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?

Dear Student


Given, L = 200(10 - t)^2where L represents the number of liters of water in the pool. differentiating both the sides w.r.t, t,
dL/dt = 200 x 2(10 - t) (-1)
= -400(10 - t)
the rate at which the water is running out

= -dL/dt = 400(10 - t)
Now, rate at which the water is running after 5 seconds will be

= 400 x (10 - 5) = 2000 L/s (final rate)
Putting T = 0 for initial rate

= 400 (10 - 0) = 4000 L/s
So, the average rate at which the water is running out is given by

= (Initial rate + Final rate)/ 2 = (4000 + 2000)/ 2
= 6000/2
= 3000 L/s
Thus, the required rate = 3000 L/s

Thanks