# What is the Cartesian form of r-theta = -2cos^3theta-cot^2theta ?

Oct 21, 2016

The presence of $\theta$ forces a split into three equations. Please see the explanation for the three equations.

#### Explanation:

Because the cotangent function has a divide by zero problem when $\theta = 0 \mathmr{and} \pi$, this translates to the Cartesian restriction $y \ne 0$

Add $\theta$ to both sides:

$r = \theta - 2 {\cos}^{3} \left(\theta\right) - {\cot}^{2} \left(\theta\right)$

Write ${\cot}^{2} \left(\theta\right)$ as $\frac{{\cos}^{2} \left(\theta\right)}{{\sin}^{2} \left(\theta\right)}$

$r = \theta - 2 {\cos}^{3} \left(\theta\right) - \frac{{\cos}^{2} \left(\theta\right)}{{\sin}^{2} \left(\theta\right)}$

Multiply both sides by ${r}^{5} {\sin}^{2} \left(\theta\right)$:

${r}^{6} {\sin}^{2} \left(\theta\right) = \theta {r}^{3} \left({r}^{2} {\sin}^{2} \left(\theta\right)\right) - 2 {r}^{3} {\cos}^{3} \left(\theta\right) {r}^{2} \left({\sin}^{2} \left(\theta\right)\right) - {r}^{5} {\cos}^{2} \left(\theta\right)$

Substitute y for $r \sin \left(\theta\right)$ and x for $r \cos \left(\theta\right)$

r^4y^2 = thetar^3y^2 - 2x^3y^2 - r^3x^2; y!=0

Use ${\left({x}^{2} + {y}^{2}\right)}^{\frac{1}{2}} = r$ to make substitutions in accordance with the power of r:

(x^2 + y^2)^(2)y^2 = theta(x^2 + y^2)^(3/2)y^2 - 2x^3y^2 - (x^2 + y^2)^(3/2)x^2; y!=0

Collect like terms:

(x^2 + y^2)^(2)y^2 = (x^2 + y^2)^(3/2)(thetay^2 - x^2) - 2x^3y^2; y!=0

The substitution for $\theta$ breaks the above into 3 equations:

(x^2 + y^2)^(2)y^2 = (x^2 + y^2)^(3/2)(tan^-1(y/x)y^2 - x^2) - 2x^3y^2; y > 0 and x > 0

(x^2 + y^2)^(2)y^2 = (x^2 + y^2)^(3/2)(tan^-1(y/x) + pi)y^2 - x^2) - 2x^3y^2; y != 0 and x < 0

(x^2 + y^2)^(2)y^2 = (x^2 + y^2)^(3/2)((tan^-1(y/x) + 2pi)y^2 - x^2) - 2x^3y^2; y < 0 and x > 0