Problem: Volume of a cubic box is $V = L^3$. How are the relative rates of change of $V$ and $L$ related?
This problem seems really simple, but I can't understand the concept of a relative rate of change. Here are my workings:
Original equation
$$V = L^3$$
I write it in form of differential equation
$$\partial V = 3L^2 \partial L$$
then I divide the 2nd line by the 1st
$$\frac {\partial V}{V} = \frac{3L^2 \partial L}{L^3}$$
$$\frac {\partial V}{V} = 3 \frac{\partial L}{L}$$
Expression of the form $\frac {\partial f(x)}{f(x)}$ is called a relative rate of change. And it can be thought of as percentage change. So as I see it by knowing the percentage change in $L$ we can work out the respective percentage change in $V$, right? Wrong. I tried to make sense of it by putting values into the equation but no success. E.g. we increase $L$ from 2 to 3 (by 50%) thus according to the last formula we should have a respective increase in $V$ of 150% (50% times 3) which is not true ($\frac{3^3-2^3}{2^3}$ is a 237.5% increase).
Can you help me out? I'm definitely missing something either in computation or more likely in understanding the concept.
Best Answer
You have set $\frac{\delta L }{L} =$ 50 percent which is quite large, as the equation holds in the limit as $\delta L \rightarrow 0$.
That is, calculating the relative change in volume as you did $$ \frac{\delta V}{V} = \frac{(L + \delta L)^3 - L^3}{L^3} = 3\frac{\delta L}{L} + 3\frac{(\delta L)^2}{L^2} = 3\frac{\delta L}{L}\left( 1 + \frac{\delta L}{L} \right) $$ So as you make $\delta L$ smaller, the approximation to $\frac{\partial V}{V}$ improves. In the limit of $\delta L \rightarrow 0$, we have equality (invoking a large dose of handwavium).