# What the is the polar form of y = x^2y-x/y^2 +xy^2 ?

May 20, 2018

$\textcolor{b l u e}{{r}^{2} = \frac{- \cos \theta}{{\sin}^{3} \theta - {r}^{2} \cdot {\sin}^{3} \theta \cdot {\cos}^{2} \theta - {r}^{2} \cos \theta \cdot {\sin}^{4} \theta}}$

#### Explanation:

Note that

$\textcolor{red}{y = r \cdot \sin \theta}$

$\textcolor{red}{x = r \cdot \cos \theta}$

$y = {x}^{2} y - \frac{x}{y} ^ 2 + x {y}^{2}$

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

$r \sin \theta = {r}^{3} \sin \theta \cdot {\cos}^{2} \theta - \frac{\cos \theta}{r {\sin}^{2} \theta} + {r}^{3} \cos \theta \cdot {\sin}^{2} \theta$

$\frac{r \sin \theta - {r}^{3} \sin \theta \cdot {\cos}^{2} \theta - {r}^{3} \cos \theta \cdot {\sin}^{2} \theta}{1} = - \frac{\cos \theta}{r {\sin}^{2} \theta}$

${r}^{2} {\sin}^{3} \theta - {r}^{4} \cdot {\sin}^{3} \theta \cdot {\cos}^{2} \theta - {r}^{4} \cos \theta \cdot {\sin}^{4} \theta = - \cos \theta$

$\textcolor{b l u e}{{r}^{2} = \frac{- \cos \theta}{{\sin}^{3} \theta - {r}^{2} \cdot {\sin}^{3} \theta \cdot {\cos}^{2} \theta - {r}^{2} \cos \theta \cdot {\sin}^{4} \theta}}$