# Plotting a polar curve

The question is, to generate a polar graph using a graphing utility, and to choose parameter interval so that the complete graph is generated.

$$r=\cos\frac{\theta}{5}$$

To find such an interval, we are looking for smallest number of complete revolutions until value of $r$ begins to repeat. Algebraically,this amounts to

$$\cos\frac{\theta}{5}=\cos\frac{\theta+2n\pi}{5}$$

For this equality to hold,$\frac{2n\pi}{5}$ must be an even multiple of $\pi$,the smallest n for which it occurs is $n=5$.Therefore, the graph will be traced completely in $5$ revolutions ($10\pi$).

But when I draw it the graph is completely traced in $5\pi$, where have I gone wrong?

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Try graphing $r=cos\theta$ by hand. It creates a circle between 0 and $\pi$. It re-traces the same circle between $\pi$ and $2\pi$. The same thing occurs in your graph. – Paul Sundheim Aug 23 '14 at 14:54
@PaulSundheim how is that same with this graph, this graph does not trace itself before $5\pi$, and i am getting $10\pi$. – SHOBHIT GAUTAM Aug 23 '14 at 14:56
As you said, "the graph is completely traced in 5π" but now you say "this graph does not trace itself before 5π"? Which is the correct statement? – Paul Sundheim Aug 23 '14 at 14:59
I just traced it on a graphing calculator. Seemed to be complete in $5\pi$ and then repeated. – paw88789 Aug 23 '14 at 15:01
@PaulSundheim The graph is completely traced in 5pi means after that you are just tracing it again.I know 5pi is correct, do you mind telling where is the computation wrong? – SHOBHIT GAUTAM Aug 23 '14 at 15:04

$r(5\pi+t)=\cos(\pi+\frac{t}5)=-\cos(\frac{t}{5})=-r(t)$. Also the angle at $5\pi+t$ points in the opposite direction of the angle at $t$. Hence you get repetition.

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$r(5\pi+t)=$?? what do you mean. – SHOBHIT GAUTAM Aug 23 '14 at 15:13
He means $r$ evaluated at $5\pi+t$ – Paul Sundheim Aug 23 '14 at 15:15
got it thanks for the help. – SHOBHIT GAUTAM Aug 23 '14 at 15:18
@Paul S. Yes. Thanks. – paw88789 Aug 23 '14 at 15:30

Since the question has been already answered, I post here a visual answer that I hope may help you to understand the problem better:

Cheers!

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very helpful and cool, which utility is this. – SHOBHIT GAUTAM Aug 23 '14 at 15:26
Hi @Shobhit. This was entirely done with Matlab. – Dmoreno Aug 23 '14 at 15:26
thank you again its cool – SHOBHIT GAUTAM Aug 23 '14 at 15:28
You're welcome mate. – Dmoreno Aug 23 '14 at 15:31