Find all points of intersection

I'm asked to find all points of intersection between

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

$\displaystyle r=1+\sin{\theta}$.

I've graphed both polar equations. The first one looks like a heart rotated 90 degrees to the right. The second like a heart flipped vertically.

I know that I'm supposed to set the equations equal to each other:

$\displaystyle 1-\cos{\theta} = 1+\sin{\theta}$

$\displaystyle \cos{\theta} + \sin{\theta} = 0$

So, I need something that either makes both cos and sin zero, or makes them opposites of each other.

So, $\displaystyle \theta = \frac{3\pi}{4}$. But now that I have that, I'm not really sure how to proceed. Do I integrate one of the functions from 0 to $\displaystyle \frac{3\pi}{4}$?

Confused...

Re: Find all points of intersection

Quote:

Originally Posted by

**centenial**

$\displaystyle \cos{\theta} + \sin{\theta} = 0$

So, I need something that either makes both cos and sin zero, or makes them opposites of each other.

$\displaystyle \cos{\theta} + \sin{\theta} = 0$

$\displaystyle -\cos{\theta} = \sin{\theta} $

$\displaystyle -1 = \frac{\sin{\theta}}{\cos{\theta}} $

$\displaystyle -1 = \tan{\theta} $

Solve this? Then sub in for $\displaystyle r$

Re: Find all points of intersection

Quote:

Originally Posted by

**centenial** I'm asked to find all points of intersection between

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

$\displaystyle r=1+\sin{\theta}$.

I've graphed both polar equations. The first one looks like a heart rotated 90 degrees to the right. The second like a heart flipped vertically.

I know that I'm supposed to set the equations equal to each other:

$\displaystyle 1-\cos{\theta} = 1+\sin{\theta}$

$\displaystyle \cos{\theta} + \sin{\theta} = 0$

So, I need something that either makes both cos and sin zero, or makes them opposites of each other.

So, $\displaystyle \theta = \frac{3\pi}{4}$. But now that I have that, I'm not really sure how to proceed. Do I integrate one of the functions from 0 to $\displaystyle \frac{3\pi}{4}$?

Confused...

What has integration got to do with the question **as posted**.