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

## y=2x^2 r(theta)= ?

Apr 8, 2018

$r \left(\theta\right) = \sin \frac{\theta}{2 {\cos}^{2} \left(\theta\right)}$

#### Explanation:

$y = 2 {x}^{2}$

$2 {y}^{2} + y = 2 {x}^{2} + 2 {y}^{2}$

$2 {\left(r \left(\theta\right) \sin \left(\theta\right)\right)}^{2} + r \left(\theta\right) \sin \left(\theta\right) = 2 {\left(r \left(\theta\right)\right)}^{2}$

$2 {\left(r \left(\theta\right)\right)}^{2} {\sin}^{2} \left(\theta\right) + r \left(\theta\right) \sin \left(\theta\right) = 2 {\left(r \left(\theta\right)\right)}^{2}$

$2 r \left(\theta\right) {\sin}^{2} \left(\theta\right) + \sin \left(\theta\right) = 2 r \left(\theta\right)$

$2 r \left(\theta\right) {\sin}^{2} \left(\theta\right) - 2 r \left(\theta\right) = - \sin \left(\theta\right)$

$r \left(\theta\right) \left(2 {\sin}^{2} \left(\theta\right) - 2\right) = - \sin \left(\theta\right)$

$r \left(\theta\right) = - \sin \frac{\theta}{2 {\sin}^{2} \left(\theta\right) - 2}$

$r \left(\theta\right) = \sin \frac{\theta}{2 - 2 {\sin}^{2} \left(\theta\right)}$

$r \left(\theta\right) = \sin \frac{\theta}{2 {\cos}^{2} \left(\theta\right)}$

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