Question

How do you graph `r=2-2costheta`?

Answer 1

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`