How do you graph #r=6costheta#?

Question

4868 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-22T14:41:29

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

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]}