Question

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 seconds? What is the average rate at which the water flows out during the first 5 seconds?

Answer 1

Given that L = 200(10 – t)²

Where L represents the number of litres of water in the pool.

Differentiating both sides w.r.t, t, we get

${\displaystyle \frac{\text{dL}}{\text{dt}}}$ = 200 × 2(10 – t)(– 1)

= – 400(10 – t)

But the rate at which the water is running out

= ${\displaystyle -\frac{\text{dL}}{\text{dt}}}$ = 400(10 – t)   .....(1)

Rate at which the water is running after 5 seconds

= 400 × (10 – 5) = 2000 L/s  .....(Final rate)

For initial rate put t = 0

= 400(10 – 0)

= 4000 L/s

The average rate at which the water is running out

= ${\displaystyle \frac{\text{Initial rate}+\text{Final rate}}{2}}$

= ${\displaystyle \frac{4000+2000}{2}}$

= ${\displaystyle \frac{6000}{2}}$

= 3000 L/s

Hence, the required rate = 3000 L/s.