How do you graph r=1+2costheta?

1 Answer

The graph of this limacon is obtained using the cartesian form
(x^2+y^2)=sqrt(x^2+y^2)+2x. Look at the dimple, at the pole r = 0..

Explanation:

Use the conversion formula r(cos theta, sin theta) = (x, y), that

gives cos theta =x/r and r=sqrt(x^2+y^2).

Hence r=1+2costheta=>r^2=r+2rcostheta

or x^2+y^2=sqrt(x^2+y^2)+2x

Then use the Socratic graphic facility.

The period for r(theta) is 2pi. You can choose #theta in [0,

2pi]#.

graph{x^2+y^2-sqrt(x^2+y^2)-2x=0 [-5, 5, -2.5, 2.5]}