Question

Question #a86e3

Answer 1

`r =(cos(theta)- sin(theta))/(cos(theta)sin(theta))`

Explanation:

Let's begin by displaying a graph of the original equation `y=x/(x+1)`:

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

`xy + y = x`

Subtract x from both sides:

`xy + y - x = 0`

Substitute `rcos(theta)` for `x` and `rsin(theta)` for `y`:

`(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`:

`rcos(theta)sin(theta) + sin(theta) - cos(theta) = 0`

Add `cos(theta)-sin(theta)` to both sides:

`rcos(theta)sin(theta) =cos(theta)- sin(theta)`

Divide both sides by `cos(theta)sin(theta)`:

`r =(cos(theta)- sin(theta))/(cos(theta)sin(theta))`

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

Answer 2

`r=1/(sintheta)-1/(costheta)`

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 `PO` 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 `costheta=x/r` and `sintheta=y/r`.

Therefore, `x=rcostheta` and `y=rsintheta`.

Now, we just plug it into `y=x/(x+1)` and simplify!

`rsintheta=(rcostheta)/(rcostheta+1)`

`sintheta=(costheta)/(rcostheta+1)`

`costheta=sintheta(rcostheta+1)`

`rcostheta=(costheta)/(sintheta)-1`

In terms of `r` we have:
`r=1/(sintheta)-1/(costheta)`

Therefore the equation is `r=1/(sintheta)-1/(costheta)`