Question

How do you graph `r=1+2costheta`?

Answer 1

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]}