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

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Answer 1 · 2018-05-02T10:38:32

#r^2(rcos^2theta+rcosthetasin^2theta-sintheta)=cotthetacsctheta#

For this we will use:
#x=rcostheta#
#y=rsinthetra#

#rsintheta=(rcostheta)^2-(rcostheta)/(rsintheta)^2+r^2costhetasin^2theta#

#rsintheta=r^2cos^2theta-(cotthetacsctheta)/r+r^2costhetasin^2theta#

#r^2sintheta=r^3cos^2theta-cotthetacsctheta+r^3costhetasin^2theta#

#r^3cos^2theta+r^3costhetasin^2theta-r^2sintheta=cotthetacsctheta#

#r^2(rcos^2theta+rcosthetasin^2theta-sintheta)=cotthetacsctheta#

This cannot be simplified further and has to be left as an implicit equation.