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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

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प्रश्न

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?

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उत्तर

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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अध्याय 6: Application Of Derivatives - Exercise [पृष्ठ १३६]

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एनसीईआरटी एक्झांप्लर Mathematics Exemplar [English] Class 12
अध्याय 6 Application Of Derivatives
Exercise | Q 9 | पृष्ठ १३६

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