What the is the polar form of #y = y/x^2+(2x-2)(y-5) #?

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Answer 1 · 2016-04-15T18:45:36

#r^4cos^2thetasin2theta-3/2 r^3costhetasin2theta-10r^3cos^3theta+10r^2cos^2theta+rsintheta =0#

#y=y/x^2 +(2x-2)(y-5)#

#y=(y+x^2(2x-2)(y-5))/x^2#

#x^2y=y+x^2(2x-2)(y-5)#

#x^2y=y+x^2(2xy-10x-2y+10)#

#x^2y=y+2x^3y-10x^3-2x^2y+10x^2#

#0=y+2x^3y-10x^3-2x^2y+10x^2-x^2y#

#0=2x^3y-3x^2y-10x^3+10x^2+y#

Now use the formula
#x=rcostheta and y=rsin theta#

#0=2r^3cos^3thetaxxrsintheta-3r^2cos^2thetaxxrsintheta-10r^3cos^3theta+10r^2cos^2theta+rsintheta#

#0=2r^4cos^3thetasintheta-3r^3cos^2thetasintheta-10r^3cos^3theta+10r^2cos^2theta+rsintheta#

#0=r^4cos^2theta(2sinthetacostheta)-3/2 r^3costheta(2sinthetacostheta)-10r^3cos^3theta+10r^2cos^2theta+rsintheta#

#0=r^4cos^2thetasin2theta-3/2 r^3costhetasin2theta-10r^3cos^3theta+10r^2cos^2theta+rsintheta#