1. ## find intersetion points

how do you find the points of intersection algebraicly?
1-cos(theta) = 1+cos(theta)

2. Originally Posted by jeph
how do you find the points of intersection algebraicly?
1-cos(theta) = 1+cos(theta)
let theta be x

1 - cos(x) = 1 + cos(x)
=> 1 = 1 + 2cos(x) ............added cos(x) to both sides
=> 0 = 2cos(x) .................subtracted 1 form both sides.

so we want all x's such that

2cos(x) = 0
=> cos(x) = 0
=> x = cos^-1(0)
=> x = pi/2 + k*pi

3. dont the curves intersect at 0 also? do i need to plug in the pi/2 into just 1 of the equations to find the intersection?

4. Originally Posted by jeph
dont the curves intersect at 0 also? do i need to plug in the pi/2 into just 1 of the equations to find the intersection?
the curves DO NOT intersect at 0

remember cos(0) is 1

so 1 - cos(0) = 1 - 1 = 0

and 1 + cos(0) = 1 + 1 = 2

so 1 - cos(x) is not equal to 1 + cos(x) at x = 0

no, pluging in is not necessary, the intersection points are the values of x (which is what i called theta) i gave you.

below is a graph that show some of the intersections

the graph intersects for pi/2 and every multiple of pi before and after that

5. Hello, jeph!

Find the points of intersection: .1 + cosθ .= .1 - cosθ
I assume these are polar curves (two cardioids).
. . Jhevon was absolutely correct.

We have: .2·cosθ = 0 . . cosθ = 0 . . θ = ±½π

The points of intersection are: .(1, ½π), (1, -½π)

6. Originally Posted by Soroban
Hello, jeph!

I assume these are polar curves (two cardioids).
. . Jhevon was absolutely correct.

We have: .2·cosθ = 0 . . cosθ = 0 . . θ = ±½π

The points of intersection are: .(1, ½π), (1, -½π)

is there any particular reason you assumed they were polar curves, Soroban?

7. Hello, Jhevon!

is there any particular reason you assumed they were polar curves, Soroban?

Well, just their appearance, especially the use of θ (theta).

It suggested an "Area between two polar curves" problem . . .
In this case: .r .= .1 + cosθ .and .r .= .1 - cosθ
. . and theor intersections (polar coordinates) must be determined.