How do you graph #r=6-5costheta#?

1 Answer
Jul 4, 2018

See graph and explanation.

Explanation:

Use #( x, y ) = r ( cos theta, sin theta)#, to get the Cartesian form

#( x^2 + y^2) = 6 sqrt( x^2 + y^2 ) - 5 x #

Graph:
graph{x^2+y^2 -6(x^2+y^2)^0.5+5x=0[-16 16 -8 8]}

Interestingly, limacons like this are from multi-loop forms

created by #r = 6 - 5 cos n theta#, n = 1.

Graph of #r = 6 - 5 cos 3 theta#, with n = 3:
graph{ (x^2+y^2)^2 -6(x^2+y^2)^1.5 +5(x^3-3xy^2)=0[-22 22 -11 11]}