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

1 Answer
Cesareo R.
Sep 28, 2016

#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))#

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