By how much does the volume of a regular square pyramid

By how much does the volume of a regular square pyramid with all edges equal change when the edge grows from $5$ to $5.2$ length units?

Step 1: The height of the pyramid

Every edge, base and lateral, has the same length $a$. The distance from the centre of the square base to a base vertex is half the diagonal: $r = \frac{a\sqrt2}{2}$.

A lateral edge of length $a$ is the hypotenuse of a right triangle with legs $h$ and $r$: \[h^{2} = a^{2}-r^{2} = a^{2}-\frac{a^{2}}{2} = \frac{a^{2}}{2} \;\Longrightarrow\; h = \frac{a}{\sqrt2}.\]

Step 2: The volume as a function of the edge

\[ V(a) = \frac13\,a^{2}\cdot\frac{a}{\sqrt2} = \frac{a^{3}}{3\sqrt2} = \frac{a^{3}\sqrt2}{6}. \]

Step 3: Differentiate

\[ V'(a) = \frac{3a^{2}\sqrt2}{6} = \frac{a^{2}\sqrt2}{2}, \qquad V'(5) = \frac{25\sqrt2}{2} \approx 17.6777. \]

Step 4: Apply the differential

With $\Delta a = 5.2-5 = 0.2$: $\Delta V \approx V'(5)\,\Delta a = \frac{25\sqrt2}{2}\cdot 0.2 = \frac{5\sqrt2}{2} \approx 3.5355$.

Step 5: Verification

\[V(5) = \frac{125\sqrt2}{6} \approx 29.4628, \qquad V(5.2) = \frac{140.608\sqrt2}{6} \approx 33.1416\]

\[\Delta V_{\text{exact}} = \frac{15.608\sqrt2}{6} = \frac{22.0730}{6} \approx 3.6788\]

\[\text{The estimate } 3.5355 \text{ is low by } 0.1433, \text{ i.e. } 3.9\% \;\;\checkmark\]