Question

Sand is being poured onto a conical pile at the constant rate of 50 cm³/ minute such that the height of the cone is always one half of the radius of its base. How fast is the height of the pile increasing when the sand is 5 cm deep.

Answer 1

$$\begin{gathered} \text{Let}\mathit{\text{r}}\text{be the radius,}\mathit{\text{h}}\text{be the height and}\mathit{\text{V}}\text{be the volume of the conical pile at any time}\mathit{\text{t}}\text{. Then,} \\ V\text{=}\frac{1}{3}\text{π}{\mathit{\text{r}}}^2\mathit{\text{h}} \\ \Rightarrow \mathit{\text{V}}\text{=}\frac{1}{3}\mathrm{\pi }{\left(2h\right)}^{2}h\left[\because h=\frac{r}{2}\right] \\ \Rightarrow \mathit{\text{V}}\text{=}\frac{4}{3}{\text{πh}}^3 \\ \Rightarrow \frac{dV}{dt}=4\mathrm{\pi }{h}^2\frac{dh}{dt} \\ \Rightarrow 50=4\mathrm{\pi }{h}^2\frac{dh}{dt} \\ \Rightarrow \frac{dh}{dt}=\frac{50}{4\mathrm{\pi }{\left(5\right)}^2} \\ \Rightarrow \frac{dh}{dt}=\frac{1}{2\mathrm{\pi }}\mathrm{cm}/\min \end{gathered}$$