The edge of a cube is measured as 10 cm with an error of 1%. The cube's volume is to be calculated from this measurement. Estimate the percentage error in the volume calculation.
Please explain to me how to do this.
The volume of a cube is
$\displaystyle V=s^3$
The error is $\displaystyle \frac{\Delta V}{V} \approx \frac{dV}{V}$
We know that $\displaystyle s=10cm \mbox{ and } \Delta s =.01(10cm)=0.1cm$
$\displaystyle dV =3s^2ds$
using the above formula we get
$\displaystyle \frac{3(10cm)^2(0.1cm)}{(10cm)^3}=\frac{30cm^3}{10 00cm^3}=\frac{3}{100}=3\%$