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?
Given, L = 200(10 – t)2 where L represents the number of liters of water in the pool.
On differentiating both the sides w.r.t, t, we get
dL/dt = 200 x 2(10 – t) (-1) = -400(10 – t)
But, 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)
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
Therefore, the required rate = 3000 L/s
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