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

Mar 20, 2016

Set:

x=rcosθ

y=rsinθ

$\sqrt{{x}^{2} + {y}^{2}} - \arccos \left(\frac{x}{\sqrt{{x}^{2} + {y}^{2}}}\right) = - 2 {x}^{2} / \left({x}^{2} + {y}^{2}\right) - {x}^{3} / {y}^{3}$

#### Explanation:

According to the following picture:

Set:

x=rcosθ

y=rsinθ

So we have:

cosθ=x/r

sinθ=y/r

θ=arccos(x/r)=arcsin(y/r)

$r = \sqrt{{x}^{2} + {y}^{2}}$

The equation becomes:

r-θ=-2sin^2θ-cot^3θ

r-θ=-2sin^2θ-cos^3θ/sin^3θ

$\sqrt{{x}^{2} + {y}^{2}} - \arccos \left(\frac{x}{r}\right) = - 2 {x}^{2} / {r}^{2} - \frac{{x}^{3} / {r}^{3}}{{y}^{3} / {r}^{3}}$

$\sqrt{{x}^{2} + {y}^{2}} - \arccos \left(\frac{x}{r}\right) = - 2 {x}^{2} / {r}^{2} - {x}^{3} / {y}^{3}$

$\sqrt{{x}^{2} + {y}^{2}} - \arccos \left(\frac{x}{\sqrt{{x}^{2} + {y}^{2}}}\right) = - 2 {x}^{2} / {\sqrt{{x}^{2} + {y}^{2}}}^{2} - {x}^{3} / {y}^{3}$

$\sqrt{{x}^{2} + {y}^{2}} - \arccos \left(\frac{x}{\sqrt{{x}^{2} + {y}^{2}}}\right) = - 2 {x}^{2} / \left({x}^{2} + {y}^{2}\right) - {x}^{3} / {y}^{3}$