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My Mathematics class has been working with error propagation, and I encountered a practice problem that I can't solve.

The problem is:

I tried using the general formula for error propagation that I was given in class :enter image description here , where c is a function of a and b. However, I keep getting a huge absolute uncertainty of 1000 or so. I would really appreciate if someone could explain how to tackle this. Much thanks.

edit:This is what I'm getting at the moment: enter image description here

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  • $\begingroup$ $\frac {\partial c}{\partial \theta}=2A \sin \theta$. It should not be squared under the square root. Only $\sigma_\theta$ should be. That is the factor $\sqrt{200}$ you are out compared to my calculation. $\endgroup$ – Ross Millikan Jun 24 '17 at 14:20
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One check you can do is make the errors go in the direction that will increase $y$ and see how much $y$ changes. The nominal value of $y$ is $100 \cos 90^\circ=0$ If we try to drive $y$ positive, we get $101 \cos 80^\circ \approx 17.54$ so we should be finding an error of about that size.

Your problem is in the $\sigma_\theta$ term. $\theta$ needs to be measured in radians when you take the derivative of the $\cos$ term and in the $\sigma^2_\theta$ term. $\frac{\partial y}{\partial \theta}=2A\sin (2\theta)=200$ $\sigma^2_\theta=\left(\frac {5\pi}{180}\right)^2\approx 0.0076$, so $\sigma_y^2=200\cdot \left(\frac {5\pi}{180}\right)^2\approx 1.52$ The disagreement with the first paragraph is because $y$ is flat at zero at the nominal value. Your equation assumes that things are linear, but here that is a bad approximation.

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  • $\begingroup$ I followed the formula that I mentioned above and got 17.46, Am I doing something wrong? I'll attach my final simplified form; I would really appreciate if you could let me know where I went wrong. $\endgroup$ – BillyBob1 Jun 24 '17 at 5:39

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