Question

Water is dripping out from a conical funnel of semi-vertical angle $\frac{\pi }{4}$ at the uniform rate of $2$ $c{m}^2$/sec in the surface, through a tiny hole at the vertex of the bottom. When the slant height of the water level is $4$ cm, find the rate of decrease of the slant height of the water.

Answer 1

Let the radius, height and the slant height of the funnel are $r,h$ and $l$ respectively.

We know the volume of funnel is $V=\frac{1}{3}\pi r^{2}h$

And it's given that semi verticle angle is $\pi /4$

From the figure, $\sin \frac{\pi }{4}=\frac{r}{l}=\frac{1}{\sqrt{2}}⟹r=\frac{l}{\sqrt{2}}$

And $\cos \frac{\pi }{4}=\frac{h}{l}=\frac{1}{\sqrt{2}}⟹h=\frac{l}{\sqrt{2}}$

Let $S$ be the curved surface area of the water cone, then we have

$S=\pi rl=\pi \frac{l}{\sqrt{2}}.l=\frac{\pi l^2}{\sqrt{2}}$

Differentiating w.r.t $t$, we get

$\frac{dS}{dt}=\frac{\pi }{\sqrt{2}}.2l\frac{dl}{dt}$

Now, $\frac{dS}{dt}=-2c{m}^2/sec$ ..... $(S$ is negative, because it is decreasing$)$

$\therefore \frac{dS}{dt}=\frac{2l\pi }{\sqrt{2}}\frac{dl}{dt}=-2c{m}^2/sec$

$\frac{dl}{dt}=-\frac{\sqrt{2}}{\pi l}$

Given $l=4cm$, then

$\frac{dl}{dt}=-\frac{\sqrt{2}}{4\pi }$ cm/sec

Hence, the slant height is decreasing at the rate of $\frac{\sqrt{2}}{4\pi }cm/sec$