A swimming pool is to be drained for cleaning. If $L$ represents the number of litres of water in the pool $t$ seconds after the pool has been plugged off to drain and $L = 200(10 - t)^2$. How fast is the water running out at the end of $5 \text{ s}$ and what is the average rate at which the water flows out during the first $5 \text{ s}$?

Instantaneous: $2000 \text{ L/s}$, Average: $3000 \text{ L/s}$

The correct answer is Option (3) → Instantaneous: $2000 \text{ L/s}$, Average: $3000 \text{ L/s}$ ##

Let $L$ represents the number of litres of water in the pool $t$ seconds after the pool has been plugged off to drain, then

$L = 200(10 - t)^2$

$∴$ Rate at which the water is running out $= -\frac{dL}{dt}$

[$-ve$ sign shows that water is drain out]

$\frac{dL}{dt} = -200 \cdot 2(10 - t) \cdot (-1) = 400(10 - t)$

Rate at which the water is running out at the end of $5 \text{ s}$

$= 400(10 - 5)$   $[t = 5\text{s}]$

$= 2000 \text{ L/s} = \text{Final rate}$

Since, $\text{initial rate} = -\left(\frac{dL}{dt}\right)_{t=0} = 4000 \text{ L/s}$

$∴\text{Average rate during 5 s} = \frac{\text{Initial rate} + \text{Final rate}}{2}$

$= \frac{4000 + 2000}{2} = 3000 \text{ L/s}$