A graduated cylinder is a vessel for measuring a volume of liquid using a ruled scale printed vertically on its side. It works because at a constant diameter, the volume can be directly related to depth by
\begin{equation*}
V = \frac{\pi d^2 h}{4} \text{,}
\end{equation*}
where \(d\) is the diameter and \(h\) is the depth (or height) in the cylinder.
When designing a graduated cylinder for manufacture, what diameter should be chosen so that the ruled markings can be printed as millimetres, with each centimetre of depth corresponding to \(10\) cubic centimetres of volume? A variation in height corresponds to a variation in volume, so that two different depths \(h_1\) and \(h_2\) with corresponding volumes \(V_1\) and \(V_2\text{,}\) we have
\begin{align*}
\change{V} \amp = V_2 - V_1 \\
\amp = \frac{\pi d^2 h_2}{4} - \frac{\pi d^2 h_1}{4} \\
\amp = \frac{\pi d^2}{4} (h_2 - h_1) \\
\amp = \frac{\pi d^2}{4} \change{h} \text{.}
\end{align*}
If we turn this into a ratio \(\inlineslope{V}{h}\text{,}\) we can directly see how a variation in depth leads to a variation in volume.
\begin{equation*}
\slope{V}{h} = \frac{\pi d^2}{4} \text{.}
\end{equation*}
We can think of this ratio as describing variation in volume per unit variation in depth. This is because if \(\change{h} = 1\text{,}\) then
\begin{equation*}
\slope{V}{h} = \frac{\change{V}}{1} = \change{V} \text{.}
\end{equation*}
So whenever the depth changes by \(1\) centimetre, the volume will change by
\begin{equation*}
\frac{\pi d^2}{4}
\end{equation*}
cubic centimetres. But if the depth changes by \(20\) centimetres, then the volume will change by
\begin{equation*}
20 \cdot \frac{\pi d^2}{4} = 5 \pi d^2
\end{equation*}
cubic centimetres.
We can now answer the original question: the requirement that each centimetre of change in depth correspond to \(10\) cubic centimetres change in volume can be stated as
\begin{equation*}
\slope{V}{h} = 10
\end{equation*}
if was assume that both the diameter \(d\) and the depth \(h\) are measured in centimetres. Solving, we obtain
\begin{align*}
\frac{\pi d^2}{4} \amp = 10 \\
d^2 \amp = \frac{40}{\pi} \\
d \amp = 2 \sqrt{\frac{10}{\pi}} \text{,}
\end{align*}
where we discard the negative solution since we want diameter to be a positive quantity. So the diameter should be approximately 3.57 cm.