Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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?

share|improve this question
    
it's a Limaçon –  john mangual Aug 12 '13 at 20:25
add comment

3 Answers

  • 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

share|improve this answer
    
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
add comment

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.)

share|improve this answer
    
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
add comment

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

share|improve this answer
    
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
show 10 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.