How do you convert 3y= 2x^2-2xy-x 3y=2x22xyx into a polar equation?

1 Answer
Oct 5, 2016

(3sin(theta)+cos(theta))/(2cos^2(theta)-2cos(theta)sin(theta))=r3sin(θ)+cos(θ)2cos2(θ)2cos(θ)sin(θ)=r

Explanation:

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

r^2=x^2+y^2r2=x2+y2

Make the necessary substitutions

3rsin(theta)=2(rcos(theta))^2-2rcos(theta)rsin(theta)-rcos(theta)3rsin(θ)=2(rcos(θ))22rcos(θ)rsin(θ)rcos(θ)

Simplify

3rsin(theta)=2r^2cos^2(theta)-2r^2cos(theta)sin(theta)-rcos(theta)3rsin(θ)=2r2cos2(θ)2r2cos(θ)sin(θ)rcos(θ)

Add rcos(theta)rcos(θ) to both sides

3rsin(theta)+rcos(theta)=2r^2cos^2(theta)-2r^2cos(theta)sin(theta)3rsin(θ)+rcos(θ)=2r2cos2(θ)2r2cos(θ)sin(θ)

Factor out rr and r^2r2

r(3sin(theta)+cos(theta))=r^2(2cos^2(theta)-2cos(theta)sin(theta))r(3sin(θ)+cos(θ))=r2(2cos2(θ)2cos(θ)sin(θ))

Isolated r^2r2

(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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