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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?
Solution
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
`"dL"/"dt"` = 200 × 2(10 – t)(– 1)
= – 400(10 – t)
But the rate at which the water is running out
= `- "dL"/"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
= `("Initial rate" + "Final rate")/2`
= `(4000 + 2000)/2`
= `6000/2`
= 3000 L/s
Hence, the required rate = 3000 L/s.
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