Question

The edge of a cube was found to be 30 cm with a possible error in measurement of 0.1 cm. Use differentials to estimate the maximum possible error, relative error, and percentage error in computing (a) the volume of the cube and (b) the surface area of the cube.

Short answer

1. For volume, the relative error is 0.01 and the percentage error is 1%.
2. For surface area, the relative error is 0.006 and the percentage error is 0.6%.

Step-by-step solution

Describe the given information

Let \(x\) be the edge length. It is given that \(x = 30\) and \(dx = 0.1\).Thus, the volume is given by \(V = {x^3}\), such that \(dV = 3{x^2}dx\). On plugging given values, \(dV = 3{\left( {30} \right)^2}\left( {0.1} \right) = 270\).Thus, minimum error which could occur in finding volume is \(270c{m^3}\).

Find the relative error in volume

The relative error in finding volume is obtained by \(\frac{{\Delta V}}{V} \approx \frac{{dV}}{V}\). Substitute the expression of \(dV\)and \(V\), and simplify to find the relative error.

\(\begin{aligned}{c}{\rm{Relative}}\,\,{\rm{error}}&= \frac{{dV}}{V}\\&= \frac{{3{x^2}dx}}{{{x^3}}}\\&= 3\frac{{dx}}{x}\\&= 3\left( {\frac{{0.1}}{{30}}} \right)\\&= 0.01\end{aligned}\)

On multiplying the relative error with 100, we get the percentage error as \(0.01\left( {100} \right) = 1\% \).

Find the surface area in surface area

The surface area of the cube is given as \(S = 6{x^2}\), such that \(dS = 12xdx\). For the given values of \(x\)and \(dx\), \(dS = 12\left( {30} \right)\left( {0.1} \right) = 36\). Thus, the maximum error which can occur, in evaluating surface area is 36cm² . Substitute the expression of \(dS\) and \(S\), and simplify to find the relative error.

\(\begin{aligned}{c}{\rm{Relative}}\,\,{\rm{error}}&= \frac{{dS}}{S}\\&= \frac{{12xdx}}{{6{x^2}}}\\&= 2\frac{{dx}}{x}\\&= 2\left( {\frac{{0.1}}{{30}}} \right)\\ &= 0.006\end{aligned}\)

On multiplying the relative error with 100, we get the percentage error as \(0.006\left( {100} \right) = 0.6\% \).