Answer
$ \frac{dV}{dt}=\pi \left[2 rh \frac{dr}{dt}+ r^2\frac{dh}{dt}\right] $
$ \frac{dV}{dt}==-20\pi $ $ in^3/s$; Decreasing
Work Step by Step
(a)$V=\pi r^2 h$ ............. eq (1)
$ \frac{dV}{dt}=\frac{dV}{dr} \frac{dr}{dt}+\frac{dV}{dh}\frac{dh}{dt}$ ........... eq(2)
Putting equation (1) in equation (2)
$ \frac{dV}{dt}=\frac{d(\pi r^2 h)}{dr} \frac{dr}{dt}+\frac{d(\pi r^2 h)}{dh}\frac{dh}{dt}$
$ \frac{dV}{dt}=\pi h\frac{d( r^2 )}{dr} \frac{dr}{dt}+\pi r^2\frac{d h}{dh}\frac{dh}{dt}$
$ \frac{dV}{dt}=2\pi rh \frac{dr}{dt}+\pi r^2\frac{dh}{dt}$
$ \frac{dV}{dt}=\pi \left[2 rh \frac{dr}{dt}+ r^2\frac{dh}{dt}\right]$ ............ eq(3)
(b) Putting $h=6$ ,$\frac{dh}{dt}=1 $ in/s. $r=10$ in , $ \frac{dr}{dt}=-1$ in/s in equation (3)
$ \frac{dV}{dt}=\pi [2 \times 10 \times 6 \times (-1)+ 10^2\times 1]=-20\pi \hspace{1mm} in^3/s $
Decreasing
