Question

How do you graph `r=6costheta`?

Answer 1

The Socratic graph for the cartesian equation
`r^2=(x^2+y^2)=6r cos theta=6x` is inserted.

Explanation:

This is the equation of the circle through the pole O( r = 0 ), with

center C at ( 3, 0 ). If P is any point ( r. theta ) and OA the diameter

through O.

`OP = r = OA cos theta = 6 cos theta.`.

graph{(x^2+y^2-6x)((x-3)^2+y^2-.005)=0 [0, 14, -3.5, 3.5]}

This circle ia a member of the family of curves

{`r = k cos (ntheta)`}, where k and n scale factors for r and

`theta`.

I let readers to see the grandeur in the graphs of the members

`r = cos (theta/3) and r = cos (theta/5)`.

Graph of `r = cos (theta/3)`:
graph{x - (x^2+y^2)(4(x^2+y^2)-3)=0[-2 2 -1 1]}

Graph of `r = cos (theta/5)`:
graph{x - (x^2+y^2)(16(x^2+y^2)^2-20(x^2+y^2)+5)=0[-2 2 -1 1]}