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]}
Explanation:
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) sox=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 usingr=1+-sqrt(1-2x)
The graph has horizontal tangent in
Alternatively one can implicitly differentiate
using that