How do you convert #y= 3x^2 -2x-xy^2 # into a polar equation?

1 Answer
Roella W. · Stefan V.
Jan 8, 2016

By substitution

Explanation:

Use the substitutions

#x= rcos(theta)" "# and #" "y= rsin(theta)#

Then

#rsin(theta) = 3r^2cos^2(theta) - 2rcos(theta) - rcos(theta)r^2sin^2(theta)#

#r^3cos(theta)sin^2(theta) -3r^2cos^2(theta)+2rcos(theta) +rsin(theta) = 0#

We know #sin^2(theta) + cos^2(theta) = 1#

#:. sin^2(theta) =1- cos^2(theta) #

Substituting into the equation above gives

#r^3cos(theta)(1-cos^2(theta))- 3r^2cos^2(theta)+2rcos(theta) +rsin(theta) = 0#

#r^3cos(theta) -r^3cos^3(theta) -3r^2cos^2(theta) +2rcos(theta) +rsin(theta) = 0#

#r^3cos^3(theta)+3r^2cos^2(theta)-(r^3+2r)cos(theta) -rsin(theta)=0#

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