Water is running in a hemispherical bowl of radius $180 \mathrm{~cm}$ at the rate of 108 cubic decimetres…
Water is running in a hemispherical bowl of radius $180 \mathrm{~cm}$ at the rate of 108 cubic decimetres per minute. How fast the water level is rising when depth of the water level in the bowl is $120 \mathrm{~cm}$ ?
$(1$ decimeter $=10 \mathrm{~cm})$
$16 \pi \mathrm{cm} / \mathrm{sec}$
$\frac{16}{\pi} \mathrm{cm} / \mathrm{sec}$
$\frac{1}{16 \pi} \mathrm{cm} / \mathrm{sec}$
$\frac{\pi}{16} \mathrm{~cm} / \mathrm{sec}$
Solution
Radius of hemispherical bowl $(\mathrm{r})=180 \mathrm{~cm}$ Rate of flow $\left(\frac{\mathrm{dV}}{\mathrm{dt}}\right)=108 \mathrm{dm}^3 / \mathrm{min}$
$\begin{aligned} & =108000 \mathrm{~cm}^3 / \mathrm{min} \\ & =\frac{108000}{60} \mathrm{~cm}^3 / \mathrm{sec} \\ & =1800 \mathrm{~cm}^3 / \mathrm{sec}\end{aligned}$
Let depth of water in bowl be $x$.
$\therefore \quad$ Volume of water in hemispherical.
$\begin{aligned} & \text { bowl }(\mathrm{V})=\frac{\pi}{3} x^2(3 \mathrm{r}-x) \\ & \mathrm{V}=\frac{\pi}{3} x^2(3 \times 180-x)\end{aligned}$
$\therefore \quad \mathrm{V}=180 \pi x^2-\frac{\pi}{3} x^3$
Differentiating w.r.t. $x$, we get
$\frac{\mathrm{dV}}{\mathrm{dt}}=360 \pi x \frac{\mathrm{d} x}{\mathrm{dt}}-\pi x^2 \frac{\mathrm{d} x}{\mathrm{dt}}$
$\left.\frac{\mathrm{dV}}{\mathrm{dt}}\right|_{x=120}=\frac{\mathrm{d} x}{\mathrm{dt}}\left(360 \pi \times 120-120^2 \pi\right)$
$\therefore \quad 1800=\frac{\mathrm{d} x}{\mathrm{dt}}\left(360 \pi \times 120-120^2 \pi\right)$
$\therefore \quad 15=\frac{\mathrm{d} x}{\mathrm{dt}}(360 \pi-120 \pi)$
$\frac{\mathrm{d} x}{\mathrm{dt}}=\frac{15}{240 \pi}$
$=\frac{1}{16 \pi} \mathrm{cm} / \mathrm{sec}$