# Question #a86e3

Oct 13, 2017

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

#### Explanation:

Let's begin by displaying a graph of the original equation $y = \frac{x}{x + 1}$:

Multiply both sides of the equation by $x + 1$

$x y + y = x$

Subtract x from both sides:

$x y + y - x = 0$

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

$\left({r}^{2}\right) \cos \left(\theta\right) \sin \left(\theta\right) + r \sin \left(\theta\right) - r \cos \left(\theta\right) = 0$

We can divide both sides of the equation by r, because this will discard the trival root $r = 0$:

$r \cos \left(\theta\right) \sin \left(\theta\right) + \sin \left(\theta\right) - \cos \left(\theta\right) = 0$

Add $\cos \left(\theta\right) - \sin \left(\theta\right)$ to both sides:

$r \cos \left(\theta\right) \sin \left(\theta\right) = \cos \left(\theta\right) - \sin \left(\theta\right)$

Divide both sides by $\cos \left(\theta\right) \sin \left(\theta\right)$:

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

Please observe that the graphs are identical. This proves that the conversion has been done properly.

Oct 13, 2017

$r = \frac{1}{\sin \theta} - \frac{1}{\cos \theta}$

#### Explanation:

Let's start by reminding ourselves on how polar coordinate systems work:

We can describe any point $P$ on the plane using the distance $r$ from that point to the center $O$. Then, we provide the angle $\theta$ of the line $P O$ in accordance with the $x$ axis.

Now to find the two parameters $r$, $\theta$ given the "rectangular" coordinates $x$, $y$, we need to use some trigonometry.

As you can see in the diagram above, the $x$, $y$ coordinates of a point represents the length of the sides of a rectangle drawn through that point - thus "rectangular" coordinates vs "polar" coordinates.

In a right-angled triangle with legs $x$, $y$, a hypotenuse $r$ and the angle adjacent to $x$ as $\theta$, we know that $\cos \theta = \frac{x}{r}$ and $\sin \theta = \frac{y}{r}$.

Therefore, $x = r \cos \theta$ and $y = r \sin \theta$.

Now, we just plug it into $y = \frac{x}{x + 1}$ and simplify!

$r \sin \theta = \frac{r \cos \theta}{r \cos \theta + 1}$

$\sin \theta = \frac{\cos \theta}{r \cos \theta + 1}$

$\cos \theta = \sin \theta \left(r \cos \theta + 1\right)$

$r \cos \theta = \frac{\cos \theta}{\sin \theta} - 1$

In terms of $r$ we have:
$r = \frac{1}{\sin \theta} - \frac{1}{\cos \theta}$

Therefore the equation is $r = \frac{1}{\sin \theta} - \frac{1}{\cos \theta}$