A stone is dropped into a quiet lake and waves move in circles at a speed of 3.5 cm per second. At the instant when the radius of the circular wave is 7.5 cm, how fast is the enclosed area increasing?

$52.5π\, cm^2/s$

The correct answer is Option (3) → $52.5π\, cm^2/s$

Let r be the radius of the circular wave and A be the area enclosed by it at any time t seconds, then

$A = πr^2$   …(i)

Diff. (i) w.r.t. t, we get $\frac{dA}{dt}=π.2г\frac{dr}{dt}$

But $\frac{dr}{dt}$ = 3.5 cm/sec = $\frac{7}{2}$ cm/sec (given)

∴ From (ii), we get $\frac{dA}{dt}=2πr.\frac{7}{2}= 7πr$.

When $r = 7.5 cm =\frac{15}{2}cm,\frac{dA}{dt}=7\pi × \frac{15}{2}= 52.5 π$.

Hence, the enclosed area is increasing at the rate of $52.5π\, cm^2/s$ when the radius of the wave is 7.5 cm.