# Water is running into an inverted cone at the rate of π cubic metres per minute.

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.

Solution:

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

Then,

$V=\frac{1}{3} \pi r^{2} h$

$\Rightarrow \frac{d V}{d t}=\frac{1}{3} \pi r^{2} \frac{d h}{d t}+\frac{2}{3} \pi r h \frac{d r}{d t}$

Now,

$\frac{h}{r}=\frac{10}{5}$ or $r=\frac{h}{2}$ and $\frac{d h}{d t}=2 \frac{d r}{d t}$

$\Rightarrow \frac{d V}{d t}=\frac{1}{3} \pi\left(\frac{h}{2}\right)^{2} \frac{d h}{d t}+\frac{2}{3} \pi\left(\frac{h}{2}\right) h \frac{1}{2} \frac{d h}{d t}$

$\Rightarrow \frac{d V}{d t}=\frac{\pi}{3}\left[\frac{h^{2}}{4} \frac{d h}{d t}+\frac{h^{2}}{2} \frac{d h}{d t}\right]$

$\Rightarrow \frac{d V}{d t}=\frac{\pi}{3} \times \frac{3 h^{2} d h}{4 d t}$

$\Rightarrow \frac{d V}{d t}=\frac{\pi h^{2}}{4} \frac{d h}{d t}$

$\Rightarrow \frac{\pi h^{2}}{4} \frac{d h}{d t}=\pi$

$\Rightarrow \frac{d h}{d t}=\frac{4}{h^{2}}$

$\Rightarrow \frac{d h}{d t}=\frac{4}{(2.5)^{2}}$

$\Rightarrow \frac{d h}{d t}=0.64 \mathrm{~m} / \mathrm{min}$