homework and exercises - The approximate uncertainty in $r$

The volume of a cylinder is given by the expression

$V=\pi r^2 h$

The uncertainties for $V$ and $h$ are as shown below

$\begin{align} V&\pm 7\%\\ h&\pm 3\% \end{align}$

What is the uncertainty in $r$?

Now, the obvious answer would be $2\%$, from $$\frac{dV}{V}=\frac{dh}{h}+2\frac{dr}{r}$$

However, rearranging to $r^2=\frac{V}{\pi h}$ gives $$2\frac{dr}{r}=\frac{dV}{V}+\frac{dh}{h}$$ which gives a different answer of $5\%$. Thus, by simply rearranging the formula, we get different values for uncertainty in $r$.

How do you explain this?

(The mark scheme lists $5\%$ as the correct answer)

Answer

You're confusing independent and dependent variables. When you propogate from uncertainties in the $x_{i}$ to some $f(x_{1},x_{2}...)$, the formula $\delta f(x_{1}...)=\sum \left|\frac{\partial f}{\partial x_{i}}\right|\delta x_{i}$ assumes that each of the $x_{i}$ is an independently measured variable and that $f$ is a dependent variable to be calculated from the $x_{i}$.

In the example you give, you have two independent measurements of $V$ and $h$ and are expected to calculate the uncertainty in $r$. Well, to use the above formula, you need to write $r$ as a dependent variable of $V$ and $h$. Therefore, it's only correct to solve for $r$ first, and then calculate the uncertainty.