how do you find the points of intersection algebraicly?

1-cos(theta) = 1+cos(theta)

Printable View

- Apr 8th 2007, 01:00 PMjephfind intersetion points
how do you find the points of intersection algebraicly?

1-cos(theta) = 1+cos(theta) - Apr 8th 2007, 01:06 PMJhevon
- Apr 8th 2007, 02:01 PMjeph
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?

- Apr 8th 2007, 02:09 PMJhevon
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 - Apr 8th 2007, 02:12 PMSoroban
Hello, jeph!

Quote:

Find the points of intersection: .1 + cosθ .= .1 - cosθ

. . Jhevon was absolutely correct.

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

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

- Apr 8th 2007, 02:16 PMJhevon
- Apr 8th 2007, 06:36 PMSoroban
Hello, Jhevon!

Quote:

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.