The side of a square is a = 8 length units. By how much

The side of a square is $a = 8$ length units. By how much does its area increase if the side increases by 1° $\Delta a = 1$; 2° $\Delta a = 0.5$; 3° $\Delta a = 0.1$ length units? Replace the increment by the differential and estimate the relative error.

Step 1: Exact increment and differential

With $S = a^{2}$: \[\Delta S = \bigl(a+\Delta a\bigr)^{2}-a^{2} = 2a\,\Delta a+\bigl(\Delta a\bigr)^{2}, \qquad dS = 2a\,\Delta a = 16\,\Delta a.\]

Step 2: A picture of the two terms

Enlarging the square adds two rectangles $a\times\Delta a$ along the two sides — that is the $2a\,\Delta a$ of the differential — plus one small square $\Delta a\times\Delta a$ in the corner, which is exactly the discarded remainder.

Step 3: The errors in closed form

\[ \bigl|\Delta S-dS\bigr| = \bigl(\Delta a\bigr)^{2}, \qquad \frac{\bigl|\Delta S-dS\bigr|}{\Delta S} = \frac{(\Delta a)^{2}}{2a\,\Delta a+(\Delta a)^{2}} = \frac{\Delta a}{2a+\Delta a} = \frac{\Delta a}{16+\Delta a}. \]

Step 4: Case 1°: $\Delta a = 1$

\[ \Delta S = 16+1 = 17, \quad dS = 16, \quad \bigl|\Delta S-dS\bigr| = 1, \quad \frac{1}{17} \approx 5.88\%. \]

Step 5: Case 2°: $\Delta a = 0.5$

\[ \Delta S = 8+0.25 = 8.25, \quad dS = 8, \quad \bigl|\Delta S-dS\bigr| = 0.25, \]

\[ \frac{0.25}{8.25} = \frac{0.5}{16.5} \approx 3.03\%. \]

Note that the denominator of the relative error is $\Delta S = 8.25$, not $2a+\Delta a = 16.5$; writing $\dfrac{0.25}{16.5}$ would halve the true figure.

Step 6: Case 3°: $\Delta a = 0.1$

\[ \Delta S = 1.6+0.01 = 1.61, \quad dS = 1.6, \quad \bigl|\Delta S-dS\bigr| = 0.01, \quad \frac{0.01}{1.61} \approx 0.62\%. \]

Step 7: Summary

\[\begin{array}{c|c|c|c|c} \Delta a & \Delta S & dS & \text{absolute} & \text{relative} \\ \hline 1 & 17.00 & 16.0 & 1.00 & 5.88\% \\ 0.5 & 8.25 & 8.0 & 0.25 & 3.03\% \\ 0.1 & 1.61 & 1.6 & 0.01 & 0.62\% \end{array}\]

\[\text{Formula check: } \frac{\Delta a}{16+\Delta a} = \frac{1}{17} = 5.88\%,\ \frac{0.5}{16.5} = 3.03\%,\ \frac{0.1}{16.1} = 0.62\% \;\;\checkmark\]

\[\text{Halving } \Delta a \text{ from } 1 \text{ to } 0.5 \text{ quarters the absolute error } (1\to0.25) \text{ and roughly halves the relative one} \;\;\checkmark\]