Question

How do you graph `r=2-4costheta`?

Answer 1

`theta in (pi/3, 5/3pi), t in (0, 6)`, symmetrical about the initial line and the shape is cardioid-like. The graph inserted is for the cartesian equivalent.

Explanation:

`r=2-4 cos theta >=0 to cos theta <=1/2 to theta in (pi/3, 5/3pi)`

`r(theta)` is periodic, with period `2pi`.

As `cos (-theta) = cos theta`, the graph is symmetrical about

`theta=0`.

So, a Table for `theta in (pi/3, pi)` is sufficient, for realizing the whole

graph.

`(r, theta): (0, pi/3) (2, pi/2) (4, 2/3pi) (6, pi)`

The inserted graph was obtained by using the cartesian equivalent

`x^2+y^2+4x=2sqrt(x^2+y^2)`

graph{x^2+y^2+4x-2sqrt(x^2+y^2)=0 [-10, 10, -5, 5]}