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$$r = 2\cos \theta -1$$

I am suppose to find the polar curve of the inner loop. Here is its graph, courtesy of Wolfram|Alpha,

enter image description here

I am having trouble working out this polar function on a cartesian graph system so my confusion comes from finding the limits of integration for the inner loop. I think if $r = 2\cos \theta -1$ that means I have $r(\theta)$ so $\theta$ is my x and r is my y. So I know that at $\theta = 0$ I have two y values, 0 and 1. How does this work though? Clearly $2\cos (0) -1$ can never be 0. So what is going on here? How does it get two values.

It seems like no one understands my question, I have two values at theta 0. How do I get an arc length if I have two arcs?

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it's a Limaçon – cactus314 Aug 12 '13 at 20:25
  • Replacing $(r,\theta)\to(r,-\theta)$, we have a same equation so the graph is symmetric with respect to the polar axe.

  • We can have the following polar point. Since the graph is symmetric so we need to draw a half of the graph:

    $$\theta=0\to r=1$$ $$\theta=\pi/6\to r=\sqrt{3}-1$$ $$\theta=\pi/3\to r=0$$ $$\theta=\pi/2\to r=-1$$ $$\theta=2\pi/3\to r=-2$$ $$\theta=5\pi/6\to r=-\sqrt{3}-1$$ $$\theta=\pi\to r=-3$$

  • If $r=0$, then $\cos(\theta)=1/2$ and so $\theta=\pi/3$ provided $0\le\theta<\pi$. This means that the point $(0,\pi/3)$ is on the graph, and an equation of the tangent line there is $\theta=\pi/3$. It is called limacon

enter image description here

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I was having trouble finding the direction since it seems like small numbers of theta give numbers around 1 and it is hard to track. – Paul the Pirate Aug 12 '13 at 21:27
Nice graph, @Babak. Pictures help sooooo much! – amWhy Aug 13 '13 at 0:36

You don't have two values. At $\theta = 0$, $r = 1$. As $\theta$ increases, up to the point where $\theta = \frac{\pi}{3}$, $r$ decreases, until at that point, $r = 0$.

Going backwards from $\theta = 0$, $r = 1$, as $\theta$ decreases, $r$ decreases, up to the point where $\theta = -\frac{\pi}{3}$, at which point $r = 0$ again.

Thus, the bounds of integration are $-\frac{\pi}{3}$ and $\frac{\pi}{3}$, as tracing the curve with these limits gives the inner loop, from $(-\frac{\pi}{3}, 0)$ around through $(0, 1)$, all the way to $(\frac{\pi}{3}, 0)$.

(Or, in cartesian coordinates, $(0,0)$ through $(0,1)$, back to $(0,0)$ again.)

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Clearly you can look at the graph and see that for theta = 0 the graph has values of 0 and 1 at the same time. – Paul the Pirate Aug 12 '13 at 22:03
The issue here is that your intuition of looking at the graph isn't quite correct. For any $\theta$, the point $(\theta, 0)$ is the same point on the cartesian plane -- $(x,y) = (0,0)$. While there is a point at $(0,0)$ in cartesian coordinates, it is not reached at $\theta = 0$, but rather at $\theta \in \{\frac{\pi}{3}, -\frac{\pi}{3}\}$. – qaphla Aug 13 '13 at 0:13
Damn, so how do I view this in that graph I have linked? That is a cartesian graph correct? How do I make sense of it? – Paul the Pirate Aug 13 '13 at 0:33
$\theta$ is the angle that you're looking at, and $r$ is the radius. While the graph is plotted on a cartesian plane, the axes are not $r$ and $\theta$, but $x = r\cos(\theta)$ and $y = r\sin(\theta)$. If you start at $\theta = 0$, this is the point $(x, y) = (0, 1)$. Then, increasing $\theta$, you go anticlockwise around the small loop until you hit $(x,y) = (0,0)$, after which you continue anticlockwise around the big loop until reaching $(x,y) = (0,0)$ again. Then, continue (still anticlockwise) around the lower part of the inner loop, coming back to $(x,y) = (0,1)$ at $\theta = 2\pi$. – qaphla Aug 13 '13 at 2:43

Note that
1) The function $r(\theta)=2\cos{\theta}-1$ is an $2\pi$-periodic function.
2) Cartesian coordinates $(x, \ y)$ can be calculated from polar by $$\left \lbrace \begin{gather} x(r,\ {\theta})=r(\theta) \cos{\theta}, \\ y(r,\ {\theta})=r(\theta) \sin{\theta}.\end{gather}\right.$$

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Hmm, so I graphed it incorrectly. The linked graph is a cartesian graph and I was attempting to use a polar equation to graph it. Is that correct? – Paul the Pirate Aug 12 '13 at 21:21
Yes, the linked graph is a cartesian graph. – M. Strochyk Aug 12 '13 at 21:37
So how do I convert? I know the formula but how do I get an r and a theta? – Paul the Pirate Aug 12 '13 at 21:38
For integration does not necessarily use Cartesian coordinates. You can integrate using polar coordinates. – M. Strochyk Aug 12 '13 at 21:55
I am trying to figure out the direction of the graph, how this function works. I can't understand it, it doesn't make sense. I have $r = 2cos\theta -1$ so I plug in 0 and get 1, how does that fit on the graph? It doesn't. – Paul the Pirate Aug 12 '13 at 21:58

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