# Where do these two polar equations meet

• Jan 23rd 2010, 06:05 PM
Jukodan
Where do these two polar equations meet
I recently came across this problem and would like some clarification as to why one of the answers to theta cannot be $\displaystyle \frac{\pi }{2}$

The problem asks to sketch two polar curves and find where they intersect, and then present those points in rectangular coordinates. I can easily see from the graph that the points are (1,0) and (-1,0). The two polar functions are,

$\displaystyle r= \cos ^{2}\theta$ and r = -1.

I set them equal to each other

$\displaystyle \cos ^{2}\theta= 1$

$\displaystyle \frac{1}{2}\left ( 1+\cos 2\theta \right )= 1$

$\displaystyle 1+\cos 2\theta = 2$

$\displaystyle \cos 2\theta = 1$

So theta should be...$\displaystyle 0,\pi,-\pi,\frac{-\pi}{2},\frac{\pi}{2}$

Yet the graph tells me only pi and -pi can be the answer. I'm confused...
• Jan 23rd 2010, 06:23 PM
tonio
Quote:

Originally Posted by Jukodan
I recently came across this problem and would like some clarification as to why one of the answers to theta cannot be $\displaystyle \frac{\pi }{2}$

The problem asks to sketch two polar curves and find where they intersect, and then present those points in rectangular coordinates. I can easily see from the graph that the points are (1,0) and (-1,0). The two polar functions are,

$\displaystyle r= \cos ^{2}\theta$ and r = -1.

I set them equal to each other

$\displaystyle \cos ^{2}\theta= 1$

Ok, is it r = 1 or r = -1??

$\displaystyle \frac{1}{2}\left ( 1+\cos 2\theta \right )= 1$

$\displaystyle 1+\cos 2\theta = 2$

$\displaystyle \cos 2\theta = 1$

So theta should be...$\displaystyle 0,\pi,-\pi,\frac{-\pi}{2},\frac{\pi}{2}$

This is wrong: $\displaystyle \cos 2\theta=1\Longrightarrow 2\theta=2k\pi\,,\,\,k\in\mathbb{Z} \Longrightarrow \theta = k\pi = 0,\,\pm \pi,\,\pm 2\pi,\ldots$

Yet the graph tells me only pi and -pi can be the answer. I'm confused...

Easier: $\displaystyle \cos^2\theta=1\Longrightarrow \cos\theta=\pm 1 \Longrightarrow \theta=k\pi ...etc$

Tonio
• Jan 23rd 2010, 11:26 PM
Pulock2009
Quote:

Originally Posted by Jukodan
I recently came across this problem and would like some clarification as to why one of the answers to theta cannot be $\displaystyle \frac{\pi }{2}$

The problem asks to sketch two polar curves and find where they intersect, and then present those points in rectangular coordinates. I can easily see from the graph that the points are (1,0) and (-1,0). The two polar functions are,

$\displaystyle r= \cos ^{2}\theta$ and r = -1.

I set them equal to each other

$\displaystyle \cos ^{2}\theta= 1$

$\displaystyle \frac{1}{2}\left ( 1+\cos 2\theta \right )= 1$

$\displaystyle 1+\cos 2\theta = 2$

$\displaystyle \cos 2\theta = 1$

So theta should be...$\displaystyle 0,\pi,-\pi,\frac{-\pi}{2},\frac{\pi}{2}$

Yet the graph tells me only pi and -pi can be the answer. I'm confused...

i am saying this very empirically. perhaps u should consider only the principal values because
-1<cos(theta)<1
• Jan 24th 2010, 01:56 AM
Jukodan
Thank you gentlemen, it seems I made this problem harder for myself than it really is. Much much thanks. :)

And also yes R = -1 but since it forms a circle I figured the radius is simply 1, hence why I set cos2theta equal to 1
• Jan 24th 2010, 04:38 PM
mr fantastic
Quote:

Originally Posted by Jukodan
I recently came across this problem and would like some clarification as to why one of the answers to theta cannot be $\displaystyle \frac{\pi }{2}$

The problem asks to sketch two polar curves and find where they intersect, and then present those points in rectangular coordinates. I can easily see from the graph that the points are (1,0) and (-1,0). The two polar functions are,

$\displaystyle r= \cos ^{2}\theta$ and r = -1.

I set them equal to each other

$\displaystyle \cos ^{2}\theta= 1$

$\displaystyle \frac{1}{2}\left ( 1+\cos 2\theta \right )= 1$

$\displaystyle 1+\cos 2\theta = 2$

$\displaystyle \cos 2\theta = 1$

So theta should be...$\displaystyle 0,\pi,-\pi,\frac{-\pi}{2},\frac{\pi}{2}$

Yet the graph tells me only pi and -pi can be the answer. I'm confused...