Uncertainty cannot be calculated?

Question

I'm doing an experiment on resonance. The phase difference between the driving force and the one oscillating is given by $$\varphi=\arcsin\left(\frac{y_1}{y_2}\right)$$ where $y_1$ and $y_2$ are measurements of voltages. At resonance $y_1=y_2$ so $φ=\frac{π}{2}$.

However, I'm trying to calculate the uncertainty of $\varphi$ , I need to take the partial derivative of $\varphi$ with respect to $y_1$ and then $y_2$. In both cases , I end up with something in the form of $$\frac{\partial\varphi}{\partial y_1}\sim\frac{1}{\sqrt{1-\left(\frac{y_1}{y_2}\right)^2}}$$ multiplied by some other constants which don't really matter. So at resonance, my denominator will be zero. Any thoughts?

Answer 1

The idea that you can use partial derivatives multiplied by the uncertainty in the independent variable(s) to estimates uncertainty in the dependent variable is just an approximation.

It is related to a Taylor series expansion with an implicit assumption that the terms involving the second derivatives and higher can be neglected.

Obviously, for a sinusoidal function this isn't going to be true near $\pi/2$ since the first derivative becomes small.

A better way to proceed in your case is to do a Monte-Carlo simulation, allow $y_1$ and $y_2$ to be generated at random values determined by their own probability distributions and uncertainties, and then calculate the consequent values of $\phi$ and build up a probability distribution of $\phi$ from which an uncertainty can be calculated.

Edit: in response to your edited function, you now have $$\frac{y_1}{y_2} = \sin \phi$$ Thus $$ \frac{\partial y_1}{\partial \phi} = y_2 \cos \phi$$ and the problem I mentioned above clearly applies at $\phi = \pi/2$.