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 $m^2$?

  • $\begingroup$ You mean what is the uncertainty in a function of r, given the uncertainty in r, no? The uncertainty as delta(y) is approxiamated by the differential,no? $\endgroup$
    – anna v
    Commented Oct 1, 2013 at 15:34
  • $\begingroup$ Hi, you can count the squares(numbers) in one dimension along the number line without bothering about area which is clearly not in the context. And a line and a curve have same dimensions. $\endgroup$
    – user28737
    Commented Oct 2, 2013 at 5:32

3 Answers 3


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?

  • $\begingroup$ I am sorry, but I believe my explanation was indeed too simplistic. I have a distance r (0.1 m through 0.4 m), where all the measurements have the uncertainty of ± 0.003m. I then, in order to linearise my graph (the number of counts (𝐶) in a time interval is proportional to the inverse square of the distance (𝑟) between the radioactive source and the Geiger counter). Therefore I have r, but I need it to be 1/r^2, in order to linearise my graph. Given r has an uncertainty of ±0.003, what would it be if I have 1/r^2, on the x-axis? $\endgroup$
    – mikkeljuhl
    Commented Oct 1, 2013 at 15:43
  • 1
    $\begingroup$ Your explanation was fine; I've shown you how to proceed. I think my answer is fairly clear; perhaps it needs some clarification but you've responded only a few minutes after I posted it so I'm inclined to think that you haven't sufficiently tried to understand it. I'm guessing you're stuck in a mindset that isn't helping you see how my answer solves your problem. Please read over my answer, take some time think about what you're trying to accomplish, and read over my answer again. $\endgroup$
    – jwimberley
    Commented Oct 1, 2013 at 15:49
  • 1
    $\begingroup$ Here is a hint: $1/r^2$ is a function of $r$. $\endgroup$
    – jwimberley
    Commented Oct 1, 2013 at 15:53

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:

$$ y = r^{-2} $$


$$\begin{align} \frac{dy}{dr} &= -2 r^{-3} \\ &= -2 y \frac{1}{r} \end{align}$$

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.


$y=1/r^2$ 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: $$\frac{1}{(0.1+0.003)^2}=94.26...\mbox{ as compared to }100-6=94$$ $$\frac{1}{(0.1-0.003)^2}=106.28...\mbox{ as compared to }100+6=106$$


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