Question

R= 12/(1-COS(THETA))

Answer 1

@SithsAndGiggles

Answer 2

What do you have to do? Plot in polar coordinates? Find the area of the region?

Answer 3

I need to complete the general form of the equation using rectangular questions, so it'd end with something like

0 = ________________________

Answer 4

Okay, so convert using
\[\begin{cases}x^2+y^2=r^2\\
x=r\cos\theta\end{cases}\]
Those two might be all you need.

Answer 5

this might be a stupid question, but which one is x and y?

Answer 6

use the picture I drew on the previous question to see

Answer 7

\[x=r\cos\theta\quad\text{and}\quad y=r\sin\theta\]
I'd suggest multiplying both sides by the denominator on the right:
\[r-\color{red}{r\cos\theta}=12\]
You can immediately replace the red term with one of the above equations. What can you do with the remaining \(r\)?

Answer 8

I'm really lost right now, I'm sorry. With the remaining r, I can find the polar coordinate? I'm not sure.

Answer 9

Use the first equation I listed, \(x^2+y^2=r^2\), and you can solve for \(r\) in terms of \(x\) and \(y\). The goal here is to replace any and all \(r\)s and \(\theta\)s with equivalent expressions of \(x\) and \(y\).

Answer 10

Okay, so what I'm going to do is replace the remaining r with 12?

Answer 11

Or would it be x= 1-cos(theta)? or something similar to that?

Answer 12

\[r=\frac{12}{1-\cos\theta}~~\iff~~r(1-\cos\theta)=12~~\iff~~r-r\cos\theta=12\]
Follow so far? Nothing fancy done here, just some algebra.

Answer 13

Yes, I'm following.

Answer 14

Great. From the set of equations I posted earlier, you have
\[\begin{cases}\color{gray}{x^2+y^2=r^2}\\
x=r\cos\theta\end{cases}~~\implies~~\begin{cases}r=\sqrt{x^2+y^2}\\x=r\cos\theta\end{cases}\]
(We ignore the negative root because \(r>0\).) Good?

Answer 15

Okay.

Answer 16

Okay, so if \(\color{red}{r=\sqrt{x^2+y^2}}\) and \(\color{blue}{x=r\cos\theta}\), and you have
\[\color{red}{r}-\color{blue}{r\cos\theta}=12,\]
what is this the same as?

Answer 17

I have no idea, but, uhm, It's late for me. i have school tomorrow morning. So, I apologize. But I have to go to sleep. Regardless, I leave you a medal for attempting to help me. Thank you.

Answer 18

\[\underbrace{\color{red}{r}}_{\sqrt{x^2+y^2}}-\underbrace{\color{blue}{r\cos\theta}}_{x}=12,\]
and so
\[\sqrt{x^2+y^2}-x=12\]