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

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Answer 1 · 2018-05-20T17:47:24

#color(blue)[r^2=(-costheta)/[sin^3theta-r^2*sin^3theta*cos^2theta-r^2costheta*sin^4theta]]#

Note that

#color(red)[y=r*sintheta]#

#color(red)[x=r*costheta]#

#y = x^2y-x/y^2 +xy^2#

#(r*sintheta)=(r*costheta)^2*(r*sintheta)-(r*costheta)/(r*sintheta)^2+(r*costheta)(r*sintheta)^2#

#rsintheta=r^3sintheta*cos^2theta-(costheta)/(rsin^2theta)+r^3costheta*sin^2theta#

#[rsintheta-r^3sintheta*cos^2theta-r^3costheta*sin^2theta]/1=-(costheta)/(rsin^2theta)#

#r^2sin^3theta-r^4*sin^3theta*cos^2theta-r^4costheta*sin^4theta=-costheta#

#color(blue)[r^2=(-costheta)/[sin^3theta-r^2*sin^3theta*cos^2theta-r^2costheta*sin^4theta]]#