Question

If water is poured into an inverted hollow cone whose semi-vertical angle is ${30}^{\circ }$ , and its depth (measured along axis) increase at the rate of 1 cm per sec, find the rate at which the volume of water is increasing when the depth is 24 cm.

Answer 1

Let the height of cone $=h$

Radius $=r$

${\theta }^{\circ }=30°$

According to question, $\frac{dh}{dt}=1cm/sec$

$tan\theta =\frac{r}{h}\Rightarrow \frac{1}{\sqrt{3}}=\frac{r}{h}\Rightarrow r=\frac{h}{\sqrt{3}}-----\left(1\right)Now,\frac{dv}{dt}=\frac{d}{dt}\left(\frac{1}{3}\pi r^{2}h\right)=\frac{1}{3}\pi \frac{d}{dt}\left(r^{2}h\right)$

Using $\left(1\right)$

$\frac{dv}{dt}=\frac{1}{3}\pi \frac{d}{dt}\left(\frac{h^2}{3}h\right)=\frac{1}{3}\pi \cdot \frac{1}{3}\frac{d}{dt}\left(h^3\right)\frac{dv}{dt}=\frac{1}{9}\pi \cdot 3h^2\frac{dh}{dt}=\frac{\pi }{3}{h}^2\frac{dh}{dt}Whenh=24cm\frac{dv}{dt}=\frac{\pi }{3}\cdot 24\cdot 24\cdot \left(1\right)=192\pi {cm}^2/sec$