Find cartesian equivalent of polar eq. r= 1 cos theta. and Find the length of cardioid r = 1 - cos theta?

Answer 1

For r = cos theta:

At first ignore complications like r = 0 or r negative. Then you can say that r = cos theta is sqrt(x^2 + y^2) = x / sqrt(x^2 + y^2) or x^2 + y^2 = x which you can manipulate to x^2 - x + 1/4 + y^2 = 1/4 or

(x - 1/2)^2 + y^2 = (1/2)^2

which is a circle with center (1/2, 0) and radius 1/2. Then check whether r = 0 or r < 0 add anything else to that (my guess would be no).

For length of r = 1 - cos theta.

Here r varies in [0, 2] so there is no r < 0 case to worry about. The length will just be the integral from theta = 0 to 2 pi of ds / d(theta) where ds^2 = (r d(theta))^2 + (dr)^2 or ds / d(theta) = ((1 - cos theta)^2 + (sin theta)^2) (d(theta))^2.

Answer 2

r=cos(theta)
r^2 = rcos(theta)
x^2+y^2=x (circle with radius 1/4 center:(1/2,0))

Using the arc length formula, I get the integral (from x=0 to 2pi (2-2costheta)^(1/2). Using the half-angle formulas for sine, and integrating, -cos (pi) +cos0 = 2

Answer 3

Solve the equation pictorially

Answer 4

Why would I want ot do that?

Answer 5

is that a real question! My god