Calculating the uncertainty of $r$, when $1/r^2$ is used I have tried to search through the forums and the Internet, however I haven't been able to find a source I was able to understand fully. 
I have a distance $r$, with an uncertainty of $\mathrm{±\ 0.003\ m}$, however in order to linearise my graph, I need to divide $1$ with $r$ squared ($1/r^2$), what would the final uncertainty for $r$ be? - Furthermore, what would the unit be? Would that still be $m$? or $m^2$?
 A: This problem is generally called propagation of error / uncertainty. You can google it and find a lot of info (I'd also recommend Taylor's "Introduction to Error Analysis"). Here's the gist of it, though. If you have independent measured quantities $x, y, z, \ldots$ with errors $
\sigma_x, \sigma_y, \sigma_z, \ldots$, then the error on a function $f(x,y,z,\ldots)$ is
$$
\sigma_f = \sqrt{ \left( \frac{\partial f}{\partial x} \right)^2 \sigma_x^2 + \left( \frac{\partial f}{\partial y} \right)^2 \sigma_y^2 + \left( \frac{\partial f}{\partial z} \right)^2 \sigma_z^2 + \cdots}
$$
As for what the unit would be, remember that you can always heuristically think of the error as a $\pm$. Ask yourself this: does the quantity $5 m^{-2} \pm 3m$, for instance, make any sense?
A: jwimberley's answer is comprehensive, and it's the one you should accept, but let me work it through for you to show how to calculate your error bars.
Your function is:
$$ y = r^{-2} $$
so:
$$\begin{align}
 \frac{dy}{dr} &= -2 r^{-3} \\
               &= -2 y \frac{1}{r}
\end{align}$$
and therefore jwimberley's expression for $\sigma_y$ gives:
$$ \sigma_y = -2 y \frac{\sigma_x}{r} $$
In your experiment $\sigma_r$ is a constant 0.003m, but the error in $1/r^2$ is not constant but depends on $r$. Take $r = 0.3$, so $1/r^2 = 0.09$, the the error bar for this point is:
$$\begin{align}
 \sigma_y &= -2 y \frac{\sigma_r}{r} \\
          &= -2 \times 0.09 \times \frac{0.003}{0.3} \\
          &= 0.0018
\end{align}$$
You might notice that the fractional error in $r$ is $1\%$, while the fractional error in $y$ is $2\%$. This isn't by chance. If you have a formula:
$$ y = x^n $$
Then for any value of $n$ if the fractional error in $x$ is $p_x$ the fractional error in $y$ is $np_x$. A quick play with the equation jwimberley gave should convince you that this is the case.
A: $y=1/r^2$ from which it follows $dy=-2/r^3 dr$.
Using the approximations $\Delta y\approx dy$ and $\Delta r\approx dr$:
$$\Delta y\approx-\frac{2}{r^3}\Delta r.$$
Now just plug in your values.
For example if $r=0.1\pm 0.003$m, then
$$\Delta y\approx-\frac{2}{(0.1\rm m)^3}0.003{\rm m}=-6.0{\rm m}^{-2}.$$
The negative result just means that y decreases as r increases, and vice-versa. If you like, you can incorporate that into expressing the uncertainty with $\mp$ instead of $\pm$:
$$y=100\mp 6{\rm m}^{-2}$$
however that is non-standard and not advisable, but included here for academic clarity.
Let's see how good this approximation is, in this particular case:
$$\frac{1}{(0.1+0.003)^2}=94.26...\mbox{ as compared to }100-6=94$$
$$\frac{1}{(0.1-0.003)^2}=106.28...\mbox{ as compared to }100+6=106$$
