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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 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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  • If $ y = f(x)$, then $ \large\frac{dy}{dx}$ measures the rate of change of $y$ w.r.t $ x $.
  • $ \bigg(\large \frac{dy}{dx} \bigg)_{x=x_0}$ represents the rate of change of $y$ w.r.t $x$ at $x=x_0$
Step 1
$ L =200(10-t)^2$
Differentiating w.r.t $t$ we get
$ \large\frac{dl}{dt}=200 \times 2(10-t)(-1)$
$ = -400(10-t)$
when $t=5$
$ \large\frac{dL}{dt}=-400(5)$
Hence the water is decreasing at the rate of $2000\: l/sec$
Step 2
When $t=5$
$ L=200(10-5)^2$
$=200 \times 25$
$ = 5000$
Hence when $ t=5$, the capacity of water is $5000$ litres.
Hence the water flowing out during the first 5 seconds
$ 5000-2000=3000 \: l$
Hence the capacity of water flowing out during the first 5 seconds is $ 3000 \: l$
answered Aug 2, 2013 by thanvigandhi_1

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