Question

Show that the relative error in computing the volume of a cubical box, due to and error in measuring the edge, is approximately equal to the three times the relative error in the edge.

Answer 1

Let length of edge of cubical box is $x$ and volume is $V$, then

$V=x^3$

Diff. w.r.t. x,

$\frac{dV}{dx}=3x^2$

$\because \Delta V=\frac{dV}{dx}\Delta x$

$\Rightarrow \Delta V=3x^2\Delta x$

$\Rightarrow \frac{\Delta V}{V}=\frac{3x^2\Delta x}{V}$

$\Rightarrow \frac{\Delta V}{V}=\frac{3x^2\Delta x}{x^3}$

$\Rightarrow \frac{\Delta V}{V}=3\frac{\Delta x}{x}$

$\Rightarrow \frac{\Delta V}{V}\times 100=3(\frac{\Delta x}{x}\times 100)$