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 $5\;{\rm{s}}$ and what is the average rate at which the water flows out during the first $5\;{\rm{s}}$ ?
Let $L$ represents the number of litres of water in the pool $t$ seconds after the pool has been plugged off to drain, then
$L = 200{(10 - t)^2}$
Therefore the rate at which the water is running out $= - \frac{{dL}}{{dt}}$
$\frac{{dL}}{{dt}} = - 200 \cdot 2(10 - t) \cdot ( - 1)$
$= 400(10 - t)$
Rate at which the water is running out at the end of $5\;{\rm{s}}$
$= 400(10 - 5)$
$= 2000\;{\rm{L}}/{\rm{s}} =$ Final rate
Initial rate $= - {\left( {\frac{{dL}}{{dt}}} \right)_{t = 0}} = 4000\;{\rm{L}}/{\rm{s}}$
Therefore the average rate during $5\;{\rm{s}} = \frac{{{\rm{ Initial rate }} + {\rm{ Final rate }}}}{2}$
$= \frac{{4000 + 2000}}{2}$
$= 3000\;{\rm{L}}/{\rm{s}}$
NCERT & Exemplar solution for CBSE Class 12 Mathematics, Application of Derivatives. Curated by Sachin Sharma. Free for all students.