Question

What the is the polar form of `y = y^3-xy+x^2`?

Answer 1

`r = (Sin(2 t)-cos(2t)-1 pm sqrt(8 - 6 Cos(2 t) + 2 Cos(4 t) - 2 Sin(2 t) -Sin(4 t)))/(3 Sin(t) - Sin(3 t))`

Explanation:

With the pass equations

`{(x=rsintheta),(y=rcostheta):}`

`y = y^3-xy+x^2 ->rsintheta=r^3sin^3theta-r^2sinthetacostheta+r^2cos^2theta` or

`sintheta=r^2sin^3theta-r(sinthetacostheta-cos^2theta)`

solving for `r` we have

`r = (Sin(2 t)-cos(2t)-1 pm sqrt(8 - 6 Cos(2 t) + 2 Cos(4 t) - 2 Sin(2 t) -Sin(4 t)))/(3 Sin(t) - Sin(3 t))`