Answer
$\frac{4\pi}{15}$ $in^2/min$
Work Step by Step
Since
$1 $hour=$60 $minutes
Minutes hand takes 60 minutes to revolve through $360^{\circ} $ (One complete revolution) $\Longrightarrow$. Then $\theta$ will be the angle swept out in $\frac{1}{60}$ minutes' Or
$60\theta =360 t$
Taking derivative with respect to $t$
$\frac{ d(60\theta)}{dt}=\frac{d(360t)}{dt}$
$ 60\frac{ d\theta}{dt}=360\frac{d(t)}{dt}$
$ \frac{ d\theta}{dt}=\frac{360}{60} =6$ .............. eq (1)
Let A be the area of sector of the circle swept out at an angle $\theta$
Let r be the radius of the circle.
Then
$A=\frac{\theta}{360} \pi r^2$
Taking derivative with respect to t
$\frac{dA}{dt}= \frac{d( \frac{\theta}{360} \pi r^2)}{dt}$
$\frac{dA}{dt}= \frac{\pi r^2}{360}\frac{d\theta}{dt}$ ................ eq (2)
Putting $\frac{d\theta}{dt}=6$, $r=4$ in equation (2)
$\frac{dA}{dt}= \frac{\pi 4^2}{360}\times6=\frac{4 \pi}{15}$ $in^2/min$
