# Finding points of intersection (polar graphs)

• Dec 14th 2009, 07:33 PM
VonNemo19
Finding points of intersection (polar graphs)
Hey guys. I'm having trouble with the rationale used in my text regarding finding points of intersection given two polar graphs. I understand that to find the points I must set the equations equal, but then, this sometimes doesn't reveal all points of intersection. If someone would be so kind as to explain

1. why doesn't solving the system yield all points of intersection;

2. What is the best way to find these points?

I've provided an example problem so that you guys can illustrate this.

Find all points of intersectionof the graphs of the equations

$r=1+\cos{\theta}$ and $r=1-\cos{\theta}$ .

Thanks.
• Dec 14th 2009, 07:57 PM
mr fantastic
Quote:

Originally Posted by VonNemo19
Hey guys. I'm having trouble with the rationale used in my text regarding finding points of intersection given two polar graphs. I understand that to find the points I must set the equations equal, but then, this sometimes doesn't reveal all points of intersection. If someone would be so kind as to explain

1. why doesn't solving the system yield all points of intersection;

2. What is the best way to find these points?

I've provided an example problem so that you guys can illustrate this.

Find all points of intersectionof the graphs of the equations

$r=1+\cos{\theta}$ and $r=1-\cos{\theta}$ .

Thanks.

Graphs should always be drawn because the solution to the simultaneous equations doe not always give all intersection points. This is because there is NOT a one-to-one correspondence between points and coordinates in the polar coordnate system.

For your example, the point (0, 0) corresponds to [0, -pi] on the first curve and [0, pi] on the second curve so this point will not show up when solving simultaneously (because the values of theta are different).
• Dec 14th 2009, 08:09 PM
VonNemo19
Quote:

Originally Posted by mr fantastic
Graphs should always be drawn because the solution to the simultaneous equations doe not always give all intersection points. This is because there is NOT a one-to-one correspondence between points and coordinates in the polar coordnate system.

Wow. I can't believe that this never entered my mind. Thanks.

Quote:

Originally Posted by mr fantastic
For your example, the point (0, 0) corresponds to [0, -pi] on the first curve and [0, pi] on the second curve so this point will not show up when solving simultaneously (because the values of theta are different).

So, did you first plot the graph, and then by inspection, discover this third point. Or did you find this point anlytically?
• Dec 15th 2009, 04:37 AM
scorpion007
Quote:

Originally Posted by mr fantastic
For your example, the point (0, 0) corresponds to [0, -pi] on the first curve and [0, pi] on the second curve so this point will not show up when solving simultaneously (because the values of theta are different).

doesn't (r, theta) = (0, -pi) => 0 = 2 on the second curve?
• Dec 15th 2009, 05:40 PM
mr fantastic
Quote:

Originally Posted by scorpion007
doesn't (r, theta) = (0, -pi) => 0 = 2 on the second curve?

Yes. Which is the whole point of my reply! The point defined by [0, -pi] is not on the second curve and therefore the origin, which is common to both curves, cannot be found using simultaneous equations.

Quote:

Originally Posted by VonNemo19
Wow. I can't believe that this never entered my mind. Thanks.

So, did you first plot the graph, and then by inspection, discover this third point. Or did you find this point anlytically?

It was obvious to me that the origin would be common to both curves.
• Dec 16th 2009, 03:18 AM
scorpion007
Sorry, I meant [0, pi] on the second curve, not [0, -pi]. That was a typo.

You said that the point (0,0) corresponds to [0, pi] on the second curve, but:

r = 1 - cos(theta)
0 = 1 - cos(pi)
0 = 1 - (-1) = 2 which is false.

Did I miss the point again? (Wondering)
• Dec 16th 2009, 03:38 AM
mr fantastic
Quote:

Originally Posted by scorpion007
Sorry, I meant [0, pi] on the second curve, not [0, -pi]. That was a typo.

You said that the point (0,0) corresponds to [0, pi] on the second curve, but:

r = 1 - cos(theta)
0 = 1 - cos(pi)
0 = 1 - (-1) = 2 which is false.

Did I miss the point again? (Wondering)

Correct, I made a typo. Here is the corrected text:

Quote:

Originally Posted by mr fantastic (corrected)
[snip]
For your example, the point (0, 0) corresponds to [0, pi] on the first curve and [0, 0] on the second curve so this point will not show up when solving simultaneously (because the values of theta are different).

My point remains unchanged.