Question

Water is dripping out at a steady rate of $1c{m}^3/sec$ through a tiny hole at the vertex of the conical vessel whose axis is vertical when the slant height of water in the vessel is $4cm$, find the rate of decrease of slant height, where the semi vertical angle of the conical vessel is $\pi /6$

Answer 1

Volume of cone $\Rightarrow v=\frac{1}{3}\pi r^{2}h$

See the figure

It is given angle $x=\pi /6$

$\cos x=h/l\Rightarrow \frac{\sqrt{3}}{2}=\frac{h}{l}$

$\cos \pi /6=h/lh=\sqrt{3}/2^l$

$\sin x=r/l\Rightarrow \frac{1}{2}=\frac{r}{l}\Rightarrow t=\frac{1}{2}$

$\sin \pi /6=r/l$

$u=\frac{1}{3}\times {\left(\frac{1}{2}\right)}^2\left(\frac{\sqrt{3}l}{2}\right)=\frac{\pi }{8\sqrt{3}}$

$\frac{dv}{dt}=1c{m}^3/s$

$\frac{d}{dt}\left(\frac{\pi }{8\sqrt{3}}{l}^3\right)=c{m}^3/s$

$\frac{\pi }{8\sqrt{3}}\frac{d}{dt}\left(l^3\right)=1c{m}^3/s$

$\frac{3l^2\pi }{8\sqrt{3}}\frac{dl}{dt}=1c{m}^3/s$

At $l=4cm$

$\frac{\sqrt{3}\left(16\pi \right)}{8}\frac{dl}{dt}=1$

$2\sqrt{3}\pi \frac{dl}{dt}=1$

$\frac{dl}{dt}=\frac{1}{2\sqrt{3}\pi }cm/sec$