How do you convert #y=x-2y+xy^2 # into a polar equation?

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Answer 1 · 2018-05-21T16:06:17

#r=(costheta-sintheta)/((2-costheta)sinthetacostheta)#

#y=x-2y+xy^2#
#x=rcostheta#
#y=rsintheta#
Thus,
#rsintheta=rcostheta-2rcosthetaxxrsintheta+rcosthetaxx(rsintheta)^2#

#rsintheta=rcostheta-2r^2sinthetacostheta+r^2sin^2thetacostheta#

#rsintheta-rcostheta=r^2(-2sinthetacostheta+sin^2thetacostheta)#

#r(sintheta-costheta)=r^2sinthetacostheta(-2+costheta)#

#r(sintheta-costheta)+r^2sinthetacostheta(2-costheta)#

#r(sintheta-costheta+rsinthetacostheta(2-costheta))=0#

#r=0#

#sintheta-costheta+rsinthetacostheta(2-costheta)=0#

#rsinthetacostheta(2-costheta)=costheta-sintheta#
#r=(costheta-sintheta)/((2-costheta)sinthetacostheta)#