How can I find the uncertainty? I need to use this formula:

The general formula for uncertanties is:

$\sigma_q = \sigma_q (a,b,c,...)$ and $\mathrm{Cov}(a,b,c,...)=0$ $\,\,$then $$\sigma_q ^2 = (\frac{\partial c}{\partial a}\sigma_a)^2+(\frac{\partial c}{\partial b}\sigma_b)^2+\,\,...$$

Is this correct: (?)

$$\sigma_{\tan(\theta)}^2 = (\frac{\partial {\tan(\theta)}}{\partial \theta}\sigma_ \theta )^2 \Rightarrow$$

$$\sigma_{\tan(\theta)} = \sec^2(\theta)\times1.2\Rightarrow$$

$$\sigma_{\tan(\theta)}=1.17 \times 1.2 = 1.4 \Rightarrow$$

$$\tan(\theta) = \tan(59.3)^{\circ} \pm \sigma_{\tan(\theta)} = 1.68 \pm 1.4 $$

thank you.

  • 1
    $\begingroup$ Recall, that derivative only works for radian measure. You need to include the conversion factor. $$\frac{\partial\tan(\theta\cdot\tfrac{\pi}{180})}{\partial \theta} = \tfrac{\pi}{180}\sec^2(\theta\cdot\tfrac{\pi}{180})$$ $\endgroup$ – Graham Kemp Aug 11 '15 at 4:05
  • $\begingroup$ @GrahamKemp Thank you!!! $\endgroup$ – Voyager Aug 11 '15 at 4:09

As $Harish has already shown, the maximum uncertainty is much smaller than what you have found. Given that you are using the general formula for error propagation (using partial derivatives, etc.) I'm guessing you are not looking for the maximum uncertainty but for the "one-sigma" uncertainty. So your answer should be a smaller than what he found. Also, he has accidentally found twice the maximum error since he subtracted the lower bound from the top bound (the full range) instead of quoting half the range.

Your problem is that you are not paying careful attention to units. In your second last line you say

$\sigma_{\tan{(\theta)}} = 1.17 \times 1.2$

But notice that your "1.2" came from the original uncertainty in $\theta$ so it is really $1.2^\circ$. So this means your answer for the uncertainty in $\tan{\theta}$ is in degrees which can't be right since $\tan{\theta}$ should be a dimensionless number. You need to put $\sigma_\theta$ into radians. This will give you a far more reasonable answer.


Notice, 1. Taking positive sign $$(\tan\theta)_{\text{max}}=\tan (59.3+1.2)^\circ=\tan 60.5^\circ$$

  1. Taking negative sign $$(\tan\theta)_{\text{min}}=\tan (59.3-1.2)^\circ=\tan 58.1^\circ$$

Hence, the maximum possible uncertainty in the value of $\tan \theta$ $$=(\tan60.5^\circ-\tan 58.1^\circ)\approx 0.160926831$$

  • $\begingroup$ Thank you. So where is my mistake above? $\endgroup$ – Voyager Aug 11 '15 at 4:00

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