Question #a86e3

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Question #a86e3

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Answer 1 · 2017-10-13T18:15:10

$r = \frac{cos (θ) - sin (θ)}{cos (θ) sin (θ)}$ $r =(cos(theta)- sin(theta))/(cos(theta)sin(theta))$

Explanation:

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

![www.desmos.com/calculator]( useruploads.socratic.org )

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

$x y + y = x$ $xy + y = x$

Subtract x from both sides:

$x y + y - x = 0$ $xy + y - x = 0$

Substitute $r cos (θ)$ $rcos(theta)$ for $x$ $x$ and $r sin (θ)$ $rsin(theta)$ for $y$ $y$ :

$(r^{2}) cos (θ) sin (θ) + r sin (θ) - r cos (θ) = 0$ $(r^2)cos(theta)sin(theta) + rsin(theta) - rcos(theta) = 0$

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

$r cos (θ) sin (θ) + sin (θ) - cos (θ) = 0$ $rcos(theta)sin(theta) + sin(theta) - cos(theta) = 0$

Add $cos (θ) - sin (θ)$ $cos(theta)-sin(theta)$ to both sides:

$r cos (θ) sin (θ) = cos (θ) - sin (θ)$ $rcos(theta)sin(theta) =cos(theta)- sin(theta)$

Divide both sides by $cos (θ) sin (θ)$ $cos(theta)sin(theta)$ :

$r = \frac{cos (θ) - sin (θ)}{cos (θ) sin (θ)}$ $r =(cos(theta)- sin(theta))/(cos(theta)sin(theta))$

![www.desmos.com/calculator]( useruploads.socratic.org )

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

Answer 2 · 2017-10-13T18:45:29

$r = \frac{1}{sin θ} - \frac{1}{cos θ}$ $r=1/(sintheta)-1/(costheta)$

Explanation:

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

![ $tutorial.math.lamar.edu$ )

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

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

As you can see in the diagram above, the $x$ $x$ , $y$ $y$ 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 $x$ $x$ , $y$ $y$ , a hypotenuse $r$ $r$ and the angle adjacent to $x$ $x$ as $θ$ $theta$ , we know that $cos θ = \frac{x}{r}$ $costheta=x/r$ and $sin θ = \frac{y}{r}$ $sintheta=y/r$ .

Therefore, $x = r cos θ$ $x=rcostheta$ and $y = r sin θ$ $y=rsintheta$ .

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

$r sin θ = \frac{r cos θ}{r cos θ + 1}$ $rsintheta=(rcostheta)/(rcostheta+1)$

$sin θ = \frac{cos θ}{r cos θ + 1}$ $sintheta=(costheta)/(rcostheta+1)$

$cos θ = sin θ (r cos θ + 1)$ $costheta=sintheta(rcostheta+1)$

$r cos θ = \frac{cos θ}{sin θ} - 1$ $rcostheta=(costheta)/(sintheta)-1$

In terms of $r$ $r$ we have:
$r = \frac{1}{sin θ} - \frac{1}{cos θ}$ $r=1/(sintheta)-1/(costheta)$

Therefore the equation is $r = \frac{1}{sin θ} - \frac{1}{cos θ}$ $r=1/(sintheta)-1/(costheta)$

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Question #a86e3

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