n*θ), where r is the radii. I'm supposed to draw the graph of this function, and calculate the area. But to calculate the area, I need to know how fast the function traverses. From the solution in my book, it says if n is even, the function traverses when θ goes from 0 to 2pi, and from 0 to pi if n is odd.

    Why is the traverse-time for this function only dependant on if n is odd or even? I do not understand this at all. Can you guys help me develop an intuition for functions in polar coordinates?
     
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  3. Feb 4, 2013 #2

    HallsofIvy

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    You will cover one lobe as r goes from 0 to 1 and back to 0. That is, as [itex]n\theta[/itex] goes form [itex]-\pi+ 2k\pi[/itex] to [itex]\pi/2+ 2k\pi[/itex] for integer k.

    The reason for the "if n is even, the function traverses when θ goes from 0 to 2pi, and from 0 to pi if n is odd." is that when [itex]cos(n\theta)[/itex] is negative, r is negative and, because r is alwas positive in polar coordinates, is interpreted as positive but with [itex]\pi[/itex] added to [itex]\theta[/itex]. When n is odd, that results in one lobe covering another. When n is even, there are 2n lobes, when n is odd there are n lobes.
     
  4. Feb 4, 2013 #3
    So you're saying when r goes from 0 to 1 to 0 (and n*θ goes from -pi/2 to pi/2) the function has fully traversed?

    Is this a general rule? That when r goes back to where it started, the function has traversed?

    I think I get you, but can you please provide an example?
     
  5. Feb 7, 2013 #4

    HallsofIvy

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    The first graph is [itex]r= 5cos(4\theta)[/itex] and the second graph is [itex]r= 5cos(5\theta)[/itex]
     

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