#### 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 represents the number of litres of water in the pool t seconds after the pool has been plugged off to drain and

$L = 200 (10 - t)^2$

To find: the rate of water running out of the pool at the last 5s. and also the average rate of water flowing at during the first 5s.

Explanation: Take the rate of Water running out given by  $-\frac{dL}{dt}$

Given $L = 200 (10 - t)^2$

Differentiation of the above-given equation results in,

$\\ \Rightarrow -\frac{\mathrm{d} \mathrm{L}}{\mathrm{dt}} = -\frac{\mathrm{d}\left(200(10 - \mathrm{t})^{2}\right)}{\mathrm{dt}}\\ Removing all the constant terms, we get \\ \Rightarrow -\frac{\mathrm{d} \mathrm{L}}{\mathrm{dt}} = -200 \cdot \frac{\mathrm{d}\left((10 - \mathrm{t})^{2}\right)}{\mathrm{dt}}\\ Using the power rule of differentiation, we get \\ \Rightarrow -\frac{\mathrm{d} \mathrm{L}}{\mathrm{dt}} = -200 \cdot 2(10 - \mathrm{t}) \cdot \frac{\mathrm{d}(10 - \mathrm{t})}{\mathrm{dt}}\\ \Rightarrow -\frac{\mathrm{dL}}{\mathrm{dt}} = -400(10 - \mathrm{t}) \cdot (-1)\\ \Rightarrow -\frac{\mathrm{dL}}{\mathrm{dt}} = 400(10 - \mathrm{t}) \ldots (i)$

Now, to find the speed of water running out at the end of 5s, we need to find the value of equation (i) with t=5

$\\ \Rightarrow\left(-\frac{d L}{d t}\right)_{t=5}=(400(10-t))_{t=5} \\ \Rightarrow\left(-\frac{d L}{d t}\right)_{t=5}=400(10-5) \\ \Rightarrow\left(-\frac{d L}{d t}\right)_{t=5}=400(5) \\ \Rightarrow\left(-\frac{d L}{d t}\right)_{t=5}=2000 \frac{L}{s} \ldots . .(i i)$

Therefore, 2000 L/s is the rate of water running out at the end of 5s

To calculate the initial rate we need to take t=o in equation (i)

$\\ \Rightarrow\left(-\frac{\mathrm{dL}}{\mathrm{dt}}\right)_{\mathrm{t}=0}=(400(10-\mathrm{t})) \mathrm{t}=0 \\ \Rightarrow\left(-\frac{\mathrm{d} \mathrm{L}}{\mathrm{dt}}\right)_{\mathrm{t}=0}=400(10-0) \\ \Rightarrow\left(-\frac{\mathrm{d} L}{\mathrm{dt}}\right)_{\mathrm{t}=0}=4000 \mathrm{~L} / \mathrm{s}$

Equation (ii) tells about the final rate of water flowing whereas equation (iii) is the initial rate.

Thus, the average rate during 5s is

$\\ = \frac{\text{initial rate} + \text{final rate}}{2}\\ Substituting \ the \ corresponding \ values, \ we \ get\\ = \frac{4000 + 2000}{2} = 3000 \, \mathrm{L} / \mathrm{s}$

After calculation, the final rate of water flowing out of the pool in 5s is 3000L/s.