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?

Home » 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?

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,

$L\text{ }=\text{ }200\left( 10\text{ }\text{ }t \right)2$

where L addresses the quantity of liters of water in the pool.

On separating both the sides w.r.t, t, we get

$dL/dt\text{ }=\text{ }200\text{ }x\text{ }2\left( 10\text{ }\text{ }t \right)\text{ }\left( -\text{ }1 \right)\text{ }=\text{ }-\text{ }400\left( 10\text{ }\text{ }t \right)$

Be that as it may, the rate at which the water is running out

$=\text{ }-\text{ }dL/dt\text{ }=\text{ }400\left( 10\text{ }\text{ }t \right)$

Presently, rate at which the water is pursuing 5 seconds will be

$=\text{ }400\text{ }x\text{ }\left( 10\text{ }\text{ }5 \right)\text{ }=\text{ }2000\text{ }L/s\text{ }\left( last\text{ }rate \right)$

$T\text{ }=\text{ }0$

for beginning rate

$=\text{ }400\text{ }\left( 10\text{ }\text{ }0 \right)\text{ }=\text{ }4000\text{ }L/s$

Along these lines, the normal rate at which the water is running out is given by

$=\text{ }\left( Initial\text{ }rate\text{ }+\text{ }Final\text{ }rate \right)/2\text{ }=\text{ }\left( 4000\text{ }+\text{ }2000 \right)/2\text{ }=\text{ }6000/2\text{ }=\text{ }3000\text{ }L/s$