Question

Water is flowing into a conical vessel 18 cm deep and 10 cm across the top at the rate of 4 cm3/min. The depth of the water is "h" cm. If the rate which the surface is rising is 0.1146 cm/min, find the value of "h"?

Answer 1

$D=2R=10 cm$ ; $H=18cm$ ; $\frac{dv}{dt}=4\frac{cm^3}{min}$ ; $\frac{dh}{dt}=0.1146\frac{cm}{min}$ .

Solution:

Filled volume:

$v=\frac13\pi r^2h$ ; $r=h\cdot \tan\theta$ ;

$v=\frac13\pi \cdot\tan^2\theta \cdot h^3$ ;

$\tan{\theta}=\frac RH=\frac{D}{2H}$ ;

$v=\frac13\pi \cdot(\frac{D}{2H})^2 \cdot h^3$

$\frac{dv}{dt}=\pi \cdot(\frac{D}{2H})^2 \cdot h^2\cdot\frac{dh}{dt}$

$h=\frac{2H}{D}\cdot\sqrt{\frac{1}{\pi \cdot\frac{dh}{dt}} \cdot\frac{dv}{dt}}=$$\frac{2\cdot 18}{10}\cdot\sqrt{\frac{1}{\pi \cdot0.1146} \cdot4}\approx\frac{2\cdot 18}{10}\cdot\sqrt{\frac{4}{0.36}}=12\ (cm)$ .

Answer: $h\approx12\ cm$ .