$$r = 2\cos \theta -1$$
I am suppose to find the polar curve of the inner loop. Here is its graph, courtesy of Wolfram|Alpha,
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?
Best Answer
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