The side of a cube is x = 5 ±0.01 m. Find the absolute
The side of a cube is $x = 5\pm0.01\ \text{m}$. Find the absolute and relative error in the computed volume.
Explanation of Method
Why a relative error triples when a quantity is cubed.
Step 1: The volume and its differential
\[ V = x^{3}, \qquad dV = 3x^{2}\,\Delta x. \]
The nominal volume is $V = 5^{3} = 125\ \text{m}^{3}$.
Step 2: Absolute error
\[ \bigl|\Delta V\bigr|_{\max} \approx 3x^{2}\bigl|\Delta x\bigr|_{\max} = 3\cdot 25\cdot 0.01 = 0.75\ \text{m}^{3}. \]
Step 3: Relative error
Dividing by $V = x^{3}$, the powers of $x$ largely cancel: \[\varepsilon_{\text{rel}} = \frac{\bigl|\Delta V\bigr|}{V} \approx \frac{3x^{2}\bigl|\Delta x\bigr|}{x^{3}} = \frac{3\bigl|\Delta x\bigr|}{x} = \frac{3\cdot 0.01}{5} = 0.006 = 0.6\%.\]
Step 4: Verification
\[5.01^{3} = 125.751501 \;\Longrightarrow\; +0.751501; \qquad 4.99^{3} = 124.251499 \;\Longrightarrow\; -0.748501\]
\[\text{Both differ from } 0.75 \text{ by only } 0.0015\ \text{m}^{3} \;\;\checkmark\]
\[\text{Relative error of the side: } \frac{0.01}{5} = 0.2\%; \qquad 3\times 0.2\% = 0.6\% \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,V = 125\pm0.75\ \text{m}^{3}, \qquad \varepsilon_{\text{rel}} \approx 0.6\%\,}\]
Remark
Remark. The general rule behind the last line: for $y = x^{n}$, $\dfrac{dy}{y} = n\,\dfrac{dx}{x}$ — the relative error is multiplied by the exponent. Squaring doubles it, cubing triples it, taking a square root halves it. It is the same fact as $d(\ln y) = n\,d(\ln x)$, which is why logarithmic differentiation is the natural tool for error propagation in products and powers.