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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 the water running out at the end of 5 seconds? What is the a

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?

Answers (1)

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}}=-200.2(10-\mathrm{t}) \cdot \frac{\mathrm{d}(10-\mathrm{t})}{\mathrm{dt}}$\\ $\Rightarrow-\frac{\mathrm{dL}}{\mathrm{dt}}=-400(10-\mathrm{t}).(-1)$\\ $\Rightarrow-\frac{\mathrm{dL}}{\mathrm{dt}}=400(10-\mathrm{t}) \ldots$(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 the 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.

 

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infoexpert22

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