If the radius of a cone is twice its height, find the approximate error in the calculation of its volume, when the radius is 10 cm and the error in radius is 0.01 cm
Let the height of cone $=h$ cm
Then radius of cone $=r = 2h$ cm $\implies$ $h = \dfrac{r}{2}$ cm
Let the error in radius $=\Delta r = 0.01$ cm
Volume of cone $= V = \dfrac{1}{3} \pi r^2 h = \dfrac{1}{3} \pi r^2 \times \dfrac{r}{2} = \dfrac{1}{6}\pi r^3$
$\therefore$ $\dfrac{dV}{dr} = \dfrac{\pi r^2}{2}$
$\begin{aligned}
\therefore \text{Error in volume } \Delta V & = \left(\dfrac{dV}{dr}\right)\times \Delta r \\
& = \left(\dfrac{\pi r^2}{2}\right) \times \Delta r \\
& = \dfrac{\pi \times 10^2}{2} \times 0.01 \\
& = \dfrac{\pi}{2} =1.571 \;\; \text{cm}^3
\end{aligned}$