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

1 Answer
Alex P.
Apr 24, 2017

Use the conversion:
#x=r cos(theta)#
#y=r sin(theta)#

Now substitute:
#r sin(theta)=(rsin(theta))/(rcos(theta))+(rcos(theta))^2(2-rsin(theta))^2#

Expand the final term:
#r sin(theta)=(sin(theta))/(cos(theta))+r^2cos^2(theta)(4-2rsin(theta)+r^2sin^2(theta))#

Multiply by #cos(theta)#:
#r sin(theta)cos(theta)=sin(theta)+r^2cos^3(theta)(4-2rsin(theta)+r^2sin^2(theta))#

At this point it is not looking like the solution can be represented 'nicely'. You could try converting everything into #sin#

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