Question

Water is dripping out from a conical funnel at the uniform rate of $4$ $c{m}^3/sec$. When the slant height of water is $3cm$, find the rate of the decrease of slant height of the water, give that vertical angle of the funnel is ${120}^o$.

Answer 1

$\alpha =\frac{{120}^{\circ }}{2}={60}^{\circ }$

$\because \frac{r}{h}=\tan \alpha$

$\Rightarrow r=\tan {60}^{\circ }\times h$

$\Rightarrow r=h\sqrt{3}$

$\therefore l=\sqrt{r^2+(h{)}^2}$

$\Rightarrow l=\sqrt{3h^2+h^2}$

$\Rightarrow l=2h\Rightarrow h=\frac{l}{2}$

$\Rightarrow \frac{dl}{dt}=\frac{2dh}{dt}$

$\because v=\frac{1}{3}\cdot \pi r^{2}h$

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

$\Rightarrow v=\pi h^3$

$\Rightarrow \frac{dv}{dt}=3\pi h^2.\frac{dh}{dt}$

$4=3\pi {\left(\frac{l}{2}\right)}^2.\frac{dh}{dt}$

$\Rightarrow \frac{dh}{dt}=\frac{4\times 4}{3\pi (3{)}^2}=\frac{16}{27\pi }$

$\Rightarrow \frac{dl}{dt}=\frac{2dh}{dt}=\frac{32}{27\pi }cm./sec$