Question

A water tank is in the shape of an inverted cone. The radius of the base is $8$ meters and the height is $12$ meters. The tank is being emptied for cleaning at the rate of $4{met}^3/minute$. Find the rate at which the water level will be decreasing, when the water is $6$ meters deep.

Answer 1

Let at any moment height and radius of a cone made by water is $h$ and $r$ respectively.
From similartiy principle of a triangle we have
$\frac{OA}{BC}=\frac{OD}{BD}$
$\therefore \frac{8}{r}=\frac{12}{h}$
$\therefore \frac{8h}{12}=r$
$\therefore r=\frac{2h}{3}$
At time $t$, volume of water in a tank is,
$V=\frac{1}{3}\pi r^{2}h=\frac{1}{3}\pi \left(\frac{4}{9}{h}^2\right)h$
$\therefore V=\frac{4\pi }{27}{h}^3$
$\therefore \frac{dV}{dt}=\frac{4\pi }{27}(3h^2.\frac{dh}{dt})$
$\frac{dh}{dt}=\frac{27}{4\pi \left(3h^2\right)}\frac{dV}{dt}=\frac{9}{4\pi h^2}\frac{dV}{dt}$
But $\frac{dV}{dt}=-4\frac{m^3}{minute}$
$\therefore \frac{dh}{dt}=\frac{9}{4\pi h^2}(-4)$
But we have $h=6m$
$\frac{dh}{dt}=-\frac{9}{\pi \left(36\right)}=-\frac{1}{4\pi }m/minute$
$\therefore {\left(\frac{dh}{dt}\right)}_{h=3}=\frac{9}{\pi \left(36\right)}=-\frac{1}{4\pi }m/minute$
$\therefore$ Rate of decrease of the height of water tank is $\frac{1}{4\pi }\frac{m}{minute}$