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

1 Answer
Oct 5, 2016

#(3sin(theta)+cos(theta))/(2cos^2(theta)-2cos(theta)sin(theta))=r#

Explanation:

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

#r^2=x^2+y^2#

Make the necessary substitutions

#3rsin(theta)=2(rcos(theta))^2-2rcos(theta)rsin(theta)-rcos(theta)#

Simplify

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

Add #rcos(theta)# to both sides

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

Factor out #r# and #r^2#

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

Isolated #r^2#

#(r(3sin(theta)+cos(theta)))/(2cos^2(theta)-2cos(theta)sin(theta))=(r^2cancel(2cos^2(theta)-2cos(theta)sin(theta)))/cancel(2cos^2(theta)-2cos(theta)sin(theta))#

#(r(3sin(theta)+cos(theta)))/(2cos^2(theta)-2cos(theta)sin(theta))=r^2#

Gather #r# to the right hand side

#((cancelr(3sin(theta)+cos(theta)))/(2cos^2(theta)-2cos(theta)sin(theta)))/cancelr=r^cancel2/cancelr#

Simplify

#(3sin(theta)+cos(theta))/(2cos^2(theta)-2cos(theta)sin(theta))=r#

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