Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2023 / Paper 1 / 14 / b / Solution

The stated total differential is that of $\pi r^{2}h$, so $V=\pi r^{2}h+C$. The zero-volume condition gives $C=0$. For positive $r,h$, divide the total differential by this volume to obtainFor the specified small changes, $dV/V\approx 2(0.01)+0.02=0.04$, so the volume increases by approximately $4\%$. The fractional expression requires $V>0$; at a degenerate zero-volume state it is undefined. For finite changes, the exact multiplier is $(1.01{)}^2(1.02)$, whose difference from $1$ is $4.0502\%$; the total differential gives its first-order approximation.