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

Question

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

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?

Answer 1

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.)

Answer 2

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 $$\left \lbrace \begin{gather} x(r,\ {\theta})=r(\theta) \cos{\theta}, \\ y(r,\ {\theta})=r(\theta) \sin{\theta}.\end{gather}\right.$$

Answer 3

• 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