Question

A water tank has the shape 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 meter 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 $4m$.

Answer 1

Let the radius and height of the cone be $r,h$ respectively.

Let the semi-vertical angle be $\alpha$


Semi- vertical angle is ${\tan }^{-1}\left(0.5\right)$
$\alpha ={\tan }^{-1}\left(0.5\right)$
$\Rightarrow \tan \alpha =0.5$
$\Rightarrow \frac{r}{h}=0.5$
$\Rightarrow r=\frac{h}{2}$

Given: $\frac{dV}{dt}=5m^3/hr$
Where $V$ is the volume of cone.
Volume of cone is $V=\frac{1}{3}\pi r^{2}h$
$\Rightarrow V=\frac{1}{3}\pi {\left(\frac{h}{2}\right)}^{2}h$

$\Rightarrow V=\frac{1}{3}\pi \left(\frac{h^2}{4}\right)h$

$\Rightarrow V=\frac{1}{12}\pi r{h}^3$
Differentiating w.r.t. $t$,
$\frac{dV}{dt}=\frac{1}{12}\pi \left(3h^2\right)\frac{dh}{dt}$

$\Rightarrow 5=\frac{1}{4}\pi h^2\frac{dh}{dt}$

$\frac{dh}{dt}=\frac{20}{\pi h^2}$

Putting $h=4$
$\Rightarrow \frac{dh}{dt}=\frac{20}{\pi (4{)}^2}$

$\Rightarrow \frac{dh}{dt}=\frac{5}{4\pi }$

$\Rightarrow \frac{dh}{dt}=\frac{5}{4}\times \frac{1}{\frac{22}{7}}$

$\therefore \frac{dh}{dt}{\mid }_{h=4}=\frac{35}{88}$

$\therefore \text{Rate of change of water level is}\frac{35}{88}m/hr$