Rewrite the rectangular equation to a polar equation, y=2x^2, what would r(theta) equal?

y=2x^2

r(theta)= ?

1 Answer
Apr 8, 2018

r(theta)=sin(theta)/(2cos^2(theta))r(θ)=sin(θ)2cos2(θ)

Explanation:

y=2x^2y=2x2

2y^2+y=2x^2+2y^22y2+y=2x2+2y2

2(r(theta)sin(theta))^2+r(theta)sin(theta)=2(r(theta))^22(r(θ)sin(θ))2+r(θ)sin(θ)=2(r(θ))2

2(r(theta))^2sin^2(theta)+r(theta)sin(theta)=2(r(theta))^22(r(θ))2sin2(θ)+r(θ)sin(θ)=2(r(θ))2

2r(theta)sin^2(theta)+sin(theta)=2r(theta)2r(θ)sin2(θ)+sin(θ)=2r(θ)

2r(theta)sin^2(theta)-2r(theta)=-sin(theta)2r(θ)sin2(θ)2r(θ)=sin(θ)

r(theta)(2sin^2(theta)-2)=-sin(theta)r(θ)(2sin2(θ)2)=sin(θ)

r(theta)=-sin(theta)/(2sin^2(theta)-2)r(θ)=sin(θ)2sin2(θ)2

r(theta)=sin(theta)/(2-2sin^2(theta))r(θ)=sin(θ)22sin2(θ)

r(theta)=sin(theta)/(2cos^2(theta))r(θ)=sin(θ)2cos2(θ)

(note: i think you can divide both sides by r(theta)r(θ) in step 5 because the function r(theta)r(θ) passes through the origin (r=0)(r=0) anyway)