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Traversetime of r=cos(n*θ) 
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#1
Feb413, 06:22 AM

P: 619

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 traversetime 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? 


#2
Feb413, 07:27 AM

Math
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PF Gold
P: 39,363

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. 


#3
Feb413, 04:21 PM

P: 619

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? 


#4
Feb713, 02:57 PM

Math
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PF Gold
P: 39,363

Traversetime of r=cos(n*θ)
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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