Question

Water is running into an inverted cone at the rate of π cubic metres per minute. The height of the cone is 10 metres, and the radius of its base is 5 m. How fast the water level is rising when the water stands 7.5 m below the base.

Answer 1

$\mathrm{Let}r\mathrm{be}\mathrm{the}\mathrm{radius},h\mathrm{be}\mathrm{the}\mathrm{height}\mathrm{and}\mathrm{V}\mathrm{be}\mathrm{the}\mathrm{volume}\mathrm{of}\mathrm{the}\mathrm{cone}\mathrm{at}\mathrm{any}\mathrm{time}t.$


$$\begin{gathered} \mathrm{Then}, \\ \mathit{\text{V}}\text{=}\frac{1}{3}\text{π}{\mathit{\text{r}}}^2\mathit{\text{h}} \\ \Rightarrow \frac{dV}{dt}=\frac{1}{3}\text{π}{\mathit{\text{r}}}^2\frac{dh}{dt}\text{+}\frac{2}{3}\text{π}\mathit{\text{rh}}\frac{dr}{dt} \\ \text{Now,} \\ \frac{h}{r}\text{=}\frac{10}{5}\text{or}\mathit{\text{r}}\text{=}\frac{h}{2}\text{and}\frac{dh}{dt}\text{=2}\frac{dr}{dt} \\ \text{⇒}\frac{dV}{dt}=\frac{1}{3}\text{π}{\left(\frac{h}{2}\right)}^2\frac{dh}{dt}\text{+}\frac{2}{3}\text{π}\left(\frac{h}{2}\right)\mathit{\text{h}}\frac{1}{2}\frac{dh}{dt} \\ \text{⇒}\frac{dV}{dt}=\frac{\mathrm{\pi }}{3}\left[\frac{h^2}{4}\frac{dh}{dt}\text{+}\frac{h^2}{2}\frac{dh}{dt}\right] \\ \text{⇒}\frac{dV}{dt}\text{=}\frac{\pi }{3}\text{×}\frac{3h^{2}dh}{4dt} \\ \text{⇒}\frac{dV}{dt}\text{=}\frac{\pi h^2}{4}\frac{dh}{dt} \\ \Rightarrow \frac{\pi h^2}{4}\frac{dh}{dt}=\mathrm{\pi } \\ \Rightarrow \frac{dh}{dt}=\frac{4}{h^2} \\ \Rightarrow \frac{dh}{dt}=\frac{4}{{\left(2.5\right)}^2} \\ \Rightarrow \frac{dh}{dt}=0.64\mathrm{m}/\min \end{gathered}$$