Question

A cone having fixed volume has semi-vertical angle of $\frac{\pi }{4}.$ At an instant when its height is decreasing at the rate of $2m/s,$ its radius increases at a rate equal to

• A. $2m/s$
• B. $4m/s$
• C. $1m/s$
• D. $8m/s$

Answer 1

Answer: (C)

As, $V=\frac{1}{3}\pi r^{2}h$
$\Rightarrow \frac{dV}{dt}=\frac{\pi }{3}\left(r^2\frac{dh}{dt}+h\cdot 2r\frac{dr}{dt}\right)=0$
$\Rightarrow r\frac{dh}{dt}=-2h\frac{dr}{dt}$

Also, $\frac{r}{h}=tan\left(\frac{\pi }{4}\right)\Rightarrow r=h$
$\therefore \frac{dh}{dt}=-2\frac{dr}{dt}\Rightarrow \frac{dr}{dt}=-\frac{1}{2}\frac{dh}{dt}$
$\Rightarrow \frac{dr}{dt}=+1$
$\Rightarrow$ radius increases at $1m/s.$