Question

A water tank has the slope of an inverted right circular cone with its axis vertical and vertex lowermost. Its semi-vertical angle is ${\tan }^{-1}\left(0.5\right)$. Water is poured into it at a constant rate of $5$ cubic metre per hour. Find the rate at which the level of the water is rising at the instant when the depth of water in the tank is $4\text{m}.$
$(\pi =\frac{22}{7})$

Answer 1

Let $r,h$ and $\alpha$ be the radius, height and the semi-vertical angle of the given conical water tank.
$\Rightarrow \tan \alpha =\frac{r}{h}\Rightarrow \alpha ={\tan }^{-1}\left(\frac{r}{h}\right)$
As $\alpha ={\tan }^{-1}\left(0.5\right)$ (given)
Therefore, $\frac{r}{h}=0.5$
$\Rightarrow r=\frac{h}{2}\cdots \left(1\right)$

Let the volume of the cone be $V$
Therefore, $V=\frac{1}{3}\pi r^{2}h$
Using $\left(1\right)$ in the above equation, we get -
$V=\frac{\pi h^3}{12}$

Differentiating the above equation with respect to time, we get -
$\frac{dV}{dt}=\frac{d}{dh}\left(\frac{\pi h^3}{12}\right)\cdot \frac{dh}{dt}$ (Using chain rule)
$=\frac{1}{4}\pi h^2\cdot \frac{dh}{dt}$
Its is given that $\frac{dV}{dt}=5{\text{m}}^3/\text{h}$ and $h=4\text{m}$
$\therefore 5=\frac{\pi }{4}(4{)}^2\cdot \frac{dh}{dt}$
$\Rightarrow \frac{dh}{dt}=\frac{5}{4\pi }\text{m/h}$
$=\frac{5\times 7}{4\times 22}\text{m/h}$
$=\frac{35}{88}\text{m/h}$

Therefore, the rate of change of water level is $\frac{35}{88}\text{m/h}$