How do you graph r=2-2costheta?

1 Answer
Nov 10, 2016

graph{(x^2+y^2+2x)^2-4x^2-4y^2=0 [-7.024, 5.46, -3.147, 3.097]}

(x^2+y^2)^2+4x(x^2+y^2)-4y^2=0

Explanation:

costheta=x/r

r=2-2x/r\ \ =>\ \ r^2+2x=2r\ \ =>\ \ r^4+4x^2+4xr^2-4r^2=0

r^2=x^2+y^2

(x^2+y^2)^2+cancel(4x^2)+4x(x^2+y^2)-cancel(4x^2)-4y^2=0

To drow the graph observe

  • the graph is simmetric over theta so over x axis
  • for theta=0, r=0 so the point (0;0)\ \ belongs to the graph
  • for theta=-pi, r=4 so the point (-4;0)\ \ belongs to the graph
  • for theta=+-pi/2, r=2 so the point (0;+-2)\ \ belongs to the graph
  • r^2-2r+2x=0, => r=1+-sqrt(1-2x) so x=1/2, r=1, y=sqrt(r^2-x^2)=+-sqrt3/2 is an extrem point
  • Using implicit differentiation on the curve equation with some calculus one can get that y'(x)=0 only when (x+1)r^2+2x^2=0 and using r=1+-sqrt(1-2x)

(x+1)(1+1-2x+-2sqrt(1-2x))+2x^2=0
2(x+1)(1-x+-sqrt(1-2x))+2x^2=0
(1-x^2)+-(x+1)sqrt(1-2x)+x^2=0
sqrt(1-2x)=1/(x+1)
(1-2x)(x+1)^2=1
x^2+2x+1-2x^3-4x^2-2x=1
2x^3+3x^2=0
x=-3/2 (=>r=3) or x=0

The graph has horizontal tangent in (0;0) and in (-3/2;+-3sqrt3/2)

Alternatively one can implicitly differentiate r^2+2x=2r
using that (r^2)'=2x+2yy' and r'=(x+yy')/r so
r(x+1)-x=(1-r)yy'\ \ \ =>\ \ \ y'=0 only when r=x/(x+1) that again is satisfied x=0 or x=-3/2