Question

How do you graph `r=4+6costheta`?

Answer 1

For making this cardioid-like graph, see explanation. The graph that is inserted is for the equivalent cartesian form `x^2+y^2-6x-4sqrt(x^2+y^2)=0`

Explanation:

`r=4+6 cos theta>=0 to cos theta >=-2/3 to`

`theta in (-131.81^o, 131.81^o)`, nearly.

As `cos(-theta)=costheta`, the graph is symmetrical with respect to

the initial line `theta = 0`.

The period of `r(theta)` is `2pi`.

So, a Table for `theta in (1. 131.61^o)` is sufficient, for making the

graph..

`(r, theta): (10, 0^o) (7, 60^o) (4, 90^o) (1, 120^o) (0, 131.81^o)`

The other half can be made using symmetry about the axis

`theta=0`.
graph{x^2 + y^2- 6x - 4sqrt (x^2+y^2) =0 [-10, 10, -5, 5]}